TITLE: Examples of Richardson orbit closures not having a symplectic resolution? QUESTION [9 upvotes]: This is a follow-up to a recent question asked by Peter Crooks here. The answer by Ben Webster includes a helpful link to the corrected arXiv version of Baohua Fu's 2003 Invent. Math. paper Symplectic resolutions for nilpotent orbits. Most of this literature is unfamiliar to me, so I may be overlooking something. I've encountered nilpotent orbits mainly in connection with various types of representation theory (sometimes in good prime characteristic, where most properties of the orbits are the same as over $\mathbb{C}$). At this point I'm still confused about some details, such as: Are there Richardson orbits whose closures fail to have a symplectic resolution (and if so, what is the lowest rank Lie algebra in which an example appears)? EDIT: Here I'm using shorthand to avoid normality questions: read "for which the normalizations of their closures fail to have ...? (Apparently the sorting out of normal orbits isn't complete yet for some exceptional types.) As Fu notes in Prop. 3.16, it follows from the main theorem of the paper that a nilpotent orbit whose closure admits a symplectic resolution must be Richardson (intersecting the nilradical of some parabolic subalgebra in a dense orbit). In turn a reviewer states: "But the converse is not always true." I don't see direct evidence of that in Fu's paper. Here the simple Lie algebras are studied case-by-case: all orbits in type $A_n$ are Richardson, with trivial component groups, forcing their closures to have symplectic resolutions. In types $G_2, F_4, E_6$, all Richardson orbits also have trivial component groups, whereas a few such orbits in types $E_7, E_8$ have component groups of order 2 and are left unsettled in the paper. (These cases were later treated geometrically here.) The discussion of types $B_n, C_n, D_n$ leaves me somewhat confused, since the explicit examples mentioned between Prop. 3.21 and Prop. 3.22 aren't Richardson orbits. This prompts the question above. ADDED: Fu reduces the problem (for a Richardson orbit) to the question of whether or not there exists a parabolic $P$ defining the orbit for which $N(P)=1$. This is the index in the full component group in $G$ (topologically, fundamental group) of the component group in $P$ of an orbit element $X$. (Here "component group" means the group $C_G(X)/C_G(X)^\circ$.) It's not clear how to compute $N(P)$ in all cases, which may be why Fu gave up on the leftover cases in types $E_7,E_8$. I'm wondering about the naive (presumably false?) statement that an orbit closure has a symplectic resolution iff the orbit is Richardson. What bothers me is that Fu seems to mention only examples of non-Richardson orbits such as minimal orbits in types other than $A_n$, etc. REPLY [2 votes]: Just adding a reference to the literature where this question seems to have cropped up : "Calculating canonical distinguished involutions in the affine Weyl groups" - Chmutova, Ostrik (pdf) They phrase the question as : "Are all Richardson orbits strongly Richardson ?" ('strongly Richardson' being their term for the existence of a resolution satisfying the necessary properties). The published version of the paper (pdf link above) also includes a family of counter examples(== one of the examples alluded to in Ben's answer).<|endoftext|> TITLE: Tokarev's theorem on Banach lattices which are Grothendieck spaces QUESTION [7 upvotes]: When browsing the literature, I have found the following theorem of E. Tokarev: Let $X$ be a Banach lattice with weakly sequentially complete dual space. Then for any Banach space $Y$, every unconditionally converging operator $T\colon X\to Y$ is weakly compact. (In other words, Banach lattices with weakly sequentially duals have Pełczyński's property (V).) (This is Theorem 1.1 here.) This theorem, if true, would have a number of fantastic consequences. For instance, a very nice theorem of W. B. Johnson would follow easily: Suppose that $X$ is a Banach space with local unconditional structure. Then either $X$ is super-reflexive or $X$ contains $\ell_\infty^n$'s uniformly or $X$ contains $\ell_1^n$'s uniformly complemented. I can produce a rather lengthy list of further applications (some of them, I believe, new and which would be useful to me for other purposes). However, I must admit I don't understand the proof. (For instance, I don't understand why the measures $\tilde{\mu}_n$ are well defined, and why we can pass to a convergent subsequence of $y_n^*$'s.) After a while, I decided to apply the so-called psychological argument: if this theorem is so strong, let us look on other results quoting it. Unfortunately, neither google nor mathscinet can find anything. I also tried to contact the author but it seems he doesn't check his mailbox. Hence my question: Does this theorem have a chance to be true? Or maybe there is an easy counter-example? REPLY [7 votes]: There is a counterexample in Figiel, T.; Ghoussoub, N.; Johnson, W. B. On the structure of nonweakly compact operators on Banach lattices. Math. Ann. 257 (1981), no. 3, 317–334.<|endoftext|> TITLE: Roots of matching polynomial of graph QUESTION [6 upvotes]: At the end of this preprint, I make the following conjecture concerning the roots of the matching polynomial: If a graph $G$ is connected and contains a cycle, then the spectral radius of $G$ strictly exceeds the largest root of the matching polynomial $\mu_G(z)$. There appears to be very little published on the roots of the matching polynomial. How might this conjecture be proved? REPLY [9 votes]: The moments (power symmetric functions, sums of powers of the roots of) the characteristic polynomial enumerate all closed walks in the graph. Chris Godsil proved that the moments of the matching polynomial count a particular type of closed walk, called tree-like. Graphs with cycles have some closed walks that are not tree-like. This proves that the even moments of the characteristic polynomial are eventually strictly greater than those of the matching polynomial if there is a cycle. That is, if the roots of the characteristic polynomial are $\lbrace \lambda_i\rbrace$ and those of the matching polynomial are $\lbrace \mu_i\rbrace$ then for large enough $k$, $$\sum \lambda_i^{2k} \gt \sum\mu_i^{2k},$$ which proves that $\lambda_{\rm max}\ge \mu_{\rm max}$. Now restrict it to the connected case. The matching polynomial has a simple largest root $\mu_{\rm max}$ and also $-\mu_{\rm max}$ as a simple root. The characteristic polynomial has a simple largest root $\lambda_{\rm max}$ and may or may not have $-\lambda_{\rm max}$ as a simple root too (iff the graph is bipartite). Let $w_{2k}$ and $t_{2k}$ be the counts of all closed walks and all tree-like closed walks, respectively. Then as $k\to\infty$ either $w_{2k}\sim \lambda_{\rm max}^{2k}$ or $w_{2k}\sim 2\lambda_{\rm max}^{2k}$, while $t_{2k}\sim 2\mu_{\rm max}^{2k}$. Now suppose there is a cycle of length $g$. An example of a closed walk that isn't tree-like is to start with a tour around the cycle twice then take any closed walk. So for $k$ large enough $w_{2k} \ge t_{2k}+t_{2k-2g}$. Therefore $w_{2k}$ is asymptotically at least $c\mu_{\rm max}^{2k}$ where $c = 1 + \mu_{\rm max}^{-2g}$. This is irreconcilable with $\lambda_{\rm max}= \mu_{\rm max}$ by the observations in the previous paragraph. Therefore $\lambda_{\rm max}\gt \mu_{\rm max}$. I bet this is known already.<|endoftext|> TITLE: When is $A$ "$L$-ish" whenever $B$ is "$L$-ish"? QUESTION [14 upvotes]: My question is about a variant of the usual notion of relative constructibility, $\le_c$ (which an earlier version of this question confusingly denoted "$\le_L$"), in set theory. Fix a countable transitive model $W\models\mathsf{ZFC+V\not=L}$. By a theorem of Barwise, given any set $A\in W$ there is a (possibly-ill-founded) end extension $W'\supseteq_{end}W$ such that $W'\models A\in L$. In light of this, we can consider the following preorder on elements of $W$: $$A\le_{L,end}B\quad:=\quad\forall W'\supseteq_{end}W[W'\models \mathsf{ZFC}+B\in L\implies W'\models A\in L].$$ Broadly speaking, I'm interested in whether there is a nice description of $\le_{L, end}$ (or an interesting subrelation, such as $\le_{L, end}$ restricted to $\mathbb{R}^W$) without referring to end extensions - and in general, anything we can say about $\le_{L, end}$, or the induced degree structure. To keep things reasonably concrete, the following specific question seems natural: Is $\le_{L, end}$ the same as $\le_c$? I'm especially interested in the situation where $V$ satisfies strong large cardinal axioms (say, "There is a proper class of Woodins) and $W$ is "far from $L$" (say, $\mathbb{R}^W$ is closed under sharps). REPLY [3 votes]: Remarks: (i) I'm interpreting the definition of $\leq_{L,\mathrm{end}}$ as quantifying over set models $W'$, not proper classes. (ii) I'm considering the main question (comparing the two orders), particularly in the case that $V$ has large cardinals, but mainly not in the case of "particular interest", i.e. where $W$ is far from $L$; some remarks on the latter case are made in the "Edit" at the bottom.) Claim: Assume ZF + there are ordinals $\kappa<\lambda$ such that $L_\kappa\models$ZFC and $L_\lambda\models$ZFC and $\kappa$ is a cardinal in $L_\lambda$. Let $\psi$ be the statement "there is a countable transitive $W\models$ZFC such that defining $\leq_{L,\mathrm{end}}$ w.r.t. $W$, then $\leq_{L,\mathrm{end}}$ is different to $\leq_{\mathrm{c}}\upharpoonright W$". Then: (i) There is a forcing extension $V[G_1]$ of $V$ such that $V[G_1]\models\psi$, and (ii) If every real has a sharp (in particular, if there is a measurable cardinal), then $\psi$ holds in $V$. In fact, we will get the two orders to disagree over $\mathbb{R}^W$. Proof: The proofs of (i) and (ii) are almost the same, so I'll deal with them simulatenously. Let $\lambda$ be least as hypothesized, and $\kappa$ the corresponding ordinal. Then $L_\lambda$ is pointwise definable by condensation etc, so $\lambda<\omega_1$, and there is a bijection $\pi:\omega\to\lambda$ with $\pi\in L_{\lambda+2}$. If every real has a sharp, then we can find a sequence $\vec{y}=\left_{\alpha<\lambda}$ of reals such that $y_\alpha<_{\mathrm{c}}y_\beta$ for $\alpha<\beta<\lambda$, and in fact with $\vec{y}\upharpoonright \beta\in L[y_\beta]$ for each $\beta<\lambda$. In any case, we can easily force the existence of such a sequence of reals. So from now on we assume that there is such a sequence. We will find a sequence $G=\left_{n<\omega}$ which is generic over $L_\lambda$ for the $\omega$-fold finite support product $\mathbb{C}^{<\omega}$ of Cohen forcing, such that setting $W=L_\kappa[G]$ (not $W=L_\lambda[G]$), we have $\leq_{\mathrm{c}}\upharpoonright X$ is a wellorder of ordertype $\lambda$, where $X=\{x_n\}_{n<\omega}$, and hence this order is not in $W$ (in fact not in $L_\lambda[G]$). However, we do have $\leq_{L,\mathrm{end}}\upharpoonright X\in W$. For this, it suffices to see it is in $L_\lambda[G]$. And for this, it suffices to see it is in $L_\lambda[G,H]$ whenever $H$ is $L_\lambda[G]$-generic for $\mathrm{Coll}(\omega,\kappa)$ (by Solovay's theorem on this business; alternatively use homogeneity of the forcing and the uniformity of the next sentence). For the latter, we observe that $\leq_{L,\mathrm{end}}\upharpoonright X$ is $\Pi^1_1(\{z\})$ whenever $z$ is a real coding $L_\kappa[G]$, and since $L_\lambda[G,H]\models$ZFC and is transitive, it is therefore in $L_\lambda[G,H]$. For the $\Pi^1_1(\{z\})$-definability, it is easy assuming ZF+DC, as if there is an uncountable counterexample $W'$ to the $\forall W'$ quantifier, we can get a countable one by taking a countable hull (cf. Remark (i) at the start). Without DC we can still do a variant of this. Suppose $W'$ is a counterexample. Let $W''=L^{W'}$, and note that $W''$ is still a counterexample. Now let $W'''$ be the definable hull in $W''$ of parameters in $W\cup\{W\}$ (recalling $A,B\in W$); since $W''$ models "$V=L$", this gives an elementary substructure, and it is countable, so $W'''$ is a countable counterexample, as desired. So we need to construct $G$. Let $T\subseteq{^{<\omega}}2$ be a perfect tree. Recall that $t\in T$ is a \emph{splitting node} of $T$ iff $t\frown(0)\in T$ and $t\frown(1)\in T$ (this is defined in the same manner for finite trees below). Given a real $x$, let $b_x$ be the infinite branch through $T$ determined by using $x(n)$ as the bit of $b_x$ following the $n$th splitting node along $b_x$. So the map ${^\omega}2\to[T]$ (the codomain is the set of branches of $T$) sending $x\mapsto b_x$ is a bijection. Note that if $T\in L$, then $x\equiv_{\mathrm{c}}b_x$. Lemma: There is a perfect tree $T\subseteq {^{<\omega}}2$, with $T\in L$, such that for every finite sequence $\vec{b}=(b_0,\ldots,b_{m-1})$ of pairwise distinct branches $b_i$ through $T$, $\vec{b}$ is $L_\lambda$-generic for $\mathbb{C}^m$ (the $m$-fold product of Cohen forcing). Proof: For $m\in[1,\omega)$, let $\mathscr{D}_m$ be the set of all open dense $D\subseteq\mathbb{C}^m$ such that $D\in L_\lambda$. Let $\mathscr{D}=\bigcup_{m\in[1,\omega)}\mathscr{D}_n$. Fix an enumeration $\vec{D}=\left_{n<\omega}$ of $\mathscr{D}$, such that each $D\in\mathscr{D}$ gets repeated infinitely often, and such that $\vec{D}\in L$. Let $m_n$ be the arity of $D_n$ (that is, $D_n\subseteq\mathbb{C}^{m_n}$). We construct a sequence $\left_{n<\omega}$ of finite trees $T_n\subseteq{^{k_n}2}$ where $k_n<\omega$, such that: (i) $T_0=\{\emptyset\}$ and $k_0=0$, (ii) for all maximal nodes $t$ of $T_n$, we have $\mathrm{lh}(t_n)=k_n$, (iii) for all maximal nodes $t$ of $T_n$, there are exactly $n$ splitting nodes $s$ of $T_n$ such that $s\subseteq t$ (so $s\subsetneq t$, since $t$ is maximal), (iv) for each $i_{\alpha<\lambda}$ and $\pi:\omega\to\lambda$ from earlier). Let $b_n=b_{z_n}$ be the branch induced by $z_n$ (as described earlier). So ($*$) for all finite $m$-tuples of distinct integers $(n_0,\ldots,n_{m-1})$, $(b_{n_0},\ldots,b_{n_{m-1}})$ is generic over $L_\lambda$ for $\mathbb{C}^m$. However, we don't know that the full sequence $\left_{n<\omega}$ is generic over $L_\lambda$ for the $\omega$-fold finite support product $\mathbb{C}^{<\omega}$ of $\mathbb{C}$. But by a standard trick, using ($*$), we can modify each $b_n$ on at most finitely many digits, producing a sequence $\left_{n<\omega}$ which is $L_\lambda$-generic for $\mathbb{C}^{<\omega}$. (Enumerate the dense subsets of $\mathbb{C}^{<\omega}$ in $L_\lambda$ as $\left_{n<\omega}$, and progressively extend conditions $p_n$ getting into $E_n$, such that, letting $p_{ni}$ be the projection of $p_n$ to the $i$th component, and letting $A_i=\mathrm{dom}(p_{ni})$, we ensure that for $n'>n$, we have that $p_{n'i}$ agrees with $b_n$ outside of $A_i$. This can be achieved using (*), since $\mathbb{C}^{<\omega}$ factors nicely.) Now let $G=\left_{n<\omega}$. So $G$ is generic over $L_\lambda$ for $\mathbb{C}^{<\omega}$, and note that since $b'_n$ eventually agrees with $b_n$, we have $b'_n\equiv_{\mathrm{c}} b_n\equiv_{\mathrm{c}}z_n=y_{\pi(n)}$. Therefore letting $X=\{b'_n\}_{n<\omega}$, the restriction $\leq_{\mathrm{c}}\upharpoonright X$ is just a wellorder of length $\lambda$, as desired. Edit: Consider now the case that $W$ is closed under sharps, assuming that $V$ is closed under sharps for reals. Then the preceding argument does not work, and in fact we have the following (related to some things Hamkins wrote in his answer above): Theorem 2: Assume ZF + for every real $x$, $x^\#$ exists. Let $W$ be a countable transitive model of ZFC which is closed under (true) sharps. Define $\leq_{L,\mathrm{end}}$ w.r.t. $W$. Let $A,B\in W$. Then $A\leq_{\mathrm{c}}B$ implies $A\leq_{L,\mathrm{end}}B$. Proof: Suppose $A\leq_{\mathrm{c}}B$, i.e. $A\in L(B)$. Then since $B^\#\in W$, it follows that $A\in L^W(B)=L_{\mathrm{OR}^W}(B)$. Let $W'$ be any end-extension of $W$ which models ZFC with $B\in L^{W'}$. If $\mathrm{OR}^{W'}=\mathrm{OR}^W$ then $L^{W'}=L^W$, so $B\in L^W$, so $L^W(B)=L^W$, so $A\in L^W=L^{W'}$. Suppose instead that $\mathrm{OR}^W\neq\mathrm{OR}^{W'}$. Let $\gamma$ be the least $W'$-ordinal such that $B\in L^{W'}$ (note $\gamma$ might be illfounded). Let $\alpha\in\mathrm{OR}^W$ be such that $A\in L_\alpha(B)$. Then $\alpha\in\mathrm{OR}^{W'}$, so $\gamma+\alpha$ makes sense in $W'$ and $\gamma+\alpha\in\mathrm{OR}^{W'}$, which easily implies that $L_\alpha(B)\subseteq L^{W'}$, so $A\in L^{W'}$, as desired.<|endoftext|> TITLE: Calculate reduction of Jacobian of hyperelliptic curve QUESTION [7 upvotes]: Suppose I have a hyperelliptic curve of genus $2$ over $\mathbb Q$. I want to get information about its Jacobian reduction at prime $p$ (especially, in case $p=2$). Also I'm interesting in the group of connected components of the Neron model of the Jacobian. Is it possible to get such information using some computer algebra system (like Magma, Sage and etc)? I know that there is Sage (and Pari) function genus2reduction (Liu algorithms implementation) that is able to give information about almost all reductions. But it doesn't give enough information about reduction at $p=2$. So I want to focus on case $p=2$ and I'm looking for other solution. UPD Gave link at genus2reduction and pointed that it is implementation of Liu algorithms. Added after Victor Miller answer. UPD2 Thanks to Michael Stoll. There is an answer for almost all cases. But for some curves his solution still doesn't work. So I'm still looking for something else. For example it is impossible to get information about reduction at $p=2$ for such curves like $y^2 = 16*x^6 + 32*x^5 + 16*x^4 + 32*x^3 + 96*x^2 + 64*x + 16$. REPLY [4 votes]: The relevant information can be obtained from a regular model of the curve over ${\mathbb Z}_p$. Such a model can be computed by repeatedly blowing up points or components of the special fiber that are singular on the arithmetic surface one obtains from the original curve. More or less detailed examples of this can be found in various places, for example here or here. The component group can be obtained from the intersection matrix of the resulting special fiber. There is an implementation of regular models in Magma by Steve Donnelly that can do this computations in many (but not yet all) cases. See the function "RegularModel" and the documentation here.<|endoftext|> TITLE: Grothendieck's Homotopy Hypothesis - Applications and Generalizations QUESTION [23 upvotes]: Grothendieck's homotopy hypothesis, is, as the $n$lab states: Theorem: There is an equivalence of $(∞,1)$-categories $(\Pi⊣|−|): \mathbf{Top} \simeq \mathbf{\infty Grpd}$. What are the applications of this hypothesis? Why is it so fundamental? Can it be "generalized", perhaps by using the following definition of spaces: a space is simply a sheaf of sets on some site $\mathbf{Loc}$ of local models with a Grothendieck topology $τ$ on it? REPLY [10 votes]: Since "Pursuing Stacks" was "written in English in response to a correspondence in English" I feel I ought to explain my own current position in relation to the Grothendieck Programme, and also to that of Whitehead, many of whose results and methods were a kind of guiding light. In particular, his theorem on free crossed modules proved in "Combinatorial Homotopy II" (CHII) led us (Phil Higgins and I) away from attempting to define higher homotopy groupoids for spaces, but instead for pairs of spaces in dimension 2, and then for filtered spaces. This suggested analogous results in discussions with Jean-Louis Loday. A general picture of the methodology is covered by the following "generic" diagram of categories and functors:      (source) with the following properties: $\rho, \Pi$ are homotopically defined, correspond under the vertical equivalence, and preserve certain colimits; $\rho \mathbb B$ is naturally equivalent to $1$; There is a natural transformation $1 \to \mathbb B \rho$ preserving some homotopical information. The purpose of the final part of 1. is to allow some calculations, and in particular some description of certain "free" objects. 3. is little vague, but is useful to bear in mind. Finally, $B= U \mathbb B$, where $U$ is the forgetful functor, is a kind of classifying space for the algebraic model. We have two examples of this situation. A. topological data = filtered spaces, broad algebraic data = (strict) cubical $\omega$-groupoids with connections, narrow algebraic data = crossed complexes. This is explained in detail in the book Nonabelian Algebraic Topology, which develops a lot of the work in CHII, and does not reach other parts. B, topological data = $n$-cubes of pointed spaces, broad algebraic data = cat$^n$-groups, narrow algebraic data = crossed $n$-cubes of groups. A survey of this may found in the expository article "Computing homotopy types using crossed N-cubes of groups", with references to the original papers, particularly Brown/Loday and Ellis/Steiner. The not-so-easy equivalences between the two kinds of algebraic data are a key to the methods. The broad model is used for intuition, conjectures and proofs; the narrow algebraic model is used for relation to classical theory and for calculations. One can hop between these at will! Criterion 1 also excludes a number of algebraic models, such as: simplicial groups; weak models; $2$-crossed modules, and others, either because they not directly defined homotopically or because they are not known to preserve certain colimits. Many people may be uncomfortable with the use of "topological data", rather than spaces, but this is how the methods work, and some thought should be given to evaluating that. There is more discussion in a talk I gave in Paris, June 5, 2014, available on my preprint page, as are talks at Galway and Aveiro. Grothendieck's analysis of "topological spaces" is referred to in another answer. My understanding of Whitehead's work is that he worked with many specific examples, or classes of examples, which had additional structure to that of a topological space. Also between the "broad" and the "narrow" there are and will be many intermediate cases, of possible use. Finally, when I told Grothendieck in 1985 that (strict) $n$-fold groupoids model homotopy $n$-types (Loday's result, in essence), he exclaimed: "But that is absolutely beautiful!" For many reasons, his thoughts on this were not developed. Actually Loday's result is for the pointed case. For more general results, see this paper, Blanc & Paoli.<|endoftext|> TITLE: Is it possible to express $\int\sqrt{x+\sqrt{x+\sqrt{x+1}}}dx$ in elementary functions? QUESTION [86 upvotes]: I asked a question at Math.SE last year and later offered a bounty for it, but it remains unsolved even in the simplest case. So I finally decided to repost this case here: Is it possible to express the following indefinite integral in elementary functions? $${\large\int}\sqrt{x+\sqrt{x+\sqrt{x+1}}}\ dx$$ REPLY [104 votes]: I'm adding a separate answer for the general question that the OP asked, which settles the question in the negative for all $n>2$ (and gives an alternate proof for $n=3$ to the one I gave above). Recall that the OP defined a sequence of algebraic functions $f_n$ by the rule $f_0(x) = 1$, $f_1(x) = \sqrt{x+1}$, and $f_{n+1}(x) = \sqrt{x + f_n(x)}$ for all $n\ge 1$. It was observed that $f_n$ has an elementary antiderivative for $n=0$, $1$, and $2$, and the problem was to determine whether $f_n$ has an elementary antiderivative for some $n>2$. I am going to show that there is no elementary antiderivative of $f_n$ when $n>2$. Assume $n>2$ (NB: This is important, because the argument below will not work for $n\le2$; the reader may enjoy finding where it breaks down), and let $K_n = {\mathbb C}\bigl(x,f_n(x)\bigr)$ be the elementary differential field generated by $x$ and $f_n(x)$. Then $K_n$ is the field of meromorphic functions on the normalization $\hat C_n$ of the algebraic curve $C_n$ defined by the minimal degree $y$-monic polynomial $P_n(x,y)$ that satisfies $P_n\bigl(x,f_n(x)\bigr) \equiv 0$. This minimal degree is $2^n$; for example, $P_2(x,y) = (y^2-x)^2-x-1$ and $P_3(x,y) = \bigl((y^2-x)^2-x\bigr)^2-x-1$, etc. Since $P_{n+1}(x,y) = (P_n(x,y)+1)^2-x-1$ for $n\ge 1$ with $P_1(x,y)=y^2-x-1$, one sees, by applying the Eisenstein Criterion to $P_n(x,y)$ regarded as an element of $D[y]$ with $D$ being the integral domain ${\mathbb C}[x]$, that $P_n(x,y)$ is irreducible for all $n\ge 1$. Hence, $\hat C_n$ is connected. It will be important in what follows to observe that $K_n$ has an involution $\iota$ that fixes $x$ and sends $f_n(x)$ to $-f_n(x)$; this is because $P_n(x,y)$ is an even polynomial in $y$. The fixed field of $\iota$ is ${\mathbb C}\bigl(x,\,f_n(x)^2\bigr)$, and the $(-1)$-eigenspace of $\iota$ is ${\mathbb C}\bigl(x,\,f_n(x)^2\bigr)f_n(x) = K_{n-1}{\cdot}f_n(x)$. Now, the curve $C_n\subset \mathbb{CP}^2$ has only one point on the line at infinity, namely $[1,0,0]$, but the normalization $\hat C_n$ has $2^{n-1}$ points lying over this point. They can be parametrized as follows: First, establish the convention that $\sqrt{u}$ means the unique analytic function on the complex $u$-plane minus its negative axis and $0$ that satisfies $\sqrt1 = 1$ and $\bigl(\sqrt{u}\bigr)^2 = u$. Let $\epsilon = (\epsilon_1,\ldots,\epsilon_{n-1})$ be any sequence with ${\epsilon_k}^2=1$ and consider the sequence of functions $g^\epsilon_k(t)$ defined by the criteria $g^\epsilon_1(t) = \sqrt{1+t^2}$ and $g^\epsilon_{k+1}(t) = \sqrt{1+\epsilon_{n-k}t g^\epsilon_k(t)}$ for $1\le k < n$. Choose, as one may, a $\delta_n>0$ sufficiently small so that, when $t$ is complex and satisfies $|t|<\delta_n$, all of the functions $g^\epsilon_k$ are analytic when $|t|<\delta_n$. In particular, one finds an expansion $$ g^\epsilon_n(t) = 1+\tfrac12\epsilon_1\,t + \tfrac18(2\epsilon_1\epsilon_2-1)t^2 + O(t^3). $$ Also, it is easy to verify that the disk in $\mathbb{CP}^2$ defined by $$ [x,y,1] = [1,\ t g^\epsilon_n(t),\ t^2]\qquad\text{for}\quad |t|<\delta_n $$ is a nonsingular parametrization of a branch of $C_n$ in a neighborhood of the point $[1,0,0]$. In the normalization $\hat C_n$, this is then a local parametrization of a neighborhood of a point $p_\epsilon\in \hat C_n$. Obviously, this describes $2^{n-1}$ distinct points on $\hat C_n$. When $x$ and $f_n$ are regarded as meromorphic functions on $\hat C_n$, it follows that there is a unique local coordinate chart $t_\epsilon:D_\epsilon\to D(0,\delta_n)\subset \mathbb{C}$ of an open disk $D_\epsilon\subset \hat C_n$ about $p_\epsilon$ such that $t_\epsilon(p_\epsilon)=0$ and on which one has formulae $$ x = \frac1{{t_\epsilon}^2} \quad\text{and}\quad f_n(x) = \frac{g^\epsilon_n(t_\epsilon)}{t_\epsilon} = \frac{1+\tfrac12\epsilon_1\ t_\epsilon +\tfrac18(2\epsilon_1\epsilon_2-1)\ {t_\epsilon}^2} {t_\epsilon} + O({t_\epsilon}^2). $$ In particular, it follows that $f_n(x)$, as a meromorphic function on $\hat C_n$, has polar divisor equal to the sum of the $p_\epsilon$ and hence has degree $2^{n-1}$. Of course, this implies that the zero divisor of $f_n(x)$ on $\hat C_n$ must be of degree $2^{n-1}$ as well. Note that the functions $g^\epsilon_k$ satisfy $g^{-\epsilon}_k(-t) = g^{\epsilon}_k(t)$, where $-\epsilon = (-\epsilon_1,\ldots,-\epsilon_{n-1})$. This implies that $\iota(p_\epsilon) = p_{-\epsilon}$ and that $t_\epsilon\circ\iota = -t_{-\epsilon}$. Now, the $2^{n-1}$ zeroes of $f_n(x)$ on $\hat C_n$ are distinct, for they are the zeros of the polynomial $q_n(x) = P_n(x,0) = (q_{n-1}+1)^2-x-1$, and the discriminant of $q_n$, being the resultant of $q_n$ and $q_n'$, is clearly an odd integer, and hence is not zero. Thus, $C_n$ is a branched double cover of $C_{n-1}$, branched exactly where $f_{n}$ has its zeros. This induces a branched cover $\pi_n:\hat C_n\to \hat C_{n-1}$ that is exactly the quotient of $\hat C_n$ by the involution $\iota$ (whose fixed points are where $f_n$ has its zeros). Since one then has the Riemann-Hurwitz formula $$ \chi(\hat C_n) = 2\chi(\hat C_{n-1}) - B_n = 2\chi(\hat C_{n-1}) - 2^{n-1}, $$ and $\chi(\hat C_1) = \chi(\hat C_2) = 2$, induction gives $\chi(\hat C_n) = (3{-}n)2^{n-1}$, so the genus of $\hat C_n$ is $(n{-}3) 2^{n-2} + 1$. (This won't actually be needed below, but it is interesting.) The only poles of $x$ and $f_n(x)$ on $\hat C_n$ are the points $p_\epsilon$, and computation using the above expansions shows that, in a neighborhood of $p_\epsilon$, one has an expansion of the form $$ f_n(x)\,\mathrm{d} x - \mathrm{d}\left(f_n(x)\bigl(\tfrac12\ x + \tfrac16\ f_n(x)^2\bigr) \right) = \left(\frac{ (1-\epsilon_1\epsilon_2) } {4{t_\epsilon}^2} + O({t_\epsilon}^{-1})\right)\ \mathrm{d} t_\epsilon\ . $$ Thus, the meromorphic differential $\eta$ on $\hat C_n$ defined by the left hand side of this equation has, at worst, double poles at the points $p_\epsilon$ and no other poles. Now, by Liouville's Theorem, $f_n$ has an elementary antiderivative if and only if $f_n(x)\ \mathrm{d} x$ and, hence, the form $\eta$ are expressible as finite linear combinations of exact differentials and log-exact differentials. Thus, $f_n(x)$ has an elementary antiderivative if and only if $\eta$ is expressible in the form $$ \eta = \mathrm{d} h + \sum_{i=1}^m c_i\,\frac{\mathrm{d} g_i}{g_i} $$ for some $h,g_1,\cdots g_m\in K_n$ and some constants $c_1,\ldots,c_m$. Suppose that these exist. Since $\eta$ has, at worst, double poles at the $p_\epsilon$ and no other poles, it follows that $h$ must have, at worst, simple poles at the points $p_\epsilon$ and no other poles; in fact, $h$ is uniquely determined up to an additive constant because its expansion at $p_\epsilon$ in terms of $t_\epsilon$ must be of the form $$ h = \frac{\epsilon_1\epsilon_2-1}{4t_\epsilon} + O(1). $$ Moreover, because $\eta$ is odd with respect to $\iota$, it follows that $h$ (after adding a suitable constant if necessary) must also be odd with respect to $\iota$. This implies, in particular, that $h$ vanishes at each of the zeros of $f_n$ (which, by the argument above, are simple zeros). This implies that $h = r\,f_n$ for some $r\in K_{n-1}$ that has no poles and satisfies $r(p_\epsilon) = (\epsilon_1\epsilon_2-1)/4$ for each $\epsilon$. However, since $r$ has no poles and $\hat C_n$ is connected, it follows that $r$ is constant. Thus, it cannot take the two distinct values $0$ and $-1/2$, as the equation $r(p_\epsilon) = (\epsilon_1\epsilon_2-1)/4$ implies. Thus, the desired $h$ does not exist, and $f_n$ cannot be integrated in elementary terms for any $n>2$.<|endoftext|> TITLE: $G_\mathbb{Z}$-homotopy type of rational Tits building $\Delta_{G, \mathbb{Q}}$ QUESTION [9 upvotes]: Take $G$ to be a standard semisimple algebraic $\mathbb{Q}$-group, e.g. $Sp_{2g}$ or $SO(h)$ for $h$ a nondegenerate quadratic form over $\mathbb{Q}$. The arithmetic group $\Gamma=G_{\mathbb{Z}}$ has an obvious simplicial action on the associated rational Tits building $\Delta_{G, \mathbb{Q}}$. It is well known (a theorem of Borel) that $G_{\mathbb{Z}}$ acts with finitely many orbits on the vertices of $\Delta_{G, \mathbb{Q}}$, i.e. finiteness of the double cosets $G_\mathbb{Z} \backslash G_\mathbb{Q} /P_\mathbb{Q}$ for $P_\mathbb{Q}$ a rational parabolic. Hence the naive quotient $\Gamma \backslash \Delta$ is some finite complex. Question: are the combinatorics of the naive quotient $\Gamma \backslash \Delta$ known? My true motivation: I would like to know more about $\Gamma$-equivariant maps with target $\Delta$. Tits-Solomon theorem says $\Delta$ has the homotopy-type of a countable wedge of spheres $S^d$ (where $d=d(G)$ is known). Question: can somebody give me a reference or explanation describing the $\Gamma$-homotopy type of $\Delta$? Is the equivariant homotopy type of the associated adele groups known, i.e. $G_\mathbb{A}(\infty)$-homotopy type of $\Delta_{G, \mathbb{A}}$? REPLY [8 votes]: $\newcommand\SL{\text{SL}}\newcommand\O{\mathcal{O}}\newcommand\cl{\text{cl}}$For $\text{SL}_n$ over a number field $K$, the rational Tits building is the classical Tits building $T_n(K)$, namely the poset of proper nonzero $K$-subspaces of $K^n$. In fact, starting here we can assume simply that $\O_K$ is a Dedekind domain with fraction field $K$. For each $d=1,\ldots,n-1$, the orbits of $d$-dimensional subspaces under $\SL_n\O_K$ can be identified with the ideal class group $\widetilde{K}_0(\O_K)$: for each $d$-dimensional $K$-subspace $V$, the intersection $V\cap \O_K^n$ is a rank-$d$ projective $\O_K$-module, and $V$ and $W$ lie in the same $\SL_n\O_K$-orbit iff $V\cap \O_K^n$ and $W\cap \O_K^n$ are isomorphic. More than this, using the fact that $\O_K$ is a Dedekind domain, you can show that one chain of subspaces is in the same orbit as another iff the ideal classes match up (i.e. iff the $i$th subspaces lie in the same orbit, for each $i$). This shows that the quotient $T_n(K)/\SL_n\O_K$ is the $(n-1)$-fold join of the discrete set $\widetilde{K}_0(\O_K)$ with itself. This is an $\ell^{n-1}$-fold wedge of $(n-2)$-spheres, where $\ell=\lvert \widetilde{K}_0(\O_K)\rvert-1$, and indeed a spherical building of type $A_1\times \cdots \times A_1$: a chamber is determined by a sequence $(c_1,\ldots,c_{n-1})$ of ideal classes $c_i\in\widetilde{K}_0(\O_K)$. If $\O_K$ is a PID (i.e. $\lvert\widetilde{K}_0(\O_K)\rvert=1$), this tells you that the quotient $T_n(K)/\SL_n\O_k$ is contractible. All of this so far is elementary, but the following theorem is not. [Edited to add reference to paper.] It is joint with Benson Farb and Andrew Putman and appears in our just-posted paper "Integrality in the Steinberg module and the top-dimensional cohomology of $\textrm{GL}_n\O_K$". Theorem: The projection map $T_n(K)\twoheadrightarrow T_n(K)/\SL_n\O_K$ is surjective on homology. An even stronger theorem is true: for every apartment in the quotient $T_n(K)/\SL_n\O_K$, there is an apartment in $T_n(K)$ whose fundamental class hits its homology class, on the nose. (This appears as Proposition 5.5 of our paper. Section 5 is self-contained, so if you are only interested in this result it suffices to read Sections 5.1 and 5.2 on pages 21-28. Still, I encourage you to check out the introduction while you're there!) To get a feeling for the strength of this result, it's a fun game to pick your favorite Dedekind domain with nontrivial class group, pick $2n-2$ ideal classes to determine an apartment in the quotient, and then try to find an apartment in $T_n(K)$ that hits it. (Warning: you probably won't be able to do this, except for $n=2$ or $n=3$! --- at least we weren't able to for a while. But the theorem we prove says that you always can.)<|endoftext|> TITLE: Log weight removal in general (weaker) prime number theorem QUESTION [5 upvotes]: Let $a_n$ be a sequence of non-negative numbers. Assume that $$\limsup _{X\to \infty}\frac{\sum_{p\leq X} a_p\log p}{X}\leq 1.$$ Can we prove that $$\limsup _{X\to \infty}\frac{\sum_{p\leq X} a_p}{X/\log X}\leq 1?$$ REPLY [5 votes]: Here is a direct argument avoiding partial summation. Let $Y:=X/\log^2 X$, then $$ \sum_{p\leq X}a_p = \sum_{p\leq Y}a_p + \sum_{Y TITLE: Convex hull of total orders QUESTION [25 upvotes]: Let $n$ be a positive integer and $\prec$ an arbitray total order on $\{1,\dots,n\}$. I associate to this order a vector $v$ with one coordinate for every pair $(i,j)$ s.t. $1\leq i\neq j \leq n$, by this definition: $$v_{ij} = \left\{\begin{array}{cc}1 & i\prec j \\ 0 & i\succ j\end{array}\right. $$ This vector satisfies the following obvious linear inequalities for every distinct triples $1 \leq i,j,k \leq n$: 1)$0\leq v_{ij}$, 2)$v_{ij}+v_{ji}=1$, 3)$v_{ik} \leq v_{ij}+v_{jk} $. Does these inequalities characterize the convex hull of all vectors associated to total orders? REPLY [8 votes]: Here's a (slightly) different write-up of David's example, combined with Emil's comment: We want to show that the partial order $1<2$, $1<6$, $3<2$, $3<4$, $5<4$, $5<6$ is not a convex combination of total orders. Now the only total orders that can contribute here are the extensions of the given partial order. We observe that there are only four extensions that make $4<1$, namely $$ 3<5<4<1<2<6 , $$ and here we may switch $3,5$ and/or $2,6$. Since initially $1, 4$ were not comparable (so $x_{14}=0$), these four total orders must get combined weight $1/2$ in the convex combination we are looking for. We can now similarly consider the (again four) total order extensions that make $6<3$, and again these must get combined weight $1/2$. We have used up all our coefficients, but all orders considered so far have $5<2$, so this will not work (since $x_{25}=0$).<|endoftext|> TITLE: Rank four quadratic Form with non trivial discriminant in I(k) QUESTION [5 upvotes]: Im sure this is a beginners question. Let $k$ be a field and $I(k)$ the fundamental ideal in the Witt-ring W(k). The Arason-Pfister-Hauptsatz states: "If $\varphi$ is any anisotropic class in $I^n(k)$, then $rank(\varphi) \geq 2^n$." It is well known that for the kernel of the discriminant $ker(e_1) = I^2$ holds. Since there is a filtration $W(k) \supseteq I(k) \supseteq I^2(k) \supseteq..$ , the set of classes of elements in $I(k)$ with non trivial discriminant is the complement $I(k)\backslash I^2(k)$. Since the Arason-Pfister-Hauptsatz only gives a lower bound on the rank of elements in $I(k)$ it might be possible that there is a class $\varphi \in I(k)\backslash I^2(k)$ with $rank(\varphi) = 4$ or even higher. Do you have an example such that this happens? In "Galois Cohomology" Serre referres to some paper of Alexander Merkurjev, in which he shows that it is possible for "every" even $N\geq1$ to find a $k$ having $u$-invariant $u(k)=N$ ,such that $I^3(k)$ vanishes i.e. $k$ has cohomological depth $cd(k)=2$. This is basically the answer to an analog of my question in case of finding some anisotropic rank $8$ class in $I^2(k)\backslash I^3(k)$. REPLY [3 votes]: The answer depends on your field $k$. For example if $k$ is the $p$-adic field $\mathbb{Q}_p$, $p\neq 2$, it is known that the only anisotropic form of dimension $4$ over $k$ is isomorphic to the norm form of the unique quaternion algebra over $k$ which is of course in $I^2$ since it is a Pfister form. By contrast you can construct many fields $k$ with your desired property, for instance consider the field $k=\mathbb{Q}(x,y,z,t)$ and the form $\varphi=\langle x,y,z,t\rangle\in I\backslash I^2$. Added: regarding your question in the comments about $I^3$: By a theorem of Merkurjev, the Hasse-Witt invariant induces an isomorphism between $I^2/I^3$ and the two torsion part of the Brauer group $Br_2(k)$. Hence it suffices to give an example of a field $k$ and an anisotropic form $\varphi\in I^2$ of dimension $8$ such that the Hasse-Witt invariant of $\varphi$ is nontrivial. For this consider an anisotropic Albert form, e.g., the form $\alpha=\langle x,y,-xy,-z,-t,zt\rangle$ over the transcendental field $\mathbb{Q}(x,y,z,t)$. Now consider the transcendental field $k=\mathbb{Q}(x,y,z,t,u,v)$ and the eight dimensional anisotropic form $\varphi=-uv\cdot\alpha\perp\langle u,v\rangle\in I^2$. The Hasse-Witt invariant of $\varphi$ is isomorphic to the Clifford algebra $Cl(\varphi)\simeq C(\langle u, v\rangle)\otimes Cl(\alpha)$ which is not split.<|endoftext|> TITLE: Volume of geodesic balls QUESTION [6 upvotes]: I have two questions (somewhat related) regarding local geometry on a SMOOTH, COMPACT Riemannian manifold. I still have a hard time getting a "good" understanding of local geometry. Question 1: It is true that there exists $\epsilon>0$ such that for all $r < \epsilon$, there exists $c_g >0$ (indep. of $x\in M$) such that $$ c_g r^n \leq Vol_g (B_x(r)), \forall x \in M, $$ where $B_x(r)$ is the geodesic ball centered at $x$. In other words, is it true that small geodesic ball are "comparable" to Euclidean balls. Question 2: I am trying to compute an integral on a "small" geodesic ball, namely the following: $$ I = \int_{(exp_{x_0}(B_{0}(R))} \rho(x_0,x)^{2-n} dV_g$$ Using normal coordinates at $x_0$, we should have that \begin{eqnarray*} I & =& \int_{(exp_{x_0}(B_{0}(R))} \rho(x_0,x)^{2-n} dV_g. \\ & \leq & C_g \int_0^{R} \rho^{2-n} \rho^{n-1} (1 + O(\rho^2)) d\rho \\ \end{eqnarray*} I really need to get that last inequality but I am very unsure about it. I think it should be true, at least for small enough $R$ (hopefully for $R$ smaller than the $\epsilon$ defined in question 1. Is that possible? Any feedback would be appreciated. REPLY [10 votes]: The answer to both questions is 'yes'. To see this, just consider the exponential map $\exp:TM\to M$ and look at the pullback of the Riemannian volume form $dV$, say $\Omega = \exp^*(dV)$ on $TM$. By the usual expansion in normal coordinates, there will be a smooth function $\phi$ on $TM$ that vanishes to order $2$ along the zero section such that, for each $x\in M$, the pullback of $\Omega$ to $T_xM$, say $\iota_x^*\Omega$ satisfies $\iota_x^*\Omega = (1+\iota_x^*\phi) dV_x$ where $dV_x$ is the ordinary Euclidean volume form on $T_xM$ (considered as a Euclidean vector space). This immediately proves what you want, since you can now use compactness to show that, when $\epsilon$ is sufficiently small, one has $|\phi(v)|\le C|v|^2$ for some constant $C>0$ and all $v\in TM$ with $|v|\le \epsilon$.<|endoftext|> TITLE: Positive primes represented by indefinite binary quadratic form QUESTION [11 upvotes]: Neil Sloane asked me about commands in computer languages to find the (positive) primes represented by indefinite binary quadratic forms. So I wrote something in C++ that works. This is for the OEIS, these primes go into sequences... Note that, within a few hours, another guy had run the tables much higher with a one-line Maple command. Some days it does not pay to get up. I thought of one I really do not understand. Discriminant $205$ has four classes of forms, $$ \langle 1, 13, -9 \rangle, \; \langle -1, 13, 9 \rangle, \; \langle 3, 13, -3 \rangle, \; \langle -3, 13, 3 \rangle. $$ The third and fourth are opposites so in the same genus, although distinct. The first two are in the principal genus, but they are not opposites, one is $-1$ times the other; in particular, they get diffeent positive primes, although both do residues $\pmod 5$ and $\pmod {41}.$ For $\langle 1, 13, -9 \rangle$ we get $$ 1,5,59,131,139,241,269,271,359,409, \ldots, $$ while for $\langle -1, 13, 9 \rangle$ we get $$ 31,41,61,251,349,379,389,401,419,431, \ldots. $$ For positive forms, low class number, there are polynomials, such as in Cox's book, such that primes represented by the principal form are those for which the polynomial factors a certain way. For a prime $p \equiv 1 \pmod 3,$ Gauss showed that $2$ is a cubic residue if an only if $p = u^2 + 27 v^2.$ Jacobi showed that $3$ is a cubic residue if an only if $p = u^2 + uv + 61 v^2.$ All I found in Henri Cohen's tables was the fact that $\mathbb Q(\sqrt {205})$ has class number $2$ and $L_K = \mathbb Q(\sqrt 5),$ appendix 12C on pages 533 and 534. See related information at IT'S A LINK. Let's see, $34$ is the smallest number where it is a surprise that there is no solution to $x^2 - 34 y^2 = -1.$ The smallest such odd number is $205,$ as there is no solution to $x^2 - 205 y^2 = -1.$ For prime $p \equiv 1 \pmod 4,$ there is always a solution to $x^2 - p y^2 = -1.$ Proof in Mordell's book. Anyway, this is why $\langle 1, 13, -9 \rangle, \; \langle -1, 13, 9 \rangle$ are distinct classes. So, that is the question, can I distinguish the represented (positive) primes by factoring some polynomial mod these primes? REPLY [2 votes]: Consider the quadratic number field $K$ with discriminant $D = pq$, where $p$ and $q$ are primes $\equiv 1 \bmod 4$ (all results below hold after a suitable modification also for $p = 2$). By results due to Dirichlet and Scholz, The fundamental unit of this field has norm $-1$ if $(p/q) = -1$ $+1$ if $(p/q) = +1$ and $(p/q)_4 (q/p)_4 = -1$ $-1$ if $(p/q) = +1$ and $(p/q)_4 = (q/p)_4 = -1$ Here $(q/2)_4$ is $+1$ or $-1$ according as $p$ is congruent to $1$ or $9$ modulo $16$. In the case $(p/q) = -1$, the Hilbert $2$-class field coincides with the genus field $K^* = Q(\sqrt{p},\sqrt{q}\,)$. This is also the Hilbert $2$-class field in the usual sense if $(p/q) = +1$ and $(p/q)_4 (q/p)_4 = -1$; in this case, however, there is a quadratic extension of the genus field unramified at finite primes, which can be constructed explicitly by solving the diophantine equation $x^2 - 4py^2 = q$ and setting $K = K^*(\sqrt{\mu})$ for $\mu = x + 2y\sqrt{p}$, where the sign of $x$ is chosen in such a way that $\mu$ is congruent to a square mod $4$, i.e., such that $x + 2y \equiv 1 \bmod 4$. In this case this means that $x$ is negative. For the smallest discriminants, the corresponding elements are $D = 2 \cdot 17$: $\mu = -5 + 2 \sqrt{2}$; $D = 5 \cdot 41$: $\mu = -11 + 4 \sqrt{5}$; $D = 13 \cdot 17$: $\mu = -9 + 2 \sqrt{17}$; $D = 5 \cdot 61$: $\mu = -9 + 2 \sqrt{5}$. Noam's generator in the case $D = 205$, by the way, is $\sqrt{\varepsilon_{205}} \cdot \omega$, where $\sqrt{\varepsilon_{205}} = \frac12(3\sqrt{5} + \sqrt{41})$ and $\omega = \frac12(1+\sqrt{5})$. It is possible to construct the corresponding polynomials, but for your purpose it is better to work with the generators, as already Noam has pointed out.<|endoftext|> TITLE: Are there any simple, interesting consequences to motivate the local Langlands correspondence? QUESTION [18 upvotes]: Let's pretend that we know local Langlands at a fairly high level of generality... i.e. we know something along the lines of: Let $G=\mathbf{G}(F)$ be the group of $F$-points of a connected reductive algebraic group $\mathbf{G}$ defined over a nonarchimedean local field $F$ with separable algebraic closure $\bar{F}$. Let $W_F'=W_F\times SL_2(\Bbb{C})$ be the Weil-Deligne group of $\bar{F}/F$ and let $^LG={}^LG^0\rtimes\mathrm{Gal}(\bar{F}/F)$ be the Langlands dual of $G$, where $^LG^0$ is the connected reductive complex algebraic group with root datum dual to that of $G$. Then there exists a natural surjective map $$\mathrm{Irr}(G)\twoheadrightarrow\mathrm{Hom}(W_F',{}^LG),$$ where $\mathrm{Irr}(G)$ is the set of equivalence classes of smooth irreducible complex representations of $G$. This map has finite fibres (the $L$-packets), is "compatible" with a list of operations: parabolic induction, twisting, etc, and is the unique such map. As far as I know, this is a theorem (or maybe very close to being one for the latter two?) for $GL_N$, $SL_N$, $Sp_{2N}$ and $SO_N$. When I'm trying to explain why I'm interested in this to someone I'll give the usual explanation along the lines of "we want to understand $\mathrm{Gal}(\bar{F}/F)$, local class field theory lets us understand the abelianisation of it in the form $F^\times\simeq W_F^{\mathrm{ab}}$, LLC generalises the dual form of this to a nonabelian setting and should tell us an awful lot about $\mathrm{Gal}(\bar{F}/F)$". Obviously you aren't going to hit them with the correspondence as stated above, but you can usually get away with saying "smooth irreps of $GL_N(F)$ naturally correspond to $N$-dim complex reps of $W_F'$, and that should generalise in a reasonable way to other groups". At this point, I'll usually have either satisfied my questioner, or they'll ask me if I can give an example of what the LLC should let us do. That's when I run in to trouble -- I don't know of a single, reasonably simple, appealing application of it. In the global case people often bring up the proof of FLT. This isn't exactly "simple", but it's at least well known and can be summarised as "if FLT doesn't hold we have a non-modular semistable elliptic curve. Wiles then uses Langlands-Tunnell as a starting point, does a lot of work and eventually shows that every semistable elliptic curve is modular, hence FLT". So... are there any such good examples of applications of the local correspondence? REPLY [13 votes]: In the case of $GL_{N}$, the $L$-packets are a non-issue, and the surjective map in the local Langlands correspondence becomes a bijection. At that point, we can think of allowing the information to flow the other way. Here's a simple application. Let $f(z)$ be a classical modular form of weight $4k+2$ for the group $\Gamma_{0}(4)$ (that is also a cusp form, in the new subspace, and is an eigenform of all the Hecke operators). If $L(f,s)$ is the $L$-function for $f(z)$, what is the sign of the functional equation for $L(f,s)$? The sign of the functional equation is always $1$, for the following reason. It is determined by the local components of the automorphic representation $\pi$ attached to $f$, and we only have to worry about the local components at $\infty$ (which is a discrete series representation that contributes a factor of $1$ to the sign because the weight is $\equiv 2 \pmod{4}$), and the local representation $\pi_{2}$ at $2$. The fact that the level of the modular form is $4$ shows that $\pi_{2}$ corresponds (under local Langlands) to a representation $\rho : W_{\mathbb{Q}_{2}} \to GL_{2}(\mathbb{C})$ that comes from a character $\chi$ of $W_{K}$, where $K = \mathbb{Q}_{2}(\omega)$ is the unramified quadratic extension of $\mathbb{Q}_{2}$, and that this character has order $6$. It follows that $\rho$ comes from an $S_{3}$ extension of $\mathbb{Q}_{2}$, and it turns out that there is a unique $S_{3}$ extension of $\mathbb{Q}_{2}$. From this, $\rho$ and hence $\pi_{2}$ is uniquely determined, and it turns out that the local root number of $\pi_{2}$ is also $1$. (This fact was also observed by Atkin and Lehner in 1970, but the explanation above gives a more conceptual reason for it to be true, in my opinion.)<|endoftext|> TITLE: Enumeration of a finite group QUESTION [23 upvotes]: Let $G=\{g_1,g_2,...,g_n\}$ be a group with $e=g_1$ and $n$ is odd, Set $$a_1=g_1$$ $$a_2=g_1g_2$$ $$a_3=g_1g_2g_3$$ $$a_n=g_1g_2...g_n$$ I am looking for example that all $a_i$ are different from each other i.e. $G=\{a_1,a_2,...,a_n\}$. By the way it is clear that $a_i\neq a_{i+1}$. Note I had asked this question there but I think it is suitable for here. I require that $n$ is odd since there are many example when $n$ is even yet I found no example in the case $n$ is odd. REPLY [49 votes]: A group $G$ with the property is called sequenceable. For a survey, see this paper by M. A. Ollis, which also tells that sequenceable groups are related to constructing row-complete latin squares. It is conjectured by Keedwell that $D_6,D_8$ and $Q_8$ are the only non-abelian non-sequenceable groups (see page 17); in particular, there should be none with odd order. It is known that an abelian group is sequenceable if and only if it has a unique element of order 2 (see page 5 for a proof). The article gives a list of groups that are known to be sequenceable. Apparently the question is not completely solved even in the case where $|G|$ has two prime factors. However, some groups of odd order are known to be sequenceable, so an example to your question would be for instance the non-abelian group of order 21 (page 5; there are other examples as well).<|endoftext|> TITLE: Positive Primes represented by an indefinite binary form, reducing poly degree from 8 to 4 QUESTION [5 upvotes]: In his lovely answer at Positive primes represented by indefinite binary quadratic form Noam found that a (positive) odd prime $p$ is represented by the indefinite form $x^2 + 13 x y - 9 y^2$ if and only if $$ x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1 $$ has a root $\pmod p;$ indeed, in that case the polynomial factors into linear factors, distinct if $p > 5.$ Now, there is the possibility (I am not sure at all) of finding a polynomial $f(x)$ of degree 4 in the sense of Cox, page 180, Theorem 9.2: if an odd prime $p$ does not divide $205,$ then we have an integral expression $p = s^2 + 13 s t - 9 t^2$ if and only if $(205|p) = 1$ and $f(x) \equiv 0 \pmod p$ has a solution. After looking again at Kronecker's result on page 88, $p = x^2 + 31 y^2$ if and only if $$ (x^3 - 10x)^2 + 31 (x^2 -1)^2 \equiv 0 \pmod p $$ has a root, compared with the cubic (from Hudson and Williams 1991) $(-31|p) = 1$ and $x^3 + x + 1$ factors completely, I looked and found this: if $u = x^4 + x^2 + 2$ and $v = x^2 + 3,$ we get $$ \color{magenta}{ u^2 + 13 u v - 9 v^2 \; \; = \; \; x^8 + 15 x^6 + 48 x^4 + 15 x^2 + 1}. $$ Despite understanding none of this, I think that easy identity means something. So, there is the question, can we go from degree 8 with no congruence conditions, down to degree 4 with $(205|p) = 1?$ Extra: here is Franz's answer, all those years ago. Note the use of the word "compositum." People don't use a word like that unless they really mean it. REPLY [2 votes]: I think the answer to the question as asked is no, but with $(205|p)=1$ it is yes, for then taking subfields of the Elkies field, look at $x^4 +x^3 + 3x^2 + 2x + 4$.<|endoftext|> TITLE: looking for reference on dihedral, tetrahedral, or octahedral forms QUESTION [5 upvotes]: I am looking for a reference on dihedral, tetrahedral, or octahedral forms. As far as I read, they are some cuspidal automorphic forms on $GL(2)$ induced from $GL(1)$. Dihedral is from $GL(1)/K$ to $GL(2)/\mathbb{Q}$, where $K$ is a quadratic extension of $\mathbb{Q}$. But is this the same for tetrahedral or octahedral forms? What's their constructions? REPLY [3 votes]: I think a basic reference is Serre's paper "Modular forms of weight one and Galois representations” in Algebraic Number Fields, ed. by A. Frohlich, ̈Academic Press, 1977, 193–268; also in OEuvres, Vol. III, Springer- Verlag, Berlin, 1986, 292–367.<|endoftext|> TITLE: Elliptic curve and Galois representation QUESTION [5 upvotes]: For an elliptic curve $E$ over ${\Bbb{Q}}$, let us consider Serre's mod $l$ representation by $\rho_{E,l} \colon {\mathrm{Gal}}({\overline{\Bbb{Q}}}/{\Bbb{Q}}) \to {\mathrm{Aut}}(\phantom{}_lE) = {\mathrm{GL}}_2({\Bbb{F}}_l)$, where $\phantom{}_lE$ is the group of $l$-torsion points on $E$. We say $\rho_{E,l}$ is ``finite" at prime $p$, which is equivalent to that $p$ is unramified if $p \not= l$. For $p = l$, it is equivalent to that there is a ${\Bbb F}_l$-vector space scheme $H$ over ${\Bbb Z}_p$ such that $H(\overline{\Bbb{Q}}_p)$ gives the representation of ${\mathrm{Gal}}(\overline{\Bbb{Q}}_p/\Bbb{Q}_p)$ when $\rho_{E,l}$ is restricted to ${\mathrm{Gal}}(\overline{\Bbb{Q}}_p/{\Bbb{Q}}_p) \subset {\mathrm{Gal}}(\overline{\Bbb{Q}}/\Bbb{Q})$. Q: Are these conditions equivalent to that $E$ has good reduction at $p$? Pierre REPLY [10 votes]: None of these conditions implies that $E$ has good reduction at $p$. Consider, for instance, the elliptic curve $E = X_0(11)$, for which $E[5] \cong \mathbb{Z}/5\mathbb{Z} \oplus \mu_5$. Then for $l = 5$, the representation $E[l]$ is unramified at every prime $p \neq l$ and is finite flat at $p = l$, but $E$ has bad reduction at $p = 11$. Your question is purely local (no need to assume that $E$ is over $\mathbb{Q}$). It is instructive to work out for oneself the case of split multiplicative reduction ("Tate curve") to get a clear picture why the answer is "no".<|endoftext|> TITLE: What goes wrong to use "bend-and-break" trick for singular varieties? QUESTION [5 upvotes]: When $X$ is a smooth projective variety, one can use Mori's bend-and-break trick to establish the cone theorem. However, when $X$ has singularity (say klt. singularity), the cone theorem is obtained by a series of hard results: vanishing theorem -> non-vanishing theorem -> rationality theorem -> cone theorem. I was wondering what would go wrong when we follow the argument in the smooth case with some simple minded modifications --- like using resolution or cyclic covering? Moreover, the bend-and-break lemma itself does not require any smoothness. REPLY [6 votes]: When we use the bend-and-break technique in the proof of the Cone Theorem, we not only need to know that under certain conditions there are rational curves through a point of our variety $X$, but we also require an upper bound on their degree (with respect to a given polarization $H$). Such a bound is only available when $X$ is smooth. See [Debarre, Higher dimensional algebraic geometry, Theorem 3.6 p. 67] for the result that we need. See also the introductions to Chapters 6 and 7 of the same book for a discussion about the different approach to the proof of the Cone Theorem in the smooth and in the singular case.<|endoftext|> TITLE: Sumsets and a bound QUESTION [8 upvotes]: Let $q$ be a positive integer. Is it true there exists a constant $C_q$ such that the following inequality holds for any finite set $A$ of reals: $$\displaystyle |A+qA|\ge (q+1)|A|-C_q\qquad (1)$$ I got this idea from the well-known inequality $|A+A|\ge 2|A|-1$, so I was thinking about the general case, but no idea about it. I experimented a bit with small values of $|A|$ and it seems to be true. This paves the road for a far more general inequality $$ |kA+\ell A|\ge (k+\ell)|A|-C_{k,\ell}\qquad (2)$$ with $k,\ell$ integers and $\text{gcd}(k,\ell)=1$. So, an answer would be welcome. A solution of $(1)$ is the thing I ask for, $(2)$ is just a bonus. And I haven't checked $(2)$ for small values yet. Anyways proving or disproving them both will be helpful. Thanks a lot. REPLY [12 votes]: Inequality (2) is true provided $(k,\ell)=1$ - this is a recent result of Balog and Shakan in 'On the sum of dilates of a set', http://arxiv.org/pdf/1311.0422.pdf. They show that for any finite $A\subset\mathbb{Z}$ and integers $k,\ell$ such that $(k,\ell)=1$. $$ \lvert kA+\ell A\rvert \geq (k+\ell)\lvert A\rvert - (k\ell)^{(k+\ell-3)(k+\ell)+1}. $$<|endoftext|> TITLE: Number of zeros of a polynomial in the unit disk QUESTION [22 upvotes]: Suppose $m$ and $n$ are two nonnegative integers. What is the number of zeros of the polynomial $(1+z)^{m+n}-z^n$ in the unit ball $|z|<1$? Some calculations for small values of $m$ and $n$ suggests the following formula: $$n-1+2\left\lfloor\frac{m-n+5}{6} \right\rfloor$$ Does this formula work for all values of $m$ and $n$? REPLY [8 votes]: Here is my solution of this problem. So we have next equation for the zeros: $$ (1+z)^{n+m}=z^n $$ We can modify it like that: $$ \Bigl(1+\frac{1}{z}\Bigr)^n \Bigl(1+z\Bigr)^m = 1 $$ Next we can mark the first factor as $re^{i\varphi}$, so the equation splits into two ones: $$ \Bigl(1+z\Bigr)^m = re^{i\varphi} \\ \Bigl(1+\frac{1}{z}\Bigr)^n = \frac{1}{r}e^{-i\varphi} $$ Let's consider the first equation. We have to extract the root: $$ 1 + z = r^{\frac{1}{m}} e^{i\frac{\varphi + 2 \pi k}{m}} $$ Next we take away $1$ and write down the module of $z$ using the condition $|z|<1$: $$ |z|^2 = |r^{\frac{1}{m}} e^{i\frac{\varphi + 2 \pi k}{m}} - 1|^2 = r^{\frac{2}{m}} - 2 r^{\frac{1}{m}} \cos \frac{\varphi + 2 \pi k}{m} + 1 < 1 $$ Finally, we divide by $r^{\frac{1}{m}}$, so: $$ r^{\frac{1}{m}} < 2 \cos \frac{\varphi + 2 \pi k}{m} $$ At the same time we know that $r^{\frac{1}{m}}<1$, so we have to understand, which inequation is stronger. Let's mark $z=\rho e^{i\psi}$ and write down the following inequation: $$ \Bigl|1 + \frac{1}{z}\Bigr|^2 = \rho^{-2} + 2\rho^{-1} \cos \psi + 1 > 1 $$ Next, $$ \rho \cos \psi + 1 > \frac{1}{2} $$ It's clear that the left part equals $\Re \sqrt[m]{re^{i\varphi}}$, so: $$ r^{\frac{1}{m}} \cos \frac{\varphi + 2 \pi k}{m} > \frac{1}{2} $$ Because of $r^{\frac{1}{m}} < 1$ we have: $$ \cos \frac{\varphi + 2 \pi k}{m} > \frac{1}{2} $$ Therefore $r^{\frac{1}{m}} < 1$ is stronger and we have $\cos \frac{\varphi + 2 \pi k}{m} > \frac{1}{2}$. Next, we perform the same procedure for the second equation, and finally obtain another inequation: $$ \cos \frac{-\varphi + 2 \pi k'}{n} < \frac{1}{2} $$ Let's transform these inequations: $$ -\frac{m}{6} < \frac{\varphi}{2\pi} + k < \frac{m}{6} \\ \frac{n}{6} < -\frac{\varphi}{2\pi} + k' < \frac{5n}{6} $$ We have to take away $\varphi$, so: $$ -\frac{m}{6} - \frac{\varphi}{2\pi} < k < \frac{m}{6} - \frac{\varphi}{2\pi} $$ And: $$ -\frac{m}{6} + \frac{n}{6} - k' < k < \frac{m}{6} + \frac{5n}{6} - k' $$ So, finally: $$ \frac{n-m}{6} < \ell < \frac{5n+m}{6} $$ where $\ell = k + k'$. The number of $\ell$'s satisfying this inequation defines the number of zeros in the unit disk.<|endoftext|> TITLE: What is a metaboliser? QUESTION [7 upvotes]: What is a "metaboliser" for a linking form? This term appears in a recent paper on the Hopf link and I cannot find any definition on the net or in any texts. I am a professional topologist trying to review a paper. REPLY [3 votes]: There is a definition on Wikipedia. The metabolizer of a bilinear form $\beta(x,y)$ on a vector space $V$ is a subspace $W$ that equals its orthogonal complement $W^{\perp}$. The bilinear form itself is then called metabolic. The metabolizer of a metabolic bilinear form is sometimes referred to as its lagrangian, see paragraph 1.C on page 17 of The Algebraic and Geometric Theory of Quadratic Forms. For applications of this concept in the context of knot theory (which I presume is the relevant context in your case), see Some examples related to knot sliceness, section 2.1, and Metabolic and hyperbolic forms from knot theory, section 2.<|endoftext|> TITLE: Obstructions to the existence of stable (and unstable?) complex structures? QUESTION [12 upvotes]: Let $V$ be a real vector bundle on a space $X$, perhaps the tangent bundle of a smooth compact manifold. I'm interested in understanding the obstructions to $V$ admitting a stable complex structure, and also understanding how much this differs from the unstable story (if we ask for $V$ to admit a complex structure). Here are some things I know. On the one hand, there are necessary conditions coming from characteristic classes. Namely, the odd Stiefel-Whitney classes $w_{2k+1}(V) \in H^{2k+1}(X, \mathbb{F}_2)$ must vanish, and the even Stiefel-Whitney classes must be in the image of the reduction map $H^{2k}(X, \mathbb{Z}) \to H^{2k}(X, \mathbb{F}_2)$, or equivalently the odd integral Stiefel-Whitney classes $\beta w_{2k}(V) = W_{2k+1}(V) \in H^{2k+1}(X, \mathbb{Z})$ must vanish, where $\beta$ is the Bockstein map $H^{2k}(X, \mathbb{F}_2) \to H^{2k+1}(X, \mathbb{Z})$. I don't know if this is sufficient in general. It certainly doesn't suffice for complex, rather than stable complex, structures, since even spheres have no odd cohomology but already $S^4$ doesn't have an almost complex structure. On the other hand, by obstruction theory we need to look at the fibration $$O/U \to BU \to BO$$ since lifting the stable classifying map $X \to BO$ of $V$ to a classifying map $X \to BU$ is equivalent to finding sections of an associated bundle with fibers $O/U$. The obstructions to doing this are cohomology classes in $H^{i+1}(X, \pi_i(O/U))$. Now, Bott periodicity implies that $O/U \cong \Omega O$, so its homotopy groups are known: they are periodic with period $8$ and the nontrivial ones are $$\pi_{8k}(O/U) \cong \pi_{8k+7}(O/U) \cong \mathbb{Z}_2$$ and $$\pi_{8k+2}(O/U) \cong \pi_{8k+6}(O/U) \cong \mathbb{Z}.$$ So there are obstructions living in $H^{8k+1}(X, \mathbb{Z}_2), H^{8k+8}(X, \mathbb{Z}_2), H^{8k+3}(X, \mathbb{Z})$, and $H^{8k+7}(X, \mathbb{Z})$. Three of these live in the same groups as the characteristic classes above so one might hope that they are in fact the same obstructions, but the obstructions living in $H^{8k+8}(X, \mathbb{Z}_2)$ don't match up. What's up with those? For the unstable picture, when $\dim V = 2n$ we need to look at the fibration $$O(2n)/U(n) \to BU(n) \to BO(2n).$$ There are induced maps $O(2n)/U(n) \to O(2n+2)/U(n+1)$ which induce isomorphisms on $\pi_k$ for (if I've calculated this correctly) $k \le 2n - 2$, so for example if we only cared about tangent bundles the stable and unstable stories almost match up except for the possibility of a mismatch involving classes in $H^{2n}(X, \pi_{2n-1}(O(2n)/U(n))$. For example, when $n = 2$ we have $O(4)/U(2) \cong S^2 \sqcup S^2$ and $\pi_0, \pi_1, \pi_2$ match up with the stable values above but $\pi_3$ is $\mathbb{Z}$ instead of being trivial. Here are some questions I have. How does the Stiefel-Whitney class story match up to the obstruction theory story? To what extent can we identify the obstructions involved in the two stories with each other? And how different are the stable and unstable obstructions? This question is closely related but I don't think it completely answers my questions. REPLY [8 votes]: I accidentally ran into this older question and noticed that the unstable situation wasn't discussed yet. So I thought I point out a couple of references discussing the obstruction classes for the existence of almost complex structures. W.S. Massey. Obstructions to the existence of almost complex structures. Bull. Amer. Math. Soc. 67 1961 559–564. (Link to paper on Project Euclid) Theorem I of Massey's paper describes a relation between the integral Stiefel-Whitney classes and the obstruction classes in integer cohomology in the stable range. The relation is that the integral Stiefel-Whitney classes are certain multiples of the obstruction classes for the existence of a complex structure. In the presence of torsion, the integral Stiefel-Whitney classes can vanish while the obstruction are non-trivial. (This would describe the relation between the Stiefel-Whitney class story and the obstruction class story in the cases $8k+3$ and $8k+7$.) Massey also describes the obstruction classes related to the first unstable homotopy group of $SO(2n)/U(n)$, living in the "mismatch group" $H^{2n}(X,\pi_{2n-1}(SO(2n)/U(n)))$ mentioned in the question. Assume that we have an almost complex structure over the $2n-1$-skeleton of the space $X$. Theorem II describes exactly the obstruction class to extend this to the $2n$-skeleton, for $n=2k$ even (which is the case where the obstruction groups have integer coefficients): the obstruction class encodes the failure of a natural relation between the Pontryagin class $p_k$ and the Chern classes of the almost complex structure on the $2n-1$-skeleton. Finally, concerning the mismatch between Stiefel-Whitney and obstruction story in the case $8k+8$. There are partial results in Theorem III of Massey's paper. Much more precise statements concerning the obstruction class in this case can be found in the following two papers. E. Thomas. Complex structures on real vector bundles. Amer. J. Math. 89 1967 887–908. (Link to paper on JSTOR) H. Yang. A note on stable complex structures on real vector bundles over manifolds. Topology Appl. 189 (2015), 1–9. (Link to paper on ScienceDirect) It seems the obstruction class for the case $8k+8$ involves discussion of quite a couple of cohomology operations. (Also check out references in these papers for further results.)<|endoftext|> TITLE: Does Turing determinacy imply full determinacy? QUESTION [13 upvotes]: The axiom of Turing determinacy is a weakening of the full axiom of determinacy, $AD$, in which only games with payoff sets which are $\equiv_T$-invariant are demanded to be determined. In "Turing determinacy and the continuum hypothesis" (published in 1989), Ramez Sami writes: "The main question so far unsettled in this particular domain can be roughly put this way: is it true that for any "reasonable" pointclass $\Gamma$ we have: Turing-Det$(\Gamma)\implies$Det$(\Gamma)$? In particular is it the case that: [over $ZF+DC$, presumably] Turing $AD$ implies $AD$?" http://link.springer.com/article/10.1007%2FBF01622874, page 153 My question is, what is the status of this question currently? Do we know whether Turing $AD$ is strictly weaker than $AD$? The only recent work I know of around Turing determinacy is from the reverse mathematical side (http://www.math.cornell.edu/~shore/papers/pdf/TDet21.pdf); I'm not at all familiar with the set theory on the subject. (I vaguely recall that Turing determinacy implies that every Suslin set is determined, but I can't remember where I supposedly learned this "fact.") REPLY [5 votes]: To show this in L(R) (or any well understood determinacy model) one apparently has to run a core model induction as in section 6.2 of https://ivv5hpp.uni-muenster.de/u/rds/core_model_induction.pdf which explains why it is open in the abstract.<|endoftext|> TITLE: Extending holomorphic functions QUESTION [10 upvotes]: Suppose $K\subset \mathbb{C}^n$ is a compact subset and $f:\mathbb{C}^n\setminus K\to \mathbb{C}$ is a holomorphic function. Then, provided $n>1$, $f$ extends to a holomorphic function defined on the whole $\mathbb{C}^n$. This is the Hartogs' extension theorem, and a proof can be found e.g. somewhere in the very beginning of Griffiths-Harris. This theorem is still true if one replaces $\mathbb{C}^n$ by a connected open subset $U$ of $\mathbb{C}^n$ (upd: and takes $K$ to be a compact subset of $U$ that does not disconnect it, or else there may be a problem even with a locally constant function), see e.g. theorem 2 in http://www.encyclopediaofmath.org/index.php/Hartogs_theorem I would like to ask if this theorem can be generalized to other complex manifolds (e.g., to Stein manifolds). If there is an answer in the algebraic case, that would be particularly interesting. REPLY [11 votes]: Yes, there are several generalizations of Hartogs Extension Theorem that hold on Stein spaces. For a good survey you can look at the paper by Øvrelid and Vassiliadou Hartogs Extension Theorems on Stein Spaces, Journal of Geometric Analysis 20 (2010), 817-836. Let me just recall Theorem 1.2 of that article, which is due to Rossi (Annals of Mathematics 78 (1963), 455–467). Theorem. Let $X$ be a connected, normal Stein space of dimension $\geq 2$. Let $K$ be a compact subset of $X$ and let $h$ be a holomorphic function on $X \setminus K$. Then there is a unique holomorphic function $H$ on $X$ such that $H=h$ on the unbounded component of $X \setminus K$.<|endoftext|> TITLE: A bound on a set QUESTION [5 upvotes]: Let $x_1,\cdots , x_n$ be a sequence of real number such that $x_i\geq 1$ for all $1\leq i\leq n$, $S=\{\alpha_1x_1+\cdots +\alpha_nx_n | \alpha_i\in\{0,+1,-1\}\}$ and $I=[a,b)$ be a Interval with length $2$. So I was wondering if there was any subsequent upper bound on $|I \cap S|$. Is there a general bound when $2$ is replaced by a positive real $\alpha$? Thanks in advance for your answers and comments. I haven't been able to guess anything, but this bound on size of a set reminded me of Sperner's Theorem but not sure about it. REPLY [2 votes]: Expanding on my comment and continuing GH from MO's answer: For $\alpha >2$, let $r:=\lceil\alpha/2\rceil+1$. Since a half-open interval $I$ of length $\alpha$ cannot contain $r$ numbers whose pairwise distance is at least $2$, for any fixed $A \subseteq \{1, \dots, n\}$, the set $A_+:=\{i \in A; \alpha_i = +1\}$ cannot contain an inclusion chain of length $r$. By Erdos's generalization of the Sperner Theorem, we have $$\lvert I\cap S_A \rvert \le \sum_{k=\lfloor |A|/2\rfloor-\lfloor (r-1)/2\rfloor}^{\lfloor |A|/2\rfloor+\lfloor r/2\rfloor} {\lvert A \rvert \choose k}, $$ with equality for $I=[-r+1, -r+1+\alpha)$ and $x_i=1$ for $i\in A$. Hence, $$|I\cap S|\leq \sum_{A\subset\{1,\dots,n\}}\sum_{k=\lfloor |A|/2\rfloor-\lfloor (r-1)/2\rfloor}^{\lfloor |A|/2\rfloor+\lfloor r/2\rfloor} {\lvert A \rvert \choose k} = \sum_{m=0}^n {n\choose m}\sum_{k=\lfloor m/2\rfloor-\lfloor (r-1)/2\rfloor}^{\lfloor m/2\rfloor+\lfloor r/2\rfloor} {m \choose k}, $$ with equality for $I=[-r+1, -r+1+\alpha)$ and $x_i=1$ for all $1\le i\le n$.<|endoftext|> TITLE: What's special about the Simplex category? QUESTION [28 upvotes]: I have been wondering lately what makes simplicial sets 'tick'. Edited The category $\Delta$can be viewed as the category of standard $n$-simplices and order preserving simplicial maps. The goal of simplicial sets is to build spaces out of these building blocks by gluing, and allow maps to be defined simplex by simplex, so it makes sense to take the free cocompletion of $\Delta$, the presheaf category $[\Delta^{op},\mathbf{Set}]$. The realisation functor $R : \Delta\to \mathbf{Top}$ can be readily extended to the cocompletion, so as to make $\hat{R}$ preserve colimits. So my questions are: How do we intuitively understand why $\hat{R}$ preserves finite products? (I understand that there are some subtleties with $k$-ification) What makes $\Delta$ special in this way, that fails for say $\Gamma=\mathbf{FinSetSkel}$, ie. "symmetric simplicial sets" and cubical sets? What is your philosophy of simplicial sets? REPLY [2 votes]: I feel one should also be somewhat eclectic and consider not only advantages but also comparative disadvantages of any candidate for a central categorical role. I admit to special pleading here since our 2001 EMS Tract on Nonabelian algebraic topology gives a major role for a homotopical foundation of algebraic topology to cubical sets. The point is that in the usual simplicial $\infty$-category theory the emphasis is on the Kan condition. In the cubical theory, the emphasis is on compositions. This allows for the replacement of the usual formal sums in basic homology theory by actual compositions of pieces for homotopically defined functors. See this mathoverflow discussion. Dan Kan's thesis and first (1955) paper were cubical, as best for intuition and conjectures, but severe disadvantages were found in the category of cubical sets by workers at Princeton. One disadvantage was that cubical groups, unlike simplicial groups, were not Kan complexes. Another was the realisation of the cartesian product of cubical sets, which had the wrong homotopy type, again unlike simplicial sets. So it was assumed that the cubical theory was quite unfixable. However work at Bangor in the 1970s with Chris Spencer and Philip Higgins, starting with the relation between double groupoids and crossed modules, found it necessary to introduce a new type of "degenerate" cube based on the monoid structures max and min on the unit interval $I=[0,1]$.The standard degeneracy in cubical sets yields cubes with opposite faces the same, where these new had some adjacent faces the same, and so made the theory a bit nearer to the simplicial theory. These new structures were called connections, because of a relation with path connections in differential geometry. As Philippe Gaucher points out, Andy Tonks proved in 1992 that cubical groups with connections were Kan complexes. In 2005, Georges Maltsiniotis proved that the geometric realisation of of the cartesian product of cubical sets with connections has the correct homotopy type. There are two main reasons for using cubical sets. One is the formula $I^m \times I^n \cong I^{m+n}$, which makes for a good tensor product of cubical sets, and also allows a convenient and direct definition of homotopies. The second and for our purposed major reason for using cubical sets is expressed in the slogan "algebraic inverses to subdivision", which is an elaboration of the idea of composition in cubical sets. The standard singular cubical set $S^{\Box} X$ of a topological space ie easily equipped with $n$ partial compositions in dimension $n$ giving $S^{\Box} X$ the structure of weak $\infty$-groupoid. A simple matrix/array notation $[a_{(r)}]$ allows for the expression of multiple partial compositions of $n$-cubes which are compatible in all the directions. This gives meaning to "algebraic inverses to subdivision". This is one of the basic intuitions behind the first proofs of higher versions of the Seifert-van Kampen Theorem. It is not at all clear to me how these ideas can be expressed in simplicial terms. There is further discussion of the intuition of these uses of cubical sets in talks I gave in Paris on June 5, 2014, and Galway, December 2014, available on my preprint page. March, 2015: I have since learned that cubical sets with connection are used in the theory of motives in this paper. It seems that the degeneracies in cubes work better here than in simplices. Another paper compares cubical and simplicial derived functors. Both these papers use cubical sets with connections, which were introduced in JPAA 1981 by Philip Higgins and me. These additional kind of degeneracies go some way to correcting some well known deficiences of the usual cubical sets, in terms of cubical groups and geometric realisations of cartesian products - see the talks referred to above. <-------><|endoftext|> TITLE: Number of Richardson orbits in simple Lie algebras of types $E_n$? QUESTION [7 upvotes]: This is a follow-up to my question about nilpotent orbits here asked in connection with an earlier discussion of symplectic resolutions. Leaving aside the connections with algebraic geometry and representation theory, I'm curious about what has been written down explicitly for the exceptional Lie algebras $E_6, E_7, E_8$. Start with a simple Lie algebra $\mathfrak{g}$ of rank $n$ over $\mathbb{C}$ (or the Lie algebra of a simple algebraic group $G$ over any algebraically closed field of good characteristic). The finitely many nilpotent orbits in $\mathfrak{g}$ (or unipotent classes in $G$) have been well studied, with expositions for example in the books by Carter (1985) and Collingwood-McGovern (1993). Carter provides detailed tables in 13.1 and partial order graphs in 13.4 (though the $E_7, E_8$ graphs have a few missing edges compared to the correct pictures in Spaltenstein's lecture notes). In particular, it's easy for exceptional types to compare the total number of orbits, labelled by Dynkin or Bala-Carter, with the number of special orbits in Lusztig's sense. While the special orbits are defined only indirectly via several kinds of representation theory, all Richardson orbits are special. The converse starts to fail in type $D_4$, where there is one special but non-Richardson orbit. Richardson orbits belong to one or more of the $2^n$ conjugacy classes of parabolic subalgebras in $\mathfrak{g}$; for instance, the subregular nilpotent orbit intersects the nilradical of each minimal parabolic $\mathfrak{p}$ in its dense orbit under the corresponding adjoint subgroup $P$. In type $F_4$ there are 16 orbits, 11 special and 9 of these Richardson. Has anyone computed the full list of Richardson orbits in types $E_6, E_7, E_8$? As noted earlier, Hirai's paper here gives "generators and relations" for producing such a list, along with explicit information about Richardson orbits with "non-even" Dynkin labelling. But for example, what is the list of Richardson orbits coming from the 256 classes of parabolics for $E_8$? One knows that there are 70 orbits, 46 of which are special; which of the latter are Richardson? REPLY [4 votes]: Isn't this question a subset of the induction question (i.e. if we induce an orbit from a Levi subalgebra up to ${\mathfrak g}$, then what orbit do we get)? This was solved by Elashvili (for exceptional types) in 1979, and his computations were verified in a 2009 joint Elashvili-de Graaf paper ("Induced nilpotent orbits of the simple Lie algebras of exceptional type", in the Georgian Mathematical Journal but also on the arXiv). By a standard result about Lusztig-Spaltenstein induction, two parabolic subalgebras with the same Levi subalgebra give the same induced orbit, so in fact we get a better bound than 2^n on the number of Richardson orbits: the number of Levi subalgebras up to conjugacy. In type F4 the Levi subalgebras are of type: maximal torus, $A_1$, $\tilde{A}_1$, $A_1+\tilde{A}_1$, $A_2$, $B_2$, $\tilde{A}_2$, $A_2+\tilde{A}_1$, $A_1+\tilde{A}_2$, $B_3$, $C_3$, $F_4$. The corresponding Richardson orbits are: regular, subregular, subregular, $F_4(a_2)$, $C_3$, $F_4(a_3)$, $B_3$, $F_4(a_3)$, $F_4(a_3)$, $\tilde{A}_2$, $A_2$, $0$. The special orbits which aren't Richardson are $A_1+\tilde{A}_1$ and $\tilde{A}_1$. In type $E_6$, the following orbits with non-even weighted Dynkin diagrams are Richardson: $D_5(a_1)$, $A_4+A_1$, $A_3$, $A_2+2A_1$, $2A_1$. These are induced respectively from the zero orbit in Levi subalgebras of type $A_2+A_1$, $A_2+2A_1$, $A_3+A_1$, $A_3+2A_1$ and $D_5$. So we get (I think) 15 Richardson orbits, and the two special orbits which aren't Richardson are $A_1$ and $A_2+A_1$. I think that the number of Richardson orbits in type $E_7$ (resp. $E_8$) is 29 (resp. 34), but I could easily have made a counting error.<|endoftext|> TITLE: Homotopy spheres with vanishing and non-vanishing $\alpha$-invariant QUESTION [12 upvotes]: I'm unsure whether this question is appropriate for mathoverflow, so feel free to criticize. All manifolds are closed, smooth and have dimensions $n\ge 5$. The Atiyah-Shapiro-Bott-Orientation gives a ring homomorphism $$\alpha\colon\Omega_*^{spin}\rightarrow KO^{-*}(pt),$$ from the spin-bordism ring to real K-theory, whose vanishing is a necessary condition for a spin manifold admitting a metric of positive scalar curvature. Recall $$KO^{-n}(pt)\cong \begin{cases} \mathbb{Z} &\mbox{if } n \equiv 0,4 (8)\\ \mathbb{Z}/2\mathbb{Z} & \mbox{if } n \equiv 1,2(8) \\ 0 &\mbox{if } n \equiv 3,5,6,7 (8). \end{cases}.$$ In dimensions $n\equiv0,4(8)$, $\alpha$ is just the $\hat{A}$-genus (respectively twice of it). Considering the spin-cobordism class of homotopy spheres, the $\alpha$-invariant induces homomorphisms $$\beta_n\colon\Theta_n\rightarrow KO^{-n}(pt),$$ ($\Theta_n$ is the group of $n$-dimensional homotopy spheres), which are zero in dimensions $n\equiv 3,5,6,7 (8)$ for trivial reasons. In the other dimensions, we have $\beta_n$ is zero in dimensions $n\equiv0,4(8)$, what is equivalent to the vanishing of the $\hat{A}$-genus for all homotopy spheres. $\beta_n$ is surjective in dimensions $n\equiv1,2(8)$. I am trying to better understand these two claims. Lawson and Michelson simply write in their book "Spin Geometry", that (2) follows from "deep work of Adams and Milnor". Since I am absolutely not an expert in this field, can someone elaborate a bit (more than "It follows from the work of Adams and Milnor.", what is not really helpful for me.) on what I really need to prove (2) and how one can place it in a wider context? How can one prove (1) and is there a reference for it? I searched the literature without good results, so simply giving references might also answer my question. REPLY [16 votes]: If $M$ is a homotopy sphere of dimension $4k>0$, then the signature is clearly zero. By the Hirzebruch signature theorem, you get $0=\langle L_k (TM); [M] \rangle = b_k \langle p_k (TM); [M] \rangle$ for a certain number $b_k \neq 0$. Therefore, the Pontrjagin classes of $TM$ are all trivial, and hence the $\hat{A}$-genus is zero. This settles part (1). Part (2) is harder. Adams proved in his $J(X)$ papers (it is one of the main results stated in the introduction to part IV) that the unit map from the sphere spectrum to $KO$ is surjective in homotopy of degrees $8k+1$ and $8k+2$. The unit map $\mathbb{S} \to KO$ factors as $$\mathbb{S} \to MSpin \to KO$$ with the first map the unit and the second the Atiyah-Bott-Shapiro orientation. All that is needed to show this claim is that the $\alpha$-invariant of a point is $1$ (this is a statement about the homomorphism $\pi_0 (MSpin) \to \pi_0 (KO)$). Therefore, there exists framed manifolds of dimension $8k+1$ and $8k+2$ whose $\alpha$-invariants (as spin manifolds) are nonzero. Now use the work of Kervaire-Milnor. They prove in ''Groups of homotopy spheres'' that each odd-dimensional framed manifold is framed cobordant to a homotopy sphere, if the dimension is at least $5$. Take a framed manifold of dimension $8k+1$ with nonzero $\alpha$ and replace it by a homotopy sphere cobordant to it. Therefore, $\beta_n$ is surjective if $n = 8k+1$ (for $k=0$, this is also true). In dimensions $8k+2$, a framed manifold is framded cobordant to a homotopy sphere if and only if its Kervaire invariant is zero. This is why I do not see a short argument in that case. EDIT: let me remark that $\beta_2$ is not surjective: the unique spin structure on $S^2$ has $\alpha=0$; to get a nonzero $\alpha$, you need the torus. EDIT II: in dimensions 8k+2, the argument (due to Milnor, I believe) is as follows. Take a homotopy sphere $\Sigma^{8k+1}$ with nonzero $\alpha$-invariant, and take the product $M=\Sigma \times S^1$, where $S^1$ has the spin structure with nonzero $\alpha$-invariant. Then $M$ has nonzero $\alpha$-invariant as well. Then do surgery on the embedded $x \times S^1$; the result is a homotopy sphere, and the $\alpha$-invariant is unchanged.<|endoftext|> TITLE: Analytic solution of a system of linear, hyperbolic, first order, partial differential equations QUESTION [5 upvotes]: In a try to solve a physical problem, I've faced a system of first-order partial differential equations of the form $$\cos\left(t\right)\partial_{x}\mathbf{u}+\sin\left(t\right)\partial_{y}\mathbf{u}+\mathbf{A}\partial_{t}\mathbf{u}+\mathbf{B}\left(t\right)\mathbf{u}=0$$ with $\left(x,y,t\right)$-independent matrix $\mathbf{A}$ and $t$-dependent / $\left(x,y\right)$-independent $\mathbf{B}$. Also, $t\in\left[0,2\pi\right]$ is not really a time, but an angle, and $\mathbf{B}\left(t+2\pi\right)=\mathbf{B}\left(t\right)$ is periodic, so I guess the solutions will be periodic as well. Actually $\mathbf{A}$ is real and diagonal, so my system of equations seems to be in a canonical form from the beginning. Note nevertheless that two eigenvalues of $$\mathbf{A}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 \end{array}\right)$$ are degenerate, so I wonder if the system is still hyperbolic (I know it is no more strictly hyperbolic according to Courant (see references below), but Wikipedia defines it as hyperbolic / I've been unable to find the Wikipedia definition in any textbook I've put an hand on). I would like (if possible) to obtain analytic (in a close form or as a series expansion) solutions of this problem, or understand a bit better which perturbation scheme I could use. It seems to me this system is extraordinary simple, and at the same time I'm paralysed by my ignorance in powerful enough methods. So my first and general question would be : which literature would you recommend to me ? about this problem. Giving more details: for the moment I've found R. Courant and P. Lax On nonlinear partial differential equations with two independent variables Commun. Pure Appl. Math. 2, 255 (1949). (beyond a paywall) where section 3 seems of interest for me, it shortly discusses an old method by O. Perron Über Existenz und Nichtexistenz von Integralen partieller Differentialgleichungssysteme im reellen Gebiet Math. Zeitschrift 27, 549 (1928). (also beyond a paywall) which seems to answer my problem, giving analytical solution of system of semi-linear partial differential equation. The book by R. Courant and D. Hilbert, Methods of Mathematical Physics, Volume II: Partial Differential Equation (John Wiley and Sons, 1962). seems too much related to non-linear systems, and I've been unable to figure out what to do with my problem. Finally a short lecture note by E. Kersalé, Analytic Solutions of Partial Differential Equations http://www1.maths.leeds.ac.uk/~kersale/teaching.html (2003) ends up with a short discussion of my problem, but I still do not figure out what to do with the system of ordinary differential equations once fond the characteristics lines / surfaces... (actually, two characteristics are circles, and two straight lines are degenerate). So are there other (perhaps better or more specific for physicists) textbook/notes you would recommend ? Thanks in advance for any remark aiming at improving this question. REPLY [3 votes]: For simplicity, consider solutions where $ u$ does not depend on $x, y$: $A u_t + B(t) u = 0$. If $y^T A = 0$, that says $y^T B(t) u = 0$, so $u$ is restricted to belong to a certain (possibly $t$-dependent) subspace. Thus for your example $$A = \pmatrix{1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & -1\cr}$$ $(B(t) u)_2 = (B(t) u)_3 = 0$. If the appropriate $2 \times 2$ submatrix of $B(t)$ is invertible, this lets you express $u_2$ and $u_3$ in terms of $u_1$ and $u_4$, and you get a periodic linear system for $u_1$ and $u_4$. The solutions are usually not periodic in $t$. Rather, the linear operator $u(0) \to u(2\pi)$ will have eigenvalues $\lambda$ corresponding to solutions where $u(2\pi) = \lambda u(0)$ (see Floquet theory). Somewhat more generally, solutions of the form $u(x,y,t) = \exp(\alpha x + \beta y) v(t)$ lead to the same type of system (with $\alpha \cos(t) + \beta \sin(t)$ added to $B$). EDIT: If the submatrix of $B(t)$ is not invertible for some $t$, you may find that some or all of the nontrivial solutions have singularities at those $t$.<|endoftext|> TITLE: Review of Tim Maudlin's New Foundations for Physical Geometry QUESTION [11 upvotes]: Tim Maudlin, a philosopher of science at NYU, has a book out called: New Foundations for Physical Geometry: The Theory of Linear Structures. The section on about the book says the following: Topology is the mathematical study of the most basic geometrical structure of a space. Mathematical physics uses topological spaces as the formal means for describing physical space and time. This book proposes a completely new mathematical structure for describing geometrical notions such as continuity, connectedness, boundaries of sets, and so on, in order to provide a better mathematical tool for understanding space-time... The Theory of Linear Structures replaces the foundational notion of standard topology, the open set, with the notion of a continuous line. The last line in the quote above caught my eye and a cursory reading of one of the chapters (7: Metrical Structures) on Google Books set off a few alarms. But, I am far from a mathematical physicist and my searches of reviews were fruitless, so my question is: Is there a link to a review of this book or what is the considered opinion about it among mathematical physicist? edit 1 I'll understand if this is closed. I suddenly realised that I am effectively indulging in what our friends in the sociology department love to call 'policing the boundaries' of our science. edit 2 Looking at how things are, I'd vote to close this question too if I could (without intending any offence to those who have participated in the discussion). However, the discussion below made me peek superficially into the history of things. Evidently, the foundations of pont-set topology as we understand it now was established by the 1920s, born from considerations in analysis, it began with Frechet's 1904 thesis, where he based an abstraction of the euclidean space on the concept of limits. It is interesting to note that Ricci and Levi-Civita's Methods de calcul differential absolu et leurs applications was published in 1901 for work done in the previous decade. REPLY [5 votes]: I have to confess that I hadn't heard about the book or its author until now, but as far as I can tell from what's available on Google, this particular volume is a book about mathematics, so I hope it gets reviewed as such. I did notice that Mathscinet has listed this book with a review pending. (Although it would best to wait for a proper review by someone who has access to the whole book, I have to say that I find some claims in the parts that I have looked at, such as "Neither Decartes nor Newton would have recognized the existence of irrational or negative numbers…", or "There may have been loose talk about irrational or negative numbers, but no rigorous arithmetical foundation for them existed. The challenge was taken in 1872 by Richard Dedekind." in the introduction a little dubious.) REPLY [4 votes]: I'm not a mathematical physicist---I work in quantum computing theory, which maybe is sort of close if you squint? FWIW, I read the first few chapters of Maudlin's new book and liked them a lot. I remember taking topology as an undergrad and thinking, "why is everything based around 'open sets,' which can be chosen totally arbitrarily except that they have to be closed under unions and finite intersections?" I mean, yes, you can build up a theory on that basis and it works very well. But the notion of open set never impressed itself on me as intuitively central, the way most other basic mathematical notions did---especially given that one can easily define "open sets" (for example, in finite spaces) that have nothing whatsoever to do with the intuitive concept of "openness" that supposedly motivated the definition in the first place. So I wondered: would it be possible to build up topology on some completely different basis? This is the main question that Maudlin sets out to answer (affirmatively) in this book. And it's a big undertaking, and one that many people will probably regard as quixotic and unnecessary even if it succeeds---which might be why no one tried it before (or maybe they did; I can't say for certain about that). In the preface, Maudlin compares his situation to that of someone who realizes that the Empire State Building would've been better if it had been built a few feet to the left: even if that's true, it's far from obvious that it's worth the effort now to move the thing! But I, for one, am happy to see someone probe the foundations of topology in this way---especially someone who writes as clearly as Maudlin, so that I can actually understand where he's going and why. Physics won't be covered until the second volume. I honestly don't know yet whether there are any real applications to physics, but if there are, one could regard them as just icing.<|endoftext|> TITLE: Finite lattices whose number of join-irreducibles does not exceed its height QUESTION [5 upvotes]: In a finite distributive lattice $L$ one has $height(L) = |J(L)|$ i.e. the size of the largest chain equals the number of join-irreducible elements. Briefly, this follows by arranging the subposet $J(L) = \{x_1,\dots,x_n\} \subset L$ so that $j > i \implies x_j \nleq x_i$, and then observing that $0 < x_1 < x_1 \lor x_2 < \dots < x_1 \lor \dots \lor x_n=1$ because join-irreducible elements are join-prime in distributive lattices. What other natural classes of finite lattices $L$ satisfy $|J(L)| \leq height(L)$? For example, do join-semidistributive lattices have this property? Many thanks. REPLY [4 votes]: Join semi-distributive lattices don't have this property because weak order on $S_n$ is join semi-distributive and doesn't have this property. (Eg, for $n=3$.) Lattices satisfying the property you are interested in are called "join-extremal" by George Markowsky, in a paper Primes, irreducibles and extremal lattices. Order 9, 265–290 (1992). The Tamari lattices, and their generalization, the Cambrian lattices, all have this property.<|endoftext|> TITLE: Comparing Krein-Rutman theorem and Perron–Frobenius theorem QUESTION [9 upvotes]: Krein–Rutman theorem is a generalization of Perron–Frobenius theorem, I know that things could be more subtle in infinite dimension, yet there's an important result in Perron–Frobenius that's missing in Krein-Rutman and I don't quite understand. In Perron–Frobenius theorem, we know that for a irreducible non-negative matrix, its positive eigenvector is unique(up to scaling), corresponding to its largest eigenvalue. the analog for positive eigenfunction is not stated in Krein–Rutman theorem. So is it possible that we have a positive operator that has two positive eigenfunctions corresponding to two distinct eigenvalues? If it helps to narrow thing down, I'm interested in integral operators in $L_2(R)$ space. Thanks in advance! REPLY [11 votes]: Beware of Wikipedia! It is true that the infinite dimensional setting makes things slightly more delicate, but actually not so much. Assuming that the positive cone $C\subset X$ under consideration is solid (i-e has non empty interior) and that your operator $T:X\to X$ is compact and strongly positive (i-e maps the positive cone $C$ into its interior $\overset{\circ}{C}$), then the following stronger conclusion holds: the spectral radius is a simple eigenvalue associated with a strictly positive eigenvector $v\in \overset{\circ}{C}$, and there is no other eigenvalue associated with (non necessarily strictly) positive eigenvectors. You can find an elementary proof here (theorem 1.2)<|endoftext|> TITLE: Is the following sum irrational? QUESTION [25 upvotes]: Is the following sum irrational? $$S = \displaystyle \sum_{n \text{ squarefree}, n \geq 1} \frac{1}{n^3}$$ The sum clearly converges, so it is bounded above by $\zeta(3) = \displaystyle \sum_{n \geq 1} \frac{1}{n^3}$. In fact, since we have $$\displaystyle \sum_{n \text{ squarefree}, n \geq 1} \frac{1}{n^3} = \prod_p \left(1 + \frac{1}{p^3}\right) = \sum_{n=1}^\infty \frac{\mu^2(n)}{n^3},$$ the sum is exactly equal to $\displaystyle \frac{\zeta(3)}{\zeta(6)}$. Hence if $S$ is irrational, then it would show that $\zeta(3)$ is not a rational multiple of $\pi^6$. Since we know that $\zeta(3)$ is not rational given Apery's work, this is a slight strengthening of the result of Apery. It seems that Apery's approach should still work for $S$, but I am not sure. Does anyone know the answer or the plausibility of Apery's approach working? Edit: one notes that if we are to sum over the reciprocals of POWERFUL numbers, i.e. those numbers $n$ such that for all primes $p$ dividing $n$, there exists an integer $k > 1$ such that $p^k || n$. In particular all powerful numbers $n$ have a unique representation as $n = a^2 b^3$, where $b$ is squarefree. Hence we have $$\displaystyle T = \sum_{n \text{ powerful}} \frac{1}{n} = \left(\sum_{a=1}^\infty \frac{1}{a^2} \right)\left(\sum_{b \text{ squarefree}, b \geq 1} \frac{1}{b^3}\right) = \frac{\zeta(2)\zeta(3)}{\zeta(6)}.$$ It would be interesting to see if Apery's methods work for the sum of the reciprocals of powerful numbers as well. REPLY [4 votes]: Unfortunately (or not), this is still an open problem. Actually, problems involving product of $\pi$ by irrational numbers seem to be out of reach. A typical one is due to Nesterenko involving algebraic independence of $\pi$ and $e^{\pi}$. In this case, Apery or Beukers methods do not work (at least in my mind).<|endoftext|> TITLE: The error in Petrovski and Landis' proof of the 16th Hilbert problem QUESTION [37 upvotes]: What was the main error in the proof of the second part of the 16th Hilbert problem by Petrovski and Landis? Please see this related post and also the following post.. For Mathematical development around this historical problem please see this paper. Added : According to their method, what of the following two statements are true?: There are uniform numbers $\tilde{H}(n)$ such that every polynomial vector field $X$ of degree $n$ satisfies: Statement 1) There are at most $\tilde{H}(n)$ real limit cycles of $X$ which lie on the same leaf. statement 2) There are at most $\tilde{H}(n)$ distinct complex leaves which contains real limit cycles. By "Leaf" I mean the leaf of the corresponding complex singular foliation of $\mathbb{C}P^{2}$. Some technical and historical aspects of these foliations are explained here. However in this linked paper there is no an explicit explianation about the "error". According to the video of lecture of Ilyashenko, provided in the answer to this question by Andrey Gogolev, we ask: What is the fate of the "persistence problem" which is mentioned by Ilyashenko? How it can be revised to become a true statement? According to the first page of the english version of the paper of Petrovski_Landis we ask "How they assume that a solution of the equation can be considered as an entire map from $\mathbb{C}$ to $\mathbb{C}P^{2}$? Can every leaf be parametrized by an entire map? Does this "entire-assumption" play an important role in their proof? Please See this related post. According to comments and answers to this question, we undrestand there is no a written paper which explains the error, explicitly.Why really this is the case? REPLY [5 votes]: • Q1 (the first yellow boxed question in the OP): The current status of the "persistence problem" has been discussed by Ilyashenko in Complex length and persistence of limit cycles in a neighborhood of a hyperbolic polycycle (2014), see also Persistence Theorems and Simultaneous Uniformization (2006). As discussed on page 285 of the 2014 paper, the persistence theorem introduced by Petrovskiĭ and Landis (P&L), without a valid proof, asks whether the real limit cycle of a real planar polynomial vector field persists as a complex limit cycle for the complexification of the corresponding differential equation. Ilyashenko notes that the theorem is still unproven for the general family of polynomial equations, but does give a proof for a particular case (and references to other special cases). In general, the persistence theorem applies if a complex singular foliation of ${\mathbb C}{\mathbb P}^2$ satisfies a certain condition of "simultaneous uniformization", explained in the 2006 paper. The error in Petrovskiĭ & Landis amounts to the absence of a proof that polynomial differential equations in the plane are simultaneously uniformizable. The 2006 paper also goes some way towards an answer of the request in the OP for "a written paper which explains the error in P&L". In connection with this recent MO question asking for Examples of incorrect arguments being fertilizer for good mathematics?, the P&L paper is a good example, as also pointed out here. • historical note: according to V.Gaiko, the error in Petrovskiĭ and Landis's proof was first noted by the Belorussian mathematician N.P. Erugin.<|endoftext|> TITLE: Are there only finitely many associative algebras of fixed dimension? QUESTION [15 upvotes]: Given an algebraically closed field $F$, for any positive integer $n$, are there always only finitely many non-isomorphic (noncommutative) associative algebras (possibly without identity) with dimension $n$ over $F$? This questions is motivated by the classification of low dimensional algebras. It seems that at least when $n$ is less than 6, the answer is yes. I'm also guessing that the number of non-isomorphic classes doesn't depend on the choice of algebraically closed fileds--I've convinced myself this is true for low dimensional cases. So far I have two ideas: 1. To compute the dimension of the variety of associative algebras of dimension n, and then consider the $GL_n(F)$ action on the variety; 2. Every algebra of dimension $n$ can be embedded as a subalgebra of $M_{n+1}(F)$. But 1. is also a difficult problem for me and I don't know how to use 2. REPLY [20 votes]: Even for commutative associative algebras it is not true. The article of Björn Poonen "Isomorphism types of commutative algebras of finite rank over an algebraically closed field" gives a classification in dimension $n\le 6$ for algebraically closed fields of arbitrary characteristic, where there are only finitely many isomorphism classes. Then there are examples given in dimension $7$ of infinitely many different algebras.<|endoftext|> TITLE: Which linear combinations of simple roots are roots QUESTION [5 upvotes]: Let $\Delta$ be the root system of a complex simple Lie algebra, $\Delta^+$ be positive roots and $\Pi$ be simple roots. We view $\Pi$ as nodes of the Dynkin diagram. Then for any two simple roots $\alpha$ and $\beta$, whether a linear combination $n\alpha+m\beta$ is a root can be judged easily from the Dynkin diagram. My question is how to do this further. Precisely, Question: Given non-negative integers $(n_\alpha)_{\alpha\in\Pi}$, is there a combinatorial criterion to judge whether $\sum_{\alpha\in\Pi}n_\alpha\alpha$ is a root? Is there a recursive way to enumerate all the $(n_\alpha)$'s such that $\sum_{\alpha\in\Pi}n_\alpha\alpha$ is a root? In particular, how to determine combinatorially the $(n_\alpha)$ corresponding to the highest root? REPLY [6 votes]: My favorite answer to #2 and #3 is Kostant's "Find the highest root game", which is written up in detail in section 5.4 of Balázs Elek's notes on reflection groups. It is not hard to show that all plays of the game (from all starting positions, i.e., simple roots) take you through all the positive roots. REPLY [2 votes]: I guess the following is not exactly what you are looking for but is somewhat revealing. A former colleague of mine tackled these type of questions in his PhD thesis. He describes something that he calls "minimal relations" i.e. relations of minimal length between roots. He gives a complete description of these minimal relations in Section 4.2.3 of his thesis which I linked below http://math.jacobs-university.de/penkov/papers/PhD_Milev.pdf As far as I know, he also proves a number of criterions of when an expression of roots is again a root.<|endoftext|> TITLE: Differentiable structure on the Gauge group of a principal bundle? QUESTION [7 upvotes]: I am currently reading this paper in which we have a map $g:I\rightarrow Gauge(P)$ for some principal bundle $P$ which is differentiated. I am looking for a reference or explanation what the "most common" differentiable structure on $Gauge(P)$ would be. Thank you very much! REPLY [5 votes]: For compact base manifold and reasonably well behaved structure groups $G$ (like a finite-dimensinoal or a Banach-Lie group) there is the structure of a Frechet-Lie group on the group $Gauge(P)$. This is worked out in Chapter 1 of Lie Group Structures on Symmetry Groups of Principal Bundles (what I have called "property SUB" there is satisfied under the above conditions on $G$). It is not stated explicitly that $Gauge(P)$ is modelled on a Frechet space, but the modelling space is a closed subspace of a finite product of Frechet spaces (Proposition 1.4) and thus a Frechet space. The same is also true for the group $Aut(P)$ of all bundle automotphisms of $P$, see Theorem 2.14. The unifying theme here is that these groups can all be seen as groups of bisections in locally trivial Lie groupoids (work in progress). If you are interested in smooth maps into $Gauge(P)$, then you can take the following criterion: let $f\colon I\to Gauge(P)$ be a map and let $f_i\colon I\to C^\infty(\overline{V_i},G)$ be the corresponding maps with respect to a chosen local trivialsation $\Phi_i\colon P|_{U_i}\to U_i\times G$ (such that $\overline{V_i}$ is a manifold with corners and $P|_{\overline{V_i}}$ is also trivial to be precise). Then $f$ is smooth if and only if all $f_i$ are smooth. This is due to the fact that $Gauge(P)$ is diffeomorphic to a submanifold of $\prod_{i=1}^n C^\infty(\overline{V_i},G)$ (which in turn is is implicitly contained in the proof of Theorem 1.11). For the smoothness of maps into mapping spaces there exist many useful criteria, like smoothness of pull-back, push-forward and composition maps.<|endoftext|> TITLE: Dennis trace map K----> THH QUESTION [9 upvotes]: I have some questions about Dennis trace map in algebraic K-Theory. I was wondering if there is some conceptual way to look at this map $K(-)\rightarrow THH(-)$ (natural transformation from K-Theory to Topological Hochschild Homology). As far as I understand, Tabuada constructed a stable model category $\mathcal{M}$ containing the category of small DG-categories (and small Spectral categories). Roughly speaking, $\mathcal{M}$ is a Bousfield localization of the category of spectral presheaves on the category of small spectral categories. The algebraic K-Theory is representable by the spectral category $\mathbb{S}-Mod^{c}$ spectra which are compact. The spectral enrichment $\mathbf{Sp}_{\mathcal{M}}$ represents the algebraic (Waldhausen) K-Theory as follows $\mathbf{Sp}_{\mathcal{M}}(\mathbb{S}-Mod^{c},A)\simeq K(A-Mod^{c})$ where $A$ is a dg (or spectral category). With this notations $\mathbf{Sp}_{\mathcal{M}}(\mathbb{S}-Mod^{c},THH(-))\simeq THH(\mathbb{S})$, the Dennis trace map is represented by the class $[1]: \pi_{0}\mathbf{Sp}_{\mathcal{M}}(\mathbb{S}-Mod^{c},THH(-))\simeq \mathbf{Z}.$ My question is the following: What happens if we restrict our self to commutative $(E_{\infty})-\mathbb{S}$-Algebras and noticing that in this case $THH(R)\simeq S^{1}\otimes R$ for any $E_{\infty}$-algebra $R$. Is it possible to construct the trace map $K(-)\rightarrow S^{1}\otimes-$ (of $E_{\infty}$-algebra?!) in purely categorical comprehensive way without using Tabuada's sophisticated machinery and Dennis-Bökstedt technical results ? PS: if there is some mathematical mistakes, please do correct me. Thank you. REPLY [12 votes]: Yes, and no. There is a "categorically comprehensive" reason for this trace map to exist, but not necessarily to construct it. And to prove that this reason is valid, one does require categorical machinery. The elevator speech answer is: THH is an algebra in some symmetric monoidal category. K is the unit in this symmetric monoidal category, so THH receives a unique algebra map from K theory. (If you are looking for a less theoretical reason, and want some explicit constructions, you might take a look at Kantorovitz-Miller, "An explicit description of the Dennis trace map.") Everything I write below, I learned from Blumberg-Gepner-Tabuada, "Uniqueness of the multiplicative cyclotomic trace." First, we note that both THH and K define functors $\infty Cat^{perf} \to Spectra$. On the righthand side is the category of spectra. (Take any model you like, so long as it's not the homotopy category of the model. You can take Lurie's oo-categorical model, or symmetric spectra if you like.) On the lefthand side is the category of perfect stable oo-categories. Roughly, these are the categories that look like modules over some ring spectrum. A different way you might describe this category is as follows: The category of spectrally enriched categories, localized with respect to Morita equivalence. Note that both categories--$\infty Cat^{perf}$ and $Spectra$--have a symmetric monoidal structure. The latter has the usual smash product, while the former has the tensor product of stable $\infty$-categories. This is given by the cocompletion of the following naive tensor product: Given two categories $A$ and $B$, the objects of $A \otimes^{naive} B$ are pairs of objects $(a,b)$, and the hom spectrum between $(a,b)$ and $(a',b')$ is given by $hom(a,a') \wedge hom(b,b')$. Moreover, we note that both THH and K satisfy the following properties: They are lax monoidal. (In fact, THH is symmetric monoidal.) This means that we have specified natural maps $K(A) \otimes K(B) \to K(A \otimes B)$, but these need not be equivalences. They are localizing: If we have a short exact sequence of categories $A \to B \to C$, we have a cofibration sequence of spectra $K(A) \to K(B) \to K(C)$, and likewise for THH. The proof of this for THH can be found in Blumberg-Mandell ("Localization theorems in topological Hochschild homology and topological cyclic homology"). Now, consider the category of all functors $\infty Cat^{perf} \to Spectra$ satisfying (2). One can construct a symmetric monoidal structure on this category. And it turns out that any functor further satisfying (1) can be made into an $E_\infty$ algebra in this category, and that K theory is in fact the unit! Since THH satsifies (1) and (2), the corresponding algebra for THH receives a unique algebra map from K theory. When $A$ is an $E_\infty$ ring, then $THH(A)$ is an $E_\infty$ ring as well; by the algebra map from $K$ to $THH$, one obtains an $E_\infty$ ring map $K(A) \to THH(A)$. More generally, if $A$ is an $E_n$-algebra, then $K(A) = K(AMod)$ is an $E_{n-1}$ ring. This is because the category of $A$-modules can be given an $E_{n-1}$-structure, and $K$ theory is lax monoidal. Moreover, you can also prove that $THH(A)$ has an $E_{n-1}$ structure as well (you can see this also using factorization homology, for instance). The fact that there is an algebra map $K \to THH$ implies that one also obtains an $E_{n-1}$-algebra map $K(A) \to THH(A)$.<|endoftext|> TITLE: Probability that $n$ is coprime to both $m$ and $m+1$ QUESTION [17 upvotes]: It is well known that the set $\{(n,m) \in \Bbb N^2 : \gcd(n,m) = 1\}$ of coprime integers has a natural density of $\zeta(2)^{-1}$ in $\Bbb N^2$. It seems reasonable to think that the density of the $\{(n,m) \in \Bbb N^2 : \gcd(n,m(m+1))=1\}$ is still positive. I am no specialist of this kind of questions so I fail to see a simple argument why. Would you have any hint? REPLY [21 votes]: The density exists and equals $$ C:=\sum_d\frac{\mu(d)\tau(d)}{d^2}=\prod_p\left(1-\frac{2}{p^2}\right)\approx 0.322634\ . $$ Note that the right hand side is the product of local densities over the primes. Indeed, the number of pairs $(n,m)\in\mathbb{N}^2$ with $1\leq n,m\leq x$ and $\gcd(n,m(m+1))=1$ equals $$ \sum_{\substack{n,m\leq x\\\gcd(n,m(m+1))=1}} 1 = \sum_{n,m\leq x} \ \ \sum_{d\mid\gcd(n,m(m+1))} \mu(d) = \sum_{d\leq x}\mu(d)\sum_{\substack{n,m\leq x\\d\mid\gcd(n,m(m+1))}} 1$$ $$ = \sum_{d\leq x}\mu(d)\tau(d)\left(\frac{x}{d}+O(1)\right)^2 = x^2\sum_{d\leq x}\frac{\mu(d)\tau(d)}{d^2}+O\left(x\sum_{d\leq x}\frac{\tau(d)}{d}\right)$$ $$ = x^2\left(C+\frac{\log(x)}{x}\right)+O\left(x\log^2(x)\right) = x^2 C+O\left(x\log^2(x)\right). $$ Hence the density of such pairs equals, as $x\to\infty$, $$ x^{-2}\sum_{\substack{n,m\leq x\\\gcd(n,m(m+1))=1}} 1=C+O\left(x^{-1}\log^2(x)\right) = C+o(1). $$<|endoftext|> TITLE: $\zeta(n)$ as a mixed Tate motive QUESTION [29 upvotes]: I am trying to understand why there exists, for each $n \geq 2$, a mixed Tate motive $M$ over $\mathbb{Q}$ such that $M \in Ext^1_{MT(\mathbb{Q})}(\mathbb{Q}(0), \mathbb{Q}(n))$ and $\zeta(n)$, the Riemann zeta function at $n$, is a period of $M$. For the easiest case $n=2$, Goncharov and Manin explain that the integral representation $$ \zeta(2)=\int_{0 \leq x \leq y \leq 1} \frac{dx}{1-x} \frac{dy}{y} $$ allows to consider $\zeta(2)$ as a period of the relative homology $$ H_2(\overline{M_{0, 5}}-A, B-A \cap B)) \quad (*) $$ where $A$ and $B$ are certain unions of irreducible components of the boundary of $M_{0, 5}$. Then one can make sense of this object in the category $MT(\mathbb{Q})$, but it does not seem to be the required $M$ (I do not think it has dimension 2...). So how to get $M$? Does one need to consider some subobject of (*)? REPLY [6 votes]: As Myshkin wrote, the existence of a motive $$M \in Ext^1_{MT(\mathbb{Q})}(\mathbb{Q}(0), \mathbb{Q}(n))$$ having $\zeta(n)$ as a period follows from work of Deligne (for compatible systems of realizations) and by Deligne and Goncharov (for motives in the sense of Voevodsky). For compatible systems of realizations, Deligne constructs this extension very explicitly in two ways: one using a rather axiomatic description (section 3 of his famous ``La droite projective...'') and a geometric construction using the unipotent fundamental group of $\mathbb P^1 \setminus \{0,1,\infty\}$ (section 16 of loc.cit.). On the other hand, an explicit geometric construction of $M$ as a motive in the sense of Voevodsky is probably very hard, and, rather intriguingly, linked with Apéry-Beukers type irrationality proofs for $\zeta$-values (see recent work of Brown: arxiv1412.6508 and of Dupont arxiv1601.00950). I'm not an expert, but it seems to me that in a nutshell, Brown constructs ``Apéry motives'' for $n=2$ and $n=3$, which reply to your question (Corollary 11.3), while Dupont constructs a large motive (Theorem 1.3), whose period matrix contains all $\zeta(n)$ for $n\geq 2$ and also a smaller motive (Theorem 1.4), which contains only the odd $\zeta(n)$ (apart from the $(2\pi i)^n$, of course). The construction of Apéry motives for $n \geq 4$ seems to remain open.<|endoftext|> TITLE: Why differential forms are important? QUESTION [50 upvotes]: Importance of differential forms is obvious to any geometer and some analysts dealing with manifolds, partly because so many results in modern geometry and related areas cannot even be formulated without them: for example if you want to learn the definition of symplectic manifold, you must first learn what is differential form. However I have heard another opinion from people with non-geometric background (say, functional analysis, probability) that differential forms are probably used only in problems where they already appear in basic definitions (like symplectic geometry). When I taught an undergraduate course "Analysis on manifold" I had a feeling that my students might have got the same impression at the end: in the second half of the semester I developed calculus of differential forms with rather few concrete applications. Personally I am completely convinced in usefulness of differential forms, but find it not so easy to explain this to people who are not used to them. I would like to make a list of concrete applications of differential forms. Also it would be interesting to know historical reasons to introduce them. To start the list, below are some examples which come to my mind. 1) The general Stokes formula $\int_M d\omega=\pm \int_{\partial M}\omega$ generalizes all classical formulas of Green, Gauss, Stokes. 2) The Haar measure on a Lie group can be easily constructed as a translation invariant top differential form. (Some experts often prefer to consider this as a special case of a more general and more difficult fact of existence of Haar measure on locally compact groups. In that generality there is no such a simple and relatively explicit description of Haar measure as there is no language similar to differential forms.) 3) The cohomology of the de Rham complex of differential forms on a manifold is canonically isomorphic to singular cohomology (with real coefficients) of the manifold. One use of this isomorphism is that even Betti numbers of a symplectic manifold are non-zero. Another non-trivial use of this fact is the Hodge decomposition of the cohomology (with complex coefficients) of a compact Kahler manifold which makes it, in particular, a bi-graded algebra (rather than just graded) and provides new information on the Betti numbers, say $\beta_{2k+1}$ must be even. REPLY [4 votes]: There is at least one more important example where people who believe that "differential forms are used only in problems where they already appear in basic definitions" are completely wrong. Differential forms and the exterior differential were created (or discovered) by Elie Cartan in the 1890's in order to solve the Pfaff problem, as follows. Let us think in $R^n$. Certainly these people will admit that a differential form of degree one, that is, a covector field $$\omega=f_1dx_1+...+f_ndx_n$$ is a natural object and deserves some attention... Here $f_1, \dots, f_n$ are real functions of $x_1, \dots, x_n$. A very natural and universal method in sciences, in order to study an object, is to consider it in the coordinates system where it gets the simplest form... Pfaff's problem is: given such a form $\omega$, in a small neighborhood of a given point, can I reduce the number of variables to some $kk$. Then, his real work is to prove that this necessary condition is sufficient, in the real-analytic case (the smooth case requires some more subtle nondegeneracy condition, if I'm correct). Also note how these simple formulas contain a lot of geometry. Foliations, symplectic and contact geometries, appear here in the consideration of the (non)vanishing of the successive derivates. Differential forms were indeed one of Elie Cartan's great discoveries.<|endoftext|> TITLE: Conjecture: for perfect graphs the fractional chromatic index rounded up equals the chromatic index QUESTION [7 upvotes]: Let $\chi'_f(G)$ be the fractional chromatic index. Based on limited experiments (up to 8 vertices and few larger graphs), I suspect: Conjecture For perfect graphs $\lceil \chi'_f(G) \rceil = \chi'(G)$ Conjecture 2 (new) For cubic claw-free perfect graphs $\lceil \chi'_f(G) \rceil = \chi'(G)$ Conjecture 3 (new) For claw-free perfect graphs $\lceil \chi'_f(G) \rceil = \chi'(G)$ Sage's fractional_chromatic_index() is not efficient for me, is there another implementation? Counterexamples or proof (especially of (2)) are welcome. Observe that the question is about edge coloring, not for vertex coloring. REPLY [6 votes]: A result of Cai and Ellis (see Theorem 5 in http://www.sciencedirect.com/science/article/pii/0166218X9190010T) implies that deciding whether a cubic perfect line-graph is $3$-edge-colorable is NP-complete. Counter-examples to Conjecture 2 can be built from their argument as follows: First, notice that every cubic bridgeless graph $G$ satisfies $\chi_f'(G)=3$. This is easily obtained using the following formula for $\chi_f'(G)$, which is derived from Edmonds' inequalities for the matching polytope of $G$: $$\chi_f'(G)=\max\left(\Delta(G),\max_{U\subseteq V(G), |U|\geq 3\, \text{odd}}\frac{|E(U)|}{\frac{|U|-1}{2}}\right).$$ Now, consider the following construction: let $H$ be a bridgeless cubic graph and $S(H)$ be the graph obtained from $H$ by subdividing each edge exactly once. Let $G$ be the line graph of $S(H)$. It is straightforward to check that $G$ is cubic, bridgeless and that: $\chi'(G)=3$ if and only if $\chi'(H)=3$. Furthermore, $G$ is perfect because $S(H)$ is bipartite. Therefore, if $H$ is a cubic bridgeless graph with $\chi'(H)=4$ (for example the Petersen graph or any other snark http://en.wikipedia.org/wiki/Snark_(graph_theory)), then $G$ is a cubic bridgeless perfect line-graph with $\chi'(G)>\lceil\chi_f'(G)\rceil$.<|endoftext|> TITLE: Overconvergent cohomology and overconvergent modular forms QUESTION [12 upvotes]: I've been reading a preprint by David Hansen (with appendix by James Newton) called Universal eigenvarieties, trianguline Galois representations and p-adic Langlands functoriality. In it he talks about using overconvergent cohomology to construct eigenvarieties. Now I was wondering if I could get a reference/ explanation as to why these eigenvarieties will be related to the ones one make coming from overconvergent modular forms (as in Kevin Buzzard's- Eigenvarieties) I don't quite see the relation between this OC cohomology and OC modular forms. I have a feeling it has something to do (as I think Hansen says on p42) with the classical Eichler--Shimura and Theorem 3.2.5 on the above paper, which is a generalization of Stevens control theorem. But it's not clear to me what the relation is. Thank you. REPLY [6 votes]: Unfortunately there isn't a simple direct relationship between OC mod forms and OC cohomology! The miracle is that they contain the same finite-slope Hecke data. As you've guessed, one proves this by using classical Eichler-Shimura and Coleman's and Stevens's control theorems, plus $p$-adic interpolation - this is a beautiful idea of Chenevier. Very roughly, OC forms and OC cohomology both spread out to coherent sheaves $M,M'$ over spectral varieties $Z,Z' \subset W \times \mathbb{A}^1$. Now the control thms + Eichler-Shimura tell you that $Z \cap Z'$ is Zariski-dense in both $Z$ and $Z'$, so $Z=Z'$ (up to nilpotents). Now $M,M'$ both have Hecke actions and the control theorems applied a second time tell you that there's a dense set of points $z \in Z$ such that for any element $t$ of the Hecke algebra, $\det(1-tX)|M_z \in k_z[X]$ divides $\det(1-tX)|M'_z$, which turns out to imply the same divisibility for any point $z \in Z$, and from here its not a big leap to get a relation between eigenvarieties. I hope this isn't too cryptic - section 5 of the paper is largely devoted to formalizing this kind of argument and giving examples. It's very tempting to expect some kind of direct relationship between OC forms and OC cohomology, but whatever form this takes, it should properly be $p$-adic Hodge-theoretic in nature. This is an active area of research; see e.g. Andreatta-Iovita-Stevens' paper "Overconvergent Eichler-Shimura isomorphisms" and the abstract for Andreatta's talk at Glenn Stevens's 60th birthday conference.<|endoftext|> TITLE: Ordinal definable sets of reals in the Solovay QUESTION [6 upvotes]: To be precise, let $\Omega$ be an inaccessible cardinal in $L$ and let N be the Solovay model defined by the Levy-collapse in this case. Then $\Omega$ is $\aleph_1$ in $N$. How many different OD (=ordinal-definable) sets of reals are there in $N$? One answers that there is exactly $\aleph_2$ of them. Now, is there a concrete, meaningful OD sequence of $\aleph_2$ different OD sets of reals in $N$? REPLY [3 votes]: We can start with a canonical sequence of $\Omega^+$ many distinct subsets of $\Omega$ in $L$, which is a sequence of $\aleph_2$ many distinct subsets of $\aleph_1$ in $N$. Then we can use the canonical partial surjection $\mathbb{R} \to \aleph_1$, which takes a real coding a well-ordering of $\omega$ to the order type of this well-ordering, to get a sequence of $\aleph_2$ many distinct subsets of $\mathbb{R}$ in $N$. I think this example is about as concrete and meaningful as we can hope for, because we will need to use somehow the fact that the ground model satisfies $2^\Omega = \Omega^+$.<|endoftext|> TITLE: Why do the projections in the Calkin algebra not form a lattice? QUESTION [14 upvotes]: Let $H$ be an infinite dimensional separable complex Hilbert space. Denote by $\mathcal{B}(H)$ the C*-algebra of bounded operators on $H$, $\mathcal{K}(H)$ the ideal of compact operators on $H$, and $\mathcal{C}(H)=\mathcal{B}(H)/\mathcal{K}(H)$ the Calkin algebra. The collection of projections $\mathcal{P}(\mathcal{C}(H))$, i.e., self-adjoint idempotents, in $\mathcal{C}(H)$ form a partially ordered set in the usual way, namely $p\leq q$ if and only if $pq=p$. It is my understanding that $(\mathcal{P}(\mathcal{C}(H)),\leq)$ does not form a lattice, i.e., meets and joins need not exist in general. (This would mean that the Calkin algebra is another example for C*-algebras with bizzarre structure of projections.) This fact is mentioned in Hadwin's paper "Maximal Nests in the Calkin Algebra" (PAMS, 1998), and there is a sketch of a proof (credited to Weaver) in Farah's notes "Set theory and operator algebras" from the Appalachian Set Theory Workshop, however I've never quite understood this proof. Would someone be able to explain/sketch this result? REPLY [3 votes]: I can't add a comment, although I'd like to clarify one point: both in Nik Weaver's Set theory and C*-algebras and Eric Wofsey's $P(\omega)/\mathrm{fin}$ and projections in the Calkin algebra it is claimed that projections in Calkin algebra do give rise to the lattice structure (see e.g. Corollary 3.4 and Wofsey's abstract, respectively). How is it possible? Both proofs (Proposition 5.26 in Farah's notes and email corresponence in this topic) seem to be general in the separable case.<|endoftext|> TITLE: A cohomology group which depends on the connection QUESTION [7 upvotes]: Warning: I am not a differential geometer, so some of the following might not make sense. Background: Let $w: (T\Omega)^k \to \mathbb{R}$ be a $k$-tensor on $\Omega$, an open subset of $\mathbb{R}^n$. We can define the "derivative" of this tensor as follows. $Dw:(T\Omega)^{k+1} \to \mathbb{R}$ is a $k+1$-tensor which enjoys the following approximation property $$ w(p+v_1)(v_2,v_3,...,v_{k+1}) = w(p)(v_2,...,v_{k+1})+Dw(v_1,v_2,...,v_{k+1})+\textrm{Error} $$ Where the error term is order $|v_1||v_2|...|v_{k+1}|$. If $f$ is a function, $Df$ is the derivative. Here is an example for $w$ a one form: $w = Pdx+Qdy$ $Dw = \frac{\partial P}{\partial x} dx \otimes dx+ \frac{\partial Q}{\partial x} dx \otimes dy+\frac{\partial P}{\partial y} dy \otimes dx+\frac{\partial Q}{\partial y} dy \otimes dy$ Notice that if you project $D$ of a one form onto the alternating one forms, you get the exterior derivative. Call a $k$-tensor $w$ closed if there are open sets $U_i$ covering $\Omega$, and $k-1$ tensors $\eta_i$ on $U_i$ with $w = D\eta_i$ on $U_i$. Call a $k$-tensor $w$ exact if there is a global $k-1$ tensor $\eta$ defined on $\Omega$ with $w = D\eta $. I am interested in the group of closed $k-$tensors mod exact $k-$tensors. To generalize this beyond subsets of $\mathbb{R}^n$, it seems clear that we would need a connection, since we need to "evaluate" $w$ at ``the same" tangent vectors, but living at different base points. I know enough about connections to know that they allow this to happen, but not much more unfortunately. I do think there should be a group defined analogously to the one I define above for any manifold with a connection. My question is simply if this group has been studied before, and if so where I should find information about them in the literature. They would possibly encode more information than the de Rham cohomology groups, including data about the geometry of the manifold instead of just the topology. Also, any information about the basic properties of these groups (really $\mathbb{R}$ -vector spaces) would be appreciated. Are they finite dimensional for reasonable spaces? Do they depend on only the topology? REPLY [2 votes]: Let me try to describe a slightly more general situation and hopefully prove my comment above. Let $V$ be a vector bundle on a manifold $M$ with connection $D : \Gamma(V) \to \Gamma(V \otimes T^*M)$. Let $Z(V)$ be the group of sections $\omega \in \Gamma(V \otimes T^*M)$ where, for an open covering $U_i$ of $M$, there exists local sections $\eta_i \in \Gamma(U_i,V)$ such that $D\eta_i = \omega |_{U_i}$. Let $B(V)$ be the group of sections $\omega \in \Gamma(V \otimes T^*M)$ such that there exists a global section $\nu\in\Gamma(V)$ such that $D\nu = \omega$. Clearly, $B(V) \subset Z(V)$. I claim that $$ Z(V)/B(V) \cong \check{H}^1(M,\Gamma_f(V,D)) $$ Here $\Gamma_f(V,D)$ is the sheaf of flat sections of $V$ with respect to the connection $D$. To get the forward, map, let $\omega \in Z(V)$. Then, as above, we can form the $\eta_i \in \Gamma(U_i,V)$. Let $\eta_{ij} = \eta_i - \eta_j \in \Gamma(U_{ij},V)$. We have $D\eta_{ij}=0$ and $\delta\eta_{ij}=0$ where $\delta$ is the Cech differential, so this is a cocyle. Any other representation of $\omega$ subordinate to the same open cover gives rise to a cocycle which differs by a $\delta$-exact term. Finally, $B(V)$ is in the kernel of this map, so we've established the forward direction. For the reverse direction, do the usual: choose a partition of unity $\sum \gamma_i = 1$ subordinate to the open cover, $U_i$. Given a cocycle $\eta_{ij}$, we can form $$ \rho_i = \sum_k \eta_{ik}\gamma_k \in \Gamma(U_i,V) $$ Let $\zeta_i = D\rho_i = \sum_k \eta_{ik} d\gamma_k$. Then, $$ \zeta_i-\zeta_j = \sum_k (\eta_{ik} - \eta_{jk}) d\gamma_k = \eta_{ij}\, d\left(\sum_k \gamma_k\right) = 0 $$ So, the $\zeta_i$ define an element $\zeta \in Z(V)$. Adding a $\delta$-exact cocycle, $\alpha_{ij} = \alpha_i - \alpha_j$ with $D\alpha_i=0$ adds to $\rho_i$: $$ \sum_k (\alpha_i - \alpha_k)\gamma_k = \alpha_i - \sum_k \alpha_k \gamma_k $$ The first term is annihilated by $D$, and the $D$ of the second gives rise to an element in $B(V)$, so we're done. To check that the composition works, note that, given $\eta_{ij} = \eta_i - \eta_k$ $$ \eta_i - \rho_i = \eta_i - \sum_k \eta_{ik} \gamma_k = \sum_k \eta_k \gamma_k $$ As above, this is a global section which gives rise to an element in $B(V)$, so the composition is the identity on $Z(V)/B(V)$.<|endoftext|> TITLE: Non Kähler blow-up of a Kähler manifold QUESTION [5 upvotes]: Is it possible to find a complete, non compact Kahler manifold $(X,\omega)$ with a closed, connected, non compact complex submanifold $Y\subset X$ of codimension at least 2 such that the blow-up of $X$ along $Y$ is not Kähler? REPLY [4 votes]: When $Y$ is compact, the blow-up is always Kahler; see e.g. Lemma 3.4 in this paper (this is a generally known folklore theorem which we had to use, and hence written down). For $Y$ non-compact the argument should be similar, but more cumbersome.<|endoftext|> TITLE: Optical methods for number theory? QUESTION [9 upvotes]: I found a paper: 'A New Method of Finding the Distribution of Prime Number', saying We stack discs and annuluses with certain rules then turn on the light to illuminate. The projection of annuluses corresponds prime number,and the projection of discs corresponds composite number. See: http://en.cnki.com.cn/Article_en/CJFDTOTAL-HNKX201201010.htm Are there other physics methods for number theory? Edit: Answers and comments here and to the corresponding meta question have shown that the answer is yes, and much broader and more recent than the sieve of Eratosthenes. Also, a more informative link to the above paper is http://wenku.baidu.com/view/1d602350be23482fb4da4cc6.html?re=view REPLY [10 votes]: Actually such optical devices were at one point in the 20th century the state of the art for doing certain number-theoretical calculations, notably factoring. See http://en.wikipedia.org/wiki/Lehmer_sieve (specifically the 1932 device, pictured in pages 18-20 of http://people.ucalgary.ca/~hwilliam/Sieve_Pictures.pdf). REPLY [7 votes]: I agree with Noam (great pictures!) Some of the links from comments are quite informative. Here are two specific cases D. N. Lehmer first used a device made of bicycle chains and rods and later devices made of gears with holes in them. This article relates in rather breathless prose that one of the gear mechanisms (combined with theory) proved in a few minutes that the great unconquered nineteen digit number $3,011,347,479,614,249,131$ , known to be a factor of $2^{95} + 1$ and suspected to be prime, is in fact prime. A few years back (maybe 2000?) there was great excitement over rumors that Adi Shamir had a breakthrough which would speed up (the sieving step of) the number theory sieve by "several orders of magnitude". The then record factoring of an RSA key was for a 465 bit integer and the breakthrough was rumored to make 512 bit keys "very vulnerable." When the details came out of the punnily named TWINKLE device it turned out to be an electro optical device using LEDs and filters. It is agreed to be quite clever, it has never been built. Enhanced (theoretical) versions might threaten 768 bit keys in under 9 months (That estimate was in 2000, for an organization willing to invest in 80,000 pentium 2 PCS and 5000 TWINKLE devices ). 1024 bit keys would probably have been well beyond that. I think that the state of the art in unbuilt (or so they say...) special purpose devices is no longer optical.<|endoftext|> TITLE: Permutation Groups Containing non-commuting $p$-cycles QUESTION [15 upvotes]: I noticed that the following is true, and that there is a reasonably elementary proof of it (in particular, the classification of finite simple groups is not needed). Let $G$ be a finite permutation group which contains two $p$-cycles which do not commute (where $p$ is any odd prime other than a Mersenne prime). Then $G$ is not solvable ( more precisely, $G$ has a non-Abelian composition factor of order divisible by $p$). Since the methods are reasonably elementary, I wonder if anyone has come across this or similar results (possibly in a Galois Theory context) and can point me to a reference? For every Mersenne prime $p,$ there is a solvable permutation group $G$ of degree $p+1$ and order $p(p+1)$ which contains two non-commuting $p$-cycles. (Later note: The analogous result is not true for $p^{2}$-cycles ($p$ prime). For $p=2$ take $G = S_{4}$ and for $p>2$ take $G$ to be a Sylow $p$-subgroup of $S_{p^{2}}.$ In each case, $G$ is a solvable (even nilpotent when $p$ is odd) permutation group containing two $p^{2}$-cycles which do not commute). REPLY [8 votes]: Fixed several inaccuracies, many thanks to Frieder Ladisch for spotting them: Let $x$ and $y$ be two non-commuting $p$-cycles, $G=\langle x,y\rangle$, and $G$ be considered as a transitive permutation group on the support $\Omega$ of $G$. We show that either $\text{AGL}_1(\mathbb F_q)\le G$ for a Mersenne prime $q$, or $G$ is simple non-abelian. Proof. $G$ is primitive (as pointed out by Frieder Ladisch already). This can be seen as follows: Let $\Delta$ be a block of a non-trivial block system. The action of the $p$-cycle $x$ on the block system is trivial, for otherwise $x$ would move $p\lvert\Delta\rvert>p$ points. On the other hand $G$ transitively moves the blocks, a contradiction. Next we show that $G$ is doubly transitive. This follows from Burnside's classical theorem if $\lvert\Omega\rvert=p$. If $\lvert\Omega\rvert>p$, then the pointwise stabilizer in $G$ of the points fixed by $x$ is transitive (via $x$) on the remaining points. By Jordan's Theorem on primitive groups with Jordan sets we again see that $G$ is doubly transitive. Let $N$ be a minimal normal subgroup of $G$. By Burnside, $N$ is either (a) elementary abelian and regular, or $N$ is (b) simple, primitive, and not regular. The Mersenne exceptions follow from looking at $p$-cycles in $\text{GL}_m(q)$ where $n=q^m$ for a prime $q$: Such a $p$-cycle fixes $q^r TITLE: Is there a effective computational criterion to all periodic points of a rational function are repelling. QUESTION [6 upvotes]: I came up with a question to know the fatou component of of some types of rational function. In some sense, I may need to give a computational criterion to existence of attracting periodic basin for a rational function, which is related to Fatou's theorem. I tried some examples , however, sometimes I can not find attracting periodic point (period from 2 to 100) with the comupters. I have no idea to know whether it has a attracting basin basin for $f$ or not. For example, $$f(z) = -2/3*z+(-2*z^3+1)/(3*z^3+5*z).$$ By the software of Ultra Fractal, and it seems that the julia set of the example is very small (may be a Cantor set), I can only see the whole screen is black. REPLY [3 votes]: The best way to study these questions is in the framework of computable analysis. I will take your rational function as being given by oracles for its coefficients and for its critical points. (That is, for a given epsilon, we expect to be able to know these values up to an error of epsilon.) In this framework, the question "Does f have non-repelling periodic points?" is undecidable. Indeed, suppose that the rational map is such that it has only repelling periodic points and is not structurally stable. (Conjecturally, every map having only repelling periodic points is structurally unstable. Under many circumstances this is known; the simplest case is the postcritically finite case where it follows from work of Thurston.) Then there are rational maps arbitrarily nearby that do have nonrepelling cycles, and hence we cannot distinguish between the two cases by considering our map only up to finite precision. However, if we replace "non-repelling" by "attracting", then the problem is semi-decidable. Indeed, we basically just iterate the critical values and see whether they converge to an attracting cycle; this can be done rigorously. (I will omit the details, but suspect they can be found e.g. in the work of Braverman and Yampolsky.) Of course, when your coefficients are given in some absolute way (e.g. as rational or algebraic numbers), then you can ask about this as a classical problem in computer science. This problem seems less natural, although I would expect it also to be undecidable. There is a result that says that the Mandelbrot set is undecidable in the sense of Shub-Smale (which assumes infinite-precision arithmetic), which suggests at least that the problem can't be solved in any naive way. Of course, there are specific examples for which it can be verified that there are no nonrepelling cycles, as mentioned by Alexandre.<|endoftext|> TITLE: About G-modules with good filtrations QUESTION [6 upvotes]: Let $k$ be an algebraically closed field of positive characteristic, and let $G$ be a reductive algebraic group over $k$ (for instance a classical group). Let $V$ be a (rational) $G$-module. We say that $V$ admits a good filtration [see for instance Jantzen, Representations of Algebraic Groups, §4.19] if there exists an ascending chain of $G$-modules $0=V_0 \subset V_1 \subset V_2 \subset ... \subset V$ such that $V=\cup V_i$ and, for each $i \geq 1$, the quotient $V_i/V_{i-1}$ is isomorphic to some $H^0(\lambda) \otimes N_i$, where $N_i$ is a (non necessary finite dimensional) vector space on which $G$ acts trivially, and $H^0(\lambda)$ is the usual induced module $ind_B^G(k_{-\lambda})$, $B \subset G$ being a Borel subgroup, $\lambda$ a dominant weight, and $k_{-\lambda}$ the one dimensional representation of $B$ corresponding to $-\lambda$. My questions are the following: 1) If $M_1$ and $M_2$ are two $G$-submodules of $M$ such that $M_1,M_2$, and $M$ admit a good filtration, does $M_1 \cap M_2$ also admit a good filtration? 2) If $M$ is a $G$-module with a good filtration and if $N \subset M$ is any $G$-submodule, does there exist a unique minimal $G$-submodule $N'$ such that $N \subset N' \subset M$ and $N'$ admits a good filtration? REPLY [7 votes]: The answer to both questions is no. For 1) let $V=H^0(\lambda)$ with $\lambda$ dominant, $U\leq V$ any submodule that doesn't have a good filtration and $\Delta(U)=\{(u,u): u\in U\}\leq V\oplus V$. Take $M_1=V\oplus0\leq (V\oplus V)/\Delta(U)$, $M_2=0\oplus V\leq(V\oplus V)/\Delta(U)$, and $M$ an injective hull of $(V\oplus V)/\Delta(U)$. Then $M_1\cap M_2\cong U$ doesn't have a good filtration. For 2) take the same example, with $N=M_1\cap M_2$. Then $M_1$ and $M_2$ are both minimal submodules of $M$ with good filtrations that have $N$ as a submodule.<|endoftext|> TITLE: Strange definition of ergodicity QUESTION [9 upvotes]: I've already asked this question on math.stack a few days ago and haven't received an answer, so I'm asking here. In an engineering course, a stationary process was defined to be ergodic if for all $k\in \mathbb{N}$ and for any bounded (measurable) function of $k+1$ variables we have $$\lim_{N\rightarrow \infty}\frac{1}{N}\sum_{n=1}^N g(X_n,\dots,X_{n+k})\overset{\text{a.s}}{=}Eg(X_n,\dots,X_{n+k})$$ From the little I've read about ergodic theory, this does not seem familiar nor does it seem to fit into the ergodic hierarchy I know, i.e ergodic, weak mixing, strong mixing etc. It seems like a different property from ergodicity (in the sense of the Birkhoff ergodic theorem). Here the boundedness of $g$ means a formulation with indicator $g$'s would be equivalent (because of DCT I think). On the other hand, any $k$-tuple of $X_i$'s is allowed.. Is there an insightful bit of intuition for this property as there are for normal ergodicity, and mixing? Where does it fit into the ergodic hierarchy? Thanks in advance! REPLY [14 votes]: Yes. That is the same thing as ergodicity. To explain it, you have probably seen somewhere that the way to understand random variables formally is functions from a (hidden) underlying space $\Omega$ to $\mathbb R$. That is: knowing the point $\omega$ of $\Omega$, one can recover $X_n(\omega)$ for each $n\in \mathbb Z$. In fact, a standard $\Omega$ to use is just the space of sequences of real numbers. Then the function $X_n$ is just the $n$th coordinate function. If you do this, then $\Omega$ is equipped with a natural map, namely the shift map, $\sigma$. You can now define a version of your function $g$ on $\Omega$, namely $G(\omega)=g(\omega_0,\ldots,\omega_k)$. The left side of your equality is now $\lim_{N\to\infty}(1/N)\sum_{n=1}^N G(\sigma^n\omega)$. The right side is $\int G(\omega)\,d\mu(\omega)$, where $\mu$ is the distribution on the set of sequences occurring. Notice: invariance of $\mu$ corresponds exactly to stationarity of the sequence of random variables.<|endoftext|> TITLE: Degree 17 number fields ramified only at 2 QUESTION [37 upvotes]: The number $17$ is the smallest odd number that occurs as the degree of a number field $K/\mathbb{Q}$ for which the only finite prime that ramifies is $2$. The non-existence for $n < 17$ follows from the paper "Number fields unramified away from $2$" by John W. Jones in the Journal of Number Theory in 2012. The existence follows from the fact that $\mathbb{Q}(\zeta_{64})$ has class number $17$ and hence if $H$ is the Hilbert class field of $\mathbb{Q}(\zeta_{64})$, then $H/\mathbb{Q}$ is Galois with $G = Gal(H/\mathbb{Q})$ of order $32 \cdot 17$. If $P$ is a Sylow $2$-subgroup of $G$, then the fixed field of $P$ is a number field $K/\mathbb{Q}$ of degree $17$ ramified only at $2$. (Actually, one of the degree $16$ subfields of $\mathbb{Q}(\zeta_{64})$ also has class number $17$, and the Galois closure of $K/\mathbb{Q}$ has degree $272$. Harbater shows that this is the unique Galois extension of $\mathbb{Q}$ of degree $272$ ramified only at $2$.) My question is the following: Is it possible to explicitly compute a degree $17$ polynomial $f(x)$ that has a root in $K$? The motivation for this question is the following. If $C/\mathbb{Q}$ is a curve of genus $g$ with good reduction away from $2$ and $J(C)$ is the Jacobian of $C$, then $\mathbb{Q}(J(C)[2^{n}])/\mathbb{Q}$ is a Galois extension, ramified only at $2$, with Galois group contained in a subgroup of $GSp_{2g}(\mathbb{Z}/2^{n} \mathbb{Z})$. Knowing something about the number fields that can arise allows one to understand the arithmetic of $C$ (and in particular, allows one to find etale covers of degree $2^{n}$). For example, since there are no $A_{3}$ or $S_{3}$ extensions of $\mathbb{Q}$ ramified only at $2$, every elliptic curve $E/\mathbb{Q}$ with good reduction away from $2$ must have a rational $2$-torsion point. (Of course, there are other ways to see this.) Here are three possible approaches to the question: $\bullet$ Use existing software that will do computations with class field theory. As a test case, I tried using Magma to compute Hilbert class fields of cubic fields with class group of prime order $p$. For $p \leq 11$, the computation was doable, but for $p > 11$, it seemed quite difficult. Maybe there are more sensible ways to attempt the computation. $\bullet$ I've seen hints in the literature at generalizations of explicit class field theory for imaginary quadratic fields to CM fields. Maybe there's a way to construct $K$ using CM abelian varieties. $\bullet$ The rigidity method (sometimes) allows one to construct a Galois extension $L/\mathbb{Q}(t)$ with a given Galois group, and has the ability to construct specializations with a small number of ramified primes. This method seems promising, although $\mathbb{Z}/17\mathbb{Z} \rtimes (\mathbb{Z}/17/\mathbb{Z})^{\times}$ does not have a rationally rigid triple of conjugacy classes. REPLY [25 votes]: Thank you for calling this problem to my attention. I computed $K$ en route to AWS (though this year's topics are a rather different flavor of number theory...). After some simplification (gp's $\rm polredabs$), it turns out that the field $K$ is generated by a root of $$ f(x) = x^{17} - 2x^{16} + 8x^{13} + 16x^{12} - 16x^{11} + 64x^9 - 32x^8 - 80x^7 $$ $$ \qquad\qquad\qquad {} + 32x^6 + 40x^5 + 80x^4 + 16x^3 - 128x^2 - 2x + 68. $$ gp quickly confirms: p = Pol([1,-2,0,0,8,16,-16,0,64,-32,-80,32,40,80,16,-128,-2,68]) poldegree(p) F = nfinit(p); factor(F.disc) returns [2 79] So the field has discriminant $2^{79}$. Let $F = {\bf Q}(\zeta_{64}^{\phantom{0}} - \zeta_{64}^{-1})$, with $F/\bf Q$ cyclic of degree $16$ of $\bf Q$. It was known that $K$ is contained in an unramified cyclic extension $L/F$ of degree $17$. Hence $L(\zeta_{17})$ is a Kummer extension of $F_{17} := F(\zeta_{17})$. Now $F_{17}$ is a $({\bf Z} / 16 {\bf Z})^2$ extension of $\bf Q$. Fortunately it was not necessary to work in the unit group of this degree $256$ field: the Kummer extension is $F_{17}(u^{1/17})$ with $u \in F_{17}^* / (F_{17}^*)^{17}$ in an eigenspace for the action of ${\rm Gal}(F_{17}/{\bf Q})$, which makes it fixed by some index-$16$ subgroup of this Galois group. Thus $u$ could be found in one of the cyclic degree-$16$ extension of $\bf Q$ intermediate between $\bf Q$ and $F_{17}$, and eventually a generator of $K$ turned up whose minimal polynomial, though large, was tractable enough for gp's number-field routines to do the rest.<|endoftext|> TITLE: Understanding of Tamagawa numbers of hyperelliptic curve QUESTION [6 upvotes]: One's can find following definition of tamagawa numbers in Dino Lorenzini paper "Torsion and Tamagawa numbers": Let $K$ be any discrete valuation field with ring of integers $O_K$ , uniformizer $\pi$, and residue field $k$ of characteristic $p \ge 0$. Let $A/K$ be an abelian variety of dimension $g$. Let $A/O_K$ denote the Neron model of $A/K$. The special fiber $A_k/k$ of $A$ is the extension $$(0) \rightarrow A^0_k \rightarrow A_k \rightarrow Φ \rightarrow (0)$$ of a finite etale group scheme $Φ/k$, called the group of components, by a connected smooth group scheme $A^0_k/k$, the connected component of 0. The order of the finite abelian group $Φ(k)$ is called the Tamagawa number of A/K. Let now $K$ be a global field, and $v$ a non-archimedean place of $K$, with completion $K_v$ and residue field $k_v$. Let $c_v$ denote the Tamagawa number of $A_{K_v}/{K_v}$, and let $c = c(A/K) := \prod_v c_v$. I have hyperelliptic curve of genus 2 over $\mathbb Q$. The questions are: Is it true that I can extract Tamagawa numbers ($c_v$ in definition) from Sage function genus2reduction (see below) output and calculate Tamagawa product (called $c$ in definition) from them? Is it ok that group of connected components is over an algebraic closure of $\mathbb F_p$? Is it true that I can use Magma's RegularModel (see below) and get tamagawa numbers $c_p$ ($=c_v$) as orders of Magma's ComponentGroup for corresponding $p$? I'm confused by group of components of the Neron model of the Jacobian of C over the **completion** at P. There is no mentioned of completition in Sage function. I'm not sure that I understand all things right. Is it true that in second case if I want to get Tamagawa product I should just calculate product of $c_p$ for all $p$ that divide discriminant of my curve? There are Necessary documentation notes about corresponding Sage and Magma functions. Sage genus2reduction function ducumentation says following: Use $R = genus2reduction(Q, P)$ to obtain reduction information about the Jacobian of the projective smooth curve defined by $y^2+Q(x)y=P(x)$. $$\dots$$ The second datum is the GROUP OF CONNECTED COMPONENTS (over an ALGEBRAIC CLOSURE (!) of $\mathbb F_p$) of the Neron model of $J(C)$. From the other side Stain in his note "What are Neron Models?" writes: When $A$ is the Jacobian of a curve $X$, there is an alternative approach that involves the "minimal proper regular model" of $X$. For example, when $A$ is an elliptic curve, it is the Jacobian of itself, and the Neron model can be constructed in terms of the minimal proper regular model $X$ of $A$ as follows. In general, the model $X → R$ is not also smooth. Let $X'$ be the smooth locus of $X → R$, which is obtained by removing from each closed fiber $X_{F_p} = \sum n_i C_i$ all irreducible components with multiplicity $n_i \ge 2$ and all singular points on each $C_i$, and all points where at least two $C_i$ intersect each other. Then the group structure on $A$ extends to a group structure on $X'$ , and $X'$ equipped with this group structure is the Neron model of A. And Magma documentation on RegularModel says: RegularModel(C, P) : Crv, Any -> CrvRegModel This computes a regular model of the curve $C$ at the prime $P$. Here $C$ is a curve over a field $F$ (the rationals, a number field or a univariate rational function field), and $P$ is a prime of the maximal order $O_F$ of $F$ (given as an element or as an ideal). ComponentGroup(M) : CrvRegModel -> GrpAb Given a regular model of a curve $C$ at a prime $P$, this returns (as an abstract abelian group) the group of components of the Neron model of the Jacobian of $C$ over the completion at $P$. (This is computed from the IntersectionMatrix of the model.) I'm quite new in this field and I'm sorry if my questions are silly. REPLY [2 votes]: (1) The answer is no for the first part of this question. This is explained in Sage's documentation you cited. The program genus2reduction only outputs the order of $\Phi(\overline{\mathbb F}_p)$, while $c_p$ is the order of $\Phi(\mathbb F_p)$. The later is a subgroup of $\Phi(\overline{\mathbb F}_p)$. I don't understand the second part of the question. (2) I do not have access to Magma. But try the following example $$ y^2=2(x(x-1)(x-2))^2+3$$ at $p=3$. According to Example 1.17 of this paper of S. Bosch and Q. Liu, $\Phi(\mathbb F_3)=\{0\}$ and $\Phi(\overline{\mathbb F}_3)=\mathbb Z/3\mathbb Z$. So if Magma gives you $c_3=1$, then it is capable of computing the Tamagawa number at least for this curve. But after reading the documentation of Magma, I would not be surprised that it will actually give an error message. For the second part of the question, it is not necessary to go to the completion of $K_v$ to compute $c_v$. (3) Yes, at any $p$ prime to a discriminant of the curve, the later has good reduction at $p$. So its Jacobian has good reduction at $p$. This implies that $\Phi$ is trivial as algebraic group and $c_p=1$.<|endoftext|> TITLE: Bouncing a ball down the stairs QUESTION [25 upvotes]: In a nutshell, the question is whether it can be faster to bounce a ball down an infinite flight of stairs than to bounce it down a ramp with the same slope. To be more specific: this is a $2$ dimensional problem, so my ball is a disk and all the action takes place in $\mathbb{R}^2$. I imagine an infinite staircase with steps having depth $1$ and height $1$, the top corner of the top step is at the origin (and the staircase is oriented down and to the right). The ball has radius $1$, so when we toss the ball on the stairs and it will impact on an outside corner of a stair every time. Gravity acts downward with constant acceleration $g$. We compute the bounces by replacing the single corner point of the stair with an imaginary line tangent to the ball at the impact point. Let $s(t)$ be the $x$-coordinate at time $t$ of the ball bouncing down the stairs, and let $r(t)$ be the $x$-coordinate at time $t$ of the ball rolling down the ramp. QUESTIONS: Are there initial conditions for which $s(t) > r(t)$ for all sufficiently large $t$? Does the answer depend on the value of $g$? IDEAS: Can we determine the distribution of the incidence angles? Is it uniform? Does the distribution govern the long-term behavior of $s$? We can replace the ball/stairs system with an equivalent one in which we track only the center of the ball. The center of the ball bounces down a "cobblestone ramp" made up of quarter circles (with radius $1$) heading down and to the right at slope $-1$. REPLY [8 votes]: Consider trajectories that start on the slope. We measure success by how long it takes to reach any point a given vertical distance below the starting point (which is not exactly the success measure in the question but may be a satisfactory replacement). Let $s$ denote distance along the trajectory. Let $t(s)$ be the time at which point $s$ is reached, and $y(s)$ be the vertical displacement from the starting point (measured downwards). Also let $u$ be the initial speed. Since the collisions are elastic, conservation of energy implies that the speed of the ball when it has gone distance $s$ along its trajectory is $\sqrt{u^2+2g\, y(s)}$. From this it follows that $$t(s) = \int_{0}^s \frac{dq}{\sqrt{u^2 + 2g\, y(q)}}.$$ Now suppose you could choose any trajectory lying on or above the smooth slope. For trajectories of given length $s$, you can minimise $t(s)$ by maximising $y(q)$ as much as possible for $0\le q\le s$, which seems a bit vague value until you realise that moving down the slope makes $y(q)$ take its maximum possible value $q/\sqrt 2$ for all $q$. Since this straight line also minimises the length of the trajectory to any given vertical distance below the starting point, we can infer a theorem: If the ball starts on the slope at speed $u$ and can be constrained by elastic collisions to take any trajectory on or above the slope, the fastest way to reach any given vertical distance below the starting point is sliding down the slope. The staircase can be drawn so that the inner corners lie on the slope. If you have to start at such an inner corner, the above theorem shows that you can't choose any initial starting speed and direction so that the ball will reach a given vertical distance below the starting point faster than it would if you used the same initial speed down the smooth slope. If you can start at places other than the inner corners, a bit more work needs to be done (starting with an exact definition of the problem). But it is still clear that the slope is better eventually. A further, much harder, question is how much worse the staircase is. As Ryan commented, repeated glancing collisions seem best but I'm not sure if such trajectories are physical. Might it be that no matter how hard you try eventually the ball will encounter hard collisions that make it bounce upwards? This would be disastrous since it can bounce up to almost the same height it started from. ADDED: In the above I treated the ball as a point mass. If it is a disk of finite radius (but still frictionless so the angular inertia is not an issue), replace what I called "the slope" by the lowest 45-degree line that the centre of the disk may touch and "the trajectory" by the path taken by the centre of the disk. Draw the staircase so that the centre of the ball is on that line when the ball is sitting on a step as far in as possible. The analysis appears to be the same.<|endoftext|> TITLE: Are there any explicit probability conserving solvers for Pauli equation? QUESTION [8 upvotes]: I know that there exist probability conserving explicit solvers for time-dependent Schrödinger's equation, for example, Visscher's one. But when I tried to take into account spin and magnetic field (so that Hamiltonian matrix is no longer real-valued) in this scheme, it appeared, although remaining stable, to not conserve total probability. I had to make time steps several orders of magnitude smaller to have total probability remain somewhat constant, and even then it visibly changed after 100 time steps. Here's what the basic algorithm looks like. Schrödinger equation $i\frac d{dt}\Psi=H\Psi$ is rewritten with $\Psi=R+iI$ as $$\left\{\begin{align} \frac {dR}{dt}&=HI,\\ \frac {dI}{dt}&=-HR. \end{align}\right.\tag1$$ These equations are then discretized as $$\left\{\begin{align} R\left(t+\frac12\Delta t\right)&=R\left(t-\frac12\Delta t\right)+\Delta tHI(t),\\ I\left(t+\frac12\Delta t\right)&=I\left(t-\frac12\Delta t\right)-\Delta tHR(t). \end{align}\right.\tag2$$ Probability density is defined as $$P(x,t)=R(x,t)^2+I\left(x,t+\frac12\Delta t\right)I\left(x,t-\frac12\Delta t\right)\tag3$$ at integer $t/\Delta t$ and $$P(x,t)=R\left(x,t+\frac12\Delta t\right)R\left(x,t-\frac12\Delta t\right)+I(x,t)^2\tag4$$ at half-integer $t/\Delta t$ and conserved provided that $H$ matrix (where Laplacian is supposed to be a finite-difference or something similar) is real-valued. My change was to make $H$ complex to allow for Pauli matrix $\sigma_y$ and magnetic field term $\propto i\partial_y A(\vec r)$. Now, with $H=H_r+iH_i,$ equations $(1)$ look like $$\left\{\begin{align} \frac {dR}{dt}&=H_rI+H_iR,\\ \frac {dI}{dt}&=H_iI-H_rR. \end{align}\right.\tag5$$ After this change probability density defined by $(3)$ and $(4)$ is no longer conserved. Apparently, Visscher's algorithm is not extensible to complex Hamiltonians, so I need to find another one. So, are there any explicit probability conserving solvers suitable for Pauli equation? NOTE There appears to be many papers related to Dirac equation, but I would like to solve the Pauli equation itself, not just Pauli regime of Dirac equation (e.g. input large potentials which would lead to Klein paradox in Dirac equation but wouldn't in Pauli equation). REPLY [3 votes]: A well-balanced and asymptotic-preserving scheme for the one-dimensional linear Dirac equation (2014) treats the effect of spin in a probability conserving way, both in the relativistic (Dirac) and non relativistic (Pauli) regime.<|endoftext|> TITLE: Discrete gradient on point clouds QUESTION [8 upvotes]: I am interested to know some ways to approximate discrete gradient if you have a function on point clouds in 2D or 3D. If you have a function defined on a grid, it well known that you can use a standard difference formula to compute the partial derivative in each direction. I came across a paper by Luo et al entitled Approximating Gradients for Meshes and Point Clouds via Diffusion Metric which uses diffusion metric. Another idea that came into mind is to defined a smoothed field using a specific kernel, then to take the gradient of the smoothed function. Is there a standard way of doing this? REPLY [4 votes]: Very many numerical methods for partial differential equations compute derivatives from values on something that is not a regular grid (they use unstructured grids). Usually these are finite element methods and they use more than just point values (they have various local integrals of the function available). A class of methods based on what you're asking for are the so-called meshfree methods. Typically, they use the point values to generate a continuous approximation by convolving with some smooth kernel (e.g., a Gaussian). Then they compute derivatives of the continuous function. Of course, if you have a lot of points this becomes expensive and there are lots of additional tricks you can play (e.g., using a kernel with compact support; employing the fast multipole method, etc.).<|endoftext|> TITLE: Normal generators of finite index subgroups in a free group QUESTION [8 upvotes]: Let $F=F(a,b)$ be the free group of rank $2$. Question 1: Given any positive integer $d$, can one always find elements $u_j,v_j,w_j \in F$, $j=1,\dots,d$, such that if $1 \le j |F/D_rF|^n$. Then, by the pigeon hole principle, there are indices $j,k$, $1 \le j < k \le d$, such that $u_{ij}$ represents the same element of the finite group $F/D_rF$ as $u_{ik}$, for all $i=1,\dots,n$. It follows that each of the elements $u_{1j}^{-1}u_{1k},\dots,u_{nj}^{-1}u_{nk}$ is contained in $D_rF$, so its Jennings-Zassenhaus degree is at least $r$. In view of (1), we can apply the theorem of Golod-Shafarevich to conclude that the group given by the presentation $$\langle a,b \,\|\, u_{1j}^{-1}u_{1k},\dots,u_{nj}^{-1}u_{nk} \rangle$$ must be infinite. Therefore no natural number $n$ from Question 2 exists.<|endoftext|> TITLE: Rational homology and finite group actions QUESTION [9 upvotes]: I'm looking for examples of the following phenomena. Let $X$ be a reasonable space (say, a CW complex) and $G$ be a finite group acting on $X$. For all $k \geq 1$, the projection map $X \rightarrow X/G$ induces a map $H_k(X;\mathbb{Q}) \rightarrow H_k(X/G;\mathbb{Q})$ which factors through the $G$-coinvariants $(H_k(X;\mathbb{Q}))_G$; let $\psi_k : (H_k(X;\mathbb{Q}))_G \rightarrow H_k(X/G;\mathbb{Q})$ be the resulting map. I want examples of $X$ and $G$ and $k$ such that $\psi_k$ is not an isomorphism. If $G$ acts freely, then the map $X \rightarrow X/G$ is a finite regular covering map and $\psi_k$ is an isomorphism by (for instance) the Cartan-Leray spectral sequence (Theorem VII.7.9 in Brown's book on group cohomology). But I have no idea what happens for non-free actions. My guess is that if it were true that $\psi_k$ were always an isomorphism, then I would have seen it somewhere, so I expect that there is a counterexample. However, I have not managed to come up with one. REPLY [13 votes]: The maps $\psi_k$ are all isomorphisms; this is a simple application of the transfer ("averaging") construction. See Theorem 2.4, Chapter III, of Bredon's book "Introduction to compact transformation groups".<|endoftext|> TITLE: Characterization of ideals in the bounded operators QUESTION [6 upvotes]: Let $\mathcal{B}(H)$ denote the C*-algebra of all bounded operators on a separable infinite dimensional complex Hilbert space $H$. It is a well-known fact that $\mathcal{B}_0(H)$, the ideal of compact operators, is the unique, nontrivial, proper closed ideal in $\mathcal{B}(H)$. Furthermore, any proper ideal in $\mathcal{B}(H)$ must be contained in $\mathcal{B}_0(H)$. Is there any reasonable characterization of all (not necessarily closed) ideals in $\mathcal{B}(H)$ which contain the ideal of finite rank operators (possibly under some definability constraints, like being Borel in the strong operator topology, etc)? Also, not being an operator theorist, are there important ideals of this type other than $\mathcal{B}_0(H)$ and the Schatten $p$-ideals (including the trace-class and Hilbert-Schmidt operators)? REPLY [4 votes]: One of my ancient papers, joint work with Gary Weiss, concerns the question whether the ideal of compact operators is the sum of two properly smaller (and therefore not closed) ideals. We proved that the answer is affirmative if one assumes the continuum hypothesis (or various weaker assumptions). The paper contains a description of ideals that "reduces" this sort of question to set theory. Later work showed that the use of assumptions (like the continuum hypothesis) beyond the usual axioms of set theory is needed; it's consistent with ZFC that $K(H)$ is not the sum of two properly smaller ideals. The joint paper with Gary Weiss is "A characterization and sum decomposition of operator ideals" [Trans. A. M. S. 246 (1978) 407-417]. For the later work, see my paper "Near Coherence of Filters II: Applications to operator ideals, the Stone-Cech remainder of a half-line, order-ideals of sequences, and slenderness of groups" [Trans. A. M. S. 300 (1987) 557-581] and the references there.<|endoftext|> TITLE: Explicit reference on generator of $H^4(BQ_8;\mathbb{Z})\cong \mathbb{Z}_8$ identified with second Chern class of standard representation QUESTION [5 upvotes]: It is extremely well-known that $H^*(BQ_8;\mathbb{Z})=\mathbb{Z}[\alpha,\beta,\gamma]$ with relations $2\alpha=2\beta=8\gamma=\alpha^2=\beta^2=\alpha\beta-4\gamma=0$, $|\alpha|=|\beta|=2$ and $|\gamma|=4$. I was wondering if anyone can given an explicit reference showing that $c_2(\sigma)$ generates $H^4(BQ_8)\cong \mathbb{Z}_8$, where $\sigma: Q_8 \hookrightarrow SU(2)$ is the standard representation (i.e., $c_2(\sigma)=c_2(E_\sigma)$, $E_\sigma= S^\infty\times_{Q_8}\mathbb{C}^2)$. Any help would be greatly appreciated, as I would like to avoid the following reasoning if possible: $E_\sigma$ is the pullback of the canonical quaternionic line bundle $E=S^\infty\times_{SU(2)}\mathbb{C}^2\rightarrow \mathbb{H}P^\infty$ by the projection $S^3/Q_8\hookrightarrow BQ_8 \stackrel{\pi}{\rightarrow}\mathbb{H}P^\infty$, and Leray-Hirsch gives $H^4(\mathbb{H}P^\infty;\mathbb{Z}_2)\cong H^4(BQ_8;\mathbb{Z}_2)\cong\mathbb{Z}_2$. Looking at $H^4(\mathbb{H}P^\infty;\mathbb{Z})\cong \mathbb{Z}\rightarrow H^4(BQ_8;\mathbb{Z})\cong \mathbb{Z}_8$ then shows that $c_2(\rho)=\pi^*(c_2(E))$ must therefore represent an odd number in $\mathbb{Z}_8$. REPLY [2 votes]: It is enough to show that the reduction mod 2 of that class is not zero. However, reducing $c_2$ mod 2 gives the Stiefel-Whitney class $w_4$ of the underlying real representation. That this class is non-zero is classical, and explicitly stated and proved for example in Quillen's "The Mod 2 Cohomology Rings of Extra-special 2-groups and the Spinor Groups", Math. Ann. 194, 197--212 (1971). (A somewhat overkill reference, proving much more).<|endoftext|> TITLE: Does this matroid have a name? QUESTION [5 upvotes]: Sorry if this question is a dumb one. I used a special family of matroids in my research. One of them, of rank 3, can be represented by the following matrix over $\mathbb F_5$ or over $\mathbb R$: $\left(\begin{array}{cccccccccccc} 1&0&0&1 &1 &0 &-1&2 &-1& 2&0 &0\\ 0&1&0&-1&0 &1 &2 &-1&0 & 0&-1&2\\ 0&0&1&0 &-1&-1&0 &0 &2 &-1&2 &-1 \end{array}\right) $ It contains 12 points, 3 lines of 5 points, 7 lines of 3 points, all the other lines are of 2 points. It is a restriction of the Dowling geometry $Q_3(\mathbb F_5^\times)$. The other matroids of this type are represented over $\mathbb R$ by the matrix, the columns of which are all possible column vectors of length $n$ (the parameter) which contain either one nonzero entry (equal to $1$) or two such entries (equal to $1$ and $-1$, or $2$ and $-1$). I would like to ask: is this class of matroids known? Any reference is welcome. Here is a geometric representation of the matroid above. The picture should be seen as a figure in projective plane, the dashed circle corresponding to the line at infinity. Thanks! REPLY [8 votes]: Matrices that contain at most two non-zero entries per column are called frame matrices. The matroids representable by frame matrices (over a finite field $\mathbb{F}$) are in fact a fundamental class in the structure theorem of Geelen, Gerards and Whittle for the class of all $\mathbb{F}$-representable matroids. I think the term frame matroids is being floated around as a possible name (at least this is what Geelen called them during a talk this week). This sort of conflicts with some earlier naming conventions though. For more information, check out the Matroid Union Blog.<|endoftext|> TITLE: Maximum of two normal random variables QUESTION [14 upvotes]: The main purpose of the following question is to get some intuition and deeper understanding why the presented method works which would hopefully help me in trying to adapt it to the setting I am dealing with in my research. Let $X,Y\sim N(0,1)$, not necessarily independent. Suppose we want to find an upper bound for $\mathbb{E}\max(X,Y)$. The most obvious approach would be something like the following $$\mathbb{E}\max(X,Y)\leq \mathbb{E}|X|+\mathbb{E}|Y|=2\sqrt{2/\pi}\approx 1.59$$ However, I've found the trick in the literature that uses Laplace transform to get something better. Although the idea is much less obvious, details are still easy. For any $\lambda >0$, Jensen's inequality gives us the following $$\mathbb{E}\max(X,Y) \leq \frac1{\lambda}\log\left(\mathbb{E}e^{\lambda\max(X,Y)}\right)\leq \frac1{\lambda}\log\left(\mathbb{E}e^{\lambda X}+\mathbb{E}e^{\lambda Y}\right) = \frac{\log(2e^{\lambda^2/2})}{\lambda}.$$ Minimizing this gives us the upper bound $\log(4) \approx 1.17$, which is better than the previous approach. Now, my question is, heuristically/intuitively, why is second method better? Or to put in a different way, is there some easy way to see that the second method should give a better bound even before doing the actual calculations that confirm this? At this stage, I don't have any intuition for why this works, and I am certainly not fine with that's the standard trick researchers in the field use. REPLY [4 votes]: Given: $X \sim N(0,1)$ and $Y \sim N(0,1)$ where $X$ and $Y$ may be correlated. If we assume that the underlying Normals are jointly Normal, then (a) the exact answer is quite simple, and (b) we can do much better, conditioning the maximum bound on the correlation coefficient. In particular, if $(X,Y)$ are bivariate Normal with correlation $\rho$, then the joint pdf $f(x,y)$ is: Then, $E\big[Max[X,Y]\big]$ is: where I am using the Expect function from the mathStatica package for Mathematica to automate the nitty gritties. Plainly, the expected maximum is contingent on $\rho$, as a quick plot illustrates: As @oferzeitouni noted, the maximum possible value is $\sqrt{\frac{2}{\pi}}$, which is attained when $\rho = -1$.<|endoftext|> TITLE: What axioms (other than choice) have a taming effect on the ordering of cardinalities? QUESTION [5 upvotes]: Axiom of choice arranges all cardinalities into a well-ordered chain but without it their ordering can be wild in general ZF models, e.g. two cardinalities may not even have inf or sup. However, according to Woodin in $ZF+DC+AD_{\mathbb{R}}$ low uncountable cardinalities form a nice lattice, and according to Caicedo and Ketchersid in $ZF+AD^++V=L\big(\mathcal{P}(\mathbb{R})\big)$ if $S$ is strictly larger than a well-ordered cardinal $\kappa$, then either $\kappa^+$ or $\kappa\cup\mathbb{R}$ embeds into $S$, and "$\omega_1$ and $\mathbb{R}$ form a basis for the uncountable cardinals" (not sure what basis means in this context). What other axioms/ZF models are known to have some taming effect on the ordering of cardinalities? Do weaker forms of choice like "every set is linearly orderable" have such effect? References appreciated. REPLY [3 votes]: I don't have anything in mind which "tames" the cardinals structure. I also think that you don't fully understand the consequences of $\sf AD$ on the cardinal structure if you say that it tames the structure down. (Note that even Woodin, in the abstract you reference, points out that the cardinal structure is very complicated.) But let me just point out the following. The axiom "Every set is linearly orderable" follows from the Boolean Prime Ideal theorem, which is shown to hold in Cohen's first model. In that model there exists a Dedekind-finite set. Therefore the example I gave you in the math.SE question about the failure of infimum of two sets holds in that model. Now, without sitting to verify the details, here's a suggestion as to how to destroy the supremum property while still having "Every set is linearly orderable". By adding $\aleph_0$ Cohen reals, and $\aleph_1$ Cohen subsets of $\omega_1$, and taking automorphisms of the forcing which look like the ones used in Cohen's first model (namely, an automorphism of a disjoint union of $\omega$ and $\omega_1$ which doesn't "mix" the two parts), and similar ideal and so on, one should be able to get a model in which there are two incomparable Dedekind-finite sets, but every set can be linearly ordered. In that case, the generalization of the example in which $\sup$ fails can be executed. (I should say that I don't remember at the moment whether or not the old question "If there is an infinite Dedekind-finite cardinal, then there are two incomparable Dedekind-finite cardinals" is answered. In case that the answer is positive, then Cohen's first model is a model for the failure of $\sup$ for cardinals as well.)<|endoftext|> TITLE: When are quotients of the diffeomorphism group Fréchet manifolds? QUESTION [8 upvotes]: Let $M$ be a compact manifold and $\text{diff}(M)$ its diffeomorphism group. Various quotients of $\text{diff}(M)$ appear in the literature, oftentimes with geometric significance. A well-known example is the space $\text{diff}(M)/\text{diff}_\mu(M)$, where $\mu$ is a measure on $M$ of total volume one and $\text{diff}_\mu(M)$ is the group of diffeomorphisms of $M$ that preserve $\mu$. (See Hamilton's Nash-Moser paper for more details.) Questions: (1) Is $\text{diff}(M)/\text{diff}_\mu(M)$ a tame Fréchet manifold? (2) Is $\text{diff}_\mu(M)$ a tame Fréchet submanifold of $\text{diff}(M)$? (Hamilton's paper says that this is an open problem. Has any progress been made on this since 1982?) (3) If $G$ is a subgroup of the diffeomorphism group, when is $\text{diff}(M)/G$ a tame Fréchet manifold? I know only of a result that addresses (3) on totally different lines. This is the Gleason-Yamabe theorem. It says that if $G$ is locally compact, then for any open neighborhood $U$ of the identity there exists a subgroup $G'$ of $G$ contained in $U$ and a compact $G'$-normal subgroup $K\subset U$ such that $G'/K$ is isomorphic to a Lie group. (Terence Tao has written about this theorem in connection with Hilbert's Fifth Problem.) Since I want to quotient out the full diffeomorphism group by a subgroup, this won't do. REPLY [6 votes]: (1) Yes, it is the set of smooth positive densities of the same total volume as $\mu$. This can be seen by the "Moser trick" directly, and it is a one dimensional affine subspace of a space of smooth post ice sections of a line bundle, that tame. (2) Yes, it is, see MR2670086 (2011m:58006) Reviewed Molitor, Mathieu(CH-LSNP-SM) The group of unimodular automorphisms of a principal bundle and the Euler-Yang-Mills equations. (English summary) Differential Geom. Appl. 28 (2010), no. 5, 543–564. for a version in a more complicated situation. The original version is in Molitor's thesis. (3) If $G$ is a tamely splitting submanifold, then one can use the Hamilton-Nash-Moser implicit function theorem. If $G$ is the isotropy group of some geometric object on which $Diff(M)$ acts, often one can identify the the quotient.<|endoftext|> TITLE: A proof for $\dim(R[T])=\dim(R)+1$ without prime ideals? QUESTION [101 upvotes]: If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$, where $\dim$ denotes the Krull dimension. If $R$ is Noetherian, we have equality. Every proof I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem. Recently T. Coquand and H. Lombardi have found a surprisingly elementary "almost" first-order characterization of the Krull dimension (see here), which in particular does not use prime ideals at all. For $x \in R$ let $R_{\{x\}}$ be the localization of $R$ at $x^{\mathbb{N}} (1+xR) \subseteq R$. Then we have $$\qquad \dim(R) = \sup_{x \in R} \left(\dim(R_{\{x\}})+1\right)\!. \qquad (\ast)$$ It follows that for $k \in \mathbb{N}$ we have $\dim(R) \leq k$ if and only if for all $x_0,\dotsc,x_k \in R$ there are $a_0,\dotsc,a_k \in R$ and $m_0,\ldots,m_k \in \mathbb{N}$ such that $$x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)=0.$$ You can use this to define the Krull dimension. A consequence of this is a new short proof of $\dim(K[x_1,\dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $\dim(R\otimes_K S)=\dim(R)+\dim(S)$ if $R,S$ are finitely generated commutative $K$-algebras. In particular $\dim(R[T])=\dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory. Question. Can we use the characterization $(\ast)$ of the Krull dimension by Coquand-Lombardi above to prove $\dim(R[T])=\dim(R)+1$ for Noetherian commutative rings $R$? Such a proof should not use the prime ideal characterization/definition of the Krull dimension. Notice that the claim is equivalent to $\dim(R[T]_{\{f\}}) \leq \dim(R)$ for all $f \in R[T]$. Maybe this question is a bit naïve. I suspect that this can only work if we find a first-order property of rings which is satisfied by Noetherian rings and prove the formula for these rings. Notice that in contrast to that the Gelfand-Kirillov dimension satisfies $\mathrm{GK}\dim(R[T])=\mathrm{GK}\dim(R)+1$ for every $K$-algebra $R$. REPLY [17 votes]: There are two kinds of primes $\mathfrak q \subset R[T]$. The first possibility is that $\mathfrak q = \mathfrak p[T]$ with $\mathfrak p = R \cap \mathfrak q$. Let us call such a prime "small". The second possibility is that the inclusion $\mathfrak p[T] \subset \mathfrak q$ is strict. Let us call such a prime "big". Suppose we have a sequence of primes $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset\mathfrak q_r$ in $R[T]$. Then we get a sequence of big/small. If the sequence has only one switch from small to big, then of course $r \leq \dim(R) + 1$. The problem comes from a sequence with multiple switches. But thinking about it for a moment we see that it suffices to prove the following. Scholium: If $\mathfrak q_0 \subset \mathfrak q_1$ in $R[T]$ lies over $\mathfrak p_0 \subset \mathfrak p_1$ in $R$ and if $\mathfrak q_0$ is big and $\mathfrak q_1$ is small, then there is a prime strictly in between $\mathfrak p_0$ and $\mathfrak p_1$. To prove this we argue by contradiction and assume there is no prime strictly in between. Observe that in any case $\mathfrak p_0 \not = \mathfrak p_1$ by our definition of big and small. After replacing $R$ by $(R/\mathfrak p_0)_{\mathfrak p_1}$ we reach the situation where $R$ is a local Noetherian domain of dimension $1$. Then $\mathfrak p_0 = (0)$ and $\mathfrak p_1 = \mathfrak m$ is the maximal ideal. Translating we have to derive a contradiction from the following: we have a nonzero prime $\mathfrak q \subset \mathfrak m[T]$ with $\mathfrak q \not = \mathfrak m[T]$. Let $K$ be the fraction field of $R$. Let $\mathfrak q_K \subset K[T]$ be the ideal generated by $\mathfrak q$ in $K[T]$. Then $\mathfrak q = \mathfrak q_K \cap R[T]$. For every $n \geq 0$ let $R[T]_{\leq n}$ be the polynomials of degree $\leq n$. Let $M_n = \mathfrak q \cap R[T]_{\leq n}$ and $Q_n = R[T]_{\leq n}/M_n$ so that we have a short exact sequence $$ 0 \to M_n \to R[T]_{\leq n} \to Q_n \to 0 $$ Now observe that $Q_n$ is a finite $R$-module, is torsion free, and has rank bounded independently of $n$. Namely, over $K$ we know that $\mathfrak q_K$ is generated by a polynomial of degree $d$ and we see that $Q_n \otimes_R K$ has dimension over $K$ at most $d$. Pick $a \in \mathfrak m$ nonzero. Then (1) $R/aR$ has finite length $c$, (2) for any finite torsion free module $Q$ of rank $r$ the length of $Q/aQ$ is $rc$, and (3) a module $Q$ with length $Q/aQ$ bounded by $rc$ is generated by $\leq rc$ elements. [Hints for elementary proofs: To prove (1) you show for any $b \in \mathfrak m$ some power of $b$ is in $aR$ otherwise $R/aR$ would have a second prime. To prove (2) you choose $R^{\oplus r} \subset Q$ and you use the snake lemma for multiplication by a on the corresponding ses. To prove (3) use Nakayama and that a finite length module is generated by at most its length number of elements.] Take $n > dc$ where $d$ is the upper bound for the ranks of all $Q_n$ found above. Then we conclude that there exists an element in $M_n$ which is not in $\mathfrak m(R[T]_{\leq n})$ because we have seen above that $Q_n$ can be generated by $\leq dc$ elements. Small standard argument omitted. This is the desired contradiction because we assumed $\mathfrak q \subset \mathfrak m[T]$. QED This answer shows that with usual commutative algebra there is a very short proof. Enjoy!<|endoftext|> TITLE: Descent of functions along finite birational morphisms QUESTION [13 upvotes]: Let $A\to B$ be a morphism of (unitary commutative) rings such that $B$ is module-finite over $A$ and there exists $f\in A$ which is a nonzerodivisor in $A$ and in $B$, with $A[1/f]\to B[1/f]$ an isomorphism. My question is: is the diagram $A\to B \rightrightarrows B\otimes_A B$ exact ? In geometric terms, this is more or less equivalent to the question whether descent of functions along a finite, scheme-theoretically birational morphism is possible. Here are some comments: Of course, I'm mainly interested in the non-flat case (otherwise it's just faithfully flat descent). I computed several examples, including the normalization of the scheme formed by $n$ lines meeting in the origin in $\mathbf{A}^2$ (i.e. the scheme $V(x^n-y^n)\subset\mathbf{A}^2$); and the resolution $k[x^n,y^n,xy]\to k[x,y]$ of a plane quotient singularity. Any time, the answer is yes. I suspect that the result might be known and maybe present somewhere in the literature. The question is related also to descent of modules for universally injective morphisms, as exposed in the Stacks Project here but I could not find there something really close to the statement I'm after. Thanks for any help! EDIT (Jan 1st, 2016): the claim that the normalization of the scheme formed by $n$ lines meeting in the origin in $\mathbf{A}^2$ is an example is not true if $n>2$. Explanations are given in this MO answer. REPLY [8 votes]: Apologies for a long comment masquerading as an answer. I think you were on the right track when you mentioned universally injective ring homomorphisms, which are also often called pure ring homomorphisms. You can find a well developed geometric theory of such homomorphisms in papers of Mesablishvili. The theory includes a generalization of the result that you and user52824 worked out in your answer. See, in particular, Theorem 6.5 of "More on descent theory for schemes," which states "A quasi-compact morphism of schemes is stable schematically dominant if and only if it is pure if and only if it is a stable regular epimorphism." Note that there is no "finite" hypothesis and no Noetherian hypothesis in the theorem. The easiest way to get a pure ring homomorphism $R\to S$ is to have an $R$-module retraction $S\to R$. One often calls such a retraction a ``Reynolds operator'' because of the representation-theory situation (invariants of a linearly reductive group) that gives rise to such retractions. In fact, an elementary argument shows that if $S$ is finitely presented as an $R$-module (which is the situation you consider) and pure, then there is such a retraction. For a proof that an $R$-algebra finitely presented as an $R$-module is pure if and only if it admits an $R$-module retraction onto $R$, one can see Corollary 5.3 of Hochster/Roberts "The purity of the Frobenius and local cohomology." The paper is now freely available online thanks to Elsevier's recent opening of access to older math articles. Although the corollary is in the middle of the paper, it belongs to a section about the basics of pure morphisms, and its proof is entirely contained in Section 5. To get endless examples of pure ring homomorphisms, you can thus use representation theory: let $k$ be a field, let $G$ be a finite group whose order is invertible in $k$, and let $V$ be a finite-dimensional representation of $G$ over $k$. The inclusion $k[V]^G \to k[V]$ is then a pure ring homomorphism, since we have a $k[V]^G$-module retraction (the Reyonlds operator mentioned earlier) $k[V]\to k[V]^G$ defined by averaging over $G$. This sort of example generalizes your quotient-singularity computation.<|endoftext|> TITLE: $6$ points lie on a conic if and only if $ABC$ and $A_0B_0C_0$ are perspective QUESTION [9 upvotes]: Let $ABC$ be a triangle with incircle $\omega$. Let $A_0,B_0,C_0$ be points outside $\omega$. The tangents from $A_0$ to $\omega$ intersect $BC$ at $A_1,A_2$. Define $B_1,B_2$ and $C_1,C_2$ similarly. Is it true that $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a conic $\Gamma$ if and only if $\Delta ABC$ and $\Delta A_0B_0C_0$ are perspective from a point? I find this to be true in all cases I could possibly check using Geogebra. I suppose it is true. Yet I have no proof or reference on this fact. So something on this topic would be helpful. Thanks a lot. REPLY [5 votes]: As noted by Todd this is a problem in projective geometry of conics. So we can use the duality principle and reformulate the "only if" part by the following statement (the other part is equivalent): Suppose $C$ and $D$ are two conics in the projective plane. For every point $A$ in $C$, we associate a line $T(A)$ in this way: Let $A_1,A_2$ are two other intersection points of tangents of $A$ to $D$ with the conic $C$. Define $T(A)$ to be the line $A_1A_2$. Then for every three points $X,Y,Z$ on $C$, $\Delta XYZ$ is perspective to the triangle with edges $T(X),T(Y), T(Z)$. To see the equivalence with the original problem, consider $X,Y,Z$ to be the dual points of edges of $\Delta ABC$, $C$ the dual of $\omega$ and $D$ the dual of the conic through $A_1,A_2,B_1,B_2,C_1,C_2$. (All conics here are nonsingular.) We use two lemmas: Lemma 1: Two triangles are perspective iff there exists a polarity which sends every vertex of one of them to the corresponding edge of the other. Lemma 2: Let $T$ ba a correlation (projective isomorphism of the plane to its dual) and $C$ a conic. If for every points $x,y\in C$, $x \in T(y) \Longleftrightarrow y\in T(x)$, then $T$ is a polarity. Proof: $T$ is equal to its dual on $C$. Now for the proof of our statement it suffices to see that $T$ is an algebraic isomorphism between $C$ and dual of a conic in the plane. (The only thing that should be checked is injectivity of $T$ which is true in the generic case. More precisely when there not exists a quadrilateral inscribed in $C$ and circumscribed about $D$.) So $T$ can be extended to a correlation in the plane which obviously satisfies the condition of Lemma 2 for $C$. Therefore $T$ is a restriction of a polarity and Lemma 1 completes the proof.<|endoftext|> TITLE: Kodaira-Spencer theory of deformation done right QUESTION [14 upvotes]: I thought in asking this question on Math StackExchange, but by my experience I don' t think anyone will notice me. Recently, I started studying deformation of complex manifolds in the sense of Kodaira and Spencer, but some things relating infinitesimal approximations are not clear for me. Let $\varpi: \mathscr{M} \twoheadrightarrow D$ be a analytic family of compact complex manifolds such that $D \in \mathbb{C}^m$ is disk centered at the origin, $M = M_0$ is covered by coordinate charts $\{ U_i \}$ and the radius of $D$ is small enough to $\{ U_i \times D \}$ cover $\mathscr{M}$. Now, let $(\zeta_i^1 (z_i , t), …,\zeta_i^n(z_i^n, t), t^1, …,t^m)$ denote the coordinates of $U_i \times D$, then, in the literature, without any formal justificative, it's stated that $$\frac{\partial \zeta_i^j}{\overline{\partial z_i}^k} = \sum_{l} \varphi_{lk}^i (t) \frac{\partial \zeta_i^j}{\partial z_i^l}$$ holds. The informal justificative usually done is by saying that if $t$ is close enough to $0$, then $pr^{(0, 1)}|_{T_t^{(0, 1)}} : T_t^{(0, 1)} \rightarrow T^{(0, 1)} $ defines an isomorphism. So it is possible to construct a map $$\varphi(t) = pr^{(1, 0)}\circ (pr^{(0, 1)}|_{T_t^{(0, 1)}})^{-1}: T^{(0, 1)} \longrightarrow T^{(1, 0)} $$ and it satisfies $(1 + \varphi(t)) (v) \in T_t^{(0, 1)}$ for all $v \in T^{(0, 1)}$. Clearly, $pr^{(0, 1)}|_{T_t^{(0, 1)}} : T_t^{(0, 1)} \rightarrow T^{(0, 1)} $ being an isomorphism is the same as the complex strutcture being the same for all $t$, so how is possible to make this construction of $\varphi$ more formal? Why the equality above holds? Maybe the assumption "t close enough to 0" should be formalized in terms of nilpotent elements, something like a kind of prorepresentable functor, such that $pr^{(0, 1)}|_{T_t^{(0, 1)}} : T_t^{(0, 1)} \rightarrow T^{(0, 1)} $ being an isomorphism makes sense. Thanks in advance. EDIT As noted by Peter Dalakov, the assertion that the splitting of $T_{\mathbb{C}}$ does not change, is totally wrong. So the remaining question is: why $$\frac{\partial \zeta_i^j}{\overline{\partial z_i}^k} = \sum_{l} \varphi_{lk}^i (t) \frac{\partial \zeta_i^j}{\partial z_i^l}$$ holds? REPLY [7 votes]: I am not sure which book you are using . But from Kodaira, Morrow's book $\it complex$ $\it manifolds$ page 149-151, the construction of vector-value $(0,1)-$form $\varphi$ will tell you why the equality holds, also you will see why $t$ is required to be close enough to 0.<|endoftext|> TITLE: Verdier localization for stable $\infty$-categories QUESTION [12 upvotes]: Verdier localization is one of the more intuitive ways to localize a triangulated category, "killing" a suitable class of objects via a functor which is universal with respect to this property. I would like to know whether it is possible to reproduce the construction of $\mathcal{T}/\mathcal C$ in a $\infty$-stable setting. It seems a well-established folklore that the category of (a model for) stable $\infty$-categories is stable (!) under this sort of operation, but I can't find a reference in Lurie HA1. In the DG model, there is a construction by Drinfeld which seems to do the job, but instead I would like to reproduce the fairly general construction of Neeman (Triangulated Categories), Ch. 2. Did anybody do this naive construction? Or rather there is a more conceptual approach? Presenting a stable $\infty$-category via a stable model category, is Bousfield (which is, as far as I understand, only a particular case of Verdier) localization enough to cover "all" the interesting cases? REPLY [12 votes]: One good source is Blumberg-Gepner-Tabuada, "A Universal Characterization of Higher Algebraic K-Theory." See Definition 5.4, which defines the Verdier quotient as the cofiber (In the oo-category of presentable stable oo-categories) of the fully faithful functor $\mathcal{C} \to \mathcal{T}$. Proposition 5.6 shows it can be characterized as a Bousfield localization, localizing at the morphisms whose cofibers are in the image of $\mathcal{C}$. Proposition 5.9 deals with Adeel's comment: It shows $Ho(\mathcal{T})/Ho(\mathcal{C}) \simeq Ho(\mathcal{T}/\mathcal{C})$.<|endoftext|> TITLE: Longest of random worm-like paths in $\mathbb{Z}^2$ QUESTION [17 upvotes]: Imagine at each lattice point of $\mathbb{Z}^2$ within $[1,3n]^2$, with coordinates $\equiv 2 \bmod 3$, we place, with equal probability, one of these six patterns:       The result is collection of disjoint "worm-like" paths, whose minimum length is $3$. For example, here is an example for $n=10$:       The longest path here starts at $(1,14)$, and has length $27$. My question is: Q. What is the growth rate of the longest path, with respect to $n$? With $10$ random trials each, the average longest path for $n=10$ is actually considerably smaller than $27$; it is in fact $18.3$. Here is a graph up to $n=50$; it appears to grow (with considerable variability) roughly proportional to $\sqrt{n}$. Is there some relatively straightforward way to see what is the expected growth rate of the longest path?       I gratefully acknowledge programming assistance from several users in response to this posting @Mathematica Stack Exchange. REPLY [16 votes]: I'll expand a bit on my comment. There are $n^2$ $3 \times 3$ tiles. From each, there are two directions you can follow the path. As you move along the path in one direction, you hit a new tile, a previously visited tile, or the edge of the region. The path only gets longer if it encounters a new tile and the tile has one of the patterns connecting it to that side, which happens with conditional probability $1/2$ since $3$ of $6$ tiles have paths connecting with any particular side. So, the number of steps you can take in each direction is dominated by a geometric random variable which has probability $1/2^n$ of being at least $n$. If there is a worm of length $L+1$ tiles, then at least one of the $n^2$ tiles has at least one direction where this geometric random variable is at least $L$. The expected count of these is $2n^2/2^L$. (Expectation is linear regardless of dependencies.) When $L=\log_2 (2n^2) = 1+2\log_2 n,$ the expected count is at most $1$. The expected number of worms of length at least $2 \log_2 n +2 + c$ is at most $1/2^c$, so the probability that there is at least one such worm has probability at most $1/2^c$. For any constant $k \gt 2$, the probability that there is a worm of length $k \log_2 n$ goes to $0$ as $n\to \infty$.<|endoftext|> TITLE: A special case of the integer Hodge conjecture QUESTION [11 upvotes]: Let $X$ be a projective complex manifold of dimension $n$. Are torsion cohomology classes in $H^{2n-2}(X,\mathbb{Z})$ algebraic? (We may assume, without loss of generality, that $n=3$, because of the Lefschetz hyperplane theorem.) I know that torsion classes (of even codimension) aren't always algebraic; the first counterexamples were found by Atiyah and Hirzebruch in 1960s. But I do not know any counterexample in this codimension. Note that by Poincare duality $H^{2n-2}(X,\mathbb{Z})\cong H_2(X)$, so the equivalent question is whether torsion classes in $H_2(X)$ are generated by algebraic curves in $X$. This looks like a "dual" to the well known fact that torsion classes in $H^{2}(X,\mathbb{Z})$ are generated by divisors (it is a torsion part of the Neron-Severi group). REPLY [3 votes]: This 2013 paper of Totaro says that it is an open question. But the integral Hodge conjecture is known to fail for dimension 1, just not via torsion.<|endoftext|> TITLE: Relationship of Euler product to coprimality densities for arbitrary sets of primes QUESTION [5 upvotes]: Continuing the curiosity of my last couple questions: Is it the case that for every set of primes $F$, the asymptotic density of the integers coprime to all of $F$ is $\displaystyle \prod_{p \in F} (1 - 1/p)$? (The case of finite $F$ is obvious. Furthermore, as the product is an upper bound on the density, this holds whenever the product goes to zero; in particular, by the divergence of the harmonic series, this holds for all co-finite $F$. But that leaves a lot of ground untouched. Ideally, there'd be some way to extend the reasoning from the finite case to the general case, but it's not clear to me how to do so given the general lack of countable additivity of density.) REPLY [11 votes]: GH from MO is answering a more difficult question where one asks for uniform estimates for the number of integers up to $x$ not divisible by primes in a set $F$ (which may also depend on $x$). This is delicate, and the answer is indeed no. The problem, as I understand it, keeps $F$ fixed, and asks for the density as $x\to \infty$. This is easier, and the answer is in fact yes. As the OP observes, if the set $F$ is finite the result is true; and also if $\prod_{p\in F} (1-1/p)=0$ then the result is true (take the restriction of $F$ to primes below some point $N$, and note that the density of the numbers coprime to $F_N$ is an upper bound for what we want; and then take $N\to \infty$). Now consider a set $F$ with $\prod_{p\in F}(1-1/p) >0$. Therefore if $G$ denotes the primes not in $F$ then $\prod_{p\in G} (1-1/p)=0$. Put $\mu_F(n)$ to be a multiplicative function defined by $\mu_F(p)=-1$ if $p\in F$ and $\mu_F(p) =0$ otherwise; put $\mu_F(p^k)=0$ for all prime powers with $k\ge 2$. So this is a small generalization of the Mobius function. Then $$ \sum_{\substack{{n\le x} \\ {(n,F)=1}} } 1 = \sum_{n\le x} \sum_{d|n} \mu_F(d) = \sum_{d\le x} \mu_F(d) \Big(\frac xd +O(1)\Big). $$ By assumption $\sum_{p\in F} 1/p <\infty$ and so the Euler product $\prod_{p\in F}(1-1/p)$ converges absolutely (to a non-zero real number) and equivalently the sum $\sum_{n=1}^{\infty} \mu_F(n)/n$ converges. Thus the main term above equals $$ x\sum_{d=1}^{\infty} \frac{\mu_F(d)}d + o(x)= x \prod_{p\in F}\Big(1-\frac{1}{p}\Big) + o(x). $$ As for the remainder term, since $\mu_F(d) =0$ unless $d$ is divisible only by primes in $F$, the remainder term is at most the number of integers up to $x$ that are coprime to all elements in $G$; but we know that the density of such integers tends to zero for large $x$. That completes the proof. More generally, Erdos asked whether for any real valued multiplicative function $f$ with $-1\le f(n)\le 1$ for all $n$ we have $$ \lim_{x\to \infty} \frac{1}{x} \sum_{n\le x} f(n) = \prod_{p} \Big(1+\frac{f(p)}{p} + \frac{f(p^2)}{p^2}+\ldots \Big) \Big(1-\frac 1p\Big). $$ Your question is a special easy case of this problem; Erdos's problem also includes the prime number theorem as the special case when $f(n)=\mu(n)$. Erdos's problem was fully solved by Wirsing, who showed that the limit does equal the expected Euler product. The extension to complex valued multiplicative functions was made by Gabor Halasz (another GH, not from MO!).<|endoftext|> TITLE: On the fundamental group of closed 3-manifolds QUESTION [5 upvotes]: I know that every finitely presented group can be realized as the fundamental group of a compact, connected, smooth manifold of dimension 4 (or higher). In dimension 2 there are strong restriction on the fundamental group of closed manifolds. I wanted to know what happen in dimension 3: which are the conditions on a finitely presented group to be realized as the fundamental group of a closed (connected) 3-manifolds? I know there are strong restriction in dimension 3, but is there an "if and only if" characterization of fundamental groups of closed 3-manifolds? In affirmative case, can you suggest me some reference? REPLY [7 votes]: Igor's suggestion of the recent paper by Aschenbrenner, Friedl, and Wilton is probably the best place to start as it has a good treatment this problem which includes both a summary of recent advances and a litany of open problems. If you want to work through a classification of geometric and non-hyperbolic manifold groups, Thurston's book "Three-Dimensional Geometry and Topology" (especially sections 4.3, 4.4. and 4.7) is very handy. I have also found the wikipedia article Seifert fiber spaces together with Scott's article: The geometries of 3-manifolds (errata here) to be especially useful in dealing with the groups of Seifert fiber spaces. Finally, Groves, Manning, and Wilton provide a theoretical algorithm for determining if a group with solvable word problem is a three manifold group in this preprint.<|endoftext|> TITLE: Arbitrarily long arithmetic progressions QUESTION [10 upvotes]: Are there arbitrarily long arithmetic progressions in which all the prime factors of all the terms are at most $N$, for some $N$? Assume all the terms are positive and the sequence of terms is increasing. I have proved that no such infinite sequence exists. Note the $N$ may vary from AP to AP. For infinite sequences let $\{a+nd\}_{n\ge 0}$ be the AP. $\text{gcd}(a,d)=s$ then $a+nd=s(x+ny)$ for some $x,y$ with $\text{gcd}(x,y)=1$ then by Dirichlet's theorem $\{x+ny\}_{n\ge 0}$ has infinitely many primes. Thus we have the prime factors of the sequence is unbounded and hence done. But I was thinking about this claim but nothing came in mind. Could someone help? Thanks a lot. REPLY [3 votes]: This is a supplement to the elementary proofs given by The Masked Avenger and Lucia. Theorem. For any $k$ there is $m$ such that no arithmetic progression of length $m$ is supported on $k$ distinct primes. Proof. We prove by induction on $k$. For $k=0$ we can clearly take $m=2$. So let $k\geq 1$ and assume the statement for $k-1$ in place of $k$. That is, there is $m$ such that no arithmetic progression of length $m$ is supported on $k-1$ distinct primes. Consider an arithmetic progression of length $m'$ supported on the $k$ distinct primes $p_1,\dots,p_k$. Without loss of generality, the terms are coprime to the difference $d$. Observe that in any consecutive block of length $m$ in the progression, each prime $p_i$ occurs as a divisor by the induction hypothesis. Hence for $m'\geq m$ we get that $p_i\nmid d$, and for $m'\geq 2m$ we get that $p_i\mid kd$ for some $0 TITLE: Equivalence of exterior forms QUESTION [6 upvotes]: Let us start with the following definition. Let $1\leqslant k\leqslant n$ and let $\omega_1,\omega_2\in\Lambda^k(\mathbb{R}^n)$. We say that $\omega_1$, $\omega_2$ are equivalent, if there exists $T\in GL(n,\mathbb{R})$ such that $$ T^\ast\omega_1=\omega_2, $$ where $T^\ast:\Lambda^k(\mathbb{R}^n)\rightarrow\Lambda^k(\mathbb{R}^n)$ is the standard lift of $T$ to $\Lambda^k(\mathbb{R}^n)$. It is known that: 1) When $k=1$, if $\omega_1,\omega_2$ are equivalent if and only if either $\omega_1,\omega_2\neq 0$ or $\omega_1=\omega_2=0.$ This is trivial to check. 2) When $k=2$, $\omega_1,\omega_2$ are equivalent if and only if, there exists $r\in\mathbb{N}$ such that $$ \omega_1^r,\omega_2^r\neq 0,\text{ and }\omega_1^{r+1}=\omega_2^{r+1}=0, $$ where power corresponds to the wedge power. A proof of this result can be found in Denis Serres' book (Matrices: Theory and Applications). QUESTION: Is there any such result when $k\geqslant 3$? Probably $k=n-1,n$ cases are easy to handle. What I'm curious to know if there are analogous results when $3\leqslant k\leqslant n-2$? REPLY [11 votes]: Normal forms for exterior forms is a classical subject. It is well-known that when $k=1,2,n{-}2,n{-}1$, or $n$ and when $(n,k)=(6,3),(7,3),(7,4),(8,3),(8,5)$, there are only a finite number of 'normal forms', i.e., orbits in $\Lambda^k(\mathbb{R}^n)$ under the action of $\mathrm{GL}(n,\mathbb{R})$. In particular, in these cases, the 'generic' orbit is an open subset of $\Lambda^k(\mathbb{R}^n)$ and these orbits (which are finite in number) are said to be stable. In the first set of cases (which the OP briefly discussed above), there is usually only one open orbit, though, in the case $(n,k) = (4m+2,4m)$ when $m>0$, it turns out that there are $2$ open orbits. For the five 'exceptional cases' listed above, one could consult Nigel Hitchin's discussion of stable forms. In these cases, typically there are $2$ open orbits, but in the cases $(n,k)=(8,3)$ or $(8,5)$, there are $3$ open orbits. In all other cases, there are continuous moduli.<|endoftext|> TITLE: Class number for binary quadratic forms discriminant $\Delta$ to class number $\mathbb Q(\sqrt \Delta)$ QUESTION [7 upvotes]: Jyrki Lahtonen has suggested I write a blog post relating binary quadratic forms to quadratic field class numbers, https://math.stackexchange.com/questions/209512/binary-quadratic-forms-over-z-and-class-numbers-of-quadratic-%EF%AC%81elds/209543#comment1727526_209543 The coincidence part of this is Jyrki bringing up this idea within a couple of weeks of Neil Sloane asking about programming for indefinite forms.... However, I remain a little uncertain about how this works with indefinite forms...from Buell, Binary Quadratic Forms, page 103, the group of binary form classes is isomorphic to the narrow class group of $\mathbb Q ( \sqrt \Delta)$ where $\Delta$ is the discriminant, where I suspect $\Delta$ must be a fundamental discriminant because multiplying by an integer square would not change a field extending $\mathbb Q.$ Then page 103, positive forms we are done, class group and narrow class group are isomorphic. Also done if there is a solution in rational integers to $u^2 - D v^2 = -4.$ Finally the problem: if there is no solution to $u^2 - D v^2 = -4,$ Buell says the class group is the squares of the narrow class group. Buchmann and Vollmer say, page 186, say the class group is a quotient of the narrow class group. Let's see., examples. I put Positive primes represented by indefinite binary quadratic form with this in mind. Cohen says that $\mathbb Q(\sqrt {205})$ has class number 2. There are four classes of indefinite binary forms of discriminant 205, and $u^2 - 205 y^2 = -4$ is impossible. So, we went from 4 to 2... In the paper with Pete Clark, he deliberately made no distinction between indefinite form $f$ and the form $-f.$ So, one possibility here is that we are just dividing by 2 to go from 4 to 2.. Maybe this is the quick version: as far as I can tell, if $1$ and $-1$ are distinct as binary forms of discriminant $\Delta,$ the principal genus has even size, call that $E.$ Suppose there are $G$ genera, so that the total number of classes of binary forms of this discriminant is $EG.$ What is the class number of $\mathbb Q ( \sqrt \Delta)?$ So, question, are the numbers the same for positive forms and when the principal form also represents $-1,$ but if indefinite and the principal form does not represent $-1,$ divide by $2?$ Perhaps I can use this to publicize an elementary trick, generally unknown: an indefinite form $\langle a,b,c \rangle$ with positive $\Delta = b^2 - 4 a c$ not a square is reduced, in the sense of Lagrange, Gauss, and Buell, if and only if: $$ ac < 0 \; \; \; \; \mbox{and} \; \; \; \; b > |a + c| $$ So, general Question: how to take the class number of binary forms of discriminant $\Delta,$ where either $\Delta \equiv 1 \pmod 4$ is squarefree, or $\Delta \equiv 0 \pmod 4$ and $\Delta/4 \equiv 2,3 \pmod 4$ and this time $\Delta/4$ is squarefree. To repeat some examples (I've got programs out the wazoo) jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle_All_Reduced 5 Sun Jun 22 20:03:44 PDT 2014 5 factored 5 1. 1 1 -1 cycle length 2 2. -1 1 1 cycle length 2 5 factored 5 1. 1 1 -1 cycle length 2 form class number is 1 jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle_All_Reduced 12 Sun Jun 22 20:03:52 PDT 2014 12 factored 2^2 * 3 1. 1 2 -2 cycle length 2 2. -1 2 2 cycle length 2 3. 2 2 -1 cycle length 2 4. -2 2 1 cycle length 2 12 factored 2^2 * 3 1. 1 2 -2 cycle length 2 2. -1 2 2 cycle length 2 form class number is 2 jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle_All_Reduced 85 Sun Jun 22 20:04:08 PDT 2014 85 factored 5 * 17 1. 1 9 -1 cycle length 2 2. -1 9 1 cycle length 2 3. 3 7 -3 cycle length 6 4. -3 7 3 cycle length 6 5. 3 5 -5 cycle length 6 6. -3 5 5 cycle length 6 7. 5 5 -3 cycle length 6 8. -5 5 3 cycle length 6 85 factored 5 * 17 1. 1 9 -1 cycle length 2 2. 3 7 -3 cycle length 6 form class number is 2 jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle_All_Reduced 136 Sun Jun 22 20:04:24 PDT 2014 136 factored 2^3 * 17 1. 1 10 -9 cycle length 4 2. -1 10 9 cycle length 4 3. 3 10 -3 cycle length 6 4. -3 10 3 cycle length 6 5. 9 10 -1 cycle length 4 6. -9 10 1 cycle length 4 7. 2 8 -9 cycle length 4 8. -2 8 9 cycle length 4 9. 3 8 -6 cycle length 6 10. -3 8 6 cycle length 6 11. 6 8 -3 cycle length 6 12. -6 8 3 cycle length 6 13. 9 8 -2 cycle length 4 14. -9 8 2 cycle length 4 15. 5 6 -5 cycle length 6 16. -5 6 5 cycle length 6 17. 5 4 -6 cycle length 6 18. -5 4 6 cycle length 6 19. 6 4 -5 cycle length 6 20. -6 4 5 cycle length 6 136 factored 2^3 * 17 1. 1 10 -9 cycle length 4 2. -1 10 9 cycle length 4 3. 3 10 -3 cycle length 6 4. -3 10 3 cycle length 6 form class number is 4 jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle_All_Reduced 205 Sun Jun 22 20:04:32 PDT 2014 205 factored 5 * 41 1. 1 13 -9 cycle length 4 2. -1 13 9 cycle length 4 3. 3 13 -3 cycle length 4 4. -3 13 3 cycle length 4 5. 9 13 -1 cycle length 4 6. -9 13 1 cycle length 4 7. 3 11 -7 cycle length 4 8. -3 11 7 cycle length 4 9. 7 11 -3 cycle length 4 10. -7 11 3 cycle length 4 11. 5 5 -9 cycle length 4 12. -5 5 9 cycle length 4 13. 9 5 -5 cycle length 4 14. -9 5 5 cycle length 4 15. 7 3 -7 cycle length 4 16. -7 3 7 cycle length 4 205 factored 5 * 41 1. 1 13 -9 cycle length 4 2. -1 13 9 cycle length 4 3. 3 13 -3 cycle length 4 4. -3 13 3 cycle length 4 form class number is 4 jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle_All_Reduced 221 Sun Jun 22 20:04:40 PDT 2014 221 factored 13 * 17 1. 1 13 -13 cycle length 2 2. -1 13 13 cycle length 2 3. 13 13 -1 cycle length 2 4. -13 13 1 cycle length 2 5. 5 11 -5 cycle length 4 6. -5 11 5 cycle length 4 7. 5 9 -7 cycle length 4 8. -5 9 7 cycle length 4 9. 7 9 -5 cycle length 4 10. -7 9 5 cycle length 4 11. 7 5 -7 cycle length 4 12. -7 5 7 cycle length 4 221 factored 13 * 17 1. 1 13 -13 cycle length 2 2. -1 13 13 cycle length 2 3. 5 11 -5 cycle length 4 4. -5 11 5 cycle length 4 form class number is 4 jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ in brief, 210 * 4 = 840, 24 reduced forms but 8 SL2 classes, 840 factored 2^3 * 3 * 5 * 7 1. 1 28 -14 cycle length 2 2. -1 28 14 cycle length 2 3. 2 28 -7 cycle length 2 4. -2 28 7 cycle length 2 5. 3 24 -22 cycle length 4 6. -3 24 22 cycle length 4 7. 6 24 -11 cycle length 4 8. -6 24 11 cycle length 4 form class number is 8 jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ REPLY [2 votes]: Alright, i have gotten the binary forms part of the story. I may or may not ever know the field side of things, that's life. From experiments with the website http://www.numbertheory.org/php/classnopos.html and comparison with my own programs, it appeared that the division in half amounted to identifying the distinct forms (whenever the principal form does not represent $-1$) $$ \langle \alpha, \beta, -\gamma \rangle \; \; \mbox{with} \; \; \langle -\alpha, \beta, \gamma \rangle, $$ where these are ``reduced'' in the sense of Gauss and Lagrange when $$ \alpha \gamma > 0 \; \; \mbox{and} \; \; \beta > |\alpha - \gamma|. $$ So it is needed to show that these really are distinct classes when $1$ and $-1$ are distinct as forms. However, this is not hard. If the two forms above are equivalent, then the opposite of $\langle \alpha, \beta, -\gamma \rangle$ in the form class group is $ \langle \gamma, \beta, -\alpha \rangle. $ Thus the hypothesis amounts to $$ 1 =\langle \alpha, \beta, -\gamma \rangle \circ \langle \gamma, \beta, -\alpha \rangle. $$ We are allowed to insist that $\gcd(\alpha,\beta ) = 1,$ important. Can always be arranged, although the result may not be ``reduced'' any longer. The algorithm of Shanks, Buell Binary Quadratic Forms, pages 64-65, tells us that, with $\gcd(\alpha,\beta ) = 1,$ $$ \langle \alpha, \beta, -\gamma \rangle \circ \langle \gamma, \beta, -\alpha \rangle = \langle \alpha \gamma, \beta, -1 \rangle. $$ In short, the hypothesis that $ \langle \alpha, \beta, -\gamma \rangle \; \; \mbox{and} \; \; \langle -\alpha, \beta, \gamma \rangle $ are equivalent implies directly that the principal form represents $-1.$ Very satisfying from my point of view. Note that we do not need to use Shanks, various books discuss the ``united forms'' approach of Dirichlet, pages 55-57 in Buell. On page 57, he confirms, with $\color{green}{\gcd(a_1,a_2,B)= 1},$ that $$ \langle a_1, B, a_2C \rangle \circ \langle a_2, B, a_1C \rangle = \langle a_1 a_2,B, C \rangle. $$ Dirichlet gives the same outcome as the Shanks method, but with no additional $\gcd$ assumptions, using $$ a_1 = \alpha, a_2 = \gamma, B = \beta, C = -1. $$ This is also Theorem 98 on page 138 of Leonard Eugene Dickson, Introduction to the Theory of Numbers. Put another way, when the principal form does not represent $1,$ we get a distinct form that does represent $-1,$ and Dirichlet says $$\color{magenta}{ \langle \alpha, \beta, -\gamma \rangle \circ \langle -1, \beta, \alpha \gamma\rangle = \langle -1, \beta, \alpha \gamma\rangle \circ \langle \alpha, \beta, -\gamma \rangle = \langle -\alpha , \beta, \gamma \rangle}. $$ As $\langle -1, \beta, \alpha \gamma\rangle$ is not the principal class, the result of the Gauss composition gives a different class from the original, therefore $\langle \alpha, \beta, -\gamma \rangle$ and $\langle -\alpha , \beta, \gamma \rangle$ must be distinct.<|endoftext|> TITLE: Fast quadratic norm algorithms QUESTION [7 upvotes]: Given rational numbers $a$ and $b$, what is the fastest way to determine whether there are any rational solutions to $a=x^2+by^2$? I am interested in the case where the numerator and denominator of $a$ have about $4$ digits, and those of $b$ have about $8$ digits. I am willing to restrict attention to solutions where $x$ and $y$ also have numerators and denominators that are not too large, if that will help. I would like to solve a large number of problems of this type, perhaps tens of millions, so efficiency is important. Both $a$ and $b$ will be different for each problem. I would also be interested in heuristics for roughly how often rational solutions exist. REPLY [6 votes]: This is a conic, so factoring the discriminant is important, then you can check local solubility at all bad primes. This gives whether it is solvable. With a little extra work (maybe negligible compared to factoring), Simon's algorithm will give you a solution. http://www.ams.org/journals/mcom/2005-74-251/S0025-5718-05-01729-1/ or alternatively Cremona-Rusin http://www.ams.org/journals/mcom/2003-72-243/S0025-5718-02-01480-1/ I would guess that in your application, the main thing is to be able to factor numbers whose size is 12 digits quite efficiently. As for heuristics, the local root number is probably $-1$ about the half the time at primes dividing the discriminant, and up to evenness of number of such primes I guess these are independent. So global solubility should depend on the number of prime factors. Doing a simple test (with small numbers) on 10000 random trials gave about 36% soluble with 3 prime factors, 20% with 4 prime factors, 10% with 5 prime factors, 5% with 6.<|endoftext|> TITLE: PBW basis and canonical basis QUESTION [8 upvotes]: Consider the example of $\mathfrak{g} = sl_3$. Then $$ \mathfrak{g} = \mathfrak{n} \oplus \mathfrak{h} \oplus \mathfrak{n}^{-}, $$ where $\mathfrak{n}$ is generated by $E_{12}, E_{13}, E_{23}$, $\mathfrak{h}$ is generated by $E_{11}-E_{22}, E_{22}-E_{33}$, $\mathfrak{n}^{-}$ is generated by $E_{21}, E_{32}, E_{31}$, $E_{ij}$ is a matrix with $1$ at $(i,j)$ and $0$ elsewhere. A PBW basis of $U(\mathfrak{n})$ is $$ B_1 = \{ E_1^a (E_1 E_2 - E_2 E_1)^b E_2^c \mid a, b, c \in \mathbb{N} \}. $$ There is another basis of $U(\mathfrak{n})$ called canonical basis which is given by $$ B_2 = \{ E_1^aE_2^bE_1^c \mid a+c \geq b \} \cup \{ E_2^aE_1^bE_2^c \mid a+c \geq b \}. $$ The basis $B_2$ has a property: given a lowest weight vector $v_0$ of a representation $V$ of $U(\mathfrak{n})$, the set $$ \{ b v_0 \mid b v_0 \ne 0 \} $$ is a basis of $V$. My question is: in general, how to compute canonical basis for $U(\mathfrak{n})$ explicitly like the above example. Can we derive a canonical basis from a PBW basis? Thank you very much. REPLY [6 votes]: It is possible to obtain the canonical basis from the PBW basis, as long as you are working in the quantum group Uq(n). The canonical basis seems to be inherently a quantum phenomenon, so it shouldn't be surprising that we want to compute in the quantum group. If Eπ is an indexing of the quantum PBW basis, then what one first proves is that applying the bar involution is unitriangular with respect to this basis. There is even a mathoverflow question about this: Convex PBW bases. Now it is a matter of linear algebra to prove that there exists a unique bar invariant basis bπ such that $$b_\pi=\sum_\sigma c_{\pi\sigma} E_\sigma$$ with cσσ=1, cσπ∈qℤ[q] if σ≠π and cσπ=0 unless σ≤π. This basis bπ is the canonical basis.<|endoftext|> TITLE: Polyhedron not circumscribed about a sphere QUESTION [10 upvotes]: Let $P$ be a polyhedron whose faces are colored black and white so that there are more black faces and no two black faces are adjacent. Show that $P$ is not circumscribed about a sphere. My teacher has conjectured this. I haven't been able to produce a counterexample, neither a solution. Is this claim true? If not a counterexample would be nice. Thanks in advance for helping. REPLY [18 votes]: This is exercise 21.3, of Mathematical Omnibus: Thirty Lectures on Classic Mathematics, by D.B. Fuks and Serge Tabachnikov. The solution is given on page 448: Assume that $P$ is circumscribed. Consider a face $A_1$, $A_2$,... $A_n$ and let $O$ be its tangency point with the sphere. Clearly, the sum of angles $A_i O A_{i+1}$ is $2\pi$. We shall sum up al these angles over all faces, taking the angles in the white faces with positive signs and the angles in the black faces with negative signs. Since there are more black faces, this sum, $\Sigma$, is negative. On the other hand, consider two adjacent faces with a common edge $AB$; see figure; The angles $AOB$ and $AO'B$ are equal. Indeed, revolve the plane $AOB$ about the line $AB$ (as if it were a hinge) until it coincides with the plane $AO'B$. This rotation takes point $O$ to $O'$ and hence the triangles $AOB$ and $AO'B$ are congruent. There are two kinds of adjacent faces; black-white and white-white; the former's contribution to $\Sigma$ cancels, and the latter contribute a positive number. Thus $\Sigma\geq 0$, a contradiction.<|endoftext|> TITLE: Convergence on a random graph QUESTION [7 upvotes]: Assume a directed graph $G = (V,E)$ is drawn from a random graph distribution, for instance Erdős–Rényi's $G(n,p)$ (but with directed edges). Let $S:V\rightarrow\mathcal{P}(V)$ be the direct successors function, that is $S(u) = (\left\{u\right\} \times V) \cap E$ Let $f_0: V\rightarrow\left\{0,1\right\}$ be an initial, arbitrary, labeling of vertices with $0$'s and $1$'s Define $$n_t(u,x) = \left|\left(\{u\} \cup S(u)\right) \cap f_{t}^{-1}(x)\right|$$ $n_t(u,.)$ is simply the number of vertices labelled one or zero among $u$ and its direct successors. Now define recursively: $$f_t(u) = \left\{ \begin{array}{cc} 0 & \textrm{if} & n_{t-1}(0) > n_{t-1}(1)\\ 1 & \textrm{if} & n_{t-1}(1) > n_{t-1}(0)\\ f_{t-1}(u) & \textrm{otherwise} & \end{array} \right.$$ Simply speaking, at each step $t$, a vertex's label is changed to reflect the majority among itself and its direct successors, or is unchanged in case of a tie. I am interested in the convergence of the sequence $f_t$, and in particular convergence to a limit that is constant over $V$ (all $1$'s or all $0$'s) for all initial conditions. How is the probability of convergence affected by the statistics of the graph? I'm sure the problem has been studied and I'd happily take some references on the topic. REPLY [2 votes]: Let $\Omega = \{0,1\}^V$ be the state space, and $T: \Omega \to \Omega$ the time evolution operator (so $f_{t+1} = T(f_t)$). Note that the evolution has a monotonicity property: if $f \le g$ (i.e. $f(u) \le g(u)$ for all $u \in V$) then $T(f) \le T(g)$. So starting from a given state we will either end up in a fixed point or a cycle that is an antichain for the partial order $\le$. It is quite possible to have a cycle: for example, consider an undirected $4$-cycle with initial configuration $f_0 = (0,1,0,1)$. Since each vertex's two neighbours have the opposite label to itself, we get $f_1 = (1,0,1,0)$, and then again $f_2 = f_0$. A configuration $f$ is a fixed point of $T$ if for each vertex $v$, its successors with opposite label to $v$ don't outnumber its successors with the same label as $v$ by more than $1$. Let's consider the Erdős–Rényi model, where $p \in (0,1)$ is the probability of any given pair $(v,u)$ being an arc. Suppose $f$ is a given configuration with $k$ $0$'s and $n-k$ $1$'s, $k = n/2 + s \sqrt{n}$. For a vertex $v$ labelled $0$, the numbers of $0$'s and $1$'s among its successors are independent binomial random variables $X_0$, $X_1$ with parameters $(k-1), p$ and $n-k, p$ respectively. $X_0 - X_1$ is then approximately normal with mean $(2k-1-n) p \approx 2 s \sqrt{n}$ and variance $(n-1) p (1-p)$. Thus $$P(T(f)(v) = f(v)) = P(X_0 + 1\ge X_1) \approx 1 - \Phi\left(\dfrac{(2k-1-n)p}{\sqrt{(n-1) p (1-p)}}\right) \approx \Phi(-s r)$$ where $\Phi$ is the standard normal random variable and $r = 2\sqrt{p/(1-p)} $. Similarly for a vertex labelled $1$, $P(T(f)(v) = f(v)) \approx \Phi(sr)$. Since the successors of different vertices are independent, $$P(f \ \text{is a fixed point}) \approx \Phi(-sr)^{n/2} \Phi(sr)^{n/2} \approx 2^{-n} \exp(-2 r^2 s^2 n/\pi)$$ It looks to me like this will imply that the expected number of fixed points that are not all $0$ or $1$ is $O(1/\sqrt{n})$. Thus almost certainly there are no such fixed points.<|endoftext|> TITLE: Equivalent Norms for the Dual of Sobolev / Bessel Spaces QUESTION [8 upvotes]: Using standard notation, we refer to $H^s(\mathbb R) = W^{s,2}(\mathbb R)$ to be the Sobolev Hilbert spaces. As is often the case, it's natural to then consider properties of functions in $H^s(\mathbb R)$ by looking at how it behaves in the Fourier domain, more specifically, we have that for any $s\in \mathbb{R}$, $ \begin{align*} H^s(\mathbb{R}) = \{ g \in L_2(\mathbb{R}) : \int (1+t^2)^{s}|\mathcal{F} g(t)|^2dt <\infty\}, \end{align*} $ where $\mathcal{F}$ is the Fourier transform. In particular to note is that this works for $s<0$ as well, and so negative order Sobolev spaces are easily defined. In particular, one can define the norm of $H^s(\mathbb{R})$ to be defined by the relation $\|g\|_s = \int (1+t^2)^{s}|\mathcal{F}g(t)|^2dt$. $\textbf{My first question:}$ I have seen, in particular the book Sobolev Spaces by Adams, Fournier (page 64), that another way to define the norm of the dual space of $H^s$, which appears to be $H^-s$ is given by the following: $$ \begin{align*} \|g\|^*_{-s} = \sup_{h\in H^{s}}\frac{\langle g,h\rangle}{\|h\|_s}, \end{align*} $$ where $\langle g,h\rangle$ is the standard $L_2$ inner product. I am having difficulty figuring out whether the norm defined by Fourier transformations or the one defined by using the dual are equivalent or not. The furthest I have been able to show is that $\|g\|^*_{-s} \leq (2\pi)^{-1}\|g\|_{-s}$ by Parceval's relation. $\textbf{My second question:}$ If there appears to be a relationship, then I would like to restrict attention to spaces of the form $H^{s}(A)$ where $A\subset \mathbb{R}$ with defined norm $\|g\|_{s,A} = \inf\{ \|g^{'}\|_s : g^{'}_{|A} = g\}$ for all $g\in H^s(A)$. Then is there a relationship between $\|g\|_{s,A}$ and $$ \begin{align*} \|g\|_{-s,A}^* = \sup_{h\in H^s(A)}\frac{\langle g,h\rangle_{A}}{\|h\|_{s,A}}, \end{align*} $$ where $\langle g,h\rangle_{A}$ is the standard $L_2$ inner product restricted to $A$? I am having difficulty trying to relate the two, because there seems to be a natural relationship globally, which I would hope to think there is a local relationship as well. REPLY [6 votes]: Q1 has been addressed by Mark in the comments (absolute values are missing in your formula), but let me quickly look at this again: Just take Fourier transforms to see that $$ \|g\|_{-s}^*=\sup_{\|h\|_s=1} |\langle g,h\rangle | = \sup \left\{ |\langle \widehat{g}, \widehat{h}\rangle | : \|(1+t^2)^{s/2}\widehat{h}\|_2 = 1 \right\} = \sup \left\{ |\langle (1+t^2)^{-s/2}\widehat{g}, (1+t^2)^{s/2}\widehat{h} \rangle | : \ldots\right\}=\|g\|_{-s} . $$ Notice that in the third $\sup$, we can simply interpret $\langle. , .\rangle$ as the standard scalar product of $L^2(\mathbb R)$; this identifies this $\sup$ as the $L^2$ norm of the first argument of the scalar product, which is $\|g\|_{-s}$. As for Q2, this formula will fail badly for restrictions because (unlike the case $U=\mathbb R$) the action of $H^{-s}(U)$ distributions on test functions does not extend to $H^s(U)$ functions. First of all, we probably want to insist on open sets $A=U$, so that we can meaningfully restrict distributions. (We restrict a distribution to an open set by only applying it to test functions with support in that set.) Then consider, for example, $g(x)=1/x$ on $U=(0,1)$. Then $g\in H^{-1/2-\epsilon}(U)$ according to your definition because $g$ is the restriction of $PV-1/x$ to $U$. However, if we test against an $h\in H^{1/2+\epsilon}(0,1)$ with $h=1$ near zero, then $\langle g, h\rangle $ isn't even defined.<|endoftext|> TITLE: Known norm varieties and the Bloch-Kato conjecture QUESTION [8 upvotes]: The Bloch-Kato conjecture states that $K_M^n(k)/l \simeq H^n(k,\mu^{\otimes n}_l)$ for every $n,l$,while $l$ is invertible in $k$. A important part in the proof of the Bloch-Kato conjecture is to proof the existence of norm varieties. In M. Rost, Norm varieties and algebraic cobordism, Proc. of the Int. Congress of Math., vol. 2 (Beijing, 2002), 77{85, Higher Ed. Press, Beijing, 2002 there are examples for norm varieties for special cases of $n,l$. For example if $l=2$ and any $n$,we write $BK(\infty,2)$, we get the Milnor conjecture and can use Pfister-Forms or better said their varieties. 1.-What newer examples of norm varieties are known? 2.-Is there hope to find a whole class of varieties,like in the case of quadratic forms,that fulfill a certain $BK(n,l)$? Fulfilling $BK(n,l)$ is also classifying feature though. 3.In the theory of quadratic forms we have the Witt-ring and the graded Witt-ring,which is also isomorphic to the galois cohomology ring. Knowing that,the Arason-Pfister Hauptsatz gives a lower bound for the rank of forms isomorphic to symbols contained in $H^n(k,\mu^{\otimes n}_l)$. Therefore it would be plausible for a general $l$ that norm varieties represented by symbols in $H^n(k,\mu^{\otimes n}_l)$ have bigger dimension the bigger $n$ is.Wouldnt it? That whole proof of BK is hard to grasp for me at the moment.So it might be that my questions are to vage and im overrating the possibilities of current motivic research by far. REPLY [3 votes]: In the paper of Rost you mentioned, there are not just examples of norm varieties - there is an outline of how to construct norm varieties in general. The preprint version of Rost's article can be found following this link. Note that in that very same article, on page 2, you find Voevodsky's characterization of norm varieties. Part of this characterization is $\dim X=p^{n-1}-1$, answering your third question. In Section 3 of the paper, Rost gives a sketch of a proof that norm varieties exist for all symbols in $K^M_n/p$, for any $p,n$. Therefore, the positive answer to your second question is also in Rost's paper. A detailed proof of the existence of norm varieties can be found in A. Suslin, J. Joukhovitski, JPAA 206, (2006), 245--276. (Note that the proof that the varieties constructed are norm varieties in weight $n$ depends on the Bloch-Kato conjecture in weight $\leq n-1$.) All the above can be found in the article on norm varieties on Wikipedia. For some general information on the structure of the proof of the Bloch-Kato conjecture, you can find an overview of a lecture of Weibel on this page. In the Milnor-Bloch-Kato section of Weibel's publications page, you can find more papers on norm varieties and the Bloch-Kato conjecture: for your question on norm varieties, you can look at the notes of Haesemeyer and Weibel; and to get an overview of the structure of the proof of the Bloch-Kato conjecture, it is probably best to look at the 2007 ICTP lecture notes.<|endoftext|> TITLE: What is the map from nodes of the E8 diagram to conjugacy classes in the binary icosahedral group? QUESTION [5 upvotes]: Let $G \subset \mathrm{SL}_2(\mathbf{C}^2)$ be a finite subgroup isomorphic to the binary icosahedral group. Let $Y$ be the minimal resolution of $\mathbf{C}^2/G$. The irreducible components of the exceptional fiber of $Y$ are naturally in correspondence with nodes of the Dynkin diagram of $\mathrm{E}_8$. Each of these components has a linking circle in $(\mathbf{C}^2 - \{0\})/G$, whose fundamental group is $G$. Thus, each component determines a nontrivial conjugacy class in $G$. There are 8 nontrivial conjugacy classes in $G$, of orders 2,3,4,5,5,6,10,10. For each node of $\mathrm{E}_8$, which conjugacy class is it labeled by? In particular, what is the order of this class? Later: It sounds like Hugh is proposing $$ \begin{array}{rrrrrrr} & & 4\\ 6 & 3 & 2 & 5 & 10 & 5 & 10 \end{array}$$ but the locations of the (6,3), and of (5,10,5,10) are just guesses. REPLY [3 votes]: I found a paper of Kirby and Scharlemann which does the computations. In figure 2 of the paper, they give the surgery diagram for the Poincar\'e homology sphere: One obtains a presentation for the fundamental group of the link complement by the usual Wirtinger method. Moreover, 2-handles are attached to each loop with framing 2 to give the 4-manifold, giving 8 more relations, appearing in the 3rd paragraph on p. 116. As indicated there, the group is generated by $a, g$ with presentation $\langle a, g | a^5 = g^3 =(ag)^2 \rangle$. We have $h=(ag)^{-1}$, and the central involution is given by $e= a^5 = g^3 = h^{-2}$ (notice $a,g,h$ are the generators of the outermost vertices of the Dynkin diagram, and $e$ represents the trivalent vertex). From their presentation, one computes $b=a^{-2}, c= a^3, d= a^{-4}, f=g^{-2}$. So indeed, the orders of these elements agree with the orders in your diagram, and essentially agrees with Hugh Thomas' answer.<|endoftext|> TITLE: Is there a "unique" homogeneous contact structure on odd-dimensional spheres? QUESTION [14 upvotes]: Let $S^{2n-1}\subset\mathbb{C}^{n}$, and denote by $\langle\,\cdot\,,\,\cdot\,\rangle$ the Hermitian product. Then $$ \mathcal{C}_p:=\{\xi\in T_pS^{2n-1}\mid\langle p,\xi\rangle=0\},\quad p\in S^{2n-1}, $$ defines a contact structure on $S^{2n-1}$, i.e., a 1-codimensional completely non-integrable distribution (in terms of contact forms, $\mathcal{C}$ may be thought of as the conformal class $[\theta]$ uniquely defined by $\mathcal{C}=\ker\theta$). I will call it standard. QUESTION: How many contact structures $\mathcal{C}$ one can equip the sphere $S^{2n-1}$ with, in such a way that the corresponding Lie group $\mathrm{Cont}\,(S^{2n-1})$ of contactomorphisms contains a finite-dimensional Lie subgroup $G$ acting transitively on $S^{2n-1}$? REMARK: By "how many" I mean, of course, up to equivalence via diffeomorphisms. Moreover, if this will facilitate the answer, some extra topological (e.g., compactness, simply-connectedness) and/or algebraic (e.g.,semi-simplicity) property can be added to $G$. Just to motivate the question, observe that, if $\mathcal{C}$ is the above standard structure, then $G$ can be taken as the unitary group $\mathrm{U}(n)$, so that there is - at least - the equivalence class of the standard contact structure. I'd like to know whether there are others. SIDE QUESTIONS. Even without an answer to the main question, perhaps some clues/references concerning the topics below will help me: how many contact structures there are on odd-dimensional spheres? in how many ways one can construct spheres as homogeneous spaces? BELOW THERE IS A LONG EDIT FOLLOWING R. BRYANT'S LAST REMARK. According to R. Bryant's remark, there is a unique "up to equivalence" homogeneous contact structure on the odd-dimensional sphere. I'm trying now to understand why. First, I lack the notion of "up to equivalence" in the context of homogeneous spaces. Here it goes my own intuition. Let $M$ be a smooth manifold which is homogeneous w.r.t. two (in principle) different Lie groups $G$ and $\widetilde{G}$, i.e., $$M=\frac{G}{H}=\frac{\widetilde{G}}{\widetilde{H}}.$$ DEFINITION. The two structures of homogeneous manifolds are equivalent iff there exists $\phi\in\textrm{Hom}\,(G,\widetilde{G})$ such that: 1) $\phi(\widetilde{H})\subseteq \widetilde{H}$ and 2) $[\phi]\in\textrm{Diff}\, M$, where $\frac{G}{H}\stackrel{[\phi]}{\longrightarrow}\frac{\widetilde{G}}{\widetilde{H}}$ is the induced map. SIDE QUESTION: is this definition correct? does it have some relevant application? does it fit in some category-theoretic approach to homogeneous spaces? Second, assuming that the above definition is the correct one, how to prove that $S^{2n-1}=\frac{SU(n)}{SU(n-1)}$ is the unique homogenous manifold structure on $S^{2n-1}$? So, given another structure $S^{2n-1}=\frac{G}{H}$, all boils down to prove that there is a group homomorphism $\phi:G\to SU(n)$, such that $\phi(H)\subseteq SU(n-1)$ and $[\phi]$ is a diffeomorphism. SIDE QUESTION: is this the right way to tackle with the problem? can infinitesimal arguments be used instead? is there any book/paper where this sort of problems are dealt with? Finally, going back to the key topic of this post, suppose that $(M,\mathcal{C})$ and $(M,\widetilde{\mathcal{C}})$ are two different homogeneous contact structure on the same manifold $$ M=\frac{G}{H}=\frac{\widetilde{G}}{\widetilde{H}}, $$ i.e., $\mathcal{C}$ is $G$-invariant and $\widetilde{\mathcal{C}}$ is $\widetilde{G}$-invariant. Infinitesimally, this means that $$ \mathfrak{g}=\mathcal{C}_o\oplus\mathbb{R}Z \oplus\mathfrak{h};\quad \widetilde{\mathfrak{g}}=\widetilde{\mathcal{C}}_o\oplus\mathbb{R}\widetilde{Z} \oplus\widetilde{\mathfrak{h}} $$ where $Z$ and $\widetilde{Z}$ are the Reeb vector fields, and $$ \mathcal{C}_o\oplus\mathbb{R}Z = \widetilde{\mathcal{C}}_o\oplus\mathbb{R}\widetilde{Z} = T_oM $$ is the tangent space at the origin. Since there a linear transformation $h\in\mathrm{Aut}\, T_oM$ such that $h(\mathcal{C}_o)=\widetilde{\mathcal{C}}_o$, the local diffeomorphism $$ \widehat{h}:=\widetilde{\exp}\,\circ h\circ \exp^{-1} $$ sends $\mathcal{C}$ to $\widetilde{\mathcal{C}}$. SIDE QUESTION: can this $\widehat{h}$ be used to prove Bryant's claim on the uniqueness of the homogeneous contact structure on $S^{2n-1}$? more generally, is there a criterion to patch together these diffeomorphisms and obtain a global diffeomorphism, thus proving the uniqueness of homogeneous contact structure on any homogeneous manifold? if not, are there examples of non-equivalent homogeneous contact structures on the same manifold? REPLY [10 votes]: Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least. Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effective and preserves a contact structure on $S^{2n+1}$. By a result of Montgomery (Simply connected homogeneous spaces, PAMS 1950), $G$ has a compact subgroup that acts transitively on $S^{2n+1}$ (and preserves the contact structure), and this implies that a maximal compact subgroup $U\subset G$ acts transitively on $S^{2n+1}$ with compact stabilizer $K = U\cap H$, so that $S^{2n+1} = U/K$ where $U$ preserves the given contact structure on $S^{2n+1}$. Without loss of generality, we can assume that $U$ is connected, which implies that $K$ is connected as well. By results of Borel, it follows that $U$ has an embedding into $\mathrm{SO}(2n{+}2)$ for which $K = U\cap \mathrm{SO}(2n{+}1)$ (i.e., $U$ acts as a transitive group of isometries of $S^{2n+1}$ endowed with its standard metric of constant sectional curvature $+1$). Examining Borel's list of the possibilities, one sees that the connected compact subgroup $U\subset \mathrm{SO}(2n{+}2)$ acts transitively on $S^{2n+1}$ and preserves a contact structure if and only if $U$ is conjugate in $\mathrm{SO}(2n{+}2)$ to one of the following subgroups $$ \mathrm{U}(n{+}1),\quad \mathrm{SU}(n{+}1),\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr)\cdot S^1,\quad \mathrm{Sp}\bigl(\tfrac{n+1}2\bigr). $$ (The latter two cases only happen when $n$ is odd.) The first three subgroups preserve a unique contact structure, namely the contact structure defined by the $1$-form $\xi$ on $S^{2n+1}$ defined by $\xi(v) = \mathrm{d}r(Jv)$, where $J:\mathbb{C}^{n+1}\to \mathbb{C}^{n+1}$ is the complex structure map and $r = |z|^2$ is the squared Hermitian norm. The fourth subgroup preserves a $2$-sphere of contact structures, namely, one identifies $\mathbb{C}^{n+1}$ with $\mathbb{H}^{(n+1)/2}$ (thought of as column vectors of height $\tfrac12(n{+}1)$ with quaternion entries) and uses the same formula as before, but now, one allows $J$ to be scalar multiplication (on the right) by any unit imaginary quaternion. Upon conjugating by an element of the subgroup $\mathrm{Sp}(1)\subset \mathrm{SO}(2n{+}2)$ consisting of multiplication on the right by a unit quaternion, any two of these contact structures can be identified, so that each of these homogeneous contact structures in the fourth case are homogeneously isometric to the contact structure identified in the first three cases. Thus, there are really only four cases to consider: When the group $G$ contains, as identity component $U$ of its maximal compact, one of the four groups listed above, and that subgroup acts on $S^{2n+1}$ preserving a metric of constant sectional curvature $+1$. This is a classification problem that can be worked out. Though I haven't done it myself, there is a routine method to do this. For example, when $U = \mathrm{U}(n{+}1)$, one could have, in addition to $G=\mathrm{U}(n{+}1)$, that $G = \mathrm{Sp}(n{+}1,\mathbb{R})$, the symplectic transformations of $\mathbb{R}^{2n+2}$, or $G=\mathrm{SU}(n{+}1,1)$, the CR-autmorphisms of $S^{2n+1}$ as a CR-manifold. (There might be others; I haven't checked.)<|endoftext|> TITLE: What morphisms induce injective/surjective maps on (Weil) cohomology? QUESTION [10 upvotes]: Let $k$ be a field, let $f \colon X \to Y$ be a morphism of $k$-varieties, and assume $X$ and $Y$ are smooth and projective. Let $H(\_)$ be a classical Weil cohomology theory (i.e. one of $\ell$-adic étale, Betti, algebraic de Rham, crystalline). Of course it depends on $k$ which of the theories are applicable, but I do not want to spell all the cases out here. What conditions can one impose on $f$ to assert that $f^{*} \colon H^{i}(Y) \to H^{i}(X)$ is injective/surjective? Currently I can only think of the rather trivial: if $f$ admits a section (resp. retraction), then $f^{*}$ is injective (resp. surjective). [Edit] For example, is it true that $f^{*}$ is injective if $f$ is dominant? [/Edit] REPLY [6 votes]: For posterity, and as it was a bit tricky to find a reference, note also that the statement holds for possibly non-projective, quasi-projective smooth varieties, or, more generally for non-compact smooth Kähler manifolds: Theorem (Wells): Let $f:X\to Y$ be a proper surjective morphism of smooth complex manifolds and assume that $X$ is Kähler. Then the following induced homomorphisms are injective $f^*:H^q(Y,\Omega^p_Y)\to H^q(X,\Omega^p_X)$ $f^*:H^k(Y,\mathbb{C})\to H^k(X,\mathbb{C})$ $f^*:H^k(Y,\Omega^p_Y(E))\to H^k(X,\Omega^p_X(f^*E))$ for a holomorphic vector bundle $E$. The reference is Wells Comparison of de Rham and Dolbeault cohomology for proper surjective mappings.<|endoftext|> TITLE: Finite union of closed convex sets is triangulable? QUESTION [17 upvotes]: I posted this question on math.stackexchange.com, but didn't get an answer. Let $A_1, \ldots, A_k \subseteq \mathbb{R}^n$ be closed convex sets. Is the union $\bigcup_{i=1}^k A_i$ triangulable, that is, homeomorphic to a simplicial complex? If so, why? This seems plausible, but hard to prove. (Would it be easier if we require the sets to be compact?) REPLY [6 votes]: This is not true even for unions of two compact convex sets. To construct a suitable counterexample, consider the unit disk $$D=\{z\in\mathbb C:|z|\le 1\}$$in the complex plane $\mathbb C$. By $\partial D:=\{z\in\mathbb C:|Z|=1\}$ we denote the boundary circle of the disk $D$. Next, choose a decreasing sequence of real numbers $(r_n)_{n\in\omega}$ such that $\lim_{n\to\infty}r_n=\inf_{n\in\omega}r_n=1$ and the points $z_n=r_ne^{i\pi/2^n}$ do not see each other behind the disk, which means that for any distinct numbers $n,m$ the interval $[z_n,z_m]:=\{tz_n+(1-t)z_m:t\in[0,1]\}$ intersects the disk $D$. Let $C$ be the (closed) convex hull of the compact set $D\cup\{z_n\}_{n\in\omega}$. It is easy to see that $C\setminus D$ has infinitely many connected components (homeomorphic to the triangle with a removed side). Now consider the compact convex sets $A_1=C\times\{0\}$ and $A_2=D\times[-1,1]$ in $\mathbb C\times\mathbb R\cong \mathbb R^3$. It can be shown that the union $A:=A_1\cup A_2$ is not homeomorphic to a simplicial complex (even to a CW-complex). Assuming that $A$ is homeomorphic to a CW-complex, we can use the domain invariance theorem to show that the boundary $$\partial A=((C\setminus D)\times\{0\})\cup (\partial D\times [-1,1])\cup (D\times\{-1,1\})$$ of $A$ in $\mathbb C\times\mathbb R$ is contained in the 2-skeleton of the CW-complex. Using the domain invariance theorem once more, it can be shown that $\partial C\cup\partial D$ is contained in the 1-skeleton of the CW-complex and hence it is homeomorphic to a finite graph, which is not true. Conclusion. The union of finitely many convex sets in $\mathbb R$ is homeomorphic to a simplicial complex. For $n\ge 3$ the union of two compact convex set in $\mathbb R^n$ can be non-homeomorphic to a CW-complex. Problem. What is the situation in dimension 2? Is the union of finitely many compact convex sets in the plane homeomorphic to a simplicial complex?<|endoftext|> TITLE: Is every separable algebra in a modular tensor category Morita equivalent to a commutative one? QUESTION [10 upvotes]: Separable algebras in modular tensor categories are interesting algebraic structures, which have received significant attention because of their connection to conformal field theories. My understanding is that it is only the Morita class of the algebra which is important for determining the conformal field theory. Also, in the more general case of a fusion category, it is known by a result of Ostrik that up to Morita equivalence, separable algebras correspond to module categories for the fusion category. Over ordinary vector spaces, every separable algebra is Morita equivalent to a commutative one, which of course makes the Morita equivalence classes easier to study. So this motivates my question: is every separable algebra in a modular tensor category Morita equivalent to a commutative one? If not, what about restricting to symmetric separable algebras? I am working always over $\mathbb{C}$ here, and assuming that the categories involved are semisimple and with a finite number of isomorphism classes of simple objects. REPLY [9 votes]: The answer to your first question is no. The category of modules over a commutative algebra has a tensor category structure such that the forgetful functor is a tensor functor, but the category of modules over noncommutative algebras need not. Since the category of modules is a Morita invariant this gives a negative answer. The usual name for this kind of situation is "Type II modular invariant" and the D_odd module categories over su(2) at roots of unity are the simplest examples. See Kirillov-Ostrik for more details on everything in this paragraph. In these examples you have a copy of super vector spaces inside su(2) at a root of unity and the algebra is the anti-symmetric one. ("Morally" super vector spaces with the group ring of Z/2 with each group element supported in its grade is the simplest answer to your question, but technically it's not modular. But you can find it inside modular categories.) It is worth noting that module categories over a modular tensor category do have a classification in terms of commutative algebras. Namely, you pick two commutative algebras A and B plus a braided equivalence between $\mathrm{Rep}^0(A)$ and $\mathrm{Rep}^0(B)$. This is proved in Davydov-Nikshych-Ostrik Cor 3.8, though the idea goes back at least to Ocneanu.<|endoftext|> TITLE: "Minimal" group C*-algebra? QUESTION [20 upvotes]: Let $\Gamma$ be a discrete group (though this could be asked for general locally compact groups) and consider the Banach $*$-algebra $\ell^1(\Gamma)$. We have two natural $C^*$-algebra completions: the reduced group $C^*$-algebra $C^*_r(\Gamma)$ which is the closure of $\ell^1(\Gamma)$ acting by left translation on $\ell^2(\Gamma)$; and the full group $C^*$-algebra $C^*(\Gamma)$ which is the universal C$^*$-algebra completion of $\ell^1(\Gamma)$. My (extremely hazy) intuition is that while $C^*(\Gamma)$ is the maximal $C^*$-completion, also $C^*_r(\Gamma)$ is the "minimal" one. But I can't really justify this. Is it right? To be precise: Let $\pi:\ell^1(\Gamma)\rightarrow B(H)$ be an injective $*$-homomorphism, for some Hilbert space $H$. It is true that $\|\pi(a)\| \geq \|a\|_{C^*_r(\Gamma)}$ for all $a\in\ell^1(\Gamma)$?? For Abelian groups, I think some Fourier analysis shows this to be true. But what about in general? REPLY [8 votes]: The following is not an answer, but records some background and some related results which suggest that the original question might be hard to answer in the generality stated. If $\Gamma$ is amenable, $\newcommand{\Cst}{{\rm C}^*}$ then the maximal and reduced $\Cst$ algebras coincide, and your question is equivalent to the following: Question. Is the maximal $\Cst$ norm the unique $\Cst$ norm on $\ell^1(\Gamma)$? We could ask the same question for a general locally compact (amenable) group $G$, not just the discrete ones. Consulting my colleague's copy of Palmer volume 2 and doing some digging online, it turns out that the answer is yes for groups of polynomial growth, but "no" for certain solvable Lie groups, see J. Boidol, Group algebras with a unique $C^{\ast}$-norm. J. Funct. Anal. 56 (1984), no. 2, 220--232. MR 86c:22006 Some related results (without proofs) and references are in Section 10.5 of Palmer vol. 2. It seems that to get positive results one has to do noncommutative versions of the argument you allude to in your question for the commutative case, only instead of dealing with regular function algebras on the dual group you have to look at a suitable theory for noncommutative star-ideals. There is a later paper of Leung and Ng C.-W. Leung, C.-K. Ng, Some permanence properties of $C^*$-unique groups. J. Funct. Anal. 210 (2004), no. 2, 376--390. MR 2005c:22014 in which, on page 2, the authors say that it is unknown whether or not a discrete amenable group must have a unique $\Cst$ norm on its $\ell^1$-group algebra. That said: none of this seems to indicate whether your original question has a positive or negative answer for, say, the free group on two generators, or SL(3,Z). Perhaps one can do something with conjugation-type representations?<|endoftext|> TITLE: Graphs of lines on del Pezzo surfaces QUESTION [6 upvotes]: Let $k$ be an algebraically closed field. To any del Pezzo surface $S$ over $k$ we may associate its graph of lines, which has one vertex for each line and an edge (with multiplicity if required) between two vertices if and only if the corresponding lines insect. For me a "line" means a $(-1)$-curve. It is well-known that the graph only depends on the degree of the del Pezzo surface, so we obtain a series of graphs $G_d$ for $d \leq 8$. For $d \geq 3$ these graphs are all well-studied by graph theorists for their own sake, for example they have their own wikipedia pages. We have $G_6$ is the cycle graph on $6$ vertices. $G_5$ is the Petersen graph. $G_4$ is the Clebsch graph. $G_3$ is the dual of the Schläfli graph. However this is where my knowledge stops. Are the graphs $G_1$ and $G_2$ similarly well-studied? e.g. Do they have special names or special properties which uniquely identify them? I appreciate that this question is slightly vague, but I hope the reader understands what I am after. For example all the above mentioned graphs are implemented in sage. Are $G_1$ and $G_2$ also implemented in sage, or perhaps another computer algebra package? Alternatively, is there a way to construct $G_2$ and $G_1$ from the $E_7$ and $E_8$ root systems? If necessary I am willing to consider the corresponding graphs obtained by removing the multiple edges. REPLY [10 votes]: (Making my comment into an answer...) Yes, these are the vertex-edge graphs of the $E_7$ and $E_8$ polytopes. This is written up in Section 8.2.5 (in the published version) of Dolgachev, Classical Algebraic Geometry.<|endoftext|> TITLE: When does a set of collinearity conditions imply collinearity of all of the points? QUESTION [13 upvotes]: Suppose we have a set of $n$ points $\{X_1,X_2,\dots,X_n\}$ in the real plane and $\mathcal{A}$ a family of subsets of $\{1,\dots,n\}$. By a "set of collinearity conditions for $\mathcal{A}$" we mean the conditions of collinearity for every subset $\{X_i:i\in A\}$ s.t. $A\in\mathcal{A}$. I'm interested in determining families for which the induced collinearity conditions imply that all of the points are on the same line. One way to propose a counterexample for a given family is to select three non-collinear points $Y_1,Y_2,Y_3$ and for each $1\leq i\leq n$ declare $X_i$ to be one of $Y_j$'s ($j=1,2,3$), in such a way that every one of $Y_j$'s have been selected at least one time and further for every $A\in \mathcal{A}$ one of $Y_j$'s doesn't appear among $\{X_i:i\in A\}$. Then this set of $n$ non-collinear points satisfies the collinearity conditions for $\mathcal{A}$. This observations leads naturally to the following conjecture: Conjecture: The collinearity conditions for a family $\mathcal{A}$ implies collinearity of all of the points if and only if there doesn't exist a partition of $\{1,\dots,n\}$ into three nonempty disjoint sets $\{B_1,B_2,B_3\}$ s.t. for every $A\in \mathcal{A}$, $A$ is contained in the union of at most two of $B_i$'s. (In fact similar conjectures can be stated in higher dimensions, if we regard the existence of a hyperplane containing all of the points.) Is this conjecture true? If yes, how the imposed equivalent condition can be checked? what's the related notion in (hyper)graph theory? Are there any related theorems to this problem? Thanks for your advice! REPLY [10 votes]: Yes, if a collection of colinearity conditions is realizable, then they are realizable using only 3 distinct points. Suppose that $x_1$, $x_2$, ..., $x_n$ in $\mathbb{R}^2$ is a collection of points realizing your colinearity conditions, and not all on a line. Choose two unequal points $x_i$ and $x_j$, and let $\overline{x_i x_j}$ be the line through them. Take $B_1$ to be the set of $x_k$ which are equal to $x_i$ (same point in $\mathbb{R}^2$); take $B_2$ to be the set of $x_k$ which are on the line $\overline{x_i x_j}$ but not in $B_1$ and take $B_3$ to be everything else. If a colinearity condition met all three of $(B_1, B_2, B_3)$, then this would correspond to a line which met $\overline{x_i x_j}$ at $x_i$ and another point, and also passed through a point not on $\overline{x_i x_j}$. Obviously, this works for configurations in $\mathbb{R}^{k-1}$ just as well: Choose points $x_1$, $x_2$, ..., $x_k$ with no linear relations between them, and let $B_d$ be the points which are in the affine linear span of $(x_1, \ldots, x_d)$ but not of $(x_1, \ldots, x_{d-1})$. Let me explain where I had seen this construction before. Work with $n$ points in $\mathbb{RP}^2$ rather than $\mathbb{R}^2$, to increase the symmetries available. Then we can encode our points as the columns of a $3 \times n$ matrix, up to rescaling of columns (because coordinates in $\mathbb{RP}^2$ are only up to scaling). Collinearity conditions say that various submatrices have rank $2$. You want to know if you can impose that various submatrices have rank $2$ without forcing the whole matrix to have rank $2$. And you only care about solutions where none of the columns are identically zero, since homogenous coordinates on projective space cannot all be zero. So your question is: Suppose we have a $3 \times n$ matrix $M$, of rank $3$, with no zero columns. Can we make there only be three distinct columns, while preserving the "rank $2$"-ness of specified submatrices. All your conditioned are unaltered by acting on the matrix by $GL_3$ on the left, so we can think on the space $GL_3 \backslash \{ \mbox{$3 \times n$ matrices of rank $3$} \}$, also known as the Grassmannian $G(3,n)$. The advantage of the Grassmannian is that is compact. If we build a family $M(t)$ of rank $3$ matrices, parametrized by $t \neq 0$ and preserving all the rank $2$-ness conditions, then it will have some limit as $t \to 0$, which we can lift back to a rank $3$ matrix. EG: $\left( \begin{smallmatrix} 1 & 0 & 0 &0 \\ 0 & t & t^2 & t^3 \\ 0 & 0 & t & t \\ \end{smallmatrix} \right)$ looks like it is approaching a matrix of rank $1$ as $t \to 0$, but it is the same family up to $GL_3$ action as $\left( \begin{smallmatrix} 1 & 0 & 0 &0 \\ 0 & 1 & t & t^2 \\ 0 & 0 & 1 & 1 \\ \end{smallmatrix} \right)$, whose limit is rank $3$. So I can approach your question by building families $M(t)$, passing through $M$, and preserving the rank $2$-ness of various submatrices, and be guaranteed that my limits will exist. An easy way to build a family of matrices that preserves the rank of all $3 \times (\mbox{whatever})$ submatrices is to look at $M \cdot X(t)$, where $X(t)$ is a one parameter subgroup of the diagonal matrices in $GL_n$. For most one parameter subgroups, some columns will become $0$ in the limit, so we can't use them. The set of one parameter subgroups for which none of the columns die is called the Bergman complex. The most common elements of the Bergman complex are indexed by "complete chains in the lattice of flats". Removing the matroid jargon, in the case of the plane, this means "a point $x_i$, and a line $\overline{x_i x_j}$ through the point". What I wrote out was the limiting $3 \times n$ matrix for that case. REPLY [4 votes]: Your conjecture is true, at least if you allow degenerate realizations where some of the points coincide (and with some form of the axiom of choice). Clearly, if there does exist a partition of this type, then there is a non-collinear realization of your collinearity conditions: just place each of your $n$ points at the vertex of an equilateral triangle corresponding to its set in the partition. In the other direction, suppose you have a realization of your collinearity conditions in which not all points are collinear. Under the axiom of choice, there exists a 3-coloring of the plane in which each line is 2-colored (see e.g. Hales and Straus, "Projective colorings", Pacific J. Math. 1982) and by an appropriate affine transformation of such a coloring you can make any three of its points, with three different colors, coincide with a chosen three non-collinear points of your realization. Then the coloring gives you your desired partition.<|endoftext|> TITLE: Is the group of isometries of a homogeneous Riemannian manifold maximal? QUESTION [7 upvotes]: I have a homogeneous Riemannian manifold X with isometry group Iso. Is Iso a maximal group? By maximal group, I mean that there does not exist another group G such that: Iso is a proper subgroup of G, G acts transitively on X by diffeomorphisms, and G has compact stabilizers Gx. I know that if such a group exists, then there is a G-invariant metric on X. So, in other words, the question is: if I have a Riemannian metric d on X with isometry group Iso which is homogeneous, can there exist another Riemannian metric d' on X with isometry group Iso', such that Iso' contains properly the group Iso? REPLY [3 votes]: (This should have been a comment, but got too long, sorry) Let us call $Iso(G)$ the full isometry group of the homogeneous space $G/H$, i.e., the largest group that acts transitively and effectively on $G/H$. Expanding on Claudio's comment, e.g., the sphere $S^{4n+3}$ can be written as $SO(4n+5)/SO(4n+4)$, $SU(2n+1)/SU(2n)$ or $Sp(n+1)/Sp(n)$, and there are (arbitrarily) small deformations of the round metric -- through so-called Berger metrics -- that make the (identity component of the) full isometry group drop from $SO(4n+5)$ to $SU(2n+1)$ or $Sp(n)$. Notice that $Iso(G/H)$ actually drops in dimension under this deformation of homogeneous metrics, not only looses some component as in item 3 below. For one more example on $\mathbb C P^n$, see my answer here; The full isometry group $Iso(G/H)$ of every homogeneous space $G/H$ with positive sectional curvature was computed by Shankar, see table given in Figure 3. Elaborating on Alex's comment, if we take $(M,g)$ any Riemannian manifold and consider $(M\times M,g\oplus\lambda g)$, then for $\lambda=1$ there is always an extra isometry, namely the one that exchanges both factors $M$.<|endoftext|> TITLE: Local inverse Galois problem QUESTION [19 upvotes]: It's a basic fact that a finite Galois extension $L/K$ of a local nonarchimedean field $K$ has solvable (in fact supersolvable) Galois group $G$. One sees this by using the ramification filtration $G_i=\{g\in G: g\beta\equiv \beta \pmod {\varpi_K^i}\}$. More nontrivially, local class field theory tells us if $L/K$ is abelian then $G$ must be a quotient of $K^\times$, whose structure as an abelian group is known explicitly. In any case, this means the inverse Galois problem over $K$ is quite circumscribed. So is it known precisely which finite groups are Galois groups over $K$? REPLY [2 votes]: I would say, it depends on what you mean by: "Is the local inverse Galois problem is solved?" One way to formulate the local inverse Galois problem is, for any given a finite group $G$ can we decide whether $G$ is the Galois group of a local field. Or more algorithmically: Does there exist an algorithm that has as input a finite group G and a local field K and outputs whether or not the case that the group G is the Galois group of an extension L/K. If you formulate it this way the answer is yes. And the reason for this maybe not as nice as one would like. The reason why the answer is yes is as follows: For a given a group $G$ there are up to isomorphism only finitely many extensions of $L/K$ of degree $\#G$. Indeed, when weighted correctly we even know explicit formula for the number of extensions and there are algorithms that explicitly enumerate these extensions. So the only thing one has to do is compute the Galois group of every extension $L/K$ of degree $|G|$, and see whether it is isomorphic to $G$. This algorithm can even be made "independent of the characteristic. I.e. for a given group $G$ on can explicitly decide for which characteristics $p$ the group $G$ can be a Galois group of $\mathbb Q_p$. The reason is that when $L/\mathbb Q_p$ is tamely ramified we know very well which groups occur (see Iwasawa's result in Jeremy's answer). So using this it is reduced to doing a finite computation as above for all the primes $p | \#G$ where there can be wild ramification. So it all depends on how you formulate the question "Is the local inverse Galois problem solved?". Cause it all depends on what way of characterizing the groups that can occur as a local Galois group you are fine with. Edit I just found out about the existence of the paper The Inverse Galois Problem for p-adic fields by David Roe. His results are only for when the residue characteristic is not $2$ since he uses the explicit description of the absolute Galois group of Jannsen and Wingberg mentioned by Jeremy. In that paper he gives an algorithm to actually count the number of extensions of K with galois group G, which is more efficient than the enumeration I described above. Additionally he gives a set of necessary and a distinct set of sufficient conditions for a group G to be realizable as a Galois group of over K, which might be of independent interest.<|endoftext|> TITLE: The lattice of covectors of an oriented matroid QUESTION [5 upvotes]: Let $M$ be an oriented matroid on the ground set $E$, and let $L(M)$ be its ranked poset of covectors. By definition, $L(M)$ is a sub-poset of the poset $\{0, \pm 1\}^E$, ordered by putting $0 < \pm 1$, with the rank being the number of nonzero coordinates. However, I want to consider $L(M)$ as an abstract ranked poset, rather than as a subposet of $\{0, \pm 1\}^E$. Question 1: Is the oriented matroid $M$ determined, up to isomorphism, by $L(M)$? Question 2: Is there a nice characterization of which ranked posets arise in this way? REPLY [4 votes]: This is an answer to question 1 (not the modified version in the comments to Ben's answer). If you consider only L(M), you have lost all the information the orientation provides (beyond the bare fact that an orientation exists). Since there can be more than one, non-isomorphic oriented matroid on the same matroid, it can't be possible to reconstruct the oriented matroid from L(M). (I don't actually have an explicit example of multiple oriented structures on one matroid in mind, but counts of the orientable and oriented matroids imply that there are orientable matroids with more than one orientation starting in rank 3 with 6 points.)<|endoftext|> TITLE: Is there a non-zero ghost map between finite suspension spectra? QUESTION [13 upvotes]: A morphism $f\colon X\to Y$ of spectra such that for every integer $n$ the induced map $\pi_n(f)\colon\pi_n(X)\to\pi_n(Y)$ on stable homotopy groups is zero is called a ghost map. Not every ghost map $f$ is the zero object of the abelian group $\operatorname{Hom}_{SH}(X,Y)$ (more precisely, its image $\bar f\in \operatorname{Hom}_{SH}(X,Y)$ in the stable homotopy category $SH$). Suppose $X=\Sigma^\infty A$ and $Y=\Sigma^\infty B$ are the suspension spectra of finite CW-complexes, let $f=\Sigma^\infty f'$ for a map $f'\colon A'\to B'$ and suppose that $f$ is a ghost map. Is $f$ necessarily zero in this case? REPLY [17 votes]: Following André's suggestion I will promote this comment to an answer. The OP's question is Freyd's Generating Hypothesis, which has been an open question for nearly 50 years. It is very hard to test, because there is no example of a nontrivial finite spectrum $X$ where all the groups $\pi_k(X)$ are known. Devinatz had a potentially interesting approach using chromatic theory and Gross-Hopkins duality, but that seems to have petered out. There is also some work on analogous questions in other triangulated categories. Many interesting and surprising consequences would follow if the Generating Hypothesis were true; Mark Hovey's paper "On Freyd's Generating Hypothesis" is a nice survey. In particular, the $p$-completion of the stable homotopy groups of spheres would be a non-Noetherian, totally incoherent ring that is injective as a module over itself. Recently Leigh Shepperson and I wrote a paper investigating the algebraic theory of such rings, in the hope of shedding some indirect light on the Generating Hypothesis.<|endoftext|> TITLE: Relativistic Control Theory QUESTION [7 upvotes]: I am looking for literature that combines General relativity and control theory. So far I found a video lecture on "Integrability meets Control Theory: Harmonic maps in GR", other than that not so much. I assume that control theory applies GR for everything that includes gravity, even if for every practical purpose Newtonian mechanics is enough. REPLY [5 votes]: There is also Inverse problems in spacetime I: Inverse problems for Einstein equations - Extended preprint version, by Yaroslav Kurylev, Matti Lassas, Gunther Uhlmann and related papers. Sounds like something different, but from a technical viewpoint inverse problems and optimal control are very similar. In optimal control you interpret some parameter as a control that you want to choose optimally to achieve a specific goal, while in inverse problems you interpret it as an unknown quantity that you like to identify from given measurements. Technically you can turn an optimal control problem into an inverse problem by replacing "control" by "unknown quantity" and "goal" by "measurements" (and vice versa).<|endoftext|> TITLE: Candidates for pure algebraic notion of cycles? QUESTION [9 upvotes]: In the interview http://www.ems-ph.org/journals/newsletter/pdf/2013-09-89.pdf, Deligne said For me, this (Hodge conjecture) is a part of the story of motives, and it is not crucial whether it is true or false. If it is true, that's very good and it solves a large part of the problem of constructing motives in a reasonable way. If one can find another purely algebraic notion of cycles for which the analogue of the Hodge conjecture holds, and there are a number of candidates, this will serve the same purpose, and I would be as happy as if the Hodge conjecture were proved. I wonder what are those "candidates" mentioned by him above, i.e. some possible pure algebraic notions of cycles? REPLY [4 votes]: Deligne's absolute Hodge classes (Deligne 1982 LNM), Andre's motivated classes (Andre 1996 IHES), Milne's rational Tate classes (Milne 2009 Moscow MJ),...<|endoftext|> TITLE: How did "normal" come to mean "perpendicular"? QUESTION [49 upvotes]: How and when did the word "normal" acquire this meaning? When I first thought of this, I couldn't really come up with any explanation that wasn't complete speculation -- pretty much all I was able to see was that it isn't any stranger than "right" in "right angle" -- the angle is probably as right as the lines are normal. But on reading the etymology note in the entry for "normal" in the American Heritage Dictionary, Middle English, from Late Latin normalis, from Latin, made according to the square, from norma, carpenter's square; I thought that was probably it -- it probably came from the perpendicular sides of a carpenter's square. Did it then? And how was it introduced? If this is indeed what's going on, either the meaning must have been introduced long ago, when people still realized what the etymology of "normal" is, or some really erudite mathematician must have introduced it, perhaps to show off their erudition. So how did it happen? REPLY [10 votes]: Perhaps it should be mentioned that while normalis did mean perpendicular in classical Latin, it was not the only possible (and perhaps not the preferred) choice of words. I asked about expressing perpendicularity in classical Latin at the new Latin Language SE site. The answer given there lists a number of expressions, mainly from Vitruvius: πρὸς ὀρθᾶς — yes, this Greek expression was used in Latin. ad normam — according to norma, a square employed by carpenters, masons, etc., for making right angles. ad perpendiculum — according to perpendiculum, a plumb-line ad perpendiculum et normam — combination of the previous two directus — "straight", essentially the same as rectus used in the medieval commentary Carlo mentions There is a finite amount of extant Latin literature and perpendicularity is not the most common of subjects, so it is difficult to tell what exactly are the differences between these phrases in usage or frequency. The examples listed above, apart from directus, mean "orthogonally". For the adjective "orthogonal" one might use directus, rectus or indeed normalis. Vitruvius does not use the word normalis at all. Notice that norma and perpendiculum are two different tools used to create straight angles, not geometrical descriptions of the angle itself. I am not aware of any ancient mathematical texts in Latin, either original or translations. It would be interesting to see which expression would be chosen in mathematical context. If you know any suitable sources, consider answering "Where to find ancient mathematics in Latin?". (The phrase mentioned in Carlo's answer can be found in the Perseus service together with an English translation in a more readable and searchable form.)<|endoftext|> TITLE: Primes dividing $2^a+2^b-1$ QUESTION [10 upvotes]: From Fermat's little theorem we know that every odd prime $p$ divides $2^a-1$ with $a=p-1$. Is it possible to prove that there are infinitely many primes not dividing $2^a+2^b-1$? (With $2^a,2^b$ being incoguent modulo $p$) 1. Obviously, If $2$ is not a quadratic residue modulo $p$ then we have the solution $a=1, b=\frac{p-1}{2}$ 2. If $2$ is a quadratic residue and the order of $2 \ modp$ is $r=\frac{p-1}{2}$ then the set $ \{2^1,2^2,...,2^{\frac{p-1}{2}}\}$ is a complete quadratic residue system modp. So,In this case, $p\mid2^a+2^b-1$ is equivalent to $p\mid x^2+y^2-1$ with $x^2,y^2$ being incogruent modp, which is always true for every $p\geq11$ . 3. It is not true that if $p \mid2^a+2^b-1$ and $q\mid2^{a'}+2^{b'}-1$ then $p\cdot q\mid 2^c+2^d-1$ . There is the counterexample: $5\mid 2^1+2^2-1$ and $17\mid 2^1+2^4-1$ but $5\cdot 17=85\not \mid2^a+2^b-1$. We can see a few examples of numbers which have the questioned property :$3,7,31,73,89,...$ (In fact,every Mersenne prime does not divide $2^a+2^b-1$) Thanks in advance! REPLY [10 votes]: This is a heuristic which suggests that the problem is probably quite hard. We have that $p | 2^{a} + 2^{b} - 1$ if and only if there is some integer $k$, $1 \leq k \leq p-1$ with $k \ne \frac{p+1}{2}$ for which $2^{a} \equiv k \pmod{p}$ and $2^{b} \equiv 1-k \pmod{p}$ are both solvable. If $r$ is the order of $2$ modulo $p$, then for each element $k$ in $\langle 2 \rangle$, the "probability" that $1-k$ is also in $\langle 2 \rangle$ is $\frac{r}{p-1}$ (assuming that $1-k$ is a "random element of $\mathbb{F}_{p}^{\times}$). So the probability that there are no solutions is about $\left(\frac{p-1-r}{p-1}\right)^{r} \approx e^{-\frac{r^{2}}{p-1}}$. (Note that if $r$ is even, then we have the trivial solution $2^{a} \equiv -1 \pmod{p}$, and $b = 1$.) Thus, in order for there to be a solution, we must have that $r$ is small as a function of $p$, no bigger than about $\sqrt{p}$. This implies that $p$ must be a prime divisor of $2^{r} - 1$ of size $\gg r^{2}$. (For example, the prime $p < 5 \cdot 10^{5}$ that does not divide a number of the form $2^{a} + 2^{b} - 1$ for which $r$ is the largest is $p = 379399$ and for this number, $r = 1709$.) However, it seems difficult to prove unconditionally that there are numbers of the form $2^{n} - 1$ with large prime divisors. Let $P(2^{n} - 1)$ denote the largest prime divisor of $2^{n} - 1$. Then, the strongest unconditional results (see the paper of Cam Stewart in Acta. Math. from 2013) take the form $P(2^{n} - 1) \geq f(n)$, where $f(n) = O(n^{1 + \epsilon})$ for all $n$. (Of course, we only need a result for infinitely many $n$, but allowing exceptions seems not to help.) Murty and Wong (2002) prove assuming ABC that $P(2^{n} - 1) \gg n^{2 - \epsilon}$ for all $n$, and Pomerance and Murata (2004) show that $P(2^{n} - 1) \gg \frac{n^{4/3}}{\log \log(n)}$ for all but a density zero subset of $n$, assuming GRH.<|endoftext|> TITLE: Curvature and Failure to return to starting point QUESTION [5 upvotes]: Assume I have a geodesic polygon $P$ in a Riemannian manifold $M$ that is given by the image of a piecewise geodesic closed curve $\gamma(t)$ (parametrized by arclength), with vertices $x_i = \gamma(t_i)$, $i=0, \dots, N$. Let $$X_i = \lim_{t \searrow t_{i-1}}\dot{\gamma}(t) \in T_{x_{i-1}}M, ~~~~~~~~~~~~~~~~i = 1, \dots, N$$ be the velocity vectors of $\gamma$ at the beginning of the respective geodesic segments. Now I can use the parallel transport the $X_i$ backwards along $\gamma$ to obtain $N$ vectors in $T_{x_0} M$ (which I also call $X_i$ by abuse of notation). Now I can look at the piecewise polygon curve given by this data, the end point of which is $$ v = \sum_{i=1}^N (t_i - t_{i-1}) X_i.$$ If $M$ is flat, then $v = 0$, as the geodesic polygon was closed. Otherwise, $v=0$ may be not equal to zero, and is somehow given in terms of the curvature of $M$. For example $$ |v|^2 = \int_0^{t_N} \!\!\!\int_0^{t_N} \bigl\langle \dot{\gamma}(t), [\gamma\|_s^t]\dot{\gamma}(s)\bigr\rangle ds dt,$$ where $[\gamma\|_s^t]$ denotes parallel transport along $\gamma$. Question(s): What are other expressions for $|v|$? Can I express this value somehow in terms of the curvature integral over surfaces whose boundary is $P$? Are there other things that come to mind regarding this situation? REPLY [2 votes]: Your question reminds me of the following result: let $X$ and $Y$ be tangent vectors in $T_p M$ such that $|X \wedge Y| = 1$ and let $\gamma$ be the exponential of a piecewise smooth closed curve in $T_p M$ based at the origin which lies in the plane spanned by $X$ and $Y$. Then: $$P_\gamma Z - Z = R(X,Y)Z \text{Area}(\gamma) + o(\text{Area}(\gamma))$$ This result has generalizations in which $\gamma$ is the boundary of a smooth surface in $T_p M$. There are formulas for the area of a surface with polygonal boundary obtained via Stokes' theorem, and I bet these formulas combined with the result above would yield something interesting.<|endoftext|> TITLE: Absolutely irreducible p-adic representation of the absolute Galois group of Q_p QUESTION [5 upvotes]: Let $p$ be a prime number, $\mathbb{Q}_p$ the field of $p$-adic numbers, $G_p$ the absolute Galois group of $\mathbb{Q}_p$ and $V$ a finite dimensional vector space over $\mathbb{Q}_p$. Assume we are given a linear representation $\rho : G_p \to GL(V)$. Can we find a closed subgroup of finite index $H$ in $G_{p}$ such that the restriction of $\rho$ to $H$ is reducible ? It is possible in the case of a representation which have coefficients in $\mathbb{F}_p$ (in this case, absolutely irreducible representations are induced). REPLY [5 votes]: The answer is no. To prove it, it suffices to show that there exists a representation $\rho: G_p \rightarrow Gl(V)$ of open image. There are several ways to do this, one is to use Chenevier prop. 1.8 in "Quelques courbes de Hecke se plongent dans l'espace de Colmez", which says that for $p=2,3,5,7$ and $\rho: G_{\mathbb Q,p} \rightarrow Gl_2(\mathbb Q_p)$ an odd representation (where $G_{\mathbb Q,p}$ is the global galois group of $\mathbb Q$ unramified outside $p$), one has $\rho(G_{\mathbb Q,p})=\rho(G_p)$. So it suffices to find now a "global" representation with open image, and this is well-known to exist (e.g. Serre's open image for elliptic curve, etc.) REPLY [4 votes]: One gets loads of crystalline counterexamples using $p$-divisible groups (and more specifically from any elliptic curve with supersingular reduction). Let $k$ be a perfect field of characteristic $p > 0$ and let $K$ be a complete discretely-valued field of characteristic 0 having residue field $k$. Let $\Gamma_0$ be a $p$-divisible group over $k$. By the unobstructedness of the infinitesimal deformation theory of $p$-divisible groups, this lifts to a $p$-divisible group $\Gamma$ over the valuation ring $O_K$ of $K$ (and even over $W(k)$). Suppose that the representation of the Galois group of $K$ associated to the generic fiber $\Gamma_K$ becomes reducible on the finite-index open subgroup, say associated to a finite extension $K'/K$. This is exactly the condition that $\Gamma_{K'}$ admits a nontrivial filtration, and by the usual schematic closure trick (and the work of Raynaud/Tate on $p$-divisible groups) that uniquely extends to such a filtration of $\Gamma_{O_{K'}}$, and hence of $(\Gamma_0)_{k'}$ for the residue field $k'/k$ of $K'/K$. In other words, as long as $\Gamma_0$ is a "geometrically isosimple" $p$-divisible group (i.e., $(\Gamma_0)_{\overline{k}}$ is simple in the isogeny category over $\overline{k}$) then all such $\Gamma_K$ furnish examples. The Dieudonne-Manin classification provides lots of concrete Dieudonne modules corresponding to geometrically isosimple $\Gamma_0$ (and then by the Fontaine-Honda formalism or other formalisms one can "write down" the data of lots of lifts over $W(k)$). And in more concrete terms, the $p$-divisible group of any supersingular elliptic curve over $k$ is of this type (so the $p$-adic Tate module of any elliptic curve over $K$ with supersingular reduction does the job) since a nontrivial composition series would (by height reasons) have to consist of terms which are either multiplicative or etale, contradicting the supersingularity hypothesis.<|endoftext|> TITLE: Analog of Newlander–Nirenberg theorem for real analytic manifolds QUESTION [16 upvotes]: It is well known that one can specify a complex structure on a real $C^\infty$ manifold in two equivalent ways: an atlas with holomorphic transition functions between charts and an integrable almost complex structure. One of the directions is straightforward. In the other direction, the celebrated Newlander–Nirenberg theorem states that an integrable almost complex structure induces a holomorphic atlas. For real analytic manifolds, I know only the atlas method (transition functions between charts are real analytic). My question: does there exist for real analytic manifolds an analog of an almost complex structure, an integrability condition and an analog of the Newlander–Nirenberg theorem? At the moment, my suspicion is that the right analog should be an almost CR structure and a corresponding integrability condition (whatever those are). But unfortunately I've not really seen this written anywhere. REPLY [4 votes]: I think this question can be addressed in a few ways. Two great answers have already been given: Real analytic is a regularity condition, while holomorphic is more algebraic, so you'd need a somewhat different condition than NN. Every $C^1$ manifold can be given a real analytic structure, so you don't even need a condition. Let me add a third, more complicated, answer where we ask the real analytic structure to be adapted to the given data. To do this we have to go back and look at what the Newlander-Nirenberg theorem gives us, and then we will be able to adapt it to give a real analytic type result. So first The Newlander-Nirenberg Theorem: Let $L_1,\ldots, L_m$ be smooth complex vector fields on a smooth manifold $M$. Suppose for each $\zeta\in M$ we have the following: $[L_j,L_k](\zeta)\in \mathrm{span}_{\mathbb{C}}\{L_1(\zeta),\ldots, L_m(\zeta)\}$. $\mathrm{span}_{\mathbb{C}} \{ L_1(\zeta),\ldots, L_m(\zeta)\} \cap \mathrm{span}_{\mathbb{C}} \{ \overline{L_1}(\zeta),\ldots, \overline{L_m}(\zeta)\}=\{0\}$. $\mathrm{span}_{\mathbb{C}}\{L_1(\zeta),\ldots, L_m(\zeta), \overline{L_1}(\zeta),\ldots, \overline{L_m}(\zeta)\} = \mathbb{C} T_{\zeta} M$. Then, there exists a complex manifold structure on $M$ (compatible with its smooth manifold structure) such that $\forall \zeta\in M$, $\mathrm{span}_{\mathbb{C}} \{L_1(\zeta),\ldots, L_m(\zeta)\} = T^{0,1}M$. Moreover, this complex structure is unique in the sense that if $M$ is given another complex manifold structure (compatible with its smooth manifold structure) such that $\mathrm{span}_{\mathbb{C}} \{L_1(\zeta),\ldots, L_m(\zeta)\} = T^{0,1}M$, then the identity map $M\rightarrow M$ is a biholomorphism between these two complex manifold structures on $M$. Now let's turn to the real setting, and address the original question. Let $X_1,\ldots, X_q$ be $C^1$ vector fields on a $C^2$ manifold $M$ such that $\forall x\in M$, $\mathrm{span}_{\mathbb{R}} \{X_1(x),\ldots, X_q(x)\} = T_x M$. Question: When is there a real analytic structure on $M$, compatible with its $C^2$ structure, such that $X_1,\ldots, X_q$ are real analytic with respect to this structure? When such a structure exists, we will see it is unique (in the strongest possible sense). We will answer this question by giving a condition which looks very similar to the Newlander-Nirenberg theorem (but is actually orthogonal to that condition, as we will see). Even though $M$ is only a $C^2$ manifold, we can define what it means for a function to be real analytic with respect to the vector fields $X_1,\ldots, X_q$. Let $V\subseteq M$ be open. For $r>0$ we define $C_{X}^{\omega,r}(V)$ to be the space of those $f:V\rightarrow \mathbb{R}$ such that the following norm is finite: $$ \| f\|_{C_X^{\omega,r}(V)} = \sum_{m=0}^{\infty} \frac{r^m}{m!} \sum_{|\alpha|=m} \| X^{\alpha} f\|_{C(V)}.$$ In defining $X^{\alpha}$ we have used ordered multi-index notation; i.e., $\alpha$ is a list of elements of $\{1,\ldots, q\}$ and $|\alpha|$ is the length of the list. For example $X^{(1,2,1,2)}= X_1X_2X_1X_2$ and $|(1,2,1,2)|=4$. We set $C_X^{\omega}(V):= \bigcup_{r>0} C_X^{\omega,r}(V)$. The space $C_X^{\omega}(V)$ is "coordinate-free": it does not depend on any choice of coordinate system or atlas. This space was originally defined in greater generality by Nelson (Ann. of Math. (2) 70 (1959), 572-615). Theorem: There is a real analytic structure on $M$, compatible with its $C^2$ structure, such that $X_1,\ldots, X_q$ are real analytic with respect to this structure if and only if: For every $x\in M$, there is a neighborhood $V_x$ of $x$, and functions $c_{j,k}^{l,x}\in C_{X}^{\omega}(V_x)$ such that $[X_j,X_k]=\sum_{l=1}^q c_{j,k}^{l,x} X_l$ on $V_x$. When this real analytic structure exists, it is unique in the sense that if $M$ is given another real analytic structure, compatible with its $C^2$ structure, such that $X_1,\ldots, X_q$ are real analytic with respect to this second structure, then the identity map $M\rightarrow M$ is a real analytic diffeomorphism between these two real analytic structures. So there's the answer to our question--and it has the same "feel" as the NN theorem: it uses an understanding of commutators of given vector fields to give a unique structure on the manifold. There's nothing special about real analytic in the above. You can replace real analytic with an appropriate space of functions which are $C^\infty$ with respect to $X_1,\ldots, X_q$ and get a corresponding result about $C^\infty$ structures. You can do it for a finite level of smoothness, too, though to obtain a sharp if and only if Zygmund spaces are used (instead of the more familiar $C^m$ spaces). This is all done in the series of papers (joint with Stovall): 1, 2, 3 No back to my comment that this was actually orthogonal to the NN theroem. One could ask the following: Question: Suppose $L_1,\ldots, L_m$ satisfy the conditions of the NN theorem (above). When are $L_1,\ldots, L_m$ real analytic with respect to the complex structure gauranteed by the NN theorem? One can answer this by giving a condition very similar to the real theory given above. You can even get a theory which unifies both the real and complex into one theorem. This is all contained in this paper.<|endoftext|> TITLE: Integrating representations of Lie algebroids QUESTION [7 upvotes]: If $A \to M$ is a Lie algebroid over a smooth manifold $M$ then a representation of $A$ is a vector bundle $E \to M$ with a flat $A$-connection $$ \nabla : \Gamma(E) \to \Gamma(E\otimes A^*). $$ If $G$ is a Lie groupoid over $M$ then a representation is a vector bundle $E\to M$ and a Lie groupoid homomorphism $G \to GL(E)$, where $GL(E)$ is the groupoid over $M$ whose arrows are linear isomorphisms of the fibers of $E$. Now suppose $G$ is a Lie groupoid that integrates $A$. Under what sort of conditions can every representation of $A$ be integrated to a representation of $G$? Two extreme cases are well-known: If $M$ is a point then we're just talking about Lie groups and algebras and every representation of the algebra integrates to a representation of the group if the group is simply connected. If $A = TM$ is the tangent Lie algebroid and $G$ is the fundamental groupoid then the Riemann-Hilbert correspondence says that every representation of $TM$ (i.e. flat vector bundle over $M$) corresponds to a representation of $G$. On the other hand, the pair groupoid also integrates $A$ but it doesn't seem like it has the same representations (I think a representation of the pair groupoid is given by a vector bundle $E\to M$ and a section of $p_1^* E^* \otimes p_2^* E \to M\times M$, where $p_j : M\times M \to M$ is the projection on the $j$th factor). I would be interested in seeing any references that talk about this. REPLY [6 votes]: The usual condition that one uses to ensure that all representations of $A$ integrate to representations of $G$ is that $G$ be "source 1-connected", meaning that all of the fibres of the source map $s : G \to M$ are connected and simply connected. That the source 1-connected condition is sufficient is a consequence of a more general theorem which asserts that if $G$ and $H$ are Lie groupoids with $G$ source 1-connected, and if $\phi : Lie(G) \to Lie(H)$ is a morphism between their Lie algebroids, then $\phi$ integrates to a unique Lie groupoid morphism $G \to H$. You can find this result, for example, in the papers "Integrations of Lie bialgebroids" by Mackenzie and Xu (Topology 39 (2000), no. 3, 445–467) "On integrability of infinitesimal actions" by Moerdijk and Mrčun (Amer. J. Math. 124 (2002), no. 3, 567–593) Your question deals with the specific case when $H = GL(E)$ is the gauge groupoid of a vector bundle $E \to M$, since a flat $A$-connection on $E$ is the same thing as a morphism $E \to \mathfrak{gl}(E)$ to the Atiyah algebroid. Regarding the specific case $A = TM$: it's a good exercise to a) convince yourself that the fundamental groupoid $\Pi_1(M)$ is source 1-connected, and b) convince yourself that a flat connection will integrate to a representation of the pair groupoid if and only if it has trivial holonomy along every loop. One way to think about b) is that the natural map $\Pi_1(M) \to Pair(M)$ expresses $Pair(M)$ as a quotient of $\Pi_1(M)$ and you're looking for representations that descend to the quotient. Finally, I should point out that, while source 1-connectedness is sufficient to conclude that every representation of $A$ integrates to a representation of $G$, it is, in general, not necessary. The example I know of is when the manifold $M$ is the Riemann sphere $\mathbb{C}P^1$ and the Lie algebroid $A$ is generated by the real and imaginary parts of the holomorphic vector fields that vanish at $0 \in \mathbb{C}P^1$. A flat connection for this algebroid is basically a flat connection in the usual sense, but it is allowed to have a singularity of logarithmic type at $0$. In this case the source 1-connected groupoid integrating $A$ is not Hausdorff, but it has a quotient that is Hausdorff, and every representation of $A$ integrates to a representation of this Hausdorff quotient.<|endoftext|> TITLE: Inherent ambiguity of the context-sensitive language $L = {a^ib^ic^id^je^jf^j \bigcap a^ib^jc^id^je^if^j} $ or $a^nb^nc^nd^ne^nf^n$ QUESTION [6 upvotes]: What is the definition of ambiguity of context-sensitive grammar?This is relevant to the definition of inherent ambiguity of context-sensitive language.And any proof for the inherent ambiguity of the language $L = {a^ib^ic^id^je^jf^j \bigcap a^ib^jc^id^je^if^j} $ or $a^nb^nc^nd^ne^nf^n$? Since another question about inherent ambiguity of the context-sensitive language leads to suspicion of definition of ambiguity of CSG and inherent ambiguity of the context-sensitive language,and has been closed and deleted,a thesis founded on web claims Inherent ambiguity of the context-sensitive language $L = {a^ib^ic^id^je^jf^j \bigcap a^ib^jc^id^je^if^j} $ or $a^nb^nc^nd^ne^nf^n$.Here comes the two questions above. REPLY [3 votes]: Let me define a context-sensitive language strongly unambiguous if it is recognizable by a context-sensitive grammar such that every string in the language has a unique derivation. (Note that in the CFL world, this condition is much stronger than plain unambiguity.) I don’t know if there is a sensible well-behaved definition of an unambiguous context-sensitive language, but any reasonable such definition should have at least the property that every strongly unambiguous language is unambiguous. (The definition in Noah S’s answer violates this condition, so I do not consider it reasonable: it may well happen that a derivable string has no leftmost derivation at all, even if it has a unique derivation; for much the same reason, that definition also apparently violates another reasonable requirement, namely that the class of unambiguous CSL should be invariant under left-to-right reversal.) As explained below in more detail, it follows that under any reasonable definition, the language given in the question is unambiguous. Furthermore, it is impossible to unconditionally prove that any context-sensitive language is inherently ambiguous without serious assumptions from computational complexity. The class of context-sensitive languages is well-known to coincide with $\mathrm{NSPACE}(O(n))$, i.e., languages recognizable by a nondeterministic linear-space Turing machine. This contains the class of deterministic linear-space languages $\mathrm{DSPACE}(O(n))$. There is a natural notion of unambiguity used in this context (e.g., classes like UL or UP): a nondeterministic Turing machine is unambiguous if every string is accepted by at most one computation path. The class $\mathrm{USPACE}(O(n))$ of languages recognized by an unambiguous linear-space Turing machine is intermediate between the other two, that is, $$\mathrm{DSPACE}(O(n))\subseteq\mathrm{USPACE}(O(n))\subseteq\mathrm{NSPACE}(O(n))=\mathrm{CSL}.$$ To put things into context, the closure of any of these classes under polynomial-time reductions is $\mathrm{PSPACE}$. Now, the point is that the reduction of $\mathrm{NSPACE}(O(n))$ to CSL preserves the number of accepting paths: that is, given a nondeterministic linear-space Turing machine $M$, one can construct a context-sensitive grammar $G$ recognizing the same language in such a way that every $G$-derivation of a string $w$ corresponds to a unique accepting path of $M$ on input $w$, and vice versa. In particular, if $M$ is unambiguous, then $G$ is strongly unambiguous. So, the class of strongly unambiguous CSL languages (and a fortiori, unambiguous CSL languages under any reasonable definition) contains $\mathrm{USPACE}(O(n))$, and therefore $\mathrm{DSPACE}(O(n))$. The language in the question can be recognized in a straightforward way by a deterministic logarithmic-space Turing machine, hence it is strongly unambiguous. Moreover, the existence of any inherently ambiguous CSL language would imply a separation of $\mathrm{USPACE}(O(n))$ and $\mathrm{DSPACE}(O(n))$ from $\mathrm{NSPACE}(O(n))$, and by padding, also a separation of (U)L from NL. These are notoriously difficult open problem in complexity, akin to separation of P from NP. Incidentally, since every CFL (or LOGCFL) language can be recognized in deterministic space $O((\log n)^2)$ (and quasi-polynomial time), this also shows that every context-free language is a strongly unambiguous context-sensitive language. Given that one might reasonably demand the opposite, this suggests that there is in fact no reasonable definition of unambiguous CSL.<|endoftext|> TITLE: Distorsion of subgroups of the mapping class group QUESTION [5 upvotes]: Let $S_{g,b}$ be an oriented surface with $b$ boundary components and $S_g^b$ be an oriented surface with $b$ punctures. Let $\mathrm{Mod}(S_{g,b})$ and $\mathrm{Mod}(S_g^b)$ their (orientation preserving) mapping class group fixing boundary/punctures pointwise. There is a short exact sequence: $$1 \to \mathbb Z^b \to \mathrm{Mod}(S_{g,b}) \to \mathrm{Mod}(S_g^b) \to 1$$ Is $\mathbb Z^b$ undistorted in $\mathrm{Mod}(S_{g,b})$ ? What is the maximal rank of an undistorted Abelian subgroup in $\mathrm{Mod}(S_g^b)$ ? Thank you in advance. REPLY [6 votes]: Farb, Lubotzky, and Minsky proved in B. Farb, A. Lubotzky and Y. Minsky, Rank-1 phenomena for mapping class groups, Duke Math. J. 106 (2001), no. 3, 581–597. that all abelian subgroups of the mapping class group are undistorted.<|endoftext|> TITLE: Combinatorics and symmetric functions QUESTION [9 upvotes]: No answers from stackexchange, so I'll try this here: (The actual questions in this posting are at the bottom.) Occasionally someone asks on stackexchange how to show that every nonempty finite set has just as many subsets of odd cardinality as of even cardinality (the empty set is the unique exception). When the set has an odd number of members, a correspondence between complements does it, but generally, one picks a distinguished element. Every set containing the distinguished element corresponds to one not containing it, and that's the bijection. Now let's take it a step further and say one adds a new element to the set, which will be "distinguished" until the next new one gets added. $$ \begin{array}{c|c} \text{odd} & \text{even} \\ \hline a & \text{---} \\ b & ab \\ c & ac \\ abc & bc \end{array} $$ Now add the new element $d$: $$ \begin{array}{c|c} \text{odd} & \text{even} \\ \hline a & \text{---} \\ b & ab \\ c & ac \\ abc &bc \\ \hline d & ad \\ abd & bd \\ acd& cd \\ bcd& abcd \end{array} $$ Those listed on the odd side get re-listed on the even side and vice-versa, and the new element is added to both. Suppose that as above we let multiplication denoted by juxtaposition represent a set of ur-elements, and addition represent a set of sets, so the first table above has $a+b+c+abc$ as its left column and $1+ab+ac+ad$ as its second column, the $1$ being the product of members of the empty set. If, further, we represent the whole table as a fraction with the odd column as the numerator and the even column as the denominator, then the operation of adding the new element is this: $$ \frac{\frac{a+b+c+abc}{1+ab+ac+bc}+d}{1+\frac{a+b+c+abc}{1+ab+ac+bc}d} = \frac{a+b+c+d+abc+abd+acd+bcd}{1+ab+ac+ad+bc+bd+cd+abcd}. $$ So we're working with a binary operation: $a\,\diamondsuit\, b=\dfrac{a+b}{1+ab}$. When the set we start with is empty, the the numerator is the sum of $0$ terms and is $0$ and the denominator is the product of $0$ terms and $\text{is }1$. (And of course $\tanh(x+y)=(\tanh x)\,\diamondsuit\,(\tanh y)$, and there's also an application to something more like ordinary trigonometric functions.) QUESTIONS: Does this algebraic way of looking at this shed light on the combinatorics, or vice-versa? What is the combinatorial meaning of the fact that the binary operation is associative? I tried this with the even terms in the numerator and odd in the denominator: $a\,\spadesuit\, b$. We get $(a\,\spadesuit\,b)\,\spadesuit\,c =a\,\diamondsuit\,b\,\diamondsuit\,c$; and $(a_1\,\diamondsuit\,\cdots\,\diamondsuit\,a_n)\,\spadesuit\, a_{n+1} = 1/(a_1\,\diamondsuit\,\cdots\,\diamondsuit\,a_{n+1})$ --- So if you apply the $\spadesuit$ operation instead of $\diamondsuit$ each time a new element is added, then you alternate between odd-in-the-numerator-and-even-in-the-denominator and even-in-the-numerator-and-odd-in-the-denominator. Does the alternation have some combinatorial significance? REPLY [11 votes]: The algebraic significance of the operation $\diamond$ is that $(a+b)/(1+ab)$ is a formal group law. See Enumerative Combinatorics, vol. 1, 2nd ed., Exercise 1.163. As for $a_1\diamond\cdots\diamond a_n$, when $n\to \infty$ we get the symmetric function $(e_1+e_3+e_5+\cdots)/(1+e_2+e_4+\cdots)$. Writing this as a power series $e_1+(e_3-e_2e_1)+\cdots$, the terms of even degree are 0, and the term of degree $2n+1$ is (up to sign) the skew Schur function $s_{\tau_n}$, where $\tau_n$ is the "staircase border strip" of Enumerative Combinatorics, vol. 2, Exercise 7.64. There is a lot of interesting combinatorics, started by Foulkes, connected with this symmetric function. See also http://math.mit.edu/~rstan/papers/altenum.pdf.<|endoftext|> TITLE: How necessary is Godel's Condensation Lemma QUESTION [12 upvotes]: It seems that the Godel's Condensation Lemma is typically used to show that certain constructible sets will appear by some stage of the construction of $L$. For example in the proof that CH holds in $L$, GCL is used to show that if $X \subset \omega$ is contructible, then $X \in L_{\omega_{1}}$. Does anyone know if there are combinatorial arguments for these sorts of facts; i.e. arguments that don't use GCL or results from model theory? How would one prove $L \models$CH to someone who did not know the Lowenheim-Skolem theorem? REPLY [5 votes]: It's been 25 years since I struggled through it, but I clearly remember that Godel's book Consistency of the Axiom of Choice and of the Generalized Continuum Hypothesis with the Axioms of Set Theory gets this result combinatorially --- exactly how, I have long forgotten. I only learned about the much, much easier Lowenheim-Skolem proof later on. I've been told that Godel wrote his book that way, in machine language, essentially, because he didn't want any philosophical doubts people might have about logic to affect their reception of his theorem.<|endoftext|> TITLE: A question on $Z^{*}$ algebras QUESTION [5 upvotes]: A $Z^{*}$ algebra is a $C^{*}$ algebra which satisfies each of the following equivalent conditions: All elements of $A$ are left zero divisor. All elements are right zero divisor. All elements are two sided zero divisors All positive elements are two sided zero divisor. The commutative(topological) interpretation for this concept is the following: For a locally compact Hausdorff space $X$, $C_{0}(X)$ is NOT a $Z^{*}$ algebra if and only if $X$ is an approximately $\sigma$- compact space (Briefly $A\sigma C$ space). that is, there are a sequence of compact subsets $K_{n}$ of $X$ such that $\cup K_{n}$ is dense in $X$. Note that the product of two $A\sigma C$ spaces is a $A\sigma C$ space. Equivalently if $X \times Y$ is not a $A\sigma C$ space then $X$ or $Y$ is not a $A\sigma C$ space. This is a motivation to ask: Assume that $A$ and $B$ are two $C^{*}$ algebras such that their minimal tensor product is a $Z^{*}$ algebra. Is it true to say that $A$ or $B$ is a $Z^{*}$ algebra? Note that if the answer to the following question were affirmative, then the answer to the above question would be affirmative, too: positive element in C* tensor product REPLY [6 votes]: Recall that a completely positive map $\phi\colon A\to B$ is said to be faithful if $\phi(a) \neq 0$ for all nonzero $a \in A^+$. Lemma: If $\phi_i\colon A_i\to B_i$ are faithful cp maps, then $\phi_1\otimes\phi_2\colon A_1\otimes_{\min}A_2\to B_1\otimes_{\min}B_2$ is faithful. Proof: Since $\phi_1\otimes\phi_2=(\mathrm{id}_{B_1}\otimes\phi_2)\circ(\phi_1\otimes\mathrm{id}_{A_2})$, we may assume one of $\phi_i$ (say $\phi_2$) is $\mathrm{id}$. Let a nonzero $a \in (A_1\otimes_{\min}A_2)^+$ be given. By definition of the minimal tensor product, there are states $f_i$ on $A_i$ such that $(f_1\otimes f_2)(a) > 0$. Observe that $(f_1\otimes f_2)(a) = f_1((\mathrm{id}_{A_1}\otimes f_2)(a))$, where $\mathrm{id}_{A_1}\otimes f_2\colon A_1\otimes A_2\to A_1\otimes{\mathbb C}\cong A_1$ is the slice map. Hence, $(\mathrm{id}_{A_1}\otimes f_2)(a) \in A^+\setminus\{0\}$ and $$(\mathrm{id}_{B_1}\otimes f_2)((\phi_1\otimes\mathrm{id}_{A_2})(a)) = \phi_1((\mathrm{id}_{A_1}\otimes f_2)(a)) \neq 0,$$ which implies $(\phi_1\otimes\mathrm{id}_{A_2})(a) \neq 0$. $\Box$ Now, if $A_i$ are non-$Z^*$ $\mathrm{C}^*$-algebras, then there are $a_i\in A_i$ which are not left zero divisors. This means that the completely positive maps $\phi_i\colon A_i\ni x\mapsto a_i^*xa_i \in A_i$ are faithful. By the above lemma, $\phi_i\otimes\phi_2$ is faithful, which means that $a_1\otimes a_2$ is not a left zero divisor.<|endoftext|> TITLE: What is the large cardinal strength of the assertion that every $\kappa$-complete filter on $\kappa$ extends to a $\kappa$-complete ultrafilter? QUESTION [21 upvotes]: It is well-known that an uncountable regular cardinal $\kappa$ is strongly compact if and only if every $\kappa$-complete filter on any set extends to a $\kappa$-complete ultrafilter on that set. The usual proof of this, in the one direction, uses $\theta$-strong compactness to handle filters of size $\theta$, even when those filters concentrate on base sets of size less than $\theta$. My question concerns the nature of the limitation of the size of the base set in this equivalence, and in particular, what is the strength of the assumption in the case of filters on $\kappa$ itself. Question. What is the large cardinal strength of the assumption that $\kappa$ is an uncountable regular cardinal for which every $\kappa$-complete filter on $\kappa$ extends to a $\kappa$-complete ultrafilter on $\kappa$? For a lower bound, every such $\kappa$ is easily seen to be a measurable cardinal, since the club filter on $\kappa$ is $\kappa$-complete and so the property gives a measure on $\kappa$. For an upper bound, if $\kappa$ is $2^\kappa$-strongly compact, then the property holds by the usual characterization of strong compactness I alluded to above. Namely, if $F$ is a $\kappa$-complete filter on $\kappa$, then let $j:V\to M$ be a $2^\kappa$-strong compactness embedding. Since $F$ has size at most $2^\kappa$, the strong compactness cover property ensures that there is some $s\in M$ with $j^{\prime\prime}F\subset s$ and $|s|^M 2^{\aleph_0}$), and the current lower bound is the existence of non-domestic mouse which implies the consistency of $\mathrm{AD}_{\mathbb{R}}$. In the paper "Equiconsistencies at subcompact cardinals", Neeman and Steel prove that under some inner model theoretical assumptions the failure of square on two consecutive cardinals, where the lower one is large (Woodin) implies that there is an inner model with ($\Pi^2_1$-)subcompact cardinal. On the other hand, less than $2^\kappa$-supercompactness of $\kappa$ is sufficient in order to get that every $\kappa$-complete filter on $\kappa$ can be extended to a $\kappa$-complete ultrafilter (indeed, Neeman-Steel's $\Pi^2_1$-subcompact is enough). Thus in a model of $\mathrm{GCH}$ and level-by-level equivalence between strong compactness and supercompactness, the least $\kappa$ which is $\kappa$-compact is strictly smaller than the least $\kappa$ which is $\kappa^{+}$-strongly compact. The proof of the failure of squares from the filter extension property on $\kappa$ relies on showing that $\kappa$ being $\kappa$-compact is equivalent to the $\kappa$-compactness of $\mathcal{L}_{\kappa,\kappa}$ for theories of size $\leq 2^\kappa$ (namely, every $\mathcal{L}_{\kappa,\kappa}$-theory $T$ of size $\leq 2^\kappa$ such that every sub-theory $T' \subseteq T$, $|T'| < \kappa$, has a model, also has a model). This shows that the filter extension property lies in the hierarchy of partial strong compactness, strictly below $2^\kappa$-strong compactness of $\kappa$. Similarly, for $\theta$ such that $\theta^{<\kappa} = \theta$, the filter extension property of $\kappa$-complete filters on $\theta$ is equivalent to $\kappa$-compactness of $\mathcal{L}_{\kappa,\kappa}$ for languages of size $\leq 2^\theta$.<|endoftext|> TITLE: Hamilton-Ivey pinching in dimension 4 QUESTION [6 upvotes]: I've heard it said (e.g., in the accepted answer to this MO question) that a major obstacle to an effective theory of Ricci Flow in dimension 4 is the absence of the Hamilton-Ivey pinching phenomenon. I'm curious about the possibilities for such a pinching in dimension 4, but I couldn't locate any information about it. I'm curious about 2 complementary questions in this regard. Are there any known partial results or indications of what such a pinching may look like in dimension 4? Are there known examples that constrain the form of or throw doubt upon such a possible pinching? Any thoughts or references to the literature are welcome. REPLY [2 votes]: http://arxiv.org/pdf/0807.1582.pdf In this paper, the author gave a local result of Hamilton-Ivey pinching on the gradient shrinking soliton ($n\ge 4$) with vanishing Weyl tensor (see prop 3.2). Also in the introduction part, the author announced a pinching result for the solution to Ricci flow in $n-$dimension provided the solution is LCF for all time, although I believe the proof has not been published yet.<|endoftext|> TITLE: Analytic maps in homotopy classes QUESTION [7 upvotes]: Let $\mathcal M$ be a compact connected real-analytic manifold. It is well known that every continuous map $f\colon\mathcal M\to\mathbb S^1$ is homotopic to a smooth map. My question is the following. Are there any sufficient conditions on $\mathcal M$ that would guarantee the existence of an analytic map $\mathcal M\to\mathbb S^1$ in every homotopy class? REPLY [10 votes]: This is always true. The Morrey-Grauert theorem says that $M$ has a real-analytic embedding in Euclidean space, so real-analytic functions $M\to\mathbb{R}$ separate points, so they are dense in the algebra of all continuous function (by the Stone-Weierstrass theorem). Thus, given a map $f:M\to S^1\subset\mathbb{C}$ we can choose a real-analytic map $g:M\to\mathbb{C}$ with $\|f-g\|<1$. The formula $h(x)=g(x)/|g(x)|$ then gives a real-analytic map $h:M\to S^1$ that is homotopic to $f$.<|endoftext|> TITLE: Symplectic Koopmanism QUESTION [5 upvotes]: Let $(M, \omega)$ be a $2n$-dimensional symplectic manifold and let $L_2(M,|\omega^n|)$ be the Hilbert space of complex-valued functions on $M$ that are square integrable with respect to the Liouville measure. A classical (but still wonderful) remark due to Koopman is that given a measure preserving map $\phi: M \rightarrow M$, the operator $$ U_\phi : L_2(M,|\omega^n|) \longrightarrow L_2(M,|\omega^n|) $$ defined by $f \mapsto f \circ \phi$ is unitary. The operator $U_\phi$ is sometimes called the Koopman operator of the map $\phi$. Since symplectomorphisms are measure preserving, it seems natural to ask the following Vague question. Is there anything at all particular about Koopman operators of symplectomorphisms? REPLY [2 votes]: The Koopman operator is bracket preserving with respect to the canonical Poisson bracket on $C^{\infty}(M)$. Formally, this implies the dual operator (the Frobenius-Perron operator) is a Poisson automorphism with respect to the Lie-Poisson structure on $\bigwedge^{2n}(M)$.<|endoftext|> TITLE: Is the crossed product $\mathcal{K} \rtimes G$ a groupoid algebra? QUESTION [5 upvotes]: Suppose G, a discrete group acting on the compact operators $\mathcal{K}$ by automorphism of C*-algebra $\mathcal{K}$. Can we view the crossed product as a groupoid C*-algebra of some groupoid? This question occurred to me when I was thinking of possible formulation of Baum-Connes conjecture for projective representation of the group. REPLY [4 votes]: Have you seen Packer-Raeburn Stabilization trick? Judith A. Packer and Iain Raeburn, On the structure of twisted group $C^*$-algebras, Trans. Amer. Math. Soc. 334 no 2 (1992), 685-718, doi:10.1090/S0002-9947-1992-1078249-7 Specially Theorem 3.4 therein establishes that the algebra is Morita equivalent to a Twisted Group $C^*$-algebra, summarizing the answer by N. Ozawa and the correction above. By considering the trick of viewing the algebra $K\rtimes G$ as a locally trivial bundle with a $G$-action over a contractible $G$-space $X$ (for instance, the total space of the universal $G$-Bundle $EG$, but also the space $\underline{E}G$ works), you can see the $C^*$-algebra in question is Morita equivalent to the algebra of sections of the Fell bundle over the transformation groupoid $X\rtimes G$. This has been documented in Kumjian's monograph Alex Kumjian, On equivariant sheaf cohomology and elementary $C^*$-bundles, J. Operator Theory 20 (1988), no. 2, 207–240 (journal pdf) Other references on this kind of tricks include Siegfried Echterhoff, Morita Equivalent Twisted Actions and a New Version of the Packer-Raeburn Stabilization Trick, Journal of the London Mathematical Society 50 Issue 1 (1994) 170–186, doi:10.1112/jlms/50.1.170 and part 5 here: Noé Bárcenas, Twisted geometric K-homology for proper actions of discrete groups, Journal of Topology and Analysis (2018) doi:10.1142/S1793525319500729, arXiv:1501.06050.<|endoftext|> TITLE: Representation varieties of 3-manifold groups in $\mathrm{SL}(n,\mathbb{C})$ QUESTION [13 upvotes]: I am looking at the variety of representations of the fundamental group of a hyperbolic 3-manifold into $\mathrm{SL}(n,\mathbb{C})$: $$\mathrm{Hom}(\pi_1(M), \mathrm{SL}(n,{\mathbb C}))$$ It is known that volume and Chern-Simons invariant of representations are constant on connected components of the representation variety. In particular, the representation variety has at least 3 components because complex conjugation of the representation changes the sign of the volume (and the trivial rep has zero volume). What else is known about the number of connected components? Are there some lower bounds, perhaps coming from other invariants? REPLY [11 votes]: These are difficult questions and very little in general is known about this. Mostly, what's known is ad hoc results for specific classes of manifolds (say, take some surgeries on 2-bridge knots...). You can find some references in answers to this related question. Here is one general construction of connected components which, in general, cannot be reduced to the volume or the CS invariant. Let $M$ be a closed oriented hyperbolic 3-manifold (I assume this is what you are asking about). Then the holonomy representation $\rho: \pi\to SL(2,C)$ of its hyperbolic structure is locally rigid in $Hom(\pi, SL(2,C))$ (and the corresponding component can be detected by maximality of the volume). WLOG, we can assume that the field $K$ generated by the matrix coefficients of $\rho(\gamma), \gamma\in \pi$, is an algebraic number field (you can always achieve this by conjugation). Then, you get other locally rigid representations $\rho^\sigma: \pi\to SL(2,C)$ by applying to $\rho$ elements $\sigma$ of the Galois group $Gal(K,Q)$. The reason is that the group $H^1(\pi, Ad\rho^\sigma)\cong H^1(\pi, Ad\rho)$ vanishes. The resulting representations $\rho^\sigma$ are (obviously) faithful but never discrete. Some of these representations will land in a compact subgroup of $SL(2,C)$ (this will always happen if $M$ is arithmetic), some will land in a subgroup conjugate to $SL(2,R)$ (and some will in neither of these groups). Such non-Zariski dense representations will have zero volume and CS invariant. Thus, these invariant cannot distinguish components of such representations, from say, the trivial one.<|endoftext|> TITLE: Fixed points of $x\mapsto 2^{2^{2^{2^x}}} \mod p$ QUESTION [33 upvotes]: Let $p$ be a prime. For how many elements $x$ of $\{0,1,\dotsc,p-1\}$ can it be the case that $$2^{2^{2^{2^x}}} = x \mod p?$$ In particular, can you find a simple proof (or, even better, several simple proofs!) of the fact that this can happen only for $< \epsilon\cdot p$ elements $x$ of $\{0,1,\dotsc,p-1\}$? (Assume, if needed, that $2$ is a primitive root of $\mathbb{Z}/p\mathbb{Z}$.) REPLY [6 votes]: Your argument for three exponentials can be simplified a bit by using the multiplicative version of van der Corput instead of the additive version. Specifically, if your equation $$2^{y^{2^k}} = 2^{y} + k \quad(\ast)$$ has many solutions then there is some bounded $l>1$ such that there are many pairs of solutions $y,ly$, and for any such pair we must have $z=2^y$ solving $$(z+k)^{l^{2^k}} = z^l + k,$$ which of course has a bounded number of solutions. (Equivalently, write the equation in terms of $y' = 2^x$ instead of $y=2^{2^x}$ and apply additive vdC.) If we're being more careful then we should define $f:\mathbf{Z}/p\mathbf{Z}\to\mathbf{Z}/p\mathbf{Z}$ by $f(x) = 2^{\bar{x}}$, where $\bar{x}$ is the representative of $x$ satisfying $0\leq\bar{x} TITLE: An identity in the free associative algebra QUESTION [7 upvotes]: Let $V$ be a finite dimensional vector space over a field of characteristic $0$, and let $T(V)$ be the tensor algebra (also called the free associative algebra) on $V$. This is actually a Hopf algebra, where the coproduct is defined on words by splitting them into two pieces in all possible ways: $$\Delta(v_{I})=\sum_{I=I'\amalg I''} v_{I'}\otimes v_{I''}.$$ Iterating the coproduct yields maps $\Delta^k:T(V)\to T(V)^{\otimes(k+1)}$ which break a word apart into $k+1$ pieces in all possible ways. Define the operator $\hat{\Delta}^k\colon T(V)\to T(V)^{\otimes (k+1)}$ which breaks a word apart into $k+1$ nontrivial pieces in all possible ways. Let $m^{k}\colon T(V)^{\otimes {k+1}}\to T(V)$ be the multiplication operator, and define $\zeta\colon T(V)\to T(V)$ by $$\zeta=\sum_{i=0}^\infty \frac{(-1)^i}{i+1}m^i\hat{\Delta}^i.$$ Because the operator $\hat\Delta^{i}$ is locally nilpotent, this will be a finite sum when applied to any given element of $T(V)$ so there is no need to worry about convergence issues. Also note that $m^0\hat{\Delta}^0$ is by convention the number of ways to split a word into one nontrivial piece, which means that it is the identity on $T^+(V)$ and trivial on the base field sitting in degree $0$. Writing everything out in terms of basic definitions, we have $$\zeta(v_I)=v_I-1/2\sum v_{I_1}v_{I_2}+1/3\sum v_{I_1}v_{I_2}v_{I_3}+\cdots, $$ where each sum is over all ways to split $I$ into nontrivial pieces $I_1,\ldots, I_k$. Low degree calculations show that $\zeta(v)=v$ $\zeta(v_1v_2)=\frac{1}{2}[v_1,v_2]$ $\zeta(v_1v_2v_3)=\frac{1}{3}[v_1,[v_2,v_3]]-\frac{1}{6}[v_2,[v_1,v_3]]$ In particular, it looks like $\zeta$ lands in the space of iterated commutators, which in this case is the same as the primitive elements in the Hopf algebra. So this is my question: How can one show that the image of $\zeta$ lies in the subspace of primitive elements? Conceivably one could either show that $\Delta(\zeta(v_I))=1\otimes \zeta(v_I)+\zeta(v_I)\otimes 1$, or directly show that $\zeta(v_I)$ is an iterated commutator. REPLY [5 votes]: $\newcommand{\id}{\operatorname{id}}$ Just adding in a couple steps missing in the answer by "an eulerian idempotent": Your map $\zeta$ can be rewritten as the logarithm of the map $\id : T(V) \to T(V)$ in the convolution algebra $\left(\operatorname{Hom}\left(T(V), T(V)\right), \star\right)$. (In fact, if we let $u$ denote the unit map and $\epsilon$ the counit map of $T(V)$, then every $i$ satisfies $\widehat{\Delta}^i = \left(\id - u\epsilon\right)^{\otimes \left(i+1\right)} \circ \Delta^i$, so that $m^i \circ \widehat{\Delta}^i = m^i \circ \left(\id - u\epsilon\right)^{\otimes \left(i+1\right)} \circ \Delta^i = \left(\id - u \epsilon\right)^{\star \left(i+1\right)}$, and thus your formula defining $\zeta$ reveals itself as the Mercator series for the logarithm of $\id$ in the convolution algebra.) So yes, this is the (first) Eulerian idempotent. To add two more references to the fact that it projects $T(V)$ onto its primitives: Gérard Henry Edmond Duchamp, Vincel Hoang Ngoc Minh, Christophe Tollu, Bùi Chiên, Nguyen Hoang Nghia, Combinatorics of $\varphi$-deformed stuffle Hopf algebras, arXiv:1302.5391v7, page 12 ("We first prove that $\pi_{1, \mathcal A}$ is a projector $\mathcal A \to \operatorname{Prim}\left(\mathcal B\right)$"; in your case it is obvious that $\mathcal B = \mathcal A$). This gives a self-contained proof Victor Reiner and your humble servant, Hopf algebras in combinatorics, June 5, 2014, Exercise 5.35 (the numbering is subject to change, so search for "e is a projection from A").<|endoftext|> TITLE: Cantor set intersecting a geometric sequence QUESTION [11 upvotes]: I was working on a problem involving finding all points in the intersection of the Cantor set $C$ and the geometric sequence $\{ (2/3)^i \}_{i=1}^\infty$. The only points I have in this intersection at the moment are $2/3$, $(2/3)^3$, and $(2/3)^9$. With the ternary representation of the Cantor set, the question can be translated to "Which powers of $2$ have no $1$'s in their ternary representations?" I have verified this in Mathematica for powers of $2$ with exponent $i \leq 10000$, and the only results I obtained was $2^1=2=(2)_3$, $2^3=8=(22)_3$, and $2^9=512=(200222)_3$. Can one actually prove (or disprove) that there is no more points in this intersection? I have worked on this seemingly simple question for a while but could not find a way to go. Thanks a lot! REPLY [11 votes]: This is an old unsolved problem. Erdos conjectured (see the first problem in that paper) that for all $n\ge 9$ the ternary expansion of $2^n$ contains the ternary digit $2$ (this is equivalent to for every $n\ge 10$ the ternary expansion of $2^n$ contains a $1$). For recent work related to this (and references) see these papers of Abram and Lagarias, and Lagarias.<|endoftext|> TITLE: Meta$^{n{-}th}$ mathematics QUESTION [9 upvotes]: Metamathematics has a reasonably clear connotation, enough to have a Wikipedia page, with Gödel, Tarski, and Turing playing leading roles; Kleene's book (Introduction to Metamathematics (Amazon link)); Chaitin's article ("Meta-mathematics and the foundations of mathematics." EATCS Bulletin, June 2002, vol. 77, pp. 167-179); etc. My question is: Q. Is there an identifiable meta-metamathematics, a scholarly study of metamathemaics, perhaps in the philosophical (rather than mathematical) literature? Or does all the literature essentially "devolve" to metamathemathics, without an identifiable line that can be drawn between metamathematics and meta-metamathematics? Citations to possibly-meta-metamathematical studies would be appreciated! Thanks! REPLY [29 votes]: My opinion is that there is no crisp distinction between mathematics, metamathematics and meta-metamathematics, and the subjects thoroughly blend one into another in such a way that prevents any coherent distinction. Furthermore, even the categorization of particular topics as mathematics or metamathematics has changed radically over time, and many topics that were formerly considered metamathematics are now just mathematics. For example, the ultrapower construction was born in metamathematics, but is now widely seen as a fundamental mathematical construction. The method of forcing was initially used only for relative consistency proofs, but is now saturated with a mixture of infinite combinatorics, ideals, Boolean algebras, topology, transfinite limits, and so on. Computability theory was born in purely philosophical speculation about what it means for a human to undertake a computable procedure, but gave birth to complexity theory and other extremely applied mathematical topics. Is the polynomial time hierarchy regarded today as metamathematics? I don't think so, but it is a part of complexity theory, which is a part of computability theory, which is traditionally considered metamathematics. The study of large cardinals is tied to fundamental issues in logic, such as definability and constructiblity, but also involves at its core essentially mathematical questions about infinite combinatorics, measure theory, complex systems of embeddings and so on. Where does the mathematics end and the metamathematics begin? It is all wrapped up together. The term metamathematics has traditionally included the entire subjects of model theory, set theory, proof theory and computability theory, but I think this kind of usage of the term is simply no longer accurate, since huge parts of these subjects are now more mathematical than metamathematical. I think that the term "metamathematics" may have made more sense as a unifying umbrella term in an earlier age, when many mathematicians were simply less familiar with these subjects than is the case today. Consider my work with Benedikt Löwe on the modal logic of forcing. The main theorem is that the ZFC provably valid principles of forcing are exactly those in the modal theory known as S4.2. Now, the principles under consideration, the principles of forcing, can themselves surely be considered as metamathematical, as they concern how truth varies in the generic multiverse, the Kripke model of possible worlds consisting of the set-theoretic universe in the context of all its forcing extensions. Since the principles are thus metamathematics, and we are proving theorems about which principles are provably valid, one could consider this to be solidly a case of meta-metamathematics. But if you look at the paper, I think you will mainly find just plain old mathematics, with detailed inductions and finite partial order combinatorics and some infinite combinatorics and forcing iterations, mixed in with some modal logic, which is essentially finite combinatorics. This example therefore illustrates my point that there is really no coherent distinction into mathematics/metamathematics/meta-metamathematics.<|endoftext|> TITLE: On an inequality among determinants QUESTION [8 upvotes]: For Hermitian matrices $X, Y$, I write $X\ge Y\ge 0$ to mean $X-Y$ and $Y$ are positive semidefinite. In Lemma 2.5 of [Linear Algebra Appl. 452 (2014) 1-6] I proved that if $X + Y\ge W + Z$, $X\ge W\ge Y\ge 0$ and $X\ge Z \ge Y\ge 0$, then $$\det X+\det Y\ge \det W+\det Z.$$ I was thinking of relaxing the condition a little bit. Is the same inequality true under only the condition $X + Y\ge W + Z$, $X\ge Y\ge 0$, $X\ge W\ge 0$, $X\ge Z\ge 0$? REPLY [2 votes]: The answer is yes. I thank fedja for discussion. WLOG, assume $X=I$ (we may do so by pre-post multiplying both sides by $\det X^{-1/2}$, with a standard continuity argument). After this, we may further assume $Y=D$ to be diagonal (by unitary similarity). So the question is equivalent to showing if $I+D\ge W+Z$ and $I\ge D, W, Z\ge 0$, then $1+\det D\ge \det W+\det Z$. Note that $I+D\ge W+Z$ implies $I+D\ge diag(W)+diag(W)$, where $diag(\cdot)$ means the diagonal part. By a simple induction on the size of the matrix, one gets $1+\det D\ge \det (diag(W))+\det (diag(Z))$. The conclusion follows by applying the Hadamard inequality, $\det (diag(W))\ge \det W$.<|endoftext|> TITLE: Maximum value of the binomial coefficient as a polynomial QUESTION [7 upvotes]: What is the maximum (absolute) value of the binomial coefficient $\begin{pmatrix}x \\ k\end{pmatrix} = \frac{1}{k!}x(x-1)(x-2)\dotsb(x-k+1)$ for real $x$ in the interval $0 \leq x \leq k-1$? REPLY [2 votes]: I'm not sure it adds anything beyond what's in Brendan McKay's answer, but I'll flesh out my comment. The absolute value of the binomial coefficient we are interested in may be written (for $x \in (0,1)$ and $k>1$ ) as $\frac{x}{k}\prod_{i=1}^{k-1}(1 - \frac{x}{i}).$ Applying the AM-GM inequality to the displayed product, this is at most $\frac{x}{k}(1 - \frac{ xH_{k-1}}{k-1})^{k-1},$ which is in turn at most $$\frac{x e^{-H_{k-1}x}}{k}.$$ As Brendan McKay noted, this is at most $\frac{1}{ekH_{k-1}},$ but perhaps the expression $\frac{x}{k}(1 - \frac{ xH_{k-1}}{k-1})^{k-1}$ will give a better estimate for small $k,$ as that takes maximum value when $x = \frac{k-1}{kH_{k-1}},$ and the maximum value attained there is $\frac{1}{kH_{k-1}}(1-\frac{1}{k})^{k}$ (thanks to Emil Jerabek for a correction here).<|endoftext|> TITLE: Primes from a Dirichlet sequence and an irrational number QUESTION [5 upvotes]: From Dirichlet's theorem on arithmetic progressions, if $\text{gcd}(a,b)=1$ we know $\{ak+b\}_{k\ge 0}$ contains infinitely many primes. Let those primes be $p_1,p_2,\cdots$. Then the real $$\alpha=0.p_1p_2p_3\cdots\tag{1}$$ is it irrational? Here the primes are placed side by side, as in $p_1=13,p_2=53,\cdots$ then the expression would be like $\alpha=0.1353\cdots$. So to rephrase my question, is $\alpha$ irrational, as defined in $(1)$? Thanks for answering. REPLY [14 votes]: It is well-known that not only does the arithmetic progression $\{ak+b\}_{k \in \mathbb{Z}^{+}}$ contain infinitely many prime numbers, but also that the series of the reciprocals of those primes diverges. The answer to the OP's question can be obtained now from the following general result: If $\{a_{i}\}_{i \in \mathbb{N}}$ is a strictly increasing sequence of natural numbers such that the series $\sum_{i=1}^{\infty} \frac{1}{a_{i}}$ diverges, then the unending decimal fraction $\alpha$ formed by juxtaposing the successive terms of the sequence $\{a_{i}\}_{i \in \mathbb{N}}$ represents an irrational number. For a proof of this theorem, see D. J. Newman, R. Breusch, and F. Herzog. Solution to problem 4494. Amer. Math. Monthly 9 (60), Nov. 1953, pp. 632-633. or N. Hegyvári. On some irrational decimal fractions. Amer. Math. Monthly 8 (100), Oct. 1993, pp. 779-780. REPLY [9 votes]: Yes, it is irrational. This is because any finite digit sequence occurs as the initial digits of a prime in your sequence (in fact you can prescribe 41% of all the digits in the beginning), hence the concatenated sequence is not periodic. Added. The OP asked for more details so here they are. Assume that $\alpha$ is rational. Then the decimal expansion of $\alpha$ is periodic after a certain digit. This implies that there is a digit $d\in\{0,\dots,9\}$ and a positive integer $n\in\mathbb{N}$ such that among any $n$ consecutive digits of $\alpha$, the digit $d$ occurs. On the other hand, by known results on the distribution of primes in arithmetic progressions, there is a $p_i$ which starts with $n+1$ digits distinct from $d$. This is a contradiction, which shows that $\alpha$ is irrational. In fact, by a result of Huxley and Iwaniec (Mathematika 22 (1975), 188-194), the interval $(x,x+x^{0.584})$ contains a $p_i$ for any sufficiently large $x>0$. So if $x$ is sufficiently large, we can find a $p_i$ whose decimal expansion agrees with that of $x$ on the initial 41% of the digits, since the error $x^{0.584}$ only affects about the last 58.4% of the digits.<|endoftext|> TITLE: Literature about metapolynomials QUESTION [5 upvotes]: We say that a function $f:\mathbb{R}^k \rightarrow \mathbb{R}$ is a metapolynomial if, for some positive integer $m$ and $n$, it can be represented in the form $$f(x_1,\cdots , x_k )=\max_{i=1,\cdots , m} \min_{j=1,\cdots , n}P_{i,j}(x_1,\cdots , x_k)$$ where $P_{i,j}$ are multivariate polynomials. Prove that the product of two metapolynomials is also a metapolynomial. This problem was given in the IMO Shortlist 2012 A7, I want to know if there is some paper on this topic metapolynomials. I searched on the web but I found nothing. I found this topic interesting. So a reference would be helpful. REPLY [5 votes]: If you put an absolute value around your polynomial, then this sort of max-min comes up in the construction of local height functions. In the terminology of Lang's Fundamentals of Diophantine Geometry (see especially Chapter 10, Theorem 3.5, page 261), the equation displayed at the top of page 262 is $$ \lambda = \sup_j \inf_i \lambda_{ij}. $$ Unsorting the definitions, one has essentially $$ \lambda_{ij}(x_1,\ldots,x_n) = \log\bigl| f_{ij}(x_1,\ldots,x_n)\bigr| $$ for a certain collection of polynomials (or maybe rational functions) $f_{ij}$. So aside from having taken logs, the function $\lambda$ is the IMO max-min with absolute values. These Weil functions are associated to divisors and do satisfy an addition formula $\lambda_{D_1}+\lambda_{D_2}=\lambda_{D_1+D_2}$, which is related to the IMO question. The IMO question is less general, but more precise because the Weil function $\lambda_D$ associated to a divisor $D$ is only well-defined up to what Lang calls an $M_K$-bounded function. On the other hand, the proof of the addition formula probably comes down to doing the IMO problem (modulo those pesky absolute value signs). Final Comment: In Lang's setup, the absolute value may be the usual one on $\mathbb R$ or $\mathbb C$, but it could also be a $p$-adic absolute value; and more generally, one really wants a collection of Weil height $\lambda_{D,v}$, one for each absolute value $v$, that have some sort of uniformity as one varies over $v$.<|endoftext|> TITLE: Combinatorics Problem: $\sum _{k=0}^{s-1} \binom{n}{k}=\sum _{k=1}^s 2^{k-1} \binom{n-k}{s-k}$ QUESTION [5 upvotes]: The question is whether the below is true. $$\sum _{k=0}^{s-1} \binom{n}{k}=\sum _{k=1}^s 2^{k-1} \binom{n-k}{s-k}$$ Mathematica can simplify as follows, but it fails to Reduce[] or Solve[]. $$2^n=\binom{n}{s} \, _2F_1(1,s-n;s+1;-1)+\binom{n-1}{s-1} \, _2F_1(1,1-s;1-n;2)$$ REPLY [25 votes]: A slightly less computational method is to note that both sides of the identity count the number of subsets of $\{1,\dots,n\}$ with fewer than $s$ elements. This is obvious for the left hand side. It's true for the right hand side because $2^{k-1}\pmatrix{n-k\\s-k}$ is the number of such subsets $S$ for which $k$ is minimal such that $|S\cup\{1,\dots,k\}|\geq s$, since such a subset $S$ is the union of an arbitrary subset of $\{1,\dots,k-1\}$ and a subset of size $s-k$ of $\{k+1,\dots,n\}$.<|endoftext|> TITLE: Linear transformation of a polyhedron QUESTION [9 upvotes]: Is there a simple proof that shows: Linear transformation of a $\mathcal{H}$-polyhedron (i.e. the intersection of finitely many closed half-spaces) is a $\mathcal{H}$-polyhedron. Minkowski sum of two $\mathcal{H}$-polyhedrons is a $\mathcal{H}$-polyhedron. I know a proof of (1.) based on Fourier-Motzkin elimination. and, I know (2.) is a simple consequence of (1.). Every different approach is appreciated. REPLY [5 votes]: I'm not familiar with Fourier-Motzkin, so I don't know how different the following argument is from what one usually does, but it's direct and elementary (and constructive, it in principle produces the new constraints from the old ones). The claim is trivial if $A\in\mathbb R^{n\times n}$ is invertible, and a general $A$ can be written as $A=PB$, with $B$ invertible and $P$ a projection, so we can focus on projections. We can in fact also assume that $P$ is a projection on a codimension $1$ subspace, say $P(y+\alpha e)=y$, for $y\perp e$ and $\alpha\in\mathbb R$. Suppose the polyhedron $Q$ is defined by the constraints $x\cdot n_j\le c_j$. We are then interested in $$ S=P(Q)=\{ y\in\{e\}^{\perp} : y\cdot n_j \le c_j + d_j\alpha \:\textrm{ for some }\alpha\in\mathbb R \textrm{ and }j=1,\ldots, N \} $$ (the same $\alpha$ for all $j$ of course). We can further assume that $d_j=0$ or $\pm 1$. Call a constraint zero, positive, or negative according to the sign of $d_j$. The zero constraints are already of the desired type and can be ignored. The case of only positive (or only negative) constraints is trivial ($S=\{e\}^{\perp}$ in both cases). In the remaining case, I claim that $y\in S$ precisely if $$ y\cdot (n_k^+ + n_j^-) \le c_k^+ + c_j^-\quad\quad\quad (1) $$ for all choices of pairs $(k,j)$ of one positive and one negative condition. Indeed, we can rewrite (1) as $$ y\cdot n_k^+ \le c_k^+ + \min (c_j^--y\cdot n_j^-) , $$ and then observe that the largest $\alpha$ that satisfies all negative constraints for a given $y$ is $\alpha=\min (c_j^--y\cdot n_j^-)$. It is now clear that (1) is equivalent to $y\in S$.<|endoftext|> TITLE: Self-dual automorphic forms on $GL(4)$ QUESTION [9 upvotes]: As is known among experts, all self-dual automorphic forms on $GL(3)$ come from symmetric square lifts from $GL(2)$. You can find this in Ramakrishnan (http://www.math.caltech.edu/~dinakar/papers/exercise-Selfdual-GL(3).pdf) or Flicker or probably Gelbart-Jacquet. I would like to know what is known for self-dual forms on $GL(4)$. An easy construction shows that symmetric cube lifts from $GL(2)$ and Rankin-Selberg lifts from $GL(2)\times GL(2)$ will give us self-dual forms on $GL(4)$. I raise the following question: Are there "many" automorphic forms on GL(4) which are NOT coming from the mentioned lifts (symmetric cube and Rankin-Selberg)? REPLY [6 votes]: As I say in my comment, a possibly useful way to think about this is from $L$-functions, and the Sato-Tate distributions for degree 4 $L$-functions. There is work of Fite, Kedlaya, Rotger, and Sutherland on this. In weight 1: http://dx.doi.org/10.1112/S0010437X12000279 http://math.mit.edu/~drew/sato-tate-g2-errata.txt In weight 3: http://arxiv.org/abs/1212.0256 I will concentrate on the second paper (weight 3). In brief: almost everything (from dimension counts) should have generic $USp_4$ Sato-Tate group, but neither ${\rm Sym}^3 A$ nor $A\otimes B$ has such a group. So the answer to the specific raised question is "no". In particular, there are a few "standard" (functorial) operations to construct a degree 4 $L$-function (here symplectic, so indeed self-dual). You listed ${\rm Sym}^3 A$ and $A\otimes B$, which are in Section 6 of the "weight 3" link above. Another example is $A\oplus B$, which is Section 5 (the $L$-functions are imprimitive here). They don't mention inductions from $GL_1$ (Grossencharacters) or $GL_2$ (Hilbert modular forms), but those are other "functorial" constructions. Most of the "smaller" Sato-Tate groups they list come about from eg complex multiplication, or a tensor/sum of "correlated" $L$-functions. But the main bulk (the "generic" case) should be Siegel modular forms. My knowledge is not that deep, but I think there are still "lifts" in the SMF class which give "generic" $L$-functions (the full $Sp_4$ Sato-Tate group), and that these are often used in the literature to exhibit examples, but that the "generic" SMF is not a lift (correspondingly, eg some small level modular forms for $GL_2$ are eta-products, they have the "generic" Sato-Tate group, but perhaps are not truly "generic" in the sense one might want). Examples of SMF lifts include the Saito-Kurokawa and (Miyawaki-)Ikeda lifts (I am no expert, there are likely more I think, I just know these words). EDIT: these are not really relevant, they give imprimitive deg 4 $L$-functions it seems. In fact, the Yoshida lift would be more relevant (it can give both the direct sum and tensor product from the spinor/standard $L$-functions). From the standpoint of what is expected, the $d$-column on page 10 of the "weight 3" preprint gives the dimension of the associated space, so indeed almost everything is thought to be $USp_4$ Sato-Tate group (dimension 10) while the ${\rm Sym}^3$ construction lands you in $D$ (dimension 3), the tensor product is generically in the normalizer of $U(2)$ (dimension 2), the direct sum (dimension 6 or 4) is $G_{3,3}$ or $G_{1,3}$, depending on whether you direct sum the same weights (wt 4 modular forms), or first Tate twist (wt 2 plus wt 4) the induction from $GL_2$ (Hilbert modular forms) is dimension 6, giving the normalizer of $G_{3,3}$ (they do not mention this one, maybe it's the normalizer of $G_{1,3}$ in non-parallel weight $[2,4]$ and with $G_{3,3}$ in parallel weight $[4,4]$, I would have to think). the induction from $GL_1$ (Grossencharacters) is in the $F_\bullet$ classes (I think, again they don't do this explicitly, and get various $F$-classes as "degenerations" as mentioned next), so dimension 2 (I don't think a full dihedral example occurs over ${\bf Q}$ in the sense they demand, but $F_{ac}$, with its $C_4$ in the fourth column, should occur for cyclic quartic fields). Everything else (the cyclic and J-cyclic data) is dimension 1 in the appropriate moduli space, as I say, coming about from having "complex multiplication" on the underlying data in one of the above constructions. For instance, tensor a CM elliptic curve with a CM modular form of weight 3, depending on how the CMs line up, you get different behavior, landing in $F$-classes, see 6.15 (the point being, you did the CM-induction before the tensor, but this can be unwound, so that's why induced $F$-classes appear). Sorry for such a sprawling answer, and indeed it is certainly not from the automorphic perspective you phrased, but I hope it helps. In brief: almost everything should have generic $USp_4$ Sato-Tate group, but neither ${\rm Sym}^3 A$ nor $A\otimes B$ has such a group. So the answer is no.<|endoftext|> TITLE: Chow motive of a fibration QUESTION [7 upvotes]: If we have a fibration of smooth projective complex varieties $F\to E\to B$, which is locally trivial in the analytic topology, and the global monodromy is trivial. Then is it true that the Chow motives satisfy $h(E)\cong h(F)\otimes h(B)$? I am pretty sure this should fail, but cannot think of a counterexample. REPLY [4 votes]: I know very little about motives, but conjecturally you should expect that $h(E) \cong h(F) \otimes h(B)$. Your hypotheses imply that $E$ and $F \times B$ have isomorphic cohomology in the category of $\mathbf Q$-Hodge structures. In fact if $E \to B$ is smooth and proper, and $B$ is smooth and connected, then Saito's theory of mixed Hodge modules implies degeneration of the Leray spectral sequence and that the Leray filtration splits in the category of mixed Hodge structures. The additional assumptions of locally trivial (I'm assuming this means in the category of complex analytic spaces) and trivial monodromy imply also that $H^p(B,R^q f_\ast \mathbf Q) \cong H^p(B) \otimes H^q(F)$. (It also implies more directly that Leray spectral sequence degenerates.) In any case, the conjectural functor (pure/mixed motives) $\to$ (pure/mixed $\mathbf Q$-Hodge structures) is conjectured to be fully faithful. This statement contains in particular the Hodge conjecture. So you should expect $E$ and $F\times B$ to have isomorphic motives, although proving this may be hard. On a sidenote, do you have any nontrivial examples of such fibrations? I have a hard time imagining one.<|endoftext|> TITLE: Differential operator simplification QUESTION [9 upvotes]: Does anyone know the explicit formulation for the $q_k$'s in, $$(x+D)^n=\sum_{k=0}^n q_k(x)D^k\ \ \ \ ?$$ I know that $e^{-x^2/2+x}$ is a fixed point of $(x+D)$. I also, know that $$(x+D)H_n(x)e^{-x^2/2} = 2n H_{n-1}(x)e^{-x^2/2},$$ where $H_n(x)$ are the Hermite polynomials. Hence, $$(x+D)^n H_k(x)e^{-x^2/2} = 0$$ for all $k TITLE: smooth affine surfaces over algebraically closed fields with trivial l-torsion of the Brauer group QUESTION [5 upvotes]: I am looking for examples of smooth affine surfaces over algebraically closed fields with trivial $\ell$-torsion of the Brauer group. Related questions: Schemes with trivial brauer group and Brauer group of projective space The affine plane $\mathbf{A}^2_k$ should be one example. Edit: I am especially interested in cases where the surface is not rational. Note that it is difficult to pass from projective surfaces to affine surfaces because of purity for Brauer groups: $0 \to \mathrm{Br}(S)(\ell) \to \mathrm{Br}(S - C)(\ell) \to H^1(C,\mathbf{Q}_\ell/\mathbf{Z}_\ell)$. Here, $H^1(C,\mathbf{Q}_\ell/\mathbf{Z}_\ell) = 0$ iff $C$ has genus $g(C) = 0$, is affine and $\bar{C} - C$ (the divisor at infinity) is a one-point set. REPLY [2 votes]: Here is an approach that should work. We know that $Br(\bf{A}_k ^2)=0$ if $k$ is an algebraically closed field. Now let $G$ be a finite group of order $n$ equipped with an action on $\bf{A}_k ^2$ such that $X$ is the quotient surface and $\bf{A}_k ^2\to X$ is flat. Now if $(\ell,n)=1$ then $Br(X)$ has no $\ell$ torsion since $Br(X)$ is split by a finite covering of order $n$ prime to $\ell$ and so must consist of torsion elements of order dividing $n$.<|endoftext|> TITLE: A geometric characterization of smooth points of a complex algebraic variety QUESTION [7 upvotes]: Let $X^m\subset \mathbb{C}^n$ be an irreducible $m$-dimensional complex algebraic subvariety. Let $\mathbb{C}^n$ be equipped with the standard Hermitian metric. Fix an arbitrary point $p\in X$. Let $V(p,\varepsilon)$ denote volume of the intersection of $X$ with the $\varepsilon$-ball centered at $p$, namely $2m$-dimensional Hausdorff measure of the intersection. Question 1. Whether there exists the limit $m(p):=\lim_{\varepsilon\to +0} \frac{V(p,\varepsilon)}{\omega_{2m} \varepsilon ^{2m}}$, where $\omega_{2m}$ is the volume of the standard $2m$-dimensional Euclidean ball? Question 2. If the limit in the Question 1 always exists, is it true that this limit $m(p)$ is a natural number? Question 3. If Question 2 has positive answer, is it true that the point $p$ is smooth if and only if $m(p)=1$? One may ask the same questions for any Kahler manifold instead of $\mathbb{C}^n$, for example for complex projective space with the Fubini-Study metric. My feeling is that it should not be very important, but I do not have a proof. A reference would be very helpful. REPLY [14 votes]: The answer to all three of your questions is yes.See the book by E M Chirka titled Complex Analytic Sets pages 189,190 and 120 .These questions are local so this is true on Kahler manifolds .<|endoftext|> TITLE: Probability over a plane QUESTION [10 upvotes]: I raise this question following the reading of Fifty challenging problems in probability with solutions. One of the problem consists in computing the probability that the quadratic equation $x^2 + 2b x+c=0$ has complex roots. The "natural way" of doing it, is to suppose that the point $(b,c)$ is randomly chosen over a large square centered at the origin, with size $2B$. By following this path, the probability is equal to $0$ when $B$ approaches $+\infty$. However, if you compute the probability over the rectangle $[-\sqrt{B},\sqrt{B}] \times [-B,+B]$, it is equal to $\frac{1}{3}$ and therefore is also the limit when $B$ approaches to $+\infty$. I agree that the first way is "more natural". My question is: how to give more mathematical background to the "natural way"? Thanks for your support on this "not so mathematical question". REPLY [12 votes]: well, to find a "natural way" to distribute the coefficients $b,c$ in the plane, you could treat this problem as the special case $n=2$ of a classic problem in random-matrix theory: take an $n\times n$ real matrix $M$ with independent identical normal distributions of the matrix elements $M_{nm}$; what is the probability that all $n$ eigenvalues are real? for $n=2$ we would then identify ${\rm tr}\,M=2b$ and ${\rm det}\,M=c$. this probability is known exactly [1] for any $n$, it equals $2^{-n(n-1)/4}$; so for $n=2$ the probability of real roots is $1/\sqrt{2}$. [1] The circular law and the probability that a random matrix has k real eigenvalues, A. Edelman (1993). (Published in Journal of Multivariate Analysis 60 (1997) 203-232.)<|endoftext|> TITLE: Free Boson Correlator $ \langle X(z)X(w) \rangle =- \ln |z - w| $ QUESTION [10 upvotes]: In physics papers, the massless free boson has a definition involving an action: $$ S(X) = \frac{1}{8\pi} \int d\sigma^2\, \partial X \overline{\partial X}$$ The random functions $X(z)$ are sampled according to the Gaussian distribution: $e^{-S(X)}$. In this case we can define the correlator to be the vacuum expectation value: $$ \mathbb{E}[X(z,\overline{z})X(w, \overline{w})] \equiv \langle \varnothing | X(z,\overline{z})X(w, \overline{w}) |\varnothing \rangle = - \tfrac{1}{2}\ln |z - w|$$ See Ginsparg, applied Conformal Field Theory. I am trying to fill in the details of this derivation... using probability theory, complex analysis or anything else. Since $X$ is a bosonic field I will just model it as a function... judging from the norm it should live in some Hardy space $X \in H^2$ - the complexification of $L^2$ - where we should fill in some Riemann surface with flat structure. I think Ginsparg uses $H^2(\mathbb{R}\times S^1)$. The equations of motion ought to be the Cauchy Riemann equations. The fields in this theory are holomorphic functions. $X(z) = \sum a_n z^n$. Then I would like to expand. $\partial X(z) = \sum n a_n z^n$ and get $$ \langle \partial X(z)\partial X(w) \rangle = \sum n\mathbb{E}[a_n^2]\,(z \overline{w})^n = \frac{1}{|1-z\overline{w}|^2}$$ There are holes everywhere in this argument - a real physicists's derivation. How can I make this more rigorous? Especially the intuition about the Gaussian random function in $L^2$ ? Thanks. REPLY [8 votes]: My answer Wiener measure and Bochner Minlos can help defining a free massless real scalar field as a random element of $S_0'(\mathbb{R}^2)$. Namely, take the Schwartz space of rapidly decaying test functions $S(\mathbb{R}^2)$ and consider the subspace $S_0(\mathbb{R}^2)=\{f| \widehat{f}(0)=0\}$ of "charge-neutral" test functions. The bilinear form $$ B(f,g)=\int_{\mathbb{R}^2} \frac{d^2\xi}{(2\pi)^2}\ \frac{\overline{\widehat{f}(\xi)}\widehat{g}(\xi)}{|\xi|^2} $$ is continuous and positive on $S_0(\mathbb{R}^2)$ and therefore by the Bochner-Minlos Theorem there exists a unique centered Gaussian probability measure $\mu$ on the topological dual $S_0'(\mathbb{R}^2)$ for which $$ \mathbb{E}(\phi(f)\phi(g))=B(f,g) $$ where $\phi$ is the corresponding random element in $S_0'(\mathbb{R}^2)$. I did not do the computation (which needs a lot of care) but I suspect that $B(f,g)$ should be a multiple of $$ \int_{\mathbb{R}^2\times\mathbb{R}^2} d^2 x\ d^2y\ f(x)(-\log|x-y|)g(y)\ . $$ The issue here is that you have both a UV and an IR problem to deal with. The slow decay of the propagator for large $\xi$ makes it so that $X$ or $\phi$ is a random distribution rather than a random function and, in particular, pointwise evaluations $X(x)$ do not make sense. Also, the divergence at $\xi=0$, or zero-mode, makes it so that there is an ambiguity of shifting the field by a constant so the field $X$ does not make sense but its "increments" do. Working with $\partial X$ is another way to circumvent this issue. Edit: I just did the computation and indeed $$ B(f,g)=\frac{1}{2\pi}\int_{\mathbb{R}^2\times\mathbb{R}^2} d^2 x\ d^2y\ f(x)(-\log|x-y|)g(y)\ . $$ This can be done by replacing $1/|\xi|^2$ by $1/|\xi|^{2-\epsilon}$ and taking the $\epsilon\rightarrow 0$ limit by dominated convergence. Then, use the representation $$ \frac{1}{|\xi|^{2-\epsilon}}=\frac{1}{\Gamma\left(\frac{2-\epsilon}{2}\right)} \int_{0}^{\infty} d\alpha\ \alpha^{-\frac{\epsilon}{2}} e^{-\alpha |\xi|^2}\ , $$ write the Fourier transforms as integrals and integrate over $\xi$. One then ends up with the computation of $$ \lim_{\epsilon\rightarrow 0^+} \int_{\mathbb{R}^2\times\mathbb{R}^2} d^2 x\ d^2y\ f(x)\left(\frac{1}{\epsilon} |x-y|^{-\frac{\epsilon}{2}}\right)g(y)\ . $$ Since $\int f(x) d^2 x=\int g(x) d^2 x=0$, the last expression is the same as $$ \lim_{\epsilon\rightarrow 0^+} \int_{\mathbb{R}^2\times\mathbb{R}^2} d^2 x\ d^2y\ f(x)\left\{\frac{1}{\epsilon} \left(|x-y|^{-\frac{\epsilon}{2}}-1\right)\right\}g(y)\ . $$ The resulting derivative at $\epsilon =0$ produces the wanted logarithm of $|x-y|$. Edit 2: A substantial elaboration on the answer I gave above has appeared recently. See the review "Log-correlated Gaussian fields: an overview" by Duplantier, Rhodes, Sheffield and Vargas. This is part of a wider program regarding the study of stationary, isotropic and self-similar Gaussian fields: "Fractional Gaussian fields: a survey" by Lodhia, Sheffield, Sun and Watson. Chapter 2 of "Construction and Analysis of a Hierarchical Massless Quantum Field Theory", Ph.D. Thesis by Ajay Chandra. "Gaussian and their Subordinated Self-similar Random Generalized Fields" by Roland Dobrushin and its follow-up " Multiple Wiener-Itô Integrals" by Péter Major.<|endoftext|> TITLE: How nilpotent is the ring of stable homotopy groups of spheres? QUESTION [5 upvotes]: Are there any known or conjectured bounds on the exponent $d(r)$ such that $x^{d(r)} = 0$ for all $x \in \pi_r^S(S^0)$? REPLY [3 votes]: If one uses the Adams-Spectral sequence based on cohmology theories other than $BP$ it is possible to say a little more. In "A vanishing line in the $BP \langle 1 \rangle$-Adams spectral sequence" Jesús González shows that when $p$ is odd the $E_2$-page of the $BP\langle 1 \rangle$-Adams spectral sequence for the sphere spectrum has a vanishing line of slope $(p^2-p-1)^{-1}$. This is used to show that the $E_\infty$-term of the classic ASS has a vanishing line of slope $(2p-1)/[(2p-2)(p^2-p-1)]$. The result you are looking for is then Corollary 4.4: In the $n$-th stable stem, any $p$-local homotopy class not detected by the $J$-spectrum is annihilated by $p^{\text{min}\{4,p\}+ \epsilon_n+S(n)}$, where $q=2p-2$, $S(n)$ is the first integer larger than or equal to $(q+1)n/[q(p^2-p-1)]$ and $\epsilon_n = 0$ unless $n+2$ is divisible by $q$, in which case $\epsilon_n$ is the highest power of $p$ dividing $n+2$.<|endoftext|> TITLE: Over a finite field, does a torsor under the component group of G lift to a torsor under G? QUESTION [6 upvotes]: Let $k$ be a finite field and $G$ a finite type smooth $k$-group scheme. Let $G^0$ and $\Gamma$ be the connected component of identity and the component group of $G$, so there is an exact sequence $1 \rightarrow G^0 \rightarrow G \rightarrow \Gamma \rightarrow 1$. Is the induced $H^1(k, G) \rightarrow H^1(k, \Gamma)$ surjective? By Lang's theorem, $H^1(k, G^0) = *$, so the map in question is injective. It is surjective if $G^0$ is central (in particular, if $G$ is commutative), since for commutative $G^0$ one has $H^2(k, G^0) = 0$ for cohomological dimension reasons. It is also surjective if the short exact sequence splits as a semi-direct product. Thus, my question is: does the surjectivity hold in general for noncommutative smooth $G$? REPLY [5 votes]: This is proved when $G$ is a linear algebraic group in Corollary III.2.4.3 of Serre's book on Galois cohomology (p. 135 in my version). I believe it is only stated here for linear algebraic groups as Serre is at that point looking at general fields of cohomological dimension at most $1$. In the special case of finite fields where one has Lang's theorem, it seems that the proof generalises to your setting.<|endoftext|> TITLE: Grothendieck group $K(\mathbb A_k^n)$ (Hartshorne Exercise III.5.4) QUESTION [7 upvotes]: This question is related with Exercise III.5.4 (Page 230) of Hartshorne's Algebraic Geometry. Here I start with recalling the definition of Grothendieck group $K(X)$ of a noetherian scheme $X$, which is the quotient of free abelian group generated by all coherent sheaves on $X$, by the subgroup generated by all expressions $\mathscr F' -\mathscr F + \mathscr F''$ whenever there is an exact sequence $0\to \mathscr F' \to \mathscr F \to \mathscr F'' \to 0$ (cf. Exercise II.6.10, Page 148 Hartshorne's Algebraic Geometry). Let $k$ be a field. This exercise essentially wants to prove $K(\mathbf P_k^r) \cong \mathbb Z^r$, generated by the images of the sheaves of regular functions of $\mathbf P_k^0, \mathbf P_k^1, \ldots, \mathbf P_k^{r}$ in $K(\mathbf P_k^r)$. We can prove this claim by induction according to the hints given in the book. It is trivial when $r =0$ which corresponds to a point. Suppose the claim is true for $r-1$. Since $\mathbf P_k^{r-1}$ can be considered as a closed subscheme of $\mathbf P_k^r$ and $\mathbf P_k^r - \mathbf P_k^{r-1} \cong \mathbf A_k^{r-1}$. Hence one has an exact sequence (cf. Exercise II.6.10, page 148): $$ K(\mathbf P_k^{r-1}) \xrightarrow{\alpha} K(\mathbf P_k^r) \to K(\mathbf A_k^{r-1}) \to 0.$$ I think I can prove $\alpha$ is injective by showing it splits. So it only remains to prove $K(\mathbf A_k^{r}) \cong \mathbb Z$ for any integer $r > 0$. But this is where I get stuck. $K(\mathbf A_k^1) \cong \mathbb Z$ is easy because we fully understand the module structure over PID ($k[x]$ is PID). I doubt this method can be extended to $r > 1$. One way I tried is to prove the surjective group homomorphism $ \gamma:\ K(\mathbf A_k^{r-1}) \to \mathbb Z$ is also injective, where $\gamma$ is induced by the ranks of coherent sheaves. To prove $\mathrm{ker}(\alpha)=0$, it is enough to prove any coherent torsion sheaf $\mathscr F$ is trivial in $K(\mathbf A_k^{r})$. Here a coherent sheaf $\mathscr F$ over an integral scheme $X$ is torsion if the stalk $\mathscr F_\eta = 0$ where $\eta$ is the generic point of $X$. But this is exactly where I get stuck. I appreciate if any one could give me some hints (answers if possible) or give a new method. REPLY [10 votes]: Hailong's answer is of course exactly right, but there's also a sense in which it's overkill. That's because it invokes two facts: Fact I: Any f.g. module over $R=k[x_1,\ldots x_r]$ has a finite resolution by projective $R$-modules. Fact II: Any projective module over $R$ is free. (This is the Quillen/Suslin theorem). From these it follows that Theorem: $K_0(R)={\mathbb Z}$. But every proof I know of Fact I shows more, namely: Fact IA: Any f.g. module over $R$ has a finite resolution by free $R$-modules. Given Fact IA (which is no harder than Fact I), the theorem follows immediately, with no need to invoke Fact II. Fact IA is a theorem of Serre (or actually a theorem of Hilbert, subsequently generalized by Serre; see comments below) that long predates Quillen/Suslin. Here is a sketch of a proof: Sketch of a proof: Let $A=k[x_1,\ldots,x_{r-1}]$, so that every f.g. $A$-module has a finite free resolution by induction. We want to show the same for $R=A[x]$. 1) If $M$ is an $A[x]$ module, we have an exact sequence $$0\rightarrow A[x]\otimes_AM\rightarrow A[x]\otimes_AM\rightarrow M\rightarrow 0$$ where the first map is $x^i\mapsto x^{i+1}\otimes m-x^i\otimes xm$ and the second is $x^i\otimes m\mapsto x^im$. 2) If $M$ is contained in a free f.g. $A[x]$-module, we can find finitely generated $A$-submodules $G,H\subset M$ so that the restriction $$0\rightarrow A[x]\otimes_AG\rightarrow A[x]\otimes_AH\rightarrow M\rightarrow 0$$ is still exact. (Proof: A free $A[x]$-module is of the form $F_0[x]$ where $F_0$ is a free $A$-module. Take $n$ large enough so that all generators of $M$ are contained in $F_0+xF_0+\ldots +x^nF_0$. Let $G=F_0+xF_0+\ldots x^nF_0$ and $H=F_0+xF_0+\ldots x^{n+1}F_0$. Check that this works.) 3) An arbitrary f.g. $A[x]$-module is a quotient of a free module. The kernel $N$ is therefore contained in a free f.g. $A[x]$-module so we have an exact sequence $$0\rightarrow A[x]\otimes_AG\rightarrow A[x]\otimes_AH\rightarrow N\rightarrow 0$$ as in 2). Use induction to construct finite free $A$-resolutions of $G$ and $H$. Tensor up to $A[x]$ and construct the mapping cone to get a finite free resolution for $N$. REPLY [8 votes]: Any module over $R=k[x_1,\dots, x_r]$ has a finite resolutions of projective modules, and projective modules are free. So, in the Grothendieck group, the class of the module is a multiple of the class of $R$.<|endoftext|> TITLE: Ordinary Generating Function for Bell Numbers QUESTION [14 upvotes]: In the OEIS entry for Bell numbers, there appears a generating function $$\sum_{k=0}^\infty B_k t^k = \sum_{r=0}^\infty \prod_{i=1}^r \frac{t}{1-it}$$ However, I could not locate any proof of reference for this formula. The contributor informs me that he discovered it by experimentation. I would appreciate any further information on this generating function. REPLY [10 votes]: As a general remark, we may see the "Exponential to Ordinary" transformation of generating functions, $$ f(x):=\sum_{r=0}^\infty a_rx^r/r!\mapsto \tilde f(t):=\sum_{r=0}^\infty a_rt^r, $$ as an operator $\mathbb{C}[[x]]\to \mathbb{C}[[t]]$ . Since $r!t^r=\int_0^\infty (tx)^r e^{-x}dx$, the transformation can be computed analytically as $$\tilde f(t)=\int_0^\infty f(tx)e^{-x}dx,$$ at least for suitably convergent $f(x)$, and for special values of $t$. If the RHS is a convergent series $\sum_{r=0}^\infty b_rt^r$, and the equality holds, for a set of values $t$ which accumulates within the disk of convergence, the identity of series $ \sum_{r=0}^\infty a_rt^r=\sum_{r=0}^\infty b_rt^r$ is then established by the principle of isolated zeros. For instance, we may compute the transform of $f_r(x):=(e^x-1)^r$ for real negative values of $t$ in terms of the Euler's Beta function by a change of variable in the integral: $$\tilde f_r(t)=\int_0^\infty (e^{tx}-1)^re^{-x}dx=(-1)^{r+1}t^{-1}\int_0^1(1-u)^r u^{-1/t-1}du=$$ $$=(-1)^{r+1}t^{-1}\frac{\Gamma(r+1)\Gamma(-1/t)}{\Gamma(r+1-1/t)}=\frac{r!t^r}{(1-t)\dots(1-rt)}\ .$$ This computation gives your identity, since for the egf $f(x)$ of the $B_r$'s we have $f(x)=e^{e^x-1}=\sum_{r=0}^\infty\frac{1}{r!}f_r(x)$ (in the sense of formal power series) so that $\tilde f(t)=\sum_{r=0}^\infty \frac{t^r}{(1-t)\dots(1-rt)}\ .$ Incidentally, note that by an analogous computation you may, more generally, compute the ordinary gf of the Stirling polynomials of the second kind, $B_r(z) :=\sum_{r=0}^nS(n,r)z^r$, starting by their egf $e^{z(e^x-1)}$. rmk. I think a more natural proof is, computing directly the ogf of the Stirling polynomials $B_r(z):=\sum_{r=0}^nS(n,r)z^r$, which consists just in translating the inductive relation $S(n+1,k)=kS(n,k)+S(n,k−1)$ in the language of formal power series, and then putting $z=1$, as done in other answers. The present answer is meant as an example of a general alternative approach.<|endoftext|> TITLE: Does the matrix exponential preserve the positive-semi-definite ordering? QUESTION [18 upvotes]: Suppose $A$, $B$, are symmetric, real valued matrices and $B-A$ is positive-semidefinite, i.e. $A≼B$. Does that imply $e^A ≼ e^B$? Would love some intuition here. I know for instance that $A≼cI \iff e^A ≼ e^cI$ and also for diagonal $A$ and $B$ the inequality is easy to show, but I'm not convinced that it translates into the general case. REPLY [21 votes]: To supplement Robert's counterexample, let me mention below some interesting facts about the matrix exponential, along with what may be regarded as the "correct" way of obtaining matrix exponential like operator inequalities. Assume throughout that $A \ge B$ (in Löwner order). The map $X \mapsto X^r$ for $0 \le r \le 1$ is operator monotone, i.e., $A^r \ge B^r$. This result is called Löwner-Heinz inequality (it was apparently originally discovered by Löwner). Now using \begin{equation*} \lim_{r\to 0} \frac{X^r-I}{r} = \log X, \end{equation*} we can conclude monotonicity of $\log X$, so that $\log A \ge \log B$. A quick experiment reveals that $A^2 \not\ge B^2$ in general (in fact $X^r$ for $r > 1$ is not monotone), which severely diminishes hopes of $e^X$ being monotone. However, though $e^A \not\ge e^B$, T. Ando (On some operator inequalities, Math. Ann., 1987) showed that \begin{equation*} e^{-tA} \# e^{tB} \le I,\qquad\forall t \ge 0. \end{equation*} Here, $X \# Y := X^{1/2}(X^{-1/2}YX^{-1/2})^{1/2}X^{1/2}$ denotes the matrix geometric mean (so basically, operator inequalities go along better with the "right" notion of a geometric mean) Additional remarks Although $A^2 \ge B^2$ does not hold, it turns out that a slightly modified version $(BA^2B)^{1/2} \ge B^2$ does hold, as does $A^2 \ge (AB^2A)^{1/2}$. These inequalities are special cases of a family of such results proved by T. Furuta, and these are called Furuta inequalities, for example, we have \begin{equation*} (B^rA^sB^r)^{1/q} \ge (B^{s+2r})^{1/q},\quad 0 \le s \le 1, r \ge 0, q \ge 1. \end{equation*}<|endoftext|> TITLE: Homeomorphism between derived sets implies homeomorphism QUESTION [6 upvotes]: (Note: I asked this question at MSE days ago and received no answer, so I'm now reposting it here.) I want to prove the following statement: Let $K_1$ and $K_2$ be two countable, compact sets of $\mathbb R$ such that $K_1'$ and $K_2'$ are homeomorphic, then $K_1$ and $K_2$ are homeomorphic. More generally, I want to prove that if $K_1$ and $K_2$ be two countable, compact sets of $\mathbb R$ have the same Cantor–Bendixson rank, $\alpha+1$, and $|K_1^{(\alpha)}|=|K_2^{(\alpha)}|$, then $K_1$ and $K_2$ are homeomorphic. I know that a proof is given by Mazurkiewicz and Sierpinski in "Contribution à la topologie des ensembles dénombrables" (1920) (http://matwbn.icm.edu.pl/ksiazki/fm/fm1/fm114.pdf). And by Millient in "A remark in Cantor derivative" (http://arxiv.org/pdf/1104.0287v1.pdf). But I want a more direct proof in $\mathbb{R}$; can anybody help me? Thanks. REPLY [3 votes]: Let me give a proof that makes use of the ordering inherited from $\mathbb{R}$. Suppose that $A\subseteq\mathbb{R}$ is a countable compact space. Then $A$ has no subset $B$ order isomorphic to the rational numbers (otherwise $A$ would be uncountable). I now claim that the order topology on $A$ is isomorphic to the subspace topology on $A$. To avoid confusion, let $\mathcal{O}$ be the order topology on $A$ and let $\mathcal{S}$ be the subspace topology on $A$. It is well known that for every ordered set, the subspace topology is finer than the order topology, i.e. $\mathcal{O}\subseteq\mathcal{S}$. However, since $(A,\mathcal{S})$ is compact, we have $\mathcal{O}=\mathcal{S}$, so the order topology and the subspace topology coincide on $A$. In particular, since $A$ is complete as an ordered set, $A$ is a complete lattice by this answer of mine. Let $\mathcal{H}_{0}$ be the set of all finite and countable successor ordinals. Let $\mathcal{H}_{\alpha}$ be the collection of all sums $\sum_{i\in B}B_{i}$ where $B_{i}\in\bigcup_{\beta<\alpha}\mathcal{H}_{\beta}$ for each $\beta<\alpha$ and where $B=\omega+1$,$B=(\omega+1)^{*}$ or $B$ is finite. I claim that $\bigcup_{\alpha<\omega_{1}}\mathcal{H}_{\alpha}$ is precisely the collection of all countable complete linear orders (this is a slight modification of a result by Hausdorff proven here). Suppose that $X$ is a countable compact linear ordering that is not in $\bigcup_{\alpha<\omega_{1}}\mathcal{H}_{\alpha}$. Then it is easy to show that there exists some $x\in X$ with $[0,x_{1/2}],[x_{1/2},1]\not\in\bigcup_{\alpha<\omega_{1}}\mathcal{H}_{\alpha}$. In other words, we can cut $X$ into two pieces not in $\bigcup_{\alpha<\omega_{1}}\mathcal{H}_{\alpha}$. Now cut each piece $[0,x_{1/2}],[x_{1/2},1]$ into pieces $[0,x_{1/4}],[x_{1/4},x_{1/2}],[x_{1/2},x_{3/4}],[x_{3/4},1]$. Continue this process until we have obtained an $x_{r/2^{n}}$ for all $n\in\omega$ and $0 TITLE: What properties of line bundles can be detected cohomologically? QUESTION [7 upvotes]: Let $X$ be a proper, finite type scheme over a field $k$. What useful properties of line bundles (e.g. amplitude, nefness) can be detected cohomologially? For example, in our setting we have the Cartan-Grothendieck-Serre theorem characterizing ample line bundles on $X$ as those line bundles $L$ such that for any coherent sheaf $\mathscr{F}$ on $X$, we have $H^i(X, \mathscr{F} \otimes L^{\otimes n}) = 0$ for all $i > 0$ and $n \gg 0$. Restricting further to projective varieties, we have that a line bundle $L$ is big iff $h^0(X, L^{\otimes n})$ grows like $n^{\dim X}$. Do other such characterizations hold for nef line bundles? semiample? other useful classes of line bundles which I've forgotten? REPLY [4 votes]: A generalization of ampleness, called $q$-ampleness, has been studied notably by Totaro, see the second paper in this volume: a line bundle $L$ on $X$ is $q$-ample if for any coherent sheaf $\mathscr{F}$ on $X$, we have $H^i(X, \mathscr{F} \otimes L^{\otimes n}) = 0$ for all $i > q$ and $n \gg 0$. See Totaro's paper and also this paper by J.C. Ottem for some interesting applications of this notion.<|endoftext|> TITLE: What are the automorphisms of a Grassmannian? QUESTION [9 upvotes]: I want to know what are the holomorphic automorphisms of a Grassmannian. Can someone tell me this? REPLY [14 votes]: Automorphisms of Grassmannians, Michael J. Cowen (1989).<|endoftext|> TITLE: The number of facets of a polyhedron under linear transformation QUESTION [5 upvotes]: Consider a (not necessarily bounded) convex polyhedron $P\subset \mathbb{R}^n$ which has $k$ facets. Let $L:\mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation. Question1: Is there a fixed constant $C$ such that the number of facets of $L(P)$ is bounded by $Ck$ ? Edit: Question2: Is there any bound on the number of facets of $L(P)$ in terms of $P$? Question3: Is there any bound on the number of (none geometrical redundant) half-spaces which defined $L(P)$ in terms of $P$? REPLY [3 votes]: No. Every bounded convex polyhedron $Q$ with $v$ vertices is an image of a simplex $P$ with $v$ vertices and $v$ facets by a linear transformation. However, $Q$ may have many facets. For example, the cross-polytopes (dual to hypercubes) have $2d$ vertices and $2^d$ facets, one for each orthant, and $2^d$ is not $O(d)$.<|endoftext|> TITLE: Quadratic Casimir of fundamental irreps of simply-laced Lie algebras QUESTION [8 upvotes]: I have the following question, motivated by the expression for the character of level 1 highest weight integrable representations of simply-laced affine algebras (in terms of the string function). It follows from the character expression (by comparing the leading "conformal dimension") that for a fundamental weight $\omega$ with mark 1 the following identity should hold in simply-laced case: $$ \frac{(\omega,\omega+2\rho)}{2(h^\vee+1)}=\frac{1}{2}|\omega|^2=\frac{1}{2}(\omega,\omega), $$ where $h^\vee$ is the dual Coxeter number of the simply-laced simple $\mathfrak{g}$. The lhs is the quadratic Casimir divided by $2(h^\vee+1)$. I checked it for $A_n$ series and $E_6,\,E_7$ explicitly, but I feel there should be some simple general argument. So the question is if there indeed is a simple argument why this should be true? Is there any generalization? REPLY [7 votes]: It turns out that it's easier to prove the following generalization: Let $\mathfrak g$ be a simple Lie algebra (not necessarily simply laced), let $\omega$ be a fundamental weight whose Dynkin mark is 1, and let $k$ be any number. Then we have $$ \frac{\langle k\omega,k\omega+2\rho\rangle}{2(h^\vee+k)} = \frac{k}{2}\|\omega\|^2. $$ (The original question is the case $k=1$) Remark: The right left hand side $\frac{\langle k\omega,k\omega+2\rho\rangle}{2(h^\vee+k)} $ is the minimal energy of the positive energy representation of the affine Lie algebra $\hat{\mathfrak g}$ of level $k$ and highest weight $k\omega$. Proof: We'll show that $$ \frac{\langle k\omega,k\omega+2\rho\rangle}{2(h^\vee+k)} \,\,\stackrel{(1)}=\,\, \frac{\| k\omega+\rho\|^2-\|\rho\|^2}{2(h^\vee+k)} \,\,\stackrel{(2)}=\,\, \frac{k\langle\omega,\rho\rangle}{h^\vee} \,\,\stackrel{(3)}=\,\, \frac{k}{2}\|\omega\|^2. $$ (1) Easy. (2) By the lemma below, the numerator $\| k\omega+\rho\|^2-\|\rho\|^2$ vanishes when $k=-h^\vee$. The function $k\mapsto \frac{\| k\omega+\rho\|^2-\|\rho\|^2}{2(h^\vee+k)}$ is therefore linear. One easily checks that it vanishes at zero and that its derivative at zero is $\frac{\langle\omega,\rho\rangle}{h^\vee}$. (3) Again by the lemma, the point $\frac \rho {h^\vee}$ is equidistant to $0$ and to $\omega$. It is on the bisecting hyperplane of the segment $[0,\omega]$, and so $\langle\omega,\frac\rho{h^\vee}\rangle=\langle\omega,\frac\omega2\rangle=\frac12\|\omega\|^2$. QED Lemma: Let $\omega$ be as above. Then $\| \rho - h^\vee\omega\| = \|\rho\|$. Proof: Let $\mathcal A$ be the Weyl alcove. Its isometry group is the automorphism group of the extended Dynkin diagram $\Gamma^e=\Gamma\cup \{\circ\}$. The vertices of $\Gamma$ whose Dynkin mark is $1$ are exactly those in the $Aut(\Gamma^e)$ orbit of the extra vertex $\circ$. Recall that $\rho$ is the unique weight in the interior of $h^\vee \mathcal A$. To finish the argument, note that the vertices $0$ and and $h^\vee\omega$ are in the same orbit under the isometry group of $h^\vee \mathcal A$, and therefore equidistant to $\rho$. QED<|endoftext|> TITLE: Principal Order Ideals in the Weak Bruhat Order QUESTION [12 upvotes]: Let $\sigma\in S_n$ be a permutation on $n$ elements, and $\mathrm{Inv}(\sigma):=\{(i,j) : 1\leq i\sigma(j)\}$ be its set of inversions. In the weak order on permutations $S_n$, we say that $\sigma \leq \tau$ if $\mathrm{Inv}(\sigma)\subseteq \mathrm{Inv}(\tau)$. The length of a permutation $l(\sigma):= |\mathrm{Inv}(\sigma)|$ is the total number of inversions of $\sigma$. Is there a formula (closed or not) for the size of a principal order ideal (principal downset) in this poset? Computing all elements in a given order ideal of this poset is easy enough for a fixed $\sigma\in S_n$ using the covering relations given by $\pi\in\{(1,2),(2,3),...,(n-1,n)\}$ and selecting permutations $\pi\sigma$ such that $l(\pi\sigma) TITLE: Abel's five terms relation from Yang-Baxter equation? QUESTION [7 upvotes]: Can the famous Abel's five terms relation satisfied by the dilogarithm be derived from (a particular case of) the theory of Yang-Baxter equations? If yes, how? Thanks for any help. REPLY [2 votes]: Quite a well-known paper Quantum Dilogarithm L. D. Faddeev, R. M. Kashaev discovers that in an appriate limit the q-exponential function degenerates to dilogarithm. (It might be suprising dilogorithm and q-exp were known for more that 100 years, but the relation between them have been uncovered in 1993). Moreover the authors show that five-term dilogarithm identity is a limit of the identity for the q-exponential functions. The last identity in contrast to the classical one, includes non-commuting variables. It is related in certain sense to Yang-Baxter (star-triangle) equation, however it seems not quite direct relation, although it is clearly from the realm of Yang-Baxter quantum group theory. I also heard that similar identities for q-exponential function appear in the proof that certain construction of universal R-matrix indeed satisfies the Yang-Baxter equation. Wikipedia article might be quite useful. The so-called "quantum dilogarithm" now appears in many papers on quantum Teichmuller theory (see papers by Fock,Rosly, Chekhov, et. al. on arXiv e.g. The quantum dilogarithm and representations quantum cluster varieties V. V. Fock, A. B. Goncharov)<|endoftext|> TITLE: Covering a set with geometric progressions QUESTION [20 upvotes]: Consider the set $S_n=\{1,2,\cdots ,n\}$. What is the minimum number of distinct geometric progressions that cover $S_n$? Let us call this number $a_n$. I was wondering about this number after doing a problem from the Allrussian MO, 1995. Can the set $\{1,2,\cdots,100\}$ be covered with $12$ geometric progressions? It becomes straightforward after observing the fact no three primes can be in a geometric progression. Hence the problem is restated to the obvious contradiction $$\pi(100)\le 24$$ Now I had searched a bit and here it is proven that $a_{100}\ge 24$. Now I would like some asymptotics, or references, or better bounds on $a_n$ . I have also found the fact that $$a_n\ge \left\lfloor{\frac{3n}{\pi^2}}\right\rfloor$$ Which is obvious since any geometric progression contains at most $2$ squarefree numbers and there are about $\dfrac{6n}{\pi^2}$ squarefree numbers less than $n$. Note it surpasses the bound given. But is something better possible? Thanks in advance. REPLY [9 votes]: We can reduce Lucia's upper bound of $3/8$ a little further as follows. Begin by taking the $n/4$ geometric progressions of common ratio $2$ beginning at each odd number at most $n/2$. Then for each odd number $x$ in the range $[n/2,25n/49]$ divisible by $25$ include the progression $\{x,(7/5)x,(7/5)^2x\}$. Pair off the remaining odd numbers. If I've added this up correctly I get $3/8 - 1/(4900)$ as an upper bound. As I explained in my comment one can improve the lower bound $3/\pi^2$ slightly by considering a set of integers $S$ larger than just the square-frees but still not containing any three terms of a geometric progression. Here is a somewhat general construction of such a set. Let $Q\subset\mathbf{Z}_{\geq0}^2$ be any set without three points on a line. Now let $p_1,p_2,\dots$ be the primes and let $$S = \left\{\text{integers }n= \prod p_i^{e_i} \text{ such that } (e_1,e_2)\in Q, (e_3,e_4)\in Q,\dots\right\}.$$ Then $S$ has no three terms of any geometric progression, and one could write down an Euler-product-like expression for the density of $S$ in terms of $Q$. Note if $Q$ contains $$\{(0,0),(1,0),(0,1),(1,1)\}$$ then $S$ contains all square-free integers, but $Q$ could be larger as well. Taking $$Q = \{(0,0),(1,0),(0,1),(1,1),(2,3)\},$$ the density of $S$ is $$\prod_{i=1}^\infty \left(1-\frac{1}{p_{2i-1}}\right) \left(1-\frac{1}{p_{2i}}\right)(1 + p_{2i-1}^{-1} + p_{2i}^{-1} + p_{2i-1}^{-1} p_{2i}^{-1} + p_{2i-1}^{-2} p_{2i}^{-3}),$$ which in any case is bounded below by $$ \frac{1 + 2^{-1} + 3^{-1} + 2^{-1} 3^{-1} + 2^{-2} 3^{-3}}{1 + 2^{-1} + 3^{-1} + 2^{-1} 3^{-1}} \prod_{p} \left(1-\frac{1}{p^2}\right) = \frac{217}{36\pi^2}.$$ Thus $\frac{217}{72\pi^2}$ is a lower bound. This problem is mentioned as problem 2014.2.1 in http://arxiv.org/pdf/1406.3558v2.pdf, though presumably others have pondered it as well.<|endoftext|> TITLE: Simplest example of non-trivial Toda bracket in spaces QUESTION [11 upvotes]: Many sources give an easy definition of a Toda bracket $\{f,g,h\}$ of appropriate maps $W \to X \to Y \to Z$ in spaces as a subset of the homotopy classes of maps $[\Sigma W, Z]$ (for example, Ravenel's Complex Cobordism.) However, the only real usages of Toda brackets that I can find are in the context of heavier lifting, in what appear to be difficult computations of stable homotopy groups of spheres. It's not difficult to construct maps satisfying the nullhomotopy criteria, but either it's not clear how to work things out or I wind up finding only a trivial bracket. Some sources give more conceptual accounts (such as this paper), but again the only examples are quite complicated from my inexperienced point of view. For instance, the only example in the linked paper is an element in $\pi_{11}(S^6)$ constructed from maps I'm not familiar with. What's the simplest natural example of a non-trivial Toda bracket in spaces? It would be especially helpful if the example illustrated the intuition or geometry behind Toda brackets, but I understand that may not be how things are. REPLY [7 votes]: Just wanted to expand @Achim Krause 's final comment: Mosher and Tangora has a nice treatment that ramps up from more to less elementary (they do get to statements in terms of Toda brackets eventually). They work stably by considering $S^n$ for $n$ large, and include the following examples involving the (suspensions of) Hopf maps $\eta, \nu, \sigma$: Assume $\eta^2 = 0$; then use part of the bracket construction and the relation $Sq^2Sq^2 = 0$ to derive a contradiction. Similar strategy to show $\eta^3 = 4 \nu$ $\eta^2 \in \langle 2, \eta, 2 \rangle$ $2 \nu \in \langle \eta, 2, \eta \rangle$ $\nu^2 \in \langle \eta, \nu, \eta \rangle$ $8 \sigma \in \langle\nu, 8, \nu\rangle$<|endoftext|> TITLE: Hodge numbers of diffeomorphic complete intersections QUESTION [10 upvotes]: Is there an example of two complex projective complete intersections that are diffeomorphic but have different Hodge numbers? Edit: as written by Daniel Loughran in the comments below, complete intersections with the same multidegree are diffeomorphic (apparently this result is attributed to R. Thom). And we know that he multidegree determines the Hodge numbers see the appendix of F. Hirzebruch's book "Topological methods in algebraic geometry". Thus we need to find two diffeomorphic complete intersections with different multidegrees such that their Hodge numbers are different. Edit 2: Oscar Randall-Williams suggested to test examples of 3-folds due to Libgober and Wood, this is a very good idea, but they have the same Hodge numbers (I made the computations via Sage macros written by Donu Arapura). I also tested examples in W. Ebeling "An example of two homeomorphic, nondiffeomorphic complete intersection surfaces." Inventiones mathematicae 99.3 (1990): 651-654 where we can encounter two homeorphic nondiffeomorphic complete intersection surfaces, and get that they have the same Hodge numbers. Edit 3: Related to this question there is this paper: "The Hodge ring of Kähler manifolds", Compositio Math. 149 (2013), 637--657 by D. Kotschick and S. Schreieder where they determine which linear combinations of Hodge numbers are birationally invariant, and which are topological invariants. REPLY [3 votes]: We got four pairs of diffeomorphic complete intersections but with Hodge numbers different. Please check the link:here<|endoftext|> TITLE: What is the fundamental group of a hypersurface? QUESTION [5 upvotes]: (A related question is this On the fundamental group of hypersurfaces). Let $X$ be a simply connected projective complex manifold of dimension at least 3. Let $Y\subset X$ be a smooth hypersurface. What is $\pi_1(Y)$ then? If $Y$ is a hyperplane section, then $\pi_2(X,Y)=0$ by the Lefschetz theorem, hence $\pi_1(Y)=0$. The question is, can we say anything about the fundamental group of a hypersurface without assuming ampleness? REPLY [4 votes]: I think that it is hard to say something is such a generality, as the following example shows. Take $X= \mathbb{P}^1 \times \mathbb{P}^2$ and let $C \subset \mathbb{P}^2$ be a smooth curve of genus $g$. Set $Y = \mathbb{P}^1 \times C$. Then $\pi_1(Y) = \pi_1(C)$. The latter group is trivial for $g=0$, it is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}$ for $g=1$ and it is a infinite nonabelian group for $g \geq 2$.<|endoftext|> TITLE: A possible extension of a determinant inequality QUESTION [19 upvotes]: It is well known that if $A, B$ are positive semidefinite matrices, then $$\det (A+B)\ge \det A+\det B.$$ I am considering a possible extension of this result. Let $\mathbb{M}_m(\mathbb{M}_n)$ denote the set of $m\times m$ block matrices with each block the usual $n\times n$ matrix. Let $\mathbf{A}=[A_{i,j}]_{i,j=1}^m, \mathbf{B}=[B_{i,j}]_{i,j=1}^m \in \mathbb{M}_m(\mathbb{M}_n)$ be positive semidefinite. Define a new matrix $M=[m_{ij}]_{i,j=1}^m$ with $m_{ij}=\det (A_{i,j}+B_{i,j})-(\det A_{i,j}+\det B_{i,j})$. Is it true that $M$ is positive semidefinite? I ran some numerical simulations, yet no counterexamples showed up (of course, numerical simulations can never be thorough). I did not find a proof even when $m=2$. Clearly, it suffices to show $\det M\ge 0$. REPLY [13 votes]: The claim is true. We prove it using a few block matrix manipulations. Note, in the proofs below $A \ge 0$ means $A$ is (symmetric) positive semidefinite. $\newcommand{\trace}{\text{trace}}$ Lemma Let $X, Y \ge 0$. Then, \begin{equation*} \otimes^k (X+Y) \ge \otimes^k X + \otimes^k Y. \end{equation*} Proof. By induction on $k$. The case $k=1$ is trivial; $k=2$ shows us the crux. Indeed, \begin{equation*} (X+Y)\otimes (X+Y) - X\otimes X - Y\otimes Y = X\otimes Y + Y \otimes X \ge 0, \end{equation*} since $X, Y \ge 0$. The general case follows similarly. - Corollary. Let $X, Y \ge 0$. Then, by restricting to the suitable symmetry class of tensors we get \begin{equation*} \wedge^k(X+Y) \ge \wedge^k X + \wedge^k Y \end{equation*} . Lemma. Let $A=[A_{ij}]$ be $mn\times mn$ with $n\times n$ blocks. Suppose $A$ is symmetric, positive semidefinite. Then, for $1\le k \le n$, the $m\binom{n}{k} \times m\binom{n}{k}$ matrix $C_k := [\wedge^k A_{ij}]$ is semidefinite. Proof. Some reflection shows that $C_k$ is a principal submatrix of $\wedge^k A$, thus, $C_k = P_k^*(\wedge^k A)P_k \ge 0$ since $A\ge 0$ and wedge products preserve positivity. Theorem. Let $A=[A_{ij}] \ge 0$ and $B=[B_{ij}] \ge 0$ be $mn\times mn$ block matrices composed of $n\times n$ blocks. Define $$M_k = [\trace(\wedge^k(A_{ij}+B_{ij}))] - [\trace(\wedge^k A_{ij})] - [\trace(\wedge^k B_{ij})],$$ for any $1\le k \le n$. Then, $M_k \ge 0$. Proof The Corollary above shows that $\wedge^k(A+B) \ge \wedge^k A + \wedge^k B$. Let $P_k$ be as in the second Lemma; then $$H_k = P_k^*(\wedge^k(A+B))P_k- P_k^*(\wedge^k A)P_k -P_k^*(\wedge^k B)P_k \ge 0.$$ The matrix $M_k$ is nothing but a (blockwise) partial trace of $H_k$, so that $H_k \ge 0 \implies M_k \ge 0$. Corollary Let $A$ and $B$ be as above. Then, \begin{equation*} M = [\det(A_{ij}+B_{ij})] - [\det A_{ij}] - [\det B_{ij}] \ge 0. \end{equation*} Proof. Observe that $\trace(\wedge^n X) = \det(X)$ for an $n\times n$ matrix $X$.<|endoftext|> TITLE: How much of GCH do we need to guarantee well-ordering of continuum? QUESTION [16 upvotes]: It's well known that, if GCH holds, then every cardinal can be well-ordered. However, I'm sure we don't need full power of GCH to prove it for specific cardinal, e.g. continuum. I have been wondering, what is the minimum amount cardinal numbers $A$ for which non-existence of cardinal $B$ with $A TITLE: Nonexistence of stable random variables QUESTION [8 upvotes]: Is there an "elementary" proof that $\alpha$-stable random variables only exist for $0 < \alpha \le 2$? By elementary I mean without using Fourier transforms. I'd be happiest with either a direct probabilistic argument, or a geometric argument, e.g., referring to embeddings of $L_p$ spaces (as long as it doesn't depend on results which are themselves proved via Fourier analysis). Note that what I'm interested in here is the fact that $\alpha$-stable random don't exist for $\alpha > 2$, not the fact that they do exist for $\alpha \le 2$. Edit: To clarify, for the purposes of this question I define an $\alpha$-stable random variable as follows. Given $\alpha > 0$, $X$ is $\alpha$-stable if, whenever $X_1, \ldots, X_n$ are independent copies of $X$, there is a real number $d=d(n)$ such that $X_1 + \cdots + X_n$ has the same distribution as $n^{1/\alpha} X + d$. And I'm quite happy to simplify to the situation when $d=0$. REPLY [6 votes]: I believe there to be a proof avoiding Fourier analysis in the literature given by Feller (1971) as part of Theorem VI.1.1. The argument can be subdivided into two parts. The first part (which was already mentioned in the comments) states, that if an $\alpha$-stable random vector has a finite non-zero variance, then $\alpha=2$. The second part is showing that variance is finite when $\alpha>2$, which is done by subdividing the integral $E(X^2)$ into a suitable series. Thus, $Var(X)=0$ when $\alpha>2$, which completes the proof.<|endoftext|> TITLE: Arveson's extension for normal completely positive maps QUESTION [7 upvotes]: My question deals with a version of Arveson's extension theorem (for the standard version, see, e.g., Paulsen's book Completely Bounded Maps and Operator Algebras). Let $\mathcal A$ be a von Neumann algebra, $\mathcal R\subset\mathcal A$ an operator system, and $\mathcal H$ a Hilbert space. If $\Phi_0:\mathcal R\to\mathcal L(\mathcal H)$ is a linear normal completely positive unital map, does there exist a linear normal completely positive unital extension $\Phi:\mathcal A\to\mathcal L(\mathcal H)$ for $\Phi_0$ (complete positivity for a linear map on an operator system defined as in Paulsen's book)? I am aware that if one gives certain restrictions for the operator system, the extension can be carried out. E.g., when $\mathcal A$ is a type-I factor and $\mathcal R$ is contained in the ultraweak closure of the set of its compact operators, the Arveson-like extension result for normal maps holds. Does anyone know whether the normal extension exists in general or whether there are some weaker versions of Arveson's theorem for normal maps? I am mostly interested in the case where $\mathcal A$ is a type-I factor. Thank you in advance. REPLY [8 votes]: The answer is NO and here's a counterexample. Let $\prod_{n\in{\mathbb N}}{\mathbb M}_n$ be the $\ell_\infty$-direct sum of the full matrix algebras ${\mathbb M}_n$, and let $X\subset\prod_{n\in{\mathbb N}}{\mathbb M}_n$ be the weak$^*$-closed subspace defined by $$X=\{ (x_n)_{n=1}^\infty : \Theta_{m,n}(x_m)=x_n \mbox{ for all }m>n\},$$ where $\Theta_{m,n}\colon{\mathbb M}_m\to{\mathbb M}_n$ is the compression to the upper-left corner. Then, $\Phi\colon X\to{\mathbb B}(\ell_2)$, $\Phi((x_n)_{n=1}^\infty) = \lim_n x_n$ is a normal unital completely positive map (in fact, an weak$^*$-homeomorphic complete order isomorphism). The map $\Phi$ does not extends to a normal ucp map on $\prod_{n\in{\mathbb N}}{\mathbb M}_n$. (In case one wants a type I factor, use the embedding $\prod_{n\in{\mathbb N}}{\mathbb M}_n\subset{\mathbb B}(\bigoplus\ell_2^n)$.) Indeed, suppose $\Psi\colon\prod_{n\in{\mathbb N}}{\mathbb M}_n\to{\mathbb B}(\ell_2)$ is any normal ucp map which maps the closed unit ball $(\prod_{n\in{\mathbb N}}{\mathbb M}_n)_1$ onto the closed unit ball $({\mathbb B}(\ell_2))_1$. Then, $x \in (\prod_{n\in{\mathbb N}}{\mathbb M}_n)_1$ such that $\Psi(x)$ is unitary is in the multiplicative domain $\mathrm{mult}(\Psi)$ of $\Psi$. Since ${\mathbb B}(\ell_2)$ is spanned by unitary elements, $\Psi$ restricted to the $\mathrm{C}^*$-subalgebra $\mathrm{mult}(\Psi)$ is a surjective $*$-homomorphism. If $\Psi$ were normal, $\mathrm{mult}(\Psi)$ would become a von Neumann subalgebra of $\prod_{n\in{\mathbb N}}{\mathbb M}_n$ and $\Psi$ would become a normal surjective $*$-homomorphism from $\mathrm{mult}(\Psi)$ onto ${\mathbb B}(\ell_2)$, which is absurd. (Although it's overkill, there's also a result of Sukochev and Haagerup--Rosenthal--Sukochev saying that the predual ${\mathbb B}(\ell_2)_*$ does not Banach embed to the predual $(\prod_{n\in{\mathbb N}}{\mathbb M}_n)_*$.)<|endoftext|> TITLE: Functions that Calculate their $L_p$ Norm QUESTION [11 upvotes]: are there any examples of functions $f:x\in\mathbb{R}_0^+\rightarrow\mathbb{R}_0^+$ and intervals $(a,b), 0\le a \lt b \le \infty$ , for which $$\Big(\int_a^b{|f(x)|^p dx}\Big)^\frac{1}{p} = f(p)$$ $$\forall p\in(a,b)$$ This question came up when thinking about $L_p$ norms as functions of $p$ and thus as a mapping of a real function to another one. In view of the comment of Mark Meckes, I would like to know, if there are also examples of non-trivial functions, i.e. that are not identical to $1$ REPLY [3 votes]: Based on Christian Remling's answer, we assume $b-a<1$. We know that $f$ has a pole on $b$. I don't know whether such an $f$ exists but I would at least like to understand the order of growth at this pole. $f(x)$ is an increasing function. For simplicity of notation let's change variables to $g(x)=f(b-x)$, then $f(x)$ is decreasing. So for any $x$, the $L^p$ norm is at least $g(x) x^{1/p}$. If we plug in $p=(b+1/\log x)$, we get $g(-1/\log x) \geq g(x) x^{ 1/ (b+ 1/\log x ) } =g(x) x^{1/b- 1/(b \log x) + 1/(b\log x^2) \dots) }= g(x) x^{1/b} e^{1/b + o(1) } $ $g(x) \leq g(-1/\log x) x^{-1/b} e^{1/b + o(1) } $ Let $h(x)= g(1/x)$, then we further simplify: $h(x) \leq h( \log x) x^{1/b} e^{1/b+o(1)}$ This gives $h(x) \leq (x \cdot \log x \cdot \log \log x \cdot \dots )^{1/b} $ I believe this asymptotic upper bound is fairly sharp, that is, if we have a bound of the form $h(x) \leq (x \cdot \log x\cdot \log \log x \cdot \dots \log^n x)^{1/b}$ then by taking the $L^p$ norm $n$ times we get a bound of the form $h(x)=O(1)$ and a contradiction, but I haven't checked this. The method of proof suggests that in the relevant regime, the operator that sends a function $f$ to the set of $L^p$ norms of $f$ is unstable, so iterating it might be a bad idea for finding a fixed point. However, some form of inverse of it might be stable?<|endoftext|> TITLE: Analytic representatives for Kahler classes QUESTION [6 upvotes]: If we are given compact complex manifold $X$ and a Kahler class $[\omega]$, can we always find a positive definite representative $\omega \in [\omega]$ that is real analytic? REPLY [10 votes]: Yes. Run the Kähler-Ricci heat flow $$ \frac{\mathrm{d}}{\mathrm{d}t}\left(\omega(t)\right) = -\mathrm{Ric}\bigl(\omega(t)\bigr) $$ with initial condition $\omega(0) = \omega$. This will exist for some time interval $[0,T)$, and the $\omega_t$ for $t>0$ will all be be real-analytic with respect to the natural real-analytic structure on $M$ given by the complex structure. Note that we have $$ \left[\omega(t)\right] = \left[\omega(0)\right] - t\,c_1(M). $$ Now, for $00$ is very small, then $\omega(t_1,t_2)$ is a positive $(1,1)$-form. It is real-analytic and, by construction, it satisfies $$ \left[\omega(t_1,t_2)\right] = \left[\omega(0)\right] = [\omega]. $$<|endoftext|> TITLE: Decomposition of $\mathrm{O}(n)$-modules coming from differential geometry QUESTION [11 upvotes]: Let $V$ be a $n$-dimensional real vector space equipped with a positively definite scalar product $g$ and let $\mathrm{O}(n)$ be the automorphism group of $(V,g)$. View $V^{\otimes k}$ as a $\mathrm{O}(n)$-module via the diagonal action. I am interested in the following specific submodules, which arise from Riemannian geometry and affine differential geometry: (1) $ \mathscr{R}(V)=\{R\in V^{\otimes 4}\mid\mathrm{Cyc}_{12}(R)=\mathrm{Cyc}_{34}(R)=\mathrm{Cyc}_{234}(R)=0\} $. Here $\mathrm{Cyc}_{234}: V^{\otimes4}\rightarrow V^{\otimes 4} $ is the $\mathrm{O}(n)$-module morphism which averages cyclic permutations of the 2nd, 3rd and 4th components, $$ \mathrm{Cyc}_{234}(x_1\otimes x_2\otimes x_3\otimes x_4)=\frac{1}{3}\left(x_1\otimes x_2\otimes x_3\otimes x_4+x_1\otimes x_3\otimes x_4\otimes x_2+x_1\otimes x_4\otimes x_2\otimes x_3\right). $$ $\mathrm{Cyc}_{12}$ is defined similarly. $\mathscr{R}(V)$ is the space where Riemannian curvature tensors live. (2) $ \mathscr{C}(V)=\{R\in V^{\otimes 4}\mid \mathrm{Tr}_{12}(R)=0,\, \mathrm{Cyc}_{34}(R)=\mathrm{Cyc}_{234}(R)=0\}. $ Here $\mathrm{Tr}_{12}: V^{\otimes 4}\rightarrow V\otimes V$ contracts the 1st and 2nd components through $g$. $\mathscr{C}(V)$ is the space where curvature tensors of torsion-free volume-preserving affine connections on a Riemannian manifold live. (3) $ \mathscr{A}(V)=\{A\in V^{\otimes3}\mid\mathrm{Cyc}_{12}(A)=A, \, \rm{Tr}_{23}(A)=0\} $, the space where the difference tensor of two torsion-free volume-preserving affine connections lives. The decomposition of $\mathscr{R}(V)$ into irreducible $\mathrm{O}(n)$-modules is well known under the name "Ricci decomposition" and is discussed briefly in the book "Einstein Manifolds" by Besse. The decomposition of $\mathscr{C}(V)$ seems obtained only recently by Blazic, Gilkey, Nikcevic, and Simon. I can't find any reference for the decomposition of $\mathscr{A}(V)$. So my question is (1) Where exactly can I find a systematic treatment of the decomposition problem for this kind of $\mathrm{O}(n)$-modules? (2) In particular, given a $\mathrm{O}(n)$-module $W$ like the above ones, i.e. a submodule of $V^{\otimes k}$ which is the kernel of some $\mathrm{Tr}$ and $\mathrm{Cyc}$ type morphisms, is there a general algorithm, or strategy, to decompose $W$ into irreducible pieces? (3) In particular, has a irreducible decomposition of $\mathscr{A}(V)$ appeared in the literature? REPLY [7 votes]: As Qiaochu Yuan wrote, these irreducible decompositions are classical. You can read about them in Hermann Weyl's The Classical Groups (for example), and the questions you are asking are basically exercises in writing out what the standard methods tell you. (Of course, there are more recent treatments, which modern readers might find more readable than Weyl. You might, for example, read about the method of Young tableau in the popular recent book Representation Theory by Fulton and Harris, which seems to be what you need.) For example, Weyl's formula gives you that $\mathscr{A}(V)$ is the direct sum of two irreducible subspaces; one of them, in your notation, is the subspace $$ \mathscr{A}_1(V)=\{A\in V^{\otimes3}\mid\mathrm{Cyc}_{12}(A)=A,\,\mathrm{Cyc}_{23}(A)=A,\,\mathrm{Tr}_{23}(A)=0\}, $$ but it is better known as the space of harmonic (i.e., traceless) symmetric cubic polynomials on $V$ and has dimension $\tfrac16 n(n{-}1)(n{+}4)$, while the other is $$ \mathscr{A}_2(V)=\{A\in V^{\otimes3}\mid\mathrm{Cyc}_{12}(A)=A,\,\mathrm{Cyc}_{123}(A)=0,\,\mathrm{Tr}_{23}(A)=0\} $$ and has dimension $\tfrac16 n(n{+}1)(2n{+}1)$ (it is usually described as the $\mathrm{O}(n)$-representation with highest weight vector $(1,1,0,\ldots,0)$).<|endoftext|> TITLE: Bound on gcd of two integers QUESTION [10 upvotes]: Well this is a problem I was fiddling with. I came up with it but it probably is not original. Suppose $a\in \mathbb{N}$ is not a perfect square. Then show that : $$\text{gcd}(n,\lfloor{n\sqrt{a}}\rfloor)\le c\sqrt n\tag{1}$$ Where $c=\sqrt[4]{4a}$ and P.S. it is the best constant, when $ax^2-y^2=1$ admits a solution. Now I have a couple questions. When $ax^2-y^2=1$ has no solution, what could be the best constant $c$ in $(1)$ then? What about when $\dfrac{\text{gcd}(n,\lfloor{n\alpha}\rfloor)}{\sqrt n}$we have a irrational number $\alpha$ which is not a quadratic surd. For example, $\pi,\sqrt[3]{2}$ etc? Any references or solutions would be helpful. Thanks. REPLY [10 votes]: Robert Israel's nice answer above already points to what happens if there are very good rational approximations to the irrational number $\alpha$. Let me just address the first question when $\alpha =\sqrt{a}$ is a quadratic irrational (with $a$ being a non-square natural number). Let $k$ be the least positive number that can be expressed as $ax^2-y^2$. Then the best constant in the bound for the gcd is $\sqrt{2\sqrt{a}/k}$. To see this, suppose $g=(n,\lfloor n\alpha \rfloor)$, and write $n=gv$ and $\lfloor n\alpha \rfloor =gu$. Then $$ \frac{1}{n} > \alpha -\frac{u}{v} > 0, $$ and multiplying both sides by $\alpha+u/v \le 2\alpha$ we get $$ \frac{2\alpha}{n} \ge \alpha^2 -\frac{u^2}{v^2} \ge \frac{k}{v^2}, $$ since $k$ is the smallest positive number of the form $a x^2- y^2$. Since $v=n/g$, we conclude that $$ g^2 \le \frac{2\alpha}{n} \frac{n^2}{k} = \frac{2\alpha n}{k}, $$ which is the claimed bound. To see that this bound is asymptotically attained, take a solution to $ax^2 -y^2 =k$ and by multiplying by a solution to the Pell equation $x^2-ay^2=1$ we can find a solution to $ax^2 -y^2 =k$ with $x$ and $y$ arbitrarily large. Further such $x$ and $y$ may be assumed to be coprime by the minimality of $k$. For such a choice we know that $$ \alpha - \frac{y}{x} \approx \frac{k}{2\alpha x^2}, $$ and choosing $n= x \lfloor (1-\epsilon) 2\alpha x/k\rfloor$ produces a large gcd for $n$ and $\lfloor n\alpha\rfloor$. Finally the negative Pell equation $ax^2-y^2=1$ has been extensively studied, and still remains not fully understood. See for example this impressive recent paper of Fouvry and Kluners in Annals. It may also be interesting to note that there do exist $a$ for which $k$ is essentially as large as possible: namely take $a$ of the form $b(b+2)$ and then the corresponding $k$ is $2b$. (One can see this from the fact that the continued fraction expansion for $\sqrt{b(b+2)}$ is $[b;1,2b,1,2b,\ldots]$.)<|endoftext|> TITLE: endomorphisms of modules over symmetric ring spectra QUESTION [5 upvotes]: I have a probably very basic question about modules over symmetric ring spectra: Let $R$ be a commutative symmetric ring spectrum and let $M$ and $N$ be module spectra over $R$. Moreover, let $\varphi \colon M \to N$ be a morphism of $R$-module spectra, which is a stable equivalence of the underlying symmetric spectra. Moreover, denote by $End_R(M)$ and $End_R(N)$ the spectra of $R$-linear endomorphisms (using the inner hom spectra). If there is any justice in the world the answer to the following should be "yes": Is $End_R(N)$ stably equivalent to $End_R(M)$? This question can be reduced to the following one: Are the maps $\varphi_* \colon End_R(M) \to Hom_R(M,N)$ and $\varphi^* \colon End_R(N) \to Hom_R(M,N)$ induced by pre- and postcomposition stable equivalences? REPLY [7 votes]: Yes, this is true. This kind of property for a model category is a consequence of what is sometimes abusively called the "SM7" axiom for the enrichment: In $R$-modules, suppose that $A \to B$ is a cofibration and $C \to D$ is a fibration. Then the natural map of $R$-modules $$ Hom_R(B,C) \to Hom_R(A,C) \times_{Hom_R(A,D)} Hom_R(B,D) $$ is a fibration of $R$-modules, which is a weak equivalence if either of the original maps is. (If the enrichment were, instead, an enrichment in simplicial sets, then this would genuinely be Quillen's SM7 axiom for simplicial model categories.) Let's suppose for the time being that $N \to M$ is a cofibration and a weak equivalence, $N$ is cofibrant, and $P$ is an arbitrary fibrant $R$-module. Applying the SM7 axiom to $N \to M$ and $P \to *$ shows that $Hom_R(M,P) \to Hom_R(N,P)$ is (a fibration and) a weak equivalence. This shows $Hom_R(-,P)$ takes acyclic cofibrations between cofibrant objects to equivalences; Ken Brown's lemma then implies that it takes arbitrary weak equivalences between cofibrant objects to equivalences. A dual proof shows that, if $Q$ is cofibrant, $Hom_R(Q,-)$ takes weak equivalences between fibrant objects to equivalences. When $N$ and $M$ are cofibrant-fibrant, we then find that all relevant maps of Hom-objects induced by pre- or post-composition with $\phi$ are equivalences. Side note: There's a subtlety here if you're using one of the "positive" model structures of Shipley's "A convenient model category for commutative ring spectra" (called the S-model structure there). There, in order for things to be homotopically well-behaved, you have to interpret the positive model structure as actually enriched in the ordinary stable model structure (Shipley proves this version of the SM7 axiom and more). Further side note: You didn't ask, but you might worry about whether you can soup this up to an equivalence of $R$-algebras. You can, but only up to a "zigzag" weak equivalence (which can't be gotten around in general) by using the endomorphism object of the map $\phi$ -- I learned this technique from Rezk's thesis. (For enrichments in spaces, an earlier paper of Dwyer and Kan gives a much more powerful tool for constructing derived endomorphism spaces.) A proof that you get equivalent endomorphism algebras (which also provides functoriality in weak equivalences, but only in the homotopy category of algebras) can be found in Section 2 of this paper.<|endoftext|> TITLE: Subgroup property stronger than being characteristic QUESTION [7 upvotes]: In what follows, all groups are assumed to be finite. Recall that if $K \leq H \leq G$ are groups, $K$ is said to be a weakly closed subgroup of $H$ in $G$ if, for all $g \in G$, $g^{-1}Kg \leq H$ implies that $g^{-1}Kg = K$. The subgroup property I am interested in is this one: what kind of subgroup of $H$ is $K$ when, for all groups $G$ with $H \leq G$, $K$ is a weakly closed subgroup of $H$ in $G$? It is trivial to verify that $1$ and $H$ are subgroups of $H$ with this property. (Here, already, the finiteness condition is important: if $H$ is infinite, one could have $g^{-1}Hg < H$ instead of having $g^{-1}Hg = H$. So I won't go there.) It is clearly necessary that $K$ be a characteristic subgroup of $H$ for this unnamed property to hold: If $H$ is finite, so is its holomorph $Hol(H) = H \rtimes Aut(H)$. So let $G = Hol(H)$. Then, regarding $Aut(H)$ as a subgroup of $G$, let $\alpha \in Aut(H)$. Then $\alpha^{-1}K\alpha \leq H$ because $\alpha$ is an automorphism of $H$. Since $K$ is, by assumption, a weakly closed subgroup of $H$ in $G$, this means $\alpha^{-1}K\alpha = K$. Since $\alpha \in Aut(H)$ was arbitrary, it follows that $K$ is a characteristic subgroup of $H$. On the other hand, merely being characteristic is not sufficient: Let $G = <(1243675), (4657)(23)>$, so that $G \cong GL_{3}(2)$. Let $H = <(4657)(23), (67)(23)>$ and $K = <(45)(67)>$. Then $K = Z(H)$, so $K$ is a characteristic subgroup of $H$. It is also true that $H \leq G$, and that $g = (13)(57) \in G$. But then $g^{-1}Kg = <(47)(65)> \leq H$, but $<(47)(65)> \neq < (45)(67) >$. I have heard of fully invariant subgroups, but this also shows that a fully invariant subgroup need not have this property: in the example above, $Z(H)$ is a fully invariant subgroup of $H$ because $Z(H)$ is the subgroup generated by the squares in $H$. What I can say is that if a subgroup $K$ of $H$ is the only subgroup of $H$ with elements having the orders they do, then $K$ is this kind of subgroup of $H$. (So, in the above example, $<(4657)(23)>$ is the unique cyclic subgroup of order 4 in $H$ and it is this kind of subgroup of $H$, unlike $Z(H)$.) Does this kind of subgroup have a name? What other sets of conditions on a characteristic subgroup imply this stronger property? REPLY [11 votes]: Instead of going via HNN extensions, one can work entirely with finite groups to show that if $X \cong Y$ are subgroups of a finite group $H$, then there exists a finite group $G \supseteq H$ such that $X$ and $Y$ are conjugate in G. The trick is to take $G$ to be the full symmetric group on he elements of $H$, with $H$ embedded as the permutations induced by right multiplication (as in Cayley's theorem). Let $\theta$ be an isomorphism from $X$ to $Y$. Now choose a set $T$ of representatives for the left cosets of $X$ in $H$, and similarly, a set $S$ of representatives for the left cosets of $Y$. Note that $|T| = |H:X| = |H:Y| = |S|$, and let $\sigma$ be an arbitrary bijection from $T$ to $S$. Extend $\sigma$ to a permutation of $H$ by setting $(tx)^\sigma = t^\sigma x^\theta$. Now $\sigma$ is an element of $G$, and I argue that conjugation in $G$ by $\sigma$ carries right multiplication by $x$ to right multiplication by $x^\theta$. This will show that the copies of $X$ and $Y$ in $G$ are conjugate in $G$. Write $y = x^\theta$ and let $r_x$ and $r_y$ be the corresponding right multiplication maps on $H$. I want to show that $\sigma^{-1}r_x\sigma = r_y$, or equivalently, that $r_x\sigma = \sigma r_y$. To check this, we apply each side to an arbitrary element $h$ in $H$. Applying $r_x\sigma$ yields $(hx)^\sigma$, and applying $\sigma r_y$ yields $h^\sigma y$. We thus want $(hx)^\sigma = h^\sigma y$. Now write $h = tu$, with $t \in T$ and $u \in X$. Then $$\ (hx)^\sigma = (tux)^\sigma = t^\sigma(ux)^\theta = t^\sigma u^\theta x^\theta = (tu)^\sigma y = h^\sigma y , $$ as wanted.<|endoftext|> TITLE: Is the leading Taylor coefficient at $s = 1$ of the $L$-series of an elliptic curve over $\mathbb{Q}$ positive, as predicted by BSD? QUESTION [9 upvotes]: Let $E$ be an elliptic curve over $\mathbb{Q}$. As proved by Wiles et al., its $L$-series $L(E, s)$ is entire. Set $r := \mathrm{ord}_{s = 1} L(E, s)$, a value conjecturally equal to $\mathrm{dim}_{\mathbb{Q}} (E(\mathbb{Q}) \otimes_{\mathbb{Z}} \mathbb{Q})$. There is a $c \in \mathbb{C}^\times$ such that $L(E, s) \sim c \cdot (s - 1)^r$ as $s \rightarrow 1$. Birch and Swinnerton-Dyer predict a great deal about $c$, in particular, that $c \in \mathbb{R}$, $c \in \mathbb{R}_{> 0}$, $c/\Omega_E \in \mathbb{Q}$ if $r = 0$, where $\Omega_E$ is the volume of $E(\mathbb{R})$ computed with respect to "the" Neron differential of $E$, and $c/\Omega_E \in \mathbb{Q}_{> 0}$ if $r = 0$. Which of these predictions are known to hold? Partial answers and answers in special cases (e.g., in the case when $E$ has potential complex multiplication) are very welcome! REPLY [12 votes]: Let me summarise the comments above that give a full answer (correct me if I am wrong). The analytic continuation of $L(E,\bar{s})=\overline{L(E,s)}$ shows that $c\in\mathbb{R}$. If $r=0$, the fact that $c>0$ is proven in On the positivity of the central value of automorphic L-functions for GL(2) Duke Math 83 1996, 1-18. It would follow from the generalised Riemann hypothesis, too. For $r=1$, it follows from Gross-Zagier and the case $r=0$. I don't know about $r>1$. That $L(E,1)/\Omega_E$ is a rational number is a consequence of the theorem of Manin-Drinfeld on modular symbols. Is just 2+3. Note that $\Omega_E$ is defined to be positive, as it is the least positive real period of a Néron differential on $E$ (or twice that depending on your normalisation).<|endoftext|> TITLE: Knot invariants in 3-manifolds that are not $\mathbb{R}^3$ or $S^3$ or $B^3$? QUESTION [5 upvotes]: This is just a reference request; I have no sharp mathematical question. Inspired by the $(3+)$-year old MO question, In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$?, I would like to ask: Q1. Are there are studies of knot polynomial invariants in the $8$ Thurston geometries? Alternatively: Q2. Are there substantive differences between the various 3-manifolds with respect to knot invariants? (Added:) Gregory Moore says, "In general [in other 3-manifolds], the knot polynomials no longer have integer coefficients,..." (p.12 of "Physical Mathematics and the Future" (2014). PDF download of 56-page paper.) REPLY [7 votes]: For question 1) Yes analogs of knot invariants exist for any manifold regardless of prime decomposition or Thurston decomposition. For example, consider the Alexander polynomial and its categorification, knot Floer homology. These can both defined for general one cusped 3-manifolds. Computations of these invariants have been used lately to answer the question: Given a homology class $\mathcal{s}$ in a manifold $M$, is there a knot $K$ such that as a curve in $M$, $K \in \mathcal{s}$ and $S^3-K' \cong M-K$ for some knot $K'$ in $S^3$. This question is equivalent to if $M$ can be realized as surgery along a knot in $S^3$. For example, Josh Greene was able to show here that if a knot in $S^3$ admits a lens space surgery then its knot Floer homology appears on a confined (but infinite) list of known examples, here. Margaret Doig offers a nearly complete classification knots in elliptic manifolds that admit an $S^3$ surgery in these two papers. And recently, Yi Ni and Xingru Zhang places significant restrictions on knots in certain Nil manifolds that could admit $S^3$ surgeries here. These are all examples of exploiting knot invariants in closed manifolds. For question 2) I think the consensus is that knot invariants are more delicate in non-simply connected manifolds. As brought up in the comments, there could be several different analogs of the unknot. One analog of unknottedness is minimal with respect to Seifert genus, i.e. a knot is minimal if for all simple closed curves in its homology class of a manifold there is no curve that bounds a lower genus (orientable) surface. However, even this question is considerably harder. There is some hope that there might be a polynomial time algorithm for recognition of the unknot in $S^3$. Along these lines, Agol, Haas, and Thurston show that just determining the genus of a knot in general 3-manifold is NP-Hard, here.<|endoftext|> TITLE: What is the difference between $\delta W^{\pm}=0$ and Einstein? QUESTION [5 upvotes]: Maybe this is a vague question. In Besse's book Einstein manifolds, $\delta W^{\pm}=0$ is considered as a generalization of Einstein metrics on four-manifolds. I was wondering what is the difference between $\delta W^{\pm}=0$ and Einstein? Are there examples of metrics with $\delta W^+=\delta W^-=0$ but not Einstein? What if $\delta W^+=\delta W^-=0$ and constant scalar curvature, is it equivalent to Einstein? Thank you very much. REPLY [8 votes]: A very simple example of a non-Einstein manifold with harmonic Weyl tensor is the product $\mathbb{S}^2\times \mathbb{R}^2$ (with the usual metrics). In Besse's book there are more examples. The example above has also constant scalar curvature. So the assumptions $\delta W=0$ and constant scalar curvature are not sufficient to say that the manifold is Einstein. Since any Einstein manifold satisfies $\delta W=0$ we can consider it as a generalization of the Einstein condition. However, as said before, they are not equivalent. One can characterize the condition $\delta W=0$ in terms of the Schouten tensor $S=\frac{1}{n-2}\left(Ric-\frac{s}{2(n-1)}g\right).$ Indeed, $\delta W=0$ if and only if $S$ is a Codazzi tensor, that is, $(\nabla_XS)(Y,Z)=(\nabla_YS)(X,Z).$ In the particular case of dimension four we can decompose the Weyl tensor $W=W^++W^-.$ A further generalization of $\delta W=0$ is $\delta W^+=0$ or $\delta W^-=0.$ If you are interested into the study of this manifolds the following papers could be useful: The geometry of weakly self-dual Kahler surfaces, by V. Apostolov, D. M. J. Calderbank and P. Gauduchon: http://arxiv.org/pdf/math/0104233.pdf Compact Kahler surfaces with harmonic anti-self-dual Weyl tensor by W. Jelonek: http://www.sciencedirect.com/science/article/pii/S0926224502000761<|endoftext|> TITLE: Mapping complexes in the simplicial localization of the category of manifolds QUESTION [5 upvotes]: Let $\mathit{Mfd}$ denote the category of smooth manifolds. Let $W$ denote all projections of the form $$M \times \mathbb{R} \to M.$$ Let $\mathit{Mfd}_W$ denote the Hammock localization of $\mathit{Mfd}$ at the class of maps $W$. Is the mapping complex $Map_W\left(M,N\right)$ in $\mathit{Mfd}_W$ between two manifolds $M$ and $N$ weakly equivalent to the space of maps from $M$ to $N$, i.e. do we have $$Map_W\left(M,N\right) \simeq \underline{Hom}\left(Sing\left(M\right),Sing\left(N\right)\right)?$$ REPLY [3 votes]: Ok, so I have an argument: First note that the homotopy coherent nerve of $\mathit{Mfd}_W$ is equivalent to the quasicategory obtained by formally inverting $W$ in $N\left(\mathit{Mfd}\right)$- this holds in generality, as shown in a recent paper of Hinich (Proposition 2.2.1 of http://arxiv.org/abs/1311.4128). Let me just be lazy and write $\mathit{Mfd}_W$ also for this quasicategory. Let $$y:\mathit{Mfd} \hookrightarrow Psh_\infty\left(\mathit{Mfd}\right)$$ be the Yoneda embedding into infinity presheaves. First, note that the localization $y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right)$ at the class of morphisms of the form $y\left(M \times \mathbb{R}\right) \to y(M)$ is canonically equivalent to $Psh_\infty\left(\mathit{Mfd}_W\right).$ To see this observe that the canonical functor $$\mathit{Mfd} \to Psh_\infty\left(\mathit{Mfd}\right) \to y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right)$$ sends $W$ to equivalences, so induces a functor $$\mathit{Mfd}_W \to y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right),$$ which one can show immediately by universal properties exhibits $y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right)$ as the free colimit completion of $\mathit{Mfd}_W$ in the world of quasicategories. Let $$y_w:\mathit{Mfd}_W \hookrightarrow Psh_\infty\left(\mathit{Mfd}_W\right)\simeq y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right)$$ denote the Yoneda embedding. Uwinding what we've done, we have that $$y_w\left(M\right) \simeq h \circ y\left(M\right)$$ for all manifolds $M.$ Let $h$ denote the localization functor $$h:Psh_\infty\left(\mathit{Mfd}\right) \to y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right).$$ By Lemma 7.5 of http://arxiv.org/abs/1311.3188, we have that $h$ can be computed as: $$h\left(F\right)\left(M\right) = \operatorname{hocolim} F\left(M \times \Delta^n_{ext}\right)$$ where $\Delta_{ext}:\Delta \to \mathit{Mfd}$ is the "standard cosimplicial manifold," with each $$\Delta^n_{ext} \cong \mathbb{R}^n.$$ Now, $$h\left(y\left(M\right)\right)\left(N\right) = \operatorname{hocolim} Hom\left(N \times \Delta^n_{ext},M\right).$$ This homotopy colimit is a simplicial diagram of sets (regarded as discrete simplicial sets), so the hocolim is simply the starting simplicial set $$Hom\left(N \times \Delta^\star_{ext},M\right).$$ Finally, (and here I guess I am being partly sloppy), this simplicial set is homotopy equivalent to $Sing(M^N),$ where $M^N$ is the compactly generated mapping space. (Incidentally, if anyone has a clean proof of this last claim, let me know). Notice that $$h\left(y\left(M\right)\right)\left(N\right)\simeq Hom\left(y(N),ih\left(y\left(M\right)\right)\right),$$ where $i$ is the inclusion of $y\left(W\right)^{-1}Psh_\infty\left(\mathit{Mfd}\right)$ into $Psh_\infty\left(\mathit{Mfd}\right),$ and $$Hom\left(y(N),ih\left(y\left(M\right)\right)\right)\simeq Hom\left(h y(N), h y(M)\right)\simeq Hom\left(y^w(N),y^w(M)\right),$$ and finally $$Hom\left(y^w(N),y^w(M)\right)\simeq Hom_{\mathit{Mfd}_w}\left(N,M\right),$$ since the Yoneda embedding $y^w$ is full and faithful.<|endoftext|> TITLE: Covering properties of strongly compact embedding QUESTION [5 upvotes]: Let $\kappa$ be a $\mu$-strongly compact cardinal, which means that there is an elementary embedding $j:V\rightarrow M$, with critical point $\kappa$ such that $M$ is well founded (even closed under $\kappa$ sequences) and there is $s\in M$ such that $j^{\prime\prime} \mu \subset s$ and $M\models |s| TITLE: Picard group generated by effective divisors: counterexample? QUESTION [9 upvotes]: Let $X$ be an integral variety defined over an algebraically closed field $k$ of characteristic 0 with finitely generated Picard group $Pic(X)$ and such that $k[X]^\times=k^\times$ (i.e. the only invertible global sections of the structure sheaf of $X$ are the constants). To my knowledge, if $X$ is quasi-projective or locally factorial, then it satisfies the following property: ($*$) $Pic(X)$ is generated by effective Cartier divisors (equivalently every Cartier divisor is linearly equivalent to a difference of two effective Cartier divisors). My question is: does there exist a variety $X$ as above that does not satisfy ($*$)? EDIT: I recall that the group of Cartier divisors of $X$ is $\text{Div}(X)=H^0(X,\mathcal{K}(X)/\mathcal{O}^*_X)$, where $\mathcal{K}(X)$ is the constant sheaf associated to the function field of $X$ and $\mathcal{O}^*_X$ is the sheaf of units of the structure sheaf of $X$. A Cartier divisor $\{(U_i,f_i)\}_{i\in I}$ is said effective if $f_i\in H^0(U_i,\mathcal{O}_X)$ for all $i\in I$. I have no experience with singular varieties, any reference related to this problem would be helpful. REPLY [3 votes]: If I understand correctly, you are asking when the natural map $\mathrm{Div}(X)\rightarrow \mathrm{Pic(X)}$ is surjective (by definition $\mathrm{Div}(X)$ is the free group generated by effective Cartier divisors). EGA IV, section 21.3 gives some general conditions under which this holds: in particular, when $X$ is reduced with finitely many irreducible components. So of course $X$ integral suffices.<|endoftext|> TITLE: When does simplicial localization commute with functor categories? QUESTION [7 upvotes]: Let $(C,W)$ be a category with a class of weak equivalences, and $D$ a small category. Then I can form the diagram category $(C^D,W^D)$ with objectwise weak equivalences, and its simplicial localization $L(C^D,W^D)$. Or I can first simplicially localize to get $L(C,W)$ and then form the $(\infty,1)$-functor category $L(C,W)^D$. If $(C,W)$ is a nice model category, so that $(C^D,W^D)$ has a model structure presenting $L(C,W)^D$, then we have $L(C^D,W^D) \simeq L(C,W)^D$. Is this true any more generally? Is there an easy counterexample where it doesn't hold? REPLY [7 votes]: For general $C$ (I will drop $W$ from the notation) this is not true even when $D$ is an infinite discrete category. In the answer to this similar question I described how to construct a sequence of relative categories $C_0, C_1, C_2, \ldots$ with objects $X_i, Y_i \in C_i$ such that $X = (X_0, X_1, \ldots)$ and $Y = (Y_0, Y_1, \ldots)$ are isomorphic as objects of $\prod_i \mathrm{Ho}(C_i)$ but not as objects of $\mathrm{Ho}(\prod_i C_i)$. We take $C = \coprod_i C_i$ and similarly we have objects $X$ and $Y$ isomorphic in $\mathrm{Ho}(C)^\mathbb{N}$ but not in $\mathrm{Ho}(C^\mathbb{N})$. Of course $\mathrm{Ho}(C^\mathbb{N})$ is the homotopy category of $L(C^\mathbb{N})$ so $X$ and $Y$ are not equivalent in $\mathrm{Ho}(C^\mathbb{N})$. I'm not sure whether $\mathrm{Ho}(C)^\mathbb{N}$ is the homotopy category of $L(C)^\mathbb{N}$ (probably not), but at least the functor $L(C)^\mathbb{N} \to \mathrm{Ho}(C)^\mathbb{N}$ reflects equivalences and hence $X$ and $Y$ are still equivalent in $L(C)^\mathbb{N}$ so $L(C^\mathbb{N}) \to L(C)^\mathbb{N}$ is not an equivalence.<|endoftext|> TITLE: When is/isn't the monoidal unit compact projective? QUESTION [8 upvotes]: I am interested in developing intuition for when the monoidal unit in a closed monoidal abelian category is or isn't compact projective. As such, my question is not looking for a yes/no answer, but rather for classes of examples in which either a yes or no applies. I will repeat my question with more detail and examples: Definitions and remarks Recall that in an abelian category $\mathcal C$, an object $X$ is compact if $\hom(X,-) : \mathcal C \to \mathrm{AbGp}$ preserves filtered colimits. It is projective if $\hom(X,-)$ is right exact. Since all $\mathbb Z$-linear functors preserve finite direct sums and since (sm)all colimits are generated by finite direct sums, cokernels, and filtered colimits, $X$ is compact projective iff $\hom(X,-)$ is cocontinuous (=preserves all colimits). In a monoidal category with monoidal unit $\mathbf 1$, the functor $\hom(\mathbf 1,-)$ is the functor of global sections or global elements or invariants. Thus to say "$\mathbf 1$ is compact projective" is the same as to say "the functor of global sections is cocontinuous." Recall that a monoidal structure on $\mathcal C$ is closed if $X\otimes : \mathcal C \to \mathcal C$ and $\otimes X : \mathcal C \to \mathcal C$ have left adjoints for all $X$; in particular, in a closed monoidal category, both $\otimes X$ and $X\otimes$ are cocontinuous. It is not hard to prove from this that in a closed monoidal category, if $\mathbf 1$ is compact projective, then so are all dualizable objects. Main questions What is a class of closed monoidal abelian categories "that appear in nature" for which the monoidal unit definitely is compact projective? What is a class of closed monoidal abelian categories "that appear in nature" for which the monoidal unit definitely is not compact projective? Examples In any semisimple category, all objects are projective. In particular, let $G$ be a reductive algebraic group over a field of characteristic $0$. Then the category of algebraic $G$-modules is semisimple. This gives a class of examples where the answer to my question is "yes". I know essentially nothing about algebraic groups in characteristic $p$. I've been told that in general reductive groups in characteristic $p$ do not have semisimple representation theory. Do they nevertheless have $\mathbf 1$ compact projective? Under what conditions? Consider the additive algebraic group $\mathbb G_a$, say over $\mathbb C$. Its category of algebraic modules is the category of vector spaces equipped with a locally nilpotent endomorphism. The only compact projective object is the $0$ object; in particular, the trivial module $\mathbf 1$ is not compact projective. Let $R$ be a commutative algebra. The category of $R$-modules is symmetric monoidal with monoidal structure $\otimes_R$. The monoidal unit is $R$ acting on itself by multiplication. It is compact projective, for somewhat stupid reasons. Let $A$ be an associative algebra. The category of $A$-$A$-bimodules is monoidal with monoidal unit $A$ acting on itself from both sides by multiplication. This is always compact. I believe, but could be mistaken, that it is projective iff $A$ is semisimple as an algebra. For most schemes $X$, the functor of global sections $\mathrm{H}^0 : \operatorname{QuasiCoh}(X) \to \mathrm{AbGp}$ is not exact. Said another way, the monoidal unit $\mathcal O_X$ is not projective. But I don't have good intuition for naturally-appearing classes of schemes for which one can say definitively that $\mathcal O_X$ is/is not projective. The finite-dimensional representation theory of a Drinfeld–Jimbo quantum group at generic $q$ is semisimple, so $\mathbf 1$ is compact projective. At roots of unity there are different possible choices for which representation theory to take, and I don't know all the answers. The Temperley–Lieb category is the monoidal $\mathbb Z[q,q^{-1}]$ category freely-generated by a self-dual object of dimension $-q^2 - q^{-2}$. It is not abelian, but its abelian envelop is. Said abelian envelop has the property that the monoidal unit is compact projective. Deligne's category $\mathrm{GL}(t)$ is the abelian envelop of the free symmetric monoidal category generated by a dualizable object of dimension $t$. It also enjoys the property that the monoidal unit is compact projective. Indeed, this is a general property of categories presented by "string diagrams" with "skein relations". There are many other categories that appear in nature. What about your favorite class? REPLY [3 votes]: I think Corollary 1.13.7 from Etingof's lectures (www.math.mit.edu/~etingof/tenscat.pdf) might be useful: Given a multiring category (= locally finite $k$-linear abelian monoidal category with biexact tensor) with right duals, then the unit object is projective iff the category is semisimple. This is a continuation of what you said yourself in the beginning. Also, $Rep(GL_t)$ is the Karoubian (not abelian) envelope of the free rigid symmetric monoidal category with one generator of dimension $t$. It is abelian when $t $ is not an integer, in which case it is also semisimple, so it's not very interesting in term of your question. When $t$ is an integer, the situation is more complicated - the category is not abelian. An interesting example is the Deligne category $Rep(S_t)$. It is the Karoubian envelope of the free rigid symmetric monoidal category generated by a Frobenius-algebra-object of dimension $t$. This category is not abelian when $t$ is a non-negative integer, but it has an abelian envelope, which is a tensor category. It is not semisimple, and of course $1$ is not projective.<|endoftext|> TITLE: Topological fundamental group of a variety QUESTION [8 upvotes]: I have an explicit question. I have a complex projective variety defined by $2\times 2$ minors of a matrix. The entries are polynomials from a weighted projective space. In fact, it's a singular 3-fold with only quotient singularities. Now I want to show that it is simply connected. Can anyone help in giving direction in this regard? REPLY [3 votes]: I am not sure this will help, but the methods we use in Fundamental Group of Moduli Spaces of Representations might be useful (if you are working over $\mathbb{C}$). In particular, these two properties are worth pointing out: Given any non-empty Zariski open subset $U \subset V$, where $V$ is normal, the homomorphism $\pi_1(U) \to \pi_1(V)$ induced by the inclusion map is surjective. The GIT quotient map $V\to V//G$ is $\pi_1$-surjective, where $G$ is a connected reductive affine algebraic group. Another tool that might be of interest is Proposition 5.8 which gives three conditions for the inclusion map $V\setminus W\hookrightarrow V$ to be 2-connected (where $W\subset V$ is a subvariety).<|endoftext|> TITLE: A sum by Ramanujan for $\coth^{2}(5\pi)$ QUESTION [23 upvotes]: Ramanujan mentions in one of his letters to Hardy that $$\frac{1^{5}}{e^{2\pi} - 1}\cdot\frac{1}{2500 + 1^{4}} + \frac{2^{5}}{e^{4\pi} - 1}\cdot\frac{1}{2500 + 2^{4}} + \cdots = \frac{123826979}{6306456} - \frac{25\pi}{4}\coth^{2}(5\pi)$$ If we put $q = e^{-\pi}$ we can see that the series is given by $$\sum_{n = 1}^{\infty}\frac{n^{5}q^{2n}}{1 - q^{2n}}\cdot\frac{1}{2500 + n^{4}}$$ While I am aware of the sum $$\sum_{n = 1}^{\infty}\frac{n^{5}q^{2n}}{1 - q^{2n}} = \frac{1 - R(q^{2})}{504}$$ and the Ramanujan function $R(q^{2})$ can be expressed in terms of $k, K$ as $$R(q^{2}) = \left(\frac{2K}{\pi}\right)^{6}(1 + k^{2})(1 - 2k^{2})\left(1 - \frac{k^{2}}{2}\right)$$ (see the derivation of this formula here). For $q = e^{-\pi}$ we have $k = 1/\sqrt{2}$ so that $R(q^{2}) = 0$ and hence $\sum_{n = 1}^{\infty}n^{5}q^{2n}/(1 - q^{2n}) = 1/504$. But getting the factor $1/(2500 + n^{4})$ seems really difficult. Any ideas on whether we can get this factor by integration/differentiation (plus some algebraic games) from the series $\sum n^{5}q^{2n}/(1 - q^{2n})$? Further Update: We have $$n^{4} + 2500 = (n^{2} + 50)^{2} - 100n^{2} = (n^{2} - 10n + 50)(n^{2} + 10n + 50)$$ so that $$n^{4} + 2500 = (n - 5 - 5i)(n - 5 + 5i)(n + 5 - 5i)(n + 5 + 5i)$$ so I believe we can do a partial fraction decomposition of $1/(n^{4} + 2500)$ but still I need to find a way to sum $\sum n^{5}q^{2n}/(1 - q^{2n})\cdot 1/(n + a)$ i.e. the problem is now simplified to getting a linear factor like $1/(n + a)$ somehow. REPLY [14 votes]: Using (see entry 24 on page 291 in Ramanujan's Notebooks II: http://www.plou) $$\frac{\pi e^{-2\pi z}}{2z[\cosh{(2\pi z)}-\cos{(2\pi z)}]}= \frac{1}{8\pi z^3}-\frac{1}{4z^2}+\frac{\pi}{4z}-\sum\limits_{n=1}^\infty \frac{1}{z^2+(z+n)^2}$$ $$+4z\sum\limits_{n=1}^\infty \frac{n}{(e^{2\pi n}-1) (4z^4+n^4)},$$ and $$\frac{n}{4z^4+n^4}=\frac{1}{4z^4}\left (n-\frac{n^5}{4z^4+n^4}\right ),$$ we get $$\sum\limits_{n=1}^\infty \frac{n^5}{(e^{2\pi n}-1) (4z^4+n^4)}=\frac{1}{8\pi}-\frac{z}{4}+\frac{\pi z^2}{4}- \sum\limits_{n=1}^\infty\frac{z^3}{z^2+(z+n)^2}$$ $$+\sum\limits_{n=1}^\infty \frac{n}{e^{2\pi n}-1}-\frac{\pi z^2e^{-2\pi z}} {2[\cosh{(2\pi z)}-\cos{(2\pi z)}]}.$$ Let us substitute $z=5i$ in this equation. Then $$\frac{\pi z^2e^{-2\pi z}}{2[\cosh{(2\pi z)}-\cos{(2\pi z)}]}= \frac{-25\pi}{2(1-\cosh{(10\pi)})}=\frac{25\pi}{4\sinh^2{(5\pi)}}=$$ $$ \frac{25\pi}{4}\coth^2{(5\pi)}-\frac{25\pi}{4}.$$ Therefore we get $$\sum\limits_{n=1}^\infty \frac{n^5}{(e^{2\pi n}-1)(2500+n^4)}= \frac{1}{8\pi}-\frac{5i}{4}+125i\sum\limits_{n=1}^\infty \frac{1}{(5i+n)^2-25}$$ $$+\sum\limits_{n=1}^\infty\frac{n}{e^{2\pi n}-1} -\frac{25\pi}{4}\coth^2{(5\pi)}. \tag{1}$$ Now (see page 6 in David M. Bradley, Ramanujan's formula for the logarithmic derivative of the gamma function: http://arxiv.org/abs/math/0505125) $$\sum\limits_{n=1}^\infty\frac{n}{e^{2\pi n}-1}= \frac{1}{24}-\frac{1}{8\pi}, \tag{2}$$ and $$\sum\limits_{n=1}^\infty\frac{1}{(5i+n)^2-25}=\frac{1}{10} \sum\limits_{n=1}^\infty\left (\frac{1}{n-5+5i}-\frac{1}{n+5+5i}\right)=$$ $$ \frac{1}{10}\left (\sum\limits_{n=-9}^\infty\frac{1}{n+5+5i}- \sum\limits_{n=1}^\infty\frac{1}{n+5+5i}\right)=\frac{1}{10} \sum\limits_{n=-9}^0\frac{1}{n+5+5i}. \tag{3}$$ Therefore, calculating the finite sum in (3), $$125i\sum\limits_{n=1}^\infty\frac{1}{(5i+n)^2-25}= \frac{5i}{4}+\frac{20594035}{1051076}. \tag{4}$$ Substituting (2) and (4) in (1), we get the Ramanujan's formula, because $$\frac{20594035}{1051076}+\frac{1}{24}= \frac{123826979}{6306456}.$$<|endoftext|> TITLE: how to find cubic polynomial that an unknown subset of a set of integers satisfies QUESTION [11 upvotes]: I have a set, $S$, of positive integers and I have reason to believe that some infinite subset of them may be parametrized by a cubic polynomial with integer coefficients evaluated at integer arguments. I have computed a finite subset of $S$, call it $S_{1}$ (all elements of $S$ less than some fixed bound), and would like to use $S_{1}$ to try to find such a cubic polynomial. Of course, there is a brute force method: for each 4-tuple in the set, $S_{1}$, I could apply Lagrange interpolation to fit the 4-tuple and then see if other integers in $S_{1}$ are also values of this cubic polynomial. However, this is not practical if the size of $S_{1}$ is say 1000 or so (in which case, there are over 41 billion such 4-tuples). So my question is whether there exist more efficient techniques for finding such a polynomial. Being able to do the same for polynomials of larger degree would also be useful for me, as here the above brute force method becomes even less practical. Does anyone know of such techniques, have relevant references,...? REPLY [2 votes]: In a comment to an earlier incorrect answer that I posted (and then deleted), Sum One made the following clarification. Here is a concrete example of the general problem. Suppose I create 950 random numbers between 1 and 10,000,000 and then I add to this set 50 numbers of the form $P(n)$ for $n=1,2,\ldots$ all less than 10,000,000 for some cubic polynomial $P(n)$. I give you this set $S$ of 1000 numbers (without any further information), tell you that some of them (at least 10, say, to avoid trivial solutions) are $P(0),P(1),\ldots$ for some cubic polynomial and ask you to find this polynomial. What follows is due to various colleagues of mine. Here is an algorithm that is cubic in the size of $S$. Write $P(n) = p_3n^3+p_2n^2+p_1n+p_0$ for some unknown integers $p_3, p_2, p_1, p_0$. For distinct $a,b,c$, we have the identity $${(b-c)P(a) + (c-a)P(b) + (a-b)P(c) \over (b-c)(c-a)(a-b)} = -(p_3(a+b+c) + p_2).$$ Now set $a=1$ and exhaust over all $1000$ possible choices from $S$ for $P(a)=P(1)$. For each such choice, exhaust over all $999\times 998$ choices of two other numbers from $S$, and form two separate lists $L_1$ and $L_2$. For $L_1$, hypothesize that the two other chosen numbers are $P(12)$ and $P(45)$, and use the above identity to compute what the value of $-((1+12+45)p_3+p_2)$ = $-58p_3-p_2$ would be if the hypothesis happened to be correct. For $L_2$, hypothesize that the two other chosen numbers are $P(23)$ and $P(34)$, and similarly compute what the value of $-((1+23+34)p_3+p_2)$ = $-58p_3-p_2$ would be if this hypothesis were correct. Then combine these two lists and sort them to look for occurrences of the same number in both lists; random collisions should be rare, so there is now a small list of candidates for 5-tuples $(P(1),P(12),P(23),P(34),P(45))$ that can be interpolated and and tested over the whole dataset. Actually I lied a bit. For what I just described to work, I only used the fact that $12+45=23+34$ and I didn't need all five numbers to be congruent modulo $11$. The reason that I chose them to be all congruent modulo $11$ is that $P(1)$, $P(12)$, $P(23)$, $P(34)$, $P(45)$ must all be congruent modulo $11$ since $P$ has integer coefficients. Therefore I can partition $S$ into congruence classes modulo $11$ and perform the above computation separately on each congruence class. The outer exhaust will then loop over only about $100$ choices for $P(1)$ and the lists $L_1$ and $L_2$ will be only about ten thousand long rather than a million long. Of course we have to do $11$ such computations but it is still a win. Congruences can be exploited in another way. The following alternative algorithm is asymptotically slower but uses less memory and should be more effective for the stated parameter sizes. We are going to have four nested loops, guessing the values of $P(1)$, $P(50)$, $P(27)$, and $P(12)$ respectively. Given a hypothesized value for $P(1)$, we know that $P(50)\equiv P(1) \pmod{49}$, so the possible values for $P(50)$ are cut down by a factor of roughly $49$. At the next level, we must have $P(27)\equiv P(1) \pmod{26}$ and $P(27)\equiv P(50)\pmod{23}$, providing further cutdown, etc. The sequence $1,50,27,12$ should provide the greatest expected cutdown, generating on average fewer than 100000 interpolations instead of 41 billion.<|endoftext|> TITLE: What is natural about the well-known bijection between conjugacy classes and irreps of a symmetric group? QUESTION [9 upvotes]: Symmetric groups possess a well-known bijection between conjugacy classes and irreducible representations. More precisely, both sets are indexed by Young diagrams. Question: To what extent is this bijection natural? This question is somewhat informal, however I hope it can be left in such a form. One of the ways to make it formal is to ask about the properties which distinguish the bijection above among all bijections between the two sets. E.g. properties like: identity class is mapped to trivial irreps... Background: It is very standard material that for any finite group the number of conjugacy classes and irreps is the same, however it is striking that no natural bijection between the two sets is known (and may not exist at all). MathOverflow has already several discussions around the subject, however they are different from the present question. A few of them are listed below. Bijection between irreducible representations and conjugacy classes of finite groups discusses bijections for general groups. The symmetric groups are not the only groups with a "natural" bijection. Some examples are collected here: Examples of finite groups with “good” bijection(s) between conjugacy classes and irreducible representations? Related questions: G/[G,G], irreps and conjugacy classes, Duality between conjugacy classes and irreducible characters for finite monoids?, among others. Conjugacy classes and irreps have many similar properties. However, in general, they do not fit each other exactly: Action of Out(G) on both sets, "reality properties" (Are there “real” vs. “quaternionic” conjugacy classes in finite groups?, If all real conjugacy classes are strongly real, then all real irreps are “strongly real”(symmetric), true ?). REPLY [4 votes]: Here is how Frobenius set up the bijection when he computed the irreducible characters of $S_n$, but I don't know whether you would consider this "natural." Let $\phi_\lambda$ denote the action of $S_n$ on the left cosets of the Young subgroup $S_\lambda = S_{\lambda_1}\times S_{\lambda_2}\times\cdots$. Equivalently, $\phi_\lambda$ is the action of $S_n$ by permuting coordinates on all $n$-tuples with $\lambda_i$ $i$'s for all $i\geq 1$. Then there is a unique irreducible character $\chi^\lambda$ appearing in $\phi^\lambda$ (necessarily with multiplicity one) that does not appear in any $\phi_\mu$ with $\mu>\lambda$ (dominance order, as defined in http://en.wikipedia.org/wiki/Dominance_order).<|endoftext|> TITLE: Approximating Markov chains by Brownian motion QUESTION [6 upvotes]: I would like a result along the following lines to be true, but haven't been able to locate it in the literature; pointers would be welcome. Let $X_{t}$ be a finite-state, irreducible, aperiodic Markov chain. Let $f$ be a bounded functional, which is mean-zero with respect to the stationary distribution of $X$. Let $$ S_{t} = \frac{1}{\sqrt{n}} \sum_{s \leq t} f(X_{s}), \quad t \leq n $$ be the partial sum process. Then there exists $\epsilon > 0$, $N_0 > 0$, and a Brownian motion $B_t$ such that, for all $n > N_{0}$, $$ g(n) := \sup_{t \leq n} |B_{t} - S_{t}| \leq n^{1/2 - \epsilon}. $$ (Interpreted either as "$t$ is an integer less than $n$" or as "interpolate $S_{t}$ linearly between integers".) In the case of partial sums of i.i.d. random variables, if I remember correctly, the Skorohod embedding gives $g(n) \sim n^{1/4}\log(n)$, and the KMT approximation gives $g(n) \sim \log(n)$. In the Markovian setting, I'm hoping for $g(n)$ to depend on the mixing time of the Markov chain. It's possible that the correct generality for this is "ergodic sequence" rather than "Markov chain". REPLY [5 votes]: It turns out that to find a whole slew of results of this shape I needed to know that (a) this sort of statement is called "strong invariance principle" more often than "strong approximation"; (b) the result as requested above is the subject of the book by Walter Philipp and William Stout, Almost sure invariance principles for partial sums of weakly dependent random variables, Memoirs of the AMS vol. 2 no. 161, July 1975.<|endoftext|> TITLE: Can R^3 be expressed as a disjoint union of pairwise linked circles? QUESTION [31 upvotes]: We can express $\mathbb{R}^3$ as a disjoint union of circles. There are some constructive ways of doing this, although it's easier to construct them sequentially by transfinite induction, applying the following step to each point $P_{\alpha}$ in the well-ordering of $\mathbb{R}^3$ by the initial ordinal of $2^{\aleph_0}$: Suppose the point $P_{\alpha}$ has not already been covered (otherwise, move on to the next point); Choose a plane passing through that point, such that the direction of the normal vector is distinct from all previous planes; This plane intersects the union of all existing circles in fewer than $2^{\aleph_0}$ points, so we can draw a circle on that plane passing through $P_{\alpha}$ and disjoint from all existing circles. We can also express $\mathbb{R}^3 - \ell$ as a disjoint union of pairwise linked circles, where $\ell$ is an arbitrary line. Specifically, we take the stereographic projection of the Hopf fibration, and remove the `circle' passing through the point at infinity. This suggests that there may be a mutual generalisation of both results, namely that $\mathbb{R}^3$ can be expressed as a disjoint union of pairwise linked circles. Can anyone prove or disprove this conjecture? REPLY [10 votes]: Whilst this doesn't answer the original question, I shall show that it's possible to partition $\mathbb{R}^3 \setminus \{ 0 \}$ into pairwise-linked disjoint circles. I'll begin with a couple of observations: Observation 1: Note that when we apply the transfinite induction to the weaker version of the problem (which doesn't require the circles to be linked), we actually have an extra degree of freedom. For example, it's possible to assert that all circles have unit radius, and the proof still works since, even after choosing a point $P$, a plane $\Pi$ incident with $P$, and a radius $R > 0$, there are $2^{\aleph_0}$ circles of radius $R$ lying on $\Pi$ and passing through $P$. Observation 1 is quite well known and is included in the literature on the subject. We won't use it directly, but the idea of being able to choose the 'sizes' of circles is key to our proof. Observation 2: If we have two disjoint great circles on $S^3$, they must necessarily be Hopf-linked. To see this, note that great circles are intersections of $S^3$ with planes, and disjointness of the circles imply that the corresponding planes are complementary (their direct sum is $\mathbb{R}^4$), in which case we can apply a linear transformation to map them to our 'standard planes' $x = y = 0$ and $w = z = 0$. This is, for example, why all of the circles in the Hopf fibration are linked. So, we would like to be able to express $\mathbb{R}^3 \setminus \{ 0 \}$ as a disjoint union of (stereographic projections of) great circles, since then we are done by Observation 2. This is indeed possible by the usual transfinite induction proof, since the appropriate analogue of Observation 1 still applies (namely that after choosing a point $P \in \mathbb{R}^3 \setminus \{ 0 \}$ and plane $\Pi$ incident with $P$, there are $2^{\aleph_0}$ (stereographic projections of) great circles lying on $\Pi$ and passing through $P$). This proof doesn't work for $\mathbb{R}^3$ since every great circle passing through $0$ is a line and therefore invalid.<|endoftext|> TITLE: A sequence of subsets of an infinite group QUESTION [11 upvotes]: Is there an infinite group $G$ such that there is not any sequence $(A_n)$ of its subsets such that always $$A_n=A_n^{-1}, \quad A_{n+1}A_{n+1}\subsetneqq A_n$$ ? link REPLY [2 votes]: This is a partial answer. There exists an infinite group $G$ having no sequence of subsets $\{A_i\}$ such that $$ A_{n+1}A_{n+1}\subsetneq A_n, \quad\hbox{and}\quad \bigcap A_i=\{1\}. $$ (We omit the inverse condition but add the the intersection-triviality condition.) Indeed, such a sequence defines a Hausdorff topology on the group where $A_i$ are neigbourhoods of $1$ and all other points are isolated. The multiplication is continuous at $(1,1)\in G\times G$ with respect to this topology. It remains to note that infinite locally non-topologizable groups do exist.<|endoftext|> TITLE: Is the homotopy category of a ring also the derived category of another ring? QUESTION [6 upvotes]: Let $R$ be an associative ring. Let $K(R)$ be the category of chain complexes of $R$-modules and chain homotopy classes of maps between them, and let $D(R)$ be its localization with respect to acyclic complexes. Is there a ring $S$ such that $K(R) = D(S)$? Is the question more likely to have a positive answer if I allow $S$ to be a differentially graded ring, or don't require that $S$ has an identity? REPLY [6 votes]: As Eric and Karol have noted it is usually not the case that there exists such an $S$ (which I'll take to be a dga; I don't know what happens off the top of my head if one asks for $S$ to be an honest ring). Indeed for $K(R) \cong D(S)$ one needs $K(R)$ to be compactly generated. But by a result of Stovicek (see Theorem 2.5 here) in order for $K(R)$ to even be well generated it is necessary that $\mathrm{Mod}\;R$ be pure semisimple i.e., every $R$-module is a direct sum of finitely presented modules. In fact this is also sufficient: $K(R)$ is well generated if and only if $R$ is right pure semisimple. This allows one to write it as a localisation of the derived category of a small dg-category (so a non-unital dga if one prefers). I don't know exactly when $K(R)$ is moreover compactly generated though.<|endoftext|> TITLE: Conjecture on irrational algebraic numbers QUESTION [6 upvotes]: Conjecture: For every irrational algebraic number $q$ and natural number $b$, the representation of $q$ on base $b$ contains all the digits $[0,\dots,b-1]$. Questions: Has this conjecture been proved, refuted or neither? If proved: Is there an estimate of the minimum length of $q_b$ containing all the digits? For example, I would expect something like $2b$ or $b^2$ for any given $q_b$. If not refuted: I suppose that it is not true for transcendental numbers. Is that correct? How can we construct a transcendental number $q_b$ which does not contain all the digits? Thanks REPLY [14 votes]: The conjecture has been neither refuted nor proved. The state of the art, as far as I know, is contained in the papers of Adamczewski and Bugeaud, in which they show that anything with a very low complexity decimal expansion cannot be an algebraic irrational. The complexity is the function $c_x(n)$ giving the number of blocks of length $n$ in the decimal expansion of $x$ (or any base). They show that if there exists a $k$ such that $c_x(n)\le kn$ for all $n$, then $x$ is either rational or transcendental. Of course, it's conjectured that $c_x(n)=10^n$ for all algebraic irrationals $x$. Your condition would be implied by the conjecture $c_x(n)>9^n$ for all algebraic irrationals $x$.<|endoftext|> TITLE: What does Freed-Hopkins-Teleman say about finite groups? QUESTION [5 upvotes]: The third in the "Loop groups and twisted K-theory" series by FHT treats compact Lie groups without any connectedness assumptions. I am trying to unwind what Theorem 2 of that paper (available here, page 5) says for a finite group. Theorem 2: "For regular $\tau$, there is a natural isomorphism $R^\tau(LG_s) \cong K_G^\tau(G)$ ..." For $G$ finite, I think that the regularity condition (page 7) is empty, as it has something to do with a maximal torus of $G$. So we can take $\tau = 0$. On the right-hand side, that means untwisted $K$-theory, which is naturally identified with $\bigoplus_g R'(Z_G(g))$, where $R'$ denotes the Grothendieck group of finite-dimensional complex representations, and $g$ runs over conjugacy class representatives. ($R'$ to avoid conflict with $R^\tau$, whose meaning I don't understand.) What is on the left-hand side, when $\tau$ is zero and $G$ is finite? It is a Grothendieck group of representations of the group of free loops in $G$, modified somehow by $\tau$ ($\tau = 0$, so not modified at all). But I don't understand what it means. For finite $G$, loops are constant and the Lie algebra is trivial, so from the definitions on page 5, $LG = LG_s = G$. But it can't simply be the Grothendieck group of complex representations of $G$, that is much smaller than the right-hand side. REPLY [5 votes]: This is maybe more of a comment. Some time ago there was a paper by Simon Willerton titled "The twisted Drinfeld double of a finite group via gerbes and finite groupoids" which tried to make sense of the statement of Freed-Hopkins-Teleman in the case of finite groups (i.e. Dijkgraaf-Witten theory).<|endoftext|> TITLE: Good ways to organize old personal mathematical resources QUESTION [27 upvotes]: I am wondering how the other Mathematicians organize their old mathematical resources, like calculation drafts, class and seminar notes etc. These old resources may be related to a wide range of areas, and if organized effectively, may be useful for future learning and research. I myself have a huge stack of old materials, and I scanned and classified some of them lately. Not sure if there is a more efficient way...... REPLY [7 votes]: Thanks to a blog post by Tim Gowers (https://gowers.wordpress.com/2013/10/24/what-i-did-in-my-summer-holidays/), I discovered TiddlyWiki (http://tiddlywiki.com/). So far, I've only used it to capture odd jottings, such as notes about references/information sources, ideas to try out, and so forth. I can see that if I devoted more effort, I could make it much more effective. In particular, the ability to tag and cross-link items in various ways is very useful.<|endoftext|> TITLE: Does forcing with recursively pointed perfect trees add a Turing degree that is minimal over $V$? QUESTION [8 upvotes]: A tree $T$ on $\omega$ is recursively pointed if it is recursive in each of its branches. We can consider a variant of Sacks forcing where the conditions are recursively pointed perfect trees ordered by inclusion. Given a $V$-generic filter $G$ for this forcing, we can define the real $x_G = \bigcap_{T\in G} [T]$. For every real $x \in V$ we have $x <_\text{T} x_G$. Given a real $x \in V[G]$ with $x <_\text{T} x_G$, must we have $x \in V$? Probably the answer to this question is well-known, but I'm afraid I don't see how the proof of the minimality property for Sacks forcing (as given by Jech) might adapt to recursively pointed trees. REPLY [7 votes]: If you two haven't worked this out already, it seems to me that $x_G$ is not minimal as a Turing degree above $V$, but is a minimal $V$ degree. For the first, since $x_G$ computes $0'$, it follows that $x_G$ is the join of two mutually 1-generic reals, neither of which computes $x_G$ and at least one of which is not in $V$. However, if $x_G$ computes $x$ and $x$ is not in $V$, then there is an $a$ in $V$ such that $x\oplus a$ computes $x_G$. This is by the usual minimality argument: $a$ is the degree of the splitting tree for the functional used to compute $x$ from $x_G$.<|endoftext|> TITLE: Examples of fundamental domains QUESTION [7 upvotes]: Everyone knows that it's difficult to compute a general fundamental domain for an arithmetic group but are there any specific examples where such domains have been calculated? I'm mostly interested in results related to orthogonal groups of lattices of type O(2, n). I've only really been able to track down two such results: Vinberg's algorithm computes a fundamental domain for reflective groups of type O(1,n); and Gottschling was able to compute a Minkowski reduced domain for the integral symplectic group Sp(4, Z), which can be considered of type O(2, 3). Would the situation improve if one were to look at a restricted problem, and study fundamental domains of groups generated by p-torsion, for example? REPLY [6 votes]: Here are some more examples of fundamental domains for arithmetic groups, although not of the type $O(2,n)$ you are most interested in. A situation in which quite something is known on fundamental domains are the Bianchi groups, i.e., the groups $PSL_2(\mathcal{O}_{\mathbb{Q}(\sqrt{-d})})$ for $d$ a positive square-free integer. These arithmetic groups act on hyperbolic space $\mathbb{H}^3$ (and therefore can be considered lattices in $SO^+(1,n)$). The first examples of fundamental domains were computed by Bianchi in 1892, whence the name. The subsequently developed technique used to compute fundamental domains in the cases of the Bianchi groups was mostly to provide an equivariant retract of hyperbolic 3-space to a 2-dimensional subcomplex on which the group acts cocompactly: D. Flöge: Zur Struktur der $PSL_2$ über einigen imaginär-quadratischen Zahlringen. Math. Z. 183 (1983) 255-279 E.R. Mendoza: Cohomology of $PGL_2$ over imaginary quadratic integers. Bonner Mathematische Schriften, 128. Computations of fundamental domains for Bianchi groups were also the basis for group cohomology computations of Bianchi groups done in J. Schwermer and K. Vogtmann: The integral homology of $SL_2$ and $PSL_2$ of euclidean imaginary quadratic integers. Comment. Math. Helv. 58 (1983) 573-598. Several algorithms have been described for computing the fundamental domain for Bianchi groups. You can have a look at the thesis of A. Rahm in which an algorithm is described which is also implemented using Pari/GP. This allows to do a couple of computations of fundamental domains for Bianchi groups: A. Rahm and M. Fuchs. JPAA 215 (2011) 1443-1472 A. Rahm. LMS Journal of Computation and Mathematics 16 (2013), 344-365. Note that only finitely many of the Bianchi groups are generated by reflections (M. Belolipetsky and J. McLeod: Reflective and quasi-reflective Bianchi groups. arXiv:1210.2759), and so Vinberg's algorithm does not apply to most Bianchi groups. The technique of retraction onto a nice subcomplex on which the group acts cocompactly has also been used by Soulé (C. Soulé: The cohomology of $SL_3(\mathbb{Z})$. Topology 17 (1978) 1-22). This technique has subsequently been improved so that today it is possible to compute fundamental domains for $SL_n(\mathbb{Z})$ for $n\leq 8$, see the works of Elbaz-Vincent, Gangl, Soulé and others. For the symplectic group $Sp_4(\mathbb{Z})$, you may also be interested in the paper A. Brownstein and R. Lee: Cohomology of the symplectic group $Sp_4(\mathbb{Z})$ I: the odd torsion case. Transactions AMS 334 (1992) 575-596. They do not compute a fundamental domain, but the relation between the $Sp_4(\mathbb{Z})$-quotient and the moduli space of genus $2$ surfaces might be interesting anyway. In a similar direction, for Hilbert modular groups, i.e., $SL_2(\mathcal{O}_{\mathbb{Q}(\sqrt{d})})$ for $d$ a positive square-free integer, a lot is known about the quotients $SL_2(\mathcal{O}_{\mathbb{Q}(\sqrt{d})})\backslash\mathbb{H}^2\times\mathbb{H}^2$ because these quotients have the structure of complex surfaces. There are several papers of Hirzebruch about this topic. This does not give precise information about the fundamental domain inside $\mathbb{H}^2\times\mathbb{H}^2$, but it does provide some information on the group action, the quotient, group cohomology etc. Of course, this can be generalized to other types of groups, but I know nothing about Shimura-varieties, so I cannot say anything about that...<|endoftext|> TITLE: Non projective hyperbolic compact complex space QUESTION [10 upvotes]: A famous conjecture by Kobayashi (perhaps slightly revisited subsequently) states that every compact hyperbolic Kähler manifold $X$ has ample canonical bundle. This implies in particular that $X$ is projective (and canonically polarized...). Here is my innocent-sounding question: is there any example of hyperbolic compact complex manifold which is not projective nor Kähler? What about if one allows singularities? In other words, is the Kähler assumption necessary in the above-mentioned conjecture? Thanks a lot in advance! REPLY [6 votes]: The non-projective Kahler surfaces are either K3, tori or elliptic surfaces, known to be non-hyperbolic. Non-Kahler surfaces are either elliptic or class VII; the elliptic surfaces are obviously non-hyperbolic. Minimal class VII surfaces with $b_2=0$ are either Hopf (obviously non-hyperbolic) or Inoue, by Bogomolov's theorem; the latter admit a holomorphic foliation with $\Bbb C$-fibers, hence non-hyperbolic. Minimal class VII surfaces with $b_2>0$ are conjecturally all Kato surfaces. The Kato surfaces (by Dlousky's theorem, I think) are obtained as limits of blown-up Hops surfaces, hence non-hyperbolic. For non-Kato surfaces, nothing is known, but they don't exist at least for $b_2=1$ by Teleman's result. I would expect that they are also non-hyperbolic, even if they exist (though everybody I know believes that non-Kato class VII${}_0$ surfaces don't exist, me included; much evidence supporting this view was collected by Teleman, Dlousky, Oeljeklaus, Toma and others).<|endoftext|> TITLE: A generalization of the Arhangelskii Theorem QUESTION [13 upvotes]: Arhangeleskii's Theorem states the following For any Hausdorff topological space $X$, $$ |X|\leq2^{\chi(X)L(X)} $$ where $\chi(X)$ is the character of $X$ and $L(X)$ is the Lindelöf degree of $X$. In the article Paracompactness of Box Products of Compact Spaces, Kenneth Kunen presents a generalization of the countable case of the previous Theorem, also by Arhangelskii. It says If $Y$ is a compact Hausdorff space and $\mathfrak{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathfrak{F}$ satisfies $$ \forall H\in \mathfrak F (|\{K\in\mathfrak F: H\cap K\neq\emptyset\}|\leq \mathfrak c), $$ then $|\mathfrak F|\leq \mathfrak c$ I would like to prove this result. Kunen says only that the proof to this Theorem is an easy modification of a proof by R. Pol of the original (on the article Short proofs of two theorems on cardinality of topological spaces). However, the steps of the original proof don't translate well to this generalization. When I try to emulate Pol's proof, I can't take the closure of each step, because that why I couldn't ensure that I take less than $\mathfrak c$ elements of $\mathfrak F$. When I tried to use something other than the closure in that step, I can't guarantee that after the induction I have closed sets. I wanted those set to be closed in order to use the compactness of the space. Maybe there is some other way to ensure the compactness, but I don't have any clue of what that could be. Does the theorem lack some extra hypothesis? There is a proof supposing that $X$ has countable cellularity, but this assumption does not help me. There is a similar result in the article Box Products, by Scott Williams, in the Handbook of Set Theoretic Topology (Lemma 4.1 on page 184). No compact Hausdorff space can be partitioned into more than $\mathfrak c$ closed $G_\delta$ sets. This result would be enough for me, but the proof in the article has a flaw. During the induction step, a function is defined in order to guarantee that each step does not exceed $\mathfrak c$ closed $G_\delta$ sets. However, that function is not always injective. I think that this can be related to the troubles I have with the other version. REPLY [4 votes]: I'll prove the theorem claimed by Kunen, assuming basic knowledge about elementary sub-models. If $(Y, \tau)$ is a compact Hausdorff space and $\mathcal{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathcal{F}$ satisfies: for every $H \in \mathcal{F}$ the set $\{K \in \mathcal{F}: K \cap H \neq \emptyset \}$ has cardinality at most continuum then $|\mathcal{F}| \leq 2^\omega$. Let $\theta$ be a large enough regular cardinal and $M \prec H(\theta)$ be a countably closed elementary submodel of $H(\theta)$ of cardinality $\mathfrak{c}$ such that $\mathfrak{c} \subset M$ and $\{Y, \tau, \mathcal{F}, \mathfrak{c}\} \subset M$. We claim that $\mathcal{F} \subset M$. Suppose this is not the case and let $K \in \mathcal{F} \setminus M$. Using compactness and regularity of $Y$, we can write every closed $G_\delta$ set $F \subset Y$ in the form $F=\bigcap \{\overline{V_n(F)}: n < \omega \}$, for a suitable decreasing family of open neighbourhoods $\{V_n(F): n < \omega \}$ of $F$. If $F \in \mathcal{F} \cap M$, then the family $\{V_n(F): n < \omega \}$ may be fixed in $M$, but then we actually have $\{V_n(F): n < \omega \} \subset M$. Note now that for every $x \in \overline{Y \cap M}$, we can find $F \in \mathcal{F} \cap M$ such that $x \in F$. Indeed, let $x \in \overline{Y \cap M} \setminus Y \cap M$. Then there is $F \in \mathcal{F}$ such that $x \in F$. Let $x_n$ be any point of $U_n(F) \cap Y \cap M$. Using compactness of $F$ and the fact that $F=\bigcap \{\overline{U_n(F)}: n <\omega \}$ we can easily see that the sequence $\{x_n: n < \omega \}$ converges to $F$, in the sense that, every open neighbourhood of $F$ contains a final segment of $\{x_n: n < \omega \}$. Now by $\omega$-closedness of $M$, we have that $\{x_i: i \geq n \} \in M$, for every $n<\omega$ and hence the set: $$\mathcal{S}=\{K \in \mathcal{F}: \{x_n: n <\omega \} \rightarrow K \}$$ is an element of $M$ (the formula defining that set has all free variables in $M$). Note that since every compact Hausdorff space is normal, $F$ can be separated from a closed set disjoint from it by a pair of disjoint open sets. Therefore, we have that if $G \in \mathcal{S}$ then $G \cap F \neq \emptyset$. Hence $|\mathcal{S}| \leq 2^{\aleph_0}$ by assumption. But then we actually have that $\mathcal{S} \subset M$ and hence $F \in M$, as we wanted. For every $F \in \mathcal{F} \cap M$ we have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \in M$, and, since $|\{H \in \mathcal{F} : F \cap H \neq \emptyset \}| \leq \mathfrak{c}$, we also have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \subset M$. So, for every $F \in \mathcal{F} \cap M$, we have $F \cap K=\emptyset$. Now, by compactness of $K$, we can find for every $F \in \mathcal{F} \cap M$, an integer $n <\omega$ such that $\overline{U_n(F)} \cap K=\emptyset$. Thus for every $x \in \overline{Y \cap M}$, we can find an open neighbourhood $U_x \in M$ of $x$ such that $U_x \cap K=\emptyset$. Now $\mathcal{U}=\{U_x: x \in \overline{Y \cap M} \}$ is an open cover of the compact space $\overline{Y \cap M}$ such that $\mathcal{U} \subset M$. So there is a finite set $\mathcal{G} \subset \mathcal{U}$ such that $\overline{Y \cap M} \subset \bigcup \mathcal{G}$. This implies that $M \models Y \subset \bigcup \mathcal{G}$ and hence $H(\theta) \models Y \subset \bigcup \mathcal{G}$. But this contradicts $K \cap \bigcup \mathcal{G}=\emptyset$. So $\mathcal{F} \subset M$ and hence $|\mathcal{F}| \leq |M| \leq \mathfrak{c}$.<|endoftext|> TITLE: Generalizations of de Franchis and function field Mordell QUESTION [6 upvotes]: The classical de Franchis theorem, as generalized by S. Kobayashi and T. Ochiai ("Meromorphic mappings onto compact complex spaces of general type," Inventiones, 1975), states that if $X$ is a complex projective varity of general type, then for any fixed complex projective variety $Y$, there are only finitely many surjective mappings $Y \to X$. (A variant of this for Kobayashi-hyperbolic complex projective manifolds $X$ was conjectured by Lang and proved by C. Horst; those manifolds are expected to be of general type anyway, but if I understand correctly, this expectation has not been proved in full generality.) I would like to ask here about the generalization of this problem whereby we drop the surjectivity assumption, and assume instead that $Y \to X$ is generically finite onto its image. To keep things simple, consider $\dim{Y} = 1$ and $\dim{X} = 2$, which is the first interesting case: Question. Let $X$ be a surface of general type and $Y$ a fixed curve. Assume that there is no curve $Z$ such that $X$ is rationally dominated by $Y \times Z$. Does $X$ contain only finitely many images of non-constant morphisms $Y \to X$? [Is it plausible that the manifestly necessary assumption on $X$ should be sufficient?] This would be a generalization of the function field Mordell conjecture. [To see this, let $X \to B$ be a fibred surface of genus $> 1$, and choose any branched covering $Y \to B$ of genus $> 1$. Then $X \times_B Y$ is a surface of general type, and if it is domianted by a product $Y \times Z$, then $X/B$ is isotrivial. ] It would likewise generalize the theorem of Miyaoka and Lu (itself a very particular case of Lang's geometric conjecture), according to which a surface of general type contains only finitely many smooth rational curves. I am aware of a 1985 paper by Y. Imayoshi ("Holomorphic maps of compact Riemann surfaces into $2$-dimensional compact $C$-hyperbolic manifolds," Math. Ann. 270), which solves the analogous question for a compact complex surface $X$ which admits a covering $\Omega \to X$ such that the bounded holomorphic functions on $\Omega$ separate points (i.e., the Caratheodory pseudo-distance is a distance on $\Omega$; such $X$ are a fortiori Kobayashi-hyperbolic.) In that case the conclusion is in fact the finiteness of the set of non-constant holomorhic maps from a fixed compact hyperbolic Riemann surface, and it is enough to assume that $X$ does not have a finite etale cover by the product of two compact hyperbolic Riemann surfaces. But presumably, $C$-hyperbolicity is a very restrictive constraint on a surface of general type. Have the above algebraic statement or similar strengthenings of function field Mordell been considered, or at least explicitly stated anywhere? REPLY [5 votes]: One of the major results in this direction, when $Y$ is either a rational or an elliptic curve, is due to Bogomolov, see F. Bogomolov, Families of curves on a surface of general type, Doklady AN SSSR 236 (1977), no. 5, 1041-1044, in Russian; English translation: Soviet Math. Dokl. 18 (1977), 1294-1277. The statement of Bogomolov's result is as follows: Theorem (Bogomolov). Let $X$ be a surface of general type with $c_1^2(X) > c_2(X)$. Then, for any $g$, the curves of geometric genus $g$ on $X$ form a bounded family (roughly speaking, a family having only a finite number of irreducible components). Since $X$ is of general type, it is not covered by rational or elliptic curves, i.e. curves of geometric genus $0$ and $1$ cannot deform. Then Bogomolov's theorem implies that there are only a finite number of rational or elliptic curves on $X$, that is we obtain the following Corollary. Let $X$ be a surface of general type with $c_1^2(X) > c_2(X)$, and let $Y$ be either a rational curve or an elliptic curve. Then there are only finitely many images of non-constant morphisms $ Y \to X$. The condition $c_1^2(X) > c_2(X)$ is rather restrictive, for instance it is never satisfied for a hypersurface $X \subset \mathbb{P}^3$. It is expected that in any case neither rational nor elliptic curves can be dense on a variety of general type $X$ (I think this is a conjecture due to Kobayashi), but in principle one could have surfaces of general type with $c_1^2(X) \leq c_2(X)$ and containing countably many rational or elliptic curves. However, I'm not aware of any example of this type.<|endoftext|> TITLE: Union of conjugates of a subgroup QUESTION [6 upvotes]: Let $G$ be a finite group, $H \leq G$ a proper subgroup. It is well known that the union of the conjugates of $H$ does not cover $G$. I would like to know of more precise results (even in special cases) saying that the union of conjugates misses some structure. For example, if $G$ is a Frobenius group, the complement of the union of conjugates is a subgroup (when $1$ is added). Is there any generalization of it? Maybe one could hope to find a submonoid (in the complement) in a more general case? I would like to be able to deduce that this union misses something more specific than just some element It is possible that there is something we can say if $G$ is a $p$-group. REPLY [3 votes]: I'm not sure whether it is really relevant, but there is a generalization by Wielandt of Frobenius's theorem which may be interpreted as follows: Let $G$ be a transitive permutation group on a set $\Omega$ , and let $G_{\alpha}$ be a point stabilizer. Suppose that $G_{\alpha} \neq \langle G_{\alpha} \cap G_{\beta} : \beta \neq \alpha \rangle.$ Then there is a proper normal subgroup of $G$ which is still transitive on $\Omega.$ This is a permutation group theoretic consequence of Wielandt's generalization of Frobenius's theorem : if $H$ is a subgroup of a finite group $G$ and there is $K \lhd H$ with $H \cap H^{g} \leq K$ for all $g \in G \backslash H,$ then there is $L \lhd G$ with $G = HL$ and $L \cap H = K.$<|endoftext|> TITLE: solvable word problem without algorithm QUESTION [5 upvotes]: Let $G$ be a finitely generated group. I wonder if there are examples where: 1) The word problem is known to be solvable in $G$ but there is no algorithm known. 2) The word problem is known to be solvable in $G$ but it is also known that no algorithm for solving the problem can be exhibite. 3) the same as 1) and 2) but with other decisional problems. $\ $ Is 2) really possible? The question relies on the difference between "$\exists x$" and "showing an $x$". It seems to me that if $G$ has solvable problem then, by definition, an algorithm $A$ that solves the problem exists. Since algorithms are build up from finite objects, in principle I can enumerate all of them and eventually find $A$ (but how can I be sure that $A$ solves the word problem for $G$?). Am I making a big confusion or the question makes sense? REPLY [8 votes]: The technique for constructing groups with unsolvable word problems applies more generally to construct groups that "simulate'' Turing machines. So, if a Turing machine halts for a recursive set of inputs, it can be arranged that the corresponding group will have a solvable word problem. However, the algorithm will depend on the set of inputs for which the machine halts. This reduces question 1) to finding a Turing machine which halts for a recursive set of inputs, but for which the set itself is not yet known. You could take, for example, a machine that halts on input 0 if there is an odd perfect number, and does not halt at all if not.<|endoftext|> TITLE: Bases for spaces of smooth functions QUESTION [16 upvotes]: Let $S$ denote the space of rapidly decreasing sequences, which means sequences $a=(a_k)_{k=1}^\infty$ such that the numbers $p_d(a)=\sup\{k^d|a_k| : 1\leq k<\infty\}$ are finite for all $d\in\mathbb{N}$. We give this space the topology generated by the family of seminorms $p_d$. Now let $M$ be a compact smooth closed manifold, and consider the space $C^\infty(M)$. For any differential operator $L:C^\infty(M)\to C^\infty(M)$ (of any nonnegative order) we have a seminorm $p_L(f)=\|Lf\|_\infty$, and we give $C^\infty(M)$ the topology determined by this family of seminorms. By a basis for $C^\infty(M)$ I mean a sequence of functions $f_k$ such that the rule $a\mapsto\sum_ka_kf_k$ gives an isomorphism $S\to C^\infty(M)$ of topological vector spaces. I think it is known that $C^\infty(M)$ always has a basis. Is this true, and if so, what is a good reference? I think I have seen it in the literature, but I cannot find it right now. Is there a reasonably effective criterion to check whether a given sequence is a basis? Suppose that $M$ is a subspace of $\mathbb{R}^m$ defined by polynomial equations (and is still compact and smooth). Is there an effective way to find a basis consisting of polynomial functions? In particular, can I just use a Gröbner basis with respect to degree-lexicographic order? Here is a little background, partly taken from some notes of Dietmar Vogt: http://www2.math.uni-wuppertal.de/~vogt/vorlesungen/fs.pdf When $M=S^1$ we can just take $f_{2k+1}(\cos(\theta),\sin(\theta))=\cos(k\theta)$ and $f_{2k}(\cos(\theta),\sin(\theta))=\sin(k\theta)$. This gives an isomorphism $S\to C^\infty(S^1)$, and of course we can precompose this with any of the many automorphisms of $S$, so $C^\infty(S^1)$ has many different bases. If $(f_j)$ is a basis for $C^\infty(M)$ and $(g_k)$ is a basis for $C^\infty(N)$ then the functions $h_{jk}(x,y)=f_j(x)g_k(y)$, enumerated in a suitable order, will give a basis for $C^\infty(M\times N)$. If $V$ is any nuclear Frechet space, then a theorem of Komura and Komura shows that $V$ is isomorphic to a subspace of $S^{\mathbb{N}}$. I do not understand all the issues here, but it seems like there is not too much difference between $S$, $S^{\mathbb{N}}$ and subspaces of $S^{\mathbb{N}}$. It is certainly known that $C^\infty(M)$ is always a Frechet space. Vogt's notes show that when $U$ is a nonempty open subset of $\mathbb{R}^m$, the space $C^\infty(U)$ is isomorphic to $S^{\mathbb{N}}$. If we choose $U$ to be a tubular neighbourhood of an embedded copy of $M$, then $M$ will be a retract of $U$ and so $C^\infty(M)$ will be isomorphic to a summand in $C^\infty(U)$, and thus to a summand in $S^{\mathbb{N}}$. REPLY [16 votes]: The paper MR0688001 Reviewed Vogt, Dietmar Sequence space representations of spaces of test functions and distributions. Functional analysis, holomorphy, and approximation theory (Rio de Janeiro, 1979), pp. 405–443, Lecture Notes in Pure and Appl. Math., 83, Dekker, New York, 1983. (Reviewer: M. Valdivia) shows that for a compact manifold $M$, the Frechet space $C^\infty(M)$ is always linearly isomorphic to the space $\mathcal s$ of rapidly deceasing sequences. Added later: If you consider the $L^2$ orthonormal basis $\phi_k$ for the Laplacian of a Riemannian metric on $M$ as suggested by Liviu Nicolaescu, then any $f\in C^\infty(M)$ is of the form $f=\sum_k f_k\phi_k$, and $f_k\in \ell^2$ initially. But $1+\Delta$ (geometric $\Delta$ here) is an isomorphism between the Sobolev spaces $H^k(M)$ and $H^{k-2}(M)$ for each $k$. By Weyl's formula the eigenvalues $\lambda_k$ of $\Delta$ satisfy $\lambda_k \sim C k^{2/\dim(M)}$ for $k\to \infty$; see page 155 of the book of Chavel, `Eigenvalues in Riemannian geometry'. Since $\bigcap_k H^k(M)=C^\infty(M)$ on a compact manifold, we see that $$(1+\Delta)^m f = \sum_k f_k(1+ \lambda_k)^{m}\phi_k$$ with coefficients again in $\ell^2$, for each $m$. Thus the coefficients $f_k(1+Ck^{2/\dim(M)})^m\in \ell^2$ for each $m$, and the $f_k$ are rapidly decreasing. Moreover, any rapidly decreasing sequence of coefficients gives a function in $C^\infty(M)$. This proves again that $C^\infty(M)\cong \mathcal s$, even with the basis of eigenfunctions for any Laplacian.<|endoftext|> TITLE: Higman's Criterion QUESTION [6 upvotes]: Higman's Criterion is a fundamental result in the representation theory of a finite group $G$ over a field $k$ of characteristic $p>0$. One version is the following (I hope the notation is standard enough): Let $M$ be a $kG$-module and let $H$ be a subgroup of $G$. Then $M$ is a direct summand of ${\rm Ind}_H^G\,{\rm Res}_H^G\, M$ if and only if there is a $kH$-endomorphism $\psi$ of $M$ such that ${\rm tr}_H^G(\psi)=1_M$. Question: suppose that $M$ has a $kH$-endomorphism $\psi$ such that ${\rm tr}_H^G(\psi)=1_M$. Does $\psi M$ contain a $kH$-submodule $L$ such that $L$ is a direct summand of ${\rm Res}_H^G\,M$ and $M$ is a direct summand of ${\rm Ind}_H^G\,L$? Motivation: Notice that if $L$ is a direct summand of ${\rm Res}_H^G\,M$ and if $N$ is a submodule of ${\rm Res}_H^G\,M$ containing $L$, then $L$ is a direct summand of $N$. As a consequence, if $N$ is a submodule of ${\rm Res}_H^G\,M$ which contains $\psi M$, then $M$ is a direct summand of ${\rm Ind}_H^G\,N$. The standard argument proving the `if' implication in Higman's Criterion appears to show this consequence directly. This seems to me to be a mysterious property of $\psi M$, at least if my Question has a negative answer. Going back to generalities, let $M$ be a $kG$-module and suppose that $M$ is a direct summand of ${\rm Ind}_H^G\,N$, where $N$ is a $kH$-module. One version of Higman's Theorem produces a $kH$-endomorphism $\phi$ of $M$ such that $t_H^G(\phi)=1_M$. This $\phi$ `factors through' $N$ in the sense that it is a composition $\phi=\phi_2\circ\phi_1$ where $\phi_1:{\rm Res}_H^G\,M\rightarrow N$ and $\phi_2:N\rightarrow{\rm Res}_H^G\,M$. Set $L=\phi M$. Then $L$ is both a submodule and quotient module of ${\rm Res}_H^G\,M$ and a subquotient ${\rm Im}(\phi_1)/{\rm Im}(\phi_1)\cap{\rm ker}(\phi_2)$ of $N$ and $M\mid{\rm Ind}_H^G\,L$. REPLY [2 votes]: This has the rather easy answer: No! Let $G$ be a finite group of order divisible by $p$ that has a $p$-block $B$ of defect $0$ and let $M$ be the unique simple $kG$-module in $B$. Take $H$ to be any nontrivial $p$-subgroup of $G$. Then $k_H$ is a submodule of Res$_H^G(M)$, as $k_H$ is the unique simple $kH$-module. So $M$ is a submodule of Ind$_H^G(k_H)$, by Frobenius reciprocity. But $M$ is a projective $kG$-module. So $M$ is a direct summand of Ind$_H^G(k_H)$. Now by the last paragraph in my posted question there is a $kH$-homomorphism $\phi$ of $M$ such that t$_H^G(\phi)=1_M$ and $\phi$ factors through $k_H$. So $\phi M\cong k_H$. But Res$_H^G(M)$ is a projective $kH$-module and $k_H$ is a non-projective $kH$-module (as $p$ divides $|H|$). So $\phi M$ has no submodule that is a direct summand of Res$_H^G(M)$.<|endoftext|> TITLE: A differentiable one-parameter family of codimension 2 subspaces of $\mathbb{C}^n$ cannot fill $\mathbb{C}^n$, right? QUESTION [9 upvotes]: Suppose that $P(t)$ is a one-parameter family of rank 2 self-adjoint projections on $\mathbb{C}^n$ that vary analytically in the real parameter $t \in [0,1]$. I claim that there must exist a vector $x \in \mathbb{C}^n$ such that $P(t)x \neq 0$ for all $t$. In other words, I am hoping to prove that $\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t) \neq \mathbb{C}^n$. This seems like it is essentially a space-filling curve type argument, and hence the requirement that $P(t)$ be differentiable is probably important (in my example, $P(t)$ is analytic in $t$). Does anyone know a reference that would provide a simple proof of this claim? REPLY [5 votes]: Let $Q(t):\mathbb C^n\to\mathbb C^n$ be the orthogonal projection onto $\text{ker}(P(t))$. Then $t\mapsto Q(t)$ is as differentiable as $P$ was. Now the mapping $(t,x)\mapsto Q(t)(x)$ has rank at most $2n-2+1<2n$. Apply Sard's theorem: The set of regular values is Lebesgue nearly everything. Take a regular value $y$. It cannot be in the image, and thus is in no kernel of $P(t)$ for any $t$.<|endoftext|> TITLE: Union of conjugates of a closed subgroup of a compact group QUESTION [6 upvotes]: Let $G$ be a compact Hausdorff group, $H \leq G$ a closed subgroup of infinite index in $G$. Is it possible that the conjugates of $H$ cover some open neighbourhood of $1$ in $G$ (or the whole of $G$)? If this is possible, I would like to know whether there are conditions on $G$ rendering this impossible. REPLY [11 votes]: The usual example is $G = {\rm SU}_2({\bf C})$ and $H$ the diagonal subgroup. Every unitary matrix is diagonalizable, and thus contained in a conjugate of $H$. REPLY [4 votes]: If your group $G$ is profinite and if $H$ is a proper closed subgroup, then $G$ cannot be a union of conjugates of $H$. This is because there must be a finite image $G_0$ of $G$ in which the image $H_0$ of $H$ is proper. Since the conjugates of $H_0$ cannot cover $G_0$ by the well-known result for finite groups, the conjugates of $H$ cannot cover $G$.<|endoftext|> TITLE: How much does the absolute value of an operator behave like an absolute value? QUESTION [14 upvotes]: Recall that the absolute value of a bounded operator $T$ on a Hilbert space $H$ is the unique positive operator $|T|$ such that $$\||T|x\|=\|Tx\|$$ for all $x\in H$. It can be defined using the continuous functional calculus, or if you have square-roots of positive operators in hand, by $|T|=(T^*T)^{1/2}$. Likewise, this can be defined in any C*-algebra. With respect to the usual ordering on self-adjoint elements, i.e., $S\leq T$ if $T-S$ is positive, does this absolute value behave like an absolute value? For instance, does it satisfy a triangle inequality like the one below? $$||T|-|S||\leq|T-S|\leq|T|+|S|$$ If $S$ and $T$ are normal and commute, then perhaps one could use the functional calculus on the commutative C*-algebra they generate to show this, but not for general operators. How about other inequalities with respect to this ordering? REPLY [3 votes]: While the above inequalities may not be valid for all operators, it is true that if and T are normal and commute, then $|T-S| \leq|T|+|S|$, see Corollary 3.6 in Mortad - on the absolute value of the product and the sum of linear operators, which also proves other related inequalities<|endoftext|> TITLE: The conjecture of Montgomery and Soundararajan on primes in short intervals: Empirical inconsistencies? QUESTION [18 upvotes]: Assume that $y/ \log x \rightarrow \infty$ and that $y/x \rightarrow 0$. Then, from a conjecture by Montgomery and Soundararajan, we expect the number of primes in the interval $[x,x+y]$ to be normally distributed with mean $y/\log x$ and standard deviation $\sqrt{y(\log x/y)/(\log x)^2}$. But numerical testing produces a slight and systematic deviation from this conjecture. Specifically, I have tested for different interval lengths $y=x^c$. For each value of $c$, I calculated $\pi(x+y)-\pi(x)$, the number of primes in $[x,x+y]$, for $N$ non-overlapping intervals. Then I normalized the data by subtracting the mean and dividing by the standard deviation corresponding to each interval, as stated by the conjecture. We should therefore expect the resulting $N$ samples to be normally distributed with mean 0 and standard deviation $\sigma=1$. However, what I get is the following: \begin{matrix} c& \sigma& N\\ \hline 0.20 & 0.967 & 240000\\ 0.25 & 0.966 & 240000\\ 0.30 & 0.965 & 240000\\ 0.40 & 0.958 & 240000\\ 0.50 & 0.947 & 240000\\ 0.55 & 0.917 & 40000\\ 0.60 & 0.899 & 20000\\ 0.65 & 0.891 & 10000\\ 0.70 & 0.889 & 5000\\ \end{matrix} While each data set indeed are normally distributed, what appears to be happening is that the standard deviation decreases with increasing $c$, as compared to the conjecture by Montgomery and Soundararajan. The numerical results seem to be reasonably consistent, so I don't think they are an artifact of sampling only a finite number of intervals (however, please point out if I did any apparent mistakes in the above). I am not fluent enough in the theory to walk through Montgomery and Soundararajan's arguments myself, so I would greatly appreciate any comments on or explanation of this finding. EDIT Including the lower order term in Montgomery and Soundararajan's conjecture, as suggested in the answer below by Lucia, we have the following revised numerics: \begin{matrix} c& \sigma& N\\ \hline 0.20 & 1.000 & 240000\\ 0.25 & 1.001 & 240000\\ 0.30 & 1.003 & 240000\\ 0.40 & 1.002 & 240000\\ 0.50 & 1.000 & 240000\\ 0.55 & 1.001 & 40000\\ 0.60 & 0.992 & 20000\\ 0.65 & 0.995 & 10000\\ 0.70 & 1.009 & 5000\\ \end{matrix} These numbers strongly support Montgomery and Soundararajan's conjecture, so it is clear that the missing lower order term was indeed responsible for the observed discrepancy. REPLY [24 votes]: There are lower order terms in the work of Montgomery and Soundararajan that may account for the discrepancies you're observing. If you look at Theorem 3 of the paper that you linked, you'll find that the standard deviation should really be $$ \frac{\sqrt{y (\log \frac xy +B)}}{\log x}, $$ where $B=1-\gamma-\log (2\pi)=-1.415\ldots$. This is asymptotically the same as what you have, but numerically the second order term can make a difference. Note also that the lower order term becomes more significant as $y$ gets larger, which is a feature that you see in your data. So this is really a question for you: whether taking the new standard deviation with lower order terms gives you values for $\sigma$ closer to $1$. I'd be curious to know the revised numerics.<|endoftext|> TITLE: New(?) reciprocity law QUESTION [18 upvotes]: Consider three functions $f, g$ and $h$ on a smooth curve $X$ over $\mathbb{C}$. I have found the following equality: $$\sum (res(f\frac{dg}{g})\frac{dh}{h}-res(f\frac{dh}{h})\frac{dg}{g})=0.$$ Here the sum is over complex points of $X$, $res$ means residue of the corresponding meromorphic form on $X$. The most important point is that object $$res(f\frac{dg}{g})\frac{dh}{h}$$ should be considered as an element of the module of absolute Kahler differentials $\Omega^1_{\mathbb{C}/\mathbb{Q}}.$ For $f=1$ it gives the Weil reciprocity law, since $$res(\frac{dg}{g})\frac{dh}{h}-res(\frac{dh}{h})\frac{dg}{g}=ord(g)\frac{dh}{h}-ord(h)\frac{dg}{g}=$$ $$=\frac{d\{g,h\}_W}{\{g,h\}_W},$$ where $\{g,h\}_W$ stands for the Weil symbol. So, $$0=\sum (res(\frac{dg}{g})\frac{dh}{h}-res(\frac{dh}{h})\frac{dg}{g})= =\sum\frac{d\{g,h\}_W}{\{g,h\}_W}=\frac{d\prod \{g,h\}_W}{\prod \{g,h\}_W}.$$ From this it follows that $\prod \{g,h\}_W$ is algebraic complex number. Since the product of Weil symbols depends continously on the functions, it should be constant. But for $g=h=1$ it is equal to one. This finishes the proof of the Weil reciprocity law. Question 1: How one can prove this formula? I have found quite a long proof with Newton polygons, following ideas of Askold Khovansky. Question 2: Can one interprete it as some kind of a reciprocity law for surfaces? Question 3: I am almost sure that it is known. I would be very grateful for any reference. REPLY [8 votes]: Look at so-called Parshin reciprocity law, for example Mazin' approach http://www.math.utoronto.ca/mmazin/thesis.pdf I guess that is exactly what you want. If I am not mistaken, your version follows from the standard version of Weil reciprocity law for the functions $f/g,f/h$, in the spirit of logarithmic differentials (cf. http://www.math.toronto.edu/askold/osaka.pdf) If not, it is very strange.<|endoftext|> TITLE: A realcompact analogue of the Baire category theorem QUESTION [5 upvotes]: Let $\frak{m}$ be the least measurable cardinal. A space $X$ is realcompact if it is homeomorphic to a closed subset of some product $\mathbb{R}^I$. Let $X$ be realcompact with $P_\frak{m}$ topology, that means - an intersection of fewer than $\frak{m}$ open subsets is again open. A subset $R\subseteq X$ is regular closed if $R=\overline{\mathop{\rm int}R}$. Question: Is the following true: Let $D_i=R_i\cap\overline{X\setminus R_i}$ be the boundaries of regular closed subsets $R_i$. Let $\kappa<\frak{m}$. Then their union has an empty interior: $$ \mathop{\rm int}\bigcup_{i<\kappa}D_i=\emptyset $$ The union is closed by the $P_\frak{m}$ topology. If $X=P_{\frak{m}}K$ is the $P_\frak{m}$ refinement of a compact space $K$ (such a refinement is always realcompact) then it is enough to assume that $D_i$'s are closed with empty interior. Otherwise it is easy to come up with a realcompact $P_\frak{m}$ space which is the union of countably many closed subsets with empty interiors. But what happens if they are boundaries of regular subsets? REPLY [3 votes]: I have a counterexample. In particular, for all regular cardinals $\kappa$ such that there is no uncountable measurable cardinal below $\kappa$, there is a space $X$ that is the union of countably many boundaries of regular open sets but where $X$ is realcompact (and much more). Let $\kappa$ be an uncountable regular cardinal. Let $D$ be a discrete space of cardinality $\kappa$ and let $D\cup\{\infty\}$ be space where $U\subseteq D\cup\{\infty\}$ is open if and only if $\infty\not\in U$ or $|D\setminus U|<\kappa$. Then $D\cup\{\infty\}$ is a $P_{\kappa}$-space. Furthermore, $D\cup\{\infty\}$ is $\kappa$-compact (by $\kappa$-compact we mean that every open cover has a subcover by less than $\kappa$-elements.). Thus $D\cup\{\infty\}$ is the one-point $P_{\kappa}$ $\kappa$-compactification of $D$. Give $(D\cup\{\infty\})^{\omega}$ the box topology and let $X\subseteq(D\cup\{\infty\})^{\omega}$ be the set of all sequences $(x_{n})_{n\in\omega}\in(D\cup\{\infty\})^{\omega}$ such that $x_{n}=\infty$ for sufficiently large $n$. Clearly $(D\cup\{\infty\})^{\omega}$ and $X$ are $P_{\kappa}$-spaces. Let $X_{n}=\{(x_{i})_{i\in\omega}\in X|x_{n}=\infty\}$. Then each $X_{n}$ is a closed subset of $X$, but $X=\bigcup_{n}X_{n}$. Now let $\{A,B\}$ be a partition of $D$ such that $|A|=|B|=\kappa$. Take note that $A,B$ are open subsets of $D\cup\{\infty\}$ and $A\cup\{\infty\}$ is a closed subset of $D\cup\{\infty\}$. Let $U_{n}=\{(x_{i})_{i\in\omega}\in X|x_{n}\in A\}$ and let $C_{n}=\{(x_{i})_{i\in\omega}\in X|x_{n}\in A\cup\{\infty\}\}$. Then $U_{n}$ is an open set and $C_{n}$ is a closed set. Furthermore, it is easy to check that $C_{n}^{\circ}=U_{n}$ and $\overline{U_{n}}=C_{n}$. Therefore $C_{n}$ is a regular closed set with $\partial C_{n}=C_{n}\setminus U_{n}=X_{n}$. We therefore conclude that $\bigcup_{n}\partial C_{n}=\bigcup_{n}X_{n}=X$. On the other hand, the space $X$ is often much more than simply realcompact. We take note that the product of finitely many $\kappa$-compact $P_{\kappa}$-spaces is $\kappa$-compact (the proof of this is a straightforward generalization of the proof of the fact that the product of finitely many compact spaces is compact). Let $Y_{n}=\{(x_{i})_{i\in\omega}\in X|x_{i}=\infty\,\textrm{for}\,i\geq n\}$. Then $Y_{n}\simeq(D\cup\{\infty\})^{n}$, so $Y_{n}$ is $\kappa$-compact. Since $X=\bigcup_{n}Y_{n}$, we conclude that $X$ is $\kappa$-compact as well. Let $X$ be a completely regular space. The Baire $\sigma$-algebra is the smallest $\sigma$-algebra such that every continuous function $f:X\rightarrow\mathbb{R}$ is measurable. If $X$ is a completely regular space and $\mathcal{M}$ is the Baire $\sigma$-algebra on $X$, then $X$ is realcompact if and only if each $\sigma$-complete ultrafilter $\mathcal{U}\subseteq\mathcal{M}$ is of the form $\{R\in\mathcal{M}|x\in R\}$ for some $x\in X$. Take note that if $X$ is a $P$-space, then the Baire $\sigma$-algebra is precisely the collection of all clopen sets on $X$. $\mathbf{Proposition}$ Suppose that $\kappa$ is an uncountable regular cardinal such that there does not exist a measurable cardinal less than $\kappa$. Then every $\kappa$-compact $P_{\kappa}$-space is realcompact. $\mathbf{Proof}$ Suppose that $\kappa$ is an uncountable regular cardinal such that there does not exist a measurable cardinal less than $\kappa$. Let $X$ be a $\kappa$-compact $P_{\kappa}$-space, and let $\mathcal{U}$ be a $\sigma$-complete ultrafilter on the Baire $\sigma$-algebra $\mathcal{M}$ (which coincides with the algebra of all clopen sets). Then since $\kappa$ is not greater than the first measurable cardinal if one exists, we conclude that the ultrafilter $\mathcal{U}$ is $\kappa$-complete. Since $X$ is $\kappa$-compact, and each intersection $\bigcap_{i\in I}C_{i}$ is non-empty whenever $|I|<\kappa$ and $C_{i}\in\mathcal{U}$ for $i\in I$, we conclude that the intersection $\bigcap\mathcal{U}$ is non-empty. Let $x\in\bigcap\mathcal{U}$. Then $\mathcal{U}\subseteq\{R\in\mathcal{M}|x\in R\}$. Therefore since $\mathcal{U}$ and $\{R\in\mathcal{M}|x\in R\}$ are ultrafilters, we conclude that $\mathcal{U}=\{R\in\mathcal{M}|x\in R\}$. Therefore the space $X$ is realcompact. $\mathbf{QED}$ We finally conclude that if $\kappa$ is an uncountable regular cardinal which does not exceed the first measurable cardinal (if one exists), then the space $X$ described above is a realcompact $P_{\kappa}$-space which is the union of the boundaries of countably many regular closed sets.<|endoftext|> TITLE: Surprising connection between linear algebra and graph theory QUESTION [13 upvotes]: What is the intuition for linear algebra being such an effective tool to resolve questions regarding graphs? For example, one can determine if a given graph is connected by computing its Laplacian and checking if the second smallest eigenvalue is greater than zero (the so called Fiedler's eigenvalue). It seems surprising to me that some questions about graphs, which are inherently discrete combinatorial structures, can be resolved by means of linear algebra. What is the intuition behind this connection? REPLY [14 votes]: I think the basic point of contact between graph theory and linear algebra is the notion of a random walk. Given an initial probability distribution $p$ on the vertex set $V$ of a graph (though of as a vector in $\mathbb{R}^{|V|}$), the probabilities of hitting different vertices after $k$ steps of a random walk are given by $W^k p$ where $W = A D^{-1}$ (with $A$ the adjacency matrix and $D$ the degree matrix). This suggests that the spectral theory of $W$ is going to be relevant to dynamical questions on graphs, and the (normalized) Laplacian is just $D^{-1/2}(I - W)D^{1/2}$. (The point of the normalization is so that eigenvalues for different graphs can be compared). You ask specifically why spectral theory for the Laplacian helps measure the connectivity of a graph. Let's first note that it is possible to use random walks to answer this question. Suppose a graph is very loosely connected, meaning it can be divided into two pieces which each have many internal connections but very few external connections. Pick any vertex and start doing random walks of various lengths starting at that vertex. Intuitively, you expect that these random walks will be much more likely to visit vertices in the same piece as the starting vertex and much less likely to visit vertices in the other piece, and this intuition can be made precise. Of course you will still probably see vertices with high degree more often then you will see vertices with low degree, so you should compare the random walk probabilities to the degree vector of the graph (normalized so that it is a probability distribution). The normalized degree vector is an eigenvector for $W$ with eigenvalue $1$, so it is not unreasonable to expect that the small eigenvalues of the Laplacian (which is conjugate to $I - W$) will carry similar information. A question you might ask is: why use the Laplacian at all if the intuition comes from random walks? The first answer is historical: the theorem relating the connectivity of graphs to the spectral theory of the Laplacian began as the Cheeger inequality, a theorem relating connectivity of Riemannian manifolds to the spectral theory of the Riemannian Laplacian operator (the two theorems have nearly identical proofs). A more substantive answer is that for any vector $f \in \mathbb{R}^{|V|}$ we have: $$\langle Lf, f \rangle = \sum_{u \sim v} (f(u) - f(v))^2$$ which is a particularly simple quadratic form on the graph, and thus linear algebra and spectral theory for $L$ is quite simple (and the estimates are pretty sharp).<|endoftext|> TITLE: Is there one binary operation foundational for set theory? QUESTION [7 upvotes]: The membership relationship "$\epsilon$" is foundational for set theory, in the sense that the axioms of any set theory are formulated in the language of "$\epsilon$". Naturally, the question arises whether there is one binary operation, which is foundational in this sense. With such an operation identified, the "algebraic set theory" would be a theory of universal algebras with this operation, and maybe with some constants (like the $0$-ary operation selecting the empty set $\emptyset$). The supports of such algebras are also a foundational issue to be addressed, but such issues are resolved in literature - these supports can be "universes", that is sets "closed" under all operations in which we are interested. This issue is not in the focus of this question - important is the binary operation in the title. I see two such operations, each of which sounds to be foundational for set theory: $f(x, y)$ = {$x$} $\cup \ y$, since ($x \ \epsilon \ y$) iff $f (x, y) = y$, $g(x, y)$ = {$x$} $\cap \ y$, since ($x \ \epsilon \ y$) iff $g(x, y)$ = {$x$}. Each of these two operations is foundational for set theory, because in any axiom of a set theory we can replace the atomic formulas of the form ($x \ \epsilon \ y$) with corresponding equivalent equations in (1) and (2). But only this formal procedure is not very interesting. Way more interesting is whether there are algebraic properties of these two operations which would allow to replace other formulas (not only the atomary formulas) with equations in terms of $f$ (or $g$) - ideally, to replace all logical formulas with algebraic equations (identities). I believe, this ideal cannot be achieved because of quantifiers, unless some infinitary generalizations of these two operations are found. Therefore, instead of the ideal goal, a realistic goal is also interesting - to replace with algebraic equations all subformulas under each quantifier (or the matrix in a prenex form). My research relates to set-theoretic modeling of natural languages and these two operations sound to model a linguistic phenomenon. I would appreciate even partial answers to my questions above, or references to something close. Here are some more concrete questions about this: Are there known any properties of the operation $f(x, y)$ — properties interesting for the foundations? I see some of such properties, but a set-theorist or algebraist will probably indicate also other properties: (a) The operation $f(x, y)$ allows to define in set theory the natural numbers (finite ordinals) by induction: $0 = \emptyset, \ n+ 1 = f(n, n)$. Here "$0$" is number "zero" and "$\emptyset$" is "empty set". (b) For any $x_1, x_2,..., x_n$, [{$x_1, x_2,..., x_n$} = $f(x_1, f(x_2,..., f(x_n, \emptyset$)..))]. Thus, the notion of finite set can be defined through operation $f$ (and the $0$-ary operation selecting $\emptyset$). Here, "defined through" is same as "is superposition of". (c) For any $x$, {$x$} $= f(x, \emptyset)$. Thus, the singleton formation operation can be expressed through $f$ (and $\emptyset$). Can the binary operation of union be expressed through $f$? If not, then are there partial cases (like finite sets) when it can? Are there any properties of operation $g(x, y)$ — properties interesting for the foundations? Is there another binary operation similar to $f$ and $g$? Could there be infinitary generalizations of $f$ or $g$ which would allow to axiomatize set theory in algebraic equations without quantifiers (except the external universal quantifiers in algebraic identities)? If I correctly phrased (in mathematical terms) this last question, then I suspect, this is a really difficult question. To make it more precise, remember that infinitary unions and intersections play the role of universal and existential quantifiers. Also, notice that if we treat comma (,) as a symbol of a binary operation in the denotation of a finite set {$x_1, x_2,..., x_n$}, then this operation is associative, commutative and idempotent (like union). This, together with 1(b), cannot help define an infinitary (in some sense) generalization of $f(x, y)$? REPLY [3 votes]: There is a whole area of algebraic set theory. The internal language of a topos can be expressed as an essentially algebraic theory, see Lambek & Scott's "Introduction to higher-order categorical logic". An essentially algebraic theory is a slight generalization of an equational theory, which you asked about. I am pretty sure that ZFC is not essentially algebraic, but can't think of a reason right now -- there must be a simple one. If you're set on classical logic for some unreasonable reason, you can always throw in one more equation, namely $p \lor \lnot p = \top$. The internal logic of a topos is more or less like bounded Zermelo set theory (where "bounded" means that only bounded separation is allowed). In any case, I am surprised you're expecting set theory to do anything useful for linguistics. The language of set theory is notoriously not at all "natural" from a human perspective (although the concept of set is). How about categorical grammar and that sort of thing?<|endoftext|> TITLE: Infinite decreasing sequence by the Turing jump QUESTION [11 upvotes]: I saw in Wikipedia the existence of an infinite sequence of Turing degrees $\bf{a_0}, \bf{a_1}, \dots$ such that $\bf{a}_{i+1}' \leq_T \bf{a}_i$ where $\bf{a}_{i+1}'$ is the Turing jump of $\bf{a}_i$. However I couldn't find references for this result. Is it a trivial consequence of one of the well-known jump inversion theorems ? REPLY [12 votes]: There is surely a proof using only classical recursion theory, but the proof I know uses second-order arithmetic. Take an $\omega$-model $M$ of $\mathsf{ATR}_0$ in which there is a countable linear order $L$ such that $M$ satisfies "$L$ is well founded", but such that $L$ is not actually well founded. Then $M$ believes that there is a set $H$ which is obtained by iterating the Turing jump along $L$ starting with $\emptyset$, because this is provable in $\mathsf{ATR}_0$. Because $M$ is an $\omega$-model, it is absolute for statements of the form "$A' = B$" when $A,B \in M$. Thus, if we take an infinite descending sequence $(a_i : i \in \omega)$ in $L$, and let $(A_i : i \in \omega)$ be the corresponding sequence of sets in $H$, then $A_{i+1}' \leq_T A_i$ for all $i \in \omega$. This method can be easily extended to stronger results -- for example we could make $A_{i+1}^{(\omega)} \leq_T A_i$. The use of $\mathsf{ATR}_0$ is not really necessary here. The key point is that there is a linear ordering $L$ of $\omega$, that is not a well ordering, for which there is a set $H^L$ that satisfies the arithmetical definition of being the iteration of the Turing jump along $L$ starting with the empty set. This kind of set $H^L$ is known more generally as a "pseudohierarchy". The existence of a suitable $L$ follows immediately from the standard fact that the set of well orderings of $\omega$ is not $\Sigma^1_1$ definable. The statement "$L$ is a linear order and $H(L)$ exists" is a $\Sigma^1_1$ formula with parameter $L$, which is satisfied by every well ordering of $\omega$, so it must also be satisfied by some other set $L$; this $L$ will satisfy the "key point" statement in the previous paragraph.<|endoftext|> TITLE: From the perspective of bordism categories, where does the ring structure on Thom spectra come from? QUESTION [24 upvotes]: To fix ideas, let's consider the Thom spectrum of framed bordism $M$, the spectrum whose homotopy groups are the framed bordism groups. $M$ has a ring spectrum structure inducing the product of manifolds on its homotopy groups. By Pontryagin-Thom, $M$ is the sphere spectrum $S$, which is even the initial ring spectrum. This fact in some sense lifts to the cobordism hypothesis as follows. A suitable version of the cobordism hypothesis should assert that the framed bordism $\infty$-category $\text{Bord}^{fr}$ is the free $E_{\infty}$ $\infty$-category with duals on a point. After inverting all of the morphisms in $\text{Bord}^{fr}$ suitably, we get an $E_{\infty}$ $\infty$-groupoid, or essentially an infinite loop space, which should be the underlying space of the Thom spectrum $M$. This tells me what $M$ looks like as a spectrum. But the higher categorical story, so far as I know, doesn't immediately tell me what $M$ looks like as a ring spectrum; I don't see how to put a familiar-looking structure on $\text{Bord}^{fr}$ inducing the ring spectrum structure on $M$. Is there such a structure? REPLY [15 votes]: Thanks to a very helpful discussion with Clark Barwick in the homotopy theory chat, I think I now understand what's going on here. In particular, the ring spectrum structure on the sphere spectrum $\mathbb{S}$ does come from a monoidal structure on $\text{Bord}$, but I was confused about how to transport this monoidal structure from the category to the spectrum. The monoidal structure can be thought of as coming from the universal property of $\text{Bord}$: since it's the free symmetric monoidal $\infty$-category with duals on a point, the $\infty$-category of symmetric monoidal functors $\text{Bord} \to \text{Bord}$ can canonically be identified with $\text{Bord}$ itself, and hence $\text{Bord}$ naturally acquires a monoidal structure, which I'll call $\circ$, coming from composition of functors $\text{Bord} \to \text{Bord}$. This is exactly analogous to how the free abelian group $\mathbb{Z}$ on a point canonically acquires a ring structure, and compatible under group completion with how the free spectrum on a point, namely $\mathbb{S}$, canonically acquires a ring spectrum structure. Here we need to know that symmetric monoidal $\infty$-categories are enriched over themselves, but this ought to be true by analogy both with the case of abelian groups and with the case of spectra. This is admittedly an indirect description. It's hard to attempt a more direct description because the resulting monoidal structure isn't all that interesting on objects, and trying to describe what it does on morphisms is what got us into this mess in the first place. So let me trudge on. In the comments I explained that I thought a monoidal structure $$\text{Bord} \times \text{Bord} \to \text{Bord}$$ couldn't induce the usual multiplication map on $\mathbb{S}$ because monoidal structures take an $n$-morphism and an $n$-morphism and return another $n$-morphism: for example, the disjoint union does provide a monoidal structure of this form (in fact it's the monoidal structure figuring in the universal property), and the induced map $$\pi_n(\mathbb{S}) \times \pi_n(\mathbb{S}) \to \pi_n(\mathbb{S})$$ is the usual abelian group structure on $\pi_n(\mathbb{S})$. The problem with this story as applied to $\circ$ is that, with the natural symmetric monoidal structures on both sides, $\circ : \text{Bord} \times \text{Bord} \to \text{Bord}$ is not a symmetric monoidal functor, so it does not induce a map $\mathbb{S} \times \mathbb{S} \to \mathbb{S}$ of spectra. This issue shows up already at the level of abelian groups: with the natural abelian group structures on both sides, the multiplication map $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ is not a homomorphism of abelian groups. As suggested by the analogy to abelian groups, $\circ$ is "bilinear": it preserves disjoint unions separately in both variables. So the fix is to use a suitable notion of "tensor product" of symmetric monoidal $\infty$-categories (suitable meaning in particular that on symmetric monoidal $\infty$-groupoids, thought of as connective spectra, it reproduces the smash product) and think of $\circ$ as instead providing a symmetric monoidal functor $$\text{Bord} \otimes \text{Bord} \to \text{Bord}$$ which reproduces the usual ring spectrum structure $\mathbb{S} \otimes \mathbb{S} \to \mathbb{S}$ (here I am using $\otimes$ for the smash product as well). Now to see how we get a multiplication map $$\pi_n(\mathbb{S}) \times \pi_m(\mathbb{S}) \to \pi_{n+m}(\mathbb{S})$$ it suffices to recall that the smash product of $S^n$ and $S^m$ is $S^{n+m}$. So, one way to answer the conceptual question "how, in this situation, did we start with an $n$-morphism and an $m$-morphism and get an $n+m$-morphism" is that the universal property of $\text{Bord}$ is extremely general: it naturally acts by endomorphisms on an object in any symmetric monoidal $\infty$-category with duals whatsoever, including the "loop spaces" of $\text{Bord}$ itself! The analogous statement in stable homotopy is that the stable homotopy groups naturally give rise to operations on the homotopy groups of any spectrum whatsoever, including the shifts of the sphere spectrum itself.<|endoftext|> TITLE: maximum sum of angles between $n$ lines QUESTION [6 upvotes]: Take $n$ lines in $\mathbb{R}^d$ (not necessary different, and all passing through the origin, though this is not important). What is maximal possible sum of angles between them for given $n$ and $d$? Conjecturally you should use only $d$ different mutually orthogonal directions, either $\lfloor n/d \rfloor$ or $\lceil n/d \rceil$ times each of them. I can not prove or disprove this even for $d=3$. Note that if you take $n$ rays instead of lines then the optimal configuration contains only two opposite directions, each $\lfloor n/2 \rfloor$ or $\lceil n/2 \rceil$ times. REPLY [5 votes]: This problem seems to be a question of Fejes Tóth posed here: L. Fejes Tóth, ``Über eine Punktverteilung auf der Kugel", Acta Math. Acad. Sci. Hungar 10 (1959), 13-19 (in German). Some recent work on this question in $\mathbb R^3$ has been done in F. Fodor, V. Vígh, and T. Zarnócz, ``On the angle sum of lines", Arch. Math. 106 (2016), 91-100. Amusingly, the above paper, in addition to the original conjecture by Fejes Tóth, references this mathoverflow post. Let $Z=\{z_1, ..., z_N\}$ be $N$ points on the sphere $\mathbb S^{d-1} \subset \mathbb R^d$. Let $F(t) = \arccos |t|$, then the ``line" (acute) angle between two vectors $z_i$ and $z_j$ is $F(z_i \cdot z_j)$. We are trying to maximize the discrete energy $$ E_F (Z) = \frac{1}{N^2} \sum_{i,j =1}^N F(z_i \cdot z_j).$$ The conjecture then states that the maximum of this energy is equal to $\displaystyle{\frac{\pi}2 \cdot \frac{d-1}{d}}$ when $N$ is a multiple of $d$ (with the necessary correction for other cases). More generally one can consider the energy integral $$ I_F (\mu ) = \int_{\mathbb S^d}\int_{\mathbb S^d} F(x \cdot y ) d\mu (x) d \mu (y) ,$$ where $\mu $ is a probability measure on the sphere. It is also reasonable to conjecture that $\max I_F(\mu ) = \displaystyle{\frac{\pi}2 \cdot \frac{d-1}{d}}$, i.e. the energy is maximized if $\mu$ is equally concentrated in vertices of an orthonormal basis. In this terms, the paper of Fodor, Vígh, Zarnócz proves that in dimension $d=3$ for any point distribution $$ E_F (Z) \le \frac{3\pi}8, $$ (with a small correction for $N$ odd), while the conjecture in $d=3$ is $\frac{\pi}3$. Ryan Matzke and I recently realized that there is a simple way to improve this bound. Here is the argument: one can easily check that for $-1\le t \le 1$ $$ F (t) = \arccos |t| \le \frac{\pi}{2} - \frac{7 \pi}{16} t^2 $$ (see the graph in the end). Therefore, $$ I_F (\mu) \le \frac{\pi}2 - \frac{7 \pi}{16} I_{t^2} (\mu) \le \frac{\pi}2 - \frac{7 \pi}{16 d} ,$$ where we have used the fact that $I_{t^2} (\mu ) \ge \frac1{d} $ (this is a well-known lower bound on the ``frame energy", it is essentially proved in fedja's reply to the aforementioned mathoverflow post: The minimum of a sum of absolute values of inner products in $\mathbb{R}^d$). Hence for $d=3$ we obtain $$ I_F (\mu ) \le \frac{17 \pi}{48},$$ which is better than $\frac{3\pi}{8}$, and is only $\frac{\pi}{48}$ away from the conjectured $\frac{\pi}{3}$. Here is a picture that ``proves" the inequality above. There is a tiny bit of room for improvement, but his will definitely not yield the conjecture.<|endoftext|> TITLE: Do's and don'ts of writing survey papers QUESTION [33 upvotes]: I am not sure if this is the appropriate forum to ask as it is not directly related to a research level (math) problem, but I figured it was worth a try. I recently attended a conference and felt that my area of research was not well represented, and simultaneously got feedback that there does not exist a single reference that summarizes the state of affairs of my topic area. I have thus decided to write an expository paper and submit it to the conference proceedings. However, I am not sure what the expectations are for expository papers. Can anyone give me their opinions on the "do's" and "don'ts" of expository writing? Thank you very much for your input. REPLY [18 votes]: There are several journals publishing mostly or only surveys: BAMS, Russian Math surveys, Expositiones math., Asterisque, Sugaku. There are many excellent surveys in these journals. Take one which you like and follow the pattern. The main thing I expect from a survey is that it should be readable by a non-specialist. By someone who wants to be introduced to the area. Perhaps a graduate student. There is a different kind of surveys: surveys for specialists. In this type of surveys, the main feature is completeness. Specialists will use it to check what is known and what is not known in your area. And what are the most important questions/conjectures. You have to decide in advance what type of a survey you want to write.<|endoftext|> TITLE: Ellipses on spheres (and other surfaces) QUESTION [25 upvotes]: Define an ellipse $E$ on a sphere as the locus of points whose sum of shortest geodesic distances to two foci $p_1$ and $p_2$ is a constant $d$. There are conditions on $\{ p_1, p_2, d \}$ for this definition to make unambiguous sense, but assume those conditions hold. (Added: Ian Algol specifies: $p \in E$ should not be a conjugate point of $p_1$ or $p_2$, i.e., the Jacobi field of a geodesic from a focus to a point $p$ of $E$ should not vanish at $p$.)           I am interested in billiard-reflection properties of ellipses on spheres, and on other curved surfaces. My initial searches for literature has turned up empty, although there seems little doubt the topic must have been studied. Here are a few questions. Q1. For an ellipse $E$ on a sphere, does a geodesic ray from $p_1$, reflecting by angle-of-incidence = angle-of-reflection from $E$, necessarily pass through $p_2$? Q2. If not, is there some other curve $C$ that has this property? In other words, could an ellipse be defined as a curve $C$ with the reflection property, rather than the sum-of-distances property? What is the relationship between the two possible definitions? Q3. What are the properties of an ellipse (defined by the sum-of-distances property) on other curved surfaces? Constant negative curvature? Arbitrary smooth surfaces? Thanks for ideas and/or pointers to the literature! Answered by Ian Algol: The answer to Q1 is Yes, not just for spheres, but—remarkably—for any smooth surface. So this answers Q2 and Q3 as well. REPLY [11 votes]: I'd like to add that the same holds not only for Riemannian, but also for Finsler metrics: E. Gutkin, S. Tabachnikov. Billiards in Finsler and Minkowski geometries J. Geom. Phys. 40 (2002), 277-301. Of course, the law of reflection should be defined appropriately. My explanation of the optical property of ellipses is as follows (tested on students). Let A and B be the foci. Consider the distance functions to A and to B. The gradients of these functions at point X are the unit vectors along the geodesics AX and BX. The sum of these unit vectors is orthogonal to the ellipse, a level curve of the sum of the functions. This implies that the angles are equal, and the ray AX reflects to the ray XB.<|endoftext|> TITLE: The metric of the expected difference of random variables QUESTION [12 upvotes]: Suppose we have a set of independent random variables $X_1,\ldots,X_n$ over $\mathbb{R}$. It is easy to see that $$d_{ij}=E[|X_i-X_j|]$$ satisfy the triangle inequality. Is there any study of such metric spaces? (Note that this metric is not the usual metric of distributions. In particular, for two identically distributed $X_1,X_2$, $d_{12}\ne 0$.) REPLY [5 votes]: I have not seen concretely your metric, but something similar is used to analyze extrema of Gaussian random fields. A very useful notion is entropy with respect to the following metric: $d(s,t)=(\mathbf{E}(X_t-X_s)^2)^{1/2}$. I think, such metrics were introduced by Dudley, but I may be mistaken. I am sure this is written in many books and papers. A minute of googling gave me http://cims.nyu.edu/~zeitouni/notesGauss.pdf, where this metric is introduced on p.18. Your independence assumption makes this metric not so useful though, it seems, and the following computation may be more elucidating for your problem: $$ E|X-Y|= E\int_R (1_{X\le x}-1_{Y\le x})^2 dx = \int_R E(1_{X\le x}+1_{Y\le x}-21_{X\le x}1_{Y\le x})dx =\int_R (P\{X\le x\}+P\{Y\le x\}-2P\{X\le x\}P\{Y\le x\})dx =\int_R ((P\{X\le x\}-P^2\{X\le x\})+(P\{Y\le x\}-P^2\{Y\le x\})+(P\{X\le x\}-P\{Y\le x\})^2)dx =\int_R F_X(x)(1-F_X(x)) dx +\int_R F_Y(x)(1-F_Y(x)) dx + \int_R(F_X(x)-F_Y(x))^2dx $$ The first term on the right-hand side depends only on $F_X$, the distribution of $X$, the second one depends only on $F_Y$, the distribution of $Y$, and the last one is the square of $L^2$ distance between the distribution functions of $X$ and $Y$. So the distance you want to consider is, in fact, a kind of $L^2$ distance on distributions with a little bit of what is called the "nugget effect" in geostatistics, where they often consider random fields $X_t$ such that $E(X_s-X_t)^2$ does not converge to zero as $s\to t$. This occurs naturally if $X_t=Y_t+Z_t$, where $Y_t$ is a very nice smooth process, and $Z_t$ has zero correlation range, i.e., $cov(Z_t,Z_s)=0$ for $t\ne s$.<|endoftext|> TITLE: Harrington's unpublished note "The constructible reals can be anything" QUESTION [10 upvotes]: Around 1974, Leo Harringto wrote an unpublished note entitled "The constructible reals can be anything", in which he proved that it is consistent that being $\Delta^1_n$ is the same as being constructible. Harrington proved his theorem using a version of almost disjoint coding of Jensen and Solovay. Is there any reference in which a proof of the above theorem is given which is similar to the Harrington's original proof. Remark. There are proofs of Harrington's theorem by Kanovei, using different methods, see for example (1) Kanovei. The independence of some propositions of descriptive set theory and second order arithmetic. Soviet Math. Dokl . 1975, 16, 4, pp. 937 – 940. (2) Kanovei. On the nonemptiness of classes in axiomatic set theory. Math. USSR Izv. 1978, 12, pp. 507 – 535. REPLY [10 votes]: It seems the Harrington's preprint is here http://logic-library.berkeley.edu/catalog/detail/2135 but to access it requires CalNet login. Maybe Harrington himself could help if you write to him: http://math.berkeley.edu/~leo/index.html P.S. Prof. Harrington kindly sent me a scan of the preprint. It can be found here http://www.inp.nsk.su/~silagadz/Harrington.pdf<|endoftext|> TITLE: If all reals are generic, is the set of reals generic? QUESTION [20 upvotes]: Let $W\subseteq V$ be two models of $\sf ZFC$ with the same ordinals. Is the following situation consistent: For every $x\in\Bbb R^V$ there is some $P_x\in W$ such that for some $G\subseteq P_x$ which is $W$-generic, $x\in W[G]$. There is no $P\in W$ and $G\subseteq P$ that is $W$-generic such that $\Bbb R^{W[G]}=\Bbb R^V$. Namely, each real is [set-]generic over $W$, but the set of reals is not. This sort of situation of course immediately exclude the case that $V$ is a generic extension of $W$; but also things like when $V=L[r]$ is obtained by coding $W$ into a real $r$. (We may assume that $\sf CH$ holds in $V$, otherwise we can force it without adding real numbers.) REPLY [4 votes]: I met Woodin recently and asked him that. He came up with a solution, modulo some technical assumption which Ashutosh showed to be consistent (although admittedly, not the same suggestion that Woodin had for solving this issue). With his kind permission, I am posting this solution here. $W$ is a model of $\sf ZFC+GCH+$"There are $\aleph_1$ ccc forcings which add independent reals" (call these forcings $\Bbb P_\alpha$). $V_1$ is a class generic extension of $W$ in which a proper class of cardinals were collapsed while preserving $\sf ZFC$ (e.g. collapse all $\aleph_{\alpha\cdot\omega+3}$ to $\aleph_{\alpha\cdot\omega+2}$). $V_2$ is coding $V_1$ into a subset of $\omega_1$ without adding reals over $W$, so $V_2=W[A]$ where $A\subseteq\omega_1$. Finally, $V$ is the finite support product of $\Bbb P_\alpha$ for $\alpha\in A$ over $V_2$. Since from $W$ to $V_2$ we didn't add any reals, and every real added to $V$ came from a countable part of the product (which is in $W$), it follows that every real number is $W$-generic for some suitable part of the product. But if you had a $W$-generic $G$ (for a set forcing) such that $W[G]$ and $V$ had the same reals, you would be able to extract $A$ and therefore compute the class generic for the now-collapsed cardinals.<|endoftext|> TITLE: How to construct a free 2-group on a groupoid? QUESTION [8 upvotes]: Let G be a groupoid. I'm wondering how to construct the free 2-group on G. By the free 2-group I mean a 2-group $\mathcal{F}\left(G\right)$ equipped with a functor $i:G\longrightarrow\mathcal{F}\left(G\right)$ such that for any 2-group $\mathcal{G}$ and any functor $F:G\longrightarrow\mathcal{G}$ there is a monoidal functor $F':\mathcal{F}\left(G\right)\longrightarrow\mathcal{G}$ and an isomorphism $\alpha:F'\circ i\Longrightarrow F$ such that the functor $F':\mathcal{F}\left(G\right)\longrightarrow\mathcal{G}$ is unique up to coherent isomorphism. Here is my attempt at a description of $\mathcal{F}\left(G\right)$ : The objects are defined inductively by requiring that $\mathcal{F}\left(G\right)$ contain an object 1 , the objects of G and for each object $g\in G$ an object $\overline{g}\in G$ such that for any two objects $x,y\in\mathcal{F}\left(G\right)$ there is an object $x\otimes y$ in $\mathcal{F}\left(G\right)$ . The 'prearrows' in $\mathcal{F}\left(G\right)$ are built as follows. We require that every arrow from G be a prearrow in $\mathcal{F}\left(G\right)$ . For each pair of prearrows $a:x\longrightarrow z$ and $b:y\longrightarrow w$ in $\mathcal{F}\left(G\right)$ there is a prearrow $a\otimes b:x\otimes y\longrightarrow z\otimes w$ . In addition we adjoin a prearrow $e_{g}:g\otimes\overline{g}\longrightarrow1$ for each $g\in G$ and we adjoin prearrows $\alpha_{x,y,z}:\left(x\otimes y\right)\otimes z\longrightarrow x\otimes\left(y\otimes z\right)$ , $\lambda_{x}:1\otimes x\longrightarrow x$ , and $\rho_{x}:x\otimes1\longrightarrow x$ for each $x,y,z\in\mathcal{F}\left(G\right)$ . The arrows are then given by equivalence classes of prearrows generated by the requirement that $\alpha,\rho,\lambda$ and e be isomorphisms, the naturality conditions on $\alpha,\rho$ and $\lambda$ , the condition that $\otimes$ is a functor, and the axioms of a monoidal category. My question is, does this construction work? If not, can anybody give an indication of a construction that does? Also, to define the free symmetric 2-group on a groupoid, does it suffice to add prearrows $\gamma_{x,y}:x\otimes y\longrightarrow y\otimes x$ for each $x,y\in\mathcal{F}\left(G\right)$ and generate the equivalence relation from the axioms of a symmetric monoidal category? REPLY [3 votes]: The notion of free crossed module was a major feature of JHC Whitehead's 1949 paper "Combinatorial homotopy II", in which he proved, using methods of transversality and knot theory, that the crossed module $$\pi_2(X \cup \{e^2_\lambda\},X,x) \to \pi_1(X,x) $$ is free on the characteristic maps of the $2$-cells. This theorem is sometimes mentioned but rarely proved in topology texts. Work of Philip Higgins and I showed how this theorem was a special case of a $2$-dim van Kampen type theorem, i.e. a colimit theorem, and the full story of this is given in Part I of the book partially titled Nonabelian Algebraic Topology, EMS 2011 (NAT). Thus the more general theorem determines $$\pi_2(X \cup _f CA,A,x) \to \pi_1(X,x)$$ for $A$ connected, in terms of the induced morphism $f_*: \pi_1(A,a) \to \pi_1(X,x)$, so that Whitehead's theorem is the case $A$ is a wedge of circles. Feb 14, 2016 This question could also be looked at in the light of the forgetful functor $$\Phi: (\text{2-groups)} \to (\text{groupoids})$$ which is a bifibration with a left adjoint say $D$, and a right adjoint $I$. As developed in Appendix B3 (Theorem B.3.2) of the book NAT, the cofibration (cocartesian) property of $\Phi$ is given by a pushout of 2-groupoids $$ \begin{matrix} D\Phi(K) &\xrightarrow{D(F)}& D(H) \\ \downarrow&& \downarrow \\ K & \to &F_*(K).\end{matrix} $$ The notion of ``free" 2-groupoid on $F: G \to H$ is the special case when $K=I(G)$. This is not stated in that Appendix but is the construction used in Part 1. For example if $F:P \to Q$ is a morphism of groups then the free crossed module on $F$ is the induced crossed module $F_*(P \to P)$, where the identity crossed module is $I(P)$. This generalises to free crossed modules over groupoids. February 25,2016 I think Fernando is right and I apologise for my misapprehension. To explain it, I am commonly looking for some algebraic explanations of how low dimensional identifications in spaces influence high dimensional homotopy invariants. The prototype was that groupoids enable the computation of 1-types by a van Kampen type theorem because, it seems, groupoids have structure in dimensions 0 and 1. Crossed modules (over groupoids, and so equivalent to 2-groupoids) have structure in dimensions 0,1,2, and give computational models of 2-types in an analogous way. So I am used to the "inducing" process coming from a bifibration of algebraic models from dimension $n$ to dimension $n-1$ and contributing to "free" structures in dimension $n$.<|endoftext|> TITLE: Differences in philosophy between Lie Groups and Differential Galois Theory QUESTION [15 upvotes]: As far as I have heard,Sophus Lie's aim was to construct an analogue of galois theory for differential galois theory. I am familiar with lie group but not with differential galois theory. What is the difference is philosophy between these 2 subjects-Lie groups and Differential Galois Theory? REPLY [18 votes]: [This is really an extended comment on the remarks of Siegel and Khavkine, but too long for a comment box.] There is an intermediate topic which straddles the world of algebra (where differential Galois theory "really" lives, in the language of differential fields and so on) and analysis (where Lie groups live), namely the theory of linear algebraic groups. Although Lie worked from an entirely analytic viewpoint (and mainly with "group chunks", as the global perspective of Lie groups as manifolds, far from just formal group laws, only took off with the work of Weyl), it was really with the advent of the theory of matrix groups and the work of Kolchin that modern differential Galois theory (building on ideas of Liouville and Ritt) really took off. For instance, the "Lie-Kolchin" theorem about (Zariski-)connected closed subgroups of matrix groups was proved by Kolchin precisely for its applications to characterizing when a "full set of solutions" to (certain kinds of) linear ODE's could be expressed in terms of iterating the operations of forming exponentials, logarithms, and solutions to algebraic equations. The analogy from the viewpoint of differential fields works very beautifully in the style of classical Galois theory (except that things can be even harder to compute in practice): to any linear ODE over a differential field $K$ with algebraically closed field of constants $C$ one constructs a "Picard-Vessiot" extension field $L$ of $K$ over which the ODE acquires a "full set of solutions", this being unique up to isomorphism of differential fields (analogous to splitting fields of polynomials in one variable), and the abstract automorphism group of the differential field extension $L/K$ faithfully represented on the finite-dimensional $C$-vector space $V$ of solutions (in $L$) to the given ODE turns out to be a Zariski-closed subgroup $G$ of ${\rm{GL}}(V)$ (the "differential Galois group" of the ODE). Moreover, this algebro-geometric structure on the automorphism group is intrinsic, and the "Galois correspondence" is that Zariski-closed subgroups of $G$ are in 1-1 inclusion-reversing correspondence with intermediate differential fields within $L/K$. In these terms, Kolchin related "solvability by successive exponentials, and logarithms, and algebraic equations" into "solvability of $G^0$". The ideas introduced by Kolchin spawned techniques with $D$-modules and other aspects of "algebraic analysis" which have become important tools in geometric representation theory. Differential algebra and differential modules remains an active field, perhaps not as popular as some other areas which make use of its ideas, but it didn't take off in quite the same way as usual Galois theory simply because the kinds of things that differential Galois theory allows one to say about differential equations don't mesh as well with the kinds of things that specialists in differential equations want to know (qualitative information, non-linear phenomena, etc.). Also, relating the abstract differential field extensions to actual concrete function spaces is a tricky matter (even within the restrictive linear setting where differential Galois theory is most relevant), much like trying to relate Galois theory to the study of specific complex numbers. Finally, and perhaps most importantly, despite the importance of compact Lie groups in particle physics, the Lie groups with the most beautiful mathematical structure (namely, semisimple ones) tend not to be the ones which are most relevant to the differential equations with symmetries that arise in a variety of physical problems. There is a beautiful book "Applications of Lie groups to differential equations" by P. Olver which does an excellent job of explaining what can be done with Lie groups in the service of symmetries in differential equations (and the Introduction explains more fully why Lie's initial dream didn't develop in quite the way he had expected).<|endoftext|> TITLE: Is formula valid for relating $\pi$ with ALL of its OEIS A002485(n)/A002486(n) convergents? QUESTION [19 upvotes]: Could anyone try to prove that the below conjectured formula is valid for relating $\pi$ with ALL of its convergents - those, which are described in OEIS via A002485(n)/A002486(n)? $$(-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n))$$ $$=(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^{2(j+2)}(k+(i+k)x^2)\big)/(1+x^2)\; dx$$ (1) and in unformatted form: (-1)^n*(Pi−A002485(n)/A002486(n))=(abs(i)*2^j)^(-1)Int((x^l(1-x)^(2*(j+2))*(k+(i+k)*x^2))/(1+x^2),x=0...1) where integer $n = 0,1,2,3,...$ serves as the index for terms in OEIS A002485(n) and A002486(n), and $\{i, j, k, l\}$ are some integers (to be found experimentally or otherwise), which are probably some functions of $n$. The "interesting" (I think) part of my generalization conjecture is that both "$i$" and "$j$" are present in both denominator of the coefficient in front of the integral and in the body of the integral itself At this time it could be shown that the formula under question is applicable for some first few convergents (of the A002485(n)/A002486(n) type) 1) For example for $\frac{22}{7}$ $$\frac{22}{7} - \pi = \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,\mathrm{d}x$$ with $n=3, i=-1, j=0, k=1, l=4$ - with regards to my above suggested generalization. In Maple notation i:=-1; j:=0; k:=1; l:=4;Int(x^l*(1-x)^(2*(j+2))*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1) yields 22/7 - Pi 2) It also works for found by Lucas http://www.math.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf formula for $\frac{333}{106}$ $$\pi - \frac{333}{106} = \frac{1}{530}\int_{0}^{1}\frac{x^5(1-x)^6(197+462x^2)}{1+x^2}\,\mathrm{d}x$$ with $n=4, i=265, j=1, k=197, l=5$ -with regards to my above suggested generalization. In Maple notation i:=265; j:=1; k:=197; l:=5;Int(x^l*(1-x)^(2*(j+2))*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1) yields Pi - 333/106 3) And it works for Lucas's formula for $\frac{355}{113}$ $$\frac{355}{113} - \pi = \frac{1}{3164}\int_{0}^{1}\frac{(x^8(1-x)^8(25+816x^2)}{(1+x^2)}$$ with $n=5, i=791, j=2, k=25, l=8$ -with regards to my above suggested generalization. In Maple notation i:=791; j:=2; k:=25; l:=8;Int(x^(2*(j+2))(1-x)^l(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1) yields 355/113 - Pi 4) And it works as well for Lucas's formula for $\frac{103993}{33102}$ $$\pi - \frac{103993}{33102} = \frac{1}{755216}\int_{0}^{1}\frac{x^{14}(1-x)^{12}(124360+77159x^2)}{1+x^2}\,\mathrm{d}x$$ with $n=6, i= -47201, j=4, k=124360, l=14$ -with regards to my above suggested generalization. In Maple notation i:=-47201; j:=4; k:=124360; l:=14;Int(x^l*(1-x)^(2*(j+2))*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1) yields Pi - 103993/33102 5) And also it works Lucas's formula for $\frac{104348}{33215}$ $$\frac{104348}{33215} - \pi = \frac{1}{38544}\int_{0}^{1}\frac{x^{12}(1-x)^{12}(1349-1060x^2)}{1+x^2}\,\mathrm{d}x$$ with $n=7, i= -2409, j=4, k=1349, l=12$ - with regards to my above suggested generalization. In Maple notation i:=-2409; j:=4; k:=1349; l:=12;Int(x^l*(1-x)^(2*(j+2))*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1) yields 104348/33215 - Pi 6) And it works as well for $\frac{618669248999119}{196928538206400}$ which, by the way, is not part of A002485/A002486 OEIS sequences: $$\frac{618669248999119}{196928538206400} - \pi = \frac{1}{755216}\int_{0}^{1}\frac{x^{14}(1-x)^{12}(77159+124360x^2)}{1+x^2}\,\mathrm{d}x$$ with $i= 47201, j=4, k=77159, l=14$ -with regards to my above suggested generalization. In Maple notation i:=47201; j:=4; k:=77159; l:=14;Int(x^l*(1-x)^(2*(j+2))*(k+(k+i)x^2)/((1+x^2)(abs(i)*2^j)),x= 0...1) yields 618669248999119/196928538206400 - Pi This question relates to my answer given in https://math.stackexchange.com/questions/1956/is-there-an-integral-that-proves-pi-333-106/127618#127618 Update: Recently Thomas Baruchel (see his answer at https://math.stackexchange.com/questions/860499/seeking-proof-for-the-formula-relating-pi-with-its-convergents ) has conducted extensive calculations and found that even the parametric formula (with four parameters) yields infinite number of solutions for each n. Thomas shared with me his calculations results and supplied me with quite a few of valid combinations of i, j, k, l values - so now I have a lot of experimentally found five-tuples {n,i, j, k, l}, which satisfy above parameterization, where n varies in the range from 2 to 26. Based on this data, of course, it would be nice to find how (if at all) i, j, k, l are inter-related between each other and with "n" - but such inter-relation (if exists) is not obvious and difficult to derive just by observation ... (though it is clearly seen that an absolute value of "i" is strongly increasing as "n" is growing from 2 to 26). RHS could be reduced (after performing integration - please let me know if I made a mistake in doing this) to: (abs(i)*2^j)^(-1)*Gamma(2*j+5)*((k+i)*Gamma(l+3)*HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2*j+l+8)+k*Gamma(l+1)*HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2*j+l+6)) May be from discussed parametric identity one could derive irrationality measure for pi, if to assume that RHS in this identity holds true, when the rational fraction on the LHS is equal to 0? Are there any {i,j,k,l}, which would satisfy such condition? Pi = (abs(i)*2^j)^(-1)*Gamma(2*j+5)*((k+i)*Gamma(l+3)*HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2*j+l+8)+k*Gamma(l+1)*HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2*j+l+6)) Update #2: Thanks to Jaume Oliver Lafont, at least one case, answering affirmatively to the last question, is identified: i=-1, j=-2, k=1, l=0 $$\pi = \int_{0}^{1}\frac{4}{1+x^2}\,\mathrm{d}x$$ Should there be infinite number of such cases? P.S. Per discussion with Jaume Oliver Lafont, depending on the value of the polynomial x degree in the integral body's numerator (while denominator stays to be the same "1+x^2"), the result varies from "Pi" to "log(2)" and also to "+/- (Pi - p/q)" as well as to "+/-(log(2)-p/q)", so perhaps now one could produce two distinct families of parameterization: one for Pi and the differences between Pi and its convergents and another for log(2) and the differences between log(2) and its convergents. P.P.S. While manipulating expressions in Wolfram Cloud Development Platform and WolframAlpha I came across the following parametric identity Sqrt[Pi] = (1/(2^j)*((k Gamma[5 + 2 j] Gamma[1 + l] HypergeometricPFQ[{1, 5/2 + j, 3 + j}, {3 + j + l/2,7/2 + j + l/2}, -1])/Gamma[6 + 2 j + l] + ((k + i) Gamma[7 + 2 j] Gamma[1 + l] HypergeometricPFQ[{1, 7/2 + j, 4 + j}, {4 + j + l/2,9/2 + j + l/2}, -1])/Gamma[8 + 2 j + l]))/(2^(-5 - 3 j -l) Gamma[5 + 2 j] Gamma[1 + l] (k HypergeometricPFQRegularized[{1, 5/2 + j,3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1] +1/2 (3 + j) (5 + 2 j) (k + i) HypergeometricPFQRegularized[{1,7/2 + j, 4 + j}, {4 + j + l/2, 9/2 + j + l/2}, -1])) (2) which indeed gave Sqrt[Pi] for each set of {i,j,k,l} given in above-listed cases 1), 2), 3), 4), 5), 6) I presume that above identity (2) will yield Sqrt[Pi] for other (infinite) number of sets of {i,j,k,l}. Is it an interesting identity? Thanks, Best Regards, Alexander R. Povolotsky I was notified that Maple simplifies the expression on the right hand side of (2) to sqrt(Pi). It seems to be true for arbitrary j,k,l,m. Based on above we could consider case: Sqrt[Pi] = (2^(5+3 j) (Gamma[5+2 j] Gamma[8+3 j] HypergeometricPFQ[{1,5/2+j,3+j},{3+(3 j)/2,7/2+(3 j)/2},-1]+2 Gamma[7+2 j] Gamma[6+3 j] HypergeometricPFQ[{1,7/2+j,4+j},{4+(3 j)/2,9/2+(3 j)/2},-1]))/(Gamma[5+2 j] Gamma[6+3 j] Gamma[8+3 j] (HypergeometricPFQRegularized[{1,5/2+j,3+j},{3+(3 j)/2,7/2+(3 j)/2},-1]+(15+11 j+2 j^2) HypergeometricPFQRegularized[{1,7/2+j,4+j},{4+(3 j)/2,(3 (3+j))/2},-1])) (3) REPLY [5 votes]: We have $$(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^{2(j+2)}(k+(i+k)x^2)\big)/(1+x^2)\; dx $$ $$= (\operatorname{sgn}(i) \cdot2^j)^{-1} \int_0^1 \big(x^{l+2} (1-x)^{2(j+2)}\big)/(1+x^2)\; dx +\frac{k}{ |i|\cdot2^j }\int_0^1x^l(1-x)^{2(j+2)}\; dx $$ Now $\frac{1}{2^j }\int_0^1x^l(1-x)^{2(j+2)}\; dx $ is just some rational number so by selecting $k$ and $i$ we can take the second term to be any rational number (positive if $k$ is required to be positive). So if we evaluate the left term and find it to be $\pm \pi$ plus a rational number, we can deduce that for any other number of the form $\pm \pi$ plus a rational number (with the same sign in the $\pm$) we can choose $k,i$ to make the integral equal that formula. This is easy to do. If we set $$\frac{x^{l+2} (1-x)^{2j+2}}{1+x^2} = \frac{a+bx}{x^2} + f(x)$$ for $a,b$ two rational numbers and $f(x)$ a polynomial in $x$ with rational coefficients, then $$x^{l+2} (1-x)^{2j+4} = a+bx +(1+x)^2 f(x)$$ $$ i ^{l+2} (1-i)^{2j+4} = a + bi$$ $$a+bi = i^{l+2} (-2i)^{j+2} = 2^2 \cdot 2^j \cdot i^{l-j} $$ As long as $j$ or $l$ are congruent modulo $2$, this is $\pm 4 \cdot 2^j$, so the leftmost term is the integral of a polynomial with rational coefficients plus $$ \pm 4 \int_0^1 \frac{1}{1+x^2} dx = \pm \pi$$ as desired.<|endoftext|> TITLE: An analogue of cabling for configuration spaces QUESTION [7 upvotes]: There is a well-known operation known as cabling for knots, and also for braid groups, where it is a homomorphism $$\beta_k \times \beta_\ell \longrightarrow \beta_{k\ell}$$ given by thickening up the $k$-strand braid and putting one copy of the $\ell$-strand braid inside each of its $k$ tubes. This arises from a map of unordered configuration spaces $C_k(\mathbb{R}^2)\times C_\ell(\mathbb{R}^2) \to C_{k\ell}(\mathbb{R}^2)$, and more generally one has an operation $C_k(M)\times C_\ell(\mathbb{R}^n) \to C_{k\ell}(M)$ for any parallelized manifold $M^n$. More generally again, if one has chosen $a$ linearly independent vector fields on $M$, there is an operation $C_k(M)\times C_\ell(\mathbb{R}^a) \to C_{k\ell}(M)$. In particular, if $M$ admits at least one non-vanishing vector field (either $M$ is non-compact or $\chi(M)=0$) there is a map $C_k(M) \to C_{k\ell}(M)$. This can be thought of as flowing each point of the configuration along the vector field and taking its image at $\ell$ different small time values to obtain the new configuration of $k\ell$ points. This map has been important in a joint project http://arxiv.org/abs/1406.4916 which I have worked on recently, where we call this the "$\ell$-replication map". We would be interested in knowing about anywhere in the literature where this has appeared before, as it seems a very natural operation to consider. We would be especially interested in the case of closed manifolds $M$. One reference which we have come across is the paper http://arxiv.org/abs/math/0701189, which is concerned with this replication map in the case $M=\mathbb{R}^2$. REPLY [2 votes]: Simpler maps $C_k(M)\to C_{k+1}(M)$ which "double a strand" appear in the work of Berrick, Cohen, Wong and Wu, where the induced maps on fundamental groups are shown to induce the degeneracies in a simplicial group structure on the braid groups of $M$. See Section 3 of http://www.ams.org/journals/jams/2006-19-02/S0894-0347-05-00507-2/<|endoftext|> TITLE: Does the symmetric group on an infinite set have a minimal generating set? QUESTION [29 upvotes]: To clarify the terms in the question above: The symmetric group Sym($\Omega$) on a set $\Omega$ consists of all bijections from $\Omega$ to $\Omega$ under composition of functions. A generating set $X \subseteq \Omega$ is minimal if no proper subset of $X$ generates Sym($\Omega$). This might be a difficult question, but perhaps the answer is known already? REPLY [4 votes]: The canonical paper on the subject (don't be fooled by the publication date, it had been around for years before then) is George Bergman's gem: Bergman, George M., Generating infinite symmetric groups., Bull. Lond. Math. Soc. 38, No. 3, 429-440 (2006). ZBL1103.20003.<|endoftext|> TITLE: Which sets occur as boundaries of other sets in topological spaces? QUESTION [24 upvotes]: This question was originally asked on MathStackExchange and is migrated here with opinion from MO meta. I am integrating the inputs from users Daniel Fischer and Emil Jerabek there into this post. Which sets occur as boundaries of other sets in topological spaces? Of course the boundary of a set is closed. But not every closed set in a topological space is the boundary of some set in that space; only the empty set occurs as a boundary in a discrete space. More generally, boundaries cannot contain isolated points of the ambient space (but can well have isolated points of itself). It is tempting to assert that boundaries have empty interiors, but this is not true, as is shown by the fact that the boundary of Q in R is R. In fact it can be seen in general that the boundary of a dense set with empty interior is the whole space. Thus the only boundary in an indiscrete space is the whole set, just as the only boundary in a discrete space is the empty set. However the intuitive feeling comes right for open sets (and then for closed sets as well): The boundary of an open set cannot contain an open set. This question concerns all subsets of a topological space. An alternative/related question shall be to characterise all topological spaces in which every closed set occurs as a boundary. Being a perfect space (i.e., one without isolated points) is a necessary condition, as a previous remark above about isolated points shows. Emil Jerabek has conjectured on meta that this (being perfect) is sufficient too, for $T_0$ second countable spaces. REPLY [13 votes]: Building on Joseph Van Name's answer A closed set is a boundary if and only if its interior is a resolvable topological space Proof: $\leftarrow$ If $A$ is a subset of $X$ whose interior $A^o$ is resolvable, i.e., $A^o=B\cup (A^o-B)$ where $B$ and $A^o-B$ are dense in $A^o$, let $C$ be the union of the complement of the set and one of the two dense parts of the interior, $C=B\cup A^c$. The interior of this union is just the complement of the set, $C^o=(B\cup A^c)^o=A^c$, as $B$ cannot contain interior points if $A^o-B$ has to be dense in $A^o$. The closure of this union, $\overline{C}=\overline{B\cup A^c} $ includes the interior $A^o$ and the complement $A^c$ of $A$, and is closed, so is the whole space, i.e., $\overline{C}=X$. Hence, $\partial C=\overline{C}-C^o=X-A^c=A.$ $\rightarrow$ For this part, note that the property of a set being a boundary is preserved by restriction to open subsets, and use Joesph Van Name's answer.<|endoftext|> TITLE: Approximate Moment Conditions QUESTION [5 upvotes]: It is known in classical probability that if two random variables $X$ and $Y$ obeys $$\mathbb{E} X^k = \mathbb{E}Y^k, \ \forall \ k \geq 1$$ with additional condition that $\mathbb{E}X^k$ does not grow too fast, then $X$ is equal to $Y$ in distribution. I was wondering that has anybody studied the same problem with the equality of moments replaced by approximation? Specifically, suppose $$|\mathbb{E} X^k - \mathbb{E}Y^k| \le \alpha_k, \ \forall \ k \geq 1$$ for some small $\alpha_k$, can we give a descent bound of $$|\mathbb{P}(X \le t) - \mathbb{P}(Y \le t)|$$ for some $t$ or uniformly in $t$ in terms of the gaps $\alpha_k$. We may assume that both $X$ and $Y$ are supported in $[0, 1]$. Many thanks! John REPLY [3 votes]: Here is one approach you can use. The differences $a_k$ provide you with bounds on the difference between the characteristic functions of the two distributions. This may be easy or not depending on your example. Then you can apply known bounds on the difference between two distributions based on the difference between their characteristic functions, for example Lemma 2 in Chapter XVI Section 3 of Feller, An Introduction to Probability Theory and its Applications, Vol II.<|endoftext|> TITLE: Lifting a Diffeomorphism to the Cotangent Bundle QUESTION [10 upvotes]: Both Abraham-Marsden and Da Silva seem to imply that given a symplectomorphism $g:T^\ast X\to T^\ast X$ which preserves the tautological $1$-form $\alpha$, it must be that $g$ is fibre preserving. In fact, this is addressed in the question Cotangent bundle lift theorem. The answer given to this question stresses very clearly that g must preserve fibres to have any chance of being a lift, even if it preserves the canonical 1-form. I would really like to see a counterexample to this. That is, a symplectomorphism which preserves the canonical $1$-form but is not the lift of a diffeomorphism. I have struggled with trying to prove both Abraham-Marsden and Da Silva's claim. Any help would be greatly appreciated. I can show the following: If $V$ is the symplectic dual of $\alpha$ (i.e. $\omega(V,\cdot)=\alpha$), then $g_\ast V=V$, or equivalently $g\circ\theta_t=\theta_t\circ g$ where $\theta_t$ is the flow of $V$. Also I can show that $\theta_t$ is fibre preserving (i.e. $\theta_t(T_p^\ast X)=T_p^\ast X)$. This last fact follows from actually computing $\theta_t$ which is $\theta_t(p,\xi_p)=(p,e^t\xi_p)$. REPLY [9 votes]: The vector field $V$ is the radial vector field in the fibers, and it vanishes along the zero locus of $\alpha$, which is the zero section $Z\subset T^*M$. Thus, $g$ must preserve $Z$ and hence be a diffeomorphism of $Z$, say $f:Z\to Z$. Since $Z\to M$ is a diffeomorphism, this can be regarded as a diffeomorphism $f:M\to M$. Since $g$ preserves $V$, it must, for each $z\in Z$, i.e., $z = 0_x$ for $x\in M$, carry the $\alpha$-limit set $S_z$ of $z$ with respect to $V$ into (and diffeomorphically onto) the $\alpha$-limit set of $g(z) = 0_{f(x)}$, but $S_{0_x}=T^*_xM$ and $S_{0_{f(x)}}=T^*_{f(x)}M$. Thus, $g(T^*_xM) = T^*_{f(x)}M$. The rest is now clear, I think.<|endoftext|> TITLE: Main Gap Phenomenon QUESTION [14 upvotes]: Shelah's Main Gap Theorem states that for all first-order, complete theories, T, in a countable language, we have that either $$I(T,\aleph_\alpha)=2^{\aleph_\alpha}$$ or $$I(T,\aleph_\alpha)<\beth_{\omega_1}(\alpha)$$ This result, while beautiful, leaves some questions. Can analogues of this result be extended to larger languages? Do other theorems exists for logics like $L_{\omega_1 \omega}$ or $L_{\omega_1 \omega}(Q)$? Do there exist any logics without the 'gap phenomenon'? Any resources would be helpful. Thanks! REPLY [9 votes]: Extending Shelah's main gap to non first-order or even to first-order theories in an uncountable language is a major hard open problem. I am familar with couple of extensions to non f.o. (both papers are available from my web page): Rami Grossberg and Bradd Hart. The classification theory of excellent classes, Journal of Symbolic Logic, 54, (1989) 1359--1381. Rami Grossberg and Olivier Lessmann. Abstract decomposition theorem and applications, Contemporary Mathematics, Vol 380, (2005), AMS, pp. 73--108. For full $L_{\omega_1,\omega}$ this seems to be super hard as it may require proving Shelah's categoricity conjecture for $L_{\omega_1,\omega}$ which despite of many hundreds of pages of approximation is still open.<|endoftext|> TITLE: Incomplete Kloosterman sum QUESTION [11 upvotes]: I am interested in an upper bound on the following incomplete Kloosterman sum $$ \sum_{\substack{x=1 \\ x+_{_{\bf Z}}x^{-1}>p}}^{p-1}e\left(\frac{x+x^{-1}}{p}\right).$$ Using the Weil's bound it is easy to show that the real part of the sum is bounded by $\sqrt p.$ It is because if $x+x^{-1}>p$ then $(p-x) + (p-x)^{-1}< p.$ Note that $x^{-1} \in \{1, \cdots , p-1\}$ and $xx^{-1} \equiv 1 \text{ mod } p$. And by $+_{_{\bf Z}}$ we mean that the sum in $\bf Z$ not in $\bf Z_p$. REPLY [12 votes]: We can estimate this using the Polya-Vinagradov method. We get a main term, which comes from the fact that two elements of $\mathbb F_p$ that sum to something greater than $p$ are more likely to sum to something a little bit greater than $p$ than a lot, and an error term. The formula is: $$ \frac{ i p}{2\pi} + O( \sqrt{p}\log{p} )$$ View the sum as a sum of the product of two characteristic functions and an exponential funcion: $$\sum_{x,y\in \mathbb F_p}\mathbf 1_{\{xy=1\} } e(x+y) \mathbf 1_{\{x+y>p\}}$$ Let $f(a,b)$ be the Fourier transform of $\mathbf 1_{\{xy=1\} }$. Let $g(a,b)$ be the Fourier transform of $\mathbf 1_{\{x+y>p\}}$. Then by Plancherel's formula, this sum is: $$\frac{\sum_{a,b\in \mathbb F_p} f(a+1,b+1) \overline{g} ( a,b)}{p^2} $$ This sum, it turns out, is easier to estimate. Our first function: $$f(a,b) = \sum_{x \in \mathbb F_p} e(ax+ bx^{-1} ) = K(ab)$$ is a Kloosterman sum, unless $a=0$ or $b=0$, in which case it is $-1$, unless both $a$ and $b$ are $0$, in which case it is $p-1$. In particular, it is bounded by $2 \sqrt{p}$, unless $a=b=0$, in which case it is $p-1$. Our second sum we may estimate by more elementary means: $$g(a,b) = \sum_{0\leq x,yp} e(ax + by) = \sum_{1\leq x 1 \}$. That integral is elementary, and is not zero; it turns out that $C = i/2\pi$. A more precise estimate using Weil's bound is $$ \sum_{x=1}^{p-1} e\Bigl(\frac{x+x^{-1}}{p}\Bigr) = \frac{i}{2\pi} p + O(p^{\frac12 + \epsilon}), $$ and if I did this right then the $p^\epsilon$ factor can be replaced by $\log^2 p$. (I see that while I was writing this up Will Sawin posted much the same answer, claiming that moreover even $\log^2 p$ has one log factor more than necessary.) [added later] My online notes on analytic number theory include a short chapter "An application of Kloosterman sums" one of whose exercises outlines an elementary proof that each Kloosterman sums is $O(p^{3/4})$; this is weaker than the Weyl bound, but still sufficient to prove the asymptotic equidistribution of $\{ (x,y) \in (0,p)^2 \mid xy \equiv c \bmod p \}$ for any $c \in ({\bf Z} / p{\bf Z})^*$. See Exercise 5 on page 3 (and also Lemma 1 on page 1 for an estimate that implies the equidistribution result).<|endoftext|> TITLE: Spencer's "six standard deviations" theorem - better constants? QUESTION [24 upvotes]: This question is about Joel Spencer's famous "six standard deviations" theorem. The theorem says that when $$ L_i(x_1,\dots,x_n) = a_{i1} x_1 + \dots + a_{in} x_n, \quad 1 \leq i \leq n, $$ are $n$ linear forms in $n$ variables with all $|a_{ij}| \leq 1$, then there exist numbers $\varepsilon_1,\dots,\varepsilon_n \in \{-1,+1\}$ such that $$ |L_i(\varepsilon_1,\dots,\varepsilon_n)| \leq K \sqrt{n} $$ for all $i$. It is stated as Theorem 1 in: Spencer, Joel. Six standard deviations suffice. Trans. Amer. Math. Soc. 289 (1985), no. 2, 679–706. Full text PDF (open access) As noted at the end of the paper, the constant $K$ which Spencer obtained is actually $K=5.32$. Question: does anybody know of a proof of the Theorem which gives a smaller value for the constant? REPLY [18 votes]: Having done the original result it is exciting to see work on the constant. My 5.32 was certainly not meant to be best possible. For the lower bounds, using Hadamard Matrices (with coefficients +1,-1 as opposed to the set system with coefficients 0,1) gives a lower bound of K=1. -- js<|endoftext|> TITLE: What recent programmes to alter highly-entrenched mathematical terminology have succeeded, and under what conditions do they tend to succeed or fail? QUESTION [49 upvotes]: I think we all occasionally come across terminology that we'd like to see supplanted (e.g. by something more systematic). What I'd like to know is, under what circumstances is it reasonable to believe that such a fight is winnable? Question. What recent programmes to alter highly-entrenched mathematical terminology have succeeded, and under what conditions do they tend to succeed or fail? Definitions. Recent = In the past fifty years. Highly-entrenched = The literature at the time had the old terminology written all over it. Succeeded = The new terminology is now used almost exclusively in research papers. REPLY [13 votes]: From around 1900 to 1970, there was a highly-entrenched practice to write maps as $f:x\to y$ with an arrow between argument and value (e.g. Weyl 1913, p. 54). Starting in the 1930s, a programme arose to write them instead as $f:X\to Y$, with the arrow now between domain and codomain (e.g. Eilenberg-Steenrod 1952, p. 4). These conflict: does $f:X\to X\cup\{X\}$ mean von Neumann’s successor map, or an unspecified map between these sets? This can get bad when dealing with sheafs, or localization. In the late 1950s, some bourbakists (Serre and Grothendieck?) attempted to lift the ambiguity by writing argument-to-value with what they called a tortillon : $x\rightsquigarrow y$. This didn't catch on. Around early 1964, more bourbakists (Serre and Borel?) tried again, this time with what they called a poussoir : $x\mapsto y$. This was a hit, and almost entirely displaced the old $\to$ in research papers. Lesson: enroll Borel?<|endoftext|> TITLE: LU factorization for $I+A$ (A skew-symmetric) QUESTION [11 upvotes]: The matrix $M=I+A\in \mathbb{R}^{n\times n}$, where $I$ is the identity and $A=-A^T$ is skew-symmetric satisfies $$ Mx\cdot x = \|x\|^2 \quad\text{for all }x\in\mathbb{R}^n. $$ Therefore, $M=LU$ has an unique $LU$-factorization. Numerical experiments with Matlab suggest that there holds the following estimate $$ \|L\| + \|U\|\leq C(1+\|A\|^2), $$ where $\|\cdot\|$ is the spectral norm and the real constant $C>0$ does not depend on the dimension $n\in\mathbb{N}$. Does anyone know of a proof for that? REPLY [4 votes]: EDIT 1. For $n=2$, the smallest $C$ is as follows. In all calculations we use the "Groebner" library of Maple. Let $C_n$ be the smallest $C$ (if there exists) in dimension $n$. Let $a\approx 0.3787$ be the number satisfying $4a^8+27a^6+34a^4-117a^2+16=0$. Note that $a$ can be calculated by radicals. Then $C_2=\dfrac{\sqrt{2a^2+4+2a\sqrt{a^2+4}}+\sqrt{2a^4+6a^2+4+2a(a^2+1)\sqrt{a^2+4}}}{2(a^2+1)}$. Moreover $C_2\approx 2.184596$ is a root of the following polynomial: $20736x^8-122832x^6+126913x^4-62224x^2+1024$. Note that $C_2$ can be calculated by radicals. EDIT 2. When $n=3$, let $A=\begin{pmatrix}0&-a_3&a_2\\a_3&0&-a_1\\-a_2&a_1&0\end{pmatrix}$ ; there are $2$ cases. Case 1. $a_1a_2a_3\geq 0$. Then the bound for $C$ seems to be $C_2$. Case 2. $a_1a_2a_3\leq 0$. Then the bound is $>C_2$ and $C_3>C_2$. Indeed, let $a_1=-0.25,a_2=0.29,a_3=0.029$ ; then $(||L||+||U||)/(1+{a_1}^2+{a_2}^2+{a_3}^2)\approx 2.18577$.<|endoftext|> TITLE: Strongly compact cardinal with bad covering properties QUESTION [8 upvotes]: This is a continuation of the question covering properties of strongly compact embedding. Recall that a cardinal $\kappa$ is $\nu$-strongly compact cardinal if there is an elementary embedding $j:V\rightarrow M$, with critical point $\kappa$, $j(\kappa) > \nu$, such that $j^{\prime \prime} \nu \subset s$ where $|s|^M < j(\kappa)$. Question 1: Is it consistent (relative to the existence of large cardinals, of course), that there is a $\nu$-strongly compact cardinal $\kappa$, $\nu > 2^{\kappa}$ regular, such that for some $j:V\rightarrow M$, (a $\nu$-strongly compact embedding elementary embedding with $crit(j)=\kappa$), $j^{\prime\prime} \nu$ cannot be covered by any set from $M$ with order-type $\nu$? Assuming that the answer to the first question is positive, how far can we get? namely: Question 2: With the same notations, how large can be the gap between the minimal $M$-cardinality of a set in $M$ that covers $j^{\prime\prime} \nu$ and $\nu$? Does there is any non trivial restriction on this $M$-cardinal? REPLY [3 votes]: The answer is yes, and indeed the situation is not merely consistent with strong compactness, but rather every strongly compact cardinal has such embeddings with no small covers. Theorem. Suppose that $\kappa$ is $\delta^+$-strongly compact, where $\delta^{\lt\kappa}=\delta$. Then there is a $\delta^+$-strong compactness embedding $j:V\to M$ such that there is no $s\in M$ with $j''\delta^+\subset s$ and $|s|^M=\delta^+$. Proof. Let $j_0:V\to M_0$ be any $\delta^+$-strong compactness embedding. Let $s_0\in M_0$ be a cover, so that $j_0''\delta^+\subset s_0$, and we may assume without loss of generality that $|s_0|^{M_0}=\delta^+$, for if there is no such cover then we are already done. Let $j_1:V\to M_1$ be any $\delta$-strong compactness ultrapower. We may consider $h=j_1\upharpoonright M_0$, so that $h:M_0\to M$ is an elementary embedding of $M_0$ into $M=j_1(M_0)=\bigcup_\alpha j_1(V_\alpha^{M_0})$. Note that we do not necessarily have the measure for $j_1$ inside $M_0$, and so this embedding $h$ is defined in $V$ rather than in $M_0$. Let $j=h\circ j_0:V\to M$ be the composition elementary embedding. (Although the argument here will not use it, in fact one can prove that $j:V\to M$ is the ultrapower by the product measure $\mu\times\nu$, where $j_0$ is the ultrapower by $\nu$ and $j_1$ is the ultrapower by $\mu$.) I claim that $j$ is a $\delta^+$-strong compactness embedding. It clearly has critical point $\kappa$. Define $s=j_0(s_0)$. It follows easily that $j''\delta^+=h''j_0''\delta^+\subset h''s_0\subset h(s_0)=s$, and so $s$ covers $j''\delta^+$. And $|s|^M=h(|s|^{M_0})=h(\delta^+)$, which is less than $h(j_0(\kappa))=j(\kappa)$. Meanwhile, since $s_0$ has size $\delta^+$ in $M_0$, it follows that $\sup j_0''\delta^+$ has cofinality $\delta^+$ in $M_0$. But now the key point is that $j_1$ and hence $h$ is continuous at ordinals of cofinality $\delta^+$. Thus, $h(\sup j_0''\delta^+)=\sup h''j_0''\delta^+=\sup j''\delta^+$. Therefore, any covering set of $j''\delta^+$ in $M$ must have size at least the cofinality of this supremum, which is $h(\delta^+)$, which is the same as $j_1(\delta^+)$, which is larger than $j_1(\kappa)$ which is significantly larger than $\delta^+$ in $M$. So every cover of $j''\delta^+$ in $M$ has size much larger than $\delta^+$. QED<|endoftext|> TITLE: How does one justify funding for mathematics research? QUESTION [142 upvotes]: G. H. Hardy's A Mathematician's Apology provides an answer as to why one would do mathematics, but I'm unable to find an answer as to why mathematics deserves public funding. Mathematics can be beautiful, but unlike music, visual art or literature, much of the beauty, particularly at a higher level, is only available to the initiated. And while many great scientific advances have been built on mathematical discoveries, many mathematicians make no pretense of caring about any practical use their work may have. So when the taxpayer or a private organization provides grants for mathematical research, what are they expecting to get in return? I'm asking not just so I can write more honest research proposals, or in case it comes up in an argument, but so I have an answer for myself. REPLY [4 votes]: Since this question has popped up to the front page, and at the risk of interpreting it over-broadly, I think it is worth reproducing a famous quote by Carl Gustav Jacob Jacobi in a letter to Legendre dated 2 July 1830: Il est vrai que M. Fourier avait l'opinion que le but principal des mathématiques était l'utilité publique et l'explication des phénomènes naturels ; mais un philosophe comme lui aurait dû savoir que le but unique de la science, c'est l'honneur de l'esprit humain, et que sous ce titre, une question de nombres vaut autant qu'une question du système du monde. (Carl Gustav Jacob Jacobi, Gesammelte Werke, band 1, Reiner, Berlin 1881, Correspondance mathématique avec Legenre, 454–455, scan here.) Translation from Wikiquote (ever-so-slightly edited): It is true that Mr Fourier had the opinion that the principal end of mathematics was the public utility and the explanation of natural phenomena; but such a philosopher as he is should have known that the unique end of science is the honor of the human mind, and that from this point of view a question of numbers is as important as a question of the system of the world. The reference to the "honor of the human mind" was used, in particular, by Dieudonné for the title of his book, Pour l'honneur de l'esprit humain (Les Mathématiques d'aujourd'hui) (1987).<|endoftext|> TITLE: Covering Spaces and Vector Bundles QUESTION [12 upvotes]: Suppose $f: Y \rightarrow X$ is a covering map between compact Hausdorff spaces $X$ and $Y$. Then $f$ induces a algebra homomorphism $f^*:C(X) \rightarrow C(Y)$ and gives $C(Y)$ the structure of a finitely generated projective module over $C(X)$. Hence $C(Y)$ can be viewed as the sections of a complex vector bundle $E \rightarrow X$. I am interested in the converse. Given a complex vector bundle $E \rightarrow X$, when is $E$ defined from a covering map as above? In particular, is this the case if the first Chern character $c_1(E) \in \check{H}^2(X ; \mathbb{Z})$ vanishes? Is there always a complex line bundle $L \rightarrow X$ such that $E \otimes L$ is defined by a covering map as above? REPLY [10 votes]: The category of flat vector bundles is equivalent to the category of local systems, see for instance this MO-question, which in turn are equivalent to representations of the fundamental group, see this MO-question. Via these correspondences, the vector bundle associated to the covering corresponds to the representation of the fundamental group of $X$ on the fibre of $f:Y\to X$ via deck transformations. In particular, the vector bundles associated to coverings are flat. A flat vector bundle has trivial characteristic classes, but triviality of characteristic classes is not sufficient for flatness, as the answer of abx shows.<|endoftext|> TITLE: Density of p-ordinary modular forms QUESTION [8 upvotes]: Fix an odd prime $p$. For concreteness, let $N$ be coprime to $p$, and let $2 \leq k \leq p$. Let $S^+(N,k)$ be the newforms in $S_k(\Gamma_1(N))$. Let $f = \sum a_n q^n \in S^+(N,k)$. We say that $f$ is $p$-ordinary if $v_p(a_p)=0$. (If $p$ splits in the field of coefficients of $f$, we require that $a_p$ not be divisible by any prime $\mathfrak{p}$ above $p$.) For a "random" $N$ and $k$ satisfying the conditions I set out at the beginning, what can be said about the proportion of $p$-ordinary forms in $S^+(N,k)$? If we let $N$ and $k$ vary, can we say anything about the density of $p$-ordinary forms in the union $\bigcup_{N,k} S_k^+(\Gamma_1(N))$? REPLY [2 votes]: An ordinary modular form needs to be $p$-stable (by definition). If you have a modular form of level $N$ co-prime to $p$, you need to associate a $p$-stabilization to this latter to get an ordinary modular form with level $Np$. If you take $T$ the Hecke algebra acting on Katz p-adic modular forms, it is known that the classical points are dense in $\operatorname{Spec}T$, and $T$ has krull dimension at least $4$, but the ordinary Hecke algebra has a dimension equal to $2$.<|endoftext|> TITLE: Morphisms every pushout of which is a weak equivalence QUESTION [6 upvotes]: Let $M$ be a category equipped with a class of weak equivalences $W$. Is there a name for a morphism $f$ such that every pushout of $f$ (including, of course, $f$ itself) is a weak equivalence? For example, if $(M,W)$ underlies a model structure, then any acyclic cofibration in the model structure is such a map. But the concept as defined depends only on the weak equivalences. For example, we may be primarily considering a particular model structure, but if it so happens that there is another model structure with the same (or smaller) class of weak equivalences, then the acyclic cofibrations in that other model structure will also be examples of this concept. A name for a the dual concept (a morphism every pullback of which is a weak equivalence) would of course be just as good, since then I could stick a "co" in front of it. REPLY [3 votes]: Maps $f$ such that every pushout along $f$ is a weak equivalence were called couniversal weak equivalences in the preprint Homotopy theory for algebras over polynomial monads by Michael Batanin and Clemens Berger. In left proper model categories such maps are characterized in Lemmas 1.5 and 1.6. I'm not sure I agree with calling these maps 'flat' as that word is already so over-used. For example, this terminology could easily cause confusion in examples like $Ch(R)$ or the stable module category. Furthermore, a common axiom for monoidal model categories is that whenever $X$ is cofibrant and $f$ is a weak equivalence, $X\otimes f$ is a weak equivalence. Motivated by the examples above, this axiom has sometimes been called the axiom that 'cofibrant objects are flat.' I discussed this axiom a bit at this mathoverflow thread. I don't know if that terminology will stick (in the paper above this axiom is called the Resolution Axiom, and in a preprint of Pavlov and Scholback it's called the left cow axiom), but it's more evidence to avoid using the already-saturated 'flat.'<|endoftext|> TITLE: Is an extension of compact Hausdorff topological groups compact? QUESTION [5 upvotes]: Let $1 \rightarrow A \xrightarrow{a} B \xrightarrow{c} C \rightarrow 1$ be a short exact sequence of topological groups (i.e., all maps are continuous, $A = \mathrm{Ker}(c)$, and $C = \mathrm{Coker}(a)$). Suppose that $A$ and $C$ are compact Hausdorff. Does it follow that $B$ is compact? REPLY [5 votes]: You have to assume that $a$ is a homeomorphism onto its image. Indeed, if you don't then you get counterexamples with $C=1$: let $A$ be a finite group with the discrete topology and let $B$ be the same group with the indiscrete topology. You also have to assume that $c$ is a topological quotient map. Indeed, if you don't then you get counterexamples with $A=1$: let $B$ be the circle with the discrete topology and $C$ be the circle with the usual compact Hausdorff topology. With these extra topological assumptions, everything works out fine. First note that $a(A) = \ker c$ is closed in $B$ since $C$ is Hausdorff, so the identity point of $B$ is closed as we can check it in the Hausdorff $A$ (using that $a$ is a homeomorphism onto a closed image). Hence, $B$ is Hausdorff (as for any topological group with closed identity point). This didn't use the compactness (and is also not at all interesting). Now that $B$ is Hausdorff, we can argue with nets: if $\{x_i\}$ is a net in $B$ then we can pass to a subnet so that $\{c(x_i)\}$ converges to some $y \in C$. Writing $y = c(x)$ and replacing $x_i$ with $x_i x^{-1}$ then reduces us to the case that $c(x_i) \rightarrow 1$ in $C$. But $C$ has the quotient topology from $B$ by hypothesis, so after passing to a subnet we can choose $a_i \in A$ such that $x_i a_i^{-1} \rightarrow 1$ in $B$. Since $A$ is compact Hausdorff, passing to a further subnet allows us to arrange that $a_i \rightarrow a$ in $A$, so $x_i a^{-1} \rightarrow 1$ in $B$; i.e., $x_i \rightarrow a$ in $B$. QED You may like to consider the more interesting/useful case when "compact" is relaxed to "locally compact".<|endoftext|> TITLE: Subset of $F_2^n$ that must contain some subspace of dimension $k$ QUESTION [6 upvotes]: This question has practical meanings in algebraic attack of stream ciphers in cryptography. It can be stated as follows: Suppose $V$ is a $n$ dimensional vector space over the field $F_2$, where $F_2$ is the simplest finite field contains only $\{0,1\}$. you can take $V$ to be $F_2^n$. $S$ be a subset of $V$. The question is, What is the minimum of the size of $S$ such that $S$ must contain some $k$ dimensional linear sub-manifold $M$ of $V$, where $k$ is a given positive integer below $n$? Here by a linear sub-manifold $M$ of dimension $k$, we mean that $M$ can be expressed as $M=\{x+W\}$, where $x$ is a fixed vector in $V$ and $W$ is a $k$ dimensional subspace of $V$. It's obvious that $|S|>2^k$，since any sub-manifold of dimension $k$ contains exactly $2^k$ points but there are subsets of size $2^k$ that are not submanifolds. I think this problem must has been considered by combinatorists, but I didn't find it in any standard textbooks on combinatorics. REPLY [8 votes]: As noted by Seva, you're looking for Szemeredi's cube lemma, which is one of the simplest results of its kind. Let $S\subset\mathbf{F}_2^n$, and let $\|S\|_{U^k}^{2^k}$ denote the number of $e_0,e_1,\dots,e_k\in \mathbf{F}_2^n$ such that $$\{e_0 + x_1 e_1 + \cdots + x_k e_k : x_i\in\{0,1\}\}\subset S,$$ normalised by $2^{n(k+1)}$. In other words define $$\|S\|_{U^k}^{2^k} = 2^{-n(k+1)}\sum_{e_0,e_1,\dots,e_k} \prod_{x\in\{0,1\}^k} 1_S(e_0 + x_1 e_1 +\cdots + x_k e_k).$$ Then $\|S\|_{U^1} = |S|/2^n$, and by the Cauchy-Schwarz inequality we have $\|S\|_{U^k} \leq \|S\|_{U^{k+1}}$ for all $k$, so $\|S\|_{U^k}^{2^k} \geq |S|^{2^k}/2^{n2^k}$. The number of $e_0,e_1,\dots,e_k$ for which $$|\{e_0+x_1e_1+\cdots + x_ke_k: x_i\in\{0,1\}\}|<2^k$$ is at most $k 2^{nk} 2^{k-1}$, so there must be at least one affine subspace of dimension $k$ in $S$ if $$|S| > 2^{n(1-2^{-k}) + 1}.$$ Upper bounds will come from random examples. Include each point of $\mathbf{F}_2^n$ in $S$ independently with probability $p$. Then any fixed $k$-dimensional affine subspace of $\mathbf{F}_2^n$ is contained in $S$ with probability exactly $p^{2^k}$. The number of such affine subspaces is at most $2^{n(k+1)}$, so the expected number contained in $S$ is at most $2^{n(k+1)} p^{2^k}$. Taking $p\sim 2^{-n(k+1)/2^k}$, you get at least one set $S$ of size at least $2^{n(1 - (k+1)2^{-k})}$ not containing such a subspace. So the exponent is somewhere between $1-(k+1)2^{-k}$ and $1 - 2^{-k}$. I think for $k>2$ the exact value is not known. EDIT: One can improve the random example a little as follows. If the expected number of subspaces is $2^{n(k+1)} p^{2^k}$ then take any set with at most this many subspaces and then just remove one point from each subspace. You get a set of size $2^n p (1 - 2^{nk} p^{2^k-1})$, so it suffices to take $p\sim 2^{-nk/(2^k-1)}$, so the exponent is somewhere between $ 1-k/(2^k-1)$ and $1-2^{-k}$.<|endoftext|> TITLE: A partition congruence modulo 13 QUESTION [10 upvotes]: In the paper "Note on certain modular relations considered by Messrs Ramanujan, Darling and Rogers" (Proceedings of London Mathematical Society (1922) s2-20 (1): 408-416) Mordell gives proofs of the identities \begin{align} \sum_{n = 0}^{\infty}p(5n + 4)q^{n} &= 5\frac{\{(1 - q^{5})(1 - q^{10})(1 - q^{15})\cdots\}^{5}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{6}}\tag{1}\\ \sum_{n = 0}^{\infty}p(7n + 5)q^{n} &= 7\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{3}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{4}}\notag\\ &\,\,\,\,+ 49q\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{7}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{8}}\tag{2} \end{align} based on theory of elliptic modular functions. He uses a slightly old notation $$\Delta(\omega_{1}, \omega_{2}) = \left(\frac{2\pi}{\omega_{2}}\right)^{12}q\prod_{n = 1}^{\infty}(1 - q^{n})^{24}$$ where $q = e^{2\pi i\omega}, \omega = \omega_{1}/\omega_{2}$. Clearly both the series given by Ramanujan lead to partition congruences modulo $5, 5^{2}, 7, 7^{2}$. Mordell shows that these results are equivalent to $$\sum_{j = 0}^{4}\left(\frac{\Delta(5\omega_{1}, \omega_{2})}{\Delta\{(\omega_{1} + j\omega_{2})/5, \omega_{2}\}}\right)^{1/24} = 5\left(\frac{\Delta(5\omega_{1}, \omega_{2})}{\Delta(\omega_{1}, \omega_{2})}\right)^{1/4}\tag{3}$$ and \begin{align} \sum_{j = 0}^{6}\left(\frac{\Delta(7\omega_{1}, \omega_{2})}{\Delta\{(\omega_{1} + j\omega_{2})/7, \omega_{2}\}}\right)^{1/24} &= 7\left(\frac{\Delta(7\omega_{1}, \omega_{2})}{\Delta(\omega_{1}, \omega_{2})}\right)^{1/6}\notag\\ &\,\,\,\,+ 49\left(\frac{\Delta(7\omega_{1}, \omega_{2})}{\Delta(\omega_{1}, \omega_{2})}\right)^{1/3}\tag{4} \end{align} Mordell goes on to mention that identities like $(3), (4)$ don't exist if we replace replace $5, 7$ by $11$. On the other hand he says that there is similar identity for modulo $13$. This seems very weird because we do have a congruence identity $p(11n + 6) \equiv 0\pmod{11}$ and we don't have a similar partition congruence $\pmod{13}$. What Mordell probably means is that we do not have an expansion like $(1)$ or $(2)$ for $p(11n + 6)$ and at the same time we do have some formula like $(1), (2)$ for $p(13n + a)$ for some integer $a, 0 < a < 13$. Is my understanding mentioned in last paragraph correct? If so, do we have an identity like $(1), (2)$ available in literature for modulo $13$? Any references regarding such an identity and its proof would be really helpful. REPLY [2 votes]: Herbert S. Zuckerman in his paper Identities analogous to Ramanujan's identities involving the partition function (published in Duke Mathematical Journal Vol 5,1939, pages 88-110) gives the following formula (which is rather cumbersome) $$\sum_{n=0}^{\infty} p(13n+6)q^n=11\prod_{n=1}^{\infty} \frac{(1-q^{13n})}{(1-q^n)^2}+468q \prod_{n=1}^{\infty} \frac{(1-q^{13n})^3}{(1-q^n)^4}+6422q^2 \prod_{n=1}^{\infty} \frac{(1-q^{13n})^5}{(1-q^n)^6}+43940q^3 \prod_{n=1}^{\infty} \frac{(1-q^{13n})^7}{(1-q^n)^8}+171366q^4 \prod_{n=1}^{\infty} \frac{(1-q^{13n})^9}{(1-q^n)^{10}}+371293q^5 \prod_{n=1}^{\infty} \frac{(1-q^{13n})^{11}}{(1-q^n)^{12}}+371293q^6 \prod_{n=1}^{\infty} \frac{(1-q^{13n})^{13}}{(1-q^n)^{14}}$$ All the coefficients on the right hand side except the first one ($11$) are divisible by $13$. Thus even though there is (not-so) nice generating function for $\sum p(13n+6)q^n$ there is no corresponding simple congruence. Oddmund Kolberg in his paper Some identities involving the partition function (published in Mathematica Scandinavica, Vol 5, 1957, pages 77-92) gave a much simpler and systematic procedure to prove Ramanujan's generating functions described in this question. The same technique of Kolberg has been successfully applied by Göksal Bilgici and A. Bülent Ekin in their paper Some congruences for modulus 13 related to partition generating function (published in Ramanujan Journal, Vol 33, 2014, pages 197-218) to prove the above mentioned formula.<|endoftext|> TITLE: Maps to projective space == line bundles; what do maps to weighted projective space correspond to? QUESTION [20 upvotes]: A map from an algebraic variety $X$ to a projective space is the same thing as a globally generated line bundle on $X$. What geometric object on $X$ corresponds to a map to a weighted projective space? REPLY [20 votes]: Let me expand my comment in a short answer. One of the most common way to build a rational map $f \colon X \dashrightarrow \mathbb{P}(a_1, \ldots, a_n)$ is to consider a line bundle $\mathscr{L}$ on $X$ together with sections $$\sigma_1 \in H^0(X, \, \mathscr{L}^{a_1}), \quad \sigma_2 \in H^0(X, \, \mathscr{L}^{a_2}), \ldots, \sigma_n \in H^0(X, \, \mathscr{L}^{a_n}),$$ and to define $f(x) := [\sigma_1(x): \ldots : \sigma_n(x)].$ Clearly, $f$ is a morphism if and only if the intersection of the zero loci of the sections $\sigma_i$ is empty. The basic example is given by hyperelliptic curves. Let $X$ be a smooth hyperelliptic curve of genus $g$ and take as $\mathscr{L}$ the hyperelliptic involution, i.e. the unique $g^1_2$ of $X$. Then it is no difficult to show that there exist $x_0, \, x_1 \in H^0(X, \, \mathscr{L})$ and $y \in H^0(X, \mathscr{L}^{g+1})$ that satisfy a polynomial relation of type $$y^2 = F_{2g+2}(x_0, \, x_1), \quad (\sharp)$$ where $F_{2g+2}$ is a homogeneous form of degree $2g+2$. This precisely means that we can write $X$ as a double cover of $\mathbb{P}^1$, branched at $2g+2$ points. Trying to homogenize the equation $(\sharp)$ in the usual way introduces a complicate singularity at infinity. Instead, it is more convenient to see $(\sharp)$ as the global equation of $X$ into the weighted projective plane $\mathbb{P}(1, \, 1,\, g+1)$. In fact, it is easy to see that the map $f \colon X \longrightarrow \mathbb{P}(1, \, 1, \, g+1)$ defined by the sections $x_0, \, x_1, \, y$ is an embedding (notice that the image does not pass through the singular point $[0: 0: 1]$ of the weighted projective plane).<|endoftext|> TITLE: Literature that helps explain what the theory of numerosities contributes with QUESTION [11 upvotes]: Since 2003 a group of Italian mathematicians (Benci, Di Nasso and Forti) has developed a new measure for infinite sets that satisfies the Euclidian principle: The whole is greater than the part. The theory has some interesting consequences. To mention some of them: It shows that the Euclidian principle can be captured in a theory of size in a consistent way. The set of numerosities has the same arithmetic as the natural numbers. We have that numerosity(even numbers)+numerosity(odd numbers)=numerosity($\mathbb N$), numerosity($\mathbb Z$)=2*numerosity($\mathbb N$)-1 and numerosity($\mathbb N\times \mathbb N$)=numerosity$(\mathbb N)^2$. It gives rise to nonstandard analysis. It can be extended to whole mathematical universes. It has applications in nonstandard probability theory, the foundation of nonstandard analysis and in number theory. But also: Sizes depend on a choice of ultrafilter in the construction of the numerosities, e.g. if odd$\in U$ then numerosity($\mathbb N$) is an even number and if even$\in U$ then numerosity($\mathbb N$) is odd. It violates translation invariance, e.g. numerosity($\mathbb N+1$) < numerosity($\mathbb N$) (more generally it violates transformation invarianse for every transformation with an infinite orbit). Makes it hard to classify "similar" sets. The existence of numerosities in some cases demand the existence of selective ultra filters (they exists if we assume the continuum hypothesis). The construction is not simple, it demands a lot of knowledge about special ultra filters and a lot of technical machinery. I have already considered Kitcher's (1984) idea of rational generalizations as a mean to explain how the contributions of the theory of numerosities differ from Cantor's theory of cardinalities (this is already done to some extend by Mancosu(2009)). But do you know any other (philosophical) theories or ideas that can help explain exactly what the new theory (more philosophical) contributes with and what the limitations of the theory may mean for these contributions? Specifically it would be nice, if there were some literature on what structure and classification means for the fruitfulness of a mathematical theory or a mathematical concept. REPLY [2 votes]: The theory of numerosities is more consistent with pre-mathematical intuitions of how collections (not to use the technical term set) should behave relative to each other. The fact that such an alternative is sorely needed is evidenced by the fact that a popular if not to say populist attempt to implement this at a naive level has apparently gained broad popularity (though perhaps not among pure mathematicians), see What is... A Grossone?<|endoftext|> TITLE: Transfers on Bloch groups and scissors congruence groups QUESTION [10 upvotes]: I have a couple of questions concerning existence and description of transfers for Bloch groups and scissors congruence groups/pre-Bloch groups. To fix notation and recall definitions: From the general algebraic K-theory machinery, we get transfers on $K_3$. In particular, for a finite field extension $E/F$, we get a map $\operatorname{tr}_{E/F}:K_3(E)\to K_3(F)$ such that the composition $K_3(F)\to K_3(E)\stackrel{\operatorname{tr}_{E/F}}{\longrightarrow}K_3(F)$ is multiplication with the degree $[E:F]$. Now by the work of Bloch, Dupont-Sah, Suslin and others, we have another description of $K_3(F)^{\operatorname{ind}}=K_3(F)/K_3^M(F)$ in terms of an exact sequence $$ 0\to \widetilde{\operatorname{Tor}}(\mu(F),\mu(F))\to K_3(F)^{\operatorname{ind}}\to B(F)\to 0. $$ In the above, the Bloch group $B(F)$ is defined as $$ B(F)=\ker\left(\mathcal{P}(F)\to \Lambda^2(F^\times):[x]\to x\wedge(1-x)\right) $$ and the group $\mathcal{P}(F)$ is the pre-Bloch group or scissors congruence group $$ \mathcal{P}(F)=\left(\bigoplus_{x\in F^\times\setminus\{1\}}\mathbb{Z}[x]\right) /\left([x]-[y]+[y/x]-[(1-x^{-1})/(1-y^{-1})]+[(1-x)/(1-y)]\right), $$ the relation coming from the five-term relation satisfied by the dilogarithm. In particular, elements of the Bloch group $B(F)$ can be written down as linear combinations of symbols $[x], x\in F^\times\setminus\{1\}$ satisfying certain relations. Now I can formulate my questions on transfers on Bloch groups and scissors congruence groups. Does the transfer on $K_3$ induce a transfer on the Bloch group? I think that this is not the case in general. The element $[x]+[1-x]\in B(F)$ is independent of $x$, and is typically denoted by $c_F$. The element $c_{\mathbb{R}}$ has exact order $6$ in $B(\mathbb{R})$, and the element $c_{\mathbb{C}}$ is trivial in $B(\mathbb{C})$. This seems to contradict transfers for the Bloch group (it does not contradict transfers for $K_3^{\operatorname{ind}}$ because the torsion moves from $B(F)$ to $\widetilde{\operatorname{Tor}}(\mu(F),\mu(F))$). Are there more torsion elements like this, in particular with other odd orders? Are there further obstructions to the existence of transfers on Bloch groups? If $F$ contains an algebraically closed fields, it follows from work of Suslin and Levine that $B(F)$ is uniquely $\ell$-divisible for $\ell$ different from the characteristic - in particular $c_F=0$. Does the $K$-theory transfer induce a transfer on the Bloch group in this situation? What would be a good reference? Is there an explicit description of what the transfer map on $K_3$ does on the Bloch group? I would be interested in a description that only uses the definition of the Bloch group via points on $\mathbb{P}^1(F)\setminus\{0,1,\infty\}$ given above. More generally, are there transfers known on scissors congruence groups/pre-Bloch groups? As written above, these groups are defined in terms of points on $\mathbb{P}^1\setminus\{0,1,\infty\}$ modulo the five-term relation. A very naive approach to the definition of transfers for pre-Bloch groups in an extension $E/F$ would be to sum over $E$-points lying over $F$-points of $\mathbb{P}^1\setminus\{0,1,\infty\}$. Has anyone ever tried to work this out, or are there known obstruction why this cannot provide a transfer? Assuming it works, how would one relate such a naive definition to the definition of transfers for algebraic K-theory? In a related direction, what torsion elements besides those in $B(F)$ are known in the scissors congruence groups over fields which are not algebraically closed (in the algebraically closed case, the scissors congruence groups are uniquely divisible)? REPLY [4 votes]: I have worked on this question for some time (in the $\mathbb{Q}$-coefficients case), trying to construct norm maps on $B_2(F)$ in a way, similar to Milnor $K-$theory. Eventually, I was able to construct norm maps on the middle cohomology groups of complexes $$ B_3(F) \otimes \Lambda^{n-3} F^{\times} \longrightarrow B_2(F) \otimes \Lambda^{n-2} F^{\times} \longrightarrow \Lambda^n F^{\times}, $$ which conjecturally coincide with $H^{n-1}_{M}(F,\mathbb{Q}(n)),$ for which existance of norm maps is known (see https://arxiv.org/abs/1511.00520). Unfortunately, I had to use the fact that the kernel of the map $$B_2(F)\longrightarrow \Lambda^2 F^{\times}$$ does not change under a simple transcendental extension of the field, so the construction of the norm is not completely explicit at the end of the day.<|endoftext|> TITLE: Why are we interested in the Fundamental Groupoid of a Space? QUESTION [26 upvotes]: The classical version of the van Kampen theorem is concerned about the fundamental group of a based space. In fact, it says that the functor $\pi_1$ preserves certain types of pushouts in $Top_*$. There is also a generalization of the van Kampen theorem that holds for the fundamental groupoid of a space $X$, where in this case it states precisely that the fundamental groupoid functor, $\Pi$, preserves certain colimits in $Top$, namely, those that arise from "nice" open coverings of $X$. The "groupoid" version of the van Kampen theorem seems to me more conceptual and more elegant than the classical version. Also, the groupoid version allows one to prove the classical version in a more or less easier way. Although, apart from the conceptual advantages of the groupoid version of van Kampen theorem, I would like to know if we are able to do any interesting calculations using the fundamental groupoid version of the van Kampen. In fact, explicitly describing a groupoid as a colimit of "simpler" groupoids is something that it is not clear at all for me. I would like to know some concrete cases where is it possible to describe the fundamental groupoid of a space, using this generalization form of the van Kampen theorem, and, if possible, to calculate the fundamental group directly from our calculation of $\Pi(X).$ REPLY [13 votes]: I'm very fond of Ronnie Brown's proof of the Jordan curve theorem using the groupoid version of the van Kampen theorem. I find it easier to understand and reproduce than other comparably elementary proofs, such as Munkres' argument using covering spaces (which can be found in his textbook Topology: a first course). I guess this application isn't really a calculation, as requested, but I hope you like it. (I guess I should mention that a small gap in the proof is filled in in this short note.)<|endoftext|> TITLE: Connection of Galois representation and arithmetic geometry QUESTION [6 upvotes]: There are lots of Galois representations which arise naturally from geometric objects, for example, Galois representations attached to elliptic curves. I know that people are interested in studying such Galois representations. There are lots of theorems which explain images of such representations are contained in special/open/closed subgroups or, form special subgroups. What geometric information can we get from such theorems? (or, why are such theorems so good for us?) REPLY [5 votes]: For example, the criterion of Néron-Ogg-Shafarevich says that if the Galois representation on the Tate module of an Abelian variety is unramified, the Abelian variety has good reduction. For this and more examples, see "Good Reduction of Abelian Varieties" by Jean-Pierre Serre and John Tate. Another example is the Tate conjecture for Abelian varieties $A, A'$ over finitely generated fields $K$: $\mathrm{Hom}(A,A')\otimes_{\mathbf{Z}}\mathbf{Z}_\ell = \mathrm{Hom}_{G_K}(T_\ell A,T_\ell A')$ The surjectivity means roughly that you can construct isogenies between Abelian varieties from $G_K$-equivariant homomorphisms between their Tate modules. Two of the equivalent formulations of modularity of elliptic curves are: 1) The Galois representation is modular [Galois representation]. 2) There is a finite morphism $X_0(N) \to E$ [geometry].<|endoftext|> TITLE: Forcing $\neg AC$ QUESTION [6 upvotes]: Sorry if this sounds like a silly reference request, but I wasn't able to track down any. I'm looking for proof, via forcing, that axiom of choice can fail in a model of $ZF$. All of papers I found where either proving something more sophisticated or, if it was some introductory paper/book, it showed that $\neg CH$ is consistent relative to $ZFC$. Again, sorry if this question sounds silly, but I would appreciate any help! Thanks in advance! REPLY [12 votes]: As Asaf wrote, the usual construction of models of $\neg AC$ consists of forcing followed by passing to an inner model. There is, however, an alternative construction that might be closer to what you're looking for. One can begin with a model of ZFC and first construct a model of ZFCA, a modification of ZFC that allows for atoms, which are not sets but can be elements of sets. With an infinite set of atoms, one can use the method of permutation models, pioneered by Fraenkel in 1922, to build submodels that satisfy ZFA but not the axiom of choice. (These are the permutation models described in Chapter 4 of Jech's "Axiom of Choice" book.) Then one can force over such a permutation model in such a way that the pure part of the resulting forcing extension (i.e., the submodel consisting of the sets that don't involve any atoms in their transitive closures) is a model of ZF (without atoms) violating choice. In other words, one can do the symmetrization before forcing, rather than afterward. Then the symmetry involves just permutations of atoms rather than automorphisms of a complete Boolean algebra. For a specific example: If you take the basic Fraenkel model (as defined in Jech's book) and force, in the usual way (finite partial functions), a family of mutually generic Cohen reals indexed by the atoms, then the pure part of the resulting forcing extension is the basic Cohen model (as defined in Chapter 5 of Jech's book). This approach to the independence of AC from ZF is used, for example, in the book "Theory of Semisets" by Vopenka and Hajek. (Unfortunately, I think the only way to read that book is straight through from the beginning; if you try to look up a particular result in it, you'll find that it depends on so much prior notation that you'll have to read practically everything that precedes the result you want.)<|endoftext|> TITLE: Multiplicative Structures on Moore Spectra QUESTION [20 upvotes]: The motivation for this question is that I want "toy examples" of how to prove/disprove the existence of multiplicative structures on examples of spectra. The class of examples I am thinking of is the Moore spectrum. For concreteness this is defined as a spectrum $X$ such that $\pi_n(X) = 0$ for $n <0$ $H_n(X)= 0$ for $n >0$ and $H_0(X) = R$ for some ring $R$. There are some curious phenomenon that happens: On one extreme, the Mod 2 Moore spectrum has no unital multiplication at all (by simple arguments in, say, Difficulties with the mod 2 Moore Spectrum) The Mod 3 Moore spectrum is not $A_{\infty}$ by Massey product arguments. The comment here on top of page 838: http://www.math.uni-bonn.de/people/schwede/rigid.pdf says that the mod $p$ Moore spectrum for $p \geq 5$ is homotopy associative by folklore (I would like to see an argument for this too!) On another extreme, since we can model the $\mathbb{Z}[q^{-1}]$ by localizing the sphere spectrum they are $E_{\infty}$. In this light, my questions are: First and foremost, I would love to see a proof of the folklore result above about $p \geq 5$ Is there a "general pattern" about multiplicative structures of the Moore spectrum as the ring/abelian group varies REPLY [5 votes]: The classical reference is as follows: @article {MR760188, AUTHOR = {Oka, Shichir{\^o}}, TITLE = {Multiplications on the {M}oore spectrum}, JOURNAL = {Mem. Fac. Sci. Kyushu Univ. Ser. A}, FJOURNAL = {Memoirs of the Faculty of Science. Kyushu University. Series A. Mathematics}, VOLUME = {38}, YEAR = {1984}, NUMBER = {2}, PAGES = {257--276}, ISSN = {0373-6385}, CODEN = {MFKAAF}, MRCLASS = {55P45}, MRNUMBER = {760188 (85j:55019)}, MRREVIEWER = {Donald M. Davis}, DOI = {10.2206/kyushumfs.38.257}, URL = {http://dx.doi.org/10.2206/kyushumfs.38.257}, } Here is the review by Don Davis: For any positive integer $q$, let $M_q$ denote the Moore spectrum whose only nontrivial homology group is $\mathbb{Z}/q$ in dimension 0. The main result is the following theorem: (a) The number of homotopy classes of multiplications on $M_q$ is $4$ if $q≡0 (4)$, 1 if $q$ is odd, and $0$ if $q≡2 (4)$. (b) These multiplications are commutative if and only if $q≡0$ (8) or $q$ is odd, and are associative if and only if $q\not\equiv 2 (4)$ and $q\not\equiv\pm 3 (9)$. A number of more technical results are proved, many involving premultiplications and regularity.<|endoftext|> TITLE: Heuristic interpretation of the 'third index' for Besov and Triebel-Lizorkin spaces QUESTION [12 upvotes]: For $p,q \in (0,\infty)$ and $s \in \mathbb{R}$, one can define certain function spaces, $B_s^{p,q}(\mathbb{R}^n)$ and $F_s^{p,q}(\mathbb{R}^n)$, the Besov and Triebel-Lizorkin spaces respectively. The definition of these spaces is a bit complicated, and it probably won't offer any new insight to anybody who hasn't already considered these spaces, but it can be found for example in Triebel's 'Theory of Function Spaces II' (section 2.3.1) or Grafakos' 'Modern Fourier Analysis' (section 6.5.1). For both of these spaces, one can rightfully refer to $p$ as the 'integrability index' and to $s$ as the 'regularity index'; for example, the spaces $F_s^{p,2}(\mathbb{R}^n)$ are equivalent to the Sobolev spaces $W_s^{p}(\mathbb{R}^n)$, and one can justify the integrability/regularity interpretations when $q \neq 2$ or for the spaces $B^{p,q}_s(\mathbb{R}^n)$. Is there a simple way of describing the role of the $q$ index? Or are stuck with referring to it as 'the third index'? REPLY [3 votes]: This is really a comment with possibly relevant references, but I would prefer to avoid creating an account. The additional parameter is sometimes referred to as the "microscopic parameter" in the literature (this may help in searching google and MathSciNet). Two references of interest are: W. Sickel, L. Skrzypczak and J. Vybiral. On the Interplay of Regularity and Decay in Case of Radial Functions I. Inhomogeneous spaces. arXiv:1201.5007 Commun. Contemp. Math 14 (2012) No. 1 (along with reference list of this article -- there is also a followup paper II. Homogeneous spaces in J. Fourier Anal. Appl. 2012). as well as the work of Brezis and Mironescu (J. Evol. Eq. 1, 2001; see Section III). As a tangential point related to the statement of the question, while $F^{p,2}_s$ is certainly the Bessel potential space $L^{s,p}$, unless I miss something this space is not the same as $B^{p,2}_s$ (note that there are many conventions for notation in the literature, and many authors use $W^{s,p}$ for the Besov space $B^{p,p}_s$).<|endoftext|> TITLE: A Weakening of the Tree Property QUESTION [8 upvotes]: If $f$ and $g$ are two functions, define $f \sim g$ if they differ only finitely often on their common domain. The following property of a large cardinal arose from a problem in model theory. I am interested in its strength. Say that a cardinal $\kappa$ has the weak tree property if the following holds: Suppose $(b_\alpha: \alpha < \kappa)$ is a sequence such that: Each $b_\alpha \in 2^\alpha$, For each $\alpha < \beta < \kappa$, $b_\alpha \sim b_\beta$. Then there is some $b \in 2^\kappa$ such that for all $\alpha < \kappa$, $b \sim b_\alpha$. The following summarizes what I know about the weak tree property: If $\kappa$ has the tree property then it has the weak tree property. (A sequence $(b_\alpha)$ such as above can also be viewed as defining an Aronszajn tree.) $\aleph_1$ does not have the weak tree property. (One of the standard constructions of an $\aleph_1$ Aronszajn tree yields this.) If $\kappa$ has countable cofinality then $\kappa$ has the weak tree property. I am interested in knowing more. For example, might $\aleph_2$ not have the weak tree property? (It is possible that $\aleph_2$ has the weak tree property since it might have the full tree property.) Any thoughts would be appreciated. PS: The example question above has been resolved in the comments. REPLY [4 votes]: I give here some upper bounds for the consistency of the weak tree property. Namely - for every successor of regular (including double successors of singulars) we can get the weak tree property by collapsing a weakly compact, and we can get the weak tree property everywhere by collapsing a strongly compact cardinal to be $\aleph_2$. Let start with some definitions: Definition: We call a sequence $\mathcal{B} = \langle b_\alpha | \alpha < \kappa \rangle$ coherent if $\forall \alpha , \beta < \kappa\,b_\alpha\sim b_\beta$. A element $b\in 2^{\kappa}$ is a thread for $\mathcal{B}$ if $b\sim b_\alpha$ for every $\alpha < \kappa$. Lemma: Let $\mathcal{B}$ be a coherent sequence with no thread. Let $\mathbb{P}$ be $\sigma$-closed forcing notion. Then after forcing with $\mathbb{P}$, $\mathcal{B}$ still has no thread. Proof: [Peter Komjath already gave a proof in his answer above. I give here a slightly more detailed one] Assume otherwise, and let $\dot{b}$ be a name for a thread. Note that since $\Vdash\dot{b} \notin V$, for every condition $p\in \mathbb{P}$ and ordinal $\alpha < \kappa$ there is ordinal $\alpha < \beta < \kappa$ and two conditions $q, q^\prime \leq p$ such that $q\Vdash \dot{b}(\beta) = 0$ and $q^\prime \Vdash \dot{b}(\beta) = 1$. Now, choose by induction ordinals $\alpha_n < \kappa$ and conditions $p_\eta \in \mathbb{P}$ for every $\eta \in 2^{<\omega}$ such that $\eta \trianglelefteq \eta^\prime \implies p_\eta \geq p_{\eta^\prime}$ and for all $\eta \in 2^n$ there is $\alpha_n < \beta_\eta$, such that $p_{\eta \frown \langle 0 \rangle} \Vdash \dot{b}(\beta_\eta )= 0$ and $p_{\eta \frown \langle 1 \rangle} \Vdash \dot{b}(\beta_\eta )= 1$. Let $\alpha_{n+1} = \max_{\eta \in 2^n} {\beta_\eta} + 1$. Let $\gamma = \sup \alpha_n$. Now, using the $\sigma$-closure of $\mathbb{P}$ pick for every $f\in 2^\omega$ a condition $p_f$ such that $p_f \leq p_{f\restriction n}$ for every $n$. By extending $p_f$, if necessary, we may assume that the finite set of difference between $b_\gamma$ and $\dot{b}\restriction \gamma$ is decides by $p_f$, and let denote it by $s_f$. Since $s_f$ is bounded for every $f$, there is $n_0 < \omega$ such that for infinite many $f\in 2^\omega$, $s_f \subset \alpha_{n_0}$. But this is impossible since for every $\eta \in 2^n$ there is only one extension $\eta^\prime \in 2^m$, $m>n$ that force that the difference between $\dot{b}$ and $b_\gamma\restriction \alpha_m$ will be all below $\alpha_n$, so there at most $2^n$ many such $f$ - a contradiction. QED Theorem: Let $\mu$ be uncountable regular cardinal and $\kappa > \mu$. $\mathbb{C} = Col(\mu,<\kappa)$. Then if $\kappa$ is weakly compact, $\mathbb{C}$ forces the weak tree property at $\mu^+$, and if $\kappa$ is $\lambda$-strongly compact $\mathbb{C}$ forces the weak tree property at $\lambda$. Proof: We start with the weakly compact case. Let $\dot{\mathcal{B}}$ be a $\mathbb{C}$-name for a coherent sequence of length $\kappa$. The elementary embedding $j: (V_\kappa,\dot{\mathcal{B}},\mathbb{C},\dots) \rightarrow M$ extends to elementary embedding $\tilde{j}:V_\kappa[G] \rightarrow M[G][H]$ in $V[G][H]$ where $H$ is a generic filter for $Col(\mu,[\kappa,j(\kappa) )$ which is $\mu$-closed. $j(\mathcal{B}) = \langle b^{j}_\alpha | \alpha < j(\kappa)\rangle$ is a coherent sequence so in particular, $b^{j}_\kappa$ is a thread. This object was added by $Col(\mu,[\kappa,j(\kappa) )$ but this is a $\sigma$-closed forcing, so by the lemma, there was some thread already in $V[G]$. The proof for the strongly compact case is the same. We extend the strongly compact embedding $j$ to $\tilde{j}:V[G]\rightarrow M[G][H]$ be forcing with $Col(\mu,[\kappa,j(\kappa) )$. Now, if $\mathcal{B}$ was a coherent sequence of length $\lambda$ in $V[G]$, (and $\sup j^{\prime\prime} \lambda < j(\lambda)$) then for $j(\mathcal{B} = \langle b^j_\alpha | \alpha < j(\lambda)\rangle$, we can pick $\delta = \sup j^{\prime\prime} \lambda$ and choose $b(\alpha) = b^j_\delta (j(\alpha)$ for every $\alpha < \lambda$. It's clear that $b\in V[G][H]$ is a thread for $\mathcal{B}$, and by the lemma, there is a thread already in $V[G]$. QED REPLY [2 votes]: Douglas: here's Yair's argument. Assume that $P$ is $\sigma$-closed and adds a $b$ as described. First, if $p$ forces that $b|\alpha=\beta$, then there are extsnions $p'$ and $p''$ of $p$ which force $b|\beta=g',g''$ such that $g'$ and $g''$ differ. (otherwise $b$ is in $V$.) Notice that you can make $\beta$ arbitrarily large. Now build the decreasing sequences $p_0\geq p_1\geq\cdots$ and $q_0\geq q_1\geq\cdots$ and increasing sequence of ordinals $\alpha_0<\alpha_1<\cdots$ such that $p_i$ forces that $b$ restricted to $[\alpha_i,\alpha_{i+1})$ is $g_i$, $q_i$ forces that $b$ restricted to $[\alpha_i,\alpha_{i+1})$ is $g'_i$ and $g_i\neq g'_i$. Let $p,q$ be the extensions of $p_i$, $q_i$, resp. Then they force that $b_\alpha$ is almost equal to two functions, which differ at infinitely many places, a contradiction.<|endoftext|> TITLE: How do we know if Vaught's Conjecture is Absolute? QUESTION [7 upvotes]: Please note that this might be some confusion on my part about the work surrounding Vaught's conjecture. First of all, Vaught Conjecture states that if a first-order complete theory $T$ in a countable language has infinitely many models of size $\aleph_0$, then either $I(T,\aleph_0)=\aleph_0$ or $I(T,\aleph_0)=2^{\aleph_0}$. We will denote this conjecture as $VC$. It has been shown that if $I(T,\aleph_0)> \aleph_1$, then $I(T,\aleph_0)=2^{\aleph_0}$. Therefore, we know that $ZFC+CH \vdash VC$. If one wants to prove that $VC$ is false, then one must construct a model $M$ such that $M\models ZFC$ and a theory $T_1$, such that $M\models ZFC+ \neg CH$ and $M \models I(T_1,\aleph_0)=\aleph_1$. But here lies my problem: Let $M_1,M_2 \models ZFC + \neg CH$. Suppose that we can construct first order theories $T_1$ and $T_2$ such that $M_1 \models I(T_1,\aleph_0)=\aleph_1$ and $M_2 \models I(T_2,\aleph_0)=\aleph_1$. Is it possible that $M_1\models I(T _2,\aleph_0)=2^{\aleph_0}$ and $M_2\models I(T_2,\aleph_0)=2^{\aleph_0}$? Better yet, we can simply ask: if $T$ is a counterexample to Vaught's Conjecture in some model $M$ of $ZFC$ must it be the case that $N\models I(T,\aleph_0)=\aleph_1$ for any $N$ such that $N\models ZFC$? Thanks! REPLY [5 votes]: The answer is yes. See section 5 of the paper "Bounds on weak scattering" by Sacks (http://www.math.harvard.edu/~sacks/bws.pdf); he cites Morley 1970 ("The number of countable models," http://www.jstor.org/stable/2271150) as the original proof.<|endoftext|> TITLE: Reading Papers in a Language you don't Speak QUESTION [14 upvotes]: First, I apologize if I'm posting this to the wrong place, but it seems correct. My adviser sent me the SGA text of Grothendieck which is in French. Though I can piece together parts of the text, I'm afraid that I'm losing significant parts of the meaning when I just have no idea what a sentence says. Google Translate was not terribly helpful. I was wondering what the standard techniques are for dealing with papers in a foreign language. REPLY [3 votes]: You can try to seek assistance in a translation site such as http://www.proz.com. This is a global site connecting clients and translators from around the world. Although there are few professional translators who understand advanced mathematics, such creatures may exist if you seek translation from a common language like French. You can register at the site and offer the job with all your conditions. Then you select the translator who suits you most. I would not expect a high-quality result, but you can help the translator by providing her/him with the terminology of the subject field. The down side is that it can cost you several hundred $, depending on the length of the paper.<|endoftext|> TITLE: Formal group law is a group object in ...? QUESTION [15 upvotes]: A formal group law over a commutative ring $R$, (by nLab) is a sequence of power srires $$ f_1,...,f_n\in R[[x_1,...,x_n,y_1,...,y_n]] $$ such that, using the notation $$ x=(x_1,...,x_n),y=(y_1,...,y_n),f=(f_1,...,f_n) $$ we have $$ f(x,f(y,z))=f(f(x,y),z) $$ $$ f(x,y)=x+y+\text{higher order terms} $$ I understand vaguely the idea of a formal group law as a power series expansion of the group law of a lie group or an algebraic group (actual or hypothetical) in the neighborhood of the identity. But I would be happy to know, if only for psychological reasons, if this definition can be recovered as simply a group object in some category. In the nLab entry about formal groups, it is written that formal group laws are one approach to formal groups, and the later is a group object in 'infinitesimal spaces', but I was unable to understand what is an infinitesimal space from the linked entry. I would appreciate if someone could explain this circle of ideas or point to the relevant literature. REPLY [14 votes]: A formal group law over a scheme $S$ is a group object in the category of framed formal schemes over $S$. Objects in this category are formal schemes $X$ over $S$ equipped with an $S$-isomorphism $X \to \operatorname{Spf} \mathscr{O}_S[[t_1,\ldots,t_n]]$ for some $n$. There is a functor from framed formal schemes over $S$ to formal schemes over $S$, given by forgetting the $S$-isomorphism. The essential image is the category of formal Lie varieties over $S$. This functor takes formal group laws to the class of formal groups that admit a framing. REPLY [5 votes]: Given a group scheme one can complete at the identity to get a group object in the category of formal schemes. Every such group object in the category of formal schemes obtained in this way is equivalent to a formal group law. The power series operations for formal group laws are induced hopf-algebra-like structure on the structure sheaf of the group scheme. If we fix the base ring then we can classify formal group laws and state that every formal group law is isomorphic to one obtained in this way. In this sense we can say that every formal group law is a group object in the category of formal schemes. The converse of this is of course not true. There are group objects in the category of formal schemes which are not just formal group laws (take a group scheme and complete along something that isn't the identity). This is talked about in "Baby Silverman".<|endoftext|> TITLE: is there a p-adic implicit function theorem? QUESTION [12 upvotes]: I am trying to find a good reference for a version of the implicit function theorem over $p$-adic manifolds. None of the texts I have consulted ( including "$p$-adic numbers, $p$-adic analysis, and Zeta functions" by Neal Koblitz and "$p$-adic analysis and Lie groups" by Peter Schneider) seem to discuss it. Of course, I can try to go through the argument in the real case and see if it works in the $p$-adic world, but it would be very desirable to have a reference. More precisely, I would like to know if a theorem of this type is true: Let $U$ and $V$ be open subsets of some $p$-adic manifolds and $f: U \to V$ a sufficiently smooth map defined on $U$ (From what I have seen, there seems to be a notion of strict differentiability in the $p$-adic setup. The function I will be interested are very nice, so most probably, I can live with a stronger assumption on $f$ too). If the rank of the derivative of $f$ at a point $x \in U$ is $r$, then after a smooth change of coordinates on possibly smaller open subsets of $U$ an $V$, $f$ can be expressed as $$f(x_1, \dots ,x_m)=(x_1, \dots , x_r, g_1(x), \dots , g_{n-r}(x)),$$ where $g_i$ are also smooth functions. I would very much appreciate if someone can confirm that this is the case and point me to a reference. REPLY [7 votes]: There are the lecture notes by Serre (Springer lecture notes 1500) of a course on Lie groups and Lie algebras. He proves the implicit function theorem for analytic functions on a $p$-adic manifold (not smooth functions, though) The following is a link: http://www.amazon.com/Lie-Algebras-Groups-University-Mathematics/dp/3540550089/ref=cm_cr_pr_product_top<|endoftext|> TITLE: A question on fixed point theory QUESTION [9 upvotes]: I asked this question in MSE, but I did not received any answer, so I repeat it here: https://math.stackexchange.com/questions/858238/a-question-on-fixed-point-property Assume that $0 TITLE: Does the category PCM (partial commutative monoids) have a closed symmetric monoidal product? QUESTION [6 upvotes]: A partial commutative monoid (PCM) is, roughly speaking, a set with a partially defined binary operation that is as associative as it can be (given that not all products are defined) and commutative. These things seem to come up in quantum logic, for example: D. J. Foulis, M. K. Bennet, Effect algebras and unsharp quantum logics, Found. Phys. 24 (1994), 1 331–1 352. and Peter Hines, A categorical analogue of the monoid semiring construction (Probably the same definition has come up in other places too - if so, please tell me about it.) Question Does the category PCM have a closed symmetric monoidal product? This might sound a bit random, so I'll try to explain a bit of background and my motivation. PCMs sit somewhere in between abelian monoids (aka $\mathbb{N}$-modules) and pointed sets (which one might call modules over the field with one element $\mathbb{F}_1$). Commutative monoids and pointed sets both form closed symmetric monoidal categories, and the commutative monoids in these categories are commutative semirings ($\mathbb{N}$-algebras) and commutative monoids with an absorbing element (the corresponding notion of $\mathbb{F}_1$-algebras) respectively. A commutative monoid in PCM should, I expect, be something like a semiring, but with a multiplication operation that is always defined and an addition operation that is only sometimes defined. Various people, including Toen-Vaquie, have studied the variants of scheme theory built from these things as their affine local models. For various reasons, I find myself wanting a category of schemes that fits in between $\mathbb{F}_1$ and $\mathbb{N}$, and so commutative monoids in PCM seem to fit the bill. The input for the general Toen-Vaquie machine is a closed symmetric monoidal category which one thinks of as the category of modules over the algebraic object we are considering. I've tagged this as reference-request in the hope that I can simply cite something, but feel free to also post an argument/construction/counter-example. REPLY [5 votes]: Yes, it does; see Theorem 11 of this paper.<|endoftext|> TITLE: Are all vector-space valued functors on sets free? QUESTION [14 upvotes]: Let $\mathbf{Set}$ be the category of finite sets and functions between them, and let $\mathbf{Vect}$ be the category of finite-dimensional complex vector spaces and linear transformations between them. There is a free functor $G \colon \mathbf{Set} \to \mathbf{Vect}$, left adjoint to the forgetful functor, given by $G(X) = \mathbb{C}^X$ and $G(f)(\phi)(y) = \sum_{f(x)=y} \phi(x)$. Is any functor $\mathbf{Set} \to \mathbf{Vect}$ of the form $H \circ G \circ F$ for $F \colon \mathbf{Set} \to \mathbf{Set}$ and $H \colon \mathbf{Vect} \to \mathbf{Vect}$? A priori one might expect lots more functors, but I'm having a hard time coming up with any. On the other hand, functoriality seems to keep $G$ from "creating chaos" to "mess up freeness" (sorry that I can't explicate this feeling better). To keep it simple, let's keep things finite(-dimensional), and not consider anything about other base fields, or monoidal structure. REPLY [19 votes]: The easiest example is the functor that sends the empty set to a 1-dimensional vector space, and all other finite sets to zero. (Any H that sends some vector space to 0 must send 0 to 0 as well). The second-easiest example is the one given by Jeremy Rickard. If you're interested in functors from the category of finite sets to finite dimensional vector spaces, you might take a look at my paper http://arxiv.org/abs/1406.0786 The main result is a structure theorem about (finitely-generated) functors in this category.<|endoftext|> TITLE: Vector bundles, Higgs bundles and the Langlands program QUESTION [15 upvotes]: This question is somewhat vaguely structured. But, I hope someone can make it more precise (or) it is indeed possible to answer it in the form that I am stating it.  Background : I recently chanced upon the following lecture note of Langlands : http://publications.ias.edu/rpl/paper/2578 . In this, he has various comments/proposals on geometrical approaches to aspects of the Langlands program.  I don't understand several things about it but the following question came to my mind.   Qn : If you are Beilinson-Drinfeld in the early 90s looking to formulate a version of Geometric Langlands. Why are moduli spaces of stable Higgs bundles (arising in Hitchin systems) on Riemann surfaces the right place to look at and not just moduli spaces of stable vector bundles on Riemann surfaces ? (the connection to the Lecture note linked above is that Langlands discusses the theory of stable vector bundles, especially the work of Harder-Narasimhan, Atiyah-Bott etc and provides certain conjectures in the language of this theory. ) PS : There is actually a 'physics answer' that I can think of.  As is well known, Atiyah-Bott were studying two dimensional Yang-Mills (gauge) theories. Such theories do not have an electric magnetic duality relating the theory for $G$ and the theory for $G^\vee$. However, such dualities can occur in four dimensional gauge theory. The most robust such duality is for the maximally symmetric $\mathcal{N}=4$ Supersymmetric Yang Mills theory in four dimensions. And in the gauge theory approach of Kapustin-Witten to the Geometric Langlands Program, this electric magnetic duality is indeed the starting point for obtaining statements about geometric Langlands. When you compactify the four dimensional theory on a Riemann surface, one naturally obtains Hitchin system(s) from the physics (I am being telegraphic here, omitting details about twists, defects etc). Plenty of MO threads do a better job of this. But, the point is that if a 4d field theory (more accurately, a TQFT built from it) is the starting point, then the Hitchin system is what one gets naturally. But, I suspect there might be another answer since Beilinson-Drinfeld's original approach was not motivated by dualities in four dimensional field theories. Hence the question. REPLY [9 votes]: I think that understanding what can be a "motivation" for Beilinson-Drinfeld work, at least several things should be kept in mind. 1) V. Drinfeld made principal contributions for functional fields case of Geometric Langlands - he established it for GL(2), as well as, creating basic constructions which lead Lafforgue to establish GL(n) case - he introduced "schutakas", and a way to use them in order to get Langlands correspondence. ( Somewhat curious, that his contribution to this field dated back to 1979, and since that time he seems to moved his principal interests to quantum groups and "math. physics". And it seems that it is kind of astonishing that work on geometric Langlands over complex numbers unified both of his interestes - "math. phys" and Langlands.) 2) A. Beilinson (with J. Bernstein) introduced so-called "Beilison-Bernstein localization" in their way to prove Kazhdan-Lusztig conjectures on Verma modules and intersection cohomology. This construction plays an important role for "understanding-motivating" Beilinson-Drinfeld work. It defines a correspondence between D-modules on flag variety and certain representations of the semisimple Lie algebra. ( In Beilinson-Drinfeld story: similar D-modules apppear as "Hitchin eigensheaves", flag manifold is substituted by moduli space of vector bundles, semi-simple Lie algebra substituted by affine Lie algebra). 3) Hitchin's paper on integrable system appeared in 1987, and pay attention that it appeared in special issued of Duke journal dedicated to 50-th anniversary of Yu.I. Manin, who was the teacher of both V.Drinfeld and A.Beilinson. So it was of course known to them. From the position of our nowdays knowledge it is easy to understand the relevance of Beilionson-Bernstein (BB) localization to Hitchin's paper: roughly speaking the center of universal enveloping of affine Lie algebra on the critical level which play crucial role in BB-localization gives exactly the quantum version of the Hitchin integrable system. However, I think in late 1980-ies, probably no one except Beilinson and Drinfeld were able to see through the clouds. (Since the center itself was not actually proved to exist in 1987, and many many other things were not yet known). Concerning the more detailed version of the question. Why are moduli spaces of stable Higgs bundles (arising in Hitchin systems) on Riemann surfaces the right place to look at and not just moduli spaces of stable vector bundles on Riemann surfaces ? I am not sure I fully understand the point to make stress on the moduli space of Higgs bundles vs. moduli space of vector bundles. In my undestanding one should stress on the following things. 1) Moduli space of Higgs bundles is (modula details) COTANGENT BUNDLE to moduli space of vector bundles. Contangent bundles to manifold appears naturally when you speak about D-modules or "quantization" - it is "classical phase space" in quantization story or the space where characteristic manifold of D-modules lives. 2) What is surpsising that in functional field case developped by V.Drinfeld in late 70-ies, he needs to work with moduli space of "schtukas" - which are vector bundles plus additional structures related to Frobenius, while Beilinson-Drinfeld story is more simple in that respect - you do not need "schtukas" and you can actually work with vector bundles.<|endoftext|> TITLE: Proof or citation? QUESTION [9 upvotes]: I'm writing an article. I suppose that I'll submit it to a more or less decent journal (in English). I have doubts about the following: I have a lemma (with quite a trivial proof). I don't want to include this lemma with complete proof due to its triviality and because it has been already proved before in my Ph.D. thesis. So I may either prove the lemma in the article (this certainly will not be very good) or simply include a citation of my dissertation instead of proof. But the thesis is in Ukrainian and it's not even available freely via the Internet. Which of the two alternatives would you choose? Do there exist other ways to do this well? REPLY [12 votes]: On my opinion, the main criterion for a reference is that it must be AVAILABLE. Either on Internet or in most university libraries. An unpublished thesis in Ukrainian which is not available on Internet does not satisfy this criterion. Such a reference is meaningless. So the real options are: a) to leave the proof to the reader or b) to include it. I do not understand why you call the second option "not very good". What's wrong with including a trivial lemma? A reader for whom it is trivial will just skip the proof. PS. Why don't you scan your thesis and put in on Internet? Even in Ukrainian.<|endoftext|> TITLE: Holomorphic Foliations having transverse sections QUESTION [6 upvotes]: In the introduction to the paper "On the Geometry of Holomorphic Flows and Foliations Having Transverse Sections" by Ito and Scardua, one reads the following "a holomorphic codimension one foliation on a compact manifold is not necessarily transverse to some compact Riemann surface. Indeed, the existence of such a compact transverse section often implies several restrictions on the foliation..." My question is simply "What restrictions?" This is not really discussed in their paper nor the references, as they concentrate on real transversals. Also, most results I have seen are local, I'm interested in the global statement as phrased by Scardua. I'd be happy to even concentrate on the case of smooth projective surfaces and smooth transverse algebraic curves. The only general result of which I am aware is the result by Bogomolov/McQuillan in "Rational Curves on Foliated Varieties" and the follow-up article by Kebekus/Conde/Toma "Rationally connected foliations after Bogomolov and McQuillan", that if the transverse curve has positive self-intersection (i.e. is an ample divisor), then the foliation is a fibration with rational leaves. (I'm phrasing it in the case of dimension 2, similar results also hold in higher dimension, giving rational connectedness of the leaves, and with weaker hypotheses as to the nature of the curve and foliation singularities, amount of tangency, etc). But how about everywhere transverse curves with non-positive self-intersection? What restrictions does this give on the foliation? REPLY [7 votes]: If a smooth curve $C$ has zero self-intersection and is everywhere transverse to a foliation on a surface then there are also strong restrictions. If $C$ is rational then the foliation is a Riccati foliation (i.e. is birationally equivalent to the projectivization of a meromorphic flat connection on rank two bundle over a curve). For an arbitrary smooth curve $C$ everywhere transverse to a foliation $\mathcal F$ on a projective surface we can argue as follows. If $T^* \mathcal F$ is not pseudo-effective then by Miyaoka's Theorem (Bogomolov-McQuillan is a generalization of Miyaoka's Theorem) then $\mathcal F$ is a foliation by rational curves. If instead $T^*\mathcal F$ is pseudo-effective then the Zariski decomposition $P+N$ ($P$ nef $\mathbb Q$-divisor and $N$ effective contractible $\mathbb Q$-divisor) of $T^* \mathcal F$ satisfies $P \cdot C = 0$. Hodge index theorem implies that $P$ is numerically equivalent to a rational multiple of $C$. This already imposes strong restriction on $\mathcal F$. It must be a foliation of special type (Kodaira dimension is not maximal). Synthesis. Assume there exists a smooth curve C of zero self-intersection which is everywhere transverse to a foliation $\mathcal F$. I believe that going through the classification of foliations of special type we can proof that at least one of the following assertions holds true: The foliation $\mathcal F$ is a foliation by rational curves. The curve $C$ is a fiber of rational fibration and the foliation is transverse to the general fiber of it (Riccati foliation). The curve $C$ is a fiber of an isotrivial elliptic fibration and the foliation is transverse to the general fiber of it (turbulent foliation). The curve $C$ is a fiber of an isotrivial fibration with fibers of genus $\ge 2$ and the leaves of the foliation are all algebraic. The main references here are McQuillan's paper Canonical models of foliations; and Brunella's book Birational geometry of foliations. Examples. In contrast, it is rather easy to produce examples of foliations tranverse to curves of negative self-intersection. Start with a reduced foliation of general type on any surface and take a sufficiently general smooth curve. The tangencies between the foliation and the curve will be simple tangencies. Blowing-up each of the tangencies points twice will give rise to a curve everywhere transverse to the resulting foliation.<|endoftext|> TITLE: Do homotopy limits compute limits in the associated quasicategory in the non-combinatorial model category case? QUESTION [6 upvotes]: Suppose that $\mathcal{M}$ is a model category which is not combinatorial, does a homotopy limit in $\mathcal{M}$ correspond to a limit in the associated $\left(\infty,1\right)$-category? How about when $\mathcal{M}$ doesn't have enough limits and colimits, but is otherwise a model category? How about when $\mathcal{M}$ is a category of fibrant objects? If there are counter-examples, under what additional assumptions (besides being combinatorial) will this be true? REPLY [18 votes]: Homotopy limits in any model category always coincide with limits in the associated $(\infty,1)$-category. To see this, you need to know the following (classical) facts: 1) given a cofibrant object $A$, the mapping space functor $Map(A,-)$ (constructed as in Hovey's book, say, using a Reedy cofibrant resolution of $A$ in the category of cosimplicial objects) is a right Quillen functor, and thus commutes with homotopy limits (up to a canonical weak equivalence); 2) furthermore, for any fibrant object $X$, the Kan complex $Map(A,X)$ canonically has the homotopy type of the space of maps from $A$ to $X$ in the simplicial localization of your model category (this due to Dwyer and Kan). 3) the localization in the sense of quasi-categories corresponds via the (homotopy coherent) nerve functor to the simplicial localization, so that you can reinterpret 2) in an appropriate way; 4) in the setting of $(\infty,1)$-categorie, the Yoneda embedding preserves limits. Therefore, by the Yoneda lemma (I mean the very usual one, applied to the homotopy category of the model category you are dealing with), to prove the property you want, it is sufficient to consider the case of the usual model category structure on simplicial sets, which obviously falls in the setting considered by Lurie (but can also be easily deduced from earlier results of Dwyer and Kan). [Edit] I realize I did not write anything about the case where $\mathcal M$ is not a bicomplete model category. In the case of a model category with only finite limits, the same is true for finite homotopy limits (same proof). More generally, finite homotopy limits in categories of fibrants objects coincide with the corresponding finite limits in the associated $(\infty,1)$-category (which is always finitely complete). For a proof, we proceed as follows. It is sufficient to consider the case of a small category of fibrant objects (this is an easy exercise, and you don't need Grothendieck universes to do so). Then Remark 3.13 in my paper [Invariance de la $K$-théorie par équivalences dérivées, J. K-theory 6 (2010), 505-546] (available here) reduces the problem to the case of a combinatorial proper simplicial model category, in which case you can apply Lurie's result. For a general discussion about homotopy limits in categories of fibrant objects, there is my paper [Catégories dérivables, Bull. SMF 138 (2010), 317-393] (available here), or Radulescu-Banu's paper.<|endoftext|> TITLE: The surjectivity of the exponential map for the isometry group QUESTION [7 upvotes]: Little is known on general conditions guaranteeing that the exponential map between a Lie algebra and an associated Lie group is surjective. Let $M$ be a noncompact connected Riemann manifold, and $G$ be its (Lie) group of isometries. Does anybody know whether the exponential map between the Lie algebra of $G$ and the connected component of $G$ is surjective? (If $M$ is compact then $G$ also is, which in this case answers my question in the affirmative.) What about replacing the isometry group with the group of conformal transformations? Edit: The question has been answered in the negative by @RobertBryant below. Does anybody know what conditions (other than compactness) should be added to get an affirmative answer? REPLY [16 votes]: No. Consider, for example, $M=\mathrm{SL}(3,\mathbb{R})/\mathrm{SO(3)}$ endowed with its $\mathrm{SL}(3,\mathbb{R})$-invariant Riemannian metric $g$ (which is unique up to a positive constant multiple). This is an irreducible symmetric space of noncompact type. The identity component of the isometry group of $(M,g)$ is $\mathrm{SL}(3,\mathbb{R})$ itself (since $\mathrm{SL}(3,\mathbb{R})$ has trivial center), and the exponential map of $\mathrm{SL}(3,\mathbb{R})$ is not surjective. To see this, note that, when all of the generalized eigenvalues of $a\in{\frak{sl}}(3,\mathbb{R})$ are real, then $A=\exp(a)$ has positive (generalized) eigenvalues, and, when $a$ has one real eigenvalue and two distinct complex conjugate eigenvalues, say, $z\not=\bar z$ and $-(z{+}\bar z)$, the eigenvalues of $A = \exp(a)$ are $\lambda = e^z$, $\bar\lambda = e^{\bar z}$, and $1/(\lambda\bar\lambda) = e^{-z-\bar z}>0$. Thus, $A = \exp(a)$, cannot have eigenvalues $(-\tfrac12, -2, 1)$ since $-\tfrac12$ and $-2$ are not complex conjugates. In particular $A = \mathrm{diag}(-\tfrac12, -2, 1)\in\mathrm{SL}(3,\mathbb{R})$ is not the exponential of anything in ${\frak{sl}}(3,\mathbb{R})$. You'll run into the same problem with the conformal group in this example, because the space of conformal transformations in this example is the same as the space of isometries. Added comment in response to OP's edit: Unfortunately, I think that, without more restrictions on the class of Riemannian metrics of interest, this is probably not a very sensible question. For example, for any connected semi-simple group $G$, the generic left-invariant metric on $G$ will have $G$ as the identity component of its isometry group, so there are many examples of such metrics for each $G$ for which the exponential map is not surjective (which happens for very many if not 'most' non-compact semi-simple Lie groups). Thus, for example, the generic left-invariant metric on $G=\mathrm{SL}(2,\mathbb{R})$ gives a $2$-parameter family of $3$-dimensional examples that are distinct up to homothety.<|endoftext|> TITLE: Progress on the standard conjectures on algebraic cycles QUESTION [38 upvotes]: What's the current state of these conjectures? Who is working on them? In "Standard conjectures on algebraic cycles" Grothendieck says: "They would form the basis of the so-called "theory of motives" which is a systematic theory of "arithmetic properties" of algebraic varieties, as embodied in their groups of classes of cycles for numerical equivalence. ... Alongside the problem of resolution of singularities, the proof of the standard conjectures seems to me to be the most urgent task in algebraic geometry." REPLY [28 votes]: For future references. Feel free to edit to include new cases, or any improvements. As for 2015, the standard conjectures on algebraic cycles is unconditionally (at lest) known for $X$: Lefschetz standard conjecture (Grothendieck conjectures $A(X)$ and $B(X)$) a curve (trivial). a surface with $H^1(X)=2\cdot\mathrm{Pic}^0(X)$ (Grothendieck). an abelian variety (Liebermann). a generalized flag manifold $G/P$ (Schubert). a smooth varieties which is complete intersections in some projective space (trivial). a Grassmannian (Liebermann?). for which $H^*(X)$ is isomorphic to the Chow ring $A^*(X)$. a smooth projective moduli space of sheaves on rational Poisson surfaces. a uniruled threefold (Arapura). a unirational fourfold (Arapura). the moduli space of stable vector bundles over a smooth projective curve (Arapura). the Hilbert scheme $S^{[n]}$ of a smooth projective surface (Arapura). a smooth projective variety of $K3^{[n]}$-type (Charles and Markman). Weak Lefschetz standard conjecture (Grothendieck conjecture $C(X)$) all of the above, since $B(X) \Rightarrow C(X)$ (Grothendieck). defined by equations with coefficients in a finite field (Katz and Messing). Hodge standard conjecture (Grothendieck conjecture $Hdg(X)$) in characteritic $0$ (Hodge). a surface (Segre, Grothendieck). Main references: Alexander Grothendieck, "Standard Conjectures on Algebraic Cycles" (1969). Steven Kleiman, "The standard conjectures" (1994). Francois Charles, Eyal Markman, The standard conjectures for holomorphic symplectic varieties deformation equivalent to Hilbert schemes of $K3$ surfaces (2011) References for proofs: Beniamino Segre, Intorno ad teorema di Hodge sulla teoria della base per le curve di una sperficie algebraica (1937) Alexander Grothendieck, "Sur une note de Mattuck-Tate" (1958) D. I. Lieberman, "Higher Picard Varieties" (1968) N. Katz and W. Messing, "Some consequences of the Riemann hypothesis for varieties over finite fields" (1973) Donu Arapura, Motivation for Hodge cycles (2005)<|endoftext|> TITLE: Pushouts of equivalences of categories QUESTION [7 upvotes]: If $f:C\to D$ is an equivalence of categories that is injective on objects, then every pushout of $f$ is also an equivalence. This follows, for instance, because such a functor is an acyclic cofibration in the canonical model structure on Cat. Suppose conversely that every pushout of $f$ is an equivalence of categories (in the terminology suggested by Karol here, $f$ is an "acyclic flat" functor). Does it follow that $f$ is injective on objects? (I'd be equally happy with an answer to the corresponding question in Gpd.) REPLY [6 votes]: Every flat functor is injective on objects. In this blog post Chris Schommer-Pries proves that there is a unique model structure on $\mathsf{Cat}$ with categorical equivalences as weak equivalences. The argument of his "Somewhat Less Trivial Lemma" also shows that if there is a flat functor that is not injective on objects, then $E \to *$ is also flat where $E$ is the walking isomorphism. All we need to know is that flat functors are closed under composition, pushout and retracts. Now, let $C_2$ be the group of order $2$ and let $E \to C_2$ classify the nontrivial element. The resulting pushout of $E \to *$ is $C_2 \to *$ which is not an equivalence and hence $E \to *$ is not flat.<|endoftext|> TITLE: Shortest path through $\sqrt{n}$ points out of $n$ QUESTION [29 upvotes]: Say I sample $n$ points uniformly at random in the unit square, and then I look for the shortest path through $\sqrt{n}$ of those points (rounding up, say). What happens to the length of this path as $n\rightarrow\infty$? Does it increase, decrease, or converge (or "none of the above")? REPLY [8 votes]: The comment by John Gunnar Carlsson and answer by Ofer Zeitouni give an upper bound. Comments by two people suggested that the actual behavior should be to decrease to $0$. In fact, there is a positive lower bound: The probability that there is a path through $\sqrt{n}$ points of length less than $0.214$ goes to $0$ as $n\to \infty$. The idea is that if we pick a random path, the probability that this path is short is extremely small. Then we use the union bound over all possible paths of length $\sqrt{n}$ out of $n$ points. It suffices to consider points chosen from a fine grid, or equivalently a large square subset of $\mathbb Z^2$, and to use the $L^1$ metric instead of $L^2$. Roughly how many paths are there in $\mathbb Z^2$ on $m$ points of $L^1$ length at most $d$, where $d \gg m$, and we start at a fixed point? We can divide the distance into $m-1$ steps plus the unused distance. The number of ways to do this is ${d+m-1 \choose m-1} \sim d^{m-1}/(m-1)! = o\left(\left(\frac{ed}{m}\right)^{m-1}\right)$. For any such division of distances, we can choose a lattice point on each "circle." There are $4r$ lattice points of distance $r \gt 0$ from the origin and for $d \gg m$ we can increase the count by making each distance nonzero. By the AM-GM inequality, the count of these choices is at most the case where they are all equal, $\left(\frac{4d}{m-1}\right)^{m-1}$. So, the number of paths is $$\begin{eqnarray} & o\left(\left(\frac{ed}{m}\right)^{m-1} \left( \frac{4d}{m}\right)^{m-1}\right) \newline &=o\left( \left(\frac{2\sqrt{e}d}{m} \right)^{2m-2}\right).\end{eqnarray}$$ The total number of paths in $\lbrace 1, 2, ..., d/c \rbrace^2$ of $m$ lattice points starting at a fixed point is $(\frac{d}{c})^{2m-2}$. So, the probability that a random path of $m$ points has $L^1$ length at most $c$ times the grid dimension is $$o\left( \left(\frac{2\sqrt{e}c}{m} \right)^{2m-2}\right). $$ Let $n=m^2$. The number of paths of length $m$ among $n$ points is $n(n-1)\cdots(n-m+1) \lt m^{2m}$. So, if we choose $c$ so that the probability is $o(m^{-2m})$ then the probability that there is a path that short in the $L^1$ norm goes to $0$ as $n\to \infty$. This is accomplished by setting $c \lt \frac{1}{2\sqrt{e}}$. Since the $L^2$ norm is smaller by at most a factor of $\sqrt{2}$, for any $c \lt \frac{1}{\sqrt{8e}} =0.2144$ the probability that there is a path whose Euclidean length is shorter than $c$ goes to $0$ as $n\to \infty$. While the constant might be improved, the probability bound is good enough to say that if we add the points one by one, then almost surely only finitely many times will there be a path on $\sqrt{n}$ points of length smaller than $c$.<|endoftext|> TITLE: Balls and bins with color QUESTION [5 upvotes]: Say I have $n$ balls each of $k$ different colors (i.e. $nk$ balls altogether), and I throw these balls independently into $N$ bins. Is there anything that can be said (in expectation, limits with respect to any of the variables, or otherwise) about the minimum number of bins that collectively contain at least one ball of each color? I can't seem to find this problem in the literature, although it seems natural to me. How about the case where $N = nk$? To clarify, I am interested in the question "how many bins do I have to look in to see one ball of every color", pursuant to Ben Barber's comment. REPLY [2 votes]: If $S$ is a set of bins, the probability that it contains at least one ball of every colour is $(1 - (1 - |S|/N)^n)^k$. There being $N \choose B$ such sets of cardinality $B$, the expected number of sets of cardinality $B$ that contain at least one ball of every colour is $E(B,N,n,k) = {N \choose B} (1 - (1 - B/N)^n)^k$. This is an upper bound on the probability that there is such a set. EDIT: Here's a little case study. I tried $N = 1000$, $n = 10$, $k = 100$. Here $E(30,1000,10,100) \approx 0.2$ and $E(31,1000,10,100) \approx 109.7$. In 100 trials (using Cplex to solve the set-covering problem of finding the minimum number of bins to get balls of all colours) the minimum numbers of bins needed were as follows: $B = 33$ in 4 trials $B = 34$ in 16 trials $B = 35$ in 41 trials $B = 36$ in 32 trials $B = 37$ in 7 trials<|endoftext|> TITLE: Geometric measures different from Hausdorff QUESTION [10 upvotes]: $\newcommand{\RR}{\mathbb{R}}\newcommand{\calF}{\mathcal{F}}\newcommand{\diam}{\mathrm{diam}}$ In geometric measure theory there are various notions of $m$-dimensional measure for sets $A\subset \RR^n$ for $m\leq n$ (some of them also for non-integer $m$, but this is not the point here). They all build on Carathéodory's general construction which works by covering $A$ with countably many sets $E_i$ from a specific base family $\calF$, measuring the sets $E_i$ with a function $\zeta$ and then building the infimum over all these coverings that are $\delta$-fine (i.e. $\zeta(E_i)\leq \delta$), and letting $\delta\to 0$. Among these specific measures are Hausdorff measure for $s>0$: This uses all sets for covering and takes $\zeta = \diam^s$, the diameter of the set raised to the $s$th power. Spherical measure is the same but restricts the family $\calF$ to consist only of balls. Dyadic net measure is again similar but uses cubes with dyadic corner points as the family $\calF$. Gross measure for $m=0,1,\dots$ is a bit different: It uses the Borel sets for $\calF$ and reuses the $m$-dimensional Lebesgue measure as follows. For some $E$ define $\zeta(E)$ as the largest $m$-dimensional Lebesgue measure you get by projecting $E$ onto any $m$-dimensional subspace. Carathéodory measure is similar to Gross measure but takes as $\calF$ only closed and convex subsets. (There are more, e.g. Federer measure or Gillespie measure…) Somehow, the Gross measure seems most natural to me as the method of covering really drives the minimizing covers to follow the set as close as possible and also directly counts the size of the covering sets in the way one wants to have in the end (slight drawback is that it only works for integer $m$). However, the Hausdorff measure seems to produce a very reasonable definition as illustrated by various examples (e.g. rectifiable curves, Cantor dust in two dimensions with Hausdorff dimension 1). In the books on geometric measure theory I considered (like Federer, Mattila, Krantz & Parks, Morgan) they describe the construction and basic estimates between them quite detailed and mainly use the Hausdorff measure afterwards. Especially for the notion of dimension it turns out that Hausdorff measure, spherical measure and the dyadic net measure all give the same notion of dimension. However, I could not find answers to these questions: Are there sets for which the Gross measure gives something unreasonable (in comparison to intuition or Hausdorff measure), e.g. a totally "wrong" size or even a "wrong" dimension? Are there sets for which the spherical or dyadic net measure gives some intuitively wrong size (or some size different from the Hausdorff measure)? I would also love to have some examples of the usefulness of the measures different from the Hausdorff measure (which appear to be handy for the analysis of self-similar sets). Hence, a third question is ${}$3. What are the spherical, dyadic net, Gross or Carathéodory measure good for? Somehow I am most interested in Gross and Carathéodory measure - they are defined in several books but basically not used. Also it is daunting to search the web for "Gross measure" and totally not helpful to search for "Carathéodory measure". REPLY [7 votes]: Hausdorff, spherical Hausdorff and dyadic net measures not only give rise to the same dimension but, for a fixed value of $m$, are comparable up to constants that depend only on the ambient dimension $d$. In particular, the property of having zero, positive and finite, or infinite measure coincides for these three measures. I am sure that there are examples (even fairly simple examples such as self-similar sets) for which the actual values of these three measures differ, but I don't have a concrete example or reference at hand. As you say Hausdorff dimension is the most used, and the reason to consider spherical or dyadic net measures is that they sometimes make calculations simpler (while yielding the same notion of dimension, and even of zero/infinite measure). One very useful way to study the size of sets is by representing them as trees, and a universal way to do so in Euclidean space is via dyadic cubes - when doing this dyadic net measures are more natural. For example, the standard proof of Frostman's lemma (one the most basic results about Hausdorff measures) uses dyadic partitions and dyadic net measures, although it is often stated for Hausdorff measure. On the other hand, covering by spheres are often easier to analyze than covering by arbitrary sets, and for this reason spherical measures is sometimes easier to compute (if one is seeking the exact value). In particular, the study of densities (the behavior of $\mu(A\cap B(x,r))/r^m$ as $r\to 0$, where $\mu$ is some measure of interest) is easier for spherical measures For nice regular sets (such as embedded manifolds) all three measures do agree, so I don't think there are examples where size according to spherical or net measures is "totally wrong". I hadn't heard of Gross or Caratheodory measures before, but here are some general remarks. I'll denote Gross measure by $G_m$, Hausdorff measure by $H_s$ and Lebesgue measure by $L_m$. Since they are defined only for integer $m$, it does not make too much sense to consider a dimension associated to Gross measure, but in some sense it agrees with the notion of Hausdorff dimension. If $\dim_H A>m$ (where $A\subset\mathbb{R}^n$ and $\dim_H$ is Hausdorff dimension), then $G_m(A)>0$. This follows from the Marstrand-Mattila projection theorem, that says that if $\dim_H A>m$, then the projection of $A$ onto almost every $m$-dimensional subspace has positive $L_m$-measure. In fact it is enough to know this for only one projection. On the other hand, $G_m$ is bounded above by a constant multiple of $H_m$ (since projecting does not increase Hausdorff measure, and on an $m$-dimensional subspace, $H_m$ is a constant multiple of Lebesgue measure). In particular, if $\dim_H(A)0$ but $G_m(A)=0$. Indeed, there exist sets of positive and finite $m$-dimensional measure such that their projection onto any $m$-dimensional subspace has zero $L_m$-measure. For example, this is the case if $A$ is a self-similar set satisfying a suitable separation condition if the orthogonal parts of the generating similitudes generate a dense subgroup of the orthogonal group (this was proved by Eroglu in the plane and Farkas in arbitrary dimension, but ad hoc examples were known long before). For my intuition, it is "correct" to say that such self-similar sets have positive $m$-measure, so Gross measure looks "wrong" to me here - but this is likely simply because I'm much more familiar with Hausdorff measure. Everything I've said about Gross measure applies equally well to Carathéodory measure.<|endoftext|> TITLE: Is every knot unavoidable in the embeddings of some graph? QUESTION [10 upvotes]: Is it the case that, for any given knot $K$, there exists some graph $G$ whose every embedding into $\mathbb{R}^3$ (or into $\mathbb{S}^3$) contains a cycle that realizes $K$? I know the famous Conway-Gordon result that the complete graph on seven vertices $K_7$ is intrinsically knotted in that every embedding contains a knotted cycle. And Joel Foisy proved that $K_{3,3,1,1}$ is also intrinsically knotted (J. Graph Theory, 2002, ACM link). Instead, I am asking a Ramsey-like question: Whether it's known that, for every particular knot, there exists some graph for which that particular knot is unavoidable. REPLY [6 votes]: The answer is no for the following silly reason. Take any graph $G$ and some embedding into $\mathbb R^3$. To any edge, we may perform the connected sum with a fixed knot $K'$ (that is, take a small straight part of the edge and connect sum to it $K'$ there). Do this operation separately to all of the edges. Then any cycle in $G$ has knot type of the form $K'\#(\text{something})$. We just need to pick a knot $K'$ so that the given $K$ is not of this form. We could modify the question and ask whether a fixed knot $K$ exists in some weaker sense in any given embedding of $G$. One such (very) weak sense could be that there exists a knotted torus $T$ in knot type $K$ and a cycle in $G$ which is contained in $T$ and cannot be made disjoint from a compressing disk for $T$. I suspect this may also have a negative answer, however. It seems one can probably always find embeddings of $G$ which are arbitrarily complicated, and in particular which are not related in any way to the given knot $K$.<|endoftext|> TITLE: Covolume of the row span of a matrix and of the kernel of a matrix QUESTION [6 upvotes]: Let $L$ be a $k$-dimensional lattice in $\mathbb{R}^n$. The covolume $\hbox{CoVol}(L)$ of $L$ is the $k$-dimensional volume of a fundamental domain for $L$, i.e., the volume of the parallelopiped spanned by a $\mathbb{Z}$-basis for $L$. Now let $m TITLE: Removal of non-isomorphic edges results in the same graph QUESTION [24 upvotes]: There exists a (simple unlabeled) graph on 6 nodes with a pair of non-isomorphic edges (i.e., there is no graph automorphism that sends one edge into the other) such that removal of either of them results in the same graph: The two non-isomorphic edges here are colored red and blue. This is the smallest example of such graph and it is unique for 6 nodes. I wonder what would be the smallest example of a graph with a triple of pairwise non-isomorphic edges and removal of any one of them resulting in the same graph. P.S. See also sequence https://oeis.org/A245246 -- extension is welcome. UPDATE: user2097 below gave a construction for a graph on $(\tbinom{n}{2}+1)\cdot n$ nodes that squares the number of such pairwise non-isomorphic edges. When applied for the aforementioned graph on 6 nodes, it gives a graph on 96 nodes with a quadruple of pairwise non-isomorphic edges, removal of each of which results in the same graph. But it is probably not the smallest such graph. REPLY [7 votes]: This is essentially a comment on user2097's excellent answer. One consequence of that answer was the construction of a graph with four non-isomorphic edges whose removals result in isomorphic graphs. The graph so constructed had 96 nodes and 293 edges. Without changing the idea of their construction, various tricks allow an improvement to a graph with the same property with 24 nodes and 47 edges. The tricks used are: Instead of connecting the nodes $i$ to nodes $v_{ec}$ whenever $i$ is in $e$, do this whenever $i$ is in $e$ and $c$ is in some set of vertices stable under the isomorphism of subgraphs. Instead of replacing all edges with a copy of $G$ or $G'$, only do this for enough edges in the complete graph to contain the two special edges and be stable under the isomorphism of subgraphs. To wit, begin with the graph described by the original poster: Replace the solid green edges with a copy of $G$ as follows: Replace the dotted green edge with a copy of $G$ without one of its special edges: The final result is as promised. It is not hard to see by starting at the 1-valent vertices that this graph has no automorphisms.<|endoftext|> TITLE: Topological transversality QUESTION [18 upvotes]: Warmup question: Let us say that two continuous functions $f,g:[0,1]\to \mathbb R$ are topologically transverse if their difference $f-g$ has only finitely many zeros, and each zero separates an interval where $f>g$ from an interval where $f TITLE: Is this weak asymptotic Goldbach's conjecture open? QUESTION [7 upvotes]: Let $\tau(x)$ be the number of even numbers $2<2n TITLE: Is the Duflo polynomial conjecture open? QUESTION [17 upvotes]: Let $G/K$ be a symmetric space. Let $\mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p}$ be a Cartan decomposition, with the odd part $\mathfrak{p}$. It is well known that the algebra of invariant differential operators in this case is commutative, and "polynomial conjecture" states that it is isomorphic to $S(\mathfrak{p})^{\mathfrak{k}}$. It was formulated by C.Torossian in a 1993 paper, but actually it is a special case of an older conjecture by Duflo (though the reference I know are proceedings of a 1986 conference, and I haven't seen them.) Is this conjecture still open? If it is, it makes me a little curious, because there aren't many symmetric spaces. What are the known and the open cases then? EDIT: The conjecture (in this form) was formulated in Torossian, C., Operateurs differentiels invariants sur les espaces symetriques I. Methodes des orbites. J. Funct. Anal. 117 (1993), no. 1, 118–173. Torossian made a reference to Duflo, M., in Open problems in representation theory of Lie groups, Conference on Analysis on homogeneous spaces, (T. Oshima editor), August 25-30, Kataka, Japan, 1986. (As I understand it, Duflo's conjecture is much more general; admittedly, I did not read this 1986 text). A more recent account is in "Quantification pour les paires symétriques et diagrammes de Kontsevich" A. Cattaneo, C. Torossian, Annales Sci. de l'Ecole Norm. Sup. (5) 2008, 787--852, available here http://www.math.jussieu.fr/~torossian/ REPLY [3 votes]: As far as I know, Duflo's conjecture is still open. Let me make several remarks: Duflo's conjecture actually says that the algebra of invariant differential operators on a symetric space is isomorphic to the $\mathfrak k$-invariant part of $S(\mathfrak g)/(h-\chi(h),h\in\mathfrak k)$, where $\chi$ is the character given by half the trace of the adjoint action of $\mathfrak k$ on $\mathfrak p$. this shift by a character did not appear in Cattaneo-Torossian paper and this was very surprising... there was indeed a mistake in that paper, which is corrected in Cattaneo-Rossi-Torossian: http://arxiv.org/pdf/1105.5973.pdf Duflo's conjecture is indeed more general. It holds for general reductive homogeneous spaces: it claims that the center of the algebra of invariant differential operators is isomorphic to the Poisson center of $\big(S(\mathfrak g)/(h-\chi(h),h\in\mathfrak k)\big)^{\mathfrak k}$. Rybnikov's result mentionned in Alexander Chervov's comment prove a localized version of it for Riemaniann reductive homogeneous spaces (i.e. it holds on the level of fraction fields).<|endoftext|> TITLE: Exactly sampling from a distribution with access to the probabilities only QUESTION [5 upvotes]: There is a discrete distribution where integers, $k$, from $1$ to $n$ occur with probability $p_{k}$, all $p_{k}$ are unknown. Rather than having access to the distribution we have access to $n$ coins, with the $k$th coin having probability $p_{k}$ of giving a result of heads. Can we sample exactly from the distribution by flipping the coins finitely many times? We can sample approximately by flipping each coin $M$ times and then randomly selecting one of the results of heads and outputting the value of $k$ for the coin that produced it. Unfortunately we require exact sampling. I'm sure this must have been studied somewhere but I really can't find it. REPLY [5 votes]: I will assume all $p_i$ are strictly between $0$ and $1$. Lemma We can use a coin with probability $p$ of coming up heads to emulate a coin with probability $p/(1+p)$ of coming up heads. Proof Flip the coin until it first comes up tails. The emulated coin is declared to be heads if we have flipped an even number of times. The emulated coin comes up heads with probability $$p(1-p) + p^3(1-p) + p^5(1-p) + \cdots = \frac{p(1-p)}{1-p^2} = \frac{p}{1+p}. \quad \square$$ Set $q_i = p_i/(1+p_i)$. So we may assume we have coins with probability of heads $q_k$. Flip all coins until exactly one coin comes up heads, and select the corresponding $k$. Let $S$ be the probability that not precisely one coin comes up heads. The probability of picking $k$ is $$q_k \prod_{j \neq k} (1-q_j) + S q_k \prod_{j \neq k} (1-q_j) + S^2 q_k \prod_{j \neq k} (1-q_j) + \cdots = \frac{q_k \prod_{j \neq k} (1-q_j)}{1-S}.$$ So the ratio of the probability of picking $k$ and the probability of picking $\ell$ is $$\frac{q_k/(1-q_k)}{q_{\ell}/(1-q_{\ell})}= \frac{p_k}{p_{\ell}}.$$ Since the probabilities must add up to $1$, the probability of picking $k$ is $p_k$.<|endoftext|> TITLE: Examples of a continuous martingale with $E[\sup\limits_{0\leq s\leq t} |M_s|]=\infty$? QUESTION [5 upvotes]: A local martingale is a martingale iff it is in the class DL. The condition: for every $t\in[0,\infty)$ $$E[\sup\limits_{0\leq s\leq t} |M_s|]<\infty\tag1$$ guarantees a local martingale $M$ is a martingale by ensuring it satisfies the condition for being in the class DL. Moreover, by Burkholder-Davis-Gundy, this means: for every $t\in[0,\infty)$, $$E[\langle M\rangle^{1/2}_t]<\infty$$ My question is: do there exist (simple?) examples of continuous martingales where these two conditions are violated? What I am really asking is that these are known to be sufficient conditions to guarantee local martingales are martingales, why are they not necessary? REPLY [4 votes]: There exist indeed a uniformly integrable martingale $X$ whose max $\sup_{k\in \mathbb{N}} |X_k|$ is not integrable. A simple example in discrete time can be found here http://www.math.fsu.edu/~nichols/martingalezoo.pdf (see the last 2 lines of point 5). A related construction is found in example 4.1 of the paper by A.S. Cherny "Some particular problems of martingale theory", which produces a uniformly integrable martingale $(X_n)_{n\in \mathbb{N}}$ and a bounded process $(H_n)_{n\in \mathbb{N}}$ such that the stochastic integral $M_n:=\sum_{k=0}^n H_k (X_{k+1}-X_k)$ is a martingale which is not uniformly integrable (by the Burkholder-Davis-Gundy inequality this implies that $\sup_{k\in \mathbb{N}} |X_k|$ is not integrable) A second (completely different) example in continuous time is sketched in Exercise 3.15 of the 2nd Chapter of Revuz-Yor's Book "Continuous Martingales and Brownian Motion, 3rd ed"<|endoftext|> TITLE: How weird can Modular Tensor Categories be over non-algebraically closed fields? QUESTION [8 upvotes]: I am trying to understand better the behaviour and character of modular tensor categories over non-algebraically closed fields. How weird can they be? The reason I am interested in this is that my collaborators and I are investigating extended 3D tqfts. These are known to be related to Modular Tensor Categories via the Reshetikhin-Turaev construction. However most of the literature seems to be focused on the case where the MTC is defined over an algebraically closed field, usually the complex numbers. I don't see a reason to restrict to this case and I can imagine that there could be some very interesting examples, and hence interesting invariants, in other cases to. For example over field $k$, a finite semisimple linear category will have simple objects whose endomorphisms rings are division algebras over $k$. If $k$ is algebraically closed, then all we get are copies of $k$. If $k$ is not algebraically closed then it is more interesting as we can have objects with different division algebras as endomorphisms. Can this happen in a modular tensor category? i.e. is there an example of a Modular Tensor Category such that the simple objects have different division algebras for their endomorphism rings? What if we drop the requirement End(1) = k? How sticky can it get? REPLY [8 votes]: An example of MTC is Drinfeld double of a finite group $G$ (over any field of characteristic zero). This category contains representation category of $G$ as a subcategory. So all endomorphisms rings that you can find in representations of finite groups, you can also find in MTC. For example the quaternions will show up in the Drinfeld double of the quaternion group.<|endoftext|> TITLE: How should I be thinking about object classifiers / universal fibrations / universes? QUESTION [9 upvotes]: I have been learning about homotopy type theory this summer. I am not a homotopy theorist but I am more comfortable with homotopy theory than I am with type theory, so the way I rationalize many of the constructions used is by thinking in terms of homotopy theory. For example, when someone says type, i think homotopy type of a space. When someone says type family, I think of the family of fibers of some fibration. $ \Sigma $ types are total spaces and dependent function types are sections of a fibration. Things like path induction and function extensionality are easy to understand from this perspective. One thing that I have never really understood is how to think about the universe $ \mathcal{U} $. It is a type, so it should be the homotopy type of some space? Assuming this, the univalence axiom says that there is one path component for every homotopy type. Moreover, we have some sort of universal fibration $$ \Sigma_{A : \mathcal{U}} A \to \mathcal{U} $$ and the fiber over the path component corresponding to the homotopy type $A$ has homotopy type $A$. I have read things like "Morally, $\mathcal{U}$ is an object classifier" and "$\Sigma_{A : \mathcal{U}} A \to \mathcal{U} $ behaves like a universal fibration", but i dont really know how to make intuitive sense of these statements. Questions: How should I be thinking about the universe $ \mathcal{U}$? Is it the homotopy type of some huge space (ignoring set theoretic problems?) What does it mean to say $ \mathcal{U}$ behaves like an object classifier? What does it mean to say $\Sigma_{A : \mathcal{U}} A \to \mathcal{U}$ is a universal fibration? REPLY [3 votes]: Zhen Lin's answer is great. Here is something I realized this morning (I believe this was one of the original motivations for the univalence axiom). You can make sense of the statement "$ \Sigma_{A : \mathcal{U}} A \to \mathcal{U} $ is a universal fibration" inside HoTT. Lets call a type family $ P : B \to \mathcal{U}$ "fibration data". The corresponding fibration is $ \Sigma_{b : B} P(b) \to B $. In the paper homotopy limits in type theory, homotopy pullbacks are constructed inside HoTT. Indeed, $$\require{AMScd} \begin{CD} \Sigma_{a:A} \Sigma_{x:\Sigma_{b:B}P(b)} \pi x = f(a) @>>> \Sigma_{b:B} P(B)\\ @VVV @VVV \\ A @>{f}>> B \end{CD}$$ it a homotopy pullback square. Note, that using the induction principle for $\Sigma$-types it is easy to construct the homotopy that makes this square homotopy commutative. We have another homotopy commutative square $$\require{AMScd} \begin{CD} \Sigma_{a:A} P(f(a))@>>> \Sigma_{b:B} P(B)\\ @VVV @VVV \\ A @>{f}>> B \end{CD}$$ where the top map and the homotopy are constructed using $\Sigma$-induction. The universal property of homotopy pullbacks gives us a map $$ \Sigma_{a : A} P(f(a)) \to \Sigma_{a : A} \Sigma_{x : \Sigma_{b : B}P(b)} \pi x = f(a) $$ and homotopies making the obvious diagram commute up to homotopy. In fact, it is easy to construct the map and homotopies directly using $ \Sigma$-induction. Again, using $\Sigma$-induction and the introduction rule for equality types in a $ \Sigma$-type we can construct a quasi-inverse to this map, so it is an equivalence. Moreover, this quasi-inverse fits into the appropriate homotopy commutative diagram, which tells us that the second commutative square above is in fact a homotopy pullback square. I worked all of this out explicitly this morning. It is a good exercise in working inside HoTT, so I won't spoil it for anyone else interested. What all of this tells us is that given fibration data $P : A \to \mathcal{U}$, the corresponding fibration is the homotopy pullback of $ \Sigma_{ A : \mathcal{U}} A \to \mathcal{U}$ along $ P $.<|endoftext|> TITLE: Continued fraction expansion of an algebraic number and its conjugates QUESTION [7 upvotes]: Let $w$ be an element of a Galois extension $L:\mathbb{Q}$ such that $\text{Gal}(L/\mathbb{Q})=\langle g\rangle$ is cyclic of order $n$ (here $\mathbb{Q}$ is rationals). Suppose we know the continued fraction of $w$. Is there a way to find the continued fraction of its conjugates? Also can someone provide me references in the direction for finding the continued fraction of $w$ itself. Please edit if things are not clear. REPLY [8 votes]: A uniform answer for all such cases is going to be too much to expect when $n > 2$. To begin with, $w$ may not even have a continued fraction expansion - it may fail to be real! (But at least the conjugates of a real algebraic number will also be real under the given conditions.) And then the relationship between $w$ and any one conjugate may be arbitrarily complicated. More can be said only when you restrict your attention to particular families, such as for example Shanks' "simplest cubic fields" defined by polynomials $X^3-TX^2-(T+3)X-1$ for integer $T \ge -1$ where the automorphisms can be expressed by a single formula independent of $T$. As to your second question: Nowadays given a defining polynomial $f$ for (real algebraic) $w$, you would use your favorite computer algebra or computational algebraic number theory software package to compute a sufficiently accurate approximation, and then turn this into the desired number of partial denominators. However, there is also a direct algorithm starting from the integer coefficients of $f$ and working only with (large) exact integers: Begin by bracketing the real roots of $f$ between successive integers using Sturm sequences. Then when you know that, for some $a\in\mathbb{Z}$ and real $\delta > 1$, one root can be written as $f(a+1/\delta)=0$, you can rewrite this as a new polynomial equation $f_1(\delta)=0$ with integer coefficients depending on $a$ and on those of $f$. Repeat as often as you like. Serge Lang and Hale Trotter were using this approach long before computer algebra software became widespread ("Continued fractions for some algebraic numbers," J. f. r. angew. Math. 255 (1972), 112-134; Addendum: ibid., 219-220). I'm not sure whether that article is available online, but you'll find a very readable introduction by E.Bombieri and A.van der Poorten here. Edit: A scan of the Lang-Trotter paper is available online, too. They do discuss the case $T=-1$ of Shanks' family in the guise of $2\text{cos}(2\pi/7)$.<|endoftext|> TITLE: Are there any integers which can't be written as a sum of two fourth powers minus a cube? QUESTION [31 upvotes]: To be precise, I am asking: Does there exist an integer $k$ such that there do not exist (possibly negative) integers $x,y,z$ satisfying $x^4+y^4=z^3+k$? Heuristically the answer must be yes, in fact, one expects that almost every $k$ should work (this is just because the sum of the reciprocals of the exponents is $\frac13+\frac14+\frac14 = \frac56<1$, so one ought to expect that up to a large bound $N$, something like $N^{5/6}$ values of $k$ are representable). For all I know, $-2$ and $4$ might already be two examples of numbers which can't be represented as sums of two fourth powers minus a cube, but I haven't been able to prove this. What I can prove is that there are no local obstructions: for any integers $n,k$ with $n\ne 0$, we can find integers $x,y,z$ with $(x,y,z)=1$ such that $x^4+y^4\equiv z^3+k\pmod{n}.$ Using the Chinese Remainder Theorem and Hensel's Lemma one can quickly reduce this claim to the case that $n=p$ is a prime. The most interesting case is when $p\equiv 1\pmod{12}$ and $p\nmid k$, and in this case we can use a trick similar to the proof of Chevalley-Warning to count the number of solutions $N_p$ modulo $p$. We start with the easy congruence $N_p \equiv \sum_{x,y,z} (1-(x^4+y^4-z^3-k)^{p-1}) \equiv -\sum_{x,y,z}\sum_{a+b+c+d=p-1} \frac{(p-1)!}{a!b!c!d!} x^{4a}y^{4b}(-z)^{3c}(-k)^d\pmod{p},$ and then note that if we fix $(a,b,c,d)$ and sum over $x,y,z$, we can only get a nonzero contribution when $p-1\mid 4a,4b,3c$ and $a,b,c>0$. From this we see that $N_p \equiv \frac{(p-1)!}{\left(\frac{p-1}{4}\right)!\left(\frac{p-1}{4}\right)!\left(\frac{p-1}{3}\right)!\left(\frac{p-1}{6}\right)!} (-k)^{\frac{p-1}{6}}\not\equiv 0\pmod{p},$ so in particular $N_p \ne 0$. REPLY [4 votes]: Section 5 of my paper http://maths.nju.edu.cn/~zwsun/179b.pdf is closely related to your question. You can find there a theorem, two conjectures and some heurisitic arguments with explanations. The heuristic arguments (not rigorous) there suggest that there are integers not of the form $x^4+y^4-z^3$ with $x,y,z$ positive integers. In view of this, I believe that there are infnitely many integers $k$ which cannot be written as $x^4+y^4-z^3$ with $x,y,z$ integers.<|endoftext|> TITLE: opposite category QUESTION [5 upvotes]: In the 2-category Cat of small categories, for each category $C$ (an object of Cat) there is also the dual category (I dare not write "dual object") $C^{op}$. Is ${op}$ the instance in Cat of a more general concept for 2-categories? (In the same sense that monads in Cat are a special case of monads-in-a-2-category $C$). REPLY [7 votes]: To make a link between Todd's answer and Qiaochu's answer: if $\mathbb{W}$ is a 2-category, then a functor: $$(-)^* \colon \mathbb{W}^{co} \rightarrow \mathbb{W}$$ is called a "duality involution" if it is self-inverse and (pseudo) naturally satisfies: $$\mathit{DFib}(A\times B, C) \approx \mathit{DFib}(A, B^* \times C)$$ where $\mathit{DFib}(X, Y)$ is the category of discrete fibrations from $X$ to $Y$ in $\mathbb{W}$. Observe that in case $\mathbb{W} = \mathbf{Cat}$ discrete fibrations are equivalent to profunctors by the generalised Grothendieck construction and its inverse; the above equivalence is then induced by pairing (evaluation) $B \times B^* \nrightarrow 1$ from autonomous (weak) 2-category of profunctors. This concept of "duality involution" was a crucial part of the definition of a 2-topos given by Mark Weber in "Yoneda structures from 2-toposes".<|endoftext|> TITLE: Without AC, which implications between the different definitions of amenability still hold? QUESTION [10 upvotes]: More precisely, I would like to know which implications between the following definitions of amenability of a discrete countable (or even finitely generated) group can be proved to hold with only ZF (and for which one it can be proved that ZF is not enough) : G admits on itself a left-invariant finitely-additive-probability-measure ; every action of G by homeomorphisms on a non-empty compact set admits an invariant Borel probability measure ; G admits a Folner sequence. I know that 3 => 1 requires more than ZF. Thank you for your time. REPLY [8 votes]: Partial answer: $1\Rightarrow2\Rightarrow3\Rightarrow (2$ for compact metrizable spaces) It will be convenient to consider also a fourth characterization, the (von Neumann-)Dixmier criterion: 4) if $K\subset G$ is finite and $f_k\in B(G)$, $k\in K$, we have $\displaystyle\inf_G\sum_{k\in K}(f_k-f_k\circ\ell_k)\le0$, where $\ell_g(x)=gx$ (left-translation). This is an easy consequence of the Følner criterion 3, and as that one it obviously holds for $G$ if and only if it holds for every finitely generated subgroup (all this in ZF). Finally, for finitely generated groups, Følner implies Dixmier using Hahn-Banach for a separable Banach space, thus in ZF. Proof of $1\Rightarrow2$ (for any discrete group). If $\mu$ is a left-invariant finitely additive probability measure on $G$, one associates to it a left-invariant mean $m(f)=\displaystyle\int_Gf(g)d\mu(g)$ defined on all bounded functions of $G$: $m(f)=\sup \displaystyle\sum_{\lambda\in\bf R}\lambda\mu(\varphi^{-1}(\lambda)\})$ over every function $\varphi\le f$ taking only finitely many values. Then if $G$ acts on a compact space $X$, we construct a Radon probability measure which is left-invariant, or equivalently a positive linear form $p$ on $C(X)$ such that $p(1_X)=1$ and $p(f\circ\ell_g)=p(f)$ where $\ell_g$ is the left-translation by $g$: choose $x_0\in X$ and set $p(f)=m(\gamma\mapsto f(\gamma.x_0))$. Then $$\eqalign{p(f\circ\ell_g)&=m(\gamma\mapsto f((g\gamma).x_0)\cr &=m(\gamma\mapsto f(\gamma.x_0))\ \hbox{by left-invariance}\cr &=p(f).\cr}$$ Proof of $2\Rightarrow 4$ (inspired by A. Paterson, Amenability, Problem 2-14 p.90, solution p. 420) We take now $B(G)$ to be the bounded complex functions on $G$. Let $A\subset B(G)$ be the $C^*$-algebra generated by $f_1,\cdots,f_n$. In other words, $A$ is the norm-closure of all polynomials in $f_1,\cdots,f_n,\overline f_1,\cdots,\overline f_n$. It is a separable commutative unital $C^*$-algebra, thus by the Gelfand representation theorem it is isomorphic to $C(X)$ for $X$ a compact metric space, this in ZF because of separability. The action of $G$ on itself by left translations gives rises to an action on $X$ such that $f(g.x)=(g^{-1}f)(x)$. By hypothesis this action has an invariant Borel (here it the same as Radon) probability measure, ie $A$ admits an invariant positive linear form $p$ such that $p(1)=1$. By invariance, $p(\displaystyle\sum_{k\in K}(f_k-f_k\circ\ell_k))=0$, which implies (by positivity) $\displaystyle\inf_G\sum_{k\in K}(f_k-f_k\circ\ell_k)\le0$. Proof of $3\Rightarrow 2$ for a compact metrizable space. If $X$ is a compact metrizable space, the space of Radon (= Borel here) probability measures $P(X)$, viewed as a subspace of $C(X)^*$ with the weak $*$-topology, is compact (and metrizable). This is true in ZF since it is contained in $[-1,1]^D$ where $D$ is a countable dense subset of the unit sphere of $C(X)$, and Tychonov's theorem in this case is true in ZF. By the Følner criterion, for every $K\subset G$ finite and $\varepsilon>0$ there existe a non-empty finite set $F\subset G$ such that $\displaystyle\max_{k\in K}|F\setminus kF|\le\varepsilon|F|$. Then, if $x_0\in X$ is fixed, $p=\displaystyle{1\over|F|}\sum_{g\in F}\delta_{g.x_0}$ belongs to the set $$F_{K,\varepsilon}=\{p\in P(X):\max_{k\in K}||p-p.g||\}\le2\varepsilon\}.$$ The family $(F_{K,\varepsilon})$ consists of closed sets and has the finite intersection property since $\displaystyle\bigcap_{i=1}^nF_{K_i,\varepsilon_i}\supset F_{K_1\cup\cdots\cup K_n,\min\varepsilon_i}$. Thus its full intersection is nonempty, and it consists of the invariant probability measures. Remarks. (i) The last proof holds whenever $P(X)$ is compact for the weak $*$-topology. I do not know if $X$ compact implies $P(X)$ compact in ZF. If not, presumably $3\Rightarrow 2$ does not hold in general. (ii) Things get better with many (all ?) other characterizations of amenability. For instance, consider the Markov-Kakutani fixed point property: 5) every $G$-space $X$ which is compact, affine and convex has a fixed point. Proof of $3\Rightarrow 5$. Fix $x_0\in X$. If $F\subset G$ is a $(K,\varepsilon)$-Følner set, the point $x=\displaystyle{1\over|F|}\sum_{g\in F} g.x_0$ belongs to the set $$F_{K,\varepsilon}=\{x\in X:(\forall k\in K)\ x-kx\in[-\varepsilon,\varepsilon]K\}.$$ The family $(F_{K,\varepsilon})$ consists of closed sets and has the finite intersection property since $\displaystyle\bigcap_{i=1}^nF_{K_i,\varepsilon_i}\supset F_{K_1\cup\cdots\cup K_n,\min\varepsilon_i}$. Thus its full intersection is nonempty, and it consists of the fixed points. Proof of $5\Rightarrow 4$. This is essentially the same proof as $2\Rightarrow 4$.<|endoftext|> TITLE: Lebesgue measure of a set of irrational numbers QUESTION [10 upvotes]: Let $I_{\lambda},$ $\lambda>0$ be a subset of all irrational numbers $\rho=[a_{1},a_{2},...,a_{n},...]\in(0,1)$ such that $a_{n}\leq \text{const}\cdot n^{\lambda}.$ Here, $[a_{1},a_{2},...,a_{n},...]$ is the continued fraction with partial quotients $a_1,a_2,\dots$. My question is: under what values of $\lambda$ the set $I_{\lambda}$ has a positive Lebesgue measure? REPLY [2 votes]: A precise solution to this problem is known. In Khintchine's book on continued fractions it says: $\mathbf{Theorem~30}$ Suppose that $\varphi(n)$ is an arbitrary positive function with natural argument $n$. The inequality $$ a_n = a_n(\alpha) \geq \varphi(n) $$ is, for almost all $\alpha$, satisfied by an infinite number of values $n$ if the series $\sum_n 1/\varphi(n)$ diverges. On the other hand, this inequality is, for almost all $\alpha$, satisfied by only a finite number of values of $n$ if the series $\sum_n 1 /\varphi(n)$ converges. (quoting from: A. Ya. Khintchine. Continued Fractions. University of Chicago Press, 1964)<|endoftext|> TITLE: Combinatorial interpretation of composition of power series? QUESTION [10 upvotes]: This is a minor curiosity that came up in a joint project recently. Consider the sequence $a_n=3\frac {(2n)!}{(n+2)!(n-1)!}$ (A000245 in OEIS). It has multiple combinatorial descriptions. One can write the corresponding generating function, but I will do it with an extra sign. $$ f(t) = -\sum_{n=0}^{\infty} a_n t^n = -t-3t^2-9t^3-28t^4-90t^5 -... $$ which has a simple though inelegant description $$ f(t) = \frac {(1-t)\sqrt{1-4t}-1+3t}{2t^2}. $$ So far it is all perfectly boring. But the interesting thing is that this $f$ satisfies the identity $$ f(f(t))=t. $$ My questions are: Has this been observed before? Are there any combinatorial interpretations of this identity in terms of the combinatorial descriptions of the sequence $a_n$? REPLY [3 votes]: When rephrased (using the positive generating series $g=-f$) as the identity $g(-g(-t))=t$, this suggests that there may exist a quadratic nonsymmetric operad with this generating series, which is Koszul and whose Koszul dual is itself. I do not know such an operad. Looking at Loday's encyclopedia of operads, a possible candidate could have been the operad of Malcev algebras, but Loday only gives the first three terms $1,3,9$, which is probably not enough to draw any conclusion. And moreover, this is not a nonsymmetric operad. It may be possible to try to find one, maybe being guided by the existing combinatorics. To illustrate, the simpler case of $g(t)=t/(1-t)$ is related to the Associative operad, which is Koszul and self-dual.<|endoftext|> TITLE: For $x,y\ge 2$ does $x^4+x^2y^2+y^4$ ever divide $x^4y^4+x^2y^2+1$? QUESTION [21 upvotes]: For a problem in combinatorics, it comes down to knowing whether there exist integers $x,y\ge 2$ such that $$ x^4+x^2y^2+y^4\mid x^4y^4+x^2y^2+1. $$ Note that $x^6(x^2-y^2)(x^4+x^2y^2+y^4)+(x^2y^2-1)(x^4y^4+x^2y^2+1)=x^{12}-1$ and so we can look at a perhaps simpler problem: Are the only solutions to $$ x^4+x^2y^2+y^4\mid x^{12}-1 $$ (where $x,y\ge 2$) given by $(x,y)=(5,6)$ and $(x,y)=(6,5)$? REPLY [6 votes]: I do not know whether there is any advantage to considering this problem in the ring of Eisenstein integers $\mathbb{Z}[\omega],$ where $\omega = e ^{\frac{2 \pi i}{3}},$ which is a PID. Then we have to ask when we can have $(x - \omega y)(x +\omega y)(x - \omega^{2}y )(x+\omega^{2}y)$ dividing $(xy- \omega)(xy +\omega)(xy- \omega^{2})(xy +\omega^{2})$ in $\mathbb{Z}[\omega],$ where $x,y$ are rational integers. I have not been able to pursue this to provide further insight myself, but someone else might. Later remark: It is easy to check that the power of $1-\omega$ dividing both expressions is the same: it is $0$ if $3$ divides $xy,$ and $2$ if $3$ does not divided $xy.$ Hence we can omit the prime $1-\omega$ from our considerations, and we only need to worry about primes in $\mathbb{Z}[\omega]$ such that $N(\pi)$ is a rational prime congruent to $1$ (mod $3$). If $\pi$ is such a prime dividing the leftmost product, we note that $\pi$ divides exactly one of the terms in the rightmost product (and, in fact, $\pi$ also divides exactly one term in the leftmost product). This leads (if the required divisibility holds ) relatively easily to the observation (already made by the OP) that the leftmost expression divides $y^{12}-1$ (and/or $x^{12}-1,$ there is symmetry in $x$ and $y$), but it is unclear to me at present whether this viewpoint provides any more useful information. Later edit: I noticed that Aaron Meyerowitz's observation (in a comment after Gerry Myerson's answer) that if the required divisibility holds, then $x^{4} + x^{2}y^{2} +y^{4}$ divides $(x^{4}-1)(y^{4}-1)$ can be derived this way. That is not particularly surprising, and the direct derivation is easier. However, perhaps less obvious is that we also have that $x^{4} + x^{2}y^{2} +y^{4}$ divides $(y^{8}+y^{4}+1)(x^{8}+x^{4}+1)$. While $(x^{4} + x^{2}y^{2} +y^{4})^{2}$divides $(x^{12}-1)(y^{12}-1),$ it is not immediately obvious to me that this last claimed divisibility is a consequence of that- for example, there might a priori be a prime $\pi$ such that $x^{4}-1$ is divisible by some higher than expected power of $\pi$- so I outline a proof: Note that if $\pi$ is a prime in $\mathbb{Z}[\omega]$ with $N(\pi) \equiv 1$ (mod $3$), then if $\pi^{m}$ divides both $x^{4}-1$ and $x^{2}- \omega y^{2},$ we have $\pi^{m}$ divides $\omega^{2}y^{4}-1,$ so that $\pi^{m}$ divides $y^{4}-\omega^{4}.$ It follows that $N(\pi)^{m}$ divides $y^{8}+y^{4} + 1.$ Hence it follows that (in $\mathbb{Z}$), ${\rm gcd}(x^{4}-1,x^{4} + x^{2}y^{2} + y^{4})$ divides $y^{8}+y^{4}+1$ (as before, the power of $3$ is taken care of). Similarly ${\rm gcd}(y^{4}-1,x^{4} + x^{2}y^{2} + y^{4})$ divides $x^{8}+x^{4}+1$. Since $x^{4}+ x^{2}y^{2} + y^{4}$ divides $(x^{4}-1)(y^{4}-1),$ the claim is established (note that ${\rm gcd}(x^{4}-1,y^{4}-1)$ has the form $2^{a}3^{b}$ if the original divisibility holds (and $(x^{2}-1)(y^{2}-1) \neq 0$). We can continue this analysis: we see (if the original divisibilty holds) that $x^{4}+x^{2}y^{2}+y^{4}$ divides ${\rm gcd}(y^{2}-1,\frac{x^{6}-1}{x^{2}-1}){\rm gcd}(y^{2}+1,\frac{x^{6}+1}{x^{2}+1}) {\rm gcd}(x^{2}-1,\frac{y^{6}-1}{y^{2}-1}){\rm gcd}(x^{2}+1,\frac{y^{6}+1}{y^{2}+1}).$ Additional edit: Conversely, it is easy to check that the rightmost product divides $3(x^{4}+y^{4}+ x^{2}y^{2})$ given that $x$ and $y$ are coprime. Also, the righmost product divides $3(x^{4}y^{4}+x^{2}y^{2}+1).$ It follows that $x^{4} + y^{4} + y^{2}x^{2}$ divides $x^{4}y^{4}+y^{2}x^{2}+1$ if and only if ${\rm gcd}(x,y) = 1$ and $x^{4}+x^{2}y^{2}+y^{4}$ is equal to ${\rm gcd}(y^{2}-1,\frac{x^{6}-1}{x^{2}-1}){\rm gcd}(y^{2}+1,\frac{x^{6}+1}{x^{2}+1}) {\rm gcd}(x^{2}-1,\frac{y^{6}-1}{y^{2}-1}){\rm gcd}(x^{2}+1,\frac{y^{6}+1}{y^{2}+1})$ when $xy$ is divisible by $3$ or $3(x^{4}+x^{2}y^{2}+y^{4})$ is equal to ${\rm gcd}(y^{2}-1,\frac{x^{6}-1}{x^{2}-1}){\rm gcd}(y^{2}+1,\frac{x^{6}+1}{x^{2}+1}) {\rm gcd}(x^{2}-1,\frac{y^{6}-1}{y^{2}-1}){\rm gcd}(x^{2}+1,\frac{y^{6}+1}{y^{2}+1})$ when $xy$ is not divisible by $3$.<|endoftext|> TITLE: weak convergence of the solutions to stochastic heat equation QUESTION [6 upvotes]: $W(t,x)=\sum_ic_ie_i(x)B^i_t$ is a Brownian motion in $L^2(R^d)$, where $\{e_i\}$ is the standard orthogonal basis and $\sum_ic_i^2<\infty$. $$\partial_t u(t,x)=\Delta u(t,x)+u(t,x)\dot{W}(t,x)$$ $$u(0,x)=f(x)\in C_C(R^d)$$ If $W$ satisfies some condition, if regarding $\{u(t,\cdot)\}$ as a process in $L^2_\rho$ $(\rho>2d)$ , $(\|u\|^2_{L^2_\rho}=\int_{R^d}u^2(x)(1+|x|)^{-\rho}dx)$, I know the distribution of $\{u(t,\cdot)\}_{t\geq 0}$ is tight. Can one prove that all the limit points are the same one? (You can add some condition on $W$ and $f$. ) REPLY [2 votes]: In linear random dynamical systems (cocycles) the notion of Lyapunov exponent usually applies. In more compact systems than yours, it is a theorem (consequence of Oseledets' multiplicative ergodic theorem and its infinite-dimensional generalizations) that there is a real number $\lambda$ such that the logarithm of the norm of most solutions is asymptotic to $\lambda t$ for large $t$. Some form of that statement should apply to your SHE (maybe you even have enough compactness since your norm gives much more weight to the bulk then to the "tails"). Then tightness would mean $\lambda\le 0$. It is unlikely that $\lambda=0$, so one should have $\lambda<0$, and then, of course, you have contraction to zero, and identical zero is a unique limit point.<|endoftext|> TITLE: Clusters of uniformly distributed random points QUESTION [8 upvotes]: This can be viewed as a toy version of this wonderful question. (I'm removing the TSP component, and I'll zoom in on the statistical part.) Let $X_1,\ldots, X_n$ be iid, with uniform distribution in $[0,1]$. Let $\ell_n$ be the length of the shortest subinterval of $[0,1]$ that contains $n^{\alpha}$ of the points $X_j$, with $0<\alpha<1$. What can we say about the distribution of the normalized length $L_n=n^{1-\alpha}\ell_n$? For example, what are the asymptotics of $EL_n$? As a warm-up of sorts, we can in fact also consider the shortest interval with $cn$ points and $L_n=\ell_n/c\,$; the answers to the linked question seem to suggest that $EL_n\to 1$ in this case. (This sounds like it should be well studied, but I couldn't locate anything on the internet.) REPLY [5 votes]: As in the $2$-dimensional problem linked, the probability that a particular interval much shorter than $n^{\alpha-1}$ contains $n^\alpha$ points is very small, and we can use the union bound over a small set of such intervals to bound the probability that the shortest interval containing $n^\alpha$ points is small. This implies $EL_n \to 1$ and more precise statements about the difference with $1$. Suppose we want to bound the probability that $L_n \gt 1-2\epsilon$. Consider intervals of length $(1-\epsilon)n^{\alpha-1}$ starting at $0, \epsilon n^{\alpha-1}, ...$. There are fewer than $n^{1-\alpha}/\epsilon$ such intervals and every interval of length $(1-2\epsilon)n^{\alpha-1}$ fits into one of these. Let us bound the probability that there are $n^\alpha$ points in a particular interval of length $(1-\epsilon)n^{\alpha-1}$ and then use the union bound. The number of points in an interval of length $(1-\epsilon)n^{\alpha-1}$ is a binomial random variable with mean $(1-\epsilon)n^\alpha$ and standard deviation under $\sqrt{n^\alpha}$. So, having $\epsilon n^\alpha$ points more than average is more than $\epsilon n^{\alpha/2}$ standard deviations above the mean. A normal approximation would suggest that the probability drops rapidly as $n$ increases. To be rigorous we can use a Chernoff bound. For $0 \lt \delta \lt 1,$ $\textrm{Prob}[\textrm{Binom} \ge (1+\delta)\mu] \le \exp(-\mu \delta^2/3).$ We can choose $\delta = \epsilon$, since $(1+\epsilon) \mu = (1+\epsilon)(1-\epsilon)n^\alpha = (1-\epsilon^2)n^\alpha \lt n^\alpha$. $\textrm{Prob}[\textrm{Binom} \ge n^\alpha] \le \exp (-c_\epsilon n^\alpha).$ So, the probability that the shortest interval containing $n^\alpha$ points is shorter than $(1-2\epsilon)n^{\alpha-1}$ is at most $\frac{n^{1-\alpha}}{\epsilon}\exp (-c_\epsilon n^\alpha).$ As $n\to\infty$ the exponential term dominates and the probability drops to $0$. As $n\to \infty$, $L_n$ is close to $1$ with high probability.<|endoftext|> TITLE: Motivation behind the definition of hochschild cohomology QUESTION [11 upvotes]: For an associative algebra $A$ one can define the Hochschild cohomology of $A$ as $ HH^n(A,A):= Hom_{\mathcal{D}(A^{op} \otimes A)}(A, [n]A)$ (this definition also works for the graded and dg cases as well). There is a lot of information contained in this gadget: when $ n=0$ it is the center and when $n=1$ it is the lie algebra $\text{Der}(A)/\text{Inn}(A)$ and when $n=2$ it says something about the deformation theory of the algebra. Also there is a cup product and Gerstenhaber bracket on $HH^*(A,A)$. My question is: What is the reason/motivation for the definition $ HH^n(A,A):= Hom_{\mathcal{D}(A^{op} \otimes A)}(A, [n]A)$, if you didnt know about all this higher structure what would lead you to study it in the first place? REPLY [8 votes]: I think you will have different answers to your questions. As you noticed there is more about Hochschild Homology in n-lab page and you also pointed out the interpretation of (positive) low dimension of Hochschild cohomology. I would say that the Hochschild cohomology is "The" cohomology theory for differential graded algebras in the following sense. Let $\mathsf{dgAlg}_{k}^{\geq 0}= \mathsf{A}$ be the category of (connective) differential graded algebras over a commutative ring $k$, which is a model category . Fix a morphism of DG $k$-algebras $f:R\rightarrow S $ with $R$ cofibrant DGA, then the homotopy groups of the mapping space $Map_{\mathsf{A}}(R,S)_{f}$ are closely related to the (negative) Hochschild cohomology $HH^{-\ast}(R,S)$ where you see $S$ as an $R$-bimodule via $f$. More precisely, for all $i>1$: $$\pi_{i}Map_{\mathsf{A}}(R,S)_{f}\cong HH^{1-i}(R,S). $$ In the particular case $id:R\rightarrow R$, we can say more i.e., $\pi_{1}Map_{\mathsf{A}}(R,R)_{id} $ is a subgroup of the units group of the algebra $HH^{0}(R,R)$ (i.e., $HH^{0}(R,R)^{\times}$). Block and Lazarev gave an interpretation of the (negative) Andre-Quillen cohomology in the case of commutative differential graded $\mathbf{Q}$-algebras denoted by $\mathsf{C}$. $$ \pi_{i}Map_{\mathsf{C}}(R,S)_{f}\cong AQ^{-i}(R,S), i>1. $$<|endoftext|> TITLE: Embeddings of forcing notions - preserve properness? QUESTION [6 upvotes]: Let $ M $ be a countable, transitive model for $ \mathsf{ZFC}^* $, i.e. for a sufficiently large finite fragment of $ \mathsf{ZFC} $. Suppose that $ \mathbb{P} := (P, {\leq_P}, \mathbb{1}_P) \in M $ and $ \mathbb{Q} := (Q, {\leq_Q}, \mathbb{1}_Q) \in M $ are forcing notions. Reminder Definition ([Kun80, VII.7.1]). A mapping $ i \colon P \to Q $ is a complete embedding of $ \mathbb{P} $ into $ \mathbb{Q} $ iff (i) $ \forall r, s \in P \ (r \leq_P s \implies i(r) \leq_Q i(s)) $, (ii) $ \forall r, s \in P \ (r \perp_P s \implies i(r) \perp_Q i(s)) $, and (iii) $ \forall q \in Q \ \exists p \in P \ \forall r \in P \ (r \leq_P p \implies i(r) \parallel_Q q) $. A condition $ p \in P $ as in (iii) is called a reduction of $ q $ to $ \mathbb{P} $. Remark. Note that in clause (ii) equivalence holds because of (i). Definition ([Kun80, VII.7.7]). A mapping $ i \colon P \to Q $ is a dense embedding of $ \mathbb{P} $ into $ \mathbb{Q} $ iff (i) $ \forall r, s \in P \ (r \leq_P s \implies i(r) \leq_Q i(s)) $, (ii) $ \forall r, s \in P \ (r \perp_P s \implies i(r) \perp_Q i(s)) $, and (iii) $ i[P] $ is dense in $ \mathbb{Q} $. Remark. Every dense embedding is a complete embedding. Theorem ([Kun80, II.3.3]). There exists a dense embedding of $ \mathbb{P} $ into $ \operatorname{RO}(\mathbb{P}) \setminus \{ \mathbb{0} \} $. Lemma ([Kun80, VII.Ex.C2]). If $ \mathbb{P} $ and $ \mathbb{Q} $ are separative and $ i \colon P \to Q $ is a complete embedding of $ \mathbb{P} $ into $ \mathbb{Q} $, then $ i $ is one-to-one, $ i(\mathbb{1}_P) = \mathbb{1}_Q $, and $ (r \leq_P s \iff i(r) \leq_Q i(s)) $ holds for all $ r, s \in P $. Now, consider the following statements: (C1) For each $ \mathbb{Q} $-generic $ H $, there exists a $ G \in M[H] $ such that $ G $ is $ \mathbb{P} $-generic over $ M $. (Then $ M[G] \subseteq M[H] $.) (C2) For each $ \mathbb{P} $-generic $ G $, there exists a $ \mathbb{Q} $-generic $ H $ such that $ G \in M[H] $. (Then $ M[G] \subseteq M[H] $.) (C3) There exists a complete embedding $ i \in M $ of $ \mathbb{P} $ into $ \mathbb{Q} $. (C4) There exists a complete embedding $ i \in M $ of $ \mathbb{P} $ into $ \operatorname{RO}(\mathbb{Q}) \setminus \{ \mathbb{0} \} $. (D1) For each $ \mathbb{Q} $-generic $ H $, there exists a $ G \in M[H] $ such that $ G $ is $ \mathbb{P} $-generic over $ M $ and $ H \in M[G] $. (Then $ M[G] = M[H] $.) (D2) For each $ \mathbb{P} $-generic $ G $, there exists a $ \mathbb{Q} $-generic $ H $ such that $ G \in M[H] $ and $ H \in M[G] $. (Then $ M[G] = M[H] $.) (D3) There exists a dense embedding $ i \in M $ of $ \mathbb{P} $ into $ \mathbb{Q} $. (D4) There exists a dense embedding $ i \in M $ of $ \mathbb{P} $ into $ \operatorname{RO}(\mathbb{Q}) \setminus \{ \mathbb{0} \} $. (Pr) If $ \mathbb{Q} $ is proper, then $ \mathbb{P} $ is also proper. Question What implications between the above statements are provable in $ \mathsf{ZFC} $? Which are not? If it is helpful, you may assume that $ \mathbb{P} $ and $ \mathbb{Q} $ are separative partial orders (in the strict sense). Main problem Suppose that (C1) holds. What additional assumptions do we need to show (Pr)? (Note that (C2) implies (Pr). So what additional assumptions does one need to show (C2) from (C1)?) Bibliography [Kun80] Kenneth Kunen: Set Theory: An Introduction to Independence Proofs. North Holland, 1980 REPLY [2 votes]: The following implications are either trivial or well known: (D$ n $) implies (C$ n $): Trivial. (C3) implies (C1): Use $ G := i^{-1}[H] $. (See [Kun80, VII.7.5].) (C3) implies (C4): The composition of two complete embeddings is a complete embedding. (D3) implies (D1): Use $ G := i^{-1}[H] $. Then $ H = \{ q \in Q : \exists p \in G \ i(p) \leq q \} $. (See [Kun80, VII.7.11].) (D3) implies (D2): Use $ H := \{ q \in Q : \exists p \in G \ i(p) \leq q \} $. Then $ G = i^{-1}[H] $. (See [Kun80, VII.7.11].) (D3) implies (D4): The composition of two dense embeddings is a dense embedding. Regarding (C1) does not imply (C3) and (C1) does not imply (C4): Let $ \mathbb{Q} $ be the Cohen forcing and let $ \mathbb{P} $ be the lottery sum of $ \mathbb{Q} $ and $ \operatorname{Col}(\omega, \omega_1) $. Then (C1) clearly holds (use $ G := H $), but (C3) and (C4) do not hold since $ \mathbb{P} $ is not c.c.c. whereas $ \mathbb{Q} $ and hence $ \operatorname{RO}(\mathbb{Q}) $ satisfy the countable chain condition. (See the comments of Yair Hayut and Andreas Blass.) Also see A common forcing extension obtained via different forcing notions by J.D. Hamkins or Lemma 25.5 in T. Jech's book Set Theory (1978 edition). Regarding (C3) implies (C2): Each $ H \subseteq Q $ which is $ \mathbb{Q} / G $-generic over $ M[G] $ is also $ \mathbb{Q} $-generic over $ M $, and $$ G \in M[G][H]_{\mathbb{Q} / G} = M[H]_{\mathbb{Q}}. $$ (See [Kun80, VII.Ex.D3] and [Kun80, VII.Ex.D4].)<|endoftext|> TITLE: Proof correctness problem QUESTION [15 upvotes]: I was watching this talk by Vladimir Voevodsky which was given at the Institute of Advanced Study in 2006. In his talk the first slide he shows has the following written on it: We need to look at the foundations again because of the Proof correctness problem Two components: $1.$ There is an accumulation of results whose proofs the math community cannot fully verify $2.$ There are more and more examples of proofs which have been accepted and later found to be incorrect This is a much more serious problem for math than it would be for any science because the main strength of mathematics is in its ability to build on multiple layers of previous constructions. Here is what he says while presenting the slide: "......As mathematics gets more and more complex, there is an accumulation of results whose correctness becomes more and more uncertain. We don't know about certain things, whether they have been proved or not. ..... every Mathematician has experienced on both sides how terrible it is nowadays to be a referee. I have a paper which is about 10 pages long and it has been lying in a journal for about 10 years now because the referee can't get through ( ? Not sure about if I understood him correctly there). I have not been much better as a referee myself. The problem is mathematics is very complex and if one wants to be responsible for a paper one referees, it takes an enormous amount of effort. It really slows things down. We do have to do something about it. From my point of view there is only one solution.... " He then goes on to talk about foundations of mathematics, automated proof verification, and so on. My question is only about the statements $1.$ and $2.$ made in the slide. Q1. I am looking for examples of such results and proofs. What evidence (if any) is there that shows the problem is on the rise? Q2. Is there a blog, article, essay, etc., which goes through or lists such results and proofs, where there is a discussion about these things ? REPLY [12 votes]: The Finnish mathematician Pertti Lounesto produced, with computer aid, a series of counterexamples to published and accepted theorems on Clifford algebras. He recorded his findings in the following two articles: P. Lounesto: Counterexamples in Clifford algebras with CLICAL, pp. 3-30 in R. Ablamowicz et al. (eds.): Clifford Algebras with Numeric and Symbolic Computations. Birkh\"auser, Boston, 1996. P. Lounesto: Counterexamples in Clifford algebras. Advances in Applied Clifford Algebras 6 (1996), 69-104. He set up a webpage where he exhibits a few of these counterexamples and offers some explanations of how these errors could arise and how he went about to find the counterexamples. After Lounesto's death the webpage was mirrored here.<|endoftext|> TITLE: Do non-subcanonical Grothendieck topologies always induce a category of fractions? QUESTION [8 upvotes]: Suppose that $\mathscr{C}$ is a (possibly higher) category and $J$ is a Grothendieck topology which is not subcanonical. Denote the composite $$\mathscr{C} \hookrightarrow \mathbf{Set}^{\mathscr{C}^{op}} \stackrel{a}{\longrightarrow} \mathbf{Sh}\left(\mathscr{C}\right)$$ by $\theta$, where $a$ is the sheafification functor. Let $W$ denote the class of morphisms in $\mathscr{C}$ which become isomorphisms after applying $\theta.$ Is the essential image of $\theta$ equivalent to the category $\mathscr{C}\left[W^{-1}\right]$ obtained from $\mathscr{C}$ by formally inverting the morphisms $W$? Question: There are certainly examples where this is true, which I provide below. Is this always the case? If not, what further conditions are needed for it to be true? Here are two examples where this is true: Example 1: Let $\mathscr{C}=\mathbf{Haus}$ be the category of Hausdorff topological spaces. Let $J$ be the Grothendieck topology whose covering families for a space $X$ consist of subspace inclusions $$\left(V_\alpha \hookrightarrow X\right)_{\alpha \in A}$$ such that for each compact subspace $K$ of $X$ there exists a finite subset $A(K) \subset A$ such that $$\left(V_\alpha\cap K \hookrightarrow K\right)_{\alpha \in A(K)}$$ can be refined by an open cover of $K$. The essential image of $\theta$ in this case is equivalent to the category of compactly generated Hausdorff spaces, and $W$ consists of all those maps $$X \to Y$$ such that for each compact Hausdorff space $C$, the induced map $$Hom(C,Y) \to Hom(C,X)$$ is an isomorphism. Formally inverting such morphisms yields a category equivalent to compactly generated Hausdorff spaces. Example 2: Let $\mathscr{C}$ be the $\left(2,1\right)$-category of Lie groupoids and smooth functors and natural transormations. Let $i$ denote the full and faithful inclusion $$i:\mathit{Mfd} \hookrightarrow \mathit{LieGpd}.$$ The adjoint pair $i^* \dashv i_*$ exhibits $\mathbf{Fun}\left(\mathit{Mfd}^{op},\mathbf{Gpd}\right)$ as a left-exact localization of $\mathbf{Fun}\left(\mathit{LieGpd}^{op},\mathbf{Gpd}\right).$ Moreover, stacks of groupoids $$\mathbf{St}\left(\mathit{Mfd}\right)$$ is a left-exact localization of $\mathbf{Fun}\left(\mathit{Mfd}^{op},\mathbf{Gpd}\right),$ so by composition is also a left-exact localization of $\mathbf{Fun}\left(\mathit{LieGpd}^{op},\mathbf{Gpd}\right).$ Therefore, there exists a unique Grothendieck topology $J$ on $\mathit{LieGpd}$ such that $$\mathbf{St}\left(\mathit{LieGpd},J\right) \simeq \mathbf{St}\left(\mathit{Mfd}\right).$$ This Grothendieck topology is not subcanonical because the assignment $$M \mapsto Hom\left(M,\mathcal{G}_1\right) \rightrightarrows Hom\left(M,\mathcal{G}_0\right)$$ does not satisfy descent unless the Lie groupoid $\mathcal{G}$ is really a manifold. $\theta$ in this case is the functor assigning a Lie groupoid $\mathcal{G}$ its stack of principal bundles. The essential image is the $\left(2,1\right)$-category of differentiable stacks, and $W$ consists of Morita equivalences. It is well known that the bicategory of fractions of Lie groupoids with respect to Morita equivalences is equivalent to differentiable stacks. REPLY [4 votes]: I don't know whether this includes your two examples, but I believe one situation where this is true is if the topology $J$ is supercanonical, i.e. includes the canonical topology. For example, if $C$ is itself a topos, then supercanonical topologies on $C$ are the same as subtoposes of $C$, which can be obtained from $C$ by inverting the local isomorphisms.<|endoftext|> TITLE: What 3-manifolds can be obtained by gluing $ S^1 \times P $ and two copies of $S^1 \times D^2$ QUESTION [5 upvotes]: Let P denote the pair of pants e.g. a sphere minus three small discs $D_1$,$D_2$,$D_3$ about marked points $x_1,x_2,x_3$. I then consider $P \times S^1$. We have boundary components $T_1$,$T_2$,$T_3$. Fix two homeomorphisms $f_1: \partial(S^1 \times D^2) \to T_1 $ and $f_2: \partial(S^1 \times D^2) \to T_2$. What oriented three manifolds with boundary can be obtained by gluing $S^1 \times D^2$ along these maps (up to homeomorphism)? Namely what are the possible homeomorphism types of manifolds: $S^1 \times D^2 \sqcup_{f_1} P \times S^1 \sqcup_{f_2} S^1 \times D^2 $ REPLY [9 votes]: You get: the solid torus, all the Seifert manifolds fibering over the disc with two exceptional fibers; the connected sum $(D\times S^1) \# L(p,q)$ of a solid torus and a lens space $L(p,q)$, for all coprime $(p,q)$. The proof goes as follows. Consider the filling meridians of the two solid tori you attach. If one meridian is attached along the fiber of $P\times S^1$, then the resulting manifold is the connected sum $(D\times S^1) \# (D \times S^1)$ of two solid tori: this is a nice exercise in 3-dimensional topology. The attaching of the second solid torus then produces $(D^2 \times S^1) \# L(p,q)$. If both meridians are not glued along the fibers, then the fibration of $P\times S^1$ extends to a Seifert fibration over the disc with at most two exceptional fibers. If at least one fiber is not exceptional, then you actually get a solid torus.<|endoftext|> TITLE: GIT over integers QUESTION [11 upvotes]: Let $G$ be a reductive algebraic group over ${\mathbb Z}$ (or a finite localization of a ring of integers $R$ in a number field) acting on an affine scheme of finite type $M=Spec(A)$ over $R$. We can take the invariants $A^G$, and then reduce modulo a prime $p$: $A^G\otimes_{\mathbb Z}{\mathbb F}_q$. We can first reduce modulo $p$, and then take invariants: $(A\otimes_{\mathbb Z}{\mathbb F}_q)^{G_p}$. Is it true that for almost all primes $p$ the results coincide? Let $M^\theta\subset M$ be the open subset of $\theta$-stable points for a choice of a character $\theta$ of $G$. Let $M^{\theta,f}\subset M^\theta$ be the open subset of $M^\theta$ formed by all the points with trivial stabilizers. Let $M//_\theta G$ be the GIT quotient (the projective spectrum of the ring of $\theta$-semiinvariants of the $G$-action on $A$). Is it true that the natural projection $M^{\theta,f}\to M//_\theta G$ is a $G$-torsor? (\'etale locally trivial?) REPLY [5 votes]: The answer to the first question is positive. One does not need normality of the generic fiber. By the universal coefficient theorem [Prop I 4.18 (a) in Jantzen Representations of algebraic groups], it suffices to know that $H^1(G,A)$ vanishes generically on $Spec(R)$. But this is implied by theorem 33 of my paper with Franjou Power reductivity over an arbitrary base. Using further suggestions by user52824 we see that the right setting has a Dedekind domain $D$ as base ring. We localize by inverting a nonzero element of $D$. We first localize to get $A$ flat over $D$, then we localize further to make $H^1(G,A)$ flat over $D$. This uses Theorem 10.5 of Good Grosshans filtration in a family as the big hammer. Then the universal coefficient theorem implies compatibility with any base change. Notice that we no longer try to kill $H^1(G,A)$. Just its $D$-torsion.<|endoftext|> TITLE: Are there workable algebraic geometry approaches for the pentagon equation? QUESTION [9 upvotes]: A pentagon equation is a system of polynomial equations of degree $3$ with several variables and integer coefficients, given by a fusion ring. A fusion ring is given by a finite set of integer matrices checking some axioms (see here p 22). A solution of its pentagon equation is the main condition for having a structure of fusion category, in which it encodes associativity (see here). For example, the irreducible complex representations of a finite group, equipped with $\oplus$ and $\otimes$, generate a fusion ring and a fusion category. The problem is that, in practice the pentagon equations we meet are huge, for example, a system of $2000000$ polynomial equations with $50000$ variables (of degree $3$ with integer coefficients), so that proving the existence of a solution (and a fortiori finding one) is very hard. Note that there are pentagon equations without solution. Question: Are there workable algebraic geometry approaches ? (for proving the existence of a solution or for finding one) In fact, the pentagon equation is more structured than just a system of scalar equations, it's a system of several invertible (unitary) matrix equations of the form $$A_1 A_2 A_3 = A_4 A_5$$ (and $A_i^* A_i = I$), such that $A_i = \tau_i B_i \tau'_i$, with $\tau_i$,$\tau'_i$ fixed permutation matrices and $B_i$ a block diagonal matrix (each block is a matrix variable), so that there are many holes (see here p 29-30). Note that in general $\nu_1\nu_2\nu_3 \neq \nu_4\nu_5$ (with $\nu_i = \tau_i\tau'_i$), so that $B_i =I$ can't be a solution. I'm interested in the equation given by this fusion ring. There are $16227$ matrix equations of dimensions from $1$ to $91$, and $2097$ invertible (unitary) matrix variables of dimensions from $1$ to $17$ (see here p 31). REPLY [6 votes]: I'm not an expert in algebraic geometry, but I can say something about methods for solving pentagon equations that will hopefully be of use. The primary way to determine whether or not there is a solution to the pentagon equations is to use Groebner basis methods. However, these begin to break down very quickly for algebraic varieties of the size that we deal with in the pentagons. However, the Fusion variety has a lot of structure that can be taken advantage of: We can fix bases on the $Hom(a \otimes b, c)$ spaces by fixing certain 10j-symbols in the pentagon equations. Ocneanu rigidity then implies that there are only finitely many solutions to the pentagon equations. When there's no multiplicity (i.e. all $Hom(a \otimes b, c)$ spaces are one dimensional), this is relatively easy. The big problem here is in trying to figure out whether or not a given 10j-symbol is identically zero or can be fixed to 1. In the case where there is multiplicity this gets much more difficult because the action of the gauge group on the $Hom(a\otimes b \otimes c, d)$ spaces is no longer one dimensional and so its extension to the $Hom(a\otimes b\otimes c, d)$ spaces can no longer be treated as multiplication by non-zero scalars. Ideally then the thing to do is find a basis for the multidimensional hom-spaces that fixes certain F-Matrices so that the gauge group only acts by scalar matrices. Some of the structure of the pentagons can be deduced from arm-bending, i.e. using the rigidity and leads to redundancy which can be summarized when solving things by hand. For solving things using computer algebra systems, this redundancy can be helpful. The pentagon equations also have the structure of a tower of varieties. By this I mean that If we consider only the matrix equations of dimension <=n, our variables can only be those 10j-symbols coming from F-Matrices of dimension <=n. Given this, one way to attack things is by finding admissible values for smaller dimensional equations and then using these as input to simplify larger ones. For example, there exists a subset of pentagon equations which are one dimensional matrix equations. By necessity, these can only include F-matrices which are themselves one dimensional, and so they form a sub-variety of the fusion variety. Solving these by themselves can sometimes be done, and this can then be used to simplify equations with the next dimension up. It's also not necessarily true that partitioning equations by dimension is the smallest division we can come up with. Given a set of solutions to equations with dimension $ TITLE: Bounds on horizontal minima of the Riemann zeta function QUESTION [8 upvotes]: It is known that $\zeta(s)$ has an infinity of zeros in the strip $0<\sigma<1$ and that those zeros become closer together as $t\rightarrow\infty$. More precisely, Littlewood showed that there is a zero $\rho=\beta+i\gamma$ with $|\gamma-t|\in O(1/\log\log\log t)$. A positive proportion of these zeros are known to have $\beta=1/2$, but $\zeta(1/2+it)$ is unbounded. In fact, $\zeta(\sigma+it)$ is bounded as $t\rightarrow\infty$ only if $\sigma>1+\delta$, $\delta>0$ fixed. On the other hand, there seem to be good reasons to expect that $$f(t)=\min_{0\leq\sigma\leq 1} |\zeta(s)|$$ approaches zero as $t\rightarrow\infty$. One reason is that zeros of $\zeta(s)$ are rather differently distributed than those of $\zeta(s)-z$ for any other value of $z$. I would like to argue that $f(t)\rightarrow 0$ here, but my approach is incomplete. In fact, I would like to conjecture that, for every $\epsilon>0$, we have $$f(t)\in O (\log^{\Theta(t) -1+\epsilon}t),$$ where $\Theta(t)=\max\{\beta:\gamma TITLE: Families of quintics in $\mathbb{Q}[x]$ with Galois group $A_5$ QUESTION [12 upvotes]: Theorem. The Galois group of a quintic polynomial $f\in\mathbb{Q}[x]$ is $A_5$ if and only if its discriminant is a rational square and its Weber sextic resolvent has no rational root. Question. What are known infinite families of of quintic polynomials in $\mathbb{Q}[x]$, each with Galois group $A_5$? I am aware of two explicit such families as follows. Example 1. $f(x)=x^5+(5t^2-1)(5x+4)$ for $t\in\mathbb{Z}$ such that $t\equiv\pm 1\pmod{21}$. This is Exercise 3.7.2 in the book Generic Polynomials by Jensen, Ledet and Yui. Example 2. $f(x)=x^5+(t^2-5^5)(x-4)$ for non-zero $t\in\mathbb{Q}$ such that $\forall u\in\mathbb{Q}: t\ne g(u)$ where $g(u)=\frac{(u^3-18u^2+8u-16)(u^3+2u^2+18u+4)}{2u^2(u^2+4)}$. This is from the paper Reducibility and the Galois group of a parametric family of quintic polynomials by Lavallee, Spearman and Williams. Note. Both are all trinomials, presumably because it is easier to construct polynomials with square discriminant when some coefficients vanish. It would be nice to see otherwise. REPLY [4 votes]: You might be interested in a paper of Hashimoto and Tsunogai on generic polynomials for quintics (https://projecteuclid.org/euclid.pja/1116443730), including one for $A_5$. It is not a trinomial as a matter of fact.<|endoftext|> TITLE: Rediscovery of lost mathematics QUESTION [119 upvotes]: Archimedes (ca. 287-212BC) described what are now known as the 13 Archimedean solids in a lost work, later mentioned by Pappus. But it awaited Kepler (1619) for the 13 semiregular polyhedra to be reconstructed.          (Image from tess-elation.co.uk/johannes-kepler.) So there is a sense in which a piece of mathematics was "lost" for 1800 years before it was "rediscovered." Q. I am interested to learn of other instances of mathematical results or insights that were known to at least one person, were essentially correct, but were lost (or never known to any but that one person), and only rediscovered later. 1800 years is surely extreme, but 50 or even 20 years is a long time in the progress of modern mathematics. Because I am interested in how loss/rediscovery might shed light on the inevitability of mathematical ideas, I would say that Ramanujan's Lost Notebook does not speak to the same issue, as the rediscovery required locating his lost "notebook" and interpreting it, as opposed to independent rediscovery of his formulas. REPLY [3 votes]: Not only can a mathematical technique be lost, but so can a whole field and even an entire science or sciences! Thom called Aristotle the only philosopher of continuity for over a millenia. So not quite lost, but a route not taken until the dawn of the 20th C. Physics is often dated from Galileo. This was Einsteins view, even if he acknowledged prior work. However, physics was worked on by ancient Greek philosophers and probably was the impetus for for Greek philosophy given that the earliest school from Miletus was concerned with it. The most comprehensive treatise that we have from that era are Aristotle's pair of books, Physics & Metaphysics. Greek philosophy was mostly forgotten in Europe by 600 AD. It was revived by Ibn Sinna (Avicenna), who by his own admission had read Aristotle's Metaphysics fifty times without understanding it until he came across a small book commentating on it by al-Farabi in a bazaar. It was this revival that triggered the scientific revolution in Europe, including that of physics. For example, the definition of force and that of inertial motion can be traced back to Aristotle. Crucially, it was Avicenna himself, going on from a modification by Philopenus who came up with implicitly, Newtons first law of motion. The importance of this law can hardly be exaggerated. Even after Einsteins revolutionary understanding of spacetime, the same law applies. Newton himself recognised this tradition, for he wrote in his Philosophical Notebooks: Plato is a friend, Aristotle is a friend but truth is a greater friend. This signalled his ambition to out-master his master which is generally the ambition of all ambitious students. The continuity of this tradition has been forgotten again with the crystallisation of physics as a science and no longer a part of natural philosophy with physicists like Susskind and countless others who are keen to show why Aristotle was so wrong. Yes, sure, but so was Newton and just as likely, Einstein too. This does not detract from their achievements and nor should it detract from Aristotle's. Feynman in his Lectures on Physics speculated if the entireity of physics was lost to future generations except for one sentence, what should that one sentence be to convey the maximum of information? He said, that one sentence should be "that all things are made of atoms". Feynman's speculation, although it seems he did not know it at the time of writing of his lectures, is actually true. Physics was forgotten and then revived through Aristotle as well as others. This post has focused on physics rather than math. But physics has been and still is, the most important application of maths and almost synonymous with it as the term geometry signifies - measurement of the earth.<|endoftext|> TITLE: Is it known which links have Seifert fibered complements? QUESTION [6 upvotes]: I believe many such links can be constructed by looking at a foliation similar to the hopf fibration, but the wrapping leaves replaced with $(p,q)$ torus knots. However, I'm interested in particular in whether there are classes of examples that don't fall under that construction. REPLY [2 votes]: I think this paper exactly answers your question http://projecteuclid.org/euclid.dmj/1077378799 . But you are basically right, the links you get are fibres (possibly degenerate) in some (possibly singular) Seifert fibration of $S^3$ and pretty much it can only be fibred in the ways you mentioned.<|endoftext|> TITLE: Orbit spaces of crystallographic groups QUESTION [6 upvotes]: In their paper "On Three-Dimensional Space Groups", Conway et al. write Although this paper was inspired by the orbifold concept, we did not need to consider the 219 orbifolds of space groups individually. We hope to discuss their topology in a later paper. Did this discussion ever materialize? Or did somebody else classify the orbit spaces for all three-dimensional crystallographic groups? REPLY [8 votes]: The orientable Euclidean orbifolds were described in section 7 of this paper of Dunbar. To complete the list, note that every crystallographic group has an index 2 orientation-preserving subgroup. The corresponding orientable orbifold quotients will be realized as 2-fold orbifold covers of the non-orientable orbifolds. So one needs to look over Dunbar's list, and find all of the orientation reversion involutions of the corresponding orientable orbifolds to find their non-orientable quotients. For the orbifolds with underlying space $S^3$, there are two possible orientation-reversing quotients: a reflection, and a suspension of an antipodal map. They admit quotients the ball and the suspension of $\mathbb{RP}^2$ respectively. So one needs to go through the list and determine which of these admit reflection or antipodal symmetries. For example, due to the well-known fact that the figure-8 knot is invertible, the label 3 figure-eight orbifold in Table 4 admits a quotient by a suspension antipodal map. For the first 4 examples in table 4, there is a reflection fixing the 1-skeleton of the simplex with quotient a Coxeter group. Some of these also admit further reflection symmetries.<|endoftext|> TITLE: A question about Dehn surgery and Brieskorn homology 3-spheres QUESTION [6 upvotes]: I have been learning about Brieskorn homology 3-spheres $\Sigma(a_1,...,a_n)$ and Seifert manifolds. My reference is the first few pages of Saveliev's "Invariants of Homology 3-spheres." If I understand correctly, it is explained that one can realize the homology three spheres by considering a trivial bundle $S^2 \times S^1$ and remove small discs $D_i$ about marked points $x_i ,..., x_n$ on the two sphere. One can then perform a rational Dehn surgery with surgery slopes $a_i/b_i$ along the links given by $\partial D_i \times pt$ to obtain the Brieskorn homology spheres. More precisely Saveliev shows that there exist $b_i$ that make this Dehn surgery construction into a rational homology sphere and then states that it follows from Kirby calculus that the resulting 3 manifolds are independent of $b_i$. I should say that, unfortunately, I don't know anything about Kirby calculus. Clearly, in this construction the $a_i$ and $b_i$ do not play symmetric roles. Nevertheless, I am interested in the following: A general question is: To what extent can we determine the homology three spheres from the $b_i$? A specific question/motivation: I am not at all a low dimensional topologist. However, a Brieskorn homology sphere naturally arises as a surgery of the type above in a construction I am working on. I know that $n=4$ and all $b_i= \pm 1$, or $n=3$ and $b_1, b_2= \pm 1.$ It is natural to ask: what manifold am I possibly looking at? REPLY [7 votes]: Seifert manifolds that are homology spheres can be determined as follows. Let $$ M =(S^2, (p_1,q_1), \ldots, (p_h,q_h)) $$ denote the Seifert manifold obtained by filling the $h$ tori in $D_h \times S^1$ (where $D_h$ is the 2-sphere minus $h$ discs) with coprime parameters $(p_i, q_i)$. Since we consider Seifert manifolds, we ask $p_i \neq 0$. We may say that $M$ fibers over the orbifold $$(S^2, p_1, \ldots, p_h).$$ The Euler number of $M$ is an important invariant, defined as: $$ e = \sum_{i=1}^h \frac {q_i}{p_i}.$$ A simple homology calculation shows that the integral homology $H_1(M)$ is finite if and only if $e\neq 0$, and in that case we have $$|H_1(M)| = ep_1\cdots p_h = \sum_{i=1}^h q_ip_1\cdots \widehat{p_i}\cdots p_h.$$ This implies that $M$ is an integral homology sphere if and only if $$\sum_{i=1}^h q_ip_1\cdots \widehat{p_i}\cdots p_h = \pm 1.$$ This in turn easily implies the following: For every set $p_1, \ldots, p_h\geqslant 2$, of $h \geqslant 3$ pairwise coprime integers there is a unique homology sphere $\Sigma(p_1,\ldots,p_h)$, which fibers over the orbifold $(S^2,p_1, \ldots, p_h)$. Every Seifert homology sphere distinct from $S^3$ arises in this way. Proof: it is easy to see that the $p_i$'s must be coprime. On the other hand, if they are, one can find easily $q_1,\ldots, q_h$ that give $|H_1(M)|=1$. Two different strings of $q_1,\ldots, q_h$ are related by ``moves'' that do not affect $M$. (The same $M$ can be described by different parameters.) This theorem is probably written somewhere, like in Orlik's book or Seifert's original paper, but I can't remember.<|endoftext|> TITLE: Negative impact of wrong or non-rigorous proofs QUESTION [38 upvotes]: The recent talks of Voevodsky (for example, http://www.math.ias.edu/~vladimir/Site3/Univalent_Foundations_files/2014_IAS.pdf), which describe subtle errors in proofs by him as well as others, as well as the famous essay by Jaffe and Quinn (http://www.ams.org/journals/bull/1993-29-01/S0273-0979-1993-00413-0/) and responses to it (http://www.ams.org/journals/bull/1994-30-02/S0273-0979-1994-00503-8/), raises for me the following question: What are some explicit examples of wrong or non-riogour proofs that did damage to mathematics or some significant part of it? Famous examples of non-rigorous proofs include Newton's development of calculus and the latter stages of the Italian school in algebraic geometry. Although these caused a lot of dismay and consternation, my impression is that they also inspired a lot of new work. Is it wrong for me to view it this way? In particular, I'm told that people proved false theorems using Newton's approach to calculus. What are some examples of this and what damage did they do? REPLY [14 votes]: A proof being wrong can mean many things: By 1932, when the Hungarian-American mathematician John von Neumann claimed to have proven that the probabilistic wave equation in quantum mechanics could have no “hidden variables” (that is, missing components, such as de Broglie’s particle with its well-defined trajectory), pilot-wave theory was so poorly regarded that most physicists believed von Neumann’s proof without even reading a translation. More than 30 years would pass before von Neumann’s proof was shown to be false, but by then the damage was done. The physicist David Bohm resurrected pilot-wave theory in a modified form in 1952, with Einstein’s encouragement, and made clear that it did work, but it never caught on. (The theory is also known as de Broglie-Bohm theory, or Bohmian mechanics.) The problems is the interpretation of what has actually been proved. I don't like the No Free Lunch Theorems for Optimization, because their assumptions are unrealistic and useless in practice, but the theorem itself certainly feels true (but in a less trivial way than what is actually proved). And the conclusion is deeply flawed. It claims that there is no difference between a buggy implementation of a flawed heuristic and a correct implementation of a reasonable solution strategy. The conclusion should rather be that we should explicitly specify what our solution strategy is supposed to achieve, not just claim that it is a good black box search strategy.<|endoftext|> TITLE: Topological relationships between family of transversal intersections of manifolds QUESTION [6 upvotes]: Let $M$ and $N$ be submanifolds of $\mathbb{R}^n$ and let $a(t)$ be a smooth path in $\mathbb{R}^n$ such that $M+a(t)$ intersects $N$ transversally for all $t \in [0,1]$. Is there a nice relationship between the submanifolds $(M+a(0)) \cap N$ and $(M+a(1)) \cap N$? Are they homotopy equivalent or better yet homeomorphic? REPLY [6 votes]: Let me rephrase the construction: you have a map $$\varphi : [0,1] \times M \to \mathbb{R}^n$$ which for each $t \in [0,1]$ is an embedding ($\varphi(t,x) = x+a(t)$ in your notation) and is transverse to $N \subset \mathbb{R}^n$. In particular, it follows that the map $\varphi$ is transverse to $N$, and so $$X:=\varphi^{-1}(N) \subset [0,1] \times M$$ is a submanifold. If we write $X_t := \varphi(t,-)^{-1}(N)$ then this is a smooth manifold for all $t$, as $\varphi(t,-)$ is transverse to $N$: thus $X$ is a cobordism from $X_0$ to $X_1$, and these are the two manifolds in your question. I claim that the projection map $\pi : X \to [0,1]$ is a submersion. To see this, note that for $(t,m) \in X$ the map $$T_{(t,m)}([0,1] \times M) \to T_t[0,1]$$ is surjective, so we can find a $v \in T_{(t,m)}([0,1] \times M)$ mapping to $d/dt$. Under the surjective map $$T_{(t,m)}([0,1] \times M) \overset{D\varphi}\to T_{\varphi(t,m)}\mathbb{R}^n \to \nu_{\varphi(t,m)}N$$ to the normal bundle of $N$, $v$ may not map to 0, but as precomposing with $$T_{(t,m)}(X_t) \to T_{(t,m)}([0,1] \times M)$$ the map stays surjective, we can modify $v$ by an element of $T_{(t,m)}(X_t)$ (which does not change its projection to $T_t[0,1]$) to get a new $v'$ mapping to 0 in $\nu_{\varphi(t,m)}N$; hence $v' \in T_{(t,m)}(X)$, and it maps to $d/dt$ under $D\pi$, as required. By picking a splitting of the bundle epimorphism $$TX \to \pi^*T[0,1]$$ we get a vector field on $X$, and integrating this shows that $X \cong [0,1] \times X_0$. Thus in particular $X_1$ and $X_0$ are diffeomorphic, but even better, $\varphi\vert_X$ gives an isotopy between them in $\mathbb{R}^n$. REPLY [2 votes]: I believe they are homeomorphic. Let $j: M\hookrightarrow\mathbb{R}^n$ denote the embedding of $M$. Your path defines a smooth homotopy $h: M\times [0,1]\to \mathbb{R}^n$ given by $h(x,t) = j(x)+a(t)$, and the assumption is that each $h_t = h(-,t): M\to \mathbb{R}^n$ is transverse to $N\subseteq \mathbb{R}^n$. This assumption implies firstly that $h$ is transverse to $N$, and therefore that $W:=h^{-1}(N)\subseteq M\times [0,1]$ is a cobordism between $h_0^{-1}(N)=(M+a(0))\cap N$ and $h_1^{-1}(N)=(M+a(1))\cap N$ (cf. the Pontryagin-Thom construction). However, it is much stronger than that. These kinds of transverse homotopies have been studied by Jon Woolf and his collaborators (see http://uk.arxiv.org/abs/0910.3322, for example). The key observation is that, under these conditions, the map $W\subseteq M\times [0,1]\to [0,1]$ which projects onto the interval has no critical points (see http://uk.arxiv.org/abs/0910.3322 Theorem 2.3, which is about maps transversal to stratifications but the same reasoning applies to maps transversal to submanifolds). Therefore $W$ is a trivial cobordism, and the two ends are homeomorphic (this uses the ideas of Morse theory). I guess this could be better explained. The slides here http://pcwww.liv.ac.uk/~jonwoolf/Slides/belfast-talk.pdf might be helpful.<|endoftext|> TITLE: Adams Operations on $K$-theory and $R(G)$ QUESTION [25 upvotes]: One of the slickest things to happen to topology was the proof of the Hopf invariant one using Adams operations in $K$-theory. The general idea is that the ring $K(X)$ admits operations $\psi^k$ that endows it more relations from which one can deduce restrictions on when a Hopf invariant one element can occur. Similarly, the representation ring $R(G)$ also admits Adams operations. One may describe its effects on characters by $\Psi^k(\chi)(g) = \chi(g^k)$. Alternatively one can parallel the construction of the Adams operations in topology by defining the Adams operations as taking the logarithmic derivative of $$\lambda_t(V) := \Sigma_k [\Lambda^k V]t^k$$ (of course we extend this formula to virtual characters as well) and defining $\Psi^n(V)$ as $(-1)^n$ times the $n$-th coefficient of the logarithmic derivative. Now, considering that my background in representation theory is minimal, my questions are: What is an application of the Adams operations in representation theory? Is there one that has the same flavor as the proof of the Hopf invariant one problem? We have the Atiyah-Segal completion theorem that gives isomorphism of rings $K(BG) \simeq \hat{R(G)}_I$ and there are Adams operations on both sides (I'd imagine that the Adams operations in group representations extends naturally to completion) - what is their relationship (if any)? REPLY [18 votes]: Adams operations played a key part in Borcherds's proof of the Conway-Norton Monstrous Moonshine Conjecture. The specific point is in section 8 of his paper (13th on the linked page), where we have a $\mathbb{Z} \times \mathbb{Z}$-graded Lie algebra $$E = \bigoplus_{m > 0, n \in \mathbb{Z}} E_{m,n}$$ equipped with an action of the Monster simple group, by homogeneous automorphisms. An important property of $E$ is that its Lie algebra homology ``lives on the boundary'', i.e., $H_0(E) \cong \mathbb{C}$, $H_1(E) \cong \bigoplus_{n \in \mathbb{Z}} E_{1,n}$, $H_2(E) \cong \bigoplus_{m > 0} E_{m,1}$, $H_i(E) = 0$ for $i \geq 3$. The Chevalley-Eilenberg isomorphism $H_*(E) \cong \bigwedge^* E$ of graded virtual Monster modules then expands to a K-theory identity: $$\mathbb{C}p^{-1} + \sum_{m >0} E_{m,1} p^m - \sum_{n \geq -1} E_{1,n} q^n = p^{-1} {\bigwedge}^* \left(\bigoplus_{m>0,n\in \mathbb{Z}} E_{m,n} p^m q^n \right).$$ A second important property of $E$ is that as a Monster module, the isomorphism type of $E_{m,n}$ depends only on the product $mn$, i.e., we have a set of modules $\{V_k\}_{k \geq -1}$ such that $E_{m,n} \cong V_{mn}$ (I recently wrote about this in a blog post). By using properties of Adams operations, we find that for any element $g$ in the Monster we get the following character identity: $$\sum_{m \geq -1} Tr(g|V_m) p^m - \sum_{n \geq -1} Tr(g|V_n) q^n = p^{-1} \exp \left(- \sum_{i>0} \sum_{m>0,n\in \mathbb{Z}} \frac{Tr(g^i|V_{mn}) p^{mi} q^{ni}}{i} \right).$$ Both identities have a left side that is a sum of power series that are pure in $p$ and $q$, while the right side has lots of mixed terms that necessarily cancel. The resulting relations constrain the graded characters $\sum Tr(g|V_n)q^n$ quite strongly, e.g., the vanishing of the $p^1 q^2$ term implies $V_4 \cong V_3 \oplus \wedge^2 V_1$, so the $\psi^2$ identity implies $$Tr(g|V_4) = Tr(g|V_3) + \frac{Tr(g|V_1)^2 - Tr(g^2|V_1)}{2}.$$ From these relations, Borcherds was able to show that the characters are Fourier expansions of modular functions that precisely match the list conjectured by Conway and Norton, by just checking the first few terms. There have been some further developments following this application. The character identity has been reinterpreted in terms of equivariant Hecke operators by Charles Thomas and Nora Ganter (and some others who I am forgetting), with a view toward elliptic cohomology operations. Gerald Höhn applied this method to solve the Baby Monster case of the Generalized Moonshine conjecture, and I've used it in my own work for showing that ``well-behaved'' actions of groups on Lie algebras can be used to form Hauptmoduln.<|endoftext|> TITLE: Can we define an "empirically generic" real number? QUESTION [12 upvotes]: Summary: My question, in a nutshell, is how we should intuitively imagine a generic real number (as opposed to a random one), and whether we can construct numbers which empirically behave like generic numbers in the same way that $e$ or $\pi$ behave empirically like random ones. I hope this is not too vague, informal or philosophical for MO. Let me explain what I'm asking in greater detail. Background: First recall the classical "duality" between (Lebesgue) measure and (topological) category: a subset of $\mathbb{R}$ is said to be "negligible" iff it is of Lebesgue measure zero, and "meager" iff it is contained in a countable union of nowhere dense closed sets (i.e., closed sets with empty interior). By Lebesgue measure theory, resp. by the Baire category theorem, a negligible, resp. meager set has empty interior. Both are "small" in a certain sense, but in an incompatible way since $\mathbb{R}$ is the union of a negligible and a meager set (as a simple example is given below). There is also a classical theorem by Erdős (refining an earlier result by Sierpiński) showing that, under the Continuum Hypothesis, there is an involution of $\mathbb{R}$ to itself which takes negligible subsets to meager ones and conversely. [What follows is badly written: jump to "edit/clarification" below for an attempt at saying things more clearly.] There are various meanings of the word "random", but the general flavor is that a real number is random iff it does not belong to a negligible Borel set which can somehow be described or coded in a simple way (e.g., random over a transitive model of set theory means that it does not belong to a negligible Borel set coded by a sequence in that model; but there are some weaker meanings of "random" where we forbid belonging only to negligible Borel sets of simpler description, e.g., those which can be described by a code computable by a Turing machine). For example, a random number will be normal in every base, because the set of real numbers which are not normal in every base is a simply described Borel set which is negligible. Now a real number like $e$ or $\pi$ is not random even in the weakest sense, because it is, well, equal to $e$ or $\pi$, and that is not random (it belongs to the Borel set $\{e\}$ or $\{\pi\}$ which is negligible and certainly computable). It does seem to be, however, "empirically random", in a sense that we (or at least, I) don't know how to make precise, but the idea being that it won't belong to any simply defined negligible Borel set which hasn't been specifically constructed to contain it. For example, it is conjectured that $e$ and $\pi$ are normal in every base: we expect their decimals to pass statistical tests of randomness. The same holds for a huge number of "naturally defined" real numbers (and not just real numbers: see this question for another case). Philosophically, it is also generally expected that the real number whose binary expansion is obtained by flipping an unbiased coin (or, better, taking some physical source of randomness) will be random in a strong sense. So, anyway, we have a good intuition of what a random number feels like. The dual notion of a "generic" number, however, is more obscure: a real number is generic iff it does not belong to a meager Borel set which can be somehow described or coded in a simple way (e.g., belonging to a transitive model of set theory, or Turing-computable, or something like this). Here is an example of something we can say about generic numbers: call a real number (between $0$ and $1$, say) an "oft-repeater in base $b$" iff its expansion in base $b$ repeats an infinite number of times all the digits up to that point. In other words, there exist arbitrarily large $n$ such that the digits $c_n$ to $c_{2n-1}$ are equal to $c_0$ to $c_{n-1}$. It is easy to see that the set of oft-repeaters in base $b$ (and therefore, in any base) is negligible but comeager (=contained in a countable intersection of open dense subsets): so a random real number is not an oft-repeater in any base, but a generic real number is an oft-repeater in every base. We don't know this either way, but I don't think anyone would seriously conjecture that $e$ or $\pi$ is an oft-repeater in any base: clearly we expect them to be empirically random and not empirically generic. Or in other words, we expect measure theory to be a better predictor of what $e$ and $\pi$ behave like than category. An "empirically generic" real number, however, would be (among many other things) an oft-repeater in every base (and it would not be normal in any base: for example, in a generic number, there are arbitrarily large $n$ such that all digits $c_n$ to $c_{n^2}$ are zero — the set of normal reals is meager). Questions: So, my questions are something like this: Are there "naturally defined" real numbers which are "empirically generic" rather than "empirically random"? Or at least, can we give some examples of (non-"naturally defined") such numbers? Is there some kind of process (physical or idealized), analog to throwing a coin, that would produce a (somewhat!) generic real number? Is there a philosophical argument explaining why measure theory predicts better than category how the numbers naturally encountered in mathematics behave? Why should we expect $e$ and $\pi$ to behave more randomly than generically (when, in fact, they are neither)? How can one intuitively visualize a generic random number? (I think I can picture a random one, and the idea of it being normal makes sense, but the fact that a generic random number has infinitely large $n$ such that all digits $c_n$ to $c_{n^2}$ are zero seems very difficult to imagine.) How would one even test empirically if a given real number is generic? (Assume you have a true generic oracle and a fake one: how would you proceed to detect which is the true one? Can we have a "genericity test" like we have randomness tests?) Perhaps the idea that genericity should behave symmetrically to randomness is naïve: please don't hesitate to tell me why this is naïve! Edit/Clarification: The above discussion was probably too messy or informal. Let me try to give a clearer restatement: Definition: If $\mathscr{T}$ is a set of Turing degrees, a real number is said to be $\mathscr{T}$-random, resp. $\mathscr{T}$-generic, iff it does not belong to any negligible, resp. meager, Borel set which can be coded by a sequence whose Turing degree is in $\mathscr{T}$ ("coding" of Borel sets being done, say, as in Jech's Set Theory). If $\mathscr{T}$ is the set of degrees belonging to some transitive model $\mathfrak{M}$ of ZFC, the reals in question are said to be random over $\mathfrak{M}$ resp. Cohen/generic over $\mathfrak{M}$ (cf. Jech, definition 26.3 and lemma 26.4 in the Third Millennium edition). If $\mathscr{T}$ is simply the degree $\mathbf{0}$ of Turing computability, I think we get a definition equivalent to Martin-Löf random numbers, and something analogous for "generic": of course, this is a much weaker property than being random, resp. generic, over a model of ZFC. We could perhaps define even weaker versions of "random", resp. "generic", by replacing $\mathscr{T}$ by a set of finer degrees, maybe primitive recursive degrees (but if the degrees are too fine, then the definition will become too sensitive on how Borel sets are coded and probably not the right way to proceed). Question number 1: While I think I have an intuitive grasp of how a random number real behaves (irrespective of what $\mathscr{T}$ is), e.g., by imagining an coin being tossed an infinite number of times, the corresponding "generic" notion is much more obscure. Is there some way to picture it intuitively? Now there is the matter of numbers like $e$ and $\pi$. Of course these numbers are not random (nor are they generic) in the sense of the above definition, or even of any reasonable weakening I can imagine. Nevertheless, $e$ and $\pi$ behave in certain ways like random real numbers, and I claim that they behave "more like random reals than like generic reals". For example, if we are to make a conjecture as to the lim.sup. and lim.inf. of the sequence $\frac{1}{n}\sum_{k=0}^{n-1} c_k$ where $c_k$ denotes the $k$-th binary digit of $\pi$, and if Pr. Eugsebel predicts "I conjecture that the limit is $\frac{1}{2}$, because the set of real numbers for which this is the case is of full measure", while Pr. Eriab predicts "I conjecture that the lim.inf. is $0$ and the lim.sup. is $1$, because the set of real numbers for which this is the case is comeager", then experimentally, it appears that Pr. Eugsebel is right and Pr. Eriab is wrong: measure theory seems to predict the empirical behavior of the decimals of $\pi$ correctly, and category does not; alternatively, $\pi$ behaves empirically like a random number (even though it is not at all random!), in this limited respect, and it does not behave empirically like a generic number. Informal definition (which probably cannot be made rigorous): Say that a real number is "empirically random" when it behaves like a random real number for this kind of simple tests. (Perhaps "pseudorandom" would be a better term for this.) For example, an "empirically random" real number should, at least, be normal in any base (note that the set of real numbers that are normal in any base is of full measure). Analogously, we want to define a number to be "empirically generic" (or "pseudogeneric") when it behaves like a generic real number. For example, an "empirically generic" real number should at least be an "oft-repeater" in any base (meaning that there exist arbitrarily large $n$ such that the digits $c_n$ to $c_{2n-1}$ are equal to $c_0$ to $c_{n-1}$); also, the lim.sup. and lim.inf. of the sequence $\frac{1}{n}\sum_{k=0}^{n-1} c_k$ where $c_k$ denotes the $k$-th binary digit of the number should be $0$ and $1$ (note that the set of real numbers satisfying these criteria comeager). Question number 2: Whereas $e$, $\pi$ and many others can reasonably be conjectured to be "empirically random", is there, dually, any real number that has been explicitly defined in mathematics that one can reasonably expect to be "empirically generic"? Or could one be defined? (By "explicitly", I mean to forbid something like "take some number outside of the union of all meager Borel sets with a computable code": this would indeed define a generic real number, not just an "empirically generic" one, but this is not explicit by any means.) The underlying philosophical question is something like this: "How come is it that randomness appears to be a much more natural notion than genericity?" But I don't really expect anyone to have an answer to that. REPLY [5 votes]: It sounds like you are talking about what in computability theory and set theory are known as Cohen generic reals (the lowest level of which in computability theory is 1-generic, then 2-generic and so on). I don't know any really natural example of a 1-generic real, but there is a fairly simple construction of one, see e.g. the book Lerman, Degrees of Unsolvability, 1983. To visualize you can imagine that we put down more and more digits in our number, and every now and then we stop and say "what kind of digits could we possibly put down?", and then we put down some digits like that. For example, you could put down 100 times as many 0s as the number of digits you've put down so far. The important thing is that you eventually cover all kinds like that. So something like: 02345234 Now let's add 23 zeroes: 0234523400000000000000000000000 Now let's add what we already have two more times: 023452340000000000000000000000002345234000000000000000000000000234523400000000000000000000000 ... and so on. But we could also have done it in a different order, so say we have 78345786345 and then add what we already have two more times: 783457863457834578634578345786345 and then add 23 zeroes 78345786345783457863457834578634500000000000000000000000 REPLY [2 votes]: You can't really have a "given" empirically-generic number, because any explicit description of $x$ can be turned into an explicit description of the meagre closed set $\{x\}$ to which it belongs.<|endoftext|> TITLE: Good properties of the $H^0$ functor (from quasi-functors to ordinary functors) QUESTION [6 upvotes]: Let $\mathcal A, \mathcal B$ be dg-categories over a field $k$. I denote by $\mathcal{RHom}(\mathcal A,\mathcal B)$ the dg-category (defined up to quasi-equivalence) which gives the internal hom in $\mathrm{Hqe}$ (see, for example, Keller's survey, Theorem 4.5). There is a natural $H^0$ functor: \begin{equation} H^0(-) :H^0(\mathcal{RHom}(\mathcal A,\mathcal B)) \to \mathcal{Hom}(H^0(\mathcal A),H^0(\mathcal B)), \end{equation} where $\mathcal{Hom}(-,-)$ here denotes the category of ordinary ($k$-linear) functors; $H^0(\mathcal{RHom}(\mathcal A,\mathcal B))$ can be identified to the category of quasi-functors $\mathrm{rep}(\mathcal A,\mathcal B)$. I've been studying the properties of this functor for a while. For example, I know that for $\mathcal A = \mathbf{1}$, the category with one object and endomorphism ring $k$, $H^0(-)$ is an equivalence; for $\mathcal A = \Delta^1$, the category freely generated over $k$ by the diagram $0 \to 1$, this functor is full and reflects isomorphisms. Assuming $\mathcal B$ is pretriangulated and taking $\mathcal A = (\Delta^1)^{\mathrm{pretr}}$ (the pretriangulated hull of $\Delta^1$) the above claim implies that $H^0(-)$ induces, in this particular case, a bijection between the isomorphism classes of objects. Do you know other situations where the $H^0$ functor exhibits some good properties as above? For example, are there other cases in which it is full and reflects isomorphisms? Or cases in which it induces a bijection between isomorphism classes of objects? (Also, I was wondering if a similar problem is studied in the framework of $(\infty,1)$-categories.) REPLY [3 votes]: In the framework of $\infty$-categories, I think it is not difficult to see that your claim for $\mathcal{A} = \Delta^1$ holds for all $\infty$-groupoids. Let $K$ be an $\infty$-groupoid (Kan complex) and $C$ an $\infty$-category (weak Kan complex). The canonical map $C \to N h C$ is an isofibration (or fibration in the Joyal model structure), where $h$ is the homotopy category functor, left adjoint to the nerve functor $N$. This implies that 1) the induced map $$ \underline{\mathrm{Hom}}(K, C) \longrightarrow \underline{\mathrm{Hom}}(K, NhC) \times_{NhC} C $$ is a trivial Kan fibration, and 2) the canonical map $$ \underline{\mathrm{Hom}}(K, NhC) \times_{NhC} C \longrightarrow \underline{\mathrm{Hom}}(K, NhC) $$ induces a full, conservative, bijective-on-objects functor on the homotopy categories. Since $K$ is a Kan complex, one can prove that $$ N \underline{\mathrm{Hom}}_{\mathrm{Cat}}(hK, hC) = \underline{\mathrm{Hom}}(NhK, NhC) \longrightarrow \underline{\mathrm{Hom}}(K, NhC) $$ is a trivial Kan fibration. Now after passing to homotopy categories one gets a full and conservative functor $$ h\underline{\mathrm{Hom}}(K, C) \longrightarrow \underline{\mathrm{Hom}}_{\mathrm{Cat}}(hK, hC). $$ I am not sure whether the assumption on $K$ Is really necessary.<|endoftext|> TITLE: Finite-dimensional inverse limits of double-dual spaces QUESTION [7 upvotes]: Let $k$ be a field and $\{V_i\}_{i \in I}$ a filtered projective system of $k$-spaces with transition maps $f_{ji}: V_j \rightarrow V_i$ for $i \leq j$ (for my purposes we may assume the index set is $\mathbb{N}$). For each $i$, let $V_i^{**}$ be the double dual of $V_i$: here this is simply the algebraic dual, $Hom_k(Hom_k(V_i, k), k)$. The $V_i^{**}$ also form an projective system. If $\varprojlim V_i^{**}$ is finite-dimensional over $k$, do we have $\varprojlim V_i \simeq \varprojlim V_i^{**}$? (Here there is no assumption that the $V_i$ themselves are finite-dimensional.) Since $V_i \hookrightarrow V_i^{**}$ canonically for each $i$ and projective limit is left-exact, it follows that $\varprojlim V_i \hookrightarrow \varprojlim V_i^{**}$. Therefore the hypothesis implies that $\varprojlim V_i$ is also finite-dimensional, and isomorphic to its own double dual $(\varprojlim V_i)^{**}$. It would suffice to show that there is an injection $\varinjlim V_i^* \hookrightarrow (\varprojlim V_i)^*$, as this would dualize to a surjection $(\varprojlim V_i)^{**} \twoheadrightarrow (\varinjlim V_i^*)^* = \varprojlim V_i^{**}$ and the proof would be complete by comparing dimensions. There is a natural map $\varinjlim V_i^* \rightarrow (\varprojlim V_i)^*$ , induced by the duals of the structural maps $\pi_i: \varprojlim V_i \rightarrow V_i$, but even with the assumption that both source and target are finite-dimensional I can't seem to show that this map is injective (nor can I think of a counterexample). REPLY [2 votes]: I think this is true. I'll try to prove the statement $$ \varinjlim V_i^* \hookrightarrow (\varprojlim V_i)^* $$ if the source is finite dimensional and $\varprojlim V_i$ is finite dimensional. I'll assume the indexing set is $\mathbb N$ but it should work for any cofiltered system. First a reduction. If we let $V_i' = \operatorname{im}(\varprojlim V_i → V_i)$ then we can choose complements $V_i = V_i' \oplus V_i''$ such that $\{V_i''\}_i$ is an inverse system itself (using that $k$ is a field). Then the tower $V'$ is levelwise finite dimensional with limit $\varprojlim V_i$, and $\varprojlim V_i'' = 0$. Now we have $$ \varinjlim V_i^* = \varinjlim (V_i')^* \oplus \varinjlim (V_i'')^* → (\varprojlim V_i')^* \oplus (\varprojlim V_i'')^* = (\varprojlim V_i)^*, $$ and on the left summands, this map is an isomorphism. So in the first statement we want to show, we can assume that $(\varprojlim V_i)^* = 0$, and hence $\varprojlim V_i = 0$. If $\{V_i\}$ is pro-trivial (this means for every $i$ there's a $j>i$ such that $V_j → V_i$ is the zero map) then $\varinjlim V_i^* = 0$ as well, so let's assume it's not. Possibly by passing to a cofinal subtower, we can choose nonzero elements $α_i\colon V_i → k$ such that $α_i|_{\operatorname{im}V_{i+1}}=0$ and $α_i$ maps nontrivially to $V_{j}$ for $j TITLE: Über theorem on unavoidable patterns? QUESTION [8 upvotes]: Let $A$ be an alphabet of $k$ symbols, and $p$ a pattern. An example of a pattern is $p=XX$, where $X$ is any finite string of symbols from $A^+$. Avoiding $p$ is avoiding any subword repeated twice in a row. Such strings are called square-free. It is well known that there are no infinite binary ($k=2$) strings of symbols that are square-free (in fact, only $0$, $1$, $01$, $10$, $010$, and $101$ are square-free), but there are infinite ternary ($k=3$) square-free strings, as proved by Axel Thue. Other examples: the patterns $X$ and $XYX$ are unavoidable on any alphabet. My question is: Q. Is there a theorem of the form: Any alphabet $A$ with $|A| \le k$ cannot avoid any patterns $p$ of the form [some description of these patterns $p$ as a function of $k$], i.e., there are no infinite strings that avoid these $p$? In other words, is there a pattern to—a characterization of—the patterns that are unavoidable, for a given $k$? Or are there, to date, only claims that specific patterns are unavoidable? REPLY [7 votes]: According to the 2013 paper "Computing the Partial Word Avoidability Indices of Ternary Patterns" by Blanchet-Sadri, Lohr, and Scott, The problem of deciding whether a given pattern is avoidable has been solved [1, 14], but the one of deciding whether it is k-avoidable has remained open. So, it seems to be an open problem.<|endoftext|> TITLE: Complex geometry text/research introduction for the analyst QUESTION [9 upvotes]: To give some background, I am mainly an analyst trained in harmonic/functional and do work on geometric pde's and spectral multipliers. Of late, I am trying to learn more about (research level) complex geometry. If I understand correctly, complex geometry means many things to many different people. To some, it is an extension of algebraic geometry, while many others would immediately think of the Newlander-Nirenberg or Yau's proof of Calabi conjecture. I am looking for some kind of research ``introduction'' to the analytic side of complex geometry. I want to develop a better appreciation of the pde theoretic tools that are relevant in complex geometry, the problems they can solve, and also some of the open problems in the field that are thought to be potentially amenable to analytic methods of attack. In other words, suppose an analyst wants to do research on complex geometry. What does he start by reading? Papers sometime down the line for sure, but initially they might be too specific/concentrated. I guess, if there were a textbook on complex geometry written by Yau, or Hormander, or Tao, amongst other people, that would be a starting point for me. If my question is too broad/unfit for this site, I apologize. I realize that it is impossible that any single book/monograph/lecture note will cover all the analytic sides of complex geometry. But even a partial answer will be appreciated. Lastly, just for example: if someone asked me what would be a good answer if someone asked the same question about real differential geometry? I would say I don't have a good answer, as the literature is just too huge. However, I would add that some of my favourites are Schoen/Yau's Lectures on Differential Geometry, Jost's Geometric Analysis, and perhaps Aubin's Nonlinear problems in Riemannian geometry. These books should certainly get one started. REPLY [7 votes]: 1) There is a great book From Holomorphic Functions to Complex Manifolds by Fritzsche-Grauert. It is very geometric and gives you the fundamentals on complex manifolds, including specialized topics, from Stein manifolds to compact Kähler manifolds, which in a sense are the two extremities in the spectrum of holomorphic geometry. An important bonus is that Grauert was arguably the deepest complex analyst in the twentieth century. It is very geometric but it will also warm your analyst's heart with sections on plurisubharmonic functions, Sobolev spaces and, Neumann operators,... 2) Another, even more analysis rich introduction, to complex geometry is Krantz's Function Theory of Several Complex Variables There you will find harmonic analysis, regularity of $\bar \partial $ operator, $H^p$ functions and all sorts of integral representations. 3) Both books contain introductions to the indispensable tool of sheaves and their cohomology, which actually had their first applications in complex analysis, and were foreshadowed in Oka's groundbreaking solution to the Levi problem. There are other good books, by Fuks, Griffiths, Hörmander (mentioned in abx's comment), Huybrechts, Ohsawa, Range, Wells,... but for me the most comprehensive introductions are 1) and 2). 4) Welcome to that enchanting land of complex geometry (full disclosure: that was where I started doing research!) and good luck!<|endoftext|> TITLE: Is the Manickam-Miklós-Singhi Conjecture solved? QUESTION [10 upvotes]: This arXiv paper is claimed to contain a proof for the MMS conjecture. But it seems that this manuscript is not yet peer reviewed by other mathematicians. I personally tried to follow the paper, but after some point I couldn't due to the lack of explanation, or my ability to understand. Given the interest of mathematicians for this problem, I think the manuscript could have been noticed by mathematicians' community. What is the status of this paper? REPLY [8 votes]: The paper has appeared in print: http://link.springer.com/article/10.1134%2FS0032946014040048 It does look strange that the author keeps posting and posting solutions to very famous problems in combinatorics every few months or so (yesterday a paper appeared claiming to have solved the famous conjecture on the singularity probability of a random Bernoulli matrix) and they all appear in 1 journal or do not appear at all. As the length of papers is 10-15 pages and they claim to solve long standing open problems, I would expect the refereeing process to be lightning fast and the best journals trying to snap those papers as soon as possible. Yet it does not happen. The papers are genuinely terribly written (I have looked at them all quite a bit as I am interested) and no effort seems to have been put in making the results more readable. It is therefore strange that the author keeps posting and posting new supposed breakthroughts without making the previous ones more accessible (if the proofs are correct, hordes of combinatorialists and probabilists would be lining up to read it and present it in seminar, but we have absolute silence instead). But this is only my opinion. I would be very glad if my gloomy outlook is misplaced.<|endoftext|> TITLE: Open problems in Federer's Geometric Measure Theory QUESTION [20 upvotes]: I wanted to know if the problems mentionned in this book are solved. More specifically, at some places, the author says that he doesn't know the answer, for example :"I do not know whether this equations are always true" p.361 4.1.8, or "I do not know..." p.189 2.10.26. Are there counterexamples or proofs of these asumptions? I add some details : in theorem 2.10.25, if $f:X\rightarrow Y$ is a lipschitzian map of metric spaces, $A\subset X$, $0\leq k<\infty$, and $0\leq m<\infty$, then $$ \int_Y^*\mathscr{H}^k(A\cap f^{-1}\{y\})d\mathscr{H}^m(y)\leq (\mathrm{Lip} f)^m \frac{\alpha(k)\alpha(m)}{\alpha(k+m)}\mathscr{H}^{k+m}(A). $$ (where $\mathscr{H}^n$ is the Hausdorff measure associated to the metrics on $X$ and $Y$, and $\int^*$ is the upper integral) provided either $\{y\in Y, \mathscr{H}^k(A\cap f^{-1})>0\}$ is the union of countable family of sets with finite $\mathscr{H}^m$ measure, or $Y$ is boundedly compact (each close bounded subset is compact). The question Federer asks is to determine if everything after "provided either..." is necessary. The other question is about currents : let $S$, $T$ be two currents on open subsets $A$,$B$ of euclidean spaces, of degree $i$ and $j$ : if $S$, $T$ are representable by integration, do we always have $\Vert S\times T\Vert=\Vert S\Vert\times\Vert T\Vert$, and $\overrightarrow{S\times T}(a,b)=(\wedge_i p)\overrightarrow{S}(a)\wedge(\wedge_j q)\overrightarrow{T}(b)$ for $\Vert S\times T\Vert$ almost all $(a,b)\in A\times B$, where $p:A\rightarrow A\times B$ and $q:B\rightarrow A\times B$ are the canonical injections (it is indeed the case if either $\overrightarrow{S}$ or $\overrightarrow{T}$ is simple $\Vert S\Vert\times\Vert T\Vert$ almost everywhere). Edit : I found the answer of the first question in a book of Yu. D. Burago and V. A. Zalgaller, Geometric inequalities, where there is a mention of a more general form for abrbitrary separable metric spaces ($X$ and $Y$ are assumed to be separable). This result was proved by Roy O. Davies, in the article Increasing Sequences of Sets and Hausdorff Measure (Proc. London Math. Soc. 20, 222-236, 1970). REPLY [7 votes]: Theorem. The following inequality is true for any metric spaces $X,Y$, any $A\subset X$, any Lipschitz map $f:X\to Y$ and any real numbers $k,m\geq 0$. $$ \int_Y^*\mathscr{H}^k(A\cap f^{-1}\{y\})d\mathscr{H}^m(y)\leq (\mathrm{Lip} f)^m \frac{\alpha(k)\alpha(m)}{\alpha(k+m)}\mathscr{H}^{k+m}(A). $$ Fereder could prove the inequality in cases when he could establish equality $\lim_{\delta\to 0^+}\lambda_\delta(g)=\int^*g\, d\psi$ (see page 187 in Federer's book [F1]). For example he could prove Theorem when $Y$ is boundedly compact i.e. bounded and closed sets in $Y$ are compact (in fact Federer's result was originally proved in [F2]). In general he could prove the inequality $$ \lim_{\delta\to 0^+}\lambda_\delta(g)\leq\int^*g\, d\psi. $$ Then he asked [F1, page 187]: The general problem whether or not the preceding inequality can always be replaced by the corresponding equation is unsolved. A positive answer to this question would imply Theorem in the general case. The problem was answered in the positive by Davies [D, page 236]: Note added 8 September 1969. H. Federer tells me that this work answers a question he raised in Geometric measure theory (Berlin, 1969) [...] However, there is no proof of Theorem in Davies' paper and one could get the proof only by reading Federer's proof, by reading Davies' paper and by understanding how to combine the two together. Surprisingly, it wasn’t until 2009 when Reichel [R] in his PhD thesis, re-wrote a complete proof of Theorem in its full generality, by following the original proof of Federer while making use of Davies’ result. Until recently Reichel’s thesis was the only place with a complete proof of Theorem, except that Reichel did not include the proof of Davies’ theorem. Davies’ theorem is however, very difficult. You can find a new and a complete proof of Theorem that avoids Davis' result in: https://arxiv.org/abs/2006.00419 This paper also contains a detailed history of the problem. [D] Davies, R. O.: Increasing sequences of sets and Hausdorff measure, Proc. London Math. Soc. 20 (1970), 222-236. [F1] Federer, H.: Geometric measure theory. Die Grundlehren der mathematischen Wissenschaften, Band 153 Springer-Verlag New York Inc., New York 1969. [F2] Federer, H.: Some integralgeometric theorems. Trans. Amer. Math. Soc. 77 (1954), 238–261. [R] Reichel, L. P.: The coarea formula for metric space valued maps. Ph.D. thesis, ETH Zurich, 2009.<|endoftext|> TITLE: Why are Witten-Reshetikhin-Turaev invariants expected to be integral? QUESTION [15 upvotes]: A Witten-Reshetikhin-Turaev (WRT) Invariant $\tau_{M,L}^G(\xi)\in\mathbb{C}$ is an invariant of closed oriented 3-manifold $M$ containing a framed link $L$, where $G$ is a simple Lie group, and $\xi$ is a root of unity. Components of $L$ are coloured by finite dimensional $G$-modules. Since Murakami in 1995, people have been proving integrality results for increasingly general classed of simple Lie groups $G$ (first $SO(3)$ then $SU(n)$, then any compact simple Lie group), roots of unity $\xi$ (first prime, then non-prime), and $3$--manifolds $M$ (first integral homology $3$-sphere, then rational homology $3$-sphere, then the general case). These results usually state that the WRT invariant is an algebraic integer- an element of $\mathbb{Z}[\xi]$- or the stronger result that it's the evaluation at $\xi$ of an element in the Habiro ring. Papers on the integrality of WRT invariants usually list wonderful things that can be done once integrality properties are established (e.g. integral TQFT or categorification or representations over $\mathbb{Z}$ of the mapping class group). But why would we expect $\tau_{M,L}^G(\xi)$ to be an element of $\mathbb{Z}[\xi]$? As far as I know that's always been the case. Is there some perhaps some not-quite-rigourous construction of WRT invariants which takes place entirely over $\mathbb{Z}[\xi]$, or maybe over the Habiro ring? Question: Why are WRT invariants expected to be algebraic integers? Is there a conceptual explanation for their integrality? Are all WRT invariants in fact expected to come from analytic functions over roots of unity (i.e. elements of the Habiro ring), and if so, why? REPLY [10 votes]: My humble point of view is that the Witten-Reshetikin-Turaev invariant (at least for $G=SU(2)$ or $SO(3)$) at the root $\xi$ is (by definition) a rational function on $\xi$ which looks very much like a polynomial with integer coefficients. (Note that the function depends on $\xi$.) When $M=S^3$, the rational function is indeed a polynomial with integer coefficients that does not depend on $\xi$: that's the (colored) Jones polynomial. So the Witten-Reshetikin-Turaev invariant of $L\subset S^3$ lies in $\mathbb Z[\xi]$ in that case. And it is natural to ask whether this holds for general $M$. When $M\neq S^3$ the rational function is no longer a polynomial and moreover it depends of $\xi$ (but this dependence is not an issue concerning integrality). The numerator of this rational function is an arbitrary element in $\mathbb Z[\xi]$, but the denominator is not arbitrary: it is just a product of quantum integers $$[n] = \frac{\xi^n - \xi^{-n}}{\xi - \xi^{-1}} = \xi^{n-1} + \xi^{n-3} + \ldots + \xi^{-n+1}.$$ Hence the invariant is "almost" a polynomial with integer coefficients: it is an element in $\mathbb Z[\xi]$ divided by a product of quantum integers like $[2]^3[4]^2[7]$. (I am probably ignoring some renormalisation factor.) As far as I remember, when $\xi$ is a $4r$-th root of unity and $r$ is prime, then every $[n]$ is actually invertible in $\mathbb Z[\xi]$, and hence the invariant indeed lies in $\mathbb Z[\xi]$.<|endoftext|> TITLE: Is there a mistake in Vapnik's "Basic Lemma"? QUESTION [6 upvotes]: I have a concern about the "Basic Lemma" which Valdimir Vapnik states and proves in his 1998 book Statistical Learning Theory (ch. 14.3, pp. 574–76): It seems like a certain coefficient should have been 2 instead of 1. I'm hoping I'm just wrong, but I'd appreciate any comments or clarifications. First, some context: Suppose a sample space $(X,\mu)$ is given, and that we have selected a system $S$ of events $A\subseteq X$. We can then define various real-valued functions over $S$. For example, the probability measure $\mu$ maps an event in $S$ into its probability; and given a fixed data set, we can also define a frequency function $f$ which maps an event into its frequency in the data set. This space of functions is equipped with a distance measure: $$||s-t|| = \sup_{A\in S} |s(A) - t(A)|.$$ This distance measure respects the triangle inequality, a property that it essentially inherits from the usual distance measure $|x-y|$ on the real number line. Then, the critical issue: Suppose that three independent and equally large data sets are drawn from $X$; we can then define the three frequency functions $f$, $f_1$, and $f_2$ corresponding to these data sets. If we fix $A$ but let the data sets be chosen randomly, the numbers $f(A)$, $f_1(A)$, and $f_2(A)$ are random variables. The probability $\mu(A)$ is deterministic. The "Basic Lemma" (p. 574) now states that $$\Pr\left(||f_1 - f_2|| > \varepsilon\right) \;\leq\; \Pr\left(||f - \mu|| >\frac{\varepsilon}{2}\right) - \Pr\left(||f - \mu|| >\frac{\varepsilon}{2}\right)^2.$$ Vapnik proves this (p. 576) by noting that the two frequencies $f_1$ and $f_2$ can only differ by $\varepsilon$ from each other if at least one of them differs by $\varepsilon/2$ from $\mu$. (Otherwise a detour over $\mu$ would reduce the distance between $f_1$ and $f_2$, violating the triangle inequality.) But the probability of a proposition is smaller than the probability of its logical consequences, so $$\Pr\left(||f_1 - f_2|| > \varepsilon\right) \;\leq\; \Pr\left( ||f_1 - \mu|| > \frac{\varepsilon}{2} \;\textrm{ or }\; ||f_2 - \mu|| >\frac{\varepsilon}{2}. \right)$$ Since $f_1$ and $f_2$ are independent, and since they are both equal to $f$ in distribution, we can reduce this to $$\Pr\left(||f_1 - f_2|| > \varepsilon\right) \;\leq\; 1 - \left(1 - \Pr\left( ||f - \mu|| > \frac{\varepsilon}{2} \right) \right)^2.$$ He then states without further comment that this proves the inequality (p. 576). This seems wrong. By expanding, $1 - (1 - a)^2 = 2a - a^2$; and $a - a^2$ is a tighter bound than $2a - a^2$, so we can't just decrease the coefficient. So where does his right-hand side come from? It seems like the inequality should in fact have read $$\Pr\left(||f_1 - f_2|| > \varepsilon\right) \;\leq\; 2\Pr\left(||f - \mu|| >\frac{\varepsilon}{2}\right) - \Pr\left(||f - \mu|| >\frac{\varepsilon}{2}\right)^2.$$ Or what? Does anybody have a thought on this? REPLY [4 votes]: You're right, it seems. Suppose $X=\{1, -1\}$; $\mu(\{1\})=\mu(\{-1\})=1/2$; $S$ is the power set of $X$; our data sets have just one element each, so that $f(\{1\})$ is either 0 or 1; and $\epsilon=\frac12$. Then $$\Pr\left(||f - \mu|| >\frac{\varepsilon}{2}\right)\ge \Pr\left(|f(\{1\})-\mu(\{1\})|>\frac14\right)=1$$ so the stated inequality, if true, would give $$\frac12=\Pr\left(||f_1 - f_2|| > \varepsilon\right) \;\leq\; \Pr\left(||f - \mu|| >\frac{\varepsilon}{2}\right) - \Pr\left(||f - \mu|| >\frac{\varepsilon}{2}\right)^2=1-1^2=0,$$ a contradiction. (If the first mistake in the book is on page 574, that's pretty good though.)<|endoftext|> TITLE: elementwise functions of positive definite matrix QUESTION [11 upvotes]: The fact that the Schur (that is, element wise) product of two positive definite (symmetric) matrices is positive definite immediately implies (using the convexity of the positive semi definite cone) that if $A$ is PSD, then so is $B=A_f,$ such that $b_{ij} = f(a_{ij}),$ where $f$ is an analytic function all of whose coefficients are positive. The question is, is this result sharp, or are there other $f$ which maps the PSD cone into itself by being element wisely applied? REPLY [4 votes]: Schoenberg's theorem was first shown without the continuity assumption by Rudin in this paper in Duke Math. J., in 1959. In particular, prior to the work of Herz cited above.<|endoftext|> TITLE: Expectation of a generalization of Dirichlet distribution QUESTION [6 upvotes]: For the standard Dirichlet, the expectation of $X_i$ is $\alpha_i/\alpha_0$, where $\alpha_0 = \sum_i \alpha_i$ [http://en.wikipedia.org/wiki/Dirichlet_distribution]. I am considering the following generalization. Suppose we are playing a simple poker game as follows. We are player 1 and observing player 2's plays. Player 2 can be dealt one of two hands with probability 1/2 (or more generally probability p) -- K or Q (player 2 sees this and player 1 does not). Then player 2 selects one of i actions. Let $X_i$ denote the rv for the probability player 2 plays action i with a K, and let $Y_i$ denote the rv for Q. Suppose that he is following a static probability distribution for all rounds, and that the $X_i$'s and $Y_i$'s are independent (his strategy for a K is independent of his strategy for a Q). Player 1 only observes the action i of P2, and not his card. I am trying to find a closed form for $E[X_i]$ and $E[Y_i]$. This generalization differs from the "Generalized Dirichlet distribution," which has a relatively simple closed-form solution for the expectations (http://en.wikipedia.org/wiki/Generalized_Dirichlet_distribution). I think the pdf of this distribution is the following, where p denotes the probability P2 is dealt a K, $\alpha_i$ is the observed number of times he has taken action $i$ so far, and $N(\alpha)$ is some normalization constant I'm not sure how to compute: $$f(x,y;\alpha) = N(\alpha) * \prod_i [p x_i + (1-p)y_i]^{\alpha_i}$$ where $x_i, y_i \geq 0, \sum_i x_i = 1, \sum_i y_i = 1.$ One can apply the binomial theorem here, but I am not sure how to proceed and if that will even help. Somehow the gamma/beta function will need to come in to play. Thanks a lot. REPLY [2 votes]: Let me try to reformulate your question, and then you can tell me whether I have misunderstood you: You are assuming that the player in question can have $2$ different hands, and can choose between $K$ different actions. For each hand $H=h$, the distribution over actions is a Dirichlet distribution with parameter vector $$ \alpha_h = (\alpha_{h,1}, \alpha_{h,1}, \ldots, \alpha_{h,K}). $$ So in a sense you are given a rectangular table $\alpha$ of parameters which indirectly tell you how the "hand" variable interacts with the "action" variable. The rest of your model is described by the following generative procedure: Choose a hand $h$ according to a multinomial distribution with parameters $\theta = (1/2, 1/2)$; Choose a probability vector $\pi$ according to a Dirichlet distribution with parameters $\alpha_h$; Choose an action $a$ according to a multinomial distribution with parameters $\pi$. Is this roughly what you had in mind? If it is, then it is a kind of Dirichlet mixture model. It has the following properties: Given that $H=h$, the expected value of $\pi_a$ (or marginal probability of $a$) is $$ \Pr(a\,|\,h) \ =\ E[\pi_a\,|\; H=h] \ =\ \frac{\alpha_{h,a}}{\sum_b \alpha_{h,b}}. $$ By the linearity of expectations, the unconditional version of this probability is the weighted sum of the conditional expectations: $$ \Pr(a) \ =\ \sum_h \Pr(a\,|\,h) \Pr(h) \ =\ \sum_h \left( \frac{\alpha_{h,a}}{\sum_b \alpha_{h,b}} \right) \theta_h . $$ Given $h$, the model is a multinomial-Dirichlet model. In such a model, the predictive probability of observing a data set in which the actions $(a_1,a_2,\ldots,a_K)$ occur with counts $n=(n_1,n_2,\ldots,n_K)$ is given by $$ \Pr(n\,|\,h) \ = \int \Pr(n\,|\,\pi) \Pr(\pi\,|\,h)\; d\pi \ =\, \frac{Z(\alpha_h + n)}{Z(\alpha_h)} Q(n), $$ where $Z$ is the normalizing constant for a Dirichlet distribution, $$ Z(v) \ =\ \frac{\sum_i\Gamma(v_i)}{\Gamma(\sum_i v_i)}, $$ and $Q$ is the multinomial coefficient $$ Q(v) \ =\ \frac{\Gamma(1 + \sum_i v_i)}{\prod_i\Gamma(1 + v_i)}. $$ You can compute the unnormalized posterior probabilities of the hands by multiplying their prior probabilities by the likelihood they assign to the data. The normalizing constant will then be the marginal probability assigned to the data set, avering over all parameters in the model: $$ \Pr(n) \ =\ \sum_h \Pr(n\,|\,h) \Pr(h) \ =\ Q(n) \sum_h \frac{Z(\alpha_h + n)}{Z(\alpha_h)} \theta_h. $$ Without further information about what the $\alpha_h$'s look like, this expression cannot be reduced any further. (If they were all somehow derived from a shared hyperprior, then perhaps; but you probably have something else in mind.) I don't know if I'm missing the mark completely here. Does any of this help?<|endoftext|> TITLE: Borel Sets in Sacks Generic Extension QUESTION [6 upvotes]: Let $\mathbb{S}$ denote Sacks forcing. This is forcing with perfect trees or equivalently forcing with uncountable Borel subsets of ${}^\omega 2$ with the relation $\subseteq$. Let $G \subseteq \mathbb{S}$ be a generic filter over $V$. Suppose $B$ is a Borel set coded in $V[G]$. This means $B$ is a subset of $({}^\omega 2)^{V[G]}$ definable by a $\Delta_1^1$ formula using a parameter in $({}^\omega 2)^{V[G]}$. Furthermore, suppose $B$ is an uncountable Borel set. The question is: Does $B$ have a ground model coded uncountable Borel subset? I believe that if the above is answered yes then in $V[G]$, the identity map is a dense embedding of $(\mathbb{S})^V$ and $\mathbb{S}$. Does this lead to any problems? A one further question: Is there some forcing property that can be used to show that the product $\mathbb{S} \times \mathbb{S}$ is different (not equivalent in the forcing sense) from the iteration $\mathbb{S} * \dot{\mathbb{S}}$? Or to differentiate between countable support products and countable support iterations of Sacks forcing? Thanks for any information or clarification. REPLY [6 votes]: Here is a property that differentiates between the product of two copies of Sacks forcing and the iteration. In the iterated Sack model (countable support iteration, the length any ordinal) the following is true: For any two reals $x,y\in 2^\omega$ there is a continuous function $f:2^\omega\to 2^\omega$ coded in the ground model such that $f(x)=y$ or $f(y)=x$. After forcing with a product of two copies of Sacks forcing there is no Borel map coded in the ground model mapping one of the two generics to the other, because then the other generic would already be contained in the extension generated by the first generic.<|endoftext|> TITLE: Constructing a "geometric" model structure on Cat by localizing the "categorical" model structure QUESTION [11 upvotes]: Let $\text{Cat}$ be the category of (small) categories and functors. There is a "categorical" (also called "canonical" or "folk") model structure on $\text{Cat}$ in which the weak equivalences are the usual equivalences of categories (and as explained here, it turns out that there is a unique choice of cofibrations and fibrations that come with them). In this model structure, all categories are fibrant and cofibrant. There is also a natural question of constructing a "geometric" model structure on $\text{Cat}$ for which the weak equivalences are functros that induce weak equivalence of nerves. i.e. there is the nerve functor $N:\text{Cat}\to \text{sSet}$, where $\text{sSet}$ is the category of simplicial sets with the standard model structure, and we would like a model structure on $\text{Cat}$, for which a functor $F:C\to D$ is a weak equivalence if and only if $N(F):N(C)\to N(D)$ is a weak equivalence. There is such a construction by Thomason, in which the model structure on $\text{Cat}$ is "transported" from $\text{sSet}$ along the adjunction $$ \tau_1 \circ Sd^2 :\text{sSet}\leftrightarrows\text{Cat}: Ex^2\circ N $$ Where $\tau_1$ is the left adjoint of $N$ (sometimes called "fundamental category") and $Sd\dashv Ex$ are the barycentric subdivision / Extension adjunction of Kan. In Thomason's model structure, the cofibrations are quite complicated (they are all "Dwyer maps"). In particular a cofibrant category must be a poset. My question is whether there exist a model structure on $\text{Cat}$, with "geometric" weak equivalences as in Thomason's, but with the same cofibrations as the "categorical" one. In other words, can we obtain a "geometric" model structure on $\text{Cat}$ by localizing the categorical one? REPLY [9 votes]: Karol's answer is excellent, however, I want to suggest another argument. I claim that a model category on $\mathbf{Cat}$ whose weak equivalences are the Thomason equivalences can never be a localization of a model structure whose weak equivalences are the categorical equivalences. Indeed, on the one hand the canonical model structure is a simplicial model category whose mapping space is given by $Map(C,D)=N(i \mathrm{Fun}(C,D))$ where $i$ sends a category to its subcategory of isomorphisms. From this description, we see that the mapping spaces are all nerve of groupoids which implies that they are $1$-truncated (i.e. that their homotopy groups of degree at least $2$ are trivial). The important point is that this fact is intrinsic to the weak equivalences in the canonical model structure because these mapping spaces can also be computed using for instance the hammock localization which only depends on the weak equivalences. On the other hand, if $M\leftrightarrows L_SM$ is a Bousfield localization, the fibrant objects of $L_SM$ are a subclass of the fibrant objects of $M$. Moreover, between fibrant objects of $L_SM$, the derived mapping spaces (computed using a simplicial enrichment or the hammock localization) are the same in $M$ or $L_SM$. This means that the $\infty$-category $L_SM$ is a full sub $\infty$-category of $M$. The category of categories with the Thomason weak equivalences is a model for the $\infty$-category of spaces. In particular, some of its derived mapping spaces are not $1$-truncated (for instance $Map(*,S^2)$is not $1$-truncated) which means that it cannot embed as a full sub $\infty$-category of the $\infty$-category of categories with the canonical weak equivalences.<|endoftext|> TITLE: Erdős multiplication problem revisited QUESTION [12 upvotes]: This is a well-known problem and is about counting the number of distinct numbers in the $n \times n$ multiplication table. The very problem has been discussed in-depth and, as such, I require no further input on it by itself. There has been, however, a significant amount of debate about it on StackOverflow, namely this question: https://stackoverflow.com/questions/24614798/find-the-number-of-distinct-numbers-in-multiplication-table As far as I understand, the problem has currently only $O(n^2)$ computational solutions (strictly speaking, $kn^2$ iterations, with $k=0.5$), while the asymptotic size of the set is equal to $$\left|\lbrace a\cdot b:\ a,b\leq N\rbrace\right|\asymp \frac{N^2}{(\log N)^c(\log\log N)^{3/2}}$$ where $$c=1-\frac{(1+\log \log 2)}{\log 2}.$$ (Ford, 2008). As far as my knowledge goes, there is no explicit way to generate a set of size $A(n)$ and to calculate its cardinality without at least $A(n)$ operations. Also, there currently exists no solution to determine the exact value of $A(n)$ without generating the set and counting its unique elements. There has been significant amount of dispute about it by certain individuals, convinced there is an $O(n)$ solution to the problem [calculating $A(n)$], and that they have found it. Although such solutions are usually disproven, I'm interested if it's at all possible for this problem to be solved strictly below $O(n^2)$, either with explicitly generating the set or using some functional relationship between $n$ and $A(n)$. Currently, both the reference solutions and the one sent by David are $O(n^2)$. Edit. For clarity, let us split this into two questions: a) Can exact $A(n)$ for a specific $n$ be actually calculated without generating the set itself (i.e. without any need to know and possibly without any method to tell if a number is in the set, or not) and if so, how? If not, possible reasons for practical/theoretical possibility/impossibility of creating such solution would be perfect. b) Can $A(n)$ be computed by generating the set in strictly below $O(n^2)$ complexity (e.g. $O(n^2/\log \log n)$ or similar)? If so, how would that be possible? related: How many different numbers can be obtained as product of first $n$ natural numbers? Distinct numbers in multiplication table Number of elements in the set $\{1,\cdots,n\}\cdot\{1,\cdots,n\}$ REPLY [7 votes]: The answer to both your questions are (essentially) yes, and are given in a recent paper of Brent, Pomerance, Purdum, and Webster. Regarding (b), they show that $A(n)$ can be computed in subquadratic time. Their algorithm also tabulates the set of products. Regarding (a), they note that for large values of $n$, exact algorithms are too slow, so they also present two Monte Carlo algorithms for approximating $A(n)$. These Monte Carlo algorithms do not tabulate the set of products (doing so would be too expensive). This answers (a) in the positive (if you don't mind Monte Carlo algorithms). They have implemented their algorithms and give exact values of $A(n)$ for all $n \leq 2^{30}$, and Monte Carlo computations for all $n \leq 2^{100000000}$.<|endoftext|> TITLE: Progressively measurable vs adapted QUESTION [23 upvotes]: I often see in stochastic calculus books the terms 'adapted process' and 'progressively measurable process'. I know there is a small difference between them (every progressively measurable process is adapted but the converse is not necessarily true) but I can't get it from the mathematical definitions. The definition of progressively measurable process is the following: A stochastic process $X$ defined on a filtered probability space $(\Omega ,{\mathcal F},{({{\mathcal F}_t})_{t \ge 0}},P)$ is progressively measurable with respect to ${({{\mathcal F}_t})_{t \ge 0}}$, if the function $X(s,\omega):[0,t]\times \Omega \rightarrow \mathbb{R}$ is $\cal{B}([0,t]) \times \cal{F}_t$ measurable for every $t\ge 0$. The definition of adapted process: A stochastic process $X$ on $(\Omega ,{{\mathcal F}},{({{\mathcal F}_t})_{t \ge 0}},P)$ is adapted to the filtration ${({{\mathcal F}_t})_{t \ge 0}}$ (or ${\mathcal F}_t$-adapted) if $X(t)\in {\mathcal F}_t$ for each $t \ge 0$. Can someone explain me the difference in simple words? I think I understand the definition of the adapted process quite well but I'm probably confused of the role of the Borel sigma algebra in the definition of the progressively measurable process. REPLY [4 votes]: Theorem T46 in Meyer's book "Probability and Potentials" says: For any $(X_t)$ a measurable real-valued process adapted to the family of $\sigma$-fields $(\mathcal{F}_t)$, there exists a modification of the process $(X_t)$ progressively measurable with respect to the same family $(\mathcal{F}_t)$. In the above example, you can easily see this is true. Change $A$ to a modification $\bar{A}=\{(x,0),x\in[0,\frac{1}{2}]\}$. There are important relations between them: (Continuous+adapted)$\rightarrow$(mean-square continuous+adapted)$\rightarrow$predictable$\rightarrow$optional$\rightarrow$progressive$\rightarrow$adapted.<|endoftext|> TITLE: forcing square with small conditions QUESTION [5 upvotes]: In the paper, Large cardinals and definable counterexamples to the continuum hypothesis, Foreman and Magidor mention a way to force $\square_{\omega_1}$ with countable conditions. (This is used in Theorem 2.17.) It not hard to guess what this is and prove it is countably closed and $\omega_2$-c.c. under CH. Questions: (1) Are a definition of this forcing and proofs of its basic properties explicitly written up anywhere? (2) Does this generalize to higher cardinals, for example forcing $\square_{\omega_2}$ with $\omega_1$-sized conditions? It seems complications could arise. REPLY [5 votes]: The appropriate reference for a forcing for adding a square sequence with countable conditions is: "S-forcing I. A "black-box" theorem for morasses, with applications to super-Souslin trees", Shelah and Stanley, Israel J. Math, vol. 43, no. 3, 1982. See section 4 example 1 pages 203-204.<|endoftext|> TITLE: If there is a dense geodesic, are almost all geodesics equidistributed? Dense? QUESTION [18 upvotes]: Let $M$ be a complete finite volume Riemannian manifold and $\gamma : \mathbb{R}^{\geq 0} \to M$ a geodesic. Suppose that $\mathrm{im}(\gamma)$ is dense. Is it equidistributed in the Riemannian measure? That is, does $$ \lim_{T \to +\infty} \frac{1}{T} \int_0^T f(\gamma(t)) \, dt = \frac{1}{\mathrm{vol}(M)} \int_M f \, \mathrm{d vol} $$ for every $f \in C_0(M)$? [False in general; true for Nilmanifolds. True a.e. in negative curvature, where the geodesic flow is ergodic. ] Let now $N \subset M$ be an (immersed) submanifold and $\gamma$ a geodesic of $M$ which is contained densely in $N$. Is the submanifold $N \subset M$ totally geodesic? [False in general, though true for some variants in constant negative curvature. But what if "totally geodesic" is weakened to "minimal"?] Added. Asaf's answer nonethtless begs a follow-up question to 1: (Revised). If there is a dense geodesic, must there also be an equidistributed one? Could it in fact be that almost every geodesic is then equidistributed? Does a single dense geodesic imply ergodic geodesic flow? And in particular: does one dense geodesic imply almost all geodesics dense? [A similar revision of 2 would instead involve the condition that almost every geodesic of $M$ that is tangent to $N$ at some point is contained in $N$; but then it should follow trivially (I think) that $N$ is totally geodesic. ] Note: There is an analogy with the equidistribution and Manin-Mumford theorems, due to Szpiro, Ullmo, and Zhang, for torsion points in abelian varieties $A/\bar{\mathbb{Q}}$: For a sequence of torsion points which is eventually outside of every torsion translate of an abelian subvariety, the Dirac masses at the Galois orbits converge to the normalized Haar measure on $A(\mathbb{C})$ (where an embedding $\bar{\mathbb{Q}} \hookrightarrow \mathbb{C}$ has been fixed). Here, I would be tempted to think of a geodesic as corresponding to a Galois orbit of torsion points (either minimizes an energy functional -- or a canonical height); and of a totally geodesic subvariety as corresponding to a Galois orbit of a translate of an abelian subvariety by a torsion point (note that in the basic case of a flat torus, the totally geodesic submanifolds are precisely the subtori). The analogy is probably only superficial, but I thought it could be worth pointing out (if only because it led me to asking this question). Added later. One more (final) question along the line of 2. In algebraic geometry, we have the following general fact: For $L$ a nef line bundle on a projective variety $X$, if $\deg_LC =0$ for a Zariski-dense set of curves $C \subset X$, then $\deg_LX = 0$. (Nef =non-negative intersection numbers, =non-negative on every curve). For if $\deg_LX > 0$, Riemann-Roch and the almost vanishing of the higher cohomology of powers of nef line bundles imply that $L$ is big, hence a power of $L$ is effective. We may also do this in an arithmetic setting. In the analogy of the preceding note which led me to consider totally geodesic submanifolds, I was misled by the Manin-Mumford theorem, which is specific to commutative group varieties and fails even for algebraic dynamical systems. Instead, subvarieties of minimal height ought to be analogous to minimal immersed submanifolds: the images of harmonic isometric immersions (which include totally geodesic ones as a particular case, and coincide with the geodesics in dimension one). Considering the previous paragraph, then, does the following question make any sense: If the closure of a minimal submanifold happens to be an immersed submanifold, is this submanifold still minimal? In the same vein: If we have a sequence of complex algebraic curves in $\mathbb{CP}^n$ (images of non-constant holomorphic maps from compact Riemann surfaces) whose supports converge to a compact real-analytic immersed submanifold $M \subset \mathbb{CP}^n$, must $M$ be a complex (algebraic) submanifold? REPLY [3 votes]: The answer to question 2 is negative. Let $M$ be the suspension of the round sphere $S^2$ by an irrational rotation $\phi$ (whose fixed point should be called the poles); i.e. $M$ is the quotient of $S^2\times \mathbb{R}$ by the relation $(x,t+1)\sim (\phi(x),t)$. Since $\phi$ is an isometry of $S^2$, $M$ inherits a well-defined Riemannian metric, which is locally isometric to a Riemannian product of $S^2$ by a small interval.In particular, the "vertical lines" $\{x\}\times\mathbb{R}/\sim$ are geodesics of $M$. Now, consider the vertical line issued from a point $p=(x,0)$; its closure is the (quotient of the) product of a circle by the $\mathbb{R}$ factor. If $x$ is neither a pole nor lying on the equator, this submanifold is clearly not totally geodesic.<|endoftext|> TITLE: Is there any progress on Problem 12 (from Schoen and Yau)? QUESTION [15 upvotes]: I saw the following question from the "Problem Section" in Schoen and Yau, page 281, problem 12: Let $M_1, M_2$ each have negative curvature. If $\pi_1 (M_1)=\pi_1 (M_2)$, prove that $M_1$ is differeomorphic to $M_2$. The authors then commented: " There is some progress due to Cheeger, Gromov, Farrell and Hsiang. Cheeger proved that $\pi_1(M)$ determines the second Stiefel bundle of $M$. Gromov proved $\pi_{1}(M)$ determines the unit tangent bundle of $M$. Farrell-Hsiang proved $\pi_{1}(M)$ determines $M\times \mathbb{R}^{3}$. Farrel-Hsiang have only to assume that one of the manifolds has negative curvature." Here the paper due to Cheeger&Gromov is this one. I am unable to find any reference on Farrell&Hsiang's work as the book says it was to be published. May I ask if there is any significant progress on this problem? This is obviously some strong rigidity theorem (if turned out to be true). A quick "scan" of the papers involved in the end of Gromov's paper did not reveal anything particularly relevant. REPLY [16 votes]: Counter-example was constructed by T.Farrell and L.Jones, in "Negatively curved manifolds with exotic smooth structures", JAMS 1989, volume 2, number 4.<|endoftext|> TITLE: Canonical Sheaf of Projective Space QUESTION [15 upvotes]: I am stuck on one step that occurs without explanation in several Algebraic geometry books. Starting from the exact sequence $$0\rightarrow \Omega_{\mathbb{P}^n}\rightarrow \mathcal{O}_{\mathbb{P}^n}(-1)^{\oplus n+1}\rightarrow \mathcal{O}_{\mathbb{P}^n}\rightarrow 0$$ it is concluded that $$\omega_{\mathbb{P}^n}=\wedge^n \Omega_{\mathbb{P}^n}\cong \mathcal{O}_{\mathbb{P}^n}(-n-1)$$ How does this follow and in particular how does $\Omega_{\mathbb{P}^1}\cong \mathcal{O}_{\mathbb{P}^1}(-2)$ follow ? Thanks in advance. REPLY [4 votes]: One could see this in the following way. We have $$\omega_{\mathbb{P}^n} = \mathcal{O}_{\mathbb{P}^n}(c_1)$$ where $c_1 = c_1(\omega_{\mathbb{P}^n}) = c_{1}(\bigwedge^n\Omega_{\mathbb{P}^n}) = c_1(\Omega_{\mathbb{P}^n})$ is the first Chern class. Now, by the Euler's exact sequence $$0\mapsto\Omega_{\mathbb{P}^n}\rightarrow\mathcal{O}_{\mathbb{P}^n}(-1)^{\oplus (n+1)}\rightarrow \mathcal{O}_{\mathbb{P}^n}\mapsto 0$$ we get $$c_1(\omega_{\mathbb{P}^n}) = c_1(\mathcal{O}_{\mathbb{P}^n}(-1)^{\oplus (n+1)})-c_1(\mathcal{O}_{\mathbb{P}^n}) = -n-1.$$ Therefore $$\omega_{\mathbb{P}^n} = \mathcal{O}_{\mathbb{P}^n}(-n-1).$$<|endoftext|> TITLE: Weil's Riemann Hypothesis for dummies? QUESTION [12 upvotes]: Weil's Riemann Hypothesis is a deep result that I don't fully understand, but it has understandable corollaries which interest me. For example: (a) For any projective curve $X$ satisfying certain conditions, the number $N$ of points in $X$ with coordinates in $\mathrm{GF}(q)$ satisfies $|N-(q+1)|\leq\mathrm{const}\cdot\sqrt{q}$. (The deviation is $0$ when $X$ is a projective line.) (b) For any nontrivial multiplicative character $\chi$ on $\mathrm{GF}(q)$ and any polynomial $f$ of degree $n$ satisfying certain conditions, we have $$\bigg|\sum_{x\in\mathrm{GF}(q)}\chi(f(x))\bigg|\leq(n-1)\sqrt{q}.$$ Questions: Is there a reference (legible to an English-speaking non-expert in the field) which gives the rigorous statements of these corollaries? In particular, I would like conditions which one can verify without a background in algebraic geometry. Are there other corollaries of Weil's Riemann Hypothesis which are also widely understandable? EDIT: I'm mostly interested in the Riemann Hypothesis, but I'm also happy to learn understandable consequences of the other Weil conjectures and related results. REPLY [9 votes]: This is an explicit version of part (a) of Gerry Myerson's answer. If $f(x,y)$ is an absolutely irreducible polynomial in $\mathbf{F}_q[x,y]$ of total degree $d>0$, and $N$ is the number of zeroes of $f(x,y)$ in $\mathbf{F}_q\times\mathbf{F}_q$, then $$ q+1-(d-1)(d-2)\sqrt{q}-d\le N\le q+1+(d-1)(d-2)\sqrt{q}. $$ Likewise, if $f(x,y,z)$ is an absolutely irreducible homogeneous polynomial in $\mathbf{F}_q[x,y,z]$ of total degree $d>0$, and $N$ is the number of zeroes of $f(x,y,z)$ in $\mathbf{P}^2(\mathbf{F}_q)$, then $$ \lvert N-(q+1)\rvert\le (d-1)(d-2)\sqrt{q}. $$ These results comprise Corollary 2 in the paper "The number of points on a singular curve over a finite field" by Leep and Yeomans (Arch. Math. 63 (1994), 420-426).<|endoftext|> TITLE: Is this graph of reciprocal power means always convex? QUESTION [5 upvotes]: Let $$ p = (p_1, \ldots, p_n) $$ be a finite probability distribution, which for convenience I'll assume to have no zeroes: thus, $p_i > 0$ for all $i$ and $\sum_i p_i = 1$. Is the function $$ q \mapsto \biggl( \sum_{i = 1}^n p_i^q \biggr)^{1/(1 - q)} $$ ($q \geq 0$) necessarily convex? Now let me give some context. For each $t \in \mathbb{R}$ and $x = (x_1, \ldots, x_n) \in (0, \infty)^n$, we can form the power mean of $x_1, \ldots, x_n$, weighted by $p_1, \ldots, p_n$, of order $t$. When $t \neq 0$, this is defined by $$ M_t(p, x) = \biggl( \sum_{i = 1}^n p_i x_i^t \biggr)^{1/t}. $$ We define $M_0(p, x) = \lim_{t \to 0} M_t(p, x)$, which works out to be $$ M_0(p, x) = \prod_{i = 1}^n x_i^{p_i}. $$ It's a well-known classical fact that $M_t(p, x)$ is increasing in $t$, for fixed $p$ and $x$. (I mean "increasing" non-strictly; e.g. it's constant in $t$ if $x_1 = \cdots = x_n$.) For instance, the fact that $M_0(p, x) \leq M_1(p, x)$ is the famous theorem that the geometric mean is less than or equal to the arithmetic mean. My question is, in some sense, at one level higher. For reasons that probably aren't relevant here, I've seen plots of the function $$ t \mapsto 1/M_t(p, p) \qquad (t \geq -1) $$ for many different distributions $p$. The fact above tells us that the graph is always decreasing. But in every case I've seen, it has also looked as if it's convex. Write $q = t + 1$ ($q \geq 0$), so that $$ 1/M_t(p, p) = \begin{cases} \Bigl( \sum p_i^q \Bigr)^{1/(1 - q)} &\text{if } q \neq 1 \\ \prod p_i^{-p_i} &\text{if } q = 1. \end{cases} $$ Then in every case I've seen, the graph has looked something like this: (source: ed.ac.uk) Notes: For $q \geq 0$, write $f(q) = \bigl( \sum p_i^q \bigr)^{1/(1 - q)}$. We know that $f \geq 0$ and $f' \leq 0$. I'm asking whether $f'' \geq 0$. If that's true, a natural conjecture is that $(-1)^k f^{(k)} \geq 0$ for all $k$: that is, $f$ is completely monotone. A theorem of Bernstein states that $f$ is completely monotone if and only if it's the Laplace transform of some finite measure on $[0, \infty)$. For an arbitrary $x \in (0, \infty)^n$, it's not necessarily true that $t \mapsto 1/M_t(p, x)$ is convex in the region $t \geq -1$. There are counterexamples. The quantity $\bigl( \sum p_i^q \bigr)^{1/(1 - q)}$ is the exponential of the Rényi entropy of $p$ of order $q$. That's why I've given this an "information theory" tag. REPLY [7 votes]: To expand on Dirk's example: for $n=3$ and $p = [1/4, 1/4, 1/2]$ we have $$ f''(0) = 3\, \left( \ln \left( 3 \right) \right) ^{2}+6\,\ln \left( 3 \right) -10\,\ln \left( 2 \right) \ln \left( 3 \right) +9\, \left( \ln \left( 2 \right) \right) ^{2}-10\,\ln \left( 2 \right) < 0$$. Added by Tom Leinster As a check, and to insure against any error in computing the 2nd derivative, I computed this: $$ \frac{1}{2}\bigl( f(0) + f(0.6) \bigr) - f(0.3) = -0.00018332\ldots < 0, $$ again proving non-convexity. (I chose $0.6$ because that's roughly the value that illustrates the non-convexity most vividly. But even so, notice how close to zero this is.)<|endoftext|> TITLE: limiting empirical spectral distribution of the Laplacian matrix on an Erdos-Renyi graph? QUESTION [8 upvotes]: Let $G$ be an Erdos-Renyi random graph (i.e. an edge ($ij$) exists with probability $0 < p < 1$ and all edges are independent). Let $L$ be the Laplacian matrix of this graph (i.e $L=D-A$, where $A$ is the adjacency matrix of the graph and $D$ is diagonal with $D_{ii} = $ degree of node $i$). Let finally $0 = \lambda_0 \leq \lambda_1 \leq \lambda_2 \leq \cdots$ be the eigenvalues of $L$. Is an analytic expression known for the limiting empirical distribution of these eigenvalues, i.e. for $\lim_{n \to \infty} \frac{1}{n} \#\{ 1 \leq j \leq n : \lambda_j \leq t \}, t \in \mathbb{R}?$ REPLY [3 votes]: As far as I know, the spectra of Laplacian of a graph follows a distribution that is the free additive convolution of a Gaussian and the semicircle law. Take a look, for example at the article: "Spectra of random graphs with arbitrary expected degrees" Raj Rao Nadakuditi, M. E. J. Newman https://arxiv.org/abs/1208.1275 In your case the arbitrary expected degree is constant, therefore the computation should result simpler than in the more general case.<|endoftext|> TITLE: Did Cauchy think that uniform and pointwise convergence were equivalent? QUESTION [21 upvotes]: I've heard that Cauchy thought he'd proved that pointwise and uniform convergence are equivalent. Is this a historical fact? If it is indeed true, I was wondering if anyone had a reference. REPLY [4 votes]: I recommend the reading of a wonderful 15-page text of Imre Lakatos, "Another case study of the method of proofs and refutations", which discuss precisely what Cauchy knew and didn't know and did and didn't do about this. It is published as an appendix in "Proofs and Refutations". Google Book page of "Proofs and Refutations":<|endoftext|> TITLE: Non-vanishing of elements in cohomology of full Flag varieties QUESTION [5 upvotes]: Consider the full flag variety $F_n$ consisting of full flags in $\mathbb C^n$. There is a collection of tautological bundles on $F_n$: $0=U_0\subset U_1\subset ...\subset U_{n-1}\subset U_n=\mathbb C^n\otimes O_{F_n}$. Recall that for $i=1,...,n-1$ the classes $\sigma_i=c_1(U_i^*)\in H^2(F_n)$ form a base in $H^2(F_n)$ and moreover generate multiplicatively the cohomology ring of $F_n$. Question. Is it true that the cohomology classes $(\sigma_1\cdot\sigma_2\cdot...\cdot \sigma_{n-1})^k$ and $(\sigma_1\cdot\sigma_2\cdot...\cdot \sigma_{n-2})^k$ are non-zero for $k$ small enough with respect to $n$ (say for $k<\log(n)/100$)? Or at least, say for $k\le 10$ for large $n$? REPLY [5 votes]: Yes. I claim that $(\sigma_1 \sigma_2 \cdots \sigma_{n-1})^{\lfloor n/2 \rfloor} \neq 0$, which is better than anything you ask for. (Here $\lfloor x \rfloor$ is $x$ rounded down to the nearest integer.) Recall that a basis for $H^{\ast}(FL_n)$ is the Schubert classes $[X_w]$, indexed by permutations $w$ in $S_n$. We use $\ell$ for the length function on $S_n$ and write $(i j)$ for the transposition that interchanges $i$ and $j$. Monk's formula states that $$\sigma_r [X_w] = \sum_{\begin{matrix} \ell( w(ij)) = \ell(w)+1 \\ i \leq r < j \end{matrix}} X_{w (ij)}.$$ The identity of $H^{\ast}(FL_n)$ is the identity permutation. Thus, we see that the coefficient of $[X_w]$ in $\sigma_{r_1} \sigma_{r_2} \cdots \sigma_{r_N}$ is the number of chains $$e = w_0,\ w_1,\ w_2,\ \ldots,\ w_N=w$$ so that $\ell(w_k)=k$ and $w_k$ is of the form $w_{k-1} (i j)$, with $i \leq r_k < j$. In particular, if the word $(r_1 \ r_1+1) (r_2\ r_2+1) \cdots (r_N \ r_N+1)$ is reduced, meaning that the product of the first $k$ transpositions has length $k$, then $\sigma_{r_1} \sigma_{r_2} \cdots \sigma_{r_N} \neq 0$. Now notice that $${\Big (}(1 \ 2)(3\ 4)(5\ 6) \cdots (2\ 3)(4\ 5)(6\ 7) \cdots {\Big )}^{\lfloor n/2 \rfloor}$$ is reduced.<|endoftext|> TITLE: Constructing quintic number fields with certain splitting behaviour QUESTION [6 upvotes]: I am looking for number fields $K$ which satisfy the following properties: $[K:\mathbb{Q}]=5$. The Galois closure of $K$ has Galois group $S_5$. For each prime $p$ which ramifies in $K$, there exists a prime ideal $\mathfrak{p}$ of $K$ of inertia degree and ramification index $1$ above $p$, i.e., we have $$(p) = \mathfrak{p} \cdot \prod_{i=1}^r \mathfrak{p}_i^{e_i},$$ with $N(\mathfrak{p}) = p$. It is well-known that one can construct infinitely many fields satisfying conditions 1 and 2 using Hilbert's irreducibility theorem. It is the last condition 3 which is the most important one, and says something like the ramification of $K$ is very mild. I have been able to find such fields by looking at databases of number fields. For example, one such field is given by $$t^5-t^4-5t^3+5t^2+2t-1 =0.$$ This has discriminant equal to $101833$, which is prime. One checks that we have the factorization $$(101833) = \mathfrak{p}_1^2 \mathfrak{p}_2\mathfrak{p}_3\mathfrak{p}_4,$$ where each ideal has norm $101833$. Do there exist infinitely many such fields? I'm possibly willing to weaken condition 2 to ask instead that the Galois group is a solvable (but still non-abelian) subgroup of $S_5$, if it helps. However, in my application I cannot remove condition 1. REPLY [5 votes]: The point of this answer, building on a comment by "so-called friend Don", is to point out that behavior has density $0$ among all quintic extensions. My guides here are the introductory portions of The geometric sieve and the density of squarefree values of invariant polynomials and Mass Formulae for Extensions of Local Fields and Conjectures on the Density of Number Field Discriminants, both by Bhargava. Let $k$ be a field. Recall that an etale $k$-algebra is a direct sum of finitely many finite separable extension fields of $k$. (We'll be in characteristic zero, so you can ignore the adjective separable.) We'll write $\mathrm{Aut}(K/k)$ for the automorphism group of $K$ preserving $k$; this is a finite group. Let $K$ be a degree $p$-adic extension of $\mathbb{Q}_p$ of degree $n$. As I understand the philosophy of these papers, the probability of that the $p$-adic completion of a degree $n$ number field will be isomorphic to $K$ is supposed to be $$\frac{p-1}{p} \frac{1}{\mathrm{Disc}_p(K)} \frac{1}{\# \mathrm{Aut}(K/\mathbb{Q}_p)}.$$ Moreover, these probabilities for distinct primes are supposed to be independent. For $K$ an etale $\mathbb{Q}_p$ algebra, we write say that $K$ has symbol $(f_1^{e_1}, f_2^{e_2}, \ldots, f_r^{e_r})$ if $K \cong \bigoplus_{i=1}^r K_i$ where $K_i/\mathbb{Q}_p$ is a field extension of ramification degree $e_i$ and residue field extension of degree $f_i$. A lemma in the Mass Formula paper (Prop 2.1) allows us to group together any place where we sum over the set of all etale $\mathbb{Q}_p$-algebras with a given symbol. So, our desired Euler factor is: The unramified extensions These are symbols where all the $e_i$ are $1$; namely $(1,1,1,1,1)$, $(2,1,1,1)$, $(3,1,1)$, $(4,1)$, $(5)$. We compute: $$\frac{p-1}{p} \left( \frac{1}{120} + \frac{1}{12} + \frac{1}{8} + \frac{1}{6} + \frac{1}{4} + \frac{1}{5} \right) = \frac{p-1}{p}.$$ It is not a coincidence that the sum came out to $1$; see equation (2.4) in the Mass formula paper. The other symbols It is also okay to have symbol $(1^2, 1,1,1)$, $(1^2, 2,1)$, $(1^2, 1^2, 1)$ or $(2^2, 1)$. Each of these symbols corresponds to several possible etale $\mathbb{Q}_p$-algebras (for example, for $p$ an odd prime, there are $2$ ramified quadratic extensions of $\mathbb{Q}_p$, so $(1^2, 1,1,1)$ describes two options), but Prop 2.1 in the Mass formula paper means we don't have to think about this. I compute that the respective terms are $$\frac{p-1}{p} \left( \frac{1}{p} \left( \frac{1}{6} + \frac{1}{2} \right) + \frac{1}{p^2} \left( \frac{1}{2} + \frac{1}{2} \right) \right)$$ Putting everything together, our Euler factor is $$\frac{p-1}{p} \left( 1+\frac{2}{3p} + \frac{1}{p^2} \right) = 1-\frac{1}{3 p} + \frac{1}{3 p^2} - \frac{1}{p^3}.$$ The point is that $$\prod_p \left( 1-\frac{1}{3 p} + \frac{1}{3 p^2} - \frac{1}{p^3} \right) =0.$$ Now, Bhargava's paper doesn't directly allow us to us the prudct in this manner because his best theorem, Theorem 1.3 in the sieve paper, only applies when, for all sufficiently large $p$, we include all the terms with discriminant $1$ or $p$. However, let $p_N$ be the probability that the ramification is as you wish for all $p < N$. Then Bhargava's Theorem 1.3 does show that $p_N$ exists and equals $\prod_{p TITLE: heat kernel on n-sphere QUESTION [15 upvotes]: I'm interested in diffusion, a.k.a. the heat kernel driven by the Laplace-Beltrami operator, on the $n$-dimensional sphere. There are lots of bounds showing that, for small times, it behaves in a way close to the heat kernel in $\mathbb{R}^n$: that is, the probability $p_t(\theta)$ that we have moved an angle $\theta$ away from the starting point, at time $t$, is bounded by a Gaussian of variance $t$. But I need some control on how the errors or leading constants in these bounds depend on $n$. The most precise estimates I've found (where the sphere has radius $1$) is due to Molchanov, $$ p_t(\theta) \sim \frac{\mathrm{e}^{-\theta^2/2t}}{(2\pi t)^{n/2}} \left( \frac{\theta}{\sin \theta} \right)^{(n-1)/2} $$ This is the first term of an asymptotic series, and if I understand his paper correctly, for fixed $n$ the next term would give a multiplicative error of $1+O(t)$. But does the constant hidden in $O(t)$ grow rapidly with $n$? Another family of bounds gives $$ p_t(\theta) \le C \,t^{-n/2} \,\mathrm{e}^{-\theta^2/(4+\delta)t} $$ Any constant value of $\delta$ would be fine with me. But we need $C$ to decay roughly as $(4\pi)^{-n/2}$ to match the normalization of the flat-space Gaussian. All we really need for our application is the following. If we think of the heat kernel as a stochastic process, let $\theta$ be the angular distance from the initial point. We know that for small enough $t$ we have $$ \mathbb{E}(\theta) \le C' \sqrt{nt} $$ for some constant $C'$, as it would be in flat space; but we need to know this holds for all $t$ up to $1/n$ or so. In other words, we need to know that the error term doesn't do something horrible like $$ \mathbb{E}(\theta) = C' \sqrt{nt} + O(2^n t) \, . $$ This would follow, for instance, from a bound on the error term in Molchanov's estimate above, although this might be overkill. Thanks! Cris REPLY [7 votes]: Yes, it is true that $\theta$ on $S_n$ is dominated by $\theta$ on $\mathbb{R}^n$. Let $(B_t)$ be a Brownian motion on the sphere. The radial process $\theta_t=d(x,B_t)$ is a Jacobi process, that is a Markov process with generator $ L=\frac{n-1}{2} \text{cotan} (r) \frac{d}{dr} +\frac{1}{2} \frac{d^2}{dr^2 } $ Since $ \text{cotan} (r) \le \frac{1}{r}$, we deduce from the comparison theorem for stochastic differential equations that $ \theta_t \le \beta_t $ where $\beta_t$ is a Bessel process, that is a Markov process with generator $ L=\frac{n-1}{2r} \frac{d}{dr} +\frac{1}{2} \frac{d^2}{dr^2 } $ The Bessel process is the radial part of the Euclidean Brownian motion in $\mathbb{R}^n$, so you get the desired stochastic domination.<|endoftext|> TITLE: Noncommutative HKR theorem QUESTION [6 upvotes]: What is the analog of HKR theorem in the noncommutative world? Recall that the well-known theorem by Hochschild-Kostant-Rosenberg says that for a smooth commutative algebra $A$ of finite type over a field $k$ (let's assume of characteristic zero) there is an isomorphism $$HH_\bullet(A)\simeq \Omega^\bullet(A)$$ where $HH_\bullet(A)$ denotes the Hochschild homology of $A$ with coefficients in itself, and $\Omega^\bullet(A):=\Lambda^\bullet(\Omega^1_k(A))$ is the space of algebraic differential forms on $A$, and $\Omega^1_k(A)$ denotes the module of Kahler differentials. There is a notion of smoothness for any associative algebra $A$, not necessarily commutative (formal smoothness). Moreover, one can define noncommutative differential forms $\Omega^\bullet_{nc}(A)$. So we have all the ingredients for noncommutative HKR, but I can't find the precise statement. I found the page on nLab, but the section is missing. Thank you very much for your help! Edit: As Mariano pointed out in the comments, $HH_\bullet(A)$ is not the same as $\Omega^\bullet_{nc}(A)$ for formally smooth $A$. I was indeed stupid here. Formal smoothness is equivalent to saying that $\Omega^1_{nc}(A)$ is a projective bimodule. But then the complex $0\to \Omega^1_{nc}(A)\to A\otimes A\to 0$ is a projective resolution of $A$, and so Hochschild homology $HH_n(A)$, defined as $Tor_n^{A^e}(A,A)$, vanish for $n>1$, while noncommutative differential forms do not vanish. But the question is still valid. What is the NC analog that the nLab page is talking about? REPLY [2 votes]: Maybe the following paper helps: Andreas Cap, Andreas Kriegl, Peter W. Michor, Jiri Vanzura: The Frölicher-Nijenhuis bracket in non commutative differential geometry, Acta Math. Univ. Comenianae 62(1993), 17--49 pdf<|endoftext|> TITLE: nth term in the Baker-Campbell-Hausdorff formula QUESTION [10 upvotes]: I am trying to prove a result for which I need the nth term of the Baker-Campbell-Hausdorff formula. I came at this particular result (which is not of significance for the question, but mentioning for context) by hypothesizing and using the first few terms of the Baker formula to verify. In order to prove my result rigorously, I think I need the nth term of the Baker formula. Is there an expression for that - I could not find it through online research. ? It would also help if I could find a proof of the BCH formula which is based on recursion i.e. if I could see how the nth term relates recursively to the (n-1)th term. REPLY [10 votes]: The Dynkin formula is somewhat cumbersome. Maybe a better choice is Goldberg's version http://projecteuclid.org/euclid.dmj/1077466673 In the commutator form Goldberg's result is reformulated in http://www.ams.org/journals/proc/1982-086-01/S0002-9939-1982-0663855-0/ (Cyclic relations and the Goldberg coefficients in the Campbell-Baker-Hausdorff formula, by Robert C. Thompson). By the way an interesting early history of the Baker-Campbell-Hausdorff-Dynkin formula can be found in http://link.springer.com/article/10.1007%2Fs00407-012-0095-8 (The early proofs of the theorem of Campbell, Baker, Hausdorff, and Dynkin, by R. Achilles and A. Bonfiglioli). REPLY [6 votes]: A completely explicit formula is obtained in a nice way in Bourbaki, Lie Groups and Lie Algebras, ch. 2, §6.<|endoftext|> TITLE: When is a Newton basin fractal continuously determined by the roots of its polynomial? QUESTION [9 upvotes]: Newton basin fractals are visualizations of the Julia sets of functions of the form: $$f_p(z) = z - p(z)/p'(z)$$ where $p$ is a complex polynomial. My question is: When is the Julia set, $J(f_p)$, continuously determined by the roots of $p$, in the sense of the Hausdorff metric? I have convinced myself that it is a necessary condition that the roots of $p$ be simple, and I'm happy to elaborate on my reasoning, but I suspect that the statement is either obvious (or obviously false) to experts. (Update: It turns out that it is obviously true.) B. Krauskopf and H. Kriete proved that: If a sequence of meromorphic functions $f_n$ converges uniformly on compact sets to a limit function $f$, and $F(f)$ is a union of basins of attracting periodic orbits, and $\infty\in J(f)$, then $J(f_n)$ converges to $J(f)$ in the Hausdorff metric. And at first glance that theorem looks like a great starting point for showing that the simple roots condition is sufficient. Unfortunately it is not the case that $F(f_p)$ is a union of attracting basins whenever $p$ has simple roots. Is anything more known about this? Perhaps there is some other condition (not merely simple roots) that makes $F(f_p)$ a union of attracting basins. I expected for this to be a well studied problem. An answer to the related question of "When is convergence of Newton-Raphson iteration not sensitive to perturbation of roots?" seems like it would have useful applications. But I have not found much beyond results like Krauskopf and Kriete's theorem, above, which don't quite apply. Edit: To give the question a little more context and motivation: Here is a program to demonstrate that in some sense, perhaps not exactly the way I've phrased it, a Newton basin fractal may transform continuously under changes to the roots of its polynomial. In the demonstration (which requires a WebGL enabled browser) the Newton basin fractal for the polynomial $z\to (z^2 - 1)(z - \lambda)$ is rendered in a 10x10 box centered at the origin. Clicking and dragging updates $\lambda$ and the resulting basins. Edit 2: I've updated the demonstration in an attempt to get more information onto the screen. Now the usual Newton basin rendering is blended with a rendering of the parameters $\lambda$ such that the sequence of iterates starting at the critical point converges very slowly. These are the parameters for which the Fatou set might include components that are not basins of attracting periodic orbits, preventing application of Krauskopf and Kriete's theorem. The roots are highlighted in yellow and the critical point is highlighted in cyan. It's easy to see that the Julia set experiences a discontinuous transformation when any two roots overlap, and that when $\lambda$ is in the "problem area" the critical point is sort of pressed towards the Julia set (which makes sense given how the problematic parameters are determined in the first place). But even as I move $\lambda$ through the problematic region I don't see the Julia set transforming in a discontinuous way, certainly nothing as obviously discontinuous as the transformations that occur when two roots overlap. I wonder if the Julia set actually is continuously determined by $\lambda$ in this region, even if the theorem above isn't quite enough to prove it. REPLY [7 votes]: The question of when the Julia set of a function depends continuously on the parameters is well-studied. For the case of polynomials, see Douady, Does a Julia set depend continuously on the Polynomial? Proceedings of Symposia in Applied Mathematics, 49, 91–135, 1994. This implies that the answer for your question is negative. More precisely, consider the bifurcation locus of the family that you are looking at; i.e., the set of unstable parameter value. (It is well-known that such parameters exist; e.g. take points on the boundaries of little Mandelbrot set copies in the parameter plane.) Now, in this locus, the set of parameters having a linearizable irrationally indifferent periodic point is dense. For these parameters, the periodic point is in the Fatou set, and hence has positive distance from the Julia set. On the other hand, arbitrarily small perturbations will lead to parameters with parabolic points, where the periodic point is in the Julia set. This proves that the Julia set does not depend continuously near such points. (The answer is also negative near parabolic parameters.) Of course, if you are in the complement of the bifurcation locus (say, with all critical points tending to attracting orbits), then the Julia set depends continuously, and even moves holomorphically. There is always a semicontinuous dependence of the Julia set, due to the fact that repelling periodic orbits are dense, and that these are stable under perturbations. In other words, under small perturbations the Julia set can get larger in a discontinuous way, but not smaller.<|endoftext|> TITLE: Detection of stable homotopy by K-theory spectra QUESTION [9 upvotes]: This is primarily a reference request. Does anyone know of any writing about algebraic K-theory spectra picking up elements in the stable homotopy groups of spheres in their Hurewicz image coming from the map $\mathbb{S}\to K(R)$? I have heard that $K_1(\mathbb{Z})$ is generated by $\eta$ but don't have any idea of even where to start. Is this something people have investigated at all? REPLY [8 votes]: I would say that the historically correct place to start is Quillen's letter to Milnor on the image of $(\pi_i O \to \pi_i^s \to K_i\mathbb{Z})$, published in Springer LNM 551 (1976). There Quillen proved that most of the image of $J$ in $\pi_*^s$ is detected in $K_*\mathbb{Z}$, including the honorary classes $\mu_{8k+1}$ and $\eta\mu_{8k+1}$ in degrees $8k+1$ and $8k+2$ that are detected in $\pi_* KO$ but are not in the image from $\pi_iO$. The cases not covered by Quillen were the multiples $\eta \alpha_{8k-1}$ and $\eta^2 \alpha_{8k-1}$ in degrees $8k$ and $8k+1$, where $\alpha_{8k-1}$ generates the image-of-$J$ in degree $8k-1$ for $k\ge1$. Waldhausen showed that these classes map to zero in $K_*\mathbb{Z}$, in Corollary 3.6 of his paper "Algebraic $K$-theory of spaces, a manifold approach" CMS Conf. Proc. 2 (1982). The cokernel-of-$J$ maps to zero in $K_*\mathbb{Z}$; see Mitchell's "The Morava $K$-theory of algebraic $K$-theory spectra" $K$-Theory (1990). Later on, these results became direct consequences of the proven Lichtenbaum-Quillen conjecture, as formulated in terms of etale $K$-theory by Dwyer-Friedlander in "Étale K-theory and arithmetic" Bull. AMS (1982). See also the discussion about $JK(\mathbb{Z})$ in Bokstedt's "The rational homotopy type of $\Omega Wh^{Diff}(*)$" in Springer LNM 1051 (1984). For any other (discrete) ring $R$, the unit map $S \to K(R)$ factors through $K(\mathbb{Z})$, so nothing more can be detected that way. For strict ring spectra $R$ a larger part of $\pi_*^s$ can be detected, up to all of $\pi_*^s$ in the case $R = S$, since $S$ splits off from $K(S) = A(*)$. As an intermediary example, for $p\ge5$ and $R = \ell$, the $p$-local Adams summand in connective topological $K$-theory, $\pi_*( K(\ell) ;\mathbb{Z}/p )$ detects $\beta'_1 \in \pi_*( S; \mathbb{Z}/p )$, as Ausoni and I showed in "Algebraic $K$-theory of topological $K$-theory" Acta Math (2002).<|endoftext|> TITLE: Examples of manifolds whose second Stiefel-Whitney satisfies a nontriviality condition QUESTION [6 upvotes]: I'm looking for examples of pairs $(M,L)$ where $M$ is a symplectic manifold, $L$ a (closed, connected) Lagrangian submanifold, such that the second Stiefel-Whitney of $L$, $w_2(TL)$, evaluates nontrivially on the image $G \subset \pi_2(L)$ of the boundary homomorphism $\pi_3(M,L) \to \pi_2(L)$. Note that such $L$ must have dimension at least $4$, since any $L$ of dimension at most $3$ is $\text{Pin}^-$, which means in particular that $w_2(TL)$ vanishes on $G$. More generally, if $L$ is so-called relatively $\text{Pin}^\pm$, then $w_2(TL)$ vanishes on $G$ as well ($L$ is called relatively $\text{Pin}^+$ if $w_2(TL)$ is in the image of the restriction morphism $H^2(M;\mathbb Z_2) \to H^2(L;\mathbb Z_2)$; it is relatively $\text{Pin}^-$ if the same is true of $w_2(TL) + w_1^2(TL)$). Indeed, suppose $L$ is relatively $\text{Pin}^+$. Then $w_2(TL)|_G$ factors through $\pi_2(L) \to \pi_2(M)$ and therefore it vanishes since of course $\pi_3(M,L) \to \pi_2(L) \to \pi_2(M)$ is zero. The same argument works if $L$ is relatively $\text{Pin}^-$ since $w_1^2(TL)$ vanishes on spheres in $L$. Lastly, I would really like to obtain such examples where $L$ is moreover assumed to be monotone with minimal Maslov at least $2$. REPLY [4 votes]: Audin, Lalonde and Polterovich have a construction (Corollary 1.2.5 in this book) which produces a Maslov-two monotone Lagrangian embedding of $M^n \times T^m$ into $\mathbb{C}^{n+m}$ for any $m \ge 1$, given that $M^n$ admits a Lagrangian immersion into $\mathbb{C}^n$. The latter condition is equivalent to $TM^n \otimes \mathbb{C}$ being a trivial unitary bundle. All we have to do is thus to find our favourite non-spin manifold with trivial complexified tangent bundle. I would go for $M=\mathbb{C}P^2 \sharp \overline{\mathbb{C}P^2}$.<|endoftext|> TITLE: function field analogy and global/absolute geometry QUESTION [22 upvotes]: The "function field analogy" seems to be a topic that is considerably bigger than any one existing writeup conveys. There are several old question on MO and and MathSE that ask for details. One of the most prominent replies has been given here and consists simply of a pointer to what is maybe the one single place that sets out to produce a table listing some key statements, namely Poonen's lecture notes (pdf, see section 2.6). While such a table is much needed (as witnessed conclusively by 35 upvotes, and counting) it seems to me that there'd eventually be much more to put in it in order to do any justice to the topic. Notably the table ought to have a third column besides arithmetic geometry over number fields and function fields, namely the column for complex curves/Riemann surfaces, which brings the geometric Langlands correspondence into the picture. And it would be good for such a table to be hyperlinked, since there'd be so much to say on each single one of its entries. In short, that motivated me to start to try to compile a function field analogy -- table on the $n$Lab. I got somewhere, but there is still some way to go. I have some questions, too. (And I would like to stress that nothing in this table is meant as claim of mine, all is me trying to reproduce what is known. If there is anything that seems outrageous, then this is a mistake on my part and I will do my best to fix it.) So in general one question is: does this look about right? And: what seem to be glaring omissions. (I am aware of some, but I hope to hear of those that I am not aware of yet.) But I also have this slightly more concrete question: From one point of view the search for $\mathbb{F}_1$ is the search for a systematic theory that would promote the function field analogy from an analogy to a well controled base change away from $\mathrm{Spec}(\mathbb{F}_1)$. That idea is expressed for instance here in another previous MO discussion of this point. However, when I scan the literature on $\mathbb{F}_1$ then I see plenty of discussion of zeta functions over these various bases, but little about the bulk rest of the function field analogy table. Does existing $\mathbb{F}_1$-theory have much to say here? For instance the very first line of the (either!) function field analogy table states that $\mathbb{Z}$ is analogous to $\mathbb{F}_q[x]$ and to some extent so "as $q \to 1$". So from the point of view of the function field analogy it would seem that the central request on any theory of $\mathbb{F}_1$ would be to give rise to a truth that reads in symbols like "$\mathbb{Z} \simeq \mathbb{F}_1[x]$", whatever it is that makes this true. I see that some people do expect just this from a theory of $\mathbb{F}_1$ (for instance in the first footnote here). However, what I have seen as actual proposals for $\mathbb{F}_1[x]$ seems to be headed in a rather different direction. Unless I am missing something, of course, and my question is: am I? Which approach to $\mathbb{F}_1$ should I look at for function field analogy purposes beyond (and that probably means: prior to) zeta functions? Finally to close an already long and vague question with something even broader, just for those who might enjoy it (all others please stop reading): what I am eventually after is an answer to my old MO question p-Adic String Theory and the String-orientation of Topological Modular Forms (tmf). Namely there are so many hints already in string theory that a function field analogy base-changing us from complex curves to arithmetic geometry over $\mathbb{F}_1$ plays a role, that I'd like to have a good enough mathematical theory of the analogy that would allow to put these hints together to a nice statement. For instance looking at application or mirror symmetry to geometric Langlands as in Gerasimov-Lebedev-Oblezin 09 makes one want to ask: "What is a sigma model in $\mathbb{F}_1$-geometry?" I am wondering how far $\mathbb{F}_1$-theory -- or maybe global analytic geometry? -- may have gotten in this respect, or what the prospect seems to be. Is this even in the line of sight of present research into $\mathbb{F}_1$? If not: what is, if anything? REPLY [2 votes]: I think the biggest thing missing from this table is the geometric picture over function fields. Almost everything under "complex Riemann surface" on your table makes sense over for algebraic curves over an arbitrary field. But of course if you go so far as to split the function field case into two columns, the first and second column of your table would be almost identical, as would the third and fourth column, which would be wasteful. However this does represent how the function field analogy actually works. Usually passing from number fields to function fields is a simple bookkeeping step (changing notation for the same concepts), and moving geometry on curves from one field to another is again bookkeeping, but passing from arithmetic to geometry over a single field requires some insight, although usually a simple one. Second, I think many mathematicians working in number theory are skeptical of $\mathbb F_1$-theory and prefer to keep it an analogy. One reason is that the analogy can fail if you look in the wrong places. If $\mathbb Z= \mathbb F_1[t]$, what is $\mathbb F_1[t^2]$? Also consider the zeta functions of these fields. The Riemann zeta function has infinitely many zeroes, while the zeta function of $\mathbb F_q(t)$ has none. To study the zeros of the Riemann zeta function in the function field model, mathematicians pass to the limit of infinitely large $g$. James Borger also pointed out the issues with the discriminant. So clearly in translating questions between the function field and number field worlds some discretion is necessary. Similar to this all currently existing $\mathbb F_1$-theories have some kind of problem where they don't fit with our intuitive idea of what an $\mathbb F_1$-theory should look like - in fact I believe that is known to be contradictory. Certainly good mathematical work can be done by finding clever workarounds and deftly avoiding these problems, and it might go so far as to solve completely or make progress on otherwise intractable number field problems. Third let me say that, not knowing much about string theory, it seems to me that your other question should not run into these problems. Indeed you seem to be concerned only with (various sophisticated forms of) analysis and integration over these fields. Identities of integrals and things like that tend to translate very well among different contexts once you've found the right way of looking at them - I have seen many examples of this. But I don't have any idea what to do in your particular problem.<|endoftext|> TITLE: Hecke-module structure implicit in definition of automorphic forms in Borel-Jacquet's Corvallis article QUESTION [12 upvotes]: Let $G$ be a connected reductive group over a number field $F$, $G_\infty=\prod_{v\mid\infty} G(F_v)$, $\mathbf{A}$ the adèles of $F$, $\mathbf{A}_f$ the finite adèles of $F$. Fix a maximal compact subgroup $K$ of $G_\infty$. We have Hecke algebras $H_\infty$ of $G_\infty$ (relative to $K$) and $H_f$ of $G(\mathbf{A}_f)$. The former is the one used by Flath and Borel-Jacquet in their Corvallis Volume 1 articles (not the one from Jacquet-Langlands) and is described in great detail in Chapter 1 of Knapp-Vogan's giant book on cohomological induction. It is the convolution algebra of left $K$-finite distributions on $G_\infty$ with support in $K$ (see Appendix B of Knapp-Vogan for distributions on manifolds). The algebra $H_f$ is the convolution algebra of $\mathbf{C}$-valued, locally constant, compactly supported functions on $G(\mathbf{A}_f)$. The important facts about these $\mathbf{C}$-algebras are that they have an "approximate identity" (a certain collection of idempotents) and a resulting notion of "smooth module" (see Chapter 1 of Knapp-Vogan for the definitions, although they use the term "approximately unital module"). One can prove that (1) smooth $\mathbf{C}$-representations of $G(\mathbf{A}_f)$ are the same as smooth $H_f$-modules (which are automatically $\mathbf{C}$-vector spaces because $H_f$ has an "approximate identity"), and (2) smooth $H_\infty$-modules are the same as $(\mathfrak{g},K)$-modules, where $\mathfrak{g}=\mathrm{Lie}(G_\infty)$, and the definition of $(\mathfrak{g},K)$-modules I'm using is from Knapp-Vogan's book, or equivalently, Wallach's Corvallis Volume 1 article. In Borel-Jacquet's Corvallis article, they define an automorphic form on $G(\mathbf{A})$ by a list of properties (see the top of page 195 of Corvallis Volume 1 for the properties). In particular, automorphic forms are complex-valued functions on $G(\mathbf{A})$ which are smooth in the sense that they are continuous, $C^\infty$ in the Archimedean variable, and locally constant in the finite adelic variable. My question is as follows. The conditions imposed on such a function by B-J to make it an automorphic form involve a right action of the Hecke algebra $H:=H_\infty\otimes_\mathbf{C}H_f$. For this to make sense, automorphic forms should live in some large $\mathbf{C}$-vector space which is simultaneously an $H_\infty$-module and an $H_f$-module, i.e., simultaneously a $(\mathfrak{g},K)$-module and a smooth $G(\mathbf{A}_f)$-representation. What space is this? I think that the right regular representation of $G(\mathbf{A})$ on the space of all $\mathbf{C}$-valued functions on $G(\mathbf{A})$ preserves the space $C^\infty(G(\mathbf{A}))$ of smooth functions defined above, and we must at least take the subspace of $K$-finite vectors (in the sense of Chapter 1 of Knapp-Vogan, which includes a continuity condition for the $K$-action) to get a $(\mathfrak{g},K)$-module structure (property (b) of B-J implies that automorphic forms are right $K$-finite anyway). We should also take the $G(\mathbf{A}_f)$-smooth vectors (I don't think smoothness is automatic, despite the terminology, again property (b) forces smoothness of automorphic forms in the sense of representations of locally profinite groups). Because the $K$ and $G(\mathbf{A}_f)$-actions commute, I think the order in which we take these subspaces doesn't matter, and we end up with a $K$-finite representation which is also a smooth representation of $G(\mathbf{A}_f)$, hence an $H(K)\otimes_\mathbf{C}H_f$-module, where $H(K)$ is the Hecke algebra of $K$ (left and right $K$-finite distributions on $K$). We'll have a differentiated action of $\mathfrak{k}=\mathrm{Lie}(K)$, but how do we get one of $\mathfrak{g}$? It doesn't quite make sense to ask if $K$-finite vectors in our space are $C^\infty$-vectors for the $G$-action, because there is no topology on the space (if all the vectors were $G$-finite, we could differentiate the $G$-action, but this is not going to be the case). If I'm on the right track, my question then boils down to: how do we get an action of $\mathfrak{g}$ on the space of $K$-finite, $G(\mathbf{A}_f)$-smooth vectors in $C^\infty(G(\mathbf{A}))$? This kind of thing has often confused me because, for example, $K$-finiteness of automorphic forms is part of the definition, but the way it is phrased in B-J (in terms of being fixed by a standard idempotent in $H$) only makes sense if we have an action of $H$, and if we need to take $K$-finite vectors to get the $H$-action in the first place, the logic is out of order! EDIT: It occurs to me that maybe we can differentiate the $G_\infty$-action from the very beginning (in $C^\infty(G(\mathbf{A}))$), using the usual formula. Namely, for $g=(g_\infty,g_f)\in G_\infty\times G(\mathbf{A}_f)$, and $X\in\mathfrak{g}$, if $f\in C^\infty(G(\mathbf{A}))$, then the function $$t\mapsto f(g_\infty\exp(tX),g_f):\mathbf{R}\to\mathbf{C}$$ will be smooth, and we can define $(Xf)(g_\infty,g_f)$ to be the derivative of this function at $t=0$. Is this the correct way to get the $\mathfrak{g}$-action and hence the $(\mathfrak{g},K)$-module structure on the subspace of $K$-finite vectors? If so, is there a less ad hoc way to get the $\mathfrak{g}$-action? The fact this isn't an actual vector-valued derivative, like what you'd have with a smooth vector in e.g. a Banach space representation of $G_\infty$, is somehow unsatisfying to me, but perhaps I shouldn't expect anything like that since the point of view here does not involve a locally convex topology on the space $C^\infty(G(\mathbf{A}))$...of which I'm aware. REPLY [4 votes]: The definition on the top of page 195 in B-J's article (in Corvallis 1) is ok, because $f\ast\xi$ is well-defined for any smooth $f:G(\mathbf{A})\to\mathbf{C}$ and any $\xi\in H$. It suffices to verify the claim for pure tensors $\xi=\xi_\infty\otimes\xi_f$, in which case $f\ast\xi$ is $f\ast\xi_\infty$ convolved with $\xi_f\in H_f$ in the usual sense. The function $f\ast\xi_\infty$ exists by the smoothness assumption, while the integral defining the convolution converges, because $\xi_f:G(\mathbf{A}_f)\to\mathbf{C}$ is compactly supported and locally constant by definition.<|endoftext|> TITLE: Serious introduction to the Langlands program for nonspecialist QUESTION [5 upvotes]: I recently became interested in the Langlands program and hope to learn more. For context, I am an analytic number theorist but have some light background in algebraic number theory and modular forms. In particular from Neukirch's Algebraic Number Theory, Diamond-Shurman's A First Course in Modular Forms, and Shimura's Modular Forms: Basics and Beyond. Aside from these, I have very little in way of prerequisites except the usual ones one would expect of any analytic number theorist. What are the best introductions to the Langlands program for someone with limited prerequisites? REPLY [2 votes]: Gelbart's AMS article Introduction to the Langlands program is a fairly standard place to start, for a broad overview of what the global conjectures look like and why we're interested in them. There's also a book by the same title by Bernstein, Gelbart et al, which gives more detail. There are also some notes by Knapp that I don't seem to have anymore, but you should be able to find them easily enough. There's an awful lot of different topics that all converge in the "Langlands program" -- to understand "everything" you're going to have to have good command of a pretty intimidating list of topics -- so the best thing to do (at least, what I found to be the best) is try and get a broad overview while taking a lot for granted, and then learn more about the things that particularly interest you. However, if you by "Langlands program" you really just mean "Langlands reciprocity for Galois representations and the local correspondence" then those references will be just fine. Once you've got a basic understanding of what the conjectures say (or at least the global conjecture for $GL_2$) the answer depends on what you mean by a "serious" introduction. If you mean some level of formality, but skipping over lots of technical details and proofs then the two references above should be fine. If you want to actually understand things at a technical level (and be able to follow the proofs where they exist) then you've got a lot of work to do. I don't really know much about the global conjectures at a technical level, but for the local conjectures a good start is Bushnell-Henniart's Local langlands conjecture for $GL_2$.<|endoftext|> TITLE: Not-lonely runners QUESTION [17 upvotes]: The lonely runner conjecture has several formulations. They all involve a number $n$ runners running on a circular track, each with a different speeds, and the conjecture is that each runner is eventually "lonely" in a technical sense. My question is essentially the obverse, and hopefully much easier; and so perhaps already answered: Q. Let $n$ runners start at random positions on a circular track, running at different speeds. Is it the case that at some time $t$, it is guaranteed that all the runners will lie within some semicircle? Or within some three-quarters circle? Or within some $(1-\epsilon)$ of a circle for $\epsilon > 0$? Informally, will the runners eventually cluster? Surely they will "cluster" within 99% of the track length? (Added 23Jul14). I see now that a version of this question is posed as an open problem by A. Dumitrescu and C.D. Toth in SIGACT News 45(2) p.71 (ACM link). REPLY [8 votes]: I'll use Christian Remling's notation: We want to show that we can always find an empty interval of size $\epsilon_n$ among any set of $n$ runners. The Masked Avenger gives an argument for $\epsilon_n =1/n$, and suggests that $\epsilon_n=2/n$ may be close to optimal by an argument I can't follow. Assuming the lonely runner conjecture for $n+1$, we get an argument for $\epsilon_n =2/n+1$ - just add one more runner, randomly placed, and wait until they're lonely. In terms of actually provable statements, I was only able to get a very slight improvement - $\epsilon_n = 1/n + c/n^{3/2}$ for some $c$. For this argument, we may assume without loss of generality that the speeds are integers. Consider each runner's position on the circle as a unit complex number function of time, and add them up. This gives a periodic function whose Fourier series $\hat{f}(n)$ is $0$ if $n$ is not the speed of the runner and, if $n$ is the speed of a runner, the starting position of that runner. So $\hat{f}(n)$ has $L_2$ norm $\sqrt{n}$, and $f(t)$ has $L_2$ norm $\sqrt{n}$. This means at some time it must take a value at least $\sqrt{n}$. We can easily see that if all the gaps are at most $1/n+\delta$, each runner can be no further than $n/\delta$ from a perfectly even position, meaning summing over the runners gives $\sqrt{n}\geq n^2/\delta$, with some constant thrown in there, which gives our result. On the other hand we get $\epsilon_n < 2/\sqrt{\log n}$ from Christian Remling's argument. So we have an exponential gap to close. EDIT: I might be able to improve Christian Remling's argument by the probabilistic method. Let the runners have speeds $1$ through $n$, with random start times. We need to rule out the gaps of size $\epsilon$ starting at the point $x$ and the time $y$ for $x,y$ in the unit square. Cover the unit square in $a \times b$ bricks. What is the probability of the runner of speed $i$ ruling out a whole brick? To rule it out at a given time they need to be in an area of size $\epsilon-a$, which they are in for a time of $(\epsilon-a)/i$ on $i$ different occasions in the unit interval. Each of these occasions has a probability of $(\epsilon-a)/i-b$ of filling the whole brick, for a probability of $\epsilon-a-bi \geq \epsilon-a-bn$. Let $p=\epsilon-a-bn$. Then each runner has probability of $p$ of covering the whole brick, so each brick has a probability of $(1-p)^n$ of going uncovered. To have a positive probability of all bricks covered, we need: $$(1-p)^n a^{-1} b^{-1} <1$$ $$ n (\log ( 1-p) ) < \log a + \log b$$ $\log(1-p)$ is about $- p$ so we have $$ p< \frac {- \log a - \log b}{ n} $$ $$ \epsilon -a - nb < \frac{ - \log a - \log b }{n} $$ Putting $a=1/n$, $b=1/n^2$ we get $\epsilon_n < 3 \log n/n+ 2/n = (3 + o(1) )\log n/n$. This gives only a logarithmic distance between upper and lower bounds.<|endoftext|> TITLE: Borel-Serre compactification of $\mathbb{H}^3 / SL_2(\mathcal{O}_K)$ QUESTION [16 upvotes]: Let $\mathbb{H}^3$ be the three-dimensional hyperbolic space. Let $K$ be an imaginary quadratic number field and $\mathcal{O}_K$ its ring of integers. Then $SL_2(\mathcal{O}_K)$ acts on $\mathbb{H}^3$ and the quotient $$ M=\mathbb{H}^3 / SL_2(\mathcal{O}_K) $$ is a non-compact hyperbolic manifold. I have heard several times that the Borel-Serre compactification of $M$ consists of adding elliptic curves with complex multiplication, as much as the class number of $K$. Does anybody know a good reference for this or can provide more details? I would be grateful REPLY [14 votes]: I do not know of any specific reference, but I will try to explain the situation a bit. For most of the stuff, one does not actually need references besides the paper of Borel-Serre defining the Borel-Serre compactification. Maybe the book of Elstrodt, Grunewald, Mennicke "Groups acting on hyperbolic space" contains some useful information, I currently do not have access to it to check. To figure out the structure of $\mathbb{H}^3/SL_2(\mathcal{O}_K)$, one can first compactify $\mathbb{H}^3$ by a $\mathbb{CP}^1$ at infinity. The action of $SL_2(\mathbb{C})$ on $\mathbb{H}^3$ can be extended to the $\mathbb{CP}^1$ at infinity; the action of $SL_2(\mathbb{C})$ on $\mathbb{CP}^1$ is given by Möbius transformations. Then there is a bijection between the cusps of $M=\mathbb{H}^3/SL_2(\mathcal{O}_K)$, the orbits of $SL_2(\mathcal{O}_K)$ acting on $\mathbb{CP}^1$, and the ideal class group of $\mathcal{O}_K$. A short explanation of this can be found in Tom Church's answer to this MO-question. A slightly longer explanation of this bijection between orbits and ideal classes is given in these notes of Keith Conrad (the link on Keith Conrad's page of notes did not seem to work). This explains why the boundary components of the Borel-Serre compactification are in bijection with the ideal class group. Now, to explain what the right boundary components should be, look at the definition of the Borel-Serre compactification. The cusps are related to the intersection of parabolic subgroups with $SL_2(\mathcal{O}_K)$. For example, the cusp for the trivial ideal class is stabilized by the subgroup of upper triangular matrices: $$ \left\{\left(\begin{array}{cc} \pm 1&\alpha\\0&\pm 1\end{array}\right)\mid \alpha\in\mathcal{O}_K\right\}. $$ [Edit: Note that, as pointed out by Ian Agol this is not true for $K=\mathbb{Q}(\sqrt{-1})$ and $K=\mathbb{Q}(\sqrt{-3})$. These number fields have additional units, therefore the stabilizers of cusps are bigger than written above.] The Borel-Serre compactification (of $\mathbb{H}^3$) adds for this cusp a copy of $\mathbb{R}^2\cong \mathcal{O}_K\otimes_{\mathbb{Z}}\mathbb{R}$ on which the above stabilizer of the cusp acts via translations by elements of $\mathcal{O}_K$. The Borel-Serre compactification of the quotient $\mathbb{H}^3/SL_2(\mathcal{O}_K)$ then adds the quotient of $\mathbb{R}^2$ by the translation group $\mathcal{O}_K$ - but this is exactly a $2$-torus. Of course, there is a natural identification of $\mathbb{R}^2$ with $\mathbb{C}$, such that the corresponding $2$-torus has an induced complex structure which makes it an elliptic curve with complex multiplication by $\mathcal{O}_K$. The same happens at all the other cusps, but the stabilizer groups of these cusps are more difficult to write down explicitly. But up to conjugacy in $SL_2(K)$, everything is as in the case of the trivial ideal class. [Edit: As in the comment of Ian Agol, for $K=\mathbb{Q}(\sqrt{-1})$ and $K=\mathbb{Q}(\sqrt{-3})$ the boundary is the orbifold quotient of $\mathbb{C}$ modulo the corresponding stabilizer group $$ \left\{\left(\begin{array}{cc} \beta&\alpha\\0&\beta^{-1}\end{array}\right)\mid \alpha\in\mathcal{O}_K,\beta\in\mathcal{O}_K^\times\right\}.] $$ However, I would consider the elliptic curve structure as somewhat more of an accident. Just like the quotients of symmetric spaces modulo arithmetic group actions may or may not be varieties (in the case at hand, the locally symmetric space fails to be an algebraic variety simply because it is $3$-dimensional and hence can not be a complex variety), the same is also true for the boundary of the Borel-Serre compactification. I think the natural thing to say is that the Borel-Serre compactification adds $2$-torus for each ideal class. [Edit: As explained in the comment of Ben Wieland, the complex structure is natural, induced from hyperbolic space. What I wanted to say is that for other types of arithmetic groups acting on other types of symmetric spaces, the Borel-Serre boundary does not necessarily have a complex structure (sometimes just for dimension reasons).]<|endoftext|> TITLE: The Jones polynomial at specific values of $t$ QUESTION [32 upvotes]: I've been calculating some Jones polynomials lately and I was just curious if there was a "physical" (or, rather, geometric) meaning to evaluating the Jones polynomial at a particular value of $t$. For example, if I take the Jones polynomial for the (right) Trefoil knot, I have $J(t) = t + t^3 - t^4$. Is there some way I can interpret $J(0)$? $J(1)$? I understand that the Jones polynomial is a laurent polynomial, so I don't expect $J(0)$ to make sense for a lot of knots (for example the left trefoil has $J(t) = t^{-1} + t^{-3} - t^{-4}$), but I thought it was worth asking. I also know that $J(t^{-1})$ gives the Jones polynomial of the mirror image knot. Is there a way to interpret $J(-t)$? $J(t^2)$? How about $J(t) = 0$? Edit to clarify what I mean when I say "physical meaning": Since the Jones polynomial is a link invariant, $J(0)$ is also a link invariant (if it exists). Does this invariant correspond to a property of the knot that you can visualise, such as, say, the linking number or the crossing number? REPLY [7 votes]: The volume conjecture predicts the existence limit of (a certain normalization of) the colored Jones polynomials evaluated at roots of unity (which is not known to exist), and that this limit is equal to the hyperbolic volume of the knot complement. This volume is uniquely defined and in some sense is a "physical" quantity. (Note: This doesn't literally answer the question, since the answer generalizes "Jones polynomial" to "colored Jones polynomial.")<|endoftext|> TITLE: History of the connection between Riemann surfaces and complex algebraic curves QUESTION [21 upvotes]: As noted in the question "Links between Riemann surfaces and algebraic geometry", there are strong connections between Riemann surfaces and algebraic geometry - for example, compact Riemann surfaces are equivalent to smooth complex algebraic curves. I wondered if anyone could give insights into the history of this relationship - for example, which was studied first, who first realised the connection, did the theories develop concurrently for any period of time, etc. Any individual insights, or links or book recommendations to find out more, would be greatly appreciated. REPLY [16 votes]: I have always felt this was due to Riemann himself: especially in: Theory of Abelian Functions, 1857. Of course the association of a Riemann surface to an algebraic curve is generally attributed to him, and there he also proves conversely there is a plane curve associated to a (compact) Riemann surface. He proves there too the Riemann part of the Riemann Roch theorem both for a Riemann surface, and explains how to prove it algebraically for a plane curve. The Roch part, provided later, is merely the addition of the residue theorem to the Riemann theorem, to compute a period integral as a residue evaluation. There are excellent translations of Riemann's works in a volume by Kendrick Press. http://www.kendrickpress.com/Riemann.htm for my account of Riemann's proof of his theorem, see pages 21-34 of: http://www.math.uga.edu/%7Eroy/rrt.pdf Let me clarify my use of the word "prove" in regard to Riemann's work. It is apparently true he made unjustified use of the Dirichlet principle in his argument that a compact Riemann surface has on it non zero entire differential forms and non constant meromorphic functions, and hence in his proof of the Riemann theorem for Riemann surfaces. However he explains how to simply write down such objects for plane curves, which by virtue of their representation allow the use of coordinate functions in this regard. The purely algebraic work of others, is sometimes pointed to as filling a gap in the rigor of Riemann's arguments for Riemann - Roch, but in the case of curves already assumed to be algebraic, such as those associated to algebraic function fields, Riemann's arguments do not require the Dirichlet principle for the existence theorems, and hence are already completely rigorous. Purely algebraic formulations of Riemann Roch, although unnecessary to render Riemann's arguments rigorous in the case of algebraic function fields over the complex numbers, do however seem important for allowing the theory to be generalized to fields other than the complex numbers. Briefly, the original Riemann Roch theorem over C, requires existence of g holomorphic differential forms where g is the topological genus, and the existence of meromorphic differentials of second kind associated to each point. Then the Riemann Roch formula for the number of meromorphic functions with given bounds on their poles is deduced from an argument via path integrals which are evaluated by Roch using residues. This original argument is completely rigorous for curves derived from algebraic functions fields over C. (I admit that although Riemann writes down the holomorphic forms on a plane curve, he only states that he could as easily write down the meromorphic ones, but does not do so.) In the algebraic setting one has the existence of forms and functions essentially by hypothesis, and usually one defines the genus in terms of the space of entire forms, and deduces the Riemann Roch formula by algebraic means, avoiding the use of path integrals. In one approach, a judicious use of duality allows one to exploit just the holomorphic differentials and to deduce the existence of meromorphic differentials of second kind after the fact. In the analytic case of a compact Riemann surface, in which the existence of either a non zero differential form or a non constant meromorphic function is non trivial, purely algebraic arguments do nothing to shore up the situation. So I feel the actual gaps in Riemann's work are on the analytic side, and were apparently filled in only later by those who supplied the analysis, such as Koebe, Hilbert, and Weyl.<|endoftext|> TITLE: Tight prime bounds QUESTION [16 upvotes]: This is a cross-post of this question on MSE. I would not usually do this, but have decided to in this case since it has had no responses having been posted as a bounty question. I did not delete the MSE question since so many users have starred it (and are apparently interested in seeing it answered) and not all of those users will have an MO account. If it receives a positive response here, I intend to link it to the MSE question for the benefit of those users. If however, this is objected to by MO users, or is not deemed 'research level', I shall respectfully remove it. The question is: Is it likely that $$\left|\operatorname{li}(n)-\sum_{k=1}^{\lfloor\log(n)\rfloor}\dfrac{\pi(n^{1/k})}{k}-\log(2)-\dfrac{1}{2}\right|<\dfrac{2\sqrt{n}}{e\log(n)}?$$ This is of course, almost identical to saying $$|R(n)-\pi(n)|<\dfrac{2\sqrt{n}}{e\log(n)}$$ where $R$ is the Riemann prime counting function, but this is a little too tight since this doesn't hold for $n=113$. Since $$ \sum_{k=1}^{\lfloor\log(n)\rfloor}\dfrac{\pi(n^{1/k})}{k}\approx\operatorname{li}(n)-\sum_{k=1}^{\infty}2\ \Re\left(\operatorname{Ei}\left(\rho_k\log\left(n\right)\right)\right)-\log(2) $$ and since $\pm\dfrac{2\sqrt{n}}{\Im(\rho_1)\log(n)}$ bounds $2\ \Re\left(\operatorname{Ei}\left(\rho_1\log\left(n\right)\right)\right)$ does it follow that $\pm\dfrac{2\sqrt{n}}{C\log(n)}$ will bound $\sum_{k=1}^{\infty}2\ \Re\left(\operatorname{Ei}\left(\rho_k\log\left(n\right)\right)\right)$ for some $C$ (assuming RH)? $C=e$ seems particularly tight. Is this a realistic bound to propose? REPLY [27 votes]: The inequality you propose would imply that $$ |\operatorname{li}(n)-\pi(n)| \ll \frac{\sqrt{n}}{\log n}. $$ On the other hand, Littlewood (1914) showed that this relation is false, in fact he proved that the left hand side divided by the right hand side is $\Omega(\log\log\log n)$. So the inequality you propose is false. Added 1. Littlewood's original paper is "Sur la distribution des nombres premiers", C. R. Acad. Sci. Paris, 158 (1914), 1869-1872. For a modern proof, see Theorem 15.11 in Montgomery-Vaughan's book "Multiplicative number theory I." (which is apparently the last result in the book). Added 2. The OP asked in a comment how the inequality above follows from his proposed bound. Here are the details. We clearly have (for $n>100$) $$ \operatorname{li}(n)-\pi(n) = S_1+\frac{\pi(n^{1/2})}{2} + S_2,$$ where $$S_1:=\operatorname{li}(n)-\sum_{k=1}^{\lfloor\log(n)\rfloor}\dfrac{\pi(n^{1/k})}{k}-\log(2)-\dfrac{1}{2}$$ $$S_2:=\sum_{k=3}^{\lfloor\log(n)\rfloor}\dfrac{\pi(n^{1/k})}{k}+\log(2)+\dfrac{1}{2}.$$ The second expression $S_2$ is a sum of less than $\log n$ terms, each less than $n^{1/3}$. Hence, by the triangle inequality, $ |S_2|\leq n^{1/3}\log n \ll \sqrt{n}/\log n$. Using also the OP's conjecture $S_1\ll \sqrt{n}/\log n$ we get, again by the triangle inequality, $$ |\operatorname{li}(n)-\pi(n)|\leq |S_1|+|\pi(n^{1/2})|+|S_2|\ll \frac{\sqrt{n}}{\log n}.$$ This, on the other hand, contradicts Littlewood's result from 1914.<|endoftext|> TITLE: In the definition of the Heegard Floer surgery exact triangle, what exactly is the correspondence between Whitney triangles and periodic domains? QUESTION [7 upvotes]: I'm reading Osváth-Szabó's notes on Heegard Floer homology, in particular about the surgery exact triangle. On page 14 (numbered 42 on the document), they describe an isomorphism between the space of homology classes of Whitney triangles $\pi_2(\mathbf{x},\mathbf{y},\mathbf{z})$ in $\mbox{Sym}^g(\Sigma)$ and $\mathbb{Z} \times \mathcal{P} $, where $\mathcal{P}$ denotes the group of periodic domains in $\Sigma$. I'm not sure if I understand this isomorphism correctly, or if there are some typos, or both. Here is how I understand the isomorphism works: Given two elements $\psi, \psi_0 \in \pi_2(\mathbf{x},\mathbf{y},\mathbf{z})$, we can associate domains $\mathcal{D}(\psi), \mathcal{D}(\psi_0)$, by taking $$\mathcal{D}(\psi) = \sum n_{z_i}(\psi) D_i,$$ where $D_i$ are the components of $\Sigma - \{\mathbf{\alpha} \cup \mathbf{\beta}\cup\mathbf{\gamma}\}$, and $z_i \in D_i$; and $n_{z_i}(\psi)$ is the algebraic intersection of $\psi$ with $z_i \times \mbox{sym}^{g-1}(\Sigma)$. It follows that if $n_z(\psi) = n_z(\psi_0)$, the domain $E = \mathcal{D}(\psi) - \mathcal{D}(\psi_0)$ is periodic: i.e. it satisfies $n_z(E) = 0$ (here, $n_z(E)$ denotes the coefficient of the component of $\Sigma - \{\mathbf{\alpha} \cup \mathbf{\beta}\cup\mathbf{\gamma}\}$ containing $z$ in $E$). So all we have to do to define the isomorphism is pick a $\psi_0$ such that $n_z(\psi_0) = 1$: then we can subtract off $n \mathcal{D}(\psi_0)$ from $\mathcal{D}(\psi)$ to get a periodic domain. Given a fixed $\psi_0$, the isomorphism is then given by $$\psi \mapsto (n, \mathcal{D}(\psi) - n\mathcal{D}(\psi_0)).$$ The inverse of the isomorphism should then be given by sending $(n, E)$ to something like $n \psi_0 + \phi$, where $\mathcal{D}(\phi) = E$; but this seems dependent on choice and I'm not entirely sure how to make sense of $n$ times a holomorphic triangle. Is this right? How do I fix the last part? Thanks very much. REPLY [4 votes]: If I had to guess, I'd go for the typo and a small misunderstanding. I think that the isomorphism should indeed depend on the choice of a triangle $\psi_0$ with $n_z(\psi_0) = 0$ (careful here: this has to be 0 and not 1, as you wrote above), but that it should be defined as $$\psi\mapsto (n_z(\psi), \mathcal{D}(\psi)-\mathcal{D}(\psi_0)-n_z(\psi)\cdot\Sigma)$$ whose inverse is given by (notice that now you only sum a triangle and some periodic domain, plus multiples of the Heegaard surface) $$(n,P)\mapsto \psi_0 + P + n\Sigma$$ The isomorphism I gave above agrees with the isomorphism in the notes only for domains with $n_z = 0$ (which are the only ones you consider in the hat version). I don't think that you need a canonical choice for $\psi_0$ (hence for the isomorphism), nor that it's possible to have one. One final remark: here you're not really summing holomorphic triangles, but rather domains, so you shouldn't worry about the "holomorphic meaning" of the sum of domains.<|endoftext|> TITLE: Is the Tate conjecture known for etale covers of products of curves QUESTION [8 upvotes]: Let $X$ be a (smooth projective geometrically connected) surface over a finitely generated field $k$. The Tate conjecture predicts that, for $l$ a prime number invertible in $k$, the Chern class map from Pic$(X) \otimes \mathbb Q_l$ to the Galois invariants of the second etale cohomology group of $X$ is surjective (and thus bijective). This conjecture is known if X is dominated by a product of curves. Is it known if X is an etale cover of a product of curves? (We can restrict our attention to the case that X is of positive Kodaira dimension.) Etale covers of a product of curves are not necessarily a product of curves. We would be done if etale covers of a product of curves are dominated by a product of curves. Is that the case? REPLY [8 votes]: The answer to the last question (and therefore to the others) is yes. An étale cover of a variety $X$ is dominated by an étale Galois $G$-cover, for some finite group $G$; and this is given by a homomorphism $u:\pi _1(X)\rightarrow G$. If $X=X_1\times \ldots \times X_p$, this gives homomorphisms $u_i:\pi _1(X_i)\rightarrow G$, and there are étale coverings $\tilde{X}_i\rightarrow X_i $ such that the composition $\pi _1(\tilde{X}_i )\rightarrow \pi _1(X_i)\rightarrow G$ is trivial. This implies that your étale cover is dominated by $\tilde{X}_1\times \ldots \times \tilde{X}_p $.<|endoftext|> TITLE: Graphs where each edge belongs to the same number of 1-factors QUESTION [5 upvotes]: Let $G$ be a simple connected graph that has at least one 1-factor. We'll define: $G$ has property A iff it is edge-transitive. $G$ has property B iff each edge belongs to the same number of 1-factors. Is it obvious that $A\implies B$ ? Is it possible to characterize the graphs that have B but not A? I think that such graphs exist, but haven't been lucky yet. The "natural" candidates like a $C_{2k}$ along with its diagonals of a certain length don't seem to work but some come close to property B. REPLY [7 votes]: Here is an example for you to enjoy. Take two disjoint copies of $K_{2,3}$, and form a cubic graph on 10 vertices by joining each of the three vertices of degree 2 in one copy to one of the vertices of degree 2 in the second copy. Here is Sage's picture of this graph. The graph is not edge-transitive because the three "cross edges" are not equivalent to the 12 other edges. Each one-factor of this graph must use exactly one of the three cross-edges, e.g. 2-5, and a pair of opposite edges from each of the two four-cycles induced by the vertices on each side other than 2, and 5, making a total of four one-factors using 2-5. Similarly there are four using 3-6 and four using 4-7. So we have twelve one-factors, giving a total of 60 edge-occurrences in a one-factor. Twelve of those occurrences were cross edges, leaving 48 occurrences shared between the remaining 12 edges, meaning that each non-cross-edge also appears in four one-factors.<|endoftext|> TITLE: If $G$ is compact, $H \leq G$ open, $V$ an irreducible $H$-rep, is $\text{Ind}_H^G$ semisimple? QUESTION [6 upvotes]: Let $G$ be a compact group, $H$ a normal open subgroup, and $K$ a $p$-adic field (so that not all $G$-reps with coefficients in $K$ are semisimple). Let $V$ be a finite-dimensional topological $K$-vector space with continuous action of $H$, such that $V$ is irreducible as an $H$-representation. Is the induced representation $\text{Ind}_H^G(V)$ semisimple? In this specific context, I'm thinking of $G$ as the Galois group of a field, if that's helpful. Thanks in advance! REPLY [6 votes]: The answer is yes. I think it should be an exercise in any book on representation theory. Since $H$ has finite index in $G$, and $V$ is finite dimensional, so is the representation $W=Ind _H^G (V)$ induced to $G$. In characteristic zero, this is equivalent to proving that the Zariski closure of $G$ in $GL(W)$ is reductive. The connected component ${\mathcal G}^0$ of the Zariski closure ${\mathcal G}$ of $G$ in $GL(W)$ is unchanged if we replace $G$ by any finite index subgroup $K$. We take $K$ to be the intersection $\cap gHg^{-1}$ as $g$ varies over $G$; this is a finite intersection since $G/H$ is finite. Restricted to $K$, the representation is semi-simple since $W$ restricted to $K$ is the span of the restriction to $K$ of the semi-simple $gV$ as $g$ varies over $G/H$. Therefore, the unipotent radical of ${\mathcal G}^0$ acts trivially on $gV$ for every $g$ and hence on $W$. Thus, ${\mathcal G}^0$ has no unipotent radical. [Edit] I should have said that I am assuming $Char (K)=0$.<|endoftext|> TITLE: Multiplicative domains and conditional expectations QUESTION [8 upvotes]: Let $M$ be an injective von Neumann subalgebra of $B(H)$. For a completely positive map $\phi:B(H)\to B(H)$, let $Mult(\phi)$ denote be the multiplicative domain of $\phi$. For any conditional expectation (CE) $E:B(H)\to M$, the range of $E$ is contained in $Mult(E)$ because $E$ is a bimodule map. Indeed, the image of $E$ is contained in $\cap\{Mult(E): E \text{ is a CE from }B(H)\text{ onto }M\}$. Question: Is $M=\cap\{Mult(E): E \text{ is a CE from }B(H)\text{ onto }M\}$? I am mainly interested in the answer to this question when $M$ is Type I, or even when $M$ is a masa. Evidence: not much. If there is a faithful CE $E$ onto $M$ then $M=Mult(E)$; in this case the answer is "yes". So, for example, the claim is true when $M$ is a purely atomic masa. I believe the answer is also "yes" when $M$ is purely atomic (not necessarily abelian). Beyond that I have no idea. REPLY [7 votes]: This is not true in general. For example, consider an infinite amenable group $\Gamma$ and let $M$ be the group von Neumann algebra $L\Gamma \subset \mathcal B(\ell^2\Gamma)$ endowed with its usual trace $\tau(x) = \langle x \delta_e, \delta_e \rangle$. For $\gamma \in \Gamma$ let $P_\gamma$ denote the rank one projection onto $\mathbb C \delta_\gamma$. Then $\{ P_\gamma \}_{\gamma \in \Gamma}$ forms an orthogonal family of projections and hence for any conditional expectation $E: \mathcal B(\ell^2\Gamma) \to L\Gamma$, and each finite set $F \subset \Gamma$ we have $$ 1 \geq \tau \circ E( \sum_{\gamma \in F} P_\gamma) = \sum_{\gamma \in F} \tau \circ E( \lambda(\gamma) P_e \lambda(\gamma^{-1})) = | F | \tau \circ E(P_e). $$ Since $\tau$ is faithful it then follows that $E(P_e) = 0$ and hence $P_e$ must be in the multiplicative domain of $E$. A slight generalization of the above argument shows in fact that if $M \subset \mathcal B(\mathcal H)$ is any injective finite von Neumann algebra without minimal projections then the compact operators are always contained in the multiplicative domain of any conditional expectation $E: \mathcal B(\mathcal H) \to M$. If one considers only type I von Neumann algebras, then the argument above can be adapted to show that $M$ being completely atomic is a necessary condition to have $M = \cap \{ Mult(E) : E {\rm \ is \ a \ CE \ from \ } \mathcal B(\mathcal H) {\rm \ to \ } M \}$. This condition should also be sufficient. When $M$ is completely atomic then there in fact exists a normal conditional expectation and the result shouldn't be too difficult from there.<|endoftext|> TITLE: Topology on the dual of a Frechet space QUESTION [5 upvotes]: If $F$ is a Frechet space, is there any locally convex space topology on the dual $F'$, such that for each local diffeomorphism $f$ from an open subset $U$ of $F$ to $F$, the map $U \times F' \longrightarrow F'$ is smooth? REPLY [4 votes]: This map is not smooth for every vector space topology on the dual space, see Remark I.3.9. in Neeb, K.-H. "Towards a Lie theory of locally convex groups" 2006 for an explicit counterexample. Thus the cotangent bundle does not carry a (natural) smooth structure for Fréchet manifolds and one has to specify what one understands under smooth differential forms more directly. The usual way is to require the chart representation of the differential $k$-form $\alpha$ to be smooth as as a map $\alpha: U \times F^k \to \mathbb{R}$ (see Definition I.4.1 in the above paper).<|endoftext|> TITLE: Newton polygons of modular polynomials QUESTION [10 upvotes]: This is pretty much straightforward curiosity. Is there anything known about Newton polygons of classical modular polynomials (polynomial relations between $j(\tau)$ and $j(n\tau)$)? I understand that modular polynomial is not really a practical way of approaching modular curves, still it would be interesting to see if properties of the Newton polygon had some relevance to the geometry and arithmetic of the curve. REPLY [6 votes]: The polynomial $\Phi_n(x,y)$ has a degree of $n \prod_{p|n } \frac{p+1}{p}$ in each variable, so the Newton polygon is contained in a square of that side length. Call this $n'$. For each fixed value of $j_1$, there are $n'$ possible values of $j_2$ with multiplicity, so plugging in any value of $x$ or $y$ gives a degree $n'$ polynomial. This means $(n',0)$ and $(0,n')$ are in the Newton polygon. If there is no cyclic isogeny of degree $n$ from a curve of $j$ invariant $0$ to itself, then $(0,0)$ is in the polygon, which determines the bottom half. If there is a unique such isogeny that is a point of multiplicity $1$, so $(1,0)$ and $(0,1)$ are in the polygon. If there are multiple isogenies, I am not sure exactly how to describe the bottom half. I think the different isogenies always form distinct branches of the modular curve so if $k$ is the number of isomorphism classes of isogenies then $(k,0)$ and $(0,k)$ are the only vertices in the bottom half. Edit: At each point $x$ of the abstract modular curve $X_0(N)$ such that $j_1(x)=0$, $j_1$ has a zero of order $1$ or $3$. Similarly if $j_2=0$ then $j_2$ has a zero of order $1$ or $3$. We will show that these two orders are the same. The order is $1$ if and only if the point is an elliptic point - that is, there is an automorphism of order three of the curve that preserves the cyclic subgroup. Modding out by this subgroup, we get an automorphism of order three of the isogenous curve that fixes the dual subgroup. So $j_1$ has degree $1$ if and only if $j_2$ does. What this means is that locally the equation relating $j_1$ and $j_2$ looks like $aj_1+bj_2+$ higher-order terms with $a\neq 0$, $b\neq 0$. Multiplying together one of these equations for each point with $j_1=j_2=0$, we a polynomial of degree $k$ with the coefficients of $x^k$ and $y^k$ nonzero as the lowest-order term, so the Newton polygon is as I described. The divisor corresponding to this side of the polygon should just be the sum of all the points with $CM$ by $Z[\omega]/(\omega^2+\omega+1)$ where the isogenous curve also has CM by the same ring, with multiplicity $1$ for the elliptic points and $3$ for the CM points. I'm not sure exactly how this relates to the Heegner point construction. Understanding the top half is basically understanding the behavior as the $j$ invariants of both isogenous curves aprroach $\infty$. How this works out is that one $j$ invariant looks like a power of the other: $j_1 = j_2^{a/b}$, where $ab=n$. This means that the slopes of the top half of the Newton polygon are exactly these rational numbers $a/b$. To make it add up I'm guessing the part of length $a$ travels a distance $(a,b) * \phi ( gcd(a,b))/gcd(a,b)$. I believe there is an explicit combinatorial way of computing this with the modular group. I figured something like this out once. There's probably also a nice reference somewhere. Basically the idea is that one can view cusps as double cosets $\Gamma_0(n) \backslash \Gamma / \langle T \rangle$. Then the order of the pole of $j$ on that cusp is the size of the double coset (viewed as a $T$-orbit in $\Gamma/\Gamma_0(n)$). The isogeny acts by viewing $\Gamma/\Gamma_0(n)$ as cyclic subgroups of order $n$ of $(\mathbb Z/n )^2$ and sending them to a dual subgroup. You can use this to count the orbits of each degree that are sent to each other degree, which gives the top of the Newton polygon. The divisors on the top are sums of cusps - basically you're summing all the cusps of a certain type. The divisors on the sides are summing all the points with $j$ invariant $0$, so they're basically the pullback of $O(1)$ from the classical modular curve. The divisors on the bottom corner near $0$ are sums of CM points, so they may be related to Heegner points.<|endoftext|> TITLE: Update on list of open problems for Cherednik/Symplectic Reflection Algebras QUESTION [8 upvotes]: Background: There are two lists of open problems about Cherednik or Symplectic Reflection Algebras from 2007: Ian Gordon's Problems, Chapter 9 in Symplectic Reflection Algebras, and Ginzburg & Etingof's list given at a workshop in 2007 here. Gordon titled these "Problems for next year". Now, more than years later, some of them are certainly solved. For example 2) in 1 or 0.1 in 2: Aspherical locus, conjugation-invariant functions $c$ for which $U_{1,c}$ and $H_{1,c}$ are Morita equivalent. There were characterized by a theorem of Bezrukavikov and Etingof in this paper as precisely those for which $U_{1,c}$ has finite homological dimension. Those aspherical values were then characterized in this paper by Dunkl and Griffeth. 6) in 1: Derived equivalences of the category $\mathcal{O}_c$ and $\mathcal{O}_{c'}$ where $c-c'$ is an integer conjugation-invariant function on the set of relections. This conjecture by Rouquier was settled in a recent paper by Losev. My questions are: A) Which of the questions posted in 1 or 2 still remain open? (Or even questions from the original paper, Chapter 17). I am particularly interested in the following ones: 1) in 1: Has there been progress in studying symplectic reflection algebras for symplectic reflection groups that are neither complex nor wreath products? 3) in 1 or 0.4 in 2: Is the Bernstein inequality for finitely generated $H_{1,c}$-modules that the Gelfand-Kirillov dimension is at least $1/2$ that of the reflection representation still open? 8) in 1: Is $\mathcal{O}_c$ Koszul? Disclaimer: Due to my unfamiliarity with this vast field, I may be picking out the wrong ones or being imprecise in my review on things that are now proved. Apologies for that. Thank you! REPLY [4 votes]: My paper with Charles classifies only aspherical values for the monomial groups $G(r,p,n)$; for most exceptional reflection groups the calculation of the aspherical locus is still open (and our technique, based on explicit norm calculations with orthogonal functions, doesn't easily generalize). Looking only at the list of problems in section 9 of Iain's paper, I believe that problem 12 has been partially solved (Bezrukavnikov and Finkelberg used the Cherednik algebra to generalize Macdonald positivity to wreath Macdonald polynomials), and problem 6 has been solved by Losev. Essentially none of the (other) problems dealing with Cherednik algebras for arbitrary groups have been solved, though I am not sure about the status of the problems dealing with GK dimension or a hyper-kaehler structure on the symplectic resolution for the $G_4$ singularity. A particularly embarrassing unsolved problem that does not appear on Iain's list: we do not have a classification of the finite dimensional irreducibles (except for the symmetric groups, and arguably, the monomial groups $G(r,p,n)$, though even there the combinatorics involved is of the Kazhdan-Lusztig type, rather than something simple like dominant weights in classical Lie theory). REPLY [4 votes]: For G(r,1,n), category $\mathcal{O}$ is Koszul. This is a consequence of a comparison theorem with a version of parabolic category $\mathcal{O}$ of affine type, recently proven by Rouquier-Shan-Varagnolo-Vasserot and by Losev when combined with slightly older work of Shan-Varagnolo-Vasserot. This technique doesn't have much hope of generalizing to other types (except maybe G(r,p,n)).<|endoftext|> TITLE: What is the number of noncrossing acyclic digraphs? QUESTION [7 upvotes]: A noncrossing graph on $n$ vertices is a graph drawn on $n$ points numbered from $1$ to $n$ in counter-clockwise order on a circle such that the edges lie entirely within the circle and do not cross each other. I am interested in the integer sequence for the number of noncrossing acyclic digraphs on $n$ vertices. I computed the first few terms of this sequence using a brute-force method. Starting with $n = 1$, they are: 1, 3, 25, 335, 5521 I was not able to find this sequence in the OEIS, which does contain integer sequences for many other classes of noncrossing graphs. Does anybody know the sequence, or has an idea for how to derive it? Thanks for your time! REPLY [11 votes]: You can split noncrossing graphs and get equations for the generating functions. I like to take the minimal root edge (if it exists). $$N=1+xN+2xN\bar N$$ This says that an acyclic noncrossing graph is either empty ($1$), or has an isolated root ($xN$), or has a minimal root edge ($2$ counts either direction) with a noncrossing graph on one side and an edge-rooted noncrossing graph on the other side (counted by $\bar N$). But now we have to deal with $\bar N$. $$\bar N = \frac{1+xN}{(1-2x\bar N)^2}+\frac{4x\bar N^2}{1-2x\bar N}-\frac{x\bar N^2}{1-x\bar N}$$ It's easier to explain this one with a picture. Note that $x$ is not counting the root(s), so if you need to weight the root you'd get $x\bar N$. If you need to include the (truly) empty case too you'd need $1+xN$, as in both equations above. You could easily modify these equations to count by components with $z$ by replacing $1+xN$ with $1+xzN$ in both equations. Then we can compute: $$N(x)=1 + 3x + 25x^2 + 335x^3 + 5521x^4 + 101551x^5 + 1998753x^6 + 41188543x^7 + 877423873x^8 + 19166868607x^9 + \dots$$ $$\bar N(x)=1 + 8x + 106x^2 + 1740x^3 + 31944x^4 + 628040x^5 + 12932888x^6 + 275367248x^7 + 6013043424x^8 + 133923593728x^9 + \dots$$ This method works in general for counting types of noncrossing graphs weighted by different things but the equations can get monstrous.<|endoftext|> TITLE: Examples of famous 'workhorse' theorems QUESTION [34 upvotes]: I use the term 'workhorse' to describe a theorem which is technically challenging to prove, perhaps very deep, but the statement is either uninteresting at first glance or too imposing to be understood by non-experts. I will give some examples of what I feel are 'workhorse' theorems: 1) Bombieri-Vinogradov theorem. This theorem, which is believed (according to Jean-marie De Koninck and Florian Luca's book) to be the reason for Bombieri's Fields Medal in 1974, asserts basically that the Generalized Riemann Hypothesis is true 'on average' over an impressive range of primes. The exact statement, however, is likely quite obtuse to non-number theorists. That said, Bombieri-Vinogradov has an impressive list of consequences, including the most recent results on bounded gaps between primes (due to Maynard). 2) Heath-Brown's 'Theorem 14'. Proved by Heath-Brown in his 2002 paper "The density of rational points on curves and surfaces", this theorem generalized the Bombieri-Pila determinant method to the $p$-adic setting. Its statement is long and difficult to understand at first glance, but it has enormous consequences including (ultimately shown by Salberger) the so-called dimension growth conjecture. It also provided uniform estimates for curves (the best result on this is a preprint due to Miguel Walsh Edit: I just found out that Walsh's paper has now appeared in print and can be found here: http://imrn.oxfordjournals.org/content/early/2014/06/29/imrn.rnu103.refs) and surfaces. It has consequences for concrete diophantine problems, including power-free values of polynomials and power-free values of $f(p)$ where $p$ ranges over the primes only (previous error bounds only provided $\log$ power savings, which are insufficient for this case). So roughly speaking a workhorse theorem is one where the statement of the theorem is not intuitive nor easy to understand, its proof is difficult and perhaps not very enlightening, but nonetheless it has extraordinary consequences and can be used to prove results which are much easier to understand or seemingly unrelated. REPLY [7 votes]: In homotopy theory, a typical workhorse theorem is of the form 'there is a model structure on XY with such an such properties'. While there are some standard techniques producing them, most examples need some extra effort. Moreover, even the very definition of a model structure will appear obscure to non-experts and the existence of one does not appear to very interesting in itself. To be a bit more concrete: One of the most often used examples of a model structure is the Kan-Quillen model structure on simplicial sets, for which there is still no entirely easy proof. It is the basis of any modern treatment of the homotopy theory of simplicial sets. This model structure is also the basis of countless other model structures (like Cauchy-Schwarz produces countless other estimates). Or the various model structures on categories of spectra (S-modules, symmetric spectra, orthogonal spectra), which form the basis of modern stable homotopy theory (unless you want to use $\infty$-categories, where one often uses other work-horse theorems like straightening-unstraightening!).<|endoftext|> TITLE: Max flow, min cut on manifolds QUESTION [12 upvotes]: If a graph has some half edges marked "input" and some half edges marked "output", it is well known that the smallest number of edges which must be cut to disconnect input from output is equal to the number of edge-disjoint paths from input to output. In the Riemannian context, say on a closed $d$-manifold, I believe the generalization is that the infimum $(d-1)$-volume of an integral $(d-1)$-cycle ( say submanifold) in a fixed homology class is equal to the supremum of closed $(d-1)$-forms integrated on that class, subject to the constraint that the $L^\infty$ norm of the form is at most 1. This is an $L^1$-$L^\infty$ duality. I do not know a reference for this and would like to. Does this generalize to all $p < d$ cycles and closed $p$-forms? REPLY [2 votes]: It's good to read a clearly-stated geometric version of max-flow min-cut. I remember reading that in the thesis of John Sullivan, A Crystalline Approximation Theorem for Hypersurfaces. A linear optimization textbook usually only states the case for graphs. Since I am here, read the thesis, perhaps I should get on a soapbox and discuss it some when I can.<|endoftext|> TITLE: A conjecture about the entropy of matrix vector products QUESTION [5 upvotes]: Consider a random $m$ by $n$ partial circulant matrix $M$ whose entries are chosen independently and uniformly from $\{0,1\}$ and let $m < n$. Now consider a random $n$ dimensional vector $v$ whose entries are also chosen independently and uniformly from $\{0,1\}$. Let $N = Mv$ where multiplication is performed over the reals. The matrices $M$ and $N$ are discrete random variables. Recall that the Shannon entropy for a discrete random variable $Z$ is $H(Z) = -\sum_z P(Z=z)\log_2{P(Z=z)}$. In the case where $P(Z=z)=0$ for some values $z$, the corresponding term in the sum is taken to be $0$. We therefore know that the (base $2$) Shannon entropy $H(M) = H(v) = n$. The fact that $H(M) = n$ is a direct result of the fact that the entire matrix is defined by its first row. If $m = \lfloor 10n/\ln{n} \rfloor$ then I would like to make the following conjecture. Conjecture: For all sufficiently large $n$, $H(N) \geq n/10$. The value $10$ is chosen somewhat arbitrarily to be a sufficiently large constant. Is this a known problem and/or can anyone see a way to approach it? Is it in fact true? REPLY [2 votes]: This is not an answer, just a long comment. As pointed out by John Mangual, this game is similar to Mastermind. In fact, it is even more similar to the game defined by Erdos and Renyi here: http://www.renyi.hu/~p_erdos/1963-12.pdf. In your language their game is the following: We are given the first row of the matrix $M$, denoted by $M'$, (so $m=1$) but what we multiply it with, $X$, is not a vector, but an $n\times a$ matrix. Our goal is to reconstruct $M$ from $N=M'X$. They show that this can be done whp if $X$ is a random matrix and $a\ge 10n/\log n$. Of course if $M$ can be reconstructed whp, that also implies that $H(N)\ge (1-\epsilon)n$. So one possible way to answer your question would be to show that their theorem holds even in the case when $X$ is a circulant matrix, as this is equivalent to $X$ being a vector and $M$ being circulant. new part So suppose we want to compute the probability that for two different random vectors, denoted by $v$ and $u$, multiplying them with the rotations of a random vector $r$ we get the same values, i.e., if the rotations of $r$ are denote by $r_1,\ldots,r_k$, then what is the chance that for all $i$ we have $vr_i=ur_i$. For any $i$, this is like a random walk, as $v$ and $u$ are also random, so $Pr[vr_i=ur_i]\approx 1/\sqrt n$. Now I claim that this statement is true even in the following conditional form if $k$ is small enough: $Pr[vr_i=ur_i\mid \forall j\ne i: vr_j=ur_j]\approx 1/\sqrt n.$ I don't think this is that hard to prove, but I cannot see a solution now.<|endoftext|> TITLE: The definition of < in Robinson's Q QUESTION [7 upvotes]: I recently had to explain how the basic axioms in Simpson's Subsystems of Second Order Arithmetic were interpretable in Robinson's Q. Most of the axioms are actually the same, except that Simpson includes an ordering relation $\lt$ with the axioms $$\lnot(m \lt 0)$$ and $$m \lt n + 1 \leftrightarrow m \lt n \lor m = n.$$ (Perhaps I shouldn't call it an "ordering relation" since the axioms are too weak to prove that it is transitive.) Though some flavors of Robinson's Q include an ordering relation, it is usually simply defined in terms of addition. In my explanation, I instinctively defined $m \lt n$ to mean $\exists k(n = m + (k + 1))$ and proceeded to show that this relation satisfies Simpson's two axioms. I later tried to find a reference for this simple fact. To my dismay, it appears that the usual way to define the ordering relation in Robinson's Q is the other way around. For example, Hájek and Pudlák define $m \leq n$ as $\exists k(n = k + m)$ in Metamathematics of First-Order Arithmetic. Sadly, this variant doesn't work for the purpose above. In fact, I've convinced myself that while they all satisfy the first axiom, mine is essentially the only variation on the theme that satisfies the second axiom. I haven't checked every source but I haven't found any that use my version of the ordering relation. This is not a serious problem since all variants work equally well for the ordering of standard natural numbers and all variants agree once a very mild amount of induction is added to the picture. Nevertheless, I'm puzzled: Is there any source that uses my definition of the ordering relation? Is there any technical reason to prefer the other way of defining the ordering relation? REPLY [3 votes]: I can’t give you a source from the top of my head, but I’m pretty sure that all obvious variants of the definition of ordering in Q appear somewhere. The technical advantage of the definition in Hájek and Pudlák is that it provably satisfies $$x\le\overline n\lor\overline n\le x$$ for every standard $n$, which is used in the canonical proof of representability of recursive functions in Q (though this can actually be circumvented) and in various Rosser-style arguments. The reverse definition does not have this property, as Q does not even prove that every nonzero $x$ is of the form $0+y$.<|endoftext|> TITLE: $RO(G)$-Graded Cohomology Theories QUESTION [6 upvotes]: Let $G$ be a compact Lie group with real representation ring $RO(G)$. Recently, I have been learning about some aspects of $RO(G)$-graded cohomology theories (for a precise definition, see Chapter XIII of May's book, Equivariant Homotopy and Equivariant Cohomology). One impetus for introducing this definition is that it allows one to properly formulate an equivariant analogue of the suspension isomorphism. Indeed, if $E_G^*$ is an $RO(G)$-graded theory with reduced counterpart $\tilde{E}_G^*$, then we have natural $\mathbb{Z}$-module isomorphisms $$\tilde{E}_G^{\alpha}(X)\cong\tilde{E}_G^{\alpha+V}(\Sigma^V(X)).$$ Here, $\alpha\in RO(G)$, $V$ is a real orthogonal $G$-module, $X$ is a $G$-space, and $\Sigma^V(X)$ is its equivariant reduced suspension with respect to $V$. To recover an underlying $\mathbb{Z}$-grading, one identifies a non-negative integer $n$ with the trivial $G$-module $[\mathbb{R}^n]\in RO(G)$ so that $$\tilde{E}_G^n(X):=\tilde{E}_G^{[\mathbb{R}^n]}(X).$$ The usual suspension isomorphism then determines the negative grading components of $\tilde{E}_G^*$. $\textbf{Question:}$ Suppose that $V$ is a finite-dimensional orthogonal $G$-representation. What is the relationship between $E_G^{n+V}(\text{pt})$ and $E_G^{n+\dim(V)}(\text{pt})$? I suspect this is well-known and understood. Yet, I am having difficulty finding references that directly address this issue. I would therefore appreciate any and all advice concerning references. REPLY [5 votes]: There is no relationship in general. If $E$ is a module over the $G$-equivariant $MU$ spectrum then $E_G^{n+V}(\text{pt})$ is naturally isomorphic to $E^{n+\text{dim}_{\mathbb{R}}(V)}(\text{pt})$ for all complex representations $V$. The same works for real representations if $E$ is a module over the equivariant $MO$.<|endoftext|> TITLE: When are isotrivial families split by a finite base-change? QUESTION [6 upvotes]: A well-known theorem of Grauert and Fischer states that a smooth proper family of complex manifolds is a locally trivial fibration as soon as all the fibers are isomorphic. It is also easy to obtain an algebro-geometric variant of this; cf. Lemma 1.3 in Buium's Differential Algebra and Diophantine Geometry. For a complete (generically smooth) family of curves $X/B$, isotriviality (take this to mean that all smooth fibers are isomorphic) is thus a priori equivalent to having a quasi-finite dominant map $B' \to B$ over which $X \times_B B'$ is a product family. Now, for families of positive genus, it is often taken for granted that $B' \to B$ may be taken to be surjective (that is, finite). Why is this, or where is it proved? This of course is false for $\mathbb{P}^1$-bundles: if there were a finite cover $f: C \to \mathbb{P}^1$ under which $\mathbb{P}(\mathcal{O}_{\mathbb{P}^1} \oplus \mathcal{O}_{\mathbb{P}^1}(1)) \to \mathbb{P}^1$ pulled back to the product bundle $C \times \mathbb{P}^1$, then the line bundles $f^* \mathcal{O}_{\mathbb{P}^1}$ and $f^*\mathcal{O}_{\mathbb{P}^1}(1)$ would be isomorphic, which they are not: one is ample, the other is not. For elliptic surfaces $E/B$ we can see it as follows. First, following a finite base change attaining stable reduction, we may assume that the family is smooth. Consider then the line bundle $\omega := 0^* \Omega_{E/B}^1$ on $B$. We know that $\omega^{\otimes 12}$ is trivial: it has the nowhere vanishing global section $\Delta$. It follows that there is a finite etale covering $p :B' \to B$ of degree dividing $12$ with $p^*\omega \cong \mathcal{O}_{B'}$ trivial. Then it is easy to see that $E \times_B B' \to B'$ splits (cf. section 3.2 of Ulmer's survey Elliptic curves over function fields). The question: If $X/B$ is a family of abelian varieties or of curves of genus $> 0$, with all members isomorphic (a smooth isotrivial morphism), is there a finite etale covering $B' \to B$ such that $X \times_B B'$ is a product family? Does the same statement hold for smooth isotrivial families of projective varieties of non-negative Kodaira dimension? REPLY [4 votes]: This is false for abelian varieties, as eluded by Ari. The moduli of polarized abelian varieties is a DM stack, as they have only finitely many automorphisms preserving the polarization, but some abelian varieties have infinite order automorphisms, which must permute the polarizations. In particular, the product of a non-CM elliptic curve with itself has automorphism group $GL_2(\mathbb Z)$. An arbitrary automorphism can be realized as monodromy over the base a nodal curve by taking the trivial family over the normalization and gluing by the automorphism. But you may consider a nodal curve pathological. A normal variety has profinite étale fundamental group and cannot have such exotic automorphisms as monodromy. An abelian variety over a normal base must support a polarization.<|endoftext|> TITLE: non-negative random variable QUESTION [5 upvotes]: Let $X$ be a real-valued random variable with $X \geq 0$ and $\mathbb E X >0$. I would like to bound $\mathbb P(X >0)$ from below using information about the first few moments of the variable. From Cauchy-Schwarz inequality we know $\mathbb P(X>0) \geq \frac{(\mathbb E X)^2}{\mathbb E X^2 }$. Unfortunately, this is not a good enough bound (for instance, if you apply it to $X$ being the absolute value of a standard Gaussian you get $\mathbb P(X>0) \geq \frac{2}{\pi}=0.636..$). Is there a refined lower bound? REPLY [9 votes]: Let $f(x) = \sum_{j=0}^d a_j x^j$ be any polynomial of degree $d$ such that $f(0) = 0$ and $f(x) \le 1$ for all $x \ge 0$. Then $$P(X > 0) \ge E[f(X)] = \sum_{j=0}^d a_j E[X^j]$$ Your Cauchy-Schwarz bound is the case $f(x) = 1 - (x - c)^2/c^2 = 2 x/c - x^2/c^2$ where $c = E[X^2]/E[X]$. If you want a bound that scales properly (so the bound for $X$ is the same as the bound for $kX$ for any $k > 0$), you can take $a_j = b_j/E[X]^j$. These bounds are best possible in the sense that they are exact for any $X$ whose distribution is concentrated on $0$ and the points where $f(x) = 1$. If $X$ is a random variable supported on $[0, L]$, you can take a polynomial $f$ such that $f(0) = 0$, $0 \le f(x) \le 1$ for $0 \le x \le 1$, and $f(x) \ge 1-\epsilon$ for $\delta \le x \le L$, and then $E[f(X)] \ge (1-\epsilon) P(X \ge \delta)$. So you can tailor the estimate to be as close to $P(X > 0)$ as you want for this particular distribution.<|endoftext|> TITLE: $K$-homology of $BG$ QUESTION [5 upvotes]: Let $G$ be a finite group. Atiyah proved that the $K$-cohomology of $BG$ vanishes in odd degrees and in even degrees is the completion of the representation ring of $G$ at the augmentation ideal. What is the $K$-homology of $BG$? I mean the stable homotopy groups of the complex $K$-theory spectrum smash $BG$. REPLY [3 votes]: For finite-type torsion spectra $X$ there is a natural isomorphism $$ KU_{n-1}(X) \simeq \text{Hom}_c(KU^n(X),\mathbb{Q}/\mathbb{Z}). $$ Here $\text{Hom}_c$ denotes the group of homomorphisms that are continuous with respect to the skeletal topology on $KU^*(X)$ and the discrete topology on $\mathbb{Q}/\mathbb{Z}$. The isomorphism can be obtained in a fairly straightforward way using $X\wedge S\mathbb{Q}=0$ (so $X=\Sigma^{-1}X\wedge S(\mathbb{Q}/\mathbb{Z})$) and standard exactness arguments. If $G$ is a finite group then we can take $X$ to be $\Sigma^\infty BG$ (without disjoint basepoint) to get reduced $K$-homology groups $KU_{2n}(BG)=0$ and $$ KU_{2n-1}(BG) = \text{Hom}_c(\widehat{J}(G),\mathbb{Q}/\mathbb{Z}). $$ Here $J(G)$ is the augmentation ideal in the representation ring, and $\widehat{J}(G)$ is the completion of $J(G)$ with respect to itself. If $G$ is a $p$-group then we just have $\widehat{J}(G)=J(G)\otimes\mathbb{Z}_p$. The above answer is of course the same as you get via local cohomology, but this argument is more elementary.<|endoftext|> TITLE: Why do Lie algebras pop up, from a categorical point of view? QUESTION [30 upvotes]: Groups pop up as automorphism groups in any category. Rings pop up as endomorphism rings in any additive category. Is there a similar way to attach a Lie algebra to an object in a category of a certain sort? Maybe even such that the attachment of a Lie algebra to a Lie group becomes a special case? And yes, I have searched through the answers to Why study Lie algebras?. REPLY [10 votes]: Lie algebras are equivalently groups internal to "infinitesimal geometry". For instance when formalized in a topos for synthetic differential geometry then a Lie algebra of a Lie group is just the first-order infinitesimal neighbourhood of the unit element (e.g. Kock 09, section 6). More generally in geometric homotopy theory, Lie algebras, being 0-truncated L-∞ algebras are equivalently "infinitesimal ∞-group geometric ∞-stacks" also called formal moduli problems (Lurie). See the link there for pointers to the proof of the equivalence, also (Pridham 07).<|endoftext|> TITLE: A strengthening of base 2 Fermat pseudoprime QUESTION [11 upvotes]: If $n$ is a prime then for all $k$ with $1 \le k \le [n/2]$, $k$ divides ${n-1 \choose 2k-1}$ because of the identity ${n-1 \choose 2k-1} \frac{n}{k}=2{n \choose 2k}$. My question is whether an integer $n$ satisfying the divisibility conditions must be prime ? This may be interesting because of the identity $ \sum_{k=1}^{[n/2]} {n-1 \choose 2k-1} \frac{1}{k}=(2^n-2)/n,$ which shows that any such composite $n$ must be a Fermat pseudoprime to the base 2 and hence must be quite rare. The sum formula which holds for all $n$ can be deduced from halving the expansion $2^{n}=\sum_{k=0}^{n} {n \choose k}$. Any such $n$ must be odd, square-free, and at least $10^7$. What is probably more interesting than this simple result is the way we are led to it by considering the iteration of the map $f(x)=x-1/x$. For each $n$ there are $2^n-2$ points with periods $n$ and since there is no fixed point, for $n$ prime they must form union of $n$ cycles and we know for each $k$ with $1 \le k \le [n/2]$, exactly ${n-1 \choose 2k-1} \frac{1}{k}$ of the cycles intersect the interval $(0,1)$ $k$ times. Since this accounts for all the cycles, we are led to the divisibility condition and the sum formula which also gives a combinatorial proof. [Added July 31: Extension to base $m>2$] We still have for $n$ prime, $m$ divides ${n-1 \choose mk-1} \frac{1}{k}$ since ${n-1 \choose mk-1} \frac{n}{k}=m {n \choose mk}$ but the sum $\sum_{k=1}^{[n/m]}{n-1 \choose mk-1} \frac{1}{k}$ grows much slower than $(m^n-m)/n$ for $m>2$. The added question is how can one relate the sum to some form of Fermat quotient for $m>2$ ? It is possible to consider a more general degree $m$ rational map of the form $f(x):=x-\sum_{j=1}^{m-1} \frac{a_j}{x-b_j}$ where $a_j>0$ and $b_j$ are distinct real numbers which has $m^n-2$ points of period $n$. Adding inclusion exclusion, there will be exactly $q_m(n):=\sum_{d|n} \mu(n/d)\frac{(m^n-m)}{n}$ primitive $m$ cycles, where $\mu$ is the Moebius function. However it is not clear if all cycles intersect $(b_1,b_2)$ for $m>2$. Incidently $q_2(n)$ starts as $1,2,3,6,9,18...$ matches some sequences in OEIS which gives many other combinatorial interpretations. REPLY [10 votes]: Such an $n$ must be prime. If $\frac{1}{k} \binom{n-1}{2k-1}$ is an integer for all $1 \leq k \leq \lfloor \frac{n}{2} \rfloor$, then $n$ divides $\binom{n}{2k}$ for all $1 \leq k \leq \lfloor \frac{n}{2} \rfloor$. Suppose that $n$ is odd and squarefree and $p$ is a prime dividing $n$. Then $p$ is odd, and so $n-p$ is even, and we have that $n$ divides $\binom{n}{n-p}$. Assuming that $n$ is odd and squarefree, then for all primes $p | n$, $n-p$ is even. However, we have that $n$ divides $\binom{n}{n-p} = \binom{n}{p} \equiv \frac{n}{p} \pmod{p}$, where the congruence follows from Lucas's theorem. This is a contradiction since $n/p$ is not a multiple of $p$.<|endoftext|> TITLE: Vanishing patterns of minors of matrix QUESTION [6 upvotes]: Let $M$ be a $m\times n$ matrix with entries in, say $\mathbb{C}$; assume $n\leq m$. Denote by $I\subseteq\{1,2,\ldots, n\}$ a subset of the columns of M. I am interested in positive results to the following (admittedly not very precise) problem: Let $M_I$ be the submatrix of $M$ formed by its columns labelled by $I$. Suppose we know which maximal minors of $M_I$ vanish; what are the possible vanishing patterns for the maximal minors of submatrices $M_{I'}$ where $I\subset I'\subseteq \{1,2,\ldots, n\}$. I would also be satisfied with answers relating just the number of vanishing maximal minors of $M_I$ and $M_{I'}$ (as opposed to the precise pattern of vanishing). I know the answer is related to relations among minors of a generic matrix, as partially discussed here, for instance, but I am looking for precise references or information concerning my question. REPLY [2 votes]: This seems like a matroid question but I'll take an elementary approach. It may be that my remarks just knock off the easy observations. I don't think anything changes if you switch to $\mathbb{Z}$ although fields of finite characteristic are different. We may as well assume that $M_{I'}=M$ and that $M_I$ is the $j$ leftmost columns. So we have a set $S=\{{v_1,v_2,\cdots,v_m\}}$ of $m$ symbols and a family $\mathcal{F}$ of $n$-subsets from $S$. We wonder when there is a way to assign the $v_i$ to be $m$ row vectors from $V=\mathbb{C}^n$ so that $\mathcal{F}$ is exactly the bases for $V$ made from $S.$ That is an interesting question in and of itself. But you to think of the $v_i$ as rows of a matrix $M$ (so the question is equivalent to which minors vanish and which don't) and wonder what more can we say given the additional information from $M_I?$ Whatever is true for $M_I$, it could be that there are no bases from rows of $M$ (i.e. all minors vanish). If so, that is the end of it. Otherwise we can call a subset of $S$ independent if it is a subset of a basis. Then it is required that given an independent set $T$ of size $k$ and a basis $B$, there is a basis made up of $T$ and some $n-k$ elements from $B$. So that is a non-trivial restriction and it applies to the information that we are provided about $M_I.$ However there are families $\mathcal{F}$ which meet this requirement and can be realized by matrices over $\mathbb{Z}_2$ but not over $\mathbb{C}$ and also families which can not be realized by a matrix over any field. Assume $M_I$ is the first $j$ columns. It will follow from below that (*) if every minor for $M_I$ vanishes, the same is true for $M$ and that (* *) otherwise we may assume that the top minor of $M_I$ is the $j \times j$ identity matrix and (* * *) that we know which entries of $M_I$ are zero and that the leading non-zero entry of each row is a $1.$ I don't actually make use of these last two but they are helpful to think about. If $j=1$ then we know which entries in column $1$ are zero. A minor of $M$ with all zero in the first column will vanish. Any other first column might or might not lead to a non-zero minor. In general, a minor for $M$ with all leftmost $j \times j$ minors vanishing also vanishes (hence (*)). What else can one say? I argue below that the answer is nothing more for the case $n=2.$ Otherwise I can think of a bit more, but not much. Column operations on $M_{I'}$ will affect the values of the minors but not which ones vanish. And permuting the rows is just a relabelling which does not change the problem. Hence (* * ). And then just the minors which use $j-1$ of the top $j$ rows will tell you the rest of (* * *) since multiplying a row by a (non-zero) constant does not change anything. What about the case $n=2?$ We can represent the family $\mathcal{F}$ by a graph with $m$ nodes labelled $v_1,\cdots,v_m$ (or unlabeled) and an edge for each putative basis. Then before including the information from column $1$ (aka $M_I$) we can say that the graph must be a complete multipartite graph possibly with some isolated vertices (call two non-zero rows equivalent if each is a multiple of the other, then the edges connect precisely the non-equivalent pairs.) Now the additional information from $M_I$ is which entries in column one are zero and which are not. The isolated vertices (if any) must come from the rows with a leading zero and the remaining rows (if any) with a leading zero form an equivalence class for the graph. Other than that the remaining classes can be anything. Note too that for $j=1$ and $n=2$ we can do column operations on $M$ without changing which minors of $M$ vanish. Each all $0$ row , if any, will always leave a $0$ in column $1$ but otherwise we can arrange to have no other $0$ entries or to have the $0$'s indicate the all $0$ rows along with any other one equivalence class. For $j \gt 1$ I will just say that any $j$ rows of $M$ with span of dimension less than $j$ give a vanishing minor in $M_{I'}.$ I think one could show that by column operations in $M$ (which have no effect on which minors of $M$ vanish) we could arrange to have no other minors of of $M_I$ vanish.<|endoftext|> TITLE: The (un)decidability of Robinson-Arithmetic-without-Multiplication? QUESTION [11 upvotes]: I asked this over at math.stackexchange, and though a number of people were interested enough to vote up the question, I didn't get an answer -- which makes me wonder whether it isn't quite so trivial/dumb as I originally feared it was. So let me try again in this more turbo-charged forum ... Take our old friend Robinson Arithmetic, and cut it down to a theory of successor and addition. To spell that out (just to ensure that we are singing from the same hymn sheet), take the first-order theory with $\mathsf{0}$ as the sole constant, and $\mathsf{S}$ and $+$ as the built-in function signs, with the five axioms $\mathsf{\forall x\ 0 \neq Sx}$ $\mathsf{\forall x\forall y\ Sx = Sy \to x = y}$ $\mathsf{\forall x(x \neq 0 \to \exists y\ x = Sy)}$ $\mathsf{\forall x\ (x + 0) = x}$ $\mathsf{\forall x\forall y\ (x + Sy) = S(x + y)}$ and whose deductive system is your favourite classical first-order logic with identity. Since this cut-down theory doesn't represent the recursive functions, you can't use the usual proof of undecidability for an arithmetic. Since this cut-down theory doesn't even know that addition is commutative, you can't do the kind of manipulations inside the theory involved in a quantifier-elimination proof of decidability (cf. what happens when we add induction to this theory to get Presburger arithmetic, i.e. Peano Arithmetic minus multiplication). Ermmmm .... so .... Drat it, I ought to know how to prove that this cut-down theory is decidable or that it is undecidable. But I seem to have forgotten, assuming I ever knew, and searching around hasn't helped me out. OK folks, I'm more than likely to be having a senior moment here [well, given the lack of answers on math.se maybe a forgivable senior moment?] -- so be gentle! -- but how do we show the theory is (un)decidable? [My bet is on undecidable, for what little that it is worth ...] REPLY [21 votes]: The theory $T'$ with axiom $Sx\ne x$ in place of $Sx\ne0$, as it was originally written, is undecidable, because the theory of groups with a distinguished nonidentity element $a$ is a conservative extension of the theory of nontrivial groups, shown undecidable by Tarski. Since this is a finite extension of $T'$ (modulo the translation of $Sx$ by $x+a$), the latter is also undecidable. The theory with the proper axiom $Sx\ne0$ is undecidable as well. The reduction above doesn’t quite work as the offending axiom is incompatible with groups, nevertheless one can apply a minor modification of Tarski’s original argument. Let $M$ be the set of affine functions $f\colon\mathbb Z\to\mathbb Z$ of the form $f(x)=ax+b$, where $a\in\mathbb N^{>0}$, $b\in\mathbb Z$, and if $a=1$, also $b\ge0$. It is easy to see that $M$ is closed under composition, and contains the identity and the function $s(x)=x+1$. The structure $$\mathcal M=\langle M,\mathrm{id},\circ,S\rangle,$$ where $S(f)=f\circ s$, satisfies axioms 1–5: in particular, $f\circ s=g\circ s$ implies $f=f\circ s\circ s_{-1}=g\circ s\circ s_{-1}=g$ where $s_n(x)=x+n$, which implies 2, and axioms 1 and 3 are consequence of the fact that $f\circ s_{-1}\in M$ iff $f\ne\mathrm{id}$. It thus suffices to prove that $\mathrm{Th}(\mathcal M)$ is hereditarily undecidable, and we can do this by interpreting $\langle\mathbb N,+,\cdot\rangle$ in $\mathcal M$. We embed $\mathbb N$ in $\mathcal M$ via $n\mapsto s_n$. The range of the embedding is definable, as $f\in M$ is of the form $s_n$ for some $n$ if and only if it commutes with $s$. Addition on $\mathbb N$ is definable in $\mathcal M$ as $s_{n+m}=s_n\circ s_m$. Finally, notice that if $f(x)=ax+b$, we have $f\circ s_n=s_{an}\circ f$, hence $$nm=k\iff\forall f\in M\,(f\circ s=s_n\circ f\to f\circ s_m=s_k\circ f)$$ for $n>0$. This shows that multiplication on $\mathbb N$ is definable in $\mathcal M$.<|endoftext|> TITLE: Homotopy excision for structured ring spectra -- reference? QUESTION [11 upvotes]: I'm looking for a reference for analogues of the Blakers-Massey triad connectivity theorem (and its higher-order generalization) for ring spectra. That is: Suppose that $A\to A_1$ is a $k_1$-connected map of (associative) ring spectra and $A\to A_2$ is a $k_2$-connected map of ring spectra, and that the maps are cofibrations so that the pushout, call it $A_{12}$, is a homotopy pushout. Then as long as all of the rings are connective the map of spectra $$ A\to holim (A_1\to A_{12}\leftarrow A_2) $$ is $(k_1+k_2)$-connected. Has anyone worked out a detailed proof of this? I would be happy to see this in any reasonable theory of structured ring spectra. EDIT: I know how a proof should go: Filter the spectrum $A_{12}$ by "word length" and examine the sequence of subquotients (the "associated graded object"), which are wedges of "tensor products" of the spectra $A_i/A$ regarded as bimodules over $A$. But I do not want to delve into technicalities if the details are already out there somewhere. REPLY [9 votes]: Theorems 1.4–1.11 in Ching and Harper's paper “Higher homotopy excision and Blakers-Massey theorems for structured ring spectra” (arXiv:1402.4775) give higher homotopy excision and Blakers-Massey (and their dual versions) for structured ring spectra and more generally, for algebras over operads.<|endoftext|> TITLE: Representation of finite groups in a compact Lie group QUESTION [9 upvotes]: Let $H$ be a finite $p$-group, and let $G$ be a compact connected Lie group. Then it is well-known that $[BH,BG]\cong Rep(H,G)$, where $BH$ and $BG$ are classifying spaces and $Rep(H,G)$ is the set of representations of $H$ in $G$. What happens if we assume that $H$ is any finite group -- do we also have $[BH,BG]\cong Rep(H,G)$? REPLY [10 votes]: In "Maps from $B\pi$ into $X$" Quart. J. Math. Oxford Ser. (2) 39 (1988), no. 153, 117–127., Wojtkowiak proves that the natural map $Rep(H,G)\rightarrow [BH,BG]$ is not surjective when $H=\Sigma _3$, $G=U(2)$.<|endoftext|> TITLE: Compute an arbitrary decimal place of $\pi$ QUESTION [8 upvotes]: Is there a method to find the value of the $n$-th decimal place of $\pi$ which is more efficient than having to compute all decimal places before as well? REPLY [11 votes]: Yes, there are such algorithms. -- See e.g. Xavier Gourdon: Computation of the $n$-th decimal digit of $\pi$ with low memory. There also is the Bailey-Borwein-Plouffe formula already mentioned by Geoff Robinson in a comment -- see https://www.math.hmc.edu/funfacts/ffiles/20010.5.shtml.<|endoftext|> TITLE: Software tools for medium-scale systems of polynomial equations QUESTION [9 upvotes]: I am attempting to find all real solutions of a system of 12 polynomial equations in 12 unknowns. The equations each have total degree 6 and contain up to 1700 terms. I am only interested in real solutions. The equations were derived as the gradients of a sum-of-squares cost function, which I am attempting to find all global optima of. I believe there are a finite number of real solutions but I have not confirmed this yet. I have floating point coefficients and I'm looking for numerical solutions (as opposed to symbolic solutions). Which software packages (and which functions specifically) are generally most promising to solve such a problem? I am aware of various functions in Maple, Matlab, and Mathematica that can solve systems of polynomial equations but there are a large number of options in each software package and I'm interested in advice on where I should be looking first for problems of this scale. A numerical dump of the cost function is here: https://raw.githubusercontent.com/alexflint/polygamy/master/out/epipolar_accel_bezier3/cost.txt REPLY [6 votes]: In Maple you can just do with(Optimization): g := (your function): Minimize(g,iterationlimit = 200); On my machine this takes only about 1.5 seconds to return the following: [2.35579022955789696*10^(-9), [x0 = .696531801759957, x1 = .286105658731833, x10 = .342973444356395, x11 = .728732510532874, x2 = .226824733028582, x3 = .551288843437034, x4 = .719479494442298, x5 = .423120942717389, x6 = .980635895595386, x7 = .684727337329935, x8 = .480773372241607, x9 = .391860913480735]] If you ask for 30 digits of accuracy (so that Maple cannot just use hardware floats) then it gets a lot slower, but still only 40 seconds. Incidentally, I tried this first as just Minimize(g), which gave me an answer with a warning that the maximum iteration limit had been reached. The default limit is 50 iterations, so I tried 200 and the warning went away. Also, there is another function called LSSolve which is specifically for the case where your objective function is a sum of squares. I did not use that because I do not know how to express your objective in the relevant form.<|endoftext|> TITLE: Are shortest halving curves simple closed geodesics? QUESTION [6 upvotes]: Let $S$ be a smooth convex surface in $\mathbb{R}^3$ (although my question may as well be asked for the surface of a polyhedron). Say that $\gamma$ is a shortest halving curve if (a) it partitions the surface area of $S$ into two equal-area halves, and (b) it is the shortest such curve (under the Euclidean metric).   Q. Is $\gamma$ necessarily a simple (non-self-intersecting) closed geodesic on $S$? On a polyhedral surface, the equivalent would be a simple closed quasigeodesic (in Alexandrov's sense). Addendum. Douglas Zare's counterexample to that vain hope:                       (Not metrically accurate.) The red circle seems to be a shortest halving curve, but it is not a geodesic. REPLY [3 votes]: I just located this relevant reference: Engelstein, Max, Anthony Marcuccio, Quinn Maurmann, and Taryn Pritchard. "Isoperimetric problems on the sphere and on surfaces with density." New York J. Math 15 (2009): 97-123. (PDF download link) "the short equator gives a least perimeter partition of the ellipsoid into two regions of equal area"   Note the similarity in Fig.7 below to Will Sawin's disc-bubble (although making a different point):<|endoftext|> TITLE: Properties of the Burnside kernel QUESTION [5 upvotes]: Let $p$ be a prime such that the free 2-generator group $B(2,p)$ of exponent $p$ is infinite. Consider the short exact sequence $$ 1\to K \to B(2,p) \to B_0(2,p) \to 1, $$ where $B_0(2,p)$ is the biggest finite $2$-generator group of exponent $p$, which exists by RBP. Question. What is known about the normal structure of the kernel $K$? More specifically, besides the obvious facts that $K$ is perfect, finitely generated, and of exponent $p$, Are there any known proper normal subgroups of $K$? Is $Z(K)=1$? Could it be that $K$ is in fact simple? REPLY [6 votes]: The answer to question 2. is yes, since the centralizer of elements in a free Burnside group (of exponent $p > 665$) are finite cyclic of order $p$. For question 3., the answer is no (for sufficiently large $p$ at least). A theorem of Olshanskii implies that any non-elementary torsion-free hyperbolic group has periodic quotients of period $p$ for sufficiently large odd numbers $p$. If we apply this to a 2-generator hyperbolic group $G$, then there is a period $p$ quotient. Moreover, for $p$ large enough, the quotient $G/G^p$ will not be isomorphic to $B(2,p)$, since it is known that any element in $F_2$ will be non-trivial in $B(2,p)$ for large $p$ (so apply this to any non-trivial relator in a presentation of $G$). Then one has a homomorphism $B(2,p) \to G/G^p$, with infinite index kernel. Intersect this with $K$ to get an infinite index normal subgroup.<|endoftext|> TITLE: Norm of triangular truncation operator on rank deficient matrices QUESTION [5 upvotes]: Let $T_{n\times n}$ be a triangular truncation matrix, i.e. $$T_{i,j}=\begin{cases}1 & i\ge j\\ 0 & i\sqrt r$$ Let $n$ grow in someway but keep $r$ fixed, is it possible to construct an example to show $$\frac{\|T\circ A\|}{\|A\|}\sim O(\sqrt r)$$ or to show the ratio is actually $O(\ln r)$? REPLY [5 votes]: The ratio is of order $O(\ln r)$. This follows from the fact that the triangular truncation is bounded on the Schatten class $S^p$ (=the operators $A$ on $\ell^2$ such that $\|A\|_p:= (Tr (A^*A)^{p/2})^{1/p}$ is finite) and has norm $O(p)$ as $p \to \infty$. Indeed, for $A \in M_n$ of rank $r$ and operator norm $1$, then for all $p \geq 2$ $$ \|T \circ A\| \leq \|T \circ A\|_p \leq C p \|A\|_p \leq C p r^{1/p}.$$ I suffices to take $p = \ln r$. The fact that the triangular truncation has norm $O(p)$ on $S^p$ is well-known. One reference that might contain a proof of this fact (or contain a reference) is the paper E. Davies, Lipschitz continuity of functions of operators in the Schatten classes, J. London Math. Soc. 37 (1988), 148–157. This is also related to the fact that the Hilbert transform is completely bounded on $L^p$ with norm $O(p)$ for $p \geq 2$. Edit I had a look at the paper by Davies that I mention above, and indeed he uses that the triangular truncation $T$ is bounded on $S^p$ with norm $O(p)$. For this fact, he refers to a book by Krein and Gohberg Theory and Applications of Volterra operators in Hilbert spaces, where they attribute the result to Krein and Gohberg, On the theory of triangular representations of nonselfadjoint operators (1961). I am not sure I understand exactly how their result formally implies that $T$ is bounded, but here is a very simple proof that is probably their proof (and that is very close to the standard proof of the boundedness of the Hilbert transform on $L^p$). Denote by $C_p$ the max of the norms of $T$ and $(1-T)$ on $S^p$ (actually it is not hard to see that $T$ and $(1-T)$ have the same norms, but this is of no use here). Then $C_2=1$, and if we prove that $C_{2p}\leq 2 C_p$ we have by interpolation that $C_p \leq p$ for all $p\geq 2$. To prove that $C_{2p} \leq C_p$, remark that for every $A$, $$ (TA)^* (TA) = T( (TA)^* A) + (1-T)( A^* TA )$$ Hence $$ \|TA\|_{2p}^2 = \| (TA)^* (TA) \|_p \leq \|T( (TA)^* A)\|_p + \|(1-T)( A^* TA )\|_p,$$ and each of the two terms is less than $C_p \|A^* TA\|_p \leq C_p \|A\|_{2p}\|TA\|_{2p}$ by Hölder's inequality. If we divide both sides by $\|TA\|_{2p}$ and take the supremum over $A$ we get the result.<|endoftext|> TITLE: Algebraic $K$-theory of algebras in symmetric spectra: reference QUESTION [10 upvotes]: I want to use the technicalities of structured ring spectra for the first time in my life, and I am not really familiar with the relevant literature. I am looking for a reference that defines algebraic K-theory for associative unital algebras in symmetric spectra. REPLY [7 votes]: Tom, not precisely sure what you want. I'm guessing you want to think of a commutative symmetric ring spectrum $R$ and then an $R$-algebra $A$. Without the extra layer, just using an $S$-algebra $R$ (not necessarily commutative), the original source for the algebraic $K$-theory of $R$ is Chapter VI of EKMM, but of course in terms of the $S$-algebras there. The essential point is to have a homotopically well-behaved symmetric monoidal category of spectra in which to work. We now have several: symmetric spectra, orthogonal spectra, and EKMM $S$-modules. Since we have Quillen equivalences that preserve smash products among them, for many foundational purposes it doesn't matter in which category you work. Since that was clear early on, there are many things in EKMM that have not been repeated in full detail in later sources. The differences in detail are not especially significant and comparison theorems are not hard to come by showing that it doesn't matter where you work. (That is emphatically not true when doing multiplicative infinite loop space theory, which is relevant but I think not what you are asking.) Probably there is a good more recent source focusing on symmetric spectra, but I don't know of one. Here is a relevant quote from a nice paper of Blumberg and Mandell (The localization sequence for the algebraic $K$-theory of topological $K$-theory): "We work in the context of EKMM $S$-modules, $S$-algebras, and $R$-modules. Since other contexts for the foundations of a modern category of spectra lead to equivalent $K$-theory spectra, presumably the arguments presented here could be adjusted to these contexts, but the EKMM categories have certain technical advantages that we exploit and that affect the precise form of the statements below."<|endoftext|> TITLE: Asymptotics for algebraic numbers of height less than one QUESTION [14 upvotes]: The question. Is an asymptotic equivalent known or conjectured for the number $N(d)$ of $\alpha \in \bar{\mathbb{Q}}$ with $h(\alpha) < 1$ and $[\mathbb{Q}(\alpha):\mathbb{Q}] \leq d$? The rather crude proof of Northcott's theorem bounds $\log{N(d)}$ by $O(d^2)$. On the other hand, upon extracting $\lfloor d/m \rfloor$-th roots from numbers of degree $m$ and logarithmic height $< d/m$, where $m$ is fixed but arbitrarily large, and using a generalization of Schanuel's theorem due to Masser and Vaaler ("Counting algebraic numbers with large height," Trans. Amer. Math. Soc. 2007), it is easily seen that $\log{N(d)} > Ad$ for $d \gg_A 0$ and any given $A < \infty$. The mentioned paper of Masser and Vaaler determines the dominant term for the opposite count involving points of large height and bounded degree. The problem considered here, involving bounded height and large degree, is likely to be more delicate. The basic lower bound $\log{N(d)} \gg d$ of the previous paragraph is likely to be improved by taking an optimal $m = m(d)$ and examining the error term (rate of convergence) in the Masser-Vaaler estimate. Perhaps such an improvement could point to the right asymptotics for $\log{N(d)}$? Has this asymptotic been studied? The answer. Indeed $\log{N(d)} \asymp d^2$, and this is not hard to see modulo a certain irreducibility hypothesis, which at the very least is implied by a conjecture of Poonen (cf. Thm. 1 in Bertini theorems over finite fields). By the Jensen-Mahler formula, the height $h(\alpha)$ of an algebraic number of degree $d$ with minimal polynomial $f \in \mathbb{Z}[x]$ coincides with the normalized logarithmic Mahler measure: $$ h(\alpha) = d^{-1} \log{M(f)} := d^{-1} \int_{S^1} \log{|f(z)|} \, \frac{d\theta}{2\pi}. $$ By the triangle inequality and a term-by-term integration, this is bounded above by $d^{-1} \log{\ell_1(f)}$, where $\ell_1(f)$ is the $L^1$-norm of the vector of coefficients of $f$. Now of course the inequality $\ell_1(f) < e^d$ has $\exp((1-o(1))d^2)$ solutions $f \in \mathbb{Z}[x]$ of $\deg{f} = d$, and it should not be too difficult to show that as many of them are irreducible (perhaps I am underestimating this problem). This supplies the lower bound $\log{N(d)} \geq (1-o(1))d^2$. Northcott's bound gives $\log{N(d)} < (1+\log{2})d^2 + O(d)$ in the other direction, and we certainly have $\log{N(d)} \asymp d^2$. Apart from verifying the claim on irreducibility, what remains to be seen is whether equality holds in $\log{N(d)} \geq (1-o(1))d^2$. But I suspect this could be a delicate issue. REPLY [5 votes]: Dubickas, Algebraic numbers with bounded degree and Weil height, Bull Aust Math Soc 98 (2018) 212-220, writes, For a positive integer $d$ and a nonnegative number $\xi$, let $N(d,\xi)$ be the number of $\alpha\in\overline{\bf Q}$ of degree at most $d$ and Weil height at most $\xi$. We prove upper and lower bounds on $N(d,\xi)$. For each fixed $\xi>0$, these imply the asymptotic formula $\log N(d,\xi)\sim\xi d^2$ as $d\to\infty$, which was conjectured in a question at Mathoverflow [Asymptotics for algebraic numbers of height less than one ].<|endoftext|> TITLE: Why polarization of abelian varieties? QUESTION [31 upvotes]: Maybe this question is not suitable for here, but I don't think I would receive a satisfactory answer in Math StackExchange. I could never understand the intuition behind polarization of abelian varieties and how it arises. I know that there is an analogy roughly with a prequantum line bundle, but I think the concept of polarization of abelian varieties came first. Furthermore, I know that an embedding in the projective space is equivalent to a Riemann form (in the complex analytic case). Why the word "polarization"? Is there any partition of the abelian variety (as in the case of a geometric quantization, for instance)? How can I think geometrically (in the lattice) about fixing a polarization? Thanks in advance. REPLY [25 votes]: Weil introduced the term "polarization" in connection with his study of abelian varieties with complex multiplication. His definition is slightly different from what one sees today; one might call it a polarization up to isogeny instead of a polarization. One can find a discussion in Weil's article "On the theory of complex multiplication" ([1955d] in volume 2 of his Oeuvres Scientifiques). There are a few notes on the context of that article (and its relation to the work of Shimura and Taniyama) at the end of the volume. Weil says in the article itself (and again in the notes) that "the word 'polarization' is chosen so as to suggest an analogy with the concept of 'oriented manifold' in topology." As Weil mentions in the article, Matsusaka (who might be described as a student of Weil) also did some important early work on polarized varieties apparently around the time that Weil invited him to Chicago (1954). Kollár, a student of Matsusaka, wrote a nice memorial article for the Notices in 2006, which discusses the work of Matsusaka, especially in relation to the theory of moduli. One can find Matsusaka's early work on polarizations in his "Polarized varieties, fields of moduli, and generalized Kummer varieties of polarized abelian varieties" in American J. of Math. vol. 80 No. 1, 1958. The introduction starts as follows: "In studying theta-functions and abelian functions, we sometimes fix the set of scalar multiples of a principal matrix attached to the given Riemann Matrix. This implies that we fix one divisor class and scalar multiples of it with respect to homology on the corresponding complex torus. In this paper we shall introduce the corresponding notion in the abstract case, not only to abelian varieties, but also to arbitrary varieties as done in Weil [22]." (Reference [22] is Weil's paper on complex multiplication mentioned above.) For independent confirmation of Weil and Matsusaka as the origin of the notion of polarization, one may see various articles of Shimura from around the same time, where he repeatedly cites these two. For example, at the end of "Modules des variétés abéliennes polarisées et fonctions modulaires," he writes, "Les notions d'une variété polarisée, d'un corps du module et d'une variété de Kummer sont dues à Weil et Matsusaka; ce dernier a traité le cas général de variétés quelconques." For intuition about polarizations, it may be helpful to think like this: the category of abelian varieties over a field $k$ (say up to isogeny) acts like a full subcategory of the category of representations of a group $G$ on $\mathbb{Q}$-vector spaces. (To show the shift in thinking we can write $V(A)$ when we think of $A$ as like a vector space.) The category of abelian varieties is too small to admit anything like the tensor products that exist in the larger category of $G$-representations, but one can see some multilinear structures. In particular, the dual abelian variety is like a dual representation---or at least a dual representation twisted by a character. A divisorial correspondence between abelian varieties $A$ and $B$ (a sufficiently rigidified line bundle on $A\times_k B$---a biextension of $(A,B)$ by $\mathbf{G}_m$) is like a $G$-invariant bilinear form on $V(A)\times V(B)$. A polarization on $A$ is a divisorial correspondence on $A\times_k A$ satisfying a symmetry and positivity condition (ampleness of pullback along the diagonal), which is like a $G$-invariant bilinear form on $V(A)\times V(A)$ satisfying a symmetry and positivity condition. There are some subtleties in the translation; for example, the symmetry on side of $A$ corresponds to antisymmetry on the side of $V(A)$. We know from the early pages of books on representation theory that $G$-invariant bilinear forms satisfying positivity conditions are useful---they give us complete reducibility, for example---and Weil used polarizations in a similar way in [1955d]. The fact that abelian varieties are polarizable corresponds to something like the fact that $G$ is a compact group, but the aforementioned subtleties mean that statement is not quite right. These vague remarks can be made precise when $k = \mathbb{C}$ in the way that Francesco Polizzi suggests: take $V(A)$ to be $H_1(A^{\rm an},\mathbb{Q})$ equipped with its standard Hodge structure, and the above remarks correspond to some of the theory of theta functions (which is what Matsusaka had in mind in his introduction). A final remark: I've always been a bit suspicious of how Weil justifies the polarization terminology, since there is a notion in classical projective geometry with a similar flavor called a "polarity." I wonder if Weil also had this classical terminology in mind when coining the term "polarization." (A polarity on the projective space associated to a vector space $V/k$ is a geometric structure related to a non-degenerate symmetric bilinear form $B:V\times V\to k$. A polarization on an abelian variety $A/\mathbb{C}$ is a geometric structure related to a certain type of alternating bilinear form on the period lattice of $A$. Both geometric structures are symmetric divisorial correspondences.)<|endoftext|> TITLE: The word problem of the free left distributive algebra on one generator QUESTION [5 upvotes]: A left distributive algebra is a set $A$ together with a binary operation, $\cdot$, satisfying $a\cdot(b\cdot c)=(a\cdot b)\cdot(a\cdot c)$. One important example of left distributive algebras arises in set theory. For $\lambda$ a limit cardinal let $\mathcal{E}_\lambda$ be the set of elementary embeddings $j: V_\lambda\rightarrow V_\lambda$. Suppose $\lambda$ is such that $\mathcal{E}_\lambda\not=\emptyset$ (this is a very large cardinal axiom, called "rank-into-rank" or more precisely "$I_3$"). Then for $j, k\in \mathcal{E}_\lambda$, we can define $$j\cdot k=\bigcup_{\alpha<\lambda}j(k\cap V_\alpha)$$ where we view $k$ as a set of ordered pairs, so that $k\subset V_{\lambda+1}$. Then it turns out that $j\cdot k\in\mathcal{E}_\lambda$ and that $(\mathcal{E}_\lambda, \cdot)$ is a left distributive algebra. Fixing $j\in\mathcal{E}_\lambda$ and letting $\mathcal{A}_j$ be the closure of $\{j\}$ in $\mathcal{E}_\lambda$, Laver (1992, http://www.sciencedirect.com/science/article/pii/000187089290016E#) proved that $\mathcal{A}_j$ is the free left distributive algebra on one generator. He also showed (under the assumption that such a $j$ exists, that is, $I_3$) that the word problem for the free left distributive algebra on one generator is decidable. There has been extensive work on the strength of various results around left distributive algebras. My question is about this last point: Is it known (within ZFC) if the word problem for the free left distributive algebra with one generator is decidable? Since the only proof of decidability I know of uses the normal form theorem, I suspect the answer is no, but I haven't been able to find out myself. Note: I've tagged this "universal algebra" as that seemed the most relevant algebraic tag. If there is a better tag, or if this tag is just too irrelevant, feel free to replace/remove it. REPLY [9 votes]: Yes. If by the word problem for left-distributive algebras, we mean determining if two elements of a free left-distributive algebra (with any number of generators) are equal. In the Handbook of Set Theory, Theorem 2.11 in the chapter on algebras of elementary embeddings states that in ZFC left division in the free left distributive algebra with one generator has no cycle (i.e. we cannot have $(...((x*x_{1})*x_{2})*...)*x_{n}=x$ where the operation is application of elementary embeddings). Theorem 2.1 in the same chapter states that if there is a left-distributive algebra where left division has no cycle, then the word problem for left distributive algebras is decidable. Combining these results we conclude that the word problem form left distributive algebras on one generator is decidable without using the I3 axiom. $\textbf{Added some time later}$ I should also mention that there are ways to represent the free LD-system with one generator without using large cardinal embeddings. From these representations, one can solve the word problem for LD-systems with one generator. Let $B_{\infty}$ denote the braid group on infinitely many strands. The braid group $B_{\infty}$ is the direct limit of the braid groups $B_{n}$ on finitely many generators. Then the braid group $B_{\infty}$ can not only be endowed with a group operation, but $B_{\infty}$ can also be given a self-distributive operation as well as follows. Let $\sigma_{i}$ be the canonical generators of $B_{\infty}$. In other words, $\sigma_{i}$ is the braid that twists the $i$-th strand with the $i+1$-th strand. Then let $\textrm{sh}:B_{\infty}\rightarrow B_{\infty}$ be the homomorphism mapping where $\textrm{sh}(\sigma_{i})=\sigma_{i+1}$. In other words, $\textrm{sh}$ is the shift map. Define an operation $*$ on $B_{\infty}$ where $a*b=a \textrm{sh}(b)\sigma_{1}\textrm{sh}(a^{-1})$. Then $*$ is a self-distributive operation on $B_{\infty}$. Furthermore, left division in $B_{\infty}$ has no cycles. Therefore for each $b\in B_{\infty}$, the LD-system $\langle b\rangle$ generated by $b$ is freely generated by $b$. Therefore the word problem for LD-systems generated by one element reduces to the word problem for braid groups. Not only is the word problem for braid groups solvable, but there are several known algorithms for the word problem on braid groups. An algorithm for the word problem for braid groups was formulated in 1969 by Garside. Presumably, the best algorithm for determining if two braid words $w_{1},w_{2}$ represent the same braid has complexity $O(l^2n)$ where $l$ is the length of the longest braid word and $n$ is the number of strings that one twists. These facts may be found in Patrick Dehornoy's book Braids and Self-Distributivity.<|endoftext|> TITLE: Motivic L-function vs motivic zeta function QUESTION [12 upvotes]: Let $M$ be a pure motive over a field $k$. Roughly speaking, the L-function of $M$ is the product over all primes $p$ of $$L_p(M,s)=\det(I-Fr_p|_{M_\ell^I} N(p)^{-s})^{-1}$$ where $Fr_p$ is a Frobenius element and $M_\ell^I$ is the $\ell$-adic realization of $M$ stable under inertia $I_p$. On the other hand, the zeta function of $M$ is the infinite series $$Z(M,t)=\sum_{n=0}^\infty[Sym^nM]t^n$$ taking values in the ring of pure motives over $k$. Is there a precise relation between the L-function and zeta function? Both seem to generalize the Hasse-Weil zeta function in different ways, but it is tempting (if nonsensical) to ask for an euler product for the series and vice versa. [EDIT: I believe this 'product' should in fact be the $\det(\dots)^{-1}$ factor, and then the global L-function defined as a product over almost all $p$] Indeed, Dhillon and Minac (1991) define a 'motivic Artin L-function' as $L(M,\rho,t)=Z((V\otimes M)^G,t)$ where $\rho$ is a representation of a finite group $G$ in a $\mathbb Q$ vector space $V$, but it is not obvious whether $L(M,\rho,t)$ and $L(M,s)$ are directly related in any way. REPLY [8 votes]: The idea is to rewrite the $L$-function as the sum of $\sharp [Sym^nM](F_p)t^n$; here "the number of $F_p$-points" of a motif is a natural homomorphism from the Grothendieck group of motives to abelian groups that extends the "usual" number of points over the field $F_p$ for varieties. So, the $L$-function can be obtained from the motivic zeta via the application of this homomorphism. Possibly, Kapranov explians this somewhere; I have read about this in a survey on motivic integration.<|endoftext|> TITLE: Cases where the number field case and the function field (with positive characteristic) are different QUESTION [30 upvotes]: In number theory there is often an analogue between statements which holds over a number field (that is, a finite field extension $K/\mathbb{Q}$) and function fields (that is, finite extensions of the form $K/\mathbb{F}_q(t)$ where $q = p^k$ for some prime $p$ and $k \geq 1$). One of the most famous examples of such an analogue is the Riemann hypothesis. Weil showed that the Riemann Hypothesis holds for the function field case. However, progress on the function field case has not shed much, if any, light on the corresponding case for number fields. An example of where the analogy breaks down (conjecturally) is the Brauer-Manin obstruction. For hypersurfaces defined over $\mathbb{P}^n(K)$ for some number field $K$, it is expected that most hypersurfaces $X$ of degree $d \geq n+1$ will be of general type, and hence neither satisfy the Hasse principle nor have its failure to satisfy the Hasse principle accounted for by the Brauer-Manin obstruction. Harari and Voloch showed that for function fields of positive characteristic, this is not the case: indeed in this paper (http://www.ma.utexas.edu/users/voloch/Preprints/carpobs3.pdf) they showed that the Brauer-Manin obstruction is the only reason Hasse principle can fail in this case. Another case where the analogy is not convincing is ranks of elliptic curves. It is known that elliptic curves over a function field of positive characteristic may have arbitrarily large rank. However, this question is not known even conjecturally in the number field case. Indeed it seems that many experts disagree on this question. At a recent summer school on counting arithmetic objects in Montreal, Bjorn Poonen gave a take on a heuristic suggesting that elliptic curves over number fields should have bounded rank. Andrew Granville, one of the organizers, agrees with this assertion. However, other experts present including Manjul Bhargava disagreed. My question is, what are some other situations where one expects genuinely different behavior between the function field setting and the number field setting? REPLY [10 votes]: I answered this question once already, but another example where $\mathbf Z$ and $k[x]$ behave differently when $k$ is a finite field was brought to my attention recently by Jeff Lagarias and it deserves being mentioned here as a separate answer. Theorem 1. The group ${\rm SL}_2(\mathbf Z)$ is finitely generated. Theorem 2. For each finite field $k$, the group ${\rm SL}_2(k[x])$ is not finitely generated. Theorem 1 goes back to the 19th century, with ${\rm SL}_2(\mathbf Z)$ having generators $S = (\begin{smallmatrix}-1&0\\ \, 0&1\end{smallmatrix})$ and $T = (\begin{smallmatrix}1&1\\0&1\end{smallmatrix})$. Theorem 2 is due to Nagao ("On GL(2,$K[X]$)," J. Inst. Polytech. Osaka City Univ.Ser. A10 (1959), 117-121). The two theorems of course also hold using ${\rm GL}_2$ rather than ${\rm SL}_2$. To appreciate the special nature of the $2 \times 2$ matrix setting, Nagao points out at the start of his paper that for $n \geq 3$ the groups ${\rm SL}_n(k[x])$ and ${\rm GL}_n(k[x])$ are finitely generated. Since a finitely generated group has only finitely many subgroups of each index, ${\rm SL}_2(\mathbf Z)$ has countably many subgroups of finite index. In contrast to that, ${\rm SL}_2(k[x])$ for finite $k$ has uncountably many subgroups of finite index. Generalizations of Theorem 2 are in A. W. Mason, "Serre's generalization of Nagao's theorem: an elementary approach," Trans. Amer. Math. Soc. 353 (2001), 749-767 and H. Behr, "Arithmetic groups over function fields. I. A complete characterization of finitely generated and finitely presented arithmetic subgroups of reductive algebraic groups," J. Reine Angew. Math. 495 (1998), 79-118.<|endoftext|> TITLE: Fibrations and Cofibrations of spectra are "the same" QUESTION [27 upvotes]: My question refers to a folklore statement that I have now seen a couple of times, but never really precise. One avatar is: "For spectra every cofibration is equivalent to a fibration" (e.g. in the accepted answer here https://mathoverflow.net/a/56575/18744), Another one is: "For spectra, fibrations and cofibrations sequences are the same" (which is stronger than the statement above, because it works both ways) At this point I am already quite happy with a suitable reference. However I do have some follow-up questions: To what extent can I switch between spaces and spectra? The suspension functor does not preserve fibrations, correct? So what can be said about the relation of the homotopy fibre (cofibre) of spaces and the homotopy fibre (cofibre) of their suspension spectra? What if the spaces involved are infinite loop-spaces? The resulting $\Omega$-spectra don't carry much different information from the spaces itself, do they? Concretely: I have a sequence $X\to Y\to Z$ of group-like H-spaces, and know that they form a homotopy fibration. I would like to make statements about the homotopy cofibre of $X\to Y$. I got the hint to 'work in spectra', where the two are "the same", but don't know what to make of it. Edit: Thank you all for your answered. Together they cover quie a number of different points of view. Initially I hoped to be able to return back to spaces after doing an excursion through spectra (having fairly explicit $\Omega$-spectra for group like $H$-spaces). It seems that there is no totally generic way to do this and I think I have now enough material to think about the particularities. REPLY [15 votes]: I'd like to expand on Matthias's answer a little bit. There is some unfortunate terminology going around. Being a cofibration resp. fibration is a particular property a map can have; for spaces Hurewicz cofibrations and Serre cofibrations (relative CW-complexes) are ones that get most use, and they have associated notions of fibrations; the concept of a model category mentined by Dan Ramras is in the background of this but I don't think your question warrants, or requires, getting out the big guns. For spectra, in the usual model structure, a cofibration will be a relative CW-spectrum, and $X → *$ is a fibration iff $X$ is an $\Omega$-spectrum. Every map can be turned into a cofibration or into a fibration if you allow yourself to change your target resp. source by a weak equivalence. For spaces, the mapping cylinder construction will turn $f\colon A → X$ into a Hurewicz cofibration $A → (A \times I) \sqcup X/\sim$, and the path space construction will turn $f\colon E → B$ into a Serre fibration. Similar constructions exist for Serre cofibrations and Hurewicz fibrations, and in fact it's an axiom of model structures that such replacements must always exist. Also in spectra. In other words, in the homotopy category, it doesn't make sense to say that a map is a cofibration or fibration. A cofibration sequence, or fibration sequence, is a completely different animal, although there's of course a relation. The usual definition of a cofibration sequence is as follows: if $A → X$ is a cofibration, then $A → X → X/A$ is a cofibration sequence, and any $K → L → M$ weakly equivalent to it is a cofibration sequence, too. So the first map need not be a cofibration. In fact, any $A → X$ can be extended to a cofibration sequence (by the mapping cone, to be explicit in spaces). Analogously for fibration sequences. It is true in spectra that if $A → B → C$ is a cofibration sequence then it is also a fibration sequence, and vice versa. As a special case (choose $B$ to be a point), the loop functor is inverse to the suspension functor. Morally, in spectra, you've inverted the suspension functor $\Sigma$, so both $\Sigma^{-1}$ and $\Omega$ want to be right adjoint to $\Sigma$, so they have to agree. That's not a proof, though, but you'll find explicit proofs in any textbook on stable homotopy theory, e.g. Adams's "Stable homotopy and generalised homology", part III. The most conceptual reason of this is the Blakers-Massey theorem, which is a theorem about pushouts/pullbacks of spaces, which roughly says that a pushout diagram with highly connected maps is also a pullback diagram in a range of dimensions. The Freudenthal suspension theorem mentioned by Matthias is a corollary of this. To address your more specific questions: $\Sigma^\infty$ preserves cofibration sequences, $\Omega^\infty$ preserves fibration sequences, but not vice versa. In fact, $\Omega^\infty$ turn cofibration sequences of spectra into fibration sequences of spaces, which is not surprising since cofibration sequences of spectra are fibration sequences after all. And for your concrete question: you have a natural map from the homotopy cofiber of $X → Y$ to $Z$ whose connectivity depends on the connectivity of $X → Y$; the Blakers-Massey theorem will tell you the details.<|endoftext|> TITLE: Small remarkable matroids QUESTION [6 upvotes]: Working on phased matroids (a generalization of oriented matroid to the complex case) I've found an interesting formula for computing the inner Tutte group (and, hence, all the Tutte groups) associated to a matroid $M$ with a ground set $\sharp(E)\leq7,$ without minors of Fano or dual-Fano type. I also wrote a sage programme which explitly compute this group. I've already do the computations for the matroids: $F_{7}^{-},$ $O7,$ $P6,$ $P7,$ $Q6$ $R6.$ In fact, these are all the matroids with $\sharp(E)\leq7$ I've found in the sage library. My questions are the following: 1) Are there some other interesting matroids $M$ with the condition $\sharp(E)\leq7$ (without minors of Fano or dual-Fano type)? 2) Maybe this matroids arise from some interesting geometric or topological construction. Could you please explain where these matroids come from? REPLY [2 votes]: Other than those you've listed, the most obvious "interesting" matroids I can think of on up to 7 elements are: $M(K_4)$, the cycle matroid of $K_4$, $W^3$, the "whirl" obtained from $M(K_4)$ by relaxing a 3-element circuit, $U_{r,n}$, the rank-r uniform matroid of size n. $(F_7^-)^*$, the dual of the non-Fano I'm pretty sure that your list plus $M(K_4)$, $W^3$, $U_{2,6}$, $U_{4,6}$, and $U_{3,6}$ covers all the 3-connected 6-element matroids. For 7 elements, there are many other matroids but I don't think there are any more that are interesting enough to have standard names.<|endoftext|> TITLE: String diagrams for bimodules over noncommutative algebras? QUESTION [6 upvotes]: I'm trying to do some calculations with bimodules (over Azumaya algebras, as it happens), and I need a string diagram notation that mixes the tensor product over the base ring (a symmetric monoidal product) and the tensor product of bimodules. I haven't seen any diagrammatic calculus that allows for this mixing, but I haven't searched exhaustively. It seems that if I do something naive I could get something completely wrong. Has anybody used/seen such a thing? REPLY [5 votes]: In general, the correct generalization of the "stringy" notation for bicategories — in which objects label regions in $\mathbb R^2$, 1-morphisms label codimension-1 defects (whose projection to $\mathbb R$ has no critical points), and 2-morphisms label codimension-2 defects, and composition is a "pushforward" of defects — to monoidal bicategories is to put everybody in $\mathbb R^3$ (giving no label to the ambient 3-space, and letting the 2-dimensional regions wiggle around such that they are parameterized by their projections to $\mathbb R^2$). If you want braided monoidal bicategories, you should use $\mathbb R^4$ (but no 3-dimensional defects), and if you want sylleptic monoidal, you use $\mathbb R^5$. In bicategories, if you go to $\mathbb R^6$, you already get symmetric monoidal, but more general $(\infty,2)$-categories do not, I think. In general, the "true" topological setting for symmetric monoidal bicategories is of surfaces-with-defects that are not embedded into any $\mathbb R^n$, or (better?) that are embedded into $\mathbb R^\infty$. Again you should include some framing data to keep the surfaces from swinging around on themselves. I think it works to give every 2-dimensional region an embedding as a domain in $\mathbb R^2$, such that restricting the 2-dimensional parameterization to any 1-dimensional boundary component commutes (up to specified homotopy) with embedding in $\mathbb R^2$ and then projecting to $\mathbb R$. I'm not aware of any paper that writes these pictures up carefully and proves coherence theorems comparable to Ross–Street for plane string diagrams, although that doesn't mean there aren't any. Lurie's TQFT paper contains essentially all of these ideas, but it's not the focus — but he does address the tangle hypothesis in the last section. The consensus among people I know who've studied the paper (I am not one of them) is that it is complete and correct: what Lurie calls an "outline" most people would call a "detailed proof". (See for example Section 2 of Scheimbauer's thesis, where Lurie's definition of the bordism category is filled in and all facts proved; but I don't think Scheimbauer has posted her thesis publicly yet.) For monoidal (rather than symmetric monoidal) bicategories, you can use Gray categories: a monoidal bicategory is a tricategory with one object, and Gray proved a coherence result that every tricategory is weakly equivalent to a Gray category. (But be careful with functors.) I haven't read Hummon's thesis, but it seems to provide such "surface diagrams" for arbitrary Gray categories. See also arXiv:1211.0529 and the n-lab page on Surface Diagrams.<|endoftext|> TITLE: Extending vector bundles from subvarieties QUESTION [8 upvotes]: Let $X$ be a smooth projective variety and let $Y\subset X$ be a smooth subvariety. Given a vector bundle $E$ on $Y$, when can $E$ be extended to a vector bundle $\tilde E$ on $X$? I.e., are there cohomology groups containing the obstructions to extending $E$? REPLY [5 votes]: An obvious obstruction comes from topology: the Chern classes of your bundle should be obtained from restriction of Hodge classes on an ambient variety. This is (more or less) enough to extend a smooth bundle $B$ from $Y$ to $X$. To be precise, you need the classifying map from $Y$ to the space $BU(r)$ to be extendable to a continuous map from $X\supset Y$ to $BU(r)$, where $r$ is rank $B$. This is not the only obstruction, because a way to find a holomorphic bundle with prescribed $(p,p)$-Chern classes on $X$ amounts to a result which is much stronger than the Hodge conjecture (and false, generally speaking). The easiest obstruction to finding a bundle with prescribed $(p,p)$-Chern classes comes from the Bogomolov inequality: for any stable $B$, one has $$\int_M [2rc_2(B) - (r - 1)c_1(B)^2]\wedge \omega^{n-2}> 0,$$ where $n$ is dimension of your manifold $M$, and $\omega$ its Kahler form (case of unstable bundles is considered separately using the Jordan-Holder filtration). Also, if this inequality is non-strict, $B$ admits a projectively flat connection, and therefore $c_2$ and the rest of Chern classes are powers of $c_1$.<|endoftext|> TITLE: Why does this antisymmetric product factor out a determinant? QUESTION [10 upvotes]: Consider a generic $n \times n$ matrix $M$. Define the $(n-1) \times n$ matrix $M_q$ to be $M$ with the $q$th row omitted, and assume that $M_q$ possesses a right inverse, $R_q$: $$R_q = M_q^T (M_q M_q^T)^{-1}$$ The components of $R_q$ will be rational functions of the components of $M$, with the determinant $\det(M_q M_q^T)$ as their denominator. Now, consider the $n \times n$ matrix $C$ of cofactors of $M$. That is, $C_{i j}$ is $(-1)^{i+j}$ times the determinant of the submatrix of $M$ obtained by omitting the $i$th row and the $j$th column. I have found by direct computation for low values of $n$ that the antisymmetric products of the $q$th row of $C$ and the components of $R_q$: $$(A_q)_{i j k} = C_{q i} (R_q)_{j k} - C_{q j} (R_q)_{i k}$$ are polynomials of degree $n-2$ in the components of $M$, with the determinant that one might expect to be present in the denominator, inherited from $R_q$, factoring out. To give an example, for $n=3$: $$M_1= \left( \begin{array}{ccc} m_{2,1} & m_{2,2} & m_{2,3} \\ m_{3,1} & m_{3,2} & m_{3,3} \\ \end{array} \right)$$ $$C= \left( \begin{array}{ccc} m_{2,2} m_{3,3}-m_{2,3} m_{3,2} & m_{2,3} m_{3,1}-m_{2,1} m_{3,3} & m_{2,1} m_{3,2}-m_{2,2} m_{3,1} \\ m_{1,3} m_{3,2}-m_{1,2} m_{3,3} & m_{1,1} m_{3,3}-m_{1,3} m_{3,1} & m_{1,2} m_{3,1}-m_{1,1} m_{3,2} \\ m_{1,2} m_{2,3}-m_{1,3} m_{2,2} & m_{1,3} m_{2,1}-m_{1,1} m_{2,3} & m_{1,1} m_{2,2}-m_{1,2} m_{2,1} \\ \end{array} \right)$$ $$ R_1 = M_1^T (M_1 M_1^T)^{-1} = \\ {\scriptsize \frac{ \left( \begin{array}{cc} m_{2,1} \left(m_{3,2}^2+m_{3,3}^2\right)-m_{3,1} \left(m_{2,2} m_{3,2}+m_{2,3} m_{3,3}\right) & m_{3,1} \left(m_{2,2}^2+m_{2,3}^2\right) -m_{2,1} \left(m_{2,2} m_{3,2}+m_{2,3} m_{3,3}\right) \\ m_{2,2} \left(m_{3,1}^2+m_{3,3}^2\right)-m_{3,2} \left(m_{2,1} m_{3,1}+m_{2,3} m_{3,3}\right) & m_{3,2} \left(m_{2,1}^2+m_{2,3}^2\right) -m_{2,2} \left(m_{2,1} m_{3,1}+m_{2,3} m_{3,3}\right) \\ m_{2,3} \left(m_{3,1}^2+m_{3,2}^2\right)-m_{3,3} \left(m_{2,1} m_{3,1}+m_{2,2} m_{3,2}\right) & m_{3,3} \left(m_{2,1}^2+m_{2,2}^2\right) -m_{2,3} \left(m_{2,1} m_{3,1}+m_{2,2} m_{3,2}\right) \\ \end{array} \right)} {\left(m_{3,2}^2+m_{3,3}^2\right) m_{2,1}^2-2 m_{2,3} m_{3,1} m_{3,3} m_{2,1}+m_{2,3}^2 \left(m_{3,1}^2+m_{3,2}^2\right)-2 m_{2,2} m_{3,2} \left(m_{2,1} m_{3,1}+m_{2,3} m_{3,3}\right)+m_{2,2}^2 \left(m_{3,1}^2+m_{3,3}^2\right)} } $$ $$(A_1)_{i j k}= C_{1 i} (R_1)_{j k} - C_{1 j} (R_1)_{i k} = \left( \begin{array}{ccc} \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) & \left( \begin{array}{c} m_{3,3} \\ -m_{2,3} \\ \end{array} \right) & \left( \begin{array}{c} -m_{3,2} \\ m_{2,2} \\ \end{array} \right) \\ \left( \begin{array}{c} -m_{3,3} \\ m_{2,3} \\ \end{array} \right) & \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) & \left( \begin{array}{c} m_{3,1} \\ -m_{2,1} \\ \end{array} \right) \\ \left( \begin{array}{c} m_{3,2} \\ -m_{2,2} \\ \end{array} \right) & \left( \begin{array}{c} -m_{3,1} \\ m_{2,1} \\ \end{array} \right) & \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) \\ \end{array} \right)$$ My question is: why are the components of $A_q$ polynomials rather than rational functions? Can this be proved for general $n$? And can $A_q$ be reduced, for general $n$, to a simpler expression that makes no reference to $R_q$? (Edited to change question from sum to individual components). REPLY [8 votes]: $\let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits} $ I will write $R$ for $R_{q}$, because $q$ is constant. In the following, I am going to assume that $R$ is an arbitrary right inverse of $M_{q}$, rather than the specific right inverse $M_{q}^{T}\left( M_{q}M_{q}^{T}\right) ^{-1}$ which you have suggested. This makes the statement a tad more general and rids us of a red herring. I shall prove that for any $i\in\left\{ 1,2,...,n\right\} $, $j\in\left\{ 1,2,...,n\right\} $ and $k\in\left\{ 1,2,...,n-1\right\} $, the value $C_{qi}R_{jk}-C_{qj}R_{ik}$ is a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$ (independently, and independent, of the choice of right inverse $R$). For every $\ell\in\left\{ 1,2,...,n\right\} $, we abbreviate $C_{q\ell}$ by $v_{q}$. Thus, $v_{\ell}=C_{q\ell}=\left( -1\right) ^{q+\ell}\det\left( \underbrace{M\text{ without row }q}_{=M_{q}}\text{ and column }\ell\right) $ (1) $=\left( -1\right) ^{q+\ell}\det\left( M_{q}\text{ without column }\ell\right) $. From this it easy to obtain (2) $\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}v_{\ell}=0$ for every $j \in \left\{1,2,...,n-1\right\}$. [Proof of (2): Let $j \in \left\{1,2,...,n-1\right\}$. Let $G$ be the result of inserting a copy of row $j$ of the matrix $M_{q}$ between the rows $q-1$ and $q$ of this matrix. Then, $G$ is an $n\times n$-matrix with two equal rows, and hence has determinant $\det G=0$. But Laplace expansion of $\det G$ along the row we have inserted yields $\det G=\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}\left( -1\right) ^{q+\ell}\det\left( M_{q}\text{ without column }\ell\right) $, since the entries of this row are $\left( M_{q}\right) _{j\ell}$ and the cofactors corresponding to this row are $\left( -1\right) ^{q+\ell} \det\left( M_{q}\text{ without column }\ell\right) $. Now, (1) yields $\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell }v_{\ell}$ $=\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell }\left( -1\right) ^{q+\ell}\det\left( M_{q}\text{ without column } \ell\right) $ $=\det G=0$, so that (2) is proven.] Since $M$ is generic, its cofactors $v_{1}$, $v_{2}$, $...$, $v_{n}$ are nonzero. Fix $i\in\left\{ 1,2,...,n\right\} $. Let $N_{i}$ denote the matrix $M_{q}$ without column $i$. Thus, $\det\left( N_{i}\right) =\det\left( M_{q}\text{ without column }i\right) =\left( -1\right) ^{q+i}v_{i}$ (because (1) yields $v_{i}=\left( -1\right) ^{q+i}\det\left( M_{q}\text{ without column }i\right) $). Let $S_{i}$ denote the matrix $\left( R_{jk}-\dfrac{v_{j}}{v_{i}} R_{ik}\right) _{j\in\left\{ 1,2,...,n\right\} ;\ k\in\left\{ 1,2,...,n-1\right\} }$. The $i$-th row of this matrix $S_{i}$ is zero; let $S_{i}^{\prime}$ denote the matrix $S_{i}$ without row $i$. Now, we claim that $N_{i}S_{i}^{\prime}=I_{n-1}$ (the $\left( n-1\right) \times\left( n-1\right) $ identity matrix). Indeed, for every $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$, the $\left( j,k\right) $-th entry of $N_{i}S_{i}^{\prime}$ is $\left( N_{i}S_{i}^{\prime}\right) _{jk}=\sum_{u\in\left\{ 1,2,...,n-1\right\} }\left( N_{i}\right) _{ju}\left( S_{i}^{\prime }\right) _{uk}$ $=\sum_{\ell\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} }\left( M_{q}\right) _{j\ell}\underbrace{\left( S_{i}\right) _{\ell k} }_{=R_{\ell k}-\dfrac{v_{\ell}}{v_{i}}R_{ik}}$ (since $N_{i}$ is the matrix $M_{q}$ without column $i$, while $S_{i}^{\prime }$ is the matrix $S_{i}$ without row $i$) $=\sum_{\ell\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} }\left( M_{q}\right) _{j\ell}\left( R_{\ell k}-\dfrac{v_{\ell}}{v_{i} }R_{ik}\right) $ $=\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell }\left( R_{\ell k}-\dfrac{v_{\ell}}{v_{i}}R_{ik}\right) $ (here, we added an $\ell=i$ addend to the sum; this did not change the sum because this addend is $0$) $=\underbrace{\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}R_{\ell k}}_{\substack{=\left( M_{q}R\right) _{jk}=\delta _{jk}\\\text{(since }R\text{ is a}\\\text{right inverse of }M_{q}\text{)} }}-\underbrace{\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}\dfrac{v_{\ell}}{v_{i}}R_{ik}}_{=\dfrac{1}{v_{i}}R_{ik}\sum_{\ell \in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}v_{\ell}}$ $=\delta_{jk}-\dfrac{1}{v_{i}}R_{ik}\underbrace{\sum_{\ell\in\left\{ 1,2,...,n\right\} }\left( M_{q}\right) _{j\ell}v_{\ell}} _{\substack{=0\\\text{(by \textbf{(2)})}}}=\delta_{jk}$. This shows that $N_{i}S_{i}^{\prime}=I_{n-1}$, and thus $S_{i}^{\prime} =N_{i}^{-1}$ (since $N_{i}$ and $S_{i}^{\prime}$ are $\left( n-1\right) \times\left( n-1\right) $-matrices). Now, we recall Cramer's rule for the inverse of a matrix. It essentially says that the inverse of a square matrix is obtained by dividing its adjoint by its determinant. In other words, if $X$ is an invertible $p\times p$-matrix for some $p\in\mathbb{N}$, and $j$ and $k$ are elements of $\left\{ 1,2,...,p\right\} $, then (3) $\left( X^{-1}\right) _{jk}=\dfrac{1}{\det X}\left( -1\right) ^{j+k}\det\left( X\text{ without row }k\text{ and column }j\right) $. Now, recall that $S_{i}^{\prime}=N_{i}^{-1}$. Hence, $\left( S_{i}^{\prime}\right) _{jk}=\left( N_{i}^{-1}\right) _{jk} =\dfrac{1}{\det\left( N_{i}\right) }\left( -1\right) ^{j+k}\det\left( N_{i}\text{ without row }k\text{ and column }j\right) $ (by (3)) $=\dfrac{1}{\left( -1\right) ^{q+i}v_{i}}\left( -1\right) ^{j+k} \det\left( N_{i}\text{ without row }k\text{ and column }j\right) $ (since $\det\left( N_{i}\right) =\left( -1\right) ^{q+i}v_{i}$) $=\dfrac{1}{v_{i}}\left( -1\right) ^{i+j+q+k}\det\left( N_{i}\text{ without row }k\text{ and column }j\right) $ for all $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$. In other words, (4) $v_{i}\left( S_{i}^{\prime}\right) _{jk}=\left( -1\right) ^{i+j+q+k}\det\left( N_{i}\text{ without row }k\text{ and column }j\right) $ for all $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$. Let $\mathbf{B}$ be the (unique) increasing bijection from $\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} $ to $\left\{ 1,2,...,n-1\right\} $. Then, for all $j\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} $ and $k\in\left\{ 1,2,...,n-1\right\} $, we have (5) $v_{i}\left( S_{i}\right) _{jk}=\left( -1\right) ^{i+\mathbf{B} \left( j\right) +q+k}\det\left( M_{q}\text{ without row }k\text{ and columns }i\text{ and }j\right) $. (Indeed, this follows from (4), applied to $\mathbf{B}\left( j\right) $ instead of $j$, because $\left( S_{i}^{\prime}\right) _{\mathbf{B}\left( j\right) ,k}=\left( S_{i}\right) _{jk}$ (since $S_{i}^{\prime}$ is the matrix $S_{i}$ without row $i$) and because column $j$ of $M_{q}$ is column $\mathbf{B}\left( j\right) $ of $N_{i}$ (since $N_{i}$ is the matrix $M_{q}$ without column $i$).) Now, fix $j\in\left\{ 1,2,...,n\right\} $ and $k\in\left\{ 1,2,...,n-1\right\} $. We need to show that $C_{qi}R_{jk}-C_{qj}R_{ik}$ is a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$. We WLOG assume that $j\neq i$ (since otherwise, $C_{qi}R_{jk}-C_{qj}R_{ik} =0$). Thus, $j\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} $. Since $C_{qi}=v_{i}$ and $C_{qj}=v_{j}$ (by the definitions of $v_{i}$ and $v_{j}$), we have $\underbrace{C_{qi}}_{=v_{i}}R_{jk}-\underbrace{C_{qj}}_{=v_{j}}R_{ik} =v_{i}R_{jk}-v_{j}R_{ik}=v_{i}\underbrace{\left( R_{jk}-\dfrac{v_{j}}{v_{i} }R_{ik}\right) }_{\substack{=\left( S_{i}\right) _{jk}\\\text{(by the definition of }S_{i}\text{)}}}$ (6) $=v_{i}\left( S_{i}\right) _{jk}=\left( -1\right) ^{i+\mathbf{B}\left( j\right) +q+k}\det\left( M_{q}\text{ without row }k\text{ and columns }i\text{ and }j\right) $ (by (5)), which is obviously a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M_{q}$ (and independent of the choice of $R$), qed. Thanks for a very nice question, and sorry for this mess of an answer... PS. I believe the genericity of $M$ is not required for (6) to hold. Does anyone see a nice proof of this? I don't. PPS. Here is a more general statement which, I think, is true. Let $A$ be a commutative ring. Let $N$ be an $\left(n-1\right) \times n$-matrix over $A$. Let $s \in A^n$ be a vector. For every $\ell \in \left\{1,2,...,n\right\}$, let $p_\ell$ denote the scalar $\left( -1\right) ^{\ell}\det\left( N \text{ without column }\ell\right)$. If $w$ is a vector and $i$ is an integer, we denote by $w_i$ the $i$-th coordinate of $w$ (whenever this makes sense). Then, every two distinct $i \in \left\{1,2,...,n\right\}$ and $j \in \left\{1,2,...,n\right\}$ satisfy (11) $p_i s_j - p_j s_i = \sum_{k=1}^{n-1} \pm \left(Ns\right)_k \det\left(N \text{ without row } k \text{ and columns } i \text{ and } j\right)$, where $\pm$ is something like $\left(-1\right)^{i+j+\left[i TITLE: Maximal ideals are prime (history answer please!) QUESTION [14 upvotes]: Please can someone tell me the history of the simple argument that any maximal ideal of a commutative ring or distributive lattice is prime? (It is understood that we have found the maximal one using Zorn's Lemma.) How were prime ideals found before the enactment of the Axiom of Choice? How do constructive algebraists find them now? What similar arguments (directly referring to polynomials or algebraic numbers) were used before ideals were invented? My own interest is really in distributive lattices and locales, from a constructive point of view, but I am aware that these notions appeared much earlier for commutative rings. Incorporating my comments: For locales/frames, (the relevant analogue of) prime ideals are (formal) points. To be precise, these are completely coprime filters. In the draft paper [http://www.paultaylor.eu/ASD/loccbv] on which I am working (and a propos of which I asked this question) I have a partly constructive argument that I also use to find points of inhabited overt subspaces. I am really more interested in the history of the arguments than the definitions (or even concepts). Maybe factorisation is the relevant idea to pursue in order to answer my history question. As Mamuka says, it all comes from prime numbers and therefore probably from the argument by infinite descent in Elements VII 31. REPLY [5 votes]: It is perhaps worth mentioning that you only need a choice function for nonempty subsets of the ring that you are working with. Indeed, given such a function $c$ and an ideal $I$ you can define $d(I)=\{x\not\in I : I + Rx \neq R\}$ and then $$ e(I) = \begin{cases} I + R c(d(I)) & \text{ if } d(I)\neq\emptyset \\ I & \text{ if } d(I) = \emptyset \end{cases} $$ You can then iterate transfinitely by the rule $$ e^\alpha(I) = \begin{cases} e(e^\beta(I)) & \text{ if } \alpha = \beta + 1 \\ \bigcup_{\beta<\alpha}e^\beta(I) & \text{ if } \alpha \text{ is a limit ordinal}. \end{cases} $$ Now $e^\alpha(I)$ will be independent of $\alpha$ when $\alpha$ is large enough, equal to $I^*$ say; and $I^*$ will be maximal provided that $I$ is proper. If you prefer a description without ordinals, we can say that $a\in I^*$ if for each family $\mathcal{J}$ of ideals such that $I\in\mathcal{J}$ $e(J)\in\mathcal{J}$ whenever $J\in\mathcal{J}$ The union of any nonempty chain in $\mathcal{J}$ is also in $\mathcal{J}$ there exists $J\in\mathcal{J}$ such that $a\in J$. Moreover, the class of rings for which one can write down a choice function is quite large. It certainly contains all rings for which we have an explicit enumeration $\mathbb{N}\to R$, and is closed under taking quotients, products, tensor products, finitely generated algebras and so on. Of course we do not have choice functions for nontrivial algebras over $\mathbb{R}$ or $\mathbb{Q}_p$. However, in some cases like this the ring will have a topology and it will be sufficient to have a choice function for nonempty open subsets of nonempty closed subsets, which can be arranged explicitly.<|endoftext|> TITLE: Hypercovers of sheaves in classical and quasi-categories QUESTION [8 upvotes]: I am interested in relating the definition of hypercovers in the $\infty$-topos of sheaves on an $\infty$-Grothendieck site to the classical definition of hypercovers of presheaves on a Grothendieck site. The definition for hypercovers in an $\infty$-topos that I am using is from Higher Topos Theory: In an $\infty$-topos $\mathfrak{X}$ a hypercovering of an element $X$ is the structure map $|U_\bullet|\to X$ of the geometric realization of a simplicial object $U_\bullet\in s\mathfrak{X}_{/X}$ such that the map $$U_n\to (cosk_{n-1}U_\bullet)_n$$ is an effective epimorphism in $\mathfrak{X}_{/X}$ (its Cech nerve is a simplicial resolution of the target) for all $n\geq 1.$ My question now is if this definition gets any easier if we restrict to an $\infty$-topos that is given by the $\infty$-sheaves on a (small) quasi-category with Grothendieck topology. Ideally, I want to relate this to the classical definition that a hypercovering on the presheaves $\mathcal{P(C)}$ of a Grothendieck site is an augmented simplicial object $U_\bullet\in s_+\mathcal{P(C)}$ (with $X\cong U_{-1}$) such that the maps $$U_n\to (cosk_{n-1}U_\bullet)_n$$ are local epimorphisms. So it all boils down to the question if there is some way of viewing the effective epimorphisms of the $\infty$-sheaf topos on an $\infty$-site as local epimorphisms. Is there any source that does this example? Lastly, I just want to note that we get $\infty$-sheaves from $\infty$-presheaves by localizing along those local epimorphisms the target of which is representable. But I don't see how this would imply my question. REPLY [6 votes]: Local epimorphisms are precisely those morphisms in $\mathcal{P}\left(\mathcal{C}\right)$ which become effective epimorphisms after applying the sheafification functor. In particular, if $f$ is an effective epimorphism in $\mathfrak{X}=Sh_\infty\left(\mathcal{C}\right),$ then when regarded as a morphism in $\mathcal{P}\left(C\right)$ via the full and faithful embedding $$Sh_\infty\left(\mathcal{C}\right) \hookrightarrow \mathcal{P}\left(\mathcal{C}\right)$$ it is a local epimorphism. That is to say, you can rephrase the second definition you give (the "classical one") as an augmented simplicial object $$U:\Delta_+^{op} \to \mathcal{P}\left(\mathcal{C}\right)$$ such that, $a \circ U$ is a hypercover in the first sense (where $a$ denotes sheafification). More precisely, $$\left(\Delta^{op}\right)^{\triangleright}\cong \Delta_{+}^{op}$$ so the augmented simplicial object $a \circ U$ of $Sh_\infty\left(\mathcal{C}\right)$ corresponds to a simplicial object in $Sh_\infty\left(\mathcal{C}\right)/a\left(X\right)$ which is a hypercover (this correspondence between an augmented simplicial object and a simplicial object in the slice category uses the definition of the Joyal's join construction as a left adjoint). Note we are also using that sheafification is a left exact functor, and the functors $cosk_n$ etc. are all computed using finite limits. In response to commented question: Why are local epimorphisms precisely those morphisms which become effective epimorphisms after sheafification? One way of phrasing what a local epimorphism is as follows: Given $f:F \to G,$ one can form its Cech nerve $C(f):\Delta^{op} \to \mathcal{P(C)}.$ $f$ is said to be an effective epimorphism if the canonical map $\operatorname{hocolim} C(f) \simeq G.$ Being a local epimorphism is equivalent to the canonical map $\operatorname{hocolim} C(f) \to G$ (which is a sieve always) to be a covering sieve. But, being a covering sieve is equivalent to being a subobject whose sheafification becomes an equivalence, i.e. $a\operatorname{hocolim} C(f) \to aG$ is an equivalence. But since $a$ is left exact and a left adjoint, one has $$a\operatorname{hocolim} C(f) \simeq \operatorname{hocolim} C(af),$$ so one sees directly that $f$ is a local epimorphism if and only if its sheafification is an effective epi.<|endoftext|> TITLE: Cut and Fold Polyhedron! QUESTION [10 upvotes]: I have two convex polyhedra such that their sums of side areas are equal. It is true that I can cut one of them and flatten it on the plane, then fold the flattened polygon to reach the other polyhedron? For special case in which these two polyhedra are cube and regular tetrahedron, The answer in Yes I think. But I think the answer is "No, not necessarily" in general case, and I think the problem is difficult! REPLY [12 votes]: The general answer is sometimes Yes, various unfold/refold pairs of convex polyhedra exist, but not always, depending on what is meant by "cut." Here is an example that Erik Demaine, Marty Demaine, Anna Lubiw, and I worked out carefully: The Latin-cross unfolding of the cube can refold into precisely 23 distinct convex polyhedra, as displayed below (all of the same surface area): the cube, two doubly covered flat quadrilaterals, seven tetrahedra, three pentahedra, each with one or more quadrilateral faces, four hexahedra, and six octahedra:     Fig. 25.30, p.408 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra, 2007. Here is a "movie" of one of the refoldings to a tetrahedron:                     But more generally, in the 2012 paper Refold Rigidity of Convex Polyhedra, we showed that every convex polyhedron can be cut open and refolded to an incongruent convex polyhedron. But if the cutting is restricted to follow edges of the convex polyhedron ("edge unfoldings"), then there are "refold-rigid" polyhedra. For example, each of the 43,380 edge unfoldings of a dodecahedron may only fold back to the dodecahedron. Added. I apologize for not initially understanding Morteza's precise question, which asks whether or not, for any two given pair of convex polyhedra $P$ and $Q$ with the same surface area, is it always possible to cut open $P$ and refold to $Q$. This remains an open question as far as I know. It is a version of Open Problem 25.31 (Fold/Refold Dissections) in Geometric Folding Algorithms.<|endoftext|> TITLE: Poisson ideals vs. ideals generated by Poisson central elements QUESTION [5 upvotes]: Let $R$ be a Poisson algebra (over $\mathbb C$, say) with Poisson center $Z = \{c \in R : \{c,R\} = 0\}$ and consider two types of ideals $I \leq R$: $I = \langle (c_i) \rangle$ is generated by Casimirs $c_i$, i.e. each $c_i\in Z$ $I$ is a Poisson ideal, i.e. $\{I,R\} \leq I$. I'm having trouble distinguishing these geometrically: both seem to be about subschemes of $Spec\ R$ that are unions of symplectic leaves. But that seems too good to be true (or I would hope someone would have told me already). Still, I will ask: If $I$ is a Poisson ideal, is $\sqrt{I} = \sqrt{ \langle I \cap Z \rangle}?$ If so, what is a reference? If not, what is a counterexample? REPLY [5 votes]: I am not that familiar with the algebraic approach but the intersection on the rhs seems quite hard to me. Take the Poisson structure on the plane given by $\{x,y\}=xy$. Then $Z$ contains only constant functions and therefore for any Poisson ideal $I$ you have $I\cap Z$ equal to $Z$ or $0$ depending whether the ideal contains constants or not. Take $I=\langle x,y\rangle$ and it looks to me your condition is not fulfilled... To put it another way there seems to me to be many situations in which the Poisson center is given only by constants and still the symplectic foliation is quite rich, thus many Poisson ideals.<|endoftext|> TITLE: Reference request: Continuity of unique maximizer of linear functional on convex set QUESTION [5 upvotes]: Does anyone know reference for a theorem of the following sort: Proposition: Let $K \subset\mathbb {R}^n$ be a compact convex set, and assume that $$f(w):=\operatorname{argmax}_{x\in K}w(x) $$ is unique for each nonzero linear functional $w:\mathbb{R}^n\rightarrow \mathbb{R}$. Then the function $f$ is continuous at nonzero $w$. ? I don't want a proof, just a citeable reference or the name of this theorem or some generalization. REPLY [3 votes]: Let $\sigma(w) = \max_{x\in K} w(x)$ be the support function of set $K$. This function is convex and its subdifferential is exactly your function $f$, i.e. $f(w) = \partial \sigma(w)$. Your assumption that $f(w)$ is unique for all $w \neq 0$ is equivalent to the differentiablity of $\sigma$ on the set of nonzero $w$. We now use the following (which can be found for example in Convex Analysis and Minimization Algorithms I: Part 1: Fundamentals by Hiriart-Urrut and Lemaréchal, Remark 6.2.6) If a convex function is differentiable on an open set, then it is continuously differentiable on that set. to conclude that $\sigma$ is continuously differentiable on the region $w \neq 0$, which is the same as saying that $f$ is continuous.<|endoftext|> TITLE: Multiplicative Identity for all elements in SU(n) QUESTION [11 upvotes]: Let $\{P_i\}$ be a subset of $SU(n)$ such that for any $U$ in another subset (or perhaps subgroup) $H$ of $SU(n)$: $$P_1UP_2U\cdots P_mU=I$$ where $I$ is the identity element. Is there a sequence $\{P_i\}$ such that $H$ be enlarged to the whole $SU(n)$? Also (perhaps more interesting): how large can $H$ be made? Perhaps a similar question can be asked about groups in general. For $SU(n)$, I suspect that $U$'s must be generated by a single Lie Algebra element [which I understand to mean $U=exp(-i \theta A)$ for a fixed Hermitian $A$ and any real number $\theta$]. Notice that a similar question can be asked about rotation matrices in $\mathbb{R}^3$ which I believe to be sufficient for a more general proof. A direction that I pursued was to note that the transformation of the generator (logarithm) of $U$ under the sequence has to be a linear map (identically zero) and then try to prove that it (or its restriction) can have a dimension of at most 1. As a non-$SU(n)$ example consider the group of affine transformations of the 3-dimensional Euclidean space and take ${P_i}$ to be 4 alternating 180 rotations (flips) along $x$ and $y$. Then $U$ can be any translation. Now It is highly unlikely that the set of $U$'s can be extended to the whole affine group by using a larger set of $P_i$'s but I can't prove it. Excuse my quantum physicist notation and I hope that this is not too elementary although I would appreciate an elementary answer. REPLY [7 votes]: I can show that it is impossible to achieve $H=SU(n)$. More generally, if $G$ is a compact connected Lie group, I will show that the map $U \mapsto P_1 U P_2 U \cdots P_m U$ is surjective, and therefore the preimage of the identity can't be all of $G$. Proof Such a $G$ is a compact connected orientable manifold so we can talk about the degree of self-maps of $G$. If $\phi: G \to G$ is a continuous function which misses a point, then $\phi$ has degree $0$. We will compute the degree of $U \mapsto P_1 U P_2 U \cdots P_m U$ and see that it is nonzero. Since $G$ is connected, $U \mapsto P_1 U P_2 U \cdots P_m U$ is homotopic to $U \mapsto U^m$. A generic matrix in $SU(n)$ has $m^{n-1}$ many $m$-th roots in $SU(n)$ (and $m^n$ many $m$-th roots in $U(n)$). At some point in my life, I worked out that all of these $m$-th roots contribute to the degree with the same sign. More generally, let $G$ be a compact connected Lie group of rank $r$ (meaning that maximal torii have degree $r$). For a generic element $x$ of $G$, there is a unique maximal torus $T$ containing $x$ and all $m$-th roots of $x$ are in that torus; there are $m^r$ of them. For our purpose, the only important fact is that $m^r \neq 0$, so $U \mapsto U^m$ has nonzero degree and $U \mapsto P_1 U P_2 U \cdots P_m U$ must be surjective. $\square$ ADDED Robert Bryant proves that, if $G$ is a compact connected Lie subgroup and $T$ is a connected subgroup so that $P_1 U P_2 U \cdots P_m U=1$ for $U \in T$, then $T$ is contained in a maximal torus. The point of this edit is to point out that, if $G$ is semisimple, and $T$ is any maximal torus, then there are $P_i$ which achieve this. Robert Bryant already does this for $SU(n)$ in his answer. Let $N(T)$ be the normalizer of $T$, so we have a short exact sequence $1 \to T \to N(T) \to W \to 1$, where $W$ is the Weyl group. Let $c$ be any element of the Weyl group which acts with no eigenvalues of $1$ on the Lie algebra of $T$ (for example, a Coxeter element) and let $h$ be the order of $c$. Let $P_1$ be a lift of $c$ to $N(T)$. Let $P_1^h = z$ (an element of $T$). Set $P_1=P_2 = \cdots = P_{h-1}$ and $P_h = P_1 z^{-1}$. Then I compute $\def\Ad{\mathrm{Ad}}$ $$P_1 \exp(t) P_1 \exp(t) \cdots P_1 \exp(t) P_1 z^{-1} \exp(t) = \exp\left( (\Ad(P_1) + \Ad(P_1)^1 + \cdots + \Ad(P_1)^h ) t \right) P_1^h z^{-1} = \exp(0) \cdot 1.$$<|endoftext|> TITLE: Horn's inequalities for n matrices QUESTION [8 upvotes]: Where I can find necessary and sufficient conditions on eigenvalues of Hermitian matrices with the relation $$A_1 + A_2 + ... + A_n = A_0 ,$$ i.e. Horn's inequalities for n matrices? Can such inequalities be obtained just from Horn's inequalities for 3 matrices if we couple matrices in this way $$(((A_1 + A_2) + A_3) + A_4) + ... = A_0 ?$$ Thanks. REPLY [10 votes]: 1) Section 7 of [Knutson-Tao-Woodward], http://arxiv.org/pdf/math/0107011.pdf 2) Yes. To obtain the (minimal set of) inequalities, you can just glom the puzzles together.<|endoftext|> TITLE: Do direct limits (filtered colimits) commute with pullbacks, in C*-algebras? QUESTION [11 upvotes]: I asked this question already on math.stackexchange, but maybe it is also useful to ask this here, since it was not answered there. Suppose we have three directed sequences of $C^*$-algebras, say $(A_n,\varphi_n)$,$(B_n,\psi_n)$ and $(C_n,\theta_n)$ and $*$-homomorphisms $\alpha_n:A_n\rightarrow C_n$ and $\beta_n:B_n\rightarrow C_n$, then we can take the pullback $A_n\times_{C_n}B_n$ for all $n\in\mathbb{N}$ and can also take the direct limit, thus $\lim_{n\rightarrow\infty}{A_n\times_{C_n}B_n}$. My question is: Does the following hold: $$\lim_{\rightarrow}{A_n\times_{C_n}B_n}=\lim_{\rightarrow}{A_n}\times_{\lim_{\rightarrow}{C_n}}\lim_{\rightarrow}{B_n}$$ or in other words: do direct limits preserve pullbacks? From my point of view this does not hold in general but I can not find a good argument. Someone an idea? Thank you very much. REPLY [12 votes]: I may be showing my ignorance of category theory, but don't think this is true. Work on $l^2$. Let $\mathcal{A}_n$ consist of the operators $A$ satisfying $\langle Ae_i, e_j\rangle = 0$ if $\max(i,j) > n$ (so $\mathcal{A}_n$ is isomorphic to the $n\times n$ matrices). The direct limit of the $\mathcal{A}_n$ is the compact operators on $l^2$. Let $P$ be a rank one projection that is not contained in any of the $\mathcal{A}_n$, let $\mathcal{B} = \mathbb{C}\cdot P$ be the C*-algebra it generates, and let $\mathcal{B}_n = \mathcal{B}$ for all $n$. Also let $\mathcal{C}_n = B(l^2)$ for all $n$. All the morphisms are inclusion maps. Each pullback $\mathcal{A}_n\times_{\mathcal{C}_n} \mathcal{B}_n$ is zero, so the direct limit of the pullbacks is zero, but the pullback of the direct limits is (isomorphic to) $\mathcal{B}$.<|endoftext|> TITLE: Is the ordinary locus affine? QUESTION [5 upvotes]: Let $p$ be a prime number and let $Y$ over $\mathbb F_p$ be a Siegel modular variety, with minimal compactification $X$. It is well known that $X^{\operatorname{ord}}$, the ordinary locus of $X$ is affine (since it is cut out by the Hasse invariant, that is a section of an ample line bundle). But what about $Y^{\operatorname{ord}}$, the ordinary locus of $Y$? So the question is the following. Is the ordinary locus of $Y$ an affine scheme? Thank you! REPLY [8 votes]: No. It's well known that the complement of a closed subset of codimension two or more in an affine variety is never affine. The minimal compactification has codimension g.<|endoftext|> TITLE: Escape the zombie apocalypse QUESTION [57 upvotes]: Consider zombies placed uniformly at random over $\mathbb{R}^2$ with asymptotic density $\mu$ zombies/area. You are placed at a random point and can move with speed $1$. Zombies move with speed $v\leq 1$ straight towards you, what is the probability $P(\mu,v)$ you can escape to infinity without a zombie catching you? What if the zombies can move in any direction and might collude to set up a wall of high density or similar tactics? Lets call it a win for zombies if for every $d>0$, one of them can get within a distance $d$ in finite time. Addendum: Is there a finite collection of colluding zombies and a player placement, from which escape is impossible? What is the least number of zombies? REPLY [3 votes]: Simple pursuit Zombies moving towards you will always catch you, but due to their lack of intelligence, your survival time increases exponentially with your relative speed. In $k=O(1)$ dimensional space ($k=2$ in the problem), the expected survival time is $d⋅(1/Θ(μd^k))^{(1+1/v)(1±o(1))/(k-1)}$ if $v$ is bounded below 1 and $μd^k→0$. I conjecture that for fixed $v$, the above $o(1)$ is unnecessary. If $μ$ depended on the distance $r$ from the origin, the threshold value for survival (at large $r$ and constant $v<1$) is $μ(r) = r^{-(k-1)v/(1+v)±o(1)}$ (the $o(1)$ is likely negative and necessary here). Consider first the continuous field version of the problem: The initial density is $μ$ and the player loses when the mass within distance $d$ of the player reaches (or exceeds) 1. Let $r(t)$ be the trajectory of the player; $r(0) = 0$. If $a$ and $b$ are trajectories of two possible zombies, we have: - $a'(t) = v \frac{r(t)-a(t)}{|r(t)-a(t)|}$ - $|b(t)-a(t)|$ is nonincreasing - For small $|b(t)-a(t)|$, $b'(t) = a'(t) - \frac{v}{|a(t)-r(t)|} (b(t)-a(t))_⊥ + O(\frac{|b(t)-a(t)|}{|r(t)-a(t)|})^2$ where the orthogonal projection $(b(t)-a(t))_⊥ = b(t)-a(t)-(b(t)-a(t))⋅(r(t)-a(t)) \frac{r(t)-a(t)}{|r(t)-a(t)|^2}$. - If $f_T(a(0)) = a(T)$, then the density at time $T$ at $a(T)$ is $μ/\det J_{f_T}$, and by integrating the above $b'$ equation, we get $\log \det J_{f_T} = -(k-1)v \int_0^T \frac{dt}{|r(t)-a(t)|}$. Now among all $a$ and $r$ with $|a(T)-r(T)|≤d$, the integral (and thus the density) is minimized if $|a(0)|=(1+v)T+d$, which requires moving in a straight line from the origin at maximum speed. Furthermore, in this case, the density at $a(T)$ matches the average density within distance $d$ up to a constant factor, and the precise bounds in the first paragraph follow. Lower bounds For the nonfield version, we can get the lower bounds by escaping in a nearly straight line while avoiding traps. This is possible here even if the player velocity is always within $ε$ (if $ε$ is $Θ(1)$) of moving with speed 1 in the positive $x$ direction. Intuitively, on a straight line, the average distance between traps of size $O(d)$ is $O(1/(μ_1 d^{k-1}))$ (where $μ_1$ is the relevant field density; $1-v=Ω(1)$), and using the (approximate) independence, the frequency of larger traps drops exponentially with trap size, with the lower bounds (in the first paragraph) reached with high probability unless the initial position is a trap. However, formalization of traps is a bit tricky, so we instead observe that we can trace out a trajectory of speed $1+v$ of sufficient clearance against the initial configuration, and then evolve it the same way as the zombies to get the escape trajectory. As long as all tangents of the initial trajectory are at an angle $≤α$ (for $α≤45°$) to the $x$ axis, this property will hold throughout the trajectory evolution, allowing us to ensure that the clearance will not shrink too much. For fixed $v$ ($μ$ does not depend on $r$), we can remove the $o(1)$ from the lower bounds for survival time by choosing a trajectory with variable $α$ with, at each point, $1/α$ at least polynomial in the distance between the point and the final destination. Also, for $v=1$ (and small $d$) and variable $μ(r) = r^{-k/2+1/6-ε}$, you can survive by making your trajectory increasingly smooth: Zombies following you at a small distance gain on you at a speed proportional to the square of the path curvature (and the square of the distance to you), so with curvature $r^{-0.5-ε}$ you can avoid $d→0$ as $r→∞$ for those zombies. In a straight line path and $d=Θ(1)$, you will encounter zombies at typical intervals $s = Θ(1/(μ(r) r^{(k-1)/2}))$. Avoiding an incoming zombie from a distance $s$ uses correction $O(\sqrt s)$, corresponding to curvature $O(s^{-1.5})$, which is $O(r^{-0.5-ε_2})$ for the above $μ(r)$. Also, for $v≤1$ and $d=0$, you can survive indefinitely (i.e. not lose at finite time) by simply moving at speed 1 in a straight line in any unoccupied direction — or even, with probability 1, by moving with speed 1 along any curve with bounded curvature, with the curve chosen independently of zombie positions. Upper bound For the upper bound, it suffices to consider a single point and a linear approximation to the problem. To escape, for every $R>0$, the player would have to cross (at time $O(R/v)$) an $a(t)$ that starts at the sphere $|a(0)|=R$. A player cannot approach $a(t)$ to within distance $d$ without increasing the field density at $a(t)$ in $ρ =(R/d)^{(k-1)/(1+1/v)}$ times. Furthermore, as long as the increase in density at $a(t)$ is $o(ρ)$ times, the cumulative relative nonlinearity within distance $O(d)$ of $a(t)$ is $o(1)$. (Proof outline: If the remaining density increase is $ρ_1$ times, then the distance of the relevant points to $a$ is $D=O(ρ_1^{1/(k-1)} d)$, and with the player far enough compared to $D^2/d$, the nonlinearity is small enough.) From there, for large enough $R$, $\{b(0):|b(t)-a(t)|≤d|\}$ contains a volume $ω(\log R)$ ellipsoid (contained within distance $O(R)$ from the origin). By a counting argument, with probability $1-o(1)$ all such ellipsoids contain at least one relevant point, as required. For the bounds without $o(1)$, we are off by a factor of $\log(1/(μd^k))$ inside the $Θ(μd^k)$. However, I expect that the $\log$ factor can be eliminated by using the player path rather than a single point $a(t)$ and by analyzing the impact of nonlinearities. Intelligent pursuit If zombies (in $k = O(1)$ dimensional space) could strategize and cooperate, and letting $D=\frac{1}{μd^{k-1}v}$, they could surround you with gaps $0$ dependent on $v$ (and small enough $d$ dependent on $v,ε$). At $v=1$, $k+1$ well-placed zombies (i.e. 3 for the plane) can win in finite time even at $d=0$. The reason is that a single pursuer can guard a half-space. They can even (deterministically) win if they have nonzero but small enough reaction time proportional to distance (i.e. the speed of light is finite). A single pursuer can guard a slowly receding half-space, allowing 1 (effectively 2) out of $k+1$ pursuers to advance.<|endoftext|> TITLE: Prove that ..., f(x-2), f(x-1), f(x), f(x+1), f(x+2),... is algebraically linearly independent without the Fourier transform QUESTION [7 upvotes]: Suppose that $f$ is in $L^2(\mathbb{R})$ and consider the set of integer translates of this function, $V=\{f(x-k):k\in\mathbb{Z}\}$. This set is linearly independent: taking the Fourier transform of the finite sum $\sum a_k f(x-k)$ one gets $p(e^{2\pi i\xi})\widehat{f}(\xi)$ for some polynomial $p$. If the sum is zero and $f$ is nonzero, then $p$ must be zero on some set of positive measure; this is an infinite set, implying that $p$ must be the zero polynomial and so each $a_k$ must be zero. I find this to be an especially nice application of the Fourier transform. My question is this: does there exist a proof of this fact which does not use the Fourier transform? The $L^2$ condition could be modified, but obviously one needs some kind of integrability condition to disallow the constant functions. One can prove this using a variant of the Fourier transform, so I should say that I'm really looking for a proof where you don't integrate against complex exponentials. As for why $V$ would be an interesting thing for mathematicians to look at: the closure of the span of $V$ (in $L^2$) is one of the fundamental objects in wavelet theory --- a principal shift-invariant space. REPLY [11 votes]: Suppose there is some linear dependence. If the set is linearly dependent, space $V$ should be finite dimensional. Fix a finite subset $S$ of $\mathbb{Z}$ so that supposedly the set of $f(x-k)$ where $k\in S$ would span. Note that as $k_0 \rightarrow \infty,$ $\langle f(x-k_0), f(x-k) \rangle \rightarrow 0,$ which would imply $\|f(x-k_0)\|$ went to $0,$ but $\|f(x-k_0)\|$ is constant, which is a contradiction.<|endoftext|> TITLE: About an identity which gives immediate proof of the permanent lemma QUESTION [7 upvotes]: Let $A$ be a $n \times n$ matrix over field $F$. Let $a_1, \cdots, a_n$ be the column vectors of $A$. For any subset $S \subseteq [n] = \{1, 2, \cdots, n\}$, let $a_S = \sum_{i \in S} a_i$. Alon's celebrated permanent lemma states that if the permanent of $A$ is nonzero, then for any $b \in F^n$ there is some set $S$ such that $a_S$ and $b$ differs in all coordinates. Let $v_S = \prod_{i=1}^n (a_{S,i} - b_i)$. Then $v_S \neq 0$ iff $a_S$ and $b$ differ in all coordinates. By attempting to prove the lemma by my own, I discovered this identity. $$\sum_{S \subseteq [n]} (-1)^{n-|S|} v_S = \operatorname{perm} A$$ So $\operatorname{perm} A \neq 0$ immediately implies the existence of some $v_S \neq 0$. Note that the formula of left-hand side depends on $b$. If we are working on a field with absolute value, then the absolute value of $v_S$ measures the difference between $a_S$ and $b$ in coordinates. For example, there is a set $S$ with $|v_S| \geq (\operatorname{perm} A) / 2^n$. The bound is tight for $A = I_n$ and $b = (1/2, \cdots, 1/2)$. This gives a quantitative version of permanent lemma. Given the simplicity of the identity, I believe that it should had been noticed by other mathematicians. In what literature can I find observations about this identity? Are there any generalizations or applications around this idea? REPLY [8 votes]: This identity is a particular case of theorem 3 in "A generalization of Combinatorial Nullstellensatz" by Michał Lasoń. Indeed your proof is essentially reinventing the combinatorial nullstellensatz. :) The usual proof of Alon's lemma looks at the polynomial $\prod_{i=1}^n (\sum_j a_{ij}x_j - b_i)$. And your identity follows by computing the coefficient of $x_1x_2\cdots x_n$ in two different ways, like in the paper above.<|endoftext|> TITLE: Whittaker models for $GL_n$ and Fourier coefficients QUESTION [7 upvotes]: Let $G$ be a compact abelian group. Then we know, because of the Peter-Weyl theorem, that $L^2(G)$ decomposes as a Hilbert space direct sum of 1 dimensional representations of $G$. Let $\mathbb{A}$ denote the adeles for $\mathbb{Q}$. Suppose we are given an automorphic funtion $\phi:GL_2(\mathbb{A})\to \mathbb{C}$ (I am following the book of Gelbart, Automorphic forms on adele groups, definition 3.3). In particular, $\phi(\gamma g)=\phi(g)$, for $\gamma\in G(\mathbb{Q})$. Fix $g\in G(\mathbb{A})$, then we get a function on $\mathbb{Q}\backslash\mathbb{A}$ defined as $$x\mapsto \phi\left(\left(\begin{array}{cc}1 & x\\ 0 & 1\end{array}\right)g\right)$$ This being a smooth function on the compact abelian topological group $\mathbb{Q}\backslash \mathbb{A}$ we can write its Fourier series, and the Fourier coefficient for the character $\psi$ will be $$W_{\phi,\psi}(g)=\int_{\mathbb{Q}\backslash \mathbb{A}}\phi\left(\left(\begin{array}{cc}1 & x\\ 0 & 1\end{array}\right)g\right)\psi(x)^{-1}dx$$ Using the Fourier coefficients we may write $$\phi(g)=\sum_{\psi}W_{\phi,\psi}(g).$$ For $GL_n$ where $n>2$ one defines analogous objects $$W_{\phi,\psi}(g)=\int_{N(\mathbb{Q})\backslash N(\mathbb{A})}\phi(xg) \psi(x)^{-1}dx$$ However, the space $N(\mathbb{Q})\backslash N(\mathbb{A})$ is no more a topological group, though it is compact. Instead of doing this, why don't we simply repeat the process for the $n=2$ case, that is, for $g\in GL_n(\mathbb{A})$ consider, for example, the inclusion of $\mathbb{A}$ in $GL_n(\mathbb{A})$ given by $x\mapsto U(x):=I+xE_{12}$ where $E_{12}$ is the matrix whose $12$ entry is 1 and all other entries are 0. The function $x\mapsto \phi(U(x)g)$ is a smooth function on $\mathbb{Q}\backslash \mathbb{A}$ and we can define the Fourier coefficients as before, that is, $$W_{\phi,\psi}(g)=\int_{\mathbb{Q}\backslash \mathbb{A}}\phi(U(x)g)\psi(x)^{-1}dx$$ EDIT Then one has $$\phi(U(x)g)=\sum_{\psi}W_{\phi,\psi}(g)\psi(x)$$ Once again, put $x=0$ to get $$\phi(g)=\sum_{\psi}W_{\phi,\psi}(g).$$ I have not checked this very carefully, but, I guess, as mentioned in the response below we will also have $$ \phi(g)=\sum_{\gamma\in \mathbb{Q}} W_\phi\left(\begin{pmatrix}\gamma&0\\0&1\end{pmatrix}g\right).$$ My question is why do we need to work with the more complicated object $N(\mathbb{Q})\backslash N(\mathbb{A})$. Moreover, if we have to work with this object, then is there an analogous theory of Fourier series behind this. REPLY [8 votes]: It is indeed reasonable to wonder what's going on with "Fourier expansions" along non-abelian (sub-) groups... since, among other things, any one-dimensional representation has to factor through the maximal abelian quotient, so must lose some information. For contrast: the unipotent radical $N$ of the standard minimal parabolic (upper-triangular matrices) in $GL_3$ is just barely not abelian. It is an example of a sort of Heisenberg group, and the representation theory of Heisenberg groups is well understood, via forms of the Stone-vonNeumann theorem (or, more generally, "Mackey imprimitivity"...) That is, there are one-dimensional repns which are trivial on the center $\pmatrix{1 & 0 & * \cr 0 & 1 & 0 \cr 0 & 0 & 1}$, and for each non-trivial character on the center a unique isomorphism class of irreducible. But this is not what we are doing in writing "Fourier expansions" of automorphic forms. The way to obtain the Fourier expension on $GL_n$ is iterative, and does use the fact that Fourier expansions along abelian subgroups lose no information. Thus, the literal Fourier expansion along $N^{n-1,1}(\mathbb Q)\backslash N^{n-1,1}(\mathbb A)$ loses no information (where $P^{i,n-i}$ is the standard maximal proper parabolic with diagonal blocks of sizes $i$ and $n-i$ and $N^{i,n-2}$ is its unipotent radical). For cuspforms, the $0$th Fourier coefficient is $0$. Now comes the trick: the upper $n-1$ block of the Levi component of $P^{n-1,1}$ acts transitively on non-trivial characters on $N^{n-1,1}$, and the isotropy subgroup of a suitable "base choice" is close to being the $(n-2,1)$ parabolic in that block of the Levi component. So the $n-1,1$ Fourier expansion of a cuspform is a sum of translates of a single Fourier component (note, not just "coefficient"). Then there's an induction, which I won't re-type here... Leading to $$ f(g) \;=\; \sum_{\gamma \in U(\mathbb Q)\backslash H(\mathbb Q)} W_{f,\psi}(\gamma g) $$ where $H$ is the upper left $n-1$ block of the Levi component of $P^{n-1,1}$ and $U$ is the unipotent radical of the standard minimal parabolic in $H$, and $$ \psi\pmatrix{1 & x_{12} & x_{13} & \ldots \cr 0 & 1 & x_{23} & \ldots \cr & & \ddots & \cr & & & 1 & x_{n-1,n} \cr & & & & 1 } \;\;=\;\; \psi(x_{12}+x_{23}+\ldots+x_{n-1,n}) $$ is a character on the unipotent radical of the minimal parabolic in $GL_n$, obviously vanishing on the derived group of the unipotent radical. The induction/iterative procedure to obtain this expansion uses the assumption that $f$ is a cuspform repeatedly to know that no information is lost in successively dropping the zero-th Fourier component in literal Fourier expansions along various abelian unipotent radicals. As GH noted, Jim Cogdell's various notes on $GL_n$ go through this process in some detail. Edit: to comment on "why not expand along this abelian subgroup..." as in the edited version of the question: first, it is certainly legitimate to do so. The merit of the slightly labyrinthine version (after Shalika and P-S) is that (because of "uniqueness of (local) Whittaker models") the resulting Whittaker function _factors_over_primes_ (for cuspform $f$ generating an irreducible repn of the adele group). That is, up to a single global normalizing constant, $W_{f,\psi}$ is entirely determined by the local data attached to $f$. This factoring-over-primes feature will not hold for arbitrary or capricious decompositions-along-subgroups, although it is very important to appreciate the situations where it does. So, again, the virtue of the Shalika-PS version of a "Fourier expansion" is that the Fourier components are essentially products of local (Whittaker) functions.<|endoftext|> TITLE: Whats the prime-to-$p$ Etale fundamental group of the affine line minus two points over $\mathbb{F}_p$? QUESTION [6 upvotes]: Background: I've often heard that the fundamental group of $\mathbb{P}^1/\mathbb{Q}-\{0,1,\infty\}$ is extremely hard to understand. First of all, it has a surjective map to the galois group $\rm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$, and then you further have a map $$\rm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\to\rm{Out}(\widehat{\rm{F}_2})$$ to try and get at. I'm trying to look at the simpler case of working over $\mathbb{F}_p$ instead of $\mathbb{Q}$. Of course, this acquires problems of its own, as degree $p$ extensions in char $p$ are hard. In fact, I'm aware that Tamagawa and many others have results that say you can use the absolute galois group of curves (possibly other varieties too) in char. $p$ to recover the curve, and they heavily use degree $p$ extensions. For this reason, I'm restricting my question to the prime-to-$p$ part of the fundamental group. Question: What is the prime-to-$p$ part of $\pi_1\left(\mathbb{P}^1/\mathbb{F}_p-\{0,1,\infty\}\right)$? If, as I suspect, the above is too difficult, I would be interested in any theorems pertaining to the structure of this group. If, however, this question turns out to be doable, I am equally interested in the similar questions with more points removed from $\mathbb{P}^1$. Thank you! For reference: In case people are unfamiliar, the prime-to-$p$ part of a profinite group $G$ is the inverse limit over all finite groups $H$ that have order prime to $p$ and which are continuous quotients of $G$. REPLY [8 votes]: The general exact sequence: For any scheme $X$ over a field $k$, there is a sequence $$ 1\to \pi_1^{et}(\overline{X})\to \pi_1^{et}(X)\to\operatorname{Gal}(\overline{k}/k)\to 1. $$ Here $\overline{X}=X\times_k\overline{k}$ in slight abuse of notation, and I am a bit sloppy about the base points here. This sequence can be found in SGA1, Exposé IX, Thm 6.1. Since $\mathbb{P}^1\setminus\{0,1,\infty\}$ is defined over $\operatorname{Spec}\mathbb{Z}$, you can compare the geometric part of the fundamental group in characteristic $0$ to characteristic $p$. I think this is an application of the specialization theory of the etale fundamental group in SGA1. In characteristic $0$, you can compare the etale fundamental group to the profinite completion of the topological fundamental group, SGA1, Exposé XII, Corollary 5.2. The result is that the prime-to-p part of the fundamental group is (independent of the characteristic) the group $\widehat{F_2}$, i.e. the pro-prime-to-p completion of the free group on $2$ generators. In the case where the base field has at least $3$ points, there is a $k$-rational point of $\mathbb{P}^1\setminus\{0,1,\infty\}$, in which case the sequence splits. The result is a split extension $$ 1\to \widehat{F_2}\to \pi_1^{et}(\mathbb{P}_{\mathbb{F}_q}^1\setminus\{0,1,\infty\})\to \widehat{\mathbb{Z}}\to 1. $$ This leaves the action to be identified. The same then applies for $\mathbb{P}^1$ with $n$ points removed. Topologically, the fundamental group is a free group on $n-1$ generators, so the geometric fundamental group (resp. its prime-to-p part) is the pro-prime-to p-completion of the free group on $n-1$ generators. The étale fundamental is an extension of this by $\operatorname{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)\cong \widehat{Z}$. Partial description of the action: At the moment, I can not write down an easy description of the action. Simply translating the definition of the étale fundamental group, the action is essentially described as follows: take an étale covering $C\to\mathbb{P}^1_{\overline{\mathbb{F}_q}}\setminus\{0,1,\infty\}$, this is defined over $\mathbb{F}_{q^n}$, and pullback along Frobenius gives another covering $C'\to\mathbb{P}^1_{\overline{\mathbb{F}_q}}\setminus\{0,1,\infty\}$. The action (and hence the extension) are then given by a homomorphism $\operatorname{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)\to\operatorname{Out}(\widehat{F_2})$. A more precise description of the action can be obtained using Grothendieck-Teichmüller theory (which attempts describing the action of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on $\widehat{F_2}$) and specialization of the étale fundamental group. For a survey on Grothendieck-Teichmüller theory can be found e.g. in this survey paper of Leila Schneps. In §3.1 of this article, you can find the description of Ihara's morphism which describes the action of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on $\widehat{F_2}$: $$ \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \operatorname{Aut}(\widehat{F_2}): \sigma\mapsto \left\{x\mapsto x^{\chi(\sigma)},y\mapsto f_\sigma^{-1}y^{\chi(\sigma)}f_\sigma\right\} $$ where $x$, $y$ are chosen generators for $F_2$, $\chi:\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \widehat{\mathbb{Z}}^\times$ is the cyclotomic character and $f_\sigma$ is the pro-loop given by composing the straight path from $0$ to $1$ with its image under $\sigma$. The specialization of the étale fundamental group should now imply that the action of $\operatorname{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p)$ is given by choosing a Frobenius element and using the following composition $$ \operatorname{Gal}(\overline{\mathbb{F}_p}/\mathbb{F}_p)\to \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \operatorname{Aut}(\widehat{F_2}). $$ This yields a well-defined extension since the Frobenius elements are conjugate, and it incidentally explains how the Grothendieck-Teichmüller theory for $\mathbb{Q}$, $\mathbb{Q}_p$ and $\mathbb{F}_p$ are related. I still do not know how to get a more explicit description of the extension; what we see from the above is that the action of Frobenius on the first loop is by raising the first loop to the $p$-th power. The action on the second loop is more complicated, because I do not know of explicit formulas for $f_\sigma$. Something along the lines of studying Grothendieck-Teichmüller theory locally is discussed in section 5 of these notes on open problems in Grothendieck-Teichmüller theory. Maybe some expert on Grothendieck-Teichmüller theory knows how to describe the action in the function field case? Or maybe this is complicated, after all the Frobenius elements are dense? A final note: to get a more precise description of the action, you can do the same thing that is done in characteristic $0$: the function field analogue of Belyi's theorem (about which you can read here) shows that the above is in fact an action on the algebraic curves over $\overline{\mathbb{F}_q}$ because all these are covers of $\mathbb{P}^1$ ramified at at most 3 points.<|endoftext|> TITLE: Is there an algorithm to write down the 27 lines of a cubic surface? QUESTION [19 upvotes]: Let $S$ be a smooth cubic surface defined by $f\in \mathbb Q[x,y,z,w]$. Is there an algorithm to write down the 27 lines on $S$? Or at least find a field extension of $\mathbb Q$ over which these lines are defined? REPLY [7 votes]: Here is yet another place to look for an algorithm: you can find a description of how to compute homogeneous equations for the Plücker coordinates of the lines on a cubic surface in the Macaulay 2 tutorial at http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.6/share/doc/Macaulay2/Macaulay2Doc/html/___Tutorial_co_sp__Fano_spvarieties.html. What you will find there is essentially is the same procedure as in Jim Carlson's Sage code mentioned in the comments above, but perhaps it is a bit easier to read. The tutorial also discusses the general problem of computing equations for the Fano variety (scheme) of a projective variety.<|endoftext|> TITLE: Fundamental group of $\mathbb{R}^3-F$ where $F\subseteq \mathbb{R}\times \{0\} \times \{0\}$ QUESTION [8 upvotes]: Maybe not research level. Let $Z\cong \mathbb{R}$ be the $z$-axis of $\mathbb{R}^3$. Clearly $\pi_1(\mathbb{R}^3-Z)\cong \mathbb{Z}$. Now if $F\subset Z$ is a closed non-empty subset, then one easily sees that $\pi_1(\mathbb{R}^3-(Z-F))=0$. What is if $F$ is not closed (for example $F=\mathbb{Q}$)? Is for any nonempty subset $F\subseteq \mathbb{R}$ the space $\mathbb{R}^3-(Z-F)$ simply connected? REPLY [7 votes]: Given a continuous pointed map $\gamma: (S^1,1) \to (\mathbb{R}^3 \setminus (Z-F), x)$, compactness of $S^1$ implies the intersection $\gamma(S^1) \cap Z$ is closed in $Z$. Thus, $\gamma$ represents an element of $\pi_1(\mathbb{R}^3 \setminus (Z-F'), x)$ for some closed nonempty subset $F' \subset F \subset Z$, and as you mentioned, this group is trivial. In particular, there is a contracting homotopy that lies in this smaller space.<|endoftext|> TITLE: Is every compact monothetic group metrizable? QUESTION [6 upvotes]: If $G$ is a compact (Hausdorff) topological group with a dense cyclic subgroup, is it necessarily true that $G$ is first countable? This claim seems to be implicit in a paper that I am reading at the moment (the paper asserts the existence of a translation-invariant metric which induces the topology of $G$, which is equivalent by the Birkhoff-Kakutani theorem). Previously posted at https://math.stackexchange.com/questions/852364/is-every-compact-monothetic-group-metrizable REPLY [4 votes]: No, a compact separable group need not be second countable and may contain ${\bf Z}$ as a dense subgroup. A simple example is a torus of infinite uncountable dimension. So let $(1, \theta_i)_I$ be an algebraic basis of ${\bf R}$ over ${\bf Q}$. We consider the compact group $G = ({\bf R / Z})^I$ endowed with the product topology. $G$ is separable but not second countable. This follows from the following classical results (see e.g. Dugundji, "Topology"). Theorem A product of Hausdorff spaces (each with more than a point) is second countable if and only if the product is at most countable and all spaces are second countable. Theorem A product of Hausdorff spaces (each with more than a point) is separable if and only if the product is at most the cardinality of the continuum and all spaces are separable. Now let us consider the element $\theta = (\theta_i)_{i\in I} \in G$. The group ${\bf Z}\theta$ generated by $\theta$ is dense in $G$. Indeed this comes from the following facts. For all finite $F \subset I$, the family $(1, \theta_i)_{i\in F}$ is free over ${\bf Q}$ hence the group generated by $(\theta_i)_{i\in F}$ is dense in $({\bf R / Z})^F$. An open set in $G$ contains a pullback (via the canonical projection) of an open set in $({\bf R / Z})^F$ for some finite $F\subset I$. Beware that this does not mean that any $y\in G$ can be approximated by a sequence of multiples of $\theta$. The group $G$ is neither second countable, first countable (By the Birkhoff-Kakutani theorem, as pointed out by YCor) nor sequentially compact. Being a cluster point does not allow to extract some converging subsequence.<|endoftext|> TITLE: K3 surfaces that correspond to rational points of elliptic curves QUESTION [10 upvotes]: In his work on mirror symmetry (http://arxiv.org/pdf/alg-geom/9502005v2.pdf) Igor Dolgachev has considered families of K3 surfaces of Picard rank at least 19 with the base given by $X_0(n)^+$, the quotient of the modular curve of level $n$ by Fricke involution. Consider a modular parametrization $\mu:X_0(n)^+ \to E$ for some rational elliptic curve $E$, say of rank two. A rational point $p$ on $E$ gives rise to a collection of points $\mu^{-1}(p)$ on $X_0(n)^+$ and therefore to a finite set of the aforementioned K3 surfaces. Question: What is special about these K3 surfaces? For example, is their Picard group of rank 19 or does it jump to 20? Do they have some other algebra-geometric characterization, such as existence of some sections of some line bundle on them? Has this been studied at all? P.S. My guess is that the jump in the rank of K3 would be some CM condition, and thus not terribly interesting. But then what is special about these K3-s that correspond to points of rank 2 curve? REPLY [4 votes]: You can ask similar questions for elliptic curves, and as far as I know, the answer to your question of "What is special..." is not much. Thus let $\mu:X_0(n)\to E$ be a modular parametrization of an elliptic curve of conductor $n$, and let $P\in E(\mathbb{Q})$. Then is there anything special about the elliptic curves parametrized by the points in $\mu^{-1}(P)$? (Aside from the obvious fact that they are Galois conjugates of one another and have marked cyclic subgroups of order $n$ that respect the Galois conjugation.) For something further to ponder, here's something that I've thought about over the years and never come up with anything interesting: How are the elliptic curves parametrized by $\mu^{-1}(kP)$ related (if at all) for $k=1,2,\ldots,\,$?<|endoftext|> TITLE: How much of homotopy theory can be done using only finite topological spaces? QUESTION [22 upvotes]: Let $X$ be a finite simplicial complex and let $B$ denote the set of barycenters of the simplices of $X$. McCord constructed a $T_0$ topology on $B$ with the property that the inclusion $B \to X$ is a weak homotopy equivalence. Clader showed that the geometric realization of $X$ can be recovered up to homeomorphism by iterating McCord's construction on the $n$th barycentric subdivision of $X$, taking an inverse limit, and passing to a certain quotient space. Moreover she proved that the inverse limit deformation retracts onto the quotient space and thus every finite simplicial complex is homotopy equivalent to an inverse limit of finite spaces. This is kind of a shocking result (though the shock wears off quickly once you stop and think about it), and it leads me to wonder: Would it be possible to dispense with the language of simplicial complexes entirely and rewrite the foundations of homotopy theory using finite topological spaces and their inverse limits? I am mainly interested in whether or not this is possible, though you may feel free to include in your answer remarks about whether or not this is desirable. Some possible complications that come to mind: Many important topological spaces (e.g. classifying spaces for most groups) do not have the homotopy type of a finite simplicial complex. One does not directly have access to certain tools, such as a Morse theory. Perhaps one can hope to approximate whatever is lost using finite models? REPLY [17 votes]: Vidit, thanks for the advertisement; Paul I'll answer your email shortly. As a minor point, there is a small but subtle mistake in Clader's work that is corrected in Matthew Thibault's 2013 Chicago thesis, which goes further in that direction. I do intend to finish the advertised book, but it is too incomplete to circulate yet. There is actually a large and interesting picture that connects mainstream algbraic topology to combinatorics via finite spaces. However, the right level of generality is $T_0$-Alexandroff spaces, $A$-spaces for short. These are topological spaces in which arbitrary rather than just finite intersections of open sets are open, and of course finite $T_0$-spaces are the obvious examples. One can in principle answer Paul's question in the affirmative, but the finiteness restriction feels artificial and the connection between $A$-spaces and simplicial complexes is far too close to ignore. The category of $A$-spaces is isomorphic to the category of posets, $A$-spaces naturally give rise to ordered simplicial complexes (the order complex of a poset) and thus to simplicial sets, while abstract simplicial complexes naturally give rise to $A$-spaces (the face poset). Subdivision is central to the theory, and barycentric subdivision of a poset is WHE to the face poset of its order complex. Categories connect up since the second subdivision of a category is a poset, which helps illuminate Thomason's equivalence between the homotopy categories of $\mathcal{C}at$ and $s\mathcal{S}et$. Weak and actual homotopy equivalences are wildly different for $A$-spaces. In the usual world of spaces, they correspond to homotopy equivalences and simple homotopy equivalences, respectively, a point of view that Barmak's book focuses on. The $n$-sphere is WHE to a space with $2n+2$ points, and that is the minimum number possible. If the poset $\mathcal{A}_pG$ of non-trivial elementary abelian $p$-subgroups of a finite group $G$ is contractible, then $G$ has a normal $p$-subgroup. A celebrated conjecture of Quillen says in this language that if $\mathcal{A}_pG$ is weakly contractible (WHE to a point), then it is contractible and hence $G$ has a normal $p$-subgroup. There are many interesting contractible finite spaces that are not weakly contractible. These facts just scratch the surface and were nearly all previously known, but there is much that is new in the book, some of it due to students at Chicago where I have taught this material in our REU off and on since 2003. This is ideal material for the purpose. (Obsolete notes and even current ones can be found on my web page by those sufficiently interested to search: Minian, Barmak's thesis advisor in Buenos Aires, found them there and started off work in Argentina based on them.) I apologize for this extended advertisement, but perhaps Paul's question gives me a reasonable excuse.<|endoftext|> TITLE: A divisor sum congruence for 8n+6 QUESTION [9 upvotes]: Letting $d(m)$ be the number of divisors of $m$, is it the case that for $m=8n+6$, $$ d(m) \equiv \sum_{k=1}^{m-1} d(k) d(m-k) \pmod{8}\ ?$$ It's easy to show that both sides are 0 mod 4: the left side since two primes appear to odd order in the factorization of $m$, and the right side since $m$ is neither a square nor the sum of two squares and so even products all appear twice. But they also seem to match mod 8 for small values. Does this keep going, or does it fail somewhere? REPLY [17 votes]: The congruence you state is true for all $m \equiv 6 \pmod{8}$. The proof I give below relies on the theory of modular forms. First, observe that $$ \sum_{k=1}^{m-1} d(k) d(m-k) = 2 \sum_{k=1}^{\frac{m-2}{2}} d(k) d(m-k) + d\left(\frac{m}{2}\right)^{2}. $$ Noting that $d(m) \equiv d\left(\frac{m}{2}\right)^{2} \pmod{8}$ if $m \equiv 6 \pmod{8}$, it suffices to prove that for every $m \equiv 6 \pmod{8}$ that $$\sum_{k=1}^{\frac{m-2}{2}} d(k) d(m-k)$$ is a multiple of $4$. The only terms in the sum that are not multiples of $4$ are those where $k$ is a perfect square, and $m-k = py^{2}$, where $p$ is prime, and $p$ divides $y$ to an even power (or $k = py^{2}$ where $p$ divides $y$ to an even power, and $m-k$ is a square). It suffices therefore to show that if $m \equiv 6 \pmod{8}$, then $m$ has an even number of representations in the form $m = x^{2} + py^{2}$ with $x, y \in \mathbb{Z}$ with $x, y > 0$ and $p$ a prime number (and the $p$-adic valuation of $y$ is even). If $m = x^{2} + py^{2}$, then either $x$ is even, which forces $p = 2$ and $y$ odd, or $x$ is odd, in which case $y$ is odd and $p \equiv 5 \pmod{8}$. The function $F(z) = \sum_{n=0}^{\infty} \sigma(2n+1) q^{2n+1}$, $q = e^{2 \pi i z}$ is a modular form of weight $2$ for $\Gamma_{0}(4)$. Here $\sigma(k)$ denotes the sum of the divisors function. A simple calculation shows that if $n \equiv 5 \pmod{8}$, then $\sigma(n) \equiv 2 \pmod{4}$ if and only if $n = py^{2}$ for some prime $p \equiv 5 \pmod{8}$ and a some perfect square $y$ (and moreover, the power of $p$ dividing $y$ is even, a condition which will remain in effect). It follows from this that $$ \frac{1}{2} \sum_{n \equiv 5 \pmod{8}} \sigma(n) q^{n} \equiv \sum_{\substack{p \equiv 5 \pmod{8} \\ y \geq 1}} q^{py^{2}} \pmod{2}, $$ where by this statement I mean that the power series on the left and right hand sides have integer coefficients and the coefficient of $q^{k}$ on the left side is congruent (modulo $2$) to the coefficient of $q^{k}$ on the right hand side. By twisting $F(z)$ by Dirichlet charaters mod $8$, one can see that $G(z) = \frac{1}{2} \sum_{n \equiv 5 \pmod{8}} \sigma(n) q^{n}$ is a modular form of weight $2$ on $\Gamma_{0}(64)$. Now, observe that $F(z) \equiv \sum_{\substack{n \geq 1 \\ n \text{ odd }}} q^{n^{2}} \pmod{2}$. Now, we find that $$ F(z) G(z) + F(4z) F(2z) \equiv \sum_{x, y \geq 1, p \equiv 5 \pmod{8} \text{ prime }} q^{x^{2} + py^{2}} + \sum_{x,y} q^{x^{2} + 2y^{2}} \pmod{2}. $$ where in the second term on the right hand side we have $x \equiv 2 \pmod{4}$ and $y \equiv 1 \pmod{2}$. The left hand side is now a modular form of weight $4$ on $\Gamma_{0}(64)$, and a computation shows that the first $1000$ Fourier coefficients are even. Indeed, Sturm's theorem proves that if the first $32$ coefficients are even, then all of them are. It follows that the number of representations of $m$ in the form $x^{2} + py^{2}$ when $m \equiv 6 \pmod{8}$ is always even, and hence the desired congruence is true.<|endoftext|> TITLE: Are open immersions in analytic geometry transverse? QUESTION [6 upvotes]: lately I have been interested in functional analysis, especially with a view towards its applications in the world of (complex) analytic geometry. I have been using R. Taylor's book Several complex variables with connections to algebraic geometry and Lie groups (as it is suggested in this MO question) and J.P. Demailly excellent book [D]: http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf. However, my poor background in functional analysis doesn't allow to me to answer to a very natural question I have: in [D], definition 5.32 the notion of transverse nuclear $A$-modules over a nuclear Fréchet algebra $A$ is introduced. In the subsequent Proposition 5.33, it is shown (in particular) that if $X$ is a (finite dimensional) Stein space and $\mathcal F$ is a coherent sheaf over $X$, then for every pair of open Stein subspaces $U' \subset U \Subset X$, $\mathcal F(U)$ and $\mathcal O_X(U')$ are transverse over $\mathcal O_X(U)$. So far so good, but the proof leaves me astonished: I would expect that $\mathcal O_X(U')$ is "analytically flat" over $\mathcal O_X(U)$ (in the sense that the functor $\mathrm{Tôr}^{\mathcal O_X(U)}_q(\mathcal O(U'), -)$ is identically zero. The reason I would expect that is simply that $U'\to U$ is an open immersion in analytic geometry, and I would expect it to be flat "in the correct sense" (of course it is flat stalkwise, but I would expect more). Can someone explain to me if the intuition coming to me from Algebraic Geometry leads me to some mistake or if it is possible to prove the above statement? REPLY [3 votes]: Hope this answer still will be useful: For an open embedding of smaller polydisk into a bigger polydisk $U \rightarrow V$, the $\mathcal{O}(V)$-module $\mathcal{O}(U)$ is not Frechet flat. The problem is that there are much more Frechet modules over $\mathcal{O}(V)$ than one would expect from comparison with the algebraic situation. For example, there are modules of ($L^2$, smooth, ...) functions on a boundary of a smaller polydisk. One reference to this is Pirkovskii, On certain homological properties of Stein algebras. He shows that there's a nontrivial first Frechet Tor over the algebra of entire functions on a line $\mathcal{O}(\mathbb{C})$ between functions on a unit disk $\mathcal{O}(D)$ and the module $L^2/H^2$, where $L^2$ is the space of $L^2$-functions on the unit circle, and $H^2$ is the Hardy space ($L^2$-functions holomorphically extendable into the inside of a circle).<|endoftext|> TITLE: Is any particular algebraic number known to have unbounded continued fraction coefficients? QUESTION [24 upvotes]: The continued fraction $$[1;1,2,3,4,5,\dots]=1+\cfrac{1}{1+\cfrac{1}{2+\cdots}}, $$ for instance, is known explicitly as a ratio of Bessel function values and is (I believe - SS) known to be transcendental. Similarly, $[1,2,2^2,2^3,2^4,2^5,\dots] $ is surely transcendental (and likely related to Liouville numbers). Are there any explicitly known algebraic numbers with unbounded continued fraction coefficients? (Of course, such a number must have degree greater than $2$, by all the usual theorems on continued fraction representations of quadratics.) Any references would be greatly appreciated. REPLY [4 votes]: I have one thought on this. But it would take some work to carry out. (*) Perron (Die Lehre von den Ketterbrüchen) in Kap. 11 evaluates certain continued fractions with polynomial entries in terms of hypergeometric functions. (**) There is an algorithm to determine when a hypergeometric function is an algebraic function. Maybe these can be combined to get an example of an algebraic number with polynomial (and thus unbounded) denominators. added: Examples of these two parts... (*) The Perron method evaluates the (non-simple) continued fraction $$ b_0+\frac{a_1}{b_1+\frac{a_2}{b_2}+\ddots}\qquad\text{with}\qquad a_n=18n^2-33n, b_n=3n-2 $$ as $$ \frac{-\frac{9}{2}}{{}_2F_1\left(-\frac{5}{6},1;\frac{1}{2};\frac{1}{3}\right)} $$ Here, $a_n$ and $b_n$ are polynomials in $n$ which is what I meant by "polynomial entries" above. (**) Unexpectedly, $$ {}_2F_1\left(\frac{1}{4},\frac{3}{4};\frac{2}{3};z\right) $$ is an algebraic function of $z$, see LINK So, the (somewhat remote) hope would be that a case of continued fraction with $a_n=1$ evaluates to something in terms of hypergeometric functions on the list of known algebraic cases.<|endoftext|> TITLE: Existence of a family of elliptic curves with large torsion subgroup QUESTION [10 upvotes]: Andrej Dujella's excellent web-site "High rank elliptic curves with prescribed torsion", at http://web.math.pmf.unizg.hr/~duje/tors/tors.html gives example of the (current) largest known rank of an elliptic curve over $\mathbb{Q}$ having each of the fifteen possible torsion subgroup structures. The related web-site "Infinite families of elliptic curves with high rank and prescribed torsion", at http://web.math.pmf.unizg.hr/~duje/tors/generic.html, gives information on the existence of infinite families. There are no known families with torsion $\mathbb{Z} / 9\mathbb{Z}$, $\mathbb{Z} / 10\mathbb{Z}$, $\mathbb{Z} / 12\mathbb{Z}$, and $\mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 8\mathbb{Z}$. I have already spent a reasonable amount of time hunting for such a beast. To save me wasting my time, I would like to ask experts whether they would expect such a family to exist, and if so, why? Alternatively, if they do not expect such a family to exist, could they also give a plausible reason? REPLY [2 votes]: It might be worth noting that in the following paper, which is also in Dujella's list of references, A. O. L. Atkin and F. Morain, Finding suitable curves for the elliptic curve method of factorization, Math. Comp. 60 (1993) (available here: http://www.ams.org/journals/mcom/1993-60-201/S0025-5718-1993-1140645-1/S0025-5718-1993-1140645-1.pdf) A. O. L. Atkin and F. Morain exhibits families of rank $ \geq 1$ and torsion group $\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z$, $\mathbb Z/5\mathbb Z$, $\mathbb Z/7\mathbb Z$, $\mathbb Z/9\mathbb Z$ and $\mathbb Z/10\mathbb Z$. However, their families are not $1$-parameter families, i.e. curves over $\mathbb Q(T)$, but rather parametrized by points on elliptic curves. Whether or not that adds anything to the likelihood of finding $1$-parameter families of rank $\geq 1$ for those less frequent torsion groups, I do not know.<|endoftext|> TITLE: Diagonalization via the Toda flow QUESTION [10 upvotes]: According to some almost indecipherable notes of a graduate Linear Algebra class, a symmetric matrix $A\in\mathbb R^{n\times n}$ can be diagonalised via the Toda flow. More specifically, if $X=X(t)\in\mathbb R^{n\times n}$ is the solution of the initial value problem ($n^2\times n^2$) \begin{equation} \frac{dX}{dt} \,=\, B(X)X-XB(X), \quad X(0)\,=\, A, \end{equation} where $B(X)=X_+-(X_+)^T=-B(X)^T$, and $X_+$ is the strict upper part of $X$: $$ X_+ \,=\, \left(\! \begin{array}{cccccccc} 0 & x_{12} & x_{13} & x_{14} &\cdots & x_{1n}\\ & 0 & x_{23} & x_{24} &\cdots & x_{2n}\\ & & 0 & x_{34} & \cdots & x_{3n} \\ &&&\ddots&& \vdots \\ & & & &0&x_{n-1,n} \\ & & &&& 0 \end{array} \!\right), $$ then $$ \lim_{t\to\infty} X(t)=\varLambda, $$ where $\varLambda$ is a diagonal matrix containing the spectrum of $A$. Is there any reference for this fact? Update. This is due to Moser: J. MOSER, Finitely many mass points on the line under the influence of an exponential potential-an integrable system, in “Dynamic Systems Theory and Applications” (J. Moser, Ed.), pp. 467-497, Springer-Verlag, New York/Berlin, 1975. I was wondering if its proof is accessible anywhere. REPLY [17 votes]: $\def\Tr{\mathrm{Tr}}$This proof is short enough that I thought I'd just write it out. On a skim, this looks like the same proof that Christian Remling pointed you to, and which Deift-Li-Tomei say is the same as the proof of Moser. Disclaimer: all signs in this argument have at best a 55% chance of being right. First of all, let $C$ be any continuous function at all from $n \times n$ matrices to $n \times n$ matrices and define $Y(t)$ by the ODE $$\frac{dY}{dt} = C(Y) Y - Y C(Y).$$ Then $$\frac{d \Tr(Y^m)}{dt} = \Tr \left( \frac{dY}{dt} Y^{m-1} + Y \frac{dY}{dt} Y^{m-2} + \cdots + Y^{m-1} \frac{dY}{dt} \right)$$ $$=\Tr \left( C(Y) Y^m - Y C(Y) Y^{m-1} + Y C(Y) Y^{m-1} - Y^2 C(Y) Y^{m-2} + \cdots + Y^{m-1} C(Y) Y - Y^m C(Y) \right) = \Tr\left( C(Y) Y^m - Y^m C(Y) \right)=0.$$ So $\Tr(Y^m)$ is constant and all the $Y(t)$'s have the same spectrum. Also, if $Y$ is symmetric and $C(Y)$ is skew-symmetric, then $C(Y) Y - Y C(Y)$ is symmetric, so symmetric matrices stay symmetric. Now, we specialize to the case of Toda flow. Let $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$ be the spectrum of $X$. A quick computation shows that $$\frac{d X_{kk}}{dt} = - \sum_{i< k} X_{ik}^2 + \sum_{i>k} X_{ik}^2.$$ So $$\frac{d (X_{11}+X_{22} + \cdots + X_{kk})}{dt} = \sum_{i \leq k,\ j > k} X_{ij}^2.$$ So all the quantities $X_{11}+X_{22} + \cdots + X_{kk}$ are increasing. Since $X$ is symmetric we have $X_{ii} \leq \lambda_1$ (an inequality of Schur), so $X_{11} + \cdots + X_{kk}$ is bounded above and we conclude that $\lim_{t \to \infty} X_{11} + \cdots + X_{kk}$ exists. As a result, $\lim_{t \to \infty} X_{kk}$ exists, call it $\mu_k$. Also, we see that $\lim_{t \to \infty} \sum_{i \leq k,\ j > k} X_{ij}^2 =0$ and we thus deduce that $\lim_{t \to \infty} X_{ij} =0$ for each $i \neq j$. So $\lim_{t \to \infty} X$ is a diagonal matrix, with diagonal entries $\mu_i$, and the same spectrum as $X$. So the $\mu$'s are a permutation of the $\lambda$'s. Finally, we want to know in what order the $\lambda$'s occur. We can't answer this in general: all the diagonal matrices are fixed points of the flow. However, I claim that $\mu_1 \geq \mu_2 \geq \cdots \geq \mu_n$ is the only stable fixed point. Proof: If $\mu_i < \mu_{i+1}$, then a tiny perturbation in direction $e_{i,i+1} + e_{i+1, i}$ is magnified, where $e_{i,j}$ is the matrix whose unique nonzero entry is a $1$ in position $(i,j)$. So almost all matrices flow to $\mu_1 \geq \mu_2 \geq \cdots \geq \mu_n$. On Toda flow and Morse flow The exact same proof works if $$B(X)_{ij} = c_{ij} X_{ij}$$ for any skew symmetric matrix $c$ with positive entries above the diagonal. In another answer, I work out that the Morse flow for the function $\psi(X) = \sum a_i X_{ii}$ is given by this equation with $c_{ij} = a_i - a_j$. (The metric on the set of matrices with fixed spectrum is induced by the $SO(n)$ action, and the inner product on $\mathfrak{so}(n)$ is the standard one.) So Toda flow would be Morse flow if we could arrange that $a_i -a_j = 1$ for all $ik$), but not in general. Still, I can imagine a fake history where Toda flow was discovered by writing down Morse flow for $\psi$ and then noticing that it still worked for any $c_{ij}$.<|endoftext|> TITLE: Origins of the Jacobi matrix QUESTION [7 upvotes]: I have several questions concerning history of Jacobi matrices. Does anybody know why the Jacobi matrix (=symmetric tridiagonal matrix) is named by Carl Gustav Jacob Jacobi? What was his contribution to the topic? What is the origin and the history of methods of the investigation of spectral properties of Jacobi matrices? Any suitable reference concerning the above questions would be helpful. Thanks. REPLY [4 votes]: In 1845 paper http://www.ima.umn.edu/preprints/April92/951.pdf (On a New Way of Solving the Linear Equations that Arise in the Method of Least Squares) Jacobi introduced a new iterative method to solve the matrix equation $Ax=b$, where the matrix $A$ is diagonally dominant. As an example, he considers the case $$A=\left(\begin{array}{ccc} 27 & 6 & 0 \\ 6 & 15 & 1 \\ 0 & 1 & 54\end{array} \right),\;\;\;b=\left(\begin{array}{c} 88 \\ 70 \\ 107\end{array}\right).$$ Jacobi uses a rotation to eliminate the biggest off-diagonal element $A_{12}= A_{21}=6$ and then solves the transformed system in three iterations each adding about one digit of accuracy. Perhaps this was a starting point of the theory of tridiagonal (Jacobi) matrices. In 1950, Lanczos introduced a method for the successive transformation of a given matrix to a tridiagonal matrix which turned to be very important for solving linear systems of equations and eigenvalue problems. For historical overview of these developments see http://www.cs.umd.edu/~oleary/reprints/j28.pdf (Some history of the conjugate gradient and Lanczos algorithms : 1948-1976, by Gene H Golub and Dianne P O'Leary). Some historically important references about the spectral theory of Jacobi operators can be found in http://annals.math.princeton.edu/wp-content/uploads/annals-v158-n1-p05.pdf (Sum rules for Jacobi matrices and their applications to spectral theory, by Rowan Killip and Barry Simon).<|endoftext|> TITLE: Linear systems on bielliptic surfaces QUESTION [6 upvotes]: A bielliptic surface is a surface of type $S=E_1 \times E_2/G$ where $E_1, E_2$ are elliptic curves and $G$ is a finite group of translations of $E_1$ acting on $E_2$ such that $E_2/G=\mathbf{P}^1$. It is easy to classify these surfaces, and it turns out that there are $7$ families. Is there a classification of the base point free and of the very ample linear systems on a bielliptic surface? REPLY [6 votes]: The structure of $\textrm{Num}(S)$ for a bielliptic surface $S$ is given in the paper by F. Serrano Divisors of bielliptic surfaces and embeddings in $\mathbb{P}^4$, Mathematische Zeitschrift 203 (1990), 527-533. First, Serrano proves that a basis for $\textrm{Num}(S) \otimes _{\mathbb{Z}} \mathbb{Q} \cong H^2(S, \, \mathbb{Q})$ is given by the fibres $A$and $B$ of the two natural, isotrivial elliptic fibrations $f_1\colon S \to E_1/G$ and $f_2 \colon S \to E_2/G$. They clearly satisfy $$A^2=B^2=0, \quad AB = |G|.$$ This allows the author to describe the ample cone of $S$: in fact, it follows by Nakai-Moishezon criterion that a divisor in the numerical class $\alpha A + \beta B$ is ample if and only if $ \alpha >0$ and $\beta >0$, see Lemma 1.3. An explicit description of a basis of $\textrm{Num}(S)$ over $\mathbb{Z}$ must take into account the multiplicities of the singular fibres of $f_1$ and $f_2$. These depend on which of the seven families we are considering, and a complete case-by-case analysis is given in Theorem 1.4. Finally, some results on very ample divisors are given in Theorem 2.1: in fact, Serrano shows that the bielliptic surfaces in one of the seven families admit a very ample linear system $|L|$ such that $H^0(S, \,L)=5$ and $L^2=10$. Consequently, these surfaces can be embedded in $\mathbb{P}^4$ as surfaces of degree $10$.<|endoftext|> TITLE: Geometry description of the GSR riffle shuffle model QUESTION [8 upvotes]: In 1992 Diaconis and Bayer announced their famous result which is now a well-known folklore: Seven shuffles is enough to randomize a deck of cards. One of the key ingredients in their proof is that an $a$-shuffle, followed by a $b$-shuffle, is equivalent to a single $ab$-shuffle. Here an $a$-shuffle is an operation consists of two steps: the first step is called a cut: cut the deck of cards into $a$ packets of size $p_1,p_2,\cdots,p_a$ according to the multinomial distribution $\begin{pmatrix}n\\p_1,p_2,\ldots,p_a\end{pmatrix}/a^n$, here $p_i\geq0$ and $p_1+\cdots+p_a=n$. The second step is called interleave: drop the cards one by one from these packets with probability proportional to their current size. (for more precise explanations, see the paper by Diaconis above or the paper by Mann ) when all cards are dropped, we get a permutation $\pi\in S_n$. Note that there may be many different pairs (cut,interleave) that result in the same permutation of the deck of the cards. In fact the number of the different pairs (cut, interleave) that result in a given permutation $\pi$ only depends on the number of raising subsequences of $\pi$: Diaconis and Bayer proved that this number is $\begin{pmatrix}n+a-r\\n\end{pmatrix}$, assuming that $\pi$ has $r$ raising subsequnces. Now we come to the key point: Diaconis found a nice geometry description of the a-shuffle, which enables him to prove his assertion economically. This description goes as follows: Drop $n$ points into the interval $[0,1]$ independently with the uniform density. Label them from left to right as $x_1<\cdots TITLE: Why should affine lie algebras and quantum groups have equivalent representation theories? QUESTION [36 upvotes]: Let $\mathfrak{g}$ be a simple lie algebra over $\mathbb{C}$ and let $\hat{\mathfrak{g}}$ be the Kac-Moody algebra obtained as the canonical central extension of the algebraic loop algebra $\mathfrak{g} \otimes \mathbb{C}[t,t^{-1}]$. In a sequence of papers, Kazhdan and Lusztig constructed a braided monoidal structure on (a certain subcategory of) the category of representations of $\hat{\mathfrak{g}}$ of central charge $k - h$ where $k \in \mathbb{C}^* \;\backslash\; \mathbb{Q}_{\geq 0}$ and $h$ is the coxeter number of $\mathfrak{g}$. They then showed that the resulting braided category is equivalent to the braided category of finite dimensional representations of the quantum group $U_q(\mathfrak{g})$ for $q = e^{\frac{\pi i}{k}}$. My question then is this: is there any conceptual explanation as to why these two braided categories should be equivalent (which does not resort to computing both sides and seeing that they are same)? The representations of $\hat{\mathfrak{g}}$ of various central charges can be considered as twists of the representation theory of the loop algebra $\mathfrak{g} \otimes \mathbb{C}[t,t^{-1}]$. On the other hand, the representation theory of $U_q(\mathfrak{g})$ is a braided deformation (which can be thought of as a form of twisting) of the representation theory of $\mathfrak{g}$ itself. Moreover, the equivalence above only holds for non-trivially deformed/twisted cases. The limiting case of the representations of $\mathfrak{g}$ is recovered by (carefully) taking $q=1$, which corresponds to $k \rightarrow \infty$ and hence does not participate in the game. On the other hand, to obtain central charge $0$ we would need to take $k=h$ which is also excluded (as the proof Kazhdan-Lustig assumes $k \notin \mathbb{Q}_{\geq 0}$). Is there any reason why these two lie algebras would have the same twisted/deformed representations, but not the same representations? REPLY [6 votes]: (Written on my phone - apologies for any typos.) A few comments: a) First, as to the source of the braided monoidal structure on the Kazhdan-Lusztig category. The category of integrable affine Lie algebra reps is naturally a factorization category, which is close morally to an E2/braided monoidal category, see DBZ's answer to mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebra/54008#54008. In some cases this can be made precise, and this gives you your braided monoidal category here. b) Now as to why one might expect these two braided monoidal categories to agree, I think the key is exactly the $\kappa\rightarrow\infty$ case you mention (corresponding to $q=1.$) A general "limiting to infinity" procedure is described in https://arxiv.org/abs/1708.05108, but here I will proceed in a more ad hoc way. Before taking this limit, let me restate the category we are interested in. The Kazhdan-Lusztig category is the category of finitely-generated $U_{\kappa}(g((t)))$-modules equipped with an action of $G$, with the conditions that the two induced $g$ actions agree and that elements of $tg[[t]]$ act nilpotently. For our $\kappa\rightarrow\infty$ limit, we need to figure out how to degenerate $U_{\kappa}(g((t)))$. Writing $\kappa=c\kappa'$ for a fixed nondegenerate $\kappa'$, we can describe $U_{\kappa}(g((t)))$ as the free algebra on $g((t))$, mod the relations $[s,t]=[s,t]_0+c\kappa'(s,t).$ We can't directly limit $c$ to infinity, but note that we can rescale the generators and rewrite the relations as $[s,t]=\frac{1}{c}[s,t]_0+\kappa'(s,t),$ which we can limit to $[s,t]=\kappa'(s,t).$ So we can reasonably set $U_{\infty}(g((t)))$ to be the tensor product of $\operatorname{Sym} g$ and a Weyl algebra on $tg[[t]]\oplus (tg[[t]])^*$ (recall $\kappa'$ induces a perfect pairing between $g[[t]]$ and $t^{-1}g[t^{-1}]$.) Now our original integrability condition limits to the conditions that $tg[[t]]$ acts nilpotently, that $g$ acts by zero (not that it agrees with the action of $G$ - recall our rescaling!), and that the $G$ actions on our representation and on the Weyl algebra are compatible. The only representations satisfying these conditions are of the form $V\otimes W$, where $V$ is a finite dimensional representation of $G$ and $W$ is the standard rep of the Weyl algebra. So our final category is indeed the category of $G$-reps. c) A more highbrow way to limit $\kappa$ to infinity is via geometric Langlands. GL switches level infinity with critical level, and the Kazhdan-Lusztig category with the Whittaker category on the affine Grassmannian for the Langlands dual group (since here we are working with abelian categories and critical level, you can think instead of the category of spherical D-modules.) Now geometric Satake tells you this gives you back $Rep(G)$. This is kind of perverse, in that you're using Langlands duality twice. The point here though is that this suggests one can prove the Kazhdan-Lusztig equivalence via similar methods to geometric Satake, by describing both sides via root datum. I believe Dennis Gaitsgory has a new proof following these lines.<|endoftext|> TITLE: Smallest Connected Graph for Given Degree Sequence QUESTION [8 upvotes]: For a given integer sequence $(d_1, d_2,...,d_n)$, a natural question is if such a sequence is graphical, i.e. is a degree sequence of some graph. According to Erdős–Gallai theorem, A sequence of non-negative integers $d_1\geq\cdots\geq d_n$ can be represented as the degree sequence of a finite simple graph on $n$ vertices if and only if $d_1+\cdots+d_n$ is even and $\sum^{k}_{i=1}d_i\leq k(k-1)+ \sum^n_{i=k+1} \min(d_i,k)$ holds for $1\leq k\leq n.$ My questions are (1) If Erdős–Gallai theorem holds, what is the condition that this graph is unique? (2) If those graphs are not unique, how to find a connected graph with smallest connectivity among them? REPLY [8 votes]: A theorem of Hakimi says that any pair of degree-equivalent graphs can be obtained one from the other by a sequence of "elementary $2$-switchings" (probably known under many other names), which involve the subgraph switch on the subgraph induced by four vertices, as illustrated in one instance below.   So whatever you seek to minimize (cf. the comments), likely it could be pursued by searching for the minimum via these $2$-switchings. Hakimi, S. Louis. "On realizability of a set of integers as degrees of the vertices of a linear graph. I." Journal of the Society for Industrial & Applied Mathematics. 10.3 (1962): 496-506. Hakimi, S. Louis. "On realizability of a set of integers as degrees of the vertices of a linear graph II. Uniqueness." Journal of the Society for Industrial & Applied Mathematics. 11.1 (1963): 135-147.<|endoftext|> TITLE: Circles avoiding rational points of height $\le h$ QUESTION [10 upvotes]: Q. Which origin-centered circles $C(r)$ (or spheres in dimension $d$) of radius $r < 1$ avoid all rational points of height $\le h$? A rational point is a point all of whose coordinates are rational numbers. A rational number $x= a/b$ in lowest terms (i.e., gcd$(a,b)=1$) has height $\max \lbrace |a|,|b| \rbrace$. A rational point of height $h$ is a rational point all of whose coordinates are of height $\le h$. For example, let $h=10$, and $r=\frac{7}{11}$ [Thanks for corrections by NAME_IN_CAPS and Noam Elkies]. Then $C(\frac{7}{11})$, unless I am mistaken, avoids all rational points of height $\le 10$, despite some near misses:           Again fix $h=10$, but for $r=\frac{1}{\sqrt{2}}$, $C(r)$ passes through $$\lbrace{ (\frac{1}{10},\frac{7}{10}), (\frac{1}{2},\frac{1}{2}), (\frac{7}{10},\frac{1}{10}) }\rbrace \;:$$           Perhaps I should phrase the question in the obverse sense: Q'. Which origin-centered circles $C(r)$ (or spheres in dimension $d$) of radius $r < 1$ include at least one rational point of height $\le h$? So I am seeking $r$ as a function of $h$. I am unfamiliar with this type of reasoning, but I seek circles that avoid low-height rational points. Thanks! REPLY [7 votes]: I'm going to interpret the question as: What is the positive rational number of least height that cannot be written as the sum of $d$ squares of rational numbers of height $h$. So $r^2=1/p$ works quite well. You might find $r^2=a/p$, for $a$ slightly less than $p$ and $a$ a sum of two squares, more aesthetically pleasing.<|endoftext|> TITLE: Convexity of a certain sublevel set QUESTION [11 upvotes]: Consider the polynomial of degree $4$ in variable $r$ $$ r^4 + (x^2 + y^2)\ r^2 - 2 x y\ r + x^2 y^2 $$ The discriminant of this polynomial in $r$ is the following expression (obtained using Mathematica) $$ \Delta := 16 x^2 y^2 \left( -x^6 + x^8 - 27 x^2 y^2 + 33 x^4 y^2 - 4 x^6 y^2 + 33 x^2 y^4 + 6 x^4 y^4 - y^6 - 4 x^2 y^6 + y^8 \right) $$ Consider the subset $D$ of $\mathbb{R}^2$ defined by the polynomial inequality $\frac{\Delta}{16 x^2 y^2} \le 0$ $$ D : -x^6 + x^8 - 27 x^2 y^2 - 4 x^6 y^2 - 4 x^2 y^6 + 33 x^4 y^2 + 33 x^2 y^4 + 6 x^4 y^4 - y^6 + y^8 \le 0 $$ I would like to prove that $D$ is convex. Any help would be appreciated. Thanks! REPLY [12 votes]: You can parametrize the zero locus of $\Delta$ (other than the origin) in the first quadrant by $$ (x,y) = \left(\frac{(3{-}t)\sqrt{(3{+}t)(1{-}t)}}{8}, \frac{(3{+}t)\sqrt{(3{-}t)(1{+}t)}}{8}\right) \qquad\qquad -1\le t\le 1 $$ and then compute that the curvature is $$ \frac{(x'y''-y'x'')}{((x')^2+(y')^2)^{3/2}} = \frac{2^{3/2}(3{+}t^2)}{\bigl(3{-}t^2\bigr)^{5/2}}. $$ Since this is always positive, it follows that the curve is strictly convex (since it is symmetric under $x$ and $y$ reflections.<|endoftext|> TITLE: Factor a sum of products of cofactors QUESTION [10 upvotes]: Let $M$ be any $n\times n$ matrix. We define the usual cofactors: $C_{i,j}$ is $(-1)^{i+j}$ times the determinant of the submatrix obtained by deleting row $i$ and column $j$ of $M$. We can write the determinant of $M$ using Laplace expansion along column $p$ as: $$\det M = \sum_{q=1}^n M_{q,p} C_{q,p}$$ Now, for any $k, p\in \{1,...,n\}$ consider the sum: $$W_{n,p}(k) = \sum_{q=1}^n {M_{q,p} C_{q,p}^2 \prod_{j\ne q}{C_{j,k}}}$$ Clearly $W_{n,p}(p)$ can always be factored, with $\det M$ as one factor: $$W_{n,p}(p) = \sum_{q=1}^n{M_{q,p} C_{q,p}^2 \prod_{j\ne q}{C_{j,p}}} = \sum_{q=1}^n{M_{q,p} C_{q,p} \prod_{j=1}^n{C_{j,p}}} = \left(\det M\right) \prod_{j=1}^n{C_{j,p}}$$ But in all the specific cases I've examined, $\det M$ appears as a factor of $W_{n,p}(k)$ even when $k\ne p$. For example, with $n=3$, $p=1$ and $k=2$: $$W_{3,1}(2)=\sum_{q=1}^3{M_{q,1} C_{q,1}^2 \prod_{j\ne q}{C_{j,2}}}=\\ \left(\det M\right)\left(M_{2,2} M_{2,3} M_{3,1}^2 M_{1,3}^2+M_{2,1} M_{2,3} M_{3,1} M_{3,2} M_{1,3}^2-M_{2,1} M_{2,2} M_{3,1} M_{3,3} M_{1,3}^2-M_{2,1}^2 M_{3,2} M_{3,3} M_{1,3}^2-M_{1,2} M_{2,3}^2 M_{3,1}^2 M_{1,3}+M_{1,2} M_{2,1}^2 M_{3,3}^2 M_{1,3}+M_{1,1} M_{2,1} M_{2,2} M_{3,3}^2 M_{1,3}-M_{1,1} M_{2,3}^2 M_{3,1} M_{3,2} M_{1,3}-M_{1,1} M_{1,2} M_{2,1} M_{2,3} M_{3,3}^2-M_{1,1}^2 M_{2,2} M_{2,3} M_{3,3}^2+M_{1,1} M_{1,2} M_{2,3}^2 M_{3,1} M_{3,3}+M_{1,1}^2 M_{2,3}^2 M_{3,2} M_{3,3}\right) $$ It might be worth noting that the second factor here cannot be written as a linear combination of products of the cofactors, with purely numeric coefficients. This is in stark contrast to the case $k=p$, when the quotient is simply a product of cofactors. I am seeking a proof that $\det M$ always divides $W_{n,p}(k)$, and a general formula for the quotient in the non-trivial case $k\ne p$. REPLY [2 votes]: This is a proof that $\det M$ divides $W_{n,p}(k)$. I can assume that the entries of $M$ are variables of polynomial ring $\mathbb{C}[M_{11},\ldots,M_{nn}]$; then, it suffices to prove that points from the (irreducible) variety $\det M=0$ satisfy $W_{n,p}(k)=0$. Indeed, let $M'$ be a generic point on this variety; denote by $v^i$ the $i$th column vector of $M'$. Since $M'$ is generic, there is $l\notin\{p,k\}$ such that $v^l=\sum_{t\neq l} \lambda_t v^t$ with all $\lambda_t$ nonzero. Then, the ratio of cofactors $C_{qp}/C_{qk}$ equals $\alpha=\lambda_p/\lambda_k$ and, in particular, it does not depend on $q$. We conclude $$W_{n,p}(k)=\sum\limits_{q=1}^{n}\,M'_{qp}C_{qp}^2\prod\limits_{j\neq q}C_{jk}=\sum\limits_{q=1}^{n}\,M'_{qp}C_{qp}\alpha\prod\limits_{j=1}^nC_{jk}=\alpha\prod\limits_{j=1}^nC_{jk}\cdot\left(\sum\limits_{q=1}^{n}\,M'_{qp}C_{qp}\right)=0.$$<|endoftext|> TITLE: Bloch group, hyperbolic manifolds and rigidity QUESTION [13 upvotes]: I have some questions concerning the hyperbolic geometry side of the rigidity question for $K_3$ which asks if the natural map $K_3^{\operatorname{ind}}(\overline{\mathbb{Q}})\to K_3^{\operatorname{ind}}(\mathbb{C})$ is surjective. Question 0, a historical aside: Recently I grew a little uncertain about the correct attribution of this question/conjecture. I have seen this conjecture attributed to Bloch somewhere, and in Dupont's book "Scissors congruences, group homology and characteristic classes" it is attributed to Sah. What's the exact history? Now for the actual mathematical questions. Recall that one can associate classes in $K_3$ (or versions of the Bloch group) to hyperbolic $3$-manifolds. Roughly, this is done by choosing a triangulation of the manifold $M$ by ideal tetrahedra, and the ideal tetrahedra naturally give classes in the scissors congruence (or pre-Bloch) group. There are several papers by Neumann and Yang about this. My main question is now: what, if anything, corresponds to the rigidity question for $K_3$ on the hyperbolic $3$-manifold side? More detailed questions are formulated below: Question 1: What are possible reasons for believing or disbelieving the conjecture? I guess the rigidity of the Cheeger-Chern-Simons invariants is one reason for believing rigidity, maybe another is that we have no clue what else could be in $K_3^{\operatorname{ind}}(\mathbb{C})$? Any other reasons? Or maybe there is an argument why hyperbolic geometry can not possibly see non-rigidity of $K_3$? I would be mainly interested in intuition from the hyperbolic geometry side, about which I know almost nothing. Question 2: Assume, just for the fun of it, that the conjecture is false, i.e., there are classes in $K_3^{\operatorname{ind}}(\mathbb{C})$ which do not come from $K_3^{\operatorname{ind}}(\overline{\mathbb{Q}})$. Would there be a hyperbolic geometry interpretation of these classes? Something related to Question 2 was discussed in Ian Agol's answer to this MO-question. Apparently, one would not see the "new" classes as manifolds with strange triangulations, but rather the new classes would come from deformations of $\operatorname{SL}_2(\mathbb{C})$-representations of the fundamental group of $M$. The representations would correspond to flat rank $2$ vector bundles. Is it possible to say that a failure of the rigidity conjecture would imply the existence of strange deformations of flat rank $2$ vector bundles on hyperbolic $3$-manifolds? Are there other things/objects in hyperbolic geometry that would deform in a strange way if the rigidity conjecture was false? I guess it is called rigidity conjecture for a reason. Question 3: Have people considered a way of going from classes in the Bloch group to hyperbolic manifolds? An element of the Bloch group can be represented as a linear combination of ideal tetrahedra, but it is not obvious to me how I could get a hyperbolic $3$-manifold from that? Is the relation defining the Bloch group ($\sum x\wedge (1-x)=0$) enough to make sure that the ideal tetrahedra can be glued to a manifold in some way? Provided Question 3 has a positive answer, then I would have a way of interpreting elements in $K_3^{\operatorname{ind}}(\mathbb{C})$ which do not come from $\overline{\mathbb{Q}}$ (assuming these exist). For simplicity, let $C$ be a smooth projective curve over $\overline{\mathbb{Q}}$. Then I would interpret elements in $K_3^{\operatorname{ind}}(\overline{\mathbb{Q}}(C))$ as (linear combinations of) deformations of ideal tetrahedra with boundary points in $\overline{\mathbb{Q}}$ with parameter space some open subcurve of $C$. If the simplices can be glued, that would provide a hyperbolic $3$-manifold together with a deformation of a triangulation (with parameter space a subcurve of $C$). The fact that the corresponding element in $K_3$ does not come from $\overline{\mathbb{Q}}$ would say that the deformation of the triangulation can not be made constant by the obvious operations on ideal tetrahedra. Is it true that I can view non-rigid elements in $K_3$ as deformations of ideal triangulations? Question 4: Now this is only meaningful if the questions above have a reasonably positive answer. Assuming the failure of the rigidity conjecture, and assuming that it is possible to represent non-rigid elements by deformations of hyperbolic-geometric objects (vector bundles or triangulations or some such thing), would there be invariants (other than their classes in $K_3$) that one could use to show that such objects yield non-rigid classes in $K_3$? The Cheeger-Chern-Simons invariants are rigid, but are there other suitably analytic invariants, maybe related to regulators, that could do the job? Any help, comments and explanations are most welcome. REPLY [3 votes]: In answer to question 3 it is not possible to go from an element of the Bloch group to a hyperbolic manifold as there are different three manifolds with the same element in the Bloch group. That is different hyperbolic manifolds which are Scissor congruent. (See Quantum Content of Gluing Equations section 1.2 https://arxiv.org/pdf/1202.6268.pdf) I'm not sure if the image of the hyperbolic three manifolds generate the Bloch group but see the introduction of section 4 of Extended Bloch Group and Cheeger-Chern-Simons Class (https://projecteuclid.org/euclid.gt/1513883371) for a description of the additional data on closed three manifolds that would generate it or even better its lift.<|endoftext|> TITLE: Existence of solutions of a polynomial system QUESTION [11 upvotes]: Fix $k \in \mathbb{N}$, $k \geq 1$. Let $p \in [0,1]$ and $x = (x_0, \ldots, x_k)$ be a $(k+1)$-dimensional real vector, and define $$S(p,x) = -x_0^2 + \sum_{i=0}^k {k \choose i} p^i (1 - p)^{k - i} \cdot (x_i - p)^2.$$ Experiments show that for small values of $k$ $$\exists x \in \mathbb{R}^{k+1} \,.\, \forall p \in [0,1] \,.\, S(p,x) = 0.$$ In other words, there are $x_i$'s such that $S(x,p)$ is identically zero as a polynomial in $p$. For a given $k$ we can expand $S(x,p)$ as a polynomial in $p$ and equate the coefficients to $0$. For $k = 2$ we get \begin{align*} 0&=0 \\ -x_0^2-2 x_0+x_1^2&=0 \\ 2 x_0-2 x_1+1&=0 \\ \end{align*} and this has two solutions: $$x = (\frac{1}{2} (-1-\sqrt{2}),\frac{1}{2},\frac{1}{2} (3+\sqrt{2}))$$ and $$x = (\frac{1}{2} (-1+\sqrt{2}),\frac{1}{2},\frac{1}{2} (3-\sqrt{2})).$$ For $k = 1, 2, 3, 4, 5, 6, 7$ there are $1, 2, 4, 8, 14, 28, 48$ solutions respectively, according to Mathematica. According to OEIS this is A068912, "the number of $n$ step walks (each step $\pm 1$ starting from $0$) which are never more than $3$ or less than $-3$." This is kind of interesting because the problem arises in statistics, see John Mount's blog post for background. Question: Is there a solution for every $k$? Addendum: John says he wants soltions in $[0,1]^{k+1}$... Here is the relevant Mathematica code: s[k_, p_, x_] := Sum[Binomial[k, i] * p^i* (1 - p)^(k - i)* (Subscript[x, i] - p)^2, {i, 0, k}] Subscript[x, 0]^2 xs[k_] := Table[Subscript[x, i], {i, 0, k}] system[k_, p_, x_] := Thread[CoefficientList[s[k, p, x], p] == 0] solutions[k_] := Solve[system[k, p, x], xs[k], Reals] To see the system of equations for $k = 4$, type system[4, p, x] // ColumnForm To see the solutions for $k = 4$, type solutions[4] To make a table of counts of solutions up to $k = 7$, type Table[{k, Length@solutions[k]}, {k, 1, 7}] // ColumnForm REPLY [6 votes]: The solutions described via the link http://winvector.github.io/freq/explicitSolution.html (posted in one of the earlier answers) can be given by the following formula: $$ x_i=\frac{(k-2i)\sqrt{k}+(2i-1)k}{2k(k-1)}=\frac{1}{2(1+\sqrt{k})}+\frac{i}{\sqrt{k}(1+\sqrt{k})}. $$ Note that (when $k$ is fixed): $x_i$ is an increasing function of $i$, and we have $$ x_0=\frac{1}{2(1+\sqrt{k})}, \quad x_k=\frac{1+2\sqrt{k}}{2+2\sqrt{k}}, $$ so all these numbers are between 0 and 1. Moreover, we have $x_i=a+bi$, so $S(p,x)$ can be represented as $$ -x_0^2+\sum_{i=0}^k\binom{k}{i}p^i(1-p)^{k-i}(U+Vi+Wi^2), $$ where $U$, $V$ and $W$ depend on $k$ and $p$ but not on $i$. It remains to use formulas \begin{gather} \sum_{i=0}^k\binom{k}{i}p^i(1-p)^{k-i}=1,\\ \sum_{i=0}^k i\binom{k}{i}p^i(1-p)^{k-i}=kp,\\ \sum_{i=0}^k i(i-1)\binom{k}{i}p^i(1-p)^{k-i}=k(k-1)p^2 \end{gather} (which are obvious) to check directly that the formulas for $x_i$ as above give a solution. This solution also simplifies to $x_i = (\frac{1}{2}\sqrt{k} + i)/(\sqrt{k}+k)$ which is exactly the smoothed estimate of the win-rate of a coin flipped $k$ times showing $i$ wins with $\sqrt{k}$ "pseudo-observations" (half wins, half losses) added first (or Bayesian inference starting with $\beta(\sqrt{k}/2,\sqrt{k}/2)$ priors, $\beta(1/2,1/2)$ being Jeffreys priors, and $\beta(1,1)$ being standard Laplace smoothing).<|endoftext|> TITLE: Is a connected graph uniquely determined by its weighted 2-step graph? QUESTION [5 upvotes]: This is an extension of a previous question: https://math.stackexchange.com/questions/876336/is-a-graph-uniquely-determined-by-its-weighted-2-step-graph/876357#876357. In that question I asked about arbitrary graphs; in this question I restrict to connected graphs. Here are the details: Let $G$ be an undirected graph. Define the 2-step graph $G^{(2)}$ of $G$ to be the weighted graph whose vertices are the same as those of $G$ but whose edges correspond to 2-step paths in $G$. Thus the weight of an edge $(u,v)$ is the number of distinct vertices $w$ such that $(u,w)$ and $(w,v)$ are both edges in $G$. (In particular, the weight of $(u,u)$ is the degree of $u$ for every vertex $u$.) My question: Are there two connected, non-isomorphic graphs $G$ and $H$ such that $G^{(2)}$ is isomorphic to $H^{(2)}$? My intuition says that the answer should be "yes", but I'm unable to construct an example. REPLY [3 votes]: A counterexample can be constructed as follows. Let $G$ be the graph of a hexagonal pyramid and $H$ be that of a pair of tetrahedra sharing one common vertex. Then, $G^2$ and $H^2$ are isomorphic while $G$ and $H$ themselves are not. Indeed, $G$ has a simple $6$-cycle but $H$ has not.<|endoftext|> TITLE: Easiest example where field of definition is not field of moduli QUESTION [11 upvotes]: There are many examples of varieties over $\overline{\mathbb Q}$ whose field of moduli is $\mathbb Q$ but which can't be defined over $\mathbb Q$. What is the easiest such example? It should be a genus two curve probably, but which one? Also, what is the easiest example of a variety over $\mathbb C$ whose field of moduli is $\mathbb R$ but which can't be defined over $\mathbb R$? REPLY [9 votes]: Here is a recipe for constructing some examples. Suppose that $f$ is a polynomial of degree 5 over $\Bbb C$ without repeated roots. Suppose that $f$ has the following two properties: The only linear fractional transformations in $x$ fixing the union of the zeros of $f(x)$ with $\infty$ is the identity. We have $-\overline{f(-\overline x)} = f(x)$. (For example, you could take $f(x) = (x^2-1)(x^2-2)(x-i)$.) Consider the resulting hyperelliptic curve $C$ of genus two of the form $y^2 = f(x)$. The projection $C \to \Bbb P^1$ determined by the canonical bundle on $C$ is projection onto the $x$-coordinate, and the first condition we've imposed on $f$ guarantees that any automorphism of $C$ must fix the $x$-coordinate (as it must fix the six ramification points). As a result, the only possible such automorphisms are the identity and $y \mapsto -y$. Thus we have a hyperelliptic curve whose automorphism group is exactly $\Bbb Z/2$. Now we also find that the second condition on $f$ means that there is an isomorphism of $C$ with its complex conjugate $\overline{C}$, given by $y \mapsto iy$, $x \mapsto -x$. This means that the field of moduli has to be strictly smaller than $\Bbb C$, and so it must be $\Bbb R$. However, this curve cannot be defined over $\Bbb R$. The obstruction, as is typical when the automorphism group is abelian, shows up as an element in $H^2(Gal(\Bbb C/\Bbb R), Aut(C))$. In terms of group cohomology, this is determined by an extension $$ 1 \to Aut(C) \to G \to Gal(\Bbb C/\Bbb R) \to 1. $$ Here $G$ is the collection of automorphisms of $C$ which are either $\Bbb C$-linear or conjugate-linear. In this case, both groups on the outside are $\Bbb Z/2$, and so we want to show that the extension is $\Bbb Z/4$. To calculate the extension, we take the nontrivial automorphism of $\Bbb C$, lift it to an isomorphism $g: C \to C$ which is conjugate-linear (which we already calculated above), and then form the composite $g^2: C \to C$. This automorphism is the hyperelliptic involution, and so the extension is nontrivial. I believe that the same method works over $\Bbb Q(i)$, but you need a sturdier Galois cohomology computation.<|endoftext|> TITLE: Geodesics on $SU(4)$ QUESTION [12 upvotes]: Are the geodesics of the following metrics on $SU(4)$ known or easy (in a way not known to me!) to find? In the adjoint representation, one can express the Killing form as a matrix and consider it as an inner product on $\mathfrak{su}(4)$. This matrix is some multiple of the identity (i.e. a scalar matrix). The geodesics of this metric are well known to be the one parameter sub-groups of $SU(4)$. Consider the basis of $\mathfrak{su}(4)$ given by $\{i\sigma^n \otimes \sigma^m\}$ where $n,m =0, x,y,z$ but not both are $0$. $\sigma^0$ is the identity matrix and the others are the usual Pauli matrices. I want to know the geodesics of the metrics formed by right translating an inner product at the identity which is given by diagonal matrix (it would have dimension $15$ by $15$) in this basis for $\mathfrak{su}(4)$. Ideally, I'd like to know the geodesics in terms of the 15 diagonal elements of the matrix defining the inner product on $\mathfrak{su}(4)$. A way to represent the solution as a matrix exponential would be especially helpful! REPLY [19 votes]: In the OP's particular case, the situation is somehwat simpler than the general case that José discusses. That's because the family of left-invariant metrics on $\mathrm{SU}(4)$ that the OP wants to consider has special properties, although just how special does not become apparent until one looks at the problem from a rather different viewpoint, using the fact that $\mathrm{SU}(4)$ is $\mathrm{Spin}(6)$. (In fact, one has $\mathrm{SU}(4)/\{\pm I_4\}=\mathrm{SO}(6)$, and the problem is much easier to describe and treat as a problem on $\mathrm{SO}(6)$, as will be seen.) First, though, a quick review of the geodesic equations for a left-invariant metric on a compact, semi-simple Lie group $G$: If $\kappa:{\frak{g}}\times{\frak{g}}\to\mathbb{R}$ is the Killing form on ${\frak{g}} = T_eG$, and $\omega:TG\to{\frak{g}}$ is the canonical left-invariant form, then the standard bi-invariant metric on $G$ is given by $\mathrm{d}s^2 = -\kappa(\omega,\omega)$. Any other left-invariant metric on $G$ can be written uniquely in the form $\mathrm{d}\bar s^2 = -\kappa(B\omega,\omega)$, where $B:{\frak{g}}\to{\frak{g}}$ is a positive definite $\kappa$-symmetric linear isomorphism. To find the $\mathrm{d}\bar s^2$-geodesic passing through $g_0\in G$ with initial velocity $L'_{g_0}(v_0)\in T_{g_0}G = L'_{g_0}\bigl({\frak{g}}\bigr)$, one has a $2$-step procedure: First, one finds the curve $v:\mathbb{R}\to{\frak{g}}$ that satisfies the Euler equation (a nonlinear ODE initial value problem) $$ v'(t) = B^{-1}\bigl[v(t),Bv(t)\bigr],\qquad v(0) = v_0 $$ and then the curve $g:\mathbb{R}\to G$ satisfying the Lie equation $$ \omega\bigl(g'(t)\bigr) = v(t),\qquad g(0) = g_0\,. $$ (When $G$ is a matrix group, this latter equation is just $g'(t) = g(t) v(t)$, with initial value $g(0) = g_0$.) Note that, when $v_0$ is an eigenvector of $B$, the solution of the Euler equation is $v(t) = v_0$, and so the geodesic is just $g(t) = g_0 \exp(tv_0)$ (i.e., the left-translation of a $1$-parameter subgroup). More generally, if $B$ preserves a subalgebra ${\frak{s}}\subset {\frak{g}}$ that contains $v_0$, then the problem reduces to finding the geodesic in the corresponding subgroup $S\subset G$ (which is totally geodesic in $G$ with respect to the metric $\mathrm{d}\bar s^2$). Next, in the OP's specific case, one has ${\frak{g}} = {\frak{su}}(4) = {\frak{so}}(6)$ and the OP has prescribed an orthogonal basis $\mathbf{b}$ consisting of 15 elements in ${\frak{su}}(4)$ and wants to consider, all together, the $15$-dimensional cone of metrics determined by the set of positive definite symmetric transformations $B:{\frak{su}}(4)\to {\frak{su}}(4)$ that preserve the $15$ lines spanned by the elements of $\mathbf{b}$. What is not apparent in the OP's description is the great deal of symmetry that the basis $\mathbf{b}$ possesses. This is much more apparent when one, instead, uses the alternative form ${\frak{so}}(6)$, i.e., the skew-symmetric linear transformations of $\mathbb{R}^6$ with its standard inner product. In this form, one can describe the OP's basis $\mathbf{b}$ as follows: Let $e_1,\dots,e_6$ be an orthonormal basis of $\mathbb{R}^6$ and let $E_{ij}\in {\frak{so}}(6)$ for $1\le i TITLE: continuum many mutually generic filters QUESTION [6 upvotes]: Given a countable model $M$ of set theory and an atomless, separative partial order $\mathbb{P} \in M$, can we construct (in the real universe) $2^\omega$ many pairwise mutually $\mathbb{P}$-generic filters $\{ G_r : r \in \mathbb{R} \}$? If CH holds, then the answer is yes. Recursively on the countable ordinals, we construct models $M_\alpha = M[G_\alpha]$, where $G_\alpha \subseteq \mathbb{P}$ meets all dense sets in $\bigcup_{\beta<\alpha} M_\beta$. We can also construct $2^\omega$ many distinct generic extensions. We can build (externally) a tree $T \subseteq \mathbb{P}$, where $T = \{ p_s : s \in 2^{<\omega} \}$. Make $T$ such that: (a) If $|s| = n$, then $p_s \in D_n$, where $\{ D_n : n \in \omega \}$ lists the dense subsets of $M$. (b) For $s$ a prefix of $t$, $p_s \geq p_t$. (c) For all $s$, $p_{s0} \perp p_{s1}$. Every real $r$ determines a distinct branch through $T$ and an associated $\mathbb{P}$-generic filter $G_r$. So we have $2^\omega$ many distinct filters, and since the models are countable, a pigeonhole argument shows that we must have $2^\omega$ distinct models. But this argument does not show that the filters are mutually generic. REPLY [8 votes]: I think you can handle this by slightly modifying the construction in the last paragraph of your question. In addition to what you did there, enumerate all the dense subsets of $\mathbb P\times\mathbb P$, say as $E_n$ ($n\in\omega$). Do this in such a way that every dense open set occurs infinitely often in the enumeration. Then at the stage of your recursion where you handle nodes $s$ of length $n$, extend your $p_s$'s so that every two of them, considered as a pair, are in $E_n$. That's only finitely much additional work at stage $n$, because there are only $2^n$ $s$'s to take care of. At the end of the construction, if $G$ and $G'$ are two of the generic filters you've produced, corresponding to paths $r$ and $r'$ through $T$, these will be mutually generic because any dense subset of $\mathbb P\times\mathbb P$ occurs as $E_n$ for infinitely many $n$, in particular for some $n$ beyond the level where $r$ and $r'$ branched apart.<|endoftext|> TITLE: Cubic graphs whose 2-factors all have the same cycle type QUESTION [9 upvotes]: Let $G$ be a bridgeless cubic graph. I am interested in such graphs where all 2-factors are isomorphic (as graphs), i.e. have the same partition as cycle type. We'll say that this partition is realized by the graph. Apart from the trivial $K_4$ and $K_{3,3}$, an example is the Petersen graph realizing $(5,5)$, as all of its 2-factors are of type $(5,5)$. We can easily derive from it a 2-connected graph with a unique 2-factor, of type $(12, 5,5)$, by taking two copies and adding the blue vertices and edges. It is in fact straightforward to see here that both "inner pentagrams" must belong to any 2-factor. In view of this kind of construction, it doesn't seem of high interest to ask which partitions can be realized by 2-connected graphs. But it is much more interesting to look at 3-connected graphs. It is not at all obvious for instance that there are 3-connected graphs which have Hamilton cycles as their only 2-factors. The following construction gives one for $n=26$: In the Petersen graph $P$, there are triples of edges with pairwise distance $3$. If we add vertices $a,b,c$ on each of the edges of such a triple (see picture), now take a second instance of this graph and add three edges $aa',bb',cc'$, we'll come up with a 3-connected graph $G$ whose only 2-factors are H-cycles, i.e. it realizes the partition $(26)$. This is because $a,b,c$ cannot be all contained in a 2-factor of $P$, so there must be two of $aa',bb',cc'$ contained in any 2-factor of $G$, and from there it is easy to check that such a 2-factor cannot contain a $C_5$ of $P$, i.e. it must pass through all vertices. EDIT: We can realize other types $(n)$ by replacing $P'$ by certain other graphs: $n=14$ by just adding one vertex at which $a',b',c'$ concur. $n=16$ by using instead a $K_3$ with vertices $a',b',c'$ $n=20$ by using a $K_4$ with $a',b',c'$ added on the edges of one of its triangles ${n=22}$ by using a $K_{3,3}$ with $a',b',c'$ added on the edges of one of its 1-factors [edited again - this one isn't hamiltonian, but has all its 2-factors of type $(17,5)$] for $n=36$ we can realize type $(n)$ as follows: start with two copies $P$ and $P'$ as above, join the three outgoing edges of $P$ (call them "blue edges") to the blue vertices and the three outgoing edges of $P'$ (call them "red edges") to the red vertices of this graph: As above, any 2-factor of $G$ must contain exactly two of the red edges, and it is clear that there cannot be a smaller cycle in it, by checking all paths between two red vertices. On the other hand, several H-cycles obviously exist. Replacing the central vertex by a small triangle doesn't change anything in terms of 2-factors, so the new graph realizes the partition (38). In fact, it looks like many (or, to be cautious, let's say several) graphs can be inscribed here instead, as long as they have a H-path between some two of the three 'corners', and as no other path exists between any two 'corners' such that the (graph induced on the) remaining vertices have a 2-factor. Believe it or not... inserting another Petersen graph here (with the three outgoing edges as above) would do $\implies$ a graph for $ n=48$ ! The general question: Which partitions (= types) can be realized by 3-connected graphs? Are there constructions that yield infinite families of such graphs (e.g. Hamiltonian graphs like the above one) ? REPLY [2 votes]: The examples given above already contain all the ingredients for a construction that yields indeed infinitely many graphs whose only 2-factors are H-cycles. Start with an arbitrary rooted binary tree. We'll think from the leaves to the root. Replace each leaf by the above graph $P$, leaving three "outgoing" edges. Replace each node by the graph $T$ shown below, which accommodates a red and a blue triple of "incoming" edges, one for each branch/leaf, and has three "outgoing" edges (in green). Finally, replace the root by one of the four small graphs as above ($K_1,K_3,K_4$ or $P$, maybe some others are possible, too), accommodating the three "incoming" edges. At each stage, two of the incoming red edges must belong to a 2-factor, and the same argument as for $n=36$ applies to show that the cycle (which must be, a forteriori, the same cycle as the one containing two of the blue incoming edges) must "continue beyond $T$". If the tree has $k$ leaves, this will result in a graph realizing $(n)$, where $n=13+(13+9)(k-1)+r=22k-9+r$ with $r\in\{1,3,7,13\}$ depending on the small graph chosen for the root.<|endoftext|> TITLE: Subgroups of Nilpotent groups with prescribed center QUESTION [7 upvotes]: Let $G$ be a torsion-free, finitely-generated, nilpotent group of nilpotency class at least 3. Does there exist a normal subgroup $N\leq G$ such that $G/N\cong \mathbb{Z}$ and $Z(G)=Z(N)$? (By $Z(H)$ I mean the center of the group $H.$) The basic examples I've played with have this property, but I'm no group theorist (this question arose in an operator algebraic setting) so that's the only evidence I have one way or the other. REPLY [4 votes]: Here are two examples. I describe it as Lie algebras (over any field $K$). (1) The 7-dimensional, 3-step nilpotent Lie algebra with basis $(X_1,\dots,X_7)$ and nonzero brackets $$ [X_1,X_2]=X_4,[X_1,X_3]=X_5,[X_2,X_3]=X_6,[X_1,X_4]=[X_1,X_5]=[X_2,X_4]=[X_3,X_6]=X_7$$ (2) The 6-dimensional, 4-step nilpotent Lie algebra with basis $(X_1,\dots,X_6)$ and nonzero brackets $$[X_1,X_2]=X_3, [X_1,X_3]=X_4, [X_2,X_3]=X_5, [X_1,X_5]=[X_2,X_4]=X_6.$$ Here's a common proof. In (1), let $V$ be the subspace generated by $X_1,X_2,X_3$, $W$ the subspace generated by $X_4,X_5,X_6$, and $Z$ the subspace generated by $X_7$. In (2), let $V$ be the subspace generated by $X_1,X_2$, $W$ the subspace generated by $X_4,X_5$, and $Z$ the subspace generated by $X_6$. Then in each $\mathfrak{g}$ of these two: $Z$ is the 1-dimensional center, and the bracket $V\times W\to Z$ defines a non-degenerate pairing. Thus for each hyperplane $X$ in $V$, its centralizer in $W$ is equal to its orthogonal $X^\bot$ with respect to this pairing. Each codimension 1 ideal of $\mathfrak{g}$ has the form $I_X=X\oplus \mathfrak{g}$ where $X$ is some hyperplane, and the center of $I_X$ is equal to the 2-dimensional ideal $X^\bot\oplus Z$. Both Lie algebra being defined over $\mathbf{Q}$, we can consider the corresponding lattice, which therefore answers your question.<|endoftext|> TITLE: Derived categories of curves equivalent then the curves are isomorphic QUESTION [6 upvotes]: I am a beginner at derived categories and I'm looking for a proof of the following fact: If $X$ and $Y$ are smooth projective curves such that $D^b(Coh\,X)$ is equivalent to $D^b(Coh\,Y)$ then $X$ and $Y$ are isomorphic. Can anyone give me the explanation of this fact or provide a good reference for this? As I understand, the crucial point here is that any object in $D^b(Coh \, X)$ splits as a sum of its cohomology sheaves but I don't know how to finish the proof. Thank you. REPLY [5 votes]: The non-elliptic case is easier. By a theorem of Bondal-Orlov, see [Huybrechts, Fourier-Mukai transforms in algebraic geometry, Theorem 4.11]: Theorem. Let $X$ and $Y$ be smooth projective varieties with equivalent derived category $D^b(X)$ and $D^b(Y)$. If the (anti)-canonical bundle of $X$ is ample, then $X$ and $Y$ are isomorphic. For elliptic curve, we need a generalized theorem by Kawamata, See [Kawamata, D-equivalence and K-equivalence] or [Huybrechts, Fourier-Mukai transforms in algebraic geometry, Theorem 6.15]: Theorem. Let $X$ and $Y$ be smooth projective varieties with equivalent derived category $D^b(X)$ and $D^b(Y)$. Then the (anti)-canonical bundle of $X$ is nef if and only if the (anti)-canonical bundle of $Y$ is nef. This theorem implies if $X$ is elliptic curve, the so is $Y$. By [Huybrechts, Fourier-Mukai transforms in algebraic geometry, Theorem 5.39], the hodge structuces of $X$ and $Y$ are the same, which yeilds that $X$ and $Y$ are isomorphic. REPLY [2 votes]: I don't know of a uniform way to prove it for all curves. For elliptic curves it follows from Hodge theory and for the rest it's a consequence of the Bondal-Orlov theorem. It's all explained in Huybrechts's excellent book on Fourier-Mukai transforms.<|endoftext|> TITLE: Does every compact manifold exhibit an almost global chart QUESTION [15 upvotes]: Let $M$ be a compact connected manifold. Is there a chart $\Psi:U \to \mathbb{R}^n$ such that the closure of $U$ is $M$? This is true for $S^n, T^n, K$, all compact surfaces, etc. If it is not true in general, what is the obstruction? REPLY [6 votes]: The exponential map for any Riemannian metric on your compact manifold $M$, based at any point $p$ of $M$, maps the tangent space $T_p M$, an ${\mathbb R}^n$, onto $M$ and is a diffeo inside the cut locus. Back on the tangent space, this `inside' of the cut locus is a star shaped domain relative to the origin, so defines a domain $V$ in ${\mathbb R}^n$ which is mapped diffeomorphically onto an open set $U$ whose closure is $M$. (The closure of the domain $V$ is homeomorphic to the closed ball in the tangent space, so this same argument shows that every compact manifold is the quotient of the n-ball by some identification of points on its boundary, the n-sphere. )<|endoftext|> TITLE: Put P inside Q! polygons/polyhedra QUESTION [5 upvotes]: We have two Polygons/Polyhedra P and Q. Does there exist a polynomial time algorithm to decide if we can put P (using translation and rotation) inside Q or not? First think about the case of which P and Q are convex and the input is the list of vertices, edges and faces of P,Q. REPLY [3 votes]: There is quite a bit of literature specifically on polygon containment, much of deriving from robotics applications. The paper below solves the containment problem, allowing translations and rotations, with a polynomial-time algorithm (roughly $O(n^6)$). Avnaim, Francis, and Jean Daniel Boissonnat. "Polygon placement under translation and rotation." Informatique theorique et applications. 23(1). 1989. p.5-28. Here is a snapshot from their introduction. The two polygons, named $E$ and $I$, have $n$ and $m$ edges respectively. This Intro reviews related, prior work:       Their algorithm extends to handle the case of polyhedra in $\mathbb{R}^3$, but for translation only (no rotation).<|endoftext|> TITLE: Comparison of etale and singular cohomology for varieties over number fields QUESTION [7 upvotes]: Whilst reading Hartshorne's appendix C I came across the comparison theorem for etale cohomology and singular cohomology: Let $X$ be a smooth projective variety over a number field $K$ and $\ell$ a prime number. Fix an embedding of $K$ into $\mathbb C$. Then there is a "natural" isomorphism of $\mathbb C$-vector space $$\mathrm H^n(X_{\bar K,et},\mathbb Q_\ell)\otimes \mathbb C \cong \mathrm H^n(X(\mathbb C),\mathbb C).$$ Do we have the slightly stronger "natural" isomorphism of $\mathbb Q_\ell$-vector spaces $$\mathrm H^n(X_{\bar K,et},\mathbb Q_\ell) \cong \mathrm H^n(X(\mathbb C),\mathbb Q)\otimes \mathbb Q_\ell?$$ I would appreciate references to the literature. REPLY [9 votes]: Yes. In fact what Artin proves in SGA4 exp XI thm 4.4 is that étale cohomology and singular cohomology agree for smooth schemes over $\mathbb{C}$ with finite coefficients. The statement you want will follow from this by taking inverse limits to get to $\mathbb{Z}_\ell$ and then extending scalars to $\mathbb{Q}_\ell$. If you don't feel like looking at SGA, you can find treatments of this in the books by Freitag-Kiehl, Milne,… Added (in response to comment). The isomorphism $H_{et}^*(X_{\bar K}, \mathbb{Q}_\ell)\cong H_{et}^*(X_{\mathbb{C}}, \mathbb{Q}_\ell)$ follows from the smooth base change theorem (cf. Milne, Etale cohomology, p 231 cor 4.3).<|endoftext|> TITLE: Die hard nilpotent spaces QUESTION [5 upvotes]: Let $V\subset\mathbb{C}^{n\times n}$ be a linear space consisting of $n\times n$ complex matrices. Say that $V$ is nilpotent if every matrix $v\in V$ is nilpotent; denote by $V^k$ the subspace spanned by all possible products $v_1\ldots v_k$ with $v_i\in V$. The conjecture is: Assume that both $V^k$ and $V^{k+1}$ are nilpotent spaces. Then, $\dim V^k\geqslant \dim V^{k+1}$. Some experiments with upper-triangular nilpotent spaces make me hope that this conjecture is true in general. I would be very happy if someone was able to help me with either a proof or a counterexample, even for $k=1$. REPLY [15 votes]: You conjecture is not true. Let $P$ be the $3\times 3$ matrix $$ \left( \begin{array}{ccc} 0&1&0\\ 0&0&1\\ 0&0&0\end{array} \right) $$ which is nilpotent with $P^2\neq 0$. Consider the subspace of $9 \times 9$ uppertriangular matrices spanned by $$ A=\left( \begin{array}{ccc}P&0&0\\0&P&0\\0&0&0\end{array} \right),~~ B= \left( \begin{array}{cc}0&0&0\\0&P&0\\0&0&P\end{array} \right). $$ Then $A^2$, $AB$ and $B^2$ are linearly independent.<|endoftext|> TITLE: A model structure on marked simplicial sets QUESTION [6 upvotes]: Do you have a reference for the following fact? And before that, is it true? The Joyal model structure on simplicial sets "lifts" to a model structure on the category of marked simplicial sets, having as fibrant objects precisely the objects sent to fibrant objects by the obvious forgetful functor $\mathbf{sSet}^+\to\mathbf{sSet}$. If it is not true, what should be a best approximation to it and where can I find it? Thanks! REPLY [2 votes]: I'm still unclear on what the motivation for looking at such a model structure might be, so I'm going to go ahead and guess that what you're ultimately interested in, Fosco, is an alternate construction of the usual model structure on $\mathrm{Set}_\Delta^+$ constructed in HTT 3.1.3.7 (where we take $S$ to be a point). Apologies if I'm guessing wrong! Verity's approach: Verity almost does this. To see this, recall that a stratified simplicial set in Verity's sense is a simplicial set equipped with certain "thin" simplicies which can have arbitrary dimension $\geq 1$ (and it's required that every degenerate simplex is thin). Verity constructs a model structure on the category $\mathrm{Strat}$ of stratified simplicial sets whose fibrant objects are what Verity calls weak complicial sets. A marked simplicial set can be considered as a stratified simplicial set by taking a 1-simplex to be thin iff it is marked, and taking all simplices of dimension $\geq 2$ to be thin. Then Verity's model structure on $\mathrm{Strat}$ restricts to a model structure on $\mathrm{Set}_\Delta^+$. The only caveat is that Verity's model structure needs to be further localized, by a "Rezk completeness condition". That is, in his model structure, every fibrant marked simplicial set will have the property that every marked simplex is an equivalence, but will not necessarily have the converse property that every equivalence is marked. This needs to be enforced by a Bousfield localization. This is not hard to do, but I'm not sure it's been done in the literature. The Cisinski/Olschok approach: It's notable that Verity's construction of his model structure is very much analogous to Joyal's construction of the Joyal model structure, which in turn is essentially an application of Cisinski's theory of model structures on Grothendieck topoi. The only reason that Cisinski's theory can't be directly applied to marked simplicial sets is that marked simplicial sets don't actually form a topos! But that's okay -- Olschok has extended Cisinski's theory to general locally presentable categories. Olshok's theory is a very general method for constructing model structures from a set of generating cofibrations, a set of elementary anodyne extensions, and a functorial cylinder object. In order to apply Olschok's theory, we must simply identify a class of elementary anodyne extensions and a functorial cylinder object (it's easy to write down a set of generating cofibrations such that the cofibrations are exactly the monomorphisms). The elementary anodyne extensions from Verity's paper (supplemented by a morphism corresponding to Rezk completeness), and the functorial cylinder given by $X \mapsto X \times \Delta[1]_t$ (where $\Delta[1]_t$ is the marked 1-simplex) yield the usual model structure. Olschok's theory allows us to identify the fibrant objects and fibrations between fibrant objects in this model structure, via lifting against the elementary anodyne extensions, if we additionally verify that the anodyne extensions are closed under certain pushouts involving the functorial cylinder; This is a straightforward combinatorial exercise.<|endoftext|> TITLE: Probability that a positive integer is the euler phi function of another positive integer QUESTION [6 upvotes]: Define $f(n) = |\{m : m\le n, \exists k \text{ s.t. }\phi(k) = m\}|$. Clearly, $f(n)\le \left\lfloor \frac{n}{2}\right\rfloor + 1$ since $\phi(n)$ is even for all $n > 2$. Is $\limsup_{n\rightarrow\infty} \frac{f(n)}{n} > 0$. It is less than $\frac{1}{2}$ by my previous statement, but I don't know how to proceed. REPLY [14 votes]: See Erick Wong's response here. In particular, Kevin Ford proved (in more precise form) that $$ f(n) = \frac{n}{\log n} \exp\left(O(\log \log \log n)^2\right),$$ whence $f(n)/n$ tends to zero. The same consequence also follows from an earlier result of Pillai (1929), available online here.<|endoftext|> TITLE: 3n+1 problem and cycles QUESTION [5 upvotes]: Just to make sure I am up to date with this problem. I know (or I think I do) that it is not yet proven that there are no non-trivial cycles for the collatz sequence (please correct me if I am wrong). But have we already reduced this problem to finding a non-trivial cycle? i.e. if we suppose that there are no non-trivial cycles at all, is it already proven that with this assumption the Collatz conjecture holds? As far as I can see in the literatures, even this is not shown (i.e. we can reduce the problem to proving whether there are no non-trivial cycles). But I may be wrong, so please correct me if I am wrong. Edit: Most of my knowledge are on par with Lagarias annotated bibliography and I believe it has been a few years since this literature review. REPLY [9 votes]: It might be a nice illustration of the general behaviour of increasing/decreasing by iterations from a purely statistical view. Consider some number odd number $a_0$ Then in the $mx+1$-problem, $a_1 = (ma_0+1)/2^A$ and the likelihood of $ma_0+1$ having $2^A$ as factor is just $1/2^A$. From this the likelihood, $ p(a_1 \ge a_0)$ can be computed. This can then be iterated to, say, $N$-iterations and the probability $ \mu_N = p(a_N \ge a_0)$ can be computed the obvious way. Probabilities $ \mu_N $ are between $0..1$; and the curves for different problem-parameters $m$ are not much detailful. So I also did a transform $y_{m,N} = \tanh^{-1}( 2\mu_{m,N}-1)$. Here are the curves for the transformed probabilities: I've inserted also the fictive "4x+1"-curve although that problemparameter $m=4$ does not make sense in reality. We see, that all curves except that of $m=3$ and $m=4$ increase, that means, the probability that $a_{m,N} \ge a_{m,0}$ for increasing number of iterations tends to $1$ for all problemparameters $m \ge 5$ Moreover, I find it amazing, that the increase of the $y$-values are asymptotically linear with the number of iterations (but didn't investigate this further). Important remark: Of course, the statistical view does not "see" exceptional cases like occurence of cycles, if their number is small. For instance, in the $5x+1$-problem we actually we know a few cycles, but without that few exceptions statistically the trajectories of most numbers $a_{5,0}$ seem to diverge. This view can only give the "big picture".<|endoftext|> TITLE: Factorization of a finite group by two subsets QUESTION [11 upvotes]: I want to write a GAP program for checking the following question. Let $G$ be a given finite group with order $n$. Is it true that for every factorization $n=ab$ there exist subsets $A$ and $B$ such that $|A|=a$, $|B|=b$ and $G=AB$? We have many candidates for a counterexample such as $PSL(2,8)$, $PSL(2,11)$, $PSL(2,13)$, $SL(2,11)$, $SL(2,13)$, $PSL(2,17)$ and $PSL(2,19)$, and need to check them. About $PSL_2(8)$, if it has a subgroup of index $21$ with the property (in the above question), then we can say that the answer for $PSL_2(8)$ is positive. Can some body check it by GAP? Any help would be highly appreciated. Now, can any one write a GAP code for $G=PSL_2(8)$, $a=21$ and $b=24$? (i.e. Are there subsets $A$ and $B$ such that $|A|=21$, $|B|=24$ and $PSL_2(8)=AB$? or equivalently, are there subsets $A$, $B$ of $PSL_2(8)$ such that $|A|=21$, $|B|=24$ and $|A^{-1}A\cap BB^{-1}|=1$?) Considering the answer from Peter Mueller, we need to try some other cases for getting a counterexample. With do attention to https://math.stackexchange.com/questions/885778/subgroups-of-groups-of-order-36 and https://math.stackexchange.com/questions/882859/groups-of-order-n2-that-have-no-subgroup-of-order-n?lq=1 , I propose the following groups: Groups of order $n=m^2$ with no subgroup of order $m$ (so $|G|=mm, a=b=m$). As one can see in the second link $n\geq 24^2$ ($m\geq 24$). Now, is it true for every group $G$ of order $n=24^2$ and $a=b=24$? What about $n=28^2$, $n=30^2$, etc.? REPLY [5 votes]: There are subsets $A$ and $B$ of orders $21$ and $24$ of $G=PSL(2,8)$ with $G=AB$. A naive search of course does not work. However, in trying to find first $A$ with $A^{-1}A$ being small, a good candidate to work with is $A=G_7G_3$, where $G_3$ and $G_7$ are subgroups of order $3$ and $7$. The smallest possible size of such a set $A^{-1}A$ is $48$. Finding $B$ with $AB=G$ is an exact cover problem: For each $g\in G$ set $S_g=\{ag\;|\;a\in A\}$. Then a set $B$ has the required property if and only if the sets $S_b$, $b\in B$, yield a disjoint union of $G$. I can't offer a GAP code, rather a quick and naive Sage code which finds such good sets $A$ and $B$: g = PSL(2,8) n = g.order() l = {i:x for i,x in enumerate(g)} li = {y:x for x,y in l.items()} g7 = g.sylow_subgroup(7) s3 = g.subgroup([[z for z in g.conjugacy_classes_representatives()\ if z.order() == 3][0]]) n3 = g.normalizer(s3) tv = [z[0] for z in g.cosets(n3)] for g3 in [s3.conjugate(z) for z in tv]: A = [g(zz)*g(z) for zz in g7 for z in g3] AA = set([z1^(-1)*z2 for z1 in A for z2 in A]) print len(AA) if len(AA) > 48: continue M = matrix(n,n,0) for i, x in l.items(): for j in [li[a*x] for a in A]: M[i,j] = 1 c = sage.combinat.matrices.dlxcpp.OneExactCover(M) if c == None: continue B = [[l[i] for i in range(n) if M[i] == row][0] for row in c] AB = set([a*b for a in A for b in B]) print len(AB) == n # Check if AB = G break One solution we get is $A=G_7G_3$ with $G_7$ and $G_3$ generated by $(3,8,6,4,9,7,5)$ and $(1,3,4)(2,7,8)(5,9,6)$, and $B=\{(), (1,2,4,3,5,6,7,9,8), (2,5,8,3,7,6,4), (1,2,5,7,3,6,8), (2,6,9,4,8,7,5), (1,2,7,8,6,4,5), (2,8)(3,7)(4,9)(5,6), (1,6,5,2,4,8,9,3,7), (1,4,6)(2,5,7)(3,8,9), (1,5,7,4,6,2,3,9,8), (2,7,6,3,4,9,5), (2,4)(3,6)(5,7)(8,9), (1,5)(3,6)(4,9)(7,8), (1,4,8,5,2,3,9,7,6), (2,9,5,7,4,3,8), (1,7)(2,5)(3,8)(4,9), (1,5,8,2,4,6,3,7,9), (2,3)(4,6)(5,9)(7,8), (1,2,6,8,4,7,9), (1,4,5,6,2,9,7,3,8), (1,4,8,6,9,3,5), (1,3,7,8,5,2,6,4,9), (1,2)(3,5)(6,9)(7,8), (1,3,9,8,2,4,7)\}$. Added: (in view of the modification of the question) The groups of orders $24^2$ and $28^2$ are solvable by Burnside's $p^aq^b$-Theorem. Hence they don't give counterexamples, as pointed out in the answers to this related question by the OP. These answers, by the way, give more promising indications where possible counterexamples may reside. (As to $PSL(2,8)$, the two potential candidates $(a,b)=(21,24)$ and $(a,b)=(12,42)$ are given there. The first doesn't exist by the result from above, and a slight modification of the above code gives $G=AB$ with $|A|=12$, $|B|=42$, so the other one doesn't exist either.)<|endoftext|> TITLE: Self-duality in a lattice QUESTION [7 upvotes]: Is there any self-dual lattice $(X,\le)$ such that there is not any self-duality $f:X\to X$ such that $f\circ f = 1_X$? REPLY [8 votes]: Yes. Let $L$ be the lattice structure on $\mathbb Z$ with the following Hasse diagram: -6 <----- -2 <---- 2 <---- 6 <--- \ / \ / \ / \ ... -5 -3 -1 1 3 5 7 ... \ / \ / \ / \ ---> -4 -----> 0 ----> 4 ----> 8 where all the diagonal arrows go upwards. It is easy to see that the only selfdualities of $L$ are of the form $f(n)=n+c$ for $c\equiv2\pmod4$, and in particular, they are never involutive.<|endoftext|> TITLE: Prime factorization "demoted" leads to function whose fixed points are primes QUESTION [21 upvotes]: Let $n$ be a natural number whose prime factorization is $$n=\prod_{i=1}^{k}p_i^{\alpha_i} \; .$$ Define a function $g(n)$ as follows $$g(n)=\sum_{i=1}^{k}p_i {\alpha_i} \;,$$ i.e., exponentiation is "demoted" to multiplication, and multiplication is demoted to addition. For example: $n=200=2^3 5^2$, $f(n) = 2 \cdot 3 + 5 \cdot 2 = 16$. Define $f(n)$ to repeat $g(n)$ until a cycle is reached. For example: $n=154=2^1 7^1 11^1$, $g(n)=20$, $g^2(n)=g(20)=9$, $g^3(n)=g(9)=6$, $g^4(n)=g(6)=5$, and now $g^k(n)=5$ for $k \ge 4$. So $f(154)=5$. It is clear that every prime is a fixed point of $f(\;)$. I believe that $n=4$ is the only composite fixed point of $f(\;)$. Q1. Is it the case that $4$ is the only composite fixed point of $f(\;)$, and that there are no cycles of length greater than $1$? (Yes: See EmilJeřábek's comment.) Q2. Does every prime $p$ have an $n \neq p$ such that $f(n) = p$, i.e., is every prime "reached" by $f(\;)$? (Yes: See JeremyRouse's answer.) There appear to be interesting patterns here. For example, it seems that $f(n)=5$ is common. (Indeed: See მამუკა ჯიბლაძე's graphical display.) REPLY [4 votes]: Was thinking about this old question again, and made another image (sorry it is unreadable), in the style of user @მამუკა ჯიბლაძე, of the numbers $\le 2000$ that ultimately map to $5$ ($5$ and $6$ are in the center, with $8$ right and $9$ left, each connecting to $6$). For example: \begin{eqnarray} n = 1788 &=& 2^2 \; 3^1 \; 149^1\\ g(1788) &=& 2 \cdot 2 + 3 \cdot 1 + 149 \cdot 1 = 156\\ n = 156 &=& 2^2 \; 3^1 \; 13^1 \\ g(156) &=& 2 \cdot 2 + 3 \cdot 1 + 13 \cdot 1 = 20\\ n = 20 &=& 2^2 \; 5^1\\ g(20) &=& 2 \cdot 2 + 5 \cdot 1 = 9\\ n = 9 &=& 3^2\\ g(9) &=& 3 \cdot 2 = 6\\ n = 6 &=& 2^1 \; 3^1 = 5\\ g(5) &=& 5 \end{eqnarray} The overall structure of the $5$-sink remains the same as far as I can take the computations.<|endoftext|> TITLE: is there any heuristics suggesting that the number of Fibonacci primes below $x$ is equivalent to $\log_{\phi}\log_{\phi}x$? QUESTION [6 upvotes]: The question of knowing whether there are infinitely many Fibonacci primes is an open question. As $F_p$ is prime only if $p$ is prime, one has $\pi_{FP}(x)\le \pi(\log_{\phi} x+0.5\log 5)$, but numerical computations seem to show that this quantity is roughly equal to $\log_{\phi}\log_{\phi}x$, where $\pi_{FP}(x)$ is the number of Fibonacci primes below $x$. Is there any heuristics suggesting this estimation should be true? Thanks in advance. REPLY [9 votes]: We need to find a reasonable-sounding answer for the following question: for a fixed prime $p$; what is the probability that $F_p$ is prime? A prime $q TITLE: Undecidable puzzles QUESTION [13 upvotes]: There are plenty of popular NP-hard puzzles, for example, generalized Sudoku ($n^2 \times n^2$-board), Flow (I cannot give a source for this), Minesweeper, etc. Recently, I read a bit about aperiodic tilings of the plane, and it is undecidable whether a set of tiles can tile the plane or not, since there might be aperiodic tilings. I do not really consider this problem a popular game you would find in a newspaper. So, my question is this: what puzzles/games are there that are undecidable in general? The Post correspondence Problem, (PCP), is undecidable and this has a very "puzzle"-feel to it, so I would say this game could qualify. Perhaps some generalization of the board game Roborally on an infinite board, would lead to some type of undecidability, that is, there is no algorithm that given a RoboRally configuration produces the resulting configuration after all rules have been applied? As a funny note, the sand-box game Minecraft allows the user to make circuits, making the game (in theory) Turing complete. REPLY [4 votes]: A recent (after the question was posted) arxiv preprint proves that Magic the Gathering is Turing-complete. The abstract states: This shows that even recognising who will win a game in which neither player has a non-trivial decision to make for the rest of the game is undecidable.<|endoftext|> TITLE: K-theory for the $C^*-$algebra of the continuous functions on the $2-$torus and the Bott projection QUESTION [5 upvotes]: I am trying to understand the K-theory for the $C^*-$algebra of the continuous functions on the $2-$dimensional torus $T^2$. In particular I am interested on the $K_0-$group. I have read that the generators of the group $K_0(C(T^2))$ are two elements: the unit $[1]$ and the Bott Projection $[Bott]$. Unfortunately, I cannot find the definition of the Bott projection for the torus, I have only seen the definition of the Bott projection for $\mathbb{R}^2$. Can someone tell me the definition of this projection? Or can someone give me another generator (in terms of projections)? I thank you all for the attention and the help. REPLY [5 votes]: If you want an explicit projection, you can form a variation of a Rieffel projection as follows. First take any function $f$ from $[-\pi/2,\pi/2]$ to $[0,1]$ that sends $-\pi/2$ to $1$, dips down to $0$ at $0$ and then goes back up to take value $1$ at $\pi/2$. Now define two more functions $$ g=\begin{cases} 0 & x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\\ \sqrt{f-f^{2}} & x\notin\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \end{cases} $$ and $$ h=\begin{cases} \sqrt{f-f^{2}} & x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\\ 0 & x\notin\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \end{cases} . $$ The projection then is then $$ p(\phi,\theta) = \left[\begin{array}{cc} f(\phi) & g(\phi)+h(\phi)e^{i\theta}\\ g(\phi)+h(\phi)e^{-i\theta} & 1-f(\phi) \end{array}\right] . $$<|endoftext|> TITLE: Is $x^{n}-x-1$ irreducible? QUESTION [17 upvotes]: Is it true that for every $n \in \mathbb{N}$, $x^{n}-x-1$ is irreducible in $\mathbb{Z}[x]$? The standard irreducibility criteria seem to fail. REPLY [29 votes]: This is true; it is due to Selmer. Ljunggren (On the irreducibility of certain trinomials and quadrinomials, Math. Scand. 1960) has obtained the complete list of reducible trinomials with $\pm 1$ coefficients. For more details see Gerry Myerson's answer to this question, which contains a review of Ljunggren's paper: About irreducible trinomials . To this I may add Prasolov's monograph Polynomials as a (probably) more accessible reference; there, you will find a complete treatment of Ljunggren's result. Added. As Greg Martin notes, the trinomial $x^n-x-1$ in fact has Galois group $S_n$. Osada's paper to which he refers is The Galois groups of the polynomials $X^n + aX^l+b$ (J. Number Theory 25, pp. 230-238, 1987), although the result already appears in an earlier paper by E. Nart and N. Vila, quoted [7] in loc.cit. However, the proof of the maximality of the Galois group uses Selmer's result that the trinomial is irreducible, and does not yield a new proof of irreducibility. The argument [that Selmer's result implies full Galois group] is presented in section 4.4 of Serre's 1988 course Topics in Galois Theory, see Remark 2 on page 42. You may find this text on the web: http://www.msc.uky.edu/sohum/ma561/notes/workspace/books/serre_galois_theory.pdf To summarize the argument, which also works for a wider class of irreducible trinomials, one checks directly that $x^n-x-1$ has at most one double root at each ramified prime. Consequently, all non-trivial inertia subgroups are generated by a transposition. Since $\mathbb{Q}$ has no unramified extensions, the Galois group of any polynomial over $\mathbb{Q}$ is generated by its inertia subgroups, and hence we see that our $G \subset S_n$ is generated by transpositions of $S_n$. Selmer's result implies that $G$ is also transitive, and it is a standard fact (proved in every course in Galois theory) that a transitive subgroup of $S_n$ generated by transpositions must be the full $S_n$.<|endoftext|> TITLE: Spectrum of this ODE QUESTION [5 upvotes]: I noticed something interesting studying this Sturm-Liouville Problem: $$ \frac{d}{dx}\left(\sqrt{(1-x^{2})}\frac{df}{dx} \right)+\frac{\left(n \alpha x+\alpha^2 x^{2} + \lambda\right)f}{\sqrt{(1-x^{2})}}= 0,$$ where $\alpha \in \mathbb{R}$ with periodic boundary conditions on $[-1,1]$. (Lambda is the eigenvalue) From basic St.-Liouville theory we know that the spectrum will be discrete cause the Sturm-Liouville operator is self-adjoint and the problem is regular. Furthermore, we will have that $-\infty <\lambda_0 < \lambda_1 \le \lambda_2 < \lambda_3 \le .. \ .$ Now I noticed by extensive numerical calculations that for odd $n$ we will have that $\infty < \lambda_0 < \lambda_1 < \lambda_2 <...< \lambda_n,$ but after $\lambda_n$ all eigenvalues occur pairwise $\lambda_{n+1}=\lambda_{n+2}$. Is there a way to explicitely show this? Edit: Due to two good questions, I want to add some information to the question: The eigenfunctions seem to a double eigenvalue seem to be even/odd functions respectively. I assume $\alpha >0$, but maybe it is worth studying $\alpha=0$ first (is similar to Legendre's ODE). Edit2: I think I could reduce my problem to a Linear Algebra problem. Due to the very different kind of question this observation includes I asked a new question about this: https://mathoverflow.net/questions/177814/explain-eigenvalue-structure-of-these-sequences-of-matrices . REPLY [3 votes]: Let's rewrite your equation as a Schrödinger equation, as follows: Introduce the new variable $t\in (-\pi/2, \pi/2)$ by $x=\sin t$. Then if $f$ solves your boundary value problem (with periodic boundary conditions), then $y(t)=f(x(t))=f(\sin t)$ satisfies $$ -y'' +V(t)y = \lambda y , \quad\quad y(-\pi/2)=y(\pi/2),\quad y'(-\pi/2)=y'(\pi/2) , $$ with the potential $$ V(t) = -n\alpha\sin t -\alpha^2\sin^2 t = \frac{\alpha^2}{2}\cos 2t - n\alpha\sin t-\frac{\alpha^2}{2}. $$ (We can do such a transformation for any SL equation; see for example here.) The additional change of variable $t=2s-\pi/2$ (so $0 TITLE: non-Borel set which intersects every compact in a Borel set QUESTION [5 upvotes]: I remember hearing some time ago that there is a locally compact Hausdorff space $X$ and a non-Borel subset $E$ which intersects every compact set in a Borel set. (This would contradict Lemma 13.9 of Royden, Real Analysis 3rd edition 1988, which is stated without proof). Is there a reference for this? Can this happen if the space $X$ is perfectly normal? REPLY [8 votes]: Take the ordinal $\omega_{1}$. Then every subset of $\omega_{1}$ intersects each compact subset of $\omega_{1}$ in a Borel set. However, not every subset of $\omega_{1}$ is Borel. If $\mathcal{M}$ is the collection of all sets which are either non-stationary or contains a closed unbounded set, then $(\omega_{1},\mathcal{M})$ is a $\sigma$-algebra that contains each closed set and hence each Borel set. However, $(\omega_{1},\mathcal{M})$ is a proper $\sigma$-subalgebra of $P(X)$. REPLY [4 votes]: If I understand the question, then you are correct, there is such a space. I'll sketch what I hope is a correct argument. Take $X=\coprod_AY$ for some fixed locally compact Hausdorff space $Y$ and some index set $A$. As long as $Y$ is sufficiently complicated (probably $Y=\mathbb R$ would work) and $A$ is sufficiently large ($\left|A\right|\geq\aleph_1$ is enough), then you can find a collection of Borel sets $\{B_\alpha\subseteq Y\}_{\alpha\in A}$ which are not all contained in any countable "stage" towards the Borel $\sigma$-algebra of $Y$ (I'm sorry I don't know the standard terminology for this; using the notation of the wikipedia entry on Borel sets, I mean that for any countable ordinal $m$, not all of the $B_\alpha$ are contained in $G^m$). Now the subset: $$\coprod_{\alpha\in A}B_\alpha\subseteq\coprod_{\alpha\in A}Y$$ is not a Borel set. If it were, it would be contained in some $G^m(X)$ for some countable ordinal $m$, but this would imply that every $B_\alpha$ is in $G^m(Y)$, a contradiction. If we take $Y$ metrizable, then $X$ will be metrizable as well, and hence perfectly normal.<|endoftext|> TITLE: Kan condition in simplicial homotopy theory QUESTION [6 upvotes]: I know Kan condition (see http://en.wikipedia.org/wiki/Kan_fibration) is something like homotopy extension condition, and I know this condition ensures homotopy defined by the naive idea to be an equivalence condition. But this condition rules out some easy-constructed simplicial sets, like those obtained from the standard simplex, and simplicial sets satisfying this condition are always quite huge. My question is: can we avoid Kan condition by considering ALL simplicial sets and define homotopy to be an equivalence condition by ARTIFICIALLY adjoining all transitivity? Can we define well-behaved homotopy groups then for all simplicial sets and still get the equivalence between its homotopy category and that of CW complex? REPLY [3 votes]: You asked: 'can we avoid Kan condition by considering ALL simplicial sets and define homotopy to be an equivalence condition by ARTIFICIALLY adjoining all transitivity?' Why not do things as follows? To turn homotopy into an equivalence relation you have to not only generate under transitivity but also to add inverses, so to do that you have to form the free groupoid ... doh! This has been done systematically by Dwyer and Kan (and also by Joyal and Tierney at about the same time) by forming the loop groupoid on the simplicial set, cf. https://ncatlab.org/nlab/show/Dwyer-Kan+loop+groupoid. Of course this ends up with a simplicially enriched groupoid (aka simplicial groupoid) and yes, of course, you can use this to calculate homotopy groups, etc. via the Moore complex construction. ... and yes of course, you can go back to simplicial sets via the loop-groupoid functors left adjoint $\overline{W}$... and low and behold that yields a Kan complex (and the fillers have the sort of properties you would like in answer to your question ... (although this is rarely shown in texts). This has an advantage in as much as it comes with a lot of fibre bundle / fibration intuitions analogous to those for the loop group in topology. These are classical, very neat and very useful!<|endoftext|> TITLE: Question about higher inductive types and computational rules QUESTION [5 upvotes]: I have been trying to make my way through the homotopy type theory book, slowly but surely, and I just finished reading this introductory series of 3 articles on hott on ScienceForAll. http://www.science4all.org/le-nguyen-hoang/homotopy-type-theory/ At some point, he describes identity proofs and higher inductive types, he shows how you could construct the integers starting with a base element 0 and two constructors, up (u) and down (d), such that udA=A, for any "integer" A. Now he says that one way of reducing the complexity of the type and flattening it is to use higher order inductive types and have two identity constructors: id_ud (n) : n = u(d(n)), id_du (n) : n = d(u(n)) Now, my question is simply: Why can't we just make up this type by playing around with computation rules? Couldn't we just posit an induction principle that would say something like: ind_Z (C,0) := C(0), ind_Z (C, u(d(n))) := ind_Z (C, n), ind_Z (C, d(u(n))) := ind_Z (C, n) I'm aware that we'd get stuck at something like uudd(0), but then I'm sure we could have more rules to swap the ups and downs around or something. Then, having those equalities at the definitional level, u(d(n))=n : Z (3-bar equality symbol), we would get the equality type from above u(d(n))=n (as a type). Is the problem that it's too strong? Thank you very much REPLY [5 votes]: If I understand the question correctly, one answer is that the rules of type theory are not (supposed to be) arbitrarily chosen independently of each other like the axioms of set theory are. They come in "packages", one for each "type-forming operation", and each package has the same general shape: it consists of a Formation rule, some Introduction rules, some Elimination rules, and some Computation rules. A Formation rule tells you how to introduce a type, e.g. "if $A$ and $B$ are types, so is $A\times B$". An Introduction rule tells you how to introduce terms in that type, e.g. "if $a:A$ and $b:B$, then $(a,b):A\times B$". An Elimination rule tells you how to use terms in that type to construct terms in other types, e.g. "if $f:A\to B\to C$, then $rec(f):A\times B\to C$". And a Computation rule tells you what happens when you apply an Elimination rule to an Introduction rule, e.g. "$rec(f)((a,b)) \equiv f(a)(b)$". These four groups of rules that pertain to any type former can't be chosen arbitrarily either; they have to be "harmonious". There's no formal definition of what this means, but the idea is that the Introduction and Elimination rules should determine each other, and the Computation rules should tell you exactly how to apply any Elimination rule to any Introduction rule and no more. A bit more specifically, there are two kinds of type formers: positive ones and negative ones. For a positive type, you choose the Introduction rules, and then the Elimination rules are essentially determined by saying "in order to define a function out of our new type, it suffices to specify its value on all the inputs coming from some Introduction rule". For a negative type, you choose the Elimination rules, and then the Introduction rules are essentially determined by saying "in order to construct an element of our new type, it suffices to specify how all the Elimination rules would behave on that element". In both cases, the Computation rules then say that these "specifications" do in fact hold (as definitional equalities). So, you can't just arbitrarily postulate Computation rules. I mean, you can, but you won't end up with a well-behaved theory. Thus, we have to regard the equalities postulated by higher inductive types as Introduction rules, with correspondingly determined Elimination and Computation rules. (We could try to make them Elimination rules instead, yielding a notion of "Higher Coinductive Type", but there's no consensus yet on what such a thing should look like.) Why do we require this sort of harmony between the rules? From a computational point of view, it's so that we can actually compute with the Computation rules. If you didn't have that sort of harmony, then you might end up with "stuck" terms with an Elimination form applied to an Introduction form but no applicable Computation rule, or conversely if there were too many Computation rules then you might have some terms that try to "compute" to many different things. From a category-theoretic point of view, it's because we're specifying objects by universal properties: a positive type former has a "left" universal property like a colimit, while a negative type former has a "right" universal property like a limit. I wrote a blog post about this here.<|endoftext|> TITLE: The shortest interval for which the prime number theorem holds QUESTION [6 upvotes]: It is well known that the prime number theorem on the form \begin{align*} \pi(x+y) - \pi(x) \sim \frac{y}{\log (x+y)} \end{align*} breaks down for short enough intervals, e.g. taking $y=(\log x)^\lambda$ for any $\lambda>1$, as shown by Maier. As to what is short enough (or long enough), both Granville, p. 7 and Soundararajan, p. 79 conjecture that the prime number theorem holds for all $x$ as long as $y\geq x^\epsilon$. However, I believe to have a heuristic argument for the following conjecture: Conjecture The choice of $y= \sqrt{x}$ is necessary and sufficient for \begin{align*} \pi(x + y)-\pi(x) \sim \frac{y}{\log (x+y)} \end{align*} to hold for all $x$ as $x\rightarrow \infty$. I understand that this can be considered a bold claim, so my question is: Q: Considering the heuristic below, is the conjecture as stated reasonable? Heuristic: To understand why this conjecture should hold, we need to look at the short intervals between consecutive primes squared, defined by $s_k:=\{p_{k}^2, \dots p_{k+1}^2-1\}$ for $k\geq 1$. These intervals naturally occur in the context of the sieve of Eratosthenes, and in particular, each $s_k$ has the specific quality of being fully sieved by the $k$ first primes; any element in $s_k$ is either divisible by some $p \in \mathcal{P}_k:=\{p_1, \dots,p_{k}\}$ or else is a prime $p \notin \mathcal P_k$. In addition, the exact distribution of primes in $s_k$ is in its entirety build up of the periodic sequences \begin{align*} \rho_{k}(n):=\begin{cases} p_k & \text{if } p_k \mid n,\\ 1 & \text{otherwise}, \end{cases} \end{align*} which we visualise for the specific example of $s_3$ by the following table: \begin{matrix} n & 25 & 26 & 27 & 28 & \bf{29} & 30 & \bf{31} & 32 & 33 & 34 & 35 & 36 & \bf{37} & \cdots & 48\\ \hline \\ \rho_1(n) & 1 & p_1 & 1 & p_1 & 1 & p_1 & 1 & p_1 & 1 & p_1 & 1 & p_1 & 1 & \cdots & p_1\\ \rho_2(n) &1 & 1 & p_2 & 1 & 1 & p_2 & 1 & 1 & p_2 & 1 & 1 & p_2 & 1 & \cdots & p_2\\ \rho_3(n) &p_3 & 1 & 1 & 1 & 1 & p_3 & 1 & 1 & 1 & 1 & p_3 & 1 & 1 & \cdots & 1 \end{matrix} Observe that the lengths of the intervals $s_k$ are of the form $|s_k|=2 p_{k+1} g_k-g_k^2$, where $g_k:=p_{k+1}-p_k$, and hence lie on the curves $2 \sqrt{x} g - g^2$, with $g=2n$, $n\geq 1$. Necessary part As $k\rightarrow \infty,$ Any interval growing slower than $\sqrt{x}$ will eventually be infinitesimal compared to arbitrarily many primes smaller than $p_k$, and cannot be expected to accurately sample the distribution of primes in $s_k$, which derives from the underlying periodic sequences $\rho_j(n)$, $1\leq j \leq k$, and where the largest period is $p_k$. What this suggests is that $y=\sqrt{x}$ is the sharp barrier below which the prime number theorem breaks down. Sufficient part On the other hand, any interval growing faster than $\sqrt{x}$ will eventually cover arbitrarily many intervals $s_k, s_{k+1}, \dots, s_{m}$. But this results in an underestimate of $\pi(x + y)-\pi(x)$, since $y/\log (x+y)$ assumes constant density of primes across $[x,x+y]$, equal to the density in the final interval $s_m$ covered, hence suggesting that $y= \sqrt{x}$ is also sufficient. At its most extreme, the latter argument is exemplified by the estimate $\pi(x) \sim x/\log x$, which is well known to be an inferior guess of the number of primes up to $x$ compared to $\pi(x) \sim \textrm{li}(x)$. (I should add that the heuristic argument is presented in greater detail in a draft manuscript I recently added to arXiv, titled Primes in the intervals between primes squared). ADDED: CLARIFICATION OF HEURISTIC ARGUMENT In an attempt to make the heuristic clearer, consider the table above. If we move across this with intervals smaller than $p_3=5$, there will be some places where we underestimate the number of primes and some places where we overestimate. This effect magnifies for larger $k$ and intervals growing slower than $\sqrt{x}$, and suggests the necessary part of the heuristic. (Consider even measuring the density across $\rho_3(n)$ only. It should be even more obvious then.) REPLY [8 votes]: I'm not an analytic number theorist, so take this all with many grains of salt. Let $S(x,y,p)$ be the set of integers in the interval $(x,x+y)$ which are NOT divisible by $p$. The argument you are imagining for the prime number theorem is $$\pi(x+y)-\pi(x) = \left| \bigcap_{p \leq \sqrt{x}} S(x,y,p) \right| \approx y \prod_{p \leq \sqrt{x}} \frac{|S(x,y,p)|}{y} \approx y \prod_{p \leq \sqrt{x}} \left( 1 - \frac{1}{p} \right).$$ Your point is that the second $\approx$ is not good on a term by term basis if $p>y$, because the true value of $|S(x,y,p)|/y$ will be either $1$ or $1-1/y$, not $1-1/p$. You therefore suggest that the whole composite approximation should also not be good. I see two immediate issues: A product of inequalities is not an inequality While it is true that $|S(x,y,p)|/y$ is not $1-1/p$, this formula can be wrong in either direction. So it is possible that the errors cancel and the products are close to equal. The sketched proof doesn't work for large $y$ Even when $y$ is as large as $x$, there is a huge issue: $\prod_{p \leq \sqrt{x}} (1-1/p) \approx e^{- \gamma}/\log \sqrt{x} = 2 e^{-\gamma}/\log x$, not $1/\log x$ as we want. So there is already something sketchy here. I don't have an intuition for why the right constant in the PNT is $1$, not $e^{-\gamma}$ or $2 e^{- \gamma}$, but since I already know that there is an issue with this sort of argument, I wouldn't take it too seriously in predicting exactly when PNT would fail. (To clarify, I know many arguments why the constant must be $1$: For example, $\sum_{p \leq n} \frac{n}{p} \log p$ should be $\approx \log n! \approx n \log n$, and this only works if the constant is $1$. And I know why the sieve argument doesn't rigorously prove that $\pi(x) \approx e^{-\gamma} x / \log x$. What I don't have is a gut level understanding of why the sieve formula is right up to the constant factor, but not actually right.) None of this amounts to an argument FOR the conjectures of Granville and Soundararajan, it just argues that I wouldn't take your heuristic particularly seriously.<|endoftext|> TITLE: Squarefree numbers $n$ such that $432n+1$ is also squarefree QUESTION [11 upvotes]: This is a second attempt (see Primes $p$ such that $432 p +1$ is prime) Is the set of squarefree numbers $n$ such that $n(432 n+1)$ is also squarefree known to be infinite? Fact: the number of such numbers $n$ such that $n\leq 10^6$ is precisely $553095$. Do we expect that $$ \lim_{x\to \infty}\frac{\# \{n \leq x\ : \ n \ \textrm{is squarefree integer, and } 432n+1 \ \textrm{is squarefree}\}}{x} = 1/2?$$ REPLY [21 votes]: Here is a quick proof that the density in question exists and equals $$ c:=\frac{2}{3}\prod_{p\geq 5}\left(1-\frac{2}{p^2}\right)\approx 0.553087\ . $$ Let $f(d)$ denote the number of solutions of the congruence $n(432n+1)\equiv 0\pmod{d}$. Note that, for $p$ prime, $f(p^2)=1$ when $p<5$ and $f(p^2)=2$ when $p\geq 5$. Let $P\geq 2$ be fixed. By a simple inclusion-exclusion sieve combined with the Chinese remainder theorem, we see that the number of $n\leq x$ such that $n(432n+1)$ is not divisible by the square of any prime $p\leq P$, equals $$ x\prod_{p\leq P}\left(1-\frac{f(p^2)}{p^2}\right)+o_P(x)=x\prod_p\left(1-\frac{f(p^2)}{p^2}\right)+o_P(x)+O(x/P).$$ Observe that the infinite product on the right equals $c$. On the other hand, the number of $n\leq x$ such that $n(432n+1)$ is divisible by the square of some prime $p>P$, is at most $$ \sum_{p>P}O(x/p^2)=O(x/P).$$ Altogether we see that the number of $n\leq x$ such that $n(432n+1)$ is square-free, equals $cx+o_P(x)+O(x/P)$. In particular, both the lower and the upper density of these numbers equals $c+O(1/P)$. These quantities are independent of $P$, hence upon letting $P\to\infty$, we see that they both equal $c$. So the density exists and also equals $c$. For related comments see the introduction of this 1953 paper by Erdős.<|endoftext|> TITLE: Rank of Elliptic Curves QUESTION [10 upvotes]: Recently, I have heard of some heuristics that would suggest that the rank of elliptic curves are bounded (specifically in the congruent number family). I always though that the best way to prove something about the rank of an elliptic curve is to look at the rank of the 2-Selmer group. Indeed, I was able to find a reference that says that the rank in the congruent number family of elliptic curves is bounded by the 2-Selmer rank. Clearly it is not known that the 2-Selmer rank is bounded, but is it known that it is unbounded in the congruent number family? What about in general? Is this equivalent to asking if the rank of elliptic curves is unbounded? I would appreciate any answers or references. REPLY [11 votes]: Cassels showed in "Arithmetic on curves of genus 1 VI" that there is no bound on the size of the 3-Selmer group for the curves of $x^3+y^3+dz^3=0$. Well, he actually showed that the 3-torsion of Sha can be arbitrarily large. Similar results are known for other primes. Tom Fisher found arbitrarily large 5-torsion for instance. He refers to Bölling for the fact that the 2-torsion can be arbitrarily large. Kloosterman and Schaefer show that the $p$-torsion can grow over certain number fields. I have little hope that the $p$-Selmer group for a prime $p$ is bounded, because the Tate-Shafarevich part of it can grow. I would not know about $p=2$ for the congruent number curves.<|endoftext|> TITLE: On a.e. approximate differentiability of certain continuous real functions QUESTION [5 upvotes]: I have the following question: If $f:[0,1]\to \mathbb{R}$ is a bounded continuous function of $\sigma$-finite variation in sense 1, then is it true that $f$ is approximately differentiable a.e. on $[0,1]$? The $\sigma$-finite variation in sense 1 is obtained in the following way. For any measurable set $E\subset [0,1]$, we define $$v_f^\delta(E)=\inf\{\sum_{i=1}diam f([a_i,b_i]):E\subset \cup_i[a_i,b_i], b_i-a_i<\delta\}$$ and $$v_f(E)=\lim_{\delta\to0}v_h^\delta(E).$$ We say that $f$ has finite variation on $E$ if $v_f(E)<\infty$. We say that $f:[0,1]\to \mathbb{R}$ has $\sigma$-finite variation in sense 1 if there exists a partition $\{E_i\}$ of $[0,1]$ such that $f$ has finite variation on each $E_i$. Note that the answer to the question is positive if the following statement is true: Let $f:[0,1]\to \mathbb{R}$ be a bounded continuous function. Then $f$ has has $\sigma$-finite variation in sense 1 implies that $f$ has $\sigma$-finite variation in sense 2. For $E\subset [0,1]$, one defines another variation as $$v_f'(E)=\sup\sum|f(b_i)-f(a_i)|,$$ where the supremum is taken over all non-overlapping intervals $\{[a_k,b_k]\}$ whose end points belong to $E$. We say that $f$ has finite variation on $E$ in sense 2 if $v_f'(E)<\infty$. A bounded function $f:[0,1]\to \mathbb{R}$ is said to have $\sigma$-finite variation in sense 2 if $[0,1]=\bigcup_{i=1}^\infty E_i$ such that $f$ has bounded variation on each $E_i$ in sense 2. Thanks for any suggestions and answers! REPLY [4 votes]: There exists a continuous bounded function $f \colon [0,1] \to \mathbb R$ with $\sigma$-finite variation in the sense 1 and with positive set of points where the function is not approximately differentiable. An illustration of one such function is below. (I will only briefly comment on it. Please (come and) ask more details if you need.) Construct an oscillating function so that the set where it oscillates has positive measure, i.e. the length of the intervals where we define the function to be affine sum to something less than $1$.      (source) This function has $\sigma$-finite variation in the sense 1: On each of the countably many intervals where the function is affine, the variation is clearly finite. On the remaining set where we "oscillate", the variation is also easily seen to be finite since on any given scale we have removed the set where we jump. Notice that this remaining set does not have finite variation in the sense 2 because for it removing the affine part does not remove the variation. The function is clearly not approximately differentiable in the oscillating part.<|endoftext|> TITLE: Generalization of Frobenius groups QUESTION [9 upvotes]: Frobenius group is a transitive permutation group on a finite set, such that no non-trivial element fixes more than one point and some non-trivial element fixes a point. In other words, if in a transitive permutation group, each element that fixes a point, fixes exactly one point, and there is at least one such element we call it Frobenius. Let us define a $t$-Frobenius group to be a transitive permutation group, such that each element that fixes a point fixes exactly $t$-points, and there is at least one such element. Has this concept been studied already? Perhaps under different name? Does $t$-Frobenius group exist for every $t$? Does every $t$-Frobenius group have a regular subgroup? REPLY [14 votes]: Yes, these groups exist for all $t$. To see that, let $G$ be a Frobenius group with complement $H$ of size $tu$ for some $u>1$, where $H$ has a normal subgroup $K$ of order $u$. We could, for example, choose $H$ to be a cyclic group of order $tu$, and we can do that for any $t$ and $u$. Then the image of the permutation representation of $G$ on the cosets of $K$ is a $t$-Frobenius group. But note that not all examples of $t$-Frobenius groups arise in this way from Frobenius groups. For example, $D_8$ on $4$ points is $2$-Frobenius, but does not arise in this way. No, they do not all have regular subgroups. With above construction with $G$ Frobenius of degree $5$ and order $20$, and $u =2$, the resulting $2$-Frobenius group has no regular subgroup of order $10$. More generally, if the subgroup $H$ of $G$ is a trivial intersection set, which means that it is disjoint from its distinct conjugates, then the permutation representation of $G$ on the cosets of $H$ is a $t$-Frobenius group, where $t = |N_G(H):H|$. This gives rise to lots more examples. But this still does not include all examples. An example not of this form is the simple group ${\rm PSL}(2,7)$ acting on the cosets of a Klein $4$-group, which is a $6$-Frobenius group of degree $42$. This is because all nontrivial elements of the stabilizer are conjugate in $G$.<|endoftext|> TITLE: A stronger version of Van der Waerden's theorem? QUESTION [9 upvotes]: Let $W$ be an infinite word over a finite alphabet $\{1,\dots,n\}$ and $k$ a positive integer. An easy application of Van der Waerden's theorem implies the existence of $k$ disjoint and consecutive intervals in $W$ such that the sum of letters in each interval are equal (Edit. See Barber's answer). On the other hand, Van der Waerden's theorem for the coloring by two colors (which easily implies the general case of every finite coloring) can be derived by this statement without much effort (Edit. See the comments). So I think of it as an equivalent form of original Van der Waerden theorem. I had conjectured a stronger version which I was unable to prove or disprove it: Conjecture: If $W$ is an infinite word over finite alphabet. Then for every positive integer $k$, there exists $k$ disjoint and consecutive intervals in $W$, with the same number of occurrence of each letter. I also know that a weaker version of this conjecture for binary alphabet and "the same" replaced by "proportional" is correct. Does there exist a similar known result? Can someone give a prove or counterexample? REPLY [18 votes]: Your claim is not true. Two consecutive blocks having the same number of occurrences of each letter (such as the English word "reappear") form what is called an "abelian square". A google search will easily produce a large literature on this subject. The best result (in terms of alphabet size) is due to the Finnish mathematician V. Keranen in 1992: he proved the existence of an infinite word over a 4-letter alphabet having no abelian squares. 4 is best possible, as an easy backtracking argument shows there is no such word over a 3-letter alphabet. Three consecutive blocks with the same property form an "abelian cube". Dekking proved in 1979 that there is an infinite word over a 3-letter alphabet avoiding abelian cubes. Again, 3 is best possible for alphabet size. Four consecutive blocks with the same property form an "abelian 4th power". Dekking also proved in 1979 that there is an infinite word over a 2-letter alphabet avoiding abelian fourth powers. You can easily find references with a google search.<|endoftext|> TITLE: Is the category of $G$-spaces a model category? QUESTION [5 upvotes]: Let $G$ be a compact Lie group and $\mathcal{C}_G$ the category of $G$-spaces (ie. topological spaces endowed with continuous left $G$-actions). Is there a model category structure on $\mathcal{C}_G$ for which (i) weak equivalences are the morphisms $f:X\rightarrow Y$ such that for all closed subgroups $H$ of $G$, $f^{H}:X^H\rightarrow Y^H$ is a weak homotopy-equivalence, and (ii) cofibrations are the morphisms $f:X\rightarrow Y$ that have the expected $G$-homotopy extension property? What I have read thus far strongly suggests that this is the case. Nevertheless, I would appreciate a reference that makes this explicit. REPLY [7 votes]: This result has nothing special to do with compact Lie groups: it works for arbitrary topological groups $G$. And as Karol gently points out, the "expected $G$-homotopy extension property" actually means "retracts of relative $G$-cell complexes". The fibrations, like the weak equivalences are created by the fixed point functors: $p$ is a $G$-fibration iff $p^H$ is a nonequivariant Serre fibration for all closed subgroups $H$. This is the ``Quillen model structure''; it was understood in the 1980's. There is also a classical or Hurewicz model structure whose weak equivalences are the $G$-homotopy equivalences and whose $G$-fibrations are the Hurewicz $G$-fibrations, defined by the $G$-CHP, and whose $G$-cofibrations really do have the $G$-HEP, which I imagine may be what the question had in mind. There is also a mixed model structure in which the weak equivalences are those of the Quillen model structure and the fibrations those of the Hurewicz model structure. Its cofibrations are the Hurewicz cofibrations that factor as composites of Quillen cofibrations and $G$-homotopy equivalences. The proofs are really no different than in the nonequivariant case. Some references on my web page are http://www.math.uchicago.edu/~may/BOOKS/alaska.pdf (Section VI.5), http://www.math.uchicago.edu/~may/PAPERS/MMMFinal.pdf (Theorem 1.8 there gives the same result as Schwede's A.1.18, with a little more detail about the properties of the Quillen model structure), and http://www.math.uchicago.edu/~may/TEAK/KateBookFinal.pdf (Chapter 17)<|endoftext|> TITLE: Why is geometric quantization (esp. Berezin-Toeplitz quantization) interesting for a symplectic geometer/topologist? QUESTION [8 upvotes]: I know that many symplectic geometers are interested in quantization as well. From what I understood, quantization isn't expected to be used as a tool to answer symplectic questions (as in translating the problem to the quantum world, solve it there, and somehow go back), unlike the relation, for example between complex geometry and tropical geometry. Rather, it is supposed to be interesting and have appeal for reasons intrinsic to quantization. As a person whose main interests are symplectic geometry, topology and algebraic geometry, why study quantization? What aspects or problems there may I find appealing? For those who approach mathematics from the physics realm, I guess the answer is straightforward, as those ways of quantizing symplectic manifolds are attemps at formalization of some physical theories and beliefs. I am looking for a more mathematically oriented answer, since I personally derive my interest and motivation from the intrinsic beauty of mathematics. Thanks REPLY [2 votes]: The method of coadjoint orbits suggests that irreducible unitary representations of a Lie group $G$ are something like quantum mechanical systems. In fact finding all irreducible unitary representation of nilpotent lie groups correspounds to geometric quantization of coadjoint orbits. In fact quantiz ation problem comes down to a finite set of coadjoint orbits for each reductive group: the nilpotent orbits. The method was carried over to geometric quantization where you want to deform the pointwise multiplication of functions with the Poisson bracket to a noncommutative product with the Poisson bracket as a first order approximation. Ultimately this lead to star product quantization. In symplectic geometry: quantization commutes with reduction : states that the space of global sections of a line bundle satisfying the quantization condition on the symplectic quotient of a compact symplectic manifold is the space of invariant sections of the line bundle. In topology, Geometric quantization is related to K-theory: About the conjecture quantisation co mmutes with reduction for noncompact simple groups the only finite dimensional unitary representations are direct sums of the trivial one which by Landsman's idea we can replace the representation ring of a group by the K-theory of its $C^∗$-algebra, and the $K$-index by the analytic assembly map.<|endoftext|> TITLE: When is $(q^k-1)/(q-1)$ a perfect square? QUESTION [18 upvotes]: Let $q$ be a prime power and $k>1$ a positive integer. For what values of $k$ and $q$ is the number $(q^k-1)/(q-1)$ a perfect square, that is the square of another integer? Is the number of such perfect squares finite? Note that $(q^k-1)/(q-1)$ is the number of points in a finite projective space of dimension $k-1$. The above question is related to the following one: How many non-isomorphic finite projective spaces are there whose numbers of points are perfect squares. Preliminary calculation shows that $(q^k-1)/(q-1)$ is a perfect square when $(k,q)$ takes on one of the values $(2,3)$, $(5,3)$, $(4,7)$, $(2,8)$, in which cases it is equal to $2^2$, $11^2$, $20^2$, $3^2$, respectively. When $k=2$, $(q^k-1)/(q-1)$ is a perfect square if and only if $q=3$ or 8. When $k=3$, $(q^k-1)/(q-1)$ cannot be a perfect square. When $q=2$, $(q^k-1)/(q-1)$ cannot be a perfect square. REPLY [31 votes]: The equation $$ \frac{x^k-1}{x-1}=y^m$$ is known as the Nagell-Ljunggren equation. It is conjectured that for $x\geq 2$, $y\geq 2$, $k\geq 3$, $m\geq 2$, the only solutions are $$ \frac{3^5-1}{3-1}=11^2,\qquad \frac{7^4-1}{7-1}=20^2,\qquad \frac{18^3-1}{18-1}=7^3.$$ For $m=2$, the equation was solved by Ljunggren (Norsk. Mat. Tidsskr. 25 (1943), 17-20). Hence for your problem (where $m=2$ and $x$ is a prime power) we can assume that $k=1$ or $k=2$. In the first case, $y=1$ and $x$ is arbitrary, which we can regard as trivial solutions. In the second case, $x=y^2-1$ is a prime power, whence it is easy to see that either $x=8$ and $y=3$, or $x=3$ and $y=2$. To summarize, your equation only has the four solutions listed in your post, besides the trivial solutions ($k=1$ and $y=1$).<|endoftext|> TITLE: Comparison of finite field extensions of $\mathbb{C}(t)$ QUESTION [6 upvotes]: Let K be a finite field extension of $\mathbb{C}(t)$. Then $K$ is isomorphic to the field of meromorphic functions on a compact Riemann surface $X$ with genius $g$. By an argument similar to the proof of Douady's theorem for $\mathbb{P}^1(\mathbb{C})$ ( cf. chapter3 of Szamuly's Galois groups and fundamental groups) one can show that $$\mathrm{Gal}(\bar{K}/K) \cong \widehat{\mathrm{F}}(X\backslash {x_0} \cup \{ \gamma_1,\dots,\gamma_{2g}\}). $$ ($\widehat{\mathrm{F}}(S)$ is the profinite completion of the free group on generators from a set $S$, $x_0$ arbitrary point in $X$ and $\gamma_i$s the standard generators of $\pi_1(X,x_0)$.) It's obvious that the topology and group structure of $\widehat{\mathrm{F}}(S)$ depends only on the cardinality of $S$ and the cardinality of compact Riemann surfaces are the same. So we can conclude that all function fields over $\mathbb{C}$ have isomorphic absolute Galois groups. But are these fields really isomorphic? How one can prove or disprove this? ( Note that we are interested only in the field structure not in the structure over $\mathbb{C}$. ) REPLY [5 votes]: For any such $K$, we can recover the subfield of constants $\mathbb{C}\subset K$ as the elements of $K$ that have $n$th roots for all $n$. Indeed, if a rational function on a curve has roots of all orders, it must have valuation $0$ at every point and hence be constant. Thus any isomorphism between two such fields $K$ and $K'$ must fix $\mathbb{C}$ setwise, and so can be described as an automorphism of $\mathbb{C}$ followed by an isomorphism of curves. That is, two such fields are isomorphic iff the corresponding curves are conjugate under some automorphism of $\mathbb{C}$. In particular, for example, this means that the genus is invariant under all such isomorphisms (as any of the usual algebro-geometric definitions of genus are preserved by automorphisms of the base field).<|endoftext|> TITLE: Bounds on the moments of the binomial distribution QUESTION [10 upvotes]: I'm looking for simple and reasonably tight bounds on the k-th moment of the Binomial distribution $B(n,p)$, namely, $E[B(n,p)^k]$. I'm interested in the case when k is large (say on the order of $\sqrt{n}$) and in exact bounds (not asymptotic). Using a relatively simple calculation based on the concentration of the binomial sum I got: $$E[B(n,p)^k] \leq (pn)^k + k \ln n \cdot k^k$$ but I suspect that better bounds are known. There is an extensive literature with bounds on moments of sums of independent r.v.'s but I could not find something suitable. For example, as far as I understand, in this case Latala's paper gives a bound of: $\left(O\left(\frac{k}{\ln k}\right)\right)^k \cdot max \{pn,(pn)^k\}$ which is both larger and asymptotic. Any references or simple alternative bounds would be appreciated. REPLY [6 votes]: For any $\beta>0$, $$\mathbb{E}B(n,p)^k\leq k!\beta^{-k}\mathbb{E}e^{\beta B(n,p)}= k!\beta^{-k}(1-p+pe^{\beta})^n.$$ Now you can plug various $\beta$, e.g. $\beta=\frac{k}{np}$ which yields $$\mathbb{E}B(n,p)^k\leq (np)^k k!k^{-k}\left((1-p)+pe^{\frac{k}{pn}}\right)^n.$$ I assume that you got your estimate by elaborating on this expression, although in the case $1\ll k\ll n$ the above is slightly better. It feels unlikely that you can get much better universal bounds. Indeed, you can write $$ \mathbb{E}B(n,p)^k=k!\int\frac{(1-p+pe^z)^n}{z^{k+1}}dz, $$ integral being over a contour around zero. When you transform it to pass through a saddle point, you are essentially solving the optimization problem on $\beta$ as above, hence the maximum of the integrand is of the same order. In the limit, Laplace's method gives you at best some polynomial corrections. Another room for small improvements would be a more careful choice of $\beta$; e. g. for $1\ll k\ll n$ you can find a second-order approximation to the extremum.<|endoftext|> TITLE: A parity counting problem for subsets over finite fields QUESTION [6 upvotes]: Let ${\mathbb F}_p$ be the prime field of $p$ elements and $b$ be an element in ${\mathbb F}_p$. For a subset $T\subseteq {\mathbb F}_p$, define $$Bias(T)=|N_e( {\mathbb F}_p,b)-N_o( {\mathbb F}_p,b)|,$$ where $$N_e( {\mathbb F}_p,b)=\# \{ D\subseteq {\mathbb F}_p | \sum_{x\in D}x=b, |D\cap T|\equiv 0 \bmod 2 \}$$ and $$N_o( {\mathbb F}_p,b)=\#\{ D\subseteq {\mathbb F}_p| \sum_{x\in D}x=b, |D\cap T|\equiv 1 \bmod 2 \}.$$ My question: is there any method to prove $$Bias(T)\leq 2^{(1/2+o(1))p}, \forall T \ne \emptyset$$ for $|T|\sim p/2$? Thank you very much. The exponential sum approach by Shparlinski can be used to show $$Bias(T)\leq 2^{0.8413p}.$$ When $|T|$ is very small, say for example, $|T|=o(p)$, or very large $(|T|=p-o(p))$, this conjecture can be solved by our sieving counting argument. Take the simplest example, $T={\mathbb F}_p$, and we count that $$N_e({\mathbb F}_p, b)=\sum_{k\ne 0, 2\mid k} {1\over p}{p\choose k}\pm 1=\frac { 2^{p-1}-1}p\pm 1,$$ $$N_o({\mathbb F}_p, b)=\sum_{k\ne p, 2\not\mid k} {1\over p}{p\choose k}\pm 1=\frac {2^{p-1}-1}p\pm 1,$$ and thus $$Bias({\mathbb F}_p) \leq 2.$$ Generally we may define $$Bias_S(T)=|N_e(S,b)-N_o(S,b)|$$ similarly for $S\subseteq {\mathbb F}_p$ and propose the same conjecture $Bias_S(T)=2^{(1/2+o(1)|S|}, \forall T\ne \emptyset$. REPLY [6 votes]: Let's identify the elements of $\mathbb F_p$ with $\lbrace 0,1,2,\dots ,p-1\rbrace$. After fixing $T\subset \mathbb F_p$ and $b$, the Bias can be read off the following generating function $$f_{T,b}(x)=x^{p-b}\prod_{i=0}^{p-1} \left(1+(-1)^{\chi(i)}x^i\right),$$ where $\chi(i)=1$ if $i\in T$, and $\chi(i)=0$ otherwise. In fact $\operatorname{Bias}(T,b)=\frac{1}{p}\left|\sum_{i=0}^{p-1} f_{T,b}(\omega^i)\right|$, where $\omega$ is a primitive $p$th root of unity. Notice for example that if $0\in T$, then the Bias evaluates to zero (adding/removing zero gives a bijection between even/odd sets). To disprove your conjecture we can look at the set $T=\lbrace t | \frac{p}{4} \leq t \leq \frac{3p}{4}\rbrace$, and set $b=0$. Using that $\prod_{i=0}^{p-1}(1+\omega^i)=2$, we can say $$\operatorname{Bias}(T,0)=\frac{2}{p}|\sum_{i=0}^{p-1}\prod_{t\in T} \frac{(1-\omega^{it})}{(1+\omega^{it})}|.$$ From here you can check that: (1) every term in the sum is nonnegative, so we may ignore the absolute value (2) the term for $i=1$ is $\left(\prod_{t\in T} \frac{1-\cos(2\pi t/p)}{1+\cos(2\pi t/p)}\right)^{1/2}>(1+\sqrt{2})^{p/8}=2^{O(p)}$.<|endoftext|> TITLE: Is there a finitely presented group with infinite homology over $\mathbb{Q}$? QUESTION [8 upvotes]: Suppose $G$ is a discrete group given by finitely many generators with finitely many relations. Can the homology groups $H_i(G, \mathbb{Q})$, or equivalently $H_i(BG, \mathbb{Q})$ (topological homology of the classifying space) be infinite-dimensional? Can they be nonzero for infinitely many $i$? For any finitely presented groups I've seen, the answer is a surprising "no" (all finitely presented groups I know act on a finite-dimensional contractible space with finite stabilizers, and it follows that above the dimension of this space, homology vanishes). But it really should be the case that a "general" finitely-presented group has infinite homology... does anyone know of an example? REPLY [6 votes]: This answers the other part of your question, not answered by Thompson's group. For each $i\geq 3$ there is a finitely presented group $G_i$ with the property that $H_i(G_i\mathbb{Q})$ is infinite dimensional. The first such examples were due to Stallings ($i=3$ and Bieri $i>3$). Take a direct product of $i$ copies of the free group $F$ on two generators, and define $G_i$ to be the kernel of the homomorphism $F^i\rightarrow \mathbb{Z}$ that sends each of the $2i$ standard generators to $1\in \mathbb{Z}$. These are the Bieri-Stallings groups. For $i=2$ the analogous group is finitely generated but not finitely presented, and for $i=1$ it is free of infinite rank. These groups can be viewed as special cases of the Bestvina-Brady construction.<|endoftext|> TITLE: Are countable FC-groups maximally almost periodic? QUESTION [7 upvotes]: An FC-group is a group in which every element has a finite conjugacy class. A group G is said to be maximally almost periodic if there is an injective homomorphism from G into a compact Hausdorff group. My question is whether every discrete countable FC-group G is maximally almost periodic. In general an FC-group G is isomorphic to a subgroup of the direct product of a locally finite FC-group and a torsionfree abelian group. Since discrete abelian groups are maximally almost periodic, this reduces the general question to the case where G is additionally assumed to be locally finite. REPLY [7 votes]: Here's a counterexample. Consider a non-abelian finite, 2-step nilpotent group $F$. Let $Z$ be its derived subgroup. Define $G$ as the quotient of the direct sum $F^{(\mathbf{N})}=\bigoplus F_n$ of copies of $F$ indexed by $\mathbf{N}$ by identification of all copies of $Z$. Thus, writing $H=F/Z$, the group $G$ lies in a central extension, with central kernel $Z$ and quotient the infinite direct sum $H^{(\mathbf{N})}$. I claim that $G$ is not maximally almost periodic, and more precisely that any homomorphism of $G$ into a compact Hausdorff group kills $Z$. 0) $G$ is an FC-group: this is because it's quotient of a direct sum of finite groups. 1) I claim that for every finite index subgroup $M$ of $G$ we have $Z\subset [M,M]$. Indeed, let $M_n$ be the image of $M$ in the $n$-th copy $H_n$ of $H$ in $H^{(\mathbf{N})}$. Then the image of $M$ in $H$ is contained in $\bigoplus M_n$. Since it has finite index, this implies that for all but finitely many $n$, we have $M_n=H_n$. Let $M'$ be the inverse image of $M$ in $F^{(\mathbf{N})}$ and $M'_n$ its projection on $F_n$. Then for all but finitely many $n$, we have $M'_nZ_n=F_n$. Since $F_n$ is nilpotent and $Z_n=[F_n,F_n]$, this implies that for those $n$, we have $M'_n=F_n$, and hence for those $n$, we have $Z_n\subset [M'_n,M'_n]$. By fixing such an $n$ and projecting into $G$, we deduce that $[M,M]$ contains $Z$. 2) To conclude, it's enough to show that every homomorphism $f$ from $G$ to a compact Lie group $L$, with dense image, kills $Z$. Fix $f$. Since $G$ is nilpotent, $L^0$ being a compact nilpotent connected Lie group, it is abelian. Define $M=f^{-1}(L^0)$. Then $M$ has finite index, hence $Z\subset [M,M]$. Since $f$ maps $M$ into the abelian group $L^0$, we deduce that $[M,M]\subset ker(f)$, hence $Z\subset ker(f)$. Besides, as I mentioned, the result is true for FC-groups $G$ with trivial center: indeed these groups act faithfully on themselves by conjugation, and since the orbits are finite, the closure of the image of $G\to Sym(G)$ is compact. This is more generally true for residually finite FC-groups. However since as you mention, it is also true for abelian groups (which may fail to be residually finite), it's unclear what it the greatest generality.<|endoftext|> TITLE: Do runs of every length occur in this sequence? QUESTION [40 upvotes]: This is a repost from user r.e.s's unsolved Math Stack Exchange question: Do runs of every length occur in this string? That question was derived from my original question on the subject: Does this sequence have any mathematical significance? There is also a related Programming Puzzles & Code Golf question: Where are the runs in this infinite string? (CCCCCC Found!) I am posting it here in hopes that some of you will be able to shed more light on the problem or even solve it. I apologize if does not fall into the "research question" category, but I imagine many of you may find it quite intriguing. The Problem Starting with the sequence $\text{001}$, consider the infinite sequence $s$ generated by repeatedly appending the last half of the current sequence to itself, using the larger half if the length is odd: $$\begin{align} \quad &\text{001}\\ &\text{00101}\\ &\text{00101101}\\ &\text{001011011101}\\ &\cdots\\ &\text{______________________________}\\ s = \ &\text{0010110111010111011010111011110...} \end{align} $$ Does every sized run of ones occur in $s$? The first few runs are easy to find but then their indices grow astronomically as we've found in the programming contest: $$\begin{align} &\text{run} \quad & \text{first index}\\ &\text{1} &\text{2}\\ &\text{11} &\text{4}\\ &\text{111} &\text{7}\\ &\text{1111} &\text{26}\\ &\text{11111} &\text{27308}\\ &\text{111111} &\approx 10^{519}\\ &\text{1111111} &? \ (\gt 10^{40501})\\ \end{align} $$ What can be said about the runs of ones in $s$? Why is the index growth so extraordinary? Can one prove or disprove that all runs of ones occur in $s$? Note that user r.e.s has done much more analysis on this problem in his/her original question. The formatting above is his/hers; I've only changed $\text{abc}$ to my original $\text{001}$ starting sequence. It is r.e.s. and the other users in the programming contest that helped generate the substring indices. I've had little to do with the analysis of this sequence except as discussion starter. REPLY [4 votes]: Some observations, (some were pointed out at other exchanges): The only way to get a run of length $n$, is to cut off the partial sequence in the middle, such that the cut-off part starts with a run of length $n-1$. A point of attack is therefore to just keep the index of the longest observed run, iterate, (which give indices to copies of this run) in the longer sequences. What has to be done is therefore so look for a form of relatively prime-ness of the indices of the copies, and the cut-off index. Conjectured lemma: Assume there is a run of $n$ ones starting at index $j$. Then, for any $k>0$ and $0\leq r TITLE: Surjective morphism from $X$ to itself is finite QUESTION [8 upvotes]: Let $X$ be a projective variety. Why is any surjective morphism from $X$ to itself finite? REPLY [10 votes]: Let $f$ be your surjective endomorphism; it is generically finite, say of degree $d$. Then $f_*:H^*(X,\mathbb{Q})\rightarrow H^*(X,\mathbb{Q})$ is surjective, because $\ f_*f^*=d.\mathrm{Id}\ $ (use $\mathbb{Q}_\ell$ instead of $\mathbb{Q}$ in characteristic $p$). Therefore $f_*$ is bijective. But then $f$ cannot contract any positive dimensional subvariety $Z$, because we would have $f_*[Z]=0$.<|endoftext|> TITLE: Partitioning $\omega_1$-branching trees of size and height $\omega_1$ QUESTION [7 upvotes]: Is it possible, in ZFC, to find an $\omega_1$-branching tree $(T,\leq)$ of size and height $\omega_1$ such that whenever $T$ is partitioned into countably many sets $T=\bigcup_{n<\omega} T_n$ one of the sets $T_n$ contains a subset $S$ which is again an $\omega_1$-branching tree of size and height $\omega_1$ (with the tree order on $S$ being just $\leq$ restricted to $S$). REPLY [12 votes]: The existence of such trees is independent of ZFC. On one hand, if $CH$ holds then $\omega_1^{<\omega_1}$ (and in fact - any $\sigma$-closed $\omega_1$-branching tree) cannot be partitioned into $\aleph_0$ many subsets, each does not contain an $\omega_1$-branching tree. Proof: Assume otherwise and let $T_n$ be such partition. Claim: For each $n<\omega$, each $t\in T$ can be extended to a node $s\in T$ such that $s$ has no extension in $T_n$. Proof: If this failed for some $n$ then $T_n$ would be $\omega_1$-branching of height $\omega_1$. $\blacksquare$ Using the claim we can construct, by recursion on $n$, a strictly increasing sequence of $s_n$'s such that $s_n$ has no extension in $T_n$. Then, by $\sigma$-closedness of $T$, all of them would have a single extension $s_\omega\in T$. However, $s_\omega$ would have no extension in any $T_n$ contradicting the fact that $T=\bigcup T_n$.$\square$ On the other hand, it is consistent (relative to the existence of an inaccessible) that $2^{\aleph_0} = \aleph_2$ and every tree of height and size $\omega_1$ can be partitioned into countably many subsets, each not containing an $\omega_1$-branching subtree. Definition: Let $T$ be a tree of height $\omega_1$, $|T|=\omega_1$. $T$ is special if there is a function $f:T\to \omega$ such that for every $x,y,z\in T$ if $f(x)=f(y)=f(z)$ and $x\leq y,z$ then $y\leq z$ or $z\leq y$. If $T$ is special, as witnessed by $f$, then $T_n = f^{-1}(n)$ is the desired partition: for every $x\in T_n$, $x$ doesn't have any two incompatible successors in $T_n$, since if $x\leq y,z$ and $y,z\in T_n$ then $f(y)=f(z)=f(x)=n$ and therefore $y$ and $z$ are comparable. Theorem (Todorcevic/Baumgartner): The statement "Every tree of size and height $\omega_1$ is special" is equiconsistent with the existence of an inaccessible cardinal. See "Some Combinatorial Properties of Trees" by Todorcevic, and "Aronszajn trees and the independence of the transfer property" by Mitchell for more details.<|endoftext|> TITLE: Examples of research on how people perceive mathematical objects QUESTION [18 upvotes]: What examples are there on research related to human perception and mathematical objects? For example, the shape of a beer glass influences drinking habits, since people are bad at integrating. There have been studies conducted, to determine what fractal dimension people prefer. There is also a lot of research done on how people choose numerical passwords, or what people consider as "random numbers" and random sequences of numbers, but I am not looking for examples on how people handles randomness, or deal with strategies in a game. What I am looking for is where math appears in relation to peoples' perception, such as in this example: there is a an interesting article on a mathematical model for "generating annoying, scratching sounds". The conclusion (conjecture) is that the boundary between order (sine wave) and chaos (white noise) is what make nails on a chalkboard so distinctively unpleasant. The researchers found that listening to the graph of the logistic map is highly unpleasant, while other more random, and more ordered functions did not have this property. The authors try to explain this experience by properties of the function they examine. Another example is the mathematical model of the Droste effect. This model (based on complex analysis) was used to complete one of M.C Escher's famous paintings. I believe this is related to Escher's map. Yet another example is the following 20-page paper, titled A mathematical theory of illusions and figural aftereffects, published in a Springer journal. Also, a while back, a physicist posted The proof of innocence on arXiv, giving a mathematical model on why he (erroneously) got a traffic ticket, due to an optical illusion explainable by mathematics. From the abstract: The paper was awarded a special prize of $400 that the author did not have to pay to the state of California. REPLY [2 votes]: The golden ratio is often cited in research on pleasing shapes to the human eye in art and architecture. For example, see http://plus.maths.org/content/os/issue22/features/golden/index<|endoftext|> TITLE: Finite extension of local fields QUESTION [6 upvotes]: Can a (higher) local field have uncountably many finite (seperable) extensions? REPLY [3 votes]: Here is a proof that $K=\mathbf{F}_q((t))$ admits only countably many extensions in each finite degree. Here $q$ is a power of a prime $p$. It is enough to prove this countability result for minimal extensions $K'\subset K$, i.e., those with no intermediate field between $K'$ and $K$. Indeed, suppose the result is proved for minimal extensions and let $d$ be minimal such that for some $q$, $K$ admits uncountably many non-isomorphic extensions $(K_i)$ of degree $d$. Then all but countably of those admit a minimal subextension $L_i$ of some degree $1 TITLE: Distribution of dropped objects QUESTION [7 upvotes]: Consider small perfectly elastic spheres being dropped from a fixed height in R^3, bouncing and coming to rest on the horizontal R^2. Assuming a reasonable distribution of minor perturbations of the initial velocity and minor perturbations of perfect elastic scattering when bouncing can anything be said about the distribution of final resting places ? Make further simplifying assumptions without making it trivial. REPLY [7 votes]: What you're asking about seems fairly close to that of what's called a ``freely-cooling granular gas'' which is of great interest in granular physics, geophysics, and even the study of the large scale structure of the universe. I haven't found any literature on the specific situation you described of spheres falling under gravity, but since it seems you're interested in what happens when a lot of inter-sphere collisions occur, I think the physical phenomena described below should be quite relevant. Accordingly, in this answer I'll give a brief summary of some of the physics literature on this topic. Freely-cooling granular gases are systems of hard sphere particles that interact via inelastic collisions, initially at a high "temperature" (particles with random velocities with some large average kinetic energy) and are studied as they dissipate energy and gradually come to a rest. At short times the behavior is described by "Haff's law" where the particles remain homogeneously distributed and energy decreases in a power law fashion as $t^{-2}$. At longer times, the system enters a coarsening regime where it becomes inhomogeneous (clustering of particles) and the exponent of the energy decay changes to $t^{-\theta}$ where $\theta$ is dimension dependent. A lot of attention has been given to dimension 1, where $\theta$ is known to be 2/3. Ben-Naim, Chen, Doolen and Redner proposed that one-dimensional systems approach at long times the "sticky limit" (where the system behaves as if the coefficient of restitution is 0). The situation in 2 and 3 dimensions is still not clear although there have been many simulations and proposed theories. Here is a recent paper by Pathak, Jabeen, Das and Rajesh with at least this nice image illustrating a simulation of the gas on the left, and another model called "ballistic aggregation" on the right. Most of the literature regarding cluster sizes and such tends to be about the dynamics, i.e. how the distribution of clusters grows with time, etc. In fact, it is not obvious to me that the notion of "final resting places" is well-defined in these systems. Inelastic collapse is an interesting phenomenon in systems of inelastically colliding particles, which is a singularity in the dynamics due to an infinite number of collisions occurring in finite time. The existence of this also means that fining "reasonable distributions of minor perturbations" can be a very subtle affair. Here's a paper by McNamara and Young describing simulations with this phenomenon in two and three dimensional systems. They do have some discussion of how one might continue the dynamics through such singularities, but I haven't read it very carefully yet. Note that the other papers above implement their simulations in a way so as to avoid these singularities altogether. This is done by replacing inelastic collisions with elastic ones whenever the relative velocity between two particles is sufficiently small (see the paper of Ben-Naim et al).<|endoftext|> TITLE: "The" kronecker foliation or "a" kronecker foliation? QUESTION [11 upvotes]: Consider the following two foliations of torus: 1)The Kronecker foliation with slope $\sqrt{2}$ 2)The Kronecker foliation with slope $\pi$ As I learn from the literature, these two foliations are not topological equivalent. The proof is that the K theory of their corresponding $C^{*}$ algebras are not isomorphic.In fact two Kronecker foliations with slopes $\alpha$ and $\beta$ are not equivalent if $\alpha$ and $\beta$ are not on the same orbit of action of $Sl_{2}(\mathbb{Z})$. See Non Commutative Geometry by Alain Connes. But intuitively it is difficult to imagine that these two foliations are different. because in both foliations all leaves are dense! one can not distinguish these two foliations via visible topological behavior. Is there an intuitive and geometric proof for this fact(without using K theory, $C^{*}$ algebras of foliation,etc.)? I think, if there is no an intuitive proof, this shows the deep power of the role of $C^{*}$ algebra of foliations, at least in this example. REPLY [7 votes]: Just what the doctor ordered, a proof using diffeology instead of $\mathrm C^*$-algebras: MR0799609 Donato, Paul(F-CNRS-T); Iglésias, Patrick(F-CNRS-T) Exemples de groupes difféologiques: flots irrationnels sur le tore. (French. English summary) [Examples of diffeological groups: irrational flows on the torus] C. R. Acad. Sci. Paris Sér. I Math. 301 (1985), no. 4, 127–130. Author summary (translated from the French): "We illustrate J.-M. Souriau's technique of `diffeological spaces and groups' in the case of the quotient $T_α$ of the standard torus by the irrational flow with slope $α$. Computing the universal covering $\mathbf R$ and the fundamental group $\mathbf Z^2$ of $T_α$ allows us to classify these tori diffeologically: $T_α$ and $T_β$ are diffeomorphic if and only if $α∼β$ modulo $\mathrm{GL}(2,\mathbf Z)$; moreover, the computation of $\mathrm{Diff}(T_α)$ reveals a difference between quadratic irrationals and other irrationals.''<|endoftext|> TITLE: What exact number of domino tilings cannot be realizable? QUESTION [12 upvotes]: Inspired by some other questions, (this and this), I wonder what numbers $n$ there are that satisfy $$p(n)=\text{there is no region that admits exactly } n \text{ domino tilings}.$$ If this is true, $n$ is non-realizable, otherwise, it is realizable. A region in the plane is a union of squares in the unit square grid, and a domino is of course two adjacent unit squares. Clearly, if $m$ and $n$ are realizable, then so is $mn$. Note that all powers of $2$ are easily realizable. $2 \times k$-regions have a Fibonacci-number of tilings, so all Fibonacci-numbers are realizable. Note that there is no restriction on the subset of the plane (it can be disconnected, or have holes). Will the answer be different if the region is simply connected? In the latter case, there is a theorem stating that we can reach each tiling from any other using "flips". Update: Let $F_k$ be the number of ways to tile a $2 \times k$-rectangle. This is a Fibonacci number, $F_1=1,F_2=2,F_3=3,\dots$. Consider the Young diagram given by the partition $(k,k,2)$. We can either choose to have a horizontal domino in the third row, which give $F_k$ tilings of the remaining, or the third row is covered by two vertical dominos. The remaining part can then be tiled in exactly $F_{k-2}$ ways. Hence, all numbers of the form $F_k + F_{k-2}$ are realizable. In particular, $7 = 2+5$, and $11 = 3+8$. We can do a more general construct on the "other end" of a long $2\times k$-shape, and see that all $F_k + F_{k-2}+F_{k-4}$ are realizable whenever $k\geq 4$. Thus, we are now incredibly close to applying Zeckendorf's theorem. REPLY [17 votes]: To follow up on the answer of dhy, there is a simple construction that works with simply connected regions: Take a $k\times k$ square with $k\geq 2$. Remove from the bottom left corner a staircase region with rows of $1,2,3,\ldots k-2$ cells, and from the top right corner a (suitably rotated) staircase region with rows of $1,2,3,\ldots k-3$ cells. The remaining region is some sort of fat diagonal of the square having rows with $3,4,4,\ldots,4,3,2$ cells from top to bottom. It is easily seen that this region has $2k-2$ domino tilings, while the same region with the $2$ cells of the last row removed has $2k-3$ possible tilings.<|endoftext|> TITLE: Can there be a power basis for a totally real field of high degree? QUESTION [10 upvotes]: A number field $K$ is said to have a power basis if there is an $\alpha \in K$ such that the full ring of integers $O_K$ is the $\mathbb{Z}$-linear span of $1,\alpha,\alpha^2,\ldots,\alpha^{\deg{K}-1}$. In other words, $O_K = \mathbb{Z}[\alpha]$; another term for this is monogenic. This happens for example for the quadratic and the cyclotomic fields. The monogenic number fields are presumably very rare; results of Bhargava imply that they are a negligible fraction (zero density) among the number fields of degree $3, 4$, or $5$, and this is conjectured to hold in every fixed degree $d > 2$. It is easily seen that totally $p$-adic number fields of degree $d$ are never monogenic for $d \gg p$. (To illustrate this, consider that the split primes in the cyclotomic field of level $N$ are the ones $\equiv 1 \mod{N}$; so they are in particular $> N > \phi(N)$.) Question. Are there monogenic totally real number fields of arbitrarily high degree? To put it differently: for totally real algebraic integers of arbitrarily high degree, may the ring $\mathbb{Z}[\alpha]$ be integrally closed? Dummit and Kisilevsky [Indices in cyclic cubic fields, in "Number Theory and Algebra," 1977] have shown that infinitely many totally real cubic fields have a power basis. There exist totally real monogenic sextic fields; an example, taken at will from page 116 of [I. Gaal, Diophantine Equations and Power Integral Bases: New Computational Methods] is the field generated by a root of $$ X^6 - 5X^5 +2X^4 +18X^3 - 11X^2 -19X+1. $$ Are there examples of higher degree? REPLY [7 votes]: Kedlaya proves that there are infinitely many irreducible integer polynomials of every degree with real roots and square free discriminant; if $P$ has square free discriminant then $\mathbb{Z}[x]/P(x)$ is integrally closed. More precisely, he constructs $c N^{1/(n-1)}$ different fields of this sort with discriminant in $[-N,N]$. According to Ash-Brakenhoff-Zarrabi, Hendrik Lenstra made the following conjecture in private communication: Let $n \geq 2$. If a degree $n$ monic integer polynomial $P$ is chosen at random, then $\mathbb{Z}[x]/P(x)$ is integrally closed with probability $6/\pi^2$. So, in some sense, monogenic fields are not that rare. We talked about these questions before here and here.<|endoftext|> TITLE: A topologist is not a mathematician - a small question QUESTION [45 upvotes]: Years ago I read about a topologist who was to enter the states as an immigrant and was asked a question about his profession. He indicated he was a topologist, but as this was not included on the officer's list, he wanted to check him in as a "mathematician", which was on his list. But, the topologist refused being classified as an ordinary mathematician, and insisted on "topologist". The argument escalated and finally they put the topologist in a psychiatric institute, out of which he was freed by a colleague mathematician who explained the situation to the police. The question is now: what's the name of this topologist; I can't remember his name anymore. Some years ago (must have been between september 2000 and october 2004) I read his biography, I believe in the Notices of the AMS, but it may have been another publication by AMS, MAA or EMS as well. Some details in the story might be slightly wrong due to failing memories. Can someone please provide me his name, and maybe the reference to the biography as well? That would be very friendly! Thanks in advance for this! REPLY [53 votes]: It seems to me you are referring to Egbert Rudolf van Kampen (but the problem at the immigration office was quickly solved by a phone call to the university, the Johns Hopkins I believe). The story is told by Mark Kac, in his book Enigmas of Chance: An Autobiography. rmk The point of Mark Kac's enjoyable anecdote is not that v.K. refused the label "mathematician", but that he was so immersed in his topic, and considered his activity so natural, that naively believed that even an immigration officer should have known what it was. According to Mark Kac's story, at the officer's request "what's a topologist", v.K. would have started explaining fundamental groups and exact sequences (possibly a humorous exaggeration).<|endoftext|> TITLE: Inaccessible becomes successor of singular QUESTION [8 upvotes]: Is it possible, starting from any large cardinal assumption, to find a countably closed forcing $\mathbb{P}$ such that for some inaccessible $\kappa$, $\Vdash_\mathbb{P} "\kappa = \lambda^+$ and $\lambda$ is singular"? REPLY [11 votes]: No. The following theorem is from a work in progress by Yair Hayut and myself. Theorem. If $\Bbb P$ is a proper forcing, and it changes the cofinality of $\kappa$ to $\mu>\omega$, then $\Bbb P$ adds a surjection from $\mu$ onto $\kappa$. Now suppose that you had such countably closed $\Bbb P$, it is certainly proper. And it changes the cofinality of $(\lambda^+)^V$ to be something which is smaller than $\lambda$, and therefore collapses $(\lambda^+)^V$, and as a consequence it must collapse $\lambda$ as well. (It should be remarked that a countably closed forcing cannot change the cofinality of something $\omega$ anyway.)<|endoftext|> TITLE: What's the state of affairs concerning the identification between quantum group reps at root of unity, and positive energy affine Lie algebra reps? QUESTION [28 upvotes]: In his paper [1], Finkelberg used Kazhdan-Lusztig's massive work [4,5,6,7,8] to prove that $Rep^{ss}(U_q\mathfrak g)$ (the semisimplification of the category of finite dimensional reps of $U_q\mathfrak g$) is equivalent to the category $Rep_k(\widetilde{L\mathfrak g})$ of level $k$ integrable highest weight modules over the affine Lie algebra $\widetilde{L\mathfrak g}$. But then, I recently learned (from Section 3 of [3]) that there was an erratum [2] where an error was discovered and corrected, and that there are cases (namely $E_6$, $E_7$, $E_8$ level 1, and $E_8$ level 2) where the Kazhdan-Lusztig story [4,5,6,7,8] cannot be applied... Question 1: Is the equivalence $Rep^{ss}(U_q\mathfrak g)\cong Rep_k(\widetilde{L\mathfrak g})$ known for all $\mathfrak g$ and $k$, or are there exceptions? Question 2: Is the equivalence $Rep^{ss}(U_q\mathfrak g)\cong Rep_k(\widetilde{L\mathfrak g})$ known just at the level of fusion categories, or have the braidings also been compared? How about the ribbon structures? References: [1] M. Finkelberg, An equivalence of fusion categories, Geom. Funct. Anal. 6 (1996), 249–267. [2] M. Finkelberg, Erratum to: An equivalence of fusion categories, Geom. Funct. Anal. 6 (1996), 249–267; Geom. Funct. Anal. 23 (2013), 810–811. [3] Y.-Z. Huang and J. Lepowsky, Tensor categories and the mathematics of rational and logarithmic conformal field theory, ArXiv:1304.7556 [4] D. Kazhdan and G. Lusztig, Affine Lie algebras and quantum groups, Duke Math. J., IMRN 2 (1991), 21–29. [5] D. Kazhdan and G. Lusztig, Tensor structures arising from affine Lie algebras, I, J. Amer. Math. Soc. 6 (1993), 905–947. [6] D. Kazhdan and G. Lusztig, Tensor structures arising from affine Lie algebras, II, J. Amer. Math. Soc. 6 (1993), 949–1011. [7] D. Kazhdan and G. Lusztig, Tensor structures arising from affine Lie algebras, III, J. Amer. Math. Soc. 7 (1994), 335–381. 25 [8] D. Kazhdan and G. Lusztig, Tensor structures arising from affine Lie algebras, IV, J. Amer. Math. Soc. 7 (1994), 383–453. REPLY [7 votes]: The following answer is from Yi-Zhi Huang. I post it here with his permission. Finkelberg's approach cannot be used to give a tensor functor in the exceptional case ($\mathfrak g=E_6$, $k=1$, $\mathfrak g=E_7$, $k=1$ and $\mathfrak g=E_8$, $k=1$ or $2$). Finkelberg's approach can be sketched as follows: Kazhdan-Lusztig constructed an equivalence between a quantum group module category at a root of unity and an affine Lie algebra module category of a negative integral level. On the quantum group side, one can take a subquotient to obtain a semisimple rigid braided tensor category (in fact a modular tensor category). The same procedure certainly works on the affine Lie algebra side. So one also has a semisimple subquotient of the affine Lie algebra module category of the negative integral level. The main idea of Finkelberg is to use the contragredient functor for affine Lie algebra modules. This functor sends modules of negative levels to those of positive levels (since the functor is the Lie algebra contragredient functor, not the vertex algebra contragredient functor). The crucial step is to prove that this functor is a tensor functor. This is the place where Finkelberg had a gap. One needs the Verlinde formula to fill the gap. From this description, you can see that Finkelberg's work has nothing to do with quantum groups. It is purely a result on the affine Lie algebra side. It is the work of Kazhdan-Lusztig that gives the connection with the quantum group. Now since Kazhdan-Lusztig's work does not cover the exceptional cases, Finkelberg's approach fails completely. To construct a tensor functor, one has to do this directly. The abelian categories are obviously equivalent and the fusion rules are known to be the same. But these are far from a construction of a tensor category equivalence. One possibility is to see whether Kazhdan-Lusztig's method can be adapted. But in this case one has to work directly with the semisimple subquotient of the quantum group category, since on the affine Lie algebra side the category is not a subquotient of a nonsemisimple rigid braided tensor category. It does not seem to be easy to directly adapt the method of Kazhdan-Lusztig since the representation theory of affine Lie algebras at positive integral levels are very different from the representation theory of affine Lie algebras at negative integral levels. I think that if there is a construction in the exceptional cases (mainly the case $E_8$, level $2$), it should also work in the general positive level case. Finkelberg's construction is natural as a construction of an equivalence between a positive integral level category and a negative integral level category. But it is not natural as a construction of an equivalence between a quantum group category and a positive integral level category. The failure in the exceptional cases is an indication. There should be a direct and natural construction that works in all the cases and provides a true understanding of the connections between the modular tensor categories from quantum groups and from affine Lie algebras.<|endoftext|> TITLE: Is there a simple proof that there is no five mutually orthogonal Latin squares of order 6? QUESTION [7 upvotes]: It is well known that there is a projective plane of order $n$ if and only if there exist a set of $n-1$ mutually orthogonal Latin squares. The first nontrivial case is $n=6$, which fails because of Bruck-Ryser theorem. The history of the problem mentions Thomas Clausen and Gaston Tarry who proved that there are no two mutually orthogonal Latin squares of order $6$. Their proof consists of a lot separate cases and Stinson gives some short proof of that in 1984, but it also contains some cases discussion. Is there a simple proof (unlike long cases discussion) that there is no five (instead of two) mutually orthogonal Latin squares of order $6$? REPLY [4 votes]: I agree with the OP that Stinson's paper, while short, doesn't give a clean and conceptual proof of the non-existence of a pair of 2 MOLS of order $6$. If the OP is happy with another proof of the non-existence of planes of order $6$, there is an alternative to Bruck-Ryser suggested by Assmus: One can show that there is no plane of order $n$ for $n\equiv6\pmod{8}$. (That's a special case of Bruck-Ryser.) The proof is contained in the second edition of Lineare Algebra by Huppert/Willems. They, however, rely on Gleason's Theorem about the weight enumerator of binary doubly-even selfdual codes. A slightly different treatment avoiding this theorem can be found in this script on coding theory.<|endoftext|> TITLE: The category of categories and adjunctions QUESTION [11 upvotes]: What is known about the category that has small categories as objects and adjunctions as morphisms? Obviously, it has neither terminal nor initial objects. But what about other kinds of limits? Are there interesting facts about this category? I am particularly interested in the special case of the category of posets and Galois connections. REPLY [5 votes]: You will have to make an arbitrary choice for the direction of morphisms: is the left adjoint "forward" or "backward"? To prevent that, you can add involutions. The resulting category $\mathbf{InvAdj}$ of involutive categories and adjunctions, that I'll define below, has a lot of interesting structure. It is a dagger category, and in fact the `mother of all dagger categories', as it universally embeds any dagger category. In particular, the full subcategory of (ortho)posets and Galois connections has dagger kernels, dagger biproducts, an an opclassifier. See these two papers. Now for the definition (from 3.1.8 of my thesis): A functor $* \colon \mathbf{C}^\mathrm{op} \to \mathbf{C}$ is called involutive when $* \circ * = \mathrm{Id}$. Define a category $\mathbf{InvAdj}$ as follows. Objects are pairs $(\mathbf{C},*)$ of a category with an involution. A morphism $(\mathbf{C},*) \to (\mathbf{D},*)$ is functor $F\colon \mathbf{C}^\mathrm{op} \to \mathbf{D}$ that has a left adjoint, where two such functors are identified when they are naturally isomorphic. The identity morphism on $(\mathbf{C},*)$ is the functor $*\colon \mathbf{C}^\mathrm{op} \to \mathbf{C}$; its left adjoint is $*^\mathrm{op} \colon \mathbf{C} \to \mathbf{C}^\mathrm{op}$. The composition of $F\colon \mathbf{C}^\mathrm{op} \to \mathbf{D}$ and $G\colon \mathbf{D}^\mathrm{op} \to \mathbf{E}$ is defined by $G \circ *^\mathrm{op} \circ F \colon \mathbf{C}^\mathrm{op} \to \mathbf{E}$; its left adjoint is $F' \circ * \circ G'$, where $F' \dashv F$ and $G' \dashv G$. (It is not arbitrary to require a left adjoint instead of a right one. A contravariant functor from $\mathbf{C}$ to $\mathbf{D}$ can be written both as a (covariant) functor $F\colon \mathbf{C}^\mathrm{op} \to \mathbf{D}$ or as a (covariant) functor $F^\mathrm{op}\colon \mathbf{C} \to \mathbf{D}^\mathrm{op}$. The latter version has a right adjoint precisely when the former version has a left adjoint.)<|endoftext|> TITLE: A general theory of quasi-functors, generalizing from dg-categories to $\mathcal V$-categories, with $\mathcal V$ monoidal model category QUESTION [7 upvotes]: I employ the vast majority of the post to develop the notion of quasi-functor between dg-categories: I think it is important to get the idea. Let $k$ be a field, and let $\mathcal V =\mathbf C(k)$ be the category of cochain complexes of $k$-vector spaces and chain maps. $\mathcal V$ is a symmetric closed monoidal category, and also has a structure of model category (the weak equivalences are the quasi-isomorphisms, the fibrations are the surjective chain maps). Categories enriched over $\mathcal V$ are called dg-categories. $\mathcal V$ is also enriched over itself. Let $\mathcal A, \mathcal B$ be categories enriched over $\mathcal V$. The category $\mathrm{Fun}_{\mathcal V} (\mathcal A, \mathcal B)$ of $\mathcal V$-functors and $\mathcal V$-natural transformations is also enriched over $\mathcal V$. The (enriched) Yoneda lemma holds, and we have the (fully faithful) Yoneda embedding: \begin{equation} \mathcal A \hookrightarrow \textrm{mod-}\mathcal A =: \mathrm{Fun}_{\mathcal V}(\mathcal A^{\mathrm{op}},\mathcal V). \end{equation} The category $\textrm{mod-}\mathcal A$ of (right) $\mathcal A$-modules inherits a (levelwise) model structure from $\mathcal V$: for example, quasi-equivalences are given by levelwise quasi-equivalences. Now, let me give the following definition: given $\mathcal A, \mathcal B$ two $\mathcal V$-categories, a $\mathcal V$-functor $F: \mathcal A \to \textrm{mod-}\mathcal B$ is called a quasi-functor if for any $A \in \mathcal A$, $F(A)$ is isomorphic to a representable right $\mathcal B$-module in the homotopy category $\mathrm{Ho}(\text{mod-}\mathcal B)$. Now, observe that we can define a "tensor product" $\mathcal A \otimes \mathcal B$ of $\mathcal V$-categories. Moreover, $\mathcal V$-functors $\mathcal A \otimes \mathcal B^{\mathrm{op}} \to \mathcal V$ correspond exactly to $\mathcal V$-functors $\mathcal A \to \textrm{mod-}\mathcal B$. They are precisely the bimodules (or profunctors). Let me denote by $\mathcal A\text{-mod-}\mathcal B$ the $\mathcal V$-category of bimodules (covariant in $\mathcal A$, contravariant in $\mathcal B$). Since it is a category of modules, it is a model category, with the levelwise structure discussed above. Now, I can define $\mathrm{rep}(\mathcal A,\mathcal B)$ as the full subcategory of $\mathrm{Ho}(\mathcal A\text{-mod-}\mathcal B)$ whose objects are the quasi-functors. Moreover, I can define $\mathrm{rep}_{\mathcal V}(\mathcal A,\mathcal B)$ as the full $\mathcal V$-subcategory of $\mathcal A\text{-mod-}\mathcal B$ whose objects are the quasi-functors which are also cofibrant as bimodules. Where can we go from here? Well, the category $\mathcal V$-$\mathbf{Cat}$ of categories enriched over $\mathcal V$ has itself a model structure: it is a known result by G. Tabuada about dg-categories, more recently generalized. The homotopy category $\mathrm{Ho}(\mathcal V\text{-}\mathbf{Cat})$ is monoidal, and $\mathrm{rep}_{\mathcal V}(\mathcal A,\mathcal B)$ gives the internal hom. In the world of dg-categories, $\mathrm{rep}(\mathcal A,\mathcal B)$ is (equivalent to) $H^0(\mathrm{rep}_{\mathcal V}(\mathcal A,\mathcal B))$. Finally, here is my question: generalize from this particular case $\mathcal V = \mathbf C(k)$ to a more general setting, possibly letting $\mathcal V$ be a monoidal model category (with some assumptions). Every definition given above should work without problems. Has someone developed a "general theory" of those quasi-functors? I've studied it in the case of dg-categories, but I guess it is just a particular case. REPLY [2 votes]: In some sense, the "universal" version of this fact was proved by Blumberg-Gepner-Tabuada as Proposition 3.3 in this paper. That is, they proved the analogue for stable $\infty$-categories, which is the description of the $\infty$-category $Fun^{ex}(A, B)$, the internal hom in the $\infty$-category of idempotent complete stable $\infty$-categories and exact functors, via what they call "right-compact" $A$-$B$-bimodules, which are the analogue of right quasi-representable functors or quasi-functors. In fact, the proof is much nicer in this setting. I claim that any "reasonable" version of Toën's theorem should follow from this version. By that I mean for any $\mathcal{V}$ such that $\mathcal{V}$-categories model some version of stable $\infty$-categories (so that my claim is basically tautological). For example, for spectral categories ($\mathcal{V}$-categories where $\mathcal{V}$ is the symmetric monoidal model category of symmetric spectra), this follows from the equivalence in Theorem 1.10 between idempotent complete stable $\infty$-categories, and spectral categories up to Morita equivalence. For dg-categories, it should follow similarly by using the equivalence between dg-categories and linear stable $\infty$-categories. (Note that the internal hom in the $\infty$-category of dg-categories is the same as the derived internal hom in the model category of dg-categories.)<|endoftext|> TITLE: Gaps between roots of trigonometric polynomials QUESTION [8 upvotes]: [Cross-posted from Math.SE because I got no responses there.] Given a polynomial in $e^{\mathrm{i}k t}$ of the form $$ p(t) = \sum_{-n\leq k\leq n} c_k e^{\mathrm{i}k t} $$ with $\bar c_{-k} = c_k$, is there a good way of characterising how close its roots can be in terms of its coefficients? Clearly this depends on coefficients, for example $1-\epsilon-\cos x$ has roots $\sqrt{\epsilon}$ apart, but for general coefficients, is there any good result? I am secretly hoping for minimum distance of something like $A n^{-1}$ (which is the case for $\sin n x$) for non-almost-degenerate coefficients. If you know a result for any kind of orthogonal polynomials (like Chebyshev polynomials), that would also be very helpful. Also, I know there are some results for asymptotic distributions of eigenvalues of Toeplitz matrices, but I am interested in a result for a fixed polynomial $p$ and its coefficients, and not just $n\to\infty$. Edit. I am mainly interested in an algorithm that, given a polynomial as above, computes the shortest gap between two adjacent roots of the polynomial. The motivation is as follows. The simplest method for calculating a root of a function numerically is the bisection method, which only works so long as you know an interval that brackets a root (an interval on which the function changes sign). In general, calculating all roots of a black-box function is really hard. So without going to more involved algorithms, the simplest algorithm to find all (simple) roots must be this one: find the shortest distance $\delta$ between adjacent roots, sample the function in intervals of length $\delta$, find the root in each interval with a sign change in it. Of course, this doesn't work without knowing $\delta$. This is not a brilliant algorithm, there are better ones out there, but I am interested in finding out how I could make it work. So the question is: given a trigonometric polynomial with only simple roots, how can I calculate the shortest distance between adjacent roots, so that the above scheme would succeed? Example. If $t_{1,2}$ are two roots, I can write $$ 2\max_{[t_1,t_2]}|f(t)| \leq \int_{t_1}^{t_2}|f'(t)|\,dt \leq |t_2-t_1|\max_{[t_1,t_2]}|f'(t)|. $$ So at least $$ |t_1-t_2| \geq 2\frac{\max_I |f|}{\max_I |f'|}. $$ Then by Bernstein's inequality, $$ \max_I |f'| \leq \|f'\|_\infty \leq n \|f\|_\infty, $$ so $$ |t_1-t_2| \geq \frac2m \frac{\max_I|f|}{\max_{[-\pi,\pi]}|f|}. $$ But then I don't know quite how to deal with the right-hand side. REPLY [5 votes]: The technical term for what you want to do is root isolation or root bracketing. One way to approach this to find the minimal distance between the roots, like you are suggesting, and also a large enough bounded interval to contain all the roots. This idea was in fact used early on in the history of root isolation for real polynomials. However, these techniques have gotten more sophisticated with time. I imagine the situation would be similar for trigonometric polynomials. Here's a reference that seems to discuss root isolation precisely for trigonometric polynomials: Real zero isolation for trigonometric polynomials, by Achim Schweikard ACM Transactions on Mathematical Software 18 350-359 (1992) http://dx.doi.org/10.1145/131766.131775<|endoftext|> TITLE: Examples of unexpected mathematical images QUESTION [265 upvotes]: I try to generate a lot of examples in my research to get a better feel for what I am doing. Sometimes, I generate a plot, or a figure, that really surprises me, and makes my research take an unexpected turn, or let me have a moment of enlightenment. For example, a hidden symmetry is revealed or a connection to another field becomes apparent. Question: Give an example of a picture from your research, description on how it was generated, and what insight it gave. I am especially interested in what techniques people use to make images, this is something that I find a bit lacking in most research articles. From answers to this question; hope to learn some "standard" tricks/transformations one can do on data, to reveal hidden structure. As an example, a couple of years ago, I studied asymptotics of (generalized) eigenvalues of non-square Toeplitz matrices. The following two pictures revealed a hidden connection to orthogonal polynomials in several variables, and a connection to Schur polynomials and representation theory. Without these hints, I have no idea what would have happened. Explanation: The deltoid picture is a 2-dimensional subspace of $\mathbb{C}^2$ where certain generalized eigenvalues for a simple, but large Toeplitz matrix appeared, so this is essentially solutions to a highly degenerate system of polynomial equations. Using a certain map, these roots could be lifted to the hexagonal region, revealing a very structured pattern. This gave insight in how the limit density of the roots is. This is essentially roots of a 2d-analogue of Chebyshev polynomials, but I did not know that at the time. The subspace in $\mathbb{C}^2$ where the deltoid lives is quite special, and we could not explain this. A subsequent paper by a different author answered this question, which lead to an analogue of being Hermitian for rectangular Toeplitz matrices. Perhaps you do not have a single picture; then you might want to illustrate a transformation that you test on data you generate. For example, every polynomial defines a coamoeba, by mapping roots $z_i$ to $\arg z_i$. This transformation sometimes reveal interesting structure, and it partially did in the example above. If you don't generate pictures in your research, you can still participate in the discussion, by submitting a (historical) picture you think had a similar impact (with motivation). Examples I think that can appear here might be the first picture of the Mandelbrot set, the first bifurcation diagram, or perhaps roots of polynomials with integer coefficients. REPLY [5 votes]: The Hasse diagram of a (partially) ordered set is often used to graphically represent an order relation; one may think of it as the minimal directed acyclic graph inducing this relation (its transitive reduction), with all arcs drawn from top to bottom. The partitions of an integer $n$ (non-increasing sequences of integers of sum $n$) are ordered by the dominance order. A similar order may be defined on the different ways to write an integer $n$ as a sum of powers of integer $b$. Both have the lattice structure, and have a striking self-similar structure. This self-similar structure was first noticed by observing the following drawings of Hasse diagrams. The first one is defined over the integer partitions of $40$ and its self-similarity was studied in this paper.                               The second one is defined over the partitions of $80$ into powers of $2$ and its self-similarity was studied in this paper. They are also mentioned in this discussion.<|endoftext|> TITLE: How many integers divide a number that involves just three non-zero digits? QUESTION [18 upvotes]: Just to be concrete, consider the digits to be binary. Hasse showed that among all the primes, only a fraction of $17/24 < 1$ divide a number of the form $2^n+1$. As a result, the integers that divide a number with just two non-zero digits have zero density. On the other hand, since $A + A = \mathbb{F}_p$ for a typical set $A \subset \mathbb{F}_p$ of cardinality, say, $> p/\log{p}$ (and much lower than that), and because under GRH and with probability $1$ there are at least as many powers of $2$ mod $p$, we expect a full density of the primes to divide a number of the form $2^m+2^n+1$. [Added: As shown by Skalba in the reference provided by "so-called friend Don," we can actually prove that this is true for all primes $p \gg_{\epsilon} 0$ having $r := \mathrm{ord}_p^{\times}2 > p^{\frac{3}{4}+\epsilon}$: consider Weil's bound for the Fermat curve of degree $(p-1)/r$. This applies also to the linked question. ] So it would seem legitimate to ask if the probability might be positive for a random integer to have a multple composed of only three non-zero digits. Note that $2^k-1$ does not have this property for $k > 3$ (all its multiples have digit sum exceding $ck$), and this family already furnishes an infinite set of pairwise co-prime integers without the property. Yet I thought it would be hard to believe that a random integer, with positive probability, will have a multiple with bounded digit sum. Is there a (better) heuristic for the expected number up to a given bound $X$ of the positive integers having a multiple of the form $2^m+2^n+1$? On the opposite extreme, wouldn't most $N$ require, for all their multiples, as many as $(1+o(1))\frac{\log{N}}{2\log{2}}$ binary ones? REPLY [7 votes]: For an integer $n$, call a prime divisor $p$ good, if the multiplicative order of $2$ modulo $p$ exceeds $p^{2/5}$, $p^2\nmid n$, and $(p-1, \varphi(n/p))p^{1/5}$, by the sum-product results of Bourgain there exists some $k$, such that for any $a$ there exist integers $e_{1,p}, \ldots, e_{k,p}$, such that $2^{e_{1,p}\varphi(n/p)}+\dots+2^{e_{k,p}\varphi(n/p)}\equiv a\pmod{p}$. Put $e_i=\sum_p e_{i,p}$, the sum taken over all good prime divisors of $n$. Then $e_i\equiv e_{i,p}\pmod{p-1}$ for all good $p$, thus $2^{e_1}+\dots+2^{e_k}\equiv a\pmod{\prod p}$. If $p$ is bad, then $2^{e_i}\equiv 1\pmod{p}$. We conclude that for a certain fixed $k$ the $k$-fold sum of the multiplicative subgroup generated by 2 contains a complete coset of the subgroup of $(\mathbb{Z}/n\mathbb{Z}, +)$ which is isomorphic to $(\mathbb{Z}/P\mathbb{Z}, +)$, where $P$ is the product of good prime divisors of $n$. Next we give an upper bound for the average product of bad prime divisors. Since the number of primes such that 2 has small order is small, and almost all integers are almost squarefree, the only relevant part is the condition on $(p-1, \varphi(n/p))$. For a prime $p$ the number of $n\leq x$ such that $n$ has $<2\log\log x$ prime divisors and $p$ is a bad prime divisor of $n$ is at most $$ \underset{d>p^{1/(10\log\log x)}}{\sum_{d|p-1}}\underset{p\neq q}{\sum_{d|q-1}}\#\{n\leq x: pq|n\} \ll \underset{d>p^{1/(10\log\log x)}}{\sum_{d|p-1}} \frac{x\log\log x}{pd^{1-\epsilon}} \ll \frac{xd(p-1)\log\log x}{p^{1+1/(12\log\log x)}} $$ If $p>e^{40(\log\log x)(\log\log\log x)}$, this becomes $\ll\frac{xd(p-1)}{p\log^3 p}$. The sum over $p$ converges, hence we find that almost all $n\leq x$ have no bad prime divisor $p>e^{40(\log\log x)(\log\log\log x)}$. Disregarding a set of density 0, the average logarithm of the product of the bad prime divisors of $n$ is at most the average logarithm of the product of the small prime disivors, which is $$ \ll \sum_{p\leq e^{40(\log\log x)(\log\log\log x)}} \frac{\log p}{p} \ll \log\log x\log\log\log x $$ We conclude that there is a constant $k$, such that for almost all $n$ have a divisor $d<(\log\log n)^2$, such that the $k$-fold sum of the subgroup generated by 2 covers modulo $n$ the arithmetic progression $k\pmod{d}$. For such $n$ every integer in the interval $[1, d]$ can be written as the sum of $\log d\ll\log\log\log n$ powers of 2, hence for almost all $n$ all residue classes modulo $n$ can be written as the sum of $\ll\log\log\log n$ powers of 2. In particular almost all $n$ divide some number with sum of digits $\ll\log\log\log n$.<|endoftext|> TITLE: Quotients by the additive group $\mathbb G_a$ QUESTION [7 upvotes]: Geometric invariant theory doesn't work so well for non-reductive groups, since invariant rings are not generally finitely generated. However, in many cases the action of a non-reductive group has a finitely generated invariant ring (e.g. when the additive group acts on a polynomial ring). How do I think of quotient varieties in this case? I have a specific example in mind: Let $G=\mathbb G_a$ act on the polynomial ring $k[x_0,\ldots,x_r,y_0,\ldots,y_r]$ by $$tx_i=x_i+ty_i, \text{ and } ty_i=y_i.$$ How should one think of a quotient $\mathbb P^{2r+1}/G$? Does it make sense to consider $\mathrm{Proj}$ of the invariant ring (i.e. the subring generated by $y_0,\ldots,y_r$ and the minors of the $2\times(r+1)$-matrix of variables)? If so, what does the resulting projective variety look like? REPLY [3 votes]: The following response is not really an answer, as the second (and probably most pertinent) part of your question (the specific example) has already been addressed. However, with respect to your general question: How do I think of quotient varieties in this case? I would recommend: Towards non-reductive geometric invariant theory by Doran and Kirwan. They work out many good examples, and rather precisely address the case when the invariant ring is finitely generated.<|endoftext|> TITLE: How to fit the parameters of differential equations with known data? QUESTION [6 upvotes]: I have the following data from chemical kinetics research to fit the parameters of ordinary differential equations: $$ \left[ \begin{array}{ccccccc} \text{No.}& t & y_1(t)&y_2(t) & y_3(t) & y_4(t) & y_5(t)\\ 1&30.0000 & 9.1300 & 0.0931 & 0.0899 & 0.1000 & 0.0000 \\ 2&60.0000 & 8.9300 & 0.1270 & 0.1230 & 0.2270 & 0.0049 \\ 3&90.0000 & 8.6000 & 0.1510 & 0.1390 & 0.4920 & 0.0153 \\ 4&120.0000 & 8.2800 & 0.1540 & 0.1490 & 0.7780 & 0.0249 \\ 5&150.0000 & 7.9700 & 0.1540 & 0.1570 & 1.0700 & 0.0329 \\ 6&180.0000 & 7.8600 & 0.1540 & 0.1600 & 1.1700 & 0.0348 \\ 7&210.0000 & 7.8100 & 0.1530 & 0.1530 & 1.2100 & 0.0404 \\ 8&240.0000 & 7.7700 & 0.1400 & 0.1420 & 1.2800 & 0.0432 \\ \end{array} \right] $$ The ordinary differential equations to fit have $k_1,k_2,k_3,k_4,k_5,k_6$ to be determined. $$ \left\{ \begin{array}{l} {y_1}'(t)=-{k_1} {y_1}(t)-{k_2} {y_1}(t),\\ {y_2}'(t)={k_2} {y_1}(t)-{k_3} {y_2}(t),\\ {y_3}'(t)={k_1} {y_1}(t)+{k_3} {y_2}(t)-{k_4} {y_3}(t),\\ {y_4}'(t)={k_4} {y_3}(t)-{k_5} {y_2}(t) {y_4}(t)+{k_6} {y_5}(t),\\ {y_5}'(t)={k_5} {y_2}(t) {y_4}(t)-{k_6} {y_5}(t)\\ \end{array} \right.$$ In order to solve it from conventional numerical optimization methods, my original thoughts are: first convert it into least square problems, then apply numerical optimization to it, but this requires symbolically solve a nonlinear system of ordinary differential equations into explicit solutions first, which seems difficult. My questions are: (1)Is it possible to determine the global (least square or similarly converted) solution to all the parameters $k_i(i=1,\cdots, 6)$ ? (2)Is the solution unique? (3) are there general approaches to solve such problems (globally if possible)? Update: Further data from replicates: $$ \begin{array}{ccccccc} \text{No}&t& y_1(t) &y_2(t) &y_3(t) & y_4(t) &y_5(t)\\ 9&30.0000 & 9.0400 & 0.1190 & 0.1040 & 0.1390 & 0.0044 \\ 10& 60.0000 & 8.8000 & 0.1640 & 0.1120 & 0.3210 & 0.0097 \\ 11&90.0000 & 8.5300 & 0.1640 & 0.1140 & 0.5630 & 0.0219 \\ 12&120.0000 & 8.1800 & 0.1600 & 0.1250 & 0.8730 & 0.0369 \\ 13&150.0000 & 7.9700 & 0.1550 & 0.1380 & 1.0600 & 0.0459 \\ 14&180.0000 & 7.7900 & 0.1580 & 0.1510 & 1.2000 & 0.0545 \\ 15&210.0000 & 7.4900 & 0.1480 & 0.1430 & 1.5000 & 0.0636 \\ 16&240.0000 & 7.0800 & 0.1390 & 0.1380 & 1.9100 & 0.0756 \\ \end{array} $$ Update 2: I am still seeking for a working approach to obtaining solutions to the problem. REPLY [2 votes]: If you have access to Maple and the Global Optimization Toolbox, you could try something like this. data:=[[0.0000,9.1300,0.0931,0.0899,0.1000,0.0000], [30.0000,8.9300,0.1270,0.1230,0.2270,0.0049], [60.0000,8.6000,0.1510,0.1390,0.4920,0.0153], [90.0000,8.2800,0.1540,0.1490,0.7780,0.0249], [120.0000,7.9700,0.1540,0.1570,1.0700,0.0329], [150.0000,7.8600,0.1540,0.1600,1.1700,0.0348], [180.0000,7.8100,0.1530,0.1530,1.2100,0.0404], [210.0000,7.7700,0.1400,0.1420,1.2800,0.0432]]: des:=[diff(y1(t),t) = -k1*y1(t) - k2*y1(t), diff(y2(t),t) = k2*y1(t) - k3*y2(t), diff(y3(t),t) = k1*y1(t) +k3*y2(t) - k4*y3(t), diff(y4(t),t) = k4*y3(t) - k5*y2(t)*y4(t) + k6*y5(t), diff(y5(t),t) = k5*y2(t)*y4(t) - k6*y5(t)]: ics := seq((y || i)(0) = data[1, i+1], i = 1 .. 5): res := dsolve({ics, des[]}, numeric, parameters = [k1, k2, k3, k4, k5, k6]): timeList := [0, 30, 60, 90, 120, 150, 180, 210]; sse := proc (k1, k2, k3, k4, k5, k6) res(parameters = [k1, k2, k3, k4, k5, k6]); add( (rhs(select(has, res(timeList[i]), y1)[])-data[i, 2])^2+ (rhs(select(has, res(timeList[i]), y2)[])-data[i, 3])^2+ (rhs(select(has, res(timeList[i]), y3)[])-data[i, 4])^2+ (rhs(select(has, res(timeList[i]), y4)[])-data[i, 5])^2+ (rhs(select(has, res(timeList[i]), y5)[])-data[i, 6])^2, i = 2 .. 8); end proc; c := GlobalOptimization:-GlobalSolve('sse'(k1, k2, k3, k4, k5, k6), k1 = 0 .. 1, k2 = 0 .. 1, k3 = 0 .. 1, k4 = 0 .. 1, k5 = 0 .. 1, k6 = 0 .. 1, timelimit = 10); [0.219132447080011505, [k1 = 8.52482740113834e-4, k2 = 5.2683998680924474e-5, k3 = 0.0, k4 = 0.05113239298267808, k5 = 0.004363021255887466, k6 = 0.0]] res(parameters = c[2]): p:=Array(1..5): for n from 1 to 5 do p[n]:=plots:-display(plots:-odeplot(res, [t, (y || n)(t)], t = 0 .. 210),plots:-pointplot([seq([data[i, 1], data[i, n+1]], i = 1 .. 8)])); end do; plots:-display(p) The code is a parameterized numeric solution of the differential equations (I shifted all the data down by 30s, so I could get the ICs at t=0) followed by a global optimization of the least-squares problem. I haven't tested the code for correctness, and if you let the optimizer run for longer or change the optimization bounds, you may get better results.<|endoftext|> TITLE: Which singularities of log pairs do not depend on the resolution? QUESTION [5 upvotes]: Let $(X,\Delta)$ be a log pair (we assume the coefficients of $\Delta\leq 1$, but could be negative rationals), and I use the definitions in the book of "Birational Geometry of Algbebraic varieties" (Page 56, Def. 2.34; Page 58, Def. 2.37) for "terminal, canonical, klt, plt, lc, dlt" singularities. My first question is: which of these six singularities do not depend on the resolution of singularities? To be precise, this means: if one happen to have a log resolution: $Y \to X$, and the discrepancy for this resolution satisfy the requirement for the corresponding singularities, then for all the log resolutions, the discrepancy satisfy the requirement. I try to use the standard argument that one dominant two resolutions by the third one. However, I always feel I might miss something (because the argument goes through when the coefficients of $\Delta > 1$, in which case, all the statement are false), below is what I did: Suppose we try to "show" lc doesn't depend on the resolution. Let $\pi: Y_1 \to X$ be the log resolution, and the coefficients of the exceptional divisors $E_1$ are all $\geq -1$ in the log reamification formula. Let $\theta: Y_2 \to X$ be another log resolution. Finally, let $p:W \to Y_2, q: W \to Y_1 $ be a common resolution where the exceptional divisors and the strict transform of $\Delta$ (which I denote by $\Delta_W$) have snc. Then, we have $$K_W + \Delta_W \sim q^*\pi^*(K_X+\Delta) + q^* E_1 + F_1,$$ where $E_1, F_1$ are some exceptional divisors of $\pi, q$ respectively. We can write $$q^*E_1 = q_*^{-1}E_1 + Exc_q(q^*E_1)$$ where $ q_*^{-1}E_1$ is the strict transform of $E_1$, and $Exc_q(q^*E_1)$ belongs to the exceptional divisor of $q$. One can do the same thing for $W \to Y_2 \to X$, and putting them together we have $$q_*^{-1}E_1 + Exc_q(q^*E_1) + F_1 \sim \theta_*^{-1}E_2 + Exc_p(p^*E_1) + F_2.$$ Because every divisor involved in the above expression are exceptional divisors of $p \circ \theta = q \circ\pi$, the numerically equivalence implies equality of divisors. Moreover, because the coefficients on the left are all $\geq -1$ (I am a little worry about $Exc_q(q^*E_1)$, but I think the assumption on the snc of divisors would imply its coefficients $\geq -1$ ), this implies the same thing on the right, and particularly for $E_2$. My second question is: Assume the coefficients of $\Delta$ is in $[0,1]$, then what are the differences between "plt" and "dlt"? I didn't realize the subtlety of $dlt$ compared to other type of singularities... REPLY [7 votes]: I guess I can take a stab at this. Certainly we are talking about log resolutions, ie that $\pi : Y \to X$ is proper and birational, $Y$ is regular, and $\text{exc}(pi) \cup \pi^{-1}_* \Delta$ has simple normal crossings. Then: KLT is independent of the log resolution LC is independent of the log resolution. PLT is independent of the log resolution as long as you assume that the resolution also separates the birational transform of all the components of $\lfloor \Delta \rfloor$. In other words it's independent of the resolution as long as that resolution is also an embedded resolution of the components of $\Delta$. DLT is definitely not independent of the resolution, however see the recent book on singularities by Kollár (with contributions by Kovács), for discussions of types of resolutions that one can read off DLT from, also see the work of Szábo. Canonical is not independent of the resolution. Consider the pair $(X, \Delta)$ where $X = \mathbb{A}^2$ and $\Delta = (1-\varepsilon) L_1 + (1 - \varepsilon) L_2$. Then $(X, \Delta)$ is its own resolution, and so you might suspect it is canonical (or terminal) but of course blowing up the origin gives you a divisor with discrepancy $1 - 2 \cdot ( 1 - \varepsilon) = -1 + 2 \varepsilon < 0$ for $0 < \varepsilon \ll 1$. Of course, in the case that $\Delta$ is an integral Cartier divisor and $K_X$ is Cartier, then it is independent of the choice of resolution. Now, what is true is that there are only finitely valuations corresponding to divisors with discrepancy $\leq 0$ for a KLT (and also canonical) pair. Hence it is always enough to go to log resolutions where every divisor, exceptional or not, with discrepancy $\leq 0$ is disjoint from every other divisor with discrepancy $\leq 0$. Terminal is not independent of the log resolution (indeed, the same example works). The same argument should also work here in terms of finding sufficiently big resolutions as we did for canonical. Your computation I think you are making it too hard. The point for KLT / LC is that if $\pi: Y \to X$ is any birational map from normal $Y$ then if $(X, \Delta)$ is KLT/LC, then $(Y, -K_Y + \pi^*(K_X + \Delta) )$ is also KLT/LC. (KLT and LC are rigged so that this happens, and the converse works too). This means that if you take a further resolution $\rho : W \to Y$ then we see that the pair $$ \big(W, -K_W + \rho^*(K_Y - K_Y + \pi^*(K_X + \Delta)) \big) = \big(W, -K_W + \rho^* \pi^*(K_X + \Delta) \big) $$ In particular, the fact that $(W, -K_W + \rho^* \pi^*(K_X + \Delta) )$ has KLT/LC singularities implies that $(Y, -K_Y + \pi^*(K_X + \Delta))$ also has KLT/LC singularities. And so it's easy to see that things are independent of the resolution. PLT vs DLT A simple normal crossings pair with coefficients in $[0,1]$ always has DLT singularities, it only has PLT singularities if each divisor with coefficient $=1$ is disjoint from every other such divisor. Basically, the point of DLT is that it is a more controlled version of LC (many objects which you produce via the MMP which a priori have LC singularities also have DLG singularities). In particular, the worst singularities of a DLT variety occur where the pair is a SNC pair, and so while these are technically more severe, they can be very easy to deal with. Hopefully this helps.<|endoftext|> TITLE: Szemeredi's theorem in the Gaussian integers QUESTION [14 upvotes]: Suppose that $S\subseteq\mathbb{Z}[i]$ has the following properties: For convenience, let $A_n = \{z : z\in\mathbb{Z}[i], \text{Nm}(z)\le n\}$ $$\limsup_{n\rightarrow\infty} \frac{|S\cap A_n|}{|A_n|} > 0$$ Then is it the case that $S$ contains arbitrarily long arithmetic progressions. REPLY [26 votes]: This is true, and follows from one-dimensional Szemeredi. Fix $\delta > 0$ such that $|S\cap A_n| \geq \delta |A_n|$ infinitely often. Let $r = \lfloor \sqrt n \rfloor$, so $A_n$ is contained in the square $S_r: \{x+iy \in {\bf Z}[i] \colon |x|,|y| < r\}$. Define the additive homomorphism $h_r: {\bf Z}[i] \rightarrow {\bf Z}$ by $h_r(x+iy) = 4rx+y$. Then $h_r$ is injective on $S_r$, and any arithmetic progression in $h_r(S_r)$ lifts to $S_r$. [The point is that $h_r$ regularly lays the horizontal segments in $S_r$, each of length $2r-1$, on ${\bf Z}$, but separated by gaps of length $2r+1$, which is short enough to reduce the density by only a finite factor but long enough that any identity $2a_2=a_1+a_3$ in the image holds on each coordinate of the $h_r^{-1}(a_i)$.] Then the image $h_r(S_r)$ is a subset of $(-4r^2,4r^2)$ of density at least $\frac\pi8 \! \delta - O(1/r)$, and by Szemeredi is guaranteed to contain arbitrarily long arithmetic progressions as $r \rightarrow \infty$. The same technique works with ${\bf Z}[i] \cong {\bf Z}^2$ replaced by ${\bf Z}^k$ for any fixed $k$. [Added later: Naturally this stratagem is far from new. This 2008 entry from Terry Tao's blog reminds me that it's called the "Ruzsa projection trick", and the map $h_r$ (and more generally the injection from $(-r,r)^k$ to ${\bf Z}$ taking $(x_1,\ldots,x_k)$ to $\sum_{i=1}^k (4r)^{k-i} x_i$) is called a "Freiman isomorphism of order $2$" to its image ("order $2$" because coincidences between sums of $2$ elements of $S$ suffice to detect arithmetic progressions of arbitrary length).]<|endoftext|> TITLE: Reference Request: "Neck Stretching Procedure" (In Symplectic Field Theory) QUESTION [6 upvotes]: I've been reading some papers in Symplectic Geometry which refer to something called "Stretching the neck", and give reference to Eliashberg, Givental and Hofer's SFT paper (http://arxiv.org/abs/math/0010059) I ran a search after the words "neck" and "stretch" and they do not appear in the paper, I guess the theory is there under a different name. Since it's a long and complicated paper and I just want to understand the general idea and results of this "neck stretching" so I can continue reading those other papers i'm currently reading, Could someone please refer me to the part in the the SFT paper which deals with neck stretching? or alternatively, refer me to some other paper/survey/someone's thesis/book in which it is clearly explained? Thank you REPLY [10 votes]: Neck-stretching is a deformation of an almost complex structure in a neighbourhood of a hypersurface. In the Eliashberg-Givental-Hofer paper, neck-stretching is called "splitting along a contact submanifold" (see Section 1.3) - the description of the almost complex structure is sketched at the end of Section 1.4. A more detailed account of neck-stretching (and the behaviour of punctured pseudoholomorphic curves under this deformation) is given in Bourgeois-Eliashberg-Hofer-Wysocki-Zehnder (e.g. Section 3.4) http://arxiv.org/abs/math/0308183 and in Cieliebak-Mohnke (e.g. Section 2.7) http://projecteuclid.org/euclid.jsg/1154467631<|endoftext|> TITLE: Mapping graphs to ordinals QUESTION [7 upvotes]: Robertson-Seymour theorem implies that graph minor relation is a well-quasi-ordering, which means (among other things) that this relation can be extended to a well-order, and other result says that this order can have order type $\psi(\Omega_\omega)$, if I remember correctly. I've been wondering - has anyone actually analysed this ordering? For example, in such ordering, what place would $K_{3,3}$ take? $K_5$? Petersen graph? Only such analysis I know of is on this page: http://googology.wikia.com/wiki/User_blog:Hyp_cos/SCG(n)_and_some_related but it only concerns subcubic graphs, because of their connection to SCG function. Thanks in advance for any help! (If you think this fits better to Math StackExchange, please move the question) REPLY [2 votes]: There are now results bounding the order type of the set of finite graphs when ordered by the graph minor relation, which is bounded between $\psi_0(\Omega_\omega)$ and $\psi_0(\Omega_\omega^{\omega^\omega})$ with respect to (presumably) Buchholz's ordinal collapsing function. In Rathjen and Krombholz's paper "Upper bounds on the graph minor theorem", this bound is cited from unpublished research, with suggestions for possible future improvements of the lower bound by relating the problem to a generalization of Kruskal's problem of well-ordering trees.<|endoftext|> TITLE: Do regular conditional distributions almost surely assign trivial measure to all members of the conditioning $\sigma$-algebra? QUESTION [8 upvotes]: Let $(X,\Sigma)$ be a standard measurable space, let $\rho$ be a probability measure on $(X,\Sigma)$, and let $\mathcal{E}$ be a sub-$\sigma$-algebra of $\Sigma$. We will say that a stochastic kernel $(\rho_x^\mathcal{E})_{x \in X}$ on $X$ is a regular conditional distribution of $\rho$ with respect to $\mathcal{E}$ if the map $x \mapsto \rho_x^\mathcal{E}(A)$ is $\mathcal{E}$-measurable for all $A \in \Sigma$; for every $E \in \mathcal{E}$ and $A \in \Sigma$, $\rho(A \cap E) = \int_E \rho_x^\mathcal{E}(A) \, \rho(dx)$. Is it necessarily the case that $\rho$-almost every $x \in X$ has the property that for all $E \in \mathcal{E}$, either $\rho_x^\mathcal{E}(E)=0$ or $\rho_x^\mathcal{E}(E)=1$? Remark: In order for the above to be satisfied, I believe it is sufficient that there exists a family $(\rho_{x,y})_{x,y \in X}$ of probability measures on $X$ such that for $\rho$-almost every $x \in X$, $(\rho_{x,y})_{y \in X}$ is a rcd of $\rho_x^\mathcal{E}$ with respect to $\mathcal{E}$; the map $(x,y) \mapsto \rho_{x,y}(A)$ is $(\mathcal{E} \otimes \mathcal{E})$-measurable for all $A \in \Sigma$. (Specifically, if we can find such a family, then I think we can show that for $\rho$-almost every $x$, for $\rho_x^\mathcal{E}$-almost every $y$, $\rho_{x,y}=\rho_x^\mathcal{E}$.) Some important remarks: The disintegration theorem guarantees that a rcd of $\rho$ with respect to $\mathcal{E}$ exists and is unique modulo $\rho$-null sets. So of course (at least if we assume AC) there exists a family $(\rho_{x,y})_{x,y \in X}$ satisfying (1); but the question is whether there necessarily exists a family $(\rho_{x,y})_{x,y \in X}$ satisfying both (1) and (2). It is worth emphasising: we do not necessarily have that for $\rho$-almost every $x \in X$, for all $E \in \mathcal{E}$, $\rho_x^\mathcal{E}(E)=\mathbf{1}_E(x)$. (For counterexamples, see the examples of "maximally improper rcd's" in the paper "Improper Regular Conditional Distributions" linked to by Jochen below.) Motivation - ergodic decomposition: I'm keen to have a "nice" proof of the ergodic decomposition theorem for stationary probability measures of stochastic semigroups (jointly measurable in space and time) on standard measurable spaces. If I understand correctly, one can reduce the question of finding an ergodic decomposition of a stationary measure of a stochastic semigroup to the question of finding an ergodic decomposition of an invariant measure of a (deterministic) dynamical system, by considering the time-shift dynamical system on the space of $X$-valued functions of time. Already I'm not sure I'd deem this "nice", but even for dynamical systems I wonder whether there's a nicer proof of the ergodic decomposition theorem than the ones I've seen. For an invariant measure $\rho$ of a measurable dynamical system, the proofs that I've seen involve using Birkhoff's ergodic theorem to show that $\rho_x^\mathcal{I}$ is ergodic for $\rho$-almost all $x$, where $\mathcal{I}$ is the $\sigma$-algebra of invariant sets. But if the answer to my question is yes, then the ergodicity of $\rho_x^\mathcal{I}$ for $\rho$-almost all $x$ is immediate (once we have established the invariance of $\rho_x^\mathcal{I}$ for $\rho$-almost all $x$---but that is easy). I guess one could argue that Birkhoff's theorem is "nice enough" as it is, but if the answer to my question is yes, then the same proof will work directly for stochastic semigroups (so that we don't have to invoke the theorem of equivalence between ergodicity with respect to a stochastic semigroup and ergodic of the corresponding Markov measure under the time-shift dynamical system). A possible approach? Perhaps I should mention a possible starting point that I've thought of, but have been unable to make into a full solution: The difficulty behind the problem is that $\mathcal{E}$ might not be countably generated; however, as hinted at by Yuri below, perhaps it is possible to use the fact that $\mathcal{E}$ is countably generated $\bmod \rho$ to help. Of course, this fact cannot mean that all arguments for the countably generated case remain valid in the general case, since as we have said already, it is not necessarily the case that for $\rho$-almost every $x \in X$, for all $E \in \mathcal{E}$, $\rho_x^\mathcal{E}(E)=\mathbf{1}_E(x)$. Nonetheless, perhaps we can proceed as follows: Let $\{E_n\}_{n \in \mathbb{N}} \subset \mathcal{E}$ be such that $\mathcal{E}$ is contained in the $\rho$-completion of $\sigma(E_n:n \in \mathbb{N})$. For each $n$, let $\mathcal{G}_n:=\sigma(E_i : 1 \leq i \leq n)$. Then I believe we have that for $\rho$-almost all $y \in X$, $(\rho_x^\mathcal{E})_{x \in X}$ is a rcd of $\rho_y^{\mathcal{G}_n}$ with respect to $\mathcal{E}$ for all $n$; [by the result mentioned in (2) of Conditional law as a random measure and convergence of random measures, combined with Levy's Upward Theorem] for any fixed Polish topology on $X$, for $\rho$-almost all $y \in X$, $\rho_y^{\mathcal{G}_n} \to \rho_y^\mathcal{E}$ in the narrow topology as $n \to \infty$. If I can somehow show that (1) and (2) together imply that $\hspace{7mm}$ for $\rho$-almost all $y \in X$, $(\rho_x^\mathcal{E})_{x \in X}$ is a rcd of $\rho_y^\mathcal{E}$ with respect to $\mathcal{E}$ then I'm done! Any ideas?? REPLY [4 votes]: I've found the answer - it's NO! The paper I found addressing the question is the following: http://projecteuclid.org/euclid.aop/1175287757 ("0-1 Laws for Regular Conditional Probabilities") A simple counterexample (which I've just slightly adapted from the counterexample in Example 2 of the above paper) is the following: Let $X=[0,1] \times \{0,1\}$ (with $\Sigma=\mathcal{B}([0,1]) \otimes 2^{\{0,1\}}$), let $\rho$ be a probability measure on $X$ whose projection onto $[0,1]$ is atomless, and let $\mathcal{E} \subset \Sigma$ be the $\rho$-completion of $\mathcal{B}([0,1]) \otimes \{0,1\}$ relative to $\Sigma$. Then given any non-trivial probability measure $m$ on the binary set $\{0,1\}$, the stochastic kernel $\hspace{5mm} \rho_{(x,i)}^\mathcal{E} \ := \ \delta_x \otimes m$ is a rcd of $\rho$ with respect to $\mathcal{E}$. Clearly, for any $x \in [0,1]$, $\{(x,0)\} \in \mathcal{E}$; and yet, for all $(x,i) \in X$, $\hspace{5mm} \rho_{(x,i)}^\mathcal{E}(\{(x,0)\}) \ = \ m(0) \, \in \, (0,1).$ Regarding my motivation: Theorem 12 of the above paper claims to be a generalisation of the ergodic decomposition theorem (for measurable maps). However, I haven't yet managed to work out how to derive the ergodic decomposition theorem from Theorem 12 of the above paper.<|endoftext|> TITLE: Salvaging Leibnizian formalism? QUESTION [7 upvotes]: Can one justify Leibniz's formalism in a suitable algebraic or topological context? We have published some papers recently where we argue that Leibniz's formalism for the calculus wasn't inconsistent as Berkeley claimed. For an insightful review see http://www.ams.org/mathscinet-getitem?mr=3053644 Berkeley claimed that Leibniz wanted to have it both ways: both $dx\not=0$ so as to form the differential ratio, and also $dx=0$ so as to get the right answer (i.e., a "standard" one). Starting about 140 years ago, Berkeley's claim of inconsistency of Leibnizian calculus acquired the status of dogma to such an extent that Robinson himself felt compelled to speak of Berkeley's "brilliant critique" of the calculus, and referred to the hyperreal framework as "a small price to pay for the removal of an inconsistency"--the implied assumption being that such an "inconsistency" was real. The reason Berkeley was wrong is that Leibniz repeatedly emphasized that he is working with a generalized notion of equality. For example if $y=x^2$, the desired formula $\frac{dy}{dx} = 2x$ does not mean that the residual $dx$ is set equal to zero but rather that it is absorbed into the generalized relation of equality "up to" a negligible term, in an exact sense to be specified. Leibniz called this principle the transcendental law of homogeneity. The principle is mentioned, for example, in the title of his 1710 paper, as reported already in 1974 by Bos. Berkeley did not take this into account and merely misunderstood Leibniz. Now this is fine for showing that Leibniz was not inconsistent (refuting Berkeley's claim). However, it is not quite enough for showing that Leibniz was actually consistent, or more precisely for formalizing Leibniz's approach. This is because it is not completely clear what the generalized relation is exactly. I will refer to such a generalized notion as "adequality" so as not be have to write "generalized equality up to" every time. In other words, if we want to be able to work with adequality as we work with the ordinary equality, we need to explain how this is done and why this works and why whenever A=B one can replace A by B in computations. Robinson circumvented the problem by using the standard part function but this isn't completely faithul to Leibniz's formalism. One attempted solution is to take the adequality 2x+dx=2x to mean that the difference of the two sides is infinitesimal. But if this is our notion of adequality, then this allows us to write down things like dx=0, as well, and if we are allowed to replace dx by 0 in calculations then we end up dividing by zero. In Lawvere's approach (also Kock, Bell) they replace the ratio formula $f '(x)=dy/dx$ by the multiplication formula $dy=f '(x)dx$. Then they get equality on the nose by working with nilsquare infinitesimals. Thus their adequality is true equality on the nose. In this way they implement (some of) Leibniz's procedures. However, this is not entirely faithful to Leibniz because Leibniz worked with arbitrary order infinitesimals, and also divided by them freely. Euler worked with what he called a generalized "geometric" equality where A=B means that the ratio of A to B is infinitely close to 1, but this does not automatically allow us to add such relations. Of course if all expressions involved are appreciable, this does work. On the other hand, we can't always assume both sides to be appreciable because this would disallow critical points, certainly a disturbing loss. How does one address this problem? The idea is to continue working with Euler's "geometric equality" and somehow to make both sides appreciable by working globally rather than at a specific point; or perhaps evaluating the expressions at a generic (or perhaps nonstandard?) point. This would hopefully allow one to manipulate an adequality between expressions as an ordinary equality. In the specific case of $y=x^2$ the problem is the zero of the derivative where Euler's geometric equality does not work, but at any other point we are OK. Could one define such a relation in an algebraic (or algebraic-geometric) context? One needs to specify the sort of expressions one is allowed to work with, i.e. introduce a limitation on the objects one is allowed to use. Or perhaps it is enough to declare expressions adequal if they are geometrically equal at a nonstandard point. Can one redefine the relation "=" in a suitable context, so that for example one could read the chain rule as literally saying $\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}$ ? The relation can be rewritten in a simpler form in terms of differentials: $dz = \frac{dz}{dy}dy$ but this still depends on the "cancellation" of $dy$ in the numerator and denominator. This works with Euler's "geometric equality" but the problem is you can't add such equalities. Already chain rule with 2 variables requires addition. Aside from the pedagogical value of being able to cancel out the two $dy$'s in $dz/dx =dz/dy\; dy/dx$, an application would be the solution of differential equations by separation of variables, etc. More generally, one would like to be able to take any argument using derivatives, and be authorized to replace $f ′ (x)$ by $dy/dx$ whenever it occurs without changing the argument otherwise. In nested arguments involving multiple derivatives this could be a significant simplification. Differential geometry is a rich source of examples where salvaging Leibniz's formalism could simplify many proofs. Beyond being able to say that the center of curvature is the point of intersection of two infinitely close normals, numerous arguments in Gauss and Riemann are simpler in their original infinitesimal form (including formula for curvature) than their modern reformulations; see discussion in Spivak, Differential geometry, volume 2, and related MSE thread. REPLY [3 votes]: This is a little long for a comment, but not really an answer. One way I think you could argue that Leibniz "got it wrong" is his treatment of higher order differentials. As far as I know he would write $$ d^2f = f''(x)dx^2 $$ When in fact we should be writing something like $$ d^2f = f''(x)(dx)^2+f'(x)d^2x $$ This new formula satisfies the chain rule, and so gives an invariant definition on manifolds. It is related to Andrej Bauer's answer as well: This is related to the Weil Algebra $W = \mathbb{R}[x,y]/(x^2,y^2)$. $M^{Spec(W)}$ is the tangent bundle to the tangent bundle of $M$, and the above formula (basically) gives the second derivative as a map from that second order tangent bundle (in the case $M = \mathbb{R}$). For a reference developing this kind of stuff without reference to intuitionistic logic, you might want to check out Kolar, Michor, and Slovak's "Natural Operations in Differential Geometry". Also, for first order differentials, what is wrong in your opinion with differential forms?<|endoftext|> TITLE: Hyperfinite type II_1 factor as the Clifford algebra QUESTION [5 upvotes]: In Connes' book Noncommutative geometry, there is a presentation of all hyperfinite factors. He reffers to type $II_1$ as the Clifford algebra of infinite dimensional Euclidean space. This factor can be constructed as a von Neumann algebra of the i.c.c (discrete) group of all permutations of $\mathbb{Z}$ leaving fixed all integers except of finite number of them. From this construction it is not clear for me what it has to do with Clifford algebra. Could anybody give me a good reason for such a name? REPLY [4 votes]: First, note that there are many manifestations of the hyperfinite $II_{1}$ factor, but it turns out that they are all isomorphic. I will thus interpret your question as asking how the Clifford algebra of infinite-dimensional Euclidean space is a hyperfinite $II_{1}$ factor. The book Spinors in Hilbert space by Plymen & Robinson answers this question in paragraph 1.3. Of course, the authors of the book give a far better explanation of the story, but let me try to give a sketch anyway. Let $V$ be a separable infinite dimensional real Hilbert space, endowed with a unitary structure $J:V \rightarrow V$. (That is, an orthogonal map that squares to $-1$). The complex Clifford algebra $C(V)$ is endowed with a complex anti-linear involution, and can be completed into a $C^{*}$-algebra, called the Clifford $C^{*}$-algebra and denoted $C[V]$. The Clifford algebra is equipped with a canonical trace $\tau: C(V) \rightarrow \mathbb{C}$, and thus with an inner product defined by $\langle \xi | \eta \rangle = \tau(\eta^{*} \xi)$. Completing the space $C(V)$ with respect to this inner product gives us a Hilbert space $\mathbb{H}_{\tau}$. The Clifford algebra $C(V)$ acts via the left regular representation on $\mathbb{H}_{\tau}$, and this representation can be extended to the Clifford $C^{*}$-algebra $C[V]$. Write $\lambda: C[V] \rightarrow B(\mathbb{H}_{\tau})$ for this representation. The von Neumann Clifford algebra $\mathcal{A}[V]$ is now defined as the weak closure of $\lambda(C[V])$, or equivalently $\mathcal{A}[V] := \lambda(C(V))''$, where a prime denotes the commutant. Finally, one may show that the von Neumann Clifford algebra $\mathcal{A}[V]$ has a unique central normal faithful tracial[*] state $\tau: \mathcal{A}[V] \rightarrow \mathbb{C}$. One then uses this to show that $\mathcal{A}[V]$ is type $II_{1}$. The fact that it is hyperfinite follows from the fact that $V$ is seperable. *I am not sure if there is any redundancy between these adjectives.<|endoftext|> TITLE: How were formulas / images added to books in post-printing-press / pre-digital times? QUESTION [19 upvotes]: I have seen that Euclid's Elements was written 300 BC and first set in type in 1482. Are there scans of that old versions available? How were formulas / images added to the books created with printing presses? Can / could formulas / images also be printed automatically or had they to draw every single image again for every single new book? Related Writing papers in pre-LaTeX era? How was Euclid's Elements likely written? REPLY [14 votes]: The following references describe mathematical printing in the 1950s, before it became too expensive and degenerated into illegible typescript: The Printing of Mathematics by T. W. Chaundy, P. R. Barrett and Charles Batey, Oxford University Press, 1954. Setting Mathematics by Arthur Phillips, in The Monotype Recorder 40 (4), Winter 1956. Printing Mathematics by Peter Basnett, in Eureka 23 (1960) 11-13. The Mathematician and the Printer. Hints on the preparation of mathematical and technical manuscripts. by R. G. Hitchings, Hodgson, London, 1961. Copy-Editing, The Cambridge Handbook for Editors, Authors, Publishers, by Judith Butcher, CUP. First edition 1975, second 1981, third 1992. At that time, ordinary text (novels and newspapers) was set using the Linotype system, in which the compositor would use a huge keyboard to select characters and then they would be set in molten lead a whole line at a time. However, this was unsuitable for mathematics and the "state of the art" was the Monotype Four Line System, which had movable type. The following piece of nonsense is a block provided by Monotype and used in the Eureka article: Each letter is set on a block (shown in grey) that occupies space on one of the four lines. However, the $g$, $X$, $b$ and $Y$ overhang their blocks and are supported by spacer blocks (white) in the other line. You could add big integral or summation signs in front of such a fractional expression, and (I think) a $d x$ vertically centered after it. PS on OCR (Optical Character Recognition) There is commercial software for OCR of maths but apparently no good FOSS. For ordinary text, Tesseract-OCR does a very good job, but its "equation" recognition seems to be limited. This article by Ray Smith of Google gives a good description of how it works. There are comments on the Web asking for recognition of general LaTeX output or handwritten mathematics. These are fanciful because they are over-ambitious, not clearly defined and not what is actually required. On the other hand there are huge numbers of old journal papers (increasingly available online as scans) that were printed with the technology described above, or less. It would be extremely valuable to be able to put these in digital form, for example for machine translation. Anyone interested in a programming project (maybe contributing to tesseract) should therefore aim to recognise the Monotype "four line" formulae above. Of course it would also be necessary to recognise symbols, fonts, sub- and superscripts. All of these would appear to be within the design concept of tesseract. Such a project should be regarded as a tool to help a human digitise the text, ie to reduce the amount of hand-editing that is needed but not necessarily eliminate it. This post should be enough on its own to start such a project. More details of the old printing technologies are in the books and articles mention at the top of this post. I have copies of these that were kindly given to me (a long time ago) by David Tranah of CUP.<|endoftext|> TITLE: Divisor sums over values of binary forms of primes QUESTION [7 upvotes]: Let $\tau$ be the divisor function, that is $$ \tau(n)=\sharp\{d \in \mathbb{N}, d|n\}. $$ I was wondering if anyone has ever proved an asymptotic estimate for the sum $$S(x):=\sum_{p,q\leq x}\tau(p^2+q^2),$$ where the summation is taken over pairs of primes. One obviously expects $$S(x)\sim c\frac{x^2}{\log x}$$ as $x \to \infty,$ where $c$ is a positive constant which is an infinite product of Euler factors. This is based on the heuristic that each of the $\pi(x)^2$ terms present in $S(x)$ is approximated by a constant multiple of $\log x$ on average. Brun-Titchmarsch and Bombieri-Vinogradov can be used to prove the upper and the lower bound $$ c\frac{x^2}{\log x} (\frac{1}{2}+o(1)) \leq S(x) \leq c\frac{x^2}{\log x} (2+o(1)), $$ as $x\to \infty$ respectively. But the question remains, $\textit{can we prove an asymptotic?}$ REPLY [4 votes]: An answer regarding the use of large sieve suggested by Lucia (too long for a comment). I guess her/his thought was along the following lines: The divisors $d$ of $p^2+q^2$ are of order $x^2$ and the hyperbola trick reduces to estimating sums of the form $$\sum_{d\leq x}\sum_{\lambda^2=-1 (d)} \sum_{\substack{p,q \leq x \\ p= \lambda q (d)}} 1. $$ Inserting multiplicative characters in the sum over primes we will be left with an error term coming from the primitive characters, which looks like $$ \sum_{d\leq x}\frac{1}{\phi(d)} \sum^*_{\chi(d)} \sum_{\lambda^2=-1(d)}\overline{\chi(\lambda)} \ \Big| \sum_{p\leq x}\chi(p)\Big|^2 ,$$ which behaves as $$\frac{1}{x} \sum_{d\leq x}\tau(d)\frac{d}{\phi(d)} \sum^*_{\chi(d)} \ \Big| \sum_{p\leq x}\chi(p)\Big|^2.$$ Ideally we would like to show that this is $o(x^2/\log x)$. However the large sieve in the form of [Th.4,p.g. 160, multiplicative Davenport] gives a bound for this quantity which is of order $O(x^2)$ and is therefore inadequate. Essentially this is the level of distribution problem. What we described can show that the contribution of $d\leq x/(\log x)^A$ to the sums with primitive characters is indeed $o(x^2/\log x)$ for an appropriate value of $A>0$ but I am not sure whether the micro logarithmic savings coming from Hooley's-Delta function can be used to say something about the remaining range. REPLY [2 votes]: When computing e.g. an asymptotic for $\sum_{p\leq x}d(p-1)$ you would like to estimate the number of primes $p$ such that $n$ divides $p-1$ by the prime number theorem as $\sim\frac{x}{\varphi(n)\log x}$. We do not know GRH, so we can't use this estimate for all $n TITLE: AdicCompletion$\dashv$Torsion adjunction on spectra? QUESTION [5 upvotes]: It seems to me that in slight paraphrase the central result of the article Marco Porta, Liran Shaul, Amnon Yekutieli, On the Homology of Completion and Torsion (arXiv:1010.4386) (theorems 6.11 and 6.12) means that for $\mathfrak{a} \subset A$ a suitably nice ideal inside a commutative ring $A$, then the total derived functors of 1) adic completion of modules at $\mathfrak{a}$ and 2) of taking $\mathfrak{a}$-torsion submodules form an adjoint pair of (co-)reflections of homotopy theories (i.e. an adjoint pair of idempotent $\infty$-(co-)monads on the $\infty$-category of chain complexes of $A$-modules). I am wondering if an analogous result would not also hold for spectra in the case that $\mathfrak{a} = (p)$ is a prime. If so that would yield a nice enhancement of the story of the arithmetic fracture square. Is forming $p$-completion of spectra adjoint to forming universal $\mathbb{Z}[p^{-1}]$-acyclic spectra (hence adic completion to $\mathbb{Q}$-acyclification), maybe at least on suitably small spectra? And how about lifting either statement to commutative monoids, i.e. to dg-algebras and further to $E_\infty$-rings, is anything known? REPLY [4 votes]: If I understand what you are asking, then yes. p-completion of p-local spectra is $X \mapsto F(M, X)$, where $M=$ fiber of $S\to S\mathbb{Q}$, while the p-torsion approximation is $X\mapsto X\wedge M$. The same story holds for any "smashing" localization. Added. A "smashing localization" $L$ gives a map of spectra $\eta\colon S\to T:=LS$ such that $T\wedge \eta$ is an equivalence. Consider the cofiber sequence $$ M\xrightarrow{\epsilon} S\xrightarrow{\eta} T. $$ Then we obtain a couple of idempotent monads $T\wedge-$ and $F(M,-)$ on spectra, and a couple of idempotent comonads $M\wedge-$ and $F(T,-)$ on spectra. Clearly, these come as two adjoint pairs of functors on spectra. Then $T\wedge -$ is just the original smashing localization $L$, and $F(M,-)$ is a "cosmashing localization". The other two functors are the corresponding acyclizations. In the case of $$ M=\Sigma^{-1} S\mathbb{Q}_p/\mathbb{Z}_p \to S \to S\mathbb{Z}[\tfrac{1}{p}]=T $$ we get the situation you described, where $M\wedge -$ is the $p$-torsion-approximation idempotent-comonad, and $F(M,-)$ is the $p$-completion idempotent-monad. (Note: as a functor from spectra to spectra, $p$-torsion approximation $M\wedge-$ is a left adjoint, but is also right adjoint to the inclusion functor of $p$-torsion spectra into all spectra.)<|endoftext|> TITLE: Proofs of the Chevalley-Warning Theorem QUESTION [17 upvotes]: A well known proof of the Chevally-Warning Theorem is the one listed on wikipedia: http://en.wikipedia.org/wiki/Chevalley%E2%80%93Warning_theorem Are there any other proofs of this, or generalizations of it? REPLY [2 votes]: It is worth to mention that in his original paper, E. Warning also finds a lower bound for the number of solutions: Apart from the references provided in other answers here, an excellent survey about Chevalley-Warning theorem, its history and applications and related problems about solving polynomial equations over finite fields, written by Jean-René Joly, is accessible from Équations et variétés algébriques sur un corps fini (L'Enseignement Mathématique (1973))<|endoftext|> TITLE: Heegaard genera of arithmetic 3-manifolds QUESTION [10 upvotes]: UPDATE: Because I was hoping that state the question as concisely as possible, the original post did not include a precise definition of arithmetic 3-manifold only a reference to Maclachlan and Reid's book where it is defined. However, this lead to an ambiguity that caused conflicting answers. With the included definition, Ian Agol's answer is correct. However, given a more restrictive definition of arithmetic 3-manifold, which is (less standard, but) used in a paper of Gromov and Guth, John Pardon's answer is correct. I apologize for the confusion and appreciate the thoughtful responses of the MO community. Following a definition was taken from Maclachlan and Reid, ``The Arithmetic of Hyperbolic 3-manifolds'', Chapter 8.2: Let $k$ be a number field with exactly one complex place and $A$ a quaternion algebra over $k$ which is ramified at all real places. Furthermore assume $A\subset M_2(\mathbb{C})$ via a complex embedding of $k$. If $\mathcal{O}$ is an order in $A$ and $\mathcal{O}^1$ is the set of elements of $\mathcal{O}$ of determinant 1, then a subgroup $\Gamma$ of $SL(2,\mathbb{C})$ is arithmetic if it is commensurable with $\mathcal{O}^1$ and a subgroup $\Gamma \subset PSL(2,\mathbb{C})$ is arithmetic if it is commensurable with $P\mathcal{O}^1$. If we consider the natural action of $PSL(2,\mathbb{C})$ on the upper-half space model of $\mathbb{H}^3$, a finite volume hyperbolic 3-manifold (or 3-orbifold) $Q \cong \mathbb{H}^3/\Gamma$ is arithmetic precisely when $\Gamma$ is arithmetic. Arithmetic 3-manifolds tend to be atypical many ways. For example, one non-generic property of arithmeticity observed by A.Borel in "Commensurability classes and volumes of hyperbolic 3-manifolds" is the following: for a fixed $V>0$, there are only finitely arithmetic 3-manifolds of volume less than $V$. Along those lines, I was wondering about the following: Question: For a fixed Heegaard genus $g$, are there only finitely many arithmetic manifolds of Heegaard genus $ TITLE: Finiteness as a motivation for compactness QUESTION [8 upvotes]: Another history question, and I am not sure if I will get any answers. (If anyone knows of a good history of math list to use for this question I would be happy for any tips. The one I used to post to is now closed.) This question deals with the motivation for compactness. I wrote a paper on this topic some time back, and one of the reviewers posed a very hard question that I am trying to answer, namely whether finiteness was a motivation for compactness. We know Frechet coined the term compact in 1904. We also know compactness related ideas (sequential, open cover, etc) were around for many decades before this. We also know that in current thinking of compactness, there are quite strong and obvious ties between compactness and finiteness (including the joke, attributed to H. Weyl: what is a compact city? it is a city that can be guarded by finitely many near-sighted policemen!) This topic has been discussed on MO and some examples given in a classic paper of Hewitt. But neither of these discussions gets at the question of historical origin. Does anyone know whether (and how and at what stage and to what extent) finiteness became a motivation for compactness? Is this just an after-the-fact phenomena, or was the idea of finiteness around before compactness became formally defined? Thanks! REPLY [2 votes]: This is not strictly an answer to the history question, but I would like to take this opportunity to record my dissent from the common assertion that compactness is about finiteness. I claim instead that compactness is about covering. Consider the relation $K\subset U$ as a function (predicate) of the argument $U$ ranging over open subspaces. This function is continuous in $U$ (with respect to the Scott topology) iff $K$ is compact. If you haven't heard of the Scott topology before, you can take this equivalence as its definition (assuming that you know the usual one for compactness!). When we also consider membership $x\in U$ as a predicate $\phi(x)$, the containment $K\subset U$ becomes $\forall x:K.\phi(x)$. So a compact subspace is one over which we may perform universal quantification. The dual treatment of the existential quantifier is described, using the novel notion of overtness in this answer by @Andrej Bauer, and the additional one by me tries to relate the idea to familiar ones such as the Newton-Raphson algorithm. In print, compactness and overtness in $\mathbb R$ are explored in the paper by Andrej and me and the further one of mine.<|endoftext|> TITLE: Local geodesics in uniquely geodesic spaces QUESTION [6 upvotes]: A while ago I asked this question in Math Stackexchange. Since I didn't receive an answer so far, I thought I'd ask it here. Suppose $Y$ is a proper length space, where every pair of points $x,y\in Y$ can be joined by a unique length minimizing geodesic (i.e. global geodesic). Can there still be more than one local geodesic joining some two points, or is it necessarily so that the space is also uniquely locally geodesic? If a uniquely geodesic space which is not uniquely locally geodesic exists, can one even cook up Riemannian manifold as an example? REPLY [7 votes]: Without further hypotheses it is easy to cook up a counterexample. Start with a standard torus of revolution, choose a point $p$ on the "outer" circle $\alpha$ and follow this circle until you reach the first conjugate point $q$ from $p$. It is well known that for points $x$ past $q$, the geodesic is no longer minimizing, and in fact there is a pair of geodesics $\beta$ and $\gamma$ joining $p$ to $x$. Now simply "trim" your manifold by removing everything outside the digonal region bounded by $\alpha$ and $\beta$. Here every point is joined to $p$ by a unique geodesic but for points $x$ on $\alpha$ beyond $q$, the geodesic $\alpha$ is no longer minimizing.<|endoftext|> TITLE: Constructing a space with prescribed cohomology ring QUESTION [8 upvotes]: The most general way I can formulate my question is the following: Question 1: Given a Gorenstein quotient ring $S$ of a polynomial ring over a field $K$, can one construct a (topological) space $X$ such that the (even degree part of the) singular cohomology ring of $X$ with coefficients in $K$ is isomorphic to $S$? Edit: as mentioned in the comments, the answer depends on the grading of the variables. I would be most interested in a uniform grading (i.e all variables have the same degree d, without restriction on the value of d). For the specific cases I have in mind, $S$ is a quotient of a polynomial ring over $\mathbb{C}$ by an ideal generated by monomials and binomials. Below is an example of such a ring $S$ that I would like to realize as the cohomology ring of some space: $$ S = \mathbb{C}[x_1,\ldots,x_7]/(x_7^2, x_3x_7-x_4x_7, x_2x_7-x_5x_7, x_1x_7-x_6x_7, x_6^2, x_3x_6-x_5x_6, x_2x_6-x_4x_6, x_1x_6-x_6x_7, x_5^2, x_3x_5-x_5x_6, x_2x_5-x_5x_7, x_1x_5-x_4x_5, x_4^2, x_3x_4-x_4x_7, x_2x_4-x_4x_6, x_1x_4-x_4x_5, x_3^2, x_1x_3-x_2x_3, x_2^2, x_1x_2-x_2x_3, x_1^2).$$ The ring $S$ above is Gorenstein and all rings that I am considering for my question are also Gorenstein (i.e satisfy Poincare duality). Edit 2: I will put this here rather than leaving the additional question in the comments because it goes deeper into what I was really interested in. Question 2: Since the answer to Question 1 seems to be affirmative, can the space X be chosen to be "nice" (e.g. compact manifold) under the assumption that $S$ (the cohomology ring of $X$) satisfies Poincare duality? REPLY [3 votes]: To answer your second question, you can apply the Sullivan-Barge realization theorem. Here is a description and references. The theorem is usually stated over $\mathbb{Q}$, but I see no reason why it shouldn't apply also over $\mathbb{C}$. In the easiest case, if the formal dimension $n$ of your ring $S$ is not a multiple of $4$, then yes, it can be realized as the complex cohomology ring of some closed $n$-manifold.<|endoftext|> TITLE: lower and upper bound for $\sum_{k=1}^n \frac{(-1)^{\Omega(k)}}k$? QUESTION [8 upvotes]: Are there known any lower and upper bounds for $$ \sum_{k=1}^n \frac{(-1)^{\Omega(k)}}k, $$ where $\Omega(n)$ is the number of prime factors counting multiplicities of $n$? Or at least is it known if it is always positive? REPLY [15 votes]: Let $\lambda(n) = (-1)^{\Omega(n)}$ and \[T(x) = \sum_{n \leq x} \frac{\lambda(n)}{n}.\] The conjecture that $T(x) \geq 0$ for all $x \geq 1$ is called Turán's conjecture (though Turán took objection to this labelling, as he did not conjecture that it was true). As Jeremy Rouse stated, Haselgrove disproved this in 1958 using methods of Ingham. The first counterexample occurs at $x = 72\,185\,376\,951\,205$ and was found by Borwein, Ferguson, and Mossinghoff in 2008. This latter paper also proves the fact that \begin{align*} \liminf_{x \to \infty} \sqrt{x} T(x) & < 0, \\ \limsup_{x \to \infty} \sqrt{x} T(x) & > 0, \end{align*} It is not known if \begin{align*} \liminf_{x \to \infty} \sqrt{x} T(x) & = -\infty, \\ \limsup_{x \to \infty} \sqrt{x} T(x) & = \infty, \end{align*} though both are conjectured to be true and, by Ingham's work, would follow from the linear independence hypothesis, which states that the positive ordinates of the imaginary parts of the nontrivial zeroes of the Riemann zeta function are linearly independent over the rationals. As to the true size of $T(x)$, it is hard to say. Unconditionally, it is not difficult to prove that $T(x) = o(1)$ (which is in fact equivalent to the prime number theorem), while the Riemann hypothesis implies that $T(x) = O_{\varepsilon}\left(x^{-1/2 + \varepsilon}\right)$ for every $\varepsilon > 0$. A probabilistic argument of Ng gives a conjectured size of maximal growth of the summatory function of the Möbius function, and the same methods would suggest that \begin{align*} 0 > \liminf_{x \to \infty} \frac{\sqrt{x} T(x)}{\left(\log \log \log x\right)^{5/4}} & > -\infty, \\ 0 < \limsup_{x \to \infty} \frac{\sqrt{x} T(x)}{\left(\log \log \log x\right)^{5/4}} & < \infty. \end{align*}<|endoftext|> TITLE: What is the analogue of simple prime closed geodesic for prime numbers? QUESTION [29 upvotes]: The prime geodesic theorem (of Margulis?) states that on a compact surface of (constant?) negative curvature, the number of prime closed geodesics of length at most $L = \log x$ is approximately $e^L/L = x/\log x$ as $x$ grows. This is commonly viewed as an analogue of the prime number theorem. Reading the report on Mirzakhani's work in relation to her receiving the Fields medal, I learned that she proved that the number of simple (i.e. non-self-intersecting) prime closed geodesics of length at most $L$ is of the order of $L^{6g-6}$, where $g$ is the genus of the surface, as $L$ grows. What would be the analogue of simple prime closed geodesic in the context of prime numbers and what would one expect the analogue of Mirzakhani's result to be? Edit: My colleague Alan Reid suggested to look at the universal cover to distinguish simple from non-simple closed geodesics. Since I want to keep the analogy with arithmetic, instead it's better to look at finite covers. Is it fair to say that simple closed geodesics are more likely to become non-prime in finite covers? That would suggest that "simple" primes would be those more likely to split in finite extensions. Edit 2: My history is sloppy. See GH's comment and Igor's answer. REPLY [14 votes]: First of all, the prime geodesic theorem is due to Sarnak, in Margulis' result the geodesics need not be prime. Secondly, for constant curvature, the result goes back to Huber and Selberg, with precise error terms. Margulis extended this to variable negative curvature (with no error term), and that was sharpened (at least theoretically) by Dolgopyat to show that there is an exponentially decaying error term, but the exponent is not explicit at all. Thirdly, the order of growth result for simple geodesics is due not to Maryam, but to yours truly: %0 Journal Article %A Rivin, Igor %T Simple curves on surfaces %J Geom. Dedicata %V 87 %D 2001 %N 1-3 %P 345--360 %@ 0046-5755 %L MR1866856 (2003c:57018) %R doi:10.1023/A:1012010721583 %U http://dx.doi.org/10.1023/A:1012010721583 Maryam proved an asymptotic result, which (as is not surprising to number theorists) is weaker in practice (my result has explicit bounds, hers does not), but certainly conceptually better (with modern technology - mixing of Teichmuller geodesic flow - one can presumably show the existence of error term a la Dolgopyat, but nothing explicit). Lastly, and getting back to the actual question, the connection between prime numbers and closed geodesics goes via the Selberg zeta function, and its analogy to the Riemann zeta function. So, I will not speak of prime numbers, but the lattice point counting. It turns out that the counting of all closed geodesics and simple closed geodesics is conceptually very similar, despite completely different result that one gets: counting closed geodesics corresponds to counting points of an orbit of a Fuchsian group (for example, $PSL(2, Z)$) in a disk in the hyperbolic plane (that was what Selberg counted initially). The insight is that simple closed curves lie in a finite set of orbits of the mapping class group, and so counting simple closed geodesic reduces to lattice point counting for the mapping class group on Teichmuller space. I am not sure what the number-theoretic analogue would be (that is, arithmetic geometers certainly study moduli, but I am not sure what easy-to-understand things this counts). Notice that the way Huber/Selberg count lattice points in symmetric spaces is via harmonic analysis, and so what is needed for a parallel theory for Teichmuller space is developing tools for analysis on Teichmuller/moduli spaces. Mirzakhani made the first step in this, by combining Greg McShane's amazing identity (or an extension thereof) with a simple idea of Siegel. This allowed her to integrate the constant function. A baby step, to be sure (an anecdote: I applied for an NSF grant once to develop analysis on Teichmuller space. It was turned down as "too narrow").<|endoftext|> TITLE: ICM 2014 streaming video QUESTION [11 upvotes]: Is there a possibility to watch ICM 2014 opening ceremony and the big talks online? I hope there is since it was possible for the previous meeting. REPLY [12 votes]: Perhaps a bit late at this point, but the opening ceremony is being broadcast live on http://www.ebs.co.kr/onair?channelCodeString=tv , with the caveat that so far there's been Korean translations over top of the English from the actual speakers. I think the talks will be uploaded to the ICM's Youtube account later on, as well: https://www.youtube.com/user/ICM2014VOD<|endoftext|> TITLE: Does every mathematics article have a DOI (Digital Object Identifier)? QUESTION [6 upvotes]: Most articles nowadays have DOI's. I am looking for a list of mathematics journals in which some (or all) articles lack this piece of metadata. I don't have access to MathSciNet, but even if I had, a lack of DOI in a MathSciNet entry wouldn't prove that no DOI exists. (Or would it? Maybe MathSciNet really has complete DOI metadata for all of the mathematics literature??) As examples, here are three journals where I suspect that not all articles have DOI's. But how can I know for sure? And what other journals lack DOI's? Journal of Algebraic Geometry Comptes Rendus Theory and Applications of Categories. REPLY [13 votes]: The short answer to your title question is no. To be able to assign DOIs, one has to pay a fee to the International DOI Foundation and not all publishers are willing to pay such fees (especially diamond-model open access publishers where both reading and publishing are free and the journal is run on a volunteer basis). So, for instance, I'm pretty sure that the Electronic Journal of Combinatorics doesn't have DOIs. Most commercial and society publications do have them, and occasional low-budget ones do as well; for instance, the Journal of Graph Algorithms & Applications is a diamond-access volunteer journal but apparently Brown University was willing to cover its fees. It's probably also the case that some journals have DOIs for some of their papers but not all of them, although I don't have examples at hand. If you want to know whether an individual paper has a DOI, you can tell by searching the DOI foundation's guest query service. (If a paper has a formula in its title, you should probably only search for a part of the title without the formula, to avoid encoding issues.) No MathSciNet access is needed. I've several times seen papers that do have DOIs but where the DOI was not listed in MathSciNet, so I don't think their data is complete.<|endoftext|> TITLE: What is the status of the extreme value theorem in forms of constructive mathematics, such as Smooth Infinitesimal Analysis? QUESTION [5 upvotes]: In certain intuitionistic frameworks the extreme value theorem cannot be proved. Depending on the exact framework, counterexamples can be constructed as well; see for example pp. 294-295 in Troelstra, A. S.; van Dalen, D. Constructivism in mathematics. Vol. I. An introduction. Studies in Logic and the Foundations of Mathematics, 121. North-Holland Publishing Co., Amsterdam, 1988. xx+342+XIV pp. ISBN: 0-444-70266-0; 0-444-70506-6 The Smooth Infinitesimal Analysis (SIA) following Lawvere, Kock, J. Bell, and others is based on a category-theoretic framework where the background logic is intuitionistic. What is the status of the extreme value theorem in SIA? REPLY [4 votes]: 1) Yes, in SIA the maximum value $y$ exists. Looking at $R^{[-1,1]}$, the space of all smooth maps, there is a smooth map $\max : R^{[-1,1]} \rightarrow R$ where $y \ge \max(f) \leftrightarrow \forall x\ f(x) \le y$ 2) No, in SIA the maximum need not be attained. Let $a,x,w$ range over $[-1,+1]$. Suppose $\forall f\in R^{[-1,1]} \ \exists x \ \forall w\ \ f(x) \ge f(w)$. Then $\forall a\ \exists x\ \forall w\ \ ax \ge aw$. Then $\forall a\ \exists x\ \forall w\ \ a(x-w) \ge 0$. If $x< 1$, then choose $w=1$, so $x-w<0$, and dividing by $x-w$ gives $a \le 0$. If $x> -1$, then choose $w=-1$, so $x-w>0$, and dividing by $x-w$ gives $a \ge 0$. Since $x< 1 \vee x>-1$, we conclude $a \le 0 \vee a \ge 0$. Since this conclusion does not hold in SIA, the original claim that the maximum is attained can not hold in SIA either.<|endoftext|> TITLE: Saturated Ultrapowers QUESTION [9 upvotes]: I posted it on MSE and didn't receive any comments or answers, so I thought I would post it here. (See https://math.stackexchange.com/questions/895549/keisler-order-saturated-ultrapowers) Keisler's paper "Ultraproducts which are not Saturated" states the following theorem as a corollary to a (much more) generalized theorem. However, I cannot, for the life of me, figure out how to prove it for the specific case. Furthermore, Malliaris's Thesis cites Shelah's Classification Theory for the proof. I looked there and found a discussion of games (which I couldn't quite follow). Theorem: Let $D$ be a regular ultrafilter over $I$ (where $|I| =\alpha$) and let $\mathfrak{A} \equiv \mathfrak{B}$. Then the ultrapowers $\prod_D\mathfrak{A}$ is $\alpha^+$-saturated iff $\prod_D \mathfrak{B}$ is $\alpha^+$-saturated. Motivation: Proving this theorem shows that the Keisler Order is a well-defined order on theories (as opposed to an order on models). Thanks REPLY [4 votes]: W.W. Comfort, S. Negrepontis, The Theory of Ultrafilters, section 13, in particular Theorem 13.7 and Corollary 13.8, might be useful to you. It contains a textbook presentation of the relevant proofs (without games).<|endoftext|> TITLE: Matching number and chromatic number QUESTION [5 upvotes]: If $G$ is a (finite) graph, denote with $\mu(G)$ the size of any maximum matching in $G$ (this number is also called the "matching number" of $G$). For odd integers $n$ we have $n=\chi(K_n) = 2\cdot\mu(K_n) + 1$. Question: is there a non-complete graph $G$ with $\chi(G) = 2\cdot\mu(G) + 1$? REPLY [9 votes]: Yes there are, but they only differ from the complete graph by adding isolated vertices. The proof goes as follows: Let $G$ be a graph with $\chi(G) = 2\cdot\mu(G) + 1$ we will show that it is a complete graph plus some isolated vertices. Consider the vertices of a maximal matching: $x_1, x_2, \ldots , x_{2\mu(G)}$. Since it is a maximal matching the other vertices form an independent set. Let us try to color the graph with $2\mu(G) \mbox{}$ colors. Lets color $x_i$ with color $i$. the remaining independent set can be colored without additional color as long as there is no vertex adjacent to all of the $x_i$. As this would be a contradiction we conclude that there is a vertex $\alpha$ outside of the matching which is adjacent to all the $x_i$. As this is a maximal matching, all the other vertices must be isolated, otherwise we could make a larger matching. So far we have a matching of size $\mu(G)$ and $\alpha$ which is adjacent to every vertex in the matching, and all the other vertices are isolated. We only have to show that the graph induced by these $\mu(G)+1$ vertices is complete. This can be done as we can form another matching of the same size by replacing an $x_i$ with $\alpha$, and the same reasoning says that $x_i$ is adjacent to every vertex of the matching, thus this subgraph is complete. REPLY [3 votes]: We have an odd number of colours. There is an edge between any pair of colours, or else those colours could be combined. So we can easily generate a matching with $\frac{\chi - 1}{2}$ edges. We are given that we cannot add an edge. Take the colour $C$ not currently involved in any matching. Suppose that some vertex $v \in C$ is not adjacent to every other colour. Then let $v$ be adjacent to some vertex of colour $A$, but not to any vertex of colour $B$. In the initial matching, pair $A$ with $B$, to get a matching of $\frac{\chi -1}{2}$. Now delete the edge between $A$ and $B$ and add the $v$ to $A$ edge and one $B$ to $C$ edge. Now the matching number is too big. Hence $v$ does not exist - every vertex in $C$ is adjacent to a vertex of every other colour. Since $C$ was arbitrary, every vertex is adjacent to a vertex of every other colour. Now we simply pick a vertex not yet in the matching and not in the colour $C$ and match it with something in $C$ to break the rule. Hence everything not in $C$ is in the matching, so each colour has only one vertex. But $C$ was chosen arbitrarily, so it too has just one member. Hence the graph is complete. EDIT: another answer just pointed out that isolated vertices can be admitted, which I missed out when I assumed that the vertex $v$ is adjacent to some other vertex (whose colour I called $A$). Apologies.<|endoftext|> TITLE: Integers with a large prime divisor in short intervals QUESTION [11 upvotes]: For an integer $n$, denote by $P^+(n)$ the largest prime divisor of $n$. Then we have the following: There exists some $c>0$, such that for all $x$ sufficiently large the number of integers $n\in[x, x+0.1\sqrt{x}]$ with $P^+(n)>x^{1/2+c}$ is $\geq c\sqrt{x}$. The proof is quite standard: You essentially need a non-trivial bound for $$ \sum_{n\leq x^{1/2+c}}\Lambda(n)\left(\left[\frac{x+0.1\sqrt{x}}{n}\right]-\left[\frac{x}{n}\right]-\frac{0.1\sqrt{x}}{n}\right). $$ Now express $\Lambda$ by Vaughan's identity, approximate $[\cdot]$ by exponential sums, and bound the latter by Weyl-van der Corput. However, while conceptually simple, the computations take at least two pages, which I would rather avoid since a) journal space is precious b) the computations are likely to distract the reader from the more algebraic main arguments, c) the question is so natural that someone probably spent a lot of time on some quantitative version of it. So my question is: Can anyone give me a reference for such a result? Note that for the application the value of $c$ is of no importance, while 0.1 should not be replaced by something much larger. Thank you in advance! REPLY [8 votes]: See the very first paper on the subject by Ramachandra (A note on numbers with a large prime factor; JLMS 1969). Ramachandra's Theorem 1 states that there is a constant $\alpha <1/2$ such that for all large $x$ the interval $[x,x+x^{\alpha}]$ contains an integer with a prime factor larger than $x^{\frac 12+\frac{1}{13}}$. Subsequent works focus on the interval $[x,x+\sqrt{x}]$ and on improving the size of the largest prime factor. It should be noted that in the proof Ramachandra focuses for simplicity on the case $\alpha=\frac 12$, but he does indicate that only small changes to the argument are needed to get some $\alpha<\frac 12$.<|endoftext|> TITLE: orthogonal group in characteristic 2 QUESTION [6 upvotes]: Let $O(2,\mathbb{Z}_2)$ be the orthogonal group of order two matrices. On $\mathbb{Z}_2$ there should exist just one odd quadratic form, hence the stabilizer subgroup $O^-$ of an odd quadratic for should be the whole group. Is it really so or I am confused? On the other hand the even stabilizer $O^+$ should have index 3, so - since the order of $O$ is 6 - I guess $O^+$ is just $\pm Id$. Can someone confirm this? REPLY [5 votes]: Assuming that, by $\mathbb{Z}_2$, you mean the finite field of order $2$, then $$ O_2^{\pm}(q) \cong D_{2(q\mp 1)}$$ where $D_{2(q\mp 1)}$ is the dihedral group of order $2(q\mp 1)$. Taking $q=2$, one obtains that $O_2^+(2)$ is cyclic of order $2$ (although not equal to $\pm I$ you suggest, because $-I=I$), while $O_2^-(2)$ is dihedral of order $6$. One can calculate all this directly, or you can refer to Proposition 2.9.1 of Kleidman and Liebeck's book.<|endoftext|> TITLE: Relationship between fragments of the axiom of choice and the dependent choice principles QUESTION [8 upvotes]: The dependent choice principle ${\rm DC}_\kappa$ states that if $S$ is a nonempty set and $R$ is a binary relation such that for every $s\in S^{\lt\kappa}$, there is $x\in S$ with $sRx$, then there is a function $f:\kappa\to S$ such that for every $\alpha<\kappa$, $f\upharpoonright\alpha R f(\alpha)$. The axiom of choice fragment ${\rm AC}_\kappa$ states that every family of size $\kappa$ has a choice function. There are several classical theorems (see Jech's "Axiom of Choice", chapter 8) concerning the relationship between the dependent choice principles and fragments of the axiom of choice. Theorem 1: Over ${\rm ZF}$, ${\rm AC}$ is equivalent to $\forall\kappa\,{\rm DC}_\kappa$. Theorem 2: Over ${\rm ZF}$, $\forall \kappa\,{\rm AC}_\kappa$ implies ${\rm DC}_\omega$. Theorem 3: It is consistent with ${\rm ZF}$ that $\forall \kappa\,{\rm AC_\kappa}$ holds but ${\rm DC_{\omega_1}}$ fails (theorem 8.9). Theorem 4: It is consistent with ${\rm ZF}$ that ${\rm AC}_\kappa$ holds for some cardinal $\kappa>>\omega$ but ${\rm DC}_\omega$ fails (theorem 8.12). Jech proves theorems 3 and 4 using permutation models (and then discusses how to obtain ${\rm ZF}$-models with the same properties). But I am wondering whether there are direct symmetric model constructions for these two results. Either a reference for the arguments or the arguments themselves would be appreciated. REPLY [7 votes]: The idea is to mimic the permutation models as given in Jech. One can then ask, "Well, in Jech he chooses some set of objects in the full universe, and shows it has a support. But in forcing we don't have a simple access to names like that, since they might not be "sufficiently determined" for us to collect them into a symmetric name!" To counter the effects of this problem here is a generalized formulation of The Continuity Lemma, as Felgner called it (for the basic Cohen model). $\newcommand{\PP}{\Bbb{P}}\newcommand{\dom}{\operatorname{dom}}\newcommand{\fix}{\operatorname{fix}}\newcommand{\sym}{\operatorname{sym}}\newcommand{\forces}{\Vdash}$ Suppose that $\PP$ is a Cohen type forcing, with $p\colon A\times\kappa\to2$ such that the domain of $p$ is $<\kappa$, ordered by reverse inclusion. We write $s(p)$ as the projection of the $\dom p$ onto $A$. Let $\scr G$ be a group of permutations of $A$ acting on $\PP$ naturally: $\pi p(\pi a,\alpha)=p(a,\alpha)$. And let $I$ be an ideal on $A$ which is closed under $\scr G$, and $s(p)\in I$ for all $i\in I$. Moreover, assume that whenever $X,Y\in I$ there is a permutation in $\scr G$ such that $\pi\upharpoonright(X\cap Y)=\operatorname{id}$ and $\pi''(X\setminus Y)$ is disjoint from $Y$. Then whenever $\dot x_1,\ldots,\dot x_n$ are symmetric names with respect to the filter generated by $\{\fix(E)\mid E\in I\}$ and $E\in I$ such that $\fix(E)$ is a subgroup of $\sym(\dot x_i)$ for each $i$, and $p\forces^{\sf HS}\varphi(\dot x_1,\ldots,\dot x_n)$ then $p\upharpoonright(E\times\kappa)\forces^{\sf HS}\varphi(\dot x_1,\ldots,\dot x_n)$. (If you are uncomfortable with the notion of $\forces^{\sf HS}$ you can instead require $\varphi$ to be $\Delta_0$, and replace the relativized quantifiers by names one at a time.) Okay let me explain this for a second, since there are plenty of conditions and plenty of more conditions in the consequences. The idea is that if $p$ forces that something happens in the symmetric model, about concrete symmetric names, then we can restrict $p$ to something whose support is in the ideal, and already this decides the same value for the same statement with the same names. In our two examples, all the conditions will be easy to verify. Theorem I: Let $\kappa$ be a successor cardinal, then it is consistent that $\sf DC_{<\kappa}$ holds, $\sf W_\kappa$ fails, and $(\forall \lambda\in\rm Ord)\sf AC_\lambda$ Proof. We take $\PP$ to be functions from $\kappa\times\kappa\to2$ with domain smaller than $\kappa$. $\scr G$ here is the group of all permutations of $\kappa$ and $I$ is $[\kappa]^{<\kappa}$. So the conditions easily hold, and just to remind you here our filter of subgroups is the one generated by $\{\fix(E)\mid E\in I\}$, and it is normal since $I$ is closed under the operation of $\scr G$, and $\cal F$ is $\kappa$-complete since $I$ is $\kappa$-complete. If $G$ is a generic filter, we let $a_\alpha=\{\beta\mid\exists p\in G: p(\alpha,\beta)=1\}$, and $\dot a_\alpha$ is going to be the canonical name for this set. Additionally, $A$ is the set of all these $a_\alpha$ and $\dot A$ will be its canonical name. Let $N$ be a symmetric model defined by $\cal F$ given above, then by standard arguments $A$ is in $N$. First off, if both the forcing and the filter are $\kappa$-closed, then $\sf DC_{<\kappa}$ holds in the symmetric model. This much is easy to verify (you can look it up in my thesis, or in some of the written notes in my homepage). So we have this for almost free. Secondly, $\sf W_\kappa$ fails since $|A|$ and $\kappa$ are incomparable. This is a standard proof, like the one in Cohen's first model with the Dedekind-finite set of real numbers. The big trick is to show that given a function $X\colon\lambda\to N\setminus\{\varnothing\}$ then we want to find a function $g$ with domain $\lambda$ such that $g(i)\in X(i)$. Suppose that $p\in G$ and $\dot X$ is a hereditarily symmetric name for $X$ such that $p$ forces $\dot X$ has the above properties. Let $E\in I$ be a support for $\dot X$, namely if $\pi\in\fix(E)$ then $\pi\dot X=\dot X$. Without loss of generality $s(p)\subseteq E$ and $|E|^+=\kappa$. Pick some $E'$ disjoint to $E$ and $|E'|=|E|$. We will find a choice function with support $E\cup E'$. For each $\gamma<\lambda$ let $q\leq p$ such that for some symmetric $\dot y_\gamma$ we have that $q\forces\dot y_\gamma\in\dot X(\check\gamma)$. Let $F$ be a support for $\dot y_\gamma$, now we can find some $\pi\in\fix(E)$ such that $\pi''F\subseteq E\cup E'$, define $\dot x_\gamma=\pi\dot y_\gamma$. Now we have that $\pi q\leq p$ (since $\pi p=p$), $E\cup E'$ is a support for $\dot x_\gamma$, and $\pi q\forces\dot x_\gamma\in\dot X(\check\gamma)$. So $\pi q\upharpoonright(E\cup E'\times\kappa)$ also forces that. For each $\gamma<\lambda$ pick a maximal antichain below $p$ of conditions as above (and we can assume they all have domain within $E\cup E'\times\kappa$), $D_\gamma$ and names $\dot x_\gamma(q)$ as above. Then $\{(q,(\check\gamma,\dot x_\gamma(q))^\bullet)\mid q\in D_\gamma,\gamma<\lambda\}$ is a choice function and $E\cup E'$ is clearly a support for it. $\square$ Theorem II: If $\kappa$ is uncountable, then it is consistent that $\sf DC$ fails, $\sf W_{<\kappa}$ and $\sf AC_{<\kappa}$ both hold. Proof. Let me skimp out on most of the details. We take $A$ in this case to be $\kappa^{<\omega}$ (you can replace $\omega$ here by the least cardinal for which you want $\sf DC$ to fail). Our automorphism group is going to be the automorphisms of the tree $\kappa^{<\omega}$ and the ideal of supports is the ideal of subtrees of cardinality less than $\kappa$ with no branches (in the case of $\sf DC$ these are really the subtrees which are well-founded). For $t\in\kappa^{<\omega}$ define $a_t$ as the Cohen set defined when fixing $t$ and $A$ as the set of all $a_t$'s. Then the structure of $\kappa^{<\omega}$ is fixed trivially by the automorphisms, so $A$ has a tree structure but no branches (since a branch would require a support with an unbounded tree). Therefore $\sf DC$ fails. To show that $\sf W_\lambda$ or $\sf AC_\lambda$ hold, for $\lambda<\kappa$, we perform a trick similar to the previous proof. However here we need to be slightly more careful. But we can also notice that the union of trees whose intersection is without branches is also without branches. Therefore the union of any less than $\kappa$ "almost disjoint" supports is a support. So here we take a name and by induction we construct a sequence of conditions and names which witness $\sf W_\lambda$ or $\sf AC_\lambda$. Simply by ensuring that the next name we take has a support which extends the previously chosen names "sideways" and not "up". This will guarantee that the union of the symmetric names for the functions at limit steps is a function. And again the generalized continuity lemma ensures we can always restrict back to smaller conditions as we progress, to ensure that their support is in the ideal.<|endoftext|> TITLE: Quasi-affineness of the base of a $\mathbb{G}_a$-torsor QUESTION [8 upvotes]: Let $\mathbb{G}_a$ be the additive group over an algebraically closed field $k$ of any characteristic. Let $X \to Y$ be a $\mathbb{G}_a$-torsor of $k$-schemes (of finite type - in case that is relevant). Suppose that $X$ is quasi-affine. Is it true that $Y$ is quasi-affine? EDIT: As David pointed out below, one should also assume that $Y$ is separated. ADD: Maybe I should add an explanation. The analogue assertion is not true for "affine" instead of "quasi-affine". The standard counterexample is $X = SL_2$ and $Y = SL_2/U$, where $U$ is the subgroup of unipotent upper triangular matrices ($Y$ is the $\mathbb{G}_m$-torsor corresponding to the tautological line bundle over $\mathbb{P}^1$). On the other hand, it is true that if $G$ is an affine group scheme of finite type and $U$ is a unipotent subgroup, that $G/U$ is always quasi-affine. In fact the quotient is quasi-affine if and only if $U$ is obervable in $G$, and $U$ is observable if it has no non-trivial characters (Referenz (not original): W. Ferrer Santos, A. Rittatore: Actions and Invariants of Algebraic Groups, Chapman & Hall (2005)). REPLY [3 votes]: It's false! Take $\mathbb{A}^3$ with coordinates $(x,y,z)$. Blow up the origin, and let $E$ be the exceptional divisor. Delete the line $E \cap \{ z=0 \}$. The resulting quasi-projective variety will be our $Y$. Since $Y$ is quasi-projective, it is separated. $Y$ is an example of a quasi-projective variety with no complete curves that is not quasi-affine. I wrote an earlier answer showing that any $Y$ must have no complete curves and suggesting that might imply $Y$ quasi-affine; I have now deleted that answer since it was a dead end except to point me towards which $Y$ to investigate. First, let us see that $Y$ is not quasi-affine. Since $B \ell_{(0,0,0)} \mathbb{A}^3$ is smooth, and $Y$ is obtained by deleting a codimension $2$ locus from it, any global function on $Y$ extends to $B \ell_{(0,0,0)} \mathbb{A}^3$. But any function on $B \ell_{(0,0,0)} \mathbb{A}^3$ contracts $E$. So any global function on $Y$ contracts $E \setminus \{ z=0 \}$, and global functions on $Y$ do not separate points. Now, we build our torsor. We will cover $Y$ with two open charts: $U$ is the complement of the proper transform of $\{ z = 0 \}$. We have $U \cong \mathbb{A}^3$, with coordinates $(x z^{-1}, y z^{-1}, z)$; we define $p = x z^{-1}$ and $q = y z^{-1}$. The other chart will be $V:= Y \setminus E$. So $V \cong \mathbb{A}^3 \setminus \{ (0,0,0) \}$. The open set $V$ is not affine, but the global functions on $V$ are $k[x,y,z]$. Note that $U \cap V \cong \mathbb{A}^2 \times (\mathbb{A}^1 \setminus \{ 0 \})$, with coordinate ring $k[x,y,z^{\pm}]$. Take $U \times \mathbb{A}^1$ and $V \times \mathbb{A}^1$, with coordinates $u$ and $v$ on the respective $\mathbb{A}^1$ factors, and let $\mathbb{G}_a$ act by translation on each $\mathbb{A}^1$. Glue these trivial torsors by $v = u + z^{-1}$; this makes sense since $z^{-1} \in \mathcal{O}(U \cap V)$. This will be $X$. We write $\pi$ for the map $X \to Y$ and $\psi$ for the map $Y \to \mathbb{A}^3$, contracting $E$. To repeat, $X = \pi^{-1}(U) \cup \pi^{-1}(V)$. We have $\pi^{-1}(U) = \mathrm{Spec}\ k[p,q,z,u]$ and $\pi^{-1}(V) = \mathrm{Spec}\ k[x,y,z,v] \setminus \{ x=y=z=0 \}$ To show that $X$ is quasi-affine, consider the following $7$ functions: $$pz=x,\ qz=y,\ z,$$ $$a:=pzu+p=xv,\ b:= qzu+q = yv,\ c:=zu+1=zv,$$ $$d := zu^2+u = z v^2 - v .$$ For each function, I have given one formula which displays it as an element of $\mathcal{O}(\pi^{-1}(U))$ and another which displays it as an element of $\mathcal{O}(\pi^{-1}(V))$, so these are global functions on $Y$. Let $R$ be the ring generated by these functions. I claim that the map from $X$ into $\mathrm{Spec}(R)$ is an embedding (at which point it is an open embedding, since both varieties have dimension $4$). We first check that the map separates points. Suppose that $\alpha_1$ and $\alpha_2$ are sent to the same point of $\mathrm{Spec}(R)$. Since $(x,y,z) \in R$, we see that $\psi(\pi(\alpha_1)) = \psi(\pi(\alpha_2))$. First, suppose this common point is not $(0,0,0)$. So $\alpha_1$ and $\alpha_2$ are in $\pi^{-1}(V)$ and are of the form $(x,y,z,v_1)$ and $(x,y,z,v_2)$. But we then have $x v_1 = x v_2$, $y v_1 = y v_2$ and $z v_1 = z v_2$, and $x$, $y$ and $z$ cannot all be $0$ on $V$, so $\alpha_1 = \alpha_2$ in this case. Now, assume that $\psi(\pi(\alpha_1)) = \psi(\pi(\alpha_2)) = (0,0,0)$. So $\alpha_1$ and $\alpha_2$ are in $U$, and are of the form $(p_1, q_1, 0, u_1)$ and $(p_2, q_2,0,u_2)$. But then $p_1 \cdot 0 \cdot u_1 + p_1 = p_2 \cdot 0 \cdot u_2 + p_2$, $q_1 \cdot 0 \cdot u_1 + q_1 = q_2 \cdot 0 \cdot u_2 + q_2$ and $0 \cdot u_1^2 + u_1 = 0 \cdot u_2^2 + u_2$, so we again conclude $\alpha_1 = \alpha_2$. This shows that $X \to \mathrm{Spec}(R)$ is injective on points, but we want to know it is an embedding of schemes. Now that we know injectivity on points, we just need to do a local computation. The following table gives inverses on each of the open sets $\{ z \neq 0 \}$, $U \cap \{ zu+1 \neq 0 \}$, $V \cap \{ x \neq 0 \}$ and $V \cap \{ y \neq 0 \}$; we leave it to the reader to check that these sets form a cover. $$\begin{array}{@{\mbox{On}\ }r@{,\ }rcl} \{ z \neq 0 \} & (x,y,z,u) &=& (x,y,z, (c-1) z^{-1}) \\ U \cap \{ zu+1 \neq 0 \} & (p,q,z,u) &=& (ac^{-1}, b c^{-1}, z, d c^{-1}) \\ V \cap \{ x \neq 0 \} & (x,y,z,v) &=& (x,y,z,a x^{-1}) \\ V \cap \{ y \neq 0 \} & (x,y,z,v) &=& (x,y,z,b y^{-1}) \\ \end{array}$$ I suspect that a complete list of relations for $R$ is $$\mathrm{rank} \begin{pmatrix} a & b & c & d \\ x & y & z & c-1 \end{pmatrix} \leq 1$$ and I suspect that the image of $X$ in $\mathrm{Spec}(R)$ is everything except the line $\{ a=b=c=x=y=z=0 \}$. (On this line, $d$ is unconstrained.) Written this way, the $\mathbb{G}_a$ action is $$\begin{pmatrix} 1 & t \\ & 1 \end{pmatrix} \begin{pmatrix} a & b & c & d \\ x & y & z & c-1 \end{pmatrix} \begin{pmatrix} 1 & & & \\ & 1 & & \\ & & 1 & t \\ & & & 1 \end{pmatrix}.$$ But I doubt I will get around to checking this.<|endoftext|> TITLE: All Kähler metrics on a complex manifold? QUESTION [7 upvotes]: Let $M$ be a complex manifold of complex dimension 2. What do we know about the set all Kähler metrics on $M$ in general and in the case of 4-torus $C^2/Z^4$? For the case of surfaces ($dim_C=1$), any compatible metric is Kähler and by the uniformization theorem, the answer is that every two such metrics are conformally equivalent and the set all Kähler metrics is nonempty. REPLY [8 votes]: adding my comment as an answer In general, Kahler metrics in $[ω_0]$ can also be parametrised as metrics of the same volume conformally equivalent to $ω_0$ by $$\{\varphi\in C^\infty(X,\mathbb R)|\; \int_Xe^\varphi\omega_0^n=\int_X\omega_0^n=vol(X,[\omega_0]) \}$$ Moreover if two metric be comformally equivalent $\omega_\varphi^n=e^u\omega_0^n$ then conformal factor and Kahler potential are related by $(1+\Delta_{\omega_0}\varphi)=e^u$ In Mirror symmetric language If $X$ and $\hat X$ be mirror to each other and be CY, then the Kahler moduli space of $\hat X$, denoted by $\mathcal M_{kah}(\hat X)$ can be identified with $K_\mathbb C(\hat X)/Aut(\hat X)$ where $$K_\mathbb C(\hat X)=\{\omega\in H^2(\hat X,\mathbb C)|Im(\omega)\in K(\hat X)\}/im H^2(\hat X,\mathbb Z)$$ We can identify the moduli space of Kahler spaces of $\hat X$ with moduli space of complex space of $X$ via Yukawa couplings See Moduli Spaces of Hyperkahler Manifolds and Mirror Symmetry by Daniel Huybrechts<|endoftext|> TITLE: Dropping altitudes to achieve nonobtuse planar triangulations: finite or infinite? QUESTION [7 upvotes]: Given a planar triangulation of (say) a convex region, imagine the following process to convert it to a triangulation with no obtuse angles: Pick an arbitrary obtuse angle at vertex $a$ of $\triangle abc$ and drop an altitude $ax$ from $a$ to the altitude foot $x$ on $bc$. If $x$ also lies in another triangle $\triangle dcb$, then add the segment $dx$ to regain a proper triangulation. Repeat. An example is shown below. The triangulation has two obtuse angles, at vertices $3$ and $4$. The left sequence starts by splitting the angle at vertex $3$ with altitude $35$, next adding segment $15$, then dropping altitude $46$ onto $15$, and so on as illustrated. The sequence depicted never terminates. The right sequence starts by splitting the angle at vertex $4$ with altitude $45$, and quickly terminates.   My question is: Q1. Is it the case that, for any planar triangulation of (say) a convex region, there is some ordering of the splittings that terminates in a finite number of splittings? Or is there a triangulation for which every splitting sequence never terminates? Q2. If the answer to Q1 is Yes (there exists some terminating sequence): Is there an algorithm (more efficient than try-all-possibilities) that finds such a terminating sequence? Incidentally, I know other methods of obtaining a nonobtuse triangulation. Here I am concentrating on this simple procedure. REPLY [7 votes]: There is an example for a triangulation with just one obtuse triangle in each step. In the image, only triangle $A_1A_2O$ is obtuse. With your algorithm we draw altitudes and reach $H_1,H_2,H_3,...$ and always we have only one obtuse triangle (in step i, triangle $A_rH_iO$ where $r$ is the remainder of $i+2$ modulo $6$), so we are forced to draw it's altitude. Red segments are the altitudes and green segments are the added segments to regain a proper triangulation. So there is a triangulation that never terminates. Now I introduce an algorithm for a triangulation without any cycle in its adjacency graph. Consider the adjacency graph of the initial triangulation. (Vertices are triangles and edges are between two triangles with common segment) If this graph is a tree, i.e. there is not a hole in the region or a vertex with 360 degrees (like $O$ in the image) in the triangulation, then we can choose the order of splitting such that the algorithm terminates. Here is the algorithm: Choose an obtuse triangle ($abc$) and drop its altitude ($ax_1$), add the other segment ($dx_1$), now probably $x_1$ is the obtuse vertex of a new triangle namely $dx_1c$, drop the altitude form $x_1$ in the new triangle ($x_1x_2$), now continue with $x_2$ and drop its altitude, and continue dropping altitudes from new vertices, you never come back to a segment because there is no cycle in the graph, so this procedure ends, now ($abc$) is done and the number of obtuse triangles are less than the the number of initial ones. Repeat this procedure for each obtuse triangle until you are done.<|endoftext|> TITLE: In a fibration, can a deformation retraction of the base be lifted to the total space? QUESTION [10 upvotes]: Given a fibration $p:E \rightarrow B$ and if $A$ is a deformation retract of $B$. Is it true that $p^{-1}(A)$ is a deformation retract of $E$?. If this is not true, can some conditions be imposed on $p$, $E$ or $B$ to make that statement true?. If so, the theorem is still valid for a strong deformation retract? REPLY [4 votes]: There is another situation when we can lift a strong deformation retraction (SDR). If your fibration $p:E\to B$ is a discrete covering ($\pi^{-1}(b)$ has the discrete topology for every $b\in B$), then no additional assumptions are needed. Indeed, by definition, the fibration $p$ has the homotopy lifting property from arbitrary spaces. To be concrete, in the diagram, we can lift the the bottom map $H$ to the dashed arrow $\tilde{H}$, so that $H\circ(id_{I}\times p)=p\circ\tilde{H}$: $$ \begin{eqnarray*} I\times E & \stackrel{\tilde{H}}{\dashrightarrow} & E\\ id_{I}\times p\downarrow\quad & & \downarrow p\\ I\times B & \stackrel{H}{\to} & B \end{eqnarray*} $$ In general, $\tilde{H}$ is only a weak retraction. However, when the fibers of $p$ are discrete, and $H$ is an SDR to $A\subset B$, then $\tilde{H}$ also an SDR from $p^{-1}(B)=E$ to $p^{-1}(A)$. To prove this, using $H(t,a)=a$ for all $t$, and all $a\in A\subset B$, we get: $$ \tilde{H}(t,c)\in p^{-1}(H\circ(id_{I}\times p)(t,c))=p^{-1}(H(t,p(c)))=p^{-1}(p(c)). $$ Hence, $\tilde{H}(t,c)=c$, for all $c\in p^{-1}(a)$, as wanted, by discreteness and continuity.<|endoftext|> TITLE: The ten martini problem - reason for name QUESTION [6 upvotes]: Why is the problem called the ten martini problem? Sounds like an interesting name for people who drink. REPLY [8 votes]: The name was coined by Barry Simon in this 1982 article (page 487): The Ten Martini Problem: The almost Mathieu operator has a Cantor spectrum. The name comes from the fact that Mark Kac* has offered ten martinis to anyone who solves it. [...] Actually, Kac said "has all its gaps there", so perhaps one should solve instead: The Ten Martini Problem: (Strong Form, or should it be Dry Form)... [*] Marc Kac, public communication at 1981 AMS Annual Meeting.<|endoftext|> TITLE: Uniruled degenerations of abelian varieties QUESTION [5 upvotes]: Suppose I have a smooth projective variety $X$ over $\mathbb{C}$ with $K_X$ semiample, and consider the fiber space $f:X\to Y$ given by $|\ell K_X|$, for some $\ell>0$ large, where $Y$ is a normal projective variety of general type. Assume that the general fiber of $f$ is an Abelian variety, I would like to say that either there is a component of some singular fiber of $f$ which is uniruled, or otherwise all singular fibers (if any) are just smooth Abelian varieties with multiplicity $m>1$. If $\mathrm{dim} X=2$ this statement is true thanks to Kodaira's classification of singular fibers of elliptic surfaces. In this case the singular fibers are either $mI_0, m>1$, or else have a component which is a rational curve. I am looking for a higher-dimensional analog of this statement, any reference (or counterexample) is greatly appreciated. REPLY [3 votes]: Some more comments: If you allow semi-stable reduction and run Minimal Model program then we have the following result due to Fujino which can be applied to semi-stable degenerations of Abelian varieties, Calabi-Yau varieties Let $f : X \to Y$ be a proper surjective morphism from a smooth quasi-projective variety $X$ to a smooth quasi-projective curve $Y$ with connected fibers. Let $P \in Y$ be a point. Assume that $Supp f^*P$ is a simple normal crossing divisor on $X$ and $f$ is smooth over $Y \setminus P$. We further assume that $K_{f^{-1}Q} \cong 0$, for every $Q\in Y\setminus P$. Then there exists a sequence of flips and divisorial contractions $$X=X_0 \to K_{X_1} \to K_{X_2} \to \cdots \to K_{X_k} \cdots \to K_{X_m}$$ over $Y$ such that $X_m$ has only $\mathbb Q$-factorial terminal singularities and $K_{X_m} \cong_{Q,Y}0$. Let $S=Supp f^*_mP $ be the special fiber of $f_m : X_m \to Y$ . If $S$ is reducible, then every irreducible component of $S$ is uniruled. If $S$ is irreducible, then $S$ is normal and has only canonical singularities if and only if $S$ is not uniruled. Note that $K_S\cong_ Q 0$ when $S$ is irreducible and has only canonical singularities.<|endoftext|> TITLE: Does OCA imply $2^{\aleph_0}=\aleph_2$? QUESTION [7 upvotes]: Is it known whether Todorcevic's Open Coloring Axiom implies $2^{\aleph_0}=\aleph_2$? The only consistency proofs for OCA that I know are the following: 1) PFA implies OCA (and also $2^{\aleph_0}=\aleph_2$). 2) Constructing a finite support ccc iteration of length $\omega_2$ starting from a model satisfying CH. (CH at intermediate stages is needed to ensure the ccc of the forcing notions, so we end with $2^{\aleph_0}=\aleph_2$). 3) Starting from a model of OCA and forcing with a Suslin tree (which does not add reals). On the other hand, I know that MA+OCA imply $2^{\aleph_0}=\aleph_2$, as well as OCA plus the Abraham-Rubin-Shelah version of OCA. The last fact is proved in Moore, J.T. Open colorings, the continuum, and the second uncountable cardinal, Proceedings of the American Mathematical Society, 130 (2002), n 9, pp. 2753-2759. where it is also asked whether OCA implies $2^{\aleph_0}=\aleph_2$. So my question is if this question has been answered since 2002. REPLY [7 votes]: Whether $OCA$ implies $2^{\aleph_0}=\aleph_2$ appears as problem 7.2 in J.T. Moore´s The Proper Forcing Axiom in the Proceedings of the ICM, 2010. Apparently it was still open by the end of 2012 according to T. Yorioka´s abstract for the Fields Institute Set Theory seminar.<|endoftext|> TITLE: Construction of the Lie functor: left vs. right invariant vector fields on Lie groups and Lie groupoids QUESTION [14 upvotes]: When constructing the Lie algebra $L(G)$ of a Lie group $G$, one usually uses the identification of the tangent space $T_1 G$ with left invariant vector fields $\mathcal{V}^l(G)$ to construct the Lie bracket on $L(G)=T_1 G$. However, in the literature on Lie groupoids and Lie algebroids some authors use left-invariant and some use right-invariant vector fields (and I think that right-invariant vector fields are more natural here). There are several reasons for using one convention or the other, for instance using the right invariant vector fields yields the usual Lie bracket on the vector fields, which are the sections of the Lie algebroid of the pair groupoid. This can be rephrased by saying that the natural action of the bisections (which are just diffeomorphisms in the case of the pair groupoid) yield a Lie algebra homomorphism from the right invariant vector fields on the Lie groupoid to the vector fields on the manifold. However, in finite-dimensional Lie theory one often wants to make Lie algebra elements act on functions on the Lie group, which then involves a $^{-1}$ in the left regular representation and thus $\mathcal{V}^l(G)$ does act naturally on $C^\infty(G,\mathbb{R})$. This has the unfortunate effect, that if we consider $G$ as a Lie groupoid with one object $(G\rightrightarrows *)$, then the Lie bracket on its Lie algebroid $L(G\rightrightarrows *)$ is not the same as the Lie bracket on the Lie algebra $L(G)$, considered as a Lie algebroid $L(G)\to *$. Now I have two questions on this: What are further reasons for using one or the other convention (for instance, calculations that are substantially easier in one or the other)? In your opinion, is this is a historical accident that one should stick to and endure the resulting signs or do you see good reasons for breaking with the conventions in favour of a unified construction of the Lie functor on Lie groups and Lie groupoids? Note: One cannot simply use left-invariant vector fields on Lie algebroids, what is frequently done in the literature. One would also have to change the way how these objects act naturally, for instance that diffeomorphisms act on a manifold naturally from the left by evaluation. REPLY [13 votes]: My view is that one needs both, left and right invariant vector fields. Some reasons: For $X\in\mathfrak g=T_eG$ the left invariant vector field $L_X$ has a flow consisting of right translations: $$ \text{Fl}^{L_X}_t(x) = x.\exp(tX) $$ and conversely. For a left action $\ell:G\times M\to M$ we need the right invariant vector field $R_X$ so that $R_X\times 0_M$ on $G\times M$ is $\ell$-related to the fundamental vector field (infinitesimal left action by $X$) $\zeta_X$ on $M$. Etc. The situation is very transparent for the left action of the diffeomorphism group $\text{Diff}(M)$ on a compact manifold $M$ itself. The Lie algebra of the diffeomorphism group is the space $\mathfrak X(M)$ of smooth vector fields on $M$ with the negative of the usual Lie bracket as bracket. Here one sees, that both left and right make their appearance at the same time. As for mechanics of a rigid body with configuration space a Lie group, as well as for continuum mechanics of a fluid, say, with configuration space the group of (volume preserving) diffeomorphisms, one uses both trivializations: Let $G$ be the group in both cases, with $\mu:G\times G\to G$ the multiplication, $\mu(x,y)=x.y=\mu_x(y)=\mu^y(x)$. Let $g(t)$ be a curve in the group acting on a material point $x$ in space by $g(t).x$. Then: The material or Lagrangian velocity is in the tangent bundle of group itself: $\partial_t g(t)\in TG$ The spatial or Eulerian velocity is right trivialized in the Lie algebra, the right logarithmic derivative: $\partial_tg(t).g(t)^{-1} = T(\mu^{g(t)^{-1}})(\partial_tg(t))$. The body or convected velocity is left trivialized in the Lie algebra, the left logarithmic derivative: $g(t)^{-1}.\partial_tg(t) = T(\mu_{g(t)^{-1}})(\partial_tg(t))$. All 3 are used: see 15.2 of `Marsden, Ratiu: Introduction to Mechanics and Symmetry'<|endoftext|> TITLE: Is a projective module of constant finite rank finitely generated? QUESTION [8 upvotes]: If $R$ is a (commutative) ring and $P$ is a projective $R$-module, then every localization of $P$ at a prime of $R$ is free by Kaplansky's theorem, and has a well-defined rank. If these ranks are all finite, must $P$ be finitely generated? The answer is no: one can take $R=k[x_1,x_2,\dots]/(x_1^2-x_1,x_2^2-x_2,\dots)$ and $P=\bigoplus_{n=1}^\infty R/\bigl(1-x_1x_2\cdots x_{n-1}(1-x_n)\bigr)$; then $P$ is projective, not finitely generated, and at each prime at most one of its summands is nonzero. At each prime $\mathfrak{p}$ of $R$, then, $P_\mathfrak{p}$ is a free rank-$1$ $R_\mathfrak{p}$-module, unless $\mathfrak{p}=(1-x_1,1-x_2,\dots)$, for which $P_\mathfrak{p}=0$. Or here's another example: let $R$ be the ring of continuous functions $[0,1]\to\mathbb{R}$, and let $P\subseteq R$ be the ideal consisting of those functions that vanish on a neighborhood of zero. Then again $P_\mathfrak{p}$ is free of rank 1 at every prime $\mathfrak{p}$ of $R$, except for those $\mathfrak{p}$ consisting of functions vanishing at $0$; at such $\mathfrak{p}$ we have $P_\mathfrak{p}=0$. Such examples must be reasonably complicated due to this result of Bass's: Proposition 4.2. Suppose $R$ has only finitely many primes minimal above $0$... If $P$ is a locally finitely generated projective $R$-module, then $P$ itself is finitely generated. ("Big Projective Modules are Free", 1963) However, I'm wondering if it's possible to remove the conditions on $R$ if we assume not only that $P$ is projective, but that its rank is a continuous function $\mathrm{Spec}(R)\to \mathbb{N}$. Since continuity of the rank function is equivalent to the rank being locally constant, we can work locally on affine opens where the rank is constant. Thus the question reduces to: If $P$ is a projective $R$-module of constant finite rank, is $P$ finitely generated? Thanks for your input. REPLY [7 votes]: For any commutative ring $R$ with 1, any projective $R$-module of constant finite rank is finitely generated. This is the content of Exercises I.2.13 and I.2.14 of Weibel's $K$-book. The argument goes as follows (all the relevant hints are in Weibel's book), I hope I did not introduce any tacit finiteness assumptions. 1) In the constant rank one case, we have the dual $\check{P}=\operatorname{Hom}_R(P,R)$ and the evaluation map $\operatorname{ev}:\check{P}\otimes_R P\to R$ whose image $\tau_P=\operatorname{im}\left(\operatorname{ev}:\check{P}\otimes_RP\to R\right)$ is called the trace. By a result of Kaplansky, $P_{\mathfrak{p}}$ is free for any prime ideal $\mathfrak{p}\subseteq R$, and we can choose a generator $x\in P_{\mathfrak{p}}$. Then $\check{x}\otimes x\mapsto 1$ under $\check{P_{\mathfrak{p}}}\otimes_{R_{\mathfrak{p}}}P_{\mathfrak{p}}\to R_{\mathfrak{p}}$. In particular, $\tau_P\not\subseteq\mathfrak{p}$ since otherwise $P\otimes_R R/\mathfrak{p}=0$ contradicts the constant rank one assumption (as pointed out in the comments of Owen Biesel). Since $\tau_P$ is an ideal, it contains $1$ and we can write $1=\sum f_i(x_i)$. For each $\mathfrak{p}$, some $x_i$ will be a generator of $P_{\mathfrak{p}}$, so the $x_i$ are generators for $P$. 2) If $P$ has constant rank $r$, then $\bigwedge^r P$ has constant rank one and by Step 1 has finitely many generators. These are of the form $\sum a_i y_{1i}\wedge\cdots\wedge y_{ri}$ and the (finitely many) $y_{ij}$ appearing generate $P$.<|endoftext|> TITLE: Levenberg's original article "A method for the solution of certain problems in least squares" QUESTION [5 upvotes]: Does there exist any digital copy of the original article (or a transcript) K. Levenberg, A method for the solution of certain problems in least-squares, Quart. Appl. Math. 2 (1944): 164-168? It is (obviously!) cited in numerous works, but the paper itself appears to have completely vanished. There seems to be not even any publisher, library, etc. proposing to buy a hard copy or anything alike. Apart from the citations all over the places, the paper has ceased to exist?! (The question has also been posted on Mathematics SE) REPLY [6 votes]: This link might be better. The paper is now open access and the pdf is searchable.<|endoftext|> TITLE: Duration and critical groups order in sandpile models and chip firing games QUESTION [5 upvotes]: The famous chip firing game (which is closely related to sandpile models) goes like this: Place chips at the vertices of a graph. REPEATEDLY: If a vertex $v$ of degree $d_{v}$ has at least $d_{v}$ chips, then it "fires" a chip along each of its incident edges. If there is vertex which can fire, the game is over. Remarkably, whether a game terminates only depends on the graph and the initial configuration of chips (and not on the vertices chose to fire during the game). Even more remarkably, the length of the game (the number of fired vertices) does not depend on the actual play but rather only on the graph and the initial configuration. Tardos and Bjorner-Lovasz-Shor have given well-known bounds for the length of the game. On the other hand, there is a vast literature which studies the "critical group" (aka "sandpile group") of a graph, which is strictly speaking for a slightly different but related game (the so-called dollar game). QUESTION: I am wondering if there is some connection between the duration of the chip firing game and the order of the critical group. This is based on the vague intuition that the larger the group, the more opportunities the game has to terminate and so it can be expected to end rather sooner than later. But I may be completely off the mark here (for instance, I am mixing both games in my reasoning...). Any thoughts on this highly appreciated. REPLY [2 votes]: Here is an example that ought to convince you that no bound like this can exist. Let $\ell(G,N)$ (for $N$ sufficiently small) denote the maximum length of a game on the graph $G$ that terminates and involves $N$ chips. Let $P_n$ denote the path graph on $n$ vertices and $S_n$ the star graph on $n$ vertices. It's not hard to see that $\ell(S_n,N) = N$ while $\ell(P_n,N) = \Theta(N^2)$. However, because they are trees, both $P_n$ and $S_n$ have trivial critical groups.<|endoftext|> TITLE: Distribution of zeroes of lacunary functions QUESTION [11 upvotes]: In a recent Math Stack Exchange question I asked about the function $$f(z)=\sum_{n=0}^\infty z^{2^n},$$ and was informed of its status is a canonical example of a lacunary series with natural boundary at $|z|=1$. A phenomenon observed by the accepted answer was that this function has a multitude of zeroes within the unit disk; it was speculated but not proven that that this set is in fact infinite. That raises the following questions, for which I've not been able to find appropriate literature: Does $f(x)$ have an infinitude of zeros within the unit circle? How can this be proven? How are the zeros distributed? (e.g. how many zeros are found within an annulus $0 TITLE: For integers $a \ge b > 1$ is $f(a,b) = a^b + b^a$ injective? QUESTION [32 upvotes]: Given two integers $a \ge b >1$, can we encode them as a unique integer $a^b + b^a$? I asked this question on math.SE, and after surviving a week with a bounty, it seems that this question is harder than I initially thought. Apparently these things have been named Leyland Numbers, but none of the literature I've been able to find on them provides proof that there are no repeats. REPLY [24 votes]: I will show that if we assume the $abcd$ conjecture (which is the case $n=4$ of Browkin and Brzezinski's $n$-conjecture that generalizes the $abc$ conjecture), then $a^b+b^a=c^d+d^c$ has only finitely many solutions with $\{a,b\}\neq \{c,d\}$. The $abcd$ conjecture claims that if $a,b,c,d$ are integers with $a+b+c+d=0$ and nonzero subsums (i.e. $a+b+c\neq 0$ etc.), $\gcd(a,b,c,d)=1$ and $\varepsilon>0$, then $$\max\{|a|,|b|,|c|,|d|\}\leq K_{\varepsilon} \text{rad}(abcd)^{3+\varepsilon}$$ (it is enough for us to assume the conjecture with any absolute constant in place of $3+\varepsilon$). Here $\text{rad}(m)$ is the product of the prime divisors of $m$. Given that Mochizuki seems to have proved the $abc$ conjecture and some generalizations of it, perhaps the $abcd$ conjecture is not that distant an assumption. If $a^b+b^a=c^d+d^c$ and $\gcd(a,b,c,d)=1$, we obtain $$\max\{a^b,b^a,c^d,d^c\}\leq K_{\varepsilon}\text{rad}(a^bb^ac^dd^c)^{3+\varepsilon}=K_{\varepsilon}\text{rad}(abcd)^{3+\varepsilon}\leq K_{\varepsilon}(abcd)^{3+\varepsilon},$$ unless some subsum of $a^b+b^a-c^d-d^c=0$ is zero, which would give $a^b=c^d$ and $b^a=d^c$ (or vice versa), but then $a$ and $c$ have the same prime factors, and writing $a=\prod_{i=1}^s p_i^{\alpha_i}, b=\prod_{i=1}^s p_i^{\beta_i}$, we see from $a^b=c^d$ that $\frac{\alpha_i}{\beta_i}=\frac{d}{b}$, which is independent of $i$, so $a\mid c$ or $c\mid a$. Similarly $b\mid d$ or $d\mid b$. After this it is easy to see that $a^b=c^d$, $b^a=d^c$ has no nontrivial solutions. In fact, if for instance $d=kb,a=\ell c$, then after simplification the equations become $\ell=c^{k-1},k=b^{\ell-1}$, so $k\geq 2^{\ell-1}$ and then $\ell\geq 2^{2^{\ell-1}-1}$. Hence $\ell=2$ and similarly $k=2$ (or $\{a,b\}=\{c,d\}$), but then $b=c$, and thus $a=d$. Now we may assume that the subsums are nonzero. Choose $\varepsilon=1,$ say, and let $d=\max\{a,b,c,d\}$. Then $2^d\leq c^d\leq K_1\text{rad}(abcd)^4\leq K_1\cdot d^{16}$, so $d$ is bounded by an absolute constant, and hence $a,b,c,d$ are all bounded. Now assume $r=\gcd(a,b,c,d)>1$. The next step is to show that $r$ is bounded. Now $a^b+b^a=c^d+d^a$ is of the form $x^r+y^r=z^r+w^r$, where $x=a^{\frac{b}{r}},...,w=d^{\frac{c}{r}}$. We will show that if $r$ is large and $(x,y,z,w)$ is any quadruple of positive integers satisfying $x^r+y^r=z^r+w^r$, then $\{x,y\}=\{z,w\}$. By homogeneity, it suffices to show that the coprime solutions satisfy $\{x,y\}=\{z,w\}.$ Since we were allowed to make the assumption $\gcd(x,y,z,w)=1$, the $abcd$ conjecture implies $$\max\{x^r,y^r,z^r,w^r\}\leq K_1\text{rad}(x^ry^rz^rw^r)^4\leq K_1(xyzw)^4,$$ unless a subsum of $x^r+y^r-z^r-w^r$ vanishes, which leads to $\{x,y\}=\{z,w\}$. If the subsums are nonzero and $w=\max\{x,y,z,w\}$, then $w^r\leq K_1w^{16}$, so $r$ is bounded or $w=1$. The last case leads to $x=y=z=w=1$. Therefore, for large $r$, the only solutions to $x^r+y^r=z^r+w^r$ are those where $\{x,y\}=\{z,w\}$. Thus also $\{a^b,b^a\}=\{c^d,d^c\}$, which was already seen to give no nontrivial solutions. Finally, let $M$ be an absolute constant that is an upper bound for $r$. Let $R=\gcd(a^b,b^c,c^d,d^c)$. We apply the $abcd$ conjecture once again to see that $\max\{\frac{a^b}{R},\frac{b^a}{R},\frac{c^d}{R},\frac{d^c}{R}\}\leq K_1\text{rad}(abcd)^4$. Now if $abcd$ has no prime divisor greater than $M$, we have $$\min\{a^b,b^a,c^d,d^c\}\geq R\geq c_0\max\{a^b,b^a,c^d,d^c\}$$ for some absolute constant $c_0>0$. Now if there exists a prime $p_1$ that divides some of $a,b,c,d$ but not all of them, then $$R\leq \prod_{p\leq M, p\neq p_1}p^{\min\{v_p(a^b),v_p(b^a),v_p(c^d),v_p(d^c)\}}\leq \frac{\max\{a^b,b^a, c^d, d^c\}}{2^{\min\{a,b,c,d\}}}\quad \quad (1).$$ If no such $p_1$ exists, all the numbers $a,b,c,d$ have the same prime factors. Let $a=\prod_{i=1}^s p_i^{\alpha_i},...,d=\prod_{i=1}^s p_i^{\delta_i}$. The condition $\gcd(a,b,c,d)\leq M$ tells $\min\{\alpha_i,\beta_i,\gamma_i,\delta_i\}\leq M$. Let $P^{\delta}$ be the largest prime power dividing $d$. Then $\min\{v_{P}(a),v_{P}(b),v_{P}(c)\}\leq M$ if $d$ is large. For example, let $v_{P}(c)\leq M$. Write $d=P^{\delta}D,c=P^{\gamma}C$, where $P\nmid C,D$. Then $$R=\gcd(a^b,b^a,c^d,d^c)\leq \gcd(c^d,d^c)\leq P^{Md} D^c=P^{Md-c\delta}d^c\leq M^{Md}d^{-\frac{c}{s}}d^c,$$ where $s\leq M$ is the number of prime factors of $d$. The last quantity is at most $\left(\frac{d}{2}\right)^c$ if $\frac{1}{2}d^{\frac{1}{s}}\geq M^{\frac{Md}{c}},$ which holds for large enough $d$ if $\frac{d}{c}$ is bounded. It must indeed be bounded since we had $\frac{c^d}{d^c}\geq c_0$. Therefore $(1)$ holds again for all large $d$. Next we show that $(1)$ holds also if $a,b,c,d$ have some prime factors greater than $M$, and then use $(1)$ to deduce a contradiction. Now one of $a,b,c,d$ has a prime factor $p_0>M$. For example, let $p_0\mid d$. Then $$R\leq \prod_{p\leq M}p^{v_p(d^c)}\leq \left(\frac{d}{p_0}\right)^c\leq \frac{\max\{a^b,b^a,c^d,d^c\}}{2^{\min\{a,b,c,d\}}},$$ so $(1)$ must always hold for large $d$. However, we had $R\geq \frac{\max\{a^b,b^a,c^d,d^c\}}{K_1\text{rad}(abcd)^4}\geq \frac{\max\{a^b,b^a,c^d,d^c\}}{K_1 d^{16}}$. Thus $K_1d^{16}\geq 2^{\min\{a,b,c,d\}}$. But $$\min\{a,b,c,d\}^d\geq \min\{a^b,b^a,c^d,d^c\}\geq R\geq \frac{\max\{a^b,b^a,c^d,d^c\}}{K_1d^{16}}\geq \frac{c^d}{K_1d^{16}},$$ implies $\min\{a,b,c,d\}\geq k_0c$ for some constant $k_0$,so $K_1d^{16}\geq 2^{k_0c}$. Still we have $\frac{d^c}{c^d}\geq \frac{1}{K_1d^{16}}$. In particular, $d^c\geq \frac{2^d}{K_1d^{16}}$, so $c\geq k_2 \frac{d}{\log d}$. But then $K_1d^{16}\geq 2^{k_0c}\geq 2^{k_3\frac{d}{\log d}}$ shows that $d$ is bounded and hence all $a,b,c,d$ are bounded.<|endoftext|> TITLE: Looking for history on a theorem of clique intersections QUESTION [9 upvotes]: I have a short paper I'm working on where I prove: Theorem: Every graph on (2t-1) vertices with no (t+1)-clique has a vertex that is contained in every t-clique. By "t-clique", I mean a complete subgraph with t vertices. It's actually a lemma for the main result in the paper, but it's where most of the work is done. It feels like something that has been done before, but I haven't come up with anything in my searches. I think it might be related to clique graphs or clique-helly graphs, but I haven't found a direct link. Does anyone know of a paper with this result? Or another result that would imply this one? Or at least a related paper that might be worth looking into? Thanks! REPLY [11 votes]: Hajnal's Clique Collection Lemma implies your theorem. See Corollary 2.10 from this paper http://arxiv.org/pdf/1101.4564v5.pdf or Hajnal's original paper http://cms.math.ca/cjm/v17/cjm1965v17.0720-0724.pdf .<|endoftext|> TITLE: Canonical Metric on Grassmann Manifold QUESTION [10 upvotes]: I was curious and quite clueless as to how we can equip the Grassmann Manifold with a canonical metric - I have yet to find anything upon this subject. REPLY [7 votes]: In fact, $Gr(n,m)$ with its canonical metric induced from an Euclidean structure is one of the few spaces where you can write down solutions of the geodesic equations explicitly by a formula and write down a formula for the geodesic distance. See the following paper for this MR1856419 Neretin, Yurii A. On Jordan angles and the triangle inequality in Grassmann manifolds. Geom. Dedicata 86 (2001), no. 1-3, 81–92.<|endoftext|> TITLE: Special linear groups contained in symplectic groups QUESTION [5 upvotes]: Let $q$ be a power of prime $p$, and $n, m, k$ positive integers such that $mk=2n$ and $2\leq m<2n$. Let $\mathrm{Sp}(2n,q)$ be the symplectic group of dimension $2n$ over $\mathrm{GF}(q)$ and $\mathrm{SL}(m,q^k)$ the special linear group of dimension $m$ over $\mathrm{GF}(q^k)$. For what values of $m,k$ does $\mathrm{Sp}(2n,q)$ contain a subgroup isomorphic to $\mathrm{SL}(m,q^k)$? It is not difficult to show that $\mathrm{Sp}(2n,q)$ contains $\mathrm{SL}(2,q^n)=\mathrm{Sp}(2,q^n)$. If $\mathrm{Sp}(2n,q)$ contains $\mathrm{SL}(m,q^k)$ then the the greatest common divisor of $m$ and $p-1$ divides 2. When $p$ is odd, consideration of representations of the lowest degree shows that $\mathrm{Sp}(2n,q)$ cannot contain $\mathrm{SL}(m,q^k)$ when $m\geq3$ (assuming $mk=2n$). Is there a simple group theoretic argument? What is the conclusion when $q=2$? REPLY [5 votes]: The question is this: When can $SL(m,q^k)$ be a subgroup of $Sp(2n,q)$ with $mk=2n$? As you point out, this is possible if $m=2$. There are many particular cases that can be ruled out by order considerations - using Zsigmondy primes. However to give a complete answer, one should observe that if such an embedding exists, then the subgroup $SL(m,q^k)$ contains a Singer cycle of $Sp(2n,q)$, i.e. a maximal irreducible cyclic subgroup. The maximal subgroups of $Sp(2n,q)$ that contain a Singer cycle were found by Bereczky: A. Bereczky, Maximal Overgroups of Singer Elements in Classical Groups, Journal of Algebra, Volume 234, Issue 1, 1 December 2000, Pages 187–206 I don't have access to Bereczky's paper but (a version of) the statement can be found at this great blog. As you will see, there are a bunch of exceptional situations and there are the field-extension subgroups. One can step recursively through the field-extension subgroups and one will either end with a subgroup $Sp(2,q^k)=SL(2,q^k)$ as you describe, or else you will end up in the exceptional cases which can be checked by hand. Thus I believe that you are correct in asserting that, in general, $m=2$ - although there might be a finite number of exceptions. (In fact one of the exceptional cases is an infinite family - but it is just the $O^-$ type orthogonal groups inside $Sp$ and, since Bereczky's result applies to orthogonal groups too, recursion can be applied here too.) Added, thanks to comment below of the OP: In fact the result can be proved much more easily: $SL(m,q^k)$ contains a Singer cycle of order $((q^k)^m-1)/(q^k-1)$. This Singer cycle must act irreducibly on the vector space associated with $Sp(2n,q)$ and so must lie in a Singer cycle of $Sp(2n,q)$. But a Singer cycle of $Sp(2n,q)$ has order $q^n+1$. We conclude that $m=2$ as required.<|endoftext|> TITLE: Tate-Shafarevich groups over finitely generated fields QUESTION [7 upvotes]: Let $G$ be an algebraic group over a number field $k$. One defines the Tate-Shafarevich set of $G$ to be $$Ш(k,G) = \ker\left(H^1(k,G) \to \prod_{v} H^1(k_v,G)\right),$$ where the product is over all places of $k$. Note that if $G$ is non-abelian, then this will only be a pointed set in general. It is known that $Ш(k,G)$ is finite if $G$ is a linear algebraic group. It is conjectured that $Ш(k,G)$ is finite if $G$ is an abelian variety. This is known in some special cases, but is open in general. Let now $G$ be an algebraic group over a finitely generated field extension $k$ of $\mathbb{Q}$. Is there a natural analogue of $Ш(k,G)$ in this setting? Is it moreover known that $Ш(k,G)$ is finite when $G$ is linear algebraic? Part of my motivation is the observation that results over number fields often generalise to finitely generated field extensions of $\mathbb{Q}$ (e.g. the Mordell-Weil theorem). So I would like to know if $Ш(k,G)$ makes sense in this more general setting. I have a vague idea of how to proceed. Namely, to choose a model for $k$ (given as a proper flat scheme of finite type over $\mathbb{Z}$ with function field $k$, say), then take our notion of place to be a point of codimension one on this model. But I'm not really sure where to go from there, nor whether finiteness should hold when $G$ is linear algebraic. REPLY [3 votes]: If $X/\mathbf{F}_q$ is a smooth projective scheme (a model of a function field in positive characteristic) and $\mathscr{A}/X$ an Abelian variety, then $H^1(X,\mathscr{A}) = \ker\Big(H^1(K,\mathscr{A}) \to \bigoplus_{x \in S}H^1(K_x^{nr},\mathscr{A})\Big)$ is an analogue of the Tate-Shafarevich group. Here, $S$ can be $X$, the set of closed points $|X|$ or the set of codimension-$1$ points $X^{(1)}$; $K_x^{nr}$ is the quotient field of the strict Henselisation of $\mathscr{O}_{X,x}$; in the case of $x \in X^{(1)}$, you can also use the completion. I have proved this in my PhD thesis, which will be available soon.<|endoftext|> TITLE: Iteration of Proper Forcing and Support of Master Conditions QUESTION [6 upvotes]: Suppose $\mathbb{P}$ is a definable proper forcing (for instance Sacks forcing). Let $\alpha$ be some ordinal. Let $\mathbb{P}_\alpha$ be the countable support iteration of $\mathbb{P}$ of length $\alpha$. It is well known that $\mathbb{P}_\alpha$ is also proper. Hence for any countable elementary structure $M \prec H_\Theta$ for sufficiently large $\Theta$ such that $\mathbb{P}_\alpha \in M$ and any $p \in \mathbb{P}_\alpha \cap M$, there exists some $q \leq_{\mathbb{P}_\alpha} p$ which is a $(M, \mathbb{P}_\alpha)$-generic condition. Let $X = \alpha \cap M$. The question is whether a $(M, \mathbb{P}_\alpha)$ generic condition $q \leq_{\mathbb{P}_\alpha} p$ can be found such that $\text{supp}(q) \subseteq X$. Thanks for any information that can be provided. REPLY [5 votes]: See also my recent paper Understanding preservation theorems, Chapter VI of Proper and Improper Forcing, I for what I've been told is the clearest exposition of iterated proper forcing.<|endoftext|> TITLE: The Mordell and Bogomolov problems in linear groups QUESTION [11 upvotes]: Many things in the arithmetic of abelian varieties have counterparts not only in linear tori, but also for semisimple linear groups. Two examples are the Tamagawa number and the conjectured finiteness of the Shafarevich-Tate group. I wanted to ask about the Mordell conjecture (proved by Faltings and Vojta), which for a complex semiabelian variety $A$ and a finitely generated subgroup $\Gamma \subset A(\mathbb{C})$ states that the Zariski closure of any subset of $\Gamma$ is the union of finitely many cosets of algebraic subgroups of $A$. For $\mathrm{SL}_2$ the obvious translation of this statement is false: the arithmetic subgroup $\mathrm{SL}_2(\mathbb{Z})$ is generated by two elements and contains a Zariski-dense subset from the algebraic set $\mathrm{tr}(A) = 0$. (One could still ask if there is any description of the possible Zariski closures $V$ of subsets of arbitrary finitely generated subgroups $\Gamma \subset G(\mathbb{C})$ of the complex points of a linear group $G$. Taking cue from this example I had asked whether, for instance, such a $V$ has to be defined over a number field. As Venkataramana explains in his answer below, this question is meaningless as it stands: at the very least, I must remove from $V$ a finite union of algebraic subgroup cosets, thereby focusing on the counterexamples to the literal Mordell-type statement.) Perhaps the analogs of the Manin-Mumford and Bogomolov problems (which concern the Zariski closures of sets of points of finite order) could make more sense in linear groups. Instead of attempting to make any more hasty guesses of how the structure of the possible Zariski closures might look like, it would be more prudent to just record this as an open-ended problem: Problem. 1. Describe the possible Zariski closures $V$ of sets of elements of finite order in $G(\mathbb{C})$. 2. Whatever these $V$ are, do they coincide with the set of subvarieties of $G$ which possess a Zariski-dense sequence of semisimple elements of $G(\bar{\mathbb{Q}})$ all of whose eigenvalues have canonical Weil heights approaching zero? REPLY [4 votes]: The answer to whether $V$ is defined over a number field is no. If $G$ is defined over the algebraic closure of $\mathbb Q$, then the number of possible choices for $V$ is countable, so, for measure theory reasons, we can choose an element $a$ in $G({\mathbb C})$ which does not lie in any $V$. So the Zariski closure of the group generated by the element $a$ is not contained in any $V$ defined over a number field. There are too many other questions which I did not follow. Borel density says that a lattice is Zariski dense; none of the statements in the question seem to contain the Borel density as a special case<|endoftext|> TITLE: Power operations and Lambda-structure-like lifts of Frobenius in $E_\infty$-geometry? QUESTION [9 upvotes]: A $\Lambda$-structure on a commutative ring $R$ is a ring endomorphism wich restricts to the $p$-Frobenius homomorphism after localizing at $(p)$. One may think of this as a "flow" $\Phi \colon Spec(R) \longrightarrow Spec(R)$ in arithmetic geometry, which lifts the Fermat p-derivation on the base $Spec(\mathbb{Z})$. If we allowed ourselves to denote derivations as endomorphisms, then with slight but very suggestive abuse of notation we have the picture $$ "\array{ Spec(R) &\stackrel{\Phi + p \cdot \partial_p^{\Phi}}{\longrightarrow}& Spec(R) \\ \downarrow && \downarrow \\ Spec(\mathbb{Z}) &\stackrel{(-)^p = id + p\cdot \partial_p}{\longrightarrow}& Spec(\mathbb{Z}) } " $$ See on the $n$Lab at Borger's absolute geometry -- Motivation for more on what I have in mind here, following ideas famously promoted by James Borger and Alexadru Buium. I would like to know if there is a sensible generalization of this from arithmetic geometry to $E_\infty$-arithmetic geometry, hence from commutative rings $R$ to $E_\infty$-rings. Via discussion which is clearly articulated for instance starting from remark 2.2.9 in Jacob Lurie's DAGXIII Rational and p-adic homotopy theory, the $E_\infty$-analog of "this" are the power operations in multiplicative cohomology theory. I am a little shaky on some details though. Therefore my question: what would be the good generalization of the concept of $\Lambda$-rings to $E_\infty$-algebra in the sense of Frobenius lifts and with an eye towards absolute geometry, as above? Can one say anything? REPLY [9 votes]: Although I don't have much to say here, perhaps the following example is worth pointing out. Let $R$ be a $K(1)$-local $E_\infty$-ring under ($p$-adic) complex $K$-theory $KU$. Then there exists a power operation $\theta: \pi_0 R \to \pi_0 R$ such that: $\psi(x) \stackrel{\mathrm{def}}{=} x^p + p \theta(x)$ defines a ring homomorphism from $\pi_0 R \to \pi_0 R$. $\theta$ satisfies all the identities needed to make $\psi$ a ring-homomorphism after "division by p." For instance, $\psi(x+y) = \psi(x) + \psi(y)$ implies that $$\theta(x+y) = \theta(x) + \theta(y) + \frac{x^p - y^p - (x+y)^p}{p},$$ where the last term is an integral polynomial in $x,y$ and is interpreted as such. $\theta$ is the basic power operation for $K(1)$-local $E_\infty$-ring spectra, as explained in these notes of Hopkins, and the algebraic structure it gives is called a "$\theta$-algebra." (These also seem to be called $p$-derivations by algebraists.) Note in particular that $\psi$ is a lift of the Frobenius. There are generalizations of $\psi, \theta$ at higher chromatic levels, too, and there (as I understand) a modular interpretation of the resulting algebraic structure in this paper of Rezk.<|endoftext|> TITLE: When does a cone contain its dual cone? QUESTION [6 upvotes]: Let $V$ be a finite-dimensional vector space with an inner-product $(,)$ and let $C\subset V$ be a cone in $V$. Let $C^\vee$ denote the dual of $C$ with respect to $(,)$, i.e., the set of vector $v\in V$ having non-negative intersection with all of $C$. Of course there is a natural identification of $V$ with its dual $V^\vee$, given by $v\mapsto (w\mapsto (v,w))$. In this way it makes sense to compare the two cones $C$ and $C^\vee$. Is there some condition on $C$ that would allow one to conclude that $C^\vee\subseteq C$?. For example, if $C$ is generated by finitely many vectors $v_1,\ldots,v_s$ and I know the matrix $M=[(v_i,v_j)]$ of all intersection numbers, can I see the above containment in terms of some kind of property of $M$? REPLY [3 votes]: The cone $C$ contains its dual cone if and only if for every $x\in\partial C$ there exists $y\in C$ so that $(x,y)\leq0$. (This is quite obvious in two dimensions.) Let $S\subset V$ denote the unit sphere. The cone $C$ is uniquely determined by $A=C\cap S$ and similarly $C^V$ by $B=C^V\cap S$, so let us restrict our attention to $A$ and $B$. If $B\neq\emptyset$, then $B\cap A\neq\emptyset$. Let $f:S\to\mathbb [-1,1]$, $f(x)=\min(x,A)$, where $(x,A)=\{(x,y);y\in A\}$. Now $f$ is continuous and reaches the value $-1$ in $S\setminus A$. The function $f$ reaches its maximum in $A$ and $B=f^{-1}([0,1])$ is connected (the dual cone is a cone). Thus $B\subset A$ is equivalent with $f|_{\partial A}\leq0$.<|endoftext|> TITLE: Learning roadmap to TQFT from a mathematics perspective QUESTION [13 upvotes]: I had asked a question on math.stackexchange but did not receive any answers. I hope that this question is appropriate for this website as it is about an advanced subject. Hence I am posting it below. I want to learn TQFT's and am looking for review articles or books. My mathematics knowledge is limited to one year of graduate course in Algebra (Groups,Rings,Fields,Categories, Modules and Homological Algebra), self study of Geometry (Manifolds, Differential Geometry, Riemannian Metrics, Curvature and Connections), elementary representation theory of finite dimensional semisimple Lie Algebras and complex analysis. Before I study TQFTs I will try to read Algebraic Topology, Characteristic Classes and K-Theory. On the physics side I am comfortable with scalar field theory and renormalization. My first question is : 1.) Are there any other subjects that I should learn before embarking on a study of TQFTs from a Mathematics perspective ? In particular should I study Algebraic Geometry (e.g. Hartshorne or Huybretchs) before or concurrently with TQFTs ? Also there are good reviews of this subject from the Physicists' perspective such as the famous review here and lots others listed here, however I wish to approach the subject from the Mathematics side. One way is to perhaps start with the original papers of Witten but as I mentioned I am more keen on learning the Mathematics side and preferably from a book or review article. 2.) Is there a good book or detailed review article useful for beginners interested in learning TQFTs from a mathematical perspective ? TQFT is now a very vast subject and I would like a book or article dealing with an overview of the subject rather than a specialized sub area. I am aware of the book by Turaev but am not sure whether it is good for beginners or satisfies the above criteria. I am more keen on learning the subject for its applications in manifolds rather than knot theory (for which a helpful answer is already available in this answer.) It would be ideal if someone could suggest a roadmap to learning this subject. Thanks ! Note : I realized that there is a related question but the answers there mostly contain references to original papers rather than books or comprehensive reviews which is what I am looking for. REPLY [5 votes]: The recent articles below since 2016 contain systematic and useful introductions to the new development in 3d and 4d TQFT or more general arbitrary d-TQFT, with an eye toward applications for strongly coupled condensed matter system and topological quantum matter in 3-dimensions, 4-dimensions and any dimension. The authors point out the relations between quantum Hamiltonian lattice models, the continuum TQFTs and group cohomology/cobordism theory. "Braiding Statistics and Link Invariants of Bosonic/Fermionic Topological Quantum Matter in 2+1 and 3+1 dimensions" by Pavel Putrov, Juven Wang, Shing-Tung Yau https://arxiv.org/abs/1612.09298 Annals of Physics 384C, 254-287 (2017) DOI:10.1016/j.aop.2017.06.019 "Tunneling Topological Vacua via Extended Operators: (Spin-)TQFT Spectra and Boundary Deconfinement in Various Dimensions" by Juven Wang, Kantaro Ohmori, Pavel Putrov, Yunqin Zheng, Zheyan Wan, Meng Guo, Hai Lin, Peng Gao, Shing-Tung Yau https://arxiv.org/abs/1801.05416 Prog. Theor. Exp. Phys. 053A01 (2018) DOI:10.1093/ptep/pty051 "Fermionic Finite-Group Gauge Theories and Interacting Symmetric/Crystalline Orders via Cobordisms" by Meng Guo, Kantaro Ohmori, Pavel Putrov, Zheyan Wan, Juven Wang https://arxiv.org/abs/1812.11959 Some results and systematic Tables provided by the above articles include: Ref 1: Ref 2: Ref 3:<|endoftext|> TITLE: Reference request: Book of Linear algebra from categorical point of view QUESTION [9 upvotes]: Is there any book of Linear algebra in the modern language of Category theory? I refer to the (systematic, formalist) study of the category whose objects are vector spaces and whose morphisms are linear maps and its consequences. REPLY [3 votes]: Filip Bár's master thesis, "On the Foundations of Geometric Algebra" might be a beginning (I don't know if this is online, but perhaps you can ask the author). This thesis develops some ideas by Grassmann in modern language, especially concerning affine spaces and affine algebras, but Chapter 2 deals with vector spaces from a basic categorical point of view. Meanwhile there are some accounts on commutative algebra from a category-theoretic point of view (Toën-Vezzosi, Lurie, B.).<|endoftext|> TITLE: Characterizing the newforms s.t. the associated symmetric square $L$-function has a pole QUESTION [8 upvotes]: I have a straightforward question. Let $f$ be a holomorphic cusp form of weight $k$, level $N$, and nebentypus $\chi$ that is new in the sense of Atkin-Lehner theory. Write its Fourier expansion at $\infty$ as $$f(z) = \sum_{n=1}^\infty \lambda_f(n)n^{(k-1)/2}e(nz)$$ and form the $L$-function $L(f,s)$ by the Dirichlet series $$L(f,s)=\sum_{n=1}^\infty \frac{\lambda_f(n)}{n^s}$$ for $\Re(s)>1$. Then $L(f,s)$ can be continued to an entire function on $\mathbf{C}$, and, by the above normalization of Fourier coefficients, obeys a functional equation relating $s$ to $1-s$. Let $\alpha,\beta$ be the Satake parameters associated to $f$; i.e. $$L(s,f) = \prod_p \left(1-\frac{\alpha(p)}{p^s}\right)^{-1}\left(1-\frac{\beta(p)}{p^s}\right)^{-1}$$ and then define the symmetric square $L$-function $L(\operatorname{sym}^2f,s)$ associated to $L(f,s)$ by $$L(\operatorname{sym}^2f,s) := \prod_p (1-\alpha(p)^2p^{-s})^{-1}(1-\alpha(p)\beta(p)p^{-s})^{-1} (1-\beta(p)^2p^{-s})^{-1}.$$ My question is, when exactly is $L(\operatorname{sym}^2f,s)$ entire, and if it is not entire, what poles can it have? Pages 136–137 of Iwaniec-Kowalski's book seem to answer this question. We know that $L(\operatorname{sym}^2f,s)$ factors like $$L(\operatorname{sym}^2f,s) = L(f\otimes f,s)L(s,\chi)^{-1},$$ where the Rankin-Selberg convolution $L(f\otimes f,s)$ has a simple pole at $s=1$ iff $f$ is self-dual ($f=\overline f$) and is entire otherwise. $L(s,\chi)$ has a simple pole at $s=1$ iff $\chi$ is principal. Therefore, my understanding is that $L(\operatorname{sym}^2f,s)$ is entire unless $f$ is self-dual with non-principal character. This can happen if $f$ is a `CM form' arising from a Hecke grössencharacter; see https://mathoverflow.net/a/164126/37110 for details. In the special case when $f$ has real Fourier coefficients and $\chi$ is not principal, $L(\operatorname{sym}^2f,s)$ has a simple pole at $s=1$. Is my understanding correct? Is it complete? Thanks in advance. REPLY [7 votes]: Yes, your understanding is correct. Here's a little bit more detail. If $f$ is a CM form, $f$ is associated to a Hecke Grössencharacter $\xi$. If $k \geq 2$, then $\xi$ is associated to an imaginary quadratic field and then $\xi$ is a homomorphism $\xi : I(\Lambda) \to \mathbb{C}^{\times}$ is a homomorphism from the group of all fractional ideals coprime to the modulus $\Lambda$, and satisfies $\xi(\alpha \mathcal{O}_{K}) = \alpha^{k-1}$, provided $\alpha \equiv 1 \pmod{\Lambda}$. Let $\omega_{\xi}(n) = \xi\left(n \mathcal{O}_{K}\right)/n^{k-1}$. This function is a Dirichlet character, and the modular form corresponding to $\xi$, which is $$ f(z) = \sum_{\mathfrak{a} \subseteq \mathcal{O_{K}}} \xi(\mathfrak{a}) q^{N(\mathfrak{a})}, q = e^{2 \pi i z} $$ has Nebentypus $\chi = \chi_{K} \omega_{\xi}$, where $\chi_{K}$ is the Kronecker character associated to $K$. Now, the symmetric power $L$-functions of $L(s,\xi)$ factor as products of degree $1$ and degree $2$ $L$-functions. In particular, $L({\rm Sym}^{2} f, s) = L(s, \xi^{2}) L(s, \omega_{\xi})$. Self-duality of $f$ is (by the question you linked to) equivalent to the statement that $\chi_{K} = \chi$, and hence $\omega_{\xi}$ is the trivial character, which shows that $\zeta(s)$ is a factor of $L({\rm Sym}^{2} f, s)$ and so $L({\rm Sym}^{2} f, s)$ has a pole. On the other hand, if $f$ is a CM form with trivial character, then $\omega_{\xi} = \chi_{K}$. Then, the first symmetric power $L$-function with a pole is $$ L({\rm Sym}^{4} f, s) = L(s, \xi^{4}) L(s, \xi^{2} \otimes \omega_{\xi}) L(s, \omega_{\xi}^{2}). $$<|endoftext|> TITLE: Asymptotic density of finite abelian and solvable groups QUESTION [6 upvotes]: For every natural number n, let: Gn be the number of distinct group structures with at most n elements; An be the number of distinct abelian group structures wit at most n elements; Sn be the number of distinct solvable group structures with at most n elements. Question 1: Is there a known limit for the quotient An/Gn ? Question 2: Is there a known limit for the quotient Sn/Gn ? REPLY [8 votes]: The number of abelian groups of order at most $n$ is $O(n)$, whereas if $n=2^k$, the number of class $2$ nilpotent groups of order $n$ is $2^{(2/27)k^3+O(k^{8/3})}=n^{\Omega(\log^2n)}$ by a result of Sims, hence the answer to question 1 is $0$. It is conjectured that the global asymptotic density of $2$-groups of nilpotent class $2$, and a fortiori of solvable groups, is $1$, but as far as I know, this has not been proved. REPLY [5 votes]: (Edited following Emil Jerabek's coment below) From results of L. Pyber (and implicitly, C. Sims) it appears likely that $\frac{f(n)}{g(n)} \to 1$ as $n \to \infty,$ where $f(n)$ is the number of isomorphism types of nilpotent groups of order $n$ and $g(n)$ is the number of isomorphism types of all groups of order $n,$ so minor modifications should yield the same answer for question 2 (which is a cumulative version- note also that all nilpotent groups are solvable). Also, the asymptotic behaviour of the number of isomorphism types of Abelian groups of order $n$ and the number of isomorphism types of nilpotent groups of order $n$ are known: both are multiplicative, so it suffices to consider the case of $p$-groups. The number of isomorphism types of Abelian groups of order $p^{k}$ is $p(k),$ the number of partitions of $k,$ which behaves like $e^{c \sqrt{k}}$ for some (known!) constant $c.$ The number of isomorphism types of groups of order $p^{k}$ is asymptotically around $p^{\frac{2k^{3}}{27}}$ (proved by C. Sims and G. Higman). This suggests that the limit of question 1 should be zero, though again you ask for a cumulative version.<|endoftext|> TITLE: Why considering schemes over discrete valuation rings? QUESTION [7 upvotes]: For many times, I find people working on schemes over DVRs, and prove theorems on such setting. For example, my latest experience is the "semi-stable reduction theorem" by Kempf, Knudsen, Mumford and Saint-Donat: they proved semi-stable reduction theorem first over $\mathbb{C}$, then turned to the situation over DVR. It is "natural" for me to work over $\mathbb{C}$, or over character $p$, or even over arbitrary schemes. But why people like to choose DVRs in particular? This question would break to two parts: (1) Is there any technical advantage to work over DVR? (2) What is the importance to consider this special case? Does this come from the interests of number theory? I have no clue of that. REPLY [5 votes]: (1) What are the technical advantages of working over a DVR? They are many. DVR are very simple rings, they are both local and principal. The principality in particular means that the flatness (which is a fundament and not sue say to grasp property in algebraic geometry) of a module or algebra over that sort of ring becomes something much simpler to understand, to check and to use, namely the propriety of being torsion-free. Also, a DVR is integrally closed, and that's a property that is very useful for certain type of reasoning in algebraic geometry. (2) The interest of working with a DVR in algebraic geometry does not come primarily from number theory. For example, you may have seen the valuative criterion of properness: to check if a morphism $f: X \rightarrow Y$ is proper, under certain mild hypothesis, you only have to check that the pull-back of this morphism over any spectrum of DVR $Y'$ is. This reduces the verification of an important relative property over an arbitrary base to the case of much simpler base, see (1), and is a crucial tool in algebraic geometry.<|endoftext|> TITLE: C*-bimodules: the mess with definitions QUESTION [6 upvotes]: I used to participate in a seminar that taught students about foundations of non-commutative geometry. It isn’t very complicated to define a C*-module $\mathcal E$ (also known as C* Hilbert module) over a C*-algebra $\mathcal A$. This structure comprises the pairing, a 1½-linear mapping ${\mathcal E}\times{\mathcal E} \to {\mathcal A}$ with some properties analogous to ones of a Hilbert space. But the mess begins when one tries to define, in a similar fashion, a C*-bimodule; one needs it to define Morita equivalence of C*-algebras in a way other than category-theoretical. Namely, it is natural to expect from a ${\mathcal B},{\mathcal A}$-bimodule $\mathcal E$: To be truly a bimodule, that means representations of $\mathcal B$ and $\mathcal A$ must commute, or the $b({\mathbf x}a) = (b{\mathbf x})a$ identity must be respected. To have two pairings, one $\mathcal B$-valued and another $\mathcal A$-valued. Shouldn’t the definition be symmetric? Two pairings should obey the exchange identity $\{{\mathbf x}|{\mathbf y}\}\,{\mathbf z} = {\mathbf x}\,({\mathbf y}|{\mathbf z})$ (that virtually defines one 2½-linear operation on $\mathcal E$, where $\mathbf y$ is the antilinear argument). It is the least obvious requirement, but it is necessary for correct behaviour of tensor product operations. I presented these algebraic properties (and some other) at this HTML page as diagrams, to be more illustrative. Alain Connes proposed to define a C*-bimodule over ${\mathcal B},{\mathcal A}$ as a representation of $\mathcal B$ in a C*-module over $\mathcal A$, i.e. a *-homomorphism from $\mathcal B$ to $\operatorname{End}_{\mathcal A} {\mathcal E}$, saying nothing about $\mathcal B$-valued pairing. The definition of $\operatorname{End}_{\mathcal A} {\mathcal E}$ implies commutation of representations. But it appears that this asymmetric definition is very weak on another side: examples, where any $\mathcal B$-valued pairing cannot be exchangeable with given $\mathcal A$-valued one, are almost trivial. There is a book “Elements of Noncommutative Geometry” by José Gracia Bondía, Joseph C. Várilly, and Héctor Figueroa, where another definition is proposed: exchange identity is required, but authors missed the commutation of representations condition. A “C*-bimodule” in their sense is not proven to be a true bimodule. One can prove that if a “C*-bimodule” is full at least on one side, then $b({\mathbf x}a) = (b{\mathbf x})a$ follows from the definition. It appears that $b({\mathbf x}a) = (b{\mathbf x})a$ is necessary for many proofs in the book, but is it unknown whether it follows from it in the general case (there is neither counterexample nor an idea to prove it). In the lectures 3 both exchange of pairings and $b({\mathbf x}a) = (b{\mathbf x})a$ were required for a C*-bimodule. But certain problem plagues even this definition. If we know that for any $b\in{\mathcal B}$ the respective left multiplication operator belongs to $\operatorname{End}_{\mathcal A} {\mathcal E}$ (that means it is bounded in the norm induced by the C*-$\mathcal A$-module structure), then we see a *-homomorphism of C*-algebras (that, as a general fact, must be bounded and have the norm 1 or 0), and subsequently can prove that left and right C*-module structures on $\mathcal E$ define the same norm (and hence, topology). Another approach is to define $\mathcal E$ as a topological vector space (and require continuity of both representations) that does the same thing, but it is unknown whether a plain vector space $\mathcal E$, without topology, has necessarily the same norm from two different structures of a C*-module. Again, counterexamples (satisfying aforementioned conditions) are unknown. It seems that Morita equivalence, the main “user” of the concept of bimodule, is indifferent to some of these glitches because only few C*-bimodules can specify a Morita equivalence of two respective algebras. But the question persists: which should be the correct definition of a C*-bimodule? REPLY [2 votes]: As far as I'm concerned $C^*$ bimodules generally denote those you attributed to A.Connes. Such a bi-module define (by tensorization over B) a functor from $B$ $C^*$-modules to $A$ $C^*$-modules. Any $C^*$ functor between these categories which preserve orthogonal sums is of this form (essentially because of the stabilization theorem). There is indeed a lot of them, they are more general than morphisms of $C^*$ algebra, but the general idea in non-commutative geometry is that we don't really have a good notion of "morphism" between non-commutative spaces, but $C^*$ bi - module form a really nice notion of "correspondences" between non-commutative space. The relation with the bi-modules that arise in the theory of Morita equivalence (which I will call equivalence bimodule) is quite simple: A $B,A$ bi-module $H$ induce an equivalence of category between $C^*$ A module and $C^*$ B module if and only if it is an equivalence bimodule. In this case $B$ is exactly the algebra of compact operator of the $A$-module $H$ (that is the closure of the spam of the operator of the form $|x\rangle \langle y|$ for $x,y \in H$ and the $B$ valued parring is simply $[x,y]=|x\rangle \langle y|$. I don't have access right now to the book you mentioned, and I don't read Russian, so I can't really comment on the other two definitions, but it seem to me that (unless they contains a small mistake) they should by equivalent to the previously mentioned definition (either of bimodule or of equivalence bimodule). Also note that equivalence bimodules are supposed to be full, which might answer a part of your questions. Take a look at Bruce Blackadar's "Operator algebras" chapter II.7. It contains all the details about this.<|endoftext|> TITLE: Birkhoff Ergodic Theorem and Ergodic Decomposition Theorem for Continuous-Time Markov Processes QUESTION [9 upvotes]: I have a couple of questions regarding ergodicity for Markov processes in continuous time. (In particular, the first question seems like it should be particularly basic, and yet I haven't managed to find a proof [or counterexample!].) We have a family $(P_x^t)_{x \in \mathbb{R},t \geq 0}$ of Borel probability measures on $\mathbb{R}$ such that for all $A \in \mathcal{B}(\mathbb{R})$, the map $(x,t) \mapsto P_x^t(A)$ is Borel-measurable; for all $x \in \mathbb{R}$, $P_x^0=\delta_x$; for all $A \in \mathcal{B}(\mathbb{R})$ and $s,t \geq 0$, $P_x^{s+t}(A)=\int_\mathbb{R} P_y^t(A) \, P_x^s(dy)$. [We can refer to $(P_x^t)_{x \in \mathbb{R},t \geq 0}$ as a "measurable stochastic semigroup". In general, "stochastic semigroups" only need to be measurable in $x$ for each $t$.] We will say that a probability measure $\rho$ on $\mathbb{R}$ is stationary if $\rho(A)=\int_\mathbb{R} P_x^t(A) \, \rho(dx)$ for all $A \in \mathcal{B}(\mathbb{R})$ and $t \geq 0$. We will say that a probability measure on $\mathbb{R}$ is ergodic if it is an extremal point of the convex set of stationary probability measures. Q1. Let $(\Omega,\mathcal{F},(\mathcal{F}_{t \geq 0}),\mathbb{P})$ be a filtered probability space, and let $(X_t)_{t \geq 0}$ be a progressively measurable real-valued homogeneous Markov process with transition probabilities given by $(P_x^t)_{x \in \mathbb{R},t \geq 0}$ -- that is to say, $P_{X_s(\cdot)}^t(A)$ is a conditional probability of $X_{s+t}^{-1}(A)$ with respect to $\mathcal{F}_s$ (for all $s,t,A$). Suppose also that $\rho:=X_{0\ast}\mathbb{P}$ is stationary. Fix any bounded measurable $f:\mathbb{R} \to \mathbb{R}$; is it the case that $\hspace{5mm} \lim_{T \to \infty} \frac{1}{T} \int_0^T f(X_t(\omega)) \, dt$ exists for $\mathbb{P}$-almost all $\omega \in \Omega$? (Please note that we do not assume any kind of continuity of $(X_t)$, but only that it is progressively measurable.) Now in terms of my motivation, what I am really after is an ergodic decomposition theorem for the setting that I'm currently working with; I think that a positive answer to Q1 will be enough for me to prove this. However, I would ideally like to know if ergodic decompositions exist more generally: Q2. Suppose $\rho$ is a stationary probability measure. Does there exist a probability measure $Q$ on the set $\mathcal{M}$ of probability measures on $\mathbb{R}$ (equipped with the usual $\sigma$-algebra, which is known to be standard) such that $Q$-almost every $\mu \in \mathcal{M}$ is ergodic; for all $A \in \mathcal{B}(\mathbb{R})$, $\rho(A) = \int_\mathcal{M} \mu(A) \, Q(d\mu)$? The following might be useful: Equivalent definitions of ergodicity: Given a stationary probability measure $\rho$, we will say that a set $A \in \mathcal{B}(\mathbb{R})$ is $\rho$-almost stationary if for all $t \geq 0$, $\rho(x \in A: P_x^t(A)=1)=\rho(A)$. (1) In analogy to Proposition 7.2.4 of books.google.co.uk/books?isbn=0521515971 (p378) for deterministic systems, we have that a stationary probability measure $\rho$ is ergodic if and only if every $\rho$-almost stationary set has $\rho$-trivial measure: If $\rho(A) \in (0,1)$ and $A$ is $\rho$-almost stationary, then $\rho$ conditioned on $A$ and $\rho$ conditioned on $\mathbb{R} \setminus A$ are stationary probability measures which can be linearly combined in the obvious way to give $\rho$. In the other direction, it suffices to show that if every $\rho$-almost stationary set has trivial measure and $\tilde{\rho}$ is a stationary probability measure that is absolutely continuous with respect to $\rho$, then $\rho=\tilde{\rho}$. Take a density $g$ of $\tilde{\rho}$ with respect to $\rho$. For each $t$, define the probability measure $\rho_t$ on $\mathbb{R} \times \mathbb{R}$ by $\rho_t(A \times B) = \int_A P_x^t(B) \, \rho(dx)$. The stationarity of $\tilde{\rho}$ implies that $\hspace{5mm} \int_{A \times (X \setminus A)} g(x_1) \, \rho_t(d(x_1,x_2)) \ = \ \int_{(X \setminus A) \times A} g(x_1) \, \rho_t(d(x_1,x_2))$ for any $A \in \mathcal{B}(\mathbb{R})$ and $t \geq 0$. Setting $A:=\{x \in X : g(x) \geq 1\}$, the above equation (combined with the stationarity of $\rho$) implies that $A$ is $\rho$-almost stationary, so $A$ has trivial measure. It follows that $\tilde{\rho}=\rho$. (2) We will say that a set $A \in \mathcal{B}(\mathbb{R})$ is invariant if for all $t \geq 0$ and all $x \in A$, $P_x^t(A)=1$. Given a set $A$ that is $\rho$-almost stationary, there exists a set $A'$ that is invariant, with $\rho(A \triangle A')=0$. Namely, set $\hspace{5mm} A' \ := \ \{ x \in X : \textrm{Leb}(t \geq 0 : P_x^t(A)<1) = 0 \}$ where $\textrm{Leb}$ denotes the Lebesgue measure. So a stationary probability measure $\rho$ is ergodic if and only if every invariant set has $\rho$-trivial measure. (It is perhaps worth pointing out that (1) does not rely on the stochastic semigroup $(P_x^t)$ being a "measurable" stochastic semigroup, but the construction in (2) does rely on this.) Update: I'm pretty sure the answer to Q2 is yes, because I think I can prove it using an ergodic theorem for measurable stochastic semigroups; namely, letting $\rho$ be a stationary probability measure, I think I can first prove that for any bounded measurable $f:\mathbb{R} \to \mathbb{R}$, $\hspace{5mm} \lim_{T \to \infty} \frac{1}{T}\int_0^T \int_\mathbb{R} \! f(y) P_x^t(dy) \; dt$ exists for $\rho$-almost all $x \in \mathbb{R}$, with the limit (as a function of $x$) being a conditional expectation of $f$ over the probability space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\rho)$ with respect to the $\sigma$-algebra of $\rho$-almost stationary sets. (As mentioned in Kifer's book "Ergodic Theory of Random Transformations", the discrete-time analogue of the above statement can be obtained as a special case of the Chacon-Ornstein ergodic theorem.) Using this fact, it should be possible to prove the ergodic decomposition theorem (by a similar approach as in the proof for deterministic dynamical systems). However, I suspect that the answer to Q1 is no (although I do not have a counterexample!!). More precisely, I suspect that the answer to Q1 is the same as the answer to my question Is it true that all stationary measurable stochastic processes are "measurably stationary"? - and I expect that the answer to that question is no (although again, I do not have a counterexample). If the answer to Q1 is no, I wonder whether perhaps it becomes yes in the particular case that $(X_t)_{t \geq 0}$ is a strong Markov process. REPLY [3 votes]: I have the answers to my two questions. (I've actually had them for a while; apologies for the delay in posting.) I will give them in reverse order: The answer to Q2 is yes; the structure of the proof is exactly as I outlined in the update. Details can be found in section 5 (in particular, Corollary 100) of my notes http://wwwf.imperial.ac.uk/~jmn07/Ergodic_Theory_for_Semigroups_of_Markov_Kernels.pdf. The answer to Q1 is also yes: Since $f$ is bounded, it is sufficient just to consider the limit as $T$ tends to $\infty$ in the integers. By the positive answer to the question Is it true that all stationary measurable stochastic processes are "measurably stationary"?, the discrete-time stochastic process $\left(\int_n^{n+1} f(X_t(\cdot)) \, dt \right)_{n \geq 0}$ is stationary, and therefore Birkhoff's ergodic theorem (applied to the shift map on $\mathbb{R}^{\mathbb{N}_0}$) gives the desired convergence.<|endoftext|> TITLE: Is there an arXiv for Beamer presentations of scientific work? QUESTION [12 upvotes]: When I give a "Beamer talk" I put in a lot of effort making the slides, and trying to give an efficient presentation of my work. The end product is often around 20 pages of figures, definitions, theorems, ideas, all color coded, and thoughtfully placed in the frame. After the talk I frequently am asked if my slides will be available, and I might be handed a thumb drive so that I will do the transfer right then. Because of this, I usually post my slides on my personal website. So, given all the extra work that goes into making this document that will accompany my public presentation of my work, I wonder two things. Does there exist a Beamer arXiv? Is there a place to add the Beamer file to the arXiv? Is there a journal that publishes slide presentations? If not, why not? Doesn't it seem like we should create a journal-like entity to publish and promote the work we are already doing? It could be something of a "Scientific American" for Mathematics. REPLY [5 votes]: Slideshare.net is your friend.<|endoftext|> TITLE: How to solve such an optimization problem QUESTION [8 upvotes]: I encounter the following optimization problem, but I can't solve it. Given $N$ variables satisfying $0 \leq x_1 \leq x_2 \leq x_3 \leq ... \leq x_N \leq 1$ and an integer $K$ no large than $N$, find the values of $\{x_i\}$ that maximize the following function. $$\sum_{S \subset \{1,2,..., N\},\\ |S| = K} \prod_{i TITLE: Faster formula to compute sum over partitions QUESTION [11 upvotes]: Let $f$ be a function from the positive integers to the real numbers (or some ring...). Let $$(\star) \quad F(n) = \sum_{n_1 \leq \cdots \leq n_j\atop n_1 + \cdots + n_j = n} f(n_1) \cdots f(n_j), $$ for each positive integer $n$, where the sum runs over all the partitions of $n$, i.e., $n_1 \leq \cdots \leq n_j$ and $n_1 + \cdots + n_j = n$. Computing $F(n)$ directly from $(\star)$ is computationally expensive since it require to sum $p(n) \sim e^{\pi\sqrt{2n/3}} / (4n\sqrt{3})$ addends, as $n \to \infty$, where $p(n)$ is the partition function. My question is: Assume that we have computed $F(1), \dots, F(n-1)$, there is some (recursive) formula to find $F(n)$ in a way faster than $(\star)$? I think the answer is affermative and that $F(n)$ can be expressed as a sum involving $F(k)$, $k < n$ and $f(h)$, $h \leq n$ with fewer addends than $(\star)$, but I am unable to figure out it. Thank you in advance for any suggestion. REPLY [12 votes]: The identity $np(n) = \sum_{m=1}^n p(n-m)\sigma(m)$, where $\sigma(m)$ is the sum of divisors of $n$ generalizes to this setting. The proof I sketched here shows that $$ nF(n) = \sum_{r=1}^n F(n-r) g(r) $$ where $$ g(r) = \sum_{m \mid r} f(m)^{r/m} m. $$ This should give a more efficient algorithm: first compute the values of $g(r)$ for $r \le N$. Then use the first formula to compute $F(n)$ iteratively for $n \le N$.<|endoftext|> TITLE: Does the right adjoint of the category of simplices functor is "homotopicaly inverse" to the category of simplices functor? QUESTION [5 upvotes]: Short Version (the question) Let $\text{Cat}$ be the category of (small) categories and $\text{sSet}$ the category of simplicial sets. There is a functor $\Gamma:\text{sSet}\to \text{Cat}$ that takes every simplicial set to its category of simplices. This functor has a right adjoint, say, $T$. Question 1: Given a simplicial set $X$, is it true that $T(\Gamma(X))$ is weakly equivalent to $X$? Question 2: Is the unit $\epsilon : 1\to T\circ \Gamma$ of the adjonction a weak equivalence of simplicial sets? Long Version (background, motivation, partial results etc.) To fix notation, let $\Delta$ be the simplex category whose objects are the non-empty finite ordinals $$[n]=(0<1<... TITLE: Kernel of the character of congruence groups QUESTION [5 upvotes]: Let $\Gamma$ be a congruence subgroup of $SL_2(\mathbb Z)$ and $\chi:\Gamma \to \mathbb{C}^*$ a character of $\Gamma$ with finite image. Is then $\ker(\chi)$ also a congruence group? If not, can someone give me a counterexample? REPLY [7 votes]: The answer is NO, in general. For a specific counterexample, let $\Gamma$ be the principal congruence subgroup of level two in $SL(2,{\mathbb Z})$. Then, $\Gamma$ modulo $\pm 1$, is the free group on two generators, and hence there is a homomorphism from $\Gamma$ onto ${\mathbb Z}/5{\mathbb Z}$ (the latter realised as the quotient of the free abelian group on two generators modulo a suitable subgroup). The congruence closure of $\Gamma$ may easily be shown to be the product $$SL_2(2{\mathbb Z}_2)\times SL(2,{\mathbb Z}_3) \times \prod _{p\neq 2,3} SL(2,{\mathbb Z}_p)$$where ${\mathbb Z}_p$ denotes the ring of $p$-adic integers. Using the fact that for $p\geq 5$, the group $SL_2({\mathbb Z}_p)$ is its own commutator, it is easy to see that any Abelian quotient of this congruence closure consists only of two and three torsion. Hence the kernel of $\Gamma$ to ${\mathbb Z}/5{\mathbb Z}$ cannot be a congruence subgroup. [Edit] I see that the link provided by Matthias Wendt answers this question completely.<|endoftext|> TITLE: Full-rank factorization of the graph Laplacian QUESTION [7 upvotes]: Is there a combinatorially meaningful full-rank factorization of the Laplacian matrix of a graph? The usual factorization $L=BB^{T}$, where $B$ is an oriented incidence matrix, is full-rank if and only if the graph is a tree (because then $B$ has $n-1$ columns and rank $n-1$). Is there a nice way to to factorize a connected graph with cycles? REPLY [5 votes]: I believe you are looking for a sparse Cholesky factorization (performed after column-row pivoting). In this sense, the view of the graph begins to intertwine with a sparse-matrix interpretation. The graph cycle appears as a "feedback loop", in the linear operator interpretation of Laplacian matrix, and is eliminated by Gaussian elimination. It is worth noting that for a tree, the Cholesky factorization coincides with the oriented incidence matrix of the corresponding graph, assuming that factorization is performed in the correct order, from the leaves to the root (EDIT: corrected!). For graphs containing many cycles, e.g. the grid graph, the factorizations are fully dense, despite originating from a sparse matrix. For everything in-between, the density of the factorization is a good gauge to how interconnected the graph is. I'm happy to provide refs / more info / detail if you can elaborate on exactly what you're looking for.<|endoftext|> TITLE: Cohen-Lenstra heuristics for totally complex fields QUESTION [7 upvotes]: If a number field $K$ is a Galois extension of $\mathbb{Q}$, and $G = \operatorname{Gal}(K/\mathbb{Q})$, then the class group of $K$ is a $\mathbb{Z}[G]$-module, and since $N = \sum_{g \in G} g$ acts as the norm on ideals, sending every ideal to a principal ideal, the class group is in fact a $\mathbb{Z}[G]/\langle N \rangle$-module. Roughly speaking, the Cohen-Lenstra heuristics predict that the class group of a totally real Galois number field $K$ should behave as a ``random" finite $\mathbb{Z}[G]/\langle N \rangle$-module modulo a random cyclic submodule. My question is this: Can a similar kind of statement be made about totally complex number fields $K$ that are Galois over $\mathbb{Q}$? For example, would it simply be a random $\mathbb{Z}[G]/\langle N \rangle$-module, without quotienting out a random cyclic submodule, as in the case for imaginary quadratic fields? Or is that too naive? Along the same lines, what can be said about the Cohen-Lenstra heuristics for the relative class group of a Galois extension of totally complex number fields? REPLY [5 votes]: The general heuristic goes as follows (see the original paper by Cohen-Martinet, but I am being a bit more conservative, since some primes that Cohen-Lenstra-Martinet called "good" seem to not be all that "good", and I am leaving them out here): Fix a base field $K$, a Galois group $G$, any prime $p$ that is coprime to $\#G$ and to the order of the group of roots of unity $\mu(K)$, any central idempotent $e$ of $\mathbb{Q}[G]$ (which automatically, by assumption on $p$, lives in $\mathbb{Z}_{(p)}[G]$) that is orthogonal to the trivial idempotent $\frac{1}{|G|}\sum_{g\in G} g$. and finally a $\mathbb{Z}_{(p)}$-free $\mathbb{Z}_{(p)}[G]$-module $\Gamma$. Let $F_i$ be the sequence of those $G$-extension of $K$ for which $\mathcal{O}_{F_i}^\times\otimes_{\mathbb{Z}}\mathbb{Z}_{(p)}$ is isomorphic to $\Gamma$ as a $G$-module, ordered by absolute value of discriminant (and arbitrarily between fields of equal discriminant). Let $A_i$ be the $p$-primary part of the class group of $F_i$. Then the sequence $eA_i$ behaves like a random sequence of finite $e\mathbb{Z}_{(p)}[G]$-modules of $p$-power order, with the probability weight of such a module $A$ inverse proportional to $\#{\rm Hom}_G(\Gamma,A)\cdot\#{\rm Aut}_G(A)$.<|endoftext|> TITLE: How short can we state the Axiom of Choice? QUESTION [5 upvotes]: How short can we state a principle which is equivalent with the Axiom of Choice under $ZF$? The principle should be a sentence in the language of set theory with only $\in$ and$=$ as extralogical relation signs; I thus disregard solutions that appeal to selectors as the epsilon operator. My motivation is to extend an interpretation of $ZF$ to one of $ZFC$, and a short sentence schema will make my work - simpler and shorter. Update: On the basis of comments I have developed an answer with a challenge as to whether we may improve. REPLY [20 votes]: The following paper by Kurt Maes is focused on a version of the question at hand here, namely, finding an equivalent formulation of AC in the language of set theory using the fewest number of quantifiers, rather than merely the shortest length. In his main result, Maes finds a 5-quantifier assertion equivalent to the axiom of choice. The statement is built on the same statement as in François's answer, but modified to use fewer quantifiers (Maes has five, in comparison with ten for François; but of course François wasn't trying to minimize that quantity). Kurt Maes, A 5-quantifier (\in,=)-expression ZF-equivalent to the Axiom of Choice. Maes's result refuted a conjecture of Harvey Friedman, which in the introduction the author mentions was stated on F.O.M., that it would not be possible to state a formulation of the axiom of choice using only five quantifiers. Please see Maes's solution in his paper. When I first heard about the Maes result (August 2004, apparently an earlier draft of his paper—I haven't checked the differences), I naturally set myself the task of proving the main result myself, without looking at Maes's argument. I would encourage the same of all of you---before reading further, try to express AC in the language of set theory using only five quantifiers! Here is what I had come up with (retrieved after digging around in my old computer files): Theorem. AC is equivalent (in ZF) to the following assertion: $$\forall A\exists B\forall a\in A\, \exists x\forall z$$ $$(x \in a \cap B) \wedge (z \in a \cap B \implies z=x) \wedge (a \neq B)$$ $$\text{or }\quad(B \in x) \wedge (x \in A) \wedge (a \neq x)$$ $$\text{or }\quad(B \in A) \wedge (z \notin B).$$ Proof. The point is that in order to get down to only five quantifiers, you have to essentially reuse the quantifiers to cover the various cases. The idea is that clause 1 expresses that $B$ is a selection set for $A$, when $A$ is a family of disjoint nonempty sets (plus something extra useful when $A$ is not like that). Clause 2 expresses that $A$ has elements that are not disjoint (at least two contain $B$). Clause 3 expresses that $A$ contains the emptyset ($B=\emptyset$). AC easily implies the assertion. If $A$ is a family of disjoint nonempty sets, then we can let $B$ be a selection set for $A$, and verify clause 1. (note: in order to get $(a \neq B)$ in the case that $A$ is a singleton, we can freely add irrelevant elements to $B$ outside of $\bigcup A$.) If $A$ contains non-disjoint sets, we let $B$ be any element which is in at least two elements of $A$, and then we can always be in clause 2, since for any element of $A$ we can find another element of $A$ containing $B$. Finally, if $A$ contains the empty set, we can set $B=\emptyset$, and verify always clause 3. Conversely, suppose that the stated principle holds. To prove AC, it suffices to construct a selection set for a family $A$ of disjoint non-empty sets. By replacing $A$ if necessary with the isomorphic copy $\{\{w\}\times a\mid a \in A\}$, where $w$ has high rank (such as $w=A$ itself), we may assume that every element of $\bigcup A$ has the same rank. Thus, every element of $A$ has rank one higher than this, and every element of $\bigcup\bigcup A$ has rank lower than this. It follows that no element of $\bigcup A$ is in $A$, and no element of $\bigcup A$ has itself elements in $\bigcup A$. For such an $A$, we get $B$ by the stated principle. Note now that Clause 2 implies $B \in\bigcup A$, and clause 3 implies $B \in A$. Meanwhile, clause 1 implies both that $B$ has an element in $\bigcup A$ and also that $B$ is not in $A$ (since it implies that $B\cap a$ is nonempty for some other $a\in A$, while sets in $A$ are disjoint). By our assumptions on $A$, these possibilities are mutually exclusive. It follows that $B$ must always be in clause 1, or always in clause 2, or always in clause 3, regardless of $a$, $x$, and $z$. If clause 3 always occurs, then $\emptyset\in A$, a contradiction. If clause 2 always occurs, then $B$ must be in more than one element of $A$, since otherwise we could let $a$ be that element, and this would contradict the disjointness of the elements of $A$. Thus, it must be that clause 1 always occurs. In this case, $B$ is a selection set, and so we have established AC. QED Although I am not aware of any utility flowing from the fact that AC can be exprssed in this manner, it is nevertheless true that proof theory has sometimes made advances by investigating the resource-limited expressive powers of languages.<|endoftext|> TITLE: A group allowing exactly 7 group topologies QUESTION [32 upvotes]: Is there a group $G$ allowing exactly 7 group topologies on $G$: $\mathcal T_{\text{trivial}}, \mathcal T_{\text{discrete}}, \mathcal T_1, \mathcal T_2,\mathcal T_3,\mathcal T_4, \mathcal T_5$ with $$\mathcal T_i\nsubseteq \mathcal T_j,~~ \mathcal T_j\nsubseteq \mathcal T_i$$ for every distinct $i,j\in \{1,2,3,4,5\}$? comments: The question above is a rephrased form of this question: Is there a group with the lattice of all group topologies on it order-isomorphic to the lattice $M_5$? It can be proved that $G$ cannot be finite. For a similar question with $p+3$ group topologies (instead of $7$) where $p$ is prime, there is a simple proof for existence of such group. This is why the number 7 is selected for this question. There is an infinite group with exactly 2 group topologies which may be used in constructing $G$. This is why the question above can be interesting. REPLY [31 votes]: Edit (Oct 12, 2015): My original answer contained an error, as noticed by user47958. I now record only those parts of what I wrote which might be relevant to a future complete solution. (What I am writing here is NOT a complete solution.) I order topologies so that $\mathcal T\subseteq \mathcal T'$ means that $\mathcal T'$ is finer than $\mathcal T$. I let $M_n$ denote the lattice of height three with $n$ atoms. Background info: If $N$ is any normal subgroup of a group $G$, then the set $\mathcal T_N$ consisting of those subsets of $G$ that are unions of cosets of $N$ is a group topology on $G$. The discrete and indiscrete topologies have the form $\mathcal T_N$ for $N =\{e\}$ or $N=G$ respectively. If $M\neq N$, then $\mathcal T_M\neq \mathcal T_N$. Since $\mathcal T_M\vee \mathcal T_N = \mathcal T_{M\cap N}$ and $\mathcal T_M\wedge \mathcal T_N = \mathcal T_{MN}$, the lattice of normal subgroups of $G$ is dually embedded in the lattice of group topologies via $N\mapsto \mathcal T_N$. For any group topology $\mathcal T$ on $G$ the intersection of the neighborhoods of the identity in $\mathcal T$ is a normal subgroup $N\lhd G$. Any set in $\mathcal T$ is a union of cosets of $N$. The topology $\mathcal T_N$ from Item 1 contains the starting topology $\mathcal T$. In particular, if $\mathcal T$ is coatom in the lattice of group topologies, and the intersection $N$ is nontrivial, then $\mathcal T = \mathcal T_N$. (Briefly, non-Hausdorff coatoms in the lattice of group topologies have the form $\mathcal T_N$ for some normal subgroup $N$.) The intersection $N$ from Item 2 equals $\{e\}$ iff $\mathcal T$ is $T_0$ (equivalently Hausdorff, equivalently Tychonoff). If $G$ is finite, then all group topologies on $G$ have the form $\mathcal T_N$. The lattice of group topologies is therefore dually isomorphic to the lattice of normal subgroups of $G$. If a group $G$ has at least three pairwise complementary normal subgroups, then $G$ is finite and in fact $G\cong \mathbb Z_p\times \mathbb Z_p$ for some prime $p$. In this case the normal subgroup lattice of $G$ is $M_{p+1}$. The problem asks if there is a topological group whose lattice of group topologies is $M_5$. Since $5$ is not a prime plus one, such a group cannot be finite. In fact, no more than two of the (co)atoms in the topology lattice can have the form $\mathcal T_N$ for $N\lhd G$. This means that (i) $G$ has at most two proper nontrivial normal subgroups, each of which is an atom and a coatom in the normal subgroup lattice of $G$. Therefore (ii) at most two of the coatoms in the group topology lattice can be non-Hausdorff. Hence (iii) at least three of the atoms in the lattice of group topologies on $G$ are pairwise complementary Hausdorff topologies. Item (i) splits the full problem into these cases: Case I. $G$ is simple and has 5 pairwise complementary proper nontrivial Hausdorff group topologies. Case II. $G$ has exactly one nontrivial proper normal subgroup $N$, $G/N$ has no nontrivial proper group topology, but $G$ has four pairwise complementary Hausdorff topologies that are nontrivial and proper. Case III. $G=S\times T$ is a product of two simple groups, where neither $S$ nor $T$ has any nontrivial proper group topology, but $G$ has three pairwise complementary Hausdorff topologies that are nontrivial and proper. I have learned that there is no known group that has even one pair of complementary proper Hausdorff group topologies. No abelian group can have complementary proper Hausdorff group topologies, as proved in Relationship between the meet and join operators in the lattice of group topologies, Bradd Clark and Victor Schneider, Proceedings AMS 98(4) (1986) 681-682. The general problem of whether there can exist complementary proper Hausdorff group topologies is posed as Question 7.1(a) in On transversal group topologies, Dikran Dikranjan, Mikhail Tkachenko, Ivan Yaschenko, Topology and its Applications 153 (2005) 786-817. Here is a construction of a group topology that might be applicable to derive properties of groups in Cases I, II or III. It is a generalization of the $\mathcal T_N$ topology. Let $H$ be a subgroup of $G$. Let $S_H$ be the set of subgroups of $G$ that are finite intersections of conjugates of $H$. The groups in $S_H$ form a system of neighborhoods of $e$ for a group topology $\mathcal T_H$ on $G$. The topology $\mathcal T_H$ is nontrivial if $H$ is proper and is proper if $\{e\}\notin S_H$. $\mathcal T_H$ is Hausdorff if $H$ is core-free. Here is an example that hints about how this construction might be applicable to the problem. Suppose that $G = A_{\omega}$ is the alternating group on $\omega$, $E\subseteq \omega$ is (say) the set of even natural numbers, and $H$ is the stabilizer of $E$. Every group in $S_H$ is the stabilizer of some set that is the symmetric difference of $E$ and a finite subset of $\omega$. Therefore $\mathcal T_H$ is a nontrivial proper Hausdorff group topology that is tied to the set $E$. You can produce lots more Hausdorff group topologies on $A_{\omega}$ by varying the set $E$. For this construction to fail to produce nontrivial proper Hausdorff group topologies, it must be that whenever $H\leq G$ is proper and core-free, then $\{e\}$ is the intersection of finitely many conjugates of $H$.<|endoftext|> TITLE: Notion of infinity in categories QUESTION [9 upvotes]: Please excuse me if the question is too vague or uninteresting. Let $\mathcal{C}$ be a category and $A$ an object in $\mathcal{C}$. Motivated by the equivalence of Dedekind-finiteness and finiteness for sets (assuming Choice), let us define that an object $A$ is $\mathcal{C}$-infinite if there is an object $A'$ in $\mathcal{C}$ corresponding to a proper subobject of $A$ and an isomorphism $f: A \to A'$ in $\mathcal{C}$. In the category Set, we have the property that ($B$ is Set-infinite and $B \subseteq A) \implies A$ is Set-infinite. The analogous property does not hold (with this definition of infinity) over arbitrary categories. For instance, the additive group $\mathbb{Q}$ has $\mathbb{Z}$ as a proper subgroup, is finite in the category Grp (since any homomorphism $f: \mathbb{Q} \to \mathbb{Q}$ must actually be an isomorphism), but $\mathbb{Z}$ is Grp-infinite by $g:\mathbb{Z} \to 2\mathbb{Z}$ with $z \mapsto 2z$. In which categories would a version of this transitivity property hold? What about other properties depending on a notion of infinity? Or determining whether a given object in a category is infinite or not? A result somewhat related to this notion is that a Banach space isomorphic to all its infinite-dimensional closed subspaces is isomorphic to a separable Hilbert space. Any such Banach space is obviously Ban-infinite. I don't know much Category Theory or Logic, but this has been troubling me for a while (I'm an undergraduate student). Thanks in advance for your contributions. REPLY [4 votes]: This is an enlarged comment. See section $8.2$ of $[1]$. Transitivity Property: Let $B$ be $\mathscr{C}$-infinite, and let $B\hookrightarrow A$. Then $A$ is $\mathscr{C}$-infinite. This is not satisfied for all $\mathscr{C}$; e.g., let $\mathscr{C}$ be the category of presheaves on $(\mathbb{N},\leq)$ (Counterexample $8.2.4$, $[1]$). The following theorem gives a sufficient condition (there might be many more) for the transitivity property to hold. Theorem 1 (Proposition $8.2.3$, $[1]$): Let $A$ be an object of a topos $\mathscr{C}$ such that there is a complemented subobject $B$ of $A$ that is $\mathscr{C}$-infinite. Then $A$ is itself $\mathscr{C}$-infinite. Proof (taken from $[1]$): Let $C\hookrightarrow B$ be a subobject of $B$ not containing the global element $b:\mathbf{1}\to B$ and an isomorphism $C\to B$. As $\models b\in B$, it follows that $\models\neg(b\in \overline{B})$; hence $\models\neg(b\in S\cup\overline{B})$. Thus $S\coprod\overline{B}$ is a proper subobject of $A$. There is an isomorphism $A\to S\coprod \overline{B}$. Hence $A$ is $\mathscr{C}$-infinite. $\square$ Bibliography $[1]$ Francis Borceux, Handbook of Categorical Algebra: Volume $3$, Sheaf Theory. $522$ pp. Cambridge University Press. Published $1994$.<|endoftext|> TITLE: Generic absence of non-trivial first integrals of geodesic flows QUESTION [6 upvotes]: Is it true that given a smooth manifold M (with or without boundary), a "generic" metric g on M does not possess any non-trivial (non-constant) first integral for the geodesic flow induced by g on the unit sphere bundle? What if we restrict ourselves to considering the first integrals of the geodesic flow which is a polynomial in the momenta? I did some search, it seems that for the case that the first integrals are linear in momenta, then a generic metric does not possess such first integral. This somehow related to the fact that a generic metric does not possess Killing vector fields. What about for higher order (Killing tensors)? Thanks a lot. REPLY [6 votes]: This is an amplification of my comment on Vladimir's answer. It's actually not at all hard to see the generality of the surface metrics that admit a $k$-th degree polynomial first integral of their geodesic flow. Here is a sketch: The question is local, so we can look at metrics on an open set $U\subset\mathbb{R}^2$. Moreover, modulo diffeomorphism, we can assume that the metric is conformal, i.e., $g = e^{2u}\bigl(dx^2+dy^2\bigr)=e^{2u}dz\circ d\bar{z}$, with cometric $$ \hat g = e^{-2u}\,\frac{\partial}{\partial z}\circ\frac{\partial}{\partial \bar{z}}, $$ which is a function on the symplectic manifold $T^*\mathbb{R}$. Now, a polynomial first integral of degree $k$ is a function $p$ on the tangent bundle of $\mathbb{R}^2$ of the form $$ p = v_0(x,y)\,dx^k + v_1(x,y)\,dx^{k-1}dy + \cdots + v_k(x,y)\,dy^k. $$ Let $\hat p:T^*\mathbb{R}^2\to\mathbb{R}$ be its $g$-dual, considered as a function on $T^*\mathbb{R}$. The condition that $p$ be constant on the geodesic flow of $g$ is simply that $\hat g$ and $\hat p$ Poisson commute, i.e., $$ \left\{\hat g, \hat p\right\} = 0. $$ Since the expression $\left\{\hat g, \hat p\right\}$ is polynomial of degree $k{+}1$ in the momenta, this is $k{+}2$ first-order equations for the $k{+}2$ unknowns $u, v_0,\ldots, v_k$. It is not difficult to see that this quasilinear first order system can locally be placed in Cauchy-Kowalewskaya form, so analytic solutions are determined by specifying these $k{+}2$ functions analytically along an appropriately non-characteristic curve. (Not all solutions are real-analytic, however.) Now, there is still too much symmetry in this formulation, namely the conformal transformations of the complex plane. Generically, one can get rid of this as follows. If one writes $\hat p$ in the form $$ \hat p = h_0(z,\bar z)\ \left(\frac{\partial}{\partial z}\right)^k + h_1(z,\bar z)\ \left(\frac{\partial}{\partial z}\right)^{k-1} \circ \frac{\partial}{\partial {\bar{z}}} + \cdots + h_k(z,\bar z)\ \left(\frac{\partial}{\partial {\bar{z}}}\right)^k $$ where $\overline{h_j} = h_{k-j}$, one finds that the vanishing of the Poisson bracket implies that $h_0$ is actually holomorphic. Now, one can always assume that $h_0$ is not identically vanishing because, otherwise, one could factor out a number of copies of $\hat g$ from $\hat p$ and so reduce the order of the integral. Now, use a holomorphic transformation to make $h_0\equiv1$. This reduces the number of unknowns by $2$ (as it fixes the real and imaginary parts of $h_0$) and reduces the number of equations by $2$ (because two of the equations are now identities), and the resulting first order system of $k$ equations for $k$ unknowns now has exactly the right generality and can still be put in C-K form locally, showing that the general (analytic) solution depends on $k$ functions of one variable. All of what I have written was known more than a century ago, and one can find an account of it in Volume 3 of Darboux', Leçons sur la Théorie Générale des Surfaces et les Applications Géométriques du Calcul Infinitésimal. Possibly, what Vladimir is referring to about the difficulties is something else: Because we now know that there are restrictions on a sufficiently high order jet of a metric in order for it to admit a polynomial geodesic first integral of degree $k$, we might want to know what those conditions are, explicitly in terms of some known invariants. However, this turns out to be extremely complicated once one goes beyond degrees $1$ and $2$. What one can say is this (which gives a more precise version of the Theorem that Vladimir states in his answer): Consider the space $\mathsf{G}_\ell$ consisting of $\ell$-jets of surface metrics modulo diffeomorphism. This is a (singular) space that is finite dimensional, of dimension $1$ when $\ell=2$ and of dimension $\tfrac12(\ell{-}2)(\ell{+}1)$ when $\ell\ge 3$. (The singular locus of $\mathsf{G}_\ell$ has properly smaller dimension, and is nonempty for $\ell\ge 3$.) Carefully applying the above analysis shows that, when $\ell\ge2k{+}2$, the locus $\mathsf{F}_\ell(k)\subset \mathsf{G}_\ell$ of diffeomorphism classes of $\ell$-jets of metrics that admit a nontrivial polynomial geodesic first integral of degree $k$ or less has codimension $C_\ell(k) = \tfrac12(\ell{-}2k{-}2)(\ell{+}1)+1$ in $\mathsf{G}_\ell$.<|endoftext|> TITLE: Applications of anabelian geometry to Galois representations? QUESTION [13 upvotes]: One aspect of anabelian geometry is the study of the action of the absolute Galois group of a field $K$ on the etale fundamental group $\pi_1(X_\overline K)$, where $X$ is a (anabelian) variety and $X_\overline K=X\times_K \overline K$. This yields a representation by outer automorphims of the fundamental group. Having been recently exposed to some of the ideas of Grothendieck in his Esquisse and letter to Faltings that this representation seems to be quite a natural action, given the definition, it seems to beg the (perhaps naive) question: how have the ideas of anabelian geometry, or homotopical algebra in general, been applied to the study of Galois representations? REPLY [9 votes]: The standard theory of Galois representations is concerned with the actions of absolute Galois groups of number fields over abelian groups (in particular, vector spaces). Anabelian geometry is about the actions (by outer automorphisms) of the same absolute Galois group over étale fundamental group of varieties which are "very far from abelian" (hence the name "anabelian".) So the two theories seem go in rather opposite directions. To obtain applications of anabelian methods to Galois representation, you therefore have to lose aside much of the "anabelianness". The outer action of the absolute Galois group $G_K$ over the fundamental group $\pi_1(X_{\bar K})$ induces an action on the quotient of $\pi_1$ by any of its characteristic subgroups. If you quotient $\pi_1(X_{\bar K})$ by its derived subgroup, you obtain $H_1(X_{\bar K})$, and basically your are back to the study of the action of Galois groups on the cohomology of varieties -- you have completely lost any anabelian aspect. Instead you can quotient $\pi_1(X_{\bar K})$ by the smallest characteristic subgroup such that the quotient is nilpotent. The resulted quotient, $\pi_1^{nil}$ is a nilpotent group with action of $G_K$, and you can study it. This was done in the case where $X$ is the projective line minus three points (arguably the simplest anabelian curve) by Deligne in his paper "le groups fundamental de la droits projective moins trots points", in the book "Galois groups over $\mathbb Q$", from a conference held in 1987. The group $\pi_1^\nil$ is much more complicated and richer than the abelian $H_1$ (just a free abelian group with two generators in whatever category you want to consider it) but still is nilpotent, so, as Deligne puts it, "far from the anabelian dream of Grothendieck". The reward you get about Galois representations from this work and subsequent papers is a geometric construction of all "good reductions" extensions of Tate representations, which is quite important. I believe that many of those representations can be proved to exist by traditional Galois representations method (involving Galois deformations for instance) but I 'm not sure for all of them, and anyway it is better to have a geometric construction.<|endoftext|> TITLE: A query about Hatcher flow on arc complex QUESTION [5 upvotes]: In the paper "Triangulations of Surfaces" Hatcher proved that the arc complex associated to a punctured surface is contractible. The main proof is divided into two parts. In the first part he assumes each boundary curve to have exactly one end point and constructed a flow (Hatcher flow) to show it is contractible. In the second part he used induction and showed that for multiple points in the boundary the arc complex is the suspension of the of the previous arc complex with one less boundary point. My question is Why can't we use the flow for multiple endpoints directly? The definition works perfectly well for the flow. Will there be any problem with the continuty of the flow? Thanks in advance. REPLY [10 votes]: The flow consists of a sequence of surgeries using one fixed oriented arc $\alpha$ to cut (and isotope) all other arcs $\beta$ to remove one point of $\alpha\cap\beta$ at a time. Each surgery cuts one arc into two arcs. Say $\beta$ is cut into $\beta'$ and $\beta''$. It can happen that one of these two arcs, say $\beta'$, is trivial, cutting off a disk $D$ from the surface such that $D$ intersects the given set $V$ of allowable endpoints of arcs only in $\partial\beta'$. In this case one discards $\beta'$ and keeps $\beta''$. The bad situation would be that this happens for both $\beta'$ and $\beta''$, so both would be discarded. This can't happen if $V$ contains at most one point in each boundary component of the surface, but it can happen when some boundary component has two or more points in $V$. The surgery flow could then produce an empty arc system, which is not allowed. There are two situations where the arc complex is homeomorphic to a sphere and hence not contractible, so the surgery process must fail, namely when the surface is a disk with $V$ contained in its boundary, or an annulus with $V$ contained in one of its boundary circles. The simplest case is when the surface is a disk with $V$ consisting of four boundary points. The arc complex is then $S^0$, two points, represented by two arcs connecting the two nonadjacent pairs of points in $V$. Surgering one arc by the other then produces only trivial arcs. On my webpage is an updated version of the original 1991 paper containing the construction of this flow. The update adds the exceptional cases when the arc complex is a sphere, and it also contains a few small improvements in the exposition. The update is titled "Triangulations of surfaces" and can be found under the heading "Updates of older papers".<|endoftext|> TITLE: Integer valued polynomial through some points with rational coordinates QUESTION [9 upvotes]: I asked this question on MSE about 5 months ago, but, even after offering a bounty, I didn't receive any answer, I hope this question isn't too easy for MO. If we have a set of points $(x_i,y_i)$ with rational coordinates and distinct $x$-coordinates, such that for every (if any) $x_i\in\Bbb{N}$ the corresponding $y_i$ is also an integer, is it always possible to build an integer valued polynomial with rational coefficents passing through those points? REPLY [21 votes]: No. The integer-valued polynomials have a basis (over $\mathbb{Z}$) given by the Newton polynomials $$\displaystyle {x \choose n} = \frac{x (x - 1)\dots(x - (n-1))}{n!}$$ and as such if $$\displaystyle f(x) = \sum a_n {x \choose n}, a_n \in \mathbb{Z}$$ is an integer-valued polynomial, then, for example, $f \left( - \frac{1}{2} \right)$ must have denominator a power of $2$, since $$\displaystyle {-\frac{1}{2} \choose n} = \frac{(-1)(-3)\dots(-2n+1)}{2^n n!} = (-1)^n \frac{ {2n \choose n} }{2^n}$$ and hence no such polynomial $f$ can pass through a point like $\left( - \frac{1}{2}, \frac{1}{3} \right)$.<|endoftext|> TITLE: Endomorphisms of a maximal ideal of a local ring QUESTION [8 upvotes]: Let $R$ be a commutative local ring with maximal ideal $\mathfrak{m}$. Is it true in general that $\text{Hom}_R(\mathfrak{m},\mathfrak{m})\cong \text{Hom}_R(\mathfrak{m}, R)$? What if the Krull dimension of $R$ is equal to one? REPLY [10 votes]: Yes, this is always true. Let us assume $\text{Hom}_R(\mathfrak m,\mathfrak m)\neq\text{Hom}_R(\mathfrak m,R)$, so we have a homomorphism $f:\mathfrak m\rightarrow R$ with image not contained in $\mathfrak m$: say $f(x)\in R^\times$ for some $x\in\mathfrak m$. We can w.l.o.g. assume that $f(x)=1$ (since $f(x/f(x))=1$). Now, we must have $y=yf(x)=f(xy)=xf(y)$ for all $y\in\mathfrak m$. Hence, $\mathfrak m$ is generated by $x$ and multiplication by $x$ is injective (since $xy\neq0$ for $0\neq y\in\mathfrak m$ follows from this equation and obviously $xy\neq0$ for $y\in R^\times$). But then we have $\mathfrak m\cong R$ as $R$-modules.<|endoftext|> TITLE: In the category of sets epimorphisms are surjective - Constructive Proof? QUESTION [20 upvotes]: The statement that surjective maps are epimorphisms in the category of sets can be shown in a constructive way. What about the inverse? Is it possible to show that every epimorphism in the category of sets is surjective without reverting to a proof by contradiction / negation? REPLY [3 votes]: I want to write down a proof that comes naturally, in a way. This proof assumes that you can form the quotient of a set modulo an equivalence relation, but does not require powersets. (So it works in any pretopos.) We have sets $ A $ and $ B $ and a function $ f $ from $ A $ to $ B $. We know that $ f $ is an epimorphism in the category of sets, and we want to prove that $ f $ is a surjection. Consider the cokernel pair of $ f $, that is the pushout of $ f $ along itself. Just as epimorphisms in an abelian category have trivial cokernels, so epimorphisms in more general categories have trivial cokernel pairs. So this is a natural thing to look at. $$ \matrix { A & \overset { \textstyle f } \rightarrow & B \\ \llap f \downarrow & & \downarrow \rlap { k _ 1 } \\ B & \underset { \textstyle k _ 0 } \rightarrow & C } $$ The cokernel pair is a set $ C $ defined as a quotient of the disjoint union $ B \uplus B $; if $ y $ is an element of $ B $, then I'll write $ y _ 0 $ and $ y _ 1 $ for elements of $ B \uplus B $. Then we have an equivalence relation on $ B \uplus B $ according to which $ y _ 0 $ and $ y _ 1 $ are equivalent iff $ y $ belongs to the image of $ f $, and no other equivalences exist besides the reflexive ones. (More explicitly, $ y _ i $ and $ z _ j $ are equal iff $ y = z $ and $ i = j $ or $ y = z $ and $ y = f ( x ) $ for some $ x $.) Then $ C $ is the quotient of $ B \uplus B $ under this equivalence relation. The cokernel pair also comes equipped with two inclusion/quotient maps from $ B $, while I'll call $ k _ 0 $ and $ k _ 1 $; $ k _ i $ maps $ y $ to the equivalence class of $ y _ i $ in $ C $. Given an element $ x $ in $ A $, $ k _ 0 ( f ( x ) ) = k _ 1 ( f ( x ) ) $ in $ C $ because $ f ( x ) $ is in the image of $ f $. So since $ f $ is an epimorphism, $ k _ 0 = k _ 1 $. This means that for every element $ y $ of $ B $, $ k _ 0 ( y ) = k _ 1 ( y ) $, so $ y $ is in the image of $ f $. So $ f $ is surjective.<|endoftext|> TITLE: Bott's Formula for Grassmannians QUESTION [5 upvotes]: Bott's Formula gives the dimension of the cohomology $H^{q}(\mathbb{P}^{n}, \Omega_{\mathbb{P}^{n}}^{p}(k))$ of the $k$-twisted sheaf of $p$-differential forms on the projective space $\mathbb{P}_{\mathbb{C}}^{n}$. I was wondering if there is a similiar formula for complex Grassmannians $\mathbb{G}(r, n)$ instead of projective spaces $\mathbb{P}_{\mathbb{C}}^{n}$. I searched for a while and I only found results that provide necessary and sufficient conditions for the vanishing of such cohomology groups. When such cohomology groups are non-zero I did not find any results providing the dimension of such cohomologies. Do you know if there is any formula for the dimension of the cohomology groups $H^{q}(\mathbb{G}(r, n), \Omega_{\mathbb{G}}(k))$ as $\mathbb{C}$-vector spaces? Where could I find it? Thank you very much! REPLY [4 votes]: Yes, there are Bott's type formulas for Grassmannians. For $r = 1$ in Lemma 0.1 of this paper http://arxiv.org/pdf/alg-geom/9306010v2.pdf In general you can look at: D. Snow, Cohomology of twisted holomorphic forms on Grassmann manifolds and quadric hypersurfaces, Math. Ann. 276 (1986), 159-176.<|endoftext|> TITLE: A conjectured formula for Apéry numbers QUESTION [20 upvotes]: A conjecture by the late Romanian mathematician Alexandru Lupas. Posted in sci.math in 2005, but no proof was found. Physicist Alan Sokal just reminded me of it, saying it was related to something he is working on. Let $P_n(z)$ be the Legendre polynomials, defined by the generating function $$ \big(1-2tz+t^2\big)^{-1/2} = \sum_{k=0}^\infty t^k P_k(z) . $$ Let $g(\alpha,\beta) = 4\cos(2\alpha)+8\sin(\beta)\cos(\alpha)+5$ be defined for $(\alpha,\beta) \in (-\pi,\pi)\times (-\pi,\pi)$ . Let $A_n$ be these Apéry numbers $$ A_n = \sum_{k=0}^n \binom{n}{k}\binom{n+k}{k}\sum_{j=0}^k \binom{k}{j}^3 . $$ examples $$ P_0(z)=1;\qquad P_1(z)=z;\qquad P_2(z)=\frac{3}{2}\;z^2-\frac{1}{2};\qquad P_3(z)=\frac{5}{2}\;z^3-\frac{3}{2}\;z; \\ A_0 = 1;\qquad A_1=5;\qquad A_2=73;\qquad A_3=1445;\qquad A_4=33001 . $$ Prove or disprove: $A_n$ is the average of $P_n\circ g$. More explicitly: for all natural numbers $n$, $$ A_n = \frac{1}{4\pi^2}\int_{-\pi}^\pi \int_{-\pi}^\pi P_n\big(g(\alpha,\beta)\big)\;d\beta\;d\alpha $$ This is surely true (Sokal says he has checked it through $n=123$). But can you prove it? additional notes Also $$ A_n = \sum_{k=0}^n \binom{k}{n}^2\binom{n+k}{k}^2 $$ The conjecture should be equivalent to $$ \frac{1}{4\pi^2}\int_{-\pi}^\pi\int_{-\pi}^\pi\frac{d\alpha\;d\beta}{\sqrt{ t^2-2t(4\cos(2\alpha)+8\sin\beta\cos\alpha+5)+1}\;} =\sum_{k=0}^\infty A_k t^k $$ In the integral we can change variables to get $$ \frac{1}{\pi^2}\int_{-1}^1\int_{-1}^1\frac{dp\;dq}{\sqrt{1-p^2} \sqrt{1-q^2}\sqrt{1-2t(8pq+8q^2+1)+t^2}\;} $$ REPLY [23 votes]: Here as an approach. It is known that: $P_{n}(x) = \sum_{k=0}^{n}\binom{n}{k}\binom{-n-1}{k}\left(\frac{1-x}{2}\right)^{k}$ (see for example http://en.wikipedia.org/wiki/Legendre_polynomials) Therefore these expressions are equal $$ \sum_{k=0}^{n}\binom{n}{k}\binom{n+k}{k}\sum_{j=0}^{k}\binom{k}{j}^{3}=\sum_{k=0}^{n}\binom{n}{k}\binom{-n-1}{k}\frac{1}{4 \pi^{2}}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\left(\frac{1-g(x,y)}{2}\right)^{k}dydx $$ if $$ \sum_{j=0}^{k}\binom{k}{j}^{3}=\frac{1}{4 \pi^{2}}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\left(\frac{g(x,y)-1}{2}\right)^{k}dydx $$ One can see that $$ \left(\frac{g(x,y)-1}{2}\right)^{k} = 4^{k} (\cos^{2}x + \cos x \sin y )^k $$ Therefore we get $$ \frac{1}{4 \pi^{2}}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\left(\frac{g(x,y)-1}{2}\right)^{k}dydx = \frac{4^k}{4\pi^{2}}\int_{0}^{2\pi}\int_{0}^{2\pi}\sum_{j=0}^{k}\binom{k}{j}\cos^{2k-j}x \sin^{j} y dxdy $$ Since \begin{align*} \int_{0}^{2\pi} \cos^{m} x dx=\int_{0}^{2\pi} \sin^{m} x dx = \frac{2\pi} {2^m}\binom{m}{m/2} \quad \text{for} \quad m \quad \text{even, otherwise =0} \end{align*} We get $$ \frac{1}{4 \pi^{2}}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\left(\frac{g(x,y)-1}{2}\right)^{k}dydx = \sum_{\ell=0}^{k/2}\binom{k}{2\ell}\binom{2\ell}{\ell}\binom{2(k-\ell)}{k-\ell} $$ and now we want to show that $$ \sum_{j=0}^{k}\binom{k}{j}^{3}=\sum_{\ell=0}^{k/2}\binom{k}{2\ell}\binom{2\ell}{\ell}\binom{2(k-\ell)}{k-\ell} $$ The last equality follows from the identity: $$ \sum_{j=0}^{k}\binom{k}{j}^{3}=\sum_{\ell=k/2}^{k}\binom{k}{\ell}^{2}\binom{2\ell}{k} $$ See for example http://arxiv.org/pdf/math/0311195v1.pdf formula (2). because $$ \binom{k}{2\ell}\binom{2\ell}{\ell}\binom{2k-2\ell}{k-\ell}=\binom{k}{k-\ell}^{2}\binom{2k-2\ell}{k} $$<|endoftext|> TITLE: Is the absolute of a compact space the projective limit of the Stone-Čech compactifications of its open dense subsets? QUESTION [5 upvotes]: Is the following statement true, and if it is, does someone have a reference? Let $X$ be a compact (i.e., compact and Hausdorff) topological space. Then the Gleason space (=Iliadis absolute, =Stone dual of the Boolean algebra of regular open sets) of $X$ is (naturally homeomorphic to) the projective limit of $\beta V$ for $V$ ranging over dense open subsets of $X$, where $\beta$ is the Stone-Čech compactification functor (and, of course, the maps $\beta V'\to \beta V$ are obtained from the inclusion $V'\to V$ by functoriality). Since every $V$ and also every $\beta V$ has the same regular open algebra, hence the same Gleason space $E(X)$, there is at least a continuous map from $E(X)$ to the projective limit in question, and I am asking whether it is a homeomorphism commuting with the obvious maps from both of these spaces to $X$. I don't suppose it should be very difficult, but I'd prefer a reference. I looked in Porter & Woods's book Extensions and Absolutes of Hausdorff Spaces with no success. There is, however, a related exercise 6Z (p.521) which describes $E(X)$ as as different projective limit (whose morphisms are finite-to-one; essentially by "splitting" finite systems of regular closed sets). The projective limit of the $\beta V$ for $V$ open dense occurs in Fine, Gillman & Lambek's Rings of Quotients of Rings of Functions, but the words "absolute" or "Gleason" don't seem to appear anywhere. Edit: It turns out that I didn't look carefully enough in Fine, Gillman & Lambek's book: the statement does appear there (even if the term "absolute" is not mentioned), by combining theorem 6.9 with theorem 11.15. I much prefer the proof that Adam Przeździecki gives below, however, because it is "direct" (describing the arrow on the spaces themselves) rather than "dual" (through maximal ideals and rings of functions), so it is more enlightening. REPLY [3 votes]: Yes, your space, let's call it $GX$, is the projective cover $EX$ of $X$ in the category of compact spaces. Recall that the projective cover $p:EX\to X$ can be characterized by the properties: $EX$ is projective, that is every epimorphism $f:Y\to EX$ has a right inverse. $p$ is irreducible, that is for every closed subset $S\subsetneq EX$ we have $p(S)\neq X$. In the category of compact spaces this is the same as the absolute you mentioned. I will compare your space to the construction by Rump ("The absolute of a topological space and its application to abelian l-groups", Appl. Categ. Structures 17 (2009), 153–174), Proposition 17: Let $\mathcal{B}X$ be the set of regular subsets of $X$ ($U$ is regular if $U={\rm int}\bar U$. Let $\Phi X$ be the family of finite subsets $A\subseteq\mathcal{B}X$ such that: $A$ is pairwise disjoint. $\bigcup A$ is dense in $X$. Let $\bar A=\coprod\{\bar U\mid U\in A\}$. The set $\Phi X$ is ordered by inclusions. Rump proves that the map $$p:\lim_{A\in\Phi X}\bar A\longrightarrow X $$ is the projective cover of $X$. The map is the projection from the limit onto the axis $A=\{X\}$ - the terminal element of $\Phi X$. We have obvious epimorphisms $\beta\bigcup A\to\bar A$ however, the indexing sets of this and your limits are not yet compatible. Let $\Phi_0X=\{\bigcup A\mid A\in\Phi X\}$ - this is a set of certain dense subsets of $X$. The terminal member $\{X\}$ of $\Phi X$ belongs to $\Phi_0X$. We have an epimorphism $\pi:\Phi X\longrightarrow\Phi_0 X$ given by $\pi(A)=\bigcup A$. If $A,B\in\Phi X$ are such that $\bigcup A=\bigcup B$ then we define $C=\{U\cap V\mid U\in A, V\in B, U\cap V\neq\emptyset\}$. Obviously $\bigcup A=\bigcup C=\bigcup B$ and we have natural homeomorphisms $\bar A\cong \bar C\cong\bar B$ so that $\pi$ induces homeomorphism of the limits: $$ \pi^*:\lim_{A\in \Phi_0X}\bar A\stackrel{\cong}{\longrightarrow}\lim_{A\in \Phi X}\bar A $$ Let $\mathcal{V}X$ be the set of dense open subsets of $X$, as in you question. Since $\Phi_0X\subseteq\mathcal{V}X$, we can now assemble our epimorphisms into a single epimorphism of the limits $$ f:GX=\lim_{V\in \mathcal{V}X}\beta V{\longrightarrow}\lim_{A\in \Phi_0 X}\beta A{\longrightarrow}\lim_{A\in \Phi_0 X}\bar A=EX $$ The left arrow is the projection induced by the inclusion. The right arrow is induced by the epimorphisms $\beta A\to\bar A$ where $\bar A$ can be defined as $\bar A_0$ where $A_0$ is any element of $\Phi X$ such that $\pi(A_0)=A$. It remains to show that $f$ is one-to-one. Suppose that $x\neq y$ in $GX$. Then there exists $V\in\mathcal{V}X$ such that their projections on $\beta V$ are different: $x_V\neq y_V$. Then there exist disjoint regular open subsets of $\beta V$, namely $W_x\ni x_V$ and $W_y\ni y_V$, we may assume that $W_x\cup W_y$ is dense in $\beta V$. Then $V_x=W_x\cap V$ and $V_y=W_y\cap V$ are regular open disjoint subsets of $V$. Then $U_x={\rm int}\bar V_x$ and $U_y={\rm int}\bar V_y$ are disjoint regular open subsets of $X$ and $U_x\cup U_y$ is dense in $X$ hence it belongs to $\Phi_0X$. The inclusions $$ V\supseteq V_x\cup V_y \subseteq U_x\cup U_y $$ induce epimorphisms $$ \beta V \stackrel{a}{\longleftarrow}\beta (V_x\cup V_y)\stackrel{b}{\longrightarrow}\bar U_x\sqcup \bar U_y $$ We have $a^{-1}(x)\subseteq\beta V_x$ and then $b(a^{-1}(x))\subseteq\bar U_x$. Similarly for $y$ hence $f(x)$ and $f(y)$ are separated on the $U_x\cup U_y$ axis of $\lim_{A\in \Phi_0 X}\bar A$ hence the injectivity of $f$. Well, it took much longer than I expected. Let me mention that we may have a much smaller limit: $EX\cong\lim_{A\in\Phi X}A$ where $A$'s are viewed as finite sets.<|endoftext|> TITLE: Lower bound on the irrationality measure of $\pi$ QUESTION [6 upvotes]: There seems to be a lot of work on the upper bound for the irrationality measure of $\pi$, but I could not find anything on a lower bound except the general $\mu(\pi)\geq2$. Looks like it is the best one known, is it? What is the conventional wisdom on what $\mu(\pi)$ should be? Does a bound follow from some grand conjecture (Schanuel's, Riemann hypothesis, etc.)? I am interested because Alekseyev proved (Theorem 2) that $\frac{1}{n^u\sin{n}}$ converges to $0$ if $\mu(\pi)u+1$. If we replace $\sin{n}$ by a sequence of independent random variables $\xi_n$ uniformly distributed on $[-1,1]$ then $\frac{1}{n^u\xi_n}$ converges to $0$ almost surely iff $\sum\frac{1}{n^u}<\infty$. Thus, if $\mu(\pi)>2$ then $\sin{n}$ is not 'uniformly distributed' on $[-1,1]$. On the other hand, if $\mu(\pi)=2$ then $\pi$ is just as irrational as $e$ and irrational algebraic numbers, while some people seem to believe that it should be 'more irrational' than that. REPLY [10 votes]: The expected value is $2$, simply because almost all real numbers have irrationality measure $2$. Note that this is "almost all" in the sense of measure theory. In the sense of topological category, it's the other way around: the set of reals of measure $\infty$ is residual. As I pointed out recently, it's a bit mysterious why measure theory makes better predictions than topological category for "naturally defined" numbers like $e$ or $\pi$, but experimentally, this appears to be the case, as the upper bound on the irrationality measure of $\pi$ is finite.<|endoftext|> TITLE: Is there an algebraic number that cannot be expressed using only elementary functions? QUESTION [15 upvotes]: (this is basically a repost of a question I asked at M.SE last year) Is there an explicit real algebraic number (such that we can write its minimal polynomial and a rational isolating interval) that cannot be expressed as a combination of the constants $\pi, e,$ integers and elementary functions (rational functions, powers, logarithms, direct and inverse trigonometric functions)? In case it is an open question, can you give an example of an algebraic number such that we do not know how to express it in elementary functions? REPLY [23 votes]: I addressed this exact question in my American Mathematical Monthly paper, What is a closed-form number? Corollary 1 in that paper states that if Schanuel's conjecture holds, then the EL numbers (i.e., the numbers expressible according to your list of rules) that are algebraic are precisely the numbers expressible using radicals. So any algebraic number that is not expressible in radicals would be an answer to your question. However, Schanuel's conjecture is not known to be true. I believe that your question is still open.<|endoftext|> TITLE: Invertible operator with countable spectrum QUESTION [6 upvotes]: Let $H$ be a separable Hilbert space and $A$ is an invertible bounded operator on $H$. Can we approximate $A$ with an invertible operator $B$ such that $sp(B)$ is a countable set? Motivation: If the answer is yes, this would give's us an alternative proof of connected ness of $GL(H)$. This alternative proof is identical to a short and interesting proof of connectedness of $GL(n,\mathbb{C})$, in page 19 of "Introduction to the Baums Connes conjecture" by Alain Valette REPLY [8 votes]: The question has already been answered above (by Mike Jury). Here is another way of arguing: Any operator with countable spectrum is in the closure of the invertible operators. So are their limits. But again, the unilateral shift provides an example of an operator not in the closure of the invertibles. For if $\|S-X\|<1$ then $\|I-S^*X\|<1$, and so $S^*X$ is invertible. $X$ cannot be invertible, since otherwise $S^*$ would also be invertible. So no translate $S+\lambda I$ can be approximated by operators with countable spectrum.<|endoftext|> TITLE: Classifying TQFTs with 1d vector spaces QUESTION [13 upvotes]: To what extent have people classified $n$-dimensional TQFTs that assign a 1-dimensional vector space to every compact oriented $(n-1)$-manifold? I have some vague reasons to suspect that the classification depends heavily on $n \bmod 8$, but I'm having trouble finding information on this. REPLY [4 votes]: I recently developed a non-linear $\sigma$ model approach ( http://arxiv.org/abs/1410.8477 ) to address this issue. Let $\text{iTO}_L^n$ be the set of the fully extended TQFTs in $n$-dimensions, that assign a 1-dimensional vector space to every compact oriented $(n−1)$-manifold. (Physically, $\text{iTO}_L^n$ is the set of L-type topologically-ordered phases in $n$-dimensional space-time that have no topological excitations.) Such a set, $\text{iTO}_L^n$, actually from an Abelain group. We find that $\text{iTO}_L^1=\text{iTO}_L^2=\text{iTO}_L^4=\text{iTO}_L^6=0$, $\text{iTO}_L^3=\mathbb{Z}$, $\text{iTO}_L^5=\mathbb{Z}_2$, $\text{iTO}_L^7=2\mathbb{Z}$. Here $\text{iTO}_L^3=\mathbb{Z}$ is generated by $\omega_3$ with d$\omega_3=p_1$ and $p_1$ the first Pontryagin class. Similar results are also obtained in http://arxiv.org/abs/1403.1467 using cobordism approach. But we disagree on the generators of $\text{iTO}_L^3=\mathbb{Z}$ and $\text{iTO}_L^7=2\mathbb{Z}$. For example, in http://arxiv.org/abs/1403.1467 , the generator of $\text{iTO}_L^3=\mathbb{Z}$ is $\frac13 \omega_3$ instead of $\omega_3$. In http://arxiv.org/abs/1406.7278 , the generator of $\text{iTO}_L^3=\mathbb{Z}$ is $\frac16 \omega_3$, instead of $\omega_3$ or $\frac13 \omega_3$.<|endoftext|> TITLE: Why are there so few zero-dimensional polynomial system solvers and is this because there is no real market for them? QUESTION [8 upvotes]: My questions involve the quotes below from wikipedia regarding solving polynomial systems, which given the size of the market for Big Data & Predictive Analysis applications I find puzzling: "This exponential behavior makes solving polynomial systems difficult and explains why there are few solvers that are able to automatically solve systems with Bézout's bound higher than, say 25" "There are at least three software packages which can solve zero-dimensional systems automatically" http://en.wikipedia.org/wiki/System_of_polynomial_equations: Questions: 1) Why are there so few automatic zero-dimensional solvers? 2) Are they just too hard to build? 3) Or is there no market for such a sophisticated solution? 4) Or is the market "cornered": i.e. are the handful of packages in 1) doing the job well enough that there's no need or demand for anything better? 5) Also, would there be a demand for solvers that could push past a Bézout's bound of 25? REPLY [5 votes]: Also not my field, but... 1) Define "few"? There are many available, already for homotopy methods. See http://www.issac-conference.org/2010/assets/SoftwareDemos/VerscheldeFinal.pdf "RELATED SOFTWARE: Homotopy continuation methods for polynomial systems are implemented in various computer programs, listed in alphabetic order: Bertini [1, 2], CONSOL [14], HOM4PS-2.0 [9], HomLab [18, Appendix C], NAG4M2 [10], PHoM [8], POLSYS PLP [26], and POLSYS GLP [20]. Both POL-SYS PLP and POLSYS GLP are part of the development of HOMPACK [24, 25]. Polyhedral homotopies need mixed volumes, computed by DEMiCs [13], Mixvol [4], and MixedVol [5]." As Wikipedia noted, Maple has (at least) two solving methods. Mathematica also has some. 2) They are perhaps not easy to build (problems with numerical errors, tracing homotopy paths), but this is not the critical factor. 3) One of the main "markets" is robotics. You can note that some of the notable authors in the field (Wampler most obviously) have jobs either directly in or related to (such as part-time consultant) the car industry. See also (Wampler and Sommese) http://dx.doi.org/10.1017/S0962492911000067 "While systems of polynomial equations arise in fields as diverse as mathematical finance, astrophysics, quantum mechanics, and biology, perhaps no other application field is as intensively focused on such equations as rigid-body kinematics. This article reviews recent progress in numerical algebraic geometry, a set of methods for finding the solutions of systems of polynomial equations that relies primarily on homotopy methods, also known as polynomial continuation. Topics from kinematics are used to illustrate the applicability of the methods. High among these topics are questions concerning robot motion." The slides of Verschelde's 2003 talk also have some applications listed. http://homepages.math.uic.edu/~jan/Talks/cimpa_first.pdf 4) The market is not cornered, in particular methods that exploit sparsity are of high interest. For instance, in the Math Reviews of Li's 2003 survey, it is noted that half the 100 pages are involved with sparsity. http://www.ams.org/mathscinet-getitem?mr=2009773 "More than half of this almost 100 page survey is devoted to the exploitation of sparsity, as most applications give rise to polynomials which have fewer monomials with nonzero coefficients than general polynomials of the same degree. Quite often, sparse systems then also have far fewer roots compared to systems of general polynomials of the same degree. Any homotopy continuation method which is based only on degree bounds will likely have to trace far too many diverging paths to find all solutions of a sparse polynomial system." See also Section 10 of the above Wampler and Sommese paper, for the types of improvements that are being made (or would like to be seen). 5) Yes, though Wikipedia's information here ("25") is not exactly reliable. For instance, see section 8 of the aforementioned Wampler and Sommese paper (http://dx.doi.org/10.1017/S0962492911000067): "To put some of this in perspective, consider a sequence of polynomial systems arising from discretizing the Lotka–Volterra population model that is presented in Hauenstein et al. (2011). HOM4PS-2.0 and PHCpack both implement the polyhedral homotopy method for solving sparse polynomial systems, while Bertini implements regeneration (discussed in Section 10.2) for solving such systems (Hauenstein et al. 2011). For these systems, the number of solutions over C is the number of paths tracked by the polyhedral homotopy method. Even so, the polyhedral homotopy becomes impractical as the number of variables increases due to the computational complexity of computing a start system via the mixed volume. For example, consider solving the n = 24, 32, and 40 variable instances of the Lotka–Volterra model using a single core running 64-bit Linux. Here the number of non-singular isolated solutions equals $\sqrt{2^n}$ (which equals the mixed volume). For the 24 variable system, PHCpack took over 18 days while HOM4PS-2.0 and Bertini both took around 10 minutes. For the 32 variable system, PHCpack failed to solve the system in 45 days, HOM4PS-2.0 took over 3 days and Bertini took roughly 5 hours. For the 40 variable system, PHCpack and HOM4PS-2.0 both failed to solve the system in 45 days, but Bertini solved the system in under a week. Since regeneration is parallelizable, we also solved the 32 and 40 variable systems using 64 cores (8 nodes each having dual 2.33 GH Xeon 5410 quad-core processors). In parallel, Bertini took less than 8 minutes and 4 hours to solve the 32 and 40 variable polynomial systems, respectively." In this example, Bertini thus found $\sqrt{2^{40}}$ isolated solutions, somewhat larger than the Wikipedia bound of 25. Of course the system should be "special" (likely meaning sparse in some sense) to deal with a large Bezout (or mixed volume) number, but many interesting systems do have particular properties. (The same is true for Gr\"obner basis methods, that the generic behavior might not be very relevant to any specific given instance.) I don't know if by "automatic" you want to be able to solve generic systems of a given Bezout number efficiently, probably that is a different question.<|endoftext|> TITLE: Homotopy groups of Fredholm operators QUESTION [13 upvotes]: If $X$ is separable complex Hilbert space and $\mathcal{F}$ the topological space of Fredholm operators on $X$, then it is well-known, that $$ \pi_0(\mathcal{F}) = \mathbb{Z}\, , $$ i.e. the connected components are classified by the index of the Fredholm operator. But what is about higher homotopy groups? What is known about $\pi_n(\mathcal{F})$ for $n \in \mathbb{N}$? REPLY [11 votes]: This answer is an elaboration of my comments on Paul Siegel's answer. Atiyah showed that there is a natural bijection $$[X,\mathcal{F}] \cong K(X)$$ whenever $X$ is a compact space. Here the left-hand side is the set of unbased homotopy classes of maps into $\mathcal{F}$. This result is proven in the Appendix to Atiyah's book K-theory. As I'll explain, it follows from this theorem that there is an isomorphism $$\pi_n (\mathcal{F}) \cong \widetilde{K} (S^n),$$ where $\widetilde{K}$ denotes reduced $K$-theory. So these groups are trivial for $n$ odd and infinite cyclic for $n$ even. Note that all path components of $\mathcal{F}$ are homotopy equivalent, because (as Atiyah explains) there is an associative product $\mathcal{F} \times \mathcal{F} \to \mathcal{F}$ that makes the set of path components of $\mathcal{F}$ into a group (namely the group $[\{*\}, \mathcal{F}] \cong K(\{*\}) \cong \mathbb{Z}$). So it doesn't matter what basepoint we choose for computing homotopy. Now, to deduce the claimed calculation of homotopy groups, let $\mathcal{F}_0$ denote the index 0 Fredholm operators (which form one connected component of $\mathcal{F}$), Atiyah's bijection restricts to a bijection $[X,\mathcal{F}_0]\cong \widetilde{K}(X)$. Next, I claim that $\mathcal{F}_0$ is simply connected. First, $\mathcal{F}_0$ is path connected by definition, so we just need to check that $\pi_1 (\mathcal{F}_0)=1$. For this, it's enough to check that each loop is (unbased) nullhomotopic, which follows from $[S^1,\mathcal{F}_0]\cong \widetilde{K}(S^1)=0$. Now $\pi_n(\mathcal{F})=\pi_n(\mathcal{F}_0)=\langle S^n,\mathcal{F}_0\rangle \cong [S^n,\mathcal{F}_0] = \widetilde{K} (S^n)$. Here $\langle\, ,\rangle$ means based homotopy classes of (based) maps, which is the same as unbased homotopy classes when the range is simply connected (this is proven in Section 4.A of Hatcher's book Algebraic Topology, for instance).<|endoftext|> TITLE: boundary density of the Von Koch flake QUESTION [5 upvotes]: Given a measurable set $K\subset \mathbb{R}^d$ we consider the occupation ratio $$f_r(x)=vol(K\cap B(x,r))/r^d$$ and especially the asymptotics when $r\to 0$. When $K$ has a fractal boundary and $x$ is on the boundary, it is not clear whether most $x$'s have a large occupation ratio. Has anyone already seen this quantity? Is it studied somewhere? If you want a precise question, let $K$ be a non-negligible set with self-similar boundary with dimension $s$ and positive lower Minkowski content. Do we have $$\liminf_r\int_{\partial K}f_r(x)d\nu(x)>0$$ where $\nu$ is the $s$-dimensional Hausdorff measure? For the Von Koch flake i think the answer is yes but I don't see a general scheme. EDIT: I insist on the fact that the set $K$ itself is not fractal, but its boundary is. REPLY [3 votes]: I don't have a reference for such quantities, but I think the answer to your precise question is negative. Consider the Sierpiński triangle $S$ as the boundary of $K$. Then if you define $K$ as the union $S \cup \text{blue triangle}$ (see the picture below), then $$\liminf_r\int_{\partial K}f_r(x)d\nu(x) = 0.$$ This is (for example) because the $\partial (\text{int}K)$ is the boundary of the blue triangle which has lower dimension than the dimension of $S$. (source) If you would like to require that $\partial (\text{int}K) = S$ still the answer would be negative. To see this, you can consider $K = S \cup \text{red triangles}$ where we have an infinite number of "removed" (red) triangles as shown in the second picture. By having sufficiently large gaps between the scales at which you attach the removed triangles, you still have $$\liminf_r\int_{\partial K}f_r(x)d\nu(x) = 0.$$ The next question could be what happens if you consider $$\int_{\partial K}\liminf_rf_r(x)d\nu(x),$$ but even for this I think the second example, if correctly tuned, should give zero. One could go further with inventing question and ask for example what happens if you require the boundary to be a self-similar type Jordan curve. Then the answer should be positive. As a final remark: I guess you could do the above examples in $\mathbb R$ with a Cantor set, but I preferred a planar case for easier illustration.<|endoftext|> TITLE: Embeddings of finitely generated groups into uniformly convex Banach spaces QUESTION [11 upvotes]: de Cornulier, Tessera, and Valette (Geom. Funct. Anal. 17 (2007), 770-792) conjectured that a finitely generated group $G$ with its word metric admits a bilipschitz embedding into a Hilbert space if and only if $G$ contains a subgroup of finite index isomorphic to $\mathbb{Z}^n$ for some $n$. As far as I know this conjecture is still open. I would like to know whether the situation changes if we allow any uniformly convex space as a target space. More precisely: Question: Does there exist an infinite finitely generated group $G$ such that (1) $G$ does not contain a subgroup of finite index isomorphic to $\mathbb{Z}^n$; (2) $G$, endowed with its word metric, admits a bilipschitz embedding into some uniformly convex Banach space? REPLY [5 votes]: A negative answer to (1) sounds like a natural extension of our conjecture. Some evidence: in the main two cases for which the conjecture is known to hold in the Hilbert case, the same argument also works for arbitrary uniformly convex Banach spaces (UCB): on the one hand non-virtually abelian nilpotent groups (which don't embed bilipschitz into a UCB, by a result of Pauls), and groups with a bilipschitz-embedded 3-regular tree (which don't embed into a UCB by Bourgain, and include non-amenable f.g. groups by Benjamini-Schramm, and also include non-virtually nilpotent solvable f.g. groups).<|endoftext|> TITLE: The Diophantine equation $x^p - 4y^p = z^2$ QUESTION [7 upvotes]: If $p \geq 5$ is a prime, are there any integers $x, y, z > p$ such that $(x, y) = 1$ and $$x^{p} - 4y^{p} = z^{2}$$ REPLY [10 votes]: See Theorem 1.2 of the paper by Bennett and Skinner, which settles the problem for $p\ge 7$ (take there $C=1$ and $\alpha_0=2$). Note that the Bennett-Skinner results are more general. (Earlier work of Darmon and Granville (using Faltings's theorem) showed that there are only finitely many solutions; again for more general such equations.) Finally GH from MO has kindly pointed out an earlier paper of Darmon that handles this particular equation (assuming Shimura-Taniyama) for $p=11$ or $p\ge 17$.<|endoftext|> TITLE: Clutching functions and Classifying maps QUESTION [9 upvotes]: Let $E\xrightarrow{p} \Sigma X$ be a principal G-bundle over a suspension. Write $\Sigma X= C_+X\cup_X C_-X$. Then there are trivialisations of the restrictions $E|_{C_+X}\cong C_+X\times G$, $E|_{C_+X}\cong C_-X\times G$, and the transition function between them over their intersection X is defined by a map $t:X\rightarrow G$. The homotopy class of this map completely characterises the bundle, and the process is in fact reversible. Given such a clutching function, one can construct a unique bundle over the suspension. So if $f:\Sigma X\rightarrow BG$ is a map classifying the G-bundle E, how does this map relate to the clutching function $t:X\rightarrow G\simeq \Omega BG$ ? How does one go between one and the other? The adjoint isomorphism $[\Sigma X, BG] = [X,\Omega BG]= [X,G]$ obviously has a part to play, but for all I have tried I cannot for the life of me craft a strong enough argument to really convince myself of their true relationship. REPLY [6 votes]: Here is one possible way of answering the question, using the simplicial model for $EG$. In this context, we can show very explicitly that the map $\Sigma X\to BG$ built (via adjointness) from the clutching function for $E$ classifies $E$ (and then if $X$ is a CW complex, any other classifying map for $E$ is homotopic to this one, by a standard cell-by-cell argument). If $E\to \Sigma X$ is a principal $G$-bundle, then as explained in the question, $E$ is trivial over each half of $\Sigma X$. So $E$ is (isomorphic to) the quotient space $$(C_{-} X\times G) \cup (C_+ X \times G)\,/\,(x_-, g)\sim(x_+, c(x)g)$$ where $c: X\to G$ is the clutching function, and the notation for the relation is supposed to indicate that $x_-$ and $x_+$ are the copies of $x$ in the bases of the cones $C_-X$ and $C_+ X$ (respectively). Let's assume that $G$ is a Lie group, so that $BG$ can be taken to mean the bar construction on $G$ (as in, say, "Segal's paper Classifying spaces and spectral sequences"). Then there is a natural map $\Sigma G \to BG$ given by the fact that $G$ is the space of 1-simplices in $BG$, and it has an adjoint $i: G\to \Omega BG$ (which sends $G$ to the loop that "runs around the 1-simplex labeled '$g$'," so to speak). The function $i \circ c: X \to \Omega BG$ has an adjoint $f: \Sigma X \to BG$. It sends $x$ to $[c(x), t]$, which means "the point t units along the 1-simplex labeled $c(x)$." We want to prove that $f$ classifies $E$ - that is, we want to construct a map $\tilde{f}: E\to EG$ that is $G$-equivariant and covers $f$. This can be done very explicitly. In Segal's setup, $EG$ is also a simplicial space, with 1-simplices parametrized by $G\times G$; the $G$-action on this (universal) principal bundle is induced by right-multiplication in $G$, and the map $EG\to BG$ sends the 1-simplex $(g,h)$ to $g h^{-1}$. I want to think of $\Sigma X$ as $X\times I$ with both ends collapsed to points, so I'll take $C_-$ to be parametrized by $t\in [0,1/2]$ and $C_+$ to be parametrized by $t\in [1/2, 1]$. We can define $\tilde{f}$ separately on the two halves of $E$: on the left half $$C_{-} X\times G = \left[(X\times [0,1/2])/(X\times 0)\right]\times G,$$ we set $\tilde{f}(x, t, g) = [(c(x)g, g), t]$. On the right half $$C_{+} X\times G = \left[(X\times [1/2, 1])/(X\times 1)\right]\times G,$$ we set $\tilde{f}(x, t, g) = [(g, c(x)^{-1}g), t]$. Note that in both cases, when we project down to $BG$ we get just $[c(x), t]$, and that the map is well-defined on $E$ because when $t=1/2$ (where the two cones meet) the first formula sends $(x_-, g) = (x, 1/2, g)$ to $[(c(x)g, g), 1/2]$ and the second formula sends the equivalent point $(x_+, c(x)g) = (x, 1/2, c(x)g)$ to $[(c(x)g, c(x)^{-1} c(x)g], 1/2] = [(c(x)g, g), 1/2]$. I guess something similar probably works for any topological group $G$, if one uses Milnor's join construction instead of Segal's simplicial model, but I haven't thought through it.<|endoftext|> TITLE: A conjecture about strongly regular graphs QUESTION [8 upvotes]: Let $G \neq K_{v}$ be a $(v,k,\lambda,\mu)$ strongly regular graph. After perusing through Brouwer's tables of parameters I have formed the conjecture $$\lambda-\mu \leq \frac{k}{2}.$$ So far I have not been able to prove it, though it seems like an easy statement. Have you seen something like this? EDIT: Now that the original claim is proved, we can ask: what is the best possible constant $C$ so that $\lambda-\mu \leq \frac{k}{C}$? REPLY [5 votes]: I have found a proof now. There is an old result, due originally to Taylor & Levingston, which says that for a strongly regular $G \neq K_{v}$: $$ k \geq 2\lambda+3-\mu. $$ It can be found on page 7 of the BCN book and in fact holds for a more general class (the so-called amply regular graphs) as pointed out by Neumaier. Now we can prove the claim in the question by considering two cases: (I) $\lambda \leq \frac{k}{2}$. Then $\lambda-\mu \leq \lambda \leq \frac{k}{2}$. (II) $\lambda>\frac{k}{2}$. Then $\lambda-\mu \leq (k-\lambda)-3 TITLE: Small values of a polynomial evaluated at roots of unity QUESTION [8 upvotes]: The MO answer https://mathoverflow.net/a/98176/11926 notes the following: Let $\gamma$ be an algebraic number that is not a root of unity. Then Baker's theorem implies that there is a constant $C(\gamma)>0$ such that for all $n\ge2$ we have $|\gamma^n-1|\ge n^{-C(\gamma)}$. An alternative way to phrase this is that $\gamma$ cannot be too close to a root of unity. Thus if we let $\boldsymbol{\mu}_n$ denote the $n$'th roots of unity, then the result says that $\min_{\zeta\in\boldsymbol{\mu}_n}|\gamma-\zeta|\ge n^{-C(\gamma)}$. My question is whether a higher dimensional analogue of this is known. Let $F(X_1,\ldots,X_k)\in\overline{\mathbb{Q}}[X_1,\ldots,X_k]$ be a polynomial with algebraic coefficients. Does there exist a constant $C=C(F)>0$ such that $$ \min_{\substack{\zeta_1,\ldots,\zeta_k\in\boldsymbol{\mu}_n\\ F(\zeta_1,\ldots,\zeta_k)\ne0\\}} \bigl|F(\zeta_1,\ldots,\zeta_k)\bigr| \ge n^{-C} \quad\hbox{for all $n\ge2$?} $$ Or if this is not known, for the application that I have in mind it would suffice to have a lower bound which implies that $$ \liminf_{n\to\infty} \frac{1}{n} \min_{\substack{\zeta_1,\ldots,\zeta_k\in\boldsymbol{\mu}_n\\ F(\zeta_1,\ldots,\zeta_k)\ne0\\}} \log\bigl|F(\zeta_1,\ldots,\zeta_k)\bigr| \ge 0. $$ REPLY [4 votes]: Here is a related, but different, problem, as mentioned by @Andreas. Let $f$ be a Laurent polynomial in $n$ variables with integer coefficients. Define $\log_0(t)$ to be $\log t$ if $t>0$ and 0 if $t=0$. Consider finite subgroups $F$ of the multiplicative $n$-torus and the Riemann sums of $\log_0 |f|$ over $F$ as $F$ becomes more and more uniformly distributed. Do these always converge to the integral of $\log_0 |f|$ over the torus with respect to Lebesgue measure (which is the log of the Mahler measure of $f$)? This question is motivated by a dynamical question about periodic points. Concrete case: Let $f(x,y) = 3 - x - x^{-1} - y - y^{-1}$. The zero set of $f$ on the 2-torus is an oval containing exactly four roots on unity (the intersections of the oval with the coordinate axes), so there are no subtleties about how various finite $F$s can intersect this oval. And yet, as far as I know, whether or not there is convergence of the Riemann sums is open even in this simple case.<|endoftext|> TITLE: Arbitrarily large $n$ divides $F_n$ QUESTION [5 upvotes]: Is it true that there exists $n \in \mathbb{N}$ with arbitrarily many prime factors such that $n$ divides $F_n$, where $F_n$ represents the n-th Fibonacci number? REPLY [14 votes]: Note that (see here): If $a$ and $b$ are in your sequence, then so is $\text{lcm}(a,b)$; If $n$ is in the sequence, then so is $F_n$. Now take any $n>12$ that belongs to the sequence, then $a_1=\text{lcm}(n,F_n)$, $a_2=\text{lcm}(n,F_n,F_{F_n})$, and so on. Each of $n$, $F_n$, $F_{F_n}$, ... belongs to the sequence by the second fact. Their successive least common multiples do as well by the first fact. Moreover, one has $\omega(a_k) \geq k-1+\omega(n)$ by the theorem of Carmichael, which answers your question affirmatively. A more challenging version I didn't think about yet could be: for which $n \in \mathbb N$ is there an $N$ in the sequence such that $\omega(N)=n$?<|endoftext|> TITLE: Left invertible operators of $B(X,Y)$ QUESTION [5 upvotes]: Suppose that $X$ and $Y$ are Banach spaces. Is $\{f\in B(X,Y):f\ \text{has a left inverse}\}$ an open subset of $B(X,Y)$? REPLY [8 votes]: Yes. Let $U=\{A\in B(X,Y);\text{there is }B\in B(Y,X)\text{ so that }BA=I_X\}$. (You probably want to have the left inverse in $B(Y,X)$. Otherwise you can get counterexamples from compact injections.) Take any $A\in U$. Then there is $B\in B(Y,X)$ so that $BA=I_X$. Suppose $C\in B(X,Y)$ satisfies $\|C\|\leq\frac12\|B\|^{-1}$. Then $B(A+C)=I_X+BC\in B(X,X)$. Now $\|BC\|\leq\|B\|\cdot\|C\|\leq\frac12$, so $I_X+BC$ has a continuous inverse by Neumann series. Since $(I_X+BC)^{-1}B(A+C)=I_X$, the operator $A+C$ has a continuous left inverse. Therefore $A$ is an interior point of $U$.<|endoftext|> TITLE: What is known about the Brauer group of an arithmetic surface? QUESTION [5 upvotes]: Let $X$ be an arithmetic surface over $\mathbb{Z}$, that is we have $\pi: X\rightarrow Spec(\mathbb{Z})$, $X$ is integral, two-dimensional and regular and $\pi$ is projective and flat. What is known about the Brauer group of $X$? Is there a relation between the Brauer group of $X$ and the Brauer group of the generic fiber $X_{\eta}$? Or is there a relation with the special fibers? For example can one find this group in the simplest example: what is $Br(\mathbb{P}^1_{\mathbb{Z}})$? How about if the generic fiber is an elliptic curve? Any other interesting facts about $Br(X)$ or hints to literature is welcome. REPLY [7 votes]: As Jason says you should read "Dix Exposes". Part 3 of "Le Groupe de Brauer III" uses a result of Artin to show that $R^i\pi_*G_m =0$ for $i\geq2$ for surfaces such as you are interested in. The issue, of course, is dealing with $p$ torsion when the residue field of a valuation ring has characteristic $p$. So the Leray spectral sequence reduces to a long exact sequence involving cohomology of $\pi_*G_m$ and $R^1\pi_*G_m$. This immediately shows that $Br(P^1_{ Z})=0$. In general you need to use Neron models to carry out a full calculation since $\pi$ cannot be smooth over ${ Z}$ if the genus is positive. So I don't think there is a simple answer. In the case of a number field $K$ you might be reduced to calculating $H^1({ O_K}, A)$ where $A$ is an abelian scheme over the ring of integers $O_K$ if $\pi$ is smooth.<|endoftext|> TITLE: Where to find (personal) motivation QUESTION [23 upvotes]: I think it would be appropriate to make this question CW... It is likely that this question will not survive here on MO for long, but I do hope that the community gives it a chance. I also hope to get answers that might help others feeling unmotivated, but in general not discouraged. Among mathematicians, especially the young bunch (postdocs and beginning tenure track), among many causes for lack of productivity, two seem to stand out most often: discouragement and lack of motivation, or the combination of both. I in particular fix attention on those mathematicians who otherwise are (would be) very productive -- an assertion made plausible perhaps by their previous contributions. I would go as far as to say that many, if not most, of us know of at least one such example. There are many possible reasons for discouragement, many of which have been discussed elsewhere (I do not intend to give specific references, especially since what I have in mind are such sources as blogs and discussion forms on the internet, including MO). Examples of such reasons are family arrangements, low salary, high stress (publish or perish kind), difficult personal relations with other members of the department, and other (or, very often, any combination of the above). Let us assume that discouragement is not the problem, but lack of motivation is. Namely, assume our young mathematician is tormented by questions such as: Who needs this? What is this all good for? (i.e. lack of immediate, appreciable feedback). There are many mathematicians far stronger than me contributing significantly to the field; what is my work good for? (self-deprecation) Would I be happier doing a nine-to-five job, getting my six figures, and living a simple, earthly life? Why am I even concerned with all this abstract nonsense? (resentment, retreat) I don't necessarily have a problem with academia. In fact, I like being in academia. Maybe I should change fields, but I feel I am now so deep into math, that changing now might actually hurt my career. Besides, I know nothing about other fields, and it would take me some time just to get up to speed. What should I do? I hope my colleagues do not find out about how I feel, as this is rather embarrassing; however, carrying on with my duties (especially research) is becoming more and more difficult. Should I share my feelings with anyone? My question is, what would you say if the said imaginary mathematician was your colleague and a close enough friend to come and share her feelings with you? REPLY [34 votes]: There was a panel session on the future of algebraic topology at the birthday conference for Gunnar Carlsson, Ralph Cohen, and Ib Madsen. At the beginning of the panel, Bill Dwyer raised the point that most of us are never going to do work that will be remembered hundreds of years hence by future mathematicians, and largely will be read by a small circle of our peers. Rather than despairing at this fact, he compared this work to "making a nice dinner for your friends," an intrinsically worthwhile and rewarding task. That is, being a part of this community and doing the work that we do is often its own reward. Not a perfect answer by any means, but one that I find more honest and compelling as time goes by. REPLY [8 votes]: To say "be honest" may clarify or debunk some commodification or over-hype-ing or iconicization issues that can cloud one's thinking... and apparently increase the distance between "work" and "reality". Reminding oneself of the apparent human tendency to iconify... can be helpful in discounting the weight of the icons. :) Asking "what is/was the prior issue?"... as opposed to asking what is "publishable". Similarly "What is the real confusion/unknown/difficulty...?", as opposed to "what are the questions raised "in the literature". "The literature" is, in effect, the best that we can do (if we do also nowadays include on-line things, IMHO), but, at the same time, it is "not canon", insofar as it is ... only the best we can do. That is, one's demotivation due to observing hypocrisy or pointless chatter passing for "research" should not be allowed to be a distraction for genuine issue. Easier said than done, sure, especially when one thinks in terms of tenure or salary increases and so on. But, if one can suspend those concerns, the corrupting of commodification explains many things (as in most human enterprises where people need to make a living). The far subtler question of whether, apart from the discrepancy between hype and daily life, one simply doesn't care about working on mathematics... well, yes, it's awkward when one has committed a part of one's life to it and changes one's mind. Conceivably... the ugliness of the commodification may have soured one's feelings... and, conceivably, once recognized, that could be discounted... thus "saving" one's interest in the original thing-itself... REPLY [6 votes]: Let me focus on the first couple of objections (what is my work good for, given that there are many mathematicians far stronger than me). My answer is: your young colleague's activity is good and useful, at least in a weak, yet meaningful sense. Even if one thinks that at last, only the work of the geniuses counts, what is still debatable, he should agree that geniuses stem from the mass of workers, and that great discoveries rest on a great number of easy examples. Archimedes or Gauss grew up in societies were the average maths culture was very high. The same happens in sport: there has been Pele, because virtually every kid in every village of Brazil plays soccer. There is no scientific society with isolated geniuses: it's a myth usually spread by those governments who want to cut research funding. As I see it, if one is able to do a honest cultural work, and to enjoy it, should do it with no hesitation. It has a great civil value, in a word were a lot of people instead dedicate themselves to drug trade, finance, politics, pornography, war.<|endoftext|> TITLE: Inverse cohomological isomorphisms QUESTION [8 upvotes]: Let $\ M'\ M''\ $ be simply-connected Hausdorff compact manifolds (possibly with boundary for another variant of the question). Let $\ f:M'\rightarrow M''\ $ be a continuous function which induces an isomorphism of the cohomological rings. Does there exist a continuous function $\ g:M''\rightarrow M'\ $ which induces the inverse cohomology ring isomorphism? REPLY [7 votes]: Some further comments on homology spheres: First, for any closed connected orientable $n$-manifold $M$ there is always a degree $1$ map $M\to S^n$ obtained by collapsing the complement of an open ball in $M$ to a point. In the reverse direction, any degree $1$ map $f:M\to N$ of closed connected orientable $n$-manifolds must induce a surjection on $\pi_1$, for otherwise $f$ could be lifted to the covering space $\tilde N \to N$ corresponding to the proper subgroup $f_*(\pi_1M) \subset \pi_1N$. If this covering is finite-sheeted, then the degree of the projection $\tilde N\to N$ is equal to the number of sheets (which is the index of $f_*(\pi_1M)$ in $\pi_1N$) since an orientation of $N$ lifts to an orientation of $\tilde N$, making the local degrees at all points in a fiber have the same sign. Thus the degree of $f$ is divisible by the number of sheets, so it can't be $1$. If the covering is infinite-sheeted then $\tilde N$ is noncompact so $H_n(\tilde N)=0$, forcing $f$ to have degree $0$. Applying this fact to a degree $1$ map $S^n\to N$ we see that $\pi_1N=0$ so it can't be a nonsimply-connected homology sphere. (These are classical arguments, incidentally.)<|endoftext|> TITLE: Is there a higher Grothendieck ring of varieties? QUESTION [27 upvotes]: Fix a field $k$. The Grothendieck ring $K_0(\mathrm{Var}_k)$ of varieties over $k$ is defined as the quotient of the free abelian group on isomorphism classes of algebraic varieties by the scissor relation. The notation suggests that there might be higher $K$-groups $K_i(\mathrm{Var}_k)$ as well, but naive attempt at defining such an object fails as $K_0(\mathrm{Var})$ is not defined as $K_0$ of an exact additive category. Is there a reasonable definition of these groups nonetheless? REPLY [34 votes]: Torsten Ekedahl proposed a definition of higher Grothendieck groups of varieties. Unfortunately it seems that he never wrote anything down on this topic before passing away. Torsten had quite a large number of unfinished mathematical manuscripts and projects. I don't know what happened to them, although surely someone at Stockholm University has taken care of them. I saw him give a talk about higher Grothendieck groups of varieties at Gerard van der Geer's birthday conference on Schiermonnikoog in 2010. The abstract is available online: http://www-irm.mathematik.hu-berlin.de/~ortega/schierm/ “Higher Grothendieck groups of varieties” We shall (slightly) modify the setup of Waldhausen’s definition of higher K-theory in order to introduce higher Grothendieck groups of varieties such that the 0’th group is the ordinary Grothendieck group of varieties. We then establish some basic properties of these groups that generalise basic properties for the ordinary Grothendieck groups. Edit: I found my (sketchy) notes from the 2010 talk!!! Any mistakes in what follows are my own; I know close to nothing about K-theory today and I knew literally nothing in 2010. Anyone who is more knowledgeable than me is very welcome to edit the following: He begins by recalling the definition of $\newcommand{\Var}{\mathbf{Var}}K_0(\Var_S)$ for a base scheme $S$. There is nothing new here. Then he declares his intention to define higher K-groups. He makes a list of properties that such higher K-groups should have (I might not have written down all of them): There should exist products $K_i(\Var_S) \times K_j(\Var_S) \to K_{i+j}(\Var_S)$ extending the usual product when $i=j=0$. For $f \colon S \to T$ there should be $f^\ast \colon K_i(\Var_T) \to K_i(\Var_S)$ resp. $f_\ast \colon K_i(\Var_S) \to K_i(\Var_T)$. When $i=0$ these should be given by fibered product, resp. by composing the structure morphism with $f$. Functoriality and projection formula: $(fg)_\ast = f_\ast g_\ast$, $(fg)^\ast = f^\ast g^\ast$, $x\cdot f_\ast y = f_\ast (f^\ast x \cdot y)$. So far we could just set the higher K-groups to be zero. We want a non-triviality condition. Consider the functor $\newcommand{\Finset}{\mathbf{Finset}}\Finset \to \Var_k$, $$ A \mapsto \coprod_A \mathrm{Spec}(k).$$ (DP: for this bullet point the notes change from $\Var_S$ to $\Var_k$. Perhaps at this point it becomes necessary to work over a field.) This should induce $K_i(\Finset) \to K_i(\Var_k)$. Recall that the K-groups of $\Finset$ are the stable homotopy groups of spheres. When $k= \mathbf F_q$ there is a functor $\Var_k \to \Finset$, $$ X \mapsto X(k).$$ The composition $K_i(\Finset) \to K_i(\Var_k) \to K_i(\Finset)$ should be the identity. When $k = \mathbf C$ there should exist a map $K_i(\Var_\mathbf{C}) \to K_i(\mathbf Z)$ such that the composition $K_i(\Finset) \to K_i(\Var_\mathbf{C}) \to K_i(\mathbf Z)$ is the "standard one" (DP: at this point in my notes I wrote ?!!) He goes on to discuss generally how to define algebraic K-theory. You want a suitable category, such that the homotopy groups of its nerve are the K-theory groups. He mentions Quillen's Q construction but says that he will follow Waldhausen's approach. Waldhausen's idea is to associate a "simplicial category" $\newcommand{\C}{\mathscr C}s\C$ to a category $\C$. He notes that there is a subtlety here, in that the simplicial identities $d_i d_{j} = d_{j-1}d_i$ etc. need to be strict. For $\Var_k$ he defines $(s \C)_n$ to be the category with objects $$ \varnothing \hookrightarrow X_1 \hookrightarrow X_2 \hookrightarrow \cdots \hookrightarrow X_n $$ where all injections are closed immersions of $k$-varieties, morphisms are isomorphisms of such diagrams. All the $d_i$ are "what you expect" except for $d_0$, which maps to $$ \varnothing \hookrightarrow X_2 \setminus X_1 \hookrightarrow X_3 \setminus X_1 \hookrightarrow \cdots \hookrightarrow X_n \setminus X_1. $$For each $n$, $N((s\C)_n)$ is a simplicial set. $N(s\C)$ is a bisimplicial set, so more or less a simplicial set. We define $$ K_i(\Var_k) = \pi_{i+1} N(s\Var_k).$$ He goes on to discuss Waldhausen's additivity theorem. Consider the category of pairs $X \hookrightarrow Y$ of closed immersions. There are three functors to $\Var_k$ mapping to $X, Y$ and $Y \setminus X$ respectively. These give three functors $K_i(\Var \hookrightarrow \Var) \to K_i(\Var)$ and the additivity theorem says that two of these sum to the third. Claim: He can prove this theorem for his definition of K-groups. He notes that all his constructions mirror those of Waldhausen for topological spaces. The biggest difference is that Waldhausen's uses the existence of a quotient $Y/X$ for $X \hookrightarrow Y$. In particular one needs to give a different proof of the additivity theorem but this is possible. My notes end here. Of course the first question one asks is whether this definition agrees with the one due to Inna Zakharevich, that Clark Barwick linked to in a comment.<|endoftext|> TITLE: An interesting calculation of derivative QUESTION [5 upvotes]: I was trying to get the probability distribution $p(n)$ from a generating function $G(s)$ like this: $G(s) = e^{a(s-1)^2}=\sum s^np(n)$ I need first to do Maclaurin expansion of the exponential and then get the $n$th order term for $p(n)$. My first thinking was it would be simple to calculate the derivatives. But it turns out to be much more difficult and also very interesting to generalize the $n$ order derivative. I list a table for the powers and coefficients of each derivative order, finding that the powers are odd numbers for odd $n$, even numbers for even $n$, the coefficients are associated to $(n-1)$th order's powers and coefficients. It is easy to see the association but I cannot generalize it. Anyone could give a shot and help me out? UPDATE: to give some feedbacks to the question for other users Actually the generating function I gave above cannot generate a proper probability sequence since negative values will show up. One way of modifying it is to add one more term in the power: \begin{equation} e^{a(s-1)^2+b(s-1)}=\sum s^np(n) \end{equation} with constraint $b>2a$ guaranteeing the positiveness of sequence elements. Using hermite polynomials and changing of variables, one can obtain \begin{equation} \nonumber p(n)=e^{a-b}\frac{\alpha^n}{n!}H_n(\frac{b+2\alpha^2}{2\alpha}) \end{equation} REPLY [10 votes]: Robert Israel noted that $p(n)$ is a $e^a$ times a polynomial of degree $n$ in $a$ with a zero of order $\lceil n/2 \rceil$ at $a=0$. We express this polynomial in terms of a Hermite polynomial $H_n$ evaluated at an imaginary argument $\alpha := \pm \sqrt{-a}$. Start from the generating function $$ \exp(2xt-t^2) = \sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}, $$ and set $(x,t) = (\alpha, \alpha s)$. Then $$ 2xt-t^2 = \alpha^2 (2s-s^2) = a (s^2-2s) = a(s-1)^2 - a, $$ whence $$ e^{a(s-1)^2} = e^a e^{a(s^2-2s)} = e^a \sum_{n=0}^\infty H_n(\alpha) \frac{(\alpha s)^n}{n!}, $$ and $p(n) = \alpha^n H_n(\alpha) / n!$. Known formulas and identities for the Hermite polynomials can then be used to study the sequence $\{ p_n \}$.<|endoftext|> TITLE: Characterising the adjoint representation of SU(N) QUESTION [7 upvotes]: One can show that the adjoint representation of $\mathrm{SU}(n)$, the image of the map $\mathrm{Ad}:\mathrm{SU}(n) \rightarrow \mathrm{Aut}(\mathrm{su(n)})\subset \mathrm{GL}(\mathrm{su}(n))$, is an $n^2-1$-dimensional lie subgroup of $\mathrm{SO}(n^2-1)$. I would to know if any characterisations of this subgroup are known? In particular, I would like to understand what object, apart from the inner product, is preserved under the action of elements from the lie group $\mathrm{Im}(\mathrm{Ad})\subset \mathrm{SO}(n^2-1)$. Alternatively, I am looking to understand the action of $\mathrm{Im}(\mathrm{Ad})$ on $\mathbb{R}^{n^2-1}$. I know that given a basis for $\mathrm{su}(n)$ one can construct a basis for the Lie algebra of $\mathrm{Im}(\mathrm{Ad})$ using the $\mathrm{ad}$ map and the structure constants for $\mathrm{su}(n)$ - However I cannot see how to use this to answer the question above? In addition, I am aware that $\mathrm{Im}(\mathrm{Ad}) \simeq \mathrm{SU}(n)/\mathbb{Z}_n$ - However I also cannot see how to use this to help me answer the above question? In the $n=2$ case one can use the first isomorphism theorem along with the well know group homomorphism between $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$, however for the general case it seems like this approach would require the construction of a group homomorphism $f:\mathrm{SU}(n)\rightarrow \mathrm{SO}(n^2-1)$ with kernel $\mathbb{Z}_n$? It seems unlikely one can construct such a homomorphism in general without the useful parameterizations of $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ one has for the case where $n=2$? Is there another way to exploit this quotient group structure? REPLY [8 votes]: Here is a different characterization of the subgroup $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)\subset\mathrm{SO}(n^2{-}1)$ that works when $n>2$. Define a skew-symmetric trilinear form $\kappa:{\frak{su}}(n)\times{\frak{su}}(n)\times{\frak{su}}(n)\to\mathbb{R}$ by the formula $$ \kappa(x,y,z)=\mathrm{tr}(xyz-xzy)\ \bigl(=\mathrm{tr}(yzx-yxz)=\mathrm{tr}(zxy-zyx)\bigr). $$ Then, for $n>2$, the subgroup $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)\subset\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$ is the identity component of the subgroup of $\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$ that preserves $\kappa$. (This doesn't work for $n=2$, of course. Also, you do have to restrict to the identity component because $\mathrm{SU}(n)$ has an outer automorphism (namely, conjugation) when $n>2$, and this preserves $\kappa$ as well; accordingly, the full $\kappa$-stabilizer has two components.) In a certain sense, this is the 'simplest' characterization of $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as a subgroup of $\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$. What I mean is that, if you want to describe this subgroup as the set of linear transformations that preserve some set of algebraic objects on the vector space ${\frak{su}}(n)$, then $\kappa$ is the object that sits in the smallest representation of $\mathrm{GL}\bigl({\frak{su}}(n)\bigr)$ that does the job. For example, if you want to define $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as the subgroup that preserves the Lie bracket $[,]:{\frak{su}}(n)\times {\frak{su}}(n)\to {\frak{su}}(n)$, then you have to think of $[,]$ as an element of ${\frak{su}}(n)\otimes {\frak{su}}(n)^\ast\otimes {\frak{su}}(n)^\ast$ or, if you want to build in the skew-symmetry of the bracket, as an element of ${\frak{su}}(n)\otimes \Lambda^2({\frak{su}}(n)^\ast)$. Either way, these vector spaces have a much higher dimension than $\Lambda^3({\frak{su}}(n)^\ast)$, which is where $\kappa$ lives. As another example, if you want to think of $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as the subgroup that preserves the quadratic form $Q(x) = \mathrm{tr}(x^2)$ and the ($\mathbb{R}$-valued) cubic form $C(x) = i\,\mathrm{tr}(x^3)$, you are asking that it preserve an element of $S^2({\frak{su}}(n)^\ast)\oplus S^3({\frak{su}}(n)^\ast)$, and, again, this has considerably higher dimension than $\Lambda^3({\frak{su}}(n)^\ast)$. You can cut this down a little bit by noting that $C$ is harmonic with respect to $Q$, so that you could look at characterizing $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ as the subgroup of $\mathrm{O}\bigl({\frak{su}}(n),Q\bigr)$ that stabilizes $C\in S^3_0({\frak{su}}(n)^\ast,Q)$, but this still is more total equations than the equations that say that an element fixes $\kappa$. I'm not saying that using $\kappa$ to define $\mathrm{Ad}\bigl(\mathrm{SU}(n)\bigr)$ is the most useful way to define it, I'm just saying that it's algebraically the least redundant. For example, look at the case $n=3$: The codimension of $\mathrm{Ad}\bigl(\mathrm{SU}(3)\bigr)$ in $\mathrm{GL}\bigl({\frak{su}}(3)\bigr)\simeq \mathrm{GL}(8,\mathbb{R})$ is $64-8 = 56$, which is equal to the dimension of $\Lambda^3\bigl(\mathbb{R}^8\bigr)$, so it follows that the $\mathrm{GL}(8,\mathbb{R})$-orbit of $\kappa$ is actually open in $\Lambda^3\bigl(\mathbb{R}^8\bigr)$. Thus, the condition of fixing $\kappa$ is $56$ independent equations on an element of $\mathrm{GL}(8,\mathbb{R})$, which is the absolute minimum number of equations you need to cut out $\mathrm{Ad}\bigl(\mathrm{SU}(3)\bigr)$ as a submanifold. REPLY [4 votes]: As your question already suggests, the "extra object preserved" is the Lie bracket. I.e., for any connected semisimple Lie group $G$, the image of $\mathrm{Ad}: G\to\mathrm{GL}(\mathfrak g)$ is precisely the identity component $\mathrm{Aut}(\mathfrak g)^0$ of $$ \mathrm{Aut}(\mathfrak g) = \{a\in\mathrm{GL}(\mathfrak g):a([X,Y])=[a(X),a(Y)] \text{ for all } X,Y\in\mathfrak g\}. $$ Proof. Clearly $\mathrm{Ad}$ maps $G$ into that identity component. To verify that it is onto it, it is enough to see that its derivative $\mathrm{ad}:\mathfrak{g}\to\mathrm{gl}(\mathfrak g)$ is onto the Lie algebra $$ \mathrm{aut}(\mathfrak g) = \{D\in\mathrm{gl}(\mathfrak g):D([X,Y])=[D(X),Y]+[X,D(Y)] \text{ for all } X,Y\in\mathfrak g\} $$ of $\mathrm{Aut}(\mathfrak g)$, or in other words, that every derivation $D$ of $\mathfrak g$ is inner: $D=\mathrm{ad}(Z)$ for some $Z\in\mathfrak g$. But that is precisely the content of Whitehead's Lemma (see any book on semisimple Lie algebras).<|endoftext|> TITLE: Nonunital $E_\infty$-rings QUESTION [12 upvotes]: An elementary fact of algebra is that the category of nonunital commutative rings is equivalent to that of $\mathbb{Z}$-augmented unital commutative rings, the equivalence being given by forming unitisation and, conversely, forming augmentation ideals. What may one say about the relation between the $\infty$-categories of nonunital $E_\infty$-rings and that of $E_\infty$-rings augmented over the sphere spectrum? (This may well be answered by a simple abstract argument. Nevertheless I seem to be stuck, please bear with me.) REPLY [3 votes]: In Higher Algebra Proposition 5.4.4.10 1, Lurie proves that for a coCartesian fibration of $\infty$-operads $q:\mathscr{C}^\otimes\to\mathscr{O}^\otimes$, where $\mathscr{O}^\otimes$, when viewed as an $\infty$-category, is pointed and $\mathscr{C}$ is a stable $\mathscr{O}$-monoidal $\infty$-category under the coCartesian fibration $q$. There is an equivalence of $\infty$-categories $\mathrm{Alg^{nu}}_\mathscr{O}(\mathscr{C})\to\mathrm{Alg^{aug}_{\mathscr{O}}(\mathscr{C})}$. Let $\mathscr{O}^\otimes=\mathrm{N}(\mathscr{F}\mathrm{in}_{*})$, where $\mathscr{F}\mathrm{in}_{*}$ is Segal's category of pointed finite sets (the $n$lab denotes it as $\Gamma$, if I'm not wrong). A $\mathbb{E}_\infty$-ring is a commutative monoidal object, hence $\mathbb{E}_\infty$-monoidal object, of $\mathrm{Sp}$, and the sphere spectrum $S$ is naturally a $\mathbb{E}_\infty$-ring, so we can let $\mathscr{C}$ to be $\mathrm{Mod}_S$. Then, the $\infty$-category of nonunital $\mathbb{E}_\infty$-rings is equivalent to the $\infty$-category of augmented $\mathbb{E}_\infty$-rings over the sphere spectrum. 1 This is an expansion of Haugseng's comment above.<|endoftext|> TITLE: A variation on Bulgarian solitare QUESTION [6 upvotes]: It appears that a variation on Bulgarian solitare has a fixed point regardless of the starting $n$. For example, let $n=69$, and consider this partition: $$ (8,8,7,7,5,5,5,5,5,4,3,3,2,2) $$ In Bulgarian solitare, $1$ would be removed from each "stack/pile" to form another stack. In the variation, I remove $1$ just from the $k=9=3^2$ largest stacks, $(8,8,7,7,5,5,5,5,5)$. That reduces those stacks to $(7,7,6,6,4,4,4,4,4)$ and adds a stack of size $9$. I find it easier to visualize this process with the stacks organized in an array with the largest stacks surrounding the upperleft corner. (Added: I just learned these are known as plane partitions.) Then $1$ is removed from the $3 \times 3$ square of stacks including the upperleft corner. The numbers are resorted and the same process applied again, always gathering from the largest $k=9$ stacks: $$ \left( \begin{array}{cccc} 8 & 8 & 5 & 4 \\ 7 & 7 & 5 & 3 \\ 5 & 5 & 5 & 3 \\ 0 & 0 & 2 & 2 \\ \end{array} \right) \;\rightarrow\; \left( \begin{array}{cccc} 9 & 7 & 6 & 4 \\ 6 & 7 & 4 & 4 \\ 4 & 4 & 4 & 3 \\ 0 & 2 & 2 & 3 \\ \end{array} \right) \;\rightarrow\; \left( \begin{array}{cccc} 9 & 8 & 5 & 3 \\ 6 & 6 & 5 & 3 \\ 3 & 4 & 4 & 3 \\ 2 & 2 & 3 & 3 \\ \end{array} \right) \;\rightarrow\; \cdots $$ The endpoint of this process, after $20$ steps, is: $$ \left( \begin{array}{cccccc} 9 & 8 & 5 & 1 & 1 & 1 \\ 6 & 7 & 4 & 1 & 1 & 1 \\ 1 & 2 & 3 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 \\ \end{array} \right) $$ This seems to be a fixed point regardless of $n$, as long as the list of stacks is longer than the largest $k$ being reduced at each step. (If $k$ encompasses the entire list of stacks, this just reduces to Bulgarian solitare, and cycles rather than fixed points can occur.) My question is: Is this true, that the process described leads to the fixed point $$(k, k{-}1, k{-}2, \ldots, 3,2,1,1,1,\ldots,1)$$ under those conditions? It's a bit surprising to me that it doesn't lead to cycles for non-triangular $n$. Perhaps it does? I have not explored extensively (and I've only looked at $k$ which are squares). REPLY [4 votes]: Yes, that is what happens if there are at least $k$ stacks at each step. It is quite clear that all stacks will be eventually bounded by $k$, since either the highest number of cards or the number of stacks with the highest number of cards reduces at every iteration if there is a stack of strictly more than $k$ cards. When we have reached this state, the biggest stack always has $k$ cards. Let $N_m$ denote the number of stacks of size $m$ and define the function $\phi_a(i)=1+\max\{0,i-a\}$, $a\in\mathbb N$. Suppose we have already reached the state when $N_m=0$ for $m>k$. At every iteration $N_k\mapsto\phi_k(N_k)$. (One stack of size $k$ is created, up to $k$ become smaller.) This map is strictly decreasing as long as $N_k>1$, so eventually $N_k=1$. Then suppose $N_k\equiv1$ and consider $N_{k-1}$. If $N_{k-1}=0$, then it will be one on the next turn, so we can assume $N_{k-1}\geq1$. At every iteration $N_{k-1}\mapsto\phi_{k-1}(N_{k-1})$. Again $\phi_{k-1}$ is strictly decreasing until it hits its fixed point $1$, so eventually $N_{k-1}=1$. This argument can be continued inductively, using $\phi_m$ for $N_m$. Eventually $N_k=N_{k-1}=\dots=N_2=1$. But $\phi_1$ is the identity map, so $N_1$ does not tend to $1$; it just stays where it happens to end up. Clarification: As the OP's example demonstrates, not every $N_i$ is always strictly decreasing. When we have reached the state where all stacks are bounded by $k$, the number $N_k$ is strictly decreasing (until it hits one). When $N_k$ has stabilized to one, $N_{k-1}$ is strictly decreasing (until it hits one), and so on. If $N_{k-1}$ happens to be zero, it will become one in one turn and stay there. So each $N_m$ will only start its monotone decay towards one when $N_i=0$ for $i>k$ and $N_i=1$ for $k\geq i>m$ (again with the exception when it grows one step). Assuming no zeros occur for $N_m$, $m\leq k$, the sequence $(N_i)_{i=1}^\infty$ is strictly decreasing in lexicographical order: if $m$ is the highest index for which $N_m$ changes, then $N_m$ decreases (strictly, unless we are at the final fixed point).<|endoftext|> TITLE: Pseudo-automorphisms on Fano varieties QUESTION [6 upvotes]: Is every pseudo-automorphism (self-birational map which does not contract any hypersurface) of a smooth Fano variety of Picard rank $1$ equal to a biregular automorphism? Remark: For $\mathbb{P}^n$, the answer is yes, and easy: every birational map of degree $>1$ contracts a hypersurface, given by its Jacobian. Same for any projective surface (because birational maps are sequence of blow-ups and blow-downs). REPLY [3 votes]: It seems to me that the Fano variety $X$ does not need to be smooth. It is enough to have that $-K_X$ is $\mathbb{Q}$-Cartier. Indeed, in this case we can embed $X$ in a projective space $\mathbb{P}^n$ with the linear system $|-mK_X|$ for $m\gg 0$. For any pseudo-automorphism $\phi:X\dashrightarrow X$ we have that $\phi^{*}(-mK_X) = -mK_X$, and hence $\phi$ induces an automorphism of $\mathbb{P}^n$ stabilizing $X\subset\mathbb{P}^n$.<|endoftext|> TITLE: Aperiodic set of corner Wang Tile QUESTION [5 upvotes]: There is quite some reference on aperiodicity of the edge-type of Wang Tile. But I could not yet find aperiodic corner type of Wang Tiles... Could someone provide me some instances (better with reference) of aperiodic Wang Tile of corner-type? Thank you:) REPLY [3 votes]: So I read the answer to your previous question. I believe it's correct. This gives a recipe to translate any edge-type Wang tile to a corner-type Wang tile. If you start with an aperiodic set of edge-type Wang tiles, then the corner-type Wang tiles you get from this construction are also aperiodic. What more is there to say?<|endoftext|> TITLE: Operator norm vs spectral radius for positive matrices QUESTION [5 upvotes]: I believe the following statement should be true but somehow I don't see an argument: For every integer $d>1$ there exists a constant $C=C(d)>1$ such that whenever $A$ is a $d \times d$ matrix with all entries integers $>0$, then $$ \frac{1}{C} \|A\| \le \sigma(A) \le C \|A\|.$$ Here $\sigma(A)$ is the spectral radius of $A$ (which in this case is the Perron-Frobenius eigenvalue of $A$), and $\|A\|=\sup_{\|v\|=1} \|Av\|$ is the operator norm of $A$. Does anyone see how to prove (or disprove) this statement? Many thanks, Ilya. REPLY [9 votes]: In fact, one can improve a little bit Terry's bound to show that $\|A\|\le(\sigma(A))^2$, and therefore $$ \sigma(A) \le \|A\| \le (\sigma(A))^2. $$ Indeed, this is an immediate consequence of the fact that for a matrix $A=(a_{ij})_{1\le i,j\le n}$ with positive integer entries one has $(\sigma(A))^2\ge\sum_{i,j=1}^n a_{ij}$, and the observation that if $x=(x_1,\ldots,x_n)$ is a unit-length vector, then $$ \|Ax\|^2 = \sum_{i=1}^n \Big( \sum_{j=1}^n a_{ij} x_j \Big)^2 \le \sum_{i=1}^n \Big( \sum_{j=1}^n a_{ij} \Big)^2 \le \Big( \sum_{i,j=1}^n a_{ij} \Big)^2, $$ implying $\|A\|\le \sum_{i,j=1}^n a_{ij}$. Notice that by Mike Jury's example above, with $\sigma(A)=\sqrt n+1$ and $\|A\|>n$, the estimate $\|A\|\le(\sigma(A))^2$ is essentially best possible. Also notice that the argument just presented actually works for any matrix $A$ with all entries greater than or equal to $1$; the assumption that the entries are integer is not really used.<|endoftext|> TITLE: Is the sequence of Apéry numbers a Stieltjes moment sequence? QUESTION [15 upvotes]: Consider the sequence of Apéry numbers $$ A_n = \sum_{k=0}^n \binom{n}{k}\binom{n+k}{k}\sum_{j=0}^k \binom{k}{j}^3 = \sum_{k=0}^n \binom{n}{k}^2\binom{n+k}{k}^2 . $$ In an email, physicist Alan Sokal conjectures that it is a Stieltjes moment sequence. That is, that there exists a probability measure $\mu$ on $[0,+\infty)$ so that $$ A_n = \int_{[0,+\infty)} s^n\;d\mu(s) \tag{1}$$ for $n = 0,1,2,\dots$. [Of course you can equivalently say that $\mu$ is a nondecreasing function with $\mu(0)=0$ and $\lim_{x\to+\infty} \mu(s) = 1$ and that (1) is a Stieltjes integral, rather than a "measure" integral.] Is that conjecture correct? Is $A_n$ a Stieltjes moment sequence? [This question is a follow-up to A conjectured formula for Apéry numbers , where a formula for $A_n$ was established.] REPLY [3 votes]: OK, here is my current thinking on Pietro's approach. We use the differential equation $$ (x^4-34x^3+x^2)U'''(x) + 3(2x^3-51x^2+x)U''(x)+(7x^2-112x+1)U'(x)+(x-5)U(x)=0 \tag{ODE3}$$ Multiply by $x^k$, $k \ge 0$, then integrate by parts as much as possible. The result is $$ \int \!{x}^{k-1} \left( {k}^{3}{x}^{2}-34\,{k}^{3}x +3\,{k}^{2}{x}^{2}+{k}^{3}-51\,{k}^{2}x+3\,k{x}^{2}-27\,kx+{x}^{2}-5\, x \right) U \left( x \right) {dx} = \left( {x}^{k+2}{k}^{2}-34\,{x}^{1+k}{k}^{2}+{x}^{k+2}k+{x}^{k}{k}^{2 }-17\,{x}^{1+k}k+{x}^{k+2}-10\,{x}^{1+k} \right) U \left( x \right) + \left( -{x}^{1+k}k-{x}^{3+k}k+34\,{x}^{k+2}k+{x}^{1+k}+2\,{x}^{3+k}- 51\,{x}^{k+2} \right) U' \left( x \right) + \left( {x}^{k+4}-34\,{x}^{3+k}+{x}^{k+2} \right) U'' \left( x \right) \tag{A}$$ In particular, for $k=0$, $$ \int (x-5)U(x)\;dx = \left( {x}^{2}-10\,x \right) U \left( x \right) + \left( 2\,{x}^{3}- 51\,{x}^{2}+x \right) U' \left( x \right) + \left( {x}^{ 4}-34\,{x}^{3}+{x}^{2} \right) U'' \left( x \right) \tag{A0}$$ Write $Q_k(x)$ for the right-hand-side of (A) and $\int R_k(x)U(x)\;dx$ for the left side. We will be doing differences like $$ \big[Q_k(x)\big]_a^b := Q_k(b)-Q_k(a) $$ since that will equal the integral on the left $\int_a^b R_k(x)U(x)\;dx$. We want to arrange a solution $U$ so that $\int_0^c R_k(x)U(x)\;dx=0$ for $k \ge 0$. As Pietro noted, this will give us the recurrence we want for the moments $M(k):=\int_0^c x^kU(x)\;dx$. And in particular from (A0) we would have $\int_0^c (x-5)U(x)\;dx = 0$ so that $M(1)=5M(0)$. Now consider (ODE3). First look at solutions as we approach $x=c$ from the left. Three linearly independent solutions are asymptotically: $1$, $(c-x)^{1/2}$, $(c-x)$. Other terms are higher powers (integer and half-integer). My calculations show: for the term $1$, we get $Q_k(c^-)=1/2\, \left( 24+17\,\sqrt {2} \right) \left( 24\,k+24+7\,\sqrt {2} \right) \left( 17+12\,\sqrt {2} \right) ^{k}$ for the term $(c-x)^{1/2}$, we get $Q_k(c^-)=0$, amazingly for the term $(c-x)$, we get $Q_k(c^-)=\left( -9792-6924\,\sqrt {2} \right) \left( 17+12\,\sqrt {2} \right) ^{k}$ terms $(c-x)^2$ and higher all produce $Q_k(c^-)=0$. Now look at solutions as we approach $x=c_o$ from the right. Three linearly independent soltuions are asymptotically: $1$, $(x-c_o)^{1/2}$, $(x-c_o)$. for the term $1$, we get $Q_k(c_o^+)=1/2\, \left( 17\,\sqrt {2}-24 \right) \left( -24\,k-24+7\,\sqrt {2} \right) \left( 17-12\,\sqrt {2} \right) ^{k}$ for the term $(x-c_o)^{1/2}$, we get $Q_k(c_o^+)=0$ for the term $(x-c_o)$, we get $Q_k(c_o^+)=-12\,\sqrt {2} \left( 17-12\,\sqrt {2} \right) ^{k+2}$ for higher terms, we get $Q_k(c_o^+)=0$ Now look at solutions as we approach $x=c_o$ from the left. Three linearly independent soltuions are asymptotically: $1$, $(c_o-x)^{1/2}$, $(x-c_o)$. for the term $1$, we get $Q_k(c_o^-)=1/2\, \left( 17\,\sqrt {2}-24 \right) \left( -24\,k-24+7\,\sqrt {2} \right) \left( 17-12\,\sqrt {2} \right) ^{k}$ for the term $(c_o-x)^{1/2}$, we get $Q_k(c_o^-)=0$ for the term $(x-c_o)$, we get $Q_k(c_o^-)=-12\,\sqrt {2} \left( 17-12\,\sqrt {2} \right) ^{k+2}$ for higher terms, we get $Q_k(c_o^-)=0$ Now look at terms as we approach $x=0$ from the right. Three linearly independent terms are asymptotically: $1$, $\log x$, $(\log x)^2$. for the term $1$, we get $Q_k(0^+)=0$ for the term $\log x$, we get $Q_k(0^+)=0$ for the term $(\log x)^2$, we get $Q_k(0^+)=0$ for $k \ge 1$ but $Q_0(0^+)=2$ for higher terms (such as $x$, $x\log x$, etc.), we get $Q_k(0^+)=0$ After all that, we come to the strategy for a solution. Start with the solution of (ODE3) approaching $c$ from the left that is asymptotically $(c-x)^{1/2}$. If we use only that, then $Q_k(c^-)=0$. (At the end we will multiply by a constant to get our final result.) Follow this solution to the left to the next singularity, that is approaching $c_o$ from the right. It will look like $r_1 + r_2(x-c_o)^{1/2}+r_3(x-c_o)+\dots$ for some constants $r_1, r_2, r_3$. Now consider solutions approaching $0$ from the right. Using only the two basis elements asymptotically $1$ and $\log x$, following them to the right until we approach $c_o$ from the left, we can choose a linear combination so that we get $r_1$ and $r_3$ from before: $r_1+r_2^*(c_o-x)^{1/2}+r_3(x-c)+\dots$. We want the same $r_1$ and $r_3$ as from the right, but $r_2^*$ is probably not the same as $r_2$. This way we get $Q_k(c_o^+)=Q_k(c_o^-)$ and $Q_k(0^+)=0$. So that all boundary terms cancel as desired. Now we still have a constant factor to set. If $\int_0^c U(x)\;dx$ turns out to be nonzero, choose that factor so that $\int_0^c U(x)\;dx = 1$. Then (as noted at the beginning) we will have $A_n = \int_0^c x^n U(x)\;dx$ for all $n \ge 0$. Another thing we have to hope for is that $U(x)$ turns out to have constant sign: The way we constructed it, we have $U(x)>0$ near $c$, but we have to hope it turns out to be nonnegative everywhere. I will see if I can get Maple to produce some pictures. I'm doing fine from $c$ going left to $c_o$, and from $c_o$ going both ways, but Maple seems reluctant to do (ODE3) from $0$ going up. added I am no expert on numerical ODEs. But here is what I get. Starting at the right, using the solution of (ODE3) with $U(x) = (c-x)^{1/2}+O((c-x)^{3/2})$ as $x \to c^-$, when I reach $c_o$ it looks like $$ U(x) = 1831.7 - 5769.99(x-c_o)^{1/2}-13916.8 (x-c_o)+O((x-c_o)^{3/2}) \qquad\text{as } x \to c_o^+ $$ Then, starting with the soution satisfying $$ U(x) = 1831.7 + 0.000103 (c_o-x)^{1/2}-13916.8 (x-c_o)+O((c_o-x)^{3/2}) \qquad\text{as } x \to c_o^- $$ we end up with $0$ on the $(\log x)^2$ term: $$ U(x) = O(\;|\log x|\;) \qquad\text{as } x \to 0^+ $$ The graph seems to be positive everywhere! So we merely need to divide by the integral. Here is a plot. (Not very informative.) Here it is near the point $c_o$:<|endoftext|> TITLE: Torsion-free group that is not of type F but is virtually of type F QUESTION [15 upvotes]: Recall that a group $G$ is of type F if there exists a compact $K(G,1)$. There are many examples of groups which are not of type F but which are virtually of type F, that is, they have finite-index subgroups of type F. For example, $SL(n,\mathbb{Z})$, mapping class groups, automorphism groups of free groups, etc. In all the examples I am aware of, the problem comes down to torsion (a group of type F must be torsion-free), and the required finite-index subgroup is just one that has no torsion. Question: Does anyone know of a torsion-free group that is not of type F but is virtually of type F? REPLY [11 votes]: It is conjectured that no such example exists. It is conjectured that $G$ is of type $F$, but it is hard to prove a group is type $F$ without explicitly exhibiting a classifying space. The class of groups of type $FP$ is a well-behaved proxy. Moreover, it is conjectured that every finitely presented group of type $FP$ is actually of type $F$. It is known that a finitely presented group of type $FP$ has a classifying space which is "finitely dominated" meaning that it is a retract in the homotopy category of a finite complex, which is just as good for most homotopical purposes. I will prove that if a torsion-free group $G$ has a finite index subgroup $H$ of type $FP$, then $G$ is itself of type $FP$. Applying this to your situation, since $H$ is finitely presented, so is $G$, so the conjecture implies that $G$ is of type $F$. It is possible that the hypothesis that $H$ is of type $F$ allow an unconditional proof, but I do not expect it. $G$ has type $FP$ if the trivial $G$-module $\mathbb Z$ has a finite length resolution by finitely generated projective $G$-modules. Type $F$ is equivalent to a finite length resolution by finitely generated free $G$-modules, plus finite presentation. The advantage of type $FP$ is that it is the intersection of two properties that can be treated separately, namely finite dimension and finite generation of cohomology $FP_\infty$. These can be interpreted as the existence of two resolutions, one of finite length and other by finite modules. The two properties can be combined to produce a resolution that is simultaneously finitely generated and finite length, but only if one allows projective modules and not just free modules, so type $F$ does not decompose this way. The easy part is finite generation of homology. If $G$ contains a finite index subgroup $H$, then the intersection of the conjugates of $H$ is a finite index normal subgroup $N$. The Hochschild-Serre spectral sequence for the group extension $N\to G\to G/N$ allows us to combine the $FP_\infty$ properties of $N$ and $G/N$ to conclude that $G$ is $FP_\infty$. The other part is finite homological dimension. By assumption $H$ has a finite dimensional classifying space with universal cover $EH$. Consider the space of $H$-equivariant maps from $G$ into $EH$. That is, intertwining the left action of $H$ on $G$ and the only action of $H$ on $EH$. This leaves the right action of $G$ on $G$, which makes this a $G$-space, a candidate for $EG$. For the quotient to be a model of $BG$, we need the space to be contractible and the action to be free. This space is isomorphic to $EH^{[G:H]}$, hence contractible and finite dimensional. The size of the isotropy is bounded by the index. So the isotropy is a finite subgroup of the torsion-free group $G$, so actually $G$ acts freely on this space. This shows that $G$ has a finite dimensional classifying space. In fact, the dimension of $G$ is the same as the dimension of $H$, although that does not come out of this construction. Reference: Ken Brown's book, Chapter VIII or his notes.<|endoftext|> TITLE: Approaching convex and discrete geometry from other disciplines QUESTION [6 upvotes]: I would like to learn some convex and discrete geometry (number 52 in MSC2010). I thought that it would be interesting to approach it from some other parts of mathematics - either by learning applications of convex and discrete geometry to these parts or (preferably) the other way round. Question: What are some interesting, non-trivial, preferably "important" results or problems lying in the intersection of convex and discrete geometry with other parts of mathematics? I am most interested in "other parts" being topology (including algebraic and geometric topology), order theory, logic (including model theory). However, other disciplines are also welcome. I did some MathSciNet searches, but I am not sure which papers could be interesting. REPLY [9 votes]: Here is a list of applications of other kinds of math to discrete and convex geometry (there is a little overlap with other answer already given). Some are not as important or groundbreaking as others but I consider all of them as reasonably interesting. The list comes in no particular order. Applications of topology to discrete geometry aka "Topological Combinatorics" -- Examples include the Ham Sandwich Theorem for measures, the colorful Tverberg theorem, and many more. A good starting point is the book "Using the Borsuk-Ulam theorem" by J. Matousek. Upper bound theorem for spheres and the $g$-conjecture -- This is a breathtaking application of commutative algebra and algebraic geometry to convex polytopes (and more generally spheres). If you want to learn from the master himself, have a look at Stanley's book "Combinatorics and commutative algebra". Polynomial method -- Using this (somewhat) recently developed technique, Guth and Katz were able to almost prove the conjectured lower bound of the Erdös distance problem. A good starting point to learn about this might be the blog entry by Tao. Monsky's theorem -- Monsky's theorem says that you cannot dissect a square in an odd number of triangles with all triangles having the same area. The proof makes essential use of 2-adic valuations and Sperner's lemma from topology. For this and many more results have a look at the book "Algebra and Tiling: Homomorphisms in the Service of Geometry" by Stein & Szabo. Monsky's result is also covered in one of the more recent editions of "Proofs from the book" by Aigner & Ziegler. The Kalai-Kleitman bound on the diameter of the graph of a polytope -- One of the biggest open problems in the theory of polytopes is the question of how large the diameter of the graph of a convex $d$-polytope with $n$ facets can be. Kalai & Kleitman showed that the diameter is at most $n^{\log(d) + 2}$. Very recently Todd was able to improve this bound to $(n-d)^{\log d}$. The technqiues used stem from extremal (hyper)graph theory and are not difficult to understand. Applications of differential geometry to discrete geometry -- Adiprasito's thesis contains nice examples of such applications. In particular he (together with Ziegler) was able to prove that in high enough dimension there are infinitely many projectively unique polytopes (which kills a conjecture by Shephard & McMullen). Optimization applied to discrete geometry -- de Oliveira Filho & Vallentin applied semi-definite programming to study ball packings and the chromatic number of $\mathbb{R}^d$. Have also a look at Firsching's paper Computing maximal copies of polytopes contained in a polytope which is another nice application of optimization techniques. Stochastic geometry -- Closely related to convex geometry is stochastic geometry. Schneider & Weil wrote a book on the topic. An application of the Möbius function from order theory to hyperplane arrangements -- This application is nicely describe in the new edition of Stanley's "Enumerative Combinatorics, vol I" (see Section 3.11). Eulerian posets -- These are special posets that have a lot of nice properties; a very good introduction is again Stanley's EC1. In particular, every face lattice of a polytope is a eulerian poset. This way we know that the Generalized Dehn-Sommerville equations for polytopes hold true, see the article by Bayer & Billera.<|endoftext|> TITLE: Who originated the standard symbols for Lie groups GL, SL, SU, etc.? QUESTION [19 upvotes]: Who was first to use symbols GL, SL, O, SO, U, SU, Sp and their projective versions, and how did this notation become standard? The notation appears in fairly modern form in Weyl's "The Classical Groups" except that he uses $O^+$ for SO. I know that Weyl coined the term "symplectic group", so the notation Sp certainly originates with him. Can anyone confirm or deny that Weyl was the originator of the standard symbols for Lie groups? Thanks. REPLY [14 votes]: In his 1939 book Weyl indeed did not use the notation SO but $O^+$, and called it the proper orthogonal group, but he did use the notation SU, or more precisely $^s$U with a tiny superscript s. The name "special unitary group" does not appear, it is always "unimodular". The footnote on the name "symplectic group", from the same book, is amusing:<|endoftext|> TITLE: How do I find coefficients of a product expansion QUESTION [8 upvotes]: Any power series $f(t) = 1 + t \mathbb{Z}[[t]]$ can be uniquely expanded in the following two ways: $$1 + \sum_{i=1}^\infty f_i t^i = \prod_{i=1}^\infty (1-t^i)^{-n_i}$$ Here, the $f_i$ and $n_i$ are integers. Is there an explicit formula for the $n_i$ in terms of the $f_i$? Similarly a power series $f(t, L) = 1 + t \mathbb{Z}[[t, L]]$ can be uniquely expanded $$1 + \sum_{i=1}^\infty \sum_{j=0}^\infty f_{i,j} t^i L^j = \prod_{i=1}^\infty \prod_{j=0}^\infty (1-t^i L^j)^{-n_{i,j}}$$ How do I find the $n_{i,j}$ in terms of the $f_{i,j}$? More generally, in an arbitrary $\lambda$-ring $\Lambda$, a power series $f(t) \in 1 + t \Lambda[[t]]$ can be uniquely expanded in the following two ways: $$1 + \sum_{i=1}^\infty f_i t^i = \prod_{i=1}^\infty (1-t^i)^{-n_i}$$ where the $f_i, n_i \in \Lambda$, and the exponent signifies $(1-t^i)^{-M} := \sum_{k=0}^\infty t^{ik} \mathrm{Sym}^k(M)$ Is there an explicit formula for the $n_i$ in terms of the $f_i$? REPLY [3 votes]: Robert Israel's answer works if you have access to the coefficients of $\log f(t)$, which isn't always the case (one example that comes to mind is when the $f_i$'s are classes of punctual Hilbert schemes, and you're trying to get invariants as in here.) To answer your question more specifically, here is my attempt at spelling out the strategy you mention. Let's start with the identity $$\frac{1}{1-at}=\prod_{k\geq 1}\frac{1}{(1-t^k)^{M_k(a)}},$$ where $M_k(a)=\frac{1}{k}\sum_{d|k}a^d\mu(k/d)$ are the necklace polynomials. Next, let $h_i\in \Lambda[x_1,x_2,\dots]$ be the complete homogeneous symmetric polynomials. We have the following fundamental generating function $$1+h_1t+h_2t^2+\cdots=\prod_{r\geq 1}\frac{1}{(1-x_rt)}.$$ So combining with our previous observation we have $$1+h_1t+h_2t^2+\cdots=\prod_{k\geq 1}\frac{1}{(1-t^k)^{M_k(x_1)+M_{k}(x_2)+\cdots}}.$$ Now, $n_k=\sum_r M_k(x_r)=\frac{1}{k}\sum_{d|k}p_d(x_1,x_2,\dots) \mu(k/d)$, so to finish off we need to express the power sum symmetric functions, $p_k$, in terms of the $h_k$'s. This is given by a variation of the Newton identities $$p_n(h_1,h_2,\dots)=(-1)^{n-1}\begin{vmatrix}h_1 & 1 & 0 & \cdots & \\ 2h_2 & h_1 & 1 & 0 & \cdots \\ 3h_3 & h_2 & h_1 & 1 & \\ \vdots & & \\ nh_n & h_{n-1} & \cdots & & h_1\end{vmatrix}.$$ So the polynomials you're looking for are $$n_k(f_1,f_2,\dots)=\frac{1}{k}\sum_{d|k}\mu(k/d)p_d(f_1,f_2,\dots).$$ Hope this helps.<|endoftext|> TITLE: Who defined and who coined "module"? QUESTION [6 upvotes]: The title of my Q. says it all: QUESTION:   Who defined and who coined: module? Would it be Emmy Noether? EDIT   In view of @anon's and KConrad's answers, and as it could have been expected, the situation is a bit complex. Thus it looks that while Dedekind coined module, the final notion was defined by Emmy Noether. By defined I mean that module at a minimum should be a generalization of linear spaces, abelian groups and ideals. REPLY [11 votes]: From: http://jeff560.tripod.com/m.html MODULE. A JSTOR search found the English term in E. T. Bell’s “Successive Generalizations in the Theory of Numbers,” American Mathematical Monthly, 34, (1927), 55-75. Bell was describing the work of Dedekind, basing his account on Dedekind’s French article, “Sur la Théorie des Nombres entiers algébriques” (1877) Gesammelte mathematische Werke 3 pp. 262-298. Dedekind used the French word module to translate his German term Modul. Stillwell writes in the Introduction to his English translation, Theory of Algebraic Integers (1996, p. 5), “Dedekind presumably chose the name ‘module’ because a module M is something for which ‘congruence modulo M’ is meaningful.” Curiously le module had once before been translated into English but then it went into English as the MODULUS of a complex number. [John Aldrich]<|endoftext|> TITLE: Separation of lattice points on the Mordell elliptic curve QUESTION [8 upvotes]: Consider the Mordell equation x^3 – y^2 = k, where x is a non-square positive integer and y^2 is the perfect square nearest to x^3. Noam Elkies (see http://www.math.harvard.edu/~elkies/hall.html) found in 1998 that there are only 25 integers below 10^18 for which |k| < sqrt(x) (the first three being 2, 5234 and 8158). For these 25 numbers (and for the approximately 30 larger numbers found to date) the separation is huge. I am not aware of any research results that prove a minimum separation between such numbers in general: If x is such a number, what is the minimum separation between x and the next highest such number? Is anyone aware of such research and/or able to comment on how simple/difficult this question is? The question obviously also relates to the separation between lattice points on the Mordell elliptic curve for large x (k fixed). Elkies (Rational points near curves and small nonzero |x^3 − y^2| via lattice reduction, May 2000) has proved an upper bound (order of sqrt(N).log(N)) for the number of such points not exceeding N, but it is not clear to me whether his method implies any minimum separation. I would greatly appreciate any information on the existence or otherwise of research on this question. REPLY [5 votes]: Hall's conjecture says that for every $\epsilon>0$ there is a $C_\epsilon$ such that if $x$ and $y$ are integers with $x^3-y^2\ne0$, then $$ |x^3 - y^2| \ge C_\epsilon \max\{|x^3|,|y^2|\}^{1/6-\epsilon}, $$ so this would imply that the separation indeed gets quite large as $x$ and $y$ increase. The polynomial version, i.e., when $x$ and $y$ are in $\mathbb{C}[T]$, was proven by Davenport with $\epsilon=0$, i.e., if $x^3 - y^2 \ne 0$ then $$ \deg(x^3-y^2) \ge \frac16 \max\{\deg(x^3),\deg(y^2)\} + C $$ for an absolute constant $C$ (in fact, we can use $C=1$). More generally in the polynomial (or function field) case, one can easily prove lower bounds for $\deg(x^n-y^m)$, with analogous conjectures over $\mathbb{Z}$ (or over number fields). Addendum: As Ryan D'Mello points out, the above doesn't really answer the question, which asks about gaps between $x$ values of solutions to $0<|x^3-y^2|<\sqrt{x}$. However, I think that in order to get a reasonable answer, one will need to assume something like Hall's conjecture. Alternatively, for a given (small) $\epsilon>0$, one might hope to unconditionally prove a gap estimate for the set $$ \bigl\{ x\in\mathbb{Z} : \text{there exists $y\in\mathbb{Z}$ with $|x|^{1/2-\epsilon}<|x^3-y^2|<|x|^{1/2+\epsilon}$} \bigr\}, $$ where the size of the gaps depends on $\epsilon$.<|endoftext|> TITLE: Homology theory represented by Madsen-Tillmann spectra QUESTION [17 upvotes]: The generalized homology theory of the Thom spectrum $MO=\varinjlim\Sigma^nMTO_n$ is bordism theory:\begin{equation*}\pi_k(MO\wedge X)=\Omega^O_k(X)\end{equation*}These groups form the ring of (unoriented) $X$-cobordism classes of (unoriented) manifolds. But what information do the Madsen-Tillmann spectra $MTO_n$ contain? Does the homotopy ring\begin{equation*}\pi_k(MTO_n\wedge X)=\text{ ?}\end{equation*}yield any useful classification of manifolds? REPLY [23 votes]: This is an exercise in understanding the Pontrjagin--Thom correspondence. The group $\pi_k(MTO(n) \wedge X_+)$ is in bijection with tuples of a $(n+k)$-manifold $M$, an $n$-dimensional vector bundle $V \to M$, a stable isomorphism $\varphi: V \oplus \epsilon^k \oplus \epsilon^\ell \cong TM \oplus \epsilon^\ell$ for $\ell \gg 0$, and a continuous map $f : M \to X$. This data is taken up to cobordism in the obvious way. Note that the spectrum $MTO(n)$ is not connective, which corresponds to the fact that the above makes sense for negative $k$. On the other hand, for a $d$-dimensional manifold $X$ the cohomology theory $[X,MTO(n)]$ is represented by tuples of a $(d+n)$-dimensional manifold $E$ with a proper map $\pi : E \to X$, an $n$-dimensional vector bundle $V \to E$, a stable isomorphism $\varphi : TE \oplus \epsilon^\ell \cong V \oplus \pi^*(TX) \oplus \epsilon^\ell$, again taken up to cobordism in the obvious way. The point that user43326 is referring to is that if $\pi : E^{d+n} \to X^d$ is a smooth fibre bundle with compact $n$-dimensional fibres, then we may define $V := T_\pi E = \mathrm{Ker}(D\pi : TE \to TX)$ to be the vertical tangent bundle and choose a splitting of the short exact sequence $$0 \to T_\pi E \to TE \to \pi^*(TX) \to 0$$ of vector bundles on $E$. This gives an isomorphism $\varphi : TE \cong V \oplus \pi^*(TX)$, and so the tuple $(\pi: E \to X, V, \varphi)$ represents a class in $[X, MTO(n)]$. (However, despite what user43326 said, it is not true that all classes in this cohomology theory arise in this way.) The reason that $$\mathrm{hocolim}_{n \to \infty} \Sigma^n MTO(n) \simeq MO$$ can be easily seen from the first description above. Concretely, a class in $\pi_k$ of the homotopy colimit is represented by a tuple a $k$-manifold $M$, an $n$-dimensional vector bundle $V \to M$ for some $n \gg0$, a stable isomorphism $\varphi : V \oplus \epsilon^\ell \cong TM \oplus \epsilon^{n-k} \oplus \epsilon^\ell$, for $\ell \gg0$, taken up to cobordism. By taking $n$ large enough, and destabilising the isomorphism, we get $\varphi: V \cong TM \oplus \epsilon^{n-k}$, and so the last two pieces of data cancel out: we are left with just $k$-manifolds up to cobordism.<|endoftext|> TITLE: Is there a hyperplane avoiding two independent sets? QUESTION [10 upvotes]: Let $V$ be a vector space over a field with $5$ elements, $A,B \subseteq V$ independent subsets. Must there be a subspace of $V$ of codimension 1 disjoint from $A \cup B$? REPLY [2 votes]: Not an answer, but this might get you more help by phrasing it in terms of a common combinatorics problem - finding a lower bound for the size of a transversal of a hypergraph. A hypergraph is a collection of points and edges like a graph, but each edge can contain any number of vertices. A transversal for a hypergraph is a set of vertices which intersect every edge. Let $V=\mathbb{F}_q^k$ be a $k$-dimensional vector space over the finite field of $q$ elements. The naive hypergraph construction would be to take each non-zero point as a vertex, and each hyperplane as an edge. We must exclude zero because it is on every hyperplane, and would give a trivial transversal. Finding a minimal transversal here will tell you how many points you need to make sure that every hyperplane is hit, hence if it is larger than $|A \cup B|$, which can be at most $2k$, then there will be a hyperplane disjoint from them. This may not be the best construction however, because points on a line in $V$ will always appear together on each edge, so we lose some symmetry. Two arbitrary points will be contained in a different number of hyperplanes depending on whether or not they are collinear with the origin. Instead, for a point $v \in V$, write $L(v) = \{v,2v,\ldots, (q-1)v\}$ for the line of points through $v$ and the origin. Construct the hypergraph $H$ with vertices $L(v)$, discarding duplicates, and hyperplanes for edges. Certainly if $v$ is on a hyperplane $P$, then $L(v)$ is also contained in $P$, so a transversal for the naive construction also gives us a transversal here, and conversely by choosing any point on the line $L(v)$. This construction has nice properties: it has $\frac{q^{k}-1}{q-1}$ vertices and the same number of edges; it is uniform, meaning all edges have the same number of vertices, $\frac{q^{k-1}-1}{q-1}$; it is regular, meaning every vertex is on the same number of edges, again $\frac{q^{k-1}-1}{q-1}$; and every pair of vertices is on the same number of edges, $\lambda = \frac{q^{k-2}-1}{q-1}$. These sorts of hypergraphs are often studied for small orders, and are related to block designs, but I could not find or produce anything that applied to your specific family of spaces. At this point, you can try looking through the literature with these terms (maybe Berge's book Hypergraphs), or ask a noted expert in the field like Noga Alon, who is also conveniently at Tel-Aviv University.<|endoftext|> TITLE: Arthur's refinement of parameters for unitary automorphic representations QUESTION [5 upvotes]: In his work on the classification of automorphic representations of a group $G$, Arthur has conjectured that the parameterization of such representations involves a homomorphism $\rho : SL_2 \times Gal(\bar{F}/F) \rightarrow G^\vee$. This is a refinement of earlier conjectures that involved homomorphisms $\tilde{\rho} : Gal(\bar{F}/F) \rightarrow G^\vee$ and thus the $SL_2$ appearing in $\rho$ is sometimes referred to as "Arthur's $SL_2$". My questions are the following : For arbitrary $G$, is there a bijection between certain kinds of autormorphic representations and the instances where the $SL_2$ plays a non-trivial role ? Ex : $G=GL_n$, my understanding is that the $SL_2$ does not play a non-trivial role and that this is related to the absence of non-tempered cuspidal unitary representations (I learnt the statement from Frenkel's review). But, it is not clear to me if there is a bijection between non-tempered cuspidal unitary representations and cases where $SL_2$ plays a non-trivial role on the Galois side. I would like to ask a question similar in spirit to the above one for the local case. Is there one ? (Arthur's article has a few comments about what the existence of such parameters means for the local case, but I could not use it to come up with a question for the local case). References : Arthur's paper "Unipotent automorphic representations : conjectures (1989)" is available here and the Frenkel review where I came across Arthur's $SL_2$ is here. REPLY [5 votes]: For $G=Gl_n$, the $SL_2$ factor of Arthur plays a trivial role in the classification only when you restrict yourself to cuspidal automorphic representations. But Arthur is interested with more general automorphic representations that are in the discrete spectrum, that is (essentially) the irreducible sub-representation (in the naive sense, that is not the one that are in the continuous spectrum) of $L^2(G(\mathbb A)/G(\mathbb Q),\omega)$, where $\omega$ is a central character. For the classification of this, the $SL_2$ factor is absolutely necessary. In fact proving that the classification of the discrete spectrum of $GL_n$ is, modulo the classification of the cuspidal spectrum of $GL_n$, as pre diced by Arthur conjecture with his $SL_2$ factor is a monumental theorem of Langlands (prior to Arthur, who wa motivated by it to introduce its $SL_2$-factor), which has been rewritten with more details by Moeglin and Waldspurger in a large book with evocative sub-title "une paraphrase de l'Écriture"). This answers your question about $GL_n$. In general, it is still true, according to Arthur's formalism now proved in many cases, that the representations (or rather an $A$-packet of representations) that have an Arthur parameter which is trivial on the $SL_2$ factor are exactly the one which are tempered. To understand more about Arthur formalism, besides Arthus' articles which are very deep and very hard, and Frankel's survey you quote, you can give a look to the Appendix of my book with Chenevier, Astérisque 328, and also to the new book of Chenevier and Lannes, Chapter 8.<|endoftext|> TITLE: A example on Fourier tranform of a continous compactly supported function QUESTION [5 upvotes]: I am trying to find a continuous compactly supported function $f$ such that the Fourier transform $f^{ft}$ and derivative $(f^{ft})'$ of the $f^{ft}$ decay faster than exponential rates, that is $$|f^{ft}(t)|=O(e^{-k\left|t\right|^{\gamma}}),\, \left|(f^{ft})'(t)\right|=O(e^{-k\left|t\right|^{\gamma}}).$$ So far I have not found such a function. Can someone help me? Thank so much. REPLY [4 votes]: There are no such functions if $\gamma>1$. Even more is true: Suppose $f\in L^2(\mathbb R)$ is supported in a set $S$ of finite measure and assume its Fourier transform $g=\hat f$ satisfies $|g(x)|\leq Ae^{-k|x|^\gamma}$ for $k>0$ and $\gamma>1$. Denote $I_m=[-m,m]$. By Benedick's theorem there is a constant $C>0$ so that $$ \|f\|_{L^2(\mathbb R)} \leq Ce^{2Cm|S|}\|g\|_{L^2(\mathbb R\setminus I_m)}. $$ If $m\geq\gamma^{-1/(\gamma-1)}$, then $x\geq m$ implies $x^\gamma-x\geq m^\gamma-m$. Therefore $$ \|g\|_{L^2(\mathbb R\setminus I_m)}^2 = \sum_\pm\int_m^\infty|g(\pm x)|^2dx \leq \sum_\pm\int_m^\infty A^2e^{-2kx^\gamma} dx \leq 2A^2e^{-2km^\gamma}\int_m^\infty e^{-2k(x-m)} dx = 2A^2e^{-2km^\gamma}(2k)^{-1}. $$ Inserting this to the previous estimate gives $$ \|f\|_{L^2(\mathbb R)} \leq Ck^{-1/2}Ae^{2Cm|S|-km^\gamma}. $$ Since $\gamma>1$, taking the limit $m\to\infty$ gives $\|f\|_{L^2(\mathbb R)}=0$ so $f=g=0$.<|endoftext|> TITLE: A question about "Zariski dense" arguments QUESTION [30 upvotes]: This question is a little basic, but I think it is consistent with the goals of MO. My question is about a certain type of argument in algebraic geometry which exploits the abundance of dense sets in the Zariski topology. A classical example is the Cayley-Hamilton theorem, and I will frame my question by sketching a false proof. Theorem: Let $A$ be a $n \times n$ matrix with complex entries. Then $p_A(A) = 0$ where $p_A$ is the characteristic polynomial of $A$. False proof: Step 1: The theorem is trivial for diagonalizable matrices. Step 2: The set of diagonalizable matrices is Zariski dense in $\mathbb{C}^{n^2}$ because it contains the complement of the zero locus of the discriminant polynomial. Step 3: The function $\mathbb{C}^{n^2} \to \mathbb{C}^{n^2}$ given by $A \mapsto p_A(A)$ is Zariski continuous, so it vanishes everywhere since it vanishes on a Zariski dense set. The Flaw: Step 3 uses the following "fact" (which perhaps belongs as an answer to the "common false beliefs" question): if $f$ and $g$ are continuous functions between topological spaces $X$ and $Y$ which agree on a dense subset of $X$ then they agree everywhere. But this need not be the case if $Y$ is not Hausdorff (and the Zariski topology certainly is not): Let $X = \mathbb{R}$, let $Y$ be the line with the double origin, and let $f, g \colon X \to Y$ be the maps which restrict to the identity on $\mathbb{R} - \{0\}$ and which satisfy $f(0) = 0_1$, $g(0) = 0_2$. I have seen this error in textbooks. In the case of the Cayley-Hamilton theorem, there is an easy fix: if you give $\mathbb{C}^{n^2}$ the norm topology then the diagonalizable matrices are still dense, $A \mapsto p_A(A)$ is still continuous, and now everything is Hausdorff. But Zariski dense arguments come up a lot in algebraic geometry, sometimes in contexts where there isn't a norm topology conveniently lying around. So my question is: can the problem be fixed in some sort of uniform way? REPLY [16 votes]: At least in the proof of Cayley-Hamilton you mentioned, you do not need the Hausdorff condition. All you need is for the points to be closed, which is the case for Zariski topology. The point is that you can reformulate the condition $f(x)=g(x)$ as $f(x)-g(x)=0$, so the argument goes through without any issues.<|endoftext|> TITLE: Slim Kurepa tree at a singular strong limit cardinal of uncountable cofinality QUESTION [10 upvotes]: For a strong limit cardinal $\kappa$ the notion of $\kappa$-Kurepa tree is trivial: the full binary tree is a $\kappa$-Kurepa tree. Accordingly, we consider the following strengthening: A slim $\kappa$-Kurepa tree is a tree $T$ of height $\kappa$ such that for every infinite $\alpha < \kappa$ the $\alpha$-th level of $T$ has cardinality $\left| \alpha \right|$, and $T$ has more than $\kappa$ many branches. If $\kappa$ is a strong limit cardinal of countable cofinality, it's easy to construct a slim $\kappa$-Kurepa tree. On the other hand, if $\kappa$ is measurable (or just ineffable) then there is no slim $\kappa$-Kurepa tree. If $\kappa$ is inaccessible, then my understanding from comments here is that there is a $\mathord{<}\kappa$-closed forcing to create a slim $\kappa$-Kurepa tree (but this destroys measurability.) What about the uncountable cofinality singular case? If $\kappa$ is a singular strong limit cardinal of uncountable cofinality, can there exist a slim $\kappa$-Kurepa tree? EDIT: This turned out to be fairly easy; see my answer below. However, I would like to know where I can find this result proved (or at least mentioned) in print. So I will accept the first answer that tells me this. REPLY [5 votes]: The following is proved by Erdos-Hajnal-Milner in ``On sets of almost disjoint subsets of a set. Acta Math. Acad. Sci. Hungar 19 1968 209–218'', from which the required result follows Theorem. ssume $\aleph_0 < cf(\kappa) < \kappa$ and $\forall \theta< \kappa, \theta^{cf(\kappa)} < \kappa.$ Let $F \subseteq P(\kappa)$ be such that $\{\alpha < \kappa: |F \restriction \alpha| \leq \alpha \}$ is stationary. Then $|F| \leq \kappa.$<|endoftext|> TITLE: Morava $K(n)$'s are not $E_{\infty}$ QUESTION [10 upvotes]: I am looking for a reference/proof that shows that the Morava $K$-theory spectra, $K(n)$ are not $E_{\infty}$ ring spectra. I suspect that this should be a calculation but I can't quite get it right. Thank you as always. REPLY [19 votes]: The assertion is Lemma 5.6.4 in Rognes's "Galois extensions of structured ring spectra" available on the arXiv. In fact, the $K(n)$ spectra do not even admit $E_2$-algebra structures. The reason is that the free $E_2$-algebra with $p = 0$ is known to be (by a theorem of Hopkins-Mahowald) the Eilenberg-MacLane spectrum $H \mathbb{F}_p$. If $K(n)$ admitted an $E_2$-algebra structure, it would receive a map of $E_2$-rings $H \mathbb{F}_p \to K(n)$, which is not possible (for instance, as $H \mathbb{F}_p \wedge K(n) = 0$). More generally, no $p$-power torsion $E_n$-local spectrum can admit an $E_\infty$-structure. This can be seen by playing with power operations and is worked out as the main result of joint work with J. Noel and N. Naumann. The closing remarks to that paper also outline a proof of the Hopkins-Mahowald theorem cited above (which was not formally documented in the literature for odd primes). Analogous results for $E_2$-algebras (or $E_m$-algebras for $m \geq 2$) seem to be unknown. For example, consider the free $E_2$-algebra with $p^2 =0 $. Is it true that this is $K(n)$-acyclic for each $n$? At the prime $2$, it is even true that any homotopy commutative ring spectrum with $2 =0$ is equivalent to a wedge of Eilenberg-MacLane spectra (and in particular is invisible to $L_n$-localization). This is a result of Wurgler.<|endoftext|> TITLE: What is the current state of the crystalline analogue of the Weil conjectures? QUESTION [9 upvotes]: In "F-isocrystals on open varieties results and conjectures" Faltings says: "Finally, we extend the theory of weights and show as much as possible of the crystalline analogue of the Weil conjectures." My questions are: What is the current state of the crystalline analogue of the Weil conjectures? Which are the articles and books about these conjectures? REPLY [8 votes]: As far as I can tell, Kedlaya solved it 12 years ago.<|endoftext|> TITLE: Finding integer points on elliptic curves via divisibility conditions like $(a+b)^2 \mid (2b^3+6ab^2-1)$ QUESTION [17 upvotes]: Is the following conjecture correct? Conjecture. The divisibility condition $(\alpha+\beta)^2 \mid (2\beta^3+6\alpha\beta^2-1)$ has no solutions in positive integers $1 \le \beta < \alpha < 2\beta$. This question is related to finding integer points on a Mordell curve. A computer search outside the range indicated turned up the expected single solution $(a,b)=(4,1)$ [corresponding to the single integer point on the original Mordell curve], as well as an unexpected solution $(\alpha,\beta)=(11364,46061)$, which I can't explain. Any insights would be appreciated. I believe a method of solution — particularly by descent — to this special case would be immediately applicable to a large class of elliptic curves. FWIW, I've developed a partial proof which I include below. The Vieta jump implies that, for any solution $(a,b)$, there is a rational solution $(b,\tfrac{6b^2}{k}-2b-a)$ with the same $k$; in the case of the one known solution $(a,b)=(4,1)$, we do have $k=1$ in the original equation $$ (5b-a+1)(a+b)^2 = 2(2b^3+6ab^2-1), $$ which makes the second Vieta root degenerate at $(1,0)$. Proof (incomplete). The divisibility hypothesis implies \begin{equation} k(\alpha+\beta)^2 = 2\beta^3+6\alpha\beta^2-1, \tag{1} \end{equation} for an integer $k \ge 1$. Rearranging (1) and replacing $\alpha$ with the variable $\xi$ yields \begin{equation*} k\xi^2 + 2\beta(k-3\beta)\xi + (k\beta^2-2\beta^3+1) = 0.% \label{EQ: solve this} \end{equation*} One root of this equation is $\xi_1 = \alpha$. By Vieta's formulas, the other root may be written as \begin{align} \xi_2 &= \frac{2\beta(3\beta-k)}{k} - \alpha = \frac{\beta^2(k-2\beta)+1}{k \alpha}. \tag{2} \end{align} First, assume $\xi_2$ is an integer. Since $\alpha$ is an integer, the first relation in (2) implies that the fraction must also be an integer. Hence $k \mid 6\beta^2$. But (1) implies both that $k$ is odd, and that $\gcd(k,\beta)=1$. Hence $k \mid 3$, so $k = 1$ or $3$. If $k = 3$, then the second relation in (2) implies that $\xi_2$ is positive when $\beta = 1$, and negative when $\beta > 1$. On the other hand, the first relation with $k=3$ gives $\xi_2 = 2\beta(\beta-1) - \alpha$. When $\beta = 1$, this implies $0 < \xi_2 = 2\beta(\beta-1) - \alpha = -\alpha$, contradicting $\alpha > 0$. When $\beta > 1$, we have $0 > \xi_2 = 2\beta(\beta-1) - \alpha$. Hence $\alpha > 2\beta(\beta-1)$, contradicting $\alpha < 2\beta$. END OF PARTIAL PROOF Note that the unexpected solution $(\alpha,\beta)=(11364,46061)$ implies \begin{equation*} k = \frac{2\beta^3+6\alpha\beta^2-1}{(\alpha+\beta)^2} = \frac{340107729770625}{3297630625} = 3 \cdot 31 \cdot 1109, \end{equation*} and then we have \begin{equation*} \frac{2\beta(3\beta-k)}{k} - \alpha = \frac{\beta^2(k-2\beta)+1}{k \alpha} = \frac{685486248}{34379} \end{equation*} not an integer. REPLY [20 votes]: This is a supplement to Lucia's answer. His/her heuristic analysis suggests that there are infinitely many odd $m$'s such that the largest square dividing $m^3-2$ exceeds $9m^2$, contradicting the conjecture. On the other hand, the abc conjecture implies that the largest square dividing $m^3-2$ is $\ll_\epsilon m^{2+\epsilon}$ for any $\epsilon>0$, which indicates that a slightly weaker version of the conjecture (i.e. one with a more restrictive constraint on the variables) is probably true. Some calculations with SAGE show that for $m<10^7$ there are only two instances where the largest square dividing $m^3-2$ exceeds $m^2$, namely $m=3$ and $m=100$. (Noam Elkies extended the range to $m<\ell<10^8$ with a more efficient gp-pari code.) I found five further values with a square divisor exceeding $m^2/2$, and these are $\{1244,11317,296428,714417,722428\}$. Added. The constant in Lucia's heuristic analysis (with $m$ restricted to odd numbers) equals $$ \frac{1}{6}\frac{1}{2}\frac{2}{3}\prod_{\substack{p>3\\f(p)=0}}\left(1-\frac{1}{p^3}\right)\prod_{\substack{p>3\\f(p)=1}}\left(1-\frac{1}{p^2}\right)\prod_{\substack{p>3\\f(p)=3}}\left(1-\frac{3}{p^2}+\frac{2}{p^3}\right)$$ times the residue of the Dedekind zeta function of $\mathbb{Q}(\sqrt[3]2)$. The initial factor $1/6$ accounts for Lucia's denominator $3$ and the fact that we only look for odd $m$'s. The other factors come from comparing the Euler factors of $\sum_{\ell=1}^\infty f(\ell)\ell^{-s}$ to those of the Dedekind zeta function. My calculation in SAGE shows that $ C\approx 0.0423$, which suggests that the average ratio between the consecutive $\ell$'s yielding a counterexample $(\ell,m)$ is about $1.867\times 10^{10}$. Summary of ongoing calculation. Counterexamples to the OP's conjecture correspond bijectively to pairs of odd numbers $(\ell,m)$ such that $\ell^2\mid m^3-2$ and $m/\ell<1/3$. I extended Noam Elkies' gp-pari calculation to the range $\ell<10^{12}$ and am checking higher ranges on several machines. The best pairs I found so far (allowing $m$ even) are $$(\ell,m)=(10444012561, 5062142741)\quad\text{with ratio}\quad m/\ell\approx 0.485$$ $$(\ell,m)=(22713683537, 7950843140)\quad\text{with ratio}\quad m/\ell\approx 0.350$$ In particular, these beat the basic ratio $m/\ell=0.600$ that corresponds to $(\ell,m)=(5,3)$.<|endoftext|> TITLE: Time in Girard's Geometry of Interaction QUESTION [15 upvotes]: Jean-Yves Girard writes at the end of his paper "Towards a Geometry of Interaction", page 105, that we have three intuitions about the nature of time: time is logic modulo the order of rules, time is the cut elimination process, time is the contents of noncommutative linear logic. Can anyone explain what these means? Does anyone know if Girard has developed these thoughts any further? REPLY [33 votes]: I did my PhD thesis in Girard's team in Marseille (my supervisor was Laurent Regnier, himself a student of Girard's) so I have quite a bit of experience with his "excentric" way of communicating and I can attempt an exegesis ( :-) ) of this particular sentence (besides, I am quite familiar with both the philosophical and technical contents of what Girard calls "geometry of interaction", or GoI). First of all, the concept of time Girard is talking about is computational time, i.e., the step-by-step evolution of a computational process. This is where his words make the most technical sense. Any broader interpretation of the word "time" in this context may (or may not) lead to futile and meaningless musings. Now, the three "intuitions" Girard is talking about correspond to three different views of logic, the first two belonging to the proof-theoretic tradition, the third more model-theoretic: logic as proof search; logic as functional programming; logic as a descriptive tool. In logic as proof search, one step of computation corresponds to one inference rule (read bottom-up). Certain inference rules commute, which means that they may possibly be applied in parallel, whereas others are related by causal dependencies, yielding a sequential evaluation. These latter are the ones that make the time "tick", so one may see computational time in proof search to be given by the successive application of clusters of mutually independent rules. This idea finds a technical realization in the notion of polarity and focusing proofs in linear logic, which was unknown at the time Girard wrote "Towards a GoI". Polarity in linear logic was introduced in the early 90s by Jean-Marc Andreoli and today is an essential aspect not only of linear proof search but of games semantics and the theory of programming languages in general. The second view is the simplest to explain: it is the well-known Curry-Howard correspondence, under which a proof may be seen as a program, the execution of which corresponds to cut-elimination. So computational time is just the succession of cut-elimination steps, i.e., rewriting steps leading to a result, which is fairly standard and intuitive I would say. The relationship between evaluation time in this sense ($\beta$-reduction/cut-elimination) and the usual notion of time defined using Turing machines (or other low-level machines) has been the direct or indirect subject of countless papers from the 90s onwards (google "implicit computational complexity", or take a look for example at this paper by Blelloch and Greiner or this very recent paper by Accattoli and Dal Lago). The third view is also very standard: logic is seen as a language in which we may state facts/propositions about some kind of world. There is a whole class of logical languages (known as temporal logics) which are taylored so as to be able to speak of worlds in which there is a notion of time. Here, Girard is suggesting that non-commutative logic, with its non-symmetric connectives distinguishing "left" from "right" (hence, presumably, "before" from "after") may provide a language with a built-in notion of time. This latter point is the most hand-wavy and, in fact, it is the only one that has had no technical development so far (at least as far as I know).<|endoftext|> TITLE: Rational points techniques on curves not using their Jacobian QUESTION [9 upvotes]: Let $C/K$ be a curve of genus > 2 over a number field $K$ and suppose there exists a $p \in C(K)$. Then a recurring theme in studying $C(K)$ is using the map $C \to J(C)$ normalized by sending $p$ to 0, for example Faltings proof that $C(K)$ is finite, Chabauty-Coleman, Mazurs determination of all $\mathbb Q$ rational points on $X_0(N)$. I want to know if there are also techniques for studying $C(K)$ that don't use the Jacobian. My main motivation is that I later want to explicitly apply these techniques to a curve $C/\mathbb Q$ for which I can prove that the closure of $J(\mathbb Q)$ is of finite index in $J(\mathbb Q_p)$ for all primes p so that Coleman-Chabauty does not work. The techniques don't necessarily need to fit in a nice theoretical framework, examples in the literature where rational point questions on specific curves are solved without using their Jacobian are welcome to! One example that I know of is Runge's method which was recently succesfully used to study rational points on certain modular curves by Bilu and Parent http://arxiv.org/abs/0907.3306 REPLY [8 votes]: There is a way of shifting the problem to other curves: start with an unramified covering $\pi \colon D \to C$ that is Galois over $\bar{\mathbb Q}$ and compute the associated Selmer set $\operatorname{Sel}^\pi(C)$. The set is finite and computable, at least in principle; its elements $\xi$ correspond to twists $\pi_\xi \colon D_\xi \to C$ of $\pi$ such that $$C(K) = \bigcup_\xi \pi_\xi(D_\xi(K)),$$ which "reduces" the problem to that of determining the sets of rational points on the curves $D_\xi$ (whose genus is larger than that of $C$, so they tend to be more difficult to deal with). One can then hope that other methods can be applied to these curves. In some cases, for example, these curves map to other curves of lower genus over $K$ or over larger fields, and one might be able to find their $K$-points or the images of the $K$-points on $D_\xi$. The problems with this approach in practice are: The computation of the Selmer set may be infeasible (it works reasonably well for 2-coverings of hyperelliptic curves, but not in many other cases); As already mentioned, the covering curves are "worse" than $C$, and one needs some luck to reduce the problem of determining $D_\xi(K)$ to something manageable (again chances are best in the hyperelliptic case); It is likely to be hard to get it to work for a family of curves like the $X^+_{\text{ns}}(N)$ that you mention in a comment above. The determination of $X^+_{\text{ns}}(13)({\mathbb Q})$ is a notorious open problem, even though the curve has genus only 3! EDIT: This problem is no longer open, see this paper by Balakrishnan, Dogra, Müller, Tuitman and Vonk. That said, I'd like to see you make progress on this question — this would also be helpful for a project of mine.<|endoftext|> TITLE: Repetend digit graphs for $1/n$ in base $b$ QUESTION [10 upvotes]: Here is a decimal expansion of $\frac{1}{34}$: $$(1/34)_{10}=0.02941176470588235\overline{2941176470588235}\ldots$$ And here is a graphical representation of the 16-digit "repetend," as a directed repetend digit graph (my terminology): $$(2,9,4,1,1,7,6,4,7,0,5,8,8,2,3,5)\;.$$       I was exploring the digit-expansion of $1/n$ in base $b$—fixing $n$ while letting $b$ vary—and find it puzzling. Here is an example, for $n=51$, and bases $b=5,\ldots,50$. The top row shows base $b$, and underneath, the length of the repetend for $\frac{1}{51}$ in that base: $$ \left( \begin{array}{cccccccccccccccc} 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\ 16 & 16 & 16 & 8 & 8 & 16 & 16 & 16 & 4 & 16 & 8 & 2 & 2 & 1 & 8 & 16 \\ \end{array} \right) $$ $$ \left( \begin{array}{ccccccccccccccc} 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 & 29 & 30 & 31 & 32 & 33 & 34 & 35 \\ 4 & 16 & 16 & 16 & 8 & 8 & 16 & 16 & 16 & 4 & 16 & 8 & 2 & 1 & 2 \\ \end{array} \right) $$ $$ \left( \begin{array}{ccccccccccccccc} 36 & 37 & 38 & 39 & 40 & 41 & 42 & 43 & 44 & 45 & 46 & 47 & 48 & 49 & 50 \\ 8 & 16 & 4 & 16 & 16 & 16 & 8 & 8 & 16 & 16 & 16 & 4 & 16 & 8 & 2 \\ \end{array} \right) $$ It is evident that the repetend length is a factor of $17{-}1$; and $n=3 {\cdot} 17$. I tried to understand when the repetend digit graphs were isomorphic, but a pattern is not evident. For example, for $\frac{1}{51}$, for bases $$b = 15,19,25,26,32,36,42,43,49 \;,$$ that graph is an octagon. Here are three of them:       So here is a specific question: Q. Is it possible to predict which of the base-$b$ digit-expansions of $1/n$ result in isomorphic repetend digit graphs? In particular, graphs which are cycles? Perhaps specifically when $n$ is a prime? REPLY [7 votes]: The following remarks do not answer your questions completely, but they may nonetheless be helpful. Note first that computing the base-$b$ expansion of $1/n$ is essentially the same thing as computing the powers of $b$ modulo $n$. Here is one way to write the steps of the base-$b$-expansion algorithm, which makes the connection clear: \begin{align*} 1 &= 0\times n + 1\\ b\times 1 &= q_1\times n + r_1\\ b\times r_1 &= q_2\times n + r_2\\ &\dots\\ b\times r_k &= q_k\times n + r_{k+1}\\ &\dots. \end{align*} At each stage, we divide $b$ times the previous remainder by $n$ to get the next remainder. The base-$b$ expansion is the sequence of quotients $q_k$. Since for all $k$, we have $r_{k+1}\equiv br_k\pmod{n}$, the remainders $1, r_1, r_2,\dotsc$ are the powers of $b$ modulo $n$. To avoid some distracting details, let us restrict ourselves to $b$ that are relatively prime to $n$. (In this case the ``repetend'' begins immediately.) The element $b$ belongs to the multiplicative group $(\mathbf{Z}/n\mathbf{Z})^\times$ of units modulo $n$, whose order is $\phi(n)$ (Euler totient). Thus the length of the base-$b$ repetend always divides $\phi(n)$. If $n$ is prime, then any divisor of $\phi(n) = n-1$ will occur, since $(\mathbf{Z}/n\mathbf{Z})^\times$ is cyclic, but typically there will be further restrictions on the lengths of the repetends (which one can work out using the prime factorization of $n$). The sequence of remainders $1, r_1, r_2, \dotsc$ repeats cyclically, where the cycle length is the (multiplicative) order of $b$ modulo $n$. From the algorithm, it is clear that $r_k$ determines $q_k$, and so the ``repetend digit graph'' for the base-$b$ expansion is a sort of contraction of this cycle. As long as $b$ is large compared to $n$, however, there will not be any contraction; indeed, if we have $b\geq n$, then $q_k$ determines $r_k$. (If $br = nq + s$ and $br' = nq + s'$ with $0\leq r,r',s,s'\leq n-1$, then $b(r-r') = s-s'$ and $|b(r-r')| TITLE: Second homology of mapping class group of genus 3 QUESTION [8 upvotes]: In a survey paper of Korkmaz it is stated that $H_2(\mathrm{Mod}_3)$ is either $\Bbb Z$ or $\Bbb Z \oplus \Bbb Z_2$, but I was not able to find out a precise computation of this group (resolving the ambiguity). Is this group known? Any reference? REPLY [11 votes]: In his paper "Lagrangian mapping class groups from a group homological point of view" (available here), Sakasai proves that the desired homology group is $\mathbb{Z} \oplus \mathbb{Z}/2$.<|endoftext|> TITLE: Is a generic closed orientable hyperbolic 3-manifold Haken? QUESTION [12 upvotes]: My question is as follows: "Is a generic closed orientable hyperbolic 3-manifold Haken?" Of course the word 'generic' can be interpreted in many ways, and the answer might depend on the way how one interprets 'generic'. As an example, let's consider the related question "Is a generic closed (aspherical) 3-manifold hyperbolic?" In this case there are various notions of 'generic', in each case the answer is a resounding yes. For example, almost all Dehn fillings on a hyperbolic knot result again in a hyperbolic 3-manifold. Furthermore Maher (Random Heegaard splittings) showed in a precise sense that a 'random gluing of two handlebodies of genus >1' gives rise to a hyperbolic 3-manifold. Similarly, in a precise sense, a 'random' fibered 3-manifold is hyperbolic. My hunch would have been that a generic hyperbolic 3-manifold is non-Haken (why should it have an incompressible surface?), but I just had lunch with another 3-manifold topologist and his guess was that a generic hyperbolic 3-manifold should be Haken. REPLY [3 votes]: Firstly, the random Heegard splitting model and the random fibering model are discussed at length in my preprint. (I believe the OP is familiar with it). I believe the question is asked there, though not answered - it is pointed out that the smaller genus of an incompressible surface will grow, but that seems to be neither here nor there. A related question is whether a random fibered manifold has an incompressible surface which is not the fiber of the fibration you started with. Here, the answer is YES, for genus 1 (Floyd-Hatcher), and probably YES for higher genus, which would tend to suggest that a random Dunfield-Thurston manifold might be Haken, but nobody knows for sure. As for Dylan's comment, it is certainly plausible. Any given configuration of tetrahedra (in particular one with a normal sphere which is not the link of a vertex), in fact, you might as well assume that it is topologically a thickened sphere, will appear with probability one, and it is highly unlikely that it will bound balls on both sides.<|endoftext|> TITLE: If the generating function summation and zeta regularized sum of a divergent series exist, do they always coincide? QUESTION [11 upvotes]: One could assign a value to divergent series by means of several summation methods. One summation method we could consider is the generating function method. Let's sum, for example, the fibonacci series by means of this method. We consider the generating function of the fibonacci series: $$ g(x) = \sum_{n=1}^{\infty} f_{n} x^{n}, $$ in which the $n$'th Fibonacci number is defined as: $f_{n} = f_{n-1} + f_{n-2} $ and $f_{0} = 0 $ and $f_{1} = 1$. In the following wikipedia article it is explained (amongst many other things) that it is possible to find a closed-form for $g$: $$g(x) = \frac{x}{1-x-x^2} . $$ Therefore, we could state that the divergent series summation by means of the generating function summation method $G$ amounts to taking the following limit: $$ G \Big{(} \sum_{n=1}^{\infty}f_{n} \Big{)} = \lim_{x \to 1} g(x) = -1 . $$ We could, also try to assign a value to the sum of all Fibonacci numbers by means of zeta function regularisation. A closed form of the function $$ z(x) = \sum_{n=1}^{\infty} f_{n}^{-x} $$ can be found in equation $(5)$ of the following paper by Navas. (Who mistakenly asserts that he is finding the analytic continuation of the Fibonacci Dirichlet series. He is actually doing zeta function regularization of the Fibonacci series. The two methods are confused quite often, though.) He finds that $$ z(x) = 5^{x/2} \sum_{k=0}^{\infty} \binom{-x}{k} \frac{1}{\phi^{x+2k} + (-1)^{k+1} } . $$ The zeta regularized sum $R$ of the Fibonacci series is therefore $$ R \Big{(} \sum_{n=1}^{\infty} f_{n} \Big{)} = \lim_{x \to -1} z(x) = \frac{1}{\sqrt{5}} \Big{(} \frac{1}{\phi^{-1} -1} + \frac{1}{\phi + 1} \Big{)} = -1 .$$ We have thus found that the generating function summation and the zeta regularized sum of the Fibonacci series coincide (define $F := \sum_{n=1}^{\infty} f_{n} )$ : $$ G(F) = -1 = R(F) .$$ This does not always happen, though. If we define the sum of natural numbers $ N = \sum_{n =1}^{\infty} n $, then $G (N) $ does not exist. This is because the corresponding generating function amounts to $$ p(x) = \sum_{n=1}^{\infty} n x^{n} = \frac{1}{ (1-x)^{2} } , $$ for which $\lim_{x \to 1 } p(x) $ does not exist. The zeta regularized sum does exist, however. We have the notorious equation $R(N) = - \frac{1}{12} $ (see this page). Question (1) is now: Do the $G$ and $R$ summation methods of a divergent series always coincide, if both methods lead to a finite number that can be assigned to the divergent series at hand? Question (2): Is there a closed form of the actual analytic continuation of the Fibonacci dirichlet series $ d(x) = \sum_{n=1}^{\infty} \frac{ f_{n} }{ n^{x} } $ ? Bonus question: are there any references for collections of generating function expansion, zeta function analytic continuations and analytic continuations of dirichlet series? For the first group of functions there is the book Generatingfunctionology by Wilf, but I can't find any big overview papers/books/articles on the last two groups of functions. REPLY [4 votes]: Regarding question number 2, you can write $f_k = \frac{1}{\sqrt{5}} \left( \phi^k - \left(1-\phi\right)^k \right)$, where $\phi^2 = \phi + 1$, $\phi > 0$. Hence $$\sum_{k=1}^{\infty} \frac{f_k}{k^s} = \frac{1}{\sqrt5} \sum_{k=1}^{\infty} \frac{\phi^k - \left(1-\phi\right)^k}{k^s}$$ By the definition of the Polylogarithm, we get $$\sum_{k=1}^{\infty} \frac{f_k}{k^s} = \frac{1}{\sqrt5}\mathrm{Li}_s\left( \phi \right) - \frac{1}{\sqrt5}\mathrm{Li}_s\left( 1-\phi \right)$$<|endoftext|> TITLE: Twist in identification with singular cohomology QUESTION [7 upvotes]: Let $X$ be a smooth projective variety over $\mathbb{Q}$ and $$V = H^m(X(\mathbb{C}), \mathbb{Q} \cdot (2\pi i)^r)$$ Then I've seen people write the comparison with complex cohomology (an isomorphism of $\mathbb{Q}_\ell$ vector spaces) as $$V \otimes_\mathbb{Q} \mathbb{Q}_\ell \cong H^m_{et}(X_{\bar{\mathbb{Q}}}, \mathbb{Q}_\ell)(r)$$ Why does twisting on the right hand side correspond to taking coefficients in $\mathbb{Q} \cdot (2\pi i)^r$ instead of $\mathbb{Q}$? What does the twist do, what is it used for? REPLY [7 votes]: Let me supplement Vivek's nice answer with a few additional comments. People coming from differential geometry typically normalize the $n$th Chern class -- in the Chern-Weil formula -- by a factor of $1/(2\pi i)^n$ to get it to be integral. However, if you want compatibility with algebraic definitions, then you don't want to do this, so then $c_n$ would take values in $H^{2n}(X,(2\pi i)^n\mathbb{Z})$. For example, in additional to the etale first Chern class explained above, you can use algebraic de Rham cohomology. Here $c_1$ is given by the induced map on $H^2$ associated to $$ d\log:\mathcal{O}_X^*[-1]\to \Omega_X^\bullet$$ Continued-- To expand slightly, one can see by a diagram chase that, under the Grothendieck isomorphism $H^2(X,\mathbb{C})\cong H^2(X,\Omega_X^\bullet)$, the image of this $c_1$ lands in $H^2(X,2\pi i\mathbb{Z})$. So by the usual tricks involving the splitting principle, one gets the a similar statement for higher Chern classes. In case, you prefer to avoid Chern classes, it follows (with some work), that the image of the cycle map $CH^n(X)_\mathbb{Q}\to H^{2n}(X, \mathbb{C})$ lands in $H^{2n}(X, \mathbb{Q}(n))$. Again this is consistent with what happens on the etale side.<|endoftext|> TITLE: Classes in $H^3(G; \mathbb{Z})$ that restrict to zero on abelian subgroups QUESTION [5 upvotes]: Let $G$ be a finite $p$-group. Is it possible to have a nonzero class in $H^3(G; \mathbb{Z})$ that restricts to zero in $H^3(A; \mathbb{Z})$ for every abelian subgroup $A \subset G$? If so, what is a simple example? REPLY [4 votes]: I don't know about in general, but what you're after is tied in to (integral) Essential Cohomology $Ess_\mathbb{Z}(G)$, as follows: Let $G$ be a finite group. An element $x\in H^*(G)$ is called essential if it has trivial restriction to all proper subgroups of $G$; the set of such elements make up the essential ideal $Ess_\mathbb{Z}(G)$. If $G$ is not a $p$-group, then this essential ideal is 0. Thus the main focus has been on mod-$p$ cohomology, especially when it's an easier ring to work under. But some stuff can be said for $\mathbb{Z}$-coefficients. For an example, here is a computation I did a long time ago: Take $G=\mathbb{Z}_2\times\mathbb{Z}_2$, so that $H^*(G,\mathbb{Z})\cong\mathbb{Z}[\alpha,\beta,\mu]/(2\alpha,2\beta,\alpha\beta^2+\alpha^2\beta-\mu^2)$ where $|\alpha|=|\beta|=2$ and $|\mu|=3$. Then $\text{res}^G_{\mathbb{Z}_2\times 1}\alpha=\text{res}^G_{1\times \mathbb{Z}_2}\beta=\eta$ and $\text{res}^G_{\mathbb{Z}_2\times 1}\beta=\text{res}^G_{1\times \mathbb{Z}_2}\alpha=0$, where $H^*(\mathbb{Z}_2,\mathbb{Z})\cong\mathbb{Z}[\eta]/(2\eta)$ with $|\eta|=2$. For the final maximal subgroup $M=\lbrace (0,0),(1,1)\rbrace$ we have $\text{res}^G_M\alpha=\eta$ and $\text{res}^G_M\beta=\eta$, so $\text{res}^G_M\alpha\beta=\eta^2$ and $\text{res}^G_M(\alpha+\beta)=0$. Now $\mu$ must restrict to a $3$-dimensional element on the maximal subgroups, which it cannot because the rings are generated by a $2$-dimensional element $\eta$. Thus $Ess_\mathbb{Z}(\mathbb{Z}_2\times\mathbb{Z}_2)=(\mu)$.<|endoftext|> TITLE: What does the defect of a block measure? QUESTION [5 upvotes]: In the context of decomposition matrices for Hecke algebras of finite Coxeter groups at a root of unity (such as the tables at the end of the book "Hecke algebras at a root of unity" by Geck-Jacon or in Geck-Pfeiffer), what is meant by the defect of a block? Is there a simple explanation of what the defect keeps track of? I understand that the blocks of defect 1 have a decomposition matrix that has 1's on and just below the diagonal and 0's else, and that the blocks of higher defect have a more complicated structure. But what is "defect" exactly? REPLY [15 votes]: Originally ( in the context of group algebras of finite groups, which background is necessary to put later generalizations in context), the defect of a block was defined by Brauer as an arithmetical quantity. If $F$ is an algebraically closed field of prime characteristic $p,$ and $G$ is a finite group whose Sylow $p$-subgroup has order $p^{a}$, then the defect of a block $B$ of $FG$ was defined to be $d,$ where $p^{a-d}$ is highest power of $p$ which divides all degrees of irreducible characters assigned to $B.$ To each block with defect $d,$ Brauer assigned a unique (up to conjugacy) $p$-subgroup $D$ of $G$ of order $p^{d}$, called the defect group of $B.$ The simplest case is when $d=0$ and $D = 1$. That occurs if and only if the block $B$ is a full matrix algebra over $F,$ in which case there is a unique simple module in $B,$ which is projective, and there is a unique irreducible (complex) character assigned to $B.$ Loosely speaking, the size and structure of $D$ is a good measure of the complexity of the algebra structure of $B,$ so that the larger the integer $d$ is, the more complicated the structure of the algebra $B$ (and its module category) is likely to be. The structure of blocks of defect $1$ ( of group algebras) was worked out by Brauer in the 1940s, and the structure of blocks with cyclic defect group (at least at the character level) was worked out by Dade in the mid 1960s, with substantial contributions by JG Thompson and JA Green. K. Erdmann made major contributions to understanding blocks with defect groups which were dihedral, semidihedral, or generalized quaternion. Module theoretic notions, such as relative projectivity, were linked to arithmetic properties by people such as JA Green, W. Feit, JL Alperin (and D.G. Higman). So for example, every module in a block $B$ is relatively $D$-projective, which means that it is a direct summand of an $FG$-module induced from $D.$ This means that every indecomposable $B$-module has a vertex (as defined by J.A. Green) contained (up to conjugacy) in $D.$ By specializing the block theory of certain Hecke algebras at well-chosen roots of unity, information about blocks for finite groups can be obtained ( eg specializing the theory for type A Hecke Algebras at $p$-th roots of unity yields information about the $p$-blocks of the symmetric groups). In general for other algebraic structures, blocks are usually obtained via some sort of linkage principle ( sometimes defined via characters, as Brauer and Feit did in the group case, sometimes via certain modules, as Brauer did). In these more general contexts, simple modules are usually directly linked if they both occur as composition factors of the same projective indecomposable module, and linked when the notion of directly linked is extended by transitivity. Again speaking loosely, the larger the defect ( or the bigger the defect group), the more complicated the Cartan matrix of the block is likely to be. Incidentally, if $B$ has defect $d$, all elementary divisors of the Cartan matrix for $B$ are powers of $p,$ and $p^{d},$ which occurs with multiplicity $1$, is the largest. The analogous statements for blocks of Hecke Algebras are rather more subtle.<|endoftext|> TITLE: What is an extragradient method? QUESTION [6 upvotes]: I've searched Google, but it seems that only research journal papers appear in search results, where some new, improved, or specialized extragradient method is discussed. I've also searched Wikipedia and Wolfram MathWorld. I would like to perhaps know the straightforward definition of that, instead of deducing it from a number of articles, where the basic/original method is not defined. REPLY [2 votes]: A good discussion is in this recent preprint: Proximal Reinforcement Learning. They use the same definition as in dragonxlwang's answer, and discuss its relation to the mirror-prox method.<|endoftext|> TITLE: What are TQFTs that are multiplicative under connected sums? Do bordisms with connected sum as monoidal product exist? QUESTION [9 upvotes]: In general, one extracts a manifold invariant from a TQFT by interpreting the closed manifold as a bordism from the empty set to the empty set. The TQFT sends this bordism to a homomorphism of the ground field, which is a number. Such invariants are always multiplicative under disjoint union, this is a consequence of the TQFT being a monoidal functor: $$\mathcal{Z}(M_1 \sqcup M_2) = \mathcal{Z}(M_1) \otimes \mathcal{Z}(M_2) = \mathcal{Z}(M_1) \cdot \mathcal{Z}(M_2)$$ Some TQFTs, like the Crane-Yetter invariant (but not, say, the Turaev-Viro model) give manifold invariants that are multiplicative under connected sum $\#$. One way to see this is to notice that they can be defined (for connected manifolds) with Kirby calculus: Given a manifold, choose a handle decomposition and consider its link diagram. The diagram is then labelled with morphisms from a ribbon fusion category and the whole diagram is evaluated as a morphism from the monoidal identity to itself, again a number. Now the evaluation of the disjoint union of two link diagrams must then give the product of the evaluations of the respective diagrams, since a ribbon fusion category is monoidal. But the disjoint union of two link diagrams of manifolds $M_1$ and $M_2$ is the link diagram of the connected sum $M_1 \# M_2$! Which leads me to believe that this multiplicativity secretly comes from monoidality of some functor again. Is there a category of bordisms where the monoidal product of morphisms (=bordisms) is the connected sum, and not the disjoint union? Are such TQFTs actually monoidal functors from this bordism category to $\mathrm{Vect}$? A related, noncategorical question was asked here: Monoid structure of oriented manifolds with connect sum REPLY [6 votes]: If the dimension of $Z(S^{n-1})$ is greater than 1, then the TQFT is not even approximately multiplicative under connect sum. If $Z(S^{n-1})$ is 1-dimensional, then a simple cut and paste argument shows that $$ Z(M_1 \sharp\, M_2) = Z(M_1 \sqcup M_2) \,/\, Z(S^n) = Z(M_1) \cdot Z(M_2) \,/\, Z(S^n) . $$ So in this case $Z$ is multiplicative under direct sum iff $Z(S^n) = 1$. (If $Z(S^n) \ne 1$ we might say that the theory is approximately multiplicative under connect sum.) When $n$ is even, the Euler characteristic of $S^n$ is non-zero and we can tensor with an Euler characteristic theory to arrange that $Z(S^n) = 1$. When $n$ is odd, the Euler characteristic of $S^n$ is zero and there is no way to tweak things so that $Z(S^n) = 1$. Turaev-Viro and Witten-REshetikhin-Tureav theories fall into this class. I'll also remark that if $M_i$ is not connected, then $M_1 \sharp\, M_2$ is not well-defined, even up to homeomorphism. But nevertheless the above claims about $Z(M_1 \sharp\, M_2)$ are true, for any choice of the ambiguous connect sum.<|endoftext|> TITLE: $f(x)$ is irreducible but $f(x^n)$ is reducible QUESTION [24 upvotes]: Does there exist an irreducible polynomial $f(x)\in \mathbb{Z}[x]$ with degree greater than one such that for each $n>1$, $f(x^n)$ is reducible (over $\mathbb{Z}[x]$)? REPLY [38 votes]: There is no such polynomial. It is clear that $f$ cannot be a cyclotomic polynomial (your condition $\deg{f} > 1$ excludes $x-1$). So suppose $f$ is non-cyclotomic and irreducible, of degree $d$, and consider a prime $p$ for which $f(x^p)$ is reducible. For $\alpha$ a root of $f(x^p)$, reducibility means that $[\mathbb{Q}(\alpha):\mathbb{Q}] < pd$. On the other hand $[\mathbb{Q}(\alpha^p):\mathbb{Q}] = d$ by the irreducibility of $f$, and we conclude that $[\mathbb{Q}(\alpha):\mathbb{Q}(\alpha^p)] < p$; this means that the polynomial $x^p - \alpha^p$ is reducible over $\mathbb{Q}(\alpha^p)$. By the Vahlen-Capelli theorem, this is only possible if $\alpha^p = \eta^p$ with $\eta \in \mathbb{Q}(\alpha^p)$. We conclude that for every prime $p$, either $f(x^p)$ is irreducible, or else each root $\xi$ of $f$ has the property that $\xi$ is a $p$-th power from $K := \mathbb{Q}(\xi)$. It remains to note that for any number field $K$, and any $\xi \in K^{\times} \setminus \mu_{\infty}$ not a root of unity, $\xi \in (K^{\times})^p$ for only finitely many primes $p$. This follows for instance from Dirichlet's unit theorem.<|endoftext|> TITLE: Which axioms of ZF are used for finite choice? QUESTION [8 upvotes]: Apologies if this is a silly question, not an expert in set theory but just wondering about it. ZF implies finite choice. But let's suppose one wanted to work without it. The thinking here is being uncomfortable about sets where none of the elements are computable/definable, therefore one couldn't pick one explicitly. The idea is to allow nonempty sets where it's not allowed to pick exactly one element, you may be able to define subsets but not a single element. Obviously, finite choice is allowed when it's possible to constructively name an element. Which axiom(s) of ZF would have to go in order to remove finite choice? Is there any set theory that rejects finite choice, or is this pruning away too many useful results? REPLY [5 votes]: As mentioned in the comments, one needs truly very few ZF axioms to prove the instances of of the axioms of choice for finite families. Let us denote by $\text{AC}\upharpoonright n$ the assertion that every collection $\cal A$ of $n$ nonempty disjoint sets has a selecting set $B$, that is, a set $B$ such that $A\cap B$ is a singleton for every $A\in\cal A$. (Note that this is not the same as the assertion that every collection of disjoint sets of size $n$ has a selector.) Part of the point now is that $\text{AC}\upharpoonright 0$ is basically vacuous, since the empty family has any set as a selector vacuously. Further, if $\text{AC}\upharpoonright n$ holds, then one may easily prove $\text{AC}\upharpoonright (n+1)$ as follows: given any family of size $n+1$, consisting of disjoint nonempty sets, we may delete an element $A$ from it and have a family of size $n$, which has a selector by the hypothesis, and this selector can be extended to a selector on the large family by adjoining any element of $A$. Thus, by induction, we prove $\text{AC}$ for all finite families. Note that in the inductive step of the argument, we don't need to worry about defining a particular element of $A$, the set we deleted, since we are not proving that there is a definable selector; rather, we are only proving the existence claim that every such family has at least one selector. Similarly, we don't need to worry about how to choose a particular element to delete---the proof shows that for each set $A\in\cal A$, there are selectors obtained by deleting that set. So there is at least one selector. Thus, by induction, we prove $\forall n\ \text{AC}\upharpoonright n$, or in other words, that the axiom of choice holds for finite families. How much of ZF suffices for this argument? Of course we needed some very basic things like extensionality, pairing and union in order to make sense of the empty set and adjoining an element and so on, and perhaps one doesn't want to do set theory anyway without these axioms. Secondly, we needed to be able to prove a statement by induction on the natural numbers. Thus, the argument can be undertaken in Kripke-Platek set theory, a very weak subsystem of ZF. One issue is that one might think that $\Sigma_2$-separation is required for the inductive argument, in order to form the set $\{ n\in\mathbb{N}\mid \neg \text{AC}\upharpoonright n\}$, with a view to proving that this set can have no least element, since the negation of $\text{AC}\upharpoonright n$ appears to be a $\Sigma_2$-assertion as it asserts the existence of a family of size $n$ with no selector. But KP includes the $\in$-foundation scheme, which allows one to prove statements of any complexity by $\in$-induction, and this implies natural-number induction. But as noted, we really only needed natural number induction for $\Pi_2$-assertions, since $\text{AC}\upharpoonright n$ has complexity $\Pi_2$, and this would be a weakening of KP. I expect that one can go much lower than KP. Lastly, as I mentioned in the comments, we don't need any induction scheme at all to prove every instance of $\text{AC}\upharpoonright n$ for meta-theoretically finite $n$, that is, to prove $AC$ for families of size $1$, of size $2$, of size $3$ and so on, as a theorem scheme. The reason is that for such meta-theoretically finite families, the induction can be undertaken in the meta-theory, rather than in the object theory, and so one needs hardly any set theory at all to prove these instances.<|endoftext|> TITLE: What arrangement of unit cubes minimizes surface area? QUESTION [21 upvotes]: For each of these two questions, one can assume that the arrangements are polycubes (for which a definition can be found in the excerpt-image below). Question A. How does one arrange $n$ unit cubes to minimize surface area? Question B. How does one arrange $n$ unit cubes to form a rectangular prism of minimal surface area? Various curricular materials discuss this problem for a specific number of cubes, but I have not seen the general case broached. For an example of an exploration at the pre-secondary level, see this instructional plan from the National Council of Teachers of Mathematics. A recent example of the problem posed in generality can be found on this math education blog-post as well as here. My own attempt at combing the literature did not produce anything of import. From the two dimensional perspective, I see a question in a somewhat similar vein on MSE about unit squares, and another 2D question on MO, but neither seems applicable to the questions above. Moreover, there are some related results (recently relayed to me by a geometer-mathematician) in: Williams, W., & Thompson, C. (2008). The Perimeter of a Polyomino and the Surface Area of a Polycube. The College Mathematics Journal, 233-237. For example, see the excerpt below: One might also check the OEIS sequence recommended by Robert Israel (or found by googling), though I cannot vouch for its validity (cf. e.g. the remarks of its contributor here). Non-trivial bounds, algorithmic approaches, or other relevant remarks are all welcome! REPLY [12 votes]: One geometric remark (which is reminiscent of a symmetrization principle used in isoperimetric problems): a minimizing polycube may be chosen to lie in the first octant, with three sides tangent to the coordinate planes, and so that the number of cubes in each column is monotonic in each coordinate, like a lozenge tiling: The point is that you can choose some direction to be the surface of the earth, and let gravity pull all of the cubes down. This does not increase the surface area: the result ends up with at most one bottom and top face in each column. In adjacent columns of height $m$ and $n$, the number of exposed faces is at least $|m-n|$, which is minimized when they are sitting flush. Repeat this in each direction, until the configuration is monotonic (like a 3-dimensional Young diagram, but I don't know if there's a mathematical label). Added: This argument gives an interpretation for the desired minimization problem. One has a sequence of Young diagrams in three directions such that the total number of squares adds up to $n$, and which are monotonic in the number of squares in each diagram (with some compatibility conditions). Let the number of squares in the young diagram adjacent to each coordinate plane be $x_1, y_1, z_1$ respectively. Then the surface area is $2(x_1+y_1+z_1)$. This is counting twice the number of blue, red, and green squares (excluding the corners) in the picture (pictured are the top squares, to which there is a corresponding bottom square in each column). We have $x_1\geq x_2 \geq \cdots \geq x_k$, and $x_1+x_2+\cdots+x_k=n$, and corresponding formulae for $y_i$ and $z_i$ counting the number of squares in each tableau parallel to a given coordinate plane. It shouldn't be too hard to solve this combinatorial optimization problem, but I don't have time to work it out now. Addendum 2: I think I can describe the optimal solution, but I don't have time to write down the complete argument now. It is helpful to consider the 2-dimensional case of Young diagrams. A length $n$ Young diagram is a sequence $x_1\geq x_2 \geq \cdots \geq x_{y_1}$, such that $\sum x_j=n$. The dual diagram is $y_1 \geq y_2 \geq \cdots y_{x_1}$, where $y_i=\max\{ j | x_j\geq i\}$. The length of the boundary is $2(x_1+y_1)$. Then the symmetrization principle says that the minimal "polysquare" is a Young diagram. Let $P(2k)$ be the maximal size of a diagram with perimeter $2k$ (the "isoperimetric function"). Since a Young diagram fits inside a rectangle of sides $x_1, y_1$ with the same perimeter, clearly $P(2k) = \max_{a+b=k} ab = \lfloor \frac{k}{2} \rfloor \lceil \frac{k}{2}\rceil$. Also, the Young diagram which minimizes perimeter for a given $n$ clearly has $P(2(x_1+y_1-1))< n \leq P(2(x_1+y_1))$. Let $k=x_1+y_1$. Then the minimal perimeter will be realized by a Young diagram which, if $k$ is even, is a $k/2(k/2-1)$ rectangle with a row of $n-k(k-2)/4$ squares added on (so fitting in a $(k/2)^2$ square and containing a $k/2(k/2-1)$ rectangle). If $k$ is odd, then the Young diagram will be obtained from an $((k-1)/2)^2$ square with a row of $n-(k-1)^2/4$ squares added (so containing a $(k-1)/2\times (k-1)/2$ square, and contained in a rectangle of size $(k-1)/2\times (k+1)/2$). Note: The perimeter minimizer is far from unique. For example, if $n=8$, then the minimizer will be realized by the Young sequences $3,3,2$ and $4,4$. We are only describing a preferred choice of minimizers. Now we may use this description of perimeter minimizing Young diagrams to describe minimal 3-D Young diagrams. As described above, we have a monotonic sequence of Young diagrams, each fitting inside the other, of sizes $x_1\geq x_2 \geq \cdots \geq x_k$. Then one checks that the perimeter is $2x_1$ plus twice the sums of the perimeters of the Young diagrams. Replace each Young diagram with a perimeter-minimizing one described above (one may observe that the choices are nested for $x_i \geq x_{i+1}$). Repeat this in the $y-$ and $z-$ directions to obtain a 3-D Young diagram in which the cross sections are perimeter minimizing 2-D Young diagrams. Then the 3-D Young diagram fits into an $a\times b \times c$ box, with $a\leq b\leq c$, and $|c-a|=1$ (so $b= a$ or $c$). A similar isoperimetric argument to the above implies that it contains an $a^3$ cube and is contained in a $c^3$ cube, where $a=\lfloor n^{1/3} \rfloor$. If $ n < a^2(a+1)$, then it is obtained from an $a^3$ cube with a $n-a^3$ optimal Young diagram attached to one face. If $a^2(a+1)\leq n < a(a+1)^2$, then it is obtained from an $a^2\times (a+1)$ box by attaching an optimal $n-a^2(a+1)$ Young diagram to an $a\times (a+1)$ face. And if $a(a+1)^2 \leq n$, then it is obtained from an $a\times (a+1)^2$ box by attaching an $n-a(a+1)^2$ Young diagram to an $(a+1)^2$ face. Again, these are not the unique perimeter minimizers, but a description of optimal ones. It's possible that I've made a mistake, but at least the symmetrization principles underlying the argument seem solid.<|endoftext|> TITLE: Determinant of a $k \times k$ block matrix QUESTION [15 upvotes]: Consider the $k \times k$ block matrix: $$C = \left(\begin{array}{ccccc} A & B & B & \cdots & B \\ B & A & B &\cdots & B \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ B & B & B & \cdots & A \end{array}\right) = I_k \otimes (A - B) + \mathbb{1}_k \otimes B$$ where $A$ and $B$ are size $n \times n$ and $\mathbb{1}$ is the matrix of all ones. It would seem that the formula for the determinant of $C$ is simply: $$\det(C) = \det(A-B)^{k-1} \det(A+(k-1) B)$$ Can anyone explain why this seems to be true or offer a proof or direct me to a proof? REPLY [3 votes]: Let us assume that $A-B$ is invertible. Write $$\begin{array}{rl} C &= \begin{bmatrix} A & B & \ldots & B\\ B & A & \ldots & B\\ \vdots & \vdots & \ddots & \vdots\\B & B & \ldots & A\end{bmatrix}\\\\ &= \begin{bmatrix} A-B & O_n & \ldots & O_n\\ O_n & A-B & \ldots & O_n\\ \vdots & \vdots & \ddots & \vdots\\ O_n & O_n & \ldots & A-B \end{bmatrix} + \begin{bmatrix} B \\ B\\ \vdots \\ B\end{bmatrix} \begin{bmatrix} I_n \\ I_n\\ \vdots \\ I_n\end{bmatrix}^T\\\\ &= (I_k \otimes (A-B)) + (1_k \otimes B) (1_k \otimes I_n)^T\\\\ &= (I_k \otimes (A-B)) \left(I_{nk} + (I_k \otimes (A-B)^{-1}) (1_k \otimes B) (1_k \otimes I_n)^T\right)\end{array}$$ Using Sylvester's determinant identity, $$\begin{array}{rl} \det (C) &= \det\left((I_k \otimes (A-B)) \left(I_{nk} + (I_k \otimes (A-B)^{-1}) (1_k \otimes B) (1_k \otimes I_n)^T\right)\right)\\\\ &= \det(I_k \otimes (A-B)) \cdot \det \left( I_{nk} + (I_k \otimes (A-B)^{-1}) (1_k \otimes B) (1_k \otimes I_n)^T \right)\\\\ &= \det(I_k \otimes (A-B)) \cdot \det \left( I_{n} + (1_k \otimes I_n)^T (I_k \otimes (A-B)^{-1}) (1_k \otimes B) \right)\\\\ &= \left(\det(A - B)\right)^k \cdot \det \left( I_{n} + k (A-B)^{-1} B \right)\\\\ &= \det((A-B)^k) \cdot \det \left( I_{n} + k (A-B)^{-1} B \right)\\\\ &= \det((A-B)^{k-1}) \cdot \det \left( A-B + k B \right)\\\\ &= (\det(A-B))^{k-1} \cdot \det \left( A + (k-1) B \right)\end{array}$$<|endoftext|> TITLE: When is the profinite completion a pro-$p$ group? QUESTION [9 upvotes]: My research area is mainly pro-$p$ groups and profinite groups. However, in the last few year I became also interested in discrete groups. Therefore, it seems to me a natural problem to look for examples of finitely generated groups such that their profinite completion is a pro-$p$ group. One trivial example is when the profinite completion is trivial or a finite $p$-group, e.g. an infinite simple group. Hence, it makes sense to require the group to be infinite and residually finite. But then you can take residually finite, $p$-groups, e.g., the Grigorchuk group. In such case tha profinite completion is pro-$p$ for an easy reason. So here are my questions: Is there an infinite finitely generated residually finite non-torsion group such that its profinite completion is pro-$p$? Is there an infinite finitely generated residually finite torsion free group such that its profinite completion is pro-$p$? (Notice that profinite completion being pro-$p$ means that all subgroups of finite index are of $p$-power index.) REPLY [5 votes]: Gustavo A. Fernández-Alcober, Alejandra Garrido and Jone Uria-Albizuri gave a positive answer to question 2 and therefore also to question 1 in On the congruence subgroup property for GGS-groups.<|endoftext|> TITLE: What is the arithmetic Nullstellensatz? QUESTION [13 upvotes]: The only precise statement (coming from a reliable source) of the "arithmetic Nullstellensatz" I can find is in Gowers's book, stating that two polynomials with integral coefficients have the same roots mod every $m$ iff they differ by a sign. I would like to know the general form of this result, and see some reference where I can read about it and some applications (perhaps). All help is appreciated. REPLY [12 votes]: One "arithmetic version" of the Nullstellensatz states that if $f_1, ..., f_s$ belong to $\mathbb{Z}[X_1,...,X_n]$ without a common zero in $\mathbb{C}^n$, then there exist $a \in \mathbb{Z} \setminus {0}$ and $g_1,...,g_s$ in $\mathbb{Z}[X_1,...,X_n]$ such that $a = f_1g_1 + ... + f_sg_s$. Finding degree and height bounds for $a$ and $g_1, ..., g_s$ has received some attention, see for example here.<|endoftext|> TITLE: Markov chain Monte Carlo: why is non-reversible MC MC not as popular? QUESTION [9 upvotes]: I am new to methods for simulating Markov chains in order to sample from the target, unknown distribution. After a couple days of reading, I found out that even though people have realized that non-reversible (or, irreversible) Markov chains usually converge faster to their invariant distributions. There are not many papers on irreversible Markov chains for the purpose of generating the unknown distribution; instead, reversible Markov chains dominate the literature. Could you please please explain why the irreversible Markov chains are so less frequently used to induce the unknown distribution one wants to sample from? Thanks in advance. Regards, Chee REPLY [10 votes]: I will take a stab at this, even though I no longer have the full eloquence of a working mathematician. First of all reversible Markov chains are really self-adjoint operators in disguise, whose spectral properties are therefore well-understood at least in theory. But even in this nice setting, many seemingly simple questions cannot be answered. For instance, take a card shuffling example, where you swap a randomly chosen pair of adjacent cards in a deck with 1/2 probability, and ask about the rate of convergence in say the most natural total variation distance. Some of its eigenvalues that correspond to irreducible characters of the symmetric group are rational, but the majority of them are irrational. Of course this does not prevent us from getting good rates of convergence in theory. The current best record of upper and lower bounds have a factor of two in between, and nobody has any clue how to close the gap. Another more serious example is the random walk on proper coloring, for which the right order of magnitude is not known in many cases of number of colors and maximum graph degree. It was conjectured the a uniform lower bound of $n log n$ should hold for all bounded degree graph, which is reasonable in light of coupon collector's phenomenon, but it is still open. A similar assertion for Glauber dynamics on Ising model has been resolved a few years ago; you can see how hard it can be to prove such a simple statement. Even if we know all the eigenvalues of the chain, we are still not guaranteed an easy path to its rate of convergence. One such example is the Glauber dynamics on Ising lattice, which has only recently been resolved in the high temperature case, and still open for the critical and low temperature regime. I can't think of other good examples at the moment, but there should be plenty in this book and this. Finally there are definitely some non-reversible chains that have been extensively and successfully studied, such as the riffle shuffle and Thorpe shuffle. The fact that one can get exact answer for the first one and the right order of magnitude (only recently) for the second one is nothing short of miracles. But from a mathematical point of view, it is more natural to exhaust the study of "easier" reversible cases before tackling the wild west.<|endoftext|> TITLE: Exponentiation and Dedekind-finite cardinals QUESTION [7 upvotes]: It is known that the sum and the product of two Dedekind-finite cardinals are also Dedekind-finite cardinals. What about cardinal exponentiation ? Question: Let A and B be two Dedekind-finite cardinals, let C be the cardinal A power B (i.e:let x be a set with cardinal A and y be a set with cardinal B and let C be the cardinal of the set of functions with domain y and range a subset of x). Is it true that C is a Dedekind-finite cardinal ? Gérard Lang REPLY [12 votes]: The answer is no, not necessarily, because if there are infinite Dedekind finite sets, then the class of Dedekind finite sets is not closed under power set, and hence not closed under $A\mapsto 2^A$. To see this, simply note that if $A$ is any infinite set, then $P(A)$ has the singletons, the doubletons, the subsets of size $3$, and so on. So we can find a countably infinite subset of $P(P(A))$, and so $P(P(A))$ is not Dedekind finite. In particular, if $A$ is Dedekind finite but infinite, then $2^{2^A}$ is not Dedekind finite, and so it is consistent with ZF that the Dedekind finite sets are not closed under exponentiation. REPLY [11 votes]: It is a theorem of Kuratowski that the following holds: There exists a surjection $f\colon A\to\omega$ if and only if there exists an injection $F\colon\omega\to\mathcal P(A)$. As Joel says, the power set of an infinite set can always be mapped onto $\omega$, therefore the second-power set is always Dedekind-infinite. So if the first one is not, then the second one is. It should be remarked that both cases can be realized in terms of consistency. Consider the case there $S=\bigcup P_n$ where $P_n$ are pairwise disjoint sets, and there is no infinite family $I\subseteq\omega$ such that a choice function for $\{P_i\mid i\in I\}$ exists. We call such $S$ a Russell set, and its consistency has been shown by Fraenkel (using atoms) and Cohen (using forcing). Then $S$ is Dedekind-finite, since otherwise pick any countably infinite subset it must either meet infinitely many pairs on a single point (thus defining a choice function for an infinite family of pairs), or include infinitely many pairs and then by induction on a fixed enumeration of this set we can choose from these pairs. But clearly $S$ can be mapped onto $\omega$. Therefore its power set, which is the exponentiation of two Dedekind-finite cardinals, is Dedekind-infinite. We say that $A$ is amorphous if $A$ cannot be written as a disjoint union of two infinite set. We can show that an amorphous set cannot be linearly ordered, and in fact if $P$ is a linearly ordered set, then any function $f\colon A\to P$ must have a finite range (which means it is almost everywhere constant). In particular this means that an amorphous set cannot be mapped onto $\omega$, and therefore its power set is also Dedekind-finite. But in fact $A^A$, all the functions from $A$ to itself is a Dedekind-finite set, because we can show that $A\times A$, while not amorphous, cannot be mapped onto $\omega$ and therefore $\mathcal P(A\times A)$ is Dedekind-finite, so $A^A$ which is a subset of $\mathcal P(A\times A)$ must be Dedekind-finite as well. Amorphous sets have been shown to be consistent by Fraenkel using atoms, and this result can be easily transferred to a context without atoms (either by forcing or by using the Jech-Sochor embedding theorem). So we can say that Dedekind-finite cardinals are closed under exponentiation if and only if every Dedekind-finite set is finite. But given two infinite Dedekind-finite sets, we cannot necessarily say that their exponentiation is Dedekind-infinite.<|endoftext|> TITLE: Non-zero smooth functions vanishing on a Cantor set QUESTION [6 upvotes]: It is easy to give examples of continuous functions $f:[0,1]\to \mathbb R_+\cup\{0\}$ non-zero but vanishing on a Cantor set (ex: Can Cantor set be the zero set of a continuous function?). It is clearly non-true for analytic functions. My question is: Are there uniformly continuous non-zero functions vanishing on a Cantor set? Are there α-Hölder continuous non-zero functions vanishing on a Cantor set? Are there continuously differentiable non-zero functions vanishing on a Cantor set? REPLY [6 votes]: This answer is to provide a precise statement for a result mentioned by Peter Michor. Theorem (Calderón-Zygmund). If $F\subset\mathbb{R}^n$ is closed, then there is a function $f$ such that $c_1(n)d(x,F)\leq f(x)\leq c_2(n)d(x,F)$, for all $x\in\mathbb{R}^n$, $f\in C^\infty(\mathbb{R}^n\setminus F)$ and $$ \left|\frac{\partial^\alpha f}{\partial x^\alpha}(x)\right|\leq c_3(n,\alpha) d(x,F)^{1-|\alpha|} \quad \text{for all multiindices $\alpha$.} $$ This result says that there is a smooth regularization of a distance function. Originally it was proved in: A.P.Calderón, A.Zygmund, A. Local properties of solutions of elliptic partial differential equations. Studia Math. 20 1961 171–225. You can also find this result in E.M. Stein, Singular integrals and differentiability properties of functions. Princeton Mathematical Series, No. 30 Princeton University Press, Princeton, N.J. 1970 (Theorem 2 in Chapter VI.2).<|endoftext|> TITLE: Dynamics in the integers - Floor function QUESTION [5 upvotes]: Let $\alpha$ be an irrational with $0<\alpha<1$. Consider the function given by \begin{align*} f: &\mathbb{N}\longrightarrow \mathbb{N}\\ &x\longmapsto [ \alpha\cdot x]\end{align*} where $[r]$ is the largest integer that is less than or equal to $r$ for $r\in\mathbb{R}$. Let $N=\left[\frac{1}{\alpha}\right]$ be the integer part of $\frac{1}{\alpha}$, and consider the following numbers: $A_1(n)=\#\{k\leq n : f(k) TITLE: A metric space of geometric shapes QUESTION [11 upvotes]: My research involves geometric shapes in $R^2$, and I need a metric with several properties such as: Families of similar shapes, such as squares, are closed in this metric. Also more general families, such as the family of 2-fat objects, are closed in this metric. Converging sequences of interior-disjoint shapes, converge to interior-disjoint limits. Every continuous measure on a converging sequence, converges to the measure of the limit. I tried the Hausdorff distance, which is a metric on the space of closed sets, but found out that it doesn't say much about measures. I tried the Symmetric distance (defined as the area of the symmetric difference), but found out that it is only a pseudo-metric. I tried to make it a metric by restricting the underlying space of shapes, but found out that it is tricky even when only polygons are considered. I thought of converting the pseudo-metric to a metric on equivalence classes and then selecting a representative shape from each equivalence class, but found no simple way to do this selection. I thought of defining a new metric which is the maximum of the Hausdorff distance and the Symmetric distance and enjoy the best of the two worlds, but at that point, it began to feel like reinventing the wheel. Surely I am not the first who needs a metric between plain geometric shapes. So my question is: Is there a paper or a book that explicitly studies the topic of metrics between shapes in the plane, not in the context of general topology but with attention to the specific geometric properties? REPLY [4 votes]: The following paper gives an overview on Riemannian geometries on shape spaces and diffeomorphism group. Martin Bauer, Martins Bruveris, Peter W. Michor: Overview of the Geometries of Shape Spaces and Diffeomorphism Groups. Journal of Mathematical Imaging and Vision, 50, 1-2, 60-97, 2014. (pdf) Edit: A metric on the space of plane shapes that it somewhat more easy to use (since it allows for explicit solutions of the PDE which is the geodesic equation, but it does not see translations) is in the following paper: Martin Bauer, Martins Bruveris, Stephen Marsland, Peter W. Michor: Constructing reparametrization invariant metrics on spaces of plane curves. Differential Geometry and its Applications 34 (2014), 139–165. (pdf) REPLY [2 votes]: I can't say for certain if this satisfies all of your requirements, but this paper of Sharon and Mumford studies 2D shapes by comparing the (suitably normalized) Riemann maps from its interior and exterior to the unit circle, by comparing these maps on the circle one represents the shape by an equivalence class of diffeomorphisms of the circle, and these can be made into a metric space. (They require smooth shapes; I don't know if this can be weakened.)<|endoftext|> TITLE: What is the maximal number of distinct values of the product of n permuted ordinals QUESTION [9 upvotes]: Because addition and multiplication of two order types are non-commutative operations, we have that for every integer n, given n ordinals, there are at most n! distinct possible values for the sum (or the product) of these n ordinals when all permutations are considered. In fact, A. Wakulic (Fund. Math., XXXVI, pp 255-260) was able to establish the effective function F(n) giving the maximal possible number of distinct values for the SUM of n ordinals, when considering all permutations. QUESTION: What is the effective function G(n) giving the maximal possible number of distinct values for the PRODUCT of n ordinals, when considering all permutations ? Gérard Lang REPLY [13 votes]: Simply take the ordinals $\omega+1,...,\omega+n$ and one obtains $n!$ distinct products (This solution was taken from Chapter 8 Problem 39 and Chapter 9 Problem 66 in the book Problems and Theorems is Classical Set Theory). This follows from the easily provable fact that for natural numbers $r_{1},...,r_{n}$ we have $$(\omega+r_{1})\cdots(\omega+r_{n})=\omega^{n}+\omega^{n-1}r_{n}+\omega^{n-2}r_{n-1}+...+\omega r_{2}+r_{1}.$$<|endoftext|> TITLE: Is it possible to sum the divergent series with prime coefficients? QUESTION [8 upvotes]: It is known that the series $$ P := \sum_{n=1}^{\infty} p_{n} \qquad \text{where } p_{n} \text{ is the n'th prime} $$ cannot be summed by means of (prime) zeta function regularization. (The result was originally due to Landau and Walfisz, see this paper. Froberg later showed it as well.) However, there are loads of other summation methods. I am wondering whether any of the following summation methods can sum the divergent series of primes. For example: Abel summation/analytic continuation of power series (what is the difference?): Does $\lim_{x \to 1^{-} } \sum_{n=1}^{\infty} p_{n} x^{n} $ exist? Lindelöf summation: Does $\lim_{x \to 0} \sum_{n=1}^{\infty} p_{n} n^{-nx} $ exist? Analytic continuation of Dirichlet series: Does $\lim_{s \to 0} \sum_{n=1}^{\infty} \frac{p_{n}}{n^{s}} $ exist? Do any of these methods or another summation method for assigning a number to the sum of primes work? If so, please also indicate what the closed form of the corresponding function (for which the limit exists) is. REPLY [5 votes]: This question is poorly formulated, so I am not sure if it can be properly answered. I have some comments, rather then an answer. First, the methods 1,2 (Abel and Lindelöf summation), as well as any other linear regular summation method with positive matrix, are useless when dealing with a series with positive summands: if the original series is not convergent, then the summation does not help. Note that almost all of the standard methods are of this kind. Analytic continuation methods, including zeta function regularization, is another story. But this is fishy: even if the continuation exists, it may well depend on the path. Besides, which function to continuate? I am at the opinion that a question like "Do any of these methods work?", in the context of analytic continuation, simply does not make any sense. What makes sense is "Does THIS method work?", and the choice of the method should be well motivated. Of course, motivation here may depend on the purpose of calculating the sum of a series, not only on the series itself: for example, zeta function regularization is a reasonable choice for a calculation of the Casimir force, because it can be shown to produce a correct physical result.<|endoftext|> TITLE: Counterexamples to Elkik's theorem in the non-Noetherian case QUESTION [8 upvotes]: Elkik in Solutions d'equations a coefficients dans un anneu Henselian, Theorem 7 proves that: Let $A$ be a Noetherian ring that is Henselian with respect to a principal ideal $(a)$. That is, if $f(x)$ is a monic polynomial with a root $\alpha \in A/(a)$ such that $df/dx (\alpha)$ is a unit in $A/(a)$, then $\alpha$ lifts to a root of $f$ in $A$. Let $\hat{A}$ be its $a$-adic completion, and let $\hat{B}$ be a formally finitely generated algebra over $\hat{a}$ that is formally smooth over $\hat{A}[a^{-1}]$. Then there exists a finitely generated algebra $B$ over $A$ that is smooth over $A[a^{-1}]$ and such that its $a$-adic completion is isomorphic to $\hat{B}$. Is this theorem true in the non-Noetherian case? If not, what is a counterexample? I would ideally prefer counterexamples of relative dimension $0$ such as finite etale covers. REPLY [3 votes]: If I understand the notation of Gabber and Romero corectly, the theorem is true in the non-Noetherian case for finite etale covers. Finite etale torsors on $X$ of degree $n$ are classified by elements of $H^1(X,S_n)$. Thus we apply Theorem 5.8.14 of Almost Ring Theory by Gabber and Ramero. To convert from my notation to theirs, take $R=A$, $t=a$, $I=(1)$, so that $R^ = \hat{A}$. This satisfies the conditions of Proposition 4.21. Let $G$ be $S_n$. Then $G$ satisfies the conditions of 5.8.4 by explicit construction, or because of Lemma 5.8.5 and the fact that it is defined over the Dedekind domain $\mathbb Z$. So we may apply the theorem and get an equality $$H^1( \operatorname{Spec} A[a^{-1}], S_n) = H^1( \operatorname{Spec} \hat{A}[a^{-1}],S_n)$$ giving what I asked for (in the finite etale case).<|endoftext|> TITLE: Are linear algebraic groups rigid? QUESTION [5 upvotes]: The underlying variety of a linear elgebraic group (say, over an algebraically closed field) is affine, so doesn't have nontrivial (infinitesimal) deformations. I'm curious to know whether it's possible to deform the group structure on a fixed variety (that admits at least a structure of an algebraic group). Edit: I should have added, though when I wrote the question I mistakenly didn't expect reductivity made any difference, that I was mostly interested in reductive groups (in which case the comment of user54268 shows my question was not that naive after all..). REPLY [10 votes]: A most excellent example of "non-rigidity" beyond the reductive case (depending on how loose one wants to be about the meaning of "rigidity") is given in 5.2--5.10 of Exp. XIX of SGA3: a smooth affine group scheme $G$ over $k[t]$ for any field $k$ of characteristic 0 such that $G|_{t \ne 0}$ is a form of ${\rm{PGL}}_2$ (reductive!) but the fiber $G_0$ is solvable with 2 geometric connected components. By definition, $G$ is the automorphism scheme of the Lie algebra over $k[t]$ whose underlying $k[t]$-module is free on a basis $\{X,Y,H\}$ that satisfy $$[H,X]=X,\,\,\,[H,Y]=-Y,\,\,\,[X,Y]=2tH.$$ (It is clear that $G|_{t\ne 0}$ becomes ${\rm{Aut}}_{\mathfrak{sl}_2} = {\rm{PGL}}_2$ over the degree-2 finite etale cover defined by $\sqrt{t}$, but proving that $G$ is $k[t]$-smooth requires some cleverness.) REPLY [5 votes]: How about this: for each $t \in F$, define a group structure on $F^3$ by $$(a,b,c) \cdot (d,e,f)=(a+d,b+e,c+f+tae).$$ When $t=0$ this is just $F^3$ with coordinate-wise addition, while for $t \neq 0$ it is $3$ by $3$ unipotent matrices.<|endoftext|> TITLE: Generalization of a theorem of Øystein Ore in group theory QUESTION [13 upvotes]: Theorem (Øystein Ore, 1938): A finite group $G$ is cyclic iff its lattice of subgroups $\mathcal{L}(G)$ is distributive. Proof: see below. Let $(H \subset G)$ be an inclusion of finite groups and $\mathcal{L}(H \subset G)$ its lattice of intermediate subgroups. I would like to generalize the above theorem of Øystein Ore to the inclusions of finite groups, i.e. find an equivalent formulation of the following property $(D)$, in term of "cyclic" notions. $$(D) \ \ \ \ \mathcal{L}(H \subset G) \text{ is distributive } $$ Definition: An inclusion of finite groups $(H \subset G)$ is cyclic if it checks $(C_0)$. $$(C_0) \ \ \ \exists g \in G \text{ such that } \langle H,g \rangle = G$$ Examples: The maximal inclusions are cyclic, and $(\{e\} \subset G)$ is cyclic iff $G$ is cyclic. Question: $(D)$ $\Rightarrow$ $(C_0)$ ? Remark: I've checked that it's true for index $[G:H]<32$ and $\vert G \vert \le 5.10^5$ with GAP. The converse is false: $(S_2 \subset S_4)$ is a cyclic inclusion with a non-distributive lattice. Bonus question: How complete $(C_0)$ for having an equivalence with $(D)$? I've tried all the following completions of strictly increasing strongness: $(C_1) \ $ $\forall K$, $H \le K \le G$, $\exists g \in G$ such that $\langle H,g \rangle = K$ $(C_2) \ $ $\forall K$, $H \le K \le G$, $\exists g \in G$, $\exists n \ge 0$ such that $\langle H,g \rangle = G$ and $\langle H,g^n \rangle = K$ $(C_3) \ $ $\exists g \in G$, $\forall K$, $H \le K \le G$, $\exists n \ge 0$ such that $\langle H,g^n \rangle = K$ Conjecture: $(D)$ is strictly between $(C_1)$ and $(C_3)$. I've checked it for index $[G:H] < 32$ and $\vert G \vert \le 10^4$. Moreover $(D)$ is not orderable with $(C_2)$. Remark: All these notions and questions can be extended to the theory of subfactors by using the coproduct of minimal projections on the $2$-boxes space (see here and there). Remark: We give here the proof of the stronger version of the theorem, for the locally cyclic groups (i.e. every finitely generated subgroup is cyclic), because it's more relevant for an attempt of generalization. Proof of the theorem of Øystein Ore (coming from this book p12-13): Suppose first that $\mathcal{L}(G)$ is distributive and let $a,b \in G$. We have to show that $\langle a,b \rangle$ is cyclic. Because $\langle a \rangle \wedge \langle b \rangle$ is centralized by $a$ and $b$, $(\langle a \rangle \wedge \langle b \rangle) \leq Z(\langle a,b \rangle)$. Also,$\langle ab \rangle \vee \langle a \rangle = \langle a,b \rangle = \langle ab \rangle \vee \langle b \rangle$, and then by distributivity $$ \langle ab \rangle \vee (\langle a \rangle \wedge \langle b \rangle) = (\langle ab \rangle \vee \langle a \rangle) \wedge (\langle ab \rangle \vee \langle b \rangle) = \langle a,b \rangle $$ So, $\langle a,b \rangle / (\langle a \rangle \wedge \langle b \rangle) \simeq \langle ab \rangle /(\langle ab \rangle \wedge (\langle a \rangle \wedge \langle b \rangle) )$ is cyclic, and then $\langle a,b \rangle$ is abelian, as cyclic extension of a central subgroup. By the structure of finitely generated abelian groups, there are $c,d \in G$ such that $\langle a,b \rangle = \langle c \rangle \times \langle d \rangle$. As we've already shown, $\langle c,d \rangle / \langle c \rangle \wedge \langle d \rangle$ is cyclic. Because $\langle c \rangle \wedge \langle d \rangle = 1$, $\langle a,b \rangle = \langle c,d \rangle $ is cyclic. Now suppose that $G$ is locally cyclic and let $A,B,C \in \mathcal{L}(G)$. Because $G$ is abelian, we just need to verify that $A(B \wedge C) = AB \wedge AC$. Clearly, $A(B \wedge C) \leq AB \wedge AC$. Let $x \in AB \wedge AC$, then $x=ab=a'c$ with $a,a' \in A$, $b \in B$ and $c \in C$. Because $G$ is locally cyclic, $\exists g \in G$ such that $\langle a,a',b,c \rangle = \langle g \rangle$. Next, $ab=a'c$ implies that $\langle g \rangle = (A \wedge \langle g \rangle)(B \wedge \langle g \rangle)=(A \wedge \langle g \rangle)(C \wedge \langle g \rangle)$. If one of the three subgroups $A \wedge \langle g \rangle$, $B \wedge \langle g \rangle$, $C \wedge \langle g \rangle$ is trivial, then $x=b=c \in B \wedge C$ or $x \in A$. In each case, $x \in A(B \wedge C)$. So suppose that all these subgroups are non-trivial and let $n,r,s$ the respective indices of $A \wedge \langle g \rangle$, $B \wedge \langle g \rangle$, $C \wedge \langle g \rangle$ in $\langle g \rangle$. So $(n,r) = 1 = (n,s)$, and then $(n,rs) = 1$ and so $\langle g \rangle = \langle g^n \rangle \langle g^{rs} \rangle = (A \wedge \langle g \rangle) (B \wedge C \wedge \langle g \rangle) \leq A(B \wedge C)$. Again, it follows that $x \in A(B \wedge C)$. So $A(B \wedge C) = AB \wedge AC$ as expected. $\square$ Remark: For a finite cyclic group $G$, the second part of the proof can be reduced to the facts that $G$ has exactly one subgroup of order $d$ for every divisor $d$ of $ord(G)$, and that lcm and gcd are distributive. REPLY [5 votes]: The answer of the main question is yes. Let $G$ be a finite group and $H$ a subgroup. Definition: The group $G$ is called $H$-cyclic if $\exists g \in G$ such that $\langle H,g \rangle = G$. Note that: $\langle H,g \rangle = G \Leftrightarrow \langle Hg \rangle = G$. Ore's theorem for intervals (1938): If the interval $[H,G]$ is distributive, then $G$ is $H$-cyclic. proof: see here, Theorem 7 p269. The following is a new proof. Lemma 0: The top interval of a distributive lattice, is boolean. proof: see here. Lemma 1: If $H$ is a maximal subgroup then $G$ is $H$-cyclic. proof: immediate. Lemma 2: Let $[K,G]$ be the top interval of $[H,G]$. If $G$ is $K$-cyclic then it is $H$-cyclic. proof: straightforward. So the proof of Ore's theorem for intervals reduces to the following Theorem. Theorem: If the interval $[H,G]$ is boolean, then $G$ is $H$-cyclic. proof: Let $M$ be a coatom of $[H,G]$, and $M^{\complement}$ its complement. By induction on the rank of the lattice (and Lemma 1), we can assume $M$ and $M^{\complement}$ both $H$-cyclic, i.e. there are $a, b \in G$ such that $\langle H,a \rangle = M$ and $\langle H,b \rangle = M^{\complement}$. Let $g=a b$ then $a=g b^{-1}$ and $b=a^{-1}g$, so $\langle H,a,g \rangle = \langle H,g,b \rangle = \langle H,a,b \rangle = M \vee M^{\complement} = G$. Now, $\langle H,g \rangle = \langle H,g \rangle \vee H = \langle H,g \rangle \vee (M \wedge M^{\complement})$, but by distributivity $\langle H,g \rangle \vee (M \wedge M^{\complement}) = (\langle H,g \rangle \vee M \rangle) \wedge (\langle H,g \rangle \vee M^{\complement} \rangle)$. So $ \langle H,g \rangle = \langle H,a,g \rangle \wedge \langle H,g,b \rangle = G$. The result follows. $\square$ The converse of Ore's theorem is false because $\langle S_2,(1234) \rangle = S_4$ but $[S_2, S_4]$ is not distributive.<|endoftext|> TITLE: Interpolation between L^1 and Sobolev Space QUESTION [8 upvotes]: Suppose $D^\alpha$ is fractional differentiation of order $\alpha$ on the real line. Is it true that $||D^\alpha f||_{L^\frac{2 \beta}{2 \beta - \alpha}({\mathbb R})} \leq C_{\alpha,\beta} ||f||_{L^1({\mathbb R})}^{1-\frac{\alpha}{\beta}} ||D^\beta f||_{L^2({\mathbb R})}^{\frac{\alpha}{\beta}}$ for $0 < \alpha \leq \beta$ arbitrary (i.e., fractional)? This is a special case of a Gagliardo-Nirenberg-Sobolev inequality, but Nirenberg's 1959 proof in Ann. Scuola Norm. Sup. Pisa only holds for integer $\alpha, \beta$. There are many authors who prove slightly different GNS inequalities, but I need this exact one and it won't budge. One can play with Nirenberg's idea to remove the constraint that $\beta$ is an integer. Knowing that, you can work even harder and get $1 \leq \alpha \leq \beta$. But for $0 < \alpha < 1$, the proof breaks seemingly irreparably. It's also unfortunately true that the $L^1$ breaks Littlewood-Paley proofs (at least without some replacement for Nirenberg's magic). Has anyone seen this particular family of inequalities anywhere? If it's a folk theorem, what's the trick? REPLY [2 votes]: The inequality is valid. The precise answer to your questions (in much more general form) is given in the resent article Brezis, H., & Mironescu, P. (2017). Gagliardo-Nirenberg inequalities and non-inequalities: the full story. Annales de l'Institut Henri Poincare (C) Non Linear Analysis. DOI:10.1016/j.anihpc.2017.11.007<|endoftext|> TITLE: Can the boundedness of $A^2$ imply the boundedness of $A$? QUESTION [5 upvotes]: Given an operator $A \in \mathcal L(B)$, $B$ being a Banach space, I came across the following question: assume $\mathrm{dom}(A)=\mathrm{range}(A)$, $\mathrm{dom}(A)$ dense in $B$. Under which conditions is it possible to obtain the boundedness of $A$ from the boundedness of $A^2$. It is clear that "in most" cases this is not possible but maybe it works for something more than the identity? REPLY [4 votes]: Your first query has been answered, but not the second. We make the simple remark that it is the case when $A$ is a self-adjoint (even normal) operator on Hilbert space by the spectral theorem. Presumably this can be extended to operators on general Banach spaces with good spectral properties (spectral operators, operators of scalar type---see the third volume of Dunford and Schwartz or Dowson's monograph for this).<|endoftext|> TITLE: Do compact groups acting irreducibly have finite subgroups which do the same? QUESTION [8 upvotes]: Let $G$ be a closed subgroup of $U(n,{\bf C})$, not necessarily connected. Regard ${\bf C}^n$ as a complex $G$-module $M$. Q. Suppose $M$ is irreducible as a $G$-module (equivalent, I think, to the condition that the linear span of $G$ is all of $M_n({\bf C})$). Does there always exist a finite subgroup $H < G$ such that $M$ is irreducible as an $H$-module? Perhaps one should view the question as one about faithful irreducible representations of "abstract" compact Lie groups and their finite subgroups, but the more concrete picture is easier for me to think about. It also seems possible to me that one could hammer the question by using structure theory for compact Lie groups in some way, and then looking down the lists to check everything works for A, B, C, D etc, but I'd prefer to have something that used the basic ideas that go into such structure theory rather than the final classification itself, since the original motivation for the question actually comes from an infinite-dimensional context. (The original motivation for the question is probably not relevant here; but I include it in case it is of interest. The question is a generalization of one raised en passant in some work of A. Blanco, and if the answer is yes, it seems that improvements can be made to some results he's obtained concerning finite-dimensional Banach spaces $X$ for which $B(X)$ is "amenable with constant $1$". Moreover, a proof which was somehow "global" in nature might be adaptable to the study of $K(X)$ for certain well-behaved infinite-dimensional $X$.) REPLY [8 votes]: No. Consider an irreducible representation of $\mathrm{SU}(2)$ of large dimension, say $n\ge n_0$. Then it cannot be irreducible for a finite subgroup, because these are either abelian or dihedral (with only 1 or 2-dimensional irreducible reps) or in a short finite list, with finitely many possible dimensions of irreducible reps for each. Maybe the best $n_0$ here is $n_0=7$ but I haven't checked. REPLY [6 votes]: The answer is already no for $G=U(2)$ acting on $Sym^6(\mathbb C^2)$. Given a subgroup $\Gamma \leq U(n) \cong SU(n) \times U(1) / Z_n$, we can enlarge it by projecting to $SU(n)/Z_n$ and $U(1)/Z_n$, take the preimages in $SU(n),U(1)$, and multiply together. The $U(1)$ isn't going to help hold together an irrep so we've reduced to subgroups of $SU(n)$. There aren't so many of these for $n=2$; the cyclic, binary dihedral, and the $E_{6,7,8}$ ones. Their largest irreps are 1, 2, 3, 4, 6-dimensional, i.e. the largest coefficient of a simple root in the expansion of the highest root of the $A,D,E$ root systems.<|endoftext|> TITLE: Coloring summands of given n-partition with given weights of colors QUESTION [6 upvotes]: Let $\lambda$ and $\sigma$ be partitions of $n$: $\lambda_1+\lambda_2+\cdots+\lambda_l=n$ and $\sigma_1+\sigma_2+\cdots+\sigma_s=n$ Let $M_{\lambda \sigma}$ be the number of ways to colour the parts of lengths $\lambda_1,\lambda_2,\cdots,\lambda_l$ with $s$ colours such that 1) each part is coloured with exactly one colour 2) the sum of the lengths of the parts coloured with colour $i$ is $\sigma_i$. I am intersted in 1) are these numbers known, for example related to the Kostka numbers? 2) is there any other information about these numbers $M_{\lambda \sigma}$ available? (reccurences, gf's) These numbers are needed to find the generating function of $n$-ary commutative rooted trees. In case the definition is not clear, here is an example. For $\lambda=(2,2,1)$ and $\sigma=(3,2)$ we have $M_{\lambda \sigma}=2$. This means that there are two ways to colour the parts of lengths $2$, $2$ and $1$ with two colours such that the sum of the lengths of the parts with first colour is $3$ and the sum of the lengths of the parts with the second colour is $2$. 1st way: 1st part and 3rd part are coloured with the 1st colour, 2nd part with the 2nd color. 2nd way: 2nd part and 3rd part are coloured with the 1st colour, 1st part with the 2nd colour. REPLY [6 votes]: Your number $M_{\lambda\sigma}$ seems to be the same as $L_{\lambda\mu}$, defined in (6.9) page 103 in Macdonalds book on symmetric functions (second edition) if I am not mistaken. These numbers are coefficients in the transition matrix between power sum polynomials $p_\lambda$ and monomial symmetrix polynomials, $m_\lambda$. The Kostka numbers give the matrix between $m_\lambda$ and $s_\lambda$, the Schur polynomials. See figure on page 104.<|endoftext|> TITLE: Tensor product of certain Sobolev spaces on non-compact manifolds QUESTION [6 upvotes]: Let $M$ be a non-compact Riemannian manifold of bounded geometry (i.e., its injectivity radius is uniformly positive and the curvature tensor and all its covariant derivatives are bounded in sup-norm). (For simplicity we may just consider $M = \mathbb{R}^n$ with the usual Euclidean metric. Assuming bounded geometry implies that different possible definitions of Sobolev norms are equivalent.) We equip $C_c^\infty(M)$, the space of all compactly supported smooth functions on $M$, with the countable family of Sobolev norms $\{\| \cdot \|_{W^{k,1}}\}_{k \ge 0}$ (it is important to me that we use $L^1$-integrability here). So we get a locally convex topological algebra which we still denote by $C_c^\infty(M)$. Do we have $C_c^\infty(M) \ \hat{\otimes} \ C_c^\infty(N) \cong C_c^\infty(M \times N)$, where $\hat{\otimes}$ denotes the projective tensor product? If we complete $C_c^\infty(M)$ in its family of norms, we get the infinite Sobolev space $W^{\infty,1}(M)$, which is a Frechet space. Do we have $W^{\infty,1}(M) \ \hat{\otimes} \ W^{\infty,1}(N) \cong W^{\infty,1}(M \times N)$? If the answers to the above questions are negative, what if we just use the four norms $\{W^{0,1}, W^{1,1}, W^{0, \infty}, W^{1, \infty}\}$ instead the countably many ones and ask the analogous two questions now? An affirmative answer here would be also good enough for my application though not very convenient. REPLY [4 votes]: The answer to the first question is no for a simple reason: I guess you do not want to complete the tensor product, but then then the left hand side is only dense in the right hand side. For your second question, I think that the answer is yes, because $L^1(M)\hat\otimes L^1(N) \cong L^1(M\times N)$. But I have not yet seen a proof of this. Maybe, the following paper which uses these (and other) spaces (but only on $\mathbb R^n$) might be helpful to you: Andreas Kriegl, Peter W. Michor, Armin Rainer: An exotic zoo of diffeomorphism groups on ℝn. 45 pages. arXiv:1404.7033. (pdf) Edit (twice): With regard to you first question: If you the locally convex direct limit topology on $C^\infty(M)$ with respect to the embeddings of spaces $C^\infty_K(M)$ of smooth functions with support contained in a fixed compact $K$, then equality holds, since the projective tensor product respects direct limits in the category of locally convex spaces. Thus the algebraic tensor product is dense in the right hand side. Moreover, all tensor products between the projective and the injective coincide since the space is nuclear. But you ask for another topology, the one induced by $W^{\infty,1}$. Density of the algebraic tensor product follows, since it is dense also in the finer topology. This Frechet space is linearly isomorphic to $\ell^1\hat\otimes \mathcal s$ which is no longer nuclear, by a result of Dietmar Vogt, namely in the paper: MR0688001 Reviewed Vogt, Dietmar Sequence space representations of spaces of test functions and distributions. Functional analysis, holomorphy, and approximation theory (Rio de Janeiro, 1979), pp. 405–443, Lecture Notes in Pure and Appl. Math., 83, Dekker, New York, 1983. (Reviewer: M. Valdivia) 3rd Edit: All your points with the exception of (iv) are correct. (iv) needs a proof. The space $W^{\infty,1}(\mathbb R^n)$ is called $\mathcal D_{L^1}$ in the book on Distributions of Laurent Schwartz. Density follows via the Stone Weierstrass theorem. The tensor product of the $C^\infty_c(M)$ spaces for the l.c. inductive limit topology is treated in: MR2296978 Reviewed Trèves, François Topological vector spaces, distributions and kernels. Unabridged republication of the 1967 original. Dover Publications, Inc., Mineola, NY, 2006. xvi+565 pp. The original version is in: MR0075539 Grothendieck, Alexandre Produits tensoriels topologiques et espaces nucléaires. (French) Mem. Amer. Math. Soc. 1955 (1955), no. 16, 140 pp. (Reviewer: J. Sebastião e Silva) As for other tensor products, I think that the $\ell^p$-tensor product satisfies $W^{\infty,p}(M)\hat\otimes^p W^{\infty,p}(N) = W^{\infty,p}(M\times N)$. But this needs a proof, too. Many different tensor products are described in: Peter W. Michor: Functors and categories of Banach spaces. Springer Lecture Notes 651, (1978), vi+99 pp.(pdf)<|endoftext|> TITLE: construction of nonmeasurable sets QUESTION [23 upvotes]: I have a history question for which I've had trouble finding a good answer. The common story about nonmeasurable sets is that Vitali showed that one existed using the Axiom of Choice, and Lebesgue et al. put the blame squarely on this axiom and its non-constructive character. It was noticed however that some amount of choice was required to get measure theory off the ground, namely Dependent Choice seemed to be the principle typically employed. But the full axiom of choice which allows uncountably many arbitrary choices to be made is of a different character, and is the culprit behind the pathological sets. This viewpoint was not really justified until Solovay showed in the 1960s that ZF+DC could not prove the existence of a nonmeasurable set, assuming the consistency of an inaccessible cardinal. My question is, in the many years before Solovay's theorem, was there any effort aimed at showing the existence of a nonmeasurable set without the use of the full AC? Was something like the following question ever posed or worked on: "Can constructions similar to those of Vitali, Hausdorff, and Banach-Tarski be done without appeal to the Axiom of Choice?" REPLY [7 votes]: I would like to point out that the question on the existence of a non-measurable set without the use of the full AC was technically established in the literature before Solovay's result (which afaik goes back to March-July 1964). In 1938 Sierpinski had established (cf. "Fonctions additives non complètement additives et fonctions non mesurables", Fund. Math.) that a non-measurable set could be constructed from the assumption (in modern terminology) that there is a prime ideal on the power set of the natural numbers extending the ideal of finite sets. He explains that the existence of such prime ideal was proven by Tarski (cf. "Une constribution à la théorie de la mesure", Fund. Math.) with the aid of the Axiom of Choice (he proved it by transfinite induction). But it remained an open question, especially in the 50's after Henkin's results, whether the existence of such prime ideals in the power sets, or, more generally, in Boolean algebras, was or not weaker than the Axiom of Choice. This was eventually settled by Halpern, who proved that it was, and his results first appeared on his doctoral dissertation, submitted in the spring 1962. Sierpinski construction is quite simple: define a function $f$ on a real number $x$ to be either $1$ or $0$, depending on whether the subset of ones in (the non integer part of) its dyadic expansion (choosing the finite development for the rationals) is or not in the prime ideal defined on the powerset of the natural numbers extending the ideal of finite sets. It follows that $f$ has arbitrarily small periods (all numbers of the form $2^{-n}$) and that $f(1-x)=1-f(x)$. From this and the fact that $f$ only takes the values $0$ and $1$ it is not hard to show that it cannot be measurable, and the preimage of $0$ provides a non-measurable set.<|endoftext|> TITLE: Maximum length of a chain of topologies on $\Bbb R$ QUESTION [12 upvotes]: Let $\frak T$ be a totally ordered set of topologies on $\Bbb R$. Is $|\frak T|\le |\Bbb R|$? REPLY [11 votes]: The answer is no. As Ashutosh notes in the comments, a modification of my original answer shows in ZFC that there is a chain longer than $|\mathbb{R}|$. Theorem. There is a linearly order chain of topologies on $\mathbb{R}$ of size larger than $\mathbb{R}$. Proof. Let $\kappa$ be the smallest cardinal so that $2^\kappa>|\mathbb{R}|={\frak c}=2^\omega$. It follows that the tree $T=2^{<\kappa}$ has size at most continuum, since $\kappa\leq\frak{c}$ and so $2^{<\kappa}$ is a $\kappa$-sized union of sets of size at most $\frak{c}$, and hence of size at most $\frak{c}$. Thus, we may label the nodes in $2^{<\kappa}$ with reals, and get $2^\kappa$ many paths through this tree. For each path $p$, let $X_p$ be the set of reals appearing on $p$ or to the left of $p$. So we have a linearly ordered subset of $P(\mathbb{R})$ of size $2^\kappa$, which is larger than the continuum. For each $X\subset\mathbb{R}$, let $\tau_X$ be the topology with the discrete topology on $X$ and the indiscrete topology on $\mathbb{R}-X$. Since $X\subset Y\iff \tau_X\subset \tau_Y$, the chain in $P(\mathbb{R})$ translates to a chain of topologies on $\mathbb{R}$ of size $2^\kappa$, and so we have a chain larger than $|\mathbb{R}|$, as desired. QED So in ZFC we may find a chain of topologies longer than $|\mathbb{R}|$. Previous answer: The answer is no, not necessarily. In particular, if the continuum hypothesis holds (or more generally, if merely $|\mathbb{R}|<2^{\omega_1}$, which is to say that Luzin's hypothesis fails), then there is a very long chain. Theorem. There is a linearly ordered chain of topologies on $\mathbb{R}$ of size $2^{\omega_1}$. This is strictly larger than $|\mathbb{R}|$ if the continuum hypothesis holds, or more generally, if Luzin's hypothesis fails. Proof. Consider the tree $T=2^{<\omega_1}$ consisting of countable ordinal length binary sequences. This tree has size continuum, and so we may label each node with a real. Consider the paths through this tree, ordered from left-to-right, which is the lexical order. There are $2^{\omega_1}$ many such paths. For each path, consider the set of reals appearing on the path or to the left. This gives a linearly ordered chain in $P(\mathbb{R})$ of size $2^{\omega_1}$. For each subset $X\subset\mathbb{R}$, let $\tau_X$ be the topology consisting of the discrete topology on $X$, and the indiscrete topology on $\mathbb{R}-X$. Thus, $X\subset Y\implies \tau_X\subset \tau_Y$. So our linearly order chain in $P(\mathbb{R})$ thus transforms to a chain in the set of topologies on $\mathbb{R}$, of size $2^{\omega_1}$. QED<|endoftext|> TITLE: Elliptic curves with trace of Frobenius values always congruent to 0 modulo 2 QUESTION [9 upvotes]: Let $E/\mathbb{Q}$ be an elliptic curve and suppose that the trace of Frobenius values are such that $a_{p}(E) \equiv 0 \pmod{2}$ for all odd primes avoiding the conductor. Is it the case that $E$ contains a nontrivial rational two torsion point? Why? REPLY [6 votes]: Alternative method without using the cubic equation (of course not fundamentally different): the Galois representation $\rho: G_{\mathbb Q} \rightarrow GL_2(\mathbb Z/2 \mathbb Z)$ on the points of $2$-torsion of $E$ satisfies tr $\rho(Frob_p)=0$ for every odd prime $p$ by hypothesis, hence tr $\rho=0$ by Chebotarev. (edited:) It has obviously determinant $1$. Hence the semi-simplification of $\rho$ is $1 \oplus 1$ (a semi-simple representation of dim 2 is characterized by its trace and determinant, even in characteristic 2). Therefore $\rho$ fixes a line, and the non-zero point of this line is a rational 2-torsion point of $E$.<|endoftext|> TITLE: Can Suslin (or Aronszajn) lines ever be orderings of abelian groups? QUESTION [15 upvotes]: I am interested in realizing linear orders as orderings of abelian groups. In particular, can Suslin lines (and other classes of line) be realised in this way? Let $\mathcal{C}$ be a class of (torsion-free) abelian groups, and $L$ be the first-order language of linear orderings. Let $LO(\mathcal{C})$ be the theory of ordered abelian groups belonging to the class $\mathcal{C}$. Recall an abelian group $A$ is ordered by the binary relation $<$ if $<$ is a linear order on $A$ such that whenever $g < h$ then $g+k < h+k$ for all $g, h, k \in A$. Let $\mathcal{K}_{LO(\mathcal{C})} = \lbrace A \mid L : A \models LO(\mathcal{C}) \rbrace$ be the class of all linear orderings which are orderings of members in $\mathcal{C}$. Suppose $\mathcal{C}$ is the class of all abelian groups. So $\mathcal{K}_{LO(\mathcal{C})}$ is a pseudo-elementary class. From A.C. Morel's work (\textit{Structure and order structure in Abelian groups}, Colloq. Math. 19 (1968), 199-209), the $\aleph_{1}$ countable members of $\mathcal{K}_{LO(\mathcal{C})}$ are known up to isomorphism. What uncountable lines can turn up in $\mathcal{K}_{LO(\mathcal{C})}$? Is it consistent that $\mathcal{K}_{LO(\mathcal{C})}$ contains a Suslin line (or some but not all Suslin lines, or all Suslin lines if these exist)? If one adds a Cohen real, does $\mathcal{K}_{LO(\mathcal{C})}$ contain a Suslin line? Does $\mathcal{K}_{LO(\mathcal{C})}$ contain an Aronszajn line, i.e. a line obtained by squashing an Aronszajn tree? Thank you for any help. REPLY [6 votes]: This answer might be late, but I found the following papers related: (A) Koppelberg, Sabine, ``Groups cannot be Souslin ordered'', Arch. Math. (Basel) 29 (1977), no. 3, 315–317. In this paper it is shown that there are no Suslin ordered groups. I may also mention that the problem of whether a Suslin order may be the underlying order of an ordered field or of some more general algebraic structure is also discussed in the following paper: (B) Felgner, Ulrich, ``Das Problem von Souslin für geordnete algebraische Strukturen''. Set theory and hierarchy theory (Proc. Second Conf., Bierutowice, 1975), pp. 83–107. Lecture Notes in Math., Vol. 537, Springer, Berlin, 1976.<|endoftext|> TITLE: Completely positive maps-equivalent definition QUESTION [12 upvotes]: The most usual definition of the completely positive map (c.p.) between two C*-algebras (say, unital) is the following: $\sigma: A \to B$ should satisfy $\sigma(1)=1$ and for each $n \in \mathbb{N}$ the map $\sigma_n: M_n(A) \to M_n(B)$ defined ``cordinatewise''should be positive. However I met also the following definition: $\sigma(1)=1$ and for each $n \in \mathbb{N}, a_1,...,a_n \in A, b_1,...,b_n \in B$ we have $\sum_{i,j}b_i^*\sigma(a_i^*a_j)b_j \geq 0$. I was able to show that the first definition implies the second: how to prove that the converse is also true (if it is so)? REPLY [13 votes]: The converse follows from two observations: 1) If $a=[a_{ij}]$ is a positive element in $M_n(A)$, then $a=x^*x$ for some $x=M_n(A)$, and if we write this out in matrix entries we have $$ a_{ij} = \sum_k x_{ki}^*x_{kj}. $$ 2) The matrix $[a_{ij}]$ is positive in $M_n(A)$ if and only if it is positive in every GNS representation $\pi$ of $A$; and this is implied by the condition \begin{equation} \sum_{ij}c_i^*a_{ij}c_j\geq 0 \end{equation} for all $c_1,\dots c_n\in A$. To see this fix a state $\rho$ on $A$, let $H$ denote the GNS space associated to $\rho$ and $\pi:A\to B(H)$ the GNS representation. $H$ is spanned by equivalence classes $[c]$ with $c\in A$. Note that since $A$ is unital, by the GNS construction we have $[c]=\pi(c)[1]$. To check that $[a_{ij}]$ is positive in the representation $\pi$, fix vectors $[c_1], \dots [c_n]$ in $H$ and complute: \begin{align} \sum_{ij}\langle \pi([a_{ij}])[c_j],[c_i]\rangle_H &= \sum_{ij}\langle \pi(c_i^*)\pi(a_{ij})\pi(c_j)[1],[1]\rangle_H \\ &=\sum_{ij}\langle \pi(c_i^*a_{ij}c_j)[1].[1]\rangle_H\\ &= \rho \left(\sum_{i,j} c_i^*a_{ij}c_j\right)\\ &\geq 0 \end{align} by assumption. Now, for the proof: suppose $\sum_{ij} b_i^*\sigma(a_i^*a_j)b_j\geq 0$ for all $a's$ and $b's$. Fix a positive element $[a_{ij}]\in M_n(A)$, factor it as in the first observation. To check the positivity of $\sigma([a_{ij}])$, we check the condition of the second observation; so fix $b_1, \dots b_n$; we have \begin{equation} \sum_{ij}b_i^*\sigma(a_{ij})b_j = \sum_k\sum_{ij}b_i^*\sigma(x_{ki}^*x_{kj})b_j\geq 0, \end{equation} where the last inequality follows by hypothesis.<|endoftext|> TITLE: Adams e-invariant QUESTION [9 upvotes]: In "On the Groups J(X) - IV", Adams introduces the $e$-invariant, which turns out to be closely connected to the image of the $J$-homomorphism, but he introduces it in a more general setting. He has $d$- and $e$-invariants of a map $X\rightarrow Y$ with respect to some exact functor $k$ from Top into an abelian category. They have the property that $d\in\text{Ext}^0(kY,kX)$ and $e\in\text{Ext}^1(kY,kX)$. He then mentions (page 27) that we might "hope" to be able to construct further invariants in higher Ext groups, but it's not really clear to me how to go about doing this. How do you construct these "higher $e$-invariants" and has anyone put them to any use? REPLY [11 votes]: Yes, this has been done. Laures has defined an f-invariant using elliptic cohomology in The topological q-expansion principle. The explanation in Section 2 of his more recent paper on Toda brackets might be a bit easier to read. Essentially the idea is the following: Look at the Adams-Novikov spectral sequence. A stable homotopy element $x$ has non-vanishing degree iff it is in exact Adams-Novikov filtration 0, i.e. detected in the 0-line. Suppose the degree vanishes. Then $x$ is in Adams-Novikov filtration (at least) 1. The most naive e-invariant is just the reduction of $x$ to the 1-line of the Adams-Novikov SS (which is an $\mathrm{Ext}^1$-term). To make it a bit easier, we map it into the 1-line of the Adams-Novikov spectral sequence with respect K-theory; this does not loose any information, essentially because K-theory carries a height-1 formal group law. Then we go one step further and map it into $\mathbb{Q}/\mathbb{Z}$. This is the Adams e-invariant. If the e-invariant of $x$ vanishes, $x$ has Adams-Novikov filtration 2. (This happens automatically if the degree of $x$ is positive and even). We map now $x$ into the second line of the Adams-Novikov $E^2$-term for elliptic cohomology -- this does not loose information (essentially) because the formal group of elliptic cohomology has height 2. Laures maps this then further into a much more complicated analogue of $\mathbb{Q}/\mathbb{Z}$, essentially a quotient by the ring of divided congruences (of modular forms) by (certain) integral modular forms. For concrete calculations with the $f$-invariant, von Bodecker's papers (e.g. as referenced in the Toda bracket paper by Laures) contain a wealth of information. I guess, there are at least two reasons why the f-invariant is less known than the e-invariant: Firstly, the 1-line of the ANSS has a (very important!) geometric interpretation as the image of $J$. Secondly, the Adams-Novikov spectral sequence was not known at the times Adams invented the e-invariant; in particular, its first line was not computed yet. In contrast, since 1981 (by work of Miller, Ravenel, Wilson and Shimomura) the Adams-Novikov 2-line is completely computed. The f-invariant can therefore not longer help with its computation. On the other hand, the f-invariant can be still useful: Say you want to compute a Toda bracket (or a product) whose result lies in Adams-Novikov filtration 2. Then you can use the f-invariant to identify it with other classes already known. This is what happens in Laures' paper referenced above.<|endoftext|> TITLE: Probing the generalization of the abc conjecture to more than 3 variables QUESTION [17 upvotes]: Browkin and Brzezinski, in "Some remarks on the $abc$-conjecture", Math. Comp. 62 (1994), no. 206, 931–939, state the following generalization of the $abc$ conjecture to more than three variables: Given $n\ge3$, let $a_1,\dots,a_n \in \Bbb Z$ satisfy $\gcd(a_1,a_2,\dots,a_n)=1$ and $a_1+\dots+a_n=0$, while no proper subsum of the $a_j$ equals $0$. Then for every $\varepsilon>0$, $$ \max\{|a_1|,\dots,|a_n|\} \ll_{n,\varepsilon} R(|a_1\cdots a_n|)^{2n-5+\varepsilon}, $$ where $R(m)$ denotes the radical of $m$ (the product of the distinct primes dividing $m$). I have two questions about this conjecture. First, the authors give constructions of $n$-tuples that show that the exponent $2n-5$ on the right-hand side cannot be improved. For example, when $n=4$, one takes any $abc$ triple $(a,b,c)$ and chooses $(a_1,a_2,a_3,a_4)=(a^3,b^3,3abc,-c^3)$, so that $$ R(|a_1a_2a_3a_4|)^{2\cdot4-5} \le (3R(abc))^3 \le 27c^3 = 27\max\{|a_1|,|a_2|,|a_3|,|a_4|\}. $$ On the other hand, a probabilistic argument suggests (I believe) that the exponent $1+\varepsilon$ should suffice. Is the exponent $2n-5$ present only because of integer points on certain lower-dimension varieties like $y^3=-27wxz$, on which the examples $(a^3,b^3,3abc,-c^3)$ all live? Second, note that the $4$-tuples given above are only relatively prime, not pairwise relatively prime. Has anyone mulled over whether the exponent $2n-5$ can be reduced if one simply strengthens the hypothesis to pairwise coprimality? (An earlier version of this question wondered why the "no vanishing subsums" condition was present, but that has been answered to my satisfaction.) REPLY [7 votes]: In A more general abc conjecture, p. 7 Paul Vojta conjectures exponent $1 + \epsilon$ outside a proper Zariski-closed subset Your question appears in The abc-conjecture and the n-conjecture,Coen Ramaekers p. 23 5.1 Strong $n$-conjecture. Adding pairwise coprimality, the exponent is $1$ without additional restrictions. Searching the web for it returns only 3 hits.<|endoftext|> TITLE: Are the quaternions not uncountably categorical? QUESTION [15 upvotes]: Boris Zilber has argued that the field of the complex numbers is "logically perfect". For one thing, the theory of an algebraically closed field of characteristic zero is uncountably categorical: it admits a unique model, up to isomorphism, of some uncountable cardinality - and thus a unique model of each uncountable cardinality, by Morley's theorem! The field of real numbers is not "logically perfect" in Zilber's sense. More precisely, the theory of a real closed field is not uncountably categorical. This led Terry Bollinger to ask about the quaternions. Has anyone studied this? For starters, we need to settle on a 'theory of quaternions'. I want a theory where the only operations are + and ×, and the only constants are 0 and 1. Here's my guess about how to proceed. First we impose the axioms of a division ring (including associativity for multiplication, to rule out the octonions!). We impose a 'characteristic zero' axiom schema that says $1 + 1 + \cdots + 1$ can never equal zero. Then we impose an 'algebraically closure' axiom saying that any polynomial whose coefficients all commute has a root. (Any commuting set of quaternions lies in a subring isomorphic to the complex numbers.) Finally, we impose an axiom saying that multiplication is not commtuative, to rule out the complex numbers. I haven't thought about this much, but I'm hoping these are enough, with the help of the Frobenius theorem. Then we can ask if the theory of quaternions is uncountably categorical. My guess is that it's not. Here's my rough reasoning. The center of the quaternions forms a copy of the real numbers. Moreoever, I'm guessing that we can prove using our 'theory of quaternions" that the center of the quaternions is a real closed field. Conversely, starting from any real closed field, we can build a model of the theory of quaternions. So, I believe we're in trouble, since the theory of a real closed field is not uncountably categorical. Perhaps someone can say if this sort of argument is valid: if in some structure we can define a subset and this subset, equipped with some operations in our theory, is a model of a second theory that's not uncountably categorical... and any model of that second theory arises in this way... then the original theory could not have been uncountably categorical. Or, if this doesn't work, maybe someone can straighten things out! REPLY [7 votes]: I don't think the axioms suffice. Here is some background: Theorem 16.16 of TY Lam's "First Course in Noncommutative Rings" states: "The centrally finite noncommutative division rings which are right algebraically closed are precisely the division rings over real closed fields. These division rings are also left algebraically closed." Lam goes on to say that, without the "centrally finite" assumption, not much seems to be known. In http://eprints.biblio.unitn.it/1626/1/ghiloni-octonions-preprint.pdf, published in J. Algebra Appl., 11, 1250088 (2012), Ghiloni seems to prove the octonionic analogue. Also, Ghiloni attributes the result above to Niven, Baer, and Jacobson, based on I. Niven, Equations in quaternions. Amer. Math. Monthly 48, (1941), 654–661. For a noncommutative ring $D$, the "right algebraically closed" condition states that for any nonconstant polynomial $a_0 + a_1 X + \cdots + a_n X^n$ with coefficients in $D$, there exists an element $d \in D$ such that $a_0 + a_1 d + \cdots + a_n d^n = 0$. This is stronger than the condition you give, in which the coefficients are required to commute with each other. The "centrally finite" condition (required in the quaternionic and octonionic results) states that the whole algebra is a finite-dimensional vector space over the center of the algebra. Lam's remark suggsts that, without the "centrally finite" assumption, there may be right/left algebraically closed division rings which are not quaternion algebras over a real closed field. Since his axioms for algebraic closure are stronger (or at least as strong) as yours, but you omit the central finiteness condition, I think your axioms could allow some exotic examples. Now, I think an exotic example is given in the paper "Algebraically-closed skew-fields" by L Makar-Limanov, Journal of Algebra Volume 93, Issue 1, March 1985, Pages 117–135. Makar-Limanov gives an even stronger definition of "algebraically closed" (using polynomials in the amalgamated product $A \ast Z[X]$ (amalgamated over $Z$) ). There Makar-Limanov constructs a centrally infinite, algebraically closed (in his strong sense) division algebra (a.k.a. skew-field). So Makar-Limanov's algebra will satisfy the axioms of the question, without being a quaternion algebra over a real closed field. Add an axiom which implies "centrally finite" and I think it's golden. Unfortunately, I can't think of such an axiom which doesn't explicitly bound the dimension over the center, for example: "For every five elements $a,b,c,d,e$ of the algebra, there exist five elements $u,v,w,x,y$ of the center of the algebra, not all zero, such that $ua + vb + wc + xd + ye = 0$."<|endoftext|> TITLE: Trapping a particle QUESTION [10 upvotes]: A particle starts a brownian walk in the middle of a long tunnel in the plane, at one end of the tunnel is a region Y of given area A. Does the shape of region Y affect average time for the particle to exit through the other end of the tunnel? If yes, what shapes (qualitatively) Y would maximize or minimize this time? REPLY [2 votes]: The region $Y$ introduces a delay time $\tau$, the average time between entry and exit. Let me first consider the case that the dynamics in the two-dimensional region $Y$ is a Brownian motion (diffusion) with diffusion coefficient $D$. The region $Y$ has area $A$, perimeter $P$, and $W$ is the width of the tunnel. The mean time between collisions with the perimeter of $Y$ can be estimated as $A/D$, and on average about $P/W$ collisions occur between entry and exit, so we have the estimate $$\tau=C\times\frac{AP}{DW}.\quad\quad[1]$$ The shape dependence enters in the coefficient $C$ of order unity, but different shapes with the same area and perimeter will have the same $\tau$ in order of magnitude. We can do better and obtain a precise expression for $\tau$ including the numerical coefficient, if we assume that $Y$ is a billiard, in which the particle moves along a straight line (velocity $v$), bouncing from the walls until it finally exits. (See diagram.) This problem has been analyzed by Lewenkopf and Vallejos under the assumption of ``weak ergodicity'', which basically means that the particle explores the whole phase space of the billiard before exiting. (This assumption is weaker than that of chaotic dynamics, it also applies to, say, a circular billiard which has regular dynamics.) Under this assumption one has simply $$\tau=\frac{\pi A}{vW}\quad\quad [2]$$ without any shape dependence. The assumption of a constant velocity is actually not needed, if only the energy $E$ of the particle is constant but its velocity varies, then one has $$\tau=\frac{1}{\tilde{\Omega}}\frac{d\Omega}{dE}\quad\quad [3]$$ where $\Omega$ is the area of phase space in the billiard and $\tilde{\Omega}$ the area of phase space connected to the tunnel. The simpler expression $\tau=\pi A/vW$ for constant velocity (momentum $p=mv$) follows from $\Omega=\pi p^2 A=2\pi mE A$ and $\tilde{\Omega}=2pW$. In this way one also obtains the generalization to a three-dimensional billiard (volume $V$) connected to a tunnel with cross-sectional area $S$: $\Omega=\frac{4}{3}\pi p^3 V=\frac{4}{3}\pi(2mE)^{3/2}V$ and $\tilde{\Omega}=\pi p^2 S$, so $$\tau=\frac{1}{\tilde{\Omega}}\frac{d\Omega}{dE}=\frac{4V}{vS},\quad \quad[4]$$ again without any shape dependence.<|endoftext|> TITLE: Erdős cardinals and ineffable cardinals QUESTION [11 upvotes]: In Cantor's Attic it is stated that an $\omega$-Erdős cardinal is a stationary limit of ineffable cardinals, and Jech book is given as a reference, but I cannot find this result in that book. I have seen proofs in other sources using that Erdős cardinals are subtle, but since Jech does not define subtle cardinals, I wonder whether there is a more direct proof. Jech proves (Theorem 17.33) that there is an ineffable cardinal below $\eta_\omega$ by constructing an elementary submodel of $V_{\eta_\omega}$ with a set of indiscernibles and a non-trivial elementary embedding. Then the critical point in ineffable. By adding a set of constants to the formal language I can ensure that the critical point is large, so I obtain an unbounded set of ineffable cardinals below $\eta_\omega$ (and the proof works for any $\eta_\alpha$), but how can I ensure a stationary set of ineffable cardinals? REPLY [13 votes]: To see that the set of ineffables is stationary, let $C\subseteq \eta_\omega$ be any closed and unbounded set. It suffices to show that the good indiscernibles arising in the definition of $\omega$-Erdos for the structure $(V_{\eta_\omega}, \in, C)$ are in $C$ since by what you have already shown such indiscernibles can be critical points of a non-trivial elementary embedding, and hence are ineffable. However if $\gamma$ is such an indiscernible then $$(V_{\gamma}, \in, C\restriction \gamma)\prec (V_{\eta_\omega}, \in, C).$$ By elementarity, as $C$ is unbounded in $\eta_\omega$, then $C\restriction \gamma$ is unbounded in $\gamma$. That is, $\gamma \in C$. [ Suppose $\mathcal{A} = \langle L_\delta[A],\in,A,B\rangle $ is a structure. $I\subseteq \delta$ is a good sequence of indiscernibles for $\mathcal{A}$ if for all $\gamma\in I$: (i) $\langle L_\gamma[A\cap \gamma],\in,A\cap \gamma,B\cap \gamma\rangle \prec \mathcal{A}$; (ii) $I\backslash \gamma$ is a set of indiscernibles for the structure $\langle L_\delta[A],\in,A,B, \langle \xi\rangle_{\xi < \gamma}\rangle$. We say that $\delta$ is $\tau$-Erdos if every first order structure $\mathcal{A}=\langle L_\delta[A],\in,A,B\rangle $ has a good sequence of indiscernibles of o.t. $\tau$. It can be shown by a combinatorial argument that each of the $\eta_\alpha$ as defined by Jech are $\alpha$-Erdos. (For this see: J. Baumgartner, Ineffability properties of cardinals II, in: R.E. Butts and J. Hintikka, eds., Logic, Foundations of Mathematics and Computability Theory, Reidel, Dordrecht, 1977, pp 87-106. However it is not the case that $\delta \rightarrow (\tau)^{<\omega}$ implies that $\delta$ is $\tau$- Erdos in general, but the least such $\delta$ (which equals $\eta_\tau$ in Jech's notation) is so.) An equivalent definition of being $\tau$-Erdos is the following: let $C\subseteq \delta$ be c.u.b. Let $F:[C]^{<\omega}\rightarrow \delta$ be any regressive function (i.e. $F(a) TITLE: Why is the section conjecture important? QUESTION [16 upvotes]: As in the title, I want to know the reason for importance of the section conjecture. Of course, the statement of conjecture is important as itself, even I cannot fully grasp the soul of it. However, what I really want to know is applications of the section conjecture. For example, can we derive properties on the set of rational points through the section conjecture such as finiteness? REPLY [19 votes]: You can start here "Fermat's last theorem" and anabelian geometry?? In particular, I mention there that: At some point Deligne thought he had a proof that the section conjecture implied Mordell, but the proof doesn't work. This is all explained in an appendix by Deligne to a paper of Stix: http://arxiv.org/abs/0910.5009 Finally, for an actual application of the section conjecture. Its truth implies the existence of an algorithm to decide whether a curve of genus bigger than one has a rational point.<|endoftext|> TITLE: Proof of the Dunford-Pettis theorem QUESTION [6 upvotes]: I would like to know where to find a complete proof of the Dunford-Pettis theorem: A sequence $(f_n)_{n\geq 0} \subset L^1$ is uniformly integrable if and only if it is relatively compact for the weak topology $\sigma(L^1,L^\infty)$. REPLY [5 votes]: This is surely in many places, but a reasonably complete proof is given in Section III.2, Theorem 15 of J. Diestel, J. J. Uhl, Vector measures. American Mathematical Society 1977.<|endoftext|> TITLE: Realizing algebraic curves as complete intersections QUESTION [11 upvotes]: I have two related questions about smooth complete algebraic curves (over $\mathbb{C}$). Does there exist a smooth complete algebraic curve $X$ that cannot be embedded as a complete intersection in $\mathbb{P}^n$ for any $n$ (nb: this is different from asking if every smooth curve in $\mathbb{P}^n$ is a complete intersection, which is of course false; e.g. the twisted cubic)? I expect that the answer is "yes", though it might be "no" for specific genera (and I'd be interested in known these genera). If you bounded $n$, then probably you could use the fact that the moduli space of curves is of general type for large genus to prove this. Fix a genus $g$. Does there exist some $n$ such that $\mathbb{P}^n$ contains a smooth genus $g$ curve as a complete intersection? I'm not really sure if the answer should be yes or no; if it is no, then I'd be interested in knowing which $g$ satisfy this. REPLY [23 votes]: 1) The genus of a complete intersection of multidegree $(d_1,\ldots ,d_{n-1})$ in $\mathbb{P}^n$ is $g=1+\frac{1}{2} d_1\cdot \ldots \cdot d_{n-1}(\sum d_i-n-1)$ (just compute the degree of the canonical bundle). This gives very particular values for $g$: $0, 1, 3, 4, 5, 6, 9, 10, 13, 15, 16,\dotsc $ . Any curve whose genus is not in this list cannot be realized as a complete intersection. 2) Even if $g$ is in that list, for $g>5$ a general curve of genus $g$ cannot be realized as a complete intersection, since the number of moduli of such complete intersection is smaller than $3g-3$ (the number of moduli of a general curve of genus $g$).<|endoftext|> TITLE: If $B\subseteq A$ are free & finite rank $R$-algebras, is $R\to A \otimes_B R$ injective? QUESTION [18 upvotes]: (In this question, all rings and algebras are commutative with identity.) I have a situation that boils down to the following data: a ring $R$, an $R$-algebra $A$ with a subalgebra $B$ such that $A$ and $B$ are free of finite rank as $R$-modules, and an $R$-algebra homomorphism $B\to R$. (Update: If it helps, you can assume that the quotient $R$-module $A/B$ is also free.) In this way, $A$ and $R$ are both $B$-algebras, and we can form their tensor product: $$\begin{array}{ccc}A\otimes_B R & \leftarrow & R\\ \uparrow & & \uparrow \\ A & \leftarrow &B \end{array}$$ The question is: Must the resulting homomorphism $R\to A\otimes_B R$ be injective? In other words, if $I$ is the kernel of $B\to R$, must $IA\cap R=0$? I know that the kernel of $R\to A\otimes_B R = A/IA$ must consist of nilpotents by the following argument: since $A$ is finite as an $R$-module it is integral as an $R$-algebra, and hence also as a $B$-alebra. Since integral extensions have the Lying Over property, the map of schemes $\mathrm{Spec}(A)\to \mathrm{Spec}(B)$ is surjective. Surjectivity is preserved by base extension, so the map of schemes $\mathrm{Spec}(A\otimes_B R) \to \mathrm{Spec}(R)$ is surjective. And a morphism of affine schemes has dense image if and only if the kernel of the corresponding ring homomorphism consists of nilpotents. So under the weaker assumption that $A$ merely be integral over $R$, I already find that the map $R\to A\otimes_B R$ is injective modulo nilpotents in $R$. Does the extra assumption that $A$ and $B$ be free of finite rank as $R$-modules imply that the kernel is $0$? Update: Here are a couple of similarly plausible statements with counterexamples, in case it helps find a counterexample to the original question: (False) Suppose $R$ is a ring, $A$ is an $R$-algebra, and $B\subseteq A$ is a subalgebra of $A$, such that $A$ and $B$ are both free as $R$-modules. Let $B\to R$ be an $R$-algebra homomorphism. Then $R\to A\otimes_B R$ is injective. A counterexample is given by setting $B=R[x]$ and $A=R[x,x^{-1}]$. If we choose the homomorphism $B\to R$ sending $x\mapsto 0$, then the tensor product $A\otimes_B R$ is the zero ring. The original question asks whether an example of such $A$ and $B$ exists with $A$ and $B$ free of finite rank as $R$-modules. (False) Suppose $A$ is a ring and $B$ is a subring of $A$ such that $A$ is finitely generated as a $B$-module. For any $B$-algebra $R$, the homomorphism $R\to A\otimes_B R$ is injective. Here's a counterexample (due to Hendrik Lenstra): let $A=\mathbb{Z}[x]/(x^2-x)$ and $B=\mathbb{Z}[y]/(y^2-2y)$; the homomorphism $B\to A: y\mapsto 2x$ is injective. However, tensoring with the $B$-algebra $B/(2)$, we obtain a non-injective map $B/(2)\to A/(2): y\mapsto 0$. The original question asks for an example in which the map $B\to R$ is a section of an $R$-algebra structure on $B$ making $B$ and $A$ free of finite rank as $R$-modules. Update: For a positive result (also from conversations with Hendrik Lenstra), the statement holds if $B$ is generate by a single element as an $R$-module, which (since $B$ is also free of finite rank as an $R$-module) implies that $B$ is of the form $R[x]/(f(x))$ with $f$ monic. The argument goes like this: Say $f$ has degree $n$ and the homomorphism $B\cong R[x]/(f(x))\to R$ sends $x$ to $r$; then by choosing the generator $x-r$ of $B$ instead of $x$, we may assume $x\mapsto 0$. In that case, the kernel $I$ of $B\to R$ is $(x)$, so we find that $f(x)\in R[x]$ is divisible by $x$; write it as $f(x) = x\cdot g(x)$ for some monic polynomial $g$ of degree $n-1$. Now suppose we have something in $IA\cap R$: it will be an equation $xa = s$ for some $a\in A$ and $s\in R$. Multiplying both sides by $g(x)$, we obtain $s\cdot g(x) = xg(x) a = f(x) a = 0$ in $A$, since $f(x)=0$ in $B$. But $s\cdot g(x)=0$ is an $R$-linear relation among $\{1,x,\dots,x^{n-1}\}$, which form an $R$-basis of $B$, so all the coefficients must be zero. But $g$ is monic, so the coefficient of $x^{n-1}$ in $s\cdot g(x)$ is $s$, so $s=0$. Thus $IA\cap R=0$, so $R\to A/IA \cong A\otimes_B R$ is injective. However, finite algebras are generally not monogenic, so there may still be counterexamples. REPLY [7 votes]: The answer to the question is "No": here is a counterexample due to Bas Edixhoven: Let $k$ be a field and $R=k[t]/(t^2)$. Define $B = R[x,y]/(x^2,xy,y^2)$ and $A =R[x]/(x^2) \times R[y]/(y^2)$; the injection $B\hookrightarrow A$ sends $x\mapsto(x,0)$ and $y\mapsto(0,y)$. Then $B$ and $A$ are free $R$-modules of rank $3$ and $4$, but if we define a homomorphism $B\to R$ sending $x\mapsto t$ and $y\mapsto t$, the map $R\to A\otimes_B R$ is not injective. Indeed, let $I=(x-t,y-t)\subset B$ be the kernel of $B\to R$, so that $R = B/I$ and $A\otimes_B R = A/IA$. Then to check whether $R\to A/IA$ is injective, we need only check whether $R\cap IA = 0$ inside $A$. But $IA$ contains $(x-t)(0,1) = (0,-t)$ and $(y-t)(1,0)=(-t,0)$, so also contains their sum $(-t,-t)=-t$ which is a nonzero element of $R$.<|endoftext|> TITLE: Three old questions on the Sacks forcing QUESTION [18 upvotes]: I came across the two following Qs in 1970. Find reals $a,b$ such that $a$ is Sacks over $L[b]$ and vice versa $b$ is Sacks over $L[a]$. Note that a Sacks $\times$ Sacks generic pair definitely does not work. Suppose that $a$ is Cohen over $L$ and $b$ is Sacks over $L[a]$. Does $a$ belong to $L[b]$? Comment: this Q answers in the positive if $a$ is Sacks or Solovay-random over $L$. The next Q is perhaps not that dead lock, but still I have no clue. Let $\mathbf P$ consist of all perfect sets $P$ on the real plane $\mathbb R\times\mathbb R$ such that all vertical and all horisontal cross-sections of $P$ are perfect (in particular, non-countable). What does it force? REPLY [9 votes]: For (2), I suggest writing $a$ as the Turing join of its even and odd parts, call them $a_0$ and $a_1$. Use $a_0$ to compute a perfect subtree $T$ of $2^{<\omega}$ such that every infinite path in $T$ is mutually Cohen generic with $a_1$. When considered as a Sacks condition in $L[a]$, $T$ forces that $a$ is not an element of $L[b]$.<|endoftext|> TITLE: Differential Geometric Aspects of Rubber Bands QUESTION [5 upvotes]: What happens, if a rubber band ( of length $l_0$ that has been stretched to length $l_1:=l_0+\Delta l\;$ and brought into the shape of a closed curve in $\mathbb{R}^3$ ) is released and if the only force at work is due to the elongation and obeys Hooke's law? Clarification in response to Andreas' comment: The force due to stretching shall be the same in each point of the rubberband. Clarification in response to Hansen's comments: for the mathematical discussion of the problem, it shall be assumed that $l_0=0$, so that the contraction doesn't stop as long as the rubber band has positive length. Furthermore, I would like the mathematical question to be discussed on basis of points and vectors of $\mathbb{R}^3$; physical phenomena like moment of inertia, or bending energy, etc., shall not play a role in this context. I acknowledge however, that trying to model real-world rubber bands is also an interesting question to be tackled, after the questions related to the (over-)simplified model have been solved. By "what happens", I mean what kind of surface is traced out by an infinitely thin rubber band of infinitesimal small initial length $l_0$? what are the coordinates of the point, to which the rubber band contracts, as its length tends to 0? what are the trajectories of the points on the rubber band during contraction? REPLY [5 votes]: Let me give a shot at a partial answer, and provide a model of the problem that I hope is physically sensible. First, since we're actually dealing with a simplification of elasticity theory, and Hooke's Law can be viewed as a specific instance of linearized elasticity, this means that the stored energy function (i.e. potential energy density) is quadratic in strains away from the rest state of the rubber band. Let me here assume that the band has rest length $l_0 = 2\pi$ and that it is parametrized by $\phi\colon S^1 \to \mathbb{R}^3$. Then the stored energy is reasonably given by $$ W(s) = \frac{k}{2}(\|\phi'(s)\| - 1)^2. $$ Assuming a uniform mass density $\rho$, we have a Lagrangian $$ L(\phi,\dot{\phi}) = \int_0^{2\pi} \frac{\rho}{2}\|\dot{\phi}(s)\|^2 - \frac{k}{2}(\|\phi'(s)\| - 1)^2 \;ds. $$ Since $L$ is invariant under translations (and rotations), the total momentum $$ P = \rho \int_0^{2\pi} \dot{\phi}(s) \;ds, $$ is a conserved quantity. Since $P = 0$ initially, and $P/\rho$ is the change of the center of mass, it follows that the center of mass is constant in time. In particular, if at any time the rubber band contracts to length zero, it will be at its center of mass point. This answers your second question. (Note that this derivation only depended on $L$ being invariant under translations, not the specific stored energy function.)<|endoftext|> TITLE: Reference request: Book of topology from "Topos" point of view QUESTION [12 upvotes]: Question: Is there any book of topology in the modern language of topos theory? Motivation: In "Sheaves in Geometry and Logic" Mac Lane and Moerdijk say: "For Grothendieck, topology became the study of (the cohomology of) sheaves, and the sheaves "sited" on a given Grothendieck topology formed a topos - subsequently called a Grothendieck topos". my question is about a book of the study of this idea. Relations between geometry and logic. REPLY [19 votes]: "Topology via Logic" is only half way there. It is firmly rooted in classical mathematics and makes no connections with toposes. Mac Lane and Moerdijk is a good suggestion. As for reading it backwards: that is pretty much the aim of my "Locales and Toposes as Spaces" (Chapter 8 in "Handbook of Spatial Logics" (ed. Aiello, Pratt-Hartman, van Bentham), Springer, 2007, pp. 429-496; ISBN 978-1-4020-5586-7). I wanted to guide the reader through the results in Mac Lane and Moerdijk in an order that brings out the "generalized space" idea of toposes. I think it's fair to say that all those fall short of bringing out some of the ideas of algebraic topology that motivated Grothendieck in the first place. I don't know any books to recommend that cover that. Steve Vickers.<|endoftext|> TITLE: Is there an analog of the Barratt-Eccles construction for group-like E_∞-spaces and E_∞-ring spaces? QUESTION [5 upvotes]: The Barratt-Eccles operad is an operad in simplicial sets that provides a particularly nice model of an E∞-operad; algebras in spaces over the Barratt-Eccles operad model E∞-spaces, i.e., homotopy coherent commutative monoids in spaces. It can be described concretely by applying the nerve functor componentwise to an operad in groupoids, which itself is obtained by applying the codiscrete groupoid functor componentwise to an operad Σ in sets such that Σ(n) is the symmetric group of order n and the operadic composition Σ(n)×(Σ(a₁)×⋯×Σ(aₙ))→Σ(a₁+⋯+aₙ) is given by stacking the permutations in Σ(aᵢ) together and composing them with the block permutation in Σ(a₁+⋯+aₙ) induced by the permutation in Σ(n). Here the codiscrete groupoid functor sends a set X to the groupoid with X as the set of objects and exactly one morphism between any pair of objects; it is the right adjoint to the forgetful functor from groupoids to sets that sends a groupoid to its underlying set of objects. I am interested in similarly spirited constructions for various cousins of E∞-spaces. Specifically, I am interested in group-like E∞-spaces, which can be thought of as homotopy coherent commutative groups and are a model for connective spectra. Another interesting case is E∞-ring spaces, which can be thought of as homotopy coherent commutative rings, and are a model for connective E∞-ring spectra. As pointed out by Peter May in his answer, operads cannot model such structures because they do not allow for operations with multiple outputs, e.g., diagonal maps, so a part of the question is what type of structure one should use. For example, simplicial algebraic theories (see http://ncatlab.org/nlab/show/(%E2%88%9E,1)-algebraic+theory) seem to be a viable option. In particular, the Barratt-Eccles construction admits a particularly elegant formulation in terms of a (2,1)-algebraic theory (i.e., a groupoid-valued algebraic theory), see http://ncatlab.org/nlab/show/(2,1)-algebraic+theory+of+E-infinity+algebras. Is there an analog of the Barratt-Eccles construction for group-like E∞-spaces and E∞-ring spaces? REPLY [11 votes]: There are serious problems making your ideas coherent here! The notion of operad was in large part intended to model kinds of algebras whose laws do not involve repeated variables, do not involve diagonal maps. Operads do not even model groups, in particular, and that was intended. Any $E_{\infty}$ operad models all connective spectra, not just connected ones, the essential point being that the zeroth space of the spectrum associated to an $E_{\infty}$-space $X$, no matter how constructed, must be a group completion of $X$. There is no "the" $E_{\infty}$ operad, rather there are many interesting ones. (There is an axiomatization of infinite loop space machines, due to Thomason and myself, that makes this precise.) If you want to model $E_{\infty}$ ring spectra using operads only, you can make a mistake by trying to use just one operad, as I did over 40 years ago, or you can do it right by using two interrelated operads, one for the addition and one for the multiplication, as I also did over 40 years ago. See http://www.math.uchicago.edu/~may/PAPERS/Final1.pdf for a modern recapitulation of that early theory.<|endoftext|> TITLE: Smoothness in Ecalle's method for fractional iterates QUESTION [5 upvotes]: Some four years ago I answered my own question on fractional iteration, concluding that there is a half iterate of sine, that is $f(f(x)) = \sin x,$ which is real analytic for $0 < x < \pi$ but cannot be extended to the complexes in a neighborhood of the origin, there is a stubborn logarithm term. Does the formal power series solution to $f(f(x))= \sin( x) $ converge? Yesterday I did the same thing for $x + x^2,$ where the conclusion is analyticity for $x > 0$ and continuity at $0.$ Having had practice, i made a better tutorial of this answer. How to obtain $f(x)$, if it is known that $f(f(x))=x^2+x$? Let's see, I put a bunch of related material at WEB PAGE. In particular, Jean Ecalle's method is on pages 346-347 and 351-352, references page 517; he did publish as part of Theorie iterative: Introduction a la theorie des invariants holomorphs, J. Math. Pures Appl. vol. 54 (1975), pages 183-258, I have not seen that one... So, here is the question, very natural I think. I have some holomorphic target function $g(z)$ which is real valued along the real line, then has fixpoint $g(0) = 0$ with $g'(0) = 1,$ which is what makes the problem difficult. Then I have a real $C^1$ function $f(x)$ with $f(0) = 0, f'(0) = 1$ and $f(f(x)) = g(x)$ along the real line, constructed with Ecalle's method, so $f$ is holomorphic on segments containing $0 < x < A$ and $-B < x < 0.$ IS IT TRUE that the real function $f(x)$ is $C^\infty?$ I would like that. Put another way, one may easily find the formal power series for a half iterate around $0,$ that is what I did four years ago. It turns out that this series has radius of convergence $0,$ this is Theorem 8.5.3 on page 347 of the KCG book. It would be nice if the formal power series displayed derivatives of all orders for the half iterate, and if Ecalle's solution matched all of that at $0.$ I guess I should not paste these here, but I just put all the directly relevant bits from the book by Kuczma, Choczewski, and Ger onto two jpegs, I can email those. Came out a little smaller than I wanted, don't know why the scanner did that; clear, though. REPLY [6 votes]: I heard back from Prof. Ecalle. If we have real analytic $f(x)$ with $A \neq 0$ and $$ f(x) = x + A x^{p+1} + o(x^{p+1}), $$ then there are natural fractional iterates $g(x),$ any rational or real order, and these are $C^\infty$ at the origin and of Gevrey class $1/p, $ which result from $$ g\left( x^{1/p} \right) $$ being the Laplace transform of an analytic function with no more than exponential growth at $\infty.$ The Gevrey bound is: $$ \left| \frac{g^{(n)}(0)}{n!} \right| < C_0 \; \; C_1^n \; \; (n/p)! $$ where the last item might need the Gamma function if $n/p$ is not an integer. For references, he gave item 7 at http://www.math.u-psud.fr/~ecalle/publi.html and then Example 2 (with $\nu = 1$ in (2.5.18)) on pages 106-107 of item 19, editor Dana Schlomiuk,1993. Google books show page 106 but hides page 107. Not sure the name Gevrey comes up in either publication.<|endoftext|> TITLE: On the positivity of matrices QUESTION [6 upvotes]: For given $n$, the following $n\times n$ real matrix $M=M^{T}$ is called positive, if $x^{T}M x\geq 0$ holds for all non-negative real $x_1,x_2,\cdots,x_n$, where $x=(x_1,x_2,\cdots,x_n)^T$. Notice here, the definition of positive is not the same as usual defination of positive semidefinite. Now the goal is to characterize the set of all such $M$. In particular, I would like to show the following statement: Such $M$ can always written as positive combination of some positive semidefinite matrix and some $E_{ij}$s, where $E_{ij}$ is the $n\times n$ matrix with the only non zero element 1 lies in the position $(i,j)$. REPLY [16 votes]: Such matrices are called copositive in the literature. Moreover, the statement you want to show is known to be false. While I don't recall a counterexample right away, an intuition for this is that it's NP-hard to check copositivity, but computationally easy to check the condition in your statement, via semidefinite programming, see e.g. this text by Pablo Parrilo. EDIT: here is an explicit counterexample: the matrix $$ M=\begin{pmatrix} 1&-1& 1& 1&-1\\ -1& 1&-1& 1& 1\\ 1&-1& 1&-1& 1\\ 1& 1&-1& 1&-1\\ -1& 1& 1&-1& 1 \end{pmatrix} $$ is copositive, but not equal to the sum of a positive semidefinite matrix $P$ and a nonnegative matrix $N$. It is taken from an old paper by P.Diananda in 1962 (Proc. Cambridge Phil.Soc., vol 58(1962)), where it is also shown that $n=5$ is minimal size for which one has such a counterexample. A quick way to see that $M\neq P+N$ is as follows. First of all notice that $N$ can be assumed to have 0s on the main diagonal. This the condition $M=P+N$, that we want to bring to a contradiction, is equivalent to the existence of a positive semidefinite $P$ with all 1s on the diagonal, and satisfying the conditions $M_{ij}-P_{ij}\geq 0$, for all $1\leq i TITLE: Can the graph Laplacian be well approximated by a Laplace-Beltrami operator? QUESTION [8 upvotes]: It seems rather well known that given a Laplace-Beltrami operator $\mathcal{L}_{M}$ on a manifold $M$ we can approximate its spectrum by that of a graph Laplacian $L_{G}$ for some $G$ (where $G$ is usually a triangulation of $M$). See here or here for details. What I am interested in is going in the opposite direction. That is: Given a fixed (finite) graph $G$ is there a way to approximate its Laplacian $L_{G}$ by the Laplace-Beltrami operator $\mathcal{L}_{M}$ of some surface $M$? The motivation for this is that if $G$ is a sufficiently dense grid, then I can take $M=\mathbb{R}^{2}$. Now suppose I add an edge to $G$ - it feels right that there should be a way to modify $M=\mathbb{R}^{2}$ into some new $M^{'}$ by somehow folding the manifold appropriately. REPLY [6 votes]: Yes, embed the graph into a compact Riemannian surface with a Riemannian metric which is concentrated near the embedded praph and induces the graph distances on the graph. Away from a tubular neighborhood of the graph the metric should be uniformly small. Then the Laplace Beltrami operator of the the surface approximates the graph Laplacian. This method was used by Yves Colin de Verdiere many times. See for example: MR0932800 (90d:58156) Colin de Verdière, Yves(F-GREN-F) Construction de laplaciens dont une partie finie du spectre est donnée. (French) [Construction of Laplacians for which a finite subset of the spectrum is given] Ann. Sci. École Norm. Sup. (4) 20 (1987), no. 4, 599–615.<|endoftext|> TITLE: A random variation on Polya's orchard problem QUESTION [10 upvotes]: Polya's orchard problem is as follows: "How thick must the trunks of the trees in a regularly spaced circular orchard grow if they are to block completely the view from the center?" See, e.g., Allen, Thomas Tracy. "Polya's orchard problem." American Mathematical Monthly (1986): 98-104. JSTOR link or this earlier MO question. This problem has been well-studied. One crude way to phrase what is known is that, if the trees/disks centered at each lattice point have radius $r$, the furthest distance one can see from the origin is $R \approx 1/r$. Here is my question. Suppose instead of disks centered on each lattice point, we have a randomly oriented segment (a $1$-dimensional disk) centered on each lattice point, of length $2r$, i.e., of radius $r$. Q. Is it still the case that the furthest one can see from the origin is expected to be $R=c/r$ for some constant $c$? Here is an example, with $r=3/8$. The "Polya radius" is about $2.47$, but in this one random instance, visibility extends about $3$ times further, $R \approx 8$:   In $\mathbb{R}^3$, the same question can be asked with now $2$-dimensional randomly oriented disks centered on each lattice point. And of course the question generalizes to $\mathbb{R}^d$. REPLY [5 votes]: Probably this problem was not considered so far, but there are some related articles. "The distribution of free path lengths in the periodic Lorentz gas and related lattice point problems" by Jens Marklof, Andreas Strömbergsson. In partiqular they give the number of spheres in a random direction. "Visibility and directions in quasicrystals" by Jens Marklof, Andreas Strömbergsson. (In each direction points which are closed to a straight line form a "quasicrystal".) 3."Perfect Retroreflectors and Billiard Dynamics" by Pavel Bachurin, Konstantin Khanin, Jens Marklof, Alexander Plakhov See also Marklof's talk at ICM-2014 (lecture I09_04) for an introduction. I think that this authors can solve your problem in arbitrary dimension.<|endoftext|> TITLE: Quotient of Projective line over rationals with an infinite subgroup of PGL(2,Q) QUESTION [5 upvotes]: I am looking for references for the following; how to calculate quotient of the projective line over the field of rationals with an infinite subgroup of PGL(2,Q), e.g, of the form $ \left( \begin{array}{cc} a+b & -a \\ a & b% \end{array}% \right), $ where a and b are rational integers. By calculate I mean calculation of the coordinate ring of the quotient or any other insight into the situation. REPLY [3 votes]: The matrix in the question is the representing matrix of multiplication with $b-a\zeta_3$ on $\mathbb{Q}(\zeta_3)$, for the basis $\{\zeta_3,1\}$. Therefore, the group of matrices in the question is $R_{\mathbb{Q}(\zeta_3)/\mathbb{Q}}(\mathbb{G}_m/\mathbb{Q}(\zeta_3))$, the Weil restriction of the multiplicative group from $\mathbb{Q}(\zeta_3)$. The action of the matrix group in the question on $\mathbb{P}^1(\mathbb{Q})$ is transitive, via direct computation. After base change to $\mathbb{Q}(\zeta_3)$, the action is conjugate to the standard action of $\mathbb{G}_m^2/\mathbb{Q}(\zeta_3)\cong R_{\mathbb{Q}(\zeta_3)/\mathbb{Q}}(\mathbb{G}_m/\mathbb{Q}(\zeta_3))\times_{\mathbb{Q}}\mathbb{Q}(\zeta_3)$ on $\mathbb{P}^1/\mathbb{Q}(\zeta_3)$. So the quotient does not exist as a variety. The same will be true for all other algebraic groups of the form $R_{K/\mathbb{Q}}(\mathbb{G}_m)$ with $[K:\mathbb{Q}]=2$.<|endoftext|> TITLE: A question on certain elliptic PDE QUESTION [5 upvotes]: Consider the elliptic PDE $$(CR)\;\;\;\;\;\;\begin{cases} U_{xx}=V_{yy}\\U_{yy}=-V_{xx} \end{cases}$$ And its consequence $$(LAP)\;\;\;\;\;\;U_{xxxx}+U_{yyyy}=0$$. Somehow, these equations are similar to the classical Cauchi Riemann and Laplace equation. 1.Assume that $U$ satisfies the "LAP" equation. Is there a function $V$ such that the pair $(U,V)$ satisfies the "CR" equations? 2.Is there an infinite dimensional algebra of complex functions $f(z)=U+iV$ which satisfies the "CR" equations?If it is the case, can one find such algebra which is invariant under the operator $D(U+iV)=U_{xx}+iV_{xx}$?(Motivated by classical holomorphic differentiation $U_{x}+iV_{x}$) 3.Is there a geometric way (in term of Reimannian metric) to introduce the above equations on an arbitrary Riemannian manifold. (Similar to the situation of conformal maps and Laplace operator=the composition of div and gradient)?If the answer is yes, what would be the index of "LAP" operator? In the other word, is there a geometric meaning for these equations/ REPLY [2 votes]: All linear functions satisfy your CR equations, so it seems fairly natural that they should be included in any related function algebra $A$. But there is no algebra of $C^1$ functions satisfying the CR equations and including a function apart from the algebra of constant functions. (I assume that the algebra is supposed to be closed under pointwise multiplication.) Let us write $z=x+iy$. Suppose $F(z)=U(z)+iV(z)$ belongs to $A$. Now also $z\mapsto x$ is in $A$, so $f(z)=xF(z)$ is in it. Thus we have $$\begin{cases} U_{xx}=V_{yy}\\U_{yy}=-V_{xx} \end{cases}$$ and $$\begin{cases} 2U_x+xU_{xx}=xV_{yy}\\xU_{yy}=-2V_x-xV_{xx} \end{cases}.$$ We thus have $U_x=V_x=0$ and we can get the same result for $x$ replaced by $y$, whence $F$ is constant. Remark: Suppose $F=U+iV$ and $f=u+iv$ solve the CR equations. If their product solves it too, we get $$\begin{cases} U_xu_x-V_xv_x=U_yv_y+V_yu_y\\-U_yu_y+V_yv_y=U_xv_x+V_xu_x \end{cases}.$$ Therefore any pair of functions in your algebra should satisfy these equations. I don't know if there is a nontrivial algebra if linear functions are excluded.<|endoftext|> TITLE: Examples of continuous differential equations with no solution QUESTION [10 upvotes]: Several theorems regarding differential equations are true not only in finite dimension but also in infinite dimension Banach spaces. This is in particular the case for the Cauchy–Lipschitz theorem. On the other end the Peano existence theorem is false for Banach space with infinite dimensions. See here for a counterexample. Do you know other counterexamples in "classical" Banach spaces that are different from $c_0$ (the space of sequences of reals converging to $0$)? In particular, is there "an easy example" in the space $C([0,1],\mathbb{R})$ with $\sup$ norm? REPLY [11 votes]: Here is an example in $C([-1,1],R)$, which is a continuous analogue to the discrete example you pointed to: $$ {du(t,x)\over dt} = \operatorname{sign}(u(t,x))\sqrt{|u(t,x)|} + x\;,\qquad u(0,x) = 0\;. $$ For any $t > 0$, the solution (in $L^\infty$) develops a discontinuity at the origin, so that it doesn't belong to $C([-1,1],R)$.<|endoftext|> TITLE: A question on BSD conjecture QUESTION [7 upvotes]: If $E$ is an elliptic curve over $\mathbb{Q}$, and $K$ is an imaginary quadratic field. If $rank E(K)\leq 1$ and both $E$ and the quadratic twist of $E$ by $K$ satisfy the full BSD conjecture, does the base change of $E$ to $K$ satisfy the full BSD conjecture? REPLY [10 votes]: Yes. This follows from the fact that BSD is invariant under Weil restriction and isogeny (the Weil restriction of $E/K$ to $\mathbb{Q}$ is isogenous to the product of $E$ with its quadratic twist). Note that you don't need to assume here that $K$ is imaginary nor anything about the rank of $E/K$.<|endoftext|> TITLE: Optimal lower bounds for the sum of digits in base $b$ QUESTION [6 upvotes]: Let $b \geq 2$ be an integer and let $s_b(n)$ be the sum of the digits of the base-$b$ representation of the nonnegative integer $n$ (e.g., $s_{10}(726)=7+2+6$). From the weak law of large numbers, it follows that $$s_b(n) > \left(\tfrac{0+1+\cdots + (b-1)}{b}-\varepsilon\right) \log_b n$$ for almost all the positive integers $n \leq x$, with at most $o(x)$ exceptions, as $x \to +\infty$. Therefore, if $\{a_k\}_{k=1}^\infty$ is a "generic" monotone increasing sequence of positive integer, it is quite reasonable to expect that $s_b(a_k) > C \log(a_k)$ for almost all $k$, i.e., with the exception of a set of asymptotic density 0, where $C$ is a positive constant (depending on the sequence $\{a_k\}_{k=1}^\infty$). By "generic" I means, informally speaking, that $\{a_k\}_{k=1}^\infty$ has not some trivial properties that makes the statement false (e.g., $a_k := b^k$). So for example, we can conjecture that $$s_b(n!) > C_1 n \log n$$ $$s_b(a^n) > C_2 n$$ $$s_b(F_n) > C_3 n$$ for almost all $n$, where $a$ is a positive integer coprime to $b$ and $F_n$ is the $n$-th Fibonacci number. In [1] and [3] they have been shown the weaker lower bounds $s_b(n!) > C_1 \log n \log \log \log n$ (see also [2]) and $s_b(F_n) > C_2 \log n / \log \log n$, for all integers $n$, respectively. Moreover, in [4] the bound $s_b(a_n) > C \log n$ for almost all $n$, has been proved for any sequence of integers $$a_n = e^{f(n)} (1 + O(n^{-\alpha}))$$ where $\alpha > 0$ and $f(x)$ is a two times differentiable function satisfying $f(x) \asymp 1/x$ for large x. My question is: Are there some non(too)trivial example of $\{a_k\}_{k=1}^\infty$ such that the optimal lower bound $s_b(a_k) > C \log(a_k)$, for almost all $k$, has been proved? Thank you very much for any suggestion/reference. [1] C. Sanna, On the sum of digits of the factorial, J. Number Theory 147, 2015, 836--841. [2] F. Luca, The number of nonzero digits of n!, Canad. Math. Bull., 45, 2002, 115--11. [3] F. Luca, Distinct digits in base b expansions of linear recurrence sequences, Quaest. Math., 23, 2000, 389--404. [4] J. Cilleruelo, F. Luca, J. Rué and A. Zumalacárregui, On the sums of digits of the some sequences of integers, Cent. Eur. J. Math., 11, 2013, 188--195. REPLY [2 votes]: The first paper you reference concerns simply the number of non-zero digits of $n!$ and not their sum. Of course that just changes the constant in front. That paper does note that the bound is quite weak compared to what one might expect. For example in the case $b=2$ his bound assures us that $n!$ has at least $7$ non-zero digits provided that $n$ is at least $767$ (by which point $n!$ has many thousand binary digits.) That is enough to allow one to establish that $n=9$ is actually the last case with as few as $6$ bits equal to $1.$ Then $9!=1011000100110000000_2$ So what can be proved is certainly far weaker than what one might expect. However the case of $n!$ is certainly not generic. It is entirely predictable that there are $\lfloor 9/2 \rfloor+\lfloor 9/4 \rfloor+\lfloor 9/8 \rfloor=4+2+1$ $0$'s on the right. The other $12$ leading digits might be expected to split roughly equally and in fact split exactly equally.<|endoftext|> TITLE: $RUCar^{V}$-semiproperness implies properness QUESTION [6 upvotes]: This is a claim in Shelah's Proper and Improper Forcing, more specifically Claim 2.3(1) of Chapter X (p. 484). The proof of the claim is "Easy" but I cannot quite figure it out. There must be some trick that I am missing. A forcing notion $P$ is $RUCar^{V}$-semiproper iff there is a cardinal $\kappa$, such that for all cardinal $\lambda \geq \kappa$, for all $N$ countable elementary submodel of $(H(\lambda), \in)$ with $P \in N$, if $p \in P \cap N$ then there is $q$ ($\in P$) extending $p$ satisfying the following: for every regular uncountable cardinal $\theta \in N$ and $P$-name $\pi \in N$ of an element of $\theta$, $q \Vdash_{P}$ "there is an ordinal $\alpha \in N$ with $\pi < \alpha < \theta$". Shelah claims every $RUCar^{V}$-semiproper forcing notion is proper. Is it really easily observable? REPLY [8 votes]: Given such an $N$ and $q$, you want to see that $q$ is $(N, P)$-generic. This is equivalent to $q$ forcing $N[G]\cap On = N\cap On$. (Corollary 2.13 page 106 of Proper and Improper Forcing) Suppose $\dot\alpha$ is a $P$-name for an ordinal with $\dot\alpha\in N$. Then $\{\xi\in On: \text{ some $r\in P$ forces $\dot\alpha$ to equal $\xi$}\}$ is in $N$, and its supremum is in $N$ as well. Thus, there is an ordinal $\xi\in N$ such that $\Vdash_P \dot\alpha<\xi$. Now let $q$ be as in your question, and let $G$ be a generic subset of $P$ containing $q$. Assume by way of contradiction that $N\cap On $ is a proper subset of $N[G]\cap On$. Let $\alpha=\min(N[G]\cap On\setminus N)$, and let $\beta$ be $\min(N\cap On\setminus\alpha)$. Notice that $\beta$ exists by our discussion above. We know that $N\cap\alpha = N[G]\cap\alpha$ by choice of $\alpha$ and using this it is not hard to see that $\beta$ must be a regular cardinal. Now our assumption on $q$ will give us an ordinal in $N\cap\beta$ above $\alpha$, contradicting the choice of $\beta$.<|endoftext|> TITLE: Finite subgroups of mapping class groups QUESTION [14 upvotes]: Given a closed, oriented surface $\Sigma$ of genus greater than 1, let $Mod(\Sigma)$ denote the mapping class group of orientation preserving diffeomorphisms of $\Sigma$ up to isotopy. Given any finite subgroup $F\subset Mod(\Sigma),$ the affirmative solution to the Nielsen Realization problem states that there exists a hyperbolic metric $h$ on $\Sigma$ such that $F$ is isomorphic to the isometry group of $h.$ Now, by Hurwitz's Theorem, the order of $F$ is bounded above by $84(g-1).$ I've always wondered if there was a topological/algebraic method to deduce a uniform bound on the order of finite subgroups of the mapping class group without going through the above mentioned argument. If $Mod(\Sigma)$ were a linear group over a field of characteristic zero, I suppose this would follow immediately from Selberg's Lemma, but I vaguely recall hearing that $Mod(\Sigma)$ has no faithful, finite dimensional linear representations. So, my question is: is there a way to see that finite subgroups of the mapping class group have a uniform bound on their order without using the above argument, and if so how sharp can one make the bounds in this manner. I've long been curious about this question and finally remembered to post it on mathoverflow. As always, thanks for your input. REPLY [12 votes]: You can also use Serre's theorem which says that kernel of the natural homomorphism from the mapping class group of $\Sigma$ to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$ is torsion free, and therefore every finite subgroup injects to $\text{Sp}(2g;\mathbb{Z}/3\mathbb{Z})$. But that gives a polynomial bound of degree $g^2$ compared to $84(g-1)$.<|endoftext|> TITLE: Is the fixed point property for posets preserved by products? QUESTION [31 upvotes]: Recall that a partially ordered set (poset) $P$ has the fixed point property (FPP) if any order preserving function $f:P\longrightarrow P$ has a fixed point. Theorem. Suppose $P$ and $Q$ are posets with the FPP and at least one of them is finite. Then $(P\times Q)$ has FPP. Note: $(a,b)\le(c,d)$ if and only if $a\le c$ and $b\le d$. Question. Suppose $P$ and $Q$ are two infinite posets with the FPP. Does $(P\times Q)$ have the FPP ? REPLY [12 votes]: The question you asked is one of the main long-open problems in fixed point theory of posets. Commenting on Paul Taylor's interesting answer: the property he describes is called strong fixed point property in order theory. In fact, if one of the posets $P,Q$ has the strong fixed point property and the other has usual fixed point property, then the product $P\times Q$ has the fixed point property. (If both of them have the strong fixed point property, then $P\times Q$ also has the strong fixed point property.) This was proved by Duffus and Sauer in 1980. Fixed point property is not equivalent to strong fixed point property even for finite posets (Pickering, Roddy 1992). Some approaches that are not mentioned in Schröder's survey mentioned by Gejza Jenča have been presented by Josef Niederle here and here. EDIT (made following one of Paul Taylor's comments): Unfortunately all of the papers cited above are behind a paywall. However, one can find free conference versions of Niederle's work: here and here.<|endoftext|> TITLE: Internal categories in an endofunctor category QUESTION [6 upvotes]: Here we see the definition of an internal category in a monoidal category. We also know that endofunctor categories support a monoidal product which is actually functor composition. It is the case that a monoid in an endofunctor category is a monad, and so it makes sense that a comonoid is a comonad. We could take the route that an internal category here is a monad on the category of internal comonoids. It also looks to me like, if there is a comonad that is also a monad, we are done, we have an internal category. Does this make sense? Is it the case that if we have Frobenius monads (a monad that can be turned into a comonad) these are the internal categories? REPLY [2 votes]: It seems that there was a bit of truth to this question. Here we see Spivak defining Categories as comonoids in the monoidal category $(Poly, \circ, y)$, section 2.5. As I point out, comonoids in an endofunctor category are comonads. I would guess that if you add a monad structure to your comonoids, you still have a category, but you have extra structure. It would be interesting to know what extra stuff you are endowing the category with if you add monad structure.<|endoftext|> TITLE: Are Anderson $T$-motives motives for the function field analogy? QUESTION [5 upvotes]: this question is related to this one Geometry for Anderson's motives?, though the previous one doesn't answer exactly my question. Let $\mathbb{C}_{\infty}$ be the function field analog of $\mathbb{C}$ for $\mathbb{F}_q (\theta) = \mathbb{Q}_{\infty}$ and $A = \mathbb{C}_{\infty} [T, \tau]$ the Anderson ring ( i.e., $T$ is central and $\tau$ acts as the Frobenius endomorphism). An Anderson $T$-motive is simply a left $A$-module $M$ which is free and finitely generated over $\mathbb{C}_{\infty} [\tau]$, and satisfies $$(T - \theta)^n M/\tau M = \{ 0 \}$$ for some $n > 0$. If, furthermore, $M$ is finitely generated over $\mathbb{C}_{\infty}[T]$ (or equivalently, free of finite rank), then it's called an abelian $T$-motive or $t$-motive. First of all, why the word "motives" in "Anderson $T$-motives"? Are $T$-motives analogous to motives of arithmetic schemes? What's its number field analogue? Thanks in advance. REPLY [6 votes]: This is just a small supplement to Wendt's (as usual) excellent response. Drinfel'd modules (or elliptic modules as Drinfel'd called them) are the function field analogues of elliptic curves. Note that elliptic curves make sense also over function fields. So elliptic modules are exotic analogues of elliptic curves. Similarly, Anderson t-motives are the function field analogues of the Tannakian category of (mixed) motives generated by abelian varieties. The Tannakian word is because the category of t-motives has a tensor product and duals, abelian varieties because the category contains elliptic modules. But even this is not quite accurate. Because only Drinfeld modules of rank two correspond to elliptic curves, and those of rank one correspond to Tate motive (or its twists), and (it is tempting to think) those of rank >2 as corresponding to non-commutative tori (these are not connected to motives, as far as I know). So Anderson t-motives is the function field analogue of the Tannakian category of mixed motives generated by abelian varieties, Artin-Tate motives, and ... Here motives would mean under homological/numerical equivalence. There is no underlying geometry for Anderson's t-motives (varieties whose Chow groups etc could give t-motives). @Wendt: it is only over finite fields that abelian varieties generate the category of motives. this does not hold over number fields.<|endoftext|> TITLE: Ore's Conjecture for perfect groups QUESTION [9 upvotes]: We know that Ore's Conjecture is a theorem now. Does anyone know a counterexample to Ore's conjecture for perfect groups? I believe there should be one, but I dont have a counterexample. Thanks. REPLY [5 votes]: An old paper of mine (Amer. Math. Monthly 1977) gives an easy way to construct groups where not every element of the derived subgroup is a commutator. In particular, if $U$ is a large enough abelian group and $H$ is a simple group, then the derived subgroup of the wreath product of $U$ by $H$ is perfect and contains non-commutators.<|endoftext|> TITLE: What is the geometric fixed points of an (equivariant) Eilenberg Maclane Spectrum? QUESTION [17 upvotes]: The following was posted to math.stackexchange to no avail: https://math.stackexchange.com/questions/908756/an-exercise-in-homology-computation-what-is-the-geometric-fixed-points-of-an-e The question I want to ask has a reasonably elementary formulation and I think there is a good chance it can be answered in this form (by someone more computationally skilled than me, or perhaps by someone recognising the construction). It is, however, motivated by equivariant stable homotopy theory and certainly insights can be drawn from this picture as well. I'm going to state the elementary form first, and then explain background and motivation. For a pointed simplicial set $X_\bullet$ I shall write $\tilde{\mathbb{Z}}[X]$ for the chain complex with $\tilde{\mathbb{Z}}[X]_n = \mathbb{Z}\{X_n\}/*$ by which I mean the free abelian group on $X_n,$ except that we treat the base point as zero. Now let $G$ be a finite group. I write $X^{\wedge G}$ for the simplicial $G$-set $X \wedge X \wedge \dots \wedge X$ ($|G|$ copies of $X$) with $G$ acting by permuting factors. Note that if $X_\bullet$ is a pointed simplicial $G$-set then $\tilde{\mathbb{Z}}[X]$ has a natural $G$-action. Finally let $S^n = \Delta^n/\partial \Delta^n$ be the $n$-sphere (naturally pointed!). Question: What is $H_{n,k} := H_{n+k}\left( \left[\tilde{\mathbb{Z}}[(S^n)^{\wedge G} \right]^G \right)$? Here the outer-most $[]^G$ means $G$-invariants in the chain complex, and $H_*$ is just homology groups. One may show that there is a stabilisation map $H_{n,k} \to H_{n+1,k}$. Write $H_{\infty, k}$ for the colimit. This is the group I'm really after. Some background: Recall that there is a tensor triangulated category $SH(G)$, the $G$-equivariant stable homotopy category. This has two structures I am interested in here: firstly there is a tensor triangulated functor $\Phi^G: SH(G) \to SH$, the geometric fixed points functor. Secondly, the category has a natural $t$-structure whose heart is equivalent to the category of mackey functors. If we write $\underline{\mathbb{Z}}$ for the constant mackey functor with value $\mathbb{Z}$ and $H\underline{\mathbb{Z}}$ for the corresponding Eilenberg-MacLane spectrum, then what I would really like to understand is $\Phi^G(H\underline{\mathbb{Z}}).$ If I interpret correctly example 2.13 and paragraph 7.3 in [1], then $H_{\infty, k} = \pi_k \Phi^G(H\underline{\mathbb{Z}})$, hence my question. Some results: Let $H$ be a proper subgroup of $G.$ Then $\Phi^G (G/H_+ \wedge S_G) = 0$. The transfer $H\underline{\mathbb{Z}} \to G/H_+ \wedge H\underline{\mathbb{Z}} \to H\underline{\mathbb{Z}}$ is multiplication by $|G:H|.$ Hence multiplication by $|G:H|$ is zero on $\Phi^G(H\underline{\mathbb{Z}})$ and so also on $H_{\infty, k}.$ In particular $H_{\infty, k} = 0$ unless $G$ is a $p$-group, in which case $H_{\infty, k}$ is $p$-torsion. One may work out $\tilde{\mathbb{Z}}[(S^n)^{\wedge G}]$ fairly explicitly, but the combinatorics of the resulting chain complex is a big mess. Using some standard results about lifting homotopies, one may prove that in the definition of $H_{n,k}$ we can replace $\left[\tilde{\mathbb{Z}}[(S^n)^{\wedge G} \right]^G$ by $(C_\bullet^{\otimes G})^G$ for any simplicial free abelian group $C_\bullet$ with $H_* C = H_* S^n,$ for example the Dold-Kan inverse to the chain complex $\mathbb{Z}[n]$, i.e. $C_k = 0$ for $k < n$, $C_n = \mathbb{Z}$ and the higher degree groups are just freely added degeneracies. Hence in particular $H_{n,k} = 0$ for $k < 0.$ But the resulting complex still does not seem particularly amenable to computation (it is related to the orbits of $G$ on $G$-sets of the form $X^{\times G}$ where $X$ has trivial $G$ action). I believe I can use this description to show that $H_{n,0} = \mathbb{Z}/p.$ Some wishful thinking led me to ask in the stackexchange post if in fact $H_{\infty, k} = 0$ for all $k > 0.$ I'm thinking now that this may be wrong. In fact using paper [2] on "derived mackey functors" I seem to have convinced myself that for $G = C_2$ (the unique group of order two), $H_{\infty,*} = \mathbb{Z}/2[t]$ where $t$ is placed in degre $2$ (i.e. the stabilised homology is two-periodic). Better take this computation with a grain of salt, though. [1] http://www.math.uni-bonn.de/people/schwede/equivariant.pdf [2] http://arxiv.org/abs/0812.2519 REPLY [17 votes]: Your identification with the geometric fixed points and the calculation for $C_2$ is correct. As Akhil remarked, one obtains the same answer for $C_{2^n}$ and this is calculated in Hill-Hopkins-Ravenel. In fact, for $G=C_{p^n}$, the same method shows $\Phi^G H\mathbb{Z}=\mathbb{Z}/p[t]$ where $|t|=2$. The key ingredients of this calculation are: All proper subgroups of $G$ are contained in the kernel of a non-trivial map $$\psi\colon C_{p^n}\rightarrow C_p.$$ The geometric fixed points $\pi_* \Phi^{C_p}H\mathbb{Z}=\pi_*^GE\tilde{P}\wedge H\mathbb{Z}$ can be calculated as $H_*^{C_p}(S^{\infty \rho};\mathbb{Z})$. Here $\rho$ is a non-trivial one dimensional complex representation of $C_p$. This is because $S^{\infty\rho}\simeq E\tilde{P}_{C_p}$, (which follows from calculating the fixed points and that $E\tilde{P}_{C_p}$ is determined up to weak equivalence by its fixed points). By pulling back along $\psi$ and applying 1. We see that $\psi^*S^{\infty \rho}\simeq E\tilde{P}_{C_{p^n}}$ and because we are using constant Mackey functor coefficients we have $H_*^{C_{p^n}}(\psi^*S^{\infty\rho};\mathbb{Z})\cong H_*^{C_p}(S^{\infty \rho};\mathbb{Z})$. This follows explicitly from the characterization of the homology you have given. The unit sphere space $S(\infty\rho)$ is a model for $EC_p$ (again look at the fixed points) and $H_*^{C_p}(S(\infty\rho)_+;\mathbb{Z})\cong H_*(BC_p;\mathbb{Z})$. This follows from the definition of the Mackey functor homology as a coend (See May: Equivariant homotopy and cohomology theory). Finally the cofiber sequence $$S(\infty\rho)_+ \rightarrow S^0\rightarrow S^{\infty\rho}$$ induces a long exact sequence in homology. In degree 0 the left hand map induces a transfer map on chain complexes, because it is induced by a projection map $C_p\rightarrow C_p/C_p$. The transfer induces multiplication by the index $$\mathbb{Z}\xrightarrow{|C_p|}\mathbb{Z}$$ and we see that $\pi_0 \Phi^G H\mathbb{Z}$ is $\mathbb{Z}/p$. These arguments determine the graded group structure. To get the ring structure one analyzes what happens in homology when you smash copies of $S^\rho$ together: the orientation class $t$ in $H_2^{C_p}(S^\rho;\mathbb{Z})$ gives a polynomial generator. For more general groups, I can say the following. For non-trivial $p$-groups one can apply a similar argument but one needs to make some changes. First replace $\rho$ with $V$ where $V=\oplus_{i\in I} V_i$ is a direct sum of representations indexed over the set $I$ of non-trivial maps $G\rightarrow C_p$. The representation $V_i$ is obtained by pulling back the representation $\rho$ from above along the corresponding map. If we use all such maps then $S^{\infty V}$ is a model for $E\tilde{P}$. The calculation of the degree 0 term above can again be applied to show that $\pi_0 \Phi^G H\mathbb{Z}=\mathbb{Z}/p$. More generally I believe one gets $\pi_0 \Phi^G HR=R(G)/(\mathrm{Ind}_H^G R(H))_{H TITLE: Are there superexponential Pfaffian functions? QUESTION [6 upvotes]: This question is motivated by model theory, but it's really an analysis question (which means it may have an easy analysis answer that I just don't have the background for). Here's the main question, followed by some definitions and motivation: Is there a partial function $f:\mathbb R\to \mathbb R$, defined on an unbounded interval $(a,\infty)$, which is both superexponential and Pfaffian? I suspect if there is a solution to the above, there is also a totally defined function, but that's the form of the question as it's useful to me. A function $f$ is superexponential if, for all $n$, there is an $a$ where for all $x>a$, $f(x)>\exp(\exp(\cdots \exp(x)\cdots))$, where $\exp(x)$ is $e^x$, and there are $n$ iterates of $exp$ in the preceding expression. Equivalently (for model theorists), a function is superexponential if and only if it is not eventually bounded by any function definable in $(\mathbb R, +,\cdot,0,1,\exp)$. It is possible to construct (real)-analytic superexponential functions without too much trouble; see this answer by fedja to a Math.StackExchange question to this effect. A Pfaffian chain is a sequence of unary functions $(f_1,\ldots,f_n)$ such that for each $k$, there is a real polynomial $p_k(x,y_1,\ldots,y_k)$ where $f_k'=p_k(x,f_1,\ldots,f_k)$ on a tail of $\mathbb R$. A function is Pfaffian if it is part of a Pfaffian chain. So for example, all polynomials are Pfaffian, as is the exponential function. However, $\sin$ and $\cos$ are not Pfaffians, since each one would need to "reference the other." For motivation see another question, where this question was raised in a comment by Joel David Hamkins; if such a function exists, it would generate a superexponential o-minimal theory, resolving an interesting open question. However, a negative answer, while interesting, would not resolve that question. I fiddled with this formulation for a few months but eventually gave up. REPLY [7 votes]: No, [1] proved more generally that the Pfaffian closure of any o-minimal polynomially bounded structure is exponentially bounded. [1] Jean-Marie Lion, Chris Miller, and Patrick Speissegger, Differential equations over polynomially bounded o-minimal structures, Proc. Amer. Math. Soc. 131 (2003), 175–183.<|endoftext|> TITLE: Vanishing eigenvalues of Jacobian QUESTION [13 upvotes]: Let $f: \mathbb{R^2}\to \mathbb{R^2}$ be a Schwartz function. If the eigenvalues of $Df$ vanish everywhere, must $f$ be constant? Does an analogous result hold when we replace $2$ by $n$? Any properties of $f$ would be of interest. For example, clearly $\nabla \cdot f = 0$. REPLY [11 votes]: For some reason, the 'edit' button didn't appear for my earlier answer, maybe because it was already accepted. Thus, I'm adding the general $n$ argument as a separate answer. In fact, there is a stronger result: Suppose that $f:\mathbb{R}^n\to\mathbb{R}^n$ is smooth and has the properties that (i) the Jacobian $Df(x)$ has no negative real eigenvalues for any $x\in \mathbb{R}^n$ and (ii) there exist $\epsilon>0$ and a constant $K$ such that $$ \bigl| f(x)\bigr| \le \frac{K}{\bigl(1+|x|\bigr)^\epsilon} \qquad \forall x\in\mathbb{R}^n. $$ Then $f\equiv0$. Here is the argument: For $t\ge 0$, consider the smooth mapping $\Phi_t:\mathbb{R}^n\to\mathbb{R}^n$ defined by $$ \Phi_t(x) = x + tf(x). $$ Of course, $\Phi_0(x) = x$. Moreover, $$ \det(D\Phi_t(x)) = \det\bigl(I_n + t Df(x)\bigr) > 0 $$ by the hypothesis that $Df(x)$ has no negative real eigenvalues for any $x$. Thus, $\Phi_t:\mathbb{R}^n\to\mathbb{R}^n$ is a local diffeomorphism for all $t$. In particular, all values of $\Phi_t$ are regular values. Moreover, when $|x|=R>0$, $$ \bigl|\Phi_t(x)-x\bigr| \le \frac{tK}{(1+R)^\epsilon}, $$ and, by taking $R$ very large, it follows easily that $\Phi_t$ must be surjective (since it barely moves the sphere $|x|=R$ for $R>>0$). Then, by degree theory (and the fact that $\Phi_t$ is an orientation-preserving local diffeomorphism), it follows that $\Phi_t$ must be injective as well. I.e., $\Phi_t$ is a diffeomorphism of $\mathbb{R}^n$ with itself for all $t\ge 0$. Suppose now that $f$ were not identically zero. By translation, I can assume, without loss of generality, that $f(0)\not=0$, say $\bigl|f(0)\bigr| = M > 0$. Now, choose an $R>>0$ satisfying $$ \frac{K}{(1+R)^\epsilon} < M, $$ which is possible because $\epsilon>0$, and then choose $t>>0$ so that $$ t\left(M-\frac{K}{(1+R)^\epsilon}\right) > R. $$ Then, for all $x\in\mathbb{R}^n$ with $|x|=R$, $$ \left|\Phi_t(x)\right| = \left|x + tf(x)\right| \le R + \frac{tK}{(1+R)^\epsilon} < t M. $$ In particular, $\Phi_t$ carries the sphere $|x|=R$ into the ball $|x|\le R'$ for some $R' \lt tM$. Thus, because $\Phi_t$ is a diffeomorphism of $\mathbb{R}^n$ with itself, it must carry the ball $|x|\le R$ into the ball $|x| \le R'$. On the other hand, $\left|\Phi_t(0)\right| = \left|tf(0)\right| = tM > R'$, and this is a contradiction.<|endoftext|> TITLE: A "good scale" that is not really a scale QUESTION [6 upvotes]: I don't know much about singular cardinal combinatorics, so I apologize in advance if I write something that is wrong or looks funny. First let me recall some basic definitions. Let $\lambda$ be a singular cardinal and let $(\kappa_i : i < \text{cf}(\lambda))$ be a sequence of regular cardinals cofinal in $\lambda$. A sequence $(f_\alpha : \alpha < \lambda^+)$ of functions $f_\alpha \in \prod_{i < \text{cf}(\lambda)} \kappa_i$ is called a scale if it is: (1) increasing modulo $<^*$, where $f <^* g$ means that $f(i) < g(i)$ for all but a bounded set of $i < \text{cf}(\lambda)$, and (2) cofinal in $\prod_{i < \text{cf}(\lambda)} \kappa_i$ modulo $<^*$. A scale is said to be good if it satisfies the additional property (3) for club many $\alpha < \lambda^+$, if $\text{cf}(\alpha) > \text{cf}(\lambda)$ then $\alpha$ is a good point, meaning that there is an unbounded set $S \subset \alpha$ and an ordinal $i < \text{cf}(\lambda)$ such that the sequence $(f_\alpha \restriction [i,\text{cf}(\lambda)): \alpha \in S)$ is pointwise increasing. The proof of the fact "if $\lambda$ is a singular cardinal and there is a strongly compact cardinal between $\text{cf}(\lambda)$ and $\lambda$, then there is no good scale of length $\lambda^+$" seems to use only properties (1) and (3) but not property (2). So my questions are: (A) Do sequences of length $\lambda^+$ satisfying (1) and (3) have a name (e.g. "good pseudo-scale")? (B) Is their existence equivalent to the existence of good scales? REPLY [5 votes]: Your pseudo-good scale is indeed equivalent to a good scale, although the good scale possibly lives on a different product. This follows easily from the construction of a scale given in Cummings' Notes on Singular Cardinal Combinatorics. A basic theorem of Shelah: If a $<^*$-increasing sequence of functions has stationary many good points of cofinality $\kappa$ for some $\kappa<\lambda$, then it has an exact upper bound $g$ such that $\mathrm{cf}(g(i))>\kappa$ for but boundedly many $i$. Taking such an exact upper bound, you can thin to $X\subseteq \mathrm{cf}(\lambda)$ so that $i\mapsto \mathrm{cf}(g(i)): i\in X$ is increasing and unbounded in $\lambda$. Then you can fix cofinal subsets of $g(i)$ of order-type $\mathrm{cf}(g(i))$ for each $i\in X$ and modify the functions to live on $\prod_{i\in X} \mathrm{cf}(g(i))$. You can check that this preserves (3), and the resulting sequence of functions is cofinal in the new product.<|endoftext|> TITLE: Group structure on an arbitrary completely regular topological space that makes $(x,y)\mapsto xy^{-1}$ continuous at $(1,1)$ QUESTION [7 upvotes]: Let $(G,\mathcal T)$ be a completely regular topological space. Is there a group structure on $G$ such that the function $$f:G\times G\to G$$ $$f(x,y)=xy^{-1}$$ is continuous at $(1,1)$? REPLY [3 votes]: Here is a partial positive answer, going in the opposite direction of Terry´s comment. Let $X$ be an infinite topological space. Suppose $X$ is first countable at $p \in X$ and for every open neighborhood $U$ of $p$ there is a smaller neighborhood $V \subseteq U$ such that $|U \setminus V|=|U|$. Then there is a group structure on $X$ with identity $p$ such that the operation $(x,y) \mapsto xy^{-1}$ is continuous at $(p,p)$. Just note that (starting with a $U_0$ of minimal cardinality) we can find $\{U_n : n \in \omega\}$ an open local base at $p$ such that for every $n \in \omega$ we have $U_{n+1} \subset U_n$ and $|U_n \setminus U_{n+1}|=|U_0|$. Now find (e.g. using compactness of first-order logic) a group $G$ containing subgroups $H_0 \supset H_1 \supset H_2 \supset \cdots$, such that $|G|=|X|$ and $|H_n \setminus H_{n+1}|=|H_0|=|U_0|$ for every $n \in \omega$, and then transfer the group structure to $X$ in the obvious way. Note that we can also change the hypothesis to: there is an open neighborhood of $p$ with no smaller neighborhoods (e.g. $p$ is isolated) and the conclusion still holds; just call $U_0$ such neighborhood, find $G$ and $H_0$ as before and ignore the rest. Edit (Some details as to why such $G$ exists): Let $\kappa=|X|$ and $\mu=|U_0|$. Consider the language of first order logic consisting of a binary function symbol (for the group operation), countably many unary predicate symbols $\{P_n : n \in \omega\}$ and constant symbols $\{c^n_\alpha : \alpha \in \mu, n \in \omega\}$. In this language consider the first order theory that includes the group axioms, for each $n$ the axioms saying that $P_n$ is a subgroup and $P_{n+1} \subseteq P_n$, axioms saying that all the $c^n_\alpha$'s are distinct and $c^n_\alpha \in P_{n} \setminus P_{n+1}$. This theory is consistent because each finite fragment of it can be satisfied, e.g. using $\mathbb{Z}$ and finitely many of its subgroups. Since the language has size $\mu$, there is (by Lowenheim-Skolem) a model for this theory of size $\mu$. Inside this model we can find our $H_n$´s as the interpretations of the $P_n$'s. Finally we let $G=H_0 \times K$ where $K$ is any group of size $\kappa$.<|endoftext|> TITLE: Does this property of a first-order structure imply categoricity? QUESTION [6 upvotes]: Let $\mathfrak{A}$ be a first-order structure over a relational language and let $\kappa$ be an infinite cardinal. Lets say that $\mathfrak{A}$ has the $\kappa$-property if for every structure $\mathfrak{B}$ of size $\kappa$ over the same language as $\mathfrak{A}$ we have: $$\mathfrak{B} \mbox{ embeds into }\mathfrak{A} \Longleftrightarrow \mbox{ Every finite } \mathfrak{B}_0\subseteq\mathfrak{B} \mbox{ embeds into } \mathfrak{A}.$$ Has this property been studied before? If so, how is it called? Note that if the theory of $\mathfrak{A}$ is $|\mathfrak{A}|$-categorical, then a simple compactness argument shows that $\mathfrak{A}$ has the $\kappa$-property for every $\kappa\leq|\mathfrak{A}|$. I don´t expect the other implication to be true, but no examples come to my mind. Hence the question in the title. Edit: I apologize because I have broken two of the rules for good MO questions. One is Make your title your question and the other is Do your homework. Thanks to Joel, Andreas and Noah for their examples, I should really have put more effort into finding them. But what I´m really interested in is the question in the body: Has this property been studied before? If so, how is it called? REPLY [6 votes]: The answer is no for uncountable cardinals $\kappa$. Let $\mathfrak{A}=\langle A,U\rangle$ be a set $A$ of size $\kappa$ with a unary predicate $U\subset A$, where $U$ and $A-U$ both have size $\kappa$. It follows that $\mathfrak{A}$ has your property, since every structure $\mathfrak{B}=\langle B,U^B\rangle$ of size at most $\kappa$ in that language embeds into $\mathfrak{A}$. But the theory of $\mathfrak{A}$ is the theory of an infinite/co-infinite predicate, which is complete because it is countably categorical, but which is not uncountably categorical because there are models of uncountable size $\kappa$ where the predicate is interpreted in a set of size less than $\kappa$, and such a model is not isomorphic to $\mathfrak{A}$. REPLY [6 votes]: The following seems to be sort of a reversed version of Joel's example, categorical in uncountable cardinals rather than in $\aleph_0$. I'll use the theory of the set $\mathbb Z$ of integers with only the immediate-successor relation. Any model of this theory looks like the disjoint union of some non-zero cardinal number of copies of $\mathbb Z$, so the theory is categorical in uncountable powers but not in $\aleph_0$. Let $\mathfrak A$ be the model consisting of $\aleph_0$ copies of $\mathbb Z$. It easily satisfies your hypothesis, essentially because any $\mathfrak B$ whose finite substructures embed in $\mathfrak A$ must satisfy that the successor relation and its inverse are functions without finite cycles. Yet the theory is not categorical in the cardinality $\aleph_0$ of $\mathfrak A$. REPLY [5 votes]: For $\kappa=\aleph_0$, maybe I'm missing something, but: Let $\Sigma$ be the language of undirected graphs Let $R$ be the random graph; note that $R$ is connected, and that every countable graph embeds into $R$ (I'm taking "$\mathcal{A}$ embeds into $\mathcal{B}$" to just mean "$\mathcal{A}$ is isomorphic to a substructure of $\mathcal{B}$"; in particular, I'm assuming embeddings need not be elementary). Let $Z$ be the "$\mathbb{Z}$-chain," that is, points in $Z$ are integers and there is an edge between $x$ and $y$ iff $\vert x-y\vert=1$. Now, consider the structure $\mathfrak{A}=R\sqcup Z$. Clearly every countable graph embeds into $\mathfrak{A}$, but I believe $\mathfrak{A}$ is not $\aleph_0$-categorical by a compactness argument: let $T=Th(\mathfrak{A})$, $\Sigma'=\Sigma\sqcup\{a, b, c\}$, and for $n\in\omega$ let $$\varphi_n\equiv \text{"No distinct pair of points from $\{a, b, c\}$ are connected by a chain of size $\le n$."}$$ Clearly $T'=T\cup\{\varphi_n: n\in\omega\}$ is consistent, and the reduct of any model of $T'$ to $\Sigma$ will be elementarily equivalent to $\mathfrak{A}$; but any model of $T'$ has at least 3 connected components, and so is not isomorphic to $\mathfrak{A}$.<|endoftext|> TITLE: $L^2$ discrepancy bound for sequences in $[0,1)$ QUESTION [5 upvotes]: Given a sequence $x_1,x_2,\dots$, let $D_n$ be the $L^2$-norm of the function $f_n$ whose value at $t \in [0,1)$ is $nt$ minus the number of $1 \leq i \leq n$ with $x_i \leq t$. What can be said about the rate at which $D_n$ must go to infinity, regardless of the choice of $x_1,x_2,\dots$? That is, what theorem lower-bounds the $L^2$ norm of $f_n$ in analogy with the way Schmidt’s discrepancy theorem (see http://mathworld.wolfram.com/DiscrepancyTheorem.html) lower-bounds the $L^\infty$ norm of $f_n$? REPLY [6 votes]: This problem has been addressed in work of Roth and Davenport. Roth showed that for any sequence there must be $n$ with $D_n$ larger than a constant times $\sqrt{\log n}$, and Davenport constructed sequences for which $D_n$ grows like at most a constant times $\sqrt{\log n}$. More precisely, for any set ${\mathcal P}$of $N$ points in $[0,1)^2$, Roth showed that $$ \int_{\alpha,\beta=0}^{1} \Big| |{\mathcal P}\cap [0,\alpha)\times [0,\beta)| - N\alpha\beta \Big|^2 d\alpha d\beta \gg \log N. $$ Apply this to the points $(n/N,x_n)$ with $n=1$, $\ldots$, $N$. It then follows that $$ \frac{1}{N} \sum_{n=1}^{N} D_n^2 \gg \log N, $$ which proves the lower bound for $D_n$. As for the upper bound, Davenport showed that Roth's result above is best possible by looking at the set $(n/N,\{n \alpha\})$ where $\alpha$ is an irrational number with bounded partial quotients (e.g. $\alpha=\sqrt{2}$). If you look at Davenport's argument (see page 133 of the paper), he really shows that for this sequence (i.e. $x_n=\{n\sqrt{2}\}$), the $L^2$ discrepancy is of size $\sqrt{\log n}$. Roth's paper is in Mathematika vol 1, 1954, and Davenport's in Mathematika vol 3, 1956.<|endoftext|> TITLE: (Smooth) Borel Conjecture for 4-dimensional torus QUESTION [6 upvotes]: Given an aspherical 4-dimensional closed manifold $M$ with fundamental group $\mathbb{Z}^4$, it is homotopy-equivalent to $T^4 = S^1 \times \ldots S^1$, the 4-dimensional torus. Question 1: Since I am no expert and could not dig out a reference I would be interested if it is open/known that under the circumstances above, $M$ must be homeomorphic (diffeomorphic) to $T^4$? Question 2: Can one say more if one knows that $M$ is smoothly covered by $\mathbb{R} \times T^3$? Remark: in all other dimensions it seems to be true (due to Wall et al.) that for dimensions $n \geq 5$ the manifold $M$ is finitely coverey by a manifold diffeomorphic to $T^n$ and for $n \leq 3$ it is even diffeomorphic to $T^3$. REPLY [10 votes]: Freedman and Quinn's "Topology of $4$-manifolds", Chapter 11.5, contains the following statement: Let $f: M\to N$ be a homotopy equivalence of compact aspherical $4$-manifolds whose fundamental groups are poly-(finite or cyclic), restricting to a homeomorphism of boundaries. Then $f$ is homotopic rel $\partial M$ to a homeomorphism. This implies a positive answer to (1) in the topological category. I would guess that the answer in the smooth category is unknown, since it appears to be unknown if there exist exotic $4$-tori, see Do there exist exotic 4-tori?<|endoftext|> TITLE: Are Besov spaces $B^{s}_{p,q}$ invariant under Fourier transform? QUESTION [5 upvotes]: (This may be very easy question for MO; as I am just trying to understand Besov spaces) Let $\phi \in C^{\infty}(\mathbb R^{n})$ with $ \operatorname{supp} \phi \subset \{\xi \in \mathbb R^{n}: |\xi|\leq 2\} , \phi(\xi)=1$ if $|\xi|\leq 1.$ We put, $$\phi_{j}(\xi)= \phi(2^{-j}\xi)- \phi(2^{-j+1}\xi), (\xi \in \mathbb R^{n}, j \in \mathbb N).$$ Then we have $$\operatorname{supp} \phi_{j} \subset \{\xi\in \mathbb R^{n}: 2^{j-1}\leq |\xi| \leq 2^{j+1} \}, j\in \mathbb N $$ and, with $\phi_{0}=\phi,$ $$\sum_{k=0}^{\infty} \phi_{k}(\xi)=1, \text{if} \ \xi\in \mathbb R^{n}.$$ Perhaps there different ways to introduce Besov spaces; we define in the following way. Let $00$ $$ \Vert \hat u\Vert_{L^1}\ge c\sum_{l\ge 1}a_{2l} 2^{2ln}. $$ Choosing $ a_{2l}=2^{-2ln} $ we find that $u\in B^0_{1,1}$ and $\hat u\notin L^1$, which implies from (1) that $\hat u\notin B^0_{1,1}$.<|endoftext|> TITLE: Binary relations as the topological closure of the diagonal QUESTION [8 upvotes]: If $(X,\tau)$ is a topological space, we can consider the product topology on $X\times X$ and take the closure of the diagonal $\Delta_X = \{(x,x): x\in X\}$, which we denote by $\mathrm{cl}(\Delta_X)$. Obviously, $\mathrm{cl}(\Delta_X)$ is a symmetric binary relation. Now we can take things upside down: Let $X$ be a set and let $R\subseteq X\times X$ be a reflexive and symmetric relation. Is there a topology $\tau$ on $X$ such that $\mathrm{cl}(\Delta_X)=R$? REPLY [4 votes]: Let $\ |X|\ge 4,\ $ e.g. $\ X := \mathbb Z/4.\ $ Let $\ j : X\rightarrow X\ $ be an involution without any fixed point, i.e. $\ (j\circ j)(x) = x \ne j(x)\ $ for every $\ x\in X;\ $ e.g. $\ j(x) = x+2 \mod 4\ $. Let $$ R\ :=\ \{ (x\ y)\in X^2\ :\ y\ne j(x) \} $$ Then there does not exist a topology $\ T\ $ in $\ X\ $ such that $\ R\ $ is the closure of the diagonal $\ D := \{ (x\ x):\ x\in X\}\ \subseteq X^2$. PROOF   By a contradiction, let $\ T\ $ be a topology in $\ X\ $ as described above. The functions $\ I_X : X\rightarrow X\ $ and $\ \imath_x : \{x\} \rightarrow X\ $ such $\ I(t):=t\ $ for every $\ t\in X\ $ and $\ \imath_x(x):=x\ $ are continuous. Thus the diagonal product $\ I_X\Delta\imath_x : X\rightarrow X^2\ $ is continuous; and so is the projection $\ \pi_1:X^2\rightarrow X\ $ given by $\ \pi_1(t\ u) := t.\ $ It follows that the canonical bijection $$\ b_x := \pi_1|X\times\{x\} :\ X\times \{x\} \rightarrow X$$ is a homeomorphism. Thus $\ X\setminus\{\imath(y)\} = \pi_1(R\ \cap\ X\times \{y\}\ $ is closed in $\ X\ $ for each $\ y\in X,\ $ hence $\ \{x\}\ $ is open in $\ X\ $ for $\ y:=j(x);\ $ this means that $\ \{x\} $ is open for every $\ x\in X,\ $ i.e. $\ X\ $ would be Hausdorff. However the closure $\ R\ $ of the diagonal is different from the diagonal--a contradiction. (so much for an intro to Gen. Top.; sorry to be boring).<|endoftext|> TITLE: Error in Maurins proof for the nuclear spectral theorem? QUESTION [13 upvotes]: I am currently studying the nuclear spectral theorem as presented in K. Maurins Monograph "General Eigenfunction Expansions and Unitary Representations of Topological Groups", second chapter or alternatively his paper "Allgemeine Eigenfunktionsentwicklungen. Spektraldarstellung abstrakter Kerne. Eine Verallgemeinerung der Distributionen auf Lie'schen Gruppen" (Bullentin de L'Academie Polonaise des Sciences. Serie des sci. math. astr. et phys. 7 (8 1959), pp. 471-479. ) which contains basically the same proof. Let $\Phi\subset H\subset \Phi'$ be a rigged Hilbert space, $\Phi$ nuclear. We want to prove, that there exists a complete set of generalized eigenfunctions in $\Phi'$. The basic idea is the following: We use the Integral formulation of the Neumann spectral theorem, i.e. every normal operator on a separable Hilbert space $H$ is unitary equivalent to a multiplication operator on a direct integral of Hilbert Spaces (which is basically an $L^2$-type complex-valued function space), the latter shall be called $\hat{H}$ and the unitary transformation $F:\varphi\rightarrow \varphi(x)$. The key element in the proof is now, that we can prove, that evaluating some $F\varphi$ at a point of the spectrum, say $\lambda$, is a continuous mapping into the complex numbers and therefore associated with some element of the dual space. Concretely, Maurin "shows" that that the mapping $F(\lambda):\varphi\rightarrow\varphi(\lambda)$ is continuous for every $\lambda$ up to a set of measure zero. I highly dout that this mapping is even well-defined, as the functions in the direct integral of Hilbert spaces are only defined up to an equivalence class, i.e. they can differ on a set of measure zero. Am I overlooking something trivial? Is there a simple way to patch this up? Is there some reference which solves this issue? I have heard that the proof given by Gelfand and Vilekin also has a similar issue. I would greatly appreciate any help, I am very stuck at the moment. REPLY [9 votes]: In this discussion, there is a pointer to a paper by G. G. Gould (J. LMS, 1968) which supposedly gives a correct but hard to read proof. This is the paper: \bib{MR0230148}{article}{ author={Gould, G. G.}, title={The spectral representation of normal operators on a rigged Hilbert space}, journal={J. London Math. Soc.}, volume={43}, date={1968}, pages={745--754}, issn={0024-6107}, review={\MR{0230148 (37 #5713)}}, }<|endoftext|> TITLE: Recognize this strange expression from linear algebra? QUESTION [19 upvotes]: I've come across an odd-looking expression and oh how I wish I had a more elegant description of it. Maybe someone who enjoys symmetric bilinear forms in characteristic two will recognize it? Or someone who chooses bases and recognizes sums with indices instantly? Here's the setup: $X$ is a finite-rank free ${\mathbb Z}$-module, of rank $r$, with basis $\{ x_1, \ldots, x_r \}$. $g \in GL_r({\mathbb Z})$ is a matrix (change of basis matrix). Finally, $C \in (X \otimes X) \otimes {\mathbb Z / 2 \mathbb Z}$ is a symmetric tensor, modulo $2$. Concretely, with respect to the basis, $C = (c_{ij})$ with $c_{ij} \in {\mathbb Z / 2 \mathbb Z}$ and $c_{ij} = c_{ji}$ for all $1 \leq i,j \leq r$. From $g$ and $C$, I am encountering the vector $a = \sum_j a_j x_j \in X \otimes {\mathbb Z / 2 \mathbb Z}$, where $$a_j = \left( \sum_{i=1}^r c_{ii} \frac{ g_{ij} (g_{ij} - 1)}{2} \right) + \sum_{k < \ell} c_{k \ell} g_{k j} g_{\ell j}.$$ I'm encountering this vector in two very specific settings related to metaplectic groups. It looks like this vector $a$ should come from something "natural" in linear algebra or combinatorics. Has anyone seen a vector like $a$ before? Contexts? REPLY [11 votes]: You never use that $g\in \mathrm{GL}_r(\mathbb Z)$, but only that it is a matrix. So I will consider $g \in \mathrm{Mat}_r(\mathbb Z) = \mathrm{End}(X)$. I never know whether to write $\mathrm{Sym}^2(X)$ or not when in characteristic $2$. Your thing is in the invariant, as opposed to coinvariant, subspace. So I guess the $2$ should be raised. Your choice of basis determines a map $\mathrm{diag} : \mathrm{Sym}^2(X) \to X$, which records the diagonal entries of the symmetric matrix (i.e. $\mathrm{diag}(c)_i = c_{ii}$). Note that this map is not canonical without a choice of basis: at least when you replace $\mathbb Z$ by $\mathbb C$, say, $\mathrm{Sym}^2(X)$ is an irrep of $\mathrm{SL}(X)$ for $\dim X \geq 2$. (Or consider how the action of $-1$ on $X$ lifts to $\mathrm{Sym}(X)$.) Recall that $\mathrm{Sym}$ is functorial, so that $\mathrm{End}(X)$ acts also on $\mathrm{Sym}^2(X)$. This action is given by $(g\cdot c)_{ij} = \sum_{kl} c_{kl}g_{ki}g_{lj}$. So, following Steve Huntsman's suggestion, the first thing to look at is $2a$. It is nothing but $$ 2a = \mathrm{diag}(g\cdot c) - g\cdot \mathrm{diag}(c). $$ Of course, your expression makes clear that this vanishes mod $2$. So perhaps the first question to ask is: Question: Why do $g\cdot$ and $\mathrm{diag}$ commute in characteristic $2$? An answer: Consider the canonical maps $\mathrm{Sym}^2(X) \to X^{\otimes 2} \to \mathrm{Sym}_2(X)$, where the last term is the coinvariant space $X^{\otimes 2}/(\mathbb Z/2)$. When $2$ is invertible, this composition is an isomorphism. But when $2=0$, this map vanishes on the off-diagonal terms, and is nothing but a version of $\mathrm{diag}$. Its image is the image in $\mathrm{Sym}_2(X)$ of $X$ under the "Frobenius" map $x \mapsto x\otimes x$ (which is linear in characteristic $2$ and not otherwise). Ok, but that's not the whole story, because $\mathrm{diag}(g \cdot c)$ and $g\cdot \mathrm{diag}(c)$ depend only on the image of $g$ in $\mathrm{Mat}_r(\mathbb Z/2)$, whereas your $a$ depends on $g \in \mathrm{Mat}_r(\mathbb Z/4)$. So what's happening? You have a map $$ \mathrm{End}(X) \to \mathrm{Hom}\bigl(\mathrm{Sym}^2(X),X\bigr) : g \mapsto \mathrm{diag}\circ g - g\circ \mathrm{diag} = [\mathrm{diag},g]. $$ Let's mod out by $4$ and $2$: $$ \begin{matrix} \mathrm{End}\otimes \mathbb Z/4 & \overset{[\mathrm{diag},-]}\longrightarrow & \mathrm{Hom} \otimes \mathbb Z/4 \\ \downarrow & & \downarrow \\ \mathrm{End}\otimes \mathbb Z/2 & \overset{[\mathrm{diag},-]}\longrightarrow & \mathrm{Hom} \otimes \mathbb Z/2 \\ \end{matrix} $$ But the bottom arrow is $0$. So the top arrow factors through $$ \ker \bigl( \mathrm{Hom}\otimes \mathbb Z/4 \to \mathrm{Hom}\otimes \mathbb Z/2\bigr) \cong \mathrm{Hom} \otimes \mathbb Z/2 $$ where the congruence is because $\mathrm{Hom}$ is free. This is one version of your map $a$. I mean, the diagram chase gave us a linear map $\mathrm{End}(X) \to \mathrm{Hom}(\mathrm{Sym}^2(X),X) \otimes Z/2$, which is nothing but a linear map $\mathrm{End}(X) \otimes \mathrm{Sym}^2(X) \otimes \mathbb Z/2 \to X \otimes \mathbb Z/2$. I think this is $(g,c) \mapsto a$. Also, I'm sure that I'm supposed to have said the word "cocycle" somewhere in this answer. So pretend I did. Edit: Except for fixing a $0$ to a $2$ somewhere, what's above is what I wrote when this answer was accepted. But it can't be correct, at least not starting with the sentence "So what's happening?". The problem is that the action of $\mathrm{End}(X)$ on $\mathrm{Sym}(X)$ is not linear except when $2=0$, so my whole "you have this commuting square" thing is nonsense. I'm sure that the hint "cocycle" is correct. I'm not sure how to make it so. Anyway, what is true is that $\mathrm{diag}$ is a natural transformation (and $\mathrm{Sym}$ a linear functor) when $2=0$ and not otherwise. With $\mathbb Z/4$ coefficients, linearity fails, but its failure is in the kernel of $\mathbb Z/4 \to \mathbb Z/2$, which is a copy of $\mathbb Z/2$. So in that sense what I wrote is correct: the vector $a$ is the failure of linearity, after recognizing that this failure is necessarily in the kernel and identifying this kernel with $\mathbb Z/2$.<|endoftext|> TITLE: Which finite simple groups can be characterized by their action on a small set? QUESTION [14 upvotes]: It is well known that a finite 4-times transitive permutation group is Matthieu, symmetric, or alternating. Another way of stating this is that the set $$ \Omega = \{(x_1, x_2, x_3, x_4), 1\leq x_i\leq n, \mbox{$x_1, \ldots, x_4$ pairwise different}\} $$ has the property that $S_n$ acts transitively on $\Omega$, and every subgroup $U TITLE: Is there a straightedge and compass construction of incommensurables in the hyperbolic plane? QUESTION [9 upvotes]: In other words, given a segment in the hyperbolic plane is there a straightedge and compass construction of a segment incommensurable with it? In the Euclidean plane one can take the diagonal of the square built on the segment, as Pythagoreans discovered. However, equivalence of this incommensurability and irrationality of $\sqrt{2}$ relies on the Euclidean Pythagorean theorem. Pythagoreans originally believed that any two segments have a common measure, how hard would it have been for them to discover their mistake if we happened to live in a hyperbolic space? The "straightedge" of course has to be hyperbolic. More precisely, a construction can use all Hilbert's axioms of the hyperbolic plane (including the axiom of Archimedes) except the Cantor's axiom of continuity. Equivalently, the question asks if there is a pair of incommensurable segments in every subset of the hyperbolic plane closed under straightedge and compass constructions, but not necessarily metrically complete. This may not be as easy as it looks. There are no squares in the hyperbolic plane, and the hypotenuse of an equilateral right triangle can be commensurable with its leg. In fact, it follows from the hyperbolic Pythagorean theorem that any number in $(\sqrt{2},2)$ can be the hypotenuse/leg ratio depending on the size of the triangle. If the ratio is rational for the given segment the Pythagorean construction won't work. One could try doubling/halving the segment multiple times and then taking hypotenuses on various concatenations, but it is conceivable that all of them remain commensurable since there do exist non-rational analytic functions that map rationals into rationals. Perhaps there is a construction more taylored to the hyperbolic plane. But standard constructions of hyperbolic parallels, and therefore of ideal triangles, do use the axiom of continuity. REPLY [14 votes]: Yes. the fundamental theorem is that the constructible angles in the non-Euclidean plane are exactly the constructible angle in the Euclidean plane. Lengths come from the two laws of cosines and the law of sines. In particular, length 1 is not constructible. As i recall, positive length $x$ is constructible if and only if $\sinh x$ is constructible in the Euclidean plane. See my 1995 Intelligencer article at http://zakuski.utsa.edu/~jagy/bib.html Let's see, in the fourth edition of his book, Marvin Jay Greenberg included Robin Hartshorne's proof of this, very different sort of language. EDIT: it seems part of the question was constructible lengths with irrational ratio. $\log 3$ and $\log 2$ should work.<|endoftext|> TITLE: Is this problem on weighted bipartite graph solvable in polynomial time or it is NP-Complete QUESTION [8 upvotes]: I encounter this problem recently and I want to know whether it is NP-Complete or solvable in polynomial time: Given a undirected weighted bipartite graph $G = (V, E)$ where $V$ can be partitioned into two sets $A$ and $B$ and $E$ is a set of edges connecting $A$ and $B$. The weight of an edge $(v, u)$ is denoted as $w(v, u)$. All weights are positive integers. Do the following sequentially: Pick a node $v \in A$, remove all node $u \in B$ for every $(v, u) \in E$ and, add the weight $w(v, u)$ to the total score for every edges deleted. The goal is to find the sequence of nodes $v_1, \dots, v_n \in A$ that maximizes the total score. I have searched for the bank of NP-Complete problems to find something possible can reduce to this problem but I haven't found anything useful yet. Any suggestions would be extremely helpful! REPLY [4 votes]: The problem is NP-hard. Without loss of generality we shall prove this in the case that all the weights are positive integers. First let us rephrase the problem as follows. Given an integer-valued matrix, find the reordering of the columns that maximizes the score of the resulting reordered matrix, where the score of a matrix is defined as follows: It is the sum of the scores of the rows, where the score of a row is the value of the leftmost nonzero entry in that row (or zero if the row is all zero, but clearly we may assume that there are no all-zero rows). To see the equivalence, let the columns correspond to the vertices of $A$ and let the rows correspond to the vertices of $B$; zeros correspond to non-edges and the integers correspond to the edge weights. The following reduction from SAT to (the decision version of) the above problem is due to my colleague Sandy Kutin. Suppose we are given a SAT instance with $m$ clauses and $n$ variables. Without loss of generality we shall assume that for each variable $x$, one of the clauses in the system is $x\vee\bar x$. We now construct a matrix with $2n+1$ columns and $m+2n$ rows as follows. For each variable $x$ we have two columns, one labeled $x$ and the other labeled $\bar x$. We also have a special column labeled $z$. For each clause, we create a row; for each literal in the clause, we put the value $2n+2$ in the corresponding column, and we put a 1 in the $z$ column. In the other columns we put zero. This gives us the first $m$ rows. Then for each of the $2n$ literals, we create a row with a 1 in the column corresponding to that literal and a 2 in the $z$ column. Now we ask the question: Is there a reordering of the columns with score at least $m(2n+2) + 3n$? If the SAT instance is satisfiable, then the answer is yes: For each literal $x$ that is TRUE, put the $x$ column to the left of the $z$ column and the $\bar x$ column to the right of the $z$ column. Then we score $2n+2$ for each of the first $m$ rows and a total of $3n$ for the remaining $2n$ rows. Conversely, in order to score as high as $m(2n+2)+3n$, we need to score $2n+2$ for each of the first $m$ rows, since scoring only $1$ for one of those rows sacrifices $2n+1$, which is more than we can make up for using the final $2n$ rows. Therefore for every clause, at least one of the literals in it must have its column to the left of the $z$ columns. To pick up an additional $3n$ from the remaining $2n$ rows, it must be the case that for each literal $x$, either the $x$ or the $\bar x$ column must appear to the right of the $z$ column (they cannot both appear to the right because then we would fail to score $2n+2$ from the $x\vee\bar x$ row). Thus by picking all the literals to the left of the $z$ column, we obtain a satisfying assignment.<|endoftext|> TITLE: Compactness of adelic quotients for unipotent groups over global fields QUESTION [5 upvotes]: Let $K$ be a global field, $\mathbb{A}_K$ the ring of adeles, and $U$ a unipotent algebraic group over $K$. Why is $U(\mathbb{A}_K)/U(K)$, when endowed with the quotient topology, compact? REPLY [7 votes]: You don't mention if $U$ is assumed to be smooth or connected, but it doesn't matter. In general, if $H$ is any affine group scheme of finite type over a global field $K$ and if $H$ does not contain ${\rm{GL}}_1$ as a $K$-subgroup scheme (as is the case when $H$ is a unipotent $K$-group scheme) then $H(K)\backslash H(\mathbf{A}_K)$ is compact. In effect, the non-compactness of the idele class group is the source of "all" non-compactness for such adelic coset spaces. This is Theorem A.5.5 in the paper "Finiteness theorems for algebraic groups over function fields" in Compositio 148. But that generality is quite hard in positive characteristic, since imperfectness forces the use of the full-blown structure theory of pseudo-reductive groups (with its extra wrinkles in characteristics 2 and 3). For just the unipotent smooth connected case the idea is way easier: directly show that a closed immersion of $K$-group schemes induces a closed embedding on such adelic coset spaces (see Lemma 4.2.5 in loc. cit.), and then apply it to $U \hookrightarrow {\rm{R}}_{K'/K}(U_{K'})$ and use that Weil restriction interacts as expected with adelic topologies, so in that way it suffices to solve the problem after a finite extension on $K$. But $U_{\overline{K}}$ has a composition series whose successive quotients are vector groups, and that descends to some finite extension, so you thereby reduce to the case when $U$ is $K$-split. That case is then readily reduced to the case of a vector group, which is clear. QED<|endoftext|> TITLE: reference for Levelt-Turritin QUESTION [5 upvotes]: Can anybody recommend a good reference to learn the Levelt-Turritin decomposition theorem of formal connections? An intuitive description of what it says would also be very appreciated. REPLY [3 votes]: For a new approach and intuitive explanation see this article. In the older literature, Levelt's Original paper is the best reference I have found! In particular, it is more understandable than many references that come after. The statement is very simple: every differential operator over the field $\overline{\mathbb{C}((t))}$ has a Jordan canonical form.<|endoftext|> TITLE: explicit uniformizer for the false Tate extension QUESTION [6 upvotes]: Let $p$ be an odd prime and let $n\geq 1$. Set $K=\mathbb{Q}_p(\zeta_{p^n})$, $L=\mathbb{Q}_p(\sqrt[p^n]{p})$, and $M=KL$. I claim that $M$ is totally ramified of degree $\phi(p^n)p^n$ (the proof simultaneously shows that $K$ and $L$ are linearly disjoint). Assume not, then there exists an unramified extension $W\subseteq M$ with $[W:\mathbb{Q}_p]=f>1$ ($f$ being necessarily a power of $p$). By the structure theorem for unramified extensions of $\mathbb{Q}_p$, we know that $W=\mathbb{Q}_p(\zeta_{r})$ where $r=p^f-1$. The extension $K$ is totally ramified over $\mathbb{Q}_p$ and therefore linearly disjoint from $W$. We thus have $$ \mathbb{Q}_p\subseteq K \subsetneqq KW \subseteq M=K(\sqrt[p^n]{p}) $$ By Galois theory, since $f>1$, it follows (this also holds true if $K$ and $L$ are not linearly disjoint) that $\sqrt[p]{p}\in KW$ and therefore $\mathbb{Q}_p(\zeta_p,\sqrt[p]{p})\subseteq KW$. But $KW$ is abelian over $\mathbb{Q}_p$ and $\mathbb{Q}_p(\zeta_p,\sqrt[p]{p})$ is not. Contradiction. Q1: How to construct (systematically in $p$ and $n$) an explicit uniformizer of $M$ (for $n>1$ of course)? Q2 Is it possible to find some uniformizer $\pi'$ of $K$ (so $v_p(\pi')=\frac{1}{\phi(p^n)}$) such that $K(\sqrt[p^n]{\pi'})=M$ (for $n>1$ of course) ? Remark: Note that Q2 is equivalent to find (by Lagrange's resolvent) $\beta\in M$ such that $$ \sum_{j=0}^{p^n-1} \sigma^j(\beta)\zeta_{p^n}^j, $$ is a uniformizer of $M$. Here $\sigma$ is a generator of $Gal(M/K)$. This seems to boil down to some difficult linear algebra. REPLY [4 votes]: I gave this question to a student as a summer project. Thanks to Mercio's answer to a similar question posted on StackExchange (link: https://math.stackexchange.com/questions/954731/ ) we have worked out how to find explicit uniformizers of $\mathbb{Q}_p(\zeta_{p^2},\sqrt[p]{p})$ for all odd p. We have written a short article on our calculations, which can be found here: http://arxiv.org/abs/1912.01656 Please let us know if you have any suggestions and/or comments for us.<|endoftext|> TITLE: What are reasons to believe that e is not a period? QUESTION [74 upvotes]: In their 2001 paper defining periods, Kontsevich and Zagier (pdf) without further comment state that $e$ is conjecturally not a period while many other numbers showing up naturally (conjecturally) are. The former claim is repeated at many other internet sources including Wikipedia but nowhere could I find a heuristic making the conjecture that $e$ is not a period more plausible than its negation. Does anyone here know of such an argument? EDIT: I figured it would be good measure (i.e. 'shows research effort') to write what was the best I could come up with myself. I don't find it very convincing however so feel free to ignore. The number $e$ is more or less defined as the value at a rational number (1) of a function that is a solution to a ordinary differential equation ($y' = y$) with rational boundary condition ($y(0) = 1$). Now K & Z point out that all periods arise in this way (replace a rational number in the defining integral with a parameter and it will satisfy an ODE). However they also warn us that the differential equations are really special and (conjecturally) satisfy a lot of criteria among which having at most regular singularities. Now the singularity at infinity of $y'= y$ is not regular as it has order 2 (while the equation is of order 1) but of course this proves nothing since nothing is stopping $e$ from being the value at some rational number of a solution to a much more complicated differential equation which might be of the right class. So what is missing from an argument along these lines is some way of making precise that $y'= y$ really is the simplest equation which produces $e$ and that 'therefore' more complicated equations can be 'reduced' to it by a series of simplifications innocent enough to preserve the regularity of the singularities if it exists (quod non). Now personally I would not buy such a claim if it wasn't for the fact that it is a bit akin to conjecture 1 from K & Z. However this line of reasoning requires a lot of 'making precise' and perhaps is an entirely wrong way of looking at it, so better ideas are welcome! REPLY [43 votes]: Periods arise from the comparison between Betti and de Rham cohomology for an algebraic variety. The Period Conjecture, due to Grothendieck, is a transcendence conjecture for periods which says that every algebraic relation between periods arises from geometry (in a certain precise sense). More generally, there is a wider class of complex numbers called exponential periods arising from the comparison of rapid decay cohomology and de Rham cohomology. The number $e$ is an example of an exponential period. There is an analogue of the Period Conjecture in the setting of exponential periods, and the truth of this conjecture would imply that $e$ is transcendental over the ring of ordinary periods (see Proposition 10.1.5 of the paper Exponential Motives by Javier Fresán and Peter Jossen). So the exponential Period Conjecture provides a heuristic coming from algebraic geometry that $e$ is not a period.<|endoftext|> TITLE: Sums of degrees of irreducible complex characters QUESTION [11 upvotes]: The sum of the degrees of the irreducible complex characters (not the square sum which is the group order) is relevant to determine the dimension of a maximal torus of the Lie algebra associated to the group algebra of a finite group over the complex field. I have investigated certain group classes as extra-special groups, abelian groups, direct products, quaternion groups, (semi-)dihedral groups, A4, SL(2,q), GL(2,q), meta-cyclic groups, meta-abelian-groups, central products, frobenius groups and determine this sum. Also symmetric groups are now known counting involutions (a case of groups for which all irre. characters are real valued). Nilpotent groups are reduced to p-groups using direct products. My question is whether there is a nice sum-formula (maybe with recursion) for groups like: alternating groups, p-groups (maybe special classes), simple groups, general linear, special linear groups, groups with united factor group, groups with normal p-Sylow-subgroup and abelian Hall-Complement, soluble groups, M-groups, ... as well. ` Maybe there is an argument or idea how to determine this sum for all groups by reduction on certain classes of groups. In the literature there is a nice inequality saying that this sum is at least the double of the linear characters for a non-soluble groups (using the classification atlas of simple groups). Maybe there are some more upper and lower bounds to derive here. REPLY [6 votes]: I just realised that a result of mine ("On the minimal norm of a non-regular generalized character of a finite group" Bull LMS 2010) can be used to give a slightly better upper bound for the sum of the irreducible character degrees of finite non-Abelian group $G$. For such a group $G,$ with irreducible characters $\chi_{1},\chi_{2}, \ldots , \chi_{k},$ labelled so that $\chi_{k}(1)$ is maximal, we have $\sum_{i = 1}^{k} \chi_{i}(1) \leq \sqrt{k|G|+ 1 - \frac{|G|}{\chi_{k}(1)}}$ (and the $``+1"$ inside the square root sign can be omitted as long as $G$ is not a Frobenius group with Abelian Frobenius kernel). This follows from the main theorem of that paper, since $\sum_{i=1}^{k} \chi_{i}$ is not a multiple of the regular character when $G$ is non-Abelian. Notice that when $G$ is dihedral of order $2h$ with $h$ odd, we have $\sum_{i=1}^{k} \chi_{i}(1) = 1 + 1 + 2\frac{h-1}{2} = h+1,$ while the above inequality gives $\sum_{i=1}^{k} \chi_{i}(1) \leq \sqrt{ 2h(2 + \frac{h-1}{2}) - h + 1} = h+1,$ so the stated inequality becomes equality for all such dihedral groups. By an earlier paper in that series, we have $\sum_{i = 1}^{k} \chi_{i}(1) \leq \sqrt{(k-1)|G| + \chi_{k}(1)^{2}}$ when $G$ is nilpotent, but non-Abelian, (where again we label so that $\chi_{k}(1)$ is maximal).<|endoftext|> TITLE: Can a sum of roots of unity be an integer? QUESTION [13 upvotes]: Let $n \geq 2$, $H \lneq (\mathbb{Z}/n\mathbb{Z})^*$, $\zeta_k$ a primitive $k$-th root of unity. Is it possible that $$\sum_{h \in H} \zeta_k^{h} \in \mathbb{Z}$$ for every $k$ dividing $n$ such that $n/k \leq |H|$? I do not know the answer even if $k=n$. That is, can it be that $$\sum_{h \in H} \zeta_n^{h} \in \mathbb{Z}$$ REPLY [12 votes]: We suppose towards a contradiction that all the sums in the first question are integers. Set $$f(x) = \prod_{h \in H}(x - \zeta_n^h)$$ We claim that all symmetric polynomials in the roots of $f$ are integers, for which it is enough to show that for each $1 \leq d \leq |H|$ $$\sum_{h \in H} \zeta_n^{dh} \in \mathbb{Z}$$ but this is true by our assumption since $\zeta_n^d = \zeta_k$ for some $k$ dividing $n$ with $n/k \leq |H|$ from the inequality for $d$. This shows that $f \in \mathbb{Z}[x]$ is of degree smaller than $\phi(n)$ (since $H$ is proper) having $\zeta_n$ as a root, contradiction.<|endoftext|> TITLE: Generating function of $p(25n + 24)$ QUESTION [8 upvotes]: In a paper titled "RAMANUJAN’S UNPUBLISHED MANUSCRIPT ON THE PARTITION AND TAU FUNCTIONS WITH PROOFS AND COMMENTARY" by Bruce C. Berndt and Ken Ono, it is mentioned that Ramanujan derived the formula \begin{align}\sum_{n = 0}^{\infty}p(25n + 24)q^{n} &= 5^{2}\cdot 63\frac{(q^{5};q^{5})_{\infty}^{6}}{(q;q)_{\infty}^{7}} + 5^{5}\cdot 52q\frac{(q^{5};q^{5})_{\infty}^{12}}{(q;q)_{\infty}^{13}}\notag\\ &\,\,\,\,+\,5^{7}\cdot 63q^{2}\frac{(q^{5};q^{5})_{\infty}^{18}}{(q;q)_{\infty}^{19}} + 5^{10}\cdot 6q^{3}\frac{(q^{5};q^{5})_{\infty}^{24}}{(q;q)_{\infty}^{25}}\notag\\ &\,\,\,\,+\,5^{12}q^{4}\frac{(q^{5};q^{5})_{\infty}^{30}}{(q;q)_{\infty}^{31}}\tag{1}\end{align} via the use of the formulas $$\frac{(q;q)_{\infty}^{6}}{(q^{5};q^{5})_{\infty}^{6}} = A^{5} - 11q + q^{2}B^{5}\tag{2}$$ and $\dfrac{(q^{5};q^{5})_{\infty}}{(q^{1/5};q^{1/5})_{\infty}}$ $=\dfrac{(A^{4} + 3Bq) + q^{1/5}(A^{3} + 2B^{2}q) + q^{2/5}(2A^{2} + B^{3}q) + q^{3/5}(3A + B^{4}q) + 5q^{4/5}}{A^{5} - 11q + q^{2}B^{5}}\text{ (3)}$ and $$\sum_{n = 0}^{\infty}p(5n + 4)q^{n} = 5\frac{(q^{5};q^{5})_{\infty}^{5}}{(q;q)_{\infty}^{6}}\tag{4}$$ This is done by replacing $q$ by $q^{1/5}$ in $(4)$ and writing the formula as $$\sum_{n = 0}^{\infty}p(5n + 4)q^{n/5} = \frac{5}{(q;q)_{\infty}}\left(\frac{(q^{5};q^{5})_{\infty}}{(q^{1/5};q^{1/5})_{\infty}}\right)^{6}\frac{(q;q)_{\infty}^{6}}{(q^{5};q^{5})_{\infty}^{6}}$$ This effectively requires us to raise equation $(3)$ to $6^{\text{th}}$ power and then take the coefficient of $q^{4/5}$. I wonder how this leads to a beautiful result like equation $(1)$. If we use multinomial theorem to calculate $6^{\text{th}}$ power then it leads to a huge number of terms containing $q^{4/5}$ (42 terms and each term a polynomial in $A, B, q$) and I don't know if this can be done via hand calculation. Is there is any other method to find the coefficient of $q^{4/5}$ or some alternative way to derive the complicated formula $(1)$? Update: Here $A, B$ are given by $$A = \frac{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}}{(q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}},\,\, AB = -1$$ REPLY [12 votes]: Hirschhorn and Hunt have published a note M. D. Hirschhorn; D. C. Hunt, A simple proof of the Ramanujan conjecture for powers of 5, J. Reine Angew. Math., 326 (1981), 1-17., where they give a proof (what they call "a simple proof") for the formula for the generating function of $p(5^{\alpha}n+\delta_n)$, with $\alpha\ge 1$ and $\delta_n$ being the reciprocal modulo $5^{\alpha}$ of 24; see Theorem $1.4$.<|endoftext|> TITLE: existence of a finite group which is the union of self normalizing subgroups QUESTION [10 upvotes]: Can a finite group G be the union of self normalizing subgroups such that the intersection between any two of these subgroups is equal to the unit of the group G? I don't think so but I can't prove it. Thank you for your help. REPLY [7 votes]: To expand a little on Nick Gill's answer: it was a common theme in some early classification-type theorems that a simple group might be a disjoint union of Hall subgroups with trivial intersection of any two of them. This occurs for example, in Suzuki's classification of simple CA-groups ( groups in which all non-identity elements have Abelian centralizers)- I do not know whether Suzuki was thinking about groups with a partition before he did that classification. The Suzuki groups are simple CN-groups ( nonidentity elements have nilpotent centralizers)-there are simple CN-groups with no such partition, eg ${\rm PSL}(2,7), {\rm PSL}(2,9), {\rm PSL}(2,17)$ - in these last groups, Sylow $2$-subgroups can have non-trivial intersections. Quite often, delicate character theory was needed in these classifications. Also, by Frobenius's theorem, if $G$ is a finite simple group, and $M$ is a nilpotent Hall subgroup of $G$ disojoint from its other conjugates, then $M$ can't be self normalizing (there are finite simple groups which have a Sylow $2$-subgroup which is a maximal subgroup, eg ${\rm PSL}(2,17)$). Hence it follows for general reasons(without CFSG but as Nick says, with some quite difficult group theory) that in a finite non-Abelian simple group which admits a partition by nilpotent Hall subgroups, these can't all be self-normalizing and disjoint from their other conjugates ( apart from the identity). However, at the moment I don't see how to tackle directly ( ie without CFSG, and with easier results than Nick quotes ) partitions of the type asked about here (even just for simple groups). Later edit: In a completely different direction, (though this may be covered by Baer's work, which predates Carter subgroups), a finite solvable group $G$ is never a union of proper nilpotent self-normalizing subgroups : for $G$ has a unique conjugacy class of nilpotent self-normalizing subgroups, the Carter subgroups, and no finite group is the union of a single conjugacy class of proper subgroups (if $G$ itself is nilpotent, it has no proper self-normalizing subgroups at all).<|endoftext|> TITLE: How much can one say about this differential equation? QUESTION [11 upvotes]: Consider the ODE $y^{\prime \prime}(x) = \cos(x) y(x)$ with boundary value conditions $y(0)=1$, $y(1)=2$. Solving it results in a linear combination of Mathieu functions, but what I find more interesting is its graph. So, the questions are: Is there really a phase transition around 300? What is the envelope of the graph? That is, are the maxima/minima growing exponentially, or does something else occur? Is the oscillating part actually periodic (which one would guess from the cos term)? In general, what qualitative methods are there for answering these questions? I am sure the actual Mathieu function has been studied extensively, but just using its specific properties seems too "rigid". REPLY [16 votes]: I would guess that you are just seeing the effects of a dynamical system with a hyperbolic fixed point. Consider the matrix equation $$ A'(x) = \begin{pmatrix} 0 & 1\\ \cos(x) & 0\end{pmatrix} A(x) $$ with initial condition $A(0) = I_2$. Because of the $2\pi$-periodicity of $\cos(x)$, this fundamental solution clearly satisfies $$ A(x+2\pi) = A(x)A(2\pi) $$ where $$ A(2\pi)\approx \begin{pmatrix} -8.065 & -8.273\\ -7.742 & -8.065\end{pmatrix}, $$ a matrix that has eigenvalues $\lambda\approx -16.068$ and $1/\lambda\approx -0.062$. (Note that the ratio of the eigenvalues is about $\lambda^2 \approx 258$.) The action of $A(2\pi)$ on $\mathbb{R}^2$ is hyperbolic, contracting along the $1/\lambda$-eigenspace and expanding along the $\lambda$-eigenspace. Now, the general solution of your equation satisfies $$ \begin{pmatrix}y(x)\\y'(x)\end{pmatrix} = A(x)\begin{pmatrix}y(0)\\y'(0)\end{pmatrix}, $$ so $$ \begin{pmatrix}y(x+2k\pi)\\y'(x+2k\pi)\end{pmatrix} = A(x)A(2\pi)^k\begin{pmatrix}y(0)\\y'(0)\end{pmatrix}, $$ and if the initial condition is very near (but does not lie in) the $1/\lambda$-eigenspace of $A(2\pi)$, it will take a few cycles of $2\pi$ for the very small $\lambda$-eigenvector component to grow, but once it does, it will become exponentially dominant. I'm guessing that your boundary values just happen to have hit on such an initial condition. REPLY [11 votes]: All these questions are basically answered by Floquet theory. The envelope is an exponential, and there is nothing special going on at 300, what you are seeing is just exponential growth, and on the scale on which you are plotting everything less that $10^{49}$ just looks like zero.<|endoftext|> TITLE: Percentage of elements that can be written as commutators QUESTION [7 upvotes]: Let $G$ be a finite perfect group. Are there any results on the number of elements of $G$ which can be written as a commutator? When $G$ is finite non-abelian simple group, then every element can be written as a commutator. REPLY [8 votes]: The proportion of elements of a finite group $G$ which are commutators is at least $\frac{1}{k(G)}$ (and the inequality is strict for non-Abelian groups) where $k(G)$ is the number of conjugacy classes of $G,$ because there are $|G|^{2}$ expressions of the form $[a,b]$ and every element of $G$ has at most $k(G)|G|$ expressions of the form $[a,b]$ (and only at the identity can that bound be achieved). Hence the number of distinct commutators in $G$ is at least $\frac{|G|^{2}}{k(G)|G|} = \frac{|G|}{k(G)}.$ On the other hand (apart from the obvious example of Abelian groups where the boound is clearly sharp), a non-Abelian extraspecial $2$-group $G$ of order $2^{2n+1}$ has only $2$ commutators, and has $k(G) = 2^{2n} + 1,$ so $\frac{|G|}{k(G)} \to 2$ as $n \to \infty,$ while we always have $\frac{|G|}{k(G)} < 2.$ Hence the proportion of commutators can get arbitrarily close to $\frac{1}{k(G)}$ for non-Abelian groups<|endoftext|> TITLE: Moments of the trace of orthogonal matrices QUESTION [18 upvotes]: Let $O_n$ be the (real) orthogonal group of $n$ by $n$ matrices. I am interested in the following sequence which showed up in a calculation I was doing $$a_k = \int_{O_n} (\text{Tr } X)^k dX$$ where the integral is taken with respect to the normalized Haar measure on $O_n.$ I noticed some work of Dianconis et al. covers a more general set of moments, but essentially the formulas I saw only seem to hold for small $k.$ For the case of $O_2$ the sequence looks like $a_{2k} = \frac{1}{2}\binom{2k}{k},$ which can be calculated by hand since $O_2$ has some special structure. I was hoping there would be some general known formula (since, if I recall correctly, this calculates the multiplicity the trivial representation in a tensor power the standard representation of $O_n$), or that it might be in the OEIS, but I don't know exactly what to look for, and don't know enough small terms to search. REPLY [7 votes]: I am writing more than a year after the question was posted, only to spell out some more details regarding the calculation implicit in the solution presented by Suvrit, and to clarify the dependence on $n$ in that solution. Write $[{\rm Tr}(AX)]^{2k}=p_{1^{2k}}(AX)$, using power sum symmetric functions. Next, expand into Schur functions, $p_{1^{2k}}(AX)=\sum_{\lambda\vdash 2k}d_\lambda s_\lambda(AX)$. Notice that in this expansion we must have $\ell(\lambda)\le n$, otherwise the Schur function is identically zero. Now use the definition of zonal polynomials (average of irreducible characters of $U(n)$ over the subgroup $O(n)$), $$ \int_{O(n)}s_\lambda(AX)dX=\frac{Z_\tau(AA^T)}{Z_\tau(1^n)},$$ which holds if $\lambda=2\tau$ (the integral vanishes otherwise) to get $$ \int_{O(N)}[{\rm Tr}(X)]^{2k}dX=\sum_{\tau\vdash k}d_{2\tau}, $$ with the condition $\ell(\tau)\le n$ (which is the only dependence of the answer on $n$).<|endoftext|> TITLE: Are there number-theoretic graphs that are far from being isomorphic QUESTION [5 upvotes]: I say that two graphs $G_1=(V_1,E_1)$, $G_2 = (V_2,E_2)$ with the same number of vertices, edges, are $\epsilon$-far from being isomorphic, if for any bijection between $V_1$ and $V_2$, the fraction of pairs $(x,y)$ for which $(x,y)\in E_2$ conditioned on $(x,y)\in E_1$ is at most $1-\epsilon$. I am looking of for an infinite family of $\epsilon$-nonisomorphic graph pairs, for which $\epsilon$ is lower-bounded by a positive constant such that these graph pairs are not distinguishable by first-order criteria, like degree distribution, spectra, etc. I am wandering whether it is reasonable to look for such graph pairs from number-theoretic constructions: For example, it is known that there are Paley graphs (say, $P(p^2)$ for prime $p$) for which there exist Peisert graphs (on $GF(p)\times GF(p)$) that are non-isomorphic to them, but with the same parameters, and even co-spectral. Is it possible to show that they are in fact "highly" non-isomorphic as well? REPLY [4 votes]: Combinatorics might be a good place to look. I give below a huge family of pairwise non-isomorphic $57$ point graphs all regular of degree $24.$ I suspect that some of them are fairly far from being isomorphic in your sense but have no idea how easy it would be to find the distance given two specific graphs. There are $11,084,874,829$ isomorphism classes of STS(19) - a steiner triple system with 19 points (and 57 blocks). Each leads to a unique strongly regular graph with parameters SRG(57,24,11,9) (There might be other SRG(57,24,11,9), I'm not sure. ) One forms such a graph with a vertex for each block and an edge joining blocks with a common point.These graphs certainly have much in common. They all have the same spectrum and are regular. Every edge is in $11$ triangles and any pair of non-adjacent vertices is connected by $9$ paths of length $2$. All these graphs have exactly 19 $K_9$ subgraphs which correspond to the points. There are also ways in which they can be quite different. One of them has an automorphism group of order $432$. The vast majority, $11,084,710,071$, have no automorphisms at all. Three of them have $84$ Pasch configurations (the maximum number) and $2,591$ of them have none at all. As I said, I do not actually know how far a pair of these can be from being isomorphic (in your sense), nor how one would prove that the distance is large. But the Pasch configuration information leads me to speculate that some might be far apart. A Pasch configuration in a STS is a set of four blocks and six points of the form $uvw,wxy,uxz,vyz$ These are exactly the cases of four pairwise intersecting blocks with no point in common to any three. In the graphs this corresponds to a $K_4$ which overlaps no $K_9$ in a $K_3.$ Whatever is true here is no doubt reflected more extremely in the case of STS(v) for $v \gt 19$ points. I just don't know if these have been exhaustively enumerated for $v \gt 19.$ There are also great numbers of strongly regular graphs which have equal parameters but are pairwise non-isomorphic arising from sets of $j$ pairwise orthogonal $k \times k$ latin squares. LATER I do like the example above and it seems in the spirit that you want. But here is another case where it might be easier to prove distance. In a precise sense, most trees are not determined by their spectrum. There are many constructions. Here is a small pair which share the spectrum and degree sequence. It is not so hard to see that they are not fully isomorphic and there is a vertex bijection which preserves 3/4 of the edges (So they are $1/4$ far from being isokmorphic?) The condition there is a vertex of degree $2$ whose neighbors have degrees $3$ and $4$ . does distinguish them. I imagine that there are more intricate trees which are harder to immediately tell apart and which are almost $1/2$ far from being isomorphic. Perhaps even $1-\varepsilon$.<|endoftext|> TITLE: Infinite dimensional 2-Hilbert spaces QUESTION [15 upvotes]: Is there a definition of an infinite dimensional 2-Hilbert space? Finite dimensional 2-Hilbert spaces have been discussed by Baez in http://arxiv.org/abs/q-alg/9609018 In the more recent paper by Baez, Baratin, Freidel and Wise http://arxiv.org/abs/0812.4969 a notion of infinite dimensional 2-vector space is discussed, building on work by Crane, Sheppeard and Yetter. They also have a few proposals for what an infinite dimensional 2-Hilbert space should be, but "the details still need to be worked out". Is that the current state of knowledge? Another way of asking this question is the following. If I want to see a field theory that has an infinite dimensional Hilbert space of states as an extended field theory, what kind of objects should I assign to codimension 2 manifolds? REPLY [6 votes]: An answer to the question in the second paragraph (what kind of objects should a quantum field theory assign to codimension 2 manifolds?) can be found in my paper http://arxiv.org/abs/1304.7328v2. I will argue that this also provides an answer to the OP's main question (what is a possibly infinite dimensional 2-Hilbert space?) The answer is: it's a von Neumann algebra (or, possibly better, the module category of a von Neumann algebra). Note that the kind of von Neumann algebras I have in mind are type $III$ factors and that for such an algebra, there is very little difference between the algebra and its representation category. If one excludes the zero module (and modules on non-separable Hilbert spaces), then the von Neumann algebra itself, viewed as a one object category, is equivalent (in the usual sense of equivalence of categories) to its representation category. In other words, such a von Neumann algebra has only one non-zero module up to isomorphism (excluding non-separable modules), and the endomorphism algebra of this unique module is the von Neumann algebra you started with. Note that this unique module is never irreducible: if you direct sum it with itself, you get again itself. One can actually be more precise: the von Neumann algebras that show up in quantum field theory are always hyperfinite type $III_1$ factors (unless they are Morita equivalent to something finite dimensional). By a famous theorem of Connes and Haagerup, there is only one hyperfinite type $III_1$ factor (subject to appropriate separability assumptions) up to isomorphism. Thus, if one takes the point of view that an ``infinite dimensional 2-Hilbert space'' is the same thing as a hyperfinite type $III_1$ factor, then the Connes-Haagerup theorem says that there's a unique infinite dimensional 2-Hilbert space, up to isomorphism. This is the analog of the well known and easy fact according to which there exists only one separable infinite dimensional Hilbert space, up to isomorphism. Now one might ask: what is the 2-inner product on $M$-Mod? (here, $M$ is any von Neumann algebra). For that, note that there is a complex-antilinear functor $M$-Mod $\to$ $M^{op}\!$-Mod given by sending a Hilbert space $H$ with action of $M$ to its complex conjugate $H\mapsto \overline H$. Here, the right action of $m\in M$ on $\overline H$ is given by $\overline\xi m:=\overline{m^*\xi}$. Given $H\in M$-Mod and $K\in M^{op}\!$-Mod, one can form the relative tensor product $K\boxtimes_M H$ (also explained here), which is some kind of fancy version of the tensor product over a ring. That's the 2-inner product. Namely, for two $M$-modules $H_1$ and $H_2$, we define $$\langle H_1,H_2\rangle := \overline{H_1}\boxtimes_MH_2 \in \text{Hilb}.$$ To contrast with Turion's answer, let me emphasise that a 2-Hilbert space is not enriched in Hilbert spaces. It is enriched in von Neumann algebras. Also, a 2-Hilbert space doesn't have a product with $\text{Vect}$ for scalar multiplication. It has a product with $\text{Hilb}$. As a final remark, let me point out that my ``2-Hilbert spaces'' are suitable for representing 2-Groups (topological groups with a $U(1)$-valued 3-cocycle in the sense of Segal), just like Hilbert spaces are good for representing centrally extended groups (topological groups with a $U(1)$-valued 2-cocycle).<|endoftext|> TITLE: Laplacian eigenfunction $L^p$ norms QUESTION [6 upvotes]: Suppose I have a compact surface (possibly with boundary), and consider the eigenfunctions of the Laplacian, normalized so that their $L^2$ norms are $1.$ Is there some general result or conjecture on how the $L^p$ norm of $f_\lambda$ (where $\lambda$ is the eigenvalue) behaves as function of $\lambda?$ Presumably this depends on the geometry of the surface (and, of course, nothing about this question is special to two dimensions, except that this case might be easier). REPLY [5 votes]: As pointed out already by the comments, Sogge has indeed made a lot of contributions in this area. Consider a 2-dimensional compact Riemannian manifold without boundary, then the $L^2$ normalized eigenfunction of Laplacian $e_{\lambda}$ which sastisfy $$ -\Delta e_{\lambda}=\lambda^2e_{\lambda} $$ has the following estimate $$ \|e_{\lambda}\|_{L^p}\leq C\lambda^{\sigma(p)},~~\lambda\ge 1, $$ here $\sigma(p)=\frac{1}{2}(\frac{1}{2}-\frac{1}{p})$, if $2\leq p\leq 6$, and $\sigma(p)=2(\frac{1}{2}-\frac{1}{p})-1$, if $6\leq p\leq \infty$. For dimension $n>2$, there are similar results. However, for manifold with boundary, the problem becomes more difficult, for dimension 2, see also the Acta paper by Smith and Sogge http://arxiv.org/pdf/math/0605682.pdf. It's also interesting to improve the bound above. The above estimates is sharp in general. The $L^{\infty}$ norm is saturated by the Zonal function (which is concentrated near the pole) on the round sphere. For lower $p$(below the critical point, which is 6 here), it's saturated by the highest weight hamonics (which is concentrated near the equator ). However, for manifold with nonpositive curvature, on can get some improvement($\log \lambda$), and for flat torus $\mathbb{T}^n$, one can get a further improvement ($\lambda^{\epsilon(n)})$.<|endoftext|> TITLE: The necessary and sufficient condition for $\textbf{global}$ conformal flatness of a n-dim (pseudo-)Riemannian manifold QUESTION [6 upvotes]: There is a theorem : 1) 2-dim (pseudo-)Riemannian manifold must be local conformal flat; 2) 3-dim (pseudo-)Riemannian manifold is local conformal flat iff the Cotton tensor vanishes. 3) n-dim (n>3) (pseudo-)Riemannian manifold is local conformal flat iff the Weyl tensor vanished. Then I'm curious about the necessary and sufficient condition for $\textbf{global}$ conformal flatness of a n-dim (pseudo-)Riemannian manifold $(M,g)$, i.e. there exist a function $\Omega(x)$ defined in the whole manifold such that $g=\Omega^2 \eta$, where $\eta$ is the flat metric. Are there some literature or textbooks covering this question? Thanks! REPLY [5 votes]: I add a small $\varepsilon$ to Robert's answer, which is a simple explanation and a simple example to what he said concerning the 2 dim case. Conformal structure of signature (1,1) on the surface is essentially the same as a pair of everywhere transversal foliations. Well, up to a double cover, to be precise. Indeed, for a metric, the lightline cone at every point is two straight lines in the tangent space intersecting at the origin. The foliations are given by condition that they are tangent to these straight lines. It is known (see for example sect. 5.1 of http://lanl.arxiv.org/abs/1002.3934) that for globally flat metrics these foliations are somehow standard: moreover, any flat torus is a quotient of $(R^2, g=dxdy)$ by a lattice. It is also easy to construct foliations that are not ``standard'', say we easily construct a foliation such that it has a Reeb component. The conformal structure corresponding to this foliation is conformally flat, since in dimension 2 any metric is conformally flat, but is not globally conformally flat<|endoftext|> TITLE: $t$-structure on modules over highly structured ring spectra QUESTION [5 upvotes]: Let $Sp$ be a suitably nice model of a suitably nice category of spectra. By this I mean that $Sp$ is a symmetric monoidal closed stable model category, where we can make sense of homotopy "groups" and so on. I'm not sure how to formalise this, but I'm thinking about the cases where $Ho(Sp)$ is the classical stable homotopy category, or the $G$-equivariant stable homotopy category for some (finite) group $G,$ or the Morel-Voevodsky $\mathbb{A}^1$-stable category of $\mathbb{P}^1$-spectra over a perfect field. Now let $A$ be a commutative ring object in $Sp.$ Then by general theory the category $A-Mod$ of module objects in $Sp$ over $A$ has essentially the same formal properties as $Sp.$ However one notion which I have not seen established for $A-Mod$ is that of a $t$-structure. So Question Does the category $A-Mod$ admit a (nice) $t$-structure? Is this written down anywhere? In a bit more detail: The category $Ho(Sp)$ affords a $t$-structure. Let's concentrate on the $G$-equivariant case, $G$ finite, since this seems to be middle ground in difficulty between ordinary stable homotopy and motivic stable homotopy. We can define homotopy presheaves $\underline\pi_n: Ho(G-Sp) \to Fun(\mathcal{O}, Ab),$ where $\mathcal{O}$ is just the subcategory of $Ho(G-Sp)$ consisting of the suspensions of $G/H_+$ for subgroups $H$ of $G.$ One proves that $(G/H_+)$ generate $Ho(G-Sp)$ and so the functors $\underline\pi_*$ form a conservative system. We tentatively put $Ho(G-Sp)^{\le 0} = \{X | \underline\pi_n(X) = 0 \text{ if } n>0\}$ and similarly for $Ho(G-Sp)^{\ge 0}.$ Unless I'm mistaken one can prove this specifies a $t$-structure. Namely we can use the fact that the sphere spectrum is connective to kill off homotopy groups using cones (and homotopy colimits I guess) thus giving one of the truncation functors, and the rest should follow. One may moreover prove that $Ho(G-Sp)^{\ge 0}$ is generated under homotopy colimits and extensions by $\mathcal{O}.$ In particular the smash product of connective spectra is connective. So here is a more refined question: Let $A$ be a connective commutative highly structured ring spectrum in the $G$-equivariant stable homotopy category, for $G$ a finite group. Do $Ho(A-Mod)^{\ge0} = \{X : \underline\pi_n(X)=0 \text{ for } n < 0\}$ and $Ho(A-Mod)^{\le 0} = \dots$ specify a $t$-structure on $Ho(A-Mod)$? Is the derived smash product over $A$ right (t-)exact? A more philosophical question to end: It seems to me that if this is true, it should be rather formal and so valid in much greater generality. What is the natural setting of such an observation, and where can I learn about it? REPLY [4 votes]: I don't have time to think carefully. However, I answered this question for the sphere $G$-spectrum here: A heart for stable equivariant homotopy theory. I see no problem in generalizing that answer to any connective commutative ring $G$-spectrum $A$. Using cell $A$-modules, one can kill the higher homotopy groups of an $A$-module, etc. The Mackey functor $\underline{\pi}_0 A$ is a Green functor, so one can form module Mackey functors over it, and the heart should be the Eilenberg-MacLane $G$-spectra of such modules, with appropriate structure as $A$-module $G$-spectra.<|endoftext|> TITLE: $\ell$-adic monodromy theorems (over $\mathbb{C}$) QUESTION [8 upvotes]: This question is about $\ell$-adic monodromy theorems for families over a number field. ($\ell$-adic analogues of Corollaries 6.2.8 and 6.2.9 in [BBD].) Notation $H$ denotes étale cohomology. Let $f \colon X \to Y$ be a proper morphism of finite type schemes over $\mathbb{C}$. Let $\mathcal{F}$ be in $\mathrm{D}^{\mathrm{b}}_{\mathrm{c}}(X, \mathbb{Q}_{\ell})$ (bounded, constructible), semi-simple of geometric origin. Let $V \subset Y$ be a (Zariski) open over which $H^{i}f_{*}\mathcal{F}$ is locally constant. Take $y \in Y(\mathbb{C})$. Question The suggested global invariant cycle theorem: Q1. If $V$ is connected, and $y \in V$, do we have a surjection $$ H^{i}(X, \mathcal{F}) \twoheadrightarrow H^{i}(X_{y}, \mathcal{F})^{\pi_{1}(V, y)}$$ And then the part that I am least sure about. (I am not even sure the formulation makes sense.) If I am not mistaken, one should replace open balls by henselian traits. Here is my try. The suggested local invariant cycle theorem: Q2. Let $B$ be the hensel localisation of $Y$ at $y$, and let $z$ be the generic point of $B$. Do we have a surjection $$ H^{i}(X_{y}, \mathcal{F}) \twoheadrightarrow H^{i}(X_{z}, \mathcal{F})^{\pi_{1}(B, z)} $$ Remarks I have the feeling that something like this should be true. Given the theorems in [BBD], I think this should follow from their §6.1 “Principes”. Yet, I don't see how. Maybe this is because I don't understand the proof of the decomposition theorem, nor how [6.2.8 and 6.2.9, BBD] are easy consequences of it. I am even more confident, because right after the decomposition theorem [6.2.5, BBD] there is a remark that there is an étale analogue. For personal applications, it is Q2 that I am most interested in. [BBD] — Beilinson, Alexander A.; Bernstein, Joseph; Deligne, Pierre (1982). “Faisceaux pervers”. Astérisque (in French) (Société Mathématique de France, Paris) 100. REPLY [8 votes]: The sages (for whom I am but an unworthy mouthpiece) say: Tell the OP to look at 6.2.9 in Deligne’s Weil II paper, it’s quite close to answering what he is asking in his Q2.<|endoftext|> TITLE: $H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group QUESTION [50 upvotes]: Recently, prompted by considerations in conformal field theory, I was lead to guess that for every compact connected Lie group $G$, the fourth cohomology group of it classifying space is torsion free. By using the structure theory of connected Lie groups and a couple of Serre spectral sequences, I was quickly able to prove that result. However, this feels unsatisfactory: as I hinted on the first paragraph, the fact that $H^4(BG,\mathbb Z)$ is torsion free seems to have a meaning. But what that meaning exactly is is not quite clear to me... In order to get a better feeling of what that meaning might be, I therefore ask the following: Question: Can someone come up with a non-computational proof of the fact that for every connected compact Lie group $G$, the cohomology group $H^4(BG,\mathbb Z)$ is torsion free? [Added later]: My paper on WZW models and $H^4(BG,\mathbb Z)$ has recently appeared on the arXiv. In it, I present a proof of the torsion-freeness of $H^4(BG,\mathbb Z)$ which is slightly different from the one below. I also show that $H^4(BG,\mathbb Z)=H^4(BT,\mathbb Z)^W$. For the reader's interest, I include a proof that $H^4(BG)$ is torsion-free [all cohomology groups are with $\mathbb Z$ coefficients, which is omitted from the notation]. Let $\tilde G$ be the universal cover of $G$, and let $\pi:=\pi_1(G)$. Then there is a Puppe sequence $$ \pi\to\tilde G\to G \to K(\pi,1)\to B\tilde G \to BG \to K(\pi,2) $$ It is a well known fact that $\pi_2$ of any Lie group is trivial: it follows that $B\tilde G$ is 3-connected and that $H^4(B\tilde G)$ is torsion-free (actually $H_4(B\tilde G)$ is also torsion-free, but that's not needed for the argument). Now, here comes the computation: $H^*(K(\mathbb Z/p^n,2)) = [\mathbb Z, 0,0,\mathbb Z/p^n,0,...]$ from which it follows that for any finite abelian group $A$ $H^*(K(A,2)) = [\mathbb Z, 0,0,A,0,...]$ from which it follows that for any finitely generated abelian group $\pi=\mathbb Z^n\oplus A$ $H^*(K(\pi,2)) = [\mathbb Z, 0,\mathbb Z^n,A,\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)},...]$ The Serre spectral sequence for the fibration $B\tilde G \to BG \to K(\pi,2)$ therefore looks as follows: $$ \begin{matrix} \vdots & \vdots\\ H^4(B\tilde G) & 0 & \vdots & \vdots & & \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots\\ \mathbb Z & 0 & \mathbb Z^n & A &\mathbb Z^{(\!\begin{smallmatrix} \scriptscriptstyle n+1 \\ \scriptscriptstyle 2 \end{smallmatrix}\!)} & H^5(K(\pi,2)) & \cdots\\ \end{matrix} $$ and the $d_5$ differential $H^4(B\tilde G)\to H^5(K(\pi,2))$ cannot create torsion in degree four. QED PS: By a result of McLane (1954), $H^5(K(\pi,2))$ is naturally isomorphic to the the group of $\mathbb Q/\mathbb Z$-valued quadratic forms on $\pi$ modulo those that lift to a $\mathbb Q$-valued quadratic form... I wonder what the above $d_5$ differential is. REPLY [35 votes]: I try to give an argument without spectral sequences, not sure if this can be considered non-computational though. At least, there is a non-computational syllabus: torsion classes in $H^4(BG,\mathbb{Z})$ would be characteristic classes of torsion bundles over $S^3$ but the latter have to be trivial. Now for a slightly more detailed argument: first, the coefficient formula tells us that torsion in $H^4(BG,\mathbb{Z})$ comes from torsion in $H_3(BG,\mathbb{Z})$ because torsion in $H_4(BG,\mathbb{Z})$ would not survive the dualization. Since $G$ is connected, the Hurewicz theorem provides a surjection $\pi_3(BG)\to H_3(BG,\mathbb{Z})$, so any torsion class in $H_3(BG,\mathbb{Z})$ comes from a map $S^3\to BG$ classifying a $G$-bundle over $S^3$. The corresponding clutching map $S^2\to G$ is homotopic to the constant map because $\pi_2(G)$ is trivial. So we find that any $G$-bundle on $S^3$ is trivial, hence $H_3$ and therefore $H^4$ of $BG$ do not have non-trivial torsion.<|endoftext|> TITLE: Injectivity under flat base change of the Picard group on smooth projective curves QUESTION [5 upvotes]: Let $K$ be a field of characteristic $0$, $X_K$ a smooth projective curve over $K$. Denote by $\bar{K}$ the algebraic closure of $K$. The base change morphism $X_{\bar{K}} \to X_K$, induces via the pull-back map, a morphism of the Picard groups, $\mbox{Pic}(X_K) \to \mbox{Pic}(X_{\bar{K}})$. Is this map injective? If not true in general, is there a known condition on $K$ under which this holds true? REPLY [7 votes]: Yes this is true. It can be proved using the Hochschild-Serre spectral sequence plus Hilbert's theorem 90 (it is true more generally for any geometrically connected projective variety $X$). The Hochschild-Serre spectral sequence yields the exact sequence: $$0 \to H^1(K, \bar{K}[X_\bar{K}]^*) \to \mbox{Pic} X \to \mbox{Pic} X_\bar{K}.$$ See for instance Lemma 6.3 of Sansuc - Groupe de Brauer et arithmétiques des groupes algébriques linéaires sur un corps de nombres where this sequence is extended further to the right. As $X$ is projective we have $\bar{K}[X_\bar{K}]^* = \bar{K}^*$, hence the first cohomology group vanishes by Hilbert's Theorem 90, thus proving the required injectivity. Note that the result is false in the affine case in general; for example over $\mathbb{Q}$ there exist algebraic tori with non-trivial Picard groups (again see Sansuc). REPLY [7 votes]: This map is injective. There is a Hochschild-Serre spectral sequence with $E^{pq}_2=H^p(\mathrm{Gal}(\bar{K}/K), H^q(X_{\bar{K}},\mathbb{G}_m))$ converging towards $H^{p+q}(X_{K},\mathbb{G}_m)$. This gives a first terms exact sequence $$0\rightarrow H^1(\mathrm{Gal}(\bar{K}/K), \bar{K}^*)\rightarrow \mathrm{Pic}(X_K)\rightarrow \mathrm{Pic}(X_{\bar{K}})\ ,$$but $\ H^1(\mathrm{Gal}(\bar{K}/K), \bar{K}^*)$ is zero by Hilbert theorem 90.<|endoftext|> TITLE: Rational distance from vertices of an equilateral triangle QUESTION [8 upvotes]: A colleague in my department posed the following question... Let $A=(0,0)$, $B=(1,0)$, and $C=(1/2,\sqrt{3}/2)$. Then $\Delta ABC$ is an equilateral triangle with sides of length 1. Let $B_{\epsilon}({\bf x}) = \{ {\bf y} \in \mathbb{R}^2 \;|\; \mbox{distance}({\bf y},{\bf x})<\epsilon \}$ be an epsilon ball centered at ${\bf x}$ using the usual Euclidean distance. Given a point ${\bf x} \in \mathbb{R}^2$ and $\epsilon>0$, is there a point $P \in B_{\epsilon}({\bf x})$ such that the distances from $P$ to the vertices of $\Delta ABC$ are all rational? It seems like it ought to be true. It is if you just demand 2 of the distances to be rational or if you allow $P$ to live in a 3D-ball (i.e. if we spill out of $\mathbb{R}^2$ into $\mathbb{R}^3$). Thanks for your thoughts! REPLY [15 votes]: Yes. In fact, more is true. T. G. Berry ( "Points at rational distance from the vertices of a triangle", Acta Arith. 62 (1992), no. 4, 391–398) proved: Theorem. Let $ABC$ be a triangle such that the length of at least one side is rational and the squares of the lengths of all sides are rational. Then the set of points $P$ whose distances $PA$, $PB$, $PC$ to the vertices of the triangle are rational is dense in the plane of the triangle. Berry says that the case where all the sides of the triangle are rational was previously proved by J. H. J. Almering ("Rational quadrilaterals," Indag. Mat. 25 (1963), 192–199).<|endoftext|> TITLE: Ternary quadratic form theta series as Hecke eigenforms and class number one QUESTION [5 upvotes]: At Simple comparison of positive ternary quadratic form representation counts Jeremy answered: "The reason is that the theta series for the sums of three squares form is an eigenfunction for all the Hecke operators (which is encoded by the relations $R(np^{2}) = (p+1 - \left(\frac{-n}{p}\right)) R(n) - p R(n/p^{2})$ and $R(4n) = R(n)$), while in general, the theta series for other ternary forms will not be. It should be easy to show that if $f(x,y,z)$ is such that its theta series is a Hecke eigenform, then $f(x,y,z)$ must be regular." I did some experiments last night, and very comprehensive experiments just now, and I have a little conjecture... Is it the case that Jeremy's condition implies class number one? I wrote a little program, all it does is test, for a form and a prime $p<15,$ and the first $20$ numbers $n$ prime to $p$ that are represented, do we always have $$ R(n p^2 ) \geq p R(n)? $$ The result: yes for the 794 class number one positive ternaries, no for the other 119 regular (or probably regular) forms, and no for evidently irregular forms. If this is all true, do we always get ( with primes for which the form is isotropic in $\mathbb Q_p$) relations of the sort $$R(np^{2}) = (p+1 - \left(\frac{-\Delta n}{p}\right)) R(n) - p R(n/p^{2})$$ where, with $$ f(x,y,z) = a x^2 + b y^2 + c z^2 + r y z + s z x + t x y $$ we have Lehman's discriminant $$ \Delta = 4abc+rst - ar^2 - b s^2 - c t^2. $$ This is so cool. Saturday, output showing the 119 regular (or probably regular) forms of class number larger than one fail my simplified test: =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= counted 913 regular ================================== 11 : 1 1 3 0 1 0 count is 3 ; n = 5 12 ; 20 16 p = 2 ; R(n p^2 ) - p R(n) = -8 15 : 1 2 2 1 0 0 count is 5 ; n = 9 22 ; 36 42 p = 2 ; R(n p^2 ) - p R(n) = -2 17 : 1 2 3 2 1 1 count is 8 ; n = 15 20 ; 60 36 p = 2 ; R(n p^2 ) - p R(n) = -4 21 : 1 2 3 0 0 1 count is 6 ; n = 11 20 ; 44 36 p = 2 ; R(n p^2 ) - p R(n) = -4 24 : 1 3 3 3 1 1 count is 6 ; n = 9 22 ; 225 102 p = 5 ; R(n p^2 ) - p R(n) = -8 27 : 1 1 7 0 1 0 count is 1 ; n = 1 4 ; 4 4 p = 2 ; R(n p^2 ) - p R(n) = -4 27 : 1 1 9 0 0 1 count is 1 ; n = 1 6 ; 4 6 p = 2 ; R(n p^2 ) - p R(n) = -6 27 : 1 2 4 1 0 1 count is 4 ; n = 7 10 ; 175 38 p = 5 ; R(n p^2 ) - p R(n) = -12 27 : 1 3 3 3 0 0 count is 8 ; n = 25 38 ; 100 50 p = 2 ; R(n p^2 ) - p R(n) = -26 32 : 1 3 3 1 0 1 count is 12 ; n = 25 22 ; 225 62 p = 3 ; R(n p^2 ) - p R(n) = -4 44 : 1 3 4 0 0 1 count is 16 ; n = 31 36 ; 279 96 p = 3 ; R(n p^2 ) - p R(n) = -12 45 : 1 3 4 0 1 0 count is 10 ; n = 27 22 ; 108 42 p = 2 ; R(n p^2 ) - p R(n) = -2 48 : 1 3 5 3 1 0 count is 10 ; n = 17 20 ; 425 84 p = 5 ; R(n p^2 ) - p R(n) = -16 50 : 1 2 7 2 1 0 count is 7 ; n = 13 16 ; 117 40 p = 3 ; R(n p^2 ) - p R(n) = -8 56 : 1 3 5 1 1 0 count is 5 ; n = 9 10 ; 225 46 p = 5 ; R(n p^2 ) - p R(n) = -4 63 : 1 3 6 3 0 0 count is 9 ; n = 25 22 ; 100 42 p = 2 ; R(n p^2 ) - p R(n) = -2 64 : 1 1 16 0 0 0 count is 1 ; n = 1 4 ; 9 4 p = 3 ; R(n p^2 ) - p R(n) = -8 72 : 1 3 7 2 1 1 count is 1 ; n = 1 2 ; 25 6 p = 5 ; R(n p^2 ) - p R(n) = -4 72 : 1 3 7 3 1 0 count is 7 ; n = 13 20 ; 325 84 p = 5 ; R(n p^2 ) - p R(n) = -16 75 : 1 4 5 0 0 1 count is 9 ; n = 29 20 ; 116 36 p = 2 ; R(n p^2 ) - p R(n) = -4 80 : 1 3 7 1 1 0 count is 4 ; n = 9 10 ; 441 66 p = 7 ; R(n p^2 ) - p R(n) = -4 81 : 1 3 7 0 1 0 count is 6 ; n = 12 14 ; 300 50 p = 5 ; R(n p^2 ) - p R(n) = -20 100 : 1 2 13 2 0 0 count is 9 ; n = 17 16 ; 153 40 p = 3 ; R(n p^2 ) - p R(n) = -8 108 : 1 1 36 0 0 1 count is 1 ; n = 1 6 ; 25 6 p = 5 ; R(n p^2 ) - p R(n) = -24 108 : 1 3 10 3 1 0 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 108 : 1 4 7 0 1 0 count is 7 ; n = 13 16 ; 325 32 p = 5 ; R(n p^2 ) - p R(n) = -48 108 : 1 5 7 5 1 1 count is 3 ; n = 7 10 ; 175 38 p = 5 ; R(n p^2 ) - p R(n) = -12 120 : 1 3 11 3 1 0 count is 9 ; n = 17 16 ; 833 80 p = 7 ; R(n p^2 ) - p R(n) = -32 121 : 1 3 11 0 0 1 count is 1 ; n = 1 2 ; 4 2 p = 2 ; R(n p^2 ) - p R(n) = -2 135 : 1 3 12 3 0 0 count is 5 ; n = 12 10 ; 588 62 p = 7 ; R(n p^2 ) - p R(n) = -8 135 : 2 2 9 0 0 1 count is 5 ; n = 17 20 ; 68 36 p = 2 ; R(n p^2 ) - p R(n) = -4 144 : 1 3 13 3 1 0 count is 9 ; n = 19 20 ; 475 84 p = 5 ; R(n p^2 ) - p R(n) = -16 147 : 1 2 21 0 0 1 count is 8 ; n = 29 20 ; 116 36 p = 2 ; R(n p^2 ) - p R(n) = -4 189 : 2 3 8 0 1 0 count is 4 ; n = 11 10 ; 44 18 p = 2 ; R(n p^2 ) - p R(n) = -2 216 : 1 3 19 3 1 0 count is 13 ; n = 31 20 ; 775 84 p = 5 ; R(n p^2 ) - p R(n) = -16 216 : 3 5 5 2 3 3 count is 13 ; n = 39 20 ; 975 84 p = 5 ; R(n p^2 ) - p R(n) = -16 225 : 2 2 15 0 0 1 count is 5 ; n = 23 20 ; 92 36 p = 2 ; R(n p^2 ) - p R(n) = -4 240 : 1 5 13 2 1 1 count is 1 ; n = 1 2 ; 49 10 p = 7 ; R(n p^2 ) - p R(n) = -4 243 : 1 7 9 0 0 1 count is 1 ; n = 1 2 ; 4 2 p = 2 ; R(n p^2 ) - p R(n) = -2 243 : 2 3 11 3 1 0 count is 9 ; n = 35 12 ; 140 20 p = 2 ; R(n p^2 ) - p R(n) = -4 256 : 1 2 32 0 0 0 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 256 : 1 4 16 0 0 0 count is 1 ; n = 1 2 ; 9 2 p = 3 ; R(n p^2 ) - p R(n) = -4 256 : 1 5 13 2 0 0 count is 11 ; n = 29 14 ; 261 34 p = 3 ; R(n p^2 ) - p R(n) = -8 289 : 3 5 6 1 2 3 count is 6 ; n = 23 10 ; 92 18 p = 2 ; R(n p^2 ) - p R(n) = -2 297 : 1 6 13 3 1 0 count is 1 ; n = 1 2 ; 4 2 p = 2 ; R(n p^2 ) - p R(n) = -2 360 : 1 3 31 3 1 0 count is 14 ; n = 37 20 ; 1813 124 p = 7 ; R(n p^2 ) - p R(n) = -16 392 : 3 3 12 -2 2 1 count is 3 ; n = 13 10 ; 117 26 p = 3 ; R(n p^2 ) - p R(n) = -4 400 : 3 3 12 2 2 1 count is 6 ; n = 23 10 ; 207 26 p = 3 ; R(n p^2 ) - p R(n) = -4 405 : 2 5 11 2 2 1 count is 1 ; n = 2 2 ; 98 8 p = 7 ; R(n p^2 ) - p R(n) = -6 432 : 1 3 36 0 0 0 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 432 : 1 3 37 3 1 0 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 432 : 1 4 28 4 0 0 count is 1 ; n = 1 2 ; 25 6 p = 5 ; R(n p^2 ) - p R(n) = -4 432 : 1 12 12 12 0 0 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 432 : 3 5 9 3 0 3 count is 5 ; n = 17 10 ; 425 42 p = 5 ; R(n p^2 ) - p R(n) = -8 441 : 3 6 7 0 0 3 count is 7 ; n = 31 20 ; 124 36 p = 2 ; R(n p^2 ) - p R(n) = -4 484 : 1 3 44 0 0 1 count is 14 ; n = 53 16 ; 477 40 p = 3 ; R(n p^2 ) - p R(n) = -8 576 : 1 4 36 0 0 0 count is 1 ; n = 1 2 ; 25 6 p = 5 ; R(n p^2 ) - p R(n) = -4 600 : 5 7 7 6 5 5 count is 10 ; n = 37 20 ; 1813 124 p = 7 ; R(n p^2 ) - p R(n) = -16 648 : 1 7 25 5 1 1 count is 1 ; n = 1 2 ; 25 6 p = 5 ; R(n p^2 ) - p R(n) = -4 675 : 1 4 45 0 0 1 count is 12 ; n = 69 20 ; 276 36 p = 2 ; R(n p^2 ) - p R(n) = -4 675 : 5 6 6 3 0 0 count is 5 ; n = 29 20 ; 116 36 p = 2 ; R(n p^2 ) - p R(n) = -4 720 : 3 5 15 3 3 3 count is 1 ; n = 3 2 ; 147 10 p = 7 ; R(n p^2 ) - p R(n) = -4 768 : 1 8 24 0 0 0 count is 1 ; n = 1 2 ; 25 6 p = 5 ; R(n p^2 ) - p R(n) = -4 896 : 3 6 14 4 2 2 count is 5 ; n = 19 6 ; 171 14 p = 3 ; R(n p^2 ) - p R(n) = -4 972 : 1 7 36 0 0 1 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 1024 : 1 8 32 0 0 0 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 1024 : 1 16 16 0 0 0 count is 1 ; n = 1 2 ; 9 2 p = 3 ; R(n p^2 ) - p R(n) = -4 1024 : 3 3 32 0 0 2 count is 12 ; n = 59 20 ; 531 52 p = 3 ; R(n p^2 ) - p R(n) = -8 1080 : 3 9 11 3 3 0 count is 5 ; n = 17 8 ; 833 40 p = 7 ; R(n p^2 ) - p R(n) = -16 1125 : 1 10 29 5 1 0 count is 1 ; n = 1 2 ; 4 2 p = 2 ; R(n p^2 ) - p R(n) = -2 1125 : 2 7 22 -6 1 1 count is 1 ; n = 2 2 ; 98 12 p = 7 ; R(n p^2 ) - p R(n) = -2 1296 : 3 4 28 4 0 0 count is 1 ; n = 3 2 ; 75 2 p = 5 ; R(n p^2 ) - p R(n) = -8 1323 : 2 8 21 0 0 1 count is 5 ; n = 29 10 ; 116 18 p = 2 ; R(n p^2 ) - p R(n) = -2 1600 : 3 3 51 -2 2 2 count is 11 ; n = 59 16 ; 531 40 p = 3 ; R(n p^2 ) - p R(n) = -8 1620 : 5 8 11 -4 1 2 count is 2 ; n = 8 2 ; 392 8 p = 7 ; R(n p^2 ) - p R(n) = -6 1728 : 1 12 36 0 0 0 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 1792 : 5 8 12 0 4 0 count is 12 ; n = 53 16 ; 477 32 p = 3 ; R(n p^2 ) - p R(n) = -16 1800 : 5 11 11 7 5 5 count is 8 ; n = 41 16 ; 2009 80 p = 7 ; R(n p^2 ) - p R(n) = -32 2048 : 1 8 64 0 0 0 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 2160 : 5 9 15 9 3 3 count is 2 ; n = 9 2 ; 441 10 p = 7 ; R(n p^2 ) - p R(n) = -4 2304 : 3 8 24 0 0 0 count is 1 ; n = 3 2 ; 75 6 p = 5 ; R(n p^2 ) - p R(n) = -4 2592 : 5 9 17 6 5 3 count is 3 ; n = 17 6 ; 425 26 p = 5 ; R(n p^2 ) - p R(n) = -4 3072 : 1 16 48 0 0 0 count is 1 ; n = 1 2 ; 25 6 p = 5 ; R(n p^2 ) - p R(n) = -4 3375 : 2 15 32 15 1 0 count is 3 ; n = 23 4 ; 92 4 p = 2 ; R(n p^2 ) - p R(n) = -4 3840 : 5 8 24 0 0 0 count is 1 ; n = 5 2 ; 245 6 p = 7 ; R(n p^2 ) - p R(n) = -8 4032 : 7 8 20 0 4 4 count is 3 ; n = 19 2 ; 2299 18 p = 11 ; R(n p^2 ) - p R(n) = -4 4096 : 3 11 32 0 0 2 count is 8 ; n = 59 10 ; 531 26 p = 3 ; R(n p^2 ) - p R(n) = -4 4500 : 7 8 23 6 7 2 count is 1 ; n = 8 2 ; 392 12 p = 7 ; R(n p^2 ) - p R(n) = -2 4536 : 5 9 27 0 3 3 count is 8 ; n = 41 6 ; 1025 22 p = 5 ; R(n p^2 ) - p R(n) = -8 5120 : 7 12 16 0 0 4 count is 12 ; n = 79 16 ; 711 32 p = 3 ; R(n p^2 ) - p R(n) = -16 5184 : 5 8 36 0 0 4 count is 7 ; n = 41 6 ; 1025 26 p = 5 ; R(n p^2 ) - p R(n) = -4 5400 : 7 7 28 -2 2 1 count is 4 ; n = 37 10 ; 1813 62 p = 7 ; R(n p^2 ) - p R(n) = -8 6912 : 1 16 112 16 0 0 count is 1 ; n = 1 2 ; 25 6 p = 5 ; R(n p^2 ) - p R(n) = -4 6912 : 1 24 72 0 0 0 count is 1 ; n = 1 2 ; 25 6 p = 5 ; R(n p^2 ) - p R(n) = -4 6912 : 1 48 48 48 0 0 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 6912 : 4 13 37 2 4 4 count is 1 ; n = 4 2 ; 100 2 p = 5 ; R(n p^2 ) - p R(n) = -8 6912 : 5 5 72 0 0 2 count is 11 ; n = 77 16 ; 1925 64 p = 5 ; R(n p^2 ) - p R(n) = -16 6912 : 8 9 24 0 0 0 count is 2 ; n = 9 2 ; 225 6 p = 5 ; R(n p^2 ) - p R(n) = -4 8232 : 5 13 33 -6 3 1 count is 14 ; n = 73 6 ; 1825 24 p = 5 ; R(n p^2 ) - p R(n) = -6 8448 : 7 15 23 -6 2 6 count is 8 ; n = 47 6 ; 1175 26 p = 5 ; R(n p^2 ) - p R(n) = -4 9216 : 3 16 48 0 0 0 count is 1 ; n = 3 2 ; 75 6 p = 5 ; R(n p^2 ) - p R(n) = -4 10125 : 9 11 29 -4 3 6 count is 2 ; n = 14 2 ; 1694 18 p = 11 ; R(n p^2 ) - p R(n) = -4 11520 : 8 15 24 0 0 0 count is 2 ; n = 15 2 ; 735 6 p = 7 ; R(n p^2 ) - p R(n) = -8 11520 : 11 16 19 8 2 8 count is 2 ; n = 19 4 ; 2299 36 p = 11 ; R(n p^2 ) - p R(n) = -8 12544 : 3 19 56 0 0 2 count is 11 ; n = 83 10 ; 747 26 p = 3 ; R(n p^2 ) - p R(n) = -4 19200 : 1 40 120 0 0 0 count is 1 ; n = 1 2 ; 49 6 p = 7 ; R(n p^2 ) - p R(n) = -8 20736 : 3 16 112 16 0 0 count is 1 ; n = 3 2 ; 75 2 p = 5 ; R(n p^2 ) - p R(n) = -8 20736 : 7 15 55 -6 2 6 count is 8 ; n = 79 8 ; 1975 36 p = 5 ; R(n p^2 ) - p R(n) = -4 24696 : 11 15 39 -3 6 3 count is 7 ; n = 53 4 ; 1325 16 p = 5 ; R(n p^2 ) - p R(n) = -4 25344 : 5 20 68 -8 4 4 count is 12 ; n = 101 10 ; 2525 34 p = 5 ; R(n p^2 ) - p R(n) = -16 25600 : 3 27 80 0 0 2 count is 9 ; n = 107 10 ; 963 26 p = 3 ; R(n p^2 ) - p R(n) = -4 27648 : 1 48 144 0 0 0 count is 1 ; n = 1 2 ; 25 2 p = 5 ; R(n p^2 ) - p R(n) = -8 27648 : 5 20 77 20 2 4 count is 6 ; n = 77 8 ; 1925 32 p = 5 ; R(n p^2 ) - p R(n) = -8 27648 : 9 17 48 0 0 6 count is 1 ; n = 9 2 ; 225 6 p = 5 ; R(n p^2 ) - p R(n) = -4 32000 : 11 16 51 8 2 8 count is 9 ; n = 91 8 ; 819 20 p = 3 ; R(n p^2 ) - p R(n) = -4 34560 : 13 24 28 0 4 0 count is 3 ; n = 37 6 ; 1813 30 p = 7 ; R(n p^2 ) - p R(n) = -12 57600 : 3 40 120 0 0 0 count is 1 ; n = 3 2 ; 147 6 p = 7 ; R(n p^2 ) - p R(n) = -8 57600 : 7 23 92 12 4 2 count is 13 ; n = 143 8 ; 7007 48 p = 7 ; R(n p^2 ) - p R(n) = -8 172800 : 9 41 120 0 0 6 count is 1 ; n = 9 2 ; 441 6 p = 7 ; R(n p^2 ) - p R(n) = -8 jagy@phobeusjunior: grep " p = " Class_Number_NOT_One.txt | wc 119 2737 9684 jagy@phobeusjunior: date Sat Sep 6 10:51:20 PDT 2014 =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= Later Saturday: I really, really like this one. In case anyone bites, related questions include: for ternary forms, to what extent does class number one (the 794) give these nice formulas as in Hecke eigenforms? I am capable of gathering evidence for and against that myself, I am guessing the only wiggle room is primes dividing the discriminant. In higher dimension: what can one say about higher theta series? For (positive) quaternary forms, we would be talking about counting representations of numbers, but then we would count representations of binary forms, for convenience i would have them Gauss reduced which is easy. Andrew Earnest initiated the study of quaternaries representing binaries and showed there are finitely many "2-regular" quaternaries. Meanwhile, Watson showed that class number one occurs only up to dimension 10. Quite recently, Nebe's students Lorch and Kirschmer compiled a reliable list of all class number one forms. Can we make a theory of theta series for, especially, codimension 2 representations, that includes the Hecke operator stuff, and relate that to class number one? Certainly seems to be a nice theory of such higher theta series, the key phrase seems to be "Siegel modular forms," see http://arxiv.org/abs/1202.4909 and other stuff by Schulze-Pillot, also for variety http://www.math.snu.ac.kr/~mhkim/thesis/thesis_33.pdf Sunday; upon further reflection, I suspect that Siegel's weighted average representation over a genus always obeys these sorts of laws, not something I knew. With class number one, the theta series for the form agrees with that of the genus, hence explaining one direction of this pretty well. It also justifies the prevalent practice, for even dimension 4 or larger, of identifying forms with improper ( determinant $-1$) equivalence, because they represent the same numbers (or lower dimension quadratic forms) with the same representation counts. Also, there is no real problem replacing a number $n$ with an equivalence class $N$ of lower dimensional quadratic forms/lattices; this is how Schulze-Pillot writes it. I have enough software written to experiment with quaternaries representing binaries. I am not sure about other codimension; quaternaries representing numbers is codimension 3, maybe the standard relation (from Jeremy's answer) changes form in codimension other than 2. Found a nice summary pdf by Winfried Kohnen. Don't see any explicit laws on representation counts, maybe reading between the lines. There are quite a number of relevant items by Lynne Walling REPLY [5 votes]: There are many specific questions raised in this post. I will address the main one and show that if a $Q$ is a ternary quadratic form and the theta series $\theta_{Q}$ is a Hecke eigenform, then $Q$ has class number one. The form $\theta_{Q}$ is a weight $3/2$ modular form and any such form has a decomposition as $E + H + C$. Here $E$ is an Eisenstein series whose coefficients are specified by the local densities of $Q$. The form $H$ is a linear combination of "unary theta series": forms of the shape $\sum \psi(n) n q^{tn^{2}}$, where $\psi(n)$ is an odd Dirichlet character. The form $C$ is a weight $3/2$ cusp form whose Shimura lifts are also a cusp forms. Each piece is orthogonal to the other two under the Petersson inner product. Because the Hecke operators are self-adjoint (or close to it) with respect to this inner product, we have that $\theta_{Q} | T_{p^{2}} = \lambda \theta_{Q}$ implies $E | T_{p^{2}} = \lambda E$, $H | T_{p^{2}} = \lambda H$, and $C | T_{p^{2}} = \lambda C$. By looking at the formulas for local densities, I've convinced myself that if $p$ does not divide the discriminant of $Q$, then $E$ is an eigenfunction for the half-integer weight Hecke operator $T_{p^{2}}$ with eigenvalue $p + 1 - \left(\frac{-\Delta}{p}\right)$. On the other hand, $C$ has a decomposition into Hecke eigenforms each of whose Shimura lifts are weight $2$ newforms. These Hecke eigenvalues are bounded in absolute value by $2 \sqrt{p}$. Finally, if $g(z) = \sum_{n=1}^{\infty} \psi(n) n q^{tn^{2}}$, then $g(z) | T_{p^{2}} = \psi(p) (p+1) g(z)$. It follows that since each of the pieces $E$, $H$ and $C$ must be a Hecke eigenform with the same eigenvalue $\lambda_{p}$, the Hecke eigenvalues are incompatible unless $H = C = 0$. The upshot is that if $\theta_{Q}$ is a Hecke eigenform then $H = C = 0$. Hence $\theta_{Q} = E$, which implies immediately that $Q$ is regular. Based on Will's computations for the regular ternary quadratic forms $Q$ the only $Q$ with $\theta_{Q}$ a Hecke eigenform are those $Q$ with class number one. I don't see an a priori reason why $H = C = 0$ implies class number one. It seems possible to me that there might be forms $Q$ in more variables that are regular, with class number greater than one, for which $C = 0$. (The $H$ term in the decomposition only appears for ternary forms.) I'm not sure what to expect for higher theta series - I would guess that the Hecke action is more complicated.<|endoftext|> TITLE: Boundary regularity of Dirichlet Eigenfunction on bounded domains QUESTION [5 upvotes]: Consider a bounded, connected and open subset $\Omega\subset \mathbb{R}^d$ and the Dirichlet Laplacian $-\Delta$ acting in $L^2(\Omega)$. Then we know that the eigenvalues of $-\Delta$ form an increasing sequence of eigenvalues $0<\lambda_1<\lambda_2\leq \lambda_3\leq....$ and we have a corresponding orthonormal basis of eigenfunctions in $L^2(\Omega)$. Furthermore the eigenfunctions are smooth functions on $\Omega$. If now $\Omega$ is sufficiently regular (e.g. smooth), then the eigenfunctions will vanish pointwise for all $x\in\partial \Omega$. I'm wondering if this is still true when no regularits assumptions are made? Do one has pointwise $\varphi(x)=0$, $\forall x\in\partial \Omega$ and all eigenfunctions $\varphi$? Best regards, REPLY [2 votes]: In case this helps, the answer is yes for a Lipschitz domain. From McLean's book Strongly Elliptic Systems and Boundary Integral Equations: If $\Omega$ is a Lipschitz domain, there is a bounded trace operator $\gamma\colon H^s(\Omega) \to H^{s-1/2}(\partial\Omega)$ for any $1/2 < s < 3/2$. Moreover, if $1/2 < s \leq 1$, one has $$H^s_0(\Omega) = \{u \in H^s(\Omega) : \gamma u = 0\}.$$ In particular, this implies that $H^1_0(\Omega)$ is precisely the set of $H^1(\Omega)$ functions that vanish on the boundary.<|endoftext|> TITLE: What are smallest finite images of triangle groups? QUESTION [11 upvotes]: Given integers $a, b, c$ all at least 2, I would like to identify the smallest group with an element $x$ of order $a$, element $y$ of order $b$ such that the product $yx$ has order $c$. For example, if $(a,b,c)=(2,3,7)$ the smallest group with elements $x, y, yx$ of orders $2,3,7$, respectively, is the simple group of order 168. This problem is related to triangle groups but here I want to concentrate on finite groups, especially the smallest such. This question arose in a discussion on permutation groups in a graduate algebra class. One can always create finite permutations $x, y$ with the desired properties but looking for small groups with these properties seemed to quickly become complicated. (This question may overlap with this MathOverflow question.) REPLY [8 votes]: Here's an elaboration of my comment above in the pairwise coprime case, in an even more special case (the fact that in a finite solvable group we can't have three elements of pairwise coprime orders whose product is the identity is, I believe, due to P. Hall). If one insists, as in the question, on the minimality of the order, then if $a,b,c$ are distinct primes, any finite group $G$ of minimal order subject to containing elements $x,y,z$ of respective orders $a,b,c$ such that $xyz = 1_{G}$ (equivalently, $xy = z^{-1}$), is necessarily a finite simple group. For let $N$ be a proper non-identity normal subgroup of $G.$ Then if none of $x,y,z$ is in $N,$ the minimality of $G$ is contradicted, as $xN yN zN = N$ in $G/N.$ If two of $x,y,z$ are in $N,$ then so is the third, and the minimality of $G$ is contradicted again.If (say) $x \in N,$ but neither $y$ nor $z$ lies in $N,$ we have a contradiction since $yNzN = N$ in $G/N$, but $yN$ and $zN$ have coprime orders. Hence there is no such normal subgroup $N$ and $G$ is simple. Later edit: Here is a general argument which may be relevant, and admits various generalizations: Let $a>5$ be an odd prime, $b>1$ be an odd divisor of $a-1$ and $c>1$ be an odd divisor of $a+1$ (note that $a$ is then neither a Fermat nor a Mersenne prime). Then $G = {\rm PSL}(2,a)$ (which is isomorphic to a subgroup of $S_{a+1}) $ contains three elements $x,y,z$ of respective orders $a,b,c$ with $xyz = 1.$ This is because if $u$ is an element of order $a,v$ an element of order $b$ and $w$ an element of order $c$ in $G,$ then each non-trivial irreducible character $\chi$ of $G$ vanishes at one of $u,v,w.$ Hence $\sum_{\chi \in {\rm Irr}(G)} \frac{\chi(u) \chi(v) \chi(w)}{\chi(1)}> 0.$ By a standard character-theoretic formula,this means that we may choose conjugates $x,y,z$ of $u,v,w$ respectively such that $xy = w^{-1},$ or $xyw = 1.$ It is also easy to check that $G = \langle x,y \rangle$ for any such $x,y$, since $\langle x,y \rangle$ must have more that one Sylow $a$-subgroup, so has $a+1$ Sylow $a$-subgroups and hence is all of $G.$<|endoftext|> TITLE: "Economic" CW-structure for Eilenberg-MacLane spaces? QUESTION [15 upvotes]: The only really "economic" cell structures for $K(\pi,n)$'s that I know is the one with a single cell in each dimension for $K(\mathbb Z/n\mathbb Z,1)$ and the one with a single cell in each even dimension for $K(\mathbb Z,2)$. Vaguely I remember attending a talk where some lower bounds on numbers of cells in each dimension for Eilenberg-MacLane spaces of cyclic groups were given. Now that I needed this again, I only could find the text "Small CW-models for Eilenberg-MacLane spaces" by Clemens Berger, which contains among other things a cell complex for $K(\mathbb Z/2\mathbb Z,2)$ with 1,0,1,1,2,3,5,8,13,21,... cells (probably the Fibonacci sequence) and a cell complex for $K(\mathbb Z/2\mathbb Z,3)$ with 1,0,0,1,1,2,4,7,13,24,... cells. What is the current state of the art? Are there for example any manageable dimensionwise finite cell structures for $K(\mathbb Z,3)$ or $K(\mathbb Z,4)$ or $K(\mathbb Z/n\mathbb Z,2)$ known? (By manageable I mean... well it is up to you :) ) Is there a geometric construction similar to the real/complex/quaternionic projective spaces known for any other spaces aside of $BO(1)=\mathbb R P^\infty=K(\mathbb Z/2\mathbb Z,1)$, $BU(1)=\mathbb C P^\infty=K(\mathbb Z,2)$ and $B(\textrm{unit quaternions})=\mathbb H P^\infty$? (The latter is of course not any Eilenberg-MacLane space but...) (Well there are also Grassmanians with their Schubert cells but I mean something as Eilenberg-MacLaneish as possible :) ) Are there any interesting lower bounds on the numbers of cells of each given dimension in a $K(\pi,n)$ known? And yes of course there is the whole ocean of nonabelian groups with very nice finite-dimensional classifying spaces but I mostly mean $n>1$ and, respectively, abelian groups... (added after two answers below) As Jeff Strom and Will Sawin indicate in their answers, homology groups provide the lower bounds, and it is more or less straightforward to arrange for a cell complex with prescribed homology, with minimal possible numbers of cells. Still the question remains (for me) whether such an "absolutely minimal" CW-complex with the correct homotopy type exists. REPLY [8 votes]: You can build them using a homology decomposition and read off the number of cells in each dimension from the homology groups. For any simply-connected space, this will give you a construction by iterated cofiber sequences $M_n \to X(n-1) \to X(n)$ where $M_n$ is a Moore space and $X = \mathrm{colim} X(n)$. If you construct $M_n$ efficiently and give $X(n)$ the inherited CW structure, then this will be the absolute fewest cells possible in each dimension to construct a space with the required homology. By an efficient construction of $M = M(G,n)$ where $G$ can be generated by $k$ elements but not fewer, and an exact sequence $0\to F_1\to F_0\to G\to 0$ where $F_0$ is free of rank $k$. Then we topologize this to get a cofiber sequence $W_1 \to W_0 \to M$, where $W_0$ is a wedge of $k$ $n$-spheres and $W_1$ is also a wedge of $n$-spheres. REPLY [6 votes]: Some quick observations on the paper you linked, that would not fit into a comment: (1) It mentions that the generating function of the number of cells for the $CW$ complexes they construct is always a rational function. So this always gives you at worst an exponential upper bound. Also it looks like your Fibonacci guess is correct, since that has generation a rational function. In fact, it looks like the ones they listed correspond to the sequence of rational functions $$\frac{1}{1-T}, \frac{1+T}{1+T-T^2}, \frac{1+T+T^2}{1+T+T^2-T^3}, \dots$$ (2) It mentions that the number of cells is equal to the number of pruned level-trees of height $n$. So presumably if one knew what that was, one would be able to compute the rational function and hence an upper bound. (3) The obvious place to look for a lower bound on the number of cells is the cohomology. In fact, all the ones you mention as economical meet exactly the lower bound coming from cohomology. I think the cohomology of the Eilenberg-Maclane space is fairly well-studied, so you could look there. In the paper, it mentions one computation of the cohomology ring, which with sufficient understanding of the kind of trees they're talking about should tell you a lower bound.<|endoftext|> TITLE: An inequality on representations and subgroups of general linear groups over finite field QUESTION [6 upvotes]: Let $q$ be a power of $p$, let $l$ be a prime different from $p$, and let $H_1$ and $H_2$ be two subgroups of $GL_n(\mathbb F_q)$ that are $l$-groups. If for all characteristic $0$ representations $V$ of $GL_n(\mathbb F_q)$, we have $$\dim V^{H_1} \leq \dim V^{H_2}$$ does it follow that: $$\dim \left(\mathbb F_q^n\right)^{H_1} \leq \dim \left(\mathbb F_q^n\right)^{H_2}$$? I'm thinking about a problem that would be significantly simplified if this were true, although I suspect it is false. I would do a large finite search for counterexamples before asking about this question, but I don't know a way to efficiently search for counterexamples. REPLY [6 votes]: The following is a series of counterexamples: Let $q$ be a prime power with $q\equiv1\pmod{8}$, and pick $\omega\in\mathbb F_q^\star$ with order $8$. Set $G=\text{GL}_2(\mathbb F_q)$, and let $H_1$ and $H_2$ be the cyclic subgroups generated by $\begin{pmatrix}\omega & 0\\0 & 1\end{pmatrix}$ and $\begin{pmatrix}\omega^2 & 0\\0 & \omega^4\end{pmatrix}$, respectively. Of course, the second inequality does not hold. The characteristic $0$ inequalities can be verified by the explicit form of the character table of $G$, as given for instance in Theorem 28.5 in this book by James and Liebeck (see also this write up for another readable account). Note that if $m$ is a multiple of the order of $g\in G$, then the dimension of the fixed space of $g$ in the representation afforded by $\chi$ is $\frac{1}{m}\sum_{k=1}^{m}\chi(g^k)$. Using this and a simple computation yields the claim.<|endoftext|> TITLE: Why is the supersingular locus the zero locus of a modular form? QUESTION [5 upvotes]: This question is related to my other question here: Examples of subspaces singled out by modular forms. Here I am wondering if there is a philosophical explanation about why the supersingular locus on the modular curve mod p is the zero locus of a modular form. Even better, are there sufficient conditions ensuring that a subset of the modular curve is the zero locus of a modular form? Thank you for your help. REPLY [3 votes]: I'm not sure I buy Felipe Voloch's answer, since there are many choices for his modular form; however, there is a natural modular form cutting out the supersingular locus. This is something special about elliptic curves; in general (e.g. for surfaces) there is not a known scheme structure on the supersingular locus, at least to my knowledge. Let $X=X(N)$ be a modular curve with $N$ prime to the characteristic $p$, and let $\pi: E\to X$ be the universal curve. Then there is a natural map of line bundles $$f: R^1\pi_*F^*\mathcal{O}_E\to R^1\pi_*\mathcal{O}_E$$ where $F$ is Frobenius. By cohomology and base change, the supersingular locus is exactly where this map vanishes (that is, an elliptic curve is supersingular if Frobenius acting on $H^1(\mathcal{O})$ is $0$). Now $\mathcal{O}_E\simeq T_{E/X}$ (slightly non-canonically) using the group structure on $E$, so $R^1\pi_*\mathcal{O}_E$ is $T_X$ (using the moduli description of $X$; $H^1(T_E)$ classifies deformations of $E$). Similarly $$R^1\pi_*F^*\mathcal{O}_E=F^*R^1\pi_*\mathcal{O}_E=T_X^{\otimes p}.$$ So we may view $f$ as a map $$f: T^{\otimes p}_X\to T_X,$$ or in other words a section to $\omega_X^{p-1}$, also known as a modular form of weight $2p-2$. Hopefully I got that all right; if not, perhaps an expert can correct me.<|endoftext|> TITLE: Maximal compact subgroup of p-adic lie groups QUESTION [11 upvotes]: Let $k$ be a number field and $S$ be a finite set of places of $k$. Let $G$ be a connected semisimple algebraic group over $k$. Let $k_S=\prod_{v\in S}k_v$ where $k_v$ is the completion of $k$ at $v$. Question: Is maximal compact subgroup of $G(k_S)$ unique up to conjugation? If it is not unique, are there finitely many of them up to conjugation? REPLY [10 votes]: Since the maximal compact subgroup question has a complicated history, and is treated at very different levels of generality in the literature (Bruhat-Tits papers in particular), it may be helpful to fill in Paul's answer a bit. There was early work in special cases by Bruhat over half a century ago, in the aftermath of Chevalley's uniform 1955 construction of split groups over an arbitrary field coming from simple Lie algebras over $\mathbb{C}$. But the clearest picture began to emerge from the important 1965 paper by Iwahori and Matsumoto (see in particular their Prop. 2.32): here. In this approach and the further work of Bruhat-Tits one considers in particular a simple, simply connected algebraic group $G$ such as $\mathrm{SL}_n$ defined over a complete non-archimedian field $K$, obtaining a $(B,N)$-pair structure and Bruhat decomposition which leads eventually to a determination of the conjugacy classes of maximal compact subgroups of $G(K)$ (which are maximal "parahoric" subgroups): the number of these is $\ell +1$, where $\ell$ is the $K$-rank of $G$. The minimal "parahoric" subgroups are themselves all conjugate. Here the usual Weyl group is expanded to an affine (or extended affine) Weyl group. The later papers by Bruhat and Tits develop such ideas in vast generality, but as early as 1966 their announcements of results show clearly the direction in which they were going. To state the technical results for $G(k_S)$ in the question here takes some care, but the basic example cited by Paul shows how the $\ell+1$ arises (the rank in his split example being $n-1$). As Paul indicates, a direct computation can be done in the smallest case $\mathrm{SL}_2$. See for example the end of $\S15$ in my old Springer Lecture Notes 789 on arithmetic groups, where the Bruhat-Tits building appears as simply a tree and Serre's ideas about groups acting on trees can be used.<|endoftext|> TITLE: Different approaches to forcing QUESTION [8 upvotes]: There are many different approaches to the forcing method, and I am looking for all known such approaches. So my question is: Question 1. Which different approaches to set theoretic forcing are available, and who first introduced them. Giving (original) references for each approach is appreciated. On the other hand it seems the approach to forcing using posets is more comfortable in forcing arguments. So my second question is: Question 2. What are the benefits of using each of these approaches with respect to other approaches? REPLY [9 votes]: At the risk of making things worse rather than better, let me point out that what appears to be one approach, say the Scott-Solovay version using complete Boolean algebras, is really a whole family of approaches, according to what universes you extend and how you extend them. Once you've defined a particular Boolean algebra $B$, you can do (at least) any of the following: (1) Build a cumulative hierarchy of $B$-valued sets, extending the usual cumulative hierarchy of ordinary (2-valued) sets. (2) Give a syntactic $B$-valued interpretation of ZFC in ZFC. [Intuitively, this is "the same" as (1), but technically it's a way of getting finitary proofs of relative consistency.] (3) Start with a countable transitive model $M$ of ZFC, interpret the definition of $B$ within $M$ to get an $M$-complete Boolean algebra $B^M\in M$, choose an $M$-generic ultrafilter $G$ in $B^M$, and form the countable transitive model $M[G]$, the smallest transitive model that includes $M$ and contains $G$. (4) Start with any model $M$ of ZFC (not necessarily well-founded), interpret the definition of $B$ in $M$ to get $B^M$, choose any ultrafilter $G$ in $B^M$ (not necessarily generic), and use it to form a 2-valued quotient $M^{(B^M)}/G$ of the Boolean-valued model $M^{(B^M)}$.<|endoftext|> TITLE: Idea and intuition behind Penrose transform QUESTION [8 upvotes]: I would like to know what a Penrose transform is, or more precisely, what is it intended to be - I'm interested in ideas, intuition and some examples of application. My knowledge of differential geometry and homological algebra are basic (at the moment, I am working on a graduation paper on derived categories), so please be rather broader than technically specific in the answer if possible. REPLY [5 votes]: well, if you are interested in developing some basic intuition on the Penrose transform, you could go back to Harry Bateman's 1904 paper The solution of partial differential equations by means of definite integrals, where he derived what is essentially the same representation of a harmonic function $\phi(w,x,y,z)$ of four variables as a contour integral of a holomorphic function $f(p,q,\zeta)$ of three complex variables: $$\phi(w,x,y,z)=\oint f[w+ix+(iy+z)\zeta,iy-z+(w-ix)\zeta,\zeta]\,d\zeta$$ An instructive geometric interpretation of Bateman's formula, which connects it to Penrose's work, is given by Michael Eastwood in his "Introduction to Penrose Transform". (You can read most of it at books.google.com and find a brief summary here.) It is also interesting to read Roger Penrose's own account of the relation between Bateman's formula and the Penrose transform.<|endoftext|> TITLE: Failing of heuristics from circle method QUESTION [12 upvotes]: The heuristic from circle method for integral points on diagonal cubic surfaces $x^3+y^3+z^3=a$ ($a$ is a cubic-free integer) seems to fit well with numerical computations by ANDREAS-STEPHAN ELSENHANS AND JORG JAHNEL. The only known exception is the surface $x^3+y^3+z^3=2$. Circle method predicts that the number of integral points $(x,y,z)$ with $\max(\vert x\vert,\vert y \vert,\vert z\vert)1$ should imply that every sufficiently large integer $n$ is either representable as $n=x^2+y^2+z^4$, or there is some local constraint, i.e. a modulus $q$ such that $n\equiv x^2+y^2+z^4\pmod{q}$ is unsolvable. This theorem is one of those cases where formulating the theorem is much more difficult than proving it. Write $x^2+y^2+z^4=p^2$ as $x^2+y^2=(p-z^2)(p+z^2)$. Now $p-z^2$ is $3\pmod{4}$ or $6\pmod{8}$, hence divisible by some prime $q\equiv 3\pmod{4}$ to the first power. But a sum of two squares cannot be divisible by such a prime with odd exponent, thus $q$ divides $p+z^2$ as well. Now $q|(p-z^2, p+z^2)$, thus $q$ divides $2p$ and $2z^2$, hence $p=q$ and $p|z$. But then $z^4$ is larger than $p^2$.<|endoftext|> TITLE: Decomposition vs filtration vs stratification QUESTION [5 upvotes]: Are there accepted/standard definitions of "decomposition", "filtration", and "stratification" of a topological space (or of a manifold, or of an algebraic variety) $X$? I tend to understand "decomposition" simply as the partition of $X$ into disjoint subspaces, perhaps with some qualification (e.g. "locally closed decomposition"). I understand "filtration" as a realization of $X$ as an increasing/decreasing (perhaps required to be finite) union of closed subspaces. It's not very clear to me what a "stratification" (whithout any further specification) is supposed to be; I think in many case it could be useful to define it as a decomposition such that the collection of the unions of the "pieces" of index $\leq i$ form a filtration (by closed subspaces). Are there contexts in which the definitions differ significantly from what I've suggested? REPLY [5 votes]: At the risk of sounding (oxy?)moronic, I'd say that the term "stratification" is locally standard. Meaning, there exist (at least) three communities which agree internally on what the term means, but these definitions are not consistent across these communities. When you see the term "stratification of $X$" without further qualifications (have you ever?) it is most likely to be one of these three. Filtrations In each case, one first needs to define a "filtration" of $X$, which the category theory types will want to define as a functor. Perhaps this is not a bad idea: the indexing category $I$ should be a well-ordered set (usually $\mathbb{N}$ or $\mathbb{R}$), and the target $T(X)$ is a suitable subcategory of the category of subsets of $X$ (always ordered by inclusion). Here there are many choices: if $X$ is a smooth manifold, maybe you'd like the target category to be smooth sub-manifolds. But maybe $X$ is any old topological space and you only care about the open sets, or only the closed sets, etc. So already, there are tons of choices, and even when $X$ is a manifold it is not clear whether a filtration of $X$ must only involve closed sub-manifolds. So, a filtration of $X$ is a functor $F: I \to T(X)$ whose colimit equals $X$. Whenever $i \leq j$ we get at least an inclusion $F(i) \hookrightarrow F(j)$ with whatever additional structure you've decided to tack on into $T(X)$ (for instance, it is typical but not universal to assume that these inclusions are cofibrations). And the choice of $T(X)$ is not even standard when $X$ is a compact manifold (closed subsets, open subsets or both?). If $X$ is a simplicial complex, then of course things are more straightforward because sub-complexes must be closed. In general people are careful to say things like "let blah be a filtration of $X$ by closed smooth sub-manifolds" or some such phrase. Stratifications A stratification is a filtration $F:I \to T(X)$ where one imposes some additional regularity on the complements $F(j) \setminus F(i)$ whenever $j \geq i$. You might ask each one to be a smooth or PL sub-manifold of $X$ for instance. Here are the three types of stratifications that you're most likely to encounter, in increasing order of generality and (therefore?) decreasing order of popularity. Whitney Stratifications: If $X$ is a smooth variety, then typically a stratification of $X$ is a Whitney stratification. Here, the indexing category if $\mathbb{N}$ and each $F(i+1) \setminus F(i)$ is a smooth manifold subject to Whitney's Condition B which you can find in Qiaochu's Wikipedia link (see comment under the question). Thom-Mather Stratifications: If $X$ is a space which shows up in singularity theory or intersection homology, then its stratifications are likely to be Thom-Mather. The precise definition, as you can see here, is wicked since it inducts on dimension. Quinn Stratifications: I've only ever seen these in the surgery theory literature (eg work of Weinberger, Cappell-Shaneson etc). Weinberger's book in particular contains a nice exposition. Here's the basic idea: for each pair of topological spaces $Z \subset Y$, we let $[Y,Z]$ be the class of stratified maps from $([0,1],\{0\})$ to $(Y,Z)$. There is an obvious map $[Y,Z] \to Z$ given by evaluating at zero. The Quinn conditions ask that each such map $[F(i) \cup F(j), F(i)] \to F(i)$ is a fibration whenever $i \leq j$.<|endoftext|> TITLE: Singular points of algebraic varieties and parametrization by Puiseux series QUESTION [8 upvotes]: Let $V\subset \mathbb{R}^n$ (or $\mathbb{C}^n$ if that makes anything easier) be an algebraic variety and $p\in V$ a possibly singular point. Let $U\subset V$ be a sufficiently small neighborhood of $p$ and $q\in U$ a nearby point. Can every such point $q$ be reached by a curve $\gamma\colon [0,1] \to U\subset \mathbb{R}^n$, where $\gamma(0) = p$, $\gamma(1) = q$ and $\gamma(t) = \gamma_1 t + \gamma_2 t^2 + \cdots$ is a convergent Taylor series with $\gamma_1 \ne 0$? I have a feeling that the answer might be No, because some $q$ may only be reachable if fractional powers of $t$ are included along with integer powers. In that case, for a given $q$ (as above) does there always exist an integer $r>0$ and a convergent Puiseux series $\gamma(t) = \sum_{k=r}^\infty \gamma_k t^{k/r}$ with $\gamma_r \ne 0$ such that $\gamma(0) = p$, $\gamma(1) = q$ and $\gamma([0,1]) \subset U$? If so, is it possible to express the integer $r$ (for a given $q\in U$ or the maximal such integer over all $q\in U$) in terms of algebraic invariants associated to the variety $V$ or the tangent cone of $V$ and $p$? Also, with point $q$ and leading coefficient $\gamma_r$ fixed, are the higher order coefficients $\gamma_k$ ($k>r$) allowed to vary arbitrarily or perhaps must they belong to some special subsets of $\mathbb{R}^n$? If so, how to characterize these subsets? Please note that my background is not in algebraic geometry, so apologies if this question is elementary. However, I've not been able to find it, at least not posed in this form, after scanning through some textbooks. My interest in it comes from the same context as this question. REPLY [4 votes]: (This is essentially a very long comment...) What is r? If your original $V$ is a curve, or alternatively after (as per Filipe) cutting your space down to a curve by taking hyperplane sections, then one has recourse to the following basic fact of algebraic geometry normalization of curves: given any (possibly singular) curve algebraic curve $C \subset \mathbb{C}^n$, there is a unique map $\nu: C^\nu \to C$ such that $C^\nu$ is smooth, $\nu$ is surjective, and is a bijection away from the singularities of $C$. So locally at some $p^\nu \in C^\nu$ mapping to your point of interest $p \in C$ you can choose a coordinate $z$ and write $$\nu(z) = (\nu_1(z), \cdots, \nu_n(z)) = (z^{a_1} + \ldots, z^{a_2} + \ldots, z^{a_3} + \ldots)$$ Filipe was explaining above that if you wanted something with a leading linear term, you have no choice but to write $z^{\mathrm{min}(a_i)} = w$. Note that this minimum doesn't depend on the coordinate $z$ you chose. But what is this minimum? If $V$ is a unibranch* curve, then $r = \mathrm{mult}_p(V)$, the multiplicity of $V$ at $p$. The multiplicity of a variety at a point has various characterizations. If $V$ were cut out by one equation and $p=0$, then it's just the degree of the minimal degree monomial in that equation. In general, it's The number of intersections, near $p$, of $V$ with a general hyperplane passing near $p$ The same for the tangent cone of $V$ at $p$ The number of copies of the exceptional divisor when you blowup $V$ at $p$ Take $\mathcal{O}_p, \mathfrak{m}_p$ to be the local (or complete local) ring of functions of $V$ at $p$, and its maximal ideal. Then the quantity $HS(n) = \mathrm{dim}_\mathbb{C} \mathcal{O}_p / \mathfrak{m}_p^{n+1}$ is called the Hilbert-Samuel function; it's eventually a polynomial in $n$ with leading term $n^{\mathrm{dim}(V)}/\mathrm{dim}(V)!$ times the multiplicity. We can use that last characterization to see $\mathrm{mult}_p(C) = \mathrm{min}(a_i)$: the point is that the completed ring of functions $\mathcal{O}_p$ is the subring of $\mathbb{C}[[z]]$ generated by the $\nu_i(z)$. One sees from the fact that the normalization is generically 1-1 that $\mathrm{gcd}(a_i) = 1$, from which it follows that sufficiently high powers $\mathfrak{m}_p^N$ will just be $z^{N \cdot \mathrm{min}(a_i)} \mathbb{C}[[z]]$. What is this word unibranch? Well, when you normalize a curve, the preimage of the point $p$ might be several points $p^\nu_i$, the neighborhood of each of which maps to a different analytically locally irreducible component of the (reducible) analytic local curve $C$. One would like to say: just treat each local irreducible component separately. However one could for example have a connected rational curve with a unique singularity $p$ such that $C$ at $p$ is analytically locally of the form $x(y^2- x^3) = 0$. This has two branches, one smooth ($x = 0$), and the other singular ($y^2 - x^3 = 0$). For any point on the curve, including points arbitrarily close to $p$ along the singular branch, one can get to them by traveling only along the smooth branch. Thus for this curve, the minimal $r$ is $1$, despite the presence of the singularity. You demanded your paths stay in a small open set $U$, which would fix the problem if $V$ is a curve. But I don't see why it couldn't reappear for higher dimensional $V$ with the property that for any neighborhood $U$ of $p$, there exists some slice of $V$ through $p$ (depending on the neighborhood) such that this slice is a curve with an analytically locally reducible singularity at $p$ while however the branches meet at a smooth point inside $U$. I suspect this means you've asked slightly the wrong question, and should demand something like that your paths `never travel too far away from the shortest path from $p$ to $q$', but I'm not exactly sure what's the best way to make that precise. What if I just want to bound $r$? Then Filipe's answer tells you the following: $r$ is less than the maximal multiplicity of any curve obtained by cutting the variety by an appropriate-dimensional linear slice. (I see no a-priori reason to believe that linear slices give better bounds than some other kind of slices, but also as mentioned above I don't know how to correctly formalize the question so that `what is $r$' actually makes sense.) I do not know the name of this number -- i.e., the maximal multiplicity of a linear-slice-down-to-a-curve -- in algebraic geometry. A related thing which does have a name is the answer to this question for a general $q$ near $p$, i.e. the multiplicity of a general linear-slice-down-to-a-curve. Such a curve is called a local polar curve and its multiplicity is called a local polar multiplicity; people study these things. I do not think either of these things can be computed from the tangent cone.<|endoftext|> TITLE: Is a Galois extension of the rationals determined by its set of completely split primes? QUESTION [8 upvotes]: apologies if this is a naive question. Consider two Galois extensions, K and L, of the rational numbers. For each extension, consider the set of rational primes that split completely in the extensions, say Split(K) and Split(L). If Split(K) = Split(L), then is it necessarily true that K and L are isomorphic as Galois extensions of the rationals? If so, for a given set of rational primes, S, is there a way to construct the extension over which S is the set of completely split primes? References welcomed! Thanks, Martin REPLY [16 votes]: This result is due to Bauer and dated to 1916: $\textbf{Theorem}:$ Let $K$ be an algebraic number field, $L/K$ and $M/K$ finite Galois extensions, and $\text{Spl}(L/K)$, $\text{Spl}(M/K)$ the set of prime ideals of $K$ which split completely in $L$ and $M$, respectively. Then $L \subseteq M$ if and only if $\text{Spl}(M/K) \subseteq \text{Spl}(L/K)$. The case of $K=\mathbb{Q}$ answers your first question in the affirmative. As for the second question, not every subset of rational primes is $\text{Spl}(K/\mathbb{Q})$ for a number field $K$, simply by cardinality arguments. Class field theory gives a description of $\text{Spl}(L/K)$ when $\text{Gal}(L/K)$ is abelian, but the general problem is open and very hard. An excellent source for this material is Keith Conrad's notes on the history of class field theory, which you can find under the Expository Notes section on his website here: http://www.math.uconn.edu/~kconrad<|endoftext|> TITLE: Nearby matrices have nearby leading eigenvectors? QUESTION [10 upvotes]: Suppose I have a symmetric positive semidefinite matrix $A$ with leading eigenvalue $1$ of multiplicity $1$ and remaining eigenvalues $\leq\epsilon$. I am told that another symmetric positive semidefinite matrix $B$ is close in some sense to $A$. I wish to conclude that the leading eigenvector of $B$ is therefore close in some sense to that of $A$. Question: What notions of closeness correspond to existing theorems of this sort? One solution: Let $a$ and $b$ denote leading eigenvectors of $d\times d$ matrices $A$ and $B$, respectively, scaled so that $\|a\|_2^2=1$ and $\|b\|_2^2$ is the leading eigenvalue of $B$. Then by triangle, Eckart–Young–Mirsky, and triangle again, we have $$\|bb^\top-aa^\top\|_F\leq\|bb^\top-B\|_F+\|B-aa^\top\|_F\leq2\|B-aa^\top\|_F\leq2\big(\|B-A\|_F+\|A-aa^\top\|_F\big)\leq2\|B-A\|_F+2\sqrt{d-1}\epsilon.$$ For this attempt, I don't like the additive loss in $\epsilon$, and I doubt it's necessary. Can I get better performance with a different norm? Perhaps the spectral norm? REPLY [4 votes]: The spectral projection of $A$ for eigenvalue $1$ can be realized as $P_A = \dfrac{1}{2\pi i} \oint_{\Gamma} (z I - A)^{-1} \; dz$ where $\Gamma$ is a contour that encloses $1$ but none of the other eigenvalues. In particular, take $\Gamma$ to be a circle of radius $r$ centred at $1$, where $r = (1-\epsilon)/2$. For $z \in \Gamma$ we have (using the $\ell^2$ operator norm) $\|(z I - A)^{-1}\| \le 1/r$. Then if $\|B - A\| < r$, $z I - B$ is invertible for $z \in \Gamma$, with $$\|(z I - B)^{-1} - (zI-A)^{-1} \| \le \dfrac{\|(z I - A)^{-1}\|^2 \|B-A\|}{1 - \|B - A\| \|(zI - A)^{-1}\|} \le \dfrac{\|B-A\|}{r(r - \|B-A\|)}$$ and $P_B = \dfrac{1}{2\pi i} \oint_{\Gamma} (z I - B)^{-1} \; dz$ is a spectral projection of rank $1$ with $\|P_B - P_A\| \le \dfrac{\|B-A\|}{r - \|B-A\|}$.<|endoftext|> TITLE: Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)? QUESTION [8 upvotes]: Let $G$ be a $n\times n-$symmetric matrix with integral coefficients and determinant $1$ (i.e. unimodular) such that the associated quadratic form is positive-definite. I am interested in having an algorithm to find a rational basis of a lattice $L$ such that $G$ is the Gram matrix of $L$. Concretely, this consists in finding a square matrix $M$ (with rational coefficients) such that $G$ factors as $^tM \cdot M$. Note that: 1) The unimodular condition and Hasse-Minkowski theory predict that this is indeed possible (this is a remark in Serre's course in arithmetics, §1.3.6 p. 86 in the French edition). 2) I demand the matrix $M$ to be square. (For non-square $M$, this is much easier: first by the Gram-Schmidt process, one can assume that $M$ is diagonal (with positive rational entries); then write each diagonal entry as a sums of (at most 4) squares). REPLY [3 votes]: Here is a method. Let $\mathrm{V}$ be the bilinear space $(\mathbf Q^n, b)$ with Gram Matrix $G$ in the canonical basis. Let $M_0$ be the lattice $\mathbf Z^n$ on $\mathbf Q^n$. Find a primitive vector $v_0\in M_0$ such that $b(v_0,v_0)$ is a square, say $b(v_0,v_0)=m^2$ in $M_0$ (note that this can be done quite efficiently : use LLL to get a basis of vectors with small norms, take the two* first vectors, study the lattice they generate). ( * Edit 1: in fact you need to take the first 4 vectors ... indeed, there exists (many) $3\times 3$ positive definite Gram matrices that don't represent any square over $\mathbf Q$ ... The simplest example is $7.I_3$.) ( Edit 2: If you don't use this trick, you can arrange so as to find a $v_0$ such that $m=2^s$ ... this will allow you to find an orthonormal basis inside $\mathbf Z[\frac 12]^n$, whose existence is predicted by the Strong Approximation Theorem.) Let $K_1$ be the lattice made of vectors $w\in M_0$ such that $(v,w)\equiv 0$ mod. $m$. Let $L_1$ be the lattice $K_1+\mathbf Z\cdot\frac{v_0}{m}$. Then $L_1$ is integral, unimodular, and has a vector of length $1$, namely $w_0:=\frac{v_0}{m}$. Let $M_1$ be the orthogonal complement of $w_0$ in $L_1$. Then $M_1$ is an $n-1$-dimensional unimodular lattice (contained in the $n-1$-dimensional subspace $^\perp$ of $\mathrm{V}$). Go on to find $v_1\in M_1$, construct $M_2$ and so on ... you're done (the matrix $M$ you were looking for is the inverse of the matrix whose columns are $(w_0,\dots,w_{n-1})$. [The classification of unimodular lattices shows that at least when $n\leq 7$, you can always arrange so as to have $(v_0,v_0)=1$ (so that $M_1\subset M_0$), and for $n\leq 24$, you can find some $v_0$ with $(v_0,v_0)=4$.]<|endoftext|> TITLE: In what sense is the classification of all finite groups "impossible"? QUESTION [25 upvotes]: I think there is a general belief that the classification of all finite groups is "impossible". I would like to know if this claim can be made more precise in any way. For instance, if there is a subproblem of the classification problem that is already equivalent to an already agreed-upon wild problem. REPLY [16 votes]: One can make the argument by wildness much more concrete than in the previous answer: Sergeichuk ["Classification of metabelian p-groups", in: Matrix problems, Inst. Mat. Ukrain. Akad. Nauk, Kiev, 1977, pp. 150-161, in Russian] showed that isomorphism of 2-step nilpotent p-groups is already wild (over $\mathbb{F}_p$), whenever the center is not cyclic of order p. To answer David Harden's question from the comments, which also gets at Timothy Chow's point about computational complexity: simultaneous conjugacy of k-tuples of matrices can be solved in polynomial time [Sergeichuk; Brooksbank-Luks; Chistov-Ivanyos-Karpinski] (or, I believe, even in $\mathsf{NC}$, depending on the field and model of computation; an even simpler algorithm puts it in $\mathsf{RNC}$). However, the problem of isomorphism of 2-step nilpotent $p$-groups is Tensor Isomorphism-complete G-Qiao. Another TI-complete problem is conjugacy of $k$-dimensional subspaces of matrices. Belitskii and Sergeichuk showed that the latter classification problem is strictly harder than $k$-tuple conjugacy (that is, subspace conjugacy contains k-tuple conjugacy but not conversely). When the subspaces are given by a spanning set, subspace conjugacy is at least as hard as Graph Isomorphism [Chapter 4 here contains a freely available and more complete version] (that is, Graph Iso reduces to subspace conjugacy in polynomial time), which is not known to be in $\mathsf{P}$. It is even harder than Code Equivalence (which itself is thought to be harder than Graph Iso).<|endoftext|> TITLE: Amenability as a geometric property QUESTION [6 upvotes]: Let $G$ be a discrete, finitely generated, and amenable group. Let $H$ be a group which is quasi- isometric to $G$. Is $H$ amenable? REPLY [11 votes]: Yes. This is essentially an immediate consequence of the Folner sets definition of amenability. You can find references at Theorem 10.23 of the article by Ghys and de la Harpe, "Infinite groups as geometric objects".<|endoftext|> TITLE: Naturality of a Kunneth formula for cohomology QUESTION [8 upvotes]: Let $X,Y$ be CW complexes. By Kunneth formula, we have a group isomorphim $$ H^n(X\times Y;G) \cong \oplus_{p+q=n} H^p(X;H^q(Y;G))$$ Is there a natural map realizing this isomorphism? REPLY [8 votes]: I came across this question in my 1961 DPhil Thesis; this was written up in two papers which are available from my Publication List, [3]. ``Cohomology with chains as coefficients'', Proc. London Math. Soc. (3) 14 (1964), 545-565. [4]. ``On K\"{u}nneth suspensions'', Proc. Camb. Phil. Soc. 60 (1964) 713-720. It is shown in the first paper that the isomorphism could be chosen to be natural with respect to maps of $X$ but not with respect to maps of either $Y$ or of $G$. The naturality was important for the second paper, which was poorly titled: it should have been something like "$k$-invariants of function spaces". Paper [3] also contains as an Appendix formulae for this isomorphism in some special cases The aim of this work was to investigate the Postnikov system of $Y^X$ by induction on the Postnikov system of $Y$. Michael Barratt suggested this as another way of determining in some cases the extensions involved in his work on exact sequences of track groups; such a determination in his paper "Track Groups II" used Whitney tube systems! .<|endoftext|> TITLE: A method for making a graph bipartite QUESTION [16 upvotes]: Given any graph $G$, can we find a bipartite subgraph of $G$ with at least $e(G)/2$ edges ($e(G)$ is the number of edges in $G$) by sequentially deleting the edge belonging to the most number of odd cycles? That is, if we take a graph and sequentially delete the edge which belongs to the most odd cycles until we have a bipartite graph, will at least half the edges remain when the graph is bipartite? The motivation is to find a a simple method for making a graph bipartite making use of the odd cycle-free interpretation of bipartiteness rather than the $2$ color class viewpoint. Note: I asked this question on M.SE. I expected I was missing some simple argument or counterexample, but no one has given an answer so perhaps it is more tricky: https://math.stackexchange.com/questions/913366/a-method-of-making-a-graph-bipartite Edit: clarity REPLY [21 votes]: A quick check with Mathematica seems to suggest that a (smallest) counterexample is given by a graph with 6 vertices and 13 edges: Here the red edges are contained in the most odd cycles (16, 12, 6, 5, 2, 2, 1, 0 respectively) and one possible sequence of edge removals is shown. However, choosing different edges one can end up with a bipartite graph with 7 edges. Edit: Any sequence of edge removals of the following graph with 8 vertices and 18 edges seems to result in a bipartite graph with at most 8 edges: Edit: For those interested: here's the Mathematica code to check that the last graph is really a counterexample. Edit: The first complete graph for which the algorithm fails is the complete graph on 10 vertices. Indeed, it seems that the largest bipartite graph one can obtain from an $n$-vertex ($n>4$) complete graph is the complete bipartite graph on $(n-3)+3$ vertices, which has $3(n-3)$ edges, i.e. smaller than $n(n-1)/4$ for $n\geq 10$.<|endoftext|> TITLE: Counting cyclic binary sequences of length $n$ where ones appear in blocks of length at least $k$ QUESTION [6 upvotes]: How many binary cyclic sequences of length $n$ exist, where ones only appear in blocks of length at least $k$? We do not consider sequences that result from each other by a cyclic shift equivalent. Example: Let $n=6$ and $k=2$, i.e. we have no isolated one. Then [0,0,0,0,0,0] [1,1,0,0,0,0] and 5 cyclic shifts of it [1,1,1,0,0,0] and 5 cyclic shifts of it [1,1,1,1,0,0] and 5 cyclic shifts of it [1,1,0,1,1,0] and 2 cyclic shifts of it [1,1,1,1,1,0] and 5 cyclic shifts of it [1,1,1,1,1,1] are all of the 29 possible sequences. The case $k=2$ is covered by https://oeis.org/A109377 REPLY [6 votes]: Consider the monoid $M_k$ in the generators $A_k=\lbrace 0, 01^k, 01^{k+1}, 01^{k+2},...\rbrace$, where $1^r$ denotes a string of $r$ ones. This contains all binary sequences with no strings of 1's of length less than $k$, except for the words $1^r$ for $r\geq k$. The monoid $M_k$ is freely generated by $A_k$. Hence if $$ F_k(x) = x+x^{k+1}+x^{k+2}+\cdots = x+\frac{x^{k+1}}{1-x}, $$ then the generating function for the number of words of length $n$ in $M_k$ is $$ G_k(x) = \frac{1}{1-F_k(x)}. $$ It is easy to check that $M_k$ is in fact very pure in the sense of Section 4.7.4 of Enumerative Combinatorics, vol. 1, second ed. Thus by Proposition 4.7.13, the generating function for cyclic words is \begin{eqnarray*} H_k(x) & = &\frac{xF'_k(x)}{1-F_k(x)}\\ & = & \frac{x(1-2x+x^2+(k+1)x^k-kx^{k+1})}{(1-x)(1-2x+x^2-x^{k+1})}. \end{eqnarray*} To account for the words $1^r$, we merely need to add $x^k/(1-x)$ to $H_k(x)$.<|endoftext|> TITLE: $L^p$ norm means QUESTION [12 upvotes]: Consider the unit sphere $S_p^{n-1}$ of an $L^p$ normin $\mathbb{R}^n.$ The question is: what is the expected value of the $L^q$ norm on $S_p^{n-1}?$ Since (I assume) this is intractable in closed form, what are the asymptotics in $q, n$ (for $p$ fixed)? REPLY [7 votes]: For $p=2$: up to multiplicative universal constants, the average $M$ of $\|\cdot\|_q$ over $S^{n-1}$ is equal to $M \simeq \sqrt{q} \cdot n^{1/q-1/2}$ when $1 \leq q \leq \log n$, $M \simeq \sqrt{\log n}/\sqrt{n}$ for $q \geq \log n$. This can be checked most easily after switching to a Gaussian integral as Yemon mentions. For the lower bound in 1, it may be useful to consider using concentration of measure. If it is a matter of reference, it can probably be extracted from Chapter 5.4 in Milman-Schechtman, "Asymptotic theory of finite-dimensional normed spaces". Indeed, the value of this average is closely related to the dimension of almost Euclidean sections of the space $\ell_q^n$. Edit: let me add more detail. First, (this is true for any norm of $\mathbb{R}^n$, just by rotational invariance of the Gaussian measure $\gamma_n$), we have $$ M = \frac{1}{\alpha_n} \int_{\mathbb{R}^n} \|x\|_q \, \mathrm{d} \gamma_n(x), $$ where $$\alpha_n = \int_{\mathbb{R}^n} \|x\|_2 \, \mathrm{d} \gamma_n(x) $$ is a constant very close to $\sqrt{n}$. Now write $$ M \leq \frac{1}{\alpha_n} \left(\int_{\mathbb{R}^n} \|x\|^q_q \, \mathrm{d} \gamma_n(x) \right)^{1/q} \simeq \sqrt{q} \cdot n^{1/q-1/2} $$ (use the fact the $L^q$ norm of a standard Gaussian variable is or order $\sqrt{q}$). This upper bound is sharp when $q \leq \log n$, this follows from concentration of measure. Finally for $q \geq \log n$, the norms $\|\cdot\|_q$ and $\|\cdot\|_{\infty}$ are equivalent, and the question reduces to estimating the expected maximum of $n$ i.i.d. standard Gaussian variables.<|endoftext|> TITLE: History of the analytic class number formula QUESTION [17 upvotes]: The (general) analytic class number formula gives a value for the residue of the Dedekind zeta function of a number field at the point $s=1$ (or, as I prefer, the leading Taylor coefficient at $s=0$). To whom should this formula be attributed? My usual go-to place for such history questions is Narkiewicz's book Elementary and analytic theory of algebraic numbers, but it only discusses the abelian case. It lists references as late as 1929 for the result in the abelian case. Does one have to wait for Artin or Hasse? REPLY [10 votes]: KConrad's answer is correct, and the analytic class number formula is due to Dedekind. Yet the whole story is a little bit more complex and it is fair to say that Dedekind's analytic class number formula is essentially due to Dirichlet and Kummer, in the following sense. Dirichlet (who had already proved the class number formula for quadratic forms in his proof on the theorem of primes in arithmetic progression) determined the number of equivalence classes of forms that split into linear factors over ${\mathbb Q}(\zeta_p)$ (a key idea of which he had during a mass in the Sistine chapel), but did not publish his results as he knew that Kummer was working out a similar formula using his new language of ideal numbers. The only reason why Kummer did not (and could not) prove the general class number formula was that his theory of ideal numbers did not seem to work in general number rings. What Kummer was missing was the definition of an algebraic integer, and at this point Dedekind enters the story. He worked out a theory of ideals in algebraic rings of integers; transferring Kummer's analytic class number formula to the general case did not require any really new idea. In no way should these comments belittle Dedekind's contributions to algebraic number theory.<|endoftext|> TITLE: Proving that the Jones polynomial is q-holonomic QUESTION [8 upvotes]: The Jones polynomial is known to have many different interpretations or definitions, by now. There are connections with QFT, quantum groups, Hilbert schemes, Cherednik algebras, etc. My question is whether any of these approaches provide a new proof of the fact that the colored Jones polynomials of a knot satisfy a linear recurrence relation with coefficients that are polynomials in $q^n$. This is a theorem of Garoufalidis and Le, in their paper "The colored Jones function is q-holonomic". My understanding is that some of these approaches mentioned above are a lot more understood for the special case of torus knots, and i would be fine restricting to this special class. Are there any more proofs of the fact that the sequence of colored Jones polynomials is q-holonomic? REPLY [11 votes]: One statement that would imply that the colored Jones polynomials are q-holonomic involves the Kauffman bracket skein module $S_q(K)$ of the knot complement. This is a module over the skein module of the torus $T^2$, and q-holonomicity follows from the statement "$S_q(K)$ is finitely generated over the subalgebra $\mathbb C [m]$," where $m$ is the meridian of $K$. (This follows from papers of Frohman and Gelca, starting with this one. See also Cor. 1.3 here.) I'm not sure if the statement in quotes is true, although for 2-bridge knots it was proved by Le, and in special cases by Gelca and others. There is also a conjecture about $S_q(K)$ that can be viewed as a quantization of the statement "$L-1$ divides the $A$-polynomial of $K$," and this conjecture implies q-holonomicity. (This is Thm 5.10 here.) Unfortunately, as far as I know, people don't know good techniques that can prove statements about skein modules for arbitrary knot complements (although I would love to be proven wrong). So it doesn't seem like these statements will lead to a new proof of q-holonomicity (at least anytime soon). For torus knots (or, more generally, iterated cables of the unknot), the colored Jones polynomials can be written in terms of a cabling formula involving the double affine Hecke algebra and its polynomial representation. (I think at the moment this is just proved for $sl_2$, or for torus knots and $sl_n$, but it seems likely to be true in general. Some refs are here, here, here). This might imply q-holonomicity because the polynomial representation is holonomic, but I don't know a proof at the moment (I might be able to add one later). This cabling formula uses the DAHA at $t=q$, but it deforms to arbitrary $t$, and produces polynomials depending on 2 variables $q,t$. I don't know whether these polynomials are $q$-holonomic, though. One might have to change to $(q,t)$-holonomic, which could be defined in terms of the polynomial representation of the DAHA. But this probably wouldn't be necessary if you allow rational functions in the "variable" $q^n$. (I'm not sure if you want to allow this or not.)<|endoftext|> TITLE: Morava modules and completed $E$-homology QUESTION [9 upvotes]: Let $E = E_n$ be the $n$-th Morava $E$-theory and let $\{ M_{I} \}$ be a tower of generalised Moore spectra. Then (see this previous question) there is a Milnor exact sequence $$0 \to \varprojlim_I {}^1 (E \wedge M_I)_{\ast+1}(X) \to E_*^\vee X \to \varprojlim_I (E \wedge M_I)_*X \to 0. $$ Here we write $E_*^\vee X$ for $\pi_*L_{K(n)}(E \wedge X)$. The functor $\mathcal{K}_*X := \varprojlim_I (E \wedge M_I)_*X$ is sometimes called the Morava module of $X$, and seems to have first been introduced by Hopkins, Mahowald and Sadofsky whilst studying the $K(n)$-local Picard group. In Hopkins-Mahowald-Sadofksy, under the condition that $\mathcal{K}_*X$ is finitely-generated, the authors show that the $E_2$-term of the $K(n)$-local $E_n$-based Adams spectral sequence can be given by $H_c^\ast(\mathbb{G}_n,\mathcal{K}_*X)$, where $\mathbb{G}_n$ is the $n$-th (extended) Morava stabilizer group. It is known that the $\varprojlim{}^1$ term of the Milnor exact sequence vanishes under suitable conditions; for example if $E_*^\vee X$ is finitely-generated, or pro-free, as an $E_*$-module. In these cases then $E^\vee_*X \simeq \mathcal{K}_*X$. I suspect that the above spectral sequence should have $E_2$-term $H_c^\ast(\mathbb{G}_n,E^\vee_*X)$. This leads to the following question: If $\mathcal{K}_*X$ is finitely-generated is there an isomorphism $\mathcal{K}_*X \simeq E^\vee_*X$? By the comments above it would suffice to show that $E^\vee_* X$ is finitely-generated. Hovey-Strickland Proposition 8.6 gives a number of equivalent conditions, including that $X$ is $K(n)$-locally dualisable or that $K_*X$ is finite. REPLY [4 votes]: Here are some initial steps, which may be useful. First, consider $A=L_{K(n)}(E\wedge X)$, which is a $K(n)$-local $E$-module with $\pi_*A=E^\vee_*X$. It is not hard to see that $A\wedge M_I=E\wedge X\wedge M_I$. Given this, we see that we have a short exact sequence $$ \lim_I{}^1\pi_{*+1}(A\wedge M_I) \to \pi_*(A) \to \lim_I\pi_*(A\wedge M_I), $$ and we want to know whether the middle term is finitely generated, assuming that the last term is. All this makes sense for an arbitrary $K(n)$-local $E$-module $A$, and I suspect that it is easiest to work in that generality. In particular, we can replace the operation $A\mapsto A\wedge M_I$ with $A\mapsto A\wedge_EE/I$. This is advantageous because we can construct $E/I$ for any ideal of the form $(u_0^{i_0},\dots,u_{n-1}^{i_{n-1}})$, and it is easy to analyse the maps between these objects. Next, recall the category of $L$-complete $E_0$-modules described in Appendix A of Hovey-Strickland. Results given there suffice to show that all $E_0$-modules arising in any plausible approach to the above problem will be $L$-complete. Note in particular that if $P$ is $L$-complete then the map $P=L_0P\to\widehat{P}=\lim_IP/IP$ is always surjective. I am not sure exactly where to go from here, but I would suggest doing the case $n=1$ first. There we need to show that if $\lim_k\pi_*(A/p^k)$ is finitely generated, so is $\pi_*(A)$. The obvious approach is to compare the tower $\{\pi_*(A/p^k)\}$ with the towers $\{\pi_*(A)/p^k\}$ and $\{\text{ann}(p^k,\pi_{*-1}(A))\}$.<|endoftext|> TITLE: Poincare lemma for non-smooth differentiable forms QUESTION [18 upvotes]: The Poincare lemma is almost always formulated for differential forms with smooth coefficients (or sometimes for currents that have distributional coefficients). I would like to have it for $C^k$-coefficients. For one dimension everything is fine: $$0\to \mathbb R \to C^1(\mathbb R) \to C(\mathbb R)\to 0$$ (where the first map assigns the constant function and the second is the derivative) is exact because of the fundamental theorem of calculus. For $\mathbb R^2$ the analogous sequence would be $$0\to \mathbb R \to C^2(\mathbb R^2) \to C^1(\mathbb R^2)^2\to C(\mathbb R^2)\to 0$$ where the first map again assigns the constant function, the second is the gradient and the third is $d: C^1(\mathbb R^2)^2\to C(\mathbb R^2)$, $(g_1,g_2)\mapsto \partial_2 g_1 - \partial_1 g_2$. This sequence is exact at the first three spots (the usual proof of the poincare lemma) but I don't see whether it is exact at the last spot, that is, whether $d$ is onto. (For the same complex with $C^\infty$ (or $\mathscr D'$) instead of $C^k$ surjectivity of $d$ is trivial: For $h\in C^\infty(\mathbb R^2)$ one can just integrate one variable: $g_1(x,y)=\int_0^yh(x,t)dt$ and $g_2=0$ yield $d(g_1,g_2)=h$. However, if $h$ is just continuous there is no reason why $g_1$ should be differentiable w.r.t. $x$.) EDIT. The discussion with Igor (and his updated answer) lead to the simple observation that it would be sufficient to show the surjectivity of $\Delta: C^2(\mathbb R^2) \to C(\mathbb R^2)$. I have therefore added the tags pde and fa. This last question looks so natural that the answer should be known (and I believe it to be negative). REPLY [10 votes]: The answer to the question is in fact NO. It is quite surprising for me that it is relatively recent. Theorem 5 in the article Additional regularity for solutions of PDE by David Preiss (J. reine angew. Math. 485 (1997), 197—207) states that there is a continuous function $\psi:\mathbb R^2 \to\mathbb R$ with compact support such that the equation div$\Psi(x) = \psi (x) $ for almost every $x\in\mathbb R^2$ possesses no locally Lipschitz solution $\Psi:\mathbb R^2\to\mathbb R^2$. I thank Pedro Lauridsen Ribeiro and Igor Khavkine for their very informative comments. In particular, the book mentioned by Pedro eventually lead me to the article of Preiss. (Unfortunately, I cannot split the bounty.)<|endoftext|> TITLE: Stability of real polynomials with positive coefficients QUESTION [20 upvotes]: Say that a polynomial in an indeterminate $x$ with real coefficients of degree $d$ has positive coefficients if each of the coefficients of $x^d,\ldots,x^1,x^0$ is (strictly) positive. For $f$ a monic univariate polynomial with real coefficients, if there exists a positive integer $m_0$ such that its $m_0$th power $f^{m_0}$ has positive coefficients, then does there exist a positive integer $m_1$ such that for each positive integer $m$ at least $m_1$, the $m$th power $f^m$ has positive coefficients? REPLY [9 votes]: Yes. This is a part of my answer to another MO question, Zeros of polynomials with real positive coefficients, @David Handelman: thanks for the reference! I will insert it to my paper. EDIT. In fact, Theorem 1 in the preprint cited above is not new. This was found as a result of David's answer. The revised preprint is posted on the arxiv and here http://www.math.purdue.edu/~eremenko/dvi/saddle13.pdf<|endoftext|> TITLE: Simple, closed geodesics in $\mathbb{S}^3$ manifold QUESTION [18 upvotes]: Lyusternik and Shnirel'man were the first to prove Poincaré's conjecture that any Riemannian metric on $\mathbb{S}^2$ has at least three simple (non-self-intersecting), closed geodesics. See, e.g., p.466 of Berger's A Panoramic view of Riemannian Geometry, or this Encyc.Math. article. (Apparently details in the L.-S. 1929 proof were not resolved until 1978 1993.) Q. Is there an extension to any Riemannian metric for $\mathbb{S}^3$? E.g., there exist at least $k > 1$ simple, closed geodesics in $\mathbb{S}^3$ under every Riemannian metric? (Added.) As Igor Rivin points out, $k=1$ is known for any smooth metric on $\mathbb{S}^n$ via a 1927 proof of Birkoff. Added a bounty. Much remains unclear to me, despite Igor's (extremely) useful citations. The bounty is offered for clarifying the situation (comments cited in several instances): (1) Klingenberg proves (Mathematische Zeitschrift}) in 1981 there are 4 closed geodesics on $\mathbb{S}^3$. (2) Long&Duan prove (Advances in Mathematics) in 2009 that there at least 2 closed geodesics on $\mathbb{S}^3$—What happened in the 28 yrs between? Was Klingenberg's proof not accepted, or did he prove something nuancedly different? (3) And it seems that none of these authors are addressing simplicity, as @alvarezpaiva noted. The L.&S. theorem explicitly proves non-self-intersection on $\mathbb{S}^2$. REPLY [6 votes]: Just to set the terminology, a closed geodesic is called simple when it is a smooth embedded circle in the Riemannian or Finsler manifold. If a closed Riemannian manifold is not simply connected one can easily show that the shortest non-contractible curve is a simple closed geodesic. Existence and multiplicity results for simple closed geodesics on simply connected manifolds are currently available only for 2-dimensional manifolds. Indeed, as it has been already mentioned, in higher dimension being simple is a generic condition: given a closed geodesic $\gamma$, with a tiny perturbation of the metric one can produce a simple closed geodesic close to $\gamma$ for the new metric. For this reason, it is not obvious how to obtain a simple closed geodesic in higher dimension by means of variational methods: a generic metric has at least one (and often more), but a very unlucky metric in principle might have none. The existence of closed geodesics possibly with self-intersections was proved by Birkhoff for Riemannian 2-spheres, and later by Lusternik-Fet in full generality (as it was mentioned above by Rivin and Alvarez Paiva). These proofs remain valid for general Finsler metrics. Beside the already mentioned case of Riemannian $S^2$, the existence of a second closed geodesic is already hard business: for suitably non-degenerate Riemannian metrics, it was a theorem of Fet from the 1960s. The theorem should hold as well for reversible Finsler metrics (i.e. Finsler metrics $F$ with the property that $F(v)=F(-v)$ for all vectors $v$). For general, non-reversible, Finsler metrics on $S^2$ it was a theorem of Bangert-Long from the late 2000s. For higher dimensional simply connected closed manifolds equipped with non-degenerate and possibly non-reversible Finsler metric, the existence of a second geodesic is a recent theorem of Duan-Long-Wang (JDG, 2015). For simply connected closed manifolds whose cohomology ring $H^*(M;\mathbb Q)$ is not generated by one element, a celebrated result of Gromoll-Meyer from 1971 implies that there are always infinitely many Riemannian or Finsler closed geodesics (possibly with self-intersections). The only simply connected manifolds not covered by this theorem are those who have the rational cohomology of a compact rank-one symmetric space, that is, the rational cohomology of $S^n$, $\mathbb RP^n$, $\mathbb CP^n$, $\mathbb HP^n$, or $\mathrm{Ca}P^2$. The case of $S^2$ was settled in a combination of celebrated papers by Bangert, Franks, and Hingston. For the other CROSSes, Hingston and Rademacher proved that a suitably generic Riemannian metric has infinitely many closed geodesics. On non-simply connected closed Riemannian or Finsler manifolds, it is very easy to find infinitely many closed geodesics when, for instance, the fundamental group is abelian and has rank larger than 1. The statement is still true, but non-trivial, if the fundamental group is only assumed to be infinite abelian (the most difficult case being when it is $\mathbb Z$); it was proved by Bangert-Hingston in the 1980s. In his book "Lectures on closed geodesics" from the 1980s, Klingenberg has a proof of the existence of infinitely many closed geodesics for any closed Riemannian manifold (of dimension at least two). However, the proof was crucially based on a divisibility lemma that was later found to be wrong. There are unfortunately other mistakes in the literature, notably by Alber (his so-called Alber Lemma) and Klingenberg. The results that I mentioned above are universally accepted and have a solid proof, but one should be careful with older results in the published literature which claim stronger conclusions. The unconditional existence of infinitely many closed geodesics, or even of a second closed geodesic on a general closed Riemannian manifold of arbitrary dimension, are still open problems. The above are among the most significant results in the field, but I certainly forgot to mention many others.<|endoftext|> TITLE: Stromquist's 3 knives procedure QUESTION [13 upvotes]: (copied from math.SE) BACKGROUND: A cake has to be divided among 3 people with possibly different tastes, such that each person receives a single connected piece, and no person prefers another person's piece. In other words, no participant should end up being envious of any other participant. Symbolically, let $\ v_{jk}\ $ be the value of the $k$-th piece to participant $\ j\ $ (the values for each player are based on a continuous measure). We would like: $$\forall_{j=1\ 2\ 3}\ \ \ v_{jj}\ \ =\ \ \max(v_{j1},\ v_{j2},\ v_{j3})$$ This problem was unsolved for several tens of years, until Stromquist (1980) suggested the following division protocol: A referee moves a sword from left to right over the cake, hypothetically dividing it into a small left piece and a large right piece. Each player holds a knife over what he considers to be the midpoint of the right piece. As the referee moves his sword, the players continually adjust their knives, always keeping them parallel to the sword. When any player shouts "cut", the cake is cut by the sword and by whichever of the players' knives happens to be the middle one of the three. The player who shouted "cut" receives the left piece. He must be satisfied, because he knew what all three pieces would be when he said the word. Then the player whose knife ended nearest to the sword, if he didn't shout "cut", takes the centerpiece; and the player whose knife was farthest from the sword, if he didn't shout "cut", takes the right piece. The player whose knife was used to cut the cake, if he hasn't already taken the left piece, will be satisfied with whatever piece is left over. If ties must be broken - either because two or three players shout simultaneously or because two or three knives coincide - they may be broken arbitrarily. Note that all knives are visible to all players. It is clear that, if all players play truthfully (according to their own value function), the resulting division is indeed envy-free. My question is: what happens if two players play untruthfully, against their own interest - can they make the third player envious? In most protocols for cake-cutting among $\ n\ $ participants, the answer is "no", i.e., every player that plays truthfully is guaranteed to receive an envy-free share, regardless of what the other players do. For example, consider the classic protocol for 2 players: "I cut, you choose". I (the cutter) have to cut the cake to two pieces that I consider to be of equal value, but, even if I cut the cake in a very strange manner to two very unequal pieces (even against my own interest), you still have a safe strategy - you just pick the piece that you consider to be more valuable, and you are guaranteed to feel no envy. In other words, the cut-and-choose protocol (and most other cake-cutting protocols) is safe for truthful players. So, my question is: is Stromquist's procedure indeed safe for truthful players? I.e. does it guarantee that every single player playing by the rules feels no envy, regardless of what the other players do? EDIT: Here is the problematic scenario I had in mind when asking. Suppose there are two evil players - Kunning (K) and Liar (L), who want to hurt the good player Marge (M). Initially K and L put their knives close to the sword (S), like this: |----------SKL------M--------| Now Marge knows that if she remains quiet, she will get the piece between L and the right border, which is larger than the piece to the left of S, so she remains quiet. But then L moves his knives discontinuously to the right, like this: |----------SK------LM--------| And at the same moment, K shouts "cut". Now the piece to the right of L is smaller than the piece to the left of S, so Marge should shout "cut" now, but there is 50% chance that the protocol will give the piece to the left of S to K, who shouted at the same time, and Marge will envy him. This scenario is possible only if L is not truthful. Why? Because a cake is assumed (as usual) to be non-atomic - the value of every subset of zero length is zero. Hence the value to the left of the sword is a continuous function of time, and the middle point of the part to the right of the sword is also a continuous function of time. But, if a player is not truthful, he may decide to make the position of his knife a discontinuous function. So my more specific question is: is my description of the above scenario correct? If it is, how can the procedure be corrected? Maybe by requiring that the location of each knife is a continuous function of time? EDIT 2: In second reading, I see that Stromquist indeed mentioned that "the players continually adjust their knives". REPLY [8 votes]: Edit: The below does rely on the assumption that knives move continuously, see the comments. I think the procedure is "safe": Each player can guarantee not to envy either of the others by following the suggested protocol. I am assuming that the rules are that players must keep their knives to the right of the sword. Intuitively, the straightforward proof of envy-freeness, following, does not use that the other players are playing truthfully (they can be playing arbitrarily). Suppose a player "Marge" wants to guarantee herself an envy-free piece. Marge follows the suggested strategy: She keeps her knife at her perceived midpoint of the section to the right of the sword, and yells "cut" if at any point she prefers the piece left of the sword to both other pieces that will result. If neither of the other players ever yell cut, then eventually Marge will yell cut and get an envy-free allocation. So the players can only disrupt her by yelling cut before she does. However, since she has not yet yelled cut, Marge must prefer one of the other two resulting pieces to the left piece. So we just need to check that, when one of the other players yells cut, Marge will get her more-preferred piece between the center and the right. If her knife was the one used to cut, then since she was following protocol, she is indifferent between the center and right pieces (preferring both to the left piece), and she gets one of them, so she is not envious. If her knife was not used to cut, then either it is nearest to the sword or it is farthest from the sword. Suppose her knife is nearest to the sword. Then she gets the center piece. But because her knife is located to make her indifferent between (a) the section between the sword and her knife and (b) the section between her knife and the rightmost edge, and because she gets section (a) and more besides, she prefers the center piece that she gets to the right piece. The analogous argument works if her knife is farthest from the sword. ... I would guess that there is another argument which says that any strategic disruptive player could be simulated by an honest player with a certain preference, implying that it would be sufficient to show the protocol is envy-free, when players are truthful, for all possible preferences.<|endoftext|> TITLE: Does there exist a Fano variety with torsion in $H^3$? QUESTION [10 upvotes]: Let $X$ be a (smooth) Fano variety over $\mathbb{C}$. If $\dim(X)=3$, inspection of the Iskovskikh-Mori-Mukai lists seems to indicate that $H^3(X,\mathbb{Z})$ is torsion free. Is there a theoretical reason for that? What happens in higher dimension? REPLY [13 votes]: There are smooth Fano $5$-folds $\widetilde Y$ with non-zero $2$-torsion in the Brauer group (i.e., $2$-torsion in $H^3(\widetilde Y,\mathbb Z)$); the examples that follow are directly inspired by Beauville's exposition of the Artin--Mumford examples in terms of Reye congruences [Springer LNM $997$, pp. $28-30$]. Start with a general $5$-dimensional linear system $L=\mathbb P^5$ of quadrics in $\mathbb P^3$ (Beauville uses a $3$-dimensional system). The discriminant locus in $L$ is a quartic $4$-fold $S$ defined by the vanishing of a symmetric $4\times 4$ determinant. The singular locus of $S$ is the scheme $C$ that is defined by the vanishing of the $3\times 3$ minors. It is known that $C$ is a smooth surface. By construction, $C$ is the scheme-theoretic intersection of $10$ cubics. Let $\pi:\widetilde{L}\to L$ denote the blow-up $Bl_CL$, $\widetilde S$ the strict transform of $S$, $H$ the class of a hyperplane in $L$ and $E$ the exceptional divisor. Since $S$ has ordinary double points along $C$, $\widetilde S$ is smooth and $\widetilde S\sim \pi^*S-2E\sim 4\pi^*H-2E$. Now let $Y\to L$ be the double cover branched along $S$ and $f:\widetilde Y\to\widetilde{L}$ the double cover branched along $\widetilde S$. Note that $Y$ has ordinary double points along the inverse image $C_1$ of $C$ while $\widetilde Y$ is smooth. Note also that $C_1\to C$ is an isomorphism and that $\widetilde Y= Bl_{C_1}Y$. Moreover, $K_{\widetilde Y}\sim f^*(-4H+E)$, from the adjunction formula (or Riemann--Hurwitz). Lemma: $4H-E$ is ample on $\widetilde{L}$ and $\widetilde Y$ is Fano. Proof: The linear system $\vert H\vert$ defines $\pi:\widetilde{L}\to L$ and $\vert 3H-E\vert$ defines a morphism $\widetilde{L}\to\mathbb P^9$, from the description of $C$. It is then clear that the induced morphism $$\widetilde{L}\to L\times\mathbb P^9$$ contracts no curves, and the lemma is proved. Now let $G$ denote the Grassmannian of lines in $\mathbb P^3$ and consider the incidence variety $$\widetilde G=\{(l,q)\vert l\in G,\ q\in\Pi,\ l\subset q\}.$$ Then $pr_1:\widetilde G\to G$ is a Zariski $\mathbb P^2$-bundle, so that $\widetilde G$ is a smooth rational $6$-fold, while $pr_2:\widetilde G\to L$ factors through $Y$. In fact, $\widetilde G\stackrel{r}{\to} Y\to L$ is the Stein factorization of $pr_2$. Notice that the fibres of $r$ are exactly the connected components of the scheme of lines in a member $q$ of $L$, so that $r^{-1}(y)$ is a smooth curve of genus zero (a conic) whenever $y$ does not lie on $C_1$. That is, $r$ is an etale $\mathbb P^1$-bundle over $Y-C_1$. On the other hand $Z:=r^{-1}(C_1)\to C_1$ is an etale fibre bundle whose fibre is the union of two copies of $\mathbb P^2$ meeting transversely in a single point. It is then an elementary local calculation to check that the induced morphism $$\tilde r:G^*=Bl_Z\widetilde G\to \widetilde Y=Bl_{C_1}Y$$ is an etale $\mathbb P^1$-bundle. (In terms of etale local co-ordinates, $r$ is the morphism $$\mathbb A^6=Spec\ k[x,y,z,t,w_1,w_2]\to Spec\ k[v_1,v_2,v_3,v_4,w_1,w_2]/(v_1v_4-v_2v_3)$$ given by $v_1=xz,\ v_2=xt,\ v_3=yz,\ v_4=yt$. Then we make a verification chart by chart. Maybe the only slightly surprising thing here is that the blow-up of the smooth variety $\widetilde G$ along the singular subscheme $Z$ should itself be smooth.) At this point we have a smooth Fano $5$-fold $\widetilde Y$ and an etale $\mathbb P^1$-bundle $\tilde r:G^*\to \widetilde Y$ whose total space is a rational $6$-fold. Lemma: $\tilde r$ does not have a generic section. Proof: Note that restricting to a general linear space $H$ of codimension $2$ in $L$ leads exactly to the picture described by Beauville; namely, a smooth rational $4$-fold $G'$ that is an etale $\mathbb P^1$-bundle over a smooth $3$-fold $X$ which is one of the Artin--Mumford examples. It follows that $G'\to X$ has no generic section, so the same holds for $\tilde r$. Hence $G^*$ represents a non-trivial $2$-torsion element of the Brauer group $Br(\widetilde Y)$. [Clearly this is cumbersome. It would be much better to find a smooth Fano variety $\widetilde Y$ of dimension at least $4$ and a smooth hyperplane section $X'$ with non-trivial Brauer group. Then weak Lefschetz would show that $\widetilde Y$ had the same property.] "Is there a theoretical reason?" I'm not sure that there is; it's remarkable that such a precise classification of Fano $3$-folds is possible.<|endoftext|> TITLE: Jordan-Holder vs Harder-Narasimhan QUESTION [5 upvotes]: Let $M$ be a module over an algebra or a group. I am interested the following decreasing filtration: $F^0M=M$; $F^iM$ is the smallest sub-module of $F^{i-1}M$ such that the quotient is semi-simple. This filtration is unique. Does it have a specific name?? Harder-Narasimhan?? Any reference?? (Jordan-Holder is when the quotients are simple, so it is a refinement of this one) Thanks REPLY [2 votes]: I usually refer to this as the "radical filtration," since the module $F^iM$ is the radical (or Jacobson radical) of the module $F^{i-1}M$. While not explicitly stated, this paper seems to suggest it could also be called the "upper Loewy series" (with the lower Loewy series being the dual thing I would usually call the "socle filtration").<|endoftext|> TITLE: Field of definition of Galois representations of weight 1 modular forms QUESTION [8 upvotes]: Let $f$ be a weight 1 modular form (let's say cuspidal, new, normalized, and a Hecke eigenform). Then there's an associated Artin representation $\rho_f: \operatorname{Gal}(\overline{\mathbf{Q}} / \mathbf{Q}) \to GL_2(\mathbf{C})$. The character of this representation is determined by $f$, so it takes values in the number field $L = \mathbf{Q}(a_n(f): n \ge 1)$. Are there examples where $\rho_f$ is not definable over $L$, i.e. isn't conjugate to a morphism $\operatorname{Gal}(\overline{\mathbf{Q}} / \mathbf{Q}) \to GL_2(L)$? (There are lots of examples of even representations in the LMFDB where this happens, coming from even Artin representations with image $Q_8$ or $S_4$, but I can't find any odd ones.) REPLY [9 votes]: Here is another proof of the result from Joel's answer, using a little more theory: a two dimensional irrep. of a finite group $G$ corresponds to a simple algebra factor $A$ of the group ring $\mathbb Q[G]$, where $A$ is $4$-dimensional over its centre (which is some number field $F$). Thus $A$ is either $M_2(F)$, or a non-split quat. alg. over $F$. In particular, in the second case $A$ is a division algebra. If $c \in G$ has order $2$, with image $\bar{c}$ in $A$, then since $c^2 = 1$, we see that $(\bar{c} -1)(\bar{c} + 1) = 0$ in $A$. If $A$ is a division algebra, we conclude that either $\bar{c} = 1$ or $\bar{c} = -1$. Thus if the image of $c$ is not scalar, we see that $A = M_2(F)$, which is to say that the given irrep. is defined over its trace field.<|endoftext|> TITLE: Pulling back quasi-coherent sheaves from a quotient stack QUESTION [5 upvotes]: In a problem I am trying to solve, the following situation occurs. $X$ is a smooth variety and $G$ is a reductive group acting transitively on $X$. We have the stack $X/G$ and a morphism $\pi : X \to X / G$. Fix a basepoint $* \in X$ and let $H$ be the stabilizer of $*$. I don't know much about stacks but I was wondering if the following statements are correct and if someone could give me a reference where I can read about them: The quasi-coherent sheaves on $X / G$ are the same as $H$-representations. Here I am thinking of the fact that an equivariant vector bundle on a homogeneous manifold is determined by its fiber over some chosen basepoint. I guess here we could also talk about the action groupoid, but since the action is transitive i don't think that is needed Let $K$ be the fraction field of $X$ and $V$ an $H$-representation. The rational sections of $\pi^* V$ are $ (K \otimes_{\mathbb{C}} V)^H$. I must admit, this is not geometrically obvious to me even in the case of manifolds, but it is all I could think of writing down. I also asked some people in my department and they seemed to think that this was reasonable. If $X$ is affine with coordinate ring $A$, then the global sections of $\pi^*V$ are $(A \otimes_{\mathbb{C}} V)^H$. Again, this is not obvious to me, but it seems like the obvious thing to write down. All of this is probably easy if you know about stacks, so hopefully someone will share some of their stacky wisdom! REPLY [4 votes]: For the first question, there is an equivalence of stacks between $X/G$ and $*/H$, and Theorem 4.46 of Vistoli's notes gives an equivalence between $H$-equivariant quasicoherent sheaves on a point (i.e., representations of $H$) and quasicoherent sheaves on $*/H$. Your conjectures about rational and global sections seem fail when $X$ is a point. In particular, pullback from $*/H$ to a point $*$ is what gives an equivalence between quasicoherent sheaves on $*/H$ and $H$-equivariant sheaves on $*$, and taking fixed points will exclude all non-trivial $H$-modules from your category. In general, you get the $G$-modules $K \otimes_{\mathbb{C}} V = K \otimes_A \operatorname{Ind}_{H}^{G} V$ and $A \otimes_{\mathbb{C}} V = \operatorname{Ind}_{H}^{G} V$, because tensoring an $H$-module $V$ with $A$ over $\mathbb{C}$ is the same as tensoring the $\mathcal{O}(H)$-comodule $V$ with $\mathcal{O}(G)$ over $\mathcal{O}(H)$.<|endoftext|> TITLE: Are bounds known for the maximum determinant of a (0,1)-matrix of specified size and with a specifed number of 1s? QUESTION [14 upvotes]: The problems of determining the maximum determinant of an $n \times n$ $(0,1)$-matrix and the spectral problem of determining exactly which other determinants can possibly occur are both reasonably well studied. What I'd like to know is whether there are bounds known if the number of 1s in the matrix is specified. In other words, if we let $f(n,k)$ denote the maximum determinant of an $n \times n$ $(0,1)$-matrix with exactly $k$ ones (in total, not per row), then are there any known bounds on the values for $f(n,k)$? I've searched Math Reviews and Googled it, but all the papers that I have found refer to the maximum over the entire set of $n \times n$ matrices. Edit: Here is a plot of the maximum determinant of a binary $6 \times 6$ matrix with $k$ 1s, where $0 \leq k \leq 36$ (as long as I haven't messed up the computation). Edit 2: Here is the analogous plot for $7 \times 7$ matrices. Edit 3: Here is a plot of the bound given by Peter Mueller's post, together with the actual values for n=7. REPLY [7 votes]: In this paper Gasper, Pfoertner and Sigg prove an interesting upper bound for a real $n\times n$ matrix $M$: Let $n\alpha$ and $n\beta$ be the sum of the entries or the squares of the entries of $M$, respectively. Set $\delta=(\alpha^2-\beta)/(n-1)$. If $\delta\ge0$, then $\lvert\det M\rvert\le \alpha(\beta-\delta)^{(n-1)/2}$. So $\alpha=\beta=k/n$ for a $(0,1)$-matrix. In contrast to the Hadamard bound, this bound is attained quite often for $(0,1)$-matrices, namely for the incidence matrices of symmetric block designs. Looking at their (somewhat involved) proof, this result is not surprising, for they show that upon fixing $\alpha$, $\beta$ and $n$, the maximum of the determinant is attained if $MM^t$ is a linear combination of the identity matrix and the all $1$ matrix.<|endoftext|> TITLE: Harish-Chandra isomorphism for compact symmetric spaces QUESTION [8 upvotes]: I would be interested to have an explicit description of the algebra of invariant differential operators on functions on a compact symmetric space $G/K$. A reference would be especially useful. Below I am making it more precise what kind of description I need. It is a generalization of the famous Harish-Chandra isomorphism which I recall first. Let $\mathfrak{g}$ be a complex reductive Lie algebra. Fix a Cartan subalgebra $\frak{h}$ and its positive roots. Then there exists an algebra isomorphism $$\gamma: Z(U(\mathfrak{g}))\mathrel{\tilde{\to}} Sym(\mathfrak{h})^W$$ of the center of the universal enveloping algebra $Z(U(\mathfrak{g}))$ and the algebra $Sym(\mathfrak{h})^W$ of $W$-invariant polynomials on $\mathfrak{h}^*$ where $W$ is the Weyl group, which is uniquely characterized by the property that any element $x\in Z(U(\mathfrak{g}))$ acts on an irreducible representation $V_\lambda$ with highest weight $\lambda\in \mathfrak{h}^*$ by the scalar $\gamma(x)(\lambda+\rho)$ where $\rho$ is the half sum of positive roots. The algebra $Z(U(\mathfrak{g}))$ can alternatively be viewed as the algebra of bi-invariant differential operators on functions on a Lie group $G$ such that $\mathfrak{g}$ is the complexification of $Lie(G)$, or the algebra of invariant differential operators on the symmetric space $(G\times G)/G$ where $G\subset G\times G$ is the diagonal subgroup. I have heard that there exists a generalization of the above isomorphism, but in the context of compact symmetric spaces (the relevant description for symmetric spaces of non-compact type is contained in Helgason's book "Groups and Geometric Analysis"). I would be happy to have a precise statement and a reference. Let $(G,K)$ be a symmetric pair of compact type. Assume that $G$ is connected, but $K$ does not have to be (it is the standard convention in the literature, though I will need the case of disconnected $K=O(n)$). I have heard that the algebra of invariant differential operators on $G/K$ is isomorphic to the algebra of $W'$-invariant polynomial on the Cartan space of the symmetric space $G/K$, where $W'$ is the Weyl group of $G/K$. Moreover this isomorphism is somehow uniquely characterized by the action of the differential operator on all irreducible subspaces of $L^2(G/K)$ in terms of their highest weights (all of them are known exactly for a compact symmetric space $G/K$). REPLY [2 votes]: I believe the isomorphism you want is Theorem 8.2 of Lepowsky's Generalized Verma modules, the Cartan-Helgason theorem, and the Harish-Chandra homomorphism (1977). Not sure about a characterization in terms of highest weights in $L^2(G/K)$...<|endoftext|> TITLE: How did height in algeb. number theory/elliptic curves started? QUESTION [9 upvotes]: Maybe this is obvious but it isn't to me yet. What is the history of heights used in say points of the project plane over a number field or of elliptic curve over a number field? I would guess people working with valuations (like Ostrowski, Krull, Dedekind, Hensel et al) would have started this idea. Just an idea.. there is a nice article on the history of valuation written by Peter Roquette, maybe I can find this there? Edit: Maybe Keith is right. In fact I took a look at the text by Roquette and as far as I could see (I didnt read the whole text though) height was not mentioned. REPLY [16 votes]: I think that Keith Conrad is correct and heights via maxs of valuations are in Weil's "Mordell-Weil" paper. But note that Weil's paper generalized Mordell's work in two ways. First, he extended from $\mathbb{Q}$ to number fields, and second from elliptic curves to abelian varieties (or at least Jacobians). So he would have needed some sort of way of measuring the arithmetic complexity of points on varieties, which he presumably viewed as embedded in projective space. Note that there's also the more general theory of heights that associates to each divisor $D$ on a variety $X$ (everything defined over $\overline{\mathbb{Q}}$) a height function $h_{X,D}$, or more properly, an equivalence class of functions modulo bounded functions. I believe that this, too, is due to Weil, although this belief comes primarily from the fact that everyone calls them Weil height functions. (If I had to guess, the name "Weil height function" is due to Lang.) Weil would have fixed a number field $K$ and defined a height function $h_K$ on $\mathbb{P}^n(K)$, and used that to define a height on $X(K)$ for any given embedding $X\hookrightarrow\mathbb{P}^n(K)$. I believe that it was Northcott (Annals paper in 1950) who observed that one can define an absolute height on $\mathbb{P}^n(\overline{\mathbb{Q}})$ by taking $$ h(x) = \frac{1}{[K:\mathbb{Q}]} h_{K}(x) $$ for any number field $K$ such that $x\in\mathbb{P}^n(K)$. Then $h(x)$ is independent of the choice of $K$. So now people often prove theorems saying that "such and such a subset of $\mathbb{P}^n(\overline{\mathbb{Q}})$ is a set of bounded height", which implies in particular that it has only finite intersection with $\mathbb{P}^n(K)$ for any number field $K$, but even more, that it contains only finitely many points defined over all number fields of a given bounded degree. For example, the set of all torsion points on an abelian variety is a set of bounded height. Note that one can also define height on projective space mimicking the "largest coordinate after clearing denominators" approach, which doesn't involve valuations per se. So, let $a=[a_0,\ldots,a_n]\in\mathbb{P}^n(K)$ for some number field $K$. WLOG, we may assume that all of the $a_i$ are in $O_K$. Then we more or less want to set $$ h(a) = \sum_{\sigma:K\hookrightarrow\mathbb{C}} \log\max_{0\le i\le n} |\sigma(a_i)|, $$ but there's a problem if the $a_i$ have a nontrivial common factor. Since $O_K$ isn't a PID, we don't have gcds, but we do if we consider ideals. So let $$ \mathfrak{A} = a_1O_K+a_2O_K+\cdots+a_nO_K $$ be the ideal generated by the coordinates of $a$. Then the correction factor is more-or-less to divide by $\mathfrak{A}$. Precisely, an alternative formula for what we now call the Weil height is $$ h(a) = \log\left(\frac{1}{N_{K/\mathbb{Q}}\mathfrak{A}} \prod_{\sigma:K\hookrightarrow\mathbb{C}} \max_{0\le i\le n} |\sigma(a_i)|\right). $$ Now, if you write the norm as a product of prime powers and write the log of the product as a sum, you'll get to usual definition as a sum over all absolute values.<|endoftext|> TITLE: Higher coherent multiplicative structures on S-algebras QUESTION [8 upvotes]: In their book, Elmendorf, Kriz, May and Mandell describe a useful category of spectra, called S-modules, where S is the sphere spectrum. Ring objects in this category can be identified with spectra with an action of an $A_\infty$-operad (if I understand correctly) and commutative rings can be identified with spectra with an action of the $E_\infty$-operad. Has anyone written about $E_n$-rings in this category? In particular, are there nice model category structures on categories of $A$-modules and $A$-algebras if $A$ is only an $E_n$-algebra? Can this perhaps be shown by somehow trapping this model structure between the nice model category structures on commutative rings and associative rings? Thanks! REPLY [10 votes]: You can pick a model for the operad $E_n$ which receives a map from the associative operad. For instance, the Boardman-Vogt tensor product of the associative operad with $E_{n-1}$ has this property. Then, if you have an algebra $A$ over that operad, it is in particular an associative algebra and you can put a model structure on its category of modules using EKMM's method. The only problem with this method is that it is not clear that the resulting model category has a monoidal structure. According to Lurie, it should have and $E_{n-1}$-monoidal structure but this is quite hard to prove using model categories. To put a model structure on $E_n$-algebras over $A$ is easy. It suffices to observe that an $E_n$-algebra over $A$ is an $E_n$-algebra in spectra together with a map of $E_n$-algebra from $A$. You can put a model structure on that category by taking the model structure on $E_n$-algebras and then taking the under-category model structure. Constructing the model structure on $E_n$-algebras is a bit technical and I don't know of a reference for EKMM spectra. However, it has been worked out in details by John Harper in symmetric spectra. In fact John Harper proves it for any operad (see http://www.msp.warwick.ac.uk/agt/2009/09-03/p057.xhtml).<|endoftext|> TITLE: When is a continous $\epsilon$-isometry of the sphere surjective? QUESTION [9 upvotes]: Equip $\mathbb S^n$ with the standard round metric. Let $f : \mathbb S^n \to \mathbb S^n$ be a continous map satisfying $\vert d(f(x),f(y)) - d(x,y)\vert \leq \epsilon$. Is $f$ is surjective for all $0 \leq \epsilon < \epsilon_0$ for some positive $\epsilon_0$? My guess would be that the answer is yes and maybe $\epsilon_0 = \pi$. REPLY [15 votes]: Your guess seems to be true. If a map $f$ is not surjective then $f$ can be considered as a continuous map from $S^n$ to $R^n$. Hence there exist two opposite points on $S^n$ which maps to the same point by Borsuk-Ulam theorem.<|endoftext|> TITLE: (co)homology of symmetric groups QUESTION [30 upvotes]: Let $S_n=\{\text{bijections }[n]\to[n]\}$ be the n-th symmetric group. Its (co)homology will be understood with trivial action. What are the $\mathbb{Z}$-modules $H_k(S_n;\mathbb{Z})$? Using GAP, we calculate this for small $n$ and $k$: This seems to be an infinite amount of data with no apparent patterns, just the stabilization for $n\geq2k$. In Stable homology of automorphism groups of free groups (Galatius - 2008) p.2 there is written: "The homology groups $H_k(S_n)$ are completely known" referring to Nakaoka's articles Decomposition Theorem for Homology Groups of Symmetric Groups, Homology of the Infinite Symmetric Group, Note on cohomology algebras of symmetric groups from 1960, 1961, 1962. I haven't found any such table in those articles, or in Cohomology of Finite Groups (Adem, Milgram - 1994). My questions are: 1) Does $H_k(S_n;\mathbb{Q})$ and $H_k(S_n;\mathbb{Z}_p)$ for all prime $p$ determine $H_k(S_n;\mathbb{Z})$? 2) How does the above table look for larger n and k, e.g. what is $H_k(S_{2k};\mathbb{Z})$ for $k=1,...,30$? 3) Is for every prime $p$ and $k\geq1$ the module $\mathbb{Z}_{p^k}$ a direct summand of some $H_k(S_n;\mathbb{Z})$? 4) Does $H_k(S_n;R)\cong H_k(S_{2k};R)$ as $R$-modules for $n>2k$ hold over any ring $R$? REPLY [8 votes]: Here are some comments, including an answer to (3). Firstly, if you want an actual explicit computation of the mod-2 cohomology of symmetric groups $S_n$ for as large an $n$ as possible, you should look at M. Feshbach's `The mod-2 cohomology rings of the symmetric groups and invariants' Topology volume 41 (2002) 57-84. This contains explicit computations of the cohomology rings $H^*(S_n;\mathbb{F}_2)$ for $n\leq 16$, including correcting a minor error in a calculation in the book of A Adem and J Milgram. The advantage of cohomology is that it is a ring, also that the cohomology of a finite group with coefficients in either $\mathbb{Z}$ or a field is a finitely presented algebra over the coefficients. Here is a non-constructive solution to (3). By the universal coefficient theorem, it suffices to show the same thing for cohomology. Now let $n$ be sufficiently large that the symmetric group $S_n$ contains a cyclic subgroup of order $p^k$; of course the least such $n$ is $n=p^k$. For any such $n$, there will be elements of order $p^k$ in $H^j(S_n;\mathbb{Z})$ for infinitely many values of $j$. To see this one uses the Evens-Venkov theorem. For any finite group $G$ and any subgroup $H$, the map from $H^*(G;\mathbb{Z})$ to $H^*(H;\mathbb{Z})$ makes the ring $H^*(H;\mathbb{Z})$ into a module for the ring $H^*(G;\mathbb{Z})$. The Evens-Venkov theorem tells us that $H^*(H;\mathbb{Z})$ is finitely generated as an $H^*(G;\mathbb{Z})$-module. Now to apply this. The cohomology ring of the cyclic group of order $p^k$ is isomorphic to a polynomial ring $\mathbb{Z}[c]/(p^kc)$, where $c$ is a generator for $H^2$. If $R$ is a (graded) subring of this ring such that the whole ring is a finitely generated $R$-module, then $R$ contains $c^m$ for some $m$, and hence $H^{2mj}(S_n;\mathbb{Z})$ contains an element of order $p^k$ for every $j$. The universal coefficient theorem then tells you that $H_{2mj-1}(S_n;\mathbb{Z})$ contains an element of order $p^k$ for all $j$. You asked about summands of order exactly $p^k$. In fact, $H^*(S_n;\mathbb{Z})$ will contain these whenever $n\geq p^k$, although I can't provide a quick argument. A quicker thing to see is that for $p^k \leq n < p^{k+1}$, the exponent of the $p$-local cohomology of $S_n$ is exactly $p^k$. I'll give the argument just for $n=p^k$. There is a subgroup of $S_{p^k}$ isomorphic to the direct product of $p$ copies of $S_{p^{k-1}}$. Furthermore, the index of this subgroup is divisible by $p$ but not divisible by $p^2$. For any finite group $G$ and subgroup $H$, there is a transfer map in cohomology $H^*(H;\mathbb{Z})\rightarrow H^*(G;\mathbb{Z})$ with the property that the composite map from $H^*(G;\mathbb{Z})$ to itself given by first mapping to $H^*(H;\mathbb{Z})$ and then transferring back up is equal to multiplication by the index $|G:H|$. Going back to the symmetric group, we know by induction on $k$ that the exponent of the $p$-part of $H^*(S_{p^{k-1}};\mathbb{Z})$ is $p^{k-1}$, and by the Kunneth formula the same holds true for the direct product of $p$ copies of this group. The restriction from $S_{p^k}$ down to this subgroup, followed by the transfer map back up is, up to units, multiplication by $p$ on the $p$-local cohomology $H^*(S_{p^k};\mathbb{Z}_{(p)})$. Hence the exponent of this group is at most $p^k$. Combined with the Evens-Venkov lower bound this gives the claim.<|endoftext|> TITLE: Distributivity of group topologies on $\Bbb Z$ QUESTION [14 upvotes]: Let $\mathcal L$ be the set of all group topologies on $\Bbb Z$. It is known that $(\mathcal L,\subseteq)$ is a modular complete lattice [1]. Is $(\mathcal L,\subseteq)$ distributive? $$~$$ [1] Lamper, Milan. Complements in the lattice of all topologies of topological groups. Arch. Math. (Brno) 10 (1974), no. 4, 221--230 (1975). REPLY [15 votes]: For each $\alpha \in S^1$, the map $\varphi_{\alpha} \colon \mathbb Z \to S^1$, given by $n \mapsto \alpha^n$, induces a topology $\tau_{\alpha}$ on $\mathbb Z$. A basis of neighborhoods of $0$ for $\tau_{\alpha}$ is given by the sets $$U_{n,\alpha} := \left\{k \in \mathbb Z \mid |\alpha^k-1| < \frac1n \right\}, \quad n \in \mathbb N.$$ I denote by $\tau_{\alpha} \wedge \tau_{\beta}$ the largest group topology that is contained in $\tau_{\alpha} \cap \tau_{\beta}$ - which is in general different from $\tau_{\alpha} \cap \tau_{\beta}$. Claim 1: If $\alpha,\beta \in S^1$ are irrational and such that $\alpha/\beta \in S^1$ is also an irrational angle, then $\tau_{\alpha} \wedge \tau_{\beta}$ is trivial. Proof: Let $U \in \tau_{\alpha} \wedge \tau_{\beta} \subset \tau_{\alpha} \cap \tau_{\beta}$ and $0 \in U$. There exists some $V \in \tau_{\alpha} \wedge \tau_{\beta}$ such that $V-V \subset U$ and $0 \in V$. Then, there exists $n$, such that $U_{n,\alpha} - U_{n,\beta} \subset V - V \subset U$. However, $\mathbb Z=U_{n,\alpha} - U_{n,\beta}$. Indeed, for any $k \in \mathbb Z$, there exists some $m$, such that $|\alpha^m-1|<\frac1n$ and $|\beta^{k+m}-1|=|\beta^{m} - \beta^{-k}|<\frac1n$, because $(\alpha,\beta)$ generates a dense subgroup of $S^1 \times S^1$ and thus can approximate the point $(1,\beta^{-k})$ arbitrarily well. q.e.d. Claim 2: $\tau_{\alpha} \vee \tau_{\beta}$ is the topology induced from the map $\mathbb Z \to S^1 \times S^1$, $n \mapsto (\alpha^n,\beta^n)$. In particular, applying various automorphisms of $S^1 \times S^1$, we see that $$\tau_{\alpha} \vee \tau_{\beta} = \tau_{\alpha} \vee \tau_{\alpha\beta} = \tau_{\beta} \vee \tau_{\alpha\beta}.$$ Let $\alpha,\beta \in S^1$ be irrational angles, such that $\alpha/\beta$ is also irrational and consider the topologies $\tau_{\alpha},\tau_{\beta},\tau_{\alpha \beta}$. Any pairwise meet (in the lattice of group topologies) is the trivial topology (by Claim 1) and any pairwise join yields the same topology (by Claim 2). Hence, embeds as a sub-lattice in the lattice of group topologies. Hence, the lattice of group topologies cannot be distributive.<|endoftext|> TITLE: Can a division algebra have degree divisible by its characteristic? QUESTION [10 upvotes]: I apologize in advance if this is easy, but I've tried Googling, and had no luck. I'm currently working on a proof, and I realized in the course of writing that this proof will break if out there in the world there exists a division algebra $D$ with the following properties: The characteristic of $D$ (as a normal ring) is $p$. The degree of $D$ (the square root of its dimension over its center) is divisible by $p$. I would really appreciate if anyone knows an example, or a proof that there are none. I could easily add hypotheses to rule out this case, but that would make things a bit messier, so I don't want to do that unless it's really necessary. Unfortunately, I know zippo about division algebras in characteristic p, other than that every finite one is actually just a field. EDIT: Thanks for the references. Doing a little reading based on Mikhail's suggestion, I found a result which is good enough for my purposes: no such division algebra exists whose center is a perfect field, which was proven in 1934 by the remarkably named Abraham Adrian Albert. Those wanting more details on his life can read this detailed and impressive obituary by Jacobson. The best detail is that his Ukrainian father decided to ditch his last name (which is now unknown!) in Victorian England, and adopt the prince consort's name instead. REPLY [15 votes]: There are counterexamples for each $p$. The easiest maybe is the following: Let $F$ be the field of order $p^p$, and $\sigma$ be an automorphism of $F$ of order $p$. Let $D=F((t))$ be the set of Laurent series of the form $\sum a_it^i$ with the usual addition. Define multiplication by $ta=a^\sigma t$. Then $D$ is a division algebra with center $Z=\mathbb F_p((t^p))$, so $[F:Z]=p^2$.<|endoftext|> TITLE: Is every elementary absolute geometry Euclidean or hyperbolic? QUESTION [20 upvotes]: Absolute geometry is any one that satisfies Hilbert's axioms of plane geometry without the axiom of parallels. It is well-known that it is either the Euclidean or a hyperbolic plane. For an elementary version we also drop the (Cantor's) axiom of continuity, Greenberg calls such geometries Archimedean H-planes in his survey paper. We can still define a metric on them in the usual way. Repeatedly bisecting a picked "unit segment" and laying its pieces off of any other gives a binary fraction (possibly infinite) that is assigned as the segment's length. The distance between two points is defined to be the length of the line segment connecting them, unique by incidence. Incidence, order and congruence imply that perpendicular is shorter than oblique, so the triangle inequality holds. Without the continuity however Archimedean H-planes may not be metrically complete. Will their completions still always be Euclidean or hyperbolic? In other words, if we remove the axiom ensuring completeness, and then take the completion, do we end up with what we started with? Here is why I am having doubts. It is easy to check by limit arguments that most axioms still hold in the completion, but not that two lines intersect at no more than one point, for example. With extra points added perhaps some lines intersect more than once. Also, there is an algebraic classification of Archimedean H-planes due to Pejas described in the Greenberg's paper (p.760). One of them, called the semi-elliptic plane, is quite peculiar. It satisfies the Lambert's hypothesis of the acute angle (in any quadrilateral with three right angles the fourth angle is acute), but any two non-intersecting lines in it have a unique common perpendicular. In other words, no two lines are asymptotically parallel. If the semi-elliptic plane is isometric to a subset of a hyperbolic one then we should be able to obtain it by removing some points from the latter. But if we remove all lines asymptotically parallel to some line in the hyperbolic plane, nothing will be left other than that line itself. On the other hand, if the completion is elliptic, as the name suggests, then how does it square with the hypothesis of the acute angle? Is the metric completion of an Archimedean H-plane always Euclidean or hyperbolic? If so than what subplane of a hyperbolic plane is semi-elliptic? If not then what is the completion of the semi-elliptic plane? Unfortunately, Pejas's original works weren't translated from German. Here is the passage from Greenberg describing the semi-elliptic plane. "Pejas gave the following example of an Archimedean H-plane in which the acute angle hypothesis holds but which is not hyperbolic; it is an example of a semi-elliptic plane, defined by the property that any two parallel lines have a unique common perpendicular (in a hyperbolic plane, two asymptotically parallel lines have no common perpendicular, whereas two divergently parallel lines have a unique one): Let $K_0$ be an Archimedean ordered field with two distinct orderings $<$ and $<'$ (for example, $K_0 = \mathbb{Q}(\sqrt{2})$). Let $L$, $L'$ be the real closures of $K_0$ with respect to these orderings within a given algebraic closure. Set $K = L\cap L'$. Then $K$ is Pythagorean, Archimedean, and contains an element $k$ such that $k<0$ and $0<'k$. We take $k$ as metric constant and the points of $\mathcal{K}$ to be all $(x,y)$ in the affine plane over $K$ for which $k(x^2 +y^2)+1>0$. $\mathcal{K}$ is the interior of the "absolute conic" $x^2 +y^2=-k^{-1}$, which is empty because $\sqrt{-k}\notin K$. Since $\mathcal{K}$ is maximal, that conic is the locus of all ideal points, so asymptotic parallels do not exist and the plane is semi-elliptic." REPLY [11 votes]: Your answer is correct, except that in the example you gave, one has to adjoin far more numbers than you described in order to get to that ordered Pythagorean not-Euclidean field K. That example is described on p.594 of the fourth edition of my book Euclidean and Non-Euclidean Geometries: Development and History (W.H. Freeman, 2008). There I called such a plane where parallel lines have a unique common perpendicular an HE-plane, abbreviating "halb elliptisch" from Pejas' classification article. Since you are only interested in the Archimedean case in order to get a natural metric once a unit segment is chosen, your HE-plane satisfies the acute angle hypothesis (by the Saccheri-Legendre theorem). You then describe an HE-plane accurately as the interior of a virtual circle in the affine plane over an ordered Archimedean Pythagorean not-Euclidean field K. An Archimedean ordered field is a subfield of R (up to isomorphism), so when you metrically complete it, you get R. To make your argument completely rigorous, you would have to prove that the metric completion of an Archimedean H-plane is again an H-plane. Then, since it is complete, it must be either the real Euclidean or the real hyperbolic plane (in the HE-plane case it is the real hyperbolic). As I pointed out on p.594 of my book and as you indicated, if the line-circle axiom holds, then an Archimedean H-plane must be either Euclidean or hyperbolic, but its coordinate field could be any Euclidean subfield of R, such as the field of constructible numbers or the field of real algebraic numbers. See also my March 2010 MONTHLY survey paper mentioned by Will Jagy entitled "Old and New Results in the Foundations of Elementary Plane Euclidean and Non-Euclidean Geometries." Section 2 is all about Will Jagy's results about regular-polygoning circles in the hyperbolic plane (you can't always "square" them), and Section 3 is about the undecidability and consistency of elementary geometry. If you email me at mjg0@pacbell.net I will send you my latest updating of that article. (The terminology for all this is confusing. Yes, Janos Bolyai did introduce the term "absolute geometry" for the common part of real Euclidean and real hyperbolic geometries. I have argued - and it has generally been accepted by other writers - that "neutral geometry" is a better name, because one remains neutral about which parallel postulate to assume. I also argued that "absolute geometry" should be the name for Bachmann's geometry based on reflections, since it includes not only neutral geometry but also elliptic and other geometries - see p.588 of my book. Furthermore, even for neutral geometry, why restrict to real geometries? A model of Hilbert's incidence, betweenness and congruence axioms we now call a Hilbert plane or an H-plane for short. Pejas classified all H-planes. His classification is described on pp. 588ff of my book.)<|endoftext|> TITLE: Flag complexes that are shellable but not vertex decomposable QUESTION [13 upvotes]: As the title suggests, I was wondering if anyone can point me to any examples in the literature to flag complexes that are shellable but not vertex decomposable. It is well-known that if a simplicial complex $\Delta$ is vertex decomposable, then $\Delta$ is also shellable. There are examples where the converse fails, but in all the examples I have seen, $\Delta$ is not a flag complex, i.e., its maximal non-faces all have cardinality two. (A flag complex is sometimes called an independence complex of a graph, since the faces of the complex correspond to the independent sets of the graph). Some context: I've been looking at the independence complexes of circulant graphs, and as part of this project, my co-authors and I discovered that the independence complex of $C_{16}(1,4,8)$ is shellable but not vertex decomposable. (This is the graph with the vertex set $\{0,1,...,15\}$ where there is an edge between $i$ and $j$ if and only if $|i-j|$ or $16-|i-j|$ is in $\{1,4,8\}$.) We are not aware of any other examples of flag complexes with this property. Even if such examples do exist, we were curious how the size of our example compared with the known examples. EDITED TO ADD: Since there has been some interest in what this graph looks like, here is another way to draw it: Graph C_16(1,4,8) http://flash.lakeheadu.ca/~avantuyl/images/c16148.jpg An easy way to construct the circulant graph $C_n(a_1,\ldots,a_t)$ is to arrange the $n$ vertices in a circle, join the vertex $0$ to the vertices $a_1,\ldots,a_t$, join the vertex $1$ to the vertices $a_1+1,\ldots,a_t+1$, and so on (of course, the addition is modulo $n$). REPLY [5 votes]: I'm very slow to respond, but have finally found some time to put your complex into GAP and examine it. Questions on $k$-decomposability on flag complexes There were two questions about $k$-decomposability and flag complexes that seemed natural, and that I didn't know the answer to. Is every flag complex that is shellable also vertex-decomposable? (Your example says that the answer is "no"!) Is every flag complex that is shellable also 1-decomposable? I think this is still open. The 1st question seemed interesting just because all of the main constructions (barycentric subdivision, CL-shellings) that give a shellable flag complex seemed to give a vertex-decomposable complex. On the 2nd question: A complex $\Delta$ is 1-decomposable if we can find either a shedding vertices $v$ or shedding edges $e$ such that respectively ($\Delta \setminus v$ and $\operatorname{link}_\Delta v$) or ($\Delta \setminus e$ and $\operatorname{link}_\Delta e$) are both shellable. The conditions for a shedding edge are similar to those for a shedding vertex — see e.g. Provan and Billera [2] or Jonsson [1] or my own [3]. In a pure $d$-dimensional complex, it suffices that deleting the edge preserves pure $d$-dimensionality. There are also definitions of $k$-decomposability for higher $k$. Since deleting either a vertex or an edge from $\Delta$ preserves the property of being flag, a positive answer to Question 2 would say that you can always shell flag complexes in the framework of $k$-decomposability, without needing to leave the world of flag complexes. (This would be some kind of partial dual result to the fact that every shellable pure $d$-dimensional complex is $d$-decomposable.) Your complex I found the details of how $\Delta = \Delta(C_{16}(1,4,8))$ failed to be vertex-decomposable somewhat surprising. Any vertex (and thus every vertex) is a shedding vertex, and of course links are shellable (and presumably vertex-decomposable). But deleting a vertex leaves a non-shellable complex. Perhaps the easiest way to see the non-shellability of the deletion is to compute the $h$-vector, which (assuming I didn't make a mistake) is $[ 1, 11, 31, 18, -1 ]$. Since a Cohen-Macaulay complex must have a positive $h$-vector (and since $\Delta \setminus v$ is pure), this gives that $\Delta \setminus v$ is not shellable. The same doesn't immediately give a proof that $\Delta = \Delta(C_{16}(1,4,8))$ isn't 1-decomposable. Indeed, deleting the edge $\{0,3\}$ from $\Delta$ (or equivalently adding the edge $\{0,3\}$ to $C_{16}(1,4,8)$ and taking the independence complex) preserves pure 3-dimensionality, and leaves a shellable complex. You can continue this process with edges $\{3,6\}$ and $\{6,9\}$, but once you've deleted these 3 edges, $\{9,12\}$ is not a shedding edge. I suspect that you can continue this process somehow to show the complex is 1-decomposable, but I haven't checked it and don't know for sure. (I hear you have an undergraduate working on this, and it might be an interesting project for her/him to check this carefully). [1] Jakob Jonsson, Simplicial complexes of graphs, Lecture Notes in Mathematics, vol. 1928, Springer-Verlag, Berlin, 2008. [2] J. Scott Provan and Louis J. Billera, Decompositions of simplicial complexes related to diameters of convex polyhedra, Math. Oper. Res. 5 (1980), no. 4, 576–594. [3] Russ Woodroofe, Chordal and sequentially Cohen-Macaulay clutters, Electron. J. Combin. 18 (2011), no. 1, Paper 208, 20 pages, arXiv:0911.4697.<|endoftext|> TITLE: Category of modules over commutative monoid in symmetric monoidal category QUESTION [7 upvotes]: Let $\left(\cal{C},\otimes ,I\right)$ be a symmetric monoidal category (not necessarily closed) and $A$ a commutative monoid in $\cal{C}$. In his DAG III (page 95), Lurie writes: In many cases, the category $\cal{M}od_A\left(\cal{C}\right)$ of $A$-modules in $\cal{C}$ inherits the structure of a symmetric monoidal category with respect to the relative tensor product over $A$. Where can I find conditions, details and proofs of this (seemingly) elementary fact? (I didn't find it in the book I searched - MacLane, Barr & Wells, or in books about operads etc.) I also need the facts that "extension of scalars" $-\otimes A$ is the left adjoint of the forgetful functor and it commutes with the tensor product. And the case when $\otimes$ distriute over the coproduct (or the bi-product). I think I can prove most of these facts, but I want to be sure about them, and it would also be much easier to refer to them. REPLY [6 votes]: One needs that $C$ is cocomplete and that $\otimes$ preserves colimits in each variable (and then $\mathrm{Mod}_C(A)$ will have the corresponding properties). More precisely, you only need that $C$ has reflexive coequalizers and that $\otimes$ preserves them in each variable. This has been known for decades, but the first clear write-up of this, at least I know of, is Florian Marty's thesis. You can also find a discussion on this in my thesis, Section 4.1 (and Chapter 6 for the issue on reflexive coequalizers). See also MO/114457 for a discussion of the internal homs by Todd Trimble. Edit. Here are some other references, which even discuss the case where $- \otimes A$ is replaced by a (suitable) symmetric monoidal monad: H. Lindner. Commutative monads. In Deuxiéme colloque sur l'algébre des catégories. Amiens-1975. Résumés des conférences, pages 283-288. Cahiers de topologie et géométrie différentielle catégoriques, tome 16, nr. 3, 1975. R. Guitart. Tenseurs et machines. Cahiers de Topologie et Géométrie Différentielle Catégoriques, 21(1):5-62, 1980. A. Kock. Closed categories generated by commutative monads. Journal of the Australian Mathematical Society, 12(04):405-424, 1971. G. J. Seal. Tensors, monads and actions. Theory Appl. Categ., 28:No. 15, 403-433, 2013. If I recall correctly, some of these references restrict the monads in such a way that they are of the form $- \otimes A$ anyway.<|endoftext|> TITLE: Is Gauss sum a p-adic measure? QUESTION [7 upvotes]: Let $\Gamma$ be Galois group of cyclotomic $\mathbb{Z}_p$ extension over $\mathbb{Q}$. Consider the function $G$ which sends each finite order character $\chi$ of $\Gamma$ to the Gauss sum $G(\chi)$, view $G(\chi)$ as an element of $\overline{\mathbb{Q}}_p$. My question is if $G$ is a $p$-adic measure on $\Gamma$? REPLY [4 votes]: No. The Gauss sum is not a $p$-adic measure. One cheap way to see this is as follows: if $\chi$ has conductor $p^n$, the $p$-adic valuation of $G(\chi)$ is $n/2$. But if $\mu$ is a measure, the asymptotics of $\int \chi \mathrm{d}\mu$ for $\chi$ of increasing $p$-power conductor, are governed by the $\lambda$ and $\mu$ invariants of $\mu$, and in particular the valuations of these numbers must tend to a limit -- they cannot tend to $\infty$.<|endoftext|> TITLE: Ham sandwich theorem for discrete measures - reference request QUESTION [5 upvotes]: A discrete version of the ham sandwich theorem states as follows (see for instance "Common Hyperplane Medians for Random Vectors" - Hill): For every $\mu_1,...,\mu_n$ discrete (i.e., purely atomic) probability measures on $\mathbb{R}^n$, there is a hyperplane $H$ defined by $\sum_{i=1}^n a_i x_i =b$ such that for every $1 \leq j \leq n$, we have that $$ \mu_j (\lbrace (x_1,...,x_n ) : \sum_{i=1}^n a_i x_i \geq b \rbrace ) \geq \dfrac{1}{2} \text{ and } \mu_j (\lbrace (x_1,...,x_n ) : \sum_{i=1}^n a_i x_i \leq b \rbrace ) \geq \dfrac{1}{2} .$$ My question is about the error when one considers open half planes. I am looking for a result of the following nature: Let $\mu_1,...,\mu_n$ are discrete probability measures that are supported on finite sets (I don't know if the assumption on the support makes a difference). Assume there is $0 <\delta$ such that for every $x \in \mathbb{R}^n$ and for every $j$ we have $\mu_j (x) \leq \delta$, then there is $\varepsilon = \varepsilon (\delta)$ and a hyperplane $\sum_{i=1}^n a_i x_i =b$ such that for every $j$, $$ \mu_j (\lbrace (x_1,...,x_n ) : \sum_{i=1}^n a_i x_i > b \rbrace ) \geq \dfrac{1}{2} - \varepsilon \text{ and } \mu_j (\lbrace (x_1,...,x_n ) : \sum_{i=1}^n a_i x_i < b \rbrace ) \geq \dfrac{1}{2} - \varepsilon ,$$ and $\lim_{\delta \rightarrow 0} \varepsilon (\delta ) = 0$. I assume that this should be a known result, but I haven't found a reference yet. REPLY [5 votes]: You could not find a reference because the statement is not true. Suppose that we are in two dimensions but both are measures are concentrated on a line. The first measure is uniform on $1,\ldots,k$ while the second measure is uniform on $-1,\ldots,-k$. In this case any open halfplane has measure $0$ or $1$ with respect to one of the metrics. If you add the condition that the points in the support are in a general position, then the statement follows trivially from the continuous version.<|endoftext|> TITLE: Prove that the Dirichlet eta function is monotonic QUESTION [10 upvotes]: Let us consider $\eta(p):= \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n^p}$ for $p>0$. Has anyone come along with an elementary proof that $\eta(x)$ is monotonically increasing on this set? By elementary I mean that this is a problem of real analysis, and there must be a solution without using analytic continuation for $\zeta(p)$ and its relations to $\eta(p)$. Another notable remark is that function $g(p):=\frac{1}{a^p}-\frac{1}{(a+1)^p}$ ($a\geq1$) can be both increasing or decreasing at $p=p_0$, depending on $a$, which makes this problem more complex than it seems at first glance. Thanks in advance! REPLY [4 votes]: This is Theorem 3 (p. 10) in the Report: J. van de Lune, Some inequalities involving Riemann's zeta-function, CWI Report ZW 50/75, CentruM Wiskunde & Informatica, Amsterdam,1975, in which the proof does not use complex variables. The pdf of this Report can be found in the CWI repository https://repository.cwi.nl/noauth/search/fullrecord.php?publnr=6895<|endoftext|> TITLE: A hypergeometric puzzle QUESTION [19 upvotes]: $$ 143\,\sqrt {3}\;{\mbox{$_2$F$_1$}\left(\frac{1}{2},\frac{1}{2};\,1;\,{\frac {3087}{8000}}\right)}= 40\,\sqrt {5}\; {\mbox{$_2$F$_1$}\left(\frac{1}{3},\frac{2}{3};\,1;\,{\frac {2923235}{2924207}}\right)} $$ While working on another problem, I found two different ways of expressing my solution. Plugging in a value, I get the above equation. Are there methods/references for proving something like this? Maple does not know them, it seems. added More generally, I have $$ \frac{\displaystyle 3\,\sqrt {2}\; {\mbox{$_2$F$_1$}\Bigg(\frac{1}{2},\frac{1}{2};\,1;\,{\frac { 64\left( 1-y \right) ^{3}}{ \left( -{y}^{3/2}\sqrt {y+8}-8\,\sqrt {y}\sqrt {y+8}+{y}^{2}-20\,y-8 \right) ^{2}}}}\Bigg)} {{\sqrt {-3\,{y}^{2}+60\,y+24+3\,{y}^{3/2}\sqrt {y+8}+24\, \sqrt {y}\sqrt {y+8}}}} = \frac{\displaystyle {\mbox{$_2$F$_1$}\bigg(\frac{1}{3},\frac{2}{3};\,1;\,{\frac { \left( y+8 \right) ^{2} \left( 1-y \right) }{ \left( 4-y \right) ^{3}}}\bigg)}} {( 4 - y)} $$ for $0 < y < 4$ ... and for the above problem I found a value of $y$ so that both $y$ and $y+8$ were perfect squares. But I was hoping someone knew methods to do it other than the round-about way I came to it. REPLY [6 votes]: The OP's general identity is only half equivalent to the one on page 258 of Ramanujan's second notebook. (Also Theorem 5.6 of Berndt's Ramanujan's Notebooks Vol 5, p.112.) It needs a second one, namely Theorem 6.1 on p.116 to complete the puzzle pieces. I. Piece 1: Theorem 5.6 If $$u_1 = \frac{p^3(2+p)}{1+2p},\quad\quad v_1 = \frac{27p^2(1+p)^2}{4(1+p+p^2)^3}$$ then for $0\leq p <1$, $$(1+p+p^2)\,_2F_1\big(\tfrac12,\tfrac12;1;u_1\big)=\sqrt{1+2p}\;_2F_1\big(\tfrac13,\tfrac23;1;v_1\big)\tag1$$ II. Piece 2: Theorem 6.1 If $$u_2 = \frac{q(3+q)^2}{2(1+q)^3},\quad\quad v_2 = \frac{q^2(3+q)}{4}$$ then for $0\leq q <1$, $$\,_2F_1\big(\tfrac13,\tfrac23;1;u_2\big)=(1+q)\;_2F_1\big(\tfrac13,\tfrac23;1;v_2\big)\tag2$$ III. Piece 3: $v_1=v_2$ It turns out the OP's was the special case $v_1=v_2$, hence, $$\frac{27p^2(1+p)^2}{4(1+p+p^2)^3} = \frac{q^2(3+q)}{4}$$ While a cubic in $q$, it conveniently factors linearly. Choosing the correct factor such that $0\leq p,q <1$, we can aesthetically express the OP's general identity as, $$\large \tfrac{(1+p+p^2)(1+q)}{\sqrt{1+2p}}\,_2F_1\Big(\tfrac12,\tfrac12;1;\tfrac{p^3(2+p)}{1+2p}\Big)=\;_2F_1\Big(\tfrac13,\tfrac23;1;\tfrac{q(3+q)^2}{2(1+q)^3}\Big)\tag{3a}$$ where, $$q = \frac{3p}{1+p+p^2}\tag{3b}$$ For example, the OP used $y=\tfrac1{36}$. But using $p=\tfrac7{10}$ on $(3)$, we recover the original equality, $$\tfrac{143\sqrt3}{40\sqrt5}\;{\mbox{$_2$F$_1$}\left(\tfrac{1}{2},\tfrac{1}{2};\,1;\,{\tfrac {3087}{8000}}\right)}={\mbox{$_2$F$_1$}\left(\tfrac{1}{3},\tfrac{2}{3};\,1;\,{\tfrac {2923235}{2924207}}\right)}$$ and puzzle solved.<|endoftext|> TITLE: Is it consistent with ZFC (or ZF) that every definable family of sets has at least one definable member? QUESTION [9 upvotes]: I consider definability to mean one of either cases: Definability without parameters (in the language of set theory), or Definability from ordinals and a real (in the same language). So my question is: Is there a model $M$ of ZFC (or at least of ZF) such that every definable family of sets (not necessarily of reals) contains at least one definable member (in the sense 1. or 2. respectively) but such that $M$ contains nonetheless many non-definable members (i.e. $M$ is not pointwise definable)? REPLY [16 votes]: The following theorem seems to express how the various definability witness properties are connected with each other and with $V=\text{HOD}$. Theorem. The following are equivalent in any model $M$ of ZF: $M$ is a model of $\text{ZFC}+\text{V}=\text{HOD}$. $M$ has a definable well-ordering of the universe. Every definable nonempty set in $M$ has a definable element. Every definable nonempty set in $M$ has an ordinal-definable element. Every $\Pi_2$-definable nonempty set in $M$ has an ordinal-definable element. Every ordinal-definable nonempty set in $M$ has an ordinal-definable element. Proof. ($1\to 2$) The usual HOD order is a definable well-ordering of the universe. ($2\to 3$) Select the least element with respect to the definable order, as in Bjorn's answer. ($3\to 4$) Immediate. ($4\to 5$) Immediate. ($4\to 1$) If $M$ thinks there is a non-OD set, then the set $A$ of all non-OD sets in $M$ of minimal rank is a definable nonempty set in $M$ with no ordinal-definable elements. ($5\to 1$) The stronger implication has now undergone a few improvements, so let me discuss it. I had proposed considering as above the set $A$ of all minimal-rank non-OD sets, which is definable and nonempty in any model of $V\neq\text{HOD}$, but has no ordinal-definable elements. I had guessed that $\Sigma_5$ would be sufficient to define $A$. In the comments, François refined this, arguing that this set was actually $\Sigma_3$-definable and indeed $\Delta_3$-definable. Using his idea, I was able to push this down to show that $A$ is $\Sigma_2\wedge\Pi_2$ definable, by the properties: $A$ is not empty; all elements of $A$ have the same rank; every element of $A$ is not in OD; every set of rank less than an element of $A$ is in OD; every set not in $A$, but of the same rank as an element of $A$, is in OD. Each of these properties is either $\Sigma_2$ or $\Pi_2$, making the set $A$ to be $\Sigma_2\wedge\Pi_2$-definable. Specifically, the first two requirements are $\Sigma_2$, being witnessed in a rank-initial segment of the universe; the third is $\Pi_2$; the fourth and fifth are both $\Sigma_2$, since they are true just in case there is a large $V_\theta$ which believes them to be true. I also noted that $A$ is not provably $\Sigma_2$-definable. Meanwhile, over at my question Can $V\neq\text{HOD}$ if every $\Sigma_2$-definable set has an ordinal-definable element?, Emil made a suggestion leading to the observation that if $V\neq\text{HOD}$, then there is a $\Pi_2$-definable set with no ordinal-definable elements. The set is simply $U=A\times V_\theta$, where $A$ is as above and $\theta$ is least such that $V_\theta$ thinks $A$ is the set of minimal-rank non-OD sets. So I refer the reader to theorem 2 in that answer, which provides the content of the implication ($5\to 1$). ($1\to 6$) Immediate, since under statement $1$, every set in $M$ is ordinal-definable in $M$. ($6\to 4$) Immediate. QED Conclusion. Thus, case (1) of the question occurs in exactly the models of $V=\text{HOD}$ that are not pointwise definable. There are such models, if ZFC is consistent, since one may take any uncountable model of $\text{ZFC}+V=\text{HOD}$. Meanwhile, case (2) of the question — ignoring the issue of real parameters — does not occur at all, since if a set has sets that are not ordinal-definable, then it will have a definable set with no ordinal-definable members, namely, the set of all non-OD sets of minimal rank, as in the implication of statement 4 to statement 1. Update. I edited to the improved statement 5, which we've now got down to the case of mere $\Pi_2$-definability, using the answer to my question Can $V\neq\text{HOD}$ if every $\Sigma_2$-definable set has an ordinal-definable element?. Update. This answer and those of the related questions have known grown into the following paper: F. G. Dorais and J. D. Hamkins, When does every definable nonempty set have a definable element? (arχiv:1706.07285) Abstract. The assertion that every definable set has a definable element is equivalent over ZF to the principle $V=\newcommand\HOD{\text{HOD}}\HOD$, and indeed, we prove, so is the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying $V\neq\HOD$ in which every $\Sigma_2$-definable set has an ordinal-definable element. Similar results hold for $\HOD(\mathbb{R})$ and $\HOD(\text{Ord}^\omega)$ and other natural instances of $\HOD(X)$. Read more at the blog post. REPLY [3 votes]: The answer in case 1. is also yes. In fact, a stronger assertion is true: there exist models of set theory in which every set is definable without parameters. Such models are called pointwise definable, and (as a first observation) are necessarily countable. A collection of results surrounding pointwise definable models of ZFC and GBC (in which every set and every class are definable without parameters) are presented in "Pointwise Definable Models of Set Theory," joint work by Joel Hamkins, David Linesky and me. Here's a link to Joel's blog post on the paper, which gives an overview & link to the paper itself: http://jdh.hamkins.org/pointwisedefinablemodelsofsettheory/<|endoftext|> TITLE: Higher dimensional generalization of: Any quadrilateral tiles the plane? QUESTION [18 upvotes]: Any (non-self-intersecting) quadrilateral tiles the plane.     (MathWorld image.) Q. What is the strongest known generalization of this statement to higher dimensions? I.e., $\mathbb{R}^d$ filling with combinatorial cuboids? Is Michael Goldberg's 37-yr-old paper the latest in $\mathbb{R}^3$? Goldberg, Michael. "On the space-filling hexahedra." Geometriae Dedicata 6.1 (1977): 99-108.     (Snippet from Goldberg.) REPLY [8 votes]: I just stumbled over this, maybe you find it useful: As Douglas Zare already discussed in his comments, a polytope tiling $\mathbb{R}^n$ must have Dehn invariant zero. This statement has appeared at least twice in the literature: Once here: H. Debrunner: Über Zerlegungsgleichheit von Pflasterpolyedern mit Würfeln. Arch. Math. 35 (1980) 583-587. And once here: J.C. Lagarias and D. Moews: Polytopes that fill $\mathbb{R}^n$ and scissors congruence. Discrete and computational geometry 13 (1995), 573-583 A similar problem for tilings by translations has been solved by H. Hadwiger: Mittelpunktspolyeder und translative Zerlegungsgleichheit. Math. Nachr. 8 (1952) 53-58. This was also reproduced by Lagarias and Moews in the paper mentioned above. See also the acknowledgement of priority by Lagarias and Moews, which discusses some of the history, all the literature references are taken from there. I have not found papers discussing if triviality of Dehn invariants is sufficient for tiling (at least in dimensions 3 and 4, where Dehn invariant and volume completely characterize scissors congruence). (Maybe it is possible to use the scissors congruence to the cube together with the cube tiling to prove the converse and construct a tiling?)<|endoftext|> TITLE: Union of conjugates in free groups QUESTION [11 upvotes]: Let $F$ be a (finitely generated) free group, $H \leq F$ of infinite index. Is it possible that $$ \bigcup_{g \in F} gHg^{-1} = F?$$ REPLY [11 votes]: The problem of characterizing groups that are union of conjugates of a proper subgroup was considered in some papers by Wiegold and others. In particular, you can look at Transitive groups with fixed point free permutations, Archiv der Mathematik 27 (1976), 473-475. Groups covered by conjugated of proper subgroups, Journal of Algebra 293 (2005), 261–268. It turns out that the answer to your question is yes. In fact, already the free group on two generators $F_2 = \langle a, b \rangle$ can be covered by the conjugates of one of its proper subgroups. This is shown in the first paper linked above, see Example 3.1 page 474.<|endoftext|> TITLE: Mathematical research papers in general science journals QUESTION [25 upvotes]: I am interested in collecting a list of research papers with a mainly mathematical focus that appeared in high-reputation general science journals without a dedicated mathematics section. This would include things like Nature or Science, but exlcude, for example, PNAS. By a "papers with a mainly mathematical focus" I mean publications whose purpose is the announcement/description/proof of a new mathematical method or result, rather than the mere use of (sophisticated) mathematical techniques to solve a scientific problem from another discipline. I would imagine that often such papers are complemented by a publication in a more traditional mathematical venue that expands on technical details maybe omitted in the general science journal, and it would be nice to collect those as well to see how the two publications differ from each other. I vaguely remember coming across one or two of such papers in the past, but I cannot find them anymore and cannot, therefore, give an example of what I am looking for. I would be grateful if someone could convert this question to Community Wiki mode, I seem to be unable to find the right check-box. REPLY [3 votes]: Evans function and Fredholm determinants by Issa Karambal, Simon J. A. Malham is in Proc. R. Soc. A., February 2015 Volume: 471 Issue: 2174. Proceedings of the Royal Society A does not have a dedicated mathematics section. It features plenty of applied mathematics papers and mathematical physics papers of course, but interestingly, sometimes also pure research mathematics.<|endoftext|> TITLE: Minimizing deep holes in sphere packings QUESTION [5 upvotes]: What's the current state of knowledge regarding packings of spheres in $n$-space that minimize the supremum of the sizes of the holes? This notion of tightness is more rigid than asymptotic density. I would expect tightest packings to coincide with densest periodic packings in low dimensions but not when $n$ is sufficiently large. REPLY [5 votes]: The problem you are asking about is sometimes known as the packing-covering problem, since it asks for a configuration with a fixed packing radius that minimizes the covering radius, irrespective of mean density. The lattice version of the problem is solved in some dimensions (see Table 3 of arXiv:math/0412320), and indeed the answer is different than the optimal packing lattice in dimensions $d>2$ and from the optimal covering lattice in dimensions $d>3$. A surprisingly open problem according to the paper linked is whether in some dimension, the optimal packing-covering lattice has a ratio greater or equal to 2 between the covering radius and the packing radius. That would mean that no lattice packing in that dimension can be saturated, since there must be a hole where another translate of the lattice can fit.<|endoftext|> TITLE: Solving a non linear equation QUESTION [15 upvotes]: I've been trying to prove that the following equation has a unique solution in interval 0 < x < 1 : $$ x = \Big(\frac{1 - (1-x^2)^K}{1 - (1-x)^K}\Big)^2 $$ Where K is a number (integer, if it helps) greater than 1. I have checked it numerically and, in addition to x=0 and x=1, there is always a solution in interval (0,1). (for instance, see this for K=6: http://goo.gl/a7OSSn) Does anyone have any idea of how I can prove the existence and uniqueness of such a fixed point? REPLY [19 votes]: Consider the function $f(x):=(1-(1-x)^k)^2 $, a strictly increasing homeo of $ (0,1)$ onto itself, with $f'(0)=f'(1)=0$. Looking at the sign of $f''$ we see that $f$ is initially strictly convex, then strictly concave (the inflection point being at $1-\big(\frac{k-1}{2k-1}\big)^{1/k}$). This implies that $f(x)/x=f'(x)$ has a unique solution $x_0$ in $(0,1)$, and in fact this means that $g(x):=f(x)/x$ is unimodal (with $g(0)=0$, $g(1)=1$, and maximum point at $x_0$). Finally, the initial equation can be written as $g(x)=g(x^2)$, which of course has a unique solution $x=c\in(0,1)$ due to the unmorality of $g$, in fact $0< c^2 < x_0 < c <1 $. $${*}$$ edit. Incidentally, the same computation works for the generalization suggested by guest in a comment : $x=\frac{ (1-(1-x^{a+1})^b )^c }{ (1-(1-x^a)^b )^c }$, with $a>0, b>1, c>1:$ we can write it in the form $g(x^{a+1})=g(x^a)$ with $g(x):=f(x)/x$ and $f(x):=(1-(1-x )^b )^c$, and the same conclusion follows.<|endoftext|> TITLE: Can $V\neq\text{HOD}$ if every $\Sigma_2$-definable set has an ordinal-definable element? QUESTION [6 upvotes]: This question arises from an issue arising in user38200's recent question concerning models of set theory in which every definable set has a definable element. In my answer to that question, with François's help, it turned out that $V=\text{HOD}$ is equivalent to the assertion that every $(\Sigma_2\wedge\Pi_2)$-definable set has an ordinal-definable element. The question here is whether this level of complexity can be reduced to $\Sigma_2$ or not. Question 1. Is there a model of $\text{ZFC}+V\neq\text{HOD}$ in which every $\Sigma_2$-definable set has an ordinal definable element? In the equivalence to $V=\text{HOD}$, the $\Sigma_2\wedge\Pi_2$ level of complexity arises because in any model of $\text{ZFC}+V\neq\text{HOD}$, the set of all minimal-rank non-OD sets is defined by a formula of complexity $\Sigma_2\wedge\Pi_2$, but has no ordinal-definable elements. I know that this set cannot in general be defined provably by a $\Sigma_2$ formula, since we could preserve the truth of any $\Sigma_2$ formula by forcing up high so as to code sets into the GCH pattern (or some other desired method of coding), and thereby change which sets down low were in OD without affecting that given $\Sigma_2$ assertion. But it isn't clear to me how that observation answers the question here, since one would seem to need to construct a counterexample model of $V\neq\text{HOD}$, where all $\Sigma_2$-definable sets had OD members, and forcing to change that one set doesn't seem to do this. I am also interested in the case of $\Pi_2$-definability. Question 2. Is there a model of $\text{ZFC}+V\neq\text{HOD}$ in which every $\Pi_2$-definable set has an ordinal-definable element? REPLY [8 votes]: Update. (June, 2017) François Dorais and I have completed a paper growing out of this answer and our others on related posts. F. G. Dorais and J. D. Hamkins, When does every definable nonempty set have a definable element? (arχiv:1706.07285) Abstract. The assertion that every definable set has a definable element is equivalent over ZF to the principle $V=\newcommand\HOD{\text{HOD}}\HOD$, and indeed, we prove, so is the assertion merely that every $\Pi_2$-definable set has an ordinal-definable element. Meanwhile, every model of ZFC has a forcing extension satisfying $V\neq\HOD$ in which every $\Sigma_2$-definable set has an ordinal-definable element. Similar results hold for $\HOD(\mathbb{R})$ and $\HOD(\text{Ord}^\omega)$ and other natural instances of $\HOD(X)$. Read more at the blog post. $\newcommand\ZFC{\text{ZFC}} \newcommand\HOD{\text{HOD}} \newcommand\P{\mathbb{P}} \newcommand\Q{\mathbb{Q}}$ Original answer. It turns out that question 1 has a positive answer, while question 2 has a negative answer. The reader may find it useful to know of the characterization of the $\Sigma_2$ properties as the semi-local properties, those which are equivalent to an assertion of the form $\exists\theta\ V_\theta\models\psi$, where $\psi$ can have any complexity. In particular, whenever a $\Sigma_2$ property $\varphi(A)$ is true of a set $A$, it is because there is some $V_\theta$ satisfying something about $A$. We may therefore preserve that $\Sigma_2$ fact about $A$, while forcing over $V$, provided that we only force up high and preserve $V_\theta$, the rank-initial-segment of the universe up to $\theta$. First, let's consider the positive answer to question 1. Theorem 1. Every model of $\ZFC$ has a forcing extension satisfying $V\neq\HOD$, in which every $\Sigma_2$-definable set has an ordinal-definable element. Proof idea: perform a forcing iteration, considering each $\Sigma_2$ formula in turn, where we try to freeze the set defined by that formula and then code one of its elements (if any) into the GCH pattern high above the witness to that $\Sigma_2$ property. In the end, every nonempty $\Sigma_2$-definable set will contain an ordinal-definable element. Proof. Start with $V$ as a ground model. Without loss, by forcing if necessary, we may assume that $V$ satisfies $V=\HOD$, so that there is a definable well-ordering of the universe. Enumerate the $\Sigma_2$ formulas $\varphi_0,\varphi_1,\ldots$, and so on. Note that we may refer to $\Sigma_2$-truth since there is a universal truth predicate for truth of bounded complexity (so there will be no issues with Tarski's theorem on the non-definability of truth). We define a full-support forcing iteration $\P$ of length $\omega$, where the forcing at each stage will become progressively more highly closed. At the first stage, we consider the formula $\varphi_0$, and ask: is there a forcing extension $V[g_0]$ in which $\varphi_0$ holds of a nonempty set $A_0$? If so, we perform such a forcing (choose the least poset forcing this), and let $\lambda_0$ be the smallest $\beth$-fixed point above the size of that forcing so that also $\varphi_0$ is witnessed in $V_{\lambda_0}^{V[g_0]}$. Next, perform additional $\leq\lambda_0$-closed forcing over $V[g_0]$ to an extension $V[g_0][h_0]$, where $h_0$ forces to code one of the elements of $A_0$ into the GCH pattern above $\lambda_0$. This preserves the definition of $A_0$ by $\varphi_0$, while ensuring that $A_0$ has an ordinal definable element. Now, let $\theta_1$ be well above this coding, and continue. At stage $n$, we have the partial extension $V^{(n)}=V[g_0][h_0]\cdots[g_{n-1}][h_{n-1}]$, which performed forcing below the cardinal $\theta_n$. We ask whether we can perform $\leq\theta_n$-forcing so that $\varphi_n$ holds of a nonempty set $A_n$ in the extension. If so, we do that forcing, let $\lambda_n$ be large enough to witness the $\Sigma_2$ property for $\varphi_n$, and then perform GCH coding above that so as to make an element of $A_n$ ordinal-definable, and let $\theta_{n+1}$ larger than all that. (Otherwise, we ignore $\varphi_n$ and let $\theta_{n+1}=\theta_n$.) Consider the corresponding extension $V[G]$, where $G\subset\P$ is $V$-generic. Finally, we force to add a Cohen subset $H\subset\delta$, where $\delta$ is a regular cardinal above $\sup_n\theta_n$, since this will force $V\neq\HOD$ in $V[G][H]$. The final desired model is $V[G][H]$. Because we used full support, it follows that the tail forcing in $\P$ after stage $n$ is $\leq\theta_n$-closed, as is the forcing to add $H$, and so preserves sets of rank below $\lambda_n$. Thus, if $\varphi_n$ defines a nonempty set in $V[G][H]$, then at stage $n$ we would have observed that it was possible to force it to hold of a nonempty set (with forcing that was sufficiently closed), and so we would have treated it at stage $n$. That is, we would have forced to code one of its elements into the GCH pattern, afterwards always preserving that definition and this coding. So in the case that $\varphi_n$ does define a nonempty set in $V[G][H]$, then the stage $n$ forcing exactly ensured that one of the elements of this set was coded into the GCH pattern of $V[G][H]$ and was therefore ordinal-definable there. The later stages of forcing were arranged so as to preserve all these definitions. Thus, $V[G][H]$ is a model of $V\neq\HOD$, as the forcing to add $H$ is weakly homogeneous, such that every $\Sigma_2$-definable nonempty set has an ordinal-definable element. QED The argument reminds me of the forcing iteration proof of the maximality principle, where one forces a given statement to be true, if it is possible to force it in such a way that it remains true in all further forcing extensions. The end result is a model where any statement that could become necessarily true by forcing, is already true, and this is precisely what the maximality principle asserts. Meanwhile, Emil's insightful suggestion in the comments leads to a negative answer to question 2. Theorem 2. If $V\neq\HOD$, then there is a nonempty $\Pi_2$-definable set (with no parameters) containing no ordinal-definable element. Proof. Let $A$ be the set of minimal-rank non-OD sets. That is, $A$ consists of all non-OD sets of rank $\alpha$, where $\alpha$ is minimal such that there is any non-OD set of rank $\alpha$. In my answer to this related question, I had proved that $A$ is characterized by a $\Sigma_2\wedge\Pi_2$ property. Emil's idea was to consider not $A$, but a related set, namely the set $U=A\times V_\theta$, where $\theta$ is the smallest ordinal such that $A\in V_\theta$ and $V_\theta\models A$ is the set of minimal-rank non-OD sets. The set $U$ is defined by the following property: $U$ consists of the cartesian product $U=A\times B$ of two sets $A$ and $B$ such that the set $B$ has the form $B=V_\theta$ for some ordinal $\theta$ such that $A\in V_\theta$ and $V_\theta\models "A$ is the set of minimal-rank OD sets and there is no $\theta'<\theta$ for which $V_{\theta'}\models A$ is the set of minimal-rank non-OD sets"; and finally, the elements of $A$ really are not in OD. This property altogether has complexity $\Pi_2$, due mainly to the last clause. The first part, requiring that $U$ has the form $A\times B$, is $\Delta_0$. The next part, asserting that $B$ has the form $B=V_\theta$ for some ordinal $\theta$ has complexity $\Pi_1$, essentially because one need only assert that $B$ is transitive and satisfies some minimal set theory such that it thinks it is a $V_\theta$, and such that $B$ contains all subsets of any of its elements, so that it is using the true power set operation. The properties asserting that $V_\theta$, that is, $B$, satisfies certain complication assertions has complexity $\Delta_0$, since all quantifiers are bounded by $B$ and hence ultimately by $U$. And finally, asserting that the elements of $A$ are really not ordinal-definable has complexity $\Pi_2$, since "$x\in\text{OD}$'' has complexity $\Sigma_2$, as any instance of ordinal-definability reflects to some $V_\theta$ and hence is locally verifiable; thus, the assertion $\forall x\in A\ x\notin\text{OD}$ has complexity $\Pi_2$. So altogether, the set $U=A\times V_\theta$ is $\Pi_2$-definable, but it can have no ordinal-definable elements, since every element of $U$ has the form $(a,b)$ for some $a\in A, b\in V_\theta$, and if the pair $(a,b)$ were ordinal-definable, then $a$ would be ordinal-definable, contradicting $a\in A$ and the fact that every member of $A$ is not ordinal-definable. QED Note that the proof is completely uniform, in that the definition of the set does not depend on the model in any way. Rather, we have a $\Pi_2$ definition that $\ZFC+V\neq\HOD$ proves is a nonempty set disjoint from OD.<|endoftext|> TITLE: What are the exact holomorphic Lagrangians in complex 2-space? QUESTION [8 upvotes]: In an exact symplectic manifold, i.e. where the symplectic form can be written $\omega = d \lambda$, it's natural to look for exact Lagrangians, i.e. $L$ on which $\lambda_L = df$. One reason is that, any disk ending on such a Lagrangian has vanishing area: $$ \int_D \omega = \int_{\partial D} \lambda = \int_{\partial \partial D} f = 0$$ On the other hand, the search for such manifold-without-boundary Lagrangians in cotangent bundles of manifolds-without-boundary is expected to be fruitless -- Arnol'd conjectured that all such are homotopic to the zero section, and Abouzaid has proven that they're at least homotopy equivalent to the zero section. I am interested in understanding how hard such things are to come by in the cotangent bundle of an open manifold. For cotangent bundles of one dimensional manifolds, the question is trivial -- a Lagrangian encloses area if it's a closed curve, and otherwise has no topology. So the first case is $T^* \mathbb{C}$. What I really want to know is what the holomorphic exact Lagrangian manifolds are; this is motivated in part by the article by Xin Jin which explains that such exact holomorphic Lagrangians give rise to perverse sheaves under the Nadler-Zaslow correspondence. All this is the motivation for the following entirely elementary question: Characterize all algebraic curves $C \subset \mathbb{C}^2$ on which the one-form $y dx$ is exact. REPLY [5 votes]: I'm not sure what you mean by 'characterize'. For example, here is a trivial characterization: Let $C$ be any Riemann surface, with meromorphic functions $a$ and $b$ with $da\not=0$. Then let $(x,y) = (a, db/da)$. The image of $C$ (minus a polar divisor) in $\mathbb{C}^2$ will be an algebraic curve on which $y\ dx = db$. Thus, there are tons of these.<|endoftext|> TITLE: Random Voronoi Diagrams QUESTION [10 upvotes]: I'm interested in what research has already been done with regards to the statistics of random voronoi diagrams. I have had a look on google scholar and results are a little inconclusive. I'm interested in things like the expected size of Voronoi cells etc. REPLY [10 votes]: The number of $(d{-}1)$-facets of a Poisson-process Voronoi cell in $\mathbb{R}^d$ is: $6$ for $d{=}2$; $\approx 15.5$ for $d{=}3$; and $\approx 37.8$ for $d{=}4$. For references and other stats, see: Masaharu Tanemura, "Random Voronoi Cells of Higher Dimensions." (link). Here is an unrelated but attractive image from www.qhull.org:     Voronoi image by KOOK Architecture, Silvan Oesterle and Michael Knauss. REPLY [9 votes]: The expected number of sides and size of a typical cell (typical is interpreted in the sense of Palm measures which basically means what you see on average in a large ball) for the Voronoi cells of a Poisson process with constant intensity in the Euclidean plane (as well as higher dimensional Euclidean spaces) are known. See for example the book by Moller "Lectures on Random Voronoi Tesselations" (it's in the series by Springer "Lecture notes in statistics"). There are other references (which I forgot but you should be able to find easily) with quite explicit calculations for the distribution of the size of the cells and also for these things when the underlying space is the hyperbolic plane. Some other (a bit more exotic) things are known in the Euclidean case such as the asymptotic shape of the ball, first passage time (see this article) Some things are known about percolation on the results adjacency graph (see this article where the hyperbolic case is treated). And there are people (including myself :)) currently thinking about random walks on these random graphs.<|endoftext|> TITLE: Software for computing Thurston's unit ball QUESTION [11 upvotes]: Is there any software which can be used for computing Thurston's unit ball (for second homology of 3-manifolds) of link complements? In particular can I do that with SnapPy? PS: even a table for Thurston's ball of two component links would be helpful for me. REPLY [9 votes]: "Better late than never." Stephan Tillmann and William Worden have produced the software package tnorm. This can be found here: https://pypi.org/project/tnorm/ The software should be able to deal with hyperbolic two-component links easily. William replies to emails, as well. :)<|endoftext|> TITLE: Wrong asymptotics of OEIS A000607 (number of partitions of an integer in prime parts)? QUESTION [23 upvotes]: Sequence A000607 in the Online Encyclopedia of Integer Sequences is the number of partitions of $n$ into prime parts. For example, there are $5$ partitions of $10$ into prime parts: $10 = 2 + 2 + 2 + 2 + 2 = 2 + 2 + 3 + 3 = 2 + 3 + 5 = 3 + 7 = 5 + 5.$ The OEIS gives an asymptotic expression $$A000607(n) \sim \exp\left(2 \pi \sqrt{\frac{n}{3 \log n}}\right). $$ Numerically, this seems to be wrong even if you take the logarithm of both sides. My conjecture is that $$\lim_{n \to \infty} \log\left(A000607(n)\right) \bigg/ \left( 2 \pi \sqrt{\frac{n}{3 \log n}} \right) \ne 1.$$ See the following graph: How might one prove or disprove this conjecture? For more references please see http://oeis.org/A000607. REPLY [41 votes]: Your data is compatible with the more refined estimates proved by Vaughan in Ramanujan J. 15 (2008), 109–121. His Theorems 1 and 2 (together with his (1.9)) reveal that $$\log(A000607(n)) = 2 \pi \sqrt{\frac{n}{3 \log n}}\left(1+\frac{\log\log n}{\log n}+O\left(\frac{1}{\log n}\right)\right). $$ For $n=50000$, we have $$\log(A000607(n)) \approx 252.663 $$ $$ 2 \pi \sqrt{\frac{n}{3 \log n}} \approx 246.601$$ $$ 2 \pi \sqrt{\frac{n}{3 \log n}}\left(1+\frac{\log\log n}{\log n}\right)\approx 300.877$$ So if you use the secondary term that is present in Vaughan's formula, the approximation (without the error term) is not below but above the actual value. We also see that in this particular instance the error is $\approx 48.214$, which is very compatible with the fact that the error term above is $O(1)$ times $$ 2 \pi \sqrt{\frac{n}{3 \log n}}\cdot\frac{1}{\log n}\approx 22.792.$$ In short, your conjecture is probably false, while Vaughan is right. The numeric anomaly is caused by a secondary term that is rather large for the $n$'s you considered.<|endoftext|> TITLE: If the direct image of f preserves coherent sheaves on noetherian schemes, how to show f is proper? QUESTION [14 upvotes]: The other direction is well known. I think it is true and I was told by several other guys doing algebraic geometry that it is indeed true but they did not know how to prove. I am also wondering whether there is a general nonsense style proof. It looks like the statement can be proved by playing adjunctions in several categories. Note that a scheme $X$ is quasi compact iff the structure sheaf $O_X$ is a compact object in category of quasi coherent sheaves $Qcoh(X)$. We also know that if $X$ is noetherian scheme, then coherent sheaves are exactly the compact objects in $Qcoh(X)$, so it looks like this statement is equivalent to say if $f_*:Qcoh(X)\rightarrow Qcoh(Y)$ preseves compact objects, then $f$ itself is proper morphism of scheme. But this somehow means that $f^{-1}$ preserves compactness in the topological sense. (I think this can also be phrased into general nonsense language,say the compact objects in category of topological space.) Then one might be able play the commutativity of $Hom(M,?)$ functor and filtered colimits game to get proof. All the comments are welcome Thanks REPLY [10 votes]: If $f:X\to Y$ is separated of finite type between noetherian schemes and $f_*$ preserves coherence, then $f$ is proper. Here is a proof that follows the geometric idea (given in the comments of Piotr Achinger and Karl Schwede) to use a curve in $X$ that is not proper over $Y$ : Since it is possible to extend a coherent sheaf on an open of $X$ to the whole of $X$ [Stacks project, Tag 01PD], we see that the statement is local on $Y$, and we are reduced to the case where $Y$ is the spectrum of a local ring. By Nagata's compactification theorem, we may find a compactification of $f$: it is the composition of an open immersion with dense image $i:X\to \overline{X}$ and a proper morphism $\overline{f}:\overline{X}\to Y$. We suppose for contradiction that $Z=\overline{X}\setminus X$ is not empty. Let us reduce to the case where $Z$ is finite and for every $z\in Z$, $\mathcal{O}_{\overline{X},z}$ is of dimension $1$. We do it by applying iteratively the following procedure. Choose a closed point $z\in Z$ at which $\mathcal{O}_{\overline{X},z}$ is not of dimension $1$. Using prime avoidance, choose a function $\phi\in \mathcal{O}_{\overline{X},z}$ that vanishes on $z$, but does not vanish identically on any irreducible component of $\overline{X}$ through $z$ nor on any positive-dimensional irreducible component of $Z$ through $z$. Replace $\overline{X}$, $X$ and $Z$ by their intersection with the closure of $\{\phi=0\}$ in $\overline{X}$. The hypothesis of preserving coherence still holds because closed immersions preserve coherence. The Hauptidealsatz ensures that $X$ is still dense in $\overline{X}$. Using noetherian induction on $Z$, and induction on the dimensions of the local rings $(\mathcal{O}_{\overline{X},z})_{z\in Z}$, we see that the procedure terminates. Hence, from now on, we suppose that $Z$ is finite and for every $z\in Z$, $\mathcal{O}_{\overline{X},z}$ is of dimension $1$. Replacing moreover $\overline{X}$ by one of its components through $z$, we may assume that it is integral of generic point $\eta$. Now we distinguish two cases. If $f(\eta)$ is the closed point of $Y$, then $X$ is an affine curve over the residue field of $Y$, and $f_*\mathcal{O}_X$ is not coherent: a contradiction. Otherwise, consider the Stein factorization $\overline{X}\to S\to Y$ of $\overline{f}$. Up to replacing $Y$ by $S$ (we do not lose the hypothesis of preserving coherence, because $S\to Y$ is finite), we may suppose that $\overline{f}$ satisfies $\overline{f}_*\mathcal{O}=\mathcal{O}$. In particular, it has connected fibers. Localizing $Y$ at a point in the image of $Z$ as in the first step, we may still assume that $Y$ is local. The points of $Z$ are closed points, hence sent to the closed point $y$ of $Y$ by properness of $\overline{f}$. Since their only generization in $\overline{X}$ is $\eta$, and since the fibers of $\overline{f}$ are connected, $\overline{f}^{-1}(y)$ consists of a single point $z=Z$. It then follows that $\overline{X}$ has only two points $z$ and $\eta$, and that $Y$ has also only two points: their images. Consequently, $\overline{f}$ is quasi-finite and proper, hence finite, and in fact an isomorphism because $\overline{f}_*\mathcal{O}=\mathcal{O}$. At this point, the hypothesis that $f_*\mathcal{O}$ is coherent means exactly that $\textrm{Frac}(\mathcal{O}_{Y,y})$ is of finite type over $\mathcal{O}_{Y,y}$, which is absurd.<|endoftext|> TITLE: How to approximate volumes of m-dimensional manifolds with m-dimensional polyhedra QUESTION [5 upvotes]: The areas of a sequence of polyhedra approaching a surface need not approach the area of the surface, but there are theorems guaranteeing that this be so. (T. Rado, On the Problem of Plateau, Chapter 1.) What is known about analogous situations for higher dimensional manifolds in $R^n$? I'd like an answer in terms that an advanced undergraduate can understand. REPLY [6 votes]: Check out my old paper with the late, lamented Fred Almgren, and especially the references there in (e.g. to the paper of Allard). The magic word is "varifold". On second thought, if you look at Frank Morgan's Geometric Measure Theory book, there is something called the Approximation Theorem, which might be just what you need. (page 77 is the beginning of the chapter on the subject, in the fourth edition of Morgan's book).<|endoftext|> TITLE: Representing a number close to 1 with a sum of reciprocals of natural numbers QUESTION [21 upvotes]: For positive integers $n_1, \ldots, n_k$, let $H(n_1, \ldots, n_k)$ denote $1/n_1 + \ldots + 1/n_k$. Let $V(N)$ be the largest possible value of $H(n_1, \ldots, n_k)$ that is less than 1, subject to the condition that $n_1 + \ldots +n_k \le N$. So $V(5) = 5/6$, realized as $1/2 + 1/3$. My question is, how does $1/(1-V(N))$ grow as a function of $N$? In particular, is there a $C$ and a $k$ such that $$ \frac 1{1-V(N)} \le C N^k\ $$ for all $N$? REPLY [4 votes]: Some greater detail about the context for the question. Thurston (see Doaudy and Hubbard) considers a branched cover $f\colon S^2 \to S^2$ such that $P_f \equiv \{ f^n(c) \mid c \in C_f$ is finite (where $C_f$ is the set of critical points of $f$). He defines a map $\sigma_f$ from $\operatorname{Teich}(S^2, P_f)$ to itself by pulling back complex structures (up to maps isotopic to the identify rel $P_f$) by $f$. The fixed points for $\sigma_f$ are exactly the rational maps that are "combinatorially equivalent" to $f$. Moreover, he proves that in the generic case for $f$, the map $\sigma_f$ is uniformly contracting on compact subsets of $M(S^2, P_f)$, which is just the space of maps of $P_f$ into the Riemann sphere, up to Moebius transformation. Moreover we understand the map $\sigma_f$ up to a bounded error as follows. Let $WA(S^2, P_f)$ be the space of real-weighted sums of disjoint curves (nontrivial and nonperipheral, and up to homotopy) on $S^2 \setminus P_f$. We can define a map $W \colon \operatorname{Teich}(S^2, P_f) \rightarrow WA(S^2, P_f)$ by weighting each curve by the modulus of the associated covering annulus, and then discarding the curves of weight less that 1. Moreover we can prove the fundamental estimate that $$ W(\sigma_f(X)) = f^*W(X) + O(1) $$ for all $X \in \operatorname{Teich}(S^2, P_f)$, where $f^*\colon WA(S^2, P_f) \to WA(S^2, P_f)$ is defined by $$ f^*\gamma = \sum_{f(\eta) = \gamma} \frac1{\operatorname{deg} f|_\eta} \eta. $$ Here the constant for the O(1) depends only on $\operatorname{deg} f$ and $|P_f|$. Now suppose that $\alpha$ is an "invariant multicurve", which means that $f^{-1}(\alpha) \subset \alpha$. Then $f^*\colon \mathbb R^\alpha \to \mathbb R^\alpha$ is a positive linear transformation (or at least nonnegative) and so has a positive leading eigenvalue $\lambda_\alpha$. If $\lambda_\alpha > 1$ it is easy to prove that $\sigma_f^nX \to \infty$ as $n \to \infty$, for any starting point $X$. It turns out that same also holds if $\lambda_\alpha = 1$. If $\lambda_\alpha = 1 - 1/K$ then we expect to take about $K$ iterations of $\sigma_f$ to get to a place where the covering annuli have modulus about $K$. In particular if $\alpha$ is just a single curve then we will have $$ W_\alpha(\sigma_f X) = (1-1/K) W_\alpha(X) + O(1) $$ where the $O(1)$ term is positive and converges in the course of the iteration. Now we should be able to construct a branched cover $f$ of degree about $N$ (say with $|P_f| = 4$) and a single curve $\alpha$ on $S^2 \setminus P_f$ such that $f^*\colon \mathbb R^\alpha \to \mathbb R^\alpha$ is multiplication by $1 - 1/K(N)$, and then by Noam Elkies' estimate we find that the number of iterations of $\sigma_f$ to get near the fixed point (from a reasonable starting point) can grow superpolynomially in $\operatorname{deg} f$ and $|P_f|$. So the iteration of $\sigma_f$ is not a polynomial-time algorithm to find the fixed point of $\sigma_f$ (which is the unique rational map representing $f$).<|endoftext|> TITLE: Is it easy to prove that $\sum_n |X(\mathbb{F}_{q^n})| t^n$ is rational? QUESTION [12 upvotes]: Background: Let $X$ be an algebraic variety over a finite field $\mathbb{F}_q$. One of the successes of Etale cohomology - previously achieved by Dwork- was proving the rationality of the Zeta function of X, defined as $$Z(X,t)=\textrm{exp}\left(\sum_{n\geq 1}\frac{|X(\mathbb{F}_{q^n})|t^n}{n}\right).$$ As a corollary, we get the rationality of $tZ'(X,t)/Z(X,t)$, which is the sum $P(X,t)=\sum_n |X(\mathbb{F}_{q^n})| t^n.$ Question: Is there an elementary way to see that $P(x,t)$ is a rational function? Note that this is strictly weaker than asking whether $Z(X,t)$ is a rational function. REPLY [11 votes]: Not an answer to the question: in a 1976 Transactions of the AMS paper, Catarina Kiefe shows that the logarithmic derivative of zeta is rational for sets definable over finite fields (it is NOT necessarily true that zeta itself is, as she points out in the paper). So, the good news is that the rationality of $Z^\prime/Z$ is strictly weaker than rationality of $Z,$ the bad news is that she uses Dwork's result (and a bunch of model theory, though the paper is quite short). The fact that she makes no reference to a direct proof of such a result for varieties may be a sign that none was known at the time.<|endoftext|> TITLE: Groups and pregeometries QUESTION [7 upvotes]: Definition. For an infinite structure $\mathcal{A}$ and $cl : P(dom(\mathcal{A})) \longrightarrow P(dom(\mathcal{A}))$ , we say that $(\mathcal{A}, cl)$ is a structure carrying an $\omega$-homogeneous pregeometry if the following holds: (a) $(\mathcal{A}, cl)$ is a pregeometry, (b) $dim(\mathcal{A})$ is the same as the cardinality of $\mathcal{A}$, (c) If $A \subseteq \mathcal{A}$ is finite and $a, b \in \mathcal{A} \setminus cl(A)$ , then there is $f\in Aut(\mathcal{A}/A)$ such that $f(a) = b$ and for all $B \subseteq \mathcal{A}$, $f(cl(B)) = cl(f(B))$. Theorem. Suppose that $(G, cl)$ is a group carrying an $\omega$-homogeneous pregeometry. Then either $G$ is commutative or unstable. Question. Are there non-commutative groups that carry an $\omega$-homogeneous pregeometry? Giving references is appreciated. REPLY [6 votes]: This question has been around for some time. Connections of homogeneous pregeomtries, quasiminimal structures and regular types have been studied in a recent article of Pilay and Tanovic. They show that the generic type of a homogeneous pregeometry is strongly regular (and generically stable) and conversely a global strongly regular type induces a homogeneous pregeomtry (Theorem 3). Earlier they analyse regular groups and ask if every regular group is commutative (which a variant of your question), see the question after Theorem 2. As far as I know, the question is still open.<|endoftext|> TITLE: Random Vornoi Diagrams (particular measures) QUESTION [5 upvotes]: This is my second question about Random Voronoi diagrams, in my first question was given some excellent advice but i was not clear in explaining what i was looking for. I'm interested to know whether there are any known results about the following measures in 2-dimensional (preferably) random Vornoi diagrams. I think i should state also that i'm working in a circular disk so if the setting of any currently known results is different would the nature of the results change if i was to change the setting? The measures are a) The variance of the number of neighbours of each cell in the Vornoi Diagram. b) The variance of the length of the edges in the random Delaunay triangulation. c) The variance of the smallest, medium, and largest angles in the Delaunay triangulation. I know it's perhaps not feasible to get information on the exact measures i am looking for, i'm therefore interested in anything related to the measures. Thanks REPLY [4 votes]: Page 104 of Moller's book (which I mentioned in my answer to your previous question) has the heading "On the distribution of the typical Poisson-Delaunay cell and related statistics". It gives an exact distribution for the distance between the nucleus and vertices of the cell plus the joint distribution for the angles to the vertices as seen from the nucleus. He then goes on to deduce the moments (e.g. variance) of some of these quantities. Of course this is for the tesselation associated to a homogeneous Poisson point process on the entire plane. It may not be exactly what you're looking for... but I think you should at least take a look and try to understand what's done there. He also gives a reference to a 1992 paper of Rathie for some of the results and the planar case is discussed separately.<|endoftext|> TITLE: Besides the tracioid are there other surfaces of revolution that have a constant negative curvature? QUESTION [9 upvotes]: There is no surface in $ R^3 $ that can represent the complete hyperbolic plane (Hilberts theorem) so we always have to do with a surface that is not completely equivalent, has a cusp somewhere, but in most publications on hyperbolic geometry, it is almost given that the tracioid (tractrix rotated about its asymptope) is a surface that has a constant negative curvature, and in many publications "tracioid" and "pseudosphere" are used interchangable. But I am wondering are there other surfaces of revolution that have a constant negative curvature? I did some searching and did find: In Klein's "Vorlesungen uber Nicht-Euclidische Geometrie" (1928) $4, page 286, figure 218 219 and 220, Klein gives three surfaces for hyperbolic surfaces: figure 218 looks like an hill figure 219 the tracioid and figure 220 looks like an single sheet hyperboloid or catenoid Unfortunedly Klein doesn't give equations of these surfaces. In Sommerville "The elements of non euclidean geometry" it says (Dover edition page 167) Furtunedly we do not require to take the imaginary circle a the type of surfaces of constant negative curvature. There are different forms of such surfaces, even of revolution, but the simplest is the surface called pseudosphere, which is formed by revolving a tractrix about its asymptope. Again a hint that other surfaces exist but no equations, but maybe he refers to surfaces that are not surfaces of revolution. In https://math.stackexchange.com/a/666101/88985 there is a link to http://www.dm.unibo.it/~arcozzi/beltrami_sent1.pdf and this publication says at page 6: Gauss published his Theorema egregium in 1827 and it was already clear that, if figures could be moved isometrically, cuvature had to be constant. Minding observed that the converse was true in the 30's, and he found various surfaces of constant negative curvature in Euclidean space, the tractroid among them. Again sadly there is no reference to the publication of Minding. I did ask a similar question at the mathematics stack exchange site https://math.stackexchange.com/q/930847/88985 and one answer refered to the virtual math museum, the gallery of famous surfaces http://virtualmathmuseum.org/Surface/gallery_o.html And under Pseudospherical Surfaces (K = -1) It mentions three surfaces of revolution with a constant negative curvature: The tracioid : http://virtualmathmuseum.org/Surface/pseudosphere/pseudosphere.html The Conic K =-1 Surface of revolution : http://virtualmathmuseum.org/Surface/conic_k-1_sor/conic_k-1_sor.html looks like Klein's hill and The hyperbolic K =-1 Surface of revolution : http://virtualmathmuseum.org/Surface/hyperbolic_k1_sor/hyperbolic_k1_sor.html looks like Klein's hyperbolioid These pages sadly do not give a lot of information. Now I am stuck: What are those other surfaces of revolution that have a constant negative curvature? and what are their (parametric) equations? Or don't they have a constant negative curvature and can it be proved that the tracioid is the only surface of revolution with an constant negative curvature? REPLY [5 votes]: The topic is quite old. There are three and only three types of rotationally symmetric surfaces for constant $K = -1/a^2$ where $a$ is the cuspidal radius of the central pseudosphere. These are the central pseudosphere (parameter $m =1$) or tractricoid, Conic type ($m < 1$) and Ring type ($m > 1$). The descriptions Ring, Conic type etc. are given by Klein in: Felix Klein," Vorlesungen über nicht-euklidische Geometrie" 3rd ed. (Berlin, 1928). with a reference/reprint iirc from Crelle's Journal. Asymptotic lines of a Chebyshev Net on these three surfaces are given by the Sine-Gordon Equation. Present day English translations of Vorlesungen may be available on googling, else one can contact German newsgroups e.g., de.sci.mathematik. Although the central pseudosphere is often referred to as "the" Beltrami surface, the physical paper model he made is isometrically equivalent to the Conic type ( m > 1) that you can readily verify in Daina's blog. I am discussing it the concurrent thread here In 1868 when making his hyperbolic plane model Beltrami was a professor of mathematics in Bologna. It had to be important to him to take his paper models with him when Beltrami returned to University of Pavia 1876.<|endoftext|> TITLE: Diophantine equation QUESTION [5 upvotes]: I would like to ask the broad community what is known about the solutions of diophantine equation $$\frac{u}{v} +\frac{v}{w} +\frac{w}{u} =t$$ where $t,v,u,w\in \mathbb{N}.$ I read a book of W. Sierpinski, $\text{ 250 Problems in Elementary Number Theory}$ and there is that this is still an open problem. Is this true? Could anyone give me some references? I know from MSE that that equation has a solutions of the form $(u,v,w,t) =(k,k,k,3)$ and $(u,v,w,t)=(k,2k,4k,5)$ and also I know that solutions of this equation need not be of the form $(u,v,w) =(a^2 b, b^2 c ,c^2 a).$ I would like to know what is known about solutions of this equation? Any references to this problem will be welcome. REPLY [9 votes]: The values of $t$ for which a solution exists are tabulated at the Online Encyclopedia of Integer Sequences. There are also references there and links to further information. In particular, there is a reference to the paper, Andrew Bremner and Richard K. Guy, Two more representation problems, Proc. Edinburgh Math. Soc. (2) 40 (1997), no. 1, 1–17, MR1437807 (98d:11037). The review by Michel Olivier says solutions to $n=x/y+y/z+z/x\,(*)$ correspond to rational points on the elliptic curves $Y^2=X^3+(nX+4)^2$; the paper tabulates $n$ for which $(*)$ has a solution. The OEIS entry also links to this page of Dave Rusin. That in turn links to these two discussions, which take up where Bremner & Guy left off.<|endoftext|> TITLE: Different notions of convergence of complex subvarieties QUESTION [8 upvotes]: Let $X$ be a smooth complex algebraic variety (or, better, complex analytic manifold). Let $\{C_i\}$ be a sequence of compact algebraic subvarieties (resp. analytic reduced subspaces) which converges to a compact reduced subspace $C$ in the sense of the analytic topology on complex points of the Hilbert scheme (resp. Douady space). Question. Is it true that $\{C_i\}$ converges to $C$ in the sense of currents, namely for any smooth differential form $\omega$ on $X$ one has $$\lim_{i\to\infty}\int_{C_i}\omega=\int_C\omega?$$ A reference would be helpful. Added: In the case when $C$ is not reduced, the right hand side in the last equality should probably be multiplied by an appropriate multiplicity. REPLY [4 votes]: For Kahler manifold this is true. First, notice that limits in the sense of Barlett or Douady clearly (up to multiplicity) coincide with the limits in Hausdorff topology on the set of compact subsets. This can be proven using the following argument: the Douady moduli space $D$ with the usual topology is Hausdorff, compact, and the map to the same space $H$ with the Hausdorff topology is continuous. By Bishop's theorem, $H$ is also compact. Bijective continuous maps between compacts are homeomorphisms. Now, the map from $H$ to currents is continuous in currents topology (also by construction, or, if you like, by semicontinuity of Lelong numbers). The same argument proves that it is also a homeomorphisms (currents are compact for obvious reasons). The same argument for complex manifold would work if the volume of the subvarieties is bounded, because this is the condition under which the Bishop's theorem remains true.<|endoftext|> TITLE: Generalizing "variation of parameters" QUESTION [11 upvotes]: I'm stuck on generalizing an ODE formula and could use your help! One way to think about "variation of parameters" is that it bakes the solution $z(t)=e^{At}z_0$ of $z'=Az$ (here $z(t)\in\mathbb{R}^n$, $A\in{\mathbb R}^{n\times n}$) into formulas for nonlinear problems. In particular, to solve $y'=Ay+G[y]$ for some $G:\mathbb{R}^n\rightarrow\mathbb{R}^n$, we can write $$y(t)=e^{At}y_0+\int_0^t e^{A(t-\tau)}G[y(\tau)]\,d\tau.$$ The integral compensates between the closed-form solution of the linear ODE and the solution of the ODE including $G[\cdot]$. Suppose instead that we wish to solve $y'=F[y]+G[y]$ for two nonlinear functions $F,G:\mathbb{R}^n\rightarrow\mathbb{R}^n.$ Furthermore, let's say we know how to solve $z'=F[z]$ in closed form via some $\Phi(z,t)$ so that $z(t)=\Phi(z_0,t).$ Is there an analogous formula to variation of parameters in this case? E.g. something of the form: $$y(t)=\Phi(y_0,t)+\int_0^t\left[\textrm{something involving $G$}\right]\,d\tau$$ PS: If it helps, we can assume both $F$ and $G$ come from a Hamiltonian dynamics problem. So, $n$ is even and contains both velocity and momentum variables, $\Phi_t$ is inverted by $\Phi_{-t}$, $\Phi_t$ is area-preserving, and so on. REPLY [13 votes]: Yes, this is called the nonlinear variation of constants formula due to Alekseev: “An estimate for the perturbations of the solutions of ordinary differential equations”, in: Vestnik Moskov. Univ. Ser. I Mat. Meh. 2 (1961), pp. 28–36. I don't think that that article is available in English. It can also be found in the book by V. Lakshmikantham and S. Leela "Differential and integral inequalities: Theory and applications" Vol. I: Ordinary differential equations.Mathematics in Science and Engineering, Vol. 55-I. New York: Academic Press, 1969, pp. ix+390. The formula is $$ \tilde{\Phi}(y_0,t) = \Phi(y_0,t) + \int_0^t D\Phi(\tilde{\Phi}(y_0,\tau),t-\tau) G(\tilde{\Phi}(y_0,\tau)) \;d\tau $$ where $\tilde{\Phi}$ denotes the flow of $F+G$. See also Appendix E in my book for this, a proof, and a few more details.<|endoftext|> TITLE: Eigenstates of Fourier transformation QUESTION [7 upvotes]: Let $\gamma$ be defined on $\mathbb R^n$ by $\gamma (x)=e^{-π x^2}$. With $\mathcal F$ standing for the Fourier transformation defined on the Schwartz space by $$ (\mathcal F u)(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx, $$ we have $ \mathcal F\gamma =\gamma. $ We can also define $\mathcal F$ for the tempered distribution ($\mathscr S'$) with the duality formula $$ \langle \widehat T,\phi\rangle_{\mathscr S',\mathscr S}=\langle T,\widehat{\phi}\rangle_{\mathscr S',\mathscr S}. $$ For instance, the Poisson summation formula is $\widehat S=S$ with $S=\sum_{k\in \mathbb Z^n}\delta_k$. Finally the question: determine all the tempered distributions $T$ such that $$ \mathcal F T=T. $$ REPLY [4 votes]: In case the specific distribution-theoretic argument is not clear... as it hadn't really been overtly mentioned in comments or answers, and is not really suggested by the classical argument as in Titchmarsh and such: E.g., for Poisson summation on $\mathbb R$: observe that the distribution "sum over integers" is annihilated by multiplication by $e^{2\pi ix}-1$, and is translation invariant. Observe that these two conditions are interchanged by Fourier transform. Show that the space of such distributions is one-dimensional: the multiplication annihilation shows that any such distribution is of order $0$ and supported at integers. By classification of distributions supported at a point, it is a sum of Dirac deltas at integers. By translation-invariance, it is (a scalar multiple of) sum-over-evaluation-at-integers. A just-slightly-more-complicated version applies in $n$-dimensions.<|endoftext|> TITLE: A class of matrix determinants between Wronskians and Vandermondes QUESTION [10 upvotes]: Update: see below Let $M$ be an $n\times n$ matrix that's constructed as follows. Construct the right-most column of $M$ as $[\alpha_1(x_1),\cdots,\alpha_n(x_n)]^T$ for some class of fixed functions $\alpha_i(x)$. Now, moving left, each column of $M$ is the derivative of the last. So the next column is $[\alpha_1'(x_1),\cdots,\alpha'_n(x_n)]^T$, and the very first column is $[\alpha_1^{(n-1)}(x_1),\cdots,\alpha^{(n-1)}_n(x_n)]^T$. I'm curious if there's a name for such matrices. Specifically, I'm wondering about the determinant of such matrices: $G(x_1,\cdots,x_n)=\det(M(x_1,\cdots,x_n))$. As Jose rightfully pointed out when all variables are set equal we get the usual Wronskian. I'm particularly curious about $\alpha_i(x)= x^{d_i}/(d_i)!$ for some decreasing positive integer sequence $d_i$. A similar example would be when $M_{ij}=\frac{x_i^{d_i+j-i}}{(d_i+j-i)!}$, with the understanding that $M_{ij}$ is zero once we are taking negative factorials. This is starting to look like an offset Vandermonde matrix, or a confluent Vandermonde. The determinant isn't quite a Schur polynomial but perhaps a known generalization? Is there a connection of these determinants to some well studied class of polynomials (it seems unlikely they will be symmetric)? Note that when written as $M_{ij}=\frac{x_i^{d_i+j-i}}{(d_i+j-i)!}$, we have that if $|d|=\sum_{i=1}^nd_i$, then when $x_1=x_2=\cdots=x_n=1$, $|d|!\det(M)$ is equal to the number of standard Young tableau of shape $(d_1,d_2,\cdots,d_n)$. Alternatively, is there a way to write $\det(M)=C\prod_{i=1}^dx_{i}^{d_i}\prod_{i\neq j}(x_i-x_j)^{a_ib_j}?$ (or possibly more complicated). Unfortunately numerical evidence suggests that fixing a variable, say $x_1$ and thinking of the determinant as a polynomial in $x_1$ gives wildly complex expressions for its roots. So the $(x_i-x_j)$ factors aren't quite right either. I've seen "generalized Vandermondes" here but unfortunately it looks like the structure in the link is simpler. As an example, take $n=4$, and $d_i=9-2i$. Then the resulting matrix is: \begin{pmatrix} \frac{x_1^4}{4!} & \frac{x_1^5}{5!} & \frac{x_1^6}{6!} & \frac{x_1^7}{7!}\\ \frac{x_2^2}{2!} & \frac{x_2^3}{3!} & \frac{x_2^4}{4!} & \frac{x_2^5}{5!}\\ 1 & x_3 & \frac{x_3^2}{2!} & \frac{x_3^3}{3!}\\ 0 & 0 & 1 & x_4 \end{pmatrix} The determinant equals: $$\frac{1}{302400}x_1^4 x_2^2 \left(10 x_1^3 x_2-30 x_1^3 x_3-70 x_1^2 x_2 x_4+210 x_1^2 x_3 x_4-21 x_1 x_2^3+105 x_1 x_2^2 x_4+210 x_1 x_3^3-630 x_1 x_3^2 x_4+105 x_2^3 x_3-525 x_2^2 x_3 x_4-350 x_2 x_3^3+1050 x_2 x_3^2 x_4\right)$$ and it looks like if we single out $x_1$ to make a polynomial in $x_1$, we get complex roots involving $x_2,x_3,x_4$. So, if there is a way to write the determinant as the sum over something nice (not involving signs of permutations), I would also be very interested in see that. Context: These kinds of determinants come up when counting Young Tableau via integrals over sub-simplices, (specifically (4.1),(4.2)). It's a surprisingly powerful way for calculating the number of non-standard Young tableau shapes. I'm just wondering if there's a neater (representation-theoretic?) interpretation of where these determinants arise from. Edit: Consider the matrix $\mathcal{M}=\left[\frac{t_i^{x+j-2}}{(x+j-2)!}\right]_{i,j=1^d}$. It's determinant is: $$\mbox{det}(\mathcal{M})=\frac{1}{\prod_{j=1}^n(x+j-2)}\prod_{i=1}^d t_i^{x-1} \prod_{1\leq i< j\leq d}(t_j-t_i),$$ where we recognize the vandermonde determinant. Then my matrix $M$, and it's determinant can be calculated from this by taking an appropriate amount of derivatives of $t_i$ in each row. So I think this does have something to do with Schur functions afterall. The Jacobi Trudi identity states that: $s_{\lambda}=\det(h_{\lambda_i+j-i})_{i,j=1}^d.$ The exponential specialization of $h_k$ to parameter $t$ gives $t^k/k!$. With $t=1$, one gets that the exponential specialization the Schur polynomial above gives the identity: $$f^{\lambda}=\det\left[\frac{1}{(\lambda_i+j-i)!}\right]_{i,j=1}^d.$$ So the question is, is it sensible to generalize exponential specialization to get an identity of the above form for $t_1,t_2,\cdots,t_d$ instead of just $t_1=\cdots=t_n=t=1$? REPLY [4 votes]: Your general determinant has to do with the Wronskian isomorphism for $SL_2$ representations namely a map $$ \wedge^m({\rm Sym}^n(\mathbb{C}^2))\rightarrow {\rm Sym}^m({\rm Sym}^{n-m+1}(\mathbb{C}^2)))\ . $$ If you follow it by the natural map $$ {\rm Sym}^m({\rm Sym}^{n-m+1}(\mathbb{C}^2))\rightarrow {\rm Sym}^{m(n-m+1)}(\mathbb{C}^2) $$ which corresponds to restricting to the diagonal, then you get the usual Wronskian or rather a homogenized version of it where polynomials in one variable are seen as binary forms. You can find more details in this paper and that one. Great question by the way and thank you for pointing out the article by Ping Sun.<|endoftext|> TITLE: Using Discrete Morse Theory to represent hom classes QUESTION [6 upvotes]: In "vanilla" Morse theory, you can construct cycles representing integral homology classes of a smooth manifold from a Morse function on the manifold (by looking at the flow into/out of the critical points and gluing/compactifying appropriately). Has something similar been proven for Discrete Morse Theory? REPLY [7 votes]: In section 11 of this paper I show that a discrete Morse function on a simplicial complex leads to a dynamical description of Forman's theory. More precisely there is a canonical flow associated to the function such that the (open) faces of the barycentric subdivision are invariant sets. The stationary points of this flow are the barycenters of the original simplicial complex. The discrete Morse function defines a continuous function on topological space defined by the simplicial complex which is affine on the faces of the barycentric subdivision and is a Lyapunov function for the above flow, i.e., it decreases along the trajectories of the flow. The key property of the flow, as far as its connection with Forman's theory is concerned, is the following: a barycenter of a face $F$, that is, a stationary point of the flow, has nontrivial Conley index if and only if the corresponding face is critical in Forman's sense. The unstable manifold of this point is the interior of the face $F$ and the (homotopic) Conley index is the homotopy type of the pair $[F,\partial F]$. We see that this is the same as the homotopy type of a pointed sphere of the same dimension as $F$. Using the finite volume flow technology of Harvey-Lawson one can then obtain a chain homotopy from the simplicial chain complex to the Forman complex.<|endoftext|> TITLE: Go I Know Not Whither and Fetch I Know Not What QUESTION [9 upvotes]: Next day: apparently my original question is harder, by far, than the other bits. So: it is a finite check, I was able to confirm by computer that, if the polynomial below satisfies $$ f(a,b,c,d) \equiv 0 \pmod {27}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 3, $$ and if $$ f(a,b,c,d) \equiv 0 \pmod {125}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 5, $$ ORIGINAL: $f$ is a polynomial in four variables. Take matrices $$ 1 = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right), $$ $$ i = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right), $$ $$ j = \left( \begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right), $$ $$ k = \left( \begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right), $$ Then take $$ f(a,b,c,d) = \det (a \cdot 1 + b \sqrt 3 i + c \sqrt 5 j + d \sqrt{15} k), $$ $$ =a^4-6 a^2 b^2+9 b^4-10 a^2 c^2-30 b^2 c^2+25 c^4+120 a b c d-30 a^2 d^2-90 b^2 d^2-150 c^2 d^2+225 d^4$$. Note that everything is commutative; $$ i^2 = 1, j^2 = 1, k^2 = 1; \; ij=ji=k, ki=ik=j,jk=kj=i. $$ It is also possible to re-write this with the square roots absorbed into the definitions of $i,j,k.$ So, questions include: does it make sense to anyone that, as I checked by brute force, that if $$ f(a,b,c,d) \equiv 0 \pmod {81} $$ then $a,b,c,d \equiv 0 \pmod 3?$ Same for $625$ and $5.$ Need to think about how to check $5$ completely. Finally, is it true that this thing represents the same numbers as $x^2 - 15 y^2,$ and what is such a thing called anyway? It might be a field norm, I dunno. Oh, from a closed question at https://math.stackexchange.com/questions/931769/integer-solution-to-diophantine-equations which I found interesting. http://en.wikipedia.org/wiki/Go_I_Know_Not_Whither_and_Fetch_I_Know_Not_What EDIT: It turns out we may use $27$ in place of $81.$ Evidently explaining this is the hard part. Confirmed, anyway. See what I can do with $125$ instead of $625.$ EDIT 2: Figured out how to program it; if the polynomial is divisible by $125,$ each variable is indeed divisible by $5.$ REPLY [12 votes]: Yes, this is a field norm; it is the norm of $a + b \sqrt{3} + c \sqrt{5} + d \sqrt{15}$, from $K = \mathbb{Q}(\sqrt{3}, \sqrt{5})$ down to $\mathbb{Q}$. Note that $a+b \sqrt{3} + c \sqrt{5} + d \sqrt{15}$ acts on the basis $(1, \sqrt{3}, \sqrt{5}, \sqrt{15})$ by $$a \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} + b \begin{pmatrix} 0 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 \\ 0 & 0 & 1 & 0 \end{pmatrix} + c \begin{pmatrix} 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 & 0 & 15 \\ 0 & 0 & 5 & 0 \\ 0 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}.$$ Now conjugate by the matrix whose diagonal entries are $(1, \sqrt{3}, \sqrt{5}, \sqrt{15})$ to get your matrix. The entries are no longer rational, so I can't think of the result as describing the action on $K$, but the determinant is the same. $\mathbb{Q}(\sqrt{15})$ has class number $2$ and $K$ is the class field. So, for a prime $p$ other than $2$, $3$, $5$, we have that $\pm p$ is a value of $x^2-15 y^2$ if and only if $p$ splits principally in $\mathbb{Q}(\sqrt{15})$ if and only if $p$ splits in $K$ if and only if $\pm p$ is a value of $f$. Also, neither $x^2-15 y^2$ nor $f$ can be $3 \bmod 4$, so the sign is the same in the two cases. However, they don't take the same set of composite values. Look at $-119 = 7 \times 17$. We have $61^2 - 15 \cdot 16^2 = -119$, but, if $7 | f(a,b,c,d)$ then $7^2 | f(a,b,c,d)$. I found this by hunting for two primes which are non-principally split in $\mathbb{Q}(\sqrt{15})$. In terms of quadratic forms, which I know you love, I needed primes of the form $3 x^2 - 5 y^2$, and I found $7=3 \cdot 2^2 - 5$ and $-17 = 3 \cdot 6^2 - 5 \cdot 5^2$. Then their product was of the form $x^2-15 y^2$. Since these primes split non principally in $\mathbb{Q}(\sqrt{15})$, they don't split further in the class field. (We can also directly compute $\left( \frac{3}{7} \right) = \left( \frac{3}{17} \right) = -1$.) So things divisible by one power of $7$ or $17$ are not norms from $K$.<|endoftext|> TITLE: Geodesic flow on infinite surfaces QUESTION [6 upvotes]: The geodesic flow on a compact hyperbolic surface (i.e. a surface with a riemannian metric of constant curvature $-1$) has been well-studied, in particular it has been known for a long time that it is ergodic (in fact mixing). On an hyperbolic surface with infinite volume however, I am not aware of any result about dynamical properties of this flow (there seems to be some results about infinite translation surfaces however). I welcome any reference to results in this direction as an answer (or perhaps more appropriately as a comment) to this question. Here are some more specific queries: let $S$ be an infinite-volume complete hyperbolic surface (for simplicity let us assume that $S$ has no cusps; if necessary one could also assume that $S$ has positive injectivity radius, though I would rather avoid making this hypothesis). If $S$ has an open subset $U$ of ends which is a Cantor set (i.e. a neighbourhood of $U$ is diffeomorphic to the boundary of a regular neighbourhood of a tree embedded in $\mathbb{R}^3$), does the geodesic flow of $S$ starting at some basepoint return to any neighbourhood of an end in $U$ infinitely many times with positive probability (depending uniformly on the basepoint in a compact subset of $S$)? If $E$ is an isolated end of $S$ with linear or quadratic volume growth, does the geodesic flow leave any neighbourhood of $E$ with probability one? Note that these questions are analogues for surfaces of the corresponding facts for random walks on infinite graphs (of bounded valency). For convenience let me recall here the definition of an end of a surface. For this one needs to fix a sequence $K_i$ of compact subsets, with $K_i\subset K_{i+1}$ and $\bigcup_i K_i = S$. An end of $S$ is then determined by a sequence $C_i$ where $C_i$ is a connected component of $S\setminus K_i$ and $C_{i+1}\subset C_i$. The topology on the set of ends has for a basis of open sets the sets of ends starting with a given finite sequence $C_1\supset\ldots \supset C_n$, for all such sequences. REPLY [6 votes]: Dynamics of the geodesic flow for infinite volume manifolds has been studied a lot, but with respect to the most relevant measure, that is the Bowen-Margulis-Patterson-Sullivan measure. A complete reference could be Roblin, Mémoires SMF. But I guess that you are interested only in the Lebesgue/Liouville measure ? In the examples you have in mind, the geodesic flow will typically be dissipative. First, the case of abelian covers of compact hyperbolic surfaces is well understood. The geodesic flow is ergodic in the case of Z or Z^2 covers, and totally dissipative in the other cases. This is due to Mary Rees and others before her. This let believe that the answer to your question 2 could be positive. But to my knowledge, nothing precise is known, for both questions. If the surface you are interested in is a regular cover of a compact hyperbolic surface, I guess that lifting the coding of the geodesic flow of the compact surface to the cover could allow to use the probabilistic results on random walks. (In their common work, Ledrappier-Sarig use such coding lifted to the cover.) However, if the surface is not a cover, I don't know any method to prove it. Bests Barbara<|endoftext|> TITLE: Does iterating a certain function related to the sums of divisors eventually always result in a prime value? QUESTION [34 upvotes]: Let define the following function for integers (from 2): $f(x)=\sigma(x)-1$, where $\sigma$ is the sum of the divisors of $x$. For example $f(6)=6+3+2=11$, $f(5)=5$. Note that $x$ is a fixed point for $f$ if and only if, $x$ is prime. If we iterate starting at any integer $x$ we get a dynamical system. Computations with Maple showed that for all integer $x$ in $[2,2000000]$ there exists an integer $n$ such that $(f \circ f \circ \dots \circ f)(x)$ is prime; that is for each integer $x$ the sequence generated is stationary. The question is: can anyone help proving the conjecture ? Thank you for any help on the subject. REPLY [5 votes]: Every prime occurs at least once (obviously) as the end of a trajectory. However many occur only once and some occur quite often. Since $f$ is non-decreasing it is easy by simple computation to find all $x$ which end up at a particular prime $p$. Primes such that $p+1$ has many small prime factors are likely to occur often. For example, there are $125$ starting points which end at $5039$ but the nearby primes $5011,5023,5051, 5059$ and $5057$ can be reached only from themselves while $5021$ can be reached starting from itself and one other place, $2650=2\ 5^2\ 53$ with $\sigma(2650)=(2+1)(25+5+1)(53+1)=2\ 3^4\ 31=5022.$ It is not too hard to establish that $\sigma(x)=2651$ has no solutions. On the other hand, $5040=2^33^25\ 7$ can be factored in may ways as the product of several factors, which must have only small prime factors. Here are all the ways so that each factor is one less than a prime. $[3, 4, 14, 30], [3, 6, 14, 20], [3, 4, 420], [3, 12, 140], [3, 20, 84], [4, 14, 90], $$[4, 30, 42], [4, 1260], [6, 14, 60], [6, 20, 42], [12, 14, 30], [6, 840],$$ [14, 18, 20], [12, 420], [14, 360], [20, 252], [30, 168], [60, 84]$ These, with $5039$ itself, give the $19$ square free integers with $\sigma(x)-1=5039.$ There are others which are not square free, such as $x=4y$ for odd $y$ with $\sigma(y)=720$ (also examples coming from $\sigma(3^3)=40$ and from $\sigma(2^5)=63$.) There are as well starting points which land at $5039$ after several steps. I'd expect $6719=(2^6\ 3\ 5 \ 7)-1$ to occur quite often, but I haven't checked.<|endoftext|> TITLE: On the conformal removability of Jordan curves QUESTION [10 upvotes]: We say that a compact subset $E$ of the Riemann sphere $\mathbb{C}_\infty$ is (conformally) removable if every homeomorphism of $\mathbb{C}_\infty$ conformal outside $E$ is actually conformal everywhere, i.e. is a Mobius transformation. My question is the following : Suppose $\Gamma$ is a non-removable Jordan curve in the plane. Does $\Gamma$ necessarily contains a non-removable proper closed subset? Thank you, Malik EDIT As mentioned by Igor Rivin, in the paper Some homeomorphisms of the sphere conformal off a curve, Chris Bishop raises the above question. However, the paper is 20 years old and I talked to Chris about it, and even back then he really wasn't sure that the problem was open. The question simply did not seem trivial to him. EDIT 2 It is a direct consequence of the measurable Riemann mapping theorem that every set of positive area is non-removable, so the answer to the question is yes if $\Gamma$ has positive area. What if the area of $\Gamma$ is zero? REPLY [2 votes]: In 1994 this was open, as evidenced by this paper by Chris Bishop (which has a nice survey). (Some homeomorphisms of the sphere conformal off a curve).<|endoftext|> TITLE: What's the current state of one-rule semi-Thue system termination problem? QUESTION [6 upvotes]: What's the current state of one-rule semi-Thue system termination problem? Search produces a lot of references, but it's hard to find out if decidability of this problem has been proven or not. REPLY [3 votes]: An important part of this problem is, as Ben points out in his answer, the word problem for one-relation monoids. This problem is still open, but it has been reduced to a number of special cases. I'll summarise these developments; there are a few gaps in proofs floating around, so sometimes one sees claims that cases are solved when they remain open. Here's an overview. Consider the one-relation monoid $M = \operatorname{Mon}\langle A \mid U = V \rangle$. The case $U=1$ is called the special case, and was solved in 1960 by Adjan [1] (sometimes this proof is attributed to his famous 1966 monograph, in which the proof also appears). He reduces the word problem in such monoids to the group case $\operatorname{Gp}\langle A \mid U=1 \rangle$, in which the word problem is famously decidable by Magnus. In fact, Adjan's student Makanin extended this result in his 1966 PhD thesis, and proved that the word problem in $k$-relator special monoids reduces to the word problem in $k$-relator groups. A word is called hypersimple if it is self-overlap free, e.g. $abb$ is hypersimple but $aba$ is not. In our one-relation monoid, if the relation $U=V$ is of the form $\alpha U \alpha = \alpha V \alpha$ or $\alpha U \alpha = \alpha$ for some hypersimple word $\alpha \in A^\ast$, then we say that the relation $U=V$ is compressible with respect to $\alpha$. To every compressible one-relation monoid $M$ there is an associated one-relation monoid $M'$ with a shorter defining relation. Adjan and his student Oganesjan [2] proved in 1978 that the word problem in a compressible one-relation monoid $M$ reduces to the word problem for $M'$. Further reductions are possible. Adjan [3] showed in 1960 that our one-relation monoid $M$ above is cancellative if and only if $U$ and $V$ have different initial and different final letters (this is an obvious necessary condition), and that in this case $M$ embeds in the group $\operatorname{Gp}\langle A \mid U=V \rangle$. Thus in this case the word problem is again decidable by Magnus' result. As with the special case, this result is also occasionally incorrectly attributed to the 1966 monograph. Thus the word problem for one-relation monoids can be reduced to the case when the words $U = V$ either begin in the same letter or end in the same letter (but not both, as otherwise it is not hard to see that we could compress). This leaves us with the right-cancellative cases $\operatorname{Mon}\langle A \mid aUb = aVa \rangle$ and $\operatorname{Mon}\langle A \mid aUb = a \rangle$, as well as the symmetrical left-cancellative cases $\operatorname{Mon}\langle A \mid aUb = bVb \rangle$ and $\operatorname{Mon}\langle A \mid aUb = b \rangle$. It is not difficult to consider all words in reverse and reduce the left-cancellative to the right-cancellative case, and vice versa. In 1987, Adjan and Oganesjan [4] (note that this paper is named very similarly to the 1978 paper, both in the original Russian as well as the English translation!) reduced these remaining cases to the two-generated case (in the excellent survey [5], this is incorrectly attributed to the 1978 paper [3]). Thus the remaining cases are $$ \operatorname{Mon}\langle a,b \mid aUb = aVa \rangle \qquad \text{and} \qquad \operatorname{Mon}\langle a,b \mid aUb = a \rangle. $$ In 1982, Oganesjan [6] claimed a proof that the word problem is decidable in $\operatorname{Mon}\langle a,b \mid aUb = a \rangle$, i.e. the second of the cases above. However, this proof depends on a result by O. A. Sarkisjan (another one of Adjan's students, as you might have guessed at this point) from the year before [7]. She claimed to have proved that the left and right divisibility problems are decidable in all cancellative one-relation monoids. However, Adjan discovered a gap in this proof in the 90s, and the gap remains unfilled. I am not certain what the gap is. In any case, this gap means Oganesjan's result is not yet proved (though apparently Adjan believed that the result was nevertheless correct). Thus we are up to speed -- those two cases remain the open ones. If the word problem is decidable in the two classes specified above, then it is decidable for all one-relation monoids. A number of special cases are known to be decidable of the above (and, in fact, it is known that almost all, in a well-defined sense, one-relation monoids have decidable word problem), but writing out all the known partial cases would be a bit ridiculous for the scope of this answer. Even small cases can cause large headaches. For example, even the "tiny" case $\operatorname{Mon}\langle a,b \mid abaab = a \rangle$ was beyond the reach of the methods of Howie and Pride [8], though this, as it turns out, can be solved using Adjan's pseudo-algorithm $\mathfrak{A}$ for solving the right divisibility problem (in this case $\mathfrak{A}$ turns out to be an algorithm) -- but that's a topic for a different post/question! References: [1] Adjan, S. I. The problem of identity in associative systems of a special form. Dokl. Akad. Nauk SSSR 135 (1960), 1297–1300. [2] Adjan, S. I.; Oganesjan, G. U. On problems of equality and divisibility in semigroups with a single defining relation. Izv. Akad. Nauk SSSR Ser. Mat. 42 (1978), no. 2, 219–225, 469. [3] Adjan, S. I., On the embeddability of semigroups in groups, Dokl. Akad. Nauk SSSR 133 (1960), 255–257. [4] Adjan, S. I.; Oganesjan, G. U.; On the word and divisibility problems for semigroups with one relation. Mat. Zametki 41 (1987), no. 3, 412–421, 458. [5] Adjan, S. I.; Durnev, V. G.; Algorithmic problems for groups and semigroups., Uspekhi Mat. Nauk 55 (2000), no. 2(332), 3–94. [6] Oganesjan, G. U.; Semigroups with one relation and semigroups without cycles., Izv. Akad. Nauk SSSR Ser. Mat. 46 (1982), no. 1, 88–94, 191. [7] O. A. Sarkisjan; On the word and divisibility problems in semigroups and groups without cycles, Izv. Akad. Nauk SSSR Ser. Mat. 45 (1981), 1424–1440. [8] Howie, James; Pride, Stephen J.; The word problem for one-relator semigroups. Math. Proc. Cambridge Philos. Soc. 99 (1986), no. 1, 33–44.<|endoftext|> TITLE: Does an equivalence of fusion categories depend on choice of simple objects within isomorphism classes? QUESTION [7 upvotes]: Let $C$ be a fusion category with simple objects $X_1,...,X_n $, and let $Y_1,...,Y_n$ be objects with each $Y_i$ isomorphic to $X_i$. Is there a monoidal auto-equivalence $F:C \rightarrow C $ which takes each $X_i$ to $Y_i$, and such that $F$ is naturally isomorphic (as a monoidal functor) to the trivial auto-equivalence? It seems like this should be true since the structure of the fusion category is determined by $6j$-symbols and one should be able to get the same $6j$-symbols for every choice of representatives of simple objects by appropriately choosing representative morphisms between their tensor products. But how can one write down such a functor? REPLY [7 votes]: Let's forget about the monoidal structure for a moment. $C$ is semisimple, and semisimplicity means that to give a functor out of $C$ is the same thing as specifying where it sends simple objects. So that means that saying $\mathcal{F}(X_i)=Y_i$ determines a unique functor $\mathcal{F}$. Now, what does it mean for this functor to be naturally isomorphic to the identity functor? Well, again by semisimplicity, to give a natural transformation between functors is just to give its components on simple objects. So fixing isomorphisms $\eta_i: X_i \cong Y_i$ determines a unique natural isomorphism $\eta: \mathcal{F} \rightarrow \mathcal{Id}$. Now let's bring the monoidal structure back into the picture. We have a functor $\mathcal{F}$ and a natural transformation $\eta: \mathcal{F} \rightarrow \mathcal{Id}$. But $\mathcal{Id}$ has a canonical structure as a monoidal functor! So you can just use "transport de structure" to move the monoidal structure on $\mathcal{Id}$ across the natural isomorphism to determine a monoidal structure $\mathcal{F}$. In other words, there's no additional content to the monoidal statement. Once you know that $\mathcal{F}$ is naturally isomorphic via $\eta$ to the identity functor, then there's a unique way to make $\mathcal{F}$ into a monoidal functor such that $\eta$ becomes a monoidal natural transformation. This is completely the same argument as saying if you have a set X and a group G and a bijection $f: X \rightarrow G$ then there's a unique way to make X into a group such that $f$ is an isomorphism of groups.<|endoftext|> TITLE: higher-order reflection QUESTION [6 upvotes]: In the first-order context, "reflection" of a formula $\varphi(x)$ below $\kappa$ refers to the the following situation: There are many ordinals $\alpha<\kappa$ such that for all $a \in V_\alpha$, $V_\alpha \models \varphi(a)$ iff $V_\kappa \models \varphi(a)$. If $\kappa$ is inaccessible, then this holds for any expansion of $V_\kappa$ in a countable language. "Many" can be taken to mean on a club set. When we move to higher-order logic, talk of reflection usually shifts to talk of indescribability. A cardinal is $\Pi^m_n$ indescribable if for any $A \subseteq V_\kappa$ and any $\Pi_n$ sentence $\sigma$ in $(m+1)$-order logic with a predicate for $A$, if $(V_\kappa, \in, A) \models \sigma$, then there is $\alpha<\kappa$ such that $(V_\alpha,\in,A\cap V_\alpha) \models \sigma$. It is a standard fact that if $\kappa$ is measurable, then there is a measure-one set of $\alpha< \kappa$ that are $\Pi^m_n$-indescribable for every $m,n$. One can also show something stronger: If $\kappa$ is measurable, there is a measure-one set of $\alpha < \kappa$ such that if $A \subseteq V_\alpha$, then there is $\beta < \alpha$ such that $(V_\alpha,\in,A)$ and $(V_\beta,\in,A\cap V_\beta)$ have the same $\omega$-order theory. Now this is not completely analogous to reflection because we're no longer talking about elementary substructures, but just elementarily equivalent structures, albeit with a common interpretation of a particular predicate. So my question is, what kind of large cardinal $\kappa$ is needed to get the following statement? For any $A \subseteq V_\kappa$ and any $n \in \omega$, there are many ordinals $\alpha < \kappa$ such that $(V_\alpha,\in,A \cap V_\alpha) \prec^n (V_\kappa,\in,A)$, where $\prec^n$ means elementary in $(n+1)$-order logic. It happens at an $\omega$-strong cardinal, but this is clearly not optimal. REPLY [7 votes]: It depends on how we define elementarity for the $(n+1)^{th}$-order language of set theory, $\mathcal L^{n}_\in$. For $X\subseteq V_\kappa$, two salient definitions are: $V_\alpha \prec^n_X V_\kappa$ iff $\forall \vec{x}\in V_\alpha(V_\alpha\vDash \phi(X\cap V_\alpha,\vec{x}) \leftrightarrow V_\kappa\vDash \phi(X,\vec{x}))$, for all $\phi\in \mathcal L^{n}_\in$ with parameters among $X, \vec{x}$. $V_\alpha \prec^{*n}_X V_\kappa$ iff $\forall \vec{x}\in V_\alpha\forall \vec{Y}\subseteq V_\alpha(V_\alpha\vDash \phi(X\cap V_\alpha, \vec{Y}, \vec{x}) \leftrightarrow V_\kappa\vDash \phi(X,\vec{Y},\vec{x}))$, for all $\phi\in \mathcal L^{n}_\in$ with parameters among $X,\vec{Y}, \vec{x}$. ${\bf Theorem}$ ${\bf 1:}$ There are no $\alpha<\kappa$ such that $V_\alpha\prec^{*n} V_\kappa$ for $n\geq 1$. ${\it Proof.}$ If $Y = V_\alpha$, $V_\alpha\vDash \forall x(x\in Y)$ but not $V_\kappa\vDash \forall x(x\in Y)$. ${\bf Theorem}$ ${\bf 2:}$ If $\kappa$ is $\Pi^{n}_1$-indescribable for $n\geq 1$, then for any $X\subseteq V_\kappa$ there are unbounded $\alpha<\kappa$ such that $V_\alpha\prec^{n-1}_X V_\kappa$. ${\it Proof.}$ In $n+1^{th}$-order Zermelo set theory we can define a $\Delta^{n}_1$ satisfaction predicate for $\mathcal L^{n-1}_\in$. Denote it $Sat$. Then we can define $Y\subseteq V_\kappa$ such that $V_\kappa$ satisfies: (*) $\langle \vec{x},\phi\rangle\in Y \leftrightarrow Sat(\phi, X, \vec{x})$, for all $\phi\in \mathcal L^{n-1}_\in$ with parameters among $X, \vec{x}$. Clearly, (*) is $\Delta^n_1$ and since $\kappa$ is $\Pi^{n}_1$-indescribable, it follows that there are unbounded $\alpha<\kappa$ which satisfy: (**) $\langle \vec{x},\phi\rangle\in Y\cap V_\alpha \leftrightarrow Sat(\phi, X\cap V_\alpha, \vec{x})$, for all $\phi\in \mathcal L^{n-1}_\in$ with parameters among $X, \vec{x}$. For any such $\alpha$ it is easy to see that $V_\alpha\prec^{n-1}_X V_\kappa$, as required. (Note that we can easily strengthen "unbounded" to "stationary" but not "club" - any $\alpha$ for which $V_\alpha\prec^1 V_\kappa$ will be inaccessible.) ${\bf Corollary:}$ $\kappa$ is totally indescribable just in case for any $X\subseteq V_\kappa$ and $n<\omega$ there are unbounded $\alpha$ such that $V_\alpha\prec^n_X V_\kappa$.<|endoftext|> TITLE: Bases of surface groups QUESTION [5 upvotes]: Let $\Gamma_g$ be a surface group of genus $g \geq 2$. A $2g$-tuple $(x_1,y_1, \dots,x_g,y_g) \in \Gamma_g^{2g}$ will be called a Surface Basis if we have the presentation $$\Gamma_g = \langle x_1, y_1, \dots, x_g, y_g \vert \prod_{i = 1}^{g}[x_i,y_i] = 1\rangle$$ Take some $1 \leq k \leq g$ and $H \leq \Gamma_g$ of finite index such that $x_1, \dots, x_k \in H$. One can show that $H$ is a surface group of genus $(g-1)[\Gamma_g : H] + 1$, and that it has a surface basis. My question is: Is there a surface basis for $H$ containing $x_1, \dots, x_k$? The analogous question for free groups has a positive answer as shown in Bases of free groups. REPLY [3 votes]: There is indeed a surface basis for $H$ containing $x_1,\ldots,x_k$. I'll give a topological proof, basically the same as the proof suggested in the comment of @HJRW. First, I'll give a topological re-interpretation of your problem, by formulating a topological property which is equivalent to the property "there is a surface basis of the fundamental group containing such-and-such a list of elements". Then I'll say why this topological property is preserved under the kind of covering maps in your question. For each genus $g$ fix a ``standard'' surface $S_g$ with base point $p$ and an isomorphism $\pi_1(S_g,p) \approx \Gamma_g$, and with an embedded rank $2g$ rose consisting of loops $\xi_1,\eta_1,\ldots,\xi_g,\eta_g$ representing $x_1,y_1,\ldots,x_g,y_g$. When $S_g$ is cut open along the rose, the result is a $2g$-gon whose boundary is attached to the rose using the word given by the standard relator in your question. Let $N_k$ be a regular neighborhood of $\xi_1 \cup \cdots \cup \xi_k$. Note that $N_k$ is a connected planar surface with connected complement; it follows that $N_k$ has $k+1$ boundary components, and that $S_g - N_k$ is a connected surface of genus $g-k$ with $k+1$ boundary components. Consider now an abstract surface $S'$ of some genus $p'$, and a rank $k$ rose embedded in $F$ consisting of loops $\gamma_1,\ldots,\gamma_k$ all with common base point $q$. Then the following are equivalent: there exists a surface basis for $\pi_1(S',p')$ containing $[\gamma_1],\ldots,[\gamma_k]$ the regular neighborhood of the rose $\gamma_1 \cup \ldots \cup \gamma_k$ is a connected planar surface with connected complement. I've already proved 1$\implies$2. For 2$\implies$1, it follows the regular neighborhood of the rose is a planar surface with $k+1$ boundary components and that its complement has genus $g'-k$ with $k+1$ boundary components, and this allows us to construct a homeomorphism between $S'$ and the standard surface $S_{g'}$ that takes $\gamma_1,\ldots,\gamma_k$ to $\xi_1,\ldots,\xi_k$. Okay, so, consider a finite index subgroup of $\pi_1(S,p)$ containing $x_1,\ldots,x_k$. Let $f : S' \to S$ be the corresponding finite degree covering map with base point $p'$ lifting $p$. The loops $\xi_1,\ldots,\xi_k$ lift one-to-one to loops $\xi'_1,\ldots,\xi'_k$ based at $p'$. The regular neighborhood $N_k$ of $\xi_1\cup\cdots\cup\xi_k$ lifts one-to-one to a regular neighborhood $N'_k$ of $\xi'_1\cup\cdots\cup\xi'_k$. Since the restricted covering map $N'_k \to N_k$ is a homeomorphism, this proves that $N'_k$ is planar. It remains to prove that $S'-N'_k$ is connected. Arguing by contradiction, assuming that $S' - N'_k$ is disconnected, consider a component $F$ of $S'-N'_k$ which meets $N'_k$. It follows that $\partial F = F \cap N'_k$ consists of a proper nonempty subset of the $k+1$ components of $\partial N'_k$. Let those components be enumerated $c'_1,\ldots,c'_j,c'_{j+1},\ldots,c'_{k+1}$ so that $\partial F = c'_1 \cup \cdots \cup c'_j$. Push this enumeration down to $N_k$ whose boundary circles are therefore enumerated $c_1,\ldots,c_j,c_{j+1},\ldots,c_{k+1}$. Let the surface $F$ be decomposed as $F = F_1 \cup F_2$ where the covering map $S' \to S$ restricts to a covering map $F_1 \to S - N_k$ of some degree $\delta_1$, and it restricts to a covering map $F_2 \to N_k$ of some degree $\delta_2$. Let $\partial F_1$ be decomposed into two subsets $\partial F_1 = \partial_{\le j} F_1 \cup \partial_{>j} F_1$, with restricted covering maps $\partial_{\le j} F_1 \to c_1 \cup \cdots \cup c_j$ and $\partial_{>j} F_1 \to c_{j+1} \cup \cdots \cup c_{k+1}$ each of degree $\delta_1$ (the base spaces of these covering maps are not connected, nonetheless these covering maps each have constant degree $\delta_1$ over each component). It follows that the restricted map $\partial F_1 - (c'_1 \cup \cdots \cup c'_j) \to c_1 \cup \cdots \cup c_{k+1}$ is a covering map of unequal degrees over different components: it has degree $\delta_1-1$ over each component of $c_1,\ldots,c_j$ and it has degree degree $\delta_1$ over each component of $c_{j+1},\ldots,c_{k+1}$. However, we have an equation $$\partial F_1 - (c'_1 \cup \cdots \cup c'_j) = \partial F_2 $$ and the degree $\delta_2$ covering map $F_2 \to N_k$ restricts to a map $\partial F_2 \to c_1 \cup \cdots \cup c_{k+1}$ of equal degree $\delta_2$ over each component. This is a contradiction.<|endoftext|> TITLE: Functions $f$ on $\mathbb{Z}/N\mathbb{Z}$ with $|f|$ and $|\widehat{f}|$ constant QUESTION [20 upvotes]: Let $N$ be a positive integer; for simplicity I'm happy to assume it's an odd prime but I'm interested in the general case too. Let $f \colon \mathbb{Z}/N\mathbb{Z} \to \mathbb{C}$ and let $\widehat{f}$ denote its Fourier transform, $\widehat{f}(\xi) = \frac{1}{N} \sum_x f(x) e^{-2 \pi i \xi x / N}$. I'm interested in functions $f$ with $|f|$ constant and $|\widehat{f}|$ constant; with the above conventions, WLOG taking $|f| = 1$ we have $|\widehat{f}| = 1/\sqrt{N}$. My question is: is anything known about such functions? Do they have a name? Is there perhaps even a precise classification of them? Any pointers or references would be appreciated. It's maybe worth saying that, although apparently a question about analysis, it is actually surely strongly algebraic in nature. Indeed, the conditions can be phrased as an algebraic set over $\mathbb{R}$ which I believe has dimension $0$ when $N$ is prime (if we add the condition $f(0) = 1$ to remove the degeneracy), which would mean the collection of such $f$ is finite and the coefficients are algebraic numbers. Furthermore all the examples I've computed are exceedingly structured, e.g. $$ f(x) = e^{2 \pi i (\alpha x^2 + \beta x + \gamma) / N} $$ where $\alpha \in (\mathbb{Z}/N\mathbb{Z})^\times$, $\beta \in \mathbb{Z}/N\mathbb{Z}$, $\gamma \in \mathbb{R}$, as well as more complicated variants. EDIT: given Sean Eberhard's partial answer below, I should maybe sketch an example to show that these quadratic examples are not the only ones, i.e. there are such functions $f$ whose values are not $N$th roots of unity. Very briefly: let $f(x)$ be $1$ when $x$ is a square modulo $N$ and $\alpha$ otherwise for some fixed $\alpha \in \mathbb{C}$. It suffices to show that there is an $\alpha$ that works. Whenever $N \equiv 3 \pmod{4}$ is prime, this is the case, and in fact $$\alpha = \frac{ -(N-1) \pm 2 \sqrt{-N}}{N+1} \ .$$ REPLY [10 votes]: $\def\Z{\mathbf{Z}}$If $N$ is prime and $f(x)$ has the form $\zeta^{g(x)}$, where $\zeta$ is an $N$th root of unity and $g$ is some function from $\Z/N\Z$ to $\Z$, then $g$ must be quadratic. The following argument is very similar to the argument given in the answer https://mathoverflow.net/a/136046/20598, so I feel at liberty to be brief on details. For each fixed $t\in\Z/N\Z$ the sum $$\sum_x \zeta^{g(x) + tx}$$ has absolute value $\sqrt{N}$ by assumption. This implies (Cavior, S.: Exponential sums related to polynomials over GF(p), Proc. A.M.S. 15 (1964) 175-178) that $$\sum_x \zeta^{g(x) + tx} = \sum_x \zeta^{ax^2 + s}$$ for some $a,s\in\Z/N\Z$ depending on $t$, and this implies that $g(x)+tx$ takes the same values and multiplicities as $ax^2+s$. In particular $g(x)+tx$ takes each value at most twice. By Segre's theorem $g$ must be quadratic.<|endoftext|> TITLE: Inverse of partial differential operator as a smooth tame map QUESTION [7 upvotes]: Tameness for maps is one of the main ingredients for the Nash-Moser inverse function theorem. A linear map $f: X \to Y$ between Fŕechet spaces with fixed seminorms is called tame if we have an estimate of the form $$ ||f(x)||_k \leq C ||x||_{k+r}$$ for some $C$ and $r$ (and all $k$), where $|| \cdot ||_k$ denotes the $k$-th seminorm. Hamilton (1982) shows that every partial differential operator is a tame map and, moreover, he claims that many inverses of partial differential operators are also tame. A few pages later, he shows that the Green's operator of an elliptic differential operator is tame. What are examples of other differential operators with tame inverses? (In particular, is there any hope that inverses to hyperbolic operators are tame?) REPLY [6 votes]: An addendum to Deanne Yang's answer. Actually, if a linear hyperbolic equation is degenerate, meaning that the characteristic roots are real but not distinct, then the inverse operator presents a loss of derivatives, meaning that it takes $H^s$ to $H^{s-m}$ for some $m>0$. This is an actual phenomenon and not a defect of the estimates. Indeed, the solution of the 1D equation $u_{tt} − t^2 u_{xx} = (4k+1) u_x $ can be written explicitly for integer $k$ and loses exactly $k$ derivatives (this was proved by Qi Min-you in 1958). Thus the inverse of a degenerate hyperbolic equation is a perfect example of a smooth tame map, and the Nash-Moser theorem can be used to handle the nonlinear equations. I used myself this idea in a couple of papers. EDIT: I lost track of the hardcopy of QMY's paper, which I had at some point, anyway from my notes the solution with initial data 0 and $\phi(x)$ should be $$u(t,x)=\sum_{j=1}^{k} \frac{\sqrt{\pi}t^{2j}}{j!(k-j)!\Gamma(j+1/2)} \phi^{(j)}(x+t^{2}/2).$$ EDIT 2: Wait: in section 6 of this paper the example is extended to a more general class of equations.<|endoftext|> TITLE: What is the level of a positive energy loop group representation? QUESTION [8 upvotes]: I am trying to learn a bit about loop group representation theory to understand its role in string geometry. Let $G$ be a Lie group. I am thinking of $\text{Spin}(n)$, so you may assume $G$ to be simple and simply connected and compact and... By the loop group $LG$ of $G$, I mean the group of smooth loops in $G$. Now, in Pressley-Segal's book "Loop Groups", they give a definition of the level of a positive energy representation in terms of pure representation theory. [page 177]. However, in the paper "What is an elliptic object?", the authors talk about a level as an element in $H^4(BG)$.[the article can be found at: http://people.mpim-bonn.mpg.de/teichner/Math/Papers.html] Now, my question is the following: In which way do these two notions connect, or even coincide? Thanks for any answer; even a reference where this is worked out would be great! REPLY [6 votes]: Let's first recall the case of finite groups $G$. A projective representation of $G$ is a homomorphism $\rho : G \to PGL_n(\mathbb{C})$. In trying to lift this to a genuine representation $G \to GL_n(\mathbb{C})$ we find an obstruction given by a class $c \in H^2(BG, \mathbb{C}^{\times})$. An invariant way to describe this obstruction is that the short exact sequence $$1 \to \mathbb{C}^{\times} \to GL_n(\mathbb{C}) \to PGL_n(\mathbb{C}) \to 1$$ gives rise to a coefficient exact sequence ending $$\cdots \to H^1(BG, GL_n(\mathbb{C})) \to H^1(BG, PGL_n(\mathbb{C})) \to H^2(BG, \mathbb{C}^{\times})$$ (where every group involved has the discrete topology). The class $c$ can be thought of as the level of the projective representation $\rho$. If $G$ is perfect, meaning that $G/[G, G] \cong H_1(BG, \mathbb{Z})$ vanishes, then universal coefficients gives an identification $$H^2(BG, \mathbb{C}^{\times}) \cong \text{Hom}(H_2(BG, \mathbb{Z}), \mathbb{C}^{\times})$$ and we can equivalently think about levels as follows: $G$ admits a universal central extension $$1 \to H_2(BG, \mathbb{Z}) \to \widetilde{G} \to G \to 1$$ and we can identify projective representations of $G$ of level $c$ with ordinary representations of $\widetilde{G}$ where the central $H_2(BG, \mathbb{Z})$ acts by the character in $\text{Hom}(H_2(BG, \mathbb{Z}), \mathbb{C}^{\times})$ corresponding to $c$. In particular, by Schur's lemma, every irreducible representation of $\widetilde{G}$ has a well-defined level. A similar but more complicated thing is happening in the case of loop groups, although I can't claim to know the details. Here is a guess at the details. With suitable hypotheses on a Lie group $G$, the loop group $LG$ should admit a universal central extension $$1 \to S^1 \to \widetilde{LG} \to LG \to 1$$ and we should be able to think of projective representations of $LG$ in terms of representations of $\widetilde{LG}$ (I am ignoring the energy circle here) where the central $S^1$ acts by a fixed character; moreover, these characters should be identified with levels in a cohomological sense as above. The identification should go something like this: given a level $k \in H^4(BG, \mathbb{Z})$ we can think of it as a class in $H^3(BG, \mathbb{C}^{\times})$ (where $\mathbb{C}^{\times}$ now has the usual topology) and then transgress it to a class in $H^2(LBG, \mathbb{C}^{\times})$. The group I actually wanted to land in is $H^2(BLG, \mathbb{C}^{\times})$, which at least looks like it has something to do with projective representations of $LG$, but in fact $$\Omega LBG \cong L \Omega BG \cong LG$$ so $LBG \cong BLG$ as long as $LBG$ is connected, which should be ensured by $G$ being connected. If $G$ is compact, simple, connected, and simply connected, then it's well-known that $\pi_2(G) \cong \pi_3(BG) \cong H_3(BG, \mathbb{Z})$ vanishes and that $\pi_3(G) \cong \pi_4(BG) \cong H_4(BG, \mathbb{Z}) \cong \mathbb{Z}$. Universal coefficients now gives an identification $$H^4(BG, \mathbb{Z}) \cong \text{Hom}(H_4(BG, \mathbb{Z}), \mathbb{Z}) \cong \mathbb{Z}$$ which gets identified above with characters of the central $S^1$, although I'm not sure how; there should be some nice functorial description of this $S^1$ but I'm not sure what it is.<|endoftext|> TITLE: Holomorphic Hoffstein-Lockhart QUESTION [9 upvotes]: In the article Hoffstein, Jeffrey; Lockhart, Paul "Coefficients of Maass forms and the Siegel zero." Ann. of Math. (2) 140 (1994), no. 1, 161–181, it is stablished a good bound for the Petersson norm of a Hecke Mass newform. As they indicate themselves, their method also works for classical holomorphic newforms of arbitrary level and weight. Is this version explicitly written somewhere? If $N$ is the level and $k$ is the weight, the bound is a function of $N$ and $k$, which is NOT exactly the same as in the Maass case (even if we replace $k$ by the corresponding eigenvalue). It would be useful to see the explicit form of the bound neatly written. I know it follows from the Rankin-Selberg method, but the important point is to control a special value of an automorphic form on $GL_3$ and I am not familiar with such forms. REPLY [5 votes]: The reference pointed out by GH gives an excellent general write-up. For those interested in completely explicit bounds in the classical case of $GL(2)$ over $\mathbb{Q}$, I've worked these out in a few cases, though none in complete generality. For example if $f$ is a newform in $S_{2}(\Gamma_{0}(D), \chi_{D})$ that does not have complex multiplication and $D$ is a fundamental discriminant, then Proposition 11 of this paper shows that $$ \frac{3}{\pi [{\rm SL}_{2}(\mathbb{Z}) : \Gamma_{0}(D)]} \iint_{\mathbb{H} / \Gamma_{0}(D)} |f(x+iy)|^{2} y^{2} \, \frac{dx \, dy}{y^{2}} > \frac{3}{208 \pi^{4} \log(D)} \prod_{p | D} \left(\frac{p}{p+1}\right). $$ The appendix to Hoffstein and Lockhart's paper explains how one gets a zero-free regions for $L(f \otimes \overline{f}, s)$ near $s = 1$, and a good reference for the classical problem of turning a zero-free region into a lower bound is Hoffstein's paper "On the Siegel-Tatuzawa theorem" (in Acta Arithmetica, 1980/1981).<|endoftext|> TITLE: Wrapped Fukaya categories of Stein manifolds QUESTION [12 upvotes]: By the work of Abouzaid, we know that the wrapped Fukaya category of $T^\ast Q$ with $Q$ a closed smooth manifold is generated by a cotangent fiber. Basically, this is an application of Abouzaid's generation criterion for wrapped Fukaya categories of Liouville manifolds, which says that a full subcategory $\mathscr{B}\subset\mathscr{W}(X)$ of the wrapped Fukaya category of a Liouville manifold $X$ generates the wrapped Fukaya category $X$ if the image of the open-closed map $\mathscr{OC}:HH_\ast(\mathscr{B},\mathscr{B})\rightarrow SH^\ast(X)$ contains the identity of $SH^\ast(X)$, where $SH^\ast(X)$ is the symplectic cohomology of the Liouville domain $X_0\subset X$, which is also an invariant of $X$ by Viterbo functoriality. Since there is a Stein a structure on $T^\ast Q$, it's natural to ask which Lagrangians generate $\mathscr{W}(X)$ for a general Stein manifold $X$. However, it seems hard to apply Abouzaid's criterion for explicit candidates which are expected to generate $\mathscr{W}(X)$. The case for $T^\ast Q$ is relatively easier mainly because in this case, the geometric information of $\mathscr{W}(T^\ast Q)$ comes entirely from the loop space $\mathscr{L}X$. For example, someone considered Lagrangian sections of certain (local) Lagrangian fibrations on log Calabi-Yau surfaces or $T^\ast S^3$ in some papers concerning mirror symmetry, but we still don't know whether these Lagrangians generate $\mathscr{W}(X)$. (These considerations are of course motivated by $\mathbb{R}_+\subset\mathbb{C}^\ast$.) On the other hand, for the study of homological mirror symmetry for affine varieties, the superpotential $W:X\rightarrow\mathbb{C}$ doesn't seem to provide a natural Lefschetz fibration on $X$ in dimensions $n\geq3$. For example, $W=z_1z_2z_3$ on $\mathbb{C}^3$ is not even a Morse-Bott fibration. Therefore, we need other techniques to find natural candidates to generate $\mathscr{W}(X)$. Recently, Auroux proposed in his IHES talk that for certain affine varieties (which I believe should be affine conic bundles), there is a single Lagrangian homeomorphic to $\mathbb{R}^n$ which is expected to generate the so-called fiberwise wrapped Fukaya category $\mathscr{F}(X,W)$. To my understanding, this category should be an $A_\infty$ subcategory of $\mathscr{W}(X)$. Combining all these I come up with the following questions. For a Stein manifold $X$, Will a single Lagrangian $L$ generate $\mathscr{W}(X)$? Is it true that $L$ must be diffeomorphic to $\mathbb{R}^n$? If we have counterexamples for the above two questions, how about restricting ourselves to the case when $X$ is an affine variety? REPLY [5 votes]: By now we understand quite well generators of such categories. The co-core disks for any relative skeleton of the sector associated to the superpotential will do. (The argument for this is geometric, and does not invoke Abouzaid's criterion.) For the notion of relative skeleton, see e.g. section 1.3 of this.<|endoftext|> TITLE: Pictures of the von Neumann polytope QUESTION [9 upvotes]: Are there any graphic portrayals of von Neumann polytopes in low dimensions? REPLY [3 votes]: The set of doubly stochastic matrices is a subset of the set of nonnegative $n \times n$ matrices whose sum of entries is $n$. These matrices form an $(n^2-1)$-dimensional simplex in $\mathbb{R}^{n^2}$, whose vertices are the $n^2$ matrices (written as row vectors of length $n^2$) $(n,0,0,...,0)$, $(0,n,0,...,0)$, ..., $(0,0,... 0,n)$. The vertices of the Birkhoff polytope are the arithmetic means of $n$ of these - suitably chosen - matrices. For $n=2$ the Birkhoff polytope lies within a tetrahedron, between the midpoints of the edges that connect points $1 = (2, 0, 0, 0)$ and $4 = (0, 0, 0, 2)$ and points $2 = (0, 2, 0, 0)$ and $3 = (0, 0, 2, 0)$.<|endoftext|> TITLE: Solution to generalized Sylvester equation QUESTION [5 upvotes]: I am interested in solving generalized Sylvester equations (for $X$) of the form: $$ \sum_{j=1}^k A_j X B_j^T = F, $$ where $A_j,B_j,X,F\in\mathbb{C}^{n\times n}$ and $k$, $n$ are integers. I will assume that there is a unique solution matrix $X$. In the case $k=1$ the matrix equation can be solved in $\mathcal{O}(n^3)$ operations by Gaussian elimination. For $k=2$, it can be solved in $\mathcal{O}(n^3)$ operations by the generalized Bartels-Stewart algorithm (for example). For $k\geq3$ the fastest algorithm I know is the naive $\mathcal{O}(n^6)$ approach, where the matrix equation is written out as a dense $n^2\times n^2$ linear system. In the case $k=3$ is there a faster known algorithm? Or, perhaps, there is a paper proving that one cannot do better than $\mathcal{O}(n^6)$ in general. Anything along these lines would be wonderful. Thank you very much in advance. REPLY [2 votes]: Not an expert in this problem, but I try to suggest something nevertheless. A practical algorithm could be an iterative method like GMRES. Each step requires a matrix-vector product which costs $O(kn^3)$; convergence holds in exact arithmetic in $n^2$ steps, and you may get a good approximation with much less than that. You could get a decent preconditioner by dropping all but two terms of the sum. Even if you are just interested in a theoretical big-O cost, this strategy should beat a Strassen-like fast multiplication method on the Kronecker product form ($O(n^{4+\omega})$ rather than $O(n^{4+2\omega})$).<|endoftext|> TITLE: Locally Closed Orbits in Real Algebraic Geometry QUESTION [5 upvotes]: Let $G$ be a real algebraic group, and let $X$ be a real affine $G$-variety. I am looking for conditions on $G$ and $X$ for which the $G$-orbits are known to be locally closed in the Zariski topology on $X$. Please feel free to offer any conditions you desire. References would also be much appreciated. Of particular interest to me is the case in which $G$ is semisimple and $X$ is a finite-dimensional $G$-representation. Over $\mathbb{C}$, one usually establishes that orbits are locally closed by appealing to Chevalley's Theorem. It tells us that the an orbit contains a non-empty open subset of its closure. If we act on this open set by elements of $G$, then we realize the orbit as being open in its closure. Unfortunately, I believe Chevalley's Theorem fails over non-algebraically closed fields. ADDED: I would be perfectly happy with conditions under which a $G$-orbit is the unique orbit of maximal dimension in its Zariski closure. This is weaker than being locally closed. REPLY [6 votes]: Let $x_0\in X(\mathbb{R})$, and consider the orbit map $f:g\mapsto g.x_0$, as a morphism of $\mathbb{R}$-schemes. Denote by $S\subset G$ the stabilizer of $x_0$. Chevalley's therorem asserts that $f$ factors as $$G\longmapsto G/S \xrightarrow{\sim} Y \hookrightarrow X$$ where the first map is the (faithfully flat) canonical projection, the second is an isomorphism, and the third is a (locally closed) immersion. The $G(\mathbb{R})$-orbit of $x_0$ is $\Omega:=f(G(\mathbb{R}))\subset Y(\mathbb{R})\subset X(\mathbb{R})$. Of course, $Y$ is the "algebraic" orbit, and $Y(\mathbb{C})=G(\mathbb{C}).x_0$. Proposition. If $Y$ is connected (in particular if $G$ is), then $\Omega$ is Zariski-locally closed in $X(\mathbb{R})$ if and only if it is equal to $Y(\mathbb{R})$. The "if" part is trivial. Conversely, if $\Omega$ is Zariski-locally closed in $X(\mathbb{R})$, the same holds in $Y(\mathbb{R})$, so we may forget about $X$ from now on. The key point is that (without any locally closed assumption) $\Omega$ is always open and closed in $Y(\mathbb{R})$, for the real topology. More generally: Lemma. Let $S$ be a real algebraic group, and let $f:U\to V$ be an $S$-torsor (= principal homogeneous $S$-bundle), where $U$ and $V$ are $\mathbb{R}$-varieties. Then $f(U(\mathbb{R}))$ is open and closed in $V(\mathbb{R})$, for the real topology. (This is certainly well known, but I don't have a reference; in fact, it also works over $p$-adic fields). Proof of Lemma: since $f$ is automatically a smooth morphism, the induced map on real points is open (implicit function theorem), so $f(U(\mathbb{R}))$ is open. To see that it is closed, consider the characteristic map $c:V(\mathbb{R})\to H^1(\mathbb{R},S)$ sending $v\in V(\mathbb{R})$ to the class of $f^{-1}(v)$ as an $S$-torsor. Now $f(U(\mathbb{R}))=c^{-1}(e)$ where $e$ is the trivial class, so it suffices to see that $c$ is locally constant. For $\xi\in H^1(\mathbb{R},S)$, consider the twist $f^\xi:U^\xi\to V$ of $f$ by $\xi$: this is a torsor under the inner twist $S^\xi$ of $S$, and it has the property that $c^{-1}(\xi)=f^\xi(U^\xi(\mathbb{R}))$, which is open as we have seen. QED Now let us finish the proof of the proposition. Note that $Y$ is a smooth connected (hence irreducible) variety. Since $\Omega$ is open for the real topology it is Zariski-dense in $Y(\mathbb{R})$, so if it is Zariski-locally closed it must be Zariski-open. But then its complement in $Y(\mathbb{R})$ is Zariski-closed, hence has empty interior for the real topology (again because $Y$ is smooth) so it must be empty since it is (strongly) open. QED Remark. If $Y$ is not connected and $\Omega$ is locally closed, the conclusion is just that $\Omega=Y_1(\mathbb{R})$ where $Y_1$ is a union of components of $Y$.<|endoftext|> TITLE: What is known about the distribution of eigenvectors of random matrices? QUESTION [23 upvotes]: Let $A$ be a real asymmetric $n \times n$ matrix with i.i.d. random, zero-mean elements. What results, if any, are there for the eigenvectors of $A$? In particular: How are individual eigenvectors distributed (probably zero-mean multi-variate Normal, but what is the covariance)? If $u_i$ and $u_j$ are eigenvectors of $A$, what is the distribution of $|u_i^*u_j|$, or, even better, the $n^2$-d joint distribution $P(u_1, ... u_n)$? What is the joint distribution of eigenvalues and their corresponding eigenvectors (or, perhaps more in line with my application described below, the conditional distribution of an eigenvector given an eigenvalue)? Numerically, I've found that every eigenvector corresponding to a complex eigenvalue has a single real element. (Naturally, real eigenvalues have corresponding real eigenvectors.) Has this been proven? What is the distribution of the number of real eigenvalues of $A$? Note: I'm not a mathematician, but a physicist working in dynamical systems, and I skipped nuclear so my knowledge of GUE/GOE results is limited to basically the circular laws. I'm really interested in constructing random real matrices $A = VDV^{-1}$ where $D$ is a diagonal matrix of eigenvalues drawn from a distribution that I control and differs from the one given by the various circular laws, and $V$ is the matrix of eigenvectors drawn from the conditional distribution of eigenvectors of random matrices given their corresponding eigenvalue. So this question can be summarized: how do I draw $V$? I don't imagine that there are complete answers to my questions yet, but any insights along those lines that will help me draw $V$ "realistically" would be appreciated. Heck, I just realized bullets two and three may have somewhat incompatible assumptions: bullet three (or, rather, my proposed application) assumes that $P(u_1, ... u_n | \lambda_1, ..., \lambda_n) = P(\lambda_1,...,\lambda_N) \prod_i P(u_i | \lambda_i)$ where $i$ ranges over a single member of each complex conjugate pair of eigenvalues, where two makes no such assumptions and just jumps to $\int P(u_1, ... u_n, \lambda_1, ..., \lambda_n) d\vec{\lambda}$ where $\vec{\lambda}$ is circular law distributed. If that seems like a weird application, my motivation is to study the influence of only the eigenvalues of the adjacency matrix of a dynamical process that takes place on a random network. A simple first attempt at this by drawing $A$, performing SVD on it to get $V$, and mucking with $D$ only gives either interesting or disastrous results depending on how you look at it. A still simple but second attempt (which to my mind seems like it should work if the conditional independence assumption holds) of drawing several random $A$'s and choosing eigenvalues and corresponding eigenvectors from them according to my desired distribution is even more disastrous (but no more interesting, I think). REPLY [3 votes]: Another update: in the paper https://arxiv.org/abs/1608.04923 it is shown how one can use methods from free probability theory to go also beyond the Ginibre ensemble and treat eigenvector correlations for random matrix models corresponding to the single ring theorem<|endoftext|> TITLE: Contracting a rational curve in a Calabi-Yau threefold QUESTION [8 upvotes]: Let $X$ be a Calabi-Yau threefold and $C \subset X$ be a rational curve with $N_{C/X}\cong \mathcal{O}\oplus \mathcal{O}(-2)$. Can one contract the curve $C$? Assuming the answer is yes, what kind of singularity does one get? Explicit equation? REPLY [5 votes]: Given a rational curve $C$ in a threefold $X$ with normal bundle $N_{C/X}\cong{\mathcal O}\oplus{\mathcal O}(-2)$, what Miles Reid proves is that, at least locally, "$C$ either contracts or moves". $C$ certainly moves infinitesimally to order one, since $\dim H^0(N_{C/X})=1$. Reid proves that if $C$ moves to order $k$ but no further, then locally at least it is contractible to the hypersurface singularity in Francesco's answer. Alternatively, if $C$ deforms to arbitrary order, then there is a small analytic disc over which $C$ deforms in a ruling inside $X$. In this case, any contraction that contracts $C$ will also have to contract all other fibres of the ruling, and locally the singularity is an $A_1$ surface double point times the small disc, given by the equation $x^2+y^2+z^2=0$ inside affine four-space. Whether a contraction exists inside a projective Calabi-Yau threefold, in either case, is a global question, and seems quite subtle. I don't really have anything else to say in the isolated case. It is very easy to produce non-isolated examples: let $\bar X$ be a singular Calabi-Yau threefold containing a smooth curve $S$ of $A_1$ singularities; many such examples exist. Blowing up $S$ gives a smooth Calabi-Yau threefold $X$ with an exceptional divisor which is a ruled surface over $S$; the fibres of the ruling have the normal bundle you want. Just to see that the problem is really global, consider the following, related example. Let $\bar Y$ be a singular Calabi-Yau threefold which contains a smooth curve $Z$ of $A_2$ singularities, in such a way that when you blow up $Z$ inside $\bar Y$, you get an irreducible exceptional divisor $D$ which fibres over $Z$. Over any point $p\in Z$ the fibre in $D$ will of course be a line pair, but it can happen that $D$ is irreducible because of monodromy. This can also be achieved inside a projective Calabi-Yau threefold $Y$. Either one of the lines in the fibre still has the normal bundle you want, and the numerical class of the curve can be contracted, but only by contracting both fibres along the whole of $Z$, so the contraction has to give $\bar Y$ back, with a curve of $A_2$ singularities.<|endoftext|> TITLE: Is $\mathbb{R}$ a $\mathbb{C}$-module without AC? QUESTION [11 upvotes]: Assuming ZFC. We can make $(\mathbb{R},+)$ into a nontrivial (scalar multiplication is not identically zero) $\mathbb{C}$-module. Now my questions are? 0.Is it consistent with $ZF$ that $\mathbb{R}$ is not a $\mathbb{C}$-module? 1.Does $AD$ (Axiom of determinacy) implies that $\mathbb{R}$ is not a $\mathbb{C}$-module? We know $AD$ leads many amazing theorems about reals, both algebraic and analytic. How does Solovay model affect Algebraic construction of reals? More precisely and especially 2.If the answer of question 1 is yes. Is it possible that $\mathbb{R}$ is not a $\mathbb{C}$-module in Solovay model? Any suggestion and reference is appreciated. REPLY [13 votes]: If all sets of reals have the Baire Property (as holds under $\sf AD$ and in Solovay's model, and in Shelah's model whose consistency strength does not require any large cardinals), then the answer is negative. Let us denote by $\sf BP$ this principle. To see that, first we use the fact that under $\sf BP$ if $\varphi\colon\Bbb R\to\Bbb Q$ is linear, then it is continuous (in general any homomorphism from a Polish group to a normed group is continuous under $\sf BP$). This means that there are no discontinuous solutions for Cauchy's functional equation: $f(x+y)=f(x)+f(y)$ with $f\colon\Bbb R\to\Bbb R$. But if $\Bbb R$ is a $\Bbb C$-module, then we can reinterpret it over $\Bbb R$ as a module of dimension $\geq 1$, but since $\Bbb C$ has dimension $2$, this gives us a nontrivial decomposition of $\Bbb R$ as a vector space over $\Bbb Q$ and therefore it gives us a discontinuous solution to Cauchy's functional equation (to see that this is the case, just note that continuous solutions are zero or have trivial kernel, here the projection on one of the direct summands is neither zero nor has a trivial kernel). Another option, which again appeals to Pettis' theorem is to note that if $\Bbb R$ is a $\Bbb C$-module, then $z\mapsto z\cdot 1_\Bbb R$ is a group homomorphism, which is therefore continuous, but therefore the range of this function is a connected subgroup of $\Bbb R$ which has to be $\Bbb R$ itself. Now the previous paragraph holds, since $\Bbb R$ is a $2$-dimensional vector space over itself and we can find a discontinuous endomorphism as before. Some references: The fact that $\sf BP$ implies that any homomorphism from a Polish group into a normed group appears in Kecrhis' book about descriptive set theory. It's a consequence of Pettis' theorem. Shelah constructed a model of $\sf ZF+DC+BP$ from a model of $\sf ZFC$. Thus eliminating the need for an inaccessible cardinal to get the Baire property. He also showed that in order to get Lebesgue measurability of all sets of reals in $\sf ZF+DC$ one has to have that there is an inaccessible cardinal in an inner model. Saharon Shelah, Can you take Solovay’s inaccessible away?, Israel J. Math. 48 (1984), no. 1, 1--47.<|endoftext|> TITLE: Immersion of $S^1$ in $\mathbb{R}^2$ that can be extended to $\mathbb{D}$ QUESTION [8 upvotes]: I was wondering about $\mathcal{C}^{\infty}$ immersions $S^1 \longrightarrow \mathbb{R}^2$ which are the restriction to $\partial \mathbb{D}$ of an immersion $\overline{\mathbb{D}} \longrightarrow \mathbb{R}^2$ ? Especially, is a computable invariant for such immersed closed curves ? REPLY [18 votes]: Here is a cool example (due I believe independently to Eliashberg, Milnor, and Blank) related to your question. There is an immersion of $S^1$ into $\mathbb{R}^2$ that extends to two different immersions of $D^2$ into $\mathbb{R}^2$. Best done with a picture but I don't know how to up load one. A reference is page 150 of "Topology of Spaces of S-Immersions" by Eliashberg and Mishachev. Glueing these two together gives rise to a map from $S^2$ to $\mathbb{R}^2$ with only fold singularities which is not homotopic through such maps to the standard quish the 2-sphere onto the plane, which is the topic of the E and M paper.<|endoftext|> TITLE: transcendence of beta values QUESTION [5 upvotes]: (1) Can anybody suggest a readable reference for Schneider's theorem that the number $$ \beta(a, b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} $$ is transcendental for $a, b \in \mathbb{Q}$ such that none of $a, b, a+b$ is an integer? (2) Fix some integer $n \geq 3$ Is the degree of transcendence of the field generated over $\mathbb{Q}$ by $$ \left(\beta(\tfrac{i}{d}, \tfrac{j}{d})\right)_{i, j=1, \ldots, d-1} $$ known? REPLY [2 votes]: Scheider's original paper is available online. The theorem you quote is proved at the end of Section 1, on Page 114. I don't know the answer to your second question, but my guess is "no".<|endoftext|> TITLE: How do small changes in a filtered complex affect the associated spectral sequence QUESTION [7 upvotes]: I have recently been learning about spectral sequences, not in the context of any problem, but mostly out of curiosity. There are two questions that have occurred to me which I've been unable to answer, even after computing a few examples. For the purposes of both questions, suppose I have two filtered chain complexes: $0=C_{0}\subset C_{1}\subset\ldots\subset C_{n}$ and $0=K_{0}\subset K_{1}\subset\ldots\subset K_{m}$. For the sake of these questions one can assume that the modules making up the complexes are finite dimensional vector spaces. If $n=m$ and there exists $q$, such that $0 TITLE: Commuting nets for commuting projections QUESTION [9 upvotes]: I think this should not be too difficult, but I am not an expert. I did not get an answer on stackexchange. Let $A$ be a $C$*-algebra and let $p,q\in A^{**}$ be two commuting projections. Then there exist self-adjoint nets $(x_i)_i$ and $(y_j)_j$ in $A$ such that $x_i\to p$ and $y_j\to q$ in the weak$^*$-toplogy. Can these nets be chosen to be commutative, that is, $x_iy_j=y_jx_i$ for all $i,j$? Extra question: Is this true if $A$ is a $JB$-algebra? REPLY [5 votes]: I think the following provides a counterexample, though the bidual of a $C^*$-algebra always makes me nervous. Let $A=M_2\otimes C[0,1]$. Any bounded, Borel measurable, $M_2$ valued function on $[0,1]$ will give an element of $A^{**}$; for the projection $p\in A^{**}$ we take the function \begin{equation} p(t)=\begin{cases} \begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix} & 0\leq t\leq \frac12, \\ \begin{pmatrix} \frac12 & \frac12 \\ \frac12 & \frac12 \end{pmatrix} & \frac12 TITLE: If any open set is a countable union of balls, does it imply separability? QUESTION [54 upvotes]: If a metric space is separable, then any open set is a countable union of balls. Is the converse statement true? UPDATE1. It is a duplicate of the question here https://math.stackexchange.com/questions/94280/if-every-open-set-is-a-countable-union-of-balls-is-the-space-separable/94301#94301 UPDATE2. Let me summarize here the positive answer following Joel David Hamkins and Ashutosh. It is a matter of taste, but I omit using ordinals and use Zorn lemma instead, which may be more usual for most mathematicians (at least, it is for me). Lemma 1. If $(X,d)$ is non-separable metric space, then for some $r>0$ there exists an uncountable subset $X_1\subset X$ such that $d(x,y)>r$ for any two points $x\ne y$ in $X_1$. Proof. For each $r=1/n$ consider the maximal (by inclusion) subset with such property. If it is countable, then $X$ has a countable $1/n$-net for each $n$, hence it is separable. Now consider two cases. Define $X_2\subset X_1$ as a set of points $x\in X_1$ for which there exist a point $y_x\in X$ such that $02d(x,y_x)\geq d(x,z)+d(x,y_x)\geq d(z,y_x)>a$ since $y_x\notin B(z,a)$. It implies that $B(z,a)$ is contained in a unique ball $B(x,d(x,y_x))$, hence we need uncountably many such balls to cover whole $U$. 2) $X_3=X\setminus X_2$ is uncountable. For any $x\in X_3$ define $R(x)>0$ as a radius of maximal at most countable open ball centered in $x$. Clearly $R(x)\geq r/10$ for any $x\in X_3$. For any $x\in X_3$ define a star centered in $x$ as a union $D=x\cup C$, where $C=\{z_1,z_2,\dots\}\subset X_3$ is a countable sequence of points with $d(x,z_i)\rightarrow R(x)+0$. Choose a maximal disjoint subfamily of stars. Clearly it is uncountable, else we may easily increase it. Denote by $U$ the set of centers of chosen stars. It is open (as any subset of $X_3$), assume that it is a countable union of balls $U=\cup_{i=1}^{\infty} B(x_i,r_i)$, $x_i\in U$. We have $r_i> R(x_i)$ for some $i$, else $U$ is at most countable. But then $B(x_i,r_i)$ contains infinitely many points of the star $D$ centered in $x_i$, while by our construction $U\cap D=\{x_i\}$. A contradiction. REPLY [37 votes]: Towards a contradiction, let us assume that we have a metric space $X = \{x_i : i < \omega_1\}$ in which any two points are at least unit distance apart and every subset of $X$ is the union of a countable family of open balls. Let $r_i$ be the supremum of all $r > 0$ such that $B(x_i, r)$ is countable. Construct $\{C_i : i < \omega_1\}$ such that (1) Each $C_i$ is countable and the infimum of $\{d(x_i, y): y \in C_i\}$ is $r_i$ (2) If $i < j$, then $x_i \notin C_j$ Now let $I \in [\omega_1]^{\omega_1}$ be such that whenever $i < j$, $x_j \notin C_i$. Suppose $Y = \{x_i : i \in I\}$ can be covered by a family $F$ of countably many balls. Let $i \in I$ be least such that $B(x_i, r) \in F$ for some $r > r_i$. Pick $y \in C_i \cap B(x_i, r)$. So $y \in \bigcup F = Y$ so that $y = x_j$ for some $j \in I$ which is impossible.<|endoftext|> TITLE: Is there a von Koch-type theorem for the generalized Riemann hypothesis? QUESTION [5 upvotes]: Helge von Koch proved in 1901 that the Riemann hypothesis is equivalent to the error term in the prime number theorem having the bound $$ \mid\pi(x)-\textrm{li}(x)\mid=O(\sqrt{x} \log x). $$ Q1: Is von Koch's result (and proof) also valid for the generalized Riemann hypothesis (GRH)? That is, if a and d are coprime, and $\pi(x,a,d)$ denotes the number of primes in the arithmetic progression $a, a+d, a+2d, \dots$, is GRH equivalent to $$ \left|\pi(x,a,d)-\frac{1}{\varphi(d)}\textrm{li}(x)\right|=O(\sqrt{x} \log x)? $$ The Wikipedia page for GRH states that GRH implies this error term, but is short on references and says nothing about whether the given error term implies GRH. Q2: Does anyone know if von Koch's paper has been translated to English; alternatively a reference to the best modern version of proving his result; or perhaps both? REPLY [13 votes]: As GH from MO and Felipe Voloch have already indicated it is standard to show that $\psi(x;q,a) = x/\phi(q) +O(x^{\frac 12+\epsilon})$ for all reduced residue classes $a\pmod q$ is equivalent to GRH for the characters $\pmod q$. I want to make the following small (but amusing) refinement: it is enough to know that $\psi(x;q,1) = x/\phi(q) + O(x^{\frac12+\epsilon})$ and $\psi(x) =x+ O(x^{\frac 12+\epsilon})$ and from these two pieces of information (rather than all $\phi(q)$ residue classes) we can get GRH for all the characters $\pmod q$. To see this, note that (starting with Re$(s)>1$) $$ -\frac{1}{\phi(q)} \sum_{\chi\neq \chi_0} \frac{L^{\prime}}{L}(s,\chi) = \int_1^{\infty} \frac{s}{x^{s+1}} \Big( \psi(x;q,1) - \frac{\psi(x,\chi_0)}{\phi(q)} \Big) dx, $$ and so by hypothesis, the RHS extends analytically to Re$(s)>1/2$. Therefore $L(s,\chi) \neq 0$ for all $\chi \neq \chi_0$ and Re$(s)>1/2$. Finally $\psi(x)=x+O(x^{\frac 12+\epsilon})$ implies RH; thus GRH follows for all the characters $\pmod q$. Put differently, this shows that the asymptotic $\psi(x;q,1) = \psi(x)/\phi(q) + O(x^{\frac 12+\epsilon})$ implies $\psi(x;q,a) =\psi(x)/\phi(q)+ O(x^{\frac 12+\epsilon})$ for all $(a,q)=1$. REPLY [8 votes]: The bound you state follows from GRH, see Corollary 13.8 in Montgomery-Vaughan: Multiplicative number theory I. Conversely, the bound you state implies GRH. I provide the proof below. Assume that for all coprime pairs $a$ and $d$ we have $$ E(x,a,d):=\pi(x,a,d)-\frac{\mathrm{li}(x)}{\varphi(d)} = O(\sqrt{x}\log x). $$ Let us introduce $$ \theta(x,a,d):=\sum_{\substack{p\leq x\\p\equiv a\pmod{d}}} \log p \qquad\text{and}\qquad\psi(x,a,d):=\sum_{\substack{n\leq x\\n\equiv a\pmod{d}}} \Lambda(n). $$ Then $$\theta(x,a,d) - \frac{x-2}{\varphi(d)} = \int_{2-}^x \log t\ d E(t,a,d) = E(x,a,d)\log x-\int_2^x\frac{E(t,a,d)}{t}dt,$$ whence $$\psi(x,a,d)=\theta(x,a,d)+O(\sqrt{x}\log x) = \frac{x}{\varphi(d)}+O(\sqrt{x}\log^2 x).$$ As a result, for any nontrivial Dirichlet character $\chi$ modulo $d>1$ we have $$\psi(x,\chi):=\sum_{n\leq x}\chi(n)\Lambda(n)=\sum_{\substack{1\leq a\leq d\\(a,d)=1}}\chi(a)\psi(x,a,d)=O(\sqrt{x}\log^2 x).$$ This implies that the Dirichlet $L$-function $L(s,\chi)$ has no zero in the half-plane $\Re(s)>1/2$, because its logarithmic derivative is analytic there: $$ -\frac{L'(s,\chi)}{L(s,\chi)}=\sum_{n=1}^\infty \frac{\chi(n)\Lambda(n)}{n^s}=\int_{2-}^\infty t^{-s}\ d\psi(t,\chi)=s\int_2^\infty \frac{\psi(t,\chi)}{t^{s+1}}\,dt.$$ Analycity follows from the local uniform convergence of the last integral in the half-plane $\Re(s)>1/2$.<|endoftext|> TITLE: History of powers beyond squares and cubes QUESTION [25 upvotes]: The ancient Babylonians understood squares:       Plimpton 322 The ancient Athenians understood cubes, if we can take doubling the cube, i.e., the Delian problem, as evidence. My question is: Q. When were 4th, 5th, $\ldots$, $n$-th powers contemplated/understood/used? I am wondering how tied was the understanding of powers/exponentiation to geometry, to spatial dimensions. Did the ancients generalize their explorations to arbitrary integer exponents? REPLY [18 votes]: The cuneiform tablet MS 2351 from the 19th century BC contains the 15-digit sexagesimal number 13 22 50 54 59 09 29 58 26 43 17 31 51 06 40, which happens to equal $20^{20}$. I also seem to remember they constructed a table of reciprocals for numbers of the form $125 \cdot 2^n$ for exponents up to $19$. Edit. Last year some fragments of a large table have been identified: Ossendrijver discovered that the complete table contained the powers of 9 up to $9^{46}$.<|endoftext|> TITLE: p-adic L-function of curves QUESTION [6 upvotes]: Given a smooth projective curve $C$ over $\mathbb{Q}$ one has the $L$-function $L(C, s)$ and the Beilinson conjectures predict its values at integers $s=n$ in terms of regulators. Is there a p-adic analogue of the story, that is a p-adic L-function $L_p(C, s)$ and a conjecture about the special values? I have seen people study p-adic L-functions of modular forms, so I guess for elliptic curves the answer is positive, at least to the first question. What about more general curves? Is there a construction of $L_p(C, s)?$ REPLY [5 votes]: To complete Chris's answer in comments, yes, we do expect such a $p$-adic $L$-function to exist, but we are far from being able to prove it. There are two problems: first we very likely need to show that the curve $C$, or equivalently its Jacobian, or equivalently its $L$-function $L(C,s)$ is automorphic, because all $p$-adic $L$-functions construction we have so far use in some fundamental way the deep properties of archimidean automorphic $L$-function. And proving the automorphy of a curve $C$ over $\mathbb Q$ (or for that matter, of an elliptic curve over a general number field) is wide open. Second, even when we know that $L(C,s)$ is $L(\pi,s)$ for nice automorphic representation $\pi$, current technologies does not allow yet to prove that the special values of $L(\pi,s)$ at integers are, conveniently normalized, algebraic numbers, let alone to interpolate those values $p$-adically into a $p$-adic $L$-function. There has been some progresses when $\pi$ is attached to a unitary group, and $p$ an ordinary prime for $\pi$ be the general case is still out of reach...<|endoftext|> TITLE: Isotropic Riemannian manifolds QUESTION [12 upvotes]: Let $M$ be a Riemannian manifold and $G$ a closed connected subgroup of isometries of $M$. Call the pair $(M,G)$ an isotropic pair if $G$ acts transitively on the sphere bundle $SM$. As an example, the pair $(S^6,SO(7))$ is isotropic, but also $(S^6,G_2)$. I am looking for a reference for the classification of all isotropic pairs. In several places in the literature, the possible manifolds $M$ are given (in the non-flat case these are symmetric spaces of rank one), but not the corresponding groups $G$.The best reference seems to be J. Tits paper Espaces homogènes et isotropes, et espaces doublement homogènes from 1954, but it is not so easy to understand his terminology. Questions: Is the following list complete? There seems to be no negatively curved versions of the pairs $(S^6,G_2)$ and $(S^7,Spin(7))$. What is the reason for this? Flat case: $\mathbb{R}^n$ with the group generated by translations and $SO(n), U(n/2), SU(n/2), Sp(n/4), Sp(n/4) U(1), Sp(n/4) Sp(1)$ $\mathbb{R}^7$ with the group generated by translations and $G_2$ $\mathbb{R}^8$ with the group generated by translations and $Spin(7)$ $\mathbb{R}^{16}$ with the group generated by translations and $Spin(9)$ Symmetric space case (each with $G$ equal to the connected component of the trivial element of the isometry group): $S^n$ sphere $\mathbb{RP}^n$ real projective space $\mathbb{H}^n$ real hyperbolic space $\mathbb{CP}^n$ complex projective space $\mathbb{CH}^n$ complex hyperbolic space $\mathbb{HP}^n$ quaternionic projective space $\mathbb{HH}^n$ quaternionic hyperbolic space $\mathbb{OP}^2$ octonionic projective plane $\mathbb{OH}^2$ octonionic hyperbolic plane Other cases: $(\mathbb{RP}^6,G_2)$ and $(S^6,G_2)$ $(\mathbb{RP}^7,Spin(7))$ and $(S^7,Spin(7))$ REPLY [5 votes]: The reason there are no 'negatively curved' analogs of $(S^6,\mathrm{G}_2)$ or $\bigl(S^7,\mathrm{Spin}(7)\bigr)$ is that, in each of these cases of homogeneous Riemannian manifolds $G/H$, the corresponding $H$-structure ($H=\mathrm{SU}(3)$ in the first case, $H=\mathrm{G}_2$ in the second) is not torsion-free. In fact, there is, up to scale, only one possible torsion tensor in each of these cases that is $H$-invariant, and the Cartan structure equations imply that the $H$-invariant part of the Riemann curvature tensor must be a quadratic expression in the torsion tensor. Since $x^2 = (-x)^2 >0$, changing the sign of the torsion tensor won't change the sign of the $H$-invariant part of the curvature tensor. That's why you can't just reverse it and get compatible structure equations (which is what works and gives rise to duality in the case of symmetric spaces). Your list does look complete to me. I think you could prove it relatively easily if you are willing to accept the known list of Lie groups acting transitively on spheres.<|endoftext|> TITLE: Hilbert's Hotel QUESTION [35 upvotes]: Hilbert's Hotel is a famous story about infinity attributed to David Hilbert (1862-1943). Is it documented that Hilbert's Hotel is in fact due to Hilbert, and if yes, where? REPLY [45 votes]: The True (?) Story of Hilbert's Infinite Hotel, by Helge Kragh (2014) What is known as "Hilbert's hotel" is a story of an imaginary hotel with infinitely many rooms that illustrates the bizarre consequences of assuming an actual infinity of objects or events. Since the 1970s it has been used in a variety of arguments, some of them relating to cosmology and others to philosophy and theology. For a long time it has remained unknown whether David Hilbert actually proposed the thought experiment named after him, or whether it was merely a piece of mathematical folklore. It turns out that Hilbert introduced his hotel in a lecture of January 1924, but without publishing it. The counter-intuitive hotel only became better known in 1947, when George Gamow described it in a book, and it took nearly three more decades until it attracted wide interest in scientific, philosophical, and theological contexts. The paper outlines the origin and early history of Hilbert's hotel paradox. At the same time it retracts the author's earlier conclusion that the paradox was originally due to Gamow. The relevant quote from Hilbert's 1924 lecture is as follows (my translation from German): An application of this fact is provided by the hotel manager, who has a hotel with a finite number of rooms. All these rooms are occupied by guests. When the guests exchange their room in any way, so that again no room has more than one guest, then that will not free a room, and the hotel manager cannot in this way make space for a newly arrived guest. We can also say: A part of a finite quantity is never equal in number to the whole. [...] How is this for an infinite quantity? Let us take as the simplest example the quantity of integer numbers. Here already this law "a part is smaller than the whole" no longer holds. We can explain this important fact easily using our example of the occupied hotel. This time we assume that the hotel has infinitely many numbered rooms, $1,2,3,4,5,\ldots$, in each of which there lives a guest. When a new guest arrives, all the manager has to do is to allow each of the old guests to occupy the room having one number higher, and this will free room number 1 for the new arrival. Of course, in this way space can be made for any finite number of new guests, and in this world of an infinite number of houses and occupants there will be no housing shortage. [...] Indeed, it is even possible to make space for an infinite number of new guests. For example, each of the old guests, originally occupying room number $n$, just has to move to number $2n$. Then the infinite number of rooms with odd numbers become free for new guests.<|endoftext|> TITLE: Why the Szpiro conjecture over number fields doesn't depend on the discriminant of the number field? QUESTION [5 upvotes]: According to Hindry p.7 Conj 3.1 and Stein Szpiro's conjecture states that the Szpiro ratio is: $$ \sigma_{E/K}=\frac{\log{|N_{K/Q}\Delta_{E/K}|}}{\log{|N_{K/Q} f_{E/K}}|}$$ Given $ \varepsilon >0$ there are only finitely many $ E/K$ with $ \sigma_{E/K}\geq 6+\varepsilon $. In particular, $ \sigma_{E/K}$ is bounded. $\Delta_{E/K}$ is the minimal discriminant. Szpiro's conjecture implies abc with a bit higher exponent over the integers. Over number fields, the uniform abc conjecture depends on the discriminant of the number field. Maybe this is because every integer can be arbitrary large power in some number field. To make $d$ a $k$-th power, work with defining polynomial $x^k -d$. So start with EC over the rationals with discriminant $\Delta$ and compute $\sigma_{E/K}$ for $K$ with defining polynomial $x^k - \Delta$ for $k$ sufficiently large. Unless the minimal discriminant takes care of this case, this will give unbounded Szpiro ratio over some number fields. Why the Szpiro conjecture over number fields doesn't depend on the discriminant of the number field while the uniform abc conjecture depends? Over $K$ with defining polynomial $x^{16} + 22384$ take $E/K : y^2 = x^3 + 7x -1$. Not sure if the discriminant is minimal, but the norm is $2^{64} \cdot 1399^{16}$ while the norm of the conductor is $2^{10} \cdot 1399$ giving Szpiro ratio $11.305664847$ Some experiments in degree 14 and sage's global_minimal_model() suggest the global minimal model preserves the high powers in the discriminant: a4,a6= -2 -1 global minimal model= Elliptic Curve defined by y^2 + (1/14*w^12-1/28*w^10+1/7*w^8+1/4*w^7-1/14*w^6+2/7*w^4-1/7*w^2-3/7)*y = x^3 + (243/56*w^13+17/16*w^12-297/28*w^11+51/2*w^10-561/14*w^9+111/2*w^8-1581/28*w^7+81/2*w^6+255/28*w^5-99*w^4+1530/7*w^3-374*w^2+3330/7*w-527)*x + (-531/112*w^13-9419/56*w^12+2945/7*w^11-40829/56*w^10+55889/56*w^9-31119/28*w^8+47939/56*w^7-1209/28*w^6-10470/7*w^5+53021/14*w^4-90715/14*w^3+125729/14*w^2-69170/7*w+107855/14) over Number Field in w with defining polynomial x^14 - 80 Delta= 2^32 * 5^14 f= 2^32 * 5 d_q= 2^4 * 5 ratio= 1.87946880927217 a4,a6= -2 1 global minimal model= Elliptic Curve defined by y^2 + (1/14*w^12-1/28*w^10+1/7*w^8+1/4*w^7-1/14*w^6+2/7*w^4-1/7*w^2-3/7)*y = x^3 + (243/56*w^13+17/16*w^12-297/28*w^11+51/2*w^10-561/14*w^9+111/2*w^8-1581/28*w^7+81/2*w^6+255/28*w^5-99*w^4+1530/7*w^3-374*w^2+3330/7*w-527)*x + (533/112*w^13+4709/28*w^12-5891/14*w^11+40819/56*w^10-55887/56*w^9+31125/28*w^8-47933/56*w^7+1185/28*w^6+10460/7*w^5-26504/7*w^4+90725/14*w^3-125767/14*w^2+69150/7*w-107857/14) over Number Field in w with defining polynomial x^14 - 80 Delta= 2^32 * 5^14 f= 2^18 * 5 d_q= 2^4 * 5 ratio= 3.17425556409935 The last two examples give the abc triples $c_4^3-c_6^2=w^{14}$ Curves of Szpiro ratios $ > 66$ are in this question and they appear to contradict boundedness of the ratio. REPLY [2 votes]: A simple example: Choose arbitrarily $a,c\in \mathbb{Z}$ with $c\neq 0$. Let $K/\mathbb{Q}$ be a number field such that $b=\sqrt{a^3-c}\in \mathcal{O}_K$ and the elliptic curve $E$ defined by $y^2=x^3-3ax-2b$ has semistable reduction over $\mathcal{O}_K$. $E$ has the $j$-invariant $j(E)=a^3/c$. Then a computation as by Silverman/Pellarin gives for the stable Faltings height $$\log|a|\le \tfrac{1}{3}\log|c|+4h_F(E)+2\log\max\lbrace 1,h_F(E)\rbrace+7.$$ On the other hand side we have $$\tfrac{1}{[K:\mathbb{Q}]}\log|N_{K/\mathbb{Q}}\Delta_{E/K}|\le \log|2^4\cdot \mathrm{Disc}(x^3-3ax-2b)|=\log|c|+\log 3^3\cdot 2^6.$$ Since $K/\mathbb{Q}$ can be chosen with bounded degree, we obtain for $c=1$ and $a$ arbitrary large an Elliptic curve of arbitrary large Faltings height $h_F(E)$ and bounded discriminant $\log|N_{K/\mathbb{Q}}\Delta_{E/K}|$. That shows, that in Szpiro's conjecture $$h_F(E)\le (\tfrac{1}{2}+\epsilon)\log|N_{K/\mathbb{Q}}\mathcal{F}_{E/K}|+C_{\epsilon,K},$$ the constant $C_{\epsilon,K}$ should really depend on $K$ and not only on $[K:\mathbb{Q}]$.<|endoftext|> TITLE: Compositional inversion and generating functions in algebraic geometry QUESTION [5 upvotes]: The exponential generating function of the graded dimension of the cohomology ring of the moduli space of n-pointed curves of genus zero satisfying the associativity equations of physics (the WDVV equations) (cf. OEIS-A074060) is the compositional inverse of the generating function for the Betti numbers for $M_{0,n}$ (cf. OEIS-A049444 and A143491). And, Brown and Bergstrom in "Inversion of series and the cohomology of the moduli spaces of $M_{0,n}^\delta$" state in the abstract: For $n \geq 3$, let $M_{0,n}$ denote the moduli space of genus $0$ curves with $n$ marked points, and $\overline{M}_{0,n}$ its smooth compactification. A theorem due to Ginzburg, Kapranov and Getzler states that the inverse of the exponential generating series for the Poincare polynomial of $H^\bullet(M_{0,n})$ is given by the corresponding series for $H^\bullet(\overline{M}_{0,n})$. In this paper, we prove that the inverse of the ordinary generating series for the Poincare polynomial of $H^\bullet(M_{0,n})$ is given by the corresponding series for $H^\bullet(M^{\delta}_{0,n})$, where $\overline{M}_{0,n}\subset M^{\delta}_{0,n} \subset \overline{M}_{0,n}$ is a certain smooth affine scheme. Is there an intuitive principle underlying this relationship of compositional inversion between the generating functions? (Getzler makes use of a generalized Legendre transform-compositional inversion in disguise-to relate the generating functions.) REPLY [6 votes]: I think you would enjoy reading Curt McMullen's paper "Moduli spaces in genus zero and inversion of power series". In some sense there is nothing there that isn't already in Getzler's paper, but everything is stated in a down-to-earth and combinatorial fashion. Let me summarize the story, first for the spaces $\overline M_{0,n}$ and $M_{0,n}$. The space $\overline M_{0,n}$ has a stratification where a stratum corresponds to a tree with no vertices of valence two. The stratum itself is isomorphic to $\prod_v M_{0,\mathrm{val}(v)}$ where $v$ runs over interior vertices of the tree and $\mathrm{val}(v)$ denotes the number of incident edges. Since the virtual Poincaré polynomial is additive over stratifications, this shows that the virtual Poincaré polynomial of $\overline M_{0,n}$ is given by a sum over trees involving the virtual Poincaré polynomials of $M_{0,n'}$ for $n' \leq n$. Now using the relationship between compositional inversion and summing over trees, well-known to combinatorists, one can thus show that the exponential generating series of virtual Poincaré polynomials of $\overline M_{0,n}$ and $M_{0,n}$ are compositional inverses of each other. (If you don't know virtual Poincaré polynomials, think about any other invariant additive under stratification, e.g. Euler characteristic.) Finally, both the spaces $\overline M_{0,n}$ and $M_{0,n}$ have pure cohomology in every degree: $H^k (\overline M_{0,n})$ is pure of weight $k$, and $H^k(M_{0,n})$ us pure of weight $2k$. Thus in both cases, the virtual Poincaré polynomial concides with the usual Poincaré polynomial (in the latter case up to a substitution $t \mapsto t^2$). This explains the second sentence in Bergström-Brown's abstract. The story for $M_{0,n}^\delta$ and $M_{0,n}$ is completely similar, the only difference being that $M_{0,n}^\delta$ has a stratification indexed by trees without vertices of valence two and with a cyclic ordering of the edges incident to each vertex. In the same way as compositional inversion of exponential generating functions corresponds to sums over trees, compositional inversion of ordinary generating functions corresponds to sums over trees with such cyclic structure. McMullen touches upon something very similar at the very end of his paper. He doesn't consider $M_{0,n}$ and $M_{0,n}^\delta$, but instead considers the choice of a connected component of $M_{0,n}(\mathbf R)$ and its closure. Combinatorially this amounts to exactly the same thing: $M_{0,n}^\delta$ is defined by choosing a connected component of $M_{0,n}(\mathbf R)$ and taking the union of all strata meeting the closure of this component. A final remark is that the duality between $H^\bullet(\overline M_{0,n})$ and $H^\bullet(M_{0,n})$ can be upgraded to a Koszul duality of two cyclic operads, the "Hypercommutative" and "Gravity" operads. This is a much stronger result than just that their generating series are compositional inverses, and this is what Getzler proves. On the other hand the cohomologies of $M_{0,n}^\delta$ and $M_{0,n}$ give rise to nonsymmetric cyclic operads (this notion is not defined in the literature, but it's not hard to give the definition). However, it turns out that they are not in any natural sense Koszul dual of each other, but it is still true that they are interchanged with each other under bar-cobar-duality, up to homotopy. (But first one needs to define a bar transform of nonsymmetric cyclic operads...) This is an operad-theoretic statement that improves on what Bergström-Brown proved. I worked this out with Johan Alm at one point but we never wrote it down properly.<|endoftext|> TITLE: High dimensional topological field theory QUESTION [5 upvotes]: In the article Topological Field Theories in 2 dimension, Constantin Teleman has the following commentary " By constrast, in higher dimension, there seem to be no interesting theories: all examples are built from characteristic classes." There is any material where I can find these examples? REPLY [2 votes]: In three dimensions there are Chern-Simons theories which are indeed built from characteristic classes. As a physicist I find them extremely interesting. In four dimensions, we have Dijkgraaf-Witten TQFT and also the Crane-Yetter TQFT, which are constructed using the algebraic data of a premodular ribbon fusion category.<|endoftext|> TITLE: Endomorphism Ring of Simple Abelian Varieties QUESTION [5 upvotes]: I know that if $A$ is a simple abelian variety over a number field $k$ with all endomorphisms defined over $K$ then $\mbox{End}(A_K)\otimes \mathbb{Q}$ is a division algebra with a positive involution. Albert's classification tells us that these types of division algebras come in four types Type I: Totally real number field Type II/III: Central simple algebra $L$ over a totally real field such that simple components of $L\otimes \mathbb{R}$ are isomorphic to $M_2(\mathbb{R})$ or $\mathbb{H}$ (depending on Type II or III respectively) Type IV: Central simple algebra over a CM-field. My question is, for a fixed dimension, $g$, can we find simple abelian varieties of dimension $g$ such that it's endomorphism ring falls into each of these types? I ask because in the paper Sato-Tate distributions and Galois endomorphism modules in genus 2 (by Fite, Kedlaya, Rotger and Sutherland), they say that for simple abelian varieties of dimension 2, then $\mbox{End}(A_K)\otimes \mathbb{R}$ can be one of $$\mathbb{R}, \mathbb{R}\times\mathbb{R}, \mathbb{C}\times\mathbb{C}, M_2(\mathbb{R})$$ If it were true that for a fixed dimension all types appear, shouldn't it be possible to get an abelian variety with endomoprhism ring $\mathbb{H}$? If it is the case that not all types can appear for a fixed dimension, is it known when they can appear? REPLY [5 votes]: Albert's classification works/is good enough for (algebraically closed) fields of characteristic zero. For the complete list of all possibilities for a given $g$ (including the case of prime characteristic) see a survey of Frans Oort: ``Endomorphism algebras of abelian varieties". Algebraic geometry and commutative algebra, Vol. II, 469–502, Kinokuniya, Tokyo, 1988. (MR0977774 (90j:11049) ). In particular, if $g=2$ then you cannot get type III.<|endoftext|> TITLE: Trees with a maximal convex hull: are the only optimal solutions Steiner trees? QUESTION [7 upvotes]: For given $n\geqslant 3$, I'm looking for a connected set composed of $n$ equal segments in the plane such that the convex hull of it has maximal area $A(n)$. To simplify notation, we'll take $\dfrac{2}{\sqrt[4]{3}}$ as the length of each segment, so the unit triangle has area $1$. It turns out that for small $n$, all the optimal solutions are Steiner trees, in the sense that each point belonging to more than one segment is the common endpoint of either two segments forming a straight line, or of three segments forming angles of $\dfrac{2\pi}{{3}}$. We'll call points of the latter kind branching points. Using Steiner trees with only one branching point and the three legs of almost same length, we get the trivial lower bounds $ \ \ \,\quad A(3k)\geqslant 3k^2$, $A(3k+1)\geqslant k(3k+2)$, $A(3k+2)\geqslant (k+1)(3k+1)$. Equally, we can write $A(n)\geqslant \left[\frac{n^2}3\right]= $ A000212 $(n)$. The oeis entry contains several interesting comments but nothing which immediately applies here. For a given $n$, there can be other Steiner trees of same length with more than one branching point which yield the same area. It might be an interesting question how many non-isomorphic ones exist, but before that, I want to ask the following: Is this bound sharp for all $n$? Are there optimal solutions which are not Steiner trees? (EDIT: in fact, for $n=2,3,4$ taking $n$ consecutive edges of a regular hexagon yields also optimal solutions, which I would however consider marginal.) This is somewhat converse to the problem of Steiner minimal trees for convex polygons but not exactly. The problem, the above construction and both questions can be immediately generalized to trees spanning volumes in $\mathbb R^d$ instead of $\mathbb R^2$. Conjecturally the answers are the same as for the plane. REPLY [5 votes]: One may ask an analogous continuous problem: Which connected set composed of simple arcs of total length $1$ has the largest convex hull? If my computations are correct, the star formed by three segments of length $1/3$ and forming the angles $2\pi/3$ gives a triangle of area $\frac{1}{4\sqrt{3}} \sim \frac{1}{6.93}$, whereas a half-circle of length $1$ gives a half-disc of area $\frac{1}{2\pi} \sim \frac{1}{6.28}$, which is larger. This implies that for large enough $n$, the Steiner tree with one branching point is not optimal. Edit: I just found that this problem has been solved in the special case that the connected set is a curve; see the following question: Largest convex hull of a unit length path The three-dimensional case is discussed here: Largest possible volume of the convex hull of a curve of unit length<|endoftext|> TITLE: Strong divisibility of Lucas sequences QUESTION [6 upvotes]: Let $a$ and $b$ be relatively prime integers and let $u_n$ be their associate Lucas sequence, i.e., the second order linear recurrence sequence satisfying $u_0 = 0$, $u_1 = 1$ and $u_{n+2} = au_{n+1} + bu_n$, for each nonnegative integer $n$. It is well know that $(u_n)_{n=0}^\infty$ is a strong divisibility sequence, i.e., it holds $$(\bullet) \quad \gcd(u_m, u_n) = u_{\gcd(m,n)} ,$$ for all the integers $m,n \geq 0$ (put $\gcd(0,0) := 0$). This in turn implies that $$(\star) \quad m \mid n \Rightarrow u_m \mid u_n ,$$ for all the integers $m,n \geq 0$. My question is: Are there some nice hypothesis under which also the reverse implication holds in ($\star$) ? Note that from ($\bullet$), we get $$ u_m \mid u_n \Rightarrow u_m = \gcd(u_m, u_n) = u_{\gcd(m,n)} ,$$ so if $(u_n)_{n=0}^\infty$ is injective then $m = \gcd(m,n)$ and thus $m \mid n$. So my question can be also answered if one gets some nice hypothesis under which $(u_n)_{n=0}^\infty$ is injective. Thank you in advance for any suggestion. REPLY [4 votes]: I found a method to solve this problem. We recall the primitive prime factor theorem Theorem 2.3.1 (Florian Luca, Effective methods for diophantine equations) If $k \notin \{1,2,3,4,6\}$, then $u_k$ has a primitive prime factor except when $(a,\Delta,k)$, where $\Delta = a^2 + 4b$, is one of the following triples: $$(1, 5, 5), (1, -7, 5), (2, -40, 5), (1, -11,5), (1, -15, 5), (12, -76, 5), (12, -1364, 5),$$ $$(1, -7, 7), (1, -19, 7),$$ $$(2, -24, 8), (1, -7, 8),$$ $$(2,-8,10), (5, -3, 10),$$ $$(1, 5, 12), (1, -7, 12), (1, -11, 12), (2, -56, 12), (1, -15, 12), (1, -19, 12),$$ $$(1, -7, 13),$$ $$(1, -7, 18),$$ $$(1, -7, 30).$$ Now, let $m$ and $n$ be positive integers such that $u_m \mid u_n$ but $m \nmid n$. From $(\bullet)$ we have that $$u_m = \gcd(u_m, u_n) = u_d ,$$ where $d = \gcd(m,n) < m$. It follows that $u_m$ has not a primitive prime factor. From Theorem 2.3.1 then or $m \in \{1,2,3,4,6\}$ or $(a,\Delta,m)$ is one of the triples listed above. In each of these cases, we can (patiently) check if actually there exist or not a divisor $d$ of $m$ such that $u_m = u_d$.<|endoftext|> TITLE: Square filling self avoiding walk QUESTION [11 upvotes]: I want to create an algorithm that fills a square grid with a random Hamiltonian path starting at a particular point. See this example. One approach is to try a free direction as a next step, and then validate whether it is still possible to complete the current path to visit each square exactly once. This step will be undone if the extension is impossible and one of the other free directions will be tried. How can we determine if it is possible to extend the path to a Hamiltonian path? Here is an example where no simple invariant seems to detect the lack of a Hamiltonian extension: We are at cell 25 and we have three possibilities: 17, 24 and 33. The path will eventually fail if you go to cell 17. (In the linked page, you can mark the white cells by clicking on them, if you want to try things out). REPLY [5 votes]: The following paper by Umans and Lenhart gives a polynomial-time algorithm for finding a Hamiltonian cycle in "solid" grid graphs (grid graphs with no holes with area larger than $1$): http://users.cms.caltech.edu/~umans/papers/LU97.ps For general grid graphs, the problem is NP-complete. Even though they search for cycles and not paths, the algorithm might be useful, since a complement of a path in a grid is either disconnected (which is easy to detect) or has at most one large hole.<|endoftext|> TITLE: Are the Szpiro ratios of 37b1 over certain number fields {33,39,42,48,51,66}? QUESTION [5 upvotes]: Related to this question. According to Hindry p.7 Conj 3.1 and Stein's notes Szpiro's conjecture over number fields states that the Szpiro ratio is: $$ \sigma_{E/K}=\frac{\log{|N_{K/Q}\Delta_{E/K}|}}{\log{|N_{K/Q} f_{E/K}}|}$$ Given $ \varepsilon >0$ there are only finitely many $ E/K$ with $ \sigma_{E/K}\geq 6+\varepsilon $. In particular, $ \sigma_{E/K}$ is bounded. $\Delta_{E/K}$ is the minimal discriminant and $f_{E/K}$ is the conductor. To my knowledge over the rationals the largest known Szpiro ratio is about $9.01$. Consider the elliptic curve with cremona label 37b1 and ainvariants $[0, 1, 1, -23, -50]$. Over the rationals the discriminant is $37^3$. Over the number field with defining polynomial $x^{22} - 37^3$ the norm of the minimal discriminant is $37^{66}$ and the norm of the conductor is $37$ while the global minimal model is the same as over the rationals, giving Szpiro ratio $66$. If this pattern continue for number fields of the form $x^n - 37^3$, the Szpiro ratio probably will be unbounded. Computation with sage, possibly assuming GRH since the number fields are not certified suggest the following large Szpiro ratios {33,39,42,48,51,66}: a_i= [0, 1, 1, -23, -50] ratio= 33.00000 Delta= 37^33 f= 37 d_E/Q= 37^3 global minimal model= Elliptic Curve defined by y^2 + y = x^3 + x^2 + (-23)*x + (-50) over Number Field in w with defining polynomial x^11 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 39.00000 Delta= 37^39 f= 37 d_E/Q= 37^3 global minimal model= Elliptic Curve defined by y^2 + y = x^3 + x^2 + (-23)*x + (-50) over Number Field in w with defining polynomial x^13 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 42.00000 Delta= 37^42 f= 37 d_E/Q= 37^3 global minimal model= Elliptic Curve defined by y^2 + y = x^3 + x^2 + (-23)*x + (-50) over Number Field in w with defining polynomial x^14 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 48.00000 Delta= 37^48 f= 37 d_E/Q= 37^3 global minimal model= Elliptic Curve defined by y^2 + y = x^3 + x^2 + (-23)*x + (-50) over Number Field in w with defining polynomial x^16 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 51.00000 Delta= 37^51 f= 37 d_E/Q= 37^3 global minimal model= Elliptic Curve defined by y^2 + y = x^3 + x^2 + (-23)*x + (-50) over Number Field in w with defining polynomial x^17 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 66.00000 Delta= 37^66 f= 37 d_E/Q= 37^3 global minimal model= Elliptic Curve defined by y^2 + y = x^3 + x^2 + (-23)*x + (-50) over Number Field in w with defining polynomial x^22 - 50653 Q1 Are the computations correct? Q2 Does increasing the degree of the number field continue to increase the ratio? Q3 (New) Is this a counterexample this formulation of Szpiro's conjecture? My concern is bounded, not infinitely many? Working will large degree fields is not fast for me. Small search found 46 ratios greater than $24$. The largest is $134.6199$ from ainvariants $[1, 1, 1, -110, -880]$ a_i= (1, 1, 1, -110, -880) ratio= 134.6199 Delta= -1 * 3^304 * 5^19 f= 3 * 5 d_E/Q= -1 * 3^16 * 5 global minimal model= Elliptic Curve defined by y^2 + x*y + y = x^3 + x^2 + (-110)*x + (-880) over Number Field in w with defining polynomial x^19 + 215233605 Added Addressing Jamie Weigandt answer. With his code and possibly assuming GRH, I get the same ratio. To assume GRH, replace K. = NumberField(x^11 - 50653) with K. = NumberField(x^11 - 50653,check=False). This avoids certification of the NF with pari's bnfcertify. Here is the output of Nf.=NumberField(x**D-d) Nf.=NumberField(x**D-d,check=False) #assume GRH print 'a_i=',ai,' ratio=',szpiro_ratio(E2),Nf a_i= [0, 1, 1, -23, -50] ratio= 30.0 Number Field in w with defining polynomial x^10 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 33.0 Number Field in w with defining polynomial x^11 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 39.0 Number Field in w with defining polynomial x^13 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 42.0 Number Field in w with defining polynomial x^14 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 48.0 Number Field in w with defining polynomial x^16 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 51.0 Number Field in w with defining polynomial x^17 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 57.0 Number Field in w with defining polynomial x^19 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 60.0 Number Field in w with defining polynomial x^20 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 66.0 Number Field in w with defining polynomial x^22 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 69.0 Number Field in w with defining polynomial x^23 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 75.0 Number Field in w with defining polynomial x^25 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 78.0 Number Field in w with defining polynomial x^26 - 50653 a_i= [0, 1, 1, -23, -50] ratio= 84.0 Number Field in w with defining polynomial x^28 - 50653 According to pari's documentation, the comment: #Don't do this. It will take forever probably will mean GRH fails for this NF, since in this case pari's bnfcertify doesn't return. Regarding his comment about Q2 and 32a1: 32a1 has $a_6=0$, so the ratio degrades for trivial reasons. I didn't claim high ratio for it. The claim that the ratio is bounded over number fields is from here and William Stein' notes Similar definition of the Szpiro's conjecture is in Silverman's The Arithmetic of Elliptic Curves, p. 275. With the above notation, there is constant $k$, depending on $\epsilon$ and $K$ s.t.: $$ N_{K/Q} \Delta_{E/K} \le k(N_{K/Q} f_{E/K})^{6+ \epsilon}\qquad (2) $$ Silverman asks "It is very interesting to ask how the constants $k$ appearing in these conjectures depend on the field $K$." In analogy with the uniform abc conjecture, probably Silverman should take $k$ the discriminant of $K$. REPLY [7 votes]: The correct discriminant ideal to compute is a locally defined ideal $$\mathcal D_{E/K}^\text{min} = \prod_{\frak {p}} {\frak{p}}^{v_\frak{p}}$$ where $v_{\frak p}$ is the valuation of the discriminant of a local minimal model of $E$ at $p$. For general number fields, global minimal models do not exist. They exist if and only if the class number is 1. This may or may not bring down your Szpiro ratios. If it doesn't, I still don't think this is inconsistent with Szpiro's conjecture since the finite set of counterexamples depends both on $\epsilon$ and the number field $K$. For every number field $K$ there is a finite (and sometimes non-empty) set of elliptic curves over $K$ with everywhere good reduction. These curves have conductor ideal $(1)$ so the Szpiro ratio can arguably be taken to be $\infty$. I can't say whether your discriminant calculations are correct because I'm not sure how exactly you computed them, and working in these number fields is a bit too hard for me to do the calculations myself today. Here's code to compute the honest Szpiro ratio: def szpiro_ratio(E): #Only works over number fields. N = E.conductor() nN = N.norm() S = N.prime_factors() local_norms =[p.norm()^E.local_minimal_model(p).discriminant().ord(p) for p in S] nD = prod(local_norms) return log(RDF(abs(nD)))/log(RDF(abs(nN))) To do these calculations, I would try: sage: E = EllipticCurve('37b1') sage: K. = NumberField(x^11 - 50653) sage: EK = E.base_extend(K) sage: szpiro_ratio(EK) #Don't do this. It will take forever. Then I would give up after 10 minutes because it would be clear that this is not a feasible approach for such high degree number fields. There might be a way around this difficulty with high degree number fields. I'd look at the code for local_minimal_model?? and try doing it by hand for primes in $K$ above $37$ to understand what's going it. Here's an example where the code does work, and gives a negative answer to Q2. sage: E = EllipticCurve('32a1').quadratic_twist(97) sage: K. = QuadraticField(97) sage: L. = NumberField(x^4 - 97) sage: szpiro_ratio(E.base_extend(K)) 2.4 sage: szpiro_ratio(E.base_extend(L)) 2.25 EDIT: (Fixed some inaccuracies, and added a more reasonable explanation for the $2-2g$ term. Look at the Shioda-Tate formula: $$ N = \frac{D}{6} + 2 - 2g + r + (h^{1,1} - \rho).$$ This is the elliptic surface analog of Szpiro's conjecture. Here $N$ is the conductor degree, $D$ is the discriminant degree, $r$ is the Mordell-Weil rank, $(h^{1,1} - \rho)$ is a non-negative number we don't need to worry about. We do have to worry about $2-2g$ where $g$ is the genus of the base curve. There are two things this could be reflecting in the number field case. (1) The negative $2g$ could be accounting for the difference between the minimal discriminant ideal and the discriminant ideal attached to a global model, since it measures the size of the Picard group of the base. (2) (2-2g) could be correspond to something like $(1 + \epsilon)\log|\text{Disc}_{K/\Bbb Q}|$. Maybe both are related to the degree of some canonical divisor. I suppose this would be an Arakelov divisor in the number field case. I think the second interpretation seems more reasonable, and agrees with what you're trying to say.<|endoftext|> TITLE: Fast computation of a Groebner basis. What is possible? QUESTION [18 upvotes]: I need to compute a Groebner basis of 18 polynomials in 19 variables the terms of which have degree at most 3. My aim is to exploit a symmetry in a PDE problem and I am not an expert in algebra or computer algebra. Mathematica is now running for 10 days without answer. A similar question description here was asked a couple of years ago but because there was no satisfying answer I raise this question again. I would appreciate if anyone could tell help me with the following two questions if there is any hope to compute the Groebner basis? if so are there fast at best parallel ready to use programms for computing the Groebner basis? Any answers, comments and reference are appreciated. REPLY [2 votes]: I have been trying to answer that question myself for some time and haven't found anything definitive in the literature. I can only point to the paper where M4GB is presented where there are some benchmarks comparing its performance against OpenF4, Magma, and FGb (Faugere algorithm that is used in Maple) for the case of $m$ polynomials of $n$ variables with $m=2n$ and $m=n+1$. It would be a very nice resource for the community to have an independent benchmark of the different libraries similar to the Mathematical Programming community benchmark service offered by Hans Mittelmann. Exploring a little more, I found this dicussion in the Google group for the Sage developers where there is a comparison of several codes out there to compute Groebner basis. Keep in mind that the performance depends heavily on your ideal.<|endoftext|> TITLE: Separability of the C*-algebra in the definition of K-homology QUESTION [7 upvotes]: There are (at least) two approaches to K-homoology: one is via the so called dual algebra which is due to Paschke. The second is via the Fredholm modules and is due to Kasparov. In Nigel Higson's book there is the note that the second approach is more flexible. I wonder what exactly does it mean: to be more precise, in the first treatment there is a assumption that the underlying $C^*$-algebra is separable and as far as I understood some properties and proofs, this assumption is heavily used. Namely, one needs the so called ample representation in order to prove that the dual algebra (essentially) doesn't depend on the representation. When it comes to discussing Fredholm modules, author assumes also that the underlying $C^*$-algebra is separable however he points out that the definition of Fredholm modules can be stated for general $C^*$-algebras. So my question is the following: Question: is it true that one can define $K$-homology groups for all $C^*$-algebras but only via Fredholm modules. So in other words, the dual algebra aproach is not proper to define $K$-homology groups for general $C^*$-algebras? REPLY [7 votes]: The main problem with dual algebras for non-separable C*-algebras is that they need not be functorial. Given representations $\rho_A \colon A \to \mathcal{B}(H_A)$ and $\rho_B \colon B \to \mathcal{B}(H_B)$ and a $*$-homomorphism $\phi \colon A \to B$, one wants to define the induced map $\phi_* \colon \mathcal{D}(B,\rho_B) \to \mathcal{D}(A, \rho_A)$ by $\phi_*(T) = VTV^*$ where $V \colon H_B \to H_A$ is an isometry such that $V^* \rho_A(a) V$ is equal to $\rho_B(\phi(a))$ modulo compact operators. Such isometries are constructed using Voiculescu's theorem, which requires that $A$ is separable. Without functoriality the whole theory pretty much collapses; for instance, one proves that K-homology is independent of the ample representation used to define the dual algebra by arguing that the K-theory map induced by $\phi_* \colon \mathcal{D}(B, \rho_B) \to \mathcal{D}(A, \rho_A)$ is independent of the isometry $V$. For Kasparov's model of K-homology, you don't run into this problem: if $(\rho, H, F)$ is a Fredholm module over $B$ then $(\rho \circ \phi, H, F)$ is a Fredholm module over $A$. Still, there are problems: without Voiculescu's theorem you don't have the excision theorem in either model (at least the standard proof doesn't work) and without excision you don't have long exact sequences. None of this should be too much of a concern, however - most C*-algebras for which K-theory is useful are separable (e.g. $C(X)$ for $X$ compact, crossed products of $C(X)$ by a locally compact group action). The main source of non-separable C*-algebras is the theory of von Neumann algebras, and K-theory/homology doesn't tell you anything about von Neumann algebras anyway.<|endoftext|> TITLE: Minimal number of generators for $GL(n,\mathbb{Z})$ QUESTION [19 upvotes]: $\DeclareMathOperator{\gl}{GL}\DeclareMathOperator{\sl}{SL}$From de la Harpe's book "Topics in Geometric Group Theory" I learnt that $\gl(n,\mathbb{Z})$ is generated by the matrices $$s_1 = \begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 1 & 0 \end{bmatrix}$$ $$s_2 = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 1 \end{bmatrix}$$ $$s_3 = \begin{bmatrix} 1 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 1 \end{bmatrix}$$ $$s_4 = \begin{bmatrix} -1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 1 \end{bmatrix}$$ whenever $n \geq 2$. It can be shown that $s_1, s_3, s_4$ suffice (a result by Hua and Reiner). The matrices $s_1$ and $s_3$ generate $\gl(n,\mathbb{Z})$ when $n$ is even and $\sl(n,\mathbb{Z})$ when $n$ is odd. But what about the odd case? For odd $n$, is $\gl(n,\mathbb{Z})$ generated by two elements? REPLY [7 votes]: Though it's a good exercise to work things out directly (as YCor does), the ideas here are all fairly old and can be extracted without much difficulty from the literature as Dietrich's reference indicates. An old paper of Hua-Reiner, for instance, is freely available online here. To supplement this with more details about generators, there is a discussion over more general rings than $\mathbb{Z}$ (but specialized frequently to $\mathbb{Z}$) in the first five sections of Chapter VII in the textbook by M. Newman, Integral Matrices, Academic Press, 1972.<|endoftext|> TITLE: Ways to show a system of polynomial equations has no solution QUESTION [12 upvotes]: I came across the following system of polynomial equations on $X_1,\dots,X_{m-2}$: $$ \begin{cases} 2X_{2s}+\sum\limits_{t=1}^{2s-1}(-1)^tX_tX_{2s-t}=0,\quad s=1,\dots,\frac{m}{2}-1,\\ X_sX_{m-s}+(-1)^s=0,\quad s=2,\dots,\frac{m}{2},\\ X_1X_{\frac{m}{2}}-2X_{\frac{m}{2}+1}=0, \end{cases} $$ where $m\ge10$ is an even integer. By Groebner basis computation, I verified up to $m=20$ that $1$ is in the ideal generated by the above polynomials in $\mathbb{Q}[X_1,\dots,X_{m-2}]$, whence the system has no solution in $\mathbb{C}^{m-2}$. Also, the number of equations is one more than the number of variables. Thus it is reasonable to conjecture that the system has no solution in $\mathbb{C}^{m-2}$ for any even $m\ge10$, and so the question arises naturally as how to prove this. Remark: For a given value (not too lagre) of $m$, there are ways such as Groebner basis and various kinds of multipolynomial resultants to show that the system has no solution. But it seems to me that these algorithmic ways give not much insight for general $m$. A further question: If we have known that the system has no solution in $\mathbb{C}^{m-2}$, we immediately deduce that it has no solution in $\mathbb{F}_p^{m-2}$ for all sufficiently large prime $p$. However, how can I then get a bound $N$ such that the system has no solution in $\mathbb{F}_p^{m-2}$ for prime $p>N$? I guess a bound like $p>m$ holds, but achieving this would be ad hoc to the equations. REPLY [4 votes]: For the Further question, See Pascal Koiran's paper (which, I believe, is the last word in the subject, even if almost 20 years old).<|endoftext|> TITLE: A perturbation question for the intersection of C*-subalgebras QUESTION [5 upvotes]: This feels like something I may have asked before (in which case, apologies) and it also might be some kind of "standard counterexample in a book on C* algebras". Let M be a unital C*-algebra and let A, B be unital, closed star-subalgebras (so they are C*-algebras in their own right). Let q be the quotient map of Banach spaces from M onto M / B. Is q(A) necessarily closed in M / B ? (Of course if B were an ideal then the answer is yes, because any star-homomorphism between C*-algebras has closed range.) If it makes any difference, I'm primarily interested in the case where M is a von Neumann algebra and A and B are sub- von Neumann algebras. REPLY [6 votes]: If $q(A)$ is closed in $M/B$, then its preimage $A+B$ is closed in $M$. Here is an example when it is not: Let $M=C([0,1],M_2)$, let $p,q\in M$ be the rank one projections $$ p= \begin{pmatrix} 1 & 0\\ 0&0 \end{pmatrix}, q(t)= \begin{pmatrix} 1-t & t^{1/2}(1-t)^{1/2}\\ t^{1/2}(1-t)^{1/2} & t \end{pmatrix}. $$ Let $A$ be the elements of the form $fp$, with $f\in C[0,1]$ and $B$ the element of the form $gq$, with $g\in C[0,1]$ (i.e., $A=pMp$, $B=qMq$). Then the elements of $A+B$ have the form $$ fp+gq= \begin{pmatrix} f+g\cdot (1-t) & gt^{1/2}(1-t)^{1/2}\\ gt^{1/2}(1-t)^{1/2} & gt \end{pmatrix}. $$ Observe that the rate of decay near zero of the bottom right corner is at least linear. So choosing functions $g_n\in C[0,1]$ such that $g_nt^{1/2}\to t^{1/4}$ and $f_n=-g_n\cdot (1-t)$ we get the matrix $$ \begin{pmatrix} 0 & t^{1/4}(1-t)^{1/2}\\ t^{1/4}(1-t)^{1/2} & t^{3/4}, \end{pmatrix} $$ in the closure but not in the algebraic sum. The subalgebras $A$ and $B$ don't share $M$'s unit but this can be fixed by adding the unit to them: the elements of $A+B+\lambda 1_2$ have $\lambda+gt$ in the bottom right corner, so the same matrix as before does not belong to it.<|endoftext|> TITLE: Stalks of étale sheaves QUESTION [5 upvotes]: I want to prove that $0 \to F\to G\to H \to 0$ is an exact sequence of étale sheaves. I understand that it is enough to show that $0\to F_{\bar{x}}\to G_{\bar{x}}\to H_{\bar{x}}\to 0$ is exact at every geometric point $\bar{x}\to X$. Why is it then enough to show that $0\to F(U)\to G(U)\to H(U)\to 0$ is exact for every strictly Hensel local scheme $U$? I ask this because I don't understand the proof of proposition 6.12 of Voevodsky's "Lecture notes on motivic cohomology". Specifically I'm trying to understand the first two paragraphs here http://www2.maths.ox.ac.uk/cmi/library/MotivicCohomologyold.pdf#page=52 and use that argument to understand the proof of theorem 7.20 ( Suslin's rigidity theorem ) here http://www2.maths.ox.ac.uk/cmi/library/MotivicCohomologyold.pdf#page=63. It seems the arguments are quite similar in that the book states that it is enough to show an isomorphism on stalks and then goes to show said isomorphism on the sheaf evaluated on some henselian schemes. Is this correct? REPLY [5 votes]: Your statement is correct, in both cases, the exactness of the sequence is established by looking at the stalks. Strictly henselian local rings show up because the stalk of the structure sheaf on the étale site at a geometric point is the strict henselization of the corresponding local ring, cf. Section 4 of Milne's lecture notes on étale cohomology, or EGA IV. To identify stalks of the sheaf with sections over the local ring, you need some condition - the limit needs to commute with the functor as mentioned in anon's comment. (This can be seen as a continuity condition). In both cases mentioned in the question, continuity works: in the case of Prop 6.12 of the lecture notes on motivic cohomology, $F=\mathbb{Z}_{\operatorname{tr}}(T)$ for $T$ a scheme of finite type. Because of the finite type assumption, every correspondence $\operatorname{Spec}\mathcal{O}_{X,x}^{\operatorname{sh}}\to T$ is in fact represented by a correspondence $U\to T$ where $U$ is an étale neighbourhood of $x$ in $X$; therefore the stalk of $\mathbb{Z}_{\operatorname{tr}}(T)$ at $x$ can be computed as $\mathbb{Z}_{\operatorname{tr}}(T)(\operatorname{Spec}\mathcal{O}_{X,x}^{\operatorname{sh}})$. (See also Exercise 1.13 in the lecture notes.) in the argument for the rigidity theorem, the functors that are being considered (i.e. presheaves with transfers) live on a category of schemes of finite type - in some sense, there is actually no definition of $F(\operatorname{Spec}\mathcal{O}_{X,x}^{\operatorname{sh}})$. So the functors are extended to local rings (or more general, essentially finite type schemes) by the definition $F(\operatorname{Spec}\mathcal{O}_{X,x}^{\operatorname{sh}})=\operatorname{colim}F(U)$ where $U$ runs over all étale neighbourhoods of $x$ in $X$. The necessary continuity problem is defined away... (As a side note: this is not possible in the original application of the rigidity theorem to K-theory because K-theory is defined for all rings. Fortunately, K-theory commutes with directed colimits as in the definition of the local rings, so the continuity requirement is again met.)<|endoftext|> TITLE: Did differential geometry undergo a notation change? QUESTION [16 upvotes]: As a graduate student, I found the old books of differential geometry used a different set of notation from modern textbooks. For example, Chern and Milnor defined the curvature 2-form by $d\omega-\omega\wedge \omega$, because they assumed that $Ds_{\alpha}=\sum_{i, \beta}\Gamma^{\beta}_{\alpha i}du_{i}\otimes s_{\beta}, \omega^{\beta}_{\alpha}=\sum_{i}\Gamma^{\beta}_{ai}du^{i}$, hence their connection matrix $\omega$ is the transpose of what we used nowadays. In today's notation, we have $\Omega=d\omega+\omega\wedge \omega$ instead. This notation transition is not so difficult, until one actually encounters calculations using the two sets of notation, and the "translation process" is not easy. For example, the torsion free condition $\nabla_{X}-\nabla_{Y}-\nabla_{[X,Y]}=0$ is equivalent to $d\theta^{i}-\theta^{j}\wedge \theta^{i}_{j}=0$ using a local coframe field and connection matrix $\theta^{i}_{j}$. While this may be well-known to experts in the field, it certainly caused confusion to a new guy like me when I tried to read old papers (for example, Chern's paper on Chern-Gauss-Bonnet theorem). When I was learning differential geometry, "modern style" textbooks (De Carmo, Taubes, John Lee, Jurgen Jost, etc.) seem contented not to introduce the old notation. This problem can be solved by extensive googling around and reading old-style textbooks (like Chern's Lectures on differential geometry), but it takes some time at least. I want to ask historically, what is the reason for this notation change? If I am not mistaken, this is a change of notation as well as a change of philosophy. In Chern's textbook there are a lot of explicit complicated computations to prove a trivial result (like connection exists on any manifold), and one gets the feeling that differential geometry is close to some kind tensor analysis. But I doubt if any modern reader (say, of John Lee's book) will feel the same way. What has happened since 60s-70s? Reference: Milnor: Morse Theory, Characteristic Classes (appendix) Chern: Lectures on Differential Geometry, Chapter 4-5. REPLY [13 votes]: I don't know of a systematic change in notation, but Chern definitely used a different convention than what most other differential geometers did. The decision stems from homogeneous spaces and deciding whether you want to work with left or right cosets. Most people use left cosets, which has the consequence that the action of the isotropy group on the fiber on the principal bundle of orthonormal frames is a right action. Chern chose to do it the other way around (just as Herstein does in his algebra text). It should be noted that there is no standard notation used by everybody in differential geometry. We all have our individual preferences. My experience is that no matter what you do, you run into situations where your choice of notation doesn't work so well. This is probably a false memory, but I remember my advisor asking me if I had created my own notation yet. Someone who has really given this a lot of thought and designed an idiosyncratic but quite versatile notation is Roger Penrose. My advice is to set up your own preferred conventions and learn to convert whatever notation is being used in a paper or text into yours. It is also good to be comfortable with working with either vector fields directly (which for me is easier to interpret geometrically) or differential forms (where calculations are often a lot easier). And although it is usually better to work with respect to an orthonormal frame of tangent vectors and the corresponding dual frame, it is also useful to be to work with arbitrary frames. And, if you want to do PDE estimates, you usually need to convert everything into local co-ordinates. ADDED: In terms of explicit formulas for a matrix group, there are two possible choices for the Maurer-Cartan form, either $A^{-1}\,dA$ or $dA\,A^{-1}$, depending on whether you want to work with left or right invariant vector fields. Again, most people use left invariant vector fields and set $\Omega = A^{-1}\,dA$, so the Maurer-Cartan equations are $$ d\Omega + \Omega\wedge\Omega = 0. $$ If instead you define $\Omega = dA\,A^{-1}$, you get $$ d\Omega - \Omega\wedge\Omega = 0. $$<|endoftext|> TITLE: Tori in Compact Riemannian Symmetric Spaces QUESTION [6 upvotes]: The closure of a 1-parameter subgroup in a compact Lie group is a torus. To what extent does this result generalize to compact Riemannian symmetric spaces? In other words, is the closure of a geodesic in a compact Riemannian symmetric space necessarily a flat totally geodesic submanifold? REPLY [6 votes]: I think that this is treated in Helgason's Differential Geometry, Lie Groups and Symmetric Spaces. The point is that, for any Riemannian symmetric space $G/K$, one has the notion of the rank $r$ of the symmetric space, which is the dimension $r$ of a maximal abelian subspace $\frak{a}$ in $\frak{k}^\perp\subset \frak{g}$, where $\frak{k}\subset\frak{g}$ are the respective Lie algebras of $K\subset G$. The exponential of $\frak{a}$ gives a torus that then maps down to a totally geodesic torus in $G/K$. (The images of this torus under the action of $G$ are the so-called 'flats' in $G/K$.) Then, because the $K$-orbit of $\frak{a}$ in $\frak{k}^\perp$ contains everything in $\frak{k}^\perp$, it follows that every geodesic in $G/K$ is a geodesic in such a 'flat', and hence its closure is itself a closed geodesic sub-torus in a flat. The case of a compact Lie group $K$ is the special case in which $G = K\times K$ and we regard $K$ as embedded diagonally in $G$.<|endoftext|> TITLE: co-dimension one minimizing verifolds QUESTION [12 upvotes]: It is known that a minimizing co-dimension verifolds within a manifold may need to be singular. I think a famous example first partially analyzed by Jim Simons is the cone on in the 8-ball of the product of two 3-spheres embedded in S^7, where the 3-spheres are the quaternions of norm root(2)/2 in each factor. All submanifolds with that boundary condition turn out to have more 7-volume than the cone. My question is: Is this phenomena stable to perturbation of metric? That is, is it always possible to perturb the ambient metric so that the co-dimension minimizer is a submanifold? REPLY [9 votes]: I think that the phenomena is not stable under perturbation of the metric, i.e., a small perturbation can cause the minimizer to be smooth. First, let me explain how it is not stable under perturbation of the "link." If you have a smooth submanifold $\Gamma^{n-1} \hookrightarrow \mathbb{S}^n \hookrightarrow \mathbb{R}^{n+1}$ so that the cone over $\Gamma$, $C_1(\Gamma)=\{tx : x\in \Gamma,t\in [0,1]\}$, is area-minimizing, then there exist arbitrarily small perturbations $\Gamma_\epsilon \hookrightarrow \mathbb{S}^{n}$ of $\Gamma$ so that the solution to the Plateau problem (i.e., the area minimizer with boundary $\Gamma_\epsilon$) for $\Gamma_\epsilon$ is regular. In general, this is a consequence of the work of Hardt and Simon, who showed that if $C(\Gamma)=\{tx : x\in\Gamma, t\geq 0\}$ is area minimizing (this is equivalent to $C_1(\Gamma)$ solving the Plateau problem for $\Gamma$), then $\mathbb{R}^{n+1}\setminus C(\Gamma)$ is foliated by smooth, area minimizing hypersurfaces which are asymptotic to $C(\Gamma)$ at infinity. For the Simons' cone you mentioned, this fact was proven earlier by Bombieri, De Giorgi, and Giusti as part of their proof that the Simons' cone is area minimizing. Now, because these smooth surfaces foliate $\mathbb{R}^{n+1}\setminus C(\Gamma)$, and are asymptotic to $C(\Gamma)$, we can find one, say $\Sigma_\epsilon$, which intersects $\mathbb{S}^n$ in a (smooth) surface, say $\Gamma_\epsilon$ which is arbitrarily close to $\Gamma$. Because $\Sigma_\epsilon$ is area minimizing, $\Sigma_\epsilon\cap B_1(0)$ must be the solution to Plateau's problem. Moreover, it is the unique solution thanks to the fact that the $\Sigma_\epsilon$'s form a foliation (e.g., by the maximum principle or a calibration argument). Now, via the above facts, I claim we can construct a small perturbation of the Euclidean metric so that $\Gamma$ bounds a unique area minimizing surface which is smooth. To do so, simply construct a diffeomorphism $\phi:\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$ so that $\phi(\Gamma_\epsilon)=\Gamma$. Clearly we can arrange that $\phi$ is close to the identity in whatever sense we want. Now, for the metric $\phi^*\delta$, the solution to the Platau problem for $\Gamma$ is unique and smooth, because it is $\phi(\Sigma_\epsilon)$. EDIT: I realize that I should have referenced the following paper of N. Smale http://www.ams.org/mathscinet-getitem?mr=1243523 "Generic regularity of homologically area minimizing hypersurfaces in eight-dimensional manifolds." where he uses this sort of argument to prove that a $7$-dimensional minimizer in an $8$-manifold will be smooth for generic metrics. I should also point out (as Smale does in the article) that this is very much unresolved in higher dimensions. This has a lot to do with the poorly understood nature of the structure of the singular set of minimal hypersurfaces in higher dimensions.<|endoftext|> TITLE: curvature flow for loops in S^2 QUESTION [12 upvotes]: Consider the unit 2-sphere and a smooth simple closed curve c embedded in it. I would guess that under the well studied parabolic equation which evolves the curve according to its curvature vector, that c shrinks through embedded sccs to a "round" point iff it bounds disks in S^2 of unequal area, and the "side" it shrinks to (which given the shrink can be unambiguously be identified as the "inside" or the "outside") is the one of smaller area. If the two sides have the same area, I would similarly guess that the curve shrinks to an equator. Does anyone know if these guesses are correct? REPLY [8 votes]: Actually, this result seems to be due to Mike Gage, and is so attributed by Grayson: @article {MR1046497, AUTHOR = {Gage, Michael E.}, TITLE = {Curve shortening on surfaces}, JOURNAL = {Ann. Sci. \'Ecole Norm. Sup. (4)}, FJOURNAL = {Annales Scientifiques de l'\'Ecole Normale Sup\'erieure. Quatri`eme S\'erie}, VOLUME = {23}, YEAR = {1990}, NUMBER = {2}, PAGES = {229--256}, ISSN = {0012-9593}, CODEN = {ASENAH}, MRCLASS = {53C22 (35K55 58G11)}, MRNUMBER = {1046497 (91a:53072)}, MRREVIEWER = {Dennis M. DeTurck}, URL = {http://www.numdam.org/item?id=ASENS_1990_4_23_2_229_0}, }<|endoftext|> TITLE: Did Leibniz really get the Leibniz rule wrong? QUESTION [41 upvotes]: A couple of posts ([1], [2]) on matheducators.SE seem to suggest that Leibniz originally got the wrong form for the product rule, perhaps thinking that $(fg)'=f'g'$. Is there any actual historical evidence for this? It seems particularly hard to believe that he would have made the hypothesis $(fg)'=f'g'$. We would then have $x'=(1x)'=(1')(x')=0$. And presumably anyone inventing calculus would take $(x^2)'$ to be a prototypical problem, and would realize pretty early on that $(x^2)'\ne (x')(x')=1$. It's also pretty trivial to disprove this conjecture based on dimensional analysis or scaling. There is some discussion on this Wikipedia talk page, with some sources cited, but it appears to be inconclusive. REPLY [49 votes]: In the manuscript "Determinationum progressio in infinitum" (pp. 668-675 of Sämtliche Schriften und Briefe, Reihe VII, Band 3, Teil C, available in pdf here), Leibniz writes on p. 673 (with "$\sqcap$" in place of "$=$"): $$ \odot = \overline{dt}\int\frac{a^2}{a^2 + t^2}. \quad\text{Hence}\quad \overline{d\odot} = \frac{a^2}{a^2 + t^2}\overline{d\overline{dt}} $$ This amounts to asserting that $d[uv] = dv\,du$ where $u=dt$ and $v=\int\frac{a^2}{a^2+t^2}$; and thus differentiating the product wrong, as the editors comment in footnote 14. On p. 668 they take this as grounds to date the manuscript early November 1675, since by November 11 he was pointing out this error (in "Methodi tangentium inversae exempla", quoted by Edwards in KConrad's comment above). Addendum: The first time Leibniz gets his general rule right appears to be in "Pro methodo tangentium inversa et aliis tetragonisticis specimina et inventa" (dated 27 November 1675; pp. 361-371 of the same Sämtliche Schriften, Reihe VII, Band 5, Teil B; English translation here), where he writes on p. 365: Therefore $d\overline xy = d\overline{xy}-xd\overline y$. Now this is a really noteworthy theorem and a general one for all curves.<|endoftext|> TITLE: Contractibility of space of embeddings of a disc QUESTION [8 upvotes]: I'm pretty sure that both of the following spaces are contractible. However, I can't seem to find a proof or a reference. Can anyone provide one? Let $D^2$ be the unit disc in $\mathbb{R}^2$. The space of smooth embeddings $f\colon D^2 \rightarrow \mathbb{R}^2$ such that $f(0)=0$ and $(Df)_0 = \text{id}$, with the $C^{\infty}$ topology. The space of topological embeddings $f\colon D^2 \rightarrow \mathbb{R}^2$ such that $f(0)=0$, with the compact-open topology. REPLY [10 votes]: For details I recommend looking at the papers of Yagasaki on arXiv especially the paper Homotopy types of homeomorphism groups of noncompact 2-manifolds, Topology and its Applications 108 Issue 2 (2000) 123-136, https://doi.org/10.1016/S0166-8641(99)00130-3, arXiv:math/0010223. Let me answer 1. Consider be the diffeomorphism group of $\mathbb R^2$ that fixes $0$ and acts as the identity at the tangent space to $0$. The group acts on your space of embeddings by postcomposition. By the parametrized isotopy extension theorem the action is transitive on each path-component and the orbit map is a fiber bundle whose fiber is the subspace of diffeomorphisms that are identity on the inclusion of the disk. The latter space is contractible by the Alexander trick towards infinity. Note that so far the two-dimensionality has not been used and the argument generalizes to the embedding of a compact manifold onto an open manifold obtained by attaching a collar to the compact one. We are left to understand the homotopy type of the above diffeomorphism group. The group is contractible as proved in the last mentioned paper of Yagasaki (or even easier, look at the Anton Petrunin's answer which really proves contractibility of the diffeomorphism group). Thus each component of the embedding space is contractible. In fact, the embedding space is path-connected essentially because any two embedded circles in $\mathbb R^2-\{0\}$ are isotopic via a compactly supported ambient isotopy. The same strategy can be used to find the homotopy type of the space of embeddings in 2. As Allen Hatcher says in his comment, the space is 2 is not contractible, and one way to explain is that the homeomorphism group of $\mathbb R^2$ that fixes $0$ is not contractible. I think the group is homotopy equivalent to $O(2)$. See the paper above of Yagasaki.<|endoftext|> TITLE: On a weak tree property for inaccessible cardinals QUESTION [8 upvotes]: Suppose that $\kappa$ is inaccessible and consider a tree of height $\kappa$ whose levels have size strictly below some cardinal $\gamma < \kappa$. Does this type of tree always have a $\kappa$-branch? The answer is known to be positive when $\gamma=\omega$, i.e., when the levels are finite. On the other hand, if the levels are only known to have size strictly below $\kappa$ itself, then the answer is negative in general since the existence of a $\kappa$-branch implies that $\kappa$ is weakly compact. I would be interested in knowing about references/proofs on whether this weak tree property holds for every inaccessible or if in fact it constitutes a large cardinal axiom whose strength is somewhere between inaccessibility and weak compactness. Ideally, I would expect a proof of a positive answer in ZFC (one can assume that the inaccessible is given). REPLY [12 votes]: The following theorem of Kurepa, proved in his thesis of 1935, seems to address your question. Theorem Suppose that $\kappa = cf(\kappa) > \gamma$, and $(T, <_{T})$ is a $\kappa$-tree each of whose levels has cardinality less than $\gamma$. Then $(T, <_{T})$ has a cofinal branch. The proof uses the Pressing Down Lemma (Fodor's theorem) and the Pigeonhole Principle with the bound $\gamma$ to find the branch. A good reference on the general area is A. Kanamori, The Higher Infinite, which contains the proof of the above theorem. You might also wish to consult J. Cummings, M. Foreman, The tree property, Advances in Mathematics, 131, 1998, 1-32.<|endoftext|> TITLE: If every nonseparable metric space contains a sequence of subsets with no convergent subsequence, does the Continuum Hypothesis hold? QUESTION [6 upvotes]: If every nonseparable metric space contains a sequence of subsets with no convergent subsequence, does the Continuum Hypothesis hold? The answer is negative, and in the interests of self-contained items, I write Ashutosh's solution below, recast in terms of the splitting number. This question was submerged in the discussion of Fedor Petrov's much-read question whether writing open sets as countable unions of balls implies separability in a metric space. The problem was solved by Joel and Ashutosh. As Timothy Chow observed, Sierpinski proved that if the continuum hypothesis is true, then every nonseparable metric space contains a sequence of subsets with no convergent subsequence (Wacław Sierpiński, "Sur l'inversion du théorème de Bolzano-Weierstrass généralisé," Fund. Math. 34 (1947), 155–156). Sierpinski's proof invokes CH strongly in a classic enumeration argument, given in Timothy's comment. Can the appeal to CH be removed? REPLY [4 votes]: This answer is the proof given by Ashutosh, but formulated in terms of the splitting number. Proposition If the splitting number $s$ is $\aleph_{1}$, then every nonseparable metric space contains a sequence of subsets with no convergent subsequence. Proof: Following Sierpinski, since the metric space $M$ is non-separable, there exist $d > 0$ and a sequence $\{p_\xi\}_{\xi<\omega_{1}}$ of points in $M$ such that $\varrho(p_\xi,p_\eta)\ge d$ for $\xi<\eta<\omega_{1}$, where $\varrho(x,y)$ is the metric on $M$. Let $S$ be a splitting family (for $[\omega]^{\omega}$) of size $\aleph_{1}$, $S = \lbrace s^\xi : \xi < \omega_{1} \rbrace$, where $s^\xi = \langle n_1^\xi,n_2^\xi,n_3^\xi,\ldots\rangle$; for a given $k \in \mathbb{N}$, let $E_k$ be the set of all $p_\xi$ such that $k\in \{n_1^\xi,n_2^\xi,\ldots\}$. The sequence $E_1,E_2,E_3,\ldots$ does not contain any convergent subsequence. For, if $E_{k_1}, E_{k_2},\ldots$ where $k_1 TITLE: Random points on the unit sphere QUESTION [25 upvotes]: Suppose you have $n$ points picked uniformly at random on the surface of $\mathbb{S}^d,$ and let the volume of the convex hull of these points be $V_{n, d}.$ Clearly, $V_{n, d}$ converges to the volume of the unit ball in $\mathbb{R}^{d+1}$ as $n$ goes to infinity, but what is the distribution (or at least the expectation) of the difference? For $d=1,$ it is a simple computation that the $V_{n, 1} - \pi = O(1/n^3)\dots$ (there seem to be a number of questions and references about points in polygons, but I seem to be failing to find anything on spheres...) REPLY [17 votes]: As mentioned in the comments, this question has been answered for random pointes on the boundary of convex bodies and even better for all intrinsic volumes. Let me offer some references: A good refernce is: Matthias Reitzner, Random points on the boundary of smooth convex bodies, Trans. Amer. Math. Soc. 354, 2243-2278, 2002 Abstract: The convex hull of $n$ independent random points chosen on the boundary of a convex body $K \subset \mathbb{R}^d$ according to a given density function is a random polytope. The expectation of its $i$-th intrinsic volume for $i=1, \dots, d$ is investigated. In the case that the boundary of $K$ is sufficiently smooth, asymptotic expansions for these expected intrinsic volumes as $n \to \infty$ are derived. By Ross M. Richardson, Van H. Vu and Lei Wu there are two papers, which are very simlilar: Random inscribing polytopes, European Journal of Combinatorics. Volume 28, Issue 8, Pages 2057–2071, November 2007 and An Inscribing Model for Random Polytopes, Discrete & Computational Geometry, Volume 39, Issue 1-3, pp 469-499, March 2008 With the following abstract: For convex bodies $K$ with $\mathcal{C}^2$ boundary in $\mathbb{R}^d$ , we explore random polytopes with vertices chosen along the boundary of $K$. In particular, we determine asymptotic properties of the volume of these random polytopes. We provide results concerning the variance and higher moments of this functional, as well as an analogous central limit theorem. Another more recent reference is Károly J. Böröczky, Ferenc Fodor, Daniel Hug, Intrinsic volumes of random polytopes with vertices on the boundary of a convex body, Trans. Amer. Math. Soc. 365, 785-809, 2013, arxiv link Let $K$ be a convex body in $\mathbb{R}^d$, let $j\in\{1, ..., d-1\}$, and let $\varrho$ be a positive and continuous probability density function with respect to the $(d-1)$-dimensional Hausdorff measure on the boundary $\partial K$ of $K$. Denote by $K_n$ the convex hull of $n$ points chosen randomly and independently from $\partial K$ according to the probability distribution determined by $\varrho$. For the case when $\partial K$ is a $C^2$ submanifold of $\mathbb{R}^d$ with everywhere positive Gauss curvature, M. Reitzner proved an asymptotic formula for the expectation of the difference of the $j$th intrinsic volumes of $K$ and $K_n$, as $n\to\infty$. In this article, we extend this result to the case when the only condition on $K$ is that a ball rolls freely in $K$.<|endoftext|> TITLE: Variety of nilpotent Lie algebras or $p$-groups QUESTION [13 upvotes]: Here's a couple of analogous questions, one in terms of finite-dimensional complex Lie algebras and one in terms of finite $p$-groups; I'd be interested in an answer to either: 1) Let $\mathcal{L}$ be an isomorphism-closed class of finite-dimensional nilpotent complex Lie algebras. Assume $\mathcal{L}$ is closed under taking finite direct products, subgroups, and quotients. Assume that it does not satisfy any common identity (i.e., any free Lie algebra is residually-$\mathcal{L}$). Does $\mathcal{L}$ necessarily consist of all finite-dimensional nilpotent complex Lie algebras? 2) Let $p$ be prime. Let $\mathcal{C}$ be an isomorphism-closed class of finite $p$-groups. Assume $\mathcal{C}$ is closed under taking finite direct products, subgroups, and quotients. Assume that it does not satisfy any common identity (i.e., any free group is residually-$\mathcal{C}$, or still equivalently the free group on 2 generators is residually-$\mathcal{C}$). Does $\mathcal{C}$ necessarily consist of all finite $p$-groups? (Note: I ask both questions in the positive but I don't particularly expect a positive answer!) REPLY [6 votes]: Here is some idea, it is not very precise. Let $F$ be the free group on two generators and let $F_p$ be its pro-$p$ completion. Let $w$ be an infinite word in $F_p$, i.e., an element of $F_p$ which is not in $F$. Let $W$ be the closed verbal subgroup of $F_p$ generated by $w$. I think that it is possible to choose $w$ so that $W \cap F$ is trivial. For example, the first congruence subgroup of $SL_2(\mathbb{Z}_p)$ ($p>2)$ satisfies a pro-$p$ identity due to Zubkov, but I think it has a dense free group. Let $D_n$ be the $n$-dimension subgroups of $F_p$. Then if I recall correctly there is a canonical way to write $w=w_nu_n$, where $u_n \in D_n$. Take your variety to be all the groups that satisfy $w_n$ for some $n$. I would guess they will satisfy your requirement. I am not sure this idea will work, but if you like it and need more reference, then please contact me.<|endoftext|> TITLE: Nonnegativity conditions for a polynomial in two variables? QUESTION [11 upvotes]: Let $$P(X,Y)= c_{22}X^2Y^2 +c_{21}X^2Y +c_{12}XY^2 +c_{20}X^2 +c_{11}XY+c_{02}Y^2+c_{10}X+c_{01}Y+c_{00}$$ be a polynomial of two variables $X$ and $Y$ with real coefficients $c_{ij}$. What are the necessary and sufficient conditions on the coefficients $c_{ij}$ such that $P(X,Y) \geq 0$ for all pairs $(X,Y)$ with $X\geq 0$, $Y\geq 0$? REPLY [4 votes]: Here's a trick (which sometimes works), but it might be contained in the comments above. As stated, the Newton polyhedron (convex hull of the exponents, the latter viewed as lattice points in ${\bf R}^2$), $K$, of $P$, is contained in the double unit square in ${\bf R}^2$. First, an obvious necessary condition for nonnegativity: the coefficients of the extreme points of $K$ must be positive (as opposed to negative). Next, for each edge of $K$, there is a corresponding polynomial (just take the exponents appearing in the edge, and use the coefficients from $P$); if $P$ is nonnegative, each of these edge polynomials must be nonnegative---but after a change of variables, the edge polynomials are really just polynomials in one variable, for which testing for nonnegativity is just the quadratic formula in your example). This gives some necessary conditions. Now a sufficient condition for positivity: If $P(x,y) >$ on the interior of the positive orthant, and each of the edge polynomials is positive (which implies the coefficients at the extreme points of $K$ are all positive), then for any $Q$ with the same supporting monomials as $P$ and $Q$ having no nonnegative coefficients, there exists an integer $M$ such that the product, $Q^M P$, has no negative coefficients. (This is a generalization of Polya and Meissner's theorems; when $K$ is the double unit square, this is Meissner's theorem.) The simplest (but probably not the best) choice for $Q$ is $\sum x^w$ where $w$ varies over the lattice points in $K$ (where $x^w = x_1^{w(1)}x_2^{w(2)}$ is the usual monomial, with $w = (w(1),w(2))$. Another choice is to replace the negative coefficients in $P$ by their absolute value. In any event, you can try testing $Q^m P$ for large (whatever that means in this context) choices of $m$. And a variation: if you suspect your thing is only nonnegative, add $\epsilon Q = \epsilon \cdot \sum_{w \in K} x^w$ to $P$, where $0 <\epsilon$ is much smaller than all the absolute values of the nonzero coefficients of $P$, and test $Q^m \cdot (P + \epsilon Q)$ for nonnegative coefficients. If succcessful, reduce $\epsilon$. If unsuccessful, increase $\epsilon$; this might give estimates of lower bounds. The problem, as usual, is that it can't give a negative answer, although it might suggest where to look.<|endoftext|> TITLE: Does the Pfaffian have a geometric meaning? QUESTION [34 upvotes]: While reviewing the proof of Gauss-Bonnet in John Lee's book, I noticed the following paragraph: " ...In a certain sense, this might be considered a very satisfactory generalization of Gauss-Bonnet. The only problem with this result is that the relationship between the Pfaffian and sectional curvature is obscure in higher dimensions, so no one seems to have any idea how to interpret the theorem geometrically! For example, it is not even known whether the assumption that $M$ has strictly positive sectional curvatures implies that $\chi(M)>0$.... (page 170) " May I ask if this "interpretation problem" has been resolved? I felt much the same way when I read Milnor's proof of Chern-Gauss-Bonnet using Chern classes, and Chern's statement in his own book using Lipschitz-Killing curvature. Neither has a geometric meaning that is "self-transparent" to me. When I had a class in the index theorem, our proof basically showed Gauss-Bonnet is a special case of Atiyah-Singer using an appropriate Dirac operator. And I do not recall that it involved much geometry, but instead a lot of algebraic manipulations. Chern suggested the following way to look at it in his book: Consider the exterior $2n$-form $$ \Omega=(-1)^{n}\frac{1}{2^{2n}\pi^{n}n!}\delta^{i_1 \cdots i_{2n}}_{1\cdots 2n}\Omega_{i_1i_2}\cdots \Omega_{i_{2n-1}i_{2n}}, \Omega=K d\sigma $$ Then the "key" to prove Chern-Gauss-Bonnet is to represent $\Omega$ on the sphere bundle of $M$ so that one has $\Omega=d\prod$, where $\prod$ is a $2n-1$-form. However, I still do not know how this shed any light on the picturesque side of the equation so that I can visualize it. So I decided to ask. I suppose that this might be one of those topics well known to experts but not written down in introductory level textbooks. Reference: John M. Lee: Riemannian Manifolds, page 170 Chern: Lectures on Differential Geometry, page 171 Milnor & Stasheff: Characteristic Classes, appendix A? For a definition of Pfaffian, see here from wikipedia. REPLY [43 votes]: The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$. Once one establishes this, Chern's proof of the Gauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$, and let $$U = \frac{X}{|X|}:M\setminus\{z_1,\ldots,z_k\}\to \mathsf{S}(M)$$ be the corresponding unit vector field, defined and smooth away from the $z_i$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $U:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $U^*\Omega = U^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, $$ \int_{M_{\epsilon}}U^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} U^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ while the $i$-th summand on the right-hand side converges to the index of $X$ at $z_i$. (This is because $U^*\Pi$ on $\partial B_\epsilon(z_i)$ differs by a term vanishing with $\epsilon$ from the pullback of the unit volume form of $\mathsf{S}_{z_i}(M)$ to $\partial B_\epsilon(z_i)\simeq \mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $U$ at $z_i$, whose degree is, by definition, the index of $X$ at $z_i$.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), one obtains Chern's proof of the Gauss-Bonnet Theorem. As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at Chern's definition of $\Pi$, which uses the Pfaffian, particularly its algebraic properties. This comes out of the computation that Chern does, and it is essentially a geometric fact, but it amounts to an explicit formula for the transgression operator defined in Chern-Weil theory for the Euler class. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question A question on Generalized Gauss-Bonnet Theorem.<|endoftext|> TITLE: The angular distribution of the $(a,b)$ in $p = a^2+b^2$, and the distribution of the lattices corresponding to prime ideals QUESTION [6 upvotes]: Here is a really basic question which I wished I understood better about the primes of the Gaussian field $\mathbb{Z}[i]$. But I was curious about the possibility of generalizing it to other (real quadratic or higher degree) number fields, see below. Consider the primes $p < X$ which split $\mathbb{Q}(i)$, i.e., the ones $p \equiv 1 \mod{4}$. Then $p = a^2+b^2$ in an essentially unique way: one can switch the signs of $a$ or $b$ or interchange them, giving a total of $8$ points $(a,b)$ on the circle $|z| = \sqrt{p}$ in the Gaussian plane. This corresponds to looking at the two primes $(a+ib)$ and $(a-ib)$ above $p$ in $\mathbb{Z}[i]$, and noting that the unit group is just $\{\pm 1, \pm i\}$. Mark all these $8$ points (so that there is no ambiguity), for all the split primes $p < X$. As $X$ grows, how do these marked points distribute in angular sectors? This sounds like Sato-Tate involving a curve $y^2 = x^3-x$ with CM by our Gaussian ring, particularly since $a \pm i b$ is a Weil number of weight one. But so also is its reflection $-i(a \pm ib) = b \mp ia$, and the two are not conjugate, hence would correspond to different elliptic isogeny classes / $\mathbb{F}_p$ in Honda's construction. (And I am confused.) Question: What is the relation, and how to solve in the most direct way possible the elementary problem here about sums of two squares? And the second part of my question, which was also my motivation, concern the possibility of extending this to general number fields $K/\mathbb{Q}$. The prime ideal theorem describes the distribution of the norms of primes of $K$. But these prime ideals sit as lattices in the real vector space $K \otimes_{\mathbb{Q}} \mathbb{R}$, with points representing specific algebraic numbers (elements) from $K$. One could ask about the distribution of other qualities of these lattices than just their covolume (which gives essentially the norm $N (\mathfrak{p})$, as their index in the integer lattice $O_K$). As before, for each prime with $N(\mathfrak{p}) < X$, we could mark the first $\deg{K} = \dim(K \otimes_{\mathbb{Q}} \mathbb{R})$ shortest vectors in the corresponding lattice (if there are repetitions, mark them all), and ask how the marked points distribute as $X \to \infty$. The primes of degree $> 1$ make a negligible contribution, and so for $\mathbb{Q}(i)$ the question reduces to the above elementary problem. Has anything been written on this type of question (the distribution of lattices of prime ideals, rather than just their covolumes), or is this question uninteresting for some reason that I do not see? REPLY [2 votes]: Try chapter 12 in the great book "Prime detecting sieves" by a frequent commentator in mathoverflow, Glyn Harman.<|endoftext|> TITLE: Examples of non isometric surfaces having the same curvature function QUESTION [10 upvotes]: I think it is really natural to believe, after doing Riemannian geometry for a little time, that sectional curvature encodes the all local geometry of a Riemannian manifold. One of the first thing one learns is that having constant sectional curvature implies that you are locally isometric to either the sphere, the euclidian space or the hyperbolic space. The paper http://www.jstor.org/stable/1970580 gives a satisfying answer to the question in dimension higher than 4 (Any diffeomorphism preserving the sectional curvature is an isometry, except in the constant case, which we already know to be locally determined by the curvature). (See also the MO question : Determining a surface in $\mathbb{R}^3$ by its Gaussian curvature for an interesting discussion on the question.) Can anyone give a example of two smooth metric on a surface having the same Gaussian curvature function ? One could say : take two different hyperbolic compact surfaces of the same genus, their curvature functions are both -1, and yet are not globally isometric. But still those surfaces are locally isometric. So can one find two metric $g_1$ and $g_2$ on the disk inducing the same curvature function such that no diffeomorphism of the disk induces an isometry on a neighborhood $(U_1,g_1)$ of $0$ on $(U_2=f(U_1),g_2)$. Which are essentially different is this sense. REPLY [2 votes]: As an example the two surfaces with embeddings $( u \cos v , u \sin v , \log u )$ - surface of revolution with exponential curve as meridian and $ ( u \cos v , u \sin v , v ) $ - helicoid have the same Gauss curvature at corresponding points equal to $-1/(1+u^2)$, even though there is no isometric correspondence at corresponding (u,v) points. Also the helicoid is isometric to the catenoid. $ ( u \cos v , u \sin v , \cosh^{-1} u ) $ page 120 Thus even two surfaces of revolution ( catenoid and exponential horn) share same Gauss curvature without isometrically mapped correspondence at (u,v). Ref page 176 DJ Struik, Isometric and geodesic mapping.<|endoftext|> TITLE: Constructing sums of squares identities QUESTION [6 upvotes]: Recall that a sum of squares formula for $[r,s,n]$ over a field $F$ is an identity of the form $$ ( x_{1}^{2} + \cdots + x_{r}^{2})( y_{1}^{2} + \cdots + y_{s}^{2}) = ( z_{1}^{2} + \cdots + z_{n}^{2}),$$ where the $z_{k}$ are bilinear expressions in the $x_{i}$ and $y_{j}$. The simplest non-trivial example is $$(x_{1}^{2} + x_{2}^{2})(y_{1}^{2} + y_{2}^{2}) = (x_{1}y_{1} - x_{2}y_{2})^{2} + (x_{1}y_{2} + x_{2}y_{1})^{2}$$ coming from the complex numbers. Similar examples can be given from the multiplication on the quaternions and octonions. Various conditions for the existence of sums of squares formulae are know, but there seem to be very few examples of such formulae. Are there any known algorithms for generating identities? What about in the special cases where identities are known to exist? For example, the case $[r,n,n]$ was solved by Hurwitz and Radon (independently). There exists a sum of squares formula with signature $[r,n,n]$ if and only if $r = \rho(n)$, where rho is the Radon--Hurwitz number. Are these identities constructible? The theorem indicates the existence of an identity with signature $(9,48,48)$ for example. How might we go about finding these? REPLY [2 votes]: There are three passages in LAM or LIBRARY of interest; Radon-Hurwitz is pages 127-131. A generalization owing largely to Pfister is pages 323-328. Finally, there is some relevance to the discussion of the field Stufe, pages 379-384. On page 328, he draws attention to cases where Radon-Hurwitz say there is no such formula where the $z$ are bilinear forms in the $x,y,$ but Pfister gives rational function solutions. Meanwhile, he recommends SHAPIRO highly, also LIBRARY<|endoftext|> TITLE: What justification can you give for the fact that "most ODEs do not have an explicit solution"? QUESTION [14 upvotes]: What justification can you give for the fact that "most ODEs do not have an explicit solution"? REPLY [10 votes]: On the non-linear side, there is a theorem of Hudai-Verenov MR0147699, which says that for a generic (open dense set) equation $y'=R(x,y)$ the graph of a generic solution is dense. See also Ilyashenko, MR0247176. This implies that this equations do not have first integrals, so they are not "solvable" in the sense that we teach in the elementary courses. This is an example of a general theorem on the subject. But of course it is well known from the last 300 years of experience with differential equations. For example, all integrable cases of the rotating top in uniform gravity field are explicitly known, and they are exceptional. There are many other examples like this.<|endoftext|> TITLE: Adelic open image for modular forms? QUESTION [8 upvotes]: There's a famous theorem of Serre that if $E$ is a non-CM elliptic curve over $\mathbf{Q}$, and $\rho_{E, \ell} : Gal(\overline{\mathbf{Q}}/{\mathbf{Q}}) \to GL_2(\mathbf{Z}_\ell)$ is its $\ell$-adic Galois representation, then the product $\rho = \prod_{\ell} \rho_\ell: Gal(\overline{\mathbf{Q}}/{\mathbf{Q}}) \to GL_2(\widehat{\mathbf{Z}})$ has open image. In particular, this implies that $\rho_{E, \ell}$ is surjective for almost all $\ell$; but is much stronger than this, as it shows that the $\rho_{E, \ell}$ for different $\ell$ are "independent" in some sense. If one works instead with general non-CM modular forms $f$ of weight $k \ge 2$, then I know of theorems (due to Ribet and Momose) describing the images of the $\rho_{f, \ell}$, showing that they are "as large as possible" for almost all $\ell$ (cf. this earlier question of mine). (The notion of "as large as possible" is much more delicate in this generality, because the coefficient field can be nontrivial, and there can be "inner twists".) Are there analogues of Serre's adelic open image results for higher-weight modular forms? EDIT. I'll just put up a guess of mine, just to show that there is a reasonable conjectural formulation which is compatible with the known results for individual $\ell$. Momose has shown that there is a subfield $F$ of the coefficient field $E = \mathbf{Q}(f)$, a quaternion algebra $B$ over $F$, and an open subgroup $H$ of $Gal(\overline{\mathbf{Q}}/{\mathbf{Q}})$, all independent of $\ell$, such that for any $\ell$ the representation $\rho_{f, \ell}: Gal(\overline{\mathbf{Q}}/{\mathbf{Q}}) \to GL_2(E \otimes \mathbf{Q}_\ell)$ sends $H$ to an open subgroup of the group $\{ x \in B(F \otimes \mathbf{Q}_{\ell}) : \operatorname{norm}(x) \in \mathbf{Z}_{\ell}^{\times(k-1)}\}$. For all $\ell$ coprime to the discriminant of $B$, we have $B(F \otimes \mathbf{Q}_{\ell}) = GL_2(F \otimes \mathbf{Q}_{\ell})$, and Ribet has shown that for all but finitely many such $\ell$ the image of $H$ is all of $\{ x \in GL_2(O_F \otimes \mathbf{Z}_{\ell}) : \operatorname{det}(x) \in \mathbf{Z}_{\ell}^{\times(k-1)}\}$. Conjecture: The image of $H$ in $GL_2(L \otimes \mathbf{A})$ contains an open subgroup of $\{ x \in B(O_F \otimes \widehat{\mathbf{Z}}) : \operatorname{norm}(x) \in \widehat{\mathbf{Z}}^{\times(k-1)}\}$. This is visibly consistent with (and implies) Momose and Ribet's results. REPLY [5 votes]: As I mentioned in my comment, an adelic open image theorem should follow in a purely group-theoretic way from the knowledge that the $\ell$-adic representations are surjective (for an appropriately specified codomain) for all sufficiently large $\ell$ and have open image for all $\ell$. The group theory necessary is Aaron Greicius's classification of maximal closed subgroups of a direct product of profinite groups (see Proposition 2.5 of this paper - the link is correct this time). Greicius's result is the following: Assume that $G_{\alpha}$, for $\alpha \in \Lambda$ are profinite groups with the property that there is no non-abelian finite simple group $M$ that is a quotient of $G_{\alpha}$ and $G_{\alpha'}$ for $\alpha \ne \alpha'$. Then, every maximal closed subgroup of $G = \prod_{\alpha} G_{\alpha}$ comes from a maximal closed subgroup $H_{\alpha} \subseteq G_{\alpha}$, or from a maximal subgroup of $G/G'$. In the case at hand, $\Lambda$ can be taken to be the set of primes $\ell \ne 2, 5$, and $G_{\ell} = {\rm im}~\rho_{f,\ell}$. The only possible non-abelian finite simple quotients of $GL_{2}(O_{F} \otimes \mathbf{Z}_{\ell})$ are the $PSL_{2}(\mathbf{F}_{\ell^{r}})$, and it's not possible to have $PSL_{2}(\mathbf{F}_{\ell_{1}^{r_{1}}})$ isomorphic to $PSL_{2}(\mathbf{F}_{\ell_{2}^{r_{2}}})$ for odd primes $\ell_{1} \ne \ell_{2}$. The proof of Greicius's proposition implies that the adelic image contains $G'$. Moreover, $SL_{2}(O_{F} \otimes \mathbf{Z}_{\ell})$ has no abelian quotients for $\ell \geq 5$, and the commutator subgroup of $SL_{2}(O_{F} \otimes \mathbf{Z}_{\ell})$ has finite index for $\ell = 3$. Greicius's result doesn't quite apply if you include $\ell = 2$ and $\ell = 5$ because $PSL_{2}(\mathbf{F}_{4}) \cong PSL_{2}(\mathbf{F}_{5})$, but it is easy to see that if $K_{2}$ is the fixed field of $\rho_{f,2}$ and $K_{5}$ is the fixed field of $\rho_{f,5}$, then $K_{2} \cap K_{5}/\mathbb{Q}$ is a finite extension.<|endoftext|> TITLE: Recursions for some binary theta series in characteristic 3 QUESTION [6 upvotes]: Define $A(0), A(1), A(2) \dots$ in ${\bf Z}/3[[x]]$ as follows. For $n$ in $\bf N$ let $s=3^{2n+1}$. Then $A(n) = \sum a_kx^k$ where $a_k$ is the mod 3 reduction of the number of representations of $k$ by the principal positive binary quadratic form of discriminant $-s$, and the sum runs over all $k$ prime to 3. Example 1__ When $n=0$, we take the form to be $u^2+uv+v^2$. The number of representations of any non-zero $k$ by this form is a multiple of 6, and so $A(0)=0$. Example 2__ $-A(1)$ is the mod 3 reduction of (the expansion of) the usual weight 12 cusp form for the full modular group. Experiment suggests a recursion for the $A(n)$ which would allow one to write each $A(n)$, $n>0$, as a polynomial of degree $(s-3)/24$ in $A=A(1)$. Explicitly I ask if the following holds: (*)_ $A(n+2)=(A^{3s}+A^{2s}+1)A(n+1)-A^{2s}A(n)$ If (*) holds, then for example: $A(2)=A^{10}+A^7+A$ $A(3)-A(2) = \sum A^k$, $k$ in $\{91,88,82,64,61\}$ $A(4)-A(3) = \sum A^k$, $k$ in $\{820,817,811,793,790,739,736,730,577,574,568,550,547\}$ I've experimentally verified the above 3 identities, and I'm sure that modular forms could be used to settle the truth of (*) for any fixed $n$. EDIT: As far as the weaker question of finding a proof that each $A(n)$ is a polynomial in $A$, there is a possibly relevant article by Skoruppa. (arXiv:0807.4694v2---Reduction mod $\ell$ of theta series of level $\ell^n$). But his main result is restricted to characteristic >3. I quote the result: It is proved that the theta series of an even lattice whose level is a power of a prime $\ell$ is congruent modulo $\ell$ to an elliptic modular form of level one. (Note that every elliptic modular form over $\bf Q$ of level 1 has as its modulo 3 reduction a polynomial in $A$. So one would want to extend Skoruppa's result to $\ell=3$). EDIT: It seems one can use an earlier result of Serre, Theorem 5.4 of "Divisibilite ..." appearing in L'Enseignement Mathematique (1976) to quickly show that the A(n) are polynomials in A. Namely let m=(s+1)/4, and consider the quaternary form uu+uv+mvv+ww+wt+tt. It's classical that the theta series attached to this form is a weight 2 modular form for Gamma_0 (N), where N is a power of 3. The Serre theorem just cited shows that the mod 3 reduction of such a modular form is the reduction of a modular form for Gamma(1), and so is a Z/3 linear combination of powers of delta= -A. Now the reduction of the theta series for ww+wt+tt is 1, so the reduction of the theta series for uu+uv+mvv is also a Z/3 linear combination of powers of A. It follows that A(n) is a Z/3 linear combination of A, A^4, A^7, A^10,... REPLY [5 votes]: We establish the recursion for all $n$ by writing the rank-2 theta series $A(n)$ in terms of the rank-1 thetas $$ S(q) := \sum_{m \in \bf Z} q^{m^2} = 1 + 2q + 2q^4 + 2q^9 + \cdots, $$ $$ T(q) := \sum_{m \in \bf Z} q^{(m+\frac12)^2} = 2q^{1/4} + 2q^{9/4} + 2q^{25/4} + \cdots. $$ The lattice corresponding to the principal positive binary form of discriminant $-s = -3^{2n+1}$ is the union of the rectangular lattice ${\bf Z} \oplus {\bf Z}\langle s \rangle$ and its translate by $(1/2, 1/2)$. The quadratic form is a multiple of $3$ iff the ${\bf Z}$ or ${\bf Z} + \frac12$ term is a multiple of $3$. Hence $$ A(n) = (S(q)-S(q^9)) \, S(q^s) + (T(q)-T(q^9)) \, T(q^s). $$ Because $9$ and $s$ are powers of $3$, an equivalent formula in characerstic $3$ is $$ A(n) = (S-S^9) S^s + (T-T^9) T^s $$ where $S=S(q)$, $T=T(q)$. We claim that the recursion is already satisfied by $A_1(n) := S^s$ and $A_2(n) := T^s$ separately, from which it will follow by linearity for $A(n) = (S-S^9) A_1(n) + (T-T^9) A_2(n)$. For both $i=1$ and $i=2$ each side of the recursion $$ A_i(n+2) = (A^{3s} + A^{2s} + 1) A(n+1) - A^{2s} A(n) $$ is the ($3^{2n}$)-th power of its $n=0$ case $$ A_i(2) = (A^9 + A^6 + 1) A_i(1) - A^6 A_i(0), $$ so we need only verify this last identity for both $i$. Now $S$ and $T$ are related by $S^4 + T^4 = 1$, because in characteristic zero $S^4 + T^4$ is the theta series of the $D_4$ lattice, whose automorphism group contains a $3$-cycle that acts freely on nonzero vectors. [Check: $A(0)$ vanishes because it equals $$ S^4-S^{12} + T^4-T^{12} = (S^4+T^4) - (S^4+T^4)^3 = 1 - 1^3 = 0.] $$ Thus $$ A(1) = S^{28} - S^{36} + T^{28} - T^{36} = S^{28} - S^{36} + (1-S^4)^7 - (1-S^4)^9, $$ which comes to $S^{24} - S^{16} + S^{12} - S^4$, and by symmetry also $A(1) = T^{24} - T^{16} + T^{12} - T^4$. Then $A_i(2) = (A^9 + A^6 + 1) A_i(1) - A^6 A_i(0)$ is just an identity in $({\bf Z}/3{\bf Z})[S]$ or $({\bf Z}/3{\bf Z})[T]$, which we verify by direct computation to complete the proof.<|endoftext|> TITLE: on the center of a Lie group QUESTION [13 upvotes]: I'm trying to set straight my various pieces of knowledge about the center of a compact Lie group, and I'm running in circles... First some definitions: • Let $G$ be compact, simple, and simply connected Lie group, with Dynkin diagram $\Gamma$. • Let $\{\alpha_i\}_{i=1...n}$ be the simple roots of $G$; they form by definition the vertex set of $\Gamma$ (sometimes, I'll write just $i$ instead of $\alpha_i$ for a vertex of $\Gamma$). • Let $\Gamma^e=\Gamma\cup\{\alpha_0\}$ be the extended Dynkin diagram, where one adds the lowest root to $\Gamma$. • Let $Z(\Gamma^e)$ be the subset of $\Gamma^e$ defined as the orbit of $\alpha_0$ under the automorphism group of $\Gamma^e$. • Let $Z(G)$ be the center of $G$, which is also the quotient $\Lambda_{coweight}/\Lambda_{coroot}$. • Let $\omega^\vee_i\in\Lambda_{coweight}$ be the fundamental coweights, defined by $\alpha_i(\omega^\vee_j)=\delta_{ij}$. Let us also agree that, by convention, $\omega^\vee_0:=0$. Then my question: Is it true (and if so, where can I find a proof of this fact) that $$\{\omega^\vee_i \,|\, i\in Z(\Gamma^e)\}$$ forms a complete set of representatives of $\Lambda_{coweight}/\Lambda_{coroot}$. REPLY [4 votes]: I have found a reference that contains the exact statement which I wanted: It is Theorem 3, on page 14 of this paper of Stephen Sawin. The vertices of the Weyl alcove which are in bijection with $Z(G)$ are called sharp corners (for self-explanatory reasons). The other corners are called dull corners.<|endoftext|> TITLE: Is there an $(\infty,2)$-category with morphisms given by $D^b\text{Coh}$? QUESTION [9 upvotes]: My question is: Has anyone constructed an $(\infty,2)$-category whose objects are (projective, maybe smooth, ...) varieties, and where the 1-morphisms from $X$ to $Y$ are given by $D^b_\infty\text{Coh}(X \times Y)$ (an $(\infty,1)$-enhancement of $D^b\text{Coh}(X \times Y)$)? By "$(\infty,1)$-enhancement" I mean some $(\infty,1)$-category whose homotopy category is $D^b\text{Coh}$. I would hope that binary composition would descend to the functor $$D^b\text{Coh}(Y\times Z) \times D^b\text{Coh}(X \times Y) \to D^b\text{Coh}(X \times Z)$$ that sends $(Q,P)$ to $\pi_{02,*}(\pi_{01}^*P \otimes \pi_{12}^*Q)$ a la the Fourier--Mukai transform (where tensor and the projections are the derived functors). If the answer is "yes", I wonder whether this is special to derived categories of coherent sheaves or whether it's a more general algebraic result about collections of $(\infty,1)$-categories of a certain kind? I'm new to infinity infinity stuff and I might have misused a technical term, so it might be best not to interpret my words too literally. REPLY [14 votes]: For smooth projective varieties it is known that $$D^b_\infty Coh(X\times Y)\simeq Hom(D^b_\infty Coh(X), D^b_\infty Coh(Y))$$ compatibly with the composition you describe (note here $D^b_\infty Coh$ coincides with the $\infty$-category $Perf$ of perfect complexes). Hence your desired $(\infty,2)$-category is a full subcategory of the $(\infty,2)$-category of dg categories (or small idempotent-complete stable $\infty$-categories) where the above Homs take place. The analogous result for $QCoh$ holds in much greater generality (without smoothness or projectivity - eg for quasicompact quasiseparated schemes). Edit: Without smoothness, you find (here) that $D^b_\infty Coh(X\times Y)$ represents functors from $Perf(X)$ to $D^b_\infty Coh(Y)$, and that functors from $D^b_\infty Coh(X)$ to $D^b_\infty Coh(Y)$ are represented by kernels which are coherent relative to the first factor. Note you have to be careful in the formulation of the question since $D^b_\infty Coh$ is not preserved by tensor product on a singular variety! (e.g. self-tor of a skyscraper at a singular point is not bounded, i.e., not in $D^b_\infty Coh$)<|endoftext|> TITLE: Bounds re Asymptotic Formula for the Sum of Largest Prime Factors QUESTION [6 upvotes]: I have a reference request related to the result : $\sum_{n=2}^{x} P(n)$ ~ $\frac{\pi^2}{12}\frac{x^{2}}{log(x)}$ as $x \rightarrow \infty$ where $P(n)$ is the largest prime factor of the positive integer $n$. Theorem 1.1 in [1] below appears to be the first proof of this result and the result was generalized in Theorem 3.1 in [2] below. I am looking for upper and lower bounds on the ratio L/R where L and R are the left and right sides respectively of the asymptotic relation above. Thanks for any help. References [1]. K.Alladi and P.Erdos. Pacific J. Math. 71(1977) 275-294 [2]. J.De Konnick and R.Sitaramachandrarao. Indian J. Pure Appl Math. 19(10) 990-1004 Oct 1988 Disclosure: This question was first posted on Math Stack Exchange where it has languished for the last three weeks without an answer (or even a comment). REPLY [6 votes]: I am not exactly sure what you mean by bounds for $L/R$, but I assume that this translates to the rate of convergence and the next terms in the asymptotic. In this short note, a more precise asymptotic for $\sum_{n\leq x} P(n)$ is given and it is shown that $$\frac{1}{x}\sum_{n\leq x}P(n)=\text{li}_g(x) +O_\epsilon \left(x e^{-c(\log x)^{3/5-\epsilon}}\right),$$ where $$\text{li}_g(x)=\int_2^x \frac{t}{x}\frac{[x/t]}{\log t}dt$$ is an integral function that shares some properties in common with $\text{li}(x)=\int_2^x \frac{1}{\log t}dt.$ In particular, we have the asymptotic expansion $$\text{li}_g(x)= \frac{c_0 x}{\log x}+\frac{1!c_1 x}{\log^2x}+\cdots+\frac{(k-1)!c_{k-1} x}{\log^{k}x}+O_k\left(\frac{x}{\log^{k+1}x}\right)$$ where $$c_k=\frac{1}{2^{k+1}}\sum_{j=0}^n \frac{2^j(-1)^j\zeta^{(j)}(2)}{j!},$$ which yields an asymptotic expansion for $\sum_{n\leq x} P(n).$<|endoftext|> TITLE: Multiprecision numerical evaluation of integral: Sage vs. PARI/GP vs. mpmath QUESTION [9 upvotes]: I am trying to compute thousands of integrals of the below type, that comes up in a conformal mapping problem, to as many accurate digits as possible (preferably 50+): $$ \int_{-1}^1\textrm{d}t \frac{\mathcal{Re}\{\log[(\cos{(\pi/130)} - t)]\}}{\sqrt{1 - t^2}} $$ The results from PARI/GP, Sage and Python's mpmath library respectively are: -2.1770705767584673426214016567105099553, (-2.1775860588840983, 1.2746272565903925e-05), -2.1774410877577223893132496923831596284 Clearly $-2.177$ is correct, but what's the best way to find a more accurate answer, accurate to 50+ digits? I've tried splitting intervals from $[-1, \tau] \cup [\tau, 1]$, where $\tau = \cos{(\pi/130)}$; that doesn't improve things but actually makes it worse. I am working at much higher decimal precision than 3. UPDATE: Following the suggestion of IgorRivin, the below is the Mathematica attempt: tau = N[Cos[Pi/130], 50]; epsilon = 2^-10; limit = N[ArcCos[1 - epsilon], 50]; T1 = NIntegrate[Re[Log[(tau - t)]]/(Sqrt[1 - t^2]), {t, -1, 1 - epsilon}, WorkingPrecision -> 50, AccuracyGoal -> 50] T2 = NIntegrate[Re[Log[tau - Cos[theta]]], {theta, 0, limit}, WorkingPrecision -> 50, AccuracyGoal -> 50] answer = T1 + T2 gives the results: Out[1]= -1.7968036050143567231750633164742621583459497767361 During evaluation of In[881]:= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in theta near {theta} = {0.024170580099064656300274166159163078224443688377727}. NIntegrate obtained -0.38078136551592029350719560689304843811203900465893 and 6.9050153103011684224977203192777088512613553501207`50.*^-7 for the integral and error estimates. >> Out[2]= -0.38078136551592029350719560689304843811203900465893 Out[3]= -2.1775849705302770166822589233673105964579887813950 As may be seen the error is still in the seventh decimal place; increasing the working precision or accuracy does not really improve things. UPDATE: The integral is exactly soluble; the result is given in my comment to the accepted answer of Emil Jeřábek, and compares well to the exact value. REPLY [10 votes]: Expanding my comment above: putting $\alpha=\pi/130$, the integral equals $$\int_0^\pi\log\left|\cos\alpha-\cos\theta\right|d\theta=\int_0^\pi\log\frac{\cos\alpha-\cos\theta}{(\theta-\alpha)\sin\alpha}\,d\theta+\int_{-\alpha}^{\pi-\alpha}\log\left|\theta\sin\alpha\right|d\theta$$ $${}=\int_0^\pi\log\frac{\cos\alpha-\cos\theta}{(\theta-\alpha)\sin\alpha}\,d\theta+\pi\log\sin\alpha+\alpha(\log\alpha-1)+(\pi-\alpha)(\log(\pi-\alpha)-1).$$ The new integrand is regular on $[0,\pi]$, hence it has a better chance to be accurately approximated by numerical integration. I leave it to someone knowledgeable with such tools to try it. In case it helps with cancellation errors, one can further write $$\int_0^\pi\log\frac{\cos\alpha-\cos\theta}{(\theta-\alpha)\sin\alpha}\,d\theta=\int_{-\alpha}^{\pi-\alpha}\log\left(\frac{\sin\theta}\theta+\frac{1-\cos\theta}\theta\cot\alpha\right)d\theta.$$<|endoftext|> TITLE: Books on the analysis of hyperbolic partial differential equations QUESTION [12 upvotes]: Most of the present books on pde analysis deal with the elliptic partial differential equations. Is there some book related to rigorous analysis with hyperbolic pdes, and especially hyperbolic systems of pdes? I want to some great books on this subject for research. Thank you. REPLY [10 votes]: It's funny that a similar question hasn't already appeared on MO. Other answers already give some good suggestions. Here's a bunch more. Note that the older ones may not be considered very pedagogical or rigorous by today's standards. Hadamard, J. Lectures on Cauchy's Problem in Linear Partial Differential Equations (Yale University Press, 1923) Courant, R. and Hilbert, D. Methods of mathematical physics. Vol. II: Partial differential equations (Interscience, 1962; German original 1937) Petrovsky, I. G. (also as Petrowsky) Lectures on Partial Differential Equations (Interscience, 1954; Russian original 1950--51) Leray, J. Hyperbolic differential equations (Institute for Advanced Study, Princeton, 1953) John, F. Partial differential equations (Springer, 1971) Lax, P. D. Hyperbolic Partial Differential Equations (AMS, 2006; original notes 1963) Garabedian, P. R. Partial differential equations (Wiley, 1964) Friedlander, F. G. The Wave Equation on a Curved Space-time (CUP, 1975) Günther, P. Huygens' Principle and Hyperbolic Equations (Academic Press, 1988) Hörmander, L. Lectures on Nonlinear Hyperbolic Differential Equations (Springer, 1997; original notes 1987) A bunch more have appeared in more recent years, some of which have already been mentioned. Here are a few other noteworthy ones. Christodoulou, D. The Action Principle and Partial Differential Equations (PUP, 2000) Bär, C., Ginoux, N. and Pfäffle, F. Wave Equations on Lorentzian Manifolds and Quantization (EMS, 2007) Dafermos, C. M. Hyperbolic Conservation Laws in Continuum Physics (Springer, 2010) Klainerman, S. Lecture Notes in Analysis (lecture notes, Princeton, 2011) Rauch, Hyperbolic PDEs and Geometric Optics (AMS, 2012)<|endoftext|> TITLE: Conjectured integral for Catalan's constant QUESTION [7 upvotes]: Numerical evidence suggests: $$ \int_0^{\frac12}\int_0^{\frac12}\frac{1}{1-x^2-y^2} dy \, dx= \frac{G}{3}\qquad (1)$$ Couldn't find the indefinite integral, though maple simplifies (1) to $$ \int _{0}^{1/2}\!-\arctan \left( 1/2\,{\frac {1}{\sqrt {-1+{x}^{2}}}} \right) {\frac {1}{\sqrt {-1+{x}^{2}}}}{dx}$$ Is (1) true? REPLY [5 votes]: The proof of (1), sketched in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf, goes as follows. With the substitution $x=x^\prime/2,\,y=y^\prime/2$, the integral (1) takes the form (we have omitted the primes) $$I_2=\int_0^1\int_0^1\frac{dx\,dy}{4-x^2-y^2}.$$ Due to the $x\leftrightarrow y$ exchange symmetry of the integrand, $$I_2=2\iint\limits_\Delta \frac{dx\,dy}{4-x^2-y^2},$$ where the triangular integration domain $\Delta$ is the lower half of the unit square $0\le x\le 1,\,0\le y\le 1$. Let us introduce the polar coordinates, as suggested by Aeryk, $x=r\cos{\theta},\,y=r\sin{\theta}$. Then $$I_2=\int\limits_0^{\pi/4}d\theta\int\limits_0^{1/cos{\theta}}\frac{2r}{4-r^2}dr=\int\limits_0^{\pi/4}\ln{\frac{4\cos^2{\theta}}{4\cos^2{\theta}-1}}\,d\theta.$$ Now, by using $\sin{3\theta}=\sin{\theta}(3\cos^2{\theta}-\sin^2{\theta})=\sin{\theta}(4\cos^2{\theta}-1)$, we cen rewrite the above integral in the form $$I_2=\int\limits_0^{\pi/4}\ln{\frac{4\cos^2{\theta}\sin{\theta}}{\sin{3\theta}}}\,d\theta=\frac{\pi}{2}\ln{2}+2I_C+I_S-I_{3S},$$ where the integrals $$I_S=\int\limits_0^{\pi/4}\ln{\sin{\theta}}\,d\theta=-\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2},\;I_C=\int\limits_0^{\pi/4}\ln{\cos{\theta}}\,d\theta=\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}$$ were calculated in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf, while for $I_{3S}$ we have $$I_{3S}=\int\limits_0^{\pi/4}\ln{\sin{3\theta}}\,d\theta=\frac{1}{3} \int\limits_0^{3\pi/4}\ln{\sin{\theta}}\,d\theta=\frac{1}{3}\left[ \int\limits_0^{\pi/2}\ln{\sin{\theta}}\,d\theta+ \int\limits_{\pi/2}^{3\pi/4}\ln{\sin{\theta}}\,d\theta\right].$$ The substitution $\alpha=-(\pi/2-\theta)$ shows that the second integral equals to $I_C$, while for the first integral the following result can be found in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf $$\int\limits_0^{\pi/2}\ln{\sin{\theta}}\,d\theta= \int\limits_0^{\pi/2}\ln{\cos{\theta}}\,d\theta=-\frac{\pi}{2}\,\ln{2}.$$ Therefore $$I_{3S}=\frac{1}{3}\left[-\frac{\pi}{2}\,\ln{2}+\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}\right]=\frac{1}{6}\,G-\frac{\pi}{4}\,\ln{2},$$ and $$I_2=\frac{\pi}{2}\ln{2}+2\left(\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}\right)+\left(-\frac{1}{2}\,G-\frac{\pi}{4}\,\ln{2}\right)-\left(\frac{1}{6}\,G-\frac{\pi}{4}\,\ln{2}\right)=\frac{G}{3}.$$ By the similar method, the following result was proved in http://www.maths.lancs.ac.uk/~jameson/catalan.pdf $$I_1=\int_0^1\int_0^1\frac{dx\,dy}{2-x^2-y^2}=G.$$<|endoftext|> TITLE: Counting norms on an infinite dimensional vector space QUESTION [10 upvotes]: It is known that whenever E is a finite dimensional real vector space, there is only one norm on E up to equivalence (actually one non discrete vector space topology). Is it known what happens when E is infinite dimensional? for sure, one can create two (infinitely many) non equivalent norms by using Hamel bases(*), but what about the precise cardinality (up to equivalence, or without taking into account equivalence at all) ? (*) E.g. Let (e_i) a Hamel basis, write x = sum x_i e_i and put ||x|| = sum of |x_i| N(x) = sum of i|x_i| REPLY [9 votes]: Take a look at the papers "Über Normtopologien in linearen Räumen" (Link to the article) as well as "Über vollständige Normtopologien in linearen Räumen" (Link to the article) by D. Laugwitz. He proves the following. Let $E$ be a vector space, denote by $a(E)$ the cardinality of a Hamel basis of $E$ and denote by $n(E)$ the number of mutually non-equivalent norms on $E$. Then $$n(E) = 2^{a(E)}, \quad\text{if}~\mathfrak{c}\leq a(E),$$ $$ \mathfrak{c}\leq 2^{a(E)} \leq n(E) \leq 2^{\mathfrak{c}}, \quad\text{if}~\aleph_0\leq a(E)<\mathfrak{c},$$ and (of course) $$ n(E) = 1 \quad\text{if}~1\leq a(E) < \aleph_0.$$ Here $\aleph_0$ denotes the cardinality of countable sets and $\mathfrak{c}$ the cardinality of the continuum. In the second article you can find the answer for complete norms.<|endoftext|> TITLE: "Forms" of quadrics QUESTION [9 upvotes]: The theory of Severi-Brauer varieties is well-known. Let $k$ be a (perfect) field. There may exist varieties not isomorphic to $\mathbf{P}^n$ over $k$, which are isomorphic to $\mathbf{P}^n$ over $\overline{k}$. They are classified by $H^1(k,\mathrm{PGL}_n)$. How about quadrics? Say $k$ is a (perfect) field and $X$ is a smooth, projective $k$-variety of dimension $n$. Assume that $X \otimes_k \overline{k}$ is isomorphic to a quadric. Is $X$ necessarily an $n$-dimensional quadric itself? If not, can you give some nice examples (e.g. over number fields) which show that this need not be the case? REPLY [6 votes]: This is a comment to Daniel Loughran's answer; I would like to add some more details on obstruction to lifting $PO(n)$-cocycles to $O(n)$-cocycles (at least in the case where $k$ has characteristic $\neq 2$). As discussed in the Wikipedia article, there is an extension of algebraic groups $$ 1\to \mathbb{Z}/2=\{\pm I\}\to O(Q)\to PO(Q)\to 1, $$ which is split in the odd case and non-split in the even case. Then there is an exact sequence in group cohomology $$H^1(k,O(Q))\to H^1(k,PO(Q))\to H^2(k,\mathbb{Z}/2), $$ see e.g. the Galois cohomology book of Serre. Since the sequence is split in case $Q$ is odd-dimensional, the element $\sigma\in H^1(k,PO(Q))$ classifying the form maps trivially to $H^2(k,\mathbb{Z}/2)$ and so the map $H^1(k,O(Q))\to H^1(k,PO(Q))$ is surjective. In the case where $Q$ is even-dimensional (which corresponds to the case $\mathbb{P}^{2n+1}$ mentioned in Daniel Loughran's answer), the extension is non-split but it is split locally in the étale topology. Therefore, the extension class lives in $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)$. This cohomology group has several interpretations. By the Merkurjev-Suslin theorem (a special case of the Milnor conjecture), there is an isomorphism $H^2_{\operatorname{et}}(k,\mathbb{Z}/2)\cong K^M_2(k)/2K^M_2(k)$. By a theorem of Merkurjev, $K^M_2(k)/2K^M_2(k)\cong {}_2Br(k)$, explaining the appearance of $2$-torsion in the Brauer group in Daniel Loughran's answer. (A possible reference for these would be the book on central simple algebras by P. Gille and T. Szamuely; alternatively, check out survey papers on the Milnor conjecture.) So for each form of the quadric $Q$ (parametrized by an element $\sigma\in H^1(k,PO(Q))$) there is an associated obstruction class in $H^2(k,\mathbb{Z}/2)\cong K^M_2(k)/2\cong {}_2Br(k)$ whose triviality is equivalent to the form being a quadric.<|endoftext|> TITLE: Mirror Symmetry for Quaternionic-Kähler Manifolds QUESTION [7 upvotes]: I take the following quote from Huybrecht's notes on hyperkähler manifolds and mirror symmetry: Mirror symmetry in a first approximation predicts for any Calabi-Yau manifold (M,g) the existence of another Calabi-Yau manifold $(M^v,g^v)$ together with an isomorphism $M^{\text{cpl}}(M) \simeq M^{\text{khl}}(M^{v})$. Here $M^{\text{cpl}}(M)$, and $M^{\text{khl}}(M^{v})$, refer to the moduli spaces of complex, resp. symplectic, structures on $M$, resp $M^v$. Mirror symmetry is supposed to be much simpler for hyperkähler manifolds, as usually the mirror manifold $M^v$ as a real manifold is $M$ itself. If one looks at quaternionic-Kähler manifolds, is there any analogous simplification that holds. Discussion of some simple examples (ie the Grassmannian $Gr(4,2)$) would be appreciated. REPLY [7 votes]: Let me summarize and supplement my remarks above. A quaternion-Kahler manifold $X$ is a Riemannian manifold with holonomy $Sp(n)Sp(1)$, its definition of course includes hyperkahler manifolds as a special case. For a hyperkahler manifold, mirror symmetry can sometimes be realized as a hyperkahler rotation, e.g. elliptic $K3$ surfaces. However, such an understanding fails in general, as was pointed out by Huybrechts in his lecture notes: http://arxiv.org/pdf/math/0210219.pdf. For the method to handle mirror symmetry in the general case, see the work of Gross-Siebert: http://arxiv.org/abs/math/0703822. Now let's exclude hyperkahler manifolds by assuming the scalar curvature is nonzero. For simplicity, let's further assume that $X$ is symmetric, then it can be written as $X=G/H$ with $G$ a simple Lie group and $H=K\cdot SU(2)$ , $K$ is the centralizer of $SU(2)$ in $G$. In this case, as far as I konw, only the mirror symmetry for the case when $G=SU(n+2)$ and $H=S\big(U(n)\times U(2)\big)$ is systematically studied. In general, for the homogeneous space $X=G/P$, where $G$ is a semisimple Lie group and $P$ a parabolic subgroup, its mirror is given by the Landau-Ginzburg model $(R,W)$, where $R\subset G^L/P^L$ is the Richardson variety, and $W:R\rightarrow\mathbb{C}$ is a holomorphic function called superpotential. Here $G^L$ and $P^L$ denote their Langlands dual. The main reason for using $X^\vee=R$ to partially compactify $(\mathbb{C}^\ast)^N$ is to get the following isomorphism between the quantum cohomology ring of $X$ and the Jacobi ring of $W$: $QH^\ast(X)\cong Jac(W)$. This isomorphism is generally expected to hold for mirror symmetry for Fano manifolds. For the case when $X$ is toric Fano, such an isomorphism is proved by Fukaya-Oh-Ohta-Ono: http://projecteuclid.org/euclid.dmj/1262271306. The general mirror construction for $X=G/P$ is done by Rietsch: http://arxiv.org/abs/math/0511124. The isomorphism $QH^\ast(X)\cong Jac(W)$ should be compared with Kontsevich's conjecture, which asserts the following: $QH^\ast(X)\cong HH^\ast\big(\mathcal{F}(X)\big)$, where $\mathcal{F}(X)$ is the Fukaya category of $X$. The interesting thing is that when $X$ is Grassmannian (e.g. $X=Gr(2,n+2)$, in which case $X$ is quaternion-Kahler), $W:X^\vee\rightarrow\mathbb{C}$ has only isolated critical points, therefore it should provide a Lefschetz fibration on $X^\vee$, so Seidel's theory (http://www.ems-ph.org/books/book.php?proj_nr=12) can be applied and one may expect that all the geometric information of the Fukaya category $\mathcal{F}(X^\vee,W)$ is contained in the superpotential $W$. This suggests that $\mathcal{F}(X)$ and $\mathcal{F}(X^\vee,W)$ should be related to each other. From an SYZ point of view, there is no such thing as mirror symmetry for Fano manifolds. Because everytime we talk about mirror symmetry for a Fano manifold $X$ we regard it as a Calabi-Yau manifold $X_0$ together with some boundary divisor $D$, i.e. $X_0=X\setminus D$. The Landau-Ginzburg mirror manifold $X^\vee$ is constructed from $X$, and $D$ determines only the superpotential $W$. For the SYZ mirror construction for some of the Fano examples, see Auroux's fundamental work: http://arxiv.org/abs/0706.3207. In fact, such a philosophy coming from SYZ mirror symmetry applies also to the case when $X=G/P$. There is a Richardson variety $R^\vee$ which is dual to $R$ inside $G/P$. The open part $R^\vee\subset X$ should be regarded to be SYZ mirror to $R\subset G^L/P^L$. But when we add back the boundary divisor $X\setminus R^\vee$, we need to use the superpotantial $W$ to correct $R$, so we end up with the Landau-Ginzbug model $(R,W)$. From such a point of view, the meaning of the following mirror dualities should be clear: $R^\vee\leftrightarrow R$ $(R,W)\leftrightarrow G/P$ $G^L/P^L\leftrightarrow (R^\vee,W^\vee)$ Let me also mention that the philosophy that $T$-duality should be realized as the Langlands duality in some special cases dates back to the work of Hausel and Thaddeus on mirror symmetry for Hitchin moduli spaces: http://arxiv.org/pdf/math/0205236v1.pdf. As the quaternion-Kahler case (with non-zero Ricci curvature) should be regarded as parallel to the hyperkahler case, it's not strange that the same philosophy works here. To get a deep understanding of mirror symmetry in the quaternion-Kahler case, we will need to study the geometric structures of a quaternion-Kahler manifold, and understand how the geometric structures exchanges when passing from $X$ to $X^\vee$. Unfortunately, personally I don't know much development in these directions. However, there is the paper of Leung (http://arxiv.org/pdf/math/0303153v1.pdf) which aims to use normed division algebras to unify the geometric structures coming from different holonomy groups. I personally believe this unified picture should be fundamental in the study of mirror symmetry for quaternion-Kahler case.<|endoftext|> TITLE: List of integers without any arithmetic progression of n terms QUESTION [16 upvotes]: Let's consider a positive integer $n$ and the list of the $n^2$ integers from $1$ to $n^2$. What is the minimum number $f(n)$ of integers to be cancelled in this list so that it is impossible to form any arithmetic progression of $n$ terms with the remaining integers? REPLY [12 votes]: EDIT: The original values of $f$ that I reported were the result of a buggy program, as noted by Wolfgang in the comments. I have fixed the bug and corrected the results. This problem is easily coded up as an integer linear program. For $1\le i\le n^2$, let $x_i$ be a 0-1 variable indicating whether $i$ is chosen. Minimize $\sum_i x_i$ subject to the constraints $\sum_{a\in A} x_a \ge 1$ for every arithmetic progression $A$ of interest. Assuming no bugs in my program, it took CPLEX only a few minutes to extend JiK's list of small values to: $$ \begin{align} f(8)&=13 &&\{6,13,18,23,28,36,37,38,39,40,43,50,57\} \\ f(9)&=15 &&\{8,15,22,29,32,41,42,43,44,45,46,54,61,68,75\} \\ f(10)&=16 &&\{10,15,22,29,36,43,53,55,56,57,58,68,73,74,84,91\}\\ f(11)&=18 && \{11,22,33,44,55,60,66,68,72,76,77,85,92,97,99,102,106,111\} \end{align} $$ The last of these took less than 15 minutes on a six-core machine (about 400,000 "ticks" in CPLEX parlance). I'm sure that cleverer techniques could push the computation further, but this is already enough to show that the OEIS doesn't already know about this sequence.<|endoftext|> TITLE: A continuous function for defining unique values to a 1024x1024 image with a 24 bit 3 color channel image QUESTION [8 upvotes]: I need to generate a color map which I am not sure exist. I have a 1024x1024 image which would contain 2^20 pixels. I have 3 color channels which each have 8 bits which would leave us with 2^24 possible colors. This means that there would actually be enough different pixel values for a unique value for a 4096x4096. This problem is easy to solve with non continuous colors where you simply use 4 bits of the final channel on both of the first two channels to create two 12 bit channels. Here is an example of the non-continuous version: Each of the individual sub-squares has a different blue value which gives us a unique value. This may be hard to see with the eye, as they are only changing by a very small amount. Using this technique, it is easy to fill up an entire 4096x4096 with unique, mathematically predictable colors. Unfortunately, I have a new constraint where all three channels of the map must remain continuos. What I mean by this is that each individual neighboring pixels channel value does not change by more than one in value. For instance, a pixel with a red value of 10 may have direct neighbors with a red value of either 9, 10 or 11 The reason for this constraint is that when sampling from this texture, individual neighboring pixels may be sub-sampled and linear interpolated together and when going along the edge of the sub-boxes, this would result in inaccurate values. To put it in a slightly different way, I need a function f and f^-1 f(x, y) = r, g, b f^-1(r, g, b) = x, y (only existing in the original x,y range) with r, g, b, being 8 bit numbers (the integers 0 - 255) and x and y being 10 bit numbers (the integers 0 - 1023). All neighboring r,g,b values must be continuous. By continuous, I mean that each individual neighboring pixels channel value does not change by more than one in each channel. Do such functions exist, and if so, what are they? REPLY [7 votes]: This problem is very roughly analogous to having a square piece of paper and you want to fold it so that it fits into a cube, except both the paper and cube are discrete, and the folds of the paper can pass through themselves under certain conditions. I'll just post the image now and maybe later post a more thorough explanation if there is interest. The image may be wrong, I've only proved it correct, not checked it. Here's the python code, including tests and image creation. I think it displays the transpose of the image pasted into this answer. The logic is not so complicated; basically you fold the image like a map to fit into the rgb color box. from itertools import islice import numpy as np import Image def first_fold(): r""" x x / \ / \ x x x x x | | | | | x x x x x | | | | | . . . . . . . . . . . . . . . | | | | | x x x x x | | | | | x x x x x \ / \ / x x """ segment_length = 204 q = 0 while True: for i in range(segment_length): yield q, 1 + i q += 1 yield q, 1 + segment_length q += 1 for i in range(segment_length): yield q, segment_length - i q += 1 yield q, 0 q += 1 def second_fold(): """ aabbccddeeff a bacbdcedfe f a b cadbecfd e f a b c daebfc d e f a b c d eafb c d e f a b c d e fa b c d e f a b c d e f a b c d e f """ segment_length = 136 k = 6 q = 0 while True: for i in range(segment_length): yield q, k + i q += 1 for i in range(k): yield q + i, k + segment_length + i q += k for i in range(k): yield q + i, k + segment_length + (k-1) - i q += k for i in range(segment_length): yield q, k + segment_length - 1 - i q += 1 for i in range(k): yield q + i, (k - 1) - i q += k for i in range(k): yield q + i, i q += k def good_image(image): arr = np.asarray(image) n = arr.shape[0] if arr.shape != (n, n, 3): return False colors = set(tuple(x) for x in arr.reshape(n*n, 3)) if len(colors) != n*n: return False if (arr < 0).any(): return False if (arr > 255).any(): return False e = np.abs(arr[1:, :, :] - arr[:-1, :, :]) if (e > 1).any(): return False e = np.abs(arr[:, 1:, :] - arr[:, :-1, :]) if (e > 1).any(): return False return True def bad_image(image): return not good_image(image) def main(): assert(good_image([ [[1, 1, 1], [1, 2, 1]], [[1, 1, 2], [2, 2, 2]]])) assert(good_image([ [[0, 0, 0], [0, 1, 0]], [[0, 0, 1], [1, 1, 1]]])) assert(good_image([ [[2, 20, 200], [1, 19, 199]], [[1, 19, 200], [2, 19, 200]]])) assert(bad_image([ [[1, 1, 1], [1, 2, 1]], [[1, 2, 1], [2, 2, 2]]])) assert(bad_image([ [[0, 0, 0], [0, -1, 0]], [[0, 0, -1], [-1, -1, -1]]])) assert(bad_image([ [[255, 255, 255], [255, 256, 255]], [[255, 255, 256], [256, 256, 256]]])) assert(bad_image([ [[2, 20, 200], [1, 19, 199]], [[1, 19, 200], [2, 19, 201]]])) assert(bad_image([ [[2, 20, 200], [1, 19, 200]], [[1, 19, 199], [2, 19, 201]]])) n = 1024 arr = np.zeros((n, n, 3), dtype=int) for u, (q, x) in islice(enumerate(first_fold()), n): for v, (a, b) in islice(enumerate(second_fold()), n): y = q + a z = b arr[u, v, :] = (x, y, z) assert(good_image(arr)) img = Image.fromarray(arr.astype(np.uint8)) img.show() main() By the way, you can get a 2048x2048 solution by changing only three constants: segment_length 136->204, k 6->11, n 1024->2048. This still uses only 1/4 of the available colors, and to exceed 1/2 you will need deeper changes than changing constants. The reason for the 1/2 barrier is that this method uses twice the color density near creases of the second fold than at less interesting parts of the image.<|endoftext|> TITLE: Generalized Hardy-Littlewood-Sobolev Inequality QUESTION [6 upvotes]: The Hardy-Littlewood-Sobolev Inequality says that $$\text{for $p,q,r\in (1,+\infty)$ such that }\quad 1-\frac1p+1-\frac1q=1-\frac1r,\tag {$\sharp$} $$ $$ \exists C, \forall u\in L^p(\mathbb R^n),\quad \Vert{u\ast\vert\cdot\vert^{-n/q}}\Vert_{L^r(\mathbb R^n)}\le C \Vert u\Vert_{L^p(\mathbb R^n)}. $$ Setting $v_q(x)=\vert x\vert^{-n/q}$, we see that $v_q\in L^q_w(\mathbb R^n)$, which is also the Lorentz space $L^{q,\infty}(\mathbb R^n)$ (the latter space is a Banach space when $q\in (1,+\infty)$). Is there a generalization of Young's inequality such as $$ \exists C, \forall u\in L^p(\mathbb R^n), \forall v\in L^{q,\infty}(\mathbb R^n),\quad \Vert{u\ast v}\Vert_{L^r(\mathbb R^n)}\le C \Vert u\Vert_{L^p(\mathbb R^n)}\Vert v\Vert_{L^{q,\infty}(\mathbb R^n)}, $$ with $p,q,r$ satisfying $(\sharp)$? REPLY [3 votes]: The weak type Young inequality $$ \Vert{u\ast v}\Vert_{L^r(\mathbb R^n)}\le C_{p,q,r} \Vert u\Vert_{L^p(\mathbb R^n)}\Vert v\Vert_{L^{q,\infty}(\mathbb R^n)}, \quad 1 TITLE: Euclidean algorithm for differential operators QUESTION [5 upvotes]: While perusing through the article "Algorithms for solving linear ordinary differential equations" by Winfried Fakler (a pdf can be found through a google search), I noticed Faker mentioning on page 2 that From a mathematical point of view linear differential operators generate a left skew polynomial ring of derivation type. The elements of such a ring are called skew polynomials or Ore polynomials. For Ore polynomials the usual polynomial addition holds. Only the multiplication is different. It is declared as on extension of the rule $a\in k$ $$Da=aD+a'$$ to arbitrary Ore polynomials. Multiplication of Ore polynomials is in fact operator composition. Therefore, $k[D]$ is not a commutative ring. This means there is a left and right division. Indeed, there exists an extended Euclidean algorithm and it is possible to determine for any two nontrivial elements a smallest nontrivial common left multiple. I'm curious to know whether anyone knows the details about this "extended Euclidean algorithm" for differential operators? REPLY [4 votes]: You may find the algorithm in: Li, Noncommutative Gröbner Bases and Filtered-Graded Transfer, Springer Lecture Notes in Mathematics, Vol. 1795. Chapter I.2<|endoftext|> TITLE: What is the best way to construct an Aronszajn Tree? QUESTION [7 upvotes]: What is the best definition of Aronszajn tree? And, what is the best proof that it exists? So I write the question to learn more about Aronszajn trees, any further detail is my intention to appreciate. REPLY [3 votes]: There is also Shelah's very enjoyable construction using descending sequences of infinite subsets of $\omega$, close in spirit to Aronszajn's tree of rational sequences, and described in Judith Roitman's book here: https://www.math.ku.edu/~roitman/stb3fullWeb.pdf.<|endoftext|> TITLE: Galois groups and prescribed ramification QUESTION [7 upvotes]: What is known about finite groups $G$ for which there exists a Galois extension $K$ of $\mathbb{Q}$ ramified only at $2$ such that $\text{Gal}(K/\mathbb{Q}) \cong G$ ? More generally, which groups can be realized over $\mathbb{Q}$ with no ramification outside a given (finite) set of primes? I am thus interested in results of two kinds: Realization of specific groups. Examples of groups which are not realizable in this manner, and restrictions on groups which are. REPLY [8 votes]: Concerning the question of Pablo that follows Joël's answer: If $k$ is an algebraically closed field, then the situation is completely understood, thanks to work of Grothendieck (in characteristic 0) completed by the proof of Abhyankar's conjecture by Raynaud and Harbater. Precisely, let $C/k$ be a smooth affine curve, complement of $r\geq1$ points in a projective smooth connected curve of genus $g$, over the alg. closed field $k$. Let $F=k(C)$ be the field of rational functions on $C$. Extensions of $C$ that are unramified on $C$ correspond to connected étale covers of $C$. If $\mathop{\rm char}(k)=0$, then Grothendieck showed (this is the main result of SGA 1) that a finite group $G$ is the Galois group of a connected étale cover of $C$ if and only if it is generated by $2g+r-1$ elements. More precisely, the Galois group of the maximal extension of $k(C)$ unramified on $C$ is a free profinite group on $2g+r-1$ elements. It is worth observing the fundamental group of a compact connected Riemann surface of genus $g$ deprived from $r\geq1$ points is a free group on $2g+r-1$ generators. (The case $r=0$ is also treated by Grothendieck, $G$ then needs to be generated by $2g$ elements $a_1,b_1,\dots,a_g,b_g$ satisfying the relation $(a_1,b_1)(a_2,b_2)\dots(a_g,b_g)=1$.) If $\mathop{\rm char}(k)=p>0$, then $C$ has « more » étale covers. For example, if $C=\mathbf A^1_k$, then $f\colon C\to C$ given by $f(t)=t-t^p$ is a connected étale cover of degree $p$. Raynaud (when $C=\mathbf A^1_k$) and Harbater (in general) showed that a finite group $G$ is the Galois group of a connected étale cover of $C$ if and only if the quotient of $G$ by the (normal) subgroup generated by its $p$-Sylow subgroups is generated by $2g+r-1$ elements. Specific examples had been exhibited by many mathematicians, such as Abhyankar itself (he has a long series of papers in this style) and Nori (when $G$ is a finite group of Lie type).<|endoftext|> TITLE: The unit tangent bundle of 2- or 4-manifolds as a principal $S^{1}$- or $S^{3}$-bundle QUESTION [5 upvotes]: What type of obstructions have been studied so that the unit tangent bundle of a Riemannian 2-(4-)manifold have a structure of a principal $S^{1}$-($S^{3}$-)bundle? REPLY [2 votes]: Here are some details for the statements in Robert Bryant's answer, mostly as an exercise for myself. A priori the unit tangent bundle of a Riemannian $n$-manifold has structure group $\text{O}(n)$. The questions in the OP can be interpreted to mean the following (if the OP means something else it would be good to clarify): when $n = 2$ what are the obstructions to reducing the structure group from $\text{O}(2)$ to $\text{SO}(2) \cong S^1$? when $n = 4$ what are the obstructions to reducing the structure group from $\text{SO}(4)$ to $\text{SU}(2) \cong S^3$? Presumably the intended map $\text{SU}(2) \to \text{SO}(4)$ comes from the natural action of $\text{SU}(2)$ on itself by, say, left multiplication. Somewhat more explicitly, this map can be written as a composite $$\text{SU}(2) \xrightarrow{\text{id} \times 1} \text{SU}(2) \times \text{SU}(2) \cong \text{Spin}(4) \to \text{SO}(4) \to \text{O}(4).$$ In the $n = 2$ case reduction of the structure group of the tangent bundle from $\text{O}(n)$ to $\text{SO}(n)$ is one of the several equivalent definitions of orientability, so that's that. In the $n = 4$ case, the factoring of the map $\text{SU}(2) \to \text{O}(4)$ above shows that we first need to reduce structure group to $\text{SO}(4)$ (so we need a choice of orientation), then to $\text{Spin}(4)$ (so we need a choice of spin structure). Now, $\text{Spin}(4) \cong \text{SU}(2) \times \text{SU}(2)$ has two distinguished $4$-dimensional spin representations given by the underlying real representations of the defining complex representations of each copy of $\text{SU}(2)$. The associated bundles of these spin representations are the two spinor bundles attached to a spin structure when $n = 4$, and the final reduction to the first copy of $\text{SU}(2)$ is equivalent to a trivialization of the second spinor bundle. $B \text{SU}(2) \cong B \text{Spin}(3)$ is $3$-connected with $\pi_4 \cong \mathbb{Z}$, hence $H_4 \cong H^4 \cong \mathbb{Z}$; the pullback of one of the generators of $H^4$ to our manifold $X$ along the classifying map $X \to B \text{Spin}(3)$ is presumably the Euler class of the corresponding spinor bundle. It is an obstruction to the trivialization of the corresponding spinor bundle, and in this case it is the only one: it vanishes iff the map $X \to B \text{Spin}(3)$ lifts to the homotopy fiber of the universal Euler class $B \text{Spin}(3) \to B^4 \mathbb{Z}$, which is $4$-connected, hence maps into it from a $4$-manifold are nullhomotopic.<|endoftext|> TITLE: Reference request: Optimal $L^p$-decay for nonhomogenous heat equation in $\mathbb R^d$ QUESTION [5 upvotes]: Let $u$ be a classical solution for the nonhomogeneous heat equation in $\mathbb R_+ \times\mathbb R^d$: $$ \begin{cases} \partial_tu(t,x)-\Delta u(t,x) = f(t,x), \\ u(0,x)=u_0(x). \end{cases} $$ Suppose $f(t,\cdot) \in L^q(\mathbb R^n), 1 \le q \le \infty$, for each $t > 0$, and suppose $f$ has the decay behavior $$ \|f(t, \cdot)\|_{L^q} \le \frac{C}{1+t^\gamma}, \quad \gamma >0. $$ Now my question is that, given $\gamma$ and $q$, what is the decay rate of $\|u(t, \cdot)\|_{L^p(\mathbb R^d)}$ for each $1\le p \le \infty$? Probably there is no decay for all values of $p$ in this range, but I am interested in the optimal results. In particular, I would be interested in a reference (book or research paper), where this has been studied in detail. I tried to derive the results from the representation formula for $u$: $$ u(t,x)=\int_{\mathbb R^d} H(t, x-y)u_0(y)\, dy + \int_0^t\int_{\mathbb R^d} H(t-s, x-y)f(s,y) \, dy \, ds, $$ where $ H(t,x)=(4\pi t)^{-d/2}e^{-\frac{|x|^2}{4t}} $ is the heat kernel. Here the problem comes with the latter term for large dimensions $d$. Simple arguments by using Minkowski's (or Hölder's) inequalities give some decay estimates, but then one ends up with some strange looking integrability conditions in the time integral. In particular, in this manner one only obtains decay for some values of $p$ and I doubt the estimates obtained are optimal in this sense. So I am wondering what is the correct approach and what would be the reference for this. REPLY [2 votes]: Suppose that $u_0=0$, otherwise $L^p$ decay estimates are well-known. Let $\{T(t)\}_{t\geq0}$ denote the heat semigroup, i.e., $T(t)$ for $t>0$ is convolution with $(4\pi t)^{-d/2}\,e^{-|x|^2/(4t)}$. Notice that, for $p\geq q$, $$ \|T(t)\|_{L^q\to L^p} = \left\|\frac1{(4\pi t)^{d/2}}\,e^{-\,\frac{|x|^2}{4t}}\right\|_{L^r} = C_{d,r}\,t^{-\,\frac{d}2\left(1-\frac1r\right)}, $$ where $1+\frac1p=\frac1q+\frac1r$ (here, $p,\,q,\,r\in[1,\infty]$). Let $p\geq q$ and $\displaystyle \frac1q-\frac1p<\frac2d$. Further let $\displaystyle \|f(t)\|_{L^q} \lesssim (1+t)^{-\gamma}$ for some $\gamma\in \mathbb R$. Then $\displaystyle \|u(t)\|_{L^p} \lesssim \alpha_\gamma(t)(1+t)^{-\,\frac{d}2\left(\frac1q-\frac1p\right)}$, where $$ \alpha_\gamma(t) = \begin{cases} (1+t)^{1-\gamma}, & \gamma<1, \\ \log \left(e+t\right), & \gamma=1, \\ 1, & \gamma>1. \end{cases} $$ Indeed, $u(t)=\int_0^t T(t-s)f(s)\,ds$, so $$ \begin{aligned} \|u(t)\|_{L^p} &\leq \int_0^t \|T(t-s)\|_{L^q\to L^p} \|f(s)\|_{L^q}\,ds \lesssim \int_0^t \frac{(t-s)^{-\,\frac{d}2\left(1-\frac1r\right)}}{(1+s)^\gamma}\,ds \\ & \lesssim t^{1-\,\frac{d}2\left(1-\frac1r\right)}\sideset{_{2\,}}{_1}F\left(1,\gamma;2-\, \frac{d}2\left(1-\frac1r\right);-t\right). \end{aligned} $$ To find the asymptotics of $\displaystyle \sideset{_{2\,}}{_1}F\left(\dots;-t\right)$ as $t\to\infty$, use $$ \begin{aligned} \sideset{_{2\,}}{_1}F\left(a,b;c;-t\right) & = \frac{\Gamma(b-a)\Gamma(c)}{\Gamma(c-a)\Gamma(b)} \,t^{-a}\sideset{_{2\,}}{_1}F\left(a,a-c+1;a-b+1;-\,\frac1t\right) \\ &\qquad +\,\frac{\Gamma(a-b)\Gamma(c)}{\Gamma(c-b)\Gamma(a)}\,t^{-b}\, \sideset{_{2\,}}{_1}F\left(b-c+1,b;-a+b+1;-\,\frac1t\right) \end{aligned} $$ in case $a-b\notin\mathbb Z$; and similarly for $a-b\in\mathbb Z$. The estimates above are best possible (as $t\to\infty$) provided that $q=1$ or $\gamma\neq1$. In case $\gamma<1$ or $(q,\gamma)=(1,1)$, this is shown by the example of $$ u(t,x) = \frac1{(4\pi t)^{d/2}}\,e^{-\,\frac{|x|^2}{4t}}\,h(t), \quad f(t,x) = \frac1{(4\pi t)^{d/2}}\,e^{-\,\frac{|x|^2}{4t}}\,h'(t), $$ where $h\in\mathscr C^\infty(\overline{\mathbb R}_+)$, $h(t)=0$ near $t=0$, upon choosing $h$ appropriately. For instance, $\displaystyle h'(t)=t^{-\delta}$ for $t\geq1$ (and $h'(t)=0$ for $0\leq t\leq1/2$) with $\displaystyle \delta=\gamma-\frac{d}2\left(1-\frac1q\right)$ will do. For $\gamma>1$, it is enough to remark that $\alpha_\gamma(t)\gtrsim 1$, with this lower bound being implied by the Cauchy problem (e.g., take $f$ such that $f=0$ for $t>1$).<|endoftext|> TITLE: Ends of quotients of Coxeter Groups QUESTION [5 upvotes]: Let C(p,q) be the Coxeter group: $C(p,q):= \langle a,b,c\hspace{1mm}|\hspace{1mm} a^2,b^2,c^2,(ac)^2,(ab)^p, (bc)^q \rangle$ for integers $p,q$ s.t. $\frac{1}{p}+\frac{1}{q}<\frac{1}{2}$. This group is infinite and one ended. Let $G$ an infinite group obtained from $C(p,q)$ by adding some relations to the presentation above. Can $G$ have 2 or infinitely many ends? REPLY [10 votes]: No. Your group has Serre's Property FA, meaning that any action on a tree has a global fixed point. (This can be deduced from the fact that it has a generating set such that every element is torsion, and some product of each pair of elements is also torsion.) Suppose now that some quotient $G$ has more than one end. By Stallings' Theorem, $G$ acts on a tree without a global fixed point, so $C(p,q)$ does too. This is a contradiction.<|endoftext|> TITLE: Is every continuous function measurable? QUESTION [18 upvotes]: This question has already been asked on Math StackExchange here, but was too old to be migrated, and I think will be more appropriate to MathOverflow. In non-Hausdorff topology it is standard to define the Borel algebra of a topological space $X$ as the $\sigma$-algebra generated by the open subsets and the compact saturated subsets. Recall that a subset is saturated if it is an intersection of open subsets, and that compact saturated subsets play the role of compact subsets when the space $X$ is not $T_1$ (which is typically the case for a partially ordered set equipped with the Scott topology for instance). In this situation, one may ask whether every continuous function $f : X \to Y$ between topological spaces is measurable, or equivalently whether every continuous function $f : X \to Y$ is such that $f^{-1}(\uparrow y)$ is measurable for all $y \in Y$, where I write $\uparrow y$ for the intersection of all open subsets containing $y$, which happens to be compact saturated. I do not expect this to be true, but I am rather looking for sufficient conditions on $X$, $Y$ or $f$ in order for $f$ to be measurable. For instance: If $Y$ is $T_1$, then every subset $\uparrow y$ is closed (it coincides with the singleton $\{ y \}$ which itself coincides with its closure), so the continuous map $f : X \to Y$ is measurable. If $Y$ is first-countable, then every subset $\uparrow y$ can be written as a countable intersection of open subsets, so again $f$ is measurable. If $f$ is open and bijective, one can show that the inverse image of $\uparrow y$ is of the form $\uparrow x$, $x \in X$, so $f$ is measurable. Do we know other such situations? Thank you for your help. REPLY [4 votes]: If $X, Y$ are topological spaces such that for every continuous map $f: X\to Y$ and any $K\subseteq Y$ compact, $f^{-1}(K)$ is a compact of subset of $X$, then every continuous function is measurable.<|endoftext|> TITLE: Is there an analog of the Birch/Swinnerton-Dyer conjecture for abelian varieties in higher dimensions? QUESTION [15 upvotes]: I am wondering if there is a multi-dimensional analog of the Birch/Swinnerton-Dyer (BSD) conjecture. The recent famous result inching toward resolution of that conjecture is: Bhargava, Manjul, and Christopher Skinner. "A positive proportion of elliptic curves over $\mathbb{Q}$ have rank one." 2014. arXiv link. As I understand it: "66.48% of elliptic curves satisfy the (rank part of the) BSD Conjecture." I see there are books such as Miyaoka, Joichi, and Thomas Peternell, eds. Geometry of higher dimensional algebraic varieties. Vol. 26. Springer, 1997. that extend elliptic curves to "higher dimensional algebraic varieties." So, Q. Is there an analog of the Birch/Swinnerton-Dyer conjecture for abelian varieties in higher dimensions? Obviously this is a naive question. Thanks for educating me! REPLY [26 votes]: Yes, there is a (well-known) analogue for abelian varieties of all dimensions. I was going to suggest looking at the Wikipedia article on the Birch/Swinnerton-Dyer conjecture and was surprised to see that it only talks about the elliptic curve case. (Clearly an opportunity for someone to add a section on generalizations.) There are also further major generalizations due to Tate (concerning the order of vanishing), Beilinson (concerning the transcendental factor in the leading coefficient, analogous to the real period) and Bloch and Kato (concerning the form of the algebraic part of the leading coefficient, analogous to the SHA*Regulator/(torsion)$^2\cdot (c_p$ factors)). [And I'm probably missing some other names here.] The abelian variety version is undoubtedly described in many places, but since I have a copy handy, I'll mention the statement over $\mathbb Q$ is in my book with Hindry, Diophantine Geometry: An Introduction, Conjecture F.4.1.6 (page 462). It says: Conjecture Let $A/\mathbb Q$ be an abelian variety, and assume that $L(A/\mathbb Q,s)$ has an analytic continuation to $\mathbb C$. Then $$ \operatorname{ord}_{s=1} L(A/\mathbb Q,s) = \operatorname{rank} A(\mathbb Q)$$ and if the rank is $r$, then $$ \lim_{s\to1} \frac{L(A/\mathbb Q,s)}{(s-1)^r} = \Omega_A \frac{\#\text{SHA}(A/\mathbb Q,s)\cdot\text{Reg}(A/\mathbb Q,s)} {\#A(\mathbb Q)_{\text{tors}}\cdot\#\hat A(\mathbb Q)_{\text{tors}}} \cdot \prod_{p} c_p. $$ Most of the terms have the same meaning as for elliptic curves. Notice the appearance of the torsion in the dual $\hat A$. An elliptic curve is self-dual, that's why $\#(E(\mathbb Q)_{\text{tors}})^2$ appears in the elliptic curve version. Also, the regulator is the pairing between a basis for the free parts of $A(\mathbb Q)$ and $\hat A(\mathbb Q)$ relative to the Poincare bundle on the product $A\times\hat A$. The fudge factors $c_p$ are for the primes of bad reduction, with $c_p=\#A(\mathbb Q_p)/A^0(\mathbb Q_p)$, where $A^0$ is the identity component of the Neron model. Alternatively, as suggested by user52824, if one lets $\mathcal A_p$ be the reduction of the Neron model of $A$ at $p$ and $\mathcal A_p^0$ its identity component, then $c_p = \#(\mathcal A_p/ \mathcal A^0_p)(\mathbb F_p)$.<|endoftext|> TITLE: Is there a generalized Birkhoff ergodic theorem? QUESTION [7 upvotes]: Is there a Birkhoff ergodic theorem for two measure preserving transformations $T$ and $S$ where $S\circ T= T \circ S$ so that $\frac{1}{n+1}\frac{1}{m+1}\sum_{i=0}^{n}\sum_{j=0}^{m}f \circ T^{i}\circ S^{j} \to E(f|C)$ for $\mu$-a.e. points where $C$ is $\sigma$-algebra of $S,T$-invariant sets and $E(f|C)$ is conditional expectation of $f$ with the respect to the $\sigma$-algebra $C$? REPLY [15 votes]: The classical Birkhoff ergodic theorem considers a map $T$ acting on a probability space $(X,\mathcal{F},\mu)$. This could alternatively be thought of as an action of $\mathbb{N}$ on $(X,\mathcal{F},\mu)$ by the transformations $T,T^2,T^3,\ldots,T^n,$ et cetera, where $n \in \mathbb{N}$ is identified with the transformation $T^n$. Since at least the 1960s there has been a general effort to extend the Birkhoff theorem to more general groups and semigroups of transformations. A recent high point of this program was a version of the pointwise ergodic theorem due to Lindenstrauss which treats amenable groups of measure-preserving transformations acting on $(X,\mathcal{F},\mu)$. The literature for semigroups of transformations as opposed to groups (since you did not mention whether $T$ and $S$ are invertible) is slightly more sparse but seems to be adequate for your purposes. In your case, since the transformations commute, you are considering the action of $\mathbb{N}^2$ on $(X,\mathcal{F},\mu)$ given by $(i,j)\mapsto T^iS^j$. For this purpose you could use the results of Bewley ("Extension of the Birkhoff and von Neumann ergodic theorems to semigroup actions", Ann. Inst. H. Poincaré Sect. B (N.S.) 7 (1971), 283–291) with $G=\mathbb{N}^2$, $\gamma$ being counting measure, and $A_n$ being some sort of rectangle in $\mathbb{N}^2$ containing $(1,1)$. Note that you must be precise about the manner in which $n$ and $m$ separately tend to infinity: to apply Bewley's theorem I think it is sufficient to assume that $n/m$ is bounded away from zero and infinity.<|endoftext|> TITLE: Derivation of Blattner's conjecture in the Beilinson-Bernstein picture QUESTION [10 upvotes]: On the last page of Schmid's article "Discrete Series", he says "In the Beilinson-Bernstein picture, discrete series modules are attached to closed $K$-orbits in $X$... the $K_{\mathbb R}$-structure of discrete series modules [i.e. Blattner's conjecture, which Schmid proved] is almost obvious from this point of view." I do indeed see how to prove Blattner's conjecture from this point of view, though there is enough battle with $\rho$-shifts that I didn't find it "almost obvious". (Added: I posted this proof on the arXiv.) Is there a place where Blattner's conjecture is derived in this algebraic setting of $(\mathfrak g,K)$-modules, rather than Schmid's original analytic setting? (I did write Schmid, though I suspect if he knew of such a reference he would have referenced it. Of course one may have come out in the intervening time, and if he has an answer I will include it here.) ADDED: Schmid's derivation sounded (on the phone) pretty isomorphic to mine, except where I write out a page (being an outsider to the field, and not clear on which parts are considered easy and which hard) he often puts just a sentence. Nothing wrong with that, of course, if you know your audience and know what will suffice for them. Anyway that doesn't answer my reference-request. REPLY [3 votes]: "Geometric Methods in Representation Theory", by Gregg Zuckerman p.283, in 'Representation Theory of Reductive Groups' ed. Peter C Trombi, Progress in Math. 290 Birkhauser(1983) Procedings of University of Utah conference may have what you are looking for.<|endoftext|> TITLE: sum of binary and ternary digits QUESTION [6 upvotes]: A problem in group theory (indices of imprimitive groups) gives rise to the following conjectures in number theory. Suppose a positive integer $n$ has binary and ternary expansions $n=\sum_{k\geqslant0}b_k2^k=\sum_{k\geqslant0}t_k3^k$. For example, $6=2^2+2=2\cdot 3$; $81=2^6+2^4+1=3^4$. Conjecture 1. If $\sum_{k\geqslant0}b_k=\sum_{k\geqslant0}t_k=2$, then $n\in\{6,10,12,18,36\}$. Conjecture 2. If $\sum_{k\geqslant0}(b_k+t_k)=4$, then $n\in\{6,10,12,18,36,81\}$. Terry Tao discusses the separation of powers of 2 and powers of 3; see his blog: http://terrytao.wordpress.com/2011/08/21/hilberts-seventh-problem-and-powers-of-2-and-3/ Conjectures 1 and 2 are related to separation problems. For example, the four $n$ with $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,2)$ or $(2,1)$ are related to the solutions to $|3^p-2^q|=1$, namely $(p,q)=(1,1),(2,3),(0,1),(1,2)$. More trivially, $(\sum_{k\geqslant0}b_k,\sum_{k\geqslant0}t_k)=(1,1)$ is related to the solution to $|3^p-2^q|=0$, namely $(p,q)=(0,0)$. I wonder whether one needs the values for the `effective constants' in Tao's blog, or whether elementary arguments suffice to prove these conjectures. Answers to the question " $3^n - 2^m = \pm 41$ is not possible. How to prove it? " may help. REPLY [9 votes]: You can prove your conjecture by combining local methods, linear forms in logarithms, and computations (either brute force or continued fractions). Suppose you have $\sum t_k=\sum b_k = 2$, i.e. $2^x+2^y=3^u+3^v$, $x>y$, $u\geq v$. Then $2^{2(x-y)}-1$ is divisible by $3^v$. Since 2 is a primitive root modulo 9, it is primitive modulo $3^k$, hence $2\cdot 3^{v-1}|2(x-y)$. In particular, $3^{v-1}\leq x$. In the same way we find that $2^{y-2}\leq u$. Baker's method gives lower bounds for linear forms in logarithms. In this case the most appropriate result google gave me is due to Bugeaud ( http://www.birs.ca/workshops/2012/12ss131/files/bugeaud_LFL.pdf , Theorem C): Let $a_1, a_2, b_1, b_2$ be rational integers, $a_1, a_2$ multiplicative independent. Let $A_1, A_2$ be real numbers satisfying $A_i\geq\max(3, a_i)$. Put $B=\frac{b_1}{\log A_2}+\frac{b_2}{\log A_1}$. Then for $\Lambda := |b_1\log a_1 + b_2\log a_2|$ we have the bound $$ \log \Lambda \geq -30.9\log A_1\log A_2\left(\max\left(21, 0.66+\log B\right)\right)^2. $$ In our case we have $a_1=2$, $a_2=3$, $b_1=x$, $b_2=u$. Then Baker gives us a lower bound for $\Lambda$, on the other hand we have $$ \Lambda = \log\frac{2^x}{3^u} = \log\frac{2^x+2^y}{3^u+3^v} + \log\frac{2^x}{2^x+2^y} + \log\frac{3^u+3^v}{3^u}, $$ and the first of the three terms on the right is 0, while the other two are very small and differ in sign, hence $\Lambda<\max(\frac{6u}{2^x}, \frac{3x}{3^u})$. I don't optimize $A_1$ and $A_2$, and just put $A_1=A_2=3$ and get $$ \max\left(\frac{6u}{2^x}, \frac{3x}{3^u}\right) > \exp\left(-37.3(0.57+\log(x+u))^2\right). $$ For $x\geq 10000$ this is impossible. You could now check all quadruples $(x,y,u,v)$ with $x<10000$ by a computer, which is easy, since $u$ is uniquely determined by $x$, and $y$ and $v$ are small, or check that $2^x-3^u$ is large for all $x,u$ in this range by computing the continued fraction expansion of $\frac{\log 3}{\log 2}$.<|endoftext|> TITLE: Finite Dimensional Simple nonunital associative Algebras QUESTION [7 upvotes]: I have the following problem: Let K be any field. An finite dimensional associative non-unital algebra A is a vector space A, togeter with a K-biliniear associative operation such that there is NO identity element for this operation. I call such an algebra simple if it has no nontrivial proper ideals. (Edit: I did not define the term ideal: An ideal is supposed to be a vector subspace, closed under the multiplication with elements of the alebgra (from left and from right)) An easy example would be the ground field K as a 1dimensional K-vector space, tohether with the zero-multiplication, sending everything to zero. This is finite dimensional, associative and it certainly has no identity-element. Furthermore, it is simple since it is one-dimensional. Now, my question is: Are there any other examples? The canonical examples of simple associative algebras are matrix algebras but since they contain the identity matrix, they are unital. Another idea would be to take the non-unital algebra A and adjoint a unit to get a unital algebra and then use the known classification there, but this so obtain unital algebra will never be simple since it contains the original non-unital abgera as an ideal. So, another formulation of the problem would be to classify the unital associative algebras with exaclty 3 ideals, where the nontrivial of the 3 has codimension 1 or something like that. I would be very grateful if someone could help me out here, Tom REPLY [6 votes]: Here is a direct argument avoiding adjoining a unit and explicit use of the Wedderburn theorem. In what follows left ideals and ideals are always subspaces. If $A^2=0$, then any subspace is an ideal. So $A$ is 1-dim and is the example you have. So assume $A^2\neq 0$. Then $A^2=A$ since $A^2$ is an ideal. If $L$ is a nilpotent left ideal and $a\in A$ then $La=0$ or $La$ is nilpotent. Thus the sum of all nilpotent left ideals of $A$ is two-sided and hence $A$ or $0$. But $A=A^2$. We conclude that $A$ has no nilpotent left ideals. Let $L$ be a minimal left ideal. Then $L^2\neq 0$ implies $L=La$ for some $a\in L$. Suppose $ea=a$ with $e\in L$. Then $e^2a=ea$ and $e,e^2\in L$. Since $L$ is finite dimensional and $La=L$ we have right multiplication by $a$ is injective on $L$. Thus $e=e^2$ and $L=La=Lea$ implies $e\neq 0$. Thus $L=Ae$ with $e$ idempotent. Next we claim by induction on dimension that each left ideal $L$ is of the form $Ae$ for some idempotent $e$. If $L$ is minimal, we are done by above. Else let $L'$ be a minimal left ideal contained in $L$ and write $L'=Ae$ with $e$ idempotent as above. Let $L''$ be the set of elements $x$ of $L$ with $xe=0$. Then $L''$ is a proper submodule of $L$ and hence $L''=Af$ with $f$ idempotent by induction. Then we claim $L=A(e+f-ef)$ and that $e'=e+f-ef$ is idempotent. If $a\in L$, then $a-ae\in L''=Af$ and so $a-ae=(a-ae)f$. Thus $ae'=a(e+f-ef)=ae+(a-ae)f=ae+a-ae=a$ and therefore $L=Ae'$. Also taking $a=e'$ we see $e'e'=e'$. This completes the induction. Thus $A=Ae$ for some idempotent. Therefore $A$ has a right identity. A dual argument shows $A$ has a left identity and so $A$ has an identity.<|endoftext|> TITLE: Number of edges in linklessly embeddable graphs QUESTION [7 upvotes]: What is the maximum number of edges of an $n$-vertex linklessly embeddable graph? A more general question is the following. What is the maximum number of edges of an $n$-vertex graph with Colin de Verdière number $\mu$? A related question would be the following. Is there a fixed space (say a 2-manifold) such that graphs which embed linklessly into $\mathbb{R}^3$ are characterized by embedding into that space? This is purely out of curiosity. Thanks. REPLY [8 votes]: The reference given in the Wikipedia article on linkless embedding for the $4n-10$ bound on the number of edges in a linkless embeddable graph is Mader, W. (1968), "Homomorphiesätze für Graphen", Mathematische Annalen 178 (2): 154–168, doi:10.1007/BF01350657. Apparently Mader proves this bound more generally for $K_6$-minor-free graphs. As the Wikipedia article also states, the example of apex graphs shows that this is tight. As for the Colin de Verdière invariant: it is known to be at least the size of the largest clique minor minus one – e.g. see http://homepages.cwi.nl/~lex/files/cdvsurvey_new2.pdf – and combining this with the Kostochka/Thomason results cited by Tony Huynh shows that the edge density can grow at most slightly superlinearly as a function of the CdV invariant.<|endoftext|> TITLE: Tannakian formalism for topological Hopf algebras QUESTION [10 upvotes]: Tannaka-Krein duality allows, under the appropriate assumptions, to reconstruct a Hopf algebra from its category of modules. This method was found to be powerful for instance in the work of Etingof-Kazhdan on quantization of Lie bialgebras. Briefly, the coproduct of a Hopf algebra $H$ (say, in vector spaces $Vect_{\mathbb{K}}$) defines a symmetric monoidal structure on its category of modules $Mod_H$. We have a forgetful functor $U:Mod_H\rightarrow Vect_{\mathbb{K}}$ called the fiber functor, so that if $U$ is equipped with a symmetric monoidal structure, then one can recover $H$ via an isomorphism $H\cong End(U)$ (the linear endomorphisms of $U$). My question is the following: is there a Tannaka duality for topological Hopf algebras, e.g. Hopf algebras in Fréchet spaces, Banach spaces, etc...(equipped with the appropriate tensor product) ? If so, what are the main results and good references about this ? REPLY [4 votes]: I believe the best answer to your question at the moment is in this paper (which only treats the case of bialgebras) http://adsabs.harvard.edu/abs/2014arXiv1411.3183L Indeed, in topological setting one has different tensor products. But if one looks carefully on the proofs on Tannaka duality in algebraic setting - they use universal properties of objects of the form $X\otimes X^\wedge$, that are dual to the universal properties of endomorphism objects. Instead of requiring the structure of rigid category (that fails in topological setting due to the necessity to use different tensor products), one can only require the existence of the above-mentioned objects, that are called coendomorphisms in the paper, and it appears that all the proofs can be carried out in this new setting (although it is probably not entirely new). At least for topological vector spaces over $p$-adic fields, the coendomorphisms are precisely the inductive tensor products of the corresponding space and it's dual, which fits into the pattern of the construction.<|endoftext|> TITLE: Preservation Results for Iterations of Non-Proper Forcing QUESTION [7 upvotes]: Suppose $\mathbb{P}$ is a forcing with the following properties: Let $G \subseteq \mathbb{P}$ be filter generic over $V$, then there exists $A \in V[G]$ such that $V[G]$ thinks $A$ is countable and $A \subseteq {}^\omega 2 \cap V$, but $A$ is not covered by any ground model countable set. That is, in the generic extension there is a countable set of ground model reals that can not be covered by a countable set from the ground model. Certainly $\mathbb{P}$ is not a proper forcing. Is it possible that $\mathbb{P}$ can preserve $\aleph_1$? As $A$ need not be a set in $V$, it seems at least theoretically possible that $\mathbb{P}$ does not need to collapse any cardinals. Now I would like to iterate this forcing by itself. However, since it is not proper, I do not know what properties can be expected to be preserved. Although this may be a bit too open ended, my next question is : What classes of forcing (by which I means things like c.c.c., proper, semi-proper, etc) could $\mathbb{P}$ potentially belong to given that it adds a countable set of ground model real that can not be covered by a countable ground model set. (Certainly $\mathbb{P}$ can not be c.c.c. or proper.) Do any of these classes have preservation theorems for iterations (countable support or perhaps some other type of iterations)? I am most interested in perserving $\aleph_1$, preserving ${}^\omega\omega$-bounding, or $\aleph_2$-chain conditions. If it is applicable, one may assume $\mathbb{P}$ has size $\aleph_1$ if $\mathsf{CH}$ holds. I am looking for some class of forcing that can help handle iterations of non-proper forcings like $\mathbb{P}$. Thanks for any information that can be provided. REPLY [7 votes]: Chapters X, XI and XV (possibly also others) in Shelah's book "Proper and Improper Forcing" deal with the problem of iterating nonproper forcing notions. (By this I mean very non-proper. E.g., not even $S$-proper for any stationary $S$.) Chapter X gives a definition of revised countable support iteration (RCS). (It is somewhat difficult to read, but there are other definitions in the literature which could be used instead. Chapter XI deals with nonproper iterations not adding reals over models of CH. There is a property of forcing notions $Q$ that I will call $Pr_{X}(Q)$ that satisfies: Any forcing which has property $Pr_X(Q)$ preserves $\omega_1$ and moreover does not add reals. AND: The limit of an RCS iteration (that uses lots of collapses) of forcing notions with $Pr_X$ will itself have property $Pr_X$. Chapter XV introduces a property that I will call $Pr_{XV}$ and proves a similar statement: Any forcing which has property $Pr_X(Q)$ preserves $\omega_1$ AND: The limit of an RCS iteration (that uses lots of collapses) of forcing notions with $Pr_{XV}$ will itself have property $Pr_{XV}$. A main point is that Namba forcing $Nm$ satisfies both of these properties. (Regardless of whether $Nm$ is semiproper or not. There are several versions of $Nm$: Laver-like or Miller-like, using the club filter or the cobounded filter; I am not sure if all of them have these properties.) Roughly speaking, $Pr_{XV}(Q)$ is this: whenever you have a sufficiently nice tree $(N_\eta: \eta\in \omega_2^{<\omega})$ of countable elementary submodels of the universe, where niceness in particular implies that the intersections of the models with $\omega_1$ converge to the same ordinal $\delta$ along every branch, then $Q$ forces that there exists a branch $\nu\in \omega_2^\omega$ such that $N_\nu[G]\cap \omega_1=N_\nu\cap \omega_1=\delta$. As I recall, if $Q=Nm$ then the generic branch $\nu$ (the union of all stems of conditions in the generic filter) will satisfy the requirement.<|endoftext|> TITLE: $k$-Disk algebras versus $E_k$ algebras QUESTION [10 upvotes]: Background: The little $k$-cubes operad is the $(\infty,1)$-operad defined by embedding disjoint unions of $k$-dimensional open cubes rectilinearly into one another, that is using maps $(0,1)^k\rightarrow (0,1)^k$ of the form $(x_i)\mapsto (a_i i_k+b_i)$ for $x_i\in (0,1)$ and $a_i\geq 0$ for $i=1,...,k$. 2-morphisms are given by isotopies of embeddings, 3-morphisms are isotopies of isotopies etc. Call this operad $\square^k$. Disjoint union equips $\square^k$ with a symmetric monoidal structure. The $(\infty,1)$-category of $E_k$-algebras with values in a symmetric monoidal $(\infty,1)$-category $(C,\otimes)$ is defined as the $(\infty,1)$-category of symmetric monoidal functors $\text{Fun}^{\otimes}(\square^k,C)$. The little $k$-disks $(\infty,1)$-operad is similarly defined as the operad of framed embeddings of open(?) disks into one another, with (higher) isotopies as (higher) morphisms. Call this operad $\text{Disk}_k^{fr}$. The $(\infty,1)$-category of $k$-disk algebras with values in a symmetric monoidal $(\infty,1)$-category $(C,\otimes)$ is again the $(\infty,1)$-category of symmetric monoidal functors $\text{Fun}^{\otimes} ( \text{Disk}_k^{fr},C)$ The question On the nLab, it is written (see here and here) that the little $k$-cubes operad and the little $k$-disks operad are distinct objects, and the latter is a generalization of the former. The main difference I notice is that the little $k$-disks operad allows one to rotate disks when embedding them, while the little $k$-cubes operad does not. Still, rotations are homotopic to the identity, so it seems (to me) safe to assume that the operads are equivalent as $(\infty,1)$-categories. Am I mistaken? Also, according to Ginot's notes (page 27, Example 12), the $(\infty,1)$-categories of algebras of $\text{Disk}_k^{fr}$ and $\square^k$ are equivalent, and this leads me to believe that the operads themselves should be equivalent. I would not be suprised of the above reveals a severe lack of understanding on my part. I am just starting to try to understand these gadgets. Any help will be greatly appreciated. REPLY [10 votes]: There is an unfortunate clash of terminologies here. Traditionally, the little discs operad comes in two variants: the "usual" $\mathtt{D}_n$: the space of arity $r$ operations consists of embeddings of that do not allow rotations (with some other conditions). In other words, such embeddings $D^n \hookrightarrow D^n$ must preserve the framing. the "framed" version $\mathtt{fD}_n$: here the embeddings are allowed to rotate the disks, and do not necessarily preserve the framing. Basically it is $\mathtt{D}_n$ together with an action of $\mathrm{SO}(n)$ (in fact it's a semi-direct product $\mathtt{D}_n \rtimes \mathrm{SO}(n)$, see P. Salvatore and N. Wahl, Framed discs operads and Batalin-Vilkovisky algebras. Q. J. Math., 2003, 54, 213-231"). These two operads are not weakly equivalent, and their categories of algebras are different. To give you an idea, $H_*(\mathtt{D}_2) = \mathtt{Ger}$ is the operad of Gerstenhaber algebras, whereas $H_*(f\mathtt{D}_2) = \mathtt{BV}$ is the operad of BV-algebras -- morally we have a circle action in addition. More generally, $\mathtt{D}_n(1)$ is contractible, whereas $\mathtt{fD}_n(1) \simeq \mathrm{SO}(n)$ is non-contractible, so the operads cannot be weakly equivalent. The first operad $\mathtt{D}_n$ is actually equivalent to $\mathrm{Disk}_n^{\mathrm{fr}}$. This makes perfect sense in this context: $\mathtt{D}_n$ is equivalent to the endomorphism operad of $\mathbb{R}^n$ in the category of framed manifolds and embeddings, and you can take the factorization homology of a $\mathtt{D}_n$-algebra only on a framed manifold. On the other hand, $\mathtt{End}_{\mathbb{R}^n} = \mathrm{Disk}_n$ in the category of unoriented manifolds and embeddings is equivalent to an operad slightly larger than $\mathtt{fD}_n = \mathtt{D}_n \rtimes SO(n)$, I think it is $\mathtt{D}_n \rtimes O(n)$. Its endomorphism operad in the category of oriented manifolds $\mathtt{End}^{\mathrm{or}}_{\mathbb{R}^n}$ is weakly equivalent to $\mathtt{fD}_n$. Unfortunately, as you can see, the two occurrences of "framed" refer to different things, and are applied in opposite manners. As far as I know, a recent trend in some circles is to do away with the terminology "framed little discs operad" altogether. With all that being said, it is indeed true that $\mathtt{D}_n \simeq \mathrm{Disk}_n^{fr}$ is equivalent to the operad $\square^n$ of little $n$-cubes (where you don't allow rotations, of course). A few possible references: R. Steiner, A canonical operad pair, Math. Proc. Cambridge Philos. Soc. 86 (1979), 443–449. C. Berger, Opérades cellulaires et espaces de lacets itérés. Ann. Inst. Fourier 46 (1996), 1125–1157. It's actually not so easy to prove. It's easy to see that they are arity-wise equivalent: in arity $r$, both spaces are equivalent to the configuration space $\operatorname{Conf}_r(\mathbb{R}^n)$ of $r$ ordered points in $\mathbb{R}^n$. It's finding an equivalence that respects the operad structure that is difficult (but possible).<|endoftext|> TITLE: Do the terms of an iid sequence whose law has infinite expected value necessarily exceed the partial sums of the sequence infinitely often? QUESTION [6 upvotes]: Let $\mu$ be a probability measure on $(0,\infty)$, and let $(\mathbf X_n)_1^\infty$ be a sequence of independent $\mu$-distributed random variables. Fix $\kappa > 0$, and consider A) $\int x \; d\mu(x) = \infty$ B) There almost surely exist infinitely many $n$ such that $$ \mathbf X_{n + 1} > \kappa \sum_{i = 1}^n \mathbf X_i. $$ Then (B) implies (A) (this follows from the law of large numbers together with the Borel-Cantelli lemma.) My question is: does (A) imply (B)? According to the generalized Borel-Cantelli lemma, condition (B) is equivalent to B$'$) The series $$ \sum_{n = 1}^\infty \mu\left(\left(\kappa\sum_{i = 1}^n \mathbf X_i,\infty\right)\right) $$ diverges almost surely. This fact was used to show that (B) holds when $d\mu(x) \sim x^{-(1 + \varepsilon)} dx$ ($\varepsilon\in (0,1)$.) I can also use it to show that (B) holds when $d\mu(x) \sim 1/(x^2\log(x))$, $d\mu(x) \sim 1/(x^2\log(x)\log\log(x))$, etc., using complicated calculations. This is highly suggestive that the result holds in general... REPLY [8 votes]: In 1970, Harry Kesten proposed essentially this question in the Advanced Problems section of The American Math Monthly. Let $X_1, X_2, \ldots, X_n$ be iid random variables and $S_n = \sum_{i=1}^n X_i$. Show that $$ \limsup_{n\to\infty} \frac{|X_n|}{|S_{n-1}|} = \infty \quad \text{with probability 1,} $$ whenever the $\mathbb E |X_i|$, the expectation of $|X_i|$ is infinite. The solution, also by Kesten, is in the March 1971 issue (The American Math Monthly, vol. 78, no. 3, 305–308). The proof relies on first reducing to the case where the $X_i$ take values in $\{2^k: k \in \mathbb N\}$. He further uses it to prove a result of Chow and Robbins, If $\mathbb E|X_i| = \infty$ and $(b_n)$ is any sequence of positive numbers, then either $\liminf_n |S_n|/b_n = 0$ almost surely or $\limsup_n |S_n|/b_n = \infty$ almost surely.<|endoftext|> TITLE: Factors of automorphy from Chern connection QUESTION [6 upvotes]: This question is inspired by a recent question about holomorphic bundles and factors of automorphy. Suppose $X$ is a compact, complex manifold whose universal cover $\widetilde{X}$ is Stein (the Stein condition is just to ensure all topologically trivial holomorphic vector bundles over $\widetilde{X}$ are holomorphically trivial). Then I've just recently come to appreciate the theorem that every holomorphic vector bundle (whose pullback to $\widetilde{X}$ is topologically trivial) \begin{align} E\rightarrow X \end{align} arises from a factor of automorphy, this is a holomorphic map: \begin{align} f:\pi_1(X) \times \widetilde{X}\rightarrow GL_n(\mathbb{C}) \end{align} satisfying the cocycle condition $f(\gamma\eta, x)=f(\gamma,\eta(x))f(\eta, x).$ The map $f$ defines a $\pi_1(X)$-action on the trivial bundle $\widetilde{X}\times \mathbb{C}^n\rightarrow \widetilde{X}$ and the quotient defines a holomorphic vector bundle over $X.$ Now suppose that the holomorphic structure on $E\rightarrow X$ is given as the $(0,1)$-part of a linear connection $\nabla$ on $E.$ If this connection is flat, then the factor of automorphy is nothing more than the monodromy representation $\rho:\pi_1(X)\rightarrow GL_n(\mathbb{C})$ of this flat vector bundle. $\textbf{My question is:}$ if the connection is not flat, is there a gauge theoretic construction of the factor of automorphy defining the holomorphic structure on $E.$ Ideally this would be something involving parallel transport using the connection $\nabla,$ but in my first fumbling attempts I haven't found a way to do this. My motivation for this question arises from something that has befuddled me for some time. I won't define everything in this section but a reference for anything I say is in Hitchin's paper "The self-duality equations on a Riemann surface." Let $X$ be a compact Riemann surface of genus at least 2. Let $\rho:\pi_1(X)\rightarrow SL_2(\mathbb{R})$ be a Fuchsian representation. Then via the non-abelian Hodge correspondence, the Higgs bundles $(E,\phi)$ corresponding to any Fuchsian representation all have the same underlying holomorphic bundle, namely \begin{align} E=K^{-\frac{1}{2}}\oplus K^{\frac{1}{2}} \end{align} where $K$ is the canonical bundle over $X.$ This has always been mysterious to me, starting from the flat bundle side; namely you form the flat bundle $E_{\rho}=\widetilde{X}\times_{\rho}\mathbb{C}^2$ and let's call the flat connection $B.$ Then by Hitchin, there exists a unique unitary connection $\nabla$ such that \begin{align} B=\nabla+\psi, \end{align} and $d_{\nabla}^{*}\psi=0.$ The holomorphic structure on $E_{\rho}$ in the Higgs bundle picture is then the one induced by the $(0,1)$-part of the unitary connection $\nabla,$ which is not the same as the one induced by the flat connection $B$ on $E_{\rho}.$ But, why should this always be $K^{-\frac{1}{2}}\oplus K^{\frac{1}{2}}.$ I've never reached a satisfying answer to this issue. I wonder if some insight into the question I asked above could clarify this situation. Thank you for any help. REPLY [2 votes]: I will not answer your question, but tell you something about the holomorphic structure of a Fuchsian representation. I guess you are aware of the paper "Twisted harmonic maps..." of Donaldson (cited in Hitchin's paper). The thing is that if you start with a Fuchsian representation, your corresponding twisted harmonic map is just the developing map of you constant curvature metric into hyperbolic 2-space sitting in hyperbolic 3-space. This leads to some kind of symmetry of your connection $B$ and its decomposition into $B=\nabla+\psi$. Basically, $\nabla$ must be diagonal and $\psi$ off-diagonal. To work things out, you can also take a look into Hitchin's paper "Harmonic tori in $S^3$" (Proposition 1.9). There Hitchin also studied the case of harmonic maps into $S^2$ which led to the same kind of symmetry as in your case (but different signs...). The next thing you need to understand is, why the entries of the diagonal part are a spin structure and its dual. To see this, look at the $(1,0)$ $\Psi$ part of $\psi$ which is a holomorphic off-diagonal 1-form with values in the endomorphism bundle: If your Fuchsian representation is the one of your Riemann surface, than the harmonic map is conformal and therefore $\det \Psi=0.$ But the developing map has no branch point, hence $\Psi$ is nowhere vanishing. From this and trace-freeness you see that, with respect to your decomposition, one of the off-diagonal terms of $\Psi$ vanishes whereas the other is constant to $1.$ This implies that the holomorphic structure is $K^{1/2}\oplus K^{-1/2}.$ If you deform your Fuchsian representation slightly, everything depends real analytically on the deformation. In particular, there is locally one off-diagonal entry (the upper right one) of $\Psi$ which is not vanishing, but still holomorphic. So in a neighborhood of the Fuchsian representation of your Riemann surface $E=K^{1/2}\oplus K^{-1/2}.$ To deduce the general case observe, that the upper right entry of $\Psi$ either vanishes everywhere or nowhere (as it is constant), but the first case can only happen at the Fuchsian representation of the Riemann surface.<|endoftext|> TITLE: what-if.xkcd.com: stabbing (simply connected) regions on the 2-sphere with few geodesics QUESTION [51 upvotes]: In the latest what-if Randall Munroe ask for the smallest number of geodesics that intersect all regions of a map. The following shows that five paths of satellites suffice to cover the 50 states of the USA: A similar configuration where the lines are actually great circles is claimed by the author: They're all slightly curved, since the Earth is turning under the satellites, but it turns out that this arrangement of lines also works for the much simpler version of the question that ignores orbital motion: "How many straight (great-circle) lines does it take to intersect every state?" For both versions of the question, my best answer is a version of the arrangement above. There has been quite some work on similar sounding problems. For stabbing (or finding transversals of) line segments see as an example Stabbing line segments by H. Edelsbrunner, H. A. Maurer, F. P. Preparata, A. L. Rosenberg, E. Welzl and D. Wood (and papers which reference it.) or L.M. Schlipf's dissertation with examples of different kinds. Is there an algorithmic approach known to tackle this problem (or for the simpler problem when all regions of the map are convex)? In the case of the 50 states of the USA, it is of course easy to see that one great circle does not suffice: take two states (e.g. New York and Louisiana) such that all great circles that intersect those do not pass through a third state (e.g. Alaska). Similarly one can show that we need at least 3 great circles. Maybe it would be helpful to consider all triples of regions that do not lie on a great circle and use this hypergraph information to deduce lower bounds. What are good methods to find lowers bounds? Randall Munroe's conjectures that 5 is optimal: I don't know for sure that 5 is the absolute minimum; it's possible there's a way to do it with four, but my guess is that there isn't. [...] If anyone finds a way (or proof that it's impossible) I'd love to see it! REPLY [3 votes]: It is known to be NP-hard to cover regions (or even just points) with a minimum number of lines. For the Euclidean plane, see Megiddo, Nimrod and Tamir, Arie (1982), "On the complexity of locating linear facilities in the plane", Oper. Res. Lett. 1 (5): 194–197, doi:10.1016/0167-6377(82)90039-6. Their construction is flexible enough that, at least for the region version, it should extend to the approximations to Euclidean geometry that one gets in small patches of spherical geometry.<|endoftext|> TITLE: CAT(0) groups that does not act on CAT(0) cubical complex QUESTION [10 upvotes]: CAT(0) groups are groups that act on a CAT(0) space properly and cocompactly. If a group acts on a CAT(0) cubical complex properly and cocompactly, then of course it is a CAT(0) Group. I am wondering the other direction, Quesion: Given a CAT(0) group G , will one always be able to find a CAT(0) cubical complex such that G acts properly and cocompactly. I assume the answer is no, but I could not find any counter example by google. If the answer is indeed no, one can further ask does there exists a CAT(0) group that can not act on a CAT(0) cubical complex properly. REPLY [10 votes]: Many CAT(0) groups cannot act geometrically on CAT(0) cube complexes. For instance: CAT(0) groups satisfying Kazhdan's property (T), eg. uniform lattices in simple Lie groups of higher rank or in quaternionic hyperbolic spaces (as mentioned by Jean Raimbault in the comments). Some Kähler groups (as mentioned by Misha in the comments), including uniform lattices in complex hyperbolic spaces. Some crystallographic groups, as proved by Mark Hagen. Some Coxeter groups (see Niblo and Reeves' article for more details). Some braid groups, eg. $B_n$ for $n=4,5,6$ (see Haettel's article Virtually cocompactly cubulated Artin-Tits groups and references therein). The group $\mathrm{Aut}(B_4) \simeq \mathrm{Aut}(\mathbb{F}_2)$ (see Piggott, Ruane and Walsh's article The automorphism group of the free group of rank two is a CAT(0) group). Probably, the easiest example of such a group is the triangle group $$T=\langle a,b,c \mid a^2=b^2=c^2=(ab)^3=(bc)^3=(ac)^3=1 \rangle.$$ It coincides with the symmetry group of the regular tilling of the Euclidean plane by equilateral triangles. So it acts geometrically on the Euclidean plane: it is a CAT(0) group. As observed by Wise, if a virtually $\mathbb{Z}^n$ group $G$ acts geometrically on a CAT(0) cube complex, then it acts geometrically on $\mathbb{E}^n$ endowed with its usual cubulation. (See Lemma 16.12 in his long paper The structure of groups with a quasiconvex hierarchy. Essentially, the idea is the following: apply the flat torus theorem to find a $G$-invariant flat $\mathbb{R}^n$ inside your cube complex $C$, and look at the CAT(0) cube complex obtained by cubulating the wallspace defined by the intersections of the hyperplanes of $C$ with your flat.) Consequently, if $T$ were cubulable then it would act geometrically on the square complex $\mathbb{E}^2$. Notice that $\mathrm{Isom}(\mathbb{E}^2)= (D_{\infty} \times D_{\infty}) \rtimes \mathbb{Z}_2$ does not contain elements of order three, so, for any action $T \curvearrowright \mathbb{E}^2$ by isometries, the elements $ab$, $bc$ and $ac$ must be trivial. Consequently, such an action must factorise through the quotient $T \twoheadrightarrow \mathbb{Z}_2$ sending all the generators to $1$. A fortiori, the action cannot be geometric (and even proper).<|endoftext|> TITLE: Holomorphic cusp forms and cohomology of GL(2,Z) QUESTION [13 upvotes]: Let $V_{k}$ denote the complex representation of $\mathrm{GL}(2)$ given by $\mathrm{Sym}^k(V)$, where $V$ is the defining 2-dimensional representation. Assume that $k$ is even. I would like to compute the cohomology group $$ H^1(\mathrm{GL}(2,\mathbf Z),V_{k}). $$ By the Lyndon-Hochschild-Serre spectral sequence, this is the same as the $\mathbf Z/2$-invariants in $H^1(\mathrm{SL}(2,\mathbf Z),V_k)$. The latter cohomology group is a direct sum of three terms, each of which corresponds to a space of modular forms of weight $k+2$: one to holomorphic cusp forms, one to antiholomorphic cusp forms, and one to Eisenstein series. I think that the $\mathbf Z/2$-invariants consist exactly of the holomorphic cusp forms, but I can't seem to figure out why. Can someone give an argument and/or a reference that explains this, preferably in geometric terms? Thanks to Joël's great answer I think I understand what's going on. Let me try to make really explicit where my confusion came from. If you write the Eichler-Shimura isomorphism $$ H^1_{\mathrm{cusp}}(\mathrm{SL}(2,\mathbf Z),V_k) \cong S_{k+2} \oplus \overline S_{k+2}$$ then it's not true as I thought (and which led to nonsensical answers) that the subspace $S_{k+2}$ is $\mathbf Z/2$-invariant and $\overline S_{k+2}$ is anti-invariant. Instead the $\mathbf Z/2$-action interchanges these two subspaces according to the involution $$ f(z) \mapsto -f(-\overline z),$$ which maps holomorphic modular forms to antiholomorphic ones and vice versa. (I guess depending on how you've defined the Eichler-Shimura isomorphism there might be a sign difference here.) Still, there is of course a natural way to identify the invariant subspace with the space of holomorphic cusp forms. REPLY [6 votes]: Let me try an answer. Instead of working with $H^1(SL(2,{\bf Z}),V_k)$, I'll work with a space which is naturally isomorphic to it, but more concrete, the space of modular symbols $Symb(V_k)$, defined as $Hom(\Delta^0,V_k)^\Gamma$. Here, $\Delta^0$ is the abelian group of divisors of degree $0$ on the projective rational line $\mathbb P^1(\bf Q)$, and $\Gamma = SL_2(\bf Z)$ for shortness. The isomorphism between those two spaces is standard, follows from the long exact sequence of cohomology attached to the pair (Poincaré half-plane with cusps $P^1(\bf Q)$ attached, $P^1(\bf Q)$), and commute with the action of Hecke operator on both spaces, in particular, what matters to us, with the action of the "Hecke operator at $\infty$" $\iota$ defined by the matrix $\left( \begin{matrix} 1 & 0 \\ 0 & -1\end{matrix}\right)$ (a generator of $GL_2({\bf Z})/SL_2({\bf Z})$). Also I see $V_k$ as the dual of the space of polynomial in one variable z of degree $\leq k$, with the natural homographic action of $\Gamma$. That's equivalent to your definition as a $Sym^k$. Now we need to describe explicitly the relation you mention between modular forms and cohomology/modular symbols. Let us begin by cusp forms. In your post, the $\Gamma$-module $V_k$ is a complex vector space, but of course it is naturally defined over $\bf Z$, and I will call $V_k(\bf R)$ the version over $\bf R$. Let $S_{k+2}=S_{k+2}(\Gamma,\bf C)$ be the space of cusp forms. Then there is an injective $\bf R$-linear map $$S_{k+2} \rightarrow Symb(V_k({\bf R})),\ \ f \mapsto \phi_f,$$ where $\phi_f$ is the modular symbol defined by $$\phi_f(D)=(P \mapsto \Re( \int_D f(z) P(z) dz)).$$ Here, for $D=[a] - [b]$, $\int_D$ means the integral along any geodesic in the Poincaré half plane relying the cusp $a$ to the cusp $b$, for more general $D \in \Delta^0$, $\int_D$ is defined by linearity, and we see the right hand side is a linear form on the space of polynomials $P$ of degree $\leq k$ with real coefficients, hence an element of $V_k(\bf R)$ as it should be. Now if $f = \sum a_n q^n \in S_{k+2}$, let $f'=\sum \bar a_n q^n \in S_{k+2}$. The $\bf R$-linear involution $f \mapsto f'$ is of course the complex conjugation on $S_{k+2}$ associated with its canonical real structure given by the forms defined over $\bf R$ (= with real coefficients). Note that $\overline{f(z)} = \sum \bar a_n \overline{e^{2 i \pi n z}} = \sum \bar a_n e^{-2 i \pi n \bar z}=f'(-\bar z)$. If $D=[\infty]-[a] \in \Delta^0$, $a \in \bf Q$, one has $\iota(D)=[\infty]-[-a]$ one has: $$\Re \left( \int_D f(z) P(z) dz \right) = \Re \left( \int_0^\infty f(a+iy) P(a+iy) i dy \right) \\ = \Re \left( \overline{ \int_0^\infty f(a+iy) P(a+iy) i dy } \right) \\ = - \Re \left( \int_0^\infty f'(- a+iy) P(a-iy) i dy \right) \\ = -\Re \int_{\iota D} f'(z) P(-z) dz, $$ hence $$\phi_{f'} = - \iota (\phi(f)),$$ that is the $\iota$ involution on modular symbols corresponds to minus the complex conjugation on modular cusp forms. Now we complexify this real linear map, getting an injective map $$S_{k+2}(\Gamma,{\bf C}) \otimes_{\bf R} {\bf C} \hookrightarrow Symb(V_k({\bf C}),$$ where the left hand side can be identified with two copies of $S_{k+2}(\Gamma,{\bf C})$ and clearly from the above and easy linear algebra, one copy is identified with the $+1$-eigenspace, an other with $-1$ eigenspace for $\iota$ on the RHS. So this justifies your claim that the subspace invariant by $\iota$ in $Symb(V_k)$ contains one copy of the holomorphic modular cusp forms. There is also the question of Eisenstein series. Working in level 1 as you do, there is only one of them for each k, and the question if whether the modular symbol corresponding to it is of eigenvalue $+1$ or $-1$ for $\iota$. This implies unfortunately that to have it right we need to be very careful in choosing a consistent system of conventions, which is a little bit too hard for me. If I interpret correctly my own computations with Samit Dasgupta however in our paper THE p-ADIC L-FUNCTIONS OF EVIL EISENSTEIN SERIES (Prop. 2.6), to appear in compositio, I think this sign is +1, which would mean that the Eisenstein series is also in the invariant subspace of $Symb(V_k)$, in other words also appears in $H^1(GL_2({\bf Z}),V_k)$. but maybe I'm wrong. What makes you think the latter is constituted of only cuspidal nodular forms?<|endoftext|> TITLE: Classification of rings satisfying $a^4=a$ QUESTION [19 upvotes]: We have the famous classification of rings satisfying $a^2=a$ (for each element $a$) in terms of Stone spaces, via $X \mapsto C(X,\mathbb{F}_2)$. Similarly, rings satisfying $a^3=a$ are classified by pairs of Stone spaces via $(X,Y) \mapsto C(X,\mathbb{F}_2) \times C(Y,\mathbb{F}_3)$. (This kind of classification works for all rings with $a^n=a$ where $n$ is such that every prime power $q$ with $q-1|n-1$ is a prime number. Do these $n$ have a concise description or a name?) Question. How to classify the rings satisfying $a^4=a$ for each element $a$? They are commutative and embed into some product of copies of $\mathbb{F}_2$ or $\mathbb{F}_4$. A typical example would be $\{f \in C(X,\mathbb{F}_4) : f(E) \subseteq \mathbb{F}_2\}$ for some Stone space $X$ and a closed subset $E$. But because of $\mathrm{Aut}(\mathbb{F}_4)=\mathbb{Z}/2$ there cannot be a purely topological classification. I suspect that the sheaf cohomology group $H^1(X \setminus E,\mathbb{Z}/2)$ might enter here. In fact, if $A$ is a ring satisfying $a^4=a$, then $E:=\{\mathfrak{p} : A/\mathfrak{p} \cong \mathbb{F}_2\}=V(a^2-a : a \in A)$ is a closed subset of the Stone space $X:=\mathrm{Spec}(A)$ and for every compact open subset $V$ of $X \setminus E$ we have an isomorphism of sheaves $\mathcal{O}_X |_{V} \cong \underline{\mathbb{F}_4}|_V$, because the sheaf of isomorphisms between them is a $\mathbb{Z}/2$-torsor and $H^1(V,\mathbb{Z}/2)=0$. But the potential failure of $\mathcal{O}_X |_{X \setminus E} \cong \underline{\mathbb{F}_4}|_{X \setminus E}$ should be encoded in $H^1(X \setminus E,\mathbb{Z}/2)$. REPLY [3 votes]: Here is a proof for one part of Neil Strickland's answer. If $R$ is a $4$-ring, then $R \to \hom_{C_2}(\hom_{\mathsf{Ring}}(R,\mathbb{F}_4),\mathbb{F}_4)$, $r \mapsto (f \mapsto f(r))$ is an isomorphism. Proof. If $r \in R$ lies in the kernel, this means that $f(r)=0$ for all $f \in \hom(R,\mathbb{F}_4)$. Hence, $f$ lies in every prime ideal, and therefore in the nilradical, which is zero. This proves that injectivity.. For surjectivity, let $X:=\hom(R,\mathbb{F}_4)$ and let $\alpha : X \to \mathbb{F}_4$ be a $C_2$-equivariant continuous map. Write $\mathbb{F}_4=\{0,1,u,u^2\}$. We may write $\alpha$ as $\alpha = \chi_{Y} + u \chi_{Z} + u^2 \chi_{\sigma Z},$ where $Y = \alpha^{-1}(\{1\})$ and $Z = \alpha^{-1}(\{u\})$ and hence $\sigma Z = \alpha^{-1}(\{u^2\})$. Notice that $Y,Z,\sigma Z$ are disjoint clopen subsets of $X$ and that $\sigma Y = Y$. In the decomposition of $\alpha$, we have to be a bit careful since $u \chi_Z$ does not lie in $\hom_{C_2}(X,\mathbb{F}_4)$, but $\chi_Y$ and $u \chi_{Z} + u^2 \chi_{\sigma Z}$ do. We show that both functions are in the image. First, let us treat $\chi_Y$. It suffices to find some idempotent element $r \in R$ such that $Y = \{f \in X : f(r)=1\}$, because then $(f \mapsto f(r))$ equals $\chi_Y$. Consider the map $X \to \mathrm{Spec}(R)$, $f \mapsto \ker(f)$. It is easily seen to be continuous. Since $f$ and $\sigma f$ have the same kernel, we get a continuous map $X/C_2 \to \mathrm{Spec}(R)$, which is clearly bijective. Since $X/C_2$ is compact and $\mathrm{Spec}(R)$ is Hausdorff, the map is a homeomorphism. It follows that there is a 1:1 correspondence between clopen subsets of $X/C_2$ and the clopen subsets of $\mathrm{Spec}(R)$, which in turn correspond to idempotent elements. Since $\sigma Y = Y$ and $Y$ is clopen, the image of $Y$ in $X/C_2$ is clopen, and we are done. Now let us treat $u \chi_{Z} + u^2 \chi_{\sigma Z}$. Since $Z$ is open, we may write $Z$ as a union of non-empty sets of the form $\{f \in X : f(r_1)=i_1,\dotsc,f(r_n)=i_n\}$ for some $r_k \in R$ and $i_k \in \mathbb{F}_4$. Since $Z \cap \sigma Z = \emptyset$, not all $i_k$ can be contained in $\mathbb{F}_2$. Since $f(r)=u^2$ is equivalent to $f(r^2)=u$, we may therefore assume that $i_1=u$. If $i_k=0$, rewrite the relation $f(r_k)=0$ as $f(r_1-r_k)=u$. If $i_k=1$, replace $r_k$ by $r_k-1$ and reduce to $i_k=0$. If $i_k=u^2$, replace $r_k$ by $r_k^2$ and reduce to $i_k=u$. Hence, each set may be written as $\{f \in X : f(r_1)=\dotsc=f(r_k)=u\}.$ We claim that this already equals $\{f \in X : f(r)=u\}$ for some $r \in \langle r_1,\dotsc,r_k \rangle$. By induction, it suffices to assume $k=2$. Then, it suffices to find a polynomial $p \in \mathbb{F}_2[x,y]$ such that for $a,b \in \mathbb{F}_4$ we have $p(a,b)=u$ if and only if $a=b=u$. A possible choice is $p := x (1 - (x-y)^3).$ In fact, if $a \neq b$ in $\mathbb{F}_4$, then $p(a,b)=0$, and otherwise $p(a,b)=a$. This implies the desired property. We have proven that $Z$ is a union of sets of the form $\{f \in X : f(r)=u\}$ for various $r \in R$. Since $Z$ is assumed to be closed and therefore compact, finitely many $r$ suffice. We claim that there is some $r \in R$ such that $Z = \{f \in X : f(r)=u\}$. By induction, we may assume that $Z = \{f \in X : f(r_1)=u\} \cup \{f \in X : f(r_2)=u\}$. Since $Z \cap \sigma Z = \emptyset$, we see that $f(r_1)=u \Rightarrow f(r_2)^2 \neq u$ for all $f \in X$, and likewise $f(r_2)=u \Rightarrow f(r_1)^2 \neq u$. Therefore, similar to the technique before, it suffices to find a polynomial $p \in \mathbb{F}_2[x,y]$ such that for $a,b \in \mathbb{F}_4$ with $(a,b) \notin \{(u,u^2),(u^2,u)\}$ we have $p(a,b)=u$ if and only if ($a=u$ or $b=u$). We define $p_1:=x^2+x$, $p_2:=(x^2+x+1)(y^2+y)$ and finally $p := x p_1 + y p_2$. As polynomial functions on $\mathbb{F}_4 \times \mathbb{F}_4$, we have $p_1 = \chi_{\{u,u^2\} \times \{0,1,u,u^2\}}$, $p_2 = \chi_{\{0,1\} \times \{u,u^2\}}$. It follows that $p = x \chi_{\{u,u^2\} \times \{0,1,u,u^2\}} + y \chi_{\{0,1\} \times \{u,u^2\}}.$ Hence, $p(a,b)=u \Leftrightarrow a=u \vee b=u$ except for $p(u^2,u)=u^2$. Since we don't have to care for $(u^2,u)$, that's enough. We have proven that $Z = \{f \in X : f(r)=u\}$ for some $r \in R$. It follows that $\sigma Z = \{f \in X : f(r)=u^2\}$. Thus, if we substract $(f \mapsto f(r))$ from $u \chi_{Z} + u^2 \chi_{\sigma Z}$, we obtain a function which has only values in $\{0,1\}$. We already know that such a function lies in the image of the counit, hence $u \chi_{Z} + u^2 \chi_{\sigma Z}$ lies in the image, too. $\square$ Thanks to Jyrki Lahtonen for pointing out the polynomials in the proof above. In some sense, we don't really have to write these polynomials down since their existence is guaranteed by the finite case of the claim which is easy to deal with since every finite $4$-ring is a direct product of copies of $\mathbb{F}_2$ or $\mathbb{F}_4$.<|endoftext|> TITLE: When do you go hunting for Lagrangian submanifolds? QUESTION [12 upvotes]: Similar to this question, I'm trying to figure out why one would be interested in Lagrangian submanifolds. But from a more geometric point of view. My best find so far is Exercise 12.4 in McDuff, Salamon, which is to ... construct a symplectic embedding $$B^{2n+2}(r)\hookrightarrow \mathbb{T}(1)\times\mathbb{R}^{2n}$$ in the following way. Find a linear Lagrangian subspace $L$ of $\mathbb{R}^2\times \mathbb{R}^{2n}$ whose $\epsilon$-neighbourhood $L_\epsilon$ projections injectively into $\mathbb{R}^2/\mathbb{Z}^2\times \mathbb{R}^{2n}=\mathbb{T}^2(1)\times \mathbb{R}^{2n}$. Then consider a composite mapping $$B^{2n+2}(r)\hookrightarrow L_\epsilon\hookrightarrow \mathbb{T}^2(1)\times\mathbb{R}^{2n}.$$ A bit earlier they explained that one may embed $B^{2n}(1)\hookrightarrow B^n(\delta)\times \mathbb{R}^n$ by $(x,y)\mapsto (\delta x,\delta^{-1}y)$. I suspect that the reason people care about Lagrangian submanifolds is that there are more situations where one may move a construction from one place to another by finding appropriate Lagrangian submanifolds. So my question is: What are other geometric problems in which the existence of a certain Lagrangian manifold is used in a manner similar to the above? That is, when do you go hunting for Lagrangian submanifolds? Edit: To clarify, I'm not that much interested in generalities, but in concrete geometrical problems where a Lagrangian does the trick. REPLY [4 votes]: Another place where Lagrangian submanifolds arise naturally is in cotangent bundles. If $M\subset N$ is a submanifold then you can define the conormal bundle of $M$ to be the set of covectors which annihilate $TM$. This is Lagrangian (in fact an exact Lagrangian). You also get a Legendrian submanifold when you intersect this with the unit cosphere bundle (Legendrians are the contact-geometric cousins of Lagrangian submanifolds). Legendrians and Lagrangians are very intimately related and you might as well go hunting for both. Now contact geometric invariants of this Legendrian (or symplectic invariants of the Lagrangian) conormal bundle are smooth invariants of the embedded submanifold $M$. In particular, Ng's knot contact homology (when $M\subset N$ is a knot in a 3-manifold) gives extremely interesting knot invariants related (see recent work of Aganagic-Ekholm-Ng-Vafa based on ideas of Witten) to Chern-Simons theory. Just as in Jean-Claude Sikorav's answer (though in a completely orthogonal direction) this indicates that (exact) Lagrangian submanifolds of cotangent bundles remember much about smooth topology in the base.<|endoftext|> TITLE: polynomials in many variables and Hasse principle QUESTION [5 upvotes]: I was wondering whether there exists any result of the form "if $f \in \mathbb{Z}[x_1, ..., x_k]$ is a polynomial (not form! I don't require homogeneity) of total degree $n$, with $k \geq \delta n$ for some explicit $\delta$, then $f(x) = 0$ satisfies the Hasse principle." I'm particularly interested in the case $n = 4$. Thanks! EDIT: I'm interested in integral solutions. REPLY [10 votes]: The integral Hasse principle can fail for polynomials of degree $2$ (or $4$), even if the number of variables is huge! For example, if $k \equiv 1 \pmod{8}$, then $$(2x_1-1)^2 + \cdots + (2x_k-1)^2 = 1$$ has the solution $(1,1/2,\ldots,1/2)$ over $\mathbf{R}$ and $\mathbf{Z}_p$ for all odd $p$, and has a solution $(a,1,\ldots,1)$ over $\mathbf{Z}_2$ for some $a$ near $1$ by Hensel's lemma, but it has no solution over $\mathbf{Z}$ for $k>1$ since each square on the left is at least $1$. For a much more detailed study of the integral Hasse principle, see Colliot-Thélène and Xu, Brauer-Manin obstruction for integral points of homogeneous spaces and representation by integral quadratic forms, Compos. Math. 145 (2009), no. 2, 309–363.<|endoftext|> TITLE: cardinality of perfect sets in generalized Baire space QUESTION [9 upvotes]: I've been unable to find an answer to the following question in the literature on generalized descriptive set theory. Consider Baire space $\kappa^{\kappa}$ where $\kappa$ is inaccessible. The basic open sets are the $U_f$ where $f\in\kappa^{<\kappa}$. A perfect set is a nonempty closed set with no isolated points. Does a perfect set have cardinality $2^{\kappa}$? Does it have to have cardinality $>\kappa$? A variation of an observation in "Generalized Descriptive Set Theory and Classification Theory" (arxiv.org/abs/1207.4311) states that if $T$ is a slim Kurepa tree (as defined in Devlin's "Constructibility") then there is no continuous (in fact Borel) injection of $2^{\kappa}$ into $[T]$, but this doesn't answer my question. Of course, if the existence of a slim Kurepa tree was consistent with $2^{\kappa}>\kappa^+$ then it would be consistent that my first question had a "no" answer; I have not been able to find any references on this question, but I wouldn't be surprised if it were so. REPLY [10 votes]: For every $\kappa$ of uncountable cofinality there is a tree $T\subseteq 2^{<\kappa}$ such that $[T]=\kappa$. The tree $T$ is the tree of all binary sequences $f\colon \alpha \to 2$, $\alpha <\kappa$ such that $f^{-1} (1)$ is finite. It is clear that for every $f \in T$, $f\frown (1), f\frown (0)$ are both in $T$, so this tree is prefect. Any branch $b$ of $T$ contains finitely many $1$-s, since $\text{cf }\kappa > \omega$ and therefore is there were infinitely many $1$-s, there was some $\alpha$ such that $b\restriction \alpha$ contains already infinitely many $1$-s. On the other hand, it is clear that for every $b\colon \kappa \to 2$, with $b^{-1}(1)$ finite, $\{ b\restriction \alpha \mid \alpha < \kappa \}$ is a branch in $T$. There are $\kappa^{<\omega} = \kappa$ such branches, as wanted. Edit: I argue that the $ZFC$ doesn't prove that there are prefect sets in $^\kappa\kappa$ of size $\kappa^{+}$. The proof is similar to the proof of the consistency of "there are no Kurepa trees". Theorem: Assume $GCH$. Let $\kappa < \eta < \mu$ be regular cardinals, $\eta$ inaccessible. Then after forcing with $\mathbb{Q} = Add(\kappa,\mu)\times Col(\kappa,<\eta)$, for every $\kappa$-tree $T$, $|[T]| \in \kappa \cup \{\kappa, \mu\}$. In this generic extension $\eta = \kappa^+$, $\mu = 2^\kappa$ and every cardinal $\geq \eta$ is preserved. Proof: We need the following well known fact: Fact: Let $T$ be a tree of height $\kappa$. If there is a $\kappa$-closed forcing that adds a branch to $T$ then $|[T]| = 2^\kappa$. Sketch of proof: Let $\mathbb{P}$ be a $\kappa$-closed forcing that adds a branch for $T$ and let $\dot{b}$ be the name of this new branch. Define an embedding of $2^{<\kappa}$ into $T$ by building a tree of conditions in $\mathbb{P}$, $\langle p_\eta \mid \eta \in 2^{<\kappa}\rangle$ such that for every $\eta$, $p_{\eta \frown (0)}, p_{\eta \frown (1)} \leq p$ give contradictionary information about the branch. Then for every $f\in 2^\kappa$, $b_f = \{ t\in T \mid \exists \alpha < \kappa,\,p_{f\restriction \alpha}\Vdash t\in \dot{b}\}$ is a cofinal branch, and $f\neq g\implies b_f \neq b_g$. Q.E.D. Let's return to the proof of the theorem. Let $G$ be a $\mathbb{Q}$-generic filter and let $\dot{T}$ be a $\mathbb{Q}$-name for a tree in $V[G]$. Note that $(\kappa^{<\kappa})^{V[G]} = (\kappa^{<\kappa})^{V}$, so we may assume that $\Vdash \dot{T}\subseteq \check{\kappa^{<\kappa}}$. By the $\eta$.c.c. of $\mathbb{Q}$, we can find a model $M\prec H_\chi$ such that $|M|<\eta$, $\dot{T}, \mathbb{Q} \in M$, $^{<\kappa}M\subseteq M$ and for every $t\in \kappa^{<\kappa}$ there is a maximal antichain $\mathcal{A} \subseteq M$ that decides whether $t\in \dot{T}$ or not (so $T\in V[M\cap G]$). Note that $M\cap G$ is the restriction of the generic filter to the coordinates that are ordinals of $M$, so it is a generic filter for the forcing $\mathbb{Q}_M := Add(\kappa, \mu\cap M)\times Col(\kappa, <(\eta\cap M))$. Let $\mathbb{P}$ be the restriction of $Add(\kappa,\mu)\times Col(\kappa,<\eta)$ to the ordinals that don't appear in $M$, so $\mathbb{Q}= \mathbb{Q}_M \times \mathbb{P}$. In $V[G\cap M]$, $2^\kappa$ is less than $\eta$ (since $|\mu \cap M| < \eta$), so if all the branches of $T$ in $V[G]$ are already in $V[G\cap M]$, we have that $V[G]\models |[T]|< \eta = \kappa^+$, and we're done. Otherwise, let $\dot{b}$ be a name for a new branch. Since $\mathbb{P} \cong \mathbb{P} \times \mathbb{P}$, $\mathbb{Q} \cong \mathbb{Q}_M \times \mathbb{P} \times \mathbb{P}$. Therefore also in $V[G]$, $\dot{b}$ is a $\mathbb{P}$-name for a new branch (by the mutually generity of the two copies of $\mathbb{P}$). Moreover, $\mathbb{P}$ is $\kappa$-closed in $V[G]$, so by the fact above - $V[G] \models |[T]|=2^\kappa$, as wanted.<|endoftext|> TITLE: How naturally can functions defined by parametric integrals be interpolated from $\mathbb N$ to $\mathbb R^+$? QUESTION [5 upvotes]: It is well known that several definite integrals $I_n$ containing a parameter $n\in\mathbb N$ can be expressed recursively (e.g. doing integration by parts) in terms of $I_{n-1} $ or $I_{n-2} $, and thus written as some expression containing factorials, e.g. the gamma function itself $$\int_0^\infty t^{n}e^{-t}dt=n\int_0^\infty t^{n-1}e^{-t}dt=\cdots=n!$$ Or take the formula $I_n:=\int_0^\pi \sin^nx\;dx=\dfrac{n-1}nI_{n-2}$, allowing to obtain $$\int_0^\pi \sin^{2n}x\;dx=\dfrac{(2n)!}{2^{2n}n!^2}\pi\ \ \text{ and }\ \int_0^\pi \sin^{2n+1}x\;dx=\dfrac{2^{2n+1}n!^2}{(2n+1)!}$$ (which BTW easily yields the Wallis product). As long as for such integrals the LHS is also defined for non-integer (say all positive real) $n$, is it "automatically" guaranteed that replacing on the RHS $n!$ by $\Gamma(n+1)$ yields a valid formula? We all know that the interpolation of the gamma function between the factorials to $\mathbb R$ is not unique but that the Bohr-Mollerup theorem assures a unique extension when adding the mild (?) condition of log-convexity. The problem: the LHS expressions above don't "know" anything about log-convexity... We can ask a similar question (even though there is no simple recursion formula in this case) about the Riemann zeta function with, for $n>1$, $$\int_{0}^{\infty} \frac{t^n}{e^t - 1} \; \frac{dt}{t}=\zeta(n) \; \Gamma(n) $$ or, closely related, this interpolation of the Bernoulli numbers $$ 4n\int_{0}^{\infty} \frac{t^{2n}}{e^{2\pi t}-1} \frac{dt}{t}=4n\frac{2^{2n-1}}{2^{2n-1}-1}\int_{0}^{\infty} \frac{t^{2n}}{e^{2\pi t}+1} \frac{dt}{t}=(-1)^{n+1}B_{2n}$$ or this one of the Euler numbers $$ \int_{0}^{\infty} \frac{t^{2n}}{\cosh\frac{\pi t}2}\; dt =(-1)^{n}E_{2n}.$$ The two last ones are somewhat intriguing because for half-integers $n$, the integrals are obviously not $0$, unlike the odd Bernoulli and Euler numbers. For Euler numbers, this can of course be explained by the fact that the "entire" integral from $-\infty$ to $\infty$ does vanish in this case (odd function), but for the Bernoulli numbers the situation is different. What is going on here? REPLY [4 votes]: I don't have a full answer to this question, but I do have a rather good semi-answer. Considering Carlson's theorem there is something that can be said about this. Now Carlson's theorem is a rather niche result that doesn't pop up as much as I think it should but it has some rather unwieldy results. To be short it gives an identity theorem for natural numbers. The result can be stated a few ways, so I'll choose a modest version of it which is a consequence of Ramanujan's Master theorem (Carlson's theorem is a stronger version of this). Let $z = x+iy$ for $\Re(z) > 0$ Let $\phi$ be holomorphic. If $|\phi(z)| < C e^{\kappa|y| + \rho|x|}$ for $C,\kappa, \rho \in \mathbb{R}^+$ where $\kappa < \pi$. If $\phi\Big{|}_{\mathbb{N}} = 0$ then $\phi = 0$. This helps us in your current situation through the following If $\phi(z) = \int_0^\pi \sin^{2z}(x)\,dx - \pi\frac{\Gamma(2z+1)}{2^{2z}\Gamma(z+1)^2}$ is bounded as above, then since it equals zero on the naturals, they must equal. Successfully interpolating your values. Now this isn't the case here, but very often it does pop up; and it quickly solves a result like this. For your Bernoulli number identity it should work through some manoeuvring. Similarly with the $\Gamma\zeta$ case, however one should divide by the Gamma function first and then apply the identity. It also works with the Euler numbers, in this case.<|endoftext|> TITLE: A specific Vandermond matrix QUESTION [5 upvotes]: Consider the Vandermond matrix $$ V (x_1, x_2, \ldots , x_n) = \begin{pmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} & x_1^n & x_1^{n+1} & \cdots \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} & x_2^n & x_2^{n+1} & \cdots \\ 1 & x_3 & x_3^2 & \cdots & x_3^{n-1} & x_3^n & x_3^{n+1} & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \cdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} & x_n^n & x_n^{n+1} & \cdots \end{pmatrix} $$ with the specific choice of $x_1 = e^{\lambda_1}, \dots x_n = e^{\lambda_n}$, where $\lambda_1, \dots, \lambda_n$ are distinct real numbers. I was wondering, if one will always get by picking arbitrary (not necessary consecutive) $n$ columns in the above "infinite" Vandermonde matrix linearly independent vectors. REPLY [8 votes]: This works, as pointed out by Samuel in a comment. Here's an easy direct argument: We are claiming that no non-trivial polynomial $$ p(x) = \sum_{j=1}^N a_j x^{n_j} $$ with $N$ terms (let me call $N$ the pseudo-degree) has $N$ or more distinct positive zeros. This is immediate from an induction on $N$. First of all, we can assume that $p$ starts with a constant term. Then, if $p$ had $N$ distinct positive zeros, its derivative would be a polynomial of pseudo-degree $N-1$ with at least $N-1$ positive zeros.<|endoftext|> TITLE: Compute only selected components of an eigenvector QUESTION [7 upvotes]: I am wondering whether it is possible to compute portions of the eigenvectors of a given (possibly very big) matrix. More formally, consider the eigenvalue problem $\mathbf{Ax} = \lambda \mathbf{x}$, where $\mathbf{A}$ is $n \times n$ Hermitian. For a fixed eigenvector $\mathbf{x}$, I am only interested in the values $\mathbf{x}_k$ for some choices of $k \in \{1,\dots,n\}$. Is it possible to restrict the computation as above? If not, is it possible to obtain an approximate solution, and under which conditions? REPLY [4 votes]: Here is some information that you might find useful. The context is that of PageRank computation, for the purpose of updating only some of ranks (essentially entries of the eigenvector). Y. Y. Chen, 2004. Local methods for estimating PageRank values Fast incremental and personalized page rank In general, chasing references for "incremental" pagerank or just "top few ranks" should help you find more related work and ideas. Alternatively, you could try to solve the problem approximately by using randomization to subsample the matrix $A$, and then using perturbation analysis to estimate the error of approximation --- some ideas for this can be found by following work related to the so-called "CUR" decomposition.<|endoftext|> TITLE: What is the most useful non-existing object of your field? QUESTION [98 upvotes]: When many proofs by contradiction end with "we have built an object with such, such and such properties, which does not exist", it seems relevant to give this object a name, even though (in fact because) it does not exist. The most striking example in my field of research is the following. Definition : A random variable $X$ is said to be uniform in $\mathbb{Z}$ if it is $\mathbb{Z}$-valued and has the same distribution as $X+1$. Theorem : No random variable is uniform in $\mathbb{Z}$. What are the non-existing objects you have come across? REPLY [8 votes]: The core model. To a large extent, inner model theory (my area of set theory) is about building the core model K which then under any reasonable hypothesis is shown to not exist.<|endoftext|> TITLE: Sets computable from enough hints QUESTION [10 upvotes]: Is there a non-computable set $X\subset\omega$ such that, for some $Y\subset\omega$, any infinite subset or cosubset (=subset of the complement) of $Y$ computes $X$? More generally, call a set $X$ $n$-hintable if there is some $f\in n^\omega$ such that, whenever I am told infinitely many bits of $f$, I can compute all of $X$. Then: What are the $n$-hintable sets? (I ran into this question while thinking about stable ramsey's theorem: given a stable $k$-coloring of pairs - that is, a $k$-coloring of pairs $c$ such that $\lim_yc(x TITLE: The impact of large cardinals in mathematics QUESTION [9 upvotes]: What are the main applications of large cardinals in ordinary mathematics, and what is the philosophy behind using them. In particular: Question 1. What is the philosophy behind accepting large cardinals in mathematics? Question 2. Who first introduced large cardinals, in which paper and for what reasons? Question 3. Do large cardinals appear in finite or discrete mathematics? Would you please give examples if there are any. I think the work of Harvey Friedman is related to this question. By large cardinal, I mean a cardinal which is at least strongly inaccessible. Thanks in advance. REPLY [7 votes]: Re: 6: You're right, Friedman certainly has something to say on this topic! Friedman has discovered a number of $\Pi^0_1$ sentences of Ramseyish flavor which have large cardinal strength. For example, say that a function $f: \omega^k\rightarrow \omega$ is good if there are rational $p, q>1$ such that $$p\vert \overline{x}\vert TITLE: Why is the inverse of a bijective rational map rational? QUESTION [6 upvotes]: Let $f:X\to Y$ be a bijective rational map of an open dense subset $X$ of $\mathbb{C}\times\mathbb{C}$ onto an open dense subset $Y$ of $\mathbb{C}\times\mathbb{C}$. How to prove that the inverse map $f^{-1}:Y\to X$ is rational as well? Could you recommend any exact reference to a theorem which guarantees this? edit Oct 30-31 I have looked carefully through the provided answers and references and still did not find any indication how to prove that the inverse map is rational. Theorem 12.83 in page 355 from [1] mentioned below does not contain the part of Proposition 2 below starting from 'hence an isomorphism onto its image'. Definition of an open immersion, see Definition 3.40 in page 83 from [1], does not say that an open immersion is an isomorphism onto its image. If there is a theorem that an open immersion in the sense of [1] is an isomorphism onto its image, then could you provide an exact reference to it (such a theorem seems to be more or less equivalent to the initial question)? Related question: Isomorphism between varieties of char 0. [1] U. Görz and T. Wedhorn, Algebraic Geometry I, Vieweg+Teubner Verlag-Springer Fachmedien Wiesbaden GmbH 2010. REPLY [11 votes]: Since you are talking about rational maps, I assume you mean "open dense" in the Zariski topology, so that $X$ and $Y$ are algebraic varieties. Therefore we have a particular case of the following well-known statement in algebraic geometry. Proposition 1. Let $k$ be an an algebraically closed field of characteristic zero and $f \colon X \to Y$ a morphism between integral $k$-schemes of finite type over $k$. Then if $f$ is bijective and $Y$ normal the morphism $f$ is an isomorphism, that is $f^{-1} \colon Y \to X$ is a morphism as well. This is in turn a consequence of the following version of Zariski's Main Theorem, see [Görz-Wedhorn, Algebraic Geometry I, Theorem 12.83 page 355]. Proposition 2. Let $f \colon X \to Y$ be a separated morphism of finite type such that $f^{\flat} \colon \mathscr{O}_Y \to f_* \mathscr{O}_X$ is an isomorphism. Let $V$ be the open set of $X$ given by the points $x$ such that $\dim f^{-1}(f(x))=0$. Then the restriction $f_{|V} \colon V \to Y$ is an open immersion, hence an isomorphism onto its image. In particular, if $f$ is dominant and all fibres of $f$ are finite then $f$ is birational. In fact, given a proper morphism $f \colon X \to Y$ with integral fibres, if $Y$ is normal then $f^{\flat} \colon \mathscr{O}_Y \to f_* \mathscr{O}_X$ is an isomorphism, see again [Görz-Wedhorn, Exercise 12.29 page 365], hence Proposition 2 implies Proposition 1. In your situation, $k= \mathbb{C}$ and $Y$ is smooth (hence normal), because it is an open dense subset of $\mathbb{C}^2$. So the previous results apply. A related discussion can be found in this MathOverflow question: Isomorphism between varieties of char 0. Edit October 31, 2014 (see comments). In Görz-Wedhorn's book an open immersion is defined as a morphism $j \colon V \to Y$ such that the underlying continuous map is a homeomorphism of $V$ onto the open set $U:=j(V)$ and the sheaf homomorphism $\mathscr{O}_Y \to j_* \mathscr{O}_V$ induces an isomorphism $\mathscr{O}_{Y|U} \cong j_* \mathscr{O}_V$ (of sheaves on $U$). This definition is equivalent to requiring that $j \colon Y \to X$ is an isomorphism onto the open subscheme $U:=j(V)$, see [EGA I, Chapitre I Proposition 4.2.2 a), page 122]. A related discussion is in this MSE question, see in particular Georges Elencwajg's answer.<|endoftext|> TITLE: Connected components of the complement of a degree-d affine hypersurface QUESTION [7 upvotes]: Let $n$ and $d$ be positive integers, and $f\in\mathbb{R}[x_1,\dots,x_n]$ be a polynomial of degree $d$. Let's consider the zero-set $M = \{x \in \mathbb{R}^n: f(x) = 0\}$ of $f$. Can we estimate the number of connected components of $\mathbb{R}^n\setminus M$? I throw out a guess: no more than $2^d$. An realisation of the bound $2^d$ is given by the example: $f(x) = x_1x_2\cdots x_d$ (when $n \ge d$). I can prove it for the case $d = 2$. UPDATE: is it possible to find an estimate that does not depend on $n$? REPLY [2 votes]: You can do a bit better than the bound in Dan's reply. For a sufficiently small $\varepsilon$, consider the polynomial $g= f^2-\varepsilon$. Note that the zero set of $g$ has at lest one connected component inside of every cell of ${\mathbb R}^n\setminus M$. Thus, the Milnor-Thom theorem gives the upper bound $d^n$. The above bound is asymptotically tight. By taking $f$ to be the product of $d$ linear expressions of the form $a_1x_1+a_2x_2+\cdots+a_nx_n+b$ you get that $M$ is the union of $n$ hyperplanes. When these hyperplanes are generic, it is not difficult to show that ${\mathbb R}^n\setminus M$ consists of at least $cd^n$ compoenets, for some constant $c$. If you would like to have a more well-behaved polynomial, you can again replace $f$ with $g= f^2-\varepsilon$.<|endoftext|> TITLE: complexity of proof of p(n) grows greater with n if for all x P(x) is unprovable? QUESTION [5 upvotes]: Is it true that if "for all x P(x)" is unprovable in pA then the complexity of the proof of P(n) becomes greater as n grows bigger? REPLY [9 votes]: You didn’t specify what you mean by “complexity”. If one interprets it as the number of lines in the proof, this is a famous conjecture of Kreisel (usually stated contrapositively: if there is a constant $c$ such that $\phi(\overline n)$ has a PA proof with at most $c$ lines for all $n$, then PA proves $\forall x\,\phi(x)$). Kreisel’s conjecture is still open for the most natural formulation of PA, however it is extremely sensitive to the choice of the language, and various variants of it have been proved and disproved. See Hrubeš for a recent paper in the positive direction, where you can also find pointers to other known results related to the conjecture. REPLY [2 votes]: Consider the assertion $P(x)$ that asserts "$x$ is not the Gödel code of a proof of a contradiction in PA". If PA is consistent, then we can prove $P(n)$ for any particular natural number $n$, since no such $n$ codes a proof. I don't know what is your measure of proof complexity, but the proof of $P(n)$ in every case is mundane: it is because the sequence coded by $n$ violates one of the syntactic requirements of being a proof. Namely, one of the sequents is not an axiom, or does not follow by modus ponens from the earlier sequents, or the conclusion is not a contradiction, and so on. In particular, the formulas appearing in the proofs of $P(n)$ have bounded complexity (although these proofs do get longer, since one must check more cases to cover the earlier sequents). But meanwhile, the assertion $\forall x\, P(x)$ is independent.<|endoftext|> TITLE: How to "lift" a transitive group action on a manifold? QUESTION [8 upvotes]: Let $M=G/H$ be a homogeneous manifold, with $G$ connected Lie group. Suppose that $\widetilde{M}$ is a covering of $M$. QUESTION: is there a general prescription to obtain a Lie group $\widetilde{G}$, starting from $G$, in such a way that $\widetilde{M}=\widetilde{G}/\widetilde{H}$? Using the case of $M=G=S^1$ and $H=1$ as a toy model (and, for instance, $\widetilde{M}=\widetilde{G}=\mathbb{{R}}$) we see that $\widetilde{G}$ has two remarkable properties: 1) it contains the group $\Gamma=\mathbb{{Z}}$ of "gauge symmetries" of $\widetilde{M}\to M$; 2) its factor by $\Gamma$ returns the original group $G$. So, I guess that the two properties above are enough to characterise $\widetilde{G}$ but I'm not able to prove it. I'm sure it's a well-known result, but I can't find any reference (I could not get which book is this "Bredon" mentioned here: lifting group action). In the case that my guess is correct, I'd like to understand if there is a constructive way to obtain $\widetilde{G}$, e.g., by realising the Lie algebra $\frak{g}$ as vector fields on $M$, lifting them to $\widetilde{M}$, and then take the group generated by their flows. REPLY [3 votes]: The Palais theorem assumes that the manifold $\tilde M$ is compact. The positive answer gives proposition 6 of the Onishchik book "Topology of transitive transformation groups". It states : For any action of a Lie group $G$ on a manifold and any covering $\pi : N \to M$ there is an action of the universal cover $\tilde{G}$ on $N$ which cover the action of $G$ on $M$, i.e. such that the projection $\pi : N \to M$ is $\tilde{G}$-equivariant.<|endoftext|> TITLE: Closure of random rotations QUESTION [6 upvotes]: Are matrix Fisher random variables closed under multiplication? For those unfamiliar with the jargon, let me unpack the terms above and repose my question. This is a question about probability distributions on rotations (i.e. on $SO(n)$). We can represent a rotation as an $n\times n$ real-valued matrix $M$. Observe that if know the first $n-1$ columns of $M$, and that the matrix is orthogonal, then the last column is determined. Therefore, to define a probability distribution on $SO(n)$, it is sufficient to consider a probability distribution on the $n \times (n-1)$ matrices whose columns are unit length and orthogonal. The set of these matrices for a fixed $n$ form an object called a Stiefel manifold, denoted $V_{n-1}(R^n)$. The matrix Fisher distribution provides a probability distribution on $V_{k}(R^n)$ (so I'm interested in the case where $k=n-1$). In particular, consider a fixed $n \times k$ matrix $F$ (not necessarily in $V_{k}(R^n)$). Then X is a matrix Fisher random variable with parameter $F$ if its pdf $\Pr(X|F)$ is proportional to: $$ \exp(Tr(F^TX)) $$ where $Tr$ is the trace. Now, suppose that $X$ is a matrix Fisher random variable with parameter $F$, $Y$ is a matrix Fisher random variable with parameter $G$, and $X$ and $Y$ are independent. Consider the (matrix) product: $$Z=XY$$ $Z$ is a random variable on $SO(n)$. So, to restate my original question: Does $Z$ have a matrix Fisher distribution? If so, what is its parameter as a function of $F$ and $G$? In case it makes any difference, I'm primarily interested in $n=3$. I asked this question on Cross Validated, but didn't get any answers or comments. As I mention in that post, there's a comment in Suvorova, "Bayesian Recursive Estimation on the Rotation Group", 2013 pointing out that since rotation matrices do not, in general commute (for $n\geq 3$), then the mean of the product of matrix Fisher random variables is not necessarily the product of the mean. For (a little) more information on this subject, see Section 13.2 of Mardia and Jupp, "Directional Statistics", 2000. REPLY [2 votes]: Yoav Kallus' comment above is correct; I'm going to sketch a few of the details below for the sake of completeness. Yoav points out that if we choose $X$ with parameter $$F=\begin{bmatrix}a&0&0\\0&0&0\end{bmatrix}$$ and $Y$ with parameter $$G=\begin{bmatrix}0&0&0\\0&a&0\end{bmatrix},$$ then for $a>0$ the probability of $X$ is maximized when it is of the form $$X=\begin{bmatrix}1&0&0\\0&\cos(\alpha)&\sin(\alpha)\end{bmatrix}$$ for any $\alpha$, and the probability of $Y$ is maximized when it is of the form $$Y=\begin{bmatrix}\cos(\beta)&0&\sin(\beta)\\0&1&0\end{bmatrix}$$ for any $\beta$. Therefore, as $a$ grows large, the $X$ and $Y$ converge to (the Stiefel manifolds corresponding to) uniform independent rotations about the $x$ and $y$ axes, respectively. Let $\tilde{X}$ and $\tilde{Y}$ be independent rotations around the $x$ and $y$ axes, respectively. Then we can compute the product $\tilde{Z}$: $$\tilde{Z}=\tilde{X}\tilde{Y}=\begin{bmatrix}1&0&0\\0&\cos(\alpha)&\sin(\alpha)\\0&-\sin(\alpha)&cos(\alpha) \end{bmatrix}\begin{bmatrix}\cos(\beta)&0&\sin(\beta)\\0&1&0\\-\sin(\beta)&0&cos(\beta)\end{bmatrix}$$ $$=\begin{bmatrix}\cos(\beta)&0&\sin(\beta)\\-\sin(\alpha)\sin(\beta)&\cos(\alpha)&\sin(\alpha)\cos(\beta)\\-cos(\alpha)sin(\beta)&-\sin(\alpha)&\cos(\alpha)\cos(\beta)\end{bmatrix}$$ The independence of $\tilde{X}$ and $\tilde{Y}$ and their symmetry around their respective axes imply that: $$E[Z]=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$ Suppose that $Z$ were a matrix Fisher random variable. It would then have some parameter $H$. Now, $E[Z]$ is a sufficient statistic for $H$ (since the matrix Fisher distribution is an exponential family), but then $H$ must be identically zero. In that case, $Z$ would have a uniform distribution. However, the entry in the top row, middle column is identically zero, so clearly this distribution is (highly) non-uniform. Therefore, $Z$ does not have a matrix Fisher distribution.<|endoftext|> TITLE: Smooth and $GL(n)$-equivariant implies algebraic? QUESTION [5 upvotes]: Context: Let $B_n$ be the space of symmetric bilinear forms on $\mathbb{R}^n$ and $L_n\subset B_n$ be the subset of non-degenerate forms of Lorentzian signature $(-,+,\ldots,+)$. Let $T$ be a finite dimensional real vector space. Both $B_n$ and $T$ carry linear representations of $GL^+(n,\mathbb{R})$, where $+$ means the connected component of the identity, with the action on $B_n$ restricting to $L_n$. We assume that, for $g\in GL^+(n,\mathbb{R})$, the corresponding linear maps acting on $B_n$ and $T$ are polynomial in the components of $g$ up to a multiple of $\left|\det g\right|$ to a rational power. For $h\in L_n$, consider its components $h_{ab}$ (using some fixed basis on $\mathbb{R}^n$, say) as functions on $L_n$. I can also define another set of functions which are the components $\varepsilon_{a_1\cdots a_n}$ of the Levi-Civita tensor, defined as a solution to the equation $$\varepsilon_{a_1\cdots a_n} \varepsilon_{b_1\cdots b_n} = \text{(antisymmetrization over $a_1\cdots a_n$ of)} ~ h_{a_1 b_1} \cdots h_{a_n b_n}.$$ Question: If I have a (non-linear) smooth equivariant map $A\colon L_n \to T$, can I conclude that $A$ is actually polynomial in the functions $h_{ab}$ and $\varepsilon_{a_1\cdots a_n}$ (up to a multiple of a rational power of $\left|\det h_{ab}\right|$) on $L_n$? I believe that the answer is Yes. Unfortunately, I don't know which (probably well-known) result implies that. In the absence of a direct answer, some hints about how to find one in the vast literature on Invariant Theory would also be appreciated. Note: Most of the literature in Invariant Theory seems to already make the assumption that all maps involved are polynomial, rational or algebraic. On the other hand, my question is about reducing to one of the contexts starting with smoothness and some other assumptions. REPLY [5 votes]: If I understand you correctly, the answer is 'no'. Because the open set $L_n\subset B_n$ is an orbit of $\mathrm{GL}^+(n,\mathbb{R})$ under the natural representation of $\mathrm{GL}^+(n,\mathbb{R})$ on $B_n$, it follows that, if $\rho:\mathrm{GL}^+(n,\mathbb{R})\to \mathrm{GL}(T)$ is the representation that defines the $\mathrm{GL}^+(n,\mathbb{R})$-action on $T$, then any $\mathrm{GL}^+(n,\mathbb{R})$-equivariant mapping $A:L_n\to T$ must satisfy $$ A(g\cdot b) = \rho(g)\bigl(A(b)\bigr) $$ for all $g\in\mathrm{GL}^+(n,\mathbb{R})$ and $b\in L_n$. In particular, since $\rho$ is smooth, it follows that $\mathrm{GL}^+(n,\mathbb{R})$-equivariance already implies that $A$ is smooth. Conversely, given any such representation $\rho$ and any fixed $b_0\in L_n$, any choice of an element $A(b_0)\in T$ defines a $\mathrm{GL}^+(n,\mathbb{R})$-equivariant mapping $A:L_n\to T$ by the above formula. (Note that, if one regards $L_n$ as the space of $n$-by-$n$ symmetric matrices of index $n{-}2$, then $g\cdot b = g b g^T$.) Thus, this is really a question about the representation $\rho$. Consider the representation $\rho:\mathrm{GL}^+(n,\mathbb{R})\to \mathrm{SL}(2,\mathbb{R})$ defined by $$ \rho(g) = \begin{pmatrix} 1 & \log\bigl(\det(g)\bigr)\\0&1\end{pmatrix}. $$ Then, for, say, $b_0 = -{x_1}^2 + {x_2}^2 + \cdots + {x_n}^2$ and $A(b_0) = \begin{pmatrix}0\\1\end{pmatrix}$, the above formula gives the smooth equivariant map $$ A(b) = \begin{pmatrix} \log\bigl(-\det(b)\bigr)\\1\end{pmatrix}, $$ for $b\in L_n$, and this mapping is not polynomial in the functions that you name in your question.<|endoftext|> TITLE: Ising model on lattices with (vertical side length) $\neq$ (horizontal side length) QUESTION [7 upvotes]: Consider the Ising model with nearest neighbours interactions on a rectangular lattice $L\times M$. If $L=M$ ($2$-dimensional square lattice), it is known (e.g., by Peierls' argument or Onsager's explicit solution) that the model exhibits a phase transition when $L=M\to\infty$. If instead we fix $L=1$ ($1$-dimensional line) and let $M\to\infty$, the model does not exhibit a phase transition. My question is: Which relations among the side lengths $L,M$ guarantee the presence/absence of a phase transition? For example, what about the case $L=\log M$ ? REPLY [5 votes]: Any increasing sequence $(\Lambda_n)_{n\geq 1}$ of finite subsets of $\mathbb{Z}^d$, $d\geq 2$, such that $\bigcup_{n\geq 1} \Lambda_n =\mathbb{Z}^d$ will do. All sequences $(\mu_{\Lambda_n}^+)_{n\geq 1}$ of finite-volume Gibbs measures in $\Lambda_n$ with $+$-boundary condition converge to the same infinite-volume Gibbs measure $\mu^+$, under which there is spontaneous magnetization as soon as the inverse temperature $\beta$ is large enough. This can be proved easily using the FKG inequality (see, for example, the chapter on the Ising model here). REPLY [2 votes]: Peierls's argument (in one of its modern forms, e.g. using chessboard estimates in the case of periodic boundary conditions) should work as long as $M$ and $L$ both go to $\infty$.<|endoftext|> TITLE: (Very) High dimensional manifolds QUESTION [16 upvotes]: Usually one regards manifolds up to dimension 4 as a part of low dimensional topology. There are plenty of various results which work only in low dimensional topology; especially in dimension 4. However still there are phenomena which occur only up from certain dimensions above 4: For example the famous result of Milnor, which states that each $PL$ manifold of dimension $n$ is in fact smooth provided that $n \leq 7$. My question is the following: Could you give an example of the (reasonable) theorem of the type "each manifold of dimension $n$ have some property $P$ provided that $n \leq K$ (and for $n>K$ there are counterexamples)," where $K$ is some large number? REPLY [28 votes]: The smallest example of a manifold that is homotopy equivalent to a topological group, but not rationally equivalent to a Lie group has dimension 1254.<|endoftext|> TITLE: Classification of countable posets? QUESTION [5 upvotes]: Is there a classification of countable posets where between each two comparable elements there is a third element between them? REPLY [3 votes]: Here is the dummies version of why countable dense posets (meaning countable posets in which every maximal chain is densely ordered) are as complicated as countable binary relations in general, so that no reasonable classification of them is to be expected. To any countable poset $(P,\lt)$ let us associate a countable set $A$ and an irreflexive binary relation $R\subseteq A\times A$ as follows: $A$ is the set of all minimal elements of $P$, and $$R=\{(a,b)\in A\times A:\exists x,y\in P\ (a\lt x\lt y\wedge b\not\lt x\wedge b\lt y)\}.$$ Let's call $(A,R)$ the BRSC of $(P,\lt)$. Proposition. Given a countable set $A$ and an irreflexive binary relation $R\subseteq A\times A$, we can construct a countable dense poset $(P,\lt)$ such that the BRSC of $(P,\lt)$ is isomorphic to $(A,R)$. Proof. We assume that $A$ is disjoint from $A\times A\times\mathbb Q$. Let $$P=A\cup\{(a,b,x):(a,b)\in R\wedge x\in\mathbb Q\}.$$ The ordering of $P$ has the following comparisons and no others: $$(a,b,x)\lt(a,b,y)\text{ when }(a,b)\in R\wedge x,y\in\mathbb Q\wedge x\lt y;$$ $$a\lt(a,b,x)\text{ when }a,b\in R\wedge x\in\mathbb Q;$$ $$b\lt(a,b,x)\text{ when }(a,b)\in R\wedge x\in\mathbb Q\wedge x\gt0.$$<|endoftext|> TITLE: Dubins car shortest paths: Decidable? QUESTION [8 upvotes]: A Dubins car follows a Dubins path in $\mathbb{R}^2$, with constant wheel speed and limited turning radius. It is known that the shortest Dubins path in the absence of obstacles follows circular arcs and straight segments, in fact, in no more than six patterns:                     Figure from Huifang Elizabeth Wang's PhD thesis The literature is vast, and I am having difficulty determining: Q. Is there an algorithm to determine the shortest Dubins path between any two points in $\mathbb{R}^2$, in the presence of polygonal obstacles? Assume that the start and end points, and the $n$ polygonal vertices, are specified by integer coordinates using at most $L$-bits. An algorithm polynomial-time in $L$ & $n$ is perhaps too much to hope for, but Q asks if the problem is decidable. Answered (11 Nov 2014). user3097732 pointed me to the paper, Curvature-Constrained Shortest Paths in a Convex Polygon, which says in the Introduction Reif and Wang[29] confirmed that the problem of deciding whether there exists a collision-free curvature-constrained path for $B$ between two given configurations amid polygonal obstacles is NP-hard. where [29] is: J. Reif and H. Wang, "The complexity of the two-dimensional curvature-constrained shortest-path problem," in Robotics: The Algorithmic Perspective (Houston, TX), A. K. Peters, 1998, pp. 49–57. NP-hard could still be undecidable, but Fortune and Wilfong proved the problem is decidable, by providing an algorithm with time complexity $2^{\mathrm{poly}(n, L)}$, where $n$ is the number of vertices of the obstacles and $L$ the number of bits in the coordinates. (Their algorithm does not actually find the path!) "Planning constrained motion," Ann. Math. Artificial Intelligence, 3 (1991), pp. 21–82. REPLY [4 votes]: You might want to take a look at "Curvature-constrained shortest paths in a convex polygon" (Agarwal et al. 2002). Use another method to decompose your environment into convex polygons, e.g "A navigation mesh for dynamic environments" (Van Toll, W. G., Cook, A. F., & Geraerts, R. 2012). Then, run a global planning algorithm, to find the subset of convex polygons that are to be traversed. For each polygon, use the 1st method to plan between polygon edges.<|endoftext|> TITLE: When is the Ad (Adjoint Representation) Morphism a Closed Map QUESTION [7 upvotes]: Given a Lie group $\mathfrak{G}$ with finite centre and with Lie algebra $\mathfrak{g}$, I am looking at a simple proof that negative definite Killing form implies compactness. This proof is given here. A sketch of the proof is as follows: The author claims that $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G})$ is a closed map; The Killing form $K:\mathfrak{g}\times\mathfrak{g}\to\mathbb{R}$ is $\mathrm{Ad}$-invariant, so a negative definite $K$ allows us to define an inner product $-K$such that $\mathrm{Ad}(\gamma)$ is orthogonal wrt $-K$ i.e. $\mathrm{Ad}(\mathfrak{G})\subseteq SO(\dim \mathfrak{g}, -K)$. $SO(\dim \mathfrak{g}, -K)$ being compact, and $\mathrm{Ad}(\mathfrak{G})$ a closed subgroup by 1., we conclude that $\mathrm{Ad}(\mathfrak{G})$ is itself compact; $\mathfrak{G}$, being an $M$-fold cover of $\mathrm{Ad}(\mathfrak{G})$ (where $M$ is finite by dint of the finite centre), is thus also compact. Crucial to this proof is the assertion that $\mathrm{Ad}(\mathfrak{G})$ is closed in $SO(\dim \mathfrak{g}, -K)$. I can prove this given nondegeneracy of the Killing form, (e.g. with Lemma 1 of G. Hochschild, "Complexification of Real Analytic Groups") but the author of the first document I linked seems to be saying that this is a much more general and well known property of $\mathrm{Ad}$. What am I missing here?: I think I'm making this harder than it should be through overlooking a simple fact. So, as in my title: When is $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G}))$ a closed map and why? Edit: It seems that this is not as trivial as I thought. Hence answers less than a full answer are helpful and acceptable to me. For example, interesting counterexamples (showing when $\mathrm{Ad}:\mathfrak{G}\to GL(\mathrm{Lie}(\mathfrak{G})$ is not closed) or sufficient conditions for it to be closed (such as semisimplicity of $\mathfrak{G}$). REPLY [5 votes]: Let $G$ be a connected Lie group. Equivalences: (i) every linear representation of $G$ has a closed image (ii) $\mathrm{Hom}(G,\mathbf{R})=0$ (iii) $G/\overline{[G,G]}$ is compact. Here homomorphisms are meant continuous, and reps are in $\mathbf{GL}_n(\mathbf{R})$ for $n$ not fixed, or equivalently in $\mathrm{GL}_n(\mathbf{C})$ in view of the closed inclusions $\mathrm{GL}_n(\mathbf{R})\subset \mathrm{GL}_n(\mathbf{C})\subset \mathrm{GL}_{2n}(\mathbf{R})$. Note that the equivalence between (ii) and (iii) is obvious. That (i) implies (ii) is clear since if $\mathrm{Hom}(G,\mathbf{R})\neq 0$, $G$ admits $\mathbf{R}$ as a quotient, and we can use a 2-dimensional complex representation of $\mathbf{R}$ whose image is dense in a 2-dimensional real torus. The implication (ii)$\Rightarrow$(i) goes by reduction to the semisimple case. Thus first assume that $G$ is semisimple. Fix a representation $\rho:G\to\mathrm{GL}_n(\mathbf{C})$. In the latter, the image of the representation is a (a priori not closed) Lie subgroup $H_1$, with Lie algebra $\mathfrak{h}$ (a real Lie aubalgebra of $\mathfrak{gl}_n(\mathbf{C})$. Let $H_2$ be the closure of $H_1$ (in the real Zariski topology, or in the real topology, it does not matter). Then $H_2$ normalizes $\mathfrak{h}$ and $H_1$. Since up to finite index, the automorphisms of $H_1$ are inner, and since $H_2$ is connected, we deduce that $H_2=H_1C$ where $C$ is the centralizer of $H_1$ in $H_2$. Decompose $\mathbf{C}^n=\bigoplus V_i$, a sub of $H_1$-irreducible complex subspaces. Then since $C$ is in the closure of $H_1$, it preserve each $V_i$. Since the $H_1$-representation on the complex space $V_i$ is irreducible, its commutant is reduced to scalars. Moreover, since the restriction of $H_1$ on $V_i$ has determinant 1, it is also the case for $C$. We deduce that $C$ is a finite abelian group. Thus $H_1$ has finite index in $H_2$. Since $H_1$ is $\sigma$-compact, a Baire argument then shows that $H_1$ has non-empty interior in $H_2$, and hence $H_1$ is the unit component of $H_2$ in the real topology. Thus $H_1$ is closed. (Another argument goes by using the fact that perfect Lie subalgebras of $\mathfrak{gl}_n(\mathbf{R})$ are always Lie algebras of some Zariski closed subgroup of $\mathrm{GL}_n(\mathbf{R})$.) Now assume $G$ arbitrary satisfying (ii). Write $G=RS$ with $R$ the connected solvable radical and $S$ a semisimple Levi factor. A simple argument based on taking the complexification at the Lie algebra level shows that $[G,R]$ is normal and that each representation of $G$ maps $[G,R]$ to unipotents. Thus let $\rho$ be a representation of $G$, say in $\mathrm{GL}_n(\mathbf{R})$, and let $H$ be its image. Then $N=\rho\Big(\overline{[G,R]}\Big)$ is a connected, unipotent normal subgroup of $H$. Being a unipotent connected Lie subgroup, it is necessarily closed in $\mathrm{GL}_n(\mathbf{R})$, actually Zariski closed. If $M$ is the normalizer of $N$, then $H\subset M\subset \mathrm{GL}_n(\mathbf{R})$ and $M$ is Zariski closed. Moreover $M/N$ stands a Zariski closed subgroup of $\mathrm{GL}_m(\mathbf{R})$ for some $m$ (indeed, if $M=\mathbb{M}_\mathbf{R}$ and $N=\mathbb{N}_\mathbf{R}$ for some $\mathbf{R}$-subgroups of $\mathrm{GL}_n$, then $\mathbb{M}/\mathbb{N}$ is a linear algebraic group and hence is isomorphic to an $\mathbf{R}$-closed subgroup in some $\mathrm{GL}_m$, and $\mathbb{M}_\mathbf{R}/\mathbb{N}_\mathbf{R}\to (\mathbb{M}/\mathbb{N})_\mathbf{R}$ is a closed map in the real topology.) Therefore, we are reduced to the case when $[G,R]=1$. In this case $G=RS$, with $S$ semisimple Levi factor, $R$ abelian, and $[R,S]=1$. Since every linear representation of $S$ factors through some finite index subgroup of its center, it is no restriction to assume that $S$ has a finite center, and in particular $S$ is closed. By (ii), we thus deduce that $R$ is compact. Hence since by the first case, $\rho(S)$ is closed, we deduce that $\rho(G)$ is closed as well.<|endoftext|> TITLE: Is the identification between symmetric tensors and homogeneous polynomials useful? QUESTION [7 upvotes]: The general question: Given an $n$-dimensional vector space $V$ over a field $k$, there exists an identification $$\mathrm{Sym}^d(V) \sim k[x_1, \dots, x_n]_d$$ between the space of symmetric order $d$ tensors on $V$ and the space of order $d$ homogeneous polynomials in $n$ indeterminates over $k$. I am wondering whether this identification is ever of much use in the study of tensors and tensor fields. In particular, I am interested in ways in which this identification might simplify problems in differential geometry, but uses in other fields would be interesting also. A potential example of what I'm asking about: This question is inspired by the following observation I made, which I would also like to confirm is valid: Let $\omega \in \mathrm{Sym}^1(V)$ be some unknown. Suppose we have a map $$\phi \colon \mathrm{Sym}^1(V) \longrightarrow k$$ along with a few known elements $\eta, \theta, \nu_1, \nu_2 \in \mathrm{Sym}^1(V)$. Additionally, we have the following system of equations: $$ \begin{align} \phi(\omega) \cdot \eta + \phi(\eta) \cdot \omega &= \nu_1 \\ \phi(\omega) \cdot \theta + \phi(\theta) \cdot \omega &= \nu_2 \end{align} $$ Our goal is to solve for $\omega$. In light of the above, we can identify $\mathrm{Sym}^1(V)$ with $k[x_1, \dots, x_n]_1 \subset k[x_1, \dots, x_n]$, form the field of fractions $k(x_1, \dots, x_n)$ and solve the above system for $\omega$ using linear algebra. Doing so, we will arrive at an expression of the form: $$ \left(\phi(\theta)\eta - \phi(\eta)\theta\right)\omega = \theta \cdot \nu_1 + \eta \cdot \nu_2 $$ where all the elements are now considered to be in $k(x_1, \dots, x_n)$, so all the products make sense. Dividing, we can obtain an expression for $\omega$ expressed entirely in terms of objects we know. We can now evaluate this expression on the appropriate vectors to obtain an expression for $\omega$ in terms of a basis for $\mathrm{Sym}^1(V)$. Note: I originally posted this question here on math.se. At the time I was unsure whether it would better to post it on mathoverflow. Given that it has remained unanswered over there for some time, I figured I would try posting it here. I am not as familiar with mathoverflow as some other stack exchange sites, so please feel free to edit my post in order to make things fit better with this site if necessary. REPLY [5 votes]: I suggest an example showing that sometimes the object we need is actually the homogeneous polynomial in velocities but in order to work with it it is convenient to view it as a symmetric tensor since it allows us to use invariant ways to work with the tensors. My example is the Killing tensor. In a simplest situation it is a symmetric $(0,2)$ tensor $K$ satisfying the equation $$symmetrisation \ of (\nabla K)=0.$$ If your prefer indices, the equation above looks $$ K_{(ij,k)}=0. $$ The tensor plays improtant role in physics and differential geometry since it is corresponds to conservative quantaties of the geodesics flow: for a symmetric tensor $K$ the function $K(\dot \gamma, \dot \gamma)$ is constant along geodesics if and only if it is a Killing tensor. We see that the geometric condition that is equavalent to the property of tensor $K$ to be Killing is actually a condition about the symmetric polynomial $K(\xi, \xi)$. Though the Killing equation above is equivalent to any other equation that is equivalent to the fact that $K(\dot\gamma, \dot\gamma)$ which is quadratic polynomial in $\dot \gamma$ is preserved along geodesics, writing it in the tensorial form above allows one to use the tensorial mashinery to work with this equation. For example, one can use the tensorial mashinery to obtain conditions on the curvature that prevent a metric to have a Killing tensor. One can also use it in order to understand how many Killing tensors may exist on a manifold. Also projective invariance of the Killing equations is better seenable on the level of Killing tensors and I did use it very extencively in my research. Everything what I said is also true for higher degree $d$ of the polynomial, i.e., for higher valency of the tensor.<|endoftext|> TITLE: Modular polynomials for elliptic curves point counting QUESTION [7 upvotes]: The Schoof-Elkies-Atkin (SEA) algorithm (for counting points on elliptic curves over a finite field) performs computations over polynomials modulo some modular polynomials. Originally the "classical" modular polynomials (the minimum polynomial of the modular function $\tau \mapsto j(\ell \tau)$ for a given prime $\ell$). It turns out that the coefficients of this polynomial increase very fast with the prime $\ell$, so in practice other polynomials should be used. In the Handbook of Elliptic and Hyperelliptic Curve Cryptography (Cohen and Frey), they present the "canonical" modular polynomials, and apparently people also use the Weber modular polynomials (Genus 1 point counting records over prime fields), which have even smaller coefficients, but I cannot find any reference on how to use them. And people also talk about other polynomials (Atkin polynomials ?). In all the descriptions of SEA I could find, they just use the classical version and add a side note "in practice we would use other polynomials". So here is the question: is there any literature dealing with those alternatives of the classical modular polynomials? (comparing them, which one is the best in the context of SEA, and how to use them even though they don't give as much information as the classical one). Which ones are used in the state-of-the-art implementations of SEA and why? (e.g. in Magma). REPLY [3 votes]: Andrew Sutherland (in his paper On the evaluation of modular polynomials, where he documents the point counting record) refers to Section 7 of the paper by himself, Broker and Lauter (Modular polynomials via isogeny volcanos) as a good source for properties of other modular polynomials. It looks like Weber modular polynomials are the most popular in general, since they are "smaller" by a factor of 1728 than the others. Sutherland's modification of the SEA algorithm (which was used to compute $\# E(\mathbb{F}_{p})$, where $p$ has $5011$ digits) used Weber modular polynomials. I would refer to Sutherland's computation as "state-of-the-art". I've had a hard time figuring out whether the implementations of SEA in Magma, PARI/GP, and Sage use which modular polynomials. It looks to me like PARI/GP primarily uses Atkin modular polynomials.<|endoftext|> TITLE: What is the (mixed strategies) equilibrium of this game? QUESTION [5 upvotes]: Given a weight vector $w\in [0,1]^d$ such that $\sum w_i=1$, the game goes as follows: Two players, $X,Y$ choose strategies $x,y\in [0,1]^d$ such that $\sum x_i = \sum y_i = 1$. The utility (profit) for player $X$ is given by $$u_X=\sum_{i:x_i>y_i} x_i\cdot w_i$$ That is summing over all coordinates in which $x$ is larger than $y$, and the profit for the coordinate is given by $x_i\cdot w_i$. Y's utility is symmetric ($u_Y=\sum_{i:y_i>x_i} y_i\cdot w_i$). This also interests me for low dimension (say $d=3$) if it makes it easier. We assume w.l.o.g that $i>j\implies w_i\geq w_j$. If $w_1\geq \frac{1}{2}$ then this is trivial, so assume this is not the case. If needed we may assume that if $x_i=y_i$ then they both get $\frac{x_i\cdot w_i}{2}$ utility for that coordinate. Is there a known name for this game? Is the equilibrium of this game computable or is it $PPAD$-hard? REPLY [6 votes]: If you change the payoff to $$\sum_{i:x_i>y_i} w_i $$ then this type of game is called a Colonel Blotto game or a Blotto game. It was studied by Borel. Your assumption that $\sum x_i = \sum y_i$ is not always a requirement. In a 1950 paper, Gross and Wagner analyzed a few variants and contributed the name. They covered the cases of $d=2$ in your notation, $n=2$ in theirs. $d=3$, $\sum x_i = \sum y_i.$ $\sum x_i = \sum y_i$ and $w_j=1/d.$ Blotto games have been identified and analyzed in a few other papers.<|endoftext|> TITLE: Does one real radical root imply they all are? QUESTION [50 upvotes]: Is there an example of an irreducible polynomial $f(x) \in \mathbb{Q}[x]$ with a real root expressible in terms of real radicals and another real root not expressible in terms of real radicals? REPLY [2 votes]: WARNING: Wrong answer follows, as per comments. Kept here to deter others from making the same mistake. Perhaps $$x_1=\sqrt{2+2^{1/4}}+\sqrt{2-2^{1/4}},\quad x_2=\sqrt{2+i2^{1/4}}+\sqrt{2-i2^{1/4}}$$ are real numbers, solutions of $\bigl((x^2-4)^2-16\bigr)^2=32$, one expressed using real radicals, the other, not.<|endoftext|> TITLE: Hlawka inequality for determinants of positive definite matrices QUESTION [17 upvotes]: It is mentioned here that if $A, B, C\in M_{n}(\mathbb C)$ are positive semidefinite, then $$\det (A+B+C)+\det C\ge \det (A+C)+\det (B+C)$$ (quoted from this article) and the special case ($C=\bf 0$) $$\det (A+B)\ge \det (A)+\det (B).$$ The latter one has many proofs. The proof of the former one given in the paper uses tensor products which are decomposed into parts that mostly vanish due to orthogonality. It is valid not only for determinants, but for all generalized matrix functions (a.k.a. immanents). Now: I have numerical evidence that a sharper Hlawka inequality $$\det (A+B+C)+\det (A)+\det (B)+\det (C)\ge \det (A+B)+\det (A+C)+\det (B+C)$$ also holds. But this one becomes wrong if determinants are replaced by permanents (and supposedly for the other immanents, too). This makes me think that the decomposition of the corresponding tensor products can very probably not be used to prove this one. Note the formal similarity with Popoviciu's inequality, though I don't think that is of any help. Is there a way to prove the Hlawka inequality for determinants? EDIT: (to include my comment from below) More generally, if $A_1,...,A_r\in M_{n}(\mathbb C)$ are psd matrices and $\chi\vdash n$ (or even: $\chi$ is an irreducible character of a subgroup of $S_n$), define for $k=1,...,r$ $$s_k:=\sum_{i_1<\cdots TITLE: Limiting probabilities for two-player game drawing random uniform numbers QUESTION [10 upvotes]: Consider this simple 2-person game I just made up: Player A goes gets to draw a uniform U[0,1] number up to X times. At any time, he may either keep his number, or draw a brand new uniform number. However, if he draws all X times, he must keep his last draw as his score. Player B goes second, and gets to draw up to Y times. Same rules apply to him. The winner is the player with the highest score. Assuming both A and B play optimally, the probability that player A wins the game is given by F(X,Y). F(X,Y) can be given by the following recurrence relationship (I'll leave it as a separate exercise to derive this): $F(0,Y) = 0$ $\forall$ Y>0 $F(X,Y) = \left(\frac{Y}{Y+1}\right)*\left(F(X-1,Y)\right)^{\frac{Y+1}{Y}} + \frac{1}{Y+1}$ Here's the question: Find the limit $\lim_{n\rightarrow \infty}$ $F(n,n)$. Obviously, this can be approximated numerically directly from the recurrence definition, but I'm wondering if this limit has an elegant closed-form solution (or perhaps can be derived as a unique root of an implicit equation). The solution has a nice interpretation to it. Basically, as you let each player have more and more turns, how much does this neutralize B's advantage of getting to go second? REPLY [2 votes]: Not an answer, but here are some numerical results to show that Julian Rosen's asymptotic lower bound is not too far from optimal. $$\matrix{F(10^1,10^1) &= 0.4284225780\cr F(10^2,10^2) &= 0.4212694133\cr F(10^3,10^3) &= 0.4205898646\cr F(10^4,10^4) &= 0.4205226511\cr F(10^5,10^5) &= 0.4205159409\cr F(10^6,10^6) &= 0.4205152700\cr}$$<|endoftext|> TITLE: What are $( \infty , n)$-categories useful for? QUESTION [9 upvotes]: I know that mathematicians are trying to construct adequate models for $( \infty, n)$-categories. Although, it seems to be an interesting task, I would like to know some explicity examples where this theory can be helpful. In fact, for me there is no doubt that $( \infty, 1)-$categories are really useful. For example, the work of Lurie/Toën/Vezzosi in Derived Algebraic Geometry, or the Cobordism hypothesis. I have also the idea that $( \infty, 2)-$categories are used in the new advances in Langlands. However, I do not know any useful application of $(\infty, n)$-categories, for $n \geq 3$. I can presume that they provide the necessary formalism to fill the missing details in Lurie's proof of the Cobordism Hypothesis, but if this is so people do not seem to care too much. REPLY [13 votes]: In my humble opinion, the fact that such a structure appears naturally on bordisms between manifolds (eventually with some additional geometric structure like framings) is already a very good motivation to care about it. Moreover, with such a strong relationship between fully dualizable objects in $(\infty,n)$ categories and extended TQFTs, there are strong expectations to better understand the latter (which produces invariants of manifolds) via the former and vice-versa. I think an interesting example of such a bridge is the recent work of Douglas, Schommer-Pries and Snyder, investigating this relationship for $n=3$ to study fusion categories: http://arxiv.org/abs/1312.7188<|endoftext|> TITLE: Has philosophy ever clarified mathematics? QUESTION [143 upvotes]: I've recently been reading some standard textbooks on the philosophy of mathematics, and I've become quite frustrated that (surely due to my own limitations) I don't seem to be gleaning any mathematical insights from them. My naïve expectation would be that philosophy might take a difficult construction or proof, and clarify it by isolating the key ideas behind it. Having isolated the key ideas, philosophy might then highlight their relevance and thus point the way forward. Beyond this, I would hope that philosophy might elucidate the `true meaning' of axioms and of definitions by examining their ontology in a wider context. In reality, to the best of my knowledge (please prove me wrong!) both of the above tasks seem to be carried out exclusively by mathematicians, physicists, computer scientists, and other natural scientists, as far as I can see. To play the devil's advocate, philosophy seems to me like it might historically have largely played an opposite role, labeling certain objects as "unreal" and "unnatural" which in fact later turned out to be fruitful to study (negative numbers, irrational numbers, complex numbers...). Question: Has it ever happened that philosophy has elucidated and clarified a mathematical concept, proof, or construction in a way useful to research mathematicians? Philosophers have created much new mathematics (e.g. the work of C.S. Peirce, much of which is bona fide mathematical research), but the question is not about this, but rather about philosophy as practiced by philosophers providing elucidation, explanation, and clarification of existing mathematics. REPLY [5 votes]: Bernhard Riemann's famous 1854 Habilitation lecture "On the hypotheses…" owed more to the influence of the anti-Kantian philosopher Johann Friedrich Herbart than to Kant himself. Here one could mention an illuminating article by Nowak: G. Nowak, in The history of modern mathematics, Vol. I (Poughkeepsie, NY, 1989), 17–46, Academic Press, Boston, MA, 1989. Nowak argues that Herbart's constructive approach to space, already cited, mirrored the content of Riemann's reference to Gauss in that both discussed construction of spaces rather than construction in space. Riemann followed Herbart in rejecting Kant's view of space as an a priori category of thought, instead seeing space as a concept which possessed properties and was capable of change and variation. Riemann copied some passages from Herbart on this subject, and the Fragmente philosophischen Inhalts included in his published works contain a passage in which Riemann cites Herbart as demonstrating the falsity of Kant's view. Riemann took from Herbart the view that the construction of spatial objects was possible in intuition and independent of our perceptions in physical space. Riemann extended this idea to allow for the possibility that these spaces would not obey the axioms of Euclidean geometry. We know from Riemann's notes on Herbart that he read Herbart's Psychologie als Wissenschaft. It is possible that, had Kant's ideas about infinite space not been challenged by Herbart, the field we know today as Riemannian geometry may have developed much later. This is meant to complement Thomas Klimpel's answer on Riemann. For those with access to mathscinet, more information can be found here. Note. Peter Heinig pointed out that the historian David Rowe similarly holds that Riemann was more of a Herbartian than Kantian; see comments below and also Rowe's review.<|endoftext|> TITLE: Fixed sets of orbit spaces QUESTION [5 upvotes]: I've run across something that surprises me, so I'm wondering (1) Is it true? and (2) Is it well known? (And if the answers are affirmative, why didn't I know this already?) Let $G$ be a compact Lie group and let $H$ and $K$ be closed subgroups. So that the question isn't trivial, you can assume that $H$ is subconjugate to $K$. Then we can consider the $WH = NH/H$ space $(G/K)^H$. The claim is that this is a disjoint union of orbits of $WH$. The first proof I found depends on the Mongtomery-Zippin theorem via a consequence that appears as Lemma 1.1 in Peter May's "Equivariant Orientations and Thom Isomorphisms": If $j\colon \alpha\to\beta$ is a homotopy between $G$-maps $G/H\to G/K$, then $j$ factors as the composite of $\alpha$ and a homotopy $c\colon G/H\times I \to G/H$, where $c(eH,t) = c_tH$ for a path $c_t$ in the centralizer $C_GH$ of $H$, starting at the identity. Reinterpreting in terms of $(G/K)^H$, this implies that, if $\alpha$ and $\beta$ are two points in the same path component of $(G/K)^H$, then $\beta = c\alpha$ for some $c$ in the identity component of $C_GH$. In particular, they are in the same $WH$-orbit. Hence, each $WH$-orbit consists of a union of path components of $(G/K)^H$. I think another proof could be found by looking at the tangent plane at an $H$-fixed point in $G/K$, as an $H$-representation, and observing that the only $H$-trivial directions are those in the direction of the $NH$-action. (But I haven't worked this out completely.) So, is this true? Is it already well known? If it is true, it has some interesting implications for equivariant ordinary homology. REPLY [2 votes]: To answer my own question: I happened to run across the answer just now. Yes, it's true and known. tom Dieck quotes it as II.5.7 in Bredon's Introduction to compact transformation groups (which I don't happen to have handy).<|endoftext|> TITLE: Lower bound of Hecke eigenvalues of Maass form QUESTION [6 upvotes]: If $f$ is a Maass form and $p$-Hecke eigenvalue (i.e. Hecke eigenvalue of usual Hecke operator $T_p$) of $f$ is $\lambda_f(p)$, do we know anything about lower bound of the sum$$S(x) = \sum_{x\le p\le 2x}|\lambda_f(p)|^2?$$ To avoid Confusion $$(T_pf)(z)=\frac{1}{\sqrt{p}}\left[\sum_{b=0}^{p-1}f\left(\frac{z+b}{p}\right)+f(pz)\right]$$ Any referecne would be highly helpful. REPLY [3 votes]: Following up on @Idoneal's comment, one can prove something a bit more precise. In particular, $S(x)\asymp\frac{x}{\log x}$, provided that $\log x\gg\log(\lambda N)$, where $f$ has level $N$ and Laplace eigenvalue $\lambda$. I'm assuming Matt Young's normalization and that the central character is unitary; also, I'm writing $f\asymp g$ to mean that both $f=O(g)$ and $g=O(f)$. This result follows from work of Motohashi https://arxiv.org/abs/1209.4140. A little extra effort is needed to get the claimed range, but all the necessary tools are in this paper.<|endoftext|> TITLE: Is there a bijection of permutations onto mathematical objects that preserve information about descents? QUESTION [5 upvotes]: $\omega \in S_n$ is an FPFI (fixed point free involution) (also called a matching) if $\omega^2=1$ and $\omega(i) \neq i$ for all $i$. For $\omega \in S_n$, a descent occurs at $i$ if $\omega(i+1) < \omega(i)$. For example, $(1 \, 3)(2 \, 4) \in S_4$, when written as $3412$, has one descent at $i=2$. I'm curious as to what kind of mathematical objects there are that FPFIs be bijected onto that preserves the descent set, if any? Or in general, if there are objects that permutations can be bijected onto that preserve the information about descents. REPLY [7 votes]: In general, to find maps transporting statistics in a well-behaved way, it is useful to try FindStat. In the case at hand, go to http://www.findstat.org/StatisticsDatabase/St000021/ (which is the statistic "number of descents of a permutation") and click on "Search for values". After a short while, you will be presented with a list of candidates, each of the following type: a statistic $stat$ on (possibly different) combinatorial objects, and a map $\phi$ such that $$ des(\pi) = stat(\phi(\pi)) $$ (possibly $\phi$ is in fact a composition of several maps) You then only have to check which of candidates have maps that are bijective. Furthermore, you will have to check that not only the number of descents but also the descent set itself is preserved. In the case at hand, Ira's example of standard Young tableaux is found, there is possibly a well behaved bijection to increasing trees, to ordered trees,... As Christian points out in the comment below, it is also possible to provide values only for a subset of permutations, in your case for fixed point free involutions. Yet another possibility is to use a collection of objects built into FindStat that fits your problem better, namely http://www.findstat.org/StatisticFinder/PerfectMatchings. The drawback of the latter two approaches is that you have to enter the values manually (or generate them with a computer program as below and use the "free" box). for n in range(1,4): for pi in PerfectMatchings(2*n): print pi, "=>", pi.to_permutation().number_of_descents()<|endoftext|> TITLE: Can we normalize a complex analytic space in a covering of an open subset? QUESTION [7 upvotes]: Let $X$ be a normal connected complex analytic space, $x\in X$ a point, $f$ a nonzero holomorphic function vanishing at $x$. Denote by $U\subseteq X$ the locus where $f$ is nonzero. Suppose that $\pi:U'\to U$ is a finite degree covering space. This gives $U'$ the structure of an analytic space. Let $A$ be the ring of pairs $(V, h)$ where $V$ is a neighborhood of $x$ in $X$ and $f$ is a bounded holomorphic function on $\pi^{-1}(V\cap U)$, where for $V'\subseteq V$ we identify $(V, h)$ with $(V', h|_{V'})$. This is a subring of the stalk of $j_* \pi_* \mathcal{O}_{U'}$ at $x$, where $j:U\to X$ is the inclusion. It contains $\mathcal{O}_{X, x}$ (the ring of germs of holomorphic functions at $x$). Question. Is $A$ a finitely generated $\mathcal{O}_{X, x}$-module? In other words, can we "normalize $X$ inside $U'$", that is, extend $U'\to U$ to a finite map $X'\to X$? REPLY [5 votes]: See Theorem 3.4 in Dethloff and Grauert's "Seminormal Complex Spaces" (observe that in your setting $A$ is empty, so the theorem becomes rather easier). This seems to be exactly what you're looking for, if I understand your question correctly. I believe that this is a modern version of very classical theorems of Grauert, Remmert, and Stein. For convenience I include the statment of the theorem, which is actually slightly stronger than what is required. Dethloff and Grauert prove: Theorem 3.4. Let $N$ be a normal complex space, and let $B\subset N$ be a nowhere dense analytic subset. Let $\pi: Y\to (N\setminus B)$ be an analytically branched covering with a critical locus $A \subset (N\setminus B)$. Assume that $A\cup B\subset N$ is analytic. Then $\pi: Y\to (N\setminus B)$ can be extended uniquely to an analytically branched covering $\pi: X\to N$, which is uniquely determined up to equivalence of analytically branched coverings. Note that Theorem 1.3 and Theorem 3.3 and the remarks following it in the same notes shows that in this case $X$ must be normal, so this does indeed answer the question in its entirety.<|endoftext|> TITLE: Boolean Valued Models of PA QUESTION [14 upvotes]: O.K, a massively naive question. I've never really studied any non-standard models of PA before. I was just wondering if there's ever been any attempt to use the kind of Boolean valued model theory familiar from set theory in the context of arithmetic. Obviously, the actual structure of the model would have to be very different, but I mean specifically the idea of using elements of a complete Boolean algebra as truth-values for sentences of PA. If this hasn't been done, has anyone got any ideas about if/how it could be done? REPLY [9 votes]: This belated answer is prompted by: (1) Joel Hamkins' answer, in which he brings attention to Boolean ultrapowers of models of arithmetic. (2) Emil Jeřábek's comment "Using sophisticated tools to construct elementary extensions is simply a waste of effort" (in the exchange with Hamkins following Jeřábek's answer). First a definition: full arithmetic is the first order theory of the full expansion of the standard model of arithmetic, i.e, it is the first order theory of the model $ (\Bbb{N}, X)_{X \subseteq{\omega}}$, where $\Bbb{N}$ is the standard model of arithmetic $(\omega, +, \times)$. Note that full arithmetic has an uncountable vocabulary. Next, two results of Andreas Blass from the mid-1970s: Theorem 1. Every model $\cal{A}=$ $(A, \cdot \cdot \cdot) $ of full arithmetic has a conservative elementary end extension $\cal{B}=$ $(B, \cdot \cdot \cdot) $, i.e., if X is a subset of B that is parametrically definable in $\cal{B}$, then $X \cap A$ is parametrically definable in $A$. Theorem 2. Assuming CH (the continuum hypothesis) there is a model of full arithmetic that possesses a nonconserative elementray end extension. And, finally, the point of this posting: back in 1990, I used the method of Boolean ultrapowers to improve theorem 2 above by eliminating the CH assumption. My result appears as Theorem 3.4 of the paper below, which includes references for the aforementioned results of Blass, as well as other applications of the method of Boolean ultrapowers. A. Enayat, Minimal elementary extensions of models of set theory and arithmetic. Arch. Math. Logic 30 (1990), no. 3, 181–192.<|endoftext|> TITLE: Independent domination number for grid graphs QUESTION [5 upvotes]: Let $G_{n,m}$ be the $n \times m$ grid graph, i.e. $G= P_n \Box P_m$, and $T_{n,m}$ the $n\times m$ torus grid graph, i.e. $G= C_n \Box C_m$, where $P_n$ and $C_n$ indicate the path graph of length $n$ and the cycle graph of length $n$, respectively. The independent domination number $i(G)$ is defined to be the minimum cardinality among all maximal independent sets of vertices of $G$. Is there any explicit formula for $i(G_{n,m})$ and $i(T_{n,m})$? If not, is there a lower bound sharper than the one given by corresponding domination numbers $\gamma(G_{n,m})$ and $\gamma(T_{n,m})$? Or is there any asymptotic results for $i(G_{n,m})/mn$ and $i(T_{n,m})/mn$ as $m,n \to \infty$? REPLY [2 votes]: The independent domination number of grid graphs is known. You can find what you are looking for in the following paper. S. Crevals and P.R.J. Ostergard, Independent domination of grids, Discrete Math. 338 (2015), 1379-1384.<|endoftext|> TITLE: How many maximal triangulations of a rectangle? QUESTION [7 upvotes]: I posted the following question on MathStackExchange, but I didn't any answer. So please let me post it on MathOverflow. Let $L_{m,n}\subset\mathbb R^2$ be a rectangle given by $[0,m]×[0,n]$ with $m,n$ positive integers. Define $N(m,n)$ to be the number of subdivisions of $L_{m,n}$ into lattice triangles of area $1/2$. Here a lattice triangle is a triangle with vertices on the lattice $\mathbb Z^2\subset\mathbb R^2$. Is there an explicit formula for $N(m,n)$? For example, $N(1,1)=2$, $N(1,2)=6$... Thank you very much in advance. REPLY [6 votes]: Have you seen this paper by Kaibel and Ziegler ? Their notion of unimodular triangulation appears to be the same as what you ask for. They do not present explicit formulas, but rather discuss upper and lower bounds which are exponential in $mn$.<|endoftext|> TITLE: Do there exist sparse graphs with large crossing number? QUESTION [13 upvotes]: Does there exist a sequence of graphs $\{ G_n \}$ such that $G_n$ has $n$ vertices, the number of edges of $G_n$ is $O(n)$, and the crossing number of $G_n$ is $\Omega(n)$? In particular, do random $k$-regular graphs satisfy this? Motivation: the crossing number inequality gives a lower bound on the crossing number $cr(G)$, just in terms of the number of vertices $v$ and edges $e$. In particular, if $e \ge 4v$ then $$ cr(G) \ge \frac{e^3}{64v^2}.$$ This is tight, up to a constant factor, meaning there are graphs whose crossing numbers are approximately this small. But this still raises the question of how large the crossing number can be, particularly for sparse graphs. REPLY [5 votes]: Actually, even the crossing number of the random 3-regular graph on $n$ vertices has crossing number $\Theta(n^2)$. This follows from the lower bound obtained by Bollobás for the isoperimetric number of random regular graphs (European Journal of Combinatorics, Vol. 9 (1988), pp. 241-244); the random $k$-regular graph (for any fixed $k$) thus has large bisection width, guaranteeing a large (i.e., $\Theta(n^2)$) crossing number.<|endoftext|> TITLE: Elliptic operators corresponds to non vanishing vector fields QUESTION [11 upvotes]: Added, June 19, 2019: The main motivation of this post is to associate an index to differential operator associated to a dynamical system such that the index has an interesting dynamical interpretation. For example we hope that the index can help us to find an upper bound for the number of attractors of a dynamical system. According to comment conversations in this post we realize that ellipticity or hypoelipticity is a very relevant or perhaps a necessary conditions for existence of "Index". Now the subject and materials of this recently hold conference, "Fredholm theory of Non elliptic operatores seems to be related to this post. Let $X$ be a non vanishing vector field on a compact manifold $M$. The only differential operator associated with $X$ which I am aware of, is the derivational operator $D(g)=X.g$. Unfortunately this operator is not an elliptic operator. From the dynamical view point,what type of elliptic operators, or at least Fredholm diff. operators, can be associated with $X$? I mean, for a given non vanishing vector field $X$, what interesting elliptic operator $D$ can be constructed such that its fredholm index contains some information about the dynamical behavior of $X$. For example: the number of attractores, or the number of isolated compact invariant sets, etc.. EditL: For a possible related post see the following: How to compute the index of such operator? REPLY [10 votes]: Perhaps you would be interested in Witten's proof of the Poincare-Hopf theorem. Given a smooth nondegenerate vector field $V$ on a smooth closed manifold $M$, the theorem asserts that the Euler characteristic of $M$ is equal to the sum of the signs of the critical points of $V$. Perhaps this isn't as interesting as the dynamical behavior that you mentioned in your question, but it's a start. Witten's approach is to use $V$ to perturb the de Rham complex by replacing the de Rham differential $d$ with the operator $$d_t = d + t i_v \colon\: \Omega^*(M) \to \Omega^*(M)$$ where $t$ is a real number and $i_V$ is the interior product with $V$. He looked at the corresponding perturbed de Rham operator $D_t = d_t + d_t^*$ (where the adjoint is defined using a choice of Riemannian metric) and as usual viewed it as a graded Dirac-type operator on the graded Clifford module $\Omega^*(M)$. $D_t$ is elliptic and hence Fredholm, and since the index of an operator is determined by its symbol class the index of $D_t$ is just the index of the usual de Rham operator $D$ which is the Euler characteristic of $M$. On the other hand, one can calculate that $$D_t^2 = D^2 + t^2 ||V||^2 + t T$$ where $T$ is some bundle map. For large values of $t$ the potential term $t^2 ||V||^2$ becomes very large except in a tiny neighborhood of the critical set of $V$, so one can show that the eigenvectors of $D_t$ concentrate near the critical set. Combining this observation with the McKean-Singer formula for the index of $D_t$ and some asymptotic analysis proves the Poincare-Hopf theorem. There are a variety of generalizations of this result in the literature - perturbing other operators, relaxing the nondegeneracy assumption, etc. I don't know this literature too well and so I don't quite know how much dynamics to expect, but it's worth a look.<|endoftext|> TITLE: Higher Cerf Theory QUESTION [15 upvotes]: Morse functions on a manifold $M$ are defined as smooth maps $f:M \rightarrow \mathbb{R}$, such that at the critical points we can find local coordinates so that $$f(x_1,\dots,x_n)=-x_1^2-x_2^2-\dots-x^2_{i}+x^2_{i+1}+\dots+x_n^2.$$ There is an interpretation of Morse functions as "generic" functions. More precisely, one gives a stratification of $J^2(M,\mathbb{R})$---the second jet space of maps from $M$ to $\mathbb{R}$, and Morse functions are those functions whose second jet is transverse to the strata. By the Thom transversality theorem, the resulting space of functions is dense and open in $C^\infty(X,\mathbb{R})$. The space of Morse functions is not connected. To make it connected one can introduce generalized Morse functions, where the critical points have a local description as above or as follows $$f(x_1,\dots,x_n)=-x_1^2-x_2^2-\dots-x^2_{i}+x^2_{i+1}+\dots+x_{n-1}^2+x_n^3.$$ We call the latter a birth-death singularity. The result in this case is that a "generic" smooth homotopy $M\times I\rightarrow \mathbb{R}$ is a Morse function at the endpoints and it is a Morse function at all values of the parameter, $t\in I$, except for finitely many values, where the function is a generalized Morse function. What I am interested in is a generalization of Cerf's result, where the parameterizing space $I$ is replaced by an n-simplex. Here is my question. Is there a "nice" description of a generic family of smooth functions of the form $M\times \Delta^n\rightarrow \mathbb{R}$? By "nice" I mean having finitely many critical points and the critical points have local descriptions similar to the ones given above. The goal roughly is to construct a simplicial set, where $n$-simplices are the generic smooth maps $M\times\Delta^n\rightarrow \mathbb{R}$. The hope is that given the transversality results one should be able to show that the simplicial set is contractible, since it would amount to extending a generic function $M\times\partial\Delta^n\rightarrow \mathbb{R}$ to a generic functions $M\times\Delta^n\rightarrow \mathbb{R}$, which we would do by first extending by arbitrary smooth function (the space of all smooth functions is contractible) and then deform it something generic relative to a neighborhood of the boundary. REPLY [5 votes]: This is what catastrophe theory does, at least for small $n$, $n\leq 10$. Volume 1 of the book by Arnold, Gussein-Zade and Varchenko on singularities has a nice description of this theory; see especially Part 2 of that book.<|endoftext|> TITLE: Why a nilpotent Lie group must be a matrix group? QUESTION [5 upvotes]: The question may be a little naive (or even appear as a duplicate) as I guess the result is well known. I saw on the other thread that " c) A solvable Lie group G is linear iff its commutator subgroup G′ is closed, and G′ has no non-trivial compact subgroup (Thm 3.2 in Chap. XVIII) " But I cannot find the reference book anywhere. May I ask someone to give a hint why this is true? Background: Try to find a non-trivial example of infranil manifold which is not a matrix Lie group. Then the speaker visiting my university claiming every nilpotent Lie group must be a matrix Lie group. Since nilpotent is defined entirely algebraically, I feel there should be a proof based on algebraic techniques. But I do not know how to prove the above statement. REPLY [5 votes]: Another reference which was not yet mentioned, I think, is the article of M. Moskowitz, "Faithful Representations and a local property of Lie groups", Math. Z. $143$, 1975. There the question is discussed when all analytic groups with a given Lie algebra $\mathfrak{g}$ have a faithful linear representation. He proves among other things the following result: Theorem All connected Lie groups $G$ with Lie algebra $\mathfrak{g}$ have a faithful linear representation if and only if $Z(\mathfrak{g})\cap [\mathfrak{g},\mathfrak{g}]=0$ and $\tilde{S}$ has a faithful representation. Here $\tilde{G}=rad(\tilde{G})\cdot \tilde{S}$ is the Levi decomposition for the simply connected group $\tilde{G}$ with Lie algebra $\mathfrak{g}$. Corollary Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. If $ad(\mathfrak{g})$ has nontrivial center then there is a locally isomorphic group without any faithful representation.<|endoftext|> TITLE: Translative packing constant strictly larger than lattice packing constant QUESTION [8 upvotes]: Simply put, my question is this: what is the smallest dimension, if any, where we can know for sure that a convex body exists whose translative packing constant is strictly larger than its lattice packing constant? Some relevant facts I know: In two dimensions, the translative packing constant is always equal to the lattice packing constant for convex bodies. For non-convex bodies, already in two dimensions there are counterexamples. Bezdek and Kuperberg give a good exposition of this. In ten dimensions, the best packing of spheres seems to be non-lattice. In lower dimensions, the best lattice packing seems to be also the best packing. The best known lattice packing in dimensions above 8 is not actually known to be best (except in 24 dimensions). Therefore, even for spheres, there is no dimension where lattices are proved to be suboptimal. See this MO question for more on that. Convex bodies that tile by translation can also tile by lattice translations. So the example cannot be a tiling body. This is famous result due to Venkov, Alexandrov, and McMullen, and Gruber's book on Convex and Discrete Geometry gives a nice treatment. Except for tiling bodies, which are ruled out be the previous point, very few bodies have known translative packing constants. However, all we need to have an example is a lower bound for the translative packing constant and an upper bound for the lattice packing constant. There are known methods to compute the lattice packing constant for polytopes, and in 3D one such method has been implemented by Betke and Henk. So, it seems that in dimension 10, there are convex bodies, namely spheres, that pack better by translation than by lattice translation. However, this is not rigorously known. If true, this example can probably be extended to higher dimensions by forming cylinders. However, it seems to me that if we allow nonspherical bodies, the dimension where translative packing starts to beat out lattice packing should be siginificantly lower. Is there an example? In dimensions low enough (e.g. 3D), candidates can be checked computationally and rigorously established if they check out. REPLY [5 votes]: This is not an answer, but rather a documentation of a failed attempt to obtain an answer. One problem for which we know there is a significant difference between unrestricted sets of translations and lattices is in tilings with equal-size cubes. Every lattice tiling with n-cubes has a pair of cubes sharing an entire (n-1)-dimensional face (Hajós). On the other hand, in dimensions $n\ge8$, there are translational tilings with n-cubes in which no two cubes share an entire (n-1)-dimensional face ($n\ge10$ due to Lagarias and Shor, improve to 8 by Mackey, n=7 case open, see Keller's conjecture). Moreover, a stronger property holds in these tilings: the center of each (n-1)-face does not lie in the relative interior of any other (n-1)-face. I call such a tiling a center-to-boundary tiling. Now consider a slightly stellated n-cube: the convex hull of a unit n-cube and a point for each face lying a height $\epsilon$ above the center of the face. It is easy to perturb a center-to-boundary tiling of n-cubes to a packing with mean volume $(1+\epsilon)^n$. I wanted to show that since lattice tilings have shared faces along at least one direction, and a lattice packing of slightly stellated cubes will be near a lattice tiling, that the mean volume (i.e. lattice determinant) of the packing will have to be $\ge 1+(n+1)\epsilon + o(\epsilon)$, but this turns out to be false. Here is an example of a family of lattices that pack $\epsilon$-stellated 5-cubes, but have determinant $1+(47/8)\epsilon+O(\epsilon^2)$: $L=A\mathbb{Z}^5$, where $$A=\left(\begin{array}{ccccc} 1+2\epsilon&-\tfrac12&0&0&0\\ 0&1+\epsilon&-\tfrac12&\tfrac12&0\\ 0&0&1+\epsilon&0&-\tfrac14\\ 0&0&0&1+\epsilon&\tfrac12\\ 0&0&-\tfrac12\epsilon&0&1+\epsilon \end{array}\right)\text.$$ Now, still $47/8>5$, but this example does not bode well for the program. For large $n$, I believe similar constructions are possible that achieve determinants $\le 1+n\epsilon+o(\epsilon)$. It might still be that this is impossible for $n=8$, but if so, it could be delicate to show.<|endoftext|> TITLE: Why is the Dynamical Mordell-Lang conjecture interesting? QUESTION [15 upvotes]: The gist of the Dynamical Mordell-Lang conjecture is: Let's say you have a set $S$. Inside $S$, you have a point $x$ where you start off at. Now, you also have a function $f: S \rightarrow S$ (is that how I correctly write "maps from something in $S$ to something else in $S$"?). So each time you apply $f$ to $x$, you land at another point inside $S$ ("duh"), but what if you keep landing inside a closed subset of $S$, $Y$? Not just once ("duh"), a few times ("hrm"), or a billion times ("hmmm"), but an infinite number of times? Clearly, $Y$ must have some special relationship to $f$? Some special "structure"? Recently, it was proved that that there is some number $k \in \mathbb{N}$ such that $Z \subset Y$ is invariant under the "$k$th iterant" of $f$, $f^{\circ k}$ (i.e. $f$ applied $k$ times to $x$). Why is Dynamical Mordell-Lang interesting? Which open questions might eventually benefit from having such a result? What new mathematical vistas does it expose? Or, to what long program of work does it add? I recall that the speaker mentioned that there was is an interesting application of this conjecture to the theory of differential equations, and that in fact, someone working on differential equations about 30 or 40 years ago conjectured that there should be a "Dynamical Mordell-Lang"...perhaps this tidbit, if it isn't too vague, might be a helpful starting point for answers? See this meta.MO question for background on this MO question REPLY [22 votes]: Maybe a brief history would provide an answer. The original Mordell conjecture asserted that if a curve $C:f(X,Y)=0$ given by a polynomial with rational coefficients has genus at least 2, then it has only finitely many points with rational coordinates. (More generally, this should be true for any number field $K$.) This was a bold general conjecture about solutions to Diophantine equations, a subject dating back to antiquity. Algebraic geometers and complex analysts studying the geometry of (smooth projective) curves discovered that a curve $C$ of genus $g$ can be embeded into an abelian variety $J$ of dimension $g$ called its Jacobian variety. So one might hope to study $C(K)$ by first analyzing the group $J(K)$ and then seeing which of those points lie on $C$. Enter the famous Mordell-Weil Theorem: The abelian group $J(K)$ is finitely generated. So now we have the group $J(\mathbb C)$, which looks like $\mathbb C^g/\text{(lattice)}$, a finitely generated subgroup $\Gamma=J(K)\subset J(\mathbb C)$, and a curve $C(\mathbb C)\subset J(\mathbb C)$. If $g\ge2$, then just by dimension count, it seems reasonable to suppose that $C(K) = \Gamma \cap C(\mathbb C)$ is finite. Lang realized that when phrased as an intersection in this way, there was no reason to restrict to curves. So let $A$ be an abelian variety (which is a variety that's also a group), let $Y\subset A$ be a subvariety, let $\Gamma\subset A(\mathbb C)$ be a finitely generated group. Mordell-Lang Conjecture The intersection $\Gamma\cap Y$ is finite unless $Y$ contains a translate of an abelian subvariety of $A$. Let's consider the special case that $\Gamma$ has rank $1$, say $\Gamma=nP_0$ for some fixed $P_0\in A$ and $n\in\mathbb Z$. Then we can describe $\Gamma$ by iterating the "translation-by-$P_0$-map" $T:A\to A$ defined by $T(P)=P+P_0$. Thus a special case of the Mordell-Lang conjecture says if the forward orbit $\{T^{\circ n}(O) : n\ge1\}$ of the iterates of $T$ intersects $Y$ infinitely often, then $Y$ contains a translate of an abelian subvariety of $A$. And now you can begin to see how to generalize the original Mordell-Lang conjecture to a more general dynamical system. Thus let $X$ be an algebraic variety, let $f:X\to X$ be a morphism, let $x_0\in X$ be a starting point, let $Y\subseteq X$ be a subvariety, and let $\mathcal O_f(x_0)=\{f^{\circ n}(x_0):n\ge0\}$ be the forward orbit of $x_0$ under iteration of $f$. Dynamical Mordell-Lang Conjecture (Version 1) If $\mathcal O_f(x_0)\cap Y$ is infinite, then there is a positive dimensional subvariety $Z\subset Y$ such that $f(Z)\subseteq Z$. Here is an alternative formulation, which is often useful for proofs: Dynamical Mordell-Lang Conjecture (Version 2) The set $\{n : f^{\circ n}(x_0)\in Y\}$ is the union of a finite set and a finite collection of one-sided arithmetic progressions. The original Mordell conjecture for curves was proven by Faltings. Vojta then gave a very different proof, and Faltings adapted and extended Vojta's ideas to prove the original Mordell-Lang conjecture for abelian varieties. There are various special cases of the dynamical Mordell-Lang conjecture that are known, but it is still open (as far as I'm aware) in full generality, even for morphisms $\mathbb P^2\to\mathbb P^2$.<|endoftext|> TITLE: Is homology finitely generated as an algebra? QUESTION [17 upvotes]: If a differential graded algebra is finitely generated as an algebra, is its homology finitely generated as an algebra? Is it easier if we impose any of the three conditions: characteristic zero; free as an algebra; generated in positive degree? generated in negative degree? What about commutative algebras or Lie algebras? (is anything else sensible to ask?) What if we reduce the grading from $\mathbb Z$ to $\mathbb Z/2$ (ie, differential super-algebras)? What I'm really interested in is the case of $\mathbb Q$-DGLAs freely generated in homologically positive degree, because that corresponds to homotopy groups of finite complexes, but the commutative case is probably more familiar. I expect the answers to be the same, heuristically from Koszul duality. REPLY [15 votes]: Another counterexample: let $A$ be the algebra $\mathbb{Q}[y,z]/(y^2) \otimes \bigwedge(x)$ with $x$ in degree 1, $y$ and $z$ in degree 2. Put a differential on this by $z \mapsto xy$. This is a commutative dga in characteristic 0 generated in positive degrees, but of course it's not free. Its homology is spanned by the classes $xz^i$ and $yz^i$ for all $i \geq 0$, and the product on the homology algebra is trivial. So it is infinitely generated as an algebra.<|endoftext|> TITLE: (non-)existence of the aperiodic monotile QUESTION [5 upvotes]: The aperiodic monotile problem asks whether there exists a single tile that every tiling of the plane made with it results non-periodic. What is known about this problem? If this tile exists, how can it be/not be? how could it (not) be constructed? EDIT: I'm thinking about the most restrictive case where the single tile should be homeomorphic to an euclidean closed disk. EDIT: The kind of answer I'm looking for is, for example, this result implies that if the monotile is to exist, it cannot be convex. REPLY [5 votes]: In three dimensions there is a known aperiodic tile, the Schmitt–Conway–Danzer. I am pretty sure that this idea can easely be extended to any dimension $d \geq 3$, if anyone cares. In one dimension, it is known that such tiling cannot exist. The problem was open in two dimensions until few years ago. And it may be solved, depending on what you understand by aperiodic monotile. The Socolar-Taylor tile is a tile which tiles the plane only aperiodically. BUT, depending on which version you pick, it is either disconnected or not the closure of its interior (you can make it connected by adding few curves to connect pieces). If you want a tile which is connected and closure of its interior, the problem is still open as far as I know...<|endoftext|> TITLE: Are Brown representable functors determined by restriction to finite complexes? QUESTION [9 upvotes]: Assume two $CW$ complexes $X,Y$ give two functors $h_X=[-,X], h_Y=[-,Y]$ on the homotopy category of $CW$ complexes whose restrictions to the full subcategory of finite $CW$ complexes are naturally equivalent. Does this imply that the two functors are naturally equivalent (i.e. that X, Y are homotopy equivalent)? REPLY [12 votes]: The answer depends on which category of CW-complexes you have in mind. It is true in the category of based connected CW-complexes simply by the Whitehead's Theorem. EDIT: The concern raised by Matthias in the comment above is addressed in Brown's paper (Theorem 2.8 of Abstract Homotopy Theory). This is only proven in the based case, but even if it works in the unbased case too, then the following still applies. However, dropping any of these assumptions makes the answer negative. In fact, Heller proves in Corollary 2.3 of On the representability of homotopy functors that there is no set of CW-complexes such that functors represented by them detect weak equivalences. (This is closely related to the failure of Brown's representability in the unbased case, discussed here based on another paper by Freyd and Heller.) In particular, finite CW-complexes do not suffice since a map inducing bijections on homotopy classes of maps out of CW-complexes might not be a $\pi_1$-isomorphism. In general, it will only induce a bijection on conjugacy classes of $\pi_1$s. On the other hand, this answer shows that this works for maps between CW-complexes with finitely generated fundamental groups.<|endoftext|> TITLE: Power series with funny behavior at the boundary QUESTION [8 upvotes]: Consider a power series $$ \sum_{n=0}^{\infty}a_nz^n $$ where $a_n$ and $z$ are complex numbers. There is radius $R$ of convergence. Let us assume that is a positive real number. It is well known that for $|z|R$ it does not converge. On the other hand, when $|z|=R$, the series can have very different behaviors. This has been discussed in many posts, e.g. Behaviour of power series on their circle of convergence , Seeking a Geometric Proof of a Generalized Alternating Series' Convergence . I am looking for some relatively easy explicit examples of power series with funny behavior at the boundary. The only I know is $$ \sum_n\frac{1}{n}z^n $$ or small variations, e.g. replacing $z$ with $z^k$. thanks REPLY [18 votes]: The series $$f(z) = \sum_{n=1}^\infty \dfrac{z^{2^n}}{n}$$ converges almost everywhere on the unit circle by Carleson's theorem (it is the Fourier series of an $L^2$ function). However, it diverges on a dense set, including all the $2^k$'th roots of unity: in fact at each of those points the real parts of the partial sums $S_k = \sum_{n=1}^k z^{2^n}/n$ are unbounded above. Now for each positive integer $N$, the set $U_N$ of points $z$ such that $\text{Re}(S_k) > N$ for some $k$ is open and dense in the unit circle. The intersection $G$ of the $U_N$ is a dense $G_\delta$ by the Baire category theorem, and the series diverges at every point of $G$. Putting these facts together, we find that the set of points of the unit circle where the series diverges is negligible in the sense of Lebesgue measure but the set where it converges is negligible in the sense of Baire category. I'd call that funny...<|endoftext|> TITLE: Convergence in energy of bounded (semi)subharmonic functions QUESTION [5 upvotes]: Consider a sequence $(f_n)$ of functions in the flat torus $T^d$ converging Lebesgue-a.e. to a limit function $f$. Assume that: 1) $|f_n|(x)\leq 1$ for every $n,x$ 2) $\Delta f_n\geq -1$ in the sense of distributions for every $n$ Can we deduce that $$ \lim_{n\to\infty}\int|\nabla f_n|^2 d\mathcal L^d=\int|\nabla f|^2 d\mathcal L^d $$ ? Notice that $(1)+(2)$ grant that $\int|\nabla f_n|^2 d\mathcal L^d=-\int f_nd\Delta f_n\leq \mathcal L^d(T^d)<\infty$ for every $n$ so that the sequence is bounded in $W^{1,2}$ and thus, given the a.e. convergence, also weakly converging. The question is then if it is strongly converging in $W^{1,2}$. In fact, I'm interested in this problem in a much more general setting, the functions being defined on a converging sequence of metric measure spaces with Ricci curvature bounded from below, but after some thinking I realized that I don't know the answer not even in this simplified setting. Some comments. The assumption on the Laplacians ensures that they are measures with uniformly bounded mass, and the uniform bound on the $W^{1,2}$-norms of the functions grant that $|\Delta f_n(E)|$ can be bounded from above in terms of the capacity of $E$ only. Therefore to conclude it would be sufficient to show that for every $\epsilon$ there exists $\delta,N$ such that for $n>N$ the set $\{|f_n-f|>\delta\}$ has capacity smaller than $\epsilon$ (here it matters to pick the upper semicontinuous representatives of the functions). I've searched in the literature and found results which are very close to this one, but not really sufficient for my purposes, see for instance the book `Growth Theory of Subharmonic Functions' by Vladimir Azarin. Yet, I found no reference for the question I'm asking. Any help would be appreciated, thanks in advance. REPLY [3 votes]: I'm reporting an example, presented to me by Bozhidar Velichkov, showing that the answer to my question is no. The example is based on a construction by Cioranescu-Murat given in their paper "Un terme étrange venu d'ailleurs". They show that there exists a sequence $\Omega_n$ of open subsets with closure contained in the unit ball $B$ with the following property. Let $u_n$ be the solution of $$ \left\{\begin{array}{ll} \Delta u_n =-1 & on\ \ \Omega_n\\ u_n=0 & on\ \ B\setminus\Omega_n \end{array} \right. $$ and notice that: 1) $u_n$ is non-negative 2) the Laplacian of $u_n$ is $\geq -1$ in the sense of distributions in $B$ 3) $u_n$ is bounded from above by the solution $v$ of $\Delta v=-1$ on $B$ with 0 boundary condition. In particular they are uniformly bounded in $L^\infty$. 4) $\int_B |\nabla u_n|^2=-\int_{\Omega_n}u_n\Delta u_n=\int_B u_n$ In particular the sequence $(u_n)$ is bounded in $W^{1,2}$ and thus compact in $L^2$. All of this is true regardless of the choice of $\Omega_n$. What Cioranescu-Murat did was to show that one can choose the $\Omega_n$'s so that $u_n\to u$ in $L^2$ for $u$ such that $\Delta u+1$ is a positive measure $\mu$ (i.e. non-negative and with strictly positive mass). This provides a counterexample because on one hand we have $$ \int_B|\nabla u_n|^2=\int_Bu_n\to \int_B u $$ and on the other $$ \int_B|\nabla u|^2=-\int u\Delta u=\int u(1-\mu)<\int u $$ (I've been a bit sloppy in the way this last line is written and also in not justifying why $u$ is strictly positive in a set of positive $\mu$-mass, but this should give the idea). Roughly said, the sets $\Omega_n$ are the whole $B/2$ minus a lot of tiny disjoint balls spread around which decrease in size and increase in number as $n\to\infty$<|endoftext|> TITLE: On the induced matrix norm $\| \cdot \|_{2,\infty}$ QUESTION [5 upvotes]: The induced norm of the matrix $A$ as a map from $(\mathbb R^n , \| \cdot \|_p)$ to $(\mathbb R^m, \| \cdot \|_q)$ is given by $$ \| A \|_{p,q} = \sup_{x\in\mathbb{R}^n\setminus \{0\}} \frac{\|Ax\|_q}{\|x\|_p}.$$ I would like to compute $\| \cdot \|_{2,\infty}$. In this paper: On the Calculation of the $l_2\to l_1$ Induced Matrix Norm, the authors presented the results for any $p,q\in\{1,2,\infty\}$, except for $p=2$ and $q=\infty$. So I would like to know if this is still an open problem. And if it is, then is there any result on finding a tight and easy-to-compute upper bound on the norm? Thank you in advance. REPLY [4 votes]: Computing such induced norms is a hard problem. For the case of $p=2$ and $q \ge 2$, have a look at this paper by Barak et al. to see how tricky the problem is. Typically, for other than the nice cases of $1,2, \infty$ style, these norms are NP-hard to compute, with well-known results for the case $p \ge q$ (see also e.g.: "Matrix norms are NP-Hard to approximate"). The paper of Barak et al focuses on the hypercontractive case of $p < q$. For the specific case of $p=2$ and $q=\infty$ mentioned above, I think the paper of Barak et al cited above mentions that the norm is just the largest 2-norm of any row of the matrix (thanks to N. Johnston for pointing this out).<|endoftext|> TITLE: Deligne's exterior power QUESTION [8 upvotes]: In "Catégories Tannakiennes", Deligne defines the $n$th exterior power of an object $A$ of an abelian tensor category $\mathcal{C}$ as the image of the morphism $$p : A^{\otimes n} \to A^{\otimes n}, a_1 \otimes \dotsc \otimes a_n \mapsto \sum_{\sigma \in \Sigma_n} \mathrm{sgn}(\sigma) \cdot a_{\sigma(1)} \otimes \dotsc \otimes a_{\sigma(n)}.$$ Let's call this $\Lambda^n_{Deligne}(A)$. Notice that $p^2=n! p$. If $n!$ acts invertibly on $\mathcal{C}$, this is certainly the correct definition of the exterior power (but then we could also define it as suggested here). But what happens in general? If $\mathcal{C}=\mathsf{Mod}(R)$ for some commutative ring $R$, then it is easy to see that the map $(a_1,\dotsc,a_n) \mapsto p(a_1 \otimes \dotsc \otimes a_n)$ is alternating and thus induces a surjective homomorphism $$\Lambda^n(A) \to \Lambda^n_{Deligne}(A).$$ Here, $\Lambda^n(A)$ is the usual exterior power. Is this map an isomorphism? In other words, is the map $\Lambda^n(A) \to A^{\otimes n}$ induced by $p$ injective? The answer is yes if $A$ is free. It follows formally that it also holds when $A$ is locally free. By Lazard's Theorem we get the result if $A$ is flat. It also holds when $A$ is cyclic, because then $\Lambda^n(A)$ vanishes for $n>1$. I've also tried some other examples. Are there any counterexamples? REPLY [2 votes]: It is not. Here is a counterexample. (I have essentially copied the setting from https://mathoverflow.net/a/87958/2530 , which in turn goes back to a PBW counterexample by P. M. Cohn.) Let $k$ be the commutative ring $\mathbb F_2 \left[\alpha,\beta,\gamma\right] / \left(\alpha^2,\beta^2,\gamma^2\right)$. Let $M$ be the $k$-module $\left\langle x,y,z\right\rangle$. Let $L$ be the $k$-module $M / \left\langle \alpha x + \beta y + \gamma z\right\rangle$. We shall abusively write $x$, $y$ and $z$ both for the elements $x$, $y$ and $z$ of $M$ and for their projections on $L$ as well. Define a $k$-linear map $f : M \otimes M \to k$ by $f\left(y \otimes z\right) = \alpha$, $f\left(z \otimes y\right) = \alpha$, $f\left(x \otimes x\right) = 0$, $f\left(z \otimes x\right) = \beta$, $f\left(x \otimes z\right) = \beta$, $f\left(y \otimes y\right) = 0$, $f\left(x \otimes y\right) = \gamma$, $f\left(y \otimes x\right) = \gamma$, $f\left(z \otimes z\right) = 0$. It is easy to see that $f$ vanishes on $\left\langle \alpha x + \beta y + \gamma z\right\rangle \otimes M + M \otimes \left\langle \alpha x + \beta y + \gamma z\right\rangle$. Hence, $f$ descends to a $k$-linear map $L \otimes L \to k$ (since $L \otimes L \cong \left(M \otimes M\right) / \left(\left\langle \alpha x + \beta y + \gamma z\right\rangle \otimes M + M \otimes \left\langle \alpha x + \beta y + \gamma z\right\rangle\right)$ as a consequence of the right-exactness of the tensor product). This latter map is also alternating, and therefore descends to a $k$-linear map $\wedge^2 L \to k$. This latter map sends $\beta y \wedge \gamma z + \gamma z \wedge \alpha x + \alpha x \wedge \beta y$ to $3 \alpha \beta \gamma = \alpha \beta \gamma \neq 0$. Hence, $\beta y \wedge \gamma z + \gamma z \wedge \alpha x + \alpha x \wedge \beta y \neq 0 \in \wedge^2 L$. But in $L \otimes L$, we have $p \left( \beta y \wedge \gamma z + \gamma z \wedge \alpha x + \alpha x \wedge \beta y \right)$ $= \beta y \otimes \gamma z - \gamma z \otimes \beta y + \left(\text{cyclic permutations}\right)$ $= \underbrace{\alpha x \otimes \beta y - \alpha x \otimes \gamma z}_{= - \alpha x \otimes \beta y - \alpha x \otimes \gamma z = \alpha x \otimes \left( -\beta y - \gamma z\right)} + \left(\text{cyclic permutations}\right)$ $= \alpha x \otimes \underbrace{\left( -\beta y - \gamma z\right)}_{=\alpha x \text{ (since } \alpha x + \beta y + \gamma z = 0 \text{ in } L \text{)}} + \left(\text{cyclic permutations}\right)$ $= \underbrace{\alpha x \otimes \alpha x}_{=\alpha^2 x\otimes x} + \left(\text{cyclic permutations}\right)$ $= \underbrace{\alpha^2}_{=0} x \otimes x + \left(\text{cyclic permutations}\right) = 0$. So $p$ fails to be injective.<|endoftext|> TITLE: On the definition of the $\alpha$-iterable cardinals QUESTION [10 upvotes]: I am reading the paper Ramsey-like cardinals II by Victoria Gitman and Philip Welch (Journal of Symbolic Logic, vol. 76, no. 2. pp. 541-560, 2011) and maybe I am missing something. According to the deffinitions provided there: An $M_0$-ultrafilter $U_0$ is $0$-good if its ultrapower $M_1$ is well-founded. $U_0$ is $1$-good if it is weakly amenable and $0$-good. $U_0$ is $2$-good if $U_1=j_{01}(U_0)$ determines a well-founded ultrapower $M_2$ of $M_1$. Hence, if $U_0$ is $2$-good, we have two well-founded ultrapowers, $M_1$ and $M_2$ and, in general, if $U_0$ is $n$-good we have $n$ well-founded ultrapowers, $M_1,\ldots, M_n$. However, $U_0$ is $\omega$-good if it provides a sequence of well-founded ultrapowers $M_1,M_2,\ldots$, but the inductive limit $M_\omega$ is not necessarily well-founded. When this happens, $U_0$ is said to be $\omega+1$-good, but the ultrapower $M_{\omega+1}$ must be ill-founded. $U_0$ is $\omega+2$-good if $M_{\omega+1}$ is well-founded, and so on. So, if I have understood the definition, for $n<\omega$, when $U_0$ is $n+1$-good the ultrapower $M_{n+1}$ is well-founded, but for $\beta\geq \omega$, when $U_0$ is $\beta+1$-good the ultrapower $M_\beta$ is well-founded, but $M_{\beta+1}$ may be ill-founded. I insist on this because, for instance, in Theorem 4.1 we have an $\alpha$-good ultrafilter, with $\alpha=\beta+1$, and so, if I am right, the ultrapower $M_\beta$ is well-founded, but we cannot say that $M_\alpha$ is also well-founded. However, the proof requires to make a complex construction within $M_\alpha$, and it seems strange to me that nowhere in the text is said something like "notice that $M_\alpha$ maybe ill-founded". In fact, I am not sure about how to handle this situation, since until now I have worked with well-founded ultrapowers only, but before trying to adapt the proof of Theorem 3.11 as indicated, I have preferred to ask here, just in case $M_\alpha$ is well-founded and I am missing why it is so. REPLY [6 votes]: The objection is correct. The argument of Theorem 4.1 uses that $M_\alpha$ is well-founded, which may not be the case if $\beta$ is a limit, according to the definition of $\alpha$-iterable cardinals given in the paper. There are two ways to address this. It does make sense to redefine the limit stages of the hierarchy to say that the direct limit is well-founded. I believe that in this case, Theorem 4.1 is correct. The new definition would say that a cardinal is $\alpha$-iterable if for every $A\subseteq\kappa$, there is a weak $\kappa$-model $M$ with $A\in M$ and a weakly amenable $M$-ultrafilter on $M$ whose $\alpha^{\text{th}}$-iterate model is well-founded. I am not sure how satisfied I am with changing the definition. I am still hoping that the increasing strength of the hierarchy can be proved with the original definition. The question is this: suppose that $\beta$ is a limit cardinal and $M_\beta$ is the well-founded direct limit of an iteration of some $M$ by an $M$-ultrafilter $U$. Does it hold in $M_\beta$, that every $a\subseteq\kappa$ is contained in a weak $\kappa$-model $m$ for which there is a weakly amenable $m$-ultrafilter $u$ producing an iteration of length $\beta$ (with a possibly ill-founded direct limit)? Currently, I don't know how to do it, although I have some ideas that I will write about later if they work out. This is related to another erroneous claim made later in the paper that we can assume in the definition of $\alpha$-iterable cardinals for $\alpha\geq 2$ that $M\models{\rm ZFC}$ (as opposed to ${\rm ZFC}$ without powerset). It is my conjecture that this assumption actually increases the consistency strength of the large cardinal notion. It seems that the argument of Theorem 4.9, which shows the statement in the case of $1$-iterable cardinals should generalize to all $\alpha$-iterable cardinals particularly with the revised definition. Update: I now think that Theorem 4.1 is correct for the definition of $\alpha$-iterable cardinals given in the paper. To fill the gap in the original argument, we need to show the following. Suppose that $M$ is a weak $\kappa$-model with an $M$-ultrafilter $U$ and $\beta$ is a limit ordinal such that $U$ can be iterated $\beta+1$-many times, meaning that the direct limit $M_\beta$ of the first $\beta$-many steps of the iteration is well-founded. We must show that in $M_\beta$, every subset of $\kappa$ can be put into a weak $\kappa$-model $M$ for which there is a $M$-ultrafilter producing $\beta$-many well-founded iterates with a possibly ill-founded direct limit. For the tree construction of the proof of Theorem 4.1 to go through, it suffices to show that all embeddings $j_{\xi\gamma}^i:M^i_\xi\to M^i_\gamma$ are elements of some set in $M_\beta$ (so that we can construct the tree on a set, whose branch will gives the desired directed system). Let $\{\kappa_\xi\mid\xi\leq\beta\}$ be the critical sequence of the iteration and assume that $V_\kappa\in M$. I want to argue that all $j_{\xi\gamma}$ are in $V_{\kappa_\beta}^{M_\beta}$. For clarity, I will give the argument for $\beta=\omega$. First, consider $f=j_{01}^i:M^i_0\to M^i_1$. We know $f\in M_1$ and has size $\kappa_1$ there (because $M^i_0$ has size $\kappa$ and $M^i_1=j_{01}(M^i_0)$ has size $\kappa_1$). Thus, $f\in V_{\kappa_2}^{M_2}$. The map $j_{2\omega}:M_2\to M_\omega$ has critical point $\kappa_2$, from which it follows that $j_{2\omega}(f)=f\in j_{2\omega}(V_{\kappa_2})=V_{\kappa_\omega}$. Next, consider the general case $f=j_{mn}^i:M^i_m\to M^i_n$. We know that $f\in M_n$ and has size $\kappa_n$ there. Thus, $f\in V_{\kappa_{n+2}}^{M_{n+2}}$, and rest of the argument is the same.<|endoftext|> TITLE: Weight multiplicity formulae for $(\mathfrak g,B)$-irreps QUESTION [5 upvotes]: Let $G$ be a complex reductive Lie group, $B$ a Borel subgroup, with which to define "dominant weight". Let $\lambda$ be an integral weight, not necessarily dominant, but nonetheless giving a one-dimensional $B$-rep $\mathbb C_\lambda$. Then we can define the Verma module $U{\mathfrak g} \otimes_{U\mathfrak b} \mathbb C_\lambda$, which has compatible actions of $\mathfrak g$ and $B$ (it's a "$(\mathfrak g,B)$-module"), and a unique irreducible quotient $V_\lambda$, again a $(\mathfrak g,B)$-module. If $\lambda$ is dominant, then $V_\lambda$ is actually a (finite-dimensional) $G$-irrep, so we know how to compute its weight multiplicities in manifestly positive ways, e.g. counting Littelmann paths. For $\lambda$ not dominant, $V_\lambda$ is infinite-dimensional, but its weight multiplicities are still finite (since they're bounded by those of the Verma module). Are there combinatorial formulae for the weight multiplicities of $V_\lambda$, when $\lambda$ is not dominant? If so, references please! REPLY [3 votes]: Maybe it should be emphasized that the question here concerns only Lie algebra representations, so the initial group language (about Lie groups or perhaps algebraic groups) is unnecessary. The framework is the 1976 BGG category $\mathcal{O}$ for a semisimple Lie algebra over $\mathbb{C}$ or other algebraically closed field of characteristic 0. In general, each linear functional $\lambda \in \mathfrak{h}^*$ (for a fixed Cartan subalgebra $\mathfrak{h}$ of the given Lie algebra $\mathfrak{g}$) defines a Verma module and its unique simple quotient, while other possible highest weight modules occupy intermediate positions. Here the weights are assumed to be integral. Such weights form an $\ell$-dimensional lattice $\Lambda \subset \mathfrak{h}^*$ (where $\ell = \dim \mathfrak{h}$), which can be viewed also as the abstract weight lattice of the underlying root system or as the character group $X(T)$ of a maximal torus $T$ having Lie algebra $\mathfrak{h}$ in a simply connected algebraic group with Lie algebra $\mathfrak{g}$. (But since $G$ usually doesn't act analytically or rationally on modules in category $\mathcal{O}$, the potential rational $T$-action doesn't add anything to the given action of $\mathfrak{h}$.) As Victor observes, there are "trivial" cases in which there exists a combinatorial formula for the weight multiplicities of a given (integral) weight $\mu$ in a simple highest weight module of highest weight $\lambda$. A Verma module is in fact simple if and only if its highest weight is antidominant (see section 4.4 in my book on category $\mathcal{O}$). For a parabolic (= generalized) Verma module, Jantzen developed a rather complicated criterion (see sections 9.12-9.13). But even in such "trivial" cases, it's necessary to compute Kostant's partition function. Other than this case, and the narrow case in type $A$ which Victor cites, I'm unaware of any infinite dimensional simple modules admitting such combinatorial formulas. Naturally it's impossible to prove that such formulas can't exist elsewhere. To approach this experimentally would be extremely challenging, since the combination of Kostant's partition function (for weight multiplicities in Verma modules) and the Kazhdan-Lusztig polynomial values at 1 will lead to lengthy calculations involving many cancellations even in fairly small ranks. The resulting raw data might or might not suggest any patterns, so a clever a priori theoretical approach would probably be needed to get new formulas. While the formulation and proof of the Kazhdan-Lusztig conjecture was a major breakthrough in terms of theoretical understanding of the infinite dimensional modules in question, the applications (to unitary representations of Lie groups in particular) continue to be quite indirect.<|endoftext|> TITLE: A learning roadmap to the Schramm-Loewner evolution (SLE) for the complex analyst QUESTION [13 upvotes]: I would like some good references to learn about the Schramm-Loewner evolution (SLE), for a complex analyst with no background in probability. A quick google search gave a lot of references on SLE that are exactly the opposite of what I'm looking for, in the sense that they assume strong background in probability and no knowledge of complex analysis. EDIT (In response to Timothy Chow's comment) : I guess what I'm looking for is a reference that (a) does not assume that the reader is familiar with the probability theory needed for SLE (except for the basics), so it should contain the required material on stochastic calculus, brownian motions, etc. (b) describes in details the classical (non-stochastic) case (c) contains an introduction to the stochastic case, which should focus more on the complex-analytic aspects than the probabilistic ones. Thank you, Malik REPLY [5 votes]: I recall seeing this post half a year ago while I was searching for knowledge myself regarding the very same topic. I thought I would throw my own master's dissertation into the picture. It can be found by clicking here. The aim while writing was to make a rigorous, self-contained and readable introduction to the SLE, for anybody who has obtained knowledge in advanced complex analysis and stochastic calculus. It is my guess that you will never find a rigorous text that does not assume familiarity with at least stochastic calculation on application-level. It treats compact hulls, and their relation to the Loewner equation. The overall method is taken directly from the brilliant notes written by Nathanael Berestycki and James Norris which may be found here. Then I introduce randomness through the Brownian motion, and present the proof of existence of the SLE trace, which was introduced in the article by Schramm & Rohde Annals Math. 161 (2005) 879-920. This may be the only part of my dissertation where the level of detail between my references and my own work is considerably different. Thus this may be the part that can save you some time. I also include proofs of a "weak" phases characterization, which can be found in the Cambridge notes, among others. Finally, I discuss the simulation method presented by Tom Kennedy in his article J.Statist.Phys.128:1125-1137,2007 NB: I should of course stress that this is a master's dissertation and not an article from a journal, which means that it has not been through any approval-procedure apart from grading.<|endoftext|> TITLE: Subgroups of one-relator groups QUESTION [10 upvotes]: I know that not every finitely-presented group may be embedded into a one-relator group, for example because of a theorem of Magnus stating that the word problem is solvable in one-relator groups. But does there is a great amount of finitely-generated groups embeddable into a one-relator group? For instance, is there any (hopefully "large") class of groups known to be embeddable into a one-relator group? REPLY [3 votes]: It is known that a right-angled Artin group $A(\Gamma)$ embeds in some one-relator group if and only if $\Gamma$ is a forest. In fact, any such $A(\Gamma)$ embeds in the one-relator group $\operatorname{Gp}\langle a,t \mid [a, a^t] = 1\rangle$, which contains a copy of $A(P_4)$ (and hence also the group corresponding to any forest by certain embeddability results). This is all proved in Gray, Robert D., Undecidability of the word problem for one-relator inverse monoids via right-angled Artin subgroups of one-relator groups, ZBL07160162. Note that in the torsion case, the set of subgroups possible is significantly reduced, and explicit analysis can be done to a greater extent. For example, every $2$-generated subgroup of a $2$-generated one-relator group with torsion is either free or is itself a one-relator group with torsion. This is all possible by the B. B. Newman Spelling Theorem, which provides a significant degree of control. In particular, this theorem implies that one-relator groups with torsion are hyperbolic, so the only right-angled Artin groups embedding in some one-relator group with torsion are free. For detailed combinatorial analysis, and some more examples in the torsion case, see much of the work by Stephen Pride, e.g. Pride, Stephen J., The two-generator subgroups of one-relator groups with torsion, Trans. Am. Math. Soc. 234, 483-496 (1977). ZBL0366.20022.<|endoftext|> TITLE: Explicit examples of undetermined games QUESTION [10 upvotes]: Suppose we have a game between two players in which they take alternating turns. The game can have finite length, length $\omega$ or any transfinite number of steps (however, I'm not concerning games which are continuous). Every game has a winning condition, which can be interpreted as the set of all game histories possible which result in a win for player 1, and other such set which specifies winning for player 2. Here we are only interested in games which always end (possibly after transfinite time) and which have no ties (if player 1 doesn't win, then player 2 wins). We define a strategy for player 1 as a function from the possible partial (i.e. up to some move of 1) history of the game to an allowed move which 1 can make. We say that strategy for player 1 is winning if it guarantees him a win in the game provided that he follows the strategy. Same for strategy and winning strategy for player 2. The game is determined if one of the players has a winning strategy. Otherwise, the game is undetermined. It's a well-known result that, if we assume axiom of choice, then there is an undetermined game of length $\omega$ with allowed moves being elements of $\Bbb N$. This is necessarilly a non-constructive proof. It's also known that there is an undetermined game in which players play for time $\omega_1$ with moves being bits, or they play countable ordinals for time $\omega$, or they play elements of $\mathcal{P}(\Bbb R)$ for time $\omega$, without assuming AC. However, only proofs of these facts which I know of are by contradiction. A game which is somewhat closer to being constructive example is the following game which uses non-principal ultrafilter on $\Bbb N$: players alternatingly form increasing sequence of numbers $a_1,a_2,...\in\Bbb N$. Then we partition $\Bbb N$ into two sets, $[0,a_1)\cup[a_2,a_3)\cup...$ which belongs to player 1, and $[a_1,a_2)\cup[a_3,a_4)\cup...$ which belongs to player 2. The winner is the player whose set is in the non-principal ultrafilter. Simple strategy-stealing shows that neither player has a winning strategy. All other examples of games which are undetermined are heavily non-constructive, which made me ask the following question: Are there any explicit examples of games which are not determined? With "explicit" I mean that we can provide the example without assuming existence of any sets beyond these which ZF can prove are existing. REPLY [10 votes]: Here is an amusing concrete non-determined game, under the assumption that the dependent choice principle fails. Assume DC fails. This means that there is a set $X$ and a binary relation $R$ on $X$, such for every $x\in X$ there is $y$ with $x\mathrel{R} y$, but there is no sequence $\langle x_n\mid n\in\omega\rangle$ with $x_n\mathrel{R} x_{n+1}$ for all $n$. In other words, the tree of finite sequences that accord with $R$ has no leaves — every node can be extended one more step — but there is no way to iterate these steps and the tree has no infinite branch. Consider the game on $X$ where player I plays a point $a$ from $X$ and player II responds with $b$, and player II wins just in case $a\mathrel{R} b$; otherwise player I wins. So the game is over very quickly, after just one move for each player. (If you insist that games should have infinitely many plays, then you can continue with infinitely many irrelevant moves.) Player I obviously cannot have a winning strategy, since whatever point $a$ is played, there is a way for II to defeat it by playing some $b$ with $a\mathrel{R} b$. But player II also cannot have a winning strategy, since any such strategy would provide a choice function on the $R$-successors of the points in $X$, and any such function could be iterated to thread $R$, whereas there is no such infinite thread. Since the game is an open game (in fact clopen), this argument shows that one cannot prove open determinacy without DC. The usual proof of open determinacy uses DC, when showing that if a position does not have an ordinal rank, then the closed player can avoid losing by maintaining rank. But in order to do so, that player needs DC to find the infinite thread of rank-maintaining moves. Update. As noted in wojowu's comment below, this example can be adapted to prove the following theorem. Theorem. ZF proves that there is a non-determined set. Specifically, in ZF we can prove that $\text{AD}_{P(\mathbb{R})}$ fails. Proof. First, let $F$ be any family of non-empty sets, and consider the game where player I selects some $A\in F$ and player II selects some $a\in A$, and the first player to violate either requirement loses. Clearly, there can be no winning strategy for player I, since if such a strategy directed player I to play a particular $A$ in $F$, then it would be defeated by the strategy for player II to always play some particular element $a\in A$. Conversely, observe that if player II has a winning strategy for the game, then we would have a choice function for $F$. So any violation of AC leads immediately to a non-determined game. In particular, if choice fails for families of nonempty sets of reals, then we get a violation of $\text{AD}_{P(\mathbb{R})}$. On the other hand, if choice holds for such families, then the reals are well-orderable, and we may construct a non-determined set for binary play. So in either case, $\text{AD}_{P(\mathbb{R})}$ fails. QED Regarding the remarks below about constructing an "explicit" game, one can easily do this as follows. Corollary. In ZF we can write down an explicit definition of a non-determined set. Proof. Consider the game where player I chooses either (1) a well-ordering of the reals, and then the rest of the payoff set is as would be constructed for a non-determined set by the usual argument given such a well-ordering; or (2) a family $F$ of nonempty sets of reals having no choice function, with player II then picking an element of the family and player I playing an element of that set, as in the theorem (but with the players reversed). This gives altogether one big payoff set, explicitly defined. Note that player I cannot have a winning strategy, since he cannot have a strategy for the resulting play in case (1), as the game is undetermined, and he cannot have a strategy that wins by placing him in case (2), since any such strategy would have to provide a choice function for $F$. Conversely, player II cannot have a winning strategy, since player I can play either a family with no choice function, in which case it is hopeless for II to play with particular element of the family, or else AC holds for all such families in which case player I can play a well-ordering and a non-determined game, for which player II can have no responding strategy. QED<|endoftext|> TITLE: Intermediate Jacobians of intersections of two quadrics QUESTION [7 upvotes]: Let $$X: \quad Q_1(x)=Q_2(x) = 0 \quad \subset \mathbb{P}^{2n+1},$$ be a smooth complete intersection of two quadrics of odd dimension over a field $k$, not of characteristic $2$. Let $J(X)$ denote the intermediate Jacobian of $X$. When $k$ is algebraically closed, I know that $J(X)$ is isomorphic to the Jacobian of the hyperelliptic curve $$y^2 = \mathrm{det}(xQ_1 + Q_2). \quad (*)$$ Is there a similar description when $k$ is non-algebraically closed? The reason I ask is that when $k$ is non-algebraically closed, it seems quite possible that we may need to take a quadratic twist of $(*)$. However the only "canonical" twists are either $(*)$ itself or the multiplication by $-1$. So my guess is that $J(X)$ should be isomorphic to the Jacobian of $$\pm y^2 = \mathrm{det}(xQ_1 + Q_2),$$ however I don't know which sign to take, and it seems quite possible that the choice of sign could even depend on $n$. Edit: As pointed out by Sasha, I should have explained which definition of the intermediate Jacobian I'm using. Firstly, in Deligne - Les intersections complètes de niveau de Hodge un. Deligne showed how to construct the intermediate Jacobian for complete intersections of Hodge level $1$ over fields of characteristic $0$, using a clever trick coming from certain universal properties (in particular, it is not explicit). In the special case of complete intersections of two quadrics $X$ as above, I believe that one can define $J(X)$ to be the albanese variety of the Fano variety of $(n-1)$-planes inside $X$. This construction should work over fields of characteristic not equal to $2$, and moreover recover Deligne's construction when $\mathrm{char}(k)=0$, though unfortunately I know neither a proof nor a reference for this fact. I would be happy with an answer which uses either definition, restricting to characteristic $0$ if necessary. REPLY [5 votes]: This is not an answer, but this is too long for a comment. The hyperelliptic curve (let me call its $C$) associated to an intersection of quadrics $X$ has an intrinsic meaning as the moduli space of some vector bundles (the restrictions of spinor bundles from the quadrics in the pencil to $X$). Over complex numbers it is a fine moduli space and the universal bundle gives a fully faithful functor $D^b(C) \to D^b(X)$ extending to a semiorthogonal decomposition $$ D^b(X) = \langle D^b(C), O, O(1), \dots, O(2n-3) \rangle. $$ Over other fields the difference is that the moduli space is not fine, the curve $C$ comes with an Azymaya algebra $A$ (given by components of even parts of Clifford algebras associated to quadrics in the pencil) and the decomposition one has is $$ D^b(X) = \langle D^b(C,A), O, O(1), \dots, O(2n-3) \rangle. $$ So, I believe that if one can twist the Jacobian by a Brauer class, then I think it might be the answer. To prove this one could consider the functor $D^b(X) \to D^b(C,A)$ given by the above decomposition and apply it to the universal family of $(n-1)$-planes. This would give a morphism from the Fano variety $F(X)$ of $(n-1)$-planes to some moduli space of objects in $D^b(C,A)$ which probably is the appropriate twisting of $J(C)$. At least over an algebraically closed field of characteristic $0$ this is an isomorphism $F(X) \cong Pic^0(C)$.<|endoftext|> TITLE: Does the limit in the Volume conjecture converge? QUESTION [8 upvotes]: The Volume conjecture says that if $J_n(q)$ are the colored Jones polynomials of a knot $K \subset S^3$, then $$\lim_{N \to \infty} \frac{ 2 \pi}N \left\vert J_N(e^{2\pi i / N})\right\vert = vol(K)$$ Here we are assuming that the complement of $K$ is hyperbolic, and $vol(K)$ is its hyperbolic volume. (The normalization is that $J_N(q) = 1$ for the unknot.) I have an ignorant question about this: Q: Is it obvious that this limit exists? One thing that might be helpful for showing the limit exists is that the $J_n(q)$ satisfy a (generalized) recursion relation (this is a theorem of Garoufalidis and Le). More precisely, there are Laurent polynomials $a_i(-,-)$ such that $$ \sum_i a_i(q,q^N) J_{N+i}(q) = 0$$ (This is true for any $N$, and the $a_i$ don't depend on $N$.) But the existence of this recursion relation was proved several years after the volume conjecture. REPLY [5 votes]: No, this is unknown. There are heuristic arguments for convergence based on the stationary phase approximation, but as far as I know, no one has made the argument precise in general. The closest I know of to a proof of convergence is an upper bound on the limsup in terms of the crossing number given in Theorem 1.3 of this paper by Garoufalidis and Le, although I should add the caveat that I haven't been following the developments on this topic closely recently.<|endoftext|> TITLE: Difference between modularity and clustering in graphs QUESTION [5 upvotes]: The definition of modularity given on wikipedia is as follows. http://en.wikipedia.org/wiki/Modularity_(networks) Modularity is one measure of the structure of networks or graphs. It was designed to measure the strength of division of a network into modules (also called groups, clusters or communities). Networks with high modularity have dense connections between the nodes within modules but sparse connections between nodes in different modules. And clustering coefficient. http://en.wikipedia.org/wiki/Clustering_coefficient In graph theory, a clustering coefficient is a measure of the degree to which nodes in a graph tend to cluster together. Evidence suggests that in most real-world networks, and in particular social networks, nodes tend to create tightly knit groups characterised by a relatively high density of ties; this likelihood tends to be greater than the average probability of a tie randomly established between two nodes (Holland and Leinhardt, 1971; I am facing a doubt here. For a given graph, if the value of modularity is high, shouldn't the value of clustering coefficient be high as well? Is it possible that modularity is extremely high and clustering coefficient is very low? Also, I have certain graphs for which this is happening. Why is it? Modularity is a measure of nodes forming clusters. And clustering coefficient is similar as well. Then, why the difference? Thanks! REPLY [4 votes]: Both concepts are heuristics to measure clustering. One looks at edge densities in given clusters compared to edge densities between clusters, the other looks at density of triangles compared to induced density of 2-paths. Very different concepts, and both may work very well in many applications. But it is very easy to construct graphs with very high modularity and very low clustering coefficient: Just take a number of complete balanced bipartite graphs with no edges between each other, and make each their own cluster. No triangles, so clustering coefficient 0. Very dense (1/2) in each cluster, so very high modularity (1/2 or so if I read the definition correctly). Upshot: if you want to measure sexual clustering in a strictly heterosexual network, better not use the clustering coefficient...<|endoftext|> TITLE: Is Every New Real in the Silver Extension a Silver Generic Real? QUESTION [5 upvotes]: Let $\mathbb{V}$ denote Prikry-Silver forcing. That is, $\mathbb{V}$ is forcing with partial functions $\omega \rightarrow 2$ with coinfinite domain or forcing with uniform trees. Let $\dot x$ denote the canonical name for the unique real given by the generic filter. A real $r \in V[G]$ is a $\mathbb{V}$ generic real if and only if there exists some $H \subseteq \mathbb{V}$ and $H \in V[G]$ such that $V[H] \models r = \dot x_G$. The question is: Is every new real in $V[G]$ a $\mathbb{V}$-generic real. This meaning, if $p \in \mathbb{V}$ and $\tau \in V^{\mathbb{V}}$ and $p \Vdash_{\mathbb{V}} \tau \in {}^\omega 2 \wedge \tau \notin \check V$, then there exists a $q \leq_{\mathbb{V}} p$ such that $q \Vdash_{\mathbb{V}} \tau = \dot x$. It is known that every new real in the Sacks extension, is a Sack generic real. Both Sacks forcing and Prikry-Silver forcing add reals of minimal degree (although Sacks actually is a minimal extension and Silver forcing is not). Does having all new reals being generic follow from adding reals of minimal degree? If not, what is a counterexample of a forcing adding minimal degree such that a new real is not a generic real for that forcing? Thanks. REPLY [9 votes]: No, not every new real in the Silver extension is generic for Silver forcing. To see this, take any new real $x$ in the extension, and form a new real $y$ by doubling every digit of $x$. So if $x=011010\ldots$, then $y=001111001100\ldots$. The real $y$ cannot be Silver generic, since it is dense in Silver forcing to violate the digits-doubling property. Given any condition, look at the first bit not specified, and then fill that bit (and possibly the next) in such a way that the double-digit property is violated.<|endoftext|> TITLE: Reference request for a treatment of Schwinger–Dyson equations QUESTION [10 upvotes]: Is there a treatment of Schwinger–Dyson equations with no mention of Green's functions? Is there perhaps a purely algebraic analog? REPLY [4 votes]: In the formulation of QFT using formal functional integrals, as mentioned by Igor in his answer, the Schwinger-Dyson equation becomes an infinite-dimensional differential equation for the partition function. As such, since the partition function is the generating functional for the Green functions, all you are doing is simply grouping the Schwinger-Dyson equations for all Green functions together in a single formula. I also share Igor's viewpoint that the Schwinger-Dyson equations written in terms of the partition function is more natural, because it makes the structure of the equation much more evident. That being said, there are two nice references that have not been mentioned by the answers given so far. One is the book Path Integral Methods in Quantum Field Theory by R. J. Rivers (Cambridge University Press, 1987), where the Schwinger-Dyson equations are a central concept and are formulated in many different ways. Another, which I strongly recommend, is the paper by M. Dütsch and K. Fredenhagen The Master Ward Identity and Generalized Schwinger-Dyson Equation in Classical Field Theory, Commun. Math. Phys. 243 (2003) 275-314, arXiv:hep-th/0211242. In this second reference, it is shown that the Schwinger-Dyson equation is actually a particular case of the quantum correction (due to perturbative renormalization) to all identities that follow from the classical equations of motion, called the Master Ward Identity (MWI for short). The quantum BV-equation mentioned in Urs's answer is also a particular case of the MWI, which has a nice algebraic interpretation - to wit, one can see the classical limit of the MWI as the simple fact that the (left-hand sides of) all identities that follow from the classical equations of motion are elements of the ideal of off-shell functionals of field configurations (satisfying certain conditions which we do not need to recall here) which is generated by the Euler-Lagrange operator. At the quantum level, not all identities survive due to renormalization - the ensuing violations are known as anomalies. This means that the "classical" ideal of formal power series in $\hbar$ of functionals generated by the Euler-Lagrange operator is no longer an ideal with respect to the quantum product. However, the most general form for the anomalies is dictated precisely by the MWI. In the case of local gauge (or BRS) symmetries, this leads through the quantum BV equation to the Wess-Zumino consistency conditions and so on. Just a final word of warning: it must be said that all of the above can be made rigorous only at the formal perturbative level. In constructive QFT, the Schwinger-Dyson equations are usually not used as a tool to construct a model but are rather derived rigorously as a consequence after the model has been constructed, since only then can they be made sense of. Analysis of QFT models using the Schwinger-Dyson equations is usually deemed as "non-perturbative" for the equations make no explicit mention of perturbation theory, but making sense of this equations outside perturbation theory is a messy business.<|endoftext|> TITLE: Maximum number of general-position points with mutual rational distances? QUESTION [5 upvotes]: Richard Guy has shown that there are six points in the plane—no three collinear, no four cocircular—such that all interpoint distances are rational. Guy, Richard. Unsolved Problems in Number Theory. Vol. 1. Springer, 2004. D20. Six general points at rational distances. p.185ff:           My question is: Q. Is it known that six is the maximum possible? Or have examples been found in the intervening decade that supersede Guy's result? Update. This question is essentially a duplicate of "Integer-distance sets." Apologies. REPLY [6 votes]: There is a heptagon with all integer distances here.<|endoftext|> TITLE: An Intriguing Tapestry: Number triangles, polytopes, Grassmannians, and scattering amplitudes QUESTION [6 upvotes]: What are the roles that the classic number arrays-- Eulerian, Narayana--play in the application of totally non-negative Grassmannians, or amplituhedrons, to string / twistor scattering theory? (This is a severely reduced version of the original post, which can be found at my blog https://tcjpn.wordpress.com/2016/11/20/an-intriguing-tapestry-number-triangles-polytopes-grassmannians-and-scattering-amplitudes/, with numerous references to the combinatorics of the associahedra and permutohedra and relations to scattering theory and Grassmannians.) REPLY [4 votes]: In this recent preprint of Postnikov https://arxiv.org/abs/1806.05307 (which is for an upcoming ICM talk in Rio) he explains a beautiful direct connection between the geometry of the positive Grassmannian and the combinatorics of hypersimplices (polytopes whose normalized volume are the Eulerian numbers).<|endoftext|> TITLE: Is a function of complete statistics again complete? QUESTION [5 upvotes]: suppose $T$ is a complete stats for a parameter $\theta$. Is any function $f(T)$ again complete? It sounds weird but the definition seems to confirm that $f(T)$ is indeed complete.. REPLY [3 votes]: Geometrically, completeness means something like this: if a vector $g(T)$ is orthogonal to the p.d.f. $f_\theta$ of $T$ for each $\theta$, $$\mathbb E_\theta g(T) = \langle g(T),f_\theta\rangle=0$$ then $g(T)=0$ i.e., the functions $f_\theta$ for varying $\theta$ span the whole space of functions of $T$. So in a way it would be more natural to say that $\theta$ is complete for $T$ than what we do say, $T$ is complete for $\theta$. This way it is not so strange that a constant function would be "complete"! Maybe an example helps. Suppose $X$ and $Y$ are independent and identically distributed Bernoulli($\theta$) random variables taking values in $\{0,1\}$, and $Z=X-Y$. Then $Z$ is incomplete for $\theta$, because taking $g=\text{identity}$, $$\mathbb E_\theta(Z)=0$$ for all $0<\theta<1$, but nevertheless $\mathbb P_\theta(Z=0)\ne 1$.<|endoftext|> TITLE: Soft question: beginners reference to moduli spaces QUESTION [17 upvotes]: What is a geometrically intuitive yet reasonably general first introduction to the theory of Moduli spaces? (Possibly introducing stacks also)? I'm looking for something which really gets the pictures across and helps build my beginners intuition. However it should be something that is not a totally trivial read either (i mean i'd like to be able to actually use the results in that paper/ short book also). If it helps: the direction I'm looking into the subject will primarily be from the point of view of analytic varieties and secondarily from the point of view of (schemes, or generalizations thereof (provided there is an initial introductory portion on that generalization)). REPLY [4 votes]: For moduli I would second Geometry of Algebraic Curves, Volume II. For a clear introduction to stacks, I like Martin Olsson's book "Compactifying Moduli Spaces for Abelian Varieties". A good motivation for studying stacks is Mumford's beautiful paper "Picard groups of moduli problems". Also there is a book being written on stacks by several authors (Behrend, Conrad, Edidin, Fulton, Fantechi, Göttsche, Kresch) that is extremely clear: http://www.math.uzh.ch/index.php?pr_vo_det&key1=1287&key2=580&no_cache=1<|endoftext|> TITLE: Are there nontrivial involutions of $S^7\times S^7$ with fixed point set homeo to $S^7$? QUESTION [10 upvotes]: The group $\mathbb{Z}_2$ acts on $S^7\times S^7$ by switching the coordinates with fixed point set $\Delta(S^7\times S^7)\cong S^7$. I want to know whether there are some other $\mathbb{Z}_2$ actions on $S^7\times S^7$ with fixed point set homeomorphic to $S^7$. We regard two actions as the same if they are conjugate in $\text{Aut}(S^7\times S^7)$. I guess this question could be discussed in different categories, say TOP or SMOOTH, so i put $\text{Aut}$ instead of $\text{Homeo}$ or $\text{Diff}$. The space $S^n×S^n$ is topologically rigid for odd $n$ (i.e. every homotopy equivalence from $M$ to $S^n×S^n$ is homotopic to a homeomorphism). You may replace $7$ by any other odd number. That would still be interesting to me. REPLY [7 votes]: First, a warning. This kind of question, classifying structures up to the automorphisms of a manifold, is usually quite difficult. It is easier to classify them up to the connected component of the automorphism group, that is, up to isotopy. It is possible to classify involutions of $S^n\times S^n$ whose fixed points are the diagonal, up to conjugacy by automorphisms in the connected component of the identity. There are finitely many equivalence classes. If $n$ is sufficiently large (7 is probably adequate), there are multiple classes. A strategy for classifying actions by $F$ on $M$ that are free away from fixed set $M^F=X$: (1) Classify homotopy classes of maps $X\to M$; immersions up to isotopy; embeddings up to isotopy. (2) Classify appropriate actions on the normal bundle. (3) Remove a tubular neighborhood of $X$ to produce a manifold with boundary, where the putative action is already known on the boundary. Classify (free) homotopy actions on the space, compatible with the boundary, ie, free $F$-spaces that are homotopy equivalent, but may not be manifolds. (4) Apply surgery theory to the quotient to produce a manifold with fundamental group $F$ and boundary the quotient of the sphere bundle with the prescribed action. Taking the universal cover gives a manifold with boundary with a free $F$-action that can be glued to $X$ to produce an action of $F$ on a manifold homotopy equivalent to $M$ with prescribed fixed set. Surgery can be applied again to determine whether it is $M$. All of these steps are individually computable, at least if both $M$ and the complement of $X$ are simply connected, so it's useful for $X$ to be codimension at least 3. The surgery step requires dimension at least 5. However, a step will often produce infinitely many possibilities; for any particular one, the next step is computable, but there is no guarantee that the infinite family has a nice description that can be dealt with simultaneously. For example, there are infinitely many homotopy classes of maps from $S^7$ to $S^7\times S^7$, each with their own embeddings. For the smooth case, I believe that this procedure classifies actions up to conjugacy by the connected component of the diffeomorphism group. In the PL case, I am not sure how to do the bundle step. In the topological category, the fixed set can have a wild embedding and the procedure breaks down. But if you restrict to group actions with tamely embedded fixed point set, it is no worse than the PL case. Let us pursue step 4. This makes sense in any category. For sufficiently large $n$, there are smooth involutions with fixed set the diagonal that are not conjugate to the usual by a diffeomorphism in the connected component of the identity, or even by a homeomorphism in the connected component of the identity. They are detected by deleting a tubular neighborhood of the diagonal to get a manifold with boundary and a free involution. The diffeomorphism or homeomorphism type of the quotient is an invariant of the involution. Recall the warning from the beginning: such classification is difficult. What is accessible to surgery theory is automorphisms within a fixed class, marked by its cover being homotopy equivalent to a subset of $M$, and other data from earlier steps. Classification up to this marking can be computed explicitly, but I will not do the calculations. I will wave my hands and assert that a group is large, hence nontrivial. It grows with $n$, but is probably nontrivial already at $n=7$. Surgery theory says that the choice of manifold structures on a Poincaré duality space $Y$ is first approximated by a choice of a stable tangent bundle. More precisely, it a reduction of the structure group of the homotopy-invariant Spivak normal fibration (with structure "group" $G$) to the structure group $CAT$ for bundles for the appropriate category. If such a reduction exists (eg, if $Y$ is already a manifold), the choice of reduction is a torsor for the space of maps into the coset space $[Y,G/CAT]$. If $Y$ is already a manifold with boundary and we do not want to change the manifold structure on the boundary, the new bundle must restrict to the boundary to be the original, so we consider maps $[Y/\partial,G/CAT]$. The deviation of this first approximation from the answer is controlled by $L$-theory, which depends only on the dimension modulo 4 and the fundamental group. This is a bounded deviation, while $[Y,G/CAT]$ can become arbitrarily large for complicated $Y$. In our example, $[Y,G/CAT]$ becomes large as the dimension increases. We apply surgery theory to two manifolds. One is $Y_1$, the complement of a tubular neighborhood of the diagonal, a deformation retract of the complement of the diagonal. It is homotopy equivalent to $S^n$ and has boundary the bundle of unit tangent vectors of $S^n$. The other is $Y_2$, the quotient by the involution, with boundary a $\mathbb RP^{n-1}$ bundle over $S^n$. Surgery yields manifolds homotopy equivalent to $Y_2$, with the same boundary. Taking the double cover and gluing back in the tubular neighborhood yields an involution on a manifold homotopy equivalent to $S^n\times S^n$. If $n$ is odd and we are in the PL or topological categories, there is only one such manifold, as you said, so this yields exotic involutions on $S^n\times S^n$. So exotic involutions in the PL or topological case are close to $[Y_2/\partial,G/CAT]$. Maps into $G/CAT$ are a generalized cohomology theory, computable from ordinary cohomology by the Atiyah-Hirzebruch spectral sequence. $Y_1$ is homotopy equivalent to $S^n$ and $Y_2$ is homotopy equivalent to $\mathbb RP^n$. So the larger the dimension, the more total cohomology $Y_2$ has, and the larger $[Y_2/\partial,G/CAT]$, and eventually it overwhelms the bounded obstructions and exotic examples appear. In a little more detail: for 2-torsion, $G/TOP$ is basically just ordinary cohomology in specific dimensions, with period 4. So it does detect much of the cohomology of $\mathbb RP^n$. (Actually, we should work not with $Y_2$, but with $Y_2/\partial$, but by Poincaré duality, its cohomology is close to that of $Y_2$.) If we want to deal with the smooth category or even $n$, we must only use exotic $Y_2$ whose double cover is standard as a $Y_1$. That is, we must restrict to the kernel $[Y_2/\partial,G/CAT]\to [Y_1/\partial,G/CAT]$. (Also, we must worry about the bounded indeterminacy coming from $L$-theory, but that is zero in our case of simply connected even dimensional manifolds.) $Y_1$ is homotopy equivalent to $S^n$, so it does not become more complicated as $n$ increases. In particular, $[Y_1/\partial,G/TOP]$ is just two homotopy groups of $G/TOP$, which are periodic and do not become more complicated with $n$. Whereas, $[Y_2/\partial,G/TOP]$ has increasing amounts of 2-torsion in the kernel. In the smooth category, the homotopy groups of $G/O$ reflect the homotopy groups of spheres and do become more complicated with $n$, but this affects both groups and is independent of $Y_2$ becoming complicated. The computation of the kernel is more complicated, but it grows and thus there are exotic smooth involutions. Indeed, that the group is large because $Y_2$ is complicated means that there are exotic smooth involutions represented by elements of $[Y_2/\partial,G/O]$ that are still nontrivial in $[Y_2/\partial,G/TOP]$, and thus remain exotic as continuous involutions. These exotic involutions are not isotopic to the original, that is, not conjugate by automorphsims in the connected component of the identity, but it is not immediately clear that they are exotic in the way you asked, that they are not conjugate by arbitrary automorphisms. The invariant used to distinguish them is the tangent bundle, but this is relative to a homotopy equivalence. There might be a homotopy self-equivalence of $Y_2$ that switches the two tangent bundles; and such a homotopy equivalence might be induced by an automorphism of $M$. Ruling out this possibility requires understanding the group of homotopy equivalences of $Y_2$ and its action on the set of bundles. It is probably tractable in this case.<|endoftext|> TITLE: Iwasawa's mu-invariant for noncyclotomic $\mathbf{Z}_p$ extensions of cyclotomic fields? QUESTION [6 upvotes]: Let $p$ be an odd prime number, $m$ a positive integer with $p\mid m$. Put $k=\mathbf{Q}(\mu_m)$. (1) Is there any example where certain noncyclotomic $\mathbf{Z}_p$-extension $k_\infty/k$ has positive $\mu$-invariant? (2) Let $n$ be an integer such that $m\mid n$ and $p$ does not divide $n/m$. Put $k'=\mathbf{Q}(\mu_n)$, $k'_\infty=k_\infty\cdot k'$. It is well-known that $\mu(k_\infty/k)\le \mu(k'_\infty/k')$. I wonder is it possible that the two sides differ by exactly 1. Here, for this part, I allow $\mu(k_\infty/k)=0$. I know that Iwasawa and Ozaki has constructed examples from cyclotomic fields which have positive $\mu$-invariant. But their examples all have degree divisible by $p$ over the base field. Their method does not apply to the above question. REPLY [2 votes]: I think I have a positive answer for (1). Let $p > 3$ be a prime, let $\mathbb{K}_1 = \mathbb{Q}[ \zeta_p ]$ be the $p$-cycotomic extension and let $q = 1 \bmod p$ be some rational prime. Let $\mathbb{F} \subset \mathbb{Q}[ \zeta_q ]$ be the subfield of degree $p$ over $\mathbb{Q}$. Obviously, $\mathbb{K}_1$ has $r_2+1 = (p+1)/2$ independent $\mathbb{Z}_p$-extensions. Let $\mathbb{M}$ be the composite of all $\mathbb{Z}_p$-extensions of $\mathbb{K}_1$ and $D(q) \subset X = Gal(\mathbb{M}/\mathbb{K}_1)$ be the decomposition group of some prime of $\mathbb{K}_1$ above $q$ - it is isomorphic to $\mathbb{Z}_p$. So we can choose some $\mathbb{Z}_p$ - extension of $\mathbb{K}_1$ which is fixed by $D(q)$; let this be $\mathbb{L}$. Then $q$ is totally split in $\mathbb{L}/\mathbb{Q}$. Let now $\mathbb{K}' = \mathbb{K}_1 . \mathbb{F}$ and let $\mathbb{L}' = \mathbb{L} . \mathbb{F}$. It is an exercise in the application of the ambig ideal lemma of Chevalley, to adapt Iwasawa's seminal proof to the present situation and show that $\mu(\mathbb{L}') > 0$. Of course, $\mathbb{K}'$ is a (quite simple) cyclotomic field, which answers your question in the affirmative. For extensions $\mathbb{K}'/\mathbb{K}$ of degree not divisible by $p$, the method of Iwasawa fails. So the question (2) appears to be difficult ...<|endoftext|> TITLE: Classifying space for fibrations with Eilenberg-MacLane space as fibers QUESTION [5 upvotes]: The following result seems to be frequently quoted: Consider the fibration $K(\pi,n)=\Omega K(\pi,n+1)\to PK(\pi,n+1)\to K(\pi,n+1)$. Let $B$ be any topological space (which is not too pathologic). Then the fiberations on $B$ with fiber $K(\pi,n)$ are classified by the space $K(\pi, n+1)$. In other words, the pull-back map $$ [B,K(\pi,n+1)]\to \{\textrm{fibrations on } B \textrm{ with }K(\pi,n)\textrm{ as fibers, up to fiber homotopy}\}$$ is a bijection. Does anyone know a proof of this result, or a reference? Thanks a lot. REPLY [9 votes]: There is a very careful analysis of this question in Lemma 3.4.2, page 57, of More Concise Algebraic Topology, by Kate Ponto and myself. Assuming that $E$ and $B$ are connected, a fibration $E\longrightarrow B$ with fiber $K(A,n)$ for an abelian group $A$ is a pullback of of the path space fibration over $K(A,n+1)$ if and only if $\pi_1(B)$ acts trivially on $A$.<|endoftext|> TITLE: Multivariate polynomial interpolations QUESTION [6 upvotes]: I have a multi-variate, continuous function from $R^n$ to $R$, which I can query for its output for any input. I would like to create an interpolation of that function by sampling a subset of the points and create a polynomial representation. I've read literature on univariate function interpolation and it seems using the Chebyshev nodes: http://en.wikipedia.org/wiki/Chebyshev_nodes will do the job, as it converges to the true function as the number of interpolant points increases. I would like an analogous result for multi-variate functions, where one can choose the samples of the function at the right locations and just interpolate, and the result converges to the function as we increase the number of samples. However I am not able to find any literature on that, is it because it is impossible to extend the result to a multi-variate case? I heard about "chebyshev grids" which is a higher dimensional chebyshev nodes, but does interpolation on the grids retains the convergence property that as the number of samples increase? REPLY [3 votes]: The answer to your question highly depends on the domain on which you plan to approximate your function. For products of compact intervals, e.g. rectangles in the plane, some kind of tensor product construction can be used. You should take a look at [1], which describes how bivariate functions are handled in Chebfun [2], a Matlab package for representing and computing with univariate and bivariate functions. There are a few references in the aforementioned paper about convergence results. For more complicated domains, I would take a look at [3]. [1] A. Townsend & L. N. Trefethen, An extension of Chebfun to two dimensions, SISC, 35 (2013), pp. C495-C518, https://people.maths.ox.ac.uk/trefethen/cheb2paper.pdf. [2] Chebfun, http:// www.chebfun.org/. [3] B. N. Ryland & H.Z. Munthe-Kaas, On Multivariate Chebyshev Polynomials and Spectral Approximation on Triangles, http://www.ii.uib.no/~hans/Chebyshev/Welcome_files/munthekaas09mcp.pdf.<|endoftext|> TITLE: Generalized Cauchy-Binet sum over a fixed subset of indices QUESTION [7 upvotes]: I originally posted this on math.stackexchange, but it quickly got buried. I removed it not too long after, thinking of rewriting it for MO, but I didn’t have a chance to post it until now. Apologies if you were one of the lucky 24 people who already saw it there. Background The Cauchy-Binet formula states that $$ \det(AB) = \sum_{S\in\tbinom{[n]}m} \det(A_{[m],S})\det(B_{S,[m]}), $$ where $A$ is an $m$ by $n$ matrix, $B$ is an $n$ by $m$ matrix, $[n]$ is notation for the set $\{1,\dots,n\}$ (similarly for $[m]$), $\binom{[n]}{m}$ is the set of all $m$-element subsets of $[n]$, and if $M$ is a $k$ by $l$ matrix and $J,K$ are subsets of $[k],[l]$, respectively, then $M_{J,K}$ denotes the submatrix of $M$ consisting of rows indexed by $J$ and columns indexed by $K$. Question Is there a version of the Cauchy-Binet formula to evaluate the following sum for general integers $0\leq j\leq m$? $$ ? = \sum_{S\in\tbinom{[n-j]}{m-j}} \det(A_{[m],S\cup T})\det(B_{S\cup T,[m]})\quad\quad (*)$$ Here $T=\{n-j+1,\dots,n\}$ so that the columns labeled by $T$ are forced to appear in the submatrices of $A$ included in the sum, and similarly for rows $n-j+1,\dots,n$ in the submatrices of $B$. When $A^T=B$ is the incidence matrix for a graph $G$, then Kirchhoff's matrix-tree theorem and effective resistance formula imply that I can compute this sum as the determinant of the Laplacian matrix of the graph $G'$, where $G'$ is $G$ where all the edges indexed by $n-j+1,\dots,n$ have been contracted. The motivation of this question is thus to try to understand contraction in a slightly more general setting. This paper by Konstantopoulos gives a nice coordinate-free version of Cauchy-Binet, but I couldn't wrangle it into what I wanted. It may also be possible that evaluating this is equivalent to evaluating the permanent or something so I shouldn’t expect anything nice after all. But I would be quite disappointed if that were so. Other special cases When $j=0$, $(*)$ reduces to ordinary Cauchy-Binet, and when $j=m$ it is of course just $\det(A_{[m],T}B_{T,[m]})$. When $j=1$, $(*)$ is equal to $\det(AB)-\det(A_{[m],[n-1]}B_{[n-1],[m]})$, and by using inclusion-exclusion I can extend this to all the other $j$. However, for large $j$, there will be a mess of terms, and it will be a rather inefficient way of evaluating this sum. Is there a simpler formula just involving one determinant? Update (13 Oct 2014) While I quite like the answer of Igor Khavkine, I would still like to know if there are any other quick ways to evaluate this. Is there any faster way? If the polynomial trick is the best, then are there some efficient ways to extract the leading coefficient of a polynomial? Interpolation can get me all $j+1$ coefficients, and thus requires $j+1$ evaluations of the determinant. It seems to me that if I have a bound on the other coefficients, I could just evaluate the determinant once with a large value of $x$ then divide by $x^j$ and then round off everything else. REPLY [6 votes]: This is probably a bit late, but the following formula is proved in this paper (Eq. S4 in the Supplementary Material): \begin{equation} \sum_{S \in \binom{[n - j]}{m - j}} \det(A_{[m], S \cup T}) \det(B_{S\cup T, [m]}) = (-1)^{j} \det \left(\begin{array}{c c} \mathbf{0}_{j \times j} & B_{T, [m]} \\ A_{[m], T} & A_{[m], [n - j]}B_{[n - j], [m]} \end{array} \right) \rm{,} \end{equation} where $0_{j \times j}$ is the $j \times j$ matrix of all zeroes. I'll leave the detailed proof to the paper, but the essential idea is to apply the Laplace expansion by complementary minors formula to the columns $T$ of $A$ and rows $T$ of $B$, rearrange sums to apply the ordinary Cauchy-Binet formula over $S$, and then recognize what remains as another expansion by complimentary minors of the above matrix on the r.h.s. The $j \times j$ block of zeros is to ensure that the minors you don't want to count end up vanishing, much along the same lines as Adam Przeździecki's answer. Notice that the special case for $j = 0$ is the ordinary Cauchy-Binet formula.<|endoftext|> TITLE: The "Rolle theorem" for sections of a vector bundle QUESTION [5 upvotes]: 1) Assume that $E\to M$ is a smooth real vector bundle and $\nabla$ is a connection. (We do not assume any metric compatibility since we do not fix a metric on $E$). Assume that for every vector field $X$ on $M$ with a periodic orbit $\gamma$ and for every $s\in \Gamma^{\infty} (E)$, the section $\nabla_X s$ vanishes on at least one point of $\gamma$. Does this implies that $E$ is a (trivial) line bundle? The motivation: The Rolle theorem is not valid in dimension greater than one. 2) Let $\ell$ be the canonical line bundle on $\mathbb{C}P^{1}\simeq S^2$, thought of as a smooth complex line bundle. Is there a connection $\nabla$ such that for every section $s$ and every periodic orbit $\gamma$ of $X$, the section $\nabla_{X}^{s}$ vanishes on at least one point of $\gamma$? I ask the above question because I searched for some unusual differential operator associated with a vector field such that these operators can count the number of attractors of a vector field $X$. As a related post, please see: Elliptic operators corresponds to non vanishing vector fields REPLY [2 votes]: As a first observation, if $(E,\nabla)$ has this Rolle property then if $i: S^1 \to M$ is any embbeding, $(i^*E,i^*\nabla)$ has this property as well (this is just a tubular neighborhood + cutoff argument). In particular we need to understand what possible such vector bundles with connection appear over $S^1$. It is fairly clear that the only possibilities are line bundles. If $i^*E$ is the Mobius bundle, then any connection has this property. If it is the trivial bundle, then the connection must have trivial holonomy (in other words there must be a trivialization for which the connection $i^*\nabla$ is the canonical one associated to that trivialization). In particular the connection must be flat. For $\dim M > 2$ this classifies all such bundles rather satisfactorily: they are precisely the line bundles with a metric, equipped with the unique connection that preserves this metric. This is because any embedded circle in $\dim \geq 3$ admits a family of embeddings $i_n:S^1 \to M$ converging to the immersion of multiplicity two $z \mapsto i(z^2)$. The holonomy along these converges to the holonomy along the multiplicity two embedding, so the holonomy along $i$ must be $-1$ (a priori it could have been any negative scaling). In dimension two, the above argument breaks down, and indeed there are line bundles on $S^1 \times \mathbb{R}$ that have holonomy a negative number $\neq 1$. However, there is clearly some reasonable casework that could be done: on $\mathbb R P^2$ all line bundles satisfying the condition are metric compatible: all homotopically nontrivial embedded circles have nontrivial normal bundle, so we may find embeddings $i_n$ converging to $i$ with multiplicity two. In the torus, we simply see that there are no holonomy maps $\mathbb{Z}^2 \to \mathbb{R}^{\times}$ which take on values besides $\{ \pm 1\}$ but also take on negative values or $1$ at every class representable by an embedding (namely if some class takes on a value beside $\pm 1$, some class representing an embedding takes on a positive value not equal to $1$), so again all line bundles satisfying our condition are metric compatible. (this section was edited)<|endoftext|> TITLE: How to prove an elementary functional equation for polylogarithms? QUESTION [6 upvotes]: Let $Li_s(z)$ denote the usual polylogarithm. The elementary functional equation $$Li_{-n}(z)=(-1)^{n-1}Li_{-n}(1/z)$$ holds for $n\geq 1$. I remember only that the proof used some reproducing property of the Stirling numbers of the second kind. This functional equation is rather useful because, taking linear combinations, it amounts to saying that a meromorphic function having a power series on the open unit disc whose coefficients are the values of a polynomial $f$ of degree $d$ evaluated at integer arguments: $$\sum_{k=0}^{\infty}f(k)z^k=\frac{1}{(1-z)^{d+1}}\sum_{k=0}^{d}\left(\sum_{j=0}^k{d+1\choose j}(-1)^jf(k-j)\right)z^k$$ is defined for $|z|>1$ by the series $$-\sum_{k=1}^{\infty}f(-k)z^{-k}.$$ I would like to know how this functional equation (or the continuation by sum over negative powers) can be proved? Please note that the use of statements like $$\sum_{k=-\infty}^{\infty}k^nz^k=\left(z\frac{d}{dz}\right)^n\sum_{k=-\infty}^{\infty}z^k=\left(z\frac{d}{dz}\right)^n 0=0$$ is acceptable only if you can also prove that the sums of positive and negative powers converge on a common arc of the unit circle. In his answer below, Fedor Petrov has pointed out that the use of divergent series as formal generating functions is justified in some cases, so the last statement is incorrect. REPLY [13 votes]: Let's try to understand in which sense this equality $$\sum_{k=-\infty}^{\infty}k^nz^k=\left(z\frac{d}{dz}\right)^n\sum_{k=-\infty}^{\infty}z^k=\left(z\frac{d}{dz}\right)^n 0=0$$ may be understood. Of course, there is no convergence in usual sense on the unit circle (because terms do not tend to 0). Actually, we do not use analysis at all, as it is purely combinatorial-algebraic statement. For two-sided Laurent series like $\sum_{k=-\infty}^{\infty} a_k z^k$ there are following well-defined operations: taking derivative and multiplying by polynomials, and we have $(fg)'=f'g+fg'$ for polynomial $f$ and two-sided series $g$. For $H(z)=\sum_{k=-\infty}^{\infty} z^k$ we have $(z-1)H(z)=0$. Hence $(z-1)H'(z)+H(z)=0$, multiplying by $(z-1)$ we get $(z-1)^2H'(z)=0$. Take a derivative again to get $(z-1)^3H''(z)=0$ and so on. This implies that $$0=(z-1)^{n+1}\left(\left(z\frac{d}{dz}\right)^n H(z)\right)=(z-1)^{n+1}(Li_{-n}(z)+(-1)^nLi_{-n}(1/z))$$ But we know that both expressions $(z-1)^{n+1}Li_{-n}(z)$ and $(z-1)^{n+1}Li_{-n}(1/z)$ are actually polynomials in $z$. What we have proved is that the sum of those two polynomials does vanish, as desired.<|endoftext|> TITLE: If all balls at $x$ and $y$ are isometric is there an isometry sending $x$ to $y$? QUESTION [20 upvotes]: Let $(X,d)$ be a metric space and $x,y \in X$. Assume that for all $r > 0$ the balls $B_r(x)$ and $B_r(y)$ are isometric. Is it true that there exists an isometry of $X$ sending $x$ to $y$? REPLY [20 votes]: There is a 5-point example $\ X := \{x\ y\ a\ b\ c\},\ $ with (symmetric) metrics $\ d\ $ as follows: $$d(x\ y) = d(a\ b) = 1$$ $$d(x\ a) = d(y\ b) = 2$$ $$d(x\ b) = d(y\ a) = 3$$ $$d(x\ c) = d(y\ c) = 6$$ $$d(a\ c) = 5\qquad\qquad d(b\ c) = 4$$ Oviously, $\ B_r(x)\ $ and $\ B_r(y)\ $ are isometric for every $\ r>0,\ $ while there is no isometry $\ f:X\rightarrow X\ $ for which $\ f(x)=y$. REMARK 1  This example is ironic because while the respective balls are isometric, the isometry of the balls doesn't respect the centers. In this sense every bounded (and especially--finite) required example would be ironic. REMARK 2  Number $\ 5\ $ is minimal.<|endoftext|> TITLE: The distribution of the shortest path through $n$ points QUESTION [9 upvotes]: In the big picture, I'd like to know: if I sample $n$ points uniformly at random in the unit square, what is the probability that the shortest path that visits each one of them is very small? More precisely: if I sample $n$ points uniformly at random in the unit square, then it is known that as $n$ becomes large, the shortest path $L_n$ through these points satisfies $$L_n/\sqrt{n} \to \beta \approx 0.71~,$$ with probability one, where $\beta$ is the "Euclidean TSP constant". My question is: for a given (small) length $\ell$, say $\ell = c\sqrt{n}$ for small $c<\beta$ , does a non-trivial lower bound for $\Pr(L_n \leq \ell)$ exist? REPLY [2 votes]: The distribution of the shortest path through $n$ points The following 4 diagrams plot the empirical distribution of the shortest path through $n$ points, when $n = 3, 20, 50$ and 100 (each simulated 100,000 times). The vertical red line denotes $0.71 \sqrt{n}$ on each plot. $n = 3$: (source) $n = 20$: (source) $n = 50$: (source) $n = 100$: (source) Even when $n = 100$, the stated asymptote of $0.71 \sqrt{n}$ leaves a substantial component of the distribution in the left tail, to the left of the asymptote. Depending on whether you are interested in large or small values of $n$, an exercise such as the above will provide a simple way to select your desired $c$, to minimise any left-tail probability. It is also apparent that using the distribution (or simulated distribution) is more helpful than just looking at expected values (or asymptotic expected values, or bounds on expected values of shortest paths), which is what most of the literature seems to do. For instance, the following diagram compares: Marks (1948) lower bound for the expected shortest path: $\sqrt{\frac{1}{2}} \left(\sqrt{n}-\frac{1}{\sqrt{n}}\right)$ Mahalanobis estimate of the expected shortest path: $\sqrt{n}-\frac{1}{\sqrt{n}}$ The actual expected shortest path [ round dots ] The OP's stated asymptote: $.71 \sqrt{n}$ (source) ... but, for your problem, you really should be looking at the distribution ... not just the first moment.<|endoftext|> TITLE: If all balls around two points are isometric... -- manifold version QUESTION [16 upvotes]: This question is a natural follow-up of this other question, asked earlier today by wspin. Let's say that a metric space $(X,d)$ has two poles if: there are two distinct points $x$, $y$ such that for every $r\ge 0$ the open balls $B_r(x)$ and $B_r(y)$ are isometric; there is no isometry $f$ of $(X,d)$ such that $f(x)=y$. Bjørn Kjos-Hanssen gave a very pretty example of a space with two poles (it's the metric space associated to an infinite graph). I've tried to turn this example into a surface by "inflating" it (imagine it in the 3-space, and take the boundary of a regular neighourhood), but failed. Hence my question: Are there connected Riemannian manifolds with two poles? If one drops the connectedness assumption, then Włodzimierz Holsztyński's answer to the question linked above is a 0-dimensional example. One can also ask for more: compactness, trivial $\pi_1$, curvature restrictions, etc., but I'll stick with the broader question for now. The rough meta-question in the back of my mind is: "How nice can a bipolar space be?", so answers/comments in this direction are very welcome. REPLY [12 votes]: The answer is no if $M$ is assumed to be complete: Let $x, y \in M$ such that for all $r > 0$ the balls $B_r(x)$ and $B_r(y)$ are isometric via $$f_r : B_r(x) \to B_r(y).$$ Set $R := \inf \{ r \mid B_r(x) = M\}$ (here $R = \infty$ if $M$ is unbounded). First I claim that For all $0 < r < R$ we have $f_r(x) = y$: Assume on the contrary that $f_r(x) \neq y$. Let $z \in M \setminus \overline B_r(y)$ and $\gamma_1 : [0,l] \to M$ be a minimal arc length geodesic from $y$ to $z$. Then $\gamma_1(r) \in \partial B_r(y)$. Also let $\gamma_2$ be a minimal geodesic from $f_r(x)$ to $\gamma_1(r)$. Since $f_r^{-1}(\gamma_1(r)) \in \partial B_r(x)$ (here we extend $f$ to $\overline B_r(x)$ by metric completion), $f_r^{-1}(f_r(x)) = x$, $\gamma_2$ is minimal and $f_r$ is an isometry it follows that $\gamma_2$ has length $r$. But then also the curve obtained by concatenating $\gamma_2$ and ${\gamma_1}_{\vert [r,l]}$ is minimal between $f_r(x)$ and $z$. This is a contradiction, since geodesics do not branch. Now, consider a sequence $f_{r_n}$ of such isometries with $f_{r_n}(x) = y$ and ${r_n} \to R$ for $n \to \infty$. After passing to a subsequence we may assume that $f_{r_n}$ converges to an isometry $f : M \to M$ with $f(x) = y$ (the isometries are determined by their differentials at $x$, which converge after some subsequence by Ascoli-Arzelà (or simply by linearity)). Two remarks: This argument holds more generally for complete length spaces with curvature bounded below. I am not sure whether the completeness assumption is necessary.<|endoftext|> TITLE: On the cohomology ring of the Hilbert scheme of points on k3 or abelian surfaces QUESTION [5 upvotes]: There are many results on the cohomology of the Hilbert scheme of points of a surface. Gottsche calcaluted the Betti numbers and Nakajima got the generators of the cohomology. Also there are results on the ring structure of the cohomology. In Manfred Lehn and Christoph Sorger's paper "The cup product of the Hilbert scheme for K3 surfaces", they have an explicit description of the ring structure when the surface has trivial canonical bundle. I am trying to calculate the cup product of some particular cohomology classes in the cohomology ring. Even though there is the explicit description by Lehn and Sorger, I found that it's very hard to do this by hand. So I think I should use Mathematica programming to do this. I am just wondering whether there is someone else who has already written the code for this ring structure so that I don't need to do reprogramming. REPLY [2 votes]: I did do this recently. Have a look here: http://arxiv.org/abs/1410.8398 and here for the source code: https://github.com/s--kapfer/HilbK3<|endoftext|> TITLE: Central limit theorem via maximal entropy QUESTION [35 upvotes]: Let $\rho(x)$ be a probability density function on $\mathbb{R}$ with prescribed variance $\sigma^2$, so that: $$\int_\mathbb{R} \rho(x)\, dx = 1$$ and $$\int_\mathbb{R} x^2 \rho(x), dx = \sigma^2$$ Fact: the density function which maximizes the entropy functional $$S(\rho) = -\int_\mathbb{R} \rho(x) \log \rho(x)\, dx$$ with the constraints above is the normal distribution $$\rho(x) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{x^2}{2\sigma^2}}$$ This can be proved using basic techniques from the calculus of variations. My question: can this be used to prove the central limit theorem? In other words, can one show directly that the limiting distribution of the average of a sequence of i.i.d. random variables maximizes entropy? Actually, I don't care too much about entropy. I'm mainly interested in the possibility of a variational proof of the central limit theorem. REPLY [16 votes]: There is a book on the subject: "Information Theory and The Central Limit Theorem" by Oliver Johnson. The article by Anshelevich mentioned by Yemon considers the operator $T$ acting on probability densities and corresponding to going from the law of a random variable $X$ to that of $(X+Y)/\sqrt{2}$ where $Y$ is an independent copy of $X$. The entropy is a Lyapunov function for this transformation which is the simplest example of a renormalization group transformation. The $N(0,1)$ is a fixed point and it is easy to diagonalize the linearization of $T$ near this fixed point using Wick monomials, i.e., Hermite polynomials. The directions corresponding to 0-th, 1-st and 2-nd moments are expanding (relevant operators) or neutral (marginal operators) while all others are contracting (irrelevant operators). Therefore if one makes the necessary arrangements (renormalization conditions) to fix these moments (e.g. subtracting $N$ times the mean and dividing by $\sqrt{N}$) then one lies on the stable manifold of the Gaussian fixed point. See the textbook on probability theory by Koralov and Sinai for more details. The generalization of the $T$ map for joint probability distributions of dependent variables, i.e., the renormalization group is explained in the book "A Renormalization Group Analysis of the Hierarchical Model in Statistical Mechanics" by Collet and Eckmann. The issue with using this type of nonlinear transformations is that the above diagonalization at a fixed point only gives information about the vicinity of that fixed point. To get results far away, having a Lyapunov function like the entropy is of great importance. This is an active area in physics which investigates generalizations of Zamolodchikov's $c$-"theorem" in conformal field theory. See for instance this article for a recent review. Entanglement entropy seems to be the Lyapunov function in this setting.<|endoftext|> TITLE: Variation of curvature with respect to immersion? QUESTION [7 upvotes]: Let $M$ be a smooth surface and let $f: M \to \mathbb{R}^3$ be a family of immersions given by $$ f(t) = f_0 + tuN_0, $$ where $f_0$ is some initial immersion, $N_0$ is the associated Gauss map, and $u: M \to \mathbb{R}$ is an arbitrary smooth function. One can easily work out, e.g., the first variation in the induced area form, but I am having a surprising amount of difficulty deriving (or tracking down known expressions for) variations of higher-order quantities. Question: What are the first variations of the mean curvature $H$ and the Gauss curvature $K$ with respect to the immersion $f$? Bonus question: What are the first variations of the principal curvatures? If these results are not immediately available, I am also appreciative of strategies and techniques for obtaining them. Finally, my apologies if these questions have well-established answers - I am having no luck finding them in the literature! REPLY [3 votes]: Have a look at this paper. Chapter 4 contains all that you are asking for, in a more general setting.<|endoftext|> TITLE: Characterization of closed convex set QUESTION [7 upvotes]: Suppose that $C$ is a bounded convex subset of $\mathbb{R}^n$ such that the optimum value, over $C$, of any linear functional is attained at some point of $C$. Does this imply that $C$ is closed? If so, is there a simple proof? If not, a counterexample would be nice. REPLY [14 votes]: No, as a counterexample take a closed "stadium" in $\mathbf{R}^2$ (the convex hull of two half-circles, I hope the word stadium makes it clear) and remove from it one of the endpoints of the half-circles. This is somehow related to the difference between extreme points and exposed points.<|endoftext|> TITLE: Homeo-Fixed point property QUESTION [12 upvotes]: Edit: According to comment of Michał Kukieła I revised the question A topological space $X$ satisfies "Homeo-fixed point" property if every homeomorphism $f$ on $X$ possess a fixed point. Is there an example of a connected manifold with this property but does not satisfies fixed point property? Edit Now i found a related paper, so I add the link to this question: http://link.springer.com/article/10.1007%2FBF02771655 REPLY [2 votes]: Let me construct a non-paracompact counterexample. Of course, a paracompact counterexample is better in many respects since differential topology is usually restricted to paracompact manifolds and because the manifolds that naturally occur and the ones sold here are all paracompact. Let $L=([0,1)\times\omega_{1})\setminus\{(0,0)\}$ be the long line. Let $M$ be any paracompact manifold with the fixed point property for homeomorphisms. For instance, $M$ could be a projective space over the reals, the complex numbers, or over the quaternions. Let $N=L\times M$. I claim that $N$ has the fixed point property for homeomorphisms but not for continuous maps. Clearly, there is a retraction $f:N\rightarrow S$ where $S\simeq M\times(0,1]$. The set $M\times(0,1]$ does not have the fixed point property since the map $M\times(0,1]\rightarrow M\times(0,1],(x,y)\mapsto(x,y/2)$ has no fixed point. Therefore there is some map $g:S\rightarrow S$ without a fixed point. Thus $g\circ f:N\rightarrow S$ has no fixed point. Now assume that $h:N\rightarrow N$ is a homeomorphism. If $x\in M$, then let $\Gamma_{x}:L\rightarrow L,\Delta_{x}:L\rightarrow L$ be the maps where $\Gamma_{x}(y)=\pi_{1}(h(y,x)),\Delta_{x}(y)=\pi_{1}(h(y,x))$. Let $C_{x}=\{\alpha\in\omega_{1}|\Gamma_{x}(\alpha,0)=(\alpha,0)\},D_{x}= \{\alpha\in\omega_{1}|\Delta_{x}(\alpha,0)=(\alpha,0)\}.$ I claim that the sets $C_{x},D_{x}$ are club sets (club stands for closed unbounded). It is clear that the sets $C_{x},D_{x}$ are closed, so we now need to show $C_{x},D_{x}$ are unbounded. It is easy to see that for all $\alpha\in\omega_{1}$, we have $\Gamma_{x}(z)>(0,\alpha)$ for sufficiently large $z$. Therefore, for all $\alpha\in\omega_{1}$ there is a sequence $(y_{n})_{n}$ with $(0,\alpha) TITLE: Does this construction yield the surreal numbers? QUESTION [10 upvotes]: There are two simple constructions for creating arbitrarily large non-Archimedean ordered field extensions of the reals. First given such a field one may consider rational functions over that field with $f(x)$ declared positive if $f(x)>0$ for all sufficiently large $x$. Secondly one can Dedekind complete the field by filling in all good/narrow cuts in the field as detailed in this discussion. I would like to know if iterating these two constructions over the ordinals produces the surreal numbers. Specifically we define $F_0$ to be the real numbers. For an ordinal $\lambda$ with predecessor $\lambda-1$, define $F_\lambda$ to be the ordered field obtained by Dedekind completing the rational functions over $F_{\lambda-1}$. Otherwise if $\lambda$ is a limit ordinal, define $F_\lambda$ to be the union of all $F_\kappa$ with $\kappa<\lambda$. REPLY [2 votes]: I've written up an expository paper incorporating ideas from this discussion. Here is a link: https://drive.google.com/file/d/0B1G4mOmOYMtha1JLc2tTY1d2SnM/view I welcome any comments, suggestions, etc.<|endoftext|> TITLE: Cuspidal modular forms - toroidal or minimal compactification? QUESTION [6 upvotes]: Let $Y$ be a Siegel variety and let $X$ be a toroidal compactification of $Y$. For any tuple of integers $\underline k$ we have the usual sheaf $\omega^{\underline k}$. The space of modular forms of weight $k$ is $H^0(Y,\omega^{\underline k})= H^0(X,\omega^{\underline k})$ (let us suppose that the genus of $Y$ is greater than $1$). The space of cuspidal forms is defined as $H^0(X,\omega^{\underline k}(-D))$, where $D$ is the divisor of the boundary of $X$. Let $X^\ast$ be the minimal compactification of $Y$ and let $\pi \colon X \to X^\ast$ be the canonical morphism. Of course we have $H^0(X,\omega^{\underline k}) = H^0(X^\ast,\pi_\ast\omega^{\underline k})$, so we can define modular forms as section of some sheaf on the minimal compactification (of course the sheaf need not to be locally free). Let $I$ be the sheaf of ideals on $X^\ast$ that gives the boundary of $X^\ast$ (that is not a divisor in general). Is it true that we have $$ H^0(X,\omega^{\underline k}(-D)) = H^0(X^\ast,\pi_\ast\omega^{\underline k} \otimes I) ? $$ In other words, can we define cuspidal forms using the minimal compactification? A related question is the following: can we consider the space $H^0(X,\omega^{\underline k} \otimes \pi^\ast I)$? Note that $I$ is not locally free, so we cannot use the projection formula. Thank you! REPLY [3 votes]: Let $i \colon \Delta \to X^\ast$ be the closed immersion corresponding to the boundary of $X^\ast$. You have an exact sequence $$ 0 \to I \to \mathcal O_{X^\ast} \to i_\ast \mathcal O_\Delta \to 0 $$ but, as you said, $\pi_\ast \omega^{\underline k}$ is not locally free in general, so, after tensoring with $\pi_\ast \omega^{\underline k}$ we only have the exact sequence $$ I \otimes \pi_\ast \omega^{\underline k} \to \pi_\ast \omega^{\underline k} \to i_\ast \mathcal O_\Delta \otimes \pi_\ast \omega^{\underline k} \to 0 $$ so the correct sheaf to consider is $$ \ker(\pi_\ast \omega^{\underline k} \to i_\ast \mathcal O_\Delta \otimes \pi_\ast \omega^{\underline k}). $$<|endoftext|> TITLE: Must a map on a compact space be surjective on $\cap_{n=1}^\infty f^n(X)$? QUESTION [8 upvotes]: Maybe I'm just being a bit dense here, but this has me stumped right now. A fairly well-know thm is the following: Let $X_0$ be a compact metric space and $f:X_0\to X_0$ be continuous. For each $n\ge1$, let $X_n=f(X_{n-1})$, i.e., the range of the nth "iterate" of $f$. Since $X_0,X_1,\dots$ is a decreasing sequence of compact sets, $X_\infty=\cap_{n=1}^\infty X_n$ is a non-empty compact set. It is easily seen that $f$ maps $X_\infty$ into itself. But, in fact, $f(X_\infty)=X_\infty$. The only argument that I know for this last fact uses sequential compactness. Briefly: fix $x\in X_\infty$, choose a sequence such that each $x_n\in X_n$ and each $f(x_n)=x$, and pick a convergent sub-sequence. So, this theorem ($f(X_\infty)=X_\infty$) actually holds for any compact and sequentially compact topological space. My question is whether or not sequential compactness is needed. There may be a simple argument that I'm missing (as I said at the start). Or, maybe there is a counterexample I'm not seeing - perhaps a mapping on $\beta\omega$? In summary: QUESTION: If $f$ is a continuous map of a compact (but not sequentially compact) space $X_0$ into itself, and $X_\infty$ is defined as above, must $f$ map $X_\infty$ onto itself? Thanks in advance for any help or pointers. -Jeff Norden PS: by "compact" I really mean "compact and Hausdorff". REPLY [2 votes]: $\bf{Method\ 1:}$ For $x\in X_{\infty}$ consider the projective system $$(f^{-n}(x) \overset{f}{\leftarrow} f^{-n-1}(x))_{n\in \mathbb{N}}$$ The projective limit $\lim_{\leftarrow} f^{-n}(x)$ is nonvoid ( $(f^{-n}(x))_{n \in \mathbb{N}}$ nonvoid compacts), so take an element in it : $(x, x_1, x_2, \ldots )$. We have $x_1 \in X_{\infty}$ and $f(x_1) = x$ $\bf{Method\ 2:}$ Let $x \in X_{\infty}$. The sets $f^{-(n+1)}(x)$ are nonvoid, therefore $f^{-1}(x) \cap f^n(X) \ne \emptyset$ and so $$\cap_{n\ge 0} (f^{-1}(x) \cap f^n(X)) \ne \emptyset$$ therefore $$f^{-1}(x) \cap (\cap_{n\ge 0} f^n(X) ) = f^{-1}(x) \cap X_{\infty}\ne \emptyset$$<|endoftext|> TITLE: Minimum word length for an unusual set of generators of the symmetric group QUESTION [5 upvotes]: Problem. Let $n\geq 2$ and let $T$ be the set of all permutations in $S_n$ of the form $$t_k:=\prod_{1\leq i\leq k/2}(i,k-i) \qquad \hbox{for $k=3,4,\ldots,n+1$}.$$ Find the least integer $f_n$ such that every $x \in S_n$ can be written as a product of at most $f_n$ elements from $T$. This is part (ii) of Exercise 1.2.16 in Chapter 1 of Permutation Groups by Dixon and Mortimer (Springer 1996) and is labeled as an unsolved problem. It has been 18 years since the book was published (I guess there hasn't been any further edition yet), so I was wondering whether it is solved or not; if it is can somebody give me a hint? It looks very tough. I have solved part (i) of the exercise, showing that $T$ generates $S_n$, and showing that $2n-3$ is always an upper bound. I asked it on math.stackexchange but did not get any answers there. I am an active member there, but here it is my first question; I hope it meets the standards here. REPLY [3 votes]: This problem is known as ''pancake sorting'' or ''sorting by prefix reversal''. Imagine you have a stack of pancakes numbered $1,2,3,\ldots$ starting from the top. Then $t_k$ corresponds to taking the top $k-1$ pancakes and ``flipping them'' (as a stack). The minimum number of flips needed to reach any given arrangement (alternatively, the diameter of the corresponding Cayley graph) is a well-known problem. The exact number is not known, but some decent bounds are. More information can be found at : http://en.wikipedia.org/wiki/Pancake_sorting https://oeis.org/A058986<|endoftext|> TITLE: Why the Dold-Thom theorem? QUESTION [42 upvotes]: Dold-Thom Theorem: $$\pi_i(SP(X))\cong\tilde{H}_i(X)$$ It's pretty miraculous, no? I've seen its proof, where you show that the composition of the functors on the left-side satisfies the axioms of a homology theory. I've also seen many uses of the theorem, to explain features about Eilenberg-MacLane spaces and other (categorical) phenomena which relate homotopy and homology. But, Is there an intuitive reason (geometrically?) why it's true? Is the Dold-Thom theorem to be expected? Why would one come to think of this? It is very intuitive and clear in low degrees, but the geometry might stop after this. The $i=0$ case is the connectedness of $X$. The $i=1$ case is the ability to lift and commute loops, when analyzing the compositions $\pi_1(X)\to \pi_1(X)^d\to \pi_1(\text{Sym}^d(X))\to H_1(\text{Sym}^d(X))\to H_1(X)$. Perhaps I can argue similarly in higher degrees when $X$ is a closed Riemann surface. This is clear for the sphere, since $\text{Sym}^d(\mathbb{C}P^1)\approx\mathbb{C}P^d$ and $\pi_i(\mathbb{C}P^\infty)\cong\tilde{H}_i(\mathbb{C}P^1)$. Note that this is also clear in the 1-dimensional case when $X\simeq S^1$, as $\text{Sym}^d(\mathbb{C}-\lbrace0\rbrace)\approx \mathbb{C}^{d-1}\times(\mathbb{C}-\lbrace 0\rbrace)$. The content of the theorem is reduced to low degrees for these simple examples. REPLY [15 votes]: This won't involve any geometry, but here is a model-independent description of the situation as I understand it. I will not prove anything. The very short summary is that The infinite symmetric product and singular chains are both models of the free $\mathbb{Z}$-module spectrum on a space $X$, where $\mathbb{Z}$ is regarded as a ring spectrum, and "homotopy groups of the free $\mathbb{Z}$-module spectrum on $X$" is a model-independent definition of the (ordinary) homology groups of $X$. First, here's the simplest version of a general definition. Let $X$ be a set and let $R$ be a commutative ring. Then the $R$-homology of $X$ is equivalently one of the following $R$-modules: the $R$-module $R[X]$ of formal $R$-linear combinations of elements of $X$, the direct sum $\displaystyle \bigoplus_{x \in X} R$, or equivalently the colimit of the constant diagram $X \ni x \mapsto R \in \text{Mod}(R)$, the value on $X$ of the left adjoint to the forgetful functor $\text{Mod}(R) \to \text{Set}$, the value on $X$ of the unique cocontinuous functor $\text{Set} \to \text{Mod}(R)$ sending $\text{pt}$ to $R$. Intuitively, $R$-homology is the canonical covariant way to linearize a set $X$ into an $R$-module: above I've just given four different ways of saying "the free $R$-module on $X$." Moreover, the forgetful functor $\text{Mod}(R) \to \text{Set}$ factors through abelian groups, and hence its left adjoint also factors through abelian groups in the other direction as the composite $$R[-] : \text{Set} \xrightarrow{\mathbb{Z}[-]} \text{Ab} \cong \text{Mod}(\mathbb{Z}) \xrightarrow{R \otimes (-)} \text{Mod}(R)$$ which is just a fancy way of saying that we can write $$R_0(X) \cong R[X] \cong R \otimes \mathbb{Z}[X].$$ This is "universal coefficients for sets": it says that to understand the free $R$-module on a set it suffices to understand the free $\mathbb{Z}$-module / abelian group on a set. The last description of $R$-homology above reflects "Eilenberg-Steenrod for sets," which says that $\text{Set}$ is the free cocomplete category on a point. Now suppose we want to linearize, not sets, but spaces, by which I mean (weak) homotopy types / $\infty$-groupoids. So let $X$ be a space and let $R$ be an $E_{\infty}$-ring spectrum. Then the $R$-homology of $X$ is equivalently (the homotopy groups of) one of the following $R$-module spectra, where $\text{Mod}(R)$ denotes the $(\infty, 1)$-category of $R$-module spectra (and "functor" means "$(\infty, 1)$-functor"): the smash product $R \wedge \Sigma^{\infty}_{+} X$, where $\Sigma^{\infty}_{+} X$ is the suspension spectrum of $X$ with a disjoint basepoint, the homotopy / $(\infty, 1)$-colimit of the constant diagram $X \ni x \mapsto R \in \text{Mod}(R)$, the value on $X$ of the $(\infty, 1)$-left adjoint to the forgetful functor $\text{Mod}(R) \to \text{Space}$, the value on $X$ of the unique homotopy cocontinuous functor $\text{Space} \to \text{Mod}(R)$ sending $\text{pt}$ to $R$. Intuitively, $R$-homology is the canonical covariant way to linearize a space into an $R$-module spectrum: above I've just given four different ways of saying "the free $R$-module spectrum on $X$." The first description above should be regarded as a direct generalization of the isomorphism $R[X] \cong R \otimes \mathbb{Z}[X]$ to spaces, except that $\mathbb{Z}$ has been replaced with the sphere spectrum $\mathbb{S}$. More precisely, the forgetful functor $\text{Mod}(R) \to \text{Space}$ factors through spectra, and hence its left adjoint also factors through spectra in the other direction as the composite $$\text{Space} \xrightarrow{\Sigma^{\infty}_{+}(-)} \text{Sp} \cong \text{Mod}(\mathbb{S}) \xrightarrow{R \wedge (-)} \text{Mod}(R).$$ In particular $\Sigma^{\infty}_{+}$, being left adjoint to the forgetful functor from spectra to spaces, should be thought of as the "free spectrum" functor, and $R \wedge (-)$, being left adjoint to the forgetful functor from $R$-module spectra to spectra, should be thought of as the "free $R$-module spectrum (on a spectrum)" functor. The last description of $R$-homology reflects an $(\infty, 1)$-categorical version of Eilenberg-Steenrod for spaces, which says that $\text{Space}$ is the free homotopy cocomplete $(\infty, 1)$-category on a point. Now, at long last, ordinary homology is the homotopy groups of the free $\mathbb{Z}$-module spectrum: $$H_{\bullet}(X, \mathbb{Z}) \cong \pi_{\bullet}(\mathbb{Z} \wedge \Sigma_{+}^{\infty} X).$$ Hopefully I've phrased things so it's clear that this story about linearizing spaces is a direct analogue of the story about linearizing sets, provided you are willing to accept (that various $(\infty, 1)$-categorical machinery works the way it ought to and) that the correct analogue of abelian groups in this setting is spectra. Here are some more things that ought to be true and that connect this story back to more model-dependent considerations. By a suitable version of the stable Dold-Kan theorem, the $(\infty, 1)$-category of $\mathbb{Z}$-module spectra should be equivalent to the $(\infty, 1)$-category presented by unbounded chain complexes of $\mathbb{Z}$-modules. This should restrict to an equivalence between connective $\mathbb{Z}$-module spectra and connective chain complexes. By the usual Dold-Kan theorem, the category of connective chain complexes of abelian groups is equivalent to the category of simplicial abelian groups (and there should be model structures on both sides making this a Quillen equivalence presenting an equivalence of $(\infty, 1)$-categories, and so forth). This equivalence more or less sends singular chains on a topological space $X$ to the free simplicial abelian group on the singular simplicial set of $X$, and modulo technical details this gives rise to the relationship between singular homology and the homotopy groups of the free $\mathbb{Z}$-module spectrum on $X$, which is connective since $\mathbb{Z}$ and suspension spectra are connective. The analogue of the free simplicial abelian group on a simplicial set for topological spaces is the free topological abelian group; this is roughly what the infinite symmetric product attempts to be, and modulo technical details (in particular, niceness hypotheses on $X$) this gives rise to the relationship between the homotopy groups of the infinite symmetric product and the singular homology of $X$.<|endoftext|> TITLE: Reverse Hausdorff Young for nonnegative functions QUESTION [6 upvotes]: The classical Hausdorff-Young inequality states that $$ \Vert \widehat{f} \Vert_{p'} \leq \Vert f \Vert_p \text{ for } 1 \leq p \leq 2. $$ For $p=2$, we even have equality due to Plancherel. If we additionally assume that $f \geq 0$, we also get $$ \Vert \widehat{f} \Vert_\infty = \widehat{f}(0) = \int f(x) \, dx = \Vert f \Vert_1, $$ i.e. we get equality in the Hausdorff-Young inequality for $p=1$ also. My question is, wether this generalizes to $1 \leq p \leq 2$ (at least asymptotically), i.e. do we have $$ \Vert \widehat{f} \Vert_{p'} \asymp \Vert f \Vert_p \text{ for } 1 \leq p \leq 2 \text{ and } f \geq 0 \text{?} $$ We can not use interpolation here (at least I do not see it), because the estimate on the "boundary" points (at least at $p=1$) is only valid for $f \geq 0$ (and the whole inequality can also only be valid for those $f$), so that the usual "splitting" (for real interpolation) can not be applied. Similarly, complex interpolation does not seem to work. But also the classical method for constructing a counterexample does not work, i.e. one can not take something like $$ f = \sum_{m=1}^n M_{\xi_m} g, $$ where $M_\xi g (y) = e^{2\pi i \xi y} g(y)$ denotes modulation, because this will violate the non-negativity. Taking $$ f = \sum_{m=1}^n T_{x_n} g $$ does not violate this assumption and I can asymptotically calculate $\Vert f \Vert_p$ in this case (for $\min_{n \neq m} |x_n - x_m| \to \infty$), but I am unable to calculate the integral $$ \Vert \widehat{f} \Vert_{p'} = \bigg( \int \big| \sum_{m=1}^n e^{\pm 2\pi i x_n \xi }\big|^{p'} \cdot |\widehat{g}(\xi)|^{p'} \, d\xi \bigg)^{1/p'}. $$ precisely enough. Any ideas would be appreciated. REPLY [5 votes]: It is an elementary exercise in Banach space theory to show that there is NO operator from $L_p$ into $L_{q}$ that satisfies the inequality when $1\le p \not= 2$ with $p TITLE: Is there a general notion of a subobject generated by a subset in any concrete category? QUESTION [5 upvotes]: Let $A$ be an object of a concrete category $C$ with a forgetful functor $F\rightarrow Set$ and let $S \subseteq F(A)$. Is there a general construction that gives us a subobject $\left$ of $A$ in $C$? $S$ should, for example, satisfy the universal property that if $B$ is a subobject of $A$ in $C$ such that $S \subseteq F(B)$, then $\left< S \right>$ is a subobject of $B$ in $C$. I know that in the category of groups it's true that if $f$ is a homomorphism of groups, then $f(\left< S \right>)=\left< f(S) \right>$. Is there a corresponding theorem here? REPLY [4 votes]: Let's denote the forgetful functor by $U : C \to \mathsf{Set}$. (Usually its left adjoint is denoted by $F$.) If $C$ is well-powered, has intersections (which are special limits), and $U$ preserves monomorphisms and intersections (for example when $U$ preserves limits), we may define $\langle S \rangle = \bigcap_{i : B \to A \text{ mono}, S \subseteq \mathrm{im}(U(i))} B$ with the evident monomorphism $\langle S \rangle \to A$. Now for the second question: If $f : A \to B$ is a morphism in $C$ and $S \subseteq U(A)$, then let $T:=U(f)(S) \subseteq U(B)$. We ask ourselves if $f(\langle S \rangle)=\langle T \rangle$ holds. But what does $f(A')$ mean for a monomorphism $A' \to A$? It should be the smallest subobject of $B$ such that $A' \to A \to B$ factors through $f(A')$ (this exists again under the mentioned hypothesis). But when $A'=\langle S \rangle$, the condition reduces by definition of $\langle S \rangle$ to that $S \to U(A) \to U(B)$ factors through $U(f(A'))$, i.e. that $T \subseteq U(f(A'))$. This shows $f(\langle S \rangle)=\langle T \rangle$.<|endoftext|> TITLE: Generators of invariant polynomials of semisimple Lie algebra QUESTION [14 upvotes]: Suppose $\mathfrak{g}$ is a complex semi-simple Lie algebra. By a theorem of Chevalley, we know that $S(\mathfrak{g})^\mathfrak{g}$, i.e. the $\mathfrak{g}$ invariant polynomials, is generated by $l$ homogeneous polynomials where $l$ is the rank of $\mathfrak{g}$. The degrees of the generators are known, namely they are the primitive exponents. I am wondering if the generators can be canonically chosen. For example, when we are considering $\mathfrak{sl}(2,\mathbb{C})$, we can take the generator to be the Casimir. Does such kind of canonical choice exist in general? REPLY [6 votes]: Here are some topological considerations that privilege a choice of generators up to scale. First, Chern-Weil theory gives an isomorphism $$S(\mathfrak{g}^{\ast})^{\mathfrak{g}} \cong H^{\bullet}(BG, \mathbb{C})$$ sending an invariant polynomial $f$ to the characteristic class of principal $G$-bundles given by evaluating $f$ on the curvature of a connection. Here $G$ is the corresponding simply connected complex Lie group, or if you prefer its maximal compact. Second, rational homotopy theory gives an isomorphism $$H^{\bullet}(BG, \mathbb{C}) \cong S( \left( \pi_{\bullet}(BG) \otimes \mathbb{C} \right)^{\ast} )$$ which singles out some particularly nice choices of generators of $H^{\bullet}(BG, \mathbb{C})$, namely dual bases of graded bases of $\pi_{\bullet}(BG) \otimes \mathbb{C}$. If $\pi_{\bullet}(BG) \otimes \mathbb{C}$ has rank at most $1$ in each degree then this graded basis is unique up to scale, and hence the corresponding dual basis is also unique up to scale. To fix scales up to sign you can take a basis of $\pi_{\bullet}(BG) \otimes \mathbb{C}$ which lifts to a basis of the torsion-free part of $\pi_{\bullet}(BG)$. In fact $\pi_{\bullet}(BG) \otimes \mathbb{C}$ has Poincare series $\sum t^{2e_i}$ where the $e_i$ are the exponents of $G$, so the above condition is satisfied whenever the exponents are distinct, and this happens for all simple $\mathfrak{g}$ except sometimes in type $D$ (and this ambiguity can be resolved). I believe the resulting choice of generators for classical types is given, up to scale, by invariant polynomials of the form $$X \mapsto \text{tr}(X^i)$$ as in Humphreys' answer. To see how this works for type $A$ let me work first with the groups $U(n)$ rather than $SU(n)$ or $SL_n(\mathbb{C})$. The sequence of inclusions $U(1) \hookrightarrow U(2) \hookrightarrow \dots$, which has colimit $U$, induces a sequence of maps $$BU(1) \to BU(2) \to \dots$$ with colimit $BU$. $BU$ has a natural infinite loop space structure given by taking direct sums of complex vector bundles, and so its cohomology is naturally a cocommutative Hopf algebra. Moreover the isomorphism $$H^{\bullet}(BU, \mathbb{C}) \cong S( \left( \pi_{\bullet}(BU) \otimes \mathbb{C} \right)^{\ast} )$$ is an isomorphism of Hopf algebras, where the RHS should be interpreted as the universal enveloping algebra of an abelian Lie algebra. Hence the distinguished choice of generators above corresponds to a choice of primitive elements in this Hopf algebra, of which there is one in each even degree. On the other hand, $H^{\bullet}(BU, \mathbb{C})$ is known to be isomorphic, as a Hopf algebra, to the Hopf algebra of symmetric functions, with the isomorphism given explicitly by sending the $i^{th}$ Chern class $c_i$ to the elementary symmetric function $e_i$. The primitive elements of the Hopf algebra of symmetric functions, over a field of characteristic $0$, are given by linear combinations of the power symmetric functions $p_i$, and tracing back through all of the isomorphisms we've written down these correspond to the trace functions $X \mapsto \text{tr}(X^i)$. The corresponding characteristic classes are the coefficients of the Chern character. This is all a long-winded way of saying "among the characteristic classes of complex vector bundles we can single out the ones that behave as nicely as possible with respect to direct sum. The Chern character is additive with respect to direct sum and so it's reasonable to single out its coefficients." In the other classical types suitable choices of embedding into $U(n)$ should also reproduce this result. The ambiguity in type $D$ comes from the classifying spaces $BSO(4n)$, where the Euler class has the same degree as a Pontryagin class. The Pontryagin class can be distinguished by the fact that it is stable, or in other words by the fact that it is the pullback of a class on $BSO(4n+1)$, and similarly the Euler class can be distinguished by the fact that it vanishes when pulled back to $BSO(4n - 1)$.<|endoftext|> TITLE: Some polytopes in $\mathbb R^n$ whose vertices have coordinates 1, -1 or 0 QUESTION [6 upvotes]: Let $n$ and $k$ be positive integers with $k\leq n$. Let $P(n,k)$ be the convex hull in $\mathbb R^n$ of the $2^k {n \choose k}$ vectors whose exactly $k$ coordinates belong to $\{\pm 1\}$ all the others being equal to $0$. Examples: $P(n,n)$ is the $n$-hypercube; $P(n,1)$ is the $n$-cross-polytope. Viewed the two classical examples above, I expect that the $P(n,k)$'s have been studied already. Some references would be welcome. I'm interested in the $f$-vector and more specifically to the facets (1-codimensional faces) of $P(n,k)$: how many and what kind of polytopes they are? Added: thinking to this question, it comes to my mind that $2n$ of the facets of $P(n,k)$'s are $P(n-1,k-1)$ (assuming that $k>1$) and that all the other ones are $(n,k)$-hypersimplexes (assuming that $k TITLE: Coloring of the plane QUESTION [8 upvotes]: I would like to know the minimum number k such that the plane R^2 can be coloured with k colors such that no colour contain all the possible distances. In other words, a colouring such that each color forbids a distance (not necessarily the same as it is the case for http://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem) Has this problem been studied? It is easy to prove that 2 colors isn't enough, but I don't know for 3 colors... We also know that this number k is at most 7 (using http://en.wikipedia.org/wiki/File:Hadwiger-Nelson.svg for example), but I did not find any better bound yet. Bests, REPLY [10 votes]: Seems to be the polychromatic number of the plane. According to my knowledge, the value is at least 4 (due to Raiskii) and at most 6 (due to Stechkin). See Chap. 4 and 6 of The Mathematical Coloring book.<|endoftext|> TITLE: Learning roadmap: 'combinatorial' probability QUESTION [10 upvotes]: I am about to finish working through Williams's Probability With Martingales. I have studied analysis up to the first five chapters of Folland's text but have not studied any combinatorics yet. It seems like 'combinatorial' probability topics like percolation, probability on graphs and networks, finite Markov chains and random walks are currently very active and I would like to be able to read the current research in at least some of these areas. While I can find many interesting texts on Amazon etc. I am not sure how well they reflect current work. I would greatly appreciate a reading list or learning roadmap for this area. [I am cross-posting this from Math.SE where it did not get an answer despite a bounty.] REPLY [4 votes]: Geoffrey Grimmett's Probability on Graphs is an excellent introduction to a variety of current active research areas in discrete probability theory, and is probably at about the level you want (you may find that you need to consult other books if there is background you need to fill in, but that isn't a bad thing). It also has quite a number of very stimulating exercises. It's available free from his web site.<|endoftext|> TITLE: Why are the angular differences of these random complex polynomial coefficients almost constant? QUESTION [5 upvotes]: This is based on მამუკა ჯიბლაძე's (not-)answer here. I guess it is better to make up a new thread for it. Let me repeat the setup here: We consider polynomials whose complex roots are randomly distributed in various senses (say, uniformly in a square around zero). Can something be said about the distribution of the coefficients? I have taken 500 random complex numbers chosen uniformly in a unit square around O, i.e. $[-\frac12,\frac12 ]\times [-\frac12,\frac12]$, and have PARI display the coefficients of the polynomial with those 500 roots. Consecutive ones are joint, and the precision \p1000 I chose should yield accurate graphics. Doing this for several dozens of polynomials, here are some kinds of patterns that arise repeatedly. Most of them are different from those displayed by მამუკა ჯიბლაძე, though I did get cycloid-like shapes (but less "smooth") from time to time. The three rightmost ones are more irregular, which seems to occur less frequently. What they mostly seem to have in common is the fact that the arguments of consecutive coefficients tend to be quite equally spaced, which is why linking them yields spirals (like #2 and #5) or star-like constellations (like #1 and #8). So I have displayed the arguments of the coefficients in order. (The argument is only defined modulo $\pi$, so if the difference of two consecutive ones is $>\dfrac\pi2$ in absolute, I have corrected it by $\pm\pi$ to get rid of unnatural jumps in the display. Some of the "folds" (or "reflections") are probably still due to that, e.g. in #4 and #8 below, at places where the slope is close to $\pm\frac\pi2$.) Note that the following graphs don't correspond to the same polynomials as above, but are again chosen to give a quite representative idea of what can happen. I have included here the y-axis graduation for convenience. It turns out that more often than not, the curve of arguments is approximatively "piecewise linear". Can this statistically be explained? Note that much can be explained in a similar way as David Speyer's answer to an earlier question, but for the angles (arguments) it seems less obvious to me. It is a good idea to divide all zeros of this "raw polynomial" $\sum c_kx^k$ by $\sqrt[n]{|c_0|}$. This planar scaling doesn't change the above angles (arguments) and makes the polynomial more "symmetric". BTW, as for the (decimal logarithms of) absolute values of the coefficients, all curves, once scaled, are close to the parabola-like one given by მამუკა ჯიბლაძე. REPLY [3 votes]: I suspect both numerical issues and issues with the random number generator to be at work simulaneously. I also get crazy values (i.e. very large ones and also some structure as shown above) for the coefficients even when I put the zeros at the $N$-th root of unity. This is seen better, the larger $N$. This suggests that the calculation of the coefficients from the roots is performed in a numerically unstable way. My pictures of coefficients did not change qualitatively when I changed the random number generator. However, I observed a different outcome when I permuted the roots randomly and then calculated the coefficients again. They should be the same, but they were not. Moreover, the pictures looked qualitatively different. This suggests again that the computation of the coefficients is unstable and also that "permuting random coefficients randomly gives another type of random". For example with z = exp(2*pi*1i*(rand(N,1))); a = poly(z); subplot(1,2,2),plot(a,'-.'),axis equal, grid on z2 = z(randperm(N)); z2 = z2(randperm(N)); a2 = poly(z); subplot(1,2,1);a2 = poly(z2);plot(a2,'-.');axis equal, grid on in MATLAB I got this (For some reason I plotted the coefficients after double repermutation of the root on the left and the coefficients of the original polynomial on the right.) In other examples the difference was not that drastically but a pointwise error between the coefficients above 10% was seen frequently.<|endoftext|> TITLE: Can knot diagrams be monotonically simplified using under moves? QUESTION [48 upvotes]: It is well known that knot diagrams cannot be monotonically simplified using Reidemeister moves. For instance, the Goeritz unknot cannot be directly simplified. On the other hand, there is a stronger move that 3-manifold topologists sometimes use, called the "under move": take a section of the knot that has only under crossings, and replace it with another under-strand connecting the same two endpoints. We can use this to simplify the Goeritz unknot one step: Question: Can every diagram of the unknot be monotonically simplified using only under moves (or maybe under and over moves)? Probably one needs to allow level moves as well. This is related to an earlier question: Are there any very hard unknots? My move is more precise than that one, and Haken's "Gordian knot" can be simplified at least one step using a few level under-moves and then a reducing under-move on the right-hand side. (I was wondering about this during a talk by Prasolov on his amazing work with Dynnikov on similar questions in the context of grid diagrams. Surely someone has considered this before.) REPLY [11 votes]: I recall a discussion on this question on math.sci.research quite a few years ago. John Conway had tried various moves to supplement Reidemeister moves. If I recall correctly, his guess was that no finite collection of moves will give monotonic simplification. The Goeritz unknot below admits a simplifying move along the dashed arc shown. If one takes its untwisted double (as Ryan suggested) one again gets an unknot. (I hope I got the twists right below, but if not then adding twists shouldn't hurt). I don't see any obvious simplifying moves, or even any helpful lateral moves, for the bottom diagram. Added: Marco correctly points out that while the doubling construction destroys the simplification at top left, it does not kill the one at bottom right. There is nothing to stop one from doing an additional untwisted doubling, with the clasp put at lower right. Perhaps that will do the trick.<|endoftext|> TITLE: eigenvalue estimate of the adjacency matrix QUESTION [5 upvotes]: The adjacency matrix of a nonempty (undirected) graph has a strictly positive largest eigenvalue $\lambda_\max$. A very easy upper estimate for it can be obtained directly by Gershgorin's theorem: $$ \lambda_{\max}\le \Delta\ , $$ where $\Delta$ is the maximal degree of the graph. Are any further estimates known? And are there known lower estimates on the lowest eigenvalue $\lambda_\min$? REPLY [5 votes]: A classic estimate is due to Constantine: $$ \lambda_{\min} \geq -\sqrt{\lceil \frac{n}{2} \rceil \lfloor \frac{n}{2} \rfloor}. $$ If $m$ is the number of edges, then $$ \lambda_{\min} \geq - \sqrt{m}. $$ A common generalization is $$ \lambda_{\min} \geq -\sqrt{MaxCut(G)}, $$ where $MaxCut(G)$ is the size of a maximal bipartite subgraph. You can find these results, with references, in the 2008 paper by Bell et al.. There are more complicated results as well, in particular using the eigenvector. If you would like to discuss such topics, I am always interested.<|endoftext|> TITLE: What happens if we rotate the kernel of an integral operator? QUESTION [10 upvotes]: Given an integral operator $K$ on $L^2(\mathbb R)$ with kernel $k(x, y)$, consider the integral operator $L$ on $L^2(\mathbb R)$, whose kernel has the form $k(\alpha x+\beta y, \gamma x+\delta y)$, where $\alpha\delta-\beta\gamma\ne 0$. It is well known that an integral operator belongs to the Hilbert-Schmidt class $\mathfrak S_2$ if and only if its kernel is square-summable; therefore, $K\in\mathfrak S_2$ if and only if $L\in\mathfrak S_2$. Is the same true if we replace the class $\mathfrak S_2$ by $\mathfrak S_1$, or by $\mathfrak S_p$ with $p\ge 1$? Can such an equivalence be proved for other properties? (References to known results would be appreciated.) This question was not answered at SE. I was not able to use hints suggested in the discussion there. In particular, this contains another question as a special case: does the integral operator on $L^2(\mathbb R)$, whose kernel is the indicator of the rhombus $\{|x|+|y|<1\}$, belong to the trace class? REPLY [8 votes]: Consider the characteristic function of the rhombus $\{|x|+|y| <1\}$ as a kernel on $L^2(I)$, with $I:=[-1,1]$. Since this kernel is symmetric, the trace norm of the corresponding compact operator $T$ is just the $\ell^1$ norm of the sequence of its eigenvalues. Let's find them. We have for any $x\in I$ $$Tu(x)=\int^{1-|x|}_{-1+|x|}u(y)dy.$$ Let $v(x):=\int_0^xu(y)dy$ and $\lambda\neq0$. The eigenvalue equation $Tu=\lambda u$ then writes $$v(1-x)-v(-1+x)=\lambda v'(x)\quad \mathrm{for\quad } x >0\; , $$ $$v(1+x)-v(-1-x)=\lambda v'(x)\quad \mathrm{for\quad } x <0\; , $$ with $v(0)=0$. Equivalently, $v(x)$ is an odd function satisfying $$2v(1-x)=\lambda v'(x).$$ If we derive both sides and use the equation again we find $$v''(x)=-(2/\lambda)^2v(x),$$ with boundary conditions $v(0)=0=v'(1).$ Hence the eigenfunctions are cosine functions, with eigenvalues $\lambda_n\sim C/n$, and $T$ is not in the trace class.<|endoftext|> TITLE: Gromov-Hausdorff convergence for non-compact metric spaces QUESTION [8 upvotes]: Let $(X_i,p_i)$, $(X,p)$ be pointed connected proper metric spaces (i.e. the closures of balls are compact). Are the following two statements equivalent? $\forall r > 0: \bar{B}_r(p_i) \stackrel{GH}{\to} \bar{B}_r(p)$. $\forall r > 0: (\bar{B}_r(p_i),p_i) \stackrel{GH}{\to} (\bar{B}_r(p),p)$. In 2, the pointed Gromov-Hausdorff distance is defined as usual but with respect to the pointed Hausdorff distance $d_H((A,a),(B,b)) := d_H(A,B) + d(a,b)$. Obviously, 2 implies 1, so the question is whether or not/under which conditions the other implication holds. REPLY [4 votes]: Here is a counterexample constructed from the $5$-point example given by Włodzimierz Holsztyński here. Consider the set $Z = \{x,y,a,b,c\}$ made into a metric space via $$d(x\ y) = d(a\ b) = 1,$$ $$d(x\ a) = d(y\ b) = 2,$$ $$d(x\ b) = d(y\ a) = 3,$$ $$d(x\ c) = d(y\ c) = 6,$$ $$d(a\ c) = 5,\qquad\qquad d(b\ c) = 4.$$ First we construct a connected metric space $Z'$ simply by connecting each two points by edges of lengths given by the respective distances and taking the metric on $Z'$ the induced intrinsic one (i hope this is clear). Now we form a noncompact space $Z''$ by adding a ray (that is the interval $[0,\infty[$) starting at $c$. Finally consider the subset $X \subset Z''$ given by $Z''$ without the edge connecting $a$ to $c$ and the one connecting $b$ to $c$. Equip $X$ with the induced restricted metric $d$ (not the intrinsic one). Then for all $r > 0$ the balls $B_r(x) \subset X$ and $B_r(y) \subset X$ are isometric. Therefore, taking $X_i = X$ and $p_i = x$ we find that $B_r(p_i) \to B_r(p)$ in Gromov-Hausdorff sense for all $r > 0$, but $(B_r(p_i),p_i) = (B_r(x),x)$ does not converge to $(B_r(p),p) = (B_r(y),y)$ for all $r \geq 6$ since there is no isometry of $X$ taking $x$ to $y$. Maybe a sufficient condition might be that the metrics of the $X_i$ are intrinsic (and complete), i.e. distances are given by infimal (minimal) lengths of continuous curves connecting points.<|endoftext|> TITLE: Is the following module over a group ring necessarily infinitely generated? QUESTION [6 upvotes]: Suppose $\Gamma$ is a (finitely presented, but this is probably irrelevant) group, and $M$ is a finitely generated (EDIT: finitely presented) module over $\mathbb{Q}\Gamma$ which is infinite-dimensional (as a vector space over $\mathbb{Q}$.) Define a module $M \wedge_{\mathbb{Q}} M$ as either the symmetric or the alternating tensor product over $\mathbb{Q}$, with the module structure given by $\gamma(m \wedge n)=\gamma m \wedge \gamma n$. Is it true that $M \wedge_{\mathbb{Q}} M$ cannot be finitely generated? My intuition is that there will always be some $\alpha \in M$ and a sequence of $\gamma_i \in \Gamma$ such that every element of the form $\alpha \wedge \gamma_i\alpha$ can't be generated by a finite set of elements. The simplest example is when $\Gamma=\mathbb{Z}$ and $M=\mathbb{Q}[\mathbb{Z}]$. Then it's easy to show that a finite set in $M \wedge_{\mathbb{Q}} M$ can only generate a finite-dimensional set of elements of the form $\alpha \wedge m$, where $\alpha$ generates $M$. [In fact, this argument generalizes to any cyclic module over any group. This also means that the statement is true for groups $\Gamma$ for which $\mathbb{Q}\Gamma$ is Noetherian, but these are rather rare.] <- the argument I had in mind here doesn't actually work, never mind To prove the general statement it's enough to prove it for modules over the free group rings, $\mathbb{Q}F_n$, though this is unlikely to make it any easier. This question, though reduced to pure algebra, came up in a topological setting. I'm wondering if there's a ring-theoretic reason that the statement is true. REPLY [5 votes]: If $\Gamma$ acts $2$-transitively on an infinite set $X$, then the permutation module $\mathbb{Q}[X]$ will be a counterexample. For example, take an action of the free group of rank $2$ on a countable set, where one generator acts transitively and the other generator has one fixed point and acts transitively on the remaining elements. Edit: In fact, $2$-transitivity is not necessary for this to work: it's sufficient that a point stabilizer acts on $X$ with only finitely many orbits. Also, if the point stabilizer is finitely generated then $\mathbb{Q}[X]$ will be a finitely presented $\mathbb{Q}[\Gamma]$-module. An example is Thomson's group $F$ acting on the set $X$ of dyadic rationals strictly between $0$ and $1$. It is a finitely presented group, as is a point stabilizer (which is isomorphic to $F\times F$), and a point stabilizer acts on $X$ with three orbits. [I have no reason to think there couldn't be a far simpler example, except for the fact that I didn't think of one.]<|endoftext|> TITLE: Thinnest 2-fold coverings of the plane by congruent convex shapes QUESTION [11 upvotes]: It is an unsolved problem to determine the "thinnest" $2$-fold covering of the plane by disks. The $2$-fold coverage problem by disks is to find the minimum number of congruent (unit-radius) disks such that each point of the plane is covered by at least $2$ disks. Here is a paper from 1997 on the topic, including their Fig.1 showing the thinnest $1$-fold covering and a potential thinnest $2$-fold covering: Bezdek, András, and Włodzimierz Kuperberg. "Circle covering with a margin." Periodica Mathematica Hungarica 34.1-2 (1997): 3-16. journal link And here is a paper from just a few days ago on this topic, which says that thinnest $k$-fold disk coverage for $k>1$ is "hopelessly difficult": Jingchao Chen. "Two-Fold Circle-Covering of the Plane under Congruent Voronoi Polygon Conditions." 3 Oct 2014. arXiv Abstract link My question is: Q. For which shapes $K$ (planar convex bodies) is the thinnest $2$-fold covering (by congruent copies of $K$) known? I would allow $K$ to be rigidly transformed: translated, rotated. Certainly if $K$ tiles the plane, the question is settled. So for all triangles, and for all quadrilaterals (see this earlier MO question), matters are settled. Some pentagons tile the plane. How about for $K$ a regular pentagon? Etc. (Added:). I am especially interested to learn if some thinnest $2$-fold covering is known for a shape that does not perfectly tile the plane. REPLY [14 votes]: [Edited to add a two-parameter family of hexagons] Here's a family of convex pentagons $K_\epsilon$, each of which does not tile the plane but does have a perfect $2$-covering by rotated translates. Start with a regular hexagon $\,\widetilde{\!H}$. Its six shorter diagonals form a "star of David" with a smaller hexagon $H$ at its center whose area is $1/3$ the area of $\,\widetilde{\!H}$. Now tile the plane with parallel copies of $\,\widetilde{\!H}$, and remove the corresponding copy of $H$ from the center of each translate. The resulting region, call it $R$, is shown shaded in the first part of the image below (with $\,\widetilde{\!H}$ and its translates outlined in blue, and the star of David in light blue). This picture shows how $R$ is also the hexagonal tiling by copies of $H$ with every third hexagon removed. Hence three parallel copies of $R$ constitute a perfect $2$-covering of the plane. Now each copy of $\,\widetilde{\!H} - H$ is tiled by six rotations of a pentagon $K_0$ with angles $120^\circ, 90^\circ, 120^\circ, 90^\circ, 120^\circ$. So $K_0$ has a perfect $2$-covering. This is not yet our pentagon: it tiles the plane, because three copies fill $H$ exactly. But we can deform $K_0$ by rotating its shortest two sides by $\epsilon$ degrees to a pentagon $K_\epsilon$, with angles $(120+\epsilon)^\circ, (90-\epsilon)^\circ, 120^\circ, (90+\epsilon)^\circ, (120-\epsilon)^\circ$, six copies of which still tile $\,\widetilde{\!H} - H$ as shown in the second part of the image. For small but nonzero $\epsilon$ this $K_\epsilon$ cannot tile the plane, not even with reflections allowed $-$ a contradiction is soon reached starting with a neighborhood of a $(120\pm\epsilon)^\circ$ angle. But $K_\epsilon$ still perfectly $2$-covers the plane via $\,\widetilde{\!H}-H$ and $R$, so we have our family of convex pentagons with a minimal tiling thickness of $2$. P.S. I later noticed a further modifiction to a two-parameter family of convex hexagons $K = K_{\epsilon,\delta}$ that do not tile the plane but do tile $R$ and thus perfectly $2$-cover the plane. Deform each edge of $\,\widetilde{\!H}$ to the same Z-shaped path ("Z-shaped": centrally-symmetric concatenation of three line segments). Call the resulting polygon $\,\widetilde{\!H}'$; it still tiles the plane. Do not change $H$. Then $\,\widetilde{\!H}' - H$ is tiled by six rotations of the same convex hexagon $K$, as seen in the first part of the next image: Hence $K$ tiles $R$, as seen in the second part, and therefore $K$ perfectly $2$-covers the plane. Compared with the pentagons $K_\epsilon$, this variation has the advantage not just of an extra parameter but also that it's much easier to check there's no perfect $1$-covering: If a convex hexagon tiles the plane then we can assume that any two adjacent tiles must share a complete edge, and that no more than three tiles meet in every corner; and that's soon seen to be impossible here.<|endoftext|> TITLE: Erdős cardinals and $0^\sharp$ QUESTION [8 upvotes]: It is well-known that if the Erdős cardinal $\kappa(\omega_1)$ exists, then $0^\sharp$ exists, but what if $\kappa(\lambda)$ exists for a limit ordinal $\omega_1^L\leq \lambda<\omega_1$? Does this still impliy that $0^\sharp$ exists? Or, alternatively: it is known that the existence of $\kappa(\lambda)$ for all countable $\lambda$ is consistent with $0^\sharp$ does not exist (with $V=L$, in fact), but, does it imply that $\omega_1^L=\omega_1$? REPLY [6 votes]: Similarly to Asaf's comment, let $\kappa = \kappa(\omega_1^L)$ (we assume that it exists). Then in $L_\kappa$, there is $\alpha$-Erdős cardinal for every $L$-countable $\alpha$ (since being $\alpha$-Erdős cardinal is downward absolute between transitive models for countable ordinals $\alpha$). So $L_\kappa$ is a model for $\forall \lambda < \omega_1,\, \kappa(\lambda)$ exists and $\omega_1 = \omega_1^L$. It is interesting to note that even when $\omega_1$ is larger then $\omega_1^L$, still the existence of $\lambda$-Erdős cardinal for every $\lambda < \omega_1$ does not imply the existence of $0^{\#}$. For example, the existence of $\kappa(\omega_1^L)$-Erdős cardinals has only mild effect on the cardinals of $L$: Although it does imply that $\omega_1^L < \omega_1$ (otherwise $0^{\#}$ exists, and then $\omega_1^L < \omega_1$), it is consistent with $\omega_1 = \omega_2^L$. The reason is the if $V \models \kappa(\omega_1^L)$ exists, and $\omega_2^L$ is countable, then one can construct in $V$ a $L$-generic filter for $Col(\omega, \omega_1^L)$, $G$, and since $L[G] \subseteq V$ and both agree that $\omega_1^L$ is countable, the fact that $\kappa(\omega_1^L)$ is $\omega_1$-Erdős in $V$, implies that it is also $\omega_1^L$-Erdős in $L[G]$, but $\omega_1^{L[G]} = \omega_2^L$. The same argument shows that if in $V$, for every ordinal $\lambda < \omega_2^L$ we had a $\lambda$-Erdős cardinal, then there is a model in which $\omega_1 = \omega_2^L$, and for every countable ordinal $\alpha$ there is a $\alpha$-Erdős cardinal.<|endoftext|> TITLE: Can monomial representations induced from nonmonomial representations? QUESTION [7 upvotes]: Let $H$ be a subgroup of $G$. Let $\rho$ be an irr representation of $G$ induced from an irr representation $\theta$ of $H$. It is well known that $\rho$ is monomial if $\theta$ is monomial. Is it possible that $\rho$ is monomial but $\theta$ is not monomial? In other words, can monomial representations induced from nonmonomial representations? REPLY [8 votes]: As Amritanshu Prasad points out, the $6$-dimensional irreducible complex representation of $S_{5}$ is indeed monomial (with respect to a suitable basis), and thinking about how to prove this directly led me to a general observation: let $G$ be a finite group, and $\chi$ be a non-linear complex irreducible character of $G$ of minimal degree. Let $H$ be a proper subgroup of $G$ of index less than $2\chi(1).$ Let $\lambda$ be a linear character of $H$ of order $m$ ( that is, $\lambda^{m}$ is the trivial character, but no smaller positive integer power of $\lambda$ is trivial). Then if $m$ does not divide $[G:G^{\prime}],$ the induced character $\theta = {\rm Ind}_{H}^{G}(\lambda)$ is irreducible. For otherwise, (by Frobenius reciprocity), there must be a linear constituent $\mu$ of $\theta$ such that $\langle {\rm Res}^{G}_{H}(\mu),\lambda \rangle \neq 0.$ But then the order of $\mu$ must be divisible by $m,$ whereas the order of $\mu$ divides $[G:G^{\prime}],$ a contradiction. To apply this result to $G = S_{5},$ take $H$ to be the normalizer of a Sylow $5$-subgroup. Then $[G:G^{\prime}] = 2,$ and $H$ has a linear character $\lambda$ of order $4,$ so the above result can be used to see that ${\rm Ind}_{H}^{G}(\lambda)$ is irreducible (for note that $S_{5}$ has no non-linear irreducible character of degree less than $4$). Incidentally, taking $G = A_{5},$ and $K$ to be a Sylow $5$-normalizer of $G$, we see that $[G:G^{\prime}] = 1,$ that $[G:K] = 6$ and that $G$ has no non-linear irreducible character of degree less than $3.$ However, $K$ has a linear character of order $2$ such that ${\rm Ind}_{K}^{G}(\lambda)$ is not irreducible ( it is a sum of two irreducible characters of degree $3$). Hence the bound $[G:H] < 2 \chi(1)$ in the result above can't be sharpened in general.<|endoftext|> TITLE: Diffusion processes with different diffusion coefficients and absolute continuity QUESTION [6 upvotes]: I would first of all like to say that I am an analyst, and so I am familiar with probabilistic methods only on a basic level. My initial situation is the following. Consider two stochastic differential equations: \begin{align} dX_t &= f(X_t) dt + \sigma(X_t) dW_t\\ dY_t &= f(Y_t) dt + \tilde\sigma(Y_t) dW_t, \end{align} where $W_t$ is the standard 1D-Brownian motion, the initial data for $Y$ and $X$ are the same and the drift $f$ and the diffusion coefficients $\sigma,\tilde\sigma$ have sufficiently nice properties. Now assume that (for some reason) we know that the measures corresponding to this processes are mutually absolutely continuous. Question: What can I conclude about the diffusion coefficients $\sigma,\tilde\sigma?$. Do they have to be equal (maybe only up to some symmetries or something like that)? I appreciate any hint or reference. Thanks in advance. REPLY [4 votes]: Well, if there are regions on the real line where the processes never get (which can be the case, e.g. if $f\equiv 1$ and $\sigma$ has zeroes with fast enough decay), then, clearly, you cannot say anything about the $\sigma$'s there. Otherwise the quadratic variation is the invariant that ensures $\sigma=\pm\tilde{\sigma}$. Suppose that $|\sigma(x)| 0$ also $|\sigma(X_t)| b^2\epsilon$, which is a contradiction.<|endoftext|> TITLE: Does $\binom{2n}{n} \equiv 2 \pmod p$ ever hold? QUESTION [21 upvotes]: Well, the title does not tell the whole story; the complete question is: Are there any primes of the form $p=2n(n-1)+1$, with integer $n\ge 1$, such that $$ \binom{2n}{n} \equiv 2\pmod p ? $$ Primes $p=2n(n-1)+1$ are not that rare: say, of all numbers of this form up to $2\cdot10^{10}$, some 12.7% are prime; however, for none of these primes the congruence above holds true. A simple heuristic is as follows. Let's say that a prime is bad if it satisfies the congruence. If the binomial coefficients $\binom{2n}{n}$ were distributed uniformly modulo $p$, then the probability that a particular prime $p$ is bad would be about $1/p$. Also, the probability that bad primes exist does not exceed the sum of probabilities for all primes $p=2n(n-1)+1>2\cdot10^{10}$ to be "bad". Hence, this probability is at most $$ \sum_{n>10^5} \frac1{2n(n-1)+1} < 5\cdot 10^{-6}. $$ REPLY [2 votes]: Confirmed up to $n = 1e6$. No example found. But note that there is probably nothing special about the number $2$. The same holds (also up to $n = 1e6$) with $2$ replaced by $3, 4, 5, 8$.<|endoftext|> TITLE: Special Kähler normal coordinates around a point QUESTION [5 upvotes]: Let $(M,\omega)$ be a compact Kähler manifold and suppose there are holomorpic vector fields vanishing at a point $p$. As a consequence we have a group $G_{p}$ of biholomorpisms fixing $p$. Let $T_{p}$ a maximal torus of $G_{p}$. Suppose $\omega$ is invariant for the action of $T_{p}$. Can we find a neighborhood $\mathfrak{U}$ of $p$ and a set of holomorphic coordinates $z$ centered at $p$ such that the action of $T_{p}$ is linear (unitary) and coordinates $z$ are Kähler normal coordinates i.e. \begin{equation} \omega=i\partial\overline{\partial}\left(\frac{|z|^{2}}{2}+\mathcal{O}(|z|^{4})\right) \end{equation} Separately these two requests can be satisfied, but I can't prove that can hold simultaneously. Is there a reference for this? Or can someone give me a hint? REPLY [4 votes]: I don't know a reference, but this desired normal form is, indeed, attainable. Here is the argument: Assume given a Kähler form $\omega$ defined on a neighborhood of $0\in\mathbb{C}^n$ and that there is a torus (i.e., a connected compact abelian Lie group) $\mathbb{T}$ acting effectively and holomorphically on $\mathbb{C}^n$ as $\omega$-isometries and fixing $0\in\mathbb{C}^n$. As the OP points out, it is known that there exist holomorphic coordinates $z = (z^r)$ centered on $0$ in which the action of $\mathbb{T}$ is linear, i.e., that there exist $n$ characters (i.e., homomorphisms) $\chi_r:\mathbb{T}\to S^1$ such that $$ a^*(z^r) = \chi_r(a)\,z^r\qquad \text{(no sum)} $$ for all $a\in \mathbb{T}$. Now, write $\omega = \mathrm{i}\partial\bar\partial\phi$, where $\phi$ is a Kähler potential for $\omega$ in a neighborhood of $0$. Since $\omega$ satisfies $a^*\omega = \omega$ for all $a\in\mathbb{T}$, by averaging $\phi$ with respect to $\mathbb{T}$, one can assume that $\phi$ is also $\mathbb{T}$-invariant. After discarding holomorphic and anti-holomorphic terms (which are $\mathbb{T}$-invariant and don't show up in $\omega$), it can be assumed that $\phi$ has a Taylor expansion up to terms vanishing to order $4$ of the form $$ \phi \equiv_4 h_{p\bar q}\,z^p\overline{z^q} + a_{pq}^r\,z^pz^q\overline{z^r} + \overline{a_{pq}^r}\,\overline{z^p}\overline{z^q}z^r $$ where $\overline{h_{p\bar q}} = h_{q\bar p}$ is positive definite, and $a_{pq}^r = a_{qp}^r$. Now, because $\phi$ is $\mathbb{T}$-invariant, it follows that $h_{p\bar q} = 0$ unless $\chi_p = \chi_q$. Thus, by collecting the $z^j$ that share the same character and making a 'block-form' linear transformation that preserves these subsets, one can arrange that the $z^p$ have been chosen so that $h_{p\bar q} = \delta_{pq}$, i.e., that the quadratic term in $\phi$ is simply $\tfrac12\bigl(|z^1|^2+\cdots+|z^n|^2\bigr)$. Now, again because of the $\mathbb{T}$-invariance of $\phi$, one sees that $a_{pq}^r = 0$ unless $\chi_r = \chi_p\chi_q$. Now, consider the change of variables of the form $$ w^r = z^r + 2a_{pq}^r\,z^pz^q. $$ Because $a_{pq}^r$ vanishes unless $\chi_r = \chi_p\chi_q$, it follows that $a^*(w^r) = \chi_r(a)\,w^r$ for all $r$, so the $w$-coordinates also linearize the action of $\mathbb{T}$. Moreover, one clearly has $$ \phi \equiv_4 \tfrac12\bigl(|w^1|^2+\cdots+|w^n|^2\bigr). $$<|endoftext|> TITLE: Is there a non-compact Poulsen simplex? QUESTION [5 upvotes]: A Choquet simplex is a closed, convex and metrizable subset of a locally convex Hausdorff topological vector space in which every point is a barycenter of a unique probability measure supported on the set of extreme points. The Poulsen simplex is a unique nontrivial compact Choquet simplex with a dense set of extreme points. This was proved by Lindenstrauss, Olsen and Sternfeld (Ann. Inst. Fourier (Grenoble) 28 (1978), no. 1, vi, 91–114); see also http://www.ams.org/mathscinet-getitem?mr=500918. The Poulsen simplex has many remarkable properties. Is there a similar object in the category of not necessarily compact (but bounded) Choquet simplices? REPLY [4 votes]: This is not a complete answer but rather a long comment. The classical (compact) Poulsen simplex is a very homogeneous object. This heuristic observation can be indeed made precise via the theory of Fraïssé limits. Clinton Conley and Asger Törnquist proved that the Poulsen simplex is indeed a Fraïssé limit for the class of compact, finite-dimensional simplices. (This is not published yet as far as I know.) I am not sure what do you mean a non-compact simplex but the general rule of thumb to define a non-compact version of the Poulsen simplex should be as follows: consider the class of non-compact simplices that you consider small / finitary (simplices in $\mathbb{R}^n$ with removed boundary?), check whether this class has the amalgamation property, if so, think of the Fraïssé limit of this class as a variant of the Poulsen simplex appropriate for your class of finitary simplices.<|endoftext|> TITLE: Are there noncommutative extensions of $\alpha_p$ by $\mathbb{G}_m$? QUESTION [9 upvotes]: Let $k$ be a field of characteristic $p > 0$ (algebraically closed, if you want; that doesn't make a difference). Consider a finite type $k$-group scheme $E$ that is a (central) extension of $\alpha_p$ by $\mathbb{G}_m$. Is $E$ necessarily commutative? Edit: $E$ is an extension of $A$ by $B$ if it fits into a short exact sequence (which is part of the data of an extension) $$ 1 \rightarrow B \rightarrow E \rightarrow A \rightarrow 1, $$ and the extension is central if $B$ is in the center of $E$. So in the case at hand I am looking at central extensions $$ 1 \rightarrow \mathbb{G}_m \rightarrow E \rightarrow \alpha_p \rightarrow 1. $$ REPLY [18 votes]: So, bilinear maps $\alpha_p \times \alpha_p \rightarrow \mathbf{G}_m$ are classified by maps from $\alpha_p$ to itself (since $\alpha_p$ is Cartier self-dual). The collection of such maps is a $1$-dimensional vector space $V$ over $k$. The group $\mathbf{Z}/ 2 \mathbf{Z}$ acts on the vector space $V$ by "swapping'' the two factors of $\alpha_p$. This action is trivial (this follows from the fact that you can write down the Cartier self-duality on $\alpha_p$ in a symmetric way). It follows that if $p \neq 2$, there are no nonzero skew-symmetric bilinear maps from $\alpha_p$ to $\mathbf{G}_m$, so any central extension of $\alpha_p$ by $\mathbf{G}_m$ must be commutative. If $p = 2$, then the symmetric bilinear maps $\alpha_2 \times \alpha_2 \rightarrow \mathbf{G}_m$ are also skew-symmetric, so this argument does not apply. It is still true that there are no noncommutative central extensions of $\alpha_2$ by $\mathbf{G}_m$, but you have to work a little bit harder to show this. You can find an argument in Example 3.2.7 of my paper with Mike Hopkins "Ambidexterity in K(n)-Local Stable Homotopy Theory" (we need this fact to discuss "alternating powers" of 1-dimensional p-divisible groups, which arise naturally from some calculations in algebraic topology).<|endoftext|> TITLE: Forcing is intuitionistic QUESTION [9 upvotes]: The main idea of why it´s necessary a generic filter $G$ to extend a countable transitive $\epsilon$-interpretation (not necessarily a model) $M$ is given by the condition (for which $G$ being a generic filter can be derived): For each formula $\varphi(x_1, ..., x_n)$ there exists a formula $Q_{\varphi} (x, x_1, ..., x_n)$ (usually denoted by $x \Vdash \varphi (x_1, ...x_n)$), such that $$\forall m_1, ..., m_n \in M(\varphi^{ M[ G ] } (F_G (m_1), ...,F_G (m_n)) \equiv \exists p \in G; Q_{\varphi}^M (p, m_1, ..., m_n)))$$ where $F_G$ is the collapsing function from $M$ with the relation $R_G (x, y) : \exists p \in G; (p, x) \in y $. However $Q_{\neg \neg \varphi} \not\equiv Q_{\varphi}$, so it´s intuitionistic in some sense. I think that it´s what inspired the Kripke semantics. But I don´t think it can be mere coincidence. So the question is Is there a more philosophical reason for why forcing is intuitionistic? Thanks in advance. REPLY [11 votes]: It seems to me that the "more philosophical" reason why forcing is intuitionistic is that forcing and intuitionistic logic have similar interpretations of the logical connectives and quantifiers. This is also why forcing is related to Kripke models and why Kripke models give semantics for intuitionistic logic; and it is also why intuitionistic logic is related to realizability. From an informal perspective, these all have a viewpoint that aligns closely with the BHK interpretation of intuitionistic logic. For example, an intuitionistic proof of $(\exists x)\phi(x)$, by the BHK interpretation, consists of an object $c$ and a proof of $\phi(c)$. If we apply the same intuition to forcing, we would suspect that if a condition $p$ "forces" $(\exists x)\phi(x)$ to hold, there should be a term $t$ such that $p$ forces $\phi(t)$ to hold. (This intuition can be complicated, in the case of set-theoretic forcing, by the fact that the term $t$ will be a $P$-name. But for recursion-theoretic forcing, $t$ may well just be a natural number.) This intuition about forcing leads to a forcing relation known as "strong forcing", which is what we obtain if we define forcing by directly considering the intuitionistic meaning of each of the connectives, as in the BHK interpretation. This will look very similar to the definition of the forcing relation in a Kripke model. We can then obtain a "weaker" classical forcing relation by using a double-negation translation to embed classical logic into the intuitionistic logic of the strong forcing relation. This classical forcing relation is weaker in that it allows a condition to force a formula (such as a disjunction or an existential formula) without providing as much evidence about why the formula will hold. For example, in recursion theoretic Cohen forcing over a model of PA to construct a generic $G \subseteq \mathbb{N}$, with the classical forcing relation $\Vdash$, we have $\langle\rangle \Vdash [5 \in G \lor 5 \not \in G]$, even though $\langle\rangle \not \Vdash 5 \in G$ and $\langle\rangle \not\Vdash 5 \not \in G$. This is because $\langle\rangle $ strongly forces $\lnot\lnot (5 \in G \lor 5 \not \in G)$. Similarly $\langle\rangle$ forces $(\exists x)[x \in G]$ without strongly forcing any particular number to be in $G$, because $\langle\rangle$ does not strongly force $G$ to be empty. In modern accounts of forcing, when the goal is to obtain classical models, the authors often bypass strong forcing entirely, and begin with a different definition of forcing that somehow has double-negation translation embedded into its definition.<|endoftext|> TITLE: Question regarding 2-mathematics: Can you stackify a 2-functor without prestackifying it first? QUESTION [6 upvotes]: Let $C$ be a site and $CAT$ the 2-category of categories. Given a contravariant 2-functor $A:C\rightarrow CAT$, we can of course consider the associated stack. This is done by first considering the associated prestack, denoted by $\hat{A}$, (i.e. sheafifying the hom) and then considering $2$-$lim_\mathfrak{U}Des(\mathfrak{U},\hat{A})$, the descent data of $\hat{A}$ with respect to the covering $\mathfrak{U}$. EDIT: My question is, could we stackify $A$ purely with the descent data? In other words, define $A'(U):=2$-$lim_\mathfrak{U}(Des(\mathfrak{U},A))$, $A''(U):=2$-$lim_\mathfrak{U}(Des(\mathfrak{U},A'))$ and $A'''(U):=2$-$lim_\mathfrak{U}(Des(\mathfrak{U},A''))$. Would $A'''$ be equivalent to the associated stack of $A$?. Thank you in advance. REPLY [9 votes]: The three-step process is given as Theorem 3.8 in Ross Street, Two dimensional sheaf theory, J. Pure Appl. Algebra 23 (1982) 251–270 under some smallness assumptions. Without some smallness hypotheses, you may run into set-theoretic problems. For example, Waterhouse produced a presheaf with no flat sheafification for Theorem 5.5 in Basically bounded functors and flat sheaves Pac. J. Math. 57 no. 2 (1975) 597-610. There is a one-step process using hypercovers instead of covers. As Zhen Lin mentioned in a comment, this is discussed in the beginning of section 6.5.3 in Lurie's Higher Topos Theory.<|endoftext|> TITLE: Manifolds admitting flat connections QUESTION [20 upvotes]: For each Riemannian manifold one can construct the Levi-Civita connection. While this connection is unique, we can call a (Riemannian) manifold flat if the Levi-Civita connection is flat. However when one tries to think about flatness in terms of pure manifolds (not manifolds plus metric tensor/connection) the natural questions arises: Question 1 Are there manifolds $M$ with the property that each connection on $M$ is never flat? Question 2 Is it possible to find a manifold $M$ with the following property: for each Riemannian metric tensor $g$ the corresponding Levi Civita connection is not flat but still there are flat torsion-free connections on $M$? Question 3 Is it possible to find a manifold $M$ with the following property: each torsion free connection cannot be flat but still there are some flat connections on $M$? REPLY [7 votes]: I will add some detail to the previous answers by abx, Robert Bryant (in the linked question), and V. S. Matveev. Regarging question 1: A simply connected manifold $M$, admits a flat connection if an only if it is parallelizable. Indeed, you can parallel-transport a frame (a basis of the tangent space) from some initial point $p$ to every point $q$ along any curve $\gamma$. The fact that the connection is flat implies that the frame that you obtain at the endpoint $q$ won't vary if you perform a continuous variation (homotopy) of the curve $\gamma$; it only depends on the homotopy class. But there's only one homotopy class, so the frame at $q$ is independent of the choice of $\gamma$, and you get a basis of vector fields $X_i$ that are parallel with respect to the connection: $\nabla_YX_i=0$ for all $X_i$ and $Y$. For examples, observe that all the spheres $S^n$ with $n\geq 2$ are simply connected, and the only parallelizable spheres are $S^1$, $S^3$ and $S^7$. If $n$ is even, you can quickly see that there is not even one nonvanishing vector field, by the Poincaré-Hopf theorem, because the Euler characteristic is 2. Regarding question 3: You can furthermore prove that spheres of dimenison $n\geq 2$ don't admite ­­­­symmetric flat connections. Indeed, if the connection was symmetric (zero torsion), then the vector fields that you obtain in last construction commute: $[X_i,X_j]=0$. Now, since each $X_i$ gives rise to an action of $\mathbb R$ on the manifold $M$, and the actions commute, you get an action of $\mathbb R^n$ by "translations" in the directions of the vector fields $X_i$, the translation by a vector $w\in\mathbb R^n$ being the result of travelling along $M$ in the direction of the vector field $\sum_i w_iX_i$ during one unit of time. The orbits of this action are open sets because the action is locally free (you can act on a point to "wiggle" it in every direction because the vector fields form a basis of the tangent space), and there is only one such orbit because $M$ has one connected component (so the action is transitive). The only compact $n$-manifold that admits a locally free, transitive Lie group action of $\mathbb R^n$ is the $n$-torus, which is not simply connected. So a compact, simply connected manifold can't have a symmetric flat connection. Regarding question 2: To see that $S^2\times S^1$ does not admit a flat metric, observe that if it did, then by the Cartan-Hadamard theorem, its universal cover would be diffeomorphic to $\mathbb R^3$. But it's universal cover is $S^2\times\mathbb R$. As V.S. Matveev observes in his comment below, the Cartan-Hadamard argument can also be used to prove that spheres don't admit a symmetric flat connection.<|endoftext|> TITLE: Milnor-Wolf result on growth of solvable groups QUESTION [6 upvotes]: The Milnor-Wolf theorem says that any solvable group has either polynomial or exponential growth. I wonder about the existence of alternative proofs of this fact. I have an impression that the original proof is more brutal and less ideological. If you have a citation or a suggestion of an alternative proof, please post. REPLY [3 votes]: There seems to be a different proof of a sharper result by our own T. Tao: http://terrytao.wordpress.com/tag/milnor-wolf-theorem/<|endoftext|> TITLE: Concise definition of subobjects QUESTION [7 upvotes]: Higher category theory tells us that it is a bad idea to identify isomorphic things. Rather, the isomorphism should belong to some additional data. Also, categorification tells us that one should, whenever possible, look at a category directly, not just on its set of isomorphism classes. These are two well-known and accepted principles, right? However, the definition of a subobject seems to contradict these principles. Why should two monomorphisms be identified with each other when they are isomorphic? What is wrong with the following definition: A subobject of an object $B$ is a monomorphism $A \to B$. With this definition, you can do categorical algebra as usual. It also works nicely in examples. For example, a subring of a ring $B$ is just an injective homomorphism of rings $A \to B$. I think this is more abstract but also more natural than the usual set-theoretic definition; for example with $\mathbb{C}=\mathbb{R}[x]/(x^2+1)$ it is not correct that $\mathbb{R}$ is a subring of $\mathbb{C}$ in the set-theoretic sense, but rather in the sense defined above, via the canonical homomorphism $\mathbb{R} \to \mathbb{C}$. For what purpose should I now look at the class of all monomorphisms $\mathbb{R} \to \mathbb{C}$ isomorphic to that? (This is a bad example since there is only one homomorphism $\mathbb{R} \to \mathbb{C}$ anyway, but I hope that my point is clear. Otherwise consider $\mathbb{Z}[i] \to \mathbb{C}$, $i \mapsto \pm i$.) REPLY [8 votes]: Of course it's not necessary to make this identification, but it's fairly harmless since the groupoid of monomorphisms into an object $X$ is equivalent to the discrete category of subobjects, and it can be a slight technical convenience, especially in relation to smallness conditions. For example, we say that a category is well-powered if for each object the class of its subobjects is "small" (is a set), and this is convenient for example when discussing certain forms of adjoint functor theorems, etc. In topos theory, assuming that a topos $E$ is well-powered (e.g., a Grothendieck topos), one way of describing a subobject classifier $\Omega$ is that the contravariant subobject functor $Sub: E^{op} \to Set$ is representable as $\hom(-, \Omega)$. This description is conceptually convenient to some people's taste.<|endoftext|> TITLE: What is an infinite prime in algebraic topology? QUESTION [42 upvotes]: The links between algebraic topology (stable homotopy theory in particular) and number theory are nowadays abundant and fruitful. In one direction, there is chromatic homotopy theory, exploiting the theory of formal groups, local fields and moduli spaces. In another one there is a whole subject of derived algebraic geometry, as well as more esoteric objects like chiral homology and their applications. Perhaps this is wishful thinking, but I would really like to view these two fields as different facets of some deeper theory. In particular, since the days of yore does AT employ a theory of localizations and completions at ideals of $\mathbb Z$. This already makes it feel as if we're really doing algebraic geometry over $\mathrm{Spec}\ \mathbb Z$. This feeling is supported by Nishida's theorem, which tells us that the sphere spectrum can be really thought about as some very nice nilpotent thickening of $\mathbb Z$. With a bit of white magic we can even apply local class field theory to the study of homotopy groups. However, there is a glaring issue: in number theory we really should be working not over $\mathbb Z$, but over some compactification of it, and should include information at infinite primes. Surprisingly, I have never seen infinite primes mentioned in homotopy theory. An obvious guess that we should study localizations w.r.t. $H\mathbb R$ instead of $H\mathbb Q$ or $H\mathbb{F}_p$ fails for a simple reason that rational cohomology 'equal' real and complex ones. Trying to google for something like "infinite primes in algebraic topology" or "infinite prime Bousfield localization" returned zero relevant results. My knowledge about infinite primes in number theory is very limited, but as far as I understand constructions mostly revolve around completions with respect to archimedean norms on extensions of $\mathbb Q$ and studying real vector bundles with metric (well, it's mostly the same). I see no way to push either of these approaches to homotopy (metric cohomology? wut?). Thus the question as stated: is there any theory exploiting some constructions (especially some form of localization) with respect to infinite primes in number fields to gain homotopy-theoretic results? Some variant of topological Arakelov geometry would be close to the ideal result, but I don't expect that it exists, so would welcome any leading threads. More broadly, what could take the role of those infinite primes and supply the missing (in what sense?) homotopy-theoretic information? Of course, one could note that perhaps a more basic question is how the Galois extensions of $\mathbb Q$ can be generalized to spectra (because otherwise there are not many potential infinite primes around), but I don't feel like it should really be an obstruction. In any case it should be a subject of another question. REPLY [10 votes]: The only result in algebraic topology I know that explicitly involves both finite and infinite primes is Dustin Clausen's lift of Hilbert reciprocity to a statement about spectra. For each prime $p$ he introduces a $p$-adic version of the J-homomorphism and proves a product formula over all J-homomorphisms, including the usual one over $\mathbb{R}$, which reduces to Hilbert reciprocity after applying $\pi_2$.<|endoftext|> TITLE: Consistency of Weak Diamond with a Weak Version of Martin's Axiom QUESTION [6 upvotes]: If $S \subset \omega_1$ is stationary, then the weak diamond principle $\Phi(S)$ states that for any $F: 2^{<\omega_1} \to 2$, there is a $g: \omega_1 \to 2$ such that for all $f: \omega_1 \to 2$, the set $\{\alpha \in S: F(f \restriction_\alpha) = g(\alpha)\}$ is stationary. Let $\Phi^*$ be the statement: for all stationary $S \subseteq \omega_1$, $\Phi(S)$. Let $MA^*$ be the following weakening of Martin's Axiom: $\mathbb{R}$ is not the union of $\omega_1$ nowhere dense sets. (Equivalently, $MA^* = MA_{\omega_1}($countable POs$)$; or $MA^* =``\mbox{cov}(\mathcal{B}) \geq \omega_2."$) I would like to know the consistency of $\Phi^* \land MA^*$. (If consistent this would have applications in model theory.) Remark 1. It is consistent that $\Phi(\omega_1)$ and $MA^*$ holds; for example, start with $\mathbb{V} = \mathbb{L}$ and force over $\mathsf{Fn}(\omega_{\omega_1}, \omega, \omega)$, or over $\mathsf{Fn}(\omega_2, \omega, \omega) \times \mathsf{Fn}(\omega_3, \omega_1, \omega_1)$. For all I know, both of these forcing extensions witness the consistency of $\Phi^* \land MA^*$. (Here $\mathsf{Fn}(I, J, \lambda)$ is the set of all partial functions from $I$ into $J$ of cardinality less than $\lambda$.) Remark 2. A related, and probably easier problem is: suppose we start with $\mathbb{V} = \mathbb{L}$ and force over $\mathsf{Fn}(\omega_1, \omega, \omega)$ to get $\mathbb{V}[G]$. Then $\mathbb{V}[G] \models \diamondsuit$. But does $\mathbb{V}[G] \models \forall \mbox{ stationary } S \subseteq \omega_1, \diamondsuit(S)$? REPLY [2 votes]: So I was looking through related questions on this site and the book "Proper and Improper Forcing" by Shelah kept popping up. So I checked it out and the appendix actually resolves the question. I rephrase the proof there because I think this way it's simpler and also because it suggests an interesting question (see the end). Let $P_0 := \mathsf{Fn}(\omega_2, 2, \omega)$ and let $P_1 := \mathsf{Fn}(\omega_3, 2, \omega_1)$. The following lemma is basic: Lemma $P_0 \Vdash MA^*$. Proof. Let $G$ be $P_0$-generic over $\mathbb{V}$. Let $f: \omega_1 \to 2$ be the generic function. Suppose $X \subseteq \omega_1$ codes an $\omega_1$-sequence of open dense sets $(\mathcal{O}_\alpha: \alpha < \omega_1)$ of $(\,^\omega \omega)^{\mathbb{V}[G]}$. Then for some $I \subset \omega_2$ with $|I| = \aleph_1$, we have $X \in \mathbb{V}[G \cap \mathsf{Fn}(I, 2, \omega)]$. But let $\alpha = (\sup I) + 1$ and let $x \in \,^\omega \omega$ be defined by $x(n) = f(\alpha + n)$. Then $x \in \bigcap_{\alpha < \omega_1} \mathcal{O}_\alpha$. End proof of lemma. We aim to show that $P_0 \times P_1 \Vdash MA^* \land \Phi^*$ provided we start with a model of $CH$. This proof is gleaned from the proof of Theorem 2.11 in the Appendix of Shelah's "Proper and Improper forcing." Customarily we view this iterated forcing as starting with $P_1$ and following it by $P_0$, since $(P_0)^{\mathbb{V}^{P_1}} = P_0$. However to get $\Phi^*$ we have to view it the other way around. So we need to look at $P_1$ in $\mathbb{V}^{P_0}$. Lemma. Suppose $G_0$ is $P_0$-generic over $\mathbb{V} \models CH$. Then in $\mathbb{V}[G_0]$: (a) $\mathsf{Fn}(\omega_3, 2, \omega) \subset P_1 \subset \mathsf{Fn}(\omega_3, 2, \omega_1)$, and for any two elements $p, q \in P_1$, if $p$ and $q$ are compatible then $p \cup q \in P_1$, (b) $P_1$ has the $\omega_2$ c.c., (c) Forcing with $P_1$ does not add any reals, (d) The set $\mathcal{B} := \{\mbox{dom}(p): p \in P_1\}$ is closed under intersections, unions, and relative complements, and if $p \in P_1$ and $X \in \mathcal{B}$, then there is some $q \, || \, p$ with $\mbox{dom}(q) = X$, (e) If $X \subseteq \omega_3$ is countable there is some (countable) $Y \in \mathcal{B}$ containing $X$. Proof. (a) Obvious. (b) Showing that $(Q \mbox{ has the $\omega_2$ c.c.})^{\mathbb{V}[G_0]}$ is the same as showing it in $\mathbb{V}$ (just go through the proof of the $\Delta$-system lemma and see that it works for any antichain in $Q$). (c) Let $G_1$ be $Q$-generic over $\mathbb{V}[G_0]$. Then also $G_1$ is $Q$-generic over $\mathbb{V}$, and $G_0$ is $P$-generic over $\mathbb{V}[G_1]$ and $G_0 \times G_1$ is $P \times Q$-generic over $\mathbb{V}$. Now it suffices to show that if $x \in \mathcal{P}(\omega) \cap \mathbb{V}[G_0 \times G_1]$ then $x \in \mathcal{P}(\omega) \cap \mathbb{V}[G_0]$. To see this, consider a $P$-nice name $\sigma$ for $x$ in $\mathbb{V}[G_1]$; so $\sigma = \{(\hat{n}, A_n): n \in \omega\}$ where $A_n \in \mathbb{V}[G_1]$ is an antichain in $P$. So $A_n \subseteq P$ is countable, so $A_n \in \mathbb{V}$ already (since $(Q \mbox{ is $\omega$-closed})^{\mathbb{V}}$). Hence $\sigma \in V$ so $x \in \mathbb{V}[G_0]$. (d) Obvious. (e) The existence of $Y$ follows from the c.c.c. of $P_0$. End proof of lemma. Lemma. Suppose $2^{\aleph_0} \leq \aleph_2$ and $Q$ is a forcing notion satisfying $(a)$ through $(e)$ above. Then $Q \Vdash \Phi^*$. Proof. Suppose towards a contradiction that $1_Q \Vdash \lnot \Phi^*$. Then (by the maximal principle) there are names $\dot{F}, \dot{S}$ such that \begin{eqnarray*} 1_Q &\Vdash& ``\mbox{ $\dot{F}: \,^{<\omega_1} 2 \to 2$ and $\dot{S} \subseteq \omega_1$ is stationary and } \\ && \forall g: \omega_1 \to 2 \,\,\, \exists f: \omega_1 \to 2 \mbox{ such that } \{\alpha \in \dot{S}: \dot{F}(f \restriction_\alpha)= g(\alpha)\} \\ && \mbox{ is nonstationary}." \end{eqnarray*} We can choose $\dot{F}$ and $\dot{S}$ to be nice names; that is $\dot{F} = \{(\hat{(\eta, i)}, A_{\eta, i}): \eta \in \,^{<\omega_1} 2, i \in 2\}$ and $\dot{S} = \{(\hat{\alpha}, A_\alpha) : \alpha < \omega_1\}$, where each $A_{\eta, i}$ and each $A_\alpha$ is an antichain in $Q$. (This uses that $(\,^{<\omega_1} 2)^{\mathbb{V}} = (\,^{<\omega_1} 2)^{\mathbb{V}^Q}$. Let $\mathcal{A} = \bigcup_{\eta, i} A_{\eta, i} \cup \bigcup_{\alpha} A_\alpha$; so by the $\omega_2$-c.c., $|\mathcal{A}| \leq \aleph_2$. Hence $D:= \bigcup_{A \in \mathcal{A}} \mbox{dom}(A)$ has cardinality at most $\aleph_2$. Hence after relabeling we can suppose that $D \cap \omega_1 = \emptyset$. Let $\dot{c}$ be the nice $Q$-name for the generic $\omega_3$-sequence: $\dot{c} = \{(\hat{(\alpha, i)}, \{(\alpha, i)\}): \alpha < \omega_3, i \in 2\}$. Let $\dot{g} = \dot{c} \restriction_{\omega_1}$, i.e. $\dot{g} = \{(\hat{(\alpha, i)}, \{(\alpha, i)\}): \alpha < \omega_3, i \in 2\}$. Then $1_Q \Vdash ``\exists f: \omega_1 \to 2$ such that $\{\alpha \in \dot{S}: \dot{F}(f \restriction_\alpha) = g(\alpha)\}$ is nonstationary." So for some nice $Q$-name $\dot{f} = \{(\hat{(\alpha, i)}, B_{\alpha, i}): \alpha < \omega_1, i \in 2\}$, we have $1_Q \Vdash ``\dot{f}: \omega_1 \to 2$ and $\{\alpha \in \dot{S}: \dot{F}(\dot{f} \restriction_\alpha) = \dot{g}(\alpha)\}$ is nonstationary." Let $\dot{C}$ be a nice name for a club subset of $\omega_1$ such that $1_Q \Vdash ``\forall \alpha \in \dot{S} \cap \dot{C}: \dot{F}(\dot{f} \restriction_\alpha) \not= \dot{g}(\alpha)$. Say $\dot{C} = \{(\hat{\alpha}, B_\alpha): \alpha < \omega_1\}$. Let $G$ be $Q$-generic over $\mathbb{V}$; we work in $\mathbb{V}[G]$. Let $F = \dot{F}_G$, let $S = \dot{S}_G$, let $g = \dot{g}_G$, let $f = \dot{f}_G$ and let $C = \dot{C}_G$. For each $\alpha < \omega_1$, let $p_\alpha$ be the unique element of $G \cap (B_{\alpha, 0} \cup B_{\alpha, 1})$ and let $q_\alpha$ be the unique element of $G \cap B_\alpha$ if it exists, or $q_\alpha = 1_Q$ if $G \cap B_\alpha = \emptyset$. Let $D_\alpha = \mbox{dom}(p_\alpha) \cup \mbox{dom}(q_\alpha)$ and let $X_\alpha = \bigcup_{\beta < \alpha} D_\beta$. Then the set $C^* = \{\alpha < \omega_1: X_\alpha \cap \omega_1 \subseteq \alpha\}$ is club. So there is some $\delta \in S \cap \mbox{acc}(C) \cap C^*$. (In the following we use the properties $(a), (d)$ and $(e)$ heavily.) Let $I \subset \omega_3$ be countable such that $I \supseteq \delta \cup \{X_\alpha \backslash \omega_1\}$ and $I \in \mathcal{B}$ (where $\mathcal{B}$ is as in the lemma). Let $J = I \backslash \{\delta\}$ (possibly $I = J$); so $J \in \mathcal{B}$. Let $p_0$ be the unique element of $G \cap A_\delta$, so $p_0 \Vdash \delta \in \dot{S}$. Let $p_1 = g \restriction_J$, so $p_1 \Vdash \dot{f} \restriction_\delta = f \restriction_\delta$ and $p_1 \Vdash \dot{C} \cap \delta = C \cap \delta$. Since $\delta \in \mbox{acc}(C)$, $p_1 \Vdash \delta \in \dot{C}$. Let $p = p_0 \cup p_1$; note that $\delta \not \in \mbox{dom}(p)$. Back in $\mathbb{V}$, $p \Vdash ``\hat{\delta} \in \dot{S} \cap \dot{C} \land \dot{F}(\dot{f} \restriction_\delta) = i"$, for some $i \in 2$. But let $q = p \cup \{(\delta, i)\}$; then $q \vDash \dot{F}(\dot{f} \restriction_\delta) = \dot{g}(\delta)$, contrary to definition of $\dot{C}$. End proof of lemma. Theorem. Suppose $\mathbb{V} \models CH$. Then $P_0 \times P_1 \Vdash MA^* \land \Phi^*$. Proof. Let $G_0 \times G_1$ be $P_0 \times P_1$-generic over $\mathbb{V}$. By Lemma 1, $\mathbb{V}[G_0 \times G_1] = \mathbb{V}[G_1][G_0] \models MA^*$. By Lemma 2, $\mathbb{V}[G_0 \times G_1] = \mathbb{V}[G_0][G_1] \models \Phi^*$. End proof of theorem. Question. Under what conditions can we find a $Q$ satisfying conditions (a) through (e)?<|endoftext|> TITLE: Schoenberg's rational polygon problem QUESTION [11 upvotes]: "A polygon is said to be rational if all its sides and diagonals are rational, and I. J. Schoenberg has posed the difficult question, ‘Can any given polygon be approximated as closely as we like by a rational polygon?’" This is a quote from D.D. Ang, D.E. Daykin, and T.K. Sheng. "On Schoenberg's rational polygon problem." Journal of the Australian Mathematical Society 9.3-4 (1969): 337-344. (journal link.) I seek to understand the current status of this question. I know (from D.E.Daykin, "Rational polygons," (journal link)) that Mordell showed that the set of all rational quadrilaterals is dense in the set of all quadrilaterals. I am only finding ~40-yr-old papers, none of which are easily accessed online. (Added:) As Aaron Meyerowitz and Gerry Myerson point out, it is unknown if there even exists a single rational octagon. But I am interested in the dense question. For example, is it already known that the set of all rational pentagons is not dense in the set of all pentagons? If not for pentagons, then for hexagons? Heptagons?         MathWorld image of a rational quadrilateral REPLY [2 votes]: This is an answer to your last question. As far as I know, it is still open whether there exists a dense subset $S$ of the plane with all pairwise distances rational. Such a set $S$ would imply a positive answer to Schoenberg's Problem. Thus, no one has currently shown that the set of rational pentagons is not dense in the set of all pentagons. Same for hexagons and heptagons.<|endoftext|> TITLE: discrete group cohomology vs continuous group cohomology for profinite groups QUESTION [8 upvotes]: Let $G$ be a profinite group and $M$ be a finite $G$-module. I can compute the cohomology of $G$ with coefficients in $M$ either as a topological group or as a discrete group. There is an obvious map $H^p(G,M)\to H^p(G^{\delta},M)$ (where $G^\delta$ denotes the underlying discrete group of $G$) which forgets that a $p$-cochain is continuous. I would like to know if there are conditions on $G$ that insure that these maps are isomorphisms. In my case $G$ is finitely generated (as a topological group). REPLY [4 votes]: If $G$ is finitely generated, then $G$ is isomorphic to its profinite completion by a result of Nikolov and Segal mentioned by Ian Agol. Thus, what are you asking is equivalent to the goodness introduced by Serre (see J. P. Serre, Galois cohomology, I.2.6). The only good finitely generated profinite groups that I know are virually polycyclic (it can be proved in the same way as Theorem 2.10 in http://arxiv.org/pdf/math/0701737.pdf). In the same paper you can find many related results. For example, a non-abelian free pro-$p$ group is not good (this is a result of A.K. Bousfield, On the p-adic completions of non-nilpotent spaces, Trans. Amer. Math. Soc. 331 (1992), 335–359).<|endoftext|> TITLE: The quotient stack $[\mathbb{A}^n / \mathrm{GL}_n]$ QUESTION [18 upvotes]: Consider the affine space $\mathbb{A}^n$ (over some base scheme) with the usual $\mathrm{GL}_n$-action. What does the quotient stack $[\mathbb{A}^n / \mathrm{GL}_n]$ classify? If $n=1$, then we get $[\mathbb{A}^1 / \mathbb{G}_m]$, which classifies line bundles together with a global section, right? In general, $[\mathbb{A}^n / \mathrm{GL}_n]$ classifies vector bundles of rank $n$ together with some additional data - is it again just a global section? I would also be happy if someone can add a geometric picture of $[\mathbb{A}^n / \mathrm{GL}_n]$, at least over some algebraically closed field. What are the points of this stack, and what are their stabilizers? What's your geometric intuition for $[\mathbb{A}^n / \mathrm{GL}_n]$? PS: I am quite new to stack-land and hope that this question is not too trivial. PPS: In the comments it is suggested to do the same as for $n=1$, so let me better explain what I've done for $n=1$. A $T$-point of $[\mathbb{A}^1/\mathbb{G}_m]$ is a $\mathbb{G}_m$-bundle $P \to T$ with a $\mathbb{G}_m$-map $P \to \mathbb{A}^1$. Then $P$ corresponds to an invertible sheaf $\mathcal{L}$ on $T$ via $P = \mathrm{Spec}_T(\bigoplus_{z \in \mathbb{Z}} \mathcal{L}^{\otimes z})$. Assuming that $T$ is affine, the $\mathbb{G}_m$-map therefore corresponds to a section of $\bigoplus_{z \in \mathbb{Z}} \mathcal{L}^{\otimes z}$, say $\sum_z a_z$, which is compatible with the $\mathcal{O}(\mathbb{G}_m)$-coaction, which comes down to $\sum_z a_z u^z = \sum_z a_z u$, i.e. $a_z=0$ for $z \neq 1$. We end up with a section $a_1$ of $\mathcal{L}$. For $n >1$, I have tried the same, but the quasi-coherent algebra induced by some locally free sheaf of rank $n$, whose spectrum is the corresponding $\mathrm{GL}_n$-bundle, is quite complicated, at least globally. Locally, we just have $P = \mathrm{GL}_n \times T$, and the $\mathrm{GL}_n$-map $P \to \mathbb{A}^n$ corresponds to a map $T \to \mathbb{A}^n$, i.e. $n$ global sections of $T$, which is one section of the free sheaf $\mathcal{O}_T^n$ of rank $n$. But a) I am not sure if this glues properly, and b) I would prefer a global argument as for $n=1$. REPLY [4 votes]: Let $\text{G}$ be a group. Then maps $$S\ \longrightarrow \ \text{BG}$$ from a test scheme $S$ is the same thing as a $\text{G}$-bundle $P\to S$ (or rather the groupoid of such). Now let $X$ be another scheme with an action of $\text{G}$. Then to any $\text{G}$ bundle we can build its associated $X$ bundle: $$(P\to S)\ \rightsquigarrow \ (P\times_\text{G}X\to S).$$ This $X$ bundle has another description, as the pullback $\require{AMScd}$ \begin{CD} P\times_\text{G}X @>>> X/\text{G}\\ @V V V @VV V\\ S @> P> > \text{BG} \end{CD} Answer: Choosing a lift of the map $P$ to $$S\ \longrightarrow \ X/\text{G}$$ is thus the same as choosing a section of the $X$-bundle $P\times_\text{G}X$ associated to $P$. Examples: $\mathbf{A}^1/\mathbf{G}_m$ classifies $\mathbf{G}_m$ bundles (a.k.a. line bundles) along with a section of the associated line bundle. Similarly, $\mathbf{G}_m/\mathbf{G}_m$ classifies $\mathbf{G}_m$ bundles with a nonvanishing section. This trivialises the line bundle, which geometrically corresponds to the fact that $\mathbf{G}_m/\mathbf{G}_m=\text{pt}$ is a single point. $\mathbf{A}^n/\text{GL}_n$ classifies $\text{GL}_n$ bundles with a section of its associated vector bundle. $\mathbf{P}^1=(\mathbf{A}^2\setminus 0)/\mathbf{G}_m$ classifies line bundles together with two sections which do not vanish simultaneously. Similarly for $\mathbf{P}^n$. A variation on 1. If instead $\mathbf{G}_m$ acts on $\mathbf{A}^1$ with weight $n$, then $\mathbf{A}^1/\mathbf{G}_m$ classifies line bundles with a section of its $n$th tensor power. Similarly for e.g. weighted projective spaces. More generally, if $\text{P}$ is a parabolic subgroup of the reductive group $\text{G}$, then the generalised flag variety $\text{G}/\text{P}$ classifies $\text{P}$ bundles with a trivialisation of the associated $\text{G}$ bundle. This gives the functor of points for e.g. Grassmannians $\text{Gr}(k,n)$.<|endoftext|> TITLE: Is this series well known? QUESTION [8 upvotes]: I recently encountered the following function $$ f(t) = \sum_{n=0}^\infty \frac{t^{n^2}}{n^2!}. $$ It seems familiar, though I cannot remember where I might have seen it before. I would like to know in what text, if any, it has been studied. Above all, I would like to know its asymptotic properties (how fast it grows) as $t$ tends to infinity ($t$ real and positive). Thanks, G. REPLY [8 votes]: This is a slice of the taylor series for $\exp(t)$. The terms that dominate are those near $n=t^{1/2}$. Using Stirling's approximation, as $t\to\infty$ with $q$ more or less bounded, we have $$\frac{t^{(t^{1/2}+q)^2}}{(t^{1/2}+q)^2)!} = \frac{e^{t-2q^2}}{\sqrt{2\pi t}} (1 + O(q^3/t)).$$ So $$f(t) \sim \frac{e^t}{\sqrt{2\pi t}}c(t),$$ where $$c(t) = \sum_q e^{-2q^2},$$ with the sum over all $q$ for which $t^{1/2}+q$ is integer. It is clear that $c(t)$ depends only on the fractional part of $t^{1/2}$. It is nice curve that oscillates between 1.2353 (when $t^{1/2}$ is midway between two integers) and 1.2679 (when $t^{1/2}$ is an integer).<|endoftext|> TITLE: Generalized Hlawka inequality QUESTION [10 upvotes]: Let $E$ be a vector space over the real (the complex case is interesting too). We consider functions $f:E\rightarrow\mathbb R$ which satisfy the homogeneity property $$f(\lambda x)=|\lambda|\,f(x).$$ In particular, we have $f(0)=0$. I am interested in the generalized Hlawka inequality (GH$n$) at order $n$. Given $n$ vectors $x_1,\ldots,x_n$, and $I\subset[\![1,n]\!]$, define $x_I=\sum_{i\in I}x_i$. Then (GH$n$) says that $$\sum_{I\subset[\![1,n]\!]}(-1)^{{\rm card}\, I}f(x_I)\le0,\qquad\forall x_1,\ldots,x_n\in E.$$ For instance, (GH1) and (GH2) are respectively $0\le f(x)$ and $f(x+y)\le f(x)+f(y)$; thus, a function $f$ satisfying (GH1) and (GH2) is a norm over $E$. (GH3) is Hlawka inequality, which is satisfied by Euclidian norms and some others, though not by all norms. Let me point out that (GH$n$) is sharp, in the sense that the equality holds true when either one vector is $0$, or all the vectors are positively collinear. If $n$ is odd, it is also an equality when $\sum_1^nx_i=0$, while if $n$ is even, this choice in (GH$n$) yields (GH$(n-1)$). One may also derive (GH2) from (GH3) by choosing $x_3=x_1$. First question : Do Euclidian norms satisfy (GH$n$) for all $n$ ? Second question : Conversely, if a norm satisfies (GH$n$) for all $n$, is it Euclidian ? Third question : Can we derive (GH$(n-1)$) from (GH$n$) when $n$ is odd ? (True if $n=3$). Edit. Guillaume, I liked your suggestion. It does give a (new ?) proof of the classical Hlawka inequality ($n=3$). However it fails as soon as $n=4$. A counter-example is $(1,1,1,-2)$, where the sum is $2$. Therefore the answer to the very first question above is No, even in one space-dimension. REPLY [4 votes]: I studied the inequality in case $n=3$, see http://link.springer.com/article/10.1007%2Fs00010-012-0178-2 You can see from this paper that not every subadditive function ($n=2$) satisfies inequality for $n=3$. So (GH3) is not equivalent to (GH1) and (GH2). I don't know what happens for greater $n$.<|endoftext|> TITLE: Publishing an elementary proof of a less-general and less-useful version of a classic result? QUESTION [6 upvotes]: Background Let $X_t$ be a stochastic process on the state space {Working, Broken}. Let $U$ be the cumulative sojourn Working during an interval $[0,\tau]$ (the process's uptime). It is well-known [1] that if the process has a finite mean time to failure (MTTF) and mean time to repair (MTTR), then $$ \begin{align*} P\left\{ X_t = \text{Working}\ | \ X_s = \text{Working} \right\} &= p + (1-p) \cdot z(t;s) \\ P\left\{ X_t = \text{Broken}\ | \ X_s = \text{Broken} \right\} &= 1-p + p \cdot z(t;s) \end{align*} $$ for some function $z$, where $p = \frac{\text{MTTF}}{\text{MTTF}+\text{MTTR}}$. Now further suppose that the times to failure have finite variance, and so too for the times to repair. Then in a classic result [2], Takács showed $U$ is asymptotically normal as $ \tau \rightarrow \infty $, and calculated the distribution's mean $\mu_U$ and variance $\sigma^2_U$. My situation I can prove that $U$ is asymptotically normal and calculate $\mu_U$ and $\sigma^2_U$, but with two caveats: I need to make an assumption about $z$. Thus my result is less general than Takács's. Calculating $\sigma^2_U$ requires calculating $\lim_{\tau\rightarrow\infty} \int_{0}^{\tau} z(t; 0) \ dt$. Thus my result is less useful than Takács's, as solving $z$ for arbitrary processes is an open question. However, the result is (perhaps?) of interest as an elementary application of Billingsley's central limit theorem for dependent variables under strong mixing [3, Theorem 27.4]. (And yes: What happened is that I learned about Billingsley's result through Wikipedia when first reviewing literature, found a proof, and then I found Takács result. There must be Murphy's Law variant for this.) If Prof Billingsley was still alive, I'd probably write to him and offer the result as a homework exercise for a new edition of his book. However, failing that option ... My question ... what journal might be good for my result? (I know I can self-publish to arXiv or equivalent with my employer. I am interested in alternatives.) Many thanks. References [1] Kishor Shridharbhai Trivedi, 2002, Probability and statistics with reliability, queuing, and computer science applications, Second ed., John Wiley & Sons, New York. [2] Lajos Takács, 1959, On a sojourn time problem in the theory of stochastic processes, Transactions of the American Mathematical Society, 93(3), 531--540. [3] Patrick Billingsley, 2008, Probability and measure, John Wiley & Sons, New York. REPLY [7 votes]: Real Analysis Exchange has an Inroads section that specifically covers the case of "a clever new proof of an important theorem".<|endoftext|> TITLE: What is the minimal $C_k$, such that every $f\colon \{-1,1\}^n\to \mathbb{R}$ of degree at most $k$ satisfies $\|f\|_2\le C_k\|f\|_1$ QUESTION [15 upvotes]: Every $f\colon\{-1,1\}^n\to \mathbb{R}$ can be repsenented as a multilinean polynomial of the form $$f(x_1,x_2,\ldots ,x_n)=\sum _{S\subseteq [n]} \hat{f}(S)\prod_{i\in S} x_i $$ The degree of the function is defined to be $\max \{|S|\,:\,\hat{f}(S)\neq0\}$. Give $\{-1,1\}^n$ the uniform probability measure. Khintchine's inequality says that if $f$ is of the form $(x_1,x_2,\ldots ,x_n)\mapsto a_1x_1+\cdots+a_nx_n$, then $\|f\|_2 \le \sqrt 2\|f\|_1$. Theorem 22 here: http://analysisofbooleanfunctions.org/?p=1472 says that every function of degree at most $k$ satisfies $\|f\|_2\le e^k\|f\|_1$. The example $(x_1+1)(x_2+1)\cdots (x_k+1)$ shows that the constant $e^k$ in this theorem cannot be reduced bellow $\sqrt 2 ^k$. My question is: What is the minimal $C_k$, such that every $f\colon \{-1,1\}^n\to \mathbb{R}$ of degree at most $k$ satisfies $\|f\|_2\le C_k\|f\|_1$? In particular, is it true that $C_k=\sqrt 2^k$? REPLY [7 votes]: When $f : \{-1,1\}^{n} \to \mathbb{C}$ is Walsh--Rademacher chaoes of degree $k$, i.e., $$ f(x) = \sum_{1\leq j_{1}<\ldots7.3...$). It is also interesting to remark that for degree 1 chaoses (Khinchin case) the sharp bound is $\sqrt{2}$ due to Szarek, however, our theorem gives $\sqrt{e}$ for free. UPDATE 9/14/2018 I think it might still be unknown whether the constant can be reduced below $e$ With Alexandros Eskenazis we came up with an argument which improves the bound $\|f\|_{2}\leq e^{k}\|f\|_{1}$ to $\|f\|_{2}\leq (2.69075...)^{k}\|f\|_{1}$ for polynomials $f$ of degree $k$ on the Hamming cube. I suspect one can further improve the bound but then one needs to carefully play with conformal maps. Anyways, here is the proof which in a sense corresponds to "averaging" the previous argument. Proof. Take any $q>2$ and consider the following domain in the complex plane $$ \Omega := \left\{ z \in \mathbb{C}\, :\, \left|z\pm \frac{q-2}{2(q-1)}\right|\leq \frac{q}{2(q-1)}\right\}. $$ Let $p$ be the conjugate exponent to $q$. It follows from complex hypercontractivity that for any $f:\{-1,1\}^{n} \to \mathbb{C}$, $f(x) = \sum_{S \subset \{1,2,\ldots, n\}} a_{S} W_{S}(x)$, where $W_{S}(x)=\prod_{j \in S}x_{j}$, $x=(x_{1}, \ldots, x_{n})$ are Walsh functions, we have $$ \|T_{z}f\|_{q} \leq \|f\|_{p} $$ whenever $z \in \Omega$, where $T_{z}f(x) = \sum_{S \subset \{1,2,\ldots, n\}} z^{|S|}a_{S} W_{S}(x)$. Let $\pi \beta$ be the exterior angle (measured in radians) of the domain $\Omega$ at point $z=i\sqrt{p-1}$. Clearly $\beta = \frac{1}{2}+ \frac{1}{\pi}\arctan\left(\frac{q-2}{2}\right)$. Next, consider the conformal map $$ \varphi(z) = \frac{1+\left(\frac{zi\sqrt{q-1}+1}{zi\sqrt{q-1}-1}\right)^{1/2\beta}}{1-\left(\frac{zi\sqrt{q-1}+1}{zi\sqrt{q-1}-1}\right)^{1/2\beta}}. $$ Notice that $\varphi$ maps conformally the comploment of $\Omega$ onto the complement of the unit disk in $\mathbb{C}$, and it has a linear growth at infinity. Therefore for any polynomial $p(z)$ of degree $k$ on $\Omega$ we see that the map $z \mapsto \frac{p(z)}{\varphi(z)^{k}\|p\|_{C(\Omega)}}$ is regular at infinity and it is bounded by $1$ on $\partial \Omega$. Therefore by the maximum principle we have $|p(z)|\leq |\varphi(z)|^{k}\|p\|_{C(\Omega)}$ for all $z \in \Omega^{c}$. In particular this means that $$ |p(1)|\leq |\varphi(1)|^{k}\|p\|_{C(\Omega)} \qquad (***) $$ Next, let $V$ be the vector space of all polynomials of degree $k$ on $\Omega$. It is the subspace of $C(\Omega)$. Let $L$ be the linear functional on $V$ which acts as follows: for any $p \in V$, $L(p)=p(1)$. It follows from (***) that $|L(p)|\leq |\varphi(1)|^{k}\|p\|_{C(\Omega)}$. By Hahn--Banach theorem $L$ extends to a bounded functional $\tilde{L} \in C^{*}(\Omega)$ so that $\|\tilde{L}\|_{C^{*}(\Omega)}\leq |\varphi(1)|^{k}$, and $\tilde{L}|_{V}=L$. The space $C^{*}(\Omega)$ can be identified to the Banach space of complex Radon measures on $\Omega$ equipped with total vartiation norm, so that $\tilde{L}(g) = \int_{\Omega} g d\mu$ for all $g \in C(\Omega)$, and $|\mu|(\Omega):=\|\mu\|_{TV} = \|\tilde{L}\|_{C^{*}(\Omega)}\leq |\varphi(1)|^{k}$. Now pick any $f:\{-1,1\}^{n} \to \mathbb{C}$ of degree $k$, and consider the polynomial $p(z)=T_{z}f(x)$. We have $$ \left\|f\right\|_{q} = \left\|\int_{\Omega}T_{z}f d\mu(z) \right\|_{q} \leq \int_{\Omega} \left\|T_{z}f \right\|_{q}d|\mu|(z)\\ \leq \int_{\Omega} \left\|f \right\|_{p}d|\mu|(z) \leq \|\mu\|_{TV} \|f\|_{p}\leq |\varphi(1)|^{k} \|f\|_{p} $$ for all $q\geq 2$ and the conjugate $p$. Using log-convexity of $L^{p}$ norms we obtain $$ \|f\|_{2} \leq \|f\|_{1} \left(\frac{\|f\|_{q}}{\|f\|_{p}}\right)^{\frac{1}{2(1/p-1/q)}} = \|f\|_{1} |\varphi(1)|^{\frac{k}{2(1/p-1/q)}} $$ Now if my calculation is correct we have $$ |\varphi(1)|^{\frac{k}{2(1/p-1/q)}} = \left(\frac{1+\cos\left(\frac{\arctan\left(\frac{2\sqrt{q-1}}{q-2}\right)}{1+\frac{2}{\pi}\arctan\left(\frac{q-2}{2}\right)}\right)}{1-\cos\left(\frac{\arctan\left(\frac{2\sqrt{q-1}}{q-2}\right)}{1+\frac{2}{\pi}\arctan\left(\frac{q-2}{2}\right)}\right)}\right)^{\frac{k}{4(1/p-1/q)}}, $$ optimizing the latter quantity over all $q>2$, and using the fact that $p=\frac{q}{q-1}$, we see that the minimal value is attained around the point $q=2.39079...$ which gives the value $(2.69075...)^{k}$.<|endoftext|> TITLE: Discrete subsets in the topology of pointwise convergence vs. metrisability QUESTION [13 upvotes]: While reading Arkhangel'skii's Topological function spaces, I encountered an unexpected application of Martin's Axiom. This is Theorem II.5.20: Assume $\mathsf{MA}+\neg \mathsf{CH}$. Let $X$ be a compact Hausdorff space. If every discrete subset of $C_p(X)$ is countable, then $X$ is metrisable. Here $C_p(X)$ stands for the space of all real-valued continuous functions on $X$ with the topology of pointwise convergence. Is it just a matter of proof or there is a consistent counter-example to this statement? If there is a counter-example, $X$ must be necessarily hereditarily separable. REPLY [7 votes]: Yes, that statement is independent of the axioms of set theory. Let $X$ be a compact strong S-space, that is a compact space such that $X^n$ is hereditarily separable but not hereditarily Lindelof for every $n \in \mathbb{N}$ (De La Vega and Kunen constructed a homogeneous space with these features in http://www.sciencedirect.com/science/article/pii/S0166864103002153) If $C_p(X)$ contained an uncountable discrete set, this would yield an uncountable discrete set in $X^n$ for some $n$. But this is impossible, because every subspace of $X^n$ is separable. Now, $X$ is not metrizable, because separable metrizable spaces have a countable base, and hence are hereditarily Lindelof. EDIT: It remains to prove that if $C_p(X)$ contains an uncountable discrete set then some finite power of $X$ also does. I believe this is due to Tkachuk. Let $D=\{f_\alpha: \alpha <\omega_1\}$ be a discrete set in $C_p(X)$ of size $\omega_1$. Let $\mathcal{B}$ be a countable base for $\mathbb{R}$. Since $D$ is discrete, for every $\alpha < \omega_1$, we can find a positive integer $n_\alpha$, a sequence $(x^\alpha_1, \dots x^\alpha_{n_\alpha})$ of elements of $X$ and a sequence $(B^\alpha_1, \dots B^\alpha_{n_\alpha})$ of elements of $\mathcal{B}$ such that $[x^\alpha_1, \dots x^\alpha_{n_\alpha}, B^\alpha_1, \dots, B^\alpha_{n_\alpha}] \cap D=\{f_\alpha\}$, where $[x_1, \dots x_n, B_1, \dots B_n]$ denotes the following basic open subset of $C_p(X)$: $\{f \in C_p(X): (\forall i \leq n)(f(x_i) \in B_i)\}$. By the pigeonhole principle, we can find an uncountable set $E \subset \omega_1$, a positive integer $n$ and a finite sequence $(B_1, \dots B_n)$ of elements of $\mathcal{B}$ such that $n_\alpha=n$ and $(B^\alpha_1, \dots B^\alpha_{n_\alpha})=(B_1, \dots B_n)$, for every $\alpha \in E$. We claim that $S=\{(x^\alpha_1, \dots, x^\alpha_n): \alpha \in E \}$ is a discrete subset of $X^n$. Indeed, by definition of the $B_i$s and the $x^\beta_i$s, we have that $f_\beta$ is the only element of $D$ such that $f_\beta(x^\beta_i) \in B_i$ for every $i \leq n$. But this implies that the open set $\prod_{i=1}^n f_\alpha^{-1}(B_i)$ intersects $S$ in the single point $(x^\alpha_1, \dots x^\alpha_n)$, and hence $S$ is discrete in $X^n$.<|endoftext|> TITLE: Is there an English translation of Minding's 1839 paper? QUESTION [8 upvotes]: Is there an English translation of "Wie sich entscheiden lässt, ob zwei gegebene krumme Flächen auf einander abwickelbar sind oder nicht..." by Ferdinand Minding, Journal für die reine und angewandte Mathematik, (page(s) 370 - 387) Berlin; 1839 Or is there a simpler or more recent proof of the same theorem? (This states that isometric triangles on surfaces with the same Gaussian curvature are congruent, and vv if I understand the theorem correctly) REPLY [7 votes]: Is there a simpler or more recent proof of the same theorem? The theorem is given as an exercise with solution (number 1.2) in this differential geometry course by Michael Eichmair at the ETH. You can also find a proof on page 292 of A New Approach to Differential Geometry using Clifford's Geometric Algebra by John Snygg, and on page 282 of Differential Geometry by Erwin Kreyszig.<|endoftext|> TITLE: Ehresmann's fibration theorem in the C1 class QUESTION [10 upvotes]: I have seen on the French Wikipedia that Ehresmann's fibration theorem is stated with the assumption that everything is $C^2$, see Théorème de Ehresmann. (On the English Wikipedia, the assumption is smooth, which I suppose means $C^\infty$, see Ehresmann's Lemma.) Here is a translation of the French Wikipedia: Ehresmann's fibration theorem states that a $C^2$ map $f:M \to N$ where $M$ and $N$ are $C^2$ differential manifolds, and such that $f$ is a surjective submersion and $f$ is proper, is a locally trivial fibration. (what is meant is that it "is a locally trivial $C^2$ fibration", I just checked Ehresmann's statement in his article Les connexions infinitésimales dans an espace fibré différentiable, Seminaire N. Bourbaki, 1948-1951, exp. n° 24, p. 153-168.) Does anybody know of a counterexample in the case where the smoothness is only $C^1$? I mean a $C^1$ map as above which would not be $C^1$ fibration. I am specially interested in the case where the domain of the submersion has dimension 2 and the range dimension 1. I suspect that the theorem holds in this case (select one fiber and build a local fibration-trivialization around it by patching the x-coordinate of local submersion-trivializations where level curves would be horizontals in $\mathbb{R}^2$). Is that already proved or disproved somewhere? REPLY [7 votes]: I believe the paper "Foliations and Fibrations" by Earle and Eells, Jr. in J. Differential Geometry 1 (1967), pp. 33-41 can be helpful. The first proposition in their fourth section is a very general extension of Ehresmann: Theorem: If $f:X \to Y$ is a proper $C^1$-map of Finsler manifolds which foliates $X$, then $f$ is a locally trivial $C^0$-fibration. Here Earles and Eells say $f$ foliates $X$ if (i) for every $x\in X$ the differential $df_x$ maps the tangent space $T_x X$ surjectively onto $T_{fx}Y$ and (ii) the fibres $\{f^{-1}(y)\}_{y\in Y}$ are closed differentiable submanifolds of $X$ defining a foliation whose leaves are the connected components of the manifolds $f^{-1}(y)$. They remark (bottom of pp. 38) that there are theorems asserting $C^k$ maps $f$ foliating $X$ define locally trivial $C^k$-foliations provided that one can find $C^k$-partitions of unity. They refer to R. Hermann, "A sufficient condition that a mapping of riemannian manifolds be a fibre bundle", Proc. Amer. Math. Soc. 11 (1960) pp. 236-242. So while their main theorem is very general (defined for Finsler manifolds modelled on Banach spaces with continuously varying family of norms on the tangent spaces) , there seems hope that in concrete finite dimensional settings one has $C^1$ foliatings maps defining $C^1$-locally trivial fibrations.<|endoftext|> TITLE: For a Sum-of-Squares cost functions J(x) is it true that J(x)-j* is also SOS? QUESTION [6 upvotes]: For polynomial optimization problems the sum-of-squares theory and Lasserre relaxation hierarchy provides a theoretically handy way of getting the solution. There are also results saying that finite relaxation is enough for global convergence. My question is related to the unconstrained optimization of polynomials of the following form $$ J(x) = \sum_{k=1}^N q^2_k(x) $$ where $q_i(x)$ are arbitrary polynomials (they do have structure in my particular case, but I hope that wont be needed to get an answer). Since $J(x)$ is sum of squares, it is non-negative. Let its minimum value be $$ j^* = \min_{x}J(x) $$ I know that there are general polynomials $P(x)$ such that $P(x)-p^*$ is not sum-of-squares. The questions is whether the assumption that $P(x)$ is sum-of-squares helps this situation or not. Is $J(x)-j^*$ is also a sum of squares polynomial? If not, do we have a counterexample? Any pointers towards relevant literature is also appreciated, as searching "is sum-of-squares" is useless for such specific questions. REPLY [2 votes]: Motzkin polynomial in the homogeneous form, i.e. $f(x):=f(x_1,x_2,x_3)=x_1^4x_2^2 + x_1^2x_2^4 − 3x_1^2x_2^2x_3^2 + x_3^6$ becomes an SOS if you add $\frac{3}{16}x_1^6$ to it. $3/16$ is an upper bound on minimum such value (a better bound is $(3/4)^6$), as can be checked e.g. by using a nice Macaulay2 package by H.Peyrl. Thus $P(x):=f(1,x_2,x_3)+\frac{3}{16}$ is a counterexample. Below is the Macaulay code to run to find an appriximation to the minimal $a$ so that $f(1,y,z)+a$ is SOS (after installing that package), and its output, in the format ({list of squares},{list of positive coefficients for them},{value of a}) S = QQ[z,y,a]; f = y^2 + y^4 + z^6 - 3*y^2*z^2 + a; (g,d,aval)=getSOS (f,{a},a) 2 2 2 17 15 3 3 9 51 1 11 3 3 ({z - -y - --, - --z + z, z , 1, y}, {-, --, ---, ----, --}, {--}) 3 60 17 4 40 136 1600 20 16<|endoftext|> TITLE: Does a classification of simultaneous conjugacy classes in a product of symmetric groups exist? QUESTION [15 upvotes]: Let the symmetric group $S_n$ on $n$ letters act on $S_n^d=S_n\times\cdots\times S_n$ by simultaneous conjugation, i.e. $\pi\in S_n$ acts on $(\sigma_1,\ldots,\sigma_d)\in S_n^d$ by $\pi.(\sigma_1,\ldots,\sigma_d)=(\pi\sigma_1\pi^{-1},\ldots,\pi\sigma_d\pi^{-1})$. I would like to know if there is a classification (a combinatorial description) of the orbits of this action, at least in the case $d=2$. For $d=1$, a combinatorial description is given by the cycle structure of the permutation. There is a related question that deals with deciding whether two elements of $S_n^d$ are in the same orbit and the author of that question also defines a canonical representative for each orbit, but it is quite indirect. It seems to me that a classification for arbitrary $d$ is probably not known. In another related question, the number of orbits is discussed, but I am really interested in a combinatorial description that classifies them. Edit. To clarify what I mean by "combinatorial description": At the very least, I am looking for an algorithm that enumerates exactly one representative from each simultaneous conjugacy class in $S_n^d$ with polynomial delay (polynomial in $n$ and $d$ will be fine). REPLY [2 votes]: You can find representatives of the orbits algorithmically by choosing $\sigma_1$ by $S_n$ conjugacy, $\sigma_2$ as representative up to conjugacy by the centralizer of $\sigma_1$, and so on. If $C=C_{S_n}(\sigma_1)$, then the $C$-orbits on $S_n$ refine the $S_n$-orbits of $S_n$ (i.e. the conjugacy classes of $S_n$), the refinement of the class containing $\tau$ corresponds to the double cosets $C_{S_n}(\tau)\setminus S_n / C$. The following commands in GAP use this method (essentially I'm asking it to enumerate homomorphisms from the free group into $S_n^d$), i.e. you get a list of orbit representative tuples: g:=SymmetricGroup(6); # or whatever degree cl:=ConjugacyClasses(g);; MorClassLoop(g,[cl,cl],rec(),8); The number of repetitions of cl indicates the value of $d$, that is use [cl,cl,cl] for $d=3$ and so on. Please note that this is calling an internal function with limited error check. It might not work for $n<3$ or $d=1$.<|endoftext|> TITLE: Intersection of a ring class field of a quadratic field K with the cyclotomic extension of K QUESTION [11 upvotes]: Let $K$ be a quadratic field. Let $f\in\mathbb{Z}_{\geq 1}$. Let $\mathcal{O}_f=\mathbf{Z}+f\mathcal{O}_K$ be the unique order of $K$ of index $f$ in $\mathcal{O}_K$. Let $H_f^{ring}$ denote the ring class field in the narrow sense (so we allow ramification at infinite real places of $K$, if such places exist) associated to the order $\mathcal{O}_f$. For a number field $L$ let us denote by $\mu(L)$ the group of roots of unity of $L$. For an integer $n\in\mathbf{Z}_{\geq 2}$ we let $\mu_n$ be the group of $n$-th roots of unity and $\zeta_n$ a primitive $n$-th root of unity. An (not so trivial) exercise in class field theory shows that $\mu(K_f^{ring})\subseteq \mu_{12}$. Let us denote by $K^{gen}$ the genus field in the narrow sense of $K$. Recall that the genus field $M^{gen}$ (in the narrow sense) of an abelian extension $M$ over $\mathbf{Q}$ is the maximal unramified extension $M^{gen}$ over $M$ (at all finite places of $M$) which is abelian over $\mathbf{Q}$. Note that $H_K=K_{1}^{ring}$ corresponds to the Hilbert class field, in the narrow sense, of $K$. We obviously have $K_{f}^{ring}\supseteq H_K\supseteq K^{gen}$. Let $d$ be the discriminant of $K$. We also have $K^{gen}\subseteq K(\zeta_d)$. Let $L=K(\zeta_n:n\in\mathbf{Z}_{\geq 2})$. So this begs the following question: Q Do we always have that $K_{f}^{ring}\cap L\subseteq K^{gen}(\mu_{12})$ ? P.S. Note that $K^{gen}(\mu_{12})\subseteq K(\mu_m)$ where $m=lcm(d,12)$. P.S.S. An explicit description of the genus field: In genreral, if $k$ is an abelian extension over $\mathbf{Q}$, we have $k\subseteq\mathbf{Q}(\zeta_m)$ for some $m$. We may associate to $k$ a subgroup $X_k$ of the Pontryagin dual of $(\mathbf{Z}/m\mathbf{Z})^{\times}$, which we denote by $X_m$. Note that $X_m$ has a natural direct sum decomposition with respect to the inertia subgroups of $\mathbf{Q}(\zeta_m)$; which are in correspondence with the primes $p|m$. This simply corresponds to the usual decomposition of a Dirichlet charcter via the Chinese remainder theorem for the prime powers dividing $m$. It is a simple exercise to show that the "cartesian closure" of the subgroup $X_k$ (with respect to this direct sum decomposition of $X_m$) corresponds to $k^{gen}$. So in the case where $k$ is quadratic over $\mathbf{Q}$, this implies that $k^{gen}$ is a multiquadratic extension over $k$. See the wikepedia page on genus fields for an explicit description of $k^{gen}$ when $k$ is quadratic. REPLY [5 votes]: The answer is no. I'll only focus on imaginary quadratic fields since they are easier to deal with than real ones. Great reference for all this theory is Cox's book Primes of the form $x^2+ny^2$. So let $K$ be imagimary, thus $d<0$, and let $D=f^2d$. Since $L=\bigcup_{n\in\mathbb{N}}\mathbb{Q}(\zeta_n)$ is the maximal abelian extension of $\mathbb{Q}$, the field $K_f^{\text{ring}}\cap L$ is the maximal subfield of $K_f^{\text{ring}}$ which is abelian over $\mathbb{Q}$. We have $\operatorname{Gal}(K_f^{\text{ring}}/K)\cong C(D)$, where $C(D)$ is the group of classes of primitive positive definite binary quadratic forms of discriminant $D$, and $G:=\operatorname{Gal}(K_f^{\text{ring}}/\mathbb{Q})$ is isomorphic to the generalized dihedral group $C(D)\rtimes\{\pm1\}$ (which means that $-1$ acts on $C(D)$ as inverse). Under this isomorphism $[G,G]$ corresponds to $C(D)^2\times\{1\}$. Since $\operatorname{Gal}(K_f^{\text{ring}}/K_f^{\text{ring}}\cap L)=[G,G]$, we get $[K_f^{\text{ring}}\cap L:K]=[C(D):C(D)^2]$. Let $r$ be the number of odd primes dividing $D$. We define number $\mu$ in the following way: if $D$ is odd then $\mu=r$, and if $D=-4n$ is even then $$\mu= \begin{cases} r&\text{if }n\equiv3\pmod4,\\ r+1&\text{if }n\equiv1,2\pmod4,\\ r+1&\text{if }n\equiv4\pmod8,\\ r+2&\text{if }n\equiv0\pmod8.\\ \end{cases}$$ Gauss proved that $[C(D):C(D)^2]=2^{\mu-1}$, so we get $[K_f^{\text{ring}}\cap L:K]=2^{\mu-1}$ and $[K_f^{\text{ring}}\cap L:\mathbb{Q}]=2^\mu$. In fact we can describe $K_f^{\text{ring}}\cap L$ explicitly. Let $p_1,\ldots,p_r$ be odd primes dividing $D$ and let $p^*=(-1)^{(p-1)/2}p$. If $D$ is odd then $K_f^{\text{ring}}\cap L=\mathbb{Q}(\sqrt{p_1^*},\ldots,\sqrt{p_r^*})$, and if $D$ is even then $$K_f^{\text{ring}}\cap L= \begin{cases} \mathbb{Q}(\sqrt{p_1^*},\ldots,\sqrt{p_r^*})&\text{if }n\equiv3\pmod4,\\ \mathbb{Q}(\sqrt{p_1^*},\ldots,\sqrt{p_r^*},i)&\text{if }n\equiv1\pmod4,\\ \mathbb{Q}(\sqrt{p_1^*},\ldots,\sqrt{p_r^*},\sqrt{-2})&\text{if }n\equiv2\pmod8,\\ \mathbb{Q}(\sqrt{p_1^*},\ldots,\sqrt{p_r^*},\sqrt{2})&\text{if }n\equiv6\pmod8,\\ \mathbb{Q}(\sqrt{p_1^*},\ldots,\sqrt{p_r^*},i)&\text{if }n\equiv4\pmod8,\\ \mathbb{Q}(\sqrt{p_1^*},\ldots,\sqrt{p_r^*},i,\sqrt{2})&\text{if }n\equiv0\pmod8. \end{cases}$$ Since $K^{\text{gen}}=K_1^{\text{ring}}\cap L$, by applying the previous formula with $f=1$ we get $[K^{\text{gen}}:\mathbb{Q}]=2^{\mu'}$, where $\mu'$ is defined same as $\mu$ with $d$ in place of $D$. Thus $[K^{\text{gen}}(\zeta_{12}):\mathbb{Q}]\le[K^{\text{gen}}:\mathbb{Q}]\cdot[\mathbb{Q}(\zeta_{12}):\mathbb{Q}]= 2^{\mu'}\cdot\varphi(12)=2^{\mu'+2}$. In particular $[K_f^{\text{ring}}\cap L:\mathbb{Q}]$ can be arbitrarily much larger than $[K^{\text{gen}}(\zeta_{12}):\mathbb{Q}]$ (if $f$ is divisible by lot of primes that don't divide $d$). However even if $[K_f^{\text{ring}}\cap L:\mathbb{Q}]\le[K^{\text{gen}}(\zeta_{12}):\mathbb{Q}]$ that still doesn't imply $K_f^{\text{ring}}\cap L\subseteq K^{\text{gen}}(\zeta_{12})$. For example take $K=\mathbb{Q}(i)$ and $f=8$. Then $K_f^{\text{ring}}=\mathbb{Q}(i,\sqrt[4]{2})$ and $K_f^{\text{ring}}\cap L=\mathbb{Q}(i,\sqrt{2})=\mathbb{Q}(\zeta_8)$ but $K^{\text{gen}}(\zeta_{12})=K(\zeta_{12})=\mathbb{Q}(\zeta_{12})$.<|endoftext|> TITLE: What did Rolle prove when he proved Rolle's theorem? QUESTION [31 upvotes]: Rolle published what we today call Rolle's theorem about 150 years before the arithmetization of the reals. Unfortunately this proof seems to have been buried in a long book [Rolle 1691] that I can't seem to find online. (Well, maybe that's fortunate because otherwise I'd have felt obligated to comb through it with my poor knowledge of French.) Rolle was an algebraist and a prominent opponent of infinitesimals. His proof of the theorem was based on something called the "method of cascades", an evolution of techniques originated by Johannes Hudde. This seems to have been a method of manipulating polynomials that was equivalent to differentiating them [Itard 2008]. Today, we would consider Rolle's theorem to be a consequence of the extreme value theorem, which in turn depends on the completeness property of the reals -- stated long after Rolle was in his grave. However, there is a revisionist argument that people as early as the 17th century had quite a clear notion of what we would today call the real number system [Blaszczyk 2012]. So what did Rolle really prove when he published his proof of Rolle's theorem? Was it just a proof for polynomials? At the time, would a proof for polynomials have been considered sufficient, on the theory that any smooth function can be approximated by a polynomial? Related: Does Rolle's Theorem imply Dedekind completeness? Blaszczyk, Katz, and Sherry, "Ten Misconceptions from the History of Analysis and Their Debunking", 2012, http://arxiv.org/abs/1202.4153 Itard, "Rolle, Michel" in Complete Dictionary of Scientific Biography, 2008, http://www.encyclopedia.com/doc/1G2-2830903713.html Rolle, Démonstration d'une Méthode pour resoudre les Egalitez de tous les degrez, 1691. REPLY [25 votes]: In the lengthy review by Victor J. Katz of The Oxford Handbook of the History of Mathematics, edited by Eleanor Robson and Jacqueline Stedall, Oxford University Press, Oxford, 2009, MR2549261 (2011e:01001), it says, Virtually the only article in the Handbook that could be classified as "internalist'' history of a mathematical idea is June Barrow-Green's article "From cascades to calculus: Rolle's theorem'', in which she looks in detail at the theorem as stated by Michel Rolle in 1690 and then looks at how the result was treated in various textbooks up to the beginning of the twentieth century. As originally stated and proved, the theorem only concerned polynomial functions and certainly involved no calculus, especially because Rolle attacked the calculus at its beginning for its lack of rigor.<|endoftext|> TITLE: Randall Munroe's Lost Immortals QUESTION [42 upvotes]: In Randall Munroe's book What If?, the "Lost Immortals" question asks: If two immortal people were placed on opposite sides of an uninhabited Earthlike planet, how long would it take them to find each other? After an entertaining discussion in Munroe's usual inimitable style, he concludes with the following suggestion: If you have no information, walk at random, leaving a trail of stone markers, each one pointing to the next. For every day that you walk, rest for three. Periodically mark the date alongside the cairn. It doesn't matter how you do this, as long as it's consistent. You could chisel the number of days into a rock, or lay out rocks to plot the number. If you come across a trail that's newer than any you've seen before, start following it as fast as you can. If you lose the trail and can't recover it, resume leaving your own trail. I find this algorithm very intriguing and I can almost—but not quite—recall seeing it before. Has this problem been studied before? In any case, my question is, can Munroe's algorithm be improved? It may be helpful to lay down some ground rules. Munroe considers planets with terrain (oceans, deserts, coastlines, etc.) but for simplicity let's assume a uniform sphere and an unlimited ability to leave a trail behind. Let's also assume that there are no pre-existing markers on the sphere that allow the players to pre-arrange something like, "Let's meet at the North Pole." Although the original question seems to specify that the players are placed at antipodes, it seems to make more sense for their starting positions to be random. Both people have some maximum speed of travel but can choose to move more slowly than that. Finally, I'm not sure whether it makes a difference if the players are allowed to leave arbitrary messages along the trail for the other player to read; if this possibility complicates the problem too much, I'd be willing to simplify by declaring success as soon as one player intersects the other's trail. REPLY [10 votes]: Munroe's algorithm runs into trouble if one immortal (say, Charles) comes across the trail of the other (say, Marie) while Marie has already ran into and is following Charles' trail. If the two have the same maximum speed, they'll chase each other forever. You'd have to establish some rule, like one person reversing direction after a full loop, or have the strategy involve gradually cutting into and closing the circle. I think the optimal symmetric strategy, assuming full trail and marker capability but no significant radius of sight, is to walk a random geodesic, leaving markers indicating the time since the start of the walk. After intersection, an immortal will be able to determine who will intersect the other's geodesic first; the first to cross the other's path follows it forward, and the second turns around and meets him. This strategy gives a worst-case end time of 1.25*(circumference of planet)/(speed of immortals).<|endoftext|> TITLE: Fundamental class in K-theory and orientability QUESTION [6 upvotes]: In ordinary homology, the classical results give the following situation: for a compact, connected, topological manifold $M$ of dimension $n$ we have, for each ring $R$, that $H_n(M,M \setminus \{x\};R) \simeq R$. The orientation of $M$ is a section $o$ such that $o(x) \in H_n(M,M \setminus\{x\};R)$ in a generator of this homology module, for each $x \in M$. This definition heavily uses the 'machinery' of this particular homology theory: in other words, it is highly non-axiomatic. However, I met the sentence that "for each (co)homology theory there is a notion of orientability". So here is my (possibly a little bit vaque): Question 1 Is the notion of orientability in any (co)homology theory axiomatic? If so, how to interpret the above, particular definition in axiomatic terms? Recently I've got interested in $K$-theory and $K$-homology. These are homology theories for $C^*$-algebras: however, the construction doesn't remind the standard constructions in homological algebra. To be more precise, it doesn't involve chain complexes and typical cycle in the theory is not an element in kernel of some (boundary) operator modulo boudary. What happens, is that one have only two $K$-theory groups, in contrast to ordinary homology. I know that orientability in $K$-theory is the existence of $spin^c$ structure but let me ask: Question 2 What is the definition of $K$-orientable space? It happens, that the notion of orientability (via singular homology), when restricted to smooth manifolds, is the same as ordinary orientability. So it is also very natural to ask: Question 3 For what class of spaces $K$-orientability is still well defined? Finally, there is a notion of a fundamental class in singular cohomology: it is an image of the orientation under the isomorphism $\Gamma\Big(M,\bigcup_{x \in M}H_n(M,M \setminus \{x\};R)\Big) \simeq H_n(M;R)$. As before, this is very specific, so again: Question 4 is it possible to define the notion of a fundamental class axiomaticly? I apologize for a rather long question: I would be grateful if anybody could clarify this issue for me. REPLY [9 votes]: Ad question 1: Let $E$ be any ring spectrum, with associated cohomology and homology theory. An $E$-orientation of a rank $n$ vector bundle $V \to X$ is a class $u \in E^n (V,V-0)$ such that the restriction to any point $x \in X$, i.e. the image in $E^n (V_x,V_x-0) \cong E^n (R^n,R^{n}-0) \cong E^0$ is a generator. For a closed manifold $M^n$; an orientation is a class $a \in E_n (M)$, such that for all $x \in M$, the image of $a$ in $E_n (M,M-x) \cong E_n (R^n, R^n-0)=E_0$ is a generator. This is an axiomatic definition. For smooth manifolds, orientations of $TM$ and of $M$ are in bijection by the Atiyah-Milnor duality theorem. Proving that a vector bundle has an $E$-orientation in practice is a different question. This is \it{almost never} a tautology. In principle, there is a homotopy theoretic strategy; but the computations quickly become intractable. Look in Rudyak's book ''Thom spectra...'', chapter VI, to see what I mean (there he discusses K-Theory). If there is a concrete geometric model of $E$-Theory, the problem is much easier. For example, if $E$ is real $K$-Theory, then a vector bundle is $KO$-orientable iff it is spin. This depends heavily on the description of $K$-theory in terms of vector bundles, i.e. it relies on linear algebra related to Clifford algebras. There might exist a purely homotopy-theoretic proof, but I bet that it is quite hard and Rudyak's exposition does not give a homotopy-theoretic proof of that fact.<|endoftext|> TITLE: Is it possible to formulate the axiom of choice as the existence of a survival strategy? QUESTION [19 upvotes]: Consider the following situation: There is an infinite set $G$ of giraffes. A lion comes and announces a set $C$ of all possible colours and an infinite cardinal $\kappa$. The hungry lion tells the giraffes that when she comes back, the following happens: The giraffes may no longer speak to each other. The lion gives each giraffe a scarf of some colour in the set $C$. Every giraffe sees all the scarves (including their own) but may be mistaken of the colour of strictly less than $\kappa$ scarves. The giraffes do not know which scarves they see correctly, only that there is a strict upper bound on the number of false colours. Then each giraffe must guess the colour of their own scarf. The lion eats all the giraffes that guess wrong. Can the giraffes agree on a survival strategy before the lion returns so that strictly less than $\kappa$ giraffes are eaten? In the typical formulation of this problem $|G|=\kappa=\aleph_0$ and the giraffes have prior knowledge of which scarves they see. Let us ignore the practical issues related to this generalized savanna for a moment. This is what the existence of a survival strategy means in more mathematical terms: Take any infinite set $G$, any nonempty set $C$ and any infinite cardinal $\kappa$. A survival strategy is a function $S:G\times C^G\to C$ (strategy) with the following property. Take any any function (colouring) $f\in C^G$ and any $o\in (C^G)^G$ (the giraffe $g\in G$ observes the colour pattern $o(g)$) so that $\left\lvert\{h\in G;o(g)(h)\neq f(h)\}\right\rvert<\kappa$ for all $g\in G$. Let $a\in C^G$ (the answers given by the giraffes) be defined by $a(g)=S(g,o(g))$. This $a$ always satisfies $\left\lvert\{g\in G;a(g)\neq f(g)\}\right\rvert<\kappa$. Note that to make sense of this existence we need not be able to compare all cardinals; it suffices that $|A|<\kappa$ is well defined. (I am not a set theorist myself. If this formulation seems inappropriate or does not seem to correspond to the story, let me know. If you know how to add clarifying details to this question, feel free to edit.) Claim: The giraffes have a survival strategy in ZFC. Proof: In the set $C^G$ of all possible scarf colourings define an equivalence relation $\sim$ by $$ a\sim b \iff \left\lvert\{g\in G;a(g)\neq b(g)\}\right\rvert<\kappa. $$ Let $q:C^G\to C^G/{\sim}$ be the quotient map and $r$ a right inverse for it. The giraffes can agree on the map $r$ based on the information given by the lion. Suppose the scarf colour pattern chosen by the lion is $f\in C^G$. All giraffes can see the equivalence class $q(f)$ but cannot be sure about $f$ itself. The giraffe $g\in G$ the guesses $r(q(f))(g)\in C$, so that the answers form the colour pattern $r(q(f))$. Since $r(q(f))\sim f$, the number of mistaken giraffes is strictly less than $\kappa$. In terms of the more mathematical formulation, the strategy is $S(g,v)=r(q(v))(g)$. $\square$ The problem: The proof relies heavily on the axiom of choice (the existence of $r$). But is it equivalent to it? That is, if we assume that the giraffes have a survival strategy for any $G$, $C$ and $\kappa$, can we deduce the axiom of choice? My vague intuition suggests an affirmative answer, but I do not have any real arguments to support it. I have not seen such claims in lists of equivalent forms of the axiom of choice. One issue is that the strategy is somewhat indeterministic in the sense that it does not give any control over the identities of the surviving giraffes. There are of course variations of the problem: Do we really need the strategy for all $G$, $C$ and $\kappa$, or is it enough let let one or two of them vary (perhaps $|G|=\kappa$ and $C=\{0,1\}$)? What if the giraffes have prior knowledge of which scarves they don't see correctly (e.g. their own)? Note: One purpose of the animal story is to give simple and reasonably intuitive names for the objects. But my main purpose is that if the result indeed is equivalent with the axiom of choice, this could be used to demonstrate what the axiom of choice is about to non-mathematical (or choice-ignorant mathematical) audience. It is of course interesting as such that the existence of a survival strategy follows from AC, but it would be much more interesting if the existence had abstract consequences, AC itself or something else. REPLY [5 votes]: The question makes sense when $\kappa=\aleph_0$ since the statement involves no choice. But according to the interesting book by Hardin and Taylor, "The mathematics of coordinated inference" (2013), mentioned above in a comment by François, even the existence of a survival strategy for $\kappa=\aleph_0$ and arbitrary $G$ and $C$ is not known to be equivalent to the Axiom of Choice. They state this as "the single most prominent open question" of chapter 3 (Question 3.6.1). ADDENDUM There is one reasonable reformulation of your question which avoids mentioning cardinalities (which are always tricky in a choiceless context) and which is equivalent to it in ZFC. It turns out that such a reformulation is actually a particular case of a theorem in the book that is unknown to imply AC, which would indicate that your question is also an open problem. Fix any proper ideal $\mathcal{I}$ in the powerset of $G$. Then you can stipulate that the subset of scarves each giraffe sees wrongly is a member of $\mathcal{I}$, and ask whether there is a strategy under which only a member of $\mathcal{I}$ guesses wrong (when AC is available, the subsets of cardinality strictly less than $|G|$) would form such an ideal). Under this reformulation, $G$ can be turned into a topological space in which the neighbourhood filter of a point consists precisely on those subsets containing the point whose complements belong to $\mathcal{I}$. For each $g \in G$ you can define an equivalence relation $\sim_g$ on $C^G$ by $f \sim_g f'$ if and only if $f$ and $f'$ agree on a deleted neighbourhood $U - \{g\}$ of $g$. Given an assignation $f$, each giraffe $g$ can detect the equivalence class of $f$ under $\sim_g$ and predict their own colour. This is precisely what Hardin and Taylor call a $\textit{near neighbourhood predictor}$. Moreover, one can easily see that all points in a given member $i \in \mathcal{I}$ are topologically isolated, and in particular any point in a nonempty subset of $i$ has a neighbourhood intersecting only finitely many points in that subset, making $i$ something which they call $\textit{weakly scattered}$. Their corollary 7.2.4 then states: "For any space $X$ of agents and set $Y$ of colours, there exists a weakly scattered-error near neighbourhood predictor" Once again, their list of open questions includes question 7.8.1, which asks whether the above proposition, provable in ZFC, actually implies AC over ZF. REPLY [5 votes]: This is related to Nate Eldredge's answer that shows some amount of choice is required. In the case $G=\mathbb{N}$, $\kappa = \omega_0$ and $C=2$, the equivalence relation you define is eventual equality of 0-1 sequences and it has a special name, $E_0$. Choose one element from each equivalence class of $E_0$, and assume that this set is Borel. In this case, there would be a Borel map $f: 2^{\mathbb{N}} \rightarrow 2^{\mathbb{N}}$ such that $xE_0y \Leftrightarrow f(x)=f(y)$. Such Borel equivalence relations are known as "smooth" (check out this MO question) and $E_0$ is known to be not smooth. (You can see page 5 of this workshop talk by S. Thomas, or Su Gao's Invariant Descriptive Set Theory Proposition 6.1.7 on page 135) Therefore, the giraffes' strategy you define is not Borel. To see that some amount of choice is necessary to construct this strategy, or any set that intersects each $E_0$-class at a single point, you can move to a model of set theory where choice fails badly and every set of reals is Borel (for example, see here). Of course, Nate Eldredge's answers reveals more than this argument but I thought it is worth mentioning.<|endoftext|> TITLE: Is every Montel locally convex vector space compactly generated? QUESTION [5 upvotes]: Let $X$ be a Hausdorff locally convex vector space. Recall (my reference is the book of H. Jarchow, Locally Convex Spaces. B.G. Teubner, 1981) that we say that $X$ is a semi-Montel space if every bounded subset of $X$ is relatively compact (equivalently, every closed and bounded subset of $X$ is compact), and a Montel space if it is semi-Montel and satisfy one (hence all) of the following conditions (equivalent under the semi-Montel hypothesis, see Proposition 11.5.1, pp. 230 of Jarchow's book): $X$ is reflexive; $X$ is barrelled; $X$ is quasi-barrelled. It is known that the strong dual of a Montel space is also Montel (Jarchow, Proposition 11.5.4, pp. 230-231). In the proof of Theorem 4.11 (5), pp. 39-40 of the book of A. Kriegl and P.W. Michor, The Convenient Setting of Global Analysis (AMS, 1997), it is shown that if $X$ is the strong dual of a Fréchet-Montel space (hence $X$ is a Montel space), then $X$ is a compactly generated topological space (also called a k-space or a Kelley space), i.e. the topology of $X$ is the final topology with respect to the inclusions of compact subsets of $X$. However, it seems to me that the proof of this assertion uses only the fact that $X$ is Montel. Question: Are Montel spaces compactly generated, or is there a counter-example to this claim? REPLY [5 votes]: An(other) example of a Montel space which is not compactly generated is $\kern.4mm\mathscr D\kern.4mm(\kern.4mm\mathbb R\kern.4mm)$ . This follows from Theorem 6.1.4(iii) and Proposition 6.2.8(ii) on pages 190 and 195 in A. Frölicher and A. Kriegl: Linear Spaces and Differentiation Theory, Wiley, Chichester 1988.<|endoftext|> TITLE: Are Banach space norms (up to equivalence) unique? QUESTION [10 upvotes]: Here is a naive question: is a "completing" norm of a vector space unique (up to equivalence) or can one find a vector space and two non-equivalent norms $\|.\|$ and $|||.|||$ that both induce a Banach space norm? That should be a classic question, but I do not find anything in text books. Cheers, Bernhard REPLY [8 votes]: I learned the following "construction" in the article "Equivalent complete norms and positivity." from Arendt and Nittka. On a Banach space $(X,\lVert\cdot\rVert)$, take an unbounded functional $\varphi$ and a point $y\in X$ such that $\varphi(y)=1$ and define the operator $S:X\to X$ by $Sx := x - 2\varphi(x)y$. Then you can check that $S^2=I$ and $\lvert x \rvert := \lVert Sx\rVert$ defines a complete norm on $X$ which is not equivalent to the $\lVert\cdot\rVert$-norm. Of course there are many such norms, you might also have a look at this question.<|endoftext|> TITLE: Is every order type of a PA model the \omega of some ZFC model? QUESTION [11 upvotes]: Let $N$ be a model of first-order Peano arithmetic, and let $\sigma$ be its order-type. Does it follow that there is a (non-transitive, expect when $M$ is the standard model) $ZFC$-model $M$ such that $\omega^{M}=\sigma$? What if $PA$ is strengthened to true arithmetic (i.e. the theory of $(\mathbb{N},+,\cdot)$ or weakened to a subtheory such as $I\Sigma_{1}$, $I\Delta_{0}+EXP$, $I\Delta_{0}$ or $IOpen$? Or when we replace $ZFC$ with something weaker like $KP$? REPLY [10 votes]: Joel Hamkins has already answered the question in the positive for countable models of PA; indeed PA can be relaxed to IOpen in his argument. In the uncountable case, as far as I know, the problem is wide open. My reason: a closely related question is still (wide) open, namely, is it possible to have distinct consistent completions $T_1$ and $T_2$ of PA such that the collection of order-types of uncountable models of $T_1$ differs from the collection of order-types of uncountable models of $T_2$ ? This open question was first brought to world attention by Harvey Friedman in his 1975-paper "One hundred and two problems in mathematical logic" (in Journal of Symbolic Logic). For the state of the art (in relation to order-types of models of PA) see Chapter 11 of the Kossak-Schmerl graduate text on on models of arithmetic. Friedman's problem is mentioned there both in Chapter 11, and also in Chapter 12 (List of 20 open questions).<|endoftext|> TITLE: Can the Cohen forcing collapse cardinals? QUESTION [7 upvotes]: Let $\kappa$ be a regular cardinal, and let $\mathbb{P} = Add(\kappa,1)$ be the standard forcing notion for adding a new subset of $\kappa$ using partial function from $\kappa$ to $2$ with domain of size $<\kappa$. This forcing is $\kappa$-closed, so it doesn't collapse cardinals $\leq \kappa$. If we further assume that $2^{<\kappa} = \kappa$ then it is also $\kappa^+$.c.c., so it doesn't collapse cardinals at all. If we extend the universe and add new bounded subsets of $\kappa$ then $\mathbb{P}$ doesn't stay $\kappa$-closed, but it might still be $\kappa$-distributive. For example it will be the case when we extend the universe by using a $\kappa$.c.c. forcing, by a theorem of Easton. The only way that I know to force that $\mathbb{P}$ collapses cardinals is by changing the cofinality of $\kappa$. Indeed, if $\mathbb{Q}$ is a forcing such that $V^{\mathbb{Q}} \models\text{cf }\kappa = \omega < \kappa$, then any $V^{\mathbb{Q}}$-generic filter for $\mathbb{P}$ codes an enumeration of all ordinals below $\kappa$ of order type $\omega$ (take $\{\alpha_n \mid n < \omega\}$ a cofinal sequence at $\kappa$. For every $p \in \mathbb{P}$, there is $n < \omega$ such that $\text{supp }p \subseteq \alpha_n$. So we can extend $p$ to a condition $q$ in which the first $\gamma$ coordinates after $\alpha_n$ are zeros and $q(\alpha_n + \gamma) = 1$. By density arguments this defines an enumeration of all $\kappa$). Question: Is it consistent that there is regular cardinal $\kappa$ and a forcing $\mathbb{Q}$ that preserves the regularity of $\kappa$ such that $\Vdash_\mathbb{Q} \check{\mathbb{P}}$ collapses $\kappa$? Question: Is it consistent that for some regular uncountable cardinal $\kappa$ there is no such $\mathbb{Q}$? REPLY [7 votes]: The following theorem of Stanley "forcing disabled" answers your first question: Theorem. Assume there is a proper class of weakly compact cardinals. Then there is a class generic extension $V$ of $L$ such that if $P\in L$ is non-trivial and uniform, $\beta$ is the least cardinal such that forcing with $P$ over $L$ adds a new subset of $\beta,$ and $card^V(\beta)$ is not in $V$ successor of a singular cardinal, then forcing with $P$ over $V$ collapses $card^V(\beta)$, if it is $>\omega.$ In fact the above theorem is true, if instead of $L$ we start with any model of $ZFC+GCH$ which has no inner model with a measurable cardinal.<|endoftext|> TITLE: Isotypic components of the action of the symmetric group on polynomials QUESTION [5 upvotes]: The polynomial ring $\mathbb{C}[x_1,\ldots,x_n]$ decomposes as a direct sum of isotypic components for the action of the symmetric group $S_n$. The isotypic component of the trivial representation is simply the ring of symmetric functions. The complementary summand--that is, the direct sum of all the other isotypic components--is a free module of rank $n!-1$ over the ring of symmetric functions. I would like to have an explicit basis, or at least an explicit set of generators, for this module. For example, if $n=2$, we have $\mathbb{C}[x_1,x_2] = \mathbb{C}[x_1,x_2]^{S_2} \oplus \mathbb{C}[x_1,x_2]^{S_2}\cdot (x_1-x_2)$, so we have a basis consisting of the single element $x_1-x_2$. More generally, the isotypic component of the sign representation is equal to symmetric functions times the Vandermonde determinant, so the Vandermonde determinant should probably be one of the elements of my generating set. REPLY [4 votes]: At Dotsenko pointed out, the quotient of ${\bf C}[x_1,\dots,x_n]$ by the ideal of homogeneous positive degree symmetric functions is isomorphic to the regular representation of $S_n$, so you might ask about a good choice of basis for this compatible with the $S_n$-action that you can lift to ${\bf C}[x_1, \dots, x_n]$. In particular, each irreducible ${\bf M}_\lambda$ appears $\dim {\bf M}_\lambda$ times. It is a classical fact that when ${\bf M}_\lambda$ appears the first time (i.e., lowest possible degree) that it appears with multiplicity 1, so there is a canonical choice for this representation (and this was constructed by Specht(?) and generalized by Macdonald to other root systems). To construct this, let $T$ be any filling of the Young diagram of $\lambda$ with entries $1, \dots, n$. For each column, take the Vandermonde product $\prod_{i < j} (x_i - x_j)$ where $i,j$ range over distinct elements in that column. Let $p_T$ be the product of these determinants. The space spanned by all $p_T$ is this lowest-degree occurrence (for a basis, only consider $T$ that are standard Young tableaux). But for the other occurrences there will be ambiguities. The following paper of Terasoma and Yamada give some choices (though it's not canonical). Annoucement: http://projecteuclid.org/euclid.pja/1195511538 Details (joint with Ariki): http://projecteuclid.org/euclid.hmj/1206127144<|endoftext|> TITLE: Extending compact operators QUESTION [6 upvotes]: Let $X$ be a separable, infinite-dimensional complex Banach space and $Y\subseteq X$ an infinite-dimensional closed subspace. Suppose $K:Y\to X$ is an arbitrary compact operator. I would like to find an infinite-dimensional closed subspace $Z\subseteq Y$ such that the restriction $K|_Z$ has a compact extension $\widetilde{K}$ to all of $X$. In other words, I would like to find a compact operator $\widetilde{K}:X\to X$ such that $\widetilde{K}z=Kz$ for all $z\in Z$. Thanks to Lindenstrauss, it is known that if $X^*=L_1(\mu)$ for some measure $\mu$ then $K$ has a compact extension. However, what I need is weaker. It suffices for my purposes to find an infinite-dimensional restriction with a compact extension. Recall that $X$ is called subprojective just in case every infinite-dimensional closed subspace $Y$ admits a further infinite-dimensional closed subspace $Z\subseteq Y$ which is complemented in $X$. So, if $X$ is subprojective then obviously I get what I want. What if $X$ is not subprojective? Can we still get some suitable $\widetilde{K}$? Can we always get a $\widetilde{K}$? Probably this is already known, which is the reason for my question. Thanks guys! REPLY [6 votes]: You can always extend a nuclear operator even to a nuclear operator. Every compact operator is nuclear on some infinite dimensional subspace, so your question has a positive answer.<|endoftext|> TITLE: Why do $12$ and $120$ occur very often in the denominators of $\zeta(-n)$ for odd $n$? QUESTION [11 upvotes]: $\zeta(-n) = - \dfrac{B_{n+1}}{n+1}$ $\zeta(-2n) = 0$ $\zeta(-1) = - \dfrac{1}{12}$ $\zeta(-3) = \dfrac{1}{120}$ $\zeta(-5) = - \dfrac{1}{252}$ $\zeta(-7) = \dfrac{1}{240}$ $\zeta(-9) = - \dfrac{691}{132}$ $\zeta(-11) = \dfrac{1}{32760}$ $\zeta(-13) = - \dfrac{3617}{12}$ What I am interested in are the sequence of denominators of these fractions. $12, 120, 252, 240, 132, 32760, 12, 8160, 14364, 6600, 276, 65520, 12, 3480, 85932, 16320, 12, 69090840, 12, 541200, 75852, 2760, 564, 2227680, 132, 6360, 43092, 6960, 708, 3407203800, 12, 32640, 388332, 120, 9372, 10087262640$ http://oeis.org/A006953 They are alternately divisible by 12 and 120. In the first $1000$ terms, $120$ occurs $53$ times, and $12$ occurs even more often. So it seems they both occur infinitely many times. My question is: Why? What is the meaning of all this? Does that make $12$ and $120$ some kinds of special numbers? REPLY [19 votes]: Von-Staudt's Second Theorem gives the exact prime decomposition of the denominator of $B_{2k}/(2k)$; it is $$\prod_{p:(p-1)\mid2k}p^{1+\nu_p(2k)}$$ The product is over primes $p$ such that $p-1$ divides $2k$, and $\nu_p(n)$ is the largest exponent $e$ such that $p^e$ divides $n$. If $k$ is prime, and if $2k+1$ is composite, then the only divisors of $2k$ are $1,2,k,2k$, and the only divisors of the form $p-1$ with $p$ a prime are $1$ and $2$, the primes being $2$ and $3$; moreover, $\nu_2(2k)=1$ and $\nu_3(2k)=0$. Thus, in all such cases, the denominator of $B_{2k}/(2k)$ is 12. If $k\equiv1\bmod3$ (and $k>1$), then $2k+1$ is composite. As there are infinitely many primes $k\equiv1\bmod3$, there are infinitely many $k$ such that the denominator of $B_{2k}/(2k)$ is $12$. A similar argument for $k=2q$ with $q$ prime and $q\equiv1\bmod{15}$ shows that there are infinitely many $k$ such that the denominator of $B_{2k}/(2k)$ is $120$. REPLY [2 votes]: This is not an answer but rather a comment to illustrate the observation. Consider the canonical primefactor-decomposition of the denominators of that values. You'll easily observe the relevant patterns in them. For better reading I've done them in two columns, for zeta(1-1),zeta(1-2) ; zeta(1-3),zeta(1-4) ;...;zeta(1-k),zeta(1-(k+1)); for index k=1 to some m. at indexes at indexes k=1,3,5,7... k=2,4,6,8,... ============= =============================================================== 2 2^2 .3 <<---------------------------------------------------- 1 2^3 .3 .5 1 2^2 .3^2 .7 1 2^4 .3 .5 1 2^2 .3 .11 1 2^3 .3^2 .5 .7 .13 1 2^2 .3 <<---------------------------------------------------- 1 2^5 .3 .5 .17 1 2^2 .3^3 .7 .19 1 2^3 .3 .5^2 .11 1 2^2 .3 .23 1 2^4 .3^2 .5 .7 .13 1 2^2 .3 <<---------------------------------------------------- 1 2^3 .3 .5 .29 1 2^2 .3^2 .7 .11 .31 1 2^6 .3 .5 .17 The rows which contain the horizontal lines denote the denominators which are exactly $12$ (and do not only contain $12$ as a factor). This patterns contain an obvious relation of the totient-value for the involved primefactors in relation to the index and are definite-ly described by the Clausen-von Staudt-theorem as mentioned in the other answer. A tiny remark: Note, that the entry at index k=0 (zeta(1)) would contain all primefactors to infinite power.<|endoftext|> TITLE: Is there any relationship between the Euler class and the Vandermonde determinant? QUESTION [12 upvotes]: Several Wikipedia articles claim that the relationship between the Euler class $e(V)$ and the top Pontryagin class $p_k(V)$ of an oriented $2k$-dimensional real vector bundle $V$ corresponds, via the splitting principle, to the relationship between the Vandermonde determinant and the discriminant (in particular, in each case the former is the square of the latter). As far as I can tell, this can't possibly be true. Most obviously, the Vandermonde determinant and the discriminant have degrees that are much too large: they grow quadratically rather than linearly in the dimension / number of variables. More importantly, the splitting principle for oriented $2k$-dimensional real vector bundles and, say, rational characteristic classes involves invariance under the Weyl group of $\text{SO}(2k)$, which is an index $2$ subgroup of the group of signed permutations on $k$ letters rather than an alternating group as the appearance of the Vandermonde determinant would suggest. In terms of invariant polynomials on the Lie algebra we should instead think of the relationship between the Pfaffian and the determinant. So, should the Wikipedia articles be corrected? Or am I missing some less obvious connection? REPLY [3 votes]: I've taken the liberty of removing this claim about the Vandermonde determinant from all of the relevant Wikipedia articles I could find, listed below for convenience. Euler class Pontryagin class Vandermonde polynomial Characteristic class Splitting principle Alternating polynomial<|endoftext|> TITLE: Does the Gamma function preserve integers? QUESTION [7 upvotes]: Does the Gamma function $\Gamma: \mathbb{C} \to \mathbb{C}$ preserve the Kummer ring $\mathbb{Z}[\exp(2\pi\imath/m)]$? And if not, then what about the Gaussian integers $\mathbb{Z}[\imath]$ or the Eisenstein integers $\mathbb{Z}[\exp(2\pi\imath/3)]$? Is it possible to characterize holomorphic function which do preserve these lattices? REPLY [13 votes]: Note that $$\frac{\Gamma(z) \Gamma(1-z)}{\Gamma(2z) \Gamma(1 - 2z)} = 2 \cos(\pi z),$$ and the RHS is transcendental for any non-rational algebraic number $z$ (by the Gelfond–Schneider theorem). So $\Gamma$ certainly won't preserve any number field $K$. It's most likely true that $\Gamma(z)$ is transcendental for algebraic $z \notin \mathbf{Z}$, but I'm not sure if that's known. REPLY [7 votes]: I don't know about characterization, but there are lots of such functions. In fact, for any map $g$ of the lattice $L$ into itself, there are continuum-many entire functions $f$ such that $f(z) = g(z)$ for $z \in L$. This is because you can get entire functions that take prescribed values on any subset of $\mathbb C$ without limit points.<|endoftext|> TITLE: Isotropy of Apollonian disk-packing QUESTION [6 upvotes]: Is there any sense in which the "epsilon-tail" of an Apollonian disk-packing (by which I mean the union of the disks of radius less than epsilon) exhibits more and more statistical isotropy as epsilon goes to zero? Here's one example of the kind of thing I mean. Label the disks $D_1,D_2,\dots$ in order of weakly decreasing radius. It seems likely that for all $n$ in a set of density 1, there is a unique largest disk $D'_n$ tangent to disk $D_n$, and the ray from the center of $D_n$ to the center of $D'_n$ determines a unit vector $v_n$. Now we can ask whether the end-points of the unit vectors $v_1,v_2,\dots$ are uniformly distributed on the circle. I'm interested in finding out what's known about questions like this. REPLY [3 votes]: Hee Oh has replied to my question and given me permission to post her reply here. What follows are her words (so "I" means "Hee") with my reformatting. What I proved with Shah (Inventiones, 2012) says that for a given Apollonian circle packing $P$, considered as a countable union of circles in the plane, "small circles in $P$ are uniformly distributed with respect to the $\alpha$-dimensional Hausdorff measure of the residual set of $P$, which is the closure of $P$", where $\alpha=1.305...$. For any region $E$ with piecewise smooth boundary the number of circles intersecting $E$ of radius at least $t$ is asymptotic to $c t^{-\alpha} H_{\alpha}(E)$. Chapter 8 of the following article can be useful: http://gauss.math.yale.edu/~ho2/newMSRI_Oh.pdf So it seems that your question amounts to asking how the $\alpha$-dimensional Haudorff measure of the residual set of $P$ (${\rm Res}(P$)) behaves. Since $\alpha$ is bigger than 1, a theorem of Marstrand says that a typical point $x$ in ${\rm Res}(P$) -- typical in the sense of the $\alpha$-dimensional Hausdorff measure -- is a condensation point for ${\rm Res}(P$), meaning that for almost all directions $\theta$, $x$ is a limit point of the intersection of $(x, \theta)$ and Res($P$) where $(x, \theta)$ is the ray from $x$ in the direction of $\theta$. I am referring to Theorem 6.1 of the following paper http://gauss.math.yale.edu/~ho2/BR_JAMS_Final.pdf where we copied a theorem of Marstrand in 1954. So, you do see circles in almost all directions in the above sense. Link to Marstrand's paper: http://dx.doi.org/10.1112/plms/s3-4.1.257<|endoftext|> TITLE: Linearly constrained eigenvalue problem QUESTION [13 upvotes]: Suppose I'd like to: \begin{align} \mathop{\text{min}}_\mathbf{x} && \mathbf{x}^T\mathbf{A}\mathbf{x} \\ \text{subject to:} && \mathbf{x}^T \mathbf{M} \mathbf{x} = 1\\ && \mathbf{C}\mathbf{x} = \mathbf{b} \end{align} where all vector variables are known except $\mathbf{x}$, and $\mathbf{C}$ is full row rank. If I didn't have the $\mathbf{C}\mathbf{x} = \mathbf{b}$ constraint then after applying the Lagrange multiplier method I could solve this as a generalized eigenvalue problem: \begin{align} \text{solve} && \mathbf{A}\mathbf{x} = \lambda \mathbf{M} \mathbf{x}\\ \text{subject to:} && \mathbf{x}^T \mathbf{M} \mathbf{x} = 1\\ \end{align} If I try to apply the same approach to my original problem I get something that doesn't look quite like a generalized Eigenvalue problem: \begin{align} \text{solve} && \left(\begin{matrix} \mathbf{A} & \mathbf{C}^T\\ \mathbf{C} & \mathbf{0} \end{matrix}\right) \left(\begin{matrix} \mathbf{x} &\\ \mu \end{matrix}\right) = \left(\begin{matrix} \lambda \mathbf{M} \mathbf{x}\\ \mathbf{b} \end{matrix}\right)\\ \text{subject to:} && \mathbf{x}^T \mathbf{M} \mathbf{x} = 1 \\ \end{align} Ideally, I'd like to reduce my problem to an instance of the generalized eigenvalue problem so I can use an off-the-shelf numerical solver. What's the best way to solve this problem? (The title of this question is the same, but I couldn't parse the actual question to verify duplicity). REPLY [10 votes]: I think Pushpendre's answer isn't quite right, but it gets you most of the way there. Getting rid of that pesky constant term is a bit tricky relative to the homogeneous case. Let's take his suggested substitutions: $$ \begin{array}{rl} M&:=N^\top N\\ y&:=Nx\\ D&:=CN^{-1}\\ B&:=N^{-\top}AN^{-1} \end{array} $$ I don't think these are strictly necessary (e.g. this encodes an assumption that $M$ is semidefinite), but it simplifies notation quite a bit for the most realistic/common case. I'm going to be cavalier about assuming certain matrices are symmetric/invertible as convenient. Then, your problem becomes $$ \begin{array}{rl} \min_y &y^\top By\\ \mathrm{s.t.} & \|y\|^2=1\\ & Dy=b \end{array} $$ This problem has Lagrange multiplier expression (I am sprinkling in constant factors so that they simplify my arithmetic later) $$ \Lambda(y;\lambda,\mu):=\frac{1}{2}y^\top By+\frac{1}{2}\lambda(1-\|y\|^2)+\mu^\top (b-Dy) $$ Differentiating with respect to $y$ shows $$ 0=\nabla_y\Lambda(y;\lambda,\mu)=By-\lambda y-D^\top\mu. $$ Here I am assuming w.l.o.g. that $B$ is symmetric. Pre-multiplying the critical point condition by $D$ shows $$ DBy=\lambda Dy + DD^\top\mu=\lambda b + DD^\top \mu. $$ Let's further assume that $D$ has full rank. The most common case is that $D\in\mathbb{R}^{m\times n},$ where $m TITLE: On unramified p-adic groups QUESTION [10 upvotes]: Let G be a reductive group over a local field F. Let O be the ring of integers of F. The following are equivalent (and groups satisfying these conditions are called unramified): (a) G is quasisplit and split after passing to an unramified extension of F. (b) G is the generic fibre of a reductive group scheme over O. This is stated, as far as I can tell without a proof or reference to one, in Tits' Corvallis article. My question is - how does one prove this equivalence? REPLY [15 votes]: Let us start with (b) => (a). We only need to show that if $\mathcal{G}$ is a reductive $O$-group scheme, then $\mathcal{G}_F$ is quasi-split and split after an unramified extension. Let us first show that $\mathcal{G}_F$ is quasi-split. Let $\mathrm{Bor}$ be the $O$-scheme parametrizing the Borel subgroups of $\mathcal{G}$, so that $\mathrm{Bor}$ is $O$-smooth (smoothness of the scheme of Borels is true for a reductive group scheme over any base). Over finite fields, reductive groups are quasi-split (ultimately by an application of (an extension of) Lang's theorem), so choose a Borel subgroup of the special fiber of $\mathcal{G}$. Hensel's lemma (or [EGA IV, 18.5.17]) then lifts the chosen residual point of $\mathrm{Bor}$ to an $O$-point. The resulting Borel subgroup of $\mathcal{G}$ shows that $\mathcal{G}$ is quasi-split, and hence so is $\mathcal{G}_F$. Now let us show that $\mathcal{G}_F$ is split after an unramified extension. Reductive group schemes are split etale locally on the base, so $\mathcal{G}$ splits over an etale cover of $\mathrm{Spec } O$. Since $O$ is Henselian, any such cover can be refined by a finite etale cover (look at what covers the closed point), i.e., by the spectrum of the ring of integers of a finite unramified extension $K/F$. Passing to the generic fiber we conclude that $\mathcal{G}_F$ splits over $K$. Now let us prove that (a) => (b). Choose a finite unramified extension $K/F$ over which $G$ splits. Let $G_0$ be the split reductive $O$-group with the same (geometric) root datum $R$ as $G$. Then $G$ is a form of $(G_0)_F$, so $G$ corresponds to an element $x \in H^1(K/F, \mathrm{Aut}_{G_0})$. Recall that $\mathrm{Aut}_{G_0}$ is an extension $$ 1 \rightarrow G_0/Z_{G_0} \rightarrow \mathrm{Aut}_{G_0} \rightarrow \mathrm{Aut}(R, \Delta) \rightarrow 1, $$ where $\Delta$ is a choice of a base of positive roots, and that the extension splits as a semi-direct product after a choice of pinnings (we fix a choice of the latter and of $\Delta$). Let $y \in H^1(K/F, \mathrm{Aut}(R, \Delta))$ be the image of $x$. The $O$-group scheme $\mathrm{Aut}(R, \Delta)$ is constant, so $y$ comes from a unique $z \in H^1(O_K/O, \mathrm{Aut}(R, \Delta))$ (where $O_K$ is the ring of integers of $K$). Let $w \in H^1(O, \mathrm{Aut}_{G_0})$ be the image of $z$ via the chosen semi-direct product decomposition, and let $\mathcal{G}$ be a resulting form of $G_0$. Then $\mathcal{G}_F$ and $G$ are inner forms of each other because their classes in $H^1(F, \mathrm{Aut}_{G_0})$ have the same image $y$ in $H^1(F, \mathrm{Aut}(R, \Delta))$. However, both $\mathcal{G}_F$ and $G$ are quasi-split (the former by (b) => (a)), so they are isomorphic due to the following general fact: over a semi-local base scheme, every reductive group scheme has a unique quasi-split inner form.<|endoftext|> TITLE: Is the regularity of finitely generated rings decidable? QUESTION [19 upvotes]: Q: Is there an algorithm to decide whether a given finitely generated (over $\mathbb{Z}$) commutative ring is regular? I mean by regular that the localization at every prime ideal is a regular local ring. The question arose from my interest in the desingularization problem. To have a desingularization algorithm of arithmetic schemes, one first needs to know the regularity of a given scheme. The definition of the regularity is point-wise. Naively one has to check the regularity point by point of $\mathrm{Spec}\, R$ for a given ring $R$. It is not an algorithm in the sense that it never halts if $R$ is regular. If $R$ is defined over a prime filed, $\mathbb{Q}$ or $\mathbb{F}_p$, then one can use the Jacobian criterion: first compute the Jacobian ideal and then check if it is trivial by computing its reduced Gröbner basis. In positive characteristic, one may also use Kunz's criterion in terms of Frobenius maps. As far as I know, there is no such a global criterion for rings over $\mathbb{Z}$. Serre's criterion by the finiteness of global dimension looks global at the first glance. But one needs to know the projective dimensions of infinitely many modules. So, my guess is that the answer to the question would be NO. Does someone know the answer or related works? REPLY [5 votes]: OK, we may assume the ring $R$ is a domain. Using the Jacobian criterion we can get a computable Zariski open $U \subset \text{Spec}(R)$ which is regular. Let $\mathfrak p \subset R$ be a prime ideal corresponding to a generic point of $\text{Spec}(R) \setminus U$. Let us say there is an algorithm to compute the dimension of $R_\mathfrak p$ and a minimal set of generators $f_1, \ldots, f_c$ of $\mathfrak p R_\mathfrak p$. I think there are even compute algebra packages which will allow you to do so (as well as the other computations below). Of course, it is completely justified to complain that I haven't shown you that one can actually make these computations and I may have overlooked other issues as well. If $c < \dim(R_\mathfrak p)$, then $R$ is not regular and we are done. If $c = \dim(R_\mathfrak p)$, then our task is to explicitly find a Zariski open neighbourhood $V$ of $\mathfrak p$ in $\text{Spec}(R)$ which is regular. To do this we may assume (after replacing $R$ by a principal localization) that (a) $\text{Spec}(R) = U \cup V(\mathfrak p)$, (b) $R/\mathfrak p$ is regular (again using some Jacobian criterion), (c) $f_1, \ldots, f_c \in R$, (d) $\mathfrak p = (f_1, \ldots, f_c)$, and (e) $f_1, \ldots, f_c$ is a regular sequence. To achieve (d) and (e) you have to know how to compute cohomology groups of explicitly given complexes of finite module and annihilators of finite modules. Once you have (a) -- (e), then $R$ is regular at every point of $V(\mathfrak p)$ by Tag 00NU and hence $R$ is regular but this just means we have found a regular neighbourhood in the original Spec.<|endoftext|> TITLE: The continuum hypothesis for packing shapes without overlapping QUESTION [12 upvotes]: Consider the finite cross $C$ (=union of line segments $\overline{(0, -1)(0, 1)}$ and $\overline{(-1, 0)(1, 0)}$) and the unit half-circle $H$. It is easy to see that we may pack continuum-many disjoint copies of $H$ into the plane $\mathbb{R}^2$ without overlapping, whereas for $C$ we may never pack uncountably many copies without overlapping. By "copy," here, I mean "copy up to rotation and translation," or more generally "image under an element of the affine special orthogonal group." The specific choice of transformation group to use isn't important, so long as it is reasonable. For $S\subseteq\mathbb{R}^n$, let $\mathfrak{p}(S)$ be the supremum of the cardinalities of disjoint packings of copies of $S$ inside $\mathbb{R}^n$. In case the continuum hypothesis holds, either $\mathfrak{p}(S)$ is countable or $\mathfrak{p}(S)=2^{\aleph_0}$; however, in general this need not be true. For example, let $G\subseteq (\mathbb{R}, +)$ be a subgroup of index $\aleph_1$ and let $H$ be a coset of $G$; then the copies of $H$ are precisely the cosets of $G$, so $\mathfrak{p}(H)=\aleph_1$. My question is: suppose $\neg CH$. Under what assumptions - both on the shape $S$ and the ambient axioms of set theory - can we conclude that $\mathfrak{p}(S)$ is either countable or continuum? For example: Does ZFC prove that, for $S\subseteq\mathbb{R}^n$ compact, $\mathfrak{p}(S)$ is either countable or continuum? I suspect the answer is "yes," but I don't see how to prove it. REPLY [5 votes]: Suppose $K \subseteq \mathbb{R}^n$ is compact. Let $G$ be the group of isometries of $\mathbb{R^n}$. The compact-open topology on $G$ is defined by declaring the sets $W_{A. U} = \{f \in G : f[A] \subseteq U\}$ open, for each compact $A \subseteq \mathbb{R}^n$ and open $U \subseteq \mathbb{R}^n$. This is the smallest topology that makes the evaluation function $(g, x) \mapsto g(x)$ continuous. It can be shown that this makes $G$ a locally compact, Polish group. Let $d$ be a witnessing metric (Kechris' Descriptive set theory book gives a formula for this metric on page 60). Let $S = \{f \in G : f[K] \cap K = \phi\}$. Then $S = W_{K, \mathbb{R}^n \backslash K}$ is open in $G$. Claim: If $id \notin \text{cl}(S)$, then every family of pairwise disjoint copies of $K$ is countable. Proof: Suppose $X \subseteq G$ is uncountable. Since $G$ is second countable, there exists $g \in X$ such that every neighbourhood of $g$ meets $X$ at uncountably many points. Let $W$ be an open set containing $id$ such that $W \cap \text{cl}(S) = \phi$. Pick $f \in X \cap gW$, $f \neq g$. Then, $(g^{-1}f)[K] \cap K \neq \phi$ hence also $f[K] \cap g[K] \neq \phi$. Claim: Assume $id \in \text{cl}(S)$. There exists a binary tree $\langle B_{\sigma} : \sigma \in 2^{< \omega} \rangle$ of open sets with compact closure in $G$ such that for all $\sigma$, for $i = 0, 1$, $\text{cl}(B_{\sigma i}) \subseteq B_{\sigma}$ and for every $f_i \in B_{\sigma i}$, $f_0[K] \cap f_1[K] = \phi$. Hence, the family $\{f[K]: (\exists x \in 2^{\omega})(\forall n)(f \in B_{x \upharpoonright n}) \}$ is pairwise disjoint and has size continuum. Proof: To construct such a tree, choose some $f \in B_{\sigma}$. Using the assumption that the isometries that map $K$ to a set disjoint with $f[K]$ accumulate near $f$, pick $g \in B_{\sigma}$ such that $f[K] \cap g[K] = \phi$. Let $B_{\sigma i}, i = 0, 1$ be open sets such that $f \in B_{\sigma0}, g \in B_{\sigma1}$, $\text{cl}(B_{\sigma i}) \subseteq B_{\sigma}$ and for every $h_i \in B_{\sigma i}$, $h_0[K] \cap h_1[K] = \phi$. This can be done by separating the compact sets $f[K], g[K]$ by open sets $U_f, U_g \subseteq \mathbb{R}^n$ and considering the disjoint neighbourhoods $W_{K, U_f}, W_{K, U_g}$ of $f, g$ respectively.<|endoftext|> TITLE: Higher vector spaces QUESTION [9 upvotes]: As far as I know there are different ways to categorify the notion of vector space/module. These appear (for example) when trying to find extended TQFTs. There are at least two ways (presented at $2$-vector-space at nLab and more generally in the article on $(\infty,n)$-modules): The 2-category $\mathsf{Vect}_2$ whose objects are $k$-linear categories, $1$-morphisms are functors and $2$-morphims are natural transformations; The 2-category $1\text{-}\mathsf{Alg}$ whose objects are associative $k$-algebras, morphisms are bimodules and $2$-morphisms are bimodules morphisms. The two are linked by some kind of Tannaka duality: there's a functor $1\text{-}\mathsf{Alg} \to \mathsf{Vect}_2$ (described here) that maps an algebra $A$ to its category of modules $\mathsf{RMod}_A$, an $(A,B)$ bimodule $N$ to the functor $- \otimes_A N$, and a bimodule morphism to the associated natural transformation. I think this is fully faithful but not essentially surjective. Am I right? How to describe the essential image of this functor? A category in the image has to at least be monoidal and rigid, for example. Is the image possibly the fully dualizable objects? REPLY [7 votes]: As I already noted in the comments: A $k$-linear category is equivalent to a module category iff it is abelian, cocomplete, and there exists a compact projective generator. This is a version of Morita's theorem, which can be found e.g. in this article by Bernhard Keller. For the question about which functors are equivalent to tensor functors the answer is given by Eilenberg-Watts theorem stating that these are exactly the functors which are right exact and preserve small coproducts (see e.g. Athens following article in nlab: http://ncatlab.org/nlab/show/Eilenberg-Watts+theorem )<|endoftext|> TITLE: Poincaré duality for (co)homology of Lie algebras? QUESTION [8 upvotes]: Let $R$ be a commutative ring and $\mathfrak{g}$ a Lie $R$-algebra that has an $R$-module basis with $n$ elements. In Algebra, Geometry, and Software Systems by Joswig & Takayama on p.200, it says that $H_k(\mathfrak{g};R)\cong H_{n-k}(\mathfrak{g};R)$ when $R$ is a field of characteristic $0$. Does Poincaré duality $H^k(\mathfrak{g};R)\cong H_{n-k}(\mathfrak{g};R)$ hold over any $R$ (or at least any PID such as $\mathbb{Z}$)? Also, the literature for (co)homology of Lie algebras seems scarce, please list any book that deals with this topic. So far, I have Weibel (Homological Algebra), Azcarraga & Izquierdo (Lie Groups, Lie Algebras, Cohomology), Hilgert & Neeb (Structure and Geometry of Lie Groups). REPLY [10 votes]: First, let me expand on the reply of Dietrich Burde: I got hold of the paper of Hazewinkel, and can now be more precise about what is and what is not there (last time I saw it was some years ago). Hazewinkel's most general result (for not necessarily trivial coefficients) is over a field. A result that he proves over a ring is as follows: Let $R$ be a hereditary ring, $\mathfrak{g}$ an Lie algebra over $R$ which is a free $R$-module of rank $n$, and $M$ a finitely generated $\mathfrak{g}$-module which is projective as an $R$-module. Then there exists a canonical splitting exact sequence $$ 0\to\mathrm{Ext}^1(H^{s+1}(\mathfrak{g},M),R)\to H^{n-s}(\mathfrak{g},(M^{\mathrm{tw}})^*)\to H^s(\mathfrak{g},M)^*\to 0 $$ where the $\mathfrak{g}$-module $M^{\mathrm{tw}}$ has the same underlying $R$-module, but the action on it is given by $\rho^{\mathrm{tw}}(x)=\rho(x)-\mathop{\mathrm{tr}}(\mathop{\mathrm{ad}}(x))\cdot1$. In the case of $M=R$, $(M^{\mathrm{tw}})^*$ is $R$ with the $\mathfrak{g}$-action given by multiplying by trace of the adjoint action. The example of YCor in the comment to the original question, in particular, is addressed by this, of course (there, incidentally, $R^{\mathrm{tw}}\simeq R$ as $\mathfrak{g}$-modules): we have $H^1(\mathfrak{sl}_2,\mathbb{Z})=0$ and $H^2(\mathfrak{sl}_2,\mathbb{Z})=(\mathbb{Z}/2\mathbb{Z})^2$, also we have $H_1(\mathfrak{sl}_2,\mathbb{Z})=(\mathbb{Z}/2\mathbb{Z})^2$, and applying Hazewinkel's result for $s=1$ (so $s+1=2$), we see no contradiction with $\mathrm{Ext}^1(H^{s+1}(\mathfrak{sl}_2,\mathbb{Z}),\mathbb{Z})=\mathrm{Ext}^1((\mathbb{Z}/2\mathbb{Z})^2,\mathbb{Z})=(\mathbb{Z}/2\mathbb{Z})^2$. However, for the kind of information you seek (Poincaré-type duality between homology and cohomology), the answer is simpler, since the main reason for the universal coefficients-kind phenomena in the formulas of Hazewinkel is that we need to dualise something we do not want to dualise :) Let me actually first give an example similar to the one Mariano gave, it is a nice simple exercise anyway, and it shows once again that twisting coefficients is unavoidable. Let $\mathfrak{g}$ be a Lie algebra over $\mathbb{Z}$ with two basis elements $x,y$ and the bracket $[x,y]=2y$. Then $$ H_i(\mathfrak{g},\mathbb{Z})=\begin{cases}\mathbb{Z}, i=0,\\ \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}, i=1,\\ 0, i>1,\end{cases} $$ $$ H_i(\mathfrak{g},\mathbb{Z}^{\mathrm{tw}})=\begin{cases}\mathbb{Z}/2\mathbb{Z}, i=0,\\ \mathbb{Z}, i=1,\\ \mathbb{Z}, i=2,\\ 0, i>2,\end{cases} $$ $$ H^i(\mathfrak{g},\mathbb{Z})=\begin{cases}\mathbb{Z}, i=0,\\ \mathbb{Z}, i=1,\\ \mathbb{Z}/2\mathbb{Z}, i=2,\\ 0, i>2,\end{cases} $$ $$ H^i(\mathfrak{g},\mathbb{Z}^{\mathrm{tw}})=\begin{cases}0, i=0,\\ \mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}, i=1,\\ \mathbb{Z}, i=2,\\ 0, i>2,\end{cases} $$ This example makes one think that we must have $$H_i(\mathfrak{g},R)\simeq H^{n-i}(\mathfrak{g},R^{\mathrm{tw}})\quad \text{ and }\quad H_i(\mathfrak{g},R^{\mathrm{tw}})\simeq H^{n-i}(\mathfrak{g},R).$$ Indeed, this is true because the pairs of complexes computing that (co)homology are isomorphic. Basically, take some basis $e_1\ldots,e_n$ of $\mathfrak{g}$, and send each multivector $e_{i_1}\wedge\cdots\wedge e_{i_k} (i_1<\dots TITLE: About the hypothesis of Zorn's lemma QUESTION [7 upvotes]: The proofs I know of Zorn's lemma give the following refinement: Let $(X,<)$ be a partially ordered set such that every well-ordered subset of $X$ has an upperbound. Then $X$ has a maximal element. In fact, Zorn's lemma is sometimes stated as such. In comparison, the usual statement asks that every totally ordered subset of $X$ has an upper bound. Does this refinement have some application? EDIT (after comments). Your comments make me realize that I did not think enough of my question, which is what the voters-to-close probably guessed. To help future readers, let me sum up the comments. If $(X,<)$ satisfies the hypothesis of the refined Zorn lemma (RZL), then (modulo AC), it satisfies the hypothesis of classical Zorn lemma as well, since any totally ordered set contains a cofinal ordered subset. (Comment of Ramiro de la Vega). ZF+RZL implies ZF+AC (Noah S). In Cohen's first model of ZF, there exists a set which satisfies the hypothesis of RZL but not that of ZL (Asaf Karagila). REPLY [11 votes]: According to Campbell, Paul J. The origin of "Zorn's lemma''. (French summary) Historia Math. 5 (1978), no. 1, 77–89, results in topology gave motivation to the Kuratowski-Zorn lemma (as it is known where I come from), starting with the result of Zygmunt Janiszewski from 1910 that every continuum between two given points contains a minimal continuum. The version of the lemma highlighted in your question was in fact proved by Kazimierz Kuratowski in 1922 (that is, 13 years before Max Zorn came up with his), in the paper. "Une méthode d'élimination des nombres transfinis des raisonnements mathématiques", Fundamenta Mathematicae 3: 76–108. http://matwbn.icm.edu.pl/ksiazki/fm/fm3/fm3114.pdf As applications of his theorem, Kuratowski gives the proofs of the mentioned result by Janiszewski (in a generalized form), as well as other theorems in point-set topology (e.g. Cantor-Bendixson theorem), Lebesgue's theorem on the existence of an uncountable class of Borel measurable sets and the existence of some Baire classes of functions.<|endoftext|> TITLE: Geodesics in finite groups QUESTION [22 upvotes]: It seems that I can generalize a result from compact, connected Lie groups to finite groups, but in order to do so, I need to have some kind of geodesics on finite groups. Below is a proposition for the definition of a geodesic. My main question is whether such geodesics have been studied or used. Definition and justification: If $G$ is a compact, connected Lie group with a bi-invariant metric, we can construct all closed geodesics as follows. Take any nontrivial homomorphism $\phi:S^1\to G$ and an element $x\in G$. Then $S^1\ni t\mapsto x\phi(t)\in G$ is a closed geodesic. The same construction can be done for finite groups if the circle $S^1$ is replaced with a cyclic group $C_n$. More explicitly, a geodesic of length $n$ on a finite group $G$ is a mapping $C_n\ni t\mapsto x\phi(t)\in G$, where $x\in G$ and $\phi:C_n\to G$ is a nontrivial homomorphism. We require $n>1$ in order to exclude singletons as geodesics. It does not matter whether we multiply by $x$ from the left or from the right. Left and right translations (cosets) give the same geodesics since if $\phi:C_n\to G$ is a nontrivial homomorphism, so is $t\mapsto x\phi(t)x^{-1}$. Reasons why this feels like a good definition: Every subgroup is totally geodesic in both cases (Lie and finite). In every nontrivial group there exists at least one geodesic in both cases. Geodesics are invariant under left and right translations in both cases. A finite abelian group is a product of cyclic groups, and a compact, connected, abelian Lie group is a product of copies of $S^1$ (a torus). Motivation: It is a typical problem in integral geometry (a subfield of inverse problems) to ask whether a function on a closed manifold is determined by its integrals over closed geodesics. In the case of compact, connected Lie groups, this is possible if and only if the group is not the trivial group, $S^1$ nor $S^3$. I wanted to consider the corresponding problem on finite groups, where of course the integral is replaced with a sum. That is, is a function on a finite group determined by its sums over all geodesics? It seems that I can give a decent answer, but not a complete classification of groups where the answer is affirmative. My current classification covers, for example, all abelian, symmetric, alternating, dihedral and dicyclic groups. Questions: In order of descending importance, this is what I would like to know: Have geodesics in finite groups been studied before, perhaps under another name? Has this "integral geometry problem" in finite groups been studied or does it have applications (concrete or abstract)? Do these geodesics have some length minimization property analogously to the continuous case? If you can answer any of the questions with a modified definition of geodesics, please do so. I am looking for a reasonable definition — or several if there are several — and I certainly do not want to forbid other definitions than mine. One possible variant is described below. An answer about any particular finite group is most welcome. I am not asking for you to solve my "discrete integral geometry problem" here, since I already have a rather comprehensive answer; I want to know if this problem and the related geodesics are already known. References to any existing answers are, of course, most welcome. (If someone wants to know what I know about the problem, please contact me personally. There is too much to be included here.) Norms on finite groups seem to have been studied, but I have found no study of geodesics. Geodesics as dynamical systems: (For those who prefer this point of view.) Geodesics on Lie groups (and all other Riemannian manifolds) can be realized as continuous time dynamical systems. There is a Hamiltonian flow on the cotangent bundle $T^*M$ (or the cosphere bundle $S^*M$) so that the projections of flow lines to $M$ are the geodesics. Similarly, let $G$ be a finite group and consider the discrete time dynamical system on $G\times G$ that sends $(a,b)$ to $(b,ba^{-1}b)$. Note that a system starting outside (resp. on) the diagonal $\Delta\subset G\times G$ stays outside (resp. on) the diagonal. By finiteness any "discrete flow line" of this system on $G\times G\setminus\Delta$ is periodic. The projections of such flow lines to the first component of $G\times G$ are in one-to-one correspondence with discrete geodesics in the sense defined above. A variant: Peter Michor suggested a different formulation in the comments below. I defined geodesics to be translations of cyclic subgroups, but it also makes sense to define geodesics as translations of maximal cyclic subgroups. In this setting a "Lie subgroup" $H$ of the finite group $G$ is such a subgroup that any maximal cyclic subgroup on it is also maximal in $G$. Then there are also non-Lie subgroups but all Lie subgroups are totally geodesic, analogously with the positive dimensional Lie groups. Answers to my questions with this kind of geodesics are also very welcome. Notes: A recent meta discussion showed green light to asking a question of this kind. There is an earlier question about geodesics on graphs, but it does not answer my question. REPLY [6 votes]: First, my apologies for this late answer, I only found the question today. Below, I probably recall too many things, but I felt it could put some context around the short answer to question 1 saying: yes, under the names "geodesic in an involutory quandle", or "cycle in a symmetric set". 1) Recall that a homogeneous symmetric space (I borrow the terminology from Loos [2]) is a homogeneous space $G/H$ of a Lie group $G$, where $H$ is an open subgroup of the fixed points subgroup $G^\sigma$ of an involution $\sigma$ of $G$ ($\sigma\in Aut(G)$ and $\sigma^2=Id$). A homogeneous symmetric space has a canonical connection $\nabla$ for which the geodesics are of the kind $t\mapsto g\cdot \exp(tX) H$ with $g\in G$ and $X\in \mathfrak{p}$. Here $\mathfrak p=\{X\in Lie(G) \mid Lie(\sigma)(X)=-X\}$. Every Lie group $K$ (compact or not) is a homogeneous symmetric space, with $G=K\times K$, $\sigma(k,k')=(k',k)$ and $H=diag(K)$. 2) There is an intrinsic presentation of symmetric spaces due to Loos: a symmetric space is a smooth manifold with a smooth product law $(x,y)\mapsto x\bullet y=s_xy$ such that $s_xx=x$ $s_xs_xy=y$ $s_x(s_yz)=s_{s_xy}(s_xz)$ $x$ is an isolated fixed point of $s_x$ The maps $s_x:M\to M$ are called the symmetries. A morphism of symmetric spaces $M\to N$ is a smooth map $\phi:M\to N$ such that $\phi(s_xy)=s_{\phi(x)}\phi(y)$. Any homogeneous symmetric space $G/H$ is a symmetric space in this sense, with $s_{gH}(g'H) = g\sigma(g^{-1}g')H$. Conversely, any connected symmetric space $M$ (with a choice of base point $o\in M$) is a homogeneous symmetric space $G/H$, with $G$ the subgroup of $Aut(M)$ generated by $\{s_xs_y\mid x,y\in M\}$ (it is a finite-dimensional Lie group), involution $\sigma(g)=s_ogs_o$, and $H=Stab_G(o)$. In this context, it can be seen that a geodesic in $M$ is simply a morphism of symmetric spaces from the real line $\mathbb R$ to $M$. Here, $\mathbb R$ has the symmetries $s_xy=2x-y$. 3) If we remove axiom 4. in the definition of symmetric space (and forget about smoothness), we get a purely algebraic object which appears under various names in the literature: kei (Takasaki [5]), symmetric set (Nobusawa [3], for finite sets), involutory quandle (Joyce [1]), symmetric groupoid (Pierce [4]). In analogy with the smooth case, one may define a geodesic in $M$ as a morphism of such spaces, from the integers $\mathbb Z$ to $M$. Here, $\mathbb Z$ has the symmetries $s_xy=2x-y$. Joyce gives an abstract definition of involutory quandle with geodesics [1, p. 30] and shows that any involutory quandle can be seen as an involutory quandle with geodesics, essentially by defining the geodesics as is done above. Obviously, a geodesic is determined by the images of 0 and 1 (the point-and-tangent-vector datum determining a geodesic in the smooth case is replaced by a pair of points in the discrete case). Nobusawa [3, pp. 570-571] calls these geodesics cycles (symmetric subspaces generated by two points). As in the smooth case, any (finite) group $G$ can be seen as a (finite) symmetric set by setting $s_gh=gh^{-1}g$. In the finite group case, the geodesics then coincide with your definition (except that singletons are not excluded). 4) That said, to my knowledge, these geodesics have not been much studied for themselves. I don't know the answer to questions 2 and 3 (except that in the present context, no metric is involved). References [1] David Joyce, An Algebraic Approach to Symmetry with Applications to Knot Theory, Thesis. http://aleph0.clarku.edu/~djoyce/quandles/aaatswatkt.pdf [2] Ottmar Loos, Symmetric Spaces. 1: General theory, Benjamin, New York, Amsterdam, 1969. [3] Nobuo Nobusawa, On symmetric structure of a finite set, Osaka J. Math. Volume 11, Number 3 (1974), 569-575. http://projecteuclid.org/euclid.ojm/1200757525 [4] R. S. Pierce, Symmetric groupoids, Osaka J. Math. Volume 15, Number 1 (1978), 51-76. http://projecteuclid.org/euclid.ojm/1200770903 [5] M. Takasaki, Abstractions of symmetric functions, Tohoku Math. J. 49 (1943), 143-207, [Japanese]. https://www.jstage.jst.go.jp/browse/tmj1911<|endoftext|> TITLE: Guessing the larger integer: A game-theoretic twist QUESTION [9 upvotes]: The starting point for this question is the old chestnut, already discussed on MO, about a game show on which the host has chosen two distinct integers and the contestant gets to reveal one of them at random, upon which the contestant must guess whether the remaining integer is larger than or smaller than the revealed integer. I would like to suggest that there is a certain precise sense in which the commonsense intuition that the contestant cannot beat the host is correct. As explained in the answers to the other MO question, the usual statement of the puzzle does not yield a precise definition of the "success probability" of the contestant. Allow me to suggest the following interpretation. The game show is repeated a countably infinite number of times. The host strategy is defined as follows: For each $n$, there is a probability distribution $h_{n,T}$ on pairs of distinct integers, that depends on $n$ as well as on the transcript $T$ of what happened on all rounds prior to the $n$th round. In other words, the host's choice of integers in the $n$th round is governed by some probability distribution that is allowed to depend on what the host has observed happen in previous rounds, but is otherwise fixed. Similarly, the contestant strategy is determined by a family of probability distributions $c_{n,T,v}$ that depend on $n$ and $T$ as well as on the value $v$ of the randomly chosen revealed integer. Under these conditions, we have a clear notion of what it means for the contestant to win any specific round $n$, and once the host strategy and the contestant strategy are fixed, we can talk about probabilities of winning and so forth. Now let me state a key definition: A contestant strategy beats a host strategy if there exists $\delta>0$ such that $P_n$, the proportion of the first $n$ rounds won by the contestant, converges in probability to $1/2 + \delta$ as $n\to\infty$. In my opinion, if you tell people that the contestant has a strategy that wins with probability greater than 1/2, many will instinctively interpret your claim in the above sense. That is, they imagine that if the game were played over and over again, the contestant would win more than 1/2 the time in the long run. This certainly seems impossible, especially in the absence of any conditions on the host strategy. If you then clarify that you mean that for any fixed choice of integers, there is a way to beat the host, then this is unsurprising but also seems to miss the point. After all, if the host is required to choose the same pair of integers in every round, aren't there much better strategies than random guessing? Like noticing the pattern after a while and always guessing correctly thereafter? You may not buy my argument that my definition of "beats" is an intuitively reasonable one, but regardless, let me proceed to the mathematical question: Is it true that (a) for every fixed host strategy, there is a contestant strategy that beats it, and (b) for every fixed contestant strategy, there is a host strategy that it does not beat? My guess is that the answer to (a) is no and that the answer to (b) is yes, and that it can be proven by a minimax argument, but I don't actually know. Note, of course, that the "standard" contestant strategy that answers the other MO question fails to beat the simple host strategy of deterministically picking $n$ and $n+1$ in round $n$; even though the win probability in every round is strictly greater than 1/2, the probabilities cannot be bounded away from 1/2 and hence cannot yield the desired $\delta$. Note also that allowing randomized strategies is, as usual in game theory, crucial, since deterministic strategies can trivially be defeated. If I were Lance Fortnow, I would next ask about computationally bounded strategies, but let's not get ahead of ourselves. REPLY [5 votes]: Any time the game is played, the host can choose a distribution that makes the contestant's advantage as close to 0 as desired (although not actually equal to 0). In your multi-round formulation, the host could, for example, behave as follows: in round $n$, choose $k$ uniformly at random from $\{1,2,\dots, n\}$ and declare $\{k, k+1\}$. If the revealed number is $1$ or $n+1$ the contestant can win for sure, and otherwise it's a 50-50 guess. So the contestant's optimal chance of winning in round $n$ (even knowing the host's strategy) is $\frac12+\frac1{2n}$. So indeed (a) no and (b) yes.<|endoftext|> TITLE: If $ F(x,\bullet) \in {L^{\infty}}(G,B) $ for all $ x \in G $, then is $ x \mapsto F(x,\bullet) $ strongly measurable? QUESTION [5 upvotes]: Let $ (X,\Sigma,\mu) $ be a $ \sigma $-finite measure space and $ B $ a Banach space. A function $ f: X \to B $ is said to be strongly $ \mu $-measurable iff it is the almost-everywhere pointwise limit of a sequence $ (s_{n}: X \to B)_{n \in \mathbb{N}} $ of integrable simple functions, where an integrable simple function $ s: X \to B $ has the form $$ s = \sum_{(E,b) \in I} \chi_{E} \cdot b $$ for some finite subset $ I $ of $ \{ E \in \Sigma \mid \mu(E) < \infty \} \times B $. Let $ G $ be a second-countable, locally compact Hausdorff group and $ \mu_{G} $ a fixed Haar measure on the Borel $ \sigma $-algebra $ \mathscr{B}(G) $ of $ G $. The second-countability condition implies that $ G $ is $ \sigma $-compact, which ensures that $ (G,\mathscr{B}(G),\mu_{G}) $ is a $ \sigma $-finite measure space. Let $ B $ be a separable Banach space and $ {L^{\infty}}(G,B) $ the set of all (equivalence classes of) $ B $-norm essentially bounded strongly $ \mu_{G} $-measurable functions from $ G $ to $ B $. Note: $ {L^{\infty}}(G,B) $ is a Banach space, and except in trivial cases, it is always non-separable. Question. If $ F: G \times G \to B $ is a strongly $ \mu_{G \times G} $-measurable function where $ F(x,\bullet) \in {L^{\infty}}(G,B) $ for all $ x \in G $, then is it true that the mapping \begin{align} G & \to {L^{\infty}}(G,B); \\ x & \mapsto F(x,\bullet) \end{align} is strongly $ \mu $-measurable? Thank you all very much for your help! REPLY [2 votes]: I think the answer is no, even in the case of $G:=\mathbb{R}$ with the Lebesgue measure and $B: =\mathbb{R}$ as a Banach space. Let $F: \mathbb{R}\times \mathbb{R} \to \mathbb{R} $ be the characteristic function of the half-plane above the diagonal: $F(x,y):=\chi_\mathbb{{R}_+}(y-x)$. So $F\in L^\infty(\mathbb{R}\times \mathbb{R} )$; however the corresponding map $\tilde F: \mathbb{R} \to L^\infty(\mathbb{R} )$ such that $x\mapsto F(x,\cdot)=\chi_{[x,\infty)}$ is not strongly measurable, because its range is not separable, as it is the discrete uncountable set $\{\chi_{[x,\infty)} : x\in \mathbb{R}\}$ (not even if we modify $\tilde F$ on some Lebesgue null set). Rmk 1. A Banach space valued strongly measurable map $f$ on a measure space $X$ is necessarily essentially separable , that is, up to removing a null set $N\subset X$, its image $f(X\setminus N)$ is a separable subset of $B$ (indeed it is included in the closure of the countable union of the images of the simple functions that converge pointwise to $f$; and a simple function is exactly a measurable function with finite range). Rmk 2. A theorem of Pettis states that a Banach space valued map $f$ on a measure space $X$ is strongly measurable if and only if: (i) it is essentially separable; (ii) it is weakly measurable (that is, measurable w.r.to the $\sigma$ algebra generated by the weak topology; (iii) its support is $\sigma$-finite. (See e.g. the Kōsaku Yosida's Functional Analysis. The proof is easy. ).<|endoftext|> TITLE: Tangent space describes the manifold's first order characteristic. Is there something like tangent space describes higher order characteristic? QUESTION [6 upvotes]: I'm learning differential geometry. I'm curious that when we learned analysis, we learned higher order derivative, while in differential geometry, first order derivative is generalized to element of tangent space. Then my question is what's the higher order derivative in differential geometry? Are there some literatures including this area? Or maybe this is just a trivial question because there is nothing interesting in higher order derivative. REPLY [9 votes]: There are several possibilities: 1.The most naive one is the following: Consider a smooth map $f\colon M\to N.$ Then its derivative $df\colon TM\to TN$ is again smooth, and you can again differentiate. Clearly, you take a lot of useless information around, so this is usually not the method used by differential geometers to solve any problems. Another possibility is by defining so-called jet bundles in a similar way for higher order derivatives as the tangent bundle is defined for first order derivatives. There are many books on this, check for example wikipedia. The "problem" with jet bundles is, that they are no vector bundles, and one somehow loses the (mulit)-linear algebra tools.. Differential geometer often use connections on vector bundles to define higher order derivatives. As the most basic example, one should mention the Levi-Civita connection $\nabla$ (defined by a Riemannian metric on the manifold). With its help you can derivate vector fields along vector fields. A nice and simple application of its use is the following: (Locally) shortest curves are exactly those curves which satisfy the second order differential equation: $$\nabla_\gamma' \gamma'=0.$$<|endoftext|> TITLE: Euler characteristic of open varieties as degree of Chern class of logarithmic differentials QUESTION [5 upvotes]: Let $U$ be a smooth variety over a subfield $k$ of $\mathbb{C}$. Let $X$ be a smooth projective variety containing $U$ as the complement of a normal crossings divisor $D$. Denote by $\chi(U)$ the Euler characteristic of $U$, defined using either de Rham cohomology or singular cohomology of the complex analytic manifold attached to $U$. Let $\Omega^1_X(\log D)$ be the sheaf of logarithmic differentials. If $X$ has dimension $n$, then it is locally free of rank $n$ and one has the equality $$ \chi(U)=\deg c_n(\Omega^1_X(\log D)) $$ This is quite easy to prove using the Riemann-Roch theorem, but I have been unable to find a written proof. Can anybody provide a reference that I can cite in a paper? REPLY [3 votes]: A simple explaination of the equivalent dual statement, i.e. using $T_X(- \log D)$ instead of $\Omega^1(\log D),$ can be found in the book by Burt Totaro Group cohomology and algebraic cycles, page 25. The corresponding link on googlebooks is here. I am sure that it can be found in many other places, anyway this is just the first explicit reference I can remember now.<|endoftext|> TITLE: Rep-tiles of order 2 QUESTION [9 upvotes]: A 2-rep-tile is a geometric shape that can be partitioned into exactly 2 smaller (dilated) copies of itself. Although there are many rep-tiles of higher orders, the only 2-rep-tiles I could find are: A. A right-angled isosceles triangle: B. An A4 paper (a $1\times \sqrt 2$ rectangle): Are there other 2-rep-tiles? Alternatively, is there a proof that other 2-rep-tiles (or at least other 2-rep-tiles that are polygons, etc.) do not exist? REPLY [3 votes]: There is a published paper ("On 2-Reptiles in the Plane") proving that there are exactly 6 rep-2-tiles. But this restricts the definition of a rep-tile to a dissection into directly similar parts. If you allow inversely similar (reflected) parts there are more. These include an infinite number of different parallelograms, parameterised by the value of the smaller angle. If you change the question from the number of rep-2-tiles to the number of classes of rep-2-tiles interconvertible by affine transformations there are at least 10, and possibly 12 (I haven't succeeded in demonstrating that two of them tile the plane). If you relax the definition to allow irreptiles (dissections in which the parts are similar, but of different sizes) there's another infinite class - of right angled triangles. And there are at least another 8 irreptiles corresponding to the polynomials $x+x^2=1$ (the golden bee shown in Aaron Meyerowitz's answer), $x+x^3=1$ (4 tiles) and $x+x^5=1$ (3 tiles). There are papers which show that the existence of a Perron number is a necessary and sufficient condition for the existence of a self-similar tile. As the coeffecients of the corresponding polynomials correspond to the number of elements in the dissection this allows constraints of the number of irreptiles of order 2 to be deduced. I don't understand the papers, but I think that they make a restriction to directly similar dissections, so I am not confident that the seven tiles shown at http://www.meden.demon.co.uk/Fractals/dimerIRR.html plus the golden bee and the right-angled triangle are an exhaustive set. (Wayback Machine)<|endoftext|> TITLE: Center of one-point stabilizer in 2-transitive groups QUESTION [8 upvotes]: In this MO question it was mentioned that the following fact seems to be true: If $G$ is doubly transitive on $X$ and the one-point stabilizer $G_x$ has a non-trivial center, then $G$ is of affine type, that is, the socle is elementary abelian. Does anyone know if it is true? I checked it with GAP for all 2-transitive groups up to degree 2499 and I no counterexamples appeared. Here is the code for groups with maximun degree of transitivity equal to two: for gr in AllPrimitiveGroups(NrMovedPoints, [1..2499], Transitivity, 2) do stab := Stabilizer(gr, 1); if not IsTrivial(Center(stab)) then soc := Socle(gr); if not IsAbelian(soc) then Print("Counterexample of degree ", NrMovedPoints(gr), "\n"); fi; fi; od; I am mainly interested in the following question: Is it true that if $G$ is a 2-transitive group with simple socle then $Z(G_x)=1$? I need only to consider groups with maximum degree of transitivity equal to two. Further, many of the groups appearing in the classification of 2-transitive groups with simple socle are easy to handle. However, I cannot prove the fact in the following cases: Socle $PSL(d,q)$, degree $\frac{q^d-1}{q-1}$, and $d\geq3$ (two actions). Socle $PSU(3,q)$, degree $q^3+1$ and $q\geq3$. Socle $Sz(q)$, degree $q^2+1$ and $q=2^{2d+1}>2$. Socle $Ree(q)$, degree $q^3+1$ and $q=3^{2d+1}>3$. Is it true that if $G$ is a 2-transitive almost simple group with simple socle isomorphic to $PSL(d,q)$, $PSU(3,q)$, $Sz(q)$ or $Ree(q)$, then $Z(G_x)=1$? References: Cameron, Peter J. Finite permutation groups and finite simple groups. Bull. London Math. Soc. 13 (1981), no. 1, 1--22. MR0599634 (83m:20008). The table with the classification of 2-transitive groups with simple socle appears in page 157, see google books) REPLY [9 votes]: This question is connected to an extreme case of the odd analogue of Glauberman´s $Z^{\ast}$-theorem. This theorem asserts that if a finite group $G$ has no non-identity normal subgroup of order coprime to the prime $p,$ and $u$ is an element of order $p$ of $G$ which commutes with none of its other $G$-conjugates, then $u \in Z(G).$ This theorem was proved (by Glauberman) without CFSG for $p =2,$ but as far as I know all proofs to date for odd $p$ use CFSG. Now suppose that $G$ is a doubly transitive permutation group with $F(G) =1,$ and that the point stabilizer $G_{x}$ has a central element $u$ of prime order $p$ (which will certainly happen if $Z(G_{x}) \neq 1).$ Then $G_{x} = C_{G}(u)$ and the permutation action is equivalent to that of $G$ acting by conjugation on the conjugates of $u.$ Now $C_{G}(u)$ permutes the conjugates of $u$ which commute with $u.$ By the double transitivity of the permutation action, we see that either all conjugates of $u$ commute with $u,$ or else no conjugate of $u$ other than $u$ itself commutes with $u.$ In the former case, $u \in F(G),$ which is excluded by hypothesis. Hence $u$ commutes with none of its other conjugates. By the general $Z^{\ast}$-theorem, we either have $u \in Z(G)$ or else $O_{p^{\prime}}(G) \neq 1.$ The former case is excluded as $F(G) = 1.$ Suppose then that $O_{p^{\prime}}(G) \neq 1.$ Then the image of $u$ is central in $G/O_{p^{\prime}}(G)$ and a Frattini argument yields $G = O_{p^{\prime}}(G)C_{G}(u).$ Since $u \not \in Z(G),$ there is a prime $q \neq p$ such that $u$ normalizes, but does not centralize, a Sylow $q$-subgroup, $Q$ say, of $O_{p^{\prime}}(G)$. Hence there is a $Q$-conjugate $v$ of $u$ such that $\langle u,v \rangle$ is a $\{p,q\}$-group. By the transitivity of $C_{G}(u)$ on the other conjugates of $u,$ it follows that $\langle u,w \rangle$ is a $\{p,q\}$-group for every conjugate $w$ of $u.$ It then follows that $[O_{p^{\prime}}(G),u] \lhd G$ is a non-trivial $q$-group, and that $F(G) \neq 1,$ contrary to hypothesis. Later edit: I noticed that this also follows from Theorem D of :Guralnick, Robert M.; Robinson, Geoffrey R. On extensions of the Baer-Suzuki theorem. Israel J. Math. 82 (1993), no. 1-3, 281–297. One needs to use an earlier theorem of E. Shult. An outline proof is as follows: We have a doubly transitive finite group $G$ with non-Abelian simple socle. Suppose that $H$ is a point stabilizer, and that $x$ is an element of prime order $p$ in $Z(H).$ Then $G$ acts doubly transitively by conjugation on the conjugates of $x$, so $C_{G}(x)$ is transitive on the remaining conjugates of $x$. There can be no other conjugate $y$ of $x$ which commutes with $x$, since if there were, all conjugates of $x$ would commute with $x$, contrary to the simplicity of the socle of $G$. By a theorem of E. Shult, there must be a $p^{\prime}$-subgroup $T$ of $G$ which is normalized, but not centralized. Hence for some $t \in T, x^{-1}x^{t}$ is a non-identity $p$-regular element. Since $C_{G}(x)$ is transitive on the remaining conjugates of $x$, we see that $x^{-1}x^{g}$ is $p$-regular for all $g \in G$. By Theorem D of the Guralnick-Robinson paper, and the argument above ( Theorem D gives $G = O_{p'}(G)C_{G}(x)$, and the argument above gives $[G,x] = [O_{p'}(G),x]$ is a $q$-group for some prime $q$, contrary to the fact that $G$ has non-Abelian simple socle).<|endoftext|> TITLE: Take contraction wrt a vector field twice and define kernel mod image. Does that give anything interesting? QUESTION [16 upvotes]: First, we make the following observation: let $X: M \rightarrow TM $ be a vector field on a smooth manifold. Taking the contraction with respect to $X$ twice gives zero, i.e. $$ i_X \circ i_{X} =0.$$ Is there any "name" for the corresponding "homology" group that one can define (Kernel mod image)? Has this "homology" group been studied by others (there are plenty of questions that one can ask........is it isomorphic to anything more familiar etc etc). Similarly, a dual observation is as follows: Let $\alpha$ be a one form; taking the wedge product with $\alpha$ twice gives us zero. One can again define kernel mod image. Does that give anything "interesting"? If people have investigated these questions, I would like to know a few references. My purpose for asking the "name" of the (co)homology group is so that I can make a google search using the name. I was unable to do that, since I do not know of any key words under this topic (or if at all it is a topic). REPLY [3 votes]: These constructions certainly are very recognisable and meaningful in the algebraic context, that is if you think of algebraic differential forms on an affine algebraic variety. Namely, the corresponding complexes both implement versions of the Koszul complex for a sequence of elements $f_1,\ldots,f_n$ in $R$, where $R$ is the ring of functions on $M$, and $n=\dim M$. Indeed, the algebra of differential forms $\Omega^\bullet(M)$ is the exterior algebra $\Lambda_R^\bullet(R\,dx_1\oplus\ldots\oplus R\,dx_n)$, and for the given vector field $X=f_1\partial_1+\cdots+f_n\partial_n$, the differential $i_X$ on $\Omega^\bullet(M)$ is the $R$-algebra (super)derivation taking $dx_i$ to $f_i$, which essentially is a commonly used definition of the Koszul complex. For the $1$-form $\omega=f_1\,dx_1+\cdots+f_n\,dx_n$, the differential $\omega\wedge$ on $\Omega^\bullet(M)$, which as we remember, is $\Lambda_R^\bullet(R\,dx_1\oplus\ldots\oplus R\,dx_n)$, is dual to the differential corresponding to $i_X$ under the isomorphism of $R$-modules $$\mathop{\mathrm{Hom}}\nolimits_R(\Lambda_R^\bullet(R\,dx_1\oplus\ldots\oplus R\,dx_n),R)\simeq \Lambda_R^\bullet(R\,dx_1\oplus\ldots\oplus R\,dx_n).$$ In any case, the Koszul complex is acyclic if and only if $f_1,\ldots,f_n$ define a complete intersection (this I think also holds in the analytic case, I have no intuition about the smooth one). There are many references, e.g Principles of Algebraic Geometry by Griffiths & Harris, or Commutative Algebra by Bourbaki.<|endoftext|> TITLE: Associative Ring Spectra and Derived Completion QUESTION [5 upvotes]: So, I was thinking before that this might have some nice, simple topos theoretic explanation, but Jacob disabused me of that notion. However, I'm still very interested in the following question: Is there a criterion for determining when a morphism $f:A\to B$ of connective $A_\infty$-ring spectra is such that $A$ can be recovered as the (homotopy) limit of the (homotopical) Amitsur complex on $f$ in the $(\infty,1)$-category of $A_\infty$-ring spectra? For what it's worth, in the context of EKMM, such a criterion is given as an (unproven) proposition in a paper of Carlsson's (Proposition 3.3 here). There, it is stated that a morphism $f:A\to B$ of connective associative $\mathbb{S}$-algebras has the desired property if $\pi_0(f)$ is an isomorphism and $\pi_1(f)$ is onto. I'm interested in proving this in the $\infty$-categorical context (and really proving it at all). REPLY [11 votes]: If by ``effective monomorphism'' you mean the categorical dual of the condition of being an effective epimorphism, then it is not (or at least not obviously) equivalent to the statement that A can be recovered as the totalization of the cosimplicial A-module given by the tensor powers of B, because tensor product is not the same as coproduct when working with associative rings. The condition that A can be recovered in this way is often (but not always) true. It's true for example if B is faithfully flat over A, or if A admits a finite filtration by B-modules, or under the hypothesis you mention. It doesn't seem obvious to me that you can interpret this in terms of Grothendieck topologies on associative ring spectra (again because tensor product is different from coproduct for associative rings), though there are some cases where this is possible (if you're willing to assume that things are approximately commutative and you stick to etale morphisms). In any case, I doubt that such an interpretation would be useful for actually proving that a particular map A -> B had the property you are interested in; more likely, it would just give you a way of restating the problem.<|endoftext|> TITLE: Quotient of the hyperbolic plane with respect to commutator group of $\pi_1(\Sigma_g)$ QUESTION [11 upvotes]: Let $\Sigma_g$ be a Riemann surface of genus $g\geq 2$ and $G=\pi_1(\Sigma_g)$. Let $\pi\colon \mathbb{H}\to \Sigma_g$ be the universal covering map. What kind of surface is $\mathbb{H}/[G,G]$? Moreover, what is $[G,G]$; e.g. if $g=2$? REPLY [5 votes]: Let me start by interpreting the question "What kind of surface is $S$?" in the case of a general connected oriented topological surface (without boundary). (I am considering only oriented surfaces just for simplicity of discussion.) If $S$ had finite complexity, i.e., would be homeomorphic to the interior of a compact oriented surface, you probably would be satisfied by the answer of the type "$S$ is has $n$ ends and genus $g$", since this provides a complete set of topological invariants. Surfaces of infinite complexity are also classified by a certain set of invariants: Its set of ends (regarded as a topological space). Its genus. Its set of ends with positive genus. You can find more details and references in this MO post. If you look closely at the surface you are interested in, $H^2/[G,G]$, you realize that its invariants are: The surface is 1-ended (simply because the abelian group $G/[G,G]$ is 1-ended). It has infinite genus (this is easy to see and is explained in Sam's answer). In particular, its only end has positive genus. To summarize: Your surface is the unique connected oriented topological surface of infinite genus and one end. If you are looking for a different answer, you should clarify what does your question really mean.<|endoftext|> TITLE: Is fourier analysis necessary to prove this? QUESTION [5 upvotes]: I have a couple of inequalities that I want to prove. The proof is easy using fourier analysis but I am wondering whether there is a proof that does not use fourier analysis. 1) For any $c, s > 0$, $$ \sum_{x \in \mathbb{Z}} e^{-\pi x^2 s^2} \ge \sum_{x \in \mathbb{Z}} e^{-\pi (x - c)^2 s^2}\;. $$ 2) For any $c,s >0$, $$ \sum_{x \in \mathbb{Z}} e^{-\pi x^2 s^2} \cos(2\pi xc) > 0\;. $$ Can someone suggest some other simple proof? In particular, for the second inequality, is it possible to partition the sum into parts each containing finitely many terms and each is individually greater than $0$. REPLY [10 votes]: These are both tantamount to known properties of the heat kernel $K_t(c)$ for $c$ in the circle ${\bf R} / {\bf Z}$. (In (1) the heat kernel is obtained by starting at $t=0$ with a row of delta functions $K_0(c) = \sum_{x \in \bf Z} \delta_x(c)$, and in (2) it's obtained by separation of variables.) Each of the properties can be proved in various ways, not all of which involve Fourier analysis. Here's one way. (1) Let $K_t(c) = t^{-1/2} \sum_{x \in \bf Z} \exp (-\pi t^{-1} (x-c)^2)$, so the desired inequality is (after multiplying by $s$) $K_t(0) \geq K_t(c)$ for $t = s^{-1/2}$. Clearly $K_t$ is an even function of $c$; it is also $\bf Z$-periodic: $K_t(c)=K_t(c')$ when $c' \equiv c \bmod \bf Z$. Thus $K_t(0) = K_t(c)$ for $c \in \bf Z$, and we may assume $0 < c < 1$. We shall show that then $K_t(0) > K_t(c)$. Now $K_t$ satisfies the partial differential equation (heat equation) $$ \frac{\partial^2 K_t(c)}{\partial^2 c} = 4 \pi \frac{\partial K_t(c)}{\partial t} $$ (which holds for for each term $t^{-1/2} \exp (-\pi t^{-1} (x-c)^2)$ separately). Hence the function $D_t(b) := K_t(b) - K_t(c-b)$ satisfies the same equation $$ \frac{\partial^2 D_t(b)}{\partial^2 b} = 4 \pi \frac{\partial D_t(b)} {\partial t}. $$ For all $t$ we have $D_t(b) = -D_t(c-1-b) = -D_t(c-b)$, whence $D_t((c-1)/2) = D_t(c/2) = 0$ for all $t$. For small $t$ it is clear that $D_t(b) > 0$ for $b \in ((c-1)/2, c/2)$, because the $x=0$ term in the sum for $K_t(b)$ dominates everything in $K_t(c-b)$. Therefore $D_t(b)$ remains positive in $b \in ((c-1)/2, c/2)$ for all $t$. In particular $0 < D_t(0) = K_t(0) - K_t(c)$, QED. (2) We could also show that $\sum_{x \in \bf Z} e^{-\pi x^2 s^2} \cos(2\pi x c)$ is proportional to a solution of the heat equation, but the fact that it approaches a delta function as $s \rightarrow 0$ seems to be a matter of Fourier analysis. Instead we write the sum as a theta function $$ \sum_{x \in \bf Z} e^{-\pi x^2 s^2} e^{2\pi i x c} = \sum_{x \in \bf Z} q^{x^2} z^{2x} $$ for $q = e^{-\pi s^2}$ and $z = e^{\pi i c}$ (average the $x$ and $-x$ terms on the RHS side to recover the LHS). We can now use the Jacobi triple product formula $$ \sum_{x \in \bf Z} q^{x^2} z^{2x} = \prod_{m=1}^\infty (1-q^{2m}) (1+q^{2m-1} z^2) (1+q^{2m-1} z^{-2}) $$ and observe that in our setting each of the factors is positive (note that $(1+q^{2m-1} z^2)$ and $(1+q^{2m-1} z^{-2})$ are complex conjugates because $q \in \bf R$ and $|z|=1$), whence the product is positive as desired.<|endoftext|> TITLE: A good book on adeles and ideles QUESTION [18 upvotes]: Many results in number theory are stated either in a classical language or in an adelic one. I am often impressed of the efficiency and the satisfactory computational properties of the adelic setting, often brief and well-behaved. They seems omnipresent, but barely justified, almost abstruse ! Unfortunately I never found a good reference book for the adeles and ideles definitions and properties : they always are treated in appendix or in a little chapter giving most of the time only the necessary stuff for the self-sufficientness ok the book. Travelling among tens of lecture notes and books (Weil, Vignéras, Goldfeld, Lang, Milne, Tate, Bump, Gelbart, etc. : all books which have not adeles as main theme !) do not seem to be a good solution in order to have a good idea of adelic objects and properties : what are them ? for what do they exist ? are there examples and computation rules ? what are local and global properties ? splitting properties ? measures ? volumes ? general methods ? approximation theorems ? compactness of adelic groups ? are so many questions always only partially answered, often referring to an other book again... So here is the question : is there any good reference, the more comprehensive possible, starting from the beginning and treating all the major aspects and properties of adeles, but not being just an arid handbook without intuition nor motivation nor examples ? REPLY [5 votes]: A short list of references among the answers, with comments, hoping it will be of some use : Ramakrishnan & Valenza, Fourier Analysis on Number Fields, GTM. There is a short chapter constructing general restricted product of groups, giving them their topology and measures, then applying to obtain adeles and ideles groups, plus approximation theorems and others properties, and class group.General but with no computations. - S. Miller, Adeles, Automorphic Forms and Representations (available there) 30 pages with various elementary calculations, interest in characters, Fourier theory and characters P. Deligne, Formes modulaires et représentations de GL(2) (available there) Focusing on lattices, and rebuilding automorphic forms on it Goldfeld & Hundley, Automorphic Representations and L-Functions for the General Linear Group, Cambridge University Press. Just defining adeles and ideles over $\mathbf{Q}$, but developing the theory of automorphic forms for GL in this setting. The two Weil's book, Basic Number Theory & Adeles and Algebraic Groups. Seems to deal with adeles in great generality.<|endoftext|> TITLE: Can an odd map be null homotopic? QUESTION [5 upvotes]: Let $G$ be a compact Lie group with invariant measure $\mu$. An odd function is a continuous function, $\phi:G\to \mathbb{C}$, such that $\int_{G} \phi d\mu=0$. An odd map is a continuous map, $f:G\to G$, such that $\phi \circ f$ is an odd function for all odd functions $\phi:G\to \mathbb{C}$. For example, $f_{g}(h)=gh$ is an odd map. Is it true to say that every odd map is not null homotopic? A weak and indirect motivation: "Every odd map on sphere is not homotopic to a constant map" Another motivation: The above $f_{g},s$ are not null homotopic. REPLY [10 votes]: Here's a counterexample: take $G=S^1=\{z\in\mathbb{C}:|z|=1\}$, $$ f(z)=\begin{cases}z^2:\mathrm{Im}(z)\geq 0,\\ \overline{z}^2:\mathrm{Im}(z)\leq 0. \end{cases} $$ This $f$ is nullhomotopic, but is an odd map because $\int_G \varphi\circ f=\int_G \varphi$ for all $\varphi:G\to\mathbb{C}$. In the example above, $f$ is continuous, but not differentiable at $z=\pm 1$. The result is true for $C^1$ functions. Suppose $f:G\to G$ is a continuously differentiable odd map, and for simplicity assume $G$ is connected. We have $\int_G \varphi\circ f\,d\mu=\int_G \varphi\,df_*(\mu)$, where $f_*\mu$ is the pushforward measure. Since $\int_G \varphi\,d\mu=0\Rightarrow\int_G\varphi\,df_*\mu=0$, we must have that $f_*\mu$ is a scalar multiple of $\mu$, and therefore $f_*\mu=\mu$ because $\mu$ has finite total measure. For $g\in G$, let's define the Jacobian determinant of $f$ at $g$ to be the determinant of the map induced on $T_g(G)$ by $L_{gf(g)^{-1}}\circ f$, and let's write $J_f(g)$ for this determinant. If $g$ is a regular value of $f$, then there is a neighborhood $U$ of $g$ which is evenly covered by $f$, say $f^{-1}(U)=V_1\sqcup\ldots\sqcup V_k$, and on $U$ $$ f_*\mu = \sum_{i=1}^k\frac{1}{|J_f\circ (f|_{V_i})^{-1}|}\mu, $$ so that $f_*\mu=\mu$ implies that on $U$: $$ \sum_{i=1}^k\frac{1}{|J_f\circ (f|_{V_i})^{-1}|}=1 $$ on $U$. This should imply that $J_f$ doesn't vanish anywhere, and since we assume $G$ is connected, $J_f$ must have constant sign. For any $g\in G$, we can compute the action of $f$ on the top homology group of $G$ locally, and is is multiplication by $\pm(\#f^{-1}(g))$ (where the $\pm$ is the sign of $J_f$). In particular this value is non-zero for some (hence all) $g\in G$. Since $f$ induces a non-zero map on the top homology of $G$, $f$ is not null-homotopic.<|endoftext|> TITLE: Extended TFT with coefficients in spans in any $\infty$-topos QUESTION [10 upvotes]: In the TFT classification article by Jacob Lurie (arXiv:0905.0465) the (∞,n)-category of correspondences (there: $Fam_n$) plays a key role, whose $k$-morphisms are $k$-fold spans of $\infty$-groupoids sliced over a suitable coefficient object $\mathcal{C}$. Motivated by some notes we had made on "Local prequantum field theory", recently Rune Haugseng considered in detail (arXiv:1409.0837) the evident generalization of this where the collection of $\infty$-groupoids is replaced by any other $\infty$-topos. The key properties of the construction are preserved under this generalization, in particular this is still an (∞,n)-category with duals. It seems plausible that, similarly, essentially all the relevant statements regarding topological field theories with coefficients in such correspondences in any $\infty$-topos will still hold. I am specifically interested in a sanity check of this for the following statement: First of all, combining prop. 3.2.8 in arXiv:0905.0465 with theorem 2.4.18 there and applying it to the special case that $\mathcal{C}$ is an $\infty$-groupoid (with duals), yields the neat statement that unoriented "local prequantum field theories with phases in $\mathcal{C}$" $$ Z_L \;\colon\; Bord_n^\sqcup \longrightarrow Fam_n(\mathcal{C})^\otimes $$ are equivalent to the choice of an $\infty$-groupoid $F$ equipped with $O(n)$-$\infty$-action and with an $O(n)$-equivariant map $$ L \;\colon\; F \longrightarrow \mathcal{C} \,, $$ where $\mathcal{C}$ is equipped with the canonical $O(n)$-action induced via theorem 2.4.6 in the above text. An equivalent way to say this is that $L$ induces a horizontal map fitting into $$ \array{ F/\!/O(n) && \stackrel{L/\!/O(n)}{\longrightarrow} && \mathcal{C}/\!/O(n) \\ & \searrow && \swarrow_{canonical} \\ && B O(n) } $$ (where the double slash denotes homotopy quotients). This makes it very manifest that this statement has an immediate analogue with all $\infty$-groupoids replaced by objects of any $\infty$-topos $\mathbf{H}$ (with $B O(n)$ regarded as the inverse image under the terminal geometric morphism of that $\infty$-topos of the homotopy type of the usual classifying space). Now one considers $Fam_n^{\mathbf{H}}(\mathcal{C})$ being the $(\infty,n)$-category of $n$-fold $\mathcal{C}$-phased correspondences, now all inside $\mathbf{H}$, and so forth. And $O(n)$-$\infty$-actions on $F \in \mathbf{H}$ are equivalent to homotopy fiber sequences in $\mathbf{H}$ of the form $F \to F/\!/O(n) \to B O(n)$. But let me maybe restrict attention to the case that $\mathcal{C}$ is still a bare $\infty$-groupoid (similarly embedded into $\mathbf{H}$ under the terminal inverse image). Is it then still true for general $\mathbf{H}$ that monoidal $(\infty,n)$-functors $$ Bord_n^\sqcup \longrightarrow Fam_n^{\mathbf{H}}(\mathcal{C})^\otimes $$ are equivalent to diagrams of the form $$ \array{ F/\!/O(n) && \stackrel{L/\!/O(n)}{\longrightarrow} && \mathcal{C}/\!/O(n) \\ & \searrow && \swarrow_{{canonical}} \\ && B O(n) } $$ in $\mathbf{H}$? (In case it matters, I am happy to assume that the terminal inverse image of $\mathbf{H}$ is fully faithful.) REPLY [4 votes]: Here an argument using the assumption that $\mathbf{H}$ has an $\infty$-site $\mathcal{S}$ of definition all whose objects are étale contractible. The proof of prop. 3.2.8 arXiv:0905.0465 shows that for $\mathbf{H} = \infty Grpd$ the equivalence in question is natural in the choice of the unoriented bulk field theory $F/\!/O(n)$. Let then $$ \mathbf{H} \stackrel{\longleftarrow}{\hookrightarrow} Func(\mathcal{S}^{op},\infty Grpd) $$ be the reflection exhibiting $\mathcal{S}$ as an $\infty$-site of definition. Given $F \in \mathbf{H}$ equipped with $O(n)$-$\infty$-action (throughout $O(n)$ denotes the homotopy type of the topological group $O(n)$, regarded as a group object in constant $\infty$-stacks) write for each $U \in \mathcal{S}$ $$ F_U := \mathbf{H}(U,F) \in \infty Grpd $$ for its value on $U$. Observe that setting $$ F_U/\!/O(n) := \mathbf{H}(U,F/\!/O(n)) \in \infty Grpd $$ exhibits an $O(n)$-$\infty$-action on $F_U$ since $\mathbf{H}(U,-)$ preserves $\infty$-limits and since by assumption on the site $\mathcal{S}$ we have $\mathbf{H}(U,B O(n)) \simeq B O(n)$, so that the homotopy fiber sequence $$ F \to F/\!/O(n) \to B O(n) $$ which exhibits the $\infty$-action of $O(n)$ on $F$ (here I am using arXiv:1207.0248) naturally induces a system of homotopy fiber sequences $$ F_U \to F_U/\!/O(n) \to B O(n) $$ exhibiting $O(n)$-$\infty$-actions on each $F_U$. It follows that a diagram $$ \array{ F/\!/O(n) && \stackrel{L/\!/O(n)}{\longrightarrow} && \mathcal{C}/\!/O(n) \\ & \searrow && \swarrow \\ && B O(n) } $$ in $\mathbf{H}$ is equivalent to an $\infty$-sheaf of diagrams of the form $$ U \;\;\; \mapsto \;\;\; \left\{ \array{ F_U/\!/O(n) && \longrightarrow && \mathcal{C}/\!/O(n) \\ & \searrow && \swarrow \\ && B O(n) } \right\} $$ in $\infty \mathrm{Grpd}$. Now the statement of the proposition for $\mathbf{H} = \infty \mathrm{Grpd}$ applies objectwise for each $U$, and since it is natural in $F_U$ (with its action) it is also natural in $U$, and so the above is equivalent to the $\infty$-sheaf ($(\infty,n)$-sheaf) of local unoriented-topological field theories in $\infty \mathrm{Grpd}$: $$ U \;\;\; \mapsto \;\;\; \left\{ \array{ && (Fam_n(\mathcal{C}))^{\otimes} \\ & {}^{}\nearrow & \downarrow \\ Bord_n^\sqcup &\underset{F_U/\!/O(n)}{\longrightarrow}& (Fam_n)^\otimes } \right\} \,. $$ But this is equivalently a field theory $$ \array{ && (Fam_n^{\mathbf{H}}(\mathcal{C}))^{\otimes} \\ & {}^{}\nearrow & \downarrow \\ Bord_n^\sqcup &\underset{F/\!/O(n)}{\longrightarrow}& (Fam_n^{\mathbf{H}})^\otimes } $$ with coefficients in $\mathbf{H}$.<|endoftext|> TITLE: Parity of $\lfloor 1/(x y) \rfloor$ not equally distributed QUESTION [23 upvotes]: A curious puzzle for which I would appreciate an explanation. For $x$ and $y$ both uniformly and independently distributed in $[0,1]$, the value of $\lfloor 1/(x y) \rfloor$ has a bias toward odd numbers. Here are $10$ random trials: $$51, 34, 1, 239, 9, 4, 2, 1, 1, 1 $$ with $7$ odd numbers. Here are $10^6$ trials, placed into even and odd bins:       About 53% of the reciprocals are odd. If I use the ceiling function instead of the floor, the bias reverses, with approximately 47% odd. And finally, if I round to the nearest integer instead, then about 48% are odd. None of these biases appear to be statistical or numerical artifacts (in particular, it seems that the 47% and 48% are numerically distinguishable), although I encourage you to check me on this. Update. To supplement Noam Elkies' answer, a plot of $x y = 1/n$ for $n=2,\ldots,100$: REPLY [58 votes]: You're dividing the square $S = \{ (x,y) \colon 0 < x < 1, 0 < y < 1\}$ into two regions according to the parity of $\lfloor 1/(xy) \rfloor$, separated by the segments of the hyperbolas $xy = 1/n$ ($n=2,3,4,\ldots$) contained in $S$. There's no reason to expect that the two regions have the same area. If I did this right, the area between the $n$-th hyperbola and the top right corner of the square is $$ A(n) := 1 - \frac{1 + \log n}{n} $$ so the discrepancy between odd and even values of $\lfloor 1/(xy) \rfloor$ is $$ (A(2)-A(1)) - (A(3)-A(2)) + (A(4)-A(3)) - + \cdots $$ which is numerically $0.066556553635\ldots$ according to the gp calculation A(n) = 1 - (log(n)+1)/n sumalt(n=1, (-1)^n*(A(n)-A(n+1))) So we expect about 53.33% odd and 46.67% even values, which seems consistent with your experiment. P.S. Using a formula I found in MO Question 140547, I gather that this number $0.066556553635\ldots$ has the closed form $$ (\log 2)^2 + \bigl(2 (1 - \gamma) \log 2\bigr) - 1, $$ where $\gamma$ is Euler's constant $0.5772156649\ldots$. P.P.S. I see that I didn't address the end of the original question: "If I use the ceiling function instead of the floor, the bias reverses, [...] if I round to the nearest integer instead, then about 48% are odd." The first part is clear because changing $\lfloor 1/(xy) \rfloor$ to $\lceil 1/(xy) \rceil$ switches even and odd values (except in the negligible case that $1/(xy)$ is an exact integer). For the nearest-integer function, the discrepancy between odd and even values is $$\bigl(A(3/2)-A(1)\bigr) - \bigl(A(5/2)-A(3/2)\bigr) + \bigl(A(7/2)-A(5/2)\bigr) - + \cdots $$ which evaluates numerically to $-0.03500998166\ldots$ (using sumalt in gp as before), which again is consistent with observation (48.25% odd, 51.75% even). There's still a "closed form" for this discrepancy, but more complicated: $$ -3 + 4 \log(2) + \pi \bigl(1 + \log(\pi/2) - \gamma - 4 \log\Gamma(3/4) \bigr). $$ This requires evaluation of $\log(1)/1 - \log(3)/3 + \log(5)/5 - \log(7)/7 + - \cdots$, which can be achieved by differentiating the functional equation for the Dirichlet L-function $L(s,\chi_4) = 1 - 3^{-s} + 5^{-s} - 7^{-s} + - \cdots$ and evaluating at $s=1$. REPLY [13 votes]: I suspect that it has to do with the fact that the most likely outcome is $\lfloor 1/(x y) \rfloor=1$ (which happens to be odd), followed with an application of the Strong Law of Small Numbers. Here are some more details. Let $Z$ be the random variable $xy$. Note that $Z$ takes values in $[0,1]$ and $\lfloor 1/Z \rfloor$ is odd if and only if $$ Z \in (1/2, 1] \cup (1/4, 1/3] \cup (1/6, 1/5] \cup \dots $$ Note that this set has measure more than $1/2$. However, the distribution of $Z$ is of course not uniform on $[0,1]$ (it is actually skewed towards $0$ instead of $1$). I suspect what ends up happening is that the distribution of $\lfloor 1/Z \rfloor$ is almost perfectly split between even and odd for say $Z < 1/5$, and the discrepancy is thus a result of what happens for $Z \geq 1/5$ (where one easily sees that odd wins out).<|endoftext|> TITLE: Separating points in the plane II QUESTION [9 upvotes]: Let A be a set of $2m$ points on the plane so that no open set of diameter $2$ has more than m of them. Define $A+A+...+A$ ($k$ times) to be the multiset of $k$-sums from $A$. That is, we consider all possible $(2m)^k$ sums. Question. Does there exists a number $k$ independent of $m$ such that the points in the latter multiset can be paired so that the distance between the points each pair is at least 2? Conjecture. $k=2$ is enough. Comment. $k=1$ is not enough for the following reason. Take three points at distance $2$ and a point in the middle of them - then there is no such pairing. The intuition here is that the points get further and further apart when we are summing and hence we shall have the desired pairing. One remark: The claim is true if $k$ is large enough for the following reason - the number of points in any open bounded open set divided by the total number of points goes to zero (concentration function bounds from probability) and so for $k$ large enough one can make any set of diameter 4 to have at most half of the points. But now notice that the graph on the multiset of the sums formed by adding an edge between two sums iff the distance between them is at least 2 has degree at least half the number of the sums as otherwise we would have more than half of the points would fit in a set of diameter 4 (the neighbourhood of that point with low degree). But this is Dirac's condition as so we have a perfect matching and are done. REPLY [2 votes]: Let me make the next step after Mirko Swirko. In quite a beautiful paper A. Hajnal proves the following. Theorem. Let $G$ be a $m$-saturated graph --- that is, $G$ does not contain a complete $(m+1)$-subgraph, but it does after adding any new edge. Then either $G$ has a vertex connected to each vertex, or the degree of every vertex in $G$ is at least $2(m-1)$. Now take our $2m$ points and connect with an edge any two at distance $<2$. Then this graph does not contain a complete $(m+1)$-subgraph. Assume that there is no vertex connected to all other vertices. Let us add the edges to this graph making no complete $(m+1)$-graph; by Hajnal's theorem, it is possible while all the degrees are less than $2(m-1)$; thus at the end we will get a vertex $x$ of degree exactly $2(m-1)$ and still no complete $(m+1)$-subgraph (maybe, this holds at the very beginning). Thus $x$ is connected to all but one other vertices; let $y$ be this exceptional vertex. Then even the obtained graph contains no complete $m$-subgraph after deleting $x$ and $y$: otherwise this subgraph augmented by $x$ forms a complete $(m+1)$-subgraph. Thus one may use these $x$ and $y$ in the induction step (they are not connected!). The only case remaining is when there initially exists a vertex connected to all --- that is, all the points lie in an oper ball of radius 2 centered at this vertex. Maybe, this case is left for some third answer?<|endoftext|> TITLE: Canonical functions in set theory and their applications QUESTION [9 upvotes]: Given regular cardinal $\kappa>\omega,$ we can define the canonical functions $f_\alpha: \kappa\to \kappa,$ for $\alpha<\kappa^+.$ Some of their properties are presented in Chapter 22 of the book "Problems and Theorems in Classical Set Theory" by Péter Komjáth and Vilmos Totik. (A) I am looking for more references and additional properties of these functions. (B) I also want to know what are the main applications of these functions, and how they can be used in proving theorems (in particular $ZFC$ results). REPLY [5 votes]: In belated partial response to (A), canonical functions are certainly already defined in the Jech-Shelah paper: A note on canonical functions, where reference is made to the work F. Galvin and A. Hajnal, Inequalities for cardinal powers, Annals of Math. 101 (1975), 491-498. The name "canonical" appears to derive from the property that if they exist, they are unique up to equivalence modulo a club. Below $\aleph_2$, they always exist; at or above, they may or may not.<|endoftext|> TITLE: On $V$-decisive and weakly homogeneous forcings QUESTION [8 upvotes]: Suppose that $\Bbb P$ is a forcing in $V$, we say that $\Bbb P$ is $V$-decisive if whenever $\varphi(x_1,\ldots,x_n)$ is a statement in the language of forcing, and $u_1,\ldots,u_n\in V$ then $1_{\Bbb P}$ decides the truth of $\varphi(\check u_1,\ldots,\check u_n)$. (Side question, was this property of $\Bbb P$ given a name in the past?) It's a classical theorem that weakly homogeneous forcing is $V$-decisive. But we can show a bit more. The question is whether or not there exists a $V$-decisive forcing whose Boolean completion is rigid, or at least not weakly homogeneous. What we know: $\Bbb P$ is $V$-decisive if and only if whenever $G$ is $V$-generic, we have that $\bigcup\{H\in V[G]\mid H\subseteq\Bbb P\text{ is }V\text{-generic}\}=\Bbb P$. (Note that we don't require that $V[G]=V[H]$, just that $H$ is generic over $V$.) Equivalently, this means that if $p,q\in\mathcal B(\Bbb P)$ (the complete Boolean algebra that contains $\Bbb P$ as a dense subset) there are embeddings of $\mathcal B(\Bbb P)\restriction p$ into $\mathcal B(\Bbb P)\restriction q$. (Correction: earlier a density requirement was added, after Joel's answer it dawned on us that there is too much here.) Equivalently, for every $p\in\Bbb P$ there is some $q\leq p$ and a projection of $\mathcal B(\Bbb P)$ which maps $\mathcal B(\Bbb P)\restriction q$ onto $\mathcal B(\Bbb P)$, then $\Bbb P$ is $V$-decisive. Note that we do not require the projection to be injective, which would essentially mean that $\Bbb P$ is weakly homogeneous. Question. Is there a $V$-decisive forcing whose Boolean completion is not weakly homogeneous? If not, is it at least consistent that there is one? In either case, can we find one whose Boolean completion is rigid? REPLY [12 votes]: $\newcommand\B{\mathbb{B}}$ Update. The answer is no to all three questions. Theorem. The following are equivalent for any complete Boolean algebra $\B$. $\B$ is $V$-decisive. For any two conditions $b,c\in\B$, forcing with $\B$ below $b$ adds a generic filter below $c$ with the same extension. That is, in any forcing extension $V[G]$ via $G\subset\B$, and any $c\in \B$, there is $V$-generic $H\subset\B$ with $c\in H$ and $V[G]=V[H]$. $\B$ is weakly homogeneous. Proof. The implication $3\to 1$ is standard. For $1\to 2$, you noticed part of this with your statement (1), but you get more than you state there; you can actually get $V[G]=V[H]$. The reason is that every condition $c$ forces that "there is a $\check V$-generic filter containing $\check c$ and giving rise to the whole extension", and so this is forced by all conditions. So let us consider the implication $2\to 3$. For this, one can use the ideas in my blog post on Common forcing extensions via different forcing notions in order to get the desired automorphisms. Specifically, assume statement $2$. Fix any two nonzero incompatible conditions $b,c\in\B$. By the argument of my blog post, we get an isomorphism of cones $\pi:\B\upharpoonright b'\cong\B\upharpoonright c'$ for some $b'\leq b, c'\leq c$. Let $d=\neg(b'\vee c')$, so that $b', c', d$ form a maximal antichain in $\B$. Since every element of $\B$ is the unique join of its parts below $b', c'$ and $d$, it follows that $\B$ is simply the product of the respective cones below these three elements. Thus, we may extend $\pi$ to an automorphism $\pi:\B\cong\B$ of the whole algebra, with $\pi(b')=c'$. Namely, $$\pi(x)=\pi(x\wedge b')\vee\pi^{-1}(x\wedge c')\vee (x\wedge d).$$ In particular, $\pi(b)$ is thereby made compatible with $c$, and so $\B$ is weakly homogeneous. QED The proof shows that you don't really need arbitrary parameters from $V$, but rather it suffices to be able to refer to any particular element $b\in\B$, and then you also need somehow to refer to the ground model in order to state $\check V$-genericity. This can be done either by having a parameter for the collection of dense subsets of $\B$ in $V$, but in some cases like $V=L$ the ground model may be definable without parameters. Meanwhile, my answer to Miha Habič's question shows that if you fall back to ordinal parameters, then it is consistent that a Boolean algebra is Ord-decisive but not weakly homogeneous. Original answer. This answer is just about partial orders, which is not what was desired. If one considers just the partial order, then the answer is yes, because in fact every forcing notion (meaning every partial order) is equivalent to a rigid forcing notion. Rigidity is simply not invariant under forcing equivalence. Theorem. Every partial order is forcing equivalent to a rigid partial order. In particular, every forcing notion is forcing equivalent to a non-weakly homogeneous forcing notion. Proof: This is a nice exercise in partial order combinatorics, which I encourage you simply to try to prove on your own. But here is one way to do it. Suppose we are given a partial order $\mathbb{P}$. We want to construct a new partial order $\mathbb{P}'$, into which $\mathbb{P}$ will densely embed, but where $\mathbb{P}'$ is rigid. As a first step, let us associate to each node $p$ a distinct rigid and pairwise non-isomorphic partial order $A_p$, having a largest and smallest element. We construct the partial order $\mathbb{P}'$ first by replacing each node $p\in \mathbb{P}$ with a copy of $A_p$, but also adding a new stubby maximal node sticking up above the largest node of $A_p$, incomparable with everything else that was placed above $p$, and also two new stubby maximal nodes sticking up above the minimal element of $A_p$. It follows that any automorphism of $\mathbb{P}'$ must respect the copies of $A_p$, since their largest and smallest elements were marked by these stubby nodes. But since the various $A_p$'s were chosen to be non-isomorphic and rigid, it follows that $\mathbb{P}'$ can have no automorphism at all. But clearly $\mathbb{P}$ densely embeds into $\mathbb{P}'$, by associating each $p\in\mathbb{P}$ with the least element of the copy of $A_p$ inside $\mathbb{P}'$. QED I take this argument to show that most forcing homogeneity arguments are not fundamentally about automorphisms of the partial order, or even automorphisms of the Boolean algebra, but rather automorphisms of cones in the Boolean algebra, in the way that you have already described.<|endoftext|> TITLE: The complex heat kernel on a Riemann manifold QUESTION [5 upvotes]: There is a vast literature available for the heat kernel. Nevertheless, I haven't been able to find almost anything useful about the kernel of the equation $\frac{1}{\mathbb{i}} \frac{\partial u}{\partial t} - \Delta u =0$. So far, I have three questions about it: 1) if $e(t,x,y)$ is the above kernel, is it bounded in $x$ (in $\mathbb{R}^n$ it is)? 2) since $e$ is not real, it is impossible to construct a Gaussian probability starting from it; what, then, can still be done? I assume one can still construct a complex measure, but what are its "nice" properties? 3) finally, do you know of any rich monography on the subject? I have found plenty of titles about the "usual" heat kernel, yet none regarding my question - which I find strange given how important the Schrödinger equation is. REPLY [3 votes]: As far as I know, the term "Mehler Kernel" is used for the integral kernel of the heat equation corresponding the the harmonic oscillator, $$ \partial_tu + \Delta u + x^2 u = 0.$$ The equation you are talking about is the Schrödinger equation. The kernel on $\mathbb{R}^n$ is $$ e_t(x, y) = (4\pi i t)^{-n/2} e^{-\frac{1}{4 i t} |x-y|^2},$$ that is the heat kernel with imaginary values plugged in for $t$. Despite the the formal similarity to the heat kernel, it has vastly different properties: For example, the integral $$u(t, x) := \int_{\mathbb{R}^n} e_t(x, y) u(y) \mathrm{d} y$$ does not converge absolutely for arbitrary functions in $L^2$, and it is not smoothing in the sense that it maps distributions to smooth functions. However, it preserves smoothness. In particular, it does not define a complex $\sigma$-aditive measure! On manifolds, this gets way worse: For example, I know from some corner of my memory that on the sphere, it is not smooth at all; instead, it is a singular distribution at every point at most times.<|endoftext|> TITLE: O-minimal Theories with Non-Dense Order Type QUESTION [6 upvotes]: I asked this question on MSE, but I haven't received any comments or responses (also, it has a very low view count), so I thought I would also ask it here. In this paper, Knight, Pillay, and Steinhorn prove that for any O-minimal structure $\mathfrak{A}$, in which the underlying order types is dense, and if $\mathfrak{B} \equiv \mathfrak{A}$, then $\mathfrak{B}$ is also O-minimal. Is there a counterexample for when the underlying order type is not dense? I haven't been able to construct a counterexample. Thanks REPLY [6 votes]: The restriction that the ordering be dense was removed by Pillay and Steinhorn in Definable Sets in Ordered Structures III, Trans. Amer. Math. Soc. 309 (1988) 469-476. The abstract of that paper reads: "We show that any o-minimal structure has a strongly o-minimal theory."<|endoftext|> TITLE: The limit of edge-midpoint convex polyhedra QUESTION [19 upvotes]: Starting with a convex polyhedron $P_1 \subset \mathbb{R}^3$, replace that with $P_2$, the convex hull of the midpoints of the edges of $P_1$. Continuing this process, we obtain a series of polyhedra approaching a smooth body $B$ (or at least, I think it approaches a smooth body). See above for $P_1,\ldots,P_5$—not to the same scale. Q1. Is $\lim_{n \to \infty} P_n$ $C^2$-smooth, starting with non-degenerate $P_1$? Or only $C^1$-smooth? Or only $C^0$? Q2. Does $P_n$ approach an ellipsoid as $n \to \infty$, for every (non-degenerate) starting $P_1$? [My guess: No.] These are, in some sense, less sophisticated versions of my earlier question, “Derived” polyhedra and polytopes, which focussed on face centroids rather than edge midpoints. But here I am asking specific questions on smoothness and the limit shapes. Still, perhaps @GjergjiZaimi's answer there holds. The limit objects are almost subdivision surfaces, but not quite. REPLY [8 votes]: The limit can be within $\varepsilon$ of any convex polytope: Start with a polytope $P$ then replace each vertex $v$ by three vertices $v_1,v_2,v_3$ forming a triangle with sides of length (less than) $\varepsilon$ with centroid $v$ and containing no point of $P$ other than $v.$ This gives a polytope $P_1 \supset P$ with three times as many vertices but with all points within $\varepsilon$ of (the closest point of) $P.$ Then as the edge midpoint process is applied to $P_1$ one will get a nested series of polytopes all containing $P.$ $P_1$ will have many faces aside from the tiny triangles and the centroids of these faces will be in the limit body, so the limit is strictly larger than $P.$ I'd still like to know the exact limit for a regular tetrahedron (or irregular, it shouldn't matter) and other simple cases. Here are the first nine stages for $P_1$ the cube (not shown) with vertices $(\pm 2,\pm 2,\pm 2).$ The centroids of the $6$ faces, at distance $\sqrt{4}$ from the origin, will be in the limit body. One is $[2,0,0].$ Then $P_2$ is a cuboctahedron ($14$ faces, $12$ vertices) introducing $8$ new centroids all at distance $\sqrt{16/3}$. One is $[4/3 ,4/3,4/3].$ Then $P_3$ has $26$ faces and $24$ vertices. This introduces $12$ centroids all at distance $\sqrt{9/2}.$ One is $[0,3/2,3/2].$ After this is $P_4$ with $50$ faces and $48$ vertices. The $24$ new centroids are all at distance $\sqrt{147/32}.$ One is $[7/8,7/8,7/4].$ Finally in this picture is $P_5$ with $98$ faces and $96$ vertices. There are $48$ new centroids introduced. Half are like $[0,7/8,15/8]$ at distance $\sqrt{137/32}$ and half are like $[7/8,23/16,23/16]$ at distance $\sqrt{627/128}.$ I'm not sure how much this reveals other than that the limit body is not a sphere. The centroids introduced, which will remain centroids at each stage and are extreme points of the limit body are $6$ at distance $2$ then $8$ at distance$\approx 2.3094$ then $12$ at distance $\approx 2.1213$ then $24$ at distances $\approx 2.1433$ and finally $24$ each at distances $\approx 2.0691,2.2132.$ In general it is non-trivial to find the faces and edges for the convex hull of a set of points (even given that all the points are vertices.) However given the information on $P_i$ it is easy to get $P_{i+1}$: Replace each face by the midpoint polygon and for each vertex find the faces it is on, from each one take an edge joining the midpoints of the two appropriate edges and then use these edges to list these points in a valid order. Here are $P_i$ for $i=6,7,8,9.$ Each is shown twice. Once centered on a triangle and once not. Notice that the even and odd cases look different. In the top view of $P_8$ it may look as if three triangles and the diamonds meet at a degree $6$ point. However that point is actually a tiny triangle and the "triangles" are trapezoids with a very short side. The "equator" for $P_9$ looks like this REPLY [4 votes]: A variant of Robert Israel's argument: Take a convex body $K$. Slice it by planes, to make a polyhedron $K_\epsilon\subset K$ whose Hausdorff distance to $K$ is less than $\epsilon^2$ and the diameter of faces is less than $C\epsilon$. Apply the edge-midpoint process. Then the limit body has many points at distance $<\epsilon^2$ to $\partial K$; one per face. Thus the set of limit bodies is dense in the set of convex bodies. In particular, most limit bodies are not ellipsoids. Alternately, you can form a polyhedron $K^\epsilon$ containing $K$, arbitrarily close to $K$, by taking finitely many tangent planes. Then again apply the edge-midpoint process to obtain a limit body close to $K$.<|endoftext|> TITLE: Map between stacks and automorphism groups QUESTION [8 upvotes]: I know that the Torelli morphism $t_g:\mathcal{M}_g\rightarrow \mathcal{A}_g$ between the stacks of smooth curves of genus $g$ and principally polarized abelian varieties of dimension $g$ is of order 2 ramified on the hyperelliptic locus. I also know that the natural map $\sigma:Aut(C)\rightarrow Aut (J(C),\Theta)$ which to an automorphism of the smooth curve $C$ associates the induced automorphism on $(J(C),\Theta)$ is an isomorphism for $C$ hyperelliptic and is such that $Aut(J(C),\Theta)=$ otherwise. My question is: how are the two notions linked? I know this is a trivial question, but i can't see how, for a non-hyperelliptic curve $C$, $Aut(J(C),\Theta)=$ should imply that $t_g^{-1}(J(C),\Theta)$ consists of two points in the stack $\mathcal{M}_g$.. REPLY [5 votes]: Look at the definition of fiber product in the category of stacks: For a scheme $S$, the $S$-points of the fiber product $X \times_Z Y$ are triples $(x, y, \sigma)$ where $x \in X(S), y \in Y(S)$and $\sigma$ is a (iso)morphism between the images of $x$ and $y$ in the groupoid $Z(S)$. Let $c:Spec\ k \to \mathcal{A}_g$ be the morphism corresponding to the point $(J(C), \Theta)$. The $Spec\ k$ valued points of $\mathcal{M}_g \times_{t_g, \mathcal{A}_g, c} Spec\ k$ are triples $(Spec\ k \to \mathcal{M}_g, Spec \ k \to Spec \ k, \sigma)$ where $\sigma$ is an isomorphism between images in $\mathcal{A}_g$. So if the map $Spec\ k \to \mathcal{M}_g$ corresponds to $C$ (which it must, for a $\sigma$ to exist, by Torelli theorem), then composing with $t_g$ corresponds to $(J(C), \Theta)$ in $\mathcal{A}_g$, and we can choose $\sigma$ to be an automorphism of $(J(C), \Theta)$. REPLY [3 votes]: It's not so much that it's two points of the stack, it's that the point of $\mathcal{A}_g$ has inertia, that is, the map restricted to a point (corresponding to a general curve) looks like $*\mapsto */\mathbb{Z}/2\cong B\mathbb{Z}/2$. The key is that a point of a moduli stack "looks like" $*/Aut$ where $Aut$ is the automorphism group of the object corresponding to the moduli point. So, for any point, we have $*/Aut(C)\mapsto */$ which is degree 2, with general point being like I described above. You can think of it as just being that if you have a set and quotient by $\mathbb{Z}/2$ it should be a double cover if the action is free, and here it's just not free, so we also have a double cover, because the target stops being just a set, but is rather a groupoid.<|endoftext|> TITLE: What version of the wreath product embedding theorem is actually stated in the famous paper of Kaloujnine and Krasner? QUESTION [6 upvotes]: This question is inspired by Terry Tao's blog post and the comments there. I have always cited M. Krasner and L. Kaloujnine, "Produit complet des groupes de permutations et le problème d'extension de groupes III", Acta Sci. Math. Szeged 14, pp. 69-82 (1951) as the source of the result that if $G$ is a group and $H$ is a subgroup, then the regular action of $G$ on itself is inside of the action of the wreath product $H\wr (G/\mathrm{Core}(H))$ on $H\times G/H$ via the identification of $G$ with $H\times G/H$ using a transversal $T$ for $G/H$ and that replacing $T$ by a different transversal changes the embedding of $G$ by an inner automorphism of the wreath product. I am aware that this result is implicit in Frobenius's theory of induced representations and the theory of monomial representations of groups (e.g. Chapter 4 this paper of Ore although $H$ is assumed finite index there) but these authors did not use the wreath product language to the best of my knowledge. Many people refer only to the special case where $N$ is normal as the Krasner-Kaloujnine embedding. I am now wondering which version they actually stated. The original paper does not seem easy to access by the obvious Google search. REPLY [3 votes]: The original paper (part II of a three part series) is available here. Page 47 lists the theorem and discusses in a footnote the relation with Ore's earlier work. From what I understand, any series of subgroups is allowed in the embedding, they need not be normal. A more limited version of their theorem is published in part III, available here, and this has caused some confusion.<|endoftext|> TITLE: How can the Cayley table for the elements of basis of a Cayley-Dickson algebra be summarized in an algebraic expression? QUESTION [5 upvotes]: One would be able to construct a Cayley table that has all $e_i$ elements of the basis of algebra $A$ where $00$ and $e_pe_q=-e_qe_p$ for $0\ne p\ne q\ne 0$. But in addition we have the wonderful formulas $$ \text{If } 2^N\le p TITLE: cohomology of classifying space of permutation groups QUESTION [9 upvotes]: Let $\Sigma_k$ be the permutation group of order $k$. Let $r: \Sigma_k\to GL(k)$ be the regular representation by permuting the order of the standard basis of $\mathbb{R}^n$. Let $\rho: B\Sigma_k\to BGL(k)=G_k(\mathbb{R}^\infty)$ be the induced map (induced from $r$) between classifying spaces. Let $\rho^*: H^*(G_k(\mathbb{R}^\infty))\to H^*(B\Sigma_k)$ be the induced homomorphism of cohomology with coefficients in $\mathbb{Z}_2$. What is the image $\text{Im} \rho^*$? REPLY [4 votes]: I suppose by $\mathbb{Z}_2$ you mean $\mathbb{Z}/2$ and not $\mathbb{Z}^{\wedge}_2$. Then we have $$H^*(BO(k);\mathbb{Z}/2)\cong \mathbb{Z}/2[w_1,w_2\cdots w_k].$$ By definition we have $\rho ^*(w_n)=w_n(r)$ where $w_n(r)$ is the $n$-th Stiefel-Whiteney class of the representation $r$. Since the regular representation factors through $O(k-1)$, we see that $w_k=0$. We can, in theory, determine the image using the fact that the Quillen's map is injective (as a matter of fact, isomorphic) for the symmetric groups, that is elements of mod $2$ cohomology of the symmetric get detected by its elementary abelian subgroups. In the case of $k=4$, for example, if we call $V_1=\{(12),(34),(12)(34),id\},V_2=\{(12)(34),(13)(24),(14)(23),id\}$ then the restriction of $w_n(r)$'s to $BV_1$ can be computed using Whitney product formula, and the restriction to $BV_2$ can be found more or less using the definition of Dickson invariants, so by comaparing with the table in http://homepages.math.uic.edu/~bshipley/ConMcohomology1.pdf Example 4.4, we see that they coincide with standard generators of $H^*(B\Sigma _4;\mathbb{Z}/2)$. Thus for $k=4$ $\rho ^*$ is surjective. In principle, this kind of analysis can be carried out for larger $k$, but I am not aware of any explicit result.<|endoftext|> TITLE: Who first defined quantum integers? QUESTION [9 upvotes]: Who first gave the definition of quantum integers $$ [m]_q = \frac{1 - q^m}{1 - q} $$ and addition as $$ [m]_q \oplus_q [n]_q = [m]_q + q^m [n]_q $$ and multiplication as $$ [m]_q \otimes_q [n]_q = [m]_q [n]_{q^m} $$ and in which context? REPLY [5 votes]: I think q-integers were first introduced by F.H. Jackson in 1903 in the paper http://www.biodiversitylibrary.org/item/130137#page/15/mode/1up (On generalized functions of Legendre and Bessel).<|endoftext|> TITLE: A curious Gauss-Sum type identity QUESTION [9 upvotes]: Let $q=e^{2\pi i/m}$, $a\in\mathbb{R}$ and $1\leq j\leq m-1$. I would like to prove that: $$(a-1)\sum_{n=0}^{m-1} q^n\frac{\prod_{k=0}^{j-2} (q^{n+k+1}-a)}{\prod_{k=0}^{j} (aq^{n+k}-1)}=0.$$ For $j=1$ I can prove it by induction: the left-hand side of the expression factors as $$\frac{\prod_{1 TITLE: Why is it so hard to compute $\pi_n(S^n)$? QUESTION [34 upvotes]: Of course it isn't really that hard - nowhere near as hard as $\pi_k(S^n)$ for $k>n$, for instance. The hardness that I'm referring to is based on the observation that apparently nobody knows how to do the calculation within the homotopy category of topological spaces. Approaches that I'm aware of include: -Homology theory (the Hurewicz theorem) -Degree theory -The divergence theorem and each of these reduces to a calculation within some other category (PL or Diff). My question is: is there something "wrong" with hTop that precludes a computation of $\pi_n(S^n)$ within that category? Certainly if my assumption that such a proof does not exist is wrong, I would be very interested to know it. I have been thinking and reading further about this question for the past couple days, and I wanted to summarize some of the main points in the answers and questions: Some techniques - e.g. the Freudenthal Suspension Theorem via the James construction or the Hurewicz Theorem with singular homology - might actually lead to proofs without any approximation arguments. But so far I'm not sure we quite have it: the proof of the Freudenthal Suspension Theorem uses the fact that $J(X)$ is homotopy equivalent to the loop space of the suspension of $X$, but the only proofs I can find of this fact use a CW structure on $X$, and similarly for the proofs of the Hurewicz theorem. Can these results be proved for the sphere without PL or smooth approximation? Perhaps this question is entirely wrong-headed: the techniques of PL and smooth approximation are very well adapted to homotopy theory, so why try to replace them with a language which may end up adding more complications with little additional insight? Fair enough. But the goal of this question is not to disparage or seek alternatives to existing techniques, it is to understand exactly what role they play in the theory. The statement "The identity map $S^n \to S^n$ is homotopically nontrivial and freely generates $\pi_n(S^n)$" makes no mention of CW complexes or smooth structures, yet apparently the statement is difficult or impossible to prove without that sort of language (except in the case $n=1$!) To seek an understanding of this observation is different from lamenting it. REPLY [25 votes]: I suppose that the proof that $\pi_1(S^1) \cong \mathbb{Z}$ using covering spaces is homotopy-theoretic. The Freudenthal Suspension Theorem (via the James construction) tells us that $\Sigma: \pi_n(S^n)\to \pi_{n+1}(S^{n+1})$ is an isomorphism for $n \geq 2$ and surjective for $n =1$. Since $S^1$ is an H-space, the suspension map $\sigma: S^1\to \Omega\Sigma S^1$ has a retraction $r: \Omega\Sigma S^1\to S^1$. Therefore $\Sigma = \sigma_*: \pi_1(S^1)\to \pi_{2}(S^{2})$ is injective (in addition to being surjection). Now Freudenthal completes the calculation. Note that we don't just get an abstract isomorphism, we get that these groups are generated by $[\mathrm{id}_{S^n}]$. EDIT: Regarding getting the James Construction homotopically: the paper Fantham, Peter; James, Ioan(4-OX); Mather, Michael On the reduced product construction. (English summary) Canad. Math. Bull. 39 (1996), no. 4, 385–389. derives the relevant properties using the Cube Theorems (which are about the mixing of homotopy pushouts and homotopy pullbakcs) of an earlier paper of Mather's.<|endoftext|> TITLE: Integrals involving trigonometric functions and polynomials QUESTION [5 upvotes]: Can one describe all the real polynomials $P(x)$ such that the following integrals converge: $$ \int_0^{\infty} \sin(P(x))dx, \int_0^{\infty} \cos(P(x))dx ? $$ Among special cases are such celebrities as Airy and Fresnel functions. REPLY [8 votes]: Let $k$ be the degree of $P$. The case $k<2$ is trivial. Let now $k\ge2$. Let us show that then both integrals converge. Indeed, without loss of generality, the leading coefficient of $P$ is positive. Then, since $k\ge2$, there is some real $a$ such that $P(x)>0$, $P'(x)>0$, and $P''(x)>0$ for $x\ge a$, so that the differentiable functions $P$ and $P'$ are positive and increasing on the interval $[a,\infty)$, and $P$ maps $[a,\infty)$ onto the interval $[b,\infty)$, where $b:=P(a)$. Let $Q\colon[b,\infty)\to[a,\infty)$ be the function inverse to the restriction $P\big|_{[a,\infty)}$ of $P$ on $[a,\infty)$. For simplicity of writing, let $P$ and $P'$ also denote the restriction $P\big|_{[a,\infty)}$ and its derivative. Then the function $\psi:=Q'=1/(P'\circ Q)$ is positive and decreasing, since $P'$ is positive and increasing and $Q$ is increasing. Moreover, $\psi(u)\to0$ as $u\to\infty$, since $P'(x)\to\infty$ as $x\to\infty$ and $Q(u)\to0$ as $u\to\infty$. Now, for any integer $m\ge b/(2\pi)$, we have $$\int_{Q(2\pi m)}^\infty\sin (P(x))\,dx=\int_{2\pi m}^\infty\sin(u)\,\psi(u)\,du =\sum_{j=0}^\infty(-1)^j I_j, $$ where $$I_j:=\int_0^\pi\sin(t)\,\psi(t+2\pi m+\pi j)\,dt. $$ Because $\sin>0$ on $(0,\pi)$, $\psi$ is positive and decreasing, and $\psi(u)\to0$ as $u\to\infty$, it follows that $I_j$ is positive and decreasing in $j$, and $I_j\to0$ as $j\to\infty$. So, the alternating series $\sum_{j=0}^\infty(-1)^j I_j$ converges, and hence so do the integrals $\int_{Q(2\pi m)}^\infty\sin (P(x))\,dx$ and $\int_0^\infty\sin (P(x))\,dx$. Quite similarly, the integral $\int_0^\infty\cos (P(x))\,dx$ converges. Another way to finish the solution (again for $k\ge2$) is to use integration by parts. Indeed, since $P'$ is positive and increasing on $[a,\infty)$ and $P'(x)\to\infty$ as $x\to\infty$, one has $$i\int_a^\infty e^{iP(x)}\,dx=\int_a^\infty \frac1{P'(x)}\cdot e^{iP(x)}\,iP'(x)\,dx =-\frac{e^{iP(a)}}{P'(a)} -\,\int_a^\infty e^{iP(x)}\,\Big(\frac1{P'(x)}\Big)'\,dx. $$ At that, $$\int_a^\infty \Big|e^{iP(x)}\,\Big(\frac1{P'(x)}\Big)'\Big|\,dx =-\int_a^\infty \Big(\frac1{P'(x)}\Big)'\,dx=\frac1{P'(a)}<\infty. $$ It follows that the integral $\int_0^\infty e^{iP(x)}\,dx$ is converging, and hence so are the integrals $\int_0^\infty\sin (P(x))\,dx$ and $\int_0^\infty\cos (P(x))\,dx$.<|endoftext|> TITLE: Mixed Hodge structure on configuration spaces QUESTION [15 upvotes]: Let $X$ be a smooth complex projective variety. Let $F(X,n)$ be the configuration space parametrizing $n$ distinct ordered points in $X$. The cohomology groups $H^k(F(X,n),\mathbf Q)$ carry a mixed Hodge structure, by Deligne. Is there an example where this mixed Hodge structure does not split? In other words, where there is no isomorphism $ H^k(F(X,n), \mathbf Q) \cong \bigoplus_{i} \mathrm{Gr}^W_i H^k(F(X,n), \mathbf Q) $? I assume the answer is yes and that an example can be found already when $n=3$ and $X$ is a curve of genus $\geq 2$, but I don't know one. I would also be interested in the analogous question for $\ell$-adic cohomology and $X$ over (say) a finitely generated field. Motivation: I was doing some computations about cohomology of configuration spaces which led me to making an optimistic conjecture, which I then realised would imply that all mixed Hodge structures of this form must split. This made me no longer believe the conjecture, but I don't actually have a counterexample. REPLY [7 votes]: In fact the conclusion in Gorinov's paper seems to be false, see E. Looijenga, "Torelli group action on the configuration space of a surface", arXiv:2008.10556<|endoftext|> TITLE: Green's function of the Ornstein-Uhlenbeck operator QUESTION [10 upvotes]: The Ornstein-Uhlenbeck operator $L$ is given by $$ Lu = \Delta u- \frac{1}{2}x\cdot \nabla u. $$ Is there a known closed form expression of the Green's function of $L$ on $\mathbb R^d$ (for $d\geq 2$ or at least for $d=2$) ? Any references? Thanks a lot! REPLY [7 votes]: Denote by $H_k(x)$ the $k$-th Hermite polynomial in one variable, $$H_k(x) =\delta^k 1,$$ where $\delta f(x)=xf(x)-f'(x)$, $\newcommand{\bR}{\mathbb{R}}$ $\forall f\in C^\infty(\bR)$. $\newcommand{\bx}{\boldsymbol{x}}$ For $\bx=(x_1,\dotsc,x_d)\in\bR^d$ and $\newcommand{\bZ}{\mathbb{Z}}$$\alpha\in\bZ^d_{\geq 0}$ we set $$ H_\alpha(\bx):=H_{\alpha_1}(x_1)\cdots H_{\alpha_d}(x_d). $$ Denote by $\Gamma$ the standard Gaussian measure on $\bR^d$, $$ \Gamma(d\bx)=\frac{1}{(2\pi)^{d/2}} e^{-\frac{1}{2}\Vert\bx\Vert^2} d\bx. $$ We have $$\int_{\bR^d} H_\alpha(\bx)^2\Gamma(\bx)=\alpha!:=\prod_j \alpha_j!, $$ $$ L H_\alpha = -|\alpha|\, H_{\alpha},\;\;\alpha=\sum_j\alpha_j. $$ Moreover, the linear span of the set of polynomials $H_\alpha(\bx)$, $\alpha\in\bZ^d_{\geq 0}$, is dense in the Hilbert space $L^2\bigl(\,\bR^d, \Gamma(d\bx)\,\bigr)$. Thus any $f\in L^2(\bR^d,\Gamma)$ has an orthogonal decomposition $$ f(\bx)=\sum_\alpha\frac{f_\alpha}{\alpha!} H_\alpha(\bx),\;\;f_\alpha=\int_{\bR^d} f(\bx) H_\alpha(\bx)\Gamma(d\bx). $$ The range of $L$ is the codimension $1$ subspace of $L^2(\bR^d,\Gamma)$ consisting of functions $f$ such that $f_0=0$. Consider the bounded operator $$G:L^2(\bR^d,\Gamma)\to L^2(\bR^d,\Gamma), $$ given by $$ G[f](\bx)= -\sum_{\alpha\neq 0}\frac{1}{|\alpha|} \frac{f_\alpha}{\alpha!} H_\alpha(\bx). $$ Then $G[f]$ belongs to the domain of $L$ for any $f\in L^2(\bR^d,\Gamma)$ and $$ LG [f]=f-f_0. $$ The operator $G$ is an integral operator and its integral kernel has the form $\newcommand{\by}{\boldsymbol{y}}$ $$K_G(\bx,\by)=-\sum_{\alpha\neq 0}\frac{1}{|\alpha|\cdot \alpha!} H_\alpha(\bx) H_\alpha(\by). $$ This means that $$G[f](\bx)=\int_{\bR^d} K_G(\bx,\by) f(\by) \Gamma(d\bx). $$ You can take $K_G$ as your Green's function. You can simplify the description of $K_G$ a bit by using Mehler's formula $$\sum_{k\geq 0}H_k(x)H_k(y)\frac{r^k}{k!}=\frac{1}{\sqrt{1-r^2}} \exp\left(-\frac{(rx)^2-2rxy+(ry)^2}{2(1-r^2)}\right). $$ For more details you can check Malliavin's book Integration and Probability.<|endoftext|> TITLE: Splitting integers 1, 2, 3, … n to avoid least possible sum QUESTION [9 upvotes]: For each positive integer n, partition the integers 1, 2, 3, … 2n into two sets of n integers each. Let g(n) be the least integer such that there is such a partition in neither of whose parts there is a subset of numbers whose sum is g(n). (Thus g(5) = 11 because of {1,3,4,5,9} and {2,6,7,8,10}). Is anything known about g(n)? REPLY [4 votes]: Let me show that $g(n)$ grows quadratically (surely, it cannot grow faster). It follows from the following Claim. Let $t$ be a sufficiently large integer, and let $A\subset\{1,2,\dots,2t-1\}$ be a subset of cardinality $|A|\geq t$. Then every integer $s\in[O(t\log t),t^2/2(1+o(1))]$ is a sum of some subset in $A$. This easily implies that $g(n)\geq n^2/2(1+o(1))$, since for every $s\in [2n,n^2/2(1+o(1))]$ we may fins a suitable $t\leq n$; then the intersection of one of the partitioning sets with $[1,2t-1]$ will satisfy the condition of the Claim. To prove the Claim, we start with some lemmas. Lemma 1. Let $a\frac{a}{a+b}(n+b+1)$. Then $B$ contains two numbers whose difference lies in $[a,b]$. Proof of Lemma 1. Assume the contrary. Let $b_1<\dots TITLE: Teaching the fundamental group via everyday examples QUESTION [26 upvotes]: This question is a "prequel" to a similar question about homology. Both questions were inspired by seeing a talk, by Tadashi Tokieda, about the interesting physics that appears in toys. What stories, puzzles, games, paradoxes, toys, etc from everyday life are better understood after learning about the fundamental group and/or covering spaces? To be more precise, I am teaching a short course on the fundamental group and covering spaces, from chapter one of Hatcher's book. I want to motivate the material with everyday objects or experiences. Here are some examples and then some non-examples, to explain what I am after. First the examples: $\newcommand{\RR}{\mathbb{R}} \newcommand{\SO}{\operatorname{SO}} \newcommand{\ZZ}{\mathbb{Z}}$ The plate (or belt) trick; this is a fancy move that a waiter can make with your plate, but it is more likely to appear in a juggling show. It is "explained" by the fact $\pi_1(\SO(3)) = \ZZ/2\ZZ$. Tavern puzzles: before trying to solve a tavern puzzle, one should check that the two pieces are topologically unlinked. You can decide this by computing $\pi_1$ of the complement of one of the pieces, and then checking the other piece is trivial. The game of skill, the endless chain (also called fast-and-loose), is explained by computing winding number, ie computing in $\pi_1(\RR^2 - 0)$. In the woodprint Möbius Strip II the ants illustrate the orientation double-cover (an annulus) of the strip. One could also perform the usual game of cutting the Möbius strip along its core curve to demonstrate a double cover of the circle by the circle. Noticeably missing are any real life toys/puzzles/games that rely on the idea of homotopy. Now for the non-examples: Impossible objects such as the Penrose tribar that exist locally, but not globally. These can be explained via non-trivial cohomology classes. But homology and cohomology are not discussed in this course. So - no cohomology! You can find many real-life examples of cohomology discussed here. Winding number (in the form of linking number) also arises in discussions of DNA replication; see discussions of topoisomerase. However DNA is not an everyday object, so it is not a good example. There are no draws in the board game Hex. This is equivalent to the Brouwer fixed-point theorem. This example is not very good, because most people don't know the game. REPLY [15 votes]: I recently heard this puzzle from Dror Bar-Natan, and there's a nice solution using the fundamental group. There are $n$ nails arranged in a line on a wall. Find a way of hanging a picture from these nails so that if any 1 nail is removed, then the picture will fall. To solve it, you can first reformulate it as follows: nails correspond to punctures in the plane, and removing a nail corresponds to filling in a puncture. The fundamental group of such a space is freely generated by loops around each puncture, and filling in a puncture corresponds to quotienting by one generator. We'd like a loop that is killed in each of these quotients, and it's easy to write one down inductively using iterated commutators. REPLY [12 votes]: My favorite example is this. You want to hang a picture (to which a piece of rope is attached in the usual way) on a wall. There are two nails drawn in the wall, close to each other. One has to hang it with the following conditions: a) The picture must be hanging (does not fall). b) When one nail is removed (any one), the picture falls. When you solve this, generalize to n nails. I asked this question many people from 10 years olds to participants of a conference in low dimensional topology. The average time required to find a solution is about the same, and it is strictly greater than $0$. Rope and nails or some substitute were always present when this question was asked). I use this example in my teaching in two ways. When I explain the correct statement of Cauchy's theorem in Complex Variable, and the difference between the fundamental group and the first homology group, and when I teach what is a non-Abelian group. My second favorite example is the belt trick which is associated to the name of Dirac, and which is mentioned in the question. Sometimes I do it without a belt, using just my hand holding a pencil. EDIT. Sorry, it is a duplicate:-) But I have a reference: http://www.math.purdue.edu/~eremenko/problems.html I did not invent this problem myself:-( It was one German student who asked me this question when we were hiking in the mountains about 15 years ago. I don't remember his name:-(<|endoftext|> TITLE: Making $\mathbb{Q}$-cohomology integral QUESTION [29 upvotes]: Let $X$ be an algebraic variety (say, smooth and projective) over $\mathbb{C}$, and fix $$\alpha\in H^i(X^{\text{an}}, \mathbb{Q})$$ with $i>0$. Does there always exist a variety $Y$ and a smooth proper morphism $f: Y\to X$ such that $f^*\alpha$ is integral, i.e., it is in the image of $H^i(Y, \mathbb{Z})\to H^i(Y, \mathbb{Q})$? Examples: If $i=1$, $\alpha$ gives a map $\pi_1(X)\to \mathbb{Q}\to \mathbb{Q}/\mathbb{Z}$. The kernel of this map defines a (finite) covering $Y$ of $X$. Pulling back to $Y$ makes $\alpha$ integral. If $i=2$, there is a map $\mathbb{Q}\to \mathcal{O}_X\to\mathcal{O}_X^*$. Let $\gamma$ be the image of $\alpha$ under the induced map $H^2(X, \mathbb{Q})\to H^2(X, \mathcal{O}_X^*)$; this is an analytic Brauer class on $X$. I believe a result of Serre(?) tells us that it is also an etale-cohomological Brauer class. As $X$ is projective, Gabber tells us that $\gamma$ is in fact an honest Brauer class, so we may choose a Severi-Brauer variety $f: Y\to X$ with class $\gamma$. Then $f^*(\gamma)=0$. If $f^*\alpha$ is non-zero, it lives in $H^{1,1}(Y)\cap H^2(Y, \mathbb{Q})$. Let $n$ be such that $n\cdot f^*\alpha$ is integral; then by the Lefschetz $(1,1)$-theorem, there is a line bundle $\mathcal{L}$ on $Y$ with first Chern class $n\cdot f^*\alpha$. Let $\mathcal{Y}_\mathcal{L}\to Y$ be the $\mu_n$-gerbe of $n$-th roots of $\mathcal{L}$. This isn't quite a smooth projective variety over $X$, admittedly, but the pullback of $\alpha$ to $\mathcal{Y}_\mathcal{L}$ is integral. In any case, I'm fine with $Y$ being a DM stack. (There is always some construction that works involving higher stacks in the analytic category, but this is not what I'm looking for.) I guess I'd prefer $Y$ to be an actual variety, so this argument reduces the $i=2$ case to the following special case: Let $X$ be as above, and let $\mathcal{L}$ be a line bundle on $X$. Fix a positive integer $n$. Is there always some smooth proper $f: Y\to X$ so that $f^*\mathcal{L}$ has an $n$-th root? If $i>2$, I have no idea what to do. Some more remarks: For varieties so that the cup product $H^1(X, \mathbb{Q})^{\otimes n}\to H^n(X, \mathbb{Q})$ is surjective (say, Abelian varities or curves of genus at least $1$), this is certainly possible. Namely, write $\alpha$ as a linear combination of things coming from $H^1$, and then make all of those integral. Qiaochu addresses the topological version. This is already difficult, I think, but it is slightly easier than the algebraic one. For example, I don't know how to make classes in $H^2(\mathbb{P}^1, \mathbb{Q})$ integral, but the Hopf map $S^3\to S^2$ makes every class in $H^2(S^2, \mathbb{Q})$ integral, since $H^2(S^3)=0$. To do the case $i=2$, it suffices to consider the case when $X=\mathbb{P}^n$. To see this, observe that my argument above (under Examples, $i=2$) reduces to the case where $\alpha\in H^{1,1}(X)\cap H^2(X, \mathbb{Q})$. Then use that ample classes generate $H^{1,1}(X)\cap H^2(X, \mathbb{Q})$. Jason Starr suggests that every smooth projective morphism $Y\to \mathbb{P}^1$ might have a section, which would give a negative answer for $X=\mathbb{P}^1, i=2$. Is that true? I certainly don't know a counterexample... REPLY [8 votes]: As Jason Starr remarks in the comments, this answer to a question of his implies the answer to both of my questions is "no." For the latter, one may take: $X=\mathbb{P}^1, \mathcal{L}=\mathcal{O}(1)$ Then there is no smooth proper morphism $f: Y\to X$ such that $f^*\mathcal{L}$ has a square root, for example. (Jason's question is about projective morphisms, but I think it's fairly easy to deduce the proper case.) For the first question, $X=\mathbb{P}^1, i=2$ is a counterexample. I would be curious to see a simpler proof of this result. I would also be curious to know the answer with the smoothness condition relaxed to flatness.<|endoftext|> TITLE: What does the sum of the reciprocals of all the highly composite numbers converge to? QUESTION [7 upvotes]: I've calculated the sum of the reciprocals of all the $156$ first highly composite numbers up to $10^{18}$: $\sum \dfrac{1}{HCC(n)} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} + \dfrac{1}{36} + ... \approx 1.1328728$ It seems to converge to this number... but is there a mathematical formula which spawns this number? Or is this number completely random? I've noticed that it's close to $e^{\frac{1}{8}} \approx 1.133148$ but this may just be a coincidence. . I've also calculated the sum of the reciprocals of all the $69$ first superior highly composite numbers up to $10^{95}$: $\sum \dfrac{1}{SHCC(n)} = \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{60} + \dfrac{1}{120} + ... \approx 0.77839341508236$ So again same question: does this number have a mathematical formula which spawns it? I've noticed that it's close to $e^{-\frac{1}{4}} \approx 0.7788008$ but again it's also maybe a coincidence. REPLY [11 votes]: Just some thoughts I wanted to share. I guess they are random numbers, which we can not express in terms of other popular constants. However, let $Q(n)$ be the number of highly composite numbers $\leq n$. Then it is known that $(\ln(x))^a\leq Q(x)\leq(\ln(x))^b$ for some $a$ and $b$ bigger than 1. Assuming $Q(x)\approx (\ln(x))^c $ for some $c>1$ and interpreting $Q'(x)$ as the density of highly composite numbers, we can naively write \begin{equation} \sum\frac{1}{HCC(n)}\approx \int_{x=1}^\infty Q'(x)\frac{1}{x}\text{ d}x\approx \int_{x=1}^\infty \frac{c(\ln (x))^{c-1}}{x^2}\text{ d}x=\Gamma[c+1].\end{equation} Now, here http://wwwhomes.uni-bielefeld.de/achim/highly.html it is said that $c\approx \frac{5}{4}$. Finally, $\Gamma[\frac{5}{4}+1]\approx 1.133$, which is close to your numerical value.<|endoftext|> TITLE: How can I have a copy of this old paper by Frobenius? QUESTION [11 upvotes]: How can I have a copy of this old paper and a translation of it? Frobenius, G. (1902). Uber primitive Gruppen des Grades n und der Klasse n - 1. S. B. Akad. Berlin 1902, 455-459. REPLY [20 votes]: It is freely available online. You can find it here: http://goo.gl/pDkRi6 (Download the file and take a look at pages 455-459 of it.)<|endoftext|> TITLE: Meaning of $g_d^r$ in algebraic geometry QUESTION [5 upvotes]: As an editor I often encounter the symbol $g_d^r$ as a noun. I tried googling but I only get papers where the symbol is used without a definition. Can someone supply a reference to a definition? Examples of usage: "Koszul cohomology groups of $g_d^r$s on singular nodal curves"; "any other basepoint-free $g_h^1$ [...] is composed with the given $g_d^1$. REPLY [5 votes]: As I understand, a $g_d^r$ is a linear system of dimeansion $r$ and degree $d$. Basically, these give you maps to $\mathbb{P}^r$ of degree $d$. The simplest example is of course hyperelliptic curves; these are the same as a $g^1_2$. The reference that I have for this is Harsthorne, Algebraic Geometry, p. 341.<|endoftext|> TITLE: Painting $n$ balls from $2n$ balls red, and guessing which ball is red, game QUESTION [15 upvotes]: The game Lucy has $2n$ distinct white colored balls numbered $1$ through $2n$. Lucy picks $n$ different balls in any way Lucy likes, and paint them red. Lucy then giftwrap all the balls so that it is impossible to tell its color without opening its wrapping. Lucy does this before offering any of the balls to Alice. Alice would like to get one of the red balls. Alice doesn't know which balls are painted red. Lucy will offer Alice the balls one by one, starting from ball $1$, until ball $2n$. Upon being offered each ball, Alice may either take the ball or pass it. If Alice takes the ball, Alice open the wrapping. If it is red, Alice takes it and the game ends -- otherwise, Alice will continue to be offered the remaining balls (unless all balls has been offered). If Alice pass the ball, the ball remains wrapped and thus Alice does not know the color of the ball, and Alice will continue to be offered the remaining balls. Note that it is possible for Alice not to get any red ball -- for example, if Alice choose not to take any ball. Alice's goal is to Obtain a red ball with probability at least $1 - 1/n$ (that is, it is okay for her not to get a red ball as long as this probability does not exceed $1/n$), and Minimize the expected number of balls Alice take. Lucy's goal is the complete opposite of Alice, that is: If it is possible for her to make Alice unable to obtain a red ball with probability at least $1 - 1 / n$, she will do so. Otherwise, maximize the number of balls Alice take in average (that is, the expected number of balls Alice take). Question What is the optimal strategy for Alice? What is the expected number of balls that Alice would take in the optimal case? Example When $n = 2$, Alice's goal is to obtain a ball with probability at least $1/2$. Here's an example strategy, which is provably optimal: Decide which one of the four balls to open at random, and open that and only that ball. Clearly the probability that the opened ball is red is $1/2$, and the expected number of balls opened is $1$. My work Let $X$ be the expected number of balls Alice would take if both players play optimally. Upper bound on X I can proof a $O(\log n)$ upper bound on $X$ by letting Alice select $\log_2 n$ balls at random to open. The probability of Alice not finding any red ball is at most $2^{-\log_2 n} = 1/n$. Lucy would then paint the last $n$ balls red, and we have $X = O(\log n)$. Lower bound on X Edit: As domotorp pointed in chat, the proof in the pdf below for my lower bound is wrong because I wrongly use the assumption that E(x) and E(y) were independent. I don't see any immediate fix to the proof, so I will try to work on this again. I proven a $\Omega(\log \log n)$ lower bound on $X$ in this pdf. I used a different terminology in the pdf: instead of having $2n$ balls, $n$ colored red, we have $2n$ bins, $n$ of which has ball inside. Motivation In the distributed systems settings, a rather common approach to guarantee that an algorithm is fault tolerant is by utilizing a bound on the number of possible failures. One of my supervisor's paper's algorithm works by utilizing a cheap algorithm that runs very fast if the number of failures is very small. To accommodate for larger number of failures, he decides to run this multiple times -- if with high probability (at least $1 - 1/n$) one of these runs have very few errors, then this algorithm will be much faster than the previously known one. At the moment, the paper utilize the $O(\log n)$ upperbound, but we wondered whether this bound is optimal. A very nice implication is that if a strategy with expectation lower than $O(\log n)$ is found for this game, then the algorithm's complexity in the paper will decrease correspondingly. However, I strongly believe that it is more probably to prove $\Omega(\log n)$ rather than $o(\log n)$. REPLY [8 votes]: Here is a sketch of an argument by my colleague Jim Roche of an $\Omega(\log n/\log \log n)$ lower bound. The basic idea is that Lucy chooses randomly between two strategies, one of which puts all the red balls early (thereby forcing Alice to open some number of the early boxes), and the other of which puts a lot of the red balls late (thereby forcing Alice to open some number of the late boxes). Specifically, with probability 1/2, Lucy distributes the $n$ red balls randomly among the first $n(1 + 1/\log_2 n)$ boxes. With probability 1/2, she distributes $n/\log_2 n$ red balls randomly among the first $n(1 + 1/\log_2 n)$ boxes, and puts the remaining red balls at the very end. We now give a lower bound on how many boxes Alice expects to open under any randomized strategy that succeeds with probability at least $1-1/n$. The argument that follows is slightly imprecise because it treats the probabilities that the boxes contain red balls as independent, whereas they are actually dependent since Lucy is constrained to paint exactly $n$ balls red. However, the error terms are negligible, and the argument is clearer if we ignore the dependencies. For any constant $C$, define $P_C$ to be the probability that Alice chooses at most $(C\ln n)/\ln\log_2 n$ of the first $n(1+1/\log_2 n)$ boxes. Then we must have $${1\over n} \ge \Pr\{\hbox{Alice fails}\} \ge {P_C\over 2} \left({1\over\log_2 n} \right)^{(C\ln n)/\ln\log_2 n}, $$ so $$P_C \le {2\over n}\exp (C\ln n) = 2n^{C-1}. $$ In particular, for $C=1/2$, the probability that Alice chooses at most $(\ln n)/2\ln \log_2 n $ of the first $n(1+1/\log_2 n)$ boxes is at least $2/\sqrt n$. Therefore, with probability 1 as $n\to\infty$, Alice must examine at least $(\ln n)/ 2 \ln \log_2 n$ of the first $n(1+1/\log_2 n)$ boxes. But remember that with probability 1/2, Lucy places only $n/\log_2 n$ red balls among the first $n(1+1/\log_2 n)$ boxes. If she does so, then the probability that Alice's first $(\ln n)/2 \ln \log_2 n$ looks uncovers a red ball is upper-bounded by the union bound $$\left({\ln n\over 2\ln\log_2 n}\right)\left({1\over \log_2 n}\right),$$ which approaches zero as $n\to\infty$. Therefore, with probability at least $1/2 -\epsilon$, Alice must examine at least $(1/2 - \epsilon)(\ln n)/\ln \log_2 n$ boxes. This establishes the $\Omega(\log n/\log \log n)$ bound.<|endoftext|> TITLE: Dimension of Hilbert scheme? QUESTION [5 upvotes]: I have a subvariety $V \subset X$, and I want to compute the dimension of the connected component of $\textrm{Hilb}(X)$ containing $[V]$. I can give explicit deformations of $V$ showing that the dimension of this component is at least $m$. I can also show that $H^0(V, N_{V /X}) \leq m$. Do these facts together imply that the dimension of this component is exactly $m$, and that $\textrm{Hilb}(X)$ is smooth at $[V]$? I am a little worried about how the tangent sheaf works in the singular case. REPLY [6 votes]: In general, the Zariski tangent space of $\textrm{Hilb}(X)$ at $[V]$ is naturally isomorphic to $\textrm{Hom}_V(I_V/I_V^2, \, \mathcal{O}_V)$. When $X$ and $V$ are both smooth and projective, this group equals $H^0(V, \, N_{V/X})$. Therefore in your case we can write $$m \leq \dim _{[V]}\textrm{Hilb}(X) \leq \dim T_{[V]} \textrm{Hilb}(X) = H^0(V, \, N_{V/X}) \leq m.$$ This means $$\dim_{[V]}\textrm{Hilb}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$ that is, $\textrm{Hilb}(X)$ is smooth at $[V]$, of dimension $m$.<|endoftext|> TITLE: Parity of primes QUESTION [6 upvotes]: While working on a completely different (combinatorial) problem, I ran a simple program to calculate the parity of the first ~50000 primes (number of 1s in their binary representation modulo 2). The following graph summarizes the result: The number of primes having even parity seems to grow slower. Is there a math explanation ? I looked to the correspoding OEIS entry, but it doesn't provide any detail. REPLY [7 votes]: It was shown (amongst other things) by Gel'fond that the asymptotic growth rates of the primes with even and odd parity base 2 expansions are the same: Gelʹfond, A. O. Sur les nombres qui ont des propriétés additives et multiplicatives données. (French) Acta Arith. 13 1967/1968 259–265. A more recent paper on this is due to Mauduit and Rivat.<|endoftext|> TITLE: A homological criterion for collapsibility? QUESTION [8 upvotes]: On page 256 of Kirby and Siebenmann one finds the following lemma (its proof an "elementary exercise", so they only give a hint): Taking $A$ to be a point and iterating this collapsing lemma, this seems to give a criterion for collapsibility of simplicial complexes: a finite simplicial complex $B$ is collapsible if and only if $\tilde{H}^m(B;\mathbb{Z}_2) = 0$ for all $m$. This seems false to me. First of all this can't be true if $B$ is a classifying space of a finitely presented acyclic group (like Higman's group). It also contradicts well-known examples of non-collapsible simplicial complexes like the house with two rooms (or at least a triangulation of it) or Rudin's non-collapsible triangulation of the 3-disk. In fact, those examples also give contradictions to the hint. So am I misinterpreting this lemma or is it not correct? It is used in the proof for the sliced structure classification theorem for open 4-manifolds that do not admit a handle decomposition. If the lemma is not correct, can that proof be fixed? REPLY [3 votes]: This question was essentially answered in the comments. I'm collecting the information in an answer, because it might be interesting to some people. Firstly, the lemma as stated is false. There was some discussion about whether the lemma can be iterated. To see this is the case, note a simplicial collapse of B onto a subcomplex containing $A$ in particular must fix $A$ and will be homotopy equivalence rel $A$. Using iteration the following simplicial complexes $B$ with $A = \ast$ give counterexamples: A triangulation of the house with two rooms is a contractible but not collapsible simplicial complex of dimension 2. The second barycentric subdivision of the dunce hat is another contractible but not collapsible simplicial complex of dimension 2. Rudin's triangulation is a non-collapsible triangulation of the 3-disk. This question lists various other interesting examples of non-collapsible triangulations of disks. Without iterating you get a counterexample from Higman's group (or any other finitely-presented acyclic group). This is the finitely presented acyclic group $G = \langle x_i|x_ix_{i+1}x_i^{-1}\rangle_{i \in \mathbb{Z}/4\mathbb{Z}}$. There is a finite simplicial complex $BG$ of dimension 2 that is a classifying space for $G$. Taking $A = *$, we get that $H^2(BG,*;\mathbb{Z}/2\mathbb{Z}) = 0$, so the lemma would imply that $BG$ collapses onto, and thus is homotopy equivalent to, a simplicial complex of dimension 1. But every finite simplicial complex of dimension 1 has a free group as fundamental group. As Oscar Randal-Williams suggests in the comments, the lemma is true if one additionally assumes that at most two $m$-simplices in $B\backslash A$ attach to an $(m-1)$-simplex in $B \backslash A$. For suppose this is the case, but all $m$-simplices $S$ in $B \backslash A$ that have the property that each face not in $A$ is the face of another $m$-simplex (there is then exactly one of these by the additional assumption). It suffices to prove we can collapse at least one $m$-simplex in $B \backslash A$, because the additional assumption and homological condition are preserved by such a collapse. Consider the sum of all $m$-simplices in the cellular chain complex of the pair $(B,A)$: $\sigma = \sum_i S_i$. Then $\partial \sigma$ is $0$, as faces either lie in $A$ or appear twice (and we are working modulo $2$). Thus $\sigma$ is a cycle, but this contradicts $H_m(B,A;\mathbb Z/2\mathbb Z)=0$, which follows from the assumption on cohomology using the universal coefficient theorem. This additional assumption is satisfied if $B$ is a PL manifold of dimension $m$, because in that case the link of every $(m-1)$-simplex is either $S^0$ or a point (depending on wither the $(m-1)$-simplex is in the boundary or not). One cannot iterate the lemma with the additional assumption, so we do not get contradiction (e.g. from Rudin's triangulation). Coincidentally, it does prove the classical result that every triangulation of the 2-disk is collapsible (apply the lemma with additional assumption, and then note that every tree is collapsible). Finally, if one looks to Lemma (e) on page 257 of Kirby-Siebenmann and its eventual application in the proof of Theorem A.1 on page 262, part ($\delta$), we see that in the proof we can get away with only using the lemma in the case that $B$ is a PL-manifold of dimension $m$. The sliced structure classification theorem for open 4-manifolds that do not admit a handle decomposition is safe.<|endoftext|> TITLE: Characterizing the real analytic Eisenstein series QUESTION [7 upvotes]: Consider the classical real analytic Eisenstein series $$ E(z,s)=\left(\pi^{-s}\Gamma(s)\frac{1}{2}\right)\sum_{(m,n)\neq(0,0)}\frac{y^s}{|mz+n|^{2s}}, $$ where $z=x+iy$. We think of $E(z,s)$ as a function on $(z,s)\in\mathfrak{h}\times\mathbf{C}$. The function $E(z,s)$ satisfies the following properties (1) For a fixed $z\in\mathfrak{h}$, $s\mapsto E(z,s)$ is holomorphic except with poles of order $1$ at $s=1$ and $s=0$ with residues $1/2$ and $-1/2$ respectively (the knowledge of one residue implies the knowledge of the other from the functional equation in $s$, see below). (2) $E(z,s)$ is $SL_2(\mathbb{Z})$-invariant in $z$ (3) $\Delta_h E(z,s)=s(1-s)E(z,s)$ where $\Delta_h$ is the hyperbolic Laplacian. (4) $E(z,s)=E(z,1-s)$ (5) For a fixed $s\in\mathbf{C}\backslash\{\frac{1}{2}\}$, we have $E(z,s)=O(y^{\sigma})$ as $y\rightarrow \infty$ where $\sigma=\max(\Re(s),1-\Re(s))$. Q1 Do the properties (1), (2), (3), (4) and (5) characterize $E(z,s)$ ? Q2 Is there some redundancy among properties (1), (2), (3), (4) and (5)? Q3 What is a good way to characterize what $E(z,s)$ is ? (I guess that representation theorists should have something nice to say for Q3) added Note that $E(z,\frac{1}{2})$ is not square integrable. Indeed, looking at the constant term of the Fourier series in $z$ of $E(z,1/2)$ we find that $E(z,1/2)\sim Cte\cdot\log(y)\sqrt{y}$. So if one integrates in the usual fundamental domain $\mathcal{D}_{T}$ of $SL_2(\mathbb{Z})$, up to height $T$, with respect to the Poincare volume, we find that $$ \int_{\mathcal{D}_T}|E(z,1/2)|^2\frac{dxdy}{y^2}\sim \int_{1}^{T} \frac{\log(y)^{2}dy}{y}\sim \frac{1}{3}\log(T)^3. $$ So as $T\rightarrow \infty$ the integral diverges. Note though that it is "almost" square integrable in the sense that it diverges extremely slowly. REPLY [4 votes]: For $Re(s)>1$, the function $E(z,s)$ is smooth on $\mathbb{H}$ and satisfies $(2)$, $(3)$ and $$ (*) \ E(z,s)-\xi(2s) \cdot y^s=O(y^{1-s}) \ \text { as } y \rightarrow +\infty. $$ (Here $\xi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s)$ is the completed Riemann zeta function.) One can prove this by computing the Fourier expansion of $E(z,s)$. These properties characterize $E(z,s)$ since the difference of any two functions satisfying them is square-integrable on $SL_2(\mathbb{Z})\backslash \mathbb{H}$ (think of the usual fundamental domain) and every eigenvalue $\lambda$ of $\Delta$ in $L^2(SL_2(\mathbb{Z})\backslash \mathbb{H})$ is $\lambda \geq 0$. Together with having meromorphic continuation to $s \in \mathbb{C}$ for fixed $z$, this characterises $E(z,s)$. (Of course, this sort of characterisation of $E(z,s)$ is well-known: see e.g. Lemma 2.5.1 in these notes or Lemma 1 in here.)<|endoftext|> TITLE: Is a group uniquely determined by the sets $\{ab,ba\}$ for each pair of elements a and b? QUESTION [22 upvotes]: This is a cross-posted question, originally active here on math.stackexchange. For a given group $G=(S,\cdot)$ with underlying set $S$, consider the function $$ F_G:S\times S\to\mathcal P(S)\\ F_G(a,b):=\{a\cdot b,~b\cdot a\} $$ from $S\times S$ to the power set of $S$. I'd like to figure out how much information from $G$ is encoded by $F_G$. In particular, does $F_G$ determine the group $G$ up to isomorphism? Given different groups $G_1$ and $G_2$ with underlying sets $S_1$ and $S_2$, suppose that a function $\varphi:S_1\to S_2$ has the property $\varphi\bigl(F_{G_1}(a,b)\bigr)=F_{G_2}\bigl(\varphi(a),\varphi(b)\bigr)$ for all $a$ and $b$ in $G_1$; if $\varphi^{-1}$ exists and has this property as well, let's say that $F_{G_1}\cong F_{G_2}$. For one thing, if $G_1\cong G_2$ then $F_{G_1}\cong F_{G_2}$. Going the other way, if $F_{G_1}\cong F_{G_2}$ then $|G_1|=|G_2|$ and $Z(G_1)\cong Z(G_2)$. Given $F_G$, it's easy to find out which pairs of elements in $G$ commute, which subsets of $G$ constitute subgroups of $G$, and which subsets of $G$ are generating sets of $G$. Moreover, if $F_{G_1}\cong F_{G_2}$ then $G_1$ and $G_2$ must have the same cycle graph. This means that if $F_{G_1}\cong F_{G_2}$ and the order of these groups is less than 16, then $G_1\cong G_2$. Indeed, according to Wikipedia's page on cycle graphs, "For groups with fewer than 16 elements, the cycle graph determines the group (up to isomorphism)." The question that's been plaguing me is whether $F_{G_1}\cong F_{G_2}\implies G_1\cong G_2$ in the general finite case. I think that a counterexample would need to involve groups of order 16 or larger. Any ideas? REPLY [10 votes]: The group is determined but the multiplication is not quite determined. Suppose that the set $S$ is known to both of us (say with an agreed order) and that I have a specific (but unknown to you) $n \times n$ matrix $A$ giving the group operation of $G$. I give you only the function $F_G.$ Then, it turns out, you can recover the the set $\{{A,A^t\}}.$ Either one will determine $G$ uniquely, but you will not know which it was that I used.<|endoftext|> TITLE: a question about minimal non-abelian groups QUESTION [6 upvotes]: Let $G$ be a minimal non-abelian group with cyclic Sylow $p$-subgroup $P$ and normal Sylow $q$-subgroup $Q$ , see [ Huppert, Endlich Gruppen I, Aufgaben III, 5.14]. My quesion is, if there is another minimal non-abelian group $G_1$ which has the same order with $G$, then can we get $G\cong G_1$ ? Thank you. REPLY [6 votes]: The answer is no, but there are at most two isomorphism types of minimal nonabelian groups of a given order that is divisible by $2$ distinct primes. (There are other examples that are $p$-groups such as $D_8$ and $Q_8$, but we are not talking about those here.) Suppose $G$ is minimal nonabelian with $|G|=p^mq^n$, where $G$ has a cyclic Sylow $p$-subgroup $P$ and a normal Sylow $q$-subgroup $Q$. Then from the result that you quote, $Q$ is elementary abelian, and $P = \langle g \rangle$ is cyclic, with $g^p$ centralizing $Q$. Also $g$ acts irreducibly on $Q$, and so $p|(q^n-1)$ but $p$ does not divide $q^k-1$ for any $k TITLE: Probability that a sum of intependent random variables hits a point QUESTION [5 upvotes]: Let $X_1,\ldots,X_n$ be independent random elements of a normed space $X$. Suppose that $\sup_{x\in X}\mathbb{P}(X_i=x)=p_i$. What is the best known upper bound for $$\sup_{x\in X} \mathbb{P}(X_1+\cdots+X_n=x)?$$ Comments: If $p_i$ are all bounded away from $0$ and $1$ independently of $i$ and $n$ then, of course, the maximum is at most $C/\sqrt{n}$, where $C$ depends on how far the $p_i$'s are from $0$. Are there bounds that are good when the $p_i$'s are non homogeneous? Like, say, $p_i=1/i$ or even $p_i=c/n$? Are there any precise results known? Then is, can one in general obtain the exact result in some key situations? Say $p_i=1/2$? REPLY [2 votes]: What you ask is a generalization of the Littlewood-Offord problem. (If each $X_i$ can take only two values, with probability $1/2$, then you get their problem.) This problem has been investigated a lot and has several generalizations, unfortunately I am not an expert on the topic, but I would guess that your problem follows easily from known methods, probably the best is to take each $X_i$ (almost) uniformly distributed on a set of size $\lceil 1/p_i \rceil$. For a recent paper, see http://annals.math.princeton.edu/wp-content/uploads/annals-v169-n2-p06.pdf.<|endoftext|> TITLE: A linear category with objects of infinite length but which is otherwise finite? QUESTION [12 upvotes]: Fix a ground field $k$. By a linear category I will mean an Abelian category which is compatibly enriched over $k$-vector spaces. A linear category is called finite if it satisfies the following four conditions: Every hom vector space is finite dimensional; There are finitely many simple objects, up to isomorphism; There are enough projective objects (i.e. for every object $x$ there is a projective object $p$ and a surjection $p \to x$; and Every object has finite length, i.e. for each object $x$, any descending chain of proper subobjects $$ x \supsetneq x_0 \supsetneq x_1 \supsetneq x_2 \supseteq \cdots $$ is necessarily finite. It is well known that a linear category is finite if and only if it is equivalent to $Rep(A)$ the category of finite dimensional modules of a finite dimensional $k$-algebra $A$. See for example Prop. 1.4 of arXiv:1406.4204 for a proof of this fact. The proof given in this reference uses all four of these properties. But one can still wonder if each of these assumptions is absolutely necessary? Perhaps some of these conditions imply the others? The following examples show that this is not the case for the first three properties: Example 1: Let $K$ be an infinite field extension of $k$. Then the category $Vect_K$ of finite dimensional $K$-vector spaces (finite $K$-dimension, that is) is a $k$-linear category which satisfies (2), (3), and (4), but not (1). For the next examples if $A$ is a $k$-algebra, not necessarily finite, we will let $Rep(A)$ denote the category of finite dimensional $A$-modules, i.e. finite dimensional vector spaces equipped with an action of $A$. Example 2: Let $A = \oplus_\infty k$, the countable direct sum of copies of the algebra $k$. Then $Rep(A)$ satisfies (1), (3), and (4), but not (2). Example 3: Let $A = k[[x]]$ be the formal power series algebra on one generator. Then $Rep(A)$ satisfies (1), (2), and (4), but not (3). So that leaves the finial property. My question is: Question: Is there an example of a linear category which satisfies the above properties (1), (2), and (3), but which fails to satisfy property (4)? It is easy to construct examples which fail to satisfy property (4), but everything I have come up with so far also fails to satisfy one of the other properties as well. One of the reasons I am interested in this question is to better understand the Deligne tensor product. It is usually defined for finite linear categories, and I would like to understand what kinds of categories we would be dealing with if various assumptions were dropped. REPLY [10 votes]: Consider the category of functors from the poset $(\mathbb{Q}_{\geq0},\leq)$ to finite-dimensional vector spaces, and let $\mathcal{A}$ be the full subcategory consisting of functors $F$ for which there is a finite sequence $$0=x_0 TITLE: Characterize the category of rings QUESTION [19 upvotes]: (Sub)categories of many well-studied mathematical objects have been characterized purely in terms of their morphisms. Some (famous) examples: Sets and functions, due to Lawvere. Modules over some ring and module homomorphisms, due to Mitchell and Freyd. Sheaves over some site and natural transformations, due to Giraud. Topological spaces and continuous functions, due to Schlomiuk. Boolean algebras and homomorphisms, by combining the above. In a similar vein: Is there a characterization of the category of rings and homomorphisms? REPLY [11 votes]: Yves Diers has the notion of a Zariski category in [Categories of commutative algebras], which apparently suffices to carry out a lot of commutative algebra in an axiomatic fashion. I reproduce the definition: A Zariski category is a category $\mathcal{A}$ satisfying the following conditions: $\mathcal{A}$ is cocomplete. $\mathcal{A}$ has a strong generating set whose objects are finitely presentable and flatly codisjunctable. Regular epimorphisms are universal i.e. stable under pullbacks. The terminal object of $\mathcal{A}$ is finitely presentable and has no proper subobject. Binary products of objects are co-universal i.e. stable under pushouts. For any finite sequence of codisjunctable congruences $r_1, \ldots, r_n$ on any object with respect codisjunctors $d_1, \ldots, d_n$, we have $$r_1 \vee^c \cdots \vee^c r_n = \mathrm{id}_{A \times A} \implies d_1 \vee \cdots \vee d_n = \mathrm{id}_A$$ where $\vee^c$ denotes the join in the lattice of congruences on $A$, while $\vee$ denotes the co-union of quotient objects of $A$. For more details, see the cited book, or the introduction of this article.<|endoftext|> TITLE: Does every smooth, projective morphism to $\mathbb{C}P^1$ admit a section? QUESTION [56 upvotes]: Possibly this has already been asked, but it came up again in this question of Daniel Litt. Does every smooth, projective morphism $f:Y\to \mathbb{C}P^1$ admit a section, i.e., a morphism $s:\mathbb{C}P^1\to Y$ such that $f\circ s$ equals $\text{Id}_{\mathbb{C}P^1}$? Edit. As Ariyan points out, this article proves that there are at least $3$ singular fibers of $f$ if either $Y$ has nonnegative Kodaira dimension, Theorem 0.1 of Viehweg-Zuo, or if the fibration is non-isotrivial with general fiber either general type or with $\omega_f$ semi-positive, Theorem 0.2 by Möller-Viehweg-Zuo. This suggests an approach to proving the conjecture, at least assuming the uniruledness conjecture (negative Kodaira dimension implies uniruled): take the MRC quotient and then apply the Minimal Model Program to try to reduce to these theorems. Unfortunately, both formation of the MRC quotient and the Minimal Model Program are likely to introduce singularities . . . Second Edit. As Ben Wieland points out, this is false in the category of compact, complex manifolds. The examples are interesting (to me) because they also come up in showing "rationally connected" fibrations over a Riemann surface in the analytic category need not admit sections. Begin with the $\mathbb{C}^\times$ -torsor $T$ over $\mathbb{P}^1$ associated to any nontrivial invertible sheaf, e.g., $\mathcal{O}_{\mathbb{P}^1}(1)$. Now let $q:T\to Y$ be the fiberwise quotient by multiplication by some element $\lambda\in \mathbb{C}^\times$ of modulus $\neq 1$. Projection of $T$ to $\mathbb{P}^1$ factors through this quotient, $f:Y\to \mathbb{P}^1$. Although every fiber of $f$ is the same Hopf surface elliptic curve, there is no section: if there were, its inverse image in $T$ would be a disjoint union of sections of $T$ (since $\mathbb{P}^1$ is simply connected), and $T$ has no sections. A Positive Answer by Paul Seidel. 22 September 2017. I received a communication from Paul Seidel that he knows how to prove this using methods of symplectic topology. REPLY [17 votes]: Yes, using some symplectic geometry. Let's say we had $X \subset {\mathbb C}P^n \times {\mathbb C}P^1$, with projection to $\mathbb{C} P^1$ a smooth morphism (meaning, for topologists, a proper holomorphic submersion; all fibres are smooth). With the restriction of the standard Kaehler form, this becomes a symplectic fibre bundle over ${\mathbb C}P^1 = S^2$. It is known that the Gromov-Witten invariant counting sections of any such fibration (with suitable incidence conditions) is always nonzero. This is proved by reversing the orientation of $S^2$ to get another such fibration, and then considering the fibre union of those two as a degeneration of the trivial fibration. References: The UCDavis math arXiv front paper with ID 9511.5111 (ADDED: this might be Seidel's $\pi_1$ of symplectic automorphism groups and invertibles in quantum homology rings, https://arxiv.org/abs/dg-ga/9511011) François Lalonde, Dusa McDuff, Leonid Polterovich, Topological rigidity of Hamiltonian loops and quantum homology, https://arxiv.org/abs/dg-ga/9710017 Dusa McDuff, Quantum Homology of fibrations over $S^2$, https://arxiv.org/abs/math/9905092 PS: Slight edits, I hope this clarifies (and that I have not misunderstood the question, of course). I added another reference (the third one): see Theorem 1.5 there, and note that the map takes values in the invertible elements of quantum cohomology (hence is nonzero).<|endoftext|> TITLE: Universality with respect to quotients QUESTION [6 upvotes]: Is there an infinite cardinal $\kappa$ for which the following statement (S) true? (S) : There is a topology $\tau_\kappa$ on $\kappa$ such that for all topological spaces $(X,\tau)$ with $|X|\leq \kappa$ there is a binary relation $\sim$ on $\kappa$ such that $(X,\tau)\cong (\kappa,\tau_\kappa)/\sim$. (I'm transferring this question from https://math.stackexchange.com/questions/987131/universality-with-respect-to-quotients) REPLY [4 votes]: The answer is no, by the argument given by Henno in his comment. It true that there are exactly $2^{2^\kappa}$-many non-homeomorphic topologies on a set of size $\kappa \geq \aleph_0$. See for example this nice argument by Stefan Geschke. On the other hand it is obvious that there are only $2^\kappa$-many binary relations on a set of size $\kappa\geq \aleph_0$, so it follows that there are no infinite cardinals for which (S) is true.<|endoftext|> TITLE: Finite etale atlas for Deligne-Mumford stacks QUESTION [9 upvotes]: Let $X$ be a smooth finite type separated connected Deligne-Mumford stack over $\mathbb C$. Does there exist a finite etale morphism $Y\to X$ with $Y$ a scheme? What if $X$ is an algebraic space (i.e., trivial stabilizers)? Edit: I changed the old question to a different question which should be more clear. An answer to the new question would help a lot in answering the old question. REPLY [4 votes]: To give a more simple example than Daniel's, you can just consider for X a projective line with a single orbifold point. By Riemann-Hurwitz X is simply connected and so there is no non-trivial finite étale morphism Y→X. This holds over an algebraically closed field of characteristic zero say (but would work in characteristic p as well by defining precisely X as a stack of roots in the sense of Vistoli - see Charles Cadman, Using stacks to impose tangency conditions on curves, for the precise definition). Also, you may want to consider the following closely related notion, taken from Fundamental Groups of Algebraic Stacks Behrang Noohi http://arxiv.org/abs/math/0201021 "An algebraic stack being uniformizable means that it has a finite étale representable cover by an algebraic space (roughly speaking, its “universal cover” is an algebraic space)." The author proceeds to show that, roughly, a DM stack X is uniformizable iff all morphisms from the stabilizers to the fundamental group of X are injective.<|endoftext|> TITLE: The sixth power integral moment of automorphic L-function attached to Maass Forms QUESTION [5 upvotes]: It is known that the sixth power integral moment of automorphic L-function attached to Cusp Forms has been proved by M. Jutila, that is $\int_{0}^{T}|L(1/2+it,f)|^{6}dt \ll T^{2+\varepsilon}$. And it should also be true for Maass Forms philosophically, are there some references about this? REPLY [3 votes]: I have seen several papers in this direction, but don't know exactly what is known. Perhaps the following reference is helpful: M. P. Young obtains in his article The second moment of $GL(3)\times GL(2)$ L-functions at special points in $2009$ such estimates for a family of $L$-functions $L(φ×u_j,s)$ where $u_j$ runs over the family of Hecke-Maass cusp forms on $SL(2,\mathbb{Z})$, $\phi$ is a fixed $SL(3,\mathbb{Z})$ Maass form (Rankin-Selberg convolution of $\phi$ with $u_j$) ; at the special points $\frac{1}{2}+it_j$. The results are stated in Theorems $3.1$ and $3.2$. In particular, $$ \sum_{t_j\le T}|L(u_j, \frac{1}{2}+it_j|^6 \ll T^{2+\varepsilon}. $$<|endoftext|> TITLE: Second-order term of the Fedosov quantised product QUESTION [5 upvotes]: In Fedosov's version of quantisation of functions on a symplectic manifold, the product is given in terms of a symplectic connection. I have looked through Fedosov's book in deformation quantisation, and can't find the second order term for the product of two functions (which will involve the curvature), though I can find other expressions to second order. Am I being very unobservant (very likely), or if there really is no expression given there, is there another place where I can find it? Application: The commutation relations of quantum mechanics arise in constructing quantum theory in classical geometry. Now suppose that we construct quantum mechanics in a geometry that is already noncommutative? In particular, if we have a first-order in some parameter noncommutative algebra of functions, can writing quantum mechanics give information on a second order deformation? REPLY [2 votes]: Indeed neither Fedosov's book nor his original paper (A Simple Geometrical Construction of Deformation Quantization, J. Diff. Geom. 40 (1993) 213-238) have an explicit formula for the second order term of his star product. To my knowledge, the first place where the recursive formulas for the terms in Fedosov's star product to all orders are written down explicitly is the paper by M. Bordemann and S. Waldmann, A Fedosov Star Product of the Wick Type for Kähler Manifolds, Lett. Math. Phys. 41 (1997) 243-253, arXiv:q-alg/9605012, particularly Theorems 3.1, 3.2 and 3.4 therein. Since these results are all due to Fedosov, only the proof of Theorem 3.4 is briefly sketched. However, if you want more details on how to get these formulae and are more or less comfortable reading German, I recommend the excellent book by S. Waldmann, Poisson-Geometrie und Deformationsquantisierung (Springer-Verlag, 2007), specially Section 6.4, pp. 444-473 and Exercise 6.12, pp. 484. As for the applications you have in mind, I believe the answer is known if your noncommutative geometry comes from a strict deformation quantization à la Rieffel (e.g. the Moyal plane). In that case, one can use the action of translations to deform both the geometry and the (already noncommutative) C*-algebra of observables.<|endoftext|> TITLE: Intersections in almost complex manifolds QUESTION [8 upvotes]: Question: Suppose $(M,J)$ is an almost complex manifold, and $X$ and $Y$ are two almost complex submanifolds (i.e. $J(T(X)) \subset T(X)$ and $J(T(Y)) \subset T(Y)$). Then must $X \cap Y$ also be an almost complex submanifold? Clarification: If $X$ and $Y$ intersect transversally, then the answer is yes. If the intersection is not transverse, however, it might be the case that $X \cap Y$ is not even a submanifold. In this case, if we revise our notion of tangent space, e.g. if $T$ is the tangent cone, then do we still have $J(T(X \cap Y)) \subset T(X \cap Y)$? If the intersection is a submanifold, it might fail to be an almost complex manifold for other reasons, e.g. it might be of odd dimension. Is there an example of this (with odd-dimensional intersection, or otherwise)? REPLY [11 votes]: Let me answer the easy part now and then come back later for the (slightly) more delicate singular case: If $(M,J)$ is a real-analytic almost-complex manifold and $X,Y\subset M$ are almost-complex submanifolds of $M$ whose intersection $X\cap Y$ is a submanifold, then $X\cap Y$ is an almost-complex manifold. To see this, recall the following easy consequence of elliptic theory: Suppose that $(M,J)$ is a real-analytic almost-complex manifold and that $u:\mathbb{R}\to M$ is a real-analytic curve in $M$. Then there exists an open neighborhood $W\subset\mathbb{C}$ of $\mathbb{R}$ in $\mathbb{C}$ and a map $h:W\to M$ that is pseudoholomorphic (i.e., $h'(iv) = Jh'(v)$ for all tangent vectors $v\in TW$) such that $h(x) = u(x)$ for all $x\in\mathbb{R}$. Moreover, $h$ is locally unique in the sense that any two such extensions $h_i:W_i\to M$ agree on some open neighborhood of $\mathbb{R}\subset W_1\cap W_2$. This immediately implies the claim: Since $X$ and $Y$ are almost-complex manifolds and $(M,J)$ is real-analytic, they are real-analytic submanifolds. If $X\cap Y$ is a submanifold (necessarily real-analytic) and $v$ is a tangent vector to $X\cap Y$, then there will be a real-analytic $u:\mathbb{R}\to X\cap Y$ such that $u'(0) = v$. Let $h:W\to M$ be a pseudo-holomorphic extension as above. Since $u:\mathbb{R}\to X$ and $u:\mathbb{R}\to Y$ are real-analytic, and $J$ restricts to each to be an almost-complex structure, it follows by the local uniqueness that, by shrinking $W$ if necessary, it can be assumed that $h(W)\subset X\cap Y$. But now $h'(0)(\partial_x) = u'(0) = v$, so $h'(0)(\partial_y) = h'(0)(i\partial_x) = Jv$. Thus, $Jv$ is also tangent to $X\cap Y$. Thus, $J\bigl(T(X\cap Y)\bigr) = T(X\cap Y)$, so $X\cap Y$ is almost-complex. Added remark about the singular case: I still don't have time to write out a careful argument for this (and my real-analytic singularity theory is a little rusty), but I thought I'd go ahead and put in the sketch of the argument for what happens when $X\cap Y$ is not assumed to be a submanifold (but I am still assuming that $(M,J)$ is real-analytic). The point is that $X\cap Y$ is still a real-analytic variety and, as such, has a Whitney stratification into a disjoint union of real analytic submanifolds (not necessarily closed). Recall that the analytic tangent variety $\mathsf{A\!T}(X\cap Y)\subset TM$ is defined to be the closure of the set of vectors $v\in TM$ that are the velocities of real-analytic curves $u:\mathbb{R}\to M$ whose images lie in $X\cap Y$. The argument given above in the case that $X\cap Y$ is a submanifold clearly extends to the more general case, showing that $\mathsf{A\!T}(X\cap Y)$ is invariant under $J$. This fact, coupled with the properties of the Whitney stratification (i.e., that the WS satisfies Whitney's Conditions A and B), should show that each of the Whitney strata of $X\cap Y$ is an almost-complex submanifold of $M$, thus implying that $X\cap Y$ is a disjoint union of almost-complex submanifolds of $M$. I'm pretty sure that this is will go through, but it does appear that the details are somewhat tedious.<|endoftext|> TITLE: Diffeomorphisms and homotopy equivalences sliced over BO(n) QUESTION [12 upvotes]: There are some classical results stating sufficient conditions on a manifold $\Sigma$ such that every homotopy equivalence $\Pi(\Sigma) \stackrel{\simeq}{\longrightarrow} \Pi(\Sigma)$ is homotopic to a diffeomorphism $\Pi(\Sigma \stackrel{\simeq}{\longrightarrow} \Sigma)$ (here "$\Pi$" denotes the functor from (simplicial) smooth manifolds to homotopy types, i.e. to infinity-groupoids). What is known about this when the homotopy equivalences are required to respect the tangent bundle structure, up to homotopy? More specifically, I am wondering about the following: Since $\Sigma$ is a manifold (of dimension $n$, say), so $\Pi(\Sigma)$ is canonically equipped with a map $\Pi(\tau_\Sigma) : \Pi(\Sigma) \to B GL(n) \simeq B O(n)$. One may hence consider the automorphism group $\mathrm{Aut}_{/B O(n)}(\Pi(\Sigma))$ whose elements are homotopy auto-equivalences of $\Pi(\Sigma)$ that are required to fit into a diagram $$ \array{ \Pi(\Sigma) && \stackrel{\simeq}{\longrightarrow} && \Pi(\Sigma) \\ & \searrow &\Leftarrow_{\simeq}& \swarrow \\ && B O(n) } \,. $$ This is naturally a topological group, whose homotopy type reflects homotopies between such maps over $B O(n)$. How close is this to the homotopy type of the diffeomorphism group? I.e. how close is $$ B \mathrm{Diff}(\Sigma) $$ to $$ B \mathrm{Aut}_{/B O(n)}(\Pi(\Sigma)) $$ ? That's what I am wondering. For readers tolerant of more stacky language I would like to add the following thought, which may or may not have some bearing on this: Write $\mathbf{B}GL(n)$ for the smooth stack which is the homotopy quotient of the point by the action of the smooth group $GL(n)$. Then the smooth tangent bundle of $\Sigma$ is equivalently encoded in a morphism of smooth stacks $$ \tau_\Sigma : \Sigma \longrightarrow \mathbf{B}GL(n) \,. $$ Since $\Pi(\mathbf{B}GL(n)) \simeq B GL(n) \simeq B O(n)$ this is such that $\Pi(\tau_\Sigma)$ is the map on bare homotopy types considered above. Now stack homomorphisms in the higher slice topos over $\mathbf{B}GL(n)$ are equivalently local diffeomorphisms $$ \mathrm{LocalDiffeos}(\Sigma) \simeq \mathrm{SmoothStacks}_{/\mathbf{B}GL(n)}(\Sigma, \Sigma) \,. $$ Hence when considering the diffeomorphism group of $\Sigma$ we may equivalently think of the slice automorphism group of $\tau_\Sigma$ over $\mathbf{B}GL(n)$. From this perspective the question is to which extent one may "take $\Pi$ into the internalized slice automorphism group" to pass from $$ \Pi(\mathbf{Aut}_{/\mathbf{B}GL(n)}(\Sigma)) = \Pi (\mathrm{Diff}(\Sigma)) $$ to $$ \mathrm{Aut}_{/\Pi(\mathbf{B}GL(n))}(\Pi(\Sigma)) = \mathrm{Aut}_{/B O(n)}(\Pi(\Sigma)) \,. $$ REPLY [11 votes]: I wanted to say I think this is a great question, though phrasing things in terms of stacks might scare off some of the people who can best answer this question. I think in general understanding the relationship between these two spaces is a very interesting and very hard problem. Here is an example which shows that they can definitely be different. 3-dimensional Lens spaces $L(p,q)$ have trivial tangent bundles, so the group you write $Aut_{/BO(3)}(\Pi(\Sigma))$ will just be the homotopy automorphism group of the homotopy type $\Pi(\Sigma)$. So if they are homotopy equivalent, then they are homotopy equivalent over $BO(3)$. Now we look at the classification of lens spaces (see wikipedia for example) and we see that $L(7,1)$ and $L(7,2)$ are lens spaces which are homotopy equivalent but which are not diffeomorphic. So for these manifolds we have: $$BAut_{/BO(3)}(\Pi(L(7,1)) \simeq BAut_{/BO(3)}(\Pi(L(7,2))$$ Now what about their diffeomorphism groups? Well we can look at arXiv:math/0411016 and we see that as spaces: $$Diff(L(7,1)) \simeq S^1 \times S^3 \sqcup S^1 \times S^3 $$ $$Diff(L(7,2)) \simeq S^1 \times S^1 \sqcup S^1 \times S^1 $$ which implies that $BDiff(L(7,1)) \not\simeq BDiff(L(7,2))$. So this shows that for at least one of these spaces the classifying space of the diffeomorphism group differs from the classifying space of the tangential homotopy automorphism group.<|endoftext|> TITLE: How do I check that this is a Frobenius algebra? QUESTION [12 upvotes]: Let $f_1,f_2,\ldots,f_n\in \mathbb C[z_1,\ldots, z_n]$ be such that the quotient ring $$A:=\mathbb C[z_1,\ldots, z_n]/(f_1,f_2,\ldots,f_n)$$ is finite dimensional (in other words, it's a zero-dimensional complete intersection). I've heard that such a ring is always a Frobenius algbera when equipped with the counit $$ \varepsilon(g):= \int_{|f_1|=\ldots=|f_n|=1} \frac{g(z_1,\ldots,z_n)}{\prod_i f_i(z_1,\ldots,z_n)}dz_1\ldots dz_n, $$ and I'd like to check that claim. By definition, $A$ is a Frobenius algebra if the bilinear form $\langle g,h\rangle:=\varepsilon(gh)$ is non-degenerate. So I would like to know whether: If $g\in \mathbb C[z_1,\ldots, z_n]$ is such that $$\int_{|f_1|=\ldots=|f_n|=1} \frac{g(z_1,\ldots,z_n)h(z_1,\ldots,z_n)}{\prod_i f_i(z_1,\ldots,z_n)}dz_1\ldots dz_n=0$$ for every $h\in \mathbb C[z_1,\ldots, z_n]$, does is then follow that $g$ is in the ideal generated by $f_1,f_2,\ldots,f_n$? REPLY [14 votes]: Yes, this is discussed in page 659 of "Principles of Algebraic Geometry" by Griffiths and Harris. They call it the local duality theorem.<|endoftext|> TITLE: How to visualise Bollobas' 1965 theorem? QUESTION [5 upvotes]: Theorem $[n]=\{1,\ldots,n\}$. Let $\lbrace (R_i, S_i), i \in I \rbrace, R_i, S_i \subset [n]$ be such that $R_i \cap S_i = \emptyset, R_i \cap S_j \ne \emptyset (i \ne j)$. Then $$\sum_{i \in I} \frac{1}{{{r_i+s_i}\choose{s_i}}}\le 1.$$ Question: I am happy with the proof below of Bollobas' Theorem, but it seems very bashy. Is there a: More elegant way to prove the theorem, and, more importantly Is there a way to visualise the proof? By this I mean construct a picture on the $n$-cube or similar object that makes the proof somewhat obvious. An example of this is Sperner's Theorem for antichains, visualised by drawing chains on the $n$-cube. I've already seen one probabilistic proof, which is elegant but I cannot visualise it. Proof For $n=1$ it is trivial as $I= \lbrace 1 \rbrace$. We remove an element $x\in [n]$ from the construction to achieve a construction with $n-1$, which we can induct on. For each $x \in \lbrace 1,...,n \rbrace$ let $I_x= \lbrace i\in I: x \notin R_i \rbrace$, and $S_i^x = S_i \setminus \lbrace x \rbrace$. $\lbrace (R_i, S_i^x, i \in I_x \rbrace$ we cannot have $R_i \cap S^x_j = \emptyset$, as if $x$ was the only element in common between $R_i$ and $S_j$, that $R_i$ was thrown out as it contained $x$. Anyway, $$\sum_{i \in I_x} \frac{1}{{{r_i+s^x_i}\choose{s^x_i}}}\le 1.$$ Let us vary $x$, and fix $i, R_i, S_i$. Given each $i \in I$, $i \in I_x$ (i.e. $x \notin R_i$) for $n-r_i$ values of $x$. Now, if $i \in I_x$ ($x \notin R_i$), then for $s_i$ values of $x$ we have $s^x_i=s_i-1$ (i.e. $x \in S_i$), and for $n-r_i-s_i$ values of $x$ we have $s_i^x=s_i$. Hence $$n \ge \sum_{x \in [n]} \sum_{i \in I_x}\frac{1}{{{r_i+s^x_i}\choose{s^x_i}}}= \sum_{i \in I} \frac{n-r_i-s_i}{{{r_i+s_i}\choose{s_i}}}+\frac{s_i}{{{r_i+s_i-1}\choose{s_i-1}}}$$ $$n \ge \sum_{i \in I} \frac{(n-r_i-s_i)(s_i)! (r_i)!}{(s_i+r_i)!}+\frac{s_i (s_i-1)!(r_i)!}{(r_i+s_i-1)!}= n\sum_{i \in I} \frac{1}{{{r_i+s_i}\choose{s_i}}}.$$ Proof 2 (elegant, but still no visualisation) Randomly order the elements of $[n]$, then $\frac{1}{{{ri+s_i}\choose{s_i}}}$ is the probability that all elements of $R_i$ are greater than those of $S_i$ (written $R_i>S_i$), as only one of the unordered partitions of $[n]$ into $r_i, s_i$ elements satisfies the condition. For all $i$ these events are mutually exclusive. Thus $$P \left( \bigvee_{i \in I} (R_i>S_i) \right)\le 1.$$ REPLY [2 votes]: One perspective on the Bollobas theorem is that it is pretty straightforwardly equivalent to the following theorem about cloud antichains. A cloud antichain in $[n]$ is a collection of intervals $(I_i)_1^m = ([C_i,D_i])_1^m$ such that there do not exist comparable sets coming from different intervals. The theorem states that any cloud antichain in $[n]$ satisfies $$ \sum_1^m \binom{n-d_i+c_i}{c_i}^{-1} \leq 1, $$ where $c_i=|C_i|$, and similar. The proof is that no maximal chain can go through more than one interval, and it is straightforward to see that (assuming I have the statement right) the terms of the sum are the respective probabilities of a uniformly randomly chosen maximal chain going through the $i$-th interval.<|endoftext|> TITLE: Lattice model for Affine Grassmannians of non type A QUESTION [6 upvotes]: There is a Lattice model for affine Grassmannians of type A, due to Lusztig. It describes affine Grassmannians of type A as the moduli space of certain subspaces in an infinite-dimensional $\mathbb{C}-$vector space. Is there any similar pictorial models for affine Grassmannian of other types? Any references? REPLY [6 votes]: For any $ G$, you can always embed the affine Grassmannian $Gr_G$ into the space of $\mathbb C [[t]]$-lattices in $ \mathfrak{g} \otimes \mathbb C ((t))$. I'm pretty sure that this appears in Lusztig's paper. Alternatively, for any $ G $, there is an embedding $ Gr_G \rightarrow \prod_{\lambda} \{ \text{lattices in } V(\lambda) \otimes \mathbb C ((t)) \}$. The image of this embedding will be those sequences of lattices which are compatible with morphisms $ V(\lambda) \otimes V(\mu) \rightarrow V(\nu) $. See section 10.3 in Finkelberg-Mirkovic for more details. Unfortunately, neither of these embeddings are as helpful as the one for $ GL_n$.<|endoftext|> TITLE: Reinhardt cardinals and iterability QUESTION [9 upvotes]: Work in $ZF$. Let $j:V\to V$ be a non-trivial elementary embedding which is iterable, so that we can iterate it and form models $M_\alpha, \alpha\in ON,$ with $M_0=V,$ and elementary embeddings $j_{\alpha, \beta},$ for $\alpha\leq \beta.$ Let $M=\bigcap_\alpha M_\alpha.$ What can we say about $M$? It seems it is not difficult to show that $M$ is also a model of $ZF$, since it suffices to show that it is closed under Godel operations. Which large cardinals of $V$ are preserved in $M$? Of course the answer seems to be trivial for some large cardinals, in particular for those below the critical point of $j$, where their existence require information in $V_{crit(j)}$ So I am particularly interested in those large cardinals whose definitions require a proper class of information, like supercompact cardinals, ... What interesting properties $M$ can have? In particular what can we say about the relation between $V$ and $M$? -- Remark. As it is stated in the comments by Hamkins, any such embedding is iterable, so we can remove the extra assumption of iterability of $j.$ REPLY [5 votes]: Note that $M=\bigcap_{\alpha}M_\alpha$ is just the $\mathrm{OR}^{\mathrm{th}}$ iterate $M_{\mathrm{OR}}$ cut off at height $\mathrm{OR}$, so we have for example $V_\kappa\preceq V_\lambda\preceq M$ where $\kappa=\mathrm{crit}(j)$ and $\lambda=\kappa_\omega(j)$, where $\kappa_0(j)=\mathrm{crit}(j)$ and $\kappa_{n+1}(j)=j(\kappa_n(j))$ and $\kappa_\omega(j)=\sup_{n<\omega}\kappa_n(j)$. So in terms of large cardinals with a first-order description, $M$ has the same kinds as does $V_\kappa$. If there is an elementary $k:V\to V$ with $\kappa_\omega(k)<\kappa$ then arguing like in A weak reflection of Reinhardt by super Reinhardt cardinals, we get an embedding $\ell:M\to M$ witnessing that $\kappa_n(k)$ is Reinhardt for $M$, for some $n<\omega$. (That is, for some $n<\omega$, we get $k_n(\kappa)=\kappa$, and then we can just stretch out the structure $(V_\kappa,{k_n\upharpoonright V_\kappa})$ to produce such an $\ell$.) There are some facts established in Reinhardt cardinals and iterates of V on the relationship between the iterates $M_\alpha$ and $V$. In particular, it is shown that for each $\alpha\geq\omega$, every set is set-generic over $M_\alpha$, but $M_\alpha$ is not a set-ground of $V$. And $\mathrm{HOD}\subseteq M_\omega$ but $\mathrm{HOD}\not\subseteq M$.<|endoftext|> TITLE: Topological characterization of injective metric spaces QUESTION [8 upvotes]: Let $\ (X\ d)\ \,(Y\ \delta)\ $ be arbitrary metric spaces. A function $\ f:X\rightarrow Y\ $ is called a metric map (with respect to the given metrics $\ d\ \delta$) $\ \Leftarrow:\Rightarrow\ \forall_{p\ q\in X}\ \delta(f(p)\ f(q))\ \le\ d(p\ q)$. Injective metric spaces were introduced in a paper by Aronszajn and Panitchpakdi, under the hyper-convex spaces name, via the binary intersection property of closed balls. Equivalently, a metric space $\ (Z\ \rho)\ $ is called injective $\ \Leftarrow:\Rightarrow\ $ for every metric space $\ (X\ d)\ $ and arbitrary $\ Y\subseteq X,\ $ and for arbitrary metric map $\ f:Y\rightarrow Z\ $ (with respect to $\ \delta := d|Y\times Y$ and $\ \rho$) there exists a metric map $\ g:X\rightarrow Z\ $ (with respect to metrics $\ d\ \rho$) such that $\ g|Y=f$. PROBLEM   Characterize topologically the toplogical spaces which are homeomorphic to the injective metric spaces. Preferably, this should be done for the class of all metric spaces. The class of separable spaces or of metric compact spaces would be great too. In the case of $1$-dimensional compact spaces $\ X\ $it is pretty obvious that they are injective $\ \Leftarrow:\Rightarrow\ X$ is an AR (i.e. absolute retract as defined by Borsuk). Sorry, if I missed some known results (I do use Google etc, but I am terrible at searching). Please, let me know. EDIT Under a pressure from some (just one?) careful MO participants I have edited the statement of my MO-Question. I must say that in my own opinion my old statement:         Characterize topologically injective metric spaces. is much better. I would rather say characterize topologically closed interval, circle, Euclidean plane and sphere   than   characterize topologically topological spaces homeomorphic to closed interval, circle, Euclidean plane and sphere. Or one can also say simply the same using the symbols: $\ I\ S^1\ \mathbb R^2\ S^2$. The reason to me is both mathematical, as well as the simplicity of language. REPLY [3 votes]: This might be useful: J. R. Isbell, Six theorems about injective metric spaces. Commentarii Mathematici Helvetici 39 (1964), 65-76. From the intro: Aronszajn and Panitchpakdi showed [1] that topologically, every injective metric space is a complete retract, and asked whether the converse is true and then Isbell goes on to state some partial results on the converse, e.g.: In infinite 2-dimensional polyhedra, collapsibility is sufficient and free contractibility necessary I've put the first page of the paper here. This contains a list of the results.<|endoftext|> TITLE: Doing some homological algebra in triangulated categories QUESTION [5 upvotes]: It's well known that chain complexes are an abelian category, and in particular we can consider chain complexes of chain complexes, i.e. double complexes. Given a double complex $A^{\bullet\bullet} \in \mathrm{Kom}(\mathrm{Kom}(\mathcal A))$ we can form the total complex $\newcommand{\tot}{\mathrm{Tot}}\tot(A^{\bullet\bullet})$ which now lies "one level lower", in $\mathrm{Kom}(\mathcal A)$. I can also try to consider chain complexes in the derived category of $\mathcal A$, but it is no longer clear (at least to me) how to build a total complex in the best way, since $d \circ d=0$ only has to hold up to homotopy in $D(\mathcal A)$. But let me now consider more generally a triangulated category $\newcommand{\T}{\mathcal T}\T$ and a sequence of objects $A^0, \ldots, A^n$ with $d_i \colon A^i \to A^{i+1}$ such that $d_{i+1} \circ d_i = 0$. It seems to me that one can define a total complex $\tot(A^\bullet) \in \T$ as an iterated mapping cone: for instance, if $n=2$, then one can first consider $B = \mathrm{Cone}(d_0)$. Then we consider the diagram $$ \begin{matrix} A^0 & \to & A^1 & \to & B \\ \downarrow & & \downarrow & & \\ 0 & \to & A^2 & \to & A^2 & \end{matrix}$$ where both rows are distinguished triangles; by one of the axioms of triangulated categories there is a map $f \colon B \to A^2$ completing this to a map of triangles, and we define $\tot(A^\bullet)=\mathrm{Cone}(f)$. However, this has a the drawback of not being functorial. For instance, I think that one would like to say that a map $A^\bullet \to B^\bullet$ of chain complexes in $\T$ is a quasi-isomorphism if $\tot(A^\bullet) \to \tot(B^\bullet)$ is an isomorphism, but this makes no sense unless $\tot$ is a functor. And one would like to say that if $f \colon \T \to \T'$ is a triangulated functor, then there is a natural equivalence between $f \circ \tot$ and $\tot \circ f$ as functors $\mathrm{Kom}(\T) \to \T'$ (am I right?), and again $\tot$ needs to be functorial. First of all, I would like to know if what I've said so far is correct. Maybe there is a better way to set things up than this? Secondly, I've heard the slogan that stable $\infty$-categories solve all problems arising from the fact that triangulated categories don't have functorial mapping cones. Is there a better behaved notion of a chain complex in a stable $\infty$-category? I suspect that I could answer these questions myself if I started reading Lurie's work, but it's a slightly intimidating amount of text and I thought I'd ask here first. REPLY [11 votes]: There are different constructions of total complexes, using direct sums, direct products, etc. Each of them needs that the bicomplex satisfies some boundedness conditions. Your third paragraph indicates that you might be interested in cohomological positive bicomplexes $A^\bullet$ $$0\rightarrow A^0\stackrel{d_0}\longrightarrow A^1\rightarrow\cdots$$ Here each $A^i$ is an unbounded complex in the abelian category $\mathcal A$. Let us define $\operatorname{Tot}(A^\bullet)$ as $\operatorname{Tot}(A^\bullet)^{n}=\prod_{p+q=n}A^{pq}$ with the usual differential. This is a homotopy limit construction, i.e. you can consider the abelian category $\operatorname{Kom}^+(\mathcal A)$ of cohomological positive complexes in $\mathcal A$ and its derived category $D(\operatorname{Kom}^+(\mathcal A))$. The obvious degree $0$ inclusion $\mathcal A\rightarrow\operatorname{Kom}^+(\mathcal A)$ induces a functor $D(\mathcal A)\rightarrow D(\operatorname{Kom}^+(\mathcal A))$, and $\operatorname{Tot}$ can be characterized as its right adjoint. This answer is maybe not very satisfactory for you, but I think it's the best you can get. There is an obvious forgetful functor $D(\operatorname{Kom}^+(\mathcal A))\rightarrow\operatorname{Kom}^+(D(\mathcal A))$, but it is not well behaved. The moral of the previous paragraph is that you cannot fully stay in the triangulated world if you want to consider totalization, but now I'll show you that you can stay close. The abelian category $\operatorname{Kom}^+(\mathcal A)$ is equivalent to $\mathcal A^{\Delta}$, i.e. cosimplicial objects in $\mathcal A$. In particular $D(\operatorname{Kom}^+(\mathcal A))\simeq D(\mathcal A^\Delta)$. This category is part of the triangulated derivator $\mathbb{D}(\mathcal A)$ associated to $\mathcal A$. This gadget is a 2- (or pseudo-) functor $$\mathbb{D}(\mathcal A)\colon \operatorname{Dia}^{\operatorname{op}}\longrightarrow\operatorname{CAT}\colon I\mapsto D(\mathcal A^I).$$ Here $\operatorname{CAT}$ is the big category of all categories and $\operatorname{Dia}$ is an appropriate full sub-2-category containing $\Delta$. Evaluating $\mathbb{D}(\mathcal A)$ at the inclusion $\Delta[0]\rightarrow\Delta$ we obtain the functor $D(\operatorname{Kom}^+(\mathcal A))\rightarrow D(\mathcal A)\colon A^\bullet\mapsto A^0$, its right adjoint is the degree $0$ inclusion $D(\mathcal A)\rightarrow D(\operatorname{Kom}^+(\mathcal A))$, and the right adjoint of this last one is totalization. The existence of such adjoints is part of the properties of a derivator. Derivators, discovered by Grothendieck, are a user fiendly model for higher homotopy theory interpolating between $\infty$-categories and their homotopy categories. Their flavor is closer to the homotopy (triangulated) categories, the techniques are 2-categorical, but they encode higher information such as mapping spaces.<|endoftext|> TITLE: Reference for a fact (?) on homeomorphic knot complements QUESTION [6 upvotes]: Does somebody have a reference (or an argument why it should be true) for the following statement? “Let $K$ and $K'$ be knots in $S^3$. If there is an orientation-preserving homeomorphism $h : S^3 \to S^3$ which takes $K$ to $K'$, then there is also an ambient isotopy $\eta : (S^3,K) \to (S^3,K')$.” It seems to me that this statement is always assumed when citing the Gordon–Luecke theorem from [Gordon, Luecke: Knots are determined by their complements. J. Amer. Math. Soc. 2 (1989), 371-415]. Namely, the authors define knot equivalence by the existence of an $h$ as in the statement (actually they work with homeomorphisms $h$ that are not necessarily orientation-preserving, but they say that everything still works in the orientation-preserving case), while everybody else seems to use the ambient-isotopy definition, even when citing the Gordon–Luecke theorem. REPLY [24 votes]: This is a question that I remember worrying about when I first started learning about knot theory. Older books have a tendency to skim over this point rather lightly, perhaps because the resolution of the question seems to involve techniques that have little to do with standard knot theory. One book that doesn't avoid treating this issue squarely is Burde and Zieschang's "Knots" (now in a third edition with a third author, Heusener, and available in an electronic version). In Chapter 1 of this book the fact that every orientation-preserving homeomorphism of $S^3$ (or equivalently ${\mathbb R}^3$) is isotopic to the identity is attributed to a 1960 paper of G.M.Fisher. (In Kirby's landmark 1969 paper on the Stable Homeomorphism Conjecture he refers to this result just as being "classical".) The earlier work of Moise on existence and uniqueness of PL structures on 3-manifolds does not seem to say anything about homeomorphisms between PL 3-manifolds being isotopic to PL homeomorphisms. Moise just proves that such homeomorphisms can be approximated by PL homeomorphisms. However, one of his approximation theorems appears to be sufficient to deduce the desired fact about knots. Namely, suppose $f:S^3\to S^3$ is an orientation-preserving homeomorphism taking a knot $K$ to a knot $K'$. Choose an open set $U$ in $S^3-K$ and choose a continuous function $\varepsilon: U \to (0,1)$ that approaches $0$ at the frontier of $U$. Then Moise's theorem states that $f$ can be $\varepsilon$-approximated by a homeomorphism $g$ which is PL on any compact set in $U$ (which just means PL on $U$ itself) and which equals $f$ outside $U$. In particular $g$ takes $K$ to $K'$. Since $g$ is PL on some open set, it can easily be isotoped (topologically) to the identity, and this isotopy restricts to a topological isotopy of $K'$ to $K$. If $K$ and $K'$ are PL knots, one might want a PL isotopy, but this argument doesn't give it. The Burde-Zieschang book does prove that two PL knots that are equivalent via a topological homeomorphism that preserves orientation are PL isotopic. This is Corollary 3.17, and they deduce it from Waldhausen's theorem that, in the PL category, two knots are isotopic if they have isomorphic "peripheral systems". The latter are defined in terms of the fundamental group of the knot complement, so they are invariant under topological homeomorphisms preserving orientation. I don't know who first proved the general result that every homeomorphism between PL 3-manifolds is isotopic to a PL homeomorphism, but there is a very nice proof using just PL topology of 3-manifolds and the Kirby torus trick in a 1976 paper of A.J.S.Hamilton. In the smooth category Cerf's theorem ($\Gamma_4=0$) certainly implies that if two smooth knots are equivalent under an orientation-preserving diffeomorphism then they are smoothly isotopic. However, one can obtain this without using Cerf's theorem by an elementary argument that works in all dimensions, even when the analog of Cerf's theorem fails (due to the existence of exotic spheres). Suppose $f:S^n\to S^n$ is an orientation-preserving diffeomorphism taking a closed proper subset $K\subset S^n$ to another such set $K'$. Since $f$ preserves orientation, it is a standard fact that $f$ can be isotoped to be the identity on a ball $B\subset S^n-K$, so we assume this has been done. Another way of stating this standard fact is to say that the map from the diffeomorphism group $Diff(B,\partial B)$ of $B$ fixing (a neighborhood of) $\partial B$ to the orientation-preserving diffeomorphism group $Diff^+(S^n)$ of $S^n$, obtained by extending diffeomorphisms via the identity outside $B$, induces a surjection on $\pi_0$. This means that we can compose $f$ with a diffeomorphism $S^n\to S^n$ supported in $B$ representing $[f]^{-1} \in \pi_0Diff^+(S^n)$ to get a new diffeomorphism that is isotopic to the identity and still takes $K$ to $K'$. (This can be done quite explicitly in fact.) [The second half of this last paragraph has been revised to correct a misstatement.]<|endoftext|> TITLE: Is every number the sum of two cubes modulo p where p is a prime not equal to 7? QUESTION [21 upvotes]: If p is a prime other than 7, can every integer be written as sum of two cubes modulo p? Has Waring's problem mod p for cubes been proved simply and directly? Thanks for your proof. Lemi REPLY [2 votes]: The argument in this MO answer: Simple proofs for the existence of elliptic curves having a given number of points shows that the number of solutions to $y^2=f(x)$, $f$ a cubic, is at most $3p/2$. Taking a quadratic twist, it follows that the number of solutions is at least $p/2$. The equation $x^3+y^3=a$ can be put in Weierstrass form with a slightly messy but elementary calculation and the result follows.<|endoftext|> TITLE: $j$-invariants of elliptic curves over finite fields QUESTION [12 upvotes]: Let $K$ be a finite field, and $\overline{K}$ its algebraic closure. It is well known that two curves are isomorphic over $\overline{K}$ if and only if they have the same $j$-invariant. If two such curves are also $K$-isogenous, I believe we can conclude that they are $K$-isomorphic, but I cannot find any reference or elementary proof of this fact; is it easy to see, or does anyone have a reference? (it seems that this result is implicitly used in the algorithms for isogeny graphs of elliptic curves, like Kohel's, where the isogenous curves are encoded by there $j$-invariant). As a side question, given a $j \in K$, how many curves over $K$ have this $j$-invariant, up to $K$-isomorphisms? For example, a (non-supersingular) curve and its quadratic twist both have the same $j$-invariant, but are not $K$-isogenous, so not $K$-isomorphic, so we have at least two $K$-isomorphism classes; is this it? REPLY [2 votes]: It is in general $\textit{not}$ true that $\textit{if}$ $A$ and $B$ are elliptic curves over the finite field $K$, lying in the same $K$-isogeny class, and with the same $j$-invariant, $\textit{then}$ $A$ and $B$ are $K$-isomorphic (see below). However, this is true if $A$ and $B$ are ordinary. You can find a related observation in Waterhouse's thesis (Abelian varieties over finite fields, Éc. Nor. Sup. 1969, remark page 542). Let me describe a different approach, similar to that suggested by Silverman. I tried to be succinct in my answer, I failed as there are several details to verify. Sorry about this. Let me slightly reformulate the question. Let $K$ be a finite field of size $q$ and characteristic $p\geq 5$. Let $E$ an elliptic curve over $K$, denote by $\pi_E:E\to E$ the Frobenius isogeny of $E$ relative to $K$. By Honda-Tate theory, $\pi_E$ defines a Weil $q$-number whose minimal polynomial over $\mathbf{Q}$ dentifies uniquely the $K$-isogeny class of $E$. Let $E'$ be a $K$-form of $E$, with associated Weil number $\pi_{E'}$. Your first question can then can be formulated as: Can we have that $\pi_{E'}$ is conjugate to $\pi_E$ without $E'$ being $K$-isomorphic to $E$? (Here by "conjugate" I mean that the $\pi_E$ and $\pi_{E'}$ have the same minimal polynomial over $\mathbf{Q}$.) To answer the question let's start by describing the $K$-forms of $E$ in a slightly different (but equivalent!) way than that used by Silverman above. Set $W_K=\textrm{Aut}_K(E)$ and $W_{\bar K}=\textrm{Aut}_{\bar K}(E\times_K\bar K)$. Since $p\geq 5$, $W_{\bar K}$ is isomorphic, as a $G_K$-module, to $\mu_n$, where $n=2,4$ or $6$. Moreover, we can think of $W_{\bar K}$ as the roots of unity inside some imaginary quadratic field $F$ embedded in $\textrm{End}_{\bar K}(E\times_K\bar K)$, where $F$ is unique if $E$ is ordinary or if $n>2$. The set of $K$-forms of $E$ is in bijection with the (group) $H:=H^1(G_K,W_{\bar K})$. Consider the two cases i) $W_K=W_{\bar K}$; ii) $W_K\subsetneq W_{\bar K}$. In the first case the evaluation map $c\mapsto c(\varphi)$ of $1$-cocycles on the arithmetic Frobenius $\varphi\in G_K$ induces an isomorphism $H\simeq W_{\bar K}$. In the second case the action of $G_K$ on $W_{\bar K}$ is by inversion, and the evaluation map $c\mapsto c(\varphi)$ as above induces an isomorphism $H\simeq W_{\bar K}/(W_{\bar K})^2$. You can check that this description of $H$ matches the Kummer-theoretic one given above by Silverman (at least as abstract groups). Now, if $c$ is a $1$-cocyle (for $G_K$ acting on $W_{\bar K}$), then Silverman in his book explains how to construct an elliptic curve $E^c$ over $K$ $\textit{and}$ an isomorphism $E^c\times_K\bar K\simeq E\times_K \bar K$. One can then show (I owe the omitted proof to Jakob Stix) that in the ring $\textrm{End}_{\bar K}(E\times_K\bar K)$ the following relation holds $\pi_{E^c}=c(\varphi)\pi_E$, where we make an implicit use of the identification $\textrm{End}_{\bar K}(E\times_K\bar K)\simeq \textrm{End}_{\bar K}(E^c\times_K\bar K)$ induced by the isomorphism $E^c\times_K\bar K\simeq E\times_K \bar K$ encoded by $c$. In other words, the Weil numbers $\pi_{E^c}$ and $\pi_E$ differ by multiplication by a root of unity. If the cohomology class defined by $c$ is non-trivial, the above formula implies that the non-trivial twist $E^c$ and $E$ are $K$-isogenous if and only if $\bar\pi_E=-\pi_E$, where $\bar\pi_E=q/\pi_E$ is the complex conjugate of $\pi_E$. Notice that from what we saw above $c$ is non-trivial in $H$ iff [$c(\varphi)\neq 1$ in case i), and $c(\varphi)\not\in W_{\bar K}^2$ in case ii)]. This last condition on $\pi_E$ then amounts to require it being purely imaginary, hence conjugate to $\sqrt {-q}$. If $q=p^e$, then $\sqrt{-q}$ arises as $\pi_E$ for some elliptic curve $E$ iff [either $e$ is odd, or else $e$ is even and $p\not\equiv 1$ mod $4$]. (In the remaining cases where $e$ is even you get $K$-simple non-geometrically simple surfaces attached to $\sqrt{-q}$). The end of the story is that if $e$ is odd or $e$ is even and $p\not\equiv 1$ mod $4$, then every elliptic curve $E$ in the $K$-isogeny class defined by $\sqrt{-q}$ admits $|H|$-many $K$-isogenous, pairwise non-isomorphic $K$-forms. All these are examples of supersingular elliptic curves ($p$ divides the trace of $\sqrt{-q}$) with not all geometric endomorphisms defined over $K$. They exhaust the instances of the phenomenon we were after.<|endoftext|> TITLE: Number of orbits of $\mathrm{SL}_2(\mathcal O_A)$ on $\mathbf P^1(A)$ when $A$ is a quaternion algebra QUESTION [8 upvotes]: This is a reference request. Let $A$ be an anisotropic quaternion algebra over $\mathbf Q$. Let $\mathcal O_A$ be a maximal order in $A$. Then $\mathrm{SL}_2(A)$ acts transitively on the right on the set $\mathbf P^1(A)$ of one dimensional left subspaces of $A^2$. Let $h_A$ be the cardinal of the set (1) of left ideal classes of $\mathcal O_A$. It is not hard to show that when $h_A=1$, one also has $\sharp[~\mathbf P^1(A)~/~\mathrm{SL}_2(\mathcal O_A)~]=1$. More generally, computer assisted experiments for $A$ of small discriminant seem to indicate that the equality $$\sharp[~\mathbf P^1(A)~/~\mathrm{SL}_2(\mathcal O_A)~]=h_A^2$$ should hold, but after some unfruitful draft padding, I have been unable to prove it. Would someone here know where to find a proof of this undoubtedly classical fact (2), and even better, name its discoverer ? . . Notes : (0) The inclusion $\mathrm{SL}_2(\mathcal O_A)\to \mathrm{GL}_2(\mathcal O_A)$ is an equality. (1) In the simpler case when $A$ is a quadratic field, the ideal classes form a group and this point is essential in the proof of the classical equality $\sharp[~\mathbf P^1(A)~/~\mathrm{SL}_2(\mathcal O_A)~]=h_A$. (2) Some googling indicates that many authors studied automorphic cuspidal things in this quaternionic setting, so that this quotient has certainly been studied, but I could not find the statement. REPLY [6 votes]: The formula holds, and seems to be due to Krafft and Osenberg : Eisensteinreihen für einige arithmetisch definierte Untergruppen von SL2(H). (German), Math. Z. 204 (1990), no. 3, 425–449. (See here : EuDML (free), or here : Springer.) Just after Satz 2.2 : "Die Zahl der Spitzzenbahnen ist also das Quadrat der (einseitigen) Klassen-zahl von $\mathcal O$".<|endoftext|> TITLE: Polynomial recurrence relation covering the integers (and then Gaussian integers) QUESTION [6 upvotes]: Say that a polynomial recurrence relation (my terminology) for $f_i$ is: $k$ initial conditions setting $f_1,\ldots,f_k$ to integers ($\in \mathbb{Z})$. A recurrence equation of the form $f_i =$ a polynomial in $f_{i-1},\ldots,f_{i-k}$. Example 1: ($k=2$): $\;f_1=1,\; f_2=1$, and $f_i=f_{i-1}+f_{i-2}$. $\implies$ the Fibonacci sequence. Example 2: ($k=1$): $\;f_1=1$, and $f_i=f_{i-1}+1$. $\implies$ $\mathbb{N}$. Q1. Is there a polynomial recurrence relation that covers $\mathbb{Z}$? In other words, I would like every integer (positive or negative) to be "reached" by some $f_i$. This may be obvious, in which case I apologize. I arrived at this question from another direction: Q2. Is there a polynomial recurrence relation over the Gaussian integers that covers the Gaussian integers? REPLY [12 votes]: For the first question, let $f_1=0$, $f_2=1$, $f_3=-1$, and $f_i=-f_{i-1}+f_{i-2}+f_{i-3}$.<|endoftext|> TITLE: Who first noticed that Stirling numbers of the second kind count partitions? QUESTION [17 upvotes]: When the Stirling numbers of the second kind were introduced by James Stirling in 1730, it was not combinatorially; rather, the numbers ${n \brace k}$ were defined via the polynomial identity $$ x^n = \sum_{k=1}^n {n \brace k} x(x-1)\ldots (x-(k-1)). ~~(\star) $$ Most modern treatments of the Stirling numbers introduce ${n \brace k}$ combinatorially, as the number of ways of partitioning a set of size $n$ into $k$ non-empty blocks, and then (combinatorially) derive identities such as ($\star$). While preparing for some upcoming talks on topics related to Stirling numbers, I realized that I have no idea who it was who first observed that the numbers ${n \brace k}$ defined by ($\star$) have a combinatorial interpretation. The earliest reference I can find is in W. Stevens, Significance of Grouping, Annals of Eugenics volume 8 (1937), pages 57--69 (available at http://onlinelibrary.wiley.com/doi/10.1111/j.1469-1809.1937.tb02160.x/abstract). This seems to predate any mention of ``Stirling numbers of the second kind'' on MathSciNet. I imagine that this is a question that has been thoroughly researched --- does anyone know of a reference? REPLY [3 votes]: Donald E. Knuth in the second part of his article Two Notes on Notation details some of the research he did on Stirling numbers and their history. He also introduces interesting notational conventions linking binomial coefficients and Stirling numbers: $\binom{n+1}{k}$ = $\binom{n}{k}$ + $\binom{n}{k-1}$ $\genfrac{[}{]}{0pt}{}{n+1}{k}$ = $n\genfrac{[}{]}{0pt}{}{n}{k}$ + $\genfrac{[}{]}{0pt}{}{n}{k-1}$ $\genfrac{\{}{\}}{0pt}{}{n+1}{k}$ = $k\genfrac{\{}{\}}{0pt}{}{n}{k}$ + $\genfrac{\{}{\}}{0pt}{}{n}{k-1}$ where $\genfrac{[}{]}{0pt}{}{n}{k}$ = the number of permutations of n objects having k cycles $\genfrac{\{}{\}}{0pt}{}{n}{k}$ = the number of partitions of n objects into k nonempty subsets A copy of the article is available at: https://arxiv.org/pdf/math/9205211.pdf The answer is in the article at page 12 where Knuth mentions that : "Christian Kramp [28] proved near the end of the eighteenth century that ... " [28] Christian Kramp, “Coefficient des allgemeinen Gliedes jeder willkührlichen Potenz eines Infinitinomiums; Verhalten zwischen Coefficienten der Gleichungen und Summen der Produkte und der Potenzen ihrer Wurzeln; Transformationen und Substitution der Reihen durcheinander”, in Der polynomische Lehrsatz, edited by Carl Friedrich Hindenburg (Leipzig, 1796), 91–122.<|endoftext|> TITLE: Is there an $L$ like inner model for $\sf Z$? QUESTION [15 upvotes]: Godel proved the consistency of the axiom of choice with the axioms of $\sf ZF$ by showing that given any model of $\sf ZF$, there is a definable class which satisfies $\sf ZFC$. The proof uses a lot of the power of $\sf ZF$, in particular it uses transfinite recursion which is equivalent to the axiom schema of replacement. And also the fact that we can talk about the ordinals as objects in the universe by looking at the von Neumann ordinals. Let $\sf Z$ denote Zermelo set theory, which is obtained by removing replacement and foundation from $\sf ZF$ and adding the separation schema instead. This is a strictly weaker theory than $\sf ZF$. We know that if $\delta>\omega$ is a limit ordinal, then $V_\delta\models\sf Z$ (and in fact more, since it also satisfies foundation). One interesting difference is that while $\sf Z+AC$ is sufficient to prove that every set can be well-ordered, it cannot prove that every set is equipotent with a von Neumann ordinal. Since, for example in $V_{\omega+\omega}$ there are only countably many von Neumann ordinals, all of which are countable, but there are uncountable sets (and one can arrange that there are uncountably many different cardinals amongst them). Is there an inner model construction, internal to $\sf Z$, which acts like $L$ in proving the consistency of $\sf Z$ with additional axioms? If not, how about an external construction? Will the answer change if we add back the axiom of foundation? REPLY [19 votes]: This answer is based on Adrian Mathias' majestic paper The Strength of MacLane Set Theory (published in the Annals of Pure and Applied Logic, 2001). Let $\sf{M}$ be the system of set theory whose axioms consist of Extensionality, Null Set, Pairing, Union, Set Difference ($x\setminus y$ exists), Power set, $\Delta_0$-Seperation, Foundation, Transitive Containment (every set is a subset of a transitive set), and Infinity. It has long been known that Zermelo set theory plus Foundation does not prove Transitive Containment (as shown by Schröder and Jensen, and Boffa). According to Mathias (see the paragraph before Theorem 1 in the introduction), Gödel's original construction of L can be implemented within $\sf{M}$ to yield the following theorem that can be proved within Primitive Recursive Arithmetic. But the proof, according to Mathias, is quite cumbersome. To my knowledge the full proof of this theorem has not appeared in print. Theorem. If $\sf{M}$ is consistent, then so is $\sf{M}$ + KP (Kripke-Platek set theory) + $\Sigma_1$-Seperation + $V=L$. In his paper, Mathias provides the proof of a slightly weaker result, where the hypothesis of the consistency of $\sf{M}$ is strengthened to the consistency of $\sf{M}$ augmented with the axiom H (introduced in section 2 of his paper). Let me also remark that the usual construction of L that one sees in set theory texts, on the other hand, can be smoothly carried out in KP + Infinity. Addendum. See also Avshalom's comment below, which includes an outline (provided by Mathias) of how certain results in Mathias' paper can be used to demonstrate Con(Z) => Con(Z + AC),<|endoftext|> TITLE: Does Nelson try to prove PA inconsistent directly? QUESTION [7 upvotes]: Edward Nelson is known for his serious attempts to show that Peano axioms, and sometimes even weaker theories, are inconsistent. I wasn't able to find Nelson's papers anywhere, so I wanted to ask a question about structure of his proofs: Did Nelson attempts try to, for some statement $\phi$, prove both $\phi$ and not $\neg\phi$? Some of you might ask a question "What other possibility could there be?", and here is one such possibility: PA might just prove the statement "PA is inconsistent". What is the difference? The difference is that supposed "proof" of contradiction might have nonstandard length. You can think of this in terms of different theory, namely $PA+\neg Con(PA)$. This theory shows that PA is inconsistent (because one of its axioms says so), and, as extension of PA, it can show itself inconsistent. However, the theory itself is still consistent. What the reasoning above shows is that PA is not $\omega$-consistent. So, slightly restating the question, Did Nelson in his attempts really try to show PA inconsistent, or just $\omega$-inconsistent? Also, if anyone knows a source where I could find Nelson's papers, I'd be thankful; this is the reason I added "reference-request" tag. Thanks for all feedback! REPLY [4 votes]: See the discussion here: https://golem.ph.utexas.edu/category/2011/09/the_inconsistency_of_arithmeti.html. As Monroe says, he tried to prove that PA was inconsistent, not just $\omega$-inconsistent.<|endoftext|> TITLE: What would be the consequences of $\displaystyle{\lim\inf_{n\to\infty}p_{n+k}-p_{n}\sim k\log k}$? QUESTION [7 upvotes]: The question is in the title: what would be the number theoretic consequences if we managed to establish the conjectured asymptotic equality $\displaystyle{\lim\inf_{n\to\infty}p_{n+k}-p_{n}\sim k\log k}$? Thanks in advance. Edit December 16th 2014: I'd be interested in references about this topic as well. REPLY [23 votes]: The situation is similar to that in your previous question what would be the consequences on the distribution of primes of $\Lambda=\infty$? : by itself, not very much, because one only needs an arbitrarily sparse sequence of narrow prime clusters to establish the conjecture. More substantial is the Hardy-Littlewood prime tuples conjecture, which implies the stated conjecture about $p_{n+k}-p_n$ (contingent on a further conjecture as to the narrowest admissible $k$-tuple, namely that the diameter of such a tuple is asymptotic to $k \log k$); this conjecture not only asserts the mere existence of prime tuples, but also predicts an asymptotic, and is broadly applicable to many further problems than the estimation of $\lim \inf p_{n+k}-p_n$, particularly if generalised to other patterns than tuples. More generally, I don't recommend putting too much time into taking random conjectures in number theory and trying to figure out what they imply; there are a few important conjectures which have a lot of applications (e.g. RH, GUE Hypothesis, Elliott-Halberstam) but most conjectures in number theory are endpoint objectives, rather than stepping stones to other conjectures; they are useful for benchmarking the state of technical progress in the subject, but not directly applicable to other problems. This is particularly the case if the conjecture is referring directly to endpoint objects of study in number theory, such as the primes, as opposed to more technical but broadly useful objects of study (e.g. mean values of exponential sums or Dirichlet series). Ultimately, analytic number theory is primarily a "second culture" subject, in which the techniques and general principles tend to be more important than the specific results proven (excepting some fundamentally important results, such as the prime number theorem, of course). See this essay by Gowers on the distinction between the two cultures of mathematics. Much of the value that arises when a long-standing conjecture in number theory is proved arises not from the logical consequences of the conjecture itself, but rather from the new techniques and methods (or perhaps the expansion of applicability of an older technique or method) that had to be introduced to resolve the conjecture, as these methods would then be likely to be applicable to many further problems, including those that had no direct relation to the original conjecture that was proved.<|endoftext|> TITLE: Fano varieties of cubic threefolds QUESTION [5 upvotes]: Let $X$ be a smooth cubic threefold over $\mathbb{C}$. Let $F(X)$ denote the Fano variety of lines in $X$, which is a smooth surface of general type. Is this class of surfaces distingushed amoungst surfaces of general type? I appreciate that this question is slightly vague. What I am looking for is something like a geometrical characterisation of such surfaces, say in terms of certain geometric invariants (i.e. hodge numbers, chern classes,...?). If this is too naive to hope for, then perhaps they have distinguished moduli, for example by forming a connected component of the moduli of surfaces of general type. REPLY [5 votes]: So, not what you're looking for, but closely related. A nodal cubic threefold's Fano surface is isomorphic to $S^2B/\sim$ where $B\in\mathcal{M}_4$ is nonhyperelliptic, and $\sim$ is constructed as follows: Let $L,L'$ be the two $g^1_3$'s on $B$, then they give maps $B\to S^2B$ by $r\mapsto p+q$ if $p+q+r$ is in the $g^1_3$. Then identify these two copies of $B$ in the natural way. So, not numerical invariants, but $F(X)$ deforms to this singular surface, so you can read off quite a few pieces of data from it. A reference to this is Donagi's "Fibers of the Prym Map" section 5.4 but I'm sure it appeared somewhere else earlier, I just don't have that reference on hand.<|endoftext|> TITLE: Is such a map null-homotopic? QUESTION [17 upvotes]: Suppose I have (semi-infinite) chain complexes $$ \cdots \rightarrow A_i \rightarrow A_{i+1}\rightarrow \cdots$$ $$ \cdots \rightarrow B_i \rightarrow B_{i+1}\rightarrow \cdots$$ over an additive category, and $A_i = B_i = 0$ for $i>0$. Suppose that $f:A\rightarrow B$ is a chain map such that, for every $k\geq 0$, $f$ is chain homotopic to a chain map which is zero in degrees $-k,\ldots, -1,0$. Is such an $f$ always null-homotopic? REPLY [6 votes]: Probably there's a much less artificial example, and even more probably there's a much simpler one, but ... Let $A=k[\varepsilon]/(\varepsilon^2)$, where $k$ is a field of characteristic two (purely so I don't need to bother about signs). Let $\mathcal{A}$ be the category whose objects are pairs $(M,N)$ of $A$-modules and whose maps $\alpha:(M,N)\to(M',N')$ are matrices $\begin{pmatrix}\alpha_1&0\\\alpha_2&\alpha_3\end{pmatrix}$ where $\alpha_1:M\to M'$, $\alpha_2:M\to N'$, $\alpha_3:N\to N'$ are $A$-module homomorphisms. Let $\mathcal{A}_0$ be the (non-full) subcategory with the same objects, but with only those maps $\alpha$ where $\alpha_2=0$. For $k\geq0$, let $X(k)$ be the chain complex $$\dots\to 0\to (A,0)\to (A,0)\to\dots\to(A,0)\to(A,0)\to0\to\dots$$ with $(A,0)$ in degrees $k+1,\dots,0$, where all the non-trivial differentials are $\begin{pmatrix}\varepsilon&0\\ 0&0\end{pmatrix}$, and let $Y(k)$ be the complex $$\dots\to 0\to (0,A)\to (0,A)\to\dots\to(0,A)\to0\to0\to\dots$$ with $(0,A)$ in degrees $k+1,\dots,1$, where all the non-trivial differentials are $\begin{pmatrix}0&0\\ 0&\varepsilon\end{pmatrix}$. Consider the chain map (over $\mathcal{A}$) $f(k):X(k)\to Y(k)$ given by $\begin{pmatrix}0&0\\ \varepsilon&0\end{pmatrix}$ in degree $k+1$ and zero in other degrees. This is null-homotopic (for example, take the homotopy given by $\begin{pmatrix}0&0\\1&0\end{pmatrix}$ in degrees $k,\dots,0$), but there's no choice of contracting homotopy with degree zero component in $\mathcal{A}_0$. Now let $\mathcal{A}^{\mathbb{N}}$ be the category of $\mathbb{N}$-graded objects of $\mathcal{A}$, and consider the chain map $$(f(k))_{k\in\mathbb{N}}:(X(k))_{k\in\mathbb{N}}\to (Y(k))_{k\in\mathbb{N}}.$$ As a map of chain complexes over $\mathcal{A}^{\mathbb{N}}$ this is null-homotopic. However, if you regard it a map of chain complexes over the (non-full) subcategory of $\mathcal{A}^{\mathbb{N}}$ with the same objects but where maps are required to have grade $d$ components in $\mathcal{A}_0$ for all but finitely many $d$, then it isn't null-homotopic but (using contracting homotopies for finitely many $f(k)$) is homotopic to a map that is zero in any given finite set of degrees.<|endoftext|> TITLE: How many isolated roots can a polynomial in $z$ and $\overline{z}$ have? QUESTION [11 upvotes]: This is probably well-known by I can't find it now online. My guess is that if the degree is $n$ then it's $2n$ but it's just a hunch. EDIT: This is an edited version. Before I asked about roots without qualification. REPLY [12 votes]: Alexandre Eremenko already described the failure of the $2n$ bound, but I thought I'd illustrate with an example. I did an unstructured manual search on cubic polynomials, and here's an example with 8 zeroes: $2z^3 + 4\bar{z}^3 - z^2 + z\bar{z} - \bar{z}^2 + z + 0.1 + 0.1i =0$. The graph shows vanishing loci of the real and imaginary parts, and the $Im(P(z,\bar{z}))=0$ locus is made of the components that are asymptotic to the $x$ and $y$ axes together with the lower right bubble.<|endoftext|> TITLE: Clustering of periodic points for a polynomial iteration of $\mathbb{C}$ QUESTION [6 upvotes]: Let $f : \mathbb{C} \to \mathbb{C}$ be a polynomial map of degree $q > 1$. Consider $E_n \subset \mathbb{C}$ the set of periodic points with period (dividing) $n$; generally, $|E_n| = q^n$. Since all the peridic points are limited to a bounded subset of $\mathbb{C}$, we know that there are always two among them within a distance of $O(q^{-n/2})$. Question. (i) How close can two distinct elements of $E_n$ get? In the asymptotic where $n \to \infty$ along a sequence, can it happen that $E_n$ contains a pair of points within a distance of $o(1/q^n)$? (ii) Given any point $z_0 \in \mathbb{C}$ and $C$, does the disk $|z-z_0| < C/q^n$ contain only $O_{C,z_0}(1)$ points from $E_n$? REPLY [6 votes]: The answer to both of your questions is negative, even if you replace $q^{-n}$ by any other sequence $(a_n)$ of natural numbers. The reason is, broadly speaking, that you may have of course have maps having multiple periodic points, and under small perturbations these will yield maps having a cycle of potentially large period nearby. A little bit more precisely, let us consider the family of quadratic polynomials, $f_c(z) = z^2+c$. The following is well-known: Proposition. Suppose that $f_c$ has a periodic point $z_0$ of period $n$, with corresponding multiplier $\mu = (f_c^n)'(z_0) = e^{2\pi i p/q}$, for some $p,q\in\mathbb{N}$. Then, for any neighbourhood $U$ of $z_0$, there is a parameter $c'$ (arbitrarily close to $c$) such that $f_{c'}$ has a periodic point $z'$ of period $nq$ such that $f_{c'}^{kn}(z')\in U$ for $k\geq 0$, and such that $z'$ itself has multiplier $\mu' = (f_{c'}^{nq})'(z') = e^{2\pi i /q'}$ for some (large) integer $q'$. (In addition, the map $f_{c'}$ also has a (repelling) periodic point of period $n$ in $U$.) (What happens is that the parameter $c$ is on the boundary of two components of the interior of the Mandelbrot set, one consisting of points having an attracting orbit of period $n$ and one having an attracting orbit of period $nq$. The perturbation is obtained by passing a little bit along the boundary of the latter component.) Now start e.g. with the map $f_{1/4}$ and $z=1/2$, and apply this proposition inductively. Take a limit of the resulting parameters (taking care to make each perturbation small enough as to not destroy the features already constructed). Clearly in this way we can find a quadratic polynomial having periodic points $(z_j)_{j\geq 0}$ of periods $(n_j)$, where $$ n_j = \prod_{k=1}^j q_j$$. (Here $q_j$ is some rapidly increasing sequence of positive integers obtained in the construction.) Given any sequence $(\varepsilon_n)$, we can inductively carry out the construction so that $q_j$ points of the orbit of $z_j$ are within distance $\varepsilon_{n_j}$ of each other. Letting $\varepsilon_n$ tend to zero faster than $a_n$, the claim follows.<|endoftext|> TITLE: Do singular values dominate eigenvalues? QUESTION [9 upvotes]: Suppose $A$ is an $n \times n$ complex matrix with singular values $s_1 \ge s_2 \ge \cdots \ge s_n$ and eigenvalues $(\lambda_i)_{i=1}^{n}$ arranged so that $|\lambda_1| \ge |\lambda_2| \ge \cdots \ge |\lambda_n|$. Is it true that $s_1 \ge |\lambda_1|,\ s_1+s_2 \ge |\lambda_1| + |\lambda_2|,$ and so on? REPLY [10 votes]: Let $\lambda(A)$ denote the vector of eigenvalues and $s(A)$ the vector of singular values (arranged in decreasing order). The claim of the question is whether $|\lambda(A)|^{\downarrow} \prec_w s(A)$. The answer to this question is yes (the proof follows using the tensor power trick Yanqi Qui mentioned above). Theorem (Weyl's Majorant theorem). Let $f: (0,\infty)\to(0,\infty)$ be such that $f(e^t)$ is convex and monotone increasing in $t$. Then $$[f(\vert\lambda_1\vert),\ldots, f(|\lambda_n|)] \prec_w [f(s_1),\ldots,f(s_n)].$$ As a corollary, using the function $t \mapsto t^p$ for any $p \ge 0$, we obtain an affirmative answer to the OP's question (actually, by merely using the special case $p=1$). Actually, Weyl's Majorant theorem provides the following log-majorization (which implies the above result): \begin{equation*} \log |\lambda(A)| \prec \log s(A), \end{equation*} and this version also answers Ilya's comment, namely $\prod_i^n |\lambda_i| = \prod_i^n s_i$. Another closely related result is that of absolute values of the Hermitian part of a complex matrix. Let $\Re(A) := (A+A^*)/2$. Then, we have Theorem (Fan-Hoffman). For every matrix $A$, we have \begin{eqnarray*} \lambda_j^{\downarrow}(\Re(A)) &\le& s_j(A),\qquad 1\le j \le n\\ |\lambda(\Re(A))| &\prec_w& s(A). \end{eqnarray*} Reference R. Bhatia, Matrix Analysis (Springer GTM, 169).<|endoftext|> TITLE: Conditions for convergence of Euler's method QUESTION [6 upvotes]: It is known that a sufficient and necessary condition for $$\dot y(t) = f(y(t), t), \quad t > 0, \quad y(0) = y_0$$ to have a unique solution is $f$ Lipschitz in $y$ and continuous in $t$. However, I didn't find in the literature that this condition could guarantee the convergence of Euler's scheme (forward or backward) to the solution. Instead, additional condition imposed on $y$ that $y$ is $C^2$ seems needed, see for example http://persson.berkeley.edu/228A/Fall10/doc/lec05-2x3.pdf. I was wondering whether it was possible to weaken the condition for convergence of Euler's scheme. REPLY [3 votes]: Forward Euler is convergent under mild conditions on $f(t,x)$, as explained below. Let $\delta t$ be the time step size parameter (assumed to be constant for clarity's sake), let $T$ be the time span of simulation and set $t_k = k \delta t$ for any $k \in \mathbb{N}_0$. By integration by parts: $$ y(t_{k+1}) = y(t_k) + f(t_k,y(t_k)) \delta t + \int_{t_k}^{t_{k+1}} (t_{k+1} - s) \frac{d}{ds} f(s, y(s)) ds $$ Let $\epsilon_k$ denote the global error of the forward Euler scheme. Since $f$ is Lipschitz we have that: $$ \epsilon_{k+1} \le (1+ L_f \delta t) \epsilon_k + \underbrace{\| \int_{t_k}^{t_{k+1}} (t_{k+1} - s) \frac{d}{ds} f(s, y(s)) ds \|}_{\text{local error of forward Euler}} $$ where $L_f$ is the Lipschitz constant of $f(t,x)$. The usual way to bound the local error appearing in this last inequality is to assume a uniform bound on the derivatives of $f(t,x)$ that enables you to pull these derivatives out of the time integral. Let us take a slightly differently approach that requires less stringent assumptions on $f(t,x)$. By Cauchy-Schwarz inequality: $$ \epsilon_{k+1} \le (1+ L_f \delta t) \epsilon_k + (\int_{t_k}^{t_{k+1}} (t_{k+1} - s)^2 ds)^{1/2} (\int_{t_k}^{t_{k+1}} \|\frac{d}{ds} f(s, y(s))\|^2 ds )^{1/2} $$ This recursion inequality simplifies to: $$ \epsilon_{k+1} \le (1+ L_f \delta t) \epsilon_k + \delta t^{3/2} M $$ where we have introduced a constant $M>0$ which we assume satisfies $$ (\int_{0}^{T} \|\frac{d}{ds} f(s, y(s))\|^2 ds )^{1/2} \le M \tag{$\star$} $$ By induction (or discrete Gronwall's Lemma), it follows that: $$ \epsilon_{k} \le \frac{e^{L_f T} M}{L_f} \; \delta t^{1/2} $$ for all $k \in \mathbb{N}$ satisfying $t_k < T$. Note that ($\star$) may hold even if $f(t,x)$ is just Lipschitz-continuous in $x$. Indeed, Rademacher's Theorem implies that a Lipschitz function is differentiable almost everywhere. The price for this more mild assumption is that the theoretical rate of convergence drops from $\mathcal{O}(\delta t)$ to $\mathcal{O}(\delta t^{1/2})$. However, numerical evidence seems to indicate this estimate is a bit pessimistic. Consider the initial value problem $$ \dot y=|y|/2-(y-1) \;, \quad y(0)=-2.3 \;, $$ where the right hand side is Lipschitz, but not differentiable at $0$. Here is a figure of the true solution over the time interval $[0,1]$. (I selected this solution so that $y(1)$ is close to the point where the right hand side of the differential equation $|x|/2-(x-1)$ is not differentiable.) Here is a figure of the relative error of forward Euler with respect to a converged target. (This metric of convergence is commonly used in the absence of an analytical solution.) For a MATLAB function file that reproduces this last figure click here.<|endoftext|> TITLE: Does there exist harmonic function with that property? QUESTION [6 upvotes]: Can one construct a harmonic function $f$ defined in the unit disk with the condition $f(0)≥1$ such that area of $\{z∈D:f(z)>0\}$ is small enough, i. e. for every $\epsilon>0$ does there exist a function harmonic in the unit disk with the condition $f(0) \ge 1$ s.t. the Lebesgue planar measure of $\{z∈D:f(z)>0\}$ is less than $\epsilon$ ? That was asked here, but not answered. REPLY [10 votes]: Yes you can. First of all the condition $|f(0)|\geq 1$ is irrelevant: you can always multiply your function on a positive constant. Take an entire function $F$ for which the set $\{ z:|F(z)|>1\}$ is contained in the strip $\{ x+iy:|y|<\pi\}$. Such a function is constructed in Hayman's book Meromorphic functions, section 4.1.1, and it is called $E_0$ there. (Other people call it Mittag-Leffler function). Make sure that $F(0)=2$ by an appropriate shift. Then consider $F(kz)$ where $k$ is very large. Then the set in the unit disk where $|F(kz)|>1$ will be as small as you wish, but will contain $0$. Finally take real part of $F(kz)$.<|endoftext|> TITLE: Measurability for disintegration of a kernel QUESTION [6 upvotes]: Let $(x, A) \mapsto P(x, A)$ be a probability kernel whose "target" (wikipedia terminology) is a product space $Y \times Z$, and say both $Y$ and $Z$ are compact metric spaces. For every $x$ there is the disintegration $(\mu_{x,y})_{y \in Y}$ characterized by $$P(x,A_1 \times A_2)=\int_{A_1}\mu_{x,y}(A_2) P(x,\pi^{-1}(dy))$$ where $\pi \colon Y \times Z \to Y$ is the canonical map. Well, but I need that $\mu_{x,y}$ measurably depends on $(x,y)$, or in other words I need a kernel $(x,y,A_2) \to \mu_{x,y}(A_2)$. How could I justify this measurability ? REPLY [2 votes]: In my own work I needed to disintegrate a probability kernel and I found a good reference. The following is a simplified version of Theorem 1.25 in Kallenberg's 2017 book, Random Measures, Theory and Applications. Theorem: Consider a $\sigma$-finite kernel $\rho: S \to T \times U$, where $T$ and $U$ are Borel spaces. There exist $\sigma$-finite kernels $\nu: S \to T$ and $\mu: S \times T \to U$ such that $\rho = \nu \otimes \mu$. The assertion remains true for any fixed $\sigma$-finite kernel $\nu: S \to T$ such that $\nu_s \sim \rho_s(\cdot \times U)$ for all $s \in S$. Here a Borel space is any measurable space isomorphic to a Borel subset of the real line, for example, a Polish space.<|endoftext|> TITLE: A forked plane continuum QUESTION [10 upvotes]: I came up with this question while trying to solve the following MO one: Does every connected set that is not a line segment cross some dyadic square? Suppose $C$ is a plane continuum (i.e. a compact, connected set in the plane), contained in the square $S=[-1,1]^2$, but missing the closed vertical line segment $I = \{0\}\times[0,1]$. Suppose $C$ contains the points $P(0,-1)$, $Q(-1,1)$, and $R(1,1)$. Two examples of such continua are shown in the pictures, in green, and the excluded line segment $I$ is shown in red. Note that in general we do not require that $C$ be path-connected, e.g. it might contain copies of the $\sin\frac1x\,$ curve, or of the pseudo-arc (the only path-connected components of which are points), or something else: The only requirements are that $C$ is a continuum in $S$, misses $I$, and contains $P(0,-1)$, $Q(-1,1)$, and $R(1,1)$.         Question. Does $C$ necessarily contain a subcontinuum $K$ that contains $P(0,-1)$, intersects both the 1st and the 2nd Quadrants, and such that $K$ "reaches further up" in either the 1st Quadrant or in the 2nd Quadrant: That is $y_1\not=y_2$ where $y_1 = \max\{y: (x,y)\in K, x>0\}$ and $y_2 = \max\{y: (x,y)\in K, x<0\}$? Comment. It is not difficult to show that for every $z\in[0,1]$ there is a subcontinuum $K_z$ of $C$ with $P(0,-1)\in K_z$ and such that $z = \max\{y: (x,y)\in K_z\}$. To do this, for each $n>0$ take a path from $P(0,-1)$ in the $\frac1n$-neigborhood of $C$ and reaching up to the line $y=z$ (but no further up), then a subsequence of these paths would converge to some $K_z$ that works. We may ask (the contrapositive?): Does there exist a continuum $C$ (contained in $S$, missing $I$, and containing $P(0,-1)$, $Q(-1,1)$, and $R(1,1)$) such that no matter how we pick $z\in[0,1]$ and a subcontinuum $K_z$ containing $P(0,-1)$ and reaching up the line $y=z$ (but no further up), we would always have that $K_z$ intersects the line $y=z\,$ both in the 1st and in the 2nd Quadrant? Comment. I thought I would work with the specific points $P(0,-1)$, $Q(-1,1)$, and $R(1,1)$, but the condition for $C$ may be weakened to only that $C$ intersects the bottom edge of the square $S$, misses $I$, and intersects the top edge of $S$ in both the 1st and the 2nd Quadrants, and then ask if a subcontinuum $K$ necessarily exist that intersects the bottom edge of $S$, and reaches further up either into the 1st Quadrant or into the 2nd Quadrant. Thank you :) REPLY [6 votes]: I will answer my own question. The answer turned out the opposite to what I expected or hoped for, to my surprise, and I guess to Dominic van der Zypen's surprise. So, Lasse Rempe-Gillen's intuition worked better, he did guess the correct answer (see comments above, following the question), although he suggested a recursive construction without verifying the details, whereas what I came up with today is a pretty specific construction based on the Cantor set (and some $\sin\frac1x$ type curves). One could make a picture, which I do below since this is perhaps the easiest way to explain it. Let $R$ be the "right half" of the usual middle third Cantor set, that is the intersection of the latter with $[\frac23,1]$. For each $r\in R$ the continuum $C$ that I will construct will contain the lower semicircle centered at the origin, of radius $r$. If $C_r$ denotes this (closed) semicircle, then the first picture below shows $C_1$, $C_{\frac23}$, $C_{\frac79}$, $C_{\frac89}$. (Note $\frac23=\frac69<\frac79<\frac89<\frac99=1$.) Each $C_r$ will be extended with a couple of vertical line segments at its endpoints, going up in the 1st and in the 2nd Quadrant. $C_{\frac23}$ will be extended with vertical line segments of length $1$, thus reaching the horizontal line $y=1$ (and intersecting it at points $(\pm\frac23,1)$). $C_{\frac79}$ and $C_{\frac89}$ will each be extended with vertical line segments of length $\frac12$. (I will comment on a formula that gives the exact length of the vertical line segments, depending on $r$, but I feel I should do that later.) $C_1$ will be the only one that is not extended (or formally extended with vertical line segments of length $0$). What was described so far is shown on the first picture below. Corresponding to each middle-third $(a,b)$ that was removed in the construction of (the "right-half" $R$) of the Cantor set, we will add a "double $\sin\frac1x$ type" arc that "connects" the semicircles $C_a$ and $C_b$, including the attached vertical line segments. For example, for the middle-third $(\frac79,\frac89)$, this arc will alternate from points at height $\frac12$ in the 1st Quadrant to points at height $\frac12$ in the 2nd Quadrant, remaining all the time "between" $C_\frac79$ and $C_\frac89$ (with the attached vertical segments). One "end" of this arc will "approach" $C_\frac79$ (including its attached vertical segments), and the other "end" will "approach" $C_\frac89$ (with its attached vertical segments). This is illustrated in the next picture. The main idea here is that if $K$ is any subcontinuum of $C$ (where $C$ is the continuum we are about to construct from all the $C_r$, attached vertical line segments, and connecting "double $\sin\frac1x$ type" arcs), and if $K$ contains the bottom point $P(0,-1)$ and at least one point on $C_\frac79$, then $K$ must also contain the connecting arc between $C_\frac79$ and $C_\frac89$. Since the "ends" of that arc "approach" $C_\frac79$ and $C_\frac89$, it follows that $K$ must contain both $C_\frac79$ and $C_\frac89$ (including respective vertical line segments). Such a subcontinuum $K$ must intersect the horizontal line $y=\frac12$ in both the 1st and the 2nd Quadrants (at points $(\pm\frac79,\frac12)$, $(\pm\frac89,\frac12)$, plus a bunch of points on the arc). As a minor variation of this argument, $K$ might be a subcontinuum (of $C$) that does not contain any point of $C_\frac79$ (or of the respective vertical line segments), but does contain at least one point of the connecting arc between $C_\frac79$ and $C_\frac89$ (and also contains the bottom point $P(0,-1)$). In this case $K$ must contain just "one end" of the connecting arc: Namely, that end which approaches $C_\frac89$. It follows that in this case $K$ must contain $C_\frac89$ with its vertical line segments. $K$ will be contained in the closed half-plane below the horizontal line $y=\frac12$, and will meet this line at $(\pm\frac89,\frac12)$, plus a bunch of points on the arc (in both the 1st and in the 2nd Quadrant). Continuing with the construction, we now add $C_\frac{25}{27}$ and $C_\frac{26}{27}$ along with vertical line segments of length $\frac14$ (reaching up to horizontal line $y=\frac14$). Note that $\frac89=\frac{24}{27}<\frac{25}{27}<\frac{26}{27}<\frac{27}{27}=1$, and $C_\frac{25}{27}$ and $C_\frac{26}{27}$ correspond to the endpoints of the removed middle-third $(\frac{25}{27},\frac{26}{27})$. We also add a "double $\sin\frac1x$ type arc" connecting $C_\frac{25}{27}$ and $C_\frac{26}{27}$ (and attached vertical line segments), alternating between points at height $\frac14$ in the 1st and in the 2nd Quadrants. (Picture enclosed further down.) Also, add $C_\frac{19}{27}$ and $C_\frac{20}{27}$ along with vertical line segments of length $\frac34$ (reaching up to horizontal line $y=\frac34$). Note that $\frac23=\frac{18}{27}<\frac{19}{27}<\frac{20}{27}<\frac{21}{27}=\frac78$, and $C_\frac{19}{27}$ and $C_\frac{20}{27}$ correspond to the endpoints of the removed middle-third $(\frac{19}{27},\frac{20}{27})$. We also add a "double $\sin\frac1x$ type arc" connecting $C_\frac{19}{27}$ and $C_\frac{20}{27}$ (and attached vertical line segments), alternating between points at height $\frac34$ in the 1st and in the 2nd Quadrants. $C_\frac{25}{27}$ and $C_\frac{26}{27}$ (with vertical line segments and connecting arc), and $C_\frac{19}{27}$ and $C_\frac{20}{27}$ (with vertical line segments and connecting arc) are illustrated at the next picture. The continuum $C$ will consist of all lower semicircles $C_r$ for $r\in R$, together with respective vertical line segments, and with connecting arcs, as indicated above. (The precise length of the vertical line segments will be described, as was promised, further down. Hopefully the above description is clear enough, for now.) If $K$ is any subcontinuum of $C$ that contains the bottom point $P(0,-1)$ and at least one point of the connecting arc between $C_\frac{25}{27}$ and $C_\frac{26}{27}$, then $K$ must contain at least "one end" of this connecting arc: At least the end which approaches $C_\frac{26}{27}$. It follows that in this case $K$ must contain $C_\frac{26}{27}$ with its vertical line segments (reaching height $\frac14$). Such a $K$ will meet the horizontal line $y=\frac14$ at least at $(\pm\frac{26}{27},\frac14)$ (plus a bunch of points on the arc) in both the 1st and in the 2nd Quadrant. Similarly, if $K$ is any subcontinuum of $C$ that contains the bottom point $P(0,-1)$ and at least one point of the connecting arc between $C_\frac{19}{27}$ and $C_\frac{20}{27}$, then $K$ must contain at least "one end" of this connecting arc: At least the end which approaches $C_\frac{20}{27}$. It follows that in this case $K$ must contain $C_\frac{20}{27}$ with its vertical line segments (reaching height $\frac34$). Such a $K$ will meet the horizontal line $y=\frac34$ at least at $(\pm\frac{20}{27},\frac34)$ (plus a bunch of points on the arc) in both the 1st and in the 2nd Quadrant. Re the exact length of the vertical line segments, the main idea here is a well-known continuous map from the product space $2^{\mathbb N}$ (where $2=\{0,1\}$ with the discrete topology, and $\mathbb N=\{1,2,3,\dots\}$) onto the unit interval $[0,1]$. To make things precise, if $s=\langle s_1,s_2,s_3,\dots\rangle\in2^{\mathbb N}$ then one defines $\displaystyle\varphi(s)=\sum_{n\in\mathbb N}\frac{2s_n}{3^n}$. Then $\varphi$ is a homeomorphism between $2^{\mathbb N}$ and the usual middle-third Cantor set. For example $\varphi\langle 0,1,1,\dots\rangle=0+\frac29+\frac2{27}+\cdots=\frac13$ and $\varphi\langle 1,0,0,\dots\rangle=\frac23+0+0+\cdots=\frac23$. One also defines a continuous $\psi$ from $2^{\mathbb N}$ onto $[0,1]$ by $\displaystyle\psi(s)=\sum_{n\in\mathbb N}\frac{s_n}{2^n}$. Note that $\psi\langle 0,1,1,\dots\rangle=0+\frac14+\frac18+\cdots=\frac12=\frac12+0+0+\cdots=\varphi\langle 1,0,0,\dots\rangle$. That is, $(\psi\circ\varphi^{-1})(\frac13)=\frac12=(\psi\circ\varphi^{-1})(\frac23)$. The map $\psi\circ\varphi^{-1}$ is a continuous map from the usual middle-third Cantor set onto the unit interval $[0,1]$ such that, for each removed middle-third, both its end-points are sent to the same point in $[0,1]$. (As illustrated above, both endpoints of $(\frac13,\frac23)$ are sent to $\frac12$.) The length of the vertical line segments attached at each $C_r$ is, essentially, given by the map $\psi\circ\varphi^{-1}$, with some adjustment. Instead of sending $\frac13$ and $\frac23$ to $\frac12$, we would like to send $\frac79$ and $\frac89$ to $\frac12$. Also, larger $r\in R$ are sent to smaller height, like $\frac{25}{27}$ and $\frac{26}{27}$ are sent to $\frac14$, while $\frac{19}{27}$ to $\frac{20}{27}$ are sent to $\frac34$. Formally, what I think works, is to send $r\in R$ to $(\psi\circ\varphi^{-1})(3(1-r))$. For example when $r=\frac79$ we have $(\psi\circ\varphi^{-1})(3(1-\frac79))=(\psi\circ\varphi^{-1})(3\cdot\frac29)=(\psi\circ\varphi^{-1})(\frac23)=\frac12$. When $r=\frac89$ we have $(\psi\circ\varphi^{-1})(3(1-\frac89))=(\psi\circ\varphi^{-1})(3\cdot\frac19)=(\psi\circ\varphi^{-1})(\frac13)=\frac12$. When $r=\frac23$ we have $(\psi\circ\varphi^{-1})(3(1-\frac23))=(\psi\circ\varphi^{-1})(3\cdot\frac13)=(\psi\circ\varphi^{-1})(1)=\frac12+\frac14+\frac18+\cdots=1$. When $r=1$ we have $(\psi\circ\varphi^{-1})(3(1-1))=(\psi\circ\varphi^{-1})(0)=0$. All these values agree with the heights of the vertical line segments from our initial less formal description. (As a side remark, in my question I required that the continuum $C$ must contain the points $Q(-1,1)$ and $R(1,1)$. The continuum $C$ that I described in this answer does not contain these points, but to rectify this, we only need to add the horizontal line segments from the point $(\frac23,1)$ to the point $(1,1)$, and from the point $(-1,1)$ to the point $(-\frac23,1)$. I prefer instead to work with the more relaxed version as stated towards the end of the question, requiring $C$ to intersect the horizontal line $y=1$ in both the 1st and the 2nd Quadrants, but not necessarily at $Q(-1,1)$ and $R(1,1)$.) Note that for every $z\in(0,1)$ the continuum $C$ constructed in the present answer will contain a subcontinuum $K_z$ such that $P(0,-1)\in K_z$, and $K_z$ intersects the horizontal line $y=z$ and is contained in the closed half-plane below this line. Specifically, $K_z$ will consist of all $C_r$ with $r\in R$ (the right-half of the middle-third Cantor set) such that $(\psi\circ\varphi^{-1})(3(1-r))\le z$, along with respective vertical line segments, and necessary connecting arcs. (The set of these $r$ will be of the form $[r_0,1]\cap R$ for a suitable $r_0$, namely $r_0$ would be the solution of the equation $(\psi\circ\varphi^{-1})(3(1-r))=z$ for $r$ in terms of $z$. For most $z\in(0,1)$ this solution will be unique. For countably many values of $z$, there will be two solutions, so in this case $K_z$ will not be uniquely determined. More precisely, if $z=\frac p{2^k}\in(0,1)$ for some integers $p,k$ (we may assume that $p$ is odd) then there will be a middle-third $(r_0,r_1)$ such that $(\psi\circ\varphi^{-1})(3(1-r_1))=(\psi\circ\varphi^{-1})(3(1-r_0))=z$. One could come up with a precise formula, but perhaps this is unnecessary.) (Credit goes to Ivan Johansen for his Graph https://www.padowan.dk/ , which I used to make the pictures.) P.S. I am tempted to call the continuum $C$ described above the Neptune Trident. This might be a little imprecise since the middle dent is missing (but the more precise "bident" or "polydent" just don't sound right). At hindsight the construction wasn't overly difficult and perhaps $C$ does not need a special name, but it doesn't hurt to propose one.<|endoftext|> TITLE: Visibility interpretation of Riemann zeta zeros on the critical line? QUESTION [7 upvotes]: This is a long shot, but ... The fraction of $\mathbb{Z}^2$ lattice points visible from the origin $1/\zeta(2)=6/\pi^2 \approx 61$%. The fraction of $\mathbb{Z}^3$ lattice points visible from the origin is $1/\zeta(3) \approx 83$%. And this generalizes to arbitrary dimensions $1/\zeta(d)$. Q. Is there some geometric/visibility interpretation of a zero of $\zeta(s)$ on the critical line, $\mbox{Re}(s)=1/2$? I suspect not, but there are various notions of fractional dimension and complex dimension that might allow $\zeta(s)$ to have a type of geometric interpretation in dimension $s$...? REPLY [3 votes]: The zeros of the Riemann zeta function are telling you something about the error term in the counting problem. Introduce the following zeta function $$Z(s) = \sum_{ (a,b)=1 } 1/\max\{a,b\}^s.$$ Then an application of Mobius inverse shows that, modulo some constants, this is equal to $$\zeta(s-1)/\zeta(s).$$ Perons formula can now be used to give an asymptotic formula for the corresponding counting problem. There is a pole of order $1$ at $s=2$ with residue $1/\zeta(2)$, which gives rise to the main term. There are also poles occuring at the zeros of the Riemann zeta function, which give rise to lower order terms in the corresponding asymptotic formula for the counting problem. This is very similar to the fact that the zeros of the Riemann zeta function control the error term in the prime number theorem.<|endoftext|> TITLE: Can infinitely many alternating knots have the same Alexander polynomial? QUESTION [16 upvotes]: There exist many constructions of infinite families of knots with the same Alexander polynomial. However, alternating knots seem very special. While there are also many result on restricting the form the Alexander polynomial of an alternating knot can have, is it known whether infinitely many alternating knots can have the same Alexander polynomial? REPLY [24 votes]: No, there cannot exist infinitely many alternating knots with the same Alexander polynomial. To see why, suppose for the contrary that $K$ belongs to an infinite family $\{K_n\}_{n\in\mathbb{Z}}$ of alternating knots with $\Delta_{K_n}(t)=\Delta_K(t)$. Immediately we have $$ \det(K_n ) = |\Delta_{K_n} (−1)| = |\Delta_K (−1)| = \det(K ). $$ for all $n$. Because these knots are alternating, each one admits a reduced alternating diagram, say $D_n$ with crossing number $c(D_n)$. The Bankwitz Theorem implies that $c(K_n) \leq \det(K_n) = \det(K)$. However, there are only finitely many knots of a given crossing number, and in particular $c(K_n)$ grows arbitrarily large with $n$, which contradicts that $c(K_n) \leq \det(K)$. For this reason, there also cannot exist infinitely many alternating knots with the isomorphic (bigraded) $\widehat{HFK}(S^3, K)$, either. In general though, I think these types of botany questions for the Alexander polynomial are interesting both on their own, and in the context of knot Floer homology. Some nice open botany and geography questions were recently mentioned in a paper of Hedden and Watson (see http://arxiv.org/abs/1404.6913).<|endoftext|> TITLE: Lie group actions with only one orbit type, but not defining a principal bundle QUESTION [7 upvotes]: Searched-for situation: A compact connected Lie group acts effectively on a closed Riemannian manifold by isometries, such that there is only one orbit type of dimension strictly less than that of the manifold, and the orbit projection $M\to M/G$ is not a principal $G$-bundle. I could not find or construct any examples of the situation above. However, I am quite sure there must be some. The reason is that several authors study group actions with only one orbit type in their papers, and if there were no examples as above then the situation in those papers could be simplified dramatically. Q: Who knows an example, or a reference to an example, of the situation described above? Background information: It is well-known (Bredon, Introduction to Compact Transformation Groups, Theorem IV.3.3) that when there is only one orbit type, corresponding to a minimal isotropy type $(H)$, the orbit projection $M\to M/G$ is a fiber bundle with fiber $G/H$ and structure group $N(H)/H$, where $N(H)$ is the normalizer of $H$. Consequently, the situation above reduces to: A compact connected Lie group acts effectively on a closed Riemannian manifold by isometries, such that there is only one orbit type of dimension strictly less than that of the manifold, and the isotropy groups of the action are non-trivial and core-free. Dropping the dimension hypothesis, any homogeneous space $G/H$, where $G$ is a compact connected Lie group and $H$ is a non-trivial core-free closed subgroup and $G$ acts by left multiplication, is an example. In this case, there is just one orbit type for the trivial reason that there is only one orbit in total. Thanks in advance for your answers. REPLY [4 votes]: Let $G=O(n+1)$ act on the $n$-sphere $S^n\cong O(n+1)/O(n)$. Let $M=S^1\times S^n$ and let $G$ only act on the second factor The effectiveness is the problem, but this is clear, as the action on $S^{n}$ by linearity determines the action on $R^{n+1}$.The projection is not a principal $G$-bundle, as indeed no element of $G$ acts freely.<|endoftext|> TITLE: Reconstructing a string from random samples QUESTION [8 upvotes]: What is known about the following problem? Reconstruct a string $\sigma$ of known length $n$ over a known alphabet $\Sigma$ from a collection of uniformly and independently chosen $k$-long subsequences of $\sigma$ where $k$ is fixed between $0$ and $n$. Recall that a $k$-long subsequence of $\sigma=\langle \sigma_1,\ldots,\sigma_n\rangle$ is a sequence $\langle \sigma_{\varphi(1)},\dots,\sigma_{\varphi(k)}\rangle$ where $\varphi$ is an increasing function from $\{1,\ldots,k\}$ to $\{1,\ldots,n\}$. REPLY [5 votes]: Given the string $\sigma$, for any word $x$ of length $k$ the probability $P_{n,k}(\sigma, x)$ that a randomly chosen $k$-long subsequence matches $x$ can be computed, e.g. using $$ P_{n,k}(\sigma,x) = \dfrac{k}{n} \delta_{\sigma_1, x_1} P_{n-1,k-1}(\sigma',x') + \left(1 - \dfrac{k}{n}\right) P_{n-1,k}(\sigma',x)$$ where $\sigma'$ and $x'$ are $\sigma$ and $x$ respectively with the first symbol removed. If the distributions for different $\sigma$ are different, in principle we can take enough samples to classify $\sigma$ with high probability. But if two $\sigma$'s have the same distribution, we can't distinguish them. For example, take $n=4$, $k=2$ and the alphabet $\{0,1\}$, Here are all the $P_{4,2}$: $$ \left[ \begin {array}{ccccc} &[0,0]&[0,1]&[1,0]&[1,1] \\ [0,0,0,0]&1&0&0&0\\ [0,0,0,1]&1 /2&1/2&0&0\\ [0,0,1,0]&1/2&1/3&1/6&0 \\ [0,0,1,1]&1/6&2/3&0&1/6\\ [0,1,0 ,0]&1/2&1/6&1/3&0\\ [0,1,0,1]&1/6&1/2&1/6&1/6 \\ [0,1,1,0]&1/6&1/3&1/3&1/6\\ [0, 1,1,1]&0&1/2&0&1/2\\ [1,0,0,0]&1/2&0&1/2&0 \\ [1,0,0,1]&1/6&1/3&1/3&1/6\\ [1,0 ,1,0]&1/6&1/6&1/2&1/6\\ [1,0,1,1]&0&1/3&1/6&1/2 \\ [1,1,0,0]&1/6&0&2/3&1/6\\ [1,1,0 ,1]&0&1/6&1/3&1/2\\ [1,1,1,0]&0&0&1/2&1/2 \\ [1,1,1,1]&0&0&0&1\end {array} \right] $$ Note that $\sigma = [0,1,1,0]$ and $\sigma = [ 1,0,0,1]$ both have the same distribution. So these two cannot be distinguished by their distributions of $2$-long subsequences. EDIT: For another example, for $n=7$ and $k=3$, $\sigma = [1, 0, 0, 1, 1, 1, 0]$ and $\sigma = [0, 1, 1, 1, 0, 0, 1]$ can't be distinguished by their distributions of $3$-long subsequences. But for $n=6$, all $\sigma \in \{0,1\}^6$ can be distinguished by their distributions of $3$-long subsequences.<|endoftext|> TITLE: Normal Covering of a Finite Group QUESTION [7 upvotes]: Suppose $G$ is a finite group and $N_1, N_2, \cdots, N_k$ are proper normal subgroups of $G$. The set $\{ N_1, \cdots, N_k\}$ is called a normal cover for $G$, if $G = \cup_{i=1}^kN_i$. I need to the main properties of finite groups with a normal covering. Anybody knows some references on this topic. There are several published papers on covering of groups by subgroups, but I like normal cover. Any comments will be highly appreciated. Sincerely, F Moftakhar REPLY [2 votes]: You can find a complete classification here: M. Chiodo, Finitely annihilated groups, Bull. Austral. Math. Soc. 90, No. 3, 404-417 (2014). This also contains a generalisation to arbitrary coverings of groups by proper normal finite-index subgroups; such a group is said to be Finitely Annihilated (F-A). See in particular corollary 5.5: A finite group is F-A if and only if it has non-cyclic abelianisation. -Maurice<|endoftext|> TITLE: Locus of complete curves on $\mathcal M_g$ QUESTION [9 upvotes]: Is the union of the complete curves on $\mathcal M_g$ Zariski dense? ($g \gg 0$) I know it is hard to find higher-dimensional complete subvarieties of $\mathcal M_g$, but a quasiprojective variety can have lots of complete curves but nothing higher dimensional, e.g. the complement of a codimension 2 linear subspace of $\mathbb P^n$. I am curious whether the known complete curves can deform in $\mathcal M_g$. REPLY [10 votes]: Let $M_g^S$ be the Satake compactification of $M_g$. It is a singular projective variety. For $g \geq 3$, the codimension of $M_g^S \setminus M_g$ is $\geq 2$. If $p \in M_g$ is a point, then a sufficiently general linear subspace through $p$ cuts out a complete curve not meeting the Satake boundary. Conclusion: through any point of $M_g$, $g \geq 3$, passes a complete curve.<|endoftext|> TITLE: Proving that the kernel of this matrix is of dimension 2 QUESTION [6 upvotes]: (Edit : see at the bottom of the question for an additional surprising possible hint.) Using a computational software program, I found that the kernel of the following matrix is of dimension 2 when $n\geqslant 2$ but I haven't managed to prove it: \begin{equation} \text{for almost all } t_1>0,\quad \text{dim}\,\text{ker}\left(\mathbf{Q}_2\mathbf{Q}_1(t_1)-\mathbf{Q}_1(t_2)^{-1}\mathbf{Q}_2\right)\overbrace{=}^?\;2 \end{equation} where $t_2$ is chosen (assuming it exists) such that $$\text{det}\left(\mathbf{Q}_2\mathbf{Q}_1(t_1)-\mathbf{Q}_1(t_2)^{-1}\mathbf{Q}_2\right)=0$$ where $n\geqslant 2$ $\mathbf{Q}_2$ is the following matrix: \begin{equation} \mathbf{Q}_2=\begin{bmatrix} \mathbf{I}_n & \mathbf{0}_n \\ \mathbf{0}_n & \mathbf{P}^{-1}\begin{bmatrix}1 & && \\ & \ddots && \\ & & 1& \\ &&& -1 \end{bmatrix}\mathbf{P} \end{bmatrix}\in\mathbb{R}^{2n\times2n} \end{equation} where $\mathbf{P}\in\mathbb{R}^{n\times n}$ is any invertible matrix. $\mathbf{Q}_1(t)$ is defined by: \begin{equation} \forall t>0,\quad\mathbf{Q}_1(t)=\begin{bmatrix}\textbf{cos}(\boldsymbol \Omega t) & \boldsymbol \Omega^{-1}\,\textbf{sin}(\boldsymbol \Omega t) \\ -\boldsymbol \Omega\,\textbf{sin}(\boldsymbol \Omega t) & \textbf{cos}(\boldsymbol \Omega t)\end{bmatrix}\in\mathbb{R}^{2n\times2n} \end{equation} and: \begin{equation} \boldsymbol\Omega=\begin{bmatrix} \omega_1 & & 0\\ & \ddots & \\ 0 & & \omega_n \end{bmatrix}\in\mathbb{R}^{n\times n},\quad \forall i\in\lbrace 1,\dots, n\rbrace, \omega_i>0 \end{equation} and the four blocks are diagonal, for example: \begin{equation} \mathbf{cos}(\boldsymbol\Omega t)=\begin{bmatrix} \cos(\omega_1t) & &0 \\ & \ddots & \\ 0 & & \cos(\omega_n t) \end{bmatrix}\in\mathbb{R}^{n\times n} \end{equation} Few properties of $\mathbf{Q}_1$ and $\mathbf{Q}_2$: Obviously, $\mathbf{Q}_2$ is invertible and $\mathbf{Q}_2=\mathbf{Q}_2^{-1}$. Also, $\det(\mathbf{Q}_1)=1$ ($\omega_i>0$ and for proper $t>0$) and: \begin{equation} \mathbf{Q}_1(t)^{-1}=\begin{bmatrix}\textbf{cos}(\boldsymbol \Omega t) & -\boldsymbol \Omega^{-1}\,\textbf{sin}(\boldsymbol \Omega t) \\ \boldsymbol \Omega\,\textbf{sin}(\boldsymbol \Omega t) & \textbf{cos}(\boldsymbol \Omega t)\end{bmatrix} \end{equation} Also, $\forall s,t,\ \mathbf{Q}_1(s+t)=\mathbf{Q}_1(s)\mathbf{Q}_1(t)=\mathbf{Q}_1(t)\mathbf{Q_1}(s)$. (Note: I've already asked this question here but did not get any answer despite a bounty.) Edit Additional possible hint: I've found that for $n=2$ (maybe it's more general), the follow results stands, if $A=\mathbf{Q}_2\mathbf{Q}_1(t_1)-\mathbf{Q}_1(t_2)^{-1}\mathbf{Q}_2$ and $\phi(x)=\det(A-xI)$: $$\phi'(0)^2=-\phi(0)\det(B)$$ where $B$ is the $A$ matrix if $\mathbf{P}=\mathbf{I}$... ! I have no clue where this comes from, maybe it's obvious for someone (maybe related to Lie algebras?). This would conclude the proof as $\phi(0)=\det(A(t_1,t_2))=0$ for the appropriate $t_1,t_2$, so that $0$ would be a double eigenvalue of $A$. REPLY [2 votes]: Thanks to @Terry Tao's answer, I can now give a solution to the question. I am open to comments or precisions. $Q_1(t)$ can be conjugated with by $\text{diag}(\omega_1,\dots,\omega_n,1,\dots, 1)$ which transforms it into a rotation matrix without affecting $Q_2$. It is bit more tricky, but $Q_2$ can be conjugated with a diagonal matrix $\text{diag}(d_1,\dots,d_n,d_1,\dots,d_n)$ which transforms it into an orientation-reversing orthogonal matrix without affecting $Q_1(t)$. More precisely, this seems to be true for an open set of $P$ dense in the matrix vector space. The conjugations do not depend on $t$, so at the end of the day, $Q_2Q_1(t_1)Q_2Q_1(t_1)$ is similar to an orientation-preserving orthogonal matrix, whose eigenvalues lie on the unit circle and can be $1$, $-1$ or appear by pair of complex conjugates. The product of the eigenvalues is $1$, so $-1$ has to appear an even number of times. The matrix $Q_2Q_1(t_1)Q_2Q_1(t_1)$ is a $2n\times 2n$ matrix, so $1$ also has to appear with an even multiplicity. So if 1 is eigenvalue of $Q_2Q_1(t_1)Q_2Q_1(t_1)$, the corresponding eigenspace is of dimension greater than 2. I do not have the mathematical background to prove it (help is welcome), but I guess one concludes by proving that the set of eigenvalues with exactly two eigenvalues equal to 1 is dense in the set of eigenvalues with at least two eigenvalues equal to 1. But it is still possible to find some values of $P$ and $\omega_i$ such that 1 has a multiplicity of 4, or even 6, etc. The last step is to remark that the eigenspace for the eigenvalue 1 of $Q_2Q_1(t_1)Q_2Q_1(t_1)$ is the eigenspace for the eigenvalue 0 of $Q_2Q_1(t_1)-(Q_2Q_1(t_2))^{-1}$ or of $Q_2Q_1(t_1)-Q_1(-t_2)Q_2$.<|endoftext|> TITLE: Riemann's formula for the metric in a normal neighborhood QUESTION [38 upvotes]: I would love to understand the famous formula $g_{ij}(x) = \delta_{ij} + \frac{1}{3}R_{kijl}x^kx^l +O(\|x\|^3)$, which is valid in Riemannian normal coordinates and possibly more general situations. I'm aware of 2 proofs: One using Jacobi fields [cf. e.g. S.Sternberg's "Curvature in Mathematics and Physics" from which the question title and formula is stolen :-) or cf. S.Lang's "Differential and Riemannian Manifolds"]. The other proof involves computing that $\partial_k\partial_lg_{ij}(x)$ shares some symmetries of curvature [cf. M.Spivak's "A Comprehensive Introduction to Differential Geometry, Vol. 2" where it is a several page "hairy computation" or cf. H.Weyl's 1923 edition of Riemann's Habilitationsvortrag (reprinted in a recent German book by Jürgen Jost) which I find uncomprehensible.] Are you aware of any other proof? Are normal coordinates necessary? While the Jacobi fields proof is short and elegant enough, it irks me that it requires "higher technology" not involved in the endproduct. Somehow the formula should be provable by pure calculus. Indeed, it is stated as an exercise in P.Petersen's "Riemannian Geometry": From the context I guess he thinks it should follow from the expression of $\partial_lg_{ij}$ as a sum of 2 Christoffel symbols and the simplified expression for curvature at $x=0$ where the Christoffel symbols vanish. Alas my attempts at this go in circles... I find the situation quite amazing: Not many textbooks treat this fundamental and historic formula. (Estimating from the sample on my shelf it is $3/17.$ E.g. it seems it's not even in Levi-Civita's classic.) Update/Scholium: In classical language: The knackpoint seems to be a "differential Bianchi formula" for the Christoffel symbols at $0$. This follows from the geodesic equation. I see no other way yet. A more modern approach minimizing (but not eliminating) the role of geodesics is in A.Gray's Tubes book. (Noted in comments. I'm waiting for www.amazon.de to deliver this treasure.) $\bullet$ While geodesics are very geometric and normal coordinates are very practical, methinks the formula is a tad ungeometric. What I'm hoping/asking for is a coordinate-independent formula for the second derivative of $g$ in terms of a suitable "reference connection". REPLY [6 votes]: I came to this post many years later, since I too was concerned about the absence of Riemann's formula in most texts, lengthy treatment in others, or reliance on more advanced techniques like Jacobi fields. I include here a direct concise proof which I think would be well suited for beginning students. We want to show that $$ g_{ij}(x)=\delta_{ij}+\frac{1}{3} R_{k ij\ell}(o)x_kx_\ell+\mathcal{O}(|x|^3), $$ where $x=(x_1,\dots, x_n)$ are normal coordinates centered at a point $o$ in a Riemannian manifold $M$. By Taylor's theorem, we need to check that \begin{eqnarray} \tag{1} g_{ij}(o)&=&\delta_{ij}\label{eq:1},\\ \tag{2} g_{ij,k}(o)&=&0\label{eq:2},\\ \tag{3} g_{ij,k\ell}(o)&=&\frac{1}{3}\big(R_{kij\ell }(o)+R_{\ell ij k}(o)\big)\label{eq:3}, \end{eqnarray} where we use the notation $f_{,i}:=\partial_i f$, and $f_{,ij}:=\partial^2_{ij} f$. \eqref{eq:1} follows immediately from the construction of normal coordinates, and \eqref{eq:2} is not difficult to establish either. \eqref{eq:3}, which is the heart of the matter, requires a bit more work. In most sources, like Spivak, vol II or the relatively recent book by Jost, which gives a very comprehensive treatment of Riemann's lecture, \eqref{eq:3} is established via symmetry properties of $g_{ij,k\ell}$ which involve long computations; although 2011 Lecture Notes of John Douglas Moore gives a very nice and efficient proof of them. Instead I will demonstrate \eqref{eq:3} via a Bianchi type cyclic identity for derivatives of Christoffel symbols, which follows quickly from the geodesic equation (as the OP had mentioned in the "Update/Scholium" above). The only place I have seen this approach is in 2013 Lecture Notes of Christian Bär. Proofs of (1) and (2) All indices here range from $1$ to $n$, and any term which involves repeated indices stands for a sum over that index. Let $\exp_o\colon T_o M\to M$ be the exponential map, and $U\subset M$ be a ball centered at $o$ such that $\exp_o\colon \exp_o^{-1}(U)\to U$ is a diffeomorphism. Let $e_i$ be an orthonormal basis for $T_o M$, i.e., $ g(e_i, e_j)=\delta_{ij}. $ Then the normal coordinates $x_i\colon U\to \mathbf{R}$ (with respect to $e_i$) are given by $$ \exp_o(x_i(p) e_i)=p. $$ The mapping $x\colon U\to\mathbf{R}^n$, given by $x:=(x_1,\dots, x_n)$ identifies $U$ with a ball centered at the origin in $\mathbf{R}^n$, which we again denote by $U$. Let $E_i(x):=\partial_i|_x$ be the coordinate vector fields on $U$. Then $g_{ij}\colon U\to\mathbf{R}$ are given by $$ g_{ij}(x):=g(E_i(x),E_j(x)). $$ Since $E_i(o)=e_i$ we immediately obtain \eqref{eq:1}. To see \eqref{eq:2} note that, by the definition of normal coordinates $x_i$, the geodesics in $U$ passing through $o$ and another point $x$ of $U$ are given by $\gamma(t):=tx$. Since $\gamma$ is a geodesic, $$ \gamma_k''(t)+\Gamma_{ij}^k(\gamma(t))\gamma_i'(t)\gamma_j'(t)=0, $$ which yields \begin{equation}\label{eq:4} \tag{4}\Gamma_{ij}^k(t x)x_i x_j=0. \end{equation} Setting $t=0$, observing that $x_i$, $x_j$ may assume any values, and recalling that $\Gamma_{ij}^k=\Gamma_{ji}^k$, we obtain \begin{equation*}\label{eq:Gamma0} \Gamma_{ij}^k(o)=0. \end{equation*} Since $ \Gamma_{ij}^k=\frac{1}{2} g^{k\ell}(g_{\ell i,j}+g_{\ell j,i}-g_{ij,\ell}), $ and $g^{ij}(o)=\delta_{ij}$, we have \begin{equation*}\label{eq:8} 0=\Gamma_{ij}^k(o)=\frac{1}{2} \big(g_{k i,j}(o)+g_{k j,i}(o)-g_{ij,k}(o)\big). \end{equation*} Adding the above equation to itself, after a cyclic permutation of indices yields \eqref{eq:2}. Proof of (3) Since $ \nabla_{E_i}E_j=\Gamma_{ij}^kE_k, $ $$ g_{ij,k}=g(\nabla_{E_k}E_i, E_j)+g(E_i, \nabla_{E_k}E_j)=\Gamma_{k i}^\ell g_{\ell j}+\Gamma_{k j}^\ell g_{i\ell}. $$ Differentiating again, and using \eqref{eq:2}, yields \begin{equation}\label{eq:5} \tag{5}g_{ij,k\ell}(o)=\Gamma_{k i,\ell}^j(o)+\Gamma_{k j,\ell}^i(o). \end{equation} Next we differentiate \eqref{eq:4} at $t=0$ to obtain $$ \Gamma_{ij,\ell}^k(o)x_i x_jx_\ell=0, $$ a homogeneous polynomial of degree $3$ which vanishes identically. The coefficient of each term $x_i x_jx_\ell$ is the sum of all $6$ permutations of lower indices of $\Gamma_{ij,\ell}^k(o)$. Since $\Gamma_{ij,\ell}^k=\Gamma_{ji,\ell}^k$, we obtain \begin{equation*}\label{eq:cyclic} \Gamma_{ij,\ell}^k(o)+\Gamma_{j\ell,i}^k(o)+\Gamma_{\ell i,j}^k(o)=0. \end{equation*} Now note that, since $ R_{ijk}^\ell=\Gamma_{ik,j}^\ell-\Gamma_{jk,i}^\ell+\Gamma_{ik}^p\Gamma_{pj}^\ell-\Gamma_{jk}^p\Gamma_{ip}^\ell, $ \eqref{eq:2} yields that $$ R_{ijk\ell}(o)=\Gamma_{ik,j}^\ell(o)-\Gamma_{jk,i}^\ell(o). $$ Here we have also used the fact that $R_{ijk}^\ell(o)=R_{ijk\ell}(o)$ due to \eqref{eq:1}. The last two displayed equations yield \begin{eqnarray*} R_{ik\ell j}(o)+R_{i\ell kj}(o) = \Gamma_{i\ell,k}^j(o)+\Gamma_{ik,\ell}^j(o)-2\Gamma_{\ell k,i}^j(o) =-3\Gamma_{\ell k,i}^j(o). \end{eqnarray*} The last equality together with \eqref{eq:5} and symmetries of $R$ now yields \begin{eqnarray*} g_{ij,k\ell}(o) &=&-\frac{1}{3}\big(R_{\ell ik j}(o)+\require{cancel}\cancel{R_{\ell kij}(o)}+R_{\ell jk i}(o)+\require{cancel}\cancel{R_{\ell kji}(o)\big)}\\ &=&\frac{1}{3}\big(R_{\ell i jk}(o)+R_{kij\ell}(o)\big), \end{eqnarray*} as desired.<|endoftext|> TITLE: Anti-bidiagonal matrix with main anti-diagonal {1,2,3,...} and first sub-anti-diagonal {-1,-2,-3,...} has eigenvalues lambda={1,-2,3,-4,...} QUESTION [5 upvotes]: Consider the anti-bidiagonal matrix $B_6\in\mathbb{R}^{6\times 6}$, defined along its anti-diagonals as follows $$ B_6=\begin{bmatrix} & & & & & 6\\ & & & & 5 & -5\\ & & & 4 & -4\\ & & 3 & -3\\ & 2 & -2\\ 1 & -1 \end{bmatrix}. $$ Its eigenvalues are $\lambda(B_6)=\{1,-2,3,-4,5,-6\}$, a fact easily verified numerically e.g. on MATLAB. Furthermore, one can numerically verify that this pattern persist. With this, I'm looking for a proof to the following statement. Proposition 1. For a matrix $B_n$ defined as above, its eigenvalues are given $$\lambda(B_n)=\{(-1)^{i+1}i\}_{i=1}^n\equiv\{1,-2,3,-4,\ldots\}.$$ This has turned out to be surprisingly difficult. It is tempting to predict the eigenvalues by reading off the diagonals. However the matrix is genuinely not triangular, nor does it share many properties with triangular matrices. It is easy to construct counter-examples where the eigenvalues do not coincide with the antidiagonals. The ramification of this statement is that it characterizes an entire class of graphs to be "well-connected", and the corresponding class of linear algebra problems to be well-conditioned irrespective of problem size. The fact that the eigenvalues are integers is quite significant, since it implies that the solutions to integer and linear program versions of this matrix would coincide, and this can be exploited to prove several graph results. I can give more background if necessary. Some candidate approaches I have tried: Recursive characteristic polynomial. It is routine to write down a formula for $\mathrm{det}(B_n - \lambda I_n)$. However, it's difficult to find its roots without resorting to variations on Newton iterations. Golub-Kahan form. One may reorder the rows and columns to yield a tridiagonal matrix with zeros along the main diagonal. More specifically, let $e_i$ denote the i-th column of the identity, define a similarity transform $E=\begin{bmatrix} e_1 & e_n & e_2 & e_{n-1} & \ldots\end{bmatrix}$, and consider $\hat{B}=E^TBE$. However, its characteristic polynomial is a mess, and we encounter the same problem trying to prove roots of the polynomial. Matrix inverse. The matrix $B_6$ has the following inverse $$B_{6}^{-1}=\begin{bmatrix}\frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6}\\ \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5}\\ \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4}\\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\\ \frac{1}{2} & \frac{1}{2}\\ 1 \end{bmatrix}.$$ A proof of this is simply gaussian elimation of a triangle-like matrix. Thus Prop 1 can be reworded as $\lambda(B_6^{-1})=\{1,-1/2,1/3,-1/4,1/5,-1/6\}$. It does not appear to be any easier to obtain the eigenvalues of $B^{-1}_n$. Indeed, its characteristic polynomial is even harder to write down. Interpretation as difference operator. Finally, the matrix can be viewed as a difference operator, and in the limit $\lim_{n\to\infty}B_n$, its eigenfunctions are orthogonal polynomials. This approach yields a proof. However it does not say anything about the cases where $n$ is small, e.g. 4 or 5. Any advice or suggestions? REPLY [10 votes]: This problem is essentially the same as this one. In particular, let $J$ be the anti-diagonal identity matrix, and $P^{-1}$ be the matrix mentioned in the link above. Then, the matrix in the current post is nothing but \begin{equation*} B_n = JP^{-T}J^T. \end{equation*} Since $J^TJ=I$, we can recover eigenvectors and values of $B_n$ using the derivation for $P$ in the linked post.<|endoftext|> TITLE: Decidability of decidability QUESTION [15 upvotes]: The questions I'm going to ask are non formal because they concern decidability of decidability, and I couldn't find any references on that after some quick searches. I hope that this thread is still "formal enough" to be productive. So, let us suppose that we're working with an ambient logic that is classical (we believe in LEM and induction principle), and that we were able to define some formal theory in our natural language that is strong enough to have undecidable statements. Let $A$ be any proposition of this formal theory, and call $Dec(A)$ the proposition "there exists a proof of $A$ or there exists a proof of $\neg A$". $Dec(A)$ is true iff $A$ is decidable, $Dec(A)$ being false iff $A$ is undecidable. Notice that because of LEM, $Dec(A) = Dec(\neg A)$. What can we say about $Dec(Dec(A))$ ? (here begins the non formal stuff, supposing we have at our disposal some stratified ``infinite order logic'' that can speak about its own propositions and proofs (say, level $n$ propositions can speak about level $m \leq n$ propositions and proofs)). If $Dec(A)$ is right, then there either exists a proof of $A$ or a proof of $\neg A$ that we can write down as a finite string of symbols. Thus, by finding such a proof, we proved that it is possible to find a proof of $A$ or that it is possible to find a proof of $\neg A$, hence $Dec(Dec(A))$ is true. By obvious induction, $Dec^n(A)$ true implies $Dec^{n+1}(A)$ true. This leads us to the following definition of $n$-undecidability: A proposition $A$ is said to be $n$-undecidable iff there exists some positive integer $n$ such that for all $m \leq n$, $Dec^m(A)$ is false and $Dec^{n+1}(A)$ is true. We can also define the notion of $\infty$-undecidability: We call $\infty$-undecidable, any proposition $A$ such that for all $n$, $Dec^n(A)$ is false. Question: can we prove that an $\infty$-undecidable proposition exists ? Let $B$ = there exists an $\infty$-undecidable proposition. What can we say about $Dec(B)$ ? If $Dec(B)$ is true, then we can either find a proof of $B$ or a proof of $\neg B$. Finding a proof of $B$ means to find a statement $C$ and to prove that for all $n$, $Dec^n(C)$ is false. This implies that $Dec^{n+1}(C)$ is true, and hence contradiction. If we can find a proof of $\neg B$, that is, a proof that there exists no $\infty$-undecidable statement, this means that for all statements $C$ we can find an $n$ such that $Dec^n(C)$ is true, and in particular, for $C = B$. Thus there exists an $n$ such that $Dec^n(B)$ is true and it therefore becomes decidable to know if it is decidable ... that there exists an $\infty$-undecidable proposition. If $Dec(B)$ is false, what can we say about $Dec^n(B)$ for all $B$ ? It appears that there must exist some $m$ such that $Dec^m(B)$ is true too, or we would have found a proof that there exists an infinitely undecidable statement, and hence contradiction. Any comment, recommandations, references about this stuff? PS: it really looks like that one needs to have a theory that is able to speak about its own lower level propositions and proofs, say, an infinite order stratified language. REPLY [14 votes]: $\newcommand\Con{\text{Con}} \newcommand\Dec{\text{Dec}}$ Let $F$ be the formal system in which the proofs are to be carried out, when it comes to your formal assertions of the form $\Dec(\varphi)$. So we assume that $F$ is described by some computable axiomatization. For example, perhaps $F$ is simply the usual first-order PA axioms. Let me assume that $F$ is true in the standard model $\mathbb{N}$, which is probably a case that you care most about. (But actually, I believe it is sufficient in this argument to assume iterated consistency assertions about $F$.) Let $A=\Con(F)$. I claim that this statement is $\infty$-undecidable with respect to $F$. To see this, argue as follows. By the incompleteness theorem, since $\Con(F)$ is true, we know that $F$ does not prove $A$, and since $F$ and $A$ are both true, it also follows that $F$ does not prove $\neg A$. So $\neg\Dec(A)$ is true (that is, in the standard model $\mathbb{N}$). But $F$ by itself cannot prove $\neg\Dec(A)$, since $F$ proves $\neg\Dec(X)\to \Con(F)$, as an inconsistent theory has no undecidable statements, and so if it did it would violate the incompleteness theorem. Note also that $F$ cannot prove $\Dec(A)$ either, since $\neg\Dec(A)$ is true. Thus, $\neg\Dec(\Dec(A))$ is true. But $F$ cannot prove this, since then again it would prove $\Con(F)$, violating incompleteness, and it also cannot prove $\Dec(\Dec(A))$, since $\neg\Dec(\Dec(A))$ is true. So $\neg\Dec(\Dec(\Dec(A)))$ is true. And so on. For the general step, if $\Dec^n(A)$ is false, then $F$ cannot prove this, since then it would prove $\Con(F)$, contrary to the incompleteness theorem, and it cannot prove $\Dec^n(A)$ either since it was false and $F$ is true, and so $\Dec^{n+1}(A)$ is false. This reasoning shows that $\Dec^n(A)$ will be false for every $n$, and so $A=\Con(F)$ is $\infty$-undecidable, assuming that $F$ is true in the standard model. It seems likely to me that the content of what it was about "true in the standard model" that the argument used should be covered by the assumption merely that $\Con^n(F)$ holds for all $n$. But I shall leave this to the proof-theoretic experts, who I hope will shed light on things. Update. More generally, I claim the following. Theorem. Assume that the formal system $F$ is true in the standard model of arithmetic $\mathbb{N}$. Then $\Dec(B)$ and $\Dec(\Dec(B))$ are equivalent for any statement $B$. So $\Dec(B)$ is equivalent to $\Dec^n(B)$ for any particular $n$ with respect to any such true formal system $F$. Proof. Note that I am not claiming that this equivalence is provable in $F$, only that it is true in the standard model. You had already noted that $\Dec(B)$ implies $\Dec(\Dec(B))$. So assume $\Dec(\Dec(B))$ is true. Thus, it is true in $\mathbb{N}$ that either there is a proof in $F$ of $\Dec(B)$ or a proof of $\neg\Dec(B)$. It cannot be the latter, because then $F$ would prove its own consistency, as we have noted, contrary to the incompleteness theorem. Thus, it must be true in the standard model that "there is a proof of $\Dec(B)$." In this case, there really is a (standard) proof of $\Dec(B)$, and so $\Dec(B)$ is true. QED Thus, once you have an undecidable statement, it is $\infty$-undecidable, with respect to any such system $F$ that is true in the standard model.<|endoftext|> TITLE: Proving the Irrationality of this Number QUESTION [19 upvotes]: I found this problem on Math.SE: Prove that $\log_35+\log_25$ is irrational. https://math.stackexchange.com/q/986227/173397. I labored on it for a few days, and couldn't find an algebraic solution- I'm not even sure if such a solution exists. All I was able to do was prove that both components were irrational by themselves (as opposed to their sum). I am wondering if anyone has seen this problem before, and/or if anyone knows a solution. If so, I could really use a hint. So far, using the Fundamental Theorem of Arithmetic (i.e., all integers have a unique prime factorization) hasn't helped me the way one would use it to show that the individual components are irrational. Thank you in advance. REPLY [8 votes]: To resonate with Henry Cohn's comment, Schanuel's conjecture implies that the natural logarithms of the primes are algebraically independent over $\mathbb{Q}$. In particular, the statement in the original post is probably true, but proving it might be out of reach at the moment.<|endoftext|> TITLE: Which ordinals can be proof-theoretic ordinals of a reasonable theory? QUESTION [11 upvotes]: When talking to a friend recently he asked a question - are there any reasonable first-order theories which have proof theoretic ordinal equal to small or large Veblen ordinal? I have then extended his question broadly - which ordinals can be proof-theoretic ordinals of any "reasonable" theory, where by "reasonable" I suggested we should mean "extending PA", though this can be discussed. (Edit: it seems convinient to be able to work with second-order theories, so instead we can think of "reasonable" as extending $\sf ACA_0$) Another, related question is the following: what is the proof-theoretic ordinal of theory PA+axiom schema asserting transfinite induction holds up to $\varepsilon_0$? I think it might be $\varepsilon_1$, but I can't be sure. Thanks in advance for feedback! EDIT: I have decided to state an alternative version of this question, which will hopefully be less ambiguous. Let our base system be $\sf ACA_0$. Suppose that we add to this system a statement "ordinal $\alpha$ is well-founded", expressed as second order predicate. Now we know that proof-theoretic ordinal of this theory will be greater than $\alpha$. Let's call ordinal $\gamma$ bounding if, whenever $\alpha<\gamma$, then PTO of $\sf ACA_0$+"$\alpha$ is well-founded" is also $<\gamma$. Then my question is, which recursive ordinals are bounding? We know that $\varepsilon_0$ is bounding, but what is the least bounding ordinal above it? EDIT2: I've recently realized that even second order theory can't really just talk about $\varepsilon_0$ or pretty much any ordinal per se, but we need to represent the ordinal in a way (e.g. we could represent $\varepsilon_0$ as an ordering on numbers representing ordinals in Cantor's normal form). Because of this, there can be multiple ways to express ordinal, and some of them could also hide some complexity (e.g. we can have an ordering which is well-ordered with order type $\varepsilon_0$ only if Kruskal's tree theorem holds). Because of this, we can have statement "$\alpha$ is well-founded$ hide an information about well-ordering of any recursive ordinal. Is that true? REPLY [9 votes]: Regarding the small Veblen ordinal, Rathjen and Weiermann gave an analysis of theories in that range of strength in Proof-theoretic investigations of Kruskal's theorem. Working over a reasonable base theory ($ACA_0$, the second order version of Peano arithmetic), both the theory $ACA_0+$Kruskal's theorem and a somewhat technical theory ($ACA_0$ plus $\Pi^1_1$ reflection for $\Pi^1_2$-BI; $\Pi^1_2$-BI is induction along internally well-ordered sets for $\Pi^1_2$ formulas, and $\Pi^1_1$ reflection means that the statement itself doesn't hold, but all its $\Pi^1_1$ consequences do). (I'm not sure if any of their theories are exactly the large Veblen ordinal.) The more general question seems too vague to answer. As the Rathjen and Weiermann article shows, there are an awful lot of reasonable-ish theories out there.<|endoftext|> TITLE: Number of monomials of deg D where each variables has low degree QUESTION [6 upvotes]: Let $D,n,d$ be three positive integers. I am looking for the number of monomials of degree $D$ in $n$ variables where each variable appears with exponent at most $d$. As a result of an application of inclusion/exclusion principle I found the following expression \begin{equation*} \sum_j (-1)^j \binom{n}{j} \binom{D-j(d+1)+(n-1)}{n-1} \end{equation*} where the summation runs over a set of indices where the expression makes sense: $0\leq j\leq n$, $j \leq D/(d+1)$. However, I am looking for a more tractable formula, if there exists one. REPLY [2 votes]: This formula has no closed form. That means it can not be written as a sum of fixed number of hypergeometric terms when $d$ is greater than $2$. The proof can be found in the book A=B by Petkovsek, Wilf and Zeilberger.<|endoftext|> TITLE: Is the set of certain polynomials finite or infinite? QUESTION [5 upvotes]: Let us consider the set of all polynomials with the following properties: i) all coefficients are integer; ii) the leading coefficient equals one; iii) all zeros are real and simple and belonging to $[-1.99,1.99] $. Is this set finite or infinite? REPLY [13 votes]: It is finite. This follows by combining two separate results: If $p \in \mathbb{Z}[x]$ is monic integer having all its complex roots lying in $[-2,2]$, then all these roots are of the form $2\cos(2\pi q)$ with $q \in \mathbb{Q}$. For the proof of this, write $p = \prod_{i=1}^d (x - 2\cos(2\pi \, t_i))$ and observe that the sequence of degree $d$ monic polynomials $p_n := \prod_{i=1}^d(x - 2\cos(2\pi n t_i))$ is also in $\mathbb{Z}[x]$ and that, of course, it has bounded coefficients (in terms of $d$ alone, independently of $n$). Thus there must be two $p_n = p_{n'}$ with $n \neq n'$; considering their roots, we get the claim with $q(n-n') \in \mathbb{Z}$. For a given $d \in \mathbb{N}$, the monic polynomial $P_d$ with simple roots at the set $\{2\cos(2\pi m/d) \mid (m,d) = 1\}$ is in $\mathbb{Z}[x]$ and irreducible. This follows from the corresponding irreducibility property of the cyclotomic polynomial $\Phi_d(z)$, which can be written as $z^{\deg{P_d}} P_d(z+1/z)$. Combining these two statements, you can see that your set consists precisely of the products of pairwise different minimal polynomials of those $2\cos(2\pi / d)$ that do not exceed $1.99$. There are only finitely many such $d$.<|endoftext|> TITLE: Examples of topologies compatible with a given dual pair QUESTION [8 upvotes]: Let $\langle X, Y \rangle$ be a pair of vector spaces put in duality by a non-degenerate bilinear form $\langle \cdot, \cdot \rangle: X \times Y \to \mathbb{R}$. A topology $\tau$ on $X$ is called compatible with the dual pair if the dual of $X$ with respect to the topology $\tau$ equals $Y$ (where $Y$ is seen as a subspace of the algebraic dual $X^*$ by the injective map $y \mapsto \langle \cdot, y \rangle$). The important theorem of Mackey-Arens characterizes all compatible topologies as polar topologies lying between the weak and the Mackey topology. Does there exist examples where a (or all) compatible topology can be described rather concretely? I'm especially interested in the duality of $C^\infty_c(\mathbb{R})$ and $C^\infty(\mathbb{R})$ given by integration. What can we say in this case? REPLY [6 votes]: Usha Kiran showed in An uncountable number of polar topologies and non-convex topologies for a dual pair that if the Mackey topology is distinct from the weak topology, then there are at least continuum-many distinct compatible topologies for the dual pair. The construction is based on unbounded subsets of the naturals, with equivalence given by having unbounded symmetric difference. Depending on your personal sensibilities, this means that they are not reasonably classifiable.<|endoftext|> TITLE: Cusps forms for $\Gamma (N)$ QUESTION [9 upvotes]: I know how to build a basis of the vector space of cusp forms for the congruence subgroups $\Gamma_1 (N)$ and $\Gamma_0 (N)$, but I couldn't find in the literature how to build a basis for $\Gamma(N)$. Also, Sage doesn't seem to be able to do this. For instance, for $N=6$ and weight 2, the dimension of this space is 1. Is it possible to compute explicitely the Fourier coefficients of the cusp form ? REPLY [19 votes]: You can do this in Sage but "in disguise". The idea is that if $f(z)$ is a cusp form for $\Gamma(N)$, then $g(z) := f(Nz)$ is a cusp form for a certain subgroup intermediate between $\Gamma_0(N^2)$ and $\Gamma_1(N^2)$ which Sage calls $\Gamma_H(N^2, [N + 1])$; the $N + 1$ is here because it generates the subgroup of $(\mathbf{Z} / N^2 \mathbf{Z})^\times$ consisting of classes that are 1 mod $N$. If $f(z) = \sum a_n q^{n/N}$, then $g(z) = \sum a_n q^n$, so if you have the $q$-expansion of $g$ then you have the $q$-expansion of $f$ and vice versa. ---------------------------------------------------------------------- | Sage Version 5.9, Release Date: 2013-04-30 | | Type "notebook()" for the browser-based notebook interface. | | Type "help()" for help. | ---------------------------------------------------------------------- sage: G = GammaH(36, [7]) sage: G.index() == Gamma(6).index() True sage: [g.q_expansion(25) for g in CuspForms(G, 2).basis()] [q - 4*q^7 + 2*q^13 + 8*q^19 + O(q^25)] (Note that this $g$ actually has trivial character, and is of CM type, so its $q$-expansion has lots of zero coefficients.)<|endoftext|> TITLE: Freely adding degeneracies does not change the homotopy type QUESTION [8 upvotes]: Background: Let $S$ be a simplicial set. By freely adding degeneracies to $S$, I mean first applying the forgetfull functor from simplicial sets to semi-simplicial sets which forget the already existing degeneracies maps and then apply its left adjoint which freely add the degeneracies. One gets a new simplicial set $S'$ which can be described explicitly by: $$ S'_n = \left\lbrace (a,f) |{ a \in S_k \text{ and } f:\{0,\dots,n\} \twoheadrightarrow \{0,\dots,k\} \atop \text{ is an order preserving surjection}}\right\rbrace $$ The simplicial operations being applied to (a,f) as if it was a formal "$f^* a$". It is know that the canonical map from $S'$ to $S$ (which send $(a,f$) to $f^* a$) is a weak equivalence. Indeed the geometric realization of $S'$ is exactly the same as the "fat geometric realization of $S$" (see def 1 and def 2 in nlab) which is known to be weakly equivalent to the classical geometric realization of $S$. My question: is it possible to give a purely combinatorial proof of the homotopy equivalence between $S$ and $S'$. The best would be an explicit finite sequence of simplicial homotopy equivalences or trivial Kan fibrations between them. My motivation is to determine to what extent a result of this kind is likely to be true in more general situations like for example simplicial objects in nice enough combinatorial model categories. It would also be helpful if someone could point me to a proof that the fat geometric realization is homotopy equivalent to the usual one. I already know two proofs of that in the literature: The first is in Segal's "Categories and cohomology theories". It deals with the more general case of the geometric realization of a simplicial space, but it just sketches the proof; and once stripped from all the topological consideration not really relevant for my question, it just say something like "it is true when $S$ is $\Delta^n$" (which is already not really trivial I think) and then there is a not completely clear argument to extend this to an arbitrary simplicial set by colimit. An other one that I have not been able to found back but which was just saying that it is a standard exercise to check that this map induce a bijection on homology and fundamental groups. I haven't check if this exercise is easy or not, but I was hoping for a more explicit proof. The nLab also mention a "more detailed" proof due to Tammo tom Dieck, but there is no link to it and I couldn't find it on google. REPLY [6 votes]: Here is a purely combinatorial and straightforward proof. First, you can easily check that $\Delta[m]'$ is the nerve of the category $[m]'$ that is obtained from the poset $[m]$ by freely adjoining one idempotent endomorphism to each object of $[m]$. These categories $[m]'$ are contractible since we can write down a natural transformation from the constant functor at $0$ to the identity functor. Correction: these idempotents are not exactly freely adjoined. They are supposed to act as identities on the morphisms of $[m].$ Next, we use a general fact (verified by induction over skeleta) that if we have a functor $F$ from simplicial sets to simplicial sets and a natural transformation between $F$ and identity such that $F$ preserves colimits and cofibrations and sends simplices to contractible simplicial sets, then this natural transformation is a weak equivalence. The first two conditions are clear for $FS = S'$ and I have verified the third one above so the conclusion follows.<|endoftext|> TITLE: If $S\subset\mathbb R$ is a $G_\delta$, is there a function $\mathbb R\to\mathbb R$ continuous exactly on $S$? QUESTION [7 upvotes]: Let $S\subset\mathbb R$ be a $G_\delta$ set. A variation on the construction of the Thomae function (which is discontinuous on the rationals and continuous elsewhere) shows that there is a function $\mathbb R\to\mathbb R$ that is continuous exactly on $S$. I'm trying to find a published reference for this result. Note: the wikipedia page on Thomae's function mentions an equivalent result, phrased in terms of $F_\sigma$'s, without giving a reference. So it's clear the result is well-known; but I'd like a reference in an article or book (even as an exercise) rather than just the wikipedia page. REPLY [10 votes]: The result you are looking for can be found in most advanced references in topology or analysis. Since I am away from my office at the time of this writing, I will provide you with a source readily available on the internet: Theorem 7.2 (p.30) in Oxtoby's Measure and Category. Note that Theorem 7.2 shows that any $F_{\sigma}$ set is precisely the set of discontinuities of some real valued function on $\Bbb{R}$; the result you want the reference for is equivalent to Theorem 7.2 thanks to De Morgan Laws. This theorem, I suspect, is due to Lebesgue, but at the moment I do not have a reference at hand to corroborate my suspicion. Addendum: According to this source (see history of Theorem 2') the above result was first proved by William H. Young in 1903, and perhaps independently by Lebesgue in 1904.<|endoftext|> TITLE: Isotrivial fibrations over $\mathbb P^1$ QUESTION [6 upvotes]: First of all I want to say that algebraic geometry is not "my field of research" so I apologize if the notation is not standard. $S$ is a smooth complex projective surface with a fibration $f$ over $\mathbb P^1(\mathbb C)$. Moreover suppose that the following properties hold for $f$: $f$ has singular fibers (a finite number). Each singular fiber is a curve with a single knot. (In this case I think that $f$ is called a stable fibration.) Under the above hypotheses I have to prove (or disprove) that the fibration $f$ can't be isotrivial. Here isotrivial means that any two smooth fibers are isomorphic as algebraic varieties. Many thanks. REPLY [5 votes]: There are three important invariants for any relatively minimal fibration $f:\,S \to B$ from a smooth complex projective surface $S$ to a smooth curve $B$: the self-intersection $\omega_{S/B}^2$, the degree of $f_*\omega_{S/B}$, and the singular index $e_f$ of $f$, where $\omega_{S/B}:=\omega_S\otimes\omega_B^{\vee}$ is the relative canonical sheaf of $f$. Here "relatively minimal" means that there is no exceptional curves (i.e., a curve isomorphic to $\mathbb P^1$ and have self-intersection equals $-1$) contained in fibers of $f$. If $f$ is stable (or semi-stable), $e_f$ is actually equal to the number of nodes contained in the singular fibers of $f$. These three invariants are all non-negative and satisfy the Noether formula: $$12\deg f_*\omega_{S/B}=\omega_{S/B}^2+e_f,$$ and the slope inequality $$\omega_{S/B}^2 \geq \frac{4(g-1)}{g} \deg f_*\omega_{S/B},~\text{equivalently,}~ (8g+4)\deg f_*\omega_{S/B}\geq g e_f,$$ where $g$ is genus of a general fiber of $f$. An important description of these invariants for semi-stable (or stable) fibrations is as follows. Let $j:B \to \overline M_g$ be the induced morphism from $B$ to compactification of the moduli space $M_g$ of smooth curves of genus $g$. Then there exist divisors $\lambda, \kappa, \delta$ on $\overline M_g$ such that $$\deg f_*\omega_{S/B}=\deg j^*\lambda, \quad \omega_{S/B}^2=\deg j^*\kappa,\quad e_f=\deg j^*\delta.$$ Now by your assumption, $f$ is stable and we have $e_f>0$. On the other hand, if $f$ is isotrivial, then the image of $j$ is a point, hence the pulling-back of any divisor is trivial and so $e_f=\deg j^*\delta=0$, which is a contradiction. We remark that the condition "$e_f >0$" implies in particular that $g>0$, since the moduli space of genus zero is a single point. PS: from the arguments, the conclusion is also true for any base $B$, not only for $B=\mathbb P^1$.<|endoftext|> TITLE: Unknotting number of knot diagrams QUESTION [9 upvotes]: Define the "diagram unknotting number" of a knot diagram $D$ as the minimal number of crossings that need to be changed in $D$ in order to get a diagram of the trivial knot (the usual unknotting number of a knot $K$ is the minimum over the diagram unknotting numbers of its diagrams). Can you give me an example of a diagram of the trefoil knot (or any other knot having unknotting number = 1) with diagram unknotting number greater than 1? REPLY [5 votes]: You should check this paper. http://arxiv.org/abs/0805.3174 http://www.worldscientific.com/doi/abs/10.1142/S0218216509007361<|endoftext|> TITLE: Product $PVPVP$ is elementwise nonnegative? QUESTION [10 upvotes]: Let $P\in \mathbb{R}^{n\times n}$ be the inverse of a positive definite M-matrix and $V\in \mathbb{R}^{n\times n}$ be any diagonal matrix. Prove (or disprove) that $PVPVP$ is elementwise nonnegative. I know of the following: $P$ is positive definite and elementwise nonnegative. Moreover, $p_{jk}p_{ii} \ge p_{ji}p_{ik}$ for any $i,j,k$. I can verify that the statement is true for $n=2$, but I don't know how to work with $n$ large. Playing around with randomly generated matrices in Matlab seems to suggest that the statement is true. Any hint or suggestion would be greatly appreciated. I've googled out that a very similar statement was put as a conjecture in this paper: Optimization of an on-chip active cooling system based on thin-film thermoelectric coolers (http://dl.acm.org/citation.cfm?id=1870955) Edit: Perhaps someone can solve this easier question: Is there a positive semi-definite and elementwise nonnegative $P$ and diagonal $V$ such that $PVPVP$ is not elementwise nonnegative? REPLY [11 votes]: First, we repeat the arguments from this stackexchange answer. $P^{-1}$ is an $M$-matrix, and can thus be written as $s(I-A)$ for some positive $s$ and some $A$ with non-negative entries. As $P^{-1}$ is positive definite, the spectrum of $A$ lies to the left of $\{ z: \hbox{Re}(z) = 1 \}$, and hence by Perron-Frobenius the spectral radius of $A$ is less than $1$. Thus we have the absolutely convergent Neumann series $$ P = s^{-1} (I + A + A^2 + \dots )$$ and hence $$ PVPVP = s^{-3} \sum_{i=0}^\infty \sum_{j=0}^\infty \sum_{k=0}^\infty A^i V A^j V A^k.$$ It thus suffices to show that $$ \sum_{i+j+k=m} A^i V A^j V A^k \quad (1)$$ has non-negative coefficients for each $m \geq 0$ (where $i,j,k$ are understood to be non-negative integers). By change of variables, this is $$ \sum_{0 \leq q \leq r \leq m} A^q V A^{r-q} V A^{m-r}.$$ Writing $A = (a_{st})_{1 \leq s,t \leq n}$ and $V = \hbox{diag}(v_1,\dots,v_n)$, the $st$ coefficient of (1) can be expanded as $$ \sum_{s=s_0,s_1,\dots,s_m=t} a_{s_0 s_1} \dots a_{s_{m-1} s_m}\sum_{0 \leq q \leq r \leq m} v_{s_q} v_{s_r}.$$ But the quadratic form $$ \sum_{0 \leq q \leq r \leq m} x_q x_r = \frac{1}{2}(x_0+\dots+x_m)^2 + \frac{1}{2} x_0^2 + \dots + \frac{1}{2} x_m^2$$ is positive definite, and the $a_{st}$ are non-negative, and the claim follows. [For the record, I found this argument while performing a perturbative analysis in the case where $P$ was close to $I$, or more precisely $P = (I-A)^{-1}$ for some $A$ with small non-negative entries.]<|endoftext|> TITLE: Why is there no Brauer scheme? QUESTION [37 upvotes]: Let $X$ be a proper scheme over a base field $k$ (one could consider more general settings, but I am primarly interested in a "geometric" situation with $k$ being algebraically closed). Then the Picard functor of $X$ is representable by the Picard scheme $Pic(X)$ of $X$, whose set of $k$-points is the Picard group $H^1 (X, \mathcal{O}_X^*)$. A natural generalization is to replace $H^1$ by $H^i$ for $i \geq 2$. For example, for $i=2$, we obtain the Brauer functor, whose set of of $k$-points is the Brauer group $H^2 (X, \mathcal{O}_X^*)$. Given the fact that I have never seen general existence results for a "Brauer scheme" and given some facts I learned from some people, it seems that the Brauer functor is not representable. So my first question is: 1) What is known about the representability of the Brauer functor? Is it representable in some cases, under which conditions? When I asked someone who might know the answer, he told me that the non-representability of the Brauer group is something related to the result due to Mumford that the Chow group of 0-cycles of some surfaces is "too big" but I don't really understand the relation. So: 2) What is the obstruction to the representability of the Brauer functor? What is the relation with the size of some Chow groups? I am interested in the questions 1) and 2) for any $i \geq 2$ and not just the Brauer case $i=2$. If the $i \geq 2$ case is not representable in general but the $i=1$ Picard case is representable, it is natural to ask: 3) What is the difference between the cases $i=1$ and $i \geq 2$ ? What is the "miracle" which does that the potential "bad things" happening for $i \geq 2$ do not happen for $i=1$? REPLY [30 votes]: Suppose that $Br(X)$ is representable in the following sense: there exists a $k$-scheme $B_X$ such that for each $k$-scheme $S$ there is a natural bijection $B_X(S)=Br(X_S)$, or perhaps we should rigidify by asking for $B_X(S)=Br(X_S)/Br(S)$. In any case, since $B_X(k)\rightarrow B_X(l)$ is injective for all field extensions $k\rightarrow l$, we see that $Br(X)\rightarrow Br(X_l)$ or $Br(X)/Br(k)\rightarrow Br(X_l)/Br(l)$ would have to be injective. There is no reason for this to be true in general. For instance, take an elliptic curve $E$ defined over $\mathbb{Q}$. Then, there is a short exact sequence $$0\rightarrow Br(\mathbb{Q})\rightarrow Br(E)\rightarrow H^1_{et}(k,E)\rightarrow 0.$$ Taking the extension $\mathbb{Q}\rightarrow\overline{\mathbb{Q}}$ in the argument above we have $Br(E)/Br(\mathbb{Q})$ is non-zero (it's in fact typically very big), while $Br(E_{\overline{\mathbb{Q}}})=0$. The basic issue is that for $i>1$, the assignment $S\mapsto H^i_{et}(X_S,\mathbb{G}_m)$ is just not a sheaf in the étale topology over $Spec\, k$. One can go in a different direction and look at the Picard stack $Pic=K(\mathbb{G}_m,1)$ and its classifing stack, $BPic=K(\mathbb{G}_m,2)$. This is a higher stack in the sense of Simpson, which in my indexing scheme (which I hope agrees with others') would be a $2$-stack. It precisely represents the Brauer group, but sections $BPic(X)$ also have higher homotopy groups: $$\pi_iBPic(X)=\begin{cases} H^2_{et}(X,\mathbb{G}_m)&\text{if $i=0$,}\\Pic(X)&\text{if $i=1$,}\\\mathbb{G}_m(X)&\text{if $i=2$}\\0&\text{otherwise.}\end{cases}$$ This story appears in Toën's paper on derived Azumaya algebras.<|endoftext|> TITLE: Existence of certain "nondegenerate" function and manifold topology QUESTION [8 upvotes]: Let $M$ be a smooth manifold without boundary, not necessarily compact. Let $f$ be a real-valued smooth function on $M\times M$. We say $f$ is good if for any point $(x,y)\in M\times M$ with local coordinates $x_i,y_j$ nearby, the matrix $\left[\dfrac{\partial^2f}{\partial x_i\partial y_j}\right]$ is nondegenerate. It should be clear that this concept is well-defined. I am wondering for what $M$, such a good function may exist. Clearly, $\mathbb{R}^n$ admits such good function by setting $f(x,y)=x\cdot y$. It seems I can prove that if such a good function exists, then $M$ is orientable and affine (admitting a flat connection). It also occurs to be that flat torus $T^n$ cannot support such a function. Thank you! REPLY [7 votes]: The question has a negative answer if $M$ is compact. The reason for this is that any smooth function on a compact manifold has at least one critical point. However, if you impose a weaker nondegeneracy condition, namely that the above matrix is nondegenerate only along the diagonal of $M\times M$, then the question has a positive answer and some interesting geometric consequences. Here are the details. For any $f\in C^\infty(M\times M$ denote by $B_f(x,y)$ the matrix you wrote viewed as a bilinear map $\newcommand{\bR}{\mathbb{R}}$ $$ T_xM\times T_yM\to \bR. $$ Equivalently we can view this as a linear map $$T_xM\to T^*_y M . $$ Note that $$B_{f+g}= B_f+ B_g. $$ Seek $f$ as a sum of functions of the from $u(x)v(y)$, $u,v\in C^\infty(M)$. The vector space $\newcommand{\eF}{\mathscr{F}}$ $\eF$ of functions of this type is dense in $C^\infty(M\times M)$. Assume from now on that $M$ is compact. Set $m:=\dim M$. If there existed a function $f\in C^\infty(M)$ such that $B_f(x,y)$ is nondegenerate for any $x,y\in M$, then you could find $h\in \eF$ such that $B_h(x,y)$ is nondegenerate for any $x,y\in M$. Observe that $$ B_{uv}(x,y)= du(x)\otimes dv(y).$$ The linear map $B_{uv}:T_xM\to T^*_y M$ has the form $$ T_xM\ni X\mapsto (Xu)(x) dv(y)\in T_y^* M. $$ Its adjoint $B_{uv}^*: T_yM\to T_x^*M$ is also nondegenerate and it is given by $$ T_yM\ni Y\mapsto (Yv)(y)du(x)\in T_x^* M. $$ Suppose that $$f=\sum_{k=1}^N u_k(x)v_k(y). $$ Then $$ B_f=\sum_{k=1}^N du_k(x)\otimes dv_k(y) $$ If $B_f(x,y)$ is nondegenerate, we deduce that $B_f(x,y): T_xX\to T^*_y M$ is onto and we conclude that the the covectors $dv_1(y),\dotsc, dv_N(y)$ span $T_y^* M$, for any $y$. Similarly, the covectors $du_1(x),\dotsc, du_N(x)$ span $T_xM$ for any $x\in M$. These two conditions can be conveniently rephrased as follows: the smooth maps $ \Phi,\Psi: M\to\bR^N$ defined by $$\Phi(x)= \bigl(\; u_1(x),\dotsc, u_N(x)\;\bigr),\;\; x\in M, $$ $$\Psi(y)=\bigl(\; v_1(y),\dotsc, v_N(y)\;\bigr),\;\;y\in M $$ are immersions. Fix a finite dimensional vector Euclidean space $E$ (think of $E$ as the above $\bR^N$) and immersions $$\Phi, \Psi: M\to E. $$ For every $x\in M$ we have inclusions $$i_{\Phi,x}, i_{\Psi,x}: T_x M\to E, $$ and surjections $$ P_{\Phi,x}, P_{\Psi,x}: E^*\to T_x^* M. $$ More precisely $i_{\Phi_x}= D_x\Phi$, $P_{\Phi_,x}= (D_x\Phi)^*$, where $D_x\Phi$ is the differential of $\Phi$ at $x\in M$. If $\DeclareMathOperator{\Gr}{\boldsymbol{Gr}}$ if $\Gr_m(M)$ denotes the Grassmannian of $m$-dimensional subspaces of $E$, then the maps $$ M\ni X \mapsto D_x\Phi(T_xM), D_x\Psi(T_xM)\in \Gr_m(E) $$ are the Gauss maps of the immersions $\Phi$ and $\Psi$. Using the metric on $E$ we can regard $P_{\Phi,x}$ as a surjection $E\to T_xM$, the orthogonal projection onto the subspace $T_xM\subset E$. In the compact case the problem you ask is now reduced to the following problem. $\mathbf{P_1}$ Find immersions $\Phi,\Psi: M\to E$, $E$ Euclidean space such that for any $x,y \in M$ the operator $$ P_{\Psi_y}\circ i_{\Phi_x}=(D_y\Psi)^*D_x\Phi: T_xM \to T_yM $$ is onto? Denote by $g: E\times E\to\bR$ denotes the inner product on $E$. Problem $\mathbf{P_1}$ is equivalent to the following. $\mathbf{P_2}$ Find immersions $\Phi,\Psi: M\to E$ such that, if $$ f:=(\Phi\times \Psi)^* g, $$ then $B_f(x,y)$ is nondegenerate for any $x,y\in M$. The last formulation shows that the example $f=x\cdot y$ you presented in your question is in a sense universal. However, the formulation $\mathbf{P_1}$ is the key to answering your question. Suppose that $\Phi$ is an immersion of $M$ in some Euclidean space $E$ of dimension $N$ with inner product $g$. Consider the varieties $$ N_\Phi:=\bigl\{ (x,p)\in E\times M;\;\; |v|=1,\;\;v\perp D_p\Phi(T_pM)\;\bigr\}, $$ $$ T_\Phi:=\bigl\{ (w,p)\in E\times M;\;\; |w|=1,\;\;w\in D_p\Phi(T_pM)\;\bigr\},$$ Note that $\dim N_\Phi= N-1$. Denote by $S(E)$ the unit sphere in $E$. We have two maps $$F_\Phi: N_\Phi\to S(E),\;\;(v,p)\mapsto v, $$ $$G_\Phi: T_\Phi\to S(E),\;\; (w,p)\mapsto w. $$ Problem $\mathbf{P_1}$ is equivalent to finding two immersions $\Phi,\Psi: M\to E$ such that the ranges of $F_\Phi$ and $G_\Psi$ are disjoint. We will show that this impossible on acount of the fact that $F_\Phi$ is onto. To see this choose $v\in S(E)$. The linear map $L_v: E\to \bR$, $L_v(x)= g(v,x)$ ($g$=the inner product on $E$) induces the function $L_v^\Phi=L_v\circ \Phi$ on $S^1$. If $p\in M$ is a critical point of $L_v^\Phi$, then $(v,p)\in N_\Phi$ which shows that $v$ is in the range of $F_\Phi$. Remark. Suppose that we require a weaker nondegeneracy condition, namely that the bilinear from $B_f(x,y)$ be nondegenerate when $x=y$. Then, as explained in this post, this bilinear form defines a canonical torsion free connection $\nabla^f$ on $TM$. More precisely, if we set $\newcommand{\pa}{\partial}$ $$ s_{jk}(x,y)=\pa^2_{x^jy^k}f(x,y), $$ and we denote by $(s^{k\ell}(x))$ the inverse of $(s_{k\ell}(x))$, then the Christoffel symbols of this connection are given by $$ \Gamma^j_{ki}(x)=\sum_{\ell} s^{j\ell}(x)\frac{\pa^3}{\pa x^\ell\pa y^k\pa y^i}f(x,x). $$ The question is then for which manifolds $M$ there exist smooth functions $f:M\times M\to \bR$ such that, for any $x\in M$ the matrix $$\bigl(\; \pa^2_{x^jy^k}f(x,x)\;\bigr)_{1\leq j,k\leq m}$$ is invertible. Here is a universal example. Suppose that $M$ is a submanifold of $\bR^N$. Then the function $F:\bR^n\to\bR^n\to\bR$, $F(x,y)=x\cdot y$ restricts to a function $f$ on $M\times M$ satisfying the above weaker nondegeneracy condition. In fact, for $x\in M$ the bilinear form $B_f(x,x)$ on $T_xM$ is the inner product on $T_xM$ induced by the Euclidean inner product on $\bR^N$ so that $B_f(x,x)$ is the induced Riemann metric. As I explained in this paper the above connection is compatible with the metric and, since it is also torsion free, it must be the Levi-Civita connection !<|endoftext|> TITLE: What are the "correct" conventions for defining Clifford algebras? QUESTION [32 upvotes]: I have three related questions about conventions for defining Clifford algebras. 1) Let $(V, q)$ be a quadratic vector space. Should the Clifford algebra $\text{Cliff}(V, q)$ have defining relations $v^2 = q(v)$ or $v^2 = -q(v)$? 2) Should $\text{Cliff}(n)$ denote the Clifford algebra generated by $n$ anticommuting square roots of $1$ or by $n$ anticommuting square roots of $-1$? That is, after you pick an answer to 1), should $\text{Cliff}(n)$ be $\text{Cliff}(\mathbb{R}^n, \| \cdot \|)$ or $\text{Cliff}(\mathbb{R}^n, - \| \cdot \|)$? More generally, after you pick an answer to 1), should $\text{Cliff}(p, q)$ be the Clifford algebra associated to the quadratic form of signature $(p, q)$ or of signature $(q, p)$? 3) Let $(X, g)$ be a Riemannian manifold with Riemannian metric $g$. After you pick an answer to 1), should the bundle of Clifford algebras $\text{Cliff}(X)$ associated to $X$ be given fiberwise by $\text{Cliff}(T_x(X), \pm g_x)$ or by $\text{Cliff}(T_x^{\ast}(X), \pm g_x^{\ast})$? For 1), on the one hand, $v^2 = q(v)$ seems very natural, especially if you think of the Clifford algebra functor as a version of the universal enveloping algebra functor, and it is used in Atiyah-Bott-Shapiro. On the other hand, Lawson-Michelson and Berline-Getzler-Vergne use $v^2 = -q(v)$, I think because they want $\text{Cliff}(\mathbb{R}^n, \| \cdot \|)$ to be the Clifford algebra generated by $n$ anticommuting square roots of $-1$. This is, for example, the correct Clifford algebra to write down if you want to write down a square root of the negative of the Laplacian (which is positive definite). For 2), this choice affects the correct statement of the relationship between $\text{Cliff}(n)$-modules and real $K$-theory, but there is something very confusing going on here, namely that with either convention, $\text{Cliff}(n)$-modules are related to both $KO^n$ and $KO^{-n}$; see Andre Henriques' MO question on this subject. For 3), whatever the answer to 1) or 2) I think everyone agrees that $\text{Cliff}(X)$ should be given fiberwise by $n$ anticommuting square roots of $-1$, where $n = \dim X$, so once you fix an answer to 1) that fixes the signs. The choice of sign affects the correct statement of the Thom isomorphism in K-theory. Lawson-Michelson use the tangent bundle but Berline-Getzler-Vergne use the cotangent bundle. The tangent bundle seems natural if you want to think of Clifford multiplication as a deformation of a covariant derivative, and the cotangent bundle seems natural if you want to think of the Clifford bundle as a deformation of exterior forms. I'm not sure how important this choice is. Anyway, I just want to know whether there are good justifications to sticking to one particular set of conventions so I can pick a consistent one for myself; reconciling the conventions of other authors is exhausting, especially because I haven't decided what conventions I want to use. REPLY [46 votes]: This is not really an answer, but rather a meta-answer as to why there exist many conventions in the first place. The symmetric monoidal category $\mathit{sVect}$ of super-vector spaces has a non-trivial involution $J$. The symmetric monoidal functor $J:\mathit{sVect}\to \mathit{sVect}$ is the identity at the level of objects and at the level of morphisms. But the coherence $J(V \otimes W) \xrightarrow{\cong} J(V) \otimes J(W)$ is non-trivial. It is given by $-1$ on $V_{odd} \otimes W_{odd}$ and $+1$ on the rest. The image of $\mathit{Cliff}(V,q)$ under $J$ is $\mathit{Cliff}(V,-q)$. So anything that you do with one convention can equally well be done with the other convention. Over the complex numbers, $J$ is equivalent to the identity functor. The symmetric monoidal natural transformation $J\Rightarrow Id$ that exhibits the equivalence acts as $i$ on the odd part and as $1$ on the even part of any super-vector space. Over the reals, $J$ is not equivalent to the identity functor, as can be seen from the fact that $\mathit{Cliff}(\mathbb R,|\cdot|^2)\not\simeq\mathit{Cliff}(\mathbb R,-|\cdot|^2)$. One last technical comment: Over $\mathbb C$, the action of $\mathbb Z/2$ on $\mathit{sVect}$ defined by $J$ is still non-trivial, despite the fact that $J$ is trivial. A trivialization of the action isn't just an equivalence $\alpha:J\cong Id$. For such an equivalence to trivialize the action, it would need to satisfy the further coherence $\alpha\circ \alpha = 1$, which isn't satisfied by any choice of $\alpha$. (To trivialize the action of a group $G$, one needs to trivialize the actions of each $g\in G$ in such a way that the trivializations of $g,h\in G$ compose to the trivialization of $gh$.) Now, as far as practical things are concerned, I would recommend minimizing the number of minus signs that you end up writing down.<|endoftext|> TITLE: Find all faces in a graph from list of edges QUESTION [5 upvotes]: I have the information from a undirected graph stored in a 2D array. The array stores all of the edges between nodes, e.g. graph[3] might be equal to [1,8,30] and represents the fact that node 3 shares edges with nodes 1 8 and 30. As the graph is undirected, graph[8] will also contain the value 3. I want to find an algorithm that will find all of the faces of the graph (my graph-theoretical knowledge is limited, I am essentially looking for all of the cycles that don't contain a smaller cycle within them), and provide the path for the boundary of each of those faces (e.g. 1->5->9->3->1). It is safe to assume that the graph I have is both planar and connected. With limited knowledge of graph-theory concepts I'd like to avoid getting too lost halfway through implementation, so simplicity is probably more valuable than efficiency. That said, the algorithm must not be horribly inefficient. REPLY [4 votes]: Expanding a bit on the comments to Federico's answer: If the neighbors of each vertex are listed in counterclockwise order, this defines an embedding on a surface, and if this information was obtained from a given planar embedding, this is the one you get back. Essentially, for a given edge $(n_1,n_2)$ you can find one of the faces it borders (and, analogously, the other one by considering $(n_2,n_1)$ instead) by looking up $n_1$ in the list of $n_2$'s neighbors, finding the next one, say $n_3$, in the (cyclic) order, and continuing with $(n_2,n_3)$, and then $(n_3,n_4)$ etc until you return to $n_1$. Note that the non-uniqueness is only an issue if $G$ is not $3$-edge-connected. If it is, even if the ordering at the nodes is not given, you should still have a unique embedding into the sphere (and into the plane, modulo choice of the outer face). Not sure how to find it, though.<|endoftext|> TITLE: Are angles between points enough to decide the realizability? QUESTION [8 upvotes]: Let n points in the plane be given whose coordinates we don't know. Assume, however, that for any triple of the points we know the angle. Question: Can we decide whether the n points are realizable in the plane, i.e. can we determine whether there is an assignment of coordinates to the points such that the angular constraints are satisfied? REPLY [4 votes]: To answer Joonas's comment in the last paragraph: yes, it is possible to check the realizability without constructing the whole realization. It is sufficient to test whether every 5-tuple of points is realizable. (For simplicity, I am also assuming general position, that is, no three points on a line.) To see this, note that when a new point is being added in Joonas's algorithm, its location is determined by an arbitrary triple of points already drawn. The new point can be drawn if all the triples agree on the same location. If two triples disagree, then there are also two triples sharing two points that disagree. Together with the new point, this makes a 5-tuple that is not realizable. By some finite case analysis of all 5-tuples, it can be probably shown that testing 4-tuples is also sufficient.<|endoftext|> TITLE: Representations of $\text{SL}(n,\mathbb{F}_p)$ and $\text{Sp}(2n,\mathbb{F}_p)$ whose dimensions are $p^k$ QUESTION [9 upvotes]: I should preface this by saying that I am not a representation theorist, so I apologize if this can easily be found in standard sources (but sadly I cannot seem to extract it from any of the books I know of). Let $p$ be a prime number. What are all the irreducible representations (over $\mathbb{C}$) of $\text{SL}(n,\mathbb{F}_p)$ and $\text{Sp}(2n,\mathbb{F}_p)$ whose dimensions are $p^k$ for some $k$? There are the Steinberg representations, but I assume that there are also others. I'm particularly interested in the symplectic group. REPLY [8 votes]: A full classification of such representations (and much more) can be found here: Prime power degree representations of quasi-simple groups by Malle and Zalesskii You can read this paper here. The main theorem implies that, apart from the Steinberg representation, there are no representations of this form. Interestingly this is not the case if you allow quasisimple covers of such groups, for instance $2\cdot {\rm Sp}_6(2)$ has an irreducible complex representation of degree $2^9$. As I mentioned in my comment, if you want to understand the theory of these representations then I would go to Finite groups of Lie type by Carter. I should also say that it's quite possible that the fact you seek can be proved more directly than via the full classification given by Malle and Zalesskii ... but I don't know enough of D-L theory to tell you how. P.S. if anyone knows how to put the correct accent on Zalesskii's name, then I'd like to know how!<|endoftext|> TITLE: Positivity of Ehrhart polynomial coefficients QUESTION [8 upvotes]: Are there any results stating that a given family of convex polytopes have Ehrhart polynomials with non-negative coefficients? What methods are available for proving such a property for some family of polytopes? Remember, the Ehrhart polynomial $p(k)$ for a convex polytope $P$ with integer vertices is given by the property that $p(k)$ counts the number of integer lattice points in the $k$-dilation of $P$, where $k$ is a positive integer. REPLY [3 votes]: Fu Liu recently put on the arXiv a nice survey about this question: https://arxiv.org/abs/1711.09962v1<|endoftext|> TITLE: Triangle inequality for $L^1$-norm with respect to a state QUESTION [7 upvotes]: It is well-known that the naive construction of non-commutative $L^p$-spaces is performed only in tracial case. I would like to know if it is really a necessity. To wit, let $\varphi$ be a normal state on a von Neumann algebra $M$. Suppose that the triangle inequality for the $L^1$-norm induced by $\varphi$ holds, i.e. $$ \varphi(|x+y|)\leqslant \varphi(|x|) + \varphi(|y|), $$ where $x,y \in M$ and $|x|:= \sqrt{x^{\ast}x}$. Is it true that $\varphi$ is a trace ($\varphi(x^{\ast}x)=\varphi(xx^{\ast})$)? What if $\varphi$ is only a normal weight? REPLY [3 votes]: If $\varphi$ is not a trace, M contains a von Neumann subalgebra isomorphic to $M_2(\mathbf C)$ on which the restriction of $\varphi$ is not a trace. Indeed, if $x \in M$ is such that $\varphi(x^*x) \neq \varphi(xx^*)$, by the normality assumption on $\varphi$ we can assume that $|x|$ has a finite spectrum, and by linearity that $x$ is a partial isometry. Replacing the projections $p=x^*x$ and $q=xx^*$ by (the still equivalent projections) $p-p\hat{}q$ and $q-p\hat{}q$, we can assume that $p$ and $q$ are orthogonal projections. This implies that the von Neumann algebra generated by $x$ is isomorphic to $M_2(\mathbf C)$. This reduces the problem to $M_2(\mathbf C)$. In this case I guess that a variant of zanin's answer should apply.<|endoftext|> TITLE: Why care about Fourier-Mukai partners? QUESTION [5 upvotes]: Two (smooth, projective, complex?) varieties are called Fourier-Mukai partners if they have equivalent derived categories of coherent sheaves. On the other hand, my general impression is that cool people "don't care" so much about plain old derived categories anymore, preferring instead to look at souped up things like dg-categories/A-infinity categories/... (for example, triangulated vs. dg/A-infinity). At this level, FM partners are no longer equivalent. From the point of view of dg-categories/infinity categories, what does it mean for two varieties to be Fourier-Mukai partners? Why keep studying these things? REPLY [9 votes]: Basically Adeel has already answered this question. First you know that the triangulated category D(X) has an essentially unique enhancement (Lunts-Orlov). This means that you can freely interpret D(X) as a dg(or $A_\infty$ or $\infty$ or)-category. Second, we know that any functor D(X) --> D(Y) (say with X and Y smooth and projective) is given by a Fourier-Mukai transform - and hence it's also a dg-functor. So, two varieties are partners in the triangulated sense if and only if they are partners in the dg-sense. The point of the dg-world is that it makes constructions behave as you'd want them to behave (like taking cones and gluing). This of course comes at the cost of having two keep track of more data (the dg structure).<|endoftext|> TITLE: Are there irreducible ideals that are not primary in $K[X_1,\dots,X_n,\dots]$? QUESTION [11 upvotes]: I can give examples of non-noetherian rings having irreducible ideals that are not primary. Among them there are idealizations and valuation domains. But the first non-noetherian ring we are thinking about is $K[X_1,\dots,X_n,\dots]$, $K$ a field. The finitely generated ideals of this ring have primary decomposition, so if they are irreducible then are necessarily primary. My question is the following: Are there irreducible ideals that are not primary in $K[X_1,\dots,X_n,\dots]$? REPLY [2 votes]: Another example using idealization construction (i.e. $(a,b)(c,d)=(ac,ad+bc)$): start with $R=k[T]_{(T)}+k(T)$, with $k=$prime field contained in $K$; primary ideals of $R$ are $(0)+(0)$, $(0)+k(T)$, and $(T^n)+k(T)$ $(n>0)$; the ideal $I=(0)+k[T]_{(T)}$ is irreducible and not primary; $R$ is countable so it's a quotient of $k[X_1,X_2,...]$; extend scalars from $k[X_1,X_2,...]$ to $K[X_1,X_2,...]$<|endoftext|> TITLE: Are convolution algebras ever "topologically noetherian"? QUESTION [5 upvotes]: For finite groups $G$, we have the group ring $k[G]$, and we can think of $G$-representations as $k[G]$-modules. It is known that for $G$ virtually polycyclic, $k[G]$ is a Noetherian ring, which means that if $V$ is a finitely generated $G$-rep, then all its submodules are also finitely generated. If $G$ is a topological group with an interesting topology, then $k[G]$ is almost never what you actually want to look at, since it ignores the topology on $G$. Indeed, what we care about are continuous representations of $G$, and there are various continuous group algebras in this case - $C(G)$, $L^{1,2,∞}(G)$, the algebra of Radon measure $M(G)$, and many variants - with multiplication again given by convolution. Here the analogous property should be "topologically Noetherian". We want a topologically Noetherian module to be one all of whose closed submodules are topologically finitely generated, and a topologically Noetherian ring to be one all of whose t.f.g. modules are top. Noeth. I'm pretty sure this reduces to the usual property of ACC on closed ideals, but if there's any subtlety, this seems like the right definition. So, are any of the various group algebras (absolutely including any not mentioned above) topologically Noetherian in this sense for nice $G$, say for $G$ a compact Lie group? My sense is that this is probably too much to hope for. REPLY [2 votes]: The completed group ring of a compact $p$-adic Lie group with coefficients in a field of characteristic $p$ is topologically Noetherian. In fact, it is even abstractly Noetherian. This is a theorem of M. Lazard from the 1960s, and you can read more about these rings here.<|endoftext|> TITLE: Ground State Degeneracy of 2+1D U(1) Chern Simons Theory? QUESTION [6 upvotes]: I am a physics graduate student trying to understand more mathematical aspects of gauge theories. How can I understand ground state degeneracy of a simple Chern Simons Theory: 2+1D U(1) $S= \int_M kAdA$ for different k- values (integers, rationals, irrationals?) on torus or other manifolds in a more mathematics/differential geometry oriented way? I have seen the calculation in physics literature, but I can't figure out how to translate it in differential geometry language. I know it must be there somewhere in literature, but I am not able to find it. REPLY [4 votes]: The Abeian Chern-Simons theory you write down, can be written in a very generic form with a symmetric bilinear integer matrix $K_{IJ}$ with a path integral (or partition function): $$ Z=\int DA \exp[i S]=\int DA \exp[i \int_{M^3} \frac{K_{IJ}}{4\pi} A_I dA_J] $$ The $\int DA$ integrates all possible $A$ configuration, while $\int_{M^3}$ integrates over the spacetime manifold. Please make sure that the level quantization $K_{IJ}$ and $4 \pi$ are in the correct form. The ground state degeneracy ($\equiv Gsd$) on a spatial manifold $M^d$ and a time $S^1$ can be computed from the partition function as $$ Z(M^d \times S^1)=\dim(\text{Hilbert space on } M^d)\equiv Gsd_{M^d}. $$ Below I give examples for various symmetric bilinear integer matrix $K_{IJ}$, and provide their ground state degeneracy on $T^2$ ($\equiv Gsd_{T^2}$), $$ Z(T^2 \times S^1)=\dim(\text{Hilbert space on } T^2)\equiv Gsd_{T^2}. $$ For $S=\int\frac{k}{4\pi} A dA$, with $k$ is an odd integer, it is a fermionic Chern-Simons TQFT which requires the manifold with a spin structure (i.e. spin manifold) to define the theory. It is fermionic in the sense that this TQFT can be realized in a Fermionic many-body quantum system (such as electrons, like filling-fraction 1/3 Langhlin quantum Hall state). $$ Gsd_{T^2}=k. $$ For $S=\int\frac{2k}{4\pi} A dA=\int\frac{k}{2\pi} A dA$, with $2k$ is an even integer, it is a bosonic Chern-Simons TQFT where the manifold with or without a spin structure can both define the theory. It is bosonic in the sense that this TQFT can be realized in a Bosonic many-body quantum system. For example the $Z_2$ topological order of Wen and $Z_2$ toric code of Kitaev, they are basically the same (bosonic) $Z_2$ gauge theory. $$ Gsd_{T^2}=2k. $$ For the most generic $S=\int\frac{K_{IJ}}{4\pi} A_I dA_J$, their ground state degeneracy on $T^2$ ($\equiv Gsd_{T^2}$), $$ Gsd_{T^2}=|\det(K)|. $$ is determined by the determinant of $K$. (1) For a more mathematical physics calculation, you may look for Belov-Moore: Classification of abelian spin Chern-Simons theories and Refs therein (those of Witten's). I believe that they provide the mathematics/differential geometry method and cite the Refs. (2) For physics, you can check those Refs involving counting the ground states via line operators or via the distinct classes of quantum wavefunction in the Hilbert space: such as Classification of Abelian quantum Hall states and matrix formulation of topological fluids: PhysRevB.46.2290, and Degeneracy of Topological Order: PhysRevB.91.125124 See also the more involved various non-Abelian Cherm-Simons theory Gsd and dimensions of Hilbert space.<|endoftext|> TITLE: What is the first eigenvalue of $p$-Laplacian on unit sphere $S^n$? QUESTION [11 upvotes]: We know that the first eigenvalue of Laplacian on the Riemannian unit sphere $S^n$ is $n$, then what is the explicit expression for the first eigenvalue of $p$-Laplacian on $S^n$? The $p$-Laplacian eigevalue equation is $-div(|\nabla u|^{p-2}\nabla u)=\lambda| u|^{p-2}u$, for any $p>1$. REPLY [5 votes]: For the case $n=1$, it is well known that for the domain $\Omega = (a,b)$, the first eigenvalue for the $p$-Laplacian is given by $$\lambda_1 = (p-1)(\frac{\pi_p}{b-a}),$$ where $$\pi_p = 2\int_0^1 \frac{ds}{\sqrt[p]{1-s^p}}.$$ For higher dimensions (n>1) where $p>1, p \neq 2$, the explicit expression is not known even for simple domains such as the sphere and the square. However, it is known that $\lambda_1$ is positive, simple, and the corresponding eigenfunction does not change sign. Several numerical algorithms exist for computing the first eigenvalue of the $p$-Laplacian. For an iterative method, see http://arxiv.org/pdf/1011.3172.pdf. For an gradient descent algorithm based on the Rayleigh quotient formulation, see http://arxiv.org/pdf/1106.0602.pdf.<|endoftext|> TITLE: How to compute $\pi_{3}$ of $L(p,q)\# L(p',q')$? QUESTION [12 upvotes]: Let $L(p,q)$ be a 3-dimensional lens space, and let $L(p',q')$ be another. Is there any known result concerning the 3rd homotopy group of the connected sum $L(p,q)\# L(p',q')$? If not, I am interested in computing it, but I have no idea. More generally, can we find a method to compute $\pi_{3}(M\# N)$ for general prime 3-manifolds $M$ and $N$? REPLY [8 votes]: Elaborating on Alex Suciu's first comment, let me offer a second way of computing $\pi_3(M\#N)$ which works for any pair of $3$-manifolds with non-trivial fundamental groups (i.e. none of them is $S^3$, which is uninteresting since it is a unit for $\#$). For any (CW-)space $X$ with universal cover $\tilde X$ there is an exact sequence (due to JHC Whitehead) $$H_4(\tilde X)\rightarrow\Gamma \pi_2(X)\rightarrow\pi_3(X)\rightarrow H_3(\tilde X)\rightarrow 0.$$ Here $\Gamma$ is Whitehead's universal quadratic functor (see below). If $X$ is a $3$-manifold with inifinite fundamental group then $\tilde X$ is an open $3$-manifold and $H_4(\tilde X)=H_3(\tilde X)=0$, hence $$\pi_3(X)=\Gamma \pi_2(X).$$ This applies to $X=M\#N$ since $\pi_1(X)=\pi_1(M)*\pi_1(N)$ is infinite. A map between abelian groups $\gamma\colon A\rightarrow B$ is quadratic if $\gamma(a)=\gamma(-a)$ and the map $A\times A\rightarrow B\colon (a_1,a_2)\mapsto\gamma(a_1+a_2)-\gamma(a_1)-\gamma(a_2)$ is bilinear. The abelian group $\Gamma A$ is characterized as the target of the universal quadratic map $A\rightarrow \Gamma A$. If $A$ is free abelian with basis $B\subset A$ then $\Gamma A$ is free abelian with basis $$\{\gamma(b)\,;\,b\in B\}\cup \{\gamma(b_1+b_2)-\gamma(b_1)-\gamma(b_2)\,;\,b_1 TITLE: Lower regularity version of Moser's theorem on volume elements QUESTION [6 upvotes]: A theorem of Moser, published in "On the Volume Elements of a Manifold" (Transactions of the Americal Mathematical Society 120, 1965; doi: 10.1090/S0002-9947-1965-0182927-5, jstor), shows that if a $C^\infty$ compact manifold $M$ has two $C^\infty$ volume forms $\omega_1$ and $\omega_2$ with the same total mass, then there is a diffeomorphism of $M$ sending one to the other. I am interested in what is known if the manifold and volume forms have lower regularity (in particular, I really want to know about the $C^{1+\alpha}$ case. Thanks for any reference suggestions. EDITED: So having had an answer from Robert Bryant, I realized I should have been more precise about the specific question(s) that I was asking: If $M$ is a $C^{1+\alpha}$ manifold and $\omega_i$, $i=1,2$ are two $C^\alpha$ volume forms with the same mass, does there exist a $C^{1+\alpha}$ diffeomorphism sending one the other? The comment below from AlvarezPaiva suggests the answer to the above might be yes, but the context there appears to be bounded subsets of $\mathbb R^n$. If $M$ is a $C^{1}$ manifold and $\omega_i$, $i=1,2$ are two continuous volume forms with the same mass, does there exist a $C^1$ diffeomorphism sending one to the other? Final full disclosure in case this drastically simplifies things: my manifold is topologically a two-dimensional torus. REPLY [11 votes]: If $M$ is a $C^{1}$ manifold and $\omega_i$, $i=1,2$ are two continuous volume forms with the same mass, does there exist a $C^1$ diffeomorphism sending one to the other? The answer is no. The equivalent problem is: given a positive and continuous function $g$, can we find a diffeomorphism with the Jacobian equal to $g$, $\det Df=g$? A counterexample is Theorem 1.2 in [1]. If $M$ is a $C^{1,\alpha}$ manifold and $\omega_i$, $i=1,2$ are two $C^\alpha$ volume forms with the same mass, does there exist a $C^{1,\alpha}$ diffeomorphism sending one the other? The answer is yes, at least locally (see also a comment of alvarezpaiva). Whether the result has a global version on manifolds, I do not know. This is Theorem 1 in [3]. For a comprehensive treatment of related results, see [2]. Theorem. Let $k\geq 0$ and $\alpha\in (0,1)$. If $\Omega\subset\mathbb{R}^n$ is abounded domain with $C^{k+3,\alpha}$ boundary and $\omega\in C^{k,\alpha}$ is a volume form such that $\int_\Omega\omega=|\Omega|$, then there is a diffeomorphism $\varphi:\Omega\to\Omega$ that is identity on the boundary, $\varphi,\varphi^{-1}\in C^{k+1,\alpha}(\bar{\Omega})$ and such that $\varphi^*\omega=dx_1\wedge\ldots\wedge dx_n$. [1] D. Burago, B. Kleiner, Separated nets in Euclidean space and Jacobians of bi-Lipschitz maps. Geom. Funct. Anal. 8 (1998), 273–282. (MathSciNet review.) [2] G. Csató, B. Dacorogna, O. Kneuss, The pullback equation for differential forms. Progress in Nonlinear Differential Equations and their Applications, 83. Birkhäuser/Springer, New York, 2012. (MathSciNet review.) [3] B. Dacorogna, J. Moser, On a partial differential equation involving the Jacobian determinant. Ann. Inst. H. Poincaré Anal. Non Linéaire 7 (1990), 1–26. (MathSciNet review.)<|endoftext|> TITLE: Weyl group actions on 0-weight spaces QUESTION [21 upvotes]: For a complex simple Lie group G with a maximal torus T, we can take a highest-weight representation V of G and look at the 0-weight space, i.e. the subspace of V of elements invariant under T. This subspace admits a representation U of the Weyl group W of G. Is there any method for determining the character of the representation U from the character of V? I feel like this should be known at least for the classical groups, but can't find any sources on it. REPLY [18 votes]: This is a reasonable question, and there has been some relevant literature over many decades. But my impression is that no definitive answer has been given, even for the classical types. I've intended for a long time to write my own survey on this topic, but meanwhile here are a few references and links: 1) Probably the earliest published work was in a very short (and untranslated) 1973 Russian paper by E.A. Gutkin Representations of the Weyl group in the space of vectors of zero weight here. Using the tensor product framework of Schur-Weyl theory, Gutkin studied the group $G = \mathrm{SL}_n(\mathbb{C})$ and its Weyl group $W=S_n$. He concluded that all irreducible representations of $S_n$ occur as zero weight spaces of irreducible $G$-representations. Of course there are infinitely many of the latter, but it's enough to consider those whose highest weights are (in the usual way) partitions of $n$; then the dual partitions parametrize the irreducible $W$-representations. 2) Though Gutkin's paper is not well referenced in other papers, there is a similar result in $\S4.1$ of a 1976 paper by B. Kostant here. Kostant went somewhat further with other simple Lie groups here. 3) Another 1976 paper, in the compact group framework of special unitary groups, also dealt with characters of symmetric groups in the setting of Schur-Weyl duality: D.A. Gay, Characters of the Weyl group of $\mathrm{SU}(n)$ on zero weight spaces and centralizers of permutation representations, Rocky Mt. J. Math. 6 (1976), 449-455. 4) A later paper by M. Reeder Zero weight spaces and the Springer correspondence, Indag. Math. 9 (1998), 431-441, extended the earlier results on symmetric groups to Lie types $D_n, E_n$. This occurs in the framework of his study of "small" representations, as in his paper Small representations and minuscule Richardson orbits, IMRN 2002, No. 5. He also gives numerous further references. 5) Yet another approach is found in a paper by S. Kumar and D. Prasad Dimension of zero weight space: An algebro-geometric approach, J. Algebra 403 (2014), 324-344 (not yet freely accessible online but posted in preprint form on arXiv:1304.4210). The picture derived from this and related literature is still incomplete, it seems, but intriguing. ADDED: I've been motivated to organize some notes online here. This is still a preliminary version, so email comments are welcome.<|endoftext|> TITLE: expression for infinite series with powers of factorial in denominator QUESTION [7 upvotes]: The series $$\sum_{k=0}^\infty \frac{\exp(c k \beta)}{(k!)^\beta} $$ has come up when I'm trying to apply the methodology in this paper (http://www.ism.ac.jp/~eguchi/pdf/Robustify_MLE.pdf) to Poisson regression. When $\beta = 1,$ it is $e^{e^c}$ and when $\beta = 2,$ it is $I_0(2e^c),$ the modified Bessel function of the first kind, but have no idea what happens at other values, or how to approach the question. Does this series have either a closed form expression in terms of special functions, or an asymptotic expansion that is numerically useful for $\beta \in [1, 5]$ and $|c| < R$? REPLY [5 votes]: I copied this from my comments above: This sum was discussed in the book "Advanced Mathematical Methods for Scientists and Engineers" by Carl M. Bender and Steven A Orszag in section 6.7 Example 4. The asymptotic result is (with $x=\exp(c \beta)$) $(2 \pi)^{(1-\beta)/2} \beta^{-1/2} x^{(1-\beta)/(2\beta)}\exp(\beta x^{1/\beta})$. Edit: Sketch of a proof following Bender's & Orszag's lines: Let us look at a somewhat simpler sum $$ \sum_{k=0}^{\infty} \frac{x^{k}}{k!^{\beta}}. $$ By examining the ratio of two consecutive terms of this sum we find that the terms are increasing until $k=k_{max}:=\lfloor x^{1/\beta}\rfloor$ and then decreasing. Using Stirling's formula and relaxing the requirement of $k$ being integer, we can expand $k!$ around $k_{0}$ by inserting $k=x^{1/\beta}+t$ $$ k! \sim x^{k/\beta} \exp(-x^{1/\beta}+t^{2}x^{-1/\beta}/2) \sqrt{2 \pi} x^{1/(2 \beta)} $$ for large $x$ and small $t$, i.e. $t^{3}\ll x^{2/\beta}$, which let us neglect higher terms in $t$. We find that the terms are sharply peaked for $k$ around $k_{0}$. Approximate the sum over $k$ by an integral over $t$, with the integration range extended to the whole real axis. (We can do this since the additional contributions to the integral are negligible.) Evaluating the integral gives the result. Edit: Getting higher orders for any real $\beta$ is straight forward but tedious with the above (Laplace-) method. However, going after chapter 5.11 in Luke's "The Special Functions and their Approximation", Vol. 1, one can give a rather compact formula for any order, when restricting $\beta$ to positive integers. Actually, it is the asymptotic expansion of the hypergeometric function $_{0}F_{\beta-1}(;1,...,1;\exp(c \beta))$, which is the same as the OP's sum. Write $x=\exp(c \beta)$. Then $$ _{0}F_{\beta-1}(;1,...,1;x)= (2 \pi)^{(1-\beta)/2} \beta^{-1/2} \exp(\beta x^{1/\beta}) x^{\frac{1}{2 \beta}-\frac{1}{2}} \sum_{k=0}^{\infty} c_{k} \beta^{-k} x^{-k/\beta} $$ where $c_{k}$ is recursively defined as $$ c_{k}=\frac{1}{\beta k} \sum_{s=1}^{\beta-1} c_{k-s} \sum_{r=0}^{s}\frac{(-1)^{s-r}}{r!(s-r)!} (r+k+\frac{1}{2}-\frac{\beta}{2})^{\beta} $$ and $c_{0}=1$, $c_{k}=0$ for $k<0$. I checked the formula with Mathematica for several values of $\beta$ and orders and it works fine.<|endoftext|> TITLE: A question about three forcings QUESTION [8 upvotes]: Let $P_1$ be the finite support iteration of random forcings of length $\omega$. Let $P_2 = \text{Random} \times \text{Random}$ and $P_3 = \text{Random} \times \text{Cohen}$. Are $P_i, P_j$, for $1 \leq i < j \leq 3$, forcing isomorphic? All I know is that all of them add a $P_3$-generic. REPLY [3 votes]: As you mentioned in a comment, in a $P_2$-extension there are two random reals the sum of which is Cohen. However, forcing with Cohen over $V[\mathrm{Random}]$ doesn´t add any real that is random over $V$. So the sum of two random reals in a $P_3$-extension cannot be Cohen, since Random doesn´t add Cohen reals. This argument can be found in Perfect sets of random reals by Brendle and Judah, at the beginning of section 3.<|endoftext|> TITLE: Is there a proof of Warning's Second Theorem using p-adic cohomology? QUESTION [13 upvotes]: Let $\mathbb{F}_q$ be a finite field, $n \in \mathbb{Z}^+$, and $f_1,\ldots,f_r \in \mathbb{F}_q[t_1,\ldots,t_n]$ with $\operatorname{deg}(f_i) = d_i$. Put $d = \sum_{i=1}^n d_i$ and suppose $d< n$. Let $V(f_1,\ldots,f_r) = \{ x \in \mathbb{F}_q^n \mid f_1(x) = \ldots = f_r(x) = 0\}$. The papers C. Chevalley, Demonstration d’une hypothese de M. Artin Abh. Math. Sem. Univ. Hamburg 11 (1935), 73–-75 and E. Warning, Bemerkung zur vorstehenden Arbeit von Herrn Chevalley. Abh. Math. Sem. Hamburg 11 (1935), 76–-83 contain the following results: Chevalley's Theorem: We have $\# V(f_1,\ldots,f_r) \neq 1$. Warning's First Theorem: We have $\operatorname{char}(F) \mid \# V(f_1,\ldots,f_r)$. Warning's Second Theorem: If $V(f_1,\ldots,f_r)$ is nonempty, its cardinality is at least $q^{n-d}$. The first two results are often combined as the "Chevalley-Warning Theorem", and the last result seems to be too often overlooked. The Chevalley-Warning Theorem is the beginning of a long story of p-adic estimates on the number of $\mathbb{F}_q$-rational points on such $\mathbb{F}_q$-schemes, culminating in the Ax-Katz Theorem which determines the minimal $p$-adic valuation of $\# V(f_1,\ldots,f_r)$ as $f_1,\ldots,f_r$ range over all polynomials of degrees $d_1,\ldots,d_r$. Famously, Katz used p-adic cohomology to prove his theorem. Other, more elementary proofs have since been given, but this approach remains a very natural one. Question: Is there a proof of Warning's Second Theorem using $p$-adic cohomology? Does such a proof appear in the literature? Warning's proof of his second theorem is an ingenious elementary argument of the sort that nowadays goes under the moniker "Polynomial Method". See this paper of Heath-Brown which includes Warning's proof as a point of departure to further work. (To the best of my knowledge, Heath-Brown's paper is the only one which directly follows up on Warning's Second Theorem. I would be very interested to learn of others.) A more modern Polynomial Method approach to Warning's Second Theorem which leads to a different kind of generalization is discussed here. Added: To address Daniel Litt's question raised in the comments: I do not know how to deduce Warning's Second Theorem (or even his first, for that matter) from the Weil Conjectures. The fact that we are making no "geometric" assumptions on our affine subscheme would make such a deduction a forbidding endeavor for me....but I do not claim that it cannot be done. I would be happy to broaden the question to ask about a cohomological proof of Warning II: you get to pick the cohomology theory. REPLY [5 votes]: I do not think that there is such a proof in the litterature, but as you have guessed, $p$-adic cohomology has something to do with it. Let $X=V(f_1,\dots,f_r)$. Let $K$ be the field of fractions of the Witt vectors of $k$, and let $\phi$ be the unique automorphism of $K$ such that $\phi(a)\equiv a^p \pmod p$. The rigid cohomology groups with compact supports of $X$, $H^i_{\rm rig,c}(X/K)$, are $\phi$-isocrystals, namely finite dimensional $K$-vector spaces endowed with a $\phi$-linear automorphism $\Phi$. If $k=\mathbf F_q$, with $q=p^r$, the Lefshetz trace formula in rigid cohomology asserts that $$ \# X(k) = \sum (-1)^i \mathop{\rm Tr}(\Phi^r|H^i_{\rm rig,c}(X/K)). $$ (Note that $\Phi^r$ is linear, so that the trace makes sense.) The theory of slopes of $\Phi$-isocrystals shows that Warning's theorem would follow from the condition that all slopes of $H^i_{\rm rig,c}(X/K)$ are $\geq 1$. As you said, it does not seem to be known without any smoothness assumption. Moreover, there might be cancellations in the trace formula that make this expectation wrong (see however this paper of Esnault and Katz). However, there is a nice theorem of Berthelot, Esnault and Rülling in this direction: Let $X$ be a flat, regular and proper $R$-scheme whose generic fiber $X_K$ is geometrically connected and satisfies $H^i(X_K,\mathscr O_{X_K})=0$ for $i>0$. Then $\# X(k)\equiv 1\pmod q$. The hard part of the proof consists in proving that the Witt vector cohomology vanishes in positive degrees.<|endoftext|> TITLE: Bound on sum of complex summands involving binomial coefficients QUESTION [7 upvotes]: I am trying to find the asymptotic behavior of the sum: $$ \sum^n_{i=0} \begin{pmatrix} 2n \\ i \end{pmatrix} x^i y^{2n-i}$$ as $n\rightarrow\infty$. Here $x$, $y$ are complex numbers and I have $|x|\leq1$ and $|y|\leq1$. I also know that $|x+y|\leq1$ and $|4xy|\leq1$. One of the approaches I thought about was to approximate the sum by an integral. However, as it turns out, this approach runs into trouble since the terms in the sum do not give a sufficiently smooth function (I think: we have complex numbers raised to non-integer powers thus giving phase ambiguities). On the other hand this sum may be related to Hoeffding's inequality applied to binomial distribution. This inequality gives an exponential upper bound. However, since $x, y$ are complex and $|x+y|\leq1$, I don't know how to apply this inequality. Is there a way to bound this sum? I need to show it decays to $0$ as $n\rightarrow\infty$. REPLY [5 votes]: This is a comment to Lucia’s answer that doesn’t fit into the comment box. In the case $|x|=|y|=1/2$, put $\omega=x/y$, and $$S_k=\sum_{j=0}^k\omega^j=\frac{1-\omega^{k+1}}{1-\omega}.$$ Then by partial summation, $$\sum_{k=0}^n\binom{2n}kx^iy^{2n-i}=y^{2n}\left[\binom{2n}nS_n-\sum_{k=0}^{n-1}S_k\left(\binom{2n}{k+1}-\binom{2n}k\right)\right],$$ the absolute value of which is bounded by $$4^{-n}\frac2{|1-\omega|}\left[\binom{2n}n+\sum_{k=0}^{n-1}\left(\binom{2n}{k+1}-\binom{2n}k\right)\right]\le\frac{4^{1-n}}{|1-\omega|}\binom{2n}{n}\le\frac{4+o(1)}{|1-\omega|\sqrt{\pi n}}.$$<|endoftext|> TITLE: A curious property of Ramanujan's function $\tau(n)$ QUESTION [11 upvotes]: As it is well known, Ramanujan's $\tau(n)$ function can be defined through the power series expansion of the modular discriminant: $$\Delta(q)=q\prod\limits_{n=1}^\infty (1-q^n)^{24}=\sum \limits_{n=1}^\infty \tau(n)q^n=q-24q^2+252q^3-1472q^4+4830q^5+\ldots.$$ In the short paper http://arxiv.org/abs/1408.2083 (Moonshine and the Meaning of Life, by Yang-Hui He and John McKay) a curious observation was made that $$\sum \limits_{n=1}^{24}\tau(n)^2\equiv 42 \;\; (\mathrm{mod} \;70).$$ Another observation of the same paper is that $$\sum \limits_{n=1}^{24}c(n)^2\equiv 42 \;\; (\mathrm{mod} \;70),$$ where $c(n)$ are defined through the power series expansion of the $SL_2(\mathbb{Z})$ elliptic modular function $j(q)$: $$j(q)=\sum \limits_{n=-1}^\infty c(n)q^n=q^{-1}+744 + 196884q + 21493760q^2+\ldots.$$ In the abstract, the authors write that "The observation is purely for the sake of entertainment and could be of some diversion to a mathematical audience". Nevertheless, is there any deep mathematics behind these curious observations? REPLY [8 votes]: (Too long for a comment.) Since the authors point out their two observations in a "jocund" air, we can match their observations with another pair. I. The Baby monster Just like the j-function $j(\tau)$ above and the Monster, the modular function $j_{2A}(\tau)$ that can be related to the Baby monster, $$\begin{aligned}j_{2A}(\tau) &=\Big(\big(\tfrac{\eta(\tau)}{\eta(2\tau)}\big)^{12}+2^6 \big(\tfrac{\eta(2\tau)}{\eta(\tau)}\big)^{12}\Big)^2 \\ &=\sum\limits_{n=-1}^{\infty}a(n)q^n\\&= q^{-1} + \color{blue}{104} + 4372q + 96256q^2 + 1240002q^3+\cdots \end{aligned}$$ also has, $$\sum \limits_{n=1}^{24}a(n)^2\equiv 42 \;\; (\mathrm{mod} \;70),$$ Recall that $e^{\pi\sqrt{58}} =396^4-\color{blue}{104}.00000017\dots$. II. Modular lambda function Given the modular lambda function $\lambda(\tau)=\lambda$ such that, $$j(\tau) = \frac{256(1-\lambda+\lambda^2)^3}{\lambda^2 (1-\lambda)^2}$$ and, $$\begin{aligned}\lambda(\tau) &= \Big(\tfrac{\sqrt{2}\,\eta\big(\tfrac{\tau}{2}\big)\eta^2(2\tau)}{\eta^3(\tau)}\Big)^8\\ &=\sum\limits_{n=1}^{\infty}b(n)q^n\\ &=16q - 128q^2 + 704 q^3 - 3072q^4 + 11488q^5 - 38400q^6 + \dots \end{aligned}$$ then, $$\sum \limits_{n=1}^{24}b(n)^2\equiv 42 \;\; (\mathrm{mod} \;70),$$ Of course, it is well-known that, $$1^2+2^2+3^2+\dots+24^2 = 70^2$$ It should be interesting if, for these four related functions, there is a reason for the congruences other than whimsy. $\color{red}{Update}$ P.S. By a really serendipitous error almost straight from the pages of the Hitchhiker's Guide, it turns out the same $24$ coefficients $s(n)$ of all four sequences also obey, $$\sum_{n=1}^{24} \frac{s(n)^2+42}{70} = \,\text{Integer}$$ The discussion below should illustrate how the error was found.<|endoftext|> TITLE: Difference between ZFC and ZF+GCH QUESTION [10 upvotes]: I hear that the axiom of choice (AC) derives from The generalized continuum hypothesis(GCH). And also hear that both AC and GCH are independent of Zermelo–Fraenkel set theory(ZF). So, I'm just curious why don't expert mathematicians use ZF+GCH instead of ZF+AC(ZFC). REPLY [4 votes]: There are two distinct questions that you might be asking. Why has the mathematical community adopted ZFC as a standard foundation and not ZF+GCH? What mathematical and philosophical arguments can be advanced for and against adopting AC and/or GCH as a fundamental axiom? Much as we might like to believe that the reason for the sociological acceptance of ZFC is that there are strong rational arguments for it, and that mathematicians accept those strong rational arguments because mathematicians are strongly rational, I do not believe that this is true. I believe that ZFC has been adopted as the standard largely for historical reasons. At some point, the mathematical community recognized the value of having some standard, because it would demonstrate that all of mathematics could in principle be formalized in a single system that avoided all the known paradoxes, and ZFC just happened to show up at the right place at the right time. I suspect that various other candidates could have "landed the job," including ZF + V=L, or even Z + C, and that ZFC was just a bit lucky. After a choice was made, most of the mathematical community lost interest in foundations and so had no real interest in tinkering with this or that axiom to get "better" foundations. I don't think that ZF+GCH was ever a serious contender, because GCH was still considered an open problem by the time ZFC had already secured its status as "the" foundation. If a statement X is considered an "open problem" then people generally do not also consider X to be a candidate for a basic axiom. By the time the independence of GCH was established, it was too late to apply for the job. Having said all that, I should add that if by "expert mathematicians" you mean experts in set theory, then the question is a bit different, because set theorists are more interested in these sorts of questions than the mathematical community as a whole is. Some of them will tell you that they reject GCH as an axiom simply because they don't believe that GCH is true. Still, even among set theorists, a sizable proportion take what you might call a "pragmatic" approach. They care mostly about whether the standard base theory is a technically convenient one for the investigations that they are interested in. Then the considerations that Asaf Karagila mentions come into play. GCH just isn't the most natural or convenient axiom to use for most things that set theorists currently care about. If it does happen to be useful in a certain context then they won't hesitate to assume it, but such occasions don't come up that often. (By the way, often Martin's axiom turns out to be what you really need when you might think you need CH.)<|endoftext|> TITLE: Determinant of a determinant QUESTION [45 upvotes]: Consider an $mn \times mn$ matrix over a commutative ring $A$, divided into $n \times n$ blocks that commute pairwise. One can pretend that each of the $m^2$ blocks is a number and apply the $m \times m$ determinant formula to get a single block, and then take the $n \times n$ determinant to get an element of $A$. Or one can take the big $mn \times mn$ determinant all at once. Theorem (cf. Bourbaki, Algebra III.9.4, Lemma 1): These two procedures give the same element of $A$. Corollary: If $B$ is an $A$-algebra that is finite and free as an $A$-module, $V$ is a finite free $B$-module, and $\phi \in \operatorname{End}_B V$, one can view $\phi$ also as an $A$-linear endomorphism, and then $\det_A \phi = N_{B/A}(\det_B \phi)$, where $N_{B/A}$ denotes norm. Corollary: For finite free extensions $A \subset B \subset C$, we have $N_{B/A} \circ N_{C/B} = N_{C/A}$. Question: Does the theorem (or either corollary) follow from some more conceptual statement, say some exterior power identities? Is there at least a proof that does not use induction on $m$? Other references containing a proof of the theorem: http://dx.doi.org/10.2307/2589750 and http://www.ee.iisc.ernet.in/new/people/faculty/prasantg/downloads/blocks.pdf (thanks to Andrew Sutherland for pointing out the former). REPLY [3 votes]: Here's another proof of the corollary. I think it's not the kind of proof you're looking for, especially since it secretly uses induction on m, but I think it is conceptual in a way. First reduction: we're trying to prove an equality between two maps of $A$-schemes $Res_{B/A} Mat_{m\times m}\rightarrow \mathbb{A}^1$. But $Res_{B/A}GL_m$ is Zariski-dense in $Res_{B/A} Mat_{m\times m}$. Thus we can assume our $\phi$ is invertible. First expansion: every pair $(V,\phi)$ as in the statement, with $\phi$ invertible, defines an element of the Quillen K-group $K_1(B)$. Now recall that the determinant extends to a map $det_B:K_1(B)\rightarrow B^\ast$, and likewise $det_A:K_1(A)\rightarrow A^\ast$. Recall also that there is a natural "transfer" map $tr_{B/A}:K_1(B)\rightarrow K_1(A)$ coming from the obvious forgetful functor on module categories. Then we can try to prove a more general claim: for any $x\in K_1(B)$, we have $$ N_{B/A} det_B(x) = det_A tr_{B/A} (x).$$ Both sides of this equation are elements of $A$. Since $A$ injects into the product of its localizations at all maximal ideals, and all the operations above commute with base change, we can thereby reduce to the case where $A$ is local, and hence $B$ is semi-local. For a general commutative ring $R$ the determinant map $K_1(R)\rightarrow R^\ast$ has an obvious section, coming from viewing elements of $R^\ast$ as endomorphisms of the unit $R$-module. When $R$ is semi-local, these maps are actually mutually inverse isomorphisms (see http://www.math.rutgers.edu/~weibel/Kbook/Kbook.III.pdf, Lemma 1.4). Since $B$ is semi-local, we can therefore reduce to the case where $\phi$ is given by an element of $B^\ast$ acting on $B$. Then the claim is simply the definition of $N_{B/A}$. I guess the moral of the story is that, thanks to the referenced lemma, the Zariski sheafification of $K_1$ identifies with $\mathbb{G}_m$. And this identification can't help but intertwine the transfer with the norm.<|endoftext|> TITLE: Special fiber of $X(p)$ in characteristic $p$ QUESTION [17 upvotes]: Let $p \geq 5$ be a prime. Let $Y(p)$ be the fine moduli space representing elliptic curves + basis of the $p$-torsion over $\mathbb{Q}_p$ and let $Y_0(p)$ be the fine moduli space representing elliptic curves + point of $p$-torsion over $\mathbb{Q}_p$. We know that $Y_0(p)$ has a model over $\mathbb{Z}_p$ whose special fiber is a union of two copies of $Y_0(1) := \mathbb{P}^1$ meeting transversaly at supersingular points. My question : what is the analogous description of the special fiber for $Y(p)$ (for some suitable model over $\mathbb{Z}_p$) ? What is the explicit description of the natural map $Y(p) \rightarrow Y_0(p)$ at the level of special fibers ? I guess it's in Katz--Mazur, but the purpose of my question is to get an answer as self-contained as possible. I don't (necessarily) ask for the idea of the proof. REPLY [16 votes]: A bit of mastication of Katz-Mazur Theorem 13.7.6 and the surrounding text seems to yield the following description of the special fiber of $Y(p)$: It is fundamentally $p+1$ copies of $\mathbb{P}^1$ (each with a nonempty finite set of punctures corresponding to cusps) all glued together at supersingular points. The completed local ring at a $k$-rational supersingular point is isomorphic to $k[[x,y]]/(y\prod_{i=0}^{p-1}(x-iy))$, i.e., we have $p+1$ curves coming together at all possible $\mathbb{F}_p$-valued slopes. The map to $Y_1(p)$ is $p$-to-1 on one component, and 1-to-1 on the other. This is because the map to $Y(1)$ has degree $p(p-1)$ on each component, while $Y_1(p)$ has a component of degree $p-1$ and a component of degree $p(p-1)$.<|endoftext|> TITLE: Parity of the Prime Counting Function QUESTION [13 upvotes]: I am interested in the distribution of the parity of $\pi(x)$, the prime counting function, over the natural numbers. Let: $\ \ E_n := \left\{ k \in \left\{1,\dots,n\right\} : \pi(k) \equiv 0 \mod 2 \right\}\ \ $ and $\ \ O_n :=\left\{ k \in \left\{1,\ldots,n\right\} : \pi(k) \equiv 1 \mod 2 \right\}$. My question: is it true that $$\lim_{n \to \infty} \frac{|E_n|}{n} = \lim_{n \to \infty} \frac{|O_n|}{n} = \frac{1}{2}$$ Are any methods known for calculating or estimating these limits? REPLY [19 votes]: The limits you conjecture are natural, and they are currently open. I believe the best known result is by Ping Ngai Chung and Shiyu Li, who proved that $$ \liminf_{n \to \infty} \frac{|E_{n}|}{n} $$ and $$ \liminf_{n \to \infty} \frac{|O_{n}|}{n} $$ are both $\geq \frac{1}{64}$. They can improve this to $\frac{1}{8}$ if they assume the Hardy-Littlewood prime $k$-tuples conjecture. The main tool used is Selberg's sieve and their paper appears in Integers in 2013. (See http://www.integers-ejcnt.org/vol13.html - the paper is A79.) They mention their result was obtained almost simultaneously by M. Alboiu.<|endoftext|> TITLE: The real numbers object in Sh(Top) QUESTION [15 upvotes]: If $X$ is a sober topological space, the real numbers object in the topos $\mathrm{Sh}(X)$ is the sheaf of continuous real-valued functions on $X$. This is proven very explicitly in Theorem VI.8.2 of MacLane & Moerdijk Sheaves in Geometry and Logic by compiling out the definition of real numbers in Kripke-Joyal semantics. A more abstract proof is in D4.7.6 of Sketches of an Elephant: since $\mathrm{Sh}(\mathbb{R})$ is the classifying topos of the theory of a real number, maps $U \to R_D$ in $\mathrm{Sh}(X)$ from an open subset $U\subseteq X$ to the real numbers object are equivalent to geometric morphisms $\mathrm{Sh}(X)/U \to \mathrm{Sh}(\mathbb{R})$, but $\mathrm{Sh}(X)/U \simeq \mathrm{Sh}(U)$ and so these are equivalent to continuous maps $U\to \mathbb{R}$. Theorem VI.9.2 of MacLane & Moerdijk makes an analogous claim for $\mathrm{Sh}(\mathbf{T})$, where $\mathbf{T}$ is a full subcategory of topological spaces closed under finite limits and open subspaces. My question is about a glossed-over point in the proof: they construct maps back and forth between sections of $R_D$ and the continuous $\mathbb{R}$-valued functions, but they don't say anything about why the maps are inverses. I believe it in the situation of $\mathrm{Sh}(X)$, but for $\mathrm{Sh}(\mathbf{T})$ it's not obvious to me, because a continuous map $Y\to \mathbb{R}$ only "knows" about the open subspaces of $Y$, whereas a map $Y \to R_D$ in $\mathrm{Sh}(\mathbf{T})$ says something about all continous maps $Z\to Y$ in $\mathbf{T}$. The most I can see how to show is that the sheaf of continuous real-valued functions is a retract of $R_D$. In terms of the abstract proof from the Elephant, the question is this: for $Y\in \mathbf{T}$, geometric morphisms $\mathrm{Sh}(\mathbf{T})/Y \to \mathrm{Sh}(\mathbb{R})$ are equivalent to continuous real-valued functions on the locale corresponding to the frame of subterminals in $\mathrm{Sh}(\mathbf{T})/Y$, but this frame is not (as far as I can see) the same as the open-set frame of $Y$. REPLY [14 votes]: Following a suggestion of Thomas Holder, we can close the gap as follows: For each object $Y$ in $\mathbf{T}$, there is a pseudonatural local geometric morphism $\mathbf{Sh}(\mathbf{T}_{/ Y}) \to \mathbf{Sh} (Y)$. $\mathbb{R}$ is Hausdorff, so for any Grothendieck topos $\mathcal{E}$, the category of geometric morphisms $\mathcal{E} \to \mathbf{Sh} (\mathbb{R})$ is essentially discrete. Essentially by definition, any local geometric morphism is a right adjoint in the bicategory of toposes. Thus, we have a pseudonatural left adjoint $$\mathbf{Geom} (\mathbf{Sh} (Y), \mathbf{Sh} (\mathbb{R})) \to \mathbf{Geom} (\mathbf{Sh} (\mathbf{T}_{/ Y}), \mathbf{Sh} (\mathbb{R}))$$ but any adjunction between groupoids is an equivalence, so we deduce that $\mathbf{T}(-, \mathbb{R})$ is not merely a retract of $R_D$ in $\mathbf{Sh} (\mathbf{T})$ but actually isomorphic to $R_D$, at least when every object in $\mathbf{T}$ is sober.<|endoftext|> TITLE: Faltings height in short exact sequences QUESTION [12 upvotes]: Let $K$ be a number field and $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ a short exact sequence of abelian varieties over $K$. Let $h(A)$ denote the logarithmic Faltings height (normalized so that it is invariant upon base change to any finite extension $K'/K$; thus, due to this normalization, one may assume that $A$, $B$, and $C$ have semiabelian reduction everywhere). Is $h(B) = h(A) + h(C)$? REPLY [6 votes]: Proposition 3.3 of Ullmo's paper "Hauteur de Faltings de quotients de J_0(N) " (American Journal of Math., 2000) seems to answer your question.<|endoftext|> TITLE: centeredness in forcing iterations QUESTION [8 upvotes]: Suppose $A$ is a complete subalgebra of a complete boolean algebra $B$, and $B$ is $\kappa$-centered. Let $G \subset A$ be a generic ultrafilter. Is $B/G$ $\kappa$-centered in $V[G]$? Naively, we might attempt to prove it as follows. Let $\{ F_\alpha : \alpha < \kappa \}$ be a collection of filters and in $V[G]$ let $F^*_\alpha = \{ [b]_G : b \in F_\alpha, b \not=_G 0 \}$. But for $b_1, b_2 \in F_\alpha$, maybe $b_1 \wedge b_2 =_G 0$, so that $F^*_\alpha$ is not a filter. An example where the centeredness cardinal goes up when modding out by a filter is given by comparing $\scr P(\omega)$ and $\scr P(\omega)/ \mathrm{fin}$. But maybe the generic filter case is different. REPLY [6 votes]: Let $P$ be the forcing for adding a Suslin tree $\mathring{T}$ by finite conditions. Then both $P$ and $P \star \mathring{T}$ are forcing isomorphic to adding $\omega_1$ Cohen reals (so $P \star \mathring{T}$ is sigma-centered). See lemma 5.6 here.<|endoftext|> TITLE: The formula for a perhaps basic identity (move from stackexchange) QUESTION [5 upvotes]: The following question is moved from math stackexchange. It seems that this is not a popular question, but I really want to know the answer so I moved it to here. The question reads as follows. We know the expansion of the following product $\displaystyle\prod_{k=1}^n(1+x+y_k)$ can be expressed by the formula $\displaystyle\sum_{k=0}^n(1+x)^{n-k}e_k(y_1, \ldots, y_n),$ where the $e_k$'s are the elementary symmetric functions, $e_k (x_1 , \ldots , x_n )=\displaystyle\sum_{1\le j_1 < j_2 < \dots < j_k \le n} x_{j_1} \dotsm x_{j_k}.$ My question is whether we have a nice formula for the expansion of the following product $\displaystyle\prod_{1\leq k\leq n, 1\leq\ell\leq m}(1+x_\ell+y_k).$ Reference for the nice formula of the above expression will be highly appreciated. (It seems to me that it is related to generating functions, but I have no background in combinatorics.) Thanks!~ REPLY [4 votes]: I guess the first product is expanded as $$ \prod_{k=1}^n(1+x+y_k)=(1+x)^n\prod_k(1+y_k(1+x)^{-1})=\sum_{k\geq 0}e_k(y_1,\ldots,y_n)(1+x)^{-k+n}.$$ For the other products you can write $$ \prod_{j=1}^m\prod_{i=1}^n(1+y_it_j)=\sum_\lambda e_\lambda(y_1,\ldots,y_n)m_\lambda(t_1,\ldots,t_m)=\sum_\lambda e_\lambda(\mathbf{y})m_\lambda(\mathbf{t})$$ where the sum is over $\lambda$ with $\ell(\lambda)\leq m$ and $m_\lambda$ is the monomial symmetric function. It seems that the product you are interested in looks something like $$ \prod_{j=1}^m\left(\sum_k e_k(y)(1+x_j)^{n-k}\right)=\sum_\lambda e_\lambda(\mathbf{y})m_{\lambda-(n^m)}((\mathbf{1-x})^{-1}) $$ In this formula, $\lambda-(n^m)=(\lambda_1-n,\ldots,\lambda_m-n)$ and you take the obvious definition of $m_\mu$ where $\mu$ is a partition with parts not necessarily positive. I'm not sure if this is helpful, since I don't know what you want to do with it. Anyway, this is related to a bilinear form on the ring of symmetric functions given by $$\langle u,v\rangle = (u,\omega(v))$$ where $(\cdot,\cdot)$ is the standard bilinear form having Schur functions as an orthonormal basis and $\omega$ is the automorphism switching the elementary and homogeneous symmetric functions. In particular, I believe this product also can be expanded as $$\sum_{\lambda} s_{\lambda^t}(\mathbf{y})s_{\lambda-(n^m)}((\mathbf{1-x})^{-1}).$$<|endoftext|> TITLE: Is there a non-abelian version of the Torelli map? QUESTION [11 upvotes]: Let $C$ be a connected compact oriented real surface of genus $g$, let $G$ be a connected compact Lie group and let $G_\mathbb{C}$ be the complexification of $G$. One considers the moduli space $M (C,G)$ of isomorphism classes of representations of the fundamental group $\pi_1 (C)$ in $G$. It is a compact real algebraic variety (in general singular) with a natural symplectic structure (see Atiyah-Bott, Goldman,...) If one fixes a complex structure on $C$, one can consider the moduli space $M_{hol} (C,G)$ of degree zero semistable holomorphic $G_\mathbb{C}$-bundles on $C$. It is a projective complex algebraic variety (in general singular) with a natural polarization. The theorem of Narasimhan-Seshadri (for $G=U(n)$, extended by Ramanathan for general $G$) gives an isomorphism between the underlying real analytic structures of $M(C,G)$ and $M_{hol}(C,G)$, with identification of the natural symplectic form on $M(C,G)$ with the natural Kähler form on $M_{hol}(C,G)$. In other words, a choice of complex structure on $C$ gives a natural polarized complex algebraic structure on $M(C,G)$ compatible with the symplectic structure. In vague terms, my question is: how does this structure changes when the complex structure on $C$ changes? More precisely, I would like to consider the moduli space $\mathcal{M}$ of polarized complex algebraic structures on $M(C,G)$ compatible with the symplectic structure. I don't know if such moduli space exists. Q1) Is there a way to make sense of the moduli space $\mathcal{M}$ ? Locally, globally? If $\mathcal{M}$ makes sense then the Narasimhan-Seshadri theorem gives a map $f: M_g \rightarrow \mathcal{M}$, $C \mapsto M_{hol}(C, G)$, where $M_g$ is the moduli space of complex structures on $C$. Q2) What can be said on the map $f$ ? Is it injective ? What is the dimension of its image?... When $G=U(1)$, we simply have $\mathcal{M} = A_g$, the moduli space of principally polarized abelian varieties of dimension $g$ and $f: M_g \rightarrow A_g$ is simply the Torelli map $C \mapsto Jac(C)$, which is injective by the Torelli theorem. So my questions could be reformulated as: is there some non-abelian version of the Torelli map and of the Torelli theorem (where "non-abelian" refers to the replacement of $G=U(1)$ by a general $G$) ? REPLY [5 votes]: In the analytic category, this question is addressed in N.Hitchin's paper Flat connections and geometric quantization. I will write $M$ for your $M_{hol}(C,G)$, and will pretend that $M$ is smooth (you can replace $G$ by a central extension, or use the normality of $M$ plus Hartogs-type arguments). Then, if $P$ be a stable $G_{\mathbb{C}}$-bundle (with no extra automorphisms), $T_{M,[P]}= H^1(C, ad P)$. The infinitesimal deformation map $$ H^1(C,T_C) \longrightarrow H^1(M, T_M) $$ is injective. This is due to Narasimhan and Ramanan when $M$ is smooth (they treat unitary bundles with coprime rank and degree), and Hitchin gives independent proof in the singular case. Next, cup product gives a map $$ H^0(T_M^2)\otimes H^1(\Omega^1_M)\to H^1(T_M), $$ and under multiplication with the Atiyah-Bott symplectic form, the polarisation-preserving deformations correspond to sections of $H^0(M,Sym^2 T_M)\subset H^0(M, T_M^2)$. We can pull these to (the total space of) $\pi:T^\vee_M\to M$: $$ H^0(M, Sym^2 T_M)\subset H^0(T^\vee_M, Sym^2 \pi^\ast T_M). $$ Proposition 2.16 in Hitchin's paper says that the cup product pairing $$ H^1(C,T_C)\otimes H^0(C, ad P\otimes K_C)\to H^1(C,ad P) $$ gives, for any class in $H^1(C,T_C)$, the corresponding deformation of the Kaehler polarisation of the moduli space. If you're working with a classical group, the section of $Sym^2\pi^\ast T_M$, corresponding to $[\kappa]\in H^1(C, T_C)$ is $$ G(\alpha,\alpha)=\int_C Tr(\alpha^2)\wedge \kappa,\quad \alpha\in T^\vee_{M,P}=H^0(C, ad P\otimes K_C). $$ In general, you take the quadratic part of the Hitchin map. There is indeed an analogy with the abelian case. Hitchin considers a line bundle over Teichmueller space, whose fibres are the ``non-abelian theta-functions'' (of sufficiently high level), and constructs a projectively flat connection on it. The (projective) flatness is the analogue of the heat equation for theta-functions. The connection is known as the Hitchin connection, WZW-connection, or KZB-connection.<|endoftext|> TITLE: Natural examples of Reverse Mathematics outside classical analysis? QUESTION [15 upvotes]: Harvey Friedman at the 1974 ICM motivated Reverse Mathematics by the following statement: When the theorem is proved from the right axioms, the axioms can be proved from the theorem. Reverse Mathematics has had many successes in finding the "right axioms," but to date mainly for theorems of classical analysis, where real numbers (or equivalent infinite objects, such as sets of natural numbers) are involved. This may be partly for historical reasons, since the subject grew from the study of subsystems of second order number theory. It seems to me that there are classical theorems in other fields that also fit the Reverse Mathematics paradigm: add a strong axiom $A$ to a weak theory $W$ and prove $A$ equivalent, over $W$, to a strong theorem $T$ of $W+A$. The example I have in mind is where $W$ is the theory of a projective plane $P$, which has three simple axioms about objects called "points" and "lines": Any two points belong to a unique line. Any two lines have a unique common point. There exist at least four distinct points. This $W$ is a weak theory with very few interesting theorems. But if we add to $W$ the theorem of Pappus as an axiom $A$, then results of von Staudt, Hilbert, and Hessenberg enable us to prove the theorem $T$: The plane $P$ can be coordinatized by a field. Conversely, if $T$ holds then we can prove $A$. This is because the Pappus axiom $A$ states that certain points lie on a line, and $A$ can be proved once we have coordinates in a field -- by computing the coordinates of the points in question and showing that they satisfy a linear equation. There are a few variations of this result. For example, if axiom $A$ is replaced by the theorem of Desargues then the equivalent theorem $T$ of $W+A$ is that $P$ can be coordinatized by a skew field. This leads me to the following questions. Are these results reasonable examples of Reverse Mathematics? Are there other natural examples outside analysis? Edit (Nov 15, 2014). Many thanks to the logicians who have answered this question. It appears to me now that the term "reverse mathematics" is too narrow for what I had in mind, but I am still interested in examples of "finding the right axioms." An even more elementary example than the Pappus axiom/theorem is the Euclid's parallel axiom. It is needed to prove many theorems, e.g. Pythagorean theorem, angle sum of a triangle equals $\pi$, ... and these theorems also prove the axiom. REPLY [9 votes]: I agree with Henry Towsner that the terminology "reverse mathematics" is generally used only for things in or near second-order arithmetic. If, however, you're interested in situations where one tries to prove a theorem from some axioms and then show that the axiom actually follows from the theorem, regardless of whether these situations are called "reverse mathematics", then I'd point to the study of weak forms of the axiom of choice. Here the base theory is ZF (or perhaps a variant ZFA that allows atoms), the additional axioms that one considers are the axiom of choice and weakenings of it, and one tries to show that various results of ordinary (not set-theoretic) mathematics are equivalent (over ZF) to one or another of these axioms. The book "Consequences of the Axiom of Choice" by Paul Howard and Jean Rubin gives lots of examples.<|endoftext|> TITLE: When exactly and why did matrix multiplication become a part of the undergraduate curriculum? QUESTION [51 upvotes]: The story about Heisenberg inventing matrices and matrix multiplication in 1925 is very well known and well documented. A few weeks later, Born and Jordan read this work and recognized matrix multiplication, because one of them happened to take a course in "hypercomplex numbers" in his youth. This is how modern quantum mechanics was born. This story clearly shows that in the second decade of the 20th century matrix multiplication was not something that every undergraduate was familiar with. My questions: When did this change? (Nowadays, I would say that this is the MOST standard part of the undergraduate curriculum in the US. More standard than Calculus. EVERY science or math major is taught to multiply matrices in her first year. It was similar in the Soviet Union, and I suppose this is the case everywhere). When did this dramatic change occur? When did linear algebra become a mandatory undergraduate subject? But an even more interesting question is WHY? My conjecture is that this has something to do with the invention of Quantum mechanics. I have some arguments explaining this. But to test my arguments, I would like to know the answer to the first question, and other opinions on the second. I know that matrix multiplication was introduced by Cayley (correct me if I am wrong), and that Hilbert-Courant was published in 1924. What physicists were using before Hilbert-Courant is not completely clear, but probably Thomson and Tait, or something similar. Of course, there is no mention of matrices in Thomson and Tait:-) I understand that this may not be within the scope of this forum. Feel free to close. But I am asking in hope that someone by chance knows the answer to question 1, and will answer before the question is closed. EDIT. Using Mathscinet, Zentralblatt and Jahrbuch, with "linear algebra", "lineare Algebra" etc. in the title, and year $< 1950$, I found a Turkish textbook of Pasha and Tefvik (1893), a textbook by H. Bohr and J. Mollerup (1938) and Russian text by Malcev (1949), from which I studied as an undergraduate. After 1950 there are many, which suggests that this curriculum revolution probably started in the early 1950s. EDIT2: Google Ngram shows a sharp growth of the usage of the word "matrix", which begins in the middle 1940s. Against my expectation, it shows a peak in 1990 and then starts declining. A similar question is posted on HSM REPLY [17 votes]: Too long for a comment, but this premise of the question: I know that matrix multiplication was introduced by Cayley (correct me if I am wrong) is indeed wrong. Gauss in Disquisitiones Arithmeticae (1801) has something called not matrix multiplication but combination of substitutions — e.g. he writes near the end of §294: $(S)=\left\{\!\!\begin{smallmatrix} \phantom{-}7245,&\phantom{0}5,&\phantom{-}22\\ -2415,&\phantom{0}2,&-28\\ 19320,&25,&\phantom{0}-7 \end{smallmatrix}\!\right\}$ combined with $(S')=\left\{\!\!\begin{smallmatrix} \phantom{-000}3,&\phantom{-000}5,&\phantom{-00}1\\ -2440,&-4066,&-813\\ \phantom{0}-433,&\phantom{0}-722,&-144 \end{smallmatrix}\!\right\}$ produces $\left\{\!\!\begin{smallmatrix} \phantom{-}9,&11,&12\\ -1,&\phantom{0}9,&-9\\ -9,&\phantom{0}4,&\phantom{0}3\end{smallmatrix}\!\right\}$. His definition is in §270:<|endoftext|> TITLE: "a shape that ... lies halfway between a square and a circle" QUESTION [20 upvotes]: An article in the Notices of the AMS, Volume 61, Issue 10, 2014 (PDF download link), on Khot's Unique Games Conjecture, says this: Another group ... found a shape that in a certain sense lies halfway between a square and a circle (though in many more than two dimensions). Like a square, copies can be placed next to each other horizontally and vertically to fill a whole space without gaps or overlaps, forming a multi-dimensional foam. But its perimeter is much smaller than a square—it’s closer to that of a circle, the object which has the smallest perimeter for the area contained. Could someone please either explain this enigmatic description further, or provide a reference to the original work? Thanks! (This issue is related to my earlier [answered] question, "Optimal planar net for catching convex shapes.") REPLY [12 votes]: Just to supplement Carlo's answer with a figure that illustrates this intermediate cube-sphere shape in $\mathbb{R}^3$, from the 2012 paper he cites—"we provide the most efficient known cubical foam in 3 dimensions":<|endoftext|> TITLE: Cantor-Bernstein for quasi-isometric embeddings? QUESTION [15 upvotes]: Suppose that two finitely generated groups quasi-isometrically embed into each other. Does it follow that the two groups are quasi-isometric? Recall that a quasi-isometry is a quasi-isometric embedding that is quasi-surjective, see e.g. https://www.math.ucdavis.edu/~kapovich/EPR/pc_lectures3.pdf REPLY [8 votes]: 1) LAMPLIGHTER GROUPS As mentioned by Yves, lamplighter groups over $\mathbb{Z}$ provide counterexamples thanks to Eskin, Fisher, and Whyte's work. Other counterexamples are given by lamplighter groups over one-ended groups thanks to the recent preprint arxiv:2105.04878. More precisely, if $F_1,F_2$ are two finite groups and if $H$ is a finitely presented one-ended group, then $F_1\wr H$ and $F_2 \wr H$ quasi-isometrically embed into each other, but they are not quasi-isometric if $|F_1|,|F_2|$ do not have the same prime divisors. (The converse holds if $H$ is non-amenable. If $H$ is amenable, $|F_1|,|F_2|$ must be powers of a common number and $H$ must admit specific auto-quasi-isometries (which always the case if $H$ is free abelian).) The advantage of these lampligthers is that one can avoid the heavy machinery of coarse differentiation. 2) COXETER GROUPS Counterexamples can also be found among Coxeter groups. I just give one example, based on the following statement: Proposition: Let $\Gamma$ be a finite simplicial graph and $u \in V(\Gamma)$ a vertex. Let $\Gamma(u)$ denote the graph obtained by gluing two copies of $\Gamma \backslash \{u\}$ along $\mathrm{link}(u)$. Then the right-angled Coxeter groups $C(\Gamma)$ and $C(\Gamma(u))$ are commensurable. Sketch of proof. Let $u_1,\ldots, u_k$ denote the vertices of $\Gamma \backslash \mathrm{star}(u)$. Then, by a standard ping-pong argument, one proves that the reflexion subgroup $\langle u_1, \ldots, u_k,u_1^u, \ldots, u_k^u \rangle$ admits $\{u_1, \ldots, u_k,u_1^u, \ldots, u_k^u\}$ as a basis. The commutation graph of this family is precisely $\Gamma(u)$. $\square$ (Some details can be found in arxiv:1910.04230; see Proposition 3.14.) Now, we apply the proposition to the following graph: Clearly, $\Gamma(u)$ contains an induced copy of $\Gamma \sqcup \{ \mathrm{pt} \}$, so $C(\Gamma)$ quasi-isometrically embeds into the free product $C(\Gamma) \ast \mathbb{Z}/2\mathbb{Z}$. The reverse quasi-isometric embedding is clear. But $C(\Gamma)$ is one-ended, because $\Gamma$ has no separating complete subgraph, so $C(\Gamma)$ and $C(\Gamma) \ast \mathbb{Z}/2\mathbb{Z}$ cannot be quasi-isometric. 3) RIGHT-ANGLED ARTIN GROUPS Given a simplicial graph $\Gamma$, the associated right-angled Artin group is defined by the following presentation: $$A_\Gamma = \langle V(\Gamma) \mid [u,v]=1, \ \{ u,v \} \in E(\Gamma) \rangle$$ where $V(\Gamma)$ and $E(\Gamma)$ denote the vertex- and edge-sets of $\Gamma$ respectively. First, a general statement. A graph $\Gamma$ is join if there exists a partition $V(\Gamma)= A \sqcup B$ such that $A,B$ are both non-empty and such that any vertex of $A$ is adjacent to any vertex of $B$. Proposition: Let $\Gamma$ be a finite simplicial graph which is not a join and which is not reduced to a single vertex. Then $A_\Gamma$ contains a quasi-isometrically embedded subgroup isomorphic to the free product $A_\Gamma \ast \mathbb{Z}$. Sketch of proof. Let $\Gamma^e$ denote the extension graph of $\Gamma$, ie., the graph whose vertices are the conjugates $gug^{-1}$, where $g \in A_\Gamma$ and $u \in V(\Gamma)$, and whose edges link two elements when they commute. In Embedability between right-angled Artin groups, Kim and Koberda prove that, if $\Lambda$ is a finite induced subgraph of $\Gamma^e$, then $A_\Lambda$ embeds into $A_\Gamma$. It turns out that $\Gamma^e$ is unbounded since $\Gamma$ is not a join, so that $\Gamma^e$ contains an induced subgraph isomorphic to $\Gamma \sqcup \{ \text{vertex} \}$, hence $A_\Gamma \ast \mathbb{Z} \leq A_\Gamma$. I don't know if it follows from their argument that the embedding is quasi-isometric, but, in the alternative proof I gave of their theorem in my thesis (see Section 8.5), it is not difficult to show that the embedding I construct is quasi-isometric. $\square$ Corollary: If $\Gamma$ is a finite connected simplicial graph which is not a join, then $A_\Gamma$ and $A_\Gamma \ast \mathbb{Z}$ quasi-isometrically embeds into each other, but they are not quasi-isometric. The fact that $A_\Gamma$ and $A_\Gamma \ast \mathbb{Z}$ are not quasi-isometric follows from the observation that $A_\Gamma$ is one-ended (as $\Gamma$ is connected and contains at least two vertices). Now, a concrete example. Consider the right-angled Artin group $$A= \langle a,b,c,d \mid [a,b]=[b,c]=[c,d]=1 \rangle.$$ Of course, $A$ quasi-isometrically embeds into $A \ast \mathbb{Z}$. Conversely, an embedding as mentioned above shows that the subgroup $$\langle a,b,c,d^2,ada^{-1} \rangle= \langle a,b,c,d^2 \rangle \ast \langle ada^{-1} \rangle \simeq A \ast \mathbb{Z}$$ is quasi-isometrically embedded into $A$, which can also be proved directly. However, $A$ and $A \ast \mathbb{Z}$ are not quasi-isometric since $A$ is one-ended. 4) A HYPERBOLIC EXAMPLE According to arxiv:1812.07799, there exist a torsion-free one-ended hyperbolic group $G$ containing two isomorphic quasiconvex subgroups $H_1,H_2 \leq G$ such that $H_1$ (resp. $H_2$) has finite (resp. infinite) index in $G$. As a consequence of a theorem claimed by Gromov (and proved by Arzhantseva), $G$ contains a subgroup isomorphic to the free product $H_2 \ast \mathbb{Z}$. As a consequence, $G$ and $G\ast \mathbb{Z}$ quasi-isometrically embed into each other but they are not quasi-isometric.<|endoftext|> TITLE: New relator in hurwitz group QUESTION [7 upvotes]: I have found that $([a,b]^2[a,b^2])^n$ is a good relator to use in my search for quotients of $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10} \rangle$. For n<=5 $H := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10}, ([a,b]^2[a,b^2])^n \rangle$ is the trivial group, but when n=6, it is the Janko group J1, and when n=7, it is the Hall-Janko group J2. I have tried finding what the group is when n = 8, but without success (magma failed, and GAP used too much memory and crashed my computer). Is there a way of finding what this group is? REPLY [8 votes]: A brute-force calculation in GAP (searching for homomorphisms) shows that the group G has a quotient $PSL_2(41)\times J1^2\times J2^2\times G_2(5)^2$. There are no other quotients that are simple groups of order $\le 10^{10}$, and (as you surely know) the group is perfect. The order of $[a,b]^2[a,b^2]$ under the simple quotients is 21 (PSL), 6 and 10 (J1), 7 and 15 (J2) and both times 31 (G2(5)). Thus this gives you information for $n=10$, but not $n=8$. Its only $PSL_2$ quotient is $PSL_2(41)$. (Noam Elkies mentioned this already, but the statement might not have been clear about that there are no others) , this is by the Plesken/Fabianska algorithm in Magma.<|endoftext|> TITLE: how to construct an oriented double cover of a lamination? QUESTION [5 upvotes]: Suppose $\lambda$ is an non-orientable lamination on a closed orientable surface. How to construct an oriented double cover of $\lambda$? REPLY [11 votes]: the construction is the same as that for a manifold. to each point you associate the two orientations with the natural induced topology.if the connected lamination is orientable the construction has two components, otherwise only one. this distinction, orientable or not, is interesting for laminations. for example, a one dimensional orientable lamination with cantor set transversals is the boundary of a two dimensional lamination with cantor set transversals. I don't know whether or not a non orientable one dimensional lamination with cantor set transversals is the boundary of a non orientable two dimensional lamination. dennis sullivan<|endoftext|> TITLE: Failure of Mostow rigidity in dimension 2 QUESTION [19 upvotes]: I am trying to understand why Mostow rigidity fails in dimension 2. More concretely, I have the following question: (1) What is an example of a quasiisometry $f$ of the hyperbolic plane $\mathbb H^2$ to itself such that for the continuation $\partial f$ of $f$ to the boundary $\partial \mathbb H^2$, there does not exist an isometry $\varphi$ of $\mathbb H$ such that $\bar{\varphi} |_{\partial \mathbb H^2} = \partial f$? The reason I am asking this question is that one way to prove classical Mostow rigidity (two closed hyperbolic manifolds of the same dimension $n \geq 3$ with isomorphic fundamental groups are isomorphic) can be deduced from the statement that for $n \geq 3$, every quasiisometry $f$ of hyperbolic space $\mathbb H^n$ extends to a homeomorphism $\partial f: \partial \mathbb H^n \rightarrow \partial \mathbb H^n$ which is the restriction of an isometry of $\mathbb H^n$. The proof of this statement decomposes into two pieces: First, one shows that $\partial f$ is a homeomorphism, secondly, one proves that it is conformal and hence has to come from an isometry of $\mathbb H^n$. The first step also works in dimension 2, but the second does not. So more general than the above, we could even ask the following (by identifying $\partial \mathbb H^2$ with $S^1$): (2) Which self-homeomorphisms of $S^1$ arise as extensions of quasiisometries $\mathbb H^2 \rightarrow \mathbb H^2$ to the boundary? REPLY [6 votes]: The actual questions you asked have already been answered, and so I am hesitant to chime in with anything. Anyway, here is an different response to your opening line: "I am trying to understand why Mostow rigidity fails in dimension 2." Here is my first thought: The structure of the fundamental group of surfaces allows for local deformations in the moduli space of hyperbolic structures. For every hyperbolic structure on a closed $n$-manifold there is a discrete faithful representation (hence irreducible) from $\pi_1(M)\to \mathrm{Isom}(M)$, unique up to conjugacy. When $M$ is a surface, $\mathrm{Isom}(M)$ is $\mathrm{PSL}(2,\mathbb{R})$ and $\pi_1(M)$ has a presentation with $2g$ generators and 1 non-trivial relation in those generators. One can use this to conclude that: $\mathrm{Hom}(\pi_1(M),\mathrm{Isom}(M))/\mathrm{Isom}(M)$ has dimension $3(2g-1)-3=6g-6$; and locally so at any irreducible representation. So there are non-trivial infinitesimal deformations in that case. However, the infinitesimal deformations can be "integrated", since the obstructions are in second cohomology which vanishes since $n=2$. Hence there is no rigidity (local and hence not global). For $n>2$, there are no such deformations; which follows from Weil's Remarks on the cohomology of groups (see here) also using information about a presentation of $\pi_1(M)$. One concludes that there are no infinitesimal deformations and thus no local deformations. Mostow's Rigidity says there is global rigidity (for $n>2$); admittedly a stronger statement than local rigidity (although infinitesimal rigidity is stronger than local rigidity). But the difference between $n=2$ and $n>2$ is seen locally as well and mostly is a consequence of nature of $\pi_1(M)$ in those cases. So one can say: Mostow Rigidity Fails in dimension 2 since there is no local rigidity in dimension 2, which follows mostly from the structure of $\pi_1(M)$.<|endoftext|> TITLE: A "small" definition of sub-(∞,1)-topoi QUESTION [9 upvotes]: Suppose I have an $(\infty,1)$-topos $\mathcal{X}$ and a (small) set of maps $S$ in $\mathcal{X}$, which therefore generates an accessible localization $S^{-1}\mathcal{X}$. Is there any "small" condition $P$ on $S$ which implies that this localization is left exact (hence a sub-$(\infty,1)$-topos), and such that any accessible left exact localization is generated by some $S$ satisfying condition $P$? By "small" I mean expressible with only (small-set-)bounded quantification and reference to categorical operations. So, for instance, "the localization $S^{-1}\mathcal{X}$ is left exact" is not small because it is a statement about all pullbacks in $\mathcal{X}$, which form a proper class (or a set in the next higher universe, whatever). The condition of 6.2.1.1(b) in Higher Topos Theory is likewise not small. But, for instance, "$S$ consists of monomorphisms" is a small condition, since it quantifies only over the small set $S$. (Being a monomorphism is a small condition in the sense I mean; the ordinary definition of "monomorphism" quantifies over all objects of the category, but it's equivalent to the diagonal being an isomorphism, and the diagonal is a "categorical operation". I hope the intent is clear; if anyone is confused I could try to formulate a more precise definition of "smallness".) Of course, "$S$ consists of monomorphisms" is probably not an answer to the question. I suspect that there is a small condition "$S$ consists of monomorphisms and ..." which answers a modified version of the question that asks about topological localizations (although at the moment I don't even see how to prove that). But I would really like an answer applying to all accessible left exact localizations. REPLY [5 votes]: This follow from some recent (I heard about this a year ago) results by Anel, Biedermann, Finster and Joyal. Unfortunately their work is not available yet, but You have some slide of Mathieu Anel on the topic on his web page presenting this (http://mathieu.anel.free.fr/mat/doc/Anel-LexLocalizations.pdf). I don't know if other references about this are available. What Anel's slide says is that given a set of arrows $S$ in an $\infty$-topos the smallest left exact localization inverting $S$ is the cocontinuous localization inverting all pullbacks of all iterated diagonal of maps in $S$. Another way to say it is that a localization is left exact if and only if the class of all morphisms inverted is stable under pullback and taking diagonal... Which is actually also equivalent to that fact that has a full subcategory of $\mathcal{X}^{\Delta[1]}$ it is closed under finite limits. I have not seen their proof yet, but when it is formulated this way it is not too hard to give a very direct proof. Using the descent property and the pullback stability of this class, it is enough to invert all such arrows whose target belong to some generating familly of the topos, so it provides the kind of small condition you are looking for. For example, Given $\mathcal{X}$ an $\infty$-topos, $(x_i)$ a generating set of $\mathcal{X}$ (of special interest, $\mathcal{X}$ is a presheaf topos and $x_i$ are the representable), then left exact localization of $\mathcal{X}$ are all obtained as localization at a set $S$ of morphisms such that: every morphisms in $S$ has one of the $x_i$ as its target. Every pullback of an iterated diagonal of $s \in S$ whose target is one of the $x_i$ is in $S$, i.e. for all $f : X \rightarrow x_i \in S$, and for all maps $x_j$ to the target of the $n$-iterated diagonal of $f$, the pullback of the $n$-th iterated diagonal of $f$ to $x_j$ is in $S$. In the case of topological localization, i.e. when all maps in $S$ are mono, $n$-iterated diagonal for $n \geqslant 1$ are all iso, so the thing about diagonal disapear and you recover the usual pulblack stability axiom for Grothendieck topology.<|endoftext|> TITLE: Extensionality in HoTT versus extensionality in internal language of a category QUESTION [7 upvotes]: What's the extension of judgmental identity in HoTT (homotopy type theory)? The Martin-Löf intensional dependent type theory with identity types is called (definitionally) extensional if the judgmental (intensional) equality coincides with the propositional equality, i.e, $$ \frac{p:Id_A(x,y)}{x\equiv y} $$ For the name "extensional" makes sense, the judgmental equality must be determined by the extension of the terms. So what's the extension of $x \equiv y$? By the definition, it just seems that the extension are the proofs (hence $Id_A (x, y)$ is the extension of $x \equiv_A y$) and, in this case, it would not match with the extension of the relational symbol "equality" used in the internal language of categories where a relational symbol (or predicate) $\varphi$ would be a subobject of a type $A$ $$[\varphi] \hookrightarrow A$$ composed by the collection (not set, there's no set theory here) of terms such that the predicate is true $\{x : A; \varphi(x) \ \text{true} \}$ . So, more generally, what's the extension of a predicate in Martin-Löf type theory? REPLY [9 votes]: The extension is the observable behavior, where "observation" means roughly "application of an eliminator". For a predicate $p$ we may observe whether it is true for a given argument, so by collecting those arguments at which $p$ holds we observe all there is to observe. More generally, a function is "observed" by application to an argument (as opposed to examination of its "source code" or measurement of time needed to compute, etc.), which leads to the rule of function extensionality $$\frac{x : A \vdash f(x) =_B g(x)}{f =_{A \to B} g}.$$ A simpler extensionality rule is that for cartesian products: $$\frac{\pi_1(u) =_A \pi_1(v) \qquad \pi_2(u) = \pi_2(v)}{u =_{A \times B} v} \tag{1} $$ It says that two pairs are equal if they behave in the same way when we observe them by projecting their components. In set theory we observe a set by testing which elements it has, and so the extensionality axiom of set theory says that two sets are equal if they have the same elements. And since in set theory there are only sets, this gives set theory as much extensionality as it could have. In the internal language of a category, say a topos, we get extensionality from universal properties, more precisely from the uniqueness requirements. For instance, (1) is validated by the fact that there is at most one morphism $C \to A \times B$ whose composition with $\pi_1$ and $\pi_2$ are prescribed maps $C \to A$ and $C \to B$. Furthermore, because in a typical internal language we interpret (judgmental) equality on $A$ as the diagonal $A \to A \times A$, which is the least equivalence relation on $A$, there is no room for another kind of equality -- the reflection principle is automatic, and so the distinction between judgmental and propostional equality does not arise. In contrast, the intension tells us how an object is built or how it works. Thus, functions $x \mapsto x + 3$ and $x \mapsto 2 + (x + 1)$ have the same behavior, but are not intensionally equal because they compute in different ways. A typical intensional rule of equality says that two objects are equal if they are built the same way. For pairs such a rule would be $$\frac{a \equiv_A a' \qquad b \equiv_B b'}{(a,b) \equiv_{A \times B} (a',b')}.$$ Intensional equality of functions says that two functions are equal if they have the same "bodies of definition" (officially this goes under the name $\xi$-rule): $$\frac{x : A \vdash e \equiv_B e'}{(\lambda x : A \,.\, e) \equiv_{A \to B} (\lambda x : A \,.\, e')}.$$ Another kind of intensional equality explains how eliminators compute, for instance $$\pi_1(a,b) \equiv_A a$$ says that the first projection computes $a$ when applied to a pair $(a,b)$. Similarly, the $\beta$-rule for functions, $$(\lambda x : A \,.\, e_1) e_2 \equiv_B e_1[e_2/x],$$ explains how $\lambda$-abstractions compute their results, so it is again an intensional rule. We shall therefore call a notion of equaity extensional if it makes objects equal when they have equal observations. We shall call an equality intensional if it makes objects equal when they are constructed the same way from equal parts, or if it tells us how to compute. We shall call a type theory extensional or intensional according to whether its judgmental equality is extensional or intensional (this is a source of confusion, as many people think that extensional type theory is so called because it has extensionality rules for propositional equality, see Mike Shulman's answer). For the purposes of mathematics it is good to have extensional equality, but less so for the purposes of computer science, where we generally care how things are computed and not just what the results are. As far as type theory is concerned, we have a choice. We could declare judgmental equality to be extensional, for instance $$\frac{\pi_1(u) \equiv_A \pi_1(v) \qquad \pi_2(u) \equiv_B \pi_2(v)}{u \equiv_{A \times B} v}$$ looks reasonable. Indeed, Agda and (recent) Coq have such a rule. But things get complicated if we take this too far. The reflection rule which you stated makes judgmental equality undecidable, which does not appeal to the typical type theorist. If we make judgmental equality of functions extensional, things get at least very complicated. It is less unreasonable to make the propositional equality extensional. This is what homotopy type theory does. It goes farther than any other kind of type theory by making the propostional equality of a universe extensional, although at the moment I cannot think of a good explanation of the Univalence axiom which starts from "what is an observation on an element of a universe". In summary, while we have a choice to make either kind of equality extensional, it makes more sense to make propositional equality extensional and keep judgmental equality intensional.<|endoftext|> TITLE: Why higher category theory? QUESTION [85 upvotes]: This is a soft question. I am an undergrad and is currently seriously considering the field of math I am going into in grad school. (perhaps a little bit late, but it's better late then never.) I have some background in commutative algebra and algebraic topology and I am learning AG quite intensively. I understand that category theory is useful for generalizing different mathetical objects, and in the field of AT, it makes the idea of homology and homotopy so clear as they are just functors from (based) space to the category of (abelian) groups, and many things are just limits or colimts; I also do understand you need the notion of abelian categories for homological algebra and constructing sheaves like the grand project done by Grothendieck. However, I still don't understand why study such elusive things like higher category theory where many things defy intuition? I have talked to several top mathematicians in my department, many of whom apperently think it is useless. However, it should not be as celebrated mathematicians like Jacob Lurie is making a lot of effort trying to understand it. So can anyone explain to me why study higher category theory? What problems are they most concerned with in higher category theory and what connections are there with AT or AG? Thank you very much in advance. REPLY [3 votes]: I refer you to a talk I gave in Paris last June on "The intuitions of cubical sets for nonabelian algebraic topology" which explains how from my 1967 groupoid version of the Seifert-van Kampen Theorem I was led to ask about the potential relevance of groupoids in higher homotopy theory. Eventually with Philip Higgins we obtained a new formulation of basic homology/homotopy theory using actual compositions of cubes, and avoiding Poincar'e's "formal sums", now expressed in terms of free abelian groups. The details are in the EMS Tract vol 15 (2011) advertised here, where there is a pdf available. This book also discusses the history and intuitions. These methods give more understanding and also homotopical computations not available by other methods. As an indication, I mention the bibliography on the nonabelian tensor product which has 131 items, mainly in group theory. Relatively few algebraic topologists have worked in this area. Here is a short exposition related to the Blakers-Massey Theorem. These papers use higher groupoids as concrete algebraic objects better able to express higher geometric ideas than purely $1$-dimensional constructions. What is called "higher category theory" has not yet, it seems to me, accommodated the cubical approach. However for the work in which I have been involved the cubical methods have proved essential. See also this mathoverflow answer on higher homotopy groupoids.<|endoftext|> TITLE: Partial inverse of a matrix - or does it have its own name? QUESTION [8 upvotes]: In my calculations I need to use something which is "between" a matrix and its inverse. That is, I invert only some dimensions. I am interested if it has an established name. That is, a matrix (here 2x2 real, but it is more general) $$ \begin{bmatrix} u' \\ v' \end{bmatrix} = M \begin{bmatrix} u \\ v \end{bmatrix} $$ defines a hyperplane in coordinates $(u,v,u',v')$. Its inverse (if exists) can be defined as a linear operator such that $$ \begin{bmatrix} u \\ v \end{bmatrix} = M^{-1} \begin{bmatrix} u' \\ v' \end{bmatrix}. $$ I am interested in inverting only some coordinates, e.g. $$ \begin{bmatrix} u \\ v' \end{bmatrix} = M^{(-1,1)} \begin{bmatrix} u' \\ v \end{bmatrix}. $$ I know it is a relatively simple thing related to the implicit function theorem, with simple formulas. Yet, I use it a lot and I need to call it somehow. So: does it have its own name? if not, is "partial inverse" fine? (not colliding with other names, not (too) confusing, etc) If you are curious, I use it in physics (optics) to relate a scattering matrix (relating input to output) to a transfer matrix (relating left/right of an interface). REPLY [9 votes]: It is a principal pivot transform, also known as sweep operator or gyration. You can check the linked review paper. REPLY [6 votes]: Partial Inversion For Linear Systems And Partial Closure Of Independence Graphs<|endoftext|> TITLE: Regularity of random Fourier series QUESTION [6 upvotes]: The following two statements appear to be true (but do correct me if I am wrong): The coefficients of a $C^k$ function on the torus $T^n$ decay at least as fast as $x^{-k}$ (where $x$ is some norm on $\mathbb{Z}^n.)$ If the coefficients of a Fourier series decay at least as fast as $x^{-k-n},$ then the Fourier series represents a $C^k$ function on $T^n.$ It appears that 2. is not quite a converse to 1. Now, the question is: if I take a random Fourier series whose coefficients decay as $x^{-k},$ what is its degree of regularity? Is it $C^k?, C^{k-n}?$ none of the above? (the case of $n=1$ would already be of great interest). REPLY [6 votes]: I consider the case of independent Gaussian (or any light-tailed, for that matter) coefficients with variances decaying like $x^{-2k}$ - note that in this case the coefficients themselves decay essentially like $x^{-k}$, up to a logarithmic correction. For this random series the answer will be $C^{k-n/2}$, again up to a logarithmic correction. To prove this first note that the covariance function of this (stationary) process is in $C^{2k-n-\varepsilon}$ because Fourier series (of the covariance) decays like $x^{-2k}$. Then use the multidimensional version of Kolmogorov's continuity criterion, as formulated in, say, Lemma 2.1 of Scheutzow, (recall that for the Gaussians all $L^p$ norms are equivalent to the $L^2$ norm and use high moments there). It will follow that a Gaussian process with $C^{\alpha}$ covariance has $C^{\alpha/2-\varepsilon}$ sample paths, hence the result.<|endoftext|> TITLE: Image, kernel, quotient and first isomorphism theorem, in a category of monoid objects QUESTION [9 upvotes]: Let $\mathcal{C}$ be a monoidal category and Mon$_{\mathcal{C}}$ the category of monoids (also called algebra objects) on $\mathcal{C}$. Questions: are there definitions of image and kernel for a monoid morphism, and of quotient of monoids? Is there a generalization of the first isomorphism theorem of groups? Remark: if my question has a negative answer in general, I'm interested in the case where $\mathcal{C}$ is the category of bifinite (dualizable) completely reducible $R$-$R$-bimodules, with $R$ the hyperfinite ${\rm II}_1$ factor. REPLY [17 votes]: In a nonabelian setting the correct notion of kernel is given by the kernel pair, and the correct notion of cokernel is given by the cokernel pair. For example, in any category, a morphism $f : a \to b$ is a monomorphism iff its kernel pair exists and is trivial, and dually $f$ is an epimorphism iff its cokernel pair exists and is trivial. By comparison, the naive notions of kernel and cokernel already fail to have this property for monoids in $\text{Set}$. There are several reasonable notions of image depending on what you want to do. Let me highlight two that are particularly nice because they can be defined in terms of finite limits and colimits: the regular image $\text{im}(f)$ is the equalizer of the cokernel pair of $f$, while the regular coimage $\text{coim}(f)$ is the coequalizer of the kernel pair. (These are nonabelian generalizations of "kernel of the cokernel" and "cokernel of the kernel" respectively.) In a category with finite limits and colimits, both of these exist and allow us to factor a morphism $f : a \to b$ as a composite $$a \to \text{coim}(f) \to \text{im}(f) \to b$$ where the first map is a regular epimorphism and is universal with respect to that property, and the last map is a regular monomorphism and is universal with respect to that property. Example. Let $C = \text{Top}$. Then a monomorphism and an epimorphism are just an injective and a surjective map, respectively. The kernel pair of a map $f : X \to Y$ is the pullback $$X \times_Y X = \{ (x, y) \in X \times X : f(x) = f(y) \}$$ which you should think of as the equivalence relation on $X$ determined by $f$. Dually the cokernel pair is the pushout $Y \sqcup_X Y$. A regular monomorphism is an embedding while a regular epimorphism is a quotient map. The canonical factorization of a map $f : X \to Y$ as a composite $$X \to \text{coim}(f) \to \text{im}(f) \to Y$$ takes the following form: $\text{coim}(f)$ is the set-theoretic image of $f$ topologized as a quotient of $X$, while $\text{im}(f)$ is the set-theoretic image of $f$ topologized as a subspace of $Y$. The first isomorphism theorem for groups can be reintepreted in this language as the following claim: If $f : G \to H$ is a homomorphism of groups, then $\text{coim}(f) \to \text{im}(f)$ is an isomorphism. (The correspondence here is that $\text{im}(f)$ is the set-theoretic image while $\text{coim}(f)$ is the quotient of $G$ by the kernel of $f$. To really sell the first identification you need to first prove that every monomorphism of groups is regular, though.) A morphism $f$ with this property is sometimes called a strict morphism. The condition that every morphism is strict, which in particular is implicitly one of the axioms defining an abelian category, is somewhat rare: for example, it already does not hold in the category of commutative rings. Example. Let $R$ be a commutative ring and let $f : R \to S^{-1} R$ be a localization. Then the coimage-image factorization $$R \to \text{coim}(f) \to \text{im}(f) \to S^{-1} R$$ takes the following form: the first and last maps are both identities. In particular, the middle map is just $f$ itself, so it is almost never an isomorphism. And yet there is still a first isomorphism theorem for rings! The issue here is that the regular image doesn't agree with the set-theoretic image for rings; said another way, regular monomorphisms don't agree with monomorphisms for rings. If you like, you can consider a somewhat different generalization of the first isomorphism theorem as stated above, where you replace $\text{im}(f)$ with whatever you think the "correct" notion of image is. But $\text{coim}(f)$ is, to my mind, definitely the correct generalization of "quotient of the codomain by the kernel."<|endoftext|> TITLE: Can Schwartz-Zippel be formulated for commutative rings instead of fields? QUESTION [18 upvotes]: The polynomials which occur in the Schwartz-Zippel lemma could be defined for any commutative ring, yet the lemma is restricted to fields. This makes it inapplicable for $(1+x^n)=1+x^n(\operatorname{mod}n)$ and similar identities, and feels a bit "unnatural" to me. Why can't Lemma (Schwartz, Zippel). Let $P\in F[x_1,x_2,\ldots,x_n]$ be a non-zero polynomial of total degree $d \geq 0$ over a field, $F$. Let $S$ be a finite subset of $F$ and let $r_1, r_2, \dots, r_n$ be selected at random independently and uniformly from $S$. Then $\Pr[P(r_1,r_2,\ldots,r_n)=0]\leq\frac{d}{|S|}.$ be generalized to Conjecture. Let $P\in R[x_1,x_2,\ldots,x_n]$ be a non-zero polynomial of total degree $d \geq 0$ over a "nice" commutative ring, $R$. Let $S$ be a finite subset of $R$ with "$\forall s, t\in S:((\exists u\in R:(u\neq 0\land su=tu))\Rightarrow s=t)$" and let $r_1, r_2, \dots, r_n$ be selected at random independently and uniformly from $S$. Then $\Pr[P(r_1,r_2,\ldots,r_n)=0]\leq\frac{d}{|S|}.$ Here "nice" would be some property that is satisfied for common rings like $\mathbb Z$ or $\mathbb Z/n\mathbb Z$. For example being a subring of the direct product of a family of fields might work for square-free $n$. (So "nice" commutative ring might be replaced by subring of a commutative (von Neumann) regular ring.) The condition "$\forall s, t\in S:(\dots)$" tries to ensure that one can apply the normal Schwartz-Zippel lemma independently to each field from the underlying family of fields. This condition is automatically satisfied for a subring of a field (like $\mathbb Z$), and can be checked easily by verifying $\operatorname{gcd}(s-t,n)=1$ for $\mathbb Z/n\mathbb Z$, hence it is no real limitation. Question 1 Does this work, or am I overlooking something? (This question also refers to the implicit assumption that the reformulation is more useful than the original lemma, because it doesn't exclude "common rings like $\mathbb Z$ or $\mathbb Z/n\mathbb Z$". It applies to $(1+x^n)=1+x^n(\operatorname{mod}n)$, because "This condition ... can be checked easily by verifying $\operatorname{gcd}(s-t,n)=1$ for $\mathbb Z/n\mathbb Z$, hence it is no real limitation.") Question 2 Is it also possible to completely omit any "niceness" requirements for the ring, maybe by modifying the condition "$\forall s, t\in S:(\dots)$" slightly? Edit Note that Question 1 has been answered in the comments, i.e. I am indeed overlooking something. The reformulation is true for arbitrary commutative rings, as shown by Emil Jeřábek's nice proof. However, it is not at all clear whether it is really more useful than the original lemma: $q\prod^{p−1}_{i=0}(x−i)$ in $\mathbb Z/pq \mathbb Z$ is a non-zero polynomial of total degree $p$, but it evaluates to $0$ for any $x\in\mathbb Z/pq \mathbb Z$. So even the reformulation can't be used directly to probabilistically check $(1+x^n)=1+x^n(\operatorname{mod}n)$. What I overlooked here is that the ability to check whether $s-t$ is a zero-divisor for the individual witnesses from $S$ used to evaluate $P$ is not enough to ensure $|S|>d$ with sufficiently high probability to efficiently test the identity $P=0$. The reformulation also can't be used directly for identity testing over $\mathbb Z$. The problem here is that the intermediate results during evaluation of $P$ can become too big for efficient computation. The original lemma can be applied to this problem by first randomly drawing a sufficiently big prime number $p$, and then doing the computation in the field $\mathbb Z/p \mathbb Z$. The probability to draw a divisor of $P$ is quite small, if $p$ was drawn from a sufficiently big collection. One could hope now that the reformulation would allow to randomly draw a sufficiently big $p^\alpha$ instead, and then do the computation in the ring $\mathbb Z/p^\alpha \mathbb Z$. But the fact that $|S|\leq p$ makes this inapplicable for $d>=p$. What one can do is apply the original lemma in $\mathbb Q$ (or the reformulation in $\mathbb Z$), and then draw sufficiently big random numbers $n_i$ for each individual evaluation of $P$. This seems to require more random bits, and doesn't need the reformulation either. REPLY [4 votes]: See Section 4 in "On Zeros of a Polynomial in a Finite Grid" to see how Schwartz-Zippel lemma and many similar results on zeros of polynomials work for arbitrary commutative rings as long as the "grid" $S_1 \times S_2 \times \dots \times S_n \subseteq R^n$ satisfies a certain Condition (D) which is the same as the condition given in your conjecture applied to each $S_i$.<|endoftext|> TITLE: Any reason I should join ResearchGate? QUESTION [16 upvotes]: I am getting "invitations" to join ResearchGate. I am not a member of any other social network, as I consider it a waste of time. Are there good reasons for a mathematician to join ResearchGate? Can anybody provide experiences that speak for or against joining? REPLY [5 votes]: On Academia.SE there is a question ResearchGate: an asset or a waste of time?. Opinions there are mostly negative (not only not too beneficial, but also can annoy others [e.g. distinguished professors with whom you are collaborating] with emails). E.g.: As is, I have yet to hear any positive success story from my peers. All I've heard of ResearchGate are complaints about their invitation spam. or My experience with ResearchGate has been negative. I was searching for a paper online, and a ResearchGate page came up. I signed up as a member because they promised to send me a pdf of the paper. They never sent it and instead sent unrelated spam. That said, as with any social network, it can change over time. Personally, I would love to have a good profile/networking site for scientists, but it seems that we need to wait. Or as put by one comment on A.SE: My personal stance is that RG is a wonderful idea (especially the Q&A parts) which are terribly executed.<|endoftext|> TITLE: a formula about conic bundles QUESTION [7 upvotes]: Let $\pi:Z\to S$ be a conic bundle over a smooth complex surface $S$. I'd like to know how to prove that $-\pi_{*}K_{Z}^{2}=4K_{S}+\Delta$, where $\Delta$ denotes the locus in $S$ over which the fibres are singular. REPLY [2 votes]: There are a number of sources, or you could just prove this for yourself. In the context of algebraic geometry, this follows from Proposition 5.1(v) of Divisor Classes and The Virtual Canonical Bundle for Genus 0 Curves by de Jong and myself.<|endoftext|> TITLE: Removing pawns - the game QUESTION [18 upvotes]: Here is a simple game I've invented (if the idea is not fresh, then please let me know): The game is played on a board. The board has some (finite) number of lines drawn on it. A pawn is placed on each intersection point of (two or more) lines. Two players take alternate turns removing pawns. On each turn, a player removes one or more pawns. All pawns removed in a single turn have to be taken from the same line. The player who cannot make a move loses (alternatively: the player who takes the last pawn wins). Here is my question: For what values of m and n does the player who begin have a winning strategy when the game is played on an $n\times m$ rectangular grid? REPLY [18 votes]: Here is a partial answer. I'll assume $m$ and $n$ are the number of intersections rather than the number of squares. If both values are even, then the second player can always win by rotating the first player's previous move by 180 degrees. If precisely one value is odd, then the first player can win by removing one row or column, making it even by even. Obviously if $m$ or $n$ is $1$ then the first player can win. I expect the general odd by odd case to be much harder because $3\times 3$ is a second player win by case analysis and the strategy appears to have no symmetry. Also the Sprague-Grundy values of the subsets of the $3\times 3$ square do not seem to have any easy pattern. I think if you want more information the right thing to do is to write a plugin for cgsuite (easier than it sounds) and analyze the low odd cases and their subsets to see if you can find a pattern in the Sprague-Grundy values.<|endoftext|> TITLE: What is the Explicit Relationship between Coadjoint Orbits and Flag Manifolds? QUESTION [6 upvotes]: Given a complex semi-simple Lie group $G$, it acts smoothly on the dual $\frak{g}^*$ of its Lie algebra $\frak{g}$ by the coadjoint action. The orbits of that action are called coadjoint orbits. A maximal Zariski closed and connected solvable subgroup of $G$ is called a Borel subgroup; and $P$ is a parabolic subgroup of $G$ if it contains a Borel subgroup. We call the quotient $G/P$ a flag manifold. I know that these two families of spaces are related, but cannot find an exact statement of the relationship. REPLY [7 votes]: Let $G$ be a complex semi-simple Lie group. Then $$\mathcal O_\lambda\cong G/B\cong G/G_{\lambda}\cong G^{\mathbb C}/P$$ where $G^{\mathbb C}$ is the complexification of Lie group $G$ and in fact, every coadjoint orbit is projective variety with Kodiaira dimension $-\infty$. By following decomposition $$G^{\mathbb C}\cong G\times \mathfrak g^{*}\cong T^*G$$, where, $G^{\mathbb C}:=\exp \{\mathfrak g+i\mathfrak g\}$. If we take $\mu:T^*G\to \mathfrak g^*$, then by previous decomposition, $\mu^{-1}(\lambda)=G$, So, $$\mu^{-1}(\lambda)/G_\lambda\cong G/G_\lambda\cong \mathcal O_\lambda$$ Coadjoint orbits are symplectic quotient of cotangent bundle of Lie groups. Two geometric property of coadjoint orbits. They are Symplectic varieties and also Kaehler varieties. In fact if $M$ be a compact Kaehler manifold, then its symplectic quotient is also kaehler manifold. So, because $T^*G$ is Kahler manifold, so coadjoint orbit is also Kaehler. Allen Knutson, says that coadjoint orbits are birationally equivalent to its open Bruhat cells. If we complexify our coadjoint orbites, i.e, $$G^{\mathbb C}/G_\lambda^{\mathbb C}\cong \mathcal O_\lambda^{\mathbb C}$$, then the complexified of coadjoint orbits have hyper kahler structure and are Stein manifolds Hirzebruch computed the first chern class of flag varieties and coadjoint orbits He showed $c_1( \mathcal O_\lambda)=2\rho$, where $\rho=\frac{1}{2}\sum_{\alpha>0} \alpha$(half sum of positive roots of Lie group $G$) In final, computing some coadjoint orbits<|endoftext|> TITLE: Does "simplicial" commute with "Bousfield localization"? QUESTION [6 upvotes]: Let $M$ be a model category and $S \subseteq \operatorname{Mor}(M)$ a set of arrows in (the underlying strict category of) $M$. Recall that the left Bousfield localization $L_SM$ of $M$ with respect to $S$ is the model category structure on the same underlying category as $M$, with the same cofibrations, more weak equivalences, and correspondingly fewer fibrations; it is the universal model category structure with these properties in which the elements of $S$ are weak equivalences. Let $M^\Delta$ denote the category of simplicial objects in $M$. It has two related model category structures: the injective one, in which cofibrations and weak equivalences are "levelwise" (and hence fibrations are complicated), and the projective one, in which fibrations and weak equivalences are levelwise (and cofibrations are more complicated). I am interested in both structures, so in some sense this question is really two questions. Question: Is $(L_SM)^\Delta$ a left Bousfield localization of $M^\Delta$? What is an explicit description of the maps at which you localize? I expect the answer is "yes", and that you can localize at the set of maps in $M^\Delta$ that are levelwise in $S$. These seems like the type of thing that should be provable by definition unpacking. But I got stuck somewhere, suggesting that maybe there is a subtlety. The worry is that perhaps there just aren't a lot of levelwise-$S$-maps in $M^\Delta$, so that testing just against the levelwise-$S$-maps might give false positives when you are trying to find out if an object is levelwise-$S$-local. REPLY [6 votes]: Under the injective model structure or the Reedy model structure, the answer is yes: this is a left Bousfield localization. For any $X \in M$, define $\Delta[n] \otimes X$ to be the element of $M^\Delta$ given by $$ (\Delta[n] \otimes X)_k = \coprod_{(\Delta[n])_k} X, $$ with maps induced by the natural maps on coproducts. This is a functor which is a left adjoint: $$ Hom_{M^\Delta}(\Delta[n] \otimes X, Y) = Hom_M(X,Y_n). $$ Moreover, $\Delta[n] \otimes (-)$ preserves cofibrations in both the injective and Reedy model structures, and so $[\Delta[n] \otimes X,Y] = [X,Y_n]$ for fibrant $Y$. Thus, in these cases localizing with respect to maps of the form $\Delta[n] \otimes f$ for $f \in S$ gives you $(L_S M)^\Delta$. However, $\Delta[n] \otimes f$ is (except in the case $n=0$) unlikely to be levelwise in $S$ unless $S$ is closed under finite coproducts. (Quick comment: the Reedy model structure you linked to does not have its cofibrations defined levelwise.)<|endoftext|> TITLE: What are the higher homotopy groups of a K3 suface? QUESTION [31 upvotes]: All K3 surfaces have the same homotopy type. What are their higher homotopy groups? I know that $\pi_1$ is trivial, and $\pi_2$ is $\mathbb{Z}^{22}$. Even if the answer isn't known in all degrees, I'll accept an answer if someone can give me $\pi_3$. According to this question, you can equivalently tell me the higher homotopy groups of an Enriques surface. REPLY [22 votes]: Here are some details on how to compute the rank of $\pi_3$. A K3 surface $X$ is, in addition to being simply connected, a compact Kähler manifold, and such spaces are known to be formal in the sense of rational homotopy theory; this means that their rational homotopy can be computed by finding a Sullivan minimal model of their rational cohomology rings, and in particular only depends on the rational cohomology ring. (Terzić's paper linked to in Reimundo's answer uses instead that a compact oriented simply connected $4$-manifold is formal.) Here is, briefly, how this computation works out, at least if I'm not misreading something. The goal is to build a graded rational vector space $V^{\bullet} = \bigoplus_{k \ge 2} V^k$ and a differential $d$ on the exterior algebra $\Lambda^{\bullet}(V)$ such that the cohomology of $(\Lambda^{\bullet}(V), d)$ agrees with $H^{\bullet}(X, \mathbb{Q})$, $dV$ is contained in $\Lambda^{\ge 2}(V)$. The machinery of rational homotopy theory, together with the fact that $X$ is formal and has homology of finite type, then guarantees that we have a natural identification $$\pi_{\bullet}(X) \otimes \mathbb{Q} \cong \text{Hom}_{\mathbb{Q}}(V^{\bullet}, \mathbb{Q}).$$ In particular, $\dim \pi_{\bullet}(X) \otimes \mathbb{Q} = \dim V^{\bullet}$. So to compute $\dim \pi_3(X) \otimes \mathbb{Q}$ it suffices to figure out how many elements we need in $V^3$. We already know that we need $\dim V^2 = b_2$, where $b_2 = \dim H^2(X, \mathbb{Q}) = 22$. The cup product $H^2(X, \mathbb{Q}) \times H^2(X, \mathbb{Q}) \to H^4(X, \mathbb{Q})$ takes the form $$\alpha \cup \beta = Q(\alpha, \beta) \gamma$$ where $Q(\alpha, \beta)$ is the intersection form and $\gamma$ is a generator of $H^4(X, \mathbb{Q})$. The only way to impose these relations on the cohomology of $(\Lambda^{\bullet}(V), d)$ is to introduce elements in $V^3$ whose differentials will impose those relations. Explicitly, let $e_1, e_2, \dots e_{22}$ be an orthogonal basis for $H^2(X, \mathbb{Q})$ with respect to the intersection form, so that $Q(e_i, e_j)$ is some nonzero multiple of $\delta_{ij}$. For $i \neq j$ we need to introduce $\frac{b_2(b_2 - 1)}{2} = 231$ new elements of $V^3$, call them $f_{ij}, i \neq j$, so that we can impose the relations $$d f_{ij} = e_i \cup e_j.$$ For $i = j$ we need to introduce introduce $b_2 - 1 = 21$ new elements of $V^3$, call them $f_i, 1 \le i \le 21$, so that we can impose the relations $$d f_i = \frac{e_i \cup e_i}{Q(e_i, e_i)} - \frac{e_{i+1} \cup e_{i+1}}{Q(e_{i+1}, e_{i+1})}.$$ (We cannot introduce the generator of $H^4(X, \mathbb{Q})$ into $V^4$ because we cannot impose a relation that is linear in this generator, so instead we impose the relation that all of the $e_i$ square, up to a normalization, to the same thing.) Altogether, we get $$\dim V^3 = {b_2 \choose 2} + (b_2 - 1) = {b_2 + 1 \choose 2} - 1 = 252$$ as expected from the other answers.<|endoftext|> TITLE: Recent, elementary results in algebraic geometry QUESTION [26 upvotes]: Next semester I will be teaching an introductory algebraic geometry class for a smallish group of undergrads. In the last couple weeks, I hope that each student will give a one-hour presentation. The usual approach here might be to suggest some nice classical stories (e.g. stuff in this thread) and have each student pick one. I'm hoping to convince the class that algebraic geometry, even as it is currently practised, is not such a frightening field as they have been led to believe -- that there is still some low-hanging, elementary fruit. To that end, I'd like to mix in some more recent results, say papers appearing in the last year or two. I'm looking for possible topics. Question: Can anyone suggest recent papers s.t. (1) the statements are appealing to novices (2) the statements can be understood by an undergrad familiar with the material in Shafarevich's book (no schemes, no derived categories, no toric varieties, no moduli spaces... you get the idea) (3) the outline of the proofs could conceivably be understood and presented by said undergrads. These do not necessarily need to be ground-breaking papers in big deal journals. Minor results are fine, as long as they plausibly sound interesting to non-experts. A quick skim suggests the median number of suitable papers per day on math.AG is 0. Any suggestions? Self-promotion is welcomed. REPLY [15 votes]: This paper showed two century-old classification results, each of very undergrad-comprehensible things, were the same; pretty amazing! http://arxiv.org/abs/1308.0751 "Sums of squares and varieties of minimal degree" by Grigoriy Blekherman, Greg Smith, and Mauricio Velasco Let X be a real nondegenerate projective subvariety such that its set of real points is Zariski dense. We prove that every real quadratic form that is nonnegative on X is a sum of squares of linear forms if and only if X is a variety of minimal degree.<|endoftext|> TITLE: Which mapping class group representations come from algebraic geometry? QUESTION [26 upvotes]: Let $\Gamma_g$ be the mapping class group of a closed oriented surface $\Sigma$ of genus $g$. There is a natural surjection $t \colon \Gamma_g \to \mathrm{Sp}(2g,\mathbf Z)$ which sends a mapping class to the induced action on $H^1(\Sigma,\mathbf Z)$. Composing $t$ with any representation of the symplectic group produces a large number of linear representations of $\Gamma_g$. These are only a small fraction of all representations of the mapping class groups. Others can for instance be obtained from 3D TQFTs or by from different constructions involving lower central series. My question is however whether the symplectic representations are the only ones that can be defined "algebro-geometrically". Let me ask a more concrete question. A representation of $\Gamma_g$ is the same as a local system on the moduli space of curves of genus $g$, $M_g$. For a representation which factors through $\mathrm{Sp}(2g,\mathbf Z)$ this local system underlies a polarized variation of Hodge structure, since it is pulled back from a PVHS on the Shimura variety parametrizing principally polarized abelian varieties of genus $g$. Is the converse true - if a local system (say with $\mathbf Q$ coefficients) on $M_g$ underlies a PVHS, is it isomorphic to one of the symplectic local systems? REPLY [7 votes]: Kontsevich constructed a family of varieties over moduli space with interesting cohomology in the middle dimension. I don't think anyone proved that it is not the symplectic representation, but Kontsevich conjectured that is faithful.<|endoftext|> TITLE: Spirals in Apollonian circle-packings QUESTION [6 upvotes]: Given mutually (externally) tangent circles $C_1,C_2,C_3$, let $C_n$ be the unique circle externally tangent to $C_{n-1}$, $C_{n-2}$, and $C_{n-3}$ for $n \geq 4$. Let $P_{\infty}$ be the point toward which the $C_n$'s tend, let $P_n$ be the center of $C_n$, and let $Q_n$ be the point of tangency between $P_{n}$ and $P_{n+1}$. What can be said about the asymptotic distribution of the unit vectors pointing from $P_{\infty}$ to $P_n$? What about the unit vectors pointing from $P_{\infty}$ to $Q_n$? Although I would like the answer to be that these unit vectors are uniformly distributed over the unit circle, I see at least a couple of reasons to doubt this: first, I suspect that for certain special initial choices of $C_1,C_2,C_3$, the set of unit vectors that arise is finite, and second, the symmetry group governing Apollonian packings is the group of conformal maps, which does not preserve Lebesgue measure on the circle. Still, I suspect that for generic choices of $C_1,C_2,C_3$, the unit vectors are asymptotically distributed according to a measure that is uniformly continuous with respect to Lebesgue measure on the circle, and I suspect that the measures that arise in this way admit a nice characterization, e.g., the set of all measures that are conformally equivalent to Lebesgue measure. (This is a more refined version of my MathOverflow post Isotropy of Apollonian disk-packing . See also my follow-up question Three-dimensional Apollonian spirals .) REPLY [5 votes]: This configuration seems to be Coxeter's loxodromic sequence of tangent circles. According to Wikipedia: The radii of the circles in the sequence form a geometric progression with ratio $$k=\varphi + \sqrt{\varphi} \approx 2.89005\ ,$$ where φ is the golden ratio. k and its reciprocal satisfy the equation $$(1+x+x^2+x^3)^2=2(1+x^2+x^4+x^6)\ .$$ The centres of the circles in the sequence lie on a logarithmic spiral. Viewed from the centre of the spiral, the angle between the centres of successive circles is $$\cos^{-1} \left( \frac {-1} {\varphi} \right)\ .$$ Higher dimensional generalization was done by Coxeter (1968).<|endoftext|> TITLE: Asymptotic behaviour of $K$-Bessel function in transition range QUESTION [5 upvotes]: It is known that the famous mistake of Iwaniec-Sarnak in their paper of $L^\infty$ norm of eigenfunction of non-cocompact arithmetic surfaces in lemma (A1) is because of they did not consider the bump of $K$-Bessel function at transition range. First, I do not understand which part of the proof (sketched below) is flawed. An eigenfunction of $SL_2(\mathbb{Z})\backslash\mathcal{H}^2$ with eigenvalue $1/4+t^2$ looks like $$\phi(z)=\sqrt{y}\sum_{n\neq0}a_nK_{it}(2\pi|n|y)e(nx).$$ First they proved a bound of sum of Fourier coefficients (Ramanujan on average) which is $(t>0)$ $$ \sum_{n\le N}|a_n|^2\ll e^{\pi t}(t+N). $$ Then, I guess that they are using a Cauchy-Schwarz inequality and an asymptote that for $t>2\pi|n|y$ $$e^{\pi t}|K_{it}(2\pi|n|y)|^2\ll (t^2-|n|^2y^2)^{-1/2},$$to deduce final final bound. The above inequality can be found here lemma 2.3 and 2.4. Isn't the above asymptote correct? 2)It is mentioned in Sarnak's letter to Morawetz (equation 43) that $K_{it}$ has a bump of size $t^{1/6}$ near $y=t$. Do we know precise asymptote of $K_{it}$ uniformly in $y$ and $t$ so that the above argument can be rectified? REPLY [7 votes]: For a published account of the corrected proof, see Section 10 in Blomer-Holowinsky: Bounding sup-norms of cusp forms of large level, Invent. Math. 179 (2010), 645-681. See especially pages 679-680, where you can also find the precise asymptotics of $K_{it}$ in the transitional range. Actually, a few years ago, a colleague of mine asked me about the same thing. What follows is my email response in slightly edited form (I am too busy to re-read carefully): There are several problems with Lemma A.1 and its proof. The problems arose from not using correctly the asymptotics of $K_{ir}(y)$ in the transitional range $y\approx r$. For $|y-r|r^{1/3}$ it is about $e^{-\pi r/2} |y^2-r^2|^{-1/4}$ times a phase, with the additional remark that for $y$ large the exponential decay kicks in. As a consequence: The display below (A.2) is wrong: $r^{-1}$ should be lowered to $r^{-4/3}$. This is not a big problem, since if we integrate from $r/2$ (instead from $r$), then the bound is OK (I learned these things from papers by Strömbergsson). In other words, in the line before (A.3) one should set $N=r/(4\pi Y)$. Actually (A.3) is too weak for this purpose, as explained below. Instead the following should be used, which is known by deeper methods: $$ \sum_{0<|n|1$ and $y>1$, but it is valid for all $y>0$ and all forms (including forms violating the Selberg conjecture). Added. For a generalization of (2) to spherical Hecke-Maass forms over an arbitrary number field, see Lemma 9 in this paper. Note that $\phi$ in that lemma satisfies ${\|\phi\|}_2=1$. Note also that (1) above was (*) in an earlier version of this post.<|endoftext|> TITLE: Bijective proof of an Abel-Hurwitz-type identity QUESTION [7 upvotes]: Can anyone sketch for me a bijective proof of the fact that the number of spanning trees of the complete graph on $n$ vertices, $K_n$ (given by the formula $ t_n = n^{n-2}$), satisfies $ t_n = \frac{n}{2} \sum_{k=1}^{n-1} {n-2 \choose k-1} t_{k} t_{n-k} $? Sasha Postnikov suggested that I take a look at http://math.mit.edu/~apost/papers/AbelHurwitz.pdf but despite a fairly strong resemblance between his identities and mine I don't see how to link them. REPLY [8 votes]: I believe this is the proof Postnikov suggested at CCCC LXI. Let us say that $K_n$ has vertices $1,2,\ldots,n$. Imagine a spanning tree of $K_n$ as being rooted at $1$. To any spanning tree $T$, we associate $T'$, the part of the tree at or below the vertex $2$ in this tree, and $T''$, the other part of the tree (which at least contains the root vertex $1$). Say $|T'| = k$ with $1 \leq k \leq n-1$. Then there are $\binom{n-2}{k-1}$ possible sets of labels for $T'$, $t_k$ possible trees for $T'$ given such labels, $t_{n-k}$ trees for $T''$ (whose set of labels is now determined), and $n-k$ places to "stick" $T'$ into $T''$ (which amounts to a choice of which vertex in $T''$ the vertex $2$ is a child of). This proves $$t_n = \sum_{k=1}^{n-1} (n-k) \binom{n-2}{k-1} t_k t_{n-k}.$$ Your equation then follows by grouping the term $k$ with the term $n-k$ in that sum, and then redistributing.<|endoftext|> TITLE: A technical question in Feix's construction of hyperkahler metric on cotangent bundles QUESTION [6 upvotes]: I am now reading Feix's paper Hyperkahler metrics on cotangent bundles and I have a technical question to ask. In his paper, for an analytic Kähler manifold $(X,J,\omega)$, Feix considered its complexification $X^c$ which in my understanding can be thought of as a neighborhood of the diagonal in $X\times\bar{X}$, where $\bar{X}$ is the complex manifold $X$ with complex structure $-J$. One can extend $\omega$ analytically to a holomorphic symplectic form $\omega^c$ on $X^c$. This $\omega^c$ determines two natural holomorphic Lagrangian foliations $L_+$ and $L_-$. Let $z_i$ and $z'_j$ be local holomorphic coordinates of $X$ and $\bar{X}$ respectively, then the leaves of $L_+$ and $L_-$ are given by $z_i\equiv const$ and $z'_j\equiv const$. As the diagonal intersects each leaf at exactly one point, one may identify the space of leaves of $L_+$ with $X$ and the space of leaves of $L_-$ with $\bar{X}$ respectively. From now on let us focus on $L_+$ exclusively. By shrinking $X^c$ if necessary, one may assume that $\Lambda_x$ is simply connected for any $x\in X=$ "space of leaves of $L_+$", where $\Lambda_x$ is the leaf corresponding to $x$. As a consequence of Lagrangian foliation, each $\Lambda_x$ has a natural affine structure, so it makes sense to write $V_x$ to be the space of affine functions on $\Lambda_x$. By 1-connectedness of $\Lambda_x$, each $V_x$ is a vector space of complex dimension $n+1$, where $n$ is the complex dimension of $X$. These $V_x$ patch up to a complex vector bundle $V\to X$. Feix claims without further explanation that this bundle is holomorphic. I really would like to know the description of the holomorphic structure here. It occurs to me that the most natural frames one can think of actually have anti-holomorphic transition functions as follows: Let $z_i,z'_j$ be local coordinates for $X^c$, the leaves of $L_+$ are $z_i\equiv const$. Fix a leaf $\Lambda_x$, it intersects the diagonal at the point whose coordinate is $z=x,z'=\bar{x}$. One can find parallel 1-forms $\theta_i$ on $\Lambda_x$ by parallel transport with respect to the flat connection with initial value specified by $\theta_i|_{(x,\bar{x})}=\textrm{d}z'_i|_{(x,\bar{x})}$. These $\theta_i$ must be a closed form and their primitives $f_i$ along with the constant function 1 form a basis of $V_x$. However, if you work with this particular frame, then under holomorphic coordinate change of $X$, the transition matrix of these frame depends anti-holomorphically on $X$. Surely one can switch $L_+$ and $L_-$ to solve the holomorphicity problem here. But I think a bigger trouble is then introduced since in that case the map $\phi$ defined by Feix loses its holomorphicity. Thank you! REPLY [3 votes]: I just read your question. Funny, I was also trying to read this paper carefully not long ago, and I was annoyed by exactly that kind of problem. I think the problem is to say that "the space of leaves of $L_+$ is $X$". In the complex manifold $X^c$ (which we can think of as $X^c = X \times \overline{X}$ indeed), the space of leaves of $L_+$ (let's call it $B_+$ like Feix), as a complex manifold , can naturally be identified with a transverse complex submanifold, at least locally. However $X$ (embedded as the diagonal in $X^c$) is not a complex submanifold! Instead, a natural choice of a transverse complex submanifold is a leaf of $L^-$, i.e. a horizontal slice $X \times \{x_0\} \subset X \times \bar{X}$. In this sense the space of leaves $B_+$ can be identified to $X = X \times \{x_0\}$, and the projection $L_+ \to B_+ \approx X$ is holomorphic, and so is the vector bundle $V \to B_+$. Also, I think this is the only way to make sense of what Feix writes at some point later (proof of Lemma 2): The complexified manifold $X^c$ is foliated by Lagrangian submanifolds and choosing a suitable section of the quotient map onto $B_+ \approx X$ we can identify $X$ as a transverse Lagrangian section. So, problem solved, right? I don't think so. As far as my understanding goes there are at least two problems (that are related) with this, if you allow me to expand. As you write, maybe $X^c$ (or the part you are working with) is just a neighborhood of the diagonal in $X \times \overline{X}$. For instance, if you insist that the leaves are simply connected but $X$ is not. So one can only identify $B_+$ locally with $X$. A more serious problem is the question of the choice of the (local) identification of $B_+$ with $X$. How does that affect Feix's construction? Obviously, it is crucial in Feix's proof that the twistor space she constructs is the twistor space of a hyperkähler structure on... $T^*X$! In order to justify that, she claims (a few lines down): We also notice that the above argument allows us to identify a neighbourhood of the zero section in $\pi^{-1}(0) \subset V^*$ with a neighbourhood of the zero section of the cotangent bundle. If what I said earlier is correct, she is (or she should be) talking about the cotangent bundle to $B_+$, and not $X$. And what worries me is that I checked: there is no way the hyperkähler structure Feix constructs is invariant under a transformation corresponding to changing the local identification $X \approx B_+$ by choosing different leaves of $L_-$ (locally) representing $B_+$. I wish I understood better the hyperkähler structure on the cotangent bundle of a Kähler manifold, but I understand neither the intuition behind Feix's construction (why should the complex structure $I_\lambda$ correspond to the Feix's twistor space's fiber $\pi^{-1}(\lambda)$?) nor the technical details of why this construction works, and what I pointed out here is one of my main issues. I would be happy if someone explained it to me. If you figure it out please let me know!<|endoftext|> TITLE: Associated graded Lie algebra of braid groups QUESTION [5 upvotes]: Let $G$ be a group and let $\Gamma_G(k)$ be the $k$th term of the lower central series of $G$. For each $k\geq 1$, set $\mathcal{L}_G(k)=\Gamma_G(k)/\Gamma_G(k+1)$ and $$\mathcal{L}_G:=\bigoplus_{k\geq 1}\mathcal{L}_G(k).$$ Then $\mathcal{L}_G$ has a graded Lie algebra structure induced from the commutator bracket on $G$. I am looking for the known results about $\mathcal{L}_G$ where $G$ is a braid group or reduced braid group (the factor group of braid group such that each string is allowed to intersect itself). I think that F.R. Cohen has some work on this, but I am not sure in which paper. Any references will be much appreciated. REPLY [5 votes]: Take a look at the following papers F.R. Cohen - S. Prassidis: "On injective homomorphisms for pure braid groups, and associated Lie algebras", J. Algebra 298 (2006), no. 2, 363–370. (available at the link http://arxiv.org/abs/math/0404278) and F.R. Cohen - J. Wu: "On braid groups and homotopy groups", Groups, homotopy and configuration spaces, 169–193, Geom. Topol. Monogr., 13, Geom. Topol. Publ., Coventry, 2008. (available at the link http://arxiv.org/pdf/0904.0783.pdf)<|endoftext|> TITLE: When is the topology generated by countable subsets? QUESTION [5 upvotes]: Let $X$ be a topological (Hausdorff) space and let $(X_\alpha)_\alpha$ be a directed family of subsets. We say that $(X_\alpha)_\alpha$ generates the topology of $X$ if a subset $U \subseteq X$ is open iff $U\cap X_\alpha$ is open in $X_\alpha$ with respect to the induced topology. Another way of saying this is that $X$ is the direct limit of the topological spaces $(X_\alpha)_\alpha$ where each $X_\alpha$ holds the subspace topology. It is well-known that the topology of a metrizable space is generated by the family of all compact subsets (one says $X$ is a "$k$-space") since the topology is determined by convergent sequences. With exactly the same argument, we obtain the result that the topology of a metrizable space is generated by all countable subsets. My question is now: For wich non-metrizable Hausdorff spaces it is true that the topology is generated by countable subsets? Trivially, this also holds for countable spaces but I would expect that there are many more examples. Also, a counter-example would be interesting. In particular I am interested in examples of the form $\mathbb R^I := \prod_{i\in I} \mathbb R$ or $(\mathbb Z/2\mathbb Z)^I := \prod_{i\in I} (\mathbb Z/2\mathbb Z) $ for an uncountable index set $I$, but of course all other examples or counter-examples are welcome as well. Thanks in advance, Tom REPLY [7 votes]: A space $X$ is said to have countable tightness if whenever $A \subseteq X$ and $p\in \bar{A}$, there is a countable $B \subseteq A$ such that $p \in \bar{B}$. It is not hard to see that a space has countable tightness if and only if its topology is generated by countable sets (in the sense described in the question), but I find it easier to think in these terms. There are easy examples of non-metrizable countably tight spaces. For instance, the one point compactification of an uncountable discrete space. Obviously any first-countable space has countable tightness, so both $\mathbb{R}^I$ and $2^I$ have countable tightness if $I$ is countable. If $I$ is uncountable then let $A$ be the set of functions which take value $1$ at countably-many coordinates and value $0$ at the rest. Then the function with constant value $1$ (call it $p$) is in the closure of $A$ but not in the closure of any countable subset of $A$. Here we are working in either $\mathbb{R}^I$ or $2^I$; so none of these has countable tightness.<|endoftext|> TITLE: Counting valid coordinates QUESTION [6 upvotes]: We are given a matrix $D = (d(i,j))_{1 \leq i,j \leq n}$ such that $d(x,z) \leq d(x,y) + d(y,z)$ for each $1 \leq x,y,z \leq n$. It is also known that $d(x,y) \in \mathbb{N}$ (In this question $0 \in \mathbb{N}$) edit: You may assume $D$ is symmetric, and there are zeroes on the main diagonal. A coordinate $t = (t_1,t_2,\dots,t_{n}) \in \mathbb{N}^n$ is said to be "good" if for all $1 \leq a,b \leq n$: $|t_a - t_b| \leq d(a,b)$ We say that two coordinates are equivalent if one could be obtained from the other by adding a constant to all the entries. How many "good" coordinates are there, given the matrix D? (We don't count equivalent coordinates twice). Any idea of how to approach this is highly appreciated. Where does this come from? Given a graph $G=(V,E)$ (simple and undirected), and a subset $T \subseteq V$ we could get "coordinates" for every node in the graph in the following manner: For every vertex $v \in V$ we define $coord(v) := (dist(v,a))_{a \in T}$. Here $dist$ is the length of the shortest path in the graph. In the question above, the matrix $D$ is the set of distances ${dist(a,b)}_{a,b \in T}$. It is interesting to know at what point the coordinates space saturates. (Which means that the set $T$ is too small for the set $G$). One way to find out would be to count the amount of possible coordinates, given the triangle inequalities restrictions. The formulation above does not take into account the requirement for $t_a + t_b \geq d(a,b)$, however it could be obtained by adding a constant to all the entries. Example To make the question more concrete, I include here an example. For the matrix: 0 2 1 2 0 1 1 1 0 We get the list of solutions: (0, 0, 0) (0, 0, 1) (0, 1, 0) (0, 1, 1) (0, 2, 1) (1, 0, 0) (1, 0, 1) (1, 1, 0) (2, 0, 1) A total of 9 solutions. Further Progress I noticed that for a matrix $D$ where $d(i,j) = k$ for $i \neq j$ and $d(i,j)=0$ otherwise we get that the amount of valid coordinates is $(k+1)^n - k^n$. In the case of $n=3$, If the distances inside $D$ are $a,b,c$, then we could represent them as $a=y+z, b=x+z, c=a+y$, where $x,y,z \geq 0$. This helps us to get rid of the triangle inequalities constraints. Is there an equivalent for cases of $n>4$? REPLY [2 votes]: The Question considers an $n\times n$-matrix $\ d,\ $ which is in fact a pseudo-matrix in the n-space $\ X:=\{1\ldots n\}.\ $ The goal is to study vectors $\ t\in\mathbb R^n,\ $ called coordinates, which satisfy the condition: $$\forall_{w\ x\,\in\, X}\ \ |t(w)-t(x)|\le d(w\ x)$$ plus the restriction to the integer values. Even for general real values, I will still call such functions $\ t\ $ (vectors)--coordinates. Let me consider the special case of an arbitrary metrics $\ d\ $ (rather than a pseudo-metrics), and let it be defined in an arbitrary set $\ X,\ $ so that we consider arbitrary metric spaces $\ (X\ d)\ $ (in full generality). I hope that my notion of metric space $\ Aim(X),\ $ which is closely related to the coordinates, will help the task of finding all coordinates. The reference is: wh, On metric spaces aimed at their subspaces, Prace Matematyczne X (1966), pp.95-100. First observe that every superspace $\ (Y\ d')\ $ of $\ (X\ d)\ $ is a source of (generalized) coordinates; namely, every function $\ t:=t_y,\ $ defined by: $$\forall_{x\in X}\ \ t_y(x) := d'(x\ y)$$ is a coordinate. A large supply of such coordinates can be obtained from a single space $\ Aim(X),\ $defined below. Perhaps the study of all coordinates can be reduced to functions $\ t_y.\ $When possible, it should be rather more convenient to analyze just one space rather than a huge number of the. Here is the definition of $\ Aim(X)$: Space $\ Aim(X)\ $ is the space of all functions $\ t:X\rightarrow\mathbb R\ $ such that: $$\forall_{x\in X}\ \sup_{y\in X} |d(x\ y)-t(y)|\ =\ t(x)$$ This condition is equivalent to: $$\forall_{x\ y\in X}\ t(x)+t(y)\ \ge\ d(x\ y)\ \ge t(x)-t(y)$$ This last version is similar but more restrictive than the coordinate condition. The space $\ Aim(X)\ $ is equipped with the uniform metrics. This space consists of functions obtained in a way described earlier above from all super-spaces $\ (Y\ d')\ $ which are aimed at $\ X,\ $ meaning that: $$\forall_{p\ q\in Y}\forall_{\epsilon > 0}\exists_{x\in X}\,\ |d'(p\ x)-d'(q\ x)|+\epsilon\ >\ d'(p\ q)$$ This is what I meat by large supply. Actually, if we identify (isometrically) $\ X\ $ with the subspace of $\ Aim(X)\ $ of all functions $\ t_x\ (x\in X)\ $ such that $\ \forall_{w\in X}\ t_x(w):=d(w\ x),\ $ then space $\ Aim(X)\ $ is aimed at $\ X$, and it is universal in this respect (by the way, it follows that spaces aimed at any fixed space $\ X\ $ cannot be arbitrarily large). The Question asks about  Any idea of how to approach this,  and at this time this is my idea, a bit raw though. Perhaps the study of all coordinates can be reduced to $\ Aim(X)$.<|endoftext|> TITLE: Localized J homomorphism QUESTION [5 upvotes]: Let $X$ be a simply connected finite CW complex, $\xi$ and $\eta$ vector bundles over $X$ of the same dimensions and their dimension is big enough, so they are stable bundles. Let $p$ be a prime. Are the following two conditions equivalent? $J(\xi)$ and $J(\eta)$ as elements in $J(X)$ have equal $p$-primary components. (That is there is a $q$ not divisible by $p$ such that $q(J(\xi) - J(\eta) ) = 0.$) The sphere bundles $S(\xi)$ and $S(\eta)$ are fiberwise $p$-equivalent. (That is there is a fiberwise map $S(\xi) \to S(\eta)$ that induces isomorphism in homologies with $Z/pZ$ coefficients.) REPLY [3 votes]: That 2. implies 1. is a consequence of Adams' mod k Dold Theorem (Theorem 1.1 of On the groups J(X)-I. Topology 2 (1963) pp. 181-195). The following is an argument for 1. implies 2. (using ideas similar to those in Adams' proof of the above as well as Sullivan's theory of p-local spherical fibrations contained in his MIT Geometric Topology notes (Chapter 4)). If $E \to X$ is a spherical fibration ($p$-local or not) over $X$, I will write $\Sigma E$ for its fiberwise suspension. It is sufficient to produce a map $\Sigma^m S(\xi) \to \Sigma^m S(\eta)$ with fiberwise degree prime to $p$ for some large $m$, as obstruction theory and the stability assumption on $\xi$ and $\eta$ allow us to compress such a map down to a map $S(\xi) \to S(\eta)$. [Edit:] In more detail, let $n$ be the dimension of the fibers of $S(\xi)$ and $S(\eta)$. The stability assumption is that $\dim X1$. As the space of fiberwise maps $\Sigma^m S(\xi) \to \Sigma^m S(\eta)_{(p)}$ is the space of sections of a fibration with fiber $Map(S^{n+m},S^{n+m}_{(p)})$ the problem reduces to finding a long diagonal lift in the commutative diagram $$ \begin{array}{ccc} S^{k-1} & \longrightarrow & Map(S^{n+m},S^{n+m}) & \stackrel{j}{\longrightarrow} & Map(\Sigma S^{n+m}, \Sigma S^{n+m}) \\ \downarrow & & \downarrow & & \downarrow\\ D^k & \longrightarrow & Map(S^{n+m},S^{n+m}_{(p)}) & \stackrel{j}{\longrightarrow} & Map(\Sigma S^{n+m},\Sigma S^{n+m}_{(p)}) \end{array} $$ where the horizontal maps $j$ denote fiber multiplication by a natural number $j$ (using the suspension coordinate). The middle vertical map is $p$-localization on higher homotopy groups, and $j$ induces multiplication by $j$ on $\pi_{k-1}$ so we are done.<|endoftext|> TITLE: Natural associative law for a ternary "group"? QUESTION [10 upvotes]: Suppose one were to define a group-like structure based on a set $G$ with a ternary (rather than binary) operator $g( a, b, c ) = \left< a, b, c \right>$. One possible definition for the associative law is that $$ \left< \left< a, b, c \right>, d, e \right> = \left< a,\left< b, c, d \right>, e \right> = \left< a, b, \left< c, d, e \right> \right> \;. $$ Is there some algebraic structure defined along these lines that has been investigated? If so, I'd appreciate a name & pointer. Are there other natural definitions of ternary associativity? And definitions of ternary identity laws? (Not entirely relevant, but I came upon this thinking about medians and centroids, and I am just looking for possible models...) REPLY [12 votes]: What you need is the keyword "polyadic groups". A polyadic group is a non-empty set $G$ equipped with an associative $n$-ary operation $f:G^n\to G$ such that for all $a_1, \ldots, a_{n}$ and $b\in G$, the equations $$f(a_1, \ldots, a_{i-1}, x, a_{i+1}, \ldots, a_n)=b, (1\leq i\leq n)$$ have (unique) solution for $x$. The simplest examples are polyadic groups derived from ordinary groups: Let $(G, \cdot)$ be a group and define an $n$-ary operation on $G$ by $$f(x_1, \ldots, x_n)=x_1x_2\cdots x_n.$$ Then $(G,f)$ becomes a polyadic group which is called the polyadic group derived from $(G, \cdot)$ and it is denoted by $\mathrm{der}(G, \cdot)$. It is easy to see that only polyadic groups of this type have identity element. In general, the structure of a polyadic group can be described in terms of ordinary groups and their automorphisms. If $(G, f)$ is a polyadic group, then there exists a binary operation $\ast$ on $G$ such that $(G, \ast)$ is a group, and there is an automorphism $\theta\in \mathrm{Aut}(G, \ast)$ and an element $b\in G$ such that $\theta^{n-1}(x)=bxb^{-1}$ and $\theta(b)=b$ and $$f(x_1, \ldots, x_n)=x_1\ast\theta(x_2)\ast\cdots \ast\theta^{n-1}(x_n)\ast b$$ and the converse is also true. So, the polyadic group is denoted in general by $\mathrm{der}_{b, \theta}(G, \ast)$. Many problems concerning polyadic groups are solved until now; Emil Post proved that any polyadic group $(G, f)$ has a covering group $G^{\ast}$ which is generated by $G$ and inside $G^{\ast}$ we have $$f(x_1, \ldots, x_n)=x_1x_2\ldots x_n$$. Moreover $G^{\ast}$ contains the retract group of $(G, f)$ as a normal subgroup and the corresponding quotient is cyclic of order $n-1$. Here is a list of some interesting works done for polyadic groups and you can find appropriate references using keywords in the internet. 1- The variety of polyadic groups and its subvarieties is studied by Dudek, Artamonov, and ... . 2- The universal Post cover and retracts are studied by Michalisky and Dudek. 3- Free polyadic groups are studied by Artamonov. 4- Representation theory of polyadic groups are studied by me and Dudek. 5- The structure of homomorphisms, automorphisms, and an isomorphism problem for polyadic groups, studied by me and Khodabandeh. 6- Simple polyadic groups are characterized by me and Khodabandeh. Similar structures are also studied during the past decades, among them are $n$-ary generalization of Lie algebras (Fillipov algebras) and $(m,n)$-rings. Here is a list of some useful references:\ Dudek W. A. Remarks on n-groups // Demonstr. Math. 1980. V. 13. P. 165–181. Post E. L. Polyadic groups // Trans. Amer. Math. Soc. 1940. V. 48. P. 208–350. Dornte W. Unterschungen ¨uber einen verallgemeinerten Gruppenbegriff // Math. Z. 1929. Bd 29. S. 1–19. Kasner E. An extension of the group concept // Bull. Amer. Math. Soc. 1904. V. 10. P. 290–291. Galmak A. M. N-ary groups. Gomel: Gomel Univ. Press, 2003. Dudek W. A. Varieties of polyadic groups // Filomat. 1995. V. 9. P. 657–674. Galmak A. M. Remarks on polyadic groups // Quasigroups Relat. Syst. 2000. V. 7. P. 67–70. Gleichgewicht B., Glazek K. Remarks on n-groups as abstract algebras // Colloq. Math. 1967. V. 17. P. 209–219. Dudek W. A., Michalski J. On retract of polyadic groups // Demonstr. Math. 1984. V. 17. P. 281–301. Dudek W. A., Glazek K. Around the Hossz´u–Gluskin theorem for n-ary groups // Discrete Math. 2008. V. 308. P. 4861–4876. Dudek W. A., Michalski J. On a generalization of Hossz´u theorem // Demonstr. Math. 1982. V. 15. P. 437–441. Hosszu M. On the explicit form of n-groups // Publ. Math. 1963. V. 10. P. 88–92. Ushan J. Congruences of n-group and of associated Hossz´u–Gluskin algebras // Novi Sad. J. Math. 1998. V. 28. P. 91–108. Dudek W. A., Shahryari M. Representation theory of polyadic groups // Algebr. Represent. Theory. 2012. V. 15. P. 29–51. Shahryari M. Representations of finite polyadic groups // Commun. Algebra. 2012. V. 40. P. 1625–1631. Khodabandeh H., Shahryari M. On the representations and automorphisms of polyadic groups // Commun. Algebra. 2012. V. 40. P. 2199–2212. Khodabandeh H., Shahryari M. Simple polyadic groups// Siberian Math. Journal, 2014.<|endoftext|> TITLE: Is there a continuous analogue of Ramanujan graphs? QUESTION [10 upvotes]: I think it might help to think of the following definition of a Ramanujan graph - a graph whose non-trivial eigenvalues are such that their magnitude is bounded above by the spectral radius of its universal cover. By "non-trivial eigenvalues" I mean all the eigenvalues except the highest and the smallest. A universal cover of a graph is the infinite tree such that every connected lift of the graph is a quotient of the tree. The spectral radius of a graph would be the norm of its adjacency matrix. It would be helpful if people can give any pointers along these directions.. REPLY [16 votes]: In fact, the original motivation behind Lubotzky--Phillips--Sarnak's construction of Ramanujan graphs was in analogy with modular curves $Y(N)=\mathbb H^2/\Gamma(N)$ for the principal congruence subgroups $\Gamma(N)\subseteq\operatorname{PSL}(2,\mathbb Z)$. So the answer is yes, there is a continuous analogue, but in fact it came first! Let me give a few more details. The spectrum of the Laplacian $\Delta$ on hyperbolic space $\mathbb H^2$ consists of $[\frac 14,\infty)$. Selberg proved that the smallest positive eigenvalue of the Laplacian on $Y(N)$ satisfies $\lambda_1(Y(N))\geq\frac 3{16}$, and conjectured that $\lambda_1(Y(N))\geq\frac 14$. Note that $\frac 14$ is exactly the inf of the spectrum of $\Delta$ on the universal cover $\mathbb H^2$. You can read more about this in an article by Sarnak. As far as I understand things, Lubotzky--Phillips--Sarnak's examples of Ramanujan graphs are discrete analogues of modular curves.<|endoftext|> TITLE: Why is Klein's representation of $PSL_2(\mathbb{F}_7)$ hard to obtain? QUESTION [16 upvotes]: In his famous article [1] Klein constructs a representation of $G=PSL_2(\mathbb{F}_7)$ in $\mathbb{C}^3$ (of which the first invariant polynomial of three variables gives rise to the famous Klein's quartic). All other irreducible representations of $G$ are very simple or natural to obtain: the action on $\mathbb{P}^1_{\mathbb{F}_7}$ produces a representation of dimension $7$. The isomorphism $G\simeq GL_3(\mathbb{F}_2) = PGL_3(\mathbb{F}_2)$ and the action on the Fano plane $\mathbb{P}^2_{\mathbb{F}_2}$ produces a representation of dimension $6$. There's the trivial representation, one of dimension $8$ that is simply induced from the normalizer of a $7$-Sylow subgroup (which can be also thought geometrically as these subgroups are point-stabilizers of the above mentioned actions). The two missing are Klein's representation and its conjugate. However, to construct those we appeal to explicit generators and relations by writing explicit $3\times 3$ matrices that correspond to the generators of $G$ of orders $7,3$ and $2$ (this last involution is the hard one to find). Is there a geometric intuition behind this representation? and if not (as it seems to be) is there an inherent reason for this representation being "hard" to obtain? In the fantastic article Elkies explains this last involution as the discreet Fourier transform on the space of functions $\mathbb{F}_7 \rightarrow \mathbb{C}$. To my mind this appearance of a Fourier transform is indication of some deeper explanation of this action involving some more elaborate geometric tricks as Mukai transforms more than an explanation itself. Many things that you may want to know about $G$ or Klein's quartic are beautifully explained in that article of Elkies. Finally, it is worth noticing that this phenomenon occurs for other groups of the form $PSL_2(\mathbb{F}_q)$ where most irreducible representations are geometric in nature, except the few ones that you need to write by hand the generators. [1]: F. Klein, “Ueber die Transformationen siebenter Ordnung der elliptischen Funktionen”, Math. Annalen 14 (1879), 428–471. REPLY [17 votes]: I guess the answer depends on what you call "geometric"... If you accept some basic algebraic geometry, you can do the following. Consider the homographs $\ \alpha :z\mapsto z+1\ $ and $\ \beta : z\mapsto -1/z\ $ of $\ \mathbb{P}^1_{\mathbb{F}_7}$. We have $\alpha ^7=\beta ^2=(\alpha \beta )^3=1$; it is an easy exercise to show that $\alpha $ and $\beta $ generate $PSL_2(\mathbb{F}_7)$. Thus we have a surjective map $\pi _1(\mathbb{P}^1\smallsetminus\{0,1,\infty\} )\rightarrow PSL_2(\mathbb{F}_7)$, hence a Galois covering $C\rightarrow \mathbb{P}^1$ with group $PSL_2(\mathbb{F}_7)$, branched along $\{0,1,\infty\}$. The Riemann-Hurwitz formula gives $g(C)=3$. The group $PSL_2(\mathbb{F}_7)$ acts on $C$, hence on the space of holomorphic forms $\Omega _C$, and this is the Klein representation.<|endoftext|> TITLE: Is a number field uniquely determined by the primes which split in it? QUESTION [17 upvotes]: Let $K/\mathbb{Q}$ be a number field. We say that a rational prime $p$ splits in $K$ if there exists a prime $\mathfrak{p}$ of $K$ above $p$ of interia degree $1$. Is a number field $K$ uniquely determined by the set of primes which split in $K$? A well-known application of the Chebotarev density theorem (Neukirch Cor. 13.10) says that this is true when $K/\mathbb{Q}$ is Galois (note that here a prime splits if and only if it splits completely). So I am really interested in what happens in the non-Galois case. Note also that the answer to the analogous question for completely split primes is no; indeed a prime is completely split in $K$ if and only if it is completely split in the Galois closure of $K$. REPLY [27 votes]: See exercises (6.3) and (6.4) of Cassels-Frohlich book on algebraic number theory. In these exercises, an example of two number fields $E,E'$ with the same zeta function is given; therefore, the set of primes which split in $E,E'$ are the same. This amounts to constructing two subgroups $H,H'$ in a finite group $G$, which are not conjugate but meet every conjugacy class of $G$ in the same number of elements. Since the OP has asked for split (but not necessarily completely split), the fact that equality of zeta functions implies that the split primes are the same needs a small argument. Let $p$ be a prime and $f_1,\cdots, f_r$ be the residue class degrees in $E$ over $p$ (in decreasing order); simlarly $g_1,\cdots, g_s$ the residue class degrees in $E'$ over $p$ (in decreasing order). Assume $f_1\geq g_1$. The local zeta functions at $p$ are $$(1-X^{f_1})\cdots (1-X^{f_r})=(1-X^{g_1})\cdots(1-X^{g_s})$$where $X=p^{-s}$. Since the multiplicity of the root $X=1$ is the same, we get $r=s$. The cyclotomic polynomial $\Phi _{f_1}(X)$ divides the left hand side and hence the right hand side. Therefore, by the uniqueness of irreducible factors on the LHS-RHS, $g_1=f_1$. We can thus cancel the factor $1-X^{f_1}$ on both sides and use induction on $r$ to conclude that $f_i=g_i$ for all $i$.<|endoftext|> TITLE: Is the unit tangent bundle of $S^{n}$ parallelizable? QUESTION [5 upvotes]: Is the unit tangent bundle of $S^{n}$ a parallelizable manifold. This is motivated by the fact that $TS^{n}$ is parallelizable? REPLY [12 votes]: W.Sutherland. A note on the parallelizability of sphere bundles over sphere. J. London Math. Soc. 39 (1964), 55--62. The answer is yes.<|endoftext|> TITLE: Topology of surfaces and mean curvature QUESTION [5 upvotes]: The Gauss-Bonnet theorem characterizes the topology of surfaces by means of their Gaussian curvature. Do there exist results characterizing the topology of surfaces embedded in $\mathbf{R}^3$ via their mean curvature? For example, I find it hard to imagine how a topological 2-sphere could be embedded in $\mathbf{R}^3$ with, for example, everywhere negative mean curvature. However, I am not familiar with any easy results which immediately rule this out. The Gauss-Bonnet theorem seems insufficient here, because the trace of the shape operator is a priori unrelated to its determinant. REPLY [10 votes]: Mean curvature depends on the choice of the "unit normal". If you change the orientation (choose the other unit normal), the computed mean curvature for the sphere is everywhere negative. In any case. Fix $x_0\in \mathbb{R}^3$. Since the embedding $S$ of the two sphere is compact, there is a point $y$ at which $|y - x_0| = \sup_{x\in S} |x - x_0|$. Thus $S\subset \overline{B_{x_0}(|y - x_0|)}$ and $S$ is tangent to $\partial B_{x_0}(|y-x_0|)$ at $y$. So trivially you have a lower bound on the Hessian and that the mean curvature must be positive at $y$ (for the "correct" choice of orientation). In fact, this has nothing to do with topology. Let $\Sigma$ be any compact surface with no boundary embedded in $\mathbb{R}^3$, the same argument as above shows that both its mean curvature (with appropriate choice of orientation) and scalar curvature must be positive at some point.<|endoftext|> TITLE: Counterintuitive consequences of the Hahn-Banach theorem QUESTION [9 upvotes]: The axiom of choice has many counterintuitive consequences like the Banach-Tarski paradox. The Hahn-Banach theorem is a consequence of the axiom of choice, but it is weaker. I would like to know some counterintuitive consequences of the Hahn-Banach theorem. Can the Banach-Tarski paradox be derived from the Hahn-Banach theorem? REPLY [12 votes]: To aid future inquiries, let us record here the go-to reference kindly provided by Willie Wong in a comment above: http://consequences.emich.edu/conseq.htm. This provides a data base with a utility to search for known implications between weak forms of the axiom of choice.<|endoftext|> TITLE: why are motives more serious than "naive" motives? QUESTION [15 upvotes]: I know my question is a bit vague, sorry for this. Let $k$ be a field of characteristic zero. Consider the Grothendieck ring of varieties over $k$, usually denoted by $K_0(Var_k)$. This is generated by isomorphism classes of varieties over $k$ modulo the relations [X]=[Y]+[X-Y] whenever $Y$ is a closed subvariety of $X$. People usually refer to [X] as the "naive" motive of $X$. On the other hand, one has Voevodsky's "true" motives $DM_{gm}(k)$ (not as true as we would like to, I know !) and to any variety $X$ we can attach an object $M(X)$in $DM_{gm}(k)$. Why is this $M(X)$ more serious than the naive one? That is, can you give some examples of properties that cannot be read at the level of $K_0(Var_k)$ but that one sees when working in $DM_{gm}(k)$? REPLY [10 votes]: Note that the Grothendieck ring of varieties does, at least conjecturally, remember some information about varieties that the category of motives does not. Under the cut-and-paste conjecture, two varieties are equivalent in the Grothendieck ring if and only if they can both be decomposed into the same set of locally closed pieces. There are many pairs of varieties which have the same motive but cannot be cut and pasted into each other (like a $\mathbb P^2$ and a fake projective plane). So assuming the cut-and-paste conjecture, the Grothendieck group remembers a lot of extra information about the varieties. In particular, because the fundamental group is a birational invariant for smooth projective varieties, the Grothendieck group class tells you the fundamental group, which is highly non-abelian invariant. So the category of motives loses some information that is contained the Grothendieck group of varieties - in some sense, the non-abelian information. One reason that the category of motives is good to work with is that it simplifies things by getting rid of this additional structure. For tasks where you only need that structure, motives are more serious.<|endoftext|> TITLE: Eilenberg-Mac lane spaces and a generalization QUESTION [14 upvotes]: Let $G$ and $H$ be two abelian groups and let $n>1, m>1$ be two different integers. How many different spaces $X$ (up to homotopy) do we have with the property $\pi_{n} X=G$ , $\pi_{m} X=H$ and $\pi_{\ast} X=0$ otherwise? is this number finite ? REPLY [6 votes]: One way to think about how to distinguish, if not classify, such spaces is by homotopy operations. In the same way that cohomology operations are natural transformations between cohomology functors, homotopy operations are natural transformations between homotopy functors. By the Yoneda lemma, natural transformations $\pi_n \to \pi_m$ are in natural bijection with the homotopy group $\pi_m(S^n)$, so elements of the homotopy groups of spheres give unary operations on homotopy. More generally, the $k$-ary operations on homotopy groups are natural transformations $\pi_{n_1} \times \dots \times \pi_{n_k} \to \pi_m$, and by the Yoneda lemma these are in natural bijection with the homotopy group $\pi_m(S^{n_1} \vee \dots \vee S^{n_k})$. For example, the Whitehead bracket $\pi_n \times \pi_m \to \pi_{n+m-1}$ is a well-known family of binary operations coming from some distinguished homotopy classes of maps $S^{n+m-1} \to S^n \vee S^m$. They make the homotopy groups of a space into a graded Lie algebra (up to a degree shift). All of the homotopy operations together give the homotopy groups of a space the structure of a $\Pi$-algebra, analogous to but much more complicated than the structure of being a module over the Steenrod algebra. A product of Eilenberg-MacLane spaces always has trivial homotopy operations, so you can distinguish a space from a product of Eilenberg-MacLane spaces by checking to see whether any of its homotopy operations are trivial. For example, the homotopy operation $\pi_2(S^2) \to \pi_3(S^2)$ given by a generator of $\pi_3(S^2) \cong \mathbb{Z}$ must be nontrivial since it is the universal example; this shows that the $3$-truncation of $S^2$ is a homotopy type with $\pi_2 \cong \pi_3 \cong \mathbb{Z}$ and all other homotopy groups trivial but which cannot be homotopy equivalent to $B^2 \mathbb{Z} \times B^3 \mathbb{Z}$. Moreover, it's a classical result that $H^4(B^2 A, B)$, the cohomology group that classifies the $k$-invariant of a space with $\pi_2 \cong A, \pi_3 \cong B$, can be identified with the group of quadratic functions $A \to B$. Given such a $k$-invariant, the corresponding quadratic function turns out to be precisely the homotopy operation $\pi_2 \to \pi_3$. So in this case the classification by $k$-invariants and by homotopy operations agree.<|endoftext|> TITLE: Combinatorial proof of the Cauchy identity for double Schubert polynomials QUESTION [5 upvotes]: The Cauchy identity for double Schubert polynomials states $$ \mathfrak{S}_w(x;-y) = \sum_{\substack{u,v \in S_n \\ w=v^{-1}u \\ l(w) = l(v) + l(u)}} \mathfrak{S}_u(x)\mathfrak{S}_v(y).$$ Is there a combinatorial proof of this identity, akin to the proof of the Cauchy identity for Schur functions via RSK and the dual Cauchy identity via dual RSK? For example, some procedure that takes a pipe dream with $x$ and $y$ weights to a pair of pipe dreams with only $x$ or $y$ weights? REPLY [6 votes]: Yes this is proven in the paper RC-graphs and Schubert polynomials, using double rc-graphs (pipe dreams). See section 4 of the link.<|endoftext|> TITLE: Mysterious quotes (at least for me) QUESTION [13 upvotes]: I heard two quotes, one from Alain Connes and an other one from Orlov. Alain Connes was talking about noncommutative geometry and he said the following: " a noncommutative algebra creates its own internal time " In a talk by Orlov about Mirror symmetry, he was asked if he considers the monoidal structure on the derived category of a scheme. Orlov said that "the monoidal structure is not natural in this context and we should not base our theory on this structure " he added "The tensor product is a NATURAL structure to consider in the category of Motives" I will be happy if someone can put some enlightenment to what Connes and Orlov meant (if the quotations above make sense ) ? REPLY [13 votes]: Alain Connes: "a noncommutative algebra creates its own intrinsic time". First of all, as Yemon Choi commented, this quote of Alain Connes is a slogan, not a theorem. "Most of the NC algebras create their own intrinsic time" would be a bit more correct, more precisely: Theorem: a von Neumann algebra of type $\rm III$ creates its own intrinsic time up to inner automorphisms. In the rest of the answer we will see how we can generate a von Neumann algebra from a given NC algebra, we will define all the notions appearing in the above theorem and explain what does it mean. From a noncommutative (unital associative) algebra $\mathcal{A}$ (with a countable base $\mathcal{b}$) over $\mathbb{C}$ , we can generate a von Neumann algebra as follows: let $H = l^2(\mathcal{b})$ be the Hilbert space generated by $\mathcal{b}$, let $H_0 = \{v \in H \ \vert \ a.v \in H \ \forall a \in \mathcal{A} \}$ (supposed dense in $H$) and $\rho$ the left regular representation of $\mathcal{A}$ on $H_0$. If $\forall a \in \mathcal{A}, \ \rho(a)$ is bounded, then $\mathcal{M} = (\rho(\mathcal{A}) \cup \rho(\mathcal{A})^*)''$ is the von Neumann algebra generated by $\mathcal{A}$ [else, by the polar decomposition, $\rho(a) = u. \vert \rho(a) \vert$ with $u$ a partial isometry (bounded), and $\mathcal{M}$ is the vN algebra generated by these partial isometries]. Note that $a \to a^*$ is the involution and $\mathcal{E}''= (\mathcal{E}')'$, is the bicommutant of $\mathcal{E} \subset B(H)$ the algebra of bounded operators. A von Neumann algebra $\mathcal{M}$ is a factor if and only if its center is trivial: $\mathcal{M} \cap \mathcal{M}' = \mathbb{C}$. Every von Neumann algebra $\mathcal{M}$ decomposes as a direct integral of factors (Murray - von Neumann). There are three types of factors: a factor is type $\rm I$ if it admits projections with a finite dimensional range; else it is type $\rm II$ if it admits no projection equivalent to an own subprojection; else it is type $\rm III$ (and we can prove that all the projections are equivalent). Modular theory : let $\mathcal{M}\subset B(H)$ be a von Neumann algebra. Let $\Omega \in H$ be a cyclic and separating vector (i.e., $\mathcal{M}.\Omega$ and $\mathcal{M}'.\Omega$ are dense in $H$). Let $S : H \to H$ be the closure of the anti-linear map $a\Omega \to a^{*}\Omega$, it admits a polar decomposition $S = J\Delta^{1/2}$, with $J$ anti-linear unitary and $\Delta$ positive. $JMJ = \mathcal{M}'$, $\Delta^{it} \mathcal{M}\Delta^{-it} = \mathcal{M}$ and $\sigma_{\Omega}^{t}(a) = \Delta^{it} a \Delta^{-it}$ gives the modular action of $\mathbb{R}$ on $\mathcal{M}$. Connes' Radon-Nikodym theorem: let $\Omega'$ be another vacuum (i.e. cyclic-separating) vector, then there is a Radon-Nikodym map $u_{t} \in \mathcal{U} ( \mathcal{M})$ [unitary operators in $\mathcal{M}$], defined such that $u_{t+s} = u_{t} \sigma_{t}^{\Omega'} (u_{s})$ and $\sigma_{t}^{\Omega'} (x) = u_{t} \sigma_{t}^{\Omega}(x) u^{\star}_{t}$. Then, modulo $Inn(\mathcal{M})$, $\sigma_{t}^{\Omega} $ is independent of the choice of $\Omega$, i.e., there exist an intrinsic group morphism $\delta : \mathbb{R} \to Out (\mathcal{M}) = Aut(\mathcal{M})/Inn(\mathcal{M})$. What Alain Connes calls the own intrinsic time, is precisely $\delta$. For the type $\rm I$ or $\rm II$, the modular action is inner, and so $\delta$ is trivial (i.e. $\ker(\delta) = \mathbb{R}$). It's non-trivial for the type $\rm III$. A factor is type $\rm III_1$ if and only if $\ker(\delta) = \{0 \}$. The type $\rm III_1$ factors exist, moreover, in some sense, most of the factors are $\rm III_1$ (see Structure of type III factors, for more details). Advertising: there will have a Master Class in Modular Theory by Serban Stratila and Masamichi Takesaki, at Chennai (India) from 24 Nov. to 04 Dec. 2014. The videos of the lectures are available here.<|endoftext|> TITLE: When are the Dolbeault and de Rham dgas homotopy equivalent? QUESTION [10 upvotes]: Let $M$ be a compact Kahler manifold. Then the Hodge decomposition says that the Dolbeault dga (of forms of all bidegree) and the de Rham dga on $\Omega_{\mathbb C}^\bullet(M)$ have isomorphic cohomology groups. Are there any stronger relationships between these two dgas? For example, are there simple conditions on $M$ that imply that these dgas are homotopy equivalent? REPLY [6 votes]: A relevant reference might be J. Neisendorfer, L. Taylor: Dolbeault homotopy theory. Trans AMS 245 (1978), 183-210. One of the results (Theorem 8) states that compact connected Kähler manifolds are both Dolbeault formal and de Rham formal (strengthening the result of Deligne-Griffiths-Morgan-Sullivan). This implies that the isomorphism of cohomology groups can be promoted to a zig-zag of homotopy equivalences of dgas.<|endoftext|> TITLE: Asymptotic formulas for Monster-related modular functions? QUESTION [8 upvotes]: Define the following, $$j(\tau) = \Big(\tfrac{E_4(\tau)}{\eta^8(\tau)}\Big)^3 = {1 \over q} + 744 + \color{blue}{196884} q + 21493760 q^2 + 864299970 q^3 + \cdots \tag{1}$$ $$j_{2A}(\tau) =\Big(\big(\tfrac{\eta(\tau)}{\eta(2\tau)}\big)^{12}+2^6 \big(\tfrac{\eta(2\tau)}{\eta(\tau)}\big)^{12}\Big)^2 = \tfrac{1}{q} + 104 + \color{blue}{4372}q + 96256q^2 + 1240002q^3+\cdots \tag{2}$$ $$j_{3A}(\tau) =\Big(\big(\tfrac{\eta(\tau)}{\eta(3\tau)}\big)^{6}+3^3 \big(\tfrac{\eta(3\tau)}{\eta(\tau)}\big)^{6}\Big)^2 = \tfrac{1}{q} + 42 + \color{blue}{783}q + 8672q^2 +65367q^3+\dots \tag{3}$$ $$j_{4A}(\tau)=\Big(\big(\tfrac{\eta(\tau)}{\eta(4\tau)}\big)^{4}+4^2 \big(\tfrac{\eta(4\tau)}{\eta(\tau)}\big)^{4}\Big)^2 = \tfrac{1}{q} + 24+ \color{blue}{276}q + 2048q^2 +11202q^3+\dots\\ \tag{4}$$ $$j_{7A}(\tau)=\Big(\big(\tfrac{\eta(\tau)}{\eta(7\tau)}\big)^{2}+7 \big(\tfrac{\eta(7\tau)}{\eta(\tau)}\big)^{2}\Big)^2 = \tfrac{1}{q} + 10+ \color{blue}{51}q + 204q^2 +681q^3+\dots\\ \tag{5}$$ where $\eta(\tau)$ is the Dedekind eta function. (Note that $196883, 4371, 782, 276, 51$ are degrees of the irreducible representations of the Monster, Baby Monster, Fischer Fi23, Conway Co1, and Held groups, respectively.) The asymptotic formula (by Rademacher?) for the coefficient of $q^n$ of $(1)$ is given by, $$a(n) \approx \frac{e^{4\pi\sqrt{n}}}{\sqrt{2}\,n^{3/4}}\tag{for 1}$$ For example, let $a(n)$ be the coefficient, and $a'(n)$ given by the formula, then, $$\begin{array}{cccc} n&100&200&300\\ a(n)&8.38\,\text{x}\,10^{52}&2.011\,\text{x}\,10^{75}&3.293\,\text{x}\,10^{92}\\ a'(n)&8.40\,\text{x}\,10^{52}&2.016\,\text{x}\,10^{75}&3.299\,\text{x}\,10^{92}\\ \end{array}$$ Question: What are the analogous coefficient formulas for $(2), (3), (4), (5)$? $\color{blue}{\text{Edit}}$: After perusing the OEIS, it seems that the fourth has, $$d(n) \approx \frac{e^{2\pi\sqrt{n}}}{2\,n^{3/4}}\tag{for 4}$$ $$\begin{array}{cccc} n&100&200&300\\ d(n)&3.04\,\text{x}\,10^{25}&3.64\,\text{x}\,10^{36}&1.26\,\text{x}\,10^{45}\\ d'(n)&3.06\,\text{x}\,10^{25}&3.66\,\text{x}\,10^{36}&1.27\,\text{x}\,10^{45}\\ \end{array}$$ though I have no proof that this is its correct asymptotic formula. REPLY [8 votes]: In a recent paper with Ken Ono and John Duncan, we give exact formulas for the coefficients of the Monsterous Moonshine modules. I will give a link to a preprint a few days when I can, but the relevant part relies on prior work of Bringmann and Ono (http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/115.pdf) which gives exact formulas for coefficients of modular forms assuming knowledge of their poles at cusps. Each of your $j_g(\tau)=\sum a_g(n)q^n$ is a Hauptmodul for a group of the form $\Gamma_g=\left\langle \Gamma_0(N),\begin{pmatrix}0&-1\\N&0\end{pmatrix}\right\rangle,$ where $N=1,2,3,$ or $4$. In this case, the dominant term in the exact formulas for $a_g(n)$ is $$\frac{2\pi}{\sqrt{N n}}I_1\left(4\pi \sqrt{n/N}\right)\sim \frac{e^{4\pi \sqrt{n/N}}}{\sqrt{2}N^{1/4}n^{3/4}}.$$ Here $I_k$ is the usual $I$-Bessel function.<|endoftext|> TITLE: Why can't we take three loops? QUESTION [38 upvotes]: Apologies for the vague title and soft question. According to Etingof, Igor Frenkel once suggested that there are three "levels" to Lie theory, which I guess could be given the following names: No loops: here we study a simple Lie algebra $\mathfrak{g}$, a Weyl group, a braid group, or a Hecke algebra, all of which have something to do with a Lie group $G$. One loop: here we study an affine Lie algebra $\widehat{\mathfrak{g}}$, a quantum group $U_q(\mathfrak{g})$, an affine Weyl group, an affine braid group, or an affine Hecke algebra, all of which I think have something to do with the loop group $LG$ of $G$. Two loops: here we study a double affine Lie algebra $\widehat{\widehat{\mathfrak{g}}}$, an affine quantum group $U_q(\widehat{\mathfrak{g}})$, an elliptic quantum group (whatever that means), a double affine or elliptic Weyl group, a double affine or elliptic braid group, or a double affine or elliptic Hecke algebra, all of which I think have something to do with the double loop group of $G$, or maybe more precisely the space of maps of some sort from an elliptic curve $E$ to $G$. The suggestion is further that this pattern doesn't continue. Why doesn't this pattern continue? I asked around and got an answer that I interpreted as follows. The trichotomy above can be matched up to the trichotomy additive group $\mathbb{C}$ : multiplicative group $\mathbb{C}^{\times}$ : elliptic curve $E$. Here is one story about the match as I understand it, which is not very well. One-dimensional algebraic groups give rise to equivariant cohomology theories. The above theories give rise to equivariant cohomology, equivariant K-theory, and equivariant elliptic cohomology respectively. Roughly, $\text{Spec } H^{\bullet}_G(\text{pt}) \cong \mathfrak{g}/G \cong \text{Bun}_G(\mathbb{C})$, while $\text{Spec } K^{\bullet}_G(\text{pt}) \cong G/G \cong \text{Bun}_G(\mathbb{C}^{\times})$, and $\text{Spec } E^{\bullet}_G(\text{pt}) \cong \text{Bun}_G(E)$, where by $E^{\bullet}_G$ I mean the equivariant elliptic cohomology theory associated to the elliptic curve $E$. There is some yoga in geometric representation theory which I'm not all that familiar with involving building interesting algebras like group algebras of Weyl groups and Hecke algebras by computing the equivariant (co)homology or equivariant K-(co)homology of some varieties of interest, which has something to do with the construction of the affine and double affine Hecke algebras mentioned above. Since we've run out of one-dimensional algebraic groups, that would be some reason to believe that the pattern doesn't continue. But nevertheless I don't have a good sense of what, if anything, prevents us from studying and saying interesting things about "triple affine Lie algebras," "triple affine Weyl groups," "triple affine Hecke algebras," etc. at least insofar as the triple loop group of a group seems perfectly well-defined. On the geometric side it seems like there's nothing stopping us studying $G$-bundles on higher dimensional varieties. On the cohomological side, cohomology, K-theory, and elliptic cohomology should optimistically just be the first three terms of an entire sequence of cohomology theories at higher chromatic levels, or from the perspective of the Stolz-Teichner program, defined in terms of higher-dimensional field theories... REPLY [17 votes]: To elaborate on Kevin's excellent answer, one can account for the current absence of "higher loop" representation theory using physics. Namely, all of the representation theoretic structures you mention fit in very naturally into the study of gauge theory, specifically 4-dimensional $\mathcal N=2$ gauge theories. These come in two main classes (with some intersection) - the quiver gauge theories, which are the natural homes for algebras like Yangians, quantum loop algebras, and elliptic quantum groups; and the class S theories (reductions of the 6d "theory $\mathfrak X$" -- the (2,0) superconformal field theory labeled by a Dynikin diagram - on Riemann surfaces), which are the natural home for geometric Langlands, double affine Hecke algebras, Khovanov homology etc. (the theory Kevin describes associated to $U_q(\mathfrak g)$ is $\mathcal N=4$ super Yang Mills, which is the case when the Riemann surface is the two-torus). So why should this be relevant? the question of attaching interesting representation theory to maps into Lie groups is very closely linked to the question of finding interesting gauge theories in higher dimensions (the latter is strictly stronger but seems like the most natural framework we have for such questions). Specifically, we want supersymmetric gauge theories, if we want them to have any relation to topological field theory or algebraic geometry etc. However there are no-go theorems for finding gauge theories in higher dimensions. Even at the classical level it is impossible (thansk to Lie theory, namely the structure of spin representations) to have a supersymmetric gauge theory in more than 10 dimensions ---- any SUSY theory in dimensions above ten also includes fields of spin two and above (so physically is a theory of gravity), while above dimension 11 we have to have higher spin fields still (which physicists tell us doesn't make sense -- regardless it won't be a gauge theory). In any case theories with gravity and other stuff are a very far stretch to be called representation theories! At the quantum level (which is what we need for representation theory) it's much harder still -- I believe there are no UV complete quantum gauge theories above dimension 4 (in other words higher dimensional theories have to have "other nonperturbative stuff in them"). All of the representation theoretic structures you mention naturally fit into theories that come from six dimensions at best (reduced to 4 dimensions along a plane, cylinder or torus in the quiver gauge theory case to see Yangians, quantum affine algebras and elliptic quantum groups, or along a Riemann surface in the class S case). Studying in particular theory $\mathfrak X$ on various reductions gives a huge amount of structure, and includes things like ``triple affine Hecke algebras" presumably when reduced on a three-torus, but there's a clear upper bound to the complexity you'll get from these considerations. Now of course you might ask what about theories that don't come from supersymmetric gauge theory? the only interesting source I've heard of for higher dimensional topological field theories is (as you hint) chromatic homotopy theory, in particular the fascianting work of Hopkins and Lurie on ambidexterity in the$K(n)$-local category. This is a natural place to look for "higher representation theory", which is touched on I believe in lectures of Lurie -- but my naive impression is these theories will have a very different flavor than the representation theory you refer to (in particular a fixed prime will be involved, and these theories certainly don't feel like traditional quantum field theory!). But it's a fascinating future direction. For a hint of what kind of representation theory this leads to we have the theorem of Hopkins-Kuhn-Ravenel describing the $n$-th Morava K-theory of BG in terms of n-tuples of commuting elements in G --- i.e. the kind of characters you might expect for G-actions on $(n-1)$-categories.<|endoftext|> TITLE: Largest area of a compactly supported positive definite function QUESTION [5 upvotes]: Consider a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$, supported on $[-1,1]$, of positive type. Assume $f(0) = 1$; what is the "largest area" $\int f\,dx$ that can be achieved? To be more precise, let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function satisfying: $f$ is supported on $[-1,1]$, for all $x$, $0 \leq f(x) \leq 1 = f(0)$, and $f$ has positive type: for any finite family of points $x_1 < \cdots < x_n$ in $\mathbb{R}$, the matrix $(f(x_i - x_j))_{ij}$ is positive semi-definite. (Equivalently--by Bochner's theorem--the Fourier transform is pointwise nonnegative.) How large can $\int f(x)\,dx$ be? Remarks One example of such a function is the "triangle" $t(x) = \max(1 - |x|,0)$. This achieves $\int t(x)\,dx = 1$. Is that the best one can do? There are functions $g$ of positive type (and satisfying the other requirements above) for which $g(x) > t(x)$ for some $x$. (However, I do not know of any for which $\int g(x)\,dx > 1$.) I asked this question the "Mathematics" stack exchange site, but didn't have any takers. REPLY [8 votes]: As noted in my comment above, the Poisson summation formula would give $$ 1= f(0) = \sum_{n\in {\Bbb Z}} f(n) = \sum_{k\in {\Bbb Z}} {\hat f}(k) \ge {\hat f}(0), $$ since ${\hat f}(k) \ge 0$ by assumption. Since ${\hat f}(0) = \int_{-1}^{1} f(x) dx$, this proves the desired bound. If you are worried about using the Poisson summation formula on just a continuous function, use the above argument with $f$ convolved with $\phi_{\epsilon}$ where $\phi_{\epsilon}$ is smooth (or just in $C^2$, say) and non-negative, supported on $(-\epsilon,\epsilon)$, has non-negative Fourier transform, and ${\hat \phi_{\epsilon}}(0)=1$. Now use the above argument with $f*\phi_{\epsilon}$, to obtain ${\hat f}(0)\le 1+O(\epsilon)$, and then let $\epsilon \to 0$.<|endoftext|> TITLE: Why does a flat degeneration induce equality in the K-Theory? QUESTION [6 upvotes]: Let $Z_1$ and $Z_2$ be two closed subschemes of a smooth, complex, algebraic variety $X$. We will be interested in the Grothendieck ring of coherent sheaves of $X$, i.e. isomorphism classes of such sheaves modulo exact sequences and a product given by the tensor product of $\mathcal{O}_X$-modules. Denote by $[\mathcal{O}_{Z_1}]$ the class of $f^\ast\mathcal{O}_{Z_1}$ in the Grothendieck Ring of coherent sheaves on $X$, where $f:Z_1\to X$ is the inclusion. Assume now that $Z_2$ is a flat degeneration of $Z_1$, i.e. there is a morphism $\pi: Y\to \mathbb{A}^1$ of schemes such that the generic fiber of $\pi$ is isomorphic to $Z_1$ and the fiber of $\pi$ over $0$ is isomorphic to $Z_2$. I am reading a paper by David Speyer with the title "A matroid invariant via the K-theory of the Grassmannian" and on page 885, right below equation (3) he says that $[\mathcal{O}_{Z_1}]=[\mathcal{O}_{Z_2}]$ holds in this case. I don't see how to prove this implication and I'd be very grateful if someone could explain it to me. REPLY [2 votes]: As Allen says, we are talking about flat subfamilies of $X$. I.e. $Y$ is a closed subscheme of $X \times \mathbb{A}^1$, flat over $\mathbb{A}^1$, with $Z_0 \times \{ 0 \}$ and $Z_1 \times \{ 1 \}$ the fibers over $0$ and $1$. Replacing $Y$ by its closure in $X \times \mathbb{P}^1$, we may assume that we instead have $Y$ closed in $X \times \mathbb{P}^1$, flat over $\mathbb{P}^1$. (Since $\mathbb{P}^1$ is smooth and one dimensional, being flat over it just means all associated primes of $Y$ map to the generic point of $\mathbb{P}^1$.) On $\mathbb{P}^1$, we have short exact sequences $$0 \to \mathcal{O}(-1) \to \mathcal{O} \to k_0 \to 0$$ and $$0 \to \mathcal{O}(-1) \to \mathcal{O} \to k_1 \to 0$$ where $k_0$ and $k_1$ are the skyscraper sheaves of a point at $0$ and $1$ respectively. So the point $0$ and the point $1$ have the same class in $K_0(\mathbb{P}^1)$. Since $Y \to X$ is flat, we can pull this back to an equality $[\mathcal{O}_{Z_0}] = [ \mathcal{O}_{Z_1}]$ on $Y$. Since $Y$ is closed in $X \times \mathbb{P}^1$, the map $\pi: Y \to X$ is proper and we can push forward this equality to $X$. We need to check that $\pi_{\ast} \mathcal{O}_{Z_i \times \{ i \}} = \mathcal{O}_{Z_i}$ and $R^j \pi_{\ast} \mathcal{O}_{Z_i \times \{ i \}} =0$ in order to know that the $K$-theoretic push forward is the obvious thing. I'm not sure where to cite this argument to, it is something that everyone I talked to about this sort of thing seemed to view as obvious. Allen does the analogue for equivariant $K_0$, which is more interesting, in Prop 3.5 of his recent paper.<|endoftext|> TITLE: Resolving ADE singularities by blowing up QUESTION [8 upvotes]: Let's say we have a finite subgroup $\Gamma \subseteq SL(2,\mathbb{C})$ and consider the quotient variety $\mathbb{C}^2/\Gamma$, which will have one of the well-known ADE or du Val surface singularities and can be embedded into $\mathbb{C}^3$ as a hypersurface with a singular point at the origin. These singularities have crepant resolutions where the exceptional fiber is a union of $\mathbb{P}^1$s connected according to an ADE Dynkin diagram. They can be obtained by repeatedly blowing up at singular points, starting with blowing up at the origin in $\mathbb{C}^3$. When doing this it seems that after the first blowup, the exceptional curve or curves you get correspond to the natural representation of $\Gamma$ on $\mathbb{C}^2$ that is given by its inclusion into $SL(2,\mathbb{C})$. (This is under the McKay correspondence, which gives a one-to-one correspondence between nodes in the Dynkin diagram and nontrivial irreducible representations of $\Gamma$ over $\mathbb{C}$.) For instance, if you start with an $A_n$ singularity, defined by $x^2+y^2+z^{n+1}=0$ in $\mathbb{C}^3$, and blow up at the origin, the exceptional divisor will be $X^2+Y^2=0$ in $\mathbb{P}^2$ which is two $\mathbb{P}^1$s meeting at a point. These two $\mathbb{P}^1$s correspond to the outermost nodes in the $A_n$ Dynkin diagram which are dual representations of $\Gamma$, the cyclic group of order $n+1$ in this case, that you get from its natural action on $\mathbb{C}^2$. (The node or nodes for each $\Gamma$ that I'm describing are also the node(s) that you would connect to the additional node on an extended Dynkin diagram.) Although I haven't checked recently I believe this also works for all the other ADE groups. My question is, is there some kind of "natural" explanation for this? I was hoping this might be explained by one of the geometric interpretations of the McKay correspondence out there, like the Bridgeland-King-Reid "Mukai implies McKay" paper (which I don't thoroughly understand). REPLY [2 votes]: We can use inkspot's key observation that the curves appearing in the first blowup are exactly the ones having negative intersection with the fundamental cycle. Let $[\rho_i]$ be the exceptional curve corresponding to $\rho_i \in Irr(\Gamma)$ where $Irr(\Gamma)$ is the set of nontrivial irreducible representations of $\Gamma$. Also let $c$ be the natural representation of $\Gamma$ on $\mathbb{C}^2$. If we then further accept that: The fundamental cycle is given by the formula $$\sum_{\rho_i \in Irr(\Gamma)} \deg(\rho_i) [\rho_i].$$ Let $C$ be the square matrix of dimension $|Irr(\Gamma)|$ defined by $C = (a_{ij})$, where $a_{ij}$ is the multiplicity of $\rho_i$ in $c \otimes \rho_j$. For any $\rho_i, \rho_j \in Irr(\Gamma)$, not necessarily distinct, the intersection $[\rho_i].[\rho_j]$ is given by the $ij$ entry of the matrix $C-2I$. ($2I-C$ is also the Cartan matrix of the appropriate root system.) These facts are mentioned in Gonzalez-Sprinberg and Verdier's paper "Construction geometrique de la correspondance de McKay", Theorem 2.2, and they also have a natural explanation of fact 1. In Prop. 1.4 they prove the formula $$2-c = \sum_{\rho_i \in Irr(\Gamma)} (\deg \rho_i)(c-2)\rho_i$$ in the representation ring $R(\Gamma)$. Combining the facts 1 and 2, we can see that after expanding everything out in the RHS, the coefficient on $\rho_i$ will be the intersection of $\rho_i$ with the fundamental cycle, for any $\rho_i \in Irr(\Gamma)$. So the LHS says exactly that $[\rho_i]$ will have zero intersection with the fundamental cycle, unless $\rho_i$ appears in $c$, in which case it has intersection $-1$.<|endoftext|> TITLE: Homotopy groups of $MO(2)$ QUESTION [13 upvotes]: Have there been any computations of the higher homotopy groups of $MO(2)$, the Thom space of the universal $O(2)$-bundle? Thom himself noted in his landmark 1954 paper that $$ \pi_1(MO(2))=0,\quad \pi_2(MO(2))=\mathbb{Z}/2,\quad\pi_3(MO(2))=0,\quad \pi_4(MO(2))=\mathbb{Z}. $$ By the Pontrjagin-Thom construction $\pi_n(MO(2))$ is the group of cobordism classes of embeddings of closed $(n-2)$-manifolds in $\mathbb{R}^n$, where a cobordism is an embedding in $\mathbb{R}^n\times [0,1]$. REPLY [6 votes]: There are a couple of references which show that $\pi_5(MO(2))=\mathbb{Z}/2$. Suzuki, H., On the realization of the Stiefel-Whitney characteristic classes by submanifolds, Tohoku Math. J., II. Ser. 10, 91-115 (1958). ZBL0107.17001. The proof uses the Postnikov tower. The conclusion is on page 108. Zvagel’skij, M. Yu., Cobordisms of embeddings with codimension two, J. Math. Sci., New York 104, No. 4, 1276-1282 (2001); translation from Zap. Nauchn. Semin. POMI 252, 40-51 (2001). ZBL1052.57042. The proof here is geometric, and (for me) harder to follow. This paper also contains a proof that $\pi_6(MO(2))=0$.<|endoftext|> TITLE: When did "Betti cohomology" come to be used the way it is today? (and how is it used) QUESTION [24 upvotes]: This is sort of a mixture of a math and history question. First the math part: thinking about it, I do not actually know how to properly use the term "Betti cohomology". I know I should, but I don't. Betti cohomology of a variety $X$ defined over a field $k\subseteq \mathbb{C}$ refers to the singular cohomology of the associated complex space $X(\mathbb{C})$. But what coefficients, integral or rational? Or even local systems? More importantly, complex conjugation induces an involution on $X(\mathbb{C})$ and then also on the cohomology. It seems that sometimes this is part of the structure of Betti cohomology, sometimes it isn't. So this is the math part of my confusion - maybe someone can tell me how I should use the term Betti cohomology appropriately. Next (and more seriously), assuming we have clarified how the term "Betti cohomology" is supposed to be used nowadays - how did this evolve? The german Wikipedia article claims that Poincaré coined the term "Betti numbers" for the ranks of singular homology groups because these ranks agreed with numbers Betti had defined for surfaces. So, what are the possible reasons for calling singular cohomology of the associated complex space "Betti cohomology"? Which papers were instrumental in making Betti cohomology a popular term? Can anyone shed light on the history of the terminology? PS: I tagged the question ag.algebraic-geometry because the "Betti cohomology" seems to be prevalently used in algebraic geometry related communities. Feel free to retag if you consider this inappropriate. REPLY [11 votes]: Some of this stuff is explained in much detail in this paper: Charles Weibel, History of Homological Algebra (1999) In particular, after reviewing the work of Riemman and Betti, it says: Inspired by Betti's paper, Poincaré (1854-1912) developed a more correct homology theory in his landmark 1895 paper "Analysis Situs". [...] In honor of Enrico Betti, Poincaré defined the nth Betti number of V to be $b_n+1$, where $b_n$ is the size of a maximal independent family. Today we call $b_n$ the nth Betti number, because it is the dimension of the rational vector space $H_n(V;\mathbb{Q})$. The reference in question is: Henri Poincaré, Analysis situs (1895) Journal de l'École Polytechnique Also, the follow-up notes mentioned in the comments are: Henri Poincaré, Complément à l'Analysis Situs (1899) Rendiconti del Circolo Matematico di Palermo Henri Poincaré, Second complément à l'Analysis Situs (1900) Proceedings of the London Mathematical Society There's also a moder translation of Poincaré's original paper and a total of five "supplements": Henri Poincaré, Papers on Topology: Analysis Situs and Its Five Supplements (translated by John Stillwell, 2009) As for the evolution toward the modern use of Betti cohomology, this paragraph from Weibel's survey seems relevant: Homological algebra in the 19th century largely consisted of a gradual effort to define the "Betti numbers" of a (piecewise linear) manifold. Beginning with Riemann's notion of genus, we see the gradual development of numerical invariants by Riemann, Betti and Poincaré: the Betti numbers and Torsion coefficients of a topological space. Indeed, the subject did not really move beyond these numerical invariants until about 1930. And it was not concerned with anything except invariants of topological spaces unit about 1945. I still don't know where the use of Betti cohomology comes from, so this is not really an answer; hopefully someone else can help. My best guess is that it is from about the same time that the concept of Weil cohomology theory. The papers in which the classic theorems of Betti cohomology are proven never use that name, see for example: Jean-Pierre Serre, Géométrie algébrique et géométrie analytique (1956) Michael Artin, The étale topology of schemes (1968)<|endoftext|> TITLE: Obtaining non-normal varieties by pushout QUESTION [9 upvotes]: In his answer to this MO question, Karl Schwede claimed that every non-normal variety can be obtained by an appropriate pushout diagram, as sketched in that answer. This would give substance to the heuristics according to which "a normal variety is a variety that has no undue gluing of subvarieties or tangent spaces" (again, see K.Schwede's answer and the examples therein). Q. Has a proof of that fact been written down somewhere since then? If yes, where? And if not, could anybody who knows it sketch it here on MO? REPLY [13 votes]: A number of people have asked me for a reference since I wrote down that answer in the other question so I'll try to write a reference here (I'm sure some experts knew it before though). I originally wrote a complicated Noetherian induction in this answer but I just realized this is really easy. Main point: There is a canonical way to find the thickened subschemes that glue to give the non-normal variety. Indeed, the non-normal locus already has a canonical scheme structure. Let's use that! Suppose $R \subseteq S$ a finite birational extension of reduced rings. Let $I = R :_{K(R)} S = \{z \in K(R)\;|\; zS \subseteq R \}$ (a fractional ideal -- usually called the conductor). Note that $I$ is an ideal of $R$ (since in particular it multiplies $R$ back into $R$) but it is also an ideal of $S$. Indeed, take $s \in S$ and $x \in I$, then $xs$ also still multiplies other $s' \in S$ into $R$ as well. Theorem: With notation as above if $A$ is the pullback of $\big( S \to S/I \leftarrow R/I \big)$ then $A = R$. Proof: Obviously we have $R \subseteq A \subseteq S$. We want to show that $R \to A$ surjects as well. Take an element $(s, \overline{r})$ in $A$ (this is a pair of elements $s \in S$ and $\overline{r} \in R/I$ with common image in $S/I$). Let $r \in R$ be any representative for $\overline{r}$. Consider $s - r \in S$. Obviously $s-r$ is sent to zero in $S/I$ and so $s - r \in I \subseteq R$. But then $s \in R$ as well. The map $R \to A$ sends $x$ to $(x, \overline{x})$. Therefore the map $R \to A$ surjects as claimed. $\square$<|endoftext|> TITLE: When does a cubic surface pass through five lines? QUESTION [10 upvotes]: The set of 5-tuples of lines in $\mathbf{P}^3$ is parametrized by the 20-dimensional product of Grassmannians $G(2,4)^{\times 5}$. The set of cubic surfaces is parametrized by a 19-dimensional projective space. I can present a line in $\mathbf{P}^3$ as a $2 \times 4$ matrix (the projectivization of the row span) and five lines in $\mathbf{P}^3$ as a $10 \times 4$ matrix. Which rational function of this matrix vanishes on the set of tuples of lines that are contained in a cubic surface? REPLY [8 votes]: I would write it differently. First, consider the universal space $M$ consisting of tuples $(S,L_1,\dots,L_5)$, where $S$ is a cubic surface and $L_i$ are lines on $S$. This space lies inside the product $P^{19}\times Gr(2,4)^5$ and can be described as the zero locus of a canonical global section of the vector bundle $$ E := O(1) \boxtimes (S^3U_1^* \oplus S^3U_2^* \oplus S^3U_3^* \oplus S^3U_4^* \oplus S^3U_5^*) $$ on it, where $U_i$ is the tautological bundle on the $i$-th copy of the Grassmannian. Consequently, the structure sheaf of $M$ has the following Koszul resolution $$ 0 \to \Lambda^{20}E^* \to \dots \to \Lambda^2E^* \to E^* \to O \to O_M \to 0. $$ We are interested in the image of $M$ in $Gr(2,4)^5$, so let us push forward the above resolution to $Gr(2,4)^5$. Note that $E^*$ restricts as a sum of $O(-1)$ to any fiber $P^{19}$ of the projection, hence the $p$-th term restricts as a sum of $O(-p)$. Thus restriction of almost all terms are acyclic, and they do not contribute to the pushforward. The only terms which do are the first and the last. The first gives $O$, and the last gives $$ \det(S^3U_1 \oplus \dots \oplus S^3U_5) = \otimes_{i=1}^5\det(S^3U_i) = O(-6,-6,-6,-6,-6). $$ This means that the equation of the image of $M$ is a hypersurface of polydegree $(6,6,6,6,6)$ on the product of the Grassmannians. Of course its equation is the one given by David Speyer --- the determinant is a homogeneous polynomial of polydegree $(12,12,12,12,12)$ in coefficients of matrices, but if you write it in terms of the Plucker coordinates (which are quadratic in coefficients of matirces), it will have polydegree $(6,6,6,6,6)$.<|endoftext|> TITLE: Number of paths through infinite trees with given "growth rates" QUESTION [10 upvotes]: (Preface: This may be a naive or easy question for experts....) Consider an infinite tree, rooted on the left, where each node has two children; the number of nodes at each level (distance from the root) $n$ is $f(n) = 2^n$. The number of paths is $2^{\omega}$, the cardinality of the continuum (uncountable). Now consider a similar tree where only one node at each level has an extra child; the number of nodes at each level $n$ is $f(n) = n+1$. In this case, the number of paths is $\omega$; the paths here are countable. My question is, what happens in between? I would formalize the setting as follows (but maybe I have some details wrong): Suppose I have an infinite rooted tree, where each node has at least one child, and the number of nodes at distance $n$ from the root is $f(n)$ where $n \leq f(n) \leq 2^n$. Then what can we say about the number of paths in terms of the "growth rate" $f$? For example, if the "growth rate" is polynomial, must there be a countable number of paths? Are there growth rate functions for which the cardinality of the number of paths is independent of ZFC, i.e. depends on the continuum hypothesis? Any other notes/comments? REPLY [6 votes]: As explained in the various comments, every tree has countably many or $2^\omega$ branches, there are trees with $2^\omega$ branches of any unbounded admissible growth rate $f(n)$, and if nodes can have arbitrarily many children, there are countable trees of arbitrary growth rate. As Joel already mentioned, the question that remains is for which growth rates $f(n)$ there exist trees with countably many branches where every node has one or two children. (Below, I will assume the latter condition automatically, and call them just “trees”.) Now, even for this question, the absolute growth rate is largely irrelevant, what really matters is how $f(n+1)$ relates to $f(n)$, as demonstrated by the following example. First, let me define for convenience that $f\colon\mathbb N\to\mathbb N$ is admissible if there exists a tree of growth rate $f(n)$. This is equivalent to the conditions $f(0)=1$, $f(n)\le f(n+1)\le2f(n)$ for every $n\in\mathbb N$. The question also demands $f(n)\ge n$, but as Joel notes, this has no good reason, so I will state this explicitly where intended. Example: For every function $g\colon\mathbb N\to\mathbb N^{>0}$ such that $\lim_ng(n)=+\infty$, there is an admissible function $f(n)\le g(n)$ such that every tree of growth rate $f(n)$ has $2^\omega$ branches. There is an admissible function $f$ such that $n\le f(n)\le2n$ for $n>0$, and every tree of growth rate $f(n)$ has $2^\omega$ branches. Proof: For 1, we can choose an infinite sequence $0=n_0d(n)$ has children $2i-1-d(n)$ and $2i-d(n)$. Theorem: For any admissible function $f$, the following are equivalent. All trees of growth rate $f$ have $2^\omega$ branches. $T_f$ has $2^\omega$ branches. There are $n_0\le n_1d(n)$ for infinitely many $n\ge n_0$; let us enumerate them as $n_11-2^{-n_0}\ge\sum_{k=1}^\infty2^{-k}d(n'_k).$$ $4\to1$:$\let\res\restriction\let\sset\subseteq\let\Sset\supseteq $ Let $T$ be a subtree of $2^{<\omega}$ of growth rate $f(n)$, $[T]\sset2^\omega$ the set of its branches, and $T_1\sset T$ the set of nonsplitting nodes of $T$. For any $b\in[T]\cup T$, let $B(b)\sset\mathbb N$ denote the set of levels on which $b$ goes through a splitting point of $T$. $b\res n\in2^n$ denotes the restriction of $b$ to the first $n$ levels. We consider a probability measure on Borel subsets of $2^\omega$ defined by $$\Pr_{\sigma\in2^\omega}(\sigma\Sset t)=\begin{cases}2^{-|B(t)|}&t\in T\\0&\text{otherwise}\end{cases}$$ for $t\in2^{<\omega}$. That is, $[T]$ has measure 1, and if $t\in T\smallsetminus T_1$, the measure of the set of all branches going through $t$ is split evenly between the two children of $t$. Claim: If $b\in[T]$, we have $$\Pr_\sigma(\sigma=b)\le\sum_{k:b\res n_k\in T_1}2^{-k}\Pr_\sigma(\sigma=b\mid\sigma\Sset b\res n_k).$$ Proof: We can assume wlog $\Pr(\sigma=b)>0$, which implies that $b$ is isolated. Let $k_1\sum_{k=1}^\infty2^{-k}d(n_k)=\sum_k\sum_{t\in T_1\cap2^{n_k}}2^{-k}\ge\sum_k\sum_{t\in T_1\cap2^{n_k}}2^{-k}\sum_{b\Sset t}\Pr_\sigma(\sigma=b\mid\sigma\Sset t)\\=\sum_{b\in[T]}\sum_{k:b\res n_k\in T_1}2^{-k}\Pr_\sigma(\sigma=b\mid\sigma\Sset b\res n_k)\ge\sum_{b\in[T]}\Pr_\sigma(\sigma=b)=1, \end{multline} a contradiction.<|endoftext|> TITLE: Differential of homological atiyah-Hirzebruch Spectral sequence for K-homology QUESTION [6 upvotes]: The first non vanishing differential $d_3$ of the cohomological Atiyah-Hirzebruch spectral sequence for computing (Complex) Topological $K$-theory out of ordinary cohomology has a description in terms of the cohomology operation $Sq^3_{\mathbb{Z}}$. Is there an explicit formula for the differential of the homological spectral sequence converging to complex K-homology? REPLY [4 votes]: One can use the fact that the AH spectral sequence for $KU$-homology is a module over the AH spectral sequence for $KU$-cohomology. For example, let $M$ be a closed orientable manifold with fundamental class $[M]\in H_n(M,\mathbb{Z})\cong H_n(M,KU_0)$. Consider a class $x\in H_p(M,KU_q)$, which we can write as $y\cap [M]$ for some $y\in H^{n-p}(M,\mathbb{Z})\cong H^{n-p}(M,KU^{-q})$. Then, we have $$d^3(x)=d^3(y\cap [M])=d_3(y)\cap [M]\pm y\cap d^3([M])=Sq_3(y)\cap[M]\pm y\cap d^3([M]).$$ This reduces the problem to computing $d^3([M])$.<|endoftext|> TITLE: Is the heat kernel more spread out with a smaller metric? QUESTION [15 upvotes]: Suppose M is a smooth manifold, and we have two Riemannian metrics on M, say g and h, with g bigger than h (i.e. for every tangent vector at every point, the norm according to g is bigger than the norm according to h). Suppose $H_g$ and $H_h$ are the heat kernels on (M, g) and (M, h) respectively. Does ${\frac{H_g(x, y, t)} {H_g(x, x, t)} \leq \frac {H_h(x, y, t)}{H_h(x, x, t)}}$ necessarily hold for all x, y, and t? This intuitively makes sense: if you have something diffusing on M, starting at the point x, it should diffuse "faster" with the metric h because distances are smaller, and so after time t it seems you should have a greater proportion of the diffusing substance at y compared to at x. On $\mathbb{R^n}$ this is trivially true, also I've tested it out numerically for a variety of metrics on the circle and it seems to hold. But a proof eludes me. While I am curious about the general case, I'm especially interested when the manifolds in question are quotients of $\mathbb{R^n}$ by a lattice, with the corresponding flat metrics, because an affirmative answer in this special case can be used to solve this problem. $\textbf{Update:}$ I've found a proof of this for the special case of translation-invariant metrics on the circle. For these spaces, the heat kernel $H(0, x ,t)$, as a function of x, is given by a wrapped Gaussian function. The variance of the Gaussian that gets wrapped is determined by t and the metric, and a smaller metric has larger variance for a fixed t. The problem then boils down to the following assertion: If $a$, $b$, and $s$ are real numbers with $0 < a \leq b < 1$, then $$\frac{\sum\limits_{n = -\infty}^{\infty} a^{(n + s)^2}}{\sum\limits_{n = -\infty}^{\infty} a^{n^2}} \leq \frac{\sum\limits_{n = -\infty}^{\infty} b^{(n + s)^2}}{\sum\limits_{n = -\infty}^{\infty} b^{n^2}} $$ From the Poisson summation formula, the above is equivalent to: If $\alpha$ and $\beta$ are real numbers with $1 > \alpha \geq \beta > 0$, and $z$ is a complex number of norm 1, then $$\frac{\sum\limits_{n = -\infty}^{\infty} {\alpha^{n^2} z^{n}}}{\sum\limits_{n = -\infty}^{\infty} \alpha^{n^2}} \leq \frac{\sum\limits_{n = -\infty}^{\infty} {\beta^{n^2} z^{n}}}{\sum\limits_{n = -\infty}^{\infty} \beta^{n^2}} $$ From the Jacobi triple product, the above is equivalent to: $$\prod\limits_{m=1}^{\infty}{\frac{(1 + \alpha^{2m - 1}z)(1 + \frac{\alpha^{2m - 1}}{z})}{(1 + \alpha^{2m - 1})^2}} \leq \prod\limits_{m=1}^{\infty}{\frac{(1 + \beta^{2m - 1}z)(1 + \frac{\beta^{2m - 1}}{z})}{(1 + \beta^{2m - 1})^2}}$$ It's easy to verify that the expressions inside the infinite products are positive for each m (recall that $|z| = 1$, and so $\frac{1}{z} = \overline{z}$), and so it suffices to show that we have this for each $m \geq 1$: $$\frac{(1 + \alpha^{2m - 1}z)(1 + \frac{\alpha^{2m - 1}}{z})}{(1 + \alpha^{2m - 1})^2} \leq \frac{(1 + \beta^{2m - 1}z)(1 + \frac{\beta^{2m - 1}}{z})}{(1 + \beta^{2m - 1})^2}$$ For the sake of concision, let $p = \alpha^{2m - 1}$ and $q = \beta^{2m - 1}$. We are done if we can show the following: $$\frac{(1 + pz)(1 + \frac{p}{z})}{(1 + p)^2} \leq \frac{(1 + qz)(1 + \frac{q}{z})}{(1 + q)^2}$$ Or equivalently: $$0 \leq (1 + qz)(1 + \frac{q}{z})(1 + p)^2 - (1 + pz)(1 + \frac{p}{z})(1 + q)^2$$ By expanding and factoring the right side, it can be shown that the above is equivalent to: $$0 \leq (1 - \Re(z))(p - q)(1 - pq)$$ It's easy to show that each of the three terms in the product is non-negative, and this proves the inequality, and so we are done. REPLY [9 votes]: Oded Regev and I just posted a paper to the arXiv that shows that this does in fact hold over $\mathbb{R}^n/\mathcal{L}$ where $\mathcal{L}$ is a lattice. See Proposition 4.2. The general case does not hold. In particular, if the second-largest eigenvalue of the heat kernel is unique, then as $t$ approaches infinity, the point with the highest mass will be the point that maximizes the corresponding eigenfunction. I think Jeff Cheeger originally pointed this out to Oded.<|endoftext|> TITLE: Does the linear automorphism group determine the vector space? QUESTION [16 upvotes]: I was recently thinking about what it means to put structure on a set. It seems to me that, in my area (representation theory), the two main ways of imposing structure on a set $X$ are: distinguishing certain permutations of $X$ as structure preserving; and distinguishing certain test functions (here I think of $X \to \mathbb C$, but that's just because it fits my example) are structure respecting. For example, given a manifold, we can look at the diffeomorphisms among its permutations, or the smooth functions among its test functions; and, given a vector space, we can look at the linear automorphisms among its permutations, or the functionals among its test functions. Under reasonable hypotheses, the test-functions perspective (more 'analysis'-flavoured, maybe) determines the structure; but does the group perspective (more 'geometry' / Erlangen-flavoured)? A colleague pointed out that, for manifolds, the diffeomorphism group does determine the manifold, in a very strong sense: http://www.ams.org/mathscinet-getitem?mr=693972. Given that success, I was moved to ask: does the linear automorphism group determine the vector space? Here's one way of making that informal question precise. Suppose that $V_1$ and $V_2$ are vector spaces (over $\mathbb C$, say) with the same underlying set $X$, and that the set of permutations of $X$ that are linear automorphisms for $V_1$ is the same as the analogous set for $V_2$. Then are $V_1$ and $V_2$ isomorphic? Another colleague, and The Masked Avenger, both thought of the axiom of choice when I asked this question; but I'm not sure I see it. It's just a curiosity, so I have no particular investment in whether answers assume, negate, or avoid choice. EDIT: Since I think it may look like I am making some implicit assumptions, I clarify that I do not mean to assume that the vector spaces are finite dimensional, or that the putative isomorphism from $V_1$ to $V_2$ must be the identity as a set map of $X$. Thus, for example, Theo Johnson-Freyd's example (https://mathoverflow.net/a/186494/2383) of letting $X$ be $\mathbb C$, and equipping it with both its usual and its 'conjugate' $\mathbb C$-vector space structure, is perfectly OK. REPLY [8 votes]: Vladimir Dotsenko is right: $GL(V)\cong GL(W)$ iff $V\cong W$. This is true for any complex (and not only complex) vector spaces. If at least one of the dimensions is finite, then this follows from Dotsenko's argument. If $\dim V$ and $\dim W$ are infinite, this is valid for vector spaces over arbitrary division rings by Tolstykh's theorem [Theorem 8.3] (the paper contains a complete proof of the theorem and a remark that it "easily follows from general isomorphism theorems proved by O’Meara in [15, Theorem 5.10, Theorem 6.7]"). [15] O. O’Meara, A general isomorphism theory for linear groups, J. Algebra, 44 (1977) 93-142.<|endoftext|> TITLE: Interpretation for a condition in fluid dynamics QUESTION [5 upvotes]: I have been working with some mathematical models in biology and fluid mechanics. My problem is about the interpretation of a condition that I found for a vector representing the velocity of a fluid. The exact question is the next: Let $\mathbf{u=(}u_{1},u_{2},u_{3})$ be a vector field representing the velocity of a fluid. After making some accounts with models using partial differential equations, I found that the matrix \begin{equation} -A:=\frac{1}{2}\left( \frac{\partial u^{j}}{\partial x_{i}}+\frac{\partial u^{i}}{\partial x_{j}}\right) _{i,j=1,2,3}% \ \end{equation} should be positive definite. Does this condition has a physical interpretation in fluid dynamics or tensors? Any comment or reference will be highly appreciate! REPLY [2 votes]: I don't know if this is the answer that you have been looking for, but let me offer a rather trivial observation. Your tensor is the Lie derivative of the metric tensor with respect to the vector field $\textbf u$. Loosely speaking, the Lie derivative $\mathcal{L}_\textbf{u}$ has an interpretation as a derivative with respect to "dragging" a tensor along the flow defined by $\textbf{u}$. A simple manipulation shows that for any vector field $\textbf{v}$, \begin{equation} \mathcal{L}_\textbf{u} (\textbf{v} \cdot \textbf{v}) = 2 \textbf{v} \cdot \mathcal{L}_\textbf{u} \textbf{v} - 2 \textbf{v} A \textbf{v} \,, \end{equation} where the matrix $A$ is the one you have defined. A flow $\textbf{u}$ with positive definite $(-A)$ has the property that \begin{equation} \mathcal{L}_\textbf{u} (| \textbf{v} |^2 ) \geq 2 \textbf{v} \cdot \mathcal{L}_\textbf{u} \textbf{v} \end{equation} for any vector field $\textbf{v}$. I hope this observation has some use to you.<|endoftext|> TITLE: Space of Borel measurable maps QUESTION [9 upvotes]: That's a question from MSE (here) that did not receive any answer for some days. I migrate it to MO. Let $X$ and $Y$ be two standard Borel spaces and consider the set $M(X,Y)$ of measurable maps $f: X \to Y$. Is $M(X,Y)$ also standard Borel? First of all, the cardinality of $M(X,Y)$ is $\mathfrak{c} = 2^{\aleph_0}$ for uncountable $X$ and $Y$ (see Cardinality of the borel measurable functions?) - so this doesn't contradict the Borel ismorphism theorem. In Srivastava, "A course on Borel sets", he considers the space of $B(X,Y) \subseteq M(X,Y)$ of Baire functions, i.e. continuous functions and closed under pointwise limit. Then he states the Lebesgue – Hausdorff theorem that $B(X,Y) = M(X,Y)$ for metrizable $X$. But I haven't found a theorem or note in the book that says that $B(X,Y)$ is standard Borel. Moreover, he also states that any Borel measurable function can be made continuous by taking a finer topology on $X$ that doesn't change the Borel $σ$-algebra of $X$, i.e. $X$ is still standard Borel. But I don't see, how to apply this theorem. Of course, if we have a measure $\mu$ on the domain then we can for example consider the quotient space $\mathcal{L}^0$ that identifies $\mu$-a.e. equal Borel measurable maps. The corresponding Ky-Fan metric that makes $\mathcal{L}^0$ Polish can of course be seen as a pseudo-metric on $M$. I somehow doubt that $M$ can always be standard Borel, since this question is so natural, but does not seem to appear in Srivastavas book (or I just oversaw some simple implication). REPLY [4 votes]: With the $\sigma$-algebra specified to be the trace of the product $\sigma$-algebra, as in yadaddy's comment: No. For let $X=[0,1]$ and $Y=\{0,1\}$, let $\mathcal A$ be the product $\sigma$-algebra on $Y^X$, and let $\mathcal B$ denote the trace of $\mathcal A$ on $M(X,Y)$. Then $\mathcal B$ is not countably generated (and hence not standard Borel). Proof: Let $\mathcal F$ be a countable subset of $\mathcal B$, so ${\mathcal F} = \{ E \cap M(X,Y) : E \in {\mathcal E}\} $ for some countable set ${\mathcal E} \subseteq \mathcal{A}$. The definition of the product $\sigma$-algebra implies that there is a countable set $T \subseteq X$ such that we have $E = \pi_T^{-1}[\pi_T^{}[E]]$ for every $E\in \mathcal E$, where $\pi_T^{}$ denotes the coordinate projection from $Y^X$ to $Y^T$. Now take some $x_0\in X\setminus T$. Then $A := \{f \in M(X,Y) : f(x_0)=0\}\in {\mathcal B}$. On the other hand, for every $F\in\mathcal F$, we have $f\in F$ iff $g \in \mathcal F$, where $g$ denotes the function obtained from $f$ by changing its value at $x_0$; this remains true if we replace $\mathcal F$ by $\sigma(\mathcal F)$, and thus we get $ A \notin \sigma(\mathcal F)$.<|endoftext|> TITLE: Sheaf cohomology on non paracompact topological spaces QUESTION [5 upvotes]: I have some confusion on the subject of sheaf cohomology on non-paracompact topological spaces, i hope you can help me. My reference is Godement's book "Topologie algebrique et theorie dex faisceaux". I know that, given a sheaf $\mathcal{F}$ on $X$ and $\phi$ a family of supports on $X$, I can define the cohomology $H^n_{\phi}(X,\mathcal{F})$ as $H^n(\Gamma_\phi(\mathcal{C}^*(X,\mathcal{F}))$ where $\mathcal{C}^*(X,\mathcal{F})$ is the canonical sequence and $\Gamma_\phi(\mathcal{F})=\{s\in\mathcal{F}(X)|supp(s)\in \phi\}$. When Godement writes $H^n(X,\mathcal{F})$, without the $\phi$ being indicated, i think he means $H^n(\Gamma(\mathcal{C}^*(X,\mathcal{F}))$. My questions are the following: 1) What is the difference between $H^n(\Gamma_\phi(\mathcal{C}^*(X,\mathcal{F}))$ and $H^n(\Gamma(\mathcal{C}^*(X,\mathcal{F}))$? What is the utility of the family of supports? 2) I've heard that when $X$ is not paracompact then $H^n(\Gamma(\mathcal{C}^*(X,\mathcal{F}))$ (but not $H^n(\Gamma_\phi(\mathcal{C}^*(X,\mathcal{F}))$) fails to be "functorial" (i think in the sense that short exact sequences don't go to long exact sequences). Can you explain this to me? What exactly doesn't work for $H^n(\Gamma(\mathcal{C}^*(X,\mathcal{F}))$? Thank you very much REPLY [2 votes]: 2) Are you sure you are not confusing the sheaf cohomology $H^n(\Gamma(C^*(X,\mathcal F))$ with Čech cohomology$\check H^n(X,\mathcal F)$? The sheaf cohomology has long exact sequences for arbitrary topological space and short exact sequence of sheaves. For paracompact spaces the two are equal, there is a proof of it in Godement's book.<|endoftext|> TITLE: How mirror of quintic was originally found? QUESTION [22 upvotes]: In the 90-91 pager "A PAIR OF CALABI-YAU MANIFOLDS AS AN EXACTLY SOLUBLE SUPERCONFORMAL THEORY", Candelas, de la Ossa, Green, and Parkes, brought up a family of Calabi-Yau 3-folds, canonically constructed from a sub-family of quintic CY 3folds, as a "mirror" to quintic an did some calculations on the mirror family to extract the GW invariants of the quintic. I was having a discussion with a group of physicist on what argument lead them to take that particular family as the mirror, whatever mirror means; and apparently, the original point of view among physicists has been changed in past two decades and they did not know the answer either. Does any one know what (may be physical) original recipe lead them to that particular mirror family? This question should have been asked before, but a brief search did not lead me anywhere. Let me know if thats the case and I would remove this post. REPLY [10 votes]: The history of this is as follows. In the paper by Candelas, Lynker and Schimmrigk there are two weighted hypersurfaces whose cohomology is mirror to that of the quintic. These therefore are two potential candidates for the mirror quintic. The question then was how to decide whether they provide mirror partners to the quintic or not. This was addressed in a paper by Lynker and Schimmrigk (http://inspirehep.net/record/27957) by transporting the Greene-Plesser construction of quotients of conformal field theories to the level of Landau-Ginzburg theories and hence weighted hypersurfaces. This established at level of physics that the two weighted hypersurfaces in the list of Candelas-Lynker-Schimmrigk are isomorphic and that at the level of physics they are both mirrors of the 1-parameter quintic family.<|endoftext|> TITLE: Embedding proper algebraic spaces QUESTION [9 upvotes]: Does every proper algebraic space (over a field, say) admit a closed immersion into a smooth proper algebraic space? Remark: Of course, if we say "projective" instead of "proper" then the answer is tautologically "yes": we can take the ambient variety to be some projective n-space. But I'm curious about the general case. REPLY [6 votes]: I do not know the general answer to this interesting question. However, there are surely examples of proper algebraic varieties that admit no embeddeding into smooth schemes. Look for instance to this paper by Roth and Vakil, page 11. Note that in the paper the authors claim that their example admits no embedding into a smooth algebraic space; however, their proof actually only work for smooth schemes, see the corrigendum in Vakil's webpage here.<|endoftext|> TITLE: Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$ QUESTION [6 upvotes]: Let $p$ be a prime number, $q=p^e$ a power of $p$, and $G=SL_2(\mathbb F_q)$. Let $V$ be the adjoint representation of $G$, i.e. $V$ is the 3-dimensional $\mathbb F_q$-space of of (2,2)-matrices of trace $0$ with coefficients in $\mathbb F_q$, and $G$ acts on it by conjugation. So $V$ is an irreducible representation of $G$. By "restriction of scalar", $V$ can also be considered as a representation of dimension $3 e$ over $\mathbb F_p$. When is this representation $V/\mathbb F_p$ of dimension $3e$ irreducible? (Edit: it is obviously not absolutely irreducible, as remarked by Venkataramana below). In general, I would be interested by any pointer to a reference mentioning questions of irreducibility of representations obtained by "restriction of scalars". REPLY [4 votes]: There is some textbook literature which essentially covers the issues raised here, though it often deals with more general situations. (Over finite fields life is simpler, since Schur indices are 1.) See for example Curtis & Reiner (1962), Section 70, and also the book Character Theory of Finite Groups by Isaacs (reprinted by AMS), especially Chapter 9 and problem (9.6). The basic criterion deals with an arbitrary finite group $G$ and its representations over a finite splitting field $\mathbb{F}_q$. In your situation, this applies to finite Chevalley groups by the 1963 paper of Steinberg (see $\S13$ of his 1967-68 Yale lectures here). What you need to know is that an absolutely irreducible $FG$-module $M$ over a subfield $F$ of the splitting field $\mathbb{F}_q$ is irreducible over the prime field $\mathbb{F}_p$ if and only if $F$ is the "field of definition" of $M$. Here a field of definition is one generated over the prime field by the traces of representing matrices. (It's probably hard to find an explicit statement and proof of this in the textbooks, but it's implicit.) In your specific example, it's therefore necessary to be careful about where the traces of the representing matrices lie. REPLY [2 votes]: $G$ acts irreducibly on $V$: Suppose that $W$ is an $\mathbb F_p$-subspace of $V$ invariant under $G$, of dimension $d$. So $W$ has $p^d-1$ nonzero elements. Hence there is an orbit of $G$ on these elements of length prime to $p$. Let $A\in W$ be a trace $0$ matrix in this orbit. The orbit length through $A$ is the index $[G:C_G(A)]$. If $A$ has no eigenvalue in $\mathbb F_q$, then $A$ is irreducible on $\mathbb F_q^2$, so by Schur's Lemma we get that $C_G(A)$ is a subgroup of the multiplicative group of $\mathbb F_{q^2}$. In particular, $p$ does not divide $|C_G(A)|$, so $p$ divides the orbit length through $A$, a contradiction. Thus $A$ has eigenvalues in $\mathbb F_q$. First suppose that $p$ is odd. If the eigenvalues are equal, then they vanish by the trace $0$ condition. If they are however distinct, then $A$ is conjugate in $G$ to a diagonal matrix. So we may assume that there is a nonzero $a$ such that either $A=\begin{pmatrix}0 & a\\0 & 0\end{pmatrix}$ or $A=\begin{pmatrix}a & 0\\0 & -a\end{pmatrix}$. In either case, one quickly computes the conjugacy class $A^G$, and finds out that in both cases the $\mathbb F_p$ span of the classes are $V$. So $W=V$ because $A\in W$. Similarly, in the case $p=2$ we obtain that we may assume $A=\begin{pmatrix}a & b\\0 & a\end{pmatrix}$ for some $a,b\in\mathbb F_q$. Again, $\mathbb F_p[A^G]=V$.<|endoftext|> TITLE: An infinite product associated with random matrices QUESTION [7 upvotes]: Motivation Let ${\mathbb F}_q$ be the field with $q$ (a power of some prime number) elements. Then the order of $GL_n({\mathbb F}_q)$ is $$(q^n-1)(q^n-q)\cdots(q^n-q^{n-1}).$$ The fact that this order is divisible by $q^{n(n-1)/2}$, which is the order of the uni-triangular subgroup, plays a role in the proof of Sylow's Theorem, but I am interested in another topic. The order of $SL_n({\mathbb F}_q)$ is actually $$q^{n(n-1)/2}(q^n-1)(q^{n-1}-1)\cdots(q^2-1).$$ From the formula above, and the fact that $M_n({\mathbb F}_q)$ has cardinal $q^{n^2}$, we see that the probability that the polynomial $Det_n$ (the determinant) takes the value $1$ over $({\mathbb F}_q)^{n^2}$ is $$\frac1q(1-q^{-2})(1-q^{-3})\cdots(1-q^{-n}).$$ Amazingly, this probability is strictly less than $\frac1q$. The probability is the same for $Det_n=x$ if $x\ne0$, while it is larger than $\frac1q$ for $Det_n=0$. Strangely enough, this probability has a non-trivial limit $$p(q)=\frac1q\prod_{m=2}^\infty\left(1-\frac1{q^m}\right)$$ as $n\rightarrow+\infty$ (large random matrices with entries in ${\mathbb F}_q$). We have $p(q)>0$ because $$\sum_{m=2}^\infty\frac1{q^m}<+\infty.$$ Question : Does this expression has a closed form ? In other words, is there a simplest formula than this infinite product. Edit. As mentionned by Pietro and Gerald (I realized that too, but was away from my internet connection), this limit probability can be expressed in terms of Dedekind's eta function: $$p(q)=\frac1{q^{1/24}(q-1)}\eta(\tau),\qquad q=e^{-2i\pi\tau}\hbox{ and }\Im\tau>0.$$ I apologize to those who use to denote $q$ the expression $e^{2i\pi\tau}$, the inverse of the present $q$. It is really unfortunate that it is used in field theory too. Now a related question: What are the special values of Dedekind's function ? For instance, do we know a simplest expression for $q=2$ ? REPLY [4 votes]: I think you actually want to look at the probability that the determinant is nonzero, which is: $$ \prod_{m \geq 1} \left( 1- \frac{1}{q^m} \right)$$ This seems like a nicer product. Of course you may recover the probability that $\det=1$ as $(q-1)^{-1}$ times this.<|endoftext|> TITLE: Number of orders of $k$-sums of $n$-numbers QUESTION [12 upvotes]: Suppose we have a $n$-element set $S$. Denote the set of its $k$-element subsets by $K$ ($|K|=\binom{n}{k}$). If the elements of $S$ are real numbers then to each $k$-element subset we can associate its sum. If these sums $\binom{n}{k}$ numbers are all distinct, they induce an order on $K$. What is number of orders on $K$ we can obtain this way? It is clear that this is not all $\binom{n}{k}!$ orders are possible. For example, because the smallest $k+1$ elements defined by the order of numbers of $S$ and $n! < \binom {\binom{n}{k}}{k}$ for sufficient large $k$. There are many other constrains on the orders of $K$. For $n=4$, and $k=2$. $S=\{a, b, c, d\}$ and $K=\{(a,b),(a,c),(a,d),(b,c),(b,d),(c,d)\}$. If $\{a, b, c, d\}=\{0.2, 0.4, 0.8, 0.3\}$, then the correspondent sums equals $\{0.6,1.0,0.5,1.2,0.7,1.1\}$, and we have the following order on $K$: $\{(a,d),(a,b),(b,d),(a,c),(c,d),(b,c)\}$. Note that we can not get the order $\{(a,b),(a,c),(c,d),(b,d),(a,d),(b,c)\}$ because from the order of the first two elements we get that $bc$. REPLY [7 votes]: You are asking for the number of regions of the real hyperplane arrangement with hyperplanes $x_{i_1}+\cdots+ x_{i_k}= x_{j_1}+\cdots+x_{j_k}$. For $n=4$ and $k=2$ the number of regions can be computed to be 48. The computation for $n=5$ and $k=2$ should be tractable, though beyond my computational skills. Based on what is known about counting the number of regions of a real arrangement, I would be very surprised if someone could give a nice answer to your question for any $n$ and $k$.<|endoftext|> TITLE: The closed form of $\sum_{n=2}^\infty(-1)^{n+1}\frac{\psi(n)}n\log(n)$ QUESTION [7 upvotes]: The following series I'm interested in $$\sum_{n=2}^\infty(-1)^{n+1}\frac{\psi(n)}n\log(n)$$ where $\psi(n)$ is digamma function arose in the evaluation of an integral I posted on MSE, https://math.stackexchange.com/questions/857301/evaluation-of-int-01-frac-log1x1x-log-left-log-left-frac1x-rig and as it can be easily seen, after a while all gets reduced to computing the foregoing series. The investigations that were done so far led nowhere, no closed form could be found. My intuition tells me there is a closed form, and that's the main reason for that I also posted the question here. To summarize, I have the following questions: $$a). \text{Is this series known in literature? If yes, could you name some sources?} $$ $$b). \text{How would you recommend me to tackle this series? }$$ $$c). \text{I would appreciate if anyone would do some research on it. }$$ Please consider this question comes from a person with no background in mathematics. REPLY [6 votes]: It's always non-trivial to give a definite negative answer to this kind of question. However, the following methods, which tend to be used in cases when closed form answers are obtained, don't seem to lead to anything useful. Explicit summation over $n$ with a finite upper limit. Knowing an explicit formula for the finite sum and taking the upper limit to infinity would give the desired series sum. There are methods that can check whether there exists an anti-difference (like an anti-derivative, but for sums instead of integrals) in a large class of functions, like those of hypergeometric type, e.g., Zeilberger's algorithm. Such algorithms are actually already implemented in Mathematica and Maple. So, since asking one of these computer algebra systems doesn't yield an answer, this method probably doesn't work. Representation as a sum of residues. Is there a function $g(z)$ and a contour in the complex $z$-plane such that the series equals the sum of residues of $g(z)$ inside this contour? Here are a couple of candidates, $g(z) = -\frac{\pi}{\sin(\pi z)} \frac{\psi(z)}{z} \log z$ or $g(z) = -\frac{1}{2}(\psi(\frac{-z+1}{2})-\psi(\frac{-z}{2})) \frac{\psi(z)}{z} \log z$, with the contour in both cases hugging the positive real axis. The simplification in this representation could come from the ability to deform the contour to encircle an alternative set of poles, with simpler structure, without changing the value of the integral. In this case, the complicated structure of the poles that appear on the negative real axis and the necessary logarithmic branch cut starting at $z=0$, does not seem to lead to any simplification. Fourier series and Parseval's identity. This is the method that is often used to compute the sum $\sum_{n=1}^\infty \frac{1}{n^2}$. The summand is split into a product of Fourier coefficients of two known functions and the sum is transformed into an integral of the product of these functions. (One can also use a slight variation of the method with power series instead of Fourier series, as shown here.) While the integral in question can be factored in multiple ways that give Fourier series of known functions (e.g., with Mathematica's help). Their product doesn't seem to be of the type that would have a closed form itself. Since these methods don't seem to be succeeding, it's likely that there is no answer in commonly used closed forms.<|endoftext|> TITLE: Killing vector fields on sphere QUESTION [8 upvotes]: Let $u$ be a smooth function on $\mathbb S^2$, and assume that for every killing vector field $V$ on $\mathbb S^2$. $$\int_{\mathbb S^2} V(u) x_j dS=0\text{,}\forall j=1,2,3$$ Is $u$ necessarily constant? REPLY [6 votes]: The answer is "no". Choose a basis $V_1, V_2, V_3$ of Killing fields. Note that $$\int\limits_{\mathbb S^2} V_iu\cdot x_j\cdot d\,\mathrm{area} = -\int\limits_{\mathbb S^2} u\cdot V_ix_j\cdot d\,\mathrm{area}$$ Threfore you can take any $u$ which is orthogonal to each of 9 functions $s_{i,j}=V_ix_j$.<|endoftext|> TITLE: What measurable quantity can constrain the number of odors human can discriminate? QUESTION [35 upvotes]: This is not a very typical MO question, but I hope you bear with me. It concerns a recent disagreement in the biology literature about how many different odors humans can discriminate. The authors of a paper in Science from March 2014 claimed that, based on their experiments, "humans can discriminate more than 1 trillion olfactory stimuli," a number which puts our other senses to shame. A recent critique of the Science paper, posted on arXiv, takes issue with the way the Science authors interpreted their data, and claims that even a number as small as merely 10 discriminable stimuli is consistent with the data. From what I can tell, the disagreement boils down to a very clear difference in assumptions, which I will try to explain in a simplified way that probably misses a lot of the more minute details, but hopefully gives a good idea of the mathematical question which is at play. The Science authors give a subject three vials to smell. Each vial contains a random, equal-parts mixture of 30 compounds out of a repertoire of 128 compounds. Two of the vials have identical mixtures, and the subject is asked to pick the odd vial out. Think of each mixture as a binary vector $\mathbf{x}\in X=\{0,1\}^{128}$ of Hamming weight 30. The authors estimate a critical Hamming distance $D$ at which 50% of mixture pairs at this distance are discriminable. Not many tests need to be run to obtain a very good estimate of $D$. Think of each discriminable "odor" as a set $S\subset X$, and assume that the mixture space is completely partitioned into $N$ odors $S_1,\ldots,S_N$. Mixture $\mathbf{x}$ and $\mathbf{y}$ can be told apart by a subject if and only if they are not in the same odor $S$. The main task is to infer a likely value for $N$. If odors are all roughly Hamming balls of equal radius, we can directly obtain the radius from the critical distance $D$, from the radius we get the volume of the ball, and we obtain $N\approx|X|/|S|$. This analysis appears to give an estimate of $N>10^{12}$. The critique points out that there is no reason to assume that the odors are roughly spherical in the absence of detailed mechanistic knowledge about the olfactory system, which apparently we do not have. One interesting section of the critique shows that a similar analysis applied to a hypothetical experiment using color stimuli yields nonsensical results. In the color vision system, we know that if $X$ is a binary vector space describing mixtures of optical spectra, then only the projection of $\mathbf{x}$ into a real three dimensional space can be sensed, $(s,m,l)(\mathbf{x}) = \sum_i x_i (s_i,m_i,l_i)$. Therefore, the colors are far from spheres in $X$; they are more like the preimages of balls under this projection. Clearly, for this kind of highly anisotropic shape, the same critical distance $D$ corresponds to a much larger volume. I don't know enough about the biology to tell whether it is reasonable to expect that something similar could happen in the olfactory system, but the critique points out that when allowing for nonspherical shapes, the same value of $D$ is consistent with many-order-of-magnitudes-larger odors. A particular construction even yields a number $N=10$ as being consistent with the data. Now, ignoring the biology, measuring $D$ is clearly a bad way to constrain $N$, since one value of $D$ is apparently consistent with both $N>10^{12}$ and $N=10$. The question I have for the MO crowd is what extra quantity, preferably obtainable from the same kind of odd-vial-out tests (maybe even extractable from the existing data), can actually constrain $N$ to a reasonable range. REPLY [8 votes]: Suppose $\Omega = \{0,1\}^{128}$ is partitioned into $N$ different sets $S_j$, with $|S_j|/|X| = p_j$. Suppose you take $n$ random pairs of points of $\Omega$ and see how many of these pairs can't be distinguished (presumably because they are in the same partition). The probability that two given points are in the same partition is $S_2 = \sum_{j=1}^N p_j^2$, so the expected number of indistinguishable pairs is $n S_2$; the fraction of pairs that are indistinguishable is an unbiased estimator of $S_2$. By Cauchy-Schwarz, $S_2 \ge 1/N$ (which would be the value if all $p_j$ are equal), so a good estimate on $S_2$ gives you a lower bound on $N$.<|endoftext|> TITLE: Algorithm to produce random number with a gamma distribution QUESTION [8 upvotes]: I'd like to produce pseudo-random numbers with different distributions for a Monte Carlo simulation. I've got the poisson distribution working nicely with an algorithm from Knuth. I'm having trouble getting a nice easy and fast algorithm for a power distribution. The gamma distribution should do, but the article in wikipaedia gives an algorithm, but remarks that it's not a good one, without providing a link to a better one. http://en.wikipedia.org/wiki/Gamma_distribution#Generating_gamma-distributed_random_variables Is there a good, fast algorithm for a gamma distribution? REPLY [8 votes]: The difficulty mentioned in Wikipedia refers to gamma distributions with small shape parameter; this has been addressed in arXiv:1302.1884: The gamma distribution with small shape parameter can be difficult to characterize. For this reason, standard algorithms for sampling from such a distribution are often unsatisfactory. In this paper, we first obtain a limiting distribution for a suitably normalized gamma distribution when the shape parameter tends to zero. Then this limiting distribution provides insight to the construction of a new, simple, and highly efficient acceptance--rejection algorithm. Pseudo-code and an R implementation for this new sampling algorithm are provided.<|endoftext|> TITLE: Dirichlet Characters as Eigenvectors QUESTION [5 upvotes]: This was asked in Math Stackexchange here but generated no comments or answers. I have slightly edited the original question with the comment in the fourth paragraph and the explicit matrix example at the end. I have a very concrete question. First an example: If one considers the additive group $Z/(n)$ which is cyclic, the corresponding group characters are the rows of the discrete fourier transform (DFT)matrix $D$. Take any $n\times n$ circulant matrix $C$. Obviously this matrix is generated by the action of the cyclic group on the first row. Its eigenvectors are the columns of $D$ (which is symmetric) and the matrix of eigenvectors diagonalizes any circulant matrix. Corresponding to each root of unity $\omega$ of order $n,$ its consecutive powers make up the eigenvector and the inner product of the first row of $C$ is clearly the corresponding eigenvalue. Now comes the question. For simplicity, assume that $n$ is a prime. Consider the multiplicative group of $Z/(n)$ which has $n-1$ elements. The Dirichlet characters of this group form an $(n-1)\times(n-1)$ matrix $D'$. Now take a ``multiplicative matrix'', $M$ where first one specifies the first row $(a_1,\ldots,a_{n-1})$ and then obtains the $k^{th}$ row as $(a_{k},a_{2k},\ldots,a_{(n-1)k})$ where the indices are computed modulo $n. In fact, if $n$ is assumed to be prime, I'd expect that the eigenvectors to be obtained by a coordinate permutation of the eigenvectors which are the columns of the DFT matrix since both groups in question are cyclic. I'd have assumed that such a matrix would have exactly the Dirichlet characters, of which there are enough when $n$ is a prime, as its eigenvectors. A few basic computations on Mathematica seemed to contradict this hope. For example taking $n=7,$ and the first row as $(1,1,1,0,0,0)$ yielding the matrix $$ \left( \begin{array}{cccccc} 1 &1 &1 & 0& 0& 0\\ 0 &1 &0 & 1& 0& 1\\ 0 &1 &1 & 0& 0& 1\\ 1 &0 &0 & 1& 1& 0\\ 1 &0 &1 & 0& 1& 0\\ 0 &0 &0 & 1& 1& 1 \end{array} \right) $$ the expected eigenvectors materialized. However, taking the first row as $(1,1,0,0,0,0)$ yielding the matrix $$ \left( \begin{array}{cccccc} 1 &1 &0 & 0& 0& 0\\ 0 &1 &0 & 1& 0& 0\\ 0 &0 &1 & 0& 0& 1\\ 1 &0 &0 & 1& 0& 0\\ 0 &0 &1 & 0& 1& 0\\ 0 &0 &0 & 0& 1& 1 \end{array} \right) $$ did not. Why is this the case? References, pointers welcome. REPLY [4 votes]: I believe the eigenvectors are the ones you guessed, but in your second example, the dimensions of some of the eigenspaces are larger than one. I would guess that Mathematica chose a basis for those eigenspaces different than the eigenvectors coming from the multiplicative Dirichlet characters. By the way, if you view the second example as the adjacency matrix of a directed graph, the graph is two disjoint directed 3-cycles (with loops), and from this point of view the eigenvectors might be more transparent.<|endoftext|> TITLE: Reverse plane geometry, anyone? QUESTION [6 upvotes]: I refer to Greenberg's wonderful 2010 MAA article "Old and new results in the foundations of elementary plane Euclidean and non-Euclidean geometries". There, and in his book, Greenberg defines a Hilbert plane as a model of the 13 Hilbert's first-order axioms of plane geometry. How many elementary equivalence classes of Hilbert planes are there? Recall that two Hilbert planes are elementarily equivalent if they satisfy the same first-order sentences. A rough estimate from scanning through Greenberg's is that there are at least $10$ elementary equivalence classes. For example, among Euclidean planes, Greenberg already lists two other elementary equivalence classes: namely semi-Euclidean planes (where the angles in each triangle sum to two right angles) and Pythagorean planes (in which right-angled satisfy the Pythagorean identity). Sorting out this zoo of plane geometries would profit from the joint effort of many mathematicians. Polymath, maybe? REPLY [8 votes]: There are exactly $2^{\aleph_0}$ elementary equivalence classes of Hilbert planes. Let $P$ be a subset of the set of odd primes. Let $K_P$ be the smallest field extension of $\mathbb{Q}(\{2^{1/p}:p \in P\})$ in $\mathbb{R}$ that is closed under taking square roots of positive elements, and let $H_P$ be the plane $K_P^2$. For each odd prime $p$, there is a first-order sentence $S_p$ expressing the existence of a sequence of $p+1$ line segments whose lengths form a geometric progression such that the last segment is twice as long as the first; it holds in $K_P^2$ if and only if $2^{1/p} \in K_P$, which by degree considerations is equivalent to $p \in P$. Thus for subsets $P$ and $Q$ of the set of odd primes, $H_P$ and $H_Q$ are elementarily equivalent if and only if $P=Q$. The number of choices for $P$ is $2^{\aleph_0}$. On the other hand, there are only countably many first-order sentences, and the number of elementary equivalence classes is at most the number of subsets of the set of first-order sentences, so there are at most $2^{\aleph_0}$ elementary equivalence classes.<|endoftext|> TITLE: Thurston geometries in dimension 4 QUESTION [15 upvotes]: In the sense of W. Thurston here, there is 3 geometries in dimension 2 and there is 8 geometries in dimension 3. Question: How many different geometries (in the sense of Thurston) do we have in dimension 4 ? REPLY [22 votes]: The 4-dimensional geometries were classified in the unpublished thesis of Filipkiewicz, which is available here.<|endoftext|> TITLE: Pathological product space norm QUESTION [10 upvotes]: Let $X$ and $Y$ be two normed vector spaces and $n(\cdot, \cdot)$ be any norm on $\mathbb{R}^2$. Is it always possible to define a norm on the product vector space $X \times Y$ as $||(x, y)||_{X \times Y} = n(||x||_X, ||y||_Y)$? Background information: the book "Advanced Calculus" by Sternberg and Loomis says the answer is negative - you have to impose an additional requirement on $n(\cdot, \cdot)$ for that to hold. However, no example is given in the text nor the exercises. Can you give me an example of such a pathological norm $n$ that fails to induce a honest normed vector space on $X \times Y$. REPLY [7 votes]: This is a comment rather than an answer but I am not entitled. The missing requirement on the two dimensional norm for the statement to be valid is that it be increasing in the natural sense that if $|x|\leq|x_1|, |y|\leq|y_1|$, then $n(x,y)\leq n(x_1,y_1)$. Norms without this property can easily be obtained by rotating a non-circular ellipse in standard position (as in the above answer).<|endoftext|> TITLE: A coalgebra structure on compact operators QUESTION [6 upvotes]: Is there a coalgebra structure $\Delta_{n}$ on $M_{n}(\mathbb{C})$ which is compatible with the natural embedding $i_{n:}M_{n}(\mathbb{C})\to M_{n+1}(\mathbb{C})$ with $i_{n}(A)= A\oplus 0$. That is $(i_{n}\otimes i_{n})\circ \Delta_{n}=\Delta_{n+1}\circ i_{n}$. If the answer is yes, can this method be used to equip the space of compact operators, the (topological) direct limit of $M_{n}(\mathbb{C}),s$, with a coalgebraic structure? REPLY [4 votes]: For $V$ finite-dimensional, the algebra $\hom(V, V) \cong V^\ast \otimes V$ is naturally self-dual, and this duality may be used to transfer an algebra structure on $\hom(V, V)$ to a coalgebra structure on $\hom(V, V)$, and vice-versa. (Notice that the duality functor $\text{Vect}^{op} \to \text{Vect}$ on finite-dimensional spaces induces a functor from coalgebras to algebras, and vice-versa.) The map $M_n(\mathbb{C}) \to M_{n+1}(\mathbb{C})$ mapping $A \mapsto A \oplus 0$ is dual to the map $M_{n+1}(\mathbb{C}) \to M_n(\mathbb{C})$ that takes an $(n+1) \times (n+1)$ matrix to the $n \times n$ matrix made from the first $n$ rows and columns. If we choose the algebra structure on the spaces $M_n(\mathbb{C})$ to be not the usual matrix multiplication but the one given by entrywise multiplication, then these are algebra maps and they dualize to coalgebra maps. This gives an affirmative answer to the question. However, this is clearly highly dependent on basis and doesn't strike me as terribly interesting. The forgetful functor $\text{Coalg} \to \text{Vect}$ creates colimits, and so the colimit of a chain of coalgebra maps $M_n(\mathbb{C}) \to M_{n+1}(\mathbb{C})$ is created from the colimit in $\text{Vect}$. This answers the second question affirmatively.<|endoftext|> TITLE: Possible cardinality and weight of an ordered field QUESTION [8 upvotes]: Is it true (in ZFC) that for any regular infinite cardinal $\kappa$ there exists an ordered field of weight $\kappa$ and cardinality $2^\kappa$ (or at least $>\kappa$)? The field of real numbers shows that for $\kappa=\aleph_0$ the answer is trivially "yes". It seems that this question has an affirmative answer in the Constructive Universe. Is it true in ZFC? Or the converse is true: there is a model of ZFC in which all ordered fields of weight $\aleph_1$ have cardinality $\aleph_1$? REPLY [4 votes]: On the one hand, if $K$ is an ordered field of weight $\kappa$, then trivially $$|K|\le\mathrm{ded}(\kappa):=\sup\bigl\{|L|:\text{$L$ is a linear order with dense subset of size $\kappa$}\bigr\}\le2^\kappa.$$ On the other hand, we have the following lower bounds, which in particular show that there is always an OF of weight $\kappa$ and cardinality strictly larger than $\kappa$. Proposition 1: If $\lambda$ is the least cardinal such that $\kappa^\lambda>\kappa$, there exists an ordered field of cardinality $\kappa^\lambda$ with a dense subfield of cardinality $\kappa$. Proof: Note that the condition implies that $\lambda$ is regular, and $\lambda\le\kappa$. First, we construct an increasing sequence of fields $\{F_\alpha:\alpha\le\lambda\}$ of cardinality $\kappa$ as follows. Let $F_0$ be an OF of cardinality $\kappa$ such that $(0,1)$ contains $\kappa$ disjoint nonempty intervals. (Incidentally, it’s easy to show that an OF of weight $\kappa$ always contains $\kappa$ disjoint intervals, but we will not need this.) If $\alpha$ is limit, we put $$F_\alpha=\bigcup_{\beta<\alpha}F_\beta.$$ For the successor step, since $\kappa^{<\lambda}=\kappa$, there are only $\kappa$ pairs of subsets $A,B\subseteq F_\alpha$ such that $AF_{\alpha}$. Using the regularity of $\lambda$, the field $F:=F_\lambda$ thus constructed satisfies: $|F|=\kappa$, $(0,1)_F$ contains $\kappa$ disjoint (nondegenerate) intervals, $F$ has an increasing cofinal subsequence $\{u_\alpha:\alpha<\lambda\}$, $F$ has the $\eta_\alpha$ property for $\aleph_\alpha=\lambda$, that is, if $A,B\subseteq F$ are such that $A\kappa$, there exists an ordered (real-closed) field of cardinality $2^\nu$ with a dense subfield of cardinality $\kappa$. Proof: Put $\mu=2^{<\nu}\le\kappa$ and $\lambda=\mathrm{cf}(\nu)$. An exercise in cardinal arithmetic shows that $\mu^{<\lambda}=\mu$ and $\mu^\lambda=2^\nu$, hence there exists a RCF of size $2^\nu$ with a dense subset of size $\mu$, which we can enlarge to $\kappa$. Note that for a given $\kappa$, the bound in the Corollary is better than in the Proposition: clearly $\lambda\le\nu$, hence $\kappa^\lambda\le(2^\nu)^\lambda=2^\nu$.<|endoftext|> TITLE: Lindelof Hypothesis implying Selberg Eigenvalue Conjecture? QUESTION [9 upvotes]: The Generalized Lindelof Hypothesis says that for the $L$-function of an automorphic form we have $$L(1/2+it)\ll Q(t)^{\epsilon}$$ for any $\epsilon>0$ where $Q(t)$ is the conductor of $L(s)$ at $t$. For a Maass form $\phi$ of level $N>1$ and eigenvalue $1/4+\lambda^2$, will Lindelof Hypothesis imply Selberg Eigenvalue conjecture, i.e., $\lambda\geq 0$? I know that functoriality results such as Gelbart-Jacquet, Shahidi, Kim gives a very good bound, such as Kim-Sarnak bound of 7/64. Obviously functoriality for of all symmetric powers implies the Selberg eigenvalue conjecture. The proof involves some non-vanishing results for families of $L$-functions (Kim-Sarnak, Luo-Rudnick-Sarnak). I am thinking about the relation of the Selberg eigenvalue conjectures and the GRH/Lindelof. Lastly, GRH itself at a single point for one L-function probably does not say much about the Selberg eigenvalue conjecture because GRH only excludes non-trivial zeros. A violation of Selberg eigenvalue conjecture will give us a zero from Gamma factors (thus it is trivial zero). But family of L-functions with GRH or Lindelof possibly can tells more about non-vanishing (trivial zeros) of L-functions. REPLY [4 votes]: To see that this is not the case, consider that it is well known that the Generalized Riemann Hypothesis implies the Lindelöf hypothesis. On the other hand, not even the full GRH implies the Selberg Eigenvalue conjecture (or any other Generalized Ramanujan-type conjecture). In fact, the best bound we can prove under GRH for a Maass form $\pi$ is $|\mathrm{Re}\mu_\pi(\infty,j)|\leq 1/4$, which is quite far from the $|\mathrm{Re}\mu_\pi(\infty,j)|=0$ we need. There's also an argument of Sarnak that goes something like this: Take $\pi$ a Maass form, and assume we know all the general Ramanujan conjectures for $\pi$ (we know from Langlands that we can treat the eigenvalue conjecture as a Ramanujan conjecture at infinity). Then we can prove that $L(s,\pi)$ is uniformly bounded in $\mathrm{Re}(s)>1+\epsilon$, with $\epsilon >0$, but we can't push this result even to $\mathrm{Re}(s)=1$. On the other direction, we have the same problem when we try to use the Generalized Lindelöf hypothesis to give information about $\pi$ beyond $\mathrm{Re}(s)=1$.<|endoftext|> TITLE: $\pm1$-polynomials with a maximal non-real root QUESTION [12 upvotes]: For given $n$, consider a polynomial $\sum_{k=0}^na_kz^k$ with all coefficients $a_k\in\{\pm1\}$. I am interested in the following: How big can the modulus of a non-real root of such a polynomial be? Wlog we can assume $a_n=1,a_{n-1}=-1$. Then a systematic search for small $n$, looking each time at the 10 or so extremal polynomials, exhibits a clear pattern for the highest coefficients: they come in groups of $3$ or $4$ consecutive +1's or -1's, starting with $z^n-z^{n-1}+++----+++---++++\cdots$. More precisely, the group sizes are displayed here (first line +1's, second line -1's) 1 3 3 4 3 3 3 4 3 3 3 4 3 3 4 4 3 3 4 4 3 3 4 4 3 3 4 . . . 1 4 3 3 4 4 3 3 4 4 3 3 4 3 3 3 4 3 3 3 4 3 3 3 4 3 . . . Note that there are never two adjacent groups of size 4 or three of size 3, so the pattern is rather regular, as may be expected. As $n$ grows, the extremal root $z_0$ (and $\bar z_0$) of the extremal polynomials does not at all jump around, but converges quite rapidly towards $0.93757749648487973269811306454355 \pm 1.2634174429011374851417570421775\; i$, e.g. we have $n=20\implies z_0\approx 0.937537 \pm 1.26337\;i$ $n=40\implies z_0\approx 0.9375774916 \pm 1.263417437\;i$ $n=60\implies z_0\approx 0.9375774964839 \pm 1.26341744290078\;i$. $n=80\implies z_0\approx 0.93757749648487963 \pm 1.263417442901137422\;i$. $n=100\implies z_0\approx 0.937577496484879732688 \pm 1.263417442901137485132\;i$. The obvious questions: Can the above 3-4-sequences be characterized? E.g. are they periodic? or self-similar? What can be said about the limit value of $z_0$? Is there a (closed form or whatever) formula for $z_0$? Although the problem is not directly related to Garsia numbers (see the article mentioned here), it naturally leads again to the question of limit points of zeros of such polynomials, given that $z_0$ visibly yields one of those: Are there only countably many non-real limit points of zeros of $\pm1$-polynomials? REPLY [10 votes]: A closely related problem was considered by Odlyzko and Poonen who looked at the class of polynomials with $0$ or $1$ coefficients. On page 330, they discuss the question of the smallest (in size) non-real root in this family (which is of course equivalent to the largest non-real root) and observe that numerically its size is about $0.734$. In Section 6 of their paper they discuss computations plus a simple Rouche's theorem argument which would enable one to compute this as accurately as desired. The same arguments should carry over to this situation. There doesn't seem to be any more natural description of the smallest size of non-real roots in the Odlyzko-Poonen paper.<|endoftext|> TITLE: Polynomial-time algorithm for determining whether a polynomial is positive on $\mathbb{N}$ QUESTION [8 upvotes]: Does there exist a polynomial-time algorithm to determine whether a given polynomial $p(n)$ with integer coefficients is positive on $\mathbb{N}$, in the sense that $p(n) \geq 0$ for all $n\in\mathbb{N}$? This question seems to be related, but it involves multivariable polynomials and asks about positivity on all of $\mathbb{R}^n$, and the answer doesn't seem to apply to single-variable polynomials. Edit: Just to clarify, the input for the algorithm should be a string that encodes the polynomial $p$ in a reasonably compact fashion, e.g. the digits of the degree of $p$ followed by the digits of the coefficients. REPLY [11 votes]: Yes. Compute a Sturm sequence and use binary search to locate the real roots. REPLY [7 votes]: Yes, sure, this can be done. You can "find" all the roots of $f$ on the real axis in polynomial time. More precisely, you can find intervals $[a_i,b_i]$ ($i=1,\dots, d$) with rational endpoints $a_i$, $b_i$, so that $f$ is monotonic on $[a_i,b_i]$ and $f(a_i)f(b_i)<0$, and so $x_i\in [a_i,b_i]$ is a real root of $f$. Assuming these intervals are ordered from left to right, you easily test whether there is a positive integer between $x_{2j-1}$ and $x_{2_j}$ (this is for the case of even degree $f$; for the odd degree the first $x_i$ can be ignored). This will pin the finitely many (at most the degree of $f$) positive integers of which $f$ might be negative.<|endoftext|> TITLE: Gorelic's Forcing for large Lindelöf spaces with points $G_\delta$ QUESTION [6 upvotes]: I am trying to understand a step for proving that there exists large Hausdorff Lindelöf Spaces with points $G_\delta$ using forcing. I am following Isaac Gorelic's "The Baire Category And Forcing Large Lindelöf Spaces With Points $G_\delta$. (http://www.jstor.org/discover/10.2307/2160344?uid=3737664&uid=2&uid=4&sid=21105156482053) When proving that $\mathbb P$ has the $\omega_2$-c.c. if $|\mathbb Q|=\omega_2$, $\mathbb Q \subset \mathbb P$, he says that by CH, by the $\Delta$-system lemma and simple counting we may assume that there are $p \neq q$ such that $\text{tp}\, I^p=\text{tp}\, I^q$, $\text{tp}\, A^p=\text{tp}\, A^q$, that by denoting $\xi'$ the unique ordinal in $A^q$ corresponding to a $\xi \in A^p$ such that te order type of $A^p\cap \xi$ is the same of $A^q \cap \xi'$ we have $\forall \xi \in A^p\, \forall \eta \in A^p\, f^p_\xi(\eta)=f_{\xi'}^q(\eta')$. Picking these $p, q$ is easy. He also states that we may assume that $\forall x \in \Delta:=A^p\cap A^q, \, f^p_\xi(x)=f_{\xi'}^q(x)$. This is also something easy to do, but I think he made a typo here. Maybe what he wanted was $\forall x, \xi \in \Delta:=A^p\cap A^q, \, f^p_\xi(x)=f_\xi^q(x)$, but I may me wrong. We will call this last supposition (*). Then he defines a common extension $r$ of the form $r=\langle I^p\cup I^r, A^p\cup A^Q, F^r, G^r, T^p \cup T^q\rangle$. I think he made a typo here, I think $I^r$ was supposed to mean $I^q$. Then he says how to build such $G^r$, of course, the set $F^r$ will be determined by $G^r$. He divides the proof in three cases. The first case states that if $\alpha \in I^p \cup I^q$, then we set $g_\alpha^r=g_\alpha^p \cup g_\alpha^q$. I think he made another typo here, I think $I^p \cup I^q$ was supposed to mean $I^p \cap I^q$. At this point, the reader must verify that this function is well defined. So let $x \in A^p \cap A^q \setminus A^\alpha$. Let $\xi \in A_\alpha$. Then $\xi \in A^p\cap A^q$ so by (*) it follows that $g_\alpha^p(x)=f_\xi^p(x)=f_\xi^q(x)=g_\alpha^q(x)$, as intended. Question 1: is the supposition (*) really neccessary or can we show that this function is well defined only by using what Gorelic stated? For the second case, suppose $\alpha \in I^p \setminus I^q$ (the third case is analogous). So by the Baire Category Theorem, there is an $H: A^q\rightarrow 2$ with $$h \in (U_{g_{\alpha'}}|A^q)\cap \bigcap\{U_B|A^p:B\in T^p\}\setminus\{f^q_\eta: \eta \in A_{\alpha'}\}\neq \emptyset$$ Okay, no problems with this. Then we set $g_\alpha^r=g_\alpha^p\cup h$. Question 2: why is $g_\alpha^r$ well defined? Edit: In order to try to show that $g_\alpha^r$ is well defined, I noticed that we may also suppose that for every $\alpha \in I_p$ and for every $x \in \Delta\setminus A_\alpha$, $g^p_\alpha(x)=g^q_{\alpha '}(x)$. So in case $x\in A^p \cap A^q \setminus A_\alpha \setminus A_{\alpha'}$, it follows that $g^p_\alpha(x)=g_{\alpha'}(x)=h(x)$. But I still don't know what to to in case $x \in A_{\alpha'}$. REPLY [6 votes]: I find that actually this forcing is conceptually clear and not too technical. Don't get bogged down with a few typos. Fact 3 is about amalgamation of similar conditions. Only $g_\alpha^p$-s are important for amalgamation (since the rest is preassigned in the ground model). There are only $\omega_1$-many order types of sets $I^p$. Assume all $p \in Q$ have the same ordertype $(I^p) < \omega_1$. Then using CH and the $\Delta$-system lemma for $\aleph_2$ countable sets $I^p$, $p \in Q$, choose $p$ and $q$ which are isomorphic structures. Namely, (1) for $\alpha \in I^p \cap I^q$, (a) type($I^p \cap\alpha $) = type ($I^q \cap\alpha$), and (b) $g_\alpha^p \cup g_{\alpha}^q$ is a function, and (2) for $\alpha \in I^p \setminus I^q$, and $\alpha' \in I^q \setminus I^p$ such that type($I^p \cap\alpha $) = type ($I^q \cap\alpha'$), $g_\alpha^p \upharpoonright (A^p \cap A^q) = g_{\alpha'}^q \upharpoonright (A^p \cap A^q)$. Now regarding Case 2: $g_\alpha^r$ is a function, because (as you've noticed) $h\upharpoonright $ domain ($g_{\alpha'}^q$) $= g_{\alpha'}^q$, and as $A_{\alpha'} \cap A^p = \emptyset$, $x \in A_{\alpha'} \Rightarrow x\notin $ domain ($g_\alpha^p$). REPLY [4 votes]: A shorthand for the proof of Fact 5 is to take a countable elementary submodel of the ground model.<|endoftext|> TITLE: About Sylvester's determinant QUESTION [8 upvotes]: If $A$ is any $n \times m$ matrix and $B$ is any $m \times n$ matrix then one familiar form of the Sylvester's identity is $\det(I + AB) = \det(I + BA)$. Now somehow curiously this above identity is often enough used along side another statement which says that for any vector $v$ and any invertible square matrix $A$, it is true that $\det(A + v v^T) = \det(A)\det(I + v^TA^{-1}v)$. Is there some relationship between these two identities? Like can one be gotten from the other or vice versa? A particularly useful (at least in some recent big breakthrough researches!) form of the Sylvester's identity is when $A= tu, B = v^\dagger $ where ($u$ and $v$ are complex vectors and $t$ is some complex number). This then shows that $\det(I + t u v^\dagger) = 1 + tv^\dagger u$ Now this simple statement above apparently implies this more powerful formulation which is not clear to me - If $C$ is any $n\times n$ (necessarily invertible?) matrix and $A$ is a rank-$1$ matrix then is $\det( I + tCA)$ is a degree $1$ polynomial in $t$? Why? (and this is the same as saying that this is "affine-linear" in t?) REPLY [8 votes]: The common ground of those two formulas is related to the Woodbury matrix identity. This relation is a useful statement that shows what happens to the inverse when one "updates" a matrix $A\in\mathbb{C}^{n\times n}$ with a term $UCV$, with $U,V\in\mathbb{C}^{n\times m}$, $C\in\mathbb{C}^{m\times m}$. The typical situation is when $m TITLE: Vertex-primitive graphs with two vertices having almost the same neighbourhood QUESTION [9 upvotes]: Hypothesis: Let $\Gamma$ be a vertex-primitive graph with two vertices $u$ and $v$ such that $$|N(u) \cap N(v)|=|N(v)|-1$$ Question: Is it true that $\Gamma$ must either be a complete graph or have prime order? Terminology and notation: By $N(v)$, I mean the set of neighbours of $v$ in $\Gamma$. By vertex-primitive, I mean that the automorphism group acts primitively on the vertices. In other words, the automorphism group does not preserve any partition of the vertex-set apart from the trivial ones (into singletons or with just one part). Comments: It is easy to see that a vertex-primitive graph with two distinct vertices having the same neighbourhood must be edgeless. From this perspective, the question is thus about the first non-trivial case. Complete graphs clearly satisfy the hypothesis. There are indeed non-complete graphs (of prime order) satisfying the hypothesis. For example, cycles of prime order. More generally, let $p\geq 5$ be a prime, let $i\in\{2,3,\ldots,\frac{p-1}{2}\}$ and let $S=\{\pm i, \pm (i+1),\ldots, \pm(\frac{p-1}{2})\}$. Then the Cayley graph $\mathrm{Cay}(\mathbb{Z}_p,S)$ is easily seen to satisfy the hypothesis (with $u=0$ and $v=1$, for example). In fact, computer calculations show that there are no counterexamples up to order $100$, say. REPLY [3 votes]: Pablo Spiga found a proof a few weeks ago. Together, we then proved a slightly more general result which is now on the arxiv: http://arxiv.org/abs/1501.05046 It is more general in two ways: it deals with digraphs rather than graphs, and it gives some information in general when two vertices have neighbourhoods differing by say $k$, although we only get a complete classification of the graphs when $k=1$, which was the original question. (This is Corollary 4.2.)<|endoftext|> TITLE: Coloring the edges of a torus graph QUESTION [12 upvotes]: Question:Consider the $2k \times 2k$ grid graph on a torus. Is it true that for every $2$-coloring of the edges, there is an antipodal pair of vertices connected by a path that changes colors at most $k-1$ times? Edit: $k>1$ as pointed out in the comments. More formally: The $2k \times 2k$ grid graph on a torus is defined as follows. $$V(G):=\{(i,j)|i,j\in [1,2k]\cap\mathbb{Z} \}$$ Two vertices are connected if and only if on one of the coordinates they coincide, and on the other one, their values differ by exactly one, modulo $2k$. We say that a pair of vertices is antipodal if their distance is maximal. Equivalently if both their coordinates differ by $k$, modulo $2k$. The path between them which we require to change colors only $k-1$ times is not required to be $2k$ long. Additional information:The interesting part of the question is that the pairs are $2k$ away, but we only have $k-1$ color changes. If true, sometimes we need $k-1$ color changes, as the following coloring shows: Color the edges $((i,j),(i+1,j))$ red if $i$ is even, and blue otherwise. Color the edges $((i,j),(i,j+1))$ red if $j$ is even, and blue otherwise. There is a similar open problem in the hypercube. There, it is conjectured that for every coloring there is an antipodal pair of vertices, connected by a path that changes color at most once: http://theory.stanford.edu/~tomas/antipod.pdf REPLY [2 votes]: Definition: Consider the cycles of length $4$ in the torus. Let us call a subgraph of the torus a right-up diagonal if it consists of $2k$ such $4$-cycles, and the $i$-th cycle's up right corner is the $i+1$-st cycle's left-down corner. Every edge of the torus can be naturally double covered with right-up cycles. Let us call a path tight if it is contained in a right up diagonal. We will prove slightly more. Theorem: In every $2$-coloring of the torus graph, there is a tight path that changes colors at most $k-1$ times. Proof: Definition: We define the right-up diagonal of length $l$ similarly to the right up diagonal ($1 \leq l \leq 2k$), but there are only $l$ connected 4-cycles. The original right up diagonal (of length $2k$) is similar to a circle, but all the other shorter ones are similar to paths (or diagonal arcs as we are on a torus). Thus the diagonals of length $l$ have a natural beginning and endpoint, which coincides if the diagonal is of length $2k$. Lemma: Consider all the right-up diagonals of length $l$ and all the left-up diagonals of length $l$. The average number of minimal color changes needed to get from the beginning to the endpoint of a diagonal, such that we can not leave the diagonal is at most $l$. Proof: Consider a $4$-cycle. There is a right-up and a left-up diagonal containing this cycle. If (WLOG) at the right-up diagonal we have to change colors twice in this cycle, it must happened this way: We arrived to the bottom left vertex of the cycle with our path having (WLOG) the color blue. Both edges adjacent to this vertex and belonging to this cycle were red, and the other two edges blue. Thus the left-up diagonal containing this cycle does not have to change colors at this cycle, as there is a red and a blue monochromatic path between the bottom right and the top left vertices. Thus on the average a single $4$-cycle can not cause more than $1$ color change. $ \square $ Corrolary: As in a diagonal of length $k$, the beginning and endpoints of the diagonal are antipodal. We have that the average number of necessary color changes from a point to its antipodal pair using only tight paths is at most $k$. (Actually when we averaged, we took into account all the four possible diagonals connecting these points, and the best tight path in every such diagonal.) So we will have such a path with only $k-1$ color changes unless the average is exactly $k$ and every such path with minimal color changes has to change colors exactly $k$ times. Consider now diagonals of length $2k$. By the lemma we have that the average number of paths with minimal number of color changes from the beginning to the endpoint (which is the same point) is at most $2k$ (not counting that maybe we arrive with a different color when we close the cycle). but every diagonal of length $2k$ is the union of two diagonals of length $k$ where we have that we need $k$ color changes. Thus if the average number of color changes is $2k$ but we always need at least $2k$ changes, we conclude that in every right-up and every left-up diagonal, there is a $2k$ cycle that changes colors exactly $2k$ times. (now we can count all the color changes, as after an even number of color changes, in the end we always arrive with the starting color) Now we will work with these cycles. Definition: Let us call a vertex in a right-up diagonal subgraph central, if every cycle of length $2k$ in this subgraph, passes through this vertex. There are $2k$ central and $4k$ non central vertices in every right-up diagonal of length $2k$. Every vertex is central in precisely one right-up diagonal of length $2k$. Consider a right up diagonal of length $2k$ and the associated cycle with $2k$ color changes. If such a cycle has a color change at a central vertex, we have a path along this cycle to the antipodal pair of this vertex with at most $k-1$ changes. Thus we can assume that every such cycle has its color changes at non-central points. But then the cycles have to change its color at every non-central point as there are $2k$ color changes. Thus if there are antipodal non-central points on this cycle, we have exactly $k-1$ color changes between them. The only case left to examine is when we have such a cycle, and every non-central point on the cycle is such that its antipodal pair is not on the cycle. We could change the cycle along a $4$-cycle of the right-up diagonal. (As every $4$-cycle has two common edges with our $2k$ cycle and two other edges.) So if we change the cycle this way and the number of color changes is still $2k$, we are done. If for every $4$-cycle we can not change our cycle without increasing the number of color changes, we will conclude that every $4$-cycle is colored properly (adjacent edges have different colors): Let C_4 be a $4$-cycle on a right-up diagonal. The associated cycle with $2k$ color changes shares exactly two edges with this cycle. At the associated cycle, there are no color changes at the central vertices, and there is always a color change at non-central vertices. Thus the change of the cycle along this $4$-cycle increases the number of color changes, the $4$-cycle had to be colored properly. (draw a nice picture) There is only one coloring where every $4$-cycle is colored properly, and at this coloring it is easy to find tight paths connecting antipodal vertices. $\square$<|endoftext|> TITLE: Harmonic Crystal using Random Walk QUESTION [5 upvotes]: Me and my advisor were looking for a specific proof of the disorder in $2d$ harmonic crystals. We could not find a paper or a textbook with it, so I thought trying my luck here. Basically, it is a proof of the instability in an harmonic lattice crystal that uses the idea of random walk and the discrete lagrangian, and it is quite self contained. We were able to somewhat reconstruct it, but a firm reference would obviously help. If my terminology is somewhat vague, here is exactly the theorem we're trying to find its proof: Consider the lattice $\Lambda = [-L\cdots L]^2 \in \mathbb{Z} ^2 $ and a scalar field $X$ on it, i.e. $ \varphi (x) \in \mathbb{R} $. The particles outside $\Lambda $ are tied down, meaning $ \varphi (x) = 0 $, $\forall x \notin \Lambda $. Energy will be defined by $\nu (X) = \Sigma _{x \sim y} (\varphi (x) \ - \varphi(y))^2$, sum over all neighboring lattice points. The partition function in the regular way: $$Z = \int\limits_{\mathbb{R} ^ {|\Lambda|} } dX \exp(-\nu (X)) $$ The theorem is as follows: For $L \to \infty$, we have that $\langle\varphi (0)^2 \rangle = \int\limits_{\mathbb{R} ^ | \Lambda |} dX \exp(-\nu (X)) \varphi (0) ^2 $ diverges like $\log (| \Lambda | )$ REPLY [10 votes]: The object you look at is called the Gaussian Free Field (on your graph, with zero boundary conditions) in dimension $2$. There is a lot known about it. For some pointers see the Wiki page http://en.wikipedia.org/wiki/Gaussian_free_field, my lecture notes http://www.wisdom.weizmann.ac.il/~zeitouni/notesGauss.pdf and Sznitman's lecture notes https://www.math.ethz.ch/u/sznitman/SpecialTopics.pdf. Your specific question really asks about the Green function for the Laplacian in the two dimensional box, for which detailed results are available in the probabilistic literature on random walks, see Spitzer's book or Lawler's book.<|endoftext|> TITLE: Dependence of solutions on parameters in partial differential equations QUESTION [7 upvotes]: In the standard homogenization problem $$-\nabla.\left(A\left(x,\frac{x}{\epsilon}\right)\nabla u^{\epsilon}(x)\right)=f\ \mbox{in } \Omega,$$ the homogenized matrix $A_0$ is given in terms of solutions to the cell problem $$-div\left(A\left(x,y\right)\nabla w_{\lambda}(x,y)\right)=-div\left(A(x,y)\lambda\right),$$ $$w_{\lambda}\mbox{ is periodic in } y,\mbox{ and } \lambda\in\mathbb{R}^d$$ where $x$ appears as a parameter. Generally, $A$ is assumed to be bounded. What condition do we require on $A$ to guarantee some regularity of $w_\lambda$, in order to carry out the homogenization? For example, if I were to use Tartar's method of oscillating test functions, I would require $w_{\lambda}$ to be at least $H^1$ in $x$, in order to carry out some calculations. Is there a general theory of how solutions of partial differential equations depend on parameters or how they depend on the coefficients, in analogy to how there is a result for dependence of zeros of polynomials on coefficients? Cross-posted from math.se Edits: Allaire's paper deals with $A$ involving two scales. For the corrector result, Theorem 2.6 of his paper, $u_1$ is written as a linear combination of the solutions to the cell problem, $w_i$. However, in the cell problem, as above, $x$ appears as a parameter, therefore, the dependence of $w_i$, and therefore, of $u_1$ on $x$ is not known. Can this be concluded from regularity of $A$ with respect to $x$? If we were to try extending Tartar's method of oscillating test functions (adapted for periodic homogenization) to the case where $A$ involves two scales, we will require information on regularity of $w_{\lambda}$ with respect to $x$, which only appears as a parameter in the cell problem. Is there some general theory of how solutions to elliptic linear partial differential equations depend on parameters? In general, how do solutions to linear partial differential differential equations depend on coefficients; is there a general theory compared to how zeros of polynomials depend continuously on coefficients? REPLY [3 votes]: Let us be precise. Murat and Tartar's H-convergence result does not require anything on $A$ (except boundedness and ellipticity) to give the existence of an homogenized limit. So to obtain the existence of an A_0, you do not need any additional assumptions. It applies to sequences of matrices $A_\epsilon(x)$ satisfying $A_\epsilon(x)\xi\cdot\xi \geq \alpha \xi\cdot\xi$ and $A^{-1}_\epsilon(x)\xi\cdot\xi \geq \beta \xi\cdot\xi$, where $\alpha$ and $\beta$ are independent of $\epsilon$, no periodicity, or symmetry, is required. But it does not give a formula for the homogenised limit. If your question is whether, when $A$ has two scales as described, it is always given by the periodic corrector formula, in 'Homogenization and Two-Scale Convergence', by G. Allaire, SIMA 23(6) 1992, the condition stated to obtain a corrector formula alike the one you want is $$ \lim_{\epsilon\to0}\int_\Omega A_{i,j}^2\left(x,\frac{x}{\epsilon}\right) dx = \int_{\Omega}\int_{Y} A_{i,j}^2(x,y) dx dy $$ for each $i$ and $j$ in $\{1,\ldots,d\}$, where $Y$ is the unit periodicity cell. The optimality of this requirement is discussed at length in the last section of the paper. One issue for example is that if you just assume for example that $A$ is symmetric and $$\alpha \xi\cdot \xi\leq A(x,y)\xi\cdot\xi\leq \beta \xi\cdot \xi \mbox{ a.e in }\Omega\times Y, $$ for any $x\in\mathbb{R}^d$ and with $\alpha,\beta>0$, then $ x\to A\left(x,\frac{x}{\epsilon}\right) $ need not be measurable.<|endoftext|> TITLE: Does every relative curve have a Picard scheme? QUESTION [20 upvotes]: More precisely: Let $X \to S$ be a smooth proper morphism of schemes such that the geometric fibers are integral curves of genus $g$. Must the fppf relative Picard functor $\operatorname{\bf Pic}_{X/S}$ be representable by a scheme? If $g \ne 1$, then some integer power of $\omega_{X/S}$ shows that $X \to S$ is projective (Remark 2 on p. 252 of Néron models by Bosch, Lütkebohmert, and Raynaud), and then $\operatorname{\bf Pic}_{X/S}$ is a scheme by Grothendieck, FGA, no. 232, Theorem 3.1. So the question is really about the case in which $g=1$ and $X \to S$ is not projective. REPLY [10 votes]: The answer is yes: $\mathbf{Pic}_{X/S}$ is representable by a scheme. I will argue that this follows from the SGA 3 result mentioned by user27920 and from Theorem 2 (c) in section 6.6 of Neron models (which itself is based on a nonflat descent result due to Raynaud). Preliminary reductions: Since the $g \neq 1$ case is clear, let us assume that $g = 1$. It suffices to work locally on $S$, so standard limit arguments reduce to the case when $S$ is Noetherian and affine. Since $\mathbf{Pic}_{X/S} = \bigsqcup_n \mathbf{Pic}^n_{X/S}$, it suffices to show each $\mathbf{Pic}^n_{X/S}$ is a scheme. The latter is a torsor under the elliptic curve $E:=\mathbf{Pic}^0_{X/S}$, so is an $S$-algebraic space of finite presentation. Further limit arguments therefore reduce to the case when $S$ is in addition local. The main argument: We will argue that every finite set of points of $X$ is contained in an open affine. As explained by user27920, it will then follow from SGA 3, V.4.1 that each $\mathbf{Pic}^n_{X/S}$ is a scheme. Choose an affine open $Y \subset X$ that meets the closed fiber of $X \rightarrow S$. Since $X \rightarrow S$ is open and $S$ is local, $Y$ automatically meets every fiber. The action map $E \times_S X \rightarrow X$ (gotten from the identification $X = \mathbf{Pic}^1_{X/S}$) is open because it is a base change of $E \rightarrow S$, so the image $EY$ of $E \times_S Y$ is open. Then checking fiberwise, we conclude that $EY = X$. Therefore, the claim about points being in an open affine follows from the result from Neron models mentioned earlier, a simplified version of which says: Theorem. Let $S$ be an affine Noetherian scheme, let $E$ be a smooth $S$-group scheme with connected fibers, and let $E \times_S X \rightarrow X$ be a group action of $E$ on a smooth $S$-scheme $X$. If there is an affine open subscheme $Y \subset X$ such that $EY = X$, then any finite set of points of $X$ is contained in an open affine.<|endoftext|> TITLE: Uniformization of a plane minus cantor set QUESTION [6 upvotes]: Let $\mathbb{D}$ be the unit disk endowed with the Poincaré metric and $G$ be a Fuchsian group such that the hyperbolic surface $\mathbb{D}/G$ is homeomorphic to the plane minus a Cantor set. Question: Is there a conformal bijection between $\mathbb{D}/G$ and $\mathbb{C} \setminus K$ for some Cantor set $K \subset \mathbb{C}$? I'm pretty sure the answer is yes but the version of the uniformization theorem I know doesn't imply this (just that the plane minus any Cantor set can be uniformized by some Fuchsian group $G$). Any good references? Thanks in advance! REPLY [6 votes]: Koebe uniformisation theorem says that any planar Riemann surface is biholomorphic to a domain in the Riemann sphere .See George Springer's book on Riemann surfaces.At the risk of self promotion you can also look at my book with T Napier titled An Introduction to Riemann Surfaces .<|endoftext|> TITLE: A back and forth Euclidean algorithm over the integers--does it have bounded length? QUESTION [10 upvotes]: cLet $a,b,c,d\in \mathbb{Z}$ and suppose we have the equation $ac+bd=1$. One way of thinking about this equation is it expresses the fact $\gcd(c,d)=1$. It is well-known that all other similar equations expressing that fact are of the form $(a+td)c+(b-tc)d=1$ for some $t\in\mathbb{Z}$. Letting $a'=a+td,b'=b-tc$, one may ask if there is a "best" choice for $t$; or equivalently, if there is a best choice for $a',b'$. In some sense, the usual Euclidean algorithm (applied to $b,d$) gives a minimal linear combination--we'll call the resulting pair $(a',b')$ the Euclidean pair. But there are sometimes other situations where a different choice of $t$ is optimal. In any case, we can pass from $ac+bd=1$ to $a'c+b'd=1$. Now, we can view this latter expression as a linear combination showing that $\gcd(a',b')=1$. So we can repeat the process to get a new $b',d'$ so that $a'c'+b'd'=1$. Continuing in this fashion by replacing either the $(a,c)$ pair or the $(b,d)$ pair with a new pair $(a',c')$ or $(b',d')$ respectively, then after a finite number of steps we can always reach the equation $1\cdot 1+ 0\cdot 0=1$. (One way to do this is just use the Euclidean pair for each replacement, except possibly at the end when one may need to pass from $0\cdot 0+1\cdot 1=1\mapsto 1\cdot 1+0\cdot 0=1$.) My question is: Starting with any quadruple of integers $a,b,c,d\in\mathbb{Z}$ with $ac+bd=1$, is there an absolute bound $n\gg 0$ so that by performing some sequence of back-and-forth switching as described above we get to $1\cdot 1+0\cdot 0=1$ within at most $n$ steps? The greedy algorithm of simply choosing the Euclidean pair will not give an absolute bound $n$, since we can easily construct a sequence which requires an arbitrarily large number of steps this way. But using the Euclidean pair is not always the best choice. By the way, this number theory question originally arose in my study of perspective decompositions of the abelian group $\mathbb{Z}\oplus \mathbb{Z}$. The connection comes from associating nontrivial idempotents $E\in \mathbb{M}_2(\mathbb{Z})$ with 4-tuples $(a,b,c,d)$ satisfying $ac+bd=1$, by writing $E=\begin{pmatrix}ac & bc\\ ad & bd\end{pmatrix}$. [Here, $a$ is the gcd of the first column, etc...] Further, I have a very ad hoc way of showing that the question above has a negative answer if we replace $\mathbb{Z}$ with the ring $\mathbb{F}_2[x]$. In that case, one can show that using Euclidean pairs is optimal. REPLY [9 votes]: We get an isomorphic problem by switching $c$ with $d$, and replacing $b$ with $-b$. Then we are considering matrices $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ in $SL_2(\mathbb{Z})$. Passage from $(a,b)$ to $(a',b')$ amounts to multiplication by a matrix $T^t = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$ for some $t \in \mathbb{Z}$, and switching between $(a,b)$ and $(c,d)$ is (up to sign) given by multiplication by $S = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. The question then amounts to whether there is a uniform bound on the length of words in $S$ and $T^t$ (as $t$ ranges over integers) describing elements of $SL_2(\mathbb{Z})$, and the answer is that such a bound does not exist. One way to see this is by examining the geometry of Ford circles in the complex upper half-plane. $T$ translates the half-plane by integers (preserving the cusp at infinity), and $S$ more or less takes cusps to their reciprocals. Thus, a composite of $S$ with $T^t$ takes the circle at infinity to a circle tangent to it. A word describing an element $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ then necessarily has length at least as long as the chain of circles connecting the cusp at infinity with the cusp $a/c$, and such chains of circles have unbounded length (indeed, they have a sharp lower bound given by the length of signed continued fractions - the shortest word is the chain of circles touching the vertical line with real part $a/c$).<|endoftext|> TITLE: Frozen coefficient method (von Neumann stability analysis) QUESTION [6 upvotes]: Earlier it was considered that frozen coefficients method for Neumann stability analysis for finite difference scheme is more heuristic than rigorous. But I have read some information in a book by Gustafsson,Kreiss,Oliger that there are some conditions, which give us possibility to use frozen coefficients+Neumann analysis at the level of full rigor. As a book didn't give any reference, could you help me to find these conditions? REPLY [2 votes]: I believe the intended reference regarding parabolic PDEs is: Fritz, John. On integration of parabolic equations by difference methods: I. Linear and quasi-linear equations for the infinite interval. Communications on Pure and Applied Mathematics, 5(2):155--211 (1952). The paper is 57 pages long and I cannot find a later reference that neatly summarizes it. The nonlinear case is handled in the last section (Section 8). I found this paper while looking through the references in Strang, Gilbert. Accurate partial difference methods II. Numerische Mathematik 6, 37-46 (1964). which contains results for the hyperbolic case.<|endoftext|> TITLE: The properness of a submersion QUESTION [10 upvotes]: Let $M$ and $N$ be two differential manifolds and there is a surjective submersion $f$ from $M$ to $N$ such that $f^{-1}(p)$ is compact and connected for any $p$ on $N$. Can we conclude that $f$ is proper, that is, the preimage of a compact set is compact? It is also posted on https://math.stackexchange.com/q/1004543/13534 REPLY [9 votes]: This is true in greater generality. If $M$ and $N$ are locally compact, and the fibers of $f$ are compact connected and $f$ is a quotient map then f is proper. This falls under the rubric of monotone light factorization. Look at the book of G.T. Whyburn and E. Duda titled Dynamic Topology. Monotone Light factorization says if a map between locally compact Hausdorff has the property, that all connected components of all fibers are compact, then the map can be factored as a proper map onto a space with connected fibers and this space maps to the co-domain with totally disconnected fibers. Thanks to Dan Petersen and Richard Andrade for correction. Addendum: Let a connected component of the fiber of a map be called a level.Suppose X and Y are locally compact Hausdorff spaces and F a continuous map such that all levels are compact .Suppose the equivalence relation R on X is given by saying x and y are equivalent if they lie in the same level, then R is a proper equivalence relation and the map from X to X/R is proper with connected fibers and F factors through X/R to Y as a map with totally disconnected fibers . In the case of the question of the OP the equivalence relation R and the one given by f coincide and therefore the map is proper<|endoftext|> TITLE: Is ${\rm lcm}\{{\rm ord}_p(q)\colon q\mid p-1,\ q>2\}>\sqrt p\ \,$? QUESTION [12 upvotes]: The following question is "ideologically related" to the one I have recently asked. For a prime $p$, let $M_p$ denote the least common multiple of the orders modulo $p$ of all odd prime divisors of $p-1$: $$ M_p := {\rm lcm}\{{\rm ord}_p(q)\colon q\mid p-1,\ q\ \text{is an odd prime}\}. $$ I am interested in the primes $p\equiv5\pmod 8$, and I want to show that, normally, $M_p>\sqrt p$ holds for such primes. In the range $5\le p<100,000,000$, there are only three exceptions (primes $p\equiv 5\pmod 8$ with $M_p<\sqrt p$): namely, $5$, $13$, and $148,997$. Are there any more such exceptional primes and if so, is the set of all these primes finite? Notice that allowing $p\equiv 1\pmod 8$ would make every Fermat prime a bold exception. REPLY [7 votes]: We have by inclusion-exclusion and Bombieri-Vinogradov theorem, we have Lemma 1 Let $00$, we have $T:=(\log x)^{N (\log x)^N}\ll \sqrt x (\log x)^{-M}$. So, the values of $q$ are within the range of $q$ provided by Bombieri-Vinogradov. By inclusion-exclusion and Mertens' estimate, we have for any $K>0$, $$\begin{align*} b(x)&=\textrm{Li}(x) \left(-\sum_{\substack{{10$ such that $$ \mathrm{ord}_p(a)\geq \sqrt p \exp((\log p)^{\delta}) $$ for all but $O(x/(\log x)^{1+\alpha})$ primes $p\leq x$. The theorem is stated with a fixed $a$, but by modifying Erdos & Murty's proof, we can relax $a$ up to a fixed power of $\log x$. For the main problem, we have the result by combining the above theorem with Lemma 1. In Lemma 1, we take $N=\min(\alpha/2,1/2)$. By counting the exceptional primes, we obtain Theorem Let $\alpha, \delta >0$ be the numbers in Theorem 1. Let $A$ be the set of primes $p\leq x$ in the residue class $5$ mod $8$ such that there is an odd prime number $q\leq (\log x)^N$ which divides $p-1$ and $$M_p\geq \mathrm{ord}_p(q)\geq \sqrt p \exp((\log p)^{\delta})$$ Then $$ |A|=\frac{\textrm{Li}(x)}{\phi(8)}+O\left(\frac{\textrm{Li}(x) }{\log\log x}\right). $$<|endoftext|> TITLE: Images of polynomials QUESTION [16 upvotes]: Let $f,g \in \mathbb{Q}[x]$ be polynomials such that $\{f(a) : a \in \mathbb{Q}\} \subseteq \{g(a) : a \in \mathbb{Q} \}$. Must there be some $h \in \mathbb{Q}[x]$ such that $f(x) = g(h(x))$ for all $x \in \mathbb{Q}$ ? REPLY [14 votes]: We don't need Faltings theorem here. See Theorem 1 in: H. Davenport, D. J. Lewis, A. Schinzel, Polynomials of certain special types. Acta Arith. 9 (1964), 107-116.<|endoftext|> TITLE: irregular pairs in half graphs - Szemeredi regularity QUESTION [5 upvotes]: Szemeredi's regularity lemma is a well-known result about partitioning large graphs into pieces such that most pairs of pieces are "regular". The precise statement takes a bit of detail so I'll just refer to the the Wikipedia link. My question is about the part of the lemma saying "all except $\epsilon k^2$ of the pairs $V_i$, $V_j$ are $\epsilon$-pseudo random". (Note: "pseudo-random" and "regular" seem to be somewhat interchangeable in this context.) Concerning the necessity of these "irregular" pairs, a frequently cited example is the "half-graph". In particular, given $n>0$, the half-graph of height $n$, $H_n$, is described as follows: $V(H_n)=\{v_1,\ldots,v_n,w_1,\ldots,w_n\}$, given, $1\leq i,j\leq n$, $(v_i,w_j)\in E(H_n)$ if and only if $i\leq j$, no other edges are present in $H_n$ (so $(\{v_1,\ldots,v_n\},\{w_1,\ldots,w_n\})$ is a bipartition for $H_n$). The graphs $H_n$ were the first observed example of where irregular pairs in Szemeredi's Lemma are necessary. I have two questions about this. 1) What is the precise statement of this claim about irregular pairs in half graphs? 2) Is there a source where this claim is explained in detail? Remarks: Concerning question 1, the closest thing I have found is Section 2.1.2 of this survey paper, which says: "The reader is invited to prove that, for small enough $\epsilon>0$, any $(\epsilon,k)$-equitable, $\epsilon$-regular partition of $H_n$ requires at least $ck$ $\epsilon$-irregular pairs, where $c=c(\epsilon)>0$ is some constant that depends only on $\epsilon$." But I am still uncertain on exactly how $n$ and $k$ are being quantified in this statement. My best guess is: "For any small enough $\epsilon>0$, there is $c=c(\epsilon)>0$ such that for all $k$ there is some large enough $n$ such that any $(\epsilon,k)$-equitable, $\epsilon$-regular partition of $H_n$ requires at least $ck$ $\epsilon$-irregular pairs." Concerning question 2, I have only found sources that mention half-graphs as examples of irregular pairs. So it would be nice to know if this is worked out in detail somewhere. REPLY [2 votes]: I had been stuck on this same problem for a while as well and asked Dr. Rödl about it. He gave me the following solution: Suppose for $1/4>\epsilon >0$, we get $M$ as the upper bound on the number of partitions by applying the regularity lemma. We can take $N = kM$ and define the Half Graph $H_N$ that you mention here. An equivalent version of the regularity lemma allows us to form equipartition on each of the halves (call them $V$ and $W$ and denote partitions by $V_i, W_j$), such that $|V_i| = |W_j| = k$ for all $i,j$ Delete the top and bottom $\epsilon k$ elements from each $W_i$ to form $W_i'$. Do the same to form $V_i'$. Then we delete $4\epsilon kM$ elements at most (at most half the number of total vertices). Then consider $W'$ as the union of $W_i'$. And similarly $V'$. Find a vertex in $V_i'$ called $v'_h$ and select $j$ such that $w'_h$ in $W'_j$ is below it (on re-indexing both $V'$, $W'$). Then consider $(V_i,W_j)$. The part of $V_i$ below $v'_h$ with the part of $W_j$ above $w'_h$ is a 0 density graph. While the part of $V_i$ above $v_h'$ with the part of $W'_j$ below $w'_h$ is a density 1 graph. Both of these parts are not too small ($>\epsilon k$) in size because of how we constructed $V', W'$. Thus we have that $(V_i, W_j)$ is an irregular pair. For every $V_i$, we can pick some $v'_h\in V_i'$, and find an irregular pair. Thus we have at least $M$ irregular pairs.<|endoftext|> TITLE: Reference request: Systems of linear PDES with constant coefficients QUESTION [10 upvotes]: I am looking for a reference for the following statement: Assume that $P_1, \dots, P_k \in \mathbb R[x_1, \dots, x_m]$ and consider a system of PDEs \begin{align} P_i(\partial / \partial x_1, \dots, \partial / \partial x_m)u = f_i, \quad i = 1, \dots, k \end{align} where $u(x_1, \dots, x_m)$ is a scalar function to be found. Then, if this system has a solution, we should obviously have $$ \sum_{i=1}^k Q_i(\partial / \partial x_1, \dots, \partial / \partial x_m)f_i = 0 $$ for each $k$-tuple $(Q_1, \dots, Q_k)$ such that $$ \sum_{i=1}^k Q_iP_i = 0. $$ The statement is that this condition is also sufficient for local solvability of our system of PDEs. More precisely, 1) if $f_1, \dots, f_k$ are $\mathrm C^\infty$ functions in a neighborhood of $x \in \mathbb R^m$, then there exists a $\mathrm C^\infty$ solution in a (possibly smaller) neighborhood of $x$; 2) if $f_1, \dots, f_k$ are analytic, then the solution is also analytic. I suspect that this should be Ehrenpreis or Malgrange, but I was not able to find the precise statement. Another question is whether there is a geometric way to understand this result. For instance, if $P_1, \dots, P_k$ are of degree $1$, then the statement can be deduced from the Frobenius theorem. REPLY [4 votes]: Theorem 7.6.13 in Hörmander's book "An Introduction to Complex Analysis in Several Variables" (North-Holland, 1990, 3rd edition) contains the result you are looking for. Sections 7.6 and 7.7 of this book give a full prove of the Ehrenpreis Fundamental Principle of which solvability is a part. See Hörmander's notes at the end of chapter 7 for historical information.<|endoftext|> TITLE: Hodge Bundles on Tropical Spaces QUESTION [6 upvotes]: I am not sure that this question even makes sense, which I suppose is part of the questions itself. In any case, I attended a talk recently wherin there was some discussion about a "tropical Teichmuller space", as for example in this paper, associated to the tropical moduli space of abelian varieties. My understanding is that the moduli space of tropical abelian varieties $A_g^{trop}$ can be identified with the skeleton of a Berkovich analytic space associated to the moduli space of abelian varieties $A_g$. Does the Hodge Bundle on the moduli space of abelian varieties "carry over" (whatever that should mean) to the analytic space? If so, can you then "pull it back" to the skeleton, onto which the analytic space deformation retracts? My question is motivated by the fact that the speaker suggested that this "tropical Teichmuller space" gives some tropical analogy to the Siegel upper half space model for $A_g$, where you take a quotient by $GL_n$ to get the moduli space of tropical abelian varieties. I am basically wondering if there is some notion of a "tropical Siegel modular form", which arises as a section of a "tropical Hodge bundle". Some nice references for tropical geometry and Berkovich analytic spaces is also welcome. Note: I also posted this question here on StackExchange, but received no answers. REPLY [4 votes]: There is a preprint Towards a tropical Hodge bundle by Bo Lin and Martin Ulirsch, which discusses the Hodge bundle over the moduli space of genus $g$ tropical curves.<|endoftext|> TITLE: K-theory of the h-cobordism category QUESTION [9 upvotes]: I was reading through Kervaire and Milnor's "Groups of Homotopy Spheres", in which the authors begin to compute the groups $\Theta_n$ of h-cobordism classes of homotopy $n$-spheres (with group operation coming from the connected sum). It occurred to me that there ought to be a whole $(\infty,0)$-category of homotopy $n$-spheres with h-cobordisms between them, diffeomorphisms between the cobordisms, and so forth. Then we might hope that connected sum gives our h-cobordism category a symmetric monoidal / $E^\infty$-space structure, which is in fact grouplike (for the same reason that $\Theta_n$ has inverses: $M\#-M=S^n$). If I am not mistaken, then that would suggest that this h-cobordism category naturally comes equipped with the structure of a connective spectrum whose $\pi_0$ is exactly $\Theta_n$. First of all, I want to know whether this procedure actually works, since I am a bit new to this and there are details that I am not sure about. And if it does work, I am curious if anything more is known about these spectra. Are they familiar objects? Do the cohomology theories have nice geometric descriptions? REPLY [10 votes]: I don't know if this is the sort of thing you have in mind, but using surgery theory you can in fact write down a single connective spectrum / infinite loop space whose $n^{th}$ homotopy group is $\Theta_n$. This is the space $PL/O$, one definition of which is that it is the homotopy fiber of the natural map $BO \to BPL$, where $BO$ is the classifying space of stable vector bundles and $BPL$ is the classifying space of stable PL-microbundles (or something like that). The relationship to exotic spheres (same as homotopy spheres by the generalized Poincaré conjecture) comes from looking at the set of homotopy classes of lifts of the unique PL-structure on the $n$-sphere to a smooth structure. Some details can be found in this article by Davis and Petrosyan.<|endoftext|> TITLE: Meaning of fibration in Kazhdan and Lusztig's paper on affine flag manifolds QUESTION [7 upvotes]: Kazhdan and Lusztig's paper "Fixed point varieties on affine flag manifolds" has the following definition on p.143: define inductively a variety $Z$ to be an "almost affine space" if $Z$ is affine or if there exists a fibration from $Z$ to an affine space such that all the fibers are "almost affine spaces" of smaller dimension. Also on p. 144 they construct various maps and claim them to be "algebraic fibrations". I am unsure what definition of fibration they are using, and am having difficulty inferring it from context. Their maps they construct are surjective so it leads me to believe they might have meant "flat and surjective" but I am not sure since they never actually define "fibration" anywhere. Does anyone know precisely what they mean by "fibration"? REPLY [5 votes]: According to Lusztig, it means locally trivial in the Zariski topology.<|endoftext|> TITLE: For an elliptic curve $E/\mathbb{Q}$ can the cohomology group $H^1(\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q}), E[p])$ be nontrivial? QUESTION [13 upvotes]: Suppose that $E$ an elliptic curve defined over $\mathbb{Q}$ and $p$ an odd prime. Let $G=\text{Gal}(\mathbb{Q}(E[p])/\mathbb{Q})$. I am wondering whether the cohomology group $H^1(G, E[p])$ can be nontrivial. If $G=GL_2(\mathbb{F}_p)$ (which is the case for all but finitely many primes $p$ if $E$ does not have complex multiplication) then $H^1(G, E[p])$ is trivial. This can be shown by considering the homothety subgroup $Z \le G$ which has order $p-1>1$. One easily sees that $H^i(Z, E[p])=0$ for all $i \geq 0$ and so the result follows from the Hochschild-Serre spectral sequence. Now suppose that $G$ is a proper subgroup of $GL_2(\mathbb{F}_p)$. Can $H^1(G, E[p])$ be nontrivial? REPLY [9 votes]: Fix elements $\zeta$ and $\alpha$ with $\zeta$ a primitive third root of unity and $\alpha^3 = -4$. These generate a field $K = \Bbb Q(\zeta,\alpha)$ which is the splitting field of $x^3 + 4$, with Galois group $G$ the symmetric group on three letters. Consider the elliptic curve $y^2 = x^3 + 1$. Unless I have miscalculated, the $3$-torsion points on this curve are the points $(x,y)$ with $x^4 + 4x = 0$. In particular, the points $(0,1)$ and $(\alpha,2\zeta+1)$ are independent 3-torsion points on this curve, so $K = \Bbb Q(E[3])$. Taking these as a basis, the resulting image of the Galois group into $GL_2(\Bbb F_3)$ must be $$\begin{bmatrix}1 & * \\ 0 & *\end{bmatrix}$$ because $(0,1)$ is fixed and the map must be injective. Let $H < G$ be the subgroup of order three. Since the coefficient group $E[3]$ is $3$-torsion, a transfer argument implies that the restriction $H^1(G;E[3]) \to H^1(H;E[3])$ is injective with image the invariants under $G/H \cong \Bbb Z/2$. If $$A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$$ represents the generator $\tau$ of $H$, then the group $H^1(H;E[3])$ is $ker(1 + A + A^2) / Im(1 - A)$, which is generated by the column vector $\left[\begin{smallmatrix}0 \\ 1\end{smallmatrix}\right]$. (This describes an element of $H^1$ by where an associated $1$-cocycle sends a chosen generator of $H$.) As a $1$-cocycle, this is represented by the map $f:H \to E[3]$ with $$f(\tau^k) = (1 + A + \cdots + A^{k-1})\left[\begin{smallmatrix}0 \\ 1\end{smallmatrix}\right].$$ The action of the element $\sigma = \left[\begin{smallmatrix}1 & 0 \\ 0 & -1\end{smallmatrix}\right]$ on this cocycle is given by $$({}^\sigma f)(\tau) = \sigma \cdot f(\sigma^{-1} \tau \sigma) = \sigma \cdot f(\tau^2) = \left[\begin{smallmatrix}1 \\ 1\end{smallmatrix}\right]$$ which shows that the two $1$-cocycles ${}^\sigma f$ and $f$ represent the same element of $H^1$. Therefore, this element of $H^1(H;E[3])$ is invariant under $G/H$ and lifts to a nontrivial element of $H^1(G;E[3])$. REPLY [2 votes]: $\newcommand{\FF}{\mathbb{F}}\DeclareMathOperator{\SL}{SL}$ OK, let me try. Write $M=E[p]$. If the order of $G$ is coprime to $p$, then $H^1(G,M)=0$. Assume that $p$ divides the order of $G$. Now by Prop 15 in Serre's "Propriétés galoisiennes... " Invent Math 15, $G$ either contains $\SL_2(\FF_p)$ or it is contained in a Borel subgroup. According to the question, we may assume that $G$ is contained in a Borel, say upper triangular matrices. Let $H=G\cap \SL_2(\FF_p)$. Restriction shows that $H^1(G,M)$ is the $G/H$-fixed part of $H^1(H,M)$. Now $H$ is a subgroup of $(\begin{smallmatrix} a & b \\ 0 & 1/a\end{smallmatrix})$ with invertible $a$ and arbitrary $b$. By assumption $H$ contains the subgroup $K$ generated by $h = (\begin{smallmatrix} 1 & 1\\ 0 & 1\end{smallmatrix})$. Again by restriction-inflation, $H^1(H,M)$ is contained in $H^1(K,M)$. The latter can be computed as usual and we find that a cocyle is determined by its image on $h$ which has to belong to $\FF_p(1, 0)$. Now we consider again the action of $G/H$ on $H^1(K,M)$. If I am not mistaken, then the class $\bar g$ of matrices of determinant $d$ acts by $(\bar g * \xi)(h) = d \cdot \xi(h)$. If so, then there are no elements in $H^1(H,M)$ fixed by $G/H$ as soon as there is an element of determinant $\neq 1$ in $G$. By the Weil pairing, this must be the case when $p>2$. Hence $H^1(G,M)=0$. Edit: However, I am mistaken as the example of Tyler Lawson shows. The action of $G/H$ may be trivial.<|endoftext|> TITLE: Is there a nonstandard model of arithmetic having precisely one inductive truth predicate? QUESTION [13 upvotes]: $\newcommand\Tr{\text{Tr}}$My question is whether there can be a nonstandard model of PA having a unique inductive truth predicate. Background. If $\mathcal{N}=\langle N,+,\cdot,0,1,<\rangle$ is a model of the first-order PA axioms, then a truth predicate on $\mathcal{N}$, also commonly called a satisfaction class, is a predicate $\Tr\subset N$ obeying the recursive Tarskian truth conditions, that is: (atomic) The $\Tr$ predicate holds of the Gödel code of an atomic formula just in case that atomic formula is true. For example, $\Tr(\underline n+\underline k=\underline r)\iff \mathcal{N}\models n+k=r$, and so on, where $\underline n$ means the term $\underbrace{1+1+\cdots+1}_n$. (conjunction) $\Tr(\sigma\wedge\tau)$ holds if and only if $\Tr(\sigma)$ and $\Tr(\tau)$ both hold. (negation) $\Tr(\neg\sigma)$ holds if and only if $\Tr(\sigma)$ does not hold. (quantifiers) $\Tr(\exists x\varphi(x))$ holds if and only if there is some $n\in N$ such that $\Tr(\varphi(\underline n))$ holds. Such a truth predicate is said to be inductive, if the model $\langle N,+,\cdot,0,1,<,\Tr\rangle$ expanded by that predicate satisfies induction in the expanded language. For nonstandard models, the existence of an inductive truth predicate implies that the model is computably saturated, and a countable model is computably saturated if and only if there is an inductive partial truth predicate, one which is defined only on all sentences up to some nonstandard finite level of complexity. Thus, some nonstandard models of PA, the non-computably saturated ones, have no inductive truth predicate. Meanwhile, Krajewsky proved that there can be nonstandard models of PA having more than one distinct truth predicate. I gave a talk yesterday for the NY Phil Logic Group in which such matters were an important part of the discussion, and Kit Fine asked a great question there: Question. Can there be a nonstandard model of arithmetic having a unique inductive truth predicate? The standard model $\mathbb{N}$, of course, has a unique truth predicate, since one can prove by induction that every such predicate must agree with actual arithmetic truth. But for countable nonstandard models, the answer is definitely no, it does not happen. The reason is that if a nonstandard countable model of PA has any inductive truth predicate at all, then it is computably saturated, and so the usual back-and-forth constructions show that the model will have many automorphisms, and furthermore many automorphisms that take true sentences to false ones and vice versa. This is simply because the truth predicate cannot be definable in the base language of arithmetic, and so there must be two elements of the model having the same types in the model in the language of arithmetic, but one is the Gödel code of a true sentence and the other the code of a false sentence. By constructing a tree of such automorphisms, one can get continuum many distinct truth predicates this way. (See also Kossak and Schmerl, Minimal Satisfaction Classes with an Application to Rigid Models of Peano Arithmetic, NDJFL 32(3), 1991.) This back-and-forth argument provides an alternative proof of Krajewsky's result. But can there be an uncountable affirmative instance of the question? We know that there can be rigid $\omega_1$-like computably saturated models, and that kind of situation is suggestive that an affirmative instance may be possible. REPLY [12 votes]: The answer to the question is in the positive. Let $(\cal{N}^*,\textrm{Tr*})$ be a rather classless elementary extension of $(\cal{N},\textrm{Tr})$, where $\cal{N}$ is a model of PA, and $\textrm{Tr}$ is a full truth predicate on $\cal{N}$, e.g., let $\cal{N}$ be the standard model of PA, and $\textrm{Tr}$ be the usual Tarskian truth predicate on $\cal{N}$. In the above, "rather classless" means that $(\cal{N}^*,\textrm{Tr*})$ has the property that every subset $X$ of the universe $N^*$ of $\cal{N}^*$ that is piecewise coded in $\cal{N}^*$ is first order definable in $(\cal{N}^*,\textrm{Tr*})$, where $X$ is piecewise coded means that the intersection of $X$ with every topped initial segment of $\cal{N^*}$ is parametrically definable in $\cal{N^*}$ (indeed, it is coded by a single element). See Theorem 2.2.14 (and the preceding two paragraphs) of the Kossak-Schmerl monograph on models of PA for the relevant existence theorem for rather classless models. Such models exist in every uncountable cardinality $\kappa$ of uncountable cofinality and indeed can be arranged to be $\kappa$-like. On the other hand, it is easy to see that every inductive set is piecewise coded. Therefore, if $\textrm{T}$ is an inductive truth predicate on $N^*$, then $\textrm{T}$ is piecewise coded and by rather classlessness it is definable in $(\cal{N}^*,\textrm{Tr*})$, and in particular the double-expansion $(\cal{N}^*,\textrm{Tr*}, \textrm{T})$ satisfies full induction. Now, an easy induction carried out internally in $(\cal{N}^*,\textrm{Tr*}, \textrm{T})$ on the complexity of formulae shows that $\textrm{Tr*} = \textrm{T}$. QED<|endoftext|> TITLE: Are there infinitely many primes p such that both p-1 and p+1 have at most 3 prime factors, counted with multiplicity? QUESTION [7 upvotes]: This is a subsequence of primes $p$ for which $p^2-1$ has at most 6 prime divisors counted with multiplicity. This sequence described in the question is the sequence A079153 in OEIS. I could not find references on the first sequence, but I found a mention of the second sequence in another MO question: Number of prime factors of the order of a finite non-abelian simple group with a reference to a survey article by Solomon from 2001: are there infinitely many primes $p$ such that $|PSL(2,p)|=(p-1)*p*(p+1)/2$ is a product of six prime factors? In the survey article it is said this problem is akin to the twin prime conjecture, but no references are given. REPLY [11 votes]: At the present time, the best (general) result is due to James Maynard, who has proven that any admissible 3-tuple takes at most 7 prime factors infinitely often (see this paper). In particular, (using a certain admissible 3-tuple related to what Robert Israel wrote) we can get that $n(n^2-1)$ has at most 11 prime factors infinitely often. I don't know if this number can be improved for the specific case you are interested in. However, the general consensus is that Dickson's conjecture is true, so both $p-1$ and $p+1$ should consist of exactly 3 prime factors, for infinitely many $p$. But this conjecture is currently out of reach of any known techniques, and appears to be harder than the twin prime conjecture, since we must control admissible 3-tuples, but the twin prime problem deals with only admissible 2-tuples.<|endoftext|> TITLE: Optimal definition of "paving by affine spaces"? QUESTION [9 upvotes]: Cell decompositions have been used in topology for a long time as a tool in computing cohomology, but the notion in algebraic geometry and arithmetic geometry of paving by affine spaces (or "affine paving" for short) seems to be a little more recent. An early occurrence is found in the survey Barry Mazur gave in 1974: Eigenvalues of Frobenius acting on algebraic varieties over finite fields, Algebraic geometry (Proc. Sympos. Pure Math., Vol. 29, Humboldt State Univ., Arcata, Calif., 1974), pp. 231–261, Amer. Math. Soc., Providence, R.I., 1975. He considers only complete varieties (where the most natural examples arise), but the definition makes sense for any variety $X$: require $X$ to have a filtration $0=X_0 \subset X_1 \subset \dots \subset X_d = X$ with each $X_i$ closed and with $X_i - X_{i-1}$ a finite disjoint union of copies of affine space $\mathbb{A}^i$. (Though not stated explicitly, the union here might be empty.) This definition was used by Hotta and Springer in their influential 1977 paper here. But as time went on, a broader definition seems to have taken over: given a filtration of $X$ by closed subvarieties as above, one only requires that each $X_i - X_{i-1}$ be a finite disjoint union of copies of a single affine space (of some dimension). It's not clear to me whether these two definitions (narrower and broader) are actually interchangeable, but the underlying notion of affine paving is itself quite natural. The prototype is $X=\mathbb{P}^1$, with a closed point and a complementary affine line. More generally, a flag variety $G/B$ has a suitable filtration with differences given by unions of Bruhat cells $BwB/B$, each isomorphic to affine space of dimension $\ell(w)$ as $w$ ranges over the Weyl group. One can then try to form an affine paving of a Springer fiber (or more general Hessenberg variety) inside $G/B$ by simply intersecting with a Bruhat-type filtration. Work in this direction occurs in a number of places, including $\S11$ of J.C. Jantzen's monograph Nilpotent orbits in representation theory, contained in the 2004 Birkhauser volume Lie Theory (Progress in Mathematics 228). There are also research papers by T. Haines (on Kazhdan-Lusztig "purity"), J. Tymoczko, E. Sommers, M. Precup (on Hessenberg varieties), etc. Given the range of applications in the literature, should the broader or narrower definition of "covering by affine spaces" be regarded now as standard? To me it seems awkward to leave ambiguity in a widely used mathematical definition. But it also seems unnatural to use a broader-than-needed definition. REPLY [8 votes]: The definitions are not equivalent: consider $$\mathbb{A}^2 \times 0 \cup 0 \times \mathbb{P}^1 \subset \mathbb{A}^2 \times \mathbb{P}^1$$ Then you cannot see the line before you see the plane. That is, you can't take any subvariety out of $\mathbb{A}^2$ first, since you can't stratify $\mathbb{A}^2$-minus-anything by cells. You can't take $0 \times (\mathbb{P}^1 \setminus 0)$ out first, because it's not closed. So there's no dimension one closed subvariety you can remove, leaving the complement a union of affines, so there's no Barry Mazur stratification. But you can take $$\mathbb{A}^2 \times 0 \subset \mathbb{A}^2 \times 0 \cup 0 \times \mathbb{P}^1$$ which is a stratification in the other sense.<|endoftext|> TITLE: Bass' stable range for Bezout rings QUESTION [7 upvotes]: As discussed in this MO topic, every principal ideal domain has stable rank at most 2. The proof in the accepted answer uses the fact that PID is a unique factorization domain, but there can be no irreducibles in case of Bezout ring (every finitely generated ideal is principal), as shown by the example of the ring of all algebraic integers. However, if one assumes the ring to be a Bezout domain, one can show that it is Hermitian, and thus of stable rank 2 (this paper, for example). It looks like there is nothing about infinitely generated ideals in either PID or stable range conditions, so it is naturally to ask if there are Bezout rings of stable rank greater than 2. Surely they have to be non-UFD. The two basic examples I know of: algebraic integers and entire functions, but both are domains (and of stable rank 1, which makes the situation even more puzzling). Another example from the Wikipedia page is of no help either. Question. Is the stable rank of Bezout rings bounded from above? Can it be strictly greater than 2? PS. All rings are considered to be commutative. REPLY [10 votes]: In "Rings of continuous functions in which every finitely generated ideal is principal" by L. Gillman and M. Henriksen (Trans. Amer. Math. Soc. 82 (1956), 366-391 link), Example 3.4 is of a topological space $X$ such that the ring of continuous functions $C(X)$ is Bezout but not Hermitian. The space $X$ is the complement of a closed half-plane in its Stone–Čech compactification. Theorem 1 of "Reduction of Matrices over Bezout Rings of Stable Rank not Higher than 2" by B. V. Zabavs'kyi (Ukrainian Mathematical Journal, 55(4) (2003) 665-670 link) states that a commutative Bezout ring is Hermitian if and only if it has stable rank (at most) two. So putting these together gives a Bezout ring with stable rank greater than two. Because of the remark I made in comments, I guess the stable rank is infinite, but I'm not sure.<|endoftext|> TITLE: The Universality Theorem by Mnev for uniform oriented matroids of rank 4 and higher QUESTION [6 upvotes]: According to the Universality Theorem by Mnev (see below theorem 8.6.6 from [1]), for any open semialgebraic variety V there is a uniform oriented matroid of rank 3 whose realization space is stably equivalent to V. My question is whether one can transfer the result also to uniform oriented matroids of rank 4 and higher, i.e., for a fixed r $\ge$ 4 and for any open semialgebraic variety V is there a uniform oriented matroid of rank r whose realization space is stably equivalent to V? 8.6.6 Universality Theorem (Mnev 1988). Let $V \subset \mathbb{R}^s$ be any semialgebraic variety. Then there exists a rank 3 oriented matroid M. whose realization space R(M) is stably equivalent to V. If V is an open subset of $\mathbb{R}^s$, then M may be chosen to be uniform. [1] A. Björner, M. L. Vergnas, B. Sturmfels, N. White, and G. M. Ziegler, Oriented Matroids. Cambridge University Press, 1999. REPLY [8 votes]: Yes, the Universality Theorem works for any rank $r\geq3$. One way to prove this is to consider a rank $3$ oriented matroid $M$ with the desired realization space, then take the dual $M^*$, do a lexicographic extension in general position and then take the dual again. Repeat this $r-3$ times. Lexicographic extensions preserve the realization space up to stable equivalence, and so does taking the dual. A lexicographic extension increases the number of elements by one and keeps the rank. Hence, doing it to the dual increases the rank of the primal by one.<|endoftext|> TITLE: Synthetic vs. classical differential geometry QUESTION [55 upvotes]: To provide context, I'm a differential geometry grad student from a physics background. I know some category theory (at the level of Simmons) and differential and Riemannian geometry (at the level of Lee's series) but I don't have any background in categorical logic or model theory. I've recently come across some interesting surveys and articles on synthetic differential geometry (SDG) that made the approach seem very appealing. Many of the definitions become very elegant, such as the definition of the tangent bundle as an exponential object. The ability to argue rigorously using infinitesimals also appeals to the physicist in me, and seems to yield more intuitive proofs. I just have a few questions about SDG which I hope some experts could answer. How much of modern differential geometry (Cartan geometry, poisson geometry, symplectic geometry, etc.) has been reformulated in SDG? Have any physical theories such as general relativity been reformulated in SDG? If so, is the synthetic formulation more or less practical than the classical formulation for computations and numerical simulations? Also, how promising is SDG as an area of research? How does it compare to other alternative theories such as the ones discussed in comparative smootheology? REPLY [6 votes]: Tim de Laat's bachelor thesis might be of interest: Synthetic Differential Geometry: An application to Einstein’s Equivalence Principle This thesis is the result of my bachelor project in both Mathematics and Physics & Astronomy. The aim of this project was to give a satisfactory and rigorous formulation of the equivalence principle of the general theory of relativity (GR) in terms of synthetic differential geometry (SDG). SDG is a “natural” formulation of differential geometry in which the notion of “infinitesimals” is very important. Smooth infinitesimal analysis (SIA) is the mathematical analysis corresponding to these infinitesimals and it forms an entrance to SDG. Both SIA and SDG are formulated in terms of categories and topoi. As I was quite new to these subjects, I first needed to study them thoroughly before I could start studying SDG.<|endoftext|> TITLE: A nilpotency question on $C^{*}$ algebras QUESTION [5 upvotes]: What is an example of a $C^{*}$ algebra $A$ with the property that: for every nilpotent(Quasi nilpotent) $a$ and for every $n\in \mathbb{N}$, there is a $b$ with $b^{n}=a$. To what extent such algebras are classified? REPLY [3 votes]: I'm not sure if this is useful: If your $C^*$-algebra also has a non-zero nilpotent element, then it will have nilpotent elements of all orders, it is not (algebraically) of bounded index, so does not satisfy a polynomial identity. $C^*$-algebras which do satisfy a polynomial identity are the subject of the following paper: B. E. Johnson. "Near inclusions for subhomogeneous $C^*$-algebras", Proc. London Math. Soc., 68:399–422, 1994.<|endoftext|> TITLE: Can groups of twice-odd order have quaternionic representations? QUESTION [5 upvotes]: Let $G$ be a finite group and $\phi\colon G\to \mathrm{GL}_d(\mathbb C)$ be an irreducible representation, with character $\chi$. Recall that $\phi$ is complex type if $\chi$ is not real-valued, $\phi$ is real type if $\phi$ is the complexification of a representation $G\to\mathrm{GL}_d(\mathbb R)$, $\phi$ is quaternionic type otherwise, i.e. $\chi$ is real-valued but $\phi$ is not real. (Equivalently, $\phi$ is quaternionic if $\mathbb{C}^d$ has a $G$-invariant symplectic form, or if there is a $G$-equivariant, conjugate-linear operator on $\mathbb{C}^d$ whose square is $-\mathrm{id}$.) Question: Is there an example of a finite group $G$ of order $4k+2$ for some $k$, such that $G$ has a quaternionic irreducible representation? Some thoughts that may be relevant: The Frobenius--Schur indicator $\frac{1}{|G|}\sum_{g\in G}\chi(g^2)$ is $-1$ iff $\phi$ is quaternionic, $+1$ iff $\phi$ is real, and $0$ iff $\phi$ is of complex type. A group of order $4k+2$ must decompose as the semidirect product of a cyclic group of order two acting on a group of order $2k+1$. Every nontrivial irreducible representation of a group of odd order is automatically of complex type (maybe this is a theorem of Burnside). The degree $d$ has to be even, since $\mathbb{C}^d$ must have a symplectic form. A counting argument using the Frobenius--Schur indicator, and a count of the the number of elements of order 2, shows that the degree $d$ must be strictly greater than $2$. So we need an irreducible representation of degree at least $6$. Searching on GAP didn't give any examples for $|G|\leq 150$. This isn't really a research question, it's just something curious. I'm asking for a friend. REPLY [9 votes]: A more general result is true: If a Sylow $2$-subgroup of $G$ is elementary abelian, then $G$ has no irreducible character of quaternionic type. Recall that $\chi$ is quaternionic iff the Schur index $m_R(\chi) = 2$, where $R$ is the real number field. Theorem 10.9 of my character theory book asserts that for any prime $p$ and any subfield of F of the complex numbers, $p$ cannot divide $m_F(\chi)$ if the Sylow $p$-subgroups of $G$ are elementary abelian. This does $not$ generalize further to groups with abelian Sylow $2$-subgroups. There is a group of order $12$ with a cyclic Sylow $2$-subgroup that has a quaternionic irreducible character.<|endoftext|> TITLE: Two rings...are they isomorphic? QUESTION [16 upvotes]: Edit: I have reverted my question to its original version (which Bjorn Pooenen answered correctly) as requested in the comments. Consider the local rings $$R = \mathbb{C}[[x,y,z]]/\langle xy+xz+yz\rangle$$ and $$S = \mathbb{C}[[x,y,z]]/\langle xy+xz+yz+xyz\rangle.$$ Is $R$ isomorphic to $S$? Some context: I am trying to understand formal neighborhoods of points on certain varieties. I expect one answer, and I'm getting a different answer. This is the first nontrivial case where the answer that I get does not obviously agree with the answer that I expect. REPLY [18 votes]: Yes, they are isomorphic. More generally, if $k$ is any algebraically closed field of characteristic not $2$, and $n$ is given, then all $k$-algebras of the form $k[[x_1,\ldots,x_n]]/(f_2+f_3+\cdots)$, where each $f_i$ is homogeneous of degree $i$, and $f_2$ is a nondegenerate quadratic form, are isomorphic. (I.e., there is only one kind of ordinary double point.) You can construct an isomorphism to $k[[x_1,\ldots,x_n]]/(x_1^2+\cdots+x_n^2)$ with your bare hands as follows. First diagonalize the quadratic form to assume that $f_2=x_1^2+\cdots+x_n^2$. Let $m$ be the lowest degree monomial of degree greater than $2$. Then $m$ is divisible by some $x_i$, say $m=x_1 g$. Performing the analytic change of variable $x_1 \mapsto x_1-g/2$ eliminates $m$ at the expense of introducing new terms of even higher degree. By iterating, one can eventually eliminate all monomials of degree 3 to obtain $f_3=0$, and then $f_4=0$, etc. The partial compositions of this sequence of analytic coordinate changes converge to a single analytic coordinate change because they stabilize modulo any given power of the maximal ideal.<|endoftext|> TITLE: Is nfcp equivalent to stable + eliminates $\exists^\infty$? QUESTION [9 upvotes]: Let $T$ be a complete first-order theory. Recall that a formula $\phi(\overline{x},\overline{y})$ has the finite cover property (fcp) if for all $n$, there exist $\overline{a}_1,\dots,\overline{a}_n$ such that $$T\models \lnot\exists \overline{x} \bigwedge_{i = 1}^n \phi(\overline{x},\overline{a}_i),$$ but for all $1\leq j \leq n$, $$T\models \exists \overline{x} \bigwedge_{i \neq j} \phi(\overline{x},\overline{a}_i).$$ As usual, a theory has nfcp if no formula has fcp. This condition was introduced by Keisler (in the context of Keisler's order) and studied by Shelah in Classification Theory, where it is shown that nfcp implies stability. It is also easy to see that if $T$ has nfcp, then $T$ eliminates the quantifer $\exists^\infty$ (there exist infinitely many). That is, for every formula $\phi(\overline{x},\overline{y})$, there is a natural number $n$ such that for all $\overline{a}$, if $|\phi(M,\overline{a})|>n$, then $|\phi(M,\overline{a})|$ is infinite. Now I have heard that there is a converse to these two facts. In particular, a partial converse (with the additional assumption of elimination of imaginaries) is stated as Theorem 2.8 in this paper, without proof or a reference: Theorem: If $T=T^{eq}$ is stable and eliminates $\exists^\infty$, then $T$ has nfcp. Can anyone provide a proof or a reference? The issue doesn't appear to be addressed in Classification Theory. REPLY [7 votes]: In fact, this theorem is from Classification Theory: see Theorem 4.4 there (the "f.c.p. theorem"). Assertion (8) exactly says that $T^{eq}$ does not eliminate $\exists^{\infty}$ and the theorem says that for stable $T$, this is equivalent to having the f.c.p.<|endoftext|> TITLE: Are all unstable homotopy groups of $U(n)$ torsion? QUESTION [7 upvotes]: The first few unstable homotopy groups of the unitary groups $U(n)$ were calculated by Borel-Hirzebruch, Toda, and Kervaire, and they are all torsion. There is a paper by Matsunaga (details below) in which the p-primary parts of the next few homotopy groups are calculated. This paper claims to be giving a complete description of these homotopy groups, but I don't see any explanation of why there are no free abelian factors. My question is: Why $\pi_{2n+i} (U(n))$ is a torsion group for $i=3,4,5$? More generally, is it know whether or not $\pi_* (U(n))$ is torsion for all $*>2n-1$? Matsunaga's paper is: MR182010 Matsunaga, Hiromichi Unstable homotopy groups of unitary groups (odd primary components). Osaka J. Math. 1 1964 no. 1, 15–24. It's freely available here: https://www.jstage.jst.go.jp/article/kyushumfs/15/1/15_1_72/_article There's actually an errata list for the article that's not contained in the freely available .pdf. It just corrects various notational errors (though not the obvious notational error in the main theorem, in which part b should refer to the 3-primary summand). REPLY [16 votes]: Being an $H$-space, $U(n)$ has the rational homotopy type of a product of odd-dimensional spheres. As we know its cohomology, these are $S^1 \times S^3 \times S^5 \times \cdots \times S^{2n-1}$. In particular, its rational homotopy groups vanish above degree $2n-1$. REPLY [12 votes]: A Sullivan minimal model for $U(n)$ is given by $\Lambda(x_1,x_3,\ldots,x_{2n-1})$ with zero differential. In fact for a Lie group, a minimal model is always a free algebra on odd generators and in this case, you can compute the degree of those generators by computing the cohomology of $U(n)$. The graded group $\pi_*(U(n))\otimes\mathbb{Q}$ is the vector space of indecomposable elements in the minimal model. Thus you get one non-torsion factors in the $i$-th homotopy of $U(n)$ for $i=1,3,\ldots,2n-1$ and all the other homotopy groups must be torsion.<|endoftext|> TITLE: Does the weak Hadwiger conjecture imply the Hadwiger conjecture? QUESTION [6 upvotes]: For any cardinal $\kappa$, let $K_\kappa$ denote the complete graph on $\kappa$. We consider the following statements: (H) If $G$ is a graph and $\chi(G) = \kappa$ then $K_\kappa$ is a minor of $G$. (WH) If $G$ is a graph and there is no graph homomorphism $c: G\to K_\kappa$, then $K_\kappa$ is a minor of $G$. For finite graphs, (H) is just Hadwiger's conjecture. It turns out that (H) is false for $\kappa \geq \aleph_0$: the disjoint union of all $K_\lambda$ where $\lambda < \kappa$ has chromatic number $\kappa$ but does not contain $K_\kappa$ as a minor. However, (WH) is true for graphs $G$ with $\chi(G) \geq \aleph_0$ (see http://arxiv.org/pdf/1312.2829.pdf ). Let us call (WH) the Weak Hadwiger conjecture. In the finite case, (WH) translates to: "If $\chi(G) = n+1$ then $K_{n}$ is a minor of $G$." It is not known whether (WH) holds for finite graphs. Question: Does (WH) imply (H) for finite graphs? REPLY [7 votes]: Well, it is possible that both (WH) and (H) are true, in which case (WH) implies (H). If on the other hand, you are asking if there is a short proof of (H) assuming (WH), then the answer is no. For example, Hadwiger's Conjecture for $n=5$ says If $G$ has no $K_5$-minor, then $\chi(G) \leq 4$, while the Weak Hadwiger's Conjecture for $n=5$ says If $G$ has no $K_5$-minor, then $\chi(G) \leq 5$. By Wagner's characterization of $K_5$-minor free graphs, the first assertion is equivalent to the $4$-Colour Theorem, while the second assertion is equivalent to the $5$-Colour Theorem (every planar graph is $5$-colourable). So unless you think the $5$-Colour Theorem (which has an easy proof) implies the $4$-Colour Theorem (which is not known to have an easy proof), the answer is no.<|endoftext|> TITLE: Gerbes and Stacks QUESTION [8 upvotes]: The definition of a gerbe on a smooth manifold that I know is that - after fixing an open cover $U_i$, a gerbe consists of the data of line bundles $L_{ij}$ on two-fold-intersections $U_{ij}$, isomorphisms $\alpha_{ijk}: L_{ij} \otimes L_{jk} \longrightarrow L_{ik}$ on three-fold intersections that satisfy a co-cycle condition on four-fold intersections. A gerbe on a site is a stack $G$, such that for every object $U$, there exists a covering $U_i$ of $U$ such that $F_{U_i}$ is non-empty for every $i$ and for any two objects $x_1$, $x_2$ in $G_{U}$, there exists a covering $U_i$ of $U$ such that $x_1|_{U_i}$ and $x_2|_{U_i}$ are isomorphic (i.e. objects exist locally and they are locally isomorphic). My question is that if these two notions are related or if it is just the same name for completely different things. In particular: Are gerbes on a manifold a special stack on the small site of that manifold? Is there a fully faithful functor of $2$-categories that sends gerbes over $M$ to stacks over (the small site of) $M$? REPLY [2 votes]: There is a canonical equivalence of $2$-categories $$St\left(Man/M\right) \simeq St\left(Man\right)/M$$ between stacks on the large site of $M$ and stacks on the site of manifolds equipped with a map to $M$ (regarding $M$ as a representable sheaf). Given a map $\pi:\mathscr{Y} \to M$ for $\mathscr{Y}$ some stack on manifolds, it corresponds to the stack $\Gamma(\mathscr{Y})$ on $Man/M$ which assigns a map $f:N \to M$ the groupoid of sections $N \to \mathscr{Y}$ of $\pi$ over $f.$ Suppose that there is a cover of $U_i$ of $M$ such that each $U_i \times _M \mathscr{Y}\simeq U_i \times BU(1)$ (or if you prefer $U_i \times BGL(1)$). Then $\Gamma(\mathscr{Y})$ is easily seen to be a gerbe on the large site for $M$. By Dan Peterson's answer, we see that from the data of a bundle gerbe, one gets a stack $\pi:\mathscr{Y} \to M$ with this property. In fact, it is not hard to show that these are equivalent data, that is, given $\pi:\mathscr{Y} \to M$ such that there is a cover $U_i$ such that $U_i \times _M \mathscr{Y}\simeq U_i \times BU(1)$ is the same as giving a bundle gerbe on $M$. By taking each bundle gerbe $\pi:\mathscr{Y} \to M$ and sending it to $\Gamma(\mathscr{Y})$, one gets a fully faithful embedding of the $2$-category of bundles gerbes over $M$ into the $2$-category of gerbes over the large site of $M$ (which furthermore embeds fully faithfully into stacks on the large site of $M$). The essential image is precisely those gerbes on the site $Man/M$ which are banded by $U(1)$, as pointed out by Reimundo Heluani. It doesn't embed into the $2$-category of stacks on the small site of $M$ however.<|endoftext|> TITLE: Book about the history of mathematics for weather prediction QUESTION [5 upvotes]: Can someone recommend a book about the history of mathematics being used for weather prediction, preferable one which covers recent developments? REPLY [7 votes]: Invisible in the Storm: The Role of Mathematics in Understanding Weather one review (AMS) a second review (EMS), "Invisible in the Storm" recounts the history, personalities, and ideas behind one of the greatest scientific successes of modern times - the use of mathematics in weather prediction. Although humans have tried to forecast weather for millennia, mathematical principles were used in meteorology only after the turn of the twentieth century. From the first proposal for using mathematics to predict weather, to the supercomputers that now process meteorological information gathered from satellites and weather stations, the authors Ian Roulstone and John Norbury narrate the groundbreaking evolution of modern forecasting.<|endoftext|> TITLE: Realization of second Stiefel-Whitney class QUESTION [6 upvotes]: I hope this is not too trivial. Let $M$ be a compact oriented manifold and $x\in H^2(M, \mathbb{Z}/2\mathbb{Z})$. My question is that if there exists a real oriented vector bundle $V$ over $M$ such that the second Stiefel-Whitney class of $V$ is $x$. REPLY [10 votes]: Not necessarily. A necessary condition is that $x^2$ must be the reduction of an integral class. This is because if $V$ is an oriented bundle then $\rho(p_1(V))=w_2(V)^2$, where $\rho$ denotes reduction mod 2. There are closed oriented manifolds and classes $x$ for which this condition is not satisfied. Since this is also a necessary condition for the Poincaré dual of $x$ to be realized by an immersion, the examples in http://arxiv.org/abs/1111.0249 should work.<|endoftext|> TITLE: Białynicki-Birula theory for non-complete varieties QUESTION [10 upvotes]: I would like to know to which extent the theory developed for smooth projective varieties in the following articles A. Białynicki-Birula, Some theorems on actions of algebraic groups. Ann. of Math. (2) , 98:480–497, 1973. A. Białynicki-Birula, Some properties of the decompositions of algebraic varieties determined by actions of a torus. Bull. Acad. Polon. Sci. S ́er. Sci. Math. Astronom. Phys. , 24(9):667–674, 1976. extends to the case of smooth non-complete (say, quasi-projective) varieties. Assume the ground field to be $\mathbb{C}$ if you want. There are similar questions on MO, but I haven't found a satisfying answer: as Dori Bejleri comments below, my question is about "the decomposition into locally closed affine cells and the implications that has about the cohomology"; the question linked in the comment by Vivek Shende, on the other hand, "asks about the existence of an open cover by torus invariant affines, which while important to the proof of the BB decomposition, is not the same thing". REPLY [9 votes]: Everything holds for a smooth quasiprojective $X$ as long as there exists a $\mathbb{C}^*$ action so that $$ \lim_{t \to 0} t \cdot x $$ exists for every $x \in X$. This is guaranteed when $X$ is projective but holds more generally. The theory really depends on a local analysis of torus actions at fixed points so as long as you have enough fixed points for the limits to exist (and $X$ is smooth), everything holds. In the projective case, say you have a rank $r$ torus $T = (\mathbb{C}^*)^r$ acting on $X$, the one parameter subgroups $\mathbb{C}^* \cong T' \subset T$ form a lattice $N$ of rank $r$. For a projective $X$, any general enough $z \in N$ gives you a one parameter subgroup with the properties required to get a BB decomposition. Furthermore, scaling $z$ by a positive multiple does not change the decomposition. So you get a cone decomposition of $N \otimes_\mathbb{Z} \mathbb{R}$ that determine your BB decomposition. As you vary within the interior of a top dimensional cone, the induced BB decomposition stays the same, and then when you cross the faces between the top dimensional cones (which correspond to degenerate choices of one parameter subgroup that do not give you enough independent conditions for the fixed points of $T'$ to be the same as those of $T$) you change the decomposition. Of course within each cone we need to pick a point that is scaled large enough. So in the quasiprojective case, the difference is that now you have to stay within certain top dimensional cones where the limit above always exists, while the other regions of $N \otimes_\mathbb{Z} \mathbb{R}$ correspond to one parameter subgroups whose limit would be outside of the variety, say in an appropriate projective compactification. Then once you have a decomposition into locally closed affine spaces you get the usual theorems about the cohomology vanishing in odd degree and being generated in even degree by the closures of the appropriate dimension affine cells, etc. As an example consider $\operatorname{Hilb}^n(\mathbb{C}^2)$ which I will denote $H^n$ for convenience. $(\mathbb{C}^*)^2 = T$ acts on $\mathbb{C}^2$ in the obvious way and thus on $H^n$. The points of $H^n$ are ideals $I$ in $\mathbb{C}[x,y]$ so that $\dim_\mathbb{C}\mathbb{C}[x,y]/I = n$. The torus fixed points of this action correspond to the monomial ideals. If you pick a subtorus $(t^p,t^q)$ with $p,q > 0$, this gives a monomial order $w$ on $\mathbb{C}[x,y]$ and $$ \lim_{t \to 0} (t^p,t^q)\cdot I = in_w(I) $$ the initial ideal with respect to this ordering, which for general orderings is always a monomial ideal in $H^n$ and so gives a BB decomposition. The special orderings where the initial ideal is not always a monomial ideal are the faces of the cones. However, if we pick $p,q < 0$, then this limit can be an ideal supported at infinity in $\mathbb{P}^2$ and so we do not get a BB decomposition of $H^n$. See also this MO question: Cell decomposition for a variety not necessarily complete?.<|endoftext|> TITLE: Two (other) rings...are they isomorphic? QUESTION [21 upvotes]: Consider the local rings $$R = \mathbb{C}[[x,y,z,w]]/\langle xyz+xyw+xzw+yzw\rangle$$ and $$S = \mathbb{C}[[x,y,z,w]]/\langle xyz+xyw+xzw+yzw+xyzw\rangle.$$ Is $R$ isomorphic to $S$? Some context: I am trying to understand formal neighborhoods of points on certain varieties. I expect one answer, and I'm getting a different answer. This is the first nontrivial case where the answer that I get does not obviously agree with the answer that I expect. Some history: In a previous post (Two rings...are they isomorphic?), I asked a version of this question with one fewer variable, and Bjorn Poonen pointed out that the rings are isomorphic because there is only one kind of rational double point. I think this was essentially an accident, hence the new post. REPLY [10 votes]: Consider the map $\varphi:S\to R$ given by $$\varphi(x) = \frac{4x}{4-x},$$ and similarly for $y$, $z$, and $w$. One needs to check that this is well-defined; indeed, $\varphi$ maps $xyz+xyw+xzw+yzw+xyzw$ to $$\frac{4^4(xyz+xyw+xzw+yzw)}{(4-x)(4-y)(4-z)(4-w)}.$$ Since $\varphi$ obviously induces an isomorphism on the associated graded of the filtration by powers of the maximal ideal, it is itself an isomorphism. More explicitly, the inverse homomorphism $\psi:R\to S$ is given by $$\psi(x) = \frac{4x}{4+x},$$ and similarly for $y$, $z$, and $w$. Note also that this generalizes to any number of variables. Thanks David and Vladimir for your answers; that was really helpful for getting me going!<|endoftext|> TITLE: Jackson's theorem for partial sum of Fourier series QUESTION [6 upvotes]: There is a classical theorem of Jackson stating that the $N$-th partial sum $S_N f$ of the Fourier series of a Lipschitz continuous function $f$ (which is periodic with period 1) satisfies $$ |f(x) - S_N f(x)| \leq c \frac{K \log N}{N} $$ uniformly for all $x \in [0,1]$, where $c$ is an absolute constant and $K$ is the constant in the Lipschitz condition. (This is stated in a more general case in the Wikipedia article here: http://en.wikipedia.org/wiki/Convergence_of_Fourier_series#Uniform_convergence ) Question: does anybody know where I can find a numeric value for the constant $c$? (It is particularly important for me that $c$ is independent of the function $f$ - this fact cannot be seen from the statement in Zygmund's book, for example). Be careful: Jackson's theorem is often stated in the form of the error between $f$ and the "best approximation" of $f$ - this is not the same as the error in the approximation by the partial sum of the Fourier series! If you know where I can find the requested inequality, please let me know. REPLY [6 votes]: The estimate in the wikipedia article you linked does what you want because the modulus of continuity satisfies $\omega(2\pi/N)\le 2\pi L/N$ ($L$ = Lipschitz constant), but here's a derivation from scratch: For convenience, let's focus on $x=0$, and assume that $f(0)=0$. We want to bound $$ (S_Nf)(0) = \int_{-\pi}^{\pi} f(x) D_N(x)\, dx , $$ where $D_N(x)=\sin(N+1/2)x/\sin x/2$ is the Dirichlet kernel. The part of the integral over $|x|\le 1/N$ is $\lesssim L/N$, with an absolute ($f$ independent) constant because $|f(x)/\sin x/2|\lesssim L$. To estimate the contributions coming from $|x|>1/N$, we integrate by parts, integrating $\sin(N+1/2)x$ and differentiating the rest. The boundary terms are $O(L/N)$ and $$ \frac{1}{N+1/2}\left| \int_{1/N\le|x|\le \pi} \left( \frac{f(x)}{\sin x/2}\right)' \cos(N+1/2)x\, dx \right| \lesssim \frac{L\ln N}{N} , $$ because $|f'|\le L$ by assumption. (Also, when you differentiate $\sin x/2$ in the denominator, use that $|f(x)|\le L|x|$ in the estimate.)<|endoftext|> TITLE: Framed version of braided monoidal category QUESTION [6 upvotes]: The operad $\mathcal{D}_2$ of little $2$-disks is an operad whose $n$-th space is a $K(PB_n,1)$ where $PB_n$ denotes the pure braid group on $n$-strands. Algebras over $\mathcal{D}_2$ have a categorical analogue called braided monoidal category. More precisely, there is an operad in groupoids $\mathcal{P}aB$ whose algebras in the category of categories are braided monoidal category and which is such that a levelwise application of the nerve to $\mathcal{P}aB$ yields an operad equivalent to $\mathcal{D}_2$. The operad $\mathcal{D}_2$ has a variant called the framed little $2$-disks operad and usually denoted $f\mathcal{D}_2$. The operad $f\mathcal{D}_2$ also has the property that each of its spaces are $K(\pi,1)$'s. My question : Is there a known framed analogue of a braided monoidal category ? More precisely, is there an operad $\mathcal{X}$ in groupoids whose algebras in the category of categories have been defined somewhere in the literature and with the property that levelwise application of the nerve to $\mathcal{X}$ yields an operad equivalent to $f\mathcal{D}_2$. Note that the question is not about the existence of $\mathcal{X}$ (which is straightforward) but really about the existence of an $\mathcal{X}$ whose algebras have been defined somewhere. REPLY [12 votes]: First of all let me correct you: it is not true that the groups $PB_n$ assemble into an operad in groups. The point is that $PB_n$ is the fundamental group of $D_2(n)$, which requires the choice of a base point. But it is impossible to choose basepoints on the spaces $D_2(n)$ simultaneously in a way that is compatible with the operad composition and the $S_n$-action. What you can do is consider the fundamental groupoids $\Pi_1(D_2(n))$. These carry natural $S_n$-actions and are compatible with composition. Since $\Pi_1(D_2(n))$ is a bit unwieldy one can try to find a "small" full subgroupoid of $\Pi_1(D_2(n))$ which is closed under the $S_n$-action and composition. This is exactly the combinatorially defined operad $PaB$ which you mention: $PaB$ is isomorphic to the suboperad generated by a single object of $\Pi_1(D_2(2))$. Now the fundamental group of $fD_2(n)$ is $PB_n \times \mathbf Z^n$, and one can play exactly the same game. One is naturally led to an operad $PaFB$ of parenthesized framed braids which is defined in the same way as $PaB$ except the braids are framed, that is, instead of isotopy classes of embedded intervalls we are considering isotopy classes of embedded ribbons. (I hope it's clear what I mean.) The algebras over $PaFB$ are exactly balanced monoidal categories. There is in fact a standard graphical calculus for balanced monoidal categories which is based on braids whose strands are embedded ribbons.<|endoftext|> TITLE: Orthogonal polynomial under linear transformation QUESTION [11 upvotes]: Let $M_n(x) = x^n$ be the standard monomials. The binomial formula allows one to expand $M_n(ax+b)$ as a linear combination of $M_k(x)$, for $k \leq n$, giving $$ M_n(ax+b) = (ax+b)^n = \sum_{k=0}^n \binom{n}{k} a^k x^k b^{n-k} = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k} M_k(x). $$ Question: Do equivalent formulas exist for other polynomial sequences? I am mostly interested in the Chebyshev polynomials of the first kind, $T_n(x) = \cos(n \arccos(x))$, seeking a formula of the form $$ T_n(ax+b) = \sum_{k=0}^{n} C(n, k, a, b) T_k(x). $$ I believe such a formula ought to exist, because one could always transform $T_n(x)$ into monomial basis, expand terms using the binomial formula, then transform back into a linear combination of $T_n(x)$'s. However, I would like to use it as a part of an applied algorithm, and thus am concerned about the computational complexity of the formula; going through a monomial basis would be an overkill. Is anyone aware of any such formula? REPLY [2 votes]: Why not use the recurrence $T_{n+1}(x) = 2 x T_n(x) - T_{n-1}(x)?$ When applying it to the transformed argument, you get $$T_{n+1}(a x + b) = 2( a x + b) T_n(a x + b) - T_{n-1}(a x + b).$$ Now, write $$T_l(a x + b) = \sum_k C_{l, k, a, b} T_k(x).$$ This gives you a linear recurrence for the $C$s, since you can get rid of the $x,$ by writing $2 a x T_k(x) = a (T_{k+1}(x) + T_{k-1} (x)).$<|endoftext|> TITLE: Applications of the Small and Great Theorems of Picard QUESTION [8 upvotes]: I just presented the two famous theorems of Picard (Small and Great) in a graduate course, but I have not managed to discover a good number of interesting applications. List of applications (rather straight-forward though) found so far: If a meromorphic function on $\mathbb C$ misses three values, then it is constant. The equation $f^n+g^n=1$ has NO non-trivial meromorphic in $\mathbb C$ solutions if $n\ge 3$. If $f$ is entire an 1-1, then it is linear. If $f,g$ are entire and $g'=f(g)$, then $f$ is linear or $g$ is constant. Could you provide any interesting applications of these theorem? I have asked this question in Mathematics StackExchange, but I only received one response. REPLY [4 votes]: In general, holomorphic maps $f: \mathbb{C} \rightarrow \mathbb{C}$ have no fixed points; but using Picard theorem we can show that $f\circ f$ always have fixed point. Theorem (Fixed-point theorem) Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be holomorphic. Then $f \circ f: \mathbb{C} \rightarrow \mathbb{C}$ always has a fixed point unless $f$ is a translation $z \mapsto z+b, b \neq 0 .$ Proof: Suppose $f \circ f$ has no fixed points. Then $f$ also has no fixed points, and it follows that $$ g(z):=\frac{f(f(z))-z}{f(z)-z} $$ is entire function. This function omits the values 0 and 1 ; hence, by Picard, there exists a $c \in \mathbb{C} \backslash\{0,1\}$ with $$ f(f(z))-z=c(f(z)-z), \quad z \in \mathbb{C} $$ Differentiation gives $f^{\prime}(z)\left[f^{\prime}(f(z))-c\right]=1-c$. Since $c \neq 1, f^{\prime}$ has no zeros and $f^{\prime}(f(z))$ is never equal to $c$. Thus $f^{\prime} \circ f$ omits the values 0 and $c \neq 0$ by Picard, $f^{\prime} \circ f$ is therefore constant. It follows that $f^{\prime}=$ constant, hence that $f(z)=a z+b$. Since $f$ has no fixed points, $a=1$ and $b \neq 0$<|endoftext|> TITLE: What makes the amenability of Thompsons group $F$ such a tricky problem? QUESTION [20 upvotes]: An open problem that seems to get a lot of attention every once in a while is the amenability of Thompsons group $F$. The problem seems to generate both proofs and disproofs at a fairly high rate, compared to many other open problems. What is more, it seems that a big part of these are actually serious attempts by serious mathematicians, rather than the "usual" elementary attempts one sees for the more famous problems. For examples, see for instance the MO question Is Thompson's Group F amenable? as well as (what as far as I can tell is the newest attempt, but I may have missed some) http://arxiv.org/abs/1408.2188. Is there something inherent to this problem which causes this, i.e. some aspect that makes so many serious mathematicians convince themselves that they have a solution, and so many other serious mathematicians to take so long to find the errors? Note that I am specifically not asking about what the errors were in previous attempts, unless there is some general type of error that tends to come up in many of them REPLY [16 votes]: For some famous open problems, there are known "obstructions" to resolving them. For example in number theory we have Siegel zeros and the parity problem for sieves, and in computational complexity theory we have naturalization and relativization/algebrization barriers. The obstructions are well known and ideas for surmounting them are scarce, so if a proposed solution appears, experts can often zoom in quickly on where the new idea for dealing with the obstruction must lie. It's harder to be fooled in such cases. For other open problems, the conceptual territory surrounding the problem is less well understood and mapped out. Instead of a mountain range with only a few well-known passes, the obstacles form a forest with myriad trees among which it is easy to get lost. I know nothing about amenable groups, but if what you say is correct, I would hypothesize that the conjecture falls into the latter category. There are lots of different approaches one could try, and instead of running into the same difficulty over and over again, each new attempt makes a foray into uncharted territory where there is not much past experience to help detect the pitfalls.<|endoftext|> TITLE: Is an infinite-dimensional "Lebesgue measure" uniquely determined by a set of positive finite measure? QUESTION [6 upvotes]: Let $\mu$ be a probability measure on a subset $C \subset \mathbb{R}^\infty$ of the space of sequences, and assume, for simplicity, that $C$ is closed and convex. We say that $\mu$ admits shifts if for any $\xi \in \mathbb{R}^\infty_0$ (finitely many nonzero coordinates) the shifted measures $\mu_{\xi} := (\cdot + \xi)_\ast \mu$ have the property $\mu_{\varepsilon \xi} \to \mu$ in total variation, as $\varepsilon \to 0$ In other words, shifting by $\varepsilon \xi$ produces a measure that is almost $\mu$-absolutely continuous. Call $\mu$ Lebesgue (on $C$ with shift space $S$) if: It admits shifts, and those shifts act ergodically $\frac{d\mu_\xi}{d \mu} = 1$ on $C \cap (C - \xi)$ for all $\xi \in \mathbb{R}^\infty_0$, i.e. $\mu$ is "as invariant as it could be", given that it lives on $C$. Examples of such measures include the (unique) Lebesgue measure on $[0,1]^\infty$, or on any other product of finite-dimensional sets. For many other sets - e.g. all unit balls of separable Banach spaces - they don't exist. Question: Is $\mu$ uniquely determined by $C$ (if it exists)? REPLY [3 votes]: Here is a counterexample: Take any vector $v \in \mathbb{R}^\infty \setminus \ell^1$. Then on the set $C := [0,1]^\infty + \{tv, 0 \le t \le 1\}$ there are infinitely many "Lebesgue measures", namely for any $0 \le t \le 1$ the product measure $\mu_t$ on the shifted cube $[0,1]^\infty + t v$ satisfies my definition. The reason is that for a $\mu_t$-typical point $x \in C$ the $n$-dimensional section $C \cap (x + (\mathbb{R}^n \times \{0\}^\infty))$ is a cube. The proof is a straightforward application of the Borel-Cantelli lemma. Intuivitely, the product measure on $[0,1]^\infty$ doesn't "recognize" $v$ as an admissible shift direction, due to the assumption $v \notin \ell^1$, so it doesn't "feel the difference" between $C$ and the shifted cube $[0,1]^\infty + tv$ in terms of the shift invariance condition.<|endoftext|> TITLE: How many Pythagorean triples are there in which every member is triangular? QUESTION [12 upvotes]: How many Pythagorean triples $(a,b,c)$ are there such that $a, b$ and $c$ are triangular? Any two solutions with only $a$ and $b$ interchanged are considered equivalent. The question of existence was posed by Zarankiewicz and answered by Sierpinski in Sur les nombres triangulaires carrés (1961): Such a triangle is known whose sides are $t_{132} = 8778, t_{143} = 10296$ and $t_{164} = 13530$, but we do not know if there are others and if their number is finite. A partial result: Since $t_n^2 = 1^3 + \ldots + n^3$, the equation $t_x^2 + t_y^2 = t_z^2$ is equivalent to $t_x^2$ being a sum of $k = z - y$ consecutive cubes. This leads one to consider elliptic curves. In Pythagorean triples and triangular numbers (1979) by D. W. Ballew and R. C. Weger it is shown by applying a theorem of Siegel that for a given $k$ there are only finitely many solutions. Actually finding such solutions has been done by R. J. Stroeker: On the sum of consecutive cubes being a perfect square (1995), without the restriction that the square is some $t_x^2$. REPLY [3 votes]: It was proved in http://arxiv.org/abs/0811.2477 (On certain diophantine equations related to triangular and tetrahedral numbers, by Maciej Ulas) that the equation $t_x^2+t_y^2=t_z^2$ has infinitely many solutions in rational numbers. It seems the number of integer solutions of this equation is still an open problem. It is also an open problem whether there exists a right triangle all sides of which are tetrahedral numbers of the form $T_n=\frac{1}{6}n(n+1)(n+2)$ (a tetrahedral version of the question asked).<|endoftext|> TITLE: Are homotopy braid groups residually nilpotent? QUESTION [5 upvotes]: A group is called residually nilpotent if given any non-identity element, there is a normal subgroup not containing that element, such that the quotient group is nilpotent. It is known that pure braid groups are residually nilpotent. Consider the pure homotopy braid group which is a factor group of the pure braid group; namely, in the deformation of braid strands, each strand is allowed to self-intersect. Do we know if pure homotopy braid groups are residually nilpotent? REPLY [3 votes]: It was shown by M. Falk and R. Randell that the pure braid groups $P_n$ (and, more generally, fundamental groups of complements of fiber-type arrangements) are residually nilpotent (see Theorem 2.6 from Pure braid groups and products of free groups, in: Braids (Santa Cruz, CA, 1986), 217–228, Contemp. Math., 78, Amer. Math. Soc., Providence, RI, 1988). A similar argument shows that such groups, and, more generally, almost direct products of residually (torsion-free) nilpotent groups are residually (torsion-free) nilpotent, see for instance Corollary 2 from V. Bardakov and P. Bellingeri, On residual properties of pure braid groups of closed surfaces, Comm. Algebra 37 (2009), no. 5, 1481–1490.<|endoftext|> TITLE: Oldest photographed mathematician QUESTION [11 upvotes]: Who is the most ancient mathematician of which we have a photograph? (or, in the same vein, what is the oldest photograph of a mathematician) A quick search on MacTutor History of Mathematics gives Binet (b.1786) as a pretender with Cauchy (b.1789) coming close... REPLY [27 votes]: Most ancient: Wikipedia has a daguerreotype of Gauss (1777–1855) on his deathbed. Or possibly Farkas Bolyai (1775–1856) in what look like similar circumstances. Less ancient, but allegedly photographed earlier: Googling "daguerreotype mathematician" gives • c. 1840, 1840: Augustus De Morgan (1806–1871) • c. 1842–1843: Ada Lovelace (1815–1852) • c. 1843: Friedrich Wilhelm Bessel (1784–1846) • c. 1843: Joseph Plateau (1801–1883) • c. 1843, 1847–1851: Charles Babbage (1791–1871) • c. 1844: William Henry Fox Talbot (1800–1877) • c. 1845: François Arago (1786–1853) • c. 1845: William Rowan Hamilton (1805–1865) (per frontispiece) • c. 1846: William Thomson (1824–1907) • c. 1846: John Couch Adams (1819–1892) • c. 1847: Hermann von Helmholtz (1821–1894) • c. 1848: John Herschel (1792–1871).<|endoftext|> TITLE: Does the boundary of a convex body contain a regular planar pentagon? QUESTION [17 upvotes]: How to prove or disprove that the boundary of any convex body in $\mathbb{R}^3$ includes 5 points which form a regular planar pentagon? The following consideration suggests the answer "yes": if we assume the boundary is described by the equation $f(x,y,z)=0$, then we have a system of $5+5+3=13$ equations in $15$ variables. In the general case an infinite solution set is expected. However, the convexity is not used here. The question was asked, but not answered, in MSE. REPLY [11 votes]: A MathSciNet search turns up the following paper: V. V. Makeev, Polygons inscribed in a closed curve and in a three-dimensional convex body, Journal of Mathematical Sciences 161 (2009), 419–423 (translated from Russian). The final result in the paper is: Corollary. Each convex body $K\subset \mathbb{R}^3$ is circumscribed about an affine-regular pentagon whose vertices lie on planes of support of $K$ parallel to one line. This doesn't directly answer your question, but the conspicuous absence of your question from the paper perhaps suggests that it is an open problem. Regardless, this paper would seem to be a good entry point into the literature.<|endoftext|> TITLE: Origin of the numbers game QUESTION [7 upvotes]: The numbers game is a (one-player) game played on a finite graph with an initial assignment of numbers to its vertices, studied by Alon, Bj\"orner, Brenti, Donnelly, Eriksson, Krasikov, Mozes, Peres, Proctor, Wildberger, and probably others as well (see http://arxiv.org/abs/math/0610702 for a detailed bibliography). It grew out of a problem that appeared in the 1986 International Mathematics Olympiad, in which the graph is the 5-cycle. But who came up with the problem, and why? I am particularly intrigued by something Yuval Peres told me today, namely, that the original solution (unlike all the other solutions I'm familiar with) works only for the 5-cycle, and not for $n$-cycles in general. REPLY [4 votes]: The problem is due to Elias Wegert, see Relaxation procedures on graphs . This blog post discusses the problem.<|endoftext|> TITLE: Disjoint images of polynomials QUESTION [5 upvotes]: Are there any $f,g \in \mathbb{Q}[x]$ such that for every root of unity $\zeta$, and every $a,b \in \mathbb{Q}(\zeta)$, $f(a) \neq g(b)?$ REPLY [18 votes]: I guess you won't be satisfied with the answer $f=0$ and $g=1$. :) But the answer is yes even if you assume that $f$ and $g$ are nonconstant. For example, consider $f(x)=2x^3$ and $g(x)=(x^3-2)^3$. If $f(a)=g(b)$ for some $a,b \in \mathbb{Q}(\zeta)$, then one finds that $2$ is a cube in $\mathbb{Q}(\zeta)$, which is a contradiction.<|endoftext|> TITLE: Questions about the "universal elliptic curve" over the affine $j$-line punctured at 0 and 1728 QUESTION [19 upvotes]: So my question refers to families of elliptic curves over the $\mathbb{A}^1_\mathbb{C}\setminus\{0,1728\}$ whose fiber above a point $j$ has $j$-invariant equal to $j$ (I understand it's not universal). Some sources give an equation for such a family, namely $$E_1 := y^2 + xy = x^3 - \frac{36}{j-1728}x - \frac{1}{j-1728}$$ Thanks to TomChurch's comments, I'm revising my questions: Does $E_1$ admit a nontrivial section? (other than the identity section) Is there a way to see this family complex-analytically using quotients of the upper half plane? What I mean is this: Let $\mathbb{H}$ be the upper half plane, and let $\mathbb{H}^\circ$ denote $\mathbb{H}$ punctured at the $SL_2(\mathbb{Z})$-orbits of $i$ and $e^{2\pi i/3}$. Let $\mathbb Z$ act on the product $\mathbb C\times\mathbb H^\circ$ by $$(m,n)\cdot(z,\tau) := (z + m\tau + n,\tau)$$ The quotient $\mathbb Z^2\backslash(\mathbb C\times\mathbb H^\circ)$ is an elliptic curve over $\mathbb H^\circ$. Let $SL_2(\mathbb Z)$ act on $\mathbb C\times\mathbb H^\circ$ by $$\gamma.(z,\tau) := \left(\frac{z}{c\tau + d},\frac{a\tau + b}{c\tau + d}\right)$$ This action descends to an action of $SL_2(\mathbb Z)$ on $\mathbb Z^2\backslash(\mathbb C\times\mathbb H^\circ)$, but as TomChurch noted, $\gamma = -I$ sends $(z,\tau)\mapsto(-z,\tau)$, the fibers of this quotient are actually copies of $\mathbb{P}^1$ and are not elliptic curves. Is there a similar construction that will actually yield a curve like $E_1$? REPLY [20 votes]: Question 1: Yes, there is a nontrivial section, namely $s: (x,y) = (-1/36, y_0)$ where $y_0$ is either solution of $y^2-y/36=-1/36^3$ (i.e. $y = 1/72 \pm 1/18^{3/2}$). [Note that the numerator $36$ in the equation for $E_1$ is a typo should be $36x$.] Using the theory of elliptic surfaces we can show that in fact the group of section is infinite cyclic with generators $s$. The formula for $E_1$, considered as an equation in three variables $j,x,y$, defines a rational surface (one can choose any $x,y$ and solve for $j$). We can clear denominators by writing $$ (x,y) = \Bigl( \frac{X}{(j-1728)^2},\frac{Y}{(j-1728)^3}\Bigr) $$ to obtain the equivalent equation $$ Y^2 + (j-1728)XY = X^3 - 36(j-1728)^3 X - (j-1728)^5 $$ with polynomial coefficients, and then use Tate's algorithm to find that the bad fibers at $j=\infty,0,1728$ are of Kodaira types $I_1$, $II$, $III^*$ respectively. The first two fibers make no contribution to the Neron-Severi lattice, and the last contributes $E_7$. Hence the Mordell-Weil group MW of sections is the quotient $E_8/E_7$ with the canonical height given by induced quadratic form; and this is an infinite cyclic group with a generator of height $1/2$. Since $s$ has height $1/2$ our claim follows. There are three further elliptic surfaces with the same $j$-invariant and good reduction away from $j=\infty,0,1728$; they are obtained from $E_1$ by quadratic twists by $\sqrt{j}$, $\sqrt{j-1728}$, and $\sqrt{j(j-1728)}$. If I did this right, the first of these is an elliptic K3 surface of maximal Neron-Severi rank and trivial MW, and the others are rational elliptic surfaces with MW groups of rank $2$ and $1$ respectively. For the K3 surface, the fibers at $j=\infty,0,1728$ are of types $I^*_1$, $IV^*$, $III^*$ and contribute $D_5$, $E_6$, $E_7$; this already accounts for a rank-20 subgroup of Neron-Severi, with discriminant $-24$, so the MW rank is zero, and one can show in various ways that there are no torsion sections (for instance the reducible fibers cannot accommodate nontrivial torsion). The remaining twists have singular fibers of type $I^*_1$, $II$, $III$ (giving $D_5 A_1$) and $I_1$, $IV^*$, $III$ (for $E_6 A_1$), which gives trivial torsion and rank $8-(5+1)=2$ and $8-(6+1)=1$ respectively.<|endoftext|> TITLE: Subquotients in the Verma filtration on Verma modules QUESTION [15 upvotes]: Let $\lambda$ be a dominant integral weight of $\mathfrak g$, a finite-dimensional reductive Lie algebra over $\mathbb C$. Let $M(w\cdot \lambda)$ denote the Verma module with high weight $w\cdot \lambda := w(\lambda+\rho)-\rho$. Then there are known inclusions $M(w\cdot \lambda) \hookrightarrow M(\lambda)$, for each $w$, and for this question we identify them with their images. Let $I_w := \{v \in W\ :\ v\geq w\}$ denote the principal upward order ideal in $W$. Does the assignment $I_w \to M(w\cdot \lambda)$ extend to a map of lattices, taking upward order ideals to submodules? EDIT: More concretely, is $$ M(w\cdot\lambda) \cap M(v\cdot\lambda) = \sum_{x\geq w,v} M(x\cdot\lambda),$$ or is the intersection strictly bigger? I believe this should follow from some geometric construction like "take $I$ to the $\mathcal D$-module of $B$-finite distributions on $G/B$ supported on $\bigcup_{i\in I} X_i$, twisted by $\mathcal L_\lambda$" but would rather quote a reference than hash out the geometric details. Also, for each $w$ there is a quotient module $ M(w\cdot \lambda) \big/ \sum_{v > w} M(v\cdot \lambda)$, which further quotients to the irrep $L(w\cdot \lambda)$. I gather it was these that Verma erroneously thought were irreducible. Is there a Kazhdan-Lusztig-polynomial way to compute the irreps in a composition series for this module? The standard use computes the multiplicity of $L(v\cdot \lambda)$ inside $M(\lambda)$ as a whole; my question is about how that multiplicity is distributed over the various $w\geq v$. Of course references would be most appreciated. REPLY [10 votes]: Contrary to Verma's initial impression, the internal structure of a "Verma module" (Dixmier's terminology) tends to be extremely complicated. However, this complexity only shows up in ranks 3 or higher, by which time it's not easy to exhibit complete details in most examples. (Concerning Verma's subtle error in his thesis, see the overview in section 4.14 of my 2008 AMS Graduate Studies volume Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$.) Throughout the 1970s there was considerable activity by BGG, Jantzen, and others, who made serious progress in understanding the submodule structure. But the definitive conjecture, very soon proved, came from Kazhdan and Lusztig in their 1979 Inventiones paper. The "localization" approach of Beilinson-Bernstein to the proof was developed further by them in order to prove the earlier Jantzen conjecture on how composition factors are placed in layers of the Jantzen filtration of a Verma module. All of this gets arbitrarily complicated combinatorially, even in type $A_n$. In particular, the estimate you give in your large shaded box for the behavior of the intersection of two Verma submodules is probably over-optimistic in most cases. The problem is that when composition factors appear with multiplicity greater than 1 (which typically happens), the extra factors occur "higher" in the structure than the corresponding Verma submodule. This is how the interpretation of coefficients of the Kazhdan-Lusztig polynomials fits with the Jantzen filtration. Probably it's worth looking closely at the first difficult case $\mathfrak{sl}_4$, where the Weyl group $S_4$ has order 24 but there are typically 26 composition factors in a Verma module with dominant integral highest weight. Chapter 8 in my book outlines the general theory in detail, with plenty of references, and 8.6 discusses some of the data for type $C_3$ as an example. A "typical" Verma module in higher rank has a great many composition factors (with multiplicities which grow rapidly); the internal structure is almost impossible to work out explicitly when there are infinitely many submodules. That said, your follow-up question in a shaded box might be more amenable to an answer involving the Kazhdan-Lusztig polynomial interpretation of the Jantzen filtration. This probably isn't straightforward to carry out in general, but a closer look at the rank 3 and 4 examples would be a good starting point. (The old tables of polynomials here computed by Mark Goresky might be helpful.) UPDATED SUMMARY: The answer to the first question is almost certainly no, though it might take a large amount of computation to pin down a specific counterexample (probably in high rank). On the other hand, the second question seems to have a positive answer, but again the computations involved may be onerous. To streamline notation, start with a weight $\lambda$ which is dominant regular (relative to the shifted dot-action of $W$ having origin $-\rho$). Take $\mu, \nu$ to be strongly linked subweights, with $\mu = w \cdot \lambda$ and $\nu = v \cdot \lambda$. So $M(\lambda)$ includes unique copies of $M(\mu)$ and $M(\nu)$, which intersect in a submodule $M$. It's clear that $M$ includes the sum of all Verma submodules $M(\pi)$ such that $\pi \leq \mu, \nu$ is strongly linked to both. The first question is whether $M$ must equal this sum. Here the highest composition factors $\mu, \nu$ of those two Verma modules occur on certain layers of the Jantzen filtration of $M(\lambda)$, while the Jantzen layers of $M(\mu), M(\nu)$ embed in corresponding layers of $M(\lambda)$ determined by length differences in $W$ (Jantzen conjecture, proved by Beilinson-Bernstein). Say $M(\pi) \subset M$, with $\pi := u \cdot \lambda$. It may well happen that one or both of $M(\mu), M(\nu)$ has extra composition factors of type $L(\pi)$ (with some multiplicity). Necessarily such factors $L(\pi)$ occur in higher layers of the Jantzen filtration of $M(\mu)$ and/or $M(\nu)$. This happens just when a KL polynomial $P_{u,w}(q)$ or $P_{u,v}(q)$ has a nonzero coefficient of degree $>0$. Polo has shown that these polynomials (for type $A_n$, when $W = S_{n+1}$) can have arbitrarily high degrees if $n$ s large enough. In this situation $M$ will involve one or more composition factors isomorphic to $L(\pi)$ besides the single copy contained in the sum of Verma submodules. At the same time, the ideas just sketched indicate that a detailed knowledge of coefficients of the relevant KL polynomials will be enough to determine the composition factor multiplicities in any quotient of a Verma module by the sum of its canonical Verma submodules. But again the required recursive computations may become quite lengthy, and no closed formula is likely.<|endoftext|> TITLE: What is the best reference for Spectral theory? QUESTION [11 upvotes]: I'm studying Bernard Aupetit: A Primer on Spectral Theory but the textbook we are using is a little bit heavy going for me. Is there a best book to learn about these things? Thank you. REPLY [6 votes]: As nobody has mentioned Reed and Simon yet, I will do so: Reed, Michael and Simon, Barry. Methods of Modern Mathematical Physics, Academic Press, 1980. Of course "spectral theory" means different things to different people, depending on what they plan on doing with it. As the title suggests, Reed and Simon is in principle aimed at mathematical physicists (quantum mechanics, etc) but it is an honest mathematics textbook (all theorems are proved, etc). The first volume begins with the basics of functional analysis and ends with the spectral theorem, and volumes 2-4 proceed from there. I think it's a good treatment for any "working analyst"; it's well motivated and down to earth. Applications to topics such as PDE are made more explicit than in texts such as Conway. There are also a ton of really good exercises. This book is a little unusual (in what I think is a good way) in that it includes, and shows the benefits of, Halmos's "multiplication operator" version of the spectral theorem, mentioned already by Jon Bannon. Unfortunately the current edition is very expensive, so you may want to try to borrow it from a library or colleague at first.<|endoftext|> TITLE: When is a `1-form' with continuous coefficients exact? QUESTION [5 upvotes]: Let $\Omega$ be a convex, bounded open subset of $\mathbb{R}^d$, and let $C^1(\bar \Omega)$ be usual space of continuous functions on $\bar \Omega$ which are $C^1$ in $\Omega$ and whose partials in $\Omega$ are bounded and uniformly continuous in $\Omega$. My question is about the natural differentiation operator $D:C^1(\bar \Omega)\longrightarrow C(\bar \Omega)^d$, given by $$ Du=(\partial_1u,\dots,\partial_du), $$ where $\partial_i u$ is the unique extension to $\bar\Omega$ of the partial derivative of $u$ in the $e_i$ direction. My question is about the failure of surjectivity of $D$ when $d>1$: for which $f=(f_1,\dots,f_d)\in C(\bar \Omega)^d$ is there a $u\in C^1(\bar \Omega)$ such that $Du=f$? There are (I think) some obvious necessary conditions concerning the relationship between the distributional derivatives of the $f_i$s if $f$ is to be `$D$ exact', and some restriction needs to be in effect to take care of what has to happen with line integrals. However, this question is a little outside my usual area of operation, so I thought I'd see if anyone know the answer. It occurs to me that there may be a connection with this question: Poincare lemma for non-smooth differentiable forms. REPLY [4 votes]: $\def\ssp{\kern.4mm} $Here is a sketch of proof of sufficiency of $d\ssp f=0$ , i.e. of $\partial_i f_j=\partial_j f_i$ (in the distributional sense) assuming that the case where $f$ is $C^1$ is known, for which I refer e.g. to the Poincaré lemma in §V.5 on pages 124−125 in Serge Lang's Differential Manifolds, Springer 1988. Fix any $x_0\in\Omega$ , and define $u$ by $\Omega\owns x\mapsto\int_{\,0}^{\,1}f(x_0+t\,(x-x_0))\cdot(x-x_0)\,{\rm d\,}t$ . For fixed $x\in\Omega$ to prove that $d\ssp u(x)=f(x)$ , take a smooth "cut-off" function $\chi$ which has value $1$ up to "sufficiently" near to the boundary of $\Omega$ and which has value $0$ "very near" to the boundary. Further, take a "smooth bump" $\varphi$ at the origin with integral $1$ whose support is "very small". Let $\Omega_0$ be a convex open set with compact closure included in the interior of $\chi^{-1}[\{1\}]$ and $x_0,x\in\Omega_0$ and $\Omega_0-{\rm supp\,}\varphi\subset\chi^{-1}[\{1\}]$ . Then letting $g=(\chi\cdot f)*\varphi$ and defining $v$ by $\Omega\owns x\mapsto\int_{\,0}^{\,1}g\ssp(x_0+t\,(x-x_0))\cdot(x-x_0)\,{\rm d\,}t$ , by the $C^1$ or even the smooth case, noting that $\partial_i g_j=(\chi\cdot\partial_i f_j)*\varphi=(\chi\cdot\partial_j f_i)*\varphi=\partial_j g_i$ holds on $\Omega_0$ , we have $\partial_i v=g_i=(\chi\cdot f_i)*\varphi$ on $\Omega_0$ . Taking here in place of $\varphi$ the function $\varphi_n:z\mapsto n^d\ssp\varphi(n\ssp z)$ , and letting $n\to\infty$ , we get $\partial_i u=f_i$ pointwise on $\Omega_0$ since both convergences $g_i\ssp(\varphi_n)\to f_i$ and $v\ssp(\varphi_n)\to u$ as $n\to\infty$ are uniform on $\Omega_0$ . As $x\in\Omega$ here is arbitrarily fixed, we get $d\ssp u=f$ on $\Omega$ which further extends to the boundary by the assumed uniform continuity of $f$ .<|endoftext|> TITLE: On a proposition in Hartshorne's paper "Ample vector bundles on curves" QUESTION [10 upvotes]: In Prop. 4.1, p. 87 of the article "Ample vector bundles on curves" (Nagoya Math. J. 43 [1971], 73--89), R. Hartshorne states the following: Let $A$ be an abelian variety [over an alg. closed field $k$, say - my assumption]. Let $X$ be a non-singular subvariety of $A$. Assume that every curve in $X$ generates $A$. Then the normal bundle $N$ to $X$ in $A$ is ample. The proof he provides is based on a criterion of Gieseker for ampleness and boils down to the following statement (X): Let $Y$ be a smooth curve over $k$ and let $\phi:Y\to A$ be a $k$-morphism, which is birational onto its image. Suppose that $\phi_*(Y)$ generates $A$. Then the map $\phi^*:H^0(A,\Omega_{A/k})\to H^0(Y,\Omega_{Y/k})$ is injective. Hartshorne does not spell out a proof of (X) but it is likely that he had the following argument in mind: "Proof": There is a factorisation $Y\to^{i}{\rm Jac}(Y)\to^{g} A$ of $\phi$, where $i$ is the morphism of $Y$ into the Jacobian ${\rm Jac}(Y)$ of $Y$ given by the choice of some $k$-point on $Y$ and $g$ exists because the Jacobian is an Albanese variety for $Y$. Now $i^*:H^0({\rm Jac}(Y),\Omega_{{\rm Jac}(Y)/k})\to H^0(Y,\Omega_{Y/k})$ is an isomorphism (this is a classical fact) and $g^*:H^0(A,\Omega_{A/k})\to H^0({\rm Jac}(Y),\Omega_{{\rm Jac}(Y)/k})$ is injective because the morphism $g$ is a surjective morphism between abelian varieties and is thus smooth. (for this argument when $k={\bf C}$, see for instance Prop. 6.3.10 (p. 30) in R. Lazarsfeld, "Positivity in algebraic geometry. II." Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. No. 49. Springer-Verlag, Berlin, 2004.) Now the "Proof" given above is apparently flawed, because when ${\rm char}(k)>0$, surjective morphisms between abelian varieties are not smooth in general. In fact, I think that (X) is false when ${\rm char}(k)>0$, although I do not have a counterexample handy. This is is also suggested by Serre's discussion of Albanese varieties in Exp. 10 of the "Séminaire Chevalley" (1958-1959) (see proof of Th. 3). In fact it seems to me that Hartshorne's Prop. 4.1 is likely to be false when ${\rm char}(k)>0.$ My question is: is there a way to give a characteristic free proof of (X) ? Or are there classical counterexamples to (X) ? Is it well-known that Hartshorne's Prop. 4.1 is false when ${\rm char}(k)>0$ ? I would also like to underline there is no implicit assumption on the characteristic of the base field in Hartshorne's paper. This is clear from the formulation of some corollaries of Prop. 4.1 (for instance Cor. 4.3). Thank you in advance for your help. REPLY [7 votes]: The assertion (X) is false in any characteristic $p > 0$ for any $Y$ with genus at least 2. (It is true and easy for $Y$ of genus 1, and true and easy and uninteresting for $Y$ of genus 0.) To see this, we may and do choose a subgroup scheme $G \subset J := {\rm{Jac}}(Y)$ of the nonzero infinitesimal Frobenius kernel of $J$ such that $G$ is equal to $\alpha_p$ or $\mu_p$. Note that the map $q:J \rightarrow J/G =: A$ has tangent mapping with kernel equal to the line ${\rm{Lie}}(G) \subset {\rm{Lie}}(J)$ that is a proper subspace (as the genus is larger than 1). For $y_0 \in Y(k)$, the associated inclusion $i_{y_0}:Y \hookrightarrow J$ defined by $y \mapsto \mathscr{O}(y - y_0)$ carrying $y_0$ to $0$ yields a line ${\rm{Lie}}(i_{y_0})({\rm{T}}_{y_0}(Y)) \subset {\rm{Lie}}(J)$, and it is a classical fact that as we vary through such $y_0$ the resulting lines inside ${\rm{Lie}}(J)$ span this space. In particular, there must be some such $y_0$ (and hence for all but finitely many $y_0$, though we won't use this) for which this line is not contained in the proper subspace ${\rm{Lie}}(G)$. Consequently, the composition $q \circ i_{y_0}:Y \rightarrow J/G$ carrying $y_0$ to 0 is injective between tangent spaces at $y_0$ and $0$. Thus, this map is formally an immersion at $y_0$, so it is separable onto its image (by smoothness considerations for $Y$ and $J/G$), yet it is also purely inseparable onto its image since $Y$ is closed in $J$ and $q$ is a purely inseparable isogeny, so $Y \rightarrow J/G$ is birational onto its image. The pullback map $\Omega^1(J/G) \rightarrow \Omega^1(Y) = \Omega^1(J)$ is dual to the quotient map ${\rm{Lie}}(J) \rightarrow {\rm{Lie}}(J/G)$ that is not surjective (for dimension reasons, as its kernel ${\rm{Lie}}(G)$ is a nonzero subspace), so this pullback map on global 1-forms is not injective.<|endoftext|> TITLE: from a circle to higher spheres QUESTION [11 upvotes]: Question: Is there a group $G$ and a CW-complex $X$ such that 1) $X$ is homotopy equivalent to the circle $S^{1}$. 2) $G$ acts on $X$ 3) the space of fixed points $X^{G}$ is weakly equivalent to $S^{2}$ ? REPLY [13 votes]: A very general answer is given by Tony Elmendorf's paper Systems of Fixed Point Sets, http://www.ams.org/journals/tran/1983-277-01/S0002-9947-1983-0690052-0/. Very loosely speaking, it says that, if you can write down a reasonable system of fixed sets, it can be realized up to homotopy equivalence. Take, for example, $G = \mathbb{Z}/2$. Define a contravariant functor $\mathscr{X}$ on the category of orbits of $G$ with $\mathscr{X}(G/e) = S^1$ and $\mathscr{X}(G/G) = S^2$; let the self-map of $G/e$ act on $S^1$ by reflection, with two fixed points, and let the map corresponding to the projection $G/e\to G/G$ map $S^2$ to one of the fixed points in $S^1$. Elmendorf's construction then gives a $G$-CW complex $X$ whose system of fixed sets maps to $\mathscr{X}$ with $X\to \mathscr{X}(G/e)$ and $X^G\to \mathscr{X}(G/G)$ both being homotopy equivalences.<|endoftext|> TITLE: Infected square QUESTION [10 upvotes]: I saw the following problem in Mathematical Puzzles from Peter Winkler (very good book, by the way): imagine you infect k cases of a chessboard nxn and the infection spreads to a case if it has at least two neighbors infected. Then k = n is the minimum number such that it is possible to infect the whole chessboard. I would like to know if it is known what happen in higher dimensions and other values for the number of neighbors for a case to be infected? REPLY [7 votes]: There is an d-dimensional version of this problem in The Art of Mathematics - Coffee Time in Memphis by Bela Bollobas. (Problem 35) According to it, the answer is $k = \lceil d(n-1)/2 \rceil + 1$.<|endoftext|> TITLE: Peano arithmetic vs. fast-growing hierarchy with pathological fundamental sequences QUESTION [11 upvotes]: Fundamental sequence for a countable limit ordinal $\alpha$ is an increasing sequence $\{\alpha[i]\}$ of ordinals of length $\omega$ such that $\lim_{i\rightarrow\omega}\alpha[i]=\alpha$. There are many (continuum many, in fact) possible choices for fundamental sequence for any ordinal, some are quite natural, like $\omega^2[n]=\omega n$, and some are quite odd, like $\omega[n]=\Sigma(n)$. Most common definition of fundamental sequences below $\varepsilon_0$ is via Wainer hierarchy. Using these, it's known that in fast-growing hierarchy, $F_{\varepsilon_0}(n)$ is a total recursive function which outgrows all recursive functions which Peano axioms can prove total. A friend of mine posed a question, if this necessarily hold under different choices for fundamental sequences. For me, it seems like the answer would be no, because we can choose some very slow fundamental sequences for all ordinals, possible making it slower than $F_\alpha(n)$ for some $\alpha<\varepsilon_0$ in Wainer hierarchy, but my friend believes the answer to be yes. To put it into a single question: Is it true that for any choice of fundamental sequences for ordinals below $\varepsilon_0$ we have that, in fast-growing hierarchy, $F_{\varepsilon_0}(n)$ outgrows all functions provably total recursive in PA? Does it make any difference if we replace it with Hardy hierarchy? Thanks for your feedback. REPLY [11 votes]: The answer is no. Choose a fundamental sequence for $\epsilon_0$ itself in the usual way, which I think is $\epsilon_0[n]=\omega^{\omega^{{\vdots}^\omega}}$, and then modify the earlier fundamental sequences for $\alpha=\epsilon_0[n]$ by making it start with $0,1,2,\ldots,n$, before resuming with the usual values. In particular, we have thereby ensured $\alpha[n]=n$ for $\alpha=\epsilon_0[n]$. It now follows, according to the rules of the fast-growing hierarchy, that $F_{\epsilon_0}(n)$, which by definition is $F_{\epsilon_0[n]}(n)$, is the same as $F_\alpha(n)$ where $\alpha=\epsilon_0[n]$, but since this is a limit ordinal, it is equal to $F_{\alpha[n]}(n)$, which is the same as $F_n(n)$, by construction. So with these modified fundamental sequences, the top function $F_{\epsilon_0}$ is basically the same as $F_\omega$. This seems completely to confirm your intuition that slowing the fundamental sequences down could make the diagonal function at the top very small. If we dropped the requirement that the fundamental sequences must be strictly increasing, we could pad with $n$ many $0$'s instead, ensuring that $\alpha[n]=0$ for limit $\alpha=\epsilon_0[n]$, and get a more extreme situation $F_{\epsilon_0}(n)=F_{\epsilon_0[n]}(n)=F_{(\epsilon_0[n])[n]}(n)=F_0(n)$.<|endoftext|> TITLE: How to prove that a kernel is positive definite? QUESTION [5 upvotes]: For example, how to prove $\forall(x,y)\in R^N\times R^N,K(x,y) = \displaystyle\frac{1}{1+\frac{||x - y||^2}{{\sigma}^2}}\\$ where $\sigma > 0$ is a parameter, is positive definite? I have tried to construct the target kernel base on some negative definite kernel, such as the square distance $||x-y||^2$, but doesn't seems to work. REPLY [10 votes]: Here is a simple trick that relies only on elementary ideas. First, we use the observation that for $t>0$ \begin{equation*} \exp(-t\|x-y\|^2) = \exp(-t\|x\|^2)\exp(2t\langle x, y\rangle)\exp(-t\|y\|^2), \end{equation*} which is clearly positive definite (use $\langle x,y\rangle^k$ is pd for $k\ge0$). Now, just use the integral \begin{equation*} \frac{1}{1+\|x-y\|^2} = \int_0^\infty e^{-t(1+\|x-y\|^2)}dt, \end{equation*} which is basically just a nonnegative sum of positive definite functions, hence itself positive definite.<|endoftext|> TITLE: What is the BRST-anti-BRST formalism? QUESTION [9 upvotes]: What is the BRST-anti-BRST formalism? Is the Sp(2) doublet the ghost, antighost pair? Introductory accounts of this subject seem to be hard to find. I would appreciate a reference for someone who knows BRST well. REPLY [3 votes]: To complete Igor Khavkine's answer: as explained here, in the Sp(2) covariant description the BRST and anti-BRST charges form a doublet with opposite ghost number.<|endoftext|> TITLE: Between compact and locally uniform: What is the name of this convergence? QUESTION [12 upvotes]: Let $X$ be a topological space, $(Y,d)$ a metric space, $f\in Y^X$, and $(f_n)$ a sequence in $Y^X$ with the following property: For every $x_0\in X$ and every $\varepsilon>0$, there exist a neigbourhood $U$ of $x_0$ and an index $n_0$ such that we have $d(f_n(x),f(x))<\varepsilon$ for every $x\in U$ and every $n\ge n_0$. How does one call this kind of convergence? Are there any standard references? Observe that $U$ is allowed to depend on $\varepsilon$, so that the condition is weaker than locally uniform convergence. On the other hand, it easily implies compact convergence (that is, uniform convergence on every compact subset of $X$); thus if $X$ locally compact, then it is actually the same as locally uniform convergence. I haven't made this up, it happens to be exactly what I can prove in a particular problem. In that problem, the functions need not be continuous. Of course, we should also generalize the metric space to a uniform space, the sequence to a net, and then ask for the name of the corresponding topology on $Y^X$. REPLY [13 votes]: This notion of convergence is not often refered to; I think mostly because it does not come from a topology. But this an excellent notion of convergence, probably the best we can put on the space of (continuous) functions. I can tell you a few things about it - but only in the case where all the functions are continuous: First observation: This notion of convergence does not depend on the uniform structure of $Y$, and we don't even need this uniform structure to define it: One will say that a net $f_{\alpha}$ of functions $X \rightarrow Y$ tends to a function $f:X\rightarrow Y$ if: For all $U$ open subsets of $Y$, $f^{-1}(U)$ admit a covering by open subsets $V_i$ such that for each $i$ there exists a $\alpha_i$ such that for all $\alpha \geqslant \alpha_i$ one has $f_{\alpha}(V_i) \subset U$. Proposition: (if $f$ is continuous) This notion of convergence is equivalent to yours. Prof: Assume convergence in my definition, then take $x_0 \in X$ and $\epsilon >0$. Let $U=\{ y | d(f(x_0),y)<\epsilon \}$ there exists a covering of $f^{-1}(U)$ by $V_i$ such that... In particular, $x_0$ is in one of the $V_i$, and for each $x \in V_i$ one has, for all $\alpha \geqslant \alpha_i$, $d(f_{\alpha}(x),f(x_0)) <\epsilon$ which is not exactly your definition, but you can restrict $V_i$ a little more using the continuity of $f$ such that additionally for all $x\in V_i$, $d(f(x),f(x_0))<\epsilon$ and we are done. The converse is essentially the same: If $ f_{\alpha}$ converge in your sense, then for each $U \subset Y$, for each $x \in f^{-1}(U)$ there exist an $\epsilon$ such that the $\epsilon$ neighbourhood of $f(x)$ is in $U$ and applying both your definition and the continuity of $f$ one can find a "$V_i$" containing $x$. Second observation Either with your definition or mine, The space of continuous functions is closed for this notion of convergence. Also, the evaluation map: $X \times Y^X \rightarrow Y$ is jointly continuous (on continuous function of course). In particular, when this notion of convergence on continuous function comes from a topology, then this topology on $Y^X$ is the exponential in the category of topological space. As it is well know that the category of topological does not admit all exponentials, this notion of convergence will not always correspond to a topology (essentially, it will fail as soon as $X$ is not locally compact and $Y$ not too trivial). In fact if you denote $K$ to one point compactification of $\mathbb{N}$, then a sequence $f_n$ tend to $f$ for this notion of convergence exactly if the corresponding map $K \rightarrow Y^X$ is continuous in the "exponential" sense, i.e. if $ K \times X \rightarrow Y$ is continuous. Thrid observation: This will probably not speak to everyone, but this notion as a really nice interpretation in topos theory and more precisely in the philosophy of geometric logic. A function from $X$ to $Y$ can be thought of as a "point" of the topos $sh(Y)$ in the internal logic of the topos $sh(X)$. This notion of convergence is exactly the notion of convergence of points of $sh(Y)$ interpreted in the internal logic of $sh(X)$. Fourth observation: This notion of convergence (with my definition) actually has a name (althoug not widely used) it is "continuous convergence" which is defined on arbitrary convergence space by: a sequence $f_{\alpha}$ converge to $f$ if for each net $x_{\beta}$ converging to $x$ the net $f_{\alpha}(x_{\beta})$ converge to $f(x)$. There is a few paper about this (a part of which are in German...) I think This paper is a good starting point (It seems freely available and there is more reference in its bibliography) but I'm not really familiar with this literature. Finally (but this might be subjective) I'm afraid that for non continuous functions this notion does not make sense so well. I mean, you can still use the definition of course, but the proof of the equivalence of the various possible definition of this notion of convergence as well as the various theoretical interpretation that we can give seems to rely a lot on continuity...<|endoftext|> TITLE: Panning for gold nuggets: a type of isoperimetric problem QUESTION [5 upvotes]: Let $C$ be a unit-radius circle in the plane. Suppose you have a total length $L$ of string available, and your task is to connect chords of $C$ using no more than $L$ of string to minimize the largest radius $r$ disk inside $C$ that can avoid all chords. So, if you have $L=4$, it is natural to $(+)$-partition $C$, which achieves $r=\sqrt{2}-1 \approx 0.414$, as illustrated left below.       If you have as much as $L \approx 5.084$ string, then the three chords shown right above achieve $r \approx 0.309$. I have no principled reason to believe this is optimal for that $L$ (although you can see that I at least arranged the six disks to have the identical radius $r$). The general question is: Q1. For a given $L$, what is the smallest $r$ one can ensure by chords of $C$ whose total length is no more than $L$? This might be approachable for small $L$, or by exploring a fixed number $n$ of chords ($n=2,3$ above). Perhaps Q2 is more tractable? Q2. What is the optimal pattern of chords as $L$ grows large? Is it a non-equally-spaced orthogonal grid? I have asked similar questions before: "Optimal planar net for catching convex shapes" and "Chord arrangement that avoids confining small or large disks." The tension here is that Yoav Kallus showed that the former question is answered by a hexagonal tiling, but straight chords cannot create such a tiling. REPLY [4 votes]: Partial update: It is easier to find the minimum $L$ possible for a given $r.$ I take this approach more consistently in the second part. I conjecture that for fixed $r$ it is optimal to use parallel chords at $y=1-2r,1-4r,1-6r,\cdots,1-2kr$ stopping at $k=\lfloor \frac1r\rfloor.$ For $L \le 2,$ I can't see doing better than a single chord at $y=-\sqrt{1-\frac{L^2}{4}}$ allowing a radius of $$r=\frac{1+y}{2}=\frac{2+\sqrt{4-L^2}}4.$$ As L increases this moves up and at $L=2$ becomes a diameter at $y=0$ with $r=\frac12.$ As $L$ increases past $2$ we can obtain $r \lt \frac12$: A short chord at $y=1-4r$ accompanies the longer (but no longer quite as long) chord which is at $y=1-2r.$ So $$L=4(\sqrt{r-r^2}+\sqrt{2r-4r^2 }).$$ Eventually, as Mirko notes, we arrive at $r=\frac13$ with equal chords at $y=\pm \frac13$ and $$L=\frac{8\sqrt{2}}{3} \approx 3.771236.$$ For somewhat larger $L$ values we can make make $|y|$ slightly smaller ( so those two chords become slightly longer) and leave a small bit over for chords at $\pm 3y.$ By my calculation, at $L=4$ that gives $$r=y=\frac{\sqrt{2\sqrt{73}-10}}8\approx 0.3327914.$$ However, on more careful analysis, chords at $1-2r,1-4r,1-6r$ for $r\approx 0.332253809$ is better. The exact value is the root of a certain degree $6$ polynomial. In your picture for $L \approx 5.084$ the vertical diameter helps somewhat but note how focussing on $r$ helps. Removing that vertical diameter but keeping the same $r$ necessitate moving the horizontal chords up slightly and adding another short one at the bottom. I (now) suspect that, for this $L$ and some larger ones, the pattern of chords at $1-2r,1-4r,1-6r$ remains optimal until $r=\frac14$ when $L=2+2\sqrt{3}\approx 5.4641.$ So with $L=5$ one gets $r \approx 0.299533.$ Somewhat more careful analysis: We will include the possibility of single point chords (called trivial) to account for disks tangent to the unit circle. However we will ignore them except when needed to have statements make sense. A chord with center at $(a,b)$ has length $2\sqrt{1-a^2-b^2}.$ We will pick a convenient orientation without comment. For example we will use horizontal chords where possible. A set of chords partitions the interior of the unit circle into connected regions. An $(L,r)-$configuration is a set of chords of combined length $L$ and disks all of radius $r$ with these properties: The interior of each disk is disjoint from the chords. No connected region can accommodate a disk of radius larger than $r.$ Each region which can contain a disk of radius $r$, actually does so. We call this an optimal configuration if there is no $(L,r')-$configuration with $r' \gt r.$ The stated problem is to find this maximal $r$ given $L.$ It seems easier to use the equivalent condition that there is no $(L',r)$ configuration with $L' \lt L$ and hence to minimize $L$ given $r.$ Of course the optimal configurations and $(L,r)$ pairs are the same. For example, as discussed above, for $\frac{1}{4} \le r \le \frac13$, chords at $y=1-2r,1-4r,1-6r$ give an $(L,r)-$configuration for $L=\,\sqrt {r-{r}^{2}}+4\,\sqrt {2\,r-4\,{r}^{2}}+4\,\sqrt {3\,r-9\,{r}^{2}}.$ This makes $r$ the larger real root of a degree $6$ polynomial $$(37748736+9437184L^2)r^6 +\cdots+L^8.$$ But that hardly seems helpful. I believe this is optimal. Each disk $D$ in an $(L,r)-$configuration (optimal or otherwise) is tangent to two or more chords. Call this set of chords $s(D).$ Our conditions require that the points of tangency with $s(D)$ must contain the center of $D$ is in their convex hull. Note that with $3$ or more chords in $s(D)$, both the center and radius are fixed. If $\frac12 \lt r \lt 1$ then there is just one disk,$D$. If $s(D)$ contains only two chords then they are $y=t$ and $y=t-2r$ for some $r \le t \le 1.$ A quick calculation shows that $t=1$ (only one non-trivial chord) is best. Note that if $D$ is tangent to the circle then this shortest non-trivial chord tangent to $D$ is all we need. I've convinced myself that at least three non-trivial cords in $s(D)$ is even worse. I won't spell out my laborious reasoning for this seemingly evident fact. So $L=2\sqrt{1-(1-2r)^2 }=4\sqrt{r-r^2}.$ For $\frac13 \lt r \lt \frac12$ more than one non-trivial chord will be needed. Two possible configurations are chords at $y=s$, $y=s-2r$ and $y=s-4r$ for some $1-2r \lt s \le 1$ with two regions allowing disks or chords at $y=s$ and $y=s-2r$ for some $r \lt s \lt 4r-1$ with only one region allowing disks. The first option with $s=1$ is best so this at least shows that $$L \le 4\sqrt{r-r^2}+4\sqrt{2r-4r^2}.$$ Again the only gap here is disposing of the case that some disk has three non-trivial chords in $s(D)$: If any disk has two chords in $s(D)$, both non-trivial, the given configuration is optimal. If two disks $D_1,D_2$ are tangent to the circle, there is no advantage to having more than one non-trivial cord in either $s(D_i).$ I won't say more about $\frac14 \lt r \lt \frac13$ than I already have. This is a good way from a proof, but at least it gives a direction. As a moral argument, anything else wastes some length by fixing both the center and radius of some disk.<|endoftext|> TITLE: Distribution of the permanent modulo $p$ QUESTION [6 upvotes]: We know that the order of $SL_n({\mathbb F}_p)$ is $$p^{n(n-1)/2}(p^n-1)(p^{n-1}-1)\cdots(p^2-1).$$ Dividing by $p^{n^2}$, we deduce the probability that $\det$ takes the value $1$ over $M_n({\mathbb F}_p)$. What if we consider instead the permanent~? What is the cardinal of the set of $n\times n$ matrices with entries in ${\mathbb F}_p$, whose permanent equals $1$ ? Of course, it is the same as if we replace $1$ by any non-zero element. A few comments: if $n=2$, the permanent of $A$ is the determinant of $A'$ where $a_{12}$ is replaced by $-a_{12}$. The distributions are therefore identical. if $p=2$, the permanent coincides with the determinant. The distributions are the same. the first non-trivial is therefore $p=n=3$, for which there are $3^9=19683$ matrices. With the permanent being notoriously hard to calculate, at least harder than the determinant, I did not try to figure out the distribution. If the answer is unknown, I'll try to write a code. REPLY [6 votes]: $\def\char{\mathop{\rm char}}$In this paper by M. Budrevich and A. Guterman prove that for every $F_q$ with $\char F_q>2$ there are more matrices of order $n\geq 3$ with zero determinant than those with zero permanent. In fact, their estimates can be strengthened to show the following. If $D(M_n(F_q))$ and $P(M_n(F_q))$ are the numbers of the corresponding matrices, then $$ D(M_n(F_q))-P(M_n(F_q))\geq (q-1)\prod_{k=3}^n(q^n-q^{k-1}). $$ (This is shown in Budrevich's dissertation.)<|endoftext|> TITLE: Every free abelian group is slender, why? QUESTION [7 upvotes]: Wikipedia states that every free abelian group is slender. Where can I find a proof? If this is not trivial, then I will also need a reference to use in my paper. REPLY [8 votes]: This follows fairly straightforwardly from the fact that every map from $A=\mathbb{Z}^\mathbb{N}$ to $\mathbb{Z}$ factors through a finite subproduct (let me call this "Specker's theorem"). Suppose $F$ is a free abelian group and $f:A\to F$ is a homomorphism. Since a subgroup of a free abelian group is free, we may assume $f$ is surjective. If $F$ is finitely generated, the conclusion follows from Specker's theorem, so we may assume $F$ is infinitely generated. Since $F$ is free, the map $f$ splits and so $F$ is a direct summand of $A$. But there are uncountably many homomorphisms $F\to\mathbb{Z}$, and hence uncountably many homomorphisms $A\to \mathbb{Z}$. This contradicts Specker's theorem.<|endoftext|> TITLE: Bounding from below the cardinality of a set of generators of the $n$-fold cartesian product of a finite group QUESTION [6 upvotes]: Let $G$ be a nontrivial finite group. Given $n\in\mathbb{Z}_{\geq 1}$, let $G^n$ be the cartesian product of $n$ copies of $G$. Further let $S\subseteq G^n$ be a generating set of $G^n$. Question: Do we always have $|S|\geq n$? REPLY [10 votes]: For an alternative proof of Wiegold's results, and in particular for finite groups G such that the number of generators of G^n is comparable to log n, see Section 3 in Jacques Thévenaz, Maximal subgroups of direct product groups, Journal of Algebra 198 (1997) 352-361. Pierre de la Harpe.<|endoftext|> TITLE: References for $K_{4k}(\mathbb{Z})$ QUESTION [23 upvotes]: Weibel's "Algebraic K-theory of rings of integers in local and global fields" says $K_{4k}(\mathbb{Z})$ are known to have odd order, with no prime factors less than $10^7$, but are conjectured to be zero. Since the paper is from 2004 I was wondering if anything new is known about those groups. I'm interested on any new research on them. REPLY [5 votes]: $K_8(\mathbb{Z})=0$ A proof of the result announced by Elbaz-Vincent and mentioned above can now be found in: Mathieu Dutour-Sikirić, Philippe Elbaz-Vincent, Alexander Kupers, and Jacques Martinet, Voronoi complexes in higher dimensions, cohomology of $GL_N(\mathbb{Z})$ for $N\geq8$ and the triviality of $K_8(\mathbb{Z})$, arXiv:1910.11598v1 (submitted on October 25, 2019) A shorter proof, which (unlike the one above) does not require additional computer calculations, is presented in: Alexander Kupers, A short proof that $K_8(\mathbb{Z})=0$, http://people.math.harvard.edu/~kupers/files/k8zshorter.pdf (version of October 21, 2019) The very last remark in Kupers’s short note addresses the case of $K_{12}(\mathbb{Z})$: Remark 2.6. If Conjecture 2 of [CFP14] were true with coefficients in $\mathbb{Z}[1/((n+1)!)]$ instead of $\mathbb{Q}$, it could be used to prove that $K_{12}(\mathbb{Z})=0$ in a similar manner. [CFP14] Thomas Church, Benson Farb, and Andrew Putman, A stability conjecture for the unstable cohomology of $SL_n(\mathbb{Z})$, mapping class groups, and $Aut(F_n)$, Algebraic topology: applications and new directions, 55–70, Contemp. Math., vol. 620, Amer. Math. Soc., 2014, MR3290086<|endoftext|> TITLE: Is there a category whose isomorphisms are precisely the simple homotopy equivalences? QUESTION [14 upvotes]: If we start with the category of finite complexes and continuous maps, and then identify two morphisms iff they are homotopic, we get the homotopy category of finite complexes, and it is trivial to observe that a continuous map $f:X\to Y$ is an isomorphism in this category iff it is a homotopy equivalence. Does there exist an equivalence relation on continuous maps (necessarily finer than that of being homotopic) compatible with composition such that the resulting quotient category has the property that a map $f:X\to Y$ is an isomorphism iff it is a simple homotopy equivalence? I would also be interested in an answer to the following variant, which may be slightly easier. Fix a group $\pi$, and consider the category of bounded complexes of finite-dimensional free $\mathbb Z[\pi]$-modules. Again, we may identify morphisms iff they are chain homotopic, and the resulting homotopy category has the property that a chain map $f:A_\bullet\to B_\bullet$ is an isomorphism iff it is a chain homotopy equivalence. Is there a finer equivalence relation on morphisms such that in the resulting quotient category, a chain map is an isomorphism iff it is a chain homotopy equivalence and has vanishing Whitehead torsion? REPLY [13 votes]: Unfortunately not. Let's say $D$ is a category and $F$ is a functor from finite complexes to $D$ that takes simple homotopy equivalences to isomorphisms. We note that for any finite complex $X$, the inclusion $i_0: X \to [0,1] \times X$ is a simple homotopy equivalence. We can factor it as a composition of elementary expansions, one cell $e^n$ at a time, by inductively gluing in $[0,1] \times e^n \cong D^{n+1}$ along $[0,1] \times \partial e^n \cup \{0\} \times e^n \cong D^{n}_{-}$. Therefore, $F(i_0)$ is an isomorphism. This fact, all by itself, forces $F$ to factor through the homotopy category. This is a fairly common manipulation; here is how it goes. If we consider the other inclusion $i_1: X \to [0,1] \times X$ and the projection $p: [0,1] \times X \to X$, we have $$id = F(p) F(i_0) = F(p) F(i_1)$$ which implies first that $F(p) = F(i_0)^{-1}$ and then that $F(i_1) = F(p)^{-1}$, which forces $F(i_0) = F(i_1)$. If $H: [0,1] \times X \to Y$ is a homotopy between two maps $f$ and $g$, then $$ F(f) = F(H) F(i_0) = F(H) F(i_1) = F(g) $$ so $F$ factors through the homotopy category, as desired.<|endoftext|> TITLE: Soft and hard part of geometry QUESTION [18 upvotes]: While listening to some lecture of Alain Connes about noncommutative geometry, he spoke about various generalizations of the classical concepts from geometry and divided it into "soft" and "hard" part. The audience seemed to know what is all about but for me it was not clear what is the distinction. In particular he mentioned vector bundles, differential forms, connections and curvatures and characteristic classes as being a "soft" part, while Riemannian metric, Riemannian curvature and Pontryagin classes was classified as the "hard" part. The rather vaque question would be: what is the criterion of being soft/hard part of geometry? More particular question: Why Pontryagin classes are part of hard geometry and the other characteristic classes are soft? REPLY [13 votes]: In the context of geometry, a distinction between "soft" and "hard" was introduced by Gromov, as explained here and applied to Soft and Hard Symplectic Geometry. In Gromov's words, 'hard' refers to a strong and rigid structure of a given object, while 'soft' suggests some weak general property of a vast class of objects. Riemannian geometry is hard, while symplectic geometry is soft "because all symplectic forms are locally diffeomorphic". (There is no analog of curvature in a symplectic structure.) The "hard" part of a symplectic structure is Gromov's non-squeezing theorem. Local invariants are "hard", which is presumably why one calls the Pontryagin class "hard". In the context of analysis, the distinction between "soft" and "hard" seems to be more widely used and more precise, as explained in this blog, with a table of correspondences:<|endoftext|> TITLE: Removing constraints in convex optimization QUESTION [9 upvotes]: Say I have a convex optimization problem of the form $$\min_x f(x) ~~ s.t.\\ g_1(x)\leq0,\\\vdots \\g_n(x)\leq 0$$ with all functions convex. Suppose that $x^*$ is a unique optimizer to my problem and that we have $g_i(x^*)=0$ for some index $i$ and that $g_j(x^*)<0$ for all other indices. Then, it is well-understood that $x^*$ is also a solution to the problem $$\min_x f(x) ~~ s.t.\\ g_i(x)\leq0~.$$ in other words, the $(n-1)$ inactive constraints did not contribute anything to the problem. My question is: does anything change if, instead of having $n$ constraint functions, I have a continuum of functions? That is, say that I now have the problem $$\min_x f(x) ~~ s.t.\\ g_t(x)\leq0 ~ \forall t\in[0,1] ~ $$ for some parameterized family of functions $g_t$, and I can again show that there is a unique minimizer $x^*$ where there is a single value of $t^*$ such that $g_{t^*}(x^*)=0$ and $g_{t}(x^*)<0$ otherwise. Are there any extra precautions I need to take before claiming that it will suffice to consider the simpler problem $$\min_x f(x) ~~ s.t.\\ g_{t^*}(x)\leq0 ~ $$? My gut tells me no, but strange things are known to happen at infinity. REPLY [4 votes]: Part of the reason this works for a finite set of constraints, and in particular the reason that the simplified problem still has a unique minimizer, is that $x^*$ is an interior point of each of the regions $g_j(x) \lt 0,\; j \ne i$ and therefore is an interior point of their intersection. This does not work if there are infinitely many constraints, and you cannot rely on uniqueness. For a simple example consider minimizing $x_2$ under the constraint $x_1^2 - x_2 \le 0$. Clearly the unique minimizer is $(0, 0)$. We can however replace the single strictly convex constraint by an equivalent family of linear constraints $g_t(x) = 2tx_1 - x_2 - t^2 \le 0,\; t \in \mathbb{R}$. The only active constraint is then $g_0(x) \le 0$, but if we remove all the other constraints, the (over)simplified problem has every $(x_1, 0),\; x_1 \in \mathbb{R}$ as a minimizer.<|endoftext|> TITLE: Term for "uncheckable constructions" QUESTION [30 upvotes]: Is there a term for "uncheckable geometric constructions"? Say, Angle Trisection and Doubling the Cube are checkable; i.e., if the answer is given one can do finite Compass-and-straightedge construction which checks that this is a right answer. (We assume that we can see if two points coincide.) On the other hand Squaring the Circle is not checkable --- there is no way to say "yes" by performing a finite Compass-and-straightedge construction. Similarly one can not check that a given point is a center of given circle by straightedge only. (Here I assume that straightedge can not draw infinite lines, so one can not check whether two lines are parallel.) REPLY [5 votes]: Quick claims about Joel's definition: Constructibly decidable, semi-decidable, and co-semi-decidable are identical for constructions which, like most classic geometric constructions, take bounded numbers of steps. (By the same proof that those decidable, semi-decidable, and co-semi-decidable are identical with a bound on the computation time.) A constructible semi-decidable problem is exactly a countable disjunction of bounded constructible decision problems, and similarly for co-semi-decidable and conjunction. (Assuming there is no computably criterion in the definition of a geometric construction that takes unbounded time.) We can classify the constructibly decidable in bounded time problems using the Euclidean geometry - algebra correspondence. Representing each point by a pair of numbers, any geometric input can be represented as a tuple of numbers. The constructible functions are exactly the compositions of rational functions and square roots. To get a yes-or-no value rather than a number as the result of our computations, we need a test. We can test whether two points are equal, so we can test equality, and we can test whether a line is on a circle, so we can test inequality. Any finite decision procedure is thus described by a logical combination of formulas where you take a function generated by rational functions and square roots and test whether it is positive. Such a decision procedure is clearly a first-order statement in the theory of real closed fields (if we take a real closure of our number line). By quantifier elimination, every first-order statement in the theory of real closed fields is given by a decision procedure, and you don't need to mess with square roots. So a constructibly decidable problem is a partition of $\mathbb R^n$ into two sets, each a countable union of first order definable sets. Need such a partition be a partition into two first order definable sets? No. For instance the set of real numbers whose integer part is even is a countable union of the intervals $[2n,2n+1)$, which are first order definable, but it is not definable, and similarly for its complement.<|endoftext|> TITLE: Representation theory of the general linear group over a finite prime field QUESTION [11 upvotes]: I am re-posting a question I asked on math.se here because I am unsatisfied with the answers I obtained. The irreducible modules of $\operatorname{GL}_n(\mathbb C)$ over $\mathbb C$ are completely classified and well-understood via Schur-Weyl duality, the algebraic Peter-Weyl theorem and the entire theory of reductive groups in characteristic zero. I am looking for a good reference on what is known about the representation theory of $\operatorname{GL}_n(\mathbb F_p)$ over $\mathbb F_p$, i.e. the study of $\mathbb F_p$-vector spaces with an action of $\operatorname{GL}_n(\mathbb F_p)$. Here, $\mathbb F_p$ is the field with $p$ elements, $p$ a prime number. Are the irreducible modules completely classified? Can they somehow be indexed by certain partitions, similar to the characteristic zero case? I am particularly interested in whether or not there is some equivalent of the Pieri rule, i.e. decomposing the tensor product of an irreducible representation with a symmetric power of $\mathbb F_p^n$. However, I suppose that question only makes sense when it is possible to classify irreducibles in some combinatorial way. REPLY [2 votes]: I would suggest James-Kerber: Representation Theory of Symmetric Groups (Encyclopedia of Mathematics and its Applications, vol.16) Addison-Wesley, 1981 The last chapter deals with the modular representation of general linear groups, and in particular, Exercise 8.4 gives the construction of all irreducible $\mathbb{F}_p[GL_n(\mathbb{F}_p)]$-modules.<|endoftext|> TITLE: Closure of the orbits of the $SL(2,\mathbb{Z})$-action on $\mathbb{R}^2$ QUESTION [12 upvotes]: I'm coming with a very basic question for which I can't find an answer. Please forgive me if I didn't search efficiently enough. What can the closure of an orbit of an element $X$ of $\mathbb{R}^2$ under the action of $SL(2,\mathbb{Z}) $ look like ? 1) Obviously if the vector space spanned by the coordinates of X has dimension $1$, $SL(2,\mathbb{Z}) \cdot X$ is a lattice. 2) The $SL(2,\mathbb{Z}) $-action is known to be ergodic on $\mathbb{R}^2$, which implies that there is a set of full measure of which every element has a dense orbit. Is this set exactly the remaining cases ? Namely when the coordinates of $X$ are $\mathbb{Q}$-linearly independent ? This reduces to the question : If $\theta \notin \mathbb{Q}$, is $SL(2,\mathbb{Z}) \cdot (1, \theta) $ dense in $\mathbb{R}^2$ ? 3) Is the question solved in the case of the $SL(n,\mathbb{Z}) $ action on $\mathbb{R}^n$ ? Any good reference is welcome ! REPLY [6 votes]: Another interpretation of this fact is as follows. Consider the unit tangent bundle $PSL(2,\mathbb{Z})\backslash PSL(2,\mathbb{R})$ of the modular surface $PSL(2,\mathbb{Z}) \backslash \mathbb{H}^2$. The horocyclic flow has two kinds of orbits: periodic orbits turning around the cusp, and dense orbits. This is known since Hedlund's work in the 30's at least. Moreover, all dense orbits are equidistributed towards the Liouville measure, which is the unique horocyclic-invariant ergodic probability measure of full support. (Results of Dani and Dani-Smillie at the end of the 70's) And the space $\mathbb{R}^2\setminus\{0\}$ identifies naturally with the space of horocycles, and the action of $PSL(2,\mathbb{Z})$ by isometries on the horocycles becomes the usual linear action of $SL(2,\mathbb{Z})$ on $\mathbb{R}^2$. Thus, a point in $\mathbb{R}^2$ coming from a horocycle centered at a rational point (or equivalently turning around the cusp) has a $SL(2,\mathbb{Z})$-orbit which is discrete, whereas the other points have dense orbits (equidistributed towards a measure which is not exactly Lebesgue on $\mathbb{R}^2$, but $dr d\theta$ in polar coordinates) Of course with this point of view, the good group is not $SL(2,\mathbb{Z})$ but $PSL(2,\mathbb{Z})$... Barbara<|endoftext|> TITLE: What does an endomorphism in a triangulated category give rise to? QUESTION [6 upvotes]: Let $D\xrightarrow[]\varphi D\xrightarrow[]kE\xrightarrow[]j\Sigma D$ be an exact triangle in a triangulated category. I am trying to figure out what structure emerges from this on the base of the axioms. For example, the octahedron axiom gives an exact triangle $E\to E^{(2)}\to E\xrightarrow[]{\Sigma k\circ j}\Sigma E$, where $E^{(2)}$ is from the exact triangle $D\xrightarrow[]{\varphi\circ\varphi}D\xrightarrow[]{}E^{(2)}\to\Sigma D$. All this resembles a beginning of a spectral sequence obtained from an exact couple (e. g. obviously the composite $d:E\xrightarrow[]j\Sigma D\xrightarrow[]{\Sigma k}\Sigma E$ satisfies $\Sigma d\circ d=0$) but it quickly becomes too entangled for me. Can anyone give me a reference for analyzing such situations? At least in simpler situations when e. g. $\varphi\circ\varphi$ is $\varphi$ or zero or invertible? Or on the other hand in more general situations when $\varphi$ has nonzero degree (i. e. $\varphi:\Sigma^nD\to D$ for any $n\in\mathbb Z$)? I have vague feeling that something like this takes place in connection with the Devinatz-Hopkins-Smith theorem, and that Bondal-Kapranov's twisted complexes might be relevant, but cannot come up with anything definite... REPLY [5 votes]: Here's an expanded version of my comment, addressing the spectral sequence part of the question. A filtered object in a triangulated category $T$ is simply a sequence $$ \dots \to X_{n-1} \to X_n \to X_{n+1} \to \dots $$ When you apply a homological functor $H : T\to A$, where $A$ is an abelian category, filtered objects in $T$ become bigraded spectral sequences in $A$. The construction of these spectral sequences uses nothing more than the triangulated structure (but you need more structure to make them functorial in the filtered object). In fact, the spectral sequence arises from the exact couple $$ H_*(X_*) \to H_*(X_*) \to H_*(C_*) \to H_*(X_*),$$ where $C_n=\operatorname{cofib}(X_{n-1}\to X_n)$ and $H_*=H\circ \Sigma^{-*}$. All this is described in much detail here, where some references are given. The usual convergence result is the following. Suppose that $T$ and $A$ have enough sequential (homotopy) colimits and that: for every $r$, $H$ preserves the colimit of $n\mapsto \operatorname{cofib}(X_r\to X_{r+n})$, for every $r$, $H_r(X_n)=0$ for $n\ll 0$. Then the spectral sequence converges strongly to $H_*(\operatorname{hocolim}_n X_n)$. Note that this applies to $H^{op}: T^{op}\to A^{op}$ as well, which gives a dual statement with limits. Now, given an endomorphism $\phi\colon D\to\Sigma^d D$, I can think of two ways to get a filtered object: (1) Just repeat the map $\phi$, with $X_n= \Sigma^{nd}D$, (2) Let $X_0=0$, $X_{1}=\operatorname{fib}(\phi)$, $X_2=\operatorname{fib}(\phi^2)$, etc. The potential targets are: (1) $H_*$ of the colim-inversion of $\phi$, (2) $H_*$ of the "$\phi$-primary torsion" in $D$. In the dual story, we'll have potential targets: (1') $H_*$ of the lim-inversion of $\phi$, (2') $H_*$ of the "$\phi$-completion" of $D$. I don't know if the spectral sequence (1) is useful. It converges strongly if $H$ preserves sequential colimits, $d<0$, and $H_*(D)$ is bounded below. It also converges trivially if $\phi$ has order $n$, the $(n+1)$st page being zero. But if $\phi$ is idempotent, the second page is zero and convergence fails completely. The spectral sequence (2) is a Bockstein spectral sequence.<|endoftext|> TITLE: Does this equation has a closed-form solution for $t$? ($(1-p)\sum_{i=0}^{n}t^i = p\sum_{i=0}^{n}(1-t)^i)$) QUESTION [5 upvotes]: We are given $n\in \mathbb N^+$ and $p\in[\frac{1}{2},\frac{n+1}{n+2}]$. Our goal is to find $t\in[0,1]$ such that $$(1-p)\sum_{i=0}^{n}t^i = p\sum_{i=0}^{n}(1-t)^i$$ Is there a closed-form solution $t(n,p)$? How about a close formula for some non-trivial $p$, e.g. for $g(n)\triangleq t(n,0.6)$? A few observations: This is equivalent, for $t\neq0,1$, to: $$(1-p)t(1-t^{n+1})=p(1-t)(1-(1-t)^{n+1})$$ $\forall n:t(n,\frac{1}{2})=\frac{1}{2}$ $\forall p:t(1,p)=3p-1$ $\forall n:t(n,\frac{n+1}{n+2})=1$ $\forall n:t(n,p)$ is monotonically increasing in $p$. $\forall p:t(2,p)=\frac{2 p + 1-\sqrt{-3+28 p-28 p^2}}{4 p - 2}$ If this is not possible, is it possible to bound it with simple function? e.g. I think I showed $$p\leq t(n,p)\leq \frac{(n+2)p - 1}{n}$$ Which works great for large $n$, but not so much for small values. Can we give tighter bound for $t$? REPLY [2 votes]: You might find this useful. Write the equation as $$ p = \dfrac{t (t^{n+1}-1)}{(1-t)^{n+2} + t^{n+2}-1} = \dfrac{(1-s)((1-s)^{n+1}-1)}{s^{n+2}+(1-s)^{n+2}-1}$$ where $s = 1-t$. If $n \ge 3$, I get $$ p = \dfrac{n+1}{n+2} - \dfrac{(n+1)}{2(n+2)} s - \dfrac{(n^2+4n+3)}{12(n+2)} s^2 + \ldots $$ so we should have $$ t > \dfrac{2(n+2)p}{n+1} - 1 $$ for $p$ near $(n+1)/(n+2)$. Indeed this appears to hold for $1/2 \le p \le 1$ and all $n \ge 1$.<|endoftext|> TITLE: In what sense is the Bayesian posterior mean a “convex combination”? QUESTION [9 upvotes]: I asked this on math.stackexchange with no response, I'm hoping someone here might have something. Suppose I want to estimate $x \in \mathbb{R}^n$ from two signals with zero mean, normally distributed noise: $y_1 = x + \epsilon_1$ and $y_2 = x + \epsilon_2$, where $\epsilon_1 \sim N(0, \Sigma_1)$, $\epsilon_2 \sim N(0, \Sigma_2)$. Using Bayesian estimation, the posterior mean is $\hat{x} = (\Sigma_1^{-1} + \Sigma_2^{-1})^{-1} (\Sigma_1^{-1} y_1 + \Sigma_2^{-1} y_2) $, which is a "convex combination" of the data points $y_1$ and $y_2$. In the univariate case, the posterior mean is a convex combination of the data points in the usual sense. In the multivariate case, the "weights" are matrices that add up to $I$. What can we say about the posterior mean (e.g. is the set of possible $\hat{x}$ convex and in what sense, what are the extreme points, how does it vary with the parameters, etc)? A search for "matrix convex combination" gives this result: http://www.math.uni-sb.de/ag/wittstock/OperatorSpace/node73.html which seems to be talking about something different. REPLY [3 votes]: Answering my own question: the posterior mean will lie on the locus of tangency between the ellipsoids centered at $y_1, y_2$ and with shape matrices $\Sigma_1^{-1}, \Sigma_2^{-1}$ respectively. See Chamberlain & Leamer (1976), "Matrix Weighted Averages and Posterior Bounds," Journal of the Royal Statistical Society. Series B (Methodological), 38(1), 73-84 for more.<|endoftext|> TITLE: Isomorphisms between spaces of test functions and sequence spaces QUESTION [9 upvotes]: I am in the process of writing some self-contained notes on probability theory in spaces of distributions, for the purposes of statistical mechanics and quantum field theory. Perhaps the simplest approach is to follow Barry Simon's philosophy in his book "Functional Integration and Quantum Physics", namely, evacuate the complexity of spaces of test functions right from the start and replace them by more concrete spaces of sequences to which they are isomorphic as topological vector spaces. For instance the Schwartz space $S(\mathbb{R}^d)$ is isomorphic to the space $\mathcal{s}$ of rapidly decaying sequences. A very nice proof of this fact using Hermite functions can be found for instance here. For applications to QFT it is also important to consider the space $S(\mathbb{Q}_p^d)$ of locally constant compactly supported functions over a $p$-adic field. This allows one to work with hierarchical models as a toy model for Euclidean models. This hierarchical simplification is similar to the use of branching random walks or Mandelbrot cascades as a toy model for the massless free field (see for instance this article by Bramson and Zeitouni or this one by Benjamini and Schramm) in probability theory or that of Walsh series as a toy model for Fourier series in harmonic analysis (see for instance this article by Do and Lacey in the hierarchical case and this one by Lie for the Euclidean case). The relevant space $S(\mathbb{Q}_p^d)$ is easily seen to be isomorphic to the space $\mathcal{s}_0=\mathbb{R}^{(\mathbb{N})}$ of almost finite sequences with the finest locally convex topology. My question concerns the existence of similar isomorphism theorems for more complicated spaces of test functions. Let $\Omega$ be a nonempty open set in $\mathbb{R}^d$ and let $\mathcal{D}(\Omega)$ be the usual space of smooth compactly supported functions with support contained in $\Omega$. Let $S(\mathbb{A}_{\mathbb{Q}})$ and $S(\mathbb{A}_{\mathbb{Q}}^{\times})$ be the spaces of Schwartz-Bruhat functions respectively over the adeles and ideles of $\mathbb{Q}$. Finally let $\mathcal{s}^{(\mathbb{N})}$ be the space of almost finite sequences of elements in $\mathcal{s}$. The latter is equipped with the locally convex topology defined by all semi-norms which are continuous when restricted to each summand $\mathcal{s}$. I suspect the following is true: $$ \mathcal{D}(\Omega)\simeq S(\mathbb{A}_{\mathbb{Q}}) \simeq S(\mathbb{A}_{\mathbb{Q}}^{\times}) \simeq \mathcal{s}^{(\mathbb{N})} $$ as topological vector spaces. Question 1: is this true or not and if it is where can I find a proof? Question 2: is the space $\mathcal{s}^{(\mathbb{N})}$ sequential? Edit 1: The references in the answer by P. Michor were spot-on regarding the isomorphism $\mathcal{D}(\Omega)\simeq \mathcal{s}^{(\mathbb{N})}$ which now I know is true. However I would prefer a more direct proof (which does not use the result by Pełczyński about spaces being isomorphic to a complemented subspace of the other). An ideal proof for me would be via the use of something like a wavelet basis $\psi_{p,x}$ where $p$ would be the scale index and $x$ the spatial location index. Intuitively, for $\mathcal{D}(\Omega)$ one has to be almost finite in the $x$ direction and suitably decaying in the $p$ direction hence a space of the form $\mathcal{s}\otimes \mathcal{s}_0$. For $S(\mathbb{R}^d)$ one would have decay in both directions resulting in $\mathcal{s}\otimes \mathcal{s}\simeq \mathcal{s}$. As for the second isomorphism, it is rather clear how to construct a linear isomorphism. So the question really is: what is the "standard" topology on $S(\mathbb{A}_{\mathbb{Q}})$ and $S(\mathbb{A}_{\mathbb{Q}}^{\times})$? Number theory references I looked at tend to dodge the question altogether. Edit 2: A possible idea to cover the case of general open sets $\Omega$ is to use some diffeomorphisms in order to relate them to $\mathbb{R}^d$. However, this MO question does not bode well for such an approach. REPLY [4 votes]: A construction for the isomorphism not using the Pełczyński decomposition method can be found in C. Bargetz: Explicit representations of spaces of smooth functions and distributions. J. Math. Anal. Appl. 424: 149–1505. 2015 DOI: 10.1016/j.jmaa.2014.12.009 C. Bargetz: Commutativity of the Valdivia–Vogt table of representations of function spaces. Mathematische Nachrichten 287(1): 10–22, 2014, DOI: 10.1002/mana.201200258. I do not think that the basis constructed in the first of the above papers really has the properties you would like to have. The construction of the isomorphism on the other hand consists of a series of relatively simple steps and it is completely constructive. More precisely, in the one dimensional case, the construction works as follows: First, using Seeley's extension operator, the function is "split" into a sequence of smooth functions on the unit interval $[0,1]$ which together with all derivatives vanish at the right boundary point $1$. Then, using an explicit formula each of these smooth functions is mapped to an element of $s$. Some topological properties of $s^{(\mathbb{N})}$ and related spaces are dicussed in C. Bargetz: Completing the Valdivia–Vogt tables of sequence-space representations of spaces of smooth functions and distributions. Monatsh. Math. 177(1): 1–14, 2015. DOI: 10.1007/s00605-014-0650-2<|endoftext|> TITLE: When was Bounded Zermelo set theory first formulated? QUESTION [9 upvotes]: Bounded Zermelo set theory, and many variants named for MacLane in some way, are used in equiconsistency proofs for Simple Theory of Types plus infinity, and for the Elementary Theory of the Category of Sets starting in 1969 (see Finite order arithmetic and ETCS). But did anyone formulate it before that? Of course Bounded Zermelo could easily have been stated before that, and the analogy with Primitive Recursive Arithmetic could have suggested it. But was it formulated earlier in fact? REPLY [8 votes]: The Princeton thesis of John Kemeny, written in 1949, was devoted to the relation between Zermelo set theory and type theory. For example, early in the thesis, there is a proof of the consistency of the simple theory of types relative to the consistency of a small fragment (nowadays known as KF) of Bounded Zermelo set theory. The comprehension scheme in KF is limited to stratifiable instances of bounded comprehension. KF was explicitly introduced and studied by Kaye and Forster in their 1991 paper in JSL entitled "End-extensions Preserving Power Set". For more detail, see subsection 8.0 of Mathias' Strength of Mac Lane Set Theory (APAL, 2001); indeed Mathias further discusses Kemeny's thesis in detail towards the end of section 8 of his paper. Based on Mathias' account, it is clear that, conceptually, Kemeny "knew about" bounded Zermelo set theory, but it is not clear to me whether he explicitly formulated any fragments of Zermelo set theory.<|endoftext|> TITLE: Inverted pair of complex analytic families QUESTION [6 upvotes]: I read the following "problem" in an old set of notes of Morrow and Kodaira which focused on deformations of complex manifolds: Find a pair of complex analytic families $\lbrace M_t\rbrace$ and $\lbrace N_t\rbrace$ with $|t|<1$ such that $M_t=N_0$ for $t\ne 0$ and $N_t=M_0$ for $t\ne0$, with $M_0\ne N_0$. (Every manifold assumed compact.) As of 1971 there were no known examples. Up to date, is this possibly true? Naively it seems impossible. In the spirit of Lefschetz fibrations, we have $M_0$ as a critical level set amongst the $M_t$ (thinking of cylinders which get pinched via a vanishing cycle). It is then not generic, and any perturbation of this fiber (locally about the nodal point) would alter the topology. So to me the inverted family $\lbrace N_t\rbrace$ couldn't exist, because it would force a generic $M_{t_0}$ to take over the role as the critical level set amongst the $N_t$, but there is nothing "non-regular" about it. If the manifolds are allowed to be noncompact, perhaps it's easy, again via a rough sketch: take $\lbrace M_t\rbrace$ to be the family of open cylinders which pinch the vanishing cycles at $t=0$, and $\lbrace N_t\rbrace$ to be the family of "pinched" cylinders where the pinching point pushes off to infinity as $|t|\to 0$. Edit: The problem is more subtle than I thought, and my example with Lefschetz fibrations doesn't apply. From the definition of a "complex analytic family", all $M_t$ are diffeomorphic to each other (in particular, they're all nonsingular spaces), and so we can only distinguish them by their holomorphic structures. In other words, $M_0\ne N_0$ means diffeomorphic but not complex-analytically homeomorphic. REPLY [2 votes]: I'm going to assume that your objects are compact analytic spaces and show that then the situation you describe cannot arise. I don't have access to Morrow and Kodaira and cannot remember whether the "problem" concerns only compact manifolds. Suppose that $\mathcal M\to S$ is a miniversal deformation of $M_0$. Then the given family $\{M_t\}$ defines a morphism $\mu:\Delta\to S$, where $\Delta$ is a complex disc. Since $M_t=N_0\ne M_0$, the map $\mu$ is non-constant and $\mu(\Delta-\{0\})$ is contained in the stratum of $S$ that consists of points $s$ such that $\mathcal M_s$ is isomorphic to $N_0$. Now this stratum is smooth and its dimension, which we have just proved to be at least $1$, equals $\dim Aut_{M_0}-\dim Aut_{N_0}$; one reference for this, in the algebraic context (but in the current analytic context the result will be the same), is Theorem 2.7 of arXiv:1210.0342 by Hyland, Ekedahl and Shepherd-Barron. (In positive or mixed characteristic, where group schemes need not be reduced, this dimension becomes $\dim Lie(Aut_{M_0})-\dim Aut_{N_0}$.) So $\dim Aut_{N_0}<\dim Aut_{M_0}$. From the assumptions, the opposite inequality is also valid, and we have a contradiction.<|endoftext|> TITLE: Picard of the product of two curves QUESTION [11 upvotes]: Can anyone point to me where I can find the proof that the Picard group of the product of two curves is isomorphic to the product of the Picard groups times the hom among the Jacobians? Does the result work on an arbitrary field? Thank you very much for your time and attention REPLY [12 votes]: It seems likely that you are assuming the curves are smooth and geometrically connected (e.g., you don't have in mind generalized Jacobians for singular curves), but you have omitted hypotheses on the ground field and have not indicated if you are assuming the existence of rational points. Let $X$ and $Y$ be geometrically integral proper schemes over a field $k$. First assume there exist $x_0 \in X(k)$ and $y_0 \in Y(k)$, so the relative Picard functors ${\rm{Pic}}_{X/k}$ and ${\rm{Pic}}_{Y/k}$ respectively classify isomorphism classes of line bundles on $X$ and $Y$ trivialized along $x_0$ and $y_0$. Let $L$ be a line bundle on $X \times Y$, and define the line bundles $L_1 = (1_X \times y_0)^{\ast}(L)$ on $X$ and $L_2 = (x_0 \times 1_Y)^{\ast}(L)$ on $Y$, so the line bundle $$L' := L \otimes (p_X)^{\ast}(L_1)^{-1} \otimes (p_Y)^{\ast}(L_2)^{-1}$$ has trivial pullbacks along $x_0 \times Y$ and $X \times y_0$. Hence, $L'$ as line bundle on $X \times Y$ trivial along $x_0 \times Y$ is classified by a $k$-morphism $f_L:Y \rightarrow {\rm{Pic}}_{X/k}$ carrying $y_0$ to $0$. This map lands inside ${\rm{Pic}}^0_{X/k}$, and it is clear that $f_{L \otimes M} = f_L + f_M$ for $L$ and $M$ on $X \times Y$. Assume now that $X$ and $Y$ are curves, so by the Albanese property of Jacobians we see that $f$ corresponds to a $k$-homomorphism $J_Y \rightarrow J_X$. This defines a homomorphism $${\rm{Pic}}(X \times Y) \rightarrow {\rm{Mor}}_k((Y,y_0), (J_X,0)) = {\rm{Hom}}_k(J_Y, J_X)$$ that admits a natural homomorphic section via the Picard functorialty of $J_X$ with respect to $(X, x_0)$ (essentially run parts of this construction in reverse). The seesaw theorem implies that the kernel of this surjective map consists of exactly those $L$ for which $L'$ is trivial, which is to say it is the image of the visibly injective map $${\rm{Pic}}(X) \times {\rm{Pic}}(Y) \rightarrow {\rm{Pic}}(X \times Y).$$ Hence, we have built the desired isomorphism, but our construction rests on the choices of $x_0$ and $y_0$. If these construction are carried out in a more careful scheme-theoretic manner over $k$-algebras uses the schematic form of the seesaw theorem, and keeps track of trivializations then this argument provides a left exact sequence of $k$-group schemes $$0 \rightarrow {\rm{Pic}}_{X/k} \times {\rm{Pic}}_{Y/k} \stackrel{j}{\rightarrow} {\rm{Pic}}_{(X \times Y)/k} \rightarrow \underline{\rm{Hom}}(J_Y,J_X)$$ where the final term of the \'etale Hom-scheme over $k$ and the induced map on geometric points is short exact. But the initial inclusion $j$ between smooth $k$-groups locally of finite type induces on Lie algebras the natural map ${\rm{H}}^1(X,O_X) \oplus {\rm{H}}^1(Y,O_Y) \rightarrow {\rm{H}}^1(X \times Y, O_{X \times Y})$ that is an isomorphism, so ${\rm{coker}}(j)$ is \'etale. Consideration of geometric points then shows that ${\rm{coker}}(j)$ maps isomorphically onto the Hom-scheme, so the left-exact sequence of $k$-groups is short exact. What happens to the identification of the cokernel of $j$ with the Hom-scheme if we change $(x_0,y_0)$, using the identification of Jacobian as dual to Picard scheme (thereby suppressing its dependence on the base point, via Grothendieck's definition of the Picard functor through sheafification)? Since the entire construction can be carried out using sections after base change over any $k$-scheme, the dependence on $(x_0,y_0)$ is classified by a $k$-scheme morphism $$X \times Y \rightarrow \underline{\rm{Isom}}({\rm{coker}}(j), \underline{\rm{Hom}}(J_Y,J_X))$$ which must be constant since the target is etale. Hence, the identification of the cokernel of $j$ with the Hom-scheme is independent of the choice of $(x_0,y_0)$, so in general without assuming such $k$-points to exist we can carry out the construction over a finite Galois extension of $k$ where rational points can be chosen and then use Galois descent to bring it down to $k$ (independently of all choices). This solves the original question in the form of a short exact sequence of smooth $k$-groups which is split-exact (hence also split on $k$-points) when $(x_0, y_0) \in (X \times Y)(k)$ is chosen. In the absence of such base points it isn't clear that one should really expect a description of ${\rm{Pic}}(X \times Y)$ in the form requested. (Presumably the case of curves of genus 0 or 1 without $k$-points may already exhibit some obstacles, but I am too lazy to sort that out at the moment.)<|endoftext|> TITLE: C*-Algebras: Dynamics vs. Derivations QUESTION [5 upvotes]: Problem Given a C*-algebra $\mathcal{A}$. Consider dynamics $\tau:\mathbb{R}\to\mathrm{Aut}(\mathcal{A})$ and $\tau':\mathbb{R}\to\mathrm{Aut}(\mathcal{A})$. (More precisely, strongly continuous one-parameter groups.) Denote their derivations by $\delta:\mathcal{D}\to\mathcal{A}$ and $\delta':\mathcal{D}'\to\mathcal{A}$. Then one has: $$\delta=\delta'\implies\tau=\tau'$$ (Here, equality is meant in terms of operators resp. maps.) How do I check this? For dynamics over Hilbert spaces I would proceed by: $$i\frac{\mathrm{d}}{\mathrm{d}t}\|\varphi(t)\|^2=\langle H\varphi(t),\varphi(t)\rangle-\langle\varphi(t),H\varphi(t)\rangle=0$$ But for the C*-algebra case this path is not directly available. Disclaimer I hope to get a hint from here. (I haven't got any respond yet from stack exchange.) REPLY [6 votes]: As was discussed in the comments, it suffices to see that $\delta$ determines $\tau$ uniquely on $\mathrm{dom}(\delta)$ which is dense in $\mathcal{A}$. Suppose that $x_0 \in \mathrm{dom}(\delta)$. Check that $t \mapsto \tau^t(x_0)$ is a solution to the initial value problem \begin{align*} \frac{d}{dt} x(t) = \delta( x(t)) && x(0) = x_0. \end{align*} If $x$ is any solution then $$\frac{d}{dt} \left( \tau^{-t}(x(t)) \right) = -\tau^{-t}\delta(x(t)) + \tau^{-t}\left(\delta(x(t))\right) = 0$$ and it follows that $\tau^{-t}(x(t)) \equiv x_0$ so that $x(t) = \tau^t (x_0)$ is the unique solution. Basically this all works for Banach space flows too. See the answers to my own question here.<|endoftext|> TITLE: Relation between $BG$ in topology and in algebraic geometry QUESTION [8 upvotes]: This could as well have been asked in the comments to this question, but I prefer to open a new one for the sake of clarity. Say $G$ is a reductive group over the complex numbers, with compact real form $K$. On the one hand one can construct The algebraic stack $[*/G]$ over the complex numbers, on the other hand one can construct The classifying space $\mathrm{B} K$, which is a topological space well defined up to homotopy. What is, exactly, the relation between the two objects? The classifying space $\mathrm{B} G$ of the underlying topological group $G$ is well known to be homotopy equivalent to $\mathrm{B} K$. And $[*/G]$ has, I think, an underlying topological stack $[*/G]_{\mathrm{top}}$. How do $\mathrm{B} G$ and $[*/G]_{\mathrm{top}}$ compare? They live in the same (2-)category, namely possibly "infinite dimensional" topological stacks. Is there a way they are two different "models" of the same "object"? REPLY [21 votes]: I'm not sure why you're replacing $G$ with $K$. So let me compare $[*/G]$ and $B(G^{an})$. As you say, for the purposes of homotopy theory, $B(G^{an})$ is as good as $BK$. One way to understand a stack $\mathcal{X}$ is via a hypercover, which is a simplicial scheme. The easiest way to get a hypercover is to choose an atlas $U\to \mathcal{X}$ and to consider the simplicial scheme $U_\bullet$ with $$U_0=U,$$ $$U_1=U\times_{\mathcal{X}}U,$$ $$U_2=U\times_{\mathcal{X}}U\times_{\mathcal{X}}U,$$ $$U_3=U\times_{\mathcal{X}}U\times_{\mathcal{X}}U\times_{\mathcal{X}}U$$ and so on, with the evident face and degeneracy maps. Let's try this with $[*/G]$. There is an obvious atlas $*\to [*/G]$. The hypercover we obtain has $0$-th level $*$, first level $*\times_{[*/G]}*=G$, second level $*\times_{[*/G]}*\times_{[*/G]}*=G\times G$, and so on; working out the face and degeneracy maps, we see that this is exactly the usual simplicial "bar construction" model for $BG$. Let me make the exact relationship explicit, now. There is a natural functor $\text{Var}_\mathbb{C}\to \text{Top}$ given by sending a variety over $\mathbb{C}$ to the topological space underlying it, in the analytic topology. One may extend this functor to simplicial varieties over $\mathbb{C}$ in the obvious way. Now if we apply this functor to the hypercover described above, obtained via the cover $*\to [*/G]$, one gets exactly the usual bar construction of $B(G^{an})$. As a lemma, you should convince yourself that it actually doesn't matter what (smooth) hypercover we take--we get the same homotopy type. So in terms of the "underlying topological space," (defined as in the previous paragraph) the stack is weakly equivalent to the usual topological object. Of course, the stack knows a lot more than the space $BG$--it has some actual algebraic geometry associated to it as well. Added Later. I think it might also be worth thinking about another relationship between $[*/G]$ and $B(G^{an})$. This relationship is "adjoint" to the one I described above. Consider the category of simplicial schemes, with the model structure induced by the Yoneda embedding into the category of simplicial presheaves on $\text{Sch}$, with your favorite model structure (weak equivalences should be given by pointwise weak equivalences). Now, localize at the subcategory generated by smooth hypercovers. The category of simplicial sets naturally embeds in the category of simplicial schemes, just by sending a set $S$ to the scheme $\sqcup_S \text{Spec}(\mathbb{C})$. Choosing a hypercover of $BG$ (which one doesn't matter, because of the localization) we get a simplicial presheaf on $\text{sSch}$ via Yoneda. Restricting this presheaf to the essential image of $\text{sSet}$, one recovers the usual $BG$ up to weak equivalence. The sense in which the stack $[*/G]$ "knows more" than the space $BG$ is that the presheaf above is certainly not determined by what it does on the essential image of $\text{sSet}$; the "algebraic geometry" of $[*/G]$ is precisely the data of what it does on the whole category $\text{sSch}$, from a functor-of-points POV.<|endoftext|> TITLE: Is there a bounded sequence of points in the plane with pairwise distances at least $1/\sqrt{|i-j|}$? QUESTION [17 upvotes]: Previously I have mentioned the following problem in an addition to the list of Contest problems with connections to deeper mathematics. Is there an infinite bounded sequence $(P_n) \subset \mathbb{R}^2$ having $d(P_i,P_j) \cdot \sqrt{|i-j|} > 1$ for all $i \neq j$? It is not too hard to see that there is no such sequence if the square root is replaced by an exponent $< 1/2$. If on the other hand this exponent is raised by an $\varepsilon$, diophantine approximations apply and prove that $P_n := (C\{n\sqrt{2}\},C\{n\sqrt{3}\})$ with $C \gg_{\varepsilon} 0$ is an explicit sequence with the desired property. This follows easily by the two-dimensional case of Schmidt's theorem (although this does not say how large $C$ needs to be). Similarly, metric diophantine approximation yields the same with $\sqrt{2}$ and $\sqrt{3}$ replaced by Lebesgue-generic points of $[0,1]$; this is considerably easier to prove, giving a non-explicit construction, but still with an $\varepsilon$. What should the answer be with $\varepsilon = 0$? REPLY [16 votes]: It seems that such a sequence exists. 1. Firstly, we take an auxiliary sequence $a(n)=\{(n+1)\sqrt2\}$ for $n\geq 0$. For every $m>n$ the standard estimate yields, say, $$ |a(m)-a(n)|=|(m-n)\sqrt2-p|=\frac{|2(m-n)^2-p^2|}{(m-n)\sqrt2+p}>\frac1{10|m-n|}; $$ here $p=[(m+1)\sqrt2]-[(n+1)\sqrt2]\leq 8(m-n)$. 2. Now we construct the sequence of points $P_i=(x_i,y_i)$ as follows. Let $i=\overline{\dots i_2i_1i_0}$ be the binary representation of $i$. Set $m(i)=\overline{\dots i_4i_2i_0}$ and $n(i)=\overline{\dots i_5i_3i_1}$, and put $x_i=a(m(i))$ and $y_i=a(n(i))$. Now take any $i>j$. Let $s$ be the minimal integer such that $2^{2s}\geq i-j$, hence $2^s<2\sqrt{i-j}$. Set $i'=\overline{\dots i_{2s+1}i_{2s}}=[i/2^{2s}]$ and $j'=\overline{\dots j_{2s+1}j_{2s}}=[j/2^{2s}]$. Then $j'\leq i'\leq j'+1$. Assume that $i'=j'$. We have either $m(i)\neq m(j)$ or $n(i)\neq n(j)$ (w.l.o.g., $m(i)\neq m(j)$), and $|m(i)-m(j)|<2^s$ since $m(i)$ and $m(j)$ differ only in the last $s$ digits. Then we have $$ d(P_i,P_j)\geq |x_i-x_j|=|a(m(i))-a(m(j))|>\frac1{10|m(i)-m(j)|}> \frac1{10\cdot 2^s} >\frac1{20\sqrt{i-j}}. $$ Now assume that $i'=j'+1$; then it is easy to see that either $m(i')=m(j')+1$ or $n(i')=n(j')+1$ (it depend on the parity of the last index $k\geq 2s$ for which $i_k\neq j_k$; w.l.o.g. $m(i')=m(j')+1$). Then we have $0\frac1{10|m(i)-m(j)|}>\frac1{10\cdot 2^{s+1}} >\frac1{40\sqrt{i-j}}. $$ Thus in any case we have $d(P_i,P_j)\cdot \sqrt{i-j}>\frac1{40}$. Now it remains to scale the whole picture with ratio 40.<|endoftext|> TITLE: "Nyldon words": understanding a class of words factorizing the free monoid increasingly QUESTION [31 upvotes]: BACKGROUND. Let me first introduce some classical definitions, which appear, e.g., in §5 of Lothaire's Combinatorics on Words, in §5.1 of Reutenauer's Free Lie algebras, and in §6.1 of Victor Reiner's and my Hopf algebras in Combinatorics. If you are not a stranger to combinatorics on words, scroll right down to the question. Words. Let $\mathfrak A$ be a finite totally ordered set. We call $\mathfrak A$ the alphabet, and refer to the elements of $\mathfrak A$ as letters. Let $\mathfrak A^\ast$ be the disjoint union of the sets $\mathfrak A^n$ over all $n \in \mathbb{N} = \left\{0, 1, 2, \ldots\right\}$. The elements of $\mathfrak A^\ast$ are called words (over the alphabet $\mathfrak A$). The concatenation of two words $u = \left(u_1, u_2, \ldots, u_n\right) \in \mathfrak A^\ast$ and $v = \left(v_1, v_2, \ldots, v_m\right) \in \mathfrak A^\ast$ is defined to be the word $\left(u_1, u_2, \ldots, u_n, v_1, v_2, \ldots, v_m\right)$; it is denoted by $uv$. The set $\mathfrak A^\ast$, equipped with concatenation as a binary operation, becomes a monoid (the neutral element is the empty word, denoted by $\varnothing$); this is the free monoid on the set $\mathfrak A$. A word $u$ is said to be a prefix of a word $w$ if and only if there exists a word $v$ (possibly empty) satisfying $w = uv$. Lexicographic order. We define a relation $\leq$ on $\mathfrak A^\ast$ by letting two words $u = \left(u_1, u_2, \ldots, u_n\right) \in \mathfrak A^\ast$ and $v = \left(v_1, v_2, \ldots, v_m\right) \in \mathfrak A^\ast$ satisfy $u \leq v$ if and only if either there exists a positive integer $i \leq \min\left\{n,m\right\}$ such that $u_i < v_i$ and ($u_j = v_j$ for every $j < i$), or the word $u$ is a prefix of $v$. This relation $\leq$ is (the smaller-or-equal relation of) a total order on $\mathfrak A^\ast$; this order is the so-called lexicographic order (or dictionary order) on $\mathfrak A^\ast$. It satisfies many of the properties one would expect from an order on words, but one must keep in mind that while $u \leq v$ always implies $wu \leq wv$ for words $u,v,w$, it does not always imply $uw \leq vw$. Lyndon words. A Lyndon word is defined to be a nonempty word $w \in \mathfrak A^\ast$ satisfying one of the following three equivalent assertions: If $u$ and $v$ are any two nonempty words such that $w = uv$, then $v > w$. If $u$ and $v$ are any two nonempty words such that $w = uv$, then $v > u$. If $u$ and $v$ are any two nonempty words such that $w = uv$, then $vu > uv$. Notice that the equivalence of these three assertions is not trivial; if we would restrict the assertions to a particular choice of $u$ and $v$, then they would cease to be equivalent. For example, if $\mathfrak A = \mathbb N$ with the usual total order, then the words $4$, $223$, $04044$ and $0303304$ are Lyndon, while the words $\varnothing$, $10$, $11$ and $242$ are not. Chen-Fox-Lyndon factorization. Lyndon words have a multitude of curious properties as well as applications to algebra (free groups, free Lie algebras, shuffle algebras, quasisymmetric functions etc.). Probably the most important result that is "behind" these applications is the Chen-Fox-Lyndon factorization theorem, which states that for every word $w \in \mathfrak A^\ast$, there exists a unique tuple $\left(a_1, a_2, \ldots, a_k\right)$ of Lyndon words satisfying $w = a_1 a_2 \cdots a_k$ and $a_1 \geq a_2 \geq \cdots \geq a_k$. For instance, if $\mathfrak A = \mathbb N$ with the usual total order, for $w = 104029423402010110101$, this tuple is $\left(1, 04, 0294234, 02, 01011, 01, 01\right)$. This factorization can be used as an alternative way to define Lyndon words. Indeed, assume that we are given some positive integer $N$, and we have defined the notion of Lyndon words of length $< N$. Then, we can define the notion of a Lyndon word of length $N$ as being a word of length $N$ for which there exists no tuple $\left(a_1, a_2, \ldots, a_k\right)$ of Lyndon words of length $< N$ satisfying $w = a_1 a_2 \cdots a_k$ and $a_1 \geq a_2 \geq \cdots \geq a_k$. This definition is implicit and impractical, and it does not help proving the uniqueness of the factorization, but I would not be surprised if it is how people actually discovered Lyndon words. Necklaces. A necklace (also known as a circular word) means an equivalence class of words modulo cyclic rotation. A necklace of nonempty words is said to be aperiodic if the number of its representatives equals the length of each of its representatives. For instance, the necklaces $\overline{1213}$ and $\overline{24424}$ are aperiodic, while $\overline{1212}$ and $\overline{404040}$ are not. Here, $\overline{w}$ denotes the necklace containing a word $w$; so we have $\overline{1213} = \overline{2131} = \overline{1312} = \overline{3121} = \left\{ 1213, 2131, 1312, 3121 \right\}$. A known fact states that every aperiodic necklace contains precisely one representative which is a Lyndon word, whereas any non-aperiodic necklace contains no such representative. Thus, if the alphabet $\mathfrak A$ has finite cardinality $q$, then the number of Lyndon words of length $n$ (for fixed positive integer $n$) equals the number of aperiodic necklaces of $n$-letter words. This number is $\dfrac{1}{n} \sum\limits_{d\mid n} \mu\left(n/d\right) q^d$. QUESTION. Nyldon words. We can define the notion of Nyldon words by recalling our alternative definition of Lyndon words using Chen-Fox-Lyndon factorization, and replacing the condition $a_1 \geq a_2 \geq \cdots \geq a_k$ by $a_1 \leq a_2 \leq \cdots \leq a_k$. In more detail: There are no Nyldon words of length $< 1$. Assume that we are given some positive integer $N$, and we have defined the notion of Nyldon words of length $< N$. Then, we define the notion of a Nyldon word of length $N$ as being a word of length $N$ for which there exists no tuple $\left(a_1, a_2, \ldots, a_k\right)$ of Nyldon words of length $< N$ satisfying $w = a_1 a_2 \cdots a_k$ and $a_1 \leq a_2 \leq \cdots \leq a_k$. Conjecture 1. For every word $w$, there exists a unique tuple $\left(a_1, a_2, \ldots, a_k\right)$ of Nyldon words satisfying $w = a_1 a_2 \cdots a_k$ and $a_1 \leq a_2 \leq \cdots \leq a_k$. The existence part of this is obvious, but the uniqueness is interesting. I have checked it by computer for $\mathfrak A = \left\{0, 1\right\}$ and word length up to $18$. Conjecture 2. Every aperiodic necklace contains precisely one representative which is a Nyldon word, whereas any non-aperiodic necklace contains no such representative. This is checked for $\mathfrak A = \left\{0, 1\right\}$ and word length up to $15$. Each of the Conjectures 1 and 2 would imply that the number of Nyldon words of given length over a given alphabet equals the number of Lyndon words of the same data. This is, in fact, equivalent to Conjecture 1 (because one can WLOG assume $\mathfrak A$ finite, and then argue using the fact that surjections between finite sets of equal size must be injections), so Conjecture 2 can be regarded as a stronger version of Conjecture 1. EXPERIMENTAL DATA. For $\mathfrak A = \left\{0, 1\right\}$, here are the Nyldon words of lengths up to $6$: length $0$: none. length $1$: the words $0$ and $1$. (For comparison: The Lyndon words are $0$ and $1$.) length $2$: the word $10$. (For comparison: The Lyndon words are $01$.) length $3$: the words $100$ and $101$. (For comparison: The Lyndon words are $001$ and $011$.) length $4$: the words $1000$, $1001$ and $1011$. (For comparison: The Lyndon words are $0001$, $0011$ and $0111$.) length $5$: the words $10000$, $10001$, $10010$, $10011$, $10110$ and $10111$. (For comparison: The Lyndon words are $00001$, $00011$, $00101$, $00111$, $01011$ and $01111$.) length $6$: the words $100000$, $100001$, $100010$, $100011$, $100110$, $100111$, $101100$, $101110$ and $101111$. (For comparison: The Lyndon words are $000001$, $000011$, $000101$, $000111$, $001011$, $001101$, $001111$, $010111$ and $011111$.) REPLY [9 votes]: My co-authors (Émilie Charlier, Manon Philibert) and I give positive answers to Grinberg's conjectures in the paper E. Charlier, M. Philibert, M. Stipulanti, Nyldon words. So it is true that there are equally many Nyldon words and Lyndon words of a given length. In addition, we show that the set of Nyldon words is a Hall set (more precisely, we prove that it is a Lazard factorization, but the notions are equivalent). In this paper, we also compare Lyndon and Nyldon words regarding different properties such as standard and Širšov factorizations, circular and comma-free codes, and Lazard factorizations. We also leave in the paper open questions for future work. I hope you will enjoy it!<|endoftext|> TITLE: Do complex tori contain quasi-projective open subsets? QUESTION [7 upvotes]: Complex tori are not associated to projective varieties in general. But can one find an open $U$ inside a complex torus $\mathbb C^g/L$ such that $U$ is the analytification of a quasi-projective variety? What if we just ask $U$ to be the analytification of a (finite type separated) scheme? I think that if $U$ comes from a scheme, then $U$ has a compactification which should map to the complex torus $\mathbb C^g/L$. This map should be birational proper and therefore $\mathbb C^g/L$ is Moishezon (and Kahler). Therefore, it is projective. So, probably the answer is "you can find a quasi-projective open if and only if the torus is algebraic". REPLY [5 votes]: The answer is negative if the complex torus is not algebraic (or equivalently, not Moishezon). More generally, if $U$ is the analytification of a separated scheme of finite type over $\mathbf{C}$ and $Y$ is a proper complex-analytic space admitting an open immersion $j:U \hookrightarrow Y$ onto the complement of a nowhere-dense analytic set then $Y$ must be Moishzeon. However, even if $Y$ is a manifold, without knowing the geometry of the complement of $Y-U$ in $Y$ perhaps the restriction map $M_Y\rightarrow j_{\ast}(M_U)$ with sheaves of meromorphic functions is not an equality on global sections (i.e., the Levi extension theorem may not apply). Thus, to prove that $Y$ must be Moishezon it is necessary to use more serious input from complex analysis as follows. We may assume $Y$ is reduced and irreducible (by the very definition of "Moishzeon") and may shrink $U$ to be affine (if it weren't already quasi-projective) and even smooth (by reducedness). The normalization $Y'$ of $Y$ is connected and $Y' \rightarrow Y$ is an isomorphism over $U$, and $M(Y) = M(Y')$ by the theory of analytic normalization, so we may replace $Y$ with $Y'$ to arrange that $Y$ is normal. By resolution of singularities, $U$ is the complement of a strict normal crossings divisor in a connected projective manifold $X$. Borel's extension theorem extends $U\rightarrow Y$ to $\nu:X\rightarrow Y$ by normality of $Y$. But $M_Y\rightarrow\nu_{\ast}(M_X)$ is an isomorphism by 10.6.2 in "Coherent Analytic Sheaves" (or really Remark 8.1.3 in loc. cit.), which rests crucially on Grauert's coherence theorem, so $M_Y(Y) = M_X(X)$ has transcendence degree $\dim X = \dim Y$ over $\mathbf{C}$. Hence, $Y$ is Moishezon.<|endoftext|> TITLE: amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic? QUESTION [9 upvotes]: I heard from someone that the following problem is an open question. (Open Problem 1)For a countable discrete group $G$, suppose it does not contain any Baumslag-Solitar subgroups $BS(m,n):=\langle x,y|xy^mx^{-1}=y^n\rangle$ and it admits a finite $K(G,1)$, is it a hyperbolic group? I could not find the relevent stuff on this problem, so I am wondering whether the following is known. (My question)For a countable discrete amenable group $G$, suppose it does not contain any Baumslag-Solitar subgroups $BS(m,n):=\langle x,y|xy^mx^{-1}=y^n\rangle$ and it admits a finite $K(G,1)$, is it a virtually cyclic group? Note that ``Yes for problem 1 implies Yes for my question" since $G$ is amenable and hyperbolic iff $G$ is virtually cyclic(?) Any references or comments are welcome! REPLY [7 votes]: Yes, this problem is as open as Problem 1. There are no feasible obstructions to counter-examples, the main obstacle is lack of tools: All known constructions of non-hyperbolic groups with finite $K(G,1)$, yield groups which contain some $BS(p,1)$, $p\ge 1$. Edit: For elementary amenable groups of finite type, positive answer to your question probably follows from the results here, but I am too busy right now to check.<|endoftext|> TITLE: Can a parent and child node have the same type in a well-founded digraph tree? QUESTION [10 upvotes]: $\newcommand\toward{\rightharpoonup}$It would help me to understand something in a current research project if someone could provide an example of directed graph $\langle G,\toward\rangle$ with the following properties The graph is connected. Every node $x$ points at exactly one parent node $y$, via $x\toward y$. Thus, the underlying relation is a tree. The relation $\toward$ is well-founded, so that every nonempty subset $X\subset G$ has an $\toward$-minimal element $x\in X$, meaning that $y \not\!\toward x$ for all $y\in X$. In the presence of DC, this is equivalent to saying that there is no infinitely receding sequence $\cdots\toward x_3\toward x_2\toward x_1\toward x_0$. Finally, the key part, there are two elements $a$ and $b$ with $a\toward b$, such that $a$ and $b$ realize exactly the same type in $\langle G,\toward\rangle$. That is, $G\models\varphi(a)\leftrightarrow\varphi(b)$ for any assertion $\varphi$ in the digraph language. Can this happen? Note that there can be no automorphism mapping $b$ to $a$, for in this case, we could iterate it to produce a receding chain, contrary to the well-founded hypothesis. Also, since $b$ is not minimal, neither can $a$ be minimal, and if there is a path from a minimal element to $a$ of length $k$, then there is one to $b$ of length $k+1$, and so if the types are the same, there is a path to $a$ of length $k+1$, and so on. So both $a$ and $b$ will have infinite rank with respect to the well-founded relation. And I think one can carry that reasoning much further. In my actual application, there is a lot more structure, and I consider $a$ and $b$ to realize the same assertions in a much stronger language, one that includes second-order logic and more. But I realized I don't have a complete grasp even on this "easy" case. What are types like in these well-founded digraph trees? REPLY [6 votes]: What about this? Consider a graph $G$ of size $\omega_1$ such that $G$ is connected, For every node $a$ of rank $\alpha$, and every $\beta<\alpha$, $a$ has $\omega_1$-many children of rank $\beta$. Every node has a parent, and there are nodes of all ranks below $\omega_1$. Now I think an Ehrenfeucht-Fraisse game should show that - if I pick nodes $a$ and $b$ in $G$ with $a$ the parent of $b$ and $rk(b)>\omega^\omega$ (I'm not sure that's optimal) - the structures $(G, a)$ and $(G, b)$ are elementarily equivalent, so $a$ and $b$ have the same type in $G$. I'm pressed for time, so I can't write out the details - of course, if there's a mistake, here it is! - but this seems to work. We can then replace $G$ with a countable subgraph $G_0$, by taking a countable elementary substructure of $G$ containing $a$ and $b$.<|endoftext|> TITLE: Is there any relationship between the topologies of the clique complex and the independence complex? QUESTION [13 upvotes]: Let $G$ be a simple graph on a finite vertex set. The clique complex $X(G)$ is the simplicial complex whose faces are complete subgraphs of $G$, and the independence complex $I(G)$ is the simplicial complex whose faces are independent subsets of $G$. Said another way, $I(G) = X( \bar{G} )$, where $\bar{G}$ denotes the complement of $G$. My question is whether there is any relationship between the topology of these two complexes. Playing around with small examples made me optimistic that there would be some way to relate the homology of $X(G)$ with the cohomology of $I(G)$, perhaps via Alexander Duality. For example, consider $G = C_5$, the $5$-cycle. Then both $X(G)$ and $I(G)$ are homeomorphic to $S^1$. The boundary of the simplex on $5$ vertices is homeomorphic to $S^3$. In particular, we have (for reduced homology and cohomology) that $H_q (X(G)) = H^{m-q-1} (I(G))$, where $m=3$. But I don't see any proof for a general case. Even worse: one can probabilistically construct examples on $n$ vertices, where both $X(G)$ and $I(G)$ both have dimension roughly $2 \log_2 n$ and all their nontrivial homology concentrated around dimension $\log_2 n$, which also does not seem to bode well for Alexander Duality. At least it does not bode well for the most simple-minded relationship, where one might hope that $I(G)$ is homotopy equivalent to the complement of $X(G)$ in the $n-2$-dimensional sphere. (The probabilistic examples I have in mind are to let $G$ be the Erdős–Rényi random graph $G(n,p)$ with $p=1/2$. Then the clique complex $X(G)$ and independence complex $I(G)$ have the same distribution. Several papers have studied the topology of such random complexes, and we know that they have homology concentrated around middle dimension.) REPLY [17 votes]: Given any graph $G$, let $G'$ denote $G$ with a new vertex $v$ adjacent to every vertex of $G$. The clique complex of $G'$ is contractible, while the independence complex of $G$ and $G'$ are the same except for an isolated vertex. Similarly we can adjoin a new isolated vertex $w$ to $G$, obtaining a graph $G''$ with a contractible independence complex but with the same clique complex as $G$ except for an isolated vertex. Thus in general there is not much connection between the topologies of the clique and independence complexes.<|endoftext|> TITLE: Robotics, Cryptography, and Genetics applications of Grothendieck's work? QUESTION [15 upvotes]: I was reading about the passing of Alexander Grothendieck, and something caught my interest: Mr. Grothendieck was able to answer concrete questions about these relationships by finding universal mathematical principles that could shed unexpected light on them. Applications of his work are evident in fields as diverse as genetics, cryptography and robotics. New York Times After extensive googling, I haven't been able to find examples. Has Grothendieck's mathematical work been applied to robotics, cryptography, or genetics, and if so, how? REPLY [4 votes]: Genetic research has some applications of grothendieck's theory, above mentioned answers inform about grothendieck cuts, and their applications to robotics. Some cancer researchers use groupoids, so to determine a gene expression, define first a grothendieck space, and then study deformations. These (Yoneda-Grothendieck) structures are useful in studying the dynamics of the cancer gene.<|endoftext|> TITLE: Universal maps between topological spaces QUESTION [12 upvotes]: Let $X,Y$ be topological spaces. We call a continuous map $u:X\to Y$ universal if for every continous map $f:X\to Y$ there is $x\in X$ such that $f(x) = u(x)$. If $u:X\to Y$ and $v:Y\to Z$ are universal, is $v\circ u: X\to Z$ universal? (I am thinking about this question because of an inspiration from Fixed points and universal maps for posets ). REPLY [8 votes]: This seems to be answered negatively in "On the composition and products of universal mappings" by W. Holsztyński (Fundamenta Mathematicae 64(2) (1969), 181-188). I've not looked at the proof or really tried to figure out why it works, but if I understand correctly what is claimed, then one counterexample from the paper is as follows: $M$ is a Möbius band, regarded as the annulus $\left\{z\in\mathbb{C}:\frac{1}{2}\leq\vert z\vert\leq 1\right\}$ with $z$ and $-z$ identified for $\vert z\vert=\frac{1}{2}$. $Q$ is the closed unit disc $\left\{z\in\mathbb{C}:\vert z\vert\leq 1\right\}$. Apparently the map $u:M\to Q$ induced by $z\mapsto\left(2-\frac{1}{\vert z\vert}\right)z$, which identifies the inner boundary of the annulus to a point, and the map $v:Q\to Q$ given by $z\mapsto z^2$ are both universal, but $v\circ u$ is not.<|endoftext|> TITLE: Push-forward of locally free sheaves QUESTION [8 upvotes]: Let $X, Y$ be smooth projective varieties and $f:X \times Y \to Y$ be the natural projetion map. Let $\mathcal{F}$ be a locally free sheaf on $X \times Y$. Is it true that $f_*\mathcal{F}$ is locally free on $Y$? If not true in general is there any additional condition on $X, Y$ under which this will hold true? REPLY [20 votes]: Let $X = P^1$, $Y = P^3$. We will take $F$ to be an extension $$ 0 \to O(-2,1) \to F \to O \oplus O \oplus O \to 0. $$ Of course $F$ is locally free. Note that $$ Ext^1(O,O(-2,1)) = H^1(P^1\times P^3,O(-2,1)) = H^1(P^1,O(-2))\otimes H^0(P^3,O(1)) $$ is a 4-dimensional vector space, so we can take $F$ to be the extension corresponding to a 3-dimensional subspace in it. Pushing forward to $P^3$ then gives an exact sequence $$ 0 \to f_*F \to O \oplus O \oplus O \to O(1) \to R^1f_*F \to 0 $$ and the middle map corresponds to out choice of 3 linear functions on $P^3$, so $R^1f_*F$ is the structure sheaf of a point and $f_*F$ is the simplest example of a reflexive sheaf which is not locally free.<|endoftext|> TITLE: Breaking up the free Lie algebra into Gl irreps QUESTION [9 upvotes]: The free Lie algebra $L(V)$ generated by an $r$-dimensional vector space $V$ is, in the language of https://en.wikipedia.org/wiki/Free_Lie_algebra , the free Lie algebra generated by any choice of basis $e_1, \ldots , e_r$ for the vector space $V$. (Work over the field ${\mathbb R}$ or ${\mathbb C}$, whichever you prefer.) It is a graded Lie algebra $$L(V) = V \oplus L_2 (V) \oplus L_3 (V) \oplus \ldots .$$ The general linear group $Gl(V)$ of $V$ acts on $L(V)$ by gradation-preserving Lie algebra automorphisms. Thus each graded piece $L_k (V)$ is a finite dimensional representation space for $GL(V)$. (The `weight' of $L_k (V)$ is $k$ in the sense that $\lambda Id \in Gl(V)$ acts on $L_k (V)$ by scalar multiplication by $\lambda^k$.) QUESTION: How does $L_k (V)$ break up into $GL(V)$-irreducibles? I only really know that $L_2 (V) = \Lambda ^2 (V)$, which is already irreducible. To start the game off, perhaps some reader out there already is friends with $L_3 (V)$ as a $GL(V)$-rep, and can tell me its irreps in terms of the Young diagrams / Schur theory involving 3 symbols? (My motivation arises from trying to understand some details of the subRiemannian geometry http://en.wikipedia.org/wiki/Sub-Riemannian_manifold of the Carnot group whose Lie algebra is the free k-step Lie algebra, which is $L(V)$-trunated after step $k$. ) REPLY [7 votes]: Fix a primitive $k$-th root of unity $\zeta_k$, and let $\rho\in S_k$ be a $k$-cycle. (I am working over $\mathbb{C}$ here, obviously.) Klyachko [Kl] proved in 1974 that $L_k(V)\simeq \{x\in V^{\otimes k}\,|\,\rho(x) = \zeta_k\cdot x\}$. From this Frobenius reciprocity gives the following formula for the multiplicity of $\mathbb{S}_\lambda(V)$ in $L_k(V)$ (though I believe this formula is much older): $\frac{1}{k}\sum_{d|k}\mu(d)\chi_\lambda(\tau^{m/d})$ Here $\mu$ denote the Möbius function, $\chi_\lambda\colon S_k\to \mathbb{Z}$ be the character of the corresponding $S_k$-irrep, and $\tau\in S_k$ is a $k$-cycle (so $\tau^{k/d}$ is a product of $k/d$ disjoint $d$-cycles). However this formula is not so helpful, since the irreducible characters of $S_k$ are not easy to compute; in practice, there is a much more useful (positive, bijective) formula as follows. [As motivation] Since $L_k(V)\subset V^{\otimes k}$, the multiplicity of $\mathbb{S}_\lambda(V)$ in $L_k(V)$ is bounded above by its multiplicity in $V^{\otimes k}$, which is the number of standard tableaux of shape $\lambda$. (A tableau $T$ of shape $\lambda$ is a labeling of the boxes of $\lambda$ by $1,\ldots,k$; it is standard if the labels are increasing within each row and within each column.) The descent set $D_T\subset \{1,\ldots,k\}$ of a tableau $T$ is the subset $D_T=\{i\,|$ box $i$ is in a higher row than box $i+1\}$. The major index $\text{maj}(T)$ is the sum $\text{maj}(T):= \sum_{i\in D_T}i$. The multiplicity of $\mathbb{S}_\lambda(V)$ in $L_k(V)$ is the number of standard tableaux $T$ of shape $\lambda$ with major index $\text{maj}(T)\equiv 1\bmod{k}$. You should have no trouble working out the computations you want by hand for small $k$ using this result (as F.C.'s computations show, the answer gets large fast as $k$ gets bigger). [Remark on references: I always thought this formula was due to Klyachko as well; from a brief look at his paper I don't see it there (but it may be implicit there; certainly his "$\text{ind}(\sigma)$" is precisely the major index modulo $k$). It appears explicitly in Kraśkiewicz-Weyman "Algebra of coinvariants and the action of a Coxeter element" which appeared as a preprint in the late 1980s. Credit can be difficult to pin down in this period, especially due to the communication barriers between the Soviet Union and other countries; I make no guarantees about the proper credit for these results.] [Kl]: Klyachko's article was in Russian; the English translation is Klyachko, Lie elements in the tensor algebra, Siberian Mathematical Journal 15 (1974) 6, 914-920, online here<|endoftext|> TITLE: Area of the minimal surface of a non-planar quadrilateral in 3d QUESTION [14 upvotes]: Consider a non-planar quadrilateral in three dimensions, i.e. four points $x_1,\dots,x_4$ in $\mathbb{R}^3$ that do not lie on a plane and connected by straight lines. Then, by general theory of minimal surfaces and the Plateau problem there exists a surface of minimal area with this lines as boundary. The situation looks like this: (Picture from http://mathworld.wolfram.com/SkewQuadrilateral.html) but note that the obvious bilinear interpolation is not the minimal surface. There are formulae for the minimal surfaces such as the Weierstraß-Enneper formula but I haven't come across a formula for this particular case of a quadrilateral. In fact, I am not interested in a formula for the surface but only look for an answer to the question: What is area of the minimal surface of the quadrilateral in terms of the four corner points $x_1,x_2,x_3,x_4$? REPLY [5 votes]: This paper seems to give a partial answer to the posed question, for skew quadrilaterals that project to rectangles: Furui, Sadataka, and Bilal Masud. "Numerical calculation of a minimal surface using bilinear interpolations and Chebyshev polynomials." arXive: math-ph/0608043 (2006).           Abstract. "We calculate the minimal surface bounded by four-sided figures whose projection on a plane is a rectangle, starting with the bilinear interpolation and using, for smoothness, the Chebyshev polynomial expansion in our discretized numerical algorithm to get closer to satisfying the zero mean curvature condition. We report values for both the bilinear and improved areas, suggesting a quantitative evaluation of the bilinear interpolation. An analytical expression of the Schwarz minimal surface with polygonal boundaries and its 3-dimensional plot is also given."<|endoftext|> TITLE: Von Neumann's consistency proof QUESTION [15 upvotes]: In the paper Zur Hilbertschen Beweistheorie, John Von Neumann has proposed a consistency proof for a fragment of first-order arithmetic (the fragment without induction and with the successor axioms only) by a variant of Hilbert's substitution method. The paper is in German, and I so could not read it. Question 1. Can anyone explain the main ideas of Von Neumann's work in this paper? Question 2. Are there any references in English, explaining Von Neumann's work in details? REPLY [5 votes]: I'm late to the party, but hope that the following might usefully complement Ali's nice answer and references, first by elaborating slightly on the historical side, and second by briefly indicating an $\varepsilon$-substitution-style consistency proof for the first-order-logic-based theory analogous to the $\varepsilon$-caculus-based system treated by von Neumann. Ackermann's Begründung des "tertium non datur" mittels der Hilbertschen Theorie der Widerspruchsfreiheit actually set out to prove the consistency of all of analysis, and Ackermann initially believed he'd done just that. However, ... while correcting the printer's proofs of his paper, he had to introduce a footnote ... that restricts his rule of substitution. After the introduction of such a restriction it was no longer clear for which system Ackermann's proof establishes consistency. Certainly not for analysis. ... In [Zur Hilbertschen Beweistheorie] ... von Neumann criticized Ackermann's proof and presented a consistency proof that ... came to be accepted as establishing the consistency of a first-order arithmetic in which induction is applied only to quantifier-free formulas. -- From Frege to Goedel (my emphasis) One can rather quickly give a proof in the first-order logic setting that conveys the spirit of an $\varepsilon$-substitution approach to proving the consistency of the mentioned theory. Let $\mathsf{T}$ be the theory which has the usual universal axioms of Peano arithmetic, but in which induction is allowed only for quantifier-free formulas. To prove $\mathsf{T}$ consistent, it suffices to show that any finite conjunction of Skolem instances of axioms of $\mathsf{T}$ is satisfiable. (Note that here we're appealing to Herbrand's theorem, which has analogues in the Hilbert-Bernays $\varepsilon$ Theorems.) We can begin to do that here by interpreting $0$, $S$, $+$, $\times$ numerically in the obvious way; all non-induction axioms of $\mathsf{T}$ are purely universal, and so will come out true under such interpretation. All that remains is numerically interpreting the Skolem function symbols arising from induction axioms (and this is what's analogous to finding numerical interpretations of $\varepsilon$-terms in the $\varepsilon$-substitution method of Hilbert, Ackermann, von Neumann). As we only have quantifier-free induction in $\mathsf{T}$, Skolem instances of induction axioms are of the form $$ \varphi(\vec{y}, 0) \wedge [\varphi(\vec{y},f_\varphi(\vec{y},x)) \rightarrow \varphi(\vec{y},Sf_\varphi(\vec{y},x))] \rightarrow \varphi(\vec{y},x)$$ where $\vec{y}$ is, say, $y_1,\dots,y_k$. Because all $\varphi$ are quantifier-free here, there is no interaction/nesting among the $f_\varphi$ arising from induction axioms, and in fact they can all be interpreted in a uniform way: namely, each $f_\varphi$ can be interpreted by $\hat{f_\varphi} : \mathbb{N}^{k+1} \rightarrow \mathbb{N}$ given as $$ \hat{f_\varphi}(n_1,\dots,n_k,m) := \left\{\begin{array}{l} \min\{j TITLE: Configuration space like subspace of sphere product QUESTION [8 upvotes]: For $k \geq 2, n \geq 1$ let $$M^{n,k} = \{(x_1,\dots,x_k) \in S^n \times \dots \times S^n \ | \ x_1 + \dots + x_k = 0\}$$ This is a compact CW-complex and almost, but not quite, a manifold. I generally want to understand this space better. More concretely, I am interested in the following questions: (1) What is the dimension of $M^{n,k}$ in a reasonable sense? (2) For which $n,k$ is $M^{n,k}$ simply connected? (3) For $k = p$ prime, the space admits a free $\mathbb Z/p$ action by cyclically permuting the coordinates. What can be said about the cohomology of $(\mathbb Z/p)\backslash M^{n.k}$? For small $k$ and $n$, it is possible to give a "hands on" description of $M^{n,k}$: $M^{n,2}$ is obviously homeomorphic to $S^n$. It is easy to see that $M^{n,3}$ is homeomorphic to the unit sphere bundle of the tangent bundle of $S^n$. The space $M^{1,4}$ is homeomorphic to $S^1 \times X$ where $X$ is the 1-dimensional CW-complex with three 0-cells and six 1-cells, such that for any pair of two different 0-cells there are two 1-cells joining them. $M^{2,4}$ is already pretty hard to describe, but it is definitely of dimension 5. REPLY [6 votes]: Your space $M^{n,k}$ is a moduli space of closed $k$-gons in $\mathbb{R}^{n+1}$ with sides of length 1, viewed up to translation. As such it is a fairly well-studied object in the literature, see for example Farber, Michael; Fromm, Viktor The topology of spaces of polygons, Trans. Amer. Math. Soc. 365 (2013), no. 6, 3097–3114. In particular, it is a closed smooth manifold of dimension $n(k-1)-1$ when $k$ is odd, and has singularities when $k$ is even.<|endoftext|> TITLE: "Productively normal" space QUESTION [6 upvotes]: If a set $S$ is endowed with the discrete topology $\mathcal{P}(S)$, then for every normal space $N$ the product $S\times N$ is normal. Question: can we endow a set $S$ with another Hausdorff topology, such that still for all normal spaces $N$ the product $S\times N$ is normal. REPLY [11 votes]: The answer is no. It was proved by Mary Ellen Rudin in $\aleph$-Dowker spaces (1978) that for any non-discrete Hausdorff space $S$ there is a normal Hausdorff space $N$ such that $S\times N$ is not normal, thus solving in the affirmative Morita´s first conjecture.<|endoftext|> TITLE: Langlands duality and multiplying cocharacters QUESTION [13 upvotes]: Recall that there is a bijection between irreducible representations of a compact real Lie group $G$ and the cocharacters (homomorphisms $U(1) \to G$, modulo conjugation) of the Langlands dual group $^LG$. The irreducible representations of $G$ have additional structure related to tensoring representations: Given representations $\alpha, \beta, \gamma$ of $G$, we have the invariant subspace $V_{\alpha\beta\gamma}$ of the tensor product $\alpha\otimes\beta\otimes\gamma$. (Or, if you prefer, the space $V_{\alpha\beta}^{\gamma^*}$ of homomorphisms from $\gamma^*$ tp $\alpha\otimes\beta$.) Is there an intrinsic way to define a Langlands dual structure on the cocharacters of $^LG$? In other words, in a natural way (and without using Langlands duality) associate to cocharacters $a,b,c$ of $^LG$ a vector space $V_{abc}$? One possibility would be to think of the cocharacters as (equivalence classes of) geodesics in $^LG$ and replace the tensoring of $G$ representations with the splicing of $^LG$ geodesics. The resulting families of broken geodesics $a \cdot b$ could flow to actual geodesics $\{c_i\}$. Before pursuing this idea I wanted to check whether there are already known answers to the main question above. ADDED LATER: The motivation for the above question was geometric Langlands TQFTs applied to the operation of gluing two disks together to obtain another disk. I had forgotten that on the Rep($G$) side a 2-sphere gives the same monoidal category as a disk, and that therefore the geometric Satake isomorphism answers my question. I still wonder whether thinking in terms of disks instead of spheres would give a different (but presumably equivalent) construction of the monoidal product on the cocharacter side. REPLY [6 votes]: This is another version of Reimundo Heluani's answer, avoiding perverse sheaves. In general, the $H$-orbits on $H/P \times H/Q$ are indexed by $W_P \backslash W_H/W_Q$. If $H = {}^L G(\mathcal K)$, so $W_H$ is the affine Weyl group $W_G \ltimes $[the weight lattice], and $P=Q={}^LG(\mathcal O)$, then $W_P \backslash W_H/W_Q =$ dominant weights. For $L,L' \in \mathcal Gr$, write $L \stackrel{\lambda}{\to} L'$ if $(L,L')$ is in the $\lambda$-orbit, for $\lambda$ a dominant weight. Now consider the space of "triangles" $$\mathcal V_{\lambda\mu\nu} := \{(L,L') : L,L' \in \mathcal Gr, base \stackrel{\lambda}{\to} L \stackrel{\mu}{\to} L' \stackrel{\nu}{\to} base \}$$ and I believe the statement is that $V_{\lambda\mu\nu} \cong H_{top}(\mathcal V_{\lambda\mu\nu})$.<|endoftext|> TITLE: degree of polynomials in nullstellensatz QUESTION [5 upvotes]: $(A)$ If $K$ is algebraically closed field, then consider $m$ polynomials $f_i$ for $i=\{1,\dots,m\}$ in $K[x_1,\dots,x_n]$ of degree $d_i$ each with no common root. We know there exists $g_i$ for $i=1,\dots,m$ such that $$f_1g_1+\dots+f_mg_m=1$$ What can we say about the minimum degrees of $g_i$? ($(A)$ has been answered below). $(B)$ If we relax the $1$ requirement or any fixed constant requirement and ask just for polynomials such that $$f_1g_1+\dots+f_mg_m\neq 0 \mbox{ (but not a constant)}$$ which seems same as $$Z(f_1g_1+\dots+f_mg_m)\bigcap Z((0))=\emptyset$$ can we say anything about the minimum degrees of $g_i$? Is anything meaningful possible if $K$ is not closed in the relaxed problem $(B)$? Again what are the minimum degrees of $g_i$ here (both lower and upper bounds)? I am only interested in $K=\Bbb C\mbox{ or }\Bbb R$ here. related:bound on degree of certain polynomials REPLY [3 votes]: The bounds in the article by Kollár as refered in the comment of Felipe Voloch are optimal, see Example 2.3 in that article. To see this concretely, assume that $m \leq n$, and let $f_1 = x_1^{d_1},f_2 = x_1 x_n^{d_2-1}-x_2^{d_2},...,f_{m-1} = x_{m-2}x_n^{d_{m-1}-1}-x_{m-1}^{d_{m-1}},f_m = 1-x_{m-1} x_m^{d_m-1}$. By the article of Kollár, one can find $g_1,...,g_m$ such that $f_1 g_1 + ... + f_m g_m = 1$ and $\deg f_i g_i \leq d_1 \cdots d_m$. In the above example, these bounds are optimal, which can be seen as follows: If we assume that $f_1 g_1 + ... + f_m g_m = 1$, and apply this equality along the curve $x = (t^{d_2... d_m - 1},...,t^{d_m-1},1/t)$, one sees that $\deg_{x_m} g_1 \geq d_1\cdots d_m - d_1$, and thus the bound is optimal. (cf. http://www.encyclopediaofmath.org/index.php/Masser-Philippon/Lazard-Mora_example) (Edit: removed claim with incorrect proof regarding (B))<|endoftext|> TITLE: $L^p$ stability of the Beltrami equation QUESTION [5 upvotes]: Let's assume that $f$ is a quasiconformal homeomorphism of $\mathbb{C}$ with Beltrami coefficient $\mu = \frac{\bar{\partial} f}{\partial f}$. Notice that by definition $\Vert \mu \Vert _{L^{\infty}} < 1$. It is well known that if $\mu_n \rightarrow \mu$ in $L^{\infty}$ then $f_n \rightarrow f$ (the f's are all normalised to fix three points) in the compact open topology (hence pointwise at least). I believe (but am not sure) that the rate of convergence is also known. My question is - In the above setting if $\mu_n \rightarrow \mu$ in $L^p$ for some $\infty >p>1$ then can we say (preferably with effective convergence estimates) that $f_n$ still goes to $f$ in the compact-open topology (or atleast pointwise)? It is also assumed that $\Vert \mu _n \Vert _{L^{\infty}} \rightarrow \Vert \mu \Vert _{L^{\infty}}$. REPLY [5 votes]: EDIT: The old answer to the originally asked question (without the additional assumption of convergence of the $L^\infty$ norms) is below. I am not sure about effective convergence estimates, but the answer to the question is yes, even under slightly weaker assumptions. It directly follows from the Bers-Bojarski convergence theorem, which can be found in the book of Lehto and Virtanen. The statement of the theorem is that if $(f_n)$ is a sequence of $K$-qc mappings converging locally uniformly to a $K$-qc mapping $f$, and if their complex dilatations $\mu_n \to \mu$ a.e., then $\mu$ is a.e. the complex dilatation of $f$. In the situation of your question, if $(f_n)$ is a sequence of normalized $K$-qc maps with dilatations $\mu_n\to\mu$ in $L^p$, any subsequence has a subsequence which converges locally uniformly to a $K$-qc limit (by compactness of the family of normalized $K$-qc maps.) Now you can pass to yet another subsequence where $\mu_n \to \mu$ a.e. Applying the Bers-Bojarski theorem gives that any such subsequential $K$-qc limit $f$ has complex dilatation $\mu$, which implies that the original sequence converges locally uniformly to $f$, where $f$ is the normalized qc map with complex dilatation $\mu$. Note that you really only need $\mu_n \to \mu$ locally in measure, and uniform quasiconformality of the sequence $(f_n)$. Old Answer (Without the assumption that $\| \mu_n \|_\infty \to \| \mu \|_\infty$.) I assume you want to have $f$ to be normalized in some way, otherwise convergence of the maps will not hold even in the case of $L^\infty$ convergence of the complex dilatations. However, even for normalized qc maps, without additional assumptions $L^p$ convergence of the complex dilatations does not imply convergence of the maps. As an example, consider the family $f_k$ of quasiconformal mappings defined in polar coordinates as $(r,\theta) \mapsto (g_k(r),\theta)$, where $g_k$ is the identity for $r > 1$, $g_k(r) = r^{k^2}$ for $1-\frac1k \le r \le 1$, and $g_k$ is linear on $[0,1-1/k]$. Then $f_k$ is quasiconformal in the plane, $\mu_k$ is supported on the annulus $1-\frac1k \le r \le 1$, so $\mu_k \to 0$ in $L^p$ for any $p<\infty$, and $f_k$ converges to the identity outside the unit disk, and to the constant $0$ inside the disk, so the limit is not even continuous. Maybe you want to require that the $f_k$ are uniformly quasiconformal, i.e., that there exists $\kappa<1$ such that $\| \mu_k \|_\infty \le \kappa$ for all $k$?<|endoftext|> TITLE: Eichler-Shimura congruence QUESTION [10 upvotes]: I'm trying to understand the Eichler-Shimura congruence which relates the Hecke operator $T_p$ to Frobenius at $p$ in characteristic $p$. Two possible ways to compute $T_p$ mod $p$ seem to be: A) Look at the map $Div(X_0(N)) \to Div(X_0(N))$ induced by the correspondence $X_0(N) \leftarrow X_0(Np) \to X_0(N)$ where the first map forgets the subgroup at $p$ and the second mods out by it. And then just take this map and look at the induced map $Div(X_0(N)_{F_p}) \to Div(X_0(N)_{F_p})$ in characteristic $p$. B) Look directly at the correspondence $X_0(N)_{F_p} \leftarrow X_0(Np)_{F_p} \to X_0(N)_{F_p}$ in characteristic $p$ and try to compute directly the induced map $Div(X_0(N)_{F_p}) \to Div(X_0(N)_{F_p})$. In case (A), starting with a point in $X_0(N)$, it has $p+1$ lifts to $X_0(Np)$ and thus the induced map on divisors will result in divisors of degree $p+1$. Explicitly, I'm getting: $$ (E,C) \to (E,(C+\ker(F))/\ker(F)) + p (E,(C+\ker(V))/\ker(V)) $$ which looks at lot like the standard Eichler-Shimura relation. But in case (B), starting with an ordinary point in $X_0(N)_{F_p}$, it has $2$ lifts to $X_0(Np)$ --- one where we pick $\ker(F)$ as our group scheme of order $p$, and another for $\ker(V)$. Thus the induced map on divisors will result in divisors of degree $2$, and it seems to yield $$ (E,C) \to (E,(C+\ker(F))/\ker(F)) + (E,(C+\ker(V))/\ker(V)). $$ What's wrong with the argument in case B? REPLY [13 votes]: There's a little more geometry here that should be accounted for in characteristic $p$. Namely, the curve $X_0(Np)_{\mathbb{F}_p}$ is reducible -- its two components are isomorphic to $X_0(N)_{\mathbb{F}_p}$ and they intersect transversally at the supersingular points (see e.g. Ribet and Stein's online notes). So the components are disjoint on the ordinary locus and account for the two subgroups you wrote down. The left pointing map $\pi: X_0(Np)_{\mathbb{F}_p} \to X_0(N)_{\mathbb{F}_p}$ in the correspondence is an isomorphism on the component corresponding to $\ker (F)$, but it is an inseparable degree $p$ map on the second component. Roughly speaking, that this is degree $p$ and inseparable comes from the fact that there are $p$ "non-canonical subgroups" in characteristic 0 which all reduce to $\ker (V)$ in characteristic $p$. To compute the effect of the correspondence on divisors, you need to pull back divisors along an inseparable map. So your argument in Case B is forgetting to multiply by the ramification index (which is $p$) for the unique preimage of $(E,\ker(V))$ under $\pi$ in the "$\ker (V)$ component" of $X_0(Np)_{\mathbb{F}_p}$ Edit: The OP asks how to check the inseparability rigorously and without using characteristic 0. As Eric says in the comments, this is a somewhat subtle issue which can be found in Katz-Mazur (see section 13.5), probably also in Deligne-Rapoport. The Ribet-Stein book has a more approachable discussion (read the proof of Theorem 12.6.4 carefully), but they implicitly use the fact that the Frobenius map $X_0(N) \to X_0(N)$ corresponds under the moduli interpretation to the Frobenius map $(E, C) \mapsto (E^{(p)}, C^{(p)})$, which is really the heart of the matter. But if you're willing to ignore issues of representability of functors, then there's an elementary way to see this last fact: first note that base change reduces us to the case where $N = 1$. Then use the parameterization of $X_0(1)$ by the $j$-invariant to see that both the Frobenius $X_0(1) \to X_0(1)$ and the map $E\mapsto E^{(p)}$ can be identified with $j \mapsto j^p$.<|endoftext|> TITLE: Can all $L^2$ holomorphic functions on a domain vanish at a particular point? QUESTION [17 upvotes]: Let $\Omega \subset \mathbb{C}^n$. Is it possible that there is a point $p \in \Omega$ such that every $f \in A^2(\Omega) = L^2(\Omega) \cap \mathcal{O}(\Omega)$ has a zero at $p$? The space $A^2(\Omega)$ is called the Bergman Space of $\Omega$. I ask since on page 56 of the second edition of Krantz's "Function Theory of Several Complex Variables", he observes that this is not possible in the bounded case: here it is obvious that no such point can exist, since the constant functions are $L^2$ and holomorphic. This got me thinking about the unbounded case. Another way of phrasing this question is "Is there an unbounded domain $\Omega$ for which the Bergman kernel has a zero at some point on the diagonal?" Such a domain would have many strange properties. To address Lukas Geyer's comment, I would like to further add the stipulation that $A^2(\Omega)$ does not consist only of the constant function $0$. I would like to point out that there are interesting examples of finite dimensional Bergman spaces, but the examples I have looked at do not have this property. REPLY [6 votes]: Since it has already been answered I would like to add that the corresponding problem in dimension $1$ has negative answer (this is sometimes called Virtanen's theorem). In dimension higher that $1$ there are a lot of examples as above, one can have additional properties as for example pseudoconvexity so these domains are not ''so exotic''. On the other hand in those examples the set where all $A^2(\Omega)$ vanish is a subvariety of dimension $n-1$ in $\Omega$ I am unaware of any example in which such a set could be an isolated point. This would be really interesting to know at least from the perspective of algebraic geometry – the absence of such points corresponds to the property of base point freeness.<|endoftext|> TITLE: The view from inside of a mirrored tetrahedron QUESTION [50 upvotes]: Suppose you were standing inside a regular tetrahedron $T$ whose internal face surfaces were perfect mirrors. Let's assume $T$'s height is $3{\times}$ yours, so that your eye is roughly at the centroid, and that you look perpendicular to a face:           My question is: Q. What do you see? Either qualitatively; or if anyone can find an image, that would be revealing. Asking the same question for the view inside a mirrored cube is easier to visualize: the opposing parallel square-face mirrors would produce a "house of mirrors" effect (in three perpendicular directions).                                         (Image from this link.) Addendum 1 (21Nov2014). To respond to Yoav Kallus' (good) question, here is a quote from Handbook of Dynamical Systems, Volume 1, Part 1. (ed. B. Hasselblatt, A. Katok), 2010, p.194: Addendum 2 (24Nov2014). Now that we have Ryan Budney's amazing POV-ray images, I would appreciate someone making an attempt to describe his $32$-reflection image qualitatively. I find it so complex that this may be an instance where a thousand words might be superior to one picture. REPLY [4 votes]: I don't know why this question got bumped to the front page, but since it did, I might as well answer that I made a movie of this a long time ago, showing the view inside a (Euclidean) regular tetrahedron with reflective faces (tinted red, green, blue and white, and intrinsically luminous so we can see something!), as well as what it's like to move along a straight line "through" the mirrors (equivalently, a billiard ball path, but with the view flipped at each reflection so that we seem to go through the mirror). In comparison, this video, made by my friend Vincent Nesme¹, shows a rendering of the 600-cell as a regular tiling of the constantly curved 3-sphere by 600 (spherical!) regular tetrahedra, and moving in a straight line (great circle) in this geometry. I had previously made such a video by using gnomonic projection to map the lines into Euclidean space and feeding the result into POVray, but while this correctly projects the lines, it does not correctly project their thickness. Vincent was much more patient than I was and wrote his own raytracer for this.<|endoftext|> TITLE: Stability of minimal surfaces QUESTION [6 upvotes]: Let $\Gamma$ be a prescribed $n-2$ dimensional set and assume $S \subset R^n$ is a minimal hyper-surface with respect to some smooth metric $g$ on $R^n$, and $\partial S= \Gamma$. Is $S$ is stable with respect to variations in the metric $g$? I believe the stability problem for minimal surfaces has to be well understood but I can not find any references. REPLY [12 votes]: Now that your comment has clarified your question, we can answer it: The answer is 'no'. There is the following well-known example: Consider the following family of circles: $C_\lambda$ is defined as $x^2+y^2 = 1$ and $z = \lambda$. Let $\lambda>0$ be fixed and orient $C_{-\lambda}$ counterclockwise and orient $C_\lambda$ clockwise. Then for $\lambda$ sufficiently small, there will be a minimal surface of rotation, a catenoid, of the form $x^2+y^2 = c^2\cosh(z/c)$ for some $0 TITLE: Continuity in Banach space for non-linear maps QUESTION [5 upvotes]: I want to find an example of a Banach space $X$ and a continuous map $f:X\rightarrow X$ such that $f$ is not bounded on the unit ball. I do not doubt that such an example exists, but I cannot make it explicit. REPLY [5 votes]: For $X=\ell^1$, put $f(x)=\sum_{n\ge 1} nx_i^n$ (this one is even analytic if you understand that word in a not too restrictive sense).<|endoftext|> TITLE: Is the Amitsur-Levitzki identity essentially unique? QUESTION [15 upvotes]: Let us consider the matrix algebra. $Mat_n(\mathbb{C})$. The Amitsur-Levitzki identity states that for any matrices $X_1, X_2, ..., X_{2n} \in Mat_n(\mathbb{C})$ the sum $\Sigma_{\sigma \in S_{2n}} sgn(\sigma)X_{\sigma(1)}...X_{\sigma(2n)}$ vanishes identically. Is it true that any identity (noncommutative polynomial, which always vanishes) between the matrices from $Mat_n(\mathbb{C})$ follows from Amitsur-Levitzki in some sense$^*$? $*$ - I assume the following precise meaning: let us consider the oper of "noncommutative polynomial mappings" with the substitution as the composition of operations. It naturally (also non-linearly) acts on any associative algebra (as we can evaluate an $r$-tuple noncommutative polynomial of $k$ variables on the $k$-tuple of elements of an algebra to obtain an $r$-tuple). Is it true that the polynomials that act trivially on the $Mat_n(\mathbb{C})$ are precisely the ideal, generated by the Amitsur-Levitski polynomial? REPLY [3 votes]: As mentioned before (for $2\times 2$ matrices), the answer is "no" . Here I give some reference on the general case. The identity of algebraicity is defined as: $$ a_n(x,y_1,\dots, y_n)= \displaystyle \sum _{\sigma \in S_{n+1}} (-1)^\sigma x^{\sigma(0)}y_1 x^{\sigma(1)}y_2 \cdots x^{\sigma(n-1)}y_n x^{\sigma(n)} $$ In boook [1] we have: Exercise 7.1.12 Show that the identity of algebraicity for $M_k(K)$, $k > 1$, does not follow from the standard identity $St_{2k}$. In the book there is a hint for the exercise, but the original references for that are preprint [2] and paper [3]. Also in paper [4] there are other polynomial identities for $M_3(K)$ which does not follow from $St_6$ and $a_3$. There you also find a good list of references on the subject. References: [1] V. Drensky, Free Algebras and PI-algebras: Graduate Course in Algebra, Springer, Singapore, 1999. [2] G. M. Bergrnan, Wild automorphisms offree P.I. algebras and some new identities, Preprint, Berkeley (1981). https://math.berkeley.edu/~gbergman/papers/unpub/wild_aut.pdf [3] V. S. Drensky, A.K. Kasparian, Some polynomial identities of matrix algebras, C. R. Acad. Bulg. Sci. 36 (1983), 565-568. [4] M. Domokos (1995) New identities for 3 × 3 Matrices, Linear and Multilinear Algebra, 38:3, 207-213, http://dx.doi.org/10.1080/03081089508818356<|endoftext|> TITLE: Powers of finite simple groups QUESTION [9 upvotes]: I have heard about the following result: for each finite simple non-abelian group $S$ and each natural number $r\ge 2$ there exists a number $n=n(r,S)$ such that the power $S^n$ is $r$-generator but $S^{n+1}$ is not $r$-generator. What is known about the numbers $n(r,S)$? Could someone give me references to this, please? (I have posted this already on mathstackexchange.com, but did not get a response.) Edit: This question is in a sense a converse to Bounding from below the cardinality of a set of generators of the $n$-fold cartesian product of a finite group. There it is basically asked for a given (arbitrary, finite) group $G$ and a given number $n$, how small can a generating set for $G^n$ possibly (not) be. In my question the input parameters were a finite simple group $S$ and a number $r\ge 2$ and the question was how big a number $n$ can possibly be so that $r$ elements are sufficient to generate the power $S^n$. Also I was interested in how this number (the biggest such $n$) is actually computed in concrete examples (or whatever is known about the computation of these numbers). Basically I wanted to know, given a finite simple non-abelian group $S$ and a number $r$, the product of how many copies of $S$ do I need to take to get the $r$-generated free object in the formation generated by $S$. @Editors/moderators: please feel free to delete the question if it is inappropriate. REPLY [7 votes]: See Collins's thesis, Theorem 2.22, page 21. Theorem 2.22. Let $S$ be a nonabelian simple group and $h_{n-1}(S) < k \le h_n(S)$. Then $r(S^k)=n$. Here, $r(G)$ is the minimal number of generators of $G$ and $h_n(G)$ is the reduced Euler function i.e. the number of generator sequences of length $n$ of $G$ divided by $|{\rm Aut}\ G|$.<|endoftext|> TITLE: Galois group for 0-dimensional motives QUESTION [15 upvotes]: It is my understanding that in dimension 0, the theory of motives should just be Galois theory for fields. I am hoping to find a reference or two to help me get some things straightened out. One can construct a category $\mathcal{M}$ of 0-dimensional motives over $\mathbb{Q}$. If I'm not mistaken, one possible construction is to start with the category of 0-dimensional varieties over $\mathbb{Q}$, with correspondences as morphisms, and formally adjoin images of idempotents. This $\mathcal{M}$ is a neutral Tannakian category, meaning it is equivalent to the representations of some pro-algebraic group. In this case that group should be the absolute Galois group of $\mathbb{Q}$. There are several fiber functors $\omega_{dR},\omega_{\ell},\ldots$ from $\mathcal{M}$ to vector spaces (over various fields), given by the different motivic realizations (de Rham, $\ell$-adic,...). Each fiber functor $\omega_\bullet$ has an automorphism group $G_{\mathbb{Q},\bullet}$, which is an affine pro-algebraic group. As I understand it, the groups $G_{\mathbb{Q},\bullet}$ should be viewed as realizations of some motivic absolute Galois group $G_{\mathbb{Q}}$. Each functor $\omega_\bullet$ identifies $\mathcal{M}$ with the representations of $G_{\mathbb{Q},\bullet}$. What are the groups $G_{\mathbb{Q},\bullet}$, as $\bullet$ ranges over the different realizations on $\mathcal{M}$? Can someone point me to a reference where these groups are described? I would love to find a reference that also describes the construction of $\mathcal{M}$. REPLY [6 votes]: What follows is a hands-on explanation (not a complete proof!) of why the Tannakian group $$G_{\textrm{dR}}(AM(\mathbb{Q}))$$ of the category of Artin motives is a nontrivial inner form of $$G_{\textrm{B}} (AM(\mathbb{Q})) = \underline{\textrm{Gal}}_\mathbb{Q}.$$ Ironically, the source for this answer is actually the original asker, who three years later published an interesting paper that clarifies this point: https://arxiv.org/abs/1706.06573. All I've done is translate a sliver of his work into very concrete language for the benefit of posterity. I'm actually going to work with a single 0-dimensional motive $X = \mathfrak{h}(\textrm{Spec } k)$ for a finite Galois extension $k/\mathbb{Q}$. This will produce a Tannakian group $G_{X, \bullet}$ associated to the subcategory of motives $\langle X \rangle$ (motives generated from $X$ by tensor powers, duals, and direct sums). $G_{X, \bullet}$ will be a quotient of the full Galois group for 0-dimensional motives; it should be clear how to fit the different $G_{X, \bullet}$ together compatibly to build the full groups $G_{\bullet}(AM(\mathbb{Q}))$. (It's just like building the full absolute Galois group out of its action on finite Galois extensions!) For completeness, let's first see why the Betti realization is the Galois group viewed as a constant group scheme. According to the Tannakian formalism, a $K$-point of the algebraic group $G_{X,\textrm{B}}$ is a system of $K$-linear automorphisms $$T_Y: H_B(Y, \mathbb{Q}) \otimes K \to H_B(Y, \mathbb{Q}) \otimes K$$ for each 0-dimensional motive $Y \in \langle X\rangle$ such that the following diagram commutes for all maps of motives $\varphi: Y \to Y'$: $\require{AMScd}$ \begin{CD} {H_B(Y, \mathbb{Q}) \otimes K} @> {T_Y}>> H_B(X, \mathbb{Q}) \otimes K\\ @V{H_B(\varphi)} VV @V{H_B(\varphi)}VV\\ {H_B(Y', \mathbb{Q}) \otimes K} @> {T_{Y'}}>> H_B(Y', \mathbb{Q}) \otimes K\\ \end{CD} The system $T_Y$ is also required to be compatible with tensors and duals. Let's first think about our generating motive $X$. We know that $H_B(X, \mathbb{Q})$ is the dual of the vector space spanned by $\textrm{Spec }\mathbb{C}$, the set of embeddings $k \to \mathbb{C}$. Maps of Artin motives $X \to X$ all happen to be (linear combinations of maps) of the form $g: k \to k$, where $g$ is an element of $G = \textrm{Gal}(k/\mathbb{Q})$. Since $g$ acts on an embedding $k \to \mathbb{C}$ by pullback (precomposition), the $G$ action on $\textrm{Spec }k (\mathbb{C})$ is the left regular representation (left multiplication on $\mathbb{Q}[G]$); then the action on $H_B(X, \mathbb{Q})$ is the dual, also known as the right regular representation (multiplication on the right by $g^{-1}$). In other words, in the special case of a morphism $g: X \to X$ induced by a Galois element $g: k \to k$, our commutative diagram becomes \begin{CD} {\mathbb{Q}[G] \otimes K = K[G]} @> {T_X}>> K[G]\\ @V R_g VV @V R_g VV\\ K[G] @> {T_X}>> K[G]\\ \end{CD} Just as in the Tannakian formalism for a finite group $G$, we need the further constraint on $T_X$ that comes from morphisms between $X \times X$ and $X$. Consider the multiplication morphism $k\otimes k \to k$; it induces a map $f: H_B(X) \otimes H_B(X) \to H_B(X)$. Under the isomorphism $H_B(X) \cong \mathbb{Q}[G]$, this map is $$g \otimes h \mapsto \begin{cases} 0, & \textrm{if } g\neq h \\ g, & \textrm{if } g = h. \end{cases}$$ (To check this, recall that we've identified $g \in \mathbb{Q}[G]$ with the dual basis element to $\iota \circ g$ in $\textrm{Spec }\mathbb{C}$. So $g\otimes h$ is dual to the map $k \otimes k \to \mathbb{C}$ that embeds the first factor by $\iota \circ g$ and the second by $\iota \circ h$, then multiplies them in $\mathbb{C}$. To evaluate $f(g \otimes h)$ on an embedding $\iota \circ g': k \to \mathbb{C}$, we pull back to $k\otimes k \to k \to \mathbb{C}$ and check whether we got the embedding which acts as $\iota \circ g$ on the first factor and $\iota \circ h$ on the second. We will get 0 unless $g = h = g'$.) (Note: this map $f$ is actually the same as the cup product map, because $k\otimes k \to k$ by multiplication is the diagonal morphism.) Now our map $T_X$ is supposed to make the following diagram commute: \begin{CD} K[G] \otimes K[G] @> {T_X \otimes T_X}>> K[G] \otimes K[G] \\ @V f VV @V f VV\\ K[G] @> {T_X}>> K[G]\\ \end{CD} (Note that $f$ and $T_X$ are extended linearly to $H_B(X) \otimes K$. To reiterate, this is because we are looking for $K$-points of the motivic Galois group.) At this stage, we're basically reduced to some annoying calculations. It turns out that there is only one way to make these diagrams commute: $T_X$ must be left multiplication by an element of $G$! (Such maps are also compatible with all tensor products, duals, and direct sums.) This result is independent of $K$, so it shows that $$G_{X,B} = \underline{\textrm{Gal}(k/\mathbb{Q})}.$$ Note as well that the calculations we have to carry out are exactly those for the category of representations of $G$ and the universal representation $K[G]$ (where $g$ acts as right multiplication by $g^{-1}$). This situation is also completely dual to the left regular representation; for example, $f$ is dual to the comultiplication map $K[G]\to K[G]\otimes K[G]$ given by $g \mapsto g\otimes g$. Having said all that, it's time for us to think about the de Rham fiber functor. Automorphisms of this fiber functor are systems $S_X$ of $K$-linear automorphisms $$H_{dR} (X) \otimes K \to H_{dR} (X) \otimes K$$ such that, for all maps of motives $\varphi: Y\to Y'$, $Y, Y' \in \langle X\rangle$, we have a commuting diagram: \begin{CD} {H_{dR} (Y) \otimes K} @>{S_Y} >> {H_{dR} (Y) \otimes K}\\ @V H_{dR} (\varphi) VV @V H_{dR} (\varphi) VV\\ H_{dR} (Y') \otimes K @> {S_{Y'}}>> H_{dR} (Y') \otimes K\\ \end{CD} Recall that the 0th de Rham cohomology is exactly the global sections of the structure sheaf, so for $X = \textrm{Spec } k$ we have $$H_{dR} (X) = \Gamma(\textrm{Spec } k, \mathcal{O}_{\textrm{Spec }k}) = k.$$ Once again, morphisms $X \to X$ in the category of Artin motives are just pullbacks of field automorphisms $g: k \to k$, and now the induced map on $H_{dR}(X) = k$ is just $g$. So our commutative diagram specializes to \begin{CD} {k \otimes K} @>{S_X} >> {k \otimes K}\\ @V g \otimes 1 VV @V g\otimes 1 VV\\ {k \otimes K} @>{S_X} >> {k \otimes K}\\ \end{CD} Now, by the Normal Basis Theorem, there is an element $\alpha \in k$ whose Galois conjugates form a basis for $k$ as a vector space over $\mathbb{Q}$. This means that $k \otimes K$ is isomorphic to $K[G]$ with Galois action by left multiplication -- $cg$ corresponds to $g\alpha \otimes c$ -- and you might think that we were back in the situation for the Betti fiber functor. The difference comes up when we consider the second commutative diagram, that coming from the map $X \times X \to X$ induced by $k \otimes k \to k$. But the key point is that the induced map analogous to the one we called $f$ before, $f': H_{dR}(X) \otimes H_{dR} (X) \to H_{dR} (X)$, is NOT equivalent to $f$ as a map of $G$ representations over $\mathbb{Q}$ (!!!!). In fact, if $K = \mathbb{Q}$, then there are NO possible $S_X$ except for the identity (unless $G$ is abelian). To see this, note that, by the two commutative diagrams, a valid $S_X$ is a $\mathbb{Q}$-linear map $S: k \to k$ such that $S(xy) = S(x)S(y)$ and $S(gx) = gS(x)$. But if $S$ respects multiplication and addition, it's a field automorphism, so $S(x) = hx$ for some $h \in G$. But then we have $hgx = ghx$ for all $g$, so $h$ is a central element. Only abelian Galois groups can have nontrivial centers, so $T = 1$ on $H_{dR} (X)$ and thus on all motives in $\langle X \rangle$. On the other hand, if we base change to a large enough $K$, then we do get some $K$-points. Suppose $\alpha$ is a primitive element of $k$ (one whose powers generate $k$ over $\mathbb{Q}$), and call its minimal polynomial $f(\alpha)$. Then $$k \otimes_{\mathbb{Q}} K = \mathbb{Q}[x]/f(x) \otimes_\mathbb{Q} K = K[x]/f(x) = \prod_{\alpha_i} K[x]/(f(x) - \alpha_i) = \prod_{\alpha_i} K.$$ Here, $\alpha_i$ denote the Galois conjugates, and the third isomorphism is by the Chinese Remainder Theorem. The multiplication in this ring (corresponding to $f'$ linearly extended to $K$) looks exactly like the multiplication $f$ on $K[G]$, which we recall is of the form: $$f(c_1g \otimes c_2h) = \begin{cases} 0, & \textrm{if } g\neq h \\ c_1c_2g, & \textrm{if } g = h. \end{cases}$$ Here, $g$ indexes which factor of the product we are in; think of it as the idempotent with a 1 in one index and a 0 in the rest. So after this base change, the arguments in the Betti realization case go through exactly, and we still have $G_{X, dR} (K) = \textrm{Gal} (k/\mathbb{Q})$. Whew! This is all done more abstractly, more succinctly, and far more canonically in Rosen's paper, but I hope the concrete example is helpful to somebody who stumbles across this question in the future.<|endoftext|> TITLE: Injectivity of Rewrite Rule in a Free Lie Algebra QUESTION [7 upvotes]: Let $L$ be a free Lie algebra (over $\mathbb{Q}$) on generators $x_1, x_2, \ldots, x_n$, and let $V_k$ be the subspace spanned by the $k$-fold brackets. Let $U_1 = \mathrm{span}\{ x_i | i< n\}\subseteq V_1$ and let $W_n \subseteq V_n$ be the span of the $n$-fold brackets that involve $x_n$ at least once. The bracket defines a function $V_k\otimes V_\ell \to V_{k+\ell}$ (which is what I think of as bracketing and rewriting in terms of a chosen basis, but actually it's just bracketing). Question: Is the composition $$ U_1 \otimes W_m \to V_1\otimes V_m \to V_{m+1} $$ injective? ($m$ has no particular relation to $n$) I have asked MAPLE to compute the dimensions of these vector spaces and it seems that the dimension of the target is generally large enough to accommodate such an injection (though I haven't proved a dimension inequality yet). REPLY [9 votes]: Yes. Embed the free Lie algebra in the free associative algebra, then consider the monomial with $x_n$ as far right as possible (i.e., with the largest number of other $\{x_i,i\leq n\}$ before it): it won't be cancelled. Let me now give details. Consider $F$ the free (associative but non-commutative) algebra (over $\mathbb{Q}$) on generators $x_1,\ldots,x_n$. Let $[P,Q]_F = P Q - Q P \in F$ be the usual commutator for elements $P,Q\in F$. Consider $L_F$ the smallest subset of $F$ containing $x_1,\ldots,x_n$ which is closed under commutators and linear combinations. Then $L_F$ endowed with the Lie bracket $[\cdot,\cdot]_F$ is isomorphic to the free Lie algebra $L$. Under this isomorphism, $U_1$ is still the span of $x_1,\ldots,x_{n-1}$, and $W_m$ is included in the vector space $W^F_m$ of polynomials in $x_1,\ldots,x_n$ with no monomial independent of $x_n$. An element of $U_1\otimes W^F_m$ is $\sum_{i=1}^{n-1} x_i \otimes P_i$ for some $P_i\in W^F_m$. Its image under the composition you care about is $S = \sum_{i=1}^{n-1} (x_i P_i - P_i x_i)$. Since everything is linear, we just need to show that $S \neq 0$ assuming that not all $P_i = 0$. Among all monomials of all $P_i$, consider $m = x_{a_1} \cdots x_{a_k} x_n x_{b_1} \cdots x_{b_l}$ with $k$ maximal. Say it is in $P_j$. Then $S$ contains the monomial $x_j m = x_j x_{a_1} \cdots x_{a_k} x_n x_{b_1} \cdots x_{b_l}$ with some non-zero coefficient coming from $x_j P_j$, and this monomial does not appear in any other term of $S$. Indeed, the monomial cannot appear in $x_i P_i$ for $i\neq j$ since it starts with $x_j$ and not $x_i$, and the monomial cannot appear in $P_i x_i$ for any $i$ since this would require the presence in $P_i$ of a monomial $m'$ starting with $x_j x_{a_1} \cdots x_{a_k} x_n$, which contradicts the maximality of $k$. It is probably possible to extend the proof to show that $U_k \otimes W_m \to V_{k + m}$ is injective, where $U_k$ is the span of $(k-1)$-fold brackets of $\{x_i, i TITLE: Groups with a unique composition series QUESTION [8 upvotes]: Which finite groups $G$ have a unique composition series? I don't mean in the sense of the Jordan-Holder theorem, but rather actually unique. Some examples are the cyclic groups $C_{p^n}$ and the symmetric group $S_3$ (which has the unique normal subgroup $A_3$). REPLY [14 votes]: Such a group $G$ must have a unique minimal normal subgroup $M,$ which must itself be simple by the uniqueness of the composition series. If $G$ is a $p$-group, it follows by induction that $G$ is cyclic ($G/M$ is cyclic by induction, so $G$ is Abelian, and hence cyclic by the uniqueness of $M$). Suppose then that $G$ is not a $p$-group. If $G$ is solvable, then it follows from the above ( and the uniqueness of the composition series) that $F(G)$ is a cyclic $p$-group. Then $G/F(G)$ is Abelian, hence cyclic of order $q^{m}$ for some prime $q \neq p$ (and in fact $q^{m}$ divides $p-1).$ Suppose then that $G$ is not solvable. Note that $G/G^{\prime}$ is a cyclic $p$-group for some prime $p,$ by the uniqueness of the composition series. More generally, $G/G^{(\infty )}$ has one of the structures above, where $G^{(\infty)}$ is the terminal member of the derived series of $G.$ We next describe the structure of $G^{(\infty)},$ so, for ease of exposition, we assume from now on that $G$ is perfect. Then since $F(G)$ is cyclic, we have $F(G) = Z(G)$, which is a (possibly trivial) cyclic $p$-group for some prime $p.$ Furthermore, $F(G)$ is the largest solvable normal subgroup of $G$. Then, as above, $G/Z(G)$ has a unique minimal normal subgroup, which is non-Abelian simple, and has trivial centralizer. Using the Schreier conjecture, it follows that $G$ is quasi-simple in the case under consideration. In summary, a general finite group $G$ with a unique composition series has one of the following structures: $G$ may be a cyclic $p$-group for some prime $p.$ $F(G)$ may be a cyclic $p$-group, and $G$ is the semidirect product of $F(G)$ with a cyclic $q$-group for some prime $q \neq p.$ $G$ is not solvable, $F^{\ast}(G)$ is quasi-simple, and $Z(G)$ is a cyclic $p$-group for some prime $p.$ Furthermore, $G/F^{\ast}(G)$ is solvable with a unique composition series, so with one of the structures above (note also that $G/F^{\ast}(G)$ is a subgroup of the outer automorphism group of the simple group $F^{\ast}(G)/Z(G)$). (Note that any group with any of these structures does indeed have a unique composition series).<|endoftext|> TITLE: Conjecture regarding closest point inside a discrete ball to a line QUESTION [13 upvotes]: I'm a PhD student in image processing, where I've stumbled into a problem that seems to be essentially number theory. I've hunted around online and while I've found many results on similar problems, this particular problem I cannot seem to find a solution to: Given a line in the plane passing through the origin making angle $\theta$ with the $x$-axis, I am trying to determine the closest nonzero Gaussian integer $n+im$ to the line obeying $|n+im| \le r$. I have a conjecture which is backed up by numerous computer tests, but no proof. The conjecture is as follows: Let $\Theta(r) = \{\theta_1,\theta_2,...,\theta_N\}$ denote the set of angles representable using Gaussian integers of this form. That is, each $\theta_k=Arg(n+im)$ for some non-zero Gaussian integer $n+im$ of norm at most $r$. Find $\theta_k, \theta_{k+1}$ straddling $\theta$, i.e. $\theta_k \le \theta < \theta_{k+1}$. Let $n+im$ be a Gaussian integer that solves our minimization problem. Then either $Arg(n+im)=\theta_k$ or $Arg(n+im)=\theta_{k+1}$. REPLY [2 votes]: I think the proof of @domotrop is not complete. I will first present my own proof. After several revisions, it is now complete. Then I will express my concern for the proof of @domotrop, and propose a fix. In the following picture, $O$ is the origin. Let $P$ and $Q$ be the closest points with angle $\theta_k$ and $\theta_{k+1}$. The point $-Q$ is the opposite of $Q$. Without loss of generality, we assume that $P$ is closer to $\ell$ than $Q$. Up to a change of sign, any closer lattice point with larger angle from $\ell$ will be in the grey area, for example the blue point. Then with the help of a parallelogram, we can find the red lattice point in either of the green areas with smaller angle from $\ell$. This red point is definitely inside the circle, so it contradicts our assumption. I think the proof of @domotrop is not complete. My concern for @domotrop's proof was the following: For example, in the picture of @Rob, $P$ minimize the angle with $\ell$ from below, but the point $(-1,-1)$ is closer with a larger angle from $\ell$. Therefore, if $\arg(n+im)$ takes value in $(-\pi,\pi]$, the points like $(-1,-1)$, which are in the lower half-space, is something that we should be careful with. That is, either there is no such point, or the opposite of the closest is indeed the next angle. The fix consists of considering the opposite of the $\theta_{k+1}$ as in my proof. Then, in the above picture, since the green parts are empty, so is the grey part, and Q.E.D. @domotrop explained that we only needs to consider the two quadrants containing $\ell$. Then his proof is complete after repeating the same argument above $\ell$ in the first quadrant, and replacing $y$ with $x$ when constructing the parallelogram.<|endoftext|> TITLE: How large can you draw an island on a map? QUESTION [12 upvotes]: A cartographer friend asked me this question: could you classify (shapes of) islands by how much space they occupy on a map (comparatively to how much space is occupied by water) if you draw them as large as possible? He had something in mind like: if $$\frac{\mathrm{area}(\mathrm{island})}{\mathrm{area}(\mathrm{water} + \mathrm{island})} < 50\%$$ then the island is "thin", otherwise the island is "fat". I thought about it a little bit but could not come up with anything enlightening. Here is a mathematical translation: let $P$ be a bounded domain in the plane (I suppose all you need is $P$ to be a measurable set). Define the "fatness" constant $$f(P) = \max_D \frac{\mathrm{area}(P)}{\mathrm{area}(D)}$$ where the max is taken over all Euclidean disks $D$ containing $P$. I don't have a very precise question, I just wonder: is there anything interesting to be said about $f$? Has it been studied? Does it have notable properties? Is it computable in some cases? etc. Note that this "fatness constant" could be generalized in several ways. For starters one could take Euclidean rectangles instead of disks. Another example of generalization would be to ask the question on the round sphere rather than in the plane. Here is another idea that comes to my mind: what if the island lives on the sphere but you want to draw it in the plane? Could we ask that same question over all representations of the island that are conformal (or satisfy some other property)? NB: As has been pointed out, the words "thin" and "fat" I used here are probably a poor choice of terminology, but I did not have a better idea. REPLY [21 votes]: Similar issues come up in studying gerrymandering (drawing political districts with partisan objectives), where it's useful to have a measure of how "irregular" a region is. You can read about various classical irregularity measures in this political science paper: Measuring the Compactness of Legislative Districts (Young, Legislative Studies Quarterly, 1988). The author calls your measure the "Roeck Test". For more mathematically-oriented references, see the answers to this math.stackexchange question. EDIT: You asked about computability, so it's worth mentioning that there's a linear time algorithm to find the smallest circle containing a plane region.<|endoftext|> TITLE: Flat manifolds and irreducible representations QUESTION [11 upvotes]: Let $M$ be a compact Riemannian manifold with vanishing curvature of Levi-Civita connection. Such manifolds were classified by Bieberbach; sometimes they are called Bieberbach manifolds. According to this classification, a Bieberbach manifold is a quotient of a torus by a finite group freely acting on it by isometries. Crystallographic group is a fundamental group of a Bieberbach manifold. There is an exact sequence $$ 0 \rightarrow {\mathbb Z}^n \rightarrow G \rightarrow L\rightarrow 0,$$ where $G$ is a crystallographic group, ${\mathbb Z}^n$ a group of translations in $G$, and $L$ the holonomy group of the Levi-Civita connection. The group $L$ is naturally embedded to the group $GL(T_xM)$ of automorphisms of the tangent space of the Bieberbach manifold. Could $T_x M$ be irreducible as a representation of $L$? I have checked for dimension $\leq 5$, and $T_xM$ is never irreducible for dimension $\leq 5$. REPLY [4 votes]: Here is the solution (with thanks to Misha Kapovich who pointed this out). Gerhard Hiss, Andrzej Szczepanski, "On torsion free crystallographic groups", Journal of Pure and Applied Algebra Volume 74, Issue 1, 10 September 1991, Pages 39-56 Abstract: The torsion free crystallographic groups arise as fundamental groups of compact flat Riemannian manifolds. Let R be a crystallographic group with point group G and translation group T. In this paper we consider the G-module T⊗z, for which we prove: If R is torsion free, then G does not act irreducibly on T⊗z. A proof of this theorem for solvable groups G was first given by G. Cliff. The theorem proves a conjecture made by the second author. The proof of the theorem uses the classification of the finite simple groups.<|endoftext|> TITLE: Powers of traces, integrals over spheres and class functions QUESTION [10 upvotes]: I asked this on math.StackExchange a while back but got no answers. I hope I'll be forgiven for the double post. Let $V$ be a complex vector space of dimension $\operatorname{dim}_{\mathbb C} V = n$, equipped with a Hermitian inner product $\langle \,\cdot\,,\,\cdot\, \rangle$, and $A$ be an endomorphism of $V$. I am interested in calculating the functional $$ \alpha_k(A) := \int_{S^{2n-1}} \langle Av, v \rangle^k \, d\mu, $$ where $d\mu$ is the Lebesgue measure induced by the inner product and $S^{2n-1}$ is the unit sphere with respect to that product. For $k = 1$, this functional is a nonzero multiple of the trace, as shown in some older questions on math.SE. These kind of integrals pop up, for example, when we try to calculate the differential form that represents the $k$-th Segre class of a Hermitian holomorphic vector bundle. Now, I don't really know any representation theory, but by extrapolating from one of the answers to the older questions the following should make a little sense: By polarizing we see that this functional is induced by the multilinear form $$ \alpha_k(A_1, \ldots, A_k) = \int_{S^{2n-1}} \langle A_1v, v \rangle \cdots \langle Av_k, v \rangle \, d\mu. $$ This form is invariant under the action of $U(n) \times \cdots \times U(n)$ ($k$ times) on $\operatorname{Aut}(V^{\oplus k})$ (by conjugation on each factor), so $\alpha_k$ should be a class function on that representation. As such, it should be a linear combination of the characters of the representation. This should mean that $\alpha_k$ is a linear combination of the multilinear forms $$ \def\tr{\operatorname{tr}} \prod_{|J|=k} \tr(A_J), $$ where we define $A_J = A_{j_1} \cdots A_{j_k}$ for a multiindex $J = (j_1,\ldots,j_k)$. For $k = 2$, for example, these are $$ \tr(A_1^2), \quad \tr(A_1) \tr(A_2), \quad \tr(A_2)^2. $$ Of course, $\alpha_k$ is a symmetric form, so there is some redundancy here ($\tr(A_1^2)$ and $\tr(A_2^2)$ must have the same coefficient in our hypothetical linear combination). We can calculate by brute force what happens in a special case here: Suppose $k = 2$ and $A$ is diagonalizable with eigenvalues $\lambda_1, \ldots, \lambda_n$. If $(e_1,\ldots,e_n)$ is an orthonormal basis of $V$ and $v = \sum z_j e_j$ a vector, we then have $$ \langle Av,v \rangle = \sum_{j=1}^n \lambda_j |z_j|^2, \quad \langle Av,v \rangle^2 = \sum_{j,k=1}^n \lambda_j \lambda_k |z_j|^2 |z_k|^2. $$ Omitting some tedious calculations over the sphere, we then find that $$ \eqalign{ \alpha_2(A) = \int_{S^{2n-1}} \langle Av,v \rangle^2 \,d\mu &= \sum_{j,k=1}^n \lambda_j \lambda_k \int_{S^{2n-1}} |z_j|^2 |z_k|^2 \, d\mu \cr &= \frac{1}{n(n+1)}\sum_{j,k=1}^n \lambda_j \lambda_k + \frac{1}{n(n+1)} \sum_{j=1}^n \lambda_j^2 \cr &= \frac{1}{n(n+1)}( \tr(A)^2 + \tr(A^2) ), } $$ as we suspected. Unfortunately, this direct approach is completely hopeless for higher $k$, so we must look for other ways to calculate the integral. Question Can we make my representation-theoretic hand-waiving precise to (1) show the existence of the hypothetical linear combination and (2) actually determine the coefficients in that combination? REPLY [7 votes]: Begin by rewriting $\alpha_k(A_1,\ldots,A_k)$ as follows (just move integrals and traces around): $$ \alpha_k(A_1,\ldots,A_k) = \mathrm{tr}\Big( (A_1 \otimes \cdots \otimes A_k) \int_{S^{2n-1}}(vv^* \otimes \cdots \otimes vv^*) d\mu \Big). $$ Well, the integral on the right is well-known to be $P_{sym}/\binom{n+k-1}{n-1}$, where $P_{sym}$ is the projection onto the symmetric subspace (i.e., the subspace of $V \otimes \cdots \otimes V$ spanned by vectors of the form $v \otimes \cdots \otimes v$). We thus have $$ \alpha_k(A_1,\ldots,A_k) = \frac{1}{\binom{n+k-1}{n-1}}\mathrm{tr}\big( (A_1 \otimes \cdots \otimes A_k) P_{sym} \big). $$ One way to write $P_{sym}$ is as $$ P_{sym} = \frac{1}{k!}\sum_{\pi\in S_k}W_{\pi}, $$ where $W_{\pi}$ is the unitary operator that permutes the $k$ tensor factors according to the permutation $\pi$ (and $S_k$ is the symmetric group on $k$ symbols). Thus $$ \alpha_k(A_1,\ldots,A_k) = \frac{1}{k!\binom{n+k-1}{n-1}}\sum_{\pi\in S_k}\mathrm{tr}\big( (A_1 \otimes \cdots \otimes A_k) W_{\pi} \big). $$ Finally, the trace on the right can be computed in terms of traces of the $A_i$'s by breaking each $\pi$ into disjoint cycles and grouping the $A_i$'s that are within a common cycle within one trace. For example, if $\pi = (1 2)(3 4 5)$ (in cycle notation) then $$ \mathrm{tr}\big( (A_1 \otimes \cdots \otimes A_5) W_{\pi} \big) = \mathrm{tr}(A_1 A_2)\mathrm{tr}(A_3 A_4 A_5), $$ and hopefully it is clear how this generalizes. When $k = 2$, this gives $$ \alpha_2(A_1,A_2) = \frac{1}{n(n+1)}\big(\mathrm{tr}(A_1)\mathrm{tr}(A_2) + \mathrm{tr}(A_1A_2)\big), $$ (the first term comes from the identity permutation, the second term comes from the transposition permutation) which agrees with the formula you found for the $k = 2$ case in your original post. When $k = 3$, we get $$ \alpha_3(A_1,A_2,A_3) = \frac{1}{n(n+1)(n+2)}\big(\mathrm{tr}(A_1)\mathrm{tr}(A_2)\mathrm{tr}(A_3) + \mathrm{tr}(A_1)\mathrm{tr}(A_2A_3) + \mathrm{tr}(A_2)\mathrm{tr}(A_1A_3) + \mathrm{tr}(A_3)\mathrm{tr}(A_1A_2) + \mathrm{tr}(A_1A_2A_3) + \mathrm{tr}(A_3A_2A_1)\big). $$ Again, we get one term in the sum for each of the $k! = 6$ permutations, and hopefully it is clear how they arise from the cycle decompositions of those permutations, and it generalizes to higher $k$ straightforwardly.<|endoftext|> TITLE: Number of median graphs? QUESTION [8 upvotes]: What is the number of $n$-vertex median graphs? These graphs generalize hypercubes and trees, and have many applications. It seems unlikely that a closed form expression is known, so I would also be interested in asymptotics or lower bounds. For more about median graphs see the survey by Klavžar and Mulder: Sandi Klavžar, Henry Martyn Mulder, Median graphs: characterizations, location theory and related structures, J. Combin. Math. Combin. 30 (1999) 103–127. (preprint) Imrich et al. claimed to have shown (at least "intuitively") that there are as many median graphs as there are triangle-free graphs. Their argument involves composing a sequence of injections. Since the sets involved are infinite, this is unfortunately not rigorous enough to conclude anything about the actual numbers. Wilfried Imrich, Sandi Klavžar, Henry Martyn Mulder, Median graphs and triangle-free graphs, SIAM J. Discrete Math. 12 (1999) 111–118. doi:10.1137/S0895480197323494 (preprint) Given the apparent connection with triangle-free graphs, I would also be interested in asymptotics or bounds for the number of $n$-vertex triangle-free graphs, for which OEIS has the following relevant sequences: A006785, A024607. [In an effort to make the question more self-contained, I append the definition of median graph from the Wikipedia link: In graph theory, a division of mathematics, a median graph is an undirected graph in which every three vertices $a$, $b$, and $c$ have a unique median: a vertex $m(a,b,c)$ that belongs to shortest paths between each pair of $a$, $b$, and $c$.] REPLY [6 votes]: The numbers of $m$-edge triangle-free and median graphs are of similar types — at least, the logarithms of these numbers are within a constant factor of each other. In one direction every median graph is triangle-free, and in the other direction the simplex graph of an $m$-edge triangle-free graph is median and has $O(m)$ edges. However, for $n$-vertex graphs, the triangle-free and median graphs have very different numbers. A median graph, or more generally a partial cube, can be completely described by giving a spanning tree together with a partition of the edges of the spanning tree into equivalence classes for the Djokovic equivalence relation, so the number of median graphs is at most exponential in $n\log n$, because it takes $O(n)$ bits to specify a tree and $O(n\log n)$ bits to specify a partition of the edges. On the other hand, the number of triangle-free graphs, or even the number of subgraphs of $K_{n/2,n/2}$, is exponential in $\Theta(n^2)$. I'm not sure whether the number of $n$-vertex median graphs is exponential in $n$, $n\log n$, or something in between, but my intuition is $n$. (There's an easy lower bound that is exponential in $n$: every tree is a median graph.)<|endoftext|> TITLE: Koebe–Andreev–Thurston theorem - where can I find a proof? QUESTION [9 upvotes]: Koebe–Andreev–Thurston theorem (known also as the circle packing theorem) says that any planar graph can be realized by a set of (interior-) disjoint disks corresponding to vertices, such that two discs are tangent iff the corresponding vertices are connected to each other. Where can I find the/a proof of this theorem, and what should I learn to understand it? I prefer proofs which are elementary, but other proofs are welcome too. REPLY [11 votes]: There are many proofs, and I'm not claiming that the following list is complete. New references are welcome. (First proof) Paul Koebe, Kontaktprobleme der konformen Abbildung, Ber. Verh. Sächs. Akad. Leipzig 88 (1936), 141–164 (German) (Thurston's rediscovery and related) Andreev, E. M., Convex polyhedra of finite volume in Lobačevskiĭ space, Mat. Sb. (N.S.) 83 (1970), no. 125, 256–260. (see also) Roeder, Roland K.W., Hubbard, John H. and Dunbar, William D., Andreev’s Theorem on hyperbolic polyhedra, Annales de l’institut Fourier 57 (2007), no. 3, 825–882. William P. Thurston and John W. Milnor, The Geometry and Topology of Three-Manifolds (Variational principle) Yves Colin de Verdière, Un principe variationnel pour les empilements de cercles, Invent. Math. 104 (1991), no. 3, 655–669 (French). Igor Rivin, Euclidean structures on simplicial surfaces and hyperbolic volume, Ann. of Math. 139 (1994), 553–580. Alexander I. Bobenko and Boris A. Springborn, Variational principles for circle patterns and Koebe’s theorem, Trans. Amer. Math. Soc. 356 (2004), no. 2, 659–689. (see also) Günter M. Ziegler, Convex polytopes: extremal constructions and f-vector shapes, Geometric Combinatorics, 2007, pp. 617–691. (An inductive proof ?) Kenneth Stephenson, Introduction to Circle Packing: The theory of discrete analytic functions, Cambridge University Press, Cambridge, 2005. (I also recommend the following completion of the theorem) Graham R. Brightwell and Edward R. Scheinerman, Representations of planar graphs, SIAM J. Discrete Math. 6 (1993), no. 2, 214–229. REPLY [7 votes]: This is proved in sections 13.6 and 13.7 of Thurston's notes. See also Corollary 5.10.4 and section 5.11 of Thurston's book. Another short proof appears in Appendix 2 of Rodin and Sullivan.<|endoftext|> TITLE: Determinant of the oriented adjacency matrix of a tree QUESTION [6 upvotes]: Let $(V,E)$ be a finite oriented directed graph, with vertices and edges ordered, and $M$ the $|V|\times |E|$ matrix with entries $$ m_{ve} = \begin{cases} 1 &\text{if $e$ points at $v$}\\ -1 &\text{if $e$ points from $v$}\\ 0 &\text{otherwise.} \end{cases} $$ If $(V,E)$ is a tree, then this matrix has one more row than being square. If we erase the row corresponding to a vertex $v$, the resulting square matrix is easily seen to have determinant $\pm 1$ or $0$. Is there a simple, known formula for its determinant? (Surely!) Example: consider $1 \stackrel{1}{\to} 2 \stackrel{2}{\to} 3$, with matrix $ \begin{pmatrix} -1&0\\ 1&-1\\ 0&1 \end{pmatrix}$. Then the three choices $v=1,2,3$ give the determinants $1,-1,1$ respectively. REPLY [2 votes]: A matrix is said to be totally unimodular if the determinant of any square submatrix of the matrix is either $0$ or $\pm 1.$ Let $G$ be a graph with incidence matrix $Q(G)$, that is, a matrix corresponding to a finite oriented directed graph of $G$. It is easily proved by induction on the order of the submatrix that $Q(G)$ is totally unimodular. The proof is taken from the book (Lemma 2.6) Bapat RB. Graphs and matrices. New York: Springer; 2010 Jul 23. Proof: Consider the statement that any $k\times k$ submatrix of $Q(G)$ has determinant $0$ or $\pm1.$ We prove the statement by induction on $k$. Clearly the statement holds for $k=1,$ since each entry of $Q(G)$ is either $0$ or $\pm1.$ Assume the statement to be true for $k-1$ and consider a $k\times k$ submatrix $B$ of $Q(G)$. If each column of $B$ has a 1 and a $-1$, then $\det B=0.$ Also, if $B$ has a zero column, the $\det B=0.$ Now suppose $B$ has a coumn with only one nonzero entry, which must be $\pm1.$ Expand the determinant of $B$ along that column and use induction assumption to conclude that $\det B$ must be 0 or $\pm1.$ If $G$ is tree on $n$ vertices, then any submatrix of $Q(G)$ of order $n-1$ is nonsingular. Proof: Consider the submatrix $X$ of $Q(G)$ formed by the rows $1,\dots, n-1.$ If we add all the rows of $X$ to the last row, then the last row of $X$ becomes the negative of the last row of $Q(G)$. Thus, if $Y$ denotes the submatrix of $Q(G)$ formed by the rows $1,\dots,n-2,n,$ then $\det X=-\det Y.$ Thu, if $\det X=0,$ then $\det Y=0.$ Continuing this way we can show that if $\det X=0$ then each $(n-1)\times (n-1)$ submatrix of $Q(G)$ must be singular. In fact, we can show that if any one of the $(n-1)\times (n-1)$ submatrices of $Q(G)$ is singular, then all them must be so. However, rank $Q(G)=n-1$ and hence at least one of the $(n-1)\times (n-1)$ submatrices of $Q(G)$ must be nonsingular.<|endoftext|> TITLE: Unitary factor in polar decompositions QUESTION [8 upvotes]: Let $A, B$ be $n$-square (Hermitian) positive definite matrices. Let $AB=U|AB|$ be the polar decomposition of $AB$. So $U$ is unitary (called the unitary factor of $AB$). What is the optimal constant $c$ such that $\|I-U\|\le c$, where the norm is the usual spectral norm? I want to have some understanding on the behaviour of the unitary factor for certain classes of matrices (e.g. matrices with real eigenvalues). Perhaps this is well known, any pointer to the existing papers is welcome. REPLY [2 votes]: I think that the part $(a)$ of proposition $2.4$ of this paper shows that for $n$ sufficiently large one can construct a $n \times n$ unitary matrix $U=-I_{2}\oplus U'$, which can be decomposed as the product of three positive matrices. For such $U$ we have $\parallel U-I\parallel=2$. In fact one can take a unitary matrix $U'$ with $Det \;U'=1$ such that $0$ lies in the interior of the convex hull of the eigenvalues of $U$. This convex hull is equal to $W(U)$, the numerical range of $U$. In fact if $U=ABC$ for positive $A,B,C$ then $UC^{-1}=AB$ so $C^{-1}=|AB|$ is the positive factor of $U $ in its polar decomposition. Of course such $U$ has $-1$ as an eigenvalue. An explicit example is the following $$\begin{pmatrix}-1& 0&0&0\\0&-1&0&0\\0&0&\sqrt{3}/2&-1/2\\0&0& 1/2&\sqrt{3}/2 \end{pmatrix} $$ According to the aboved linked paper this matrix is a product of three positive matrices. This matrix, with distance $2$ from the identity matrix $I_{4}$, is the unitary factor of polar decomposition of $AB$ for two positive matrices $A,B$. So according to the notation $C_{n}$ in the answer by Loup Blanc, we have $C_{n}=2, \; \forall n\geq 4$.<|endoftext|> TITLE: When is a topological space the homotopy colimit of an open covering? QUESTION [19 upvotes]: Suppose that $X$ is a topological space and $\left(U_i \to X\right)$ is an open cover. We can associate to it the Cech diagram of this cover $$C_U:\Delta^{op} \to Top.$$ I know that for many good classes of topological spaces, the homotopy colimit of $C_U$ is $X$ (e.g. for manifolds). How general is this result? Does it hold e.g. for locally contractible spaces? I believe that In general $X$ is (weakly) homotopy equivalent to the fat geometric realization of $C_U$ (e.g. see Cor. 4.8 here: http://arxiv.org/abs/0907.3925) , but for a general simplicial space, this need not agree with hocolim. Any feedback or references would be appreciated. Thanks! REPLY [19 votes]: It is true in complete generality that $X$ is the homotopy colimit of $C_U$ (and hence that the fat realization computes the homotopy colimit in this case). This is a special case of Lurie's version of the Seifert-van Kampen theorem. More precisely, Proposition A.3.2 in Higher Algebra says that that the "underlying homotopy type" functor $$Sing: Open(X) \to \mathcal{S} $$ is a cosheaf, so it lifts to a colimit-preserving functor $Shv(X) \to \mathcal{S}$. Since $X$ is the colimit of $C_U$ in $Shv(X)$, $Sing(X)$ is the colimit of $Sing(C_U)$ in $\mathcal{S}$. ETA: The proof of the above result actually shows that $Sing$ is a hypercomplete cosheaf. This reminded me that Dugger and Isaksen also prove this fact in their paper Hypercovers in topology.<|endoftext|> TITLE: Underlying idea for (automorphic) L-function? QUESTION [40 upvotes]: Edit: So with a few more months of math under my belt, I recognize some of the issues with this question. I still hope for an answer, so let me say a few things. Within the Langlands philosophy, L-funtions associated to various automorphic representations are the invariant which ought to parametrize an extremely complicated and far-reaching interaction between automorphic representations of separate groups. This of course refers to Functoriality. I get that the conjecture (ie: the definition of the Selberg class) is (vaguely) that the nice L-functions found in number theory and algebraic geometry should come from automorphic obects in some way, but the definition of the automorphic L-function is still mysterious to me. I can read the definitions, and the $\mathrm{GL}_n$ theory is truly beautiful. But it all seems like a big analogy chase, having seen the useful nature of Euler products and Dirichlet series in the past. In particular, I can't see why one would expect a connection between these L-functions and a correspondence as vast as functoriality suggests. Is there a more satisfying rational for the definition of automorphic L-functions than "we get Euler products, and by analogy they ought to be important"? To preface, I am a student of automorphic representation theory, and I know full well the definition of the L-function attached to an automorphic representation. I am intending to give a talk on the question in the title to a group of graduate students and young researchers. While the history of and ubiquity of L-functions is an aspect of what I want to explain, there is a nagging question in the back of my mind that I do not know how to approach: Why is it that a Dirichlet series with analytic continuation and functional equation is such a potent idea? What is a unifying idea behind these constructions? For many (all?) instances of L (or zeta)-functions in number theory, representation theory, (Artin, Hasse-Weil, Dirichlet, etc.) and perhaps many other fields I know less about (Selberg's zeta function) , the hope is that these are all instances of automorphic L-functions and are related in deep ways to an automorphic representation. But on the automorphic side of things, I don't understand what the L-function actually is. The converse theorems gives me a partial answer in that the L-function is some local-global object (The definition as an Euler product) which encodes, along with its twists, automorphy of the representation. This then leads me to other questions for another time, and still seems more about why the L-function is useful, as opposed to what the L-function is. My question, then, is What is the (conjectural) underlying idea of what an L-function is, either in the automorphic case or more generally? Is there a sense of why such a construction gives a powerful way of connecting different areas of mathematics? I have read Bump's Book Automorphic Forms and Representations, a few articles such as Iwaniec's and Sarnak's enjoyable article Perspectives on the Analytic theory of L-functions, as well as many of the brilliant responses to related questions here on MO. As this is my first question, I apologize if my question is not clear, or is duplicate to a question I have not yet found. Thank you for your help! Edit: In terms of an answer, let me say this: I was hoping that there is a known way, perhaps in terms of the relevant group, to see why the L-function construction should be so fundamental to so many theories. If there isn't a known answer in this sense, as was indicated by @Myshkin's answer, then I will be happy with intuition or heuristic understanding that is in this direction. Please let me know if this is still too broad or unclear. Thank you! REPLY [3 votes]: I'm not really sure what you're looking for, but here are some thoughts, and if you want me to try to expand on some aspect, let me know. As an analytic gadget $L$-functions provide a way to use complex analysis to study algebraic objects. Why are they good analytic functions for number theory? Well, suppose you have some nice number theoretic object $X$. Then often you can expect some sort of local-global principle, meaning $X$ is mostly (or completely) determined by associated local objects $X_p$ at all primes $p$ (well, sometimes you should include prime powers). In fact, it often happens (Chebotarev density and the like), that you only need to know $X_p$ for a set of primes of sufficiently large density. So, if you want an analytic function to take advantage of local-global principles, it makes sense to look for a kind of function that has an "Euler product," so the objects can be studied in terms of corresponding local analytic data. In my mind, this is the first property you want. For convergence of this "Euler product," you need local factors that are close to 1 (at least in some large domain), and something like a polynomial or geometric series is one of the simplest things you could guess. For analytic reasons, you also want to be able to control the zeroes/poles of the local factors, which makes polynomials in $q^{-s}$ useful. From another point of view, if you have some number theoretic sequence $(a_n)$, and you want to make an analytic function out of it, how can you do this? Well, the most natural thing to do is use a series. What are the options? Power series? Fourier series? Dirichlet series? If $(a_n)$ is a nice arithmetic sequence (multiplicative, polynomial growth) you won't typically have much convergence with a power or fourier series, and more importantly, you don't get the Euler product. As a tool for functoriality I don't (entirely) think of $L$-functions as some magical world where everything gets tied together (though they are quite magical). I think of them as a useful tool to compare different objects which have related representations (though some people may take the opposite view). Many interesting algebraic objects naturally have associated representations to them (or already are representations themselves), and the central idea in Langlands' conjectures is that automorphic representations of GL($n$) are the mother of all "arithmetic" (Galois) representations. Meaning, to some arithmetic object $X$, we associate $n$-dimensional representations (sometimes just local ones), and these should correspond to automorphic representations of GL($n$). For the functoriality aspects, the two key properties of $L$-functions are additivity ($L(s,\rho_1 \oplus \rho_2) = L(s, \rho_1)L(s,\rho_2)$) and inductivity ($L(s,I_{E}^F(\rho)) = L(s, \rho)$), and for this it matters what the precise form of the local factors are. Roughly you want it to be a "characteristic polynomial" of the representation (at least for a finite-dimensional Galois representation---on the automorphic side, you want the analogous thing for Langlands parameter) so that additivity will hold. Recall that the correspondence between Galois and automorphic representations says that the representations should locally correspond at almost all places, so it suffices to look at unramified places, where both sides should reduce to the 1-dimensional case of local class field theory. Then the additivity of $L$-functions says the $L$-functions correspond. Of course the main problem in functoriality is showing that $\pi = \otimes \pi_v$ is (globally) automorphic---constructing the $\pi_v$ for almost all $v$ is easy. Then inductivity is useful to get compatibility with base change. Note: not all Dirichlet series formed from arithmetic objects should give automorphic $L$-functions. For instance, Koecher-Maass series (constructed with Fourier coefficients of Siegel modular forms) and zeta functions of binary cubic forms don't have Euler products. Also, Zagier's "Naive BSD" paper considers variations of $L$-series of elliptic curves which do have Euler products, but not (apparently) meromorphic continuation to $\mathbb C$. So it's better to look at things where, say $a_n$'s are multiplicative, and the $a_p$'s are traces of representations. Edit: Upon rereading my answer, I realize I may not have spelled out a couple conclusions explicitly: The above reasoning more-or-less motivates the definition of $L$-functions of Galois representations. The automorphic definition is then motivated by a combination of the result of Hecke that gives $L$-functions of modular forms as integral representations combined with the predictions of the local Langlands correspondence. My point of view is that the reason $L$-functions connect different ares of mathematics is because they reflect things like the global Langlands correspondence. Unfortunately, I don't have a good intuitive explanation for why Langlands' conjectures should be true (one can make the trite remark that it generalizes class field theory, but that doesn't answer why in my mind).<|endoftext|> TITLE: Precise density estimates for Cantor sets QUESTION [6 upvotes]: Let $C_\lambda$ be the classical Cantor set associated to a real number $0<\lambda<\frac{1}{2}$, as defined for example in the book of K. J. Falconer The geometry of fractal sets. I recall briefly the construction. Starting with the unit interval, we remote from the center of the interval an interval of length $1-2\lambda$. After $n$ steps, we obtain $2^n$ intervals equally spaced out, each one of length $\lambda^n$. For the Hausdorff measure $\mathscr{H}^s$ of dimension $s$, we use the function $g(S)=\text{diam}(S)^s$ for the Caratheodory construction and we define the upper density of a set $A\subset \mathbb{R}^n$, at a point $x\in\mathbb{R}^n$ by $$ \Theta^{*s}(A,x)=\limsup_{r\rightarrow 0}\frac{\mathscr{H}^s(A\cap B(x,r))}{(2r)^s} $$ and for the lower density $\Theta_*^s$ we replace $\limsup$ by $\liminf$. We can prove that $\dim C_\lambda=\frac{\log 2}{\log(1/\lambda)}=s_\lambda$, and even $\mathscr{H}^{s_\lambda}(C_\lambda)=1$, and I wanted to obtain precise estimates for the upper and lower density of $C_\lambda$. In fact I think that I managed to prove that for all $x\in C_\lambda$, we have $$ 2^{-(s_\lambda+1)}\leq \Theta_*^{s_\lambda}(C_\lambda,x)\leq 2^{-s_\lambda}\leq \Theta^{*s_\lambda}(C_\lambda,x)\leq 1. $$ The problem is that I know that there is a lower bound more precise for $\Theta^{*s_\lambda}(C_\lambda,x)$, namely, a constant $c>2^{-s_\lambda}$, such that $\Theta^{*s_\lambda}(C_\lambda,x)\geq c$ for all $x\in C_\lambda$. I would like to get a confirmation of the validity of the calculated estimates, and I would really appreciate a hint for the last lower bound. REPLY [3 votes]: Upper densities In the following, I freely use the well known fact that $s_\lambda$-Hausdorff measure gives mass $2^{-k}$ to all intervals that make up the stage $k$ in the construction of $C_\lambda$. I don't think it is correct that $\Theta^{* s_\lambda}(C_\lambda,x)\ge c$ for all $x\in C_\lambda$ and some $c>2^{-s_\lambda}$. For example, let $\lambda=1/3$ so that $C=C_\lambda$ is the middle-thirds Cantor set. Write $s=s_{1/3}=\tfrac{\log 2}{\log 3}$, and $$ D(x,r) = \frac{\mathcal{H}^{s}(B(x,r)\cap C)}{(2r)^s} $$ By self-similarity, $D(0,r)=D(0,r/3)$, so $$ \Theta^{*s}(C,0) = \max\{ D(0,r): r\in [1/3,1]\}. $$ Now if $r\in [1/3,2/3]$, then $\mathcal{H}^s(B(0,r))=1/2=(1/3)^s$ so $D(0,r)\le 2^{-s}$. Otherwise, write $r=2/3+t$ for some $t\in [0,1/3]$. Then $$ \mathcal{H}^s(B(0,r)\cap C) = \mathcal{H}^s([0,1/3]\cap C)+\mathcal{H}^s([2/3,2/3+t]\cap C)\le \tfrac{1}{2}+t^{-s}, $$ whence $$ D(0,r) \le \frac{\tfrac{1}{2}+t^{s}}{2^s (2/3+t)^s} \le 2^{-s}, $$ from elementary calculus. Hence $\Theta^{*s}(C,0)=2^{-s}$. There's nothing special about $\lambda=1/3$ here. Also, although the point $0$ is special, the same holds for any ternary point. Lower densities The lower density problem was essentially solved, for more general self-similar sets, in [Feng, De-Jun. Exact packing measure of linear Cantor sets. Math. Nachr. 248/249 (2003), 102--109]. In Theorem 1.1, a formula is given for the infimum of $$ \frac{\mathcal{H}^{s_\lambda}(B(x,r)\cap C_\lambda)}{(2r)^s} $$ over all $x\in C_\lambda$ such that $B(x,r)\subset [0,1]$. Note that the example after Theorem 1.1 are exactly the central Cantor sets (with $\lambda=(1-\beta)/2$). It seems this value is strictly larger than $2^{-s_\lambda-1}$ for all $\lambda\in (0,1/2)$. In Theorem 2.1, Feng shows that $\Theta_*^{s_\lambda}(C_\lambda,x)$ equals this infimum for almost all $x$. Clearly, this infimum equals the smallest possible value of $\Theta_*^{s_\lambda}(C_\lambda,x)$, at least if we exclude $x=0,1$ (and if $\lambda\le 1/3$ this restriction is not necessary, as any extreme point of a construction interval will have the same density as $0$ and $1$). Edit (after Paul-Benjamin's comment): to give an upper bound for the lower density at every point of $C_\lambda$, I consider again the case $\lambda=1/3$ for concreteness, although a similar argument should work for any $\lambda$. Suppose first that $x$ is such that it belongs to infinitely many "left intervals" and infinitely many "right intervals" in the construction (or in other words the ternary expansion of $x$ has infinitely many zeros and twos). Then there are infinitely many $k$ such that the distance from $x$ to the boundary of the level $k$ interval of the construction of $C$ containing $x$ is at least $(2/9) 3^{-k}$. Let $r=r_k=3^{-k}(1+2/9)$. Then the ball $B(x,r)$ intersects only one interval of level $k$ in the construction, so that $$ \frac{\mathcal{H}^s(B(x,r)\cap C)}{(2r)^s} = \frac{1}{2^s(11/9)^s} < 2^{-s}. $$ Otherwise, without loss of generality, from some generation on $x$ is always on the left interval of the construction. In that case, $B(x, 2 3^{-k})$ meets only one interval of the construction for all sufficiently large $k$ and we get an even smaller upper bound for the lower density. Hence $b=2^{-s} (11/9)^{-s}$ works (I'm not claiming this is optimal, though it might be).<|endoftext|> TITLE: Biquadratic reciprocity for $p\equiv 1\pmod 4$ and $q\equiv 3\pmod 4$ QUESTION [7 upvotes]: For a prime $p\equiv 1\pmod 4$, let $\left(\frac{\cdot}{p}\right)_4$ denote the rational biquadratic residue symbol; that is, $$ \left(\frac{a}{p}\right)_4 = \begin{cases} \ \ \ 1\ &\text{if $a$ is a biquadratic residue modulo $p$}, \\ -1\ &\text{otherwise.} \end{cases} $$ Also, denote by $\left(\frac{\cdot}{p}\right)$ the quadratic residue (Legendre) symbol. Burde has shown that if $p$ and $q$ are primes satisfying $\left(\frac pq\right)=1$ and both congruent to $1$ modulo $4$, then writing $p=a^2+b^2$ and $q=c^2+d^2$ with $a$ and $c$ odd (and $b$ and $d$ even), one has $$ \left(\frac{p}{q}\right)_4\left(\frac{q}{p}\right)_4=\left(\frac{ac-bd}{q}\right). $$ Is there a version of Burde's result for the case where $q\equiv 3\pmod 4$? To be more specific, Suppose that $p\equiv 1\pmod 4$ is prime, and write $p=a^2+b^2$ with $a$ odd (and $b$ even). Is there any natural way to associate with every prime $q\equiv 3\pmod 4$ satisfying $\left(\frac{p}{q}\right)=1$ integers $c$ and $d$ so that $$ \left(\frac{q}{p}\right)_4=\left(\frac{ac-bd}{q}\right)? $$ (Inserting in the right-hand side factors like $(-1)^{(q-1)/4}$ would be just fine with me.) REPLY [7 votes]: It is perhaps not a direct analogue, but the reciprocity formula of K. S. Williams, K. Hardy and C. Friesen published in 1985 gives an expression for all odd primes $q>1$. This formula comprises all known rational quartic reciprocity laws: Theorem (Williams, Hardy, Friesen): Let $p\equiv 1(4)$ be a prime, and let $A,B,C$ be integers such that $A^2=p(B^2+C^2)$, $2\mid B$, $(A,B)=(A,C)=(B,C)=1$, $A+B\equiv 1(4)$. Then for every odd prime $q>1$ with $(p/q)=1$ we have $$ \left(\frac{q}{p}\right)_4=\left(\frac{A+B\sqrt{p}}{q}\right). $$ An elementary proof and a discussion how to derive Burde's law from this is given in Franz Lemmermeyer's article Rational quartic reciprocity.<|endoftext|> TITLE: Intersection of nonzero prime ideals is zero -- does it have a name? QUESTION [5 upvotes]: The Rabinowitch trick (in Eisenbud's Commutative Algebra with a view toward Algebraic Geometry, page 132) says that $R$ (commutative unital ring) is Jacobson if and only if for every prime ideal $P \subset R$, if $(R/P)[b^{-1}]$ is a field, then $R/P$ is a field. Equivalently, each quotient domain $S = R/P$ has the property: (*): $S$ is a field or the intersection of its nonzero primes is $(0)$. Does this property (*) have a name? REPLY [10 votes]: An integral domain $R$ for which the intersection of the nonzero prime ideals is nonzero is a Goldman domain. Equivalently: the fraction field $K$ is finitely generated as an $R$-algebra (equivalently, $K = R[f]$ for some $f \in K$). The latter property is usually taken as the definition, but the equivalence is almost immediate: see e.g. $\S 12.1$ of these notes. Note also that the prominence of Goldman domains in commutative algebra is due as much to Kaplansky as to Oscar Goldman: under the name "G-domain", they play a surprisingly central role in his (perhaps slightly eccentric but very) influential text Commutative Rings. For a general ring I don't quite know the answer to your question, but in his 1966 paper The pseudo-radical of a commutative ring, Robert Gilmer defines in any commutative ring the pseudo-radical to be the intersection of all nonzero prime ideals. You can try to chase this down in the literature and see what you come up with.<|endoftext|> TITLE: Spectral sequence of a bicomplex equipped with a group action QUESTION [7 upvotes]: Let $(A_1^\bullet,\partial_1)$ and $(A_2^\bullet,\partial_2)$ be complexes of $\mathbb{C}$-vector spaces. We suppose that $(A_1^\bullet,\partial_1)$ and $(A_2^\bullet,\partial_2)$ are equipped with an action of a discrete group $G$, i.e., $G$ acts on $A_1^\bullet$ (resp. $A_2^\bullet$) by preserving the grading and commuting with $\partial_1$ (resp. $\partial_2$). Then $A_1^\bullet$ and $A_2^\bullet$ are $\mathbb{C}[G]$-module. We suppose further that $A_1$ is a free $\mathbb{C}[G]$-module. Let $(A^\bullet,\partial)$ be the product complex, i.e, $A^k= \bigoplus_{p+q=k}A_1^p\otimes_\mathbb{C} A_2^q$ and $\partial = \partial_1+(-1)^q\partial_2$. Let $F^p = \bigoplus_{p'\geqslant p}A_1^{p'}\otimes_\mathbb{C} A_2^q$ be the classical filtration. Let $(E_r)$ be the spectral sequence associated to $((A^\bullet)^G,\partial,F^\bullet)$. The question : does $(E_r)$ converge at $r=2$ ? PS I have known that the answer is positive if $G$ is a finite group. A comprehensible example for the above construction is : let $X$ and $Y$ be topological manifolds, let $G=\pi_1(X)$, we suppose that $G$ acts on $Y$, let $\widetilde{X}$ be the universal cover of $X$, let $A^\bullet_1$ be the (simplical, Cech or de Rham) complex of $\widetilde{X}$, $A^\bullet_2$ be the complex of $Y$, then $(A^\bullet)^G$ is the complex of $\widetilde X\times_GY$. REPLY [6 votes]: In the following I'm going to assume that $G$ is an infinite group. Then the answer to your question is yes, but probably not for the reason that you expect. Let's consider your filtration on $(A^\bullet)^G$. By definition we have $$ (F^p)^G = \bigoplus_{p' \geq p} (A_1^p \otimes A_2^q)^G $$ because taking invariants commutes with direct sums. The $E_0$-term of this spectral sequence is, in degree $(p,q)$, the quotient $$ ((F^p)^G/(F^{p+1})^G)^{p+q} = (A_1^p \otimes A_2^{q})^G. $$ The $d_0$-differential is induced by the boundary map on $A_2^\bullet$. By assumption, for each $p$ the group $A_1^p$ is a free $\Bbb C[G]$-module. Therefore, $$ (A_1^p \otimes A_2^q)^G = \bigoplus (\Bbb C[G] \otimes A_2^q)^G. $$ Given a $\Bbb C[G]$-module $M$, any element $x \in \Bbb C[G] \otimes M$ can be written uniquely as a sum $\sum_{g \in G} g \otimes a_g$ for some unique $a_g \in M$, all but finitely many of which are nonzero. Because the group $G$ is infinite, the only way that such an element can be invariant under the $G$-action is if it is zero. Therefore, the spectral sequence is zero at the $E_0$-term, and so it collapses pretty quickly. (Based on your description, however, it sounds like you might not actually want free modules: the cochains on $\widetilde X$ are usually not free, but instead are a product of coinduced modules $\Bbb C^G \cong \prod_{g \in G} \Bbb C$. This is isomorphic to $\Bbb C[G]$ if $G$ is finite. You have to be a little bit careful then, because there may not be an equivalence $C^*(\widetilde X \times Y) \simeq C^*(\widetilde X) \otimes C^*(Y)$ without either assuming $Y$ is equivalent to a finite complex or that you're taking a completed tensor product.)<|endoftext|> TITLE: Is it possible to define higher cardinal arithmetics QUESTION [30 upvotes]: In number theory there are several operators like ‎addition, ‎multiplication and ‎exponentiation defined from ‎$‎‎‎\omega‎‎\times‎‎\omega‎$ ‎to ‎‎$‎‎‎\omega‎$. Each ‎of ‎them ‎is defined as an ‎iteration of ‎‎‎the other. ‎The sequence of building such iterated operators can go further to define faster and faster hyperoperators‎. The first of them is tetration which is defined as iterated exponentiation. Let ‎$‎‎m\uparrow n$ ‎denote the tetration of ‎$‎‎m$ and ‎$‎n‎$‎ ‎that ‎is‎ ‎‎$‎‎\underbrace{m^{m^{m^{.^{.^{.}}}}}}_{n - times}$. This operator appears in several interesting occasions in logic, computations and combiantorics, for example see these Wikipedia articles on Graham's number, Ackermann's function, busy Beaver function and Chaitin's incompleteness theorem. ‎ Now consider the infinitary case. ‎In set theory ‎addition, ‎multiplication ‎and ‎exponentiation are defined for ‎cardinal ‎numbers. ‎ Question 1. What ‎about ‎‎$‎‎‎\kappa‎‎\uparrow‎‎\lambda‎$? ‎How should we define this? ‎ ‎ Intuitively, we expect to define ‎$‎‎\aleph_0‎\uparrow‎\aleph_0$ ‎to be ‎‎$‎\aleph_0^{‎‎\aleph_0^{‎‎\aleph_0^{.^{.^{.}}}}}.$ But this intuitive definition of tetration has some counter-intuitive properties, as then ‎we ‎expect ‎to ‎have ‎‎$‎‎‎‎\aleph_0^{(‎‎\aleph_0‎\uparrow‎\aleph_0)}=‎‎\aleph_0‎\uparrow‎\aleph_0$ which is ‎impossible ‎by ‎Cantor's ‎theorem which says ‎$‎‎‎\forall ‎‎\kappa‎\geq\aleph_0\;\;\;\aleph_0^{‎\kappa‎}>‎\kappa‎$. Note that for the cases of addition, multiplication and exponentiation, we have quite natural operations $f_+, f_\times$ and $f_e$ such that given cardinals $\kappa, \lambda$, we have $f_+(\kappa,\lambda)=\kappa+\lambda, f_\times(\kappa, \lambda)=\kappa\times \lambda$ and $f_e(\kappa,\lambda)=\kappa^\lambda.$ Question 2. Is there a natural operation $f_t$ defined so that for all natural numbers $m,n$ we have $f_t(m,n)= ‎‎‎m\uparrow n$, and so that its definition is so natural that it also works for infinite cardinal numbers? The next question is taken from Noah's answer, where an answer to it may help in defining the tetration for higher infinite. Question 3. What is $m\uparrow n$ counting? See also What combinatorial quantity the tetration of two natural numbers represents?. But note that the answers given in the above question are so that they are not suitable for treating infinite cardinals. REPLY [24 votes]: Let me begin with an observation: $2 \!\uparrow\uparrow\! n$ is equal to the number of sets of rank at most $n$. [If you followed the combinatorics tag here and have forgotten what the rank of a set is, see this article for a refresher. Roughly, the rank of a set is the "depth" of the nesting of braces when it's written out in long form. For example, we have one set of rank $0$ (namely $\emptyset = \{\}$). There are two sets of rank $0$ or $1$ (namely $\{\}$ and $\{\{\}\}$), there are $4$ sets of rank at most $2$ (namely $\{\}$, $\{\{\}\}$, $\{\{\{\}\}\}$, and $\{\{\},\{\{\}\}\}$), there are $16$ of rank at most $3$, $65536$ of rank $\leq 4$, etc.] This observation is easily proved by induction, and it gives us a natural answer to your Question 3 when we restrict ourselves to tetration with base 2. What about the unrestricted version of Question 3? For this, we want to stop viewing set membership as a binary relation (membership value of $0$ or $1$). Instead, let's allow $k$ possible membership values: a set can be a member of another with value $0$ (not a member), $k-1$ (fully a member), or anything in between. For example, this view still gives us one set of rank $0$, namely the empty set $\{\}$, but now we have $k$ sets of rank $0$ or $1$, namely the sets containing only $\{\}$, with value between $0$ and $k-1$ (of course, the set containing $\{\}$ with value $0$ is identified with $\{\}$, so that we only get $k-1$ new sets of rank $1$, for a total of $k$ with rank $0$ or $1$). It is easy to check that we get $k^k$ "sets" of rank at most $2$, $k^{k^k}$ of rank at most $3$, and $k \!\uparrow\uparrow\! n$ sets of rank at most $n$. So a plausible answer to your Question 3 is: $k \!\uparrow\uparrow\! n$ is equal to the number of sets of rank at most $n$ when we adopt a $k$-valued notion of set membership. One of the nice things about this answer is that it generalizes easily to higher cardinals. As you know, the notion of $B$-valued set membership, where $B$ is some set (usually a Boolean algebra or a partial order) is a common one in set theory. Sets with a $B$-valued notion of membership are often called $B$-names, and they are naturally equipped with a notion of rank. In light of the foregoing discussion on finite tetration, I propose the following as a fairly natural definition for transfinite tetration: $\kappa \!\uparrow\uparrow\! \nu$ is equal to the number of $B$-names of rank at most $\nu$, where $B$ is a set of size $\kappa$. Notice that this definition coincides with the inductive definition in Noah's answer. One drawback to this definition (noticed already by Noah) is that the value of $\kappa \!\uparrow\uparrow\!\nu$ is determined by treating $\kappa$ as a cardinal while treating $\nu$ as an ordinal. Right now, I can see two approaches to "fixing" this drawback: either live with it (it's not a bug, it's a feature!), or modify the definition to (something like) one of the following three possibilities (the third being my personal preference): $\kappa \!\uparrow\uparrow\! \nu$ is the number of $B$-names hereditarily smaller than $\nu^+$, where $B$ is a set of size $\kappa$. $\kappa \!\uparrow\uparrow\! \nu$ is the number of $B$-names of rank at most $\nu$, where $B$ is a set of size $\kappa$. $\kappa \!\uparrow\uparrow\! \nu$ is the number of $B$-names of rank less than $\nu^+$, where $B$ is a set of size $\kappa$. REPLY [2 votes]: A category-theoretic angle. Accordingly, more questions than answers. But at least I hope it uncovers some subtle issue. Let us ask how to define transfinite iterates of an endofunctor $F$. That is, what properties should uniquely (up to isomorphism) determine $F^\lambda$ as something like "$\underbrace{\cdots\circ F\circ F\circ\cdots}_{\text{$\lambda$ times}}$". If this is solved, then $\kappa\uparrow\lambda$ could be defined as the cardinality of $F^\lambda(\text{singleton})$, where $F$ is the functor from sets to sets given by $F(S)=X^S$, with $X$ any (fixed) set of cardinality $\kappa$. (It is true that $F$ is actually contravariant, but there still is no problem iterating it.) When $\lambda$ is an ordinal, one might try direct limit of the $\lambda$-shaped diagram, with $F$ as transition functors. In this way one gets something peculiar. Already at $\lambda=\omega$ the limiting category might be not equivalent to the category of sets. Whatever it is, $F^\lambda(\text{singleton})$ will be defined, it will be an isomorphism class of objects of a category defined uniquely up to equivalence. So it remains to describe this category, at least the class of isomorphism classes of its objects. This will then give a natural candidate for $\kappa\uparrow\lambda$ as one of the members of this class, uniquely determined by the above. A variant - we could take the (large) groupoid of sets and bijections, we don't actually need non-bijective morphisms. However I don't know whether we would obtain something essentially different at limiting stages. Still another variant - we might take the $\textit{inverse}$ limit of the backwards diagram; if it does have a terminal object, we might take the value of the limiting functor on it. This then would be a genuine set. Here however I don't know whether the terminal indeed exists in the limiting category. When $\lambda$ is just a cardinal, I am at a loss. I need some ordering to get started with the above. After a while I decided to post a separate question about the nature of these limiting categories, Infinite iterates of the contravariant hom endofunctors on sets<|endoftext|> TITLE: Isomorphic Hadwiger graphs QUESTION [6 upvotes]: Let $G$ be a graph, then we define its Hadwiger graph $\textrm{Hadw}(G)$ in the following way: $V(\textrm{Hadw}(G)) = \{S\subseteq (V(G): S\neq \emptyset\textrm{ and } S \textrm{ is connected}\}$; $E(\textrm{Hadw}(G)) = \{\{S,T\}\subseteq V(\textrm{Hadw}(G)): S\cap T = \emptyset \textrm{ and } (\exists s\in S, t\in T): \{s,t\}\in E(G) \}.$ $G$ embeds to $\textrm{Hadw}(G)$ by $v\mapsto \{v\}$, and the Hadwiger conjecture states $\chi(G) \leq \omega(\textrm{Hadw}(G))$. Question: If $G, H$ are (finite or infinite) graphs, does $\textrm{Hadw}(G) \cong \textrm{Hadw}(H)$ imply $G\cong H$? REPLY [2 votes]: $\def\Hadw{\mathop{\rm Hadw}}$If it is not a mauvais ton, I would like to add an easier proof found independently by Sergey Dolgikh and Marat Abdrakhmanov (we used this fact as a contest problem). We assume that $G$ is finite and connected, and again we find all vertices in $H=\Hadw(G)$ corresponding to the vertices of $G$. Lemma. If $S$ is a vertex of $\Hadw(G)$ with $|S|>1$ and $v\in S$, then $\deg_H S<\deg_H \{v\}$. Proof. Every neighbor $T$ of $S$ in $H$ can be augmented by a path in $S$ to provide a neighbor $T'$ of $\{v\}$; all these neighbors are distinct since $T'\setminus S=T$. Moreover, $\{v\}$ has some neighbor contained in $S$. Now, for every $T\in V(H)\setminus\{V(G)\}$ we find its neighbor of maximal degree. By the lemma, this neighbor is a one-vertex set. On the other hand, for every $v\in V(G)$, the set $\{v\}$ is the unique maximal degree neighbor of any component of $G-\{v\}$. Thus we have reconstructed all sets of the form $\{v\}$.<|endoftext|> TITLE: Are there known cases of the Mumford–Tate conjecture that do not use Abelian varieties? QUESTION [15 upvotes]: (For a formulation of the Mumford–Tate conjecture, see below.) The question As far as I know, all non-trivial known cases of the Mumford–Tate conjecture more or less depend on the Mumford–Tate conjecture for Abelian varieties. That is, we know it for: Projective spaces (trivial) [edit] Other varieties whose cohomology is generated by algebraic cycles (more or less trivial) [/edit] Abelian varieties up to dimension $7$ (I guess, maybe some edge cases left?) Curves of genus $\le 7$ (uses the above fact) K3-surfaces (using the Kuga–Satake construction, hence AV's) [edit] I am probably missing out on some results (Pink proved MT-conj for lots of AV's of certain dimensions, etc… The list is to give an idea of the type of results. [/edit] Q1: Is there an example of a variety $X$ for which the Mumford–Tate conjecture is known, but the proof does not reduce to AV's and projective spaces? Preciser motivic (up to a precise notion of motive) formulation of the question: Q2: Is there an example of a motive $M$, such that (i) the Mumford–Tate conjecture is known for $M$, and (ii) the motive $M$ is not in the category of motives generated by AV's and the Tate/Lefschetz motive. (I.e., Milne's $\mathbf{LCM}$ category, in [Milne, 1999b].) Motivation for the question Originally the Mumford–Tate conjecture was formulated for Abelian varieties. For Abelian varieties we indeed know some examples where it is true. It seems very natural to generalise the conjecture to arbitrary smooth projective varieties, and I think nowadays most people mean the general version when referring to the conjecture. However, I do not know of any evidence for the more general version apart from undeniable beauty. Background (formulation of the Mumford–Tate conjecture) Let $k$ be a finitely generated field of characteristic $0$. Let $X$ be a projective smooth variety over $k$. Let $i$ be an integer. The Betti cohomology $H^{i} = H^{i}(X(\mathbb{C}), \mathbb{Q})$ carries a Hodge structure. The category of (pure) Hodge structures is Tannakian. Therefore, $H^{i}$ generates a sub-Tannakian category $\langle H^{i} \rangle^{\otimes}$, and the associated affine group scheme over $\mathbb{Q}$ is called the Mumford–Tate group. We denote it with $\mathrm{MT}_{X}^{i}$. (Alternatively, it is the Zariski closure over $\mathbb{Q}$ of the image of the Deligne torus in $\mathrm{GL}(H^{i})$.) The $\ell$-adic étale cohomology $H_{\ell}^{i} = H_{\text{ét}}^{i}(X_{\bar{k}}, \mathbb{Q}_{\ell})$ carries a Galois representation. The category of (finite-dimensional) Galois representations is Tannakian. Therefore, $H_{\ell}^{i}$ generates a sub-Tannakian category $\langle H_{\ell}^{i} \rangle^{\otimes}$, and the associated affine group scheme over $\mathbb{Q}_{\ell}$ is called the $\ell$-adic monodromy group. We denote it with $G_{\ell}$.(Alternatively, it is the Zariski closure over $\mathbb{Q}_{\ell}$ of the image of the Galois group in $\mathrm{GL}(H_{\ell}^{i})$.) The Mumford–Tate conjecture (for $X$ and $i$) is: $\mathbf{MT}^{i}(X)$: Via the comparison isomorphism of Betti cohomology and $\ell$-adic étale cohomology, the group $\mathrm{MT}_{X}^{i} \times_{\mathbb{Q}} \mathbb{Q}_{\ell}$ is isomorphic to $G_{\ell}^{0}$, the identity component of $G_{\ell}$. REPLY [11 votes]: It follows from results of Ribet in "On l-adic representations attached to modular forms" (Invent. Math. 28 (1975), 245–275) that the Mumford-Tate conjecture holds for the motives attached to modular forms for $SL_2(\mathbb{Z})$. Blasius shows in the article "Modular forms and abelian varieties" (Séminaire de Théorie des Nombres, Paris, 1989–90, 23–29,Progr. Math., 102) that the motives attached to these forms are not in LCM (as defined in the question) if the weight is $>2$. (Similar results probably hold in much greater generality.)<|endoftext|> TITLE: Exactness of an additive left Kan extension QUESTION [6 upvotes]: Let $\phi:R\to S$ be a flat ring homomorphism and consider the induced adjoint pair $$\phi_!:R-Mod\rightleftarrows S-Mod:\phi^*,$$ where $\phi_!=(S\otimes_R -)$. The right adjoint $\phi^*$ is easily described if we see $\phi$ as an additive functor between the one-object categories $R$ and $S$ and we view a left $S$-module $M$ as an additive functor $M:S\to Ab$. In this case, $\phi^*$ is just composition by $\phi$, $(M:S\to Ab)\mapsto (M\circ\phi:R\to Ab)$. By definition of flat homomorphism, $\phi_!$ is an exact functor, while $\phi^*$ is exact for any ring homomorphism $\phi$. Consider now the categories of finitely presented modules $fp(R)$ and $fp(S)$, and restrict the functor $\phi_!$ to an additive functor $F:fp(R)\to fp(S)$. Denote also by $fp(R)-Mod$ and $fp(S)-Mod$ the categories of additive functors $fp(R)\to Ab$ and $fp(S)\to Ab$, respectively. Of course, also here there is an induced exact functor $$F^*:fp(S)-Mod\to fp(R)-Mod$$ and, using additive Kan extensions one may construct a left adjoint $$F_!:fp(R)-Mod\to fp(S)-Mod.$$ Can we prove that the functor $F_!$ is exact? Spelling this out, rember that, given a sequence of left $fp(R)$-modules $$0\to M_1\to M_2\to M_3\to 0$$ this sequence is exact if and only if, for all $P\in fp(R)$, the sequence of abelian groups $$0\to M_1(P)\to M_2(P)\to M_3(P)\to 0$$ is short exact. Thus one should prove (or disprove) that for any short exact sequence as above, $$0\to F_!M_1(Q)\to F_!M_2(Q)\to F_!M_3(Q)\to 0$$ is short exact for all $Q\in fp(S)$. REPLY [5 votes]: Daniel Schäppi's answer made me realize that I actually can say something about this. I'll keep Daniel's notation. Something more than what Daniel said is true, and it holds in a more general context: Given that $\mathcal{A}$ has finite limits, $F$ is left exact iff $\mathrm{Lan}_Y YF$ is. This is shown by Kelly (mimic Daniel's argument and use Kelly's Thm 6.11 (i) and (v) to show that a functor $\mathcal{A} \to \mathcal{V}$ is left exact iff its extension is) at the generality of categories enriched in a locally finitely presentable symmetric monoidal closed $\mathcal{V}$. This property of the class of finite limits (that if a functor $F$ between categories with finite limits preserves all finite limits, then so does $F_!$) is a weak version of what is called soundness by Adamek, Borceux, Lack, and Rosicky in the $\mathrm{Set}$-enriched context. Other classes of limits which have this property are the class of finite products, and the class of limits indexed by categories with fewer that $\alpha$ morphisms for some fixed regular cardinal $\alpha$. This criterion was used to generalize soundness to the enriched context in Day and Lack and Lack and Rosicky. A related, stronger condition, concerning the exactness of left Kan extensions of functors $F: \mathcal{A} \to \mathcal{V}$ where $\mathcal{A}$ can be arbitrary, is studied in the enriched context by Dostal and Velebil. One important reason to study soundness is that it allows a theory of categories which are "locally presentable with respect to a class of limits". The classical case of locally $\alpha$-presentable categories is an example (when the class of limits is those indexed by diagrams with $<\alpha$ morphisms), and another example is the theory of (many sorted) Lawvere theories: a category of models of a Lawvere theory is precisely a category which is locally presentable with respect to the class of finite products. In the enriched case especially, there are potentially other interesting examples waiting to be investigated.<|endoftext|> TITLE: Reference for a strong intermediate value theorem for measures QUESTION [16 upvotes]: Let $\mu$ be a finite nonatomic measure on a measurable space $(X,\Sigma)$, and for simplicity assume that $\mu(X) = 1$. There is a well-known "intermediate value theorem" of Sierpiński that states that for every $t \in [0,1]$, there exists a set $S \in \Sigma$ with $\mu(S) = t$. I would like to use the following stronger conclusion for such a measure: There exists a chain of sets $\{S_t \mid t \in [0,1]\}$ in $\Sigma$, with $S_t \subseteq S_r$ whenever $0 \leq s \leq r \leq 1$, such that $\mu(S_t) = t$ for all $t \in [0,1]$. (One can view this as the existence a right inverse to the map $\mu \colon \Sigma \to [0,1]$ in the category of partially ordered sets.) This statement appears (albeit hidden within a proof) on the Wikipedia page for "Atom (measure theory)," and even includes a sketch for the proof! However, I would like to see some mention of this in the literature. I've checked the Wiki references and they both seem to prove the weaker statement. I looked in Fremiln's Measure Theory, vol. 2, and again found the weaker version but not the stronger. Question: Can anyone provide me with such a reference? A proof. In case anyone stumbles to this page and wants to see a proof, I'll sketch one that is more constructive than the one that I linked to above. Set $S_0 = \varnothing$ and $S_1 = X$. By Sierpiński, there exists $S_{1/2} \in \Sigma$ of measure $1/2$. For each Dyadic rational $q = m/2^n \in [0,1]$ ($1 \leq m \leq 2^n$), we may proceed by induction on $n$ to construct each $S_q$. Now given $r \in [0,1]$, set $S_r = \bigcup_{q \leq r} S_q$. (This is essentially the same method of proof as the one in the reference provided in Ramiro de la Vega's answer.) REPLY [2 votes]: It is also a special case of Theorem 15 (p. 43) in: A. Fryszkowski, Fixed Point Theory for Decomposable Sets, Topological Fixed Point Theory and Its Applications 2, Dordrecht: Kluwer Academic Publishers, 2004.<|endoftext|> TITLE: Massive cancellations QUESTION [33 upvotes]: Let $A=\{a_1,\ldots,a_k\}$ be a fixed, finite set of reals. Let $S_A(n)$ be the set of all reals that are expressible as the sum of at most $2^n$ terms, where each term is a product of at most $n$ numbers from $A$ (here each element of $A$ can be reused an unlimited number of times). Finally, let $d_A(n)$ be the minimum of $|x|$, over all nonzero $x\in S_A(n)$. I'm interested in how quickly $d_A(n)$ can decrease as a function of $n$. Exponential decrease is easy to obtain (indeed, we have it whenever there's an $a\in A$ such that $|a|\lt 1$), but anything faster than that would require extremely finely-tuned cancellations, which continue to occur even as $n$ gets arbitrarily large. Thus, let's call $A$ tame if there exists a polynomial $p$ such that $d_A(n)\gt 1/\exp(p(n))$ for all $n$, and non-tame otherwise. Then here's my question: Does there exist a non-tame $A$? Note that I don't care much about the dependence on $k$ (holding fixed the approximate absolute values of the $a_i$'s), which might be triple-exponential or worse. To illustrate, if $A$ is a set of rationals, then it's easy to show that $A$ is tame. By using known results about the so-called Sum-of-Square-Roots Problem (which rely on results about the minimum spacing between consecutive roots of polynomials, from algebraic geometry), I can also show that if $A$ is a set of square roots of rationals---or more generally, ratios of sums and differences of square roots of positive integers---then $A$ is tame. More generally, I conjecture that every set of algebraic numbers is tame, and maybe this can even be shown similarly, but I haven't done so. But while thinking about this, it occurred to me that I don't have even a single example of a non-tame set---hence the question. Meanwhile, I would also be interested in results showing, for example, that every $A$ consisting of sines and cosines of rational numbers is tame. If anyone cares, the origin of this question is that, if every $A$ is tame, then it follows that given any $n^{O(1)}$-size quantum circuit over a fixed, finite set of gates (i.e., unitary transformations acting on $O(1)$ qubits at a time), the probability that the circuit outputs "Accept" is at least $1/\exp(n^{O(1)})$. Likewise, if all $A$'s that are subsets of some $S\subseteq\Re$ are tame, then the same result follows, but for quantum circuits over finite sets of gates where all the unitary matrix entries belong to $S$. (So for example, from the result mentioned above about square roots of rationals, we get that any quantum circuit composed of Hadamard and Toffoli gates satisfies this property.) This issue, in turn, is relevant to making fully precise the definition of the complexity class PostBQP (quantum polynomial time with postselected measurements), which I invented in 2004. If all $A$'s are tame, then there's no problem with my 2004 definition; if some $A$'s are not tame, then the definition needs to be amended, to restrict the set of gates to ones like {Toffoli,Hadamard} that won't give rise to doubly-exponentially-small probabilities. Update (Nov. 30): For those who are interested, I now have a blog post that discusses this MO question and where it came from (among several other things). REPLY [20 votes]: Embarrassingly, it turns out that Greg Kuperberg previously answered my question: see Theorem 2.11 of this paper. In particular, he proved that all sets of algebraic numbers of tame, and he also observed that there exist sets including non-algebraic numbers that are non-tame (though he didn't include that observation in his paper). Moreover, he did this for exactly the same reason why I was interested in it: namely, in order to fix an ambiguity in the definition of the complexity class PostBQP. The clincher is that he actually wrote to me to explain all this! But I didn't pay attention at the time, and then I forgot about it when the question came up later. My apologies to anyone else who spent time on this on the assumption that it hadn't already been answered. But at least I now really understand the answer!<|endoftext|> TITLE: On a minimal algebraic number field which satisfies the principal ideal theorem QUESTION [6 upvotes]: By an algebraic number field, we mean a finite extension field of the field of rational numbers. Let $k$ be an algebraic number field, we denote by $\mathcal{O}_k$ the ring of algebraic integers in $k$. Let $K$ be a finite extension field of an algebraic number field $k$. Suppose for every ideal $I$ of $\mathcal{O}_k$, $I\mathcal{O}_K$ is principal. Then $K$ is called a PIT(Principal Ideal Theorem) field over $k$. Let $K$ be a PIT field over $k$. We say $K$ is a minimal PIT field over $k$ if $L/k$ is not PIT for every proper subextension $L/k$ of $K/k$. (1) Let $k$ be an algebraic number field and $K/k$ be a finite extension. Is $K/k$ a minimal PIT if and only if $K/k$ is the Hilbert class field? (2) Let $K/k$ and $L/k$ be minimal PITs. Are $K/k$ and $L/k$ isomorphic? REPLY [10 votes]: The answer to the second question is also "no". Take $k= \mathbb{Q}(\sqrt{-5})$. Then the only non-trivial class capitulates in $H=k(i)$ and it also does in $K=k(\sqrt{-3})$, yet $H$ and $K$ are not isomorphic.<|endoftext|> TITLE: Are there some tables or handbooks of homology and homotopy groups of every manifold which has been calculated? QUESTION [10 upvotes]: Are there some tables or handbooks of homology and homotopy groups of every manifold which has been calculated? Or are there some tables or handbooks which list some common calculated results of differential geometry, differential topology and algebraic topology? REPLY [13 votes]: Matthias Kreck and collaborators have set out to create an encyclopaedic catalog of manifolds and their properties a while ago (http://www.map.mpim-bonn.mpg.de/Main_Page), although I don't think it's quite at a stage where one might describe it as useful. Its aim is to collect all known information about "real-world" manifolds, whatever that means. It's a wiki, so if you're interested in helping assemble such a catalog, you might consider contributing. Personally I think that the world of manifolds is too rich and, well, manifold, to be usefully catalogued in this way, but I'd love to be corrected.<|endoftext|> TITLE: Fermat's last theorem over larger fields QUESTION [47 upvotes]: Fermat's last theorem implies that the number of solutions of $x^5 + y^5 = 1$ over $\mathbb{Q}$ is finite. Is the number of solutions of $x^5 + y^5 = 1$ over $\mathbb{Q}^{\text{ab}}$ finite? Here $\mathbb{Q}^{\text{ab}}$ is the maximal abelian extension of $\mathbb{Q}$. By Kronecker-Weber, this is the field obtained from $\mathbb{Q}$ by adjoining all roots of unity. REPLY [22 votes]: This is not quite an answer, but not quite a comment either. We can at least show that $X(({\mathbb Q}^{\text{ab}})^{\text{ab}})$ is infinite (where $X$ is the quintic Fermat curve). There are in fact (at least) two ways of showing this. $X$ has an automorphism $\tau$ of order 3 defined over $\mathbb Q$ (rotate the projective coordinates). The quotient $X/\langle \tau \rangle$ is a hyperelliptic curve $Y$. As I pointed out in a related thread, $Y({\mathbb Q}^{\text{ab}})$ is infinite for every hyperelliptic curve $Y$. Now the preimages on $X$ of any ${\mathbb Q}^{\text{ab}}$-point on $Y$ are defined over a cyclic degree-3 extension of ${\mathbb Q}^{\text{ab}}$, so are in $X(({\mathbb Q}^{\text{ab}})^{\text{ab}})$. Fix any $y \in {\mathbb Q}$, then the points $(x:y:1) \in X$ have $x$ in ${\mathbb Q}(\mu_5, \sqrt[5]{1-y^5}) \subset ({\mathbb Q}^{\text{ab}})^{\text{ab}}$. The second construction obviously generalizes to arbitrary Fermat curves. In my originial post, I claimed that the first one does, too. But this seems to be wrong: the quotient of the degree $n$ Fermat curve $x^n + y^n + z^n = 0$ by the obvious $S_3$-action has positive genus as soon as $n \ge 6$, so the quotient by the cyclic group is no longer obviously hyperelliptic. (It is true, however, that this curve maps to the hyperelliptic curve $y^2 = x^n + \frac{1}{4}$, but this is via the quotient w.r.t. the action of $\mu_n$ such that $\zeta$ sends $(x:y:z)$ to $(\zeta x : \zeta^{-1} y : z)$. This would lead to a weaker conclusion, replacing the double by the triple abelian closure of $\mathbb Q$. I had mixed up the two quotients.) Note that these constructions use the fact that $X$ has (many) nontrivial automorphisms. So perhaps another interesting question is the following. Let $C$ be a nice (smooth, projective, geometrically irreducible) curve over $\mathbb Q$. Can we at least show that $C({\mathbb Q}^{\text{sol}})$ is infinite? Here ${\mathbb Q}^{\text{sol}}$ denotes the union of all finite Galois extensions of $\mathbb Q$ with solvable Galois group. (Originally the question was formulated for (sufficiently generic) curves of genus 3. However, as René pointed out, we find lots of quartic points by intersecting the plane quartic $C$ with rational lines, and all quartic fields are contained in ${\mathbb Q}^{\text{sol}}$.) EDIT: The question whether solvable points exist has been studied, see for example a recent preprint by Trevor Wooley and the references given there, in particular this paper by Ambrus Pál.<|endoftext|> TITLE: A funny factorization of the Jacobian coming from the lines on the Fermat cubic QUESTION [13 upvotes]: Here is something which came up in my algebraic geometry class, and I'm wondering if it has a deeper explanation. Let $F(w,x,y,z) = w^3+x^3+y^3+z^3$ and let $X$ be the cubic surface in $\mathbb{P}^3$ where $F$ vanishes. As is well known, there are $27$ lines on $X$ (in characteristic $\neq 3$). I had my students verify that the obvious equations on the Grassmannian $G(2,4)$ define the $27$ points corresponding to these lines as a reduced scheme. This is easiest to do in an affine cover. For example, consider the affine chart of lines of the form $$\mathrm{RowSpan} \begin{pmatrix} 1 & 0 & p & q \\ 0 & 1 & r & s \end{pmatrix}.$$ This chart contains $18$ of the $27$ points in question. The obvious equations come from setting equal to $0$ the coefficients of $t^3$, $t^2 u$, $t u^2$ and $u^3$ in $$F(t,u,pt+ru,qt+su).$$ I.e. $$1 + r^3 + s^3 = 3 p r^2 + 3 q s^2= 3 p^2 r + 3 q^2 s= 1 + p^3 + q^3=0. \quad (\ast)$$ To verify that they define a radical ideal, one must check the Jacobian condition: I.e., that $$\det \begin{pmatrix} 0 & 0 & 3 r^2 & 3 s^2\\ 3 r^2 & 3 s^2 & 6pr & 6 qs \\ 6pr & 6 qs & 3 p^2 & 3 q^2 \\ 3 p^2 & 3 q^2 & 0 & 0 \\ \end{pmatrix} \neq 0$$ at each of the $18$ roots of $(\ast)$. (Actually, I only assigned them to do one root; then the large symmetry group of $X$ does the rest.) So, here is the thing I can't explain. Out of curiosity, I computed the above determinant and factored it. It turns out that $$\det \begin{pmatrix} 0 & 0 & 3 r^2 & 3 s^2\\ 3 r^2 & 3 s^2 & 6pr & 6 qs \\ 6pr & 6 qs & 3 p^2 & 3 q^2 \\ 3 p^2 & 3 q^2 & 0 & 0 \\ \end{pmatrix} = - 81 \det\begin{pmatrix} p & q \\ r & s \end{pmatrix}^4 .$$ Is there any deep reason for this? Does that determinant even have any significance away from the $18$ points which describe lines on $X$? REPLY [7 votes]: The Jacobian matrix consists of coefficients of $t^3,t^2u,tu^2,u^3$ in the following $4$ partial derivatives $$\partial_p F(t,u,pt+ru,qt+su) = \partial_{y}F(t,u,pt+ru,qt+su)t,\\ \ldots\\ \partial_s F(t,u,pt+ru,qt+su)= \partial_{z}F(t,u,pt+ru,qt+su)u.$$ Following B. Wellington's comment, (up to permutation of columns) this matrix is by definition the Sylvester matrix of the dehomogenizations $P(t)$ and $Q(t)$ of $P(t,u) = \partial_{y}F(t,u,pt+ru,qt+su)$ and $Q(t,u) = \partial_{z}F(t,u,pt+ru,qt+su)$ respectively, regarded as homogenous polynomials in two variables $t$ and $u$. In our particular case, $P(t,u) = 3(pt + ru)^2$ and $Q(t,u) = 3(qt+su)^2$. Since the resultant of two polynomials is defined as the product of the differences of their roots, it follows immediately that the determinant of the Jacobian matrix (up to sign) is $$Res(P(t),Q(t)) = 81 Res(pt+r,qt+s)^4 = 81(pq-rs)^4.$$<|endoftext|> TITLE: Algebraic dependency over $\mathbb{F}_{2}$ QUESTION [28 upvotes]: Let $f_{1},f_{2},\ldots,f_{n}$ be $n$ polynomials in $\mathbb{F}_{2}[x_{1},x_{2},\ldots,x_{n}]$ such that $\forall a=(a_1,a_2,\ldots,a_n)\in\mathbb{F}_{2}^{n}$ we have $\forall i\in[n]:f_{i}(a)=a_{i}$. Can $f_{i}$'s be algebraically dependent over $\mathbb{F}_{2}$?. Or even, can we say something about lower bound on the transcendence degree of this set $\{f_{1},f_{2},\ldots,f_{n}\}$ of polynomials over $\mathbb{F}_{2}$? REPLY [23 votes]: Not for $n=2$. I'm afraid this answer uses a lot more algebraic geometry than the question; I spent some time trying to remove it and failed. Suppose, for the sake of contradiction, that $f_1$ and $f_2$ obey a polynomial relation $g(x,y)$. Let $X$ be the curve $g(x,y) = 0$ in $\overline{\mathbb{F}_2}^2$ (the algebraic closure of $\mathbb{F}_2$) and let $\tilde{X}$ be its normalization. So $(f_1, f_2)$ gives a map $\mathbb{A}^2 \to X$ which, since $\mathbb{A}^2$ is normal, must factor through $\tilde{X}$. This describes $\tilde{X}$ as the image of a rational variety, so $\tilde{X}$ is unirational. For curves, unirational is the same as rational. So $X$ is a genus zero curve (with some number of punctures.) But a genus zero curve defined over $\mathbb{F}_2$ can have at most three $\mathbb{F}_2$-points, so the map $\mathbb{A}^2 \to \tilde{X}$ must identify two of the four $\mathbb{F}_2$-points of $\mathbb{A}^2$. This contradicts that these points are supposed to have distinct images under the composition $\mathbb{A}^2 \to \tilde{X} \to X \subset \mathbb{A}^2$. I see no reason the result should hold for $n=3$, and have played a little with a counterexample where $\mathbb{A}^3$ maps to a cubic surface, but I haven't found an example yet. For example, $x^2 y + x y^2 + z^2 + z$ is a smooth cubic that passes through all eight points of $(\mathbb{F}_2)^3$ (and even remains a smooth cubic in $\mathbb{P}^2$ through all fifteen points of $\mathbb{P^2}(\mathbb{F}_2)$); I see no reason that we couldn't map $\mathbb{A}^3$ to it. Observation: The key question is whether there is a polynomial map $\mathbb{A}^n \to \mathbb{A}^N$, for any $N$, which is defined over $\mathbb{F}_2$, has $(n-1)$-dimensional image and is injective on $\mathbb{F}_2^n$. If so, we can easily interpolate $n$ polynomials in $N$ variables so that the composite $\mathbb{A}^n \to \mathbb{A}^N \to \mathbb{A}^n$ is the identity on $\mathbb{F}_2^n$.<|endoftext|> TITLE: Example of proof using the generic matrix QUESTION [7 upvotes]: There is a really nice proof of the Cayley-Hamilton Theorem using the generic matrix. I expose it briefly. One defines the generic matrix $G:=(X_{ij})_{ij} \in\mathcal{M}_n(\mathbb{Z}[X_{ij}]_{ij})$. The discriminant $\Delta_G$ of $\chi_G$ (characteristic polynomial of $G$) is an element of $\mathbb{Z}[X_{ij}]_{ij}$. Now $\Delta_G = 0$ is impossible, since it would imply in particular (after specialization) that any element of $\mathcal{M}_n(\mathbb{C})$ has at least one double eigenvalue (which is clearly false). Whence $\Delta_G \neq 0$. This means that $\chi_G$ has only simple roots (in some field of decomposition, say $\mathbb{K}$), which in turn means that $G$ is diagonalizable in $\mathcal{M}_n(\mathbb{K})$ and $\chi_G(G)$ follows easily. We get then the Cayley-Hamilton for any matrix after specialization. I am searching for other nice (and easy) use of this generic matrix. In fact any generic argument like this one, could be nice also. I have to precise that I am not an algebraist, so I would really appreciate a simple example. I find this proof a bit magic but at the same time very natural (meaning, after all, that $\chi_G(G)=0$ is just a general algebraic identity, just like $(a+b)^2 = a^2+ ab +ba + b^2$, and it is a way to compute it). Thanks for your help ! REPLY [9 votes]: A famous example is the existence and uniqueness of a polynomial ${\bf Pf}$ in the entries of a $2n\times2n$ alternate matrix, called the Pfaffian, such that $${\bf Pf}(A)^2=\det A,\quad\forall A\in{\rm Alt}_{2n},\qquad{\bf Pf}(J_{2n})=1,\quad J_{2n}:=\begin{pmatrix} 0_n & -I_n \\\\ I_n & 0_n\end{pmatrix}.$$ Proof: we know that the determinant of a $2n\times2n$ alternate matrix $A$ with entries in a field $k$ is a square, because $A=P^TJ_{2n}P$ for some $P$. Apply this to the matrix whose entries are indeterminates ; here $k$ is the field of rational fraction ${\mathbb Q}(X)$. Therefore the Pfaffian exists, at least as a fraction $\frac PQ$. We may suppose $P/Q$ irreducible. Then write $P^2=Q^2{\rm Det}$, where ${\rm Det}$ is a polynomial. Every irreducible factor of $Q$ divides $P^2$, hence $P$, and must be constant. Therefore $Q=1$ and ${\bf Pf}=P$ is a polynomial. The same argument gives that it has entries in $\mathbb Z$.<|endoftext|> TITLE: A game of stones QUESTION [42 upvotes]: How I arrived at this question is a rather long story having to do with the honors calculus class I am teaching. At this point it's sheer curiosity on my part. Here is the game. $\newcommand{\bZ}{\mathbb{Z}}$ We start with a finite collection of stones placed at random somewhere on the set of nodes $\newcommand{\eN}{\mathscr{N}}$ $$\eN=\{2,3,4,\dotsc\}. $$ We can view a distribution of stones as a function $s:\eN\to\bZ_{\geq 0} $ with finite support, $s(n)=$ the number of stones at $n$. Its weight is the nonnegative integer $$|s|=\sum_{n\in\eN} s(n). $$ We say that a distribution $s$ is overcrowded if $s(n)> 1$ for some $n\in\eN$. A node $n$ is called occupied (with respect to $s$) if there is at least one stone at $n$, $s(n)>0$. We are allowed the following moves: choose an occupied node $n$. Then you move one stone from location $n$ to location $n+1$ and add a new stone at location $n^2$. Note that such a move increases the weight by $1$. Now comes the question. Is it true that for any initial distribution of stones $s:\eN\to\bZ_{\geq 0}$ and any positive integer $N$ there exists a finite sequence of allowable moves such that after these moves we obtain a new distribution of stones which (i) is not overcrowded, and (ii) no node $n1$ and define $f:\eN\to\eN$, $f(n)=n+k$. Now change the definition of an allowable move as follows. Pick an occupied node $m$. Move a stone from location $m$ to $m+1$ and add a stone at the node $f(m)$. We will denote by $T_m$ this move. Let $s_0:\eN\to\bZ_{\geq 0}$ be the configuration consisting of a single stone located at $n=2$. I claim that there exists $N>0$ such that $s_0$ cannot be moved past $N$ without overcrowding. The proof is based on a conservation law suggested by David Eppstein's answer. Consider the polynomial $P(x)=x^k+x-1$. Note that $P(0)<0$ and $P(1)>0$ so $P$ has at least one root in the interval $(0,1)$. Pick one such root $\rho$. We use $\rho$ to define the energy of a configuration $s:\eN\to\bZ_{\geq 0}$ to be $$ E(s):=\sum_{n\in \eN} s(n)\rho^n. $$ If $m$ is an occupied location of a configuration $s:\eN\to\bZ_{\geq 0}$, then $$E(T_m s)= E(s)-\rho^m+\rho^{m+1}+\rho^{m+k}=E(s)+\rho^mP(\rho)=E(s). $$ Thus allowable moves do not change the energy of a configuration. Let $N$ be a positive integer such that $$\rho^{N-2}<1-\rho. \tag{1} $$ Suppose now that using allowable move we can transform $s_0$ to a configuration $s$ such that $$ s(n)=0,\;\;\forall nn+1$, $\forall n \in \eN$ and there exists a probability measure $\pi$ on $\eN$ such that $$ \pi\bigl(\; f(n)\;\bigr)+\pi(n+1)-\pi(n)\geq 0,\;\;\forall n\in\eN. \tag{2}$$ Using $f$ to define the allowable moves, one can show that there exists $N>0$ such that $s_0$ cannot be moved past $N$ without overcrowding. The proof uses the entropy $$E_\pi(s)=\sum_{n\in\eN}s(n)\pi(n). $$ Note that this entropy is precisely the expectation of $s$ with respect to the probability measure $\pi$, and it does not decrease as we apply allowable moves $$E(s)\leq E(T_m s), \;\;\forall s. $$ For $f(n)=n+k$ we can define $$ \pi(n)=(1-\rho)\rho^{n-2}. $$ This remark raises the following natural question. Find the functions $f:\eN\to \eN$ such that $f(n)>n+1$, $\forall n\in \eN$, and there exists a probability measure $\pi$ on $\eN$ satisfying (2). How fast can such a function grow as $n\to \infty$? Remark 3. (a) For $f(n)=n^2$ the condition (2) reads $$ \pi(n^2)+\pi(n+1)\geq \pi(n). \tag{3} $$ One can show that a series $$\sum_{n\geq 1}p(n) $$ with nonnegative terms satisfying (3) is divergent if not all the terms are trivial. (This was the rather tricky honors calculus problem that prompted the present question.) Hence, for the function $f(n)=n^2$, there do not exist probability measures satisfying (2), suggesting indirectly that the original question could have a positive answer. (b) If $f(n)=n +1+ \lfloor \sqrt{n}\rfloor$, $\alpha>1$ is sufficiently close to $1$ and $$\pi(n)=\frac{C}{n^\alpha}, \;\;C\sum_{n\geq 2}\frac{1}{n^\alpha}=1,$$ then the condition (2) is satisfied so moving without overcrowding is not possible if the allowable moves use the function $f(n)$. Remark 4. Have a look at Michael Stoll's superb answer. REPLY [8 votes]: This does not answer the original question, but explores a more general version. Let $\mathcal N = \{n_0,n_0+1,n_0+2.\ldots\}$ for some fixed $n_0 \in \mathbb Z$. Let $f \colon {\mathcal N} \to {\mathcal N}$ be a map such that $f(n) > n$ for all $n \in \mathcal N$. We consider the game with stones on $\mathcal N$, where a legal move replaces a stone at position $n$ with two stones at $n+1$ and $f(n)$. Let us say that $f$ is good if for any finite configuration of stones and any $N \in \mathcal N$ there is a sequence of legal moves that transforms it into one that is not overcrowded and does not occupy a position $n < N$. We say that $f$ is bad otherwise. The original question was wether $f(n) = n^2$ is good and was answered positively by Aaron Meyerowitz. The question I want to discuss here is Question. Where (in terms of growth behavior of $f$) is the "phase transition" between bad and good? I am going to argue that this is at a growth behaviour like $f(n) \sim n^2/2$. For this, I will combine the entropy approach mentioned in the OP's Remarks and the idea of the proof in Aaron's answer. Lemma 1. Assume that there exists a discrete measure $\mu$ on $\mathcal N$ such that (i) $0 < \mu(\mathcal N) < \infty$ and (ii) for all sufficiently large $n \in \mathcal N$, $\mu(\{n+1,f(n)\}) \ge \mu(\{n\})$. Then $f$ is bad. Proof. For a configuration $s$ we define $\mu(s) = \sum_{n \in \mathcal N} s(n)\mu(\{n\})$. Let $n' \in \mathcal N$ satisfy $\mu(\{n'\}) > 0$ and be large enough so that (ii) holds for all $n \ge n'$. Consider a configuration $s$ of $m$ stones at $n'$ where $\mu(s) = m \mu(\{n'\}) > \mu(\mathcal N)$. Then by (ii) any sequence of legal moves leads to a configuration $s'$ with $\mu(s') \ge \mu(s) > \mu(\mathcal N)$, whereas any non-overcrowded position $s''$ has $\mu(s'') \le \mu(\mathcal N)$. So in fact no position we can reach by legal moves from $s$ avoids overcrowding. $\Box$ Lemma 2. Assume that there is a constant $C$ such that $f(n+1) + C \ge f(n) + n$ for all sufficiently large $n \in \mathcal N$. Then $f$ is good. Proof. Fix $q \in \mathcal N$ such that $q > C+2$ and the condition on $f$ holds for $n \ge q$. Let $s_0$ be the position consisting of one stone at $q$. We let $s_{m+1}$ be the position obtained from $s_m$ by applying the legal move to every stone in $s_m$. I claim that no $s_m$ is overcrowded. This is certainly true for $m = 0$. Assume the claim is false and let $m \ge 0$ be the smallest $m$ such that $s_{m+1}$ is overcrowded. Then there are $n_1, n_2 \in s_m$ such that $n_1+1 = f(n_2)$. There are two possibilities. Either $n_1 = q+m$; then $n_1$ is the smallest occupied place in $s_m$ and we have the contradiction $n_1+1 = f(n_2) > n_2 + 1 \ge n_1 + 1$ (we use $n_2 \ge q > C+1$ here). Or we obtain $n_1$ from $q$ by a sequence of steps $n \mapsto n+1$ and $n \mapsto f(n)$ containing at least one application of $f$. In this case, at least the last $f(n_2) - f(n_2-1) - 1 \ge n_2 - (2+C)$ steps must be $n \mapsto n+1$ (since there is no value of $f$ between $f(n_2-1)$ and $f(n_2)$). So $m \ge n_2 - (2+C)$. On the other hand, $n_2 \ge q + m$, so we get a contradiction again, since $q > C+2$. This shows the defining property for good $f$ for single-stone positions at places $> C+2$. It has already been observed in this thread that this is sufficient to establish that $f$ is good. $\Box$ Theorem. (1) If $\limsup_{n \to \infty} \dfrac{f(n)}{n^2} < \frac{1}{2}$, then $f$ is bad. (2) If $\liminf_{n \to \infty} \bigl(f(n+1)-f(n)-n\bigr) > -\infty$, then $f$ is good. In particular, if $f(n) = n^2/2 + k n + O(1)$ for some $k \in \mathbb R$, then $f$ is good. Proof. Part (2) follows immediately from Lemma 2. For part (1), fix $\alpha > 1$ such that $2^{-\alpha} > \beta = \limsup f(n)/n^2$ and consider a measure $\mu$ given by $\mu(\{n\}) = (n (\log n)^\alpha)^{-1}$ for $n \gg 0$. Then $\mu(\mathcal N)$ is finite. Also, $$ \mu(\{n\}) - \mu(\{n+1\}) \sim \frac{1}{n^2 (\log n)^\alpha} ,$$ which grows more slowly than $$ \mu(\{f(n)\}) \ge \frac{1}{\beta 2^\alpha n^2 (\log n)^\alpha}(1 + o(1)), $$ so the assumptions of Lemma 1 are satisfied. $\Box$ Corollary. If $f$ is a polynomial, then $f + O(1)$ is good if and only if $\deg f \ge 3$ or $\deg f = 2$ and the leading coefficient of $f$ is $\ge 1/2$.<|endoftext|> TITLE: Non-abelian freeness of SU_2 QUESTION [10 upvotes]: The distribution of the trace of a random element of $SU_2$ is the Sato-Tate distribution. The analogue of the Gaussian distribution in free probability theory is the Wigner semicircle distribution. The Sato-Tate distribution and the Wigner semicircle distribution are the same. Combinatorially, this comes from the fact that the moments of both those two distributions are the Catalan numbers. We can categorify this as follows: If $A$ is a self-dual element of a monoidal abelian category over a field, then $Hom(1, A^{\otimes n})$ is a vector space. (By self-duality, this is also equal to $Hom(A^{\otimes a}, A^{\otimes b} )$ for any $a,b$ such that $a+b=n$.) This categorifies the $n$th moment in the following way: If $G$ is a compact Lie group, and $A$ is a self-dual representation in the category of representations of $A$, then the dimension of $Hom(1, A^{\otimes n})$ is the $n$th moment of the trace of a random element of $G$ acting on $A$. So in the case $G=SU_2$, $A$ the standard representation, the dimension is the Catalan numbers. On the other hand, if you take a free monoidal abelian category on one self-dual generator, the moments are the Catalan numbers. (One has to take dimensions, not just over the base field, but over the ring generated over the base field by the element of $Hom(1,1)$ defined by the canonical morphism $1 \to A \otimes A \to 1$.) Since the category of representations of $SU_2$ is certainly a monoidal abelian category, we have a functor from the second category to the first that sends the free generator to the standard representation. So we have a morphism from the free moment vector spaces to the moment vector spaces of $SU_2$. I believe I can show by an explicit combinatorial argument that this is an isomorphism, categorifying this identity between the two definitions of Catalan numbers. Hence the free monoidal abelian category on one self-dual generator is equivalent to the category of finite-dimensional representations of $SU_2$. Is there some conceptual reason why this is so? Note that the limit of the category of representations of $USP_{2n}$ as $n$ goes to $\infty$ is the free symmetric monoidal abelian category on one self-dual generator. REPLY [3 votes]: I think what you said is not quite right; let me explain why. The free monoidal category on a self-dual object (ignoring abelian for now) has a description involving a version of the cobordism hypothesis (namely the "tangle hypothesis": see Theorem 4.4.4 in Lurie's On the classification of topological field theories). What it works out to is that this gadget has Objects finite sets of points in $\mathbb{R}$ and Morphisms ambient isotopy classes of cobordisms in $\mathbb{R} \times [0, 1]$ between objects (living in $\mathbb{R} \times \{ 0 \}$ and $\mathbb{R} \times \{ 1 \}$ respectively). Here the monoidal structure is stacking sets of points on top of each other. This category deserves to be called something like the category of planar tangles. As you know, it's a fun exercise to work out that, letting $X$ denote the object generating this category, there are exactly $C_n$ elements of $\text{Hom}(1, X^{\otimes 2n})$ all of whose connected components intersect the boundary at least once. This is essentially a slightly disguised version of the interpretation of the Catalan numbers in terms of parentheses. In fact, $\text{End}(1)$ is freely generated by a circle, and as an $\text{End}(1)$-module $\text{Hom}(1, X^{\otimes 2n})$ is freely generated by those $C_n$ elements. Starting from here you can get something more linear by first asking for the free monoidal $k$-linear category on a self-dual object, and you can get that just by taking the free $k$-linear category on the above. Now $\text{End}(1)$ is a polynomial ring $k[x]$ but everything else I just said is still true. This category is a version of the Temperley-Lieb category. Specializing to $k = \mathbb{C}$ you get something that looks an awful lot like the monoidal subcategory of $\text{Rep}(\text{SU}(2))$ generated by the defining representation $V$, but not quite: the circle, which corresponds to $x$ in our free category, gets sent to $-2$ in $\text{Rep}(\text{SU}(2))$ (this is due to the fact that $V$ is self-dual but a map $V \otimes V \to 1$ exhibiting that self-duality can't be chosen to be symmetric; in other words, $V$ is quaternionic). If you want to get every object in $\text{Rep}(\text{SU}(2))$, it turns out you don't have to go anywhere near as far as abelian: it suffices to first require the existence of finite direct sums and second to require that idempotents split. In other words, it suffices to require Cauchy completeness. The corresponding free object gets us every object in $\text{Rep}(\text{SU}(2))$ but again there's that subtlety with $x \in \mathbb{C}[x]$ being sent to $-2$. At this point I want to convince you that you cannot recover $\text{SU}(2)$ itself from this object. There are two ways I know of to try something like this, one involving Tannaka-Krein reconstruction using a fiber functor and one involving Doplicher-Roberts reconstruction. In both situations it's crucial that you have access to not only the monoidal structure on $\text{Rep}(\text{SU}(2))$ but the braided monoidal structure, which this construction doesn't provide (and trying the corresponding free braided monoidal construction gets you extra morphisms coming from tangles in $\mathbb{R}^3$). In fact you can choose various braided monoidal structures that don't get you $\text{SU}(2)$ but that will instead get you the quantum groups $U_q(\mathfrak{su}(2))$. Here $x$ will be sent to something like $ - q^2 - q^{-2}$ depending on your conventions. So we haven't found $\text{SU}(2)$ itself but some kind of combinatorial / topological shadow of both it and its quantum group relatives. Incidentally, I have no idea what this has to do with free probability; it would be cool if someone knew.<|endoftext|> TITLE: Could there be an exact formula for the Ramsey numbers? QUESTION [14 upvotes]: Let $R(k)$ denote the diagonal Ramsey number, i.e. the minimal $n$ such that every red-blue colouring of the edges of $K_n$ produces at least one monochromatic $K_k$. Is it possible that there exists an absolute constant $C>0$ and polynomial $p\in \mathbb{Q}[x]$ such that for all $k$ (or perhaps for all sufficiently large $k$) $$ R(k) = C^kp(k)? $$ Even the asymptotic behaviour of $R(k$) is not known (it is known that $\sqrt{2}^{k(1+o(1))}\leq R(k)\leq 4^{k(1+o(1)}$.) So a proof of such an exact result is certainly out of reach. But showing that there cannot be such a formula might be possible. There are exact formulas known for related Ramsey-type functions (e.g. fractional Ramsey numbers [2] and Ramsey numbers restricted to various classes of graphs, such as planar graphs, line graphs, and perfect graphs [1,3].) As far as I can see, however, the question of an exact formula for the classical Ramsey numbers has not been considered. [1] Belmonte, Heggernes, van't Hof and Saei, Ramsey numbers for line graphs and perfect graphs, Computing and combinatorics, 204–215, Lecture Notes in Comput. Sci., 7434, Springer, Heidelberg, 2012. http://link.springer.com/chapter/10.1007%2F978-3-642-32241-9_18 [2] Brown and Hoshino, Proof of a conjecture on fractional Ramsey numbers, J. Graph Theory 63 (2010), no. 2, 164–178. http://onlinelibrary.wiley.com/doi/10.1002/jgt.20416/pdf [3] Steinberg and Tovey, Planar Ramsey numbers, J. Combin. Theory Ser. B 59 (1993), no. 2, 288–296 REPLY [3 votes]: Obviously I cannot prove it, but I think the answer is almost surely 'no'. In fact, I don't think any combination of polynomials, exponentials, and factorials will do the job: I believe you should need (at least) to have involvement of the parity mod $2$ function. My reason for this is of course quite trivial. When you try to find a lower bound example for $R(3,4)$, you observe that neighbourhoods are independent hence the maximum degree of the graph you are constructing is three. Then you note that every vertex has at most five non-neighbours, otherwise you use the value of $R(3,3)$ in the non-neighbourhood. So you think: OK, I should have a $3$-regular graph on $9$ vertices. Unfortunately this doesn't exist because $3$ and $9$ are both odd, and you are forced to use $8$ vertices. I would expect this kind of parity obstacle to show up quite often.<|endoftext|> TITLE: Averaging maps of Riemannian manifolds QUESTION [5 upvotes]: Let $M$ be a compact Riemannian manifold. We know how to average functions $f\colon M\to {\mathbb R}$; the integral $\frac{\int_M f}{\int_M 1}$ returns a value in ${\mathbb R}$. If intead $f\colon M\to {\mathbb R}^n$, then integrating each component of $f$ separately returns a value in ${\mathbb R}^n$. What happens if $f\colon M\to N$ and $N$ is again a Riemannian manifold? In which cases can one average maps to $N$ in a canonical way that depends only on the Riemannian structure of $N$? The value of such averaging should be a point in $N$. A possible way of obtaining this value could be to take the minimum in $y\in N$ of the following function $g$: \begin{equation}g(y) = \int_M d(y,f(x))dx,\end{equation} and I only expect this minimum to exist and be unique if $M$ is diffeomorphic to a convex subset of ${\mathbb R}^n$ and its Riemannian metric is close to the standard metric (which is the case in which I am interested). REPLY [5 votes]: The anwer to this question follows from Theorem 1.2 in the paper Karcher, H. Riemannian center of mass and mollifier smoothing. Comm. Pure Appl. Math. 30 (1977), no. 5, 509–541. provided by Igor Belegradek in the comments. More specifically, the function on $N$ \begin{equation} g(y) = \frac{1}{2}\int_M d(y,f(x))^2dx \end{equation} has a unique minimum when $N$ is a convex ball (i.e., for any p,q\in N, there is a unique geodesic between them, and it lies in the ball) and either of the following conditions hold: The sectional curvatures at each point of the ball are non-positive, The sectional curvatures at each point of the ball are bounded by $\Delta>0$, in which case the radius of the ball should be strictly less than $\frac{1}{2}\pi\Delta^{-1/2}$.<|endoftext|> TITLE: Busy beaver function vs low Turing degrees QUESTION [9 upvotes]: Let $BB(n)$ denote busy beaver function. It's well known that $BB(n)$ dominates all computable functions (I'm quite certain it includes partial computable functions too). However, I was wondering if we can show similar domination over functions computable from some fixed oracle $A$. In particular, I have been wondering, is it the case that: If $A$ has low r.e. Turing degree (i.e. $A'\equiv_T 0'$), then $BB(n)$ dominates all functions computable from $A$? Best I was able to show is that no $A$-computable function can dominate $BB(n)$ (this is quite simple, because if we can bound $BB$, then we can solve halting problem and compute $0'$). On the other extreme, I know that there exist uncomputable degrees for which the conclusion holds (e.g. hyperimmune-free degrees, because all functions computable from it are dominated by computable functions). However, if we restrict our degrees to be recursively enumerable, I don't know the answer, so a lot simpler version of my question is: Is there an uncomputable r.e. Turing degree such that $BB(n)$ dominates all functions computable from $A$? I expect the answer to this last question to be "yes". Thanks in advance. REPLY [9 votes]: It turns out you're asking about the Array Nonrecursive degrees. A degree $\mathbf{a}$ is called Array Nonrecursive (ANR) if for every function $f$ wtt-below $\emptyset'$, there is an $\mathbf{a}$-computable function $g$ which is not dominated by $f$. You don't need to consider every $f \leq_\text{wtt} \emptyset'$, though; it's enough to consider the modulus function for $\emptyset'$. Now note that $BB \leq_{wtt} \emptyset'$, since we can compute $BB(n)$ after checking which machines with $n$ states halt on the empty input. So every ANR degree computes a function which infinitely often exceeds $BB$. On the other hand, given a Turing machine $T$, we can effectively construct a 2-symbol machine that simulates $T$ and, if $T$ ever halts, writes in unary on the tape the number of steps $T$ performed. If $T$ has Gödel number $n$, let $h(n)$ be the number of states in this simulating machine. We can assume that $h$ is a strictly increasing function, by padding. Then if $T$ converges, it does so in at most $BB(h(n))$ steps. This means $BB(h(n))$ is greater than the modulus of $\emptyset'(n)$. Now, suppose $\mathbf{a}$ computes a function $g$ which infinitely exceeds $BB$. Assume that $g$ is non-decreasing. Then define $f(n) = g(h(n+1))$. This is an $\mathbf{a}$-computable function. For any $g(x) > BB(x)$, fix the least $n$ with $h(n+1) > x$. Then $f(n) = g(h(n+1)) \ge g(x) > BB(x) \ge BB(h(n))$, so $f(n)$ is greater than the modulus of $\emptyset'(n)$. Thus $\mathbf{a}$ is ANR. So the answer to your first question is no, because there are low, r.e., ANR degrees. On the other hand, there are noncomputable r.e. degrees which are not ANR, so the answer to your second question is yes. Edit: Here's why it's enough to escape the modulus function of $\emptyset'$. Suppose $g$ infinitely often exceeds the modulus function. Assume $g$ is nondecreasing. Fix $f \leq_\text{wtt} \emptyset'$. Let $h$ be computable such that the use of $f(n)$ is bounded by $h(n)$ (here using the definition of wtt-reduction). Assume also that $h$ is nondecreasing. Let $m$ be the modulus of $\emptyset'$. Then for every $x$ with $g(x) > m(x)$, let $n$ be least such that $h(n+1) \ge x$. Then $g(h(n+1)) \ge g(x) > m(x) \ge m(h(n))$. So $t(n) = g(h(n+1))$ infinitely often exceeds $m(h(n))$, and $t$ is computable from $g$. Now we can define a new function $u$ from $t$ as follows: on input $n$, try to compute $f(n)$ using $\emptyset'_{t(n)}\!\!\upharpoonright_{h(n)}$ in place of the oracle for $\emptyset'$ until one of two things happen: we see our guess for $f(n)$ converge, or we see $t(n) < m(h(n))$. In the first case, output our guess for $f(n)+1$. In the second case, output 0. For any $n$ with $t(n) > m(h(n))$, we will see convergence and our guess will be the true value of $f(n)$, so $u(n) = f(n)+1$. Thus $u$ infinitely often exceeds $f$, and $u$ is computable from $t$ which is computable from $g$.<|endoftext|> TITLE: Integral domains with totally ordered spectra QUESTION [5 upvotes]: In my research I ended up trying to prove some properties of integral domains such that their spectrum is a totally ordered poset. Are there some nice (ubiqitous/natural) examples of such domains, which are not valuation domains? (Of course any local 1-dimensional domain is such, so non-noetherian almost perfect domains form a class of nice examples, so of higher Krull dimension?) REPLY [5 votes]: A good source of such rings consist of the so-called "pseudo-valuation domains" of Hedstrom and Houston. A link to their original (1978) article is: http://projecteuclid.org/download/pdf_1/euclid.pjm/1102810151. Among other things in that article, they show the following facts that are relevant to your situation: Given any pseudo-valuation domain $R$, there is a unique valuation domain $V$ with $R \subseteq V \subseteq$ the fraction field of $R$, such that the map of prime spectra is an order isomorphism. If you happen to start with a valuation domain of the form $V=K+M$ ($K$ a field, $M$ the maximal ideal of $V$) -- a form that many valuation rings found in nature take -- then take any proper subfield $F$ of $K$ (assuming of course that $K$ isn't a prime field). The subring $R=F+M$ will then be a pseudo-valuation ring whose spectrum is order-isomorphic to that of $V$.<|endoftext|> TITLE: Mayer-Vietoris sequence for topological K-theory QUESTION [7 upvotes]: I'm reading the paper Loop groups and twisted K-theory I by Freed, Hopkins, and Teleman. They give some examples of computing (twisted) K groups using the Mayer-Vietoris sequence. I'm a bit confused with some of their computations, for instance $S^3$ (their example 1.4 in the first section). They take subsets $U_+ = S^3 \backslash(0,0,0,-1)$ and $U_- = S^3 \backslash(0,0,0,1)$ and then they say that $K^0(U_\pm) \simeq \mathbb Z$. I don't understand where this comes from since $U_\pm$ are non-compact so I believe $K^0(U_\pm)$ should be the reduced $K^0$ of a 1-point compactification. The compactifications of these spaces are $S^3$ so shouldn't $K^0(U_\pm) = \tilde {K^0}(S^3) = 0$? However, it seems like if you replace $U_\pm$ by shrinking it a bit to make it closed, these computations work out. So I'm wondering what the exact statement of Mayer-Vietoris is for $K$-theory (specifically, what type of covers you can take) or if Freed, Hopkins, and Teleman are using a different definition of $K^0$ for which $K^0(U_\pm)$ is indeed $\mathbb Z$. Any references would also be appreciated since I couldn't find much in the literature about a Mayer-Vietoris sequence for $K$-theory. REPLY [9 votes]: "since $U_\pm$ are non-compact so I believe $K_0(U_\pm)$ should be the reduced $K_0$ of a 1-point compactification" This is a convention often used in $K$-theory of $C^*$-algebras: by default, people take "$K$-theory" to mean compactly supported $K$-theory. But that this not the convention used for that particular Mayer-Vietoris computation in FHT. There, the version of $K$-theory that is being used is homotopy invariant (so that the $K$-theory of $\mathbb R^n$ is the same as the $K$-theory of a point).<|endoftext|> TITLE: Isomorphism problem for two radical extensions QUESTION [6 upvotes]: Let $n\geq 2$ and let $a,b\in{\mathbb Q}$. Suppose that both the polynomials $A=X^n-a$ and $B=X^n-b$ are irreducible. We want to know whether ( * ) there is a root $\alpha$ of $A$ and a root $\beta$ of $B$ such that ${\mathbb Q}(\alpha)={\mathbb Q}(\beta)$, or if you prefer whether the quotient rings $\frac{{\mathbb Q}[X]}{A(X)}$ and $\frac{{\mathbb Q}[X]}{B(X)}$ are isomorphic. An obvious sufficient condition for ( * ) is that $\frac{b^j}{a^i}$ be a perfect $n$-th power in $\mathbb Q$ for two positive integers $i,j$ coprime to $n$. Is it a necessary condition also ? I also posted this question on MSE : https://math.stackexchange.com/questions/1036372/isomorphism-problem-for-two-radical-extensions-of-the-same-degree REPLY [11 votes]: It seems probably "yes", but the case $4|n$ is a pain. The following method covers all cases except when all of the following conditions hold: $4|n$, $a = -h^2$, and $b = -g^2$ for nonzero rational $h$ and $g$. First, to save notation later, we note that it is harmless to initially replace $a$ with $a^i$ for any $i$ coprime to $n$, and likewise to replace $b$ with $b^j$ for any $j$ coprime to $n$. Such appropriate $i$ and $j$ can be found via Kummer theory over $\mathbf{Q}(\zeta_n)$; let's give that predictable argument now. By hypothesis the polynomials $X^n - a$ and $X^n - b$ over $\mathbf{Q}$ admit respective roots $\alpha$ and $\beta$ in an algebraic closure (or equivalently, in a sufficiently large finite Galois extension) $F$ of $\mathbf{Q}$ such that $\mathbf{Q}(\alpha) = \mathbf{Q}(\beta)$ as subfields of $F$. Consider the subfield $K = \mathbf{Q}(\zeta_n)$ of $F$ which contains a primitive $n$th root of unity. Then $K(\alpha) = K(\beta)$ inside $F$, yet $K(\alpha)$ is visibly a splitting field over $K$ of the (possibly reducible) polynomial $X^n-a \in K[X]$ and likewise $K(\beta)$ is visibly a splitting field over $K$ of $X^n - b \in K[X]$. Hence, by usual Kummer theory there exist $i, j$ coprime to $n$ such that $a^i/b^j$ is an $n$th power in $K$. Now taking advantage of our initial remark about adjustment of $a$ and $b$, we may and do assume that $a/b$ becomes an $n$th power in $\mathbf{Q}(\zeta_n)$. It therefore is sufficient (though not necessary!) to prove for $n \ge 1$ that if a nonzero $q \in \mathbf{Q}$ becomes an $n$th power in $\mathbf{Q}(\zeta_n)$ then $q$ is an $n$th power in $\mathbf{Q}$. This sufficient criterion fails for $n=2m$ with odd $m$, as we see via $m$th powers of non-squares whose square roots generate quadratic subfields of $\mathbf{Q}(\zeta_{n})$ for such $n$. We will prove this sufficient criterion for odd $n$, and then use that conclusion (and even method of proof) to address even $n$. Case $n$ odd Here are two methods. Firstly, this case can be handled by the method in Pace Nielsen's answer without sign constraints since (i) for odd primes $p$ the group $\mathbf{R}^{\times}$ is $p$-divisible, (ii) for $n$ odd (or twice an odd) and any nonzero rational $t$ the polynomial $X^n - t$ is irreducible over $\mathbf{Q}$ if and only if $t$ is not a rational $p$th-power for every prime $p|n$ (due to Theorem 9.1 in Chapter VI of the 3rd edition of Lang's "Algebra", which entails an extra constraint when $4|n$). The following second method has the merit that it never uses the irreducibility hypothesis on $X^n-a$ and $X^n-b$ over $\mathbf{Q}$ (but we will need such irreducibility all over the place when handling even $n$, as we must since a root of the reducible $X^4+4 = (X^2 +2x+2)(X^2-2X+2)$ generates the field $\mathbf{Q}(i)$ that is generated by a root of the reducible $(X^2+1)(X^2-1) = X^4-1$). I claim for any $n$ (allowing even $n$ too -- useful for later!) and odd positive $d|n$ that the natural map $\mathbf{Q}^{\times}/{\mathbf{Q}^{\times}}^d \rightarrow \mathbf{Q}(\zeta_n)^{\times}/{\mathbf{Q}(\zeta_n)^{\times}}^d$ is injective. Passing to $p$-primary parts for primes $p|d$, we may assume $d=p^j$ for $1 \le j \le e = {\rm{ord}}_p(n)$ with $p$ an odd prime. Since $\mathbf{Q}^{\times}$ has no nontrivial $p$th root of unity (as $p$ is odd), the right-exact sequence $$\mathbf{Q}^{\times}/{\mathbf{Q}^{\times}}^p \stackrel{x^{p^{j-1}}}{\rightarrow} \mathbf{Q}^{\times}/{\mathbf{Q}^{\times}}^{p^j} \rightarrow \mathbf{Q}^{\times}/{\mathbf{Q}^{\times}}^{p^{j-1}} \rightarrow 1$$ is also injective on the left and hence is short exact. (The informed reader will recognize this as expressing a fact about Galois cohomology formalism, but that is not logically necessary, though it certainly informs the motivation for what I am about to do.) Likewise, since the group of $p^{j-1}$th-roots of unity in $\mathbf{Q}(\zeta_n)$ is the image under $p$-power on the group of $p^j$th-roots of unity in $\mathbf{Q}(\zeta_n)$ (here we use that $p^j|n$ rather than that $p$ is odd), it follows by reasoning with the snake lemma or elementary Galois cohomology formalism that the analogous right-exact sequence for $\mathbf{Q}(\zeta_n)$ in place of $\mathbf{Q}$ is also injective on the left and hence is short exact. The two resulting short exact sequences fit into an evident commutative diagram, so by induction on $j$ the injectivity assertion is reduced to the case $j=1$. Now it remains to show that if $q$ is a nonzero rational and $q = \theta^p$ for some $\theta \in \mathbf{Q}(\zeta_n)$ then $q$ is the $p$th-power of a rational number. If $q$ is not such a $p$th-power then $X^p - q$ is irreducible over $\mathbf{Q}$ and so the root $\theta$ generates a degree-$p$ extension of $\mathbf{Q}$ which must be abelian (and in particular Galois). Hence, this primitive extension has to split $X^p-q$, so it must contain a primitive $p$th root of unity. Being a degree-$p$ extension, it follows that the degree $p-1$ of $\mathbf{Q}(\zeta_p)$ over $\mathbf{Q}$ must divide $p$. That forces $p=2$, a contradiction. This settles the case of odd $n$, and also shows for general $n$ that $b/a = c^{n'}$ for some rational $c$ with $n'$ the "odd part" of $n$ (and again we have not yet used irreducibility of $X^n-a$ and $X^n-b$ over $\mathbf{Q}$). Case $n$ even Now we may assume $n = 2^en'$ for odd $n'$ and $e \ge 1$. As we saw above, $b/a = c^{n'}$ for a nonzero rational $c$. To go further, we need a property of the field $\mathbf{Q}(\alpha)$ (which we will prove when $e=1$, and when $e > 1$ provided that $a$ and $b$ are not negatives of rational squares): P: there is a unique quadratic subfield (so $\mathbf{Q}(\sqrt{a})$ is the unique one). Here is an important consequence of P when $4|n$: for $m|n$ that is a power of 2, if $\mathbf{Q}(\zeta_m) \subset \mathbf{Q}(\alpha)$ then $m|4$. Indeed, otherwise $8|m$ yet $\mathbf{Q}(\zeta_8)$ does not have a unique quadratic subfield. Via Galois theory, P reduces to a group theory problem for subgroups $G$ of $\mathbf{Z}/n\mathbf{Z} \rtimes (\mathbf{Z}/n\mathbf{Z})^{\times}$ such that $p_2:G \rightarrow (\mathbf{Z}/n\mathbf{Z})^{\times}$ is surjective and $H := G \cap (\mathbf{Z}/n\mathbf{Z})^{\times}$ has index $n$ in $G$. Namely, is the only index-2 subgroup of $G$ containing $H$ precisely $G \cap (((n/2)\mathbf{Z}/n\mathbf{Z}) \rtimes (\mathbf{Z}/n\mathbf{Z})^{\times})$? If $e=1$ then the answer is affirmative. Indeed, in such cases $n=2n'$ with $n'$ odd, so $(\mathbf{Z}/n\mathbf{Z})^{\times} = (\mathbf{Z}/n'\mathbf{Z})^{\times}$ and $\mathbf{Z}/n\mathbf{Z} = \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/n'\mathbf{Z}$ with $\mathbf{Z}/2\mathbf{Z}$ the 2-part. Thus, $G$ is a subgroup of $$\mathbf{Z}/2\mathbf{Z} \times (\mathbf{Z}/n'\mathbf{Z} \rtimes (\mathbf{Z}/n'\mathbf{Z})^{\times}),$$ so the image $G'$ of $G$ in $\mathbf{Z}/n'\mathbf{Z} \rtimes (\mathbf{Z}/n'\mathbf{Z})^{\times}$ (corresponding to the irreducible $X^{n'}-a$) must be full (by applying Theorem 9.4 in Chapter VI of 3rd edition of Lang's "Algebra" to the odd $n'$). So either $G$ is full or of index 2. Hence, $G$ contains the odd-order subgroup $\mathbf{Z}/n'\mathbf{Z}$. Passing to the quotient by this brings us to an analogous question for the commutative group $\mathbf{Z}/2\mathbf{Z} \times (\mathbf{Z}/n'\mathbf{Z})^{\times}$ that is easy. Suppose $c$ is not a rational square, so $\mathbf{Q}(\sqrt{c})$ is a quadratic subfield of $\mathbf{Q}(\alpha)$, but the only such field is $\mathbf{Q}(\sqrt{a})$ by P (which is proved when $e=1$), so $ac$ is a rational square. The same then holds for $ac^{n'}$ since $n'$ is odd, but $ac^{n'} = b$, contradicting that $X^n - b$ is irreducible over $\mathbf{Q}$ with $2|n$. Hence, $c$ must be a rational square, so $b/a$ is a $2n'$th-power of a rational number. This settles the case $e = 1$ unconditionally, so now assume $e \ge 2$. This brings us to: Additional tedium to handle $4|n$ We shall give a general argument when $4|n$ assuming P holds for the case under consideration. (This cannot happen if $a = -h^2$ for rational positive $h$, as then $\mathbf{Q}(\alpha^{n/4})$ is the biquadratic $\mathbf{Q}(i, \sqrt{h})$ when $h$ is a nonsquare and is $\mathbf{Q}(\zeta_8)$ when $h$ is a square.) Then we will prove that P does hold when $4|n$ away from the case when $a$ is the negative of a rational square (and so likewise away from the case when $b$ is the negative of a rational square); that will then leave unsettled only the exceptional cases mentioned at the start, for which P definitely is false. I will leave it to someone else to find an argument to handle such "biquadratic" situations. Fix the choice of $\alpha$ in $\mathbf{C}$. The possibilities for $\beta$ in $\mathbf{C}$ take the form $\zeta \alpha c^{1/2^e}$ for an $n'$-th root of unity $\zeta$ and a $2^e$-th root of $c^{1/2^e}$ of $c$ in $\mathbf{C}$. Some such $\beta$ lies in $\mathbf{Q}(\alpha)$ by hypothesis, so this latter field must then contain $\beta/\alpha = \zeta c^{1/2^e}$. Hence, its $2^e$th-power $\zeta^{2^e} c$ lies in $\mathbf{Q}(\alpha)$, so $\zeta^{2^e}$ does since $c \in \mathbf{Q}^{\times}$. But $\zeta$ is a power of $\zeta^{2^e}$ since $\zeta$ is an odd-order root of unity, so $\zeta, c^{1/2^e} \in \mathbf{Q}(\alpha)$ too. We can then replace $\beta$ with $\beta/\zeta$, so now $\beta = c^{1/2^e}\alpha$ for a $2^e$-th root $c^{1/2^e}$ of $c$ that lies in $\mathbf{Q}(\alpha)$. Write $c = {c'}^{2^{e'}}$ for maximal $e' \le e$ and rational $c'$, so $1 \le e' \le e$. If $e' = e$ then we are done (as then $c^{n'} = {c'}^n$ is an $n$th power), so we assume $e' < e$. By maximality of $e'$, neither $c'$ nor $-c'$ is a rational square. Fix a $2^{e-e'}$th-root ${c'}^{1/2^{e-e'}}$ of $c'$, so the above distinguished $2^e$th root $c^{1/2^e} = \beta/\alpha$ has the form $z {c'}^{1/2^{e-e'}}$ for some $2^e$th-root of unity $z$. Raising the identity $\beta = z {c'}^{1/2^{e-e'}}\alpha$ to the $2^{e-e'}$th-power, we get $\beta^{2^{e-e'}} = \varepsilon c' \alpha^{2^{e-e'}}$ for $\varepsilon := z^{2^{e-e'}}$ a $2^{e'}$th-root of unity. But $\beta^{2^{e-e'}} \in \mathbf{Q}(\alpha)$, so the rational multiple $\varepsilon c'$ of $\varepsilon$ lies in $\mathbf{Q}(\alpha)$. Hence, $\varepsilon \in \mathbf{Q}(\alpha)$. By a consequence of P noted earlier, the $2^{e'}$th-root of unity $\varepsilon$ is equal to $\pm 1$ or $\pm i$. In the $\pm i$ case, the unique quadratic subfield $\mathbf{Q}(\sqrt{a})$ (see P) would have to equal $\mathbf{Q}(i)$, forcing $a$ to be the negative of a rational square. Let us rule out the possibility $a = -h^2$ for rational $h$. In such cases the 4th roots of $a$ lies in the field $\mathbf{Q}(\zeta_8, \sqrt{h})$ whose Galois group over $\mathbf{Q}$ is a direct product of 2 or 3 copies of $\mathbf{Z}/2\mathbf{Z}$. But that is inconsistent with the fact that the 4th root $\alpha^{n/4}$ of $a$ generates a quartic field with a unique quadratic subfield (by P). We conclude that $\varepsilon = \pm 1$. In particular, $\varepsilon c'$ is not a rational square. Since $e-e' > 0$ we see that the non-square $\varepsilon c'$ becomes a square in $\mathbf{Q}(\alpha)$, so $\mathbf{Q}(\sqrt{\varepsilon c'}) = \mathbf{Q}(\sqrt{a})$. In such cases $\varepsilon c' = a u^2$ for rational $u$. Write $\varepsilon c' = a^s w^{2^f}$ with odd $s$, rational $w$ and maximal $1 \le f \le e-e'$ (so $\pm w$ are not rational squares); equivalently, $\beta^{2^{e-e'}} = a^s w^{2^f} \alpha^{2^{e-e'}}$ with odd $s$, rational $w$, and maximal $1 \le f \le e-e'$. (Here and in some places below the exponent of $2^f$ looks like $2f$; not sure why.) If $f = e-e'$ then raising both sides to the $2^{e'}n'$th-power gives $b = a^{2^{e'}n's}w^n a$, so $b/a^{1+2^{e'}n's}$ is an $n$th power with $1+2^{e'}n's$ coprime to $n$ (since $e' > 0$). Assume instead $f < e-e'$. We will deduce a contradiction in such cases. Since $a = \alpha^n$ is a $2^{e-e'}$th-power in $\mathbf{Q}(\alpha)$, we conclude that $w^{2^f}$ is too. Writing $w^{2^f} = x^{2^{e-e'}}$ for some $x \in \mathbf{Q}(\alpha)$, the ratio $x^{2^{e-e'-f}}/w$ in $\mathbf{Q}(\alpha)^{\times}$ is a $2^f$th root of unity. We saw that such a root of unity has to be $\pm 1$ or $\pm i$, so raising to the 4th power yields $x^{2^{e-e'-f+2}} = w^4$ with $e-e'-f+2 \ge 3$. Thus, $x^{2^{e-e'-f+1}} = \pm w^2$ with $e-e'-f+1 \ge 2$. In the case of $-w^2$ it follows that $-1$ is a square in $\mathbf{Q}(\alpha)$, so the unique quadratic subfield $\mathbf{Q}(\sqrt{a})$ (see P) coincides with $\mathbf{Q}(i)$, which is to say $-a$ is a rational square, a case that we have ruled out. Hence, $x^{2^{e-e'-f+1}} = w^2$, so $x^{2^{e-e'-f}} = \pm w$ with $e-e'+f \ge 1$. Thus, one of $\pm w$ becomes a square in $\mathbf{Q}(\alpha)$, but we saw that neither is a rational square, so one of $\pm w/a$ is a rational square: $w = \pm a {w'}^2$ for some sign and some rational $w'$. It follows that $w^{2^f} = a^{2^f}{w'}^{2^{f+1}}$, so $\beta^{2^{e-e'}} = a^{s+2^f} {w'}^{2^{f+1}} \alpha^{2^{e-e'}}$. Since $s+2^f$ is odd and $w'$ is rational, this contradicts the maximality of $f$. Verifying P when $4|n$ and $a$ is not negative of a rational square: Since $4|n$ and $X^n-a$ is irreducible, so is $X^4-a$. It is classical that the Galois group for the irreducible $X^4-a$ is either $D_4 = (\mathbf{Z}/4\mathbf{Z}) \rtimes (\mathbf{Z}/4\mathbf{Z})^{\times}$ or $(\mathbf{Z}/2\mathbf{Z})^2$, with the latter happening precisely when $a = -h^2$ for rational $h$ (in which case a single root generates the biquadratic splitting field that is $\mathbf{Q}(i, \sqrt{h})$ when $h$ is a non-square and $\mathbf{Q}(\zeta_8)$ when $h$ is a square). Looking at the subgroup structure of $D_4$ and what corresponds to $\mathbf{Q}(\alpha)$ in cases with $n=4$, another intrinsic characterization is that for $n=4$ the $D_4$-case is precisely the one for which $\mathbf{Q}(\alpha)$ admits a unique quadratic subfield. Consider a general rational $c$ such that $X^n - c$ is irreducible (with $4|n$), and let $\mathbf{Q}(\gamma)$ be the field generated by a root of $X^n - c$. The Galois group $G$ for the splitting field of $\mathbf{Q}(\gamma)$ over $\mathbf{Q}$ is a subgroup of $$G(n) := (\mathbf{Z}/n\mathbf{Z}) \rtimes (\mathbf{Z}/n\mathbf{Z})^{\times}$$ (see the discussion just above Theorem 9.4 in Ch. VI of the 3rd edition of Lang's "Algebra"), where this injection $G \hookrightarrow G(n)$ is well-defined up to conjugation (given $c$). The composite map $G \hookrightarrow G(n) \rightarrow G(8) = D_4$ has image that corresponds exactly to the Galois group of the splitting field of $X^4-c$, so this map is surjective if and only if $X^4-c$ has Galois group $D_4$. We claim that in such $D_4$-cases, $\mathbf{Q}(\gamma)$ has a unique quadratic subfield. Once this is proved, it follows that if $X^4 - a$ is in the $D_4$-case (i.e., $a$ is not the negative of a rational square) then $\mathbf{Q}(\alpha)$ has a unique quadratic subfield, so likewise for $\mathbf{Q}(\beta) \simeq \mathbf{Q}(\alpha)$, and hence the subfield $\mathbf{Q}(\beta^{n/4})$ has only one quadratic subfield. But then $X^4 - b$ must also be in the "$D_4$-case" too (as otherwise $\mathbf{Q}(\beta^{n/4})$ would be a biquadratic field). In other words, once the above general claim is proved we would have unconditionally settled all cases except those for which $4|n$ and $a = -h^2$ and $b = -g^2$ for nonzero rational $h$ and $g$. The subfield $\mathbf{Q}(\gamma)$ corresponds to $G \cap (\{0\} \times (\mathbf{Z}/n\mathbf{Z})^{\times})$, so our aim is to prove that this lies inside a unique index-2 subgroup of $G$ (as is clear when $G=G(n)$). The hypothesis of being in the $D_4$-case says exactly that the reduction map $G \rightarrow G(4)$ is surjective, so $n=4$ is settled and we now assume $n > 4$. Either $n = 2^e$ with $e \ge 3$ or $n = mp$ for an odd prime $p$ with $4|m$. Consider the latter cases, so $X^m - c$ fall into the $D_4$-case (i.e., $X^4 - c$ has $D_4$ splitting field). Let $\gamma' = \gamma^p$, a root of $X^m - c$, so by induction $\mathbf{Q}(\gamma')$ has a unique quadratic subfield. Since $[\mathbf{Q}(\gamma):\mathbf{Q}] = n = mp = p[\mathbf{Q}(\gamma'):\mathbf{Q}]$, $\mathbf{Q}(\gamma)$ is a degree-$p$ extension of $\mathbf{Q}(\gamma')$. But $p$ is an odd prime, so if $F$ is a quadratic subfield of $\mathbf{Q}(\gamma)$ then it must lie inside $\mathbf{Q}(\gamma')$ and so is unique by induction. It remains to treat the case $n = 2^e$ with $e \ge 3$. First consider $e=3$. Then $G$ is a subgroup of $G(8)$ that maps onto $G(4)$ (so its index divides $[G(8):G(4)] = 4$) and onto $(\mathbf{Z}/8\mathbf{Z})^{\times}$. By inspection, the element $(2,1) \in G(4)$ of order 2 does not admit an order-2 lift in $G(8)$, so $G \twoheadrightarrow G(4)$ has nontrivial kernel. Thus, $G$ has index 1 or 2 in $G(8)$. If index 1 then we are done for $e=3$, so assume the index is 2. Thus, $G$ contains $2\mathbf{Z}/8\mathbf{Z}$. Since $G$ maps onto $G(4)=D_4$ and onto $(\mathbf{Z}/8\mathbf{Z})^{\times}$, it is not hard to deduce that $G$ is the preimage of the graph inside $\mathbf{Z}/2\mathbf{Z} \times (\mathbf{Z}/8\mathbf{Z})^{\times}$ of one of the two order-2 quotient $q: (\mathbf{Z}/8\mathbf{Z})^{\times} \twoheadrightarrow \mathbf{Z}/2\mathbf{Z}$ which is not the reduction onto $(\mathbf{Z}/4\mathbf{Z})^{\times}$. In both cases, by inspection $G$ has a unique index-2 subgroup containing $G \cap (\mathbf{Z}/8\mathbf{Z})^{\times}$. This settles $e = 3$. Suppose $e \ge 4$. We proceed separately depending on whether the image of $G \rightarrow G(8)$ is full or one of the two index-2 subgroups obtained in the study of the case $e=3$ (applied to $X^8-c$). First assume $G \rightarrow G(8)$ is surjective. In such cases we claim that $G = G(2^e)$ (so in such cases there would be a unique quadratic subfield). Proceeding by induction on $e\ge 4$, we can assume $G \rightarrow G(2^{e-1})$ is surjective (as holds when $e=4$ by hypothesis). But $\#G(2^{e-1}) = (1/4)\#G(2^e)$, so $G$ has index 1, 2, or 4 in $G(2^e)$. If the index is 4 then $G$ maps isomorphically onto $G(2^{e-1})$, so all elements of $G(2^{e-1})$ of order 2 would admit an order-2 lift in $G(2^e)$. But inspection shows that none of the elements of $G(2^e)$ of order 2 reduce to the element $(2^{e-2},1) \in G(2^{e-1})$ of order 2. Hence, the index of $G$ in $G(2^e)$ is 1 or 2. Thus, $G \cap ((\mathbf{Z}/2^e\mathbf{Z}) \times \{1\})$ has index 1 or 2 in $\mathbf{Z}/2^e \mathbf{Z}$. If the latter index is 1 then $G$ contains the entire {\em normal} subgroup $\mathbf{Z}/2^e\mathbf{Z}$ of $G(2^e)$ yet maps onto the quotient by it, so we would get $G=G(2^e)$ as desired. Suppose instead that $G$ has index 2 in $G(2^e)$; we will show that $G$ cannot then map onto $G(8)$ (and we will identify two possibilities for its image in $G(8)$). Certainly $G$ corresponds to an index-2 subgroup $H$ of the quotient $$G(2^e)/((2\mathbf{Z}/2^e \mathbf{Z}) \times \{1\}) = (\mathbf{Z}/2\mathbf{Z}) \times (\mathbf{Z}/2^e\mathbf{Z})^{\times}$$ that maps {\em onto} $\mathbf{Z}/2\mathbf{Z} \times (\mathbf{Z}/4\mathbf{Z})^{\times}$ since we are in the $D_4$-case. Likewise, since the projection $H \rightarrow (\mathbf{Z}/2^e\mathbf{Z})^{\times}$ is surjective it must be an isomorphism for size reasons. In other words, $H$ has to be the graph of a quotient map $q:(\mathbf{Z}/2^e\mathbf{Z})^{\times} \twoheadrightarrow \mathbf{Z}/2\mathbf{Z}$, so $q$ factors through the quotient $(\mathbf{Z}/8\mathbf{Z})^{\times}$ modulo squares, and $q$ is one of the two such quotients which is not the reduction map onto $(\mathbf{Z}/4\mathbf{Z})^{\times}$ (since $H$ maps onto $(\mathbf{Z}/2\mathbf{Z}) \times (\mathbf{Z}/4\mathbf{Z})^{\times}$). Hence, $$G = H_{q,e} := \{(x, u) \in (\mathbf{Z}/2^e\mathbf{Z}) \rtimes (\mathbf{Z}/2^e\mathbf{Z})^{\times}\,|\,q(u) = x \bmod 2\},$$ so the image of $G$ in $G(2^{e-1})$ has to be index 2 (since $e-1 \ge 3$ and $q(u)$ only depends on $u \bmod 8$), showing that this situation cannot arise when $G \rightarrow G(8)$ is surjective. This settles the case $e \ge 4$ when $G \rightarrow G(8)$ is surjective. Assume instead that the image of $G$ in $G(8)$ is $H_{q,3}$ for one of the two quotients $q:(\mathbf{Z}/8\mathbf{Z})^{\times} \twoheadrightarrow \mathbf{Z}/2\mathbf{Z}$ that is not the quotient $(\mathbf{Z}/4\mathbf{Z})^{\times}$ (as we have seen is the case when $e=3$). We claim that $G = H_{q,e}$, as has been shown above when $e=3$. Let's first show that this would do the job by proving $H := H_{q,e}$ has a unique index-2 subgroup containing $H \cap (\mathbf{Z}/2^e\mathbf{Z})^{\times} = 1 \times \ker q$. Since $H$ contains $(2\mathbf{Z}/2^e\mathbf{Z}) \times \{1\}$, any index-2 subgroup contains $4\mathbf{Z}/2^e\mathbf{Z}$, so we seek index-2 subgroups of $H = H_{q,e}$ containing the subgroup $$(4 \mathbf{Z}/2^e \mathbf{Z}) \rtimes \ker q \supseteq (4 \mathbf{Z}/2^e \mathbf{Z}) \rtimes (1 + 8\mathbf{Z})/(1 + 2^e\mathbf{Z}).$$ Hence, the problem is reduced to the same for the image $H_{q,3}$ in $G(8)$ which we already handled by inspection. It remains to prove that necessarily $G = H_{q,e}$ when $e \ge 4$. Proceeding by induction, we may assume that the image of $G$ in $G(2^{e-1})$ is the index-2 subgroup $H_{q,e-1}$ (as we know to hold when $e-1=3$; i.e., when $e=4$) Hence, $G \subset H_{q,e}$ and it has index 1, 2, or 4 in $H_{q,e}$; we want to show that its index is 1. If the index is 4 then $G$ maps isomorphically onto $H_{q,e-1}$, so all order-2 elements of $H_{q,e-1}$ admit order-2 lifts in $G(2^e)$. Determination of all order-2 elements shows that the image under the reduction map $G(2^e) \rightarrow G(2^{e-1})$ of the order-2 elements consists of precisely the elements $(y,-1)$ with $y \in 2\mathbf{Z}/2^{e-1}\mathbf{Z}$. But at least one of the order-2 elements $(0, \pm (1 + 2^{e-2})) \in G(2^{e-1})$ lies in $H_{q,e-1}$ (treat $e=4$ separately), and neither has the form $(y,-1)$. Hence, $G$ has index 1 or 2 in $H_{q,e}$. We will rule out index 2, so assume $G$ has index 2 in $H_{q,e}$. The quotient map $H_{q,e} \rightarrow H_{q,e-1}$ is 4-to-1 with subgroup $G \subset H_{q,e}$ of index 2 that must be 2-to-1 onto $H_{q,e-1}$. Let $C_e$ be the normal subgroup $2\mathbf{Z}/2^e\mathbf{Z} \subset H_{q,e}$, so $C_e$ is 2-to-1 onto $C_{e-1} \subset H_{q,e-1}$. The induced quotient map $H_{q,e}/C_e \rightarrow H_{q,e-1}/C_{e-1}$ is the 2-to-1 natural map $\Gamma_{q,e} \rightarrow \Gamma_{q,e-1}$ between graphs, so it is identified with the reduction map $\pi:(\mathbf{Z}/2^e\mathbf{Z})^{\times} \rightarrow (\mathbf{Z}/2^{e-1}\mathbf{Z})^{\times}$. In particular, if $G$ contains $C_e$ then $G/C_e$ provides a splitting to $\pi$, which does not exist. Thus, $G \cap C_e$ has index 2 in $C_e$, so passing to the quotient by the normal subgroup $4\mathbf{Z}/2^{e-1}\mathbf{Z}$ that is 2-to-1 onto its image in $C_{e-1}$ gives an index-2 subgroup $\overline{H}_{q,e} \subset (\mathbf{Z}/4\mathbf{Z}) \rtimes (\mathbf{Z}/2^e\mathbf{Z})^{\times}$ mapping 2-to-1 onto the analogous $\overline{H}_{q,e-1}$ and we have an index-2 subgroup $\overline{G} \subset \overline{H}_{q,e}$ that maps onto $\overline{H}_{q,e-1}$ and hence provides a splitting. Thus, it is enough to show that the natural map $\overline{H}_{q,e} \rightarrow \overline{H}_{q,e-1}$ has no splitting. Passing to the quotient of each side by the normal subgroup $2\mathbf{Z}/4\mathbf{Z}$ gives the reduction map $\Gamma_{q,e} \rightarrow \Gamma_{q,e-1}$ that we have already noted has no splitting.<|endoftext|> TITLE: Equitably distributed curve on a sphere QUESTION [11 upvotes]: Let $\gamma=\gamma(L)$ be a simple (non-self-intersecting) closed curve of length $L$ on the unit-radius sphere $S$. So if $L=2\pi$, $\gamma$ could be a great circle. I am seeking the most equitably distributed $\gamma(L)$, distributed in the sense that the length of $\gamma$ within any disk is minimized. This is something like placing repelling electrons on a sphere, but here the curve self-repels. So there should be no "clots" of $\gamma$ anywhere on $S$. I am especially interested in large $L$. A possible $\gamma$ is shown below, surely not optimal for its length:             Here is an attempt to capture more formally "equitably distributed." I find this an awkward definition, and perhaps there is a more natural definition. Around a point $c \in S$, measure the $r$-density of $\gamma$ as the total length within an $r$-disk: $$d_\gamma(c,r) = | \gamma \cap D(c,r)|$$ where $D(c,r)$ is the disk of geodesic radius $r$ centered on $c$. Then define $d_\gamma(r)$ as the maximum of $d_\gamma(c,r)$ over all $c \in S$. Finally, we can say that, for two curves $\gamma_1$ and $\gamma_2$ of the same length $L$, that $\gamma_1 \le \gamma_2$ if $d_{\gamma_1}(r) \le d_{\gamma_2}(r)$ for all $r \in (0,\pi)$, i.e., $\gamma_1$ is less concentrated than $\gamma_2$ for all $r$ up to a hemisphere. This definition provides a partial order on curves of a given length $L$. One version of my question is: Q. What do the minimal elements of this poset look like, especially as $L$ gets large? These minimal curves are in some sense nowhere densely clotted. Update. Acknowledging Gerhard Paseman's remark, I thought I would include this attractive image of a space-filling curve on a sphere:     (Image from this website). But notice it is certainly not equidistributed in any sense, crowding near the northpole. REPLY [4 votes]: This is from "Population coding under normalization": Numerical solutions of a necklace of electrons<|endoftext|> TITLE: Teichmuller geodesics vs. geodesics in the hyperbolic plane QUESTION [7 upvotes]: Geodesics in $\mathbb H^2$ have the following properties: For every two points in the plane there exists a unique geodesic joining them. Every geodesic determines exactly two points on the boundary of $\mathbb H^2$. Conversely, every pair of points on $\partial \mathbb H^2$ determine a unique geodesic Any two different geodesics that travels at bounded distance are asymptotic. How many of the above properties are true for Teichmuller geodesics on the Teichmuller space of a closed surface with Thurston's boundary? Could you also provide references, please? REPLY [5 votes]: This subject is addressed in Kasra Rafi's very nice paper (particularly relevant to point 4).<|endoftext|> TITLE: Realizing a monoid as $\mathrm{End}(G)$ for some graph $G$ QUESTION [6 upvotes]: This question relates to Realizing groups as automorphism groups of graphs. Given a monoid $M$, is there a graph $G$ such that the endomorphism monoid $\textrm{End}(G)$ is isomorphic to $M$? REPLY [12 votes]: Trying to forestall another question along these lines let me add that every locally presentable category fully embeds into the category of graphs - see Adámek, Rosický Locally presentable and accessible categories 1994. In fact, depending on set theories in which we work, every concrete category fully embeds into graphs - see Pultr, Trnková, Combinatorial, algebraic and topological representations of groups, semigroups and categories 1980. For example: the category of metrizable spaces embeds into Graphs and in some set theories the category of Hausdorff topological spaces also embeds into Graphs. Monoid is a category with a single object, hence a special case of the first result.<|endoftext|> TITLE: Localizations or quotients of categories? QUESTION [9 upvotes]: Motivation: In the classical construction of the derived category of an abelian category, one (roughly) starts with an abelian category $\mathcal{A}$, then considers the quotient category $\mathcal{K}$ of $\mathcal{A}$-chain complexes modulo homotopy and observes this is triangulated; finally, the derived category $\mathcal{D}$ is the localization of the above triangulated category with respect to some morphisms defined in terms of the triangulated structure. I have always thought as the first process of modding out by homotopy equivalences as being a way to make homotopy-equivalent maps look the same or inverting things which are homotopy equivalent to invertible arrows. Then realized that this is also the aim of localising: force some arrows to become isomorphisms. Here the question: is there a conceptual frame in which we can see both localisation and modding out by a relation as special cases? Or, are the two concepts better related that my vague interpretation that they both make some arrows invertible? I have seen someone in nLab saying that they are entirely different constructions but that left me a bit disappointed since I still believe a bit in the feeling that at the end of both days I find myself with a new category having the same objects as the old one, and some new isomorphisms. More generally: I realise that if I had a category in my hands and were willing to see some arrows become isomorphisms, I would not know which of the two approaches would be natural in my case, whence the feeling I still have some fog in my mind. I should add that a colleague of mine observed that he would expect the first construction to be reminiscent of taking closed immersions, and the second to open immersions; which I fully agree to, but would again like to know how to think about these parallels when dealing with a category. REPLY [5 votes]: The term you're looking for is 'coinverter' (dual of the notion of inverter). This is a weighted colimit in the $Cat$-enriched category $Cat$. See this page on the nLab.<|endoftext|> TITLE: Reflexive subspaces of non-separable abstract $L_1$ spaces QUESTION [6 upvotes]: An abstract $L_1$ space is a Banach lattice $E$ such that $\|x+y\|=\|x\|+\|y\|$ for disjoint $x,y\in E$. The space $L_1[0,1]$ is a separable example that contains subspaces isomorphic to $L_p[0,1]$ for $1 TITLE: a question about connected open sets in $R^2$ QUESTION [10 upvotes]: Let $U,V$ be two nonempty connected open sets in $\mathbb{R}^2$ and $U\subsetneqq V$.I want to ask if there must exist an open ball $B\subset V$ such that $B\not\subset U$ and $B\cap U$ is a nonempty connected open set. REPLY [13 votes]: Let $V$ be the complement of a point $a$, and $U$ the complement of a ray $r$ with end-point $a$. Every ball not containing $a$ and intersecting $r$ is in fact split by $r$. REPLY [6 votes]: Choose an enumeration $\{ r_n \}_{n \in \mathbb{Z}_{\geq 0}}$ of the rational numbers, and form the subset $A \subset \mathbb{R}$ given by the union of neighborhoods of radius $2^{-n}$ around $r_n$. Let $U$ be the union of $A \times \mathbb{R}$ with a suitable open half-plane that makes the set connected. Any ball $B$ that is both disjoint from the half-plane and not contained in $U$ satisfies the property that $B \cap U$ has infinitely many connected components. Thus, we may choose $V$ to be the union of $U$ and the complement of $A \times \{0\}$.<|endoftext|> TITLE: Site dependance of the Cech weak equivalences on simplicial sheaves QUESTION [6 upvotes]: Let $\mathcal{T}= sh(C,J)$ a Grothendieck topos of sheaves over a (ordinary) site. One endows the category of simplicial presheaves over $C$ with the "Rezk-Lurie" model structure: that is we start with the projective model structure on simplicial presheaves and then we take the Left Bousefield localization with respect to the map of the form $j \hookrightarrow x$ where $j$ is a covering sieve of $x$ for the topology $J$, where they are both seen as 'constant' simplicial presheaves. This is a model for the $(\infty,1)$-topos of (possibly non Hypercomplete) $\infty$-sheaves over $(C,J)$. It is proved in the appendix of Hypercovers and simplicial presheaves, by Dugger, Hollander and Isaksen that for any simplicial presheaves $S$ the map from $S$ to its levelwise sheafication is a weak equivalence. In particular, any map of simplicial presheaves is a weak equivalence for this model structure if an only if its level sheafication is a weak equivalence. This mean that one get a nice notion of weak equivalence for maps between simplicial objects of the topos $\mathcal{T}$, that we will call "Cech weak equivalence" (by opposition to the Jardin-Joyal weak equivalence which are easily describe internally as the maps inducing isomorphism on all $\pi_n$). My question is: To what extent this notion of "Cech weak equivalence" depends on the choice of a site of definition of the topos $\mathcal{T}$ ? I'm completely willing to assume that $C$ has all finite limite if it change something (see the next remarks) The key result which I think is relevant to this question is Lurie's Proposition 6.4.5.7 of Higher topos theory, which (assuming $C$ has finite limits) describe the universal property of the $(\infty,1)$-topos of $\infty$-sheaves over $(C,J)$ purely in terms of $\mathcal{T}$, and hence proves that the model categories of simplicial presheaves over different sites of definitions of $\mathcal{T}$ (admiting finite product) are all canonically Quillen equivalent. This allows to says a few thing, but unfortunately this is not completely enough to conclude that the weak equivalence between simplicial sheaves are the same (only between those which are fibrant/cofibrant) ! My area of expertise being more topos theory than model category I might be missing something, that is why I'm asking this question here. PS: it also appears that starting with the injective model structure instead of the projective one does not change the weak equivalences at all (even after localization). REPLY [4 votes]: It is independent, provided the two sites of definition, also yield the same infinity topos (so e.g. when they both have finite limits, or the infinity-topos is hypercomplete). Given two simplicial objects $F$ and $G$ of the topos of $\mathcal{E}$, so $$F,G:\Delta^{op} \to \mathcal{E},$$ consider the embedding $$\theta:\mathcal{E} \hookrightarrow Sh_\infty\left(\mathcal{E}\right),$$ where I am writing $Sh_\infty\left(\mathcal{E}\right)$ for the $1$-localic infinity-topos corresponding to $\mathcal{E}.$ I claim that a map $f:F \to G$ is a weak equivalences if and only if the induced map between $\operatorname{colim} \theta \circ F \to \operatorname{colim} \theta \circ G$ is an equivalence in $Sh_\infty\left(\mathcal{E}\right)$. Proof: $f$ is a weak equivalence if and only if it is is one when considered as a map of simplicial presheaves, which is if and only if it becomes one in the associated infinity-topos. Choose a site $C$. In simplicial presheaves on $C$ with the global model structure, the homotopy colimit of $\theta \circ F$ is $F$ itself, which can be computed as the diagonal of the bisimplicial presheaf which is constant in one direction. Hence, the colimit of $\theta \circ F$ is the infinity sheaf associated to $F$.<|endoftext|> TITLE: another question about connected open sets in $R^2$ QUESTION [5 upvotes]: Before posting this question,I just asked a similar question:a question about connected open sets in $R^2$. I got several nice answers.Now I want to ask: Let $U$ be a nonempty connected open set in $\mathbb{R}^2$ and $U\not=\mathbb{R}^2$.I want to ask if there must exist an open ball $B\subset \mathbb{R}^2$ such that $B\not\subset U$ and $B\cap U$ is a nonempty connected open set. REPLY [5 votes]: The answer is no. It is sufficient to find a simple curve from the origin to infinity with the property that every circle intersects it at least twice. The region will be the complement of this curve. Let us begin with the graph of $y=\sqrt{x}\sin(1/x), x\geq 0$. This is a simple curve, and its complement is connected. The complement almost has the desired property: most circles intersect this graph at least twice, so the intersection of the corresponding discs with our region are not connected. It is true that there are circles that intersect our graph only once: these are circles of sufficiently large radius which look almost line vertical lines near the places where they intersect our graph. But it is easy to modify our curve so that all circles (and all straight lines) with intersect it at least twice. For this we arrange appropriate zig-zag's on our graph near the $x$-axis. If this is not clear enough I will scan a picture after the holidays. REPLY [2 votes]: Yet worse example: take $U$ to be the complement of a pseudo-arc $P$ in the unit sphere. For any open disc $D$, such that $P \not\subset \bar D$ and $D\not\subset U$ we have $D\cap U$ is not connected. Take a point in $P$ as the north pole and consider the stereographic projection of $U$ to the plane. For the obtained set $U'$ and any open topological disc $D$ such that $D\not\subset U'$ the set $D\cap U'$ is not connected.<|endoftext|> TITLE: Elementary embeddings with the same critical point QUESTION [8 upvotes]: Question: Is it consistent (relative to the existence of large cardinals) that there is an elementary embedding $j\colon V\to M$ (where $M$ is transitive model) that factors as $j = j_n \circ k_n$ for $n < \omega$, such that for every $n$, $\text{crit }j_n = \kappa$ (the same ordinal), but $j_{n} (\kappa) < j_{n + 1} (\kappa)$ for every $n$? The motivation for this question is the following observation: It is possible to get from single measurable cardinal many different elementary embeddings $j_n\colon V \to M_n$ with the same critical point (by using iterated ultrapower) and a single model $M$ and elementary embedding $j \colon V \to M$ such that $j = k_n \circ j_n$ for $k_n \colon M_n \to M$. The question is whether the "dual" situation, which seems to be more problematic, is possible. REPLY [4 votes]: Yes, this situation can occur. One should simply undertake the dual of the construction you had suggested with iterated ultrapowers. Specifically, suppose that $\mu$ is a normal measure on a measurable cardinal $\kappa$ and let $j:V\to M_\omega$ be the embedding arising from iterating the ultrapower $\omega$ many times. Thus, $M_\omega$ is the direct limit of the system of embeddings $j_{n,k}:M_n\to M_k$, where $j_{n,n+1}$ is the ultrapower of $M_n$ by $\mu_n=j_{0,n}(\mu)$. The sequence $\langle\kappa_n\mid n<\omega\rangle$ is the critical sequence. For any set $S\subset\{\kappa_n\mid n<\omega\}$, we may form the seed hull $$X_S=\{j(f)(\vec s)\mid f:\kappa^{<\omega}\to V, f\in V, \vec s\in S^{<\omega}\},$$ and this is an elementary substructure of $M$, which can be seen by verifying the Tarski-Vaught criterion, and it contains the range of $j$. Further, it is a basic fact of normal ultrapowers that no seed $\kappa_n$ can be generated from the others. That is, if $\kappa_n\notin S$, then $\kappa_n\notin X_S$. You can find this and related results in section 3 of my paper, Canonical seeds and Prikry trees, JSL 62, 1997 (adapted from a chapter of my dissertation). It uses the normality of $\mu$, and this particular fact is not necessarily true without that assumption, as shown in the paper. Now, for each finite $n$, let $S_n=\{\kappa_m\mid m\geq n\}$ and let $\pi_n:X_{S_n}\cong N_n$ be the Mostowski collapse of $X_{S_n}$. Thus, rather than including all seeds up to $\kappa_n$, which is how you might have proved the situation you mentioned at the end of your question, we instead undertake the dual set, using only seeds from $\kappa_n$ and upward. Let $k_n=\pi_n\circ j:V\to N_n$, and let $j_n=\pi_n^{-1}:N_n\to M$, so that we have a commutative triangle of elementary embeddings $j=j_n\circ k_n$. Since $S_n$ contains only $\kappa_m$ for $m\geq n$, it follows that $\kappa_i\notin X_{S_n}$ for $i TITLE: What's the volume of $\{x\in[0,1]^n|\sum x_i\le t\}$ for real $t$? QUESTION [10 upvotes]: EDIT: Thanks for your answers and comments. There is indeed a classical easy formula, given by Pietro Majer (with a simple nice proof) in his answer below. Given $x\in\mathbb{R}^n$, $x_i$ denotes its $i$-th coordinate. My question is: What is $Vol(\{x\in[0,1]^n|\sum_{i=1}^n x_i\le t\})$ for $t\in\mathbb{R}$ ? Is there some kind of "easily computable" formula for it ? A colleague asked me this in relation to a problem in applied mathematics, and I thought that the answer was going to be easy to find, but after trying to compute it by hand in dimension $3$ and then generalizing, I just surrendered. I am pretty sure that it should be somehow classical, but I could not track down a formula or a reference in the general case. I should specify that I know next to nothing about section of convex bodies, and I might be missing something easy. With some browsing, I found the following partial answers (which I hope I do not mix-up): If $t$ is an integer, then the volume is the sum of the $t$ first Eulerian numbers divided by $n!$. Between and integer $t TITLE: What problem would you base your mathcoin on? QUESTION [36 upvotes]: Recently, a variant of electronic currency, based on prime sextuplets, broke the record in generating the largest known set of six primes, packed as closely as possible, that is, a sextuple $(p,p+4,p+6,p+10,p+12,p+16)$ such that all numbers are prime. I am not certain how significant this is to the mathematical community, but there are always a need for more data, which can be used to formulate conjectures. So, if the mathematical community had access to a large amount of computer power, what would be worthwhile compute, and why? Ignoring details about how to make the computation run in parallel and all that. Personally, I would like to see the Ehrhart series for the Birkhoff polytopes, these are only known up to $n=10$. Another series that would be nice to know some more entries of are the number of 1324-avoiding permutations in $S_n$ for larger $n$, (record from 2013 2014 is for $n=36$). This is the smallest instance of enumerating pattern avoiding permutations where no formula is known. Finding more Mersenne primes is also quite interesting. I wonder if I will see the 100th Mersenne prime in my lifetime. REPLY [9 votes]: Apologies if this answer is nitpicky and diverges slightly from the questions intent, but this particular soft question appears to warrant a soft answer. Implicit in your term "mathcoin" are assumptions that the problem: Allows new solutions to be found in regular tunable time intervals, Is memory hard so that specialized ASICs cannot speed up the computations, and Any significantly faster algorithms represent the sort of dramatic advancement that it'd emerge from across the mathematical community simultaneously and not from a secret project. Of course, collisions in cryptographic hashed satisfies 1 and 3, while a memory hard cryptographic hash like scrypt or argon2 satisfies 2 as well. Almost all the answers proposed here thus far fail even 1 and maybe all here fail 2. Worse, there are probably no calculations that both pass 3 and represent useful mathematical (resp. scientific) research, at least not for 2 (resp. 10) years. In short, a naive "mathcoin" as suggested here sounds impossible to do correctly. There might be less naive approaches where perhaps any problem goes and the problem's difficulty is established by previous attempts, but that's actually solving a social problem to achieve 1 with widely varied math problems as simply a source of randomness that defeats ASICs. As an aside, there are proof-of-useful-work systems that could possibly support a bitcoin like currency without wasting resources, but again they all provide useful social services like file storage (filecoin) or anonymity via onion routing. Update 2018: There is still no evidence of any suitable mathcoin problem but there are seemingly suitable science and engineering problems. Examples: Proof-of-moon: All miners must buy telescope arrays to explore the night sky looking for specific astronomical events. If you find one, then reporting it in a canonical way and hashing the report with a nonce can win a block, with the difficulty possibly adjusted by the event. Proof-of-vulnerability: All miners run advanced static analysis tools on software to discover vulnerabilities in published software. A miner wins the block with the hash specific characteristics of a vulnerability plus a nonce, so once a vulnerability is found the miner will eventually win a block. There are several challenges though: First, you must prevent a vulnerability from being used twice, potentially this could be done by resolving to a particular point in the code, but then projects with more releases become better targets. Second, static analysis tools have false positives, so you might need to test consequences somehow. Third, you'll likely need some sort of proof-of-stake system to permit humans to ultimately arbitrate the vulnerabilities. Forth, you want to tackle closed source software too, so the definition of published becomes tricky. Fifth, developers might get annoyed with vulnerabilities being posted to some blockchain without even attempting to warn them first.<|endoftext|> TITLE: How can the generators of subalgebra $\mathfrak g^{\sigma}$ of $\sigma$-stable elements be expressed through generators of Lie algebra $\mathfrak g$? QUESTION [5 upvotes]: Let $\mathfrak g$ be the semisimple Lie algebra of type $D_{4}$. Let $\sigma$ be the 3-rd order automorphism of $\mathfrak g$ induced by the triality of $D_{4}$: $$ \sigma:\alpha_{1}\mapsto\alpha_{3}\mapsto\alpha_{4},\alpha_{2}\mapsto\alpha_{2} $$ Let $\mathfrak g^{\sigma}$ be the subalgebra of $\sigma$-stable elements of $\mathfrak g$. How can Serre generators of $\mathfrak g^{\sigma}$ be expressed through Serre generators of $\mathfrak g$? I suppose that "Serre generators", "Serre-Chevalley generators", and "Chevalley generators" are equivalent terms. (In my opinion, perhaps the use of a Chevalley basis would be more straightforward?) REPLY [5 votes]: In the G$_2$ case, I think that you will find all the information you need in Levasseur, T.; Smith, S. P. Primitive ideals and nilpotent orbits in type G2. J. Algebra 114 (1988), no. 1, 81–105. Things can also be viewed in a very nice symmetric way using 4-ality. See Section 3.4 of J. Landsberg and L. Manivel, Representation theory and projective geometry. In: Algebraic transformation groups and algebraic varieties, Enc. Math. Sci., 132, Springer-Verlag, 2004, 131-167.<|endoftext|> TITLE: Bounded operator on a normed space with empty spectrum QUESTION [11 upvotes]: A bounded operator acting on a complex Banach space has non-empty spectrum, and the proof of this fact uses the completeness of the space. Is there any example of bounded operator acting on a complex non-complete normed space with empty spectrum? I understand that the spectrum of an operator $T$ is the set of all complex numbers $\lambda$ such that $T-\lambda I$ is not bijective. REPLY [14 votes]: Take an operator on a Banach space whose image is dense, whose spectrum is $\{0\}$ but that has no kernel, for example $$ T(f)(x)=\int_x^1f(y)dy $$ acting on $H:=L^2([0,1])$. Then its restriction to the dense subspace $D:=\bigcap_n Im(T^n)$ should have the property you desire. First of all, $D$ is dense in $H$ -- see the comments below by Alexander Shamov. Clearly, $T|_D$ is invertible. So all that remains to be checks is that for $\lambda\not=0$, the operator $(T-\lambda)$ is invertible on $D$. Recall that $(T-\lambda)^{-1}$ makes sense on $H$. The trick is to note that $(T-\lambda)^{-1}$ preserves $D$. Indeed it preserves each subspace $Im(T^n)$: If $x\in Im(T^n)$ write it as $x=T^ny$ and then we have $(T-\lambda)^{-1}x=(T-\lambda)^{-1}T^ny=T^n(T-\lambda)^{-1}y\in Im(T^n)$. QED<|endoftext|> TITLE: Analytical solution of diffusion PDE with Robin boundary condition QUESTION [5 upvotes]: I need to find the analytical solution of the time-independent diffusion equation with constant coefficients on the unit disk $\Omega$ with subject to Robin boundary conditions. The formulation is as follows: $ - \alpha \Delta\varphi (\mathbf{x}) + \beta \varphi(\mathbf{x}) = q(\mathbf{x}), ~ \mathbf{x}\in \Omega$ $ \varphi(\mathbf{x}) + 2\alpha (\nabla \varphi(\mathbf{x}) \cdot \mathbf{n}) = 0, ~ \mathbf{x} \in \partial\Omega$ where $\alpha$ and $\beta$ are not dependent on $\mathbf{x} \in \Re^2$, $\mathbf{n}$ is the normal to $\partial\Omega$. Is it possible to derive a closed form analytical solution (e.g. Green function) for this case at all? Is it already described anywhere in the literature? I would be very grateful for any help. Thank you very much! REPLY [3 votes]: Yes, there is an associated Green's function for your problem, which is explicitly known (at least for complex $\alpha$, $\beta$ with positive real parts). More precisely, the "Robin Green's function", in which you are interested into, can be represented by the sum of an integral expression (= Green's function on $\mathbb{R}^2$) and a Fourier series that contains a product of several modified Bessel functions (= corrector function). The expression is very complicated, but it is known, which even allows you to consider inhomogeneous Robin boundary conditions. Unfortunately, I do not know any reference where you can find the Green's function for your problem. However, the case $\alpha=1$ and $\beta=0$ has deeply been studied in the literature. The Greens function you are looking for should be called the Robin Green's function for the perturbed Poisson equation on the unit circle (disk). Good luck.<|endoftext|> TITLE: Teaching stochastic calculus to students who know no measure theory (or PDE, or...) QUESTION [19 upvotes]: I've got quite a challenge as my teaching assignment for the next Fall (not that I want to get rid of it, quite the contrary, but I still feel like asking for advice won't hurt :-)). I'm to teach the (elementary) stochastic integration and differential equations to the students with practically zero background (the question "in what?" is meaningless here: the answer is "in everything"). I do not want to take the ancient Egyptian approach however ("Do this like we have shown you and your answer will be right!", or whatever else was written on that papyrus a few thousand years ago). So, my current plan (after covering some basic probability and discrete time Markov chains, which, thanks God, is a part of the course) is to introduce countably many independent standard Gaussians, construct the Brownian motion on dyadic rationals by successive splitting of each step $\xi$ into $\frac 12(\xi+\eta)$ and $\frac 12(\xi-\eta)$ drawing independent copies from my countable independent list as I need them, show the relevant continuity (or some crude version of it), extend to reals by continuity, assume that all functions are piecewise continuous (or even constant: after all it is still a dense linear space in $L^2$), integrate with respect to $dW$ using dyadic Riemann procedure and solve the stochastic ODE by the Euler method using the same sequence of discrete dyadic approximations again and again. This seems possible in principle with the full level of rigor if you restrict the classes of random functions you consider and think in advance of how to pass to the limits without ever invoking any non-baby versions of Fubini or Dominated Convergence, but some details promise to be a lot of nuisance to figure out from scratch. So I wonder 1) Is there any decent book (or, even better, lecture notes) that follow this or similar approach? 2) Do you have any better idea of how to carry out this task? The final goal is not very lofty: just to make sure that the student eyebrows won't reach the stratosphere when someone will later talk to them about option pricing, etc. in the financial math. courses, but I still want to do the job decently by common mathematical standards. Thanks in advice for any ideas, references, etc. REPLY [2 votes]: In addition to Steve Shreve's "Stochastic Calculus for Finance II: Continuous time Models" you should definitely use Steve Shreve's "Stochastic Calculus for Finance I" about the binomial discrete time model. Then, rather than worry about existence of Brownian motion you can spend time on the Girsanov theorem and the mathematical proof of the fundamental theorems of asset pricing in volume II. The interplay between the math and the financial interpretation in terms of complete markets etc. is very interesting.<|endoftext|> TITLE: Are these three different notions of a graph Laplacian? QUESTION [11 upvotes]: I seem to see three different things that are being called the Laplacian of a graph, One is the matrix $L_1 = D - A$ where $D$ is a diagonal matrix consisting of degrees of all the vertices and $A$ is the (possibly signed) adjacency matrix. The other is to say that the matrix $L_2$ is a $\vert V \vert \times \vert V \vert$ matrix such that $(L_2)_{ii} = deg(v_i)$ and $(L_2)_{ij} = -\frac{1}{\sqrt{deg(v_i)deg(v_j) } }$ The third is to say that $L_3 = BB^T$ (where $B$ is the incidence matrix) Can someone kindly clarify what is the relation between these three pictures (and or may be all these are the same somehow!?). [..for example for which of them would the positive-semi-definiteness and the the heat-equation intuition hold?...] REPLY [2 votes]: Let me comment a bit on Chris Godsil's answer. The fact that $L_1\ne L_3$ follows from the easy-to-check fact that $L_1=\mathcal I\mathcal I^T$, where $\mathcal I$ is the incidence matrix of an arbitrary orientation of the graph. The nullity of $\mathcal I^T$ is the number of connected components of the graph, whereas the nullity of the transpose of the signless incidence matrix (which is what you presumably denote by $B^T$) is the number of bipartite connected components. So these matrices are clearly different. However, as pointed out by Chris Godsil, they yield the same information in the regular case: the difference is that the upper end of the spectrum of $L_3$ corresponds to the lower end of the spectrum of $L_1$, and vice versa: This was in fact, as far as I can judge, the main reasons for the introduction of $L_3$ in a 1994 paper by Desai and Rao. The fact that the nullities of $L_1,L_2$ coincide can be seen by an argument that might be interesting for your purposes: $L_2$ (or at least the matrix $L_2:=D^{-\frac12}L_1 D^{-\frac12} $ suggested in the comment by Aaron Meyerowitz) is similar to $\tilde{L}_2:=D^{-1}L_1$, which is often called normalized Laplacian as well. It turns out that both $L_1,\tilde{L}_2$ can be seen as Fréchet derivatives of the same energy functional $\mathcal E:f\mapsto\|\mathcal I^T f\|_{\ell^2(E)}^2$, but with respect to two different Hilbert spaces -- more precisely, to the vector space $\mathbb R^V$ endowed with the inner products $$ (f|g):=\sum_{v\in V}f(v)g(v) $$ and $$ (f|g):=\sum_{v\in V}f(v)g(v)\deg(v) $$ respectively (obvious modifications are needed in the case of infinite graphs).<|endoftext|> TITLE: Minimal surfaces + Semi-Geodesic Coordinates QUESTION [5 upvotes]: Let $(M,g)$ be a three dimensional smooth Riemannian manifold and suppose that $\Gamma$ is an embedded minimal surface in $M$. Define the Fermi or semigeodesic coordinates around this surface through the local diffeomorhism $Z:\Gamma \times \mathbb{R} \to M$ $ Z(y,z) = Exp_y (z N(y)) $ where $N$ is the unit vector along $\Sigma$ Define $\Gamma_{t} = \left\{ {x_3=t} \right\} $. and let $H_{x_3}$ denote the mean curvature of $\Gamma_{x_3}$ Does there exist a conformal factor $c$ such that if $ \hat{g} = c g$ then $\hat{H_t}$ is constant for all sufficiently small t? Thanks REPLY [4 votes]: Yes, this can always be done as long as $\Gamma$ has a positive injectivity radius. In fact, calculation yields the following formula: Let $V$ be the unit vector field (defined in a neighborhood of $\Gamma$) that is $g$-perpendicular to the level sets of $t$ (i.e., $V$ is perpendicular to each $\Gamma_t$). This $V$ is well-defined on a $\Gamma$-neighborhood $U\subset M$. Then $$ \mathrm{d}c(V) = 2 H\,c - 2\hat H\,c^{3/2}, $$ where $H$ is the function on $U$ that gives the mean curvature of the level sets $\Gamma_t$ in the metric $g$ and $\hat H$ is the function on $U$ that gives the mean curvature of the level sets $\Gamma_t$ in the metric $\hat g = c g$. Thus, one can specify any desired function $\hat H$ (in particular, a function constant on the each level set $\Gamma_t$) and choose any desired initial function $c_0>0$ on $\Gamma=\Gamma_0$, and there will be a unique solution $c>0$ to the above equation (which is a first order PDE) that satisfies $c = c_0$ along $\Gamma_0$. This solution $c$ will be defined on an open neighborhood $\hat U\subset U$.<|endoftext|> TITLE: Simplest non-constructible set of integers compatible with the nonexistence of $0^\sharp$? QUESTION [5 upvotes]: What is the simplest non-constructible set of integers (say, in the analytical hierarchy) that is compatible with the nonexistence of $0^\sharp$? In particular, can there still be a non-constructible $\Delta^1_3$ set of integers if one assumes the nonexistence of $0^\sharp$? REPLY [8 votes]: It is consistent, relative to just $ZFC$, that there exists a non-constructible $\Delta_3^1$ set of integers. Such a result was first produced in the paper ``Some allpications of almost disjoint sets'' by Jensen-Solovay, using almost disjoint forcing. In fact we can prove the following stronger result which is due to Jensen ``definable sets of minimal degree'': It is consistent that $V=L[R],$ where $R$ is minimal (in the degree of constructibility) and it is the unique solution of a $\Pi_2^1$ predicate (hence it is $\Delta_3^1$).<|endoftext|> TITLE: $\omega$-nerve versus $\Theta$-nerve QUESTION [5 upvotes]: To which extent the adjunction $F\dashv N_\omega$ generated by the $\omega$-nerve described at $n$Lab - oriental (obtained as a particular instance of the nerve-realization paradigm) is linked to the adjunction generated by the functor $O_{[\Theta]}\colon \Theta\to \textbf{Str}\text{-}\omega\text{-}\mathbf{Cat}$ (Joyal's $\Theta$-category), $$\text{Lan}_y(O_{[\Theta]})\dashv N_{[\Theta]},$$ where the functor $N_{[\Theta]}\colon \textbf{Str}\text{-}\omega\text{-}\mathbf{Cat} \to [\Theta^{op}, \mathbf{Sets}]$ sends $C\in \textbf{Str}\text{-}\omega\text{-}\mathbf{Cat}$ to the presheaf $\textbf{Str}\text{-}\omega\text{-}\mathbf{Cat}(O(-), C)$? Is there any reference to learn about, and quote properly, affinities and differences between the two? REPLY [4 votes]: Important edit: A few years ago, Dimitri Ara mentioned to me in private correspondence that he had written a bruteforce program to find objecta in $\Theta$ that do not admit an orientation that can be coherently pasted together. He told me way back then that they found an example in low dimensions very quickly. I do not know if this ended up in any of his published articles yet, but definitively, there is no such extension. He and Maltsiniotis ended up defining a non-full subcategory of $\Theta$ that was coherently orientable, but I don't know the definition. It turns out, due to a series of papers by Richard Steiner and then a later paper by Dimitri Ara and Georges Maltsiniotis, that there doesn't appear to be a meaningful definition for an 'oriental' for any given object of the category $\Theta$. What Ara and Maltsiniotis did figure out was the notion of a lax join of strict $\omega$-categories using Steiner's representation of pasting diagrams as special chain complexes with some additional structure. Essentially the construction is given on "loop-free pasting diagrams", that is to say, directed complexes in the sense of Steiner (which happen to include the pasting diagrams that define $\Theta$ and also the simplicial orientals), then extend it to the category of strict $\omega$-categories by merit of a specialized version of the Day Convolution theorem. On the chain complexes underlying the directed complexes, this is given by the desuspension of the tensor product of the suspensions of two directed complexes, that is, $$A\star B = \Sigma^{-1}(\Sigma A \otimes \Sigma B)$$. In particular, the orientals are obtained by the formula $$\mathcal{O}[n+1] = \mathcal{O}[n] \star [0]$$. This is all contained in the paper of Ara and Maltsiniotis (link) Their definition of the join can also be extended easily to the category of cellular sets using the much simpler (weighted) Day Convolution construction for presheaves. That is, for cellular sets $S,T \in \operatorname{Psh}(\Theta)$, the join $$S\star T (-) = \int^{[a],[b]\in \Theta^+} S^+_a\times T^+_b \times \operatorname{Hom}_{\operatorname{DirComp}^+}( - , [a]\star[b]) $$ where the $+$ means that we're looking at augmented Directed Complexes, augmented cellular sets, and the augmented Disk category $\Theta$. This essentially means attaching an initial object to the categories. There are several other nerves that one can take with respect to different cocellular objects $\Theta\to \operatorname{Str-\omega-Cat}$, but it is important to note that the naive version induced by the fully faithful embedding of $\Theta$ in $\operatorname{Str-\omega-Cat}$ does not induce a Quillen adjunction between strict $\omega$-categories and the various models for weak $\omega$-categories. It turns out that the 'correct' nerve to induce the Quillen adjunction is something a bit more complicated and involves first applying Metayer's resolution by polygraphs to the objects of $\Theta$. Edit: Andrea Gagna informed me by personal correspondence that it appears to be nontrivial and thusfar unproven that the lax join as described above for cellular sets will be associative (and similarly the lax tensor product's extension to cellular sets). The extension of the join is locally biclosed (similarly the extension of the tensor product is truly biclosed), but without further work, the associativity and coherence axioms for monoidal catagories remain to be demonstrated. For example, Andrea gave an interesting example that Dendroidal sets equipped with the Day Convolution-induced tensor product from the tensor product of operads fails to be associative (see the erratum to this paper of Cisinski and Moerdijk), so it's not just free. I have geometric/combinatorial reasons to expect that both will be associative for cellular sets, but it is, as far as I know, an open, if peripheral, question.<|endoftext|> TITLE: Which ordered fields are homeomorphic to their power? QUESTION [23 upvotes]: It is well known that $\mathbb{R}^2\ncong \mathbb{R}$. It is also known that $\mathbb{Q}^2\cong \mathbb{Q}$. It is a corollary to Sierpiński's theorem which states that every countable metric space without isolated points is homeomorphic to $\mathbb{Q}$. (A proof can be found here and a discussion here.) As a consequence from the theorem, for every countable subfield $\mathbb{F}\subset \mathbb{R}$ we know that $\mathbb{F}^2\cong \mathbb{F}\cong \mathbb{Q}$. My question is: what can be said about the general case? In particular, is $\mathbb{R}$ unique in this sense, i.e. is it the only ordered field which is not homeomorphic to it's power? Putting this in another way, I want know if this property is a completeness axiom (in the sense discussed for example here or here). REPLY [13 votes]: $\let\ob\overline\let\sset\subseteq\let\nsset\nsubseteq\DeclareMathOperator\dom{dom}$ There are many such fields different from $\mathbb R$. Proposition: Let $F$ be a topological field of cardinality $2^\kappa$ with a dense subfield $D$ of cardinality $\kappa$. Then there exists an intermediate field $D\sset K\sset F$ of cardinality $2^\kappa$ such that the powers $K^n$ for $n\in\mathbb N$ are pairwise non-homeomorphic. Interesting choices include: $D=\mathbb Q$, $F\subsetneq\mathbb R$. This will produce an incomplete archimedean ordered field $K$. $D=\mathbb Q(x)$ with $x>\mathbb Q$, $F=\hat D$ (the Scott completion of $D$, which has cardinality $2^\omega$, being a perfect Polish space). This will produce a nonarchimedean OF of cardinality $2^\omega$. Whenever $\kappa^{<\kappa}=\kappa$, there exists an OF $|F|=2^\kappa$ with a dense subfield $|D|=\kappa$. [Under the assumption, there exists a $\kappa$-saturated OF $D$ of cardinality $\kappa$. Then it is easy to construct a nested tree of intervals $\{(a_t,b_t):t\in2^{<\kappa}\}$ such that $b_t-a_tm_\alpha$. For any $\alpha$ and $a\in F^{n_\alpha}$, we put $$\ob f_\alpha(a)=\lim_{\substack{x\in D^{n_\alpha}\\x\to a}}f_\alpha(x)$$ if it exists, and similarly for $\ob g_\alpha$. For $a\in F^{n_\alpha},b\in F^{m_\alpha}$, we define $$h_\alpha(a)=b\iff \ob f_\alpha(a)=b\text{ and }\ob g_\alpha(b)=a.$$ Note that $h_\alpha$ is a partial injective function. Then, if $D\sset K\sset F$ and $h\colon K^n\to K^m$ is a homeomorphism, there exists an $\alpha<2^\kappa$ such that $h\sset h_\alpha$. We construct a strictly increasing sequence $\{K_\alpha:\alpha\le2^\kappa\}$ of subfields of $F$, and an increasing sequence $\{A_\alpha:\alpha\le2^\kappa\}$ of subsets of $F$, such that: $K_\alpha$ and $A_\alpha$ are disjoint, and of cardinality at most $\kappa+|\alpha|$. There is no field $K_{\alpha+1}\sset K\sset F$ disjoint from $A_{\alpha+1}$ such that $h_\alpha$ restricts to a homeomorphism $K^{n_\alpha}\to K^{m_\alpha}$. Then $K=K_{2^\kappa}$ satisfies the required conditions, hence what remains is to carry out the construction. We put $K_0=D$, $A_0=\varnothing$, and for limit $\gamma\le2^\kappa$, we define $$K_\gamma=\bigcup_{\alpha<\gamma}K_\alpha,\qquad A_\gamma=\bigcup_{\alpha<\gamma}A_\alpha$$ as usual. For the successor step, assume $K_\alpha$ and $A_\alpha$ have been already constructed. We drop the subscripts from $f_\alpha,g_\alpha,h_\alpha,n_\alpha,m_\alpha$ to simplify the notation, and we say that $K$ is a good field if it is a subfield $K_\alpha\sset K\sset F$ disjoint from $A_\alpha$. Case 1f: There is a good field $K$ such that $K^n\nsset\dom(\ob f)$. We pick $(a_1,\dots,a_n)\in K\smallsetminus\dom(\ob f)$, and define $K_{\alpha+1}=K_\alpha(a_1,\dots,a_n)$, $A_{\alpha+1}=A_\alpha$. Case 2f: There is a good field $K$ such that $\ob f[K^n]\nsset K^m$. We pick $(a_1,\dots,a_n)\in K^n$ and $i$ such that $b_i\notin K$ for $(b_1,\dots,b_m)=\ob f(a_1,\dots,a_n)$. We put $K_{\alpha+1}=K_\alpha(a_1,\dots,a_n)$, $A_{\alpha+1}=A_\alpha\cup\{b_i\}$. Cases 1g, 2g: There is a good field such that $K^m\nsset\dom(\ob g)$, or $\ob g[K^m]\nsset K^n$: similar. Case 3: There is a good field such that $\ob f\restriction K^n$ and $\ob g\restriction K^m$ are not mutually inverse. Say, $a\ne a'\in K^n$, $b\in K^m$, $\ob f(a)=b$, and $\ob g(b)=a'$. We put the coordinates of $a,a',b$ in $K_{\alpha+1}$. Case 4: None of the previous cases applies, i.e., $h$ restricts to a bijection $K^n\to K^m$ for every good field $K$. I claim that this is in fact impossible. Recall that $n>m$. Since $|K_\alpha(A_\alpha)|<2^\kappa$, we can find $a_1,\dots,a_n\in F$ algebraically independent over $K_\alpha(A_\alpha)$. Then $$K_\alpha(a_1,\dots,a_n)\cap A_\alpha=K_\alpha\cap A_\alpha=\varnothing,$$ hence $K=K_\alpha(a_1,\dots,a_n)$ is a good field, thus by assumption, $h(a_1,\dots,a_n)=(b_1,\dots,b_m)\in K^m$. Then $K'=K_\alpha(b_1,\dots,b_m)$ has transcendence degree at most $m$ over $K_\alpha$, hence it is a proper subfield of $K$. However, it is also a good field, hence using the assumption again, $h^{-1}(b_1,\dots,b_m)\in(K')^n$, which implies $K\sset K'$, a contradiction. A minor final complication is that the previous steps do not ensure that $K_\alpha\subsetneq K_{\alpha+1}$. However, this is easy to fix: by the argument in Case 4, there always exists $a\in F$ transcendental over $K_\alpha$ such that $K_\alpha(a)$ is a good field, hence we can throw it in.<|endoftext|> TITLE: $\zeta(0)$ and the cotangent function QUESTION [37 upvotes]: In preparing some practice problems for my complex analysis students, I stumbled across the following. It is not hard to show, using Liouville's theorem, that $$\pi\cot(\pi z)=\frac{1}{z}+\sum_{n=1}^\infty\left(\frac{1}{z+n}+\frac{1}{z-n}\right),$$ which implies that $$-\frac{\pi z}{2}\cot(\pi z)=-\frac{1}{2}+\sum_{k=1}^\infty\zeta(2k)z^{2k},\qquad 0<|z|<1.$$ This formula predicts correctly that $\zeta(0)=-\frac{1}{2}$, and allows to calculate $\zeta(2k)$ as a rational multiple of $\pi^{2k}$ as well (in terms of Bernoulli numbers). Is there some simple explanation why the above prediction $\zeta(0)=-\frac{1}{2}$ is valid? Perhaps there is a not so simple but still transparent explanation via Eisenstein series. Added. Just to clarify what I mean by "simple explanation". The second identity above follows directly from the first identity, i.e. from basic principles of complex analysis: $$-\frac{\pi z}{2}\cot(\pi z)=-\frac{1}{2}+\sum_{n=1}^\infty\frac{z^2}{n^2-z^2}=-\frac{1}{2}+\sum_{n=1}^\infty\sum_{k=1}^\infty\left(\frac{z^2}{n^2}\right)^k =-\frac{1}{2}+\sum_{k=1}^\infty\zeta(2k)z^{2k}.$$ I would like to see a similar argument, perhaps somewhat more elaborate, that explains why the constant term here happens to be $\zeta(0)$, which seems natural in the light of the other terms. REPLY [7 votes]: Here is an explanation based on the Euler-Maclaurin summation formula. (Or rather, since we'll only ever need two terms of the Euler-Maclaurin summation, it's really more or less just "the trapezoid rule".) I think it's a good explanation because it sticks to the general structure of the argument outlined in the question, and its "18th-century-friendly" spirit. First, let's review how to apply Euler-Maclaurin to $\zeta$. We have $$ \begin{align} \zeta(s) &= \sum_{n=1}^N \frac{1}{n^s} + \sum_{n=N+1}^{\infty} \frac{1}{n^s} \\ &= \sum_{n=1}^N \frac{1}{n^s} + \int_N^{\infty} \frac{1}{x^s} \, dx - \frac12 \frac{1}{N^s} + \mathrm{Error} \\ &= \sum_{n=1}^N \frac{1}{n^s} + \frac{1}{s-1} \frac{1}{N^{s-1}} - \frac12 \frac{1}{N^s} + \mathrm{Error} \tag{1}. \\ \end{align} $$ We can say more about the error term later - for now, all we need to know is that $\mathrm{Error} = O ( \frac{1}{N^{\operatorname{Re}(s)+1}} )$ as $N \to \infty$, for each $s$. The key thing to note is that, even though the computation initially required $\operatorname{Re}(s)>1$ in order to be valid, in fact the "equation" $$ \zeta(s) = \sum_{n=1}^N \frac{1}{n^s} + \frac{1}{s-1} \frac{1}{N^{s-1}} - \frac12 \frac{1}{N^s} + O \left( \frac{1}{N^{\operatorname{Re}(s)+1}} \right) \tag{2} $$ has a unique constant solution $\zeta(s)$, for each $s$ with $\operatorname{Re}(s) > -1$ and $s \ne 1$. This is one way to define $\zeta(s)$ for all such $s$. In particular, we can plug in $0$ and immediately find that $\zeta(0) = -\frac12$. What about the function $f(z) = -\frac{\pi z}{2} \cot(\pi z)$? We can also apply Euler-Maclaurin to $f$ in the same way: $$ \begin{align} f(z) &= -\frac12 + \sum_{n=1}^N \frac{z^2}{n^2-z^2} + \sum_{n=N+1}^{\infty} \frac{z^2}{n^2-z^2} \\ &= -\frac12 + \sum_{n=1}^N \frac{z^2}{n^2-z^2} + \int_N^{\infty} \frac{z^2}{x^2-z^2} \, dx - \frac12 \frac{z^2}{N^2-z^2} + O \left( \frac{1}{N^3} \right) \tag{3} \\ \end{align} $$ as $N \to \infty$, for each $z$. The right-hand side of (3), without the error term, is what we'll call $f_N(z)$; we can simplify it to $$ f_N(z) = \sum_{n=1}^N \frac{z^2}{n^2-z^2} + z \cdot \frac12 \left( \log\left(1+\frac{z}{N}\right) - \log\left(1-\frac{z}{N}\right) \right) - \frac12 \frac{N^2}{N^2-z^2} \tag{4}. $$ I should emphasize that (4) is just a somewhat more elaborate variant of $-\frac12 + \sum_{n=1}^{\infty} \frac{z^2}{n^2-z^2}$, much like how (2) is just a somewhat more elaborate variant of $\sum_{n=1}^{\infty} \frac{1}{n^s}$. Now when we expand (4) into power series, we recognize the expression from (2) as the coefficients, and we recognize that we can make the sum run from $k=0$ to $\infty$, not just $k=1$ to $\infty$: $$ \begin{align} f_N(z) &= \sum_{n=1}^N \sum_{k=1}^{\infty} \frac{z^{2k}}{n^{2k}} + \sum_{k=1}^{\infty} \frac{1}{2k-1} \frac{1}{N^{2k-1}} z^{2k} - \frac12 \sum_{k=0}^{\infty} \frac{z^{2k}}{N^{2k}} \\ &= \sum_{k=0}^{\infty} \left( \sum_{n=1}^N \frac{1}{n^{2k}} + \frac{1}{2k-1} \frac{1}{N^{2k-1}} - \frac12 \frac{1}{N^{2k}} \right) z^{2k}. \tag{5} \\ \end{align} $$ This is exactly what we want: take the limit as $N \to \infty$ to conclude that $f(z) = \sum_{k=0}^{\infty} \zeta(2k) z^{2k}$. (To be rigorous in the final step, we'd need to be more precise about the error term in (1). The actual bound we get from Euler-Maclaurin is $ \lvert \mathrm{Error} \rvert \le \mathrm{constant} \cdot \int_N^{\infty} \left\lvert \frac{s(s+1)}{x^{s+2}} \right\rvert \, dx $ for all $N$ and all $s$ such that $\operatorname{Re}(s) > -1$ and $s \ne 1$. This lets us control the size of the difference $\sum_{k=0}^{\infty} \zeta(2k) z^{2k} - f_N(z)$.) This proves that all the even-power coefficients of the power series of $-\frac{\pi z}{2} \cot(\pi z)$ are the corresponding values of $\zeta$, including $\zeta(0)$ as the constant term.<|endoftext|> TITLE: Action of the homotopy braid groups on reduced free groups QUESTION [8 upvotes]: Firstly some definitions: $B_n$ is the braid group with $n$ strands. $\widetilde{B_n}$ is "homotopy braid group", which is a factor group of $B_n$ by adding the relation that $A_{j,k}$ commutes with $gA_{j,k}g^{-1}$, where $A_{j,k}$ are the usual generators of the pure braid group $P_n$ defined by $$A_{j,k}=(\sigma_{k-1}\sigma_{k-2}\cdots\sigma_{j+1})\sigma_j^2(\sigma_{j+1}^{-1}\cdots\sigma_{k-2}^{-1}\sigma_{k-1}^{-1})$$ and $g$ is an element of the subgroup of $P_n$ generated by $A_{1,k},A_{2,k},\cdots,A_{k-1,k}$. $F_n$ is the free group with $n$ generators $x_1,\cdots,x_n$ $K_n$ is the factor group of $F_n$ such that each element of the form $[x_i,g^{-1}x_ig]$ is trivial, where $i=1,\cdots,n$ and $g\in F_n$ and $[,]$ is the commutator. It is known that $B_n$ acts on $F_n$. More precisely, there exists a group homomorphism $\alpha:B_n\to\mathrm{Aut}(F_n)$ defined by $\alpha(\sigma_s)(x_s)=x_{s+1}$ $\alpha(\sigma_s)(x_{s+1})=x_{s+1}^{-1}x_sx_{s+1}$ $\alpha(\sigma_s)(x_t)=x_t$ if $t\neq s, s+1$ It is routine to check that this action descends to an action of $B_n$ on $K_n$; that is, $\alpha$ induces a map $\overline{\alpha}:B_n\to \mathrm{Aut}(K_n)$ defined by the same relations above. My question is: Does $\overline{\alpha}$ induce an action of $\widetilde{B_n}$ on $K_n$? My guess is that the answer is positive. In fact, one only needs to verify the relation that for $k=2,\cdots,n$, $i,j=1,\cdots,k-1$, $r=1,\cdots,n$ $$(\alpha([A_{j,k},{A_{i,k}^{-1}A_{j,k}A_{i,k}}]))(x_r)=x_r$$ in the reduced free group $K_n$, which should be verifiable by the defining relation 1-3 above. But this is very tedious and I am not sure if it is practical. Anyone knows some tactic to tackle this difficulty? REPLY [8 votes]: The short answer is yes. I will try to explain the intuition behind appearance of the reduced free groups and the link homotopy and why they are related; for the rest, I refer to the great papers of Milnor, Goldsmith, Habegger&Lin and Bar-Natan. Pure braids which are allowed to go ''back in time'' were given a name by Habegger and Lin: an $n$-component string link $\sigma$ is a proper embedding of $n$ intervals (strings) into the cylindar $\mathbb{D}^2\times I$ so that $i$-th string starts and ends at the prescribed point $p_i\in\mathbb{D}^2$. As already mentioned, if $\sigma$ is actually a pure braid, then we can look at the induced automorphism $\xi_\sigma$ of the free group $\mathbb{F}\langle x_1,\dots,x_n\rangle=\pi_1(\mathbb{D}^2_n,*)$; here we denote $\mathbb{D}^2_n:=\mathbb{D}^2\setminus\{p_1,\dots,p_n\}$ and each generator $x_i$ corresponds to a small loop around the single puncture $p_i$, plus an arc to the basepoint. Geometrically $\xi_\sigma$ is described by noticing that the complement $\mathbb{D}^2\times I \setminus\sigma$ deformation retracts onto both ends $\mathbb{D}^2_n\times \{0\}$ and $\mathbb{D}^2_n\times \{1\}$, inducing an isomorphism on their fundamental groups. One can actually use Wirtinger presentation for the complement of the braid to write down an algorithm which determines $\xi_\sigma(x_i)$ for each generator $x_i\in\pi_1(\mathbb{D}^2_n\times \{0\},*)$. That goes roughly as follows. Drag the meridinal loop $x_i$ upwards along $i$-th strand and the first time it passes an undercrossing in the braid diagram (assuming we chose the basepoint ''above the projection plane''), transform the meridian $x_i$ according to a Wirtinger-type move: $$ x_i\mapsto \alpha y_i\alpha^{-1}:=y_i^\alpha $$ where $\alpha$ is the meridian corresponding to the overcrossing arc and $y_i$ is the new meridian of the $i$-th strand after the crossing. Now drag $\alpha$ and $y_i$ upwards in the same manner; moreover, repeat with all other meridians to arcs that show up on the way, until you get a word only in the meridians to the upmost arcs. This word is $\xi_\sigma(x_i)$. However, the group $\pi_1(\mathbb{D}^2\times I \setminus\sigma)$ for a more general string link is not isomorphic to the free group. Whenever a ''reversal in time'' happens, we are forced to drag our meridian down, so some undercrossing on our way might force us to conjugate by something we already had before. For example, for the string link in Jjm's post (read it from right to left) we get $x_4\mapsto y_4^\alpha$ and $y_4\mapsto z_4^{x_4}$, so we can't terminate the procedure! One way to get around this is to notice that problems arise when we are trying to conjugate by an element which is already a conjugate of the meridian for the same strand. Hence, if we now imagine that we are always allowed to change a crossing of a strand with itself, this problem will dissapear: we are introducing relations that any two conjugates of $x_i$ commute with each other. Definition. [Milnor] For any group G normally generated by some elements $a_1,\dots,a_k$ define $\mathbb{R}G$ as the quotient of $G$ by relations that impose that each $a_i$ commutes with any of its conjugates. For example, we have the reduced free group $\mathbb{RF}_n:=\mathbb{F}_n\big/\langle [x_i,x_i^\alpha]=1, \forall \alpha \rangle$. From this perspecitve it is natural to introduce the relation of link homotopy on string links: the homotopy of the strings relative to their boundary, such that distinct strings remain disjoint at all times. One can show that the set of equivalence classes, called homotopy string links, forms a group $\mathcal{H}_n$ with respect to the concatenation. Moreover, we now have the isomorphism of reduced groups $\mathbb{R}\pi_1(\mathbb{D}^2\times I \setminus\sigma)\cong \mathbb{RF}_n$ and every homotopy string link induces an automorphism of $\mathbb{RF}_n$ analogous to $\xi_\sigma$ above. Habegger and Lin give a characterization of these autuomorphisms analogous to the Artin's presentation. One of the most useful tools when dealing with these groups is the existence of (split) short exact sequences: $\require{AMScd}$ \begin{CD} 0@>>> \mathbb{F}_n @>>> P\mathcal{B}_n @>{\text{forget last strand}}>> P\mathcal{B}_{n-1} @>>> 0\\ @. @VVV @V{h}VV @VVV \\ 0@>>> \mathbb{RF}_n @>>> \mathcal{H}_n @>{\text{forget last strand}}>> \mathcal{H}_{n-1} @>>> 0 \end{CD} where the map $h$ is just the quotient by the link homotopy. Using them one shows that: one can comb a pure braid, one cam comb a string link, any string link is link-homotopic to a pure braid (i.e. $h$ is onto) and the kernel of $h$ (first described by Goldsmith) is normally generated by pure braids $[A_{jk},A_{jk}^g]$, for $g\in \mathbb{F}\langle A_{1k},\dots,A_{k-1k}\rangle$. J. Milnor, Link groups, Ann. of Math. 59 (1954),177-195. D. Goldsmith, Homotopy of links: in answer to a question of E. Artin, Topology Conference, Lecture Notes in Math., vol 357, Springer-Verlag, Berlin, Heidelberg (1974), pp. 91-96. N. Habegger and X. Lin, The classification of links up to link-homotopy, J. Amer. Math. Soc. 3 (1990), no. 2, 389–419 D. Bar-Natan, Vassiliev homotopy string link invariants, Journal of Knot Theory and its Ramifications (1995), 4(01), pp.13-32.<|endoftext|> TITLE: Perron-Frobenius theory for reducible matrices QUESTION [12 upvotes]: Can someone suggest some sources/references dealing with the Perron-Frobenius theory for nonnegative matrices that are reducible? Specifically, if $A\ge 0$ is a $d\times d$ matrix with no assumptions about irreducibility, I am interested the precise description of a) the set of all non-negative eigenvectors $v\ge 0$ in $R^d$ for $A$; and b) the dynamics of the action of $A$ on the non-negative quadrant $W=\{w\in R^d | w\ge 0\}$, particularly the limiting behavior (up to projectivization) of the sequences $A^nw$ where $w\in W$ and $n=1,2,3,...$. The standard sources on the Perron-Frobenius theory only deal with primitive and irreducible nonnegative matrices, and I could not find any sources that treat in detail the case of an arbitrary $d\times d$ matrix $A\ge 0$. REPLY [2 votes]: I recently ran into a similar problem. Something that helped me a lot was Chapter 9 of the first edition of the Handbook of Linear Algebra (which seems to be Chapter 10 of the second edition---just look for something written by Rothblum), edited by Professor Hogben.<|endoftext|> TITLE: The resolution of which conjecture/problem would advance Mathematics the most? QUESTION [6 upvotes]: This is an impossibly broad question, and makes the unwarranted assumption that Mathematics is a uniform field. It might make more sense to ask the same question restricted to, say, Mathematical Logic, or Differential Geometry, or Number Theory. Nevertheless, I think there may be a sense in which certain fields of mathematics are (temporarily) blocked by major unresolved conjectures, and others are not. In which case, a breakthrough in a "blocked" field might advance Mathematics—broadly construed—the most. For example, prior to 2002, I would have ventured: the Poincaré conjecture & Thurston's geometrization conjecture. But since Perelman's resolution, I see (in my limited vision) no equivalent major conjecture outstanding in Differential Geometry. Similarly, one might might view Fermat's last theorem as a key blockage within Number Theory before the 1995 Wiles-Taylor resolution; or twin primes before Zhang's 2013 breakthrough and subsequent polymath's improvements. My parochial, biased viewpoint is that resolution of the P=NP question, tomorrow, would represent the greatest advance in Mathematics. But perhaps others think that, e.g., resolution of the Riemann hypothesis would represent the larger advance? Of course, how a question is resolved makes a huge difference: A narrow proof is less an advance than a broad reconceptualization that reshapes the landscape. Q. I welcome responses specific to narrow fields, but also remarks on Mathematics (if such exists!) as a whole—from which we may all learn from one another. Editor's note: there is a meta thread about this post at Should we reopen "The resolution of which conjecture/problem would-advance mathematics the most?". Please read that before casting open/close votes. REPLY [3 votes]: I conjecture the question to be premature ... I would nominate the Riemann Hypothesis, since it is clear that something occurs that we fundamentally don't understand. But folding other things in with RH, such as the Artin Conjecture, is known to be a good idea (since Weil). And the good behaviour of L-functions should be expressed by a "geometric" principle. Such a thing, if convincingly formulated, would have a strong claim. There is presumably another major conjecture to do with how K-theory would rule geometry. Algebra rules, subordinating geometry then analysis. Such a clutch of conjectures would delineate the reach of structure, at least into the heart of the mathematical heritage of the 19th century. Anyway, top-down questions provoke top-down answers, by suggesting "incremental" progress is beside the point. But it isn't, of course.<|endoftext|> TITLE: When can we reach a real by forcing? QUESTION [7 upvotes]: I'm sure this is well-known, but: suppose I have a non-constructible real $r\in V-L$. Under what conditions is there a poset $\mathbb{P}\in L$ and a $G$ which is $\mathbb{P}$-generic over $L$, such that $r\in L[G]$? (Note: $G$ is not necessarily a real.) Obviously it is consistent (e.g., with $V=L$) that the answer is "always;" for a non-trivial example of this, the result holds if $V$ is gotten from $L$ by adding a minimal degree of constructibility. It seems unlikely that the answer would be "always" in general, but I can't come up with a counterexample. REPLY [5 votes]: See "When is a given real generic over L?" by Fabiana Castiblanco and myself for an answer to this question: https://ivv5hpp.uni-muenster.de/u/rds/fabiana_ralf.pdf<|endoftext|> TITLE: When does a Catalan number equal a Fibonacci number? QUESTION [14 upvotes]: The $n=3$'rd Catalan number (A000108) is $1,1,2,5$ : $\frac{\binom{2n}{n}}{n+1}=\frac{\binom{6}{3}}{4}=\frac{20}{4}=$ 5. The $n=4$'th Fibonacci number (A000045) is $1,1,2,3,5,...$ : 5. Q. Which other Fibonacci numbers (besides $\{1,2,5\}$) are also Catalan numbers? There seems to be no other "small" solutions, at least up to Fibonacci/Catalan numbers around $10^{60}$. REPLY [38 votes]: A result of the type you seek follows easily from Carmichael's theorem, that if $m > 12$, then there is a prime $p$ that divides $F_{m}$, but does not divide $F_{k}$ for $k < m$. Suppose $C_{n} = \binom{2n}{n}/(n+1)$ and we assume that $C_{n} = F_{m}$. All the prime factors of the left hand side are $\leq 2n$, while by Carmichael's theorem there is a prime $p | F_{m}$ that does not divide any earlier Fibonacci number. However, since for $p > 5$, $p$ divides either $F_{p-1}$ or $F_{p+1}$ we have that $m \leq p+1 \leq 2n$. However, from the asymptotic size of $C_{n}$ and $F_{m}$, we must have that $m \approx \frac{\log(4)}{\log(\phi)} n$, and this contradicts the inequality $m \leq 2n$.<|endoftext|> TITLE: Thom isomorphism from the ABGHR perspective QUESTION [7 upvotes]: In ABGHR Thom spectra are described in the following way: we start with a morphism of Kan complexes $X\to \mathbb{S}\text{-line}$, where $\mathbb{S}\text{-line}$ is an $\infty$-groupoid which is equivalent to the well known $BGL_1(\mathbb{S})$. Then the Thom spectrum $Mf$ is defined to be the colimit of the composition of functors (of $\infty$-categories) $X\to\mathbb{S}\text{-line}\to\mathbb{S}Mod$. If we have a commutative ring spectrum $R$ there is a functor $\mathbb{S}\text{-line}\to R\text{-line}$ and it is a theorem that the colimit of the composition $X\to\mathbb{S}\text{-line}\to R\text{-line}\to RMod$ is equivalent to $Mf\wedge R$. There is a fibration of $\infty$-groupoids $R\text{-triv}\to R\text{-line}$ that we should think of as being our model of $EGL_1(R)\to BGL_1(R)$. If the composition $X\to R\text{-line}$ lifts to $R\text{-triv}$ then we say that $Mf$ is $R$-oriented and we have a Thom isomorphism $Mf\wedge R\simeq X_+\wedge R$. My question is the following: we know classically that $Mf\wedge Mf\simeq X_+\wedge Mf $ (perhaps there are some conditions I need to put on $f$ to make this true?). So, if we assume $X$ is an infinite loop space and $X\to \mathbb{S}\text{-line}$ is a morphism of infinite loop spaces then we should have that $X\to\mathbb{S}\text{-line}\to Mf\text{-line}$ lifts to $Mf\text{-triv}$, yielding the classical Thom isomorphism described above. Is there an easy way to see that there is such a lift in this framework? A more general question is: we know that such lifts always give such Thom isomorphisms, but are there conditions under which we can deduce a lift from such an isomorphism? REPLY [7 votes]: Let's try to see what a lift $\tilde f : X\to R\textrm{-triv}$ is. Recall that $R\textrm{-triv}$ is the $\infty$-groupoid of $R$-lines with a specified isomorphism with $R$, so a lift $\tilde f$ corresponds to giving a natural equivalence of $f$ with the constant functor $X\to R\textrm{-line}$ with value $R$. This is the same as giving an $R$-module map from the colimit of $f$ to $R$ which is an equivalence after precomposing with all the inclusions $f(x)\to \textrm{colim }f$. But $\textrm{colim }f=Mf\wedge R$, so a lift is effectively giving an $R$-linear map $Mf\wedge R \to R$ such that for all $x$ the composition $f(x)\wedge R\to R$ is an equivalence. Moreover an $R$-module map $Mf\wedge R\to R$ is the same thing as a spectrum map $Mf\to R$. That is, a lift is the same datum as a Thom class. Specializing in your case it is easy to see that the identity $Mf\to Mf$ gives exactly the Thom class you need (because the map $f$ is multiplicative, that is a loop space map), so you're done.<|endoftext|> TITLE: When do people actually use the maximum entropy distribution? QUESTION [18 upvotes]: One of the standard problems in convex optimization is the calculation of the maximum entropy distribution that satisfies some set of criteria. For example, if $\mathbf{x} \in \mathbb R^n$ is an unknown vector of probabilities, one might solve a problem like $$\mathrm{maximize}_\mathbf{x\geq0} -\sum_{i=1}^n x_i \log x_i ~~~s.t. \\ A\mathbf{x}\leq \mathbf b \\ \sum_{i=1}^n x_i = 1~,$$which would of course find the probability distribution that satisfies a certain set of inequalities whose entropy is as large as possible. My question is: where do people actually use this (maybe not just using inequality constraints, other kinds of constraints would be possible as well of course)? I see problems of this kind arising in textbooks very often, but I don't see research papers in which one really makes use of these distributions to some specific aim. REPLY [5 votes]: Thermodynamics can be derived from this principle: \begin{align*} 0 &\overset!=\delta \Big[\underbrace{-k_B\langle\ln p\rangle}_{S} + \alpha\Big(\langle1\rangle - 1\Big) + \beta\Big(\langle E\rangle - U\Big) + \mu\Big(\langle N\rangle - N\Big)\Big] \\\text{where}\ \langle x\rangle :&= \sum_i p_i x_i \ \text{denotes the expectation value of $x$} \end{align*} $k_B$ is the Boltzmann constant, which makes the entropy $S$ equivalent to the physical one, while the Lagrangian multiplicators $\beta$ and $\mu$ which conserve (internal) energy and particle count respectively turn out to be equivalent to the thermodynamical quantities $\beta^{-1}=k_B T$ (i.e. $\propto$ inverse temperature) and chemical potential $\mu$. For the canonical ensemble (i.e. no particle exchange, $N_i\equiv N$) this yields the Boltzmann distribution $$p_i = \frac{e^{-\beta E_i}}{\sum\limits_j e^{-\beta E_j}}.$$<|endoftext|> TITLE: Poincare-like inequality on compact Riemannian manifolds QUESTION [6 upvotes]: I am looking for a Poincare Inequality on balls but instead of euclidean space, I have a compact Riemannian manifold without boundary. The inequality I am looking for is the equivalent of $$ \int_{B_{r}(x)} |f(y) - f(z)|^{p} dy \leq c r^{n+p-1} \int_{B_{r}(x)} |Df(y)|^{p} |y-z|^{1-n} dy$$ where $f \in C^{1}(B_{r}(x))$ and $z \in B_{r}(x)$. I would be grateful for any source you can point me to. I am also interested to know if there is a version of this inequalilty if the manifold has boundary. Thank you. REPLY [7 votes]: A more accurate concept of the inequality you are looking is a Sobolev-type inequality or more precisely Hardy's inequality. The proof of this inequality can be found in most books on weighted Sobolev inequality, say the excellent book by V. G. Maz'ya. If you just want a directly proof, say for instance the following paper "Kinnunen, Juha; Martio, Olli, Hardy's inequalities for Sobolev functions. (English summary) Math. Res. Lett. 4 (1997), no. 4, 489–500. " I would like to talk more about this inequality from the point view of capacity or equivalently non-linear potential. The standard proof of all (fractional) Sobolev-type inequality based on point-wise characterization of Sobolev functions, which basically reads as for a.e. $x,y$ $$|u(x)-u(y)|\leq c(n)|x-y|^{1-\alpha/p}(M_{\alpha/p}|Du|(x)+M_{\alpha/p}|Du|(y)),$$ where $M_\beta$ is the standard fractional maximal operator. In order to gain global estimates, one just integrate the above inequality and use the boundedness of the maximal operator. The equivalence of characterizations of Sobolev function can be found in the nice book of J.Heinonen, "Lectures on Analysis on Metric Spaces, Springer Verlag, Universitext 2001." or the forthcoming book of J.Heinonen et al. "Sobolev spaces on metric measure spaces, an approach based on upper gradients." Hardy-type inequalities are usually regarded as weighted Sobolev type inequality by considering the potential term $|y-z|^{k}$ or $d(y,\partial \Omega)^k$ as a general weight. I encourage you to read the nice paper by P.Koskela, P.Hajlasz, Isoperimetric inequalities and imbedding theorems in irregular domains. J. London Math. Soc. (2) 58 (1998), no. 2, 425–450 and P.Koskela, P.Hajlasz, Sobolev met Poincaré. Mem. Amer. Math. Soc. 145 (2000), no. 688, x+101 pp. The basic capacity view is to regard the right-hand side as certain capacity, and characterize the Sobolev/Hardy type inequality in terms of certain capacity. Then one could use the standard techniques from potential theory to do capacity estimates. This point of view is quite important in proving Sobolev/Hardy type inequality for irregular domains or in metric setting. As a sample, you could also read the following paper. Koskela, Pekka; Lehrbäck, Juha, Weighted pointwise Hardy inequalities. (English summary) J. Lond. Math. Soc. (2) 79 (2009), no. 3, 757–779. See also the homepage of J.Luhrback for the recent progress on Hardy type inequality for general domains: http://users.jyu.fi/~juhaleh/publ.html. If you want a global inequality with a uniform constant, then you have to impose (non-negative) lower bounded on the Ricci curvature (in metric measure spaces as well). In the smooth compact Riemannian setting, the constant is of course uniform.<|endoftext|> TITLE: Transcendence degree of the surreals over the subfield generated by the ordinals QUESTION [8 upvotes]: Consider the Grothendieck ring $K[\Omega]$ of the semiring $\Omega$ of all ordinals under the operations of natural sum and product. Its quotient field $K(\Omega)$ is naturally a subfield of the ordered field No of surreal numbers. Is the field No algebraic over the field $K(\Omega)$ (and therefore equal to the real closure of $K(\Omega)$)? If not, what is the transcendence degree of No over $K(\Omega)$? EDIT: I think on second thought that the real closure of $K(\Omega)$ intersected with $\mathbb{R}$ in No is possibly just the real closure of $\mathbb{Q}$, rather than all of $\mathbb{R}$. Perhaps then I need to consider the real closure of the compositum in No of $K(\Omega)$ and $\mathbb{R}$ and ask if No is algebraic over that. If that's not true, then is there some way of bootstrapping such a construction to some sort of algebraic construction of No from $K(\Omega)$? Related mathoverflow post: Will Sawin's answer to another mathoverflow question shows that No is a proper extension of $K(\Omega)$. See Are Conway's omnific integers the Grothendieck group of the ordinals under commutative addition? It is perhaps relevant to note that $K[\Omega]$ can be identified with the polynomial ring $\mathbb{Z}[\omega^{\omega^\alpha}: \alpha \in \Omega]$ generated by the (algebraically independent) delta numbers $\omega^{\omega^\alpha}$ for $\alpha \in \Omega$, and the field $K(\Omega)$ is then identified with the field $\mathbb{Q}(\omega^{\omega^\alpha}: \alpha \in \Omega)$ of rational functions in the delta numbers. ADDITION TO ORIGINAL QUESTION: Since in NBG the class $\Omega$, and therefore also the class $\{\omega^{\omega^\alpha}: \alpha \in \Omega\}$, maps onto every class, it follows that the ring $K[\Omega]$ maps homomorphically onto every commutative ring, even those with an underlying proper class. In particular, there is a surjective ring homomorphism $K[\Omega] \longrightarrow \operatorname{No}$, or equivalently a maximal ideal $M$ in $K[\Omega]$ such that $K[\Omega]/M$ is isomorphic to $\operatorname{No}$. Is there a nice way to define such a homomorphism and/or maximal ideal? YET ANOTHER ADDITION: Is $K[\Omega]$ isomorphic to a quotient ring of the ring Oz of omnific integers? REPLY [7 votes]: The transcendence degree of the surreals over the ordinals is proper class sized. There are many ways of seeing this; here's one. Recall that there is an order-preserving map $x\mapsto \omega^x$ from the surreals to the positive surreals which extends the usual ordinal exponentiation and satisfies $\omega^{x+y}=\omega^x \omega^y$, and the numbers of the form $\omega^x$ are linearly independent over $\mathbb{Q}$ (or $\mathbb{R}$). Now consider the elements $\omega^{\omega^{-x}}$ for all positive surreals $x$. It is easy to see that these are algebraically independent over the ordinals.<|endoftext|> TITLE: $Spin(7)$ as stabilizer of a $4$-form QUESTION [8 upvotes]: According to Bryant's work on special holonomy groups, $G_2\subset SO(7)$ may be defined as the group preserving the following 3-form: $\phi_0=\mathrm{d}x_{123}+\mathrm{d}x_{145}+\mathrm{d}x_{167}+\mathrm{d}x_{246}-\mathrm{d}x_{257}-\mathrm{d}x_{347}-\mathrm{d}x_{356}$ where $\mathrm{d}x_{ijk}=\mathrm{d}x_i\wedge\mathrm{d}x_j\wedge\mathrm{d}x_k$. In a similar fashion, $Spin(7)\subset SO(8)$ may be defined as the group preserving the following 4-form: $\Omega_0=\mathrm{d}x_{1234}+\mathrm{d}x_{1256}+\mathrm{d}x_{1278}+\mathrm{d}x_{1357}-\mathrm{d}x_{1368}-\mathrm{d}x_{1458}-\mathrm{d}x_{1467}-\mathrm{d}x_{2358}-\mathrm{d}x_{2367}-\mathrm{d}x_{2457}+\mathrm{d}x_{2468}+\mathrm{d}x_{3456}+\mathrm{d}x_{3478}+\mathrm{d}x_{5678}$ But of course, with this definition, it is not obvious at all that $Spin(7)$ is really a double cover of $SO(7)$. Is there any way to understand this fact? (that is, which is the point of defining $\Omega_0$ like this?) In the book On Quaternions and Octonions there is a definition of $Spin(7)\subset SO(8)$ related to 'octonions automorphisms' (they are not really automorphisms) which reveals some of this double-cover nature. Perhaps these could be a bridge linking these two sides, but at this moment it is not clear for me. Any suggestion will be welcomed. REPLY [11 votes]: There are at least two sources for this: First, Harvey and Lawson, Calibrated geometries (Acta Math. 1982) proves this (i.e., that the stabilizer of $\Omega_0$ is isomorphic to the nontrivial double cover of $\mathrm{SO}(7)$) using properties of the octonions. Second you can find a proof that doesn't rely on knowledge of the octonions but instead relies on the classification of Lie algebras and groups in Bryant's Metrics with exceptional holonomy (Ann. of Math, 1987, Theorem 4 of Section 2). There are even more 'direct' (i.e., elementary) proofs available, of course. If those don't satisfy you, then ask again and I'll sketch one here when I have time.<|endoftext|> TITLE: Recent ideas in Macaulayfication? QUESTION [7 upvotes]: Kawasaki has shown that a quasi-projective variety over a field has a Macaulayfication. His construction does not preserve the Cohen-Macaulay locus of the original variety, only a finite set of points in the Cohen-Macaulay locus. Have there been any recent improvements of Kawasaki's result that give Macaulayfications that preserve the entire Cohen-Macaulay locus of the original variety? REPLY [2 votes]: Regarding Gorensteinification, it is possible to start with a demi-normal (ie, G_1, S_2, and seminormal) quasi-projective variety U and find a demi-normal projective closure. This is not really a Gorensteinification, but something close to normalization without completely normalizing a given projective closure of U. Also, for Q-Gorenstein varieties at least, the cyclic covering lemma with Cohen-Macaulayfication would probably do the trick.<|endoftext|> TITLE: Base change in homotopy type theory QUESTION [8 upvotes]: Recall that with the internal language of 1-toposes, we have the nice, basic, and useful result that geometric sequents are stable under base change along geometric morphisms: If $\varphi$ and $\psi$ are geometric formulas (formulas not containing $\forall$ and $\Rightarrow$), and if $\mathcal{E} \models \forall x:X. (\varphi(x) \Rightarrow \psi(x))$, then also $\mathcal{F} \models \forall x:f^*X. (f^*\varphi(x) \Rightarrow f^*\psi(x))$, where $f : \mathcal{F} \to \mathcal{E}$ is a geometric morphism. This is useful for lots of things, for instance for comparing validity on a space with validity at all points and for using classical reasoning in proving geometric sequents. I have two naive questions relating the $(\infty,1)$-categorical generalization of this. Question 1. Is there a similar preservation statement for the internal language of $(\infty,1)$-toposes, at least in those cases where it's known that homotopy type theory serves as the internal language? (Note that, in light of propositions as types, one should probably formulate the preservation statement in slightly different terms.) Question 2. If yes, is it useful? Or are, with the prevalent use Pi types and universes (which are probably not preserved by the inverse image parts of arbitrary geometric morphisms), few formulas of the required form? REPLY [2 votes]: The answer to question 1 is yes. To see this, I think it's better to consider a "geometric sequent" to be of the form $$ (x:X), \phi(x) \vdash \psi(x) $$ since this avoids all mention of $\forall$ and $\Rightarrow$. Now we can see that the point is just that the "geometric" logical operations $\exists$, $\wedge$, $\vee$ (and more generally its infinitary version $\bigvee$), $\top$, and $\bot$ all correspond to categorical operations that are preserved by inverse image functors — finite limits and arbitrary colimits — while the meta-logical operation $\vdash$ just means the existence of a morphism, which is certainly preserved by any functor. The same argument works essentially unaltered in the homotopy/∞ case, if we replace "formulas" by what HoTT calls hprops or "mere propositions", and the logical operations by their $(-1)$-truncated versions, since these are all interpreted by $(\infty,1)$-categorical constructions built out of finite limits and arbitrary colimits. (Although a version of HoTT allowing infinitary operations such as $\bigvee$ is not yet well-studied.) As for question 2, it's true that in HoTT we usually make use of higher-order operations. This is not so different from doing mathematics in a 1-topos, although working with non-0-truncated types does mean that some things that used to be formulable using "mere relations" no longer are. So geometric logic, and its preservation by geometric morphisms, is perhaps a bit less useful than in 1-topos theory. On the other hand, one can also consider a larger sort of "geometric type theory" involving un-truncated operations such as pushouts that are nevertheless still preserved by inverse image functors. One particular application of geometric logic for 1-toposes is of course the theory of classifying topoi. The theory of classifying $(\infty,1)$-topoi is not yet as well-developed, but indications are promising. Closely related material has been developed, in different language, by Jacob Lurie under the name of geometry (see also the nlab page). Recently Andre Joyal and some collaborators have been working on showing that the $(\infty,1)$-topos of parametrized spectra is a classifying topos for "stable objects"; the latter notion involves loop-spaces of suspensions, and thus belongs to "geometric type theory" but not the more restrictive (-1)-truncated "geometric logic".<|endoftext|> TITLE: Formula for the Perimeter of a spherical triangle? QUESTION [10 upvotes]: Consider the ordinary sphere $\mathbb{S}^2\subset \mathbb{R}^3$ and a spherical triangle $T\subset \mathbb{S}^2.$ I'm looking for a formula from which the perimeter $P$ of $T$ is "computable" given the three interior angles $\gamma_1,\gamma_2,\gamma_3$ (and the area $\vert T \vert$) of the triangles, i.e. i'm searching for a formula $$P=f(\gamma_1,\gamma_2, \gamma_3),$$ where $f$ is an explicitly given function (which is hopefully not too complicated). Are there such formulas? I couldn't find any utilizable connections. For instance there is a nice formula for the planar case given by the equation: $\frac{P^2}{4\vert T \vert}= \sum_{i=1}^3 \cot \left( \frac{\gamma_i}{2}\right)$, where $P$ denotes the perimeter of the planar triangle. I hope some experts can help me. Best regards REPLY [15 votes]: First, you don't need to know the area separately, since that is given by the classic formula $$ |T| = (\gamma_1+\gamma_2+\gamma_3) - \pi. $$ Second, if $\ell_i$ is the length of the side opposite $\gamma_i$, then the standard spherical trig formula called the polar law of cosines gives $$ \ell_i = \cos^{-1}\left(\frac{\cos\gamma_i + \cos\gamma_j\cos\gamma_k}{ \sin\gamma_j\sin\gamma_k}\right) $$ where $(i,j,k)$ is a permutation of $(1,2,3)$. Now, $$P = \ell_1+\ell_2+\ell_3$$ is such a formula.<|endoftext|> TITLE: Variance of the roots of a complex polynomial QUESTION [11 upvotes]: Let $P\in\mathbb{C}[X]$ be a complex polynomial of degree $n\geq 2$ with complex roots $\alpha_1, \alpha_2,\ldots, \alpha_n$. My question is about the existence of a formula for the variance of the roots of $P$ in terms of the coefficients of $P$. First, let me fix some notations. I will denote by $m=m(P)$ the arithmetic mean of the roots $$m=\frac{\alpha_1+\alpha_2+\cdots+\alpha_n}{n},$$ and by $v=v(P)$, the variance $$v=\frac{1}{n}\sum_{i=1}^n\vert \alpha_i-m\vert^2,$$ that is, the arithmetic mean of the squared distance to the mean, which is a quantity of great interest in probability theory. Note that $v=0$ if and only if $P$ has one root of multiplicity $n$. Le us write $$P(X)=a_nX^n+a_{n-1}X^{n-1}+\cdots+a_0=a_n\prod_{i=1}^n(X-\alpha_i).$$ Clearly, the arithmetic mean can be expressed in terms of the coefficients as $$m=-\frac{a_{n-1}}{na_n}.$$ Consider now the problem of expressing the variance $v$ in terms of the coefficients by a formula involving only arithmetic operations, radicals and conjugation (in what follows, a formula means such a formula). By a standard computation, the variance can be written as $$v=(n-1)\vert m\vert^2-\frac{2}{n}Re(\sum_{i2$, the diagonal subgroup contains no normal subgroup, and so the normal closure of $M$ over $L$ is $K$. For $n>4$, this is not solvable.<|endoftext|> TITLE: Banach space modulo a one-dimensional subspace =? QUESTION [12 upvotes]: My question is the following: Given an infinite dimensional Banach space $E$ and a one-dimensional linear subspace $F\subset E$. It is well-known that this one-dimensional linear subspace is closed and always complemented. So, we have the following topological isomorphism: $$E \cong F \times E/F.$$ Now, what is the isomorphism type of $E/F$ ? In particular, is $E/F$ isomorphic to $E$ ? I know this is true for all Hilbert spaces since they admit orthonormal bases and I would expect that the existence of some sort of "topological basis" could be related to this question but so far I could not find anything. Furthermore, I did not know where to look for an answer to this question. By the way: Of course, I believe in the axiom of choice. EDIT: All isomorphisms are meant to be topological isomorphisms, i.e., linear homeomorphisms. Isometric isomorphism are too restrictive, I would assume (although in the Hilbert space case, one can even get isometric isomorphisms, but I do not need that) Thank you in advance, Tom REPLY [6 votes]: If you consider the spaces constructed by Gowers, Maurey and others exotic, then it turns out that the answer is also negative in the class of $C(K)$-spaces. Indeed, Koszmider constructed a compact, Hausdorff space $K$ so that $C(K)$ is not isomorphic to its hyperplanes: P. Koszmider, Banach spaces of continuous functions with few operators. Math. Ann. 330, No. 1 (2004), 151–183 . Of course there is a price you have to pay for this–$C(K)$ is inseparable (i.e. $K$ is non-metrisable) and it is easy to see that a separable $C(K)$-space will never have this property as it contains a complemented copy of $c_0$ which apparently has this property.<|endoftext|> TITLE: History of integral notation for coends QUESTION [16 upvotes]: I'm searching the wheres and whys about the integral notation for co/ends. Who was the first to adopt it? Can you give me a precise pointer or tell me the whole story about it? Was s/he motivated by the Fubini rule only, or there is another additional reason? Thanks a lot! REPLY [19 votes]: From Ross Street's An Australian conspectus of higher categories: Kelly developed the theory of enriched categories describing enriched adjunction and introducing the variety of limit he called end. I later pointed out that Yoneda had used this concept in the special case of additive categories using an integral notation which Brian Day and Max Kelly adopted. Following this, Mac Lane discussed ends for ordinary categories. The Day-Kelly paper referred to is B. Day and G.M. Kelly, Enriched functor categories, Reports of the Midwest Category Seminar, III, Springer, Berlin, 1969, pp. 178–191. whose introduction states What we call ends and coends were introduced in the case $V = \text{Ab}$ by Yoneda; we borrow from him the "integral notation". The Yoneda paper referred to is Yoneda, N., On Ext and exact sequences. Jour. Fac. Sci. Univ. Tokyo 8 (1960), 507 - 576. where we find written on page 546 (thanks to Francesco for the page reference) We shall write $\int_C H$ for the integration $I$ of $H$, and $\int_C^\ast H$ for the cointegration $J$ of $H$. where his "integration" is our "coend", and dually. REPLY [8 votes]: According to the book Galois Theory, Hopf Algebras, and Semiabelian Categories by George Janelidze, Bodo Pareigis and Walter Tholen, the integral notation for (co)ends, generally attributed to MacLane, is originally due to Yoneda. See in particular the interesting historical footnote 1, page 201. The precise reference given is page 546 of Yoneda, Nobuo: On Ext and exact sequences, J. Fac. Sci. Univ. Tokyo Sect. I 8 (1960), 507–576.<|endoftext|> TITLE: Quaternion Wishart matrices of half-integer dimension? QUESTION [6 upvotes]: For a physics application (quantum delay times of a chaotic scatterer) I need to generate $m$ positive random variables $\lambda_1,\lambda_2,\ldots\lambda_m$ with probability distribution $$P_\beta(\lambda_1,\lambda_2,\ldots\lambda_m)\propto \prod_{k=1}^{m}\lambda_k^{\beta m/2}e^{-\beta\lambda_k/2}\prod_{1\leq i \beta(m-1)/2$ and arbitrary $\beta> 0$ by using tri-diagonal matrices (generalizing Wishart distributions) was done by Dumitiru and Edelman: http://www.math.washington.edu/~dumitriu/JMathPhys_43_5830.pdf<|endoftext|> TITLE: A generalized mean-value theorem QUESTION [8 upvotes]: I'm pretty sure that if the function $f$ is continuous on $[x_1,x_3]$ and twice-differentiable on $(x_1,x_3)$, with $x_1 < x_2 < x_3$, then there must exist $x$ in $(x_1,x_3)$ for which $f''(x)$ equals $$\dfrac{((x_2-x_1)(f(x_3)-f(x_1)) - (f(x_2)-f(x_1))(x_3-x_1))}{ ((x_2-x_1)(x_3-x_2)(x_3-x_1))}.$$ (This complicated expression gives the constant second-derivative of the unique polynomial of degree 2 passing through the points $(x_i,f(x_i))$ for $1 \leq i \leq 3$.) A simple proof would be nice, but even nicer would be a pointer to the literature. Surely this result is known, and indeed is just a special case of a known result for curves passing through $n$ prescribed points. REPLY [3 votes]: The stated result is a special case of the Schwarz mean value theorem, which plays a crucial role in Dörge's proof of the Hilbert irreducibility theorem. References I know of are the books Diophantine Geometry by Lang (p. 148), Selected Topics on Polynomials by Schinzel (p. 174), and Generic Polynomials by Jensen, Ledet and Yui (p. 69).<|endoftext|> TITLE: rational points of a hyperelliptic curve QUESTION [12 upvotes]: I have the following hyperelliptic curve of genus $2$: $$ y^2 = 561 x^6 - 41904 x^5 + 627264 x^4 + 11860992 x^3 - 197074944 x^2 + 124416^2 $$ I need to find all the rational points on this curve. There is one obvious point at $(0, 124416)$. I have some experience in finding rational points on elliptic curves. I also have the reference Handbook of Elliptic and Hyperelliptic Curve Cryptography (Discrete Mathematics and Its Applications). Most of this work uses genus $2$ curves reduced to quintic form. So a related question is how to transform this sextic to the quintic form. Thank you Lorenz Menke A question as to where this problem originates. The nuclear electric quadrupole Hamiltonian characteristic equation for spin $9/2$ is $$ E_p^5 - 99 (3 + z)E_p^3 - 1188(1 - z)E_p^2 + 1188(3 + z)^2 E_p + 11664(3 + z)(1 - z) $$ where the energy states are $E = 2 E_p$ and $z = \eta^2$ where $\eta$ is the asymmetry parameter. Now solve this equation for $z$. The above sextic is the square root term. Thus when the sextic is a perfect square for a rational energy state the resultant asymmetry parameter squared is also a rational. This set of rational points of this sextic are the values that result in the quadrupole quintic Galois factoring into a linear times a quartic. The only other values of $z$ that reduce this quintic are $z = 0, 1, -3,$ and $9$. The goal is to find all the points $z$ such that the quintic factors over the rationals. REPLY [18 votes]: By now there is a fairly rich literature on computing the set of rational points on curves of higher genus, see for example my survey paper on "Rational points on curves". What one can do for your concrete curve is this. First, one can simplify the curve equation (even more than Noam did); your curve is isomorphic to $$y^2 = x^6 - 12 x^5 - 6 x^4 + 82 x^3 + 1461 x^2 - 630 x + 225.$$ Then most methods aiming at finding the set of rational points use the group of rational points on the Jacobian variety of the curve (which by the Mordell-Weil theorem is a finitely generated abelian group). In your case, this group turns out to be free abelian of rank 2. Which is bad news, since the method of choice ("Chabauty's method") only works when this rank is strictly less than the genus. Since (as Noam already mentioned) the defining polynomial has large (even maximal) Galois group, the next method one could try ("Elliptic curve Chabauty") will be infeasible. So despite all the progress that was made in the last couple of decades or so, a curve like yours (with rank $\ge$ genus and no special geometric or arithmetic properties [the smoothness of the discriminant does not help here]) is still likely to resist our efforts. So what can be done regarding your question? Using explicit generators of the Mordell-Weil group (which takes some computation to provably find), one can fairly efficiently show that there are no rational points beyond the 18 that Noam found up to any given (even quite large) height bound, for example by using the "Mordell-Weil sieve". And if you are interested in integral points, then one can combine the Mordell-Weil sieve with bounds coming from Baker's method (linear forms in logarithms) to prove that you know all integral points. If desired, I can add details on any or all of the methods mentioned above.<|endoftext|> TITLE: Is below every cohesive set a 1-generic? QUESTION [5 upvotes]: A set $X$ is called cohesive for $(R_i)_{i\in \mathbb{N}}$ if it is infinite and for each $i$ we have $X\subseteq^* R_i$ or $X\subseteq^* \overline{R_i}$. (Where $X\subseteq^*Y$ means that $X$ is contained in $Y$ up to finitely many exceptions.) A set $X$ is called cohesive if it is cohesive for all computable enumerable sets. Further, a set $X$ is called 1-generic if for each computably enumerable set $S\subseteq 2^{<\mathbb{N}}$ of strings there is an initial segment of $\sigma$ of $X$ such that either $\sigma \in S$ or $\sigma \not\subseteq \tau$ for all $\tau \in \sigma$. Both notions are well studies in computability theory. However, I was not able to find a answer to the following question in the literature: Is (Turing) below each cohesive set a 1-generic set? To give some background on the question: The reverse of the above question is known to be false. That is there is a 1-generic which has no cohesive set below it. For the proof one has to note that there are low 1-generic sets but no low cohesive sets. (See Jockush, Stephan: A cohesive set which is not high). Moreover it is known that below each cohesive set is a weakly 1-generic. (A set $X$ is weakly 1-generic if for each dense c.e. set $S$ there is an inital segment $\sigma$ of $X$ with $\sigma\in S$.) REPLY [6 votes]: No -- each high degree contains a cohesive set. And Cooper showed there is a high minimal degree. But no 1-generic has minimal degree.<|endoftext|> TITLE: Adjoining torsion points from abelian varieties QUESTION [8 upvotes]: Let $L/\mathbb{Q}$ be the field generated over $\mathbb{Q}$ by all of the (projective) coordinates of all of the torsion points of all abelian varieties defined over $\mathbb{Q}$. Is $L$ algebraically closed? REPLY [13 votes]: Another proof that $L = \,\overline{\bf \!Q\!}\,$: Clearly $L$ is contained in $\,\overline{\bf \!Q\!}\,$, so we need only show $L$ contains every algebraic number $x \notin \bf Q$. Let $P(X)$ be the minimal polynomial of $x$. If $\deg P$ is odd, then the class of $((x,0)) - (\infty)$ is a $2$-torsion point on the Jacobian of the elliptic or hyperelliptic curve $y^2 = P(x)$. If $\deg P$ is even, then the class of $((x,0)) - (\infty)$ is a $2$-torsion point on the Jacobian of the elliptic or hyperelliptic curve $y^2 = x P(x)$. QED<|endoftext|> TITLE: metric on ${\bf SPD}_n({\mathbb R})$ QUESTION [7 upvotes]: The cone ${\bf SPD}_n({\mathbb R})$ of symmetric positive definite matrices is endowed with a nice geometrical structure. The midpoint of the (unique) geodesic between $A$ and $B$ is the so-called geometric mean $$A\sharp B=A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}.$$ What is the associated distance ? REPLY [7 votes]: If I'm not mistaken, this midpoint comes from the structure of Riemannian symmetric space on the space of symmetric positive definite matrices. The Riemannian metric is defined by $g(S)(h_1,h_2)=trace(S^{-1}h_2S^{-1}h_2)$, where $S$ is in $SDP_n$ and $h_1,h_2$ are symmetric matrices (tangent vectors at $S$). This metric has a lot of nice properties: $Gl(n)$ acts transitively and isometrically on $SDP_n$ by $g\cdot S=gSg^{T}$, making it a homogeneous Riemannian manifold. (In particular it is complete.) $S\mapsto S^{-1}$ is an isometric involution whose differential at $Id\in SDP_n$ is minus the identity. This makes $SDP_n$ a Riemannian symmetric space of non compact type (in particular it has nonpositive curvature). In the classification of Riemannian symmetric spaces it is often described as the quotient $GL(n)/O(n)$. This metric also arises as the Fisher information metric on the space of centered Gaussian random vectors on $\mathbb{R}^n$ (which are characterized by their covariance matrix). This makes it useful in some applications, as a mean of treating covariance matrices in a basis independent manner. All these properties makes it a very nice object, and it turns out that most of the objects of Riemannian geometry (connection, geodesics, curvature) can be described explicitly in terms of matrix computations. The distance is given by: $d(A,B)=\|\log(AB^{-1})\|_F$ , where $\|\ \|_F$ is the Frobenius norm. I found the formula in this article: http://arxiv.org/pdf/0807.4462v2.pdf.<|endoftext|> TITLE: Starshapeness of polynomial tracts with respect to the (entire collection of) critical points contained in the tract QUESTION [5 upvotes]: I recently found out (Piranian, "The Shape of Level Curves") that a polynomial tract (ie a connected component of a set of the form $G=\{z:|p(z)|<\epsilon\}$ for some $\epsilon>0$) need not be starshaped with respect to the zeros of $p$ contained in $G$. This disappointed me bitterly, as that starshapeness was a pivotal step in a proposed "proof" I had of Smale's mean value conjecture. The places where $G$ is not starshaped with respect to the zeros of $p$ in $G$ are near critical points of $p$ in $G$ or in $\partial G$, so I still hold out a tiny bit of hope for the starshapeness of $G$ with respect to the critical points of $p$ contained in $G$: Conjecture: If $G$ is a tract of $p$ with smooth boundary containing more than one distinct zero of $p$, then $G$ is starshaped with respect to the critical points of $p$ contained in $G$. Intuitions/proofs/disproofs/references are all welcome. EDIT: Note that when I say that $G$ should be "starshaped with respect to the critical points", I mean that each point in $G$ can be seen by some one of the critical points of $p$ in $G$, not of course that some single critical point can see all points in $G$. Note also that I added the assumption that $G$ contains more than one distinct zero of $p$ (since otherwise $G$ will not contain any critical points of $p$. One reason I think this is plausible: If we consider the lemniscate of Bernoulli, and let $G$ be the interior of a level curve of $p$ which is a bit bigger, the critical point of $p$ is right in the center, so should be able to "see" both lobes. In the counter-example of Piranian to my desired conjecture (that tracts are star-shaped with respect to the zeros they contain), the points that killed the starshapeness were close to the boundary of $G$, so perhaps if we assume $\partial G$ is smooth, $G$ will contain enough critical points to see into all "corners". REPLY [5 votes]: There is no hope: according to a theorem of Hilbert, every analytic Jordan curve $J$ can be approximated by a lemniscate $\{z:|P(z)|=\epsilon\}$. So the set does not have to be starlike with respect to any point. For this theorem of Hilbert, see, for example J. L. Walsh, ``Interpolation and Approximation by Rational Functions in the Complex Plane,'' 5th ed., Amer. Math. Society, Providence, RI, 1969.<|endoftext|> TITLE: Fixed point of fatness QUESTION [8 upvotes]: For the purposes of this question, define the following properties of convex sets in the plane: A set is $R$-fat (for $R\geq 1$) if it contains a disc of side-length $x$ and is contained in a disc of side-length $R\cdot x$, for some positive $x$. A set is $R$-cuttable if it can be cut (using a straight line) to two non-empty convex $R$-fat shapes. For example: A disc is 1-fat. It is not 1-cuttable since it cannot be cut to two discs. But it is 2-cuttable since it can be cut to two semi-discs, and a semi-disc is 2-fat. A $1\times 10$ rectangle is $\sqrt{101}$-fat. It is $\sqrt {26}$-cuttable since it can be cut to two $1\times 5$ rectangles. Hence it is also $\sqrt{101}$-cuttable, MY QUESTION IS: what is the smallest value $R_0$ such that every convex $R_0$-fat shape is $R_0$-cuttable? The first example shows that $R_0 \geq 2$. NOTE: I asked a similar question in Math.SE, currently with no reply. REPLY [3 votes]: Still, Pietro Majer is right --- there is no such $R_0$. Let $S$ be the convex hull of the unit circle centered at the origin and the point $A=(2R-1,0)$. Then it is $R$-fat, with the unit disk inside it and the smallest enclosing disk with radius $R$ and center $(R-1,0)$. Now assume that $S$ is cut into two convex sets $T$ and $T'$, with $T$ containing $A$. Let $O$ and $r$ be the center and the radius of a largest disk $D$ contained in $T$ (we have $r<1$, otherwise $T=S$). Then the abscissa of $O$ is at most $(2R-1)(1-r)$. Now, $T$ contains the point $B$ on the ray $AO$ with $AB=AO+r\geq (2R-1)r+r=2Rr$. Notice that $B$ lies strictly inside $S$; since the cut is straight and $B$ lies on the boundary of $D$, $T$ also contains some other point $C$ on the line perpendicular to $AB$ and passing through $B$. Then $AC>AB=2Rr$, so the smallest disr enclosing $T$ has the radius greater than $Rr$, and $T$ is not $R$-fat. REMARK, edited. Notice that in this example, one may still cut $S$ into two $R$-fat non-convex but simply connected sets by a non-straight cut (namely, by a circular arc). As Erel mentions in the comment, this is always possible for every $R>1$.<|endoftext|> TITLE: Dehn algorithm and normal forms in small cancellation groups QUESTION [8 upvotes]: I found this statement in B. Cavallo, D. Kahrobaei's paper arXiv:1311.7117 Secret Sharing using Non-Commutative Groups and the Shortlex Order, page 7. "C′(1/6) continue to be an ideal platform for this protocol because reducing words to a minimal length has polynomial time complexity using a deterministic version of Dehn’s algorithm. In fact, it is the same algorithm used to solve the word problem. " Could someone provide a reference that Dehn algorithm reduces words to normal forms in C'(1/6) groups in polynomial time? If this is not true, then in which classes of groups except virtually free gps is this true? Thanks, Olga REPLY [5 votes]: The proof is not difficult, and the argument can be found in older publications if you hunt around. I think it is essentially proved in a paper by Otto and Madlener. The point is that, if there is a Dehn algorithm that reduces a word to a geodesic, then you can cary out the reduction on a single stack. So the word problem of the group as a formal language is context-free, and it follows by the Muller-Schupp Theorem that the group is context-free. (But of course the proof of that uses the deep result of Dunwoody about the accessibility of finitely presented groups.) To see that you can do the reduction on a single stack, observe that if you have a geodesic reduction of the word read so far on the stack, then the next generator you read can reduce the geodesic length by at most one, and so you will have to do at most two length reductions with the Dehn algorithm to reduce the new word to a geodesic. Since that requires only a bounded amount of space, you can carry it out on the stack.<|endoftext|> TITLE: Injection of the proper class of ordinals in every proper class QUESTION [5 upvotes]: Is it possible to prove in the set theory NBG (with local choice but without global choice) that the proper class of ordinals injects in every proper class ? REPLY [9 votes]: The answer is no. That principle is equivalent to global choice. To see this, consider the class $W$ consisting of all well-orderings of any rank-initial segment $V_\alpha$, for any $\alpha$. If we had an injection of Ord into $W$, then there must be unboundedly many $\alpha$'s that are used, since each $V_\alpha$ has only a set-sized family of well-orderings. Thus, we have a global selection of well-orderings of unboundedly many $V_\alpha$, and from this we can define a well-ordering of the entire universe. Namely, $x TITLE: Geometric interpretation of the average of two independent Cauchy distributions QUESTION [9 upvotes]: Let me state two facts: (1) It is well known that if one takes a point uniformly distributed on the unit circle, and then takes it stereographic projection, the corresponding measure induced on the real line is the standard cauchy distribution. It is also easy to see that if you take the average of two independent cauchy distributed random variables, it is also cauchy distributed. My question is the following: Taking cue from (1) above, can we have a geometric proof of (2)? REPLY [3 votes]: Maybe something like this will work. Consider $U_1$ and $U_2$ drawn uniformly at random on the unit circle. Because they are uniformly distributed, we may rotate the circle until $U_1$ is at the ``north pole" without altering their distributions. Now, stereographic projection takes $U_1$ to $Z_1 = 0$ and $U_2$ to some point $Z_2$. Then $\frac{1}{2}(Z_1 + Z_2) = \frac{Z_2}{2}$, which is Cauchy distributed with scale $\frac{1}{2}$ by your point (1). However, noting that we had two directions we could have rotated -- corresponding to two distinct projections of $U_2$ -- we recover our factor of 2, giving the result (2).<|endoftext|> TITLE: Homogeneous spaces that are homotopy tori QUESTION [13 upvotes]: Let $G$ be a compact Lie group, and let $H$ be a closed subgroup such that $G/H$ is homotopy equivalent to a torus. Is it true that $H$ is normal and $G/H$ is isomorphic to a torus as a Lie group? One line of attack would be to reduce to the case where $G$ is connected, then to the case where $G$ is the product of a torus with a simply connected group, then try to use the classification of simply connected groups. A proof like that would be OK, but I would actually prefer a more uniform proof that just works directly with $G$. REPLY [7 votes]: Reduce to the case when $G$ is connected. Because $H$ is compact, $\pi_0(H)$ is finite, so by the homotopy long exact sequence $\pi_1(G)\to \pi_1(G/H) = \mathbb Z^n$ maps onto a finite index subgroup, hence $H^1(G/H, \mathbb Z) \to H^1(G, \mathbb Z)$ is surjective. Fix an invariant metric on $G$ and, applying Hodge theory, represent a basis of $H^1(G/H, \mathbb R)$ by harmonic $1$-forms. The harmonic $1$-forms are unique, so since the cohomology classes are $G$-invariant, the $1$-forms are $G$-invariant. Hence they give a homomorphism from the Lie algebra of $G$ to the abelian Lie algebra of rank $n$. This gives a homomorphism from the universal cover of $G$ to $\mathbb R^n$ and hence a, after modding out by the lattice corresponding to the homology of $G/H$, a homomorphism from $G$ to a torus. I really wish we were done at this point, but I don't see a way to make the next part easy: The image of $H$ in this torus is a closed subgroup, hence a finite union of tori. It cannot contain a nontrivial torus or else there would be a nontrivial map from $\pi_1$ of $H$ to $\pi_1$ of that nontrivial torus to $\pi_1$ of $G/H$. So it is a finite union of points. But the map from $\pi_0$ of $H$ to the torus factors through $H_1(G/H)$, hence factors through the zero map. So the image of $H$ is trivial. If $H'$ is the kernel of the map to this torus, then $G/H \to G/H'$ is a homotopy equivalence. Its also a fibration, so the fiber is contractible. The fiber is a compact manifold, so it's a point. Hence $H=H'$ and $G/H=G/H'$ is a torus.<|endoftext|> TITLE: Hausdorff spaces with trivial automorphism group QUESTION [14 upvotes]: Is the singleton space the only Hausdorff space $X$ such that the set of automorphisms $\varphi: X\to X$ equals $\{\textrm{id}_X\}$? REPLY [8 votes]: It's easy (if a little tedious, just a little) to construct a non-trivial metric dendroid (like an infinite tree or precisely--a one-dimensional compact metric AR) such that the ramification degree of each point is finite; for each natural number $\ n=1\ 2\ \ldots\ $ there exists exactly one point which has ramification $\ n+2$; the set of all points of ramification $\ > 2\ $ is dense. Then for every homeomorphism of our dendroid onto itself each point which has ramification $\ > 2\ $ is fixed. It follows that every such homeomorphism is the identity of our dendroid. A construction is obtained by starting with an interval--that's your initial stage of the induction, then at each stage add several short intervals which have one end at the middle of each existing intervals; at the center of $n$-th interval you add $\ n\ $ interval so that ramification will be the unique $\ n+2.\ $ You iterate this at infinitum. You folow it up with ending the end-point at each combinatorially infinite branch (but their geodesic metric length is finite), thus we get a compact. REMARK   An interval of a tree is meant any maximal interval such that all its internal points have ramification $\ 2.\ $ Each interval of a tree at any next stage, which was set-theoretically contained in the previous tree, is one of the two halves of the respective larger interval of that previous tree. Regardless of the stage at which theses intervals occur, they are all numbered $\ 1\ 2\ \ldots$. With each stage there is a bunch of intervals which are consecutively indexed above the previous tree, and below the next one. (Non-triviality of a dendroid means, by definition, that it has more than one point). The above dendroid (like any non-trivial dendroid) has non-constant continuous maps into itself, different from identity. Indeed, this is true for arbitrary AR which has more than one point. Thus the given example, while rigid, is not strongly rigid.<|endoftext|> TITLE: Is there a scheme parametrizing the closed subgroups of an algebraic group? QUESTION [19 upvotes]: In the following, let $G=\operatorname{GL}_n(\mathbb{C})$ or $G=\operatorname{\mathbb PGL}_n(\mathbb{C})$, depending on whichever has a better chance of yielding an affirmative answer. One could more generally ask the question for a complex, (reductive, affine,) algebraic group - but since this generalization is not of importance to me, I leave it as a bonus objective. Question: I am asking whether there is a scheme that parametrizes the closed, algebraic subgroups of $G$ in the same way that the Hilbert scheme classifies all closed subvarieties. Now, I am quite certain that $G$ has a Hilbert scheme $\operatorname{\mathfrak Hilb}_G$. Then, I might refine the question by asking whether the closed subgroups of $G$ form a (possibly closed) subscheme of $\operatorname{\mathfrak Hilb}_G$. Finally, of course I would like to know as much as possible about how that scheme, if it exists, looks like (as in, geometric properties of any kind are welcome). My thoughts: Denoting by $\mu:G\times G\to G$ the group operation and by $\iota:G\to G$ inversion, it seems plausible to me that the $H\in\operatorname{\mathfrak Hilb}_G$ which satisfy $\mu(H\times H)=H$ and $\iota(H)=H$ should form a closed subscheme, which would then be exactly what I am looking for. However, I also feel that I might only have a naive and vague idea of what is going on. REPLY [2 votes]: If we work with algebraic varieties instead of schemes, then there is a positive answer: I described an ind-variety parametrising families of connected algebraic subgroups of an algebraic group $G$, cf. “Universal family of the subgroups of an algebraic group,” (arXiv) which I would like to outline. Please note that this work is a pre-print, and while some persons have read it with enthusiasm it never was formally peer-reviewd and a lot of inaccuracies and clumsy statements are probably still to be found there. A natural idea is to replace each connected group by its Lie algebra and use the universal property of the Grassmann variety to construct our universal family. The main difficulties in this approach lie in the complex structure of the set $\mathbf A$ of $k$-dimensional algebraic subalgebras of the Lie algebra $\mathfrak g$ of $G$, that is, the set of subalgebras of $\mathfrak g$ which are the Lie algebra of an algebraic subgroup of $G$. The complexity of this set is entirely imputable to tori. For a torus, the kernel of the exponential mapping is a lattice and the closed subgroups of that torus correspond to the linear subspaces of its Lie algebra that are generated by points in this lattice. We can view them as the rational points in the complex Grassmann variety. This is also why we need ind-varieties in the construction. Differentiating families of sub groups The well-known study of the analytic structure of complex algebraic varieties by Serre (GAGA) allow us to see that the family the family of Lie algebras $\mathop{\mathcal{L}}(\mathcal{H})$ associated to a flat family $\mathcal{H}$ of subgroups of $G$ is also flat: 3.4 Corollary. – Let $\mathcal{H}$ be a family of connected $k$-dimensional algebraic subgroups of $G$ parametrised by $Q$. If $\mathcal{H}$ is flat, then there is a morphism $\psi:Q\to\Gamma$ such that $\mathop{\mathcal{L}}(\mathcal{H})$ is obtained by pulling back the tautological bundle $\mathcal{T}\to\Gamma$ through $\psi$, where $\Gamma$ is the Grassmann variety of $k$-dimensional subspaces of the Lie algebra of $G$. In this setting, the image of $Q$ through $\psi$ is a constructible subset of $\Gamma$, contained in the set $\mathbf A$ of algebraic subalgebras of $\mathfrak g$. It is remarkable that each subset can be obtained this way, that is, that we can solve the problem of Integrating families of subalgebras Theorem. – If $P$ is a constructible subset of $\Gamma$, contained in the set $\mathbf A$ of algebraic subalgebras of $\mathfrak g$ then the closure $\bar P$ of $P$ in the Grasmmann variety $\Gamma$ is also a subset of $\mathbf A$ and the subset $$ \mathcal H(\bar P) = \left\{ (g,p) \in G \times\bar P \mid g \text{ belongs to the irreducible subgroup of $G$ with Lie algebra $p$} \right\} $$ is a flat family of algebraic subgroups of $G$ parametrised by $\bar P$ and is equal to $\overline{\mathcal H(P)}$. We can then construct a universal family by gluing together all the $\mathcal H(\bar P)$ when $P$ runs through the maximal irreducible subvarieties of the Grassmann variety $\Gamma$ which are contained in the set $\mathbf A$ of algebraic subalgebras of $\mathfrak g$. The main observations is that there is actually a closed condition recognising subgroups $H$ of an algebraic group: with $\theta(x,y) = xy^{-1}$ this condition can be written $\theta(H\times H) \subset H$. (Incidentally, I am quite confident I learnt this characterisation of subgroups of a group in @laurent-moret-bailly's lectures in my third year at the university.) This proves that $\overline{\mathcal H(P)}$ is a set-theoretic family of subgroups of $G$. Thanks to Chevalley's structure theorem for algebraic groups and the observation that families of subgroups of an abelian variety are rigid, i.e. essentially reduced to a trivial family (see 6.3 and following), the difficult part in the Theorem above is to prove in the affine case that $\bar P$ is contained in the set $\mathbf A$ of algebraic subalgebras of $\mathfrak g$. It is enough to suppose $P$ irreducible and we can approach points $p_\infty$ in the closure of $P$ by a curve $C \subset \bar P$ meeting $P$ on a dense subset of $P$. We now make the crucial remark that the Levi-Malcev decomposition is semi-continuous and that families of semi-simple subalgebras of $\mathfrak g$ are rigid, which implies that we can use $G$ operation to assume all points in $C$ to contain the semi-simple part of $p_\infty$. We are essentially left with the case where each algebra in $P$ is solvable and the exponential restricts here to a an algebraic morphism!<|endoftext|> TITLE: Small objects vs Compact objects QUESTION [10 upvotes]: Given a cocomplete category $C$, is there an example of an object which is small but not compact? I am working with the following definitions of small and compact: Given a cardinal $\kappa$ one says that an object $X$ is $\kappa$-compact, if ${\rm Hom}(X,-)$ commutes with $\kappa$-filtered colimits. One says $X$ is $\kappa$-small if the same happens if indexing category of the colimit is an ordinal. small means $\kappa$-small for some $\kappa$ compact means $\kappa$-compact for some $\kappa$ REPLY [11 votes]: There is no difference for $\kappa = \aleph_0$. The point is that you can build colimits for filtered diagrams using just colimits for chains. Every filtered category $\mathcal{J}$ admits a cofinal directed diagram, i.e. a cofinal functor $\mathcal{I} \to \mathcal{J}$ where $\mathcal{I}$ is directed. Every countable directed poset $\mathcal{I}$ admits a cofinal $\omega$-chain: just take an enumeration of the elements of $\mathcal{I}$ and repeatedly use directedness to get a cofinal chain of length $\omega$. Every directed poset $\mathcal{I}$ of cardinality $\lambda$ is the union of a $\lambda$-chain of directed subposets of cardinality $< \lambda$. (Observe that every infinite subset $S \subseteq \mathcal{I}$ is contained in a directed subposet of $\mathcal{I}$ of the same cardinality as $S$.) Thus, by induction, every directed diagram in $\mathcal{C}$ has a colimit constructed using only colimits for chains. It is tempting to try to generalise this to regular cardinals $\kappa > \aleph_0$, but the subtlety is in (3): in general, $\kappa < \lambda$ is not enough to imply that every subset $S$ of a $\kappa$-directed poset $\mathcal{I}$ of cardinality $< \lambda$ is contained in a $\kappa$-directed subposet of $\mathcal{I}$ of cardinality $< \lambda$. (For this, we need $\kappa \triangleleft \lambda$; see Theorem 2.11 in [Locally presentable and accessible categories].) I suppose the point is that, for the purposes of the small object argument, $\kappa$-smallness suffices. But one often gets $\kappa$-compactness as well.<|endoftext|> TITLE: Compactly supported functions and Sobolev spaces on manifolds QUESTION [5 upvotes]: It is well-known that if a complete Riemannian manifold has bounded curvature and injectivity radius bounded away from zero, then the space $C^\infty_c(M)$ is dense in the Sobolev spaces $W^{k, p}(M)$ for $k=0, 1, 2$. My question: Is this an if and only if? That is, if $C^\infty_c(M)$ is dense in these Sobolev spaces, does $M$ necessarily have bounded curvature and injectivity radius bounded away from zero? REPLY [3 votes]: The general answer to this question is no. Global bound on the Ricci curvature is not necessary for the density of smooth functions with compact supports. Indeed, when $(M,g)$ is a smooth complete Riemannian manifold with positive injectivity radius and lower bound for the Ricci curvature, then the smooth functions with compact support are density in the Sobolev spaces for $p$ equals 2. This can be found for instance on Emmanuel Hebey's book: Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities.<|endoftext|> TITLE: Strongly rigid Hausdorff spaces QUESTION [13 upvotes]: A space $(X,\tau)$ is called rigid if $\textrm{Aut}(X)=\{\textrm{id}_X\}$. We say $(X,\tau)$ is strongly rigid if for every continuous map $f:X\to X$ we have that $f = \textrm{id}_X$ or $f$ is constant (that is there is $x_0\in X$ such that $f(x)=x_0$ for all $x\in X$). Is there a strongly rigid Hausdorff space with more than 1 element? REPLY [17 votes]: Yes, see for example "Continua which admit only the identity mapping onto non-degenerate subcontinua" by H. Cook (Fund. Math. 60, 1967, 241-249).<|endoftext|> TITLE: In general, are 'Young symmetrisers' given by Littlewood-Richardson 'Orthogonal projection Operators'? QUESTION [7 upvotes]: Consider $V^{\otimes n}$ where $V$ is vector space and the representation of GL(V) acting in the usual way. Now if I consider tensor products or plethysms of irreducible spaces, this is not in general irreducible. As an example Let $U \equiv (\bigwedge^3 V) \otimes (Sym^2 V) \subset V^{\otimes 5}$. The Littlewood-Richardson algorithm can applied in this case to find the irreducible spaces contained in $U$. For and arbitrary element $ X \equiv ( v_1 \wedge v_2 \wedge v_3 )\otimes (w_1 \otimes_s w_2) \in (\bigwedge^3 V) \otimes (Sym^2 V)$. The algorithm generates a standard tableau, for concreteness one of them will be, $T_{\lambda} = (v_1 w_1, v_2, v_3, w_2)$ where comma separates the rows. The young symmetriser associated with this is $c_{\lambda}$ and acts in the obvious manner on $( v_1 \wedge v_2 \wedge v_3 )\otimes (w_1 \otimes_s w_2) $. Thus span of $c_\lambda X$ will generate an irreducible subspace of $U$. Question 1) Given a decomposition of a reducible space into irreducibles. Any element can be uniquely decomposed as the sum of elements each living in a irreducible space. My question is what is the projection operator $P_{\lambda}$ associated with a standard young tableau generated by Littlewood-Richardson, for a given vector in the reducible space ? I understand that young symmetriser is an idempotent and will give one of the elements in the irreducible space, the span of which generates the whole irreducible space. However I don't believe this projects the 'correct' component, given an arbitrary element. In our running example, $c_{\lambda}X$ would be one of the elements in the irrep. However in general it is not true that $c_\lambda X$ is the orthogonal component of $X$ inside the irreducible rep characterised by the standard tableau $\lambda$ generated by Littlewood-richardson algorithm. Question 2) In general is there a procedure find out the 'orthogonal components' living in irrep characterised by the standard Young tableau ? REPLY [3 votes]: For the number of boxes less than five, the young idempotents for standard tableaux of the same shape are mutually orthogonal projectors. Once $n\ge5$ the $Y=NP$ projectors are not in general mutually orthogonal --- i.e $Y_iY_j$ is not always equal to $\delta_{ij} Y_j$. There are various methods to obtain mutually orthogonal projectors. These are described in many places --- for example the book of Littlewood on group representations. Young himself introduced a recursive construction of mutually orthogonal primative idempotents in his seminormal representation. The lecture notes of Garsia are a good source and available online: http://www.math.ucsd.edu/~garsia/somepapers/Youngseminormal.pdf<|endoftext|> TITLE: Is the classifying space of a symmetric monoidal category an infinite loop space? QUESTION [6 upvotes]: Wikipedia states: The classifying space (geometric realization of the nerve) of a symmetric monoidal category is an infinite loop space. If my mind is correct, Segals delooping machine gives a functor $Sp$ from topological symmetric monoidal categories $C$ to prespectra (In this question, a presepectrum is a sequence of based spaces $X_0,X_1$,... with structure maps $\Sigma X_i\rightarrow X_{i+1}$) with $Sp(C)_0=BC$ the classifying space of the category. If I remember correctly, one of Segals results is that the adjoints of the structure maps $Sp(C)_i\rightarrow \Omega Sp(C)_{i+1}$ are weak homotopy equivalences for $i\ge1$ and for $i=1$ this is the case iff $\pi_0(Sp(C)_0)=\pi_0(BC)$ is an abelian group with the induced multiplication which comes from the fact that $BC$ is a commutative monoid up to homotopy. So the classifying space of a topological symmetric monoidal category $C$ can in this sense infinitely delooped only if $\pi_0(BC)$ is a group. I ask myself: 1.) Did I say anything stupid or wrong? 2.) Is Wikipedia wrong? REPLY [5 votes]: You need group Completion, indeed.<|endoftext|> TITLE: Fourier-Mukai functors being identity on objects QUESTION [11 upvotes]: Let $X$ be a projective variety over $\mathbb{C}$, denote by $D^b(X)$ the bounded derived category of coherent sheaves on $X$. Suppose we have a Fourier-Mukai functor $\Phi_{X\rightarrow X}^\mathcal{P}:D^b(X)\rightarrow D^b(X)$ being an auto-equivalence on $D^b(X)$, and further assume that $\Phi_{X\rightarrow X}^\mathcal{P}$ acts identically on objects, then is it possible that $\Phi_{X\rightarrow X}^\mathcal{P}$ fails to be the identity functor? How to find a simple example to illustrate this? REPLY [9 votes]: If $X$ is smooth and projective, then any such FM functor is in fact naturally isomorphic to the identity functor. This follows immediately from Corollary 5.23 of Huybrechts' book on Fourier-Mukai tranforms. Briefly, the idea is that the hypotheses ensure that $\mathcal{P}$ is a quasi-isomorphic to a sheaf on $X\times X$ that's supported set-theoretically on the diagonal and moreover is flat over $X$ via either projection map. One then argues that $\mathcal{P}$ is of the form $\mathcal{O}_\sigma\otimes L$, where $\mathcal{O}_\sigma$ is the structure sheaf of the graph of an automorphism of $X$ and $L$ is a line bundle pulled back from $X$. When $X$ is not smooth, I'm not completely sure what happens. If $P$ is in fact a perfect complex on $X\times X$, then this reasoning still goes through. Otherwise, one needs to worry about the difference between $D^b(X)$ and $D_{perf}(X)$, the derived category of perfect complexes on $X$.<|endoftext|> TITLE: Minimal "sumset basis" in the discrete linear space $\mathbb F_2^n$ QUESTION [14 upvotes]: For a set $C\subseteq \mathbb F_2^n$, let $2C=C+C:=\{\alpha+\beta\colon \alpha,\beta\in C\}$. I want to find $C$ of the smallest possible size such that $2C=\mathbb F_2^n$. Let $m(n)$ be the size of a minimal $C$. I have found the following bounds: $$m(n) \ge B(n):=\frac{1+\sqrt{2^{n+3}-7}}{2} $$ and $$ m(n) \le A(n) := \left\{\begin{array}{ll} 2^{\frac{n+2}{2}}-2 & {\rm if}\, n\ge4\,{\rm is}\, {\rm even},\\ 2^{\frac{n+1}{2}}+2^{\frac{n-1}{2}}-2 & {\rm if}\, n\ge5\,{\rm is}\, {\rm odd}. \end{array}\right. $$ It is easy to see that $$\lim_{n\rightarrow\infty}{\frac{A(2n)}{B(2n)}=\sqrt{2}},\quad\lim_{n\rightarrow\infty}{\frac{A(2n+1)}{B(2n+1)}}=\frac{3}{2}.$$The last equations show that there might be some improvements for lower or/and upper bounds. Also I have found the following results if it can help:$$m(1)=1,m(2)=3,m(3)=5,m(4)=6,m(5)=10,m(6)\in\{12,13,14\}.$$ With great probability $m(6)=14$ (actually I am using randomized algorithm to find possible solutions and it didn't find better results for $n=6$ than 14)․ This sequence could be either https://oeis.org/A099190 or https://oeis.org/A176747. But unfortunately both are excluded, because for $n=7$ they can't match with it. Also I would like to mention that upper bounds are not the best. For example for $n=7$ computer found $C$ such that $|C|=20$. Also better results were found for $n=8,9,10$. Actually this problem is related with my thesis and I understand that I should do it myself but I have been thinking about it for about two months and I can't find any clever method to get better bounds. Any hints and suggestions would be very appreciated. Thanks! REPLY [7 votes]: This is an open problem well known in coding theory. Let $m:=|C|$, and write the vectors of your set $C$ as columns of a matrix, say $M_C$. Your condition $C+C={\mathbb F}_2^n$ translates as follows: for any non-zero vector $z\in{\mathbb F}_2^n$, there exists a vector $x\in{\mathbb F}_2^m$ of weight $|x|=2$ such that $M_Cx=z$. As a result, the (linear) code $K_C$ with the parity check matrix $M_C$ (that is, $K_C:=\ker M_C<{\mathbb F}_2^m$) has covering radius $2$: for any given $y\in{\mathbb F}_2^m$, either $y\in K_C$, or one can find $x\in{\mathbb F}_2^m$ with $|x|=2$ and $M_Cx=M_Cy$, and then $M_C(x+y)=0$, showing that $y$ differs from $x+y\in K_C$ in exactly two coordinates. It is also easy to see that the rows of $M_C$ must be linearly independent, so that $K_C$ has co-dimension $m$. This argument is reversible and it shows that the quantity you are interested in is the smallest possible length of a (linear) code with covering radius $2$ and co-dimension $m$. Denote it by $\ell(m)$. Here is a selection of bounds that can be found in Covering Codes by Cohen, Honkala, Litsyn, and Lobstein (Elsevier 1997): $\ell(2m)\le 27\cdot 2^{m-4}-1$ for $m\ge 4$ (Theorem 5.4.27); $\ell(2m+1)\le 5\cdot2^{m-1}-1$ for $m\ge 1$ (Theorem 5.4.27); $\ell(2m+1) \ge 2^{m+1}+1$ for $m\ge 2$ (Theorem 7.2.16). I am not sure whether any of these estimates have been improved over the last two decades, and also whether any non-trivial lower bound for $\ell(2m)$ is known.<|endoftext|> TITLE: Concrete Examples of Shimura Surfaces QUESTION [12 upvotes]: First a disclaimer: I am at best a part-time arithmetic geometer, so please accept my apologies when I am too naive or get something wrong. From time to time I have tried to learn something about Shimura varieties. The friendliest example I came across so far are Shimura curves arising from quaternion algebras. By this, I will mean the following: Choose an indefinite quaternion algebra $B$ over $\mathbb{Q}$ with discriminant $D\neq 1$ and a maximal order $\Lambda \subset B$. Then $\mathcal{X}^B$ is the stack over $\mathbb{Z}[\frac1D]$ classifying, for $R$ an $\mathbb{Z}[\frac1D]$-algebra, projective abelian surfaces $A$ over $R$ with an action $\Lambda \to End(A)$ satisfying the following condition: If $R\to S$ is a ring map such that $\Lambda \otimes S \cong M_2(S)$, then the two summands of the $M_2(S)$-module $T_eA \otimes_R S$ are locally free of rank $1$. These kinds of Shimura curves have (among others) the following nice properties: $\mathcal{X}^B$ is a smooth and proper Deligne-Mumford stack. The deformation theory is controlled by the associated $p$-divisible group. More precisely: We have an étale map $(\mathcal{X}^B)_p \to \mathcal{M}_{p-div}$ to the moduli stack of 1-dimensional $p$-divisible groups. One has a good understanding of the coarse moduli space over $\mathbb{C}$ (quotient of upper half-plane). In many cases one can write down equations of coarse moduli spaces over $\mathbb{Q}$ or even $\mathbb{Z}$ of the curve or its quotient by the Atkin-Lehner involutions. Points (especially of maximal height of the $1$-dimensional formal group) can have interesting automorphism groups; but one can choose level structures so that points have only trivial automorphisms. Its definition does not require adèles. (This probably only an advantage due to my inexperience.) Is there anything like this for Shimura varieties of dimension $2$? I do not need huge families, but only a few examples I can touch with my hands, with a reasonably simple definition and a good moduli interpretation (deformation theory controlled by $1$-dimensional $p$-divisible group). REPLY [20 votes]: As Dylan says, the examples you are looking for are called Picard modular surfaces. So far as I know, the simplest example of a Shimura surface is from a paper "Sur des fonctions de deux variables indépendantes analogues aux modulaires" of Picard (yes, that Picard) from 1883. From the formal group perspective, it's attached to the abelian integral $$ \int \frac{dz}{\sqrt[3]{z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4}}. $$ Roughly, the development goes as follows: Given a monic degree-four polynomial $f(z)$ without double roots, the equation $w^3 = f(z)$ defines a plane curve whose closure is a smooth curve of genus 3 with an action of $\Bbb Z/3$ (on $w$). The only isomorphisms between them which are $\Bbb Z/3$-equivariant are induced by the action itself and the maps $z \mapsto z+t$. The Jacobian of said curve is 3-dimensional abelian variety with an action of $\Bbb Z/3$. It's also principally polarized -- this roughly corresponds to the cohomology pairing on $H^1$ of the curve. There is a basis $$ \frac{dz}{w^2}, \frac{z dz}{w^2}, \text{ and }\frac{dz}{w} $$ of the 1-forms. This splits the formal completion of the Jacobian into two $\Bbb Z/3$-invariant summands, one 2-dimensional, one 1-dimensional. If $p \equiv 1 \mod 3$, then this splitting extends to the $p$-divisible group in the desired manner. (The formal group is actually additive concentrated in even heights for primes congruent to 2.) This defines for you a map of moduli from the moduli of these curves to a corresponding Shimura variety of $3$-dimensional polarized abelian surfaces with an action of the ring $\Bbb Z[\zeta_3]$. The action $[-1]$ on the abelian varieties doesn't lift to any of these curves (none of them are hyperelliptic), and so the map of moduli factors through a $\Bbb Z/2$-quotient. However, I think (?) that the Torelli theorem implies that after you take this quotient, the result is an open substack. It's not the whole Shimura surface. Roughly, there are situations where the polynomial $f(z)$ degenerates, but the abelian variety does not. In terms of the Deligne-Mumford compactification, if you have only double roots then the smooth curve can be made to degenerate down to a stable curve of arithmetic genus $3$, whose "Jacobian" $Pic^0$ is still an abelian variety (but which now decomposes as the product of some 1- and 2-dimensional abelian varieties with $\Bbb Z/3$-action). If you close up under this procedure, you get the whole Shimura surface. The surface is not compact in this case; there is one "point at infinity." There are two compactifications: the Satake-Baily-Borel compactification adds points to the moduli parametrizing so-called semi-abelian schemes (this compactification is singular), and the smooth compactification expands it so that these points are actually (quotients of) elliptic curves. In this case, the moduli is pretty simple because it's described entirely in terms of the $a_i$. Away from the primes $2$ and $3$, you can eliminate the coefficient $a_1$ explicitly by a translation, and the Satake-Baily-Borel compactification is the weighted projective stack $$ Proj(\Bbb Z[1/6, a_2, a_4, a_3^2]) $$ with $|a_i| = 3i$. (The element $a_3$ gets a square because of this $\pm 1$ issue.) The moduli of curves hits the portion where the quartic discriminant of $f(z)$ doesn't vanish, and the uncompactified part is the portion where $f(z)$ has no triple zeros. This is the complement of the part defined by the ideal $(a_2^2 + 12 a_4,27 a_3^2 + 8 a_2^3)$. The largest automorphism group occuring here is for the curves $w^3 = x^4 + a_3 x$. The resulting abelian scheme has $18$ (equivariant) automorphisms. At $p=7$ this is a height-$3$ point (but there's another one with only 6 automorphisms). Unless I'm mistaken, the height-$3$ locus has mass $(p-1)(p^2-1)/1296$. For other results, many have studied this modular surface and other Picard modular surfaces in the intervening years, but most of them can't be analyzed so concisely. The book "The Zeta Functions of Picard Modular Surfaces" by Langlands et al. has a wealth of information about their general structure (in particular, about calculations with the trace formula), and can lead to a number of other useful places in the literature. Feel free to contact me if you have further interest.<|endoftext|> TITLE: Does Brownian motion immediately visit both sides of a Jordan curve? QUESTION [11 upvotes]: Let $C$ be a Jordan curve in $\mathbb{R}^2$. By the Jordan curve theorem, $\mathbb{R}^2 \smallsetminus C$ is uniquely partitioned into two connected regions $A$ and $B$ (the interior and exterior). Question 1: Let $x_0$ be an arbitrary point in $A$. Let $X$ be a planar Brownian motion starting at $x_0$. Let $$\tau_C = \inf \{t: X(t) \in C \},\qquad\tau_B = \inf \{t: X(t) \in B \}.$$ Does $\tau_C = \tau_B$ a.s.? Question 2: Let $y_0$ be an arbitrary point on $C$. Let $Y$ be a planar Brownian motion starting at $y_0$. Let $$\tau_A = \inf \{t: Y(t) \in A \},\qquad\tau_B = \inf \{t: Y(t) \in B \}.$$ Does $\tau_A = \tau_B = 0$ a.s.? Observations: A positive answer to question 2 gives a positive answer to question 1. (Let $y_0 = X(\tau_C)$.) The answers to both questions 1 and 2 are clearly true for "nice curves", for example smooth curves or curves with some kind of cone condition on each side. However, possibly some messy curves (e.g. ones with positive 2-dimensional Lebesgue measure) would present a problem. If the answer to either question is yes, it would be interesting to know the answer in higher dimensions as well (replacing Jordan curves with their higher dimensional analogues). REPLY [11 votes]: As to question 2: Planar Brownian motion started at $y_0$ will almost surely loop around $y_0$, i.e., disconnect $y_0$ from $\infty$ immediately, so it has to hit $A$ and $B$ immediately, too, and $\tau_A = \tau_B = 0$ a.s., no matter what kind of Jordan curve $C$ is. This should also imply that the answer to the first question is yes, by the strong Markov property. In higher dimensions there are easy counterexamples to question 2. E.g., if $C$ is a topological sphere, smooth except for one point where it is locally the "Lebesgue thorn", i.e., with one irregular boundary point $y_0$, then one of the stopping times is almost surely positive. I am not sure about question 1 in higher dimension.<|endoftext|> TITLE: minimizing an integral over integer-coefficient polynomials $\displaystyle \inf_{f \in \mathbb{Z}[x]} \int_a^b f(x)^2 \, dx $ QUESTION [8 upvotes]: Let's consider the space $L^2[a,b]$ of functions on the interval and the norm: $$ ||f(x)||^2 = \int_a^b |f(x)|^2 \, dx $$ Now what if we consider only polynomials with integer coefficients: $f(x) \in \mathbb{Z}[x]$? If we write $f(x) = \sum a_n x^n$ with each $a_n \in \mathbb{Z}$, this norm is a rational quadratic form over the integers with rational coefficients. $$ \inf_{f \in \mathbb{Z}[x],f\neq 0} \int_a^b f(x)^2 \, dx $$ Does it obtain a minimum value over the set of integer-valued polynomials? What is it? REPLY [11 votes]: Thanks to BigM for the link to Ofir's MO Question 121913, which cites a 120-year-old paper of Hilbert for the result that the integral can get arbitrarily small as long as $b-a < 4$: D. Hilbert: Ein Beitrag zur Theorie des Legendre'schen Polynoms, Acta Math. 18 (1894), 155$-$159 If $b-a \geq 4$ then an elementary argument using properties of Legendre orthogonal polynomials shows that there is a positive lower bound. I gave this argument an hour ago in my answer to the earlier MO question; I guess it must be an old result, perhaps known already to Hilbert (who mentions Legendre polynomials in the title of his paper!), but it's the kind of result that's easier to prove than to find in the literature. That question did not ask for the value of the minimum, but I see that the bound is an increasing function of $\deg f$, and is sharp for $\deg f = 0$, whence the minimum value is $b-a$, attained only by the constant polynomials $\pm 1$.<|endoftext|> TITLE: Is non-existence of the hyperreals consistent with ZF? QUESTION [20 upvotes]: I know that it is possible to construct the hyperreal number system in ZFC by using the axiom of choice to obtain a non-principal ultrafilter. Would the non-existence of a set of hyperreals be consistent with just ZF, without choice? Let me be conservative, and say that by a "set of hyperreals," I just mean a set together with some relations and functions such that the transfer principle holds, and there exists $\epsilon > 0$ smaller than any real positive real number. REPLY [6 votes]: In response specifically to the title of the question: "Is non-existence of the hyperreals consistent with ZF?", technically speaking the answer is NO. Kanovei and Shelah constructed a definable model of the hyperreals in ZF; see http://arxiv.org/pdf/math/0311165.pdf Therefore ZF is not "consistent with the nonexistence of the hyperreals". Of course, to prove any of their properties (such that that they are actually a proper extension, satisfy transfer, etc) one needs AC, but the same goes for many other crucial mathematical results (see below). Goldblatt in his book "Lectures on the hyperreals" takes countable additivity to be part of the definition of measure (see M1 on page 206). Then he uses hyperfinite partitions, outer measures, and transfer to show that express the Lebesgue measure in terms of the Loeb measure (page 217). In particular countable additivity of Lebesgue measure follows. Note that the failure of countable additivity of the Lebesgue measure (such countable additivity is taken for granted in analysis) is also consistent with ZF; see Is sigma-additivity of Lebesgue measure deducible from ZF? Similarly, it is consistent with ZF that the Hanh-Banach theorem (arguably foundation of functional analysis) fails.<|endoftext|> TITLE: Characteristic polynomial of Kronecker/tensor product QUESTION [5 upvotes]: This was asked before on stackexchange but no answer was given. The question is the following: Let $A$ and $B$ be matrices in $GL(n)$ and $GL(m)$ respectively. Their tensor product $A\otimes B$ is defined explicitly by the Kronecker product: $$A\otimes B=\begin{bmatrix}a_{11}B&\dots&a_{1n}B\\\vdots&\ddots&\vdots\\a_{n1}B&\dots&a_{nn}B\end{bmatrix}\in GL(nm).$$ Question: is there a known expression of its characteristic polynomial in terms of those of $A$ and $B$? It seems there might not be one, but I would like to be proven wrong. (In contrast, the direct sum $A\oplus B=\begin{bmatrix}A&0\\0&B\end{bmatrix}$ has characteristic polynomial $p_A\cdot p_B.$) EDIT: I've worked out an example for $n=m=2$: Given $2\times 2$ matrices $A$ and $B$ with characteristic polynomial $p_A=t^2-a_1t+a_2$ and $p_B=t^2-b_1t+b_2$, I find the coefficients of $p_{A\otimes B}=t^4-c_1t^3+c_2t^2-c_3t+c_4$ can be expressed as: \begin{align} c_1&=a_1b_1\\ c_2&=a_1^2b_2+b_1^2a_2-2a_2b_2\\ c_3&=b_1b_2a_1a_2\\ c_4&=a_2^2b_2^2 \end{align} Similarly for $n=2,m=3$, \begin{align} c_1&=a_1b_1\\ c_2&=a_1^2b_2+b_1^2a_2-2a_2b_2\\ c_3&=b_1b_2(a_1a_2-3a_3)+a_3b_1^3\\ c_4&=a_1a_3(b_1b_2-2b_2^2)+a_2^2b_2^2\\ c_5&=b_1b_2^2a_1a_2\\ c_6&=a_2^3b_2^2 \end{align} It is clear that in general $c_1=a_1b_1$ and $c_n=a_n^{\text{deg}(p_B)}b_n^{\text{deg}(p_A)}$, $c_2$ also seems to be consistent, but a general formula for $c_i$ eludes me still. REPLY [14 votes]: As David Handleman observed, you need (assuming you are over a splitting field) simply the polynomial that has the products of eigenvalues as roots. Using the resultant, you could calculate this polynomial as $\mbox{res}_y(P_A(y),P_B(x/y)\cdot y^m)$. (This is a polynomial in $x$.) For example, let $P_A(x)=(x-2)(x+3)=x^2+x-6$ and $P_B(x)=(x+5)(x-7)=x^2-2x-35$. Then $P_B(x/y)\cdot y^2=x^2-2xy-35y^2$ and the resultant becomes $x^4+2x^3-479x^2+420x+44100=(x-15)(x-14)(x+10)(x+21)$. REPLY [11 votes]: If the characteristic polynomials of $A$ and $B$ factor as $P_A(x) = \prod_{i=1}^n (x - \lambda_i)$ and $P_B(x) = \prod_{j=1}^m (x - \mu_j)$, then the characteristic polynomial of $A \otimes B$ is $$P_{A \otimes B}(x) = \prod_{i=1}^n \prod_{j=1}^m (x - \lambda_i \mu_j)$$ If all $\lambda_i \ne 0$ this could be written as $$ \prod_{i=1}^n \lambda_i^m \prod_{j=1}^m (x/\lambda_i - \mu_j) = (\det A)^m \prod_{i=1}^n P_B(x/\lambda_i)$$ or similarly if all $\mu_i \ne 0$ as $(\det B)^n \prod_{j=1}^m P_A(x/\mu_j)$.<|endoftext|> TITLE: Green's operator of elliptic differential operator QUESTION [16 upvotes]: Let $P:\Gamma(E)\rightarrow\Gamma(F)$ be an elliptic partial differential operator, with index $=0$ and closed image of codimension $=1$, between spaces $\Gamma(E)$ and $\Gamma(F)$ of smooth sections of vector bundles $E\rightarrow M$ and $F\rightarrow M$ on a compact Riemannian manifold $(M,g)$ without boundary. Question: What is the elliptic operator's associated Green's operator? More concretely, $(M,g)=(\mathbb{S}^m,g)$ be the unit $m$-sphere with constant curvature =1 metric $g$, so $\text{Ricc}(g)=g$. Also let $E=F=S^2\mathbb{S}^m$, the space of $2$-covariant tensors on $\mathbb{S}^m$. Consider the operator: \begin{align*} P:\Gamma(S^2\mathbb{S}^m)&\rightarrow\Gamma(S^2\mathbb{S}^m)\\ h&\mapsto Ph_{ij}=\frac{1}{2}g^{kl}(\nabla_i\nabla_jh_{kl}+\nabla_k\nabla_lh_{ij}). \end{align*} The symbol is: $$ \sigma_P(\xi)h_{ij}=\frac{1}{2}g^{kl}(\xi_i\xi_jh_{kl}+\xi_k\xi_lh_{ij}). $$ We can show that $P$ is elliptic with index $=0$ and closed image of codimension $=1$. Question: What is the Green's operator of $P$? Reference request: A good reference on Green's operators for elliptic partial differential operators would be welcomed. Addition: There is mention of Green's operator on pages 157 & 158 of Hamilton's 1982 paper The Inverse Function Theorem of Nash and Moser. In the context of the above, it goes roughly as follows: Choose finite dimensional vector spaces $N$ and $M$ and continuous linear maps $$ j:\Gamma(S^2\mathbb{S}^m)\rightarrow N\qquad\text{and}\qquad i:M\rightarrow\Gamma(S^2\mathbb{S}^m). $$ Define another map \begin{align*} L:\Gamma(S^2\mathbb{S}^m)\times M&\rightarrow\Gamma(S^2\mathbb{S}^m)\times N\\ (h,x)&\mapsto L(h,x)=(Ph+ix,jh), \end{align*} which is required to be invertible. Then Theorem 3.3.3. states that: The inverse map $$ L^{-1}:\Gamma(S^2\mathbb{S}^m)\times N\rightarrow\Gamma(S^2\mathbb{S}^m)\times M $$ is a smooth, tame, and linear. For each $k\in\Gamma(S^2\mathbb{S}^m)$ there is a unique $h\in\text{kernal}\,j$ such that $Ph-k\in\text{image}\,i$. The resulting map \begin{align*} G:\Gamma(S^2\mathbb{S}^m)&\rightarrow\text{kernal}\,j\\ k&\mapsto Gk=h \end{align*} is smooth, tame, and linear, and is called the Green's operator of $P$. New question: I'm wondering how to adapt this construction to the map $P$ above. What would the finite-dimensional vector spaces $N$ and $M$, and the maps $j$ and $i$, be to make this construction work? REPLY [10 votes]: If it exists, the inverse of an elliptic operator $P$ is its Green's operator. In general, an inverse does not exist, but a parametrix does. A parametrix is an operator $Q$ such that $PQ-I$ and $QP-I$ are compact operators. (I am assuming that $P$ is an elliptic operator on a closed manifold and acts between sections of vector bundles.) The existence of a parametrix is a basic application of pseudodifferential calculus, but it can also be deduced from elliptic estimates. The existence of a parametrix implies that $P$ is Fredholm, thus the kernel and cokernel of $P$ are finite-dimensional. Using this, one can make $P$ become part of an invertible operator $$ \begin{pmatrix} P& R_-\\ R_+& 0\end{pmatrix}^{-1}= \begin{pmatrix}E&E_+\\ E_-&E_{-+}\end{pmatrix}, $$ where $R_-$ has finite-dimensional domain and $R_+$ finite-dimensional range. This setup is presented, with many useful applications given, by Sjöstrand and Zworski in their paper Elementary linear algebra for advanced spectral problems. See, in particular, the section 2.4 on analytic Fredholm theory. The operator $P$ is invertible if and only if the finite-dimensional operator $E_{-+}$ is, and $P^{-1}=E-E_+ E_{-+}^{-1}E_-$. In the context of Theorem 3.3.3 of Hamilton's paper on the inverse function theorem set $P=L(f)$, $R_+=j$, and $R_-=i$. He calls $G(f)=E$ the Green's operator, which it is when he is allowed to ``forget'' his spaces $M$ and $N$. The finite-dimensional spaces $M$ and $N$ can be chosen as, respectively, the cokernel and the kernel of $P$. It may be useful to choose them larger if it is desired to have them idependent of $f$ in $P=L(f)$. Decomposing with respect to kernel and cokernel is standard in Fredholm theory. In Deane Yang's answer the decomposition is interior relative to the spaces while here, and in Hamilton's paper, it is exterior.<|endoftext|> TITLE: Is the free abstract group residually of rank d > 2? QUESTION [8 upvotes]: Let $d \geq 2$ be an integer, and let $\mathcal{F}_d$ be the family of finite groups such that $G \in \mathcal{F}_d$ if and only if every subgroup of $G$ can be generated by at most $d$ elements. Is there a finite nontrivial word $w = w(x_1, \dots, x_n)$ which is trivial on $\mathcal{F}_d$? That is, can it be that for any choice of $G \in \mathcal{F}_d$ and $y_1, \dots, y_n \in G$ we have $w(y_1, \dots, y_n) = 1$. Alternatively: Let $F$ be a free group of rank $\aleph_0$. Is it possible that the intersection of all finite index subgroups $N \lhd F$ with $F/N \in \mathcal{F}_d$ is nontrivial? I am also interested in the analogous question for $p$-groups. REPLY [4 votes]: No, there is not such a word, by the following two facts. The two elements $$A=\left(\begin{array}{cc}1&2\\0&1\end{array}\right),\quad B=\left(\begin{array}{cc}1&0\\2&1\end{array}\right)$$ of $\text{SL}_2(\mathbf{F}_p)$ satisfy no nontrivial relation of length less than $c\log p$. Every subgroup of $\text{SL}_2(\mathbf{F}_p)$ is generated by at most 2 elements. For 1, there is a well known ping-pong argument which proves that $A$ and $B$ generate a free subgroup of $\text{SL}_2(\mathbf{Z})$. The argument can be reproduced in $\text{SL}_2(\mathbf{F}_p)$ for words which are not too long. For 2, the subgroups of $\text{SL}_2$ are well understood. Refer for instance to Theorem 6.17 in Suzuki's book: Suzuki, Michio. Group theory. I. Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 247. Springer-Verlag, Berlin-New York, 1982. xiv+434 pp. ISBN: 3-540-10915-3 MR0648772<|endoftext|> TITLE: Matroids relaxations of a given matroid QUESTION [5 upvotes]: Let $\mathcal{M}$ be a rank-$d$ matroid on $[n]$. Say a matroid $\mathcal{N}$ is a relaxation of $\mathcal{M}$ if $\mathrm{rank}(\mathcal{N})=d$, $\mathrm{groundset}(\mathcal{N})=[n]$, and every independent set of $\mathcal{M}$ is an independent set of $\mathcal{N}$ (observe that this notion of relaxation is in the labeled sense). Has anyone seen the set of all relaxations of $\mathcal{M}$ naturally show up in some context? I'm particularly interested in the algorithmic task of listing all relaxations of a given matroid $\mathcal{M}$, and in any heuristic to make an implementation more efficient than performing the enumeration by brute force; it would be even better if such an implementation already exists. Actually, I only care about those relaxations of $\mathcal{M}$ that are realizable over a fixed field, say $\mathbb{C}$, which brings me to the next question: Does anyone know of an implementation of a (finite) algorithm that decides whether a matroid is representable and provides a representation in case it is? (that such a finite algorithm exists can be proved using Groebner bases) Here I'm explicitly disregarding any complexity issues; I just want to be able to do compute a realization for some small matroids. Note: the first question is different from the "opposite" one of listing all matroids that specialize a fixed matroid, which has been asked here. I wonder how different the task of listing all relaxations is. Update: I implemented the following brute-force algorithm in sage to enumerate the relaxations of a rank-$d$ matroid on $[n]$: For every subset $\mathcal{F}$ of the nonbases of $\mathcal{M}$, check whether $bases(\mathcal{M})\cup \mathcal{F}$is the collection of bases of a matroid. According to sage's method .is_valid() for matroids, the outcome for all the matroids I tested was that all $2^{\#\{nonbases(\mathcal{M})\}}$ subsets of $\binom{[n]}{d}$ obtained form the bases of a matroid, which wasn't what I expected. Can it be that there's a bug in sage's .is_valid() method, or is it actually to be expected that most naive relaxations of a matroid (as enumerated above) are actually matroids? (the reward is independent from the update in the question) REPLY [4 votes]: You are describing what are usually called "weak maps" between matroids. So a bijection $\varphi: E(\mathcal{N}) \to E(\mathcal{M})$ is called a weak map if for every independent set $I$ of $\mathcal{M}$ the inverse image $\varphi^{-1}(I)$ is independent in $\mathcal{N}$. In your case, you are asking when the identity function is a weak map, which just requires every independent set of $\mathcal{M}$ to be independent in $\mathcal{N}$. There is lots of stuff known about weak maps (and strong maps) and other things. Given a bunch of matroids on the same ground set, the weak-map relation is a partial order. The best place to start is around page 280 of the second edition of Oxley's book Matroid Theory, which refers to the original papers. The "minimal" relaxation that can be done is to take a single non-basis $N$ of $\mathcal{M}$ and form a matroid whose bases are ${\mathrm{bases}} (\mathcal{M}) \cup \{N\}$. This will work if (and, I think, only if) $N$ is a circuit-hyperplane of $\mathcal{M}$. So you certainly do not expect all of your relaxations to pass Sage's is_valid routine. In fact, I wanted to do a test to convince myself that it was working ok, so I ran the following code, just testing all the relaxations of a single randomly chosen matroid (in this case the matroid called J). All this does is create an array of matroids, each of which are specified by their bases, for which I am using the bases of J together with one of the non-bases. Then I ask is_valid() which of those matroids are valid. m = matroids.named_matroids.J() bases = [x for x in m.bases()] nonbases = [x for x in m.nonbases()] circuithyps = [x for x in m.circuits() if x in m.hyperplanes()] tstsets = [bases + [nonbases[i]] for i in [0..len(nonbases)-1]] tstmats = [Matroid(x) for x in tstsets] tstcheck = [m.is_valid() for m in tstmats] tstcheck and I obtained the following result: [True, False, False, False, False, False, False, False, False, False, False, False, False, True, False, False, False, False, False, False] which when we check the number of circuit-hyperplanes, is exactly what we expected. [frozenset(['c', 'b', 'e', 'd']), frozenset(['h', 'e', 'g', 'f'])] You should let us see the code that you used that produced the incorrect answers, so we can tell whether it is a Sage bug, a sage.matroids bug, or user-error! Finally, Stefan van Zwam has written a Mathematica program using Groebner bases to determine whether a matroid is representable, but the problem is hard unless you only want to know about matroids with less than 10 elements. I am not sure if he intends to add it to the Sage code-base or whether he even has it still. Ask him: https://www.math.lsu.edu/~svanzwam/programming.html<|endoftext|> TITLE: Rationality of moduli spaces of rational curves QUESTION [6 upvotes]: Let $\overline{M}_{0,n}$ be the moduli space of Deligne-Mumford stable pointed rational curves, and let us consider the quotient $\widetilde{M}_{0,n} = \overline{M}_{0,n}/S_n$. Clearly, there is a dominant rational map $\overline{M}_{0,n}\dashrightarrow \widetilde{M}_{0,n}$, and $\widetilde{M}_{0,n}$ is unirational. If $n=4,5$ then $\widetilde{M}_{0,n}$ is a curve and a surface respectively. Thus, it is rational. Is it known that $\widetilde{M}_{0,n}$ is rational for any $n\geq 3$? If so, do you know a reference for this? REPLY [4 votes]: I don't know if this is exactly the answer you are looking for but for even $n$ this is clearly related to the result by Katsylo for moduli spaces classical binary forms or hyperelliptic curves, see this paper. Another link for accessing the article by Katsylo is this one. Theorem 0.2 therein seems to give a positive answer to your rationality question for all $n$ (even or not) except possibly for $n=10$ (or genus four hyperelliptic curves). I think this exception has been covered later by Bogomolov. Note that for $n=10$ now one has an explicit description for the ring (rather than just field) of invariants, see this article by Brouwer and Popoviciu.<|endoftext|> TITLE: Good lecture notes/books on Jacobian of hyperelliptic curve QUESTION [7 upvotes]: I want to understand what the Jacobian variety is from an algebraic (or arithmetic?) perspective. I want to know: What is the definition of the Jacobian? Widely know facts about it. Why the Jacobian of an elliptic curve is the curve itself. How to map points on a hyperelliptic curve to the Jacobian. As much as possible about the torsion subgroup and torsion 2-subgroup of the Jacobian of a hyperelliptic curve What is the connection between the fundamental unit in the corresponding ring and torsion points of the Jacobian? How to get sum of points on the Jacobian (maybe the Mumford algorithm). I want to understand it as quickly as I can, so I want to read a clear monograph. It would be great if that monograph (book or some lecture notes or survey or something) requires minimal possible background. Thanks for any references. REPLY [2 votes]: An elementary introduction to hyperelliptic curves by Menezes, Wu and Zuccherato gives a good elementary overview, especially if you want to understand the arithmetic on Jacobians. If you would like to go a bit deeper into the topic, Part A of the book Diophantine Geometry: An Introduction by Hindry and Silverman is very useful if you want to understand the geometry of these objects better. The chapter is titled The Geometry of Curves and Abelian Varieties. I found it really useful, it was not an easy reference to find and I wish I knew about this book earlier.<|endoftext|> TITLE: A generalization of Jensen's Inequality QUESTION [8 upvotes]: Jensen's inequality is well known as $$E\big[f(X)\big]\le f\big(E[X]\big)$$ where $X$ is a integrable random variable and $f: R\to R$ is a bounded concave function, see also http://en.wikipedia.org/wiki/Jensen%27s_inequality Now I have a question about whether we may have a generalized result for a more abstract space. Let $\Omega:=D([0,1],R)$ be the space of all cadlag functions defined on $[0,1]$. Denote by $X$ the canonical process, i.e. $X_t(\omega)=\omega_t$. Let $f: \Omega\to R$ be a bounded concave function and $P$ be a martingale measure, i.e. $f\big(\alpha \omega+(1-\alpha)\omega'\big)\ge \alpha f(\omega)+(1-\alpha)f(\omega')$ for any $\omega, \omega'\in\Omega$, $\alpha\in [0,1]$ and $X=(X_t)_{0\le t\le 1}$ is a $P-$martingale. Could we also show that $$E^P\big[f(X)\big]\le f\big(E^P[X]\big),$$ where $E^P[X]\in \Omega$ is a constant function taking $E^{P}[X_0]$. Thx for the reply! REPLY [5 votes]: Since $X=(X_t)_{t\in[0,1]}$ is a martingale, we have $E(X_t|X_0)=X_0=X_0\,\omega_1(t)$ for each $t\in[0,1]$, where $\omega_1$ denotes the function on $[0,1]$ with constant value $1$. So, by Jensen's inequality for conditional expectations (see e.g. [cond. Jensen's ineq.]), $$E(f(X)|X_0)\le f(E(X|X_0))=f(X_0\,\omega_1), $$ whence $Ef(X)=EE(f(X)|X_0)\le Ef(X_0\,\omega_1)$, as desired.<|endoftext|> TITLE: Computing naive algebraic singular homology QUESTION [7 upvotes]: I'm not sure if this counts as research level, since it might just be an expression of my ignorance. But anyway. Let $\Delta^n_A = Spec(A[X_0, \dots, X_n]/\sum X_i - 1)$ denote the standard algebraic simplex over $A.$ These assemble into the standard cosimplicial affine scheme $\Delta_A^\bullet.$ Hence, given any scheme $X,$ one may define a simplical set $Sing_\bullet(X)(A) = Hom(\Delta^\bullet_A, X)$ and then, taking the associated chain complex, we may define the naive algebraic homology groups as $H_*(X)(A) = h_* \mathbb{Z}Sing_\bullet(X)(A),$ where we do the usual thing of taking the free simplicial abelian group on $Sing_\bullet$ and alternating sums of face maps as differentials. For more on the construction, compare: References for the "nerve of an algebraic variety" As is pointed out in that answer, this particular construction is not in general well-behaved. (The good version is to use morphisms into infinite symmetric products, which then computes singular homology.) However: can we compute these naive groups for any spaces at all? What I have come up with: $H_0(\mathbb{G}_m)(A) = A^\times,$ $H_n(\mathbb{G}_m) = 0$ for $n>1$ because $\mathbb{G}_m$ is A^1-invariant. $H_0(\mathbb{A}^n)(A) = \mathbb{Z}$ and the higher groups vanish - because $\mathbb{A}^n$ are algebraically contractible [I haven't checked this in detail.] $H_0(\mathbb{P}^1)(k) = \mathbb{Z}$ for any field $k$ But these observations are all essentially trivial. Using work of cazanave on naive homotopy classes of endomorphisms of $\mathbb{P}^1,$ one may compute $H_0(\mathbb{P}^1)(k[X]),$ but this is already fairly non-trivial. Perhaps some concrete questions: Can anyone compute $H_n(\mathbb{P^1})(k)$ or $H_n(PGL_2)(k),$ for some $k$ or $n>0$? Are these groups zero for $n$ sufficiently large? REPLY [8 votes]: There has not been so much research on $\operatorname{Sing}(X)(A)$, and most of it concentrated on its homotopy. Nevertheless, some things are known: Among the first results are those by Jardine in a series of papers at the beginning of the 80s, see the first five entries on his list of publications. You can also check Jardine's thesis. From his results you get that $H_0(G)(k)\cong\mathbb{Z}$ for a group $G$ whose $k$-points are generated by unipotent elements. He also proved $H_1(SL_n)(k)\cong K_2(k)$ for $n\geq 3$ and $H_1(Sp_{2n})(k)\cong K_2(k)$ for $k$ algebraically closed. Very classical results of Rector and Gersten identify the homotopy of $\operatorname{Sing}(GL)(R)$ with algebraic K-theory if $R$ is a regular ring (basically a consequence of homotopy invariance for algebraic K-theory). You can read about this in Chapter 4 of Jardine's thesis. The results of Jardine are now know to hold more generally. For $G$ isotropic reductive and $R$ a regular ring, $H_0(G)(R)\cong\mathbb{Z}[G(R)/E(R)]$ by unstable homotopy for $K_1$-functors, a result of Stavrova. This relates $H_0$ with Whitehead groups. For $G$ isotropic reductive and $k$ a field, $H_1(G)(k)\cong H_2(G(k),\mathbb{Z})$ is proved in my paper with Konrad Voelkel, as a consequence of homotopy invariance in one variable for homology of isotropic reductive groups. The identification $H_1(G)(k)\cong H_2(G(k),\mathbb{Z})$ can also be derived from an $\mathbb{A}^1$-homotopy point of view. This is explained in Morel's "$\mathbb{A}^1$-algebraic topology over a field", Section 6.2 and in this paper. Another result which can be deduced from the results in Morel's book: the spaces $\mathbb{A}^n\setminus 0$ have the affine Brown-Gersten property so that the $\mathbb{A}^1$-$(n-2)$-connectivity implies $H_i(\mathbb{A}^n\setminus 0)(k)=0$ for $1\leq i\leq n-2$ and $H_{n-1}(\mathbb{A}^n\setminus 0)(k)\cong K^{MW}_n(k)$. If we allow $X$ to be more general than a scheme, we can also say something about classifying spaces of algebraic groups. The very classical results mentioned above can be reformulated as follows: homotopy invariance of algebraic K-theory implies that for $R$ a regular ring $H_n(BGL)(R)\cong H_n(BGL(R),\mathbb{Z})$. On the left is the naive algebraic singular homology of the infinite Grassmannian, and on the right is group homology of the infinity general linear group over $R$, viewed as discrete group. It is a subject of current investigations if an isomorphism such as this holds for linear algebraic groups (related to weak homotopy invariance). Weak homotopy invariance and its current status are discussed in my MO-answer here and some results related to computation of $H_3(BSL_2)(k)$ can be found in my paper with Kevin Hutchinson.<|endoftext|> TITLE: Compute adjugate matrix over commutative ring QUESTION [8 upvotes]: Let $A$ be a $n\times n$ matrix over a commutative ring. I'm looking for a good method to compute its adjugate matrix. My current approach is to use the Cayley-Hamilton theorem: $$\text{adj}(A) = -(A^{n-1} + c_1A^{n-2}+\ldots +c_{n-2}A+c_{n-1}\text{I}) $$ where $$\lambda^n + c_1\lambda^{n-1} + \ldots + c_{n-1}\lambda +c_n = \det(\lambda\text{I}-A).$$ i.e. the $c_i$ are the coefficients of the characteristic polynomial, which I can determine using the Berkowitz algorithm. In particular, I never have to make divisions, so all works over an arbitrary commutative ring. However, this method needs $\mathcal{O}(n^4)$ arithmetical ring operations, so there might be better alternatives. I hope for something like $\mathcal{O}(n^3)$, as for matrix inversion over a field. REPLY [3 votes]: A reference dated 2001-2003 is http://www4.ncsu.edu/~kaltofen/bibliography/01/KaVi01.pdf Over a commutative ring, the complexities of the calculation of $\det(A)$ and $adj(A)$ are the same. With the standard multiplication, the authors obtain a bit-complexity in $n^{10/3+\epsilon}N^{1+\epsilon}$ where $N$ is the length of the entries of $A$. Yet, the programming seems complicated ; there is a simpler method with exponent $3.5$. Conclusion: If you have a robust and simple method with exponent $4$, then it is perfect.<|endoftext|> TITLE: Euler's Triangular Number closure properties QUESTION [7 upvotes]: Burton, in "Elementary Number Theory", states that the following problems are due to Euler 1775: If $n$ is a triangular number, then so are $9n+1$, $25n+3$ and $49n + 6$. R. F. Jordan in the J. of Recreational Mathematics (1991, vol.23, p.78) proves the following generalization: Let $t_k$ be the $k$-th triangular number. For all $k$ and all triangular numbers $n$, $(2k+1)^2n + t_k$ is also triangular. and wonders whether Euler actually proved this generalization. Jordan wasn't the first to prove this, see for example the following proposed solution to the 49th Putnam 1988 question B6. I looked through the Euler Archive, but didn't locate the suitable 1775 manuscript (either appearing in 1775, written in 1775, or presented in 1775). What did Euler prove and where? What does Burton refer to? EDIT: The only part of the question that remains unanswered is: As Euler didn't seem to prove this generalization, who was the first to notice it? REPLY [4 votes]: Dickson, History of the Theory of Numbers, Volume II, page 12, writes, L. Euler (pp. 264-5, about 1775) noted that $9\Delta_a+1=\Delta_{3a+1}$, $25\Delta_a+3=\Delta_{5a+2}$, $49\Delta_a+6=\Delta_{7a+3}$, $81\Delta_a+10=\Delta_{9a+4}$. The reference Dickson gives is to Opera Postuma, 1, 1862.<|endoftext|> TITLE: Formal languages with non-unique interpretations of terms QUESTION [6 upvotes]: In mathematical logic and model theory, one considers interpretations of syntactic expressions: terms without free variables are interpreted as elements of some structure, formulas without free variables have truth values, formulas with free variables can be interpreted as relations. Multiple expressions may have identical interpretations. For example, $\ulcorner 1 + 1\urcorner$ and $\ulcorner 2\urcorner$ are both interpreted as $2$. Question: does anyone ever consider formal languages where terms can have multiple interpretations? Is there some standard approach or framework? I am thinking about this because i am trying to understand the $\omicron$ and $O$ notation in analysis, like in $$ \ln(x) =\omicron(x),\quad x\to +\infty. $$ Also, when calculating an indefinite integral, on often writes $$ \int 2x\,dx = x^2 + C. $$ Update. I understand that when the equality sign is used with $\omicron$/$O$ notation, it does not represent an equivalence relation. I also know that $\omicron(g(x))$ can be viewed as a set of functions. However, this interpretation does not fit my intuition well. When i write $\sin x = x + \omicron(x^2)$, $x→0$, i do not think about sets of functions, i think that i am replacing an anonymous implicitly understood function with a placeholder. In other words, the designated object does not change (it is still a function or a number, not a set of functions or set of numbers), only the notation is abbreviated and made less explicit, a bit like when i write "$1 + 2 + 3$" instead of "$((1 + 2) + 3)$". REPLY [3 votes]: It is possible to have multiple meanings for a given piece of syntax, this goes under the general name of polymorphism. For example, projections from a pair $\pi_1 : A \times B \to A$ and $\pi_2 : A \times B \to B$ are polymorphic because $\pi_1$ can mean the first projection for any sets $A$ and $B$, whereas a monomorphic notation (the opposite of polymorphic) would require us to always write $\pi_1^{A,B}$ (note that $\pi_A$ is broken as well as $\pi^{A,B}_A$ – think of the case $A = B$). Logicians and mathematicians are the wrong people to ask about these issues. You should talk to computer scientists because they are constrained by two factors: they want usable syntax, or else people will not use it, and the syntax has to be sensible, or else computers will not understand it. There are many solutions to giving multiple meaning to a single piece of notation, for example operator overloading in C++ (bad example), notation scopes in proof assistants such as Coq, and type classes in Haskell. General mechanisms for resolving ambiguous notations can be quite involved. For instance, type classes in Haskell and Coq are really like little prolog programs which direct the machine in finding out what the user meant when he or she wrote down an expression. Unfortnately, the examples you give about intervals and the little $o$ notation are not of the kind that can be repaired without at least some changes. Let me state explicitly that I sympathize with the idea that notation should not hinder expression of ideas, and I use sloppy notation all the time as well – but a prerequisite for using sloppy notation is to be able to use non-sloppy notation as well. Non-sloppy notation is especially important in teaching and when things get tricky.<|endoftext|> TITLE: When is the diagonal inclusion a $\Sigma_2$-cofibration? QUESTION [8 upvotes]: Recall that a space $X$ is called locally equiconnected or LEC if the diagonal map $d:X\hookrightarrow X\times X$ is a cofibration. For example, CW-complexes are LEC. There is some discussion of this concept at this MO question. Let $G$ be a finite group. Recall that a $G$-map $i: A\to Y$ is called a $G$-cofibration if it has the $G$-homotopy extension property with respect to all $G$-maps $f: Y\to Z$. If $X$ is LEC, is the diagonal map $d: X\hookrightarrow X\times X$ a $\Sigma_2$-cofibration, where the symmetric group acts trivially on $X$ and by permuting factors on $X\times X$? I've tried searching the literature on equivariant homotopy theory, but this doesn't seem to fall quite in that territory since $X$ itself does not come equipped with a group action. REPLY [3 votes]: I ended up needing this again, and eventually proved that $d:X\hookrightarrow X\times X$ is a $\Sigma_2$-cofibration as long as $X$ is an ENR. Although this doesn't quite answer the original question as asked, it seems general enough to be useful, so I thought I'd post it here. More generally, if $G$ is a finite group, $X$ is a $G$-ENR, and $A\subseteq X$ is a closed sub-$G$-ENR, then the inclusion $A\hookrightarrow X$ is a $G$-cofibration. The reference: Proposition 2.7 and Corollary 2.8 of https://arxiv.org/abs/1703.07142.<|endoftext|> TITLE: $E_n$-space and n-connected pointed space QUESTION [5 upvotes]: Is it true that the homotopy category of group-like $E_n$-spaces is equivalent to the homotopy category of pointed $n$-connected spaces ? If it is true, what should be the statement when $"n\rightarrow \infty"$ ? By $n$-connected space $X$, I mean that $\pi_{i}X=0$ for $0\leq i\leq n-1$. Edit Notions: The $\infty$-category of group-like $E_n$-spaces is denoted by $\mathbf{G}_{n}$ The category of pointed $n$-connected spaces is denoted by $\mathbf{Top}_{n}$. As Peter May and Ring Spectra noticed, $$Bar^{n}:\mathbf{G}_{n}\longrightarrow \mathbf{Top}_{n}:\Omega^{n}$$ is an $\infty$-equivalence. It seems very natural that the homotpy limit $$ holim(\dots \rightarrow \mathbf{G}_{n+1}\rightarrow \mathbf{G}_{n}\rightarrow\dots \mathbf{G}_{1})$$ is the $\infty$-category of group-like $E_{\infty}$-spaces i.e. connective spectra. My question is the following: How can we see that $$ holim(\dots \rightarrow \mathbf{Top}_{n+1}\rightarrow \mathbf{Top}_{n}\rightarrow\dots \mathbf{Top}_{1})$$ is naturally equivalent to the $\infty$-category of connective spectra without using $E_{n}-spaces$? PS: As Peter May noticed there is a problem with the my definition of $n$-connectivity. But I think the idea is clear. REPLY [4 votes]: Denis and ``Ring Spectra'', thanks for the references. I did not treat non-connected spaces in "The geometry of iterated loop spaces", which is why you couldn't find that there. It should have been treated in the immediate sequel "$E_{\infty}$ spaces, group completions, and permutative categories", but that perversely and for no good mathematical reason restricts to the case $n=\infty$. The proof works the same way in general. Max, $n$-connected means $\pi_i = 0$ for $i\leq n$ (you can look it up on Wikipedia if you do not believe me; this is implicit in Ring Spectra's answer). Briefly, the essential point is to start with an $E_n$-space $X$ and construct from it an $n-1$-connected space $Y_n(X)$ and a natural map $X\longrightarrow \Omega^n Y_n(X)$ that is a weak equivalence if $X$ is connected and a group completion in general, hence a weak equivalence if $\pi_0(X)$ is a group. Conversely, if $Y$ is an $n-1$-connected space, then of course $X=\Omega^n Y$ is a grouplike $E_n$-space. With the construction of Geo, $Y_n(X) = B(\Sigma^n, C_n, X)$ where $C_n$ is the monad associated to an $E_n$-operad. The group completion property for $n\geq 2$ follows from the group completion property for the natural map $\alpha_n\colon C_nX \longrightarrow \Omega^n \Sigma^n X$ (which I didn't yet know how to prove when I wrote Geo, but did thanks to work of Fred Cohen when I wrote the cited sequel). The case $n=1 is classical and special.<|endoftext|> TITLE: Decidability of Frankl's union-closed sets conjecture QUESTION [6 upvotes]: Is it conceivable that Frankl's union closed sets conjecture is undecidable in $\mathsf{ZFC}$, or is this quite implausible, perhaps due to the "finitistic" nature of the statement, or for some other reason? REPLY [8 votes]: The statement has complexity $\Pi^0_1$, which means that it has a single universal quantifier, quantifying over the possible union-closed sets, and then making a simple assertion about those objects. Although this is a very simple level of complexity, it is the same complexity as consistency assertions, and these admit of a robust independence phenomenon. For example, even ZFC, if consistent, admits of independent statements of this level of complexity, such as the statement Con(ZFC), which asserts that ZFC is consistent. Meanwhile, if we can prove in ZFC (or some other theory) that the union-closed set conjecture was independent of PA, then it would follow that the statement was true in the standard model. The reason is that if a $\Pi^0_1$ statement is independent of PA, then it can admit of no standard counterexample, since PA would prove that this was a counterexample. So, the situation is that if we can prove in ZFC that the union-closed set conjecture is independent of PA, then we would be able to prove in ZFC that it is true.<|endoftext|> TITLE: Smallest non-trivial conjugacy classes in simple groups and classes of involutions QUESTION [5 upvotes]: I am interested in finding the size of the smallest non-trivial conjugacy class of the simple groups $PSL(d,q)$ with $d>2$, $Sz(q)$ with $q>2$ and $R(q)$ with $q>3$. My first question is related to the involutions in Suzuki and Ree groups. For $q=2^{2n+1}>2$, the Suzuki group $Sz(q)$ has a unique class of involutions of size $(q^2+1)(q-1)$. For $q=3^{2n+1}>3$, the Ree group $R(q)$ has a unique class of involutions of size $q(q-1)(q+1)$. Are these classes the smallest non-trivial conjugacy classes? With respect to the groups $PSL(d,q)$ my questions are the following: Let $G=PSL(d,q)$ with $d>2$. Is some class of involutions the smallest non-trivial conjugacy class of $G$? I did some experiments with GAP and it seems that the smallest conjugacy class of $PSL(d,q)$, $Sz(q)$, $R(q)$, is some conjugacy class of involutions. REPLY [3 votes]: Concerning your first question for the Suzuki groups in characteristic 2, it's helpful to go back to the original announcement by Suzuki: A new type of simple groups of finite order, Proc. Nat. Acad. Sci. USA 46 (1960), no. 6, 868-870 (available through JSTOR and maybe other channels). Note that the order of such a group is $q^2(q-1)(q^2+1)$, with $q= 2^{2n+1}$. Suzuki approached his groups in the setting of 2-transitive permutation groups, but later it was seen that they could be viewed efficiently as groups of Lie type: here one gets a split BN-pair of rank 1 and nice relations with an ambient finite group of type $B_2 = C_2$. Anyway, Suzuki realized that his groups are simple for $q>2$ and that each has only a few conjugacy classes of maximal (proper) subgroups. You've pointed to the Sylow 2-subgroups of order $q^2$, each of which is the full centralizer of a nontrivial central involution (a unipotent element in Lie theory). In addition there are other types of centralizers, of orders $q-1$ (split "torus"), $q+2r+1$, and $q-2r+1$ with $r=2^n$. Here all centralizers turn out to be nilpotent, which is not too surprising from the Lie-theoretic viewpoint since the BN-pair has rank 1. Given this much information, it's true that "small" classes (which have "big" centralizers) must consist of involutions here. In the case of the Ree groups in characteristic 3, contained in groups of Lie type $G_2$, the story is probably similar since you again have a "rank" 1 BN-pair (though I haven't checked all details). Note that the Ree groups in characteristic 2 are more complicated, since they are twisted subgroups of Lie type groups $F_4$ and have "rank" 2. P.S. Looking at the order of a Ree group and some partial information about centralizers in Ward's 1966 paper here, I'm unsure where the data in the question about involutions fits in. Anyway, there must be fairly complete data about centralizers in the literature. ADDED: For a complete list of subgroups, see Theorem 6.5.5 in The Classification of the Finite Simple Groups, Number 3, by Gorenstein-Lyons-Solomon, AMS 1998. These include all centralizers, and it appears that the largest subgroup in the list is the centralizer of (any) involution. If so, this answers the question for Ree groups too. (One of the two G-L-S sources is a paper by Kleidman here.)<|endoftext|> TITLE: Paper by I. N. Sanov, Solution of the Burnside problem for exponent 4 QUESTION [7 upvotes]: I have searched extensively online and for copies of printed journals containing the paper which details Sanov's solution to the Burnside Problem for exponent 4, which is widely cited in many papers and texts on the Burnside problem. The reference for the paper is I. N. Sanov, "Solution of the Burnside problem for exponent 4", Uchen. Zap. Leningrad State Univ. Ser. Mat. 10 (1940), 166-170 (i.e. Leningrad State University Annals). Could anyone please help with obtaining a pdf of this, preferably in English? Thank you REPLY [5 votes]: As it was pointed out by Alex Dugas, one can find Sanov theorem in Hall's book here, see Theorem 18.3.1. According to the author, the proof does not determine precisely the order of $B(n,4)$. However, it is quite easy to show that $B(2,4)$ is finite: $|B(2,4)|=4096$. Let me add that the following GAP code verifies that $|B(2,4)|\leq4096$. (Then it is easy to conclude that indeed one has $|B(2,4)|=4096$.) The idea is to generate a random set $w_1,w_2,\dots,w_N$ for some big $N$ and checks whether the group $F_2/\langle w_1^4,...,w_N^4\rangle$ is finite: gap> f := FreeGroup(2);; gap> a := f.1;; gap> b := f.2;; gap> rels := Set(List([1..10000],x->Random(f)^4));; gap> gr := f/rels;; gap> Order(gr); 4096<|endoftext|> TITLE: applications of C$^*$-algebras in the field of PDEs QUESTION [12 upvotes]: I know only a little bit about C$^*$-algebras and I want a to know if you know a nice apllication or the influence of them in the field of partial differential equations (it is better that it is understandable for graduate students), or maybe can explain me why they are important for pseudo-differential-operators. A similar question i found here https://math.stackexchange.com/questions/798342/application-of-calgebras but there are not enough answers to convince me. I'm not sure if it is ok to ask it here, but maybe anyone could know more about C$^*$-algebras and partial differential equations. Regards REPLY [12 votes]: I think the canonical connection between C*-algebra and differential operators is Connes' index theorem for foliated manifolds. I don't know if that counts as PDEs but it's certainly related. Every foliated manifold $M$ has an associated C*-algebra $A$ which is noncommutative (except in trivial cases) but in some way embodies the idea of "the continuous functions on $M$ that are constant on leaves". Any pseudodifferential operator $D$ on $M$ which is elliptic on each leaf has an "index" which belongs to the $K_0$ group of $A$. There is an analytic definition and a topological definition of the index, and Connes' index theorem says that they agree. It is a profound generalization of the Atiyah-Singer index theorem. Connes' notes on the subject can be found here.<|endoftext|> TITLE: Which natural numbers are a square minus a sum of two squares? QUESTION [24 upvotes]: Question: Which natural numbers are of the form $a^2 - b^2 - c^2$ with $a>b+c$? This question came up in (Eike Hertel, Christian Richter, Tiling Convex Polygons with Congruent Equilateral Triangles, Discrete Comput Geom (2014) 51:753–759), where it was shown that numbers of this form are numbers of equilateral triangles (of the same size) that can tile a convex pentagon. It was shown (and this is elementary), that any number not of this form must be an idoneal number and thus be among $$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 15, 16, 18, 21, 22, 24, 25, 28, 30, 33, 37, 40, 42, 45, 48, 57, 58, 60, 70, 72, 78, 85, 88, 93, 102, 105, 112, 120, 130, 133, 165, 168, 177, 190, 210, 232, 240, 253, 273, 280, 312, 330, 345, 357, 385, 408, 462, 520, 760, 840, 1320, 1365, 1848$$ or equal to one of at most two more unknown numbers (which are at least $\geq 2.5 \cdot 10^{10}$). Interestingly, some of these known idoneal numbers can be written as in the question, for example $7=3^2-1^2-1^2$, but most of them can't. It is known that if the Generalized Riemann Hypothesis holds, then the list above (without the two unknown numbers) is a complete list of idoneal numbers. Then one can easily check which ones can be of the form as in the question. Hence, assuming the conjecture, the question can be answered. Anyhow, my hope was that maybe the question is more elementary to answer. Question: Can one answer the above question without using the Generalized Riemann Hypothesis? REPLY [23 votes]: The numbers that are not of the form $a^2-b^2-c^2$ with $b$ and $c$ positive and $a>b+c$ are precisely the idoneal numbers apart from $7$, $28$, $112$, $15$, $60$, and $240$. As noted in the problem, the paper by Hertel and Richter shows that the numbers not of this form are necessarily idoneal numbers, and a quick calculation shows that $7$, $28$, $112$, $15$, $60$ and $240$ are represented. I now show that if $n$ is of the form $a^2-b^2-c^2$, and $n$ is not of the form $7$ or $15$ times a square, then $n$ is not an idoneal number. The only idoneal numbers that are of the form $7$ or $15$ times a square are the six exceptional numbers listed above, (see Theorem 12 of Kani's article linked below) so the proof is complete. It may be helpful to recall that $n$ is an idoneal number if and only if, for every reduced quadratic form $Ax^2+Bxy+Cy^2$ of discriminant $-4n$, we have either $B=0$ or $A=B$ or $A=C$. We call here such forms as being of idoneal type. For a good survey on idoneal numbers see Kani. Suppose $n=a^2-b^2-c^2$ with $b$ and $c$ positive, and $a>b+c$. Then $$ -4n = (2c)^2- 4 (a-b)(a+b), $$ and if $a-b>2c$, then the quadratic form $(a-b)x^2 + (2c)xy + (a+b)y^2$ is a reduced form of discriminant $-4n$ not of idoneal type, and so $n$ is not idoneal. Now suppose $a-b<2c$. Here we use $$ -4n=(2a-2b-2c)^2 -4 (a-b)(2a-2c), $$ and note that either $(a-b)x^2 + (2a-2b-2c)xy + (2a-2c)y^2$, or $(2a-2c)x^2 +(2a-2b-2c)xy+ (a-b)y^2$ is a reduced form, and it is not of idoneal type unless $a+b=2c$. Thus $n$ is not idoneal unless $a-b=2c$ or $a+b=2c$. Reversing the roles of $b$ and $c$ we must also have either $a-c=2b$ or $a+c=2b$. These equations are solvable only if $b=c$ and $a=3c$ (so that $n=7b^2$) or $a=5b$ and $c=3b$ (or $a=5c$ and $b=3c$) which leads to numbers $n$ of the form $15b^2$.<|endoftext|> TITLE: What are the outer automorphisms of a Coxeter group? QUESTION [7 upvotes]: I want to know the outer automorphisms of the Weyl group of $\mathrm{E}_8$, if any. But I might as well ask the question more generally. Suppose we have a Coxeter diagram. This gives a Coxeter group. What are the outer automorphisms of this group? It seems we get one from any symmetry of the diagram; are these all of them? (If we were forming the simply connected compact Lie group $G$ from a Dynkin diagram, we'd know every outer automorphism of $G$ comes from a symmetry of the diagram. So, I'm hoping this analogous result is true. But maybe it's too bold a generalization; I'll settle for Coxeter diagrams that come from Dynkin diagrams. The Dynkin diagram of $\mathrm{E}_8$ has no symmetries, of course.) REPLY [13 votes]: It seems we get one from any symmetry of the diagram; are these all of them? No and no. The $A_n$ diagrams have a diagram symmetry of order $2$ for every $n$, but the induced automorphism of $S_{n+1}$ is inner for every $n$ (it's conjugation by the permutation $k \mapsto n - k$ acting on $\{ 0, 1, 2, \dots n \}$, say). On the other hand, $S_6$ has an exceptional outer automorphism. Edit: The outer automorphism group of the Coxeter group $E_8$ appears to have order $2$ and is generated by $w \mapsto w_0^{\ell(w)} w$ where $w_0$ is the longest element, which is central. I found this result in this thesis by William Franszen (which has much more information about this problem), which was the second search result I got for "outer automorphisms of coxeter groups." REPLY [6 votes]: It's useful to look at a short paper by Tits Sur le groupe des automorphismes de certains groupes de Coxeter, J. Algebra 113 (1988), no. 2, 346-357, available online here. While he doesn't treat arbitrary Coxeter groups he does include the ones of interest to you. However, the results for $E_8$ need to be made more explicit. Probably there is more recent literature coming from the study of the isomorphism problem for Coxeter groups. In any case, the main point is to look at "commuting" subsets of the simple generators. The article by Tits is partly based on an earlier one that year in the same journal by L.D. James here.<|endoftext|> TITLE: Orthogonal mud cracks and Maxwell's reciprocal figures QUESTION [15 upvotes]: Is there a succinct mathematical/physical explanation of why mud cracks tend to meet orthogonally?                     Wikipedia image in this article There is considerable literature, which I have only skimmed (I will admit), but so far I have not encountered an intuitive explanation. If anyone can either provide a high-level explanation, or point me to an appropriate source, I would appreciate it. This is unjustified speculation, but I wonder if there is any connection to Maxwell's reciprocal figures:           James C. Maxwell, "On Reciprocal Figures and Diagrams of Forces," 1864. (journal link). REPLY [2 votes]: Suggest you follow this link for a discussion on similar cracking that forms columnar basalt during the cooling of lava. I have collected several samples (from the Mt Baker, WA area) that seem to confirm that the "perfect" shape is that of a hexagon, but that conditions were often not nearly perfect resulting in a variety of other shapes in cross section. Regardless of the cross section shape, the cross section at different locations in the same column may vary in size. Sometimes this variance is considerable. http://blogs.agu.org/georneys/2012/11/18/geology-word-of-the-week-c-is-for-columnar-jointing/<|endoftext|> TITLE: Uniqueness of Complex Orientation of Morava K-theory QUESTION [6 upvotes]: It is known that the $n^{\text{th}}$ Morava $K$-theory at a prime $p$, denoted $K(n)$, is complex oriented. In other words, it admits a theory of Chern classes, or equivalently a morphism of homotopy associative, commutative ring spectra $f:MU\to K(n)$. This complex orientation is related to the height $n$ Honda formal group law in the following way: the complex orientation of $K(n)$ tells us that $K(n)^\ast(\mathbb{C}P^\infty)\simeq \pi_\ast(K(n) )[[x]]$. Recall that there is an element in $x_{MU}\in MU^2(\mathbb{C}P^\infty)$ such that, for the multiplication map $t:\mathbb{C}P^\infty\times\mathbb{C}P^\infty\to\mathbb{C}P^\infty$, $t^\ast(x_{MU})$ is given by the universal formal group law $F_u\in MU_\ast[[x,y]]$. Then our map $f:MU\to K(n)$ maps $x_{MU}$ to some $x_{K(n)}\in K(n)^2(\mathbb{C}P^\infty)$. Then we obtain a formal group law by looking at $t^\ast(x_{K(n)})\in K(n)^\ast(\mathbb{C}P^\infty\times\mathbb{C}P^\infty)$ and choosing an isomorphism $$K(n)^\ast(\mathbb{C}P^\infty\times\mathbb{C}P^\infty)\cong \pi_\ast(K(n))[[x,y]].$$ If we choose any multiplicative map $MU\to K(n)$, will the above construction return the height $n$ Honda formal group law? If not, is it somehow important how we construct $K(n)$ to make sure that the right orientation shows up? Ultimately I'm interested in the moduli space of complex orientations on $K(n)$ and how this relates to formal group laws on $\pi_\ast(K(n))$. Also, would changing the complex orientation (and thus potentially the construction of $K(n)$) affect which $A_\infty$-structure on $K(n)$ with which we end up? REPLY [17 votes]: First, I claim that if we ignore the ring structure and complex orientation, then there is a unique spectrum (up to homotopy equivalence) that deserves to be called $K(n)$. I do not know whether there is a nice way to see this before setting up a lot of chromatic homotopy theory based on a particular choice of $K(n)$. However, once one has set all that up, one can say, for example, that $K(n)$ is the unique spectrum $X$ such that $X=L_nX$ and $X$ is a retract of $X\wedge F$ for some finite spectrum $X$ of type $n$ and $\pi_*(X)\simeq K(n)_*$ as graded abelian groups. Now, there are various constructions that produce different versions of $K(n)$ as ring spectra or $MU$-algebra spectra, with or without an $A_\infty$ structure. Because of the previous paragraph, these can be thought of as giving different ring structures, algebra structures or $A_\infty$ structures on the same object. Now suppose we have a formal group law $F$ of height $n$ over the ring $K(n)_*$, which is compatible with the grading in the usual way. I claim that the standard $K(n)$ admits a ring structure and orientation for which the associated FGL is $F$. Indeed, $F$ gives a ring map $\phi_F\colon MU_*\to K(n)_*$. By thinking about the $p$-series and the grading we see that this must be surjective in nonnegative degrees. Now define $I_F$ to be the kernel of $\phi_F$, and define $v_n$ to be the coefficient of $x^{p^n}$ in the standard FGL over $MU_*$. We then find that $\phi_F$ induces an isomorphism $v_n^{-1}MU_*/I_F\to K(n)_*$. One can check that $I_F$ is automatically generated by a regular sequence. At least at odd primes, it follows that there is a commutative ring object $K_F$ in the category of strict $MU$-modules with $\pi_*(K_F)\simeq v_n^{-1}MU_*/I_F$, and this is unique up to unique isomorphism in the relevant category. The underlying spectrum of $K_F$ is equivalent to the standard $K(n)$, and the claim follows. Slightly weaker statements are still available if $p=2$. One can also ask about the situation where we fix a ring structure on $K(n)$, and look at all the FGLs that we get from maps $MU\to K(n)$ that are compatible with that ring structure. Using the Atiyah-Hirzebruch spectral sequence we see that there exist elements $x\in K(n)^2\mathbb{C}P^\infty$ that restrict to the standard generator in $K(n)^2\mathbb{C}P^1$, and it follows in the usual way that there is at least one ring map $MU\to K(n)$, giving a formal group law $F$ say. After fixing this ring map, a standard line of argument shows that the full set of ring maps $MU\to K(n)$ bijects with the set of $K(n)_*$-algebra maps $K(n)_*MU\to K(n)*$, and thus with the set of formal power series over $K(n)_*$ of the form $x+O(x^2)$. It follows in turn that the full set of possible FGLs that we can get is just the set of FGLs that are strictly isomorphic to $F$. In particular, if we set things up so that $F$ is the Honda FGL, then by changing the orientation we can get all of the uncountably many FGLs that are strictly isomorphic to the Honda FGL, but not equal to it. I do not think that there is any known construction of a specific $A_\infty$ structure on $K(n)$. The original theorem of Alan Robinson just showed that there were uncountably many structures, without specifying a particular one. I think that that theorem did not say anything about the map $MU\to K(n)$ being $A_\infty$. I am not sure what there is in more recent papers. Anyway, I think that the main message is that it does not matter very much which FGL you use, so long as it has height $n$. There are some minor technical advantages in using a $p$-local FGL so that you can exploit the fact that $BP_*$ is much sparser than $MU_*$. Whichever FGL you use on $K(n)_*$, you can always use the universal deformation to get a compatible version of $\widehat{E(n)}$. A theorem of Ando says that this $\widehat{E(n)}$ has a preferred coordinate that lifts the chosen coordinate on $K(n)$ and interacts nicely with power operations; but this works perfectly well for any choice of $K(n)$ coordinate. [UPDATE: I previously attributed the $A_\infty$ structure on $K(n)$ to Andy Baker, but I misremembered that; I have corrected it to Alan Robinson. See also Lennart Meier's comment: Vigleik Angeltveit has shown that the $A_\infty$ structure is unique in a sense different from that studied by Robinson.]<|endoftext|> TITLE: Formal systems needed to formalize relative independence results QUESTION [5 upvotes]: We know that Con(ZF) implies Con(ZFC+GCH), Con(ZF+neg(AC)) and Con(ZFC+neg(CH)). But what are some weak theories in which these relative independence results are provable? In particular, are they provable in PA and even in some interesting proper subtheories of PA? Are there similar relative independence results whose proofs require more than PA or even ZFC? REPLY [14 votes]: Independence results such as the ones mentioned that can be proved using basic syntactic methods (relative interpretation, forcing) can be formalized in the theory $\mathit{PV}_1$ (also known as $T^0_2$, $\mathit{VP}$, and the $\forall\Sigma^b_1$-fragment of $S^1_2$): it has function symbols for all polynomial-time algorithms introduced by means of Cobham’s limited recursion on notation, and it can be axiomatized by definitional equations for the functions, a form of quantifier-free induction, and something to the effect of $0\ne1$ and $x\le1\to x=0\lor x=1$. This theory is as weak as it gets among usable fragments of arithmetic: it is interpretable in Robinson’s arithmetic $Q$ on a definable cut. It is included in $I\Delta_0+\Omega_1$, and a fortiori in stronger fragments of PA. I should perhaps clarify how one deals with forcing, as it is conceptually a model-theoretic method. Let me take $\mathrm{Con}(\mathit{ZFC})\to\mathrm{Con}(\mathit{ZFC}+\neg\mathit{CH})$ for concreteness. In ZFC, you can define the relevant Cohen algebra $B$, and build the Boolean-valued universe $V^B$. In particular, for each formula $\phi(x_1,\dots,x_n)$, you can construct a definable function $\|\phi\|\colon(V^B)^n\to B$ that gives the truth value of $\phi$ in the Boolean universe. Then, in $\mathit{PV}_1$, you prove $\forall\phi\,(\mathit{ZFC}+\neg\mathit{CH}\vdash\phi\Rightarrow\mathit{ZFC}\vdash\|\phi\|=1)$ by induction on the structure of the proof. There are some more demanding relative consistency results: in particular, the consistency of GB relative to ZF, or of $\mathit{ACA}_0$ relative to PA, rely on cut elimination; they are provable in $I\Delta_0+\mathit{SUPEXP}$, but not in $I\Delta_0+\mathit{EXP}$. This is still nowhere near as strong as PA or ZFC. I can’t recall any examples of the latter sort, and somewhat suspect that in most cases, if your relative consistency required such a strong theory, you could as well make it an absolute consistency result.<|endoftext|> TITLE: When do we have a bijection between a proper class A and its power set class P(A)? QUESTION [5 upvotes]: We work in the set theory NBG with the axiom of (local choice but without global (class) choice. For every class A P(A) is the class of all sets x included in the class A. We know that P(A) is a set iff A is a set and a proper class iff P(A) is a proper class. We also know that if A is a set there is no bijection between A and P(A), and that P(P(V))=V, where V is the universal class. It is clear that if there is a bijection between A and V, then there is a bijection between A and P(A). Question: Is it true that if there is a bijection between A and P(A) then there is a bijection between A and V? This question is not interesting under global choice, where all proper classes are bijective. REPLY [5 votes]: Yes, this is provable in NBG. To see this, let $F$ be a one-one function from $\mathcal P(A)$ into $A$. By transfinite recursion on $\in$, we define a function $G$ from $V$ to $A$ such that $G(x) = F(G[x])$. A simple induction then establishes that $G$ is one-one.<|endoftext|> TITLE: Different definitions of spin structures QUESTION [6 upvotes]: This is the definition of spin structure according to Wikipedia: which is supposed to be the standard definition. But in the book The Geometry of Four-Manifolds (Donaldson-Kronheimer, page 76) one finds a rather different definition, at least for a 4-dimensional vector space ($S$ is supposed to be a general two-dimensional complex vector space with Hermitian metric and compatible complex symplectic form): What is the meaning of the second definition? Everything seems quite involved and unrelated. Any idea will be helpful and welcome. REPLY [10 votes]: As it stands, the second definition is a concrete description of the spin group in dimension four. It defines an action of the simply connected group $SU(2)\times SU(2)$ on a vector space of real dimension four, which preserves a positive definite inner product, and this identifies $SU(2)\times SU(2)$ with the universal covering of the special orthogonal group of this four-dimensional Euclidean space. To view it as a definition of a spin-structure on a manifold, one has to do this in each point of the manifold. This means that the spin strucuture in this sense is given by two auxiliary complex rank two bundles $S^+$ and $S^-$ which are endowed with Hermitian bundle metrics and compatible complex symplectic forms, togehter with an isomorphism between the tangent bundle and the bundle $Hom_J(S^+,S^-)$ which respects the inner products on the two spaces (the given Riemannian metric and the inner product constructed point-wise as in definition 2). The equivalence between the two definitions is obtained as follows: To go from definition 1 to definition 2, you form the associated bundles corresponding to the two basic (complex) spin-representations of $Spin(n)$. In the opposite direction, you form the frame bundle of $S^+\oplus S^-$ (with structure group $SU(2)\times SU(2)$, which is isomorphic to $Spin(n)$) and the identification of $Hom_J(S^+,S^-)$ with $TM$ preserving bundle metrics shows that this bundle is a two fold covering of the $SO(n)$-frame bundle associated to the Riemannian metric.<|endoftext|> TITLE: A question about symmetric matrix QUESTION [6 upvotes]: Let $A= (a_{ij})_{ij}, 1 \leq i, j \leq n$ be a symmetric $n \times n$ matrix. Suppose (1) $a_{ij} \geq 0$ are real numbers; (2) The sum of each row $\sum_{j=1}^{n} a_{ij} = 1$ for $1 \leq i \leq n$. Then I want to show the following: there must exists a nonzero $\prod_{i=1}^n a_{i, \sigma(i)}$, where $\sigma \in S_n$ is an element of the symmetric group $S_n$. In other words, there must exists a nonzero summand in the expression of $\det A$. REPLY [2 votes]: An other solution (but Darij's is enough). It is classical that if all these products are zero, then $A$ contains a submatrix $k\times\ell$ of zeros, with $k+\ell>n$ (see for instance Exercise 7, Chapter 3 of my book, 2nd edition). Then an elementary calculus shows that the sums of entries in the opposite block equals $n-k-\ell$. This is negative, contradicting the fact that the entries are non-negative.<|endoftext|> TITLE: Probability that a self-avoiding walk on $\mathbb{Z}^3$ closes to a polygon QUESTION [6 upvotes]: The probability that a random walk on $\mathbb{Z}^3$ returns to the origin is about 34%. This is (part of) Pólya's theorem. I have been looking for an analogous (numerical) result for the probability that a random "myopic" self-avoiding walk on $\mathbb{Z}^3$ returns to the origin to form a self-avoiding polygon. By a "myopic self-avoiding walk," I mean a walker who takes a sequence of random (unit) steps, each restricted to avoid previously visited sites (to quote Yoav Kallus's and Vincent Beffara's comments). Any other related information—e.g., probability of the walk ending in a cul-de-sac before $n$ steps (see below)—would be welcomed, as would either small-$n$ results or asymptotics.             Origin: green. Cul-de-sac: red, reached after 335 steps. A 2D version of this question was posed earlier, but not entirely answered: "Probability that a “closable” self-avoiding random walk forms a polygon" REPLY [3 votes]: A little data from a limited Monte Carlo simulation (as suggested by Ben Crowell). Letting paths run up to $2000$ steps, I find: 56% are still free after $2000$ steps. 33% have ended in a cul-de-sac before reaching $2000$ steps. 11% close to a self-avoiding polygon. The 11% doesn't contradict Ben's $\frac{1}{6}$ because I did not count the doubly covered segment EW as a self-avoiding polygon. I wanted to restrict attention to simple polygons; for my count, a $1 \times 1$ square is the smallest polygon. Here is one self-avoiding polygon, of length $433$ steps:       Next question: What is the knot complexity (say, expected crossing number) of these self-avoiding polygons? Difficult to tell in the example above, but it may well be the unknot.<|endoftext|> TITLE: When $C(X)$ is an injective $C(X)$-module? Current answer is erroneous QUESTION [7 upvotes]: It is an old question if every injective Banach space is isomorphic as Banach space to $C(X)$-space. I would like to know if the weakened module version of this question is answered. More precisely: For which compact Hausdorff spaces $X$ the module $C(X)$ is an isomorphically injective $C(X)$-module. I know that for $X$ Stonean $C(X)$ is even an isometrically injective $C(X)$-module. REPLY [7 votes]: It's equivalent to that $C(X)$ is isometrically ijective (i.e., $X$ is stonean). We take the definition cited by Yemon in the comment and let $\iota\colon C(X)\hookrightarrow C(Y)$ be a faithful $*$-homomorphism from $C(X)$ into an injective abelian $C^*$-algebra $C(Y)$. Then, there is a $C(X)$-module projection $\Phi$ from $C(Y)$ onto $C(X)$. In particular, $C(X)$ is isomorphically injective, but Huruya'e example (Proc AMS 1984) says isomorphic injectivity need not imply isometric injectivity (though his example is isomorphic as a Banach space to an isometrically injective one). We will exploit the fact that the projection $\Phi$ is a $C(X)$-module map. Let $\Phi^*\colon X \to C(Y)^*$ be the continuous map define by $\langle\Phi^*(x),f\rangle = \Phi(f)(x)$. Here $C(Y)^*$ is equipped with the weak$^*$-topology. Note that $\|\Phi(x)\| \geq 1$ for every $x\in X$, and define a continuous map $\Psi^*\colon X\to C(Y)^*$ by $\Psi^*(x)=|\Phi^*(x)|/\|\Phi^*(x)\|$. This in turn gives rise to a unital positive contractive map $\Psi\colon C(Y) \to C(X)$, which is given by $\Psi(f)(x) = \langle\Psi^*(x),f\rangle$. We claim that $\Psi$ is a projection onto $C(X)$. Let $h\in C(X)_+$ be given arbitrary. Since $\Phi(\iota(h)f)(x)=h(x)\Phi(f)(x)$ for all $f\in C(Y)$, one has $\iota(h)\Phi^*(x)=h(x)\Phi^*(x)$ as an element in $C(Y)^*$ (which is also viewed as a complex Radon measure on Y). Since $\iota(h)\geq0$, this implies that $\iota(h)|\Phi^*(x)|=|\iota(h)\Phi^*(x)|=h(x)|\Phi^*(x)|$ and so $\iota(h)\Psi^*(x)=h(x)\Psi^*(x)$ for every $x\in X$. This means that $\Psi(\iota(h))(x)=h(x)$ for every $x \in X$.<|endoftext|> TITLE: Generalization of Giroux's Theorem for Higher Dimensions? QUESTION [5 upvotes]: Just wanted to know if Giroux's theorem for 3-dimensional contact manifolds can be generalized: In contact geometry for manifolds of dimension 3 , we have Giroux's theorem , stating that for any compact, oriented manifold there is a bijection between the set of contact structures up to isotopy and the set of open book decomposition up to the operation of positive stabilization (meaning that any two stabilized manifolds give rise under Giroux's theorem, to the same contact structure up to isotopy). Open books are known to exist for all odd-dimensional manifolds. Are there similar relationships on (2n+1)-manifolds, n>1 say compact and oriented, between contact structures and open books, or between open books and some other property? Thanks. REPLY [6 votes]: Giroux proved that for every contact manifold $(M^{2n+1},\alpha)$, there exists an open book supporting the contact form. An open book consists of a codimension two submanifold $K^{2n-1}\subseteq M^{2n+1}$ and a fibration $M\setminus K\to S^1$ (which is "standard" in some tubular neighborhood of $K$) with fibers $F_t$. Such an open book is said to support a contact form $\alpha$ if: $\alpha$ restricts to a contact form on the binding $K^{2n-1}\subseteq M^{2n+1}$. $d\alpha$ is a symplectic form on the pages $F_t$, and the associated Liouville vector field $X_\alpha$ on $F_t$ is outward pointing along $\partial F_t$ (equivalently, the orientation on $K$ induced by the contact form $\alpha$ agrees with the orientation on $K=\partial F_t$ induced by the symplectic form $d\alpha$ on $F_t$). See http://arxiv.org/abs/math/0305129 for the proof. Conversely, any exact symplectomorphism $\varphi$ of a Liouville domain $(F,d\lambda)$, gives rise to a contact structure on the resulting open book with page $F$ and monodromy $\varphi$. The question of whether there is a unique (up to positive stabilization) open book supporting a given contact structure is (as you mention) a theorem of Giroux in dimension three and is open in higher dimensions.<|endoftext|> TITLE: What is a good reference for conormal distributions? QUESTION [6 upvotes]: May I humblely ask what is a good reference for conormal distributions (for student with some rudimentary pseudo-differential operator background)? I heard from my advisor that it is useful in index theory, but most of the lecture notes I read are quite opaque (like Simanca's) by using baroque notations, and it is difficult to see the theory in transparent manner without going over all the technical details. For a concrete example, it is not clear to me how to compute the principal symbol of the product of two conormal distributions. I assume this is a subject well known to experts(like Richard Melrose, Rafe Mezzeo, etc), but I found it difficult to find a good reference that does not assume I know most of the basic theory already. Update: I know Hormander's book is the Bible in our field. It is just I do not know if I will have enough time to digest it properly. Is there any alternatives? REPLY [5 votes]: I think the third volume of Hormander's The Analysis of Linear Partial Differrential Operators is a good reference. chapter 18 is an introduction to PsDOs, especially, in section 18.2, it's devoted to conormal distributions, which I think is very useful for people with PsDO background. He starts with the fact that the wave front set of the kernel of an PsDO is contained in the Conormal distribution, and further discussed some regularity properties of such distributions, which in turn gives an invariant definition on manifolds. In addition, you will see what conormal distributions look like locally in this section as well.<|endoftext|> TITLE: Does a left basis imply a right basis, without AC? QUESTION [8 upvotes]: If $_DV_D$ is a $D$-$D$-bimodule, and we have a $D$-basis for $V_D$, do we still need AC to get a $D$-basis for $_DV$? (The original question appears below. But this shorter question gets at the heart of my question, and makes it clear it has more logical foundations.) Let $D$ be a division ring and let $_D V_D$ be a $D$-$D$-bimodule. If we temporarily forget the left module structure, and just look at the right $D$-module structure, we have $V_D= \bigoplus_{i\in I}e_iD$ for some basis $\{e_i\}_{i\in I}$. It is a well-known fact that $E:={\rm End}(V_D)\cong {\rm CFM}_I(D)$ where ${\rm CFM}_I(D)$ is the ring of $I\times I$ column finite matrices. These are the matrices where each column has only finitely many nonzero entries. If we think of the elements of $V_D$ as columns of size $I\times 1$ with only finitely many nonzero entries, then ${\rm CFM}_I(D)$ acts on the left of $V_D$ simply by matrix multiplication. (Of course, when $|I|=n$ is finite, then ${\rm CFM}_I(D)=\mathbb{M}_n(D)$ is just the usual ring of $n\times n$ matrices.) So we have a natural bimodule structure on $V$, namely $_{E}V_D$. Our original bimodule structure $_DV_D$ gives rise to a homomorphism $\varphi:D\to E\cong {\rm CFM}_I(D)$. Conversely, given any such homomorphism (and a fixed basis for $V_D$ indexed by $I$) we get a $D$-$D$-module structure on $V$. We could do all of this over again on the other side. From the left $D$-module structure $_DV$, we can fix a basis $\{f_j\}_{j\in J}$ and corresponding decomposition $_DV=\bigoplus_{j\in J}Df_j$. The right $D$-module structure then corresponds to a homomorphism $\psi:D\to {\rm RFM}_J(D)$. (The ring ${\rm RFM}_J(D)$ is the ring of $J\times J$ row finite matrices.) So given an index set $I$ and a homomorphism $\varphi:D\to {\rm CFM}_I(D)$, there is a corresponding index set $J$ and a homomorphism $\psi:D\to {\rm RFM}_J(D)$. My question is whether there is a canonical way to describe the correspondence $(I,\varphi)\leftrightarrow(J,\psi)$. If not a canonical way, given the information $I$ and $\varphi$, can we at least describe $|J|$ and $\psi$ explicitly from that data, after a choice of basis? REPLY [3 votes]: This is a very incomplete answer, but maybe others can fill in the gaps (and I'll try to). [Edit: I've not been able to make this idea work, although the ideas may lead somewhere, so I'll leave this here. Below I've added some more specific comments about the difficulties I found.] First, a reminder of the construction in Andreas Blass' beautiful proof that in ZF "every vector space over every field has a basis" implies AC. Let $k$ be a field, and suppose that the set $X$, which is the disjoint union of subsets $X_i$, contradicts the axiom of multiple choice (i.e., it's not possible to simultaneously choose finite subsets of each $X_i$). If $K$ is the subfield of $k(X)$ consisting of elements homogeneous of degree zero in each $X_i$, and $V$ is the $K$-subspace of $k(X)$ spanned by $X$, then $V$ does not have a $K$-basis. Here's my idea. Suppose: (a) $k\cong K$ as fields, and (b) $V$ has a $k$-basis. Then $V$, considered as a $k$-bimodule with standard left action, and with right action via the isomorphism of $k$ with $K$, has a left basis but no right basis. Condition (a) can be satisfied by taking $k$ to be the rational functions over some field in a countable union of copies of $X$, homogeneous of degree zero in each copy of each $X_i$, and $K$ the same but with one extra copy of $X$. I don't know if (b) is necessarily true without AC, but I think I see a proof (but haven't checked the details) that it's true if every set has a total order, which is strictly weaker than AC. If I'm right, then this shows that in ZF a $k$-bimodule with a left basis may not have a right basis, but doesn't prove that AC is equivalent to the absence of a counterexample. [Edit: I haven't been able to write down an explicit basis, but on the other hand I haven't convinced myself that it's impossible. When I tried, I kept wanting to write homogeneous rational functions in terms of homogeneous elements $y/x$ where $x,y\in X_i$ after making a choice of $x\in X_i$ for each $i$, which is not so helpful when the $X_i$ are supposed to contradict AC! Maybe this is a hint that this method can't work. I didn't see that it's obviously easier to prove that $K$ has a $k$-basis, and if it provably doesn't, then this also answers the question, using $K$ instead of $V$.] A simpler but related question that I don't know the answer to: Let $k$ be a field and $Y$ a set. In ZF without AC, must $k(Y)$ have a $k$-basis? If not, how strong a version of AC is necessary? As before, I think I see how to prove it if $Y$ has a total order.<|endoftext|> TITLE: Roots of a polynomial in a finite field related to Fermat's Last Theorem QUESTION [7 upvotes]: In my class, we proved the following condition: define the polynomial $P_l(x)$ as $$P_l(x) = \sum_{r=1}^{l-1}{\frac{1}{r}x^{l-1-r}}$$ Then if for all $a \in \mathbb{Z}/l\mathbb{Z}-\{0,1\},$ $P_l(x)$ does not vanish (mod $l$), then the First Case of Fermat's Last Theorem(1CFLT) is true for $\mathbb{Z}$ and exponent $l.$ It appears that the polynomials, $P_l(x)$ always have a solution in $\mathbb{Z}/l\mathbb{Z}-\{0,1\}$ whenever $l \equiv 1(\text{mod 3}).$ For instance, $P_7(x) \equiv x^5+ 4 x^4 + 5 x^3+ 2 x^2 + 3 x+6 (\text{mod 7})$ has roots $3,5.$ I have tested this on Sage for primes $l \equiv 1(\text{mod 3})$ less than $1000$. Does this hold for all primes $l \equiv 1(\text{mod 3})$? REPLY [20 votes]: This is true. Let $P(x) = \sum_{n=1}^{l-1} \frac{x^n}{n} \in {\mathbb F}_l[x]$ (this is $x^{l-1} P_l(x^{-1})$). Then we have $P(1-x) = P(x)$ (check that they have the same derivative and the same value at $1/2$) and $x^l P(x^{-1}) = -P(x)$. Now if $l \equiv 1 \bmod 3$, then there is a primitive sixth root of unity $\omega \in {\mathbb F}_l$. Note that $1-\omega = \omega^{-1}$. We then obtain $$ \omega P(\omega^{-1}) = \omega^l P(\omega^{-1}) = -P(\omega) $$ and $$ P(\omega^{-1}) = P(1 - \omega) = P(\omega) .$$ So $(1+\omega) P(\omega^{-1}) = 0$, which implies (since $\omega \neq -1$) that $P(\omega) = P(\omega^{-1}) = 0$. These two are actually double roots, since the derivative also vanishes. In fact, this argument shows that $x^2 - x + 1$ divides $P(x)$ whenever $l > 3$. Another remark: Since $$ P'(x) = 1 + x + x^2 + \ldots + x^{l-2} = \frac{x^{l-1}-1}{x-1} = \prod_{a \in {\mathbb F}_l \setminus \{0,1\}}(x-a), $$ a root $a \in \overline{\mathbb F}_l$ of $P$ is a multiple root (and then of multiplicity 2) if and only if $a \in {\mathbb F}_l \setminus \{0,1\}$. (Added later:) One can also give a "combinatorial" proof. The relations satisfied by $P$ imply that the roots of $P$ come in orbits of the $S_3$-action on ${\mathbb P}^1_{\mathbb F_l}$ generated by $a \mapsto 1-a$ and $a \mapsto a^{-1}$ (we think of $P$ has a polynomial of degree $l$ with "leading coefficient" zero, so that $\infty$ is a root). These orbits have size 6, with only three exceptions: $\{0,1,\infty\}$, $\{-1,2,1/2\}$ and $\{\omega, \omega^{-1}\}$ (this uses $l > 3$). We always have the first orbit, which leaves $l-3$ roots. Any orbit other than the third special orbit will contribute a multiple of 3 to the number of roots. So the contribution of the two-element orbit is $\equiv l \bmod 3$, and it can only be 0, 2 or 4. So for $l \equiv 1 \bmod 3$, it must be 4, implying that both are double roots and hence in ${\mathbb F}_l$, and if $l \equiv 2 \bmod 3$, the contribution must be 2, so we have simple roots and they are not in ${\mathbb F}_l$. (This actually proves that $\omega \in {\mathbb F}_l$ iff $l \equiv 1 \bmod 3$ without using that ${\mathbb F}_l^\times$ is cyclic.) (Added Dec 11, 2014:) As Gjergji Zaimi points out in a comment below, we have that $P(x)$ is (for $l > 2$) the image of $$Q_l(x) = \frac{(x-1)^l - (x^l-1)}{l} \in \mathbb Z[x]$$ in ${\mathbb F}_l[x]$ (use that $\frac{1}{l}\binom{l}{k} = \frac{1}{k} \binom{l-1}{k-1} \equiv (-1)^{k-1}/k \bmod l$). For any prime $l > 3$, $Q_l(\omega) = 0$ (where now $\omega \in \mathbb Q(\sqrt{-3})$), which implies that $P(\bar\omega) = 0$ (where $\bar\omega$ is the image of $\omega$ in $\mathbb F_l$ or $\mathbb F_{l^2}$). This also explains the relation to FLT. (In fact, the observation that $\omega^l + (\omega^{-1})^l = 1$ gives an $l$-adic first case solution of Fermat's equation when $l \equiv 1 \bmod 3$ gives another proof, assuming the statement on FLT in TZE's question.)<|endoftext|> TITLE: What are the current views on consistency of Reinhardt cardinals without AC? QUESTION [16 upvotes]: It's well known that Reinhardt cardinals are inconsistent, provided that we have access to axiom of choice, but, as far as I know, we are clueless about this when we don't assume choice. For me, the fact that Reihardt cardinals are inconsistent with choice already makes them, in a way, "wrong", that is, I'd believe personally that Reinhardt cardinals cannot exist in $V$ anyways. However, I have never seen any claims like this. So my question is: Is it believed that Reinhardt cardinals can be consistent without choice? Are there any solid arguments for/against them? (other than "we haven't found them inconsistent, so they should be consistent") Thanks in advance. Note: I added soft-question tag because this question is more about philosophical views. Feel free to remove it if you think otherwise. REPLY [15 votes]: The following theorems of Woodin may be related: Theorem. ($ZF$) Assume that $ZFC$ proves the $HOD$ Conjecture. Suppose $\delta$ is an extendible cardinal. Then for all $ \lambda>\delta$ there is no non-trivial elementary embedding $j : V_{\lambda+2}\to V_{\lambda+2}$ Thus (assuming that $ZFC$ proves the $HOD$ Conjecture) one nearly has a proof of Kunen's inconsistency theorem without using the Axiom of Choice. Theorem. Assume $ZF$+ there exists a Reinhardt cardinal + there exists a proper class of supercompact cardinals is consistent. Then there exists a genric extension of the universe which satisfies $ZF$ + the axiom of choice + there exists a proper class of supercompact cardinals, and such that in it Woodin's $HOD$ conjecture fails. As far as I know, Woodin believes the $HOD$ conjecture is true (at least the current methods can not be used to solve the problem), so by the above theorem we may expect to show that at least "$ZF$+ there exists a Reinhardt cardinal + there exists a proper class of supercompact cardinals" is not consistent.<|endoftext|> TITLE: Parity of self-linking QUESTION [7 upvotes]: For a class $x$ in $H_{2k}$ of a 4k-manifold $M$, the self-intersection $x.x$ agrees mod 2 with the cap product of $x$ with the Wu class $v_{2k}$. If instead $x$ is a torsion element of $H_{2k}$ of a (4k+1)-manifold, the self-linking of $x$ defines an element of order 2. Is this equal to the inner product of $x$ with $v_{2k}$? REPLY [8 votes]: Yes. For any space $M$ there is defined a torsion linking pairing on the $Q/Z$-coefficient cohomology groups $$L:H^m(M;Q/Z) \times H^n(M;Q/Z) \to H^{m+n+1}(M;Q/Z);(a,b) \mapsto \delta a \cup b = a \cup \delta b$$ with $\delta:H^p(M;Q/Z)\to H^{p+1}(M)$ the Bockstein coboundaries for $p=m,n$, such that $$\text{image}(\delta)=\text{kernel}(H^{p+1}(M)\to H^{p+1}(M;Q))=\text{torsion}(H^{p+1}(M)).$$ This pairing is $(-1)^{(m+1)(n+1)}$-symmetric $$L(a,b)=(-1)^{(m+1)(n+1)}L(b,a) \in H^{m+n+1}(M;Q/Z).$$ If $M$ is a closed connected oriented $(m+n+1)$-dimensional manifold this gives a $(-1)^{(m+1)(n+1)}$-symmetric torsion linking pairing $$L:H^m(M;Q/Z) \times H^n(M;Q/Z) \to H^{m+n+1}(M;Q/Z)=Q/Z$$ which induces the usual nonsingular linking pairing on the torsion groups $$L:{\rm torsion}(H^{m+1}(M)) \times {\rm torsion}(H^{n+1}(M)) \to Q/Z.$$ For a $(4k+1)$-dimensional manifold $M$ and $m=n=2k$ this linking pairing is skew-symmetric. The self-linking defines a linear map $$H^{2k}(M;Q/Z) \to Q/Z~ ;~ a \mapsto L(a,a)=0~\text{or}~1/2$$ which is represented by the image of the Wu class $v_{2k}(M) \in H^{2k}(M;Z_2)$ under the injection $Z_2 \to Q/Z~;~1 \mapsto 1/2$. The Poincare dual of a torsion homology class $x \in H_{2k}(M)$ is of the form $\delta a\in H^{2k+1}(M)$ for some $a \in H^{2k}(M;Q/Z)$ and $$\text{self-linking}(x)=L(a,a)= \delta a \cup v_{2k}(M) = \langle x,v_{2k}(M) \rangle \in Z_2\subset Q/Z.$$ [Subsequent editing 17.1.2015] On page 49 of the 1976 AMS Memoir "A product formula for surgery obstructions" by John Morgan it is stated without proof that for a $(4k+1)$-dimensional manifold $M$ $$L(x,x) ~=~\langle x,v_{2k}(M)\rangle \in Z_2 \subset Q/Z~{\rm for}~ x \in {\rm torsion}(H_{2k}(M))~.$$ It must be confessed that the above "proof" does not go into sufficient detail. So here is a direct proof of Morgan's formula. Let $C=C(M)$ be the chain complex of $M$, $C^*={\rm Hom}_Z(C(M),Z)$ the cochain complex, and consider the cup$_0$ chain map and the cup$_1$ chain homotopy $$\phi_0 ~=~ [M]\cap -~ :~ C^{4k+1-*} \to C~,~\phi_1 ~:~ \phi_0~ \simeq~ T\phi_0~ :~ C^{4k+1-*} \to C$$ given by the evaluation of an extended diagonal chain map on a fundamental cycle $[M] \in C_{4k+1}$, with $T$ the signed transposition involution on ${\rm Hom}_Z(C^*,C) = C\otimes_ZC$. In particular there are defined $Z$-module morphisms $$\phi_0~:~C^{2k+1} \to C_{2k}~,~ \phi'_0:C^{2k} \to C_{2k+1}~,~ \phi_1~:~C^{2k+1} \to C_{2k+1}$$ such that $$d \phi'_0~=~ \phi_0 d^* ~:~ C^{2k} \to C_{2k}~,$$ $$\phi_0 + \phi'{_0}^*~ =~ d\phi_1+\phi_1 d^*~ :~ C^{2k+1} \to C_{2k}$$ $$d\phi_0 + (d\phi_0)^*~ =~ d\phi_1d^*~ :~ C^{2k} \to C_{2k}~.$$ The intersection number of cohomology classes $x \in H^{2k+1}(M)$, $y \in H^{2k}(M)$ is given by $$\langle x \cup y,[M]\rangle ~=~\phi_0(x)(y) \in Z~ .$$ Also $$\langle Sq^{2k}(x),[M]\rangle ~ =~ \phi_1(x)(x)~=~\langle v_{2k}(M) \cup x,[M]\rangle \in Z_2~.$$ The linking number of torsion cohomology classes $x,z \in {\rm torsion}(H^{2k+1}(M))$ is $$L(x,z)~ =~ \phi_0(d^*)^{-1}(x)(z) \in Q/Z~.$$ In view of the formal identity $$\phi_0(d^*)^{-1} + (\phi_0(d^*)^{-1})^*~ =~ \phi_1$$ the self-linking number of $x=z$ $$L(x,x)~ =~ \phi_0(d^*)^{-1}(x)(x) \in Q/Z$$ is represented by a rational number $r \in Q$ such that $$2r~ =~ \phi_1(x)(x) \in Z \subset Q$$ so that $r=0$ or $1/2 \in Q/Z$, according to the value of $\phi_1(x)(x) \in Z_2$, and $$L(x,x)~ =~ \langle v_{2k}(M) \cup x,[M]\rangle \in Z_2 \subset Q/Z~.$$ For a fuller account of the generalized Wu-Thom relationship between the symmetric structure on a manifold (resp. geometric Poincare complex) and the Wu classes of the normal bundle (resp. Spivak normal fibration) see section 9 of my 1980 Proc. LMS paper The algebraic theory of surgery II. http://www.maths.ed.ac.uk/~aar/papers/ats2.pdf For a fuller account of the relationship between intersection and linking numbers and the Wu classes check out section 3.3 of my 1981 Princeton book Exact sequences in the algebraic theory of surgery http://www.maths.ed.ac.uk/~aar/books/exacsrch.pdf<|endoftext|> TITLE: Is any connected fibre of a fibration of a sphere also a sphere? QUESTION [13 upvotes]: Is it known whether, for any fibration of a sphere with connected fibres, the fibre has to be a sphere? $S^1$, $S^3$ or $S^7$? REPLY [20 votes]: It was shown in Browder, William. Fiberings of spheres and $H$-spaces which are rational homology spheres. Bull. Amer. Math. Soc. $\bf 68$ 1962 202–203. that for any fibration $F \to S^n \to B$, where $B$ is a non-trivial polyhedron and $F$ is connected, $F$ must have the homotopy type of $S^1,S^3$ or $S^7$. This builds on previous work of Spanier-Whitehead and Borel.<|endoftext|> TITLE: Why do Pell equations appear in Ramanujan's pi formulas? QUESTION [50 upvotes]: While answering this MSE question about the Pell equation $x^2-29y^2=1$, I noticed that certain fundamental solutions appeared in Ramanujan's famous pi formula. I. Given the fundamental unit $\displaystyle U_{29} =\tfrac{5+\sqrt{29}}{2}$ and, $$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$ $$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot1820^2=1$$ $$2^6\left(\big(U_{29}\big)^6+\big(U_{29}\big)^{-6}\right)^2 =\color{blue}{396^4}$$ we find those integers all over Ramanujan's, $$\frac{1}{\pi} = \frac{2 \sqrt 2}{\color{blue}{9801}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot\color{blue}{70\cdot13}\,k+1103}{\color{blue}{(396^4)}^k}$$ The same integers appear in related pi formulas in this post. II. A similar thing happens with the Chudnovsky formula. I knew that H. Chan explored $x^2-3dy^2 = 1$ in relation to $j(\tau)$. Given the fundamental unit for $n=3d=3\times163=489$, $$U_n =u+v\sqrt{489} =7592629975+343350596\sqrt{489} = \big(35573\sqrt{3}+4826\sqrt{163}\big)^2$$ so $u^2-489v^2=1$. The units had a common form for $d=19,43,67,163$, for the last being, $$U_n = \left(\tfrac{1}{18}(\color{brown}{640320}-6)\sqrt{3}+4826\sqrt{163}\right)^2$$ and some experimentation yielded, $$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+6 = \color{brown}{640320}$$ $$3\sqrt{3}\big(U_n^{1/2}-U_n^{-1/2}\big)+18 = \color{brown}{2^2\cdot3^3\cdot7^2\cdot11^2}$$ $$\sqrt{163}\big(U_n^{1/2}+U_n^{-1/2}\big) = \color{brown}{2^2\cdot19\cdot127\cdot163}$$ and the Chudnovsky formula, $$12\sum_{k=0}^\infty (-1)^k \frac{(6k)!}{k!^3(3k)!} \frac{\color{brown}{2\cdot3^2\cdot7\cdot11\cdot19\cdot127\cdot163}\,k+13591409}{(\color{brown}{640320}^3)^{k+1/2}} = \frac{1}{\pi}$$ The same thing happens with the other $d$, for example, $$U_{201} = \left(\tfrac{1}{18}(\color{brown}{5280}-6)\sqrt{3}+62\sqrt{67}\right)^2$$ so it is not a fluke. Plus, these $U_n$ are denominators in Ramanujan-Sato pi formulas of level 9. Q. Anyone knows the reason why a fundamental unit $U_{n}$ would appear in a pi formula? REPLY [6 votes]: I think is simple. All are due to algebraic units. If $z=a+b\sqrt{d}$ is a unit and $N(z)=a^2-d\cdot b^2$, then $N(z)=1$ (unit). Hence if we deal with algebraic units, then $N(z)=1=a^2-d\cdot b^2$, is a Pell equation. See paper [1] (the knowledge of elliptic singular modulus $k_r$ requires the knowledge of algebraic units). It have been proven that most of these formulas (Ramanujan formulas for $1/\pi$, $1/\pi^2$, etc) rise from elliptic functions (see [3],[5],[6]). For example: Set $$ \phi(z):={}_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,z\right)=\frac{4K^2\left(\sqrt{\frac{1}{2}\left(1-\sqrt{1-z}\right)}\right)}{\pi^2} $$ Then we ask whether exist constants $a_1,b_1$ and $g$ such that $$ \sum^{\infty}_{n=0}\frac{\left(\frac{1}{2}\right)_n^3}{(n!)^3}z^n(a_1n+b_1)=\frac{g}{\pi}\leftrightarrow b_1\phi(z)+a_1\phi'(z)=\frac{g}{\pi}. $$ Setting $w=\sqrt{\frac{1}{2}\left(1-\sqrt{1-z}\right)}$, then $1-2w^2=\sqrt{1-z}$. Also by direct calculation $$ \sum^{\infty}_{n=0}\frac{\left(\frac{1}{2}\right)_n^3}{(n!)^3}4^n(w-w^2)^n(a_1n+b_1)=\frac{4K(w)\left[a_1E(w)+(b_1-a_1(1-w)-2b_1w)K(w)\right]}{\pi^2(1-2w)} $$ Here ofcourse $K(w)$ and $E(w)$ are the complete elliptic integrals of the first and second kind. Also $$ \alpha(r)=\frac{\pi}{4K^2(k_r)}-\sqrt{r}\left(\frac{E(k_r)}{K(k_r)}-1\right) $$ where $k_r$ is the elliptic singular modulus i.e the solution of $$ \frac{K(\sqrt{1-k_r^2})}{K(k_r)}=\sqrt{r}, r>0. $$ Also $$ \frac{dK(x)}{dx}=\frac{E(x)}{x(1-x^2)}-\frac{K(x)}{x} $$ Hence setting $w=k_r$, $a_1=1$ and $b_1=\left(\frac{\alpha(r)}{\sqrt{r}}-k_r^2\right)\left(1-2k_r^2\right)^{-1}$ we arrive to the general formula $$ \sum^{\infty}_{n=0}\frac{\left(\frac{1}{2}\right)_n^3}{(n!)^3}4^n(k_rk'_r)^{2n}\left(n+\frac{\alpha(r)-k_r^2\sqrt{r}}{\sqrt{r}(1-2k_r^2)}\right)=\frac{1}{\pi\sqrt{r}(1-2k_r^2)} $$ Evaluations of $k_r$ and $\alpha(r)$ can given for positive rationals $r$. For example with $r=2$, we have $k_2=\sqrt{2}-1$, $\alpha(2)=\sqrt{2}-1$. Hence we get the next formula $$ \sum^{\infty}_{n=0}\frac{\left(\frac{1}{2}\right)_n^3}{(n!)^3}(40\sqrt{2}-56)^{n}\left(n+\frac{2}{7}-\frac{1}{7\sqrt{2}}\right)=\frac{8+5\sqrt{2}}{14\pi} $$ For $r$ positive integer we can find $k_r$ and $\alpha(r)$ using methods of [1],[2]. For a table of values of $k_r$ from r=1 to 100 see [4]. For evaluations of $\alpha(r)$ see[5] [1]: Mark B. Villarino: "Ramanujan most singular modulus". arXiv:math/0308028v4 [math.HO] 2005. [2]: D. Broadhurst. "Solutions by radicals at Singular Values $k_N$ from New Class Invariants for $N\equiv3(\textrm{mod})8$". arXiv:0807.2976 [math-ph], (2008). [3]: N.D. Bagis, M.L. Glasser. "Ramanujan type $1/\pi$ approximation formulas". Journal of Number Theory, Elsevier., (2013) [4]: J.M. Borwein, M.L. Glasser, R.C. McPhedran, J.G. Wan, I.J. Zucker. "Lattice Sums Then and Now". Cambridge University Press. New York, (2013). [5]: J.M. Borwein and P.B. Borwein. "Pi and the AGM". 1987 ed., New York: John Wiley and Sons Inc., (1987) [6]: N.D. Bagis. "A General Method for Constructing Ramanujan-Type Formals for Powers of $1/\pi$". The Mathematica Journal. Vol. 15., (2013) ...CONTINUED From Chan Huang's Unit Theorem we have that: If $n\equiv 2(mod4)$, then $k_n$ is unit. From relation $$ \frac{(k'_n)^2}{k_n}=2g_n^{12}\textrm{, }k'_n=\sqrt{1-k_n^2} $$ where $g_n$ is Weber's invariant: $$ g_n=2^{-1/4}q^{-1/24}\prod^{\infty}_{m=0}(1-q^{2m+1})\textrm{, }q=e^{-\pi\sqrt{n}}. $$ Now if $\delta_1,\delta_2,\ldots,\delta_{\nu(-8n)}$ are the distinct odd divisors of $-8n$ and $\delta'_l$ are the complimentary divisors of $\delta_l$, such $\delta'_l\delta_l=-8n$ $$ g_{2n}=\prod^{\nu(-8n)}_{l=1}\left(\frac{T_l+U_l\sqrt{\delta_l}}{2}\right)^{\nu(\delta_l)\nu(\delta'_l)/\nu(-8n)}, $$ where $\nu(n)$ is the number of properly primitive classes of discriminant of $n$ and $(T_l,U_l)$ is the minimal solution of the Pell equation $$ x^2-\delta_l y^2=4 $$ NOTES. LEMMA 1. If $uv:=g_n^6$ $2U:=u^2+u^{-2}$, $2V=v^2+v^{-2}$ $W=\sqrt{U^2+V^2-1}$, $2S=U+V+W+1$, then $$ k_n^2=(\sqrt{S}-\sqrt{S-1})^2(\sqrt{S-U}-\sqrt{S-U-1})^2(\sqrt{S-V}-\sqrt{S-V-1})^2\times $$ $$ \times(\sqrt{S-W}-\sqrt{S-W-1})^2. $$ LEMMA 2. If $\sqrt{\alpha}:=\sqrt{ab}+\sqrt{(a+1)(b-1)}$ $\sqrt{\beta}:=\sqrt{cd}+\sqrt{(c-1)(d-1)}$, then $$ x_1:=(\sqrt{a+1}-\sqrt{a})(\sqrt{b}-\sqrt{b-1})(\sqrt{c}-\sqrt{c-1})(\sqrt{d}-\sqrt{d-1}) $$ $$ x_2:=-(\sqrt{a+1}+\sqrt{a})(\sqrt{b}+\sqrt{b-1})(\sqrt{c}+\sqrt{c-1})(\sqrt{d}+\sqrt{d-1}) $$ are roots of $$ x-x^{-1}=2(\sqrt{\alpha\beta}+\sqrt{(\alpha+1)(\beta-1)}) $$<|endoftext|> TITLE: Fastest algorithm to compute the width of a poset QUESTION [15 upvotes]: An colleague recently came to me with a problem concerning the scheduling of tasks in the presence of constraints (of the kind: task $x$ can't begin until task $y$ has been completed). It turned out that the problem was equivalent to that of computing the width (cardinality of maximum antichain) of a poset. I know that the problem of computing the width of a poset can be translated into that of finding the size of a maximum matching in a certain bipartite graph, which in turn can be found using the Hopcroft–Karp algorithm, in time $O(n^{5/2})$ (where $n$ is the number of elements in the poset). To me, that's ``polynomial time; end of story''; but to my colleague, who is working with actual data sets, the degree of the polynomial is very important. My question: what is the current state-of-that art in terms of algorithms for computing the width of a general poset? REPLY [9 votes]: For the general case, I believe that there is nothing asymptotically faster than the $O(n^{5/2})$ algorithm that you alluded to. However, if for example the posets in question are known to have small width, then you can do better. See "Recognition algorithms for orders of small width and graphs of small Dilworth number," by Felsner, Raghavan, and Spinrad, Order 20 (2003), no. 4, 351–364. They show that whether a poset has width $k$ can be decided in time $O(kn^2)$. Posets of width at most 3 can be recognized in $O(n)$ time and posets of width at most 4 can be recognized in $O(n\log n)$ time.<|endoftext|> TITLE: First-order definable bijection between $P(On)$ (or $No$) and $V$? (Is this equivalent to $V = HOD$?) QUESTION [11 upvotes]: It is known that locally one can ``code'' any set in the von Neumann universe $V$ by a set of ordinals. But can one do this globally? In other words, is there a first-order definable bijection $P(On) \longrightarrow V$, where $P(On)$ is the class of all subsets of the class $On$ of all ordinals? Note that $P(On)$ is in natural bijection with the ordered field $No$ of surreal numbers. Thus, the question is equivalent to: Is there a first-order definable bijection $No \longrightarrow V$? If we assume $V = HOD$, then indeed there a first-order definable bijection $No \longrightarrow V$ (since $V = HOD$ is equivalent to the existence of a first-order definable bijection $On \longrightarrow No$ (or $On \longrightarrow V$); see Definable map from all the ordinals to the surreal numbers with a dense image?) However, I would be surprised if the converse were also true. It seems like the existence of a first-order definable bijection $No \longrightarrow V$ (or $P(On) \longrightarrow V$) should be weaker than $V = HOD$ but still perhaps not provable in ZFC. EDIT: Given Andreas Blass' answer that the statement is indeed independent of ZFC, I'm still wondering whether or not any of the implications $(1) \Rightarrow (2) \Rightarrow (3)$ are reversible, where the respective statements are (1) $V = HOD$, (2) there is a definable bijection $P(On) \longrightarrow V$, and (3) there is a definable linear ordering of $V$. REPLY [11 votes]: I will use $(1)$, $(2)$, and $(3)$ as defined in the EDIT to the question. It is known that $(1)\Rightarrow (2)$ does not reverse over ZF, indeed, by a result due to Solovay (appearing as Theorem 3.3 of my paper below) if ZF is consistent, then there is a model of ZF + $(2)$ + $\lnot$ AC, which shows that over ZF the implication $(1) \Rightarrow (2)$ does not reverse. The status of the above failure of reversibility over ZFC (as opposed to ZF), as well as the reversibility-status of $(2) \Rightarrow(3)$ over ZF or ZFC, were posed in a 2004-paper of mine (see Conjecture 4.3.2 of the paper below), and remain open. More detail: my paper The Leibniz-Myscielski Axiom in Set Theory (Fund. Math. 181, 2004, pp.215-231) shows that the existence of an injection (equivalently: a bijection) of V into P(Ord) is equivalent to a "logical" principal dubbed LM (for Leibniz-Mycieslki), where: LM:$\ \ \ \forall x\forall y$ $[x\neq y\rightarrow\exists\alpha>\max \{\rho(x),\rho(y)\}$ $Th(V_{\alpha},\in,x)\neq Th(V_{\alpha},\in,y)].$ In the above, $\rho(x)$ is the ordinal-valued set theoretic rank of $x$, and $Th(V_{\alpha},\in,x)$ is the first order theory of the structure $(V_{\alpha},\in,x)$. LM captures the spirit of Leibniz's principle on the identity of indiscernibles because, by a theorem of Mycielski, a consistent T completion of ZF includes LM iff T has a $\cal{M}$ model with no indisceribles , i.e., if $a$ and $b$ are a pair of distinct elements of $\cal{M}$, then there is a formula $\phi(x)$ in the language of set theory such that $\cal{M}\models\phi(a)\land\lnot\phi(b)$. As shown in the aforementioned paper, the existence of an injection from V into P(Ord) is also equivalent with the global version of the Kinna-Wagner Selection Principle, in other words, models of ZF + LM are precisely models of ZF in which there is a definable class function $\mathbf{F}$ that "chooses" a proper nonempty subset from each set with at least two elements, i.e., $\forall x(\left| x\right| \geq2\rightarrow \emptyset\neq\mathbf{F}(x)$ $\subsetneq x);$ REPLY [6 votes]: $\newcommand\Ord{\text{Ord}}\newcommand\HOD{\text{HOD}}$Let me point out that the existence of a definable bijection of $V$ with $P(\Ord)$ is equivalent to the assertion that the universe is Leibnizian over the ordinals, that is, that any two distinct sets $a\neq b$ satisfy different formulas with ordinal parameters. This is proved in Ali Enayat's fine paper, On the Leibniz-Mycielski axiom in set theory, where he discusses a variety of issues surrounding this axiom. For the forward implication, suppose that we have a definable bijection of $V$ with $P(\Ord)$, given by a specific definition. Thus, to every object $a$ we have definably associated a distinct set $F(a)\subset\Ord$. In particular, if $a\neq b$, then there is some ordinal $\alpha$ such that $\alpha\in F(a)$ and $\alpha\in F(b)$ get different truth values, which establishes that the types of $a$ and $b$ over the ordinals are different. We could have even allowed that the bijection was definable with ordinal parameters. Conversely, suppose that the universe is Leibnizian over the ordinals. This means that whenever $a\neq b$, then there is some formula $\varphi$ and ordinal parameters $\vec\alpha$ such that $\varphi(a,\vec\alpha)$ is true and $\varphi(b,\vec\alpha)$ is false. By reflection, there is some ordinal $\theta$ such that $a,b,\vec\alpha\in V_\theta$ and $\varphi$ is absolute to $V_\theta$. Thus, we have $V_\theta\models\varphi(a,\vec\alpha)$ and $V_\theta\models\neg\varphi(b,\vec\alpha)$. Since $V_\theta$ is $\Pi_1$ definable, this shows that $a$ and $b$ have different $\Sigma_2$ formulas with the parameters $\varphi,\vec\alpha$ and $\theta$. (In particular, because we have bounded the complexity of the formula this way, this shows that being Leibnizian over the ordinals is first-order expressible in ZFC.) For each object $a$, let $\theta_a$ be the next $\Sigma_3$-correct ordinal beyond the rank of $a$, and let $T_a$ be the type of $a$ in $V_\theta$ over ordinal parameters less than $\theta$. Thus, $T_a$ is the set of tuples $\langle\varphi,\vec\alpha\rangle$ such that $V_{\theta_a}\models\varphi(a,\vec\alpha)$. Since any $\Sigma_2$ assertion about $a$ will reflect below $\theta_a$, it follows from the Leibnizian assumption that $a\neq b$ implies $T_a\neq T_b$. Further, since $T_a$ is a set of tuples of formulas and ordinals, we may by Gödel coding view $T_a$ as a set of ordinals. Thus, we have defined an injective map $a\mapsto T_a$ of $V$ into $P(\Ord)$, as desired. In the Leibnizian property, we may fold the ordinal parameters $\vec\alpha$ into the ordinal $\theta$, by making $\vec\alpha$ definable in $V_\theta$, and say it equivalently like this: for any $a\neq b$, there is some ordinal $\theta$ and formula $\varphi$ for which $V_\theta\models\varphi(a)\wedge\neg\varphi(b)$. This is what Enayat calls the Leibniz-Mycielski axiom LM. At that the conclusion of that paper, he conjectures that the implication $V=\HOD$ to LM is not reversible.<|endoftext|> TITLE: Which universities teach true infinitesimal calculus? QUESTION [9 upvotes]: My colleague and I are currently teaching "true infinitesimal calculus" (TIC), in the sense of calculus with infinitesimals, to a class of about 120 freshmen at our university, based on the book by Keisler https://www.math.wisc.edu/~keisler/calc.html. Two of my colleagues in Belgium are similarly teaching TIC at two universities there. I am also aware of such teaching going on in France in the Strasbourg area, based on Edward Nelson's approach, though I don't have any details on that. Which schools, colleges, or universities teach true infinitesimal calculus? A colleague in Italy has told me about a conference a few months ago on using infinitesimals in teaching in Italian highschools. This NSA (nonstandard analysis) conference was apparently well attended (over 100 teachers showed up). In Geneva, there are two highschools that have been teaching calculus using ultrasmall numbers for the past 10 years. Anybody with more information about this (who to contact, what the current status of the proposal is, etc.) is hereby requested to provide such information here. Note 1 in response to Dan's comment: usually nowadays the term infinitesimal calculus is used as a dead metaphor for "the calculus". Thus, calculus courses routinely go under the name "infinitesimal calculus" for historical reasons, whereas in point of fact no infinitesimal ever appears on the blackboard. When I refer to "true infinitesimal calculus" I mean calculus with infinitesimals (as explained above) as opposed to ordinary "infinitesimal calculus" as found in Thomas-Finney and other textbooks. Note 2 in response to Pietro's comment: I should clarify that calculus using infinitesimals is not limited to Keisler's book. There are several books of this sort, including some that are unrelated to Robinson's theory, such as those by A. Kock and J. Bell. Note 3: thanks to Bjørn Kjos-Hanssen for the update on the history of TIC teaching at the University of Hawaii. REPLY [3 votes]: Since you mentioned University of Hawaii -- David Ross regularly teaches nonstandard analysis and infinitesimal calculus at University of Hawaii in various forms, for instance MATH 649K (a regular grad course), Spring 2008 MATH 699 (reading course), Spring 2011 We also usually cover the topic in our senior-level course on mathematical logic (MATH 455), although I suppose a lot of universities do that. But infinitesimal calculus is not part of our standard calculus syllabus.<|endoftext|> TITLE: How to calculate the infinite sum of this double series? QUESTION [12 upvotes]: I'm calculating this double sum: $$ \sum _{m=1}^{\infty } \sum _{k=0}^{\infty } \frac{(-1)^m}{(2 k+1)^2+m^2} $$ I know the answer is $$ \frac{ \pi \log (2)}{16}-\frac{\pi ^2}{16} $$ which can be verified by numerical calculations. I used the Taylor expansions of $log (1+x)$ and $arcsin (x)$ at x=1 to replace $log (2)$ and $\pi$. I got $$ \sum _{m=1}^{\infty } \sum _{k=0}^{\infty } \frac{(2 m+1) (-1)^{k+m}}{4 m(2 k+1) (2 m-1)} $$ I tried to play with the dummy variables but failed. Any idea? REPLY [9 votes]: Here is a proof based on Hachino's idea. We have agreed it's enough to prove that $$S=\sum_{m=1}^\infty\frac{(-1)^m\tanh(m\pi /2)}{m}=\frac{\log 2-\pi}{4}. $$ Plug in the expansion $$\tanh(m\pi /2)=\frac{1-e^{-m\pi }}{1+e^{-m\pi }}=1+2\sum_{n=1}^\infty(-1)^ne^{-nm\pi }. $$ Changing order of summation and using the Taylor series of $\log(1+x)$ gives $$S=-\log 2+2\sum_{n=1}^\infty(-1)^{n+1}\log(1+e^{-n\pi })=\log\left(\frac{\prod_{n=1}^\infty(1+e^{-(2n-1)\pi})^2}{2\prod_{n=1}^\infty(1+e^{-2n\pi})^2}\right).$$ The infinite products can be recognized as theta values. Recall that $$\theta_2(z,q)=2q^{1/4}\cos z\prod_{n=1}^\infty(1-q^{2n})(1+2q^{2n}\cos 2z+q^{4n}), $$ $$\theta_3(z,q)=\prod_{n=1}^\infty(1-q^{2n})(1+2q^{2n-1}\cos 2z+q^{4n-2}). $$ There are lots of conflicting notations for theta functions; I use the conventions of Whittaker and Watson, A course in modern analysis. It follows that $$S=\log\left(e^{-\pi/4}\frac{\theta_3(0,e^{-\pi})}{\theta_2(0,e^{-\pi})}\right). $$ It remains to show that $$\frac{\theta_3(0,e^{-\pi})}{\theta_2(0,e^{-\pi})}=2^{1/4}. $$ This is clear from generalities on elliptic functions. Again in the notation of Whittaker and Watson, the modulus $k$ is given by $$k=\frac{\theta_2(0,q)^2}{\theta_3(0,q)^2}. $$ The value $q=e^{-\pi}$ corresponds to $k=k'$, where $k'$ is the complementary modulus defined by $k^2+(k')^2=1$. Thus, $k=1/\sqrt 2$, which gives the desired result. If you had a tough childhood, you might worry that I changed the order of summation in series that are not absolutely convergent. You might want to plug in a convergence factor $x^m$, repeat the argument above and prove that both sides of the resulting identity are left continuous at $x=1$.<|endoftext|> TITLE: Is the Weil–Deligne representations coming from $\ell$-adic cohomology independent of $\ell$? QUESTION [6 upvotes]: Let $F$ be a $p$-adic field. Let $(G_{F}, W_{F}, I_{F})$ denote the (absolute Galois group, Weil group, inertia group) of $F$. Let $X/F$ be a proper smooth variety. Let $\ell$ be a prime number $\ne p$. The $\ell$-adic cohomology $H_{\ell}^{i} = H_{\text{ét}}^{i}(X_{\bar{F}}, \mathbb{Q}_{\ell})$ is naturally endowed with a continuous Galois representation $\rho$. The Weil–Deligne representation By the $\ell$-adic monodromy theorem of Grothendieck, one associates a Weil–Deligne representation with $H_{\ell}^{i}$: It gives a nilpotent “monodromy” operator $N$. One restricts the representation $\rho$ to $W_{F}$, and changes it to $$ \sigma (\Phi^{a}x) = \rho(\Phi^{a}x) \exp(-t(x)N), \qquad a \in \mathbb{Z}, x \in I_{F} $$ where $\Phi \in W_{F}$ is a Frobenius element, and $t \colon I_{F} \to \mathbb{Z}_{\ell}$ a projection onto the $\ell$-adic component of $I_{F}$. The Weil–Deligne representation associated with $H_{\ell}^{i}$ is $(\sigma, H_{\ell}^{i}, N)$. It is an object in $\mathrm{WDRep}_{\mathbb{Q}_{\ell}}(W_{F})$. Question Choose an (non-canonical, non-continuous!) embedding $i_{\ell} \colon \mathbb{Q}_{\ell} \to \mathbb{C}$. By extending scalars, we obtain an object $(\sigma, H_{\ell}^{i}, N) \otimes_{i_{\ell}} \mathbb{C}$ in $\mathrm{WDRep}_{\mathbb{C}}(W_{F})$. Let $\ell' \ne p$ be another prime. Let $i_{\ell'} \colon \mathbb{Q}_{\ell'} \to \mathbb{C}$ be an embedding. We can repeat the entire process to obtain an object $(\sigma, H_{\ell'}^{i}, N) \otimes_{i_{\ell'}} \mathbb{C}$ in $\mathrm{WDRep}_{\mathbb{C}}(W_{F})$. Q. Are $(\sigma, H_{\ell}^{i}, N) \otimes_{i_{\ell}} \mathbb{C}$ and $(\sigma, H_{\ell'}^{i}, N) \otimes_{i_{\ell'}} \mathbb{C}$ isomorphic in $\mathrm{WDRep}_{\mathbb{C}}(W_{F})$? From a motivic viewpoint this should certainly be true, but on the other hand maybe this is one of the strongest forms of $\ell$ independence that one can ask for. If there is no answer in general, I would be very happy to learn about partial cases, where this is known. REPLY [4 votes]: In a recently published paper, Rutger Noot has studied the case of abelian varieties. In 2011, he had conducted the PhD Thesis of Abhijit Laskar where the methods are applied in a motivic setting to obtain more general results depending on the Mumford-Tate group of the underlying motive.<|endoftext|> TITLE: Cauchy-Schwarz proof of Sidorenko for 3-edge path (Blakley-Roy inequality) QUESTION [21 upvotes]: Is there a "Cauchy-Schwarz proof" of the following inequality? Theorem. Given $f \colon [0,1]^2 \to [0,1]$, one has $$ \int_{[0,1]^4} f(x,y)f(z,y)f(z,w) \, dxdydzdw \geq \left(\int_{[0,1]^2} f(x,y) \, dxdy\right)^3. $$ Background. This inequality is due to Blakley and Roy (1965). In fact, they proved even more, namely when the LHS corresponds to a path of length $k$ (above $k=3$) and the RHS is $(\int f)^k$. This is a special case of a more general Sidorenko's conjecture, which claims that $t(H,W) \geq (\int W)^{e(H)}$ for any bipartite graph $H$. The general case of Sidorenko's conjecture is still open. See, e.g., this note by Conlon, Fox, and Sudakov (although there has been some other progress since then). Szegedy and Li gives a different proof of the above inequality, using convexity of the logarithm function. Also see the paper of Kim, Lee, and Lee for another approach. On page 28 of Lovasz' book on graph limits, it states this inequality without proof, and then says ... and this is already quite hard, although short proofs with a tricky application of the Cauchy–Schwarz inequality are known. So my question is: how does one prove the inequality above using Cauchy-Schwarz? Update: It has been shown that there is no vanilla sum-of-squares proof of the inequality https://arxiv.org/abs/1812.08820 REPLY [6 votes]: Let me add a possibly shortest proof of the inequality you asked, although not by Cauchy-Schwarz. Following Sasha's notation, let $g(x)=\int f(x,y) dy$. Then by Holder's inequality, \begin{align*} \int f(x,y)f(y,z)f(z,w) &= \int g(y)f(y,z)g(z)\\ &=\int g(y)f(y,z)g(z) \int \frac{f(y,z)}{g(y)} \int \frac{f(y,z)}{g(z)}\\ &\geq \left(\int f(y,z) \right)^3. \end{align*} As both of us already knew, it is proved that the flag algebra caculus does not give the answer for this inequality. Let me also add the reference for those readers who didn't know the recent result by Blekherman, Raymond, Singh, and Thomas: https://arxiv.org/pdf/1812.08820.pdf<|endoftext|> TITLE: Lens spaces and generalized Petersen graphs QUESTION [18 upvotes]: Recently I came across this mathoverflow question, in which the number of homeomorphism classes of 3-dimensional lens spaces $L(p, q)$ is computed as a function of $p$. Using the OEIS, I found a connection to so-called generalized Petersen graphs: For $m\geq 3$ and $1\leq k TITLE: Gauss--Lucas type theorem for tracts and higher derivatives of a polynomial QUESTION [10 upvotes]: The Gauss--Lucas Theorem states that all zeros of derivative of a degree $n$ complex polynomial $p(z)$ are contained in the convex hull of the zeros of $p$. By iteration, this implies that the zeros of $p',p^{(2)},\ldots,p^{(n-1)}$ are contained in the convex hull of the zeros of $p$. The Riemann--Hurwitz Theorem (among others) implies that if a tract $D$ of $p$ (namely a component of the set $\{z:|p(z)|<\epsilon\}$ for some $\epsilon>0$) contains all the zeros of $p$ in its bounded face, then all the critical points of $p$ are contained in $D$. My conjecture is that in fact, if $D$ is a tract of $p$ and contains all the zeros of $p$, then $D$ also contains all the zeros of $p',p^{(2)},\ldots,p^{(n-1)}$. This certainly does not follow by straight-forward iteration, since in general there may not exist some tract $D'$ of $p'$ which contains all the zeros of $p'$ in $D$, and is itself contained in $D$. It seems that the tracts and level curves of $p'$ do not interact very nicely with the tracts and level curves of $p$ (even worse for $p'',p''',\ldots$). This question was originally asked on M.SE, where it received no answers, however user Behavior did point out that this conjecture could be equivalently stated as: Conjecture: Let $M=\max(|p(z)|:p'(z)=0)$, then if $w$ is a zero of any derivative of $p$, then $|p(w)|\leq M$. REPLY [7 votes]: Very interesting problem. At first I thought the statement might be true for real valued polynomials, but unfortunately I found the following, Unless I am misunderstanding the problem, I believe this is the counterexample you were looking for. Although, we note that the second derivative does look suspiciously still trapped inside of the tract. Maybe, your conjecture holds for only the first and second derivative.<|endoftext|> TITLE: What is wrong with the "naive" proof of the Hauptvermutung? QUESTION [6 upvotes]: The Hauptvermutung is the statement that any two PL structures on a topological space have a common refinement. It is false in general, but (I think) true for some low dimensional manifolds. The special case I'll consider is when the topological space is a smooth 2-dimensional manifold. In this case, the Hauptvermutung (roughly) states that for any two embeddings of two graphs into the manifold (such that the resulting faces are homeomorphic to disks), there is a graph (with an embedding) that contains both graphs as a subgraph that preserves that PL structure (cue the hand-waving). It's "obvious" that if you take the union of the two graphs (again, embedded into the manifold) and add all the necessary points (such as the points at intersections of edges of the graphs), that you'll get a refinement common to both graphs (bar some subtle cases, like when the two graphs don't intersect) In this answer, it's stated that this proof is wrong for "fractal" reasons. Does this mean that the proof still goes through if we restrict to finite graphs? If not, why? This question is motivated by a problem posed a year or two back in my Differential Geometry class. The problem required that I prove that the Euler Characteristic (on a smooth, compact surface), defined by the quantity $V-E+F$ for any embedding of a graph, is well defined. I used essentially the proof above to construct a common refinement of any two graphs that preserved the quantity $V-E+F$. Since that particular homework was never returned, I never found out if my proof was correct, so now I'm wondering if I missed something. EDIT: Since this approach clearly doesn't work (thanks to Alex Degtyarev in the comments), I might ask if there's any other way of proving that the Euler characteristic for compact surfaces defined in this way is well-defined. I know you can just use the homology groups (which is the preferred approach), but since we didn't mention homology in my class, I suspect there's a more elementary method of proof. REPLY [6 votes]: As pointed out in comments, your version of the Hauptvermutung cannot be true because there is no reason why the intersection of two graphs should be nice in any way. This has nothing to do with $C^0$ vs $C^\infty$. Since any closed subset of $\mathbb{R}^n$ is the zero set of a smooth function, you can have a pair of smooth graphs where two edges intersect along a very bad closed set. Let's stick to smooth triangulations. It means there is a simplicial complex $K$ linearly embedded in some $\mathbb{R}^N$ and a homeomorphism $h$ from $K$ to your surface $S$ whose restriction to each simplex $\sigma$ is a smooth embedding (ie. if P denotes the affine subspace spanned by $\sigma$ in $\mathbb{R}^N$, there is an open neighborhood $U$ of $\sigma$ in $P$ and a smooth embedding $h_\sigma$ from $U$ to $S$ such that $h$ and $h_\sigma$ coincide on $\sigma$). Now Whitehead's theorem (which holds in any dimension) says that if $h : K \to S$ and $g : L \to S$ are two such smooth triangulations then one can perturb $h$ and $g$ and get sub-complexes $K' \subset K$ and $L' \subset L$ such that $h^{-1}\circ g$ is a linear isomorphism from $L'$ to $K'$. Again: you cannot hope to get rid of the perturbation here (but you can make it $C^1$-small). If you want to prove invariance of Euler characteristic using this path then you now have to prove a purely PL fact: if $K$ is a simplicial complex and $K'$ a subcomplex then $\chi(K) = \chi(K')$. Clearly it's enough to consider the case where $K$ is a simplex (a triangle in your case). One very nice trick is to extend any edge of $K'$ until it hits the boundary of $K$. This way you get a common refinement $K''$ which is obtained from $K$ and $K'$ by only one kind of operation: bissection (cutting one simplex using an hyperplane). This operation obviously preserves $\chi$. This trick can be used to get a very easy proof of invariance of simplicial homology. See also Siebenmann's proof of Reidemeister-Singer or Giroux's proof of the open book theorem for contact 3-manifolds for further use of this trick. Further reading: Lurie's exscellent lecture notes for a proof of Whitehead's theorem. Moise's book Geometric Topology in Dimensions 2 and 3 if you are interested in the $C^0$ case from a traditional point of view. Hatcher's great expository paper about smoothing in dimension 2.<|endoftext|> TITLE: Cartesian product of small objects QUESTION [10 upvotes]: Let's say we have a locally $\lambda$-presentable category and a pair of $\lambda$-presentable objects $A$ and $B$. Is it true that $A \times B$ is $\lambda$-presentable? REPLY [6 votes]: A simpler counterexample is given by the slice category $S/\mathrm{Set}$ for a "large" (= of cardinality to be chosen later) set $S$. This category can be viewed as the category of models of an algebraic theory with one nullary operation for each element of $S$, so it is locally finitely presentable. Its initial object $S$ (equipped with the identity map) is of course finitely presentable, but $S \times S$ is a free object on the "large" set of pairs $\{\,(a, b) \mid a \in S, b \in S, a \ne b\,\}$. By choosing the cardinality $S$ large enough, we can then make $S \times S$ not only not finitely presentable, but not $\mu$-presentable for any fixed regular cardinal $\mu$.<|endoftext|> TITLE: Hausdorff dimension of R x X QUESTION [25 upvotes]: In general, the Hausdorff dimension of a product is at least the sum of the dimensions of the two spaces. Does equality hold if one space is Euclidian? So let $X$ be a metric space and let $\mathit{dim}_H$ denote the Hausdorff dimension. Does the following hold (and why)? $$\mathit{dim}_H(X \times \mathbb{R}) = \mathit{dim}_H(X) + 1$$ (And if this does not hold in general, is it at least true for e.g. compact length spaces $X$?) REPLY [35 votes]: Yes, it holds. In general it holds for any two separable metric spaces $X, Y$ with $Y$ totally bounded that $$ \dim_H(X)+\dim_H(Y) \le \dim_H(X\times Y) \le \dim_H(X)+\dim_B(Y), $$ where $\dim_B(Y)$ is upper box counting dimension. These inequalities can essentially be found in Falconer or Mattila's textbook (for subsets of Euclidean space, but the proofs for separable metric spaces is the same), but in short the left-most one follows from Frostman's Lemma and the right-most one directly from the definitions. Frostman's Lemma in separable metric spaces was proved by Howroyd, this is also in Mattila's book. In particular, if $Y$ has equal Hausdorff and upper box counting dimension (which holds if $Y$ is a compact interval) then $$ \dim_H(X\times Y)=\dim_H(X)+\dim_H(Y). $$ Now $\mathbb{R}$ is not compact (so box counting dimension is not defined for it), but expressing $\mathbb{R}=\cup_j [-j,j]$ and using countable stability of Hausdorff dimension, it follows that the answer to your question is positive (at least when $X$ is separable - the reason I need this assumption is for Frostman's Lemma). In fact, the inequality above holds for packing dimension $\dim_P$ in place of $\dim_B$; I stated it with $\dim_B$ because upper box counting dimension is better known, the proof of the inequality with $\dim_B$ is easier, and this is enough for answering your question. But the inequality with packing dimension is also important and is sharper (since $\dim_P\le\dim_B$). Edit: here is the simple argument for $\dim_H(X\times [-k,k])\le \dim_H X+1$, which is the bound the OP was concerned about. Fix $\delta>0$. Let $B(x_i,r_i)$ be a covering of $X$ with $r_i<\delta$, such that $$ \sum_i r_i^{\dim_H(X)+\delta} < 1. $$ (Such covering exists by definition of Hausdorff measure/dimension). For each $i$ split $[-k,k]$ into intervals $I_{ij}$ of length $2r_i$ (so there are $k/r_i$ of them). Then $B(x_i,r_i)\times I_{ij}$ has radius $\sim r_i$ and is a $O(\delta)$-covering of $X\times [-k,k]$. We obtain $$ \sum_{ij} r_i^{\dim_H(X)+1+\delta} = \sum_i {k\over r_i}r_i^{\dim_H(X)+1+\delta} = \sum_i kr_i^{\dim_H(X)+\delta} < k. $$ Since $k$ is fixed we obtain the estimate $\dim_H(X\times [-k,k])\le \dim_H(X)+1$. REPLY [14 votes]: Yes, taking the product with $\mathbb{R}^n$ always increases the Hausdorff dimension by exactly $n$, and here is a reference. In the first paragraph of [1], Kaoru Hatano notes (in slightly different notation): If $E \subset \mathbb{R}^m$ and $F \subset \mathbb{R}^n$ have dimensions $\dim(E) = \alpha$ and $\dim(F) = \beta$, then the dimension of the product $E \times F$ is in the range $J = [\alpha + \beta, \min(\alpha + n, m + \beta)]$. In your case, $F = \mathbb{R}^n$, so we get $\min(\alpha + n, m + n) = \alpha + n$, and the range $J$ consists of just one point, therefore $$\dim(E \times \mathbb{R}^n) = \dim(E) + n.$$ [1] Kaoru Hatano. Notes on Hausdorff dimensions of Cartesian product sets. Hiroshima Math. J. Volume 1, Number 1 (1971), 17-25. http://projecteuclid.org/euclid.hmj/1206138139<|endoftext|> TITLE: Does there always exist a sequence of handle moves between handle decompositions that does not increase index? (+ ref. request) QUESTION [6 upvotes]: Reference request: Firstly, I'm looking for a proof of the following well-known result about handle decompositions: ($\ast$) Given two handle decompositions of a smooth $n$-manifold $M$, there exists a sequence of handle pair creations, cancellations, and handle slides (as well as isotopies) that takes one decomposition to the other. I think I broadly understand the conceptual idea behind the proof, namely that handle decompositions correspond to Morse functions on $M$, and homotopies of the Morse functions give such sequences of handle moves, but I'd like to be able to work through this more rigorously. So far the only lead I've been able to find is Gompf and Stipsicz's 4-Manifolds and Kirby Calculus (Theorem 4.2.12), where they attribute the result to Cerf and simply reference his paper La stratification naturelle des espaces de fonctions différentiables réelles et le théorème de la pseudo-isotopie. Unfortunately my French is extremely rudimentary so I haven't been able to locate the relevant part(s) of the paper. Would anyone be able to point me in the right direction, or know where I can find a proof of this theorem? Main question: If an $n$-dimensional smooth manifold $M$ has handle decompositions $\mathscr{H}, \mathscr{H}^\prime$, both in terms of $0,1,\dots, k$-handles where $k TITLE: Localizations of model categories and $\infty$-categories QUESTION [7 upvotes]: I am interested in the relation between Bousfield localizations of model categories and localizations of $(\infty,1)$-categories. According to Hirschhorn's book we can form the left Bousfield localization of a left proper cellular model category along any set of maps. According to Lurie's book we can form the (left) localization of a presentable $\infty$-category along a small collection of maps. How are those two results related? Is the simplicial nerve of a left proper, cellular model category a presentable quasi-category? Furthermore we also know that we can form the right Bousfield localization of a right proper cellular model category along any set of objects. Is there an analogous theory of right localizations on $\infty$-categories? EDIT: It has been pointed out that combinatorial model categories correspond exactly to presentable $\infty$-categories and left Bousfield localizations exist for every left proper, simplicial, combinatorial model category and correspond exactly to the localizations of the corresponding presentable $\infty$-category. This still leaves the question if there is a similar theory and an existence theorem for right localizations of $\infty$-categories corresponding to the theory of right Bousfield localizations of something like right proper, simplicial, cellular model categories. Thanks already! REPLY [6 votes]: There is also an existence theorem for right Bousfield localizations of presentable $\infty$-categories. In fact, it follows from the existence theorem for left Bousfield localizations. Let $K$ be a collection of objects in $C$ and let $D\subset C$ be the subcategory of $K$-colocal objects. Then $D$ is manifestly closed under colimits. By the adjoint functor theorem, the inclusion $D\subset C$ has a right adjoint provided that $D$ is accessible. I claim that $D$ is accessible if the colimit closure $\bar K$ of $K$ in $C$ is accessible, for instance if $K$ is small. To prove this, let $W$ be the class of $K$-colocal equivalences, viewed as a full subcategory of $C^{\Delta^1}$. Note that $W$ is the class of $E$-local objects for $$ E=\{(0\to x) \to (x\stackrel=\to x),\quad x\in \bar K_0 \}, $$ where $\bar K_0$ is small and generates $\bar K$ under colimits. By the existence theorem for left Bousfield localizations (HTT 5.5.4.15), $W$ is presentable. Consider the functor $F: C\to (\mathrm{Fun}^R(W,S)^{op})^{\Delta^1}\simeq W^{\Delta^1}$ which sends $c\in C$ and $f:a\to b$ in $W$ to $\mathrm{Map}(c,a)\to \mathrm{Map}(c,b)$. The functor $F$ is accessible since it preserves colimits. The inclusion $D\subset C$ is the homotopy pullback by $F$ of the diagonal $W\to W^{\Delta^1}$. By HTT 5.4.6.6, $D$ is accessible. We can be a bit more precise. Unraveling the previous argument, we see that the right adjoint $R: C\to D$ is computed as follows. The counit $Rc\to c$ is $L(0\to c)$ where $L$ is the left adjoint of $W\subset C^{\Delta^1}$. The usual construction of $L$ by transfinite induction shows that in fact $Rc\in\bar K$, so that $D=\bar K$. So if $C$ is $\kappa$-accessible and every object of $K$ is $\kappa$-compact, then $D$ is $\kappa$-accessible.<|endoftext|> TITLE: Semiproper but not proper QUESTION [13 upvotes]: Assume V=L. Is there a semi-proper notion of forcing that is not proper? Namba forcing isn't semi-proper in L, and Prikry forcing isn't even available there. REPLY [7 votes]: The following is proved as Claim 2.1 in Shelah's paper ``Forcing axiom failure for any $\lambda>\aleph_1$'' Theorem. There is a forcing notion of size $2^{\aleph_2}$ which is not proper but is $\{\aleph_1\}$-semi proper. In fact Shelah proves more: for every regular $κ>ℵ_1$ there is a forcing notion $P$ of cardinality $2^κ$ such that there is a stationary set $S⊆[2^κ]^{ℵ_0}$ not preserved by $P$ and if $χ>2^κ$, $p∈P∈N≺(H(χ),∈), N$ countable, then there is $q∈P$ stronger than $p$ and $q$ forces that $N∩κ$ is an initial segment of $N[G]∩κ.$<|endoftext|> TITLE: Continuity of taking collapse maps QUESTION [5 upvotes]: Let $U$ and $V$ be open subsets of $\mathbb R^n$ and let $\mathrm{OEmb}(U,V)$ denote the space of open embeddings of $U$ into $V$ with the compact-opent topology. Let $\bar{U},\bar{V}$ denote their one-point compactifications and let $\mathrm{map}(\bar{V},\bar{U})$ denote the space of maps between them again with the compact-open topology. There is a (set-theoretic) map $$\mathrm{OEmb}(U,V)\longrightarrow \mathrm{map}(\bar{V},\bar{U})$$ that sends an embedding $e$ to the map $\phi(e)$ given by $\phi(e)(y) = e^{-1}(y)$ if this value exists and $\phi(e)(y)=\infty$ otherwise. Fact: This map is continuous. More generally, it is also continuous if $U$ and $V$ are Hausdorff locally compact locally connected topological spaces. What reference should I cite for these facts? REPLY [3 votes]: I haven't found an explicit reference in the literature, but as Johannes Ebert pointed to me, in the second part of Theorem 4 of R. Arens, Topologies for Homeomorphism Groups, American Journal of Mathematics, Vol. 68, No. 4 (Oct., 1946), pp. 593-610 it is proven that the inverse mapping $\mathrm{Homeo}(X)\to \mathrm{Homeo}(X)$ that sends a homeomorphism to its inverse is continuous with respect to the compact-open topology, provided that $X$ is Hausdorff, locally compact and locally connected. The same proof works as well to prove my question (under the assumptions that $X$ and $Y$ are Hausdorff, locally compact, and $Y$ is locally connected). Therefore I think that Arens article is a good reference for this fact (together with an indication saying that the proof has to be mildly adpated). This is how the adaptation would go: Theorem: If $X$ and $Y$ are Hausdorff locally compact spaces and $Y$ is locally connected, then the collapsing map $$\phi\colon \mathrm{OEmb}(X,Y)\longrightarrow \mathrm{map}(\overline{Y},\overline{X})$$ is continuous with the compact-open topologies. Proof: Take an embedding $e$ and consider a subbasic neighbourhood $(K,U)$ of $\phi(e)$, where $K\subset \overline{Y}$ is compact and $U\subset \overline{X}$ is open. Let us denote by $U^c$ the complement of $U$ in $X$ (not in $\overline{X}$). An embedding $f$ belongs to this neighbourhood if $\phi(e)(K)\subset U$, and this holds if and only if $e(U^c)\subset K^c$. Therefore $e(U^c)\subset K^c$ and if any other embedding $f$ satisfies that $f(U^c)\subset K^c$, then it holds that $\phi(f)\in (K,U)$. We distinguish two cases: \underline{If $\infty\in U$}, then $U^c$ is compact and we are done ($(U^c,K^c)$ is an open neighbourhood of $e$ that gets mapped into the neighbourhood $(K,U)$ of $\phi(e)$. \underline{If $\infty\notin U$}, then $K\subset e(X)$, and therefore $e^{-1}(K)$ is compact. This is the situation that Arens handles. We use that $X$ is Hausdorff locally compact to construct a pair of open neighbourhoods $V,W$ of $K$ with compact closure of $e$ such that $$e^{-1}(K)\in V\subset \overline{V}\subset W\subset \overline{W}\subset U.$$ Then $N:=\overline{W}\setminus V$ is compact, and if $f\in (N,K^c)$, then $\phi(f)(K)\subset N^c$. Observe that $N^c$ is the disjoint union of $V$ and $\overline{W}^c$. Now we use that $X$ is locally connected to assume, without loss of generality that $K$ is connected (this is detailed in Arens' paper). Therefore, if $f\in (N,K^c)$, then either $\phi(f)(K)\subset V\subset U$ (this is what we want) or $\phi(f)(K)\subset \overline{W}^c$. In order to rule out the second case, we make the additional assumption (again without loss of generality) that $K$ has non-void interior, then we take a point $p$ in the interior. If $f\in (N,K^c\cap e(W))\cap (\{e^{-1}(p)\},\mathring{K})$, then $f(e^{-1}(p))\subset K$, but then $e^{-1}(p)\in \phi(f)(K)\cap V$, which is non-empty, and therefore we have ruled out the second case and $f\in (K,V)\subset (K,U)$.<|endoftext|> TITLE: String Orientation and Level Structures QUESTION [16 upvotes]: Atiyah, Bott and Shapiro defined orientations of real and complex K-theory that were later refined to maps of ($E_\infty$-ring) spectra $$MSpin \to KO$$ and $$MSpin^c \to KU.$$ Likewise, but more complicated, the Witten genus was refined in a paper by Ando, Hopkins and Rezk to the string orientation $$MString \to tmf$$ into connective $tmf$. This has also a "complex" version: for every (even periodic) elliptic spectrum, there is an orientation $$MU\langle 6\rangle \to E,$$ as already shown earlier in Elliptic Spectra, the Witten Genus and the Theorem of the Cube. This provides "string orientations" for many variants of $TMF$ with level structure; for example, $TMF(n)$ is an elliptic spectrum for $n\geq 3$. But it does not constitute a fully satisfactory theory of string orientations for topological modular forms with level structures for the following two reasons: It is not obvious that the map $MU\langle 6\rangle \to TMF(n)$ factor over its connective version $tmf(n)$. The latter is understood to be the connective cover of the "compactified" $Tmf(n)$ defined first by Goerss and Hopkins and then in greater generality by Hill and Lawson. Many level structures do not define elliptic spectra. For example, the spectra $TMF_0(n)$ are not complex oriented in general. So my question is: Is there a notion of "string bordism with level structure" and corresponding string orientations, producing for example a map into $tmf_0(3)$? Note: The composition $MString \to tmf \to tmf_0(3)$ is obviously not what I want. REPLY [10 votes]: Dylan's nice paper at https://arxiv.org/abs/1507.05116 answers this question.<|endoftext|> TITLE: Using Stokes' theorem to define "area" enclosed by a curve QUESTION [18 upvotes]: I am trying to figure out what the next calculation of the "area" (or "volume" in higher dimensional analogues) using Stokes' theorem really means. Here is my thought process: $2$-dimensional case: given a closed simple piece-wise smooth curve $C$ in $\mathbb{R}^2$ you can find out the area enclosed by $C$ using Green's theorem by choosing an orientation on $C$ and calculating $$\text{Area} = \left\vert \dfrac{1}{2} \oint_C (x dy - y dx) \right\vert .$$ $2$-dimensional case in $\mathbb{R}^n$: When you have a piece-wise smooth non self intersecting map $\phi : S^1 \rightarrow \mathbb{R}^n$ such that $\phi (S^1)$ is contained in a $2$-dimensional plane of $\mathbb{R}^n$, you can again find the area enclosed by $\phi (S^1)$ by choosing an orthonormal coordinate system $(x_1,...,x_n)$, choosing an orientation on $S^1$ and calculating using Stokes' theorem $$\text{Area}^2 = \sum_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint_{\phi (S^1)} (x_i dx_j - x_j dx_i) \right)^2 .$$ Now my question is what is the meaning of calculating the above quantity when $\phi (S^1)$ is not contained in a $2$-dimensional plane. I want to use the same computation as above, but since I want the number I get to be independent on the coordinate system, I'll have to average with respect to all coordinate systems. To be explicit: let $\phi : S^1 \rightarrow \mathbb{R}^n$ be a piece-wise smooth non self intersecting map and fix an orientation on $S^1$. Let $SO_n (\mathbb{R})$ be the special orthogonal group and let $\mu$ be the normalized Haar measure on $SO_n (\mathbb{R})$ (i.e., $\mu (SO_n (\mathbb{R}) ) =1$). Define the "area" (or the "Stokes area" in lack of a better name) bounded by $\phi (S_1)$ as $$\text{"Stokes area"}^2 = \int_{SO_n (\mathbb{R})} \left( \sum_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint_{\gamma.\phi (S^1)} (x_i dx_j - x_j dx_i) \right)^2 \right) d \mu (\gamma) .$$ It is not hard to give analogues in any dimension to get "Stokes volume" (note that the analogue in dimension $1$, i.e., for $S^0$, comes out just the usual Euclidean distance between points). My questions are: Is this quantity well-known / studied ? If so, I would very much appreciate a reference. What does this "Stokes area" represent geometrically (I have no intuition about it)? Is there a connection (say by some inequality) to the area of a minimal surface enclosed in $\phi (S^1)$? Edit: by the answer of Will Sawin below, it is clear that we can consider only $$\text{"Stokes area"}^2 = \sum_{1 \leq i < j \leq n} \left( \dfrac{1}{2} \oint_{\gamma.\phi (S^1)} (x_i dx_j - x_j dx_i) \right)^2 .$$ and that is quantity is always less or equal than the area of a surface enclosed by the curve. The question remains is there is a connection to the infimum of the areas of all the surfaces enclosed by the curve. In the case the curve is contained in a plane, the formula computes this minimum. But what about the general case: is there a constant (maybe depending on the dimension of $\mathbb{R}^n$) $c(n)$ such that $$ \text{"Stokes area"} \geq c(n) (\text{infimum of the areas of surfaces enclosed by the curve}) ?$$ REPLY [5 votes]: This is not the answer to your question, but a related result. If $\gamma(x)=(x(s),y(s))$ is a closed curve on the plane, then $$ \mathfrak{D}=\frac{1}{2}\int_0^1(\dot{y}(s)x(s)-\dot{x}(s)y(s))\, ds $$ is the oriented area. In particular if the curve has the shape of the digit $8$, the oriented area equals zero because the curve has opposite orientation in the upper and the lower part of $8$. In the case of closed curves in $\mathbb{R}^{2n}$ with coordinates $x_1,y_1,\ldots,x_n,y_n$, one can define the symplectic area as the sum of oriented areas of projections onto the $x_iy_i$-planes, $i=1,2,\ldots,n$. More precisely, the symplecic area is $\mathfrak{D}=\mathfrak{D}_1+\ldots+\mathfrak{D}_n$, where $$ \mathfrak{D}_j=\frac{1}{2}\int_0^1(\dot{y}_j(s)x_j(s)-\dot{x}_j(s)y_j(s))\, ds $$ is the oriented area of the projection of the curve on the $x_iy_i$-plane. Theorem (Isoperimetric inequality). If $\gamma=(x_1,\ldots,x_n,y_1,\ldots,y_n):[0,1]\to\mathbb{R}^{2n}$ is a closed rectifiable curve prametrized by constant speed, then the following isoperimetric inequality is true $$ L^2\geq 4\pi|\mathfrak{D}|, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*) $$ where $L$ is the length of $\gamma$ and $\mathfrak{D}=\mathfrak{D}_1+\ldots+\mathfrak{D}_n$ is defined above. Moreover the equality in (*) holds if and only if $\gamma$ is a circle of the following form: there are points $A,B,C,D\in\mathbb{R}^n$ such that form \begin{equation} \label{Eq1} \gamma(s)=(C+iD)+(1-e^{+2\pi is})(A+iB), \quad \text{when} \quad L^2=4\pi\mathfrak{D} \end{equation} and \begin{equation} \label{Eq2} \gamma(s)=(C+iD)+(1-e^{-2\pi is})(A+iB) \quad \text{when} \quad L^2=-4\pi\mathfrak{D}. \end{equation} The proof follows from a simle and straightforward adaptation of a standard proof of the isoperimetric inequality due to Hurwitz, see: P. Hajłasz, S. Zimmerman, Geodesics in the Heisenberg group. Anal. Geom. Metr. Spaces 3 (2015), 325–337.<|endoftext|> TITLE: Can a subset of the plane have nontrivial $H_2$ or $\pi_2$? QUESTION [55 upvotes]: This is a question that occurred to me years ago when I was first learning algebraic topology. I've since learned that it's a somewhat aesthetically displeasing question, but I'm still curious about the answer. Is it possible for a subset of $\mathbb R^2$ to have a nontrivial singular homology group $H_2$? What about a nontrivial homotopy group $\pi_2$? REPLY [41 votes]: Apparently the asphericity is due to Zastrow (see Cannon-Conner-Zastrow). Also apparently the result that the higher homology groups vanish is due to Zastrow, but his habilitation thesis never seems to have appeared.<|endoftext|> TITLE: Differential Algebra Book QUESTION [15 upvotes]: I'm looking for a couple good textbooks covering differential algebra. I'm a prospective Ph.D. student, and this is potentially applicable to my specialization. As such, I'm not afraid of depth; I've got a few years to work through it all. Specifically, I'm interested in the connections between differential algebra and algebraic geometry; a focus on computation would be appreciated, as well. Any solid foundational books, along with any covering the above topics would be appreciated. EDIT: I must clarify: I'm looking mostly for books covering the general theory of algebraic structures equipped with differential operators, with other books specifically supporting the topics given. When I says "connections to algebraic geometry" I speak about Grobner bases. I'm specifically looking to understand the f4 and f5 algorithms. EDIT 2: Also, I am not a pure math student. I am studying the applications of algebraic techniques to problems in statistics. I'm fine with abstract topics, but only as long as they provide good insight towards concrete problems. I try to stay away from anything that can't be implemented on a computer. REPLY [6 votes]: Except for Buium's book these suggestions mostly cover only algebraic versions of linear differential equations and this is only a limited view of the theory developed by Kolchin and others. Kaplansky remains, I think, the best introduction to the basic algebra in rings with differential operators. There is also Kolchin's book "Differential Algebra and Algebraic Groups" although the latter part of this book is an exposition of algebraic groups Kolchin developed that is hard to follow. The first three chapters are useful though. Finally I would suggest looking at the sequence of papers by Jerry Kovacic who took on, until his untimely death, the task of reformulating differential algebra geoemetry from a scheme theoretic perspective.<|endoftext|> TITLE: What is Known About the Complexity of Calculating Minimal Surface Polyhedra? QUESTION [5 upvotes]: I am currently ruminating about ways of generalizing Minimum Spanning Trees to Minimum Spanning "Hypertrees", where the cost is associated with simplex volumes and, where certain topological constraints must be met (the "Hyper" is borrowed from the use in Hyper Graphs); I also managed to devise a Binary Programming formulation, which could also be adapted for calculating Minimal Surface Area Polyhedra spanning a set of 3D points, but I couldn't find anything new on the complexity of that task; there is a paper by Joseph O'Rourke from 1981 (Polyhedra of minimal area as 3D object models), where it is stated, that it is not known whether it is an NP problem. Question: have there been improvements in determining the computational complexity of finding polyhedra of minimal surface area through a given set of 3D points? REPLY [4 votes]: The problem is NP-hard even for the restricted version when the points form contours lying in parallel planes: Althaus, Ernst, and Christian Fink. "A polyhedral approach to surface reconstruction from planar contours." Integer Programming and Combinatorial Optimization. Springer Berlin Heidelberg, 2002. 258-272. (download link for conf version). Here is an illustration of a part of their proof:   Perhaps also of interest is that there may not exist a polyhedron through given points, i.e., vertices might need be added: Carole Gitlin, Joseph O'Rourke, and Vinita Subramanian. "On reconstructing polyhedra from parallel slices." International Journal of Computational Geometry & Applications 6.01 (1996): 103-122. (link to paper download)                   Work on discrete minimal surfaces might be relevant to your interests, e.g., Polthier, Konrad. "Computational aspects of discrete minimal surfaces." Proc. of the Clay Summer School on Global Theory of Minimal Surfaces, 2002. (link to Konrad's papers).<|endoftext|> TITLE: Are there known ways to posit definable global choice in ZF without positing V=L? QUESTION [8 upvotes]: I need a global choice function defined by a formula in (a fragment of) ZF. There is no harm in assuming V=L for my purposes. But I wonder if there are any familiar alternative ways to get this? The comments make see I also want a weakened part of GCH. Namely the power set of $\aleph_n,\ n\in\mathbb{N}$ should be $\aleph_m$ for some $m\in\mathbb{N}$. I see that HOD is known compatible with some extreme failure of CH, but I have not found what. Is it compatible with failure of this weakened part of GCH? REPLY [9 votes]: Two comments/answers: (1) By an old theorem of Roguski, for any $\Sigma_2^{\text{ZFC}}$ sentence $\phi$, the theories $\text{ZFC} + \phi$ and $\text{ZFC + V=HOD} + \phi$ are equiconsistent. Roguski's result appears in his paper Extensions of models for ZFC to models for ZF+V=HOD with applications, in Set theory and hierarchy theory, pp. 241–247. Lecture Notes in Math., Vol. 537, Springer, Berlin, 1976. (2) Since there is a parenthetical reference to fragments of $\text{ZF}$ in the first line of the question: the formulation and salient consequences of $\text{V=HOD}$ heavily depend on stratification of the universe into rank initial segments of the form $V_\alpha$, and on the veracity of the Montague-Levy reflection theorem; the latter is equivalent over $\text{ZF}$ without $\{\text{Replacement, Infinity}\}$ to the conjunction of $\text{Replacement}$ and $\text{Infinity}$ (by an old result of Azriel Levy). So $\text{V=L}$, rather than $\text{V=HOD}$, is the safe way to arrange global choice, at least for fragments of $\text{ZF}$ that extend Kripke-Platek set theory.<|endoftext|> TITLE: Non-unique splittings of homotopy idempotents QUESTION [15 upvotes]: By a homotopy idempotent I mean a map $f:X\to X$, where $X$ is a space, equipped with a homotopy $f\circ f \sim f$. In contrast to the situation in stable homotopy theory (where $X$ would be a spectrum or a chain complex), not every homotopy idempotent splits; there is a counterexample in Warning 1.2.4.8 of Higher Algebra. I'm interested in the dual question of uniqueness: supposing that a homotopy idempotent splits (up to homotopy), can it have multiple inequivalent splittings? Put differently, can a space have multiple inequivalent retracts that induce the same homotopy idempotent? The meaning of "equivalent" here is somewhat subtle. A homotopy idempotent is, of course, in particular an ordinary idempotent in the homotopy category, and splittings of ordinary idempotents are unique (up to isomorphism). Therefore, any two splittings of the same homotopy idempotent must be incoherently homotopy equivalent. That is, if the first splitting consists of $s:A\to X$ and $r:X\to A$ such that $r\circ s \sim 1$ and $s\circ r \sim f$, and similarly the second consists of $A'$, $s'$, and $r'$, then we must have an equivalence $e:A\simeq A'$ such that $s' \circ e \sim s$ and $e \circ r \sim r'$. However, these homotopies might not be coherently related to the given ones. Note that Corollary 4.4.5.7 and Proposition 4.4.5.12 of Higher Topos Theory imply that to give a splitting of a homotopy idempotent is equivalent to extending it to a "fully coherent idempotent". So the question could equivalently be phrased: can a given homotopy idempotent admit multiple inequivalent coherentifications? One last note: I do mean, as I said, that a homotopy idempotent is equipped with a homotopy $f\circ f \sim f$. If the homotopy is allowed to vary, then certainly the same underlying map $f$ can admit inequivalent splittings/coherentifications. For instance, if $f$ is the identity on $S^1 = B\mathbb{Z}$, then in addition to the trivial homotopy $f\circ f \sim f$ there is a nontrivial one, which also admits a splitting, and any splitting or coherentification will remember the difference between these two homotopies. But if we fix a given homotopy $f\circ f \sim f$, can it be split/coherentified in multiple ways? I expect the answer is yes, but I would like to see an explicit example. REPLY [4 votes]: The question "can a given homotopy idempotent admit multiple inequivalent coherentifications" ought to be approachable by the standard spectral sequence machinery, so let me try to do that. I'll more or less follow the Dwyer-Kan approach. Let $M=\langle f\,|\, f^2=f \rangle = \{1,f\}$, the walking idempotent as a monoid. Let $\mathcal{S}=$ the simplicially enriched category of CW-complexes, and $h\mathcal{S}$ its homotopy category. Clearly, a "homotopy idempotent" gives rise to a functor $M\to h\mathcal{S}$. (But not quite conversely, as you've built a choice of homotopy into the definition.) A "coherentification" of it should amount to a commutative diagram $$ \begin{array}{ccc} {\widetilde{M}} & \to & \mathcal{S} \\ \downarrow &&\downarrow \\ M & \to & h\mathcal{S}, \end{array} $$ where $\widetilde{M}\to M$ is a cofibrant approximation to $M$ in simplicial monoids. Let's fix a space $X$ and $f\colon X\to X$ such that $ff=f$, thus determining a functor $\gamma\colon \widetilde{M}\to M\to \mathcal{S}$. We want to compute something about a full subspace of $$\newcommand{\Map}{\mathrm{Map}} \Map_{s\text{Monoids}}( \widetilde{M}, \Map(X,X) ) $$ whose points are $\phi\colon \widetilde{M}\to \Map(X,X)$ such that $h(\phi)\colon \pi_0\widetilde{M}\to \pi_0\Map(X,X)$ coincides with our chosen $\gamma$. There's a spectral sequence for this: $$ E_2^{s,t}= H_Q^s(M, A_t) \Longrightarrow \pi_{t-s}\Map_{s\text{Monoids}}(\widetilde{M},\Map(X,X))_{\gamma}. $$ The corner $E_2^{0,0}$ is anomalous. It is not given by cohomology; rather, it corresponds to a choice of $\gamma\colon M\to \pi_0\Map(X,X)$, which we have fixed here. We are going to be interested in the groups $E^{t,t}_2=H_Q^t(M,A_t)$ for $t\geq 1$, which potentially contribute to $\pi_0$. The cohomology is "Quillen cohomology" of the monoid $M$. The coefficients are in an abelian group object in $\text{Monoids}_{/M}$. Such an abelian group object amounts to: For each $x\in M$, an abelian group $A(x)$. For each $x,y\in M$, group homomorphisms $x\cdot A(y)\to A(xy)$ and $\cdot y\colon A(x)\to A(xy)$ which are natural (e.g., $x\cdot(y\cdot a)=(xy)\cdot a$ and $1\cdot a=a$, and similarly on the right), and which commute: $(x\cdot a)\cdot y= x\cdot (a\cdot y)$. From this you can build the monoid $A := \coprod A(x)$ with product defined by $$ a\cdot b := (a\cdot y)+ (x\cdot b) \qquad\text{for $a\in A(x), b\in A(y)$.} $$ For a monoid $M$ acting on a space $X$, we use the coefficient systems defined by $$ A_t(f) := \pi_t \Map(X,X)_f \qquad \text{for $f\in M$.} $$ The "$M$-actions" are defined by pointwise pre-or-post composition of a map $a\colon S^t\to \Map(X,X)$ with the constant map $S^t\to *\xrightarrow{f} M\to \Map(X,X)$ where $f\in M$. Quillen cohomology is in principle hard to compute, because you need to choose a cofibrant resolution of $M$. However, there is a theorem: $$ H^t_Q(M;A) \approx H^{t+1}_{HM}(M,\{1\}; A), $$ where the other side is (relative) "Hochschild-Mitchell" cohomology. Basically, Hochschild-Mitchell cohomology is a Quillen cohomology of $M$, regarded as an object in the category of spaces equipped with a two-sided action by $M$. There is a nice bar complex for computing this. Anyway, when I work through all this for $M=\{1,f\}$, I think I get the following. The Quillen cohomology $H_Q^t(M,A)$ for $t\geq0$ is equal to the cohomology of a complex: $$ B \to B \to B \to B \to \cdots. $$ The group $B=A(f)$. The first differential is $a\mapsto f\cdot a + a\cdot f - a$. The second differential is $a\mapsto f\cdot a-a\cdot f$. Then they alternate. The two actions $f\cdot, \cdot f$ are commuting idempotents on the abelian group $B$. Thus, you can write elements of $B$ as $2\times 2$-matrices, so that the actions by $f$ amount to left and right multiplication by $\begin{bmatrix} 1&0\\0&0 \end{bmatrix}$. It is then easy to show that $$ H^t_Q(M, A) =0 \qquad \text{if $t\geq 1$.} $$ Note that $H^0_Q(M,A)$ is not generally $0$. Thus we ought to conclude that the space $\Map_{s\text{Monoid}}(\widetilde{M}, \Map(X,X))_\gamma$ is connected, with higher homotopy groups $\pi_t= H^0_Q(M,A_t)$. Of course, this is not exactly right: the coefficient system $A_1$ is a group object, but may not be an abelian group object. So I need to contemplate the non-abelian cohomology $$ E_2^{1,1} = H^1_Q(M, A_1),\qquad A_1(f) = \pi_1\Map(X,X)_f. $$ That's more than I want to do right now. In any case, it seems to me the answer to your question comes down to this single group. [Roughly, this group seems to have something to do with "adjusting" a choice of homotopy $\alpha\colon f\sim ff=f$ by a choice of homotopy $\beta\colon f\sim f$; it's something like the orbits of the action $\alpha\mapsto (\beta\beta)\alpha(\beta^{-1})$ on the set $\{\alpha\,|\,\alpha\cdot f=f\cdot \alpha\}$.] (It's possible, of course, that I've misinterpreted what you mean by "inequivalent" coherentifications. The approach I described presupposes one particular notion of equivalence, but there are others.) (Or I may have screwed up some other way.)<|endoftext|> TITLE: Obstructions for $E_n$-algebras QUESTION [6 upvotes]: In Alan Robinson's paper, Classical Obstructions and S-algebras, he provides conditions for a ring spectrum to have an $A_n$ and $\mathbb{E}_\infty$-structure. Have the obstructions for an object of an $\infty$-operad $\mathcal{C}^\otimes$ from having an $\mathbb{E}_n$-algebra structure been studied? More precisely, what conditions must an object $\mathscr{C}$ of $\mathcal{C}^\otimes$ satisfy to be an object of $\mathrm{Alg}_{/\mathbb{E}_n}(\mathcal{C}^\otimes)$? REPLY [14 votes]: Let me expand a little on what Qiaochu and Craig mentioned. If you want an obstruction theory for building an uber-gadget, you'll need (i) an algebraic approximation to such gadgets, and (ii) a way to resolve every uber-gadget by special uber-gadgets where the algebraic approximation remembers everything about the thing you started with. Condition (i) usually takes the form of a functor $Uber \rightarrow Approx$ with some nice properties, and then for (ii) you hope that maybe there is an adjoint to this functor, and that $Uber$ is generated under sifted colimits by the essential image of this adjoint (I'm sweeping everything important and techncal under the rug). Once you have this, you can start trying to formulate the problem of starting with an object in $Approx$ and adding structure until you get to an object in $Uber$ (this shouldn't be too surprising: you've essentially required that $Uber$ is monadic over $Approx$ so that elements of $Uber$ are like elements of $Approx$ with extra structure). If you do this very carefully, you'll have expressed the space of objects lifting a given one. All of this is developed for $E_\infty$-ring spectra in Goerss-Hopkins and you can find write-ups here: http://www.math.northwestern.edu/~pgoerss/ . I'm pretty sure the same proofs give you an $E_n$-obstruction theory, and this appears elsewhere as well. Now, you probably already knew all that. But I said it to remind you that there's just no hope of proceeding if you don't have (ii) and for (ii) you need your algebraic approximation to be strong enough. So you shouldn't ask: "Given a spectrum, what's the space of $E_\infty$ ring structures on is?" You need to have a little bit to get going, some candidate for the Dyer-Lashof operations or (in the Goerss-Hopkins case) a commutative comodule algebra over $E_*E$ for a suitable cohomology theory, $E$. Otherwise your answer will probably be ridiculous and uncomputable (this is sort of what Qiaochu was pointing out.)<|endoftext|> TITLE: Is the homomorphism poset directed if the codomain is directed? QUESTION [12 upvotes]: Let $P,Q$ be partially ordered sets (posets). We consider the set $\text{Hom}(P,Q)$ of order-preserving functions $f:P\to Q$. (We call a function $f:P\to Q$ order preserving if $x\leq y$ in $P$ implies $f(x)\leq f(y)$ in $Q$.) There is a natural ordering relation on $\text{Hom}(P,Q)$ given by $f\leq g$ if and only if $f(p) \leq_Q g(p)$ for all $p\in P$. Let $D$ be directed and $P$ be a poset. Is $\text{Hom}(P,D)$ necessarily directed? REPLY [7 votes]: Let me give a simple necessary and sufficient condition which characterizes which directed sets $D$ satisfy the property that $Hom(P,D)$ is directed for all posets $P$. Suppose that $D$ is a poset. Then I claim that $Hom(P,D)$ is directed for every poset $P$ if and only if there is a function $L:D\times D\rightarrow D$ such that $L(x,y)\geq x,L(x,y)\geq y$ for each $x,y\in D$ and where if $x_{1}\leq x_{2},y_{1}\leq y_{2}$, then $L(x_{1},y_{1})\leq L(x_{2},y_{2})$. In other words, $Hom(P,D)$ is directed for each poset $P$ if and only if $D$ is $``$monotonely directed$"$ or $``$uniformly directed$"$. The proof is straightforward, but let me give a proof below regardless. Suppose there is such a function $L:D\times D\rightarrow D$. Then whenever $f,g\in Hom(P,D)$ are order preserving, then define $h(x)=L(f(x),g(x))$. Then whenever $x\leq y$, we have $f(x)\leq f(y),g(x)\leq g(y)$, so $h(x)=L(f(x),g(x))\leq L(f(y),g(y))=h(y)$, so $h$ is order preserving. Furthermore, $h(x)=L(f(x),g(x))\geq f(x)$ and $h(x)=L(f(x),g(x))\geq g(x)$, so $h\geq f$ and $h\geq g$, so $Hom(P,D)$ is always directed. Now suppose that $Hom(P,D)$ is directed for each poset $P$. Then $Hom(D\times D,D)$ is directed where $D\times D$ is given the partial ordering where $(x_{1},y_{1})\leq(x_{2},y_{2})$ if and only if $x_{1}\leq x_{2}$ and $y_{1}\leq y_{2}$. Let $\pi_{1},\pi_{2}:D\times D\rightarrow D$ be the projection mappings. In other words, $\pi_{1}(x,y)=x,\pi_{2}(x,y)=y$. Then $\pi_{1},\pi_{2}$ are order preserving, so there is some $L\in Hom(D\times D,D)$ with $\pi_{1},\pi_{2}\leq L$. Therefore, $L$ is an order preserving map with $L(x,y)\geq\pi_{1}(x,y)=x$ and $L(x,y)\geq\pi_{2}(x,y)=y$.<|endoftext|> TITLE: Higher order arithmetic and fragments of ZFC QUESTION [6 upvotes]: Zbierski "Models for Higher Order Arithmetics" (BULL. DE L'ACAD. POLONAISE DES SCIENCES Serie des sciences math., astr. et phys. - Vol. XIX, No. 7, 1971) defines ZF$_n$ as ZFC with the power set axiom limited to $n$ successive power sets starting from the natural numbers $\mathbb{N}$. Note this includes the axiom of choice. He proves ZF$_n$ is a conservative extension of $n+2$ order arithmetic with the axiom scheme of choice.. He notes Gandy had proved already ZF$_2$ is not conservative over $4$th order arithmetic without the axiom scheme of choice. But what is known about the case where the axiom of choice is also removed from ZF$_2$? Sochor "Constructibility in higher order arithmetics" (Arch. Math. Logic (1993) 32:381-389) shows that $n$-th order arithmetic without choice can interpret $n$-th order arithmetic with choice.though obviously it is not a conservative extension. So the issue is not equiconsistency. It is conservativity. REPLY [3 votes]: The usual way to prove something like this is to take a model of $(n+1)$-st order arithmetic $\mathcal{A}$ and interpret a model of set theory $\mathcal{V}$ in $\mathcal{A}$ such a the usual interpretation of $(n+1)$-st order arithmetic in $\mathcal{V}$ leads to a model isomorphic to $\mathcal{A}$. There are a few recipes to do this, using graph, trees and other convenient encodings of sets. See the discussion here, I will use wellfounded accessible pointed graphs (WAPGs), though Zbierski probably uses trees (I don't know since I haven't found the paper). The model $\mathcal{A}$ of $(n+1)$-st order arithmetic has sorts $N = S_0,S_1,\ldots,S_n$, where $N$ is the number sort and each $S_{i+1}$ consists of subsets of the previous sort $S_i$. The number sort $N$ has the usual arithmetic structure with full induction for the implied language. Each sort $S_i$ ($i \geq 1$) has full comprehension within the implied language. I'll refrain from making choice assumptions to see how far things go. Each sort has a definable pairing function, so we can talk about tuples, relations and functions. In particular, we can talk about WAPGs coded in $S_n$. The model $\mathcal{V}$ is obtained by collecting all WAPGs in the top sort $S_n$. Equality $G \equiv H$ between two WAPGs $G$ and $H$ is interpreted as the existence of a bisimulation between $G$ and $H$, and membership $G \in H$ is interpreted as the existence of a bisimulation between $G$ and an immediate subgraph $H/x$ of $H$. After quotienting by $\equiv$ if desired, the result is an interpretation of the language of set theory which is automatically extensional and wellfounded (in the internal sense). Furthermore, since WAPGs coded in $S_n$ and bisimulations between them are definable in $(n+1)$-th order arithmetic, every formula $\phi$ in the language of set theory has a translation $\phi^V$ in the language of $(n+1)$-th order arithmetic such that $\mathcal{V} \vDash \phi \iff \mathcal{A} \vDash \phi^V$. Furthermore, we can define canonical WAPGs for each sort $\omega,\mathcal{P}(\omega),\ldots,\mathcal{P}^{n-1}(\omega)$. Specifically, the WAPG for $\omega$ is the graph of the ordering relation on the sort $N$; the WAPG for $\mathcal{P}(\omega)$ is built on top of that one by adding a node for each $X \in \mathcal{S}_1$ and linking it to its elements; etc. Using the fact that $\mathcal{A}$ has full comprehension and the translation, we see that each $\mathcal{P}^{i+1}(\omega)$ really is the powerset of the previous one as understood in $\mathcal{V}$. So the natural interpretation of the arithmetical sorts $\omega,\mathcal{P}(\omega),\ldots,\mathcal{P}^{n-1}(\omega)$ in $\mathcal{V}$ are canonically isomorphic to the original sorts $N,S_1,\ldots,S_{n-1}$ from $\mathcal{A}$. The top sort $\mathcal{P}^n(\omega)$ is not necessarily a set in $\mathcal{V}$ but it is a definable class. Again, full comprehension can be used to argue that $\mathcal{P}^n(\omega)$ is canonically isomorphic to the original top sort $S_n$. So far, we haven't needed any choice; I only used full comprehension in $\mathcal{A}$. It remains to check which axioms $\mathcal{V}$ satisfies. Many are straightforward to check. We already checked extensionality, foundation and infinity. Pairing and union correspond to simple combinatorial manipulations of WAPGs. Comprehension follows from comprehension in $\mathcal{A}$. Replacement/collection are problematic without further assumptions about $\mathcal{A}$. There is a good reason for that since we have just shown one half of: Theorem. The theory $\mathrm{Z}^-_{n-1}$ (extensionality, foundation, pairing, union, comprehension and the existence of $\omega,\mathcal{P}(\omega),\ldots,\mathcal{P}^{n-1}(\omega)$) is a conservative extension of $(n+1)$-th order arithmetic (without choice). The missing half is the straightforward proof that $\mathrm{Z}^-_{n-1}$ is an extension of $(n+1)$-th order arithmetic. In other words, that the interpretation $N = \omega, S_1 = \mathcal{P}(\omega), \ldots, S_n = \mathcal{P}^n(\omega)$ always yields a model of $(n+1)$-th order arithmetic. Here is a reason why some choice principles might be necessary. For simplicity, I will assume we are working in $2$-nd order arithmetic. I will show that if collection holds in $\mathcal{V}$ then countable choice holds in $\mathcal{A}$. Say $\phi(n,x)$ is a formula in $2$-nd order arithmetic that relates each natural number $n$ with a set $x$ of natural numbers. We can translate $\phi(n,x)$ to a synonymous formula $\psi(n,x)$ in $\mathcal{V}$ such that $(\forall n \in \omega)(\exists x \subseteq \omega)\psi(n,x)$. Every set in $\mathcal{V}$ is countable, by construction, so if there is a set $b \in \mathcal{V}$ such that $(\forall n \in \omega)(\exists x \in b)\psi(n,x)$, then back in $\mathcal{A}$ there is a countable sequence $\langle x_n \rangle$ coded in $S_1$ such that $\phi(n,x_n)$ holds for every $n$. In a similar fashion, for any $n$, if $S_{n-1}$ is wellorderable in a model $\mathcal{A}$ of $(n+1)$-st order arithmetic, then collection in $\mathcal{V}$ gives choice for definable $S_{n-1}$-indexed families in $\mathcal{A}$.<|endoftext|> TITLE: Domains of raising and lowering operators in QM QUESTION [6 upvotes]: Let $H : \operatorname{dom}(H) \subset L^2(\Omega) \rightarrow L^2(\Omega)$, where $dom(H) \subset H^2(\Omega)$, $\Omega \subset \mathbb{R}$ should be a bounded open interval(so 1-d setting(!)) and $H$ be a self-adjoint Schrödinger operator $H = -\frac{d^2}{dx^2} + V$ and $V \in C^{\infty}$. Now assume that the Schrödinger equation has a ground-state solution $\psi$ (with $\psi > 0$- not sure if we actually need this, but it should definitely simplify quite something). By rescaling the potential, we can get that $H \psi = 0.$ Then we define $W(x) :=\frac{\psi'(x)}{\psi(x)}$ and closed(!) operators $A = \frac{d}{dx}+ W $ and $A^* = -\frac{d}{dx}+W$. Now, my question is: Given this situation, are the domains of the operators $A,A^*$ fixed? Or is there at least a standard way to construct an appropriate domain of $A$ by using the domain of $H$ ? REPLY [2 votes]: In the applications of the method that I know of, one usually doesn't make explicit use of the operators $A, A^*$; rather, this is a method to go from the original potential $V=W^2-W'$ to a new potential $W^2+W'$ that has an extra eigenvalue. However, if the original operator $H$ has Dirichlet boundary conditions $y(a)=y(b)=0$ (here I assume that $\Omega=(a,b)$), then you could take $$ D(A)= \{ y\in H^1: y(a)=y(b)=0 \} , $$ so $D(A^*)=H^1$, and $H=A^*A$ comes out right, with the correct domain. You now definitely want to assume that $0$ is below the spectrum of $H$; if $0$ is the bottom of the spectrum, then $\psi(a)=\psi(b)=0$ and $W$ is not locally integrable near $a,b$ and this doesn't work. The operator $H_2$ then has boundary conditions $y'=Wy$ at $a,b$. For more general boundary conditions, this doesn't seem to work.