TITLE: Topology on $p$-adic period rings in an article by Fontaine, part II QUESTION [6 upvotes]: This is a follow-up to this question. See that question for background and relevant notation. In the answer to that question, it is claimed, if I understand the answer correctly, that a basis of neighborhoods of $W(R) \otimes_{W(k)} K = W(R)[1/p]$ is given by sets of the forms $p^{-n}W(\mathfrak{a})+p^NW(R)$ for $\mathfrak{a}$ a non-zero ideal of $R.$ On the other hand, if I consider the following notes, pg. 65, Exercise 4.5, it is claimed that a basis of neighborhoods of $W(R)[1/p]$ is given by sets of the form $$U_{N,\mathfrak{a}} = \bigcup_{n > -N} p^{-n}W(\mathfrak{a}^{p^n})+p^NW(R), N \geq 0,$$ where $\mathfrak{a}$ is a non-zero ideal. Is it true that the two topologies thus defined are the same? If so, why is this the case? It is easy to see that each $U_{N,A}$ contains a basis element of the form $p^{-n}W(A)+p^NW(R),$ but the other inclusion is not clear to me (is it even true?). REPLY [6 votes]: Let me just make a small comment that while the question may seem subtle from the perspective of topological rings, from the perspective of condensed rings everything really is "obvious". The reduction $\mathcal{O}_{\mathbb{C}_p}/p$ is discrete, so there's no question there. Then you just place that discrete ring in the condensed world, and do all the constructions in there. So form the inverse limit over Frobenius $\mathcal{O}_{\mathbb{C}_p}^\flat$, pass to Witt vectors, then invert $p$. This gives a condensed ring: there's nothing to check. On $T$-valued points for a profinite set $T$, it's the analogous period ring construction for the perfectoid algebra $Cont(T;\mathbb{C}_p)$ replacing $\mathbb{C}_p$. From this perspective, the condensed or topological structure on period rings is just a shadow of their structure as a pro-etale sheaf on perfectoid spaces, as exploited in Scholze's old (?!) article on $p$-adic Hodge theory of rigid analytic spaces. Now, there still remains the question of comparing this condensed structure to the topological structure in Fontaine's definition. Recall that there is a fully faithful functor, $X \mapsto \underline{X}$ with $\underline{X}(T)= Cont(T,X)$, from compactly generated weak Hausdorff topological spaces to condensed sets; this passes to rings as well. The Witt vectors of $\mathcal{O}_{\mathbb{C}_p}^\flat$ is a countable inverse limit of discrete rings, so it is a sequential Hausdorff space and hence compactly generated. Now it suffices to note that multiplication by $p$ map is a closed inclusion (the quotient is $\mathcal{O}_{\mathbb{C}_p}^\flat$ as the latter is perfect), and that the fully faithful functor $X \mapsto \underline{X}$ above sends sequential unions along closed inclusions to sequential unions, by the standard fact in point-set topology that if a compact Hausdorff space is a sequential union of closed subpsaces, then it is equal to one of those closed subspaces. However, probably in general the topological structure is not so important once you have the condensed structure. So one should probably just be happy with the first paragraph of this answer.<|endoftext|> TITLE: Bounding maximum probabilities in sum of i.i.d discrete RVs QUESTION [9 upvotes]: Let $X$ be a a discrete RV with $\mathbb{P}(X=k) TITLE: Derived base change in étale cohomology QUESTION [9 upvotes]: Given a commutative square of ringed topoi $$\begin{array}{ccc}X'\!\! & \overset{f'}\to & Y'\!\! \\ \!\!\!\!\!{\small g'}\downarrow & & \downarrow{\small g}\!\!\!\! \\ X & \underset f\to & Y\end{array}\label{Dia square}\tag{1}$$ and an object $K \in D(X)$, there is a canonical base change morphism $$Lg^*Rf_* K \to R(f')_* L(g')^* K;\label{Dia base change}\tag{2}$$ see for example [Stacks, Tag 07A7]. Coherent story. Assume (\ref{Dia square}) is a pullback square of schemes (with their categories of $\mathcal O_X$-modules), where $f$ is qcqs and $K \in D_{\mathbf{Qcoh}}(X)$. If $g$ is flat, then (\ref{Dia base change}) is an isomorphism [Stacks, Tag 02KH]. This holds more generally if $X$ and $Y'$ are tor independent over $Y$ [Stacks, Tag 08IB]. There are also versions demanding only that $K = \mathscr F \in \mathbf{Qcoh}(X)$ is flat over $Y$ and of finite presentation on $X$, at least if $f$ is proper and of finite presentation [Stacks, Tag 0B91]. It seems that the failure of (\ref{Dia base change}) to be an isomorphism is mostly explained by derived phenomena. When $X$ and $Y$ are not tor independent, it seems conceivable to me that the failure of (\ref{Dia base change}) to be an isomorphism can be salvaged by taking $X'$ to be the derived base change $X \times_Y^{\mathbf L} Y'$ instead of the usual base change (are there any results of this type?). Étale setting. Now let (\ref{Dia square}) be a pullback square of schemes (or even $\mathbf C$-varieties if you like), but now equipped with the étale topoi of $\mathbf Z/n$-sheaves. If you like, you may assume $K \in D_{ctf}(X_{\operatorname{\acute et}}, \mathbf Z/n)$ (constructible with locally finite tor dimension, see e.g. [Stacks, Tags 03TQ and 08CG]). Because the sheaf of rings $\mathbf Z/n$ doesn't change, all pullbacks $g^*$ are already exact, so $Lg^* = g^*$. In particular, we already see that flatness and tor independence are not the issue. However, the base change map (\ref{Dia base change}) is not always an isomorphism: Example. Let $X = \mathbf A^2 \setminus \{0\}$ and $Y = \mathbf A^1$, where $f$ is the first coordinate projection. Let $Y' \subseteq Y$ be the origin, so that $X' = \mathbf A^1 \setminus \{0\}$. Let $K = \mathbf Z/n[0]$. Then $Rf_*K$ is a complex whose $H^0$ is $\mathbf Z/n$ and whose $H^3$ is $g_*(\mathbf Z/n)$ (the constant sheaf at the origin). On the other hand, $(g')^* = \mathbf Z/n[0]$, and $R(f')_*(g')^*K$ is a complex whose $H^0$ is $\mathbf Z/n$ and whose $H^1$ is $\mathbf Z/n$. So it differs from $g^*Rf_*K$. Question. Can the failure of cohomology to commute with base change be explained by certain derived phenomena? For example, is the situation better if we replace $X'$ with some sort of homotopy fibre (when $Y' \subseteq Y$ is a point) or more generally a homotopy fibre product? It seems that there cannot be a Grothendieck complex that computes $Rf_*K$ after arbitrary base change, because $g^*$ is always exact. So the situation is really fundamentally different from the coherent case, where the non-flatness of $g^*$ accounts for the failure of cohomology to commute with nonderived base change (at least in the situations described above where derived base change holds). REPLY [5 votes]: Here is a long comment about the étale setting. If $X\to Y$ is a universal homeomorphism of schemes (or Deligne-Mumford stacks), then it induces an equivalence of small étale sites, and thus of topoi: $$X_{\acute et}\cong Y_{\acute et} .$$ This remains true for derived schemes/Deligne-Mumford stacks: is X is a derived scheme, and if $tX$ denotes its truncation (same underlying space but with $\pi_0(\mathcal O_X)$ as structural sheaf), then the canonical closed immersion $tX\to X$ induces an equivalence of small étale sites (i.e. it is the same to determine an étale map to $X$ or an étale map to $tX$). In particular, extending the formalism of étale cohomology to derived schemes does not change the cohomology groups nor the (derived) categories of étale sheaves. This does not mean that derived geometry is not interesting in this context (on the contrary, this means that there are new cohomology classes which have a geometric interpretation if we allow derived geometry in the picture). But this means that derived geometry will not change anything about the obstruction to base change in the étale context. However, one may see the theory of sheaves as an extension of intersection theory, where (constructible) sheaves are "cycles". In fact, if $F$ is a constructible sheaf on $X$, one defines a cycle (or equivalently a constructible function with values in $\mathbf Z$) by taking the rank (or the Euler characteristic) of each fiber of $F$. If $F$ and $G$ are in $D^b_{ctf}(X,\Lambda)$, then the (derived) internal Hom $$Hom(F,G)$$ may be seen as an "intersection pairing". There is a pullback of sheaves induced by a morphism of schemes $f:X\to Y$, $$f^*:D^b_{ctf}(Y,\Lambda)$\to D^b_{ctf}(X,\Lambda).$$ There is also a pushforward $Rf_!$ (with compact support). If $f:X\to Y$ is a map of schemes over a base schemes $S$ and $F\in D^b_{ctf}(X,\Lambda)$, we may wonder if the formation of $Rf_*(F)$ is compatible with base change of the form $S'\to S$. It so, we may say that $F$ and $f$ are transversal to each other. Deligne's generic base change theorem says that there is a dense open subscheme of $S$ over which $F$ and $f$ are indeed transversal. On the other hand, if $f$ is smooth, it is transversal to any $F$. It is tempting to say that $F$ is smooth if it is transversal to any $f$; the good news is that this is exactly what we do, up to a translation between French and English: in the $\ell$-adic context, when $\Lambda=\mathbf Z_\ell$, $\mathbf Q_\ell$, $\bar{\mathbf Q}_\ell$, one can prove that a sheaf $F$ is smooth in the sense above if and only if it is lisse in the usual sense! In this sense, Deligne's generic base change theorem is a kind of sheaf-theoretic Bertini theorem. One can also say when two sheaves $F,G\in D^b_{ctf}(X,\Lambda)$ are transversal to each other: this is when $$f^*Hom(F,G)\cong Hom(f^*F,f^*G)$$ for all maps $f:X'\to X$. Another notion of smoothness for a sheaf $F$ would be that it is transversal to all sheaves $G$. One shows that this notion of smoothness relatively to sheaves is in fact equivalent to the notion of smoothness relatively to morphism of schemes, i.e. to the property of being lisse. Interestingly enough, we have a big locus of smoothness: for any $F\in D^b_{ctf}(X,\Lambda)$, there is a dense open subscheme $U$ such that $F|_U$ is lisse in $D^b_{ctf}(U,\Lambda)$. The transversality property between sheaves (or between sheaves and morphisms) can be seen through the singular support (whose behaviour is very strongly related to vanishing cycles). This is developped in full in the context of (archimedian) analytic geometry in the book of Kashiwara and Schapira "Sheaves on Manifolds" and has been developped recently in the context of étale sheaves on schemes by A. Beilinson and T. Saito. There remains the problem of the compatibility between the operations on schemes and the operations on the associated cycles (there is a more clever way to construct cycles associated to sheaves: the characteristic cycles). In particular, with the pushforward. From a curve to a point, this is the Grothendieck-Ogg-Shafarevitch formula; the general case is not fully understood yet (in positive characteristic). But the conclusion of all this is that the behaviour of base change formulas, that is the control of pushforward functors has maybe more to do with ramification theory than with derived geometry. This does not mean that derived methods have nothing to say though: the work of Toën and Vezzosi on Bloch's conductor formula goes through non-commutative motives, Voevodsky's motives and trace formulas for modules over $E_n$-algebra with $n<\infty$.<|endoftext|> TITLE: Number of irreducible representations of $SO_3(\mathfrak{o}/\mathfrak{p}^l)$ QUESTION [6 upvotes]: $\DeclareMathOperator\SO{SO}$Let $F$ be a finite extension of $\mathbb{Q}_p$, and let $\mathfrak{o}$ denote the ring of integers, with maximal ideal $\mathfrak{p}$. Let $G_l$ denote the finite group $\SO_3(\mathfrak{o}/\mathfrak{p}^l)$. Question: Is there a formula for the number of irreducible representations of $G_l$ in a given dimension $d$? Remarks: 1. I am aware of the results of Aizenbud–Avni - Representation growth and rational singularities of the moduli space of local systems (which gives an estimate $C d^{22}$ for the number I'm interested in, over $\mathfrak{o}$) and of Jaikin-Zapirain - Zeta function of representations of compact $p$-adic analytic groups (which gives a qualitative result on the representation zeta function). Both hold for more general algebraic groups. My question is whether there exists a quantitative, explicit (possibly recursive in $l$) formula in the special case of $\SO_3$. For $\operatorname{GL}_2$ in place of $\SO_3$, such a formula can be found in Onn - Representations of automorphism groups of finite O-modules of rank two (Thm. 1.4). Remark: This question was originally posed for $\SO_2$. As Paul Broussous pointed out, it is trivial in this case since $\SO_2$ is abelian. REPLY [3 votes]: As far as I know, there is currently no such explicit formula in the literature. In fact, even for $\mathrm{SO}_3(\mathbb{F}_q)$ (i.e., the case $l=1$), I have not seen a neat table of the dimensions and multiplicities of all irreps, although this could certainly be obtained from the work of Lusztig, Irreducible representations of finite classical groups, Invent. Math. 43, 125-175 (1977). For some groups of type A of low rank over $\mathfrak{o}_l$, the representation zeta function has been determined explicitly, as long as the residue characteristic of $\mathfrak{o}$ is not too small: Avni, Klopsch, Onn, Voll, Similarity classes of integral (\mathfrak p)-adic matrices and representation zeta functions of groups of type (\mathsf{A}_{2}), Proc. Lond. Math. Soc. (3) 112, No. 2, 267-350 (2016); see Theorem C. In principle, one can extract the number of irreps of dim $n$ as the $n$-th coefficient in the Dirichlet series expansion of the representation zeta function, but in practice this can be difficult. Note that the bound $Cd^{22}$ of Aizenbud--Avni is an upper bound on the number of irreps of dimension at most $d$, so it's not exactly the number you are asking about. Also, for particular types, this bound can be made stronger. In any case, an upper bound is something rather different than an exact formula.<|endoftext|> TITLE: Compactly generated vertex stabilisers in compactly generated t.d.l.c. groups acting on trees QUESTION [5 upvotes]: In the article cited below, I. Castellano gives a proof for the following result (Proposition 4.1). Let $G$ be a compactly generated totally disconnected locally compact group. Suppose that $G$ acts discretely on a tree $\mathcal T$ such that the group $G$ is acting without edge inversions; the quotient graph $G\backslash\mathcal T$ is finite; the edge stabilisers $G_e$ are compact open subgroups of $G$. Then the vertex stabilizers $G_v$ are compactly generated. Here acting discretely means that the stabilizers are open subgroups of $G$. The proof as cited is rather involved and uses cohomology arguments, while the proposition looks rather innocent. Is there a more straightforward, elementary proof for this result? Note that the proposition should also hold for trees that are not locally finite (and in this case, the topology becomes a lot more subtle). [1] Castellano, I., Rational discrete first degree cohomology for totally disconnected locally compact groups. arXiv:1506.02310 [math.GR]. REPLY [3 votes]: As YCor suggests, proceed by Bass–Serre theory. We can write $G$ in the form $\frac{G_{v_1} \ast \dots \ast G_{v_m} \ast F(E)}{\langle \langle \overline{e}\alpha_e(g)e\alpha_{\overline{e}}(g)^{-1} \; (g \in G_e), \; e\overline{e}, \; e \; (e \in E') \rangle \rangle}$ where $E$ is a set of representatives for the edges in the quotient graph, $E' \subseteq E$ is a set of representatives for the edges in a spanning tree of the quotient graph, and $\alpha_e: G_e \rightarrow G_{o(e)}$ is the natural embedding of an edge group into the vertex group of the origin of the edge. Note that $E$ is finite and the Bass–Serre relations are gluing together vertex groups along compact open subgroups $G_e$. (This also ensures that everything is fine in terms of the group topology: you can just start with the topology of one of the edge groups, and then as a group topology, it extends uniquely to the whole of $G$.) Write $F$ for the image of $F(E)$ in $G$. Let's look at $G_{v_1}$. We have a compactly generated open subgroup $H_0$ of $G_{v_1}$ generated by $\alpha_e(G_e)$ for all $e \in E$ such that $o(e)=v_1$. In other words, all the amalgamation of $G_{v_1}$ with the other vertex groups happens inside $H_0$. Now write $G_{v_1}$ as a directed union $\bigcup_{i \in I}H_i$, where each $H_i$ is compactly generated and $0$ is the least element of $I$. Let $K_i$ be the subgroup of $G$ generated by $H_i \cup G_{v_2} \cup G_{v_3}\cup\dots\cup G_{v_n} \cup F$. Then $G = \bigcup_{i \in I}K_i$; since $G$ is compactly generated, in fact $G = K_i$ for some $i$. In particular, every $g \in G_{v_1}$ can be expressed as a product of elements of $H_i \cup G_{v_2} \cup G_{v_3}\cup\dots\cup G_{v_n} \cup F$. Using the normal form theorem for graphs of groups, we conclude that $g \in H_i$. Thus $G_{v_1}$ is compactly generated.<|endoftext|> TITLE: Do Sobolev spaces contain nowhere differentiable functions? QUESTION [26 upvotes]: Does the Sobolev space $H^1(R^n)$ of weakly differentiable functions on a bounded domain in $R^n$ (or a more general Sobolev space) contain a continuous but nowhere differentiable function? REPLY [18 votes]: As Nate Eldredge pointed out, $W^{1,2}(\mathbb{R})$ functions are absolutely continuous on $\mathbb{R}$, and therefore differentiable a.e., and so the answer is no. For $n\geq 2$, the answer is yes. When n=2, this is a classical result of L. Cesari Cesari, Lamberto, Sulle funzioni assolutamente continue in due variabili, Ann. Sc. Norm. Super. Pisa, II. Ser. 10, 91-101 (1941). ZBL0025.31301. see also an explicit construction applicable to $W^{1,n}(\mathbb{R}^n)$ in J. Serrin, Serrin, J., On the differentiability of functions of several variables, Arch. Ration. Mech. Anal. 7, 359-372 (1961). ZBL0109.03904. The paper of Cesari also contains an a.e. differentiability result for $W^{1,p}$, $p>2$ (in two dimensional settings); this was generalized to higher dimensions by A. Calder\'on in Riv. Mat. Univ. Parma, 1951. For $n>2$, it looks like a suitable construction is given at the question Are functions of bounded variation a.e. differentiable?.<|endoftext|> TITLE: How to accelerate progress in mathematical research? QUESTION [23 upvotes]: After completing a Ph.D. in pure mathematics, 10 years ago I left academia for working in industry. There, a typical question is "What can we do to accelerate $x$?" when a project is slowed down, and the typical answer is "Let the people concerned with the issue focus on it and/or bring in some experts", which usually solves the issue. I wonder if mathematical research can work the same way. Say, if you had $100 million to spare and really wanted to see the Riemann hypothesis resolved, what would you do? Would it help to finance a special decade at some institution, where 25 leading researchers are free from everyday concerns (in particular, administrative and teaching duties) and can spend their entire time working on this problem together? Would it be better to use these funds to let the 25 experts each supervise 10 graduate students over a course of 20 years? Or to support some sort of crowdsourcing? Or is it just not possible to focus exclusively on one (incredibly difficult) problem and one should rather pursue whatever is doable at the moment? Is it similar to (paraphrasing Don Knuth) "Computer science is like the Great Wall of China where each workman contributes a brick"? REPLY [6 votes]: Here's one budget for spending the money over 10 years. Obviously all the numbers are only indicative. $\$$75 million for child care for mathematically trained people who want to work on these issues, an average of $\$$7,500 per year for 10 years for 1000 people each year. Household cleaning and food preparation could also be included. This would free up the time of current researchers, and open up the research to mathematically trained people, especially women, who are spending their time on work in the household instead of research. $\$$10 million as prize money for unconditional proofs of known consequences of either the Riemann or Generalized Riemann Hypothesis. Perhaps this would be 20 prizes, each worth $\$$500,000, based on the lists of consequences here, here or here. The ideas in those proofs would be good sources ideas for proving the Riemann hypothesis. $\$$10 million in travel grants to encourage global collaboration on these topics. Perhaps this would be 10 years of 400 grants per year of $\$$2,500 each, covering airfare and a week of expenses in each case. $\$$4 million to help people write up their research or their expository works in the area. Perhaps this means that for each of 10 years there are 10 people being paid an average of $\$$40,000 per year to help write up this work. $\$$1 million to make existing numerical research in the area more accessible, e.g. better access to tables of the zeroes, translations of relevant algorithms into nice packages in several languages. Note all of this money may go further in countries with high levels of mathematics but cheaper costs of living. Conditions for work on the Riemann hypothesis are already relatively good for math professors at American or European universities; to make a big step forward, it may help to involve people who are mathematically talented but not in those roles, for whatever reason.<|endoftext|> TITLE: How abelian can a non-abelian finitely generated group be? QUESTION [9 upvotes]: For a finite subset $S$ of a group write $$k(S)=\frac{|\{(g,h)\in S^2 \,|\,gh=hg\}|}{|S|^2}$$ for the propability that two elements in $S$ commute. It is known that $k(G)>5/8$ implies that a finite group $G$ is abelian. Is there any result in this direction for $G$ finitely generated? For example: If we denote by $B_r(G)$ the elements of word-length at most $r$ w.r.t. some finite generating set, is there a condition on the limit $$\lim_{r\longrightarrow\infty} k(B_r(G))$$ (or maybe on $\liminf$ or $\limsup$) that implies $G$ abelian? REPLY [7 votes]: Try not to be seduced by balls. When it comes to infinite groups, you get much better theory if you focus on random walks. For example, the most basic thing you might ask of your density function is that every subgroup $H\leq G$ have density $1/[G:H]$. This is true if you measure density with random walks, by the basic theory of Markov chains (I have heard the infinite-index case attributed to Avez, but see also these two previous MO questions.). On the other hand, if you measure density with balls then you can have an infinite-index subgroup with density nearly $1$ (see here for example). If you measure density with a random walk, then there is a very satisfactory and unsurprising answer to your question. A group has positive commuting probability if and only if it is virtually abelian, and in particular if it has commuting probability more than $5/8$ then it must be abelian. (In fact, if a group has positive commuting probability then its commuting probability is also realized by some finite group.) For all this, see this paper of Tointon. The same results hold for balls, provided the balls have subexponential growth. As mentioned in the comments, it has been conjectured by Antolin, Martino, and Ventura that the same is true of balls of exponential growth. I am a little sceptical of this conjecture, since as mentioned there can be infinite-index subgroups of positive density. It was admittedly essential in that construction that the subgroup have exponential growth, but I cannot at the moment convince myself that you can't have a positive-density infinite-index abelian subgroup, or even center.<|endoftext|> TITLE: Furthest distance half the diameter? QUESTION [7 upvotes]: Let $S$ be the surface of a convex body, polyhedral or smooth, embedded in $\mathbb{R}^3$. For a point $x \in S$, let $F(x)$ be the set of furthest points from $x$, measured by shortest paths on the surface $S$. Let $f(x)$ be the length of those shortest paths: $|x y|$ for $y \in F(x)$. It seems natural to hope that Hypothesis: For any $x \in S$, $f(x) \ge \tfrac{1}{2} \mathrm{diam}(S)$. Here $\mathrm{diam}(S)$ is the maximum distance between any two points on $S$ (again measured by shortest paths on the surface of $S$). Suppose, for example, that $\rho$ is a diameter-realizing geodesic. Then for any $x \in \rho$, $f(x) \ge \tfrac{1}{2} |\rho|$, just tracking along $\rho$. A non-comprehensive literature search has failed to uncover a relationship between $f(x)$ and $\mathrm{diam}(S)$. Itoh, Jin‐ichi and Costin Vǐlcu. "Criteria for farthest points on convex surfaces." Mathematische Nachrichten 282, no. 11 (2009): 1537-1547. Journal link. REPLY [14 votes]: Denote the diameter by $d$ and distance by $|x-y|$. Then there are $y,z$ such that $d=|y-z|$ and we have by triangle inequality for every $x$: $$d=|y-z|\leq |y-x|+|x-z|\leq 2f(x),$$ so we obtain your inequality. Notice that I did not use convexity, or any other of your assumptions, only the triangle inequality.<|endoftext|> TITLE: Evaluation of hypergeometric type continued fraction QUESTION [7 upvotes]: Is there a (possibly hypergeometric-type) explicit evaluation of the continued fraction $$a-\dfrac{1.(c+d)}{2a-\dfrac{2.(2c+d)}{3a-\dfrac{3.(3c+d)}{4a-\ddots}}}$$ Even the special case $d=0$, $a=1$ would be interesting. Note that $$\tanh^{-1}(z)=\dfrac{z}{1-\dfrac{1^2z^2}{3-\dfrac{2^2z^2}{5-\ddots}}}$$ but that does not seem to help. REPLY [5 votes]: This is found in [1] $\S 82$, Satz 5. It covers the case where the numerator $a_n$ is polynomial of degree $2$ in $n$ and the denominator $b_n$ is degree $1$. If I plugged in correctly, we get for your continued fraction: Let $a, c, d$ be complex numbers satisfying: $c \ne 0, a \ne 0, a^2 \ne 4c$, and $(a^2-4c)/a^2$ is not a negative real. Then the value is $$ {\frac { \left( d+c \right) \sqrt {{a}^{2}-4\,c}+a \left( c-d \right) }{2c} \;{\mbox{$_2$F$_1$}\left(1,{\frac {d+c}{c}};{\frac { \left( 3\,c+d \right) \sqrt {{a}^{2}-4\,c}+a \left( c-d \right) }{2c\sqrt {{a}^{2}-4\,c}}};{\frac {\sqrt {{a}^{2}-4\,c}-a}{2\sqrt {{a}^{2}-4\,c}}}\right)}^{-1}} $$ The sign of the square-root is chosen so that $\displaystyle\frac{a}{\sqrt{a^2-4c}}$ has positive real part. (The numerator ${{}_2F_1}$ has a zero in there, so it turns out to be constant.) [1] Oskar Perron, Die lehre von den Kettenbrüchen, 2 Auflage 1929<|endoftext|> TITLE: Is the volume of relative cycles at least the systole of the manifold? QUESTION [5 upvotes]: Let $M$ be a manifold with boundary $\partial M$. Suppose that $M$ is equipped with some structure for which a notion of volume for chains can be defined. For example, if $M$ is triangulated, then the volume of a simplicial chain is just the number of cells with non-zero coefficients in the formal sum (let's work over the field of 2 elements for simplicity). An analogous notion can be defined if $M$ has a Riemannian metric instead. Let $p:C_i(M)\rightarrow C_i(M,\partial M)$ be the canonical map from $i$-chains to relative $i$-chains, and let $p_*:H_i(M)\rightarrow H_i(M,\partial M) $ the induced map. Define for any $c\in C_i (M,\partial M)$: $$vol(c):=\inf\{vol(c') \mid p(c')=c\}$$ Clearly, $p$ cannot increase volume but may very well decrease it. However: Is it true that the volume of any non-contractible relative $i$-cycle representing an homology class in $Im(p_*)$ is at least $sys_i(M)$? Here $sys_i(M)$, namely the $i$-systole of $M$, is by definition the infimal volume of a non-contractible $i$-cycle in $M$. Diagram chasing at the level of chains (and not only at the level of homology) was the main type of arguments I tried to produce for proving such a claim. But I also suspect there might be some not-so-hard counter-example that I'm missing. REPLY [2 votes]: Here is a counterexample. Let $T$ be the $2$-torus obtained by identifying the opposite sides of the square on $\mathbb R^2$ with vertices $(\pm 1, \pm 1)$. Let $B_{1-\varepsilon}$ be the disk $x^2+y^2<1-\varepsilon$ and let $M$ be the complement in $T$, $M=T/B_{1-\varepsilon}$. It is clear that the systole $sys_1(M)$ is equal to $2$. Indeed, this is the length of the shortest non-contractible cycle in $M$. However, if we take such a cycle, for example given by the left side of the square, its image in $H_1(M, \partial M)$ equals the class of the segment of length $2\varepsilon$ that joins two points $(0, -1+\varepsilon)$ and $(0, 1-\varepsilon$) on $M$. Clearly $2\varepsilon<2=sys_1(M)$.<|endoftext|> TITLE: Challenge: Non-Gaussian quartic integral and path integral in Quantum field theory QUESTION [15 upvotes]: (1) It is well-known that we can get a Gaussian integral of this type, where $x$ is in $\mathbb{R}$: $$ \int_{-\infty}^{\infty} dx e^{-ax^2}=\sqrt{(2\pi)/a}. \tag{i} $$ We can generalize this integral from a number $a$ to a matrix value $A$, from a number $x$ to a vector $X=(x_1,x_2,...,x_N)$. So we generalize $$ \int_{-\infty}^{\infty} Dx \; e^{\frac{i}{2}X^T \cdot A \cdot X+ i JX } = \int_{-\infty}^{\infty} dx_1 \int_{-\infty}^{\infty} dx_2 \dots \int_{-\infty}^{\infty} dx_N e^{\frac{i}{2}X^T \cdot A \cdot X+ i JX }=\left(\frac{(2\pi i)^N}{\det(A)}\right)^{1/2} e^{-\frac{i}{2} J^T \cdot A^{-1} \cdot J}. \tag{ii} $$ (2) We can generalize again the integral to a functional integral or the so-called path integral in Quantum Field Theory. So for a real scalar field $\phi$, can be regarded as the earlier Eq(i) real valued $x$ generalized to the field $$ x \to \varphi, $$ and the matrix value $A$ is generalized to the matrix operator $M$, so we can evaluate the path integral in the $N$-dimensional spacetime manifold $\Sigma_N$: $$ \int_{-\infty}^{\infty} D \varphi e^{i \int_{\Sigma_N} (d^N x)\big( -\frac{1}{2}\varphi \cdot M \cdot \varphi + J \cdot \varphi \big)} . \tag{iii} $$ where the Green's function $G(x_i-x_f)$ is a solution of the delta function $\delta^N(x_i-x_f)$ on right hand side (r.h.s): $$ -M G(x_i-x_f)=\delta^N(x_i-x_f). $$ Therefore, the $A$ matrix in Eq.(ii) is related to $M$, while also related to $G$ via the inverse generalization $$ A \to M, $$ $$ A^{-1}_{if}=G(x_i-x_f). $$ Thus we obtain: $$ \int_{-\infty}^{\infty} D \varphi e^{i \int_{\Sigma_N} (d^N x)\big( -\frac{1}{2}\varphi \cdot M \cdot \varphi + J \cdot \varphi \big)}=Constant\cdot e^{\frac{-i}{2} \int d^N x_i \int d^N x_f J(x_i) G(x_i-x_f) J(x_f) }. \tag{iii} $$ involving the Green's function $G$. up to a not crucial constant $Constant$ which I do not care. (3) Question: Suppose I have a quartic integral from (i) to (i') $$ \int_{-\infty}^{\infty} dx \; e^{h x - a x^2-b x^4}=?. \tag{i'} $$ I suppose that we can still easily get an answer. How about the matrix value integral from (ii) to (ii'): $$ \int_{-\infty}^{\infty} Dx \; e^{\frac{i}{2}X^T \cdot A \cdot X+ i J \cdot X +(X^T \cdot X)^2 } =?. \tag{ii'} $$ Do we also have a QFT path integral analogy: $$ \int_{-\infty}^{\infty} D \varphi e^{i \int_{\Sigma_N} (d^N x)\big( -\frac{1}{2}\varphi \cdot M \cdot \varphi + J \cdot \varphi + \varphi^4 \big)} =?. \tag{iii'} $$ My question concerns the neat evaluation of (i'), (ii') and (iii'). Thank you! Here we can focus on the real valued function field $\varphi(x) \in \mathbb{R}$ REPLY [2 votes]: For your integral (i'), at least the $h=0$, $a>0$, $b>0$ case can be done analytically (or using Mathematica or Gradshteyn-Ryzhik): $$ \int_{-\infty}^\infty e^{-ax^2-bx^4}{\rm d}x = \frac{1}{2} \sqrt{\frac{a}{b}} e^{\frac{a^2}{8 b}} K_{\frac{1}{4}}\left(\frac{a^2}{8 b}\right). $$ The $h\not=0$ case seems not to permit a closed-form solution, as far as I can tell (but of course the integral exists). For symmetric positive-definite $A$, the multidimensional case can be expressed in terms of the one-dimensional integrals in an eigenbasis of $A$. I don't think it will have a closed-form solution unless $J=0$ and $A$ is a multiple of the identity. The path integral case is very different, because it only makes sense upon renormalization (the parameters of the action will depend on the regulator used to define and evaluate the integral). It is generally believed that the integral exists in a meaningful sense only for $\lambda=0$ (i.e. "$\phi^4$ theory is trivial").<|endoftext|> TITLE: Computable models of the ordinal numbers QUESTION [9 upvotes]: It's known, for example in the answer to this question: Is there a computable model of ZFC? that ZFC has no computable model. My questions is: is there a model of ZFC for which the order relation on the class of ordinals is computable? REPLY [16 votes]: Yes, this can happen: if $M$ is a countable $\omega$-model of ZF whose well-founded part has ordertype $\omega_1^{CK}$ (that is: has the shortest well-founded part possible for $\omega$-models), then $Ord^M$ as a linear order is just the Harrison order: $$\omega_1^{CK}+(\omega_1^{CK}\cdot\eta),$$ where $\eta$ is the ordertype of the rationals. This linear order does in fact have a computable copy, and is one of the basic examples/counterexamples in computable structure theory: a computable linear order which is illfounded but has no hyperarithmetic descending sequence (it has other nice properties too). The key to seeing that this must be the ordertype is the following pair of observations: Given such an $M$ and an $\alpha\in Ord^M$, there must be an interval in $Ord^M$ beginning with $\alpha$ and isomorphic to $\omega_1^{CK}$. No interval in $Ord^M$ can be isomorphic to $\omega_1^{CK}+1$. As to why such a model exists in the first place, this is trickier; depending how you phrase it, it's an application of either the Gandy basis theorem or the Barwise(-Kreisel) compactness theorem. Unfortunately, this doesn't have a one-line explanation.<|endoftext|> TITLE: How rare are unholey permutations? QUESTION [12 upvotes]: For $S\subset [n]:=\{1,2,\dotsc,n\}$, define $\delta(S)$ to be the number of $m\in S$ such that $m+1\notin S$. Given a permutation $\pi$ of $[n]$, we define the holeyness $D(\pi)$ of $\pi$ as being $$\max_{S\subset [n]} (\delta(\pi(S))-\delta(S)).$$ It is clear that $D(\pi)\geq 0$. What is (roughly or exactly) the number of permutations $\pi$ of $[n]$ such that $D(\pi)\leq k$? What if you restrict $\pi$ to be an $n$-cycle? REPLY [2 votes]: This is just a long comment. I computed $P_{n}(q)=\sum_{\pi \in S_n} q^{D(\pi)}$, and got the following polynomials: \begin{array}{l} 1 \\ 2 \\ 4 q+2 \\ 22 q+2 \\ 36 q^2+82 q+2 \\ 478 q^2+240 q+2 \\ 576 q^3+3942 q^2+520 q+2 \\ 14840 q^3+24518 q^2+960 q+2 \\ \end{array} It seems that for odd $n$, $P_{n}(q)$ has degree $d=(n-1)/2$, and the leading coefficient is $((d+1)!)^2$. Similarly, $\sum_{\pi \in S_n, type(pi)=(n)} q^{D(\pi)}$ gave me \begin{array}{l} 1 \\ 1 \\ 2 q \\ 6 q \\ 8 q^2+16 q \\ 85 q^2+35 q \\ 96 q^3+564 q^2+60 q \\ 1978 q^3+2982 q^2+80 q \\ \end{array} Also, it seems like both families of polynomials (checked for up to $n=8$), are real-rooted. One can also consider a cyclic version of holeyness, where we add $1$ to $\delta(S)$ if not both $1$ and $n$ are in $S$. Here, $n$ is the number of elements in the permutation. What happens is that $D(\pi) = D( w\pi w^{-1})$ if $w = (123\dotsc,n)$, so every polynomial is divisible by $n$. \begin{array}{l} 1 \\ 2 \\ 6 \\ 16 q+8 \\ 110 q+10 \\ 216 q^2+492 q+12 \\ 3346 q^2+1680 q+14 \\ 4608 q^3+31536 q^2+4160 q+16 \\ \end{array} These also seem real-rooted. REPLY [2 votes]: Unless I am very mistaken, there is an easy way to establish a bound of the "much better" kind mentioned in the comments above. (I don't doubt one can and should give a more precise answer.) Write $\phi$ for $\pi \sigma \pi^{-1}$ (meaning $\pi^{-1}\circ \sigma \circ \pi$), where $\sigma = (1 2 \dotsc n)$. Then $\delta(\pi(S))$ is the number of $m\in S$ such that $\phi(m) \notin S$. (Well, there could be a difference of $1$, which does not matter. Let us redefine $\delta(S)$ to be the number of $m\in S$ such that $\sigma(m)\notin S$. Perhaps we should consult OEIS again, after this redefinition?) For the sake of simplicity, we shall restrict our attention to maps $\pi$ without fixed points; that implies $\phi$ has no fixed points either. (This case covers the case of $\pi$ ranging over $n$-cycles, in particular.) Let $v=v_\phi$ be the number of $m\in [n]$ such that $\phi(m) \notin \{\sigma(m), \sigma^{-1}(m)\}$. A bit of doodling shows that the number $w=w_\phi$ of consecutive pairs $A=\{m,\sigma(m)\}$ such that $\phi(A)\cap A=\emptyset$ (called them "valid" pairs) is $\geq v_\phi/2$. Now we build a set $S$ as follows. At each step, we add to $S$ a valid pair $\{m,\sigma(m)\}$ that has not yet been marked as forbidden. We also mark eight pairs as forbidden: $$\{\phi(m),\sigma^{\pm 1}(\phi(m))\}, \{\phi(\sigma(m)),\sigma^{\pm 1}(\phi(\sigma(m)))\}, \{\phi^{-1}(m),\sigma^{\pm 1}(\phi^{-1}(m))\}, \{\phi^{-1}(\sigma(m)),\sigma^{\pm 1}(\phi^{-1}(\sigma(m)))\}.$$ In this way, we get to build a set $S$ of size at least $2 \cdot w_\phi/9 \geq v_\phi/9$ with $\delta(S)\leq |S|/2$ and $\phi(S)\cap S = \emptyset$, so that $\delta(\pi(S)) = |S|$. Hence $$D(\pi) \geq v_\phi/18.$$ Thus, the number of distinct $\phi$ without fixed points coming from permutations $\pi$ with $D(\pi)\leq k$ is at most $2^n n^{18 k}$ (in fact, at most $2^{n-18k} n^{18 k}$). Now, at most $n$ permutations $\pi$ (in fact, exactly $n$ permutations $\pi$) give rise to the same $\phi = \pi \sigma \pi^{-1}$. Therefore, the total number of permutations without fixed points $\pi$ of $[n]$ such that $D(\pi)\leq k$ is at most $$2^n n^{18 k + 1}.$$ I think (though I haven't checked yet) that the analysis for arbitrary permutations $\pi$ should require only a little more work.<|endoftext|> TITLE: Eccentricity in the number of representations for sets too large to be Sidon sets QUESTION [5 upvotes]: Let $A=\{a_10$? More concretely, let $E(A)$ denote the "eccentricity" of $A$, given by $$E(A) = \sum_n \max\{r_A(n)-2,0\}. $$ If $|A|>(1+\epsilon)\sqrt{N}$ for some $\epsilon>0$, then must there exist some $\delta>0$ such that $E(A)>\delta N$? My impetus for asking this question comes from my attempts to understand the binary digits of $\sqrt{2}$. Currently it is known that the number of $1$'s in the first $N$ binary digits of $\sqrt{2}$ is $\ge \sqrt{2N}(1+o(1))$, and for some infinite sequence of $N$'s this can be improved to $\ge \sqrt{8N/\pi}(1+o(1))$. However, this bound comes in part from assuming that the set of indices of the $1$'s behaves like a Sidon set, which it is too large to actually be. If it could be shown that $E(A)>\delta N$, then I believe a stronger lower bound could be proven. REPLY [4 votes]: There exists a set $A$ on $\{1,...,N\}$ where $ |A|\geq\frac{2}{\sqrt{3}}\sqrt{N}$ and $E(A)=o(N)$. Let $B$ be a Sidon set on $\{1,...,n\}$. Let $C = \{3n+1-b | b\in B\}$. Let $A=B\cup C$. Suppose $a TITLE: Are maps corresponding to affinoid subdomains flat in the Banach sense? QUESTION [7 upvotes]: $\newcommand{\Sp}{\mathrm{Sp}}\newcommand{\abs}[1]{\lvert #1\rvert}\newcommand{\comptensor}{\mathbin{\hat{\otimes}}}$ Let $k$ be a complete non-archimedian field and let $X = \Sp(B)$ be a $k$-affinoid space. Let $V = \Sp(B') \subseteq X$ be an affinoid subdomain. It is well-known that the corresponding map $B \to B'$ is a flat ring homomorphism; see e.g. Cor. 7.3.2/6 in Bosch-Güntzer-Remmert. Let us say that $B'$ is Banach-flat over $B$ if whenever $M \to N \to P$ is an admissible exact sequence of Banach $B$-modules then the completed tensor product sequence $M \comptensor_B B' \to N \comptensor_B B' \to P \comptensor_B B'$ is also admissible exact. (A map $f \colon M \to N$ of Banach $B$-modules is called admissible if there is a constant $C>0$ such that any $n \in f(M)$ there is a preimage $m \in M$ such that $f(m) = n$ and $\abs{m} \le C \abs{n}$. By Banach's open mapping theorem, this condition is equivalent to $f(M)$ being closed in $N$. Generally exact sequences of Banach modules can only be expected to behave well, if all mappings are admissible.) Is it true that if $V = \Sp(B') \subseteq X = \Sp(B)$ is an affinoid subdomain, then $B'$ is a flat Banach algebra over $B$? Note that for a functor being exact in the Banach sense, it suffices to consider short exact sequences $0 \to M \to N \to P \to 0$. Taking the completed tensor products is always Banach-right exact (because being admissible right exact is equivalent to being a cokernel diagram and because completed tensor product is left adjoint to Banach-Hom), so it suffices here to show that tensoring with $B'$ preserves admissible injective maps of Banach modules. One can show that fixing $M$, the association $V = \Sp(B') \mapsto M \comptensor_B B'$ is a sheaf on $X$ (more precisely, its Čech complex is admissible exact). In particular, $M \comptensor B' \to \prod_i M \comptensor B_i'$ is admissible injective, if $\Sp(B') = \bigcup_i \Sp(B_i')$. By the theorem of Gerritzen and Grauert we can therefore assume that $V = X(f_1/f, \dots, f_r/f)$ is a rational subdomain of $X$, for which we have a rather explicit description of the algebra $B'$. But so far I had no success. REPLY [6 votes]: This is not true. Assume that $X$ is the closed unit disc (given by $|T| \le 1$, with algebra $B$) and $V$ is a smaller disc (given by $|T| \le r$ for some $r \in (0,1)$, with algebra $B_V$). Consider the annulus $W$ defined by $|T|=1$ with algebra $B_W$. Then the restriction map $B \to B_W$ is injective and admissible (because function on the disc reach their maximum on the boundary). If we do the completed tensor product with $B_V$, we get the map $B_V \to B_W \hat{\otimes}_B B_V$. But $B_W \hat{\otimes}_B B_V$ is the algebra of $V\cap W = \emptyset$, that is to say 0, so the map is not injective.<|endoftext|> TITLE: Character theory and Quantum Chemistry QUESTION [9 upvotes]: Who (presumably a chemist) realized first the efficiency of character theory in calculations of orbitals of atoms? In which year? REPLY [7 votes]: Introducing groups into quantum theory, by Erhard Scholz (2006): In quantum chemistry, representations of permutation groups made their first appearance about the same time as they did in spectroscopy. The topic was opened up by a joint publication [1] of two young physicists, Walter Heitler and Fritz London, who had come to Zürich on Rockefeller grants in 1926, respectively 1927, to work with E. Schrödinger. Group theoretic methods were first applied in two papers by W. Heitler [2,3]. [1] Heitler, W., London, F. Wechselwirkung neutraler Atome und homöpolare Bindung nach der Quanten-mechanik. Zeitschrift für Physik 44, 455–472 (1927). [2] Heitler, W., Zur Gruppentheorie der homöopolaren chemischen Bindung. Zeitschrift für Physik 47, 835–858 (1928). [3] Heitler, W., Zur Gruppentheorie der Wechselwirkung von Atomen. Zeitschrift für Physik 51, 805–816 (1928). REPLY [4 votes]: I think this realization is an immediate consequence of that of the efficiency of group theory in the quantum theory of angular momentum. The early culprits for the Gruppenpest are Weyl in his 1931 book "Gruppentheorie und quantenmechanik" and B. van der Waerden in his 1932 book "Die gruppentheoretische Methode in der Quantenmechanik". You can probably also find a discussion of this in the more recent book by Biedenharn and Louck.<|endoftext|> TITLE: The spectrum of the Hodge Laplacian on a Riemannian manifold QUESTION [5 upvotes]: The Hodge Laplacian operator on differential forms on a (compact?) Riemannian manifold carries useful information about the topology of the manifold. In particular, the multiplicity of the zero eigenvalue is equal to the betti number. Does the non-zero spectrum carry similarly useful topological information (in particular, independent of the metric)? More generally, what is known about the spectrum? What is the spectrum for instance on classical examples like Euclidean space, the flat torus, the spheres or the riemann surfaces? From the wiki page, I know that on a compact manifold, the spectrum is non negative but this is about the limit of my knowledge. REPLY [2 votes]: This "book" might interest you: http://math.bu.edu/people/sr/articles/book.pdf I haven't found a book title or an author inside the book, but on page 35 they claim, that even calculating the eigenvalues/eigenforms for the non-flat 2-torus of the laplacian on 1-forms is basically impossible. I don't know whether that is something one would expect. Unfortunately all I have is the above link, as I only got it from another thread: https://math.stackexchange.com/questions/617557/relation-of-hodge-theorem-to-eigenfunction-basis-of-laplacian But maybe you will find other information inside the book. They have a "Hodge Theorem 1.30 for the Laplacian on k-forms" on page 34 that says: For a closed connected orientable Riemannian manifold the eigenvalues of the Laplace operator on k-forms are all non-negative. There exists an orthonormal basis of $L^2(\Omega^k(M))$ consisting of eigenfunctions (or lets call them eigenforms I guess) of the Laplacian on k-forms. The eigenspaces are all finite-dimensional and the eigenvalues accumulate only at infinity. So the eigenspaces of the Laplacian on k-forms are finite dimensional. Is that common knowledge? I haven't read much more of "the book". Somewhere else I read that the spectrum of the Laplacian only consists of its eigenvalues (common knowledge?): https://arxiv.org/pdf/1710.09579.pdf on page 11. This short text might also interest you: https://math.berkeley.edu/~alanw/240papers03/chen.pdf starting on page 8, he explains that one can build up Morse-homology via a deformed Laplacian, where critical points of index k of a morse function correspond to eigenforms of the Laplacian on k-forms (the same k) and the exterior differential d on the eigenforms corresponds to the dual of the connecting-flow-lines counting boundary operator in Morse Homology.<|endoftext|> TITLE: Fundamental group of $M_g^\circ$ QUESTION [7 upvotes]: Let's work over the complex numbers $\mathbb{C}$. Let $g\geq3$ be an integer. Let $\mathcal{M}_g$ be moduli stack of smooth genus $g$ curves. Let $M_g$ be the corresponding coarse moduli scheme. They share an open subscheme $M_g^\circ$ parametrizing automorphism-free smooth genus $g$ curves. What is the fundamental group $\pi_1(M_g^\circ)$? How does it compare to the following notions: (1) The orbifold mapping class group $\pi_1(\mathcal{M}_g)$, (2) The topological fundamental group $\pi_1(M_g)$, (3) The mapping class group of a genus $g$ surface, $\mathrm{Mod}_g$? REPLY [12 votes]: Except in a few trivial cases, the locus of curves which have an extra automorphism will have codimension greater than one in $\mathcal M_g$. When that happens, the fundamental group of $\mathcal M_g$ must equal the fundamental group of $\mathcal M_g^{\circ}$. (To see this, one can pass to the $3$-torsion cover of $\mathcal M_g$, which is a smooth scheme, and use the fact that removing a codimension two subscheme of a smooth scheme does not affect $\pi_1$.) So it equals cases 1 and 3.<|endoftext|> TITLE: Decay of matrix coefficients of non-tempered representation QUESTION [6 upvotes]: A theorem of Cowling--Haagerup--Howe gives an effective decay rate of the matrix coefficients of a tempered representation $\pi$ of a semi-simple algebraic $G$ in terms of Harish-Chandra $\Xi$ function, of the form $$\langle \pi(g) u,v\rangle \ll_{u,v} \Xi(g),\quad u,v\in \pi\setminus\pi^G, g\in G.$$ Does there exist a similar, possibly weaker, bound in case of non-tempered representations (possibly of the above form but $\Xi(g)$ is replaced by $\Xi(g)^{1-\delta}$ for some $0<\delta<1$)? I am, in particular, interested in $\mathrm{GL}(n)$ and $\pi$ being an irreducible automorphic representation of it. For $n=2$ it is known that such $\delta$ exists (spectral gap) due to the works of Selberg, Gelbart--Jacquet, Kim--Shahidi. Thanks in advance! REPLY [2 votes]: There is some confusion here, as literally the construction of complementary series in $SL_{2}$ will give you unitary representations with arbitrary slow decay. For any homogeneous space $G/\Gamma$, there exists a ``spectral gap'' in your definition. The statement about Selberg's work is about showing a uniform spectral gap for a tower of coverings amounting to principal congruence subgroups, not just about the dual of $SL_{2}$ say. Now for $n\geq 3$, $SL_{n},PGL_{n}$ are Kazhdan groups, hence such an estimate is immediate for all (non-trivial) unitary representations, without pertaining to a particular lattice/tower of coverings etc. You essentially can recover such a bound from the proof of Cowling-Howe-Haagrup (which implement Howe's technique to show temperedeness). The exact statement appears in Hee Oh's famous paper about decay estimates of matrix coefficients - https://gauss.math.yale.edu/~ho2/doc/mat.pdf as Theorem 2.5 (she defines the theorem for $K$-finite vectors which are usually enough, with some extra work in light of HC results you can probably recover something for smooth vectors, if you will use some Sobolev norms). See also Shalom's annals paper for related results.<|endoftext|> TITLE: Is the union of a chain of elementary embeddings elementary? QUESTION [20 upvotes]: I am a little confused about what I think must be either a standard theorem or a standard counterexample in model theory, and I am hoping that the MathOverflow model theorists can help sort me out about which way it goes. My situation is that I have a chain of submodels, which is not necessarily an elementary chain, $$M_0\subseteq M_1\subseteq M_2\subseteq\cdots$$ and I have elementary embeddings $j_n:M_n\to M_n$, which cohere in the sense that $j_n=j_{n+1}\upharpoonright M_n$. So there is a natural limit model $M=\bigcup_n M_n$ and limit embedding $j:M\to M$, where $j(x)$ is the eventual common value of $j_n(x)$. Question. Is the limit map $j:M\to M$ necessarily elementary? A natural generalization would be a coherent system of elementary embeddings $j_n:M_n\to N_n$, with possibly different models on each side. The question is whether the limit embedding $j:M\to N$ is elementary, where $M=\bigcup_n M_n$, $N=\bigcup_n N_n$ and $j=\bigcup_n j_n$. And of course one could generalize to arbitrary chains or indeed, arbitrary directed systems of coherent elementary embeddings, instead of just $\omega$-chains. I thought either there should be an easy counterexample or an easy proof, perhaps via Ehrenfeucht-Fraïssé games? REPLY [18 votes]: First I'll give a counterexample to the natural generalization, adapted from my earlier comment: Let $M_n = (\mathbb{N},<)$ for all $n$, and let $N_n = (\mathbb{N}\sqcup \frac{1}{n!}\mathbb{Z},<)$, where all elements of $\frac{1}{n!}\mathbb{Z}$ are greater than all elements of $\mathbb{N}$ in the order. Note that each $N_n$ is isomorphic to $(\mathbb{N}\sqcup\mathbb{Z},<)$, and there is a natural inclusion $N_n\subseteq N_{n+1}$ for all $n$, since $\frac{1}{n!}\mathbb{Z} \subseteq \frac{1}{(n+1)!}\mathbb{Z}$ as subsets of $\mathbb{Q}$. Now $M_n\preceq N_n$ for all $n$, but $\bigcup_n M_n\not\preceq \bigcup_n N_n$, since the former is $(\mathbb{N},<)$ and the latter is $(\mathbb{N}\sqcup\mathbb{Q},<)$, which is not discrete. Now I'll explain how to turn any counterexample to the natural generalization into a counterexample to the original question. Suppose we have a coherent system of elementary embeddings $j_n\colon M_n \to N_n$ such that the limit map $j\colon M\to N$ is not elementary. Let $L$ be the language of this counterexample, which we assume to be relational, and let $L' = L\cup \{E\}$, where $E$ is a new binary relation. For each $n$, we construct a structure $M^*_n$ as follows: $E$ is an equivalence relation with countably many classes, which we denote by $(C_i)_{i\in \mathbb{Z}}$. We interpret the relations from $L$ on each class $C_i$ so that $C_i$ is a copy of $M_n$ when $i\leq 0$ and a copy of $N_n$ when $i > 0$. There are no relations between the classes. There is a natural inclusion $M_n^*\subseteq M_{n+1}^*$ for all $n$, in which each class $C_i$ in $M_n^*$ is included in to the class $C_i$ in $M_{n+1}^*$ according to the inclusions $M_n\subseteq M_{n+1}$ and $N_n\subseteq N_{n+1}$. Let $j_n^*\colon M^*_n\to M^*_n$ map $C_i$ to $C_{i+1}$ as the identity on $M_n$ for all $i<0$, as the identity on $N_n$ for all $i>0$, and as $j_n \colon M_n\to N_n$ for $i = 0$. Then $j_n^*$ is an elementary embedding, and $j_n = j_{n+1}\restriction M_n^*$. In the limit, $M^* = \bigcup_n M_n^*$ has equivalence classes such that $C_i$ is a copy of $M$ when $i\leq 0$ and a copy of $N$ when $i>0$. And the limit map $j^*\colon M^*\to M^*$ maps $C_i$ to $C_{i+1}$ as the identity on $M$ for all $i<0$, as the identity on $N$ for all $i>0$, and as $j\colon M\to N$ for $i=0$. This $j^*$ is not elementary, since an $L$-formula whose truth is not preserved by $j$ can be relativized to the equivalence class $C_0$ to give an $L'$-formula whose truth is not preserved by $j^*$.<|endoftext|> TITLE: For algebraic $\alpha$, does $G(\mathbb Z[d\alpha])$ have a finite index in $G(\mathbb Z[\alpha])$? QUESTION [5 upvotes]: Let $\alpha$ be an algebraic number and $G$ be a connected $\mathbb Q(\alpha)$-algebraic group, and $d\in\mathbb Z^+$. We fix a faithful representation of $G$ in some $GL_n$ and identify $G$ with its image under this representation. For a subring $R$ of $\mathbb Q(α)$, $G(R):=G\cap GL_n(R)$. Does $G(\mathbb Z[d\alpha])$ have a finite index in $G(\mathbb Z[\alpha])$? What if $d\alpha$ is an integral element? REPLY [6 votes]: The answer is yes for $\alpha$ an integer and no for $\alpha$ a general algebraic number. To see the problem with algebraic numbers, take $\alpha=1/2, d=2$. Then almost any algebraic group, in particular $\mathbb G_m$ or $\mathbb G_a$, will not have its $\mathbb Z$ points a finite-index subgroup of its $\mathbb Z[1/2]$-points. If $\alpha$ satisfies a monic polynomial equation of degree $k$, then $\mathbb Z[d\alpha]$ contains all elements in $\mathbb Z[\alpha]$ that are congruent to $0$ or $1$ mod $d^{k-1}$, so $G(\mathbb Z[d \alpha])$ contains the kernel of the natural homomorphism from $G( \mathbb Z[\alpha])$ to $G(\mathbb Z[\alpha]/d^{k-1})$. Because the target is finite, its kernel has finite index. Edit: Assuming that every element of $\mathbb Z[\alpha]$ shows up as an entry of a matrix in $G(\mathbb Z[\alpha])$. Then we can actually prove a converse - if $G(\mathbb Z[d\alpha])$ is finite-index, then $\mathbb Z[d \alpha]$ contains a finite-index ideal in $\mathbb Z[\alpha]$. To do this, take coset representatives for each of the finitely many cosets, and observe that every element of $\mathbb Z[\alpha]$ arises as a matrix entry of a product of one of these representatives with a matrix over $\mathbb Z[d\alpha]$, which implies the entries are a finite basis for $\mathbb Z[\alpha]$ as a $\mathbb Z[d\alpha]$-module. Each of these basis elements can be written as a polynomial in $\alpha$, hence its multiple by some power of $d$ is a polynomial in $d\alpha$. So altogether there is some fixed power of $d$ all whose multiples in $\mathbb Z[\alpha]$ lie in $\mathbb Z[d\alpha]$.<|endoftext|> TITLE: Introductory textbook on geometry of hyperbolic space QUESTION [10 upvotes]: I am looking for an introductory textbook to the geometry of the hyperbolic space $\mathbb{H}^n$. The book should include explicit description of geodesics and horospheres in various models (hyperboloid, Poincaré, Klein). Apologies if the question is not appropriate for this site. REPLY [2 votes]: Prasolov, V. V.; Tikhomirov, V. M., Geometry. Transl. from the Russian by O. V. Sipacheva. Transl. edited by A. B. Sossinski, Translations of Mathematical Monographs. 200. Providence, RI: American Mathematical Society (AMS). xi, 257 p. (2001). ZBL0977.51001.<|endoftext|> TITLE: The Planck constant for mathematicians QUESTION [69 upvotes]: The questions Q1. What are simple ways to think mathematically about the physical meanings of the Planck constant? Q2. How does the Planck constant appear in mathematics of quantum mechanics? In particular, quantization is an important notion in mathematical physics and there are various forms of quantization for classical Hamiltonian systems. What is the role of the Planck constant in mathematical quantization? Q3. How does the Planck constant relate to the uncertainty principle and to mathematical formulations of the uncertainty principle? Q4. What is the mathematical and physical meaning of letting the Planck constant tend to zero? (Or to infinity, if this ever happens.) Motivation: One purpose of this question is for me to try to get better early intuition towards a seminar we are running in the fall. Another purpose is that the Planck constant plays almost no role (and, in fact, is hardly mentioned) in the literature on quantum computation and quantum information, and I am curious about it. Related MO questions: Does quantum mechanics ever really quantize classical mechanics? ; REPLY [9 votes]: The previous answers all give good discussions of the physical significance of $\hbar$ and the classical limit "$\hbar \to 0$" (in large quotation marks), but few of them discuss your "motivation" comment that $\hbar$ rarely appears in the study of quantum computing and information. David Mermin dedicates an entire section of his great paper "From Cbits to Qbits" to that question: Like my disapproving colleague, some physicists may be appalled to have finished what purports to be an exposition of quantum mechanics — indeed, of applied (well, gedanken applied) quantum mechanics — without ever having run into Planck’s constant. How can this be? The answer goes back to my first reason why enough quantum mechanics to understand quantum computation can be taught in a mere four hours. We are interested in discrete (2-state) systems and discrete (unitary) transformations. But Planck’s constant only appears in the context of continuously infinite systems (position eigenstates) and continuous families of transformations (time development) that act on them. Its role is to relate the conventional units in which we measure space and time, to the units in which it is quantum-mechanically natural to take the generators of the unitary transformations that produce translations in space or time. If we are not interested in location in continuous space and are only interested in global rather than infinitesimal unitary transformations, then $\hbar$ need never enter the story. The engineer, who must figure out how to implement unitary transformations acting over time on Qbits located in different regions of physical space, must indeed deal with $\hbar$ and with Hamiltonians that generate the unitary transformations out of which the computation is built. But the designer of algorithms for the finished machine need only deal with the resulting unitary transformations, from which $\hbar$ has disappeared as a result, for example, of judicious choices by the engineers of the times over which the interactions that produce the unitary transformations act. Deploring the absence of $\hbar$ from expositions of quantum computer science is rather like complaining that the $I$-$V$ curve for a $p$-$n$ junction never appears in expositions of classical computer science. It is to confuse computer science with computer engineering. So basically, quantum algorithms researchers are implicitly setting $\hbar = 1$, or more precisely they're sweeping it up in the various fixed time constants and so on that determine the gates' operation at the microscopic level, and only considering the value of the dimensionless ratios $(\Delta E) T/\hbar$ that arise from the Schrodinger equation.<|endoftext|> TITLE: Which knot invariants have no known diagram-independent descriptions? QUESTION [13 upvotes]: Many knot invariants in knot theory are discovered by finding a property of knot diagrams which is invariant under the three Reidemeister moves. Now in principle, any knot invariant can be described in a diagram-independent way, that is, as a property of the three-dimensional knot itself without reference to diagrams of the knot. But in practice, it can take years between the development of a knot invariant and the discovery of a diagram-independent description of it. So my question is, for what knot invariants is a diagram-independent description not yet known? REPLY [8 votes]: I'd be happy to be proven wrong, but I would argue that this is still the case of the Jones polynomial and its generalizations, and my, maybe naive, understanding is that it's one of the many reasons it was considered fairly mysterious when it was discovered. In fact, finding such a description was one of the motivations for Witten's Jones paper (he says as much in the introduction). As beautiful as this is, and even if like many others I'm more than happy to think about this as an actual definition, this is strictly speaking not mathematically rigorous and recovers only the values of the Jones polynomial at roots of unity. Those can, I believe, now be given a diagrammatic-free definition in the framework of TFT's but this relies on fairly recent result. I also believe the case of generic $q$ is within reach but hasn't been done yet, and is closely related to exciting recent development in low-dimensional topology, e.g. finding a rigorous construction of analytic continuation of Chern-Simons theory, factorization algebras, the AJ conjecture etc..<|endoftext|> TITLE: Relation between information geometry and geometric deep learning QUESTION [21 upvotes]: Disclaimer: This is a cross-post from a very similar question on math.SE. I allowed myself to post it here after reading this meta post about cross-posting between mathoverflow and math.SE, I did try both the 1. and 2. suggestions of the accepted answer (suggestion 2. - asking for help on a mathoverflow meta dedicated post - was impossible because I lacked the reputation necessary to answer on mathoverflow meta). I'm currently working on information geometry (IG) and geometric deep learning (GDL). As I started without specific knowledge of both, their respective names led me to believe for a short and naive period that GDL was defined by the use of IG notions in deep learning. This now appears to me as substantially inaccurate, but since there are indeed various connections, I would like to clarify the relation between them. In Geometric deep learning: going beyond Euclidean data, GDL is defined as: Geometric deep learning is an umbrella term for emerging techniques attempting to generalize (structured) deep neural models to non-Euclidean domains such as graphs and manifolds. The article later defines Riemannian manifolds and metrics, calculus on manifolds, etc to complete the toolkit needed to build a machine learning favorable environment. On the other hand, in Amari's Information Geometry and Its Applications, IG is described in the following paragraph: Information geometry has emerged from studies of invariant geometrical structure involved in statistical inference. It defines a Riemannian metric together with dually coupled affine connections in a manifold of probability distributions. These structures play important roles not only in statistical inference but also in wider areas of information sciences, such as machine learning, signal processing, optimization, and even neuroscience, not to mention mathematics and physics. In the same book, Amari mentions a neural manifold, containing all neural networks: The set of all such networks forms a manifold, where matrix $W =w_{ji}$ is a coordinate system. The underlying question is: to what extent is GDL related to IG ? If I'm not mistaken, working on a riemannian manifold with a neural network, and relying on a riemannian gradient for its training, implies one is doing geometric deep learning, but not necessarily information geometry. An example of such GDL/non-IG neural network is SPD Net, which relies on SPD matrices for intermediate representations, defines new transformations aiming at keeping representations on a manifold (hence a bilinear mapping layer $W.X.W^T$ with $W$ belonging to a compact Stiefel manifold - cf. BiMap Layer section of the previous article), and depends on a riemannian gradient. It seems similar to IG since we speak of riemannian manifold and riemannian gradient, but it doesn't match the manifold of probability distributions aspect. A similar SPD matrices neural network architecture can be found in this paper. My current understanding is that GDL and IG are closely related since they happen to rely on similar mathematical objects, and perhaps more importantly, that in both cases we try to reach the right representation of a "latent subspace". The learning mechanisms of GDL then owe their success to optimization on the well chosen manifolds using gradients possibly defined in IG, the manifolds ideally describing a subspace where all possible data is living, not only the samples available for training. However, this relation is limited to specific cases of GDL, where a neural network uses manifolds and metrics also used in IG. One can notice that this completely excludes, among others, the graph-based part of GDL (?). Finally, we could still find a direct relation between GDL and IG if we used neural networks to work on manifolds of probability distributions, which are the very objects that IG targets. That seems rather relevant for generative models, but it's a very specific case of GDL for which the high usefulness of IG is contextual. Related AI SE post: What is geometric deep learning?, my answer there could be inaccurate depending on what the community will answer here. Mathoverflow addition: this mathoverflow question looking for IG references made me discover Amari's paper "Information geometry in optimization, machine learning and statistical inference" which is closely related to my question, but predates the definition of GDL in Geometric deep learning: going beyond Euclidean data, and thus ignores its position with respect to the latter. It doesn't seem that inappropriate to post this "boundaries definition" question on this SE Q&A since GDL is a recent and active research field (I've seen a SPD neural networks paper accepted at NIPS 2019 for example). If, after all, it still is out of place, sincere apologies, I'll just wait for answers on math.SE. EDIT: related paper establishing a link between Deep Learning (not specific to GDL) and IG through the Fisher-Rao norm: "Fisher-Rao Metric, Geometry, and Complexity of Neural Networks". REPLY [3 votes]: The fields you're talking about are typically concerned with two different geometric spaces: The space of input data to a neural network (geometric deep learning) The parameter space of all neural networks with a given architecture (information geometry) Many natural applications of neural networks involve input data with a discrete Euclidean-type structure: 1D for time series, 2D for images or audio, 3D for video. That "Geometric Deep Learning" paper discusses applying neural networks to input data with other types of geometry, such as graphs and networks. A central problem is figuring out the right architecture to handle a particular type of data. On the other hand, suppose you're studying the question of training a particular neural network. That is, you have a specific architecture in mind, let's say with $n$ weight (and maybe bias) parameters, where any given set of parameters may be viewed as a point in $\mathbb{R}^n$. When you study the dynamics of training, it can be useful to think about different metrics on this space. For example, some common regularization methods rely on $L_1$ or $L_2$ norms. The "information geometry" line of work looks at other metrics, with the goal of capturing more sophisticated concepts of network capacity, invariances to certain transformations, etc. A paper with a relatively brief, self-contained exposition is Fisher-Rao Metric, Geometry, and Complexity of Neural Networks. To sum up: Geometric deep learning is concerned with problems where the domain, or input data, is far from being modeled by a standard Euclidean space. Information geometry is traditionally used to analyze dynamics on a neural network parameter space ($\mathbb{R}^n$, but with a non-Euclidean metric). So in that sense, they are conceptually distinct. However, they both use similar mathematics, and certainly both could arise in studying a particular neural network.<|endoftext|> TITLE: Is there something like a vision paper in mathematics? QUESTION [8 upvotes]: I spoke with a computer scientist a few weeks who told me that in computer science there is something called a "vision paper", which is a paper that does not contain concrete results, but rather outlines a more general vision of what one either wants to do, or how a field may develop etc. A future vision in any case is presented. Searching the internet "vision paper" for computer science actually did not return many results - lest so for mathematics. But maybe there do exist such journals that offer a home for such articles in mathematics and it just happened that I didn't find that. If you know of any, please let me know. The journal doesn't exclusively have to publish vision papers (actually, I think that would be bad, since it would probably be soon filled with all kinds of fantasies regarding how some subfields of mathematics might develop), but should also allow such papers. There are some journals that allow papers containing open problems and conjectures. I am looking for something a bit broader than that, where a vision of a field can be presented that is extrapolated from the current state. I am interested because I myself am about to write such an article, and I'm not sure where to best post it. On a blog? I'd like to make it citeable and everything that purely online may vanish at some point in time (e.g. the blog providing company may shut down). A paper solely in arXiv? Would be an opportunity, but perhaps lacks in credentials. REPLY [5 votes]: The "mother of all vision papers", the Langlands program, was simply a handwritten letter; so I would not worry too much about "where to post it", it's the content that will determine the impact. For archival and citation purposes you could see if arXiv accepts it, and otherwise you could upload it on Zenodo: that will give you a DOI, and you can update the paper (the DOI will always point to the latest version). Zenodo is backed by CERN, so it should be a secure repository.<|endoftext|> TITLE: Dimensional gap of group representations QUESTION [13 upvotes]: The problem is inspired by eigenvalue bounds of random Cayley graphs on $SL_2(q)$. Definition. An infinite series of finite groups $S$ is α-rich if the dimension of the smallest nontrivial representation of $G$ on $\mathbb{C}$ is $\Omega(|G|^\alpha)$ for every $G\in S$. For example, the series of groups $SL_2(q)$ is $\frac{1}{3}$-rich. Question. Are there any series of $α$-rich groups with $\alpha>\frac13$? Known. One only needs to consider simple groups, and by CFSG, it's just going through Lie groups. Along the lines of David E Speyer, a permutation representation on $O(|G|^\frac13)$ points rules out the possibility for the group to beat $SL_2(q)$. This holds for all classic groups of type $A_n (n\geq2)$, $B_n(n\geq3)$, $C_n(n\geq2)$, $D_n(n\geq4)$, $^2D_n(n\geq3)$, $^2A_n(n\geq4)$: just consider the action of these groups on the 1-dimensional subspaces of their defining vector space. $^2A_2$ is not in the list, but $^2A_2(q)$ has a representation on $q^2-q+1$ dimensions, hence ruled out. The same argument rules out $F_4$, $E_n$ and $^2E_6$: All of them are covered in a paper given by Derek Holt. None of them beats $SL_2(q)$. $G_2$ and $^2G_2$ do not beat $SL_2(q)$, by Derek Holt's answer. For $^2B_2(q)$, see Wikipedia: it almost beats $SL_2$, but it has two characters of dimension $O(q^{3/2})$. For $^3D_4(q)$, see Deriziotis, D. I., and G. O. Michler. "Character table and blocks of finite simple triality groups $^3D_4(q)$." Transactions of the American Mathematical Society 303.1 (1987): 39-70.: there's a character of dimension $q(q^4-q^2+1)$. The last case $^2F_4$ is found in Die unipotenten Charaktere fur die GAP-Charaktertafeln der endlichen Gruppen vom Lie-Typ. M. Claßen-Houben; Diplomarbeit, RWTH Aachen; 2005. According to the conventions of the paper, the group has size $q^{52}$, and there's a character of dimension $q^2Φ_{12}Φ_{24}$, which is $O(q^{14})$. REPLY [9 votes]: This answer is still not quite complete, but I hope to finish it soon! The smallest degrees of the faithful permutation representations of the finite simple groups have all been known for a while now. The most convenient reference is probably Table 4 of this paper, although none of the results are original to that paper. The results for the exceptional groups of Lie type were proved in a series of papers of A. V. Vasil'ev. Anyway, this approach works for $E_7(q)$ (order about $q^{133}$, minimal permutation degree about $q^{27}$), $E_8(q)$ (order about $q^{248}$, minimal permutation degree about $q^{57}$, $^{2}E_6(q)$ (order about $q^{78}$, minimal permutation degree about $q^{21}$), and ${}^3D_4(q)$ (order about $q^{28}$, minimal permutation degree about $q^9$). However, the approach fails for $G_2(q)$, which has order about $q^{14}$ and minimal permutation degree $(q^6-1)/(q-1)$, $^{2}G_2(q)$ (with $q$ an odd power of $3$), with order about $q^7$ and minimal permutation degree $q^3+1$, and $^{2}F_4(q)$ (with $q$ an odd power of $2$), order about $q^{26}$, minimal permutation degree about $q^{10}$. So I have been hunting around for results about minimal degrees of representations of $G_2(q)$,$^{2}G_2(q)$, and $^{2}F_4(q)$. For $G_2(q)$ I found them on Page 126 of G. Hiss, Zerlegungszahlen endlicher Gruppen vom Lie-Typ in nicht-definierender Charakteristik, Habilitationsschrift, Aachen 1990. For sufficiently large $q$, these are $q^4+q^2+1$, $q^3+1$, and $q^3-1$ when $q \equiv 0,1,-1 \bmod 3$, respectively. So this is less that $q^{1/3}$, and we are OK. For $^{2}G_2(q)$, the character tables are computed in this paper. The table is towards the end of the paper, and the smallest character degree is $q^2-q+1$, so again we are OK! I still need to check $^{2}F_4(q)$. Added later: i have now located a better reference for minimal degrees of characters of exceptional groups of Lie type, namely Lübeck, Frank, Smallest degrees of representations of exceptional groups of Lie type. Comm. Algebra 29 (2001), no. 5, 2147–2169, available here. In particular, for sufficiently large $q$, the smallest character degree of ${}^2F_4(q^2)$ has degree $11$ in $q$, which is easily sufficient to prove the required result.<|endoftext|> TITLE: Napkin Folding Problem / Rumpled Ruble Problem QUESTION [6 upvotes]: I am an outsider to this group. I'm a journalist and am working on a piece about theoretical math/geometry. Simply put, when a napkin is folder in such a way to increase its perimeter is that strictly theoretical? Like, the physical limitations of the world (tension and width of cloth) make it impossible to increase the perimeter. Can someone help me understand what is really being discussed when geometry people discuss the folding of a ruble or napkin? -thanks REPLY [11 votes]: I think you will find this discussion of the napkin folding problem instructive: Chapter 5 of Lectures on piecewise distance preserving maps. The problem involves "Origami moves", meaning folding and unfolding of a thin sheet of paper, and asking whether the perimeter of the flattened folded sheet can increase. The answer is no if all layers of the folded sheet are always folded simultaneously, as in this figure: You have to allow for unfolding of a folded layer, as in the figure below, which actually does increase the perimeter, however, not yet beyond the perimeter of the original square. 18th century origami masters already discovered, before this problem was posed by mathematicians in the 20th century, that you can in fact increase the final perimeter beyond the initial perimeter, for example in this crane figure: The perimeter of the flattened crane is about 0.5% larger than the perimeter of the initial square. The crane has 80 layers of folded paper. By increasing the number of folds it is possible to make the final perimeter as large as one wants, without mathematical limit. There is of course a physical limit, due to the nonzero thickness of the paper.<|endoftext|> TITLE: What is the state of research on finding all Prime Knots with 17 Crossings? QUESTION [8 upvotes]: In this 1998 journal paper, all the prime knots with 16 or fewer crossings are found (some of which were found earlier by others). There are over 1.7 million such knots. But the prime knots with 17 crossings have not yet been tabulated. Here is what this book says: This is probably hard and requires new ideas. But this book was written in 2004, so things may have changed since then. There have certainly been a lot of developments in knot theory over the past 15 years. So my question is, what is the state of research on finding all prime knots with 17 crossings? Are we relatively close to doing so, and have partial results been discovered? REPLY [11 votes]: Ben Burton has found that there are 352,152,252 prime non-trivial knots with up to 19 crossings. See here for the tables.<|endoftext|> TITLE: Trying to understand Fisher's proof QUESTION [5 upvotes]: $\newcommand{\al}{\alpha}$ For $i=1,\dots,n$, let \begin{equation} R_i:=\frac{X_i}{X_1+\dots+X_n}, \end{equation} where the $X_i$'s are iid standard exponential random variables. Let $$R_*:=\max_{1\le i\le n}R_i. $$ Fisher1 gave the formula \begin{equation} P(R_*>x)=\sum_{j=1}^n(-1)^{j-1}\binom nj(1-jx)_+^{n-1} \end{equation} for $x\in(0,1)$ (using somewhat different notation), where $u_+:=\max(0,u)$. I have a proof of this result and a certain generalization of it. My problem is that I understand almost nothing in Fisher's proof (on pages 57--58 of his paper). In particular, I don't understand the following: What does (the polynomial (?)) $f$ in $t$ (introduced (?) on page 57 of Fisher's paper) have to do with the spline (?) $\text{P}$ in $g$; Why does $f$ have to have the differential properties in a neighborhood of $t=1$ that Fisher says $f$ has to have? How does Fisher make the jump from those properties of $f$ to the (correct) final expression for $\text{P}$? Fisher seems to provide absolutely no details on this. I will appreciate any help in filling these huge gaps in my understanding. 1Fisher, R. A., Tests of significance in harmonic analysis., Proceedings Royal Soc. London (A) 125, 54-59 (1929). ZBL55.0950.16, MR2079. REPLY [3 votes]: $\newcommand{\al}{\alpha}$ I seem to finally get it. Fisher denotes $P(R_*>x)$ by $\text{P}$ (also using $g$ in place of $x$) and calls this probability "the probability integral". He says: We may therefore represent the probability integral by the form $$\text{P}=\al_1(1-g)^{n-1}+\al_2(1-2g)^{n-1}+\ldots+\al_n(1-ng)^{n-1}, $$ in which as many terms are to be taken as have positive quantities within the brackets. The last term is therefore included for no possible value of $g$, but is written above in order to utilise the condition that when $g<1/n$ the probability integral shall be unity. This condition is sufficient to determine the $n$ coefficients by equation of the coefficients of $g^0,g^1,\ldots,g^{n-1}$. In other words, Fisher says that $\text{P}$ must be of the form $$\text{P}=\sum_{j=1}^n \al_j(1-jg)_+^{n-1}, $$ and the coefficients $\al_j$ can be determined by the condition that $\text{P}=1$ when $g<1/n$. This is a clever observation. Indeed, for $g<1/n$, $\text{P}$ coincides with the genuine polynomial $\sum_{j=1}^n \al_j(1-jg)^{n-1}$, and this polynomial must be identically equal to $1$ for $g\in(0,1/n)$, by the probability meaning of $\text{P}$. This will require that, in particular, all the derivatives of $\text{P}$ in $g$ of orders $\ge1$ be identically $0$, which will of course uniquely determine the $\al_i$'s. It is still unclear to me what $f$ has to do with $\text{P}$, but this now does not seem to matter much.<|endoftext|> TITLE: Are all cuspidals induced? QUESTION [7 upvotes]: This is a follow-up to this question by Marc Palm asked 7 years ago: Let $K$ be a finite extension of $\mathbb{Q}_p$, and $G$ a reductive group over $K$. Is every irreducible cuspidal representation induced from an open, compact-mod-center subgroup? My specific questions are: Question 1: Have there been any new results on this question in the last 7 years? (Meta remark: I don't know of a different possibility to draw attention to an MO question again, asking for new results, other than asking the question again.) Question 2: Do people generally believe that this is true, or are they expecting that there are some weird counterexamples somewhere? This is of course a very hazy question, but sometimes there is a common "folklore" belief in a community about some open questions. I don't belong to the community of experts on this, that's why I'm asking. Question 3: I once heard the name Bernstein Conjecture for the conjecture that the answer is "yes". Is this a common name? REPLY [9 votes]: Question 1. Yes indeed. a) There are new results for classical groups and their inner forms (works of Shaun Stevens, Daniel Skodlerack, ...). In particular Skodlerack proved that in the case of "quaternionic forms" of classical groups, in residue characteristic not $2$, any irreducible supercuspidal representation is induced. Daniel Skodlerack, "Cuspidal irreducible representations of quaternionic forms of p-adic classical groups for odd p" arXiv:1907.02922 math.RT math.NT b) Jessica Fintzen has improved a result of J.K. Yu by proving that for general reductive groups, J.K. Yu's construction gives all supercuspidal representations as induced representations when the residue characteristic does not divide the order of the Weyl group. Jessica Fintzen, "Tame cuspidal representations in non-defining characteristics" arXiv:1905.06374 She has also corrected errors in Yu's work and has improved his construction (cf. her Arxiv papers). c) Martin Weissman gave a very short and elegant proof that for any rank $1$ reductive group, irreducible supercuspidal are induced. This gave new instances of induced supercuspidal representations. His proof is based on a deep result of Schneider and Stuhler (cf. IHES paper). Martin Weissman, "An induction theorem for groups acting on trees", arXiv:1808.08944 . Question 2. I think the community of experts does indeed think that the conjecture should hold true. Question 3. I have personnally never heard of that.<|endoftext|> TITLE: Is there an example of two strict monoidal categories which are (monoidally) equivalent, but not strictly? QUESTION [8 upvotes]: By strict equivalence, I mean a monoidal equivalence whose underlying monoidal functors are strict, and here I am looking for two monoidal categories which are not strictly equivalent. REPLY [3 votes]: Here is a very universal/abstract example: (The following is some very classical machinery, but I'm not very familiar with the literature on this so I'm not sure which reference should be quoted) Let $\textbf{Mon}$ be the $2$-category of monoidal categories and (pseudo) monoidal functor between and $\textbf{SMon}$ the category of strict monoidal category and strict monoidal functor between them. The forgetfull functor $\textbf{SMon} \rightarrow \textbf{Mon}$ has a left adjoint $F: \textbf{Mon} \rightarrow \textbf{SMon}$ defined as follow: given a monoidal category $C$, the set of objects of $F(C)$ is the free monoids on the objects of $C$. Given two objects $a,b \in F(C)$, formally written as $a_1 \otimes \dots \otimes a_n$ and $b_1 \otimes \dots \otimes b_m$ (where $\otimes$ denotes the product in the free monoid), one defines $Hom_{F(C)}(a,b)$ to be $Hom_C(\overline{a},\overline{b})$ where $\overline{a}$ and $\overline{b}$ denotes object obtain by computing the tensor product $a_1 \otimes \dots \otimes a_n$ and $b_1 \otimes \dots \otimes b_m$ in $C$, as these object are well defined up to unique isomorphisms, there is no ambiguity on what is the Hom set between them. In particular, $F$ is a comonad on $\textbf{SMon}$ whose co-Kleisli category is the category of strict $2$-category and pseudo-monoidal functor between them. Given a strict monoidal category $C$, the counit $F(C) \rightarrow C$ is a strict monoidal functor, and is always an equivalence of category. So it is an equivalence of monoidal category. But requiring it to have a strict inverse give a very strong property to $C$: Lemma: For any strict monoidal category $C$, The following are equivalent: The equivalence $F(C) \rightarrow C$ have a strict monoidal inverse. Any pseudo-monoidal functor $C \rightarrow D$ is equivalent (as pseudo monoidal functor) to a strict monoid monoidal functor $C \rightarrow D$. Any strict monoidal functor $E \rightarrow C$ which is an equivalence has a strict monoidal inverse. (Note: Is there a standard name for this kind of objects ? maybe 'flexible' has been used for this ? ) Indeed, A pseudo-monoidal functor $C \rightarrow D$ is the same as a strict monoidal functor $F(C) \rightarrow D$, hence if $\theta:C \rightarrow F(C)$ is a strict monoidal inverse to $F(C) \rightarrow C$ precomposing $F(C) \rightarrow D$ with $\theta$ gives a strict monoidal functor $C \rightarrow D$ which as a pseudo-monoidal functor is equivalent to one we started from. The third point follow from the second because any strict monoidal functor which is an equivalence has a pseudo-monoidal inverse, so if every pseudo-monoidal functor out of $C$ is equivalent to a strict one, one obtains a strict inverse. The first point immediately follow from the third. So finding an example as asked by the question is exactly the same as finding a $C$ that do not satisifes these conditions: $F(C) \rightarrow C$ will be an example, and conversely given any example as in the question, then at least one of the two monoidal category involved do not satisfies these equivalent conditions. Most strict monoidal categories works, but to give a very simple example: Take $M$ to be a monoid, seen as discrete strict monoidal category, then $F(M) \rightarrow M$ has a strict inverse if and only if the map of monoid $LM \rightarrow M$ (where $LM$ denotes the free monoid on $M$) admits a retraction. Which I think only happen when $M$ is free, and anyway do not happen for most monoids.<|endoftext|> TITLE: Haar-null union of dense subsets QUESTION [5 upvotes]: Let $\{X_i\}_{i \in \mathbb{R}-\{0\}}$ be a set of subsets of a separable infinite-dimensional Fréchet space $X$ and $I$ be uncountable. Moreover, suppose that (Dense $G_{\delta}$) $X_i$ is a dense $G_{\delta}$ subset of $X$ not containing $0$, (Almost Contains a Linear Subspace) For each $i$, there exists a dense linear subset $E_i\subset X$ satisfying $$ E_i-\{0\}\subseteq X_i $$ (Disjoint) $\bigcap_{i \in I} X_i=\emptyset$, (Not a Cover) $\cup_{i \in I} X_i \neq X-\{0\}$, Can we conclude that: $$ X - \bigcup_{i \in \mathbb{R}-\{0\}} X_i, $$ is Haar-null, or at-least it is finite-dimensional? I have never seen this type of result and am pretty new to this type of thing but I ask here since it seems beyond the level of math-stack exchange. Relevant Definitions: Haar-null set: A subset $A\subseteq X$ is Haar-null if there exists a Borel probability measure $\mu$ on $X$ and a Borel subset $A\subseteq B$ satisfying $$ \mu\left( B+x \right)=0 \qquad (\forall x \in X). $$ Facts: I do know that $X=X_i -X_i$ upon applying the Baire category theorem. (Also from the comments the Pettis Lemma). This means that every element in $X$ can be represented as a sum of elements from each $X_i$. In the case (not covered by my question) where $I$ is a singleton, this paper gives a counter-example. Intuitions: As intuition, it can be seen here, that if $X$ is locally compact, then a Borel set is Haar-null if and only if it is of Haar-measure $0$. REPLY [3 votes]: In the Frechet space $X:=\mathbb R^\omega$ consider the dense linear subspace $$L_0:=\{(x_n)_{n\in\omega}\in\mathbb R^\omega:|\{n\in\omega:x_n\ne0\}|<\omega\}.$$ Fix a countable base $\{V_n\}_{n\in\omega}$ of the topology of the space $L_0$ and in each set $V_n$ choose a point $x_n$, which is not contained in the linear hull of the set $\{x_i\}_{i TITLE: The $K$-theory homology of the Eilenberg-MacLane spectrum QUESTION [13 upvotes]: Let $KU$ be the complex $K$-theory spectrum and $H\mathbb{Z}$ be the Eilenberg-MacLane spectrum. For $n\in \mathbb{Z}$, it is known what the homology groups $KU_{n}(H\mathbb{Z})$ are? REPLY [21 votes]: We have $KU_*(H\mathbb Z) = \pi_*(KU\wedge H\mathbb Z)$; this ring is concentrated in even dimensions and carries an isomorphism between the additive and multiplicative formal group law, hence is rational. Thus the map $$ KU\wedge H\mathbb Z\to (KU\wedge H\mathbb Z)\wedge H\mathbb Q\cong KU\wedge H\mathbb Q $$ is an isomorphism. The Chern character $KU\to \sum_{i\in\mathbb Z} \Sigma^{2i}H\mathbb Q$ is a rational equivalence, so that $$ KU_{2n}(H\mathbb Z) \cong\mathbb Q, KU_{2n+1}(H\mathbb Z) \cong 0\ . $$ Alternatively, the Snaith theorem $KU\cong \Sigma^{\infty}\mathbb{CP}^{\infty}[\beta^{-1}]$, where $\beta\in \pi_2(\mathbb{CP}^{\infty})$ is the Bott element, shows that $$ H_*(KU)\cong H_*(\mathbb{CP}^{\infty})[\beta^{-1}]\ . $$ It's easy to calculate that $H_*(\mathbb{CP}^{\infty})$ is a divided power algebra on the generator $\beta\in H_2(\mathbb{CP}^{\infty})$. Since $\beta^n$ is divisible by $n!$, this ring is the Laurent series over $\mathbb Q$ generated by $\beta$. REPLY [20 votes]: Since for any two spectra $E,F$ we have $$ E_n(F)=[\mathbb{S}^n,E\wedge F]\cong [\mathbb{S}^n,F\wedge E] = F_n(E), $$ you may as well ask about $H_n(KU;\mathbb{Z})$, the integral homology of the complex $K$-theory spectrum. This calculation is carried out e.g. in Chapter 16 of the book Switzer, Robert M., Algebraic topology – homology and homotopy., Classics in Mathematics. Berlin: Springer. xii, 526 p. (2002). ZBL1003.55002. In particular, Theorem 16.25 gives the Pontrjagin algebra structure as $$ H_*(KU;\mathbb{Z})\cong \mathbb{Q}[u,u^{-1}],$$ the ring of finite Laurent series over $\mathbb{Q}$ in an element $u\in H_2(KU;\mathbb{Z})$.<|endoftext|> TITLE: Origin of the term "sinc" function QUESTION [20 upvotes]: Is the sinc function defined here, really a short form of "sinus cardinalis" as proposed by Wikipedia? This information is deleted now but it existed some time ago. Even if we search Google Books for this term, a lot of new books call sinc as sinus cardinalis without even bothering to check Woodward's original paper. The origin of sinc is attributed to Peter Woodward's work dating 1952, but he never mentions sinus cardinalis anywhere. This term is not listed in any etymological dictionary of mathematics and even unabridged Oxford English Dictionary. What is meant by cardinal sine, if we assume whoever tried to rationalize the term, thought of it as a meaningful term? Here is the paragraph from Woodward, P. M.; Davies, I. L. (March 1952). "Information theory and inverse probability in telecommunication" (PDF). Proceedings of the IEE - Part III: Radio and Communication Engineering. 99 (58): 37–44. doi:10.1049/pi-3.1952.0011 Thanks. Woodward in his book writes:"Probability and Information Theory, with Applications to Radar" pg 29. REPLY [16 votes]: While irruption of cardinal in this context must somehow relate to Whittaker’s — also unexplained — use of the word (to name the functions subject to his sampling theorem), it seems far less clear that Woodward’s $“\mathrm c”$ had anything to do with it. AFAICT, that whole notion originated in this ambiguous (and/or misread) statement of Higgins (1996, p. 4): Definition 1.2$$ \operatorname{sinc}v:= \begin{cases} \dfrac{\sin\pi v}{\pi v},&\quad v\ne0,\\ 1,&\quad v=0. \end{cases} \tag1 $$ The name $“\operatorname{sinc}”$ 1 is common in the engineering literature, and we shall make much use of it from now on. 1 The name is usually held to be short for the Latin sinus cardinalis. It was introduced by Woodward (1953, p. 29), although it is not certain whether this is the earliest occurrence. Now this just conflates talk about a notation ($\operatorname{sinc}$) with talk about a purportedly related name for $(1)$. Whereas, not only does Woodward (as you noted) nowhere write “cardinal” or even cite Whittaker, in fact hardly anyone before 1996 seems to have called $(1)$ any name at all: Year: notation: name: Woodward & Davies 1952 (p. 41) sinc — Woodward 1953 (p. 29) sinc — Jagerman & Fogel 1956 (p. 143) — cardinal series kernel Nathan 1956 (p. 788) sinc — Bracewell 1957 (p. 69) sinc — Raabe 1958 (p. 181) sinc — Ragazzini & Franklin 1958 (p. 31) — cardinal hold response Linden & Abramson 1960 (p. 26) sinc — Helms & Thomas 1962 (p. 179) sinc — Lochard 1962 (p. 714) ? sinus cardinal Petersen & Middleton 1962 (p. 303) — cardinal function Battail 1964 (p. 128) sinc sinus cardinal Bracewell 1965 (p. 62) sinc — Detape 1965 (p. 9) — sinus cardinal Burdic 1968 (p. 48) sinc — Goodman 1968 (p. 14) sinc — Robaux & Roizen-Dossier 1970 (p. 733) sinc — McNamee, Stenger & Whitney 1971 (p. 142) sinc — Oswald 1975 (p. 65) — sinus cardinal Lannes & Pérez 1983 (p. 163) sinc sinus cardinal Schempp 1983 (p. 213) sinc sinus cardinalis De Coulon 1984 (p. 23) sinc sinus cardinal Usher 1984 (p. 98) sinc — Léna 1986 (p. 111) sinc sinus cardinal Schempp 1986 (p. 193) sinc sinus cardinalis Butzer, Splettstößer & Stens 1988 (p. 2) sinc — Stenger 1993 (p. vi) sinc — The above are all I found who used a special notation and/or name for $(1)$ — please add any that I missed. Attendant observations: Even the sinc notation was rather rare among the scores who published about the sampling theorem. (250+ papers in engineering journals over 1950–1975, according to Butzer (1983, p. 186).) Bracewell 1965 is apparently the first who attributed it to Woodward. Of those who used it, only five (that I could find) concurrently used the name sinus cardinal(is), mostly after 1983. (Lochard 1962 might too, and would be interesting to get your hands on.) Stenger 1993 says that Whittaker’s series “reappeared (...) in the important papers of Hartley, Nyquist and Shannon, who illustrated [their] role in communication theory. The term “sinc function” (...) was first defined and used in these papers, where sinc is defined by” $(1)$. I couldn’t corroborate the last sentence. Butzer & al. (2011, p. 65512) write that “The sinc function (...) had been defined by Raabe’s teacher Küpfmüller”. However, it is unclear just what they mean to say he “defined” — thing, name, notation? Incidentally this paper, or the slides by Stanković & al. (2013), should dispel any preconception on what language things happened in.<|endoftext|> TITLE: An explicit formula for a cuspidal form of weight $2$ and arbitrarily large prime level QUESTION [6 upvotes]: In Miyake's book on modular forms explicit formulas for the $q$-expansions of a basis of the space Eisenstein series of arbitrary level and weight were given. I guess similar formulas for a basis of the entire space cuspidal forms do not exist. Is there at least an explicit formula for the $q$-expansion of a non-zero cuspidal holomorphic modular form of level $\Gamma_0(p)$ and weight $2$ for arbitrarily large primes $p$? REPLY [5 votes]: Borisov and Gunnells have defined and studied the so-called toric modular forms. The idea is to take products of Eisenstein series of lower weight. So in your case, (linear combinations of) products of two Eisenstein series of weight 1. More precisely, Borisov and Gunnells consider Eisenstein series $s_a$ for $a \in \mathbb{Z}/N\mathbb{Z}$, $a \neq 0$. They have weight 1 and level $\Gamma_1(N)$, and their $q$-expansions are completely explicit. They show that the pairwise products $s_a s_b \in M_2(\Gamma_1(N))$ behave very much like modular symbols. In particular, and modulo Eisenstein series of weight 2, they satisfy the nice 3-term Manin relations, one can compute the action of the Hecke operators, and so on. You are interested with $\Gamma_0(p)$ so you would need to consider \begin{equation*} f_{a,b} = \sum_{k=1}^{p-1} s_{ka} s_{kb} \end{equation*} which belongs to $M_2(\Gamma_0(p))$. It may not be cuspidal, but since $\Gamma_0(p)$ has only two cusps, there is only one Eisenstein series and you can compute the Eisenstein component of $f_{a,b}$ just from its leading term. In this way you get explicit cusp forms of weight 2 for any $p$. Borisov and Gunnells have shown that the (cuspidal components of) $f_{a,b}$ span the space of cusp forms of analytic rank zero, i.e. the space generated by the newforms $f$ satisfying $L(f,1) \neq 0$.<|endoftext|> TITLE: Is there any deep philosophy or intuition behind the similarity between $\pi/4$ and $e^{-\gamma}$? QUESTION [55 upvotes]: Here is a couple of examples of the similarity from Wikipedia, in which the expressions differ only in signs. I encountered other analogies as well. $${\begin{aligned}\gamma &=\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1-xy)\ln xy}}\,dx\,dy\\&=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-\ln {\frac {n+1}{n}}\right).\end{aligned}}$$ $${\begin{aligned}\ln {\frac {4}{\pi }}&=\int _{0}^{1}\int _{0}^{1}{\frac {x-1}{(1+xy)\ln xy}}\,dx\,dy\\&=\sum _{n=1}^{\infty }\left((-1)^{n-1}\left({\frac {1}{n}}-\ln {\frac {n+1}{n}}\right)\right).\end{aligned}}$$ $${\begin{aligned}\gamma &=\sum _{n=1}^{\infty }{\frac {N_{1}(n)+N_{0}(n)}{2n(2n+1)}}\\\ln {\frac {4}{\pi }}&=\sum _{n=1}^{\infty }{\frac {N_{1}(n)-N_{0}(n)}{2n(2n+1)}},\end{aligned}}$$ (where $N_1(n)$ and $N_0(n)$ are the number of 1's and 0's, respectively, in the binary expansion of $n$). I wonder whether is there any algebraic system where $4e^{-\gamma}$ would play a role similar to what $\pi$ plays, say in complex numbers, or a geometric system where $4e^{-\gamma}$ would play some special role, like $\pi$ in Euclidean and Riemannian geometries. REPLY [55 votes]: The intuition may be helped by considering the generalized Euler constant function $$\gamma(z)=\sum_{n=1}^\infty z^{n-1}\left(\frac{1}{n}-\ln\frac{n+1}{n}\right),\;\;|z|\leq 1.$$ Its values include the Euler constant $\gamma=\gamma(1)$ and the "alternating Euler constant" $\ln 4/\pi=\gamma(-1)$. So any general integral formula or recursion relation for $\gamma(z)$ will establish a connection of the type noted in the OP. The properties of the function $\gamma(z)$ have been studied in The generalized-Euler-constant function and a generalization of Somos's quadratic recurrence constant (2007). Somos's constant $\sigma=\sqrt{1\sqrt{2\sqrt{3\cdots}}}$ is obtained as $\gamma(1/2)=2\ln(2/\sigma)$. Another special value $$\gamma(i)=\frac{\pi}{4}-\ln\frac{\Gamma(1/4)^2}{\pi\sqrt{2\pi}}+i\ln\frac{8\sqrt\pi}{\Gamma(1/4)^2}.$$<|endoftext|> TITLE: Isometries of convex hypersurfaces QUESTION [8 upvotes]: The well known Pogorelov’s rigidity theorem says that if two convex closed surfaces in Euclidean 3-space are isometric to each other being equipped with their intrinsic metrics, then they are congruent, namely there is an isometry of the Euclidean space which maps one surface onto the other. Are there counter examples to Pogorelov’s theorem in higher dimensions? REPLY [9 votes]: There is definitely no counterexample for convex polyhedra and no counterexample for bodies with smooth boundary with non-degenerate second fundamental form. (In the latter case even small open subsets of the boundary cannot be deformed, the argument can be found e.g. in Spivak volume 5.) I do not know if the question was asked somewhere in full generality, although it sounds very natural. If one could prove a sort of stability estimate: if the intrinsic metrics of surfaces are $\epsilon$-close, then the bodies are $\delta$-close, where $\delta$ depends on $\epsilon$ and some "rough" geometric characteristics of the bodies like their diameter, then one would get the uniqueness in the general case by approximation. Such a stability estimate is claimed to be proved in Volkov's article reproduced as appendix to the new edition of Alexandrov's "Convex polyhedra", but the proof is flawed.<|endoftext|> TITLE: Explicit bounds from Tao's result on Collatz conjecture QUESTION [12 upvotes]: A new preprint by Terry Tao has recently appeared and has established some interesting results regarding the topic of Collatz conjecture. I will not cite the precise result, but rather an equivalent formulation which Tao notes in his Remark 1.4: For any $\delta>0$ there exists a constant $C_\delta$ such that $\mathrm{Col_{min}}(N)\leq C_\delta$ for all $N$ in a subset of $\mathbb N+1$ of logarithmic density at least $1-\delta$. My question concerns whether anything is known about the rate of growth of $C_\delta$ as $\delta\to 0$. I will pose two specific questions in that regard. The first one I imagine might have been known before Tao's recent result. Are there any values of $\delta<1$ for which an explicit upper bound for $C_\delta$ is known? The second essentially asks if anything sort-of-explicit can be deduced from Tao's result. Is the function $N\mapsto C_{1/N}$ upper bounded by a computable function? (Note that I don't think the answer is "obviously yes" from mere existence of $C_\delta$, since checking that a particular value works seems to be $\Sigma^0_3$ (there exists $N_0$ such that for all $N>N_0$ there exists $M$ such that in $M$ steps at least $1-\delta$ of numbers below $N$ go below $C_\delta$) and is not obviously any lower) REPLY [2 votes]: In v2 of the arXiv preprint, there is more detail, including fairly explicit dependence on $\delta$.<|endoftext|> TITLE: Homotopy fibre sequence and left Bousfield localization QUESTION [5 upvotes]: Let $\mathcal{M}$ be a pointed model category and $\mathcal{C}$ a class of maps in $\mathcal{M}$ for which the left Bousfield localization ${\rm L}_{\mathcal{C}}\mathcal{M}$ exists (see Hirschhorn, Model categories and their localizations, Ch. 3). Assume $F\to E\to B$ is a fibre sequence and all three objects are fibrant (in $\mathcal{M}$). If $F, B$ are fibrant in ${\rm L}_{\mathcal{C}}\mathcal{M}$ (i.e. they are $\mathcal{C}$-local, Hirschhorn, def. 3.1.4), do we have that $E$ is also $\mathcal{C}$-local? Or can we replace this fibre sequence (in $\mathcal{M}$) by $F\to E'\to B$ with $E'$ $\mathcal{C}$-local? REPLY [5 votes]: The idea you're looking for is called "fibrewise localization". It's defined in Dror Farjoun's book "Cellular Spaces, Null Spaces, and Homotopy Localization", and also in Hirschhorn's book (since you clearly have a copy of the latter, I'll use references from there). Hirschhorn's Definition 6.1.1 defines the fiberwise localization of a map $p: E\to B$, of pointed spaces, as a factorization $p = q\circ i$ where $q$ is a fibration, and for every $b\in B$, the induced map on homotopy fibers $HFib_b(p) \to HFib_b(q)$ is a $\mathcal{C}$-local equivalence. In general, you construct such localizations by localizing the fibers. To be fibrant in the fiberwise localization model structure means that $p$ is a fibration and that all fibers are $\mathcal{C}$-local. When the fiberwise localization exists, it tells you how to replace your $F\to E\to B$ by the weakly equivalent $F\to E' \to B$. Unfortunately, Hirschhorn's 6.1.4 proves that fiberwise localization DOES NOT EXIST in the category of pointed spaces. Further conditions are needed, as Kevin Carlson pointed out. So the general answer to your question has to be "no", or else fiberwise localization would exist for free. One place that gives such conditions is Chataur and Scherer, "Fiberwise localization and the Cube Theorem", Theorem 4.3 (note that the "cube axiom" is required). In the situation of your question, $F$ is already local, so $\eta: F\to L_C(F)$ can be taken to be the identity. The proof of 4.3 shows that $E'$ (there denoted $\overline{E}$) is constructed as a localization of $E$, so is local, as you wanted.<|endoftext|> TITLE: $Ext^1$ for some modules over the polynomial ring in one variable QUESTION [5 upvotes]: Let $M$ be a module over the polynomial ring $\mathbb C[x]$ such that $x+n$ is invertible in $M$ for every integer $n$. Let $N=\oplus _{n>0} \mathbb C_n$ where $\mathbb{C}_n := \mathbb C[x]/x+n$. $\mathbf{Question:}$ Is it true that $Ext^1(M,N)=0$? It is clear that $Ext^1(M,\mathbb C_n)=0$ but this is not enough since $M$ is not finitely generated and therefore $Ext(M,?)$ does not commute with direct sums. REPLY [5 votes]: $\newcommand{\bC}{\mathbb{C}}\newcommand{\bZ}{\mathbb{Z}}$The resolution constructed by Dylan Wilson seems to show that $Ext^1(M,N)$ is non-zero already for $M=\bC[x][\frac{1}{x+n},n\in\bZ]$. There is an identification $$Hom(\bC[x][\frac{1}{x+n},n\in\bZ],\bigoplus\limits_{n\in\bZ}\bC[x,\frac{1}{x+n}]/\bC[x])=\prod\limits_{n\in\bZ}'\bC((x+n))$$ where the restricted tensor product is taken with respect to the submodules $\bC[[x+n]]\subset\bC((x+n))$ Proof. Given a homomorphism $f:\bC[x][\frac{1}{x+n},n\in\bZ]\to\bigoplus\limits_{n\in\bZ}\bC[x,\frac{1}{x+n}]/\bC[x]$ associate to it the element of the restricted product $(a_n)_{n\in\bZ}$ given by $a_n=\lim\limits_{k\to\infty}(x+n)^k\widetilde{f(\frac{1}{(x+n)^k})_n}$ where $\widetilde{f(\frac{1}{(x+n)^k})_n}$ denotes an aribtrary lift of the $n$-th component of $f(\frac{1}{(x+n)^k})$ to an element of $\bC[x,\frac{1}{x+n}]$. For almost all $n$ the element $a_n$ lies in $\bC[[x+n]]$ because $f(1)_n$ is zero for almost all $n$. The map $f\mapsto(a_n)$ from the LHS to the RHS is injective because a homomorphism $f:\bC[x][\frac{1}{x+i},i\in\bZ]\to\bC[x,\frac{1}{x+n}]/\bC[x] $ is completely determined by its restriction to $\bC[x][\frac{1}{x+n}]$. The map is surjective because a Laurent series $a_n\in\bC((x+n))$ gives a homomorphism $\bC[x,\frac{1}{x+n}]\to\bC[x,\frac{1}{x+n}]/\bC[x]$ which then extends uniquely to a homomorphism from the whole of $M$. $\square$ This yields a formula for the desired RHom $$RHom(M,\bigoplus \bC[x]/(x+n))=\\ RHom(M,\mathrm{fib}(\bigoplus\limits_{n\in\bZ}\bC[x,\frac{1}{x+n}]/\bC[x]\to \bigoplus\limits_{n\in\bZ}\bC[x,\frac{1}{x+n}]/\bC[x]))=\\ \mathrm{fib}\left(\prod\limits_{n\in\bZ}'\bC((x+n))\xrightarrow{(x+n)_{n\in\bZ}} \prod\limits_{n\in\bZ}'\bC((x+n))\right)$$ So, for instance, the "adele" $(1,1,\dots)$ gives a representative of a non-zero class in the Ext^1.<|endoftext|> TITLE: Do Minkowski sums have anything like calculus? QUESTION [7 upvotes]: Is there anything resembling differential calculus over the space of (nicely behaved) regions in $\mathbb{R}^d$, where addition is interpreted in terms of Minkowski sums? For example, it is known that Minkowski sums act linearly on the perimeter of two-dimensional convex bodies. Is there any sense in which one could define something like a (constant) gradient for the perimeter function? I would assume that if anything, said gradient would just be a circle centered at the origin. REPLY [6 votes]: There is a theory for convex bodies. If $A, B \subset \mathbb{R}^n$ are convex bodies, whose interiors contain the origin, you can use set addition to define $A+tB$, for any $t \ge 0$, and $$ V(A,B) = \lim_{t\rightarrow 0+} \frac{A+tB}{t}. $$ This is known as the first mixed volume. $h_A$ and $h_B$ are the support functions, then $$ V(A,B) = \int_{S^{n-1}} h_B\,dS_A, $$ where $S^{n-1}$ is the unit sphere and $dS_A$ is known as the surface area measure of $A$. In this sense, the gradient of the volume function is this measure. If $B$ is the unit ball, then $V(A,B)$ is the surface area (perimeter if $n = 2$). If you differentiate again, you get higher order curvature measures. It turns out that $V(A+tB)$ is a polynomial of degree $n$, called the Steiner polynomial, and its coefficients are called mixed volumes.<|endoftext|> TITLE: Status of the extended form of the Lichtenbaum conjecture QUESTION [11 upvotes]: The extended Lichtenbaum conjecture concerns the relationship of special values of $L$-functions of number fields $K$, to the algebraic $K$-theory and etale cohomology of the ring of integers $O_K$. For example, when $K$ is totally real and $n$ is even, it is known that (cf. Section 4.7.4 of Kahn): $$\zeta_{K}(1-n) = \pm 2^{r_1} \frac{|K_{2n−2}(O_K)|}{|K_{2n−1}(O_K)|} = \pm \frac{|H^{2}(O_K;Z(n))|}{|H^1(O_K;Z(n))|}.$$ I am interested in the case $n$ odd. In that case, the conjectured value involves additionally a determinant of values of the Beilinson regulator. In particular, formula (4.34) of the aforementioned reference says that the conjectured value "should" be $$\lim_{s \to 1−n} (s+n)^{-g_n} \zeta_K(s) = \pm \frac{|H^{2}(O_K;Z(n))|}{|H^{1}(O_K;Z(n))_{tors}|} R'_n(K).$$ For certain number fields $K$, this is known to be true up to a power of 2, e.g. Theorem 4.1 of Kolster. However, this and other references go back a number of years, and it seems not impossible to me that nowadays more is known: For which number fields $K$ is the above formula for $\lim_{s \to 1−n} (s+n)^{-g_n} \zeta_K(s)$ known to be correct, without discarding powers of 2? I am particularly interested in the case $K=\mathbb{Q}$. REPLY [5 votes]: Thanks very much for the question! Looking it up a bit I found the PhD thesis "The Lichtenbaum Conjecture at the Prime 2 " by Ion Rada (a student of Kolster) which proves that for every abelian number field $K$ and every odd integer $n \geq 3$ one has $$ \zeta_K^{\ast}(1 - n) = \varepsilon_n(K) \cdot 2^{\mu_2(K^{+})} \cdot \frac{h_n(K)}{w_n(K)} \cdot R_n^{B}(K) $$ where: $\varepsilon_n(K) \in \{ \pm 1\}$ is determined by the $\Gamma$-factors in the functional equation; $h_n(K) := \prod_p \lvert H^2_{\text{et}}(\mathcal{O}_K^{S_p(K)},\mathbb{Z}_p(n)) \rvert$, where (as far as I understand) $S_p(K)$ is the set of places of $K$ consisting of the places lying above $p$ and the Archimedean places corresponding to real embeddings; $w_n(K) := \prod_p \, \lvert H^0(K,\mathbb{Q}_p/\mathbb{Z}_p(n))\rvert $; $R_n^B(K)$ denotes Beilinson's regulator; $K^{+}$ is the maximal real subfield of $K$; for every number field $F$ and every prime $p$ we set $\mu_p(F)$ to be the $\mu$-invariant of the module $U_p(F)/C_p(F)$, where $U_p(F)$ denotes the projective limits of the groups $\mathcal{O}_{F_{p^r}}^{\times}$, where $F_{p^r}$ denotes the $r$-th field in the cyclotomic $\mathbb{Z}_p$-tower of $F$, and $C_p(F)$ denotes the group of circular units. I believe that the main ingredient is the proof of the Milnor conjecture by Voevodsky.<|endoftext|> TITLE: Nowhere vanishing section implies reduction of structure group QUESTION [5 upvotes]: Description I noticed a repeating theme in vector bundle theory, and wonder if there's a theorem that describes this kind of phenomenon. Given a vector bundle $E$ over a manifold $X$. If there is a nowhere-vanishing section for some associated bundle, then the structure group for $E$ can be reduced. Example Let $M^m$ be an Riemannian manifold. If there is a nowhere-vanishing section for $\Lambda^{m} {T^*M}$, then the structure group for $TM$ can be reduced from $O(m)$ to $SO(m)$. Let $M^7$ be a compact oriented Riemannian manifold. If there is a nowhere-vanishing section for $\Lambda^3_+T^*M$, then the structure group of $M$ can be reduced to $G_2$. (Implicitly mentioned by a informal/use-at-your-risk note section 2.4 proposition 2.) Question Is there a more general theorem that depicts more examples like these? Update (2019-09-17) I think I have a more specific statement in mind, but lack of a rigorous proof. Given that the statement is correct, I feel that the proof can be found in some elementary textbooks about principle G-bundles. Thus, I apologize posting this question here rather than math-stack-exchange: I posted this here based on my ignorance about principle $G$-bundles. The statement (still need a proof!) goes as follows: Let $E\to M$ be a $G$-bundle with vector space fiber $F$, $\Phi$ a functor from ($G$-rep) to ($G$-rep), and $H$ the stabilizer group in $G$ for some fixed element $x\in\Phi(F)$. Then the followings are equivalent. The structure group of $E$ can be reduced to $H$. There exists a nowhere vanishing global section for $\Phi(F)\to M$. For example, let $v$ be a nonzero vector of $\mathbb{R}^n$, its stabilzer in $GL(n)$ is isomorphic to $GL(n-1)$, so the existence of a nonvanishing section reduces the structure group from $GL(n)$ to $GL(n-1)$. Similarly, the existence of an independent $k$-frame reduces the structure group to $GL(n-k)$. If you find more comfortable with vector bundles (rather than $G$ bundles as I did), this is simply saying that there is a rank $k$ trivial subbundle in our vector bundle. Also notice that if $v$ was taken to be a zero vector, then its stabilizer is the whole group and the statement says the structure group can be reduced to itself (a null statement!). I find this statement useful (if proved!) for it can be applied to more complicated cases when the functor $\Phi$ is not taken as an identity. For more example, If $\Phi=Sym^{2,*}$, $x$ is a nondegenerated metric, then the stabilizer is the orthogonal group. Having a nonvanishing metric thus means that the structure group can be reduced to the orthogonal group. If $\Phi=\Lambda^{top}$, similar arguments show that the existence of a nonvanishing top form is equivalent to "orientableness". For $\Phi=\Lambda^{3,*}$, the second example in the original post shows a $G_2$ structure on a $7$-manifold. In this paper, proposition $3.2$ proves that for a 7-dimensional compact manifold $M^7$, then $TM$ admitting a $Spin_7$ structure implies a $SU(2)$ structure! And the argument is to first use representation theory to create three independent spinors, whose stabilizer is $SU(2)$ at the fibre level, and to successfully reduce the structure group from the large $Spin_7$ to $SU(2)$. REPLY [2 votes]: Informally, say a type of structure $S$ which can form "bundles" over a manifold is a "symmetry structure" if all fibers are isomorphic (in an appropriate, say categorical, sense). Let $G(S)$ be the automorphism group of a single fiber (usually a Lie group). It is not hard to see that in that case an "$S$-bundle" over a manifold $X$ is given by precisely the same data as a principal $G(S)$ bundle, a.k.a. a $G(S)$ torsor. In particular, this implies that the data of an $S$-bundle is the same as the data of an $S'$-bundle provided they have the same symmetry group. Now structures like a Riemann structure are not just bundle structures: they are bundle structures which "enrich" the tangent bundle. The tangent bundle is "a priori" a $GL_n$-structure, and other important structures are "reductions of structure" from $GL_n$ to the automorphism group of a fiber of whatever structure you are considering. In particular, the pair $(V, g)$ for $V$ an $n$-dimensional vector space and $g\in S^2V^*$ a positive-definite bilinear form has automorhpism group isomorphic to $O(n).$ Thus choosing a positive-definite sectino of $S^2T^*_x$ consistently at all points $x$ produces a "reduction of structure" on the tangent bundle from a bundle of $GL_n$-symmetric objects to a bundle of $O(n)$-symmetric objects. Similarly, if you fix, in addition to $g\in S^2T^*X$ a nonzero length section of $\Lambda^n(V)$ which has length $1$ w.r.t. $g$, the symmetry group decreases to $SO(n)$ (note that any nonzero section can be positively rescaled to a length-one section, hence the observation you wrote). If you fix "too much" data on top of the metric, eventually your bundle ceases to be a "symmetry structure" and you might get infinitely many non-isomorphic possibilities for your fibers. However so long as this does not happen, you will always have a group behind the scenes describing the bundle behavior.<|endoftext|> TITLE: Beauville-Laszlo for schemes QUESTION [8 upvotes]: For a commutative ring $A$ and $f\in A$ a non-zero divisor, the Beauville-Laszlo theorem gives a gluing statement for vector bundles on $A$ in terms of a vector bundle on $A\big[\frac{1}{f}\big]$, a vector bundle on $\hat{A}$ the ($f$)-adic completion of $A$ and an isomorphism on $\hat{A}\big[\frac{1}{f}\big]$. Is there a similar statement for a scheme $X$ and an effective Cartier divisor $D\subset X$? REPLY [13 votes]: Completing the discussion under Will Sawin's answer. The question has been answered completely and affirmatively by Ben-Bassat and Temkin in their paper "Berkovich spaces and tubular descent" (Adv. Math. 2013), but one has to be careful when globalizing $\hat A$ and (especially) $\hat A[1/f]$. (1) Beauville and Laszlo treat the non-affine case in the Corollary on p. 8 of their paper. However, what they do to define $\hat X$ (by taking the relative spec over $X$ of $\varprojlim \mathcal{O}_X/I_Z^n$) seems flawed for the reason that completion does not commute with localization, e.g. $$ (\varprojlim k[[t]][x]/(t^n))[1/x] \neq \varprojlim k[[t]][x, x^{-1}]/(t^n). $$ Therefore there does not exist a quasi-coherent $\mathcal{O}_X$-algebra whose value on an affine open $U= \operatorname{Spec} A$ is $\hat A$ (completion w.r.t. the ideal of $U\cap Z$ in $U$). See also paragraph 1.3 (p. 219) in Ben-Bassat and Temkin. (2) Clearly, the global version of $\hat A$ should be $\hat X$, the formal completion of $X$ along $Z$, which is a formal scheme but not a scheme. It is more delicate to define the "tubular neighborhood" $W = "\hat X \setminus Z"$ (quotes because, taken literally, this is the empty formal scheme). The answer is that $W$ is an object of rigid analytic geometry, to which there are several approaches. Ben-Bassat and Temkin choose to work in Berkovich spaces, but discuss the other approaches as well in Section 4.6. They remark that whatever framework we choose, the category of coherent sheaves or vector bundles will be the same. (3) Their main result is: Theorem (2.4.8). Let $X$ be a scheme of finite type over a field $k$, and let $Z$ be a closed subscheme of $X$. We denote by $\hat X$ the formal completion of $X$ along $Z$, and by $W$ the corresponding Berkovich space (which the authors define). Then $$ \operatorname{Coh}(X) \stackrel{\sim}{\longrightarrow} \operatorname{Coh}(X\setminus Z) \times_{\operatorname{Coh}(W)} \operatorname{Coh}(\hat X). $$<|endoftext|> TITLE: Is the tensor product of two infinite dimensional objects in the BGG category $\mathcal{O}$ of a semisimple Lie algebra always not in $\mathcal{O}$? QUESTION [10 upvotes]: Let $\mathfrak{g}$ be a finite dimensional complex semisimple Lie algebra (according to a comment of Victor Ostrik, we need to further require that $\mathfrak{g}$ is simple) and we can consider its BGG category $\mathcal{O}$. It is well-known that $\mathcal{O}$ is not closed under tensor product, i.e. take two $\mathfrak{g}$-modules $M$ and $N$, the tensor product $M\otimes N$ is still a $\mathfrak{g}$-module but not necessarily in the category $\mathcal{O}$. In an answer to this MO question, Jim Humphreys showed that an object $M$ in the category $\mathcal{O}$ is “tensor-closed” (meaning that $M\otimes N$ is in $\mathcal{O}$ whenever $N$ is) if and only if $M$ is finite dimensional. He also wrote a note on it. My question is: if $M$ and $N$ in $\mathcal{O}$ are both infinite dimensional, then is it always true that $M\otimes N$ is no longer in $\mathcal{O}$? In particular let $L(\lambda)$ and $L(\mu)$ be the unique simple module with highest weights $\lambda$ and $\mu\notin \Lambda^{+}$, is $L(\lambda)\otimes L(\mu)$ in the category $\mathcal{O}$? REPLY [2 votes]: I will answer the question in the affirmative for $\mathfrak{sl}_n$, and in general provide a property of the Weyl group that would imply an affirmative answer in general. This property is true in type A, but not for the other rank 2 cases. The other rank 2 Lie algebras are small enough to be dealt with easily. Let me assume for now that λ is regular and integral. Write $\lambda = x\cdot \lambda'$, where x is in the Weyl group and λ' is dominant. Write ch(L(λ)) as a rational function in reduced form, and let T be the set of positive roots α such that $(1-e^\alpha)$ does not appear in the denominator. Let G be the subgroup generated by all reflections in all elements of T. I claim that if g is in G, then gx≤x in Bruhat order. Switch to indexing simples and Vermas by the Weyl group. We have $$\operatorname{ch}(L(x))=\sum_y P_{x,y}\operatorname{ch}(M(y)).$$ The claim follows from two simple observations. Firstly, for $(1-e^\alpha)$ to not appear in the denominator, we must have $P_{x,y}+P_{x,sy}=0$, where s is the reflection for the root α. Secondly, that Px,x=1 and Px,y=0 unless y≤x. Now we turn to the question. Consider two simples L(λ) and L(μ). From the above discussion we obtain two Weyl group elements x and y, together with two sets of positve roots Tx and Ty, and groups Gx and Gy generated by the corresponding reflections, so that $G_xx\leq x$ and $G_y y\leq y$. If $L(\lambda)\otimes L(\mu)$ is in category O, then $T_x\cup T_y$ is all positive roots by looking at the character. For L(λ) and L(μ) to be infinite dimensional, we must have that x and y are not the identity. Now we have a condition on the Weyl group which is necessary for a negative answer. In the symmetric group, subgroups generated by reflections are products of smaller symmetric groups. It is easy to see that for any two proper such subgroups, there is a reflection which is not contained in either. Thus in type A, since every positive root appears in either Tx or Ty, either Gx or Gy is the entire symmetric group, which then contradicts the Bruhat ordering condition as x and y are not the identity. If we move beyond regular or integral weights, then it becomes even harder to cancel factors in the denominator of the character, so the argument should be easier. Beyond type A, I haven't checked any non-type A Weyl groups of rank at least three to see if my criterion is enough to answer the question.<|endoftext|> TITLE: How are reflection groups related to general point groups? QUESTION [9 upvotes]: I always tried to understand how the finite reflection groups of $\Bbb R^d$ (of some fixed dimension $d$) relate to the point groups of the same space $\smash{\Bbb R^d}$ (finite subgroup of the orthogonal group $\smash{\mathrm O(\Bbb R^d)}$). Initially, I was under the impression that each point group is a subgroup of a finite reflection group. This turned out to be wrong, which is obvious in hindsight. Many reflections groups have symmetries in the placement of their mirrors that can be used to enlarge the group. So let's take these enlarged groups instead. From my geometric understanding, by that I mean the symmetry groups of the uniform polytopes. So I shall call them uniform point groups. Most (or all?) uniform polytopes can be generated from a reflection group, and then it has all the symmetries of this group, but might have more. Question: Is every point group a subgroup of a uniform point group? Regardles of the answer to that question, I am open for any statement that sheds light on the placement of reflections groups (or easily derived groups thereof) inside the family of general point groups. Update Sep. 2019 There seems to exist a counterexample in dimension four, namely, a point group denoted $\pm[I\times C_n]$ that is supposedly not the subgroup of a symmetry group of a uniform polytope. This was mentioned in this answer by Günter Rote. Currently I am not able to verify the claim. So, any hint is welcome. REPLY [3 votes]: Here is a more worked-out and concrete version of the proposed counterexample mentioned in the "Update" of the post. We start from a symmetric arrangement of 12 great circles $F_1,\ldots,F_{12}$ on the 3-sphere $\mathbb{S}^3$: 12 circles from the Hopf fibration, shown in stereographic projection to $\mathbb{R}^3$. They are the inverse image of the 12 corners of the regular icosahedron under the Hopf map $\mathbb{S}^3\to\mathbb{S}^2$ (fibers of the Hopf fibration). On each circle we place 70 equidistant points: 840 points on 12 circles. The supporting hyperplanes of the 3-sphere at these points form a 4-polytope $P$ with 840 equal facets: A perspective view of a facet, a 3-polytope with 40 vertices and 22 sides. (In order to ensure that all faces are equal, the regular 70-gons on the different circles cannot just be placed arbitrarily in Step 2. The points form the orbit of a specially chosen starting point under the group $\pm[I\times C_7]$.) The facets are thin plates of roughly pentagonal shape centered at points of a circle $F_i$ and lying perpendicular to $F_i$. The plates stack up to form a twisted pentagonal tube that surrounds $F_i$. Such a tube makes one full turn of $360^\circ$ as it winds around the circle. The 12 tubes fill the space around the 3-sphere and enclose it completely. The symmetry group $G$ of $P$ has 8400 elements: a given plate can be mapped to any of the 840 plates in 10 different ways. There are only rotations (determinant $+1$), no reflections (determinant $-1$). (Because of the special choice of the starting point of the orbit, the group $G$ is larger than the group $\pm[I\times C_7]$ by which the orbit was generated. I guess, with a generic orbit, the symmetry group will reduce to size 840, but the picture will be messier.) The polytope $P$ is clearly not a uniform polytope: none of its 2-faces is a regular polygon. To definitely answer the question, one would have to argue why $G$ is not a subgroup of the symmetries of a different, uniform, polytope $P'$. One could use the classification of the point groups from the book of Conway and Smith, but maybe there is a more direct argument. The group must contain all rotations of a regular 70-gon. This example and the programs (in Sage) that were used to produce the images were prepared with the help of my student Laith Rastanawi.<|endoftext|> TITLE: Relations between boundaries of groups acting on hyperbolic spaces with WPD elements QUESTION [5 upvotes]: Let $(X,d)$ be Gromov-hyperbolic space and let $\Gamma$ be a finitely generated group acting on $\Gamma$ by isometries. Recall the following two definitions. Say that the action is acylindrical if for every $\epsilon$, there exist $R,N$ such that for every two points $x,y\in X$ with $d(x,y)\geq R$, there are at most $N$ elements $g\in \Gamma$ such that both $d(x,g\cdot x)\leq \epsilon$ and $d(y,g\cdot y)\leq \epsilon$. Let $h\in \Gamma$ be a loxodromic element with respect to the action. Say that $h$ is WPD if for every $\epsilon$ and every $x\in X$, there exists $m\in \mathbb{N}$ such that the set of elements $g\in \Gamma$ satisfying both $d(x,g\cdot x)\leq \epsilon$ and $d(h^m\cdot x,gh^m\cdot x)\leq \epsilon$ is finite. Of course if the action is acylindrical, every loxodromic element is WPD. One of the main result of Osin's paper Acylindrically hyperbolic groups is that if $\Gamma$ is not virtually cyclic and acts on a hyperbolic space $X$ with a WPD element, then it acts acylindrically on a hyperbolic space $Y$. We thus have two boundaries for $\Gamma$, namely its limit set $\Lambda_X\Gamma$ in the Gromov boundary of $X$ and its limit set $\Lambda_Y\Gamma$ in the Gromov boundary of $Y$. My question is as follows. Question. Can we construct the space $Y$ such that $\Lambda_X\Gamma$ and $\Lambda_Y\Gamma$ equivariantly homeomorphic ? At least can we construct the space $Y$ such that there is an equivariant embedding $\Lambda_Y\Gamma\hookrightarrow \Lambda_X\Gamma$ ? For example, does Osin's construction of $Y$ yield an embedding $\Lambda_Y\Gamma\hookrightarrow \Lambda_X\Gamma$ ? Some motivation. On the one hand, usually, when $\Gamma$ acts on a space $X$ with WPD elements, one has a geometric interpretation of $\Lambda_X\Gamma$. This is for example the case for the group $Out(F_n)$ acting on the free factor complex or on the sphere complex. On the other hand, when one has an acylindrical action, one can derive a lot of properties from this action. One thing I'm interested in is the following. Given a random walk on $\Gamma$, the set $\Lambda_Y\Gamma$ endowed with the harmonic measure is a model for the Poisson boundary. Thus, if $\Gamma$ acts on $X$ with a WPD element and if $\Lambda_Y\Gamma\hookrightarrow \Lambda_X\Gamma$ (or even better $\Lambda_Y\Gamma\simeq \Lambda_X\Gamma$) then one has a nice geometric interpretation of the Poisson boundary. Basically, the proof of Osin's theorem goes as follows. Recall that a subgroup $H$ of $\Gamma$ is hyperbolically embedded in $\Gamma$ (with respect to a subset $S$ of $\Gamma$) if the group $\Gamma$ is generated by $S$ and $H$ and the Cayley graph $\mathrm{Cay}(\Gamma, S\cup H)$ is hyperbolic, the subgroup $H$, endowed with the induced metric $d_H$ is a proper metric space. This induced metric $d_H(h_1,h_2)$ is basically given by the smallest possible length of a path from $h_1$ to $h_2$ staying outside of $H$. Now, assume that $\Gamma$ acts on $X$ with a WPD element. First, if $h$ is WPD, then it is contained in a maximal virtually cyclic subgroup $E(h)$ and $E(h)$ hyperbolically embeds into $\Gamma$. Second, if $H$ is hyperbolically embedded into $\Gamma$ with respect to a set $S$, define a set $S'$ consisting of all elements $g$ such that a geodesic from 1 to $g$ does not stay longer than $D$ inside $H$, for some fixed constant $D$. Then, the Cayley graph $\mathrm{Cay}(\Gamma,S'\cup H)$ is hyperbolic and the action of $\Gamma$ on this Cayley graph is acylindrical. Let us look at an easy example now. Consider the free group $\Gamma=F_2$ and let $X$ be its Cayley graph with respect to the standard system of generators. Denote by $a$ and $b$ these two generators. Then, $a$ is WPD and one can take $E(a)=\langle a\rangle$. Clearly, $\langle a \rangle$ hyperbolically embeds into $\Gamma$ with respect to the set $S=\{b\}$. Now, Osin's construction yields $S'=\langle b\rangle$. Thus, the space $Y$ on which $\Gamma$ acylindrically acts is $\mathrm{Cay}(\Gamma,\langle a\rangle \cup \langle b\rangle)$. This Cayley graph $\mathrm{Cay}(\Gamma,\langle a\rangle \cup \langle b\rangle)$ is quasi-isometric to the coned-off graph of $\Gamma$, considered as hyperbolic relative to its free factors. Its Gromov boundary is the set of conical limit points and it embeds into the Gromov boundary of $\Gamma$. Precisely, the boundary of $\mathrm{Cay}(\Gamma,\langle a\rangle \cup \langle b\rangle)$ is the set of infinite words alternating elements of $\langle a \rangle$ and elements of $\langle b \rangle$. In particular, in this example, we indeed have $\Lambda_Y\Gamma\hookrightarrow\Lambda_X\Gamma$. Note however that one could take $Y$ to be the standard Cayley graph of $\Gamma$ and then one would get $\Lambda_Y\Gamma\simeq\Lambda_X\Gamma$. REPLY [2 votes]: I think the equivariant embedding you ask for is given in Theorem 3.2 of this paper: https://arxiv.org/pdf/1601.00101.pdf. Actually, there is such an embedding any time you cone off uniformly quasiconvex subspaces of a hyperbolic space. Added: Theorem 3.2 states that if $X$ and $Y$ are hyperbolic and $f \colon X \to Y$ is coarsely Lipschitz, coarsely surjective, and alignment preserving, then there is a subspace $\partial_Y X$ of $\partial X$ that is homeomorphic to $\partial Y$. The inverse of the homeomorphism is the embedding $\partial Y \to \partial X$ that you are looking for. One source of examples of such a map $f \colon X \to Y$ comes from work of Kapovich--Rafi. See Corollary 2.4 of this paper https://arxiv.org/pdf/1206.3626.pdf, where the alignment preserving property is exactly their `Moreover' statement. In fact, Corollary 2.4 is the result that Osin uses to show that his space (i.e. $\mathrm{Cay}(\Gamma,S'\cup H)$ in your notation) is hyperbolic. As for a precise statement about coning off quasiconvex subspaces, see Proposition 2.6 in the Kapovich--Rafi paper above. In that setting, the resulting map $X \to Y$ is alignment preserving and so gives an embedding $\partial Y \to \partial X$ by Theorem 3.2 of my paper with Spencer.<|endoftext|> TITLE: Descent properties of topological Hochschild homology QUESTION [10 upvotes]: Question: What is the finest topology on $\mathrm{CAlg}$ (commutative ring spectra) for which THH (Topological Hochschild Homology) satisfies descent? Adaptations of the arguments appearing in Section 3 of BMS2 show that THH has flat descent for simplicial commutative rings. However, the methods therein cannot be used to check whether THH has flat (or any weaker form of) descent for commutative ring spectra; simplicial $R$-algebras have the special property of being the nonabelian derived category of the category of finitely generated polynomial $R$-algebras. REPLY [5 votes]: In Theorem 1.2 of B. I. Dundas and J. Rognes: "Cubical and cosimplicial descent", Journal of the London Mathematical Society (2) 98 (2018) 439-460, DOI 10.1112/jlms.12141, we showed that for each $1$-connected map $\phi : A \to B$, of connective commutative $S$-algebras, the map from $THH(A)$ to the homotopy limit of the cosimplicial spectrum $$ [q] \mapsto THH(B \wedge_A \dots \wedge_A B) $$ (with $q+1$ copies of $B$) is an equivalence. Here $1$-connected means that $\pi_0(\phi)$ is an isomorphism and $\pi_1(\phi)$ is surjective. This should let you reduce the general descent question to the one for (simplcial) commutative rings.<|endoftext|> TITLE: Estimating $\sum_{n\leq x: n \in A} d(n)^a$ from below for large sets $A\subset \{1,2,\ldots,x\}.$ QUESTION [5 upvotes]: I apologise for the long-windedness of this question. Let $a$ be a positive real constant and let $d(n)$ denote the number of divisors of $n.$ Define $$ S_a(x)=\sum_{n\leq x} d(n)^a. $$ For $a=1,$ the following is well known $$ S_1(x)=\sum_{n\leq x} d(n)=x \log x + (2 \gamma -1) x +{\cal O}(\sqrt{x}) $$ while for more general $a$, one has $S_a(x) \sim C(a) x (\log x)^{2^a -1}$ where $$ C(a) = \Gamma(2^a)^{-1} \prod_p \left( 1 - \frac{1}{p} \right)^{2^a} \left( \sum_{k \geq 0} \frac{(k+1)^a}{p^k}\right). $$ More accurate estimates are available via the Selberg-Delange method, though the details get quite technical. My question is the following. Let the subset $A$ be defined by $A\subset \{1,2,\ldots,x\},$ (assume $x$ is integer or use the floor function) with $$\#A\geq \frac{x}{2}+ c \frac{\log x}{\log \log x}:=Z(x).$$ Now define $$ S^{A}_{a}(x)=\sum_{n\leq x:n \in A} d(n)^a. $$ I want to lower bound this sum, as $A$ varies subject to the size condition, i.e., to derive a lower bound to $$ M:=\min \left\{ S_a^A(x): A \subset \{1,\ldots,x\}, \#A =Z(x) \right\}. $$ For $a=1,$ I hope a lower bound of the form $M\gg x \log x$ may be possible if the Poisson approximation is good enough. Essentially this would say that half the points achieve a constant factor of the full sum, even when restricted to those points with the fewest number of divisors. From a Poisson approximation point of view, it would seem that approximately half the points $n\in \{1,\ldots,x\}$ have $\omega(n)\leq \log\log n,$ with the relevant Poisson distribution having mean $\log\log n.$ I am unsure if the known techniques are strong enough to address such a delicate size specification, compared to say $\lceil x/2 \rceil,$ but some comments related to Selberg-Delange have a discussion of how a multiplicative $O(1+(\log x)^{-c'})$ factor can be shown for various sums, which may imply that it is indeed possible. REPLY [4 votes]: Even if you take $Z(x)=(1-\varepsilon)x$ for some fixed $0<\varepsilon<1$, you are going to get $$ M=x(\ln x)^{a\ln 2+o(1)}. $$ To prove this, observe that both $\omega(n)$ and $\Omega(n)$ have normal order $\ln\ln n$ (here $\Omega$ and $\omega$ are numbers of prime factors with and without multiplicity respectively). Also note that $$ 2^{\omega(n)}\leq d(n)\leq 2^{\Omega(n)} $$ for every $n$, which implies that for any $\delta>0$ for all but $o(x)$ numbers $n\leq x$ we have $$ (\ln x)^{\ln 2-\delta}\leq d(n) \leq (\ln x)^{\ln 2+\delta}. $$ Therefore, for any $A\subset [1,x]\cap \mathbb N$ with $|A|\geq cx$ for some $c>0$ we have $$ S_a^A\geq (|A|-o(x))(\ln x)^{a(\ln 2-\delta)}, $$ so that $M\gg x(\ln x)^{a(\ln 2-\delta)}$. On the other hand, if you throw all the $n$ with $d(n)>(\ln x)^{\ln 2+\delta}$ out of your interval, you will still have $x-o(x)$ numbers left. This implies that for any $\varepsilon>0$ and $x$ large enough there is $A$ with $|A|\geq (1-\varepsilon)x$ such that $$ S_a^A\leq |A|(\ln x)^{a(\ln 2+\delta)}. $$<|endoftext|> TITLE: Applications of mathematics in clinical setting QUESTION [30 upvotes]: What are some examples of successful mathematical attempts in clinical setting, specifically at the patient-disease-drug level? To clarify, by patient-disease-drug level, I mean the mathematical work is approved to be used as part of a decision making process to prescribe a specific treatment for a specific patient? I do not mean general modeling attempts that study and simulate on a more or less theoretical level. Additionally, are these mathematical work have to be approved by the regulators like the FDA or a national board of physicians? REPLY [4 votes]: Amazingly, noone mentionned the use of the inverse Radon transform in the scanner / medical imaging.<|endoftext|> TITLE: Are locally presentable categories determined by their objects? QUESTION [14 upvotes]: Let $f: \mathcal{C} \rightarrow \mathcal{D}$ be a colimit-preserving functor between locally presentable categories. Assume that $f$ induces an equivalence between the groupoids underlying $\mathcal{C}$ and $\mathcal{D}$. Is $f$ necessarily an equivalence? What if $\mathcal{C}$ and $\mathcal{D}$ are assumed to be linear/dg/stable? REPLY [21 votes]: The answer in general is no. Let $\mathcal C$ be the category of sets, let $\mathcal D$ be the category of pointed sets (with basepoint-preserving maps), and let $f: \mathcal C \to \mathcal D$ be the functor which adds a disjoint basepoint. Then $f$ is an equivalence on underlying groupoids, but not an equivalence of categories. Moreover, the forgetful functor is a right adjoint to $f$. Of course, since any locally presentable 1-category is also presentable as an $\infty$-category, this answers the question in the $\infty$-categorical case, too. But for a more "intrinsically $\infty$-categorical" example, let $\mathcal C$ be the $\infty$-category of spaces, let $\mathcal D$ be the $\infty$-category of pointed spaces with disjoint basepoint (i.e. pointed spaces such that the connected component of the basepoint is contractible; morphisms are basepoint-preserving maps), and let $f: \mathcal C \to \mathcal D$ be the functor which adds a disjoint basepoint. Again, the forgetful functor is a right adjoint to $f$, and $f$ is an equivalence on object spaces. It's perhaps surprising that $\mathcal D$ is presentable, but it is: colimits are formed by taking the colimit in the category of pointed spaces and then collasping the connected component of the basepoint to be contractible; from this presentability is easy to verify. The answer in the stable case is yes. Assume that $\mathcal C$ and $\mathcal D$ are stable (or even just additive with suspensions), and that $f$ is exact (or even just preserves finite direct sums and suspensions). First, $f$ is faithful on the homotopy category. For if $f(\phi) = 0$, then $f\begin{pmatrix} 1 & \phi \\ 0 & 1 \end{pmatrix} = f\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Since $f$ is faithful on isomorphisms, we get $\begin{pmatrix} 1 & \phi \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, so that $\phi = 0$. Now, $f$ is full on the homotopy category. For given a map $\phi$, because $f$ is full on isomorphisms, we have $f\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & \phi \\ 0 & 1 \end{pmatrix}$ for some $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, so that $f(b) = \phi$. Thus $f$ is an equivalence on the homotopy category (since it is essentially surjective too). In this context, this implies that $f$ is an equivalence, as $\pi_n(Hom(X,Y)) = \pi_0(Hom(\Sigma^n X, Y))$. This argument (without the bit about suspensions) also works for any additive ordinary category (i.e. any additive category with discrete hom-spaces). Here's a special case where the answer is yes. Proposition: Let $f: L \to M$ be a sup-preserving map between complete lattices. Suppose that $f$ is a bijection. Then $f$ is an isomorphism. Proof: We want to show that for $x,y \in L$, we have $x \leq y \Leftrightarrow f(x) \leq f(y)$. The forward implication follows from $f$ being sup-preserving (and hence order-preserving). For the reverse implication, suppose that $f(x) \leq f(y)$. We have $f(\sup(x,y)) = \sup(f(x),f(y)) = f(y)$. Because $f$ is a bijection, we have $\sup(x,y) = y$, i.e. $x \leq y$. Two observations for the general case: Let $f_!: \mathcal C \to \mathcal D$ be a left adjoint between locally presentable categories which induces an equivalence $\iota f_!: \iota \mathcal C \to \iota \mathcal D$ of spaces of objects, as in the question statement. Claim: $f_!$ is conservative. Proof: We may factor $f_!$ as a localization $l_!: \mathcal C \to L\mathcal C$ followed by a conservative left adjoint $g_!: L\mathcal C \to \mathcal D$. Because $l_!$ is a localization, its right adjoint $l^\ast$ is fully faithful. But this means that on object spaces, $\iota f_!$ factors through a retract $\iota L \mathcal C$ of $\iota \mathcal C$. Since $\iota f_!$ is an equivalence, $\iota l_!$ must be an equivalence, and $\iota l^\ast$ must also be an equivalence. Because $l^\ast$ is fully faithful, this implies that $l^\ast$ must be an equivalence, so that $l_!$ is also an equivalence. Thus $f_!$ may be identified with the conservative functor $g_!$. Corollary: $f_!$ is faithful. Proof: This is a general fact: a conservative left adjoint between cocomplete categories is faithful, for if $f_!(\phi) = f_!(\psi)$ for some $\phi,\psi: C^\to_\to C'$, then the map $f_! C' \to coeq(f_!\phi,f_!\psi)$ is an isomorphism. Since $f_!$ is conservative and preserves colimits, the map $C' \to coeq(\phi,\psi)$ is an isomorphism, i.e. $\phi = \psi$. For the faithfulness statement, I'm not 100% sure how this translates into the $\infty$-categorical world.<|endoftext|> TITLE: Is $\sum_{k=0}^n (|\sin(k)|-2/\pi) $ bounded by a constant $M$? QUESTION [8 upvotes]: I know $\sum_{k=0}^{n} \sin(k)$ is bounded by a constant and $\sum_{k=0}^{n} \sin(k^2)$ is not bounded by a constant. Then, what about $\sum_{k=0}^{n} (|\sin(k)|-2/\pi)$? From numerical calculation, $\max_{n=0...10^8}(\sum_{k=0}^{n} (|\sin(k)|-2/\pi))=0.0900478$ which is much small compared to $\max_{n=0...10^8}\sum_{k=0}^{n} \sin(k^2)=1882.86$. So, I suppose $\sum_{k=0}^{n} (|\sin(k)|-2/\pi)$ can be bounded by a constant, but I don't know how to prove it. REPLY [2 votes]: As a partial answer, we can show that if $S_N=\sum_{k=0}^N (|\sin(k)|-2/\pi)$, then: $S_N=O(1)-\frac{4}{\pi}\sum_{k=2}^{N}\sum_{m=1}^{[k \log^2 k]}\frac{\cos 2mk}{4m^2-1}=O(\sum_{m=1}^{[N\log^2 N]}{\frac{\min(N\log^2N, ||\frac{m}{\pi}||^{-1})}{m^2}})$, where as usual $||x||$ represents the distance to the closest integer. In particular if we know that $||\frac{m}{\pi}||$ is not too "small" (e.g it is $>>m^{-1+\epsilon}$) for enough $m$ we may be able to show boundness or at least some better estimate than the uniform distribution $o(N)$ noted by @Wojowu above. Conversely, if there are infinitely many $m$ with $||\frac{m}{\pi}|| < C\frac{1}{m^2}$, the estimate doesn't work at least as boundness goes, though a cleverer estimate of the double trigonometric sum above could work. Proof: We use the Fourier series $|\sin k|=\frac{8}{\pi}\sum_{m=1}^{\infty}\frac{\sin^2 mk}{4m^2-1}$ which converges absolutely and we cut it at $[k\log^2 k]$ for $k \ge 2$ as the terms for $k=0,1$ in $S_N$ are absorbed in the $O(1)$ above, so we won't bother with them. Then the tail of the Fourier series for $|\sin k|$ is obviously bounded by $C\sum_{m > k\log^2 k}\frac{1}{m^2}=O(\frac{1}{k\log^2 k})$, hence summing on $k \ge 2$ we get the tail $O(1)$. For the rest, using $2\sin^2(mk)=1-\cos(2mk)$ and $2\frac{1}{4m^2-1}=\frac{1}{2m-1}-\frac{1}{2m+1}$, we get that $\frac{8}{\pi}\sum_{m=1}^{[k \log^2 k]}\frac{\sin^2 mk}{4m^2-1}=\frac{2}{\pi}+O(\frac{1}{k\log^2 k})-\frac{4}{\pi}\sum_{m=1}^{[k \log^2 k]}\frac{\cos 2mk}{4m^2-1}$ Putting the above together we get the first statement above, namely that: $S_N=O(1)-\frac{4}{\pi}\sum_{k=2}^{N}\sum_{m=1}^{[k \log^2 k]}\frac{\cos 2mk}{4m^2-1}$ Rearranging the terms with the same $1 \le m \le [N\log^2 N]$ and using the well known (and easy to prove) estimate: $\sum_{j=m}^{n}\cos(2\pi jx)=O(\min(n-m+1, ||x||^{-1}))$, we get our claimed estimate result: $\sum_{k=2}^{N}\sum_{m=1}^{[k \log^2 k]}\frac{\cos 2mk}{4m^2-1}=O(\sum_{m=1}^{[N\log^2 N]}{\frac{\min(N\log^2N, ||\frac{m}{\pi}||^{-1})}{m^2}})$ since for fixed $m$, the sum in $k$ has at most $N \log^2N$ terms and the corresponding $x$ is of course $\frac{m}{\pi}$<|endoftext|> TITLE: Tannaka-Krein duality in Chari-Pressley's book QUESTION [8 upvotes]: I am not sure that this was not discussed before, so excuse me in this case. This can be considered as a special case of my previous question here. V.Chari and A.N.Pressley in their "Guide to Quantum Groups" formulate the following version of Tannaka-Krein's duality (Theorem 5.1.11): Let $R$ be a commutative ring, let $C$ be an essentially small, $R$-linear, rigid, abelian, monoidal category, and let $\varPhi:C\to\operatorname{mod}_R$ be a $R$-linear exact faithful monoidal functor. Then there exists a Hopf algebra $A$ over $R$ and an equivalence of $R$-linear categories $C\to\operatorname{corep}_A$ whose composite with the forgetful functor $\operatorname{corep}_A\to\operatorname{mod}_R$ is $\varPhi$. They give an outline of the proof, but I must say that I don't understand some of its details. I think that there must exist an accurate text with complete proof. Can anybody give me a reference to it? Also if somebody could cast some light on possible generalizations of this proposition (to the case of Hopf algebras in braided monoidal categories), I would appreciate this very much. REPLY [2 votes]: In the case where $k$ is a field this should be worked out in full generality (for Hopf algebras) in Chapter 5 of 'Tensor Categories' by Etingof, Gelaki, Nikshych and Ostrik. This includes the infinite dimensional case. For the finite dimensional case this can be generalized even further to weak Hopf algebras, so that every Fusion category is monoidally equivalent to the category of representations for a weak Hopf algebra. This is stated in 'On Fusion Categories.' (Corollary 2.22) where it is there referenced as due to Hayashi in 'A canonical Tannaka duality for finite seimisimple tensor categories ' and Szlach´anyi in 'Finite quantum groupoids and inclusions of finite type.' A proof should also be provided in Ostrik's 'Module categories, weak Hopf algebras and modular invariants.' An additional reference is section 7.23 of 1. As a note, this is referenced in the nlab article, under the table 'Tannaka duality for categories of modules over monoids/associative algebras' for 'Hopf algebras.' Following the Hopf algebras link takes you to the Hopf algebras article which includes a section on 'Tannakian Duality' and includes a link to Bakke's 'Hopf algebras and monoidal categories' and Vercruysse's 'Hopf algebras---Variant notions and reconstruction theorems'. For weak hopf algebras following the links leads to 3 and 5. Finally, a gold standard reference for understanding Tannakian duality is Pierre Delign's 'Catégories Tannakiennes' and its related generalizations, linked in the nLab article on said paper.<|endoftext|> TITLE: Extension of a bilinear form to the exterior algebra QUESTION [5 upvotes]: In Serre's Local Fields, at the beginning of the chapter III section 2, he has wrote "it is known that $T$ extends to a non-degenerate bilinear form on the exterior algebra of $V$", where $T$ is a non-degenerated bilinear form over a vector space $V$. I get confused about this well-know extension. Is there any explicit formula for this extension? Thank you. REPLY [8 votes]: Let $k$ be a nonnegative integer. Let $K$ be a commutative ring, and let $V$ and $W$ be two $K$-modules. Let $\alpha : V \times W \to K$ be a $K$-bilinear form. Then, there is a $K$-bilinear form \begin{align} \alpha_k : \wedge^k V \times \wedge^k W &\to K; \\ \left(v_1 \wedge v_2 \wedge \cdots \wedge v_k , w_1 \wedge w_2 \wedge \cdots \wedge w_k\right) &\mapsto \det\left(\left( \alpha\left(v_i, w_j\right) \right)_{1\leq i\leq k, \ 1\leq j\leq k}\right) \end{align} (where $\left( \alpha\left(v_i, w_j\right) \right)_{1\leq i\leq k, \ 1\leq j\leq k}$ is the $k\times k$-matrix with entries $\alpha\left(v_i, w_j\right)$). Note that the image of $\left(v_1 \wedge v_2 \wedge \cdots \wedge v_k , w_1 \wedge w_2 \wedge \cdots \wedge w_k\right)$ under this form is often called a Gram determinant (though the name traditionally stands for some particular cases), and can be rewritten as $\sum_{\sigma \in S_k} \left(-1\right)^{\sigma} \prod_{i=1}^k \alpha\left(v_i, w_{\sigma\left(i\right)}\right)$ (where $S_k$ is the symmetric group on the set $\left\{1,2,\ldots,k\right\}$, and where $\left(-1\right)^{\sigma}$ denotes the sign of a permutation $\sigma$). Why is this form $\alpha_k$ well-defined? Here is the straightforward way to see this (there might be slicker arguments): For each $v_1, v_2, \ldots, v_k \in V$, we define a $K$-linear map \begin{align} A_{v_1, v_2, \ldots, v_k} : \wedge^k W &\to K; \\ w_1 \wedge w_2 \wedge \cdots \wedge w_k &\mapsto \det\left(\left( \alpha\left(v_i, w_j\right) \right)_{1\leq i\leq k, \ 1\leq j\leq k}\right) . \end{align} This is well-defined (by the universal property of $\wedge^k W$), since the determinant $\det\left(\left( \alpha\left(v_i, w_j\right) \right)_{1\leq i\leq k, \ 1\leq j\leq k}\right)$ depends $K$-multilinearly on the $w_1, w_2, \ldots, w_k$ and vanishes when two of $w_1, w_2, \ldots, w_k$ are equal (indeed, this follows from standard properties of determinants). Now, consider the map \begin{align} A : V^{\times k} &\to \operatorname{Hom}_K\left(\wedge^k W, K\right); \\ \left(v_1, v_2, \ldots, v_k\right) &\mapsto A_{v_1, v_2, \ldots, v_k} . \end{align} This map $A$ is $K$-multilinear and kills all $k$-tuples $\left(v_1, v_2, \ldots, v_k\right) \in V^{\times k}$ that have two equal entries (indeed, this all boils down to checking identities between $K$-linear maps, such as \begin{align} A_{v_1, v_2, \ldots, v_{i-1}, v_i + v'_i, v_{i+1}, \ldots, v_k} &= A_{v_1, v_2, \ldots, v_{i-1}, v_i, v_{i+1}, \ldots, v_k} + A_{v_1, v_2, \ldots, v_{i-1}, v'_i, v_{i+1}, \ldots, v_k} ; \\ A_{v_1, v_2, \ldots, v_{i-1}, \lambda v_i, v_{i+1}, \ldots, v_k} &= \lambda A_{v_1, v_2, \ldots, v_{i-1}, v_i, v_{i+1}, \ldots, v_k} ; \\ A_{v_1, v_2, \ldots, v_{i-1}, v, v, v_{i+2}, \ldots, v_k} &= 0 ; \end{align} but any identity between $K$-linear maps can be proven by evaluating both sides at the generators $w_1 \wedge w_2 \wedge \cdots \wedge w_k$ of $\wedge^k W$; but when evaluated this way, all these identities again follow from basic properties of the determinant). Thus, this map $A$ gives rise to a $K$-linear map \begin{align} A' : \wedge^k V &\to \operatorname{Hom}_K\left(\wedge^k W, K\right); \\ v_1 \wedge v_2 \wedge \cdots \wedge v_k &\mapsto A_{v_1, v_2, \ldots, v_k} \end{align} (by the universal property of $\wedge^k V$). This map $A'$, in turn, induces a $K$-bilinear form \begin{align} \alpha_k : \wedge^k V \times \wedge^k W &\to K; \\ \left(p, q\right) &\mapsto \left(A'\left(p\right)\right)\left(q\right) \end{align} (by uncurrying). An immediate verification reveals that this $K$-bilinear form $\alpha_k$ is precisely the form $\alpha_k$ presented above. Of course, by considering $k$ as variable, we can glue these $K$-bilinear forms $\alpha_k : \wedge^k V \times \wedge^k W \to K$ together into a $K$-bilinear form $\alpha_\wedge : \wedge V \times \wedge W \to K$. This latter form, I believe, is the form Serre wants. Why is this latter form $\alpha_\wedge$ non-degenerate when $\alpha$ is non-degenerate? Here, we say that a $K$-bilinear form $\beta : P \times Q \to K$ is non-degenerate if there exist bases $\left(p_i\right)_{i \in I}$ and $\left(q_i\right)_{i \in I}$ of $P$ and $Q$, respectively, such that $\beta\left(p_i, q_j\right) = \delta_{i, j}$ for all $i \in I$ and $j \in I$. Two such bases are called dual bases for the form $\beta$. Now, assume that $\alpha$ is non-degenerate. Thus, there exist dual bases $\left(v_i\right)_{i \in I}$ and $\left(w_i\right)_{i \in I}$ for the form $\alpha$. Consider such bases. WLOG assume that the set $I$ is totally ordered (since we can always equip $I$ with a total order). The basis $\left(v_i\right)_{i \in I}$ of $V$ induces a basis $\left(v_{i_1} \wedge v_{i_2} \wedge \cdots \wedge v_{i_k}\right)_{\left(i_1, i_2, \ldots, i_k\right) \in I_k}$ of $\wedge^k V$, where $I_k$ is the set of all strictly increasing $k$-tuples $\left(i_1 < i_2 < \cdots < i_k\right) \in I^k$. Similarly, the basis $\left(w_i\right)_{i \in I}$ of $W$ induces a basis $\left(w_{i_1} \wedge w_{i_2} \wedge \cdots \wedge w_{i_k}\right)_{\left(i_1, i_2, \ldots, i_k\right) \in I_k}$ of $\wedge^k W$. It is now easy to see that these two bases of $\wedge^k V$ and $\wedge^k W$ are dual bases for the $K$-bilinear form $\alpha_k$. Thus, the $K$-bilinear form $\alpha_k$ has dual bases, i.e., is non-degenerate. Qed.<|endoftext|> TITLE: Integrability of log of distance function QUESTION [6 upvotes]: Let $E\subset B_1(0)\subset \mathbb{R}^n$ be a compact set s.t. $\lambda(E)=0$, where $\lambda$ is the Lebesgue measure, and $B_1(0)$ is the Euclidean unit ball centered at the origin. Is the following integral finite: $$\int_{B_1(0)}-\log d(x,E)d\lambda(x)<\infty?$$ Although this question seems trivial, I have failed to find a reference to it or to variations of it in previous discussions. I was not able to come up with a counter-example nor a proof. I also asked in mathstackexchange a variation of it, but didn’t get a sufficient answer. Thanks ahead REPLY [7 votes]: The integral in question is finite for most sets of measure zero, but can diverge to $\infty$ for some sets. An example in one dimension is obtained by constructing a Cantor set where at stage $k$ the middle $1/(k+1)$ proportion is removed from each of the $2^{k-1}$ intervals obtained at stage $k-1$. Thus the $2^k$ intervals obtained at stage $k$ will each have length $2^{-k}/(k+1)$. Therefore, each of the $2^k$ middle intervals removed in the next stage will have length $2^{-k}/[(k+1)(k+2)]$, and each of these will contribute at least $k/2$ times its length to the integral. Summing over $k$ gives a harmonic series which diverges. The example can be lifted to higher dimensions by taking a Cartesian product with a $n-1$ dimensional box.<|endoftext|> TITLE: The number of quadratic forms attaining Hermite's constant QUESTION [6 upvotes]: $\require{AMScd}$ I'm considering minimum values (at non-zero integer points) of real, positive-definite, quadratic forms of determinant $1$. These are functions $f:\mathbb{R}^n\to \mathbb{R}$ which can be represented as $x\mapsto x^\intercal Ax$ for some symmetric, positive-definite matrix $A$ with determinant $1$. Call this function space $\mathcal{Q}_n$. Since I'm interested in extreme values (at integer points), I'm going to identify two such functions $f \sim f'$ if there is some invertible integer matrix $\Gamma$ such that $f'(\cdot)=f(\Gamma \cdot)$. This gives me a set $\mathcal{Q}_n/\sim$. Now (using Hermite or Minkowski) one can easily demonstrate uniform (over $f$) bounds such as $$\min\limits_{v\in \mathbb{Z}^n\smallsetminus 0}f(v) \leq \left(\frac{4}{3}\right)^{(n-1)/2}, $$ so that $$S_n:=\sup\limits_{f} \left\lbrace\min\limits_{v\in \mathbb{Z}^n\smallsetminus0}f(v)\right\rbrace $$ is finite. Moreover, one can show (by a compactness argument due to Mahler) that in each dimension $n$, this bound is achieved by some particular quadratic form. Question: In each dimension $n$, is there a finite bound on the equivalence classes $[f] \in \mathcal{Q}_n/\sim$ that can attain this maximum? That is, can we have an infinite number of inequivalent $f$ for which $$\inf\limits_{v\in \mathbb{Z}^n\smallsetminus0}f(v) = S_n? $$ For example, in dimension $2$ (resp. $3$) Hermite (resp. Gauss) showed that this max is only attained by (an appropriate multiple of) $[x^2+xy+y^2]$ (resp. $[x^2+y^2+z^2+xy+xz+yz]$). Thanks for reading! Sorry if this question is not appropriate. REPLY [8 votes]: There are only finitely many inequivalent forms which may be local maxima for the Hermite invariant. Voronoi showed (1908) that the lattices attaining a local maxima are extreme, i.e. perfect and eutactic. Voronoi also showed there are only finitely many perfect lattices in each dimension, and these must all be integral by an older result of Korkine and Zolotarev (1877). There does not appear to be any known explicit bounds on the number of inequivalent extreme lattices holding for all dimensions. The forms which attain Hermite's constant are unique for $d\leq 8$ and these are all root lattices (see Conway and Sloane, Chapter 6 for more details). For more details on this, Chapter 3 of Martinet's book Perfect Lattices in Euclidean Spaces may be useful. All of the perfect lattices/forms have been classified up to dimension eight and the number of these lattices in dimensions $d=1,2,3,4,5,6,7,8$ are $1, 1, 1, 2, 3, 7, 33, 10916$ respectively; the last number is known thanks to the work of Sikirić, Schürmann and Vallentin. The approach here is essentially algorthmic, (using Voronoi's algorithm) and in principle, given enough computational resources it is possible to enumerate all such forms in a given dimension. There is an ongoing attempt to classify all $9$-dimensional perfect lattices where the number of such found is already in the millions.<|endoftext|> TITLE: A q-rious identity QUESTION [22 upvotes]: Let $[x]_q=\frac{1-q^x}{1-q}$, $[n]_q!=[1]_q[2]_q\cdots[n]_q$ and ${\binom{x}{n}}_{q}=\frac{[x]_q[x-1]_q\cdots[x-n+1]_q }{[n]_q!}$. Computer experiments suggest that $$\det \left(q^\binom{i-j}{2}\left(\binom{i+r}{j}_{q}x+\binom{i+r-j}{j}_{q}\right)\right)_{i,j = 0}^{n - 1} = (1+x)^n$$ Any idea how to prove this? REPLY [4 votes]: Let $A_n$ be the matrix involved in the problem and let $L_n=\left((-1)^{i-j}\binom{i}{j}_q\right)_{i,j=0}^{n-1}$. Observe that $L_n$ is lower-triangular with 1's in the diagonal. Multiplying, we have: $$L_nA_n=\left(x\,u_{i,j}(r)+u_{i,j}(r-j)\right)_{i,j=0}^{n-1}$$ where $$u_{i,j}(r)=\sum_{k=0}^i(-1)^{i-k}q^{\binom{k-j}{2}}\binom{i}{k}_q\,\binom{k+r}{j}_q\,.$$ Now \begin{eqnarray*} u_{i,j}(r)&=&\sum_{k=0}^i(-1)^{i-k}q^{\binom{k-j}{2}}\binom{i}{i-k}_q\,\binom{k+r}{k+r-j}_q \\ &=&(-1)^{r+i-j}\,q^{\binom{r+1}{2}}\,\sum_{k=0}^i q^{k(k+r-j)}\binom{i}{i-k}_q\,\binom{-j-1}{k+r-j}_q \\ &=&(-1)^{r+i-j}\,q^{\binom{r+1}{2}}\,\sum_{k=0}^i q^{k(k+r-j)}\binom{i}{i-k}_q\,\binom{-j-1}{k+r-j}_q \\ &=&(-1)^{r+i-j}\,q^{\binom{r+1}{2}}\,\binom{i-j-1}{i-j+r}_q =q^{\binom{i-j}{2}}\,\binom{r}{j-i}_q\,. \end{eqnarray*} Hence $u_{i,j}(r)=0$ when $i>j$ and $u_{i,i}(r)=1$. Thus $L_nA_n$ is upper triangular with $x+1$ as elements of the diagonal.<|endoftext|> TITLE: When does $axy+byz+czx$ represent all integers? QUESTION [27 upvotes]: For which $a,b,c$ does $axy+byz+czx$ represent all integers? In a recent answer, I conjectured that this holds whenever $\gcd(a,b,c)=1$, and I hope someone will know. I also conjectured that $axy+byz+czx+dx+ey+fz$ represents all integers when $\gcd(a,b,c,d,e,f)=1$ and each variable appears non-trivially, though I'm less optomistic about finding prior results on that. Here are some results: If $\gcd(a,b)=1$ then $axy+byz+czx$ represents all integers. [Proof: Find $r,s$ with $ar+bs=1$, then take $x = r$, $y = n - crs$, $z = s$.] $6xy+10yz+15zx$, the first case not covered above, represents all integers up to 1000. Similarly $77xy+91yz+143zx$ represents all integers up to 100. [by exhaustive search] If $\gcd(a,b,c)=1$ then $axy+byz+czx$ represents all integers mod $p^r$. [proved in the above link] The literature on this is hard to search because these are not positive definite forms, and many apparently relevant papers only consider the positive definite case. For old results, the most relevant parts of Dickson's History of the Theory of Numbers (v. 2, p. 434; v. 3, p. 224) mention only the case of $xy+xz+yz=N$. Does anyone here know a general result or reference? REPLY [4 votes]: I have figured out some things; it is much quicker, as far as computing, to find a way for the Hessian matrix of the ternary quadratic form, is to have it represent the (two by two) Hessian of the form $xy;$ this form, or its quadratic space, is often called The Hyperbolic Plane; see page 15 in Cassels. Once this is done, there is just the business of adding an appropriate third row to "rows" to get a nice result. The final quadratic form is (y - 1250*z)*x + (-797*z*y - 5751*z^2) $$ xy -797yz - 1250 zx - 5751 z^2, $$ which is universal because we can take $z = 0, y = 1,$ and $x$ equal to the target number. Oh, the beginning form was your $$ 77yz + 91 zx + 143xy $$ $$ \left( \begin{array}{ccc} 830 &-3486 &-2145 \\ 616& -2587 & -1592 \\ -3& -5& 12 \\ \end{array} \right) \left( \begin{array}{ccc} 0 &143 &91 \\ 143& 0&77 \\ 91&77 &0 \\ \end{array} \right) \left( \begin{array}{ccc} 830 &616 &-3 \\ -3486& -2587 &-5 \\ -2145&-1592 &12 \\ \end{array} \right) = \left( \begin{array}{ccc} 0&1 &-1250 \\ 1&0 & -797 \\ -1250&-797 &-11502 \\ \end{array} \right) $$ Note: it turns out to be quite easy, with explicit matrices, to take the form with the visible hyperbolic plane to the form $xy - (abc) z^2,$ that GH has already proved equivalent to the original $ayz+bzx+cxy.$ $$ \left( \begin{array}{ccc} 830 &-3486 &-2145 \\ 616& -2587 & -1592 \\ 1431507& -6012097& -3699553 \\ \end{array} \right) \left( \begin{array}{ccc} 0 &143 &91 \\ 143& 0&77 \\ 91&77 &0 \\ \end{array} \right) \left( \begin{array}{ccc} 830 &616 & 1431507 \\ -3486& -2587 &-6012097 \\ -2145&-1592 & -3699553 \\ \end{array} \right) = \left( \begin{array}{ccc} 0&1 &0 \\ 1&0 & 0 \\ 0& 0 &-2004002 \\ \end{array} \right) $$ ======================================================== parisize = 4000000, primelimit = 500000 ? h = [ 0,143,91; 143,0,77; 91,77,0] %1 = [ 0 143 91] [143 0 77] [ 91 77 0] ? rows = [ 830, -3486, -2145; 616, -2587, -1592 ] %2 = [830 -3486 -2145] [616 -2587 -1592] ? columns = mattranspose(rows) %3 = [ 830 616] [-3486 -2587] [-2145 -1592] ? rows * h * columns %4 = [0 1] [1 0] ? ? ? ? rows = [ 830, -3486, -2145; 616, -2587, -1592; -3,-5,12 ] %5 = [830 -3486 -2145] [616 -2587 -1592] [ -3 -5 12] ? matdet(rows) %6 = 1 ? columns = mattranspose(rows) %7 = [ 830 616 -3] [-3486 -2587 -5] [-2145 -1592 12] ? rows * h * columns %8 = [ 0 1 -1250] [ 1 0 -797] [-1250 -797 -11502] ? x %9 = x ? y %10 = y ? z %11 = z ? g = rows * h * columns %12 = [ 0 1 -1250] [ 1 0 -797] [-1250 -797 -11502] ? vec = [ x,y,z] %13 = [x, y, z] ? vect = mattranspose(vec) %14 = [x, y, z]~ ? vec * g * vect / 2 %15 = (y - 1250*z)*x + (-797*z*y - 5751*z^2) ? ======================================================<|endoftext|> TITLE: Kähler manifold with even-only singular cohomology QUESTION [5 upvotes]: Given a simply connected smooth projective variety (hence a compact Kähler manifold) with singular cohomology generated in even degrees, do we know that there is a Morse function on it such that all its Morse indices are even? REPLY [9 votes]: Assuming that by `cohomology' you mean integer coefficients, there is a general result saying what you want for simply connected manifolds of dimension $5$ or more, without the Kähler condition. According to Smale (Generalized Poincare’s conjecture in dimensions greater than four, Ann. Math. 74, No, 2, 391-406 (1961)) for $n > 5$ and Barden (Simply connected five-manifolds, Ann, Math, 82,365-385 (1965)) if $n = 5$ a simply connected $n$-manifold has a handle decomposition (ie Morse function) with the fewest handles that you need for a chain complex representing the homology. In your example, having only even dimensional cohomology means the same for homology, and hence there's no torsion. Thus there is a chain complex with only even index generators and so, for complex dimension at least $3$, by Smale a Morse function with the same property. The case of complex surfaces is not yet established, although there are many concrete examples where there are decompositions with only even index handles. (For example, hypersurfaces in ${\mathbb C}P^3$, elliptic surfaces without multiple fibers...). See the book of Gompf-Stipsicz for some of this. A more recent striking example, due to Akbulut, is the Dolgachev surface, equal to a two-fold log transform of orders $2$ and $3$ on an elliptic surface $E(1)$. This had been previously suggested as a counterexample to your statement.<|endoftext|> TITLE: Convexity of the expectation of boolean functions QUESTION [5 upvotes]: Let $$f:\{-1,1\}^n \to \{-1,1\}$$ be a monotone, odd ($f(-x)=-f(x)$) Boolean function. Let $$F:[0,1]\to[0,1]$$ denote the probability that $f(x_1,...,x_n)$ where $x_1,...,x_n$ are i.i.d. $\pm1$ R.V. with probability $p$ to be 1. Our question is the following: Is it true that $F$ is convex in $[0,1/2]$ for any such monotone odd $f$? (It is easy to see that $F$ is symmetric around $1/2$.) Another way to view this is that $dF/dp$ is increasing in $[0,1/2]$, meaning the bits become more influential as $p$ is closer to $1/2$. So far, in the example we worked out (such as majority) this turned out to be the case, and we could not find any references, nor managed to prove it or find a counterexample. A related fact - when I encountered "graphs" of the probability that certain monotone graph properties are satisfied in $G(n,p)$ as a function of $p$, they were always "drawn" as convex till the threshold and concave afterwards, but I could not find any explicit references to this phenomena. REPLY [3 votes]: By Russo's formula, $F'(p)$ coincides with the expected number of pivotal elements at $p$ (where a an element $x_i$ is pivotal for a configuration in $\{-1,1\}^n$ if changing the sign of $x_i$ also will change the sign of $f$). So if $F'$ is increasing on $[0,1/2]$, then we expect to see most pivotal elements at $p = 1/2$. I don't think that this is true in general. Instead of 'majority', think of 'weighted majority' where $x_1$ has disproportionally large weight. Then intuitively $x_1$ will be pivotal for most values of $p$, but any other $x_i$ is more likely to be pivotal for $p$ close to $0$ or close to $1$, so we'd expect fewer pivotal elements at $1/2$ than closer to $0$ or $1$. If I'm not mistaken, the following example exhibits this behaviour. Let $$f = \mathrm{sign}(5 x_1 + \sum _{i=2}^7 x_i)$$ Then $f$ is $1$ exactly when either $x_1 = 1$ and at least one of the other $x_i$ is $1$, or $x_1 = -1$ and all other $x_i$ are equal to $1$. Hence $$F(p) = p(1-(1-p)^6)+ (1-p)p^6$$ whose second derivative assumes negative values.<|endoftext|> TITLE: Linear independence of element-wise powers of positive vectors QUESTION [7 upvotes]: Consider a vector $x$ with $0 < x_1 < \cdots < x_n < \infty$, and let $0 < \gamma_1 < \cdots < \gamma_n < \infty$. I would like to show that $x^{\gamma_1}, \ldots, x^{\gamma_n}$ are linearly independent, where $x^{\gamma_i}$ is defined as the vector $(x_1^{\gamma_i}, \ldots, x_n^{\gamma_i})$. It is clear that $x^{\gamma_1}$ and $x^{\gamma_2}$ are linearly independent, and I might have an overly complicated argument for the case $x^{\gamma_1}$, $x^{\gamma_2}$, and $x^{\gamma_3}$, but I'm really at a loss about how to tackle the general case. REPLY [10 votes]: The following proposition shows that $x^{\gamma_1}, \dots x^{\gamma_n}$ are indeed always linearly independent if $x \in \mathbb{R}^n$ has $n$ mutually distinct strictly positive entries. Proposition. For all real numbers $\gamma_1 < \dots < \gamma_n$ (be they positive or not) and each tuple $0 \not= (\alpha_1, \dots, \alpha_n) \in \mathbb{R}^n$ the function $$ f_n: (0,\infty) \ni t \mapsto \alpha_1 t^{\gamma_1} + \dots + \alpha_n t^{\gamma_n} \in \mathbb{R} $$ has at most $n-1$ distinct zeros (by "distinct" I mean that we do not count multiplicities of zeros). Proof. We show the proposition by induction. The assertion is obvious for $n = 1$, so assume that the claim has been proved for some $n \in \mathbb{N}$ and consider $n+1$ terms now. Assume that $f_{n+1}$ has at least $n+1$ distinct zeros in $(0,\infty)$. Then all the coefficients $\alpha_1, \dots, \alpha_{n+1}$ are non-zero (otherwise $f_{n+1}$ would be a non-trivial linear combination of $n$ terms with at least $n+1$ zeros). The function $t^{-\gamma_1}f_{n+1}(t)$ also has at least $n+1$ distinct zeros, and hence, its derivative $(t^{-\gamma_1}f_{n+1}(t))'$ has at least $n$ distinct zeros (as a consequence of Rolle's theorem). This is a contradiction since $(t^{-\gamma_1}f_{n+1}(t))'$ is a non-trivial linear combination of the $n$ terms $t^{\gamma_2-\gamma_1-1}, \dots, t^{\gamma_{n+1}-\gamma_1 - 1}$. Remark. It is important in the above proof that we show the result for all real numbers $\gamma_k$ (and not only for positive ones), since even if all $\gamma_k$ are positive we need the induction hypotheses for negative exponents, too. So this is a nice example where strengthening the hypothesis is needed to make induction work..<|endoftext|> TITLE: What properties of a finite group fibre functor give its endomorphisms a hopf algebra structure? QUESTION [5 upvotes]: Tannaka duality for a finite group lets us recover the group algebra $\mathbb{C}[G]$ as the endomorphisms of the forgetful functor $F:RepG\rightarrow Vect$, and taking the monoidal automorphisms recovers the grouplike elements of this hopf algebra, which we can recognise as just our group $G$. Is there a diagrammatic way of getting the comultiplication/antipode structure of this hopf algebra? By diagrammatic, I mean a description that shouldn't rely on the objects of our categories, and is expressible in the higher level structure, such as the multiplication being composition of endomorphisms. I would think such a description would apply for arbitrary functors between categories that share enough of the nice properties of $RepG$ and $Vec$, but as I am interested in $RepG$, an ad-hoc method for this fibre functor would also be of interest. REPLY [4 votes]: The tensor product functor $\otimes : \text{Rep}(G) \times \text{Rep}(G) \to \text{Rep}(G)$ is "bilinear": it preserves colimits in both variables. It can therefore be regarded as being a "linear" functor on a "tensor product" category $$\otimes : \text{Rep}(G) \otimes \text{Rep}(G) \to \text{Rep}(G)$$ which can be identified with $\text{Rep}(G \times G)$. In general we have $\text{Mod}(R) \otimes \text{Mod}(S) \cong \text{Mod}(R \otimes_k S)$ for $k$-algebras $R$ and $S$, and we have $k[G] \otimes_k k[G] \cong k[G \times G]$. All of this can be understood rigorously using either the Morita 2-category of bimodules or an enriched version of the universal property of presheaves, but if you can take it on faith that this sort of thing makes sense at least when the representation category is semisimple we don't have to go into it. Composing $\otimes$ with the fiber functor $F : \text{Rep}(G) \to \text{Vect}$ then gives a composite functor $$F \circ \otimes : \text{Rep}(G \times G) \to \text{Vect}.$$ The endomorphisms of this functor give $k[G \times G]$, and whiskering along $\otimes$ should define a map $k[G] \to k[G] \otimes_k k[G]$ reproducing the comultiplication (although I haven't checked this). I think we can also construct the antipode but the idea I have in mind involves some funny business with the dualization functor.<|endoftext|> TITLE: Spectrum of a Subspace of Matrices QUESTION [5 upvotes]: I conjecture the following: Let $U \subset \Bbb C^{n \times n}$ be an affine subspace, and let $S_U$ denote the "spectrum of $U$", that is $$ S_U = \{\lambda \in \Bbb C : \det(A - \lambda I) = 0 \text{ for some } A \in U\}. $$ Then either all elements of $U$ have an identical spectrum, or $S_U = \Bbb C$. Is this correct? Some simple examples of each case: for any fixed $\lambda_i$, $$ U_1 = \left\{\pmatrix{\lambda_1&t\\0&\lambda_2}: t \in \Bbb C\right\}, \quad U_2 = \left\{\pmatrix{\lambda_1&0\\0&t}: t \in \Bbb C\right\}. $$ Clearly, $S_{U_1} = \{\lambda_1,\lambda_2\}$ and $S_{U_2} = \Bbb C$. Are there any other possibilities? I suspect that there is a quick algebraic-geometry approach here that I am missing. An aside: I would also be interested in the case of real affine subspaces of $\Bbb C^{n \times n}$, if anyone has insights on what the possibilities are there. Note that the symmetric real matrices are an example of a subspace where we attain $S_U^{(\Bbb R)} = \Bbb R \subsetneq \Bbb C$. We can clearly attain any "line" in $\Bbb C$, but I wonder if there are other possibilities. REPLY [3 votes]: To expand on Christian Remling's answer a bit: in his example, setting $det(A(t) - \lambda) = 0$ becomes $(1-\lambda)t+\lambda^3=0$, and solving for $t$ gives $t = x^3/(x-1)$ -- so for any $\lambda \in \mathbb{C}$, we have $x \in S$ by choosing this $t$, with the exception of $\lambda=1$. In general for an affine subspace of dimension $d$, we can define the space by $A(t_1,t_2,\dots t_d)$, and all entries in $A$ are linear in the $t_i$. Then $\det(A(t) - \lambda)$ is an $n$th degree polynomial in all the $t_i$ and $\lambda$. It is possible that the determinant has no dependence on any $t_i$, in which case we have a finite spectrum. Otherwise, we have a dependence on the $t_i$. Then $\det(A(t)-\lambda)=0$ has a solution in $t_1$ whenever at least one of the non-constant coefficients is nonzero. (In the above example, as long as the linear coefficient $1-\lambda\neq 0$.) Since the leading coefficient (that is not a constant 0 polynomial) is an $n-1$th degree polynomial in $\lambda$, it is zero at a most $n-1$ different places, and this is at most $n-1$ places that the spectrum can fail to be continuous. The spectrum only fails to be continuous when all of the coefficients in the $t_1$ polynomial are zero, so this is an upper bound. So, theorem: For an affine subspace of $n\times n$ matrices, the spectrum is either finite (of size at most $n$), or it is $\mathbb{C}\setminus K$, where $K$ is a finite set of exceptions (of size at most $n-1$).<|endoftext|> TITLE: Combinatorial species and differential categories QUESTION [11 upvotes]: It is mentioned in the introduction to [1] that (Cartesian) differential categories might be the unifying framework for differentiation in various branches of mathematics including combinatorics. It is also mentioned in [2] and other papers on tangent categories that there are tangent categories of combinatorial species. I don't see how the obvious definition of the category of species can be made into a tangent category, but maybe this is true for some other category of species. Questions: What is the relationship between combinatorial species and differential/tangent categories? Is there a differential or tangent category of species? Is the operation of differentiation of species related to these structures? [1] Blute, R.; Cockett, J. R. B.; Seely, R. A. G., Cartesian differential categories, Theory Appl. Categ. 22, 622-672 (2009). [2] Cockett, J. R. B.; Cruttwell, G. S. H., Connections in tangent categories, Theory Appl. Categ. 32, 835-888 (2017). ZBL1374.18016. REPLY [5 votes]: I contacted Geoffrey Cruttwell with regards to this question. Here is his reply: There hasn’t been any published paper on a tangent category of combinatorial species. However, the ideas can be found in a talk by my co-author, Robin Cockett here: Can you Differentiate a Polynomial?. The idea there is to show that polynomial functors form a Cartesian differential category (and thus a tangent category, since any Cartesian differential category is a tangent category). A particular example of this differentiation is then differentiation of combinatorial species (eg., as in wikipedia). We haven’t yet gone any further with these ideas, but there may be interesting developments if one attempts to apply some of the tangent category theory to this example (though it may be a bit tricky, as this example is perhaps more naturally a differential/tangent bicategory). So, there is no differential category of species. Instead, there is a (putative) differential bicategory in which 1-morphisms (between particular objects) are species and the operation of differentiation on them is the usual differentiation of species. To see this, species should be represented as analytic functors. Then the construction of this bicategory should be similar to the bicategory of polynomial functors (which are a special case of analytic functors) which is described in the slides linked above.<|endoftext|> TITLE: Contributions of Mary Cartwright to the theory of entire functions QUESTION [9 upvotes]: I have seen on the Wikipedia page for the mathematician Mary Cartwright that she achieved many new results in the field of entire functions and the zeroes of entire functions and that many of these were included in her 1956 book on the subject. I do not have access to this book, and was wondering if anyone could tell me the specific main results which Cartwright contributed in this area (I already know of Cartwright's theorem from the Wikipedia page). Edit: Changed 'integral' to 'entire' REPLY [13 votes]: Mary Cartwright proved many important theorems in the theory of entire functions (too many to list them here). For a survey of her contributions I recommend her obituary: Zbl 1032.01034 Hayman, W. K. Mary Lucy Cartwright (1900–1998), Bull. London Math. Soc. 34 (2002), no. 1, 91–107. written by Hayman, who also made many important contributions to entire functions. Nowadays she is more famous for her contribution to non-linear dynamics (partially joint with Littlewood) but these papers were VERY much ahead of their time, and were not sufficiently appreciated until the late 1980s. Before that she was more famous for her contribution to the theory entire functions. For example, the class of entire functions of exponential type with the property $$\int_{-\infty}^\infty\frac{\log|f(x)|}{1+x^2}dx<\infty,$$ which plays a fundamental role in harmonic analysis, is called the Cartwright class. (The principal theorem about zero distribution of functions of this class is called Levinson's theorem, but it was proved independently by Levinson and Cartwright). Many of her results on entire functions were improved since then or absorbed into more general theories, but her book on entire functions remains one of her most cited works.<|endoftext|> TITLE: Fredholm theory of non elliptic operators QUESTION [5 upvotes]: In this question we search for a big list of non elliptic operators whose Fredholm index is finite or whose Fredholm theory is extensively discussed. The main motovation is the conference linked in the head of the following post:The subject of the conference is "Fredholm Theory of Non Elliptic Operators". I read the abstract of talks in the conference but I did not underestand that what is a precise non elliptic differential operator whose index is finite. In the above MO post it is discussed that ellipticity play a crucial role for Fredholm ness but there are some sence of index theory for non eliptic operators. I would appreaciate if you give answer, 1 item per answer, which present either a precise example of non eliptic operator of finite index or contains an extended explaination of one of the abstract listed in the above conference on "Fredholm Theory for non Elliptic operators". I ask moderators to consider this question as a wiki question. REPLY [5 votes]: I think I am in a good position to answer this. The Fredholm property of elliptic operators as maps between Sobolev spaces on compact manifolds rests on elliptic regularity properties. If an operators is not elliptic on a closed manifold this operator will not be Fredholm. This is however no longer true of you have a manifold with boundary and you impose boundary conditions, i.e. modify the domain. If the boundary conditions are carefully chosen then Hoermander's propagation of singularity theorem, replacing elliptic theory, can be used to show the Fredholm property even in non-elliptic situations. For example the Lorentzian Dirac operator is not elliptic, but it is Fredholm under some conditions (compact Cauchy surface, etc) if APS boundary conditions are imposed on spacelike hypersurfaces. Take a look at An index theorem for Lorentzian manifolds with compact spacelike Cauchy boundary by Christian Baer and Alexander Strohmaier where an index theorem is proved for such operators. You can also look at the published version American Journal of Mathematics, 1421-1455 141.5.<|endoftext|> TITLE: Ricci flow preserves almost Kahler condition? QUESTION [9 upvotes]: I have been unable to find a reference to the following (perhaps too naive) question. Suppose we have an almost Kahler manifold $(M^{2n},\omega,J,g)$ i.e. the almost complex structure $J$ is non-integrable but $\omega$ is closed. If we run the Ricci flow on the metric i.e. $$\frac{\partial g_t}{\partial t}=-2Ric(g_t)$$ while leaving $J$ unchanged, so that we also get a flow for $\omega_t:=g_t(J\cdot,\cdot)$, does this imply that $g_t$ stays compatible wih $J$? The answer is known to be yes if $J$ is integrable and this corresponds to the Kahler Ricci flow. Moreover, $\omega_t$ stays closed. Assuming the compatibility with the almost complex structure holds, does it also follow that $d\omega_t=0$ for all $t$? I fail to see the difference to proving the above statement whether $J$ is integrable or not. REPLY [13 votes]: The answer is, in general 'no': Under the Ricci-flow, a metric need not remain compatible with $J$ if $J$ is not integrable, even if the associated $2$-form $\omega$ is assumed closed. I don't see how to see this directly without doing some calculation, but the basic idea is this: Consider an almost-complex $4$-manifold $(M^4,J)$. The set of $J$-compatible metrics on $M$ is an open cone in an affine space, the sections of a positive cone in a bundle $H\subset S^2(T^*M)$ of real rank $4$ over $M$. This bundle has a linear embedding $\Phi:H\to\Lambda^{1,1}(T^*M)$ into the $\mathbb{R}$-valued 2-forms of $J$-type $(1,1)$ defined by letting $\Phi(g)(u,v) = g(Ju,v)$ for any $u,v\in T_xM$. The almost-Kähler condition is that $\mathrm{d}\bigl(\Phi(g)\bigr) = 0$, so it is a linear condition on the sections of this bundle. In order for the Ricci-flow starting at a metric $g$ to yield a family of metrics compatible with $J$, it would at least have to be true that $\mathrm{Ric}(g)$ be a section of $H$. Unfortunately, in the case that $J$ is not integrable, the condition $\mathrm{d}\bigl(\Phi(g)\bigr) = 0$ is not sufficient to guarantee that $\mathrm{Ric}(g)$ be a section of $H$. The reason is as follows: There is a canonical splitting $S^2(T^*M) = H \oplus C$, where $C$ is the bundle of quadratic forms that are the real part of a $J$-bilinear quadratic form on $M$. This $C$ is a bundle of real rank $6$ over $M$. Let $\pi_H$ (respectively, $\pi_C$) denote the canonical projection from $S^2(T^*M)$ to $H$ (respectively, $C$). By calculation one can show that, when $\mathrm{d}\bigl(\Phi(g)\bigr) = 0$, the tensor $\pi_C\bigl(\mathrm{Ric}(g)\bigr)$, which is a section of $C$, can be written in the form $$ \pi_C\bigl(\mathrm{Ric}(g)\bigr) = L\bigl(\nabla^g(N_J)\bigr) $$ where $N_J$ is the Nijnhuis tensor of $J$ (a section of $T\otimes_{\mathbb{C}}\Lambda^{0,2}(T^*)$), $\nabla^g$ is a canonical $g$- and $J$-compatible connection (with torsion, in general), and $L$ is a (surjective) canonical linear operator $L:T^*\otimes_\mathbb{C}T\otimes_{\mathbb{C}}\Lambda^{0,2}(T^*)\to C$. In particular, it is easy to show that $\pi_C\bigl(\mathrm{Ric}(g)\bigr)$ does not, in general vanish when $\mathrm{d}\bigl(\Phi(g)\bigr) = 0$. (Of course, when $N_J=0$, i.e., $J$ is integrable, the above formula shows that, indeed, $\pi_C\bigl(\mathrm{Ric}(g)\bigr)=0$.) There remains the interesting question as to whether, when $g$ satisfies both $\mathrm{d}\bigl(\Phi(g)\bigr) = 0$ and $\pi_C\bigl(\mathrm{Ric}(g)\bigr)=0$, the Ricci-flow will preserve both of these conditions. I have not done the calculations necessary to check this, but it shouldn't be all that hard to do. It is not hard to produce examples of non-integrable $J$ for which there exist compatible metrics $g$ that satisfy both $\mathrm{d}\bigl(\Phi(g)\bigr) = 0$ and $\pi_C\bigl(\mathrm{Ric}(g)\bigr)=0$ and for which the Ricci-flow does preserve both conditions.<|endoftext|> TITLE: Existence of n-axial elements in groups with at least 2 ends QUESTION [8 upvotes]: Let $G$ be a finitely generated group. Fix some symmetric finite generating set $S$ for $G$, and write $\Gamma$ for the Cayley graph of $G$ with respect to $S$. Given finite subsets $X,S,Y$ of $G$, we say that $S$ separates $X$ from $Y$, denoted $(X:S:Y)$, if $X$ and $Y$ lie in distinct connected components of $\Gamma\setminus S$. If $B_n$ denotes the $n$-ball in $G$, we say that $g\in G$ is $n$-axial if $(g^a B_n: g^b B_n : g^c B_n)$ whenever $a2$-ended groups. By geodesics I mean paths in the Cayley graph whose length is equal to the word metric between their initial and terminal vertex. If $G$ is a group with finite generating set $S$, $A$ finite and $G \setminus A$ has all components infinite and has at least three components (in the Cayley graph w.r.t. $S$). Suppose $h \notin A$, then there exists $k$ such that $A \cap kA = \emptyset$ and $kh$ is connected to $h$ in $G \setminus (A \cup kA)$. Proof. Consider the set $P \subset G$ consisting of all geodesics in $G \setminus A$ that begin from $\partial A$ (the boundary of $A$, i.e. elements adjacent to $A$ but not inside it), in the component of $h$. Write $P_n$ for the set of nodes first reached by some geodesic in $P$ after exactly $n$ steps. Since every translate of $A$ splits $G$ into at least three components, it is easy to show that for some $\alpha > 1$, we have $|P_n| \geq \alpha^n$ for all large enough $n$. For all $g \in P_n$, fix a geodesic representation $w_g \in S^n$, representing a the path from some element $e_g \in \partial A$ to $g$ arbitrarily. Now pick $h'$ in $G \setminus A$ in $P_m$ for large $m$. Suppose that for all $g \in P_n$, the set $g(h')^{-1}A$ intersects the range of the geodesic $w_g$ (in the obvious sense of partial products starting from $e_n$). Then, intuitively, we can use this fact to compress the information in the paths in $P_n$. More precisely, if $g \in P_n$ then $g = g'a^{-1}h'$ for some $a \in A$ where $g'$ lies on the range of $w_g$. Because $h' \in P_m$ has word norm at least $m - |e_g|$ and $w_g$ is a geodesic, $g'$ must be in $P_{n-m+k}$ where $k \in [-3r,3r]$ where $r$ is the maximal word norm among elements of $A$. In particular, the distance of $g$ from $\partial A$ is reduced by at least $m-3r$ when we multiply it by a suitable $h^{-1}a$. We can now describe any element of $P_n$ by listing at most $\lceil n/(m-3r) \rceil$ elements $a$, encountered when iteratively applying this observation, starting from $g$, and listing also the element of $P_{n'}$ for $n' < m+3r$ obtained in the end. Now, $C |S|^{m+3r+1} |A|^{\lceil n/(m-r) \rceil}$ (where $C |S|^{m+3r}$ is an upper bound for the number of elements obtained in the end, $C$ is the boundary size of $A$) has slower asymptotic growth in $n$ than $\alpha^n$, as soon as $|A|^{1/(m-3r)} < \alpha$, a contradiction if $m$ was picked large enough. This means we can necessarily find some $g \in P_n$ (for any sufficiently large $n$) such that $g(h')^{-1} A$ does not intersect the geodesic $w_g$. Assuming $n$ and $m$ are large enough, we have both $A \cap g(h')^{-1}A = \emptyset$, and that the connected component of $g$ in $G \setminus (A \cup g(h')^{-1}A)$ contains both $h$ and $g(h')^{-1}h$. The first fact is obvious, and for the second, observe that as $h$ and $g(h')^{-1}h$ are, respectively, in the same connected components as $h'$ and $g(h')^{-1}h' = g$ in $G \setminus A$ and $G \setminus g(h')^{-1}A$ respectively, it is suffices that $n$ and $m$ be larger than the length of a minimal path between $h$ and $h'$ in $G \setminus A$. This means we can pick $k = g(h')^{-1}$. End square. Let $G$ have at least $3$ ends. Then $G$ has an $n$-axial element for some $n$. Proof. Let $A$ finite and $G \setminus A$ has all components infinite and has at least three components, you can always find such by flood filling the finitely many finite ones. Let $h_1 \notin A$ and $h_2 \notin A$ be in distinct components of $G \setminus A$. Now, use the previous lemma to find $k_1$ such that $h_1$ and $k_1h_1$ are in the same component of $G \setminus (A \cup k_1A)$ and $A$ and $k_1A$ are disjoint. Then find $k_2$ so that $A$ and $k_2A$ are disjoint and $h_2$ is in the same component of $G \setminus (A \cup k_2A)$ as $k_2h_2$. Then $k_2^{-1}k_1$ is $n$-axial, if $n$ is bigger than the size of $A$. Now consider the sequence $B_0 = A, B_1 = k_2^{-1}A, B_2 = k_2^{-1} k_1 A, B_3 = k_2^{-1} k_1 k_2^{-1} A, B_4 = k_2^{-1}k_1k_2^{-1}k_1 A...$ i.e. you multiply the left translation element from the right alternately by $k_2^{-1}$ and $k_1$. We have $(B_0 : B_1 : B_2)$: The separation $(k_2A : A : k_1A)$ is clear, and gives $(A : k_2^{-1}A : k_2^{-1}k_1A)$ by translating. We have $(B_1 : B_2 : B_3)$: We prove that $(A : k_1A : k_1k_2^{-1}A)$ holds: suppose there is a path from $A$ to $k_1k_2^{-1}A$ that does not go through $k_1A$. Then, there is a path from $k_1h_1$ to $k_1h_2$ that does not go through $k_1A$ (which would be a contradiction). To see this, take a path from $k_1h_1$ to $h_1$ (using the assumption on $k_1$) that does not visit any of the three sets, then move to $k_1k_2^{-1}A$ and then $k_1k_2^{-1}h_2$ without visiting $k_1A$ (using the counterassumption that $(A : k_1A : k_1k_2^{-1}A)$ does not hold). Now, observe that since $h_2$ and $k_2h_2$ are in the same connected component of $G \setminus (A \cup k_2A)$, by translating by $k_1k_2^{-1}$ we get that $k_1k_2^{-1}h_2$ and $k_1h_2$ are in the same connected component of $G \setminus (k_1k_2^{-1} A \cup k_1A)$, so we have found a path from $k_1h_1$ to $k_1h_2$ that avoids $k_1A$. From $(A : k_1A : k_1k_2^{-1}A)$, we get $(B_1 : B_2 : B_3) = (k_2^{-1}A : k_2^{-1}k_1A : k_2^{-1}k_1k_2^{-1}A)$ by translating, as desired. We have shown $(B_0 : B_1 : B_2)$ and $(B_1 : B_2 : B_3)$, and by translating by $k_2^{-1}k_1$ from the left we get $(B_i : B_{i+1} : B_{i+2}) \implies (B_{i+2} : B_{i+3} : B_{i+4})$, so by induction $(B_i : B_{i+1} : B_{i+2})$ holds for all $i$. By a basic connectivity lemma (see e.g. Cohen's thesis) we have $(B_i : B_j : B_k)$ for all $i < j < k$. The sequence $B_{2i}$ proves that $k_2^{-1}k_1$ is axial. End square. Edit: The axial part had a chirality issue. That's what you get for trying to be fast, I should maybe have drawn a picture. Fixed some typos as well.<|endoftext|> TITLE: Is it possible to constructively prove that every quaternion has a square root? QUESTION [20 upvotes]: Is it possible to constructively prove that every $q \in \mathbb H$ has some $r$ such that $r^2 = q$? The difficulty here is that $q$ might be a negative scalar, in which case there might be "too many" values of $r$. Namely, $r$ could then equal any vector quaternion of magnitude $\sqrt{|q|}$. The presence of this seemingly severe discontinuity suggests that there can't be a way to constructively prove that every quaternion has a square root. The variety of constructivism can be as strong as possible. So any Choice principle, or Markov's Principle, or Bar Induction, is allowed. My thoughts were to do some kind of reduction to $LPO$ or $LLPO$ or $LEM$. But I don't see how. The way to find a square-root classically is as follows: If $q = w + xi + yj + zk$ is not a scalar quaternion, then it lies on a unique "complex plane". This is due to the fact that a vector quaternion (of the form $xi + yj + zk$) always squares to $-(x^2 + y^2 + z^2)$, which is a negative scalar. The problem then reduces to finding the square root of a complex number. The difficulty is exactly in the case when $x=y=z=0$ and $w < 0$, in which case $q$ and $r$ lie on all complex planes. REPLY [14 votes]: Reduction to LLPO (Lesser Limited Principle of Omniscience). The statement LLPO is the following (from Wikipedia): For any sequence a0, a1, ... such that each ai is either 0 or 1, and such that at most one ai is nonzero, the following holds: either a2i = 0 for all i, or a2i+1 = 0 for all i, where a2i and a2i+1 are entries with even and odd index respectively. This is considered a quintessentially non-constructive claim. The claim that every quaternion has a square root implies LLPO. Consider a sequence $(p_n)_{n \geq 1} \in \{0,1\}$, with the property that at most one element of the sequence is equal to $1$. Consider the following infinite quaternionic series $q = -1 + i\left(1 - \sum_{n=1}^\infty\frac{1 - p_{2n}}{2^n}\right) + j\left(1 - \sum_{n=1}^\infty\frac{1 - p_{2n+1}}{2^n}\right)$. The series clearly converges. Now we assume that we can get an $r$ such that $r^2 = q$. Consider the angle $\theta$ between $r$ and $i$ (considered as 4d vectors with the standard inner product), and likewise consider the angle $\phi$ between $r$ and $j$. Either $\theta > \arctan(1/2)$ or $\phi > \arctan(1/2)$, as these two open regions cover all the non-zero quaternions. If $\theta > \arctan(1/2)$ then we conclude that all $p_{2n}=0$. If $\phi > \arctan(1/2)$ then we conclude that all $p_{2n+1}=0$. This is exactly LLPO.<|endoftext|> TITLE: Testing ideal membership in the Weyl algebra: a simple example QUESTION [6 upvotes]: In Example 1.1.4 of the book Grobner Deformations of Hypergeometric Differential Equations, it is stated without proof that $$\partial^2 \in D\cdot \langle x\partial^4, x^3\partial^2 \rangle \tag{$\star$}$$ where $D$ denotes the Weyl algebra over $\mathbb{k}[x]$, and $ D\cdot \langle x\partial^4, x^3\partial^2 \rangle$ denotes the left-ideal generated by the operators $x\partial^4,x^3\partial^2$. Since Example 1.1.4 is at the very beginning of the book, and no proof is given by the authors, I'm presuming that there a simple way to verify this (i.e. without using the machinery of the main text). My current line of attack: So far, this is my thinking: since $D$ is a domain, the equation $(\star)$ is equivalent to solving the following linear equation in the non-commutative ring $D$: $$1=D_1x\partial^2 +D_2 x^3. ~~~~~~~~~~~~~~~~$$ where $D_1,D_2\in D$ are unknowns. I am able to prove (by brute force calculation) that $\text{ord}\,D_2\geq 1$. Obviously, $\text{ord}\, D_2 = \text{ord}\, D_1+2$. The brute-force calculation in the general case quickly spirals out of control. REPLY [5 votes]: Following my nose gave the following argument. Writing $I$ be the left ideal generated by $x\partial^2$ and $x^3$ and using $\cdot$ to stress multiplication we get $$ x^2 \cdot x\partial^2 - \partial^2\cdot x^3 = [ x^3, \partial^2] = -6x^2\partial - 6x\in I$$ So $$\frac{1}{6}x\partial\cdot (-6x^2\partial - 6x) + x^2\cdot x\partial^2 = x[x^2,\partial]\partial - x^2\partial + x[x,\partial] = -3x^2\partial - x\in I$$ Taking a suitable $\mathbb{k}$-linear combinations of these gives $x\in I$ and $x^2\partial\in I$. Then $$\partial^2\cdot x - 1\cdot x\partial^2 = [\partial^2,x]=2x\partial \in I.$$ Since also $2\partial x\in I$ we conclude $$\partial x- x\partial=[\partial,x]=1\in I$$ as required. The general strategy at each step is to take two elements in the left ideal with the same principal symbol with respect to the filtration with $F_0=\mathbb{k}$, $F_1=\mathbb{k}\cdot\{1,x,\partial\}$ and $F_n=F_1^n$ and then compute their difference which will live in a lower filtered part. I haven't read the book but I'd imagine that this idea is at the heart of computations throughout.<|endoftext|> TITLE: Is an integral sum of periodic vectors always a sum of integral periodic vectors? QUESTION [9 upvotes]: Update: I have found reference to this problem. It is known as "the Rédei-de Bruijn-Schönberg theorem", which is proved in the following papers: N. G. de Bruijn: On the factorization of cyclic groups, Indag. Math.15(1953), 370-377. L. Rédei: Ein Beitrag zum Problem der Faktorisation von Abelschen Gruppen, ActaMath. Acad. Sci. Hungar.1(1950), 197-207. I. J. Schoenberg: A note on the cyclotomic polynomial, Mathematika11(1964), 131-136. End of update. The background of this question is something about cyclotomic fields, but the statement doesn't involve any algebraic number theory. I just get puzzled by this (might be stupid) little question... Let $n>1$ be an integer, and consider the vector space $\mathbb{C}^n$. A vector $v=(v_1, \cdots, v_n) \in \mathbb{C}^n$ is called periodic, if there is a proper divisor $d$ of $n$, such that $v_i = v_{i + d}$ for all $i$; integral, if every $v_i$ is an integer. Question: if an integral vector can be written as a finite sum of periodic vectors, then is it true that it can always be written as a finite sum of integral period vectors? It should be clear that the field $\mathbb{C}$ could be replaced with any field of characteristic zero (e.g. $\mathbb{Q}$). I would guess that the claim is true, but cannot convince myself with a proof... So far I can only prove the case when $n$ has at most $2$ different prime factors, which doesn't help much in the general case. I've also tried to adopt a point of view from cyclotomic fields, or representation theory, or doing some Fourier transform - but again I'm not intelligent enough to morph the question to something known... Therefore I add all the possibly relevant tags. EDIT: I add here a proof when $n = pq$ is the product of $2$ different primes. The more general case $n = p^r q^s$ is morally the same. Since $n = pq$, every periodic vector $a$ either satisfies $a_i = a_{i + p}$ for all $i$, or satisfies $a_i = a_{i + q}$ for all $i$. Hence any finite sum of periodic vectors can be written as $a + b$, where $a_i = a_{i + p}$ and $b_i = b_{i + q}$. Now suppose that $v = a + b$ is such a sum, which is integral. This means that $a_i + b_i = 0\mod\mathbb{Z}$, which then gives: $$a_i = -b_i = -b_{i + q} = a_{i + q} \mod \mathbb{Z}.$$ Together with $a_i = a_{i + p}$, we conclude that all $a_i$ are equal $\mod\mathbb{Z}$, hence all $b_i$ also, and we can adjust them with a constant vector so as to make them both integral. This trick doesn't work for more than $2$ prime factors, though... REPLY [4 votes]: A beautiful question! Though I don't have the time or the space to fill in all the details, I think one can answer the question in the following way. The answer is yes. We can reformulate the question as follows: let $I \subset \mathbb{Z}[x]$ be the ideal generated by the polynomials $$ \frac{x^n-1}{x^d-1}, \quad d \mid n, \quad d \neq n. $$ We claim that $I$ is $\mathbb{Z}$-saturated, that is, within $\mathbb{Q}[x]$, $$ \mathbb{Q} I \cap \mathbb{Z}[x] = I. $$ (Attach to each polynomial $f \in \mathbb{Q}[x]$ the sequence given by the coefficients of $f$ mod $x^n - 1$. The passage from $\mathbb{C}$ to $\mathbb{Q}$ is effected by some easy linear algebra.) We claim, indeed, that $$ I = J := (\Phi_n) $$ is the ideal generated by the $n$th cyclotomic polynomial. Note that $J$ is visibly $\mathbb{Z}$-saturated. Clearly $$ (x^n-1) \subseteq I \subseteq J. $$ Also, $\mathbb{Q}I = \mathbb{Q}J$, because the only common roots of the generators of $I$ are at the primitive $n$th roots of unity and are all simple. (This step uses the Nullstellensatz, which, since we have just one variable, isn't all that deep.) So if $I \neq J$, the discrepancy can be found locally, that is, there is a prime $p$ such that the images $\bar{I}, \bar{J}$ of $I$ and $J$ in $\mathbb{F}_p[x]$ are unequal. (Pass to the finite-dimensional $\mathbb{Q}$-vector space $\mathbb{Q}[x]/(x^n-1)$. The images $L_I$, $L_J$ are $\mathbb{Z}$-lattices with the same $\mathbb{Q}$-span; if they are unequal, pick $p | [L_J : L_I]$.) If $p \nmid n$, we are done by the Nullstellensatz again, because the $n$th roots of unity remain distinct in $\bar{\mathbb{F}}_p$. Now $p \mid n$. Let $$ \bar K = \left\{\frac{f}{\Phi_n(x)} : f \in \bar I \right\}. $$ We wish to prove that $\bar K$ is the unit ideal, that is, that the polynomials in $\bar K$ have no common roots. Suppose there is a common root $\lambda \in \bar{\mathbb{F}}_p$. Then $\lambda^n = 1$, since $x^n - 1 \in \bar I$. Let $\lambda$ be a primitive $r$th root of unity; note that $r \mid n$ and $p \nmid r$, so $r \mid n'$, where $n = p^k n'$ with $p \nmid n'$. We get a contradiction as follows: If $r \neq n'$, then the element $$ \frac{x^n - 1}{\Phi_n(x) (x^{p^k r} - 1)} = \prod_{d|n,\, d\nmid p^k r,\, d \neq n} \Phi_d \in \bar K $$ has no root at $\lambda$ (because the factors $\Phi_r, \Phi_{pr}, \ldots, \Phi_{p^k r}$ have all been eliminated). If $r = n'$, then the element $$ \frac{x^n - 1}{\Phi_n(x) (x^{n/p} - 1)} = \prod_{d|n,\, p^k \parallel d,\, d \neq n} \Phi_d \in \bar K $$ likewise has no root at $\lambda$.<|endoftext|> TITLE: Obstruction to splitting an object in derived category into a sum of two-term complexes QUESTION [12 upvotes]: Let $\mathcal{A}$ be an abelian category, and $D$ its bounded derived category. An object $M \in D$ may be described as a list of cohomology objects $H^i = H^i(M)$ together with some complicated glueing data. I am interested only in the case when $\mathcal{A}$ has homological dimension two. For example, $\mathcal{A}$ can be a category of coherent sheaves on a smooth surface. In this case the glueing data amounts to a collection of classes $\xi_i \in \mathrm{Ext}^2(H^i, H^{i-1})$ between each pair of adjacent cohomology objects, with no restrictions on the choices. By definition, an object $M \in D$ is quasiisomorphic to a direct sum of complexes concentrated in a single degree (i.e., shifts of objects from $\mathcal{A}$) if and only if each $\xi_i$ vanishes. Similarly, some objects in $D$ are quasiisomorphic to direct sums of complexes concentrated in two adjacent degrees. Is it possible to characterize this property by vanishing of some obstructions built in terms of the presentation of an object as a collection $\{ (H_i, \xi_i) \}_{i \in \mathbb{Z}}$ above? REPLY [5 votes]: As pointed out in nikola karabatic's answer, a decomposition of M as a direct sum of two-term complexes induces decompositions $H^i=H^i_a\oplus H^i_b$ for each $i$. These have the property that $\xi_i\in\text{Ext}^2(H^i_a,H^{i-1}_b)$, where $\text{Ext}^2(H^i_a,H^{i-1}_b)$ is regarded as a direct summand of $\text{Ext}^2(H^i,H^{i-1})$ in the obvious way. Or equivalently, there are vanishing obstructions in the other three summands of $\text{Ext}^2(H^i,H^{i-1})$. Conversely, if there are such decompositions of the $H^i$, then the reconstruction of $M$ from the data $\{(H^i,\xi_i)\}_{i\in\mathbb{Z}}$ gives a direct sum of two term complexes with cohomology $H^i_a$ in degree $i$ and $H^{i-1}_b$ in degree $i-1$.<|endoftext|> TITLE: Average number of iterations for the Euclidean algorithm to terminate QUESTION [7 upvotes]: Let $N$ be a positive integer and $0 \leq s < N$. We try to divide $s$ into $N$ using the Euclidean algorithm: $N = q_1 s + r_1 $ $r = q_2 r_1 + r_2 $ $\vdots$ $r_{K-1} = q_{K-1} r_K$ If we choose $-r_{i-1}/2 \leq r_i < r_{i-1}/2$, I think this determines the $q_i$'s uniquely, but I don't think this matters for my question. For a fixed $N$ and $0 \leq s < N$, define $K_s$ to be the number of iterations before the Euclidean algorithm terminates, (i.e. $K$ in the instance written above). Are there existing tools to characterize $$ \frac{1}{N} \sum_{0 \leq s < N} K_s, $$ for a given $N$? REPLY [2 votes]: The complexity analysis of the Euclidean algorithm is much much older that the given reference. It goes back to at least Lamé, before 1785. The above answer should be viewed as modern version of Lamé's theorem. The average is actually a normal variable, there are several difficult proofs, see Morris - A short proof that the number of division steps in the Euclidean algorithm is normally distributed, and earlier references.<|endoftext|> TITLE: Which of the proofs of the fundamental theorem of algebra can actually produce bounds on where the roots are? QUESTION [30 upvotes]: One of the old classic MO questions is a big-list of proofs of the fundamental theorem of algebra. Here is a second big-list question about this big list: Which of the FTA proofs can, even in principle, be mined to produce bounds on where the roots of a complex polynomial $f(z) = z^n + a_{n-1} z^{n-1} + \dots + a_0 = 0$ are, as a function of its coefficients? Which of them can further be mined to produce algorithms for locating roots? Here my interest is not in actually having such bounds or algorithms, since clearly there are more direct ways to find them; it's in classifying the proofs based on how effective or explicit they are or can be made. Example. The proof via Rouche's theorem proceeds as follows: we want to find $R$ such that Rouche's theorem guarantees that $z^n$ and $f(z)$ have the same number of roots in the disk of radius $R$. Rouche's theorem says that it suffices to find $R$ such that $|z^n| > |a_{n-1} z^{n-1} + \dots + a_0|$ for all $|z| = R$. This is guaranteed if we guarantee $$R^n > |a_{n-1}| R^{n-1} + \dots + |a_0|$$ so we can pick, for example, $R > \text{max}(|a_{n-1}| + \dots + |a_0|, 1)$, and then Rouche's theorem guarantees that $f(z)$ has $n$ roots in the disk of radius $R$. (Note that the assumption that $z$ is a root of $f(z)$ immediately gives $|z^n| = |a_{n-1} z^{n-1} + \dots + a_0| \ge |a_{n-1}| |z|^{n-1} + \dots + |a_0|$, so we already know that if any roots exist they must lie in this disk; again, my interest isn't in actually having bounds, it's in knowing what bounds can be mined from what proofs!) This example is relatively straightforward; it's much less clear whether the proofs that proceed via e.g. Liouville's theorem, degree arguments, fundamental group computations, cohomology computations, the Lefschetz fixed point theorem, etc. can be mined to produce bounds, but I have hope that at least some of them can, possibly after substantial effort, and possibly with terrible bounds. For example one might hope to mine a bound from the Lefschetz fixed point argument by bounding how complicated the simplicial subdivision + homotopy to a simplicial map you need is, etc. REPLY [7 votes]: Here is an article with a proof where the abstract says "Moreover, the proof is constructive and immediately translates to an algebraic root-finding algorithm" : https://pnp.mathematik.uni-stuttgart.de/igt/eiserm/publications/roots.pdf . It also discusses your question with an historical last section.<|endoftext|> TITLE: Tensoring irreducible $B$-series representations/ Type B Littlewood-Richardson QUESTION [5 upvotes]: When tensoring finite dimensional representations of the Lie algebra ${\frak sl}_n$, we have an explicit algorithm given in terms of Young diagrams. See Section 4 of this paper. Do there exist similar pictures for the $B$ and $D$ series? I am specifically interested in simplest case, where one of the irreducible representations being tensored is the fundamental representation. REPLY [4 votes]: Littelmann used standard monomial theory to give a unified Littlewood–Richardson rule for the simple reductive algebraic groups of types $A$, $B$, $C$ and $D$ (and some others) in which the coefficients enumerate certain generalized standard tableaux. See (a) in the theorem on page 346 of Littlemann's paper, A generalization of the Littlewood-Richardson rule., J. Alg. 130 (1990) 328–368.<|endoftext|> TITLE: General topological space with closure operation as in Russian translation of Hausdorff's 1914 and 1927 Mengenlehre QUESTION [6 upvotes]: In the Russian translation (by P. Alexandroff and A. Kolmogorov) of chapter "Point Sets in General Spaces" Hausdorff (1914), the notion of a general topological space is defined as set $R$ with closure operation $\overline{\circ}$ defined for each subset of $R$. I.e. a subset $M$ of $R$ does not necessarily belongs to its closure. Moreover, mostly no more constraints are put on $\overline{\circ}$. As 6th example, he provides: 6) $R$ is an infinite set. Let $\overline{M} = R \setminus M$. My two questions: How such topological structures are called in the modern setting? What fields are they used in? UPD: Probably, the example was introduced not by Hausdorff but by P. Alexadroff or A. Kolmogorov. This "general" closure operation is used in the russian merged version of 1914 and 1927 books to axiomatize general topological spaces and show the correspondence between metric, neighbourhoods, and topological (without "general") spaces. Still, the questions persist. REPLY [11 votes]: 1. Introductory Comments and Contents/Summary First, to repeat a point that came up in the comments, we’re not talking about alternative methods of defining a topological space, such as using the interior operator or the frontier operator. Rather, we’re talking about removing some, or even all of the axioms for the topological closure operator and asking what such generalized notions of a topological space are called and where they arise. Regarding what they are called, the most common terms are probably generalized topology and extended topology and pretopology. In what follows I’ll give a progression of generalized closure notions, beginning with arbitrary operators from ${\mathcal P}(X)$ to ${\mathcal P}(X)$ (where $X$ is a set and ${\mathcal P}(X)$ is the set of subsets of $X)$ and ending with topological closure operators. Many closure operators of interest are not directly comparable to all those in this progression, but I think the notions I’ve chosen provide an appealing path to the topological closure operator. Because the literature is filled with terms having multiple meanings $-$ depending on the author and the field $-$ and a seemingly endless parade of terms for every kind of nuance, I’ve tried to pick terms that are unlikely to have multiple meanings and I've tried to minimize the introduction of terms (other than sometimes pointing out useful search terms). For example, I’ll try to be consistent in using “operator” (rather than mapping, function, operation, transformation, etc.) when mentioning a closure operator. Let $T$ be a function from ${\mathcal P}(X)$ to ${\mathcal P}(X).$ The Kuratowski closure operator axioms for $T$ are (K1) through (K4) below, where their ordering is in accordance with that of the progression of generalized closure operators I'll give. (semi-K1) $\;$For each $A,\,B \in {\mathcal P}(X),$ we have $T(A) \cup T(B) \subseteq T(A \cup B).$ (K1) $\;$For each $A,\,B \in {\mathcal P}(X),$ we have $T(A \cup B) = T(A) \cup T(B).$ (K2) $\;T(\emptyset) = \emptyset.$ (K3) $\;$For each $A \in {\mathcal P}(X),$ we have $A \subseteq T(A).$ (K4) $\;$For each $A \in {\mathcal P}(X),$ we have $T(T(A)) = T(A).$ CONTENTS/SUMMARY 1. Introductory Comments and Contents/Summary 2. Unrestricted Closure Operators Notions such as interior, neighborhood, limit point, etc. can be defined. Definition of a partial order on closure operators. 3. Monotone Closure Operators $\;$ [satisfy (semi-K1)] Two ways to define a certain “natural” monotone closure operator from an unrestricted closure operator. Each monotone closure operator has a least fixed point and a greatest fixed point, and there are two standard ways of obtaining them $-$ impredicative or “from above”, iterative or “from below”. Many common mathematical objects can be realized as such fixed points $-$ closed sets, open sets, the linear span of a set, the convex hull of a set, the deductive closure of a set of wffs in logic, the $\sigma$-algebra generated by a collection of sets, etc. A certain monotone closure operator can even be used to prove the Cantor-Bernstein theorem for cardinal numbers. 4. Union-Preserving Closure Operators $\;$ [satisfy (K1)] There exists a greatest union-preserving closure operator that is less than or equal to a given monotone closure operator (in fact, for a given unrestricted closure operator). 5. Base Closure Operators $\;$ [satisfy (K1), (K2)] The prototypical example is the derived set operator in topology and its various generalizations obtained by using different notions of “limit point”. There is a “natural” topological space associated with each base closure operator. These so-called fine topologies are important in abstract potential theory. 6. Čech Closure Operators $\;$ [satisfy (K1), (K2), (K3)] There is a “natural way” to obtain a Čech closure operator from any base closure operator. A relatively large part of general topology can be carried out in the context of Čech closure spaces, but most of the discussion here will involve the interrelationships between a base closure operator and its associated Čech and topological closure operators. 7. Topological Closure Operators $\;$ [satisfy (K1), (K2), (K3), (K4)] Each Čech closure operator can be transfinitely iterated to obtain a topological closure operator, and the resulting topological space is the same as that obtained from the collection of closed sets in the Čech closure space. 8. Selected References References are mostly restricted to those that specifically involve, entirely or partly, various generalized closure operator notions. 2. Unrestricted Closure Operators Definition: An unrestricted closure operator on the set $X$ is a function from ${\mathcal P}(X)$ to ${\mathcal P}(X).$ I’m using “unrestricted” rather than “arbitrary” because “arbitrary” is too pervasive and useful as a general descriptive term. Stadler/Stadler [62] make the following observation in the first sentence of their manuscript: “$\dots$ we explore the surprising fact that some meaningful topological concepts can already be defined on a set $X$ endowed with an arbitrary set-valued set-function, which we will interpret as a generalized closure operator.” To illustrate this observation, note that given an restricted closure operator $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ and $A \subseteq X,$ we can define the $T$-interior of $A$ by ${\text {int}}_T(A) = X – T(X – A).$ Now we can define a $T$-neighborhood of a point $x \in X$ to be any set $N \subseteq X$ such that $x \in {\text {int}}_T(N),$ and we can define $x \in X$ to be a $T$-limit point of $A \subseteq X$ if each $T$-neighborhood of $x$ intersects $A$ in at least one point other than $x.$ These and other notions for unrestricted closure operators are discussed in Stadler/Stadler [62] (pp. 1-6). The only topology text I'm aware of that at least briefly discusses unrestricted closure operators is Mamuzić [44] (mostly in §1$-$§3, pp. 13-22), but I haven't put forth much effort in looking for such texts. Mamuzić's discussion includes notions such as accumulation point, derived set, interior and exterior boundary of a set, etc. in the setting of an unrestricted closure operator. However, the nature of his treatment makes it difficult to use the first few pages of his book as a reference for this topic $-$ results involving topological spaces and his more general notion of a neighborhood space are mixed together in such a way that close reading for context is often needed to determine just what assumptions are being made for those results that specifically pertain to generalized topological spaces. Incidentally, Sierpiński [60] begins with an essentially unrestricted notion of a derived set (set of limit points of a set), but his approach does not appear to provide a natural setting for arbitrary unrestricted closure operators. Sierpiński begins by assigning to each point $x \in X$ a (nonempty?) set ${\mathcal N}_x \subseteq {\mathcal P}(X)$ of "neighborhoods" of $x$ (I believe he requires $x$ to belong to each neighborhood of $x,$ but he is not explicit about this) and then the notion of a limit point of a subset of $X$ is defined as above. Now define a closed set to be a set that contains all its limit points, and finally define the closure of $A \subseteq X$ to be the intersection of all closed sets containing $A.$ However, it is easy to see that not every unrestricted closure operator on $X$ will arise by taking closures of sets in this manner, because all of Sierpiński's generalized closure operators $cl$ obviously satisfy $A \subseteq B \implies cl(A) \subseteq cl(B).$ In fact, Sierpiński's closure operators satisfy (semi-K1), (K2), (K3), and (K4) $-$ see bottom of p. 7 of Sierpiński [60]. It will be useful to define a partial order on the set of unrestricted closure operators on a fixed set. Definition: Let $T_1$ and $T_2$ be unrestricted closure operators on $X.$ We say that $T_1 \leq T_2$ if, for each $A \in {\mathcal P}(X),$ we have $T_1(A) \subseteq T_2(A).$ It is easy to see that, for all unrestricted closure operators $T_1,$ $T_2,$ and $T_3$ on $X,$ we have (i) $T_1 \leq T_2;$ (ii) $T_1 \leq T_2$ and $T_2 \leq T_1$ implies $T_1 = T_2;$ (iii) $T_1 \leq T_2$ and $T_2 \leq T_3$ implies $T_1 \leq T_3.$ 3. Monotone Closure Operators $\;$ [satisfy (semi-K1)] Definition: We say that $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ is a monotone closure operator on $X$ if for each $A,\,B \in {\mathcal P}(X)$ we have $T(A) \cup T(B) \subseteq T(A \cup B).$ In the literature these are also called isotone and increasing and order-preserving. Lemma: Let $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ be an unrestricted closure operator on $X.$ Then $T$ is a monotone closure operator on $X$ if and only if for all $A,\,B \in {\mathcal P}(X)$ we have $A \subseteq B$ implies $T(A) \subseteq T(B).$ Monotone Modification of an Unrestricted Closure Operator: Given an unrestricted closure operator $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X),$ there is a natural way to associate with $T$ a monotone closure operator ${\overline{T}}:{\mathcal P}(X) \rightarrow {\mathcal P}(X).$ (Here I’m using “natural” in an informal descriptive sense, not in a category-theoretic sense.) For each $A \in {\mathcal P}(X),$ either of the following two equivalent formulations can be used to define $\overline{T}(A)$: $$\begin{array}{l} \overline{T}(A) & = & \bigcap\,\{T(B):A \subseteq B\} \\ \\ \overline{T}(A) & = & \{x \in X: A \cap N \neq \emptyset \;\; {\text {for each}} \; T {\text {-neighborhood}}\; N \; {\text {of}} \; x \}\end{array}$$ Recall that “$T$-neighborhood of $x$” was defined earlier for unrestricted closure operators. For the following results, see Day [14] (p. 186) and Stadler/Stadler [62] (p. 7; note that the union in equation (11) on p. 7 should be an intersection). (a) $\;\overline{T}$ is a monotone closure operator. (b) $\;\overline{T} \leq T$ (c) $\;T_1 \leq T_2\;$ implies $\;\overline{T_1} \leq \overline{T_2}$ (d) $\;$If $T'$ is a monotone closure operator on $X$ and $\;T' \leq T,\;$ then $\;T' \leq \overline{T}.$ Remarks: (1) Note that (a), (b), and (d) can be summarized by saying $\overline{T}$ is the greatest monotone closure operator that is less than or equal to $T.$ (2) If $T$ itself is a monotone closure operator, then $T = \overline{T}.$ This follows from (a), (b), (d), and antisymmetry of $\leq.$ (3) If $T$ is any unrestricted closure operator, then $\overline{T} = \overline{\overline{T}}.$ This follows from (a) and (2). (Just to keep matters straight, I'll mention that this last assertion says the "monotone modification process" is idempotent, not that $\overline{T}$ is an idempotent closure operator.) (4) An immediate consequence of (2) is that every monotone closure operator on $X$ arises from the monotone modification of some unrestricted closure operator on $X.$ Thus, the monotone modification process does not necessarily introduce any additional properties, such as (K2) or (K3). Example (complement operator): If $T^c$ is the complement operator, defined by $T^c(A) = X – A,$ then the monotone modification of $T^c$ is such that $\overline{T^c}(A) = \emptyset$ for each $A \in {\mathcal P}(X).$ I suppose this can be understood by recognizing that the monotone modification of the complement operator has to be maximally trivial to counteract the maximal way in which the complement operator fails to be monotone. Of course, more generally, this maximal trivialness of $\overline{T}$ will also occur for any unrestricted closure operator $T$ that satisfies $T(X) = \emptyset.$ The most significant application of monotone closure operators is the fact that each monotone closure operator has at least one fixed point, along with the two contrasting methods for obtaining a fixed point $-$ “from above” and “from below”. Note that an unrestricted closure operator can have no fixed points. A rather trivial example is $X = \{a\}$ with $T$ defined by $T(\emptyset) = \{a\}$ and $T(\{a\}) = \emptyset.$ Fixed Points of Monotone Closure Operators (impredicative or “from above” approach) Let $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ be a monotone closure operator on $X,$ and define subsets $E_{*}$ and $E^{*}$ of $X$ by $$\begin{array}{l} E_{*} & \triangleq & \bigcap\,\{A \in {\mathcal P}(X): \; T(A) \subseteq A\} \\ E^{*} & \triangleq & \bigcup\,\{A \in {\mathcal P}(X): \; A \subseteq T(A) \} \end{array}$$ The following results hold: (a) $\;T(E_{*}) = E_{*}$ (b) $\;T(E^{*}) = E^{*}$ (c) $\; T(E) = E\;$ implies $\;E_{*} \subseteq E \subseteq E^{*}$ Some to all of the results above can be found in several standard undergraduate level set theory texts, such as Dalen/Doets/Swart [13] (Theorem 4.1 on p. 262), Enderton [18] (Exercise 30 on p. 54 $-$ see this and this), Hrbacek/Jech [33] (Exercises 1.10 & 1.12 on p. 69), and Moschovakis [50] (Problem 6.11 on p. 85). Although a complete statement of the results above, with proof, is given in Devidé [15] (1962), the less precise result that a monotone closure operator has a fixed point was observed in Knaster [38] (p. 133, lines −5 to −3), which was joint work with Tarski presented on 9 December 1927, and the results above (and more) were obtained in 1939 by Tarski in the more general context of a non-decreasing function from a complete lattice to itself (see footnotes on pp. 285, 286 in Tarski [63]). Fixed Points of Monotone Closure Operators (iterative or “from below” approach) Let $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ be a monotone closure operator on $X.$ For each ordinal $\alpha \geq 1,$ we define (by transfinite induction) $T^{\alpha}:{\mathcal P}(X) \rightarrow {\mathcal P}(X),\,$ the $\alpha$-th iterate of $T,$ by $$\begin{array}{l} T^1 (A) & = & T(A) \\ T^{\alpha}(A) & = & T\left(\,\bigcup \,\{T^{\gamma}(A): \; \gamma < \alpha \} \,\right) \;\; {\text{if}} \; \alpha > 1 \end{array}$$ For those concerned about statements involving “all ordinals”, it is enough to define $T^{\alpha}$ for all ordinals less than or equal to the least infinite ordinal whose cardinality is greater than the cardinality of $X.$ In (c) below, $|X|^{+}$ denotes the least cardinal number greater than $|X| \,=\, {\text {card}}\,(X).$ The following results hold: (a) $\;T^{\alpha}$ is a monotone closure operator for each $\alpha.$ (b) $\; \alpha \leq \beta\;$ implies $\; T^{\alpha} \leq T^{\beta}.$ (c) $\;$For each $A \in {\mathcal P}(X)$ there exists an ordinal $\;\lambda(A) < |X|^{+}\;$ such that, for each $\;\alpha \geq \lambda(A), \;$ we have $\;T^{\alpha}(A) = T^{\lambda(A)}(A).$ (d) $\;T^{\lambda(\emptyset)}(\emptyset) = E_{*},\;$ where $E_{*}$ is from the earlier “from above” approach. (e) $\;T^{\lambda(X)}(X) = E^{*},\;$ where $E^{*}$ is from the earlier “from above” approach. The least upper bound of the (minimal) ordinals $\lambda(A)$ in (c) is often called the closure ordinal of $T,$ which we will denote by $|T|.$ (Hammer uses the term projecting order of $T$ on p. 61 of [23].) Note that $|T| \leq |X|^{+}$ and, for each $\alpha \geq |T|,$ we have $T^{\alpha} = T^{\lambda}.$ (For completeness, I suppose we would want to define the closure ordinal of the identity closure operator to be $0.)$ For example, $|T| = 1$ for the usual closure and interior operators of a topological space, as well as for the linear span, the convex hull, the deductive closure, and the $\sigma$-algebra examples below. Also, $|T| = 2$ for the usual boundary [= frontier] operator in a topological space (or even the boundary operator in certain more generalized spaces $-$ see Theorem C.6 on p. 77 of Hammer [20]). However, it is easy to see that $|T|$ can be arbitrarily large (even more, $|T|$ can have any specified ordinal value) by considering the “limit point” operation when $X$ is an ordinal (see also this). Incidentally, since $|T|$ is the least ordinal at which the iterates of $T$ “stabilize” for all subsets of $X,$ a natural question is whether $E_{*}$ and/or $E^{*}$ can be reached in fewer than (and possibly also different from each other) $|T|$ many steps from $\emptyset$ and $X.$ Again, ordinals come to our rescue. If $X$ is the ordinal ${\omega}^{\omega} + {\omega}^2$ and $T$ is the limit point operator, then I believe $|T| = {\omega}^{\omega},$ while $E_{*} = \emptyset$ is reached in $0$ steps and $E^{*} = {\omega}^{\omega}$ is reached in $3$ steps. If $T$ is continuous in the sense that $\;T\left(\bigcup\limits_{n=1}^{\infty}A_n\right) = \bigcup\limits_{n=1}^{\infty}T(A_n)\;$ for every non-decreasing sequence $\;A_1 \subseteq A_2 \subseteq \cdots\;$ of subsets of $X,$ then $E_{*}$ can be reached in at most ${\omega}_0$ many steps and, in fact, we have $\;E_{*} = \bigcup\limits_{n=1}^{\infty}T^{n}(\emptyset).$ These results about iterating monotone closure operators and their connections with monotone inductive definitions can be found in Aczel [1], Barwise [2] (Chapter VI), Barwise/Moss [3] (Chapter 15), Dalen/Doets/Swart [13] (Section III.4 on pp. 260-266), Devlin [16] (Section 7.4 on pp. 159-163), Enderton [18] (Exercise 9 on p. 78), Hinman [32] (pp. 22-26), Hrbacek/Jech [33] (Exercises 1.13 & 1.14 on p. 69), Moschovakis [48], Moschovakis [49] (pp. 404-405), Moschovakis [50] (Chapter 6), and Pohlers [52] (pp. 109-114). For generalizations of the notion of $T$ being continuous and their consequences on restricting the size of $|T|,$ see Let $\Gamma$ be a $\kappa$-based monotone operator where $\kappa$ is regular. Then the closure ordinal of $\Gamma$ is $\kappa$, Aczel [1] (pp. 746-747), and Schwarz [58]. Some Monotone Closure Operators Whose Fixed Points are Well Known (1) The topological closure, interior, derived, and other operators of a topological space. Also, various refinements of some of these operators, such as the sequential convergence operator and the analogous operators in various generalized topological spaces (e.g. see Lei/Zhang [41]). (2) The set of linear combinations of a subset of a vector space. (3) The convex hull of a subset of a normed vector space. In 1955 Preston Clarence Hammer (1913-1986), motivated by certain issues in convex geometry, published the first of his several papers on an extensive study of generalized topological notions that he called extended topology. In the references below I’ve included Hammer’s papers that seem most relevant to closure operator notions and which I happen to have a copy of, ordered by submission date. Hammer’s Ph.D. students Don Arthur Mattson (1965) and George Clifford Gastl (1966) also did some of this work (both at University of Wisconsin-Madison), but I have not seen their dissertations. Before these two Ph.D. dissertations, there was a Ph.D. dissertation at Oregon State University in 1959, Rio [54], that was largely based Hammer’s 1955 paper [19]. Although Rio does not mention Hammer in his Acknowledgments, Rio’s supervisor was Bradford Henry Arnold, who came to Oregon State University in 1947, which was about the same time that Hammer left Oregon State University (having been there since 1940). I do not know whether Hammer and Arnold knew each other sufficiently to have inspired the choice of Rio's topic, or whether the choice of Rio's topic was simply a coincidence (i.e. Arnold/Rio were simply interested in the topic after seeing Hammer's 1955 paper). Incidentally, Hammer [23] (pp. 71-74) discusses some results in Rio’s dissertation. For results more specific to viewing the convex hull as a monotone closure operator, see How to characterize the convex hull/closure operator, Bennett [4], Jamison [34], Koenen [39], and van de Vel [66]. (4) The deductive closure of a set of well formed formulas (i.e. the consequence operator) in systems of logic. For more about this example, see Properties of the deductive closure, Brown/Suszko [7], Malinowski [43], Martin/Pollard [45], Pohlers [52], and Wójcicki [68]. (5) sigma-algebras Let $Y$ be a set, and let $X = {\mathcal P}(Y),$ and define $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ by $$ T(\mathcal C) \; = \; \left\{\bigcup_{n=1}^{\infty}A_n:\; A_n \in \mathcal C \; \text{for each} \; n \right\} \; \cup \; \{Y-A: \; A \in \mathcal C\} \; \cup \; \{\emptyset, \, Y\}. $$ Then given a collection $\mathcal C$ of subsets of $Y,$ we have that $E_*$ is the $\sigma$-algebra generated by ${\mathcal C}.$ The following sentence in Rayburn [53] (middle of p. 756) caught my interest, but I don’t know whether what he does in this paper is of much significance here: “By analogy with topology, we ask about operators from $P(X)$ to $P(X)$ which generate $\sigma$-algebras.” (6) Cantor-Bernstein Theorem Let $f:X \rightarrow Y$ and $g:Y \rightarrow X$ be injective functions and define $\;T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)\;$ by $\;T(A) = X - g[Y - f[A]].$ (Here, $h[B]$ denotes the image of the set $B$ under the function $h.)$ Then $T$ is a monotone closure operator on $X,$ and $E_{*}$ can be used to give a proof of the Cantor-Bernstin theorem for cardinal numbers. Indeed, $T$ is continuous, and hence the iterative method produces $E_{*}$ as the countable union $\;\bigcup_{n=1}^{\infty}T^n(\emptyset),\;$ which can be identified with one of the iterative methods of proving the Cantor-Bernstein Theorem. This example originated with Knaster [38] (joint work with Tarski, presented 9 December 1927), which also probably gives the first published statement that a monotone closure operator has a fixed point. See also Wikipedia: Knaster–Tarski theorem, Mathworld: Tarski's Fixed Point Theorem, Devlin [16] (pp. 77-78), Hinkis [31] (Chapters 31 and 35, on pp. 317-322 and 343-355), Hrbacek/Jech [33] (Exercises 1.10−1.14 on p. 69, where the continuity of $T$ and its implications are also discussed), Mendelson [46] (Appendix D: A Lattice-Theoretic Proof of the Schröder-Bernstein Theorem on p. 200), Sierpiński [59] (pp. 145-147), Tarski [63] (footnotes on pp. 285, 286), and Willard [67] (Problem 1J on p. 15). Regarding the above applications of monotone closure operators, see also Arturo Magidin’s 28 July 2011 essay. SELECTED REFERENCES ANSWER CONTINUES HERE REPLY [5 votes]: THIS CONTINUES MY ANSWER THAT BEGINS HERE SELECTED REFERENCES 4. Union-Preserving Closure Operators $\;$ [satisfy (K1)] Definition: We say that $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ is a union-preserving closure operator on $X$ if $T$ is a monotone closure operator on $X$ such that for each $A,\,B \in {\mathcal P}(X)$ we have $T(A \cup B) = T(A) \cup T(B).$ The term additive operator is often used, but I thought “additive” creates too much mental interference with unintended meanings from other areas of mathematics. Also, closure spaces arising from closure operators that are only required to satisfy (K1) do not seem to be of enough interest for the use of “union-preserving” here to cause problems. An example of a monotone closure operator that is not a union-preserving closure operator is the interior operator on $X = {\mathbb R}.$ Note that the interior operator satisfies (semi-K1) $-$ and also (K2) and (K4), for those keeping track $-$ but not (K2) (consider $A = \mathbb Q$ and $B = {\mathbb R} – {\mathbb Q}).$ I do not know of a naturally occurring example of a union-preserving closure operator that doesn’t also satisfy at least one of the remaining axioms for a topological operator, let alone an example from which useful insight is provided by considering it in the context of a closure operator. I also do not know how to obtain a “minimally differing” union-preserving closure operator from a monotone closure operator that doesn't explicitly make use of existing union-preserving closure operators, but the following at least shows that there exists a greatest union-preserving closure operator that is less than or equal to a given monotone closure operator. In fact, the same method works for an unrestricted closure operator $-$ see Hammer [23] (definition of $f_6$ at the bottom of p. 70). Union-Preserving Modification of a Monotone Closure Operator: Given a monotone closure operator $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X),$ the greatest union-preserving closure operator $T^{\text {u-p}}$ on $X$ that is less than or equal to $T$ can be defined by letting $$ T^{\text {u-p}}(A) \; = \; \bigcup\,\{T'(A): \; T' \leq T \; {\text {and}} \; T' \; {\text {is a union-preserving closure operator on}} \; X \} $$ Regarding the assumption (K1), the following may be of interest: (from Hammer [23], top of p. 65) The classical topologists have persisted in requiring additivity of the closure function in the definition of topological spaces. From one standpoint the additivity axiom might be called the sterility axiom. That is to say, it requires that two sets cannot produce anything (a limit point) by union that one of them alone cannot produce. On the other hand, of course, as with the analogous independence of events in probability theory, the known presence of additivity produces special benefits. 5. Base Closure Operators $\;$ [satisfy (K1), (K2)] Definition: We say that $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ is a base closure operator on $X$ if $T$ is a union-preserving closure operator such that $T(\emptyset) = \emptyset.$ Sometimes (but not here) the requirement $T(X) = X$ is included (e.g. see bottom of p. 4700 of Bliedtner/Loeb [6]). I believe the name originated in Lukeš/Malý/Zajíček [42], but I don’t know why they used the specific word “base”. A rather trivial example of a union-preserving closure operator that is not a base closure operator is given by $T(A) = X$ for all $A \in {\mathcal P}(X),$ with $X \neq \emptyset.$ Most of the results I give here concerning base closure operators, and many other results not included here, can be found in Chapter 1: Base Operator Spaces (pp. 5-37) of Lukeš/Malý/Zajíček [42]. Base Modification of a Union-Preserving Closure Operator: Given a union-preserving closure operator $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X),$ we can obtain a base closure operator $T^{\text {base}}$ on $X$ by redefining $T$ to have the value $\emptyset$ at $A = \emptyset:$ $$ T^{\text {base}}(A) \; \triangleq \; \left\{\begin{array}{c} \emptyset & \text{if} & A = \emptyset \\ T(A) & \text{if} & A \neq \emptyset \end{array} \right. $$ Note that $T^{\text {base}}$ is the greatest base closure operator less than or equal to $T.$ The prototypical example of a base closure operator is the derived set operator of a topological space: Example (derived set operator): Let $(X,\tau)$ be a topological space and define $T$ by $$ T(A) \; = \; \{x \in X: \; A \cap (U - \{x\}) \neq \emptyset \;\; \text{for each} \; \tau \text{-open set} \; U \; \text{containing} \; x\}. $$ Then $T$ is a base closure operator on $X.$ By replacing “$A \cap (U - \{x\}) \neq \emptyset$” with “$A \cap U$ is not small” for various notions of “not small” (i.e. by using various stronger notions of being a limit point of a set), we obtain many other examples of base closure operators. For example, in the appropriate spaces we could require $A \cap U$ to not be finite, not be countable, not be a first Baire category set, not have Lebesgue measure zero, etc. In fact, the motivation in Lukeš/Malý/Zajíček [42] for considering the abstract notion of a base closure operator was to generalize all the various notions in potential theory for a set to be thin/not-thin at a point. Indeed, if $T$ is a base closure operator on $X$ and $x \in X,$ then $\{A \in {\mathcal P}(X): \; x \notin T(A) \}$ is an ideal of sets that we could call the $T$-thin sets at $x.$ Moreover, if $X$ is a set and if for each $x \in X$ we specify a nonempty ideal ${\mathcal I}_x$ of subsets of $X,$ then there exists a base closure operator on $X$ such that, for each $x \in X,$ we have ${\mathcal I}_x$ equal to the collection of $T$-thin sets at $x.$ Regarding the last two sentences, see Lukeš/Malý/Zajíček [42] (1.A.2 on p. 10 and 1.A.17 on p. 16). Definition ($T$-topology): Let $T$ be a base closure operator on $X$ and let $C \in {\mathcal P}(X).$ We say that $C$ is a $T$-closed set if $T(C) \subseteq C.$ It is not difficult to show, for a fixed base closure operator $T,$ that the collection of $T$-closed sets satisfies the axioms to be the closed sets for some topology ${\tau}^T$ on $X.$ Of course, the corresponding topological closure operator, which I will denote by ${\tau}^T{\text {-}}{\text {Cl}},$ might not be equal to $T,$ but we always have $\;T \leq {\tau}^T{\text {-}}{\text {Cl}}.$ [[ Proof of this last inequality: Choose $A \in {\mathcal P}(X)$ and let $C$ be any $T$-closed set containing $A.$ Since $A \subseteq C,$ we have $T(A) \subseteq T(C);$ and since $C$ is $T$-closed, we have $T(C) \subseteq C.$ Therefore, $T(A)$ is a subset of any such set $C,$ and hence $T(A)$ is a subset of the intersection of all such sets $C.$ Thus, for each $A \in {\mathcal P}(X)$ we have shown that $T(A) \subseteq {\tau}^T{\text {-}}{\text {Cl}}(A).$ ]] If $(X,\tau)$ is a topological space, then of course the topological closure operator ${\tau}{\text {-}}{\text {Cl}}$ for $(X,\tau)$ is a base closure operator on $X.$ Moreover, as is easy to check and what one would expect, the topological closure operator corresponding to ${\tau}{\text {-}}{\text {Cl}}$ (regarding ${\tau}{\text {-}}{\text {Cl}}$ as a base closure operator) is equal to ${\tau}{\text {-}}{\text {Cl}}.$ That is, we have ${\tau}^{({\tau}{\text {-}}{\text {Cl}})}{\text {-}}{\text {Cl}} = {\tau}{\text {-}}{\text {Cl}}.$ The topologies that arise in this way in potential theory $-$ the topologies ${\tau}^T$ where $T$ is a base closure operator, defined using a limit point notion in which a set is NOT $T$-thin at a point $x$ $-$ are called fine topologies (Wikipedia page and google search) because, in Euclidean spaces, these topologies are finer (i.e. have more open sets) than the usual Euclidean topology. [Note that a more stringent condition to be a limit point causes fewer points to be added to the set to get its topological closure, hence the topological closure is smaller, hence more closed sets are needed to obtain (via intersection) this smaller topological closure, hence there are more open sets. (There is probably a less circuitous way of seeing this.)] An alternative way that fine topologies like this arise in potential theory is that it is useful to consider topologies on ${\mathbb R}^n$ (or more general spaces) such that, for various collections of functions from ${\mathbb R}^n$ to $\mathbb R$ that are larger than the collection of ordinary continuous functions, all functions in the larger collection will be continuous. This is accomplished by using finer topologies on ${\mathbb R}^n$ and the usual topology on ${\mathbb R}.$ 6. Čech Closure Operators $\;$ [satisfy (K1), (K2), (K3)] Definition: We say that $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ is a Čech closure operator on $X$ if $T$ is a base closure operator on $X$ such that for each $A \in {\mathcal P}(X)$ we have $A \subseteq T(A).$ Note that a Čech closure operator differs from a topological closure operator in that (K4), which says that $T = T^{2},$ is weakened to $T \leq T^{2}.$ Thus, if a Čech closure operator $T$ satisfies $T^2 \leq T,$ then $T$ is a topological closure operator. The corresponding generalized topological spaces are also called pretopological spaces. I’m using Čech’s name because Eduard Čech (1893-1960) studied these generalized closure functions extensively in his massive 893 page book [9] (1966 English translation). Čech also published a survey study of these notions in 1937. For some historical details about this part of Čech’s work, see Koutsk [40] (especially pp. 108-109). For instance, Čech was motivated by the fact that the idempotency property of closure was not satisfied by many spaces of functions for various notions of convergence. “He supposed that this transition to the more general concept would create no difficulties for the participants of the seminar and therefore without any hesitation he went on with his lectures. However, there were some difficulties. The concept of closures of sets supposed by axiom (4) [= idempotency] had been meanwhile so deeply rooted in the considerations of some members that misunderstandings occurred too frequently.” (Koutsk [40], near bottom of p. 108) An example of a base closure operator that is not a Čech closure operator is the ordinary derived set operator $D:{\mathcal P}(\mathbb {R}) \rightarrow {\mathcal P}(\mathbb {R}),$ where for example $D(\mathbb Z) = \emptyset,$ and hence $A \subseteq D(A)$ is false for $A = {\mathbb Z}.$ Example (zero distance from a set): Let $X$ be a set and let $d:X \times X \rightarrow [0,\infty)$ satisfy $d(x,x) = 0$ and $d(x,y) = d(y,x)$ for all $x,\,y \in X.$ Thus, $(X,d)$ is what we get if we begin with a metric space and drop the triangle inequality and drop the assumption that $d(x,y) = 0$ implies $x = y.$ Čech [9] calls this a semi-pseudometric space. Then $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ is a Čech closure operator, where $T$ is defined by $$ T(A) \; = \; \left\{x \in X: \; \inf_{a \in A} d(x,a) \leq 0 \right\}. $$ The reason for using $\leq 0$ rather than $= 0$ is to accommodate the convention that $\inf \emptyset = -\infty.$ Čech [9] (p. 302) observes that if ${\lambda}^{*}$ and ${\lambda}_{*}$ are the Lebesgue outer and inner measures on $[0,1],$ then $d^{*}(A,B) = {\lambda}^{*}(A\,\triangle\,B)$ defines a pseudometric on ${\mathcal P}([0,1])$ and $d_{*}(A,B) = {\lambda}_{*}(A\,\triangle\,B)$ defines a semi-pseudometric on ${\mathcal P}([0,1])$ that is not a pseudometric on ${\mathcal P}([0,1]).$ $(A\,\triangle\,B$ is the symmetric difference of $A$ and $B.)$ Čech Modification of a Base Closure Operator: Given a base closure operator $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X),$ we can obtain a Čech closure operator $T^{\text {Čech}}$ on $X$ by letting $T^{\text {Čech}}(A) \triangleq A \cup T(A)$ for each $A \in {\mathcal P}(X).$ It is not difficult to show that $T^{\text {Čech}}$ is the smallest Čech closure operator greater than or equal to $T.$ Since every Čech closure operator is a base closure operator, the same notion of a $T$-closed set continues to apply when $T$ is a Čech closure operator and, for a fixed Čech closure operator $T,$ the collection of $T$-closed sets satisfies the axioms to be the closed sets for some topology ${\tau}^T$ on $X.$ Thus, repeating what I’ve already said in the more general case of base closure operators, for any Čech closure operator $T$ we have $\;T \leq {\tau}^T{\text {-}}{\text {Cl}},$ where ${\tau}^T{\text {-}}{\text {Cl}}$ is the topological closure operator for the topological space $(X,{\tau}^T).$ Now suppose we begin with a base closure operator $T.$ From $T$ we obtain the Čech modification $T^{\text {Čech}},$ and from $T^{\text {Čech}}$ we obtain the collection of $T^{\text {Čech}}$-closed sets, and from the collection of $T^{\text {Čech}}$-closed sets we obtain the topological closure operator ${\tau}^{T^{\text {Čech}}}{\text {-}}{\text {Cl}}$ for the topological space $(X,{\tau}^{T^{\text {Čech}}}).$ Since the collection of $T$-closed sets is identical to the collection of $T^{\text {Čech}}$-closed sets (proof in a moment), it follows that $(X,{\tau}^T) = (X,{\tau}^{T^{\text {Čech}}}).$ [[ Proof: $(\subseteq)$ Let $A \subseteq X$ be $T$-closed. Then $T(A) \subseteq A.$ Hence, $T^{\text {Čech}}(A) \triangleq A \cup T(A) \subseteq A,$ which shows that $A$ is $T^{\text {Čech}}$-closed. $(\supseteq)$ Now let $A \subseteq X$ be $T^{\text {Čech}}$-closed. Then $T^{\text {Čech}}(A) \subseteq A,$ and hence $A \cup T(A) \subseteq A,$ which implies $T(A) \subseteq A,$ so $A$ is $T$-closed. ]] Again, suppose $T$ is a base closure operator. To summarize the results above, we have $$ T \; \leq \; T^{\text {Čech}} \; \leq \; {\tau}^T{\text {-}}{\text {Cl}} $$ where $T^{\text {Čech}}$ is the Čech modification of $T$ and ${\tau}^T{\text {-}}{\text {Cl}}$ is the identical topological closure operator associated with both $T$ and $T^{\text {Čech}}.$ Lukeš/Malý/Zajíček [42] use the term strong base operator (defined on middle of p. 8 and used frequently thereafter) for a base closure operator $T$ such that $T^{\text {Čech}} = {\tau}^T{\text {-}}{\text {Cl}}$ (the main interest being, of course, when $T \neq T^{\text {Čech}}),$ which they show is equivalent (Theorem 1.2 at the bottom of p. 7) to the property $T^2 \leq T.$ [[ To prevent possible confusion, if $T$ were a Čech closure operator, then as we mentioned at the beginning of this section, the inequality $T^2 \leq T$ would imply that $T$ is a topological closure operator, and hence we would have $T = {\tau}^T{\text {-}}{\text {Cl}}.$ However, the $T$ that we're considering here is only assumed to be a base closure operator, and it happens to be the case that we can have both $T^2 \leq T$ and $T \neq {\tau}^T{\text {-}}{\text {Cl}}.$ ]] Moreover, if $T$ is a base closure operator, then $T^{\text {*base}}:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ defined by $T^{\text {*base}}(A) = {\tau}^T{\text {-}}{\text {Cl}}(T(A))$ is the smallest strong base closure operator greater than $T$ (see top half of p. 9), and thus we can consider $T^{\text {*base}}$ to be the strong base modification of the base closure operator $T.$ Finally, at least for here (see [42] for still more such results), 1.A.8 at the bottom of p. 12 states that for a base closure operator $T,$ the operators $T$ and ${\tau}^T{\text {-}}{\text {Cl}}$ commute $-$ that is, we have $T \circ ({\tau}^T{\text {-}}{\text {Cl}}) = ({\tau}^T{\text {-}}{\text {Cl}}) \circ T.$ Various results about Čech closure operators can be found in many of the references below, but probably the most helpful along with the specific citations above are Čech [9] (especially pp. 237-241, 250-254, 272-275, 555-564, 832-833, 861-863), Kannan [35] (pp. 42-43), Roth [55], Roth/Carlson [56], and Thron [65]. Dolecki/Greco [17] study, among other things, equivalence classes of Čech closure operators that generate the same topological closure operator. 7. Topological Closure Operators $\;$ [satisfy (K1), (K2), (K3), (K4)] Definition: We say that $T:{\mathcal P}(X) \rightarrow {\mathcal P}(X)$ is a topological closure operator on $X$ if $T$ is a Čech closure operator on $X$ such that for each $A \in {\mathcal P}(X)$ we have $T(T(A)) = T(A).$ Recall that if $T$ is a Čech closure operator, then $T \leq T^{2}.$ Since the composition of two Čech closure operators is a Čech closure operator, it follows that if $T$ is a Čech closure operator, then we have $$ T \; \leq \; T^2 \; \leq \; T^3 \; \leq \; T^4 \; \leq \; \cdots, $$ where $\;T^2 = T \circ T,\;$ $\;T^3 = T \circ T \circ T,\;$ etc. are the finite composition iterates of $T.$ This is exactly the same situation we had earlier when we were iterating a monotone closure operator. In this case, each Čech closure operator can be transfinitely iterated to obtain a topological closure operator. Example (Baire function hierarchy): Let $X = {\mathbb R}^{\mathbb R}$ be the set of functions from $\mathbb R$ to ${\mathbb R}.$ Given ${\mathcal A} \in {\mathcal P}(X)\;$ (thus, $\mathcal A$ is a set of functions from $\mathbb R$ to ${\mathbb R}),\;$ let $T(\mathcal A) = \{f \in X: \; f \; {\text {is a pointwise limit of a sequence of functions in}} \; \mathcal A \}.$ Then $T$ is a Čech closure operator on $X.$ Moreover, if $\mathcal C$ is the set of continuous functions, then $T(\mathcal C)$ is the set of Baire $1$ functions, $T^2(\mathcal C)$ is the set of Baire $2$ functions, etc. In this case $T$ has to be iterated ${\omega}_1$ many times before we get a topology (the topology of pointwise convergence), and ${\tau}^T{\text {-}}{\text {Cl}}(\mathcal C)$ $-$ the corresponding topological closure operator applied to the set of continuous functions $-$ is equal to the set of Borel measurable functions. See this Mathematics Stack Exchange answer by Andrés E. Caicedo for more details. Topological Modification of a Čech Closure Operator: Let $T$ be a Čech closure operator on $X.$ For each ordinal $\alpha \geq 1$ we define $T^{\alpha}$ by transfinite induction: $$\begin{array}{l} T^1 & = & T \\ T^{\alpha + 1} & = & T \circ T^{\alpha} \\ T^{\lambda} & = & \sup \,\{T^{\gamma}: \; \gamma < \lambda \}, \;\; {\text{if}} \; \lambda \; {\text {is a nonzero limit ordinal}} \end{array}$$ The following results hold: (a) $\;\alpha \leq \beta \;$ implies $\;T^{\alpha} \leq T^{\beta}$ (b) $\;$For each ordinal $\alpha \geq 1,$ we have $T^{\alpha}$ is a Čech closure operator on $X.$ (c) $\;$For each ordinal $\alpha \geq 1,$ we have $\;{\tau}^{T^{\alpha}}{\text {-}}{\text {Cl}} = {\tau}^T{\text {-}}{\text {Cl}}.\;$ That is, all the iterates of $T$ have the same corresponding topological closure operator. (d) $\;$There exists an ordinal $\lambda \leq |X|^{+}$ such that $T^{\alpha} = T^{\lambda}$ for all $\alpha \geq \lambda.$ (e) $\;$Let $|T|$ be the least of the ordinals $\lambda$ in (d) above. If $\;1 \leq \alpha < \beta \leq |T|,\;$ then $\;T^{\alpha} < T^{\beta}.$ (f) $\;$If $\alpha \geq |T|,\;$ then $\;T^{\alpha} = {\tau}^T{\text {-}}{\text {Cl}}.$ (g) $\;$Let $(X,\tau)$ be a topological space and let ${\tau}{\text {-}}{\text {Cl}}$ be the topological closure operator for $(X,\tau).$ Then ${\tau}{\text {-}}{\text {Cl}} = \sup {\mathcal C},$ where $\mathcal C$ is the collection of all Čech closure operators $T$ such that ${\tau}^T{\text {-}}{\text {Cl}} = {\tau}{\text {-}}{\text {Cl}}.$ Regarding (g), note that each of the iterates $T^{\alpha}$ belongs to ${\mathcal C},$ but since there might be other Čech closure operators, besides these, that have ${\tau}{\text {-}}{\text {Cl}}$ for their corresponding topological closure operator, (g) is not automatic from (c) alone. In fact, this possibility can actually arise, and examples can be found in Dolecki/Greco [17]. Various aspects of the above results can be found in Čech [9] (pp. 274-275, 836), Hammer [23], Hausdorff [30], Kannan [35] (pp. 42-43), Kent/Richardson [37], Lukeš/Malý/Zajíček [42] (Exercise 1.A.10 on p. 13), and Novák [51]. SELECTED REFERENCES REPLY [5 votes]: 8. Selected References The references below are mostly restricted to those that specifically involve, entirely or partly, various generalized closure operator notions. Thus, the literature on generalized topological notions dealing with the vast zoo of various semi-open set notions, various types of separation axioms in generalized spaces, various generalized continuity notions, category-theoretic connections, etc. are not included unless I thought it had specific relevance to something discussed here. [1] Peter Henry George Aczel, An introduction to inductive definitions, pp. 739-782 in Jon Kenneth Barwise (editor), Handbook of Mathematical Logic, North-Holland, 1977, xi + 1165 pages. [2] Jon Kenneth Barwise, Admissible Sets and Structures. An Approach to Definability Theory, Perspectives in Mathematical Logic #13, Springer-Verlag, 1975, xiii + 394 pages. [3] Jon Kenneth Barwise and Lawrence Stuart Moss, Vicious Circles. On the Mathematics of Non-Wellfounded Phenomena, CSLI Lecture Notes #60, 1996, x + 390 pages. [4] Mary Katherine Bennett, Convexity closure operators, Algebra Universalis 10 #3 (1980), 345-354. [5] Garrett Birkhoff, Lattice Theory, corrected reprint of 3rd edition, American Mathematical Society Colloquium Publications #25, American Mathematical Society, 1979, vi + 418 pages. [6] Jürgen Bliedtner and Peter Albert Loeb, The optimal differentiation basis and liftings of $L^{\infty}$, Transactions of the American Mathematical Society 352 #10 (October 2000), 4693-4710. [7] Donald Jerome Brown and Roman Suszko, Abstract Logics, Dissertationes Mathematicae 102 (1973), 9-41. [8] Stanley Neal Burris, Theory of Pre-Closures, Ph.D. dissertation (under Allen Seymour Davis), University of Oklahoma, 1968, iv + 47 pages. This is a study of closure operators that satisfy (semi-K1) and (K3). [9] Eduard Čech, Topological Spaces, revised and translated by Zdeněk Frolík and Miroslav Katětov, John Wiley and Sons, 1966, 893 pages. Čech published a survey of his Čech closure operator work in 1937: Topologické prostory, Časopis pro Pěstování Matematiky a Fysiky 66 #4 (1937), D225-D264. [10] Ákos Császár, Generalized open sets, Acta Mathematica Hungarica 75 #1-2 (April 1997), 65-87. In this paper and the next paper, Császár studies various topological ideas in the context of a generalized topological space defined by a monotone closure operator, although for many of the results he (explicitly) assumes additional hypotheses, as needed. [11] Ákos Császár, On the $\gamma$-interior and* $\gamma$-closure of a set*, Acta Mathematica Hungarica 80 #1-2 (July 1998), 89-93. [12] Sterling Gene Crossley, Semi-Topological Properties and Related Topics, Ph.D. dissertation (under Shelby Keith Hildebrand), Texas Technological College [after 1968: Texas Tech University], 1968, iii + 76 pages. [13] Dirk van Dalen, Hans Cornelis Doets, and Henricus Cornelius Maria de Swart, Sets: Naïve, Axiomatic and Applied, International Series in Pure and Applied Mathematics #106, Pergamon Press, 1978, xviii + 339 pages. [14] Mahlon Marsh Day, Convergence, closure and neighborhoods, Duke Mathematical Journal 11 #1 (March 1944), 181-199. [15] Vladimir Devidé, On monotone mapings [sic] of the power-set, Portugaliae Mathematica 21 #2 (1962), 111-112. [16] Keith James Devlin, The Joy of Sets. Fundamentals of Contemporary Set Theory, 2nd edition, Undergraduate Texts in Mathematics, Springer-Verlag, 1993, x + 192 pages. [17] Szymon Dolecki and Gabriele H. Greco, Topologically maximal pretopologies, Studia Mathematica 77 #3 (1984), 265-281. [18] Herbert Bruce Enderton, Elements of Set Theory, Academic Press, 1977, xiv + 279 pages. [19] Preston Clarence Hammer, General topology, symmetry, and convexity, Transactions of the Wisconsin Academy of Sciences, Arts and Letters 44 (1955), 221-255. [20] Preston Clarence Hammer, Kuratowski’s closure theorem, Nieuw Archief voor Wiskunde (3) 8 (1960), 74-80. [21] Preston Clarence Hammer, Extended topology: reduction of limit functions, Nieuw Archief voor Wiskunde (3) 9 #1 (1961), 16-24. [22] Preston Clarence Hammer, Semispaces and the topology of convexity, pp. 305-316 in Victor LaRue Klee (editor), Convexity, Proceedings of Symposia in Pure Mathematics 7, American Mathematical Society, 1963. [23] Preston Clarence Hammer, Extended topology: set-valued set-functions, Nieuw Archief voor Wiskunde (3) 10 #? (1962), 55-77. [24] Preston Clarence Hammer, Extended topology: additive and subadditive subfunctions of a function, Rendiconti del Circolo Matematico di Palermo (2) 11 #3 (1962), 262-270. [25] Preston Clarence Hammer, Extended topology: perfect sets, Portugaliae Mathematica 23 #1 (1964), 27-34. [26] Preston Clarence Hammer, Extended topology: the continuity concept, Mathematics Magazine 36 #2 (March 1963), 101-105. [27] Preston Clarence Hammer, Extended topology: Continuity I, Portugaliae Mathematica 23 #2 (1964), 77-93. [28] Preston Clarence Hammer, Extended topology: connected sets and Wallace separations, Portugaliae Mathematica 22 #4 (1963), 167-187. [29] Preston Clarence Hammer, Extended topology: structure of isotonic functions, Journal für die reine und angewandte Mathematik 213 #3-4 (1964), 174-186. [30] Felix Hausdorff, Gestufte räume, Fundamenta Mathematicae 25 (1935), 486-502. [31] Arie Hinkis, Proofs of the Cantor-Bernstein Theorem. A Mathematical Excursion, Science Networks / Historical Studies #45, Birkhäuser, 2013, xxiv + 429 pages. [32] Peter Greayer Hinman, Recursion-Theoretic Hierarchies, Perspectives in Mathematical Logic #7, Springer-Verlag, 1978, xii + 480 pages. [33] Karel Hrbacek and Thomas J. Jech, Introduction to Set Theory, 3rd edition, Pure and Applied Mathematics #220, Marcel Dekker, 1999, xii + 291 pages. [34] Robert Edward Jamison, A General Theory of Convexity, Ph.D. dissertation (under Victor LaRue Klee), University of Washington, 1974, v + 120 pages. [35] Varadachariar Kannan, Ordinal Invariants in Topology, Memoirs of the American Mathematical Society 32 #245 (July 1981), vi + 164 pages. [36] T. Kavitha, Some Problems on Čech Closure Spaces, Ph.D. dissertation (under P. T. Ramachandran), University of Calicut (Kerala, India), August 2017, viii + 139 pages. Generalized Čech closure operators $-$ closure operators that satisfy (semi-K1), (K2), (K3) $-$ are studied in Sections 5.4-5.4 on pp. 120-131. [37] Darrell Conley Kent and Gary Douglas Richardson, The decomposition series of a convergence space, Czechoslovak Mathematical Journal 23 #3 (1973), 437-446. [38] Bronisaw Knaster, Un théorème sur les fonctions d’ensembles [A theorem on functions of sets], Annales de la Société Polonaise de Mathématique [= Rocznik Polskiego Towarzystwa Matematycznego] 6 (1927), 133-134. This is not a published paper, but rather an abstract from the section titled Comptes-rendus des séances de la Société Polonaise de Mathématique Section de Varsovie [Reports of the sessions of the Warsaw Section of the Polish Mathematics Society] (session taking place 9 December 1927). The results were obtained jointly with Tarski and presented by Knaster. [39] William B. Koenen, The Kuratowski closure problem in the topology of convexity, American Mathematical Monthly 73 #7 (August-September 1966), 704-708. [40] Karel Koutský, Čech's topological seminar in Brno, 1936-1939, Časopis pro Pěstování Matematiky 90 #1 (1965), 104-117. [41] Yinbin Lei and Jun Zhang, Generalizing topological set operators, Electronic Notes in Theoretical Computer Science 345 (28 August 2019), 63-76. [42] Jaroslav Lukeš, Jan Malý, and Luděk Zajíček, Fine Topology Methods in Real Analysis and Potential Theory, Lecture Notes in Mathematics #1189, Springer-Verlag, 1986, x + 472 pages. [43] Jacek Malinowski, On generalizations of consequence operation, Bulletin of the Section of Logic (University of Łódź) 31 #2 (2002), 135-143. [44] Zlatko P. Mamuzić, Introduction to General Topology, translated by William Joseph Pervin and James Leo Sieber and Robert C. Moore, P. Noordhoff (Netherlands), 1963, 159 pages. [45] Norman Marshall Martin and Stephen Randall Pollard, Closure Spaces and Logic, Mathematics and Its Applications #369, Kluwer Academic Publishers, 1996, xviii + 230 pages. The closure spaces of this book satisfy (semi-K1), (K3), and $T^2 \leq T.$ That is, from the topological closure operator axioms we drop half of (K1), (K2), and half of (K4). [46] Elliott Mendelson, Schaums Outline of Theory and Problems of Boolean Algebra and Switching Circuits, Schaums Outline Series, McGraw-Hill Book Company, 1970, viii + 213 pages. [47] Eliakim Hastings Moore, Introduction to a Form of General Analysis, pp. vii-viii + 11-50 in American Mathematical Society Colloquium Publications #2, Yale University Press, 1910, x + 222 pages. Moore’s extensionally attainable property (defined on middle of p. 59) is often cited as an early, and perhaps the earliest, formulation of a general closure operator notion. Supposedly it is equivalent to a closure operator $T$ on a set that satisfies (semi-K1), (K3), and $T^2 \leq T$ (e.g. pp. 16-17 of Martin/Pollard [45]), but I have not attempted to verify this or consider whether such claims might be somewhat ahistorical. For a historical study of Moore’s General Analysis program, see Reinhard Siegmund-Schultze, Eliakim Hastings Moore’s “General Analysis”, Archive for History of Exact Sciences 52 #1 (January 1998), 51-89. [48] Yiannis Nicholas Moschovakis, Elementary Induction on Abstract Structures, Studies in Logic and the Foundations of Mathematics #77, North-Holland, 1974, x + 218 pages. [49] Yiannis Nicholas Moschovakis, Descriptive Set Theory, Studies in Logic and the Foundations of Mathematics #100, North-Holland, 1980, xii + 637 pages. [50] Yiannis Nicholas Moschovakis, Notes on Set Theory, 2nd edition, Undergraduate Texts in Mathematics, Springer-Verlag, 2006, xii + 276 pages. [51] Josef Novák, On some problems concerning multivalued convergences, Czechoslovak Mathematical Journal 14 #4 (1964), 548-561. [52] Horst Wolfram Pohlers, Proof Theory, Lecture Notes in Mathematics #1407, Springer-Verlag, 1989, viii + 213 pages. [53] Marlon Cecil Rayburn, On the lattice of $\sigma$-algebras, Canadian Journal of Mathematics 21 (1969), 755-761. [54] Sheldon Theodore Rio, On the Hammer Topological System, Ph.D. dissertation (under Henry Arnold Bradford), Oregon State College [after 1961: Oregon State University], June 1959, iv + 53 pages. [55] David Niel Roth, Čech Closure Spaces, Master of Arts thesis (under John Warnock Carlson), Emporia State University (Emporia, Kansas), July 1979, iv + 25 pages. [56] David Niel Roth and John Warnock Carlson, Čech closure spaces, Kyungpook Mathematical Journal 20 #1 (June 1980), 11-30. [57] Marcin Jan Schroeder, Modification of pre-closure spaces as closure/interior operations on the lattice of pre-closure operators, RIMS Kôkyûroku #1712 (Algebras, Languages, Algorithms in Algebraic Systems and Computations), Research Institute for Mathematical Sciences (Kyoto University, Japan), September 2010, 148-155. [58] Gideon Ernst Schwarz, A note on transfinite iteration, Journal of Symbolic Logic 21 #3 (September 1956), 265-266. See the half-page review by Dana Stewart Scott in Journal of Symbolic Logic 22 #3 (September 1957), p. 303. [59] Wacław Franciszek Sierpiński, Algèbra des Ensembles [Algebra of Sets], Monograe Matematyczne #23, Nakadem Polskiego Towarzystwa Matematycznego (Warszawa-Wrocaw), 1951, ii + 205 pages. [60] Wacław Franciszek Sierpiński, General Topology, translated by Cypra Cecilia Krieger, 2nd edition, Mathematical Expositions #7, University of Toronto Press, 1952, xii + 290 pages. [61] Bärbel M. R. Stadler and Peter Florian Stadler, Generalized topological spaces in evolutionary theory and combinatorial chemistry, Journal of Chemical Information and Computer Sciences 42 #3 (May 2002), 577-585. [62] Bärbel M. R. Stadler and Peter Florian Stadler, Basic properties of closure spaces, unpublished manuscript, 29 January 2002 (web release date), 20 pages. [63] Alfred Tarski, A lattice-theoretical fixpoint theorem and its applications, Pacific Journal of Mathematics 5 #2 (June 1955), 285-309. The main results were announced in A fixpoint theorem for lattices and its applications (abstract #496, received 27 June 1949), Bulletin of the American Mathematical Society 55 #11 (November 1949), 1051-1052. [64] Wolfgang Joseph Thron, Topological Structures, Holt, Rinehart and Winston, 1966, x + 240 pages. [65] Wolfgang Joseph Thron, What results are valid on closure spaces, Topology Proceedings 6 (1981), 135-158. [66] Marcel Lodewijk Johanna van de Vel, Theory of Convex Structures, North-Holland Mathematical Library #50, Elsevier Science Publishers B.V., 1993, xv + 540 pages. [67] Stephen W. Willard, General Topology, Addison-Wesley, 1970, xii + 369 pages. [68] Ryszard Wójcicki, Theory of Logical Calculi. Basic Theory of Consequence Operations, Synthese Library: Studies in Epistemology, Logic, Methodology, and Philosophy of Science #199, Kluwer Academic Publishers, 1988, xviii + 473 pages. [69] Rafał Ryszard Zduńczyk, Simple systems and closure operators, Bulletin de la Société des Science et des Lettres de Łódź, Recherches sur les Déformations, 66 # 3 (2016), 65-72.<|endoftext|> TITLE: Is min exponents of three positive integers $n$, $n+1$ and $n+2$ $=1$ true or false? QUESTION [6 upvotes]: Given a positive integer $P>1$, let its prime factorization be written $$P=p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots p_k^{a_k}$$ Define the functions $h(P)$ by $h(1)=1$ and $h(P)=\min(a_1, a_2,\ldots,a_k)$ Is the follows property true or false? The property: Let $n$ is a positive integer then $\min(h(n), h(n+1), h(n+2)) = 1$ PS: The property was checked up to $n=5.10^7$ REPLY [15 votes]: It is actually an old conjecture of Erdős, Mollin, and Walsh that the pattern you have noticed does indeed go on forever, i.e., there are no three consecutive powerful numbers.<|endoftext|> TITLE: Translated version of a Caratheodory article QUESTION [5 upvotes]: This excellent introduction to Compressive Sensing cites a couple of (seemingly) interesting Caratheodory papers from 1907-1911. These are: [46] C. Caratheodory. Uber den Variabilitätsbereich der Koeffizienten von Potenzreihen, die gegebene Werte nicht annehmen. Math. Ann., 64:95–115, 1907. [47] C. Caratheodory. Uber den Variabilitätsbereich der Fourierschen Konstanten von positiven harmonischen Funktionen. Rend. Circ. Mat. Palermo, 32:193–217, 1911. These papers keep getting cited in various other articles on the subject as well but to date, I have not been able to find them. Would anyone know where I could find a preferably English translation? I am not too bothered if it could be found in a printed book of translated works either. Alternatively, it could be a different paper in English that covers the subject sufficiently. The papers show that if you have a (positive) sum of $k$ sinusoids, you can recover the mix completely by knowing the value of the sum at $t=0$ and any $2k$ time points. Even as a subset of Compressed Sensing problems, this is a really interesting proposition and I would like to have a closer look. REPLY [3 votes]: Google translate output of the first page of the 1907 paper, no postprocessing (other than LaTeXing the formulas). Over the range of variability of the coefficients of power series that do not assume given values. Introduction. Given an analytic function $ y $ of the complex variable z, which assumes the value $ y = A_0 $ for $ z = 0 $ and is regular in the neighborhood of this point, there are certain restrictions inside the circle $ z <\rho $ is subject to the question arises whether not at the same time for the coefficients of the functional element $$y=A_0+\sum_{k=1}^\infty A_k z^k,$$ which is the function of creating restrictions that can be determined. A special case of this kind of question occurs in the well-known generalization that E. Landau has given for Picard's theorem on whole transcendental functions.*) This sentence can be expressed as follows: If the function $ y $ for $ z = 0 $ assumes the value $ y = A_0 $, inside the unit circle is regular and leaves the values ​​zero and one, and if the real and imaginary part of the coefficient $ A_1 $ as the coordinates of a point of the planes, this point must lie inside a circle whose radius you can specify. *) During the printing of this work, Mr. Landau has published a detailed account of the questions in question (quarterly of the Naturforschenden Gesellschaft in Zurich vol. 51, p. 252). In particular, in par. 15 of this work is an analogous problem to an algebraic problem recurred procedure.<|endoftext|> TITLE: Is it possible to use Feynman diagrams to represent a dot product $a \cdot b$? QUESTION [5 upvotes]: Feynman diagrams are topological entities, but they describe linear operators It has been observed that Feynman diagrams are in particular string diagrams (morphisms in monoidal categories)) in a given category of representations I want to know if exist a way to represent or explain a dot product $a \cdot b$ using Feynman diagrams An inner product space ("scalar product") is a vector space $V$ equipped with a (conjugate)-symmetric bilinear form or sesquilinear form: a linear map from the tensor product $V \otimes V$ of $V$ with itself, or of $V$ with its dual module $\bar{V} \otimes V$ to the ground ring $k$. REPLY [6 votes]: In A Not-so-Characteristic Equation: the Art of Linear Algebra by Elisha Peterson, the dot product, cross product, and many other linear algebraic operations are described in terms of diagrams. There are also many good references in the paper for further exploration. See also this MO post: How can I learn about doing linear algebra with trace diagrams?<|endoftext|> TITLE: Guessing the number of other $1$'s in a binary sequence QUESTION [6 upvotes]: I have posed the following question on math.stackexchange.com but have not received an answer. So I would like to seek experts' opinion here. Consider the set of all binary sequence of length $n+1$, $B=\big\{(b_i)_{i=0}^n\,\big| b_i\in\{0,1\}, \forall i\big\}$. Construct a function $f: \{0,\cdots,n\}\times \{0,1\}\to \{0,\cdots, n\}$, such that $\forall (b_i)_{i=0}^n\in B,\,\exists i \colon f(i,b_i)=\sum_{j\ne i}b_j$. What is a systematic way to construct this function? Putting it more colloquially, we assign $n+1$ persons one-to-one to all the digits of an arbitrary binary sequence of length $n+1$. Each person can see but the digit assigned to him. Devise a strategy so that at least one person guesses correctly the sum of the remaining digits. REPLY [6 votes]: Since every person knows his assigned digit, the problem is equivalent to guessing the sum of all digits. Label the persons with $0,1,...,n$. Let person $0$ guess $0$ when given a $0$, and $n+1$ when given a $1$. Let person $k$ guess $k$ for every $k\in \{1,...,n\}$. At least one person can correctly guess the sum of all digits. So when they subtract their given digit from the guessed sum, at least one person can correctly guess the sum of the remaining digits.<|endoftext|> TITLE: Kinematic formula for Euler characteristic QUESTION [7 upvotes]: Is there a formula for $\int \chi(K \cap gL) \: dg$ (where $\chi$ is Euler characteristic) analogous to the kinematic formula for $\int \mu(K \cap gL) \: dg$ (where $\mu$ is Lebesgue measure)? In both expressions $K$ and $L$ are compact convex bodies, $g$ varies over a group of isometries acting on the ambient space, and $dg$ signifies integration with respect to the Haar measure of that group. REPLY [9 votes]: Yes, this is called the principal kinematic formula: $$\int \chi(K \cap gL)\, dg = \sum_{k=0}^n c_{nk} V_k(K) V_{n-k}(L),$$ where $V_i$ are the intrinsic volumes, and $c_{nk}$ certain constants. See e.g. Section 4.4 in Schneider, Rolf, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and Its Applications. 44. Cambridge: Cambridge University Press. xiii, 490 p. (1993). ZBL0798.52001. At the end of that section there are historical references. Note that if $L$ is a ball of radius $r$, then there is no dependence on the "rotational part" of $g$, so one integrates over translations only, and the formula reduces to the Steiner formula for the volume of an $r$-neighborhood of $K$.<|endoftext|> TITLE: Are the coefficients of certain product of Rogers-Ramanujan Continued Fraction non-negative? QUESTION [6 upvotes]: Let $$R(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^{2}}{1 + \cfrac{q^{3}}{1 + \cdots}}}}$$ The following equality is famous: $$\cfrac{q^{1/5}}{R(q)} = \prod_{k>0} \cfrac{(1-q^{5k-2})(1-q^{5k-3})}{(1-q^{5k-1})(1-q^{5k-4})} ( = f(q))$$ The coefficients of $f(q)$ can be positive or negative. In fact, $$f(q) = 1 + q - q^3 + q^5 + q^6 - q^7 - 2 q^8 + \cdots$$ Let $$g(q) = \prod_{k>0} f(q^k) = f(q)f(q^2)f(q^3) \cdots$$ $g(q)$ $= (1 + q - q^3 + q^5 + \cdots)(1 + q^2 - q^6 + q^{10} + \cdots)\cdots$ $= 1 + q + q^2 + q^3 + 2q^4 + 3q^5 + 3q^6 + 3q^7 + 4q^8 + 6q^9 + \cdots$ The coefficients of $g(q)$ seem non-negative. Are the coefficients of $g(q)$ non-negative? REPLY [11 votes]: Notice that we can write $$f(q)=\prod_{n\geq 1} (1-q^n)^{-\left(\frac{n}{5}\right)}$$ therefore $$g(q)=\prod_{k\geq 1} f(q^k)=\prod_{n\geq 1} (1-q^n)^{-a(n)}$$ where $a(n)=\sum_{d|n}\left(\frac{d}{5}\right)$, where $\left(\frac{d}{5}\right)$ is the Legendre symbol. Now, $a(n)$ is easily seen to be multiplicative with $a(5^k)=1$, $a(p^k)=k+1$ when $p\equiv \pm 1\pmod{5}$, and $a(p^k)=\frac{1+(-1)^k}{2}$ when $p\equiv \pm 2\pmod{5}$. This means that $a(n)\geq 0$ for all $n$, so $g(q)$ is a product of series with nonnegative coefficients, and thus has nonnegative coefficients itself (or even nondecreasing ones as suspected in the comments).<|endoftext|> TITLE: Anti-holomorphic involutions of a complex linear algebraic group QUESTION [6 upvotes]: Let $G$ be a connected linear algebraic group over the field of complex number ${\Bbb C}$. Let $G({\Bbb C})$ denote the complex Lie group of ${\Bbb C}$-points of $G$. Let $\sigma$ be an anti-holomorphic involution of $G({\Bbb C})$, that is, a an automorphism of the real Lie group $$\sigma\colon G({\Bbb C})\to G({\Bbb C})$$ such that $\sigma$ is anti-holomorphic and $\sigma^2={\rm id}$. The anti-holomorphic involution $\sigma$ naturally acts on the ring of holomorphic function on $G({\Bbb C})$: $$({}^\sigma\!\! f)(g)=\overline{f(\sigma^{-1}(g))},$$ where the bar denotes complex conjugation (and, of course, $\sigma^{-1}=\sigma$). We say that $\sigma$ as above is anti-regular, if, when acting on the ring of holomorphic functions on $G$, $\sigma$ preserves the subring of regular functions (recall that $G$ is an algebraic group). Question. Are all anti-holomorphic involutions anti-regular in the following cases: (1) $G$ is a connected linear algebraic group; (2) $G$ is a (connected) reductive algebraic group; (3) $G$ is a (connected) semisimple algebraic group? Remark. An anti-regular involution $\sigma$ of $G({\Bbb C})$ defines by Galois descent a real structure on $G$. Indeed, we may put $$ G_{\Bbb R}={\rm Spec}\,({\Bbb C}[G]^\sigma),$$ where ${\Bbb C}[G]^\sigma$ is the subring of fixed points of $\sigma$ in the ring of regular functions ${\Bbb C}[G]$ on $G$. Conversely, an algebraic group $G_{\Bbb R}$ over ${\Bbb R}$ defines a complex algebraic group $G:=G_{\Bbb R}\times_{\Bbb R} {\Bbb C}$, and the complex conjugation on ${\Bbb C}$ induces by functoriality an anti-regular involution $\sigma$ on $G({\Bbb C})$. REPLY [5 votes]: (1): No; (2,3): Yes (and also for unipotent groups). On the abelian group $\mathbb{G}_{\mathrm{a}}\times \mathbb{G}_{\mathrm{m}}=\mathbf{C}\times\mathbf{C}^*$, consider the anti-holomorphic involution $$(z,w)\mapsto (\bar{z},\exp(i\bar{z})\bar{w}):$$ it is not "anti-regular". In the semisimple case, it's the same. One can reduce to $G$ simply connected, and in this case, the group of holomorphic automorphisms corresponds to the group of automorphisms of the Lie algebra, and this is the same as the algebraic automorphism group. Now, since there exists an algebraic real form, there exists at least one algebraic anti-regular automorphism, and hence the whole coset of anti-holomorphic automorphisms consists of anti-regular ones. In the case of a torus $(\mathbf{C}^*)^d$, the answer is yes, and actually every (holomorphic or anti-holomorphic) endomorphism is regular or anti-regular. For this, it is enough to prove the case of $d=1$, and indeed every (anti)holomorphic endomorphism has the form $z\mapsto z^d$ or $z\mapsto \bar{z}^d$ for some $d\in\mathbf{Z}$. The reductive case follows: every (holomorphic or anti-holomorphic) automorphism is regular or anti-regular (by acting on the derived subgroup on the one hand and the connected center on the other hand). Also for $G$ unipotent, the (holomorphic or anti-holomorphic) automorphism group is the same as the automorphism group of the Lie algebra and hence acts (anti)-regularly.<|endoftext|> TITLE: Expected supremum of normalised random walk QUESTION [5 upvotes]: Let $X^i\in \mathbb R^d$ be iid. random variables for $i=1$ to $n$. Assume $\mathbb E[X^i]=0$ and the covariance matrix $\mathbb C[X^i] = \mathbb E[X^iX^{iT}] = I$ is the identity matrix. Define $S^k=\frac{1}{\sqrt k}\sum_{i=1}^k X^i$, so that $\mathbb C[S^k] = I$. Question We would like to prove a bound on the form $$ \Pr\left[\sup_{k=1}^n\left\|S^k\right\|_2> Kt\right]<\frac{1}{t^2}\tag{1}$$ for some universal constant $K$ or perhaps some slowly increasing function in $n$, like $K(n)=\sqrt{\log n}$ or $K(n)=\sqrt{2\log\log n}$ as we would expect from the Law of the Iterated Logarithm. (Updated: Actually the Iterated Logarithm probably suggests that defining $S^k=\frac{1}{\sqrt {k\log\log k}}\sum_{i=1}^k X^i$ would be the correct normalization.) Approach 1 One approach is a union bound using the multivariate Chebyshev bound: $$\Pr\left[\left\|S^k\right\|_2> s\right]<\frac{d}{s^2}.$$ Taking $s=t \sqrt {dn}$ we get by a union bound $\Pr\left[\sup_{k=1}^n\left\|S^k\right\|_2> t\sqrt{dn}\right]<\frac{1}{t^2}.$ Slightly stronger we might take the $t$ value for each $k$ to be an increasing sequence like $t_k=\sqrt{k\log k}$. This would then give us $$\Pr\left[\sup_{k=1}^n\frac{1}{\sqrt{k}\log k}\left\|S^k\right\|_2> K t\right]<\frac{1}{t^2}$$ for some constant $K$. However this is still much worse than we would expect. Approach 2 Alternatively, since $\sqrt{k}S^k$ is a martingale, we can use Doob's inequality on the individual coordinates to get $$\Pr\left[\sup_{k=1}^n(\sqrt{k}S^k_i)^2> t\right]<\frac{\mathbb E[(\sqrt{n}S^n_i)^2]}{t}$$ which implies $\Pr\left[\sup_{k=1}^n \sqrt{\frac{k}{n}} S^k_i > t\right]<\mathbb E[(S^n_i)^2]/t^2$ and so by a union bound over the $d$ coordiantes, $$ \Pr\left[\sup_{k=1}^n\sqrt{\frac{k}{n}}\left\|S^k\right\|_2> t\right] \le \sum_{i=1}^d \Pr\left[\sup_{k=1}^n \sqrt{\frac{k}{n}} S^k_i > t\sqrt{d}\right] \le\frac{\mathbb E[\left\|S^n\right\|_2^2]}{t^2d} =\frac{1}{t^2}. $$ This is closer to what we were aiming for, but the factor of $\sqrt{\frac{k}{n}}$ still imposes a much stronger requirement on the early terms than we were aiming for. Question again Am I wrong to expect the inequality in $(1)$ given only the first and second moments are known on the $X^i$? In that case, say I also know the moment generating function of the $X^i$s, can I somehow use this knowledge with the Doob inequality or other vector-Martingale inequalities? REPLY [3 votes]: Let $X_i:=X^i$, $S_k:=\sum_1^n X_i$, $T_k:=S_k/\sqrt k$, $|\cdot|:=\|\cdot\|_2$, $n\in\{1,2,\dots\}$, and $m:=\lceil\log_2 n\rceil$, so that $2^m\ge n$. Then \begin{equation} \max_{k=1}^n|T_k|\le\max_{k=1}^{2^m}|T_k| =\max_{j=0}^m\max_{2^{j-1}0$ \begin{multline} P\big(\max_{k=1}^n|T_k|\ge t\big) \le\sum_{j=0}^m P\big(\max_{k=1}^{2^j}|S_k|\ge 2^{(j-1)/2}t\big) \\ \le\sum_{j=0}^m \frac{E|S_{2^j}|^2}{2^{j-1}t^2} =(m+1)\frac{2d}{t^2}\le\frac{2(2+\log_2 n)d}{t^2}; \end{multline} the second inequality in the latter display is an instance of Doob's inequality, which holds because $(S_k)$ is a martingale and hence $(|S_k|^2)$ is a submartingale. Equivalently, for any real $s>0$ \begin{equation} P\big(\max_{k=1}^n|T_k|\ge s\sqrt{2+\log_2 n}\big) \le\frac{2d}{s^2}. \end{equation} So, the extra factor is $\sqrt{2+\log_2 n}$, as you expected.<|endoftext|> TITLE: Set of points with a unique closest point in a compact set QUESTION [11 upvotes]: Let $K\subset\mathbb{R}^n$ be any compact set. Let $\operatorname{Unp}(K)$ be the set of points in $$ \operatorname{Unp}(K)=\{x\in\mathbb{R}^n\setminus K:\, \exists ! y\in K \ \ |x-y|=d(x,K)\}. $$ Here are some properties. The distance function is Lipschitz and hence differnetiable a.e. (Rademacher's theorem). If the distance function is differentiable at $x$, then $x\in \operatorname{Unp}(K)$. For a proof, see https://mathoverflow.net/a/299066/121665. Therefore almost all points of $\mathbb{R}^n\setminus K$ belong to $\operatorname{Unp}(K)$. For every $x\in \mathbb{R}^n\setminus K$, there is $y\in K$ such that $|x-y|=d(x,K)$ (although $y$ is not unique). Then the interior of the segment $xy$ is contained in $\operatorname{Unp}(K)$ (triangle inequality). Thus $\operatorname{Unp}(K)$ contains the union of disjoint open segments and the set $\mathbb{R}^n\setminus (\operatorname{Unp}(K)\cup K)$ is of measure zero and is contained in the endpoints of these segments. Question. Is it true that $\operatorname{Unp}(K)$ contains an open set? The interest in the study of the properties of the distance function on the set $\operatorname{Unp}(K)$ is motivated by results of Federer about sets of positive reach. H. Federer, Curvature measures. Trans. Amer. Math. Soc. 93 (1959), 418–491. REPLY [4 votes]: There are many counterexamples as the following result of Zamfirescu [1] shows. Theorem. For most of the compact sets $K\subset\mathbb{R}^n$, $\operatorname{Unp}(K)$ has empty interior, meaning that the set of points in $\mathbb{R}^n$ without a unique nearest point in $E$ is dense in $\mathbb{R}^n$. Here "most of the compact sets" is understood in the Baire category sense with respect to the Hausdorff metric on the space of compact sets in $\mathbb{R}^n$. [1] T. Zamfirescu, The nearest point mapping is single valued nearly everywhere. Arch. Math. (Basel) 54 (1990), 563–566.<|endoftext|> TITLE: Space with semi-locally simply connected open subsets QUESTION [11 upvotes]: A topological space $X$ is semi-locally simply connected if, for any $x\in X$, there exists an open neighbourhood $U$ of $x$ such that any loop in $U$ is homotopically equivalent to a constant one in $X$ or, equivalently, if the functor $$\Pi_1(U)\rightarrow\Pi_1(X)$$ induced by the inclusion $U\subseteq X$ factorizes through a groupoid in which for each pair of objects there is exactly one morphism. My question is: is it true that if a space $X$ is such that, for any open subset $U\subseteq X$ ($X$ included), $U$ is semi-locally simply connected, then $X$ must be locally simply connected? Notice that this implies that $X$ is locally path connected. I tried to post this question on stack exchange some months ago, but I didn't receive any answer. It was just something that came to my mind while I was studying for my master thesis. REPLY [7 votes]: $X$ need not be locally simply connected. Consider the following construction: Let $X_0=S^1$ and $X_n=CS^1$ for all $n\geq 1$ where $CS^1$ is the cone over the circle. Let $Y_0=X_0$, and for $n\geq 1$, let $Y_n\subseteq X_n$ be the base of the cone, $S^1\times\{0\}$ and $Z_n\subseteq X_n$ be the arc $\{b_0\}\times I$ where $b_0$ is the basepoint of $S^1$. Let $X'$ be the shrinking wedge of all $X_n$, that is, every neighborhood of the wedge point contains all $X_n$ with $n\geq k$ for some $k$. Finally, let $X=X'/\sim$, where the relation $\sim$ glues $Z_n$ along $Y_{n-1}$ by the exponential map for all $n\geq 1$.   Such a space is not locally simply connected since every neighborhood $U$ of the basepoint must include some $Y_n\approx S^1$ yet not include the cone $X_n$ necessary to nullhomotope a loop around $Y_n$. Yet it is locally semilocally simply connected since we can choose a neighborhood $V\subseteq U$ such that $Y_n\subseteq V$ only if $X_n\subseteq U$. Thus the inclusion $V\to U$ induces a trivial homomorphism on fundamental groups.<|endoftext|> TITLE: Classification of vertex-transitive zonotopes QUESTION [5 upvotes]: Zonotopes are convex polytopes that can be defined in several equivalent ways: parallel projections of cubes, Minkowsi sums of line segments, only centrally symmetric faces, ... I wonder whether there exists a calssification of all vertex-transitive zonotopes. I know only of the following examples: omnitruncations of uniform polytopes (this is probably the same as $W$-permutahedra, see comments). This already includes the interval $[0,1]$, all regular $2n$-gons, and, e.g. the following polyhedra in $\smash{\Bbb R^3}$: $\qquad\qquad\qquad\qquad\qquad$ cartesian products of any of these above. This includes $d$-cubes, prisms, duo-prisms, ... Are there any more? For that matter, are there even any more zonotopes for which all vertices are on a common sphere? REPLY [2 votes]: Update I recently uploaded a preprint in which I work out the details that are missing below. So in fact, vertex-transitive zonotopes are $\Gamma$-permutahedra. I believe to have (at least a roadmap to) a proof of the following: Theorem. If $P\subset\Bbb R^d$ is a vertex-transitive zonotope, then $P$ is a $\Gamma$-permutahedron. That is, $P$ is the convex hull of the orbit of an appropriately chosen point $\smash{v\in\Bbb R^d}$ under a finite reflection group $\smash{\Gamma\subset\mathrm{GL}(\Bbb R^d)}$. In other words, $P$ is the omnitruncation of some uniform polytope (when considered with a certain subgroup of its symmetries). I will give some thoughts about my proof, since I have not thought through every detail: Every zonotope can be uniquely written as the Minkowski sum of line segments with pair-wise trivial intersection. Let's call $r\in\Bbb R$ a root of $P$ if $\mathrm{conv}\{-r,r\}$ is one of these line segments. One then shows that the set of roots of $P$ forms a root system (without integrality condition).1 One further shows, that the zonotope $P$ has the same symmetries as its set of roots, hence that its symmetry group is a reflection group. (until here, I think, David had another approach using the normal fan of $P$). Let $\tilde \Gamma$ be the symmetry group of $P$. Since $P$ is vertex-transitive, $P$ is the orbit polytope of some point $\smash{v\in\Bbb R^d}$ w.r.t $\smash{\tilde \Gamma}$. As David observed, this group might be too large to call $P$ a $\smash{\tilde\Gamma}$-permutahedron. Consider the subgroup $\Gamma\subseteq\tilde\Gamma$ generated by all reflections in $\tilde\Gamma$ that fix no vertex of $P$. Then $\Gamma$ is a reflection group. Show that $P$ is the orbit polytope of $v$ under $\Gamma$. Then $\Gamma$ acts vertex-transitively and -regularly on $P$, hence $P$ is a $\Gamma$-permutahedron. Some notes on 1 Let $R$ be the set of roots of $P$. How to show that $R$ is a root system: Choose any two (linearly independent) $r,r'\in R$ and consider the 2-dimensional set $R':=\mathrm{span}\{r,r'\}\cap R$. Let $P'$ be the zonotope generated by $R'$. This zonotope is a 2-face of $P$, and by using the argument that $\mathrm{Aut}(P)=\mathrm{Aut}(R)$ one can conclude that from the vertex-transitivity of $P$ follows the vertex-transitivity of $P'$. (This part is sketchy right now, and makes some trouble. How to fix this? I think that that the faces of a vertex-transitive polytope do not necessarily have to be vertex transitive! Update: yes they are vertex-transitive, see the preprint) It follows that $P'$ is a $2n$-gon with possibly alternating edge lengths. One convinces oneself that the roots of $P'$ are a root system ($2n$ roots equally spaces by $\pi/n$, maybe of alternating lengths), that is, $R'$ (and hence $R$) contains the reflection of $r'$ on the hyperplane defined by $r$. Since $r$ and $r'$ were chosen arbitrarily, this shows that $R$ is a root system.<|endoftext|> TITLE: Determinantal questions on Alternate Sign Matrices QUESTION [6 upvotes]: Let $\mathcal{A}_n$ be the set of all Alternating Sign Matrices (ASM) of size $n\times n$. The cardinality $\#\mathcal{A}_n$ is well-known $$\#\mathcal{A}_n=\prod_{k=0}^{n-1} \frac{(3k+1)!}{(n+k)!}.$$ I brought up some questions to bear. QUESTION 1: What are these values on the sum of determinants? $$\sum_{M\in\mathcal{A}_n}\det(M).$$ QUESTION 2: What is known (can be said) about the determinants distribution of these matrices in $\mathcal{A}_n$? Such as, (a) what is the range of possible values? (b) how many are singular/regular? (c) how many achieve the value 1? or -1? EXAMPLES. Define the Laurent polynomial $$L_n(z)=\sum_{M\in\mathcal{A}_n}z^{\det(M)}.$$ Then, we have $L_2(z)=\frac1z+z$, $L_3(z)=\frac3z + 1+3z$, $L_4(z)=\frac{12}z+16+ 14z$, $$L_5(z)=\frac{86}z + 253 + 90z, \qquad \text{and} \qquad L_6(z)=\frac2{z^2}+ \frac{1032}z + 5368 + 1032z+ 2z^2.$$ This way, Question 1 reads $L_n'(1)$ where $'=\frac{d}{dz}$. Credit. These computations are made courtesy of Zeilberger's Maple package, ROBBINS. REPLY [3 votes]: (Half of an answer to Question 1) $M \in \mathcal{A}_n \Leftrightarrow w_0M \in \mathcal{A}_n$, where $w_0 := \mathrm{antidiag}(1,1,...,1)$ (this is just a flip across an axis of symmetry), so when $\binom{n}{2}$ is odd (i.e., $n \equiv 2,3 \mod 4$) we have $\sum_{M \in \mathcal{A}_n} \mathrm{det}(M) = 0$ since in that case $\mathrm{det}(w_0) = -1$, as observed in your data.<|endoftext|> TITLE: Solovay’s model QUESTION [9 upvotes]: Solovay proved that if $\kappa$ is inaccessible, then if we adjoin a generic $G \subseteq \mathrm{Col}(\omega,{<}\kappa)$, then in the extension, every set of reals in $L(\mathbb R)$ is Baire- and Lebesgue-measurable and has the perfect subset property. In several publications, such as “Martin’s Maximum Part 1” by Foreman-Magidor-Shelah, this is applied to a situation in which a generic for something other than the Levy collapse is taken. A lemma is invoked that says the following: Suppose $G \subseteq \mathbb P$ is generic a forcing that turns an inaccessible $\kappa$ into $\omega_1$, and every real in $V[G]$ lives in some $V[H]$, where $H$ is $\mathbb Q$-generic and $|\mathbb Q| < \kappa$. Then there is another extension $V[G’]$, where $G’$ is $\mathrm{Col}(\omega,{<}\kappa)$-generic, and $\mathbb R^{V[G]} = \mathbb R^{V[G’]}$. Thus the conclusion of Solovay’s Theorem holds in $V[G]$. Question: Do we actually need the above lemma? Can we prove Solovay’s Theorem directly about a forcing $\mathbb P$ satisfying the hypotheses of the above lemma? Is there any place in the proof where we use a special feature of the Levy collapse such as its “continuity”? REPLY [9 votes]: The usual proof (at least the proof that I know) uses the fact that in $V[H]$, the forcing $P/H$ (or $P:Q$) is again equivalent to the Levy collapse, and in particular sufficiently homogeneous, so that every formula with parameters in $V[H]$ (in the forcing language of $P/H$) has Boolean value $1$ or $0$.<|endoftext|> TITLE: Image of a map on cohomology rings QUESTION [13 upvotes]: The following seems like an extremely basic algebraic topology question, but it's not something I ever learned, nor does it look familiar to the algebraic topologists I've asked. Let $f:X\to Y$ be a map, inducing $f^*:H^*(Y)\to H^*(X)$. Hence the image $R$ of $f^*$ is a subring of $H^*(X)$. Is there a natural way to factor $f$ as $X \to Z \to Y$, such that $f^*$ factors as $H^*(Y) \twoheadrightarrow R \hookrightarrow H^*(X)?$ I have a particular $X,Y$ in mind (the inclusion of one compact complex manifold into another, each with even-degree cohomology) but I'm hoping phrases like "Postnikov tower", "cofibrant replacement", "mapping cone" will serve to give a general answer. REPLY [10 votes]: One special case of your set up is when $Y=X$ and $f^*$ is idempotent: $f^* \circ f^* = f^*$. In this case, let $Z$ be the mapping telescope of $X \xrightarrow{f} X \xrightarrow{f} X \rightarrow \dots$. This comes with a canonical map $r: X \rightarrow Z$ such that $r^*$ is monic with image equal to the image of $f^*$, and in many cases, one can show that there exists $i: Z \rightarrow X$ such that $r^* \circ i^* = f^*$. (We are basically looking to lift an idempotent in homology to an idempotent in homotopy. One sufficient condition for $i$ to exist is that $X$ be a $p$--complete CW complex of finite type: see the short paper Atomic spaces and spectra I wrote with J.F. Adams back in the late 1980's.)<|endoftext|> TITLE: Morphisms from $bstring$ to $X\otimes \mathbb{Q}$ and sequences $s_n\in\pi_n(X)\otimes \mathbb{Q}$ QUESTION [6 upvotes]: I'm currently studying Ando-Hopkins-Rezk's work Multiplicative orientations of KO-theory and of the spectrum of TMFs. At a point a presumably obvious isomorphism is mentioned, which I'm however not able to immediately see (I guess I'm missing something obvious here, forgive me if the question is too naive). I'm referring to the identification $$ s: [bstring, X\otimes \mathbb{Q}] \cong \mathbf{D}(X):=\left\{s\in\prod_{k\geq 4} \pi_{2k}X\otimes \mathbb{Q}\,\biggr\vert\, s_k=0 \text{ if $k$ is odd}\right\} $$ (equation (5.10) in the article), where $X$ is a spectrum and $bstring$ is the spectrum realizing the standard infinite loop space structure on $BString=BO\langle 8\rangle$. While I clearly see there's a natural morphism $s: [bstring, X\otimes \mathbb{Q}] \to \mathbf{D}(X)$, that fully encodes the datum of the sequence of group homomorphisms $$ \phi_{\ast,n}: \pi_n(bstring) \to \pi_n(X)\otimes \mathbb{Q} $$ associated with (the homotopy class of) a morphism $\phi\colon bstring \to X\otimes \mathbb{Q}$, I don't clearly see why any such a sequence should be induced by a unique (homotopy class of a) morphism from $bstring$ to $X_\mathbb{Q}$, i.e., I don't immediately see why $s$ should be a bijection. I guess the answer is in some property of $bstring_\mathbb{Q}$ so well known not to be explicitly mentioned in the article, or in some even more fundamental property of rational homotopy theory of infinite loop spaces (or even general spectra) I'm missing at the moment, but I have not been able to work out this answer myself. Thanks a lot for any possible hint to help me solve my puzzlement REPLY [7 votes]: This is an assemblage of known results, I'll try to put a reference for all of them. By a classical theorem of Serre all stable homotopy groups are finite in positive degree. In particular we have $\mathbb{S}_\mathbb{Q}=H\mathbb{Q}$. Rationalization is a smashing localization, so that $\pi_*(\mathbb{S}_\mathbb{Q}\otimes E)\cong \pi_*E \otimes_{\mathbb{Z}}\mathbb{Q}$. This follows immediately from the fact that we can write $$E_\mathbb{Q}=\mathrm{colim}\left(E\xrightarrow{2}E\xrightarrow{3}E\xrightarrow{4}E\xrightarrow{5}\cdots\right)$$ and that tensor (smash) products and homotopy groups commute with filtered colimits of spectra. By Schwede-Shipley Morita theory, the category of modules over $H\mathbb{Q}$ is equivalent to the derived category of $\mathbb{Q}$. (Theorem 5.1.6 in Stable model categories are categories of modules) Moreover under this equivalence homotopy groups correspond to homology groups (because $H\mathbb{Q}$ is sent to $\mathbb{Q}$ in degree 0). In the derived category $D(\mathbb{Q})$ for every object $M$ there is an equivalence $M\cong \bigoplus_{n\in\mathbb{Z}} (H_nM)[n]$. This is an easy exercise, using the fact that in $\mathbb{Q}$-vector spaces every short exact sequence splits. In particular for every spectrum $E$, the spectrum $E\otimes H\mathbb{Q}$ has homotopy groups $\pi_*(E\otimes H\mathbb{Q})\cong \pi_*E\otimes_{\mathbb{Z}}\mathbb{Q}$ and so there is an equivalence of $H\mathbb{Q}$-modules (in particular of spectra) $$ E\otimes H\mathbb{Q}\cong \bigoplus_{n\in\mathbb{Z}} \Sigma^nH(\pi_nE\otimes_{\mathbb{Z}}\mathbb{Q})\cong\prod_{n\in\mathbb{Z}} \Sigma^nH(\pi_nE\otimes_{\mathbb{Z}}\mathbb{Q})$$ (it's not necessary, but it is very convenient to observe that in this case for degree reasons the coproduct and the product coincide) So, to conclude, we have that $$[bstring, X\otimes H\mathbb{Q}]\cong \left[ bstring, \prod_{n\in\mathbb{Z}}\Sigma^nH(\pi_nX\otimes_{\mathbb{Z}}\mathbb{Q})\right]\cong \prod_{n\in\mathbb{Z}} H^n(bstring;\pi_nX\otimes_{\mathbb{Z}}\mathbb{Q})\cong \prod_{n\in\mathbb{Z}} \pi_nX\otimes_{\mathbb{Z}} H^n(bstring;\mathbb{Q})\,.$$<|endoftext|> TITLE: Why not add cuspidal curves in the moduli space of stable curves? QUESTION [21 upvotes]: Let $\mathcal{M}_{g,n}$ be the moduli space (stack) of stable smooth curves of genus $g$ with $n$ marked points over $\mathbb{C}. $ It's known that by adding stable nodal curves to $\mathcal{M}_{g,n}$, the resulting space $\overline{\mathcal{M}}_{g,n}$ is compact. But why is it so? For example, consider the following family of elliptic curves in $\mathcal{M}_{1,1}$, $$y^2=x^3+t,$$ where $t \in \mathbb{C^*}$. Then this family of elliptic curves degenerates into the cuspidal cubic curve $$y^2 = x^3.$$ So why is $\overline{\mathcal{M}}_{1,1}$ compact? REPLY [41 votes]: If you add cuspidal curves, then $\overline{\mathcal{M}}_{1,1}$ will no longer be separated, which is the scheme/stack analogue of Hausdorff. Specifically, consider the families $$y_1^2 = x_1^3 + t^6 \ \mbox{and}\ y_2^2 = x_2^3 + 1$$ (so the second family is a constant family with no $t$-dependence). For all nonzero $t$, they are isomorphic by the change of variables $y_1 = t^3 y_2$, $x_1 = t^2 x_2$. So they should give the same map from $\mathbb{C}^{\ast}$ to moduli space (namely, a constant map). If the cuspidal curve corresponded to a point of moduli space, then this map would have two limits. The situation is similar with regard to the family $y_3^2 = x_3^3 + t$ that you consider. On the level of coarse moduli spaces, this family also corresponds to a constant map $\mathbb{C}^{\ast} \to \overline{\mathcal{M}}_{1,1}$. The subtlety is that the families $y_2^2 = x_2^3 + 1$ and $y_3^2 = x_3^3 + t$ are not isomorphic over $\mathrm{Spec}\ \mathbb{C}[t^{\pm 1}]$, but only over the cover where we adjoin a $6$-th root of $t$. The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover. In general, when choosing a definition for a moduli space, if you allow too many objects, you will fail to be separated and, if you allow too few objects, you will fail to be proper (analogue of compact). So the answer to "why don't we include" is usually "that would break separatedness" and the answer to "why must we include" is "in order to be proper".<|endoftext|> TITLE: Given a zero-dimensional ideal $(f_1,...,f_n)$, is the ideal $(f_1-\alpha_1,...,f_n-\alpha_n)$ also zero-dimensional? QUESTION [6 upvotes]: Suppose you have a zero-dimensional ideal $I=(f_1,...,f_n)$ in a polynomial ring $R=k[x_1,...,x_n]$ over a field $k$, so that $\dim_k(R/I)<\infty$. Take indeterminates $\alpha_1,...,\alpha_n$ and consider the ideal $J=(f_1-\alpha_1,...,f_n-\alpha_n)$ of the ring $S=K[x_1,...,x_n]$ over the field $K=k(\alpha_1,...,\alpha_n)$. Is it true that $J$ is also a zero-dimensional ideal? I'm especially interested in the relationship between $\dim_K(S/J)$ and $\dim_k(R/I)$. In all the examples that I tried (with my very limited Macaulay2 abilities), I found that $\dim_K(S/J)=\dim_k(R/I)$. My thought is to compute (reduced) Gröbner bases $G_I$ for $I$ and $G_J$ for $J$, and then to compare the leading monomials that show up in $G_I$ and $G_J$. Since $I$ is zero-dimensional, there are finitely many standard monomials of $G_I$, and my hope is that the standard monomials of $G_J$ are a subset of these. If it helps, you may assume that $k$ is algebraically closed of characteristic 0 and each $\alpha_i$ is of the form $\alpha_i=f_i(z_1,...,z_n)$ for indeterminates $z_1,...,z_n$. Edit: If you assume that $f=(f_1,...,f_n):\mathbb{A}^n_k\to\mathbb{A}^n_k$ is quasi-finite, then $R/I$ has Krull dimension zero and hence $I$ is zero-dimensional. Since quasi-finiteness is stable under base change and composition, one can show that $J$ is a zero-dimensional ideal of $S$. However, this doesn't seem to tell us anything about the relationship between $\dim_K(S/J)$ and $\dim_k(R/I)$. One idea is to note that $\dim_K(S/IS)\leq\dim_k(R/I)$ and try to compare $\dim_K(S/IS)$ and $\dim_K(S/J)$. REPLY [3 votes]: If $f$ is not generically finite-to-one then there are easy counterexamples (e.g. take $f_1 = x_1$, $f_2 = x_1 - 1 \in k[x_1, x_2]$). So assume $f$ is generically finite-to-one. Claim 1: If $f$ is generically finite-to-one and $\dim_k(R/I) < \infty$, then $\dim_k(R/I) \leq \dim_K(S/J) < \infty$. Claim 2: It is possible in the situation of Claim 1 that $\dim_k(R/I) <\dim_K(S/J)$. For the first claim note that $\dim_k(R/I)$ is the number (counted with multiplicity) of points in $f^{-1}(0)$ whereas $\dim_K(S/J)$ is the number (counted with multiplicity) of points in a generic fiber of $f$. For the second claim you can take any $f: k^n \to k^n$ such that $f$ is generically finite-to-one, $f^{-1}(0)$ is finite, but $|f^{-1}(0)| < \deg(f)$ (where $\deg(f)$ is the cardinality of a generic fiber of $f$). For an explicit example, take $f_1 = x_1x_2 - 1$, $f_2 = x_2(x_1x_2 - 1) + x_1$. Then $\dim_k(R/I) = 0$, whereas $\dim_K(S/J) = 2$. (I knew of this map from a paper of Zbigniew Jelonek - it is one of the simplest maps from $k^2 \to k^2$ which is quasi-finite but non-surjective.)<|endoftext|> TITLE: Has this "backwards" perspective on toposes been studied? QUESTION [15 upvotes]: Topos theory can be seen as a categorification of topology via the following analogies. \begin{array}{|c|c|} \hline \text{locales}&\text{Grothendieck toposes}\\\hline \text{open sets}&\text{sheaves}\\\hline \text{continuous maps}&\text{geometric morphisms}\\\hline \text{bases}&\text{sites}\\\hline \text{topological spaces}&\text{ionads}\\\hline \end{array} But when I was first learning about topos theory I was temporarily confused by the following two results. Proposition 1 For a set $S$ there's a correspondence between topologies on $S$ and finite limit preserving (necessarily idempotent) monads on $\mathbf{2}^S$. A topology corresponds to its closure operation. Proposition 2 For a category $C$ there's a correspondence between Grothendieck topologies on $C$ and finite limit preserving idempotent monads on $\mathbf{Set}^{C^\mathrm{op}}$. A Grothendieck topology corresponds to its sheafification operation. The similarity between these propositions suggests that we can also view topos theory as a categorification of topology in such a way that sheaves are the categorification of closed sets rather than open ones. From this strange perspective the analogies would be the following. \begin{array}{|c|c|} \hline \text{topologies}&\text{Grothendieck topologies}\\\hline \text{topological spaces}&\text{sites}\\\hline \text{closed sets}&\text{sheaves}\\\hline \text{continuous maps}&\text{morphisms of sites}\\\hline \end{array} Has this point of view been studied anywhere? Does it have any use? REPLY [3 votes]: Actually, the closure operator of a topology is a finite colimit preserving monad on a powerset.<|endoftext|> TITLE: Combination topological space and locale? QUESTION [16 upvotes]: The traditional theory of topological spaces (as formalized by Bourbaki) starts with a set of points, then builds a structure on that. In contrast, the theory of locales starts with a frame of opens (open subspaces), identifying the points (if these are even needed) from that data. I once read something suggesting that we should do both: start with both a set of points and a frame of opens, then specify the relation between these. This was pretty speculative (and probably also being done constructively, but I'm not sure about that). Is there a developed theory of this, or at least a name that I can look up or ask about? To give this question some mathematical weight, I'm going to write down some definitions of the sort of thing that I'm thinking of. Nevertheless, if you have an answer to the previous paragraph that disagrees in some details with what follows, then this is still an answer to my question! (But if it's nothing at all like what follows, then it's probably not what I'm looking for.) A space $X$ consists of a set $|X|$ (the set of points of $X$), a frame $\mathcal{O}_X$ (the frame of opens of $X$), and a frame homomorphism from $\mathcal{O}_X$ to the power set $\mathcal{P}|X|$ of $|X|$. (I'm provocatively claiming the generic term ‘space’ because I hope that somebody will say that the proper term is _____ as found in _____'s paper _____, which would be a perfect answer to this question.) As a function to a power set may be reinterpreted as a binary relation, so the frame homomorphism from $\mathcal{O}_X$ to $\mathcal{P}|X|$ may be reinterpreted in a more elementary way as a binary relation $\in_X$ between $|X|$ and (the underlying set of) $\mathcal{O}_X$ with these properties: for each point $a$ of $X$, $a \in_X \top_{\mathcal{O}_X}$; for each point $a$, open $U$, and open $V$ of $X$, $a \in_X U \wedge V$ iff $a \in_X U$ and $a \in_X V$; and for each point $a$ and collection $\mathcal{U}$ of opens of $X$, $a \in_X \bigvee \mathcal{U}$ iff, for some $V \in \mathcal{U}$, $a \in_X V$. Given a space $X$ and a space $Y$, a (continuous) map $f$ from $X$ to $Y$ consists of a function $|f|$ from $|X|$ to $|Y|$ and a frame homomorphism $f^*$ from $\mathcal{O}_Y$ to $\mathcal{O}_X$ that makes the diagram $$ \matrix { \mathcal{O}_Y & \overset{f^*}\to & \mathcal{O}_X \\ \downarrow & & \downarrow \\ \mathcal{P}|Y| & \underset{|f|^{-1}}\to & \mathcal{P}|X| } $$ commute. Or in more elementary terms: for each point $a$ of $X$ and open $U$ of $Y$, $a \in_X f^*(U)$ iff $|f|(a) \in_Y U$. Spaces and maps form a category $\mathrm{Sp}$. A space $X$ is topological if the frame homomorphism from $\mathcal{O}_X$ to $\mathcal{P}|X|$ is monic. In elementary terms: for each open $U$ and open $V$ of $X$, if for each point $a$ of $X$, $a \in_X U$ iff $a \in_X V$, then $U = V$. (It follows that $U \leq V$ iff $a \in_X U$ implies $a \in_X V$. Note that $a \in_X V$ if $a \in_X U \leq V$ in any space $X$.) The topological spaces (see what I did there?) form a full subcategory $\mathrm{Top}\,\mathrm{Sp}$ of $\mathrm{Sp}$. A subset $G$ of $|X|$ is open if it is in the image of the frame homomorphism from $\mathcal{O}_X$. In this way, every space gives rise to a topological space in the usual sense, and indeed we get a functor from $\mathrm{Sp}$ to the category of such spaces and the continuous functions between them. When restricted to $\mathrm{Top}\,\mathrm{Sp}$, this is an equivalence of categories. A space $X$ is localic if the points of $\mathcal{O}_X$ (in the locale-theoretic sense) correspond to the points of $X$. Explictly: for each completely prime filter $\mathcal{F}$ in $\mathcal{O}_X$, for some unique point $a$ of $X$, for each open $U$ of $X$, $U \in \mathcal{F}$ iff $a \in_X U$. (It follows that $U = V$ iff they have the same points. Note that in any space $X$, a point of $X$ defines a completely prime filter in this way, but we want all such filters to arise in this way and that different points generate different filters.) The localic spaces form a full subcategory $\mathrm{Loc}\,\mathrm{Sp}$ of $\mathrm{Sp}$. As $\mathcal{O}_X$ is a frame, every space gives rise to a locale, and indeed we get a functor from $\mathrm{Sp}$ to the usual category of locales. When restricted to $\mathrm{Loc}\,\mathrm{Sp}$, this is an equivalence of categories. A space is sober if it is both topological and localic. The sober spaces form a full subcategory $\mathrm{Sob}\,\mathrm{Sp}$ of $\mathrm{Sp}$, which is thus equivalent both to the category of sober topological spaces in the usual sense and to the category of topological (aka spatial) locales. REPLY [13 votes]: The proper term is topological system as found in Vickers' book Topology via Logic. Vickers actually uses precisely your expanded definition. What you call "topological" Vickers calls "spatial" and you both use the word "localic" with the same meaning.<|endoftext|> TITLE: Changing tiles by swapping squares QUESTION [7 upvotes]: In an $n\times n$ table, initially there is a $1\times n$ tile in each row. A swap is an operation that involves choosing two tiles, move one square from the first to the second tile, and simultaneously move one square from the second to the first tile. Tiles can change shapes arbitrarily - the only constraint is that each tile must remain (edge-)connected throughout and does not have a hole in it. Is it possible to perform swaps and end up with an $n\times 1$ tile in each column? For $n=2,3,4$ it can be easily checked that this is possible. $n=5$ is also possible but the sequence is much more complicated, and it is not clear there is a pattern. REPLY [3 votes]: Yes, it is possible. The procedure I came up with is really tedious to describe in detail, but I'm attaching a sketch that hopefully should make things clear. For $k \leq n$ denote by $T_{n,k}$ the following tiling of a $k \times n$ grid. The first tile uses all $k$ squares of the leftmost column, and the leftmost $n-k+1$ squares of the bottom row. The $i$-th tile for $i > 1$ consists of all squares that are incident (vertically, horizontally, or diagonally) to squares of the $(i-1)$-th tile, but have not been used in any other tile. In particular, $T_{n,n}$ is the tiling where each column is a tile, and $T_{n,1}$ consists of a single tile containing all squares. It's not hard to check that each tile of $T_{n,k}$ consists of exactly $n$ squares. The top left sketch in the attached picture shows $T_{n,k}$ for $n=8$, $k=5$. I claim that there is a sequence of swaps taking us from $T_{n,k}$ to a tiling where one tile occupies the top row, and the remaining rows form a tiling $T_{n,k-1}$. Applied inductively, this allows us to go from $T_{n,n}$ (all columns) to the tiling where every row is its own tile. As I mentioned, this sequence is a bit tedious to describe, but I hope that it is clear from the attached picture: in every step, each tile is a path. We swap the green initial piece of one of the paths for the red final piece of the path starting in the top left corner (the tiles stay connected, if we do the swaps in the order indicated by the arrows). After such a sequence of swaps, the last square of the path starting at the top left corner will be incident to the next path we'd like to perform swaps on. Conversely, the first square of that next path will be incident to the path starting at the top left corner, so we can keep going until one of the tiles occupies the top row. Note that in the $i$-th step we are transforming the $(i+1)$-th path of $T_{n,k}$ into the $i$-th path of $T_{n,k-1}$ on the bottom $k-1$ rows, so the result of our swap sequence is as claimed.<|endoftext|> TITLE: Rank-one positive decomposition for a entry-wise positive positive definite matrix QUESTION [6 upvotes]: I have asked this question in math.se without any success. Let $\mathbf{A}$ be a symmetric $n\times n$ positive semi-definite matrix and also such that each of its entries is positive. Does $\mathbf{A}$ have a decomposition of the form \begin{align} \mathbf{A} \,=\,\sum_{i=1}^{k}\mathbf{y}_i\mathbf{y}_i^T \end{align} where each vector $\mathbf{y}$ is also entry-wise positive and $k\leq n$. REPLY [5 votes]: No, there exist doubly nonnegative matrices which are not completely positive, see for example The difference between 5 x 5 doubly nonnegative and completely positive matrices (2009). A graph-based characterization of doubly nonnegative matrices which are completely positive is: Every doubly nonnegative matrix realization of a graph $G$ is completely positive if and only if $G$ does not contain an odd cycle of length at least 5, see Open problems in the theory of completely positive and copositive matrices (2015). An alternative characterization is give in a 2011 MO question: A doubly nonnegative $n\times n$ matrix is completely positive if and only if the $n$ vectors making up the Gram matrix lie in the non-negative orthant of some space of dimension $>n$. REPLY [3 votes]: Not necessarily. If this would hold, all components of the $y_i$ would be bounded in terms of $A$; so, by compactness argument, the same would hold for matrices and columns with non-negative entries. We will show that this is not the case. Let $I$ and $J$ be the identty and the all-ones matrices of order $3\times3$. Set $$ A=\left[\matrix{ 100I&J\\J&100I}\right]. $$ Then each $y_i$ may contain at most one non-zero among the first three entries, the same for the last three. On the other hand, for each $a=1,2,3$ and $b=4,5,6$ there should be a $y_i$ with non-zero $a$th and $b$th entries. Thus $k\geq9$. Remark. This argument shows that, in general, $k\geq [n^2/4]$. It cannot show more, since the edges of an arbitrary graph on $n$ vertices may be split into at most $n^2/4$ cliques. It may happen that this estimate is sharp.<|endoftext|> TITLE: Extending a triangulation of the boundary of $M \times I$ QUESTION [8 upvotes]: (Sorry for what is probably a rather foundational PL-topology question.) By a triangulation of a manifold $M$, I mean a homeomorphism with the geometric realization of a simplicial complex, $h: |K| \to M$. I am wondering if I am given two triangulations $h_0 : |K_0| \to M \times \{0\}$ and $h_1 : |K_1| \to M \times \{1\}$, can I find a triangulation $h : |K| \to M \times I$ of $M \times [0,1]$ that extends $h_0$ and $h_1$? By this I mean that $K_0$ and $K_1$ both include into $K$ and composing the geometric realization of either of these inclusions with $h$ commutes with $h_0$ and $h_1$. I am happy to assume $M$ is compact, orientable, and smooth. I don't particularly care about the triangulations being PL (i.e. I don't care if the links of vertices are spheres). REPLY [8 votes]: I think, one has to assume that the triangulations are smooth (i.e. restrictions of $h_i$ to every simplex are smooth). Then the answer is yes, this is a special case of a theorem by Munkres: a $C^r$-triangulation of the boundary of a manifold extends to a $C^r$-triangulation of the manifold, see Theorem 10.6 in Munkres, J. R., Elementary differential topology. Lectures given at Massachusetts Institute of Technology, Fall, 1961. Revised ed, Annals of Mathematics Studies. 54. Princeton, N.J.: Princeton University Press. XI, 112 p. (1966). ZBL0161.20201. In the PL category (where again one should assume that the triangulations define the same PL structure) this is a corollary in the article Armstrong, M. A., Extending triangulations, Proc. Am. Math. Soc. 18, 701-704 (1967). ZBL0149.41301.<|endoftext|> TITLE: Road map to learn about $\mathrm{Out}{F_n}$ QUESTION [10 upvotes]: I'm a last year undergraduate student and I have taken a graduate course in geometric group theory. I'd like to start reading some more advanced stuff in geometric group theory and in particular about automorphisms of free groups. (In specific I'm interested in $\mathrm{Out}(F_n)$.) Could you please suggest me material (books, notes, papers)? I've started studying "The topology of $\mathrm{Out}(F_n)$" by Mladen Bestvina. Are there any similar works (or more introductory ones)? Thanks! (If this is not the place for such a question please let me know) REPLY [4 votes]: I'm obliged to add "The topology, geometry, and dynamics of free groups. Part I: Outer space, fold paths, and the Nielsen/Whitehead problems" by Lee Mosher which I find very helpful (looking forward to Parts II & III).<|endoftext|> TITLE: What's the "actual" shape of a black hole accretion disk? QUESTION [16 upvotes]: [Warning: I have no expertise in general relativity, so this question might not be very rigorous] More and more often we come across science popularization articles like this one which show beautiful images of (simulations of) a black hole surrounded by its accretion disk: Of course, what we see is the effect of how the trajectories of light rays escaping the accretion disk are bent by the effect of space-time curvature, which is very high in the proximity of the black hole and more or less zero away from it. This results in a sort of "artifact" in which there seems to be two or more rings of falling bright matter conjoined together, or one disk presented from two or more sides maybe by some gravitational lensing. This is a phenomenologically correct thing: we -physical beings- can only perceive that matter through the light it emits, and the light is subject to the laws of general relativity. But what's the "actual" shape of the accretion disk? Let me try to make sense of the above question. First of all, the topology. The accretion material can be seen as a subset of 3-space (minus the black hole singularity), or we can probably consider the level sets of a "brightness function". Second, the metric. I can imagine that in a given instant the whole picture is a leaf of a constant time foliation of spacetime (does such foliation exist in the case of a black hole?), which is a Riemannian manifold diffeomorphic to 3-space (minus the singularity). So the accretion disc is in fact a metric space with the induced distance. Can this metric space be isometrically (or with some limited amount of distortion) embedded in flat 3-space? How would it look like? What I would like to know -I guess- is how this object would be perceived by a non-physical being that is able to "see" the metric of the space slice all at once without being bounded by the need of probing things by relativistic light rays. In other words, I want to do away with the general relativity aspects, and just keep the visualization of a subset of a Riemannian 3-manifold aspect, which may not be so crazy after all: I think I also have a more precise question, whose answer may indicate that the above question does or does not make much sense: Are light rays geodesics also for the induced Riemannian metric on the constant-time slices? If they are, my previous questions are probably not very interesting (e.g. already answered by the first picture). REPLY [5 votes]: This is not really a complete answer, but would not have fit in a comment. Your description of the topology and the time slices contains some incorrect assumptions. Let's take the simplest example of a black hole spacetime, which is the Schwarzschild spacetime (a black hole with zero charge and zero angular momentum). This spacetime is split into interior and exterior regions, with an event horizon in between. The event horizon is a null surface, not a timelike surface, so it doesn't act like a stationary boundary. It would be more accurate to think of it as if it were moving at the speed of light. (In the maximal analytic extension of the spacetime, there are actually two more regions, for a total of four, but these are not present in a black hole that forms by gravitational collapse.) The exterior is stationary, meaning that it has a timelike Killing vector. An observer whose world-line is parallel to this vector observes a gravitational field that doesn't change over time, and can be considered to be at rest relative to the black hole. Because the Schwarzschild spacetime is also not rotational, such observers can synchronize their clocks, and we therefore have a preferred time foliation. However, this preferred time foliation does not exist for rotating black holes, which are what really exist in our universe. (The simulation in the OP is of a rotating black hole.) But the interior region is not stationary. Therefore the preferred time foliation can't be extended into the interior. What this tells us is that there is never any well-defined way to talk about what is happening "now" inside a black hole. You can extend your spacelike surface of simultaneity into the interior, but there is no preferred way to do so. We can say that the infalling matter never reaches the horizon, but we can also say that it does. (According to an infalling observer, the matter passes through the horizon and reaches the singularity in milliseconds in the case of a solar-mass black hole.) I would interpret this as meaning that your question doesn't actually have a meaningful answer if you try to talk about anything but the exterior region. I can imagine that in a given instant the whole picture is a leaf of a constant time foliation of spacetime (does such foliation exist in the case of a black hole?), which is a Riemannian manifold diffeomorphic to 3-space (minus the singularity). The singularity is analogous to a spacelike surface, not a timelike one. A spacelike surface of simultaneity can be chosen either to intersect the singularity or not to intersect it. Therefore there are at least two different topologies possible for a surface of simultaneity. Are light rays geodesics also for the induced Riemannian metric on the constant-time slices? This question only makes sense on the exterior (and only for a non-rotating black hole), where there is a preferred foliation. There you can project out the time dimension. In that region, I think the answer is no. Geodesics of the spacetime are in general not geodesics after this type of projection. For example, the orbit of the earth is an ellipse under this type of projection, and that's not a geodesic of the spatial metric. I think it's pretty easy to see why we shouldn't expect the geodesic property to be preserved under projection. Different spacetime geodesics can be tangent to one another after projection, and you can't have different geodesics with the same tangent vector where they intersect.<|endoftext|> TITLE: Is it a known property of positive integers $n> 2 $ that one must have $n < \mathrm{rad}(n(n-1)(n-2))$? QUESTION [5 upvotes]: Let $P(n)$ be the statement that $$n < \mathrm{rad}(n(n-1)(n-2)),$$ where $\mathrm{rad}$ is the radical of an integer, that is defined as $$\operatorname{rad}(m)=\prod_{\substack{p\mid m\\p\text{ prime}}}p$$ for integers $m>1$ with $\operatorname{rad}(1)=1$. I checked that $P(n)$ holds for $3 \le n \le 3.10^7$. My question: Is $P(n)$ true for any positive integer $n \geq 3$? Also, is this a pre-existing conjecture? REPLY [4 votes]: Suppose we have such an $n$ where $P(n)$ is false. Now define $(a, b, c) := (1, n(n-2), (n-1)^2)$ and observe that these three numbers are pairwise coprime and satisfy $a + b = c$. Then we have: $$ \textrm{rad}(abc) = \textrm{rad}(n(n-1)^2(n-2)) = \textrm{rad}(n(n-1)(n-2)) \leq n $$ Moreover, the radical of $n(n-1)(n-2)$ cannot be equal to any of $n$, $n-1$, and $n-2$ (because then that term would be squarefree, so equal to its own radical, and the other two terms in the product would make the radical strictly larger), so we actually have: $$ \textrm{rad}(abc) \leq n - 3 < n - 1 = c^{\frac{1}{2}} $$ Such a counterexample $n$ would therefore result in an ABC triple with a quality $q > 2$, which is vastly larger than the current record-holder (which has a quality of 1.6299). There are no such triples below $10^{20}$ (see here), and we can probably extend this much further by only looking for $q > 2$ instead of $q > 1.4$, but already this implies that your conjecture is true at least up to $10^{10}$.<|endoftext|> TITLE: Whitney stratification of algebraic varieties QUESTION [5 upvotes]: When do the orbits of an action on an algebraic variety make a Whitney stratification? REPLY [2 votes]: The answer is indeed always when there are fnitely many orbits. Let $Y \subset X$ be an orbit, $y \in Y$, and $U \subset X$ a $G$-invariant open such that $y \in Y \cap U$, the relative open $Y \cap U$ is dense in $Y$, closed in $U$ and $U$ embedds $G$-equivariantly in some $\mathbb{C}^N$. Let us denote by: $$ C(X) = \overline{\{(x,H) \in U_{smooth} \times (\mathbb{C}^N)^*, \ \textrm{such that} \ T_{X,x} \subset H \}},$$ where the overline is the Zariski closure in $\mathbb{C}^N \times (\mathbb{C}^N)^*$. The variety $C(X)$ is known as the conormal space and is also $G$-invariant. Note that the projection map : $\pi : C(X) \longrightarrow X$ is $G$-equivariant. Let: $$ \rho : \mathrm{Bl}_{\pi^{-1}(Y)}C(X) \longrightarrow C(X)$$ be the blow-up of $C(X)$ along $\pi^{-1}(Y)$. The map $\rho$ is again $G$-equivariant, so that the composite: $$ \pi \circ \rho : \mathrm{Bl}_{\pi^{-1}(Y)}C(X) \longrightarrow C(X)$$ is $G$-equivariant. In particular, all fibers of $\pi \circ \rho$ are isomorphic and we find : $$ \dim (\pi \circ \rho)^{-1}(y) = \dim (\pi \circ \rho)^{-1}(Y) - \dim Y = N-2 - \dim Y.$$ Hence, proposition 2.3.1 in the paper Variétés Polaires II by Bernard Teissier shows that the pair $(X,Y)$ staisfies the Whitney conditions at $y$. This is true for all $y \in Y$, so that the pair $(X,Y)$ satisfies the Whitney conditions. As hinted in the comments, if there are finitely many orbits, then you get a finite stratification of $X$ by orbits, and by the above arguments, this is a Whitney stratification. Note however that it is certainly not the minimal stratification, as illustrated by the case $X$ smooth and $G$ acts non-trivially with finitely many orbits.<|endoftext|> TITLE: Theta functions on an elliptic curve and Serre duality QUESTION [19 upvotes]: Given an elliptic curve $E$ (over $\mathbb{C}$) and line bundle $L$, one can identify $H^0(E,L)$ with a particular space of theta functions. Serre duality gives a perfect pairing between $H^0(E,L)$ and $H^1(E,L^{-1})$, does this pairing have a nice expression in terms of theta functions? I'm either looking for some pairing on the space of theta functions themselves, or a particular way to interpret $H^1(E,L^{-1})$ as objects somehow dual to theta functions. REPLY [10 votes]: Let $X$ be an elliptic curve $\mathbb{C}/\Lambda$, where $\Lambda$ is a lattice of real rank $2$ in $\mathbb{C}$. A theta function is a holomorphic section of a line bundle $L$ on $X$ whose transition from $U$ to $U + \ell$ is given by$$f(z + \ell) = e^{a_\ell z + b_\ell} f(z)$$for $f \in \Lambda$, where $a_\ell, b_\ell \in \mathbb{C}$ and $f$ is an entire function on $\mathbb{C}$. This theta function can be regarded as an element of $\text{H}^0(X, L)$ when $\text{H}^0(X, L)$ is the kernel and $\text{H}^1(X, L)$ is the cokernel of the map$$C_{0, 0}^\infty(X, L) \overset{\overline{\partial}}{\to} C_{0, 1}^\infty(X, L),$$where $C_{0, q}^\infty(X, L)$ is the space of all $L$-valued $C^\infty(0, q)$-forms on $X$ for $q = 0, 1$. Thus $\text{H}^1(X, L^{-1})$ is the cokernel of$$C_{0,0}^\infty(X, L^{-1}) \overset{\overline{\partial}}{\to} C_{0, 1}^\infty(X, L^{-1}).$$We introduce a smooth metric $H(z) > 0$ for $L$. The compatibility condition for $H(z)$ is$$H(z + \ell) \left|e^{a_\ell z + b_\ell}\right|^2 = H(z)$$for $\ell \in \Lambda$ so that the pointwise norm square$$H(z + \ell)|f(z + \ell)|^2 = H(z)|f(z)|^2$$is a well-defined function on $X$. One way to explicitly write down such a metric $H(z)$ is to choose $H(z) = e^{-\gamma|z|^2}$ so that the curvature$${{\sqrt{-1}}\over{2\pi}}\partial\overline{\partial}(-\log H(z)) = \gamma {{\sqrt{-1}}\over{2\pi}} \text{d}z \wedge \text{d}\overline{z}$$of the metric $H(z)$ of $L$ is positive when $\gamma$ is a positive constant. The compatibility condition$$H(z + \ell) \left|e^{a_\ell z + b_\ell}\right|^2 = H(z)$$corresponds to the conditions $a\overline{\ell} = a_\ell$ and ${1\over2}|\ell|^2 = b_\ell$. With the use of the metric $H(z)^{-1}$ for $L^{-1}$, the cokernel $\text{H}^1(X, L^{-1})$ of$$C_{0,0}^\infty(X, L^{-1}) \overset{\overline{\partial}}{\to} C_{0,1}^\infty(X, L^{-1})$$can be identified as the kernel of the $\overline{\partial}^*$ of the map $\overline{\partial}$ (with respect to the metric $H(z)^{-1}$ of $L^{-1}$). An element $\text{H}^1(X, L^{-1})$ can now be described by a $C^\infty$ $(0, 1)$-form $g(z)\,\text{d}\overline{z}$ on $\mathbb{C}$ such that: (i) $g(z + \ell) = e^{-a_\ell z - b_\ell}g(z)$ for $z \in \mathbb{C}$ and $\ell \in L$ and (ii) $e^{\gamma z \overline{z}} g(z)$ is antiholomorphic, which means that $\overline{\partial}^* g(z) = -H(z)\partial_z(H(z)^{-1}g(z))$ (with $H(z) = e^{-\gamma z\overline{z}}$) is identically zero. In other words, $g(z)\,\text{d}\overline{z}$ defines a $L^{-1}$-valued $(0, 1)$-form on $X$ so that (i) states the fact that $g(z)\,\text{d}\overline{z}$ is $L^{-1}$-valued and (ii) states the fact that $g(z)\,\overline{z}$ is harmonic with respect to the metric $H(z)^{-1}$ of $L^{-1}$. Note that harmonic means both $\overline{\partial}$-closed and $\overline{\partial}^*$-closed, but $\overline{\partial}$-closedness is automatic for a $(0, 1)$-form on the compact Riemann surface $X$. The Serre duality comes from the pairing$$\text{H}^0(X, L \otimes K_X) \times \text{H}^1(X, L^{-1}) \ni (f, g\,\text{d}\overline{z}) \mapsto \int_X f(z)\,\text{d}z \wedge g(z)\,\text{d}\overline{z}$$because $K_X$ is trivial so that we can identify $f \in \text{H}^0(X, L)$ with $f\,dz \in \text{H}^0(X, L \otimes K_X)$.<|endoftext|> TITLE: The work of mathematicians outside their professional environment QUESTION [14 upvotes]: As it is reasonable to think the work of mathematicians will be developed/made in their offices of universities (or in eventual seminars or conferences), here are the colleagues, books and journals, connection to databases and blackboards. My belief is that a great part of mathematicians continue, somehow, their work outside working hours of their professional environment the university. In fact I think they have enough resources in their homes for this purpose and that they communicate with collaborators or colleagues while they are in the continuation (progress/attempts) of their research in their homes. I even evoke periodic meetings of nearby collaborators to study and work in specific problems. Question. Is it reasonable to think that the professional mathematician does research in mathematics outside the office of his/her university? Typically, under what conditions? Many thanks. The secondary question is a general overview of this situation and scenario, in case that the work outside of their professional enviroment is remarkable and can be characterized. I don't know if there are well-known examples of proofs of theorems due to mathematicians having an origin at home, coffee shops...I say research sessions/working day outside their offices. Thus an answer for the question under what conditions? should be pedagogical and informative, so that your colleagues and the general public can to know how the research in mathematics is done outside of university and get good results (and if there are general advices to schedule research sessions, remarkable preferences or tips to research in mathematics outside your office of your university). REPLY [2 votes]: I almost never do research at my place of work in a University. I teach my classes there, then go home to think. It's almost impossible for me to really think in a public setting.<|endoftext|> TITLE: What are the applications of the Mazur-Ulam Theorem? QUESTION [14 upvotes]: Every bijective isometry between normed spaces is affine. This well-known and beautiful statement, the Mazur-Ulam Theorem, was proved in 1932, but the proof has been simplified and polished in years, by contributions of more people (Lyusternik, Borsuk, Vogt, Väisälä, and others I'm not aware of) until the final form of a little gem (the last version I know, by Bogdan Nica, can be explained in conversation). Reaching a linearity property from a metric assumption is something peculiar. Another result of this "from metric to linear" form, that comes to my mind, is the linearity of the metric projector on a subspace of a Hilbert space. However, the latter is a fundamental result with very important and rich consequences on the structure of Hilbert spaces, and applications in the Calculus of Variations. On the opposite, I know no application of Mazur-Ulam theorem, and I would be really glad to learn some. Applications of any generalization of it are also welcome. edit. Of course, knowing that surjective isometries on a normed space are linear has a clear foundational application, since any theory about isometries of normed spaces can assume linearity, and exploit the linear theory of operators. Then the question is: what natural sources of (a priori non linear) isometries of normed spaces are there? REPLY [6 votes]: The Mazur-Ulam theorem is used in Bader, Uri; Furman, Alex; Gelander, Tsachik; Monod, Nicolas Property (T) and rigidity for actions on Banach spaces. Acta Math. 198 (2007), no. 1, 57–105 in order to simplify the theory of isometric actions on Banach spaces. In this connection it is worth mentioning that the question on an analogue of the Mazur-Ulam theorem for convex sets (which they ask on page 88) was answered by Schechtman, see Generalizing the Mazur-Ulam theorem to convex sets with empty interior in Banach spaces<|endoftext|> TITLE: Which $\ast$-algebras are $C^\ast$-algebras? QUESTION [21 upvotes]: It's well-known that the norm on a $C^\ast$-algebra is uniquely determined by the underlying $\ast$-algebra by the spectral radius formula. Therefore there should be a way to axiomatize $C^\ast$-algebras directly in terms of the $\ast$-algebra structure, without explicitly talking about a norm. Question 1: How does one do this? That is, which $\ast$-algebras are $C^\ast$-algebras? Question 2: How does one axiomatize those $\ast$-algebras which embed into a $C^\ast$-algebra (equivalently, embed into their $C^\ast$-enveloping algebra)? Some possibilities: Perhaps a $\ast$-algebra is a $C^\ast$-algebra iff the spectral radius is a complete, submultiplicative norm? Perhaps a $\ast$-algebra embeds into a $C^\ast$-algebra iff every element has finite spectral radius? If the first guess above (or something like it) is correct, it would still be nice to break it down into more manageable chunks. EDIT: I'm currently fascinated by the following observation. Let that if $A$ be any algebra over $\mathbb C$, and $a \in A$. Let $B$ be the subalgebra of $A$ generated by $a$, and let $C$ be the subalgebra of $A$ obtained from $B$ by closing under those inverses which exist in $A$, so that $C \cong \mathbb C[a][\{(a-\lambda)^{-1} \mid \lambda \not \in Spec(a)\}]$. Writing a general element $c \in C$ as a rational function $c = \phi(a)$, we have $Spec(c) = \phi(Spec(a))$. It follows that he spectral radius is a homogenous, subadditive, submultiplicative, power-multiplicative function on $C$. If we assume that the spectral radius in $A$ of any nonzero element of $C$ is finite and nonzero, it follows that the spectral radius is in fact a submultiplicative, power-multiplicative norm on $C$. So it seems natural to stipulate that (if $A$ is a $\ast$-algebra, and maybe assuming that $a$ is normal?), every "Cauchy sequence" in $C$ should have a unique "limit" in $A$ with respect to the spectral radius. I wonder how far this condition is from guaranteeing that $A$ is a $C^\ast$-algebra? REPLY [4 votes]: TL;DR: A $*$-algebra embeds in a $C^*$-algebra iff it is archimedean with no non-trivial infinitesimals. In this case the $*$-algebra is a $C^*$-algebra iff a certain norm is complete. I do explicitly talk about a norm here, but it is explicitly constructed from a natural order structure associated with the $*$-operation. Let $A$ be a complex $*$-algebra. In what follows we assume $A$ is unital and regard accordingly $\mathbb{C}$ as subalgebra of $A$. This is not a real restriction, as one can always unitalize $A$. We define $$ A_+=\left\{\sum_{i=1}^n x_i^*x_i\mid n\in\mathbb{N},~x_1,\ldots,x_n\in A\right\}$$ and note that it is a convex cone in $A$, thus it defines on it a partial order by $$ x\leq y \quad \Longleftrightarrow \quad y-x\in A_+. $$ Next we define for $x\in A$, $$ \|x\|=\sqrt{\inf\{\alpha\in \mathbb{R}_+\mid x^*x\leq \alpha\}}\in [0,\infty] $$ (using the conventions $\inf\emptyset=\infty$ and $\sqrt{\infty}=\infty$). Claim: $A$ is embeddable in a $C^*$-algebra iff it is archimedean with no non-trivial infinitesimals iff $\|\cdot\|$ is a norm on $A$ and in this case, $A$ is a $C^*$-algebra iff this norm is complete. Let me elaborate. If $-1\in A_+$ then $\|\cdot\|=0$ identically, thus we assume from now on that $-1\notin A_+$. $A$ is said to be archimedean if $\|\cdot\|$ attains only finite values. We define the set of infinitesimal elements in $A$ to be $A_i=\{x\mid \|x\|=0\}$. The claim above follows from the following two facts. For proofs see section 2.1 here. Fact 1: $A$ has a unital $*$-representation into a $C^*$-algebra iff it is archimedean and $A_i$ is a two-sided ideal which is in the kernel of every such a representation. Fact 2: If $A$ is archimedean then $\|\cdot\|$ is a seminorm on $A$ and the corresponding norm on $A/A_i$ satisfies the $C^*$-property. In particular, $A/A_i$ $*$-embeds into the $C^*$-algebra obtained by completion.<|endoftext|> TITLE: How are characteristic classes morphisms of infinite loop spaces? (if they are) QUESTION [10 upvotes]: The direct sum of real vector bundles endows $BO=\mathrm{colim} BO(n)$ with a natural structure of abelian group up to homotopy. The same applies to the classifying spaces of all groups in the Whitehead tower of $O$, i.e., one has a natural structure of abelian group up to homotopy on $BSO$, $BSpin$, $BString$, etc. As $w(E\oplus F)=w(E)\cup w(F)$ one sees that $$ w_1\colon BO \to K(\mathbb{Z}/2\mathbb{Z},1) $$ is a morphism of abelian groups up to homotopy, and that similarly $$ w_2\colon BSO \to K(\mathbb{Z}/2\mathbb{Z},2) $$ is a morphism of abelian groups up to homotopy. One can even make a step further and see $$ BSO \to BO\xrightarrow{w_1} K(\mathbb{Z}/2\mathbb{Z},1) $$ and $$ BSpin \to BSO\xrightarrow{w_2} K(\mathbb{Z}/2\mathbb{Z},2) $$ as ``short exact sequences of abelian groups up to homotopy''. One can be more ambitious here. Not only $BO$ is an abelian group up to homotopy, but it is an $\infty$-loop space, i.e. $BO=\Omega^\infty bo$ for a certain connective spectrum $bo$. The same applies to $BSO$, $BSpin$,etc., and it also applies to $K(\mathbb{Z}/2\mathbb{Z},n)$ as $K(\mathbb{Z}/2\mathbb{Z},n)=\Omega^\infty\Sigma^n H\mathbb{Z}/2$. So one may hope that the above sequences are actually infinitely deloopable and come from fibrations $$ bso \to bo\xrightarrow{\Omega^{-\infty}w_1} \Sigma H\mathbb{Z}/2 $$ and $$ bspin \to bso\xrightarrow{\Omega^{-\infty}w_2} \Sigma^2 H\mathbb{Z}/2 $$ of connective spectra. Versions of this latter statement seem to appear in the literature, at least in the form "$BSO \to BO\xrightarrow{w_1} K(\mathbb{Z}/2\mathbb{Z},1)$ is a fibration of infinite loop spaces" which however I am only able to give a precise meaning by interpreting it as above. For instance one finds: "Recall that $BSpin^c$ participates in a fibration of infinite loop spaces $K(\mathbb{Z},2)\to BSpin^c\to BSO\xrightarrow{bw_2}K(\mathbb{Z},3)$'' in section 7 of Ando-Blumberg-Gepner's Twists of K-theory and TMF. My question is: Is it true that the above are indeed fibrations of connective spectra inducing the usual fibrations of topological spaces via $\Omega^\infty$? Where can I find a rigorous proof of this statement? REPLY [11 votes]: Your sequences are all arise in the following standard way. Suppose $x$ is an $(n-1)$--connected spectrum and let $X = \Omega^{\infty} x$. One always has a fibration sequence $$y \rightarrow x \rightarrow \Sigma^n H\pi_n(X)$$ and applying $\Omega^\infty$ to this yields a fibration sequence of spaces $$Y \rightarrow X \rightarrow K(\pi_n(X),n).$$ $Y$ is the $n$--connected cover of $X$. (In your situation, one has successive covers $bspin \rightarrow bso \rightarrow bo$.)<|endoftext|> TITLE: In what sense is SL(2,q) "very far from abelian"? QUESTION [25 upvotes]: I am far from an expert in this area. I'd be grateful if someone could explain in what sense $\mathop{SL}(2,q)$ is "very far from abelian," to quote Emanuele Viola? Why does Theorem 1 (below) justify this "far from abelian" claim?                     (Snapshot from blog here.) Earlier related MO questions: Measures of non-abelian-ness How nearly abelian are nilpotent groups? . REPLY [26 votes]: Another measure of how far a finite group $G$ is from commutative is the "commuting probability", which goes back at least as far as W.H. Gustafson, and certainly predates the work of Gowers. This is just the probability that a pair of elements of $G$ commute, where the uniform distribution is put on $G \times G$. This turns out to be $\frac{k(G)}{|G|}$, where $k(G)$ is the number of conjugacy classes of $G$. It has been noted by several authors, including in recent years P. Lescot, and long ago, (implicitly) by E. Wigner, that this is somewhat related to the smallest degree $d$ of a non-linear complex irreducible character of $G$, though as Derek Holt's example in comments illustrates, the influence can wane if $G$ is far from perfect and has many linear characters. However, when $G$ is perfect, we have $1 +(k(G)-1)d^{2} < |G|$, so that $k(G) < \frac{|G|}{d^{2}}+1,$ and the commuting probability of $G$ is bounded above by something only slightly larger than $\frac{1}{d^{2}}.$ As noted in the question the smallest non-linear complex irreducible character degree of ${\rm SL}(2,q)$ is $\frac{q-1}{2}$ when $q$ is odd, so leading to a upper bound for the commuting probability of something close to $\frac{4}{q^{2}}$ for ${\rm SL}(2,q).$ Another approach to this in the case $G = {\rm PSL}(2,q)$ is to note that ${\rm PSL}(2,q)$ always has at most $q+1$ complex irreducible characters (equality is achieved when $q$ is even). In this case, we have $|G| = \frac{q(q-1)(q+1)}{2}$ if $q$ is odd, and the commuting probability for $G$ is less than $\frac{2}{q(q-1)}$ when $q$ is odd (and is equal to $\frac{1}{q(q-1)}$ when $q$ is even). The same inequalities hold for ${\rm SL}(2,q)$. Note that this gives that the commuting probability of $G = {\rm PSL}(2,q)$ is bounded above by something like $c|G|^{\frac{-2}{3}}$ for a small fixed constant $c.$ On the other hand, Bob Guralnick and I proved (using the classification of finite simple groups) that for any finite group $G$ with $F(G) = 1$, the commuting probability of $G$ is at most $|G|^{-\frac{1}{2}}$, so the bound which holds for ${\rm PSL}(2,q)$ is significantly smaller than the general bound we obtained. Later edit: To be more precise, the arithmetic mean (say $\mu_{d}(G)$) of the complex irreducible character degrees of $G$ is quite strongly related to the commuting probability ${\rm cp}(G)$ of $G$. The Cauchy-Schwartz inequality gives $\sum_{\chi \in {\rm Irr}(G)} \chi(1) \leq \sqrt{k(G)|G|}$ so that $\mu_{d}(G){\rm cp}(G) \leq \sqrt{{\rm cp}(G)}$ and hence ${\rm cp}(G) \leq \frac{1}{\mu_{d}(G)^{2}}.$ REPLY [9 votes]: I am going to piggyback off Bullet51's answer and mention product-free sets. In particular, one might want to see a failure of Theorem 1 (from the question) for a finite abelian group. As is explained in the introduction to Gowers' paper on quasirandom groups, if $G$ is a finite abelian group then there is some set $X\subseteq G$ of "constant density", namely $|X|\geq |G|/3$, such that $X$ is "product-free" (or "sum-free" for abelian groups). Specifically, if $x_1,x_2\in X$ then $x_1+x_2\not\in X$. So, in the context of Theorem 1, take $A=B=X$ and $C=-X$. Let $g=0$. Then we have $Pr[a+b+c=g]=0$, and so $$ |Pr[a+b+c=g]-1/|G||=1/|G| $$ (which of course is not less than $1/|G|^{1+\Omega(1)}$). There are probably more sophisticated ways of handling this, but the upshot is that abelian groups have very large product-free sets, while "quasirandom" groups like $SL(2,q)$ do not. Gowers shows that if $G$ is a finite group with no nontrivial representations of degree less than $k$, then any set of size greater than $|G|/k^{1/3}$ is not product-free. For $G=SL(2,q)$ we can take $k=(q-1)/2$. So there is some absolute constant $\epsilon>0$ such that if $G=SL(2,q)$ then $G$ has no product-free set of size greater than $|G|^{1-\epsilon}$ (in contrast to abelian groups, which have product-free sets of "linear" size, namely, $|G|/3$). By the way, in this paper Green and Ruzsa determine the maximal densities of sum-free subsets of finite abelian groups.<|endoftext|> TITLE: Solving $AXB + X\odot C = D$ QUESTION [10 upvotes]: I need to solve the following equation for $X$ with $d$-by-$d$ matrices $A,B,C,D$ and Hadamard product $\odot$ $$AXB + X\odot C = D$$ Vectorizing all terms gives a solution with $O(d^6)$ complexity, which is intractable since $d\approx 1000$ in my application. Is there something I can do to get an estimate in $O(d^3)$ time? To add more information about the structure, A,B,C are moment matrices. Specifically, for random variables $X,Y$ they are $$A_{ij}=E[X_iX_j]$$ $$B_{ij}=E[Y_iY_j]$$ $$C_{ij}=E[X_iY_j]$$ $$E_t[D_{ij}]=C_{ij}$$ To give even more background, this comes up in problem of speeding up neural network training which for a single layer can be viewed as the following problem $$\text{minimize}_{W} E[(x'Wy)^2]$$ Here $x$ and $y$ are random variables with shape $(d,1)$, the gradient is $E[x_iy_j]$ and the curvature is $E[x_i x_j y_k y_l]$. The goal is to use curvature information to obtain Newton-like correction to a noisy estimate of the gradient. Applying Isserlis theorem we can approximate curvature rank-4 tensor in terms of rank-2 covariance tensors which leads to matrices $A,B,C$ above. Substituting small sample estimate of gradient into $D$ and large sample estimate of curvature into $A,B,C$, then solving for $X$, gives us a preconditioned gradient step. Incidentally, using the same sample to estimate curvature and gradient, gives slightly simplified problem: $$AXB + X\odot C = C$$ $A$ and $B$ are known to be ill-conditioned -- $n$-th eigenvalue is approximately $1/n$ and majority of eigenvalues are numerically 0, possibly 90% of all eigenvalues. Coincidentally, this ill-conditioning is what allows neural networks to generalize. This implies that $C$ is also singular. REPLY [2 votes]: I am going to reduce your problem to another form. In a truncated version, this modified problem has been discussed here and it seems there is no apparent efficient solution à la Bartels–Stewart. However, it still might be valuable to have an alternative approach at hand, especially because such equations are being actively discussed (see T. Damm, Direct methods and ADI-preconditioned Krylov subspace methods for generalized Lyapunov equations, Numer. Linear Algebra Appl. 15, no. 9 (2008): 853-871) and because iterative approaches based on the standard Lyapunov solvers do exist. Let us apply the vec-operation to the both sides of your equation using the commutativity of Hadamard product $$ \mathrm{vec}(AXB)+\mathrm{vec}(C\odot X)=\mathrm{vec}(D). $$ Now we use the following properties $$ \text{vec}(ABC)=(C^\mathrm{T}\otimes A)\text{vec}(B) $$ and $$ \text{vec}(A\odot B)=\text{vec}(A)\odot\text{vec}(B). $$ Thus $$ (B^\mathrm{T}\otimes A)\text{vec}(X)+\mathrm{diag}\!\left[\mathrm{vec} (C)\right]\mathrm{vec} (X)=\mathrm{vec}(D). $$ Let us introduce a new matrix $U$: $$ U=(B^\mathrm{T}\otimes A)+\mathrm{diag}\!\left[\mathrm{vec} (C)\right]. $$ Our goal is now to write it in the form of a Kronecker product. $$ U=\sum_i \sigma_i V_i^\mathrm{T}\otimes W_i. $$ This is known as the nearest Kronecker product problem. Using SVD decomposition on a permuted version of $U$ as Van Loan (J. Comp. Appl. Math, 123 (2000) 85) proposed, the factors $V_i$ and $W_i$ and the singular values $\sigma_i$ can be determined, and the original equation can be written as $$ \sum_i \sigma_i\left(V_i^\mathrm{T}\otimes W_i\right)\mathrm{vec} (X)=\mathrm{vec}(D), $$ which is equivalent to $$ \sum_i \sigma_i W_i X V_i=D. $$ Truncating the sum to just 2 terms, a standard Lyapunov equation is obtained.<|endoftext|> TITLE: A reinterpretation of the $abc$ - conjecture in terms of metric spaces? QUESTION [23 upvotes]: I hope it is appropriate to ask this question here: One formulation of the abc-conjecture is $$ c < \text{rad}(abc)^2$$ where $\gcd(a,b)=1$ and $c=a+b$. This is equivalent to ($a,b$ being arbitrary natural numbers): $$ \frac{a+b}{\gcd(a,b)} < \text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^2$$ Let $d_1(a,b) = 1- \frac{\gcd(a,b)^2}{ab}$ which is a proven metric on natural numbers. Let $d_2(a,b) = 1- 2 \frac{\gcd(a,b)}{a+b}$, which I suspect to be a metric on natural numbers, but I have not proved it yet. Let $$d(a,b) = d_1(a,b)+d_2(a,b)-d_1(a,b)d_2(a,b) = 1-2\frac{\gcd(a,b)^3}{ab(a+b)}$$ Then we get the equivalent formulation of the inequality above: $$\frac{2}{1-d_2(a,b)} < \text{rad}(\frac{2}{1-d(a,b)})^2$$ which is equivalent to : $$\frac{2}{1-d_2(a,b)} < \text{rad}(\frac{1}{1-d_1(a,b)}\cdot\frac{2}{1-d_2(a,b)} )^2$$ My question is if one can prove that $d_2$ and $d$ are distances on the natural numbers (without zero)? Result: By the answer of @GregMartin, $d_2$ is a metric. By the other answer $d$ is also a metric. Edit: By "symmetry" in $d_1$ and $d_2$, this interpretation also suggests that the following inequality is true , which might be trivial to prove or very difficult or might be wrong and may be of use or not in number theory: $$\frac{1}{1-d_1(a,b)} < \text{rad}(\frac{2}{1-d(a,b)})^2$$ which is equivalent to $$ \frac{ab}{\gcd(a,b)^2} < \text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^2$$ (This is not easy to prove, as the $abc$ conjecture $c=a+b < ab < \text{rad}(abc)^2$ would follow for all $a,b$ such that $a+b < ab$.) Second edit: Maybe the proof that $d_2,d$ are distances can be done with some sort of metric transformation, for example maybe with a Schoenberg transform (See 3.1, page 8 in https://arxiv.org/pdf/1004.0089.pdf) The idea, that this might be proved with a Schoenberg transform comes from the fact that: $$d_1(a,b) = 1-\exp(-\hat{d}(a,b))$$ so $d_1$ is a Schoenberg transform of $\hat{d}(a,b) = \log( \frac{ab}{\gcd(a,b)^2}) = \log( \frac{\text{lcm}(a,b)}{\gcd(a,b)})$ which is proved to be a metric (see Encyclopedia of Distances, page 198, 10.3 ) Third edit: Here is some Sage Code to test the triangle inequality for triples (a,b,c) up to 100: def d1(a,b): return 1-gcd(a,b)**2/(a*b) def d2(a,b): return 1-2*gcd(a,b)/(a+b) def d(a,b): return d1(a,b)+d2(a,b)-d1(a,b)*d2(a,b) X = range(1,101) for a in X: for b in X: for c in X: if d2(a,c) > d2(a,b)+d2(b,c): print "d2",a,b,c if d(a,c) > d(a,b)+d(b,c): print "d",a,b,c so far with no counterexample. Related: An inequality inspired by the abc-conjecture and two questions REPLY [3 votes]: This question has already very good answers. I justed wanted to highlight that it is possible to shorten the proofs, using the following: If $X_a = \{ a/k | 1 \le k \le a \}$ then $X_a \cap X_b = \gcd(a,b)$, which is straightforward to prove. Then $d_1(a,b) = 1-\gcd(a,b)^2/(ab) = 1-|X_a \cap X_b|^2 / (|X_a||X_b|)$ is the squared cosine metric (see Encyclopedia of Distances) and $d_2(a,b) = 1-2\gcd(a,b)/(a+b) = 1-2|X_a \cap X_b| / (|X_a|+|X_b|)$ is the Sorensen Metric (Encyclopedia of Distances). Hence $d_1,d_2$ are metrics of the form $d_i = 1- s_i$ where $s_i$ is a similarity. But then $s=s_1 \cdot s_2$ is also a similarity and $d=d_1 +d_2 -d_1 d_2 = 1-s=1-s_1 s_2$ is a metric.<|endoftext|> TITLE: Connected subgraphs and their sums QUESTION [11 upvotes]: Let $G=(V,E)$ be an undirected graph with $|V|\geq 4$ such that for any distinct vertices $a_1,a_2,b_1,b_2$, there is a path from $a_1$ to $a_2$ and a (vertex-)disjoint path from $b_1$ to $b_2$ (in other words, the graph is $2$-linked). Assign a nonnegative real number to each vertex in $V$. Suppose the sum of these numbers is $2s$, and a subset of them has sum exactly $s$. What is the largest constant $c$ for which (regardless of $G$ and $s$) there always exists a subset $V'\subseteq V$ such that both $V'$ and $V\backslash V'$ form connected subgraphs, and the sum of the numbers in $V'$ belongs to $[cs,s]$? The following example shows that $c\leq 5/6$: a clique $K_5$ with one edge $e$ removed, numbers $3,3$ on the two vertices adjacent to $e$, and $2,2,2$ on the remaining vertices. Another example is to take $K_{3,3}$ with $3,3,3$ on one side and $1,1,1$ on the other. Is $c=5/6$ tight? (It is easy to see that such a constant $c$ must exist: $c=0$ works since for any connected graph, its vertices can be partitioned into two parts such that both parts are connected.) REPLY [4 votes]: Here is an argument that shows a lowerbound of $c \geq \frac{2}{3}$. For this, we only need the weaker assumptions that $G$ is $2$-connected and the weight of each vertex is at most $\frac{W}{2}$, where $W$ is the total weight. Both these assumptions are clearly implied by the hypotheses of the original question. Theorem. Let $G=(V,E)$ be a $2$-connected graph and $w: V \to \mathbb{R}_{\geq 0}$ be such that $w(v) \leq \frac{1}{2}\sum_{u \in V} w(u):=\frac{W}{2}$ for all $v \in V$. Then there exists a subset $X \subseteq V$ such that $G[X]$ and $G[V \setminus X]$ are both connected and $\frac{W}{3} \leq \sum_{u \in X} w(u) \leq \frac{W}{2}$. Proof. We proceed by induction on $|E|$. If $|E|=3$, then $G=K_3$. In this case, there is a vertex $x$ with $w(x) \geq \frac{W}{3}$, so we may take $X=\{x\}$. For the inductive step, first note that we may assume $w(u) < \frac{W}{3}$ for all $u \in V$. Otherwise, we can take $X=\{u\}$, since $G-u$ is connected by $2$-connectivity. Choose a spanning tree $T$ of $G$, and let $x$ be a leaf of $T$ and $y$ be the unique neighbour of $x$ in $T$. By assumption, $w(x)+w(y) < \frac{2W}{3}$. If $w(x)+w(y) > \frac{W}{2}$, we may take $X=V \setminus \{x,y\}$. Therefore, we may assume that $w(x)+w(y) \leq \frac{W}{2}$. We use the well-known fact that in a $2$-connected graph, every edge can be either deleted or contracted to maintain $2$-connectivity. Let $e=xy$. If $G \setminus e$ is $2$-connected, we can apply induction. If $G / e$ is $2$-connected, we let the contracted vertex have weight $w(x)+w(y)$ and we apply induction. $\square$ Note that $2$-connectedness is required to prove $c>0$. To see this consider a star with $k$ leaves where each leaf has weight $1$ and the center of the star has weight $k$.<|endoftext|> TITLE: The locus of polynomials with specified root multiplicities QUESTION [9 upvotes]: Let $\mathcal{P}_d\cong\mathbb{A}^d$ denote the set of monic degree $d$ polynomials defined over an algebraically closed field of characteristic $0$, where we identify $f(x)$ with its coefficients. The multiplicities of the roots of $f(x)\in\mathcal{P}_d$ defines a partition $\pi(f)$ of $d$. For example, if $f(x)=(x-\alpha)^d$, then $\pi(f)=(d)$, and if $f(x)$ has distinct roots, then $\pi(f)=(1^d)$. For any partition $\sigma$ of $d$, the set $$ \mathcal{P}_d(\sigma) := \bigl\{ f\in\mathcal{P}_d : \pi(f)=\sigma\bigr\} $$ is a quasiprojective subvariety of $\mathbb{A}^d$. (This follows from elimination theory.) For example, $\mathcal{P}_d(d)$ is a curve, while $\mathcal{P}_d(1^{d-2},2)$ is an open subset of the discriminant locus $\bigl\{f\in\mathcal{P}_d:\operatorname{Disc}(f)=0\bigr\}$. Do these varieties $\mathcal{P}_d(\sigma)$ have a name? My best guess was discriminantal variety, but that term does not seem to be in use. Where have these varieties been studied? Specific references would be appreciated. REPLY [6 votes]: I may be wrong but I think "coincident root loci" mentioned in Gjergjji's comment was coined by my coauthor Jaydeep Chipalkatti. The ideals of such varieties are in general poorly understood. For a general study see: J. Chipalkatti, "On equations defining Coincident Root loci", J. Algebra 267 (2003), no. 1, 246-271. J. Chipalkatti, "Invariant equations defining coincident root loci", Archiv der Math. 83 (2004), no. 5, 422-428. H. Lee and B. Sturmfels, "Duality of multiple root loci", J. Algebra 446 (2016), 499-526. For particular cases, the following might be also of interest. A. Abdesselam and J. Chipalkatti, "Brill–Gordan loci, transvectants and an analogue of the Foulkes conjecture", Adv. Math. 208 (2007), no. 2, 491-520. It has a description of the ideal for a partition with two equal parts. A. Abdesselam and J. Chipalkatti, "The bipartite Brill-Gordan locus and angular momentum", Transformation Groups 11 (2006), no. 3, 341-370. It has a description of the ideal for a partition with two unequal parts. A. Abdesselam and J. Chipalkatti, "On Hilbert Covariants" Canadian J. Math. 66 (2014), no. 1, 3-30. It has several set-theoretic systems of equations for the varieties corresponding to rectangular partitions, and a conjecture about minimal degree of generators for the ideals. This is an $SL_2$ analogue/toy version of the Foulkes-Howe conjecture. There is also work about these varieties from a topological point of view, e.g., F. Napolitano, "On some topological invariants of algebraic functions associated to the Young stratification of polynomials" Topology Appl. 134 (2003), no. 3, 189-201. The relation to Hurwitz stacks is studied in: J. Bertin and M. Romagny, "Champs de Hurwitz", Mémoires SMF, no. 125-126 (2011), 219 p. See also here for the arXiv version. The relevant result is Theorem 9.16 relating coincident root loci and Hurwitz stacks of cyclic coverings of $\mathbb{P}^1$. Also, an interesting geometric approach to these varieties is in: G. Katz "How tangents solve algebraic equations, or a remarkable geometry of discriminant varieties", Expositiones Math. 21 (2003), no. 3, 219-261. Of course, this is by no means an exhaustive bibliography. There are many references I left out in this short MO post.<|endoftext|> TITLE: Lesser known examples of perseverance with a successful ending QUESTION [6 upvotes]: The stories of Wiles, of Perelman, and of Zhang, are very well-known to illustrate that sometimes great results are achieved through particularly long perseverance. What are lesser known-examples of that kind ? These need not be epoch-making results like those mentioned, but, say, they required at least 6 or 7 years of dedicated effort (not necessarily by a single person, and also not necessarily a first-time paper i.e. a paper correcting a previously incomplete attempt also would qualify as a good answer). Edit: to clarify, I do not have in mind long endeavours by many people over centuries like Galois Theory, really just a particular theorem within the lifetime of their author(s). P.s. if moderators could help make this question community-wiki, that would be great, thank you. REPLY [2 votes]: I'm not sure if you're insisting on examples in which a mathematician (or group of mathematicians) works single-mindedly on a single problem for many years and finally conquers the problem. If so, then the following might not qualify. But it might qualify if your conditions are not as stringent as that. I quote from Ilse Fischer's response to being awarded the AMS Robbins Prize. The idea of working on Robbins' last open conjecture on alternating sign matrices slowly manifested in my mind when I was writing a grant proposal about 10 years ago, when I identified it as an ultimate, albeit unrealistic, goal. In the beginning I hardly dared spend much time on it, but every now and then I discussed it with other combinatorialists. Roger Behrend and Matjaž Konvalinka were obviously among them, but I also had a particularly fruitful exchange with Arvind Ayyer back in 2012, which led us to several conjectures on the enumeration of extreme diagonally and antidiagonally symmetric alternating sign matrices of odd order. About three years later, Arvind, Roger, and I were able to prove these conjectures, and to some extent also this work paved the way for the eventual proof of Robbins' conjecture. I feel deeply honored and moved to receive, together with Matjaž and Roger, the David P. Robbins Prize.<|endoftext|> TITLE: Solution to simple mathematical game QUESTION [28 upvotes]: Consider the following game (that I made up). Two players each attempt to name a target number. The first player begins by naming 1. On each subsequent turn, a player can name any larger number that differs from the current number by a divisor of the current number, including itself. For example, if the current number were 6, the next number named could be any one of 7, 8, 9, or 12. A player that names the target number wins. A player forced to name a number larger than the target number loses. If the target number is odd, the first player can win trivially by always naming an odd number, because the difference between the current number and the target number is even, and no odd number has an even divisor. The thing I can't seem to figure out is whether the first player also has a winning strategy for every even number greater than 6, and what that strategy is if so. Obviously, one can determine the optimum course of play for any target number by working backward, and doing so seems to suggest the first player can always win, but I haven't been able to understand why. Is there any way to prove that the first player can always win for any number greater than 6? REPLY [5 votes]: I think that every even target greater than $6$ is indeed a first player win. I have no proof but present evidence and speculation on what might lead to a proof. For every even target at least up to $10000$ there is a strategy which is $2k \rightarrow 2k+1$ except for a small list of exceptions. For example Target $T=9580$ Strategy: The other player will always state even numbers. Our response should be $2k \rightarrow 2k+1$ with these four exceptions $$4786 \rightarrow 7179,\ 7184 \rightarrow 7633,\ 9100 \rightarrow 9105,9572 \rightarrow 9574,\ 9578\rightarrow 9580$$ Let me justify that. Consider the sets $$W=\{9574,9580\}$$ $$L=\{4787,7185,9101,9573,9575,9579\}$$ These are the even numbers which it is a winner to state and the odd numbers which it is a loser to state. If the opponent never states a member of $W,$ they can't win. But those are $T=2^2\cdot 5\cdot 479$ and $9574=2\cdot 4787 .$ If we never state the odd numbers $9579,9575,9101,7185=T-1,T-5,T-479,T-2395$ or $4787=9574-4787,$ then they can't state anything from $W$ , unless perhaps we state an even number. We could state $9580$ and win. And if we state $9574=2\cdot 4787$ they can only reply $9576$ to which we can reply $9577 \notin L.$ So we just need responses other than $2k \rightarrow 2k+1$ for $2k+1 \in L.$ So preferably $2k \rightarrow 2k+t\lt T$ where $t \gt 1$ is an odd divisor of $k$ and $2k+t \notin L$. And we have done this for four of the five cases.For $9573-1=9572=2^2\cdot 2393$ there is no such $t$ to use. Fortunately, $9572 \rightarrow 9574$ allows us to state a winner. OPTIONAL DIGRESSION: Here are three more examples target $T=2^j$: Strategy: $2k\rightarrow 2k+1$ except $2^k-2 \rightarrow 2^k.$ So $W=\{2^k\}\ \ L=\{2^k-1\}.$ target $T=2p$ for $p$ prime but not a Fermat prime $p=2^s+1.$ Strategy: $2k \rightarrow 2k+1$ except $2p-2 \rightarrow 2p$ and $p-1 \rightarrow p-1+t$ where $t \gt 1$ is the next smallest odd divisor. So $W=\{2p\}\ \ L=\{p,2p-1\}.$ Since $p-1$ is not a power of $2,$ there is such an odd divisor. target $T=2p$ for $p$ a Fermat prime $p=2^s+1:$ Strategy: $2k \rightarrow 2k+1$ except $2p-2 \rightarrow 2p$ and $p-3 \rightarrow p-3+t$ where $t\gt 1$ is the next smallest odd divisor (so also $t \gt 3.$). here $W=\{p-1,2p\}\ \ L=\{p,2p-3\}.$ Since $p-1=2^s$ it is actually an even winner as the only responses are $p$ and even losers. END OF DIGRESSION Question which arise are: How do we come up with the (or a) valid strategy? How can we rule out $1 \in L?$ How large might the set of exceptions be? The methods for an even target $T=2n$ is to start $W=\{2n\}$ and $L=\{\}$ where $W$ will be the even winners and $L$ will be the odd losers. While there are unexamined members of $W \cup L$, consider the largest such. If it is $2k \in W$, then add $2k-t$ to $L$ where $t$ ranges over the odd divisors of $k$. Then pick the next unexamined member. If it is $2f+1 \in L$ then either find a good response to $2f$ or, if there is none, add $2f$ to $W$: Consider $2f+t$ where $t$ goes through the divisors of $2f$, As soon as you get an even winner in $W$ or and odd number not in $L$, add that to your list of exceptional rules. If none of those happen, add $2f$ to $W.$ Then pick the next unexamined member. Stop when everything currently in $W \cup L$ has been examined. The exceptions, given an even target $T \gt 6$, are $W,$ the even numbers it is a winner to state and $L$ the odd numbers it is a loser to state. For $\mathcal{T}$ a set of even targets (such as all even integers greater than $6$), One way to prove that every $T \in \mathcal{T}$ is a first player win is to show that every $T \in \mathcal{T}$ has smallest exception $e \gt 1$. A stronger result might be easier. I would conjecture that $e \gt \frac{T}{4}.$ For even $6 \lt T \lt 10000$ there are $114$ cases where $\frac{e}{T} \lt \frac{15}{41}.$ Here are the first few pairs $[T,e]$ if we sort according to increasing ratio $\frac{e}{T}$ $$[114,31],[10,3],[30,9],[462,151],[130,43],[450,151],[1254,421],[2058,691],[1250,421],[3654,1231],[2050,691],[2530,853],[3650,1231],[4930,1663],[5730,1933],[8454,2851],[8450,2851],[9250,3121],[9570,3229]$$ So $\frac{e}{T} \gt \frac14$ (up to $T=10000$) and $\frac{e}{T} \gt \frac13$ for $T \gt 130.$ On the other hand, there are $T$ on the edge of the examined range with $\frac{e}{T} \lt 0.337$ There are $12$ cases for $4 \leq T \leq 10000$ with $2$ exceptions, The powers of $2.$ However there are $1285$ with exceptional set of size $3$, all cases of $T=2p$ and some, but not all, $T=2^sp$ for larger $s.$ Here are the first few counts: $$[2, 12], [3, 1285], [4, 28], [5, 609], [6, 187], [7, 60], [8, 525].$$ The largest are $72,75,80,83,84,94,97$ which occur once each. Here is a cumulative plot So just under $\frac23$ of the cases have exceptional sets of size $12$ or less and over $90\%$ have exceptional set of size $25$ or less. You can just say that the target is the largest number which can be stated and the first player without a move loses. I didn't consider the case of first player without a move wins, but it should be largely similar.<|endoftext|> TITLE: Minimization problem for convolution QUESTION [11 upvotes]: Let $g(x)$ be a non-negative function supported on $[0,1]$. Let $g \ast g$ denote the convolution of $g$ with itself. Question: What is the smallest possible $L^1(0,1)$ norm of $g$, if I require that $(g \ast g) (t) \geq 1$ for all $t \in [0,1]$. Clearly one needs $\|g\|_1 \geq 1$. However, $\|g\|_1=1$ cannot be achieved. But what is the best value that can actually be achieved? (Maybe the optimizing function is explicitly known? Something like $g(x) = 1/\sqrt{\pi x}$ works and gives $\|g\|_1 \approx 1.13$, but probably something smaller is possible.) REPLY [5 votes]: Rick Barnard and I looked at the same problem for the auto-convolution instead of the convolution: arXiv. Our constants are a bit worse because you basically need two square-root singularities, one on each side. These types of inequalities tend to be relevant in combinatorics and they tend to be pretty hard (we discuss some in our paper). One I like a lot can be found in this MO post. (I would have commented but you need 50 reputation for that, sorry.)<|endoftext|> TITLE: Concrete description of lift in Arens-Eells space QUESTION [5 upvotes]: Let $X$ be a compact pointed metric subspace of the $d$-dimensional Euclidean space $(\mathbb{R}^d,d_E)$ and let $AE(X)$ denote its Arens-Eells space. Then a result of Nik Weaver shows that for every Lipschitz map $f:X\rightarrow E$ into a separable Banach space, there exists a unique continuous linear extension $F:AE(X)\rightarrow E$ satisfying $$ F\circ \delta = f, $$ where $\delta$ is the canonical isometric embedding of $X$ in $AE(X)$. (See Nik's book for more details). Question: Is there a concrete description of what $F$ is or how to explicitly construct it? I would like to use it for computations... REPLY [3 votes]: This has already been answered but I would like to add some points which I hope might be of interest. The clearest expression is, in my opinion, in the general setting of a complete metric space $M$ with a base point $x_0$ and radius $1$. One then defines the Banach space $F$ consisting of the Lipschitz functions which respect the base, i.e., map $x_0$ onto $0$, with the natural norm. Then one can embed the metric space isometrically into a Banach space $E$ with the universal property that every Lipschitz map on $M$ into a Banach space $G$ which respects the base lifts to a unique linear operator on $E$ with the same norm. If one takes $G$ to be one-dimensional, then one sees that the dual of $E$ is the space of Lipschitz functions above. Now the unit ball of the latter has a natural compact topology (pointwise or uniform convergence) and so, by standard duality theory, is a dual space. One can then turn this reasoning on its head and define $E$ to be its predual. One can see this more clearly if one uses a little terminology from category theory. If we map a Banach onto its unit ball, then we define a functor from the category of Banach spaces (with linear contractions as morphisms) into that of pointed metric spaces with base-point preserving Lipschitz functions, as above, then what we have constructed is just an adjoint functor. That is, the Arens-Eells space can be interpreted as a Free-functor and $AE(X)$ is a free object over $X$. This is perhaps not really a concrete construction, but it follows from the existence that the space is just the so-called free vector space over $M$ (as a pointed set), completed under a suitable norm (basically the observation of Nik Weaver above). On the other end of the concrete-abstract spectrum, the existence of such an object (often called the free Banach space over $M$) can be deduced from the Freyd adjoint theorem.<|endoftext|> TITLE: Is there a measure on $[0,1]$ that is 0 on meagre sets and 1 on co-meagre sets QUESTION [9 upvotes]: I'm curious if there is a finite measure on the $\sigma$-algebra of subsets of $[0,1]$ with the Property of Baire, whose null sets are exactly the meagre sets. I'd also be interested how "nice" such a measure can be like can it be Radon(when restricted to Borel sets) for example. REPLY [13 votes]: The answer is no. Assume that such a measure $\mu$ exists. First, since every singleton in $[0,1]$ is closed with empty interior, $\mu(\{x\}) = 0$ for all $x \in [0,1]$. Write $B_{x,\epsilon}$ for the open ball around $x$ of radius $\epsilon$ with respect to the standard metric on $[0,1]$. By countable additivity, for all $x \in [0,1]$, $\mu(B_{x,2^{-n}}) \to 0$. If we take an enumeration of the rationals $(q_i)_{i \in \mathbb{N}}$, for each $i$ there exists an $n_i$ such that $\mu(B(q_i,2^{-n_i})) < 2^{-i}$. So $D_1 = \bigcup_{i=1}^\infty B(q_i,2^{-n_i})$ is a dense open set with $\mu(D_1) \leq 1$. By re-doing the previous construction, picking $\mu(B(q_i,2^{-n_i})) < 2^{-(i+k)}$, we can define dense open sets $D_k$ with $\mu(D_k) \leq 2^{-k}$. Now, by countable additivity $N = \bigcap_{k=1}^\infty D_k$ has measure zero. The set $[0,1]\setminus N$ is a union of closed sets with empty interior, i.e. a meagre set, so $\mu([0,1] \setminus N) = 0$ as well, so $\mu([0,1]) = 0$. I see that Nate Eldredge was a bit quicker than I was, so I'll add that it is possible to find a finitely-additive probability measure whose null sets are exactly the meagre sets -- this is most easily done using the isomorphism between the Baire property algebra modulo meagre sets and the algebra of regular open sets. REPLY [11 votes]: No. For any finite Borel measure $\mu$ on $[0,1]$, there is a comeager Borel set of $\mu$-measure zero. First note that $\mu$ has at most countably many atoms, so it will be possible to find a countable dense set $D \subset [0,1]$ containing no atoms, i.e. $\mu(D) = 0$. Now any finite Borel measure on a metric space is outer regular, so for any $n$ there is an open set $U_n$ containing $D$ and with $\mu(U_n) < 1/n$. Setting $G = \bigcap_n U_n$, we see that $G$ is a dense $G_\delta$ (hence comeager) and $\mu(G) = 0$. Relevant to your second question, the answers in the question linked above also mention that any finite Borel measure on a Polish space is Radon.<|endoftext|> TITLE: How to construct the Moore spectrum? QUESTION [6 upvotes]: I am trying to understand how the Moore spectrum is constructed. And in reading Foundations of Stable Homotopy Theory by David Barnes and Constanze Roitzheim, I see that in example 8.4.7 (pg 340) they show how to construct the Moore spectrum. The construction goes something like this: Let $G$ be an abelian group. Assume that $G$ is obtained by the following short exact sequence: $$0 \to F_2 \xrightarrow{\rho} F_1 \to G \to 0$$ where $F_2$ is the free group on the set $I_2$ and $F_1$ is the free group on the set $I_1$. The set $I_1$ is a generating set of $G$ and $\rho(I_2)$ is the corresponding set of relations. No issues here since we can always find a group presentation. Next we construct a cofiber sequence: $$\bigvee_{I_2} \mathbb{S} \xrightarrow{r}\bigvee_{I_1}\mathbb{S} \to M(G)$$ where $\pi_0 (r)=\rho$. Now I can see that $F_2=\bigoplus_{I_2} \mathbb{Z}$, $F_1=\bigoplus_{I_1} \mathbb{Z}$ and since $\pi_0(\mathbb{S})=\mathbb{Z}$ and $\pi_0$ preserves coproducts, I can see that $\rho$ sits between the following: $$\pi_0\left(\bigvee_{I_2} \mathbb{S}\right) \xrightarrow{\rho} \pi_0\left(\bigvee_{I_1}\mathbb{S}\right)$$. I don't however see that such an $r$ ought to exist such that $\pi_0(r)=\rho$. Why can we do this? REPLY [12 votes]: What you're missing is that $[\mathbb{S},\mathbb{S}]=\mathbb{Z}$. Let now $R$ be the infinite matrix of integers representing $\rho$. Note that since $\rho$ takes value in $\bigoplus_{I_1}\mathbb{Z}\subseteq \prod_{I_1}\mathbb{Z}$, all of its columns have only finitely many non-zero values. So, for every column of $R$ we can construct a map $\mathbb{S}\to \bigvee_{I_1}\mathbb{S}$ sending $1\in\pi_0(\mathbb{S})$ to the column in $\bigoplus_{I_1}\mathbb{Z}=\pi_0\bigvee_{I_1}\mathbb{S}$ (for example by composing $\mathbb{S}\to \bigvee_{I_1'}\mathbb{S}\to \bigvee_{I_1}\mathbb{S}$, where $I'_1$ is the finite subset of $I_1$ where the column is non-zero and using that finite wedges are categorical products and so $[\mathbb{S},\bigvee_{I_1'}\mathbb{S}]=\prod_{I_1'}[\mathbb{S},\mathbb{S}]=\prod_{I_1'}\mathbb{Z}$). Finally, using that the wedge is the categorical coproduct we can put it all together in a map $$\bigvee_{I_2}\mathbb{S}\to \bigvee_{I_1}\mathbb{S}$$ as required. In fact what I'm doing is essentially proving that the map $$\pi_0:\left[\bigvee_{I_2}\mathbb{S},\bigvee_{I_1}\mathbb{S}\right]\to \mathrm{Hom}\left(\bigoplus_{I_2}\mathbb{Z},\bigoplus_{I_1}\mathbb{Z}\right)$$ is an isomorphism (injectivity is quite easy to show).<|endoftext|> TITLE: Correspondence between classes of model categories and classes of $\infty$-categories QUESTION [16 upvotes]: We know by Karol Szumiło's thesis (https://arxiv.org/pdf/1411.0303.pdf) that there is an equivalence between the two fibration categories of cofibration categories on one side and cocomplete $\infty$-categories on the other. By a dual argument, we obtain an equivalence between fibration categories and complete $\infty$-categories (I'm guessing, in the form of cofibration categories). Does either of these constructions restrict to the case of combinatorial model categories vs. presentable $\infty$-categories? In other words, is it an (almost) immediate consequence of the main theorem therein that there is also an equivalence between some model-like structures of combinatorial model categories and presentable $\infty$-categories? Similar question, but considering all model categories (or maybe just categories with both a fibration and a cofibration structure) on one side and complete/cocomplete $\infty$-categories on the other? REPLY [3 votes]: As pointed out in the answer to Localizing $\mathrm{CombModCat}$ at the Quillen equivalences, the answer to Question 1 whether there is also an equivalence between some model-like structures of combinatorial model categories and presentable ∞-categories? is affirmative and is provided by the paper Combinatorial model categories are equivalent to presentable quasicategories. The model-like structure used in the paper is that of the relative category. Concerning Question 2: Similar question, but considering all model categories (or maybe just categories with both a fibration and a cofibration structure) on one side and complete/cocomplete ∞-categories on the other? there are pretty severe size issues with defining “all model categories” or “all complete/cocomplete ∞-categories”. If, however, we use universes to resolve these size issues, then the answer is positive and can be deduced from Theorem 8.2 in the cited paper, in a manner similar to Proposition 8.3.<|endoftext|> TITLE: Meaning of A-infinity relations QUESTION [8 upvotes]: I am learning A-infinity category with Fukaya category in mind, and would like to understand the meaning of A-infinity relations. In particular, as $N=1$, it means $dd=0$. As $N=2$, it means that $d$ satisfies Leibniz's rule if we regard the second operation as multiplication. As $N=3$, it means that the multiplication satisfies associativity at the homology level. Questions These are "classical" interpretations. Though, I still wish to understand the higher operations. What are their meanings? Are there ways to view it as natural in a pure operad-ic view point? As mentioned above, the second operation can be viewed as ordinary multiplication at the homology level. How about the higher ones? How do they behave at the homology level (or perhaps this is not the right question to ask..) ? I have tried some papers on Fukaya category, wikipedia, and n-lab. Any other references will be appreciated. Thank you very much. REPLY [13 votes]: For your first question. Suppose $(A,d,m,m_3,m_4\dots)$ is an $A_\infty$-algebra. The operation $m_3$ gives a homotopy between $m(-,m(-,-))$ and $m(m(-,-),-)$, which I will abusively denote as $a(bc)$ and $(ab)c$. Now consider the two operations $A^{\otimes 4} \to A$ given by $a(b(cd))$ and $((ab)c)d$. Using $m_3$, you have two ways of going from the first to the second: Use $m_3$ three times to create a homotopy: $$a(b(cd)) \to a((bc)d) \to (a(bc))d \to ((ab)c)d.$$ Use $m_3$ twice to create a homotopy: $$a(b(cd)) \to (ab)(cd) \to ((ab)c)d.$$ If you combine these two homotopies, you get two classes of degree $2$ maps $A^{\otimes 4} \to A$. These two maps have no reason to be homotopic. Well, the higher operation $m_4$ gives a homotopy between these two homotopies! The even higher operations $m_5$, $m_6$ and so on work the same way. This is very nicely encoded in Stasheff's associahedra. The first two associahedra is just a point, representing the identity and $m_2$; the next one is a segment, representing $m_3$, a homotopy between $a(bc)$ and $(ab)c$; the next one is a pentagon, whose edges are the five arrows I drew above; and so on. For your second question, the answer is Massey products. Very briefly, suppose that you have a differential graded algebra $A$ and three cycles $a,b,c$ such that $ab = d\alpha$ and $bc = d\beta$. Then the class $abc$ vanishes "in two different ways", because $abc = d(\alpha c) = d(a \beta)$. It follows that $\alpha c - a \beta$ is a homology class, called the triple Massey product $\langle a,b,c \rangle$. The operation $m_3$ can be used to represent this triple Massey product on homology. It's a bit technical to explain how, and the explanation involves the Homotopy Transfer Theorem. For both answers, I think a good reference that cites pretty much all the other possible ones is the book Algebraic Operads by Loday and Vallette. REPLY [7 votes]: I'll be informal but try to give a topologist's intuitive interpretation, which gives the original source of the idea. Historically, $A_{\infty}$ structures start with Stasheff's work determining what higher coherence homotopies are needed to ensure that an $H$-space X (a space with a product with unit) has a classifying space, given by a bar construction. This was later interpreted as meaning that $X$ has an action by the Stasheff operad in spaces, in fact CW complexes. The chain complex associated to that operad gives an operad in chain complexes and thus a notion of $A_{\infty}$ algebra in chain complexes, with its associated bar construction. Going from there to DG categories is not a big leap. Thus the idea is that the specifics of the definition are encoding relevant higher homotopies. REPLY [4 votes]: One interpretation I prefer is obtained by defining $A_{\infty}$ structures on graded space $L$ as a degree $-1$ square zero derivation $d$ on (co)free coalgebra on $L[-1]$ aka its bar construction. As it is cofree, derivation is determined by projection on cogenerators, so you have a bunch of maps $L^{\otimes k} \to L$ of degree $2-k$. Now all relations become consequences of the fact that $d^2 = 0$ on $T_*(L[-1])$ and can be obtained by collecting elements of same degree. Other way around, beginning with a ($A_{\infty}$-)coalgebra one can construct free algebra with a square zero derivation on it encoding comultiplication (and higher operations). Composing these two together, we obtain a strictly associative dg algebra which is an "unfolded model" of $\infty$-algebra. So the way I interpret higher operations is that they are just cramped up ordinary associative multiplication when underlying space is too small. Main upshot is that this kind of algebraic structure can be transferred along vector space retractions, and as every chain complex can be retracted onto its homology, it feels like a "right one" for doing algebra in chain complexes. Good refences are "Algebraic operads" by J.-L. Loday and B. Valette, "Koszul duality for operads" by Ginzburg and Kapranov and "Modules over Operads and Functors" by Fresse.<|endoftext|> TITLE: Entire function not less than $\sqrt{\sinh x/x}$ on the real line QUESTION [5 upvotes]: If the answer to this question is in affirmative, then this would yield a good answer to this question. Let $f\colon \mathbb C\to\mathbb C$ be an entire function whose values on the real line are real and bounded from below by $$ f(x)\geq \sqrt{\frac{\sinh x}{x}}, \qquad 0\neq x\in\mathbb R. $$ [EDIT] Also, we have $|f(x+iy)|\leq C\exp(|x|/2)$ and, moreover, $f(x+iy)\to0$ as $y\to\infty$ for each separate $x$. Is it true that $f(0)\geq 2/\sqrt\pi$? Remarks. 1. If so, this estimate is sharp and achieved, e.g., on the analytic continuation of $$ f(x)=e^{x/2}\sqrt{\frac8{\pi x}}\int_0^{\sqrt {x/2}}\exp(-t^2) \,\mathrm{d}t =\frac{e^{x/2}}{\sqrt{\pi x}}\int_0^{x}\frac{e^{-s/2}}{\sqrt s} \,\mathrm{d}s, \qquad x>0. $$ (The fact that this function satisfies the inequality follows from the connection to the question mentioned above.) 2. It is worth nothing to mention that one may assume $f$ to be even. 3. Clearly, $\sqrt{\sinh z/z}$ is not an entire function, as $\sinh z$ has many simple zeroes. 4. Only the first additional constraint $|f(x+iy)|\leq C\exp(|x|/2)$ does not exclude the counterexample $\cosh(z/2)$; thanks to Michael Engelhardt who noticed that I need some extra conditions. REPLY [5 votes]: The answer is no. Let $f$ be the function described under Remark 1 of the OP. By continuity, there exists a $\delta >0$ such that $f(x) - \sqrt{(\sinh x)/x} > 0.05$ on the real axis for $|x|\leq \delta $ (note that, at $x=0$, the aforementioned difference is $2/\sqrt{\pi } -1 = 0.128$). Consider furthermore the function $$ g(z)=\epsilon \left[\frac{2\sinh (z/2)}{z}-(1+\kappa )\frac{\pi^{2} \cosh (z/2)}{z^2+\pi^{2} } \right] $$ $g$ satisfies the required constraints $|g(x+iy)| \leq C\exp (|x|/2)$ and $g(x+iy)\rightarrow 0$ as $y\rightarrow \infty$ for each separate $x$. On the real axis, $g(x)\geq 0$ for $\kappa =0$; we can adjust to a $\kappa > 0$ such that $g(x)\geq 0$ for $|x|\geq \delta $ but $g(x)<0$ for $|x|<\delta $. We can furthermore adjust $\epsilon $ such that $|g(x)|<0.05$ for $|x|< \delta $. With such a choice of $\kappa $ and $\epsilon $, the function $f+g$ is a counterexample to the OP's question, $f(0)+g(0) < 2/\sqrt{\pi } $ while satisfying all the given constraints.<|endoftext|> TITLE: Applications of schemes to mathematical physics QUESTION [16 upvotes]: Could anyone cite some applications or developments in mathematical physics or string theory that use schemes? I find curious the fact that while things like derived algebraic geometry and stacks have certain applications to mathematical physics I cannot find such applications for the case of (underived) schemes. Just to clarify: I know that for example you can have Calabi-Yau threefolds that are projective algebraic varieties and that in some sense algebraic varieties ⊂ schemes ⊂ stacks. However I am looking for specific (underived) scheme constructions like Fano schemes, Hilbert schemes, scheme-theoretic interesections etc REPLY [3 votes]: In string theory, gauge symmetries on D-branes can be studied using classical scheme theory. This was introduced around the late 1990s and early 2000s, cf. this 1998 paper, this 2000 note, these 2003 notes, and this 2007 paper to start.<|endoftext|> TITLE: Comultiplication on objects in an (abelian?) category QUESTION [5 upvotes]: Looking for example at $R$-modules for some commutative $R$, we have the direct sum and the tensor product acting analogously to addition and multiplication. After studying a little bit about co-algebras and bi-algebras, I wondered if there is any way to define something that would be analogous to comultiplication in this setting. What I might imagine is an additive functor $C \to C \otimes C$ when $C$ is an abelian category, under some suitable definition for the tensor (I looked u a little about it and I saw that there are a definitions for that, which I did not fully understand). Also, I can't really imagine what would be an analog for the counit except or maybe (the) functor $C \to *$. REPLY [7 votes]: Sure, we can define such things. Let's work in the Morita 2-category $\text{Mor}(k)$ over a commutative ring $k$, which has objects $k$-algebras $A$, morphisms $k$-bimodules (with composition given by tensor product), and 2-morphisms bimodule homomorphisms. Equivalently, applying the forgetful functor $\text{Hom}(k, -)$, we can think of the Morita 2-category as having objects the cocomplete $k$-linear categories $\text{Mod}(A)$ of right modules over $k$-algebras, morphisms given by tensor product with a bimodule (equivalently, cocontinuous $k$-linear functors, by the Eilenberg-Watts theorem) 2-morphisms natural transformations. The Morita 2-category is a categorified version of modules; specifically it can be thought of as a 2-category of module categories over $\text{Mod}(k)$, which itself can be thought of as a categorified commutative ring. It has a tensor-hom adjunction $$\text{Hom}(A \otimes B, C) \cong \text{Hom}(A, [B, C])$$ where $A \otimes B$ is the ordinary tensor product over $k$ and $[B, C] = B^{op} \otimes C$ is the internal hom. This adjunction says that we can identify $(A \otimes B, C)$-bimodules naturally with $(A, B^{op} \otimes C)$-bimodules. It is moreover the case that $\otimes$ really deserves to be called the tensor product in this setting, in that $\text{Mod}(A \otimes B)$ is the universal recipient of a "bilinear" (cocontinuous and $k$-linear in each variable) functor out of $\text{Mod}(A) \times \text{Mod}(B)$. The unit of the tensor product is $\text{Mod}(k)$, so we can define a "comonoidal object" in this setting to be equipped with a comultiplication $\text{Mod}(A) \to \text{Mod}(A \otimes A)$ (that is, an $(A, A \otimes A)$-bimodule) and a counit $\text{Mod}(A) \to \text{Mod}(k)$ (that is, a left $A$-module), plus an associator and stuff like that, satisfying some axioms that look like the axioms of a monoidal category. This is the dual of a sesquialgebra structure, in the very nice sense that a structure of this sort on $A$ is exactly a sesquialgebra structure on $A^{op}$.<|endoftext|> TITLE: when is the normal cone a linear scheme? QUESTION [5 upvotes]: Let $Y$ be a nonsingular variety and $X\subset Y$ a closed subscheme. A linear scheme over $X$ is a scheme of the form $\textbf{Spec}\, ( Sym _{\mathcal{O}_X} F) $, where $F$ is a coherent sheaf on $X$. If the embedding $X\subset Y$ is regular, i.e. if every point of $Y$ has a neighborhood over which the ideal $I$ defining the above embedding is generated by a regular sequence, then it is well known that the normal cone $C_{X/Y}= \textbf{Spec}\, (\oplus _{i\geq 0} I^i/I^{i+1})$ is isomorphic to the linear scheme $ \textbf{Spec}\, ( Sym _{\mathcal{O}_X} I/I^2) $- the total space of the conormal sheaf. Question: Is the converse true? That is, suppose that $X$ is a closed subscheme of a nonsingular variety $Y$ and $C_{X/Y}$ is isomorphic to a linear scheme. Is it true that $X\subset Y$ is regular embedding? REPLY [2 votes]: I believe the answer is negative. Take $Y= \mathbb{A}^3= \textbf{Spec}\, k[x,y,z]$ and $X= V(xz,yz)$, so $I=(xz,yz)$. Note that $X$ is the union of the plane $z=0$ and the line $x=y=0$. Then $X\subset Y$ is not regular: any neighborhood of the origin contains a point in the plane and a point in the line, so their local rings have different dimensions. Now, let $A=k[x,y,z]$, $\overline{A}=\dfrac{k[x,y,z]}{(xz,yz)}$. The surjection of graded rings $$A[S,T]\to \oplus _{i\geq0} I^i$$ which sends S to $xz$ and $T$ to $yz$ has kernel $(yS-xT)$ so it induces a surjection $$\overline{A}[S,T]\to \oplus _{i\geq0} \dfrac{I^i}{I^{i+1}} $$ with kernel $R=(\bar{y}S-\bar{x}T)$. Since $R$ is defined by an equation in degree one, and the above surjection factors through $Sym I/I^2$, we actually have $$\frac{\overline{A}[S,T]}{R}\cong\oplus _{i\geq0} \dfrac{I^i}{I^{i+1}}\cong Sym I/I^2 $$ so the normal cone $C_{X/Y}$ is the linear scheme associated to the conormal sheaf of $X$ in $Y$.<|endoftext|> TITLE: Manifolds with nonwhere vanishing closed one forms QUESTION [12 upvotes]: I am trying to find examples of closed manifolds $M$ admitting a nowhere vanishing closed one form. I am wondering if there are any examples beyond $N\times S^1$. REPLY [24 votes]: If $f: M \to S^1$ is a submersion (and so a fiber bundle if $M$ is compact) then $f^*d\theta$ is a nowhere-vanishing closed 1-form. There are many more such manifolds and fibrations than just products, and the manifolds have the name mapping tori. If $(M,\omega)$ is a manifold equipped with a nowhere vanishing closed 1-form, then in fact there is a fibration over the circle $f: M \to S^1$ so that $f^*d\theta$ is arbitrarily close to $\omega$. This is a theorem of Tischler, whose short proof can be read here.<|endoftext|> TITLE: "a sign that one should be computing K-theory" QUESTION [12 upvotes]: Allen Knutson said here in comments below the question that I generally regard torsion in (co)homology as a sign that one should be computing K-theory instead, which has less of it. I know one or two things about torsion groups, examples of cohomology groups that has non zero torsion part, some definitions $K$-theory. Can some one help me to understand why torsion in cohomology should remind about computing $K$-theory? Any references are welcome. Some not very well posed question is : what are (some of the) other signs that one should be computing K-theory? REPLY [14 votes]: Usually you are computing $H^*(X)$ or $K^*(X)$ for a reason; for example if $H^*(X)\not\simeq H^*(Y)$ then you know that $X$ and $Y$ are not homotopy equivalent, but if $H^*(X)\simeq H^*(Y)$ and this object has a complicated structure, then it is a reasonable guess that $X\simeq Y$. You can do this kind of argument with ordinary cohomology or complex $K$-theory or Morava $E$-theory or any of various other invariants; you just choose whichever one seems likely to be effectively calculable and sufficiently rich as to provide the information that you need. A common case is where $X$ is equivalent or closely related to a Borel construction $(Z\times EG)/G$ for some finite group $G$ and some $G$-space $Z$. In those cases, $H^*(X)$ will be related to $H^*(BG)$, and $H^*(BG)$ is typically an ugly sort of ring with many generators and random-looking relations and lots of torsion. (In fact, if $|G|=n$ and $a\in H^i(BG)$ with $i>0$ then $na=0$.) On the other hand, the Atiyah-Segal completion theorem says that $KU^0(BG)$ is a completion of the representation ring $R(G)$, and $R(G)$ is small, nicely structured and free of torsion. So $K$-theory is likely to be a more tractable invariant in these cases. The Morava $E$-theory ring $E^0(BG)$ is often free of torsion as well, and is a richer invariant than $KU^0(BG)$.<|endoftext|> TITLE: When does a graph underlie the Hasse diagram of a poset? QUESTION [25 upvotes]: For any finite poset $P=(X,\leq)$ there is a graph $G$ underlying its Hasse diagram $H=(X,\lessdot)$, so that $V(G)=X$ and $E(G)=\{\{u,v\}:u\lessdot v\}$. With that said, is it possible to characterize all these graphs? That is, those graphs which underlie the Hasse diagram of some finite poset? It's easy to see any digraph $D$ is a Hasse diagram of a finite poset iff every directed path in $D$ is an induced path in $D$, so I'm trying to describe graphs that have an orientation with this property. Clearly they must all be triangle free since any orientation of a triangle will result in either a cycle which isn't acyclic or an arc between two vertices and a path of length two between those vertices which isn't transitively reduced. Moreover its easy to see any bipartite graph will fall into this class since if the vertices in an arbitrary graph $G$ can be partitioned into two independent sets $X$ and $Y$ then if we define $R=\bigcup_{e\in E(G)}(e\cap X)\times (e\cap Y)$ we see $P=(X\cup Y,R)$ is a poset with a height of two and that $G$ underlies the Hasse diagram of $P$. In fact if for any poset $P$ we let $h(P)$ denote the height of $P$ then whenever a graph $G$ underlies the Hasse diagram of $P$ we must have that $\chi(G)\leq h(P)+1$ as a consequence of the Gallai–Hasse–Roy–Vitaver theorem though note despite this class of graphs being closed under edge subdivisions, graphs in this class are not closed under edge contractions which rules out using the graph minor theorem. However because this class of graphs is closed under the removal of edges and vertices (since removing arcs from the Hasse diagram of any finite poset yields another Hasse diagram which is a subposet of its parent) we know there is a unique inclusion minimal family of forbidden subgraphs $F$ (unique up to isomorphisms of the graphs in $F$) which completely characterizes all graphs that can be oriented to form the Hasse diagram of a finite poset i.e. $F$ is such that a graph $G$ does not contain a subgraph isomorphic to a graph in $F$ if and only if $G$ has an orientation which is a Hasse diagram of a finite poset. Thus to characterize graphs underlying any finite Hasse diagram, it suffices to only characterize the forbiden subgraph family $F$. Further one can prove if we call any digraph a pseudocycle when it is either a directed cycle or if it can be formed by flipping one arc in a directed cycle, then since the transitive closure of any relation containing a path $(v_0,v_1,\ldots v_n)$ must have the arc $(v_0,v_n)$ this means no transitively reduced digraph has the pseudocycle consisting of all the arcs in that path and $(v_0,v_n)$ likewise all digraphs which are Hasse diagrams must be acyclic thus we see $G\in F$ iff we have: $$(1)\text{ Every orientation of }G\text{ contains at least one pseudocycle}$$ $$(2)\text{ Every proper subgraph of $G$ has an orientation containing no pseudocycles}$$ In fact every graph $G\in F$ is both $2$-vertex connected and satisfies $\chi(G)\geq \text{girth}(G)$, because if graphs $A$ and $B$ underlie Hasse diagrams of finite posets and satisfy $\small |V(A)\cap V(B)|\leq 1$ then their union $A\cup B$ must further underlie a Hasse diagram of a finite poset, likewise by the first property above if $G\in F$ then we know every orientation of $G$ must contain a pseudocycle digraph $D$, while since the underlying graph of $D$ is a cycle graph in $G$ with $|V(D)|$ edges, we see $\small |V(D)|\geq\text{grth}(R)$ yet by definition the pseudocycle $D$ must contain a directed path of length $\small |V(D)|-1$ or equivalently a directed path with $\small |V(D)|$ vertices, which means $D$ contains a directed path with $\text{grth}(G)$ vertices, thus proving every orientation of $G$ contains a directed path with $\text{grth}(G)$ vertices, therefore we see as a corollary of the Gallai–Hasse–Roy–Vitaver theorem, that we get $\chi(G)\geq\text{grth}(G)$ as required. Though these few properties alone still provide no where near enough information on graphs in $F$ to give any kind of general characterization of them. So what sort of information has been proven about the graphs in $F$? I know that the smallest two graphs in $F$ by order are the triangle graph and the Grötzsch graph, but what are some others? Are there others? I would imagine there are, in fact if I had to guess I would say $|F|=\aleph_0$, for if $F$ was finite then let $m=\max(\text{cir}(G):G\in F)$ be the maximum circumference of any graph in $F$ then this means every graph $G$ with $\text{grth}(G)>m$ must underlie a Hasse diagram of a finite poset, since if a graph $G$ has girth greater then $m$ we know that by definition no graph in $F$ can be isomorphic to a subgraph of $G$ as otherwise it would then contain a cycle whose length was $m$ or smaller contradicting the assumption $G$ had girth larger then $m$. Yet this seems like far too "nice" of a property... So to reiterate how can this set of forbiden subgraphs $F$ characterizing all graphs underlying finite Hasse diagrams, be more simply expressed then this? Can someone just list a few other graphs in $F$ for me, apart from either the triangle or Grötzsch graph? REPLY [21 votes]: This problem was proved to be NP-complete in https://link.springer.com/article/10.1007/BF00340774, but a mistake was discovered, and later corrected, for a simple proof and brief history, see https://link.springer.com/article/10.1007%2FBF01108825. I've asked the same question myself earlier this year from some people working in the field, the above links were given to me by Piotr Micek. Some papers to read are also given in https://arxiv.org/pdf/1908.08250.pdf on page 2, before Theorem 1. Also see https://link.springer.com/article/10.1007/BF00400288.<|endoftext|> TITLE: Is data science mathematically interesting? QUESTION [78 upvotes]: I have seen a plethora of job advertisements in the last few years on mathjobs.org for academic positions in data science. Now I understand why economic pressures would cause this to happen, but from a traditional view of university organization, but how does data science fit in? I would have guessed that at most, a research group having to do with something labeled "data science" could be formed as an interdisciplinary project between applied mathematicians, statisticians, and computer scientists, with corporate funding. But I don't see why it is a fundamentally distinct intellectual endeavor, prompting mathematics hires specifically in data science. The first time I heard the term “data science,” it was said that they wanted to take PhD’s who had experience in statistically analyzing large data sets, and train them for a few weeks to apply these skills to marketing and advertising. Now just a few years later people want to hire professors of this. Question: What about data science is particularly interesting from a mathematical point of view? REPLY [22 votes]: To begin, there is a family of results which are sometimes referred to as "No free lunch" theorems. Each of these results, in their own way, asserts that any optimization algorithm is just as good as any other if you average over the space of all optimization problems. On the other hand, we know that in specific domains some algorithms vastly outperform all others (that we're aware of) - for detecting objects in images, convolutional neural networks are state of the art, and in computational linguistics the best you can do for most tasks is a neural network with an LSTM or transformer architecture. In both of these cases, the state of the art algorithms perform vastly better than, say, logistic regression. How can we reconcile the "No free lunch" theorems with our empirical experience? The answer has to be that object detection in images and standard NLP tasks aren't "typical" optimization problems - some combination of the data and the task has some special structure which particular neural architectures are unusually good at detecting. What is this structure? Why are known algorithms so good at learning it? Can we generate new algorithms (neural or otherwise) that are even better? These are all essentially math problems, sitting somewhere at the intersection of optimization theory and information theory. They are pretty wide open - except in simple cases like logistic regression - there isn't much in the way of theory which characterizes an algorithm as optimal for a particular task among the space of all possible optimization algorithms. An influential paper from 2014 proposes to use the theory of renormalization groups from physics to tackle this question, and there are other attempts using gauge theory or the principle of maximum entropy. Another line of attack involves the so-called "manifold hypothesis", which asserts that real world data sets (presented as a set of points in euclidean space) tend to cluster near a high codimension submanifold. That's my answer for your main technical question, but I'll also make a remark about the academic politics. There are significantly more job openings in data science than there are people to fill them, so much so that many companies (like AirBnB) have found it cheaper and easier to start internal data science training programs than hire outside people. This problem is not likely to go away any time soon, so it's sensible to incentivize universities to start degree programs in the field, even if it's not yet a fully fleshed-out academic discipline. This has plenty of historical precedent - for instance, academic programs in forensic science and financial mathematics sprouted in the same way in the 1990's.<|endoftext|> TITLE: Universal property of $\mathbb S[z]$ and $E_\infty$-ring maps QUESTION [9 upvotes]: Let $\mathbb S[z]$ be the free $E_\infty$-ring spectrum generated by the commutative monoid $\mathbb N$. That is, $\mathbb S[z] = \Sigma^\infty_+ \mathbb N$. In Bhatt-Morrrow-Scholze II (https://arxiv.org/abs/1802.03261), they define a map out of this spectrum by choosing an element of the target. In particular, they use the uniformizer $\varpi$ of $\mathcal O_K$ to define a map $\mathbb S[z] \to \mathcal O_K$ by $z \mapsto \varpi$. This occurs in the paragraph after Theorem 11.2. By all indications, this seems to be an $E_\infty$-ring map. This is not the only place in the literature where such a property of $\mathbb S[z]$ is used. In this paper of Krause--Nikolaus (https://arxiv.org/abs/1907.03477), they define the same map at the beginning of section 3. However, we know the universal property of $\mathbb S[z]$: it is the free $E_1$-algebra on a single generator. This does not often coincide with the free $E_\infty$-algebra on a single generator. See here or the second answer here. My questions are the following: Are these maps $\mathbb S[z] \to \mathcal O_K$, $z \mapsto \varpi$ actually $E_\infty$-ring maps? Or just $E_1$? How do you obtain $E_\infty$-ring maps out of $\mathbb S[z]$? In particular, for which elements of $\Omega^\infty A \simeq \operatorname{Map}_{E_1}(\mathbb S[z], A)$ can we lift an $E_1$-map to an $E_\infty$-map? When the target is an Eilenberg--Maclane spectrum, can we obtain an $E_\infty$-map $\mathbb S[z] \to HB$ for any element of $B = \pi_0 HB$? Ideally, the answer to this second question would come in the form of a natural transformation $\Omega^\infty \to \operatorname{Map}_{E_\infty}(\mathbb S[t], -)$. But this is probably too optimistic. REPLY [6 votes]: $\newcommand{\E}{\mathbf{E}}$Dylan answered question 3 (and hence question 1) in the comments, but here's another equivalent way to see it: $\E_\infty$-maps $S^0[z]\to R$ with $R$ a discrete ring (i.e., viewed as an Eilenberg-Maclane spectrum) are the same data as $\E_\infty$-maps $\tau_{\leq 0} S^0[z] = H\mathbf{Z}[t] \to R$ of $\E_\infty$-rings, i.e., maps $\mathbf{Z}[t]\to R$ of rings. Question 2 is hard: in general, $\E_n$-maps $\Sigma^\infty_+ \Omega^n X\to A$ (where $X$ is a topological space and $A$ is an $\E_n$-ring) are the same data as maps $X\to B^n\mathrm{GL}_1(A)$ of spaces. If $n=\infty$, that's saying that $\E_\infty$-maps $\Sigma^\infty_+ \mathbf{Z}\to A$ are the same things as maps $H\mathbf{Z}\to gl_1(A)$ of spectra. (In Lurie's elliptic cohomology papers, the space of such maps is denoted $\mathbf{G}_m(A)$.) Lurie's answer in the question linked to in the comments is concerned with computing $\mathbf{G}_m(A)$ in the special case of $A$ being a Morava E-theory. In fact, that result can be made explicit at height $1$; see http://math.uchicago.edu/~amathew/notes_thursday.pdf. Here's a summary of what goes on. If $R$ is a $K(1)$-local $\E_\infty$-$K_p$-algebra, it's a little easier to study $\E_\infty$-maps $S^0[z]\to R$. Such maps factor through the $K(1)$-local spectrum $L_{K(1)} S^0[z]$, which admits a concise presentation as a $K(1)$-local $\E_\infty$-ring. Recall (e.g. from https://web.math.rochester.edu/people/faculty/doug/otherpapers/knlocal.pdf) that the $p$-adic K-theory of the free $K(1)$-local $\E_\infty$-ring $L_{K(1)} S^0\{z\}$ on a generator in degree zero is simply the polynomial ring $\mathbf{Z}_p[z, \theta(z), \theta^2(z), \cdots]$, where the $K(1)$-local power operation $\theta$ takes $\theta^{n-1}(z)$ to $\theta^n(z)$. The element $\theta(z)\in \pi_0 (K_p \wedge L_{K(1)} S^0\{z\}) = \pi_0 L_{K(1)} K_p\{z\}$ defines a map $S^0\to L_{K(1)} K_p\{z\}$ which extends (by the universal property of free $\E_\infty$-rings) to a map $L_{K(1)} K_p\{z\} \xrightarrow{\theta} L_{K(1)} K_p\{z\}$. Since this map must commute with power operations, it sends $\theta^{n-1}(z)$ to $\theta^n(z)$ on homotopy. The $\E_\infty$-cone $L_{K(1)} K_p\{z\}/\!\!/\theta(z)$ on this element is defined via the pushout of the diagram $$K_p \leftarrow L_{K(1)} K_p\{z\} \xrightarrow{\theta} L_{K(1)} K_p\{z\}.$$ A simple calculation with the long exact sequence on homotopy shows that there is in fact an equivalence $L_{K(1)} K_p\{z\}/\!\!/\theta(z) \simeq L_{K(1)} K_p[z]$ of $\E_\infty$-rings. This is the only compact presentation of any strict spectral polynomial ring that I know of. As a side remark, here's a curious observation. There is an element $\psi^p(z)\in L_{K(1)} K_p\{z\}$, so we can consider the $\E_\infty$-cone $L_{K(1)} K_p\{z\}/\!\!/\psi^p(z)$. Using the relation $\psi^p(z) = z^p + p\theta(z)$, you can show that this $\E_\infty$-ring has divided powers on $z$.<|endoftext|> TITLE: What is the smallest class of spaces closed under finite homotopy colimits, finite homotopy limits, and splitting of homotopy-coherent idempotents? QUESTION [14 upvotes]: Finite CW complexes fail spectacularly to be closed under finite homotopy limits (e.g. $\Omega S^1 = \mathbb Z$). More subtly, they fail to be closed under homotopy retracts (by the Wall finiteness obstruction). In some sense, this is why one works with $\pi$-finite spaces when one needs a notion of "finite" space which is closed under finite limits. But what happens if we bite the bullet and just close up under finite limits, colimits, and retracts? In the end, this is probably not the most natural thing to do (for instance, we might really want closure under extensions or something), but the question remains. Questions: Let $\mathcal F$ be the smallest class of spaces closed under finite homotopy colimits, finite homotopy limits, and splitting of homotopy-coherent idempotents. What is an explicit description of $\mathcal F$? Fix a prime $p$, and let $\mathcal F_p$ be the smallest class of $p$-local spaces closed under finite homotopy colimits, finite homotopy limits, and splitting of homotopy-coherent idempotents. What is an explicit description of $\mathcal F_p$? Notes: On the one hand, $\mathcal F$ is contained in the class of countable CW complexes. On the other hand, $\mathcal F$ contains all finite-dimensional countable CW complexes. To see this, note that by closure under finite limits and colimits, $\mathcal F$ contains the empty space and the one-point space, and thus by closure under retracts and finite colimits it contains all retracts of finite CW complexes. And $\mathcal F$ also contains $\mathbb Z = \Omega S^1$. Then by taking the homotopy pushout of the obvious maps $\mathbb Z \leftarrow S^0 \times \mathbb Z \to \mathbb Z$, we see that $\mathcal F$ contains $S^1 \times \mathbb Z$. Repeating, we see that $\mathcal F$ contains $S^n \times \mathbb Z$ for each $n$. By gluing, we obtain all finite-dimensional CW complexes with countably many cells. For example, plus constructions are obtained by attaching only 2-cells and 3-cells (the number of which is bounded in terms of the size of the fundamental group), so if $X \in \mathcal F$ and $P \subseteq \pi_1(X)$ is a perfect normal subgroup, then $X^+_P \in \mathcal F$, too. In particular, by taking $X = \vee^\omega S^1$ and suitable $P$, we get spaces in $\mathcal F$ with arbitrary countable fundamental group. There are also spaces in $\mathcal F$ such as $\Omega S^2$, with cells in arbitrarily large dimension. It's not clear to me exactly which infinite-dimensional countable CW complexes are in $\mathcal F$. For instance, what about $\mathbb C \mathbb P^\infty$ or $\mathbb R \mathbb P^\infty$? If we were working with just simply-connected spaces, the question would be much easier because the Eilenberg-Moore spectral sequence would always converge, and we could make some kind of argument about Serre classes. But I think the presence of non-simply-connected spaces complicates things. REPLY [5 votes]: (The discussion below is for pointed spaces.) I'll use $\mathcal{F}_*$ for the pointed version of your $\mathcal{F}$. As Nicholas Kuhn says, this is related to the closed classes studied by E. Dror Farjoun and W. Chach\'olski. Including closure under limits in your collection actually makes it closer to the dual concept, which I have studied under the name of "resolving classes" and "resolving kernels". I can't give a definitive answer to the question as posed, but I can show that the pointed version $\mathcal{F}_*$ contains no Eilenberg-MacLane spaces $K(G,n)$ for $n> 1$ or for $G$ finite abelian and $n = 1$ A resolving kernel is a collection of spaces of the form $$ \mathcal{R} = \{ Y \mid \mbox{$\mathrm{map}_*(X,Y) \sim *$ for all $X\in \mathcal{X}$}\} $$ for some collection of spaces $\mathcal{X}$. Theorem: Let $\mathcal{A}$ be a collection of pointed spaces such that $\Sigma\mathcal{A}\subseteq \mathcal{R}$ $\mathcal{A}\wedge \mathcal{A} \subseteq \mathcal{R}$ (up to weak equivalence). Write $\overline{\mathcal{A}}$ for the closure of $\mathcal{A}$ under "finite-type wedge" and extensions by cofibrations (and therefore under pushouts). If $\mathcal{R}$ is a resolving kernel and $\Sigma \mathcal{A} \subseteq \mathcal{R}$, then $\overline{\mathcal{A}}\subseteq \mathcal{R}$. (A finite-type wedge is a wedge in which the connectivities of the summands increases to infinity.) It is worth pointing (ha ha) out that the pointed case is quite different from the unpointed case, because getting even a single wedge into a resolving kernel is enough to bootstrap to the hypotheses of this theorem. Explicitly: Theorem: If $X$ is a finite-type space with $\mathrm{map}_*(X, S^n\vee S^m) \sim *$ for any two $n, m > 1$, then $\mathrm{map}_*(X,K)\sim *$ for all simply-connected finite-dimensional CW complexes $K$. (If $\pi_1(X)$ is not a perfect group, then "simply-connected" can be dropped from the conclusion). Now let's think about $\mathcal{F}_*$, the smallest class of pointed spaces containing all the spheres and closed under homotopy pushouts homotopy pullbacks homotopy retracts. This is contained in $\mathcal{S}$, Edit: Or maybe not! There are two classes and two closure properties: $\overline{\mathcal{A}}$ is closed under pushouts, while $\mathcal{R}$ is closed under pullbacks; and $\overline{\mathcal{A}}\subseteq \mathcal{R}$. the smallest resolving kernel containing all the spheres. Let $\mathcal{M}_p$ denote the resolving kernel associated to $\mathcal{X} = \{ B\mathbb{Z}/p \}$ for your favorite prime $p$. Miller's theorem (the Sullivan conjecture) tells us that $$ \mathcal{F}_*\subseteq \mathcal{S} \subseteq \mathcal{M}_p. $$ But obviously $B\mathbb{Z}/p \not \in \mathcal{M}_p$, so we can conclude $$ B\mathbb{Z}/p \not\in \mathcal{F}_* $$ Similarly, $\mathcal{F}_*$ cannot contain Eilenberg-MacLane spaces $K(G,n)$ for any finite abelian group $G$ (such spaces also cannot see spheres by an easy extension of Miller's theorem). And using closure under forming homotopy fibers, we can see that $\mathcal{F}_*$ cannot contain $K(\mathbb{Z},n)$ for any $n$. Thus we have Proposition: If $G$ is an abelian group, then $K(G,n) \not\in \mathcal{F}_*$ for $n > 1$.<|endoftext|> TITLE: Order polynomial of shifted double staircase QUESTION [11 upvotes]: This question is related to my earlier question looking for posets with product formulas for their order polynomials. Recall that the order polynomial $\Omega_P(m)$ of a finite poset $P$ is defined by $$\Omega_P(m) := \# \textrm{ weakly order preserving maps $P\to \{1,2,\ldots,m\}$}.$$ Now let $\lambda = (n,n-1,n-2,...,1) + (k,k-1,k-2,...,1)$ for $0 \leq k < n$ be a ``shifted double staircase'' shape (see e.g. Figure 6(c) in Stanley's paper "Promotion and Evacuation", which is linked below). And let $P$ be the poset corresponding to $\lambda$ (i.e., the poset on the boxes of $\lambda$ viewed as a shifted shape, with $u \lessdot v$ if the box $v$ is directly right of or directly below the box $u$). Question: Is it true that for this $P$ we have $$\Omega_P(m) = \prod_{1 \leq i \leq j \leq n} \frac{(m+i+j-2)}{(i+j-1)}\cdot \prod_{1 \leq i \leq j \leq k} \frac{(m+i+j-1)}{(i+j)}?$$ Testing some small cases it looks like this formula works, and this is not an example I have seen in the literature (but a pointer to a place where it is addressed would also definitely be appreciated!). For context, let me explain some similar formulas which are known. If $P$ is the poset associated to the (unshifted) staircase $\lambda = (n,n-1,n-2,...,1)$ (this poset is also the Type A root poset), then $$\Omega_{P}(m) = \prod_{1\leq i \leq j \leq n} \frac{i+j+2m-2}{i+j}.$$ While if $P$ is the poset associated to the shifted staircase $\lambda = (n,n-1,n-2,...,1)$ (this poset is the Type B/D minuscule poset), then $$\Omega_P(m) = \prod_{1 \leq i \leq j \leq n} \frac{(m+i+j-2)}{(i+j-1)}.$$ Both of these formulas can be seen for instance in the paper "New Symmetric Plane Partition Identities from Invariant Theory Work of De Concini and Procesi " by Proctor (linked below). Note that the shifted staircase is just the case $k=0$ of the shifted double staircase and the conjectured formula agrees with the known formula in this case. The formula for the case $k=n-1$ is also known: in this case the poset is the Type B root poset; see for instance the abstract by Hamaker and Williams linked below. Hamaker, Zachary; Williams, Nathan, Subwords and plane partitions, Proceedings of the 27th international conference on formal power series and algebraic combinatorics, FPSAC 2015, Daejeon, South Korea, July 6–10, 2015. Nancy: The Association. Discrete Mathematics & Theoretical Computer Science (DMTCS). Discrete Mathematics and Theoretical Computer Science. Proceedings, 241-252 (2015). ZBL1335.05177. Proctor, Robert A., New symmetric plane partition identities from invariant theory work of De Concini and Procesi, Eur. J. Comb. 11, No. 3, 289-300 (1990). ZBL0726.05008. Stanley, Richard P., Promotion and evacuation, Electron. J. Comb. 16, No. 2, Research Paper R9, 24 p. (2009). ZBL1169.06002. REPLY [5 votes]: Tri Lai and I proved this conjecture, using the techniques from the theory of lozenge tilings. Indeed, this result is almost already proved by Ciucu in https://arxiv.org/abs/1906.02021. We just need to allow slightly more general parameters for the "flashlight" region of the triangular lattice which he considers- and the techniques he developed there suffice to do that. What we are able to show specifically is that for the region: the number of lozenge tilings of $F(x,y,z,t)$ is $$ \prod_{1\leq i \leq j\leq y+z}\frac{x+i+j-1}{i+j-1}\prod_{1\leq i \leq j \leq z} \frac{x+i+j}{i+j} \prod_{i=1}^{t}\prod_{j=1}^{z}\frac{(x+z+2i+j)}{(x+2i+j-1)}.$$ The case $t=0$ corresponds to the order polynomial for the shifted double staircase (where $y+z=n$, $z=k$, and $x=m-1$ in the notation of the original question). EDIT: The paper with the details is now on the arXiv: https://arxiv.org/abs/2007.05381. EDIT 2: Soichi Okada has proved an algebraic extension of the SDS order polynomial product formula in: https://arxiv.org/abs/2009.14037.<|endoftext|> TITLE: Show that the Markov chain of random tiling is irreducible QUESTION [5 upvotes]: Consider a Markov chain on a state-space which is slightly weird: It is the space of all tilings of a hexagon as shown in the left-hand side of the figure below with three types of rhombi: yellow, purple and cyan(a tiling means we cover the hexagon exactly with no empty space). The Markov chain picks one hexagon consisting of all three types of rhombi uniformly at random and flips it as is shown in the right-hand side of the figure. This is a finite-state Markov chain. Show that this Markov chain is irreducible. The question does not ask for precise mathematical expression. My idea: W.L.O.G. the original tilings of a hexagon have three cases denoted by $\{1, 2, 3\}$ which the probability of each one is $1/3$. Then after flipping each one, we get three new tilings denoted by $4, 5, 6$ where state $4$ got from flipping the state $1$ and $5$ got from $2$, $3$ got from $6$. So there are two elements of the state space is $\{1, 2, 3, 4, 5, 6\}$. It is clear that state $1, 4$ communicate each other by flipping the state $1$ or $4$. Similar with $2, 5$ and $3, 6$. But how to get the $2$ from the other states except for $5$ Is there any idea? Thanks! REPLY [6 votes]: Suppose the sidelengths of your hexagon are $k, \ell, m$. A standard way of looking at this is to turn the picture by 30 degrees to the left and to view the yellow / cyan tiles as $m$ functions $f_i \colon \{1,\ldots,k+\ell\}$ into $\mathbb{Z}$ with the following constraints: $f_i(1) = i$, $f_i(k+\ell) = i-k+\ell$, $f_{i+1}(n) \ge f_i(n)+1$ and $f_i(n+1)-f_i(n) \in \{\pm 1\}$. The Markov chain then boils down to choosing $i$, $n$ and $\delta \in \{\pm 2\}$ at random and to change $f_i(n)$ into $f_i(n) + \delta$ whenever this can be done without breaking the above constraints. In other words, it turns local maxima of the $f_i$ into local minima and vice-versa, as long as this can be done by preserving the order between the functions. It is now easy to see that the minimal element of this state space can be reached from every starting point: first turn local maxima of $f_1$ into local minima until you've reached the minimal configuration for $f_1$, then do the same for $f_2$, etc.<|endoftext|> TITLE: What is the form of the $(v_0,v_1)$-pizza curve? QUESTION [7 upvotes]: Assume that there are two (competing) pizza houses situated at the points $0$ and $1$ on the complex plane. These pizza houses can deliver pizza to points of the plane with the largest velocities $v_0$ and $v_1$, respectively. Definition. A closed subset $P$ of the complex plane is called a $(v_0,v_1)$-pizza curve if $P$ is a common boundary of two open connected sets $U_0,U_1$ in $\mathbb C$ such that $0\in U_0$ and $1\in U_1$; $U_0\cap U_1=\emptyset$ and $U_0\cup P\cup U_1=\mathbb C$. for any point $z\in P$ there exists a positive real number $t_z$ such that (i) for every $\varepsilon>0$ and every $k\in\{0,1\}$ there exists a smooth curve $\gamma_k:[0,t_z+\varepsilon)\to U_k$ such that $\gamma_k(0)=k$, $\lim_{t\to t_z+\varepsilon}\gamma_k(t)=z$ and $\lvert\gamma_k'(t)\rvert\le v_k$ for every $t\in[0,t_z+\varepsilon)$; (ii) for every $\varepsilon>0$ and every $k\in\{0,1\}$ there exists no smooth curve $\gamma_0:[0,t_z-\varepsilon)\to U_k$ such that $\gamma_k(0)=k$, $\lim_{t\to t_z-\varepsilon}\gamma_k(t)=z$ and $\lvert\gamma_k'(t)\rvert\le v_k$ for every $t\in[0,t_z-\varepsilon)$. Problem. What is the form of a $(v_0,v_1)$-pizza curve? Is it unique? Remark 1. If $v_0=v_1$, then the answer to this problem is well-known: the $(v_0,v_1)$-pizza curve is unique and coincides with the line $\{z\in\mathbb C:\Re(z)=\frac12\}$. So the problem essentially concerns the case $v_0\ne v_1$. Remark 2. It can be shown that each $(v_0,v_1)$-pizza curve locally coincides with the graph of some Lipschitz function. REPLY [2 votes]: I can present an example of a pizza curve. We lose no generality assuming that $v_0=1$ and $v_1=c>1$. A $(1,c)$-pizza curve $P$ is the union $A\cup B\cup \tilde B$ of three pieces: the (shorter) arc $A$ of the circle $C$ (with equation $c^2(x^2+y^2)=(1-x)^2+y^2$), connecting the (conjugated) points $z_0$ and $\bar z_0$ such that the segments $[z_0,1]$, $[\bar z_0,1]$ lie in the tangent lines from $1$ to the circle $C$. the curve $B=\{z_0e^{t(i+1/\sqrt{c^2-1})}:t\in [0,\pi-\arg(z_0)]\}$. the curve $\tilde B$, symmetric to $B$ with respect to the real axis. The curve $B$ has the property that for any $t\in[0,\pi-\arg(z_0)]$ the length $L_t$ of the curve from the initial point $z_0$ to the point $z_t=z_0e^{t(i+1/\sqrt{c^2-1})}$ satisfies the equation $L+L_t=c\lvert z_t\rvert$ where $L$ is the length of the segment $[1,z_0]$. Observe that $\lvert z_t\rvert$ and $\frac1c(L+L_t)$ are equal to the times spent by the pizza runners to get the point $z_t$ running on its own territory (including the boundary). So, the problem is whether each $(1,c)$-pizza curve coincides with the pizza curve $P$ described above.<|endoftext|> TITLE: Connection between entropy and the set of factors of a sequence QUESTION [8 upvotes]: Let $a = (a_n)_{n=0}^\infty$ be a bounded real-valued sequence. By a factor of $a$ I mean a finite block $w \in \mathbb R^l$ that appears in $a$, that is, there exists $n \geq 0$ such that $a_n a_{n+1} \dots a_{n+l-1} = w_0w_1 \dots w_{n+l-1}$. Let $A_l \subset \mathbb{R}^l$ denote the set of factors of $a$ of length $l \geq 0$. We can always ensure that $a$ is produced by some topological dynamical system $(X,T)$ in the sense that $a_n = F(T^n x)$ where $x \in X$ and $F$ is a continuous map on $X$ (let $X$ be the orbit closure of $a$ under the shift operation, and let $T$ be the shift, $x = a$ and $F(x) = x_0$). Assume additionally that $(X,T)$ is minimal. I'm interested in the connection between the behaviour of $A_l$ and the entropy of $(X,T)$. In particular, I would like to know how the assumption that $(X,T)$ is zero-entropy translates to asymptotics on the size of $A_l$. The situation is quite simple if $a_n$ takes on only finitely many values. Then it is well-known that $(X,T)$ has entropy zero if and only if the growth of $A_l$ is subexponential: $|A_l| = \exp(o(l))$. More generally, the entropy of $X$ is given by $\lim_{l\to \infty} \log |A_l|/l$. When $a_n$ is allowed to take on arbitrary values the situation becomes less clear to me. It seems (although I'm worried I'm missing some technical issues along the way) that zero entropy should be equivalent to: For any $\epsilon > 0$, the number of boxes with sidelength $\epsilon$ in $\mathbb{R}^l$ needed to cover $A^l$ is $\exp(o_{\epsilon}(l))$. This post on Terence Tao's blog mentions a similar-looking condition, but (?) with a different metric on $\mathbb{R}^l$: For any $\epsilon > 0$, the balls with radius $\epsilon$ in $\mathbb{R}^l$ needed to cover $A^l$ is $\exp(o_{\epsilon}(l))$. Since unit cubes and unit balls in $\mathbb{R}^l$ behave rather differently as $l \to \infty$, I don't see an easy argument showing equivalence between the two (the natural idea of converting $\epsilon$-covering with balls into an $\epsilon'$-covering with boxes and vice versa fails). However, I see no reason why the two couldn't be equivalent either: after all, the sets we are covering have some additional structure. Are 1. and 2. equivalent? If so, which norms besides $\ell^2$ and $\ell^\infty$ can one take on $\mathbb{R}^l$? If not, what makes $\ell^2$ special? What is a good reference for equivalence of 2. with zero entropy? REPLY [4 votes]: Here's an attempt. Let me restrict to functions with values in $[0,1]$ and my entropies are computed with binary log. If we consider $X \subset [0,1]^{\mathbb{Z}}$ with the compact topology obtained from the product of standard topologies on $[0,1]$, then all metrics are equivalent, so they give the same notion and value of topological entropy. The correct topology in this sense is $\ell_{\infty}$: by refining partitions to be in the natural basis, refining by shifting and using compactness, it is easy to see that it's enough to consider topological entropy with respect to finite open covers of the "alphabet" $[0,1]$. This is the same as using $\ell_{\infty}$ balls, i.e. boxes (and covering by boxes and packing disjoint boxes gives the same entropy by the basic theory). Now the way I understand your question is that you want to consider a closed invariant set $X \subset [0,1]^{\mathbb{Z}}$ in the same natural topology, and just naively apply the formula of entropy but with some $\ell_p$ distance, $p \geq 1$, used to compare distance of partial orbits. Well, why not. Let's use the one you suggest: how many $\epsilon$-balls do we need to cover the partial orbits restricted to their values in $[0,1]^n$. I claim that: It is possible that all these "$\ell_p$-entropies" are positive for $1 \leq p < \infty$, even if topological entropy ("$\ell_\infty$-entropy") is zero, for a minimal subsystem of $[0,1]^{\mathbb{Z}}$. Proof. Pick for each $n \geq 1$ a finite minimal binary subshift over alphabet $\{0,1\}$ (so a single double-periodic orbit) $X_n$ where the number of words of length $f(n) = 2 \cdot 3^{n^2}$ is $2^{f(n)}$, that is, all subwords of that length occur. Clearly $X_n$ has zero entropy. Now, consider the (infinite) product $X$ of the subshifts $X_n$. Computing the usual topological entropy with respect to any $\epsilon > 0$ you see only finitely many of the subshifts $X_n$, so you see at most the sum of their entropies, $0+0+\cdots+0 = 0$, thus $X$ has zero entropy. If you pick the periods of the $X_n$ to be coprime, this system is minimal. We can realize it inside $[0,1]^{\mathbb{Z}}$ by using digits 0 and 2 in ternary expansions to code the bits from different subshifts, so that roughly we uncover one more layer $X_n$ of $X$ each time $\epsilon$ is divided by $3$ (for $\ell_\infty$ distance). Now, consider the $\ell_p$-entropy of this system, computed for $\epsilon = 1$. Consider a length $m = f(n)$ for some $n$. Let us ignore everything but the contribution of $X_n$ in the partial orbits $[0,1]^{m}$. Intuitively, this won't make the covering task any easier. Concretely, simplify our partial orbits to be in $\{3^{-n}, 2 \cdot 3^{-n}\}^m$, where each coordinate only depends on $X_n$-value. It is easy to see that this does not increase any $\ell_p$-distances, since if the $X_n$-value is different, then distance is at least $3^{-n}$ in that coordinate no matter what the other $X_j$-values are. Since distances were not increased, it's enough to show that this new set is hard to cover. Now, we have a set of partial words in $\{3^{-n}, 2 \cdot 3^{-n}\}^m$ that we have to cover by $\ell_p$-balls. By the choice of $X_n$, this set of simplified partial orbits is exactly the set $\{3^{-n}, 2 \cdot 3^{-n}\}^m$ of cardinality $2^m$. A radius-$1$ ball in $\ell_p$-norm covers at most a Hamming ball or radius $3^{pn}$ (where Hamming distance is the usual non-normalized one, considering $\{3^{-n}, 2 \cdot 3^{-n}\}$ as a binary alphabet), and to cover $2^m$ binary words with Hamming balls of radius $3^{pn}$ (which have cardinality $2^{3^{pn}}$), for any $n \geq p$ you need at least $$ 2^m/2^{3^{pn}} \geq 2^m/2^{3^{n^2}} = 2^{m/2} $$ balls, by the choice of $m = f(n)$, so the $\ell_p$-entropy, as it is lower bounded by $\log_2 2^{m/2} / m = 1/2$, is positive. QED. In fact this system should have infinite $\ell_p$-entropy for all $1 \leq p < \infty$ since I only considered the contribution of $X_n$ to the entropy, and the contributions of the $X_n$ are essentially independent. I didn't do the math so I'll just claim it and leave as an exercise. Alternatively, it should be easy to modify it slightly so the $X_n$-contributions directly tend to $\infty$ by using larger alphabets and replacing ternary expansions with $s$-adic expansions for a sequence of bases $s$. Anyway, if the above construction is correct (and I understood your questions), this answers your first question and perhaps to some extent obsoletes the others (?). Some quick observations, if some $\ell_p$-entropy is zero, then $\ell_\infty$-entropy is zero by a ball comparison argument. And as you yourself note, $\ell_p$-entropy can be zero even in interesting cases, for example when $X$ is a subshift over a finite alphabet $A \subset [0,1]$. You can also presumably do inverse limits as above, but make $X_n$ so small with respect to $X_{n-1}$ that things work out, and get systems where $\ell_p$ entropies are zero but which are not e.g. expansive.<|endoftext|> TITLE: Projecting a symmetric matrix onto the space of linear operators with a particular eigenvalue QUESTION [5 upvotes]: Specifically, I am interested in the case where one eigenvalue is exactly $0$. Given an $n \times n$ symmetric matrix, I would like to find the closest $n\times n$ symmetric matrix that has one eigenvalue that is equal to $0$. Although the $n\times n$ matrices form a Hilbert space, the set of symmetric matrices with a zero eigenvalue is probably not convex, because the eigenvalues of a sum of two matrices has a complex relationship with the original matrices. Thus, the classic projection theorem does not guarantee a unique solution. However, I only need a solution. Since the eigenvalues of a matrix are continuous in the entries, the set of $n\times n$ matrices with a zero eigenvalue should be closed, and so the set of solutions shouldn't be totally insane. In any case, how can I find the closest $n\times n$ symmetric matrix(es) with a zero eigenvalue to an arbitrary given $n\times n$ symmetric matrix? REPLY [7 votes]: Any (real) symmetric matrix may be diagonalised by an orthogonal matrix. So let your matrix be $A$ and let $O$ be an orthogonal matrix such that $D=OAO^{-1}$ is diagonal. Note that the transformation $A\mapsto OAO^{-1}$ is distance preserving. Now get $D'$ by replacing whichever diagonal entry of $D$ has least absolute value, and let $A'=O^{-1}D'O$. (Nathaniel Johnston's comment let me find the related paper "Low-Rank Positive Approximants of Symmetric Matrices" by Achiya Dax.)<|endoftext|> TITLE: Is the category of atomless Boolean algebras with complete embeddings closed under coproducts? QUESTION [6 upvotes]: Consider the category whose objects are atomless Boolean algebras (not necessarily complete) and whose arrows are complete embeddings. Does a coproduct exist in this category for any two atomless Boolean algebras $\mathbb{B}$ and $\mathbb{C}$? REPLY [4 votes]: This category does not have co-products. To see this, let $\newcommand\B{\mathbb{B}}\B$ be any atomless complete Boolean algebra with a nontrivial automorphism $\pi:\B\to\B$. For example, the forcing to add a Cohen real. I claim that $\B$ has no co-product with itself in your category. Suppose toward contradiction that $\B\sqcup\B$ is the co-product, with complete embeddings $i,j:\B\to\B\sqcup\B$ realizing the co-product universal property. Let $f_1:\B\to\B$ and $f_2:\B\to\B$ both be the identity embedding. By the universal property, there is $f:\B\sqcup\B\to\B$ making a commutative diagram. It follows that $i(b)$ and $j(b)$ are both carried by $f$ to $b$. Since $f$ is an embedding, this means in particular that $i(b)=j(b)$. Now replace $f_2$ with the automorphism $\pi$. By the universal property, there is again a complete embedding $f:\B\sqcup\B\to\B$ making the diagram commute. But now $f$ must take $i(b)$ both to $b$ and to $\pi(b)$, which is impossible if $\pi$ moves $b$. So we don't have co-products.<|endoftext|> TITLE: Groebner Bases for submodule over polynomial ring with integer coefficients QUESTION [6 upvotes]: It is well-known that there exist Groebner bases for ideals in polynomial ring $\mathbb Q[x]$ which can be found algorithmically Moreover, I don't think it is hard to show that there exist Groebner bases for ideals in $\mathbb Z[x]$. But I am having trouble defining Groebner Bases for submodule of free $\mathbb Z[x]$-modules and showing Groebner bases exist. So my question is how do we define Groebner bases for $\mathbb Z[x]$-modules and are we able to find it algorithmically? REPLY [6 votes]: There is a description of the appropriate Groebner basis algorithm in this book: Franz Pauer, Andreas Unterkircher. Gröbner Bases for Ideals in Laurent Polynomial Rings and their Application to Systems of Difference Equations. AAECC 9, 271–291 (1999) I've implemented it in the single-variable case (in the software Regina) and I've been meaning to implement it in the multi-variable case as well. But I usually get too sad to finish – when I look at how inefficient the algorithm is. Some day I'll have it fully implemented in Regina. The book is quite well written. I find it easy to read.<|endoftext|> TITLE: What is known about the eigenvectors of the $2^n \times 2^n$ Hadamard matrix? QUESTION [11 upvotes]: What is known about the eigenvectors of the $2^n \times 2^n$ Hadamard matrix defined recursively by $H_1=(1)$ and $$ H_N=\begin{pmatrix}H_{N/2} & H_{N/2} \\ H_{N/2} & -H_{N/2}\end{pmatrix}, $$ where $N=2^n$? Edit: The answer below provides a "literal" answer to the problem. However, is there a deeper meaning to the eigenvectors? For the Fourier transform operator, for example, Hermite polynomials provide an excellent and rich theory of the eigenvectors. Since the Hadamard transform is indeed a Fourier transform (over the Boolean cube as the underlying group), one could expect the eigenvectors to have a clean interpretation. REPLY [6 votes]: It seems to me that $H_{N}$ is the character table of an elementary Abelian $2$-group of order $2^{n}$ (with respect to a suitable ordering of elements). As such, its rows are orthogonal by the orthogonality relations for group characters. Also, it is clear, by induction that $H_{N}$ is symmetric.Hence we have $H_{N}H_{N}^{t} = 2^{n}I$ (since it is a character table) and $H_{N}^{2} = 2^{n}I$. Thus the eigenvalues of $H_{N}$ are $\pm \sqrt{2^{n}}$, as already noted by Carlos Beenakker. Note also that $H_{N}$ has trace zero for $N > 1,$ so that both square roots occur with equal multiplicity as eigenvalues. Note that since $H_{N}$ is a character table of an Abelian group for $N \geq 2$, its rows and columns are mutually orthogonal. Now since $T = \frac{H_{N}}{2^{\frac{n}{2}}}$ is a matrix of multiplicative order two, we have $T\frac{I+T}{2} = \frac{I+T}{2}$ and likewise $T\frac{I-T}{2} = -\frac{I-T}{2}.$ Hence the columns of $\frac{I+T}{2}$ are eigenvectors of $T$ with eigenvalue $1$ and the columns of $\frac{I-T}{2}$ are eigenvectors of $T$ with eigenvalue $-1$. We can also see that $\frac{I+T}{2}$ and $\frac{I-T}{2}$ are mutually orthogonal idempotent matrices with sum $I$. It follows that $\frac{I+T}{2}$ has rank $2^{n-1}$, as does $\frac{I-T}{2}.$ Hence the columns of $\frac{I}{2} + \frac{H_{N}}{2^{1+\frac{n}{2}}}$ are eigenvectors of $H$ with eigenvalue $2^{\frac{n}{2}}$ spanning the $2^{\frac{n}{2}}$-eigenspace and the columns of $\frac{I}{2} - \frac{H_{N}}{2^{1+\frac{n}{2}}}$ are eigenvectors of $H$ with eigenvalue $-2^{\frac{n}{2}}$ spanning the $-2^{\frac{n}{2}}$-eigenspace.<|endoftext|> TITLE: Is the barycenter of a convex plane curve Lipschitz with respect to the Hausdorff distance? QUESTION [5 upvotes]: Crossposted from Math Stack Exchange For a convex curve $C$, define its barycenter to be $$b(C) = \frac{1}{\mathcal H^1(C)} \int\limits_C x d \mathcal H^1(x)$$ Is there a constant $L$ such that for $C_1,C_2$ convex curves, $|b(C_1) - b(C_2)| \leq L d_H(C_1,C_2)$? Note that the corresponding claim fails if "convex curve" is replaced by "convex body" or "smooth curve", and is true if the barycenter of the curve is replaced by the barycenter of the measure on the curve weighted by the curvature, in which case we recover the Steiner curvature centroid. REPLY [4 votes]: The answe is "yes". Assume $d_H(C_1,C_2)<\varepsilon$. Let $\lambda_1$ and $\lambda_2$ be the length-measures of $C_1$ and $C_2$. We can assume $C_1$ is surrounded by $C_2$; the general case can be reduced to this one. Consider the closest-point projection of $C_2$ to $C_1$. Let $\mu$ be the push-forward of $\lambda_2$ for this projection. Clearly $\lambda_1\le \mu$. Therefore $$|b(\lambda_1)-b(\mu)|<\frac{|\mu-\lambda_1|}{|\mu|}\cdot \mathrm{daim}\, C_1.$$ It remains to apply the following observations: $|b(\lambda_2)-b(\mu)|\le \varepsilon$; $|\mu-\lambda_1|<2\cdot\pi\cdot\varepsilon$; $2\cdot \mathrm{daim}\, C_1\le |\lambda_1|\le \pi\cdot\mathrm{daim}\, C_1$.<|endoftext|> TITLE: How should I think about presentable $\infty$-categories? QUESTION [26 upvotes]: Let me start out with a confession. I have never cared much for set-theoretic size issues, for they seem not to cause much trouble in my day-to-day mathematical life. Despite that, I have always been uncomfortable with their existence, although this discomfort is mostly rooted in a personal aesthetic ideal: I see technical digressions on cardinals, universes, and transfiniteness as a stain on otherwise clean mathematical theories. In the literature on $\infty$-categories, a great deal of attention appears to be given to so-called presentable $\infty$-categories. As a reminder, we say an $\infty$-category $\mathcal{C}$ is presentable if it has all small colimits, and there exists a regular cardinal $\kappa$ so that $\mathcal{C}$ can be realised as the category of $\kappa$-small $\operatorname{Ind}$-objects of some small $\infty$-category. While in most mathematical areas I am interested in, size contraints are of only minor interest, in higher category theory, they appear to arise a lot, usually in the form of requiring certain $\infty$-categories to be presentable. As an example, in Lurie's Higher Algebra, the word 'presentable $\infty$-category' appears hundreds of times. Being such a common occurrence, I can no longer sweep these constraints under the rug: I have to face reality. Having swallowed this set-theoretic pill, the solution appears simple: I shall make myself remember the definition of presentability, and apply the truths as decreed by Lurie and others only to those $\infty$-categories which either Google tells me are presentable, or which I have verified to be presentable myself. Several problems arise, however, and they are the motivation for the questions (or rather, three perspectives on a single question) that I wish to ask. I fail to truly understand the definition. I have only a limited conception of what a cardinal really is, let alone 'regular cardinal'. I have neither feeling for the definition, nor for determining whether a given $\infty$-category satisfies it. Question 1. What is the intuitive idea behind presentability? What does it mean to presentable? How do I recognise if a given $\infty$-category is presentable? I fail to truly understand the motivation behind the definition. I am not well-versed enough in the concepts involved to understand what goes wrong if we do not assume presentability, nor do I understand why presentability is defined the way it is. What if we drop 'regular' in 'regular cardinal'? Why not realise $\mathcal{C}$ as $\operatorname{Pro}$-objects? Question 2. Why is the definition of presentable the way it is, and not something slightly different? I fail to see the beauty in the definition. Although this one is purely subjective, I hope someone feels what I feel. There are these beautifully clean theorems in higher category theory — theorems which have nothing to do with size — that convince me straight away that $\infty$-categories truly are natural and intrinsically simple things, but on top of that one has size constraints all over the place. They simply feel in dissonance with an otherwise flawless theory. Question 3 (optional). In your opinion, why is presentability a natural, and aesthetic definition? REPLY [29 votes]: Presentable $\infty$-categories can be understood without every having to think about cardinals. An $\infty$-category is presentable iff it is equivalent to one of the form $\mathcal{P}(C,R)$, where $C$ is a small $\infty$-category, $R=\{f_i\colon X_i\to Y_i\}$ is a set of maps in $\mathrm{PSh}(C)=\mathrm{Fun}(C^\mathrm{op}, \mathrm{Gpd}_\infty)$, and $\mathcal{P}(C,R)$ is the full subcategory of $\mathrm{PSh}(C)$ spanned by $F$ such that $\mathrm{Map}(f,F)$ is an isomorphism of $\infty$-groupoids for all $f\in R$. That's it. The conditions that $C$ is small and $R$ is a set allow you to show that the inclusion $\mathcal{P}(C,R)\to \mathrm{PSh}(C)$ admits a left adjoint, which implies that $\mathcal{P}(C,R)$ is complete and cocomplete, which is something you definitely want. "Presentable" should be thought of in terms of "presentation", analogous to presentations of a group. In some sense $\mathcal{P}(C,R)$ is "freely generated under colimits by $C$, subject to relations $R$". More precisely, there is an equivalence between (i) colimit preserving functors $\mathcal{P}(C,R)\to D$ to cocomplete $\infty$-category $D$, and (i) a certain full subcategory of all functors $F\colon C\to D$ that "send relations to isomorphisms" (precisely: those $F$ such that $\widehat{F}(f)$ is iso for all $f\in R$, where $\widehat{F}\colon \mathrm{PSh}(C)\to D$ is the left Kan extension of $F$ along $C\to \mathrm{Psh}(C)$). So its easy to construct colimit preserving functors from presentable categories (and all such functors turn out to be left adjoints). REPLY [8 votes]: Just to be a little bit contrary, let me point out one concrete reason that locally presentable categories are not aesthetic: it is unclear whether they have any analogue in constructive mathematics. Most of basic category theory is entirely constructive (the classic quip is "have you ever seen anyone prove that a diagram commutes by assuming that it doesn't and deriving a contradiction?"). But once you move into locally-presentable world, all that goes out the window: the well-ordering and transfinite iteration techniques, and the theory of ordinal and cardinal numbers that underlie them, rely relentlessly not only on the law of excluded middle but also the axiom of choice. It's not impossible that there is an analogous theory constructively, but as far as I know, no one has ever managed to write one down.<|endoftext|> TITLE: Non isomorphic two term complexes with isomorphic kernel, image and cokernel QUESTION [15 upvotes]: Let $R$ be a ring. Can we have two $R$-module maps $A, B: R^n \to R^m$ such that $\mathrm{Ker}(A) \cong \mathrm{Ker}(B)$, $\mathrm{Im}(A) \cong \mathrm{Im}(B)$ and $\mathrm{CoKer}(A) \cong \mathrm{CoKer}(B)$, but such that there is no commutative diagram $$\begin{matrix} R^n & \overset{A}{\longrightarrow} & R^m \\ \cong && \cong \\ R^n & \overset{B}{\longrightarrow} & R^m \\ \end{matrix} \quad ?$$ Comments: I have examples where any two of kernel, image and cokernel match but the third doesn't. The trickiest is to get just the image different: Take $R = \mathbb{R}[x,y,z]/(x^2+y^2+z^2-1)$ and the matrices $\left[ \begin{smallmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&0 \\ \end{smallmatrix} \right]$ and $\left[ \begin{smallmatrix} 1-x^2&-xy&-xz \\ -xy&1-y^2&-yz \\ -xz&-yz&1-z^2 \\ \end{smallmatrix} \right]$. To get just kernel or just cokernel different, use the same ring and the matrices $\left( \left[ \begin{smallmatrix} 1\\0\\0 \end{smallmatrix} \right], \left[ \begin{smallmatrix} x\\y\\z \end{smallmatrix} \right] \right)$ or $\left( \left[ \begin{smallmatrix} 1&0&0 \end{smallmatrix} \right], \left[ \begin{smallmatrix} x&y&z \end{smallmatrix} \right] \right)$ We can make short exact sequences $0 \to \mathrm{Ker}(A) \to R^n \to \mathrm{Im}(A) \to 0$ and $0 \to \mathrm{Im}(A) \to R^m \to \mathrm{CoKer}(A) \to 0$, giving classes $E_A \in \mathrm{Ext}^1(\mathrm{Im}(A), \mathrm{Ker}(A))$ and $F_A \in \mathrm{Ext}^1(\mathrm{CoKer}(A), \mathrm{Im}(A))$. One can show that the diagram exists if and only if one can choose the isomorprhisms between Ker, Im and CoKer to make $E_A = E_B$ and $F_A = F_B$. Thus, a closely related question is: Can you find a ring $R$, finitely generated $R$-modules $M$ and $N$ and two short exact sequences $0 \to M \to R^k \to N \to 0$ whose classes in $\mathrm{Ext}^1(N,M)$ are in different orbits for the action of $\mathrm{Aut}(M) \times \mathrm{Aut}(N)$? If we don't ask for the middle term to be free, there are counterexamples here and here, but I don't see how to tweak any of those counterexamples to make the middle term free. REPLY [12 votes]: Let $R=k[x,y,z]/\left((xy-1)z\right)$, for $k$ some field. Let $A:R\to R$ be multiplication by $z$. Let $B:R\to R$ be multiplication by $xz$. Then $A$ and $B$ have the same image, since $z=xyz$, and therefore isomorphic cokernels. They also have the same kernel, namely the set of polynomials $p(x,y)$ that are multiples of $xy-1$. The only units in $R$ are scalars, so there are no isomorphisms that complete a commutative square.<|endoftext|> TITLE: History of the study of Verma modules in terms of Kazhdan Lusztig Theory QUESTION [7 upvotes]: Let $\mathfrak{g}$ be a complex finite dimensional semisimple Lie algbera, $W$ be the Weyl group, $\rho$ be the half sum of positive roots, $M(\eta)$ be the Verma module of weight $\eta$ and $L(\eta)$ its unique simple quotient. Then it is well-known that $[M(w\cdot(-2\rho)):L(x\cdot(-2\rho))]=P_{w_0w,w_0x}(1)$, where $P_{u,v}$ is the Kazhdan Lusztig polynomial of $W$ and $w_0$ is the longest element in $W$. The above is an example that illustrating the fact that one can study Verma modules in terms of Kazhdan Lusztig Theory. For more evidences, we also have Kazhdan-Lusztig Conjectures for affine Lie algebra (See Masaki Kashiwara and Toshiyuki Tanisaki --- Characters of irreducible modules with non-critical highest weights over affine Lie algebras) and the paper: RONALD S. IRVING--- The socle filtration of a Verma module. I would like to have the brief/full history of the study of Verma modules in terms of Kazhdan Lusztig Theory. If possible, please provides reference(s) for each aspect of the study of Verma modules. It would be great to have history about Verma modules over different types of Lie algberas (semisimple, affine or even Kac Moody Lie algebra). REPLY [4 votes]: It's probably too soon to expect a good historical overview, but for example Steve Kleiman has already written a scholarly article (The development of intersection homology theory) emphasizing the original KL (Kazhdan-Lusztig) conjecture. This is available in arXiv versions or in the published version cited there. One contributor easily overlooked is Vinay Deodhar, who probed further in what he called "KL theory". See for example his influential joint paper Deodhar–Gabber–Kac - Structure of some categories of representations of infinite-dimensional Lie algebras (MSN), or the 1991 survey Deodhar - A brief survey of Kazhdan-Lusztig theory and related topics (MSN). He soon dropped out of the research world because of illness in his immediate family, but was starting to get involved again when he himself developed serious health problems and later died. But in the late 1970s, when I met him at IAS Princeton where Jantzen was then visiting, both he and Jantzen were getting close to the Kazhdan-Lusztig viewpoint. There is also a characteristically readable textbook by Roger Carter, Lie algebras of finite and affine type. See especially Chapters 19-20. This is only the first step toward Verma modules and simple modules for general Kac-Moody algebras, but it conveys much of the flavor of the subject. In the direction of quantum enveloping algebras, there are of course many papers and books (including the one by Lusztig, Introduction to quantum groups (MSN)) which suggest the open-ended nature of the representation theory there. But in this and other directions, there is apparently no comprehensive history yet available. Probably it's too soon for that.<|endoftext|> TITLE: Closed embeddings of monoidal categories in *-autonomous ones QUESTION [9 upvotes]: It's often very convenient for objects in a monoidal category to have duals. Hence, it's natural to wonder whether an arbitrary monoidal category can be embedded in one where all objects have duals. For the usual notion of dual in symmetric monoidal categories, a complete answer to this was given by Joyal, Street, and Verity: any full monoidal subcategory of a compact closed monoidal category (that is, a symmetric monoidal category in which all objects have duals) is a traced monoidal category, so being traced is a necessary condition for embeddability in a compact closed category — and it is also sufficient, by the "Int-construction". Now suppose we relax the notion of dual, replacing compact closed categories by *-autonomous ones (but keeping the symmetry, for simplicity). Every object of a *-autonomous category still has a "dual" in a useful sense, but since the unit and counit of the duality use different monoidal structures, no "traces" are induced. And indeed, in this case, every closed symmetric monoidal category with pullbacks can be embedded in a *-autonomous category, namely its Chu construction relative to an arbitrary choice of dualizing object. (The condition "with pullbacks" can be replaced by "with finite products" if we take the dualizing object to be the terminal one.) So far, so good. But a significant part of the value of having dual objects is that they can be used to construct the internal-homs as $[A,B] = (A \otimes B^*)^*$. So it's natural to ask whether an arbitrary closed symmetric monoidal category can be embedded in one with duals in a way that preserves the closed structure. In the compact closed case, we're out of luck: since $A^* = [A,I]$ in a compact closed category, any sub-closed-monoidal-category of a compact closed category is itself already compact closed. But in the *-autonomous case there's room for a nontrivial answer. For instance, I believe that if $\bot$ is an object of a closed symmetric monoidal category $\mathcal{C}$ such that $[[A,B],\bot] \cong A\otimes [B,\bot]$ for all $A,B$, then the embedding of $\mathcal{C}$ in ${\rm Chu}(\mathcal{C},\bot)$ preserves the internal-homs. This is a strong condition, of course, but it's satisfied for instance if $\bot$ is a zero object. Thus, any closed symmetric monoidal category with pullbacks and a zero object can be closedly-embedded in a *-autonomous one. On the other hand, this sufficient condition is certainly not necessary, since not every *-autonomous category has a zero object. Indeed, I don't know of any nontrivial necessary condition. That is, I don't know any condition that can be expressed in the language of closed symmetric monoidal categories that is true of full closed-monoidal-subcategories of *-autonomous categories but is not true of all closed symmetric monoidal categories. Hence my question: What is a necessary and sufficient condition on a closed symmetric monoidal category (feel free to assume it has any limits and colimits you like) for it to be closed-monoidally-embeddable in a *-autonomous category? Failing that, is there any nontrivial necessary condition for such an embedding to exist? REPLY [5 votes]: In my preprint *-autonomous envelopes I showed that any closed symmetric monoidal category is closed-monoidally-embeddable into a $\ast$-autonomous category, using a variation of Hyland's polycategorical envelope. Thus, there is no additional necessary condition.<|endoftext|> TITLE: Geometry of the section $X_0(N) \to X_0(pN)$ given by the canonical subgroup QUESTION [5 upvotes]: Goren and Kassaei's paper "The Canonical Subgroup: a Subgroup-Free Approach" takes the position that the canonical subgroup of order $p$ for elliptic curves over $\mathbb{Z}_p$ with $\Gamma$-level structure that are "not too supersingular" can simply be viewed as a partially-defined section of the forgetful map $X(\Gamma_0(p) \cap \Gamma) \to X(\Gamma)$. (Passing to the congruence subgroup $\Gamma_0(p) \cap \Gamma$ just adds the extra information of an order $p$ subgroup, assuming $\Gamma$ has no level structure at $p$.) My question is, what does this map look like? For simplicity, assume $\Gamma$ gives no level structure at $p$. The section as described above is a function, defined over $X(\Gamma)$ minus some disks, to $X(\Gamma_0(p) \cap \Gamma)$; it should be injective, so that it's like including this portion of $X(\Gamma)$ as a subspace of $X(\Gamma_0(p) \cap \Gamma)$. Can there be any of these missing disks where we could extend the section to the entirety of that area? Does it look like we just glue more $g$-holed tori to $X(\Gamma) \setminus \{\text{disks}\}$ along the edge of the deleted disks? For a more explicit picture, assume that $X(\Gamma)$ is just a torus, and its supersingular locus is a single disk. Could $X(\Gamma_0(p) \cap \Gamma)$ also be a torus, so that the section given by the canonical subgroup could be extended to all of $X(\Gamma)$? Or would it have to be a higher genus surface to give some obstruction to extending it? Is there any restriction on the genus of $X(\Gamma_0(p) \cap \Gamma)$? It seems like just looking at the map to $X(\Gamma)$ means that it would have to be a genus $2$ surface. If the supersingular locus were two disks, would this restriction on the genus disappear? EDIT: When we look at the canonical subgroup of order $p^n$ (which should be cyclic?), each canonical subgroup of order $p^{ TITLE: Interference between entire functions with separated indicator diagrams QUESTION [5 upvotes]: Let $F,H:\mathbb{C}\to\mathbb{C}$ be entire functions of mean exponential type and of completely regular growth. Assume further that the indicator diagrams $I_F$ and $I_H$ are on the imaginary axis and separated, e.g., $I_F=\imath[a,b]$ and $I_H=\imath[b+1,c]$. Question: Does $F+H \in L^2(\mathbb{R})$ imply $F\in L^2(\mathbb{R} )$? In other words, can $F$ and $H$ interfere on $\mathbb{R}$ in a way that $F+H\in L^2(\mathbb{R}^2)$ without $F$ or $H$ being $L^2(\mathbb{R})$? Discussion: Motivation behind this question comes from real analysis. If we assume in addition that $F,H$ are polynomially bounded on $\mathbb{R}$ then they are Fourier transforms of compactly supported distributions $u_F,u_H$ on $\mathbb{R}$, and $\mathrm{supp}\,u_F=\imath I_F$, $\mathrm{supp}\,u_H=\imath I_H$ (Paley-Wiener, Plancherel-Polya). Qualitatively, the behaviour at infinity of $F$ and $H$ reflect the local regularity of $u_F$ and $u_H$. If $F+ H\in L^2(\mathbb{R})$ then $u_F+u_L\in L^2(\mathbb{R})$, i.e., the sum of the two distributions is locally integrable. However, since $\mathrm{supp}\,u_F\cap\mathrm{supp}\,u_H=\emptyset$, adding $u_H$ to $u_F$ cannot change the local regularity of $u_F$, and we expect $u_F$ to be locally integrable as well, which entails $F\in L^2(\mathbb{R})$. In the above question we do not assume that $F$ and $H$ are polynomially bounded on $\mathbb{R}$, but we let $I_F$ and $I_H$ be at a positive distance. In this case $F$ and $H$ can be thought of as Fourier transforms of analytical functionals, but the latter objects do not have the notion of compact support, and the analogy doesn't work. Thank you. Edit: For an entire function $F$ of mean exponential type, the indicator function can be defined as $$ h_F(\theta)=\limsup_{r\to\infty}\frac{\log|f(re^{\imath\theta})|}{r} $$ and indicates the rate of growth in direction $\theta$. It can be shown that there exists a compact convex set $I_T\subset\mathbb{C}$ (called the indicator diagram of $F$) such that $h_F$ is exactly the support function of $I_F$, $$ h_F(\theta)=\sup_{z\in I_F}\mathrm{Re}\left[ze^{-\imath\theta}\right]. $$ Intuitively, the thickness of the projection of the set $I_F$ on the ray $\theta$ shows the rate of growth of $F$ along that ray. REPLY [6 votes]: The proof that you indicated for polynomially bounded case can be extended almost literally to the general case, if you use the correct generalization of Fourier-Laplace transform. Indeed, suppose that $F+G\in L^2(R)$. Then by the Paley-Wiener, this sum is a Fourier transform of some $L^2$ function, say $h$ supported on $[a,c]$ since the indicator diagram of $F+G$ belongs to $[a,c]$. On the other hand, $F+G$ is the Borel transform of an analytic function in $C\cup\{\infty\}\backslash([a,b]\cup[b+1,c])$. By the uniqueness theorem of the Borel transform, this analytic function "coincides" with $g$ in the sense of hyperfunctions. It follows that both $F$ and $G$ are in $L^2$. Ref. MR1026013 Kaneko, A. Introduction to hyperfunctions. Kluwer, Dordrecht, 1988. On my opinion, this is the best introduction for beginners. Two other good books (more advanced) are Mitsuo Morimoto, An introduction to Sato's hyperfunctions. AMS, Providence, RI, 1993. and L. Hormander, Analysis of linear differential operators with partial derivatives, volume I, chapter 9.<|endoftext|> TITLE: Maximum size of a critical set that sums to $n$ QUESTION [5 upvotes]: Say that a set $S \subset \mathbb Z^+$ can express $n$ if there is some way to add elements of $S$ (possibly more than once) to equal $n$. Call $S$ critical if moreover no proper subset of $S$ can express $n$. For instance, $\{3,4\}$ is critical for $n=11$, because $3+4+4=11$, but $11$ is not a multiple of $3$ or $4$. Given $n$, what is the maximum size of a critical set? Call this number $u_n$. Observe that if $S$ is a critical set for $n$, then $\{2n+1\}\cup\{2k \mid k \in S\}$ is critical for $4n+1$. [1] So the growth rate for $u_n$ is at least logarithmic. I hypothesise that $u_n = \Theta(\log n)$. In fact, it seems to be that $u_n \sim \ln n$. How does one prove that $u_n = \Theta(\log n)$ (assuming that it's true)? [1] - Proof: Clearly, $S' = \{2n+1\}\cup\{2k \mid k \in S\}$ can express $4n+1$. We see that we must use $2n+1$ at least once to do this, as the other elements of $S'$ are all even. What's left is to express $2n$. We can't use $2n+1$ to do this as it's greater than $2n$. So we use what's left of $S'$, which is $\{2k \mid k \in S\}$, and we see that it's critical for $2n$. QED Erratum: The claim in [1] was previously that $S' = \{n+1\}\cup\{2k \mid k \in S\}$ was critical for $3n+1$. REPLY [2 votes]: It seems that $u_n\sim \log_2 n$. To show that $2^{u_n}\leq n+1$, notice that the sums of all subsets of a critical set $S_n$ are distinct and do not exceed $n$. Indeed, if two subsets have the same sum, we may assume they are disjoint. Then we may use this equality to get rid of one of their elements in a representation of $n$. The converse estimate is similar to the approach to a question I mentioned in a comment, with some technicalities. Assume that $n\geq t^32^t$, so that $t\sim \log_2n$. Choose $k\leq t$ such that $k$ is coprime with $2^k-1$ (we may achieve $k\sim t$ even by choosing a prime $k$). Let $ 1\leq a\leq k$ satisfy $b=(2^k-1)a\equiv n\pmod k$; notice that $b(k-1)k2^k$. We claim that the set $$ S=\{D+a2^i\colon 0\leq ikD+k2^k>n$. So the sum contains exactly $k$ summands. Subtracting $kD$ and dividing by $a$, we get a representation of $2^k-1$ as a sum of $k$ powers of $2$ which is unique.<|endoftext|> TITLE: Wreath product $S_k\wr S_n$ inside $S_{kn}$ QUESTION [7 upvotes]: I want to understand wreath products a little better. Currently, my intuition about them is as follows. Take $nk$ disks, pile them in order forming $n$ piles of $k$ disks (one pile contains disks {1,...,k}, another pile disks {k+1,...,2k}, so on). The wreath product $H_{k,n}=S_k \wr S_n$ preserves this strucutre, i.e. it only permutes disks within piles or permutes whole piles. In other words, if $P=S_k\times S_k\times \cdots\times S_k$ is the direct product of groups permuting disks within piles, then $H_{k,n}$ is its normalizer inside $S_{kn}$. I believe all the above is correct. What I need help with is how $H_{k,n}$ sits inside $S_{kn}$. For example, what is the cardinality of the double coset $H_{k,n}/S_{kn}\backslash H_{k,n}$? For $k=2$ I known that this cardinality equals $p(n)$, the number of partitions of $n$. (My intuition is that it would always be $p(n)$ for any $k$, but I am not confident about it.) Suggestions of accessible sources would be welcome (I am a physicist). REPLY [4 votes]: The numbers arise in two contexts of physics: The first is counting symmetric tensor invariants without color. See Counting Tensor Model Observables and Branched Covers of the 2-Sphere, chapter 6. The second is in the counting of vacuum Feynman graphs. The paper features the enumeration of row/column permutation equivalence classes of RC-magic squares, which are $n\times n$ integral matrices with nonnegative entries whose rows and columns sums to $k$. See A combinatorial matrix approach for the generation of vacuum Feynman graphs multiplicities in φ$^4$ theory. The equivalence classes of RC-magic squares are in 1-1 correspondence with the double cosets. To see this, let $g$ be an element of $S_{kn}$, acting on $n$ piles of $k$ disks. We construct an RC-magic square $M(g)$ in the following way: $M(g)_{ij}$ is the number of elements in the $j$-th pile which were originally in the $i$-th pile. In order to prove 1-1 correspondence, we need to prove 1)Two elements in the same double coset necessarily corresponds to the same equivalence class 2)Two elements in the same equivalent class necessarily lies in the same coset. Let $P$ denote the group permuting disks within piles, and $T$ the group permuting piles. Let $H$ be the wreath product group. To prove 1), let $g'=h_1gh_2$, where $h_1,h_2\in H$. Then there exists $p_1,p_2\in P$, $t_1,t_2\in T$ such that $g'=t_1p_1gp_2t_2$. $p_1$ and $p_2$ induce no action on $M(g)$, $t_1$ permute the rows, and $t_2$ permute the columns. So $g$ and $g'$ belongs to the same equivalent class. To prove 2), let $M(g)$ and $M(g')$ lie in the same class. Then $M(g)=t_1M(g')t_2$ for some $t_1,t_2\in T$. ($t_1$ and $t_2$ induce actions on $M(g')$). So $g$ and $t_1g't_2$ moves the same number of balls from pile $i$ to pile $j$ for each $i,j$, which means they differ by elements in $P$: $p_1g=t_1g't_2p_2$ for some $p_1,p_2\in P$. (It is possible to make the point more rigorous, by assigning an order on the elements, and let $p_1$ and $p_2$ sort the elements in each pile). It follows that $g$ and $g'$ are in the same coset. For $k=3$, one can find the exact numbers in Ronald C. Read's PhD Thesis, Some enumeration problems in graph theory, page 156. The asymptotics are given as $$ \frac{(3n)!}{(n!)^26^{2n}}\exp\Bigl(2 - \frac{2}{9n} + O(n^{-2})\Bigr),$$ which can be found in Brendan Mckay's MathOverflow answer.<|endoftext|> TITLE: Permanent of the Coxeter matrix of a distributive lattice QUESTION [13 upvotes]: Let $L$ be a finite distributive lattice with $n$ elements. Let $C=(c_{x,y})$ be the $n \times n$ matrix with entry 1 in case $x \leq y$ and 0 else. The Coxeter matrix of $L$ is defined as the matrix $G_L=-C^{-1}C^T$ (this is the matrix of the Auslander-Reiten translation acting on the Grothendieck group of the derived category of the poset). I noted that for $n \leq 10$ it was always true that the permanent of $G_L$ is either $1$ or $-1$. I was able to prove it only for some small cases such as Boolean algebras and some random examples. Question 1: Is this true in general? Question 2: Does one have a nice order theoretic characterisation when it is $1$ or $-1$ in case question 1 is true? Question 3: Let $L_n$ be the set of distributive lattices with n elements. Is the sum $|\sum_{L \in L_n}^{}{\mathrm{Perm}(G_L)}|$ bounded for $n \rightarrow \infty$? For $n \leq 10$ it was at most 2. It is also interesting to note that for arbitrary finite lattices it seems that the permanent of $G_L$ can be arbitrary large. My knowledge of permanents is close to zero so I'm sorry in case this question is not suitable for MO. The values of this statistic for posets has been entered recently here: http://www.findstat.org/StatisticsDatabase/St001472 . REPLY [7 votes]: $\def\low{\mathop{\mathrm{low}}}\def\perm{\mathop{\mathrm{perm}}}$This seems to be a (hopefully correct now) answer to Q1 and Q2. But this looks a bit strange --- perhaps, it is worth it to check some consequences on small lattices. We implement Birkhoff's theorem in a dual form, identifying $L$ with the lattice of upper sets of some poset $P$ (the structure of which is known); we thus regard each $x\in L$ as an upper set in $P$, so that $x\leq y\iff y\subseteq x$ (!). Next, each element $x\in L$ is determined by the set $\min x$ of its minimal elements, and each independent set $Q\subseteq P$ determines an element $u_Q=\{x\in P\colon \exists q\in Q\quad x\geq q\}$ with $\min u_Q=Q$. 1. First, let us learn the structure of $C^{-1}=(\mu(x,y))$. As has been already mentioned in the comments, $\mu$ is the Möbius function of $L$. Fix an element $x$; for every $J\subseteq \min x$, introduce $x^J=x\setminus J$. Clearly, all the $x^J$ are pairwise distinct. Then one can easily see that $\mu(x,x^J)=(-1)^{|J|}$, and these are the only nonzero values of $\mu(x,\cdot)$ (indeed, the matrix $C'$ determined by these values satisfies $C'C=I$. (Not needed) Similarly, denoting $\low y=\max(P\setminus y)$, we may define $y_J=y\cup J$ for any $J\subseteq \low y$ and see that $\mu(y_J,y)=(-1)^{|J|}$ are the only nonzero value of $\mu(\cdot,y)$. 2. Let now $-G_L=C^{-1}C^T=(g_{xy})$ (we omit the minus sign for clarity; this changes the sign of the permanent in a clear manner). We have $$ g_{xy}=\sum_{z\geq x\vee y}\mu(x,z) =\sum_{z\subseteq x\cap y}\mu(x,z) =\sum_{\textstyle{J\subseteq \min x\atop x\subseteq J\cup y}}\mu(x,x^J) =\begin{cases} (-1)^{|\min x|}, & y\cap x=x\setminus \min x;\\ 0, & \text{otherwise.} \end{cases} $$ To state it simple, the element $g_{xy}$ is nonzero if and only if $y\cap x=x\setminus \min x$, and all such elements for a fixed $x$ are the same, namely, they equal $(-1)^{|\min x|}$. This already yields that all nonzero summands in $\perm G_L$ are equal --- namely, each of them equals $(-1)^{\sum_x|\min x|}$. This shows that the answer to Q1 is affirmative if and only if the permanent has a unique nonzero summand (which is what I called strange), and, given that, answers Q2. We are now to check that strange claim. 3. So, we need to show that there exists a unique permutation $\sigma\colon L\to L$ satisfying $x\cap \sigma(x)=x\setminus\min x$. We start with constructing such permutation $\tau$, and then show its uniqueness. For any $x\in L$, let $$ \tau(x)=\bigcup_{\textstyle{y\in L\atop y\cap x=x\setminus\min x}}y; $$ that is, $\tau(x)$ is the minimal (in $L$) element satisfying the required property. We show that $\tau$ is a permutation by indicating its right inverse $\tau^{-1}$ as $$ \tau^{-1}(y)=u_Q,\qquad \text{where } Q=\max(P\setminus y). $$ Indeed, it is clearly seen that $\tau(\tau^{-1}(y))=y$ for all $y\in L$. Finally, let $\sigma$ be a permutation satisfying the conditions above; then $\sigma(x)\subseteq \tau(x)$ for all $x\in L$. Then $$ \sum_{y\in L}|y|=\sum_{x\in L}|\sigma(x)|\leq \sum_{x\in L}|\tau(x)|=\sum_{y\in L}|y|, $$ so the middle inequality turns into equality. This may happen only if $\sigma=\tau$, which finishes the proof.<|endoftext|> TITLE: Does asymmetric fraction of finite groups tend to $0$? QUESTION [12 upvotes]: Let’s define asymmetric fraction of a finite group $G$ as the number $$\mathrm{af}(G) = \frac{|\{(g, a) \in G \times \mathrm{Aut}(G)\mid a(g) = g\}|}{|G|\cdot|\mathrm{Aut}(G)|}.$$ Equivalently it can be defined as $\mathrm{P}(A(X) = X)$, where $A$ and $X$ are independent uniformly distributed random elements of $\mathrm{Aut}(G)$ and $G$ respectively. Is it true, that $$\forall \epsilon > 0,\; \exists N \in \mathbb{N},\; \forall G \big((\lvert\,G\,\rvert > n) \to (\mathrm{af}(G) < \epsilon)\big)?$$ I know, that $$\mathrm{af}(C_{p^n}) = \dfrac{p^n + \Sigma_{i = 1}^n p^ip^{n - 1 - i}(p - 1)}{p^{2n - 1}(p - 1)} = \dfrac{(np - n + 1)}{p^n(p - 1)}$$ and, that $$\mathrm{af}(G) \leq \frac{1}{2} + \dfrac{|\{g \in G\mid \forall a \in \mathrm{Aut}(G), \; a(g) = g\}|}{2|G|}.$$ However this is clearly not enough to prove the statement. This question was already asked by me on MSE, but received no answers Also, if we restrict the question to abelian case, then the statement will be true. It was proved by Gary Sherman in "What is the Probability an Automorphism Fixes a Group Element?" REPLY [3 votes]: Here is a possible road map to a solution. Let $G$ be a finite group such that $\def\af{\operatorname{af}}\def\Aut{\operatorname{Aut}}$$\af(G)$ is bounded away from $0$. We want to show that $|G|$ is bounded. By the theorem of Peter Neumann that Geoff Robinson mentioned, $G$ is bounded-by-abelian-by-bounded. In fact, $G$ has a bounded-index $2$-step nilpotent normal subgroup $H$ such that $|H'|$ is bounded. We can moreover assume that every $h \in H$ has a bounded-size $\Aut G$-orbit. This almost reduces us to considering $H$ in place of $G$, except that automorphisms of $H$ need not extend to automorphisms of $G$. In any case, $G = H$ is an excellent model problem (i.e., $2$-step nilpotent with bounded commutator). These groups tend to have loads of outer automorphisms, so $H$ should have plenty of large $\Aut H$-orbits. (The archetypal example here is the extraspecial $p$-group of order $p^{2n+1}$ and exponent $p$, which has outer automorphism group containing $\operatorname{Sp}(2n, p)$.) Finally we have an extension problem. Given that $H$ has bounded index in $G$, we need to show that at least one of the large $\Aut H$-orbits contains a large $\Aut G$-orbit. For example, what if $G$ is virtually abelian? (Split extensions such as a dihedral group are not interesting, because it's easy to extend automorphisms in that case.)<|endoftext|> TITLE: Map from a classifying space to a stack QUESTION [6 upvotes]: Let $G$ be an algebraic group over a field $k$, and let $BG$ be its classifying space. Let $X$ be a stack over $k$ (e.g. $X$ could be the Picard stack $Pic(S)$, for some scheme $S$). I'm trying to understand what it means to have a morphism $$BG \to X$$ of stacks. Since $BG=[pt/G]$, my guess is that a map $BG \to X$ is just an object $x \in X(k)$ together with a homomorphism of groups $G(k) \to Aut(x)$. Is this correct? This doesn't seem to take into account any of the geometric structure of $G$, so I wouldn't be surprised if this is isn't right. What is the right way to think of maps out of a classifying stack? REPLY [11 votes]: You're almost there! The problem is that, as you've surmised, the group $\mathrm{Aut}(x)$ does not capture enough of the geometric structure of $G$. But that's easily solved: For every $x\in X$ we can define a sheaf of groups $\underline{\mathrm{Aut}}(x)$ such that for every $k$-scheme $S$ $$\underline{\mathrm{Aut}}(x)(S):=\mathrm{Aut}_{X(S)}(x_S)\,.$$ where $x_S$ is the image of $x$ under the functor $X(k)→X(S)$ induced by the map of schemes $S\to \mathrm{Spec}\,k$. That this is a sheaf follows from the definition of stack ("descent for morphisms"), the details might depend on your preferred way of phrasing the definition. Then, for every sheaf of groups $G$ we can define a stack $BG$, such that for every $k$-scheme $S$ $BG(S)$ is the groupoid of $G_S$-torsors over $S$ (this recovers the notion when $G$ is a group scheme). Note that $BG(k)$ has a distinguished object $\ast$ (corresponding to the trivial $G$-torsor over $k$) such that there's a canonical isomorphism of sheaves of groups $$\underline{\mathrm{Aut}}(\ast)\cong G\,.$$ Then the groupoid of maps $BG\to X$ is the groupoid of pairs $(x\in X(k),\varphi:G\to \underline{\mathrm{Aut}}(x))$, where $x$ is an object of $X(k)$ and $\varphi$ is a map of sheaves of groups. The morphisms in the groupoid are the morphisms in $X(k)$ that conjugate the maps from $G$. The map in one direction is easy: it sends $F:BG\to X$ to the pair $(F(\ast),G\cong \underline{\mathrm{Aut}}(\ast)\to \underline{\mathrm{Aut}}(F(\ast)))$. The map in the other direction is slightly trickier: suppose I have a pair $(x,\varphi)$. Then for every $S$ I have to give a functor $BG(S)\to X(S)$. Essentially this sends a $G_S$-torsor $T$ to $T\times_{G_S} x_S$, where $G_S$ acts on $x_S$ via $\varphi$. Making this precise is a bit annoying and hence I shall leave it as an exercise. Really, the easy way of conceptualizing this (and arguably proving it) is to notice that $BG$ is the stackification of the prestack $BG^{pre}$ sending $S$ to the groupoid with one objects and $G(S)$ automorphisms. Then we can compute the groupoid of maps $BG\to X$ as the groupoid of maps $BG^{pre}\to X$, and the latter is easily seen to coincide with my description above. For the homotopy theorists that might be listening: yes the above could be simply summarized by saying that the functor from groups to pointed groupoids sending $G$ to $BG$ is the left adjoint to the functor sending $X$ to $\Omega X$. All I described here works in an arbitrary topos: I did not use that the sheaf of groups $G$ came from a group scheme.<|endoftext|> TITLE: Some interesting and elementary topics with connections to the representation theory? QUESTION [17 upvotes]: I'm going to give a talk to talented high school seniors (for nearly 1.25-1.75 hours, maybe a little bit longer). They know some abstract algebra (groups, rings, modules...), linear algebra (including the Jordan normal form, nilpotent operators, etc.), a real (and basics of complex) calculus, some elementary topology (basics from the general topology, fundamental groups and maybe something about manifolds or vector fields), basics of algebraic geometry and a lot of combinatorial stuff (including the graph theory and generating functions). My main interest is the representation theory, so I'd like to discuss an algebraic topic connected with this branch of mathematics. It'll be cool if this topic contains some beautiful combinatorial constructions. Despite this, some level of abstraction is required... I think that it mustn't also be too famous (so the standard things like representations of symmetric groups (even in Vershik-Okounkov approach) or the basics of Lie theory aren't acceptable). Ideally, it has to be new to me... Can someone give me a piece of advice about which topic can be chosen in this situation? I thought about the cluster algebras (maybe in the flavor of the first pages of the paper [1]?)... But are there any elementary applications of them? (And, by the way, are there any classical monographs about this subject?) [1] -- http://ovsienko.perso.math.cnrs.fr/Publis/FriezeNew1.pdf UPD: There is a question which looks like similar to this on MO. Namely, Fun applications of representations of finite groups . But, of course, it's very different. The reason is that I'm more interested in representations of more complicated than finite groups structures like, for example, quivers or Lie algebras. So the other question isn't relevant: its topic is too narrow. REPLY [3 votes]: One example of an elementary application of cluster algebras is the proof that the Somos-4 and Somos-5 sequences, which are defined by a simple recursion, are integral. This is so because the entries of these sequences are cluster variables with the initial cluster variables specialized to 1, so it follows from the Laurent phenomenon. See Fomin and Zelevinsky's paper The Laurent Phenomenon. One wouldn't have time to prove the Laurent phenomenon, but one can demonstrate it, and it is quite striking.<|endoftext|> TITLE: Diagonal plus low-rank decomposition of symmetric matrices QUESTION [11 upvotes]: Let $r(n)$ be the smallest integer such that All symmetric $n\times n$ matrices with non-zero real entries can be written as the sum of a diagonal matrix and a matrix of rank $r(n)$ What is $r(n)$? I can show that $$r \left( \binom{k}{2} \right) \geq \binom{k-1}{2}$$ Furthermore, I can show that $r(n)\leq n-2$ and in particular $r(3) = 1$. REPLY [7 votes]: I am afraid that it is quite possible that the rank of any diagonal perturbation of a real symmetric $n\times n$-matrix $A$ with non-zero entries is always at least $n-2$. That is, $r(n)\geqslant n-2$, you say that also $r(n)\leqslant n-2$, thus $r(n)=n-2$. Moreover, it may happen that the submatrix formed by the columns $3,4,\ldots,n$ and rows $1,2,\ldots,n-2$ is always non-degenerated. Namely, take $$ A_{ij}=\begin{cases}2,&\text{if}\quad |j-i|\geqslant 2 \,\,\text{and}\,\, \min(i,j)\geqslant 2\\ 1,& \text{otherwise}\end{cases} $$ Say, it is how it looks for $n=6$: $$\pmatrix{1&1&1&1&1&1\\ 1&1&1&2&2&2\\ 1&1&1&1&2&2\\ 1&2&1&1&1&2\\ 1&2&2&1&1&1\\ 1&2&2&2&1&1}.$$ The above claim follows by subtracting the $(n-1)$-st column from the $n$-th, then $(n-2)$-nd column from the $(n-1)$-st etc.<|endoftext|> TITLE: $B(\chi), L'(1,\chi)/L(1,\chi),\dotsc$ QUESTION [5 upvotes]: Let $\chi$ be a primitive Dirichlet character of modulus $q>1$. Write, as is customary, $B(\chi)$ for the constant in the expression $$\frac{\Lambda'(s,\chi)}{\Lambda(s,\chi)} = B(\chi) + \sum_\rho \left(\frac{1}{s-\rho} + \frac{1}{\rho}\right),$$ where $\Lambda(s,\chi)$ is a completed Dirichlet $L$-function and $\sum_\rho$ is a sum over its zeros. Obviously, $B(\chi) = \Lambda'(0,\chi)/\Lambda(0,\chi)$. Since $$\frac{L'(s,\chi)}{L(s,\chi)} = \frac{\Lambda'(s,\chi)}{\Lambda(s,\chi)} - \frac{1}{2} \frac{\Gamma'((s+\kappa)/2)}{\Gamma((s+\kappa)/2)} - \frac{1}{2} \log \frac{q}{\pi},$$ where $\kappa = [\chi(-1)=-1]$, we see that $$B(\chi) = b(\chi) - \frac{\gamma}{2} - \kappa \log 2 + \frac{1}{2} \log \frac{q}{\pi},$$ where $b(\chi)$ is the constant term in the Laurent expansion of $L'(s,\chi)/L(s,\chi)$ around $s=0$. We can easily show that $$b(\chi) = \log \frac{2\pi}{q} + \gamma - \frac{L'(1,\overline{\chi})}{L(1,\overline{\chi})}$$ by taking logarithms on both sides of the functional equation. We thus obtain that $$B(\chi) = \frac{1}{2} \log \frac{4^{1-\kappa} \pi}{q} + \frac{\gamma}{2} - \frac{L'(1,\overline{\chi})}{L(1,\overline{\chi})}.$$ It seems clear to me that this expression for $B(\chi)$ must be (very) classical. Now, looking in Montgomery-Vaughan for something else, I see that, in section 10.3, it states that "The constant $B(\chi)$... was long considered to be mysterious; the simple formula (10.39) for it [namely, the expression for $B(\chi)$ right here] is due to Vorhauer (2006)." Here Vorhauer (2006) is an unpublished preprint (not accessible online). I'd gladly give credit where credit is due, but I can't help thinking that this expression must have been known long before 2006. Does anybody have an earlier reference? (And what would be so mysterious about $B(\chi)$? IMHO, it is just tricky for the same reason that $L'(1,\chi)/L(1,\chi)$, viz., the possibility of a Siegel zero. Or is it just that we don't have an expression for it as nice as the class number formula? (Do we? EDIT: for $\chi$ odd, we do; see Prop. 10.3.5 (due to...?) in Henri Cohen's Number Theory.) On the issue of bounding it, see $|L'(1,\chi)/L(1,\chi)|$.) REPLY [5 votes]: I am replying to this question “for odd, we do; see Prop. 10.3.5 (due to...?) in Henri Cohen's Number Theory”. I would be happy to insert a comment instead, but my MO-reputation is not good enough... In a paper published in 1989, Kanemitsu wrote that this formula was first published by Berger in 1883. Kanemitsu’s paper is entitled “ On evaluation of certain limits in closed form”, pages 459-474, of the volume: Proceedings of the International Number Theory Conference held at Université Laval, July 5-18, 1987 Ed. by Koninck, Jean M. de / Levesque, Claude Series:De Gruyter Proceedings in Mathematics. Berger’s paper is “sur une sommation des quelques series”, Nova Acta Reg. Soc. Sci. Ups.(3) 12 (1883).<|endoftext|> TITLE: Is $S$ a smooth submanifold of $M$? QUESTION [5 upvotes]: Let $G$ be a Lie group and $H$ a Lie subgroup of $G$. Let $M$ be a smooth manifold. Let $\theta$ be a left smooth action of $G$ on $M$. Let $S=\{p\in M| G_p=H\}$, where $G_p$ is the isotropy group of $p$. Is $S$ a smooth submanifold of $M$? REPLY [8 votes]: More details in Thomas Rot's comment. Every closed subset of any manifold is the set of zeroes of a vector field, even one which is complete (for example, bounded in a complete Riemannian metric). So the vector field generates a group action, whose fixed point set is that closed set. So if we take $G$ to be the real line, and $H=G$, we get any closed set arising as $S$, for some vector field. To be more explicit, if $f(x)$ is bounded and vanishes to all orders at the origin, and nowhere else, then $\sin(1/x)f(x)\partial_x$ can be our vector field on the real line, with fixed points $x$ at $1/x \in 2\pi \mathbb{Z}$.<|endoftext|> TITLE: Usefulness of Nash embedding theorem QUESTION [37 upvotes]: Nash embedding theorem states that every smooth Riemannian manifold can be smoothly isometrically embedded into some Euclidean space $E^N$. This result is of fundamental importance, for it unifies the intrinsic and extrinsic points of view of Riemannian geometry, however, it is less clear that it is also useful. Most if not all results in differential geometry that I am aware of seem to have been obtained using the intrinsic point of view so avoiding any recurse to Nash embedding theorem. Can you mention results that used in their proof the Nash embedding in an essential way, or results whose proof was considerably simplified by the Nash embedding result? If not can you explain why Nash theorem is less useful and powerful than expected? REPLY [14 votes]: Sobolev mappings between Riemannian manifolds. Let $N$ be closed, and $M$ is compact, possibly with boundary. A natural definition of Sobolev mappings between Riemannian manifolds $W^{1,p}(M,N)$ requires the isometric embedding ot $N$ into an Euclidean space $N\subset\mathbb{R}^\nu$ which is the Nash theorem. Then the space is defined as $$ W^{1,p}(M,N)=\{u\in W^{1,p}(M,\mathbb{R}^\nu):\, u(x)\in N \text{ a.e.}\} $$ The space is equipped with the metric inherited from $W^{1,p}(M,\mathbb{R}^\nu)$. The space does not depend on the isometric embedding of $N$. If $\iota_1:N\to\mathbb{R}^{\nu_1}$ and $\iota_2:N\to\mathbb{R}^{\nu_2}$ are two isometric embeddings and we denote by $$ W^{1,p}_{\iota_1}(M,N) \quad \text{and} \quad W^{1,p}_{\iota_2}(M,N) $$ the spaces obtained with respect to these embeddings, then for a mapping $u:M\to N$ we have that $$ \iota_1\circ u\in W^{1,p}_{\iota_1}(M,N) \quad \text{if and only if} \quad \iota_2\circ u\in W^{1,p}_{\iota_2}(M,N). $$ However, the metric in the space depends on the embedding, but the map $$ W^{1,p}_{\iota_1}(M,N)\ni u\mapsto \iota_2\circ\iota_1^{-1}\circ u\in W^{1,p}_{\iota_2}(M,N) $$ is a homeomorphism of spaces. This class of mappings appear in a natural way in the study of geometric variational problems for mappings between manifolds. Like for example, the theory of harmonic mappings. One of the early problems was the question whether smooth mappings are dense. That led to a very fruitful research showing deep connections to algebraic topology. You can find some basic information about this space as well as references in the survey paper (available on my website): P. Hajłasz, Sobolev mappings between manifolds and metric spaces. In: Sobolev Spaces in Mathematics I. Sobolev type Inequalities pp. 185-222. International Mathematical Series. Springer 2009.<|endoftext|> TITLE: Condition for two surfaces to not live inside a common threefold QUESTION [5 upvotes]: Let $Y_1$, $Y_2$ be two complex smooth projective surfaces, are there some restrictions for $Y_1$ and $Y_2$ to be embedded in a common smooth projective threefold? The first thought is to use Lefschetz hyperplane theorem, is there any example that $Y_1$ and $Y_2$ have same first Betti number but can't be embedded in a common threefold? REPLY [6 votes]: No, there are no restrictions. Here is a construction of such a threefold $X$. [I assume that the $Y_i$ are connected, and I look for a connected $X$.] For $i\in\{1,2\}$, let $C_i$ be the cone over $Y_i$ in some projective embedding of $Y_i$. Let $f_i:C'_i\to C_i$ be the blow-up of the vertex of $C_i$, with exceptional divisor isomorphic to $Y_i$. Choosing a projective embedding of $C_i$, and then a generic projection to $\mathbb{P}^3$, we obtain a finite morphism $g_i:C_i\to\mathbb{P}^3$. Composing $g_1$ with a generic element of $\textrm{PGL}_4(\mathbb{C})$, we may assume that the image by $g_i$ of the vertex of $C_i$ is included in the locus above which $g_{3-i}$ is étale (for $i\in\{1,2\}$). Consider the fiber product $(C'_1)\times_{\mathbb{P}^3}(C'_2)$ of the morphisms $g_i\circ f_i$. It contains $\textrm{deg}(g_1)$ copies of $Y_2$ and $\textrm{deg}(g_2)$ copies of $Y_1$, and it is smooth along these subvarieties. Using that $C_1'$ and $C'_2$ are irreducible, it is easily verified that one can choose one copy of $Y_1$ and one copy of $Y_2$ belonging to the same connected component $Z$ of the smooth locus of $(C'_1)\times_{\mathbb{P}^3}(C'_2)$. We let $X$ be a smooth projective compactification of $Z$.<|endoftext|> TITLE: Does almost every point in Euclidean space have unique projection on any given set? QUESTION [13 upvotes]: I am stuck at one point in my research, where I need to prove something which appears trivial to me, but cannot find a rigorous proof. I describe it below. Whenever I will say projection, I will mean the $L^2$ projection in Euclidean spaces. We all know that the projection of a point on a closed, convex subset of the Euclidean space is unique. Now, if a set $A \subseteq \mathbb{R}^n$ is closed, we can easily show that the projection of a point $p \in \mathbb{R}^n$ on $A$ exists, but can find counterexamples, such that it is not unique. So, here is my first question (assume henceforth, that $A$ is a closed set): Is it true that the set of all points $p \in \mathbb{R}^n$ that have more than one projection on the set $A$ (let us call that set $\mathcal{P}(A)$), has Lebesgue measure $0$? It seems too much for question 1 to have an affirmitive answer for any arbitrary closed set $A$ (I do not even know whether the set $\mathcal{P}(A)$ is measurable, but if not, I can work with outer measures). So, here comes a simpler question. If $A$ is the union of two convex sets, is it true that $\mathcal{P}(A)$ has Lebesgue measure $0$? Even if this seems too much, I would really be happy to have an affirmitive answer (with a proof) to the following even simpler question: If $A$ is the union of two polyhedra (a polyhedron is a finite intersection of half-spaces, and hence, is convex), does $\mathcal{P}(A)$ have Lebesgue measure $0$? I intuitively feel that 3 must be correct, the reason being as follows. If a point $p$ has two distinct projections in $A$, then these projections must lie on two different faces of the union of the polyhedra, must be projections of $p$ on the respective faces too, and must be equidistant from $p$. The set of such point seems to be a lower dimensional hyperplane, which has Lebesgue measure $0$. But I cannot make this further rigorous. The problem is, I cannot seek help from linear algebra, as these faces are flats, and not even subspaces. Any help (at least with answering question 3) will be greatly appreciated! REPLY [15 votes]: That is true. The set of points with non-unique projection has measure zero. The proof is a beautiful application of the Rademacher theorem. You can find comments and the link to a proof here: Set of points with a unique closest point in a compact set. See also: https://mathoverflow.net/a/324877/121665.<|endoftext|> TITLE: Does the mean ratio of the perimeter to the hypotenuse of right triangles converge to $1 + \dfrac{4}{\pi}$? QUESTION [17 upvotes]: Conjecture: Let $\mu_x$ be the arithmetic mean of the ratio of the perimeter to the hypotenuse of all primitive Pythagorean triplets in which no side exceeds $x$; then, $$ \lim_{x \to \infty}\mu_x = 1 + \frac{4}{\pi}$$ Based on data for $x \le 10^{11}$, the computed value agrees with the conjectured value up to $5$ decimal places. Primitive Pythagorean triplets $a^2 = b^2 + c^2, \gcd(b,c) = 1$ are given by $a = r^2 + s^2$, $b = r^2 - s^2$ and $c = 2rs$ where $r > s$ are natural numbers. Let the $n$-th primitive triplet be the one formed by the $n$-th smallest pair in increasing order of $(r,s)$. It was proved in this post MSE that the arithmetic mean $\mu_n$ of the ratio of the perimeter to the hypotenuse of first $n$ primitive Pythagorean triplets approaches $\dfrac{\pi}{2} + \log 2$ as $n \to \infty$. However the above claim is still open. Question: I am looking for a proof or disproof of this conjecture. REPLY [16 votes]: This is right. Primitive Pythagorean triples are parametrized as $(u^2-v^2, 2uv, u^2+v^2)$ with $\mathrm{GCD}(u,v) = 1$ and $u+v \equiv 1 \bmod 2$. To have $0 < a,b < c \leq R^2$, we must have $0 < v < u$ and $u^2+v^2 \leq R^2$. The ratio of perimeter to hypotenuse is $\tfrac{(u^2-v^2)+(2uv)+(u^2+v^2)}{u^2+v^2} = \tfrac{2u(u+v)}{u^2+v^2}$. Ignoring the $GCD$ and parity conditions, we want to compute $$\sum_{0 TITLE: Text on old-fashioned differential geometry QUESTION [8 upvotes]: I am seeking good books on the geometry of surfaces in Euclidean space, which would in particular discuss Darboux frames. Please explain for each suggestion why you like this book (classics are welcome). REPLY [11 votes]: Chern explains moving frames in a masterful 4 pages: (1990, pp. 210–213). Further nice book-length treatments using them: Valiron (1950, Chap XIII–XIV) — last of the classic Cours, translated, very complete. Guggenheimer (1963, Chap. 10–11) — actually says “Darboux frames” throughout. O’Neill (1966, Chap. VI–VII) — standard choice. Cartan (1967, Chap. III) — as learned from his father(?) Do Carmo (1971, Chap. 5–6) — not (1976) which mostly does without frames. Blaschke (1973, Chap. 5–6) — as taught to Chern(?) REPLY [9 votes]: There is a very general discussion of moving frames in Clelland's book, From Frenet to Cartan: The Method of Moving Frames, and this in particular includes discussion of Darboux's use of frames on surfaces. Darboux's own 4 volume Lecons sur la Theorie Generale des Surfaces at les Applications Geometriques du Calcul Infinitesimal is quite readable, although very long.<|endoftext|> TITLE: Do the following binary vectors span $\mathbb{R}^n$? QUESTION [7 upvotes]: Defining the binary vectors Let an ordered triple of natural numbers $(r, d, n)$ such that $0 \leq r < d \leq n$ be given. Consider the binary vector $v_{(r,d,n)} \in \mathbb{R}^n$ such that for all $i \in \{0\} \cup [n-1]$: \begin{align*} (v_{(r,d,n)})_i = 1 & \quad\text{if $i \equiv r \mod d$} \\ (v_{(r,d,n)})_i = 0 & \quad\text{otherwise.} \end{align*} In other words, $r$ is the remainder, $d$ is the divisor, and $n$ is the dimension. An example vector Let's take $r = 1$, $d = 3$, and $n = 14$. In this case, we have: $$ v_{(1,3,14)} = (0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1). $$ Defining the subspaces Let an ordered pair of natural numbers $(m, n)$ such that $m \leq n$ be given. Consider the subspace $V_{(m,n)} \subseteq \mathbb{R}^n$ given by $$ V_{(m,n)} = \operatorname{span} \{ v_{(r, d, n)} \; \vert \; 0 \leq r < d \leq m \}. $$ In other words, $m$ is a bound for the divisor. My Question Let a natural number $n$ be given. Consider the function $k(n)$ given by: $$ k(n) \mathrel{:=} \min \{ m \; \vert \; V_{(m,n)} = \mathbb{R}^n \}. $$ Without much effort, it can be shown that $k(n) \leq n$ and $k(n) = \Omega(\sqrt{n})$. Can we prove asymptotically tighter bounds on $k(n)$? My intuition is that $k(n) = O(\sqrt{n} \cdot \log(n))$. Update 1 The problem has be solved thanks to @Ilya Bogdanov. Below I added a snippet of Octave code in case anyone is interested in checking on smaller values of $n$. As requested, for the $n = 30$ case, we have $k(30) = 10$. % Parameters n = 30 m = 10 % Compute number of rows & cols rows = (m + 1) * m / 2 cols = n % Construct matrix M = zeros(rows, cols); count = 1; for d = 1:m for r = 0:(d-1) for c = 1:cols if r == mod(c-1, d) M(count, c) = 1; end end count += 1; end end % Print matrix M % Print rank rankOfM = rank(M) Update 2 Below is a table of values for $k(n)$ when $1 \leq n \leq 50$. | n | k(n) | ------------- | 1 | 1 | | 2 | 2 | | 3 | 3 | | 4 | 3 | | 5 | 4 | | 6 | 4 | | 7 | 5 | | 8 | 5 | | 9 | 5 | | 10 | 5 | | 11 | 6 | | 12 | 6 | | 13 | 7 | | 14 | 7 | | 15 | 7 | | 16 | 7 | | 17 | 7 | | 18 | 7 | | 19 | 8 | | 20 | 8 | | 21 | 8 | | 22 | 8 | | 23 | 9 | | 24 | 9 | | 25 | 9 | | 26 | 9 | | 27 | 9 | | 28 | 9 | | 29 | 10 | | 30 | 10 | | 31 | 10 | | 32 | 10 | | 33 | 11 | | 34 | 11 | | 35 | 11 | | 36 | 11 | | 37 | 11 | | 38 | 11 | | 39 | 11 | | 40 | 11 | | 41 | 11 | | 42 | 11 | | 43 | 12 | | 44 | 12 | | 45 | 12 | | 46 | 12 | | 47 | 13 | | 48 | 13 | | 49 | 13 | | 50 | 13 | REPLY [9 votes]: Let $v_{r,d}$ be an infinite sequence defined in the same way, and let $V_m$ be the span of all corresponding sequences with $d\leq m$. For a fixed $d$, the linear span of all $v_{r,d}$ is the set of all linear recurrences with characteristic polynomial $$ x^d-1=\prod_{k\mid d} \Phi_k(x), $$ where $\Phi_k$ is the $k$th cyclotomic polynomial. Therefore, $V_{m}$ is the set of all linear recurrences with characteristic polynomial $$ P_m(x)= \mathop{\mathrm{lcm}}\left\{x^d-1\colon d\leq m\right\}= \prod_{k\leq m} \Phi_k(x). $$ Thus, whenever $ \deg P_m\geq n$, the sequences from $V_m$ may have arbitrary first $n$ terms, so $V_{m,n}=\mathbb R^n$. Conversely, if $\deg P_m TITLE: The product of two supersingular elliptic curves is independent of which ones we pick QUESTION [5 upvotes]: In a comment on this MO question, Qing Liu says "In positive characteristic p, if you take two supersingular elliptic curves $E_1,E_2$, then $E_i×E_j$ is isomorphic to $E^2_1$ for any pair $i,j$." Why is this true? REPLY [5 votes]: See Theorem 3.5 in "Supersingular K3 surfaces" by TetsuJi Shioda, or a recent paper "Abelian varieties isogenous to a power of an elliptic curve" at https://arxiv.org/abs/1602.06237. Let $C_0$ be a supersingular elliptic curve over an algebraically closed field $k$ of char $p>0$, and $R:= \operatorname{End}(C)$ which is a maximal order in the quaternion algebra $D_{p,\infty}=\operatorname{End}(C)\otimes \mathbb Q$. Note all supersingular elliptic curves are isogenus, and there is a bijection between supersingular elliptic curves over $k$ and rank one projective right $R$ modules (both up to isomorphism) given by $C \mapsto \operatorname{Hom}(C,C_0)$. The key point for us is that if the natural right $R$ module $\operatorname{Hom}(C,C_0)$ is free i.e $\operatorname{Hom}(C,C_0) \cong R$, then $C \cong C_0$. Similar results hold for product of supersingular elliptic curves. Now the proof is finished by an old fact that any projective module of rank $g \geq 2$ over $R$ is free, see "M. Eichler, Über die Idealklassenzahl hyperkomplexer Systeme, Math. Z. 43 (1938), 481–494", which is written in old language and it seems only a few people know the proof.<|endoftext|> TITLE: Gromov hyperbolic groups which are solvable are elementary QUESTION [7 upvotes]: I have read on wikipedia that a Gromov hyperbolic group which is solvable is elementary (i.e. virtually cyclic). Where can I find a proof of this fact? There is a proof of a similar fact in Bridson-Haefliger that if a solvable group $\Gamma$ acts properly and cocompactly on a CAT(0) space, then it is virtually abelian. Is the proof similar to this one? REPLY [13 votes]: I do not know an exact reference. I think it is a folklore. Here is a proof using basic properties of hyperbolic groups which can be found in any book on hyperbolic groups. Let $G$ be solvable and hyperbolic. Then it has an Abelian normal subgroup $H$, the last nontrivial member of the derived series. If $H$ is finite, then $G/H$ is quasi-isometric to $G$, so it is hyperbolic and we can proceed by induction on the solvability class. So $H$ is infinite. It cannot contain an infinite locally finite subgroup because all finite subgroups of a hyperbolic group have uniformly bounded orders. $H$ cannot contain a subgroup isomorphic to $\mathbb Z^2$ because $G$ is hyperbolic. Hence $H$ is virtually cyclic and contains a characteristic infinite cyclic subgroup $H_0$. Then $H_0$ is normal in $G$. So $G$ has a homomorphism into $Out(\mathbb Z)$ which is a group of order 2. Hence $G$ has a subgroup $N$ of index at most 2 which centralizes $H_0$. A centralizer of an infinite cyclic subgroup in a hyperbolic group is virtually cyclic. Hence $G$ is virtually cyclic. PS. I was talking about Gromov hyperbolic, hence finitely generated discrete groups. The (solvable) group of $2 \times 2 $ upper triangular matrices with determinant 1, with Riemannian metric is quasi-isometric to the hyperbolic plane, hence is hyperbolic itself. PPS. Another proof is based on Druţu, Cornelia; Sapir, Mark Tree-graded spaces and asymptotic cones of groups. With an appendix by Denis Osin and Mark Sapir. Topology 44 (2005), no. 5. If $G$ is solvable, it satisfies a non-trivial law. Hence if $G$ is finitely generated but not virtually cyclic, its asymptotic cones do not contain cut points. But if $G$ is hyperbolic then all asymptotic cones are trees where every point is a cut-point. Hence $G$ is virtually cyclic. This proof is of course an overkill, but it works for every group satisfying a non-trivial law, not just solvable groups. Also "hyperbolic " can be replaced by "relatively hyperbolic", "acylindrically hyperbolic" and even by "lacunary hyperbolic". PPPS Of course the easiest proof is using the fact that every nonelementary hyperbolic group contains a noncyclic free subgroup (a solvable group cannot contain such a subgroup). I do not remember who proved it first. Gromov, probably. Maybe Olshanskiy or Delzant.<|endoftext|> TITLE: When does a locally symmetric space have no odd degree Betti numbers? QUESTION [11 upvotes]: Let $G$ be a semisimple real lie group, $K$ be a maximal compact subgroup of $G$, $\Gamma$ be a torsion-free cocompact discrete subgroup. The Betti number the locally symmetric space $X_{\Gamma}:=\Gamma \backslash G/K$ can be expressed in term of unitary representations of $G$ by Matsushima's formula. The formula looks quite complicated in general, the question is: when does $X_{\Gamma}$ have all vanishing odd degree Betti numbers? I am interested in some examples and some necessary conditions. Example: Consider the toroidal compactification of Siegel threefold with $3$ level structure, its Betti numbers are $1, 0, 61, 0, 61, 0, 1$, see "The Siegel modular variety of degree two and level three" by J. William Hoffman and Steven H. Weintraub (Trans AMS 2001). REPLY [3 votes]: Here is an answer of sorts. For simplicity, I restrict to cocompact torsion-free lattices. Then $b_{odd}(X_\Gamma)=0$ implies that $X$ is even-dimensional. The situation is clear if $dim(X)\le 2$, so the first interesting case is when $dim(X)=4$. There are (essentially) three 4-dimensional symmetric spaces of noncompact type (up to rescaling the metric on de Rham factors): The real-hyperbolic space, ${\mathbb H}^4$ (when $G=PO(4,1)$), the complex-hyperbolic space ${\mathbb H}^2_{\mathbb C}$ (when $G=PU(2,1)$) and the product of two hyperbolic planes ${\mathbb H}^2\times {\mathbb H}^2$ (when $G=PO(2,1)\times PO(2,1)$), up to commensuration of Lie groups. $X={\mathbb H}^4$. I do not know if there are examples with $b_1=0$. However, it is known that for every arithmetic uniform lattice $\Gamma< PO(4,1)$ there exists a congruence-subgroup $\Gamma_1<\Gamma$ such that $b_1(\Gamma_1)= b_1(X_{\Gamma_1})>0$ (Millson). I do not think anybody knows what happens with non-arithmetic manifolds. $X={\mathbb H}^2_{\mathbb C}$. Arithmetic lattices come in two types: All lattices of type I again have congruence-subgroups $\Gamma_1$ with $b_1(\Gamma_1)>0$ (Kazhdan). However, for all congruence-subgroups $\Gamma_1$ in type II arithmetic lattice, $b_1(\Gamma_1)=b_3(\Gamma_1)=0$. So all odd Betti numbers vanish. The question of positivity of 1st Betti number of finite-index non-congruence subgroups is wide-open. The non-arithmetic case is mostly open. (There is one nonvanishing result, due to Yeung.) $X={\mathbb H}^2\times {\mathbb H}^2$. The case of reducible lattices is (essentially) clear, so I will consider irreducible lattices. For all such lattices $\Gamma$, $b_1(\Gamma)=b_3(\Gamma)=0$. (This follows from Margulis' super-rigidity, but might have been known earlier.) So all odd Betti numbers vanish. I will not attempt to analyze the case when $dim(X)\ge 6$. One gets vanishing results for $b_1$, but the situation with higher odd Betti numbers are very much unclear, unless you know something about odd Betti numbers of the dual compact metric space: Nonvanishing of some of those implies nonvanishing of $b_{odd}(\Gamma)$. There are vanishing theorems going back to Matsushima, but they do not deal with all odd Betti numbers.<|endoftext|> TITLE: Points on hyperelliptic curves: $y^2=5(x^2-3)(x^2+2)(x^2-11/5)$ QUESTION [6 upvotes]: González-Jiménez and Xarles studied a problem in Diophantine number theory and they obtained several nice results via elliptic curve Chabauty's method over quadratic number fields. At page 73 in paper there are six 5-tuples mentioned where they could not determine the set of rational solutions. Trying to do something over degree 4 composite fields I was able to handle some of the remaining cases, but not the tuple $(0,2,5,7,11).$ One possible genus 2 curve to attack this case is $y^2=5x^6 - 16x^4 - 19x^2 + 66,$ so it is bielliptic the rank of the Jacobian is 2 and there are 8 torsion points. The degree six polynomial is reducible: $5(x^2-3)(x^2+2)(x^2-11/5).$ That is there are some degree 4 number fields suggested by the factors. I could not make use of those 3 fields. Is there any argument to choose a quadratic number field, let say NumberField($x^2\pm d$) and one of the 3 suggested fields by the factorization and work in the composite quartic fields? REPLY [8 votes]: You can apply the so-called Elliptic Chabauty over the biquadratic field $K:=\mathbb{Q}(\sqrt{3},\sqrt{11/5})$ (also equal to the field adjoining a root of $ 25x^4 - 260x^2 + 16$). Over this field there are two possible 2-coverings (one corresponding to the points with coordinate $x=1$, the other with coordinate $x=-1$). Both curves have a map to a genus one curve given by an equation $$ \delta_{\pm} y^2=(x-\sqrt{3})(x-\sqrt{11/5})(x^2-2)$$ with $$\delta_{\pm}=-(\pm 1-\sqrt{3})(\pm 1-\sqrt{11/5})$$ (one for every sign). Both cases we get an elliptic curve with rank 2, so we can apply Elliptic Chabauty MAGMA function, which answers that the only points with rational $x$-coordinate are the ones with $x=\pm 1$.<|endoftext|> TITLE: Encounters with partitions of unity QUESTION [7 upvotes]: Not sure how this would be received here. This question is about smooth partitions of unity. Let $M$ be a manifold. Consider an open cover $\{U_\alpha\}_{\alpha\in \Lambda}$ of $M$. A collection of smooth functions$\{p_\alpha:U_\alpha\rightarrow \mathbb{R}\}_{\alpha\in \Lambda}$ is called a smooth partition of unity subordinate to the cover $\{U_\alpha\}$ if $\text{supp}(p_\alpha)\subseteq U_\alpha$ for each $\alpha\in \Lambda$, the collection of supports $\{\text{supp}(p_\alpha)\}_{\alpha\in \Lambda}$ is a locally finite set, $\sum_{\alpha\in \Lambda}p_\alpha=1$. It is known that given any open cover $\{U_\alpha\}$, we can produce a partition of unity on $M$ subordiante to this cover. Next question is, what is the use of partitions? Suppose I am given a $n$-form $\omega$ on a (oriented) $n$-manifold $M$. I have to make sense of (give a reasonable definition for) $\int_M\omega$. Suppose that $\omega$ is (compactly supported) zero outside a chart $(U,\phi:U\rightarrow \mathbb{R}^n)$ (let $\psi:\phi(U)\subseteq \mathbb{R}^n\rightarrow U$ be its inverse). Then, pullback $\omega$ along $\psi$ to get $n$-form $\psi^*\omega$ on $\phi(U)\subseteq \mathbb{R}^n$ and we know how to integrate an $n$-form on an open subset of $\mathbb{R}^n$. So, we know what is $\int_{\phi(U)}\psi^*\omega$ is and define $\int_M\omega:=\int_{\phi(U)}\psi^*\omega$. Suppose $\omega$ is arbitrary (compactly supported), then, ask for partition of unity $\{p_\alpha\}$ and consider $p_\alpha\omega$. This is compactly supported in $U_\alpha$. So, $\int_{U_\alpha}p_\alpha\omega$ make sense and define $\int_{M}\omega=\sum \int_{U_\alpha}p_\alpha\omega$. Given a manifold $M$, how do I know that there exists a Riemannian metric on $M$? Partition of Unity. Given a manifold $M$, how do I know there exists a connection on the tangent bundle $TM\rightarrow M$? Partitions of unity. Given a principal bundle over manifold $M$, how do I know connection exists on the principal bundle $P\rightarrow M$? Partitions of unity (along with trivialization of course). My question is the following: Is partition of unity used for anything serious than making sure some structures can be glued to give a global structure? Do you, as a research mathematician, come across the necessity of using the partition of unity for any reason other than similar to what I mentioned above? REPLY [5 votes]: Partition of unity allows one to prove that the sheaf of smooth functions is acyclic - which is needed in the construction of de Rham cohomology. As such, it is also completely analogous to the (weak) Hilbert's Nullstelensatz in algebraic geometry, which allows to break 1 into sum of polynomials vanishing in some closed subsets provided the intersection of these subsets is 0.<|endoftext|> TITLE: Higher-dimensional version of the "Magic Cube Lemma" for homotopy pushouts/pullbacks QUESTION [8 upvotes]: The "Magic Cube Lemma" is a surprising (to me) relationship between (homotopy) pushouts and (homotopy) pullbacks of spaces: Consider a cubical diagram $I^3\to \mathcal{S}$ in the $\infty$-category of spaces/homotopy types (where $I=\Delta^1=\mathrm{N}\{0\to 1\}$): $\require{AMScd}$ \begin{CD} A @>>> @>>> B @. \\ @VVV @. @VVV @.\\ @. A' @>>> @>>> B'\\ @VVV @VVV @VVV @VVV\\ C @>>> @>>> D @. \\ @. @VVV @. @VVV\\ @. C' @>>> @>>> D'\\ \end{CD} where there are supposed to be arrows connecting the back layer ($A,B,C,D$) to the front layer ($A',B',C',D'$). (Maybe a kind soul can explain how I draw actual cubical diagrams with AMScd) Assume the following: The back and front layers are pushout squares The top layer ($A,B,A',B'$) and the left layer ($A,C,A',C'$) are pullback squares Then the lemma asserts that also the right layer ($B,D,B',D'$) and the bottom layer ($C,D,C',D'$) are pullback squares. A different way of stating the same lemma is as follows: Let $I^2\to \mathrm{Fun}(I,\mathcal S)$ be a square of arrows in spaces. \begin{CD} a @>f>> b\\ @VgVV (*) @VVg'V\\ c @>>f'> d\\ \end{CD} where $a\colon A\to A'$, etc. From this perspective, $f$,$f'$,$g$ and $g'$ are natural transformations of diagrams $I\to\mathcal S$. Then the lemma can be restated as follows: Assume that the square $(\ast)$ is a pushout and that the natural transformations $f$ and $g$ are Cartesian. Then also $f'$ and $g'$ are Cartesian. More generally, the following also holds (essentially by the fact that every colimit can be built from coproducts and pushouts): Let $U\colon K^\triangleright\to\mathrm{Fun}(I,\mathcal S)$ be a colimit diagram of arrows. If for every edge $e\colon I\to K$ the natural transformation $U\circ e\colon I\to\mathrm{Fun}(I,\mathcal S)$ is Cartesian, then the same is true for every edge $e\colon I\to K^\triangleright$. (Here $K^\triangleright = K\star [0]$ denotes the right cone on the category/simplicial set $K$) I am interested in the following higher dimensional version: Call a natural transformation $f\colon a\to b$ between cubical diagrams $a,b\colon I^n\to\mathcal S$ relatively Cartesian, if it is a Cartesian (i.e.\ a limit diagram) when viewed as a diagram $I^{n+1}=I\times I^n\to\mathcal S$. (Equivalently: $f$ is a $p$-Cartesian edge for the canonical Cartesian fibration $p\colon\mathrm{Fun}(I^n,\mathcal S)\to \mathrm{Fun}(I^n_\star,\mathcal S)$, where $I^n_\star$ denotes the punctured cube obtained by removing the initial vertex $(0,\dots,0)$). Let \begin{CD} a @>f>> b\\ @VgVV (**) @VVg'V\\ c @>>f'> d\\ \end{CD} be a square $I^2\to \mathrm{Fun}(I^n,\mathcal S)$ of $n$-dimensional cubical diagrams. Is it true that: if $(\ast\ast)$ is a pushout, and $f$ and $g$ are relatively Cartesian, then $f'$ and $g'$ are also relatively Cartesian? Or more generally: Given a colimit diagram $K^\triangleright\to \mathrm{Fun}(I^n,\mathcal{S})$ such that every edge in $K$ is sent to a relatively Cartesian transformation, is the same true for all edges of $K^\triangleright$? If it is true: can one deduce it formally from the original magic cube lemma (I tried, but failed) or does one maybe need an additional input about the $\infty$-category $\mathcal S$? If it holds for $\mathcal S$, does it also hold in any $\infty$-topos? REPLY [4 votes]: I don't think so. Take $n=2$, and consider a map $f\colon b\to a$ between objects of $\mathrm{Fun}(I^2, \mathcal{S})$. If $a$ and $b$ are pullback squares, then any map $f$ between them is relatively Cartesian in your sense. Also, if $a$ is a pullback and $f$ is relatively cartesian, then $b$ must also be a pullback. (Any commutative square $a$ has a collection of "total fibers", which are the fibers of the map $a(0)\to a(1)\times_{a(12)} a(2)$ over points in the target. The total fibers are all contractible iff $a$ is a pullback. A map $b\to a$ is relatively Cartesian iff it induces equivalences on all total fibers.) So let $a$ be the pullback square which displays $\Omega X$ as a pullback of $*\to X \leftarrow *$, and let $b$ and $c$ be pullback squares of contractible spaces. Then the pushout $d$ is a square with $\Sigma \Omega X$ at the initial corner, $\Sigma X$ at the terminal corner, and $*$ at the other two spots. For $d\to c$ to be relatively Cartesian, since $c$ is certainly a pullback we would have to have $d$ be a pullback to, i.e., we would have $\Sigma\Omega X\xrightarrow{\sim} \Omega\Sigma X$, which almost never happens.<|endoftext|> TITLE: Torsion in the cohomology of Fano varieties of lines QUESTION [10 upvotes]: Let $\mathrm{X}$ be a cubic $d$-fold, and $\mathrm{F}(\mathrm{X})$ its Fano variety of lines. Is the integral cohomology of $\mathrm{F}(\mathrm{X})$ torsion-free? For $d=3$ A. Collino (`The fundamental group of the Fano surface, I') proves that there exists an exact sequence $$[\pi_1,\pi_1]\rightarrow\pi_1\rightarrow\mathbf{Z}^{\oplus 10}\rightarrow 0$$ where $\pi_1=\pi_1(\mathrm{F}(\mathrm{X}))$ (Prop. 2.3.4 in the paper). Hence $\mathrm{H}_1(\mathrm{F}(\mathrm{X});\mathbf{Z})=\pi_1^{\mathrm{ab}}$ is torsion-free, which is enough. For $d=4$ Beauville and Donagi show that $\mathrm{F}(\mathrm{X})$ is deformation equivalent to $\mathrm{Hilb}^2(\mathrm{K3})$, the integral cohomology of which is torsion-free (for example by the general results of this paper of Totaro). So the only remaining case is $d>4$, where $\mathrm{F}(\mathrm{X})$ is a Fano variety in the usual sense. Higher-dimensional Fano varieties can have torsion in their integral cohomology, see for example here. REPLY [7 votes]: Here is an expanded version of my comments. Let's work over the complex numbers which I suppose is assumed in the question. Let $K_0(Var)$ be the Grothendieck ring of varieties (see e.g. https://arxiv.org/abs/1405.5154). Furthemore, let $K_0(HS)$ be the Grothendieck group of integral polarizable Hodge structures with respect to direct sum operation, and let $K_0(FinAb)$ be the Grothendieck group of finite abelian groups also with respect to direct sum operation. We have a composition of group homomorphisms $$ K_0(Var) \to K_0(HS) \to K_0(FinAb) $$ where the first map takes a class of a smooth projective variety to its total Hodge structure, and the second map takes a Hodge structure to its torsion part (forgetting the weights). Section 3 in Efimov's paper https://arxiv.org/pdf/1707.08997.pdf gives the details for the definition of the first map, from Bittner's blow up presentation of the Grothendieck ring of varieties. Second map is straightforward. Existence of these maps implies that classes of smooth projective varieties control the torsion in cohomology groups. Now let $X$ be a smooth cubic of dimension $n$. According to https://arxiv.org/abs/1405.5154, Thm. 5.1, the class of the Fano variety $F(X)$ in the Grothendieck ring is expressed as a combination of $Hilb_2(X)$ and $\mathbb{P}^n \times X$ (with a coefficient $\mathbb{L}^2$ that shifts cohomology groups). Explicitly in $K_0(FinAb)$ we have the following: $$ [H^*(F(X), \mathbb{Z})_{tors}] + (n+1) [H^*(X, \mathbb{Z})_{tors}] = [H^*(Hilb_2(X), \mathbb{Z})_{tors}] $$ In fact, since the category of finite abelian groups has unique decompositions into indecomposables, equality of classes implies isomorphism hence $$ H^*(F(X), \mathbb{Z})_{tors} \oplus H^*(X, \mathbb{Z})_{tors}^{\oplus(n+1)} \simeq H^*(Hilb_2(X), \mathbb{Z})_{tors} $$ Thus it suffices to show that $Hilb_2(X)$ has torsion-free cohomology groups. All hypersurfaces are torsion-free, and by Totaro's result https://arxiv.org/pdf/1506.00968.pdf, Thm 2.2, same is true for $Hilb_2(X)$.<|endoftext|> TITLE: Does the Cantor-Schröder-Bernstein Theorem hold in the category opposite to the category of noetherian commutative rings? QUESTION [12 upvotes]: I asked this question on Mathematics Stackexchange, but got no answer. Let $A$ and $B$ be noetherian commutative rings with one, and let $f:A\to B$ and $g:B\to A$ be epimorphisms. Are the rings $A$ and $B$ necessarily isomorphic? [In this post "ring" means "commutative ring with one", and morphisms are required to map $1$ to $1$. By definition, a morphism of rings $f:A\to B$ is an epimorphism if for all pairs of morphisms $(g,h):B\rightrightarrows C$ the equality $g\circ f=h\circ f$ implies $g=h$. Surjective morphisms are epimorphic, but the converse does not always hold: for instance the inclusion $\mathbb Z\to\mathbb Q$ is an epimorphism.] The busy reader is invited to skip the sequel. For more details about epimorphisms see $\bullet$ MathOverflow thread What do epimorphisms of (commutative) rings look like?. $\bullet$ Stacks Project Section Epimorphisms of rings. $\bullet$ Samuel Seminar . See in particular Section 2 of Exposé Number 7 by Daniel Ferrand. The answers to the following variants of the above question are known: (1) If $f:A\to B$ and $g:B\to A$ are injective morphisms of noetherian rings, are $A$ and $B$ necessarily isomorphic? The answer is No, as shown by the following example taken from a comment of Sam Lichtenstein to this question. Let $K$ be a field, $x$ an indeterminate, $f:K[x^2,x^3]\to K[x]$ the inclusion, and $g:K[x]\to K[x^2,x^3]$ the (clearly injective) morphism defined by $g(p(x))=p(x^2)$. Note that $K[x^2,x^3]$ is not isomorphic to $K[x]$ because the ideal $(x^2,x^3)$ of $K[x^2,x^3]$ is not principal. (2) If $f:A\to B$ and $g:B\to A$ are surjective morphisms of rings, are $A$ and $B$ necessarily isomorphic? The answer is No, as shown by the following example taken from the same comment of Sam Lichtenstein. Set $$ A:=\mathbb Z/(4)\times\mathbb Z/(4)\times\cdots,\quad B:=\mathbb Z/(2)\times A, $$ let $f:A\to B$ be defined by $f(x_1,x_2,\dots)=(h(x_1),x_2,\dots)$, where $h$ is the unique ring morphism from $\mathbb Z/(4)$ to $\mathbb Z/(2)$, and let $g:B\to A$ be defined by $g(x_1,x_2,\dots)=(x_2,x_3,\dots)$. The rings $A$ and $B$ are not isomorphic because the equations $2x=0$ and $x^2=x$ have no nonzero simultaneous solutions in $A$, and one such solution in $B$ (namely $x=(1,0,\dots)$). (3) If $f:A\to B$ and $g:B\to A$ are surjective morphisms of noetherian rings, are $A$ and $B$ isomorphic? The answer is Yes, because surjective endomorphisms of noetherian rings are isomorphisms. But epimorphic endomorphisms of noetherian rings are not always isomorphisms: see this answer of Eric Wofsey. REPLY [17 votes]: No because you can take $A = \mathbf{Z}[x, 1/(x - n); n\geq 0]$ and $B = \mathbf{Z}[x, 1/x, 1/(x - n); n \geq 2]$ and the maps are $B \to A$ is the inclusion and $A \to B$ sends $x$ to $x - 2$. The reason $A$ is not isomorphic to $B$ is that the gaps between the ``missing points'' are different for $A$ and $B$. More precisely, any isomorphism $A \to B$ sends $x$ to something of the form $(a x + b)/(cx + d)$ with $a, b, c, d \in \mathbf{Q}$ and there is no such function which sends the set $\{0, 2, 3, \ldots\} \cup \{\infty\}$ bijectively to $\{0, 1, 2, 3, \ldots\} \cup \{\infty\}$.<|endoftext|> TITLE: Can we choose smoothly the singular vectors of a matrix? QUESTION [6 upvotes]: $\newcommand{\GLm}{\text{GL}_n^-}$Let $A$ be a real $n \times n$ matrix with non-positive determinant. Suppose that the smallest singular value of $A$ is strictly smaller than all the others (it has multiplicity $1$). Question: Do there exist an open neighbourhood $O$ of $A$, and smooth maps $U:O \to \operatorname{SO}_n$, $V:O \to \operatorname{O}_n^-$ such that $$ X=U(X)\Sigma(X)V(X)^T$$ holds for every $X \in O$, where $\Sigma(X) = \operatorname{diag}\left( \sigma_1(X),\dots\sigma_n(X) \right)$, and $\sigma_1(X)$ is the smallest singular value of $X$? Note that I don't care about the ordering of $\sigma_2,\dotsc,\sigma_n$, but I specifically want the minimal singular value to be in a fixed position. Comment 1: If we replace the requirement that $\sigma_1 $ has multiplicity $1$ by the requirement that all the singular values are distinct, then the answer is positive: In that case the map \begin{align*} \mu: \operatorname{SO}_n\times \mathcal D\times \operatorname{O}_n^-\to Y \\ (U,\Sigma,V)\mapsto U\Sigma V^T \end{align*} is a local diffeomorphism, so is locally invertible (here $\mathcal D$ is the space of $n\times n$ diagonal matrices with strictly increasing positive entries, and $Y=\{A \,|\, \text{$\det A<0$ and all the singular values of $A$ are distinct}\}$). This works when $\det A <0$. When $\det A=0$ (and all its singular values are distinct) a slight adaptation of this argument works. Comment 2: If such maps $U$, $V$ exist, then the map $A \to \Sigma(A)$ is also smooth. In general the ordered singular values cannot be chosen smoothly when they "cross", but here I don't require to keep the ordering of $\sigma_2,\dotsc,\sigma_n$ fixed, so I think there is no obstruction, but I may be wrong. REPLY [6 votes]: Consider the set of matrices $$ \begin{pmatrix} -1&0&0\\ 0&2-a&b\\ 0&b&2+a \end{pmatrix} $$ For $a=0$ and $b$ small and positive, the singular vectors are $(1,0,0)$, $(0,1,1)/\sqrt 2$, $(0,1,-1)/\sqrt 2$. For $a>0$ small and $b=0$, the singular vectors are the coordinate directions. Hence the singular vectors fail to even be continuous in a neighbourhood of diag(-1,2,2).<|endoftext|> TITLE: Characterising natural transformations between tensor functors QUESTION [9 upvotes]: I would like to know if the following conjecture is correct and if so what's a good citation for its proof. Let $\mathsf{E}$ be the category of euclidean vector spaces, i.e. objects are finite-dimensional real vector spaces endowed with a scalar product and morphisms are isometries. The tensor powers $(-)^{\otimes a}$ are functors $\mathsf{E}\to\mathsf{Vect}$. There are some obvious natural transformations between these functors. For example taking scalar products of any two tensor factors gives $\binom{a}{2}$ many natural transformations $(-)^{\otimes a} \to (-)^{\otimes(a-2)}$. Permuting tensor functors gives $a!$ many natural transformations $(-)^{\otimes a}\to(-)^{\otimes a}$. Of course linear combinations and compositions of these are also natural transformations. The conjecture is that these are essentially all there is to have. More precisely I think that the following statement is true: Let $a,b\in\mathbb{N}$ be fixed. The only natural transformations $(-)^{\otimes a} \to (-)^{\otimes b}$ are either zero if $b-a \notin 2\mathbb{N}$ or compositions of an element of $\mathbb{R}[S_a]$ (acting by permuting the tensor factors) and $\frac{b-a}{2}$ many scalar products if $b-a\in 2\mathbb{N}$ Considering Qiaochu Yuan's comment, a second question is also interesting to ask: If one considers $Iso(\mathsf{E})$ (i.e. the category with only bijective isometries as morphisms) instead of $\mathsf{E}$, how can we characterise the natural transformations between the analogous functors then? The Casimir element $\Omega_V:=\sum_i b_i\otimes b_i$ is independent of the choice of the basis $b_1,\ldots,b_n$ of $V$ and therefore gives rise to different natural transformations like $\mathbb{R}\to V^{\otimes 2}, 1\mapsto\Omega_V$. I'd guess that that's essentially it and every natural transformation is composed of $\mathbb{R}[S_a]$, some number of traces and some number of insertions of $\Omega$. REPLY [7 votes]: The second question has a well known, affirmative answer from invariant theory. Note that $O(V)$ acts on every tensor power $V^{\otimes k}$ and by conjugation also on $Hom(V^{\otimes a},V^{\otimes b})$ and the homomorphisms commuting with $O(V)$ are exactly the fixed points of this conjugation action. We have a natural isomorphism $$Hom(V^{\otimes a},V^{\otimes b}) \cong (V^{\otimes a})^\ast \otimes V^{\otimes b} \cong (V^\ast)^{\otimes a}\otimes V^{\otimes b}$$ The scalar product gives us an isomorphism $V\cong V^\ast$ which translates this into $$Hom(V^{\otimes a},V^{\otimes b}) \cong V^{\otimes a}\otimes V^{\otimes b} = V^{\otimes(a+b)}$$ Explicitly this last one is given by $$x_1\otimes\cdots\otimes x_a \otimes y_1\otimes \cdots \otimes y_b \mapsto \left(v_1\otimes\cdots\otimes v_a \mapsto \prod_{i=1}^a \langle x_i,v_i\rangle y_1\otimes\cdots\otimes y_b\right)$$ All of these are natural w.r.t. isometries so that our problem translates into finding fixed points on $V^{\otimes k}$ for every $k\in\mathbb{N}$. Since $v\mapsto -v$ is an isometry, there are no non-zero fixed points for uneven $k$. So we are left with the case where $a+b$ is even. For the special case $a=b$, we want the centraliser of $\rho(O(V))$ inside $End(V^{\otimes a})$ where $\rho$ is the representation $O(V)\to End(V^{\otimes a}), \phi\mapsto\phi^{\otimes a}$. This centraliser turns out to be well-known: It is $\psi(B_a(d))$ where $d:=\dim(V)$, $B_a(d)$ is the so-called Brauer algebra and $\psi: B_a(d) \to End(V^{\otimes a})$ is its natural representation. This is the analogue of Schur-Weyl-duality for orthogonal groups. The Brauer algebra has a basis indexed by perfect pairings $D$ of the numbers $1,2,\ldots,2a$ each of which describes a combination of permutations (when a number in $\{1,\ldots,a\}$ is paired with a number in $\{a+1,\ldots,2a\}$) and the map $v_1\otimes v_2 \mapsto \langle v_1,v_2\rangle \Omega$, where $\Omega:=\sum_{i=1}^{\dim(V)} e_i\otimes e_i$ is the Casimir element. If we follow the isomorphisms above, we find that these basis elements give us the following spanning set of fixed points in $(V^{\otimes 2a})^{O(V)}$: $$\Omega_D := \sum_{\substack{1\leq i_1,\ldots,i_{2a}\leq \dim(V) \\ \lbrace x,y\rbrace\in D \implies i_x=i_y}} e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_{2a}}$$ where $D$ runs over all perfect pairings of the numbers $1,2,\ldots,2a$. Now the fixed points do not care that arrived at the number $2a$ by considering $a=b$. Therefore this is also the answer for the general case where $a+b$ is any even number. If we follow all the isomorphisms backward, we find that these fixed points $\Omega_D\in V^{\otimes(a+b)}$ represent the maps $\tau^D: V^{\otimes a}\to V^{\otimes b}$ that come from permuting some tensor factors (=numbers in $\{1,\ldots,a\}$ get paired with numbers in $\{a+1,\dots,a+b\}$), contracting some (=two numbers in $\{1,\ldots,a\}$ get paired with each other), and introducing Casimir elements in some factors of the result (=two numbers in $\{a+1,\ldots,a+b\}$ get paired with each other) which is what we wanted to prove. To prove the original conjecture, we have to find all the linear combinations $$\tau = \sum_D \alpha_D \tau^D$$ which are natural not only w.r.t to bijective isometries but w.r.t. all isometries. Our theorem is proven if we can prove that $$D\text{ contains a pairing }\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\} \implies \alpha_D=0$$ Because then all introductions of Casimir elements are eliminated and only traces and reordering of tensor factors remain. In particular it would follow that $a\geq b$ must hold if $\tau\neq 0$. That together with the earlier parity argument would prove our theorem. Step 1.: Consider a vector space $V$ of dimension $d$ with an orthonormal basis $e_1,\ldots,e_d$ and define for any assignment of indices $j: \{1,\ldots,a\} \to \{1,\ldots,d\}$ the tensor $v_j\in V^{\otimes a}$ as $$v_j := e_{j(1)} \otimes e_{j(2)} \otimes \cdots \otimes e_{j(a)}$$ If we have two such assignments $j: \{1,\ldots,a\} \to \{1,\ldots,d\}$ and $k: \{1,\ldots,b\} \to \{1,\ldots,d\}$, then we can compute the scalar product $\langle{\tau_V(v_j),v_k}\rangle$ as follows: \begin{align*} \langle{\tau_V(v_j),v_k}\rangle &= \sum_D \alpha_D \langle{\tau_V^D(v_j),v_k}\rangle \\ &= \sum_{D,i} \alpha_D \prod_{s=1}^a \langle{e_{i(s)},e_{j(s)}}\rangle \langle{e_{i(a+1)}\otimes\cdots\otimes e_{i(a+b)}, e_{k(1)}\otimes\cdots\otimes e_{k(b)}} \rangle \\ &= \sum_{D,i} \alpha_D \prod_{s=1}^a \langle{e_{i(s)},e_{j(s)}}\rangle \prod_{t=1}^b \langle{e_{i(a+t)}, e_{k(t)}}\rangle \end{align*} where - as above - $i$ runs over all indexing functions $i:\{1,2,\ldots,a+b\}\to\{1,\ldots,d\}$ such that $$\{x,y\}\in D \implies i(x)=i(y).\tag{*}$$ Step 2: We prove the following lemma If $d\geq \frac{a+b}{2}$, then for every perfect pairing $D$, there are assignments $j: \{1,\ldots,a\} \to \{1,\ldots,d\}$ and $k: \{1,\ldots,b\} \to \{1,\ldots,d\}$ such that $$\langle{\tau_V(v_j),v_k}\rangle = \alpha_D$$ If moreover $D$ contains a pair $\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\}$, then it is possible to find such $j$ and $k$ with $\forall 1\leq x\leq a: j(x) \leq m-1$. Proof of the Lemma: To see this, note that the factors in both products in the above expression for the scalar product are either one or zero so that if the whole summand is non-zero, then $\forall s: i(s) = j(s)$ and $\forall t: i(a+t) = k(t)$ must hold, i.e. there can only be one $i$: The one that agrees with $j$ on $\{1,\ldots,a\}$ and with $k$ on $\{a+1,\ldots,a+b\}$. Therefore most summands in the expression are zero anyway and can be ignored. The sum reduces to $$\langle{\tau_V(v_j),v_k}\rangle = \sum_{D} \alpha_D \prod_{s=1}^a \langle{e_{i(s)},e_{j(s)}}\rangle \prod_{t=1}^b \langle{e_{i(a+t)}, e_{k(t)}}\rangle$$ where $i$ is the one function that agrees with $j$ in the lower and with $k$ in the upper part of $\{1,2,\ldots,a,a+1,\ldots,a+b\}$. Additionally, there are only a limited number of pairings $D$ that are compatible with this indexing function $i$, because $i$ also has to satisfy the condition $(*)$. Now let our given pairing $D$ be $\Big\{\{x_1,y_1\}, \{x_2,y_2\}, \ldots, \{x_m,y_m\}\Big\}$. Then if we choose $j$ and $k$ such that $i(x_1)=i(y_1) = 1$, $i(x_2)=i(y_2)=2$, ..., $i(x_m)=i(y_m)=m$, then there is only a single pairing that satisfies $(\ast)$ for this indexing function $i$, namely the pairing $D$ we started with. All summands other than the $D$-th are therefore zero and the whole sum simplifies to $$\langle{\tau_V(v_j),v_k}\rangle = \alpha_D \prod_{s=1}^a \langle{e_{i(s)},e_{j(s)}}\rangle \prod_{t=1}^b \langle{e_{i(a+t)}, e_{k(t)}}\rangle = \alpha_D\cdot 1$$ We have complete freedom which numbering of the pairs in $D$ we choose to define $i$. If $\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\}$ is one of the pairs in $D$, we can choose our ordering such that $x_m=\alpha$ and $y_m=\beta$. Therefore the numbers $1,2,\ldots,a$ get mapped to something other than $m$ by $i$. That proves the lemma. Now back to the proof of the theorem. Step 3. Let's focus on a specific example: Choose $V$ to be large enough that $d\geq \frac{a+b}{2}$ and set $U:=span(e_1,\ldots,e_{m-1})$. The embedding $\phi: U\to V, v\mapsto v$ is an isometry so that our natural transformation $\tau$ must satisfy $$\forall v\in U^{\otimes a}: \tau_U(v) = \tau_V(v)$$ Now in $\tau_U(v)$ all the sums only have indices running up to $\dim(U)=m-1$ whereas in $\tau_V(v)$ the same indices run up to $d$. On the right hand side however some summands containing $e_m,e_{m+1},\ldots,e_d$ may appear. Consider the linear combination $\tau = \sum_D \alpha_D \tau^D$ we started with and let $D$ be one of the perfect pairings which contain a pair $\{\alpha,\beta\} \subseteq \{a+1,\ldots,a+b\}$. By the lemma just proven, there is an indexing function $i$ and its two parts $j$ and $k$ such that $j$ takes only values inside $\{1,\ldots,m-1\}$ and $\langle{\tau_V(v_j),v_k}\rangle = \alpha_D$. Moreover, because $j$ takes values in $\{1,\ldots,m-1\}$, $v_j$ must be in $U^{\otimes a}$ and $v_k$ contains $e_m,e_{m+1},\ldots,e_d \in U^\perp$ in the $\alpha$-th (and $\beta$-th too) tensor factor. We conclude $$\alpha_D = \langle{\tau_V(v_j),v_k}\rangle = \langle{\tau_U(v_j),v_k}\rangle = 0$$ and that proves the theorem.<|endoftext|> TITLE: Why is the inertia stack of a smooth Deligne-Mumford stacks called inertia? QUESTION [6 upvotes]: Let $\mathcal{X}$ be a smooth Deligne-Mumford stack. Then there is an associated stack $I\mathcal{X}$, called the inertia stack of $\mathcal{X}$. Why is the inertia stack called "inertia"? We can also assign a rational number to each object $(x,g)$ of $I\mathcal{X}$, called $\text{age}(x,g)$.Then we can shift the usual grading on $H^*(I\mathcal{X})$ by age to obtain a so-called age grading on the Chen-Ruan orbifold cohomology $H^*_{\text{orb}}(\mathcal{X})$. As discussed here, the age grading is motivated by orbifold GW theory, which is closely related to orbifold string theory. So another related question is: why is the age grading called "age"? REPLY [7 votes]: I think of the word "inertia" in "inertia stack" as representing the same idea as the "inertia" in "inertia group" (which presumably came first). This latter group typically comes up when one has a ramified Galois cover $X\rightarrow Y$ (say, of algebraic varieties over an algebraically closed field $k$ of characteristic 0). If the cover has Galois group $G$, then $Y = X/G$, and the ramification points are precisely those points $x\in X$ with nontrivial stabilizer in $G$. Such a stabilizer $G_x := \text{Stab}_G(x)$ is called the inertia group at $x$. (Of course such stabilizers have meaning for any group $G$ acting on a thing $X$, so we may as well define the inertia groups of a group action as the stabilizer subgroups.) As mentioned above, a Galois cover $X\rightarrow Y$ identifies $Y$ with the scheme quotient $X/G$. What happens when we take the stacky quotient instead? Well in this case the map $X\rightarrow X/G$ will factor through the stacky quotient $X\rightarrow [X/G]\rightarrow X/G$, and the second map identifies $X/G$ with the coarse scheme of $[X/G]$. Now we may consider the inertia stack of $[X/G]$. By definition it is a stack $I$ over $[X/G]$ whose fiber over a geometric point $\text{Spec }k\rightarrow [X/G]$ is the automorphism group of that point. In our case, if we let $$\pi : X\rightarrow [X/G]$$ be the quotient map, and if $x\in X$ is a geometric point, then the fiber $I_{\pi(x)}$ is precisely the inertia group $G_x$ of $x$, which to me seems to justify its name. In general, I believe the following is a correct statement: If a stack $\mathcal{X}$ is locally a quotient of a scheme by a faithful group action, then the fiber of the inertia stack of $\mathcal{X}$ above a geometric point $x$ of $\mathcal{X}$ is precisely the inertia group of a local presentation of $\mathcal{X}$ at (a point above) $x$. Here's my favorite "real world" example. Consider the covering $\mathcal{H}\rightarrow\mathcal{H}/SL(2,\mathbb{Z})$, where $\mathcal{H}$ is the upper half plane. The stacky quotient $[\mathcal{H}/SL(2,\mathbb{Z})]$ is naturally isomorphic to the moduli stack of elliptic curves, where to a point $\tau\in\mathcal{H}$, we associate the elliptic curve $E_\tau := \mathbb{C}/\langle 1,\tau\rangle$. It's a simple calculation that for $\gamma\in SL(2,\mathbb{Z})$ and $\tau'\in\mathcal{H}$, the set of isomorphisms $Isom(E_\tau,E_{\tau'})$ is precisely the set of $\delta\in SL(2,\mathbb{Z})$ satisfying $\delta\tau = \tau'$. From this we see explicitly how the fibers of the inertia stack give precisely the inertia groups of the (nonfaithful) action of $SL(2,\mathbb{Z})$ on $\mathcal{H}$, and their images in $PSL(2,\mathbb{Z})$ give precisely the inertia groups of the Galois covering $\mathcal{H}\rightarrow\mathcal{H}/SL(2,\mathbb{Z})$.<|endoftext|> TITLE: Best constant approximation in $L^p(\Omega)$ QUESTION [8 upvotes]: For $\Omega$ a bounded open set of $\mathbf{R}^d$ and $f\in L^p(\Omega)$ the infimum \begin{align*} \inf_{C\in\mathbf{R}} \|f-C\|_p \end{align*} is reached (by compactness). For $1C_p(f)} |f-C_p(f)|^{p-1} = \int_{f TITLE: Frégier and Frégier's Theorem QUESTION [5 upvotes]: A curious and interesting gem is Frégier's theorem, quoted here from David Wells: Choose any point $P$ on a conic, and make it the vertex of a right angle which rotates about $P$. Then the lines through the points of intersection, $AA$, $BB$, and so on, will all pass through a fixed point $X$ which lies on the normal at $P$, that is, on the line through $P$ perpendicular to the tangent at $P$. Who was Frégier, and where can I find the earliest publication of his theorem? REPLY [2 votes]: Frégier appears to have published the result in M. Frégier, Géométrie analitique. Théorèmes nouveaux sur les lignes et surfaces du second ordre. Annales de Mathématiques pures et appliquées, tome 6 (1815-1816), p. 229-241 Here's a PDF. and what looks like the relevant theorem The following article helped me track this down, via some citations in the introduction: LIPPS, KATHERINE SUE. “Envelopes of Certain Families of Conics.” Pi Mu Epsilon Journal, vol. 2, no. 8, 1958, pp. 359–363. JSTOR, www.jstor.org/stable/24344939. Accessed 27 June 2020. For anybody who wants to access Gergonne's Annales, there is an index at http://www.numdam.org/item/AMPA/<|endoftext|> TITLE: Is Van der Waerden's conjecture really due to Van der Waerden? QUESTION [19 upvotes]: Van der Waerden's conjecture (now a theorem of Egorychev and Falikman) states that the permanent of a doubly stochastic matrix is at least $n!/n^n$. The Wikipedia article, as well as many other sources, provides the following citation: B. L. van der Waerden, "Aufgabe 45", Jber. Deutsch. Math.-Verein., 35 (1926), 117. However, I just went to the library to look up volume 35 of Jahresbericht der Deutschen Mathematiker-Vereinigung, and page 117 lies right smack in the middle of a paper by Karl Menger, "Bericht über die Dimensionstheorie." There is nothing remotely resembling Van der Waerden's conjecture on page 117. In the book A Course in Combinatorics by J. H. van Lint and R. M. Wilson, there is a chapter on Van der Waerden's conjecture, and there is a footnote saying that Van der Waerden told one of the authors that he had never made any such conjecture. So maybe Van der Waerden never made the conjecture? Or did he make the conjecture and then forget about it, with the citation getting garbled somewhere along the line? The earliest citation to "Aufgabe 45" in Google Scholar seems to be from 1960, so maybe that's when the error was introduced. But I don't know how to search for earlier citations to a possibly non-existent reference by Van der Waerden. ADDENDUM (prompted by Mark Sapir's comments to Matt F.'s answer below): Here's what Van Lint and Wilson say (Chapter 12, 2nd edition): In 1926, B. L. van der Waerden proposed as a problem to determine the minimal permanent among all doubly stochastic matrices. It was natural to assume that this minimum is per $J_n = n!\,n^{-n}$. … The assertion [that this minimum is uniquely achieved by the constant matrix] became known as the ‘Van der Waerden conjecture’ (although in 1969 he told one of the present authors that he had not heard this name before and that he had made no such conjecture). So it sounds like Van der Waerden himself saw a distinction between asking for the minimum value, and conjecturing what the minimum value is (or conjecturing that the minimum value is uniquely attained by the constant matrix). REPLY [16 votes]: Aufgabe 45 is in the second Abteilung of volume 35, indeed on page 117, as here. He was asking about permanents, but rather than asking for a proof about $n!/n^n$, he asked for the determination of a minimum. Translation: "The Function $Q\cdots$ (summing over all permutations of the indices $1,\ldots,n$), under the conditions $a_{ik}\cdots$, can only take positive values (Dénes König, Graphs and their Applications #2, Math. Ann. 77, p. 47). The minimum of the function is (under the given conditions) to be determined."<|endoftext|> TITLE: Can we perturb a map $\mathbb{R}^n \to \mathbb{R}^n$ to have distinct singular values? QUESTION [8 upvotes]: $\newcommand{\SO}[1]{\text{SO}(#1)}$ $\newcommand{\dist}{\operatorname{dist}}$ Let $\mathbb{D}^n$ be the closed $n$-dimensional unit ball, and let $f:\mathbb{D}^n \to \mathbb{R}^n$ be smooth. Set $$X=\text{GL}^+_n \cup \{ A \in M_n \, | \text{ the singular values of } \, A \text{ are distinct }\}$$ Here $M_n$ is the space of real $n \times n$ matrices. Do there there exist $f_n \in C^{\infty}(\mathbb{D}^n, \mathbb{R}^n)$ such that $f_n \to f$ in $W^{1,2}(\mathbb{D}^n, \mathbb{R}^n)$ and $df_n \in X$ everywhere on $ \text{int}(\mathbb{D}^n) $? Can we at least perturb $f$ to make the points where the are recurring singular values isolated? We need to understand what happens to the zeroes of the discriminant of the characteristic polynomial of $df^Tdf$ under perturbation. REPLY [3 votes]: Here is a positive answer for $n=2.$ The argument doesn't seem to generalize easily to higher $n.$ The idea is to write $df\in X$ as $\star df_1 + df_2\neq 0$ and make use of Hodge decomposition. The set $X$ consists of the matrices not of the form $(\begin{smallmatrix}a&b\\b&-a\end{smallmatrix}),$ in the $(dx,dy)$ basis (for matrices with $\sigma_1=\sigma_2,$ the middle part of an SVD is scalar, so the matrix has to be a scalar multiple of an orthogonal matrix). So we want to find an approximating sequence $g^{(n)}$ with $$\star dg^{(n)}_1 + dg^{(n)}_2\neq 0$$ on the unit ball, with the usual Hodge star operator $\star(a\;dx + b\;dy)=-b\;dx+a\;dy.$. For the Hodge decomposition I want to replace $\mathbb D^2$ by a more convenient space. I'll use $\mathbb R^2,$ but using a non-compact manifold is not essential an the argument could use a compactification $\mathbb R^2/\Lambda\mathbb Z^2.$ We can assume $f$ extends to a function $\bar f$ in $C^\infty_c(\mathbb R^2,\mathbb R^2),$ for example using the Sobolev extension theorem plus a standard result on density of compactly-supported smooth functions in Sobolev spaces. The combination $\zeta=\star d\bar f_1+d\bar f_2$ is a compactly-supported smooth $L^2$ vector field. By a perturbation we can arrange that, on a neighborhood of $\mathbb D^2,$ $\zeta$ is non-zero except at isolated points Specifically... pick a bounded open neighborhood $U$ of $\mathbb D^2$ pick a $\psi$ in $C^\infty_c(\mathbb R^2,\mathbb R^2)$ that is strictly positive on $U$ define $\phi:U\times \mathbb R^2\to\mathbb R^2$ by $\phi(x,M)=(M-\zeta(x))/\psi(x)$ and consider a regular value $N\approx (0,0)$ for the restriction $\phi|_{U\times\{(0,0)\}}$ The preimage $\phi^{-1}(\{N\})$ is the graph $\{(x,\zeta(x)+N\psi(x))\}.$ The preimage $\phi|_{U\times\{(0,0)\}}^{-1}(\{N\})$ consists of isolated points $(x,(0,0))$ such that $\zeta(x)+N\psi(x)=(0,0).$ Projecting from the graph $\{(x,\zeta(x)+N\psi(x))\}\subset U\times \mathbb R^2$ to $U$ is a diffeo, so projecting a set of isolated points gives a set of isolated points. So the points $x\in\mathbb D^2$ with $\zeta(x)+N\psi(x)=(0,0)$ are isolated. By pushing these out of the unit ball - we can approximate $\zeta$ in $L^2$ by a sequence of smooth $L^2$ vector fields $\gamma_n$ such that $\gamma_n\neq 0$ everywhere in the unit ball. Each $\gamma_n$ has an orthogonal Hodge decomposition which we can write as $\gamma_n=\star dg^{(n)}_1 +dg^{(n)}_2$ where $g^{(n)}_1,g^{(n)}_2$ are determined up to additive constants. The components $\star dg^{(n)}_1$ and $dg^{(n)}_2$ are "longitudinal and transverse" fields defined by pointwise projections in Fourier space, and since $\gamma_n$ has bounded Sobolev norms $(\int(1+|\xi|^2)^k|\hat\gamma_n(\xi)|^2d\xi)^{1/2}$ (where $\hat \cdot$ is Fourier transform), so do $\star dg^{(n)}_1$ and $dg^{(n)}_2.$ So they're smooth. The functions $g^{(n)}_i$ are unique if we add the requirement $\int_{\mathbb D^2} g^{(n)}_i=\int_{\mathbb D^2} f_i$ for $i=1,2.$ Because Hodge decomposition is an orthogonal decomposition, $\star dg^{(n)}_1+dg^{(n)}_2\to \star d\bar f_1+d\bar f_2$ in $L^2$ implies $dg^{(n)}\to d\bar f$ in $L^2.$ The Poincaré–Wirtinger inequality then gives $g^{(n)}|_{\mathbb D^2}\to f$ in $W^{1,2}(\mathbb D^2,\mathbb R^2).$<|endoftext|> TITLE: K-theory for a (geometric) stack QUESTION [5 upvotes]: There is a notion of $K$-theory for a manifold $M$. Is there a notion of $K$-theory for a stack $\mathcal{D}\rightarrow \text{Man}$ that is representable by a Lie groupoid $\mathcal{G}$; that is $B\mathcal{G}\cong \mathcal{D}$? One reference I could find for $K$-theory for Algebraic stacks is Bertrand Toen‘s thesis. Unfortunately, only title and abstract are in English and everything else is in French. So, are there any references in English that introduce and discuss $K$-theory for stacks. What prerequisites would be needed to understand such theory for stacks? REPLY [2 votes]: This is written nicely here. The the paper is called Loop Groups and Twisted $K$-theory I, by Freed, Hopkins and Teleman. Have a look at Section 3 and the Appendix.<|endoftext|> TITLE: A question about the Tannaka-Krein reconstruction of finite groups QUESTION [8 upvotes]: In Chapter 15 Section 15.2.1 of Quantum Groups and Noncommutative Geometry, 2nd edition, the authors raised a question: can we reconstruct a finite group $G$ from its category of finite dimensional complex representations $\text{rep}_{\mathbb{C}}(G)$. The answer depends on how we consider $\text{rep}_{\mathbb{C}}(G)$. Actually we cannot reconstruct $G$ if we only consider $\text{rep}_{\mathbb{C}}(G)$ as an abelian category or monoidal category, but we can reconstruct $G$ if we consider $\text{rep}_{\mathbb{C}}(G)$ as a symmetric monoidal category. The proof outlined in the book is to consider the forgetful functor $$ F: \text{rep}_{\mathbb{C}}(G)\to \text{vect}_{\mathbb{C}} $$ to the symmetric monoidal category of finite dimensional $\mathbb{C}$-vector spaces. Then we show there is an isomorphism $$ G\to \text{Aut}^{\otimes}(F) $$ where $\text{Aut}^{\otimes}(F)$ denotes the group of natural isomorphisms of $F$ which are compatible with the tensor product on $\text{rep}_{\mathbb{C}}(G)$. I want to see the impact $\text{Aut}^{\otimes}(F)$ when we add the symmetric structure on the monoidal categories. But it seems to me that there is no difference between the symmetric monoidal case and merely monoidal case. For example, symmetric means the following diagram commutes for $\alpha\in \text{Aut}^{\otimes}(F)$ $$ \require{AMScd} \begin{CD} F(V\otimes W)@>F(S_{V,W})>> F(W\otimes V)\\ @V \alpha_{V\otimes W} VV @VV \alpha_{W\otimes V} V\\ F(V\otimes W) @>>F(S_{W,V})> F(W\otimes V) \end{CD} $$ where $S_{V,W}$ is the symmetry functor in $\text{rep}_{\mathbb{C}}(G)$. However since $F$ is the forgetful functor we automatically have $$ F(S_{V,W})\cong S_{F(V),F(W)} $$ and since $\alpha$ preserve the monoidal structure we have $$ \alpha_{V\otimes W}\cong \alpha_V\otimes \alpha_W. $$ Therefore since we have the commutative diagram $$ \begin{CD} F(V)\otimes F(W)@>S_{F(V),F(W)}>> F(W)\otimes F(V)\\ @V \alpha_{V} \otimes\alpha_{W} VV \circlearrowright @VV \alpha_{W}\otimes\alpha_{V} V\\ F(V)\otimes F(W) @>>S_{F(W),F(V)}> F(W)\otimes F(V) \end{CD} $$ the first diagram also commute. In conclusion there is no difference of $\text{Aut}^{\otimes}(F)$ whether we consider $\text{rep}_{\mathbb{C}}(G)$ as a symmetric monoidal category or just a monoidal category. What did I do wrong in this argument? REPLY [6 votes]: Although you already have one sufficient answer, I thought I would add here that Etingof and Gelaki, "Isocategorical groups" considered exactly this question. For finite groups, they gave a complete characterization of when two non-isomorphic finite groups can have the same monoidal categories $rep(G)$ (ignoring the symmetric structure). One consequence of their characterization is that all finite groups of odd order, or twice odd order, and all simple finite groups are determined even by $rep(G)$ considered only as a monoidal category.<|endoftext|> TITLE: Is every finite Borel measure on a locally compact Hausdorff, $\sigma$-compact and separable space automatically regular? QUESTION [8 upvotes]: The conditions stated in the question seem mouthful and a bit arbitrary, so let me provide some backgrounds. Definition Let $\mu$ be a Borel measure on a topological space. We say: $\mu$ is outer regular on a Borel set $E$ if $\mu(E) = \inf\{\mu(U) : E\subseteq U \text{ is open}\}$, $\mu$ is inner regular on a Borel set $E$ if $\mu(E) = \sup\{\mu(K) : E\supseteq K \text{ is compact}\}$ (Some authors call this tight), $\mu$ is a Radon measure if $\mu$ is finite on all compact sets, outer regular on all Borel sets and inner regular on all open sets, $\mu$ is a regular measure if $\mu$ is finite on all compact sets and both outer regular and inner regular on all Borel sets. The subtle difference between a Radon measure and a regular measure is annoying. Fortunately, every $\sigma$-finite Radon measure on a locally compact Hausdorff space is automatically regular: Theorem 1 Let $X$ be a locally compact Hausdorff space. Then every Radon measure on $X$ is inner regular on all of its $\sigma$-finite sets. In particular, every $\sigma$-finite Radon measure on $X$ is regular. Corollary 1 Let $X$ be a locally compact Hausdorff space. A finite Borel measure on $X$ is regular if and only if it is outer regular on all Borel sets and inner regular on all open sets. Next, by using Riesz representation theorem on locally compact Hausdorff spaces, one can prove the following: Theorem 2 Let $X$ be a locally compact Hausdorff space. If every open set in $X$ is $\sigma$-compact, then every Borel measure on $X$ that is finite on compact sets is regular. A special case is a separable metric space: Corollary 2 Let $X$ be a locally compact, separable metric space. Then every finite Borel measure on $X$ is regular. The proof of this corollary relies on the following general topological result: Lemma A locally compact metric space is $\sigma$-compact if and only if it is separable, in which case every open set is $\sigma$-compact. In case local compactness is not given, one still has the following result: Theorem 3 Let $X$ be a separable complete metric space. Then every finite Borel measure on $X$ is regular. Note that a locally compact metric space can be given a compatible complete metric, so Theorem 3 also implies Corollary 2. Another special case of Theorem 2 is a second-countable space: Corollary 3 Let $X$ be a second-countable and locally compact Hausdorff space. Then every finite Borel measure on $X$ is regular. Proof: A locally compact Hausdorff space is topologically regular, so by Urysohn's metrization theorem, $X$ is metrizable. On the other hand, a second-countable space is separable. Thus, $X$ is locally compact, separable and metrizable, so Corollary 2 applies. Finally, in view of the above results, my question is: Question Let $X$ be a locally compact Hausdorff, $\sigma$-compact and separable space, which may or may not be metrizable. Is every finite Borel measure on $X$ regular? If not, please give a counter-example. REPLY [4 votes]: No. Counterexamples include $X = \{0,1\}^\kappa$ or $[0,1]^\kappa$ for any $\aleph_0 < \kappa \le \mathfrak{c}$. These spaces are compact Hausdorff (Tikhonov's theorem) and are separable (Hewitt-Marczewski-Pondiczery). (This special case of H-M-P can also be proved directly. For example, with $X =[0,1]^\kappa$, identify $\kappa$ with a subset of $[0,1]$, so that $X$ is a space of functions from $[0,1]$ to itself; then it is easy to show that the set of polynomials with rational coefficients is dense in $X$.) Now the uncountable successor ordinal $\omega_1 + 1$, with its order topology, can be embedded in $X = \{0,1\}^\kappa$ or $X = [0,1]^\kappa$. See Fremlin, Measure Theory, 434K(d) (thanks to Robert Furber for the reference). One can also follow a construction similar to the Stone-Čech compactification and embed $\omega_1 + 1$ into $[0,1]^{C(\omega_1+1, [0,1])}$, noting that $|C(\omega_1 + 1, [0,1])| = \mathfrak{c}$. Either way, call the embedding $\phi$. Let $\nu$ be the famous Dieudonné measure on $\omega_1 + 1$, where $\nu(B) = 1$ if $B$ contains a closed unbounded subset of $\omega_1$ and $\nu(B) = 0$ otherwise. (See Fremlin 411Q and 4A3J for details.) Then the pushforward $\mu = \nu \circ \phi^{-1}$ is a finite Borel measure on $X$ which is not regular. Specifically, since $\phi$ is an embedding and $\omega_1$ is open in $\omega_1 + 1$, then the image $A = \phi(\omega_1)$ is a relatively open subset of the compact set $\phi(\omega_1 + 1)$, so that $A$ is the intersection of an open set and a closed set in $X$, and in particular is Borel. Moreover $\mu(A) = \nu(\omega_1) = 1$. But if $K$ is any compact subset of $A$, then $\phi^{-1}(K)$ is a compact subset of $\omega_1$, hence $\mu(K) = \nu(\phi^{-1}(K)) = 0$. Thus inner regularity fails. This example is also Exercise 7.14.130 of Bogachev's Measure Theory.<|endoftext|> TITLE: Mackey theory in the setting of locally profinite groups QUESTION [5 upvotes]: $\DeclareMathOperator\Hom{Hom}$Let $R$ be a commutative ring (not necessarily unital). Let $G$ be a finite group, and let $H_1, H_2$ be subgroups of $G$. Recall the following standard result [1, Thm. 10.23]: Theorem. Let $L_i$ be an $RH_i$-module, $i=1,2$. Then $$\Hom_{RG}(L_1^G,L_2^G) \simeq \bigoplus_{x^{-1}y \in D} \Hom_{R({^xH_1} \cap {^yH_2})}({^xL_1}, {^yL_2})$$ as $R$-modules, where the sum is taken over all $D \in H_1 \backslash G/H_2$. Is there an analogue of the above theorem for ordinary smooth representations of a locally profinite group (l.p.g. for short)? I have already checked introductory chapters of some classical references in ordinary smooth representation theory of l.p.g.'s [2, 3, 4] but without success. The closest hint I was given seems to be the general discussion in [5, Sec. 5.5], but unfortunately I cannot access it freely on the Internet or in any library from my university. Thank you in advance. Bibliography [1] C. W. CURTIS AND I. REINER, Representation Theory of Finite Groups and Associative Algebras, John Wiley and Sons Inc, 1962. [2] C. J. BUSHNELL AND G. HENNIART, The Local Langlands Conjecture for GL(2), Springer, 2006. [3] W. CASSELMAN, Introduction to the theory of admissible representations of p-adic reductive groups (1974), Preprint, University of British Columbia. [4] I.N. BERNSTEIN AND A.V. ZELEVINSKY, Representations of the group GL(n, F) where F is a local non-archimedean field, Uspekhi Mat. Nauk 31 (1976), 5–70. [5] M.-F. VIGNÉRAS, Représentations l-modulaires d'un groupe réductif p-adique avec l ≠ p, Progress in Math. 137, Birkhaüser, 1996. REPLY [2 votes]: The answer to your question is the main theorem of : Kutzko, P. C. Mackey's theorem for nonunitary representations. Proc. Amer. Math. Soc. 64 (1977), no. 1, 173–175. You've got to be careful with the induction functors. In the setting of locally profinite groups, the "good" analogue of the induction functor of finite group representations is the compact-induction functor from an open compact subgroup.<|endoftext|> TITLE: Can set-like objects obeying ZFC be constructed in Euclidean geometry? QUESTION [5 upvotes]: Is it possible to base set theory on Euclidean geometry, by carefully defining various notions from set theory in terms of geometric objects, so that the ZFC axioms can be shown to hold for them? Analytic geometry allows Euclidean geometry to be based on set theory, and I am curious about whether the same can be done in reverse. REPLY [6 votes]: No, this can't be done. The key concept here is "simulation" - when is one theory strong enough to understand, in some sense, another? There are various versions of this (in particular, the term "interpretability" is very relevant). Below I'll give one which is fairly simple and applicable to this situation; it has drawbacks in many contexts, but those aren't relevant here. Say that a theory $T$ can simulate a theory $S$ if there is some computable function $f$ from sentences in the language of $S$ to sentences in the language of $T$ such that for each $\varphi$ we have $S\vdash\varphi$ iff $T\vdash f(\varphi)$. This is an extremely broad notion: for example, (first-order) PA can simulate ZFC and vice versa. However, by the same token the dividing lines it produces are quite strong. In particular, we have: If $T$ is decidable and $S$ is not - that is, the set $\{\varphi: T\vdash\varphi\}$ is computable but the set $\{\psi: S\vdash\psi\}$ is not computable - then $T$ cannot simulate $S$. Proof: If $T$ could simulate $S$ via $f$ we would have $$\{\psi: S\vdash\psi\}=\{\psi: T\vdash f(\psi)\},$$ and the latter of these is computable since $f$ is computable and $T$ is decidable. $\quad\Box$ Now ZFC is not a decidable theory - indeed, Godel's incompleteness theorem (as strengthened by Rosser) shows that basic mathematics is in fact undecidable - but Euclidean geometry is (as a consequence of the decidability of real closed fields). So there's no real way to go from geometry to set theory. And there are of course other examples of natural theories which at first glance may look foundationally promising but turn out to be decidable, like set-theoretic mereology or arithmetic of naturals with addition alone. By contrast, even a very basic theory of addition and multiplication of naturals is already undecidable, so these "promising" theories can't simulate much at all. EDIT: Of course, this depends on what exactly we understand as "geometry." Certainly the geometry of Euclid is reducible to the theory of real closed theories, and so the above applies to it; on the other hand, arguably things like tiling problems and cellular automata are "geometric" in nature and these are of course complex enough to encode Turing machines.<|endoftext|> TITLE: Langlands Reciprocity and Fermat's Last Theorem QUESTION [6 upvotes]: Question: Can Langlands Reciprocity be used to prove Fermat's Last Theorem? Background A few years ago I was reading a book on the Langlands Program and the introduction provided a list of motivating consequences. Among the usual suspects was the claim that it would provide a simple proof to Fermat's Last Theorem. At the time, this tidbit had little to do with my purpose for reading the book; I stored it as a curiosity and moved on. At some point I casually brought it up to a researcher. The response was something along the lines of "Sure. Just apply reciprocity and it falls out." I am curious again, so I have attempted to find some explanation. Unfortunately, Wiles' proof is often designated as part of the Langlands Program itself, so searching the relevant terms yields expository resources directed at that work. My question is directed at alternative approaches to FLT which have been forgotten in the wake of Wiles' success. Unfortunately, I do not remember the book which provoked the question. REPLY [4 votes]: I think the confusion here lies in what is being called reciprocity (and perhaps the interpretation of "simple"). If by Langlands reciprocity, you mean a correspondence between classical Artin representations and automorphic representations, I think there is no known simple way to get Fermat's Last Theorem just from this. Echoing Paul Garrett's comment, I believe the only approaches to FLT known to be feasible are through elliptic curves, but there is no direct correspondence between (non-CM) elliptic curves and Artin representations. If you are a bit more liberal, and mean a suitable correspondence between $\ell$-adic representations or mod $p$ representations (or maybe motives) and automorphic representations, then one can interpret Modularity of Elliptic Curves and Serre's Conjecture as special cases of this generalized reciprocity, from which it is relatively easy to conclude Fermat's Last Theorem. For instance, Richard Taylor in this general overview considers the Shimura-Taniyama Conjecture to be a special case of Langlands' reciprocity conjectures.<|endoftext|> TITLE: How fast is the continuum changing with respect to the relative change of size of the forcing notion? QUESTION [5 upvotes]: Has there been any research done on the related rates of forcing? If I force to increase the size of the continuum $\mathfrak{c}$ by 5 $\aleph$'s, say from $\aleph_2$ to $\aleph_7$, how fast does the size of the notion of forcing $\mathbb{P}$ change from the ground model to the forcing extension? It seems they must just have a constant ratio, but perhaps there are small forcings which have a very big effect. REPLY [7 votes]: I don't know if the following results are related, but they might be interesting: Gitik an I have results, which simply say that (sometimes under the assumption of the existence of large cardinals) one can have a pair (W, V) of models of ZFC, such that adding an $Add(\omega, \kappa)$-generic over V, adds an $Add(\omega, \lambda)$-generic over W, for some $\lambda > \kappa.$ Also, there are results by Shelah and Woodin that one can have a pair (W, V) of models of ZFC, such that W satisfies GCH, V=W[R], for some real R, and In V, the continuum is arbitrary large.<|endoftext|> TITLE: Iterated limits equal? QUESTION [5 upvotes]: Consider the Banach algebra $B_2(H)$ of Hilbert Schmidt operators on a Hilbert space $H$ with Hilbert Schmidt norm. We know that $B_2(H)$ is a Hilbert space as well with $\left=tr(B^*A)$.I am looking for an example of pair of sequences $A_i,\tilde A_j$ and $T_i,\tilde T_j$ in the closed unit ball of $B_2(H)$ and $B(B_2(H))$ respectively and a $D\in B_2(H)$ such that both the iterated limits of $\left$ exists but $$\lim_i\lim_j \left\neq \lim_j\lim_i \left$$ Otherwise prove that if iterated limits exists then they are equal. REPLY [8 votes]: For $\xi,\eta\in H$ let $\theta_{\xi,\eta}$ be the rank-one operator $\theta_{\xi,\eta}(\gamma) = (\gamma|\eta) \xi$ for $\gamma\in H$. Let $(e_i)$ be an orthonormal sequence in $H$, set $S_i = \theta_{e_1, e_i} \in B(H)$ and let $R_j$ be the projection onto the span of $\{e_1,e_2,\cdots,e_j\}$. Then $$ \lim_i \lim_j (S_iR_j(e_i)|e_1) = \lim_i (S_i(e_i)|e_1) = \lim_i (e_1|e_1) = 1, $$ while $$ \lim_j \lim_i (S_iR_j(e_i)|e_1) = \lim_j \lim_i (R_j(e_i)|e_i) (e_1|e_1) = 0. $$ Now we "embed" this into your example. Let $\tilde A_j = \theta_{e_1,e_1}$ for all $j$, and set $A_i = \theta_{e_i,e_1}$ so $A_i \tilde A_j = A_i$. Define $\tilde T_j(a) = \theta_{R_j(a(e_1)), e_1}$ for $a\in B_2(H)$; a simple estimate shows that $\tilde T_j$ is bounded. Define $T_i(a) = \theta_{S_i(a(e_1)),e_1}$. Then $$ \tilde T_j (a) (e_1) = \theta_{R_j(a(e_1)), e_1} (e_1) = R_j(a(e_1)) \qquad (a\in B_2(H)). $$ Thus $$ T_i \tilde T_j(a) = \theta_{S_i R_j(a(e_1)), e_1} \implies T_i \tilde T_j (A_i \tilde A_j) = T_i \tilde T_j(\theta_{e_i,e_1}) = \theta_{S_iR_j(e_i),e_1}. $$ If we now let $D=\theta_{e_1,e_1}$ then $$ \Big\langle T_i \tilde T_j (A_i \tilde A_j), D \Big\rangle = \operatorname{Tr}\Big( \theta_{S_i R_j(e_i),e_1} \Big) = (S_i R_j(e_i) | e_1). $$ As above, this has different iterated limits.<|endoftext|> TITLE: Does there exist a genus $g$ curve over $\mathbb{Q}$ with every type of stable reduction? QUESTION [14 upvotes]: Let $g\geq3$ be an integer, let $\{\Gamma_i|i \in I\}$ be the set of all stable graphs of genus $g$. (We say a graph is stable if it is the dual graph of a stable curve.) Let $X$ be a curve defined over $\mathbb{Q}$, we say it has reduction type $\Gamma_i$ , if there is a model $\mathcal{X}_{p_i}$ over $\mathbb{Z}_{p_i}$, whose generic fiber is $X_{\mathbb{Q}_{p_i}}$ and special fiber is a stable curve with dual graph $\Gamma_i$. Given $g\geq3$, does there always exist a smooth curve over $X/\mathbb{Q}$, (or some number field $K$?) such that every $\Gamma_i$ appear as the reduction of certain prime $p_i$? (If we ask the same question over global function field (finite extension of $\mathbb{F}_p(t)$), I think the answer is "yes", as we can interpolate the loci in $\overline{\mathcal{M}}_g$ by curves. ) REPLY [9 votes]: Here is a stacky variant of Will Savin's answer, with a more precise result. The reference is my paper: Problèmes de Skolem sur les champs algébriques, Compo. Math. 125 (2001), 1—30 (1) First, note the following: Let $F$ be a local field, $Y$ an $F$-scheme of finite type, $C\to Y$ a stable curve of genus $g$ over $Y$, and $\Gamma$ a stable graph of genus $g$. Let $U$ (resp. $U'\subset U$) be the set of points $y\in Y(F)$ such that $C_y$ has (potential) reduction type $\Gamma$ (resp. has stable reduction on $O_F$, with reduction type $\Gamma$). Then both $U$ and $U'$ are open in $Y(F)$, for the valuation topology. (Exercise). (2) Now let $I'\subset I$ correspond to singular curves (i.e. we remove the one-point graph from $I$). For each $i\in I'$, fix a prime $p_i$ (they have to be pairwise distinct!). I will make the assumption that (*) there exists a smooth curve over $\mathbb{Q}_{p_i}$ with potential reduction graph $\Gamma_i$. (I am pretty sure that this is always true; otherwise, choose $p_i$ accordingly). Put $R:=\mathbb{Z}\left[(1/p_i)_{i\in I'}\right]$. Theorem. There is a number field $K$, and a stable curve $\mathscr{X}$ over $O_K$ such that: • $\mathscr{X}$ is smooth over $R\otimes_{\mathbb{Z}}O_K$; • for each $i\in I'$ and each prime $\mathfrak{p}$ of $O_K$ above $p_i$, $\mathscr{X}$ has reduction graph $\Gamma_i$ at $\mathfrak{p}$. Assume further that for each $i\in I'$ there is a stable curve over $\mathbb{F}_{p_i}$ with graph $\Gamma_i$. (This holds if $p_i$ is large enough). Then we can take $K$ totally split at each $p_i$. Proof: Consider the moduli stack $\mathscr{M}_g$ (resp. $\overline{\mathscr{M}}_g$) of smooth (resp. stable) curves of genus $g$. Let us prove the first claim. For each $i$, let $\Omega_i\subset{\mathscr{M}}_g(\mathbb{Q}_{p_i})$ be the subcategory of those smooth curves with (potential) reduction graph $\Gamma_i$. By (1), this is $p_i$-adically open in ${\mathscr{M}}_g(\mathbb{Q}_{p_i})$, in the sense of Definition 2.2 in the paper. Moreover, it is not empty, due to assumption (*). It is straightforward to check that $\mathscr{M}_{g,R}\to\mathrm{Spec}(R)$ with the local data $(\Omega_i)_{i\in I'}$ constitutes a Skolem datum in the sense of Definition 0.6. (We use the fact that $\mathscr{M}_g$ is smooth with geometrically connected fibers over $\mathrm{Spec}(\mathbb{Z})$). Now apply Theorem 0.7: this almost gives the result (with $K$ totally split at each $p_i$) except that the curve may not have stable reduction outside $\mathrm{Spec}(R)$. To fix this, just enlarge $K$, possibly losing splitness. For the second claim, we do the same, replacing $\Omega_i$ by $\Omega'_i$ consisting of curves with stable reduction of type $\Gamma_i$ over $\mathbb{Z}_{p_i}$: the extra assumption on $p_i$ guarantees that $\Omega'_i\neq\emptyset$. QED<|endoftext|> TITLE: Maximal numbers of summands in middle terms of short exact sequences QUESTION [8 upvotes]: Let $A$ be a finite dimensional algebra and $M$ and $N$ indecomposable $A$-modules. Denote by $\xi(M,N)$ the maximal number of indecomposable summands of a modules $X$ such that there is a non-split short exact sequence $0 \rightarrow N \rightarrow X \rightarrow M \rightarrow 0$. Question 1: Is there an easy example of $A$ such that $\sup \{ \xi(M,N) | M, N $ indecomposable $\}$ is infinite? Question 2: Is there an easy example of $A$ such that $\sup \{ \xi(M,\tau(M)) | M $ indecomposable $\}$ is infinite? (here $\tau$ denotes the Auslander-Reiten translate. One might also ask for an algebra where the number of middle term indecomposable summands of Auslander-Reiten sequences can be arbitrarily large) Question 3: Do we have $\sup \{ \xi(M,N) | M, N $ indecomposable $\}=\sup \{ \xi(M,\tau(M)) | M $ indecomposable $\}$? Bonusquestion: Can $\xi(M,N)$ be calculated with the GAP-package QPA? REPLY [5 votes]: For question 1: For a self-injective algebra, if $0\to X\to Y\to Z\to0$ is a short exact sequence with $Y$ and $Z$ indecomposable, then there is a short exact sequence $0\to\Omega Z\to X\oplus P\to Y\to0$, for some projective module $P$. It's easy to come up with examples for, say, $k[x,y]/(x^3,y^3)$ where $X=\text{soc }Y$ has arbitrarily many summands (and $Y/X$ is indecomposable).<|endoftext|> TITLE: 1-dimensional pure gauge theory QUESTION [6 upvotes]: I am learning TQFT from compact Lie groups by Freed, Hopkins, Lurie, and Teleman: https://arxiv.org/abs/0905.0731 , and got stuck very hard even in the first section ($n = 1$), which was "trivial but included for completeness". In particular, I have several unfamiliar terms while it discusses "1-dimensional pure gauge theory". Each term will require some explanations, if not too long. Any pointers to places where I can learn more about the terms will be highly appreciated. 1. 1-dimensional "pure" gauge theory I have an impression that gauge theory is just bundle theory in math term. But how about the adjective "pure"? 2. The standard quantization procedure Given a compact Lie group $G$, an abelian character $\lambda: G \to U(1)$, and a $G$-bundle with connection over the circle ($g$ being its holonomy), they define a 1d-TQFT by assigning to the circle the number $\lambda(g)$. Then "by the standard quantization procedure", they assign to the positively oriented point a subspace of $\mathbb{C}$, depending on if the abelian character is trivial or not. I know little about geometric quantization - all I have read is J. Baez's informal introduction (http://www.math.ucr.edu/home/baez/quantization.html). But I have know idea how these two relate, and the article even claimed that this procedure relates to "the Gauss law in physics", making it even more mysterious for me.. 3. Path integral over the groupoid $G//G$ Now the value assigned to the circle is just the dimension of the vector space assigned to the positively oriented point, and as above this value is either 0 or 1 depending on whether the character is nontrivial or not. The authors claimed that this may be understood as the result of the **path integral over the groupoid $G//G$ of connections on $S^1$ with respect to Haar measure: $$ \frac{1}{|G|} \int_G \lambda(g)dg = 0 \mbox{ or } 1$$ .. This has nothing to do with what I think a path integral is: to me, a path integral is an integral over all path/section space with a suitable weight. I hope I express my questions clear. If there's any confusion, please let me know. Thank you. REPLY [6 votes]: Just to add something about Gauss' law to the excellent previous answer: The Hamiltonian of a pure gauge theory, as initially derived from the Lagrangean, typically has the structure $H=E^2 +B^2 - A_0 \nabla \cdot E$, where $E$ and $B$ are the electric and magnetic fields, and $A_0 $ is the temporal component of the gauge field. In addition, the Lagrangean contains no time derivative of $A_0 $, and therefore there is no conjugate momentum for $A_0 $ - it is not a dynamical degree of freedom. Instead, it acts as a Lagrange multiplier enforcing $\nabla \cdot E =0$. That is Gauss' law, a constraint on the remaining dynamical fields. There are distinct ways of dealing with this. One avenue is to use the gauge freedom to choose $A_0 =0$. Then the non-dynamical degree of freedom is gone, but one has lost the knowledge about the Gauss law constraint. In $A_0 =0$ gauge, one must impose the constraint in addition to the dynamical equations generated by the Hamiltonian. Another avenue is to choose a different gauge, but then the formalism still contains $A_0 $. The "equation of motion" for $A_0 $ is, again, not dynamical, but a constraint on $A_0 $, which one can then solve to eliminate $A_0 $ (this, e.g., generates the Coulomb interaction). In this manner of speaking, Gauss' law doesn't play quite such a central role - one thinks of $A_0 $ being constrained, as opposed to the other fields being constrained. Either way, one eliminates two components of the gauge field: One by choice of gauge, one by constraint. Only the remaining components are dynamical degrees of freedom. Thus, light comes with two polarizations, even though one starts out with a 4-component field $A_{\mu } $.<|endoftext|> TITLE: Classification of shod Dyck paths QUESTION [13 upvotes]: A sequence $[c_0,c_1,...,c_{n-1}]$ with $n \geq 2$ is called a Dyck path in case $c_{n-1}=1$, $c_i \geq 2$ for $i \neq n-1$ and $c_i-1 \leq c_{i+1}$ for each $i$. For example the Dyck paths for $n=4$ are the 5 sequences [ 2, 2, 2, 1 ], [ 3, 2, 2, 1 ], [ 2, 3, 2, 1 ], [ 3, 3, 2, 1 ], [ 4, 3, 2, 1 ]. In general they are enumerated by the Catalan sequence. Note that those sequences are just the area sequence in the classical definition of Dyck paths, see https://arxiv.org/pdf/1811.05846.pdf page 6. One also has the coarea sequence associated to a Dyck path $[d_0,d_1,...,d_{n-1}]$ which can be formally defined via $d_0=1, d_1=2$ and $d_i = \min \{k \geq 2 | k \geq c_{i-k} \}$ for $i \geq 1$. A module of a Dyck path $[c_0,c_1,...,c_{n-1}]$ is a tuple $(i,m)$ with $0 \leq i \leq n-1$ and $1 \leq m \leq c_i$. We call a Dyck path $[c_0,c_1,...,c_{n-1}]$ shod in case the following is satisfied: For each tuple $(i,m)$ with $1 \leq m \leq c_i -1$ and $m < c_{i-1}$ we have (($c_i-m=c_{i+m}$) or ($d_{i-1}=d_{i+m-1}-m$)). This condition looks complicated but has a nice pictorial interpretation in a Dyck path. Question 1: Is there a nice combinatorial description of shod Dyck paths? Question 2: Is it true that the number of shod Dyck paths is equal to $\frac{(n-2)^3+2(n-2)}{3}+1$, which is also equal to the number of permutations of length n which avoid the patterns 321, 2143, 3124 , see https://oeis.org/A116731 . This is true for $n \leq 9$. So the sequence starts with 1,2,5,12,25,46,77,120.... Question 2' (equivalent to Question 2): Call a shod Dyck path special in case it additionally satisfies that there exists an $i \neq n-1$ such that $c_{i+1}>c_i-1$ and $c_{i+c_i}>c_{i+1}-c_i+1$. Then the number of special shod Dyck paths $[c_0,c_1,...,c_{n+2}]$ seems to be given by the Square pyramidal numbers $\sum\limits_{k=0}^{n}{k^2}$, see https://oeis.org/A000330. It counts for example the Number of permutations avoiding 13-2 that contain the pattern 32-1 exactly once.Is this true? Background: A finite dimensional algebra is called shod, in case each indecomposable module has projective dimension at most one or injective dimension at most one. Recently in https://www.sciencedirect.com/science/article/pii/S0001870815300219 such algebras are characterised as endomorphism algebras of 2-term silting complexes in derived categories of Ext-finite hereditary abelian categories. Dyck paths correspond to Nakayama algebras with a linear quiver (which are also exactly the admissible quotient algebras of the algebra of upper triangular matrices) and the elementary question asks for a combinatorial classification of Nakayama algebras that are shod. Note also that Dyck paths are in bijection with permutations of length n which avoid the pattern 321 and maybe viewing Dyck paths as such pattern avoiding permutations might be better for this question. But there are various bijections from Dyck paths to 321 avoiding permutations and it is not clear what is the "best" one for such questions. edit: Here is another approach to question 2 leading to a possibly much nicer enumeration result, namely the sum of all squares: Call a shod Dyck path special in case it additionally satisfies that there exists an $i \neq n-1$ such that $c_{i+1}>c_i-1$ and $c_{i+c_i}>c_{i+1}-c_i+1$. Then the number of special shod Dyck paths $[c_0,c_1,...,c_{n-1}]$ seems to be given by the Square pyramidal numbers $\sum\limits_{k=0}^{n-3}{k^2}$: https://oeis.org/A000330 , which is a much more well known sequence. They enumerate for example according to oeis "Number of rhombi in an n X n rhombus" and "Number of permutations avoiding 13-2 that contain the pattern 32-1 exactly once". I can classifly the shod non-special Dyck paths, so it would be enough to solve question 1 and 2 for special shod Dyck paths. For $n=4$ the unique special shod Dyck path is [2,2,2,1], for $n=5$ we have [ [ 3, 2, 2, 2, 1 ], [ 3, 3, 2, 2, 1 ], [ 2, 2, 3, 2, 1 ], [ 2, 3, 3, 2, 1 ], [ 3, 3, 3, 2, 1 ] ] and for $n=6$ we got [ [ 4, 3, 2, 2, 2, 1 ], [ 4, 3, 3, 2, 2, 1 ], [ 4, 4, 3, 2, 2, 1 ], [ 3, 2, 2, 3, 2, 1 ], [ 3, 3, 2, 3, 2, 1 ], [ 3, 2, 3, 3, 2, 1 ], [ 4, 3, 3, 3, 2, 1 ], [ 4, 4, 3, 3, 2, 1 ], [ 2, 2, 4, 3, 2, 1 ], [ 2, 3, 4, 3, 2, 1 ], [ 3, 3, 4, 3, 2, 1 ], [ 2, 4, 4, 3, 2, 1 ], [ 3, 4, 4, 3, 2, 1 ], [ 4, 4, 4, 3, 2, 1 ] ]. REPLY [3 votes]: Here are some conjectures each of which would answer the question. There are lots of ways to map Dyck paths to permutations as well as many Wilf equivalent pattern avoidance classes. Hence, these conjectures may not be the "right" way or the "best" way to go. But I did some computations and these conjectures fit with bijections already implemented in SageMath (see the documentation on Dyck Words). Conjecture A: Applying to_132_avoiding_permutation() takes shod Dyck paths to $Av(132, 4321, 4213)$. Conjecture B: Applying to_312_avoiding_permutation() takes shod Dyck paths to $Av(312, 1234, 1324)$. Conjecture C: Applying to_321_avoiding_permutation() takes shod Dyck paths to $Av(321, 2134, 2143)$. Here $Av(\pi_1, \dots, \pi_k)$ denotes all permutations which avoid each of the patterns $\{\pi_1, \dots, \pi_k\}$. I came to these conjectures by looking at the two non-shod Dyck paths which map to permutations of length $4$. Provided my code is correct, each conjecture is checked for permutations up to (and including) length $12$. I should also note the numbers come out to suggest Wilf equivalence between all these, but I don't know whether or not that is known. I'm not sure of the status of the literature on avoiding one pattern of length $3$ and two patterns of length $4$. (Edit: The Wilf equivalence classes for one pattern of length $3$ and two patterns of length $4$ are known from Systematic Studies in Pattern Avoidance . Hence, it is known the sets in Conjecture A, B, and C have the same number of permutations of each length. However, it would still be nice to understand them bijectively of see the connection to shod Dyck paths.)<|endoftext|> TITLE: A toy model in 0-d QFT QUESTION [12 upvotes]: Questions For any positive integer $r$, compute $$(\frac{d}{dY})^r e^{(Y^2)}| _{Y=0}.$$ The answer should directly relates to a counting problem about Feynman diagrams. Is there a tutorial for how Feynman diagrams work in this context? I look forward to an answer a lot, since the question has been reduced to the simplest form. Thank you! EDIT: 3. Turns out the answer to the first question is trivial considering its Taylor expansion. So a better question should be what's the benefit of solving the first one by the combinatorial way. Context I am a math student trying to learn QFT and Feynman Diagrams using Hori et al's Mirror Symmetry. Much to my surprise, even a toy model in 0-dimension the theory is already complicated (for me). On a point, a function is just a number, so integration over all functions reduces to an ordinary integral. I am looking at the following particular toy model: $$ \int dX e^{-S(X)}, $$ where $S(X) = \frac{1}{2}X^2 + i\epsilon X^3$, and am focusing on the perturbation with small $\epsilon$, which reduces to the computation $$(\frac{d}{dY})^r e^{(Y^2)}| _{Y=0}.$$ Basic attempts show that this is a combinatorial problem, which I have no idea how to solve. This is where the book introduces Feynman diagrams, claiming that they help computing the value at zero of the $r$-th derivative above. However, the explanation is not clear to me. I don't know what the book means by "choosing pairs", "contracting", and "propagators". I have tried other lecture notes online, but all of what I have found use physics terminologies making the situation more complicated. REPLY [12 votes]: Draw a point from which $r$ lines emanate (a "vertex"). The ends of the lines are associated with the derivatives. Now let the derivatives act. Two things can happen: a.) The derivative acts on the exponential. Represent this by attaching a dot to the end of the line (now the derivative is gone, but there's a factor $Y$ associated with the end of the line, symbolized by the dot). b.) The derivative acts on one of the dots, i.e., one of the $Y$ prefactors you've generated by previous applications of derivatives. Represent this by attaching the line you're considering to one of the dots at the end of another line, erasing the dot (that factor $Y$ is now gone, having acted with the derivative) - so altogether you now have a loop emanating from the vertex and going back into it. Finally, set $Y=0$. This means that all diagrams which still have dots are erased. So, your combinatorial problem is to find all different diagrams you can construct this way; in effect, in your example, pairing off lines emanating from the vertex to form loops. There are some numerical factors to take care of, here, just the factors 2 from taking derivatives of $Y^2 $. In the exponential, you can think of the $Y^2 $ as two dots connected by a line - then this incorporates a.) into the notion used in b.) of "derivatives erasing dots". And you can incorporate the aforementioned factors 2 this way - you have a choice of which dot you're attaching and erasing. Although it might be a bit facile of me to say, since I'm familiar with the physics parlance, don't be afraid of trying to understand the physics tutorials on this. As far as these diagrammatics go, you don't need to understand the physics behind "propagators", "vertices" and the like. You just need a rudimentary dictionary of what these things are in the graphs. I already mentioned "vertex". "Propagators" are just the lines. The dots are "sources". When you connect two lines, erasing the dot, you're "contracting" ... So let's complete the exercise: Of course, for odd $r$, the result is zero - there are no valid diagrams. For even $r$, there are $r!/(2^{r/2} (r/2)!)$ ways of pairing off lines coming from the vertex. Multiplied by a factor $2^{r/2} $ from taking $r/2$ derivatives of the exponential, as mentioned above, yields $r!/(r/2)!$. As a check, these are indeed the absolute values of the constant terms in the Hermite polynomials $H_r $. REPLY [10 votes]: Here is how I understand the appearance of diagrams in computing that integral, which may not be how the book does it. We can formally expand as a series in $\epsilon$ and write $$\int_{-\infty}^\infty dX\,e^{-S(X)} = \sum_{n \ge 0} \frac{(-i\epsilon)^n}{ n!} \int_{-\infty}^\infty dX\,e^{-\frac{1}{2}X^2}X^{3n}$$ so the problem reduces to computing the integral on the RHS. This is somewhat well-known to vanish for odd $n$ and to be equal to $\sqrt{2\pi}$ times the double factorial $$(3n - 1)!! = (3n - 1) (3n - 3) \cdots 1$$ when $n$ is even. Combinatorially, the double factorial counts perfect matchings (partitions into blocks of size 2) of a set of size $3n$. Given a perfect matching of the set $\{1, \dots, 3n\}$, we can construct a $3$-regular graph as follows: The vertices are $\{1, \dots, n\}$. To each vertex $j$, attach three "half-edges" labelled $\{3j-2, 3j-1, 3j\}$. Join half-edges together according to the perfect matching to form edges. Thus we can interpret the coefficient of $\epsilon^n$ as counting cubic graphs of size $n$. (Though the way we've labelled the half-edges means there is some overcounting to deal with.) Thinking of it this way might not seem very motivated, but for slightly more complicated integrals the benefit is easier to see. For instance, if we added an $\epsilon^4$ term as well, we'd get something counting graphs where vertices can have either degree 3 or 4. The Isserlis–Wick theorem is the generalization of that "well-known" fact about double factorials to the multivariate setting, and can be used to interpret multivariate integrals of this type as counting graphs with decorated edges.<|endoftext|> TITLE: On approximate simultaneous diagonalization QUESTION [6 upvotes]: It is well known that two $n\times n$ symmetric positive semidefinite matrices $A$, $B$ such that $AB=0$ are simultaneously diagonalizable. My question is related to the existence of a specific simultaneous diagonalization in the following sense: Let $\{A_k\}$, $\{B_k\}$ be two sequences of symmetric matrices converging to positive semidefinite matrices $A$ and $B$, respectively, such that $AB=0$. Is it the case that there exist a basis $\{v_i^k\}$ of eigenvectors of $A_k$ and a basis $\{w_i^k\}$ of eigenvectors of $B_k$, for all k, such that each $v_i^k$ and $w_i^k$ converge to some $c_i$ such that $\{c_1,\dots, c_n\}$ form a simultaneous basis of eigenvectors for $A$ and $B$? REPLY [7 votes]: The answer is no in general. For a $2\times 2$-counterexample, let $A = 0$, let $B$ be the diagonal matrix with diagonal entries $1$ and $0$ (i.e. $B$ is the projection onto the first component), choose $B_k = B$ for each $k \in \mathbb{N}$ and $$ A_k = \frac{1}{k} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$ for each $k \in \mathbb{N}$. Then each sequence $(v_k) \subseteq \mathbb{R}^2$ of eigenvectors of $A_k$ can only converge to a scalar multiple of $(1,1)$ or to a scalar multiple of $(1,-1)$. However, only (scalar multiples of) the canonical unit vectors $(1,0)$ and $(0,1)$ simultaneously diagonalize $A$ and $B$.<|endoftext|> TITLE: Grothendieck-Verdier duality without the noetherian condition QUESTION [7 upvotes]: The Grothendieck-Verdier duality: $$ Rf_*\big(R\mathcal{H}\textit{om}_X^\bullet(\mathcal{E}^\bullet,f^!\mathcal{F}^\bullet)\big) \cong R\mathcal{H}\textit{om}^\bullet_Y(Rf_*\mathcal{E}^\bullet,\mathcal{G}^\bullet) $$ is known to hold for $f:X\to Y$ being a proper map of noetherian schemes. Is there a way to get rid of this requirement on schemes to be noetherian, possibly by putting some extra conditions on the morphism $f$? For example: let $X$ be smooth (projective in necessary) and consider the projection to the second factor $f:X\times U\to U$, where $U$ is some affine scheme. Does one have the duality in this setting? Edit: After reading about the topic in Neeman's and Lipman's work, I have not managed to find an explicit construction of the right adjoint $f^\times$. In what cases is the explicit construction of $f^\times$ known? What would it be in the above example? REPLY [8 votes]: Does one have the duality in this setting? Yes, we do have duality in a very general setting. Your question is equivalent to asking for the existence of a right adjoint to the derived pushforward functor $\mathbf{R}f_*\colon \mathbf{D}_{qc}(X) \to \mathbf{D}_{qc}(Y)$. It turns out that it exists for any morphism $f\colon X \to Y$ of qcqs schemes. Look at tag 0A9D for more details. P. S. It is not a good idea to call this adjoint functor by $f^!$ unless $f$ is proper. Usually, another functor is denoted by $f^!$ that actually differs from the right adjoint $f^\times$ (SP denotes this functor by $a(f)$). UPD: If $Y$ is qcqs and $f\colon X \to Y$ is proper and smooth of (pure) relative dimension $d$, then $f^!(-)\cong f^{\times}(-)\cong \Omega^d_{X/Y}[d]\otimes_{\mathcal O_X} f^*(-)$. Look at tag 0BRT and tag 0A9U for a proof in the noetherian case. The proof in the non-noetherian setup is essentially the same. Just note that noetherianness of $Y$ is used only to show that $\mathbf{R}f_*$ preserves perfect objects. However, it is true without this assumption as it is briefly explained in Example 2.2 of the paper Quasi-perfect scheme-maps and boundedness of the twisted inverse image functor.<|endoftext|> TITLE: How much of the category of motives can be recovered from automorphisms of the Betti functor QUESTION [7 upvotes]: Say we are working with schemes over a field $k\subset \mathbb{C}.$ A motive in the sense of Voevodsky is a functor $Sch\to D^bVect$ from (an appropriate category of) schemes to the DG category of complexes of vector spaces satisfying some descent and locality conditions such as $A^1$-locality (I will be purposefully ambiguous about the specific conditions, as I don't want to commit to a specific motivic category). Let $M$ be the category of such motives. Then $M$ is a full subcategory of the functor category $Fun(Sch, D^bVect).$ Now certain motivic categories are known to be Tannakian or derived Tannakian, i.e. fully controlled (as a symmetric monoidal category) by an algebraic group $G$, which is defined to be the category of automorphisms of a (symmetric monoidal) fiber functor $$F_M:M\to D^bVect.$$ If the fiber functor $F$ is "sufficiently nice" then it is representable in the functor category by a functor $$F:Sch\to D^bVect,$$ and thus its (algebraic) group of automorphisms is equivalent to the group of automorphisms of the functor $F.$ As the notion of fiber functor is quite flexible, this seems like a very promising approach to constructing motivic categories: namely, take a "nice" symmetric monoidal functor $F:Sch\to D^bVect,$ consider its algebraic group of (symmetric monoidal) automorphisms, $G$, and take the category of representations of $G$. I haven't seen a careful construction of motivic invariants from this point of view, but it seems very plausible to me that such a description exists. For example on the level of Tate motives, the associated graded of the weight filtration is known to be a fiber functor, which seems to imply that automorphisms of the functor $X\mapsto Gr(C^*(X))$ (graded with respect to the weight filtration), which should be noncanonically equivalent to the Betti homology functor, should have automorhpism group which completely controls mixed Tate motives. Perhaps something similar can be said with the full motivic category and the algebraic $K$ group functor. Is something like this true? If true, this would imply that motivic Galois groups can be defined without the category of motives, as automorhpisms of a a suitable "generating" functor. Is this written up somewhere? REPLY [7 votes]: A precise way to say that "sufficiently nice functors" from schemes to complexes of vector spaces determine realization functors of Voevodsky's motives is the notion of mixed Weil cohomology. There are rather explicit descriptions (i.e. categories) which describe what can be determined by the automorphisms of Betti cohomology. For the derived version, you might have a look at: I. Iwanari, Tannakization in derived algebraic geometry, J. K-Theory 14 (2014), 642—700. (also https://arxiv.org/abs/1112.1761). The non derived (hence coarser) version is the theory of Nori motives (introduced by Nori in the early 2000's), which is the subject of a recent book of A. Huber and of S. Müller-Stach, published by Springer in 2017. Yves André's Bourbaki talk is a good source as well, in which Ayoub's contributions are well explained. Either way, none of these constructions determines Voevodsky's theory of motives. It is not known that the Betti realization functor is conservative on Voevodsky's motives (Ayoub is working hard on it for years, but there is no complete proof so far). Morever, the conservativity of the Betti realization would not be sufficient to prove that Voevodsky's motives form a (derived) tannakian category (conservativity only implies that the tannakian category associated to Voevodsky's motives does not depend on the choice of a realization functor such as Betti cohomology). But more optimistic conjectures would imply that. To be more precise: the approach of Voevodsky has two important features: it satisfies a reasonable universal property (in some higher categorical sense) and the main cohomology theory represented there (i.e. determined by the unit of the tensor product) is classical intersection theory (under the form of (higher) Chow groups). This latter property is what is lost (until we prove very non-trivial conjectures which go much beyond conservativity, such as the Hodge conjecture) when we go to more tannakian versions of motives. Finally, this idea of defining motives by forcing Betti cohomology to be a fiber functor is not new at all: if we restrict to pure motives (motives of smooth and projective varieties), this boils down to Deligne's theory of `cycles motivés' revisited by Yves André in 1996 (this is very well explained in terms of Nori's approach at the end of the book of Huber and Müller-Stach).<|endoftext|> TITLE: How to control Wasserstein distance in terms of characteristic function QUESTION [5 upvotes]: Let $\mathcal P(\Omega)$ be the set of probability measures supported on some compact subset $\Omega\subset\mathbb R^d$. For $\mu\in\mathcal P(\Omega)$, denote by $F_{\mu}$ its characteristic function, i.e. $$F_{\mu}(x)~:=~\int_{\mathbb R^d}e^{i\langle x,y\rangle}\mu(dy)~=~\int_{\Omega}e^{i\langle x,y\rangle}\mu(dy),\quad \forall x\in\mathbb R^d.$$ Let $W(\cdot,\cdot)$ be the Wasserstein distance of order $1$. My question is whether there exists a continuous function $c:\mathbb R_+\to\mathbb R_+$ with $c(0)=0$ s.t. it holds for all $\mu$, $\nu\in\mathcal P(\Omega)$: $$W(\mu,\nu)~\le~c\big(\|F_{\mu}-F_{\nu}\|\big),\quad \mbox{with } \|F_{\mu}-F_{\nu}\|:=\max_{x\in\mathbb R^d}|F_{\mu}(x)-F_{\nu}(x)|.$$ Any answer, comments and references are highly appreciated! Personal thought: We use Kantorovich's duality, i.e. $$W(\mu,\nu)~=~ \sup_{f\in \rm{Lip}_1(\mathbb R^d)} \int f d\mu -\int fd\nu ~=~\sup_{f\in \rm{Lip}_1(\Omega)} \int f d\mu -\int fd\nu,$$ where $\rm{Lip}_1(\mathbb R^d)$ and $\rm{Lip}_1(\Omega)$ denote respectively the collection of $1-$Lipschitz functions on $\mathbb R^d$ and $\Omega$. It remains to write the integral $\int f d\mu$ in terms of $F_{\mu}$. Denote by $\hat f$ the Fourier transform of $f$, i.e. $$\hat{f}(x)~:=~ \int_{\mathbb R^d}e^{i\langle x,y\rangle}f(y)dy.$$ We notice $\hat{f}$ is well defined for each $f\in \rm{Lip}_1(\Omega)$. Then we may apply Parseval's equality under "suitable conditions" $$\int_{\mathbb R^d}f(x)\mu(dx)~=~\int_{\mathbb R^d}\hat{f}(x) F_{\mu}(x)dx.~~~~~~~~~~~~ (\ast)$$ I am not familiar with the regularity analysis of $\hat f$. If someone knows how to proceed based on $(\ast)$, I am happy to know. REPLY [2 votes]: For measures on the real line, There is no such function $c$. Given $n$, let $\mu_n$ assign mass $1/n$ to each of $-n,n$ and remaining mass to zero. Let $\nu_n$ be Dirac measure at zero. As $n$ grows, the Fourier transforms get uniformly close, with sup norm distance of order $1/n$, yet the Wasserstein distance remains 2. For measures on a fixed compact set there is such a function $c $ since both metrics metrize the weak * topology and the space of probability measures on a compact set is itself weak * compact.<|endoftext|> TITLE: Immersion in $\mathbb R^3$ of a Klein bottle with Morse-Bott height function without centers QUESTION [6 upvotes]: Can the Klein bottle be immersed in $\mathbb R^3$ so that the associated height function be of Morse-Bott type and have no centers? That is, the height function would have only Bott-type extrema and saddle singularities. A Bott-type singularity is a non-degenerate singular circle: a circle where the derivative is zero with the function being quadratic on transverse curves. A center is a Morse-type local extremum: an isolated singularity around which the function is $\pm(x^2_1+x^2_2)$ in some local coordinates. My intuition is that no. I think such function cannot have (Morse) singularities other than Bott-type extrema (because they would increase the genus), and I cannot see how to connect an (even) number of immersions of Bott-type extrema (circles) by tubes in a non-orientable way without additional singularities (this should follow from the Whitney–Graustein theorem). For a torus, such an immersion (embedding) is a doughnut lying flat on the table. However, I can't see how this can be done for the Klein bottle. The answer here does not seem to do the trick because it also increases the genus. REPLY [7 votes]: I wanted to prove that this is impossible but instead proved that this is possible... Unfortunately, it is a bit hard to draw the picture but I'll try to explain how this should look like. Construction. In this construction the Klein bottle will be included between the planes $z=0$ and $z=1$. The curves $\{z=1\}\cap K$ and $\{z=0\}\cap K$ are both the eight figure curve (with rotation index $0$). And both are Bott circles. Let us call the first curve $S_1$ and the second $S_0$. Now, the function $z$ restricted to $K\setminus S_0\cup S_1$ has no critical points. And $K\setminus S_0\cup S_1$ is the immersed image of two cylinders $C$ and $C'$, both propagating in $\mathbb R^3$ from the plane $z=0$ to the plane $z=1$. The intersection of $C$ and $C'$ with any plane $z=c$ (where $c\in [0,1]$) is a figure eight curve. The last detail is two explain how $C$ and $C'$ look like. So we will take as $C$ just the direct product of a vertical interval with the figure eight curve. To construct $C'$ we need to do something a bit trickier. Namely to construct it we start from $S_0$ in $z=0$ and then start to rotate it so that it the plane $z=t$ it is the figure eight curve rotated by $\pi t$. Thus, for $t=1$ it will be rotated by $\pi$. Now, one can easily check that if we rotate a figure eight by $\pi$, it changes its orientation! So if we glue $C$ with $C'$, we get the Klein bottle indeed. To finish the construction one just needs to smoothen out the described surface at $S_0$ and $S_1$. But this is not hard to do.<|endoftext|> TITLE: Is there a general theory of "compactification"? QUESTION [35 upvotes]: In various branches of mathematics one finds diverse notions of compactification, used for diverse purposes. Certainly one does not expect all instances of "compactification" to be specializations of one unique general notion. But perhaps there is something to be said from a more "taxonomic" perspective? That is, can we systematically categorize what are the chief distinctions to be drawn between different types of "compactification"? Let's look at some examples. I would love to get some more examples to add to this list. Topology: one-point compactification of locally compact spaces. Stone-Cech compactification of completely regular spaces. Bohr compactification of a topological group. Algebraic Geometry: Wonderful compactification of a $G$-space. Deligne-Mumford compactification of a moduli stack of curves. Differential Geometry: end compactification of a manifold. Mathematical Physics: Various spacetime compactifications I'm getting increasingly out of my depth as I go on, but let's list some Commonalities: One needs a notion of "compact". One identifies a class of "nice" spaces and canonical maps to "compact" spaces. Such maps should have "dense image" in an appropriate sense. One is typically interested in cases where the canonical maps are "embeddings" in an appropriate sense. Distinctions: One might try to compactify in a "maximal" or "minimal" way. One may wish to have some interpretation of the new points as "ideal points" of the original space, e.g. "points at infinity". These might be equivalence classes of some kind of "line" in the old space for example. In the case where one is compactifying some kind of moduli space, one likes to have a geometric interpretation of the new points one is adding, so that the compactification is also some kind of moduli space. Sometimes one is interested in compactifying a broad class of spaces, and may want some kind of universal property. Other times, one is compactifying one or a handful of particular space(s), and the emphasis is more on the geometric interpretation of the new points one is adding. Question: Are there further commonalities between different notions of compactification? Are there further important distinctions to be drawn? To what extent is there a general theory of "compactification"? REPLY [27 votes]: There's a distinction that I find striking but don't know how to formalize usefully or how to evaluate its importance: In algebraic geometry, moduli spaces get compactified, and this involves adding a relatively small set to the original space. Roughly speaking, the original space parametrizes some nice objects, and the compactification adds points "at infinity" parametrizing some sort of degenerations of those nice objects. Typically, the part at infinity has lower dimension than the original space. In contrast, in "my world" of ultrafilters and Stone-Cech (or similar) compactifications, the part at infinity tends to be far bigger than the original space. A countable discrete space and the real line both have Stone-Cech compactifications of cardinality $2^c$ where $c$ is the cardinal of the continuum. And it's not just a matter of the size of the part at infinity; it's also a matter of the intuitive interpretation of those extra points. It's difficult for me to imagine a non-principal ultrafilter on $\mathbb N$ as a "degenerate" natural number.<|endoftext|> TITLE: Is $h^{0,k}$ a topological invariant? QUESTION [6 upvotes]: Let $X$ and $Y$ be two smooth projective varieties over $\mathbb{C}$ such that $X(\mathbb{C})$ is homeomorphic to $Y(\mathbb{C})$. Is it true that $\dim_{\mathbb{C}} H^k(X,\mathcal{O}_X)=\dim_{\mathbb{C}} H^k(Y,\mathcal{O}_Y)$ for $k\ge 2$? (Note that for $k=1$ it follows by Hodge theory). REPLY [13 votes]: The answer is no. There are counterexamples already for surfaces, due to X. Gang and F. Campana (unpublished). The link to Campana's article is here, and the relevant result is Proposition 0.1 Il existe des surfaces projectives complexes $S$, $S_0$ simplement connexes et de type général homéomorphes, mais ayant des nombres de Hodge différents. You can look at this paper and at the references cited therein for further details.<|endoftext|> TITLE: Rational Diophantine set for the non-squares QUESTION [7 upvotes]: Related to Hilbert's Tenth problem. Is there polynomial with integer coefficients $P(a,x_1,...,x_n)$ such that $P(A,X_i)=0$ has rational solutions $X_i$ iff $A$ is not the square of integer (or as another question not the square of rational)? We think if $P$ is homogeneous and ask about integer solutions, scaling the solution might cause problems: $A^2,A X_1, A X_2, ...$ Over the integers solution is trivial via Pell equation: $$ (2+x_1^2+x_2^2+x_3^2+x_4^2)^2- a x_5^2=1 $$ REPLY [10 votes]: The set $A$ of non-squares (of rationals) is Diophantine in $\mathbb{Q}$ by [1]. The set $B:=\mathbb{Q}\smallsetminus\mathbb{Z}$ is also Diophantine by [2]. The set of non-squares of integers is equal to $A\cup B$, hence Diophantine. For a generalization of [1], see also [3]. [EDIT] The paper [1] treats arbitrary (non-)$n$-th powers, but the case of (non-)squares was proved earlier by Poonen [4]. [1] Colliot-Thélène, Jean-Louis; van Geel, Jan, Le complémentaire des puissances $n$-ièmes dans un corps de nombres est un ensemble diophantien, Compos. Math. 151, No. 10, 1965-1980 (2015). ZBL1346.14066.. [2] Koenigsmann, Jochen, Defining $\mathbb Z$ in $\mathbb Q$, Ann. Math. (2) 183, No. 1, 73-93 (2016). ZBL1390.03032.. [3] Dittmann, Philip, Irreducibility of polynomials over global fields is diophantine, Compos. Math. 154, 761-772 (2018). ZBL06861881. [4] Poonen, Bjorn, The set of nonsquares in a number field is diophantine, Math. Res. Lett. 16, No. 1, 165-170 (2009). ZBL1183.14031.<|endoftext|> TITLE: Quotients of $K3$ surfaces by finite groups QUESTION [10 upvotes]: Let $G$ be a finite subgroup of the group of automorphsims of a $K3$ surface $S$. Consider the quotient $S/G$. I am interested in the collection of such qutients: $$\{ S/G \mid S\text{ is a K3 surface, }G\text{ is a finite subgroup of }Aut(S) \}$$ Is there any classification result for the collection of quotients? What if I assume that $G$ contains an involution that acts on $H^{2,0}(S)$ as multiplication by $-1$? Is the collection finite up to deformation? REPLY [7 votes]: This question can be divided into two: what is the classification result for finite group actions on $K3$ surfaces, and what is the quotient space of $K3$ surface by a finite group action. As @abx said, there is a huge literature about these, and I'll try to list some of them. Since we have global Torelli theorem for $K3$ surfaces, the classification problem for automorphism groups can be translated into lattice theoretic problem. V.V. Nikulin (1979) first systematically studied this. Mukai (1988) classified the maximal finite groups acting symplectically on $K3$. There are $11$ such groups, and they are related to the Mathieu group. The quotient of a $K3$ by a finite symplectic group action is still a $K3$ (with ADE singularities). By looking at the configurations of those singularities, Gang Xiao (1996) gave another (more geometric) approach for the classification by Mukai. Kondo (1998) simplified Mukai's approach via Niemeier lattices. A finite group $G$ acting on a $K3$ fits into a short exact sequence: $$1\longrightarrow G_s\longrightarrow G\longrightarrow \mu_n\longrightarrow 1$$ where $G_s$ is the subgroup of symplectic automorphisms in $G$, and $\mu_n$ is a cyclic group. To understand the quotient, we first quotient by $G_s$ and obtain an ADE $K3$, then quotient by the non-symplectic action of $\mu_n$. A $K3$ surface with a non-symplectic involution is called a $2$-elementary $K3$. Such $K3$ are classified by Nikulin (1981, factor groups of groups of automorphisms of hyperbolic forms with respect to subgroups generated by 2-reflections), and the fix locus of the involution in each case is described there (Theorem 4.2.2). It turns out that there is only one case when the involution has no fixed point, then the quotient is Enriques surface; for other cases, the fixed loci are nonsingular curves, and the quotients are smooth rational surfaces (Remark 4.5.4). If a $K3$ $S$ has group action by $G$ with non-symplectic part $\mu_n$ has order at least $3$, then the quotient $S/G$ must be rational. For proof of this, I recommend G. Xiao's paper https://arxiv.org/abs/alg-geom/9512007. The action of $\mu_n$ on the trascendental lattice is free, and this implies that $\varphi(n)\le 21$. Such $n$ is at most $66$. G. Xiao proved in the above paper that $n=60$ can not be realized. All other $n$ with $\varphi(n)\le 21$ can be realized, as constructed mainly by Kondo (1992, Automorphisms of algebraic K3 surfaces which act trivially on Picard groups). Therefore, the symplectic part and non-symplectic part have only finitely many possibilities. Of course there are different ways to combine the two parts to get the group $G$, and I think a classification in general is still widely open. However, we at least know that there are only finite possibilities for the group $G$, and not hard to prove that there are only finitely many deformation types of actions of $G$ on $K3$.<|endoftext|> TITLE: Unicity of the BGG complex QUESTION [7 upvotes]: A friend and I are writing a paper that uses the BGG resolution of $L(\lambda)$ (where $\mathfrak g$ is a semisimple complex Lie algebra, $\lambda \in P^+$ is a dominant integral weight, and $L(\lambda)$ the simple finite module with highest weight $\lambda$). A point is unclear for us : one needs to choose appropriate signs for each edge $w \to w'$ in the Bruhat graph. It's probably easy but we can't see why the resulting complex is independent of the choice of the signs. Of course the maps will be different, but we expect that two different choices of signs give isomorphic complexes. There is a discussion of unicity of signs in Humphrey's book "Representations of Semisimple Lie Algebras in the BGG Category $\mathcal O$" chapter 6. It seems to me the conclusion is that the signs $\sigma(w',w)$ can be actually chosen in $\{-1,1\}$. I don't understand why it clarifies the uniqueness question, i.e why two different choices of such signs give the same complex up to isomorphism. I've probably misunderstood the argument or I don't see something obvious, but in any case I would appreciate clarification about it. Thanks in advance ! REPLY [5 votes]: Theorem 33 in the preprint [1] gives the uniqueness of BGG resolutions (= direct sums of Verma modules resolving a simple module) in category $\mathcal{O}$, both in regular and singular blocks. (However, this does not apply directly to parabolic versions of category $\mathcal{O}$). [1]: Mazorchuk-Mrđen: BGG complexes in singular blocks of category $\mathcal{O}$, https://arxiv.org/abs/1907.04121<|endoftext|> TITLE: If $\ell_0$ regularization can be done via the proximal operator, why are people still using LASSO? QUESTION [5 upvotes]: I have just learned that a general framework in constrained optimization is called "proximal gradient optimization". It is interesting that the $\ell_0$ "norm" is also associated with a proximal operator. Hence, one can apply iterative hard thresholding algorithm to get the sparse solution of the following $$\min \Vert Y-X\beta\Vert_F + \lambda \vert \beta \vert_0$$ If so, why people are still using $\ell_1$? If you can just get the result by non-convex optimization directly, why are people still using LASSO? I want to know what's the downside of the proximal gradient approach for $\ell_0$ minimization. Is it because of the non-convexity and randomness associated with? That means the initial estimator is very important. REPLY [5 votes]: The application of proximal gradient to this exact problem is considered in (equations (2) and (35) in) the Uncertainty in Artificial Intelligence (UAI) 2019 paper Fast Proximal Gradient Descent for A Class of Non-convex and Non-smooth Sparse Learning Problems, Yingzhen Yang and Jiahui Yu, with algorithm supplement at http://auai.org/uai2019/proceedings/supplements/508_supplement.pdf Unfortunately, due to the non-convexity, there is no guarantee that the algorithm will converge to the global optimum, which is what is desired. (Note: the remainder of my answer is not covered at all in the above links.) The solution obtained will depend on the algorithm starting point (initial estimator as you refer to it). An obvious starting point which can be used is the solution of the LASSO problem. If the proximal gradient algorithm is implemented with a line search guaranteeing decrease of the objective function, and the LASSO solution is used as starting solution, then the proximal gradient solution will be at least as good as, and hopefully better than, the LASSO solution when evaluated relative to the $\ell_0$ regularized objective function. So you can think of this as a post-processed LASSO solution moving it in the direction of the $\ell_0$ regularized optimal solution. I don't know how good a starting solution the LASSO solution will wind up being. If you really want to find the globally optimal solution to the $\ell_0$ regularized problem, it can be formulated as a (Convex) Mixed-Integer Quadratic Programming problem (MIQP) and solved to global optimality using standard solvers such as CPLEX, GUROBI, XPRESS, or MOSEK. However, given that the problem is NP Hard, it might take a while. I don't know whether the following would be meritorious from an overall computation time standpoint, but you could use the proximal gradient solution as a starting point for the MIQP optimizer, which might at least speed up the MIQP optimizer. LASSO is much faster and easier to solve, so I don't think it's going to be out of business for a while.<|endoftext|> TITLE: Generalized Hodge-Tate weights of an arbitrary p-adic Galois representation QUESTION [5 upvotes]: Let $V$ be a continuous representation of the absolute Galois group of $\mathbb{Q}_p$ with coefficients in $\mathbb{Q}_p$. The theory of Sen attaches to $V$ generalized Hodge-Tate weights which are elements in $\overline{\mathbb{Q}}_p$. My question is: are the generalized Hodge-Tate weight always elements of $\overline{\mathbb{Z}}_p$? In his infinite fern paper from 1997, Mazur says he does not know of an example (at least in dimension 2) where the weights are not in $\overline{\mathbb{Z}}_p$. Personally, I do not recall of ever seeing an example with weights not lying in $\mathbb{Z}_p$. Thanks! REPLY [7 votes]: If $K$ is a finite extension of $Q_p$, let $G_K = Gal(Q_p^{alg}/K)$. Let $\chi: G_K \to Z_p^\times$ be the cyclotomic character. If $n \geq 1$ and $K$ is large enough, then $\chi(G_K) \subset 1+p^{n+1} Z_p$ so that $\chi(g)^{1/p^n}$ (using the binomial series) converges and makes sense if $g \in G_K$. This gives you a character of $G_K$ with HT weight $1/p^n$. Now you can induce it to $G_{Qp}$ to get a representation with HT weights $1/p^n$. For a more general construction, see lemma 2.1.3 of G. Di Matteo's "On admissible tensor products in p-adic Hodge theory". For a result in the opposite direction, see remark 4.1.3 of L. Berger & P. Colmez "Familles de représentations de de Rham et monodromie p-adique".<|endoftext|> TITLE: Which $\infty$-groupoids correspond to simplicial abelian groups? QUESTION [7 upvotes]: Kan complexes model $\infty$-groupoids, so since every simplicial abelian group is a Kan complex, every simplicial abelian group yields an $\infty$-groupoid. What sort of $\infty$-groupoids do you get? (A natural guess would be grouplike symmetric monoidal $\infty$-groupoids, but I think that's not the correct answer roughly because $\mathbb{Z}$ is not the sphere spectrum.) I realize that this question can be taken to be tautological, but I'd like an answer that I could say specialize $2$-groupoids described by the bicategory axioms. I'd also like to understand what you get if you translate the functor taking simplicial set to the corresponding simplicial abelian group across the homotopy hypothesis. REPLY [4 votes]: The paper HHA.5(1), 2003, pp.49–52 gives a few categories equivalent to chain complexes, including cubical abelian groups with connections, and cubical $\omega$-groupoids with connections. There are also the simplicial $T$-complexes of N. Ashley, referenced [Ash88] in the bibliography of NAT-book. A simplicial $T$-complex is a simplicial complex with in each dimension a spacial kind of simplex called thin satisfying Dakin's axioms: 1) degenerate elements are thin; 2) every horn has a unique thin filler; 3) if every face but one of a thin simplex is thin then so also is the remaining face. (cf Note 222 p. 476 of the NAT book). These structures of $T$-complex do not model all homotopy types, but only the "linear" ones (no quadratic information, such as Whitehead products) but, like linear algebra, it has its uses!<|endoftext|> TITLE: When are homomorphisms between Banach algebras contractions? QUESTION [5 upvotes]: When are homomorphisms between Banach algebras contractions? I recall from my student days that there are results which show that a positive answer to the above question holds under very general conditions (on the algebras). I have tried a literature search but only turned up the fact that it is true for $\ast$-homomorphisms between $C^\ast$-algebras (Proposition 3.2.2 in Dales et al., “Banach spaces of continuous functions as dual spaces”). Does any forum member have a reference for a definitive account of this theme? I should add that I know that this not true for ANY pair of algebras, and I know that there are substantial results on classes of algebras for which it IS true. As a non-expert on complete normed algebras, I am looking for some references on the present state of the art. REPLY [11 votes]: About the only positive result that comes to mind is the fact that homomorphisms cannot increase the spectral radius, so that if the range space is a uniform algebra then homomorphisms are necessarily contractive. In my view and in my experience, at the level of generality considered by this question, the assertion/hope that "a positive answer to the above question holds under very general conditions" is not backed up by evidence. The fact that the answer is negative even for bijective homomorphisms ${\bf M}_2 \to {\bf M_2}$, with both sides carrying the natural ${\rm C}^*$-norm, is one reason to doubt that much can be said in the noncommutative setting, without extra restrictions on the nature of the homomorphism. (To get such homomorphisms, let $$ s_t= \left(\matrix{ 1 & t \\ 0 & 1 } \right)$$ and consider the automorphism of ${\bf M}_2$ given by $x \mapsto s_t^{-1} x s_t$. A simple calculation shows that the norm-1 element $$p=\left(\matrix{1 & 0 \\ 0 & 0 }\right)$$ satisfies $$s_t^{-1}ps_t = \left(\matrix{ 1 & t \\ 0 & 0} \right)$$ and the latter matrix has norm $ > |t|$.) One can also find commutative unital Banach algebras with trivial Jacobson radical (and hence for which the spectral radius does at least see every element) such that there are continuous unital endomorphisms of the algebra with norm strictly bigger than $1$. For instance, take $A_+=\ell^1({\bf Z}_+)$ with convolution product (a.k.a. the completion of the polynomial ring ${\bf C}[z]$ in the natural $\ell^1$-norm). Continuous unital endomorphisms of $A_+$ are uniquely determined by where they send the generating element $\delta_1$ (thought of as the variable $z$) and conversely every power-bounded element $a\in A_+$ definess a continuous unital endomorphism of $A_+$ which sends $\delta_1\mapsto a$. It now remains to note that there exist power-bounded elements of $A_+$ which have norm $>1$; see MR0241980 (39 #3315) Reviewed D. J. Newman, Homomorphisms of $l_+$. Amer. J. Math. 91 (1969), 37–46. https://mathscinet.ams.org/mathscinet-getitem?mr=241980 which provides the example $a= (\delta_0+\delta_1-\delta_2)/\sqrt{5}$ among others. To my mind, if a property of pairs of Banach algebras ("every continuous homomorphism from $A$ to $B$ is contractive") fails for $A=B={\bf M}_2$ or $A=B=\ell^1({\bf Z}_+)$ with convolution, it is hard to justify a hope or claim that it holds for a wide class of algebras. Being by training a Banach algebraist rather than a ${\rm C}^*$-algebraist, I just don't see why one would expect homomorphisms to be automatically contractive.<|endoftext|> TITLE: Does injectivity of $\pi_1(\partial U) \to \pi_1(M)$ imply injectivity of $\pi_1(U) \to \pi_1(M)$? QUESTION [16 upvotes]: Let $M$ be a smooth compact manifold of dimension $n$, and let $U$ be a smooth compact manifold with boundary, of the same dimension $n$, embedded in $M$. The embedding induces maps on $\pi_1$. If $\pi_1(\partial U) \to \pi_1(M)$ is injective, does this imply that $\pi_1(U) \to \pi_1(M)$ is injective? If true, can you direct me to a reference or a short proof? EDIT: I reformulated the question adding compactness of M and U to rule out the counterexample given in an answer. REPLY [4 votes]: I think I have found a somewhat elementary proof for the claim, and I would like to share it in case it is of interest for anyone, and also for some feedback. The idea is to show the injectivity of $\pi_1(U)\to\pi_1(M)$ directly, by proving that every loop in $U$ that bounds a disc in $M$, bounds a disc contained in $U$. Let $\gamma\colon S^1 \to U$ be a loop that is contractible in $M$, therefore there exists a map $u\colon D \to M$ such that $u\vert^{}_{\partial D}\equiv \gamma$, (here $D$ denotes the unit disk). Without loss of generality we may assume that $\gamma$ is in the interior of $U$ (it can be homotoped inward away from the boundary), that $\gamma$ and $u$ are smooth (by Whitney's smooth approximation theorem) and that $u\pitchfork \partial U$ (by Thom's transversality theorem). Consider the preimage $C=u^{-1}\left(\partial U\right)$, this is a compact one dimensional submanifold of $D$, hence it is a disjoint union of embedded closed curves, $C=\bigsqcup_j C_j$. We denote by $D_j$ the closed topological disc whose boundary is $C_j$. Some of the curves $C_j$ may encompass others. We call a curve $C_j$ a maximal curve if it is not encompassed by any other component of $C$. More formally, $C_j$ is maximal if there exist no other component $C_k$ and a topological disc $D_k\subset D$, such that $C_j \subset D_k$ and $\partial D_k = C_k$. Restricting $u$ to each of the maximal curves $C_j$ yields a loop $u\vert^{}_{C_j}$ contained in $\partial U$ and contractible in $M$ by $u\vert^{}_{D_j}$. Since $\pi_1(\partial U) \to \pi_1(M)$ is injective, the loop $u\vert^{}_{C_j}$ is contractible inside $\partial U$, and we thus can redefine $u$ on each of the disks that the maximal curves bound, such that $u$ now gives a contraction of $\gamma$ in $U$. Does this make sense?<|endoftext|> TITLE: Do mixing homeomorphisms on continua have positive entropy? QUESTION [6 upvotes]: I am trying to understand relations between various measures of topological complexity. I have read that expansive homeomorphisms on continua, for example, have positive entropy. But I do not know whether another property called mixing also implies positive entropy. Let $X$ be a connected compact metric space. Question. If a homeomorphism $f:X\to X$ is mixing, then does $f$ necessarily have positive entropy? A homeomorphism $f:X\to X$ is mixing if for every pair of non-empty open sets $U,V$ of $X$, there exists a positive integer $M$ such that $f^m(U)\cap V\neq\varnothing$ for all $m\geq M$. That is, if $U$ is open and non-empty, then $d_H(f^n(U),X)\to 0$ as $n\to\infty$, where $d_H$ is the Hausdorff metric. See https://en.wikipedia.org/wiki/Topological_entropy for the two equivalent definitions of the topological entropy of a map $f:X\to X$. EDIT: It occurs to me that my question was essentially asked in: Kato, Hisao, Continuum-wise expansive homeomorphisms, Can. J. Math. 45, No. 3, 576-598 (1993). ZBL0797.54047. Question 6 in that paper is my question with a weaker hypothesis (it is not difficult to show that mixing implies sensitive dependence on initial conditions). I'm not sure if anyone ever published a counterexample. REPLY [6 votes]: edit I realized there is a simpler construction that achieves the same (examples of top. mixing zero-entropy homeos on $S^3$): Instead of the bi-directional flow through cuboids, have a flow from bottom to top on the cylinder $D^2 \times [0,1]$, and have it slow down to rate $0$ on the boundary. Also, instead of all the $C_i$ playing the same role, have a single special cylinder $C$ and to get mixing, thread small cylinders from its top to its bottom, making sure at all times not to introduce periods (by refining open sets similarly as in the construction below). Open sets in the existing gadget will go around these loops and a little part of them eventually ends up on a free area on the top of $C$, and you do the threading from there. (And threading through open sets not in the gadget is easy.) The details of making the flow continuous in the limit, and the calculations showing that slow enough rate in the new cylinders implies zero entropy, are the same (and I still didn't do them). original I think there are counterexamples on all path-connected closed manifolds of dimension at least $3$, or at least I don't know what special properties I could possibly be using. For concreteness, we can think about $M = S^3$. I'll describe an $\mathbb{R}$-flow whose time-$1$ map will have the desired property. Some figures I drew on the blackboard are also attached ($U_1$ and $V_1$ in the figure should be $U_2$ and $V_2$ resp.). First, let's introduce for every $\epsilon > 0$ an flow $f_\epsilon : \mathbb{R} \times C \to C$ on the solid block $C = [0,1]^2 \times [0, R]$ (think of a large $R$). Think of $[0, R]$ as being the vertical axis, and we're staring at $C$ from the front. On the boundary of $C$, there is no movement. Inside $C$ pick two vertical lines from top to bottom, say $A$ and $B$. On the line $A$, the dynamics is trivial, i.e. all points are fixed. On the line $B$, points are moving upward at some positive rate, which should be considered very slow and parametrized by $\epsilon$ (the time-$1$ map on $B$ should behave roughly like $x \mapsto x^{1 + \epsilon}$ does on $[0,1]$). On the strip $S$ between $A$ and $B$, the dynamics is also an upward flow, whose rate is interpolated between those of $B$ and $A$ in a continuous way. Of course near the bottom and top boundaries, the movement has to slow down and stop. Outside $S$, the vertical movement quickly dies off, and turns into horizontal movement parallel to the strip $S$, so that if $A$ is on the left and $B$ on the right, the dynamics moves points say from left to right, so that the closer they are to $S$, the slower they go (and very close they also shift up). We want to introduce some horizontal movement immediately after leaving $S$, so that all points close to $S$ but not on its affine hull will eventually reach the right boundary of $C$. (On the affine hull of $S$ you have to stop movement altogether when you hit $S$.) In the blackboard photo, see the leftmost figure for a front "perspective" view of $C$ and some indications of the vector field on $S$, and see the bottommost "top view" figure for indications of the horizontal flow. The point is now that if you go into (properly inside) $C$ from the bottom, somewhere between the bottom points of $A$ and $B$, then you walk up $C$, and you can control how long it takes to reach the top from the bottom of $C$. (Of course the bottom and top have no flow because they are on the boundary, so this is indeed only true if you step properly inside $C$, but that'll change later when we start embedding copies of this in $M$.) We need some niceness properties from the flow, which we call the splotch properties, because they describe how the dynamics splotches open sets to the boundary of $C$. If you take an open set inside $C$, then we want that almost all points (all but the ones in the two-dimensional affine hull of $S$) will eventually stop moving vertically and start tending towards the right side of $C$. We want that the limit of these points on the right boundary contains a relative open set, i.e. as the open set is squeezed to the right side, the splotch you get in the $\omega$-limit always contains a square (homeomorphic copy of $[0,1]^2$). For the inverse dynamics, we want the same to happen on the left. Assuming far enough from $S$ there is no vertical movement whatsoever, this should be more or less automatic, I didn't write any formulas though. Another thing we need that if $U$ tends to the splotch $U^+$ on the right hand side and we pick a small square $D$ inside $U^+$, then some small open ball inside $U$ has its splotch ($\omega$-limit) contained in $D$. Now, having $f_{\epsilon} : \mathbb{R} \times C \to C$ with these properties, let's think of $C$ as very flexible and as carrying the dynamics of $f_{\epsilon}$. So when I stretch $C$ around the manifold $M$, the conjugate dynamics follows along. Now, enumerate a sequence of pairs of open sets $(U_i, V_i)$ in $M^2$ so that for any pair of open sets $U, V$ in $M^2$ there exists $i$ such that $U_i \subset U$ and $V_i \subset V$. We may assume $U_i$ and $V_i$ are open balls whose radius tends to $0$ very quickly. We build a sequence of flows inductively. To get the first one, called $g_1$, take a path from the center of $U_1$ to the center of $V_1$ and position an elongated $C$ called $C_1$ along this path so that its bottom is in $U_1$ and its top is in $V_1$. Now, the $f_{\epsilon_1}$-flow in $C$ (pick some tiny $\epsilon_1$) turns into a flow $g_1$ on $M$ which in $C_1$ uses the flow conjugate to $f_{\epsilon_1}$, and fixes all other points of $M$. Observe that for any large enough $m$, in the time-$1$ flow of $g_1$ we can get from $U_1$ to $V_1$ in exactly $m$ steps, by picking a suitable position between the lines $A$ and $B$. Now, we continue the construction process. We have some $U_2$ and $V_2$, and want to do the same for them. If they are disjoint from $C_1$, then this can be done in exactly the same way, and if they are not entirely inside $C_1$ they can be refined to be completely disjoint from $C_1$. So consider the case where one or both are inside $C_1$; suppose for concreteness that both are inside $C_1$. Follow the dynamics forward from $U_2$. It travels according to the flow of $C_1$ until most of it gets very close to the side of $C_1$ that corresponds to the right side of $C$, and it tends to some splotch $U_2^+$ in the $\omega$-limit. Follow it also backward to obtain an $\alpha$-limit $U_2^-$ on the left side. Then do the same for $V_2$ to get $V_2^+$ and $V_2^-$. Now, it's possible that $U_2^+$ and $V_2^+$ intersect, then refine these sets to be smaller so that they don't, using the splotch properties. Do the same in the backward direction. Now, since $U_2^+$ contains a square $D \cong [0,1]^2$ on the boundary of $C_1$, we can glue another copy $C_2$ of $C$ so that $D$ becomes the bottom square of $C_2$, and on $C_2$ use the flow $f_{\epsilon_2}$. Of course, since there is no movement on the boundary of $C_1$ nor the boundary of $C_2$, the dynamics are not in any way connected. But we can distort the dynamics near the common boundary of $C_1$ and $C_2$ slightly so that the flow drags points of $C_1$ into $C_2$, in particular we want that some points are dragged into the $S$-strip of $C_2$ and start moving upward along $C_2$. Now glue the top of $C_2$ to $V_2^-$ and distort the flow on the boundary of $C_1$ and $C_2$ as we did with the bottom. Observe that the distortion can be made arbitrarily small and made to affect an arbitrarily small area, by making the flow along $C_2$ arbitrarily slow by decreasing $\epsilon_2$ and making $C_2$ very thin. Observe that in this new flow $g_2$, in the time-$1$ map we can still get from $U_1$ to $V_1$ in $m$ steps, for any large enough $m$ (as we didn't modify the dynamics on the $C_1$-copy of $S$), but now we can also get from $U_2$ to $V_2$ in any large enough number of steps by following the dynamics to the $S$-strip of $C_2$, and picking a point on the $S$-strip with a suitable rate (the speed at the beginning and the end of these orbits cannot be controlled, but we can freely control the length of time in the middle of the orbit where we go through $C_2$). Note that in the flow $g_2$, the dynamics of every point that is not on (what corresponds to) the affine hull of $S$ in $C_1$ or $C_2$ tends to the boundary of $C_1 \cup C_2$. (We made sure $U_2^+ \cap V_2^+ = \emptyset$ to ensure no periodic behavior is introduced, and every point has singleton $\alpha$ and $\omega$-limit sets.) Now, the idea is to continue by induction, keeping roughly the following characteristic: we have in $M$ a finite set of very thin elongated blocks $C_i$. Outside the sets $C_i$, there is no movement, and on $C_i$ points move according to $f_{\epsilon_i}$, with $\epsilon_i$ tending to $0$ very fast, plus some linking behavior at their common boundaries. Every point that is not on (a set correspond to) the affine hull a copy of $S$ in some $C_i$ will eventually tend to the boundary of their union, so in particular every open set will contain an open ball which has such movement and a well-defined connected splotch as its $\omega$-limit. To get the next step of the construction, for $U_{i+1}, V_{i+1}$ again follow their forward and backward iterates until you hit splotches on some boundaries (the sets may split finitely many times as you travel between the sets $C_j$, but just refine them; againmake sure $U_{i+1}^+ \cap V_{i+1}^+ = \emptyset$ and $U_{i+1}^- \cap V_{i+1}^- = \emptyset$). Now drag a very thin copy of $C_{i+1}$ between these splotches. Observe that the existing $C_j$s don't disconnect $M$ (if you made them thin enough), so it is indeed possible to position $C_{i+1}$ this way. Finally add a bit of additional flow to connect the dynamics of $C_{i+1}$ to whereever you glued them on the boundary of $\bigcup_{j = 1}^i C_j$. Observe that no periodic points are introduced at any finite level. Now, take the pointwise limit of these flows as $i \rightarrow \infty$, say $g$. This should tend to a flow assuming $\epsilon_i$ tends to zero fast enough, and everything is regular enough, and all the added $C_i$s are small enough and so on. Just do the math and you'll see (obviously I haven't done it or I'd show it). On the other hand, if $\epsilon_i$ goes to zero very fast, clearly there is no entropy. This is true on finite levels of the construction just because every point has singleton limit sets. To get it for $g$, do the calculations, make sure not to introduce too many new partial orbits with respect to any fixed resolution $\epsilon > 0$ at finite steps. REPLY [2 votes]: The simplest counterexample would be the identity map on a one-point space.<|endoftext|> TITLE: Explicit isomorphism $\pi_{n+1}(\mathbb{RP}^n) \cong \pi_1(\mathbb{RP}^{n-1})$ QUESTION [13 upvotes]: From covering space theory we know that $\pi_{n+1}(\mathbb{RP}^n) \cong \pi_{n+1}(\mathbb{S}^n)$. From wikipedia I can notice that $\pi_{n+1}(\mathbb{S}^n) \cong \pi_1(\mathbb{RP}^{n-1})$.* My question is: is there an explicit isomorphism $\pi_{n+1}(\mathbb{RP}^n) \cong \pi_1(\mathbb{RP}^{n-1})$? My question is motivated by the fact that $\mathbb{RP}^n \cong \mathbb{R}^n \cup \mathbb{RP}^{n-1}$ and $I^{n+1} = I^n \times I$ with $\mathring{I^n} \cong \mathbb{R}^n$. *I "know" the standard calculation of $\pi_{n+1}(\mathbb{S}^n)$, for example via Pontryagin construction or $J$-homomorphism. I was wondering if it's possible to compute it in the way I stated. (I posted this originally on math.stackexchange) EDIT I not accepted user51223's answer yet because since the groups are isomorphic for every $n$ (and not only at 2), I find artificial to distinguish the two cases. Moreover, I was looking for an isomorphism whose proof does not involve a previous knowledge of the isomorphism type of one of the groups. To be honest, I was looking for an isomorphism valid all $n$ valid for the whole group (not only at 2) which does not involve the knowledge neither of $\pi_{n+1}(\mathbb{RP}^n)$ nor of $\pi_1(\mathbb{RP}^{n-1})$ induced by a map (optional) Does $\lambda_n: \mathbb{RP}^{n-1} \to \Omega^n\mathbb{RP}^n$ satisfies these conditions? REPLY [21 votes]: Let $n>2$. You have a map $\lambda_n:\mathbb{R}P^{n-1}\to\Omega^n S^n$ defined using reflection maps. This is the map that leads to the Kahn-Priddy theorem. This map extends to an $n$-fold loop map $$\lambda_n:\Omega^n\Sigma^n\mathbb{R}P^{n-1}\to\Omega^n S^n$$ which according to Kahn-Priddy Theorem induces an epimorphism on ${_2\pi_i}$ for $0 TITLE: Arens regularity of Banach algebras QUESTION [6 upvotes]: I was trying to learn the concept of Arens regularity of Banach algebras from T.W Palmers book -"Banach algebras and the general theory of $*$-algebras". There he have discussed the Arens regularity of common Banach algebras like $L^1(G)$, $C^*$-algebras,$M(G)$, $K(H)$ etc. Some other primary Banach algebras that comes to my mind are Schatten p-class operators and their tensor products. So my questions are- 1) Is $S_1(H)$ (algebra of trace class operators on Hilbert space) Arens regular? 2) What about Arens regularity of projective tensor products $S_{p_1}(H)\otimes_\gamma S_{p_2}(H)$? ($1\leq p_1,p_2<\infty$). These seems to be the very first objects people might have investigated for Arens regularity. Please suggest a reference(book,papers etc) where these things have been discussed or provide some hints. REPLY [5 votes]: For the question about trace-class operators, you could look at my New York Journal article There is a survey of sorts; see Theorem 5.39 for your question. For the trace-class operators in particular, I cite Dales's book, Theorem 2.6.23, where the result is attributed to Ulger. (See the notes at the end of Section 2.6 in Dales's book. There he references Palmer's work and Ulger's work: you could chase these references to see if there is a definitive first source.) For the projective tensor product question, you could look at Ulger's paper. I must admit to being a little wary of this paper, because of the errata. Assuming there is no mistake, you could combine Theorems 3.4 and 4.5 to, possibly, show that $S_{p_1} \otimes_{\gamma} S_{p_2}$ is Arens regular. To do this, you'd need to know that certain maps from $S_{p_1}$ to $S_{p_2'}$ were compact. I'm not an expert here in the Banach space geometry (the analogous result for $\ell^p$ spaces is true, and is "Pitt's Theorem").<|endoftext|> TITLE: Optimisation: $H(X_1) + H(X_2) +H(X_3) - H(X_1+X_2+X_3)$ QUESTION [7 upvotes]: Consider the following optimisation. $$\max [H(X_1) + H(X_2) +H(X_3) - H(X_1+X_2+X_3)],$$ where H denotes the Shannon entropy, + denotes addition over real numbers, and the maximum is taken over all the random variables $X_1$, $X_2$, $X_3$ that are independent Bernoulli. Question: How to prove that the maximum is achieved when $X_1$, $X_2$, and $X_3$ follow a uniform distribution? Edit (as requested): For a discrete random variable $X$ with probability mass function $p$ with support $\mathcal{X}$, $H(X) = - \sum_{x \in \mathcal{X}} p(x) \log p(x)$. REPLY [2 votes]: Too long for a comment. Note that $\bar{\bar{x}}=1-\bar{x}=1-(1-x)=x$. Similar for $y$ and $z$. Denote $u:=(x,y,z)$ then $\bar{u}=(\bar{x},\bar{y},\bar{z})$. So your problem can be rewritten as $$\max_{u\in(0,1)^3}f(u)+f(\bar{u})$$ Note that if $u^*\in\arg\max_{u\in(0,1)^3}f(u)+f(\bar{u})$ then so is $\bar{u}^*$ because $$f(\bar{u}^*)+f(\bar{\bar{u}}^*)=f(\bar{u}^*)+f(u^*)=f_{\max}$$ It suffices therefore to prove that $f(u)+f(\bar{u})$ is a strictly concave function of $u\in(0,1)^3$ since in this case if a maximum exists it would be unique. This would force $u^*=\bar{u}^*$ or equivalently $x^*=1-x^*,y^*=1-y^*,z^*=1-z^*$ implying $x^*=y^*=z^*=1/2$.<|endoftext|> TITLE: Magic $\mathbb{Z}\times\mathbb{Z}$-square QUESTION [5 upvotes]: Is there an injective map $j:\mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}$ satisfying: For every $z\in \mathbb{Z}$ we have $$\lim_{N\to \infty}\sum_{k=-N}^Nj(k,z) = 0 = \lim_{N\to \infty}\sum_{k=-N}^Nj(z,k)\text{ ?}$$ That is, for every $z\in \mathbb{Z}$ there is $N_0=N_0(z)\in \mathbb{N}$, such that for every integer $N\geq N_0$ we have $$\sum_{k=-N}^Nj(k,z) = 0 = \sum_{k=-N}^Nj(z,k).$$ Weaker variant: For $A\subseteq \mathbb{N}=\{1,2,\ldots\}$ we let $\mu^+(A) = \lim\sup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n}.$ Given an injective map $j:\mathbb{Z}\times\mathbb{Z} \to \mathbb{Z}$, and given $z_0\in\mathbb{Z}$, we say that $N\in\mathbb{N}$ is equitable with respect to $z_0$ if $$\sum_{k=-N}^Nj(k,z_0) = 0 = \sum_{k=-N}^Nj(z_0,k).$$ We denote the set of equitable integers with respect to $z_0$ by $\text{Eq}(z_0)$. We say that an injective map $j:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ is a weakly magic square if $\mu^+(\text{Eq}(z))>0$ for all $z\in\mathbb{Z}$. Is there a weakly magic square? REPLY [3 votes]: With the unusual definition of convergence, and infinite space, there's so much flexibility that it's difficult to go wrong. Here's one way to proceed. Enumerate the rows and columns arbitrarily and suppose we are currently trying to fill some row or column $r$ having previously filled $n$ rows and columns such that the desired property holds. Fill additional cells of $r$ (with distinct integers that have not been used so far) until The filled in cells of $r$ form an interval centred about $0$. The sum of the entries of these cells is $0$. Every integer from $-n$ to $n$ appears somewhere in the part of the grid filled so far. Once this is done (and it remains to check that it can be), fill in the remaining entries of $r$ by pairs $a$, $-a$ where $a$ is a power of the $(n+1)$st prime $p_{n+1}$. (1) and (3) are easy to ensure. Once they are in place the filled in cells have some sum $s$. We seek some $b$ such that $b$ and $-(b+s)$ are both unused; placing them in the cells at either end of the filled in interval brings the sum so far to $0$. To see that there exists such a $b$, let $A$ be the set of absolute values of integers used so far. $A$ contains only finitely many elements that are not powers of the first $n$ primes, so $A$ has zero density and its complement must contain an interval of length $|s|+1$. The endpoints of this interval, with the correct signs added, can be taken as $b, -(b+s)$. Continuing in this way eventually fills the entire grid using every integer exactly once.<|endoftext|> TITLE: Can I cover a compact set by balls {B} such that {2B} has bounded overlap? QUESTION [5 upvotes]: Suppose I have a compact set $K \subset B_1(0) \subset \mathbb{R}^n$. Can I always find a family of open balls $\{B_{r_j}(x_j)\}$ such that $x_j \in K$ and $B_{r_j}(x_j) \subset B_1(0)$ for each $j$; $K \subset \bigcup_j B_{r_j}(x_j)$; and The collection $\{B_{2r_j}(x_j)\}$ has bounded overlap, that is to say that there is some number $N = N(n)$ such that each point of $\bigcup_j B_{2r_j}(x_j)$ lies in at most $N$ balls from the collection. In other words, $$ \sum_j \mathbf{1}_{B_{2r_j}(x_j)} \leq N $$ The motivation behind a question like this is a fairly common situation where at each point $y \in K$ I can prove an estimate like $$ \int_{B_{r/2}(y)} f \leq C\int_{B_r(y)} g, $$ but what I really want is an estimate of the form $$ \int_{\{x\ :\ \mathrm{dist} (x,K) < r/10\}} f \leq C \int_{\{x\ :\ \mathrm{dist} (x,K) < 10r\}} g $$ The numbers 1/2 and 10 are not important, but to do this it is natural to try to find a collection of balls as described. In specific cases I have constructed this collection myself but now I wonder if there is a general lemma that I happen not to have heard of. REPLY [6 votes]: Let $r$ be less than half the distance from $K$ to the complement of $B_1(0)$. Start with a ball of radius $r$ centered at each point of $K$. By https://en.wikipedia.org/wiki/Besicovitch_covering_theorem there is a subcover of $K$ by balls that is a union of $C_n$ collections $A_i$, where each $A_i$ consists of pairwise disjoint balls of radius $r$. For each collection $A_i$ of disjoint balls, the balls of twice the radius can overlap at most with multiplicity $3^n$ because a ball of radius $3r$ around the point of overlap can contain at most $3^n$ disjoint balls of radius $r$. So overall we conclude that $N(n) \le 3^n C_n$. PS Here is a self-contained argument that avoids using the Besicovitch covering lemma: Given $r$, let $Z=\{z_j\}_{j=1}^m$ be a maximal subset of $K$ such that the open balls $\{B(z,r/2): z \in Z\}$ are pairwise disjoint. Then the open balls $\{B(z,r): z \in Z\}$ form a cover of $K$ (If $y \in K$ was not covered, that would contradict the maximality of $Z$). Finally, the collection of balls $S=\{B(z,2r): z \in Z\}$ has overlap multiplicity at most $5^n$: Indeed, if a point $w$ is in the intersection of $L$ balls from $S$, then $B(w,5r/2)$ contains $L$ disjoint balls of radius $r/2$; comparing volumes gives $L \le 5^n$.<|endoftext|> TITLE: Extending a map from $S^n\to M^n$ to a nice map from $B^{n+1}\to M^n$ QUESTION [16 upvotes]: Let $S^n$ and $B^{n+1}$ be the unit sphere and unit ball in $\mathbb{R}^{n+1}$, and let $M^n$ be a contractible space of dimension $n$. If necessary, assume that $M^n$ is a contractible simplicial $n$-complex. Let $\sigma\colon S^n\to M^n$ be a continuous map. Question Does there exist a continuous map $\beta\colon B^{n+1}\to M^n$ such that for any $b\in B^{n+1}$, the point $b\in\mathbb{R}^{n+1}$ lies in the convex hull of the preimage $\sigma^{-1}(\beta(b))$? Edit The following stronger statement is false already if $n=1$. There exists a continuous map $\beta\colon B^{n+1}\to M^n$ and an involution $f$ on $S^n$ with $\sigma=\sigma\circ f$, such that for any $b\in B^{n+1}$ there exists some $x\in S^n$ such that $\sigma(x)=\beta(b)$, and $b$ lies on the line segment between $x$ and $f(x)$ in $\mathbb{R}^{n+1}$. The figure below illustrates a counter example to this stronger claim. $Y$ is a star graph with three edges, and $\sigma$ is indicated by the dashed lines. None of the points interior to the white triangle lie between any pair of points that map to the same point on $Y$. REPLY [7 votes]: The answer is no, essentially since higher homotopy groups of spheres are nontrivial. For example, in the $n=3$ case let $M=B^3$ and let $\sigma\colon S^3 \to M$ be the Hopf map from $S^3$ to the boundary sphere $\partial M$ of $M$. Viewing $S^3$ as the unit sphere in $\mathbb{C}^2$, the group $U(1)$ of unit complex numbers acts on $S^3$ by scalar multiplication, and the preimage of each point in $\partial M$ is a great circle in $S^3$ which is an orbit under the action of $U(1)$. Now suppose $\beta\colon B^4\to M$ is any map satisfying the given conditions. Note that $\sigma^{-1}(\beta(b))$ will be empty if $\beta(b)\notin \partial M$, so it must be the case that $\beta(b)\in \partial M$ for all $b\in B^4$, i.e. $\beta$ maps $B^4$ to $S^2=\partial M$. Note that each $\sigma^{-1}(\beta(b))$ is one of the circles described above, and the associated convex hull is a disk through the origin. Now, if $b$ is any point on the sphere $S^3\subset B^4$, then $b$ lies in the convex hull of $\sigma^{-1}(\beta(b))$ if and only if $b$ lies on the circle $\sigma^{-1}(\beta(b))$, which occurs if and only if $\beta(b)=\sigma(b)$. Thus $\beta$ restricts to $\sigma$ on the boundary sphere $S^3$. But this is impossible since $\beta$ maps $B^4$ into the $2$-sphere $\partial M$ and the Hopf map is not nullhomotopic.<|endoftext|> TITLE: 'Continuity' of the étale topos QUESTION [11 upvotes]: In certain concrete situations, I can show that the small étale topos of an inverse limit of schemes is the inverse limit of the associated toposes, for example, if $X$ is a (qcqs) scheme relative over some affine scheme $\operatorname{Spec}(R)$, and $S$ is a filtered colimit of $R$-algebras where $S=\varinjlim S_i$, I can show that there is an equivalence of small étale toposes $$\operatorname{\acute{E}t}(X\times_{\operatorname{Spec}(R)} \operatorname{Spec}(S))\simeq \varprojlim \operatorname{\acute{E}t}(X\times_{\operatorname{Spec}(R)} \operatorname{Spec}(S_i)).$$ (Note: The inverse limit here is being taken in the category of toposes, not in the category of locally ringed toposes!!) I heard from a friend that there is a general statement that given an inverse system $\{X_i\}$ of schemes with affine transition maps and limit $X$, then we have an equivalence of étale toposes: $$\operatorname{\acute{E}t}(X)\simeq \varprojlim\operatorname{\acute{E}t}(X_i).$$ I looked high and low for a reference, and I couldn't find a proof of this statement. Can someone provide a reference? REPLY [2 votes]: Milne Etale Cohomology, III Lemma 1.16: Let $I$ be a small filtered category and $i\rightsquigarrow X_{i}$ a contravariant functor from $I$ to schemes over $X$. Assume that all schemes $X_{i}$ are quasi-compact and that the maps $X_{i}\leftarrow X_{j}$ are affine. Let $X_{\infty}=\varprojlim X_{i}$, and, for a sheaf $F$ on $X_{\mathrm{et}}$, let $F_{i}$ and $F_{\in fty}$ be its inverse images on $X_{i}$ and $X_{\infty}$ respectively. Then $$ \injlim H^{p}((X_{i})_{\mathrm{et}},F_{i})\overset{\simeq}{\longrightarrow}H^{p}((X_{\infty})_{\mathrm{et}},F_{\infty}). $$ The proof is highly technical. It is based on the fact [EGA. IV.17] that the category of etale schemes of finite-type over $X_{\infty}$ is the direct limit of the categories of such schemes over the $X_{i}$ and (III, 3) that etale cohomology commutes with direct limits of sheaves. (See [SGA. 4, VII, 5.8] or Artin, Grothendieck Topologies, Harvard notes, 1962, III, 3] for the details.)<|endoftext|> TITLE: Why is the billiard problem for obtuse triangles so hard? QUESTION [30 upvotes]: This is an incredibly naive question so this may be closed. Nevertheless, I have been reading about the problem asking if every obtuse triangle admits a periodic billiard path, which has been open for a very long time. As someone who has not worked on this problem, I am wondering why what (on the surface) appears to be a "simple" problem is in fact so difficult to solve. From the little I have read, it would appear that there has indeed been progress into the problem by the likes of Schwartz, Halbeisen et al., Vorobets et al., and more, however none have actually solved this problem. I find it curious that finding periodic billiard paths for acute triangles via the Fagnano billiard orbits is so natural and even simple, yet as soon as the same question is asked about right or obtuse triangles the ease in answering the question is vanquished. Would anyone here be able to explain to me why this is (I know why Fagnano orbits do not exist in obtuse/right triangles), and how we happened upon methods such as unfoldings to be the best machinery in asking questions about this problem? REPLY [20 votes]: Theorem 1.1 in Richard Swartz's paper Obtuse Triangular Billiards I: Near the (2,3,6) triangle rules out easy proofs: He shows that, for any $\epsilon>0$ and any $N>0$, there is a triangle whose angles are within $\epsilon$ of $(\pi/2, \pi/3, \pi/6)$ and for which any closed path involves more than $N$ bounces. So we can't write down some finite list of path types that cover all obtuse triangles.<|endoftext|> TITLE: Extension of $FP_{n}$ group QUESTION [5 upvotes]: I am reading a paper (Finitely presented residually free groups by Bridson, Howie, Miller III, and Short, Theorem 5.2) where they write the following: Since $S_{0}$ is a group of type $FP_{n}(\mathbb{Q})$ and there is a series $$S_{0}\triangleleft S_{1}\triangleleft \cdots \triangleleft S_{l}=T$$ where each $S_{i+1}/S_{i}$ is finite or infinite cyclic, then by the obvious induction, $T$ is of type $FP_{n}(\mathbb{Q})$. I understand that if $S_{i+1}/S_{i}$ is finite and $S_{i}$ of type $FP_{n}(\mathbb{Q})$, then so is $S_{i+1}$, because that property is inherited in finite extensions. Nevertheless, I really can't prove the same when $S_{i+1}/S_{i}$ is infinite cyclic. Why is this true? Maybe it does not always hold, but in my case yes: I am working in direct products of limit groups, so $S_{0}$ is a full subdirect product of limit groups, $S_{0}\leq \Gamma_{1}\times \cdots \times \Gamma_{n}$. REPLY [7 votes]: If $0\to A \to B \to C\to 0$ is a short exact sequence of groups and $A$, $C$ are of type $FP_{n}$, then so is $B$. Reference: Proposition 2.2 in Homological finiteness properties of fibre products by Kochloukova and Ferreira Lima (arXiv link).<|endoftext|> TITLE: Is there an integral fusion ring which is not of Frobenius type? QUESTION [6 upvotes]: Combinatorially, a fusion ring $\mathcal{F}$ is nothing but a finite set $B=\{b_1, \dots, b_r\}$ (generating the $\mathbb{Z}$-module $\mathbb{Z} B$) together with fusion rules: $$ b_i \cdot b_j = \sum_{k=1}^r n_{i,j}^k b_k$$ with $n_{i,j}^k \in \mathbb{Z}_{\ge 0}$, satisfying axioms slightly augmenting the group axioms: (Associativity) $b_i \cdot (b_j \cdot b_k) = (b_i \cdot b_j) \cdot b_k $ , i.e., $\sum_s n_{ij}^sn_{sk}^t = \sum_s n_{jk}^sn_{is}^t$. (Neutral) $b_1 \cdot b_i = b_i \cdot b_1 = b_i$, i.e., $n_{1i}^j = n_{i1}^j = \delta_{ij}$. (Inverse/Adjoint) $\forall i \ \exists!j $ (denoted $i^*$) such that $n_{ij}^1>0$. In addition, $n_{i^*,k}^{1} = n_{k,i^*}^{1} = \delta_{i,k}$. Frobenius-Perron reciprocity: $n_{ij}^k = n_{i^*k}^j = n_{kj^*}^i$. It follows that: $*$ induces an antihomomorphism of algebra, providing a structure of $*$-algebra to $\mathbb{C}\mathcal{B}$, Frobenius-Perron theorem: $\exists!$ $*$-homomorphism $d:\mathbb{C}\mathcal{B} \to \mathbb{C}$ with $d(\mathcal{B}) \subset (0,\infty)$, with $\mathbb{C}\mathcal{B}$ is a finite dimensional von Neumann algebra given by $b_i^* = b_{i^*}$. The number $d(b_i)$ is called the Frobenius-Perron dimension of $b_i$, whereas $\sum_i d(b_i)^2$ is called the Frobenius-Perron of $\mathcal{F}$, noted $\mathrm{FPdim}(\mathcal{F})$. Let $[d(b_1), d(b_2), \dots , d(b_r)]$ be the type of $\mathcal{F}$. The fusion ring $\mathcal{F}$ is called: of Frobenius type if for all $i$, $\frac{\mathrm{FPdim}(\mathcal{F})}{d(b_i)}$ is an algebraic integer, integral if for all $i$ the number $d(b_i)$ is an integer, and then Frobenius type just means that $d(b_i)$ divides $\mathrm{FPdim}(\mathcal{F})$ for all $i$, commutative if for all $i,j$, $b_i \cdot b_j = b_j \cdot b_i$, i.e., $n_{i,j}^k = n_{j,i}^k$. It is a famous open problem whether the Grothendieck ring of a fusion category is of Frobenius type. False belief: George Kac proved in MR0304552 that the Grothendieck ring of $Rep(K)$ with $K$ a finite dimensional Kac algebra (i.e. Hopf $*$-algebra) is of Frobenius type. The fusion category $Rep(K)$ is unitary and integral. Now, there are many fusion rings which are not Grothendieck rings of a fusion category, so that perhaps fusion rings which are not of Frobenius type are already known. Consider the following three properties for a fusion ring: (1) integral, (2) commutative, (3) unitary (i.e. admits a unitary categorification). Consider a subset $S \subseteq \{1,2,3\}$, then: Question ($S$): Is there a fusion ring satisfying (i) for all $i \in S$, but not of Frobenius type? [it is a unified way to ask $2^3=8$ questions] REPLY [6 votes]: If $3 \not \in S$ then the answer to Question($S$) is yes. There are integral commutative fusion rings which are not of Frobenius type. Examples: Non-simple: rank $4$, FPdim $15$, type $[1,1,2,3]$, and fusion rules: $$ \begin{smallmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{smallmatrix} , \ \begin{smallmatrix} 0&1&0&0\\1&0&0&0\\0&0&1&0\\0&0&0&1 \end{smallmatrix} , \ \begin{smallmatrix} 0&0&1&0\\0&0&1&0\\1&1&1&0\\0&0&0&2 \end{smallmatrix} , \ \begin{smallmatrix} 0&0&0&1\\0&0&0&1\\0&0&0&2\\1&1&2&1 \end{smallmatrix} $$ Simple: rank $6$, FPdim $143$, type $[1,4,4,5,6,7]$, and fusion rules: $$ \begin{smallmatrix}1&0&0&0&0&0\\0&1&0&0&0&0\\0&0&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\end{smallmatrix} , \ \begin{smallmatrix}0&1&0&0&0&0\\1&0&1&1&1&0\\0&1&0&1&0&1\\0&1&1&1&0&1\\0&1&0&0&1&2\\0&0&1&1&2&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&1&0&0&0\\0&1&0&1&0&1\\1&0&2&0&0&1\\0&1&0&2&1&0\\0&0&0&1&2&1\\0&1&1&0&1&2\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&1&0&0\\0&1&1&1&0&1\\0&1&0&2&1&0\\1&1&2&1&0&1\\0&0&1&0&2&2\\0&1&0&1&2&2\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&1&0\\0&1&0&0&1&2\\0&0&0&1&2&1\\0&0&1&0&2&2\\1&1&2&2&1&1\\0&2&1&2&1&2\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&0&1\\0&0&1&1&2&1\\0&1&1&0&1&2\\0&1&0&1&2&2\\0&2&1&2&1&2\\1&1&2&2&2&2\end{smallmatrix} $$ Note that $15= 3 \times 5$ and $143 = 11 \times 13$. They admit no categorification because by MR2098028, any fusion category of Frobenius-Perron dimension $pq$ (with $p,q$ different odd primes) is group-theoretical, whereas by MR2735754, a (weakly) group theoretical fusion category is of Frobenius type. Now, four new simple integral (commutative) fusion rings not of Frobenius type and on which a unitary categorification cannot be excluded according to my current knowledge: rank $6$, FPdim $924 = 2^2 \cdot 3 \cdot 7 \cdot 11$, type $[1,7,8,12,15,21]$ and fusion rules: $$\begin{smallmatrix}1&0&0&0&0&0 \\ 0&1&0&0&0&0 \\ 0&0&1&0&0&0 \\ 0&0&0&1&0&0 \\ 0&0&0&0&1&0 \\ 0&0&0&0&0&1\end{smallmatrix} , \ \begin{smallmatrix}0&1&0&0&0&0 \\ 1&0&0&1&1&1 \\ 0&0&1&1&1&1 \\ 0&1&1&1&1&2 \\ 0&1&1&1&1&3 \\ 0&1&1&2&3&3\end{smallmatrix} , \ \begin{smallmatrix}0&0&1&0&0&0 \\ 0&0&1&1&1&1 \\ 1&1&1&1&1&1 \\ 0&1&1&2&1&2 \\ 0&1&1&1&2&3 \\ 0&1&1&2&3&4\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&1&0&0 \\ 0&1&1&1&1&2 \\ 0&1&1&2&1&2 \\ 1&1&2&1&3&3 \\ 0&1&1&3&3&4 \\ 0&2&2&3&4&6\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&1&0 \\ 0&1&1&1&1&3 \\ 0&1&1&1&2&3 \\ 0&1&1&3&3&4 \\ 1&1&2&3&4&5 \\ 0&3&3&4&5&7\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&0&1 \\ 0&1&1&2&3&3 \\ 0&1&1&2&3&4 \\ 0&2&2&3&4&6 \\ 0&3&3&4&5&7 \\ 1&3&4&6&7&10\end{smallmatrix}$$ rank $6$, FPdim $1320 = 2^3 \cdot 3 \cdot 5 \cdot 11$, type $[1,9,10,11,21,24]$ and fusion rules: $$ \begin{smallmatrix}1&0&0&0&0&0 \\ 0&1&0&0&0&0 \\ 0&0&1&0&0&0 \\ 0&0&0&1&0&0 \\ 0&0&0&0&1&0 \\ 0&0&0&0&0&1\end{smallmatrix} , \ \begin{smallmatrix}0&1&0&0&0&0 \\ 1&0&0&1&1&2 \\ 0&0&1&1&1&2 \\ 0&1&1&1&1&2 \\ 0&1&1&1&3&4 \\ 0&2&2&2&4&3\end{smallmatrix} , \ \begin{smallmatrix}0&0&1&0&0&0 \\ 0&0&1&1&1&2 \\ 1&1&0&0&2&2 \\ 0&1&0&1&2&2 \\ 0&1&2&2&3&4 \\ 0&2&2&2&4&4\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&1&0&0 \\ 0&1&1&1&1&2 \\ 0&1&0&1&2&2 \\ 1&1&1&1&2&2 \\ 0&1&2&2&4&4 \\ 0&2&2&2&4&5\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&1&0 \\ 0&1&1&1&3&4 \\ 0&1&2&2&3&4 \\ 0&1&2&2&4&4 \\ 1&3&3&4&7&8 \\ 0&4&4&4&8&9\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&0&1 \\ 0&2&2&2&4&3 \\ 0&2&2&2&4&4 \\ 0&2&2&2&4&5 \\ 0&4&4&4&8&9 \\ 1&3&4&5&9&11\end{smallmatrix} $$ rank $7$, FPdim $560 = 2^4 \cdot 5 \cdot 7$, type $[1,6,7,7,10,10,15]$ and fusion rules: $$ \begin{smallmatrix}1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&1&0&0&0&0\\0&0&0&1&0&0&0\\0&0&0&0&1&0&0\\0&0&0&0&0&1&0\\0&0&0&0&0&0&1\end{smallmatrix} ,\begin{smallmatrix}0&1&0&0&0&0&0\\1&0&0&0&1&1&1\\0&0&1&0&1&1&1\\0&0&0&1&1&1&1\\0&1&1&1&0&1&2\\0&1&1&1&1&0&2\\0&1&1&1&2&2&2\end{smallmatrix}, \begin{smallmatrix}0&0&1&0&0&0&0\\0&0&1&0&1&1&1\\1&1&0&1&1&1&1\\0&0&1&1&1&1&1\\0&1&1&1&1&1&2\\0&1&1&1&1&1&2\\0&1&1&1&2&2&3\end{smallmatrix} , \begin{smallmatrix}0&0&0&1&0&0&0\\0&0&0&1&1&1&1\\0&0&1&1&1&1&1\\1&1&1&0&1&1&1\\0&1&1&1&1&1&2\\0&1&1&1&1&1&2\\0&1&1&1&2&2&3\end{smallmatrix} , \begin{smallmatrix}0&0&0&0&1&0&0\\0&1&1&1&0&1&2\\0&1&1&1&1&1&2\\0&1&1&1&1&1&2\\0&1&1&1&2&3&2\\1&0&1&1&2&2&3\\0&2&2&2&3&2&4\end{smallmatrix} , \begin{smallmatrix}0&0&0&0&0&1&0\\0&1&1&1&1&0&2\\0&1&1&1&1&1&2\\0&1&1&1&1&1&2\\1&0&1&1&2&2&3\\0&1&1&1&3&2&2\\0&2&2&2&2&3&4\end{smallmatrix} , \begin{smallmatrix}0&0&0&0&0&0&1\\0&1&1&1&2&2&2\\0&1&1&1&2&2&3\\0&1&1&1&2&2&3\\0&2&2&2&3&2&4\\0&2&2&2&2&3&4\\1&2&3&3&4&4&6\end{smallmatrix} $$ rank $7$, FPdim $798=2 \cdot 3 \cdot 7 \cdot 19$, type $[1,7,8,9,9,9,21]$ and fusion rules: $$ \begin{smallmatrix}1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&0&1&0&0&0&0\\0&0&0&1&0&0&0\\0&0&0&0&1&0&0\\0&0&0&0&0&1&0\\0&0&0&0&0&0&1\end{smallmatrix} , \begin{smallmatrix}0&1&0&0&0&0&0\\1&0&0&1&1&1&1\\0&0&1&1&1&1&1\\0&1&1&1&1&1&1\\0&1&1&1&1&1&1\\0&1&1&1&1&1&1\\0&1&1&1&1&1&5\end{smallmatrix} , \begin{smallmatrix}0&0&1&0&0&0&0\\0&0&1&1&1&1&1\\1&1&1&1&1&1&1\\0&1&1&2&1&1&1\\0&1&1&1&2&1&1\\0&1&1&1&1&2&1\\0&1&1&1&1&1&6\end{smallmatrix} , \begin{smallmatrix}0&0&0&1&0&0&0\\0&1&1&1&1&1&1\\0&1&1&2&1&1&1\\1&1&2&1&1&2&1\\0&1&1&1&2&2&1\\0&1&1&2&2&1&1\\0&1&1&1&1&1&7\end{smallmatrix} , \begin{smallmatrix}0&0&0&0&1&0&0\\0&1&1&1&1&1&1\\0&1&1&1&2&1&1\\0&1&1&1&2&2&1\\1&1&2&2&1&1&1\\0&1&1&2&1&2&1\\0&1&1&1&1&1&7\end{smallmatrix} , \begin{smallmatrix}0&0&0&0&0&1&0\\0&1&1&1&1&1&1\\0&1&1&1&1&2&1\\0&1&1&2&2&1&1\\0&1&1&2&1&2&1\\1&1&2&1&2&1&1\\0&1&1&1&1&1&7\end{smallmatrix} , \begin{smallmatrix}0&0&0&0&0&0&1\\0&1&1&1&1&1&5\\0&1&1&1&1&1&6\\0&1&1&1&1&1&7\\0&1&1&1&1&1&7\\0&1&1&1&1&1&7\\1&5&6&7&7&7&8\end{smallmatrix} $$<|endoftext|> TITLE: discriminant of subfield of $\mathbb{Q}(\zeta_p)$ QUESTION [6 upvotes]: For a field $K\subset \mathbb{Q}(\zeta_p)$ $~$($\zeta_p$ a primitive pth root of unity, p a prime), it seems to be the case that the discriminant of $K$ is $p^{[K:\mathbb{Q}]-1}$ (according to Sage). How can I prove that/ is the proof implied by something written down somewhere? I have a feeling this is somewhat deep so I ask it here; if it's insufficiently deep I'm happy to close it and ask elsewhere. REPLY [11 votes]: This would also work for non-abelian, and even non-Galois extensions: if $L/\mathbb{Q}$ is totally ramified and tamely ramified at $p$ and $K$ is an intermediate field, then $K$ is also totally tamely ramified at $p$, so the exponent of $p$ in the discriminant of $K$ is $[K:\mathbb{Q}]-1$ (Serre, Corps Locaux, Proposition 13).<|endoftext|> TITLE: Subsequences of an orthonormal basis generating a strongly embedded subspace in $L_2(0,1)$ QUESTION [6 upvotes]: A closed subspace $M$ of $L_2(0,1)$ is said to be strongly embedded if the norms $\|\cdot\|_2$ and $\|\cdot\|_1$ are equivalent on $M$. Let $(f_n)_{n\in \mathbb N}$ be a orthonormal basis of $L_2(0,1)$. Suppose that $\limsup_{n\to\infty}\|f_n\|_1>0$. Is it possible to find a subsequence $(f_{n_k})_{k \in \mathbb N}$ which generates a strongly embedded subspace? The answer is positive for $(e^{i2\pi nt})_{n\in \mathbb Z}$ and for the Walsh functions (finite products of Rademacher functions). It is negative for the Haar system $(h_n)$ because $\lim_{n\to\infty} \|h_n\|_1=0$. REPLY [6 votes]: The answer is ``yes". Slightly more generally, if $(f_n)$ is an orthonormal sequence in $L_2:= L_2(0,1)$ whose $L_1$ norms are bounded away from zero, then there is a subsequence that spans a strongly embedded subspace. It is equivalent (by extrapolation) to get a subsequence that in the $L_p$ norm with $p:= 3/2$ is equivalent to an orthonormal basis in a Hilbert space. The upper $\ell_2$-estimate in $L_p$ for linear combinations of the $f_n$ comes for free because $(f_n)$ is orthonormal and $p<2$. For the lower estimate, use the fact that $L_p$ has an unconditional basis to pass to a subsequence of $(f_n)$ that is unconditional. By the cotype $2$ property of $L_p$ when $p<2$, linear combinations of that subsequence of $(f_n)$ has a lower $\ell_2$ estimate. EDIT Oct. 21, 2019: This question is closely related to the Maurey--Rosenthal [MR] problem whether there is a normalized weakly null sequence in $L_1$ that has no unconditionally basic subsequence. An example was given in [JMS]. It is natural to ask whether such a sequence can be bounded in $L_p$. The simple answer above says ``no" if $p\ge 2$. However, there are examples that are bounded in $L_p$ for all $p<2$--see the conditions (7) and (8) in [JMS]. [JMS[ Johnson, William B.; Maurey, Bernard; Schechtman, Gideon Weakly null sequences in L1. J. Amer. Math. Soc. 20 (2007), no. 1, 25–36. [MR] Maurey, B.; Rosenthal, H. P. Normalized weakly null sequence with no unconditional subsequence. Studia Math. 61 (1977), no. 1, 77–98.<|endoftext|> TITLE: How do functional equations for zeta functions arise from the structure of a homology group? QUESTION [14 upvotes]: I have read in various disparate sources that certain zeta functions satisfy functional equations as a consequence of some structure on some homology group. Here is an example of a quote in this spirit, from the Wikipedia on functional equations for L-functions. There are also functional equations for the local zeta-functions, arising at a fundamental level for the (analogue of) Poincaré duality in étale cohomology Let me be clear that I do not understand this quote. Can someone elaborate on what the author of this quote is talking about here? Are there other simple examples of this that don't come from number theory? Thanks! REPLY [19 votes]: There are examples that don't come from number theory, although it's not much simpler. Specifically, the Lefschetz zeta function. Let $X$ be a compact manifold of dimension $d$ and let $f: X\to X$ be a map. Let $L(f^n)$ be the number of fixed points of $f^n$, counted with appropriate multiplicity. Then if we define $$\zeta_f(t) = e^{ \sum_{n=1}^{\infty} L(f^n) \frac{t^n}{n} }$$ by the Lefschetz fixed point formula we have $$ \zeta_f(t) =e^{ \sum_{n=1}^{\infty} L(f^n) \frac{t^n}{n} } = e^{ \sum_{n=1}^{\infty} \sum_{i=1}^{d} (-1)^i \operatorname{tr} ( f^n, H^i(X,\mathbb Q))\frac{t^n}{n} }=\prod_{i=1}^{d} \left( e^{ \sum_{n=1}^{\infty} \operatorname{tr} ( f^n, H^i(X,\mathbb Q))\frac{t^n}{n} }\right)^{(-1)^i} $$ $$\prod_{i=1}^{d}\left( \det ( 1 - t f, H^i(X,\mathbb Q)) \right)^{(-1)^i}$$ Now using the fact that the eigenvalues of $f$ on $H^i$ are $\deg f$ divided by the eigenvalues of $f$ on $H^{d-i}$, which comes from Poincare duality, by substituting each term we can prove the functional equation (if $d$ is even) $$ \zeta_f ((\deg f)^{-1} t^{-1} ) = \prod_{i=1}^{d}\left( \det ( 1 - (\deg f)^{-1} t^{-1} f, H^i(X,\mathbb Q)) \right)^{(-1)^i}= \prod_{i=1}^{d}\left( \det ( 1 - t^{-1} f^{-1} , H^{d-i} (X,\mathbb Q)) \right)^{(-1)^i}= \prod_{i=1}^{d}\left( \det ( 1 - t^{-1} f^{-1} , H^{i} (X,\mathbb Q)) \right)^{(-1)^i}$$ $$=\prod_{i=1}^{d}\left( \det ( tf - 1, H^i(X,\mathbb Q)) \right)^{(-1)^i} t^{ (-1)^{i+1} \dim H^i(X,\mathbb Q)} / \det(f,H^i(X,\mathbb Q)) =\prod_{i=1}^{d}\left( \det ( 1- tf 1, H^i(X,\mathbb Q)) \right)^{(-1)^i} t^{ -\chi(M)} C = C t^{-\chi(M)} \zeta_f(t) $$ for some constant $C$. The number theory one is proved exactly the same way.<|endoftext|> TITLE: Converse of a result of Koblitz and Ogus on algebraic products of gamma values QUESTION [8 upvotes]: Let $n$ be a positive integer, $a_1,\ldots,a_{n-1}\in\mathbb{Z}$. Suppose that for every $u\in(\mathbb{Z}/n\mathbb{Z})^\times$, $$ \tag{$\star$} \sum_{i=1}^{n-1} i a_{(ui\!\!\!\mod n)} = 0. $$ Then the product of values of the Gamma function $$ \tag{$\star\star$} \prod_{i=1}^{n-1}\Gamma\left(\frac{i}{n}\right)^{a_i} $$ is known to be an algebraic number. This is a consequence of the reflection and multiplication formulas for $\Gamma$, and the result is proved in an appendix by Koblitz and Ogus to [1]. I would like to know if the converse of this result is expected to be true. Since the converse is almost certainly open if it is true, I will ask: Are there any examples where the product $(\star\star)$ is known to be algebraic but $(\star)$ does not hold? [1] Pierre Deligne, Valeurs de fonctions $L$ et périodes d'intégrales, Proceedings of Symposia in Pure Mathematics, Vol. 33 (1989), part 2, pp. 313-346 REPLY [6 votes]: A convenient way to formulate this kind of questions is to use the language of distributions introduced by Kubert and Lang. I gave a short account in this answer to a previous MO question. Your question is equivalent to asking whether the $\Gamma$ distribution is universal. Rohrlich has conjectured that this is the case, see Lang, Cyclotomic fields I and II, Chapter 2, Appendix, p. 66. Lang has made the stronger conjecture that every polynomial relation between the values of $\Gamma$ at rational arguments is in the ideal generated by the standard relations, see $\S$3.1.5 in this survey of Michel Waldschmidt. It seems that these conjectures are wide open. According to Waldschmidt (see $\S$5.3 in the same survey), already the case $n=5$ seems open.<|endoftext|> TITLE: Which elements of $H^2(M,\mathbb{Z}/2)$ are the $w_2(E)$ for a real bundle $E$? QUESTION [20 upvotes]: Any element of $H^1(M,\mathbb{Z}/2)$ is the $w_1(E)$ of a real line bundle $E$ over $M$. I wonder how to characterize (probably using the Steenrod squares) which elements of $H^2(M,\mathbb{Z}/2)$ are the $w_2(E)$ of a real vector bundle $E$ over $M$. Considering tensor products and tensoring by line bundles, it is clear that such elements form a subgroup of $H^2(M,\mathbb{Z}/2)$. I know that any element killed by $Sq^1$ is realizable this way, since if $Sq^1 v=0$ for $v\in H^2(M,\mathbb{Z}/2)$ then $v$ is a reduction of an element $c$ in $H^2(M,\mathbb{C})$. Then one can pick a complex line bundle whose $c_1$ is $c$, which always exists... REPLY [5 votes]: The other answers consider the case where the bundle in question is orientable. It's then natural to ask if $v \in H^2(M; \mathbb{Z}/2)$ can be realised as $w_2(E)$ for some vector bundle $E$, is it necessarily realised as $w_2(E')$ for some orientable vector bundle $E'$? It turns out that the answer is yes! To see this, let $L$ denote the determinant line bundle of $E$ and note that $w_1(E) = w_1(L)$. Consider the vector bundle $E' := (E\oplus L\oplus\varepsilon^k)\otimes L$ where $k$ is chosen such that $\operatorname{rank}(E\oplus L\oplus\varepsilon^k) = 4p + 2$ for some $p$. Using the formulae for the first and second Stiefel-Whitney classes of a tensor product (see this note for example), we have \begin{align*} w_1(E') &= w_1((E\oplus L\oplus\varepsilon^k)\otimes L)\\ &= \operatorname{rank}(L)w_1(E\oplus L\oplus\varepsilon^k) + \operatorname{rank}(E\oplus L\oplus\varepsilon^k)w_1(L)\\ &= w_1(E) + w_1(L) + (4p+2)w_1(L)\\ &= 0 \end{align*} and \begin{align*} w_2(E') &= w_2((E\oplus L\oplus\varepsilon^k)\otimes L)\\ &= \binom{1}{2}w_1(E\oplus L\oplus\varepsilon^k)^2 + \binom{4p+2}{2}w_1(L)^2 + 1^2w_2(E\oplus L\oplus\varepsilon^k)\\ &\quad\, + (4p+2)^2w_2(L) + (4p+1)w_1(E\oplus L\oplus\varepsilon^k)w_1(L)\\ &= 0 + w_1(L)^2 + w_2(E\oplus L) + 0 + w_1(E\oplus L)w_1(L)\\ &= w_1(L)^2 + w_2(E) + w_1(E)w_1(L) + w_2(L) + (w_1(E) + w_1(L))w_1(L)\\ &= w_1(L)^2 + w_2(E) + w_1(L)^2 + 0 + 0\\ &= w_2(E). \end{align*} Therefore, if $E$ is a vector bundle with $w_2(E) = v$, then $E'$ is an orientable vector bundle with $w_2(E') = v$. This observation can be used to show that the set of cohomology classes which can be realised as $w_2(E)$ for some vector bundle $E$ is a subspace of $H^2(M; \mathbb{Z}/2)$. The only condition that needs to be verified is that the sum of realisable cohomology classes is again realisable. If $v_1$ and $v_2$ are realised, then by the above, we can find orientable bundles $E_1$ and $E_2$ such that $w_2(E_i) = v_i$. By taking the direct sum with a trivial bundle if necessary, we can arrange for $E_1$ and $E_2$ to have odd ranks $m_1$, $m_2$ respectively. It then follows from the same tensor product formula that $$w_2(E_1\otimes E_2) = m_2^2w_2(E_1) + m_1^2w_2(E_2) = w_2(E_1) + w_2(E_2) = v_1 + v_2,$$ so $v_1 + v_2$ is realisable.<|endoftext|> TITLE: Sheafification of loop scheme/group QUESTION [5 upvotes]: Let $X$ be a scheme over $K = k((t))$, where $k$ is a field. We define the loop scheme $LX$ to be the functor from the category of $k$-algebras to sets by $R \mapsto LX(R) := X(Spec (R((t))))$. Do we need sheafification in order to make $LX$ a sheaf? A particular interest is when $X$ is a connected linear reductive group over $K$ and fpqc site is chosen. REPLY [4 votes]: Let me give an answer beynd the case when $X$ is affine. Upshot: Minimally restricting the setup, the answer is "LX is a sheaf, so you don't need to sheafify" for all quasi-projective schemes $X/K((t))$, and with respect to a very strong topology. Suppose $K$ is has positive characteristic, and restrict your functor LX to the category ${\rm Perf}_K$ of perfect $K$-algebras. Recall the arc-topology on schemes, introduced by Bhatt--Morrow arXiv:1807.04725. Essentially, a map $X \rightarrow Y$ of qcqs schemes is an arc-cover if it's surjective and any specialization relation between points in $Y$ lifts to $X$. E.g., you could take the spectrum of the product over all possible valuation rings at all points of $Y$, thus obtaining a (huge) affine scheme, which maps to $Y$, and this would give you an arc-cover. Note that the arc-topology is even stronger than the $v$-topology of Bhatt--Scholze arXiv:1507.06490, and "much" stronger than the fpqc-topology. E.g. if $V$ is a valuation ring of rank $2$ and $\mathfrak{p}$ is its unique prime ideal of height $1$, then ${\rm Spec} V_{\mathfrak{p}} \,\dot\cup\, {\rm Spec} V/\mathfrak{p} \rightarrow {\rm Spec} V$ is an arc-cover, but not a $v$-cover (and so in particular, not an fpqc-cover), see Corollary 2.9 of arXiv:1807.04725. This said, we have the following result, see Theorem 5.1 of arXiv:2003.04399: Theorem. Suppose $X$ is a quasi-projective $K((t))$-scheme. Then LX is a sheaf for the arc-topology on ${\rm Perf}_K$. Remark 1. I would guess that this should also extend to the case that $K$ has characteristic zero, but the proof in loc.cit. uses perfectoid spaces, and so cannot literally be carried over to the case when $K$ has characterstic zero. Remark 2. This result continues to hold in a mixed characteristic setup, i.e., when $K((t))$ is replaced by an appropriate $p$-adic field with residue field $K$ (e.g. if $K = \mathbb{F}_p$, instead of $K((t))$ you could take $\mathbb{Q}_p$, or some finite totally ramified extension of it).<|endoftext|> TITLE: Convergence of the series of Legendre polynomials QUESTION [5 upvotes]: Consider the generating function of Legendre polynomials: $$\frac{1}{\sqrt{1 - 2xt + t^2}} = \sum\limits^{\infty}_{n=0} P_n(x)t^n$$ Is it true that for $0 TITLE: Can one determine the trace map for a nonsingular projective variety explicitly? QUESTION [8 upvotes]: I've never understood how one would actually go about computing a trace map associated with the canonical sheaf on a smooth projective variety, if it's even possible. Hartshorne proves that the canonical sheaf represents the relevant functor through some very non-explicit homological algebra gymnastics. The name of the map seems to suggest that I ought to be able to be able to write down some kind of matrix, hopefully in terms of homogeneous coordinates, and take its trace. Is this true (if not, why is it called a trace map)? How does one do this? (I also posted this here: https://math.stackexchange.com/questions/3400374/can-one-determine-the-trace-map-for-a-nonsingular-projective-variety-explicitly) REPLY [7 votes]: This is a good question. It's one that my colleague Joe Lipman spent a lot of time thinking about. You can look at some of his papers for a more explicit answer for computing the trace. Probably you should start with his book "Dualizing sheaves, differentials and residues on algebraic varieties." As for name, the Grothendieck trace really is a classical trace in some cases, e.g. for finite flat maps. I don't know if that's the reason it's called that, however.<|endoftext|> TITLE: Size of sets with complete double QUESTION [19 upvotes]: Let $[n]$ denote the set $\{0,1,...,n\}$. A subset $S\subseteq [n]$ is said to have complete double if $S+S=[2n]$. Let $m(n)$ be the smallest size of a subset of $[n]$ with complete double. My questions are: Question 1: has this function been studied before? Any reference? It looks related to Sidon sets, but not quite the same. Question 2: It is easy to see that $m(n)\geq 2\sqrt{n}-1$. Can we show that $m(n)\approx 2\sqrt{n}$ asymptotically? (see the comments on how to get to $2\sqrt{2n}$). I found one similar question (for $\mathbb F_2^n$ instead of integers) here but could not find further useful references. REPLY [4 votes]: Short answer: Such a set $S$ is known as a restricted additive basis. As observed in comments, the general term to search for is [finite] additive basis: a finite set $S \subset \mathbb{N}$ such that $S+S \supseteq [2n]$. Another term is postage stamp problem (not to be confused with the Frobenius postage stamp problem which is different). The extra condition that $S \subseteq [n]$ makes this the restricted postage stamp problem and the solutions are restricted additive bases. ("Restricted" means that the elements are restricted to be in $[n]$.) It is then (almost) equivalent to ask given $k$, what is the largest $n$ ... given $n$, what is the smallest $k$ ... such that there exists a restricted additive basis of $k$ elements for the interval $[2n]$. More about that "almost" at the end of this answer. Let us write $n(k)$ for the largest $n$ when $k$ is given, and $k(n)$ for the smallest $k$ when $n$ is given. Here the question asks for $k(n)$. For $n(k)$ this is OEIS A006638, where currently the largest entry is $a(47)=734$ by yours truly (2015). Some translation needed: That "47" counts only the nonzero elements, so in fact there are 48 elements; that "734" means the target interval is $[2n]=[734]$, so in our current notation it becomes $n(48)=367$, and $k(367)=48$. To the specific questions: Q1. The OEIS entry gives some pointers to literature. The oldest reference that I know is H. Rohrbach, "Ein Beitrag zur additiven Zahlentheorie", Math. Z., 42 (1937), 1-30, which studies both the unrestricted and the restricted versions. Symmetric bases have also been studied, see for example S. Mossige, "Algorithms for computing the h-range of the postage stamp problem", Math. Comp. 36 (1981), 575–582. Q2. About asymptotics: No, one cannot achieve $k(n) \sim 2\sqrt{n}$ asymptotically. Gang Yu, "Upper bounds for finite additive 2-bases", Proc. AMS 137 (2009), 11-18, shows that $$\limsup_{n \to \infty} \frac{n}{k_r(n)^2} \le 0.419822,$$ where $k_r(n)$ is the smallest size of a restricted basis for $[n]$. I believe this is currently the best lower bound for $k_r$. Translated to our notation (with target interval $[2n]$) this means the factor in front of $\sqrt n$ is asymptotically at least $2.18264$. And about the "almost": Generally $k(n)$ increases as $n$ increases, but there are some known cases where $k(n)$ suddenly steps down. One of them was also noticed in the comments above: "What the heck is going on at n=31?". Indeed we have $k(31)=14$ but $k(32)=13$; $k(51)=18$ but $k(52)=17$; and $k(57)=19$ but $k(58)=18$. See Table 5 in this 2018 paper by me and coauthors, where the corresponding problem is studied in two dimensions.<|endoftext|> TITLE: Representation of integers by principal binary quadratic forms QUESTION [5 upvotes]: Let $k > 1$ be a positive, square-free integer. Consider the quadratic form $f_k(x,y) = x^2 - ky^2$. When $k$ is composite, is it easy to determine for which proper divisors $k^\prime$ of $k$ does the equation $$\displaystyle f_k(x,y) = k^\prime$$ solvable in integers $x,y$? Here we do allow $k^\prime$ to be possibly negative. REPLY [3 votes]: As noted in an earlier answer referring to Pall's 1969 paper in the Journal of Number Theory, it is unlikely that there is a simple way to describe all the divisors of $k$ that are represented by the form $x^2-ky^2$, for $k$ a positive nonsquare integer. There are algorithms to compute these divisors for any particular value of $k$, but a general criterion is lacking. If we interpret ''divisor of $k$'' in the broadest sense possible, with negative numbers allowed as well as $1$ and $k$ themselves, then there are either two or four divisors of $k$ that are represented. These occur in pairs with the two numbers in each pair having opposite sign and product equal to $-k$. One pair is of course $(1,-k)$. Here are a few examples of the pairs that arise for small composite values of $k$: $k=6: (1,-6),(3,-2)$ $k=10: (1,-10),(-1,10)$ $k=14: (1,-14),(2,-7)$ $k=15: (1,-15)$ $k=21: (1,-21),(7,-3)$ $k=22: (1,-22),(11,-2)$ $k=26: (1,-26),(26,-1)$ $k=30: (1,-30),(6,-5)$ $k=35: (1,-35)$ The paper by Pall says there are two pairs for each $k$, but Pall is considering divisors of the discriminant $4k$, not just divisors of $k$. Thus in the case $k=15$ with discriminant $60$ the pair $(10,-6)$ would also be included by Pall since $10$ is represented by $x^2-15y^2$ when $(x,y)=(5,1)$ and $-6$ is represented when $(x,y)=(3,1)$. In these extra cases allowed by Pall the product of the two numbers in a pair is $-4k$ rather than $-k$. In addition to allowing divisors of the discriminant $4k$ instead of just divisors of $k$, Pall also restricts attention entirely to divisors that occur as first coefficients of forms $ax^2+cy^2$ or $ax^2+axy+cy^2$ equivalent to the given form $x^2-ky^2$. This makes no difference when $k$ is squarefree, but in other cases it does sometimes make a difference. For example the form $x^2-45y^2$ represents the pairs $(1,-45)$ and $(9,-5)$ but it also represents the divisor $-20$ of the discriminant $180$, with $-20$ not appearing as the leading coefficient of any form $ax^2+cy^2$ or $ax^2+axy+cy^2$ of discriminant $180$. It is quite enlightening to look at this question from the viewpoint of Conway's topographs. In these terms what Pall is considering are mirror symmetries of the topograph. Each mirror symmetry is a reflection of the topograph across a line that passes through two regions of the topograph corresponding to two divisors of the discriminant that are represented by the given form. For positive nonsquare discriminants the topograph is periodic, and if the topograph has any mirror symmetries (the principal form $x^2-ky^2$ always has mirror symmetries) then there are exactly two lines of mirror symmetry in each period, for a total of four divisors represented.<|endoftext|> TITLE: Adjusting the definition of a well-powered category to category theory with universes: size issues QUESTION [6 upvotes]: Wikipedia and Borceux (Handbook of Categorical Algebra, Part I) give the following definitions of subobjects and well-powered categories: A subobject of an object $X$ of a category $\mathsf{C}$ is an equivalence class of the equivalence relation $\equiv$ on the class of all monomorphisms with codomain $X$ where $f \equiv g$ whenever there is an isomorphism $h$ such that $g = f\circ h$. A category $\mathsf{C}$ is well-powered if, for any $X \in \mathsf{C}$, all subobjects of $X$ form a set. There are two different approaches to handling size issues with universes, although they agree on what a (Grothendieck) universe is. One, following Grothendieck, declares that, for a universe $\mathcal{U}$, a set $X$ is $\mathcal{U}$-small if it is isomorphic to an element of $\mathcal{U}$, a category $\mathsf{C}$ is $\mathcal{U}$-category if its $\mathsf{Hom}$-sets are $\mathcal{U}$-small and it is $\mathcal{U}$-small if it is a $\mathcal{U}$-category and with $\mathsf{Ob(C)}$ being $\mathcal{U}$-small. This approach depends heavily on Bourbaki set theory and the global choice operator $\tau$. For example, a covariant $\mathsf{Hom}$-functor $\mathsf{Hom}(X,-)\colon\mathsf{C}\to\mathcal{U}\text{-}\mathsf{Set}$ doesn't actually send $Y \in \mathsf{C}$ to $\mathsf{Hom_C}(X,Y)$ but merely to a set in $\mathcal{U}$ isomorphic to it. Another approach (which I'm not sure due to whom, but the book Higher Categories and Homotopical Algebra by Cisinski and the paper Homotopy Limit Functors on Model Categories and Homotopical Categories by Dwyer-Hirschhorn-Kan-Smith use this approach) is to declare that a set $X$ is $\mathcal{U}$-small (or a $\mathcal{U}$-set, the form which I will use here to avoid confusion) if it belongs to $\mathcal{U}$, a category $\mathsf{C}$ is a $\mathcal{U}$-category if objects constitute a subset of $\mathcal{U}$ and all $\mathsf{Hom}$-sets are actually $\mathcal{U}$-sets and it is $\mathcal{U}$-small if it is a $\mathcal{U}$-category whose set of objects is a $\mathcal{U}$-sets. My question regards the latter approach. Now an obvious choice for a category to be called $\mathsf{C}$ a $\mathcal{U}$-well-powered is to have a $\mathcal{U}$-set of subobjects as in the definition above. However, this turns out to be useless, since it is not true for most of the categories that we want to be well-powered (again, in the category $\mathcal{U}\text{-}\mathsf{Set}$ of $\mathcal{U}$-sets the set $\bigcup_{Y \in \mathcal{U}} X^Y$ is generally not a $\mathcal{U}$-set). Another approach is to make an exception and relax the definition of $\mathcal{U}$-"smallness", requiring the set of equivalence classes to be $\mathcal{U}$-small in the sense of the first definition. This will now exclude any categories which should be well-powered as when we demanded that the set in question actually belong to $\mathcal{U}$. However, this is not in line with the philosophy of second approach and will probably turn out to be useless in applications if we stick with it. Now I think of redefining a subobject of $X$ to be any monomorphism with codomain $X$ (some books do this, like Riehl's Category Theory in Context) and to say that a category $\mathsf{C}$ is $\mathcal{U}$-well-powered if, for any $X \in \mathsf{C}$, there exists a $\mathcal{U}$-set of monomorphisms with codomain $X$ containing precisely one monomorphism for each equivalence class of the aforementioned equivalence relation $\equiv$. This is stronger than requiring each equivalence class to be a $\mathcal{U}$-small, but it seems to work for the usual categories such as those of sets, groups, topological spaces, etc. What I'm not sure about is how useful will it be if we work with $\mathcal{U}$-sets rather than $\mathcal{U}$-small sets (in particular, if our $\mathcal{U}$-categories have $\mathcal{U}$-sets of morphisms between every two objects rather than simply $\mathcal{U}$-small sets). I thought a little about relation of these approach to the Special Adjoint Functor Theorem, and I think it should work fine. The intuition is this: even if we need to find a monomorphism in this $\mathcal{U}$-set $S$ of distinct representative subobjects with a certain property and we find an object with said property which may not be its element, then we can "replace" it with an equivalent element $S$ which should still satisfy the same property due to the $\equiv$ relation. This is vague, but I think this is what happens in the proof of a Special Adjoint Functor Theorem. However, I don't know much about well-powered categories and their uses in mathematics, so I need an advice from experts whether this approach leads to any trouble. In particular, I'd like to know if "my" definition of well-poweredness works well with theory of presentable and accessible categories (whose definitions also need to be adjusted if we use universes a-la the paper of Low: see here). REPLY [7 votes]: One simple way is to read the traditional definitions (à la Grothendieck, etc) but replacing existence conditions with chosen structure: A category $C$ is $U$-locally-small if it is equipped with a $U$-valued map $h : C_0 \times C_0 \to U$ and isomorphisms $h(x,y) \cong C(x,y)$. (Equivalently, a $U$-valued functor naturally isomorphic to the original hom-functor.) A category $C$ is $U$-well-powered if it is equipped with a map $S : C_0 \to U$, and isomorphisms $S(x) \cong \mathrm{Sub}(x)$. (Equivalently, a $U$-valued functor naturally iso to the original shbobject functor.) Both of these can be recovered just by reading the traditional definitions, but with a “$U$-small set” take to mean a set equipped with an isomorphism to some chosen element of $U$. These are a fairly minimal tweak to the traditional definitions to make them non-choicy. They are easily seen to respect equivalence of categories, and also equivalence of universes (suitably defined); in this way, they are considerably more robust than the Borceux-style definition and similar ones. While these definitions do involve a choice of extra structure, this structure is in each case unique up to canonical isomorphism. This approach works analogously completely off-the-shelf for every smallness condition I know.<|endoftext|> TITLE: Local Hölder continuity QUESTION [5 upvotes]: Assume that $f:[0,1]\to [0,1]$ is Hölder continuous with the constant $1/3$ and assume that for every $s\in [0,1]$ we have $$\limsup_{x\to s}\frac{\lvert f(x)-f(s)\rvert}{\lvert x-s\rvert^{1/2}}<\infty.$$ Does this imply that $$\sup_{x\neq s}\frac{\lvert f(x)-f(s)\rvert}{\lvert x-s\rvert^{1/2}}<\infty?$$ REPLY [8 votes]: $\newcommand{\de}{\delta}$ $\newcommand{\fl}{\lfloor1/x\rfloor}$ The answer is no. E.g., consider the function $f$ defined by the conditions that $f(0):=0$ and $$f(x)=\frac{(x-x_{n+1})^{1/2}(x_n-x)} {(x_n-x_{n+1})^{7/6}} $$ for each natural $n$ and all $x\in[x_{n+1},x_n]$, where $x_n:=1/n$, so that $n=\fl$. Details: Let \begin{equation} \de_n:=[x_{n+1},x_n],\quad h_n:=x_n-x_{n+1}=\frac1{n(n+1)}, \end{equation} \begin{equation} g(x):=g_n(x):=(x-a)^{1/2}(b-x),\quad a:=a_n:=x_{n+1},\quad b:=b_n:=x_n, \end{equation} $AB\vee CD:=\max(AB,CD)$. Hence, \begin{equation} f(x)=\frac{g_n(x)}{h_n^{7/6}}=\frac{g(x)}{h_n^{7/6}} \end{equation} for each natural $n$ and all $x\in\de_n$. Next, for any $x,y,h$ such that $a\le x\le y=x+h\le b$ we have \begin{align} |g(y)-g(x)|&=|(y-a)^{1/2}(b-x)-(x-a)^{1/2}(b-x) \\ &+(y-a)^{1/2}(b-y)-(y-a)^{1/2}(b-x)| \\ &=\big|(b-x)\big((y-a)^{1/2}-(x-a)^{1/2}\big)-(y-a)^{1/2}h\big| \\ &\le(b-x)\big((y-a)^{1/2}-(x-a)^{1/2}\big)\vee(y-a)^{1/2}h \\ &\le(b-x)h^{1/2}\vee(y-a)^{1/2}h \\ &\le(b-a)h^{1/2}\vee(b-a)^{1/2}h \\ &=(b-a)h^{1/2}=h_n(y-x)^{1/2}, \end{align} so that $f$ is locally $1/2$-Hölder continuous on $(0,1]$ and, moreover, $|f(y)-f(x)|\le h_n^{-1/6}(y-x)^{1/2}\le(y-x)^{1/3}$ and hence \begin{equation} |f(y)-f(x)|\le(y-x)^{1/3}, \tag{*} \end{equation} for any $x,y$ such that $x_{n+1}\le x\le y\le x_n$. If now $x\in\de_m$ and $y\in\de_n$ for some natural $m$ and $n$ such that $m>n$, then $x_{m+1}\le x\le x_m\le x_{n+1}\le y\le x_n$ and \begin{align} |f(y)-f(x)|&\le f(x)\vee f(y) \\ &=(f(x)-f(x_m))\vee(f(y)-f(x_{n+1})) \\ &\le(x_m-x)^{1/3}\vee(y-x_{n+1})^{1/3}\le(y-x)^{1/3}. \end{align} So, (*) holds for all $x,y$ such that $0 TITLE: On the finite simple groups with an irreducible complex representation of a given dimension QUESTION [14 upvotes]: This answer of Geoff Robinson shows that a finite simple group admits an irreducible complex representation (irrep) of dimension $3$ if and only if it is isomorphic to $A_5$ or $\mathrm{PSL}(2,7)$. Question: Is there such a classification for every dimension $d$? [Or at least, for every small dimension, let's say less than $15$?] Weaker question: Are there finitely many finite simple groups with an irrep of dimension $d$? If yes: What are the best known lower/upper bounds for the order of a finite simple group with an irrep of dimension $d$? The following table provides the number $n$ of finite simple groups $G$ of order less than $10^6$ with an irrep of a given dimension $d < 15$, together with the minimun and maximum orders. $$\begin{array}{c|c|c|c|c} d &n&min&max & G \newline \hline 3 &2&60&168&A_5, \mathrm{PSL}(2,7) \newline \hline 4 &1&60&60&A_5 \newline \hline 5 &4 &60 &25920&A_5, A_6, \mathrm{PSL}(2,11), \mathrm{PSp}(4,3) \newline \hline 6 &4 &168 &25920&\mathrm{PSL}(2,7), A_7, \mathrm{PSU}(3,3), \mathrm{PSp}(4,3) \newline \hline 7 &5 &168 &20160&\mathrm{PSL}(2,7), \mathrm{PSL}(2,8), \mathrm{PSL}(2,13), \mathrm{PSU}(3,3), A_8 \newline \hline 8 &4 &168 &181440& \mathrm{PSL}(2,7), A_6, \mathrm{PSL}(2,8), A_9 \newline \hline 9 &4 &360 &3420& A_6, \mathrm{PSL}(2,8), \mathrm{PSL}(2,17), \mathrm{PSL}(2,19) \newline \hline 10 &5 &360 &25920& A_6, \mathrm{PSL}(2,11), A_7, M_{11}, \mathrm{PSp}(4,3) \newline \hline 11 &4 &660 &95040& \mathrm{PSL}(2,11), \mathrm{PSL}(2,23), M_{11}, M_{12} \newline \hline 12 &4 &660 &62400& \mathrm{PSL}(2,11),\mathrm{PSL}(2,13), \mathrm{PSL}(3,3), \mathrm{PSU}(3,4) \newline \hline 13 &5 &1092 &62400& \mathrm{PSL}(2,13), \mathrm{PSL}(3,3), \mathrm{PSL}(2,25), \mathrm{PSL}(2,27), \mathrm{PSU}(3,4) \newline \hline 14 &6 &1092 &604800& \mathrm{PSL}(2,13), A_7, \mathrm{PSU}(3,3), A_8, \mathrm{Sz}(8), J_2 \end{array} $$ REPLY [20 votes]: I will write this an an answer, though the answer to the basic question is provided by one of the oldest results in finite group theory. There is a theorem of C. Jordan, proved in the 19th century, that there is a function $f : \mathbb{N} \to \mathbb{N}$ such that whenever $n \in \mathbb{N}$ and $G$ is a finite subgroup of ${\rm GL}(n,\mathbb{C}),$ then there is an Abelian subgroup $A \lhd G$ with $[G:A] \leq f(n)$. Hence, in particular, for any integer $d >1$, there are only finitely many finite simple groups $G$ with a complex irreducible representation of degree $d$. Explicit bounds for Jordan's theorem were given by many people over the years, such as Frobenius and Blichfeldt. The bounds given prior to the classification of finite simple groups (CFSG) were far from optimal, but, using CFSG, B. Weisfeiler outlined a proof of a close to optimal bound in sufficiently large $n$. Unfortunately, Weisfeiler disappeared before his work was published in complete form. Recently, M.J. Collins published a complete proof with the optimal bound $f(n) = (n+1)!$ for sufficiently large $n$ ( I think $n > 72$ will do, if my memory is correct). I think Collins also gives the maximal value of $f(n)$ (with the optimal choice of function $f$) for smaller values of $n$, but this is not in general an answer to the first question. I am not sure of the largest value of $n$ such that all finite irreducible simple subgroups of ${\rm GL}(n,\mathbb{C})$ are known. I believe that people like G.Malle and G.Hiss have done this up to $n =19$ - ( Later edit- see Derek Holt's answer, which indicates a much higher bound due to those authors) ( the cases $n \leq 11$ were understood before CFSG by the work of many authors). REPLY [14 votes]: This is just to add some references to Geoff Robinson's answer for classifications of irreducible representations in low dimensions. These are all for quasisimple groups i.e. perfect groups $G$ for which $G/Z(G)$ is simple. In the paper G. Hiss and G. Malle. Low-dimensional representations of quasi-simple groups. LMS J. Comput. Math. 4 2001, 22-63. Corrigenda: LMS J. Comput. Math. 5 2002, 95-126, the authors classify all such representations of degree up to $250$, not just in characteristic $0$, but in all characteristics other than the defining characteristic of finite groups of Lie type. To complement this work, in the paper F. Lübeck, Small degree representations of finite Chevalley groups in defining characteristic. LMS J. Comput. Math. 4 (2001), 135-169, the author handles the case of groups of Lie type in defining characteristic, going up to dimension at least $250$ in all cases (and higher in many case). These papers are all freely available online.<|endoftext|> TITLE: Replacing triangulated categories with something better QUESTION [27 upvotes]: Gelfand and Manin in their 1988 book on homological algebra write that the non-functoriality of cones means that "something is going wrong in the axioms of a triangulated category. Unfortunately at the moment we don't have a more satisfactory version." Is this still a fair description of the situation? REPLY [19 votes]: I would propose a different alternative. It is the theory of (Grothendieck's) derivators, at least the stable variant. It was also developed by Heller under the name "homotopy theories" and very much related to Keller's "towers of triangulated categories" and to Franke's "systems of triangulated diagram categories". Roughly, to a base triangulated category one adds all homotopy limits and colimits, essentially adjoints (that arise as Kan extensions) to the constant diagram with values in a triangulated category of "coherent diagrams". Once you have a derivator, to be stable is a property, not a structure on it. This property is reasonably easy to check in the main examples and distinguishes stable phenomena. Stability immediately yields a collection of distinguished triangles satisfying the usual axioms. Also octahedra and higher triangles are produced by this property and they behave in a right way form the homotopical point of view, implicitly satisfying the universal properties up to homotopy that defines them. A very nice exposition is the paper by Groth "Derivators, pointed derivators, and stable derivators" (Algebraic & Geometric Topology 13 (2013), 313-374) http://www.math.uni-bonn.de/~mrahn/publications/groth_derivators.pdf For some people this notion is simpler than $\infty$-categories and encompasses the work recently done with just the axioms mentioned by Peter May in his answer together with the existence of arbitrary coproducts. The idea of Grothendieck was to express the deep meaning behind the notion of homotopy. To the extent he achieved this is debatable. But, the flexibility of stable derivators for extending homotopical constructions in triangulated categories without making recourse to model categories is one of the features some people may find useful.<|endoftext|> TITLE: Non-differentiable Lipschitz functions QUESTION [20 upvotes]: As far as I understand, there are Lipschitz functions $f:\mathbb{R}\to\ell^\infty$ that are nowhere differentiable in the Frechet sense. Where can I find such an example? REPLY [24 votes]: In addition to Anthony Quas' answer, it might be worthwhile to mention the following general observations. A Banach space $X$ is said to have the Radon-Nikodym property if every Lipschitz mapping $f: \mathbb{R} \to X$ is differentiable almost everywhere. Here are a some interesting observations concerning this property (which can all be found in [1, Section 1.2]): Every reflexive Banach space has the Radon-Nikodym property [1, Corollary 1.2.7]. If $X$ is separable and has a pre-dual Banach space, then $X$ has the Radon-Nikodym property [1, Theorem 1.2.6]. If $V$ is a closed vector subspace of $X$ and $X$ has the Radon-Nikodym property, then so has $V$ (this is obvious, of course). The space $c_0$ does not have the Radon-Nikodym property [1, Proposition 1.2.9]. (The counterexample given there is actually the same as in Anthony Quas' answer). It follows in particular that a Banach space that contains an isomorophic copy of $c_0$ as a closed subspace, does not have the Radon-Nikodym property. In particular, the space $C([0,1])$ does not have the Radon-Nikodym property. A more explicit counterexample on this space can also be found in [1, Example 1.2.8(a)]. The space $L^1(0,1)$ does not have the Radon-Nikodym property [1, Proposition 1.2.10]. Reference: [1]: Arendt, Batty, Hieber, Neubrander: "Vector-valued Laplace Transforms and Cauchy Problems" (2011). Remark The question asks for a Lipschitz function $f: \mathbb{R} \to X$ which is nowhere differentiable. As pointed out by Bill Johnson (in the comments below this answer), if a Banach space $X$ does not have the Radon-Nikodym property, then there always exists a Lipschitz continuous function $f: \mathbb{R} \to X$ that is nowhere differentiable.<|endoftext|> TITLE: Purity of Brauer group for stacks QUESTION [6 upvotes]: Let $k$ be a field, let $X$ be a smooth quasi-projective $k$-variety, let $Z\subset X$ be a closed subscheme of codimension at least $2$, it is shown that the restriction map $\mathrm{H}^2(X,\mathbb{G}_m)\to\mathrm{H}^2(X-Z,\mathbb{G}_m)$ is an isomorphism. Let $\mathcal{X}$ be a smooth Deligne-mumford stack over $k$, let $\mathcal{Z}$ be a closed substack of codimension at least $2$, does it still hold that $\mathrm{H}^2(\mathcal{X},\mathbb{G}_m)\to\mathrm{H}^2(\mathcal{X}-\mathcal{Z},\mathbb{G}_m)$? REPLY [7 votes]: The answer seems to be positive and actually at least in the context of regular (locally) noetherian Deligne--Mumford stacks. (Actually, Artin stack should also be enough as we can compute the Brauer group also as the fppf-cohomology of $\mathbb{G}_m$.) Let $p\colon X \to \mathcal{X}$ be an \'etale cover. We obtain a descent spectral sequence $$H^i(X^{\times_{\mathcal{X}}j}; \mathbb{G}_m) \Rightarrow H^{i+j}(\mathcal{X}; \mathbb{G}_m).$$ Likewise we obtain a spectral sequence $$H^i((X-p^{-1}(\mathcal{Z}))^{\times_{\mathcal{X}}j}; \mathbb{G}_m) \Rightarrow H^{i+j}(\mathcal{X}-\mathcal{Z}; \mathbb{G}_m).$$ As \'etale maps are smooth of relative dimension zero, $X - p^{-1}(\mathcal{Z})$ has still codimension at least $2$ in $X$ and likewise for all higher fiber products. Moreover, under your regularity hypotheses purity also holds for the group of units and for the Picard group. Thus, the map $\mathcal{X} - \mathcal{Z} \hookrightarrow \mathcal{X}$ induces an isomorphism of $E_2$-terms of spectral sequence for $i\leq 2$ and all $j$. Thus, it also induces an isomorphism on convergenda for $i+j\leq 2$, which is exactly the statement we want.<|endoftext|> TITLE: When is the Lie algebra of automorphisms of a geometrical structure finite-dimensional? QUESTION [7 upvotes]: Let $M$ be an $n$-dimensional smooth manifold and $\Theta$ some tensor field on $M$, so a smooth section of $TM^{\otimes r} \otimes T^*M^{\otimes s}$ for some $(r,s)$. Let $\mathfrak{g}_\Theta$ denote the Lie subalgebra of vector fields which leave $\Theta$ invariant: $$ \mathfrak{g}_\Theta = \{ X \in \mathcal{X}(M) \mid \mathcal{L}_X \Theta = 0 \} $$ If $\Theta = g$, a pseudo-riemannian metric, then it is well-known that $\mathfrak{g}_\Theta$ is finite-dimensional, with dimension bounded above by $n(n+1)/2$. On the other hand if $\Theta = \omega$, a symplectic structure, then it is again well known that $\mathfrak{g}_\Theta$ is infinite-dimensional, since it contains the hamiltonian vector fields. I expect it is not easy, given $\Theta$, to determine whether $\mathfrak{g}_\Theta$ is finite- or infinite-dimensional, but I thought I'd ask here. I'm actually interested in knowing what the generic case is. I know from examples that if we allow a metric to become degenerate, then the "isometry" Lie algebra becomes infinite-dimensional. But on the other hand, a random metric need not have any isometries at all. Hence I am not sure which way to bet. Any comments would be appreciated. REPLY [7 votes]: The general procedure for deciding the 'local' version of this question, at least in the real-analytic connected case, was certainly known to Élie Cartan and, probably known to Lie in some form. The basic idea is this: The condition that a vector field $X$ preserve $\Theta$ is a system $\Sigma_\Theta$ of linear partial differential equations on $X$. $\Sigma_\Theta$ may not be involutive, in which case, one prolongs the system to a system $\Sigma'_\Theta$ that is involutive and then one computes the Cartan characters. If the last nonzero character is $s_p$ where $p>0$, then the sheaf of local vector fields that satisfy $\Sigma_\Theta$ has infinite dimensional stalks and hence the local infinitesimal symmetries of $\Theta$ are an infinite dimensional Lie algebra. If the last nonzero character is $s_0$, then the sheaf of local vector fields that satisfy $\Sigma_\Theta$ is a finite dimensional Lie algebra of dimension $s_0$. The assumption of real-analyticity is not always needed. The global question (i.e., how many global solutions $X$ there are) is more subtle, and the answer often depends on details in the particular case, as usual. There is often a way to show that the end result is going to be finite dimensional before you actually have to compute the involutive prolongation $\Sigma'_\Theta$, and such 'practical' criteria are more often used than the full result. REPLY [7 votes]: If an analytic geometric structure is rigid in the sense of Gromov (see Gromov and d'Ambra), its symmetry vector fields form a finite dimensional Lie algebra. This is not very helpful, since we don't know too many different ways to prove rigidity, but there are lots of examples in: D'Ambra, G.(F-IHES); Gromov, M.(I-CAGL) Lectures on transformation groups: geometry and dynamics. Surveys in differential geometry (Cambridge, MA, 1990), 19–111, Lehigh Univ., Bethlehem, PA, 1991. For example, if a structure (for example, a pseudo-Riemannian metric) induces an affine connection (for example, the Levi--Civita connection), then it is rigid, and its Lie algebra of symmetry vector fields is finite dimensional. Similarly for a projective connection, or (in dimension 3 or more) for a conformal connection. You could also look at: M. Gromov, Rigid transformations groups, in Géométrie Différentielle (Paris, 1986), Hermann, 1988. A. M. Amores, Vector fields of a finite type G-structure, J. Diff. Geom., 1979. R. Quiroga-Barranco; A. Candel, Rigid and finite type geometric structures. Geom. Dedicata 106 (2004), 123–143. R. Zimmer, Ergodic theory and the automorphism group of a $G$-structure, in Group representations, ergodic theory, operator algebras, and mathematical physics (Berkeley, Calif., 1984), 1987. The paper of Quiroga-Barranco and Candel explains how to prove rigidity of many types of geometric structures, including Cartan geometries modelled on effective homogeneous spaces, so including pseudo-Riemannian geometries, affine connections, conformal connections in dimension 3 or more, and projective connections in dimensions 2 or more. REPLY [4 votes]: This article gives many examples of manifolds endowed with a structure whose group of automorphisms is finite dimensional. https://www.ams.org/journals/tran/1964-113-01/S0002-9947-1964-0164299-4/S0002-9947-1964-0164299-4.pdf In this article, it is shown that the group of automorphisms of an elliptic $G$-structure defined on a compact manifold is finite dimensional. https://projecteuclid.org/download/pdf_1/euclid.jmsj/1260541200<|endoftext|> TITLE: Can two small exotic smooth $\mathbb{R}^4$ manifolds be combined as a standard smooth $\mathbb{R}^4$? QUESTION [7 upvotes]: I just seen De Michelis and Freedman's paper Uncountably many exotic $\mathbf{R}^4$'s in standard 4-space, J. Differential Geometry 35 (1992) pp 219-254, doi:10.4310/jdg/1214447810. If I understand correctly, they said there are many small exotic $\mathbb{R}^4_\xi$ manifolds which can embedding into a standard $\mathbb{R}^4$. I can not catch all the ideas in their paper, but I would very like to know the question: Can two small exotic $\mathbb{R}^4_\xi$ and $\mathbb{R}^4_{\xi'}$ manifolds which embedding in a standard $\mathbb{R}^4$ be combined (annihilated) each other and become a global standard $\mathbb{R}^4$ manifold? Or can a global standard $\mathbb{R}^4$ manifold be continuously deformed to create two small exotic smooth $\mathbb{R}^4_\xi$ and $\mathbb{R}^4_{\xi'}$ manifolds? REPLY [4 votes]: If the "combining" you're referring to is the end-sum, it turns out that no such exotic $\mathbb{R}^4$'s exist. This was actually shown by Gompf in the appendix to "An infinite set of exotic $\mathbb{R}^4$'s" (Journal of Differential Geometry 1983). The basic idea is to suppose $R_1 \natural R_2 = \mathbb{R}^4$ and use the Eilenberg swindle to get $$R_1 = R_1 \natural (\natural_{i=1}^\infty \mathbb{R}^4) = R_1 \natural (\natural_{i=1}^\infty (R_2 \natural R_1)) = R_1 \natural R_2 \natural R_1 \natural R_2 \dots = \mathbb{R}^4 \natural \mathbb{R}^4 \natural \dots = \mathbb{R}^4 $$ [Edit: old note] I think any such exotic $\mathbb{R}^4$ would have to be standard at infinity. To see this, suppose two exotic $\mathbb{R}^4$'s $R_1$ and $R_{2}$ have end sum $R_1 \natural R_2$ diffeomorphic to the standard $\mathbb{R}^4$. The end sum $R_1 \natural R_2$ is constructed by taking smoothly properly embeddings of rays $\gamma_i: [0, \infty) \rightarrow R_i$. We then take tubular neighborhoods of ray which will be diffeomorphic to $[0,\infty) \times \mathbb{R}^3$. Delete each of these tubular neighborhoods to get $U_i \subset R_i$. We then glue $U_1$ and $U_2$ together along the new boundary to get $R_1 \natural R_2$. When Gompf introduced this in his aforementioned paper, he showed that this produces a well defined smooth manifold homeomorphic to $\mathbb{R}^4$. If $R_1 \natural R_2$ is diffeomorphic to $\mathbb{R}^4$, then there is a neighborhood of infinity $V \subset R_1 \natural R_2$ that is diffeomorphic to a neighborhood of infinity of $\mathbb{R}^4$, namely $S^3 \times \mathbb{R}$. This induces a diffeomorphism with a neighborhood of infinity of each $U_i$ with a neighborhood of infinity of $\mathbb{R^3} \times (-\infty,0]$. We can then glue the neighborhood of the rays $\gamma_i$ back in. This will induce a diffeomorphism of the neighborhoods of infinity of $R_i$ with a neighborhood of infinity of $\mathbb{R^3} \times (-\infty,0]$ glued with $[0,\infty) \times \mathbb{R}^3$. This will be the standard $\mathbb{R}^4$ and so $R_i$ is diffeomorphic to $\mathbb{R}^4$ at infinity.<|endoftext|> TITLE: How can we not know the $s$-measure of the Sierpiński triangle? QUESTION [24 upvotes]: I'm preparing a presentation that would enable high-school level students to grasp that the (self-similarity) dimension of an object needs not be an integer. The first example we look at is the Sierpiński triangle and with some effort we learn that its dimension is $$s := \log(3)/\log(2) \approx 1.585.$$ After this I thought that it would be nice to mention what the actual Hausdorff $s$-measure of the triangle is, but all I found was measure estimates for a certain class of Sierpiński carpets and some estimates of the Sierpiński triangle. I am literally shocked to learn that we apparently do not know the exact value of Hausdorff $s$-measure of the Sierpiński triangle! Especially since it's such a concrete and symmetric object. To comply to the idea of this site hosting questions instead of rants, I formulate my bafflement as follows: Why is the $s$-measure of the Sierpiński triangle and other self-similar fractals so hard to calculate? Are we missing a link to some complicated machinery or is the problem connected to some deep problem that one would expect to remain unsolved? REPLY [5 votes]: The latest I could find is Móra, Péter, Estimate of the Hausdorff measure of the Sierpinski triangle, Fractals 17, No. 2, 137-148 (2009). ZBL1178.28007. where the best values are given as $$ 0.77 \le \mathscr{H}^s(\Lambda) \le 0.81794 . $$ He also proves an upper estimate $\mathscr{H}^s(\Lambda) \le 0.819161232881177$ of which he says "everybody can check it easily". He also explains why this is harder and more technical than the reasons found in the comments here; for instance, the obvious upper bound gives only $\mathscr{H}^s(\Lambda) \le 1$.<|endoftext|> TITLE: A question regarding Kadison-Kaplansky idempotent conjecture (A nearest group element $g$ to a nontrivial self adjoint unitary element u ) QUESTION [8 upvotes]: Edit: According to answer and comments by Prof. Valette we edite the question. The Kadison Kaplansky conjecture says: Kadison-Kaplansky conjecture: If $G$ is a torsion-free discrete group then $C^*_{\mathrm{red}}(G)$ has no nontrivial projection. It is a particular case of a more general conjecture, The Baum-Connes conjecture. Obviously existence of a non-trivial projection $e$ for a $C^*$-algebra $A$ implies that $A$ has a self-adjoint unitary element $u=1-2e$ which is neither $1$ nor $-1$. Our main question is the following: Question: Let $G$ be a discrete group and $u$ be a self-adjoint unitary element of $C^*_{\mathrm{red}}(G)$ different from $1$ and $-1$. Let $g\in G \subset C^*_{\mathrm{red}}(G)$ be a nearest element to $u$ among all group elements $h\in G\subset C^*_{\mathrm{red}}(G)$. Here by "nearest" we mean the nearest according to the distance arising from the operator norm on $C^*_{\mathrm{red}}(G)$. Can one say that such a $g$ is a torsion element of $G$ which is different from the neutral element $e\in G$? Or can one find a nearest element $g$ as above and then prove that $g$ is a nontrivial torsion element? Does the main question has an affirmative answer at least in the abelian case? A refinement of the question according to comment discussion: Assume that $u$ is a non trivial self adjoint unitary element of $C^*_{\text{red}} G$ and $g\in G$ is a group element such that $g$ or $-g$ minimize the quantity $|\pm h-u|,\quad h\in G$. Does this implies that $g$ is of finite order?Can one find such a $g\neq e$? Note: By emphasising on the word "non-neutral element" one can easily check that this nearest element $\pm g$ is always non neutral for the simple case $G=\mathbb{Z}/3\mathbb{Z}$. How can this very interesting existing answer be generalized to a group which is not necessarilly abelian? REPLY [6 votes]: Let $G$ be a discrete abelian group, denote by $\epsilon$ the trivial character. Let $u\in C^*_r(G)$ be a self-adjoint unitary element such that $\epsilon(u)=1$. If $g\in G$ is such that $\|u-g\|<2$, then $g$ has finite order. Indeed, by Pontryagin duality $C^*_r(G)\simeq C(\hat{G})$, with $\hat{G}$ the Pontryagin dual of $G$. Since the Fourier transform $\hat{u}$ takes the values $\pm1$ on $\hat{G}$, and $\hat{u}$ has value 1 at the identity of $\hat{G}$, as $\hat{u}^{-1}(1)$ is clopen we see that $\hat{u}=1$ on $\hat{G}^0$, the connected component of identity of $\hat{G}$. Observe now that, denoting by $T(G)$ the torsion subgroup of $G$, the dual of $G/T(G)$ identifies canonically with $\hat{G}^0$. If $g\in G$ has infinite order, it defines a non-zero element of $G/T(G)$, so $\hat{g}$ defines a non-trivial character of $\hat{G}^0$. Hence the image of $\hat{G}^0$ is a non-trivial, closed, connected subgroup of $\mathbb{T}$, so it is $\mathbb{T}$. So there exists $\chi\in\hat{G}^0$ such that $\chi(g)=-1$, to the effect that $|(\hat{u}-\hat{g})(\chi)|=2$, hence $\|\hat{u}-\hat{g}\|=\|u-g\|=2$. This concludes the proof.<|endoftext|> TITLE: Can we characterize a periodic function by the compactness of the set of its translates? QUESTION [8 upvotes]: Given a function $f$, let us define the translates $f_t(x)=f(x-t)$. A (Bochner) almost-periodic function is a bounded continuous function on $\mathbb R^\nu$ such that the set of functions $\{f_t\vert t\in\mathbb R^\nu\}$ form a precompact set with respect to the supremum norm (a precompact set is a set whose closure is compact). This definition is taken from Appendix 1 of this paper. I was wondering if we insisted that the set is compact rather than just precompact, is this equivalent to $f$ being periodic? That is, Is it true that a bounded continuous function $f$ on $\mathbb R^\nu$ is periodic if and only if the set of functions $\{f_t\vert t\in\mathbb R^\nu\}$ form a compact set with respect to the supremum norm? REPLY [10 votes]: I would also like to focus on the case $\nu =1$ (to avoid some slightly annoying but probably trivial bookkeeping issues). Suppose that $\{f_t\}$ is compact. This shows first of all that $f$ is uniformly continuous, or else we could shift problematic points to zero, say, to obtain a sequence with no convergent subsequence. Now if $f$ wasn't periodic, then $d(s,t)=\|f_s-f_t\|$ defines a translation invariant metric on $\mathbb R$, and $(\mathbb R,d)$ is compact, by assumption. The identity map $(\mathbb R, |\cdot |)\to (\mathbb R, d)$ is continuous (since $f$ is uniformly continuous). This implies that $(\mathbb R,d)$ is still pathwise connected. Moreover, $\mathbb R$ is a topological group also with the metric $d$: for example, if $d(s_n,s), d(t_n,t)\to 0$, then \begin{align*} \|f_{s_n+t_n}-f_{s+t}\| &= \|f_{s_n} - f_{s+t-t_n}\| \le \| f_{s_n}-f_s\| + \|f_s-f_{s+t-t_n}\| \\ & = \|f_s-f_{s_n}\| + \|f_{t_n}-f_t\| \to 0 . \end{align*} A pathwise connected compact abelian metric group is a torus (see Theorem 8.46(iii) of the tome of Hoffmann and Morris for this step), but clearly this is absurd here since a torus has torsion and $\mathbb R$ doesn't. So $f$ is periodic. The other direction is of course trivial.<|endoftext|> TITLE: Does positive scalar curvature imply vanishing of the simplicial volume on a closed Riemannian manifold? QUESTION [7 upvotes]: Are there any relationship between the scalar curvature and the simplicial volume? The simplicial volume is zero (positive) on Torus (Hyperbolic manifold) and those manifolds does not admit a Riemannian metric with positive scalar curvature. What do we know about the simplicial volume of a Riemannian manifold with positive scalar curvature? REPLY [7 votes]: In a preliminary version of what would become Gromov's "A Dozen Problems, Questions and Conjectures about Positive Scalar Curvature", he writes on page 88: Neither is one able to prove (or disprove) that manifolds with positive scalar curvatures have zero simplicial volumes. Possibly, these conjectures need significant modifications to become realistic. Simplicial volume didn't make it into the published version of these notes, maybe because he thought his conjectural relationship between scalar curvature and simplicial volume was too hopeless to be worth mentioning, but at any rate this is reasonable evidence that this question was an open problem in 2017, and I haven't seen any evidence of progress since then.<|endoftext|> TITLE: Embeddability of all graphs of cardinality $\kappa$ into one graph of cardinality $\kappa$ QUESTION [7 upvotes]: Does every infinite cardinal $\kappa$ have the following property? There is a simple, undirected graph $G_0=(\kappa, E_0)$ such that every simple, undirected graph $G=(\kappa, E)$ is isomorphic to an induced subgraph of $G_0$. REPLY [10 votes]: This turns my comments (and those of Will Brian) into an answer. The summary is that the answer to the question is consistently "yes", and consistently "no". It is reasonable to treat the question one cardinal at a time: Let $P(\kappa)$ be the property that there is a graph of cardinality $\kappa$ that embeds every graph of cardinality $\kappa$ as an induced subgraph. We of course have that $P(\aleph_0)$ holds, witnessed by the countable Rado graph. In the paper "On universal graphs without instances of CH", Shelah points out that it is consistent for $P(\aleph_1)$ to fail. So it is consistent that $P(\kappa)$ fails for some infinite $\kappa$. But we do have the following general statement. Proposition. If $2^{<\kappa}=\kappa$ then $P(\kappa)$ holds. Proof. Let $T$ be the theory of the Rado graph. Assume $2^{<\kappa}=\kappa$. Then one can build a $\kappa^+$-universal model $M$ of $T$ of cardinality $\kappa$ (by 5.1.7, 5.1.8, and 5.1.16 of Chang and Keisler). Since any graph of cardinality $\kappa$ is an induced subgraph of some model of $T$ of cardinality $\kappa$, it follows that $M$ embeds any graph of cardinality $\kappa$. Remark. I also think that this can be turned into a general fact about elementary classes with the right amalgamation and joint embedding properties. If GCH holds then any infinite cardinal $\kappa$ satisfies $2^{<\kappa}=\kappa$, and so it is consistent that $P(\kappa)$ holds for all infinite $\kappa$. Remark. In the same paper mentioned above, Shelah also shows that it is consistent that $P(\aleph_1)$ holds, while also CH fails (in particular, $2^{\aleph_0}=2^{\aleph_1}=\aleph_2$). The paper "Universal structures in power $\aleph_1$" by Mekler has some related results. Thanks to Will Brian for pointing out the relevance of GCH and tracking down the references to Chang and Keisler.<|endoftext|> TITLE: Yau's conjecture on nodal sets for manifolds with boundary QUESTION [7 upvotes]: I've just read a review paper about Yau's conjecture on nodal sets of the eigenfunctions for the Laplace operator on manifolds. Briefly, if $\phi_\lambda$, $\lambda$ are an eigenpair for the Laplace-Beltrami operator on a manifold $M$, i.e. $$-\Delta\phi_\lambda = \lambda\phi,$$ then as $\lambda \to \infty$, there are constants $c$, $C$ such that $$c\sqrt\lambda \le \text{area}(\{\phi_\lambda = 0\}) \le C\sqrt\lambda.$$ I've seen several results for manifolds without boundary of varying degrees of regularity. For example, Donnelly and Feffermann proved that Yau's conjecture is true for real analytic manifolds, but as I understand, the conjecture hasn't been proven yet for $C^\infty$ manifolds. Are there extensions of the Yau conjecture to manifolds with boundary? Most of the work I've found (mainly other papers by Malinnikova and Logunov) considers only manifolds without boundary. REPLY [4 votes]: Yau's conjecture was proved for the real analytic manifolds by Donelly and Fefferman, first for manifolds without boundary, then for manifolds with boundary in the paper Nodal Sets of Eigenfunctions: Riemannian Manifolds With Boundary, Analysis, et Cetera (1990), 251--262. The main result there is stated below. Theorem 1.2 On any real analytic Riemannian manifold $M$ with boundary, the $n-1$ dimensional Hausdorff measure, $\mathcal{H}^{n-1}(N)$, of the nodal set $N$ of eigenfunction $f$, $\Delta f = -\lambda f$, (with Dirichlet or Neumann conditions on the boundary of $M$) satisfies $$c_1\sqrt{\lambda}\leq \mathcal{H}^{n-1}(N) \leq c_2 \sqrt{\lambda},$$ for some positive constants $c_1$, $c_2$. There are other papers that handle more general boundary conditions, many of which reference the above paper.<|endoftext|> TITLE: How can I improve my mathematical creativity? QUESTION [10 upvotes]: NOTE: This post has been completely rewritten, but the ideas remain the same. I've been trying to figure out the divide between "good" and "great" mathematicians, and one metric I see repeatedly is "mathematical creativity". Great mathematicians seem to be able to pull new constructs out of thin air, whereas a merely "knowledgeable" mathematician can only mull about in the immediate facts. This is where I feel I'm at personally in my mathematical career: mulling about in trivialities. I can derive maybe a few trivial implications from the assumptions, but often I need help in solving a problem, and once I see the solution's ingenuity I get the nagging feeling that I never would've been able to come up with those ideas myself. I'm learning a lot of "things", but just learning the things hasn't seemed to improve my problem solving ability. Even solving math assignments, contest problems, working through proofs again and again, memorizing lemmas using spaced repetition software, all of it seems to have only made a marginal improvement. No matter how much math I seem to cram into my head, my problem solving creativity seems to remain at just about the same level. Mathematical creativity is tricky, it's not like traditional creativity: it's paradoxically "creativity with constraint". How can one simultaneously think "outside the box", but "within the lattice" of mathematically reasonable ideas? What can be done to encourage mathematical creativity? I think this warrants deeper discussion than we're giving it credit for. There is likely no formula for creativity, otherwise it would be common knowledge by now. Some people will just be genetically more insightful than others. I think it would be unwise to end the discussion there, though. Not everyone may be genetically built for bodybuilding, not everyone may be able to achieve the Schwarzenegger physique, but this doesn't invalidate the breadth of scientific research on bodybuilding. We've extensively developed bodybuilding techniques, we have rigorous means of discussing it. I see the mind in the same way, with creativity as the muscle. Who's to say we there isn't a means of improving creativity like there's a means for building muscle, or a means for increasing our knowledge? Why shouldn't we expect that we simply lack the tools to discuss mathematical creativity? REPLY [9 votes]: It partially depends on what kind of "creative" you want to be. Gian-Carlo Rota divided mathematicians into two groups, "Problem-Solvers" and "Theorizers". The problem-solvers solve new problems and the theorizers rethink old, already-solved problems until they are so obvious that they don't really need solutions. There's some differences between the two groups, but I imagine that they are also similar in some important ways. I think one thing that stymies creativity in mathematics is the emphasis on proofs. I think that proofs are one of the things that makes mathematics great, but the hyper-focus on them actually distracts from teaching others to think great thoughts. You almost never read about how someone first started thinking about something, or how they imagined it that allowed them to write the proof in the first place. These things that take place in the imagination are rarely discussed, and, instead, mathematics focuses on the end-result, the proof, rather than the process that got there. I think the key to creativity of both types (problem-solving and theorizing) is a willingness to think about a problem in a new way, or to take a new perspective on a field. It's great to know all there is to know about a field, but sometimes this becomes a trap, such that you get used to thinking about a field in the same way as everyone else. You get sucked down that hole, and don't even realize you are there. For this reason, in Meta Math, Chaitin suggests that newcomers are sometimes the best people to advance a field. They aren't stuck in the same way of thinking as everyone else. But, I think that there is another way out of the hole - being widely read. And by this I don't just mean mathematics. Read philosophy, physics, theology, engineering, poetry. Many mathematical tools were developed by the physicists long before they were provable by the mathematicians, and they are much more likely to relay the reasoning that led to their ideas. These things can give you a new perspective - an ability to see things that others don't see. One other method I suggest for creativity is to not worry about the impossible. There are, indeed impossible things. However, most things that are "impossible" are only impossible due to certain assumptions (which are reasonable to everyone at the time, and sometimes so obvious that they are completely unstated). One can be creative by imagining a world without some combination of those assumptions, and seeing what the new world looks like. Even if removing the assumption is impossible, the practice of imagining the world like this is helpful. It helps you practice in seeing through to what alternate sets of assumptions might look like, and helps your brain to recognize situations in which changing assumptions makes the world look more correct. If you are a problem-solver, the practice of removing assumptions is extremely helpful. If you are a theorizer, I would suggest looking into computer programming. In computer programming, there is a concept called "refactoring", where you re-analyze existing code to see where there are redundancies and things can be reworked in a way that is simultaneously simpler and more powerful. These tools and techniques can often be applied to mathematics to rework existing mathematics into a form that is likewise simultaneously simpler and more powerful. See here for an example.<|endoftext|> TITLE: Delooping a fibration sequence with loopspace fiber and finite CW complexes QUESTION [5 upvotes]: The following question is somewhat similar to a previous one on MathOverflow, except that my application does not directly involve Eilenberg-MacLane spaces $K(G,n)$, and so I don't see the immediate need for $n$-simple maps as in the theory of Postnikov $k$-invariants. Suppose that $\Omega T \longrightarrow F \longrightarrow E$ is a homotopy fibration sequence, where $T$ and $F$ are connected finite CW complexes. Is there a quantification in some sort of obstruction theory for extending this sequence further to the right in the form $F \longrightarrow E \longrightarrow T$ ? This is a 1950s-era topology question. Of course, the original sequence extends further to the right as $F \longrightarrow E \longrightarrow B hAut(\Omega T)$. First, this is by Stasheff's paper for finite CW complexes as the fiber and any CW base, working with products in the full topological category [Sta63]. Later, this is by May's simplicial upgrade to infinite CW complexes as the fiber, working in the compactly generated category [9.5, 9.8][May75]. In my application, $T$ happens to have the structure of a Lie group, but this is a happy accident of its dimension, and I'm hoping for an answer that can work independent of this fact. [Sta63]: James D Stasheff, A classification theorem for fibre spaces, Topology 2, 1963 [May75]: J Peter May, Classifying spaces and fibrations, Memoirs AMS 1:155, 1975 EDIT: As suggested, I renamed the base space $B$ to $T$, in order to not confuse it with the classifying space functor. Also, I added two citations. REPLY [6 votes]: This question is addressed in the paper Ganea, T., Induced fibrations and cofibrations, Trans. Am. Math. Soc. 127, 442-459 (1967). ZBL0149.40901. A first observation is that $\Omega T\to F\to E$ extends to the right if and only if it is induced from the based path fibration $\Omega T\to PT\to T$ by a map $p:E\to T$. In Section 2, various sufficient conditions for such a fibration to be induced are given. Two sample results: Corollary 2.5: Suppose that $\pi_q(E)\neq 0$ only if $m\le q\le n + m-1$ and that $\pi_q(\Omega T) \neq 0$ only if $n\le q\le n + m-1$, where $n\ge m\ge 2$. If the Whitehead product pairing $W:\pi_n(\Omega T) \otimes \pi_m(F)\to \pi_{n + m-1}(F)$ vanishes, then $\Omega T\to F\to E$ is induced. Theorem 2.10: Suppose $\Omega T$, $F$ and $E$ all have the homotopy type of aspherical CW complexes. Then $\Omega T\to F\to E$ is induced if and only if the image of the induced map $\pi_1(\Omega T)\to \pi_1(F)$ lies in the center.<|endoftext|> TITLE: Divisibility of certain polynomials QUESTION [5 upvotes]: Consider the finite sums $$F_n(q)=\sum_{k=1}^nq^{\binom{k}2}$$ with exponents the triangular numbers $\binom{k}2$. When $n$ is odd, it appears that $F_n(q)$ does not factorize over $\mathbb{Z}[q]$. On the other hand, when $n=2m$ is even QUESTION. is it true that $F_{2m}(q)$ is divisible by the product $$\prod_{j\geq0}(1+q^{m/2^j})$$ where the product extends so long as $m/2^j$ is an integer. Examples. Here is a sample: \begin{align} (1+q^2)(1+q)\,\vert&\, F_4(q); \qquad (1+q^3)\,\,\vert\,F_6(q); \\ (1+q^4)(1+q^2)(1+q)\,\,\vert&\,F_8(q); \qquad (1+q^6)(1+q^3)\,\,\vert\,F_{12}(q). \end{align} REPLY [9 votes]: Yes. Let $n = a*2^b$ with $a$ odd. Then your question is whether $\prod_{j = 1}^{b} \left((1 + q^{\frac{n}{2^j}})\right) \, | \, F_n(q)$. Multiplying both by $q^a - 1$, the question becomes whether $F_n(q)(q^a - 1)$ is divisible by $q^{n} - 1$. Consider the multiset $\{{i(i - 1)} (\text{mod} \; 2n)\}_{i = 1}^{2n}$. I claim that this multiset is invariant under translation by $2a$, that is, the number of times the residue class $x (\text{mod} \; 2n)$ appears is the same as the number of times $x + 2a$ appears. The Chinese remainder theorem on rings allows us to analyze the multiset using the remainders modulo $a$ and $2^{b + 1}$ separately; the distribution of outcomes will be the product of the distributions on each. We first tackle $a$. As $2a \equiv 0 (\text{mod} \; a)$, the distribution of remainders modulo $a$ will be invariant under translation by $2a$. On the other hand, consider the map $i \mapsto i (i - 1): \mathbb{Z}/2^{b + 1} \rightarrow \mathbb{Z}/2^{b + 1}$. Clearly, all outcomes must be even. I claim that each even outcome appears exactly twice. Assume $i(i - 1) = j (j - 1) (\text{mod} \; 2^{b + 1})$. Then $(i - j)(i + j - 1) = 0 (\text{mod} \; 2^{b + 1})$. The two factors are of opposite parity, so one of them is odd, while the other must be divisible by $2^{b + 1}$. Therefore, each outcome appears at most twice. But by pigeonhole principle, we therefore have that each even outcome appears exactly twice - and so the distribution of remainders modulo $2^{b + 1}$ is invariant under translation by any even amount, including $2a$. Putting these together, we have that the multiset of outcomes $\{i(i - 1) (\text{mod} \; 2n)\}_{i = 1}^{2n}$ is invariant under translation by $2a$. We have that $i (i - 1) \equiv (1 - i)(1 - i - 1)$, so by restricting to $1 \leq i \leq n$, we cut the multiset exactly in half - so $\{i(i - 1) (\text{mod} \; 2n)\}_{i = 1}^{n}$ is also invariant under translation by $2a$. Then, by dividing the polynomial by $2$ (and also dividing the modulus by $2$), we get that the multiset of outcomes $\{{i \choose 2} = \frac{i(i - 1)}{2} (\text{mod} \; n)\}_{i = 1}^{n}$ is invariant under translation by $a$. This implies that $\sum_{i=1}^n q^{i \choose 2} \equiv \sum_{i=1}^n q^{{i \choose 2} + a} (\text{mod} \; q^n - 1)$, as if $x \equiv y (\text{mod} \; n)$, then $q^x \equiv q^y (\text{mod} \; q^n - 1)$. We can rewrite this as $q^n - 1 | \sum_i q^{{i \choose 2} + a} - q^{i \choose 2} = (q^a - 1) F_n(q)$, so we are done.<|endoftext|> TITLE: $d^3$ in the Atiyah-Hirzebruch spectral sequence for (twisted) $KO$ QUESTION [12 upvotes]: Cross posted from here after no responses and a bounty being placed on the question. Let $h^n(-)$ be a generalised cohomology theory. For a space $X$ there is a spectral sequence known as the Atiyah-Hirzebruch spectral sequence: $E_2^{p,q}:=H^p(X;h^q(\ast))\Rightarrow h^{p+q}(X)$. In the case of complex topological $K$-theory, i.e $KU^n(X)$, the differentials admit nice descriptions in terms of higher cohomology operations (i.e. $d^3=Sq^3$, the Steenrod square of degree $3$). For twisted $KU$ we have that $d^3(-)=Sq^3(-)+ \lambda\smile (-)$ where $H\in H^3(X;\mathbb{Z})$ is the class of the twist. This may be a naive question but is there a similar description for real topological (twisted) $K$-theory, $KO$? I am particularly interested in $d^3$. REPLY [6 votes]: Expressions for $d^2$ and $d^3$ of the twisted Atiyah-Hirzebruch spectral sequence can be found in Grady, Daniel, and Hisham Sati. "Twisted differential KO-theory." arXiv preprint arXiv:1905.09085 (2019) Proposition 18, where the twist is given by elements $(\sigma_1,\sigma_2)\in H^1(X,\mathbb Z/2)\times H^2(X,\mathbb Z/2)\cong H^*(X,\tau_{\le 1}bgl_1 KO)$: The $E_2$-page is given by twisted cohomology $H^p(X;(KO^q)_{\sigma_1})$, with $\mathbb Z/2$ acting by the sign representation on all groups The only nonvanishing $d_2$ are \begin{align*} d_2^{p,-8t}:H^p(X;\mathbb Z_{\sigma_1})\to H^{p+2}(X;\mathbb Z/2)&, x\mapsto \operatorname Sq^2(r(x)) + \sigma_1\cup Sq^1(r(x)) + \sigma_2\cup r(x)\\ d_2^{p,-8t-1}:H^p(X;\mathbb Z/2)\to H^{p+2}(X;\mathbb Z/2)&, x\mapsto \operatorname Sq^2(x) + \sigma_1\cup Sq^1(x) + \sigma_2\cup x \end{align*} The only canonically defined $d_3$ is $$ d_3^{p,-8t-2}:H^p(X;\mathbb Z/2)\to H^{p+2}(X;\mathbb Z_{\sigma_1}), x\mapsto \operatorname \beta(Sq^2(x)) + \beta(\sigma_2\cup r(x))\\ $$ Here $r: H^p(X;\mathbb Z_{\sigma_1}) \to H^p(X;\mathbb Z/2)$ and $\beta: H^p(X;\mathbb Z/2) \to H^{p+1}(X;\mathbb Z_{\sigma_1})$ are reduction mod 2 and the corresponding Bockstein. The above twists are those defined by geometry, i.e. by the inclusion $\operatorname {sLine}^\otimes\to \operatorname {Pic}(\operatorname{sVect}^\otimes)$ of superlines as tensor-invertible super vector spaces, i.e. the twists arising from ``super gerbes'' on $X$. More generally, twists for $KO$-theory are classified by $H^*(X;bgl_1 KO)\cong H^1(X,\mathbb Z/2)\times H^2(X,\mathbb Z/2)\oplus H^*(X,\tau_{> 1}bgl_1 KO)$. By degree considerations, you can see that the ones in the third group don't change the above differentials (for $d_3$, use that the multiplication $KO_2\otimes KO_{8t+2}\to KO_{8t+4}$ vanishes). The only remaining canonically defined differential is $$ d_5^{p,-8t-4}:H^p(X;\mathbb Z_{\sigma_1})\to H^{p+4}(X;\mathbb Z_{\sigma_1}) $$ I don't know anything about it besides the untwisted part, which you can find in Tyler Lawson's answer.<|endoftext|> TITLE: Are epimorphic endomorphisms of noetherian commutative rings always injective? QUESTION [9 upvotes]: This question was asked, but not answered, on Mathematics Stackexchange. [In this post "ring" means "commutative ring with one".] Let $A$ be a noetherian ring, and let $f:A\to A$ be an endomorphism which is also an epimorphism. Is $f$ necessarily injective? Eric Wofsey gave an example of an epimorphic endomorphism of a noetherian ring which is not surjective. (It is well known, and easy to prove, that surjective endomorphisms of noetherian rings are automatically bijective.) [By definition, a morphism of rings $f:A\to B$ is an epimorphism if for all pairs of morphisms $(g,h):B\rightrightarrows C$ the equality $g\circ f=h\circ f$ implies $g=h$. Surjective morphisms are epimorphic, but the converse does not always hold: for instance the inclusion $\mathbb Z\to\mathbb Q$ is an epimorphism.] For more details about epimorphisms see $\bullet$ MathOverflow thread What do epimorphisms of (commutative) rings look like?. $\bullet$ Stacks Project Section Epimorphisms of rings. $\bullet$ Samuel Seminar . REPLY [8 votes]: This is false. I will use freely that $f \colon A \to B$ is an epimorphism if and only if $B \otimes_A B \to B$ is an isomorphism; in particular this is also equivalent to $\operatorname{Spec} B \to \operatorname{Spec} A$ being a monomorphism. Example. Let $k$ be a field of characteristic $0$, and let $$R = k\left[x,\tfrac{1}{x-1},\tfrac{1}{x-2},\ldots\right].$$ Then $R$ has a self-map $x \mapsto x-1$, whose image on spectra is the complement $U$ of the origin. Note that $U \amalg \{0\} \to \operatorname{Spec} R$ is a monomorphism, so the map \begin{align*} g \colon R &\to R \times k\\ x &\mapsto (x-1,0) \end{align*} is an epimorphism. Now let $A = R \times k$, and let $f$ be the composition $A \stackrel\pi\twoheadrightarrow R \stackrel g\to A$, where $\pi$ is the natural projection. It is an epimorphism since both $\pi$ and $g$ are, and it is clearly not injective. $\square$ Picture. Here is a picture of the situation: Remark. However, the result is true if $A$ has a unique associated prime $\mathfrak p$ (i.e. $A$ is irreducible without embedded primes). Indeed, we get an ascending chain $$\mathfrak p \subseteq f^{-1}(\mathfrak p) \subseteq f^{-1}(f^{-1}(\mathfrak p)) \subseteq \ldots,$$ which stabilises by the Noetherian hypothesis. But if $\operatorname{Spec} A \to \operatorname{Spec} A$ is a monomorphism, in particular it is injective, so this forces $\mathfrak p = f^{-1}(\mathfrak p)$. Then the base change $A_{\mathfrak p} \to (f_*A)_{\mathfrak p}$ of $f \colon A \to f_*A$ along $A \to A_{\mathfrak p}$ is an epimorphism as well, hence so is the composition $A_{\mathfrak p} \to (f_*A)_{\mathfrak p} \to A_{\mathfrak p}$ since it is a further localisation. But this is an epimorphism of Artinian rings, hence surjective by this post. Since both rings are Artinian of the same length, we conclude that it's an isomorphism. Since $\mathfrak p$ is the unique associated prime, the map $A \to A_{\mathfrak p}$ is injective, which gives the result. I'm not sure what happens if $A$ is allowed to have embedded primes. It seems hard to imagine what a counterexample could look like, but I also have no argument to rule it out. Remark. Note also that there are trivial counterexamples without the Noetherian hypothesis, even if $A$ is a domain. For example, $A = k[x_0,x_1,\ldots]$ has a surjective endomorphism that is not injective.<|endoftext|> TITLE: Is there a name for this equivalence relation? QUESTION [7 upvotes]: Let $M$ be an arbitrary set and let $\mathscr{F}$ be a family of subsets of $M$. Is there a known name for the following equivalence relation or its corresponding partition? $\sim_{M,\mathscr{F}}\,=\bigl\{(x,y)\in M\times M\bigm|\forall A\in\mathscr{F}\,(x\in A\leftrightarrow y\in A)\bigr\}$. REPLY [7 votes]: $\mathscr F$-indistinguishability. In analogy with Topological indistinguishability. REPLY [5 votes]: The elements of this partition are precisely the atoms of the complete Boolean algebra generated by the family.<|endoftext|> TITLE: A hyperplane separation question QUESTION [7 upvotes]: I have a subset $K\subset X^\ast$ of the dual of a Banach space $X$. (In fact $X$ is $C^1(M)$ for some smooth compact manifold $M$.) I hope that there exists $x\in X$ such that every $k\in K$ satisfies $k(x)>0$. I know that the convex hull of $K$ does not contain the origin, and I know that $K$ is compact. Is that enough? REPLY [5 votes]: Here is a counterexample: Let $M$ be the one-dimensional unit circle, so we can identify $C^1(M)$ with \begin{align*} X = \{f \in C^1([0,1]): \, f(0) = f(1), \; f'(0) = f'(1)\}. \end{align*} For each $z \in [0,1)$ let $d_z \in X^*$ be given by $\langle d_z,f\rangle = f'(z)$ for each $f \in X$. Then the set \begin{align*} K := \{d_z: \, z \in [0,1)\} \end{align*} is weak${}^*$-compact (since $M$ is compact), but $K$ cannot be separated from $0$ by an element of $X$. Indeed, let $f \in X$. If $f$ is constant, then every functional in $K$ vanishes on $f$. If $f$ is not constant then, due to the periodic nature of $f$, there exist $z_1,z_2 \in [0,1)$ such that $f'(z_1) < 0$ and $f'(z_2) > 0$. Hence, a convex combination of $d_{z_1}$ and $d_{z_2}$ vanishes on $f$. Remark (Edited after a comment by Tom Goodwillie). The counterexample above uses that the manifold $M$ is closed. The comment by Tim Goodwillie below explains how this example can be adjusted to obtain a counterexample on $C^1([0,1])$.<|endoftext|> TITLE: Notation for the set of all injections from $A$ into $B$ QUESTION [9 upvotes]: Is there a common notation for the set of all injections from $A$ into $B$? Some set-theorists use $B^{(A)}$, e.g., A. Levy in his book Basic Set Theory. But some combinatorists use $B^{\underline{A}}$ or $(B)_A$, e.g. JMoravitz's answer in this question. Some other combinatorists also use $\mathrm{Inj}(A,B)$, e.g., M. Aigner in his book Combinatorial Theory. But I don't like a notation of this kind, since I want something similar to $B^A$ or ${}^AB$ which is commonly used to denote the set of all maps from $A$ to $B$. Any suggestions for a notation are welcome. REPLY [14 votes]: The notation suggested by cardinal equalities such as \begin{array}{l|l|l} \text{concept} & \text{notation} & \text{cardinality} \\ \hline \text{disjoint union of $A$ and $B$} & A + B & |A + B| = |A| + |B| \\ \text{Cartesian product of $A$ and $B$} & A \times B & |A \times B| = |A| \times |B| \\ \text{set of functions from $A$ to $B$, also $A \rightarrow B$} & B^A & |B^A| = |B|^{|A|} \\ \text{set of permutations of $A$, also $\text{Sym}(A)$} & A! & |A!| = |A|! \\ \text{set of $k$-element subsets of $A$} & \binom{A}{k} & \left|\binom{A}{k}\right| = \binom{|A|}{k} \\ \text{set of $k$-element partitions of $A$} & \left\{{A \atop k}\right\} & \left| \left\{{A \atop k}\right\} \right| = \left\{{|A| \atop k}\right\} \end{array} is \begin{array}{l|l|l} \text{concept} & \text{notation} & \text{cardinality} \\ \hline \text{set of injections from $A$ to $B$} & B^{\underline{A}} & |B^{\underline{A}}| = |B|^{\underline{|A|}} \end{array} because the falling factorial \begin{align*} |B|^\underline{|A|} = \frac{|B|!}{(|B| - |A|)!} \end{align*} is precisely the number of injections from $A$ to $B$.<|endoftext|> TITLE: How can I solve an orthogonal-constrained Sylvester equation? QUESTION [5 upvotes]: I am currently facing a Sylvester equation $AX+XB = C$ where $A$, $B$, $C$ are all symmetric and a special constraint here is that $X$ should be orthogonal. The Sylvester equation itself may not hold perfectly and hence the least-square solution is also ok. REPLY [2 votes]: Let me assume that $A$ and $-B$ have no common eigenvalues. Without the orthogonality constraint there is then a unique solution $Y$ of the Sylvester equation $AY+YB=C$, which you can find using known methods. Let $Y$ have the singular value decomposition $Y=U\Sigma V^T$. Then $X=UV^T$ is the orthogonal matrix that minimizes $\sum_{ij}(X_{ij}-Y_{ij})^2$. (For a proof, see here.)<|endoftext|> TITLE: Is there an algorithm for determining whether an expression involving nested radicals is rational? QUESTION [30 upvotes]: Specifically, consider expressions involving integers, addition, multiplication, division, and nth roots for any positive integer n. Is there an algorithm that can determine whether such an expression is a rational number? If the expression is a rational number, is it possible to determine which rational number (e.g. test whether an expression is 0? I've seen papers on denesting radicals, but I couldn't find anything on testing for rationality. REPLY [5 votes]: Yes, you can do so exactly, as long as you can express arbitrary rational numbers exactly. One way of doing this (though probably not the most efficient one) is to explicitly construct a field extension $E$ (as a rational vector space) through iterative adjoining of roots of equations x^k-a. (This can be done e.g. using resultants, but is hard to get to work efficiently in larger cases.) In this extension you can express the radical expression (or one interpretation of the radical expression) exactly. Then choose a basis of $E$ that has one rational basis vector (say $1$), then an element of $E$ is rational if its coefficients for all other basis vectors are zero. To take the famous Daniel Shanks example given by Esteban Crespi, a polynomial for the field $E$ is $x^8-8x^7-196x^6+1208x^5+8742x^4-43224x^3-41476x^2+227880x+8609$. (I calculated this as iterated extension.) Now, using GAP (not that it is the best tool for this, but its the one I know), we can construct the extension $E$: gap> pol:=x^8-8*x^7-196*x^6+1208*x^5+8742*x^4-43224*x^3-41476*x^2+227880*x+8609;; x^8-8*x^7-196*x^6+1208*x^5+8742*x^4-43224*x^3-41476*x^2+227880*x+8609 gap> e:=AlgebraicExtension(Rationals,pol);; gap> x:=X(e,"x");; # define variable Now we take roots of certain quadratic polynomials to get the various radical expressions (here a is a root of the degree 8 polynomial, we work in the basis $1,a,a^2,\ldots$). gap> r5:=RootsOfUPol(x^2-5)[1]; # root 5 -609/61755136*a^7+2095/30877568*a^6+133959/61755136*a^5-26101/30877568*a^4-7625627/61755136*a^3+12216013/30877568*a^2+92459893/61755136*a-67326631/30877568 gap> r29:=RootsOfUPol(x^2-29)[1]; #root 29 20639/111931184*a^7-113083/111931184*a^6-290319/111931184*a^5+13886605/111931184*a^4+206942825/111931184*a^3-340055609/111931184*a^2-1241123913/111931184*a+556675495/111931184 gap> r22:=RootsOfUPol(x^2-(22+2*r5))[1]; # root of expression starting with 22 -312563/1790898944*a^7+843909/895449472*a^6+64760293/1790898944*a^5-101635911/895449472*a^4-3089942017/1790898944*a^3+2366180495/895449472*a^2+18967544655/1790898944*a-3396381133/895449472 gap> r11:=RootsOfUPol(x^2-(11+2*r29))[1]; # root of expression starting with 11 -318289/3581797888*a^7+1674531/3581797888*a^6+66219455/3581797888*a^5-200118013/3581797888*a^4-3175703931/3581797888*a^3+4847083105/3581797888*a^2+19761660493/3581797888*a-15977491783/3581797888 gap> r55:=RootsOfUPol(x^2-(55-10*r29))[1]; # root of expression starting with 55 -587121/3581797888*a^7+2690715/3581797888*a^6+125087039/3581797888*a^5-281976997/3581797888*a^4-6353565723/3581797888*a^3+4276345801/3581797888*a^2+54570779021/3581797888*a-1796485759/3581797888 gap> r16:=RootsOfUPol(x^2-(16-2*r29+2*r55))[1]; # root of expression starting with 16 -342159/3581797888*a^7+1944125/3581797888*a^6+71070753/3581797888*a^5-244253347/3581797888*a^4-3446466469/3581797888*a^3+6034696383/3581797888*a^2+23536102611/3581797888*a-5417921945/3581797888 Finally we can calculate left and right side and e.g. subtract. gap> left:=r5+r22; -20639/111931184*a^7+113083/111931184*a^6+4290319/111931184*a^5-13886605/111931184*a^4-206942825/111931184*a^3+340055609/111931184*a^2+1353055097/111931184*a-668606679/111931184 gap> right:=r11+r16; -20639/111931184*a^7+113083/111931184*a^6+4290319/111931184*a^5-13886605/111931184*a^4-206942825/111931184*a^3+340055609/111931184*a^2+1353055097/111931184*a-668606679/111931184 gap> left-right; !0 Thu everything works exactly, but even in this tiny example the coefficients are a mess, so this in not something to do by hand.<|endoftext|> TITLE: Zero differential in Serre spectral sequence for configuration spaces QUESTION [6 upvotes]: I moved this question from Math StackExchange. I am trying to compute homology of $Conf(n, \mathbb{R}^2)$ - ordered configurations of $n$ points on the plane - using Serre spectral sequence. I know that this computations has been done for cohomology in Arnold's The Cohomology Ring of Colored Braid Group. There is one step though which is unclear for me. Namely, he claims that $d_2$ - the only differential which might be non-zero - is actually zero. The only possible differential $d_2$ is in fact zero (this easily follows from the existence of the secant of the surface). Why this is true? REPLY [13 votes]: I'll write $C_n$ for the configuration space, and $X_n$ for $\mathbb{R}^2$ with $n$ points removed. You are presumably thinking about the spectral sequence $$ E_2^{pq} = H^p(C_{n-1};H^q(X_{n-1})) \Longrightarrow H^{p+q}(C_n), $$ with differentials $d_r\colon E_r^{pq}\to E_r^{p+r,q+1-r}$. It is a key point that the projection $\pi\colon C_n\to C_{n-1}$ has a section: for example, we can use $$ \sigma(z_1,\dotsc,z_{n-1}) = \left(z_1,\dotsc,z_{n-1},(\max(\|z_1\|,\dotsc,\|z_{n-1}\|)+1).(1,0)\right). $$ This implies that the map $\pi_1(C_n)\to\pi_1(C_{n-1})$ is surjective, and thus that $\pi_1(C_{n-1})$ acts trivially on $H^*(X_{n-1})$. This is also a free abelian group by induction on $n$, so the $E_2$ term can be rewritten as $H^p(C_{n-1})\otimes H^q(X_{n-1})$. It is clear that $H^*(X_{n-1})$ is generated by classes in degree $1$, and it will suffice to show that these support no differentials. Any differentials after $d_2$ land in the lower half plane and so are zero. The differential $d_2$ on $H^1(X_{n-1})$ lands in $H^2(C_{n-1})$, and anything in the image will be killed by the map $H^2(C_{n-1})\to H^2(C_n)$ (by the basic definition and properties of the Serre spectral sequence). However, that map is injective, because of the existence of the section $\sigma$. Thus, $d_2$ is zero and the spectral sequence collapses.<|endoftext|> TITLE: Is the commensurator of a tree lattice a simple group? QUESTION [7 upvotes]: Let $T$ be an $n$-regular tree ($n\geq3$). Let $\operatorname{Aut}^+(T)$ be the subgroup of index 2 of $\operatorname{Aut}(T)$ preserving the bicoloring of the tree for which adjacent vertices have distinct colours. Let $\Gamma$ be a lattice in $G:=\operatorname{Aut}^+(T)$. There are a number of open problems concerning whether the commensurator $\operatorname{Comm}_G(\Gamma)$ is discrete, or dense. In the case where $\Gamma$ is uniform, the commensurator is known to be dense in $\operatorname{Aut}^+(T)$. This is also true for non-uniform lattices of Nagao type (Abramenko–Rémy). In any of these cases or in any specific examples, is the commensurator $\operatorname{Comm}_G(\Gamma)$ known to be simple? REPLY [2 votes]: This is only a partial answer which follows from Henry Wilton's (@HJRW) comment. In the appendix of the paper here (arXiv 1712.01091) Caprace proves: For $m\geq 3$, consider the commensurator ${\rm Comm}_{{\rm Aut}(T)}(W_m)$ of the free Coxeter group $W_m$ of rank $m$, where $T$ is the Cayley tree. Then, ${\rm Comm}_{{\rm Aut}(T)}(W_m)$ is almost simple ${\rm Comm}_{{\rm Aut}(T)}(W_m)$ contains a simple subgroup $B$ such that $[W_m : B \cap W_m] < \infty$.<|endoftext|> TITLE: Area method in Lobachevskian geometry QUESTION [8 upvotes]: There are many proofs in Euclidean geometry using the area method; for example, Ceva's theorem or the proof of Pythagorean theorem shown below. Do you know such proofs in hyperbolic geometry? I mean something where area helps, but does not appear in the statement. I suspect that the answer is no, but I hope to be surprised. REPLY [2 votes]: Well, you may get hyperbolic Ceva's theorem using area in the manner similar to Euclidean's one. Namely, using the formula for the area $E$ of a triangle with sides $a,b,c$ and angles $A,B,C$: $\sin \frac{E}2 \cosh \frac{c}2=\sinh \frac{a}2\sinh\frac{b}2\sin C$ (analogue of Euclidean $E=\frac12ab\sin C$) we get the following interpretation of cevian: a line $\ell$ through a vertex $A$ of triangle $\triangle ABC$ is a locus of points $P$ for which the ratio $$\sin \frac{E(PAB)}2\cdot\cosh\frac{BP}2:\sin \frac{E(PAC)}2\cdot\cosh\frac{PC}2$$ is a constant which I denote $\varkappa(\ell;BA,CA)$. The above areas should be oriented but let's ignore this. When are three lines $\ell_a$ through $A$, $\ell_b$ through $B$, $\ell_c$ through $C$ concurrent? When $\varkappa(\ell_a;BA,CA)\cdot \varkappa(\ell_b;CB,AB)\cdot \varkappa(\ell_c;AC,BC)=1$. (Again, I ignore the question of concurrent lines without common point which should be treated differently in hyperbolic geometry.) Now if $\ell_a\cap BC=A_1$, we have $$ \varkappa(\ell_a;BA,CA)= \sin \frac{E(A_1AB)}2\cdot\cosh\frac{BA_1}2:\sin \frac{E(A_1AC)}2\cdot\cosh\frac{CA_1}2=\\ \frac{\sinh\frac{BA_1}2\cosh\frac{BA_1}2}{\cosh \frac c2}: \frac{\sinh\frac{CA_1}2\cosh\frac{CA_1}2}{\cosh \frac b2}= \frac{\sinh BA_1}{\cosh \frac c2}: \frac{\sinh CA_1}{\cosh \frac b2} $$ that yields hyperbolic Ceva.<|endoftext|> TITLE: Infinite cyclic subgroups of $\text{SL}_2(\mathbb{Z})$ QUESTION [11 upvotes]: Let $\Gamma = \operatorname{SL}_2(\mathbb{Z})$ be the usual modular group. It is well-known that $\Gamma$ contains infinitely many distinct (non-conjugate even) subgroups which are isomorphic to the infinite cyclic group $(\mathbb{Z}, +)$. Indeed, for any square-free integer $d > 1$, the unit group of the quadratic field $\mathbb{Q}(\sqrt{d})$ will give rise to such a group. More generally, any irreducible, indefinite binary quadratic form $f(x,y) = ax^2 + bxy + cy^2$ will induce such a subgroup in $\Gamma$, with an explicit generator given by $$\displaystyle \begin{pmatrix} \dfrac{t_f + bu_f}{2} & au_f \\ \\ -cu & \dfrac{t_f - bu_f}{2} \end{pmatrix},$$ where $(t_f, u_f)$ is the fundamental (positive) solution to the Pell equation $x^2 - \Delta(f) y^2 = 4$. Is this a bijection? That is, each infinite cyclic subgroup of $\Gamma$ must arise from an irreducible binary quadratic form in this way (up to conjugacy)? REPLY [18 votes]: The maximal infinite cyclic subgroups are of the form you mentioned (to get all infinite cyclic subgroups you also need to include their finite index subgroups). This follows from Theorem 1.4 in Sarnak, Peter, Class numbers of indefinite binary quadratic forms. J. Number Theory 15 (1982), no. 2, 229--247 since the generator of such a subgroup must be a primitive hyperbolic element. Edited: Oops, I missed that $\mathrm{SL}_2(\mathbb{Z})$ also has infinite cyclic subgroups generated by parabolic elements, but those are also stabilizers of quadratic forms of discriminant $0$ (that is, that are squares).<|endoftext|> TITLE: Asymptotics of multinomial coefficients QUESTION [8 upvotes]: Binomial coefficients have a well known asymptotics (https://en.wikipedia.org/wiki/Binomial_coefficient#Bounds_and_asymptotic_formulas) given by $$\binom nk\sim\binom{n}{\frac{n}{2}} e^{-d^2/(2n)} \sim \frac{2^n}{\sqrt{\frac{1}{2}n \pi }} e^{-d^2/(2n)}.$$ Is there corresponding asymptotics for multinomials (I am more interested in what happens analogously in the denominator)? REPLY [9 votes]: Suppose that $k$ is a fixed natural number, $n\to\infty$, and \begin{equation*} a_i=\frac nk+o(n^{2/3}) \end{equation*} for each $i$; here in what follows, $i\in\{1,\dots,k\}$. Let \begin{equation*} h_i:=\frac kn\,a_i-1=o(n^{-1/3}), \tag{1} \end{equation*} so that $h_i\to0$ and \begin{equation} a_i=\frac nk\,(1+h_i). \tag{2} \end{equation} By Stirling's formula, \begin{equation*} n!\sim(2\pi)^{1/2} n^{1/2}\Big(\frac ne\Big)^n \end{equation*} and, by (2), \begin{equation*} \begin{aligned} a_i!&\sim(2\pi)^{1/2} \Big(\frac nk\Big)^{1/2}(1+h_i)^{1/2}\Big(\frac n{ke}(1+h_i)\Big)^{a_i} \\ &\sim(2\pi)^{1/2} \Big(\frac nk\Big)^{1/2}\Big(\frac n{ke}(1+h_i)\Big)^{a_i}. \end{aligned} \end{equation*} Therefore and because $\sum_{i=1}^k a_i=n$, \begin{equation*} \prod_{i=1}^k a_i!\sim(2\pi)^{k/2} \Big(\frac nk\Big)^{k/2}\Big(\frac n{ke}\Big)^n \prod_{i=1}^k(1+h_i)^{a_i}, \end{equation*} which implies \begin{equation*} \binom{n}{a_1,\ldots,a_k}=\frac{n!}{\prod_{i=1}^k a_i!} \sim(2\pi)^{1/2-k/2}\frac{n^{1/2}}{(n/k)^{k/2}}\frac{k^n}{e^u}, \end{equation*} where \begin{equation*} u:=\sum_{i=1}^k a_i\ln(1+h_i)=\frac nk\sum_{i=1}^k (1+h_i)\ln(1+h_i), \end{equation*} by (2). Note also that (i) $(1+h)\ln(1+h)=h+h^2/2+O(|h|^3)$ as $h\to0$, (ii) $\sum_{i=1}^k h_i=0$ (by the definition of $h_i$ in (1) and the condition $\sum_{i=1}^k a_i=n$), and (iii) $h_i=o(n^{-1/3})$. It follows that \begin{equation*} u=\frac n{2k}\,\sum_{i=1}^k h_i^2+o(1)=\frac k{2n}\,\sum_{i=1}^k(a_i-n/k)^2+o(1). \end{equation*} Thus, \begin{equation*} \binom{n}{a_1,\ldots,a_k} \sim(2\pi n)^{1/2-k/2}k^{n+k/2}\exp\Big\{-\frac k{2n}\,\sum_{i=1}^k(a_i-n/k)^2\Big\}. \end{equation*} In particular, when $k=2$, we get the Wikipedia result quoted in the OP: \begin{equation*} \binom na\sim\frac{2^n}{\sqrt{\pi n/2 }} e^{-2(a-n/2)^2/n} \end{equation*} if $n\to\infty$ and $a=\frac n2+o(n^{2/3})$.<|endoftext|> TITLE: An extremal property of points on the unit sphere of a 2-dimensional Banach space QUESTION [5 upvotes]: Let $(X,\|\cdot\|)$ be a 2-dimensional real Banach space and $S=\{x\in X:\|x\|=1\}$ be its unit sphere. Assume that $S$ is smooth in the sense that for any $y\in S$ there exists a unique functional $y^*:X\to\mathbb R$ such that $y^*(y)=1=\|y^*\|$. This unique functional $y^*$ will be called the supporting functional at $y$. Let $x,y\in S$ be points such that $$\|y-x\|+\|y+x\|=\max\{\|s-x\|+\|s+x\|:s\in S\}.$$ Question. Is $y^*(x)=0$? REPLY [8 votes]: The answer is no, in general. For a counterexample, consider the $\ell^p$-norm on $\mathbb{R}^2$ with $p=4$, and let $x = e_1 = (1,0)$. We first note that the vectors $e_2 = (0,1)$ and $-e_2$ do not maximize the function \begin{align*} f: S \ni s \mapsto \|s-x\|+\|s+x\| \in [0,\infty). \end{align*} Indeed, we have $f(e_2) = f(-e_2) = 2\cdot\|(1,1)\| = 2\cdot 2^{1/4}$. On the other hand, consider the vector $s_0 = 2^{-1/4} \cdot (1,1) \in S$. Then \begin{align*} f(s_0) = \|(2^{-1/4} - 1, 2^{-1/4})\| + \|(2^{-1/4}+1, 2^{-1/4})\| > 2\cdot 2^{1/4} \end{align*} (where I, admittedly, used a computer to check the latter inequality). So $f$ is neither maximized at $e_2$ nor at $-e_2$. Hence if $y$ maximizes $f$, then the first entry of $y$ is non-zero. But $y^* \in \mathbb{R}^2$ is the pointwise product of $y$ and $|y|^{p-2}$. Thus, the first entry of $y^*$ is also non-zero, which shows that $y^*(x) \not= 0$.<|endoftext|> TITLE: Duals of the spinor representations of $\frak{so}_{2n}$ QUESTION [6 upvotes]: For the $D_n$-series simple Lie algebra $\frak{so}_{2n}$ a curious phenomenon occurs for the fundamental representations corresponding to the spinor nodes of the Dynkin diagram, which is to say the spinor representations $V_{\pi_{n-1}}$, and $V_{\pi_{n}}$: In the case where $n$ is even both $V_{\pi_{n-1}}$ and $V_{\pi_{n}}$ are self-dual representations, which is to say $$ V_{\pi_{n-1}}^{\vee} \simeq V_{\pi_{n-1}}, ~~~~~ V_{\pi_{n}}^{\vee} \simeq V_{\pi_{n}}. $$ However, in the odd case, the two representations are dual to each other, which is to say $$ V_{\pi_{n-1}}^{\vee} \simeq V_{\pi_{n}}. $$ Is there a conceptual reason why this occurs, why the even and odd case behave differently? Why does the action of the longest element of the Weyl group behave differently in each case. REPLY [4 votes]: Here is one explanation, although I am not sure if this is the conceptual explanation you are looking for. Let $E = \mathbb{R}^n$ with orthonormal basis $\varepsilon_1, \ldots, \varepsilon_l$. You can realize a root system of type $D_n$ as $\Phi = \{ \pm (\varepsilon_i \pm \varepsilon_j) : i \neq j \}$. The Weyl group $W$ is the group of permutations and sign changes on $\varepsilon_1$, $\ldots$, $\varepsilon_n$ involving only an even number of sign changes. That is, for $\sigma \in W$ you have $\sigma(\varepsilon_i) = c_i \varepsilon_{\pi(i)}$ with $\pi \in Sym_n$, $c_i = \pm 1$ and $c_1c_2 \cdots c_n = 1$. From this you already see that $-1 \in W$ if and only if $n$ is even. Also, the weights occurring in the two irreducible spin representations are precisely those of the form $$\lambda_I = \frac{1}{2} (\sum_{i \in I} \varepsilon_i - \sum_{i \not\in I} \varepsilon_i)$$ for a subset $I \subseteq \{1, \ldots, n\}$. Clearly $\lambda_I$ and $\lambda_{J}$ are conjugate under the Weyl group if and only if $|I| = |J| \mod{2}$. One of the spin representations has highest weight $\lambda = \frac{1}{2}(\sum_{i = 1}^n \varepsilon_i)$ and the other one has highest weight $\mu = \frac{1}{2}(\sum_{i = 1}^{n-1} \varepsilon_i - \varepsilon_n)$. So $\lambda$ and $-\mu$ are conjugate under the Weyl group if and only if $-\mu$ involves an even number of minus signs, equivalently $n$ is odd.<|endoftext|> TITLE: Source on smooth equivalence relations under continuous reducibility? QUESTION [10 upvotes]: This question was asked and bountied at MSE, but received no answer. In the context of Borel reducibility, smooth equivalence relations (see the introduction of this paper) are rather boring since everything boils down to the number of classes. However, the situation seems more interesting when we restrict attention to continuous reducibility. I'd like to know more about the structure of smooth equivalence relations with respect to continuous reducibility; what is a good source on this topic? (I'm happy to restrict to equivalence relations on Baire space if that would help, but in general I'm interested in arbitrary Polish spaces.) REPLY [2 votes]: Seems like this structure must be pretty complicated. For example, consider Brownian motion $\{W_t\}_{t\ge 0}$ with the equivalence relations $$t\sim_\omega s\iff W_t(\omega)=W_s(\omega).$$ Here $\omega\in\Omega$ is an outcome from the sample space, giving a particular path of Brownian motion. This $\sim_\omega$ is smooth. It seems that these equivalence relations should almost surely not be continuously reducible to eachother, i.e., for almost all $(\omega_1,\omega_2)\in\Omega\times\Omega$ there should not be any continuous $f$ with $$t\sim_{\omega_1} s\iff f(t)\sim_{\omega_2}f(s).$$ However I don't have a proof of that... just a thought!<|endoftext|> TITLE: How to prove $(\phi-1)(\phi-2)...(\phi-p) = \sqrt{5} + p\left(\frac{1}{2}+A\sqrt{5}\right) \bmod p^2$? QUESTION [8 upvotes]: We consider the solution of $x^2=x+1$ and denote them as $\phi=\frac{1}{2}(1-\sqrt{5}),\bar\phi=\frac{1}{2}(1+\sqrt{5})$. Suppose $\phi \not\in \mathbb{F}_p$. In other words, $\sqrt{5} \not \in \mathbb{F}_p\Leftrightarrow p = \pm 2 \bmod 5$. Arbitrary element of $\mathbb{F}_p(\phi)$, $a$ and $b$ satisfies $ab=0\Leftrightarrow a=0 \lor b=0$. From Fermat's little theorem and factor theorem, $$(x-1)(x-2)...(x-p) = x^p -x \bmod p .$$ Then, put $x=\phi$. Since Frobenius map,$x \mapsto x^p$, transfer $\phi$ to the conjugate of itself, $\bar\phi$, \begin{align} &(\phi-1)(\phi-2)...(\phi-p) = \bar\phi -\phi \bmod p\mathbb{Z}(\phi) \\ \Leftrightarrow &(\phi-1)(\phi-2)...(\phi-p) = \sqrt{5}\bmod p\mathbb{Z}(\phi) \end{align} Then, what about $(\phi-1)(\phi-2)...(\phi-p) \bmod p^2\mathbb{Z}[\phi]$? Surprisingly, this can be expressed as $$(\phi-1)(\phi-2)...(\phi-p) = \sqrt{5} + p\left(\frac{1}{2}+A\sqrt{5}\right) \bmod p^2\mathbb{Z}[\phi]$$ empirically (I verify this rule by a code here ideone. $a$ and $b$ of the output corresponds to $a=\frac{1}{2}p$ and $b=1+pA$). Here $\sqrt{5}$ is defined as $1-2\phi$. $\frac{1}{2}$ is the inverse of $2$ at $\bmod p^2\mathbb{Z}$. Such element can be found by extended Euclidean algorithm because $2$ and $p^2$ are coprime to each other. I cannot find the rule for $A$, but everytime the coeffieicnt of $p$ of $\mathbb{F}_p$ is $\frac{1}{2}$ at $\bmod p^2$. How to prove this rule? My idea: Consider the coefficient, $c_i$ of $(x-1)(x-2)...(x-(p-1))(x-p)=\sum_{i=0}^{p-1}c_ix^i$. It seems $c_i \bmod p=0$ for all $i$ and especially, $c_{2i}=0\bmod p^2$ for $2 \leq 2i \leq p-3$ if $p$ is prime. If $\phi$ is in the form, $a\sqrt{5}$, we only need to consider the term $c_0, c_{p-1}$ because other terms are divisible by $p^2$ or in the form $a'\sqrt{5}$. REPLY [6 votes]: Start with $$(\phi-i)(\phi-(p+1-i))=\phi^2-\phi(p+1)+i(p+1-i)=p(i-\phi)+(1+i-i^2).$$ Using this for $i=1,2,\ldots,(p-1)/2$ we get $$ \prod_{j=1}^p(\phi-j)=\left(\phi-\frac{p+1}2\right)\prod_{i=1}^{(p-1)/2}\left(p(i-\phi)+(1+i-i^2)\right). $$ Expand the brackets and take it modulo $p^2\mathbb{Z}[\phi]$. We get $$ M:=\left(\phi-\frac{p+1}2\right)\prod_{i=1}^{(p-1)/2}\left(1+i-i^2\right)+p\left(\phi-\frac12\right)\prod_{i=1}^{(p-1)/2}\left(1+i-i^2\right)\sum_{i=1}^{(p-1)/2}\frac{i-\phi}{1+i-i^2}. $$ Denote $T=\prod_{i=1}^{(p-1)/2}\left(1+i-i^2\right)$. Considering $M$ modulo $p$ we get $-\sqrt{5}T/2$, thus by your calculation of $M$ modulo $p$ we get $T\equiv -2\pmod p$. Next, we should look mod $p$ for $$ \frac{M-\sqrt{5}}{p}=-\sqrt{5}\frac{T+2}{2p}+T\left(-\frac12-\frac{\sqrt{5}}2\sum_{i=1}^{(p-1)/2}\frac{i-\phi}{1+i-i^2}\right). $$ And here we should look for ``rational'' part, which must be equal to $\frac12$ modulo $p$. This rational part equals $$ T\left(-\frac12-\frac54\sum_{i=1}^{(p-1)/2}\frac{1}{1+i-i^2}\right), $$ and our claim reduces to $$ \sum_{i=1}^{(p-1)/2}\frac{1}{1+i-i^2}\equiv -\frac15 \pmod p. $$ This can not be hard and it is not. We have $1+i-i^2=5/4-(i-1/2)^2$. The guys $(i-1/2)^2$ run over the set $\mathcal{R}$ of all non-zero quadratic residues when $i$ goes from 1 to $(p-1)/2$ (indeed, they are quadratic residues for sure, non-zero and mutually distinct: if $(i-1/2)^2=(j-1/2)^2$, then either $i=j$ or $p$ divides $i+j-1$ which can non be in our range). So we should have $\sum_{r\in \mathcal{R}} 1/(5/4-r)=-1/5$. Denote $f(x)=\prod(x-r)=x^{(p-1)/2}-1$. We should prove $\frac{f'(5/4)}{f(5/4)}=-1/5$. This is true: $f(5/4)=-2$, $f'(5/4)=\frac{p-1}2\cdot (-1)\cdot \frac45=\frac25$.<|endoftext|> TITLE: Why does this matrix have zero determinant? QUESTION [14 upvotes]: This curious identity arose from studying reductions of the maximal ideal in certain monomial algebra. It can be proved "by hand", (i.e, using Macaulay 2), but I am seeking a more conceptual understanding and related references if they exist. Let $R$ be a commutative ring. For two vectors $v=(a,b,c,d), w=(A,B,C,D)\in R^4$, we define $v\star w:= (aA,aB+bA,bB, cC,cD+dC,dD)\in R^6$. Given any 3 vectors $v_1,v_2,v_3\in R^4$, we can form a $6\times 6$ matrix $M$ whose rows are $v_i\star v_j$, $1\leq i,j\leq 3$. Then: $$\det(M)=0$$ It is not clear to me how to explain this. The kernel of $M$ is a column of degree $6$ polynomials, so the relations are quite complicated. Question: Is there a way to conceptually explain the vanishing of $\det(M)$? Have you seen similar identities? REPLY [21 votes]: Three vectors $v_1,v_2,v_3$ lie in a hyperplane $H:\alpha x+\beta y+\gamma z+\delta t=0$, in this plane we have $Q(v,v):=(\alpha x+\beta y)^2-(\gamma z+\delta t)^2=0,\forall v\in H$. Thus by polarization $Q(v,w)=\frac 14 (Q(v+w,v+w)-Q(v-w,v-w))=0$ for all $v,w\in H$ that yields a relation between columns of your matrix: if $v=(a,b,c,d), w=(A,B,C,D)$, then $Q(v,w)=\alpha^2 aA+\beta^2 bB+\alpha\beta(aB+bA)-\gamma^2 cC-\delta^2 dD-\gamma\delta(cD+dC)$.<|endoftext|> TITLE: Abelianization of mapping class groups $\Gamma_{g,n}$ QUESTION [7 upvotes]: Let $S_{g,n}$ be a Riemann surface of genus $g$, with $n$ points removed. The mapping class group of $S_{g,n}$ is denoted by $\Gamma_{g,n}$. Is there a reference where the abelianization of $\Gamma_{g,n}$ calculated (or at least for $g$ sufficiently large, are they trivial)? REPLY [9 votes]: The following statement can be found in Section 5 of Low-dimensional homology groups of mapping class groups: a survey: Theorem: Let $g \geq 1$. Then $$H_1(\Gamma_{g,r}^n,\mathbb{Z}) \simeq \left\{ \begin{array}{cl} \mathbb{Z}_{12} & \text{if $(g,r)=(1,0)$} \\ \mathbb{Z}^r & \text{if $g=1,r \geq 1$} \\ \mathbb{Z}_{10} & \text{if $g=2$} \\ 0 & \text{if $g \geq 3$} \end{array} \right.$$ $\Gamma^n_{g,r}$ denotes the mapping class group of a connected orientable surface of genus $g$ with $r$ boundary components and $n$ punctures. Precise references are given in the survey.<|endoftext|> TITLE: Taylor $k$-differentiability of a real function at a point QUESTION [5 upvotes]: I am interested in the standard name for the following weak form of $k$-differentiability. Definition. A function $f:\mathbb R\to\mathbb R$ is called Taylor $k$-differentiable at a point $x_0$ if there are real numbers $a_0,\dots,a_k$ such that $$f(x)=\sum_{n=0}^k\tfrac1{n!}a_n(x-x_0)^n+o(|x-x_0|^k).$$The numbers $a_0,\dots,a_k$ are unique (if exist) and can be called the derivatives $f^{(0)}(x_0),\dots,f^{(k)}(x_0)$ of $f$ at $x_0$. The Taylor formula says that each $k$-differentiable function at a point $x_0$ is Taylor $k$-differentiable. The converse is not true: the Dirichlet-like function $$f(x)=\begin{cases}x^{k+1}&\mbox{if $x$ is rational};\\ 0&\mbox{if $x$ is irraional} \end{cases} $$ is Taylor $k$-differentiable at zero (with all derivatives $f^{(0)}(0)=\dots=f^{(k)}(0)=0$) but is discontinuous at all non-zero points, so is not $k$-differentiable in the standard sense. So, my question: Has the Taylor $k$-differentiability some standard name, accepted in the literature? If yes, what is a suitable reference? Thanks. REPLY [3 votes]: I would not recommend "Taylor $k$-differentiable", because the name "Taylor" always refers to the standard polynomial made by successive derivatives at $x_0$, and because as far as we know this generalization was not in the mind of Taylor. A quite standard and self-explanatory name is: $f$ has a polynomial expansion of order $k$ at $x_0$, where "order $k$" refers to the form of the remainder, $o(|x-x_0|^k)$. If you like a shorter name linked to a mathematician, I'd go for Peano k-differentiable, which I think is also customary, since I think Peano (I'll hopefully add a reference later) made quite an extensive and thorough study of this subject. In particular, the implication you mention and the form of the remainder $o(x-x_0)^k$ is due to Peano (it's a one-page note, Une nouvelle forme du reste dans la formule de Taylor, 1889). This polynomial expansion has a nice calculus, analogous to the calculus of derivatives, though, as soon as you want deeper results, such as e.g. the inverse mapping theorem, you need to assume the existence of this polynomial expansion everywhere, with continuously depending coefficients: but this is equivalent to being a standard $C^k$ function. REPLY [3 votes]: It's called the “de la Vallée-Poussin derivative” or “Peano derivative” (both terms seem to have been used, sometimes with a difference between them; the latter is perhaps slightly more standard): see this answer and the references it contains for a proof of the fact that if the $k$-th Peano derivative exists and is continuous on an interval then it is, in fact, a $k$-th derivative. Edit: The term “de la Vallée-Poussin derivative” seems to have been introduced in a paper by Marcinkiewicz and Zygmund, “On the Differentiability of Functions and Summability of Trigonometrical Series”, Fund. Math. 26 (1936) 1–43. The term “Peano derivative” occurs in Oliver's paper “The Exact Peano Derivative”, Trans. Amer. Math. Soc. 76 (1954) 444–456, alongside dlVP, but does not provide an explanation of where the author got this name.<|endoftext|> TITLE: First coefficient of totally positive fundamental unit modulo 3 QUESTION [19 upvotes]: Suppose $p$ is a prime number such that $p\equiv 7 \pmod{12}$. Since $p \not \equiv 1 \mod{4}$, the ring of integers of $\mathbb{Q}(\sqrt{p})$ is $\mathbb{Z}\oplus\mathbb{Z}(\sqrt{p})$ with fundamental unit of the form $a+b\sqrt{p}$, where $a, b > 0$ It is of the norm $+1$, because $a^2 - pb^2 \equiv -1 \pmod{4}$ doesn't have solutions when $p\equiv -1 \pmod{4}$. It is easy to see that $a\equiv \pm 1 \pmod{3}$, since if $3|a$, then $a^2 - pb^2 \equiv - b^2 \not\equiv 1 \pmod{3}$ A computer-checked observation. For all said $p$ such that $p<20000$ the following holds: $a\equiv -1 \pmod{3}$ That, I think, is rather surprising. I tried proving that $a\equiv -1$ for the case when the period of continued fraction for $\sqrt{p}$ is of length $4$, but got nowhere. The claim is false when $p$ is not prime, for example $\mathbb{Q}(115)$ has the fundamental unit $1126+105\sqrt{115}$. It is also false when we drop modulo conditions, say, in $\mathbb{Q}(\sqrt{11})$ the fundamental unit is $10+3\sqrt{11}$ What is known about coefficients of the continued fraction for $\sqrt{p}$ ? I only know that it is of the form $[a_o; \overline{a_1, a_2, ..., a_2, a_1, 2a_0}]$ and that when the class number of the field is 1, the sum $a_1 - a_2 + a_3 - ... \pm 2a_0$ is divisible by 3 (the latter is a corollary from a theorem by Don Zagier). All that cannot even preclude the case when all coefficients of the continued fraction are divisible by 3. Any hints are welcome. REPLY [14 votes]: By Theorem 1.1 of Zhang-Yue: Fundamental units of real quadratic fields of odd class number, J. Number Theory 137 (2014), 122-129, we have that $a\equiv 0\pmod{2}$. From here it is a simple matter to prove that $a\equiv 2\pmod{3}$, hence in fact $a\equiv 2\pmod{6}$. To see this, let us write the unit equation as $$(a-1)(a+1)=pb^2.$$ The left hand side is divisible by $3$, hence so is $b$. The factors on the left hand side are odd and coprime, so they are of the form $pb_1^2$ and $b_2^2$ in some order, where $b=b_1b_2$. Note that exactly one of the $b_i$'s is divisible by $3$. Therefore, adding the factors of the left hand side, we get modulo $3$ that $$2a=pb_1^2+b_2^2\equiv b_1^2+b_2^2\equiv 1\pmod{3}.$$ Hence $a\equiv 2\pmod{3}$ as claimed.<|endoftext|> TITLE: A finite alternating sum QUESTION [18 upvotes]: We have stumbled upon the following finite alternating sum, which we have trouble analyzing. The sum is: $$ S_n = \sum_{j=0}^n \frac{ (-1)^j e^{-j} }{j!} (n-j)^j $$ We have observed numerically that $S_n \approx 2 n e^{-n}$. We would like to establish whether this conjecture is true. More precisely, we would like to show that $S_n= \Theta(n e^{-n})$. The sum is quite hard to evaluate numerically because the summands grow in absolute value initially and then decrease toward zero. The largest term is exponential in $n$ while the sum is conjectured to converge to zero exponentially fast. Any references or ideas are welcome! REPLY [18 votes]: WhatsUp's generating function allows to get an asymptotics: we have $$\sum S_n t^n=\frac{t}{e^{t/e}-t},\sum S_n e^nt^n=\frac{t}{e^{t-1}-t}.$$ Denote $t-1=x$, then $$\frac{t}{e^{t-1}-t}=\frac{1+x}{e^x-1-x}=\frac{1+x}{x^2/2+x^3/6+\ldots}=\frac{2(1+x)}{x^2(1+x/3+\ldots)}=\frac2{x^2}+\frac{4/3}{x}+O(1)$$ for small $x$. Thus $$\frac{t}{e^{t-1}-t}=\frac2{(1-t)^2}-\frac{4/3}{1-t}+F(t)=\sum \left(2n+\frac23\right)t^n+F(t),$$ where $F$ does not have pole at 1. I claim that there is also no pole with absolute value at most 1, that is, the function $e^{t-1}-t$ does not have zeroes in the closed unit disc other than 1. Indeed, assume that $0\leqslant |r|\leqslant 1$ and $\varphi\in (-\pi,\pi]$ and $e^{re^{i\varphi}-1}=re^{i\varphi}$. Taking the arguments of both sides we get $r\sin \varphi=\varphi$ that yields $\varphi=0$, otherwise $|\varphi|>|\sin \varphi|\geqslant r|\sin \varphi|$. So $r=e^{r-1}\geqslant 1+(r-1)=r$ with equality only for $r=1$. This all implies that $F$ is analytic in a disc with radius greater than 1, therefore the coefficients of $F$ are $O(a^n)$ for certain $a\in (0,1)$ and in your question we get $$S_n=(2n+2/3)e^{-n}(1+O(a^n)).$$<|endoftext|> TITLE: Approximation of a Sobolev surface by a smooth surface QUESTION [6 upvotes]: I was quite sure that the answer to the following question is known, and was surprised not to find any reference: Let $M$ be a compact, oriented $2$-dimensional manifold with boundary. Let $f:M\to R^3$ be a $W^{2,2}$-map such that $Df$ has full rank a.e. Can $f$ be approximated in $W^{2,2}$ by smooth immersions? (As standard, one can endow $M$ with any Riemannian metric to define the Sobolev spaces.) REPLY [2 votes]: I do not know the full answer yet, but in the case in which $f:\Omega\to\mathbb{R}^3$, $\Omega\subset\mathbb{R}^2$ bounded and convex, is an isometric immersion, then it can be approximated by smooth isometric immersions in $W^{2,2}$ norm. This is Theorem I in: M. R. Pakzad, On the Sobolev space of isometric immersions. J. Differential Geom. 66 (2004), no. 1, 47–69. Perhaps you can find relevant references and results there. The result of Pakzad has been generalized in: Z. Liu, M. R. Pakzad, Rigidity and regularity of codimension-one Sobolev isometric immersions. Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 14 (2015), no. 3, 767–817.<|endoftext|> TITLE: Smooth vs regular vs non-singular QUESTION [23 upvotes]: This is a very basic question, but I can't find a clean answer anywhere. In introductory algebraic geometry books working over the complex numbers, it's usual to use these three words interchangeably. A point on a variety $X$ is smooth/regular/nonsingular if the dimension of the tangent space at the point is equal to the dimension of the variety. On the other hand, I know that people sometimes find it important to distinguish between these terms, maybe when defining smoothness of morphisms, or working over non-closed fields,... I want to make sure I know the right definitions of these terms in current use. In what contexts should each be defined? What implies what? How should I think of them? Edit: See for example https://en.wikipedia.org/wiki/Regular_scheme , which says there are regular schemes that aren't smooth. There are also these notes of Vakil, where he has crossed out "smooth" and replaced it with "nonsingular": https://math.stanford.edu/~vakil/0708-216/216class21.pdf The notes seem to suggest it's because "smooth" is reserved as a property of morphisms. Is there a reason Wiki is happy to say "smooth scheme" but Ravi isn't? Is "nonsingular" the same as "regular"? REPLY [29 votes]: In the general context, "regular" is a property of a scheme (or a ring, or local ring), and "smooth" is a property of a morphism of schemes. "Regular" means exactly that at every point, the dimension of the (Zariski) tangent space is equal to the (Krull) dimension (of the local ring at that point). A map $f: X \to Y$ is smooth if the fibers over geometric points of $Y$ are regular, and $f$ is locally of finite presentation and flat. We also use the relative point of view, so a scheme $X$ over $S$ is a smooth scheme over $S$ if the map $X \to S$ is smooth. The first potential source of confusion is that a map between two regular schemes can fail to be smooth. This is not hard to see once you realize that there's no need for the fibers to be regular - for instance $xy$ defines a map $\mathbb A^2 \to \mathbb A^1$. The second potential source of confusion is that a regular scheme over a perfect field is necessarily smooth over that field, but for an imperfect field this fails. See some examples. Usually over an imperfect field you want to consider smooth schemes and not regular ones as they are better-behaved. Going in the other direction, smooth schemes over non-regular bases can fail to be regular, but smooth schemes over regular bases will be regular. I think people rarely use "nonsingular" when they are trying to be careful about this distinction, but I think it's more likely to mean "regular".<|endoftext|> TITLE: Best introductory texts on pointless topology QUESTION [16 upvotes]: As I understand it, there are three canonical textbooks on pointless topology: the classic "Stone Spaces" by Johnstone, "Topology via Logic" by Steve Vickers, and the newer "Frames and Locales" by Picado and Pultr. I am curious for a comparison between the emphases of these three books. As far as prerequisites go, I have a reasonable background in category theory, though no exposure to sheaf theory, and have seen enough point-set topology to understand Stone-Čech compactification and Tychonoff's theorem. I have also seen a bit of commutative algebra and am willing to do additional reading to catch up on anything necessary. I have no background in lattice theory, but I think all three texts have either appendices or introductory chapters on the subject. Having glanced through their indices, I have a vague impression of the material covered in each text. For instance, Johnstone's book contains a great deal of material on representation theorems, and also a chapter devoted to applications of locale theory to representations of rings (including discussion on Zariski and Pierce spectra); as far as I can tell this is absent from "Frames and Locales". "Topology via Logic" also has some discussion of Zariski spectra as well as a number of applications to theoretical computer science. On the other hand, "Frames and Locales" gives discussion on constructive mathematics (in particular why locale theory is a nice language for topology in constructive mathematics), while (again as far as I can tell) this is missing from the other two books. In short, my question is: for which reader is each text most suitable, and which text would be the best self-contained introduction to the field for someone with my background. Apologies if this is not appropriate for the site, but I've struggled to find any extensive discussion comparing the three books. REPLY [4 votes]: Topology via Logic - theoretical computer scientist Stone Spaces - pure mathematician Both are really good. Topology via logic as it gives a good account of domain theory, including power domains. The first few chapters of Stone Spaces really have no equal. Comprehensive, starting at the basics, excellent narrative flow and take you to the heights (I would even recommend it as an introduction to the basic constructions of sheaf theory). Enjoy!<|endoftext|> TITLE: Free ordered field? QUESTION [8 upvotes]: There is no such thing as a free field, because there are no morphisms between fields of different characteristics. However, ordered fields seem to be much better behaved: There is an initial object ($\mathbb{Q}$, the rational numbers) and a terminal object (No, the surreal numbers). Does this mean a free ordered field exists, or would the need to be able to answer x > y, where x and y are independent objects in the field, make it non-free? REPLY [14 votes]: I interpret a "free ordered field" to mean the existence of a left adjoint functor to the$^1$ forgetful functor from ordered fields either to sets or to totally ordered sets. In both cases, the answer is "no". You don't even need two points x,y and compare them, a singleton $X=\lbrace\ast\rbrace$ is enough. I claim that supposing a "free ordered field over $X$" - let's denote it by $\mathbb{Q}_X$ - cannot exist. Proof: Let's have a look at where the element $\ast\in\mathbb{Q}_X$ is located in the order. There are two standard orderings on the function field $\mathbb{Q}(T)$; one making $T$ positive, but infinitesimal (i.e. monic polynomials of higher degree get progressively smaller), one making it positive and infinite (i.e. monic polynomials of higher degree get progressively larger). Now consider the (order preserving) map $\lbrace\ast\rbrace \to \mathbb{Q}(T), \ast\mapsto T$. If $\mathbb{Q}_X$ really had the universal property of a free ordered field, both maps would extend to maps of ordered fields $\mathbb{Q}_X\to\mathbb{Q}(T)$. This gives the contradiction that both $\ast < 1$ and $\ast > 1$ would have to hold inside $\mathbb{Q}_X$. $^1$ In fact this proof shows that there is no left adjoint to a number of functors originating on the category of ordered fields. The functors $K\mapsto K$, $K\mapsto K\setminus\{0\}$, $K\mapsto K\setminus\mathbb{Q}$, $K\mapsto \{\text{transcendental elements}\}$, $K\mapsto \{\text{positive elements}\}$, and many more do not possess left adjoints. And slight variations of the proof also exclude other functors one could think of like $K\mapsto \{\text{algebraic elements}\}$. REPLY [10 votes]: There is no terminal object in the category of ordered fields with non-decreasing field morphisms. Indeed morphisms are injective and there are ordered fields of arbitrary cardinality (strictly speaking $\mathbf{No}$ is not a set, hence not an object in this category). Moreover $\mathbb{R}$ embeds into $\mathbf{No}$ in more than one way so $\mathbf{No}$ is in no way terminal. No ordered field enjoys a non-empty free object, because for any such field $K$ and $\varnothing\neq X \subseteq K$, the map $X \longrightarrow \{0\}$ if $X\neq \{0\}$ or $X\longrightarrow \{1\}$ if $X=\{0\}$ cannot be extended into an embedding $K \longrightarrow K$.<|endoftext|> TITLE: Is the map sending a continuous function to its period measurable? QUESTION [12 upvotes]: Let $C(\mathbb{R})$ be the space of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ with the compact-open topology, and the associated Borel $\sigma$-algebra. Consider the function $p$ from $C(\mathbb{R})$ to $\mathbb{R}_{\geq 0} \cup \{\infty\}$ that maps a continuous function to its period, with the convention that non-periodic functions get mapped to $\infty$. Is the function $p$ a measurable function on $C(\mathbb{R})$? The only way I know of constructing measurable functions is to realize them as iterated lim, limsup, or liminf of a sequence of continuous functions. It's not clear here what continuous functions approximate the period in any reasonable manner. Questions: Is there a way to approximate the period of $f \in C(\mathbb{R})$ using a continuous map from $C(\mathbb{R})$ to $\mathbb{R}$, which on taking appropriate limits, converges to the described function $p$? Is there some other way of showing that the map $p$ is measurable? REPLY [21 votes]: Isn't the set $p^{-1}([0, T_0])$ closed for every finite $T_0$? Suppose that $f_n$ has period $T_n \leqslant T_0$ and it converges locally uniformly to $f$. By passing to a subsequence, we may assume that $T_n$ has a limit $T$. Uniform convergence of $f_n$ on $[x, x + T_0]$ implies that $$f(x + T) = \lim f_n(x + T_n) = \lim f_n(x) = f(x),$$ and hence $f$ has period at most $T$.<|endoftext|> TITLE: Is Hausdorff Measure equal to Hausdorff Content on rectifiable (metric) spaces? QUESTION [6 upvotes]: Let $(X,d)$ be an $\mathcal{H}^n$-rectifiable metric space, i.e. there exits a collection of Lipschitz maps from measurable subsets of $\mathbb{R}^n$ to $X$ such that $ \mathcal{H}^n(X \backslash \cup_i f_i(A_i)) = 0 $. Is it true that for any subset $A \subset X$, $$ \mathcal{H^n}(A) = \mathcal{H}^n_\infty (A) \ .$$ The claim is true on $X = \mathbb{R}^n$. Note that the same equality fails horribly if we consider $\mathcal{H}^k$ for $k TITLE: What is the topology on the set of field orders QUESTION [7 upvotes]: Inspired by this question I was wondering whether there is a natural topology on the set of all orders on a field (that extend a given order on a subfield)? For example for the function field $\mathbb{Q}(X)$ there are the following examples of total orders: The convention $X>q$ for all $q\in \mathbb{Q}$; the convention $0 0\}$ for every $x\in F^\times$. This is a special case of the construction of the real spectrum of a ring $A$. That is the collection of the ``partial orders'' of $A$, which are essentially couples $(p,\le)$ where $p\in\mathrm{Spec}\,A$ is a prime ideal and $\le$ is an order on the residue field $k(p)$. We can topologize it as the closed subspace of $$\mathrm{sign}:\mathrm{Sper}\,A\to \prod_{x\in A}\{+1,0,-1\}$$ where $\{+1,0,-1\}$ has the coarsest topology such that $\{0\}$ is closed and $\{+1\}$ and $\{-1\}$ are open. To give a concrete example, the points of $\mathrm{Sper}\,\mathbb{Q}(x)$ are of one of the following forms $a^+,a^-$ where $a$ is a real algebraic number. The order is given by $f\ge 0$ iff $f$ is non-negative in $(a,a+\epsilon)$, resp. $(a-\epsilon,a)$ for some $\epsilon>0$ $r$ where $r$ is a trascendental real number. Here $f\ge0$ iff $f(r)\ge0$ $+\infty,-\infty$. Here the order is given by $f\ge0$ iff $f$ is non-negative on $(M,\infty$), resp. $(-\infty,-M)$ for some $M>0$. Equivalently, this picks the sign of the leading coefficient (resp. $(-1)^{\deg f}$ times the sign of the leading coefficient). These are all distinct, as one can easily show. The topology of $\mathrm{Sper}\,\mathbb{Q}(x)$ has a basis of clopen subsets of the form $(a,b)$ where $a,b$ are either real algebraic numbers or $\pm\infty$. This is the set of orders $\ge$ such that $f\ge0$ if $f$ is non-negative on $(a,b)$. Note that $a^+\in (a,b)$, but $a^-\not\in(a,b)$. There is a more comprehensive treatment of this, with more examples, in chapter 7 of Bochnak, Jacek; Coste, Michel; Roy, Marie-Françoise, Real algebraic geometry. Transl. from the French., Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. 36. Berlin: Springer. ix, 430 p. (1998). ZBL0912.14023.<|endoftext|> TITLE: Group of parallelizations of $M^3$ finitely generated? QUESTION [5 upvotes]: Let $M^3$ be a compact orientable 3-manifold. Then $TM$ is trivial and let's go ahead and fix a trivialization $\tau : M \times \mathbb{R}^3 \to TM$. Then given a map $g : (M, \partial M) \to (SO(3), 1)$ we can consider the new trivialization $g \cdot \tau$ that is given by $g \cdot \tau (p,v) = \tau(p,g(p)(v))$. The group of homotopy classes $[(M,\partial M), (SO(3),1)]$ then parameterizes all of the homotopy classes of trivializations of $M$ that agree with $\tau$ on the boundary of $M$. I know that this group is abelian, that the degree map is a homomorphism to $\mathbb{Z}$, and that the degree map is rationally an isomorphism. Is this group finitely generated? As more of an aside, for $M$ instead an $n$-manifold is the group $[(M,\partial M), (SO(n), 1)]$ abelian? REPLY [2 votes]: I believe that for any finite $n$-complex $X$, the group $$[X, SO(n)]_*$$ is finitely generated. I will follow Igor Belegradek's approach. In fact I only think $H^*(X;\Bbb Z)$ finitely generated and maybe $\pi_1$ finitely generated is necessary. There is a fibration $B\text{Spin}(n) \to BSO(n) \to B^2(\Bbb Z/2)$ coming from the fact that the extension is central, where the last map induces an isomorphism on $\pi_2$; this loops to the usual fibration (but now where every map is a loop map) $$\text{Spin}(n) \to SO(n) \to B(\Bbb Z/2).$$ In general this implies that if $X$ is a CW complex we have $$0 \to [X, \text{Spin}(n)] \to [X, SO(n)] \to H^1(X;\Bbb Z/2)$$ is an exact sequence of groups, which gives the desired result as soon as you know that $[X, \text{Spin}(n)]$ is finitely-generated. Now we have the short exact sequences $$[X, \Omega S^n] \to [X, \text{Spin}(n)] \to [X, \text{Spin}(n+1)] \to [X, S^n].$$ A priori the last map is exact at the level of pointed sets, but in fact for $X$ a $(\leq n)$-complex this is the same as the map $[X, \text{Spin}(n+1)] \to [X, K(\Bbb Z, n)] = H^n(X;\Bbb Z)$, and this is indeed a group map, given by looping the map $B\text{Spin}(n+1) \to K(\Bbb Z, n+1)$ given by the Euler class $e$. So this is a short exact sequence of groups when $X$ is a $(\leq n)$-complex, and $[X, S^n] = H^n(X;\Bbb Z)$ is a finitely-generated group. It follows from applying the Postnikov tower carefully that $[X, \Omega S^n]$ is also finitely-generated. So we have $$A \to [X, \text{Spin}(n)] \to [X, \text{Spin}(n+1)] \to B$$ for finitely generated abelian groups $A, B$ so long as $X$ is a $(\leq n)$-complex; passing to a quotient $A'$ of $A$ and a subgroup $B'$ of $B$ we have an exact sequence $$0 \to A' \to [X, \text{Spin}(n)] \to [X, \text{Spin}(n+1)] \to B' \to 0.$$ It follows that $[X, \text{Spin}(n)]$ is finitely generated iff $[X, \text{Spin}(n+1)]$ is so long as $X$ is a $(\leq n)$-complex. For us, we conclude that since $X$ is an $n$-complex, we have $[X, \text{Spin}(n)]$ f.g. iff $[X, \text{Spin}]$ f.g. To conclude observe that from the fibration $\text{Spin} \to SO \to B(\Bbb Z/2)$ (where all maps are loop maps) we conclude there is an exact sequence of groups $$0 \to [X, \text{Spin}] \to [X, SO] \to H^1(X;\Bbb Z/2),$$ and in particular $[X, \text{Spin}]$ is a subgroup of $[X, SO] = K^1(X)$. Now it follows from the Atiyah-Hirzebruch spectral sequence that $H^*(X;\Bbb Z)$ finitely generated implies that $K^1(X)$ is finitely generated.<|endoftext|> TITLE: A very slowly diverging series QUESTION [7 upvotes]: It is known and quite easy to prove that $S_{\mathbb N}(x) = \sum_{n\in\mathbb N, n\leq x} \frac 1 n$ grows as $\ln x$. Even more, $\lim_{x\rightarrow\infty} S_{\mathbb N}(x)-\ln x = \gamma$, the Euler-Mascheroni constant. It was also proved by Mertens that $S_{\mathbb P}(x)$, the sum of reciprocals of the primes not exceeding $x$, grows as $\ln\ln x$, and actually, $\lim_{x\rightarrow\infty} S_{\mathbb P}(x)-\ln\ln x = M$, the Meissel-Mertens constant. My questions is, is there an example of a set $A\subset \mathbb N$, such that $\lim_{x\rightarrow\infty} S_A(x)-\ln\ln\ln x=C$ for some real $C$. It is quite obvious that such a set exists, however I am looking for some "natural" examples (maybe some that were not specifically constructed as an answer to this question, but rather appeared during some research). REPLY [4 votes]: A nice example was found by Erdos; "On a problem of G. Golomb". Let $p_1 = 3$, and for $i > 1$, let $p_i$ be the least odd prime exceeding $p_{i-1}$ which is not congruent to $1$ mod $p_{j}$ for any $j < i$. That this sequence of primes has the property you seek follows from eq. (37) there (take logarithms).<|endoftext|> TITLE: Combinatorial Skeleton of a Riemannian manifold QUESTION [5 upvotes]: In Chung and Yau's paper: "A combinatorial trace formula" (MSN), they proved a combinatorial version of Selberg's trace formula for lattice graphs. I learned also in the setup that it makes sense to define Laplacian, its eigenvalues, and heat kernel of a graph (could be infinite and with loops). It showed a strong analogue to Riemannian geometry! On both sides, we can extract useful information from the objects' spectra. This was not overwhelmingly surprising to me, however, in the geometric case because I can sort of feel the picture of a laplacian and heat kernel based on my experience. Questions What's curious to me are the following: Having zero experience with graph theory, is there a better way to appreciate/interpret what those eigenvalues, heat kernel, … mean? There must be much more behind it. What are some updated important results? Do people understand this much better than they did 23 years ago? Most interestingly (to me), is there a way to extract a graph $G$ from a given Riemannian manifold $M$, so that the spectral theory on both sides agree? Any pointers to related work will be highly appreciated. REPLY [6 votes]: For construction of a sequence of discrete Laplacians whose spectrum converges to that of a given Riemannian manifold, see Dodziuk, Jozef, Finite-difference approach to the Hodge theory of harmonic forms, Am. J. Math. 98, 79-104 (1976). ZBL0324.58001. A good survey on spectral graph theory is Colin de Verdière, Yves, Spectres de graphes, Cours Spécialisés. 4. Paris: Société Mathématique de France. vi, 114 p. (1998). ZBL0913.05071. (There are surely more recent surveys or lecture notes, maybe somebody else can point at them.) Eigenvalues of the graph Laplacian carry interesting combinatorial information, such as the expansion parameter and the number of spanning trees, see Wikipedia articles Expander graphs and Kirchhoff's theorem. In order to approximate the geometry of a manifold, it is appropriate to use weights on the edges. The so-called cotangent Laplacian (well-known in dimension 2 and having an analog in higher dimensions) is a good choice.<|endoftext|> TITLE: Is there a systematic theory for Gibbs measures (better if on Hilbert spaces)? QUESTION [8 upvotes]: During these first months in my PhD, I realized how my computational problems can be drastically reduced to one single problem: Find an efficient way to sample from a Gibbs measure. Let me elaborate: if $H$ is a Hilbert space, $\mu$ a gaussian measure on it, then I need to numerically approximate a probability measure of the form $\mu_1(x) = Z^{-1} e^{-G(x)} \mu(dx)$ for a "potential" $G$ and a normalization constant $Z$. I do not have a fixed class for $G$, sometimes $H$ is just $\mathbb{R}^d$ for $d >> 1$, and generally the hypothesis on it might vary a lot (say, $G$ might come from a Neural Network, or from a PDE discretization). It would be absolutely fantastic to study how the properties on $G$ afflict $\mu_1$, and how they might be exploited e.g. to project numerical methods (in my case, e.g., a good approximation of $\mu_1$ might be used for image reconstruction). By searching on the web I realized how they are commonly called "Gibbs measure". I was very excited and tried to understand more, but on the other hand I found "only" material concerning the discrete case (lattices). This is certainly a starting point, but I was a bit confused and the approaches were different. Therefore I ask: Does a systematic theory for the (generic) problem above exist? Are there classics books that you recommend to study? Can you suggest some literature / papers? (Theoretical investigations are welcome, as well as numerics-oriented results) Thanks in advance. Ps: my background is in mathematics, but I am in a Computer Science department. People around me tried to help, and my question arouse after having successfully discussed with them - we are all very curious! REPLY [8 votes]: The field of improving convergence of sample averages is known as "enhanced sampling". As Robert pointed out, this is an incredibly hard problem. In my field (theoretical chemistry), we have been struggling with it for the last-half century. The correct way to approach this problem on depends heavily on what information you have available to you. The simplest case is when you can evaluate $Z$ exactly. In this case, your best bet is likely survey sampling. Failing that, let's assume you can evaluate $G$ at all points $x$ at reasonable cost. If: you have a good idea of what $\mu$ looks like "globally" (i.e. you know where in $H$ its modes are located, their variances, and the decay of the tails as you move away from $G$) your best is likely to be importance sampling. you don't know what $\mu$ looks like, but you know it is dominated by movement along a known lower-dimensional manifold, you can try umbrella sampling or metadynamics. you can evaluate $G$ at all points $x$ at reasonable cost, and you know that $G$ increases reasonably gently, you can try parallel tempering (this is basically the same as umbrella sampling, mathematically, just applied in a different way). you can evaluate $G$ at all points $X$, and you have samples from a related distribution $\pi$ that you don't understand well, but is easy to sample, you can try annealed importance sampling or Hamiltonian replica exchange. evaluation of $G$ is possible but expensive, and the problem isn't too high-dimensional, you can look into surrogate modeling. you can't directly evaluate $G$, but you have a dynamics that preserves $\mu$, you can try Nonequilibrium Umbrella Sampling. There are many other options and algorithms and this a very much an active area of research. However, hopefully this is enough to point you in the right direction. If you are looking for a mathematical treatment of the subject, "Free Energy Computations: A Mathematical Perspective" could be a good start. It's a bit hard to recommend reading without knowing more about the type of problem you have at hand. Nevertheless, this should hopefully be a good start.<|endoftext|> TITLE: A new combinatorial property for the character table of a finite group? QUESTION [21 upvotes]: Let $G$ be a finite group and $\Lambda = (\lambda_{i,j})$ its character table with $\lambda_{i,1}$ the degree of the ith character. Consider the following combinatorial property of $\Lambda$: for all triple $(j,k,\ell)$ $$\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} \ge 0.$$ It is a consequence of a more general result involving subfactor planar algebra and fusion category (see here Corollary 7.5, see also this answer). Question: Is this combinatorial property already known to finite group theorists? If yes: What is a reference? If no: Is there a group theoretical elementary proof? In any case: Are there other properties of the same kind? To avoid any misunderstanding, let us see one example. Take $G=A_5$, its character table is: $$\left[ \begin{matrix} 1&1&1&1&1 \\ 3&-1&0&\frac{1+\sqrt{5}}{2}&\frac{1-\sqrt{5}}{2} \\ 3&-1&0&\frac{1-\sqrt{5}}{2}&\frac{1+\sqrt{5}}{2} \\ 4&0&1&-1&-1 \\ 5&1&-1&0&0 \end{matrix} \right] $$ Take for example $(j,k,\ell) = (2,4,5)$, then $\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} = \frac{5}{3} \ge 0$. REPLY [12 votes]: This is indeed well-known in the character theory literature, and goes back to Frobenius and Burnside. What you are calculating is a positive rational multiple of a class algebra constant, and class algebra constants are clearly non-negative. Using the notation in David Speyer's comment, it is well-known, and derived in most representation theory texts that $\frac{|G|}{|C_{G}(f)| |C_{G}(g)|} \sum_{\chi} \frac{\chi(f)\chi(g)\chi(h)}{\chi(1)}$ is the number of times $h^{-1}$ is expressible as a product of a conjugate of $f$ and a conjugate of $g$. The character theory formula is easily derived from the expressions of the class sums as linear combinations of primitive central idempotents of the group algebra $\mathbb{C}G$, and can be found in many texts .<|endoftext|> TITLE: Can a compact object be a nontrivial self-retract? QUESTION [6 upvotes]: Let $\mathcal C$ be a locally finitely-presentable category, and let $X$ be a finitely-presentable object of $\mathcal C$. Question: Can there exist a nontrivial idempotent on $X$ whose fixed points are isomorphic to $X$? I don't think this should be possible in reasonable cases. Instances of this question include: Can a finitely-presentable ring be a nontrivial retract of itself? Can a finitely-presentable group be a nontrivial retract of itself? Can a finitely-presentable module be a nontrivial retract of itself? et cetera. EDIT: It occurs to me that there's a "universal" example where this does occur: let $\mathcal A$ be the category freely generated by an object with a nontrivial self-retract, and let $\mathcal C$ be the presheaf category on $\mathcal A$. So I suppose I'm really looking for conditions ensuring that this doesn't happen. REPLY [2 votes]: The property just says that every split epimorphism $X \to X$ (equivalently, every split monomorphism $X \to X$) is an isomorphism. I find this is easier to analyse. I do not agree that this "should" hold in reasonable cases. I would rather say that it holds "just by accident" in some categories. There is an immediate counterexample given by the universal cocomplete category with an object $X$ and two morphisms $i,p : X \to X$ satisfying $p \circ i = \mathrm{id}_X$. I omit its construction here, but one can check from its construction that it is locally finitely presentable, that $X$ is finitely presentable and that $p$ is no isomorphism. There are non-commutative rings $R$ with $R \cong R^2$ as $R$-modules (for example take $R$ to be the endomorphism ring of an infinite-dimensional vector space), so here $\mathrm{Mod}_R$ doesn't have the property. But when $R$ is commutative, then $\mathrm{Mod}_R$ has this property. It has already been proven in the comments, but here is a different proof: Let $M \cong M \oplus N$ for some finitely presentable $M$. Then $N$ is finitely presentable as well. By Noetherian approximation, we way massume that $R$ is Noetherian*. Then $M$ is Noetherian. But then every epimorphism $M \to M$ is an isomorphism, so $N=0$. I am pretty sure that the category of groups doesn't have the property either, but it will be hard to find counterexamples. *Details: $R$ is a filtered colimit of finitely presentable, thus Noetherian commutative rings $R_i$. Then $M \cong M_i \otimes_{R_i} R$ for some $i$, the same for $N$, since $M$ and $N$ are finitely presentable (just lift the presentation). The homomorphism $M \to M \oplus N$ also lifts to $M_i \to M_i \oplus N_i$ after increasing $i$ if necessary, the same for its inverse, and since they become inverse in the colimit, they become inverse over $R_i$ if we increase $i$ again.<|endoftext|> TITLE: Delooping the quotient space $SU/SU(n)$ QUESTION [8 upvotes]: Let $SU$ denote the infinite unitary group. Does the quotient space $SU/SU(n)$ admit a delooping $X$? One could also ask that this space $X$ sit in a fiber sequence $BSU(n)\to BSU\to X$, but this is not strictly part of the question. Note that $SU/SU(n)$ is not a topological group, because $SU(n)$ is not normal in $SU$ --- but this doesn't prohibit $SU/SU(n)$ from admitting a delooping. Perhaps a geometric construction of a H-space structure can be given by viewing the space as a Stiefel manifold. Note that $SU$ is an infinite loop space by Bott periodicity. REPLY [10 votes]: I'll work with mod $2$ cohomology. Note that $H^*(BSU(2))$ is polynomial on $c_2$ (in degree $4$) and $H^*(BSU)$ is polynomial on $c_k$ for $k\geq 2$. Here $c_k$ has degree $2k$ and so $H^6(BSU)=\{0,c_3\}$. If $X$ exists then it seems we should have $H^*(X)$ polynomial on generators in degrees $6,8,10,\dotsc$. In particular, $H^6(X)$ should be generated by $c_3$ and $H^{10}(X)$ should be generated by a single element that is $c_5$ modulo decomposables. However, $H^*(X)$ should also be closed under the action of the Steenrod algebra so it should contain $\text{Sq}^4(c_3)$, which is $c_2c_3$ by a calculation with symmetric polynomials. This is inconsistent, so $X$ cannot exist. I would guess that this line of argument can be improved to show that $SU/SU(2)$ does not deloop, but I have not tried to work out the details.<|endoftext|> TITLE: Is an overring of an order reflexive as a module over the order? QUESTION [9 upvotes]: Let $R$ be a one-dimensional Noetherian domain with fraction field $K$, let $\tilde{R}$ be the integral closure of $R$ in $K$, and assume that $\tilde{R}$ is finitely generated as an $R$-module. (In this situation one sometimes says that $R$ is an "order" in the Dedekind domain $\tilde{R}$.) For a fractional $R$-ideal $I$ we have $(R:I) := \{x \in K \mid x I \subset R\} = \operatorname{Hom}_R(I,R)$. Question 1: Does it always hold that $(R:(R:\tilde{R})) = \tilde{R}$? Question 2: Same question but with an intermediate ring $R \subset R' \subset \tilde{R}$. Does it always hold that $(R:(R:R')) = R'$? Comments: 1) For any fractional $R$-ideal $I$ we have $(R:I) = \operatorname{Hom}_R(I,R)= I^{\vee}$. Therefore my questions are equivalent to asking whether $\tilde{R}$ and $R'$ must be reflexive as $R$-modules. A one-dimensional Noetherian domain $R$ is Gorenstein iff every fractional ideal is reflexive, so the answer to 2) is yes if $R$ is Gorenstein. 2) In a draft of a paper of mine, I wrote down $(R:(R:R')) = R'$ without any justification. Now I'm concerned! The main results of the paper apply to a class of orders that are in fact Gorenstein, so it's not so terrible, but I would like to extend these results to a larger class of orders if possible (and also build up background results in a graceful way). REPLY [10 votes]: The answer to Question 1 is yes by (the proof of) Proposition 2.14 in this paper. The answer to Question 2 is not. Let $R=k[t^3,t^4,t^5]$ so $\tilde R=k[t]$. I claim that for this ring any immediate ring $R\subsetneq R'\subsetneq \tilde R$ is not reflexive. Let $m=(t^3,t^4,t^5)$ be that maximal ideal of $R$. As the semigroup generated by $(3,4,5)$ contains all numbers from $3$, $mR'\subseteq m\tilde R\subset R$. It follows that $Hom_R(R',R)$ contains $m$, and since it has to be an ideal of $R$ and can not be $R$, it is $m$. So now we just need to show that $Hom_R(m,R)\neq R'$. But this is clearly equal to $\tilde R$, as multiplication by $t^i, i\geq 0$ maps $m$ to $R$ and no negative power works. In general, I think for a non-Gorenstein domain of dimension one, torsion free modules are rarely reflexive.<|endoftext|> TITLE: Colimits in the category of (not necessarily locally convex) topological vector spaces QUESTION [9 upvotes]: Do colimits in the category of (not necessarily locally convex) topological vector spaces (over R, C, respectively) exist in general? If no, is there a well-known condition of when they exist? If yes, how can I describe the topology of the colimit? (The example in my mind has following properties. First, it is a filtered colimit; it is a chain of inclusions, not necessarily linearly ordered, and not necessarily countable. Also, each inclusion is in general not a homeomorphism onto the image. Topology gets coarser and coarser as you embed into a larger space.) (I'm also interested on a condition when a subset of the colimit is contained in one of the component spaces. For example, in the test function space $\mathcal{C}_c^\infty(\mathbb{R}^n)$ (which is of course NOT a colimit in the category of topological vector spaces), any bounded set is contained in one of $\mathcal{C}^{\infty}(K)$, $K$ a compact set. When can I say something similar?) REPLY [8 votes]: The Springer Lecture Notes 639 Topological Vector Spaces of Adasch, Ernst, and Keim contain in § 4 a more or less explicit construction of inductive (=co-) limits in the category of topological vector spaces based on the notion of a string: A sequence $(U_n)_{n\in\mathbb N}$ of balanced and absorbing sets such that $U_{n+1}+U_{n+1}\subseteq U_n$. Similarly to the locally convex theory (if $U$ is absolutely convex the sequence $U_n=2^{-n}U$ is a string) you can define a vector space topology by specifying a directed family of strings. If you have an inductive spectrum $(E_\alpha,i_\alpha)_{\alpha\in I}$ where $i_\alpha:E_\alpha\to E$ are linear mappings such the union of their ranges span $E$ you consider the family of all strings in $E$ all whose preimages are strings for the given topologies. The basics of such inductive limits are then rather similar to the locally convex case. If the index set is countable and all $E_\alpha$ are locally convex the inductive limits in the categories of topological vector spaces and of locally convex spaces coincide -- but this seems to be the only case where you have "regularity results" on the behaviour of bounded or compact sets of the inductive limit like the one of Dieudonne-Schwartz or (if house advertising is permitted) in my Springer Lecture Notes Derived Functors in Functional Analysis.<|endoftext|> TITLE: Are quadrics the cones of maximal symmetry? QUESTION [5 upvotes]: A paper by Ehlers, Pirani, and Schild axiomatizes the geometry of general relativity in what seems like a nice way. However, Jacobson criticizes one aspect of the system as not natural: One deep question is why the causal cone is given by a quadric in the tangent space. After all, one can easily imagine a partial ordering relation that arises from an infinitesimal conical structure which is not a quadric. In the EPS paper, the quadratic nature of the light cone is derived from their axioms. This is not very satifying however, since one of the axioms [$\text{L}_1$] is not particularly physically natural. Aside from any axioms, there is a special property of quadrics that might underlie the fact that the causal structure is given by one. Namely, quadrics have the largest possible symmetry group of any conical subset of the tangent space. This is the Lorentz group, together with the conformal rescalings, a group with 7 continuous parameters. I should have a reference for this but I don’t know of one. Perhaps Herman Weyl proved it. Perhaps it is not even true (see Problem 11). Is this actually an open problem? In general, is it possible to characterize "proper" Finsler metrics (those not arising from a bilinear form) as being Finsler metrics that lack symmetry? It's not even obvious to me how to formalize this question, since normally I would talk about the symmetry of a metric in terms of its Killing vectors, but I don't think that machinery applies to a proper Finsler metric. Therefore I'm not sure how to describe the relevant symmetry in a way that doesn't depend on the choice of a basis. References Jürgen Ehlers, Felix A. E. Pirani, Alfred Schild, "The geometry of free fall and light propagation," republished in General Relativity and Gravitation, 2012, Volume 44, 1587, https://doi.org/10.1007/s10714-012-1353-4 T. A. Jacobson, "A spacetime primer," http://terpconnect.umd.edu/~jacobson/spacetimeprimer.pdf REPLY [4 votes]: There is a whole theory of Lorentz-Finsler spaces, and their casuality has been considerably developed in recent years. This fact alone shows that it is not compelling to work with round light cones (i.e. cones that intersected by a hyperplane not passing through the origin of $T_xM$ give an ellipsoid. Notice that the notion of ellipsoid is affine, so the notion of round cone is well posed). The indicatrix (velocity/observer space) $I_x\subset T_xM$, is the locus $\{y: 2L(x,y)=-1\}$, where $L(x,y)$ is the Finsler Lagrangian whose vertical Hessian has Lorentzian signature $(-, +, +, +)$. It can be shown to be asymptotic to the boundary of a sharp convex cone with non-empty interior (the light cone). The sharpness of the cone expresses the finiteness of the speed of light in every direction. The non-roundness the fact that the speed of light is not isotropic. What distinsuishes general relativity is the fact that the indicatrix (and hence the cone) is sent into itself by a group of endomorphisms of $T_xM$ which has maximal dimension. This is not the case in general. When this action is transitive the velocity space is said to be homogeneous. One can say that the relativity principle still holds true, in the sense that an observer cannot tell her position in $I_x$, i.e. her velocity (so velocity is only relative), by local measurements. However, the velocity space need not be isotropic (examples with affine sphere indicatrices can be constructed). Suppose that any two spacetime points have isomorphic indicatrices, i.e. the velocity space does not change its geometry from point to point. The interesting fact is that the so called Berwald spaces, i.e.\ those Lorentz-Finsler spaces for which the connection does not depend on the velocity variable $y$, that are proper in your sense (i.e. the indicatrix has no symmetry) locally are necessarily translationally invariant (flat).<|endoftext|> TITLE: Is the simplicial nerve a localization? QUESTION [7 upvotes]: Given a simplicial category $\mathcal{C}_{\ast}$ (if necessary, you may assume it's fibrant), denote as $\mathcal{C}$ its underlying ordinary category, and as $\mathcal{W}$ the class of all equivalences in $\mathcal{C}_{\ast}$ (in the simplicially enriched sense, i.e. $f: A \to B$ is an equivalence if there exists $g: B \to A$ and 1-simplices $gf \to id_A$ and $fg \to id_B$ in Map(A,A) and Map(B,B) respectively). Regard $\mathcal{W}$ and $\mathcal{C}$ as discrete simplicial categories, and form the following homotopy pushout in the Joyal model structure on simplicial sets: $\require{AMScd}$ \begin{CD} N(\mathcal{W}) @>>> N(\mathcal{C})\\ @VVV @VVV\\ Ex^{\infty}N(\mathcal{W}) @>>> \mathcal{C}^{\infty}_{\mathcal{W}} \end{CD} which induces a functor of $\infty$-categories $q: \mathcal{C}^{\infty}_{\mathcal{W}} \to N_{\Delta}(\mathcal{C}_{\ast})$. Now, my intuition tells me that $N_{\Delta}(\mathcal{C}_{\ast})$ should be the $\infty$-category obtained by localizing at the equivalences, in other words, that $q$ should be an equivalences of $\infty$-categories. Is there a self-contained, slick way to show that this is the case? If it can help, I'd like to point out that this is the same as showing that the corresponding adjoint map $\tilde{q}: \mathfrak{C}[\mathcal{C}^{\infty}_{\mathcal{W}}] \to \mathcal{C}_{\ast}$ is a weak equivalence of simplicial categories. Furthermore, we know that $\tilde{q}$ is induced by the homotopy pushout (in the Bergner model structure on simplicial categories) given by the image of the above square along the functor $\mathfrak{C}$. REPLY [9 votes]: This is not true. Here is a counter example. We let $\mathcal{C}_*$ be the following simplicial category. It has two objects 0 and 1. Their only endomorphisms are the identity. There are no morphisms from 1 to 0. The simplicial set of morphisms from 0 to 1 is the simplicial circle $\Delta[1] / \partial \Delta[1]$. if you want a fibrant version, you could just as well use any reduced model of the simplicial circle such as the Bar construction on $\mathbb{Z}$. Now $\mathcal{C}= [1]$, the free walking 1-cell. And its nerve is the one-simplex. There are no non-trivial equivalences, so $W \cong \partial [1] = 0 \sqcup 1$. Then $Ex^\infty N(W) = W$ and so your $\mathcal{C}^\infty_W = N(\mathcal{C}) = \Delta[1]$. But this is definitely not equivalent to your original. $$\mathfrak{C}(\Delta[1]) = [1] \to \mathcal{C}_*$$ is NOT a DK-equivalence.<|endoftext|> TITLE: Derived Category of the derived critical locus, is it the category of Matrix Factorizations? QUESTION [7 upvotes]: Let $W \in \mathbb{C}[x_1, \dots, x_n]=R$ be a polynomial with an isolated critical point at the origin. A Matrix Factorizations for $W$ consists a $\mathbb{Z}/2\mathbb{Z}$-graded finite free $R$-module $E$ with an odd differential $d:E \to E$ satisfying $d^2 = W \cdot Id$ instead of the usual square zero condition. Chain maps between matrix factorizations and chain homotopies are defined in the same way as for ordinary complexes. There is a category of matrix factorizations with morphisms homotopy equivalence classes of chain maps. This category is supposed to be a model for the "derived category of the singularity". For example this point of view is advocated by Orlov, see arxiv/math/0302304. Another way to get to the derived category of the singularity is via derived algebraic geometry, say in the sense of Vezzosi's Derived Critical Loci I - Basics. In this approach we let $X= spec(R)$ and consider the derived intersection of the zero section $0:X \to T^*X$ and $dW: X \to T^*X$. This will be a derived scheme, i.e. a scheme with a sheaf non-positively graded dg-algebras satisfying a few regularity conditions. Presumably we should be able to take the bounded derived category of coherent sheaves for such a derived scheme. My Question: Are these two versions of the derived category of a singularity related in any way? Are they in some sense equivalent? What is known about their relationship, if anything. Note that the first one is a $\mathbb{Z}/2\mathbb{Z}$-graded triangulated category (shift squares to the identity), while that is not true for the later. Perhaps they become equivalent if we use a 2-periodic version of the derided category? REPLY [12 votes]: These are indeed related. The first thing to know is that they both ``live'' (i.e., sheafify) over the critical locus (this is not saying much if you assume $W$ has isolated critical points, but interesting in more general cases). This is clear for $\operatorname{Coh}(\operatorname{Crit}(W))$. For matrix factorizations, we can identify $\operatorname{MF}(W)$ with the category of singularities $\operatorname{Coh}(W^{-1}(0))/\operatorname{Perf}(W^{-1}(0))$, which vanishes on the regular part of $W^{-1}(0)$. In the case where the superpotential $W$ is Morse, or Morse-Bott, the category of matrix factorizations is close to being the $2$-periodization of $\operatorname{Coh}(\operatorname{Crit}(W))$. This is the subject of Teleman's Matrix Factorisation of Morse-Bott functions. The basic case to understand is when the superpotential is $W = x^2$ defined on $\mathbb{A}^1$. The category of matrix factorizations is $\operatorname{Coh}(k[x]/x^2)/\operatorname{Perf}(k[x]/x^2)$. A Koszul duality computation tells us that this is the category of modules over the algebra $k[\beta, \beta^{-1}]\langle \eta | \eta^2 = \beta \rangle$. Here $\beta$ is a variable of cohomological degree $2$ giving the $\mathbb{Z}/2$-grading, and $\eta$ is a variable of degree $1$. We interpret this result as a $2$-periodic Clifford algebra on a generator $\eta$ of degree $1$. The derived critical locus in this case is just $\operatorname{Spec}(k)$. You can then think about $\operatorname{MF}(x^2)$ as the $2$-periodization of $\operatorname{Coh}(\operatorname{Spec}(k))$, with an extra Clifford factor. From this you can build higher dimensional Morse functions: the rule is that $\operatorname{MF}(f \boxplus g) = \operatorname{MF}(f) \otimes_{k[\beta, \beta^{-1}]} \operatorname{MF}(g)$ (see Preygel's Thom Sebastiani and Duality for Matrix Factorizations). The tensor product of $k[\beta, \beta^{-1}]\langle \eta | \eta^2 = \beta \rangle$ with itself turns out to be $k[\beta, \beta^{-1}]$. With this one can conclude that when $W = x_1^1 + \ldots x_n^2$ the category of matrix factorizations is the $2$-periodization of $\operatorname{Coh}(\operatorname{Crit}(W))$, with an added Clifford factor in odd dimensions. In the Morse-Bott case one has extra corrections arising from the topology of the normal bundle to the critical locus. When you move beyond the Morse-Bott case the two categories start diverging. Their relationship in general is roughly like the relationship between the algebra $D_X$ of differential operators on a smooth scheme $X$, thought of as the deformation quantization of $T^*X$, and $\mathcal{O}_X$. The derived critical locus of $W$ has a $(-1)$-shifted symplectic structure. If one works in the $2$-periodic context this turns into a $1$-shifted symplectic structure, which allows one to deform the symmetric monoidal category $\operatorname{QCoh}(\operatorname{Crit}(W))$ into a monoidal category. This is the monoidal category of modules over the $E_2$-algebra associated to the Gerstenhaber algebra $\mathcal{O}_{\operatorname{Crit}(W)} \otimes k[\beta, \beta^{-1}]$. This $E_2$-algebra turns out to agree with the Hoschild cochains of $\operatorname{MF}(W)$, which implies that the deformation of $\operatorname{QCoh}(\operatorname{Crit}(W))\otimes k[\beta,\beta^{-1}]$ acts on (the ind-completion of) $\operatorname{MF}(W)$.<|endoftext|> TITLE: Ethics questions concerning a referee assignment QUESTION [19 upvotes]: I recently refereed a paper that I returned to the author(s) for revision. The thrust of their argument relied on a claim whose justification I felt was lacking. I dutifully raised the issue in my report and, in addition, I corrected another portion of their proof. The author(s) have yet to revise their work and, in the interim, I came up with a justification for their claim. I now have a proof of this result (which is important for another paper I'm working on) and would like to publish it. What are the ethics/options here? Do I need to provide them with the correct proof? May I submit the result as my own after their revision? Should I recuse myself from serving as the referee? EDIT: Not sure if this makes a difference, but the conjecture the author(s) purports to resolve is one that I raised in a previous a paper and one for which I obtained partial results with similar methods. Although, I have worked on the problem and obtained partial results, I would not have obtained a proof without refereeing their work. REPLY [22 votes]: The answer to your question really depends on how much work required justifying their claim. In my personal experience: It was many times that a referee provided me with an argument that lead to a simplification of my poof or even provided me with new results that I included in my paper. I did the same many times when I refereed the papers. Therefore if the justification did not require much work, I would give it to the authors of the paper and let them publish it. Then you would be mentioned as an anonymous referee. May I submit the result as my own? If so, must I wait until their revision before I submit my own paper? That I do not really understand. Since they do not have justification of the claim, they cannot publish it. If they come up with a justification, your proof is of not much value since they already have a proof. In my opinion, you must not publish your proof before their paper is published. If proving the claim is really difficult and required a lot of work, I do not really know what to advise. In one related situation I did as follows: I was a referee of a paper $X$. However, before refereeing the paper I already knew how to prove a much better result (5 pages of elementary calculations vs 40 pages of a difficult proofs of far better results). But I only had my proof in a draft. I wrote to the author that I was a referee and I suggested that he would withdraw the paper from the journal and we would publish a joint paper. This is exactly what happened.<|endoftext|> TITLE: Why is weak Kőnig's lemma weaker than Kőnig's lemma? QUESTION [13 upvotes]: Kőnig's lemma states that any finitely-branching tree with infinitely many nodes contains an infinite path. Weak Kőnig's lemma states the same thing about binary trees. It's known that these are not equivalent over the base system $RCA_0$, but I'm struggling to see what goes wrong with the following construction: Take an arbitrary infinite finitely-branching tree $T$; Apply the Knuth transform to convert this into an equivalent binary tree $B$ on the same vertex-set; Use the statement of Weak Kőnig's lemma to find an infinite path $P = (x_1, x_2, x_3, \dots)$ in $B$. Note that for each pair $(x_i, x_{i+1})$ of consecutive terms in $P$, $x_{i+1}$ is either a child or a sibling of $x_i$ when viewed as elements of $T$. We then define a subsequence which consists of only the terms $x_i$ such that $x_{i+1}$ is a child (rather than a sibling) of $x_i$. The resulting subsequence is then an infinite path in the original finitely-branching tree $T$. Since this proof appears to be valid, my guess is that it's using something that can't be proved in $RCA_0$ (or indeed in $WKL_0$). REPLY [7 votes]: Interestingly, even Kőnig's lemma restricted to subtrees of $\omega^\omega$ for which every node has at most two children is equivalent to the full Kőnig's lemma, not weak Kőnig's lemma. This is because you can make such a tree $T$ for which an infinite branch gives you a halting oracle: For each $\sigma\in T$ of length $n$, put $\sigma\frown0$ in $T$, and if the $n$th program halts in $k$ steps, also put $\sigma\frown k$ in $T$, and cut off all descendents of $\sigma\frown0$ of length $n+k$ (that is, instead of following the previous rules for creating child nodes, don't create any child nodes). This is a computable tree, but the only infinite branch has a $0$ in the $n$th place iff the $n$th program does not halt. This can be relativized to any oracle, and hence gives you $ACA_0$, which as Bjorn Kjos-Hanssen noted, is enough to prove Kőnig's lemma. And Kőnig's lemma for subtrees of $n^\omega$ is equivalent to weak Kőnig's lemma. The proof of Kőnig's lemma from weak Kőnig's lemma that you described can be carried out in $RCA_0$ in this case. Same goes if you let nodes in deeper levels have more children, just so long as there is a computable bound for the labels of the children that any given node can have. This is all just to emphasize what Bjorn already hinted at: binary vs finitely-branching is sort of a red herring; the difference is whether or not there is a function bounding how far you have to look for child nodes.<|endoftext|> TITLE: Abstract Jordan decomposition maybe not exist QUESTION [6 upvotes]: An abstract Jordan decomposition of an element of a Lie algebra L is a decomposition of the form a = a$_{s}$ + a$_{n}$, where (a) ad a$_{s}$ is a diagonalizable (equivalently semisimple) endomorphism of L. (b) ad a$_{n}$ is a nilpotent endomorphism. (c) [a$_{s}$, a$_{n}$] = 0 . This note defines the abstract Jordan decomposition in an arbitrary Lie algebra. Abstract Jordan decomposition in a Lie algebra is unique when it exists iff its centre is zero. It seems that the abstract Jordan decomposition maybe not exist even when its centre is zero, who can show me an example? The same question is at here with no answer. REPLY [10 votes]: In $\mathfrak g=\left\{\begin{pmatrix}x&x&y\\0&x&z\\0&0&0\end{pmatrix}:x,y,z\in \mathbf R\right\}$, $\mathrm{ad}$-semisimple and $\mathrm{ad}$-nilpotent elements all have $x=0$; so they don’t span.<|endoftext|> TITLE: Geometric intuition for Mather's cube theorem QUESTION [5 upvotes]: Mather's cube theorem for the category of topological spaces says that given a homotopy-commutative cube: If one pair of opposite faces are homotopy pushouts and the two remaining faces adjecent the source vertex are homotopy pullbacks, then the final two faces are also homotopy pullbacks. What is the geometric intuition behind this theorem? REPLY [4 votes]: I'm not sure what counts as an intuitive explanation, but this is sort of how I think about it. Say that $B=B_1\cup B_2$ and $B_0=B_1\cap B_2$. This is the second pushout square. Now let $E_1$ be a bundle over $B_1$ and let $E_2$ be a bundle over $B_2$, and suppose that the restriction to $B_0$ is the same for both bundles -- call it $E_0$. These are the two given pullback squares. Now let $E$ be the union of $E_1$ and $E_2$ along $E_0$. This is the other pushout square. $E$ should be a bundle over $B$ whose restriction to $B_1$ (resp. $B_2$) is the bundle $E_1$ (resp. $E_2$).<|endoftext|> TITLE: What is the best known upper bound for G(k) in Waring's Problem? QUESTION [9 upvotes]: Waring's problem: for fixed $k, s$ write natural numbers $n$ in the form $n=x_1^k+\dots+x_s^k$. $$ G(k):= \min\{s:\text{ all sufficiently large $n$ can be written as above}\} $$ A quick search / Wikipedia / Vaughan-Wooley's survey paper all suggest that the best upper bound for $G(k)$ is (Wooley) $$ G(k)\leq k\log k+k\log\log k+O(k). $$ Recent work of Bourgain, Demeter, Guth (https://arxiv.org/abs/1512.01565) and Wooley (https://arxiv.org/abs/1708.01220) on Vinogradov's theorem seems to improve several quantities related to Waring's problem and similar questions (eg. $g(k)$, which is as $G(k)$ but for all $n$). But I can't find any evidence of improvements in $G(k)$ following from this result. Does anyone know if such improvements are possible, and if so where to find them? REPLY [4 votes]: While not exactly the classical quantity $G(k)$, there are partial improvements on asymptotic behavior for the Waring-Goldbach problem, namely for $H(k)$, defined to be the least integer $s$ such that every sufficiently large positive integer congruent to $s$ modulo $K(k)$ may be written as $$p_1^k + p^k_2 + ··· + p^k _s = n,$$ where $p_1,\dots,p_s$ are prime numbers and where $K(k)=\Pi_{(p-1)|k} p^{\gamma}$, and $\gamma$ is a parameter defined below. Shortly after Bourgain, Demeter, and Guth's resolving of the Vinogradov's conjecture Kumcev and Wooley used their results to improve Hua's estimate $$H(k)\leq k(4 \log k + 2 \log \log k + O(1)) \text{ as }k\rightarrow\infty, \text{ to }$$ $$H(k)\leq (4k − 2) \log k + k − 4 \text{ for } k\geq 3.$$ This was improved again later by Kumcev and Wooley to $$H(k) ≤ (4k − 2)\log k − (2 \log 2 − 1)k − 3$$ holding for large $k$. To compute the value $K(k)$, let natural $k$ and prime number $p$ be given, and define $\theta = \theta(k, p)$ to be the integer with $p^\theta|k$ but $p^{\theta+1}\nmid k$, and $\gamma=\gamma(k, p)$ by $$ γ(k, p) =\begin{cases} \theta + 2,\ \text{ when } p = 2 \text{ and }\theta > 0,\\ \theta + 1, \text{ otherwise.} \end{cases}$$<|endoftext|> TITLE: What does the KL being symmetric tell us about the distributions? QUESTION [9 upvotes]: Suppose two probability density functions, $p$ and $q$, such that $\text{KL}(q||p) = \text{KL}(p||q) \neq 0$. Intuitively, does that tell us anything interesting about the nature of these densities? REPLY [5 votes]: One example where this happens is when $P$ and $Q$ are "antipodal" distributions -- say, $p=(p_1,\ldots,p_n)$, $q=(q_1,\ldots,q_n)$, $q_i=1-p_i$, and $$P=\mathrm{Ber}(p_1)\times\mathrm{Ber}(p_2)\times\ldots \mathrm{Ber}(p_n)$$ and $Q$ is defined analogously. Then $KL(P||Q)=KL(Q||P)$. These play a role in optimal decision theory, see http://jmlr.org/beta/papers/v16/berend15a.html and https://projecteuclid.org/euclid.aos/1564797865 . REPLY [5 votes]: We're looking at the equation $$-\sum_{x\in\mathcal{X}} P(x) \log\left(\frac{Q(x)}{P(x)}\right) = -\sum_{x\in\mathcal{X}} Q(x) \log\left(\frac{P(x)}{Q(x)}\right). $$ In the Bernoulli case, $$-p \log\left(\frac{q}{p}\right) -(1-p) \log\left(\frac{1-q}{1-p}\right) = -q \log\left(\frac{p}{q}\right) -(1-q) \log\left(\frac{1-p}{1-q}\right) $$ The only solutions are random variables $X$ and $Y$ with $X=1-Y$ (mentioned by @Aryeh) and the trivial solution $X=Y$:<|endoftext|> TITLE: Tor functor and invertible elements QUESTION [6 upvotes]: Let $A$ be a commutative ring, $a \subset A$ be an ideal. For $A$-module $M$ let $S \subset A$ be the set of elements, which are invertible in $M$, so $M$ is actually a $S^{-1}A$-module. It is not hard to show, that If $S\cap a \neq \emptyset$, then $\operatorname{Tor}_*^A(A/a, M) = 0$. Under what conditions the converse is also true? I can prove it for PID, so I am interested if it can be extended to wider classes of rings, Noetherian for example. Also, I would be glad to receive some references on relations between Tor functors, quotient rings and multiplicative sets. UPD I’m concerned about special case, when $a$ is prime, or even maximal (because of some geometric interpretations), but the general case is interesting too. REPLY [3 votes]: Without any finiteness assumptions on $M$, the converse fails already for $A=k[x, y]$ ($k$ is a field). Take $M=k[x,y^{\pm 1}]\oplus k[x^{\pm 1},y]$ and $a=(x,y)$. The module $M$ is flat over $A$ so $M\otimes^{\mathbb{L}}_A A/a=M\otimes_{A}A/a=M/aM=0$. However $S=k\setminus\{0\}$ because the sets of elements invertible on $k[x^{\pm 1},y]$ and $k[x,y^{\pm 1}]$ are $\{ax^n|a\in k\setminus\{0\},n\in\mathbb{N}\}$ and $\{ay^n|a\in k\setminus\{0\},n\in\mathbb{N}\}$ respectively. On the positive side, for a finitely generated $M$ the vanishing of the tensor product $A/a\otimes_A M$ already implies that some element of $a$ is invertible on $M$. Indeed, consider the annihilator ideal $I=Ann(M)\subset A$ of the module $M$. Since $M$ is finitely generated, the closed subspace $V(I)\subset Spec\, A$ is the support of $M$. Claim. If $A/a\otimes_A M=0$ then $V(a)\cap Supp\, M=\emptyset$ Proof. This is immediate from Lemma 10.39.9 (1) on Stacks Project https://stacks.math.columbia.edu/tag/00L3. The intersection $V(a)\cap V(I)$ is empty if and only if $a+I=A$. Take $f\in a, g\in I$ such that $f+g=1$. The element $g$ annihilates $M$, so the multiplication by $f$ on $M$ is the identity endomorphism, hence $f\in S\cap a$.<|endoftext|> TITLE: Topological Spin manifolds in dimension 4 QUESTION [9 upvotes]: In his ICM Adress at Nice (Proceedings of the International Congress of Mathematicians Nice, September, 1970, Gauthier-Villars, editeur, Paris 6 e ,1971, Volume 2, pp. 133-163.), Robion Kirby adresses the problem, whether the fourth topological Spin bordism group is $\mathbb{Z}$ or $\mathbb{Z}\oplus \mathbb{Z}/2$. Depending on whether the triangulation obstruction $$\Delta:\Omega_{4}^{\rm Spin Top}\to \mathbb{Z}/2 $$ is zero or not. This being equivalent to the fact that Rokhlin's divisibility property of the signature still holds modulo 2, in the sense that the signature is eight times the triangulation obstruction modulo two. This should be an example of a topological spin manifold which is not bordant to a smooth spin manifold. ¿Is this problem solved? Are there newer references for this problem I am not aware of? REPLY [8 votes]: The map $\Omega_4^{\text{Spin}} \to \Omega_4^{\text{SpinTop}}$ is taken isomorphically by the signature to the inclusion $16\Bbb Z \hookrightarrow 8\Bbb Z$, so that the groups are abstractly isomorphic but that the natural map is not an isomorphism. This is obtained as Theorem 13.1 on page 325 of Kirby and Siebenmann's (1977) foundational essays. A representative for the nonzero element of $\Omega_4^{\text{SpinTop}}/\Omega_4^\text{Spin} = \Bbb Z/2$ is given by Freedman's E8 manifold: take the E8 plumbing, a smooth manifold with intersection form E8 and boundary the Poincare homology sphere $\Sigma(2,3,5)$, and cap the boundary off with a contractible topological manifold. The result is topologically spinnable because the manifold is simply connected and the intersection form is even.<|endoftext|> TITLE: When and why are Adams operations "non-negative"? QUESTION [6 upvotes]: We can think of the unary operations in a lambda-ring as integer linear combinations of Young diagrams; for example the operation $\lambda^n$ corresponds to the Young diagram with $n$ rows and one column. Some of these operations are manifestly non-negative in the following sense: they're linear combinations of Young diagrams with natural number coefficients. If we apply a manifestly non-negative operation to an element of the representation ring $R(G)$ coming from a representation of $G$, we get another element coming from a representation - not just a formal difference of such elements. Similarly, if we apply a manifestly non-negative operation to an element of the K-theory $K(X)$ coming from a vector bundle on $X$, we get another element coming from a vector bundle - not just a formal difference of such elements. I'm confused about Adams operations. For $k > 1$, the Adams operation $\psi_k$ is apparently not manifestly non-negative, since it's given by an alternating sum of hook-shaped Young diagrams with $k$ boxes. However, if we apply $\psi_k$ to an element of $K(X)$ coming from a vector bundle over $X$, I believe we get an element coming from a vector bundle. It's certainly true for line bundles: $\psi_k [L] = [L^{\otimes k}]$ when $[L]$ is the element of $K$-theory coming from a line bundle $L$. It's also true for bundles that split as a sum of line bundles, since $\psi_k : K(X) \to K(X)$ is a ring homomorphism. And I think it follows for all vector bundles using the splitting principle for K-theory (Corollary 4.3.4 here). So, it seems that the Adams operations, while not manifestly non-negative, are still non-negative in the sense that they send elements of $K(X)$) coming vector bundles to elements coming from vector bundles - not merely formal differences of such. My questions are: 1) Is this true? 2) If so, which integer linear combinations of Young diagrams give operations that are non-negative in this sense? 3) What's really going on here? In particular, I've defined "non-negative" using $K(X)$, but these should be examples of a more general phenomenon. The Grothendieck ring $K(C)$ of any symmetric monoidal Cauchy-complete $\mathbb{Q}$-linear category $C$ is a lambda-ring, in a way that generalizes this. We can define "non-negative" operations on $K(C)$ to be those sending elements coming from objects of $C$ to elements coming from objects of $C$. Are Adams operations always non-negative on $K(C)$, or this just true for certain $C$? Which $C$ are these? And which linear combinations of Young diagrams give operations that are non-negative on $K(C)$ for all symmetric monoidal Cauchy-complete $C$? REPLY [4 votes]: The Adams operations aren't always non-negative. (They are if you restrict to the subring generated by line bundles, though.) Here's a counterexample in the world of finite CW-complexes. Let $X$ be the truncated projective space $\mathbb{RP}^{2n-1}/\mathbb{RP}^{2t-1}$. It's reduced $K$-theory was calculated by Adams here to be $\mathbb{Z}\oplus\mathbb{Z}/2^{n-t}\mathbb{Z}$. Let $\mu$ and $\nu$ be generators of each factor. They are best understood as follows. The inclusion of the $2t$-skeleton gives a map $S^{2t}\hookrightarrow X$. Pulling back along this map sends $\mu$ to a generator of $\tilde{K}(S^{2t})$. Let $\eta$ be the complexified tautological line bundle over $\mathbb{RP}^{2n-1}$. Pulling back along the quotient map $\mathbb{RP}^{2n-1}\rightarrow X$ sends $\nu$ to the class $([\eta]-1)^{t+1}=(-2)^{t}([\eta]-1)$. Note that the last equality is a simple consequence of the fact that real line bundles are self dual. Adams also calculated the effect of the Adams operations on these generators. For example, if $k$ is odd then $$\psi^{k}\mu=k^{t}\mu+\frac{k^{t}-1}{2}\nu$$ (Sidenote: that 2 in the denominator is the poison dart in Adams' vector-fields-on-spheres proof. For our purposes it is only important that $\psi^{k}\mu$ has a component in both the $\mu$ and $\nu$ factor. If we add a $t$ dimensional trivial bundle to the class $\mu$, then it is represented by an honest bundle $V$, which can be constructed via clutching, by cutting apart the $S^{2t}$ that comprises the $2t$-skeleton of $X$. Therefore, $$\psi^{k}[V]=k^{t}[V]+ t -k^{t}t +\frac{k^{t}-1}{2}[\eta] -\frac{k^{t}-1}{2}$$ Multiplying the clutching function of $V$ by $k^{t}$ (as an element in $\pi_{2t-1}U(t)=\mathbb{Z}$ gives a vector bundle $V_{k^{t}}$ representing $k^{t}\mu+t$, so $$\psi^{k}[V]=[V_{k^{t}}]+\frac{k^{t}-1}{2}[\eta] -\frac{k^{t}-1}{2}$$ Now, it's a general fact that if $E$ is a vector bundle such that the class $[E]-n$ is representable by a vector bundle, then the last $n$ Chern classes of $E$ must be zero. We can achieve this obstruction in our example by choosing the right $k$, $n$, and $t$. The first term has a single non-vanishing (top) chern class, which is $k^{t}$ times a generator of $H^{2t}(X)=\mathbb{Z}$, and the chern class of the second term is easily understood in terms of the chern class of $\eta$. In particular its top chern class will be nonzero as long as $n$ is very large.<|endoftext|> TITLE: A concrete example of the deficiency of triangulated categories? QUESTION [12 upvotes]: There seems to be a general sentiment that triangulated categories are not the "correct" notion to use because mapping cones of morphisms are unique, but only up to non-unique isomorphism. Does anyone know a concrete example of a "proof" or application that we would like to make with them, but which gets hindered because of this property? REPLY [15 votes]: Since I have already given a similar answer recently, I don't want to be branded as the "anti-triangular" guy: the formalism of triangulated categories can be useful in certain settings. That said they do have some definite shortcomings, and I would like to list a few in this answer. As a homotopy theorist, most of my examples will come from homotopy theory, but I'm sure there are others from algebraic geometry as well. The first easy example is: there's no notion of algebra and module in a triangulated category such that for an algebra $A$ the category of $A$-modules is again triangulated. If you go looking at more refined approach you'll see that to define an associative algebra or a module over an associative algebra in a "well-behaved" way, you need to involve higher homotopies that you are forgetting when you are restricting yourself to the triangulated category. This alone is a huge shortcoming: we like to be able to talk about categories of modules! Another one, maybe more abstract, but important for computations is the following: There is no notion, as far as I know, of descent for triangular categories. That is, there's no way in which one could say "this family of triangular categories form a sheaf". For example, let $B/A$ be a faithful Galois extension of commutative ring spectra with finite Galois group $G$, in the sense of Rognes' monograph. A good concrete example can be the $C_2$-extension $KO\to KU$. Suppose we know the Picard group of $B$ (for example because it is even periodic, or because it has some other property that makes it easy to compute). We would like to deduce some information about the Picard group of $A$. Using a more refined approach (either ∞-categories or model categories or what have you), you can define an invariant of an algebra $C$ called the Picard space $\mathrm{Pic}\,C$, which is an $E_∞$-space such that $\pi_0\mathrm{Pic}\,C$ is the Picard group of $C$. Then you can prove that there's an equivalence $$\mathrm{Pic}\,A\cong(\mathrm{Pic}\,B)^{hG}$$ and obtain a spectral sequence relating the Picard group of $B$ to the one of $A$. In fact most of the actual computations of the Picard group of ring spectra I know are done in a similar fashion. You could also do a more advanced version when your ring spectrum of interest is the global sections of some sheaf of ring spectra. This is how the Picard group of Tmf and TMF have been computed. Another maybe more abstract way of exploiting descent is in putting symmetric monoidal structures on derived categories of schemes. If we denote the derived $\infty$-category of a scheme $\mathscr{D}(X)$, descent tells us that we can express it as a limit $$\mathscr{D}(X)\cong \lim_{\mathrm{Spec}\,R\subseteq X}\mathscr{D}(R)$$ where the limit is indexed by the affine open subsets. But now the diagram we are taking the limit of is a diagram of symmetric monoidal stable $\infty$-categories, so the limit inherits a symmetric monoidal structure. Concretely, the tensor product $P\otimes Q$ has the property that on every affine chart we have $$\Gamma(\mathrm{Spec}\,R ,P\otimes Q)\cong \Gamma(\mathrm{Spec}\,R,P)\otimes_R\Gamma(\mathrm{Spec}\,R,Q)$$ The traditional way of defining the tensor product on the triangulated category $h\mathscr{D}(X)$ is a lot more involved, and you cannot do this (natural) approach there due to the failure of Zariski descent for $h\mathscr{D}(X)$. The last example is algebraic K-theory. You cannot define the higher algebraic K-theory of a triangulated category (only the $K_0$), there are different stable ∞-categories with the same underlying triangulated category but different algebraic K-theory spaces. As someone who works a lot on algebraic K-theory, this for me is enough of a drawback to make me not wanting to use triangulated categories.<|endoftext|> TITLE: The free smooth path space on a manifold QUESTION [8 upvotes]: Let $M$ be a closed, smooth manifold and let $PM$ be the space of unbased piecewise smooth paths $[0,1] \to M$. Then restricting a path to its boundary gives a map $$ PM \to M \times M . $$ Question is this map a fiber bundle? Andrew Stacey showed that a related map, the free smooth loop fibration $LM \to M$, is a fiber bundle (see The differential topology of loop spaces, arXiv:math/0510097). However, an inspection of his method shows that it does not immediately adapt to the situation above. REPLY [8 votes]: Yes. The technical details are in Yet More Smooth Mapping Spaces and Their Smoothly Local Properties, specifically in Section 5 which establishes that smooth manifolds are smoothly locally deformable which means that there are lots of diffeomorphisms flying around. Interestingly, although I considered subspaces I didn't consider spaces over other spaces. Nonetheless, the same technology allows us to do so. Let $M$ be a smooth manifold. Section 5 of Yet More ... shows that $M$ is smoothly locally deformable. In the discussion preceding Proposition 3.12 it is shown that this means that there is a neighbourhood $M \subseteq V \subseteq M \times M$ and a smooth map $\phi \colon \mathbb{R} \times V \to \operatorname{Diff}(M)$ with the following properties: For $v \in V$, $t \mapsto \phi_{t,v}$ is a group homomorphism $(\mathbb{R},+) \to \operatorname{Diff}(M)$. For $t \in \mathbb{R}$ and $v = (x,y) \in V$, $\phi_{t,v}$ is the identity outside $V_x := \{x' : (x,x') \in V\}$. For $v = (x,y) \in V$, $\phi_{1,v}(y) = x$. Now let $T$ be a compact smooth space and $S \subseteq T$ a compact subset. We assume that there is a neighbourhood $S \subseteq U \subseteq T$ with a retraction $\tau \colon U \to S$, and a bump function $\sigma \colon T \to [0,1]$ such that $\sigma(S) \subseteq \{1\}$ and $\overline{\sigma^{-1} (0,1]} \subseteq U$. Fix a class of function that is closed under diffeomorphism and which satisfies a sheaf condition in that functions can be defined locally. Let $\alpha \colon S \to M$ be a function. Define $C\big((S,T),(V,M)\big)_\alpha$ to be the space of functions $\beta \colon T \to M$ with the property that $(\alpha, \beta\mid_S)$ maps $S$ into $V$. Define $C(T,M)_\alpha$ to be the space of functions $\beta \colon T \to M$ such that $\beta\mid_S = \alpha$. Define $C(S,V)_\alpha$ to be the space of functions $\beta \colon S \to M$ such that $(\alpha,\beta)$ maps $S$ into $V$ (I'm not sure my notation is the best here!). We define $\Phi \colon C\big((S,T), (V,M)\big)_\alpha \to C(T,M)_\alpha \times C(S,V)_\alpha$ as follows. The map to the second factor is simply the restriction to $S$. The map to the first factor takes a function $\beta \colon T \to M$ to the function: $$ t \mapsto \begin{cases} \phi_{\sigma(t), (\alpha(\tau(t)), \beta(\tau(t)))}\big(\beta(t)\big) & t \in U \\\\ \beta(t) & t \notin U \end{cases} $$ The conditions on $\phi$ mean that this patches together to give a well-defined function. The inverse of $\Phi$ takes a pair $(\beta,\gamma)$ to: $$ t \mapsto \begin{cases} \phi_{-\sigma(t), (\alpha(\tau(t)),\gamma(\tau(t)))}\big(\beta(t)\big) & t \in U \\\\ \beta(t) & t \notin U \end{cases} $$ The case in point uses piecewise-smooth functions, $T = [0,1]$ and $S = \{0,1\}$. The conditions are easily checked. Further Reading The differential topology of loop spaces, particularly Proposition 5.1. This contains the germ of the idea. Yet More Smooth Mapping Spaces and Their Smoothly Local Properties, this contains the technical results needed. Proposition 3.12 is quite close to what you need here. This would establish that $LM \subseteq PM$ has a tubular neighbourhood, which says that it is a bundle on a neighbourhood of a diagonal. Interestingly, I didn't consider fibrations of mapping spaces. Maybe I should add another section ... The Smooth Structure of the Space of Piecewise-Smooth Loops about piecewise-smooth maps.<|endoftext|> TITLE: Is a vector space naturally isomorphic to its dual? QUESTION [29 upvotes]: This question may not be as easy to answer as you think! Some tangentially-related questions have appeared on math.stackexchange but I'm not really convinced by the answers. In the sequel I will assume all vector spaces under discussion are finite dimensional. A vector space is naturally isomorphic to its double dual In an early linear algebra course we are told that "a finite dimensional vector space is naturally isomorphic to its double dual". The isomorphism in question is ${**}_V : V \to V^{**}$, $v^{**}(\phi) = \phi(v)$. We are told that this isomorphism is "natural" because it doesn't depend on any arbitrary choices. The notion of "natural", or "independent or arbitrary choice", is made precise via the concept of a category theoretical "natural transformation". Specifically, the operation $**$ on vector spaces gives rise to a functor whose action on maps is $f^{**} : V^{**} \to W^{**}$, $f^{**}(v^{**}) = f(v)^{**}$. In fact this is exactly the condition for the naturality square to commute and so ${**}_V$ is indeed a natural transformation (between the identity functor and $**$) which is an isomorphism. A vector space is naturally isomorphic to its dual! So far, so familiar. But there's something that doesn't quite hold up about all this. Let's adapt the above to show that $V$ and $V^*$ are "naturally isomorphic". We do this by following exactly the same procedure, replacing $**$ everywhere with $*$. The only change we have to make is to come up with an arbitrary isomorphism $*_V$ for each $V$. Other than that, the whole construction goes through unchanged. Specifically, once we have chosen $*_V$ we define the functorial action on morphisms to be $f^{*} : V^{*} \to W^{*}$, $f^{*}(v^{*}) = f(v)^{*}$. In particular I have a natural isomorphism between the identity functor and $*$! Objections to the construction One could make a few objections to this construction, but they seem to be circular. "But you admitted that $*_V$ depends on an arbitrary choice!" I did, but that was informal language. In what formal sense is it arbitrary? The notion of "naturality" was supposed to rule out constructions that are arbitrary! "Your definition of $f^*$ is invalid. It depends on $*_V$." So what? My definition of $f^{**}$ depends on ${**}_V$ but it is uncontroversial. "You should have defined $f^{**}(\hat{v})(\phi) = \hat{v}(\phi \circ f)$ and then it's clear that it doesn't depend on $**_V$. You can't do that for $f^*$." But your $f^{**}$ is the same as my $f^{**}$! Is there some formal way of specifying that a functor does not depend on a natural transformation? And besides, what's the problem if it does? "It's a problem because it depends on something that depends on arbitrary choice ..." "Whilst the functor $**$ is the real double dual functor, $*$ is one you just made up. It is indeed isomorphic to the identity functor but that doesn't mean anything about 'a vector space being isomorphic to its dual'". Why not? I've followed exactly the same recipe for both of them, using the notion of "natural transformation" as I was supposed to. "Sure, but the result is interesting only in the case of $**$ because your definition of $*$ depended upon arbitrary choice ..." Conclusion All attempts to explain why I haven't really shown that a finite dimensional vector space is naturally isomorphic to its dual seem to invoke circular reasoning. I can only conclude that if the notion of natural transformation is going to be used to formalise the concept of "independent of arbitrary choice" then something needs to be tightened up. My choice of $*_V$ was indeed arbitrary but it is not ruled out by the notion of natural transformation. How could we proceed? Could the absence of choice be used to rule out the construction of $*_V$? Answers to an earlier question seem to suggest that is an irrelevant issue but in light of the above I'm not convinced. More generally, does this kind condition perhaps only make sense in a constructive or intuitionistic setting? I have a clue about how to formalise this condition in type theory via parametricity, so perhaps that is the key! (This question is similar to an earlier one. I'm not convinced by the answer. It seems to be making an objection of the third form above, which seems circular to me.) REPLY [3 votes]: I think that the issue boils down to a problem with the meaning that is generally ascribed to isomorphic objects. Typically it is said that isomorphic objects are "identical" in the sense that they are interchangeable i.e. "a theorem proved about one group is true for all isomorphic groups" But isomorphic objects are not always interchangeable. Ultimately it depends on what additional structure/relationships you need to consider hence the reason that the categorical definition of natural includes functors. Natural isomorphisms are about describing higher levels of identity and interchangeability than standard isomorphisms. Saying "arbitrary choices" implies that a non-arbitrary choice would be OK but actually the phrase really means that additional data/structure is required which limits the interchangeability that is possible. In the case of the $V\rightarrow V^{**}$ isomorphism you require a specific basis for $V$ in order to define the mapping in $V^*$. However if you already have an inner product defined then the dual basis can be defined with no additional inputs and in this case $V^{*}$ is genuinely interchangeable with $V$ similar to the duality between theorems in projective geometry when swapping points and lines.<|endoftext|> TITLE: Classification of fibrations $\Bbb S^k\longrightarrow\Bbb S^d\longrightarrow B$ QUESTION [7 upvotes]: Does there exist a complete classification of all fiber bundles $\Bbb S^k\longrightarrow\Bbb S^d\longrightarrow B$, that is, fibrations of $\smash{\Bbb S^d}$ with each fiber homeomorphic to $\smash{\Bbb S^k}$ for some fixed $k\le d$. The Wikipedia page on Hopf fibrations contains a list of some real/complex/quaternionic/octonionic fibrations. In other words: is this list complete? I am then interested, which of the base spaces $B$ that appear in above classification admit a topological/Lie group structure (compatible with its present topology). Update The last part of my question about topological/Lie group structure (which should have been a separate question from the start) was partially answered here. The projective spaces listed there are exactly the base spaces of sphere fibrations by great spheres (according to "On fibrations with flat fibres" by Ovsienko and Tabachnikov). It says nothing about the general case, though. REPLY [13 votes]: I'll assume that $1\leq k TITLE: Forcing and new ordinals QUESTION [5 upvotes]: $\textbf{Question}$: In expositions to forcing, why do we insist on not adding new ordinals to a countable transitive model $M$ of ZFC? For example, after ruling out transitive proper class models as possible candidates for satisfying (ZFC plus) $\textbf{V} \neq \textbf{L}$, Kunen writes on pages 185 "We shall describe a general procedure for finding countable transitive models $N$ of ZFC such that $M \subseteq N$ and $o(M) = o(N)$." Why insist on $o(M) = o(N)$? Later, he goes on to give examples of "non-generic" objects that shouldn't be added to $M$ as they code ordinals $> o(M)$ (Exercise (A3)). Why is this an issue? REPLY [2 votes]: You might consider the following quote from Cohen's paper, "The Discovery of Forcing", Rocky Mountain Journal of Mathematics, Volume 32, number 4, 2002, pg. 1091 (found under title on the Web): So we are starting with a countable standard model $M$, and we wish to to adjoin new elements and still obtain a model. An important decision is that no new ordinals are to be created. Just as Godel did not remove any ordinals in the constructible universe, a kind of converse decision is made not to add any new ordinals. Since (as you probably know from your readings about forcing) Prof. Cohen was the creator of the forcing technique, forcing was (at its inception), designed not to add new ordinals. The natural question to ask now is, "Why?" (this, of course, is the question you already asked). A clue as to why is found in Jech's Set theory: Third Millenium Edition, Chapter 13, in Theorem 13.28 and the preceding paragraph: If $M$ is a transitive model of $ZFC$, then the Axiom of Choice in $M$ enables us to code all sets in $M$ by sets of ordinals and the model is determined by its sets of ordinals. The precise statement of this fact is: if $M$ and $N$ are two transitive models of $ZFC$ with the same sets of ordinals, then $M$ = $N$ [in which case, $M$ and $N$ have identical ordinals--my comment]. In fact, a slightly stronger assertion is true (on the other hand, one cannot prove that $M$ = $N$ if neither model satisfies $AC$.) Theorem 13.28. Let $M$ and $N$ be transitive models of $ZF$ and assume that the Axiom of Choice holds in $M$. If $M$ and $N$ have the same sets of ordinals, i.e., $P^{M}$($Ord^{M}$) = $P^{N}$($Ord^{N}$) [$P^{M,N}$ is just the power set operation restricted to the models $M$, $N$, respectively and $Ord^{M,N}$ are just the ordinals of $M$ and $N$, respectively (Kunen uses essentially the same definition for $Ord^{M, N}$)--my comment], $M$ = $N$. As regards Profs. Hamkins' and Karagila's comments to each other regarding Prof. Hamkins' answer to you, you might ask either (or both) of them for a proof of their assertion, i.e., that adding ordinals to a model of $ZFC$ + $V$ $\ne$ $L$ might (or would) make $V$ = $L$ true again. If they would design their proofs so as to educate, I'm sure you would find them very interesting (and then you would truly know and understand why one "doesn't insist on adding ordinals to a countable transitive model $M$ of $ZFC$ in expositions of forcing"). As regards Prof. Hamkins' assertion that there are "other model-theoretic methods that do add new new ordinals", you might take a look at his answer to S A's mathoverflow question, "Can 'syntactic forcing' add ordinals?". As regards the fact that forcing does not add new ordinals, you might take a look at tomasz's mathstackexchange question, "A question about the proof that forcing extensions don't add ordinals", paying particular attention to Halbeisen's text mentioned, Profs. Caicedo's and Blass's comments, and the theorem in Halbeisen's text that tomasz mentions in his question. You might find this information at least somewhat helpful in your search for an answer.<|endoftext|> TITLE: Étale fundamental group of multiplicative group over an algebraically/separably closed field QUESTION [5 upvotes]: This is a repost of my question here. Do we know the structure of the étale fundamental group $\pi^\text{et}_1(\mathbb{G}_{m,K^\text{sep}})$ of the multiplicative group, for a given field $K$? For instance what is $$\pi_1^\text{et}(\mathbb{G}_{m,\overline{\mathbb{F}_p}})?$$ REPLY [6 votes]: I am not sure if much is known about the structure of $\pi_1(\mathbf{G}_m)$ itself, but at least there is a complete characterization of its finite quotients. This is the content of the so-called Abhyankar's Conjecture proven by Raynaud and Harbater. To formulate the conjecture, let me recall the definition of $p(G)$ for a group $G$. It is defined as the minimal subgroup of $G$ generated by all Sylow $p$-subgroup. It is easily seen that $p(G)$ is normal in $G$, so it makes sense to speak of the quotient group $G/p(G)$. Abhyankar's Conjecture: Let $X$ be a smooth connected curve over an algebraically closed field $k$ of characteristic $p$. Suppose that $X$ is of genus $g$ and $S=\{s_0, \dots, s_r\}$ is a finite set of closed points of $X$. Then a finite group $G$ can be realized as a quotient of $\pi_1(X-S)$ if and only if $G/p(G)$ is generated by at most $2g+r$ elements. The idea of Raynaud's proof for $X=\mathbf{P}^1$ is explained in Chapter $9$ of the book `Rigid analytic geometry and its applications' by Fresnel and van der Put. The general case is proven in the paper ``Abhyankar's conjecture on Galois groups over curves'' by Harbater. Remark: There is a natural question if $\pi_1(X)$ can be reconstructed from the set of its finite quotients. It turns out to be false as $\pi_1(\mathbf{A}^n)$ and $\pi_1(\mathbf{A}^m)$ have the same finite quotients, but they have different cohomological dimensions. Look at Proposition $7.3.1$ in Achinger's paper for more details.<|endoftext|> TITLE: Derived categories and classical theorems in homological algebra QUESTION [10 upvotes]: So far I have studied fundamental part of derived category theory, for example, the existence of derived functors, the "composition of derived functors", and so on. Now I came up with some questions about derived functors in the sense of derived category theory. - (1) Are the derived categorical derived functors universal in the classical sense? Let $\mathscr{A, B}$ be two abelian categories, $f : \mathscr{A} \to \mathscr{B}$ a left exact functor, and assume that $\mathscr{A}$ has enough injective. Then there exists "the" right derived functor $\mathbb{R}^+f: D^+(\mathscr{A}) \to D(\mathscr{B})$ of $f$, and its $i$-th cohomology of an object $X$ in $\mathscr{A}$ (considering as the complex which has $X$ at $0$-th degree) is the classical $R^if(X)$. So $\{ H^i(\mathbb{R}^+f(X)) \}_i$ is universal, in the sense of Hartshorne's AG, chapter III, i.e., for every $\delta$-functors $\{ g^i \}_i$ from $\mathscr{A}$ to $\mathscr{B}$ (i.e., a collection of additive functors satisfying the following condition: for every short exact sequence $0 \to X \to Y \to Z \to 0$ in $\mathscr{A}$, there exists the "connection map" $g^iZ \to g^{i+1}X$, making the long sequence exact.), if we have a natural transformation $f \to g^0$, there exists a unique natural transformations $R^if \to g^i$. Now is there a derived categorical interruption of this phenomena? I.e., for such $\{g^i\}$ and $f \to g^0$, does there exist a $\delta$-functor $g: D^+(\mathscr{A}) \to D(\mathscr{B})$ (such that for every $X \in \mathscr{A}$, $H^i g(X) = g^i(X)$) and $\mathscr{Q}f \to g \mathscr{Q}$? ($\mathscr{Q}$ is the localizing functor.) If so, then by the universality (in derived categorical sense), we have $\mathbb{R}^+f \to g$. (I read this post, but it seems to be a bit different from my question.) - (2) Can we show the "Grothendieck's Tohoku" easily using derived category? This is related to (1). This theorem says that, in particular, for a $\delta$-functor $\{g^i\}$, if $g^i(I) = 0$ for every $i \gt 0$ and every injective object $I$, then this is universal, i.e., $g^i \cong R^i g^0$. If (1) is true, then I think that this theorem can be translated into the following form: Let $\{ g^i \}$ be a $\delta$-functor and $g : D^+(\mathscr{A}) \to D(\mathscr{B})$ be "the morphism" as in (1). If $g^i(I) = 0$ for every $i$ and injective $I \in \mathscr{A}$, then $g$ is the right derived functor of $g^0 : K^+(\mathscr{A}) \to K(\mathscr{B})$. Is this true? - (3) How can we use derived categorical derived functors in order to show propositions around spectral sequences? I'm studying derived categories by Hartshorne's "Residues and duality". In this text, the author says "What used to be a spectral sequence is now simply a composition of functors. And of course one can recover the old spectral sequence..." (see the remarks after the Proposition 5.4.) I think the author means that we can show "all" propositions which are used to be shown by the spectral sequence argument using derived categories. For example: a) Let $\mathscr{A,B,C}$ be abelian categories, $f : \mathscr{A} \to \mathscr{B}, g: \mathscr{B} \to \mathscr{C}$ left exact functors, and suppose that $\mathscr{A,B}$ has enough injectives and that $f$ takes every injective object of $\mathscr{A}$ to a $g$-acycic object. Let $X \in \mathscr{A}$. If $R^if(X) = 0$ for every $i \gt 0$, then $R^n(gf)(X) \cong (R^ng)(f(X))$. (This is obvious. I could show it.) b) More generally, in the situation of (a), if $R^ifX = 0$ for $0 \lt i \lt q$, then there exists an exact sequence $0 \to R^qg(f(X)) \to R^q(gf)X \to g R^q f X$. c) Let $f : X \to Y$ be a morphism of proper schemes over a field, $\mathscr{F}$ a coherent sheaf on $X$. Then $\chi(\mathscr{F}) = \sum_p (-1)^p \chi(R^p f_* \mathscr{F})$. d) Let $f : X \to Y$ be a morpshim of schemes, and $\mathscr{F}$ a sheaf of modules on $X$. Assume that for all $q$, $\dim \operatorname{supp}R^qf_*\mathscr{F} = 0$. Then $H^0(Y, R^nf_*\mathscr{F}) \cong H^n(X, \mathscr{F})$. (This proposition is used in Mumford's Abelian varieties, $\S$8, theorem1.) (All propositions are trivial if we use the Grothendieck spectral sequence.) I think there are a lot of such proposition related the Grothendieck spectral sequence, but If I understand these (a)~(d), it seems to be similar to show any other such type propositions. Related post: this (a) and this(c). - (4) Grothendieck group of a derived category This is a related question to (3). This post shows (3) (c). As we can see from this post, it seems to be very important to understand Grothendieck groups of derived categories. But I don't know any references of Grothendieck groups of derived categories, especially of algebraic geometrical objects. I've found some fundamental properties. (e.g., this post and this pdf.) Are there any other important propositions of Grothendieck groups? And would you give me some references? - Lastly, (5) Easier proof of classical homological propositions I've heard that we can show the Kunneth's formula (of schemes) more easily using derived categories. (see here and for singular cohomology here.) I want to know such propositions. Would you give me references? I also want references of derived categorical proof of cohomology and base change theorems(see III.12 of Hartshorne's AG or here. The later statement is too abstract for me. I prefer Hartshorne-like concrete statement.) Any help will be much appreciated!! REPLY [15 votes]: Your question might be compacted to someting like: Do I need derived categories to study cohomology of sheaves? Of course, the answer depends on your particular interests. Let me anyway give you some starting points to help you to make up your mind. (1) Are the derived categorical derived functors universal in the classical sense? Let $f : \mathcal{A} \to \mathcal{B}$ a left exact functor between abelian categories and denote also by $f$ its extension to the corresponding homotopy categories. $\mathbf{K}(\mathcal{A})$ denotes the category of complexes with maps up to homotopy. The derived functor $\mathbf{R}f : \mathbf{D}(\mathcal{A}) \to \mathbf{D}(\mathcal{B})$ satisfies a universal property that implies that the collection $\{\mathbf{R}^if\}_{i \in \mathbb{N}}$ is a universal $\delta$-functor such that $\mathbf{R}^0f = f$. Where $\mathbf{R}^if: = \mathrm{H}^i\mathbf{R}f$. See [L, $\S2.1$]. (2) Can we show the "Grothendieck's Tohoku" easily using derived category? This is related to the existence of acyclic objects for a certain functor. For details, I suggest you to look at [L, $\S$ 2.2]. (3) How can we use derived categorical derived functors in order to show propositions around spectral sequences? As long as you refer to the so called Grothendieck spectral sequence on the composition of functors, it is replaced by the following theorem. Let $f : \mathcal{A} \to \mathcal{B}$ and $g : \mathcal{B} \to \mathcal{C}$ left exact functors between abelian categories. If $f$ takes $f$-acyclic objects into $g$-acyclic objects we have a natural isomorphism $$ \mathbf{R}gf \cong \mathbf{R}g\mathbf{R}f $$ (there is an analogous theorem for left derived functors). Thus, every time $\mathbf{R}f$ reduces to $f$ you obtain similar formulas than the ones you obtain by the collapse of the spectral sequence. The advantage is that the argument is simpler and you don't have limitations on finiteness or boundedness of the complex involved. Also, arguments involving three or more functors are seamless, something that would require multi-graded spectral sequences. On the other hand, derived categories will never help you in computing delicate properties for certain spectral sequences like the Adams spectral sequence or similar ones. (4) Grothendieck group of a derived category In favorable cases $\mathrm{K}_0(\mathbf{D}(\mathcal{A}))$ and $\mathrm{K}_0(\mathcal{A})$ agree. A small advantage would be that the oposite of a class of an object $X$ is not a virtual object but $-[X] = [X[1]]$. A discussion of $\mathrm{K}_0$ in the geometric context is in SGA6, exposé IV, to begin with. (5) Easier proof of classical homological propositions I won't assert they are easier but they are in my opinion clearer and broader. For base-change and Künneth I suggest you to look at [L, Theorem (3.10.3)]. The Künneth formula is more general than any other I've seen in the literature. Its expression via spectral sequences, if possible, would look extremely complicated. Besides, if you really want to understand Grothendieck-Serre duality beyond Cohen-Macaulay maps and schemes, then derived categories are indispensable. Final remarks For me, [L] is a very good introduction to the use of derived categories in Algebraic Geometry. Unfortunately it is not self-contained, so you need at least the first chapter in [KS1] and looking at Spaltenstein paper on unbounded resolutions. Alternatively you have all the needed prerequisites in [KS2]. Bibliography [KS1] Kashiwara, Masaki; Schapira, Pierre: Sheaves on manifolds. Grundlehren der Mathematischen Wissenschaften, 292. Springer-Verlag, Berlin, 1994. [KS2] Kashiwara, Masaki; Schapira, Pierre: Categories and sheaves. Grundlehren der Mathematischen Wissenschaften, 332. Springer-Verlag, Berlin, 2006. [L] Lipman, Joseph: Notes on derived functors and Grothendieck duality. Foundations of Grothendieck duality for diagrams of schemes, 1–259, Lecture Notes in Math., 1960, Springer, Berlin, 2009. [SGA6] Berthelot, Pierre; Alexandre Grothendieck; Luc Illusie, eds. Séminaire de Géométrie Algébrique du Bois Marie - 1966-67 - Théorie des intersections et théorème de Riemann-Roch - (SGA 6) (Lecture notes in mathematics 225). Berlin; New York: Springer-Verlag, 1971.<|endoftext|> TITLE: Does the nerve functor preserve dependent products when they exist? QUESTION [11 upvotes]: Consider the nerve functor $N : \mathbf{Cat} \to \mathbf{sSet}$; it is fully faithful, preserves finite limits, products and exponentials, etc. I am wondering if it additionally preserves whatever dependent products exist in $\mathbf{Cat}$. For instance, even though $\mathbf{Cat}$ is not locally Cartesian closed, we do have right adjoints to pullback along product projections. In this case, I'm curious whether the nerve functor takes these to dependent products in the category of simplicial sets. If these dependent products are indeed preserved, I am interested in understanding the general conditions under which other "nerve-like" situations exhibit the same behavior, meaning functors of the form $X \mapsto \mathbb{D}[i -,X]$ for $i : \mathbb{C}\to \mathbb{D}$, primarily considering the case where $i$ is a subcategory inclusion. EDIT: I was hypothesizing that this property would follow from the following conditions: the subcategory that generates the nerve is dense, and thence the nerve is fully faithful the nerve functor has a left adjoint REPLY [2 votes]: I have (I think) a proof under the assumptions I mentioned in my question, that the nerve is built from a dense subcategory and that there is a corresponding realization functor as a left adjoint. I would very much appreciate any readers of this answer to comment or edit if they find errors in my reasoning; and if you have an idea for a better proof, to post your own answer which I will accept. Thank you! Let's fix some notation. We will consider a fully faithful, dense inclusion $K : C_K\hookrightarrow C$. This generates a nerve functor $N(c) = C[K\bullet,d] : C \to \widehat{C_K}$; because $K$ is dense, the nerve functor is fully faithful. We assume a left adjoint to $N$, the realization $|-|: \widehat{C_K} \to C$. I am only going to use this left adjoint as a crutch at some points to transpose along; it may be that there is a better proof which doesn't use it. We will now check that a dependent product in $C$ is taken to a dependent product in $\widehat{C_K}$. We fix $f : d\to c,g : e\to d$ and wish to check that $\Pi_fg : C/c$ (if it exists) is taken to $\Pi_{Nf}Ng : \widehat{C_K}/Nc$. We note that the slice of the presheaf category is in fact another presheaf category, the category of presheaves $\widehat{C_K/Nc}$ on the category of elements $C_K/Nc$. Therefore, we may check the isomorphism by probing against "representables", fixing $x : C_K/Nc$; in this case, $x$ is concretely a map $x : K(\partial_0x)\to c$ in $C$. Let us write $\mathcal{E}/X$ for each slice $\widehat{C_K/X}$. First, we transpose: \begin{align*} \mathcal{E}/Nc[Y(x),N(\Pi_fg)] &\cong C/c[|Y(x)|, \Pi_fg] \end{align*} We note that $|Y(x)| : C/c$ is actually just the map $x$ described above; this is a consequence of the density of $K$. \begin{align*} &\cong C/c[x,\Pi_fg] \\ &\cong C/d[f^*x, g] \end{align*} Now, because $N$ is fully faithful, we may just put it everywhere: \begin{align*} &\cong \mathcal{E}/Nd[N(f^*x),Ng] \end{align*} But $N$ is continuous; so we may commute it into the pullback: \begin{align*} &\cong \mathcal{E}/Nd[Nf^*Nx,Ng] \\ &\cong \mathcal{E}/Nc[Nx,\Pi_{Nf}Ng] \end{align*} But $Nx$ is isomorphic to $Yx$ (this follows again from the assumption that $K$ is fully faithful and dense), so we are done.<|endoftext|> TITLE: Open problems from antiquity solved with analytic geometry QUESTION [34 upvotes]: E. T. Bell wrote in Men of Mathematics: Though the idea behind it all is childishly simple, yet the method of analytic geometry is so powerful that very ordinary boys of seventeen can use it to prove results which would have baffled the greatest of the Greek geometers -- Euclid, Archimedes, and Apollonius. I don't necessarily believe everything in the book, but this passage sounds plausible enough to make me wonder. Are there any good examples of an open problem from antiquity which seemed inaccessible, but was later easily solved by converting it to the Cartesian plane? REPLY [2 votes]: I come across what looks like a good fit in Brianchon, Solution de plusieurs problèmes de géométrie, J. École Polytechnique 4, nº 10 (1810) 1–15, page 5: Pappus reports also that the Greek geometers had tried in vain to solve this more general problem. « Given a circle and three poles, arranged in arbitrary manner, inscribe in this circle a triangle whose sides, extended if necessary, each go through one of the given poles. » With the help of analysis applied to geometry, the moderns easily overcame the difficulty, and this once famous question now amounts to very little; Lagrange has given a beautiful analytic solution (Mémoire de Berlin, 1776) (etc.; according to Senapati (2019) Pappus had solved the case where the 3 points are aligned, in Mathematicae collectiones, Book 7, Prop. 117.)<|endoftext|> TITLE: Monoidal category that is not spacial QUESTION [9 upvotes]: Selinger ("A survery of graphical languages for monoidal categories", 2009) defines a "spacial monoidal category" as one that satisifies the following string-diagram axiom This diagram says that for every $h : I \to I$ and object $A$, $$ \rho_A \circ (\text{id}_A \otimes h) \circ \rho^{-1}_A = \lambda_A \circ (h \otimes \text{id}_A) \circ \lambda_A^{-1},$$ where $\lambda_A$ and $\rho_A$ are the left and right unitors. What is an example of a monoidal category that is not spacial by this definition? REPLY [3 votes]: An example I learned about in Khovanov's "Heisenberg algebra and a graphical calculus", 2010, is the restriction and induction functors for the infinite chain $S_0\subset S_1\subset S_2\subset\cdots$ of symmetric groups. This is also explained in Likeng and Savage, "Embedding Deligne's category $\operatorname{Rep}(S_t)$ in the Heisenberg category," 2019. Let $\mathcal{S}=\prod_{m\in\mathbb{N}}\bigoplus_{n\in\mathbb{N}}\mathrm{Bim}{(S_n,S_m)}$, where each $\mathrm{Bim}(S_n,S_m)$ is the category of $(S_n,S_m)$-modules over a field $k$. This has a monoidal structure given by tensor products of compatible bimodules, with the monoidal unit being $I=\prod_{m}k[S_m]$ with each $k[S_m]$ as an $(S_m,S_m)$-module. We may regard $(S_n,S_m)$-modules as objects of $\mathcal{S}$ by setting all non-$m$ indices to the $0$ bimodule. The induction functor $\mathrm{Ind}_{S_n}^{S_{n+1}}$ can be given as an object $$\mathrm{Ind}=\prod_{n\geq 1} {}_n(k[S_n])_{n-1},$$ where the subscript notation means $k[S_n]$ is treated as an $(S_n,S_{n-1})$-module. This gives induction in the sense that, for each $S_n$-module $M$, the object $\mathrm{Ind}\otimes M$ is $\operatorname{Ind}_{S_n}^{S_{n+1}}M$. Similarly, restriction is given by $$\mathrm{Res}=\prod_{n\geq 1}{}_{n-1}(k[S_n])_n.$$ Just as induction and restriction are biadjoint functors, the $\mathrm{Ind}$ and $\mathrm{Res}$ objects are left and right duals to each other. Two of the four associated pairings and copairings are $\mathrm{Ind}\otimes\mathrm{Res} \to I$ and $I\to \mathrm{Ind}\otimes\mathrm{Res}$. Since $$\mathrm{Ind}\otimes\mathrm{Res} =\prod_{n\geq 1} {}_{n}(k[S_n]\otimes_{S_{n-1}}k[S_n])_n,$$ we may give their definitions in the form \begin{align*} \mathrm{Ind}\otimes\mathrm{Res} &\to I\\ (g\otimes h&\mapsto gh)_{n\geq 1}\\ \end{align*} and \begin{align*} I &\to \mathrm{Ind}\otimes\mathrm{Res}\\ (g& \mapsto \sum_{i=1}^n gg_i\otimes g_i^{-1})_{n\geq 1}, \end{align*} where $g_1,\dots,g_n\in S_n$ form a set of coset representatives for $S_n/S_{n-1}$. The composition $h:I\to \mathrm{Ind}\otimes\mathrm{Res} \to I$ of these is $h=(n\operatorname{id})_{n\geq 1}$. We can calculate \begin{align*} \operatorname{id}_{\mathrm{Ind}}\otimes h &= ((n-1)\operatorname{id})_{n\geq 1}\\ h\otimes \operatorname{id}_{\mathrm{Ind}} &= (n\operatorname{id})_{n\geq 1}, \end{align*} and therefore $\mathcal{S}$ is not spacial. Graphically, this is that counter-clockwise loops cannot be dragged across an upward strand, imagining $\mathrm{Ind}$ as an up-arrow and $\mathrm{Res}$ as a down arrow.<|endoftext|> TITLE: Freiman inequality for projective space? QUESTION [10 upvotes]: This question is suggested by some results in a paper I am writing. I would like to write it down there but want to make sure that it is not known or at least MO-hard. Freiman's inequality states that for a set $A$ of vectors that span $\mathbb R^d$, we have: $$|A+A| \geq (d+1)|A| -\binom{d+1}{2} $$ My question is what happens in $\mathbb{RP}^d$? To be really concrete let $L(A)$ be the set of lines given by the vectors in $A$. Is this true that if $A$ is a set of vectors that span $\mathbb R^{d+1}$, we still have: $$|L(A+A)| \geq (d+1)|L(A)| -\binom{d+1}{2} ?$$ If not, can any inequality be proved? Remark: there seems to be some confusion in the comments, so let's give one example. Let $d=1$ and $A=\{(1,0), (-1,0), (0,1), (0,-1)\}$. Then $|L(A)|=2$ (the x- and y-axis) and $|L(A+A)|=4$ (the axes plus the diagonal lines). Also, it is clear that this projective version implies the original one. It is tempting to try to deduce it from the original version, but I can't see a way to do it. REPLY [5 votes]: As stated, the answer is no, Freiman's inequality no longer holds. The counter example is $A$ being the vertices of an equilateral triangle on the unit circle. I found this by looking at the proof of Freiman's inequality (given as Lemma 5.13 of Tao-Vu book on Additive Combinatorics) and see where the proof fails. In my particular setting, the set $A$ satisfies some additional convexity conditions, so perhaps the result may hold for such situations. UPDATE: it turned out that the proposed inequality held under one extra assumption: that the origin is outside the convex hull of $A$. That will rule out examples such as above, and still implies the classical inequality. The proof follows the same line as the original proof, with some extra care needed.<|endoftext|> TITLE: Big etale topos vs small etale topos QUESTION [5 upvotes]: Are they equivalent? That is, given a sheaf of sets $\mathscr{F}$ defined on the small etale site on $X$, is there an essentially unique way to extend it to a sheaf on the big etale site on $X$? If not, what is an example of a sheaf which cannot be extended? What about for sheaves of abelian groups? REPLY [4 votes]: There is a site morphism $$i: X_{\rm \acute{e}t}\to {\rm \acute{E}t}(X),$$ giving an adjunction (indeed, a geometric morphism of topoi—abstract nonsense) $${\rm EXT}=i^*: \mathsf{Shv}(X_{\rm \acute{e}t})\rightleftarrows\mathsf{Shv}({\rm \acute{E}t}(X)): i_*={\rm Res},$$ where ${\rm EXT}=i^*$ is given by some Kan extension, $i_*={\rm Res}$ is really the restriction. Standard text books on étale cohomology say that the unit $F\to{\rm Res}({\rm EXT}(F))$ of the adjunction $({\rm EXT}, {\rm Res})$ is an isomorphism. So every sheaf on the small étale site is always extendable to the big étale site and restricts back to the original sheaf. So your question is not a correct one. On the other hand, the counit ${\rm EXT}({\rm Res}(G))\to G$ is usually not an isomorphism. As in the above comments, different big étale sheaves can restrict to isomorphic small étale sheaves. The correct question is to ask for such examples, already given by the above comments. What is important in étale cohomology theory is, for any $A\in\mathsf{ShvAb}(X_{\rm \acute{e}t}), B\in\mathsf{ShvAb}({\rm \acute{E}t}(X)), U\in X_{\rm \acute{e}t}$, there are canonical isomorphisms $${\rm H}^n_{\rm \acute{e}t}(X; A)\cong{\rm H}^n_{{\rm \acute{E}t}(X)}(X; {\rm EXT}(A)), {\rm H}^n_{\rm \acute{e}t}(X; {\rm Res}(B))\cong{\rm H}_{{\rm \acute{E}t}(X)}^n(X; B); $$ $${\rm H}^n_{\rm \acute{e}t}(U; A)\cong{\rm H}^n_{{\rm \acute{E}t}(X)}(U; {\rm EXT}(A)).$$<|endoftext|> TITLE: Integral (co)homology of $SU/SO$ QUESTION [6 upvotes]: I would like to know the integral cohomology of $SU(\infty)/SO(\infty)$ (to degree 5 or 6, say.) Mimura-Toda says $H^*(SU/SO,\mathbb{Z}/2\mathbb{Z})=\wedge[w_2,w_3,\ldots]$ where $w_i$ is a pullback of Stiefel-Whitney classes via $SU/SO\to BSO$. I'd like to know the image of Hurewicz images, too. Bott periodicity basically says $\pi_2(SU/SO)=\pi_3(SU/SO)=\mathbb{Z}/2\mathbb{Z}$ and I would like to know their images in $H_*(SU/SO,\mathbb{Z})$. I feel stupid for asking this, because I can find many results on exceptional symmetric spaces by googling, while I do not find it for the non-exceptional symmetric spaces... REPLY [7 votes]: First, I'll mostly talk about $U/O$ rather than $SU/SO$ because $U/O$ can be descibed as $B(\mathbb{Z}\times BO)$ or as the $8k-1$'th space in the $\Omega$-spectrum for $KO$. This gives $\pi_0(U/O)=0$ and $\pi_1(U/O)=\pi_0(KO)=\mathbb{Z}$. From the Hurewicz and universal coefficient theorems this gives $H^1(U/O)=\text{Hom}(\pi_1(U/O,\mathbb{Z})$. This describes $[S^1,U/O]=\pi_1(U/O)$ and $[U/O,S^1]=[U/O,K(\mathbb{Z},1)]=H^1(U/O)$ and using this together with the $H$-space structure we obtain a splitting of $U/O$ as the product of $S^1$ with the universal cover, which is $SU/SO$. So there is not too much difference between $U/O$ and $SU/SO$. The ring $H_*(U/O;\mathbb{Z}/2)$ has generators $a_k$ of degree $k$ for $k>0$, satisfying $a_{2k}=a_k^2$. We can therefore discard the even generators and say that $H_*(U/O;\mathbb{Z}/2)$ is polynomial on the classes $a_{2k+1}$. The Bockstein operation is given by $\beta(a_{2k+1})=a_{2k}$ and $\beta(a_{2k+2})=0$. (This would be more awkward to state if we had not introduced the classes $a_k$ for all $k$.) It follows that the classes $h_{4k+1}=a_{4k+1}+a_{2k}a_{2k+1}$ satisfy $\beta(h_{4k+1})=0$. In fact one can show that the ring $H_*(H_*(U/O;\mathbb{Z}/2),\beta)=\ker(\beta)/\text{img}(\beta)$ is an exterior generated by these classes $h_{4k+1}$ (including $h_1$, which should be interpreted as $a_1$). One can also show that there are classes $b_{4k+1}\in H_{4k+1}(U/O)$ lifting $h_{4k+1}$, and that $H_*(U/O)/\text{torsion}$ is an exterior algebra generated by these elements. The torsion is all annihilated by $2$ and is the image of the Bockstein map $\beta'\colon H_i(U/O;\mathbb{Z}/2)\to H_{i-1}(U/O)$. It is awkward to give a full description of the torsion, but if you are only interested in low degrees you should be able to spell it out. I am reading all this off from my thesis, where everything is done using Hopf ring methods. Certainly some parts of the calculation appear already here: Henri Cartan, Périodicité des groupes d’homotopie stables des groupes classiques, d’après Bott, Sem. H. Cartan, vol. 60, Ecole Normale Supérieur, 1959. However, I don't have that to hand, so I don't remember precisely what is covered.<|endoftext|> TITLE: A historical mystery : Poincaré’s silence on Lebesgue integral and measure theory? QUESTION [51 upvotes]: Lebesgue published his celebrated integral in 1901-1902. Poincaré passed away in 1912, at full mathematical power. Of course, Lebesgue and Poincaré knew each other, they even met on several occasions and shared a common close friend, Émile Borel. However, it seems Lebesgue never wrote to Poincaré and, according to Lettres d’Henri Lebesgue à Émile Borel, note 321, p. 370 … la seule information, de seconde main, que nous avons sur l’intérêt de Poincaré pour la « nouvelle analyse » de Borel, Baire et Lebesgue the only second-hand information we have on Poincaré's interest in the "new analysis" of Borel, Baire and Lebesgue is this, Lebesgue to Borel, 1904, p. 84: J’ai appris que Poincaré trouve mon livre bien ; je ne sais pas jusqu’à quel point cela est exact, mais j’en ai été tout de même très flatté ; je ne croyais pas que Poincaré sût mon existence. I learned that Poincaré finds my book good; I do not know to what extent that is accurate, but I nevertheless was very flattered; I did not believe that Poincaré knew of my existence. See also note 197, p. 359 Nous ne connaissons aucune réaction de Poincaré aux travaux de Borel, Baire et Lebesgue. We do not know any reaction of Poincaré to the works of Borel, Baire and Lebesgue. To my mind this situation is totally unexpected, almost incredible: the Lebesgue integral and measure theory are major mathematical achievements but Poincaré, the ultimate mathematical authority at this time, does not say anything??? What does it mean? So, please, are you aware of any explicit or implicit statement by Poincaré on the Lebesgue integral or measure theory? If you are not, how would you interpret Poincaré’s silence? Pure disinterest? Why? Discomfort? Why? Something else? This question is somewhat opinion-based, but The true method of forecasting the future of mathematics is the study of its history and current state. according to Poincaré and his silence is a complete historical mystery, at least to me. REPLY [22 votes]: Poincaré studied with Hermite, who famously in a letter 1893 to Stieltjes wrote „I turn with terror and horror from this lamentable scourge of continuous functions with no derivative.“ Poincaré himself is often quoted „Heretofore when a new function was invented it was for some practical end; today they are invented expressly to put at fault the reasoning of our fathers; and one will never get more from them than that.“ Of course these quotes are older than the Lebesgue integral, yet they may explain why integration of pathological functions was not considered to be important by Poincaré and other French mathematicians.<|endoftext|> TITLE: Explanatory vs Non-explanatory Proofs QUESTION [19 upvotes]: In a philosophical context, I’m currently thinking about how best to explicate mathematicians’ judgements that some correct proofs are ‘explanatory’ while others are not. In this vein, I’m trying to collect examples of theorems that have two salient proofs, one of which is judged to be explanatory whilst the other is not (even better if the examples exhibit strong disagreement regarding which proof is more explanatory). Other things being equal, simpler examples are preferred, and I’m especially interested in examples from abstract algebra, order theory and topology. Pointers towards relevant debates in the history of math would also be appreciated. (Disclaimer: this question is related to but distinct from the question below, which concerns the relationship between explanation and beauty in mathematical proof: An example of a proof that is explanatory but not beautiful? (or vice versa).) REPLY [40 votes]: Paul Halmos once gave the following example in a talk for a general audience. Suppose there is a tennis tournament with 128 players. In the first round 64 of them are paired off with the other 64, they play their games, and all the losers are ejected from the tournament. In the next round the remaining 64 players are paired off and this time the 32 losers are ejected. Eventually one player is left, who wins the tournament. How many games were played in total? The most obvious way to solve this is to add up 64 + 32 + ... If you remember a formula about geometric series, you can find this sum quickly. But the explanatory proof is different. Every player aside from the eventual winner loses exactly one game, the game in which they are ejected. So the total number of games = the total number of losers = 127. REPLY [12 votes]: This is a well-known and well-documented example: the first proof of the Alternating Sign Matrix theorem was a complicated, inductive "manipulation of generating function"-style argument by Zeilberger; shortly thereafter Kuperberg gave a shorter proof based on a connection to the six-vertex model of statistical mechanics and the Yang-Baxter equation. The history of the Alternating Sign Matrix conjecture is beautifully told in the book Proofs and Confirmations by Bressoud.<|endoftext|> TITLE: Subfields of Hilbertian fields QUESTION [7 upvotes]: This question is about the Remark on the top of page 22 of Serre's Topics in Galois Theory, available here : http://www.ms.uky.edu/~sohum/ma561/notes/workspace/books/serre_galois_theory.pdf My question is as follows. Let $L/K$ be a finite separable extension of fields. Assume that $L$ is Hilbertian. Does it follows that $K$ is Hilbertian? I believe the answer is yes (and this shouldn't be too hard to show). However, Serre claims that the answer is negative, and refers in his remark to Kuyk's paper W. Kuyk. Extensions de corps hilbertiens, J. of Algebra 14 (1970), 112-124 However, Kuyk seems to only give an example of a field extension $L/K$ of infinite degree such that $K$ is not Hilbertian, but $L$ is. The example is given in the Remarque on page 2 of his article. Note that $K$ can be taken to be the maximal $2$-extension of $\mathbb{Q}$. This is not a Hilbertian field, whereas its extension $L$ given by the maximal nilpotent extension of $\mathbb{Q}$ (or $K$) is Hilbertian. REPLY [9 votes]: The maximal (pro-)solvable extension $L$ of $\mathbb{Q}$ is not Hilbertian, but every proper finite extension of $L$ is. See Fried-Jarden (third edition, 2008), Example 13.9.5.<|endoftext|> TITLE: What does the free action of a surface group on an R-tree look like? QUESTION [7 upvotes]: Morgan and Shalen "Free action of surface groups on R-trees" 1989 shows that surface groups (genus at least 2) act freely on some real trees (R-trees). Their proof seems to be non-constructive, requiring the Baire category theorem. But the object they are constructing sounds quite simple. Their Proposition 17 says: On every closed non-exceptional hyperbolic surface, there exists a measured geodesic lamination whose leaves and complementary regions are all simply connected. Does anyone have a picture of this lamination on the genus 2 surface? And, using that, a picture of the R-tree that the genus 2 surface group acts freely on? REPLY [11 votes]: Here is a picture of a train track, which approximates a lamination: Another picture: The leaf space of this lamination (obtained by collapsing each leaf and complementary polygon to a point) is an $\mathbb{R}$-tree. As you contract the leaves to points, you might get a picture looking somewhat like this: If you fill in with the boundary circle, you get a dendrite which might help you visualize an $\mathbb{R}$-tree:<|endoftext|> TITLE: Is the motivic homotopy spectrum of Hermitian K-theory $\eta$-complete? QUESTION [5 upvotes]: Theorem 1.2 of the paper "The motivic Hopf map solves the homotopy limit problem for K-theory" (see https://elibm.org/article/10011880) says that (under certain assumptions on the base field) the homotopy fixed points of the action of $C_2$ on the motivic algebraic $K$-theory spectrum $KGL$ give the $\eta$-completion of the Hermitian $K$-theory spectrum $KQ$, where $\eta$ is the (Morel's) algebraic version of the Hopf map. My question is: is it known whether $KQ$ is $\eta$-complete in this context, i.e., whether the $\eta$-completion operation does not really change anything? REPLY [3 votes]: If I understand the question correctly, you're asking if there are base fields $k$ of characteristic not $2$ and such that $k[\sqrt{-1}]$ has a finite $\mathbb{Z}/2$-cohomological dimension (the assumptions used in the above mentioned paper) for which the map $KQ \to KGL^{hC_2}$ is not an equivalence. Note that by the Tate fiber square (mentioned also in that paper), this is the same as asking whether $KW \to KGL^{tC_2}$ is an equivalence, a necessary condition for which is that the map $W(k) \to \pi_0(K(k)^{tC_2})$ is an isomorphism of abelian groups, where $W(k)$ is the Witt group of $k$ and $K(k)$ is its $K$-theory spectrum. As shown in Theorem 16 of this paper, the last map coincides in this context with the completion map $$ (*) \quad W(k) \to \lim_n W(k)/I^n $$ with respect to the augmentation ideal $I \subseteq W(k)$. An example of a field satisfying the above conditions and for which this completion map is not an isomorphism is the field $\mathbb{R}$ of real numbers, where $(*)$ is the map $\mathbb{Z} \to \lim_n\mathbb{Z}/2^n$.<|endoftext|> TITLE: Possible application of divergence Theorem? QUESTION [7 upvotes]: suppose that $f \in C^1 (\mathbb{R}^{N+1},\mathbb{R})$. It's well known that if all his points are regular points i.e. $$\nabla f (x) \neq 0 \; \; \; \forall x \in \mathbb{R}^{N+1}$$ then, for every fixed $c \in \mathbb{R}$, the level set: $$f^{-1}(c)=\{x \in \mathbb{R}^{N+1}: f(x)=c\}$$ is a $C^1$ manifold of dimension $N$. So if $\Omega:=\left \{ x \in \mathbb{R}^{N+1}: f(x)< c \right \}$, supposed bounded, with enough regularity on a vector field $G$, I can apply the classical divergence theorem: $$\int_{\Omega} \operatorname{div}(G) \,\mathrm{d}\mathcal{L}^{N+1}=\int_{\partial \Omega}\langle G,\nu\rangle \,\mathrm{d}H_N$$ where $H_N$ is the $N$-dimensional Hausdorff measure. Now suppose that not all his points are regular and let $\operatorname{crit}(f)$ be the set of critical points of $f$ i.e. $$\operatorname{crit}(f):=\{x \in \mathbb{R}^{N+1}: \nabla f (x)=0 \}$$ Now, suppose that for a fixed $\bar{c} \in \mathbb{R}$, I can prove that: $$H_N \left( f^{-1}(\bar{c}) \cap \operatorname{crit}(f) \right)=0 $$ So $f^{-1}(\bar{c}) \setminus \operatorname{crit}(f)$ remains a $C^1$ manifold of dimension $N$, while $f^{-1}(\bar{c}) \cap \operatorname{crit}(f)$ is set of measure zero. So the points of $f^{-1}(\bar{c})$ have $\nabla f(x) \neq 0$, except for a set of measure zero. My question is: is possible to apply divergence theorem to $\Omega=\left \{ x \in \mathbb{R}^{N+1}: f(x)< \bar{c} \right \}$ also in this case? I found a lot of generalization of divergence theorem but apparently none for my case. Thanks. REPLY [3 votes]: In your case the boundary has (locally) finite $H^N$ measure and such sets have finite perimeter. It is known that for sets of finite perimeter the divergence theorem is true. Later I will write a more complete answer, but now I am traveling so it will take me a week or so.<|endoftext|> TITLE: What does the image of the integer lattice under a norm look like? QUESTION [5 upvotes]: The question that I shall ask here has arisen in the context of Diophantine approximation. I find it rather interesting, and I have no idea how to answer it. Any help, advice, or suggestions for references would be much appreciated. Let $S$ be a countably infinite subset of $\mathbb{R}_{>0}$ that satisfies the following conditions. Let $n$ be an arbitrary element of $\mathbb{Z}_{\geq 2}.$ 1) For any integer $k \geq 1,$ we have $k \cdot S \subseteq S.$ (In particular, $S$ is unbounded.) 2) There exists a (necessarily unique) strictly increasing sequence $t_\bullet = (t_k)_{k \geq 1}$ such that $\{t_k : k \geq 1\} = S.$ Moreover, there exists a constant $C > 0$ such that for any $k \geq 1,$ we have $0 < t_{k+1} - t_k \leq C.$ (Notice that we clearly have $\lim_{k \to +\infty} t_k = +\infty$, in light of 1) above.) 3) There exists a constant $D > 0$ such that $$\limsup_{T \to +\infty} \frac{\mathrm{card}\{ k \in \mathbb{Z}_{\geq 1} : t_k \leq T\}}{T^n} \leq D.$$ Is it necessarily true that there exists a norm $\eta$ on $\mathbb{R}^n$ such that $\eta(\mathbb{Z}^n) - \{0\} = S$? (Clearly conditions 1), 2), and 3) are necessary in order for this to be true; I am wondering whether they are also, in fact, sufficient.) REPLY [6 votes]: I doubt that there is a simple characterization. In any case conditions (1,2,3) are not sufficient. For example, if $n=2$ then $S$ cannot be ${\bf Z}_{\geq 100}$, and there are similar counterexamples for every $n \geq 2$, as a consequence of the following observation. Proposition. If $\eta$ is a norm on ${\bf R}^n$, and $t_1$ is the smallest element of $\eta({\bf Z}^n)$, then for any $M>0$ there are at most $(2M+1)^n$ integer vectors $v$ with $\eta(v) \leq M t_1$. Proof: Consider the $\eta$-balls of radius $t_1/2$ centered on all such $v$. They have disjoint interiors, and are all contained in the $\eta$-ball of radius $(2M+1)t_1/2$ about $0$. Therefore their total volume is no larger than the volume of the ball of radius $(2M+1)t_1/2$. But the volume of a ball of radius $r$ is proportional to $r^n$. Hence the total number of radius $t_1/2$ balls is at most $\left[ ((2M+1)t_1/2) \left/ \, (t_1/2) \right. \right]^n = (2M+1)^n$, Q.E.D. Equality is attained when $\eta$ is the sup norm and $M$ is an integer. (Then we get a perfect packing of cubes.)<|endoftext|> TITLE: A Thom spectrum from "doubled" tautological bundles? QUESTION [8 upvotes]: Let us consider real vector bundles, and denote by $V_k$ the tautological bundle $V_k\to BO(k)$. From $$ Thom(V\oplus 1_{\mathbb{R}}\to X)=\Sigma Thom(V\to X) $$ and from $j^*V_{k+1}=V_k\oplus1_{\mathbb{R}}$, where $j\colon BO(k) \to BO(k+1)$ is the canonical embedding, one sees that writing $MO_k=\Sigma^{-k}Thom(V_k\to BO(k))$ one gets a sequence of morphisms $MO_k\to MO_{k+1}$ and one can define $$ MO=\lim_{\to} MO_k $$ This way (and taking infinite suspension) one defines the Thom spectrum $MO$, whcih can therefore informally be thought as the infinite desuspension of the Thom spectrum of the tautological infinite rank vector bundle over $BO(\infty)$. Direct sum of vector bundles and the fact that $j_{m,n}^*V_{m+n}=V_m\oplus V_n$, where $j_{m,n}\colon BO(m)\times BO(n)\to BO(m+n)$ is the canonical embedding makes $MO$ a ring spectrum. Apparently, one should be able to do the very same construction by "doubling" the $V_k$'s, i.e., by considering vector bundles $$ W_k=V_k\oplus V_k $$ noticing that $j^*W_{k+1}=W_k\oplus 1_{\mathbb{R}^2}$ and so $Thom(j^*W_{k+1}\to BO(k)) = \Sigma^2 Thom(W_{k}\to BO(k))$, and then considering the pointed spaces $M_2O_k=\Sigma^{-2k}Thom(W_k\to BO(k))$ and then the colimit $$ M_2O=\lim_{\to} M_2O_k $$ This again should be a ring spectrum by the same reason as for $MO$. As, for any real vector space $V$, its double $V\oplus V$ carries a canonical comple structure $J\colon V\oplus V \to V\oplus V$, and since the defining representation $\mathbb{C}^n$ of $U(n)$ restricted to $O(n)$ splits as the sum of two copies of the defining representation $\mathbb{R}^n$ of $O(n)$, one should get a morphism of ring spectra $M_2O \to MU$. By looking at how the $\hat{A}$ polynomial is derived from the Todd polynomial, one would suspect that the analogous morphism $$ M_2Spin \to MU $$ and a compatibility between this, the standard complex orientation of $KU$, the Atiyah-Bott-Shapiro orientation of $KO$ and complexification of vector bundles is what secretely lies behind the Atiyah-Singer formula for the index of a twisted spin complex, making Atiyah-Singer formula a version of the general kind of Hirzebruch-Riemann-Roch formulas one has when dealing with morphisms of generalized cohomology theories and pushforwards (see, e.g., Panin-Smirnov). However, I've not been able so far to locate something resembling $M_2O$ or $M_2Spin$ in the literature I have searched in, so I could possibly be on a false track here. Is the construction sketched above possibly correct? What is a reference to it? has it a more canonical name than $M_2O$? is it really related to the construction and relevance of $\hat{A}$ along the lines sketched above? REPLY [3 votes]: I think that both of your examples, $M_2Spin$ and $M_2O$, arise naturally in the context of Thom spectra induced by $(B,f)$-structures. Given a $(B,f)$-structure $\mathcal{B}= \{f_n: B_n \to BO(n)\}$, the associated Thom spectrum $M\mathcal{B}$ is defined componentwise as: $$ M\mathcal{B}_k = Thom(f_k^*V_k\to B_k), $$ where the maps $\Sigma M\mathcal{B}_k \to M\mathcal{B}_{k+1}$ are given by looking at the pullback square $\require{AMScd}$ $$ \begin{CD} \mathbb{R}\oplus f_k^*V_k @>>> f_{k+1}^*V_{k+1}\\ @VVV @VVV \\ B_k @>>> B_{k+1}. \end{CD} $$ One can 'double' this construction replacing the maps $f_k$ with $\tilde{f_k}$ defined to be the composition: $$ \tilde{B_{2k}}:=B_k\overset{f_k}{\to} BO(k) \overset{\Delta}{\to} BO(k) \times BO(k) \overset{j_{k,k}}{\to} BO(2k) $$ and get a $S^2$-$(B,f)$-structure, a $(B,f)$-structure indexed only on even natural numbers, denoted by $2\mathcal{B}$. By definition, the Thom spectrum $M_2\mathcal{B}$ associated to this new $S^2$-$(B,f)$-structure, is $$ (M_2\mathcal{B})_{2k} = Thom(\tilde{f_k^*}V_{2k}\to \tilde{B_{2k}}) = Thom(f_k^*V_k\oplus f_k^*V_k \to B_k) $$ $$ (M_2\mathcal{B})_{2k+1} = \Sigma(M_2\mathcal{B})_{2k}. $$ In your case, $M_2Spin$ and $M_2O$, are (as sequential spectra) the Thom spectra associated to the 'doubled' $(B,f)$-structures that classically define $MSpin$ and $MO$, i.e. the $(B,f)$-structures respectively given by the maps $BSpin(k)\to BO(k)$ and ${{\rm id}}_{BO(k)}$. All the details and references can be found in the nlab pages Thom spectrum and G-structure.<|endoftext|> TITLE: Symplectic structure vanishing simultaneously on two totally real subspaces QUESTION [7 upvotes]: Let $L_1,L_2$ be two $\mathbb{R}$-linear subspaces of $\mathbb{C}^n$ that are both totally real, that is, $$L_j \cap \bigl(i\cdot L_j\bigr) = \{0\}$$ and $$\dim_{\mathbb{R}} L_j = \dim_{\mathbb{C}} \mathbb{C}^n = n$$ or equivalently $$L_j \oplus (i\cdot L_j) = \mathbb{C}^n . $$ I would like to know under what conditions I can find a symplectic structure $\omega$ on $\mathbb{C}^n$ that tames $i$ and that vanishes when restricted to $L_1$ and $L_2$. Taming means $$\omega\bigl(v,i\cdot v\bigr) > 0$$ for every $v\in \mathbb{C}^n$ different from $0$. Remark: In general such $\omega$ might not exist, suppose for example that $L_1$ is spanned by $(1,0)$, and $(0,1)$ in $\mathbb{C}^2$ and that $L_2$ is spanned by $(1,0)$ and $(i,1)$. Then we can easily check that $\omega$ cannot vanish on $L_2$ as $$ \omega\bigl((1,0), (i,1)\bigr) = \omega\bigl((1,0), (i,0)\bigr) + \omega\bigl((1,0), (0,1)\bigr) = \omega\bigl((1,0), i\cdot (1,0)\bigr) > 0 ,$$ where we only have used that $\omega$ vanishes on $L_1$, and that $\omega$ tames $i$. It deduce that a vector $v\in L_2$ that is transverse to $L_1\cap L_2$ must not lie in $L_1 + \mathbb{C}\cdot (L_1\cap L_2)$, but I have not managed to prove that this condition is sufficient. REPLY [2 votes]: I'll give a positive answer for two generic totally real planes in $\mathbb C^2$. I believe this generalises to larger $n$, though I don't prove it - just give a possible plan of a proof with one step completed. Lemma 1. Suppose $L_1$ and $L_2$ are two generic totally real $2$-planes in $\mathbb C^2$ then the desired form exists. Proof. Let $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ be the complex coordinates in $\mathbb C^2$. Let $$\omega_1=dx_1\wedge dx_2-dy_1\wedge dy_2,\;\;\omega_2=dx_1\wedge dy_2+dy_1\wedge dx_2.$$ Note that $\omega_1=\mathrm{Re}(dz_1\wedge dz_2)$, $\omega_2=\mathrm{Im}(dz_1\wedge dz_2)$. In particular, both $\omega_1$ and $\omega_2$ restrict to zero on all complex lines in $\mathbb C^2$. Now, let $W$ be the two-dimensional vector space in $\Lambda^2({\mathbb R^4}^*)$ spanned by $\omega_1$ and $\omega_2$. Consider the evaluation map given by restricting a form $\omega\in W$ to $L_1$ and $L_2$ $$e: W\to \Lambda^2 L_1^*\oplus \Lambda^2 L_2^*.$$ I claim that for generic $L_1$ and $L_2$ this map is an isomorphism. Indeed, it is enough to prove this claim for at least one pair of real planes. So let $L_1$ be the plane spanned by vectors $e_{x_1},e_{x_2}$ and $L_2$ to be spanned by $e_{x_1}, e_{y_2}$. In this case the map is clearly an isomorphsim. Now suppose that $L_1$ and $L_2$ are generic, and such that the map $e$ is an isomorphism. Let us take $\omega=dx_1\wedge dx_2+dy_1\wedge dy_2$. Then its restriction to $L_1$ and $L_2$ gives an element $v\in \Lambda^2 L_1^*\oplus \Lambda^2 L_2^*$. Let $\tilde\omega\in W$ be such that $e(\tilde \omega)=v$. Then it is easy to see that the symplectic form $\omega-\tilde \omega$ is doing the job. Indeed $\omega$ restricts positively to all complex lines and $\tilde \omega$ is zero on all complex lines. END OF Proof. It seems to me that the same idea could work in higher dimensions as well. One takes again $W$ to be the space spanned by all real and imaginary parts of holomorphic $2$-forms on $\mathbb C^n$. The real dimension of this space is $n(n-1)=2*\frac{n(n-1)}{2}$. So it might be that the eavaluation map $e$ is still generically an isomorphism from $W$ to $\Lambda^2 L_1^*\oplus \Lambda^2 L_2^*$. But one has to check whether this is indeed so. So, to check that the map $e$ is injective it would be enough to show that there exist two real $n$-planes $L_1, L_2\subset \mathbb C^n$ such that there is no holomorphic $2$-form $\omega$ such that ${\mathrm Re}(\omega)$ restricts as zero to both $L_1$ and $L_2$. Here one can consider different cases when ${\mathrm Re}(\omega)$ has different ranks, but assume for simplicity that ${\mathrm Re}(\omega)$ is symplectic. Then the space of couples of $L_1, L_2$ on which it vanishes has dimension $n(n+1)$ which is twice the dimension of the Lagrangian grassmanian. Now, recall the the space of all forms in $W$ has dimension $n(n-1)$, however two proportional sympelctic forms define the same Lagrangian grassmanian. So we get $n(n-1)-1$. So we conclude that the dimension of pairs of $\mathbb R^n$'s in $\mathbb C^n$ on which a non-degenerate form from $W$ can vanish is $n(n+1)+n(n-1)-1=2n^2-1$. However, the dimension of the space of all pairs of real $n$-planes in $\mathbb C^n$ is $2n^2$. Since $2n^2>2n^2-1$ we conclude that there is a pair on which none of non-degenerate forms from $W$ vanish simultaneously. I would guess that one can treat the cases of degenerate forms from $W$ in a similar way.<|endoftext|> TITLE: What is the meaning of Text inside of AMS logo QUESTION [30 upvotes]: What is the meaning of the text inside this AMS logo? The image is from here, and the logo seems to have been frequently used until the 80's. The text is ΑΓΕΩΜΕΤΡΗΤΟΣ ΜΗ ΕΙΣΙΤΩ but Google translate and Googling have not helped. Added: (07-11-2019) Who is the author (designer/ suggester) of this AMS logo? REPLY [36 votes]: ἀγεωμέτρητος μη εἰσίτω - Let no one untutored in geometry enter here According to tradition this text was displayed in the entrance of Plato's Academy. (The tradition is of a late data, see this critical discussion.) Nicolaus Copernicus choose to print it on the cover page of the book De Revolutionibus. image from 1938 (left) and from 2015 (right) De Revolutionibus REPLY [28 votes]: The phrase «Μηδείς αγεωμέτρητος εισίτω μοι την θύρα» (or «Μηδείς αγεωμέτρητος εισίτω μοι την στέγη») is a phrase that supposedly was on the threshold of Plato's Academy and translates to: "do not let anyone ignorant of geometry through my door" (or under my roof). See for instance this discussion https://www.plato-dialogues.org/faq/faq009.htm The phrase in AMS' logo translates to "Do not be ignorant geometry". Edit: The above translation is not correct. The word "εισίτω" is a form of the verb "εἴσειμι" which means "to come into" (wiktionary). I initially thought that it was a form of "ειμί" which means "to be". So finally the AMS' logo says: "Do not enter if you don't know geometry" or "You will not enter if you don't know geometry".<|endoftext|> TITLE: Decision problems for which it is unknown whether they are decidable QUESTION [33 upvotes]: In computability theory, what are examples of decision problems of which it is not known whether they are decidable? REPLY [3 votes]: It remains open whether the won-position problem of infinite chess is decidable, the problem of determining whether a given finite position in infinite chess is winning for white or not. See Richard Stanley's question at https://mathoverflow.net/q/27967. Meanwhile, the mate-in-$n$ problem of infinite chess is decidable, and so the open part of the problem necessarily concerns positions with infinite game value. Also open is the optimal-play problem: given a won position, is a given move optimal?<|endoftext|> TITLE: Model theory of the complex numbers with conjugation QUESTION [10 upvotes]: Do there exist some results on the theory of $\mathbb{C}$ in the language $\{0,1+, \times, \overline{\cdot}\}$, where $\overline{\cdot}$ is the conjugation map $\overline{a+ib} = a - ib$? I'm wondering in particular if it has QE (or if a sensible expansion of the language has QE), and if it's decidable. But properties like model-completeness, (bi)-interpretability with standard, well-understood structures etc. would be helpful to know too. REPLY [16 votes]: It is a decidable theory, because it is interpretable in the real-closed field $\langle\mathbb{R},+,\cdot,0,1\rangle$, which has a decidable theory. We can interpret complex numbers $a+bi$ as pairs of real numbers $(a,b)$, and the complex structure, including conjugation, is definable in the reals. (Indeed, this is easily seen to be a bi-interpretation, since we can define $\mathbb{R}$ via conjugation in $\mathbb{C}$.) By Tarski's theorem on real-closed fields, that theory is decidable, and so we can decide the complex theory also. Basically, for any given statement in the complex field with conjugation, we can translate it to a question in the real-closed field. By Tarski's result, that question is equivalent to a quantifier-free assertion in the real-closed field $\langle\mathbb{R},+,\cdot,<\rangle$, which we can then easily decide. Since the interpretation doesn't involve any quantifiers (note that we can clear the use of unary minus in conjugation by moving negatives to the other side of any equation), it follows as Alex Kruckman notes in the comments that we will get model-completeness of the complex field with conjugation. My earlier claim that we get full QE for $\mathbb{C}$ stumbles on the fact that $\langle \mathbb{R},+,\cdot\rangle$ doesn't have QE, since you need the order in Tarski's result. As Alex mentions, you can define the positive real line in $\langle\mathbb{C},+,\cdot,\bar{}\rangle$, but this will not be quantifier-free definable. If we add the real and imaginary part operators and the order for the real line as a relation on $\mathbb{C}$, however, then we will get QE in the corresponding expansion, and this expansion will also be bi-interpretable with the real-closed field.<|endoftext|> TITLE: Automorphism group of compact hyperkähler manifolds QUESTION [8 upvotes]: Let $M$ be a compact simply-connected hyperkähler manifold, and let $$ \mathrm{Aut}(M) $$ be the automorphism group of $M$, i.e. the group of tri-holomorphic diffeomorphisms preserving the metric. Then, $\mathrm{Aut}(M)$ is a Lie group since the isometry group of a compact Riemannian manifold is a Lie group as well as the group of biholomorphisms of a compact complex manifold. Questions: (1) Is $\mathrm{Aut}(M)$ discrete, or are there examples where $\dim \mathrm{Aut}(M)>0$? (2) Can $\mathrm{Aut}(M)$ contain a positive-dimensional compact Lie group? REPLY [5 votes]: By $\mathrm{Aut}(X)$ I denote the group of holomorphic automorphisms of a complex manifold $X$. I assume that hyperkähler manifolds are compact and simply-connected. There are two standard series of deformation types of hyperkähler manifolds: $K3^{[n]}$-type. Hyperkähler manifold that is a deformation of a Hilbert scheme of $n$-points on a given projective $K3$-surface. Kummer $n$-type. Hyperkähler manifold that is a deformation of a fiber over zero of the addition morphism $T^{[n+1]}\rightarrow T$, where $T$ is an abelian surface and $T^{[n+1]}$ is its Hilbert scheme of $n+1$-points. For these series you have the following results. Lemma 7.1.3 of Monagardi's thesis. Let $X$ be a hyperkähler of $K3^{[n]}$-type. Then the map $$\mathrm{Aut}(X)\rightarrow O\left(\mathrm{H}^2(X,\mathbb{Z})\right)$$ is injective, where the right hand side is the orthogonal group with respect to Beauville-Bogmolov form on $X$. Remark. The case $n=1$ of the result above is precisely the case of $K3$-surfaces (any $K3$ surface is deformation equivalent to a projective $K3$-surface). There is even more precise celebrated result of Pyatetskii-Shapiro and Shafarevich. Theorem by K.Oguiso. Let $X$ be a hyperkähler of Kummer $n$-type. Then the map $$\mathrm{Aut}(X)\rightarrow \mathrm{Aut}_{\mathbb{Z}}\left(\mathrm{H}^*(X,\mathbb{Z})\right)$$ has trivial kernel. Since every compact smooth manifold is homeomorphic to a finite CW-complex (by virtue of Morse theory), we derive that the ring $\mathrm{H}^*(X,\mathbb{Z})$ is finitely generated as an abelian group. Thus for these two standard series groups of automorphisms are discrete. I strongly recommend Mongardi's thesis as an excellent source of various results and references in this direction. Now the results concerning the two series above are important due to the following remark by Huybrechts (page 4 of this paper). Let us state explicitly the following immediate consequence of (2.1): 2.2.Two birational hyperkähler manifolds are deformation equivalent. In particular, their Hodge, Betti, and Chern numbers coincide. The result was used to show that most of the known examples, with the exception of O’Grady’s exceptional examples in dimension six and ten, are deformations of the two standard series provided by Hilbert schemes of points on K3 surfaces and generalized Kummer varieties. You may also be interested in Huybrechts exhibition of a Torelli theorem due to M.Verbitsky for HK manifolds, which is also related to your questions. As for your questions I don't know the answer, but the results presented above and Huybrechts remark suggest that if there are continuous subgroups, then you should look for them in the realm of O'Grady's examples.<|endoftext|> TITLE: Is the invariant subalgebra of the von Neumann algebra $L(F_k)$ isomorphic to $L(F_k)$? QUESTION [8 upvotes]: Let the symmetric group $G=S_{k}$ act on the von Neumann algebra of the free group $L(F_k)$ via permuting its generators. Is the fixed point algebra under the action isomorphic to the whole algebra, i.e., $L(F_k)^{S_k}\cong L(F_{k})$? REPLY [5 votes]: No, this is not the result that you should expect. Below, I prove that $L(F_k)^{S_k} \cong L(F_N)$ where $N = 1 + (k-1) |S_k| = 1 + (k-1) \cdot k!$. Of course, we do not know if $L(F_k) \not\cong L(F_N)$, but that is independent of the question. In analogy with the Nielsen-Schreier index formula, one expects that a subfactor of index $m$ in $L(F_k)$ is isomorphic with $L(F_{1+(k-1)m})$, but proving this is an open problem. Note that this should even hold for non-integer indices $m$ by using interpolated free group factors. For the specific example in the question, this can however be worked out. Slightly more generally, one may consider a transitive and faithful action of a finite group $\Gamma$ on a finite set $I$ and prove that $L(F_I)^\Gamma \cong L(F_N)$ where $N = 1 + (|I|-1)|\Gamma|$. To prove this formula, denote $A = L(\mathbb{Z})$ equipped with its natural trace. View $L(F_I)$ as the free product of $|I|$ copies of $A$. Fix an element $i_1 \in I$ and denote by $\Gamma_1 < \Gamma$ the stabilizer of $i_1$. The crossed product $L(F_I) \rtimes \Gamma$ is then naturally isomorphic with the amalgamated free product $$M = (A \otimes L(\Gamma_1)) *_{L(\Gamma_1)} L(\Gamma) \; .$$ Since $\Gamma \curvearrowright I$ is faithful, the action $\Gamma \curvearrowright L(F_I)$ is outer and $M$ is a $II_1$ factor. By Theorem 4.2 of [DK11], $M$ is isomorphic with an interpolated free group factor $L(F_s)$ and, using the notation of [DK11], the value of $s$ can be computed as $$s = \operatorname{fdim}(A \otimes L(\Gamma_1)) + \operatorname{fdim}(L(\Gamma)) - \operatorname{fdim}(L(\Gamma_1)) \; .$$ Since $A \otimes L(\Gamma_1)$ is diffuse, one has $\operatorname{fdim}(A \otimes L(\Gamma_1))=1$. For any finite group $\Lambda$ and using the canonical trace on $L(\Lambda)$, one has that $$\operatorname{fdim}(L(\Lambda)) = 1 - \frac{1}{|\Lambda|} \; .$$ We thus conclude that $$s = 1 + \frac{1}{|\Gamma_1|} - \frac{1}{|\Gamma|} \; .$$ So we have proven that $$L(F_s) \cong M \cong L(F_I) \rtimes \Gamma \cong M_{|\Gamma|}(\mathbb{C}) \otimes L(F_I)^\Gamma \; .$$ We conclude that $L(F_I)^\Gamma \cong L(F_s)^{1/|\Gamma|} \cong L(F_N)$ where $$N = 1 + |\Gamma|^2 (s-1) = 1 + (|I|-1) |\Gamma| \; .$$ [DR11] K.J. Dykema and D. Redelmeier, The amalgamated free product of hyperfinite von Neumann algebras over finite dimensional subalgebras. Houston J. Math. 39 (2013), 1313–1331.<|endoftext|> TITLE: Given a symmetric polynomial in F_q, write it in terms as elementary symmetric polynomials. How to find out the coefficient? QUESTION [5 upvotes]: Consider the finite field $F_q$, where $q$ is a power of an odd prime and $N$ is a power of $q$. We have a homogeneous symmetric polynomial $$ E_{l,s}(x) = \sum_{\substack{l_1+l_2+\cdots +l_s=l \\ l_i\geqslant 1}}\ \sum_{1\leqslant i_1 TITLE: Geometric intuition behind this chain homotopy QUESTION [8 upvotes]: My question has to do with the chain homotopy that appears in Lee's Introduction to Topological Manifols and Rotman's Introduction to Algebraic Topology proofs that the inclusion $$C_\bullet^\mathcal{U}(X)\hookrightarrow C_\bullet(X)$$ induces an isomorphism in singular homology $$H_p^\mathcal{U}(X)\cong H_p(X)$$ For all $p\geq 0$. In both references, a chain homotopy $h:C_p(X)\longrightarrow C_{p+1}(X)$ between the barycentric subdivision operador and the identity map is given by If $p=0$, $h$ is the zero homomorphism. If we have defined $h$ up to some $p\in\mathbb{N}$, and $\sigma$ is a $p-$simplex in $X$, then $$h\sigma=\sigma_\#b_p*(i_p-si_p-h\partial i_p)$$ Where $b_p$ is the barycentre of the standard $p-$simplex $\Delta_p$ and $*$ is the cone operator. We then extend $h$ linearly to singular chains: $h\big(\sum_{i\in I}n_i\sigma_i\big)=\sum_{i\in I}n_ih\sigma_i$ In contrast with the chain homotopy that appears in the proof of the homotopy axiom, this is really less intuitive, and relies heavily on the equation $$\partial(w*c)=c-w*\partial c$$ So my questions are: How should we understand geometrically this map $h$? What is the geometric intuition that allows us to choose this a a good chain homotopy for our purposes? How should we understand the formula $\partial(w*c)=c-w*\partial c$? What is the meaning of this equation geometrically speaking? How to come up with such a map in the first place? How has this theorem developed historically? I understand perfectly both demonstrations, since the calculations are easy to follow; I am just concerned with how this map gives no intuition at first glance about the geometry involved. REPLY [13 votes]: When thinking about chain homotopies in a setting involving simplices it can be helpful to consider the product $\Delta^p\times I$ where $\Delta^p$ is a $p$-simplex and $I=[0,1]$. The formula $h\sigma=\sigma_{\sharp} b_p \ast(i_p-si_p-h\partial i_p)$ corresponds to a certain inductively defined subdivision of $\Delta^p\times I$ obtained by coning off a subdivision of $\Delta^p\times\partial I \cup \partial \Delta^p \times I$ to a point in the interior of $\Delta^p\times I$. The subdivision of $\Delta^p\times\partial I \cup \partial \Delta^p \times I$ is $\Delta^p$ itself (unsubdivided) on $\Delta^p\times \{0\}$ and the barycentric subdivision of $\Delta^p$ on $\Delta^p\times\{1\}$. These are the terms $i_p$ and $si_p$ in the formula. On $\partial \Delta^p\times I$ one uses the subdivision given by induction. This is the term $h\partial i_p$. The term $\sigma_{\sharp} b_p$ corresponds to the point in the interior of $\Delta^p\times I$ that one cones off to, with the symbol $\ast$ denoting the coning operation. What is perhaps most puzzling is that the formula says nothing about taking the product with $I$, but this is because in reality one takes the subdivision of $\Delta^p\times I$ and projects it to $\Delta^p$ before applying the map $\sigma$, whose domain is $\Delta^p$ rather than $\Delta^p \times I$. I have seen this method of subdividing $\Delta^p\times I$ in several books when they are developing homology theory, but it is more complicated than necessary. A simpler subdivision that suffices is to cone off a subdivision of $\Delta^p\times \{0\} \cup \partial \Delta^p \times I$ to the barycenter of $\Delta^p\times\{1\}$, where $\Delta^p\times \{0\}$ is unsubdivided and $\partial \Delta^p \times I$ has the subdivision given inductively. On $\Delta^p\times \{1\}$ this gives just the usual barycentric subdivision, which is also defined inductively. There is a picture of this subdivision of $\Delta^p\times I$ in the case $p=2$ on page 122 of my algebraic topology book. Perhaps other books such as the Lee book you mention don't give a picture because the picture would be more complicated for the more complicated subdivision. An advantage of the simpler subdivision is that the formula for $h\sigma$ becomes just $\sigma_{\sharp} b_p \ast(i_p-h\partial i_p)$, without the term $si_p$. The more complicated formula is given in the classic book of Eilenberg and Steenrod (page 197) without pictures or explanation. Perhaps other books are just following suit.<|endoftext|> TITLE: Does anyone know a basepoint-free construction of universal covers? QUESTION [23 upvotes]: Let $X$ be a real manifold (for simplicity). The standard construction of the universal cover $\varphi: \widetilde{X} \longrightarrow X$ involves fixing a basepoint $p \in X$ and considering homotopy classes of paths from $p$ to $x \in X$. Is there an alternative construction of $\varphi$ that avoids choosing a basepoint? REPLY [4 votes]: Here is another attempt at pinning down the meaning of "canonical" in reference to Tom's answer. Let $X$ be a nice space (connected, locally path-connected and semi-locally simply connected). Let $\pi_X$ be the fundamental groupoid of $X$: this is a category whose objects are points $x\in X$, where a morphism $x\to y$ is a homotopy class of path fixing the endpoints. Let $U_X$ be the groupoid of universal covers: an object is a universal cover $X_1 \to X$ and a morphism $X_1 \to X_2$ is an isomorphism of covers over the identity map of $X$. There is a functor $$f:\pi_X\to U_X$$ (i.e., a homomorphism of groupoids) given by the usual construction of a universal cover. Then $f$ is an equivalence of categories (covering space theory). Let $$g: U_X \to \pi_X$$ be its adjoint (which is defined up to unique isomorphism). This means for any $\tilde X\in U_X$, with $g(\tilde X) = x\in X$ we have a preferred isomorphism $$ f(x) \cong \tilde X\, . $$ In other words, a universal cover determines a basepoint and a basepoint determines a universal cover.<|endoftext|> TITLE: What is the hidden symmetry behind four generic planes in $\mathbb{R}^4$? QUESTION [10 upvotes]: Consider the action of $\operatorname{GL}(\mathbb{R}^4)$ on the Grassmannian of 2-dimensional subspaces of $\mathbb{R}^4$. In experiments, I observe that four randomly drawn points in this space are simultaneously stabilized by nontrivial members of $\operatorname{GL}(\mathbb{R}^4)$. This violates my naive attempts at counting dimensions: The stabilizer subgroup of any given plane is a generic subset of a 12-dimensional subspace of $\mathbb{R}^{4\times 4}$. If I draw three planes at random, the intersection of the corresponding subspaces is 4-dimensional, as expected. If I draw a fourth plane at random, I expect the resulting intersection to be the span of the identity matrix, but alas, I consistently obtain a 2-dimensional intersection. Only after I draw a fifth plane does the resulting intersection equal the span of the identity matrix. What's going on here? More generally, when should I expect subsets of Grassmannian spaces to have trivial stabilizer? REPLY [12 votes]: Work over any field $k$. Taking two generic planes in 4-dimensions, we can get them to our favourites by linear transformation $k^4=k^2\oplus k^2$, reducing $GL_4$ to $GL_2 \times GL_2$. A third plane, generically, is a graph of a unique linear map from one to the other $y=Ax$. The group action is by matrix similarity. We normalize to get $A=I$, reducing to $GL_2$. A 4th plane, generically, is $y=Ax$ but with $A$ having distinct nonzero eigenvalues. The action of $GL_2$ on $A$ is by conjugation. The group preserving all 4 planes is now reduced to the subgroup preserving a splitting of the plane into eigenspaces of $A$, i.e. $k^{\times} \times k^{\times}$.<|endoftext|> TITLE: Can it be that universal measurability is preserved by projections? QUESTION [5 upvotes]: I am currently discovering descriptive set theory—with much pleasure! It is something of a surprise to me that, while the Borel hierarchy is indexed by $\omega_1$, the projective hierarchy is only indexed by $\mathbf{N}$, and that classical descriptive set theory stops there… In particular, it would seem natural to “merge” the respective ideas behind Borel and projective hierarchies, to consider the smallest algebra of subsets of $\mathbf{N}^{\mathbf{N}}$ stable under complementation, countable unions and projections. Has this been done; and if yes, what is it called? I wonder in particular if it would be consistent with ZFC for every set in this algebra to be universally measurable? (and, maybe, that being universally measurable is preserved by projections…?). Clearly this cannot be a theorem of ZFC, since assuming $V=L$ proves existence of a $\mathbf{\Delta}^1_2$ set not being universally measurable; but I have read nowhere that my question would be contradictory with ZFC… (Also, subsidiary question: the notation for Borel and projective hierarchies, with respective exponents $0$ and $1$, suggest that that there would be a “$k$-th order hierarchy” for $k \geq 2$, giving rise to $\mathbf{\Sigma}^k_n$, $\mathbf{\Pi}^k_n$ and $\mathbf{\Delta}^k_n$ sets… But I have nowhere read such a definition! So, does it exist something continuing the Borel and projective hierarchies in this way?…). REPLY [7 votes]: I'm not an expert, so please correct me if I'm wrong: We can indeed continue the projective hierarchy beyond its finite levels. And like the Borel hierarchy, we can do this "from below" as follows: $\bf\Sigma^1_1$ is the class of analytic sets. $X$ is ${\bf \Pi^1_\alpha}$ iff $\omega^\omega\setminus X$ is ${\bf\Sigma^1_\alpha}$. For $\alpha>0$, a set $X$ is ${\bf \Sigma^1_\alpha}$ iff $$X=\bigcup_{i\in\omega}Y_i$$ for some family of sets $(Y_i)_{i\in\omega}$ such that each $Y_i$ is the projection of some $Z_i$ with $Z_i\in{\bf \Pi^1_{\beta_i}}$ for some $\beta_i<\alpha$. (Note that this last clause means that we don't simply take unions of pointclasses at limit stages; we could do that instead if desired.) As with the Borel hierarchy, the regularity of $\omega_1$ implies that this hierarchy stablizes at level $\omega_1$. Call the sets that appear in this hierarchy the transfinitely projective sets (I don't know if there is a standard term for them). It's not hard to show that, just like the projective sets, each transfinitely projective set $X$ is an element of the inner model $L(\mathbb{R})$. Under large cardinals, $L(\mathbb{R})$ satisfies ZF + DC + AD (indeed a bit more), and so in fact assuming large cardinals, every transfinitely projective set is "tame" for the usual meanings of the word "tame." In particular, the answer to your question is yes. (Well, the first question, anyways - obviously this doesn't address the issue of whether in universal measurability could be preserved by projections consistently with ZFC. I suspect the answer to that question is negative, although I don't see it at the moment.) It's worth noting that pointclasses of the form $\mathcal{P}(\omega^\omega)\cap\mathcal{M}$ for some "tame" (assuming large cardinals) inner model $\mathcal{M}$ are extensively studied. These, per the above, reach far beyond the transfinitely projective. I think this explains why the transfinitely projective sets aren't as well-studied: my impression is that if we want to go past the finite levels of the projective hierarchy we generally want to go way past that to one of these more set-theoretic pointclasses. (Relatedly, see this old Mathoverflow question.)<|endoftext|> TITLE: Linear suborders of $(P(\omega),\subseteq)$ QUESTION [7 upvotes]: Consider the partial order $(P(\omega),\subseteq)$. Let $L$ be a dense linear suborder. Does $L$ have a countable dense subset? (Note that it contains a copy of $\mathbb R$, via Dedekind cuts of $\mathbb Q$.) REPLY [14 votes]: For any $a\neq b\in\omega$ pick a set $A_{ab}\in L$ such that $a\in A_{ab},b\not\in A_{ab}$, if such exists, and let $D$ be the set of all $A_{ab}$ we have picked. We claim $D$ is dense in $L$. Let $X,Y\in L$ with $X\subsetneq Y$. Pick any $Z$ such that $X\subsetneq Z\subsetneq Y$ and take $a\in Z\setminus X,b\in Y\setminus Z$. Then $Z$ is an element containing $a$ but not $b$, so $A_{ab}\in D$ exists. We have $A_{ab}\not\subseteq X$, so $X\subseteq A_{ab}$, and $Y\not\subseteq A_{ab}$, so $A_{ab}\subseteq Y$. Hence $D$ contains an element $A_{ab}$ strictly between $X$ and $Y$, so $D$ is a dense countable subset of $L$. Note that countable choice was used in the proof, but it is unavoidable, since (\mathbb R and hence also) $P(\omega)$ can contain a finite Dedekind-infinite subset, and this will necessarily contain no dense countable subset. REPLY [11 votes]: Note that the lexicographic order of $2^\omega$ is a linear extension of $(\mathcal P(\omega),\subseteq)$, namely if $A\subseteq B$, then $\min(A\mathbin{\triangle}B)\in B$, which means that the first point that the sets differ must be in $B$. But $2^\omega$ is hereditarily separable, and therefore every linear suborder of $\mathcal P(\omega)$ is separable.<|endoftext|> TITLE: Unifying "cohomology groups classify extensions" theorems QUESTION [18 upvotes]: It is common for the first or second degree of various cohomologies to classify extensions of various sorts. Here are some examples of what I mean: 1) Derived functor of hom, $\text{Ext}^1_R(M, N)$. Let $R$ be a ring (not necessarily commutative, with a $1$). As is its namesake, $\text{Ext}^1_R (M, N)$ classifies extensions $0 \rightarrow N \rightarrow L \rightarrow M \rightarrow 0$ up to isomorphism. We can put an abelian group structure on this, the Baer sum. 2) Quillen cohomology, $\text{D}^1_{R} (A/B, M)$. Let $R$ be a commutative ring and let $A$ be an $R$-algebra. Let $B$ Be an $R$-algebra with a map into $A$ (so we have a sequence $R \rightarrow A \rightarrow B$). The first Andre-Quillen cohomology $D^1(A/B, M)$ is $\text{Exalcomm}(A/B, M)$, the set of $A$-algebra extensions $0 \rightarrow M \rightarrow C \rightarrow B \rightarrow 0$, where we set $a b = 0$ for $a, b \in M$. In other cohomologies, $H^2$ classifies extensions: 3) Group cohomology, $H^2(G;M)$. Let $G$ be a group, and let $M$ be a $G$-module. $H^2(G;M)$ classifies extensions $0 \rightarrow M \rightarrow L \rightarrow G \rightarrow 0$ where $L$ is a $G$-module and $M$ has the same $G$-module structure as the one inherited from $L$. 4) Lie algebra cohomology, $H^2(\mathfrak{g};M)$. Let $R$ be a ring. Let $\mathfrak{g}$ be a Lie-algebra over $R$, and let $\mathfrak{a}$ be an abelian Lie-algebra over $R$. $H^2(\mathfrak{g}, \mathfrak{a})$ classifies central extensions $0 \rightarrow \mathfrak{a} \rightarrow \mathfrak{h} \rightarrow \mathfrak{g} \rightarrow 0$. See remark 4.7. here. 5) Hoschild cohomology, $H^2(R, M)$. Let $R$ be a commutative ring. The Hoschild cohomology module $H^2(R, M)$ classifies extensions $0 \rightarrow M \rightarrow S \rightarrow R \rightarrow 0$ where $ab = 0$ for $a, b \in M$. Note that Hoschild cohomology and Quillen cohomology are related; often $D^{n+1}(R, M) = H^n(R, M)$. (1) and (2) are potentially different from (3) and (4), but there is the mentioned relationship between (2) and (5). I am wondering if there is a (possibly categorical) unification of these theorems. Probably it would be easier to unify (1) and (2), and (3) and (4) separately- I can't say for sure there is a relationship between these pairs. There is probably even a generalization to "extensions of length $n$" for many of these. For instance, $\text{Ext}^n(M, N)$ classifies exact sequences $0 \rightarrow N \rightarrow X_1 \rightarrow \cdots \rightarrow X_n \rightarrow M \rightarrow 0$ (See Weibel, $\textit{An Introduction to Homological Algebra}$, 3.4.6) REPLY [16 votes]: $\newcommand{\cA}{\mathcal{A}}\newcommand{\Ext}{\mathrm{Ext}}\newcommand{\Hom}{\mathrm{Hom}}$Let $\cA$ be an abelian category; then, $\Ext_\cA^i(A,B)$ is literally $\Hom_{D(\cA)}(A, B[i])$, where $B[i]$ denotes the shift. A good way of thinking about this is as the cohomology of the derived Hom $\mathrm{R}\Hom_{D(\cA)}(A, B)$. In the case $i=1$, an element of $\Ext^1_\cA(A,B)$ is then literally a map $A\to B[1]$. The kernel $K$ of this map (in the derived category) defines a distinguished triangle $K\to A\to B[1]$. Since we're working in a derived --- in particular, triangulated --- category, we can rotate this further back to a distinguished triangle $B\to K\to A$. This is precisely the data of an extension. Choosing $\cA$ correctly specializes to most of the examples you wrote in your question. For instance: If $\cA = \mathrm{Mod}_R$ for some commutative ring $R$, then $\Ext_\cA$ is literally the Ext-group that you wrote down. If $G$ is a group, then $\cA$ can be the category of representations of $G$ on $\mathbf{Z}$-modules (i.e., abelian groups). Recall that a representation is just a module over a group ring. In this case, the group cohomology $\mathrm{H}^i(G; M)$ is $\Ext^i_{\mathbf{Z}[G]}(\mathbf{Z}, M)$: you can think of this as derived Homs from the unit object $\mathbf{Z}$ into your $G$-representation. I'll address central extensions below. If $\cA$ is the category of modules over the universal enveloping algebra $U(\mathfrak{g})$ of a Lie algebra $\mathfrak{g}$ over $k$, then $\mathrm{H}^i(\mathfrak{g}; M) = \Ext^i_{U(\mathfrak{g})}(k, M)$. Again, these are derived Homs from the unit object $k$ into your $\mathfrak{g}$-module. If $A$ is a $k$-algebra, and $\cA$ is the category of $A$-$A$-bimodules, then Hochschild cohomology is $\Ext^i_{D(\cA)}(A, M)$. Again, these are homs from the unit object $A$ into your $A$-$A$-bimodule. Andre-Quillen cohomology fits into this formalism as well, but this time you're not taking derived Homs into an object in the heart of your derived category (i.e., a usual module). Let $A$ be a commutative ring, and $B$ be an $A$-algebra. If $M$ is a $B$-module, then the Andre-Quillen cohomology is (by definition) the cohomology of the derived Hom $\mathrm{RHom}_B(L_{B/A}, M)$. But if $P_\bullet$ is a simplicial cofibrant resolution of $B$ as an $A$-algebra, we have $\mathrm{RHom}_B(L_{B/A}, M) = \mathrm{Der}_A(P_\bullet, M)$ --- that's what the cotangent complex represents. If you write down what the first cohomology group of this thing is, like with the above examples, you'll get precisely the result you stated in the question. By the way, one relation between Andre-Quillen cohomology and Hochschild cohomology comes from the Hochschild-Kostant-Rosenberg theorem, which says that if $k$ is of characteristic $0$, then the Hochschild homology $A\otimes_{A\otimes_k A^{op}}^\mathbf{L} A$ of a $k$-algebra $A$ is given by $\Omega^\bullet_{A/k} = \mathrm{Sym}^\bullet(\Omega^1_{A/k}[1])$. One can think of the Hochschild homology as the homotopy colimit of the constant $S^1$-indexed diagram (where $S^1$ is the simplicial circle $\Delta^1/\partial \Delta^1$) in simplicial $k$-algebras with value $A$, so there's an $S^1$-action on $\Omega^\bullet_{A/k}$. This defines a map $C_\ast(S^1; k)\otimes_k \Omega^\bullet_{A/k}\to \Omega^\bullet_{A/k}$ in the derived category of $k$-modules. But $S^1$ is rationally formal, and so $C^\ast(S^1; k) = k[d]/d^2$, where $d$ lives in degree $1$; the action of $d$ on $\Omega^\bullet_{A/k}$ is precisely the de Rham differential. Before moving to central extensions, let me say that there wasn't much about derived categories of abelian categories that I used above, other than the facts that they have a notion of distinguished triangles which you can rotate (already true in triangulated categories) and that they are enriched over abelian groups (so taking cohomology of complexes makes sense). You can do all of these constructions more generally in stable model categories/stable $\infty$-categories, and in these more general cases, you recover the simple man's cohomology from algebraic topology 1. More precisely, let's replace $\cA$ with the category $\mathrm{Sp}$ of spectra, so shifts are now given by suspension. We can then still form "derived Hom", i.e., the mapping spectrum. If $X$ is a spectrum, and $E$ is a spectrum, then $\Hom_\mathrm{Sp}(X, \Sigma^i E)$ is a perfectly well-defined spectrum, and its $\pi_0$ is just the cohomology group $E^i(X)$. Of course, you can replace $\mathrm{Sp}$ with, for instance, modules over a structured ring spectrum, and the same sort of construction holds. If your structured ring spectrum is the Eilenberg-Maclane spectrum associated to a classical ring $R$, then you'd be working in the derived category of chain complexes over $R$. It is important to remark, however, that the cotangent complex in this spectral world of a discrete ring regarded as a structured ring spectrum is not the same as the usual notion of cotangent complex: this is the distinction between topological Andre-Quillen cohomology and ordinary Andre-Quillen cohomology. (If $R$ is a $\mathbf{Q}$-algebra, then all is well and everything agrees with the classical notions.) As Qfwfq states in the comments, the first, second, and fifth examples (in the order listed in my answer) can all be thought of as taking place in the category of sheaves over an algebro-geometric object. For example, $\mathrm{Mod}_R = \mathrm{QCoh}(\mathrm{Spec}(R))$, while $G$-representations on $\mathbf{Z}$-modules are $\mathrm{QCoh}(\mathrm{Spec}(\mathbf{Z})/G)$, where $X/G$ denotes the stacky quotient. For the fifth example, you have to work in the stable category of quasicoherent sheaves on the derived scheme $\mathrm{Spec}(B)$. In the third example, you'd be working in the stable category of left modules (in spectra) over $U(\mathfrak{g})$. In fourth example, you can do the same thing: the category of $A$-$A$-bimodules is the stable category of left modules (in spectra) over $A\otimes_k A^{op}$. OK, what about central extensions and $\mathrm{H}^2$? Consider the case of the group cohomology $\mathrm{H}^2(G; M)$, where $M$ has a $G$-action. An element in this is given by a map $BG\to B^2 M$, which upon looping defines a loop map $G\to BM$. Let $L$ denote the fiber of this map; then, rotating this loop map defines a fiber sequence $M\to L\to G$ of loop spaces, i.e., a group extension. Moreover, the map $M\to L$ is central. Of course, this argument is pretty general, and you can run it in the Lie algebra setting as well. As for obstruction theory: there's a lot to say here, and I've already written a bunch in this answer. However, this is the point of cotangent complexes: the recipe is to break down your obstruction theory problem into understanding extending along square-zero extensions, and then you're in the land of derivations. I can elaborate more if you'd like. REPLY [3 votes]: There is a general simplicial theory of n-torsors due in part to Duskin (see J. Duskin, 1975, Simplicial methods and the interpretation of “triple” cohomology, number 163 in Mem. Amer. Math. Soc., 3, Amer. Math. Soc., and then P. Glenn, Realization of cohomology classes in arbitrary exact categories, J. Pure. Appl. Alg., 25, (1982), 33 – 105.) That theory is quite abstract and VERY general. In other cases, for instance in group cohomology, a related set of results is found in J. Huebschmann, Crossed n-fold extensions of groups and cohomology, Comm. Math. Helv., 55, (1980), 302 – 313. Those papers are quite old now and there is more up to date treatments including work by Beke, Aldrovandi and others with varying amounts of `geometric' input and interpretation. The generalisation of this to obstruction theoretic versions in non-abelian settings can be found in papers by Breen, eg. On the classification of 2-gerbes and 2-stacks, Ast ́erisque, 225, (1994), 160, and nice notes of his: L. Breen, 2009, Notes on 1-and 2-gerbes, in J. Baez and J. May, eds., Towards Higher Cat- egories, volume 152 of The IMA Volumes in Mathematics and its Applications, 193–235, Springer.<|endoftext|> TITLE: The Floer Equation is Elliptic QUESTION [6 upvotes]: Let $(M,\omega)$ be a symplectic manifold and $H \in C^\infty(M \times \mathbb{S}^1)$. Furthermore, let $J$ be an $\omega$-compatible almost complex structure on $M$. The Floer equation is the following partial differential equation: $$\partial_s u + J(\partial_tu - X_{H_t} \circ u)=0$$ where $X_{H_t}$ is the unique vector field defined by $i_{X_{H_t}}\omega = -dH_t$ and $u \colon \mathbb{R} \times \mathbb{S}^1 \to M$, where $(s,t) \mapsto u(s,t)$. I think this is a nonlinear first-order partial differential equation. In studying solutions to this equation, one applies elliptic regularity. I do know what a second order elliptic equation is and I checked the definition on Wikipedia for the other cases than $2$, but I am confused what exactly the terminology elliptic means and how we can see this explicitely in the case of the Floer equation. REPLY [4 votes]: First of all, ellipticity is defined in terms of the principal symbol of an operator, and the Hamiltonian term is zeroth order in the derivatives of u, so let's assume wlog that we're taking about J-holomorphic maps. Moreover, in local coordinates, the difference between J and your favourite constant-coefficient complex structure also doesn't depend on derivatives of u, so the Floer equation is a zeroth order perturbation of the usual Cauchy-Riemann equation, which is elliptic in the sense explained by Denis Serre above (i.e. it implies harmonicity of the components). Here's an alternative way to see that pseudoholomorphic maps satisfy an elliptic second order equation. Consider the $L^2$-energy $\int|du|^2 dvol$ of your map u with respect to the almost Kaehler metric given by J and the symplectic form. The critical points of this are called harmonic maps and satisfy the (second order) harmonic map equation (usual harmonicity of components if the metric is Euclidean). Pseudoholomorphic maps are not only critical points, they're global minimisers of the energy: the energy can be rewritten as $\int\omega$ (constant) plus a term which vanishes if and only if u is pseudoholomorphic. In the instanton/soliton literature, I think this is called the Bogomolny trick (writing your action as a topological term plus a term which vanishes if and only if a first order equation vanishes). Finally, it is possible to write the Floer equation as a pseudoholomorphic map equation into a larger dimensional manifold with an almost complex structure depending on J and H (I think this is due to Gromov, and is explained somewhere in McDuff-Salamon "J-holomorphic curves and symplectic topology"), so ellipticity of the Floer equation follows from that of the pseudoholomorphic map equation.<|endoftext|> TITLE: Relative Ext of Avramov-Martsinkovsky as a derived Hom QUESTION [6 upvotes]: Avramov-Martsinkovsky (http://mathserver.neu.edu/~martsinkovsky/Relative.pdf) have defined an exotic version of Ext between two modules over (for simplicity) Gorenstein rings. The basic idea of their construction in the Gorenstein case is to treat "totally reflexive" modules (=modules of G-dimension zero) as the analogue of projective modules. One then takes suitable left resolutions by these modules to obtain this exotic Ext. I am wondering whether anyone had formulated a "relative derived category" where these Exts are the Hom groups in the category? As a concrete question: Positselski has defined exotic versions of derived categories in https://arxiv.org/pdf/0905.2621.pdf. Are the derived categories studied in Section 3.9 related to the [AM] construction? Remark: I am aware of "stable module categories," which I gather play the role that I want with "relative Ext" replaced by "Tate cohomology." REPLY [2 votes]: Yes, there is a Gorenstein derived category where the Ext groups relative to the Gorenstein projectives can be expressed as Homs in this Gorenstein derived categories. See https://arxiv.org/pdf/1712.04587.pdf section 3 (especially corollary 3.3.4) for a nice overview on this and https://www.sciencedirect.com/science/article/pii/S0021869310000517 for the original research article.<|endoftext|> TITLE: $\lfloor \frac{3^n}{2^n} \rfloor = \lfloor \frac{3^n-1}{2^n-1} \rfloor$ for $n\geq2$ QUESTION [14 upvotes]: I stumbled upon the following claim online: $\lfloor \frac{3^n}{2^n} \rfloor = \lfloor \frac{3^n-1}{2^n-1} \rfloor$ for all integers $n\in \mathbb{N}$, $n\geq2$. Checking with the computer, the claim seems to be true at least for integers $n$ such that $2 \leq n \leq 10000$. How could I prove this claim is true for all integers such that $n\geq 2$? REPLY [26 votes]: Mahler proved in 1957 (see here) that if $q$ is a positive rational number which is not an integer, then the distance of $q^n$ to the nearest integer is $(1-o(1))^n$. In particular, taking $q=3/2$, we have for $n$ sufficiently large that $$\lfloor q^n\rfloor+1-q^n>(q/2)^n>\lfloor q^n\rfloor/2^n.$$ Rearranging the two sides, we get for $n$ sufficiently large that $$\left\lfloor\frac{3^n}{2^n}\right\rfloor>\frac{3^n-1}{2^n-1}-1,$$ hence also that $$\left\lfloor\frac{3^n}{2^n}\right\rfloor=\left\lfloor\frac{3^n-1}{2^n-1}\right\rfloor.\tag{$\ast$}$$ Mahler's proof is ineffective (i.e. it does not produce a lower bound for $n$), because it relies on Roth's approximation theorem, which is ineffective. As far as I know, we still don't have an effective lower bound for $n$ beyond which $(\ast)$ is guaranteed to hold.<|endoftext|> TITLE: Does Peetre's theorem hold in complex analysis? QUESTION [5 upvotes]: Let $E, F$ be two smooth vector bundles over a smooth manifold $M$. Peetre's theorem states that any $\mathbb{R}$-linear morphism $D: \mathcal{E} \to \mathcal{F}$ of the sheaves of sections of $E$ and $F$ is a differential operator, i.e. there exist trivializing charts $U_i$ such that on $D|_{U_i}$ is a differential operator in the sense of real analysis. Question 1: Does this result also hold for complex manifolds? By this I mean that $M$ is a complex manifold, $E$ and $F$ are holomorphic vector bundles and $D$ is $\mathbb{C}$-linear. We then also restrict to sheaves of holomorphic sections. I don't think the proof of the real version can be adapted directly, since it relies on bump functions. This is related to my m.SE question in which I showed that the result does not hold for the sheaf of rational functions on $\mathbb{C}$. However, on any locally ringed space $(X, \mathcal{O})$ one can define differential operators of order $0$ as multiplication with a global section of $\mathcal{O}$ and differential operators of order $n$ inductively as $\mathbb{C}$-linear morphisms of sheaves $D: \mathcal{E} \to \mathcal{F}$ satisfying that $f D - D f$ is a differential operator of order $n-1$. General differential operators can then be defined as $\mathbb{C}$-linear morphisms of sheaves which locally are differential operators of finite order. This gives the expected result for the differential operators on a Scheme (EGA IV Prop 16.8.8) (i. e. on $n$-dimensional affine space they are polynomials in the $n$ partial derivatives). By Peetre's theorem this definition also coincides with the classical notion of differential operators on a smooth manifold. Question 2: If $\mathcal{E}$ and $\mathcal{F}$ are sheaves of sections of holomorphic vector bundles over a complex manifold, does the above definition of differential operators produce the classical differential operators between holomorphic vector bundles? REPLY [3 votes]: I think Peetre's theorem is false in the holomorphic category. Namely, one can construct a counter example (a differential operator of infinite order) already in the case of one complex variable. We want to exhibit a sheaf morphism $D\colon \mathcal{O} \to \mathcal{O}$, where $\mathcal{O}$ is the sheaf of holomorphic functions in one complex variable $z$ of the form $$ D[f](z) = \sum_{k=0}^\infty d_k \frac{f^{(k)}(z)}{k!} , $$ with infinitely many non-zero $d_k$. Since this is a purely local question, we can just assume that we are only dealing with functions $f$ defined on some neighborhood of $z=0 \in \mathbb{C}$. First, recall that the Taylor coefficients of any holomorphic function $f(z)$ grow no faster than exponentially (as a consequence of the Cauchy integral formula), that is $$ \left|\frac{f^{(k)}(0)}{k!}\right| < \frac{C}{r^k} , $$ for some $C, r > 0$, where $r$ is at least as large as the radius of a disk that fits into the domain to which $f(z)$ has a unique analytic continuation. Since the same argument works also about any point of the domain of $f(z)$, we can always find $0 < R \le r$ such that $$ \left|\frac{f^{(k)}(z)}{k!}\right| < \frac{C}{R^k} , $$ uniformly on some (possibly small) neighborhood of $z=0$. The point is that such a neighborhood and corresponding constants $C, R > 0$ exist for any function $f(z)$ holomorphic at $z=0$. Now, let $d_k$ be the coefficients of some non-polynomial entire function $d(z) = \sum_{k=0}^\infty d_k z^k$, meaning that infinitely many $d_k\ne 0$ and the sums $\sum_{k=0}^\infty |d_k|/R^k$ converge for any $R>0$. For example $d_k = 1/k!$. The question to answer is the following: given $f(z)$ holomorphic on some neighborhood of $z=0$, does $D[f](z)$ define a holomorphic function on some (possibly smaller) neighborhood of $z=0$? By combining the above estimates, we see that the answer is Yes, since the inequalities $$ \left|\sum_{k=0}^N d_k \frac{f^{(k)}(z)}{k!} \right| \le C \sum_{k=0}^\infty \frac{d_k}{R^k} < \infty $$ for arbitrarily large $N$ mean that the series defining $D[f](z)$ converges uniformly (and hence to a holomorphic function) on the same neighborhood of $z=0$ on which $|f^{(k)}(z)/k!| < C/R^k$ uniformly. So $D$ maps germs of holomorphic functions to germs of holomorphic functions and, given how it was defined, it also clearly satisfies all the other properties of a morphism of sheaves. And yet, $D$ is not a differential operator of finite order. By construction, $D$ is a differential operator of infinite order with constant coefficients. But clearly, we can take the coefficients $d_k(z)$ as holomorphic functions, provided that they satisfy similar locally uniform bounds. So there are lots of possibilities for constructing sheaf morphisms that are not finite order differential operators.<|endoftext|> TITLE: Covering the disk with a family of infinite total measure QUESTION [17 upvotes]: Let $(U_n)_n$ be an arbitrary sequence of open subsets of the unit disk $D(0,1)\subseteq \mathbb{R}^2$ s.t. $\sum_{n=0}^\infty \lambda(U_n)=\infty$ (where $\lambda$ is the Lebesgue measure). Does there exist a sequence $(q_n)_n$ in $\mathbb{R}^2$ s.t. $D(0,1) \subseteq \bigcup_{n=0}^\infty (q_n+U_n)$? With the notation $q_n+U_n$, I mean $$q_n+U_n:=\{x\in \mathbb{R}^2|x-q_n\in U_n\}$$ EDIT: Fedor Petrov was quick to find an easy answer to this one and I'm forced to accept it. His method doesn't hold up though if I additionally demand that all the $U_n$ are convex. So, submissions with a take on such a related question are still welcome (although I'll not be able to reward your submission with a "accepted answer" badge) UPDATE: Acting on popular request, I've reposted the revised question over here. REPLY [14 votes]: No even in dimension 1 (and multiplying the example for $\mathbb{R}$ by the small segment you get a counterexample in $\mathbb{R}^2$). Take the set $A_n\subset \mathbb{R}$ defined as $\bigcup_{k\in \mathbb{Z}} (2k\cdot 10^{-n},(2k+1)\cdot 10^{-n})$. I claim that there exists no finite family of translates $\bigcup_{n=1}^\infty (A_n+q_n)$ which covers $[0,1]$. Indeed, we may recursively find a nested family segments $[0,1]\supset \Delta_1\supset \Delta_2\ldots$ such that length of $\Delta_i$ equals $10^{-i}$ and $\Delta_i\cap (A_i+q_i)=\emptyset$. The intersection of $\Delta_i$'s is not covered by our translates. Now if $U_n=[0,1]\cap A_n$, the measures of $U_n$ are bounded from below.<|endoftext|> TITLE: What do physicists mean by a topological quantum gravity theory QUESTION [11 upvotes]: This is a jargon-like question. The fact that this is posted here rather in a physics forum indicates two things I know too little physics. An explanation with more mathematics flavors will be appreciated more.. Background I should first explain what a gravity theory is in my imagination: it seems to be a theory that governs the relation between the space and the matter. For example, Hilbert dealt with this problem by introducing a functional on the space of metrics, a consequence being Einstein's field equations that relates the curvature of the space-time and the mass-energy-momentum tensor. A quantum field theory seems (to me) to be a field theory where each field could possibly happen. In our case, the fields are the metrics, whose amplitudes can be computed by some "quantized" action weight Question This leads the confusing part: a topological theory (seems to) mean a theory that does not depend on the geometry (in particular, metric)! What does a topological quantum gravity theory mean then? REPLY [19 votes]: Physicists here. The input for a physical theory is always some topological space and some structure (such as a metric) that depends on the specific context. The dynamics are invariant under the isometries thereof. For example, the theory of Special Relativity deals with a manifold of the form $\mathbb R^n$, and with a (pseudo)metric $\operatorname{diag}(-1,+1,+1,\dots,+1)$. The dynamics are invariant under the so-called Poincaré transformations, i.e., the group of isometries of the metric above. We typically think of gravity as a manifestation of a non-trivial geometry, i.e., a generalization of Special Relativity where the manifold and the metric are no longer necessarily of the form above. There are two layers for a theory that includes gravity: Gravity as a background field, where the manifold and the metric are fixed, and the dynamics correspond to other degrees of freedom propagating in this manifold, and Gravity as a dynamical field, where the metric (and possibly the topological space itself) is determined by some dynamical equations. The system is to be determined by solving a self-consistent set of equations that include the metric, and the rest of degrees of freedom, each influencing each other. The former doesn't have a specific name as far as I know; we just call it "dynamics in curved spacetime". The latter is known as a "theory of gravity", the prototypical example being General Relativity and its extensions. Here the metric is determined by a set of PDEs. This system of equations is invariant under diffeomorphisms, as couldn't be otherwise. This is regarded as a generalization of the statement that the dynamics are to be invariant under the isometries of the metric, but now we allow any possible map, not only an isometry (because there is no fixed metric to begin with). This is also known as general covariance. The epithet "quantum" refers to the fact that the dynamics are, well, quantum. There is no perfectly convincing definition of what it means to be quantum (cf. this physics.SE post), but the general sentiment is that the state of the system is described by a vector in some Hilbert space (as opposed to a classical system, where the state is described by some point in some fibre bundle over your manifold). A "quantum theory of gravity" is, thus, a model of a system where we include gravity (non-trivial geometry/topology) in a quantum mechanical way. Whatever the model is, it is to be general covariant. A standard way to construct such a model proceeds as follows: First construct a quantum mechanical model that depends on a fixed background metric. We know how to do this, at least in a formal way (that is perfectly good for our purposes). Integrate the previous object with respect to all metrics, whatever that may mean. The latter step guarantees that the result is general covariant. Unfortunately, we don't really know how to do that in practice; any attempt has failed. Witten (https://projecteuclid.org/euclid.cmp/1104178138) proposed an alternative method to construct a quantum theory of gravity: instead of integrating over all metrics, set up a model that does not depend on a metric at all, from the very beginning. The dynamical variables are typically differential forms, and we only admit operations that do not require a metric (exterior differentiation). Models that are metric-independent are known as topological, because they only depend on the manifold as a topological space (typically with some extra structure, such as a framing or a spin structure, etc.). So, to sum up: a theory of gravity is a theory where the physical manifold is a dynamical variable itself. One can accomplish this by introducing a metric and allowing it to interact with (and feel the back-reaction from) other degrees of freedom. Another way is to not introduce a metric at all, and use degrees of freedom that can be defined without reference to a metric, such as differential forms. Making the theory "quantum" is still an open problem, and we don't really know what we want here: what does it even mean to have a quantum theory of gravity? what should we ask of such a model? Integrating over metrics is very problematic, while topological gravity is perfectly well-defined, even if very unrealistic from a physical point of view. Perhaps we should use it as a toy model to explore what are the properties of quantum theories of geometry/topology without the noise caused by other more realistic models.<|endoftext|> TITLE: Functoriality of filtered spectral sequences QUESTION [5 upvotes]: What is the appropriate functoriality statement of a filtered chain map between filtered spectral sequences? Suppose that we have two filtered chain complexes $C,C'$ and a filtered chain map $f\colon C\to C′$. This induces a map between the spectral sequences of the filtered chain complexes, and my question is what map does this induce on each page. In particular, what relation, if any, is there between these induced maps, and the map that $f$ induces on the total homology of the complexes $f_\ast\colon H(C)\to H(C′)$? Is there a way that the associated graded of $f$ is related to the above? Is there any prefered reference for these matters? REPLY [3 votes]: Let us use increasing filtrations $F^pC \subset F^{p+1}C\subset \ldots$. Then the homology of $C$ and $C'$ also inherit the structure of filtered modules by $F^pH(C) = im(H(F^p(C)\rightarrow C))$, or in words: a homology class in $H(C)$ has filtration degree $p$, if there is a representing chain in $F^pC$. With this structure $H(f):H(C)\rightarrow H(C')$ is filtration preserving; and the subquotients $F^{p}H_q(C)/F^{p-1}H_q(C)$ are exactly the entries of the $E_\infty$-page. The groups arising in the finite pages are the groups of $r$-almost cycles divided by the group of $r$-almost boundaries (that name is really misleading; they are really honest boundaries, but there are more boundaries then just them). To simplify, let us use the notation $Z^{p,r}_q$ for all the chains whose boundary lives $r$ steps further down in the filtration, i.e. $$Z^{p,r}_q=\{x\in F^{p}C_q\mid dx \in F^{p-r}C_{q-1}\}.$$ Then the groups appearing on the $r$-th page of the spectral sequence are $$\frac{Z^{p,r}_q}{Z^{p-1,r-1}_q+d(Z^{p+r-1,r-1}_{q+1})}.$$ Informally an element is represented by a chain whose boundary lives $r$ steps further down in the filtration and we divide out things one step further down in the filtration and boundaries of chains living at most $r-1$ steps further up in the filtration. Now the filtration preserving map induces maps $Z^{p,r}_q\rightarrow Z'^{p,r}_q$ and these induce map on the quotients above. This is the map that $f$ induces on finite pages. It sometimes differs from author to author (and application to application) which index convention is used. So the groups given above really arise somewhere, but where exactly depends on conventions. EDIT: I said I used increasing filtrations, but the signs in my definitions were for decreasing filtrations. I changed that.<|endoftext|> TITLE: What's an example of an $\infty$-topos not equivalent to sheaves on a Grothendieck site? QUESTION [45 upvotes]: My question is as in the title: Does anyone have an example (supposing one exists) of an $\infty$-topos which is known not to be equivalent to sheaves on a Grothendieck site? An $\infty$-topos is as in Higher Topos Theory (HTT) 6.1.0.4: an $\infty$-category which is an accessible left-exact localization of presheaves on a small $\infty$-category. A Grothendieck site is a small $\infty$-category $\mathcal{C}$ equipped with the $\infty$-categorical variant of the classical notion of a Grothendieck topology $\mathcal{T}$, as in HTT 6.2.2: a collection of sieves (subobjects $U\to j(C)$ of representable presheaves on $\mathcal{C}$) satisfying some axioms. Sheaves on $(\mathcal{C},\mathcal{T})$ are presheaves of $\infty$-groupoids on $\mathcal{C}$ which are local for the sieves in $\mathcal{T}$. Such form a full subcategory $\mathrm{Shv}(\mathcal{C},\mathcal{T})$ of the $\infty$-category of presheaves. Note: the question Examples of $(\infty,1)$-topoi that are not given as sheaves on a Grothendieck topology appears superficially to be equivalent to this one. In practice it is not exactly the same. As answers to that question show, many interesting $\infty$-topoi exist which can be described without reference to any Grothendieck site. But it is still conceivable that a suitable site exists. Also note: any $\infty$-topos $\mathcal{X}$ can be obtained as an accessible left-exact localization of some $\mathrm{Shv}(\mathcal{C},\mathcal{T})$ with respect to a suitable class of $\infty$-connected morphisms (HTT 6.2.2, 6.5.3.14), e.g., the class of hypercovers. However, this does not immediately preclude $\mathcal{X}$ being equivalent to $\mathrm{Shv}(\mathcal{C}',\mathcal{T}')$ for some other Grothendieck site $(\mathcal{C}',\mathcal{T}')$. Added remark. I asked this question because I had, for a long time, tacitly assumed that such examples were plentiful, until I thought about it and realized I had no basis for thinking that. As no answers have yet been given, and I'm not aware of any tools which would likely lead to a resolution one way or the other, it looks to me that this should be regarded as an open question. REPLY [8 votes]: Not an answer -- the question is very much open! But I think it's worth compiling together some of the observations made in the comments (this answer is community wiki; feel free to add, correct, change it): A fundamental difference between 1-topos theory and $\infty$-topos theory is that not every left exact localization of an $\infty$-topos $\mathcal E$ (even: of a presheaf $\infty$-topos) is localization with respect to a Grothendieck topology (a so-called topological localization). Rather, every left-exact localization $L$ of $\mathcal E$ factors as the topological localization at the Grothendieck topology $J$ generated by $L$, followed by a cotopological localization, so that $L\mathcal E$ lies somewhere between $J$-sheaves and the hypercompletion thereof. Thus it's tempting to think, as Charles reports doing for some time, that almost any sheaf $\infty$-topos $\mathcal E$ which is not hypercomplete should yield examples of non-sheaf-$\infty$-toposes by taking cotopological localizations of $\mathcal E$. But of course, such $\infty$-toposes might admit sheaf presentations by changing the site. Indeed, Charles gives an example of a sheaf $\infty$-topos which is not hypercomplete, but whose hypercompletion does turn out to be a sheaf $\infty$-topos (in fact a presheaf $\infty$-topos) over a different site. So it's unclear when the situation of (2) is likely to yield examples. So far, we don't seem to have any candidate property enjoyed by sheaf $\infty$-toposes but not by $\infty$-toposes which are not-obviously-sheaf-$\infty$-toposes. In 1-topos theory, we can do one better than stated in (1) above: every 1-topos is a sheaf topos over itself (or a suitable small subcategory thereof) via the canonical topology. This is known to fail for $\infty$-topoi. For example, let $C$ be a site such that representable sheaves are hypercomplete (e.g., a 1-site). If $Sh(C)$ is not hypercomplete (see below for examples), then the hypercompletion $Sh(C)^\mathrm{hyp}$ is not sheaves on itself with respect to the canonical topology. Here are some examples of non-hypercomplete sheaf $\infty$-toposes, whose hypercompletions might be candidates for non-sheaf $\infty$-toposes. Maybe folks could add more: $Sh(Q)$, where $Q$ is the Hilbert cube (HTT 6.5.4.8) $\varprojlim_n Sh(B\mathbb Z/p^n)$ (HTT 7.2.2.31) parameterized spectra, or more generally $n$-excisive functors Ironically, the classifying topos for $\infty$-connective objects is a sheaf topos.<|endoftext|> TITLE: Monomorphisms in $\mathcal{C}\!at_\infty$ QUESTION [5 upvotes]: I'm trying to work through what the $(-1)$-truncated morphisms are in $\def\Catinf{\mathcal{C}\!at_\infty} \Catinf$. BLUF: The correct characterization is that $F : C \to D$ is a (-1)-truncated map of $\infty$-categories iff, on hom spaces, $C(X,Y) \to D(FX, FY)$ is a (-1)-truncated map of spaces whose essential image contains the equivalences. I've seen it stated, e.g. at nLab, that these should be precisely the full-and-faithful functors. However, by 5.5.6.15 of Higher Topos Theory, a functor $F : C \to D$ is $(-1)$-truncated iff the diagonal $\Delta : C \to C \times_D C$ is an equivalence (i.e. $(-2)$-truncated). Consider the model given by simplicially enriched categories, and the special case that $C$ and $D$ are the ordinary categories ${\bf 1} + {\bf 1}$ and ${\bf 2}$ respectively. That is, $C$ is the discrete category with two elements, and $D$ adjoins a single morphism between them. Since all of the hom-spaces are either empty or the point, these are fibrant objects. Furthermore, $C \to D$ is a fibration on hom-sets, and has the equivalence lifting property. Thus, $F$ is a fibration of the model structure. Thus, the ordinary pullback computes the homotopy pullback, and it's easy to see that $C \to C \times_D C$ is, in fact, an isomorphism of simpicially enriched categories. But $C \to D$ is very much not a full functor. Instead, if I've worked through the details correctly, a functor $F : C \to D$ being $(-1)$-truncated is equivalent to the weaker condition $C(X,Y) \to D(FX, FY)$ is a $(-1)$-truncated map of spaces If $FX \simeq FY$, then $X \simeq Y$ This includes full-and-faithful functors, but it also includes more general examples. So I have conflicting information. Is nLab in error? Have I made an error? Have I made some other serious misunderstanding? REPLY [4 votes]: The statement at nLab is indeed incorrect, but your condition is also too strong. The first part should be replaced with a weaker condition that the map $C(X,Y) \to D(FX,FY)$ is a $(-1)$-truncated map of spaces.<|endoftext|> TITLE: Commutative algebras with modules of small complexity QUESTION [10 upvotes]: Let $A$ be a finite dimensional commutative algebra. We can assume that it is local. Question: Which such $A$ have the property that every finite dimensional $A$-module has complexity at most 1? (This should be equivalent to the simple module having complexity equal to one or equivalently bounded Betti numbers) Note that a module has complexity at most one if and only if the terms in a minimal projective resolution have bounded dimensions. Examples include $A=K[x]/(x^n)$. Do you know other examples? REPLY [11 votes]: There are no other examples. This property is equivalent to $A$ being a hypersurface (see Avramov's note "Infinite Free Resolutions"). By Cohen Structure Theorem, an Artinian local hypersurface (which is automatically complete) that contains a field must be isomorphic to $k[[x]]/(f)$, which is equal to $k[[x]]/(x^n) = k[x]/(x^n)$ with $n$ being the smallest power of $x$ in $f$.<|endoftext|> TITLE: Counterpart of cyclotomic polynomials for elliptic divisibility sequences QUESTION [18 upvotes]: Let $(U_n)_{n \in \mathbb{N}}$ be a Lucas sequences given by $$U_0 = 0,\quad U_1 = 1,\quad U_n = P U_{n - 1} - Q U_{n-2},$$ where $P,Q$ are integers with $P^2 - 4Q \neq 0$. It is well known that the following product formula holds $$U_n = \prod_{d \mid n} \Phi_d(\alpha, \beta) ,$$ where $\Phi_d(\alpha, \beta) \in \mathbb{Z}$, $$\Phi_k(X,Y) := \prod_{\zeta \,\text{ $k$th primitive root of $1$}}(X - \zeta Y)$$ denotes the $k$th homogenous cyclotomic polynomial, and $\alpha,\beta$ are the two roots of $X^2 - PX + Q = 0$. Let $(D_n)_{n \in \mathbb{N}}$ be an elliptic divisibility sequence, that is, there exists an elliptic curve $E$ over the rationals with a point $P$ of infinite order, and $D_n$ is determined by $$nP = \left(\frac{A_n}{D_n^2}, \frac{B_n}{D_n^3}\right) ,$$ where $A_n, B_n$ are integers with $\gcd(A_n, D_n) = \gcd(B_n, D_n) = 1$. As far as I understand, elliptic divisibility sequences have many properties in common with Lucas sequences. For example, like Lucas sequences, they are strong divisibility sequences and (under certain condition) satisfy a primitive divisor theorem. My question is if there exists a counterpart of cyclotomic polynomials for elliptic divisibility sequences, that is, some quantities $\Psi_d \in \mathbb{Z}$ such that a product formula $$D_n = \prod_{d \mid n} \Psi_d ,$$ holds; and, if so, what is known about $\Psi_d$. REPLY [16 votes]: The counterpart of the cyclotomic polynomials are elliptic division polynomials, which can be defined recursively by a non-linear recursion (usually presented as a pair of recursions, one for odd indices and one for even). They are classical, dating back to the 19th century, and you can find them in many sources, including for example Exercise 3.7 of my Arithmetic of Elliptic Curves. However, if you take an point $P$ on $E(\mathbb Q)$ and write $\,x(nP)=A_n/D_n^2$, then you generally don't quite get $D_1^{(n^2-1)/2}\Psi_n(x(P))$. The issue is that the generic numerator and denominator of $\,x(nP)\,$ can have some cancellation at primes of bad reduction. So assuming that you've taken a minimal Weierstrass equation you should get something like $$ D_n = \pm D_1^{(n^2-1)/2}\Psi_n(x(P)) \cdot\! \prod_{p\mid \Delta_E} p^{k_{n,p}}, $$ Okay, here $\Psi_n$ is the analogue of $x^n-1$, so now instead of taking $\Psi_n$, which contains the $x$-coordinates of all of the points of order $n$, you can just use only the points whose order is exactly $n$, just as for primitive $n$th roots of unity. Let's call that $\Psi_n^*$, and then $$ \Psi_n = \prod_{d\mid n} \Psi_n^*. $$ Evaluating this at $P$ and multiplying by $D_1$ to the appropriate power will give a decomposition of the sort you want for $D_n$ except that the primes dividing $\Delta_E$ may not work quite right. There's likely some way to adjust them so that everything works, but I don't know a reference offhand. Addendum: Let $$ E[n] = \{P\in E\!: nP = 0\} \quad\text{and}\quad E[n]^* = \{P\in E\!: \text{$nP=0$ and $mP\ne0$ for $m TITLE: Covering the disk with a family of infinite total measure - the convex sequel QUESTION [12 upvotes]: Let $(U_n)_n$ be an arbitrary sequence of open convex subsets of the unit disk $D(0,1)\subseteq \mathbb{R}^2$ s.t. $\sum_{n=0}^\infty \lambda(U_n)=\infty$ (where $\lambda$ is the Lebesgue measure). Does there exist a sequence $(q_n)_n$ in $\mathbb{R}^2$ s.t. $D(0,1) \subseteq \bigcup_{n=0}^\infty (q_n+U_n)$? With the notation $q_n+U_n$, I mean $$q_n+U_n:=\{x\in \mathbb{R}^2|x-q_n\in U_n\}$$ This question is very similar to this one, but I was encouraged in the comments over there to ask away nontheless. REPLY [3 votes]: The planar case is rather simple but I'm still struggling with dimension 3 and higher, so we'd better keep this thread open at least for a while. Lemma 1: Suppose we have finitely many infinite strips of total width at least $8$. Then we can move them (without rotations) so that they cover a disk of radius $1/4$ (the exact numbers do not matter; all that is essential is that they are two absolute constants). Proof: We can choose a subset of strips with total width at least $1$ so that all directions lie in an angle of size $\pi/4$. WLOG, that angle is $0\le\theta\le \frac \pi 4$, where the angles are counted from the vertical up direction counterclockwise. Arrange the strips so that their directions satisfy $0\le \theta_1\le\theta_2\le\dots\le\theta_n\le\frac\pi 4$. Let $w_1,w_2,\dots,w_n>0$ be their widths. We will think of our strips as going "vertically", so each strip has the "left" and the "right" edge. Now position the $j$-th strip so that its right edge passes through the point $(w_1+\dots+w_j,0)$ on the horizontal axis. By induction, we can see that then the first $j$ strips cover the triangle bounded by the horizontal axis, the vertical axis, and the line passing through $(w_1+\dots+w_j,0)$ in direction $\theta_j$ (note that this triangle doesn't need to depend on $j$ in a monotone way!). Thus, in the end we'll cover the triangle bounded by the axes and the line through $(w_1+\dots+w_n,0)$ in direction $\frac \pi 4$, which, since $w_1+\dots+w_n\ge 1$, contains a disk of the required size. WLOG all our convex sets $U_j$ are rectangles with sides $a_j\le b_j$; moreover, we can think that $b_j$ is always some negative power of $2$. Lemma 2: Assume that we have some finite collection of $U_j$ with common $b$ and various $a$'s and orientations. If $\sum_j a_j\ge 8b$, then we can use them to cover a disk of radius $\frac b4$. Proof: This is just Lemma 1 in disguise: scaling by a factor of $b$ does not matter and the piece of any strip of width $a_j$ in the disk so small is always contained in a rectangle $a_j\times b$ in the same direction. Now we just take a fixed $b$ and start dividing rectangles into finite families with sum of $a_j$ between $8b$ and $9b$. Each family can be used to cover a disk of area comparable with the total area of the family and the useless remainders will have total area of $8b^2$ or less, so the sum of remainder areas will be finite. This reduces the problem to the case when all convex sets are disks, which can be handled similarly to the classical Vitali lemma: just consider 2 times smaller disks and throw them into the unit disk $D(0,1)$ (at least their center has to be thrown in that disk), without intersection with previously thrown disks and throwing them in decreasing order of radius as long as you can. This process must terminate after a finite number $N$ of steps since at every step the yet uncovered area decreases with the area of the disk currently being thrown and those areas sum to infinity. For later purposes, recollect $0 TITLE: Pinwheel Tilings and C* algebras, K-theory QUESTION [7 upvotes]: I was reading that spaces of tilings can be related to C*-algebras and K-theory. Here is an example of the pinwheel tiling. [1] They construct a space called $\mathcal{A}\mathbb{T}_{pin}$ and show that the $K$-groups are inductive limits of abelian groups $$ \mathbb{Z} \oplus \mathbb{Z} \stackrel{\left[ \begin{array}{cc} 2 & 3 \\ 3 & 2 \end{array} \right]}{\longrightarrow} \mathbb{Z} \oplus \mathbb{Z} \stackrel{\left[ \begin{array}{cc} 2 & 3 \\ 3 & 2 \end{array} \right]}{\longrightarrow} \mathbb{Z} \oplus \mathbb{Z} \stackrel{\left[ \begin{array}{cc} 2 & 3 \\ 3 & 2 \end{array} \right]}{\longrightarrow} \mathbb{Z} \oplus \mathbb{Z} \stackrel{\left[ \begin{array}{cc} 2 & 3 \\ 3 & 2 \end{array} \right]}{\longrightarrow} \dots $$ The C* algebra $\mathcal{A}\mathbb{T}_{pin}$ is itself an inductive limit of some kind, based on a symbolic coding of the dynamics of the pinwheel tiling itself. The spaces are spanned by step functions checking that they agree on a patch, $$ e_P(T, T') = \left\{ \begin{array}{cc} 1 & T \text{ and }T'\text{ agree on some patch} \\ 0 & \text{otherwise} \end{array} \right. $$ Some translate of $T$ and $T'$ agree on a patch $P$ (up to translation and rotation). As well as a character of sorts: $$ z(T, T') = \left\{ \begin{array}{cc} e^{\angle T(0)i} & T =T'\\ 0 & \text{otherwise} \end{array} \right. $$ The tilings look approachable, these algebras look very complicated. Could anyone describe what these function spaces are like and why we need C* algebras to describe such elementary geometric shapes ? REPLY [4 votes]: The object associated to the tiling is, usually, a groupoid $C^*$-algebra: so the correct questions should be: why a groupoid? why the $C^*$-algebra of a groupoid. What I've understood of this approach is related to Penrose tiling, I doubt things will be much different for other tilings. I will speak about Penrose. The point is you should not think about one tiling but a whole family of tilings. There are infinitely many Penrose tilings, that can be described by certain infinite sequences, each sequence describing how you build the tiling starting from one tile. The space of all Penrose tilings is called the Penrose universe. A natural thing to do, to study such space, is to put a topology on it. You put a topology through a sort of metric which counts (the inverse of) how long is a sequence on which two tilings coincide. Unfortunately (or fortunately) any two P. tiling coincide at infinitely many spots; the resulting topology on the Penrose universe is highly non separating: any point is open and dense. If you look at the quotient space it consists of only one point. Continuous functions on this space are constants. Not much to work with. How to get some information out of this space? The idea is to associate to it a groupoid. Why a groupoid? Because a groupoid is a very natural and useful object each time you have "partially defined" equivalences. Given two Penrose tiling $T_1$ and $T_2$ there is a compact subset $K_1\subseteq T_1$ with an isometry to a compact subset of $K_2\subseteq T_2$. Therefore you have an "arrow" between $T_1$ and $T_2$ that can be composed with another arrow from $T_2$ to a third Penrose tiling $T_3$ only if its range is contained in the source of the second arrow. This gives you a groupoid structure on the set of Penrose tiling that turns out to be a topological groupoid. It looks like what you're having on pinwheel tilings is exactly the same (equivalences of finite patches). Why this object describes well the Penrose universe? Because it contains details on all the tilings and details on all the partial equivalences between them. The space of orbits of this groupoid is, in fact, your highly singular Penrose universe and this topological groupoid is some kind of desingularized version. Isotropy subgroups are self equivalences of a tiling. How do you study a topological groupoid of this kind? This is where $C^*$-algebra theory comes in. Associating a $C^*$-algebra to a topological groupoid gives you tools to study your groupoid. For example $K$-theory which is one of the most important invariants. Usually data from $K$-theory can be useful to recover the frequency of occurrences of certain sub patches in the tiling. There are a number of points where my description is not completely precise, but that's the idea as I've understood it. Ready to correct after comments by real experts.<|endoftext|> TITLE: Continuous functions taking uncountably many values countably often QUESTION [23 upvotes]: Let $f$ be a continuous function defined on the closed interval $[0,1]$. Clearly $f$ is bounded and attains its bounds. Then my question is how often can $f$ take a value in its range countably many times? Formally let $Z(f,p)=\{x\in \text{Image}(f):|f^{-1}(x)| = p\}$. I would like to know large $Z(f,\aleph_0)$ can be? I conjecture that we can find $f$ with $|Z(f,\aleph_0)|=2^{\aleph_0}$. Note that I can find f with $Z(f,2^{\aleph_0}) = \aleph_0$ which I think makes the conjecture plausible: We know that the zero set of a continuous function on $[0,1]$ can be any closed subset as we can take $f$ to be the distance to the closed set. If we let the closed set be the cantor ternary set $C$ on $[0,1]$ we can create a function on $[0,1]$ with uncountably infinite zero set. By taking suitable scalings of the function on intervals $[\frac{1}{2^{2n+1}}, \frac{1}{2^{2n}}]$ with ramps in between we can find a function $f$ s.t. $Z(f)$ is countably infinite. Hence we have $Z(f,2^{\aleph_0}) = \aleph_0$. REPLY [10 votes]: Just to complement the answers that were already given, one can also have $|Z(f,2^{\aleph_0})|= 2^{\aleph_0}$: just take for $f$ a typical sample of Brownian motion. More precisely, almost every realisation of Brownian motion has the property that a set of $c$s of positive Lebesgue measure has a preimage of positive Hausdorff dimension. This can be shown by combining the fact that the zero level set of BM has positive Hausdorff dimension with Fubini and the strong Markov property.<|endoftext|> TITLE: Algorithm for comparing $x + y \cdot \log(z)$ for $x,y,z$ rational? QUESTION [6 upvotes]: I'm playing with some methods of comparing two real numbers of the form $x + y \log(z)$, where $x,y,z$ are rational numbers and $z$ is positive. There are various estimates on irrationality measures of logarithms of rational numbers (e.g. Wu, below). But I wonder what the best algorithm is for comparing two such real numbers. And how does the time depend on the height of $x_1,x_2$, $y_1,y_2$, and $z_1,z_2$, in the worst case, if one wishes to know whether $$x_1 + y_1 \log(z_1) < x_2 + y_2 \log(z_2)?$$ Is something proven or conjectured here? Wu, Qiang, On the linear independence measure of logarithms of rational numbers, Math. Comput. 72, No. 242, 901-911 (2003). ZBL1099.11037. REPLY [5 votes]: We can get an algorithm by constructively recasting a standard proof of the transcendence of $e$, and simplifying it since we need only that $e$ is not a root of a rational number. The result may not be the best algorithm, but it makes the time-analysis easy. Algorithm To start, we assume without loss of generality that $x_1, x_2, y_1, y_2$ are all integers. Let $t=x_2-x_1$ and $u=z_1^{y_1}/z_2^{y_2}$, so the question is equivalent to deciding whether $e^t>u$. Let $u=a/b$, with $a$ and $b$ integers. Choose an odd $p$ with $p>3t^2$ and $2^p>b 3^t t$. (We do not need $p$ to be prime, though it was in the source.) Then let $$f(p,t,x)=\frac{x^p(x-t)^p}{p!\ e^x}$$ $$M=\int_0^\infty f(p,t,x) dx$$ $$M_t=e^t \int_t^\infty f(p,t,x) dx$$ $$\epsilon=e^t \int_0^t f(p,t,x) dx$$ Now we are looking to determine whether \begin{align} e^t &> u \\ (M_t + \epsilon)/M &> a/b \\ bM_t - aM &> -b \epsilon \end{align} where all three expressions are equivalent. On the left hand side, we can evaluate the integral for $M$ using the identity $\int_0^\infty x^n e^{-x} dx = n!$, and conclude that $M$ is an integer. Using the change of variables $y=x-t$, we conclude that $M_t$ is also an integer, and so is $bM_t - aM$. On the right hand side, $p>3t^2$ makes $|f(p,t,x)|<1/2^p$, so the second condition on $p$ gives $0<-b\epsilon<1$. Thus the inequality holds iff $bM_t-aM>0$, and the algorithm is just to compare those as integers. Time Analysis The significant time in this algorithm is in calculating $M$ and $M_t$. To calculate $M$, we added $p+1$ summands, each of which is bounded by $$\frac{(2p)!}{p!}\binom{p}{p/2}t^p < (3pt)^p$$ The time for the algorithm is roughly the time to write out all those summands, which is $O(p^2\log(pt))$ or $$O(t^4 \log t +(\log b)^2\log\log b)$$<|endoftext|> TITLE: Can you give an example of two projective morphisms of schemes whose composition is not projective? QUESTION [20 upvotes]: Grothendieck and Dieudonné prove in $EGA_{II}$ (Proposition 5.5.5.(ii), page 105) that if $f:X\to Y, g:Y\to Z$ are projective morphisms of schemes and if $Z$ is separated and quasi-compact, or if the underlying topological space $\operatorname {sp}(Z)$ is noetherian, then the composition $g\circ f:X\to Z$ is also projective. Question: Is there an example where $Z$ satisfies neither of the two sufficient conditions above and where $g\circ f:X\to Z$ is not projective? Edit I have corrected my initially wrongly stated sufficient conditions on $Z$, caused by the change in terminology: Prescheme in EGA=Scheme nowadays Scheme in EGA=Separated scheme nowadays. Thanks a lot to R. Van Dobben de Bruyn for making me aware of my initial confusion. REPLY [17 votes]: Here is a locally Noetherian separated counterexample. I also give some motivation for this construction afterwards. Definition. Let $Z$ be an infinite chain of affine lines: $Z = Z_1 \amalg_{p_1} Z_2 \amalg_{p_2} \ldots$, where $Z_i \cong \mathbf A^1_{\mathbf C}$ and $p_i$ is the point $1$ in $Z_i$ and the point $0$ in $Z_{i+1}$. Let $Y = \mathbf P^2_Z$, which is clearly projective over $Z$. Write $Y_i = Z_i \times_Z Y$ for the irreducible component of $Y$ above $Z_i$, with structure map $g_i \colon \mathbf A^1 \times \mathbf P^2 \cong Y_i \to Z_i \cong \mathbf A^1$. Finally, let $X$ be obtained by blowing up $Y_i$ at $i$ collinear points $W_i \subseteq \mathbf A^1 \times \mathbf P^2$ in the fibre $g_i^{-1}(2)$ for all $i$ (the only thing that matters is that we choose a point $2$ that is not one of the glueing points $0$ and $1$). Then $f \colon X \to Y$ is projective because $X \subseteq \mathbf P(\mathcal I_W)$ is a closed immersion, where $W = \bigcup_i W_i \subseteq Y$ is the closed subvariety we're blowing up and $\mathcal I_W$ its ideal sheaf (which is of finite type because that's a local condition). Notation. Write $h = g \circ f \colon X \to Z$, and once again write $X_i = Z_i \times_Z X$ for the irreducible component of $X$ above $Z_i$. Denote the maps $X_i \to Y_i$ and $X_i \to Z_i$ by $g_i$ and $h_i$ respectively. Proposition. The map $X \to Z$ is not projective. We will show that there is no $h$-ample line bundle on $X$ by computing all line bundles on $X$. Note that for any open $U \subseteq Z_i \cong \mathbf A^1$, we have $\operatorname{Pic}(U \times \mathbf P^2) \cong \operatorname{Pic}(\mathbf P^2)$ by pullback. Lemma. The projection $\pi \colon Y = X \times \mathbf P^2 \to \mathbf P^2$ induces an isomorphism $$\pi^* \colon \operatorname{Pic}(\mathbf P^2) \stackrel\sim\to \operatorname{Pic}(Y).$$ Proof. Indeed, $\pi^*$ is injective since $\pi$ has a section. Every $\mathscr L$ on $Y$ restricts to some $\mathcal O_{Y_i}(n_i)$ on each $Y_i$, which has to be the same $n_i = n$ for all $i$ by restricting to $Y_i \cap Y_{i+1} = p_i \times \mathbf P^2$. We can choose identifications $\mathscr L|_{Y_i} \cong \mathcal O_{Y_i}(n)$ compatibly for all $i$ starting with $i = 1$ and moving through the chain. At any stage we might have to modify the chosen identification on $Y_{i+1}$ by an element of $\operatorname{Aut}(\mathscr L|_{p_i \times \mathbf P^2}) = \mathbf C^\times$, which is harmless because there are no loops. $\square$ By the same reasoning, any line bundle on $X$ is pulled back from $\mathbf P^2$ away from $\bigcup_i h_i^{-1}(2)$, so it suffices to glue line bundles on the open covering consisting of the locus $U = X \setminus \bigcup_i h_i^{-1}(2)$ where $f \colon X \to Y$ is an isomorphism, and the opens $$U_i = X_i \setminus h_i^{-1}\Big(\{p_{i-1},p_i\}\Big)\cong \operatorname{Bl}_{W_i} \bigg(\left(\mathbf A^1 \setminus \{0,1\} \right) \times \mathbf P^2\bigg)$$ consisting of $X_i$ minus its intersections with $X_{i-1}$ and $X_{i+1}$. Corollary. We have $$\operatorname{Pic}(X) \cong \operatorname{Pic}(\mathbf P^2) \times \prod_{i=1}^{\infty} \prod_{r = 1}^i \mathbf Z E_{i,r}.$$ Proof. We have $\operatorname{Pic}(U_i) \cong \operatorname{Pic}(\mathbf P^2) \times \prod_{r = 1}^i \mathbf Z E_{i,r}$ where the $E_{i,r}$ are the exceptional divisors above the $i$ points of $W_i$. The result now from the description above by glueing on $U$ and the $U_i$, since there is no compatibility condition between the coefficients in $E_{i,r}$ and $E_{i',r'}$ if $i \neq i'$. $\square$ Proof of Proposition. Now suppose $\mathscr L = (n,n_{i,r}) \in \operatorname{Pic}(X)$ is $h$-ample. In particular this implies that $\mathscr L|_{h^{-1}(z)}$ is ample for each $z \in Z$. Apply this to $z = 2 \in Z_i$ to get $nH + \sum_{r = 1}^i n_{i,r}E_{i,r}$ ample on $h_i^{-1}(2)$. But the fibre $h_i^{-1}(2)$ contains the strict transform $\operatorname{Bl}_{W_i}(\mathbf P^2)$ of the fibre $g_i^{-1}(2) \cong \mathbf P^2$ as a closed subvariety, and the restriction of $E_{i,r}$ to $\operatorname{Bl}_{W_i}(\mathbf P^2)$ is the $r^{\text{th}}$ exceptional divisor $E_r$ of $\operatorname{Bl}_{W_i}(\mathbf P^2) \to \mathbf P^2$. So $nH + \sum_{r = 1}^i n_{i,r}E_r$ has to be ample on $\operatorname{Bl}_{W_i}(\mathbf P^2)$. Since $E_r \cdot E_{r'} = -\delta_{r,r'}$ and $E_r \cdot H = 0$, we must have $n_{i,r} < 0$. If $\ell$ is the strict transform of the line through $W_i$ (which are collinear by assumption), then $\ell \cdot E_r = 1$ and $\ell \cdot H = 1$. Hence, $(nH + \sum n_{i,r}E_r)\cdot \ell > 0$ forces $n > -\sum n_{i,r} \geq i$. Since $i$ is arbitrary, this is impossible. $\square$ Motivation. The idea behind the construction is as follows. As I commented above, if $Z$ is affine, $\mathscr L$ is $f$-ample on $X$ and $\mathscr M$ is $g$-ample on $Y$, there is some $a$ such that $\mathscr L \otimes f^* \mathscr M^{\otimes a}$ is $(g \circ f)$-ample by Tag 0C4K. For $Z$ quasi-compact, you can choose this $a$ uniformly over all affines, but in general you might need a bigger and bigger $a$. The first thing you try is $Z$ an infinite disjoint union of points, where you need a bigger and bigger $a$ on each component (for example when you blow up more and more collinear points in $\mathbf P^2$). But this doesn't work because in a disjoint union you have too much freedom to choose a completely different line bundle on $X$ that does the job ("you can choose a different $a_i$ per component"). However, if you put yourself in a situation where you can compute $\operatorname{Pic}(X)$, and there is a reason why all the $a_i$ have to be the same for any line bundle on $X$, then you can make this into an argument leading to a contradiction.<|endoftext|> TITLE: Monoids of endomorphisms of nonisomorphic groups QUESTION [17 upvotes]: Can monoids of endomorphisms of nonisomorphic groups be isomorphic ? REPLY [15 votes]: It is proved in $[$1$]$ that the tetrahedral group $A_4$ of order $12$ and the binary tetrahedral group of order $24$ have isomorphic endomorphism monoids. So this gives a finite example. It is also the smallest order example. Moreover, no other finite groups have isomorphic endomorphism monoid to the endomorphism monoid of these two groups without being isomorphic to one of them. $[$1$]$ P. Puusemp, Groups of order 24 and their endomorphism semigroups, J. Math. Sci. (2007) 144(2) 3980–3992 (link at Springerlink).<|endoftext|> TITLE: Derivations of universal enveloping algebra of Lie algebras QUESTION [6 upvotes]: We know a lot about derivations of Lie algebra. However, for the universal enveloping algebra of Lie algebra, we have few references about it. My question: describing the derivations of enveloping algebras is very hard or easy? And are there any applications about it? For example, for the one-dimensional Lie algebra, the derivations of the universal enveloping algebra form the one-sided Witt algebra. REPLY [5 votes]: Associated to any associative algebra $A$ is the Hochschild cochain complex \begin{align*} HH^n(A) &= \operatorname{Hom}(A^{\otimes n},A),\\ \mathrm d f(a_0,\dots,a_n) &= a_0f(a_1,\dots,a_n) + (-1)^{n+1} f(a_0,\dots,a_{n-1})a_n + \sum_{i=1}^n (-1)^i f(a_0,\dots,a_{i-1}a_i,\dots,a_n) \end{align*} In particular, $\operatorname{ker} d^1\subset \operatorname{Hom}(A,A)$ are exactly the derivations, and $\operatorname{im} d^0\subset \operatorname{Hom}(A,A)$ are the inner derivations. Thus there is an exact sequence $$ 0\to Z(A)= HH^0(A)\to A\to \operatorname{Der}(A)\to HH^1(A)\to 0 $$ For $A = \operatorname{Sym}(V)$, we have $HH^i(A)\cong \operatorname{Hom}(\Lambda^i V,\operatorname{Sym}(V))$ (the Hochschild-Kostant-Rosenberg isomorphism). In this case this exact sequence encodes that a derivation is uniquely and arbitrarily defined by what it does on generators. This cochain complex carries an obvious commutative multiplication and a $1$-shifted Lie bracket which uniquely extends the commutator of derivations and their action on algebra elements as a biderivation. For $A = U\mathfrak g$, the PBW filtration defines a filtration on the Hochschild complex, and the associated graded is $HH^*(\operatorname{Sym}(\mathfrak g))$. Thus you get a spectral sequence with $E^1$-page $\operatorname{Hom}(\Lambda^i \mathfrak g,\operatorname{Sym}(\mathfrak g))$, where the differential is given by taking the Lie bracket $[\Pi,-]$ with the element $\Pi\in\operatorname{Hom}(\Lambda^2\mathfrak g,\mathfrak g)$ defining the Lie structure (the fact that this squares to zero is equivalent to the Jacobi identity). In fact, this complex can be identified with the Chevalley-Eilenberg complex calculating the Lie algebra cohomology of $\mathfrak g$ with values in $\operatorname{Sym}(\mathfrak g)$. A quite deep theorem by Kontsevich implies that this spectral sequence degenerates at the $E^2$-page (compare Pevzner, Michaël, and Ch Torossian. "Isomorphisme de Duflo et la cohomologie tangentielle." Journal of geometry and Physics 51.4 (2004): 486-505.). This implies the following for the two groups we are interested in: $Z(U\mathfrak g) = HH^0(U\mathfrak g)\cong (\operatorname{Sym}\mathfrak g)^{\mathfrak g}$ (Duflo isomorphism) $HH^1(U\mathfrak g)\cong H^1(\mathfrak g,\operatorname{Sym}\mathfrak g)$ (recall that Lie derivations are given by $H^1(\mathfrak g,\mathfrak g)$) Theoretically you can also identify the maps in the above exact sequence explicitly by chasing through the proof of the formality theorem.<|endoftext|> TITLE: BISH: If a function is pointwise positive, is its infimum positive? QUESTION [5 upvotes]: Let $f:[0,1] \to \mathbb R$ be a uniformly continuous function such that each value of $f(x)$ is greater than zero. Is its infimum greater than zero in BISH? I believe that it is indeed the case if one assumes the Fan Theorem. But independent of it, I'm not sure. [edit] Note: It's possible to get around this problem in practice by interpreting $f > 0$ to mean that there exists a constant $c$ such that $f(x) > c > 0$ for all $x$. This is an example of a pseudo-order. The fact that the infimum is greater than zero is then a tautology. REPLY [10 votes]: In BISH the follwoing two statements are equivalent: (i) If $f:[0,1] \to \{y\in\mathbb R\, | \,y>0\}$ is uniformly continuous, then there is $n\in\mathbb N$ such that $\forall x \in [0,1]\ [f(x)>\frac{1}{n}]$ (ii) The Fan Theorem FT This was already proved in Julian, W.H., and Richman, F., 1984, “A uniformly continuous function on [0, 1] that is everywhere different from its infimum”, Pacific Journal of Mathematics,111: 333–340. A simpler proof, and a lot more consequences, are given in my paper On the foundations of constructive mathematics --- especially in relation to the theory of continuous functions<|endoftext|> TITLE: Rational approximation of an integer combination of two irrationals QUESTION [6 upvotes]: Let $x$ be an irrational number, and $\beta$ strictly larger than its irrationality index, which means that for some $C>0$, for all $n\in \mathbb{Z}^*$, $$d(nx,\mathbb{Z})>C n^{-\beta}.$$ It is known that for a.e. irrational number $x$, the irrationality index is $1$. It is even known that some numbers satisf the above for $\beta=1$ (for instance, $x=\sqrt{2}$). By arguments from measure theory, I have been able to prove that if $a_n$ satisfies $$\sum_{n=1}^{\infty}na_n<\infty,$$ almost every couple $(x,y)$ of $\mathbb{R}^2$ satisfies for some $C>0$ $$d(nx+my,\mathbb{Z})>Ca_{|n|+|m|},n,m\in \mathbb{Z}.$$ Ideally, I would like, as for a single number $x$, find irrational numbers $(x,y)$ such that this holds in the limiting case $a_n=n^{-2}$. Has anyone an idea? Or has anyone a useful reference for such things? REPLY [9 votes]: Yes, such $(x,y)$ exist; for example, $x = \root 3 \of 2$ and $y = x^2$. For $l,m,n \in \bf Z$, define $$ N(l,m,n) := l^3 + 2m^3 + 4n^3 - 6lmn \in {\bf Z}; $$ this is the algebraic norm $$ (l + mx + nx^2) (l + m\rho x + n \bar\rho x^2) (l + m\bar\rho x + n \rho x^2) $$ of $l + mx + nx^2$, where $\rho$ is the cube root of unity $e^{2\pi i/3} = (-1 + \sqrt{-3})/2$. But if $(m,n) \neq (0,0)$ then $l + mx + nx^2 \neq 0$, so $\left|N(l,m,n)\right| \geq 1$ and $\left|l + mx + nx^2\right| \gg (|l|+|m|+|n|)^{-2}$. Taking for $l$ the integer nearest to $-(mx+nx^2)$ we deduce that $d(mx+nx^2,{\bf Z}) \gg (|m|+|n|)^{-2}$, as claimed. The same argument (which generalizes the familiar one for $\sqrt 2$) shows that in general if $x$ is an algebraic number of degree $D$ then $$ d\Bigl(\sum_{j=1}^{D-1} n_j x^j, {\bf Z}\Bigr) \gg \left( \sum_{j=1}^{D-1} \left| n_j \right| \right)^{1-D} $$ for $n_1,\ldots,n_{D-1} \in \bf Z$ not all zero. This is best possible up to the value of the implicit constant, because Dirichlet's celebrated "pigeonhole" argument shows that conversely for any $N$ one can find integers $n_1,\ldots,n_{D-1} \in [-N,N]$, not all zero, such that $d(\sum_{j=1}^{D-1} n_j x^j, {\bf Z}) \ll N^{1-D}$.<|endoftext|> TITLE: Free abelian subgroups of $\mathrm{SL}_n(\mathbb{Z})$ QUESTION [5 upvotes]: Does anybody know what is the biggest $r$ such that $\mathbb{Z}^r$ is isomorphic to a subgroup of $\mathrm{SL}_n(\mathbb{Z})$? It cannot be bigger that the virtual cohomological dimension of $\mathrm{SL}_n(\mathbb{Z})$, which is $\frac{n(n-1)}{2}$, since the cohomological dimension respects inclusions. But I suspect it must be smaller. REPLY [7 votes]: The answer is $\lfloor n^2/4\rfloor$, namely $m^2=n^2/4$ for even $n=2m$ and $m(m+1)=(n^2-1)/4$ for odd $n=2m+1$. Indeed, one has a free abelian subgroup of rank $\lfloor n^2/4\rfloor$, consisting of upper unipotent matrices with two blocks of size $\lfloor n/2\rfloor=m$ and $\lceil n/2\rceil\in\{m,m+1\}$ in $\mathrm{GL}_n(\mathbf{Z})$: $$\begin{pmatrix} I_m& *\\0 & I_{m}\end{pmatrix}\text{ or }\begin{pmatrix} I_m& *\\0 & I_{m+1}\end{pmatrix}.$$ Next, this is the upper bound. Indeed, it is known that the largest dimension of an abelian subalgebra of matrices is $\lfloor n^2/4\rfloor+1$ (see this MO answer, attributing to Schur). Hence the maximal dimension of an abelian subgroup of $\mathrm{GL}_n$ is $\lfloor n^2/4\rfloor+1$. Given that such a subgroup contains the scalar matrices, one deduces that the largest dimension of an abelian subgroup of $\mathrm{SL}_n$ is $\lfloor n^2/4\rfloor$. Now consider a free abelian subgroup of finite rank $r$ in $\mathrm{SL}_n(\mathbf{Z})$. Then its Zariski closure in $\mathrm{SL}_n(\mathbf{R})$ has dimension $d\le \lfloor n^2/4\rfloor$, and has a discrete subgroup isomorphic to $\mathbf{Z}^r$. Hence, the quotient of its unit component by its maximal compact subgroup has dimension $\delta\le d$ so is isomorphic to $\mathbf{R}^\delta$, and still has a discrete subgroup isomorphic to $\mathbf{Z}^r$. So $r\le\delta\le d\le\lfloor n^2/4\rfloor$.<|endoftext|> TITLE: If the sum of two independent random variables is discrete uniform on $\{a, \dots,a + n\}$, what do we know about $X$ and $Y$? QUESTION [10 upvotes]: Basically I want to know whether the sum being discrete uniform effectively forces the two component random variables to also be uniform on their respective domains. To be a bit more precise: Suppose we know $X$ and $Y$ are independent and $$ X+Y \sim UNIF({1, \dots , n})$$ Does this necessarily imply that both $X$ and $Y$ are discrete uniform as well? REPLY [6 votes]: As Lutz Mattner pointed out in his comment to another question, an affirmative answer is given in: Krasner and Ranulac (1937), Sur une propriété des polynomes de la division du cercle, C.R. Acad. Sci. Paris 204, 397–399 (which, unfortunately, does not seem to be available online, except for the Russian version due to D. Raikov). Connection to the other question was observed by @user44191. REPLY [5 votes]: 1. Proof of the claim The following lemma shows that $X$ and $Y$ may also be regarded as integer-valued random variables in OP's scenario. Lemma. Assume that $X$ and $Y$ are independent random variables. Suppose that there exists a finite set $S\subset\mathbb{R}$ satisfying $$ \mathbb{P}(X+Y \in S) = 1 \qquad \text{and} \qquad \mathbb{P}(X+Y = s) > 0, \quad \forall s \in S. $$ Then there exist $x_0, y_0 \in \mathbb{R}$ such that $X' := X + y_0$ and $Y' := Y + x_0$ satisfy $$\mathbb{P}(X' \in S) = 1 \qquad\text{and}\qquad \mathbb{P}(Y' \in S) = 1.$$ Moreover, $\mathbb{P}(X' = \min S) > 0$ and $\mathbb{P}(Y' = \min S) > 0$. The proof is postponed to the end. Now write $[\![n]\!] := \{0, \cdots, n-1\}$. Then we prove Proposition.(1, Lemma 2.1) Let $X$ and $Y$ be independent random variables such that $X+Y$ is uniformly distributed over $[\![n]\!]$. Then both $X$ and $Y$ have uniform distribution. The following proof is based on the reference 1) mentioned in @Mark Wildon's comment. Proof. In light of the lemma above, we may assume that both $X$ and $Y$ are supported on $[\![n]\!]$ as well as $\mathbb{P}(X=0,Y=0)=1/n$. Using this, set $$ A(x) := \sum_{k\geq 0} a_k x^k \qquad \text{and} \qquad B(x) := \sum_{k\geq 0} b_k x^k $$ where $a_k := p_X(k)/p_X(0)$ and $b_k := p_Y(k)/p_Y(0)$. Then it follows that $a_k, b_k$ are all non-negative, $a_0 = b_0 = 1$, and $$ A(x)B(x) = 1 + x + \cdots + x^{n-1}. $$ From this, it is easy to check that both $A(x)$ and $B(x)$ are palindromic, which will be used later. Now, to establish the desired assertion, it suffices to show that all of the coefficients of $A(x)$ and $B(x)$ lie in $\{0, 1\}$. To this end, assume otherwise. Let $k_0$ be the smallest index such that either $a_{k_0} \notin \{0, 1\}$ or $b_{k_0} \notin \{0, 1\}$. We know that $k_0 \geq 1$. Moreover, $$ a_{k_0} + \underbrace{ a_{k_0-1}b_{1} + \cdots + a_1 b_{k_0 - 1} }_{\in \mathbb{N}_0} + b_{k_0} = [x^{k_0}]A(x)B(x) \in \{0, 1\}, $$ forces that $a_{k_0} + b_{k_0} = 1$. So both $a_{k_0}$ and $b_{k_0}$ lie in $(0, 1)$. But if we write $d = \deg B(x)$, then we have $b_{d-k_0} = b_{k_0}$ and $b_d = b_0 = 1$, and so, $$ 1 < a_{k_0}b_{k_0} + 1 \leq a_{d} + \cdots + a_{k_0}b_{d-k_0} + \cdots + b_{d} = [x^{d}]A(x)B(x) \in \{0, 1\}, $$ a contradiction. Therefore no such $k_0$ exists and the desired claim follows. $\square$ References. 1) Behrends, E., 1999. Über das Fälschen von Würfeln. Elem. Math. 54, 15–29. https://doi.org/10.1007/s000170050051 2. Further questions Based on some simulations as well as actual computation for small $n$, I conjecture that the followings hold: Conjecture. Let $A(x)$ and $B(x)$ be monic polynomials with coefficients in $[0, \infty)$. Assume that there exists an integer $n \geq 1$ such that $$ A(x)B(x) = 1 + x + \cdots + x^{n-1}. $$ Then there exist positive integers $1 = n_0 \mid n_1 \mid \cdots \mid n_d = n$, not necessarily distinct, such that $$ A(x) = \frac{(x^{n_1} - 1)}{(x^{n_0} - 1)} \frac{(x^{n_3} - 1)}{(x^{n_2} - 1)} \cdots, \qquad B(x) = \frac{(x^{n_2} - 1)}{(x^{n_1} - 1)} \frac{(x^{n_4} - 1)}{(x^{n_3} - 1)} \cdots. $$ This implication of this conjecture is that, up to shift, $X$ is supported on the set of the form $$ \{ (c_0 + c_2 n_2 + c_4 n_4 + \ldots) : c_k \in [\![n_{k+1}/n_k]\!] \} $$ and likewise $Y$ is supported on $$ \{ (c_1 n_1 + c_3 n_3 + c_5 n_5 + \ldots) : c_k \in [\![n_{k+1}/n_k]\!] \}. $$ This may be regarded as the converse of the fact that, if $Z$ is sampled uniformly at random from the set $[\![n]\!]$ and $Z = \sum_{k\geq 0} C_k n_k$ with $C_k \in [\![n_{k+1}/n_k]\!]$, then $C_k$'s are independent. Addendum - Proof of Lemma. First, we note that both $X$ and $Y$ are bounded. Indeed, choose $x > 0$ so that $\mathbb{P}(|X| \leq x) > 0$ and note that $$ \mathbb{P}(|Y| \geq y) = \frac{\mathbb{P}(|Y| \geq y, |X| \leq x)}{\mathbb{P}(|X| \leq x)} \leq \frac{\mathbb{P}(|X + Y| \geq y - x)}{\mathbb{P}(|X| \leq x)} $$ can be made to vanish if $y$ is chosen sufficiently large. This shows that $Y$ is bounded. A similar argument shows that $X$ is also bounded. Next, choose the smallest interval $[x_0, x_1]$ so that $\mathbb{P}(X \in [x_0, x_1]) = 1$, and likewise, choose the smallest interval $[y_0, y_1]$ so that $\mathbb{P}(Y \in [y_0, y_1]) = 1$. Then $x_0 + y_0 = \min S$. Indeed, If $x_0 + y_0 < s$, then write $s = x+y$ with $x > x_0$ and $y > y_0$. Then $$ 0 < \mathbb{P}(X \leq x, Y \leq y) \leq \mathbb{P}(X + Y \leq s) $$ shows that $s \geq \min S$. Letting $s \downarrow x_0 + y_0$, this proves $x_0 + y_0 \geq \min S$. If $x_0 + y_0 > s$, then $\mathcal{D} := \{(x, y) : x+y \leq s\} \cap ([x_0, x_1]\times[y_0, y_1]) = \varnothing$, and so, $$ \mathbb{P}(X+Y \leq s) = \mathbb{P}((X, Y) \in \mathcal{D}) = 0. $$ This implies that $s < \min S$ and thus $x_0 + y_0 \leq \min S$. Together with $\mathbb{P}(X+Y = \min S) > 0$, this implies $\mathbb{P}(X = x_0) > 0$ and $\mathbb{P}(Y = x_0) > 0$. From this, $$ \mathbb{P}(X+y_0 \notin S) = \mathbb{P}(X+Y \notin S \mid Y = y_0) \leq \frac{\mathbb{P}(X+Y \notin S)}{\mathbb{P}(Y = y_0)} = 0 $$ and hence $\mathbb{P}(X+y_0 \in S) = 1$. A similar argument shows that $\mathbb{P}(Y+x_0 \in S) = 1$, and therefore the claim follows by setting $a = y_0$ and $b = x_0$. $\square$<|endoftext|> TITLE: Attempt at applying linear programming to the partial sums of the Möbius inverse of the Harmonic numbers QUESTION [6 upvotes]: Let $a(n)$ be the Dirichlet inverse of the Euler totient function: $$a(n) = \sum\limits_{d|n} d \cdot \mu(d) \tag{1}$$ and let the matrix $T(n,k)$ be: $$T(n,k)=a(\gcd(n,k)) \tag{2}$$ It has been proven by both joriki and GH from MO that for $n>1$: $$\Lambda(n) = \sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k} \tag{3}$$ Let $M(n,k)$ be the lower triangular matrix: $$M(n,k)=\underset{m\geq k}{\sum _{m=1}^n} a(\gcd (m,k)) \tag{4}$$ Conjecture 1: For $1 < k \leq n$, for all $n$: $$\;\;\;\;-(k-1) \leq M(n,k) \leq (k-1)$$ Conjecture 2: $$ \sum_{k=1}^{k=n} M(n,k) = 1$$ $$\sum _{k=2}^n \frac{M(n,k)}{k}=\sum _{m=1}^n \left(\underset{k \mid m}{\sum _{k=1}^m} H_k \mu \left(\frac{m}{k}\right)-1\right) \tag{5}$$ I am interested in investigating: $$\sum _{k=2}^x \frac{M(x,k)}{k} < C\left\lfloor x^{1/2+\epsilon}+\frac{1}{2}\right\rfloor \tag{6}$$ Therefore we form the linear programming problem $(7)$: $$\begin{array}{ll} \text{minimize} & \displaystyle\sum_{k=1}^{k=n} \frac{y_{k}}{k} \\ \text{subject to constraints:} & n + \displaystyle\sum_{k=2}^{k=n}y_{k}=1 \\ & y_1 \geq -1 \\ \text{and for $k>1$:} & M(n,k) \leq y_k \leq M(n,k) & \tag{7} \end{array}$$ The solution to the linear programming problem $(7)$ (blue dots) will automatically coincide with LHS of $(6)$ (the red lines) as shown in this graph: Because of the answer to this question here proven by Marcus Ritt and the other parallell answer here by Maxim, I find it natural to ask whether the output (the blue lines) from the following linear programming problem $(8)$ is greater than $(7)$. In the program I made the change that I put the upper variable bound to $0$ and the lower variable bound to the negated absolute value of the entries in the lower triangular matrix $M(n,k)$. $$\begin{array}{ll} \text{minimize} & \displaystyle\sum_{k=1}^{k=n} \frac{y_{k}}{k} \\ \text{subject to constraints:} & n + \displaystyle\sum_{k=2}^{k=n}y_{k}=1 \\ & y_1 \geq -1 \\ \text{and for $k>1$:} & -|M(n,k)| \leq y_k \leq 0 & \tag{8} \end{array}$$ Can anything be said whether there exist a constant $C$ such that: $C$ times the output from LP-problem $(8)$ $\geq$ The output from LP-problem $(7)$? Or put in pictures. Is there a constant $C$ such that the irregular red curve is bounded by the irregular blue curve? That for $C$, however large, say $C=1000$ or greater. What we do know given conjecture 1 above which implies that $-|M(n,k)| \geq -(k-1)$, is that the wiggly blue curve below is bounded by the smooth continuous blue curve $f(x)$: From the answer at the operations research forum we also know that the solutions to the linear programming problem: $$\begin{array}{ll} \text{minimize} & \displaystyle\sum_{k=1}^{k=n} \frac{y_{k}}{k} \\ \text{subject to constraints:} & n + \displaystyle\sum_{k=2}^{k=n}y_{k}=1 \\ & y_1 \geq -1 \\ \text{and for $k>1$:} & -(k-1) \leq y_k \leq 0 & \tag{9} \end{array}$$ is the continuous blue curve $f(x)$ and it is asymptotic to: $$f(x)=C\left(-\left\lfloor \sqrt{2 (x-1)}+\frac{1}{2}\right\rfloor +H_{\left\lfloor \sqrt{2 (x-1)}+\frac{1}{2}\right\rfloor } + \text{Binomial term} \right) \tag{10}$$ ($C=2$ was multiplied with later). Anyways, the the solutions to $(8)$ are always bounded by the solutions to $(9)$. The question is whether the solutions to $(7)$ are bounded by the solutions to $(8)$? The binomial term can be found in the OEIS. Edit: Minor error: $f(x)$ should have been $f(n)$ to suit the linear programming problem. The answer, if it is to be found, lies in comparing this matrix from the solution of the Linear Programming tagged $(8)$ starting: $$\begin{array}{llllllllllllllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & -2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & -1 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & 0 & 0 & -4 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & -2 & -1 & -1 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & -1 & 0 & -2 & -3 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & 0 & -1 & -1 & -2 & -2 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & -2 & 0 & 0 & 0 & -4 & 0 & -2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & -1 & -1 & -4 & -1 & -1 & 0 & 0 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & 0 & 0 & -3 & 0 & -2 & 0 & 0 & -5 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & -2 & -1 & -2 & -2 & -1 & -1 & -1 & 0 & 0 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & -1 & 0 & -1 & -3 & 0 & 0 & -1 & -5 & -1 & 0 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & 0 & -1 & 0 & -2 & -6 & -1 & 0 & -2 & 0 & 0 & 0 & 0 & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & -2 & 0 & -4 & 0 & -5 & 0 & -2 & 0 & -1 & 0 & 0 & 0 & 0 & \text{} & \text{} & \text{} \\ 1 & -1 & -1 & -1 & -3 & -1 & -4 & -1 & -1 & -1 & -1 & 0 & 0 & 0 & 0 & 0 & \text{} & \text{} \\ 1 & 0 & 0 & 0 & -2 & 0 & -3 & 0 & 0 & 0 & -4 & 0 & -7 & 0 & 0 & 0 & 0 & \text{} \\ 1 & -1 & -2 & -1 & -1 & -2 & -2 & -1 & -2 & -1 & -3 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}$$ with the matrix from the solution of the Linear Programming tagged $(7)$ which essentially is the matrix $M$ except for the first column, starting: $$\begin{array}{llllllllllllllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & -2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & -1 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & 0 & 0 & -4 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & -2 & -1 & -3 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & -1 & 0 & -2 & 3 & -6 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & 0 & -1 & -1 & 2 & -5 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & -2 & 0 & 0 & 0 & -4 & 0 & -2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & -1 & -1 & -4 & -1 & -3 & -1 & -1 & 4 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & 0 & 0 & -3 & 0 & -2 & 0 & 0 & 5 & -10 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & -2 & -1 & -2 & 2 & -1 & -1 & -2 & 4 & -9 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & -1 & 0 & -1 & 3 & 0 & 0 & -1 & 5 & -8 & 3 & -12 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & -1 & 0 & -1 & 0 & 2 & -6 & -1 & 0 & 4 & -7 & 2 & -11 & 6 & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & -2 & 0 & -4 & 0 & -5 & 0 & -2 & 0 & -6 & 0 & -10 & 7 & 8 & \text{} & \text{} & \text{} \\ 1 & -1 & -1 & -1 & -3 & -1 & -4 & -1 & -1 & -1 & -5 & -1 & -9 & 6 & 9 & -1 & \text{} & \text{} \\ 1 & 0 & 0 & 0 & -2 & 0 & -3 & 0 & 0 & 0 & -4 & 0 & -8 & 7 & 10 & 0 & -16 & \text{} \\ 1 & -1 & -2 & -1 & -1 & 2 & -2 & -1 & -2 & -1 & -3 & 2 & -7 & 6 & 8 & -1 & -15 & 2 \end{array}$$ Associated Mathematica program: https://pastebin.com/rHaXxVcj Edit: 25.11.2019: I am not entirely sure but I believe the question approximately boils down to: Let: $a(n)=\sum\limits_{d \mid n} \mu(d)d$ Prove or disprove that there exists a constant $c$ such that the inequality: $$\sum\limits_{r=2}^{n} \frac{\sum\limits_{m=r}^{n} a(\gcd (m,r))}{r} \geq c\underset{\sum\limits_{k=2}^{r} -\left|\sum\limits_{m=k}^{n} a(\gcd (m,k))\right|\geq -(n-1)}{\sum _{r=2}^n} -\frac{\left|\sum\limits_{m=r}^{n} a(\gcd (m,r))\right|}{r} \tag{11}$$ holds for all $n$ I am asking because it has been proven that the right hand side is bounded from below by: $$c\left(-\left\lfloor \sqrt{2 (n-1)}+\frac{1}{2}\right\rfloor +H_{\left\lfloor \sqrt{2 (n-1)}+\frac{1}{2}\right\rfloor } + \text{Binomial term} \right)$$ Mathematica: Clear[a, b, nn]; nn = 60; a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]]; Monitor[a1 = Table[Sum[Sum[a[GCD[m, r]], {m, r, n}]/r, {r, 2, n}], {n, 1, nn}];, n] g1 = ListLinePlot[a1, PlotStyle -> {Red, Thick}]; Monitor[a2 = Table[Sum[ If[Sum[-Abs[Sum[a[GCD[m, k]], {m, k, n}]], {k, 2, r}] >= -(n - 1), -Abs[Sum[a[GCD[m, r]], {m, r, n}]]/r, 0], {r, 2, n}], {n, 1, nn}];, n] g2 = ListLinePlot[a2, PlotStyle -> {Thick}]; Show[g2, g1] It is as said conjectured that the blue curve times a constant is greater than the red curve. The blue curve is bounded from below by a function whose leading term is the floor function of a square root. Same graph as above but for a 1000 times 1000 sized matrix: Edit 30.4.2020: More efficient program and plot: (*start*) (*Mathematica*) Clear[a]; nn = 2000; constant = 2*Sqrt[2]; a[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]]; Monitor[TableForm[ A = Accumulate[ Table[Table[If[n >= k, a[GCD[n, k]], 0], {k, 1, nn}], {n, 1, nn}]]];, n] TableForm[AB = Transpose[A]/Range[nn]]; AB[[1, All]] = 0; g1 = ListLinePlot[Abs[Total[AB]], PlotStyle -> Red]; Clear[AB]; TableForm[B = -Abs[A]]; Clear[A]; B[[All, 1]] = Range[nn]; TableForm[B1 = Sign[Transpose[Accumulate[Transpose[B]]]]]; Clear[B] Quiet[Show[ ListLinePlot[ v = ReplaceAll[ Flatten[Table[First[Position[B1[[n]], -1]], {n, 1, nn}]], First[{}] -> 1], PlotStyle -> Blue], Plot[constant*Sqrt[n], {n, 1, nn}, PlotStyle -> {Pink, Thick}], g1, ImageSize -> Large]] ListLinePlot[v/(constant*Sqrt[Range[nn]])]; (*end*) The pink curve is 2*Sqrt(2)*Sqrt(x): REPLY [7 votes]: Here is a proof of Conjecture 2. First, we have \begin{split} \sum_{k=1}^n M(n,k) &= \sum_{k=1}^n \sum_{m=k}^n \sum_{d|\gcd(m,k)} d\cdot\mu(d) \\ &= \sum_{m=1}^n \sum_{k=1}^m \sum_{d|\gcd(m,k)} d\cdot\mu(d). \end{split} Second, denoting $g:=\gcd(m,k)$ and $k':=\frac{k}{g}$, we get: \begin{split} \ldots &= \sum_{m=1}^n \sum_{g|m} \sum_{k'=1\atop \gcd(k',m/g)=1}^{m/g} \sum_{d|g} d\cdot\mu(d) \\ &= \sum_{m=1}^n \sum_{g|m} \sum_{d|g} d\cdot\mu(d)\cdot\varphi(\frac{m}{g}), \end{split} where $\varphi(\cdot)$ is Euler's totient function. Introducing $g':=\frac{g}{d}$ and recalling formula (15), we finally get: \begin{split} \ldots &= \sum_{m=1}^n \sum_{d|m} \sum_{g'|m/d} d\cdot\mu(d)\cdot\varphi(\frac{m}{g'd}) \\ &= \sum_{m=1}^n \sum_{d|m} d\cdot\mu(d)\cdot\frac{m}{d} \\ &= \sum_{m=1}^n m \sum_{d|m} \mu(d) \\ &= \sum_{m=1}^n m\cdot \delta_{m,1} \\ &= 1. \end{split}<|endoftext|> TITLE: Why is Langlands functoriality usually related with period integral in a third group? QUESTION [10 upvotes]: In the introduction of "PERIODS OF AUTOMORPHIC FORMS "by HERVE JACQUET, EREZ LAPID, and JONATHAN ROGAWSKI, they said "In many cases, it should be possible to characterize the $H$-distinguished cuspidal representations as images with respect to a functorial transfer to $G$ from a third group $G'$" here $G$ is a reductive group over a number field $F$ , and $H \subseteq G$ is a subgroup obtained as the fixed point set of an involution $\theta$. The case $(GL_n(E),GL_n(F))$ and $(GL_n(E),U_n(F))$ is well-studied by Jacquet, here $E/F$ is a quadratic extension. For general Galois model, there is a conjecture by Prasad (see https://arxiv.org/abs/1512.04347). There are examples beyond the involution case, e.g the nonvanishing of Shalika period integrals is characterized in terms of functorial transfers from $GSpin(2n + 1)$. There are also local analogues for the story. The question is (at least over $p-$adic field)： Why is Langlands functoriality usually related with period integral in a third group? What do we know beyond Galois case? What's the relation with relative Langlands program by Sakellaridis and Venkatesh? Toy model: an irreducible cuspidal representation of $GL_n(\mathbb F_{q^2})$ is distinguished by $U_n(\mathbb F_q)$ iff it comes from $GL_n(\mathbb F_q)$ by Shintani lifting. REPLY [7 votes]: A lot is known--too much to try to summarize--and I think this philosophy came about after seeing numerous examples, beginning with Harder-Langlands-Rapoport (base change for GL(2)), and thinking about the connection with L-functions. Namely, in many instances one knows that images of functorial lifts can be detected by poles or nonvanishing of certain $L$-functions at special values. For instance, often images of functorial lifts can be detected via factorizations of $L$-functions for functorial lifts---e.g., to relate to Will Sawin's comment for a lift from GL(n) to GL(n) $\times$ GL(n), the zeta function divides $L(s, \pi \times \check \pi)$ so the latter has a pole at $s=1$. In addition, theta lifts are often nonvanishing only when a certain $L$-value is non-zero. Now one often has integral representations for $L$-functions, and one can often detect poles/nonvanishing of $L$-functions at special values via period integrals. In addition, one can often compare periods on different groups, as in much of Jacquet's work and the Gan-Gross-Prasad conjectures. Sakellaridis-Venkatesh is an attempt to build a general theory for how periods are related on different groups. They primarily study the local question, but do make some global conjectures at the end as well. Some places you can look for instances of what is known: surveys/papers on integral representations, poles of $L$-functions, theta correspondence and descent; Jacquet's papers; introduction to Sakellaridis-Venkatesh, ...<|endoftext|> TITLE: Formal group law for oriented bordism QUESTION [10 upvotes]: From this answer I learned that the coefficient ring $MSO^{*}[1/2]$ of oriented bordism with 2 inverted supports an odd formal group law and is infact the universal such ring. Is there a reference/proof for this fact? As motivation, I should mention that I'm trying to prove the fact that if $MU^{*}\rightarrow E^{*}$ is a map of rings where $E^{*}$ is a Landweber exact ring such that the image of the degree 2 generator $z_1\in MU^{*}\simeq \mathbb{Z}[z_1, \dots, ]$ vanishes in $E^{*}$, then this map must factor through the canonical "forget complex structure" map $MU^{*}\rightarrow MSO^{*}[1/2]$. If I can show that these conditions imply that $E^{*}$ has an odd formal group law, then I'd be able get my desired factorization $MSO^{*}[1/2]\rightarrow E^{*}$. It therefore would be very helpful to understand why $MSO^{*}[1/2]$ itself has an odd formal group law (in particular the universal such one), as it might be the case that the proof method for $MSO^{*}[1/2]$ uses only that $MSO^{*}[1/2]$ is a Landweber exact ring for which the image of $z_1$ vanishes under the map $MU^{*}\rightarrow MSO[1/2]^{*}$ and hence generalizes straightaway to my general case. Note: An odd (one dimensional) formal group law $F(X, Y)$ is one for which $F(X, -X)=0$. REPLY [5 votes]: One basic point is as follows. The usual inclusions $U(n)\to O(2n)$ give rise to a map $BU\to BSO$ of $E_\infty$ spaces and then a map $MU\to MSO$ of ring spectra, which gives a complex orientation of $MSO$. Note that a map $X\to MSO(n)$ of spaces represents a class in $MSO^n(X)$. The orientation class $x\in MSO^2(\mathbb{C}P^\infty_+)=MSO^2(BSO(2)_+)$ is just the zero section $BSO(2)_+\to BSO(2)^T=MSO(2)$, where $T$ is the tautological bundle. There is a suspension isomorphism $MSO^2(BSO(2)_+)=MSO^3(\Sigma BSO(2)_+)$ and $\Sigma x$ is represented by an evident map $\Sigma BSO(2)_+\to MSO(3)$, covering $Bi$, where $i$ is the evident inclusion $i\colon SO(2)\to SO(3)$. It is a key point that $i$ extends to give a map $O(2)\to SO(3)$, with $i(A)=A\oplus\det(A)$. Covering this, we get a map $BO(2)^{\lambda ^2T}\to MSO(3)$ extending $\Sigma x$. Now choose $g\in O(2)\setminus SO(2)$, and let $\beta$, $\gamma$ and $\delta$ be the automorphisms of $SO(2)$, $O(2)$ and $SO(3)$ given by conjugating with $g$, $g$ and $i(g)$ respectively. As $\gamma$ and $\delta$ are inner automorphisms, it is standard that $B\gamma$ and $B\delta$ are homotopic to the respective identity maps, and one can check that they also induce maps of the relevant Thom spaces that are homotopic to the identity. From this it follows that $\beta^*(\Sigma x)=\Sigma x$. However, $\beta$ reflects the suspension coordinate and acts as the inversion map on $SO(2)$ so we have $\beta^*(\Sigma x)=-\Sigma [-1](x)$. As this is equal to $\Sigma x$, we see that the formal group law is odd. I am sure that I have seen the oddness proved somewhere in the literature, but I do not remember where. I thought it was in a paper by Steve Mitchell, but if so, I have not found the right one. I think that the argument given was not the same as the one above, but I do not remember what it was.<|endoftext|> TITLE: Can the Petersson inner product $\langle f(z), f(2z) \rangle$ be zero? QUESTION [6 upvotes]: Suppose $f$ is a weight $k$ cuspidal Hecke eigenform on $\Gamma_0(N)$. Then $f(2z)$ is a weight $k$ cuspform on $\Gamma_0(2N)$. Is it possible that $f(z)$ and $f(2z)$ can be orthogonal (regarded as forms on $\Gamma_0(2N)$)? That is, can the Petersson inner product $\langle f(z), f(2z) \rangle = 0$, where the product is taken over $\Gamma_0(2N) \backslash\mathcal{H}$? More generally, can $\langle f(z), f(nz) \rangle = 0$ (where the product is regarded over the appropriate quotient of the upper half plane)? REPLY [7 votes]: Yes, the Petersson inner product can be zero. In my paper "Explicit bounds for sums of squares (see Lemma 5) I show that if $f$ is a newform of level $N$ and $p$ is a prime that does not divide $N$, then $$ \langle f(z), f(pz) \rangle = \frac{a(p)}{p^{k-1} (p+1)} \langle f(z), f(z) \rangle. $$ So if you can find a weight $k$ level $N$ newform with odd $N$ for which $a(2)$ vanishes, then this gives you an example. The newform $f(z) = q - 2q^{3} + \cdots$ of weight $2$, level $19$ and trivial character is an example of such an $f$.<|endoftext|> TITLE: What is $\mathbb{E} [\max_{\sigma \in \{ \pm 1\}^n} \sigma^T Z \sigma]$ for a random Gaussian matrix $Z$? QUESTION [5 upvotes]: Given an $n \times n$ random matrix $\mathbf{Z}$ with each entry i.i.d. $\mathcal{N} (0,1)$, what is $\mathbb{E} [\max_{\sigma \in \{ \pm 1\}^n} \sigma^T Z \sigma]$ as $n \to \infty$? If this is too much to ask, are there any good known upper or lower bounds? REPLY [7 votes]: This is related to the Sherrington-Kirkpatrick Spin Glass model. In arxiv.org/pdf/1412.0170.pdf, Dmitry Panchenko writes that $$\lim_{N\to\infty} \frac{1}{N} \mathbb{E}\big[\max_\sigma \frac{1}{\sqrt{N}} \sum_{i TITLE: Cofibrant simplicial categories QUESTION [5 upvotes]: Reading about simplicial categories, and in particular the model structure in sCat, I found various sources, among which I can mention this one or section 16.2 in Riehl's book “Categorical homotopy theory”, which seem to imply that the trivial simplicial structures on the posets $[n]$ are not cofibrant. Now, this leaves me a bit puzzled, in that I can exhibit a very explicit description of the initial maps of these objects as compositions of pushouts of generating cofibrations. For instance, $[2]$ is the smallest of these simplicial categories which is claimed not to be cofibrant. $[1]$ is already known to be cofibrant (for one thing, it is isomorphic to $\mathfrak{C}\Delta^1$). Now consider the following two diagrams: $\require{AMScd}$ \begin{CD} \emptyset @>>> [1]\\ @VVV @VVV\\ [0] @>>> [1] \coprod [0] \end{CD} $\require{AMScd}$ \begin{CD} [1]_{\emptyset} @>f>> [1] \coprod [0]\\ @VVV @VVV\\ [1]_{\Delta^0} @>>> [2] \end{CD} where $f$ selects the terminal point of $[1]$ and the adjoined point. It is clear that the first diagram is a pushout. It really seems to me that the second one is, too, both by explicit construction of a pushout in $\textbf{sCat}$ and by checking the universal property. The left vertical maps in both diagrams are generating cofibrations. I conclude that the inclusion $[1] \to [2]$ is also a cofibration. Is there a mistake that I don't see? What goes wrong in this argument? REPLY [2 votes]: Simplicial categories $[n]$ are indeed cofibrant. The Simplicial category $\mathfrak{C}\Delta^n$ is also cofibrant and weakly equivalent to $[n]$, so the former is a cofibrant replacement of the latter. Note that $\mathfrak{C}\Delta^n$ is isomorphic to $[n]$ only when $n \leq 1$ (so this is not true for $n = 2$). The point is that simplicial functors $\mathfrak{C}\Delta^n \to C$ carry more information than $[n] \to C$, so the homotopy coherent nerve of $C$ remembers enough information about $C$ as opposed to the ordinary nerve.<|endoftext|> TITLE: Proofs of Euler's characteristic formula for n-Dim polytopes QUESTION [5 upvotes]: Twenty proofs of Euler's formula V - E + F - 1 = 1, which applies to convex polyhedrons, i.e., 3-dimensional polytopes, are presented at the Geometry Junkyard. I'm interested in proofs of the more general formula for the Euler characteristic number for bounded, convex polytopes of dimension greater than three as well since the signed, refined face partition polynomials enumerating the k-dimensional faces (k=0 to n) of the n-dimensional associahedra and providing the compositional inversion of formal power series obey the extended Euler formula $$V - E + (2-D-faces) - (3-D-faces) + ... $$ $$(-1)^{n-1} ((n-1)-D-facets) + (-1)^n = 1,$$ and proofs of the formula might provide insight on derivations of the face partition polynomials. (Same applies to permutahedra and multiplicative inversion.) I'm particularly interested in proofs related to a generalized Gauss-Bonnet theorem, proofs related to differential geometry. 1) Which of the Junkyard proofs can be extended beyond three dimensions to any n-dimensional bounded, convex polytope? 2) Do you have references to other proofs for indefinite dimensions? REPLY [4 votes]: The most straightforward way to extend Euler's formula to all convex polytopes is to show that all convex polytopes are shellable, which was only proved in 1971 by Bruggesser & Mani. Their paper is available online here: https://www.mscand.dk/article/view/11045. This fills in a gap in an 1901 argument of Schläfli. See the paper for details.<|endoftext|> TITLE: Betti sequence of finite dimensional commutative algebras QUESTION [9 upvotes]: Given a finite dimensional commutative local $K$-algebra $A$ for a field $K$. Associated to $A$ is its dimension $d_A$ and the Betti-sequence $c_i=dim(Ext_A^i(S,S))$ where $S$ is the unique simple $A$-module. Question: Are there only finitely many such $K$-algebras $A$ with a fixed dimension $d_A$ and associated sequence $c_i$? If not, then can one characterise for which sequences there are only finitely many for a fixed dimension? Is there an example of such a sequence that is not constant? For example for $c_i=1$ constant and $d_A=n$, the unique ones are $A=K[x]/(x^n)$ as we saw in a previous thread Commutative algebras with modules of small complexity . REPLY [8 votes]: I will first show that the general answer is no (in fact one should expect the opposite even in the graded case) by the Pigeonhole principle. Then we shall construct some concrete examples via an old result of Tate. Let $(A,k)$ be a local artin algebra. We write $A=R/I$ as a (minimal) quotient of a regular local ring $(R,k)$. Note that $\dim R$ is bounded above by $d_A-1$. Claim 1: Fixing $d_A$, then the set of possible sequences $\{c_i\}_{i\geq 0}$ is countable. Proof: this follows from an inequality proved by Serre: $$\sum c_it^i \ll \frac{(1+t)^{\dim R}}{1-\sum_{i>0}\dim_k Tor_i^R(k,A) t^{i+1}} $$ (where $\ll$ means that the LHS is term by term dominated by the RHS). For references, look up "Golod rings", this is such a popular subject even I had a paper about it. Now, to finish the proof, note that the denominator of RHS is a polynomial of degree $\dim R$, and each $\dim_k Tor_i^R(k,A)$ is bounded above by $d_A\dim_k Tor_i^R(k,k)$ as $A$ has a filtration by $d_A$ copies of $k$ and $\dim Tor_i(k,-)$ is subadditive on short exact sequences. Finally, $\dim_k Tor_i^R(k,k) = \binom{\dim R}{i}$. Claim 2: Fixing $d_A$, and assume $k$ is algebraically closed, the set of isoclasses of $A$, even assuming $A$ standard graded, can be uncountable. Proof: this comes from standard facts that the set of isoclasses of $A$ with a fixed Hilbert function are parametrized by Hilbert schemes. Putting together Claim 1 and 2 shows that there are sequences with uncountably many $A$s. Example: For a concrete example, a result by Tate (mentioned in Eisenbud's paper on complete intersection) implies that when $A$ is a complete intersection, the $\{c_i\}$ only depends on $\dim R$, so even two general quadrics in $k[x,y]$ would give infinitely many examples.<|endoftext|> TITLE: Rates of convergence of mollifiers with Sobolev norms on manifold QUESTION [5 upvotes]: Let $M$ be a smooth compact Riemannian manifold of dimension $n$, and let $H^s_p(M)$ for $s\in \mathbb{R}$ be the fractional Sobolev space of order $s$ on the manifold (defined for instance through the Laplace-Beltrami operator). Let $k:\mathbb{R}\to \mathbb{R}$ be a radial kernel and let $k_h=h^{-n}k(\cdot/h)$ for $h>0$. For $f:M\to \mathbb{R}$, one define $k_h*f$ by $$ k_h*f(x) = \int_{T_x M} k_h(|v|)f(\exp_x(v)) d v.$$ Consider the following assertion: For $s0$ the easiest way to obtain convolution estimates on manifolds is described below. First of all, by the Whitney embedding theorem (or by Nash theorem if you want to preserve the Riemannian metric) we may assume that $M$ is $n$-dimensional submimanifold of $\mathbb{R}^k$ for some $k>n$. The trace operator is bounded as $T:H^{s+1/p}_p(\mathbb{R}^k)\to H^{s}_p(\mathbb{R}^{k-1})$ and there is an extension operator $E:H^{s}_p(\mathbb{R}^{k-1})\to H^{s+1/p}_p(\mathbb{R}^k)$. The same applies to compact submanifolds of codimension $1$. By induction, if $M\subset \mathbb{R}^k$ s a minifold of dimension $n$, there is a trace operator $T:H^{s+(k-n)/p}_p(\mathbb{R}^k)\to H^s_p(M)$ and the extension operator $E:H^s_p(M)\to H^{s+(k-n)/p}_p(\mathbb{R}^k)$. Now for $f\in H^s_p(M)$ we define convolution on $M$ as follows $$ K_h f=T(k_h*(Ef)). $$ That is, we extend $f$ to $\mathbb{R}^k$, we apply this desired convolution on $\mathbb{R}^k$, and then we restrict the resulting function back to $M$. That usually gives all estimates you want. Assuming that the convolution in $\mathbb{R}^k$ satisfies your estimate, the same estimate will hold on $M$. Indeed, \begin{equation*} \begin{split} &\Vert f-K_h f\Vert_{H^s_p(M)}= \Vert T(Ef-K_h*(Ef))\Vert_{H^s_p(M)} \leq C_1\Vert Ef-K_h*(Ef)\Vert_{H^{s+(k-n)/p}(\mathbb{R}^k)}\\ &\leq C_2 h^{r-s}\Vert Ef\Vert_{H^{r+(k-n)/p}_p(\mathbb{R}^k)}\leq C_3 h^{r-s}\Vert f\Vert_{H^r_p(M)}. \end{split} \end{equation*}<|endoftext|> TITLE: Arzela-Ascoli for L_p-norm QUESTION [7 upvotes]: Since I am from a different mathematical field and couldn't find it: Is there something which would be best called an Arzela-Ascoli version for the $L_p$-norm, namely: Let $X,Y$ be two nice measurable/normed spaces (i.e. compact, locally compact, hausdorff and whatever properties you might want), e.g. $X$ a closed interval and $Y= \mathbb R$. Let now $f_n$ be a family of piecewise differentiable maps from $X$ to $Y$ such that their $L_p$-norm converges, i.e. $||f_n'||_p \to \lambda$. Does there exist a converging subfamily $f_k \to f$ such that $f$ is piecewise differentiable and $||f'||_p=\lambda$? The question arose from the following setting: Let $X$ and $Y$ be two finite metric graphs (you can think of them as closed intervals glued together at their endpoints with metric & measure from $\mathbb R$). Given a family of piecewise differentiable maps $f_n: X \to Y$, does there exist a piecewise differentiable map $f: X \to Y$, such that the $L_p$-norm of its derivative $||f'||_p$ attains the infimum of $||f'_n||_p$? Here the family $f_n$ is the set of all piecewise differentiable maps homotopic to a given function. Hence I think we can even assume that the $f_n$ are continuous and piecewise linear, since averaging along edges decreases the $L_p$-norm of its derivative , i.e. do we have $||f'_n||_p \geq |\frac{f_n(b)-f_n(a)}{b-a}|$ for $f_n:[a,b] \to \mathbb R$? For $p=\infty$ such an $f$ exists by Arzela-Ascoli, but does this also hold for other $L_p$-norms? EDIT: In the first version of the question the $f_n$ and $f$ in the box were only continuous and $||f_n||_p$ converged to $\lambda=||f||_p$. REPLY [2 votes]: For your interest in a minimal $f$, you might want to read a beginners textbook on Sobolev-Spaces and the calculus of variations, especially on the direct method, which is all about this. The beginner-texts won't deal with metric graphs, but I am pretty sure that you can see how to generalize to this from the solution of the problem on intervals (or rather general domains in $\mathbb{R}^n$) that you will find there. Let me give you a short overview for $p>1$ (the case $p=1$ is a bit more tricky in the analysis and it turns out that any monotone function is a minimizer): The direct method in a simple example Let $I=[a,b]$ be an interval and consider the Sobolev-space $W^{1,p}(I;\mathbb{R})$, which is the closure of $C^1(I;\mathbb{R})$ in the norm $$\|f\|^p_{W^{1,p}} = \|f'\|_{L^p}^p + \|f\|_{L^p}^p.$$ In general for every $f\in W^{1,p}(I,\mathbb{R})$, which you can approximate in the norm using smooth $(f_k)_k$, the weak-derivative $f' := \lim_{k\to\infty} f_k'$ is a well defined $L^p$ function. It won't be the pointwise derivative, but in many cases you can almost treat it as such. Furthermore one can prove that $W^{1,p}$ is a reflexive space (here $p>1$ is needed). Now let's say, we are interested in the class of functions $A := \{f\in W^{1,p}: f(a) = 0, f(b)=1\}$. (One needs to make sure that $f(a)$ and $f(b)$ are well defined here, this is in fact due to the trace-theorem, that you will also find in any relevant text.) And we want to minimize $\|f'\|_p$ in this class. (This method is a bit of overkill for such a simple problem, where we could guess the solution.) Then first we note that there is an obvious bound from below and so we can pick a sequence $(f_k)_k$ such that $\|f_k'\|_{L^p} \to \inf_{g\in A} \|g'\|_{L^p}$. In particular then $\|f_k'\|_{L^p}$ is bounded. Now we need the next tool which is Poincare's inequality, which due to $f_k(a) = 0$ (which prevents shifting by constants) also implies that $\|f_k\|_{W^{1,p}}$ is bounded. But then there is the Banach-Alaoglu theorem, which is kind of the convergence theorem you were looking for. This tells us that $f_k$ has a weak* (=weak, since $W^{1,p}$ is reflexive) converging subsequence with some limit $f \in W^{1,p}(I;\mathbb{R})$. Now finally using a bit more of the toolbox: The trace-theorem also shows that the boundary conditions are stable under weak convergence, so $f \in A$. And weak convergence of $f_k$ in $W^{1,p}$ means weak convergence of $f_k$ and $f_k'$ in $L^p$. Finally the norm is lower semi-continuous under weak convergence. There is no norm convergence as you requested in your question, but the point is that lower semicontinuity is enough, as we get $$\inf_{g\in A} \|g'\|_{L^p} \leq \|f'\|_{L^p} \leq \liminf_{k\to\infty} \|f_k'\|_{L^p} = \inf_{g\in A} \|g'\|_{L^p}$$ So $f$ is indeed a minimizer in the class $A$. Now the final step would be to use the Euler-Lagrange equation to show that $f$ is in fact affine (which would translate to piecewise affine in you case). However... All this being said there might be another solution to your problem requiring far less hard analysis. In your metric graphs you only ever have to consider piecewise affine functions. Let $f:[a,b] \to Y$ be a map from an interval to a metric graph. Now for fixed $f(a),f(b)$ and a fixed path from $f(a)$ through the vertices $f(x_1) = p_1,..., f(x_n) = p_n$ to $f(b)$, the minimal $f$ is the piecewise affine one. Then since the set of $x_i$ with $a \leq x_1 \leq ... \leq x_n \leq b$ is compact and $\|f'\|_p$ continuous with respect to that choice, there is a minimal $x_1 < ... < x_n$ in there. Now there are finitely many loop free paths to choose in a finite graph, so there is a finite choice and thus a minimal $p_1,...,p_n$ as well. But then for any $f: X \to Y$, with $f(v_i)$ fixed on the vertices $v_i$, you'll get a minimal $f$ by applying the previous reasoning. And now since $Y$ is compact, and $\|f'\|_p$ should be continuous under the choice of $f(v_i)$, you'll get your minimizer. I'm no expert on metric graphs, but I am quite sure something like this has already been written down though.<|endoftext|> TITLE: Club filter basis in $\omega_1$ QUESTION [5 upvotes]: My question is about existing of basis of club filter club($\omega_1$) with cardinality $c$. Does it exist? REPLY [11 votes]: That's independent of ZFC. On the one hand, it's consistent with ZFC that $2^{\aleph_1}=\mathfrak c$, in which case the whole club filter on $\aleph_1$ has cardinality $\mathfrak c$. On the other hand, the continuum hypothesis is consistent with ZFC and implies that the club filter has no basis of size $\mathfrak c=\aleph_1$. That's because the diagonal intersection of any $\aleph_1$ club sets is again club.<|endoftext|> TITLE: Can we recover all $k$-minors of a square matrix from some of them? QUESTION [9 upvotes]: This is a cross-post. Let $k,n$ be natural numbers, $1k$. Otherwise, if $\text{rank}(A)\le k$, then even if we know all the $k$-minors of $A$ except one, we cannot recover the last one. Indeed, take $A=\pmatrix{D&0\\ 0&0}$ where $D$ is any diagonal matrix of size $k$. The $k$-minor corresponding to the first $k$ rows and columns (which is $\det D$) cannot be recovered from the other $k$-minors (which are zeroes). REPLY [3 votes]: Here is another point of view on the question. Assume that you are interested in $k$-minors, what you're going to do is focus on submatrices of $A$ of size $k\times n$ by eliminating $n-k$ rows. Such a $k\times n$ submatrix has $n\choose k$ $k$-minors, and these are subject to what is known as Plücker equations: Theorem: an ordered collection of $n\choose k$ integers is the collection of (lexicographically ordered) maximal minors of some $k\times n$ matrix if and only if these numbers satisfy a set of polynomial equations known as Plücker equations. Context: see these lecture notes by Alexander Yong. Proof: see Schubert Calculus by Kleiman and Lakso. In practice, it means that there is some maximal number of $k$-minors that can be fixed independently, after which all the others will be uniquely determined by the equations. Plücker equations for $(n,k)$ can be displayed by typing Grassmannian(k-1,n-1) into Macaulay2 (the $-1$ come from projective reasons). Here is one of these equations for $n=6, k=3$: $$p_{2,3,4} p_{1,3,6} -p_{1,3,4}p_{2,3,6} +p_{1,2,3}p_{3,4,6}=0$$ As expected, if all minors are zero except one of them, then the Plücker equations will be of no help to find that one, as the variables always come by pairs.<|endoftext|> TITLE: Lagrangian intersection Floer homology: understanding some assumptions QUESTION [7 upvotes]: Let $(X,\omega)$ be a symplectic manifold and $L\subset X$ be a Lagrangian subspace. Let $\mu_L:H_2(X,L;\mathbb{Z})\to \mathbb{Z}$ be the Maslov index homomorphism. Usual hypothesis Recall that $L$ is said to be monotone, if there exists $c>0$ such that the following identity holds for all $\beta\in \pi_2(X,L)$: $$c \mu_L(\beta)=\omega(\beta).$$ The minimal Maslov number is defined to be: $$\inf \{\mu_L(\beta) \ |\ \beta\in \pi_2(X,L), \ \omega(\beta)>0 \}.$$ Now given two monotone Lagrangians $L_0,L_1$ call assumption $A1$ $$(A1): \text{the minimal Maslov number of $L_0$ and $L_1$ are strictly greater than 2}$$ and assumption $A2$ $$(A2): \text{ $L_0$ is Hamiltonian isotopic to $L_1$}\phantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaa}$$ In defining Lagrangian intersection Floer homology groups $HF(L_0,L_1)$ usually $A1$ or $A2$ is assumed. how are these assumption exploited in the construction of the homology? and what's the role of monotonicity? Why assuming the Lagrangians to be spin it is said to simplify things? How can we use the spin assumption? REPLY [9 votes]: When you try and prove that $d^2=0$ ($d$ being the Floer differential) you need to look at the boundary of the moduli space of index 2 J-holomorphic strips with one boundary on $L_0$, one on $L_1$. Certainly one component of the boundary will consist of "broken strips", i.e. pairs of index 1 strips with a common asymptote that contribute to $d^2$. If that were everything, we'd immediately conclude $d^2=0$. However, you could bubble off a disc at the side of the strip. The index of the disc and the strip have to add up to 2, but if you have e.g. Maslov 0 discs this could happen without affecting the index of the strip. Monotonicity ensures that zero index discs have zero area, so have to be constant and so don't bubble (they wouldn't eat up any energy/wouldn't be stable). Maslov 1 discs only occur for nonorientable Lagrangians, so let's ignore those for now. Maslov 2 discs would leave index zero strips, which have to be constant (higher Maslov discs would give negative index strips, which don't exist generically, so don't contribute). This means that $d^2x$ ends up being a multiple of $x$, the multiple being the count of Maslov 2 discs through $x$ which bubble on $L_0$ minus the number of discs which bubble on $L_1$ (discs can bubble on either side of the strip). If $L_0=L_1$ these numbers are the same, so $d^2=0$. If the minimal Maslov index is bigger than 2 then these numbers are both zero, so again $d^2=0$. In general, you either restrict to Lagrangians with the same number of Maslov 2 discs through each point, or you come to terms with the fact that $d^2$ is not zero and your Fukaya category is "curved". The spin assumption allows you to orient your moduli space of J-holomorphic strips, and hence work with coefficients over fields of characteristic other than 2 (oriented moduli spaces tell you how to assign signs to your counts of strips). Edit: there are weaker assumptions (relatively pin) that suffice, but spin is (maybe?) easier to explain.<|endoftext|> TITLE: Sum of two $n$-th powers, of two $m$-th powers, but not of two $mn$-th powers QUESTION [7 upvotes]: Let $m$ and $n$ be two coprime numbers. Let $S_n$ denote the set of integers that are a sum of two $n$-th powers of integers, for example $7\in S_3$ given $7=2^3+(-1)^3$. Analogously define $S_m$ and $S_{mn}$. My question is if this set is infinite: $(S_n\cap S_m) \smallsetminus S_{mn}$. For $n=2$ and $m=3$ I could manage to prove it is infinite and the solution is elementary (constructive). However, I don't see any clue of how to prove that this is true for arbitrary $n$ and $m$ (obviously, here we are allowed to use as much machinery as needed). REPLY [2 votes]: Likely it is not known how to answer your question in general, but here are some comments that perhaps are useful. Note that $S_2$ is massive. If we put $S_2(X) = \{x \in S_2: x \leq X\}$, then $S_2(X) \sim c_0 X (\log X)^{-1/2}$, where $c_0$ is Landau's constant. Thus it is very easy for $S_2(X)$ to ``accidentally" contain elements of $S_3(X)$, which has density $\sim c_1 X^{2/3}$ for some explicit constant $c_1$ (this was first computed in a paper by T. Wooley). Note that $S_6(X)$ has density $c_2 X^{1/3}$ (see https://link.springer.com/article/10.1007%2Fs00208-019-01855-y) for some $c_2 > 0$, so heuristically most of these occurrences will not be accounted for by the second phenomenon. In general, for each $m$ there exists a constant $c_m$ such that $S_m(X) \sim c_m X^{2/m}$. When $m,n$ are both large, i.e. when $\frac{2}{m} + \frac{2}{n} < 1$, then it is unlikely that $S_m(X), S_n(X)$ intersect at all, unless there is a reason for them to. The reason of course is that $S_{mn}(X)$ is contained in both. Another comment is that when $\min\{m,n\} \geq 5$ the surface $x^m + y^m = u^n + v^n$ is of general type, so conjecturally the set of rational points is not Zariski dense. Since there is an obvious subvariety containing lots of points (corresponding to $S_{mn} \subset S_m \cap S_n$), the naive conjecture is that this subvariety contains all of the points.<|endoftext|> TITLE: Surmounting set-theoretical difficulties in algebraic geometry QUESTION [21 upvotes]: The category $\text{AffSch}_S$ of affine schemes over some base affine scheme $S$ is not essentially small. This lends itself to certain set-theoretical difficulties when working with a category $Sh(\text{AffSch}_S)$ of abelian sheaves on $\text{AffSch}_S$ with respect to some Grothendieck topology. In fact, many definitions of the notion of category would not consider this a category at all. Nevertheless, in some sense, such a category of sheaves should be somewhat like a presentable category; it should have a collection of generators indexed by $\text{AffSch}_S$. As such, I would like to be able to use arguments involving theorems like the adjoint functor theorem. For example, I would like to show that for an affine scheme $c$, the evaluation functor $\mathcal{F}_c: Sh(\text{AffSch}_S) \to Ab$ given by $\mathcal{F}_c(F)=F(c)$ has a left-adjoint. If $Sh(\text{AffSch}_S)$ were presentable, this would follow from the adjoint functor theorem for presentable categories. Even though $\text{AffSch}_S$ is not essentially small, can we still expect statements such as the adjoint functor theorem to hold for $Sh(\text{AffSch}_S)$? I know that in many cases, one is able to restrict to some sufficiently large small subcategory of the category of affine schemes, but I'm not sure how to do it in this case. To make matters worse, certain "small" sites such as the small $\text{fpqc}$ site over a scheme are not even essentially small, so when working with an arbitrary topos associated to $\text{AffSch}_S$ it seems difficult to restrict to small subcategories for many purposes. REPLY [19 votes]: Let me start by discussing a bit the option of having a large class of generators. You might be interested in the notion of locally class-presentable. To be precise here, I need to be a bit set-theoretical, thus, let me start with an informal comment. Informal comment. Indeed your category is locally class-presentable, class-accessibility is a very strong weakening of the notion of accessibility. On one hand, it escapes the world of categories with a dense (small) generator, on the other, it still allows to build on the technical power of the small object argument via a large class of generators. Locally class-presentable and class-accessible categories, B. Chorny and J. Rosický, J. Pure Appl. Alg. 216 (2012), 2113-2125. discusses a part of the general theory of class-accessibility and class-local presentability. Unfortunately, the paper is designed towards a homotopical treatment and thus insists on weak factorization systems and injectivity but a lot of techniques coming from the classical theory can be recast in this setting. Formal comment. To be mathematically precise, your category is locally large, while locally class-presentable categories will be locally small. Here there are two options, the first one is to study small sheaves $\mathsf{Shv}_{\text{small}}(\text{AffSch})$, this is a full subcategory of the category of sheaves and contains many relevant sheaves you want to study. In the informal comment, this is the locally class-presentable one. Two relevant paper to mention on this topic are: Exact completions and small sheaves, M. Shulman, Theory and Applications of Categories, Vol. 27, 2012, No. 7, pp 97-173. Limits of small functors, B. J. Day and S. Lack, Journal of Pure and Applied Algebra, 210(3):651-683, 2007. The other option is the one of being very careful with universes, in fact restricting to small presheaves might destroy sometimes your only chance of having a right adjoint. It would be too long to elaborate on this last observation here. As a general remark, small presheaves will give you the free completion under small colimits, while all presheaves will give you the free completion under large colimits, how big you need to go depends on the type of constructions that you need to perform. Coming to the adjoint functor theorem, let me state the most general version that I know of. Since it is an if and only if, I hope it provides you with a good intuition on when one can expect a right adjoint to exists. The dual version is true for functors preserving limits. Thm. (AFT) Let $f: \mathsf{A} \to \mathsf{B}$ be a functor preserving colimits from a cocomplete category. The following are equivalent: For every $b \in \mathsf{B}$, $\mathsf{B}(f\_,b): \mathsf{A}^\circ \to \mathsf{Set}$ is a small presheaf. $f$ has a right adjoint. This version of the AFT is designed for locally small categories and can be made enrichment sensitive - and thus work also for locally large categories - using the proper notion of smallness, or equivalently choosing the correct universe. Unfortunately, I do not know a reference for this version of the AFT. Indeed it can be deduced by the too general Thm 3.25 in On the unicity of formal category theories by Loregian and myself, where it appears as a version of the very formal adjoint functor theorem by Street and Walters in the language of the preprint. Finally, about to the evaluation functor, I am not an expert of the abelian world, but it looks to me that one can mimic the argument presented in the accepted answer to this question (at least if the topology is subcanonical). Thus, the left adjoint should indeed exist. https://math.stackexchange.com/questions/2187846/adjoints-to-the-evaluation-functors This answer is closely connected to this other.<|endoftext|> TITLE: How can I simplify this sum any further? QUESTION [11 upvotes]: Recently I was playing around with some numbers and I stumbled across the following formal power series: $$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)$$ I was able to "simplify" the above expression for $a=1$: $$\sum_{k=0}^\infty\frac{x^k}{k!}\cdot2^k=e^{2x}$$ I also managed to simplify the expression for $a=2$ with the identity $\sum_{i=0}^\infty\frac{x^{2k}}{(2k)!}=\cosh(x)$: $$\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\biggl(\sum_{l=0}^k\binom{2k}{2l}\biggr)=\mathbf[\cdots\mathbf]=\frac{1}{4}\cdot(e^{2x}+e^{-2x})+\frac{1}{2}=\frac{1}{2}\cdot(\cosh(2x)+1)$$ However, I couldn't come up with a general method for all $a\in\Bbb{N}$. I would be very thankful if someone could either guide me towards simplifying this expression or post his solution here. REPLY [7 votes]: The explicit formula is as follows: $$ S_a=\frac{1}{a^2}\left(\sum_{z^a=2^a}+2\sum_{p_a(z)=0}\right)e^{az} $$ where the polynomials $p_a$ are given by A244608. For example, \begin{align} p_9(z)&= 1 - 13604 z^9 - 13359 z^{18} + 247 z^{27} + z^{36}\\ p_{10}(z)&=3125-383750 z^{10}-73749 z^{20}+502 z^{30}+z^{40} \end{align} whose roots in the complex plane look like follows: The first few solutions are \begin{align} S_1&=e^{2 x}\\ S_2&=\frac{e^{-2 x}}{4}+\frac{e^{2 x}}{4}\\ S_3&=\frac{2 e^{-x}}{9}+\frac{e^{2 x}}{9}+\frac{2}{9} e^{\frac{x}{2}-\frac{1}{2} i \sqrt{3} x}+\frac{2}{9} e^{\frac{x}{2}+\frac{1}{2} i \sqrt{3} x}+\frac{1}{9} e^{-x-i \sqrt{3} x}+\frac{1}{9} e^{-x+i \sqrt{3} x}\\ S_4&=\frac{e^{-2 x}}{16}+\frac{1}{8} e^{(-1-i) x}+\frac{1}{8} e^{(-1+i) x}+\frac{1}{16} e^{-2 i x}+\frac{1}{16} e^{2 i x}+\frac{1}{8} e^{(1-i) x}+\frac{1}{8} e^{(1+i) x}+\frac{e^{2 x}}{16} \end{align} as given by \begin{align} p_1(z)&=0\\ p_2(z)&=0\\ p_3(z)&=1+z^3\\ p_4(z)&=4+z^4\\ p_5(z)&=-1+11z^5+z^{10}\\ p_6(z)&=-27+26z^6+z^{12} \end{align} etc. Quoting the OEIS entry, the coefficients are found as follows: Let $\omega$ be a primitive $j$-th root of unity. Let $L(k)=\sum_{p=0}^{j-1} c(p)\omega^{kp}$ with $c(0)=2$ and $c(i)=C(j,i)$ if $i>0$. Then $p(j,X)=(X-L(1))(X-L(2))\dots(X-L([(n-1)/2]))$. REPLY [2 votes]: Actually the true generalisation for your $a=1$, $a=2$ cases is this expression for your sum: $$\frac{1}{a^2}\sum_{s=0}^{a-1}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)}.$$ where $\omega$ is an $a$-th root of unity ($\omega^a=1$, $a \neq 1$). When $a=1$ we have $\omega=1$ and your sum is $e^{2x}$. For $a=2$ we have $\omega=-1$ and your sum is $\frac{1}{4}(e^{2x}+e^{-2x}+e^0+e^0)=\frac{1}{4}(e^{2x}+e^{-2x}+2)$. Both agreeing with your observations. Proof: Using the fact that $\sum_{l=0}^{a-1}\omega^{lr}=0$ unless $a|l$ we can show: $$\sum_{l=0}^k\binom{ak}{al}=\frac{1}{a}\sum_{s=0}^{a-1}(1+\omega^s)^{ak}$$ and$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}=\frac{1}{a}\sum_{r=0}^{a-1}e^{\omega^r x}.$$ Substituting the above expressions into your sum we obtain: $$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)=\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\frac{1}{a}\sum_{s=0}^{a-1}(1+\omega^s)^{ak}=\frac{1}{a}\sum_{s=0}^{a-1}\sum_{k=0}^\infty\frac{(x(1+\omega^s))^{ak}}{(ak)!}$$$$=\frac{1}{a}\sum_{s=0}^{a-1}(\frac{1}{a}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)})=\frac{1}{a^2}\sum_{s=0}^{a-1}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)}.$$<|endoftext|> TITLE: Is $C^{\infty}(M)$ dense in weighted Sobolev space $W_{X}^{1}(M)$? QUESTION [8 upvotes]: Let $M$ be a compact manifold without boudary and let $X_{1},\ldots,X_{m}$ be smooth vector fields on $M$. Consider the following weighted Sobolev space: $$ W_{X}^{1}(M)=\{f\in L^{2}(M)|X_{j}f\in L^2(M)， 1\leq j\leq m\}.$$ We can prove that $W_{X}^{1}(M)$ is a Hilbert space. My question is: Can we claim that $C^{\infty}(M)$ dense in $W_{X}^{1}(M)$? I found some results about the above question. For a bounded domain $\Omega$ in $\mathbb{R}^n$, the Meyers-Serrin theorems for function spaces associated with a family of vector fields were studided by N. Garofalo and D.M. Nhieu in [1], which shows that the space $$\overline{C^{\infty}(\Omega)\cap W_{X}^{1}(\Omega)}^{\|\cdot\|_{W_{X}^{1}}}=W_{X}^{1}(\Omega).$$ Does this result also hold for compact manifolds without boudary? Thank you very much! [1] Garofalo, Nicola; Nhieu, Duy-Minh, Lipschitz continuity, global smooth approximations and extension theorems for Sobolev functions in Carnot-Carathéodory spaces, J. Anal. Math. 74, 67-97 (1998). ZBL0906.46026. REPLY [9 votes]: The density result is true for any family of vector fields with Lipschitz coefficients. Theorem. Let $X_1,\ldots,X_k$ be a system of vector fields with Lipschitz coefficient on a compact Riemannian manifold (with or without boundary), If $f\in L^p(M)$ and $X_j\in L^p(M)$, $j=1,2,\ldots,k$, then there is a sequence of smooth functions $f_i\in C^\infty(M)$ such that $$ \Vert f-f_i\Vert_{L^p}+\sum_{j=1}^k \Vert X_j f- X_j f_i\Vert_{L^p}\to 0 $$ as $i\to\infty$. This result is due to Friedrichs. For a short proof as well as relevant references, see Theorem 11.9 in P. Hajlasz, P. Koskela, Sobolev met Poincare, Memoirs Amer. Math. Soc. 688 (2000). In that paper the result is proved on domains in $\mathbb{R}^n$, but as it is indicated at the beginning of the proof, partition of unity allows you to assume that the function has compact support and with that argument you can also prove the result on compact manifolds. The proof is tricky, but short (half page long) and rather elementary.<|endoftext|> TITLE: Extension of Baire's Theorem QUESTION [6 upvotes]: Let $X$ be a topological space, $\kappa$ be a cardinal number, such that there exists a dense subset $A\subseteq X$ of cardinality $\kappa$ but there does not exist a dense subset $A'\subseteq X$ of cardinality less than $\kappa$. Now, suppose that $X$ is a metric space which satisfies the above property for $\kappa\geq \aleph_0$. In this generality is there a Baire-type theorem stating that if $(A_{i})_{i \in I}$ is a collection of dense open subsets of $X$ each of Cardinality at most $\kappa$ and $I$ is of cardinality $\kappa$ then $ \cap_{i \in I} A_i \neq \emptyset? $ REPLY [14 votes]: Such a hypothetical Baire theorem is not true: for every cardinal $\kappa$ of uncountable cofinality and any cofinal subset $C\subset \kappa$ of cardinality $|C|=\mathrm{cf}(\kappa)$, the Hilbert space $\ell_2(\kappa)$ of density $\kappa$ is a complete metric space that can be written as the union $\bigcup_{\alpha\in C}\ell_2(\alpha)$ of $\mathrm{cf}(\kappa)$ many closed nowhere dense subsets $\ell_2(\alpha)=\{x\in\ell_2(\kappa):x^{-1}(\mathbb R\setminus\{0\})\subset [0,\alpha)\}$. Then the intersection $\bigcap_{\alpha\in C}U_\alpha$ of the dense open sets $U_\alpha=\ell_2(\kappa)\setminus\ell_2(\alpha)$ is empty. The same concerns the closed unit ball $B$ of $\ell_2(\kappa)$. In the weak topology, $B$ is (uniform Eberlein) compact, which can be written as the union of $\mathrm{cf}(\kappa)$ many closed nowhere dense subsets. So, the Baire Theorem does not extend even to the intersection of $\aleph_1$ many dense open sets in nice (namely, uniform Eberlein) compact spaces. On the other hand, there are some extensions of the Baire Theorem for topological spaces of countable cellularity, but a metrizable space has countable cellularity if and only if it is separable, so such extensions do not concern metrizable spaces. Namely, the Martin's Axiom is equivalent to the statement: For any compact Hausdorff space $K$ of countable cellularity, the intersection of less than continuum many dense open sets in $K$ is not empty. As a first reading of this topic, you can look at section 1 called "Topology of MA" of the survey paper "Versions of Martin's Axiom" of William Weiss in "Handbook of Set-Theoretic Topology" published by Elsevier in far 1984. Also there exists a well-known cardinal characteristic $\mathrm{cov}(\mathcal M)$ of the continuum, equal to the smallest number of open dense subsets in the real line whose intersection is empty. The value of $\mathrm{cov}(\mathcal M)$ is between $\omega_1$ and $\mathfrak c$, but its exact position in the segment $[\omega_1,\mathfrak c]$ depends on additional axioms of Set Theory. The cardinal $\mathrm{cov}(\mathcal M)$ is one of 10 cardinal characteristics of the continuum, composing the famous Cichon's Diagram.<|endoftext|> TITLE: Can we approximate a vector field on the plane with non-vanishing vector fields in $W^{1,2}$? QUESTION [6 upvotes]: Let $V$ be a compactly-supported vector field on $\mathbb{R}^2$, whose zeros inside some open neighbourhood of the closed unit disk $\mathbb{D}^2$ are isolated. Does there exist a sequence of vector fields $V_n \in C^\infty \cap W^{1,2}$ on $\mathbb{R}^2$, such that $V_n \to V$ in $W^{1,2}$ and the $V_n$ do not vanish on $\mathbb{D}^2$? If we replace the $W^{1,2}$ convergence with $L^2$ convergence, than the answer is positive. The idea is to push the zeroes out of the disk by composing $V$ with a diffeomorphism which affects a region of very small measure. REPLY [3 votes]: Going from the topological index idea: let $C$ be a circle around one of the isolated zeros, and let $D$ be a disk containing $C$. By the trace theorem, your vector fields in $W^{1,2}(D)$ restricts to vector fields $W^{1/2,2}(C)$ on the circle. Since they do not vanish, you can regard them as vector fields $W^{1/2,2}(C, \mathbb{S}^1)$. (If there's any problem with my argument, it would be this step, passing from an $\mathbb{R}^2$ valued function to $\mathbb{S}^1$ valued one.) The $W^{1/2,2}(C,\mathbb{S}^1)$ functions are continuously embedded in $VMO(C,\mathbb{S}^1)$ since $C$ is one dimensional, and one can use the $VMO$-degree theory (originally due to Boutet de Monvel and Gabber, and extended by Brezis and Nirenberg, see this survey by Brezis). The upshot is that for continuous functions the VMO degree coincides with the the usual topological degree, and the $VMO$-degree is continuous under $VMO$-convergence. So I think Pietro's argument against the $W^{2,2}$ case using degree theory should carry over also, telling you that you shouldn't be able to do your approximation. To be more precise: the chain of arguments should go something like: since your $V_n$ are in $C^\infty$ and has no zeros, their corresponding topological degree would be zero, measured either in $VMO$ or classically; but $V_n$ converges to $V$ in $VMO$ (by argument above), and this means that $V$ has to also have degree zero, a contradiction.<|endoftext|> TITLE: Continuity of volume of boundary of Riemannian manifolds in the Gromov-Hausdorff sense QUESTION [6 upvotes]: Let $\{X_i^n\}$ be a sequence of smooth compact Riemannian $n$-dimensional manifolds with boundary. Assume that this sequence has uniformy bounded below sectional curvature, and each $X_i$ is geodesically locally convex near its boundary (the latter assumption is equivalent that $X_i$ is an Alexandrov space). Assume $\{X_i\}$ converges in the Gromov-Hausdorff sense to a compact smooth Riemannian manifold $X$ with boundary and $\dim X=n$, i.e. there is no collapse. Is it true that volume of the bondary $vol_{n-1}(\partial X_i)$ converges to $vol_{n-1}(\partial X)$? The case $n=2$ is already interesting to me. Remarks. (1) By the Perelman stability theorem $X_i$ is homeomorphic to $X$ for $i\gg 1$. (2) By Burago-Gromov-Perelman one has convergence of volumes of manifolds $vol_n(X_i)\to vol_n(X)$. REPLY [3 votes]: [The statement holds for a sequence of extremal susbsets (not necessary boundary).] According to Theorem 1.2. in my "Applications of quasigeodesics and gradient curves", $\partial X_i\to \partial X$ as length spaces in the sense of Gromov--Hausdorff. Then the same argument as in Burago--Gromov--Perelman shows that $\mathrm{vol}\,\partial X_i\to \mathrm{vol}\,\partial X$.<|endoftext|> TITLE: Finiteness of birational types for targets of algebraic fibrations QUESTION [8 upvotes]: Let $X$ be a smooth projective variety. A fibration is a surjective map with connected fibers between projective varieties. Is it true that there's a finite number of birational equivalence classes of projective varieties (say, with representatives $Y_1,\ldots, Y_n$) such that if $X\to Y$ is a fibration then $Y$ is birational to some $Y_i$? REPLY [6 votes]: As Jorge already pointed out this is too much to hope for. On the other hand, if you can put some restriction on $Y$, then there are results in this direction. A theorem of Severi implies that if you restrict $Y$ to be a curve of genus at least $2$, then this is true. (Severi's theorem is slightly more general, requiring only that the map is dominant) Severi's theorem was generalized to the case when $Y$ is a surface of general type by Martin-Deschamps and Lewin-Ménégaux (two people :)... ...and it was generalized to arbitrary dimension ($Y$ is still of general type) by Hacon and McKernan. (See Corollary 1.4 of this paper)<|endoftext|> TITLE: Homotopic classification of maps $M \to \mathbb{RP}^n$ where $M$ is a compact orientable $n$-dimensional manifold QUESTION [13 upvotes]: It is well known that if $M$ is a compact orientable $n$-dimensional manifold, then $[M, \mathbb{S}^n] \cong \mathbb{Z}$, i.e the maps are classified by their degree. What is known about $[M, \mathbb{RP^n}]$ under the same hypotheses? REPLY [16 votes]: This seems to have been worked out in the 1960s by Paul Olum, see Section 1 of Olum, P., Cocycle formulas for homotopy classification; maps into projective and lens spaces, Trans. Am. Math. Soc. 103, 30-44 (1962). ZBL0135.23203. Briefly summarizing, two based maps $f,g:M\to \mathbb{R}P^n$ are based homotopic if they agree on the generators of $H^1(\mathbb{R}P^n;\mathbb{Z}/2)$ and $H^n(\mathbb{R}P^n;\mathbb{Z}^w)$ (twisted cohomology; this has to be interpreted in terms of the difference homomorphism $(f-g)^*$ in twisted coefficients, described in the Appendix), and a certain difference cocycle in $C^n(\mathbb{R} P^n,x_0;\mathbb{Z}/2)$ represents zero in $H^n(\mathbb{R}P^n,x_0;\mathbb{Z}/2)$. The result for free homotopy classes is then deduced from this. So, secondary operations are needed to describe the answer, but this seems to be one of the few cases where this can be done explicitly.<|endoftext|> TITLE: Consequences of eigenvector-eigenvalue formula found by studying neutrinos QUESTION [29 upvotes]: This article describes the discovery by three physicists, Stephen Parke of Fermi National Accelerator Laboratory, Xining Zhang of the University of Chicago, and Peter Denton of Brookhaven National Laboratory, of a striking relationship between the eigenvectors and eigenvalues of Hermitian matrices, found whilst studying neutrinos. Their result has been written up in collaboration with Terence Tao here. From what I understand, although very similar results had been observed before, the link to eigenvector computation had not been explicitly made prior to now. For completeness here is the main result, "Lemma 2" from their paper: Let $A$ be a $n × n$ Hermitian matrix with eigenvalues $\lambda_i(A)$ and normed eigenvectors $v_i$. The elements of each eigenvector are denoted $v_{i,j}$. Let $M_j$ be the $(n − 1) × (n − 1)$ submatrix of $A$ that results from deleting the $j^{\text{th}}$ column and the $j^{th}$ row, with eigenvalues $\lambda_k(M_j)$. Lemma 2. The norm squared of the elements of the eigenvectors are related to the eigenvalues and the submatrix eigenvalues, $$|v_{i,j}|^2\prod_{k=1;k\neq i}^n(\lambda_i(A)-\lambda_k(A))=\prod_{k=1}^{n-1}(\lambda_i(A)-\lambda_k(M_j))$$ I was wondering what are the mathematical consequences of this beautiful result? For example are there any infinite dimensional generalisations? Does it affect matrix algorithms or proofs therein? What about singular values? REPLY [13 votes]: I want to add to the list of places where this identity has been used previously. For a recent one, see arXiv:1710.02181 “State transfer in strongly regular graphs with an edge perturbation”. Equation (2) in Section 2 of this reads \[ \frac{\phi(X\setminus a,t)}{\phi(X,t)} = \sum_r \frac{(E_r)_{a,a}}{t-\theta_r} \] If the eigenvalue $\theta_r$ is simple. $E_r=z_rz_r^T$ and we have formula (2.3) in the pdf given in Carlos's reply. I have used it in a number of places in work on continuous quantum walks, e.g - Lemma 7.1 and Corollary 7.2 in arXiv:1011.0231 "When can perfect state transfer occur?" It is used implicitly in my paper with Brendan McKay: "Spectral conditions for the reconstructibility of a graph", JCT B (30), 1981, 285–289. We missed the connection to neutrinos :-( There is an extensive treatment in Chapter 4 of my book "Algebraic Combinatorics". Chapman & Hall, New York, 1993. Related identities appear in C. A. Coulson and H. C. Longuet-Higgins: "The electronic structure of conjugated systems I. General theory", Proc. Roy. Soc. London A191 (1947), 39–60. They prove that if $G$ is a graph with adjacency matrix $A$ and characteristic polynomial $\phi(G,t)$ and $\phi_{ij}(G)$ is the $ij$-entry of $\phi(G,t)(tI-A)^{-1}$, then \[ \phi_{ij}(G,t) = \sum_P \phi(G\setminus P,t) \] (where the sum is over all paths $P$ in $G$ from $i$ to $j$, and $G\setminus P$ is $G$ with all vertices in $P$ deleted). Further \[ \phi_{ij}(G,t) = \sqrt{\phi(G\setminus i,t)\phi(G\setminus j,t)-\phi(G,t)\phi(G\setminus {i,j},t)}. \] The connection with eigenvectors arises because if $A=\sum_r \theta_r E_r$ is the spectral decomposition of $A$, then \[ \frac{\phi_{ij}(G,\theta_r)}{\phi'(G,\theta_r)} = (E_r)_{i,j}. \] Essentially, the identity relates diagonal entries of spectral idempotents of Hermitian matrices to characteristic polynomials of principal submatrices. The off-diagonal entries are related to Green's functions, and I expect that there are papers in the physics literature where versions of the identity appear.<|endoftext|> TITLE: when is an integer sequence the trace of a monad on FinSet? QUESTION [12 upvotes]: Given $(a_n \in \mathbb{N})$, when is there a monad $T$ on $\mathrm{FinSet}$ such that $$ | T(n) | = a_n\quad\forall n\in \mathbb{N}\:? $$ REPLY [4 votes]: I can give a more categorical description of the construction in Valery Isaev's answer. Given a functor $F$ from $(\operatorname{Finite Sets}, \operatorname{Isomorphisms})$ to $\operatorname{FinSet}$ and a subset $E$ of $F(0)$, we can construct a natural functor from $(\operatorname{Finite Sets}, \operatorname{Injections})$ to $\operatorname{FinSet}$. The construction sends a functor $F$ to the functor $F'$ given by $$F'(S) = \bigcup_{A \subseteq S} F(A)$$ unless $S$ is empty, in which case it is $F(0) \setminus E$. I claim every functor from $\operatorname{FinSet}$ to $\operatorname{FinSet}$, when restricted to isomorphisms, arises this way. To prove this let $G$ be a functor from $\operatorname{FinSet}$ to $\operatorname{FinSet}$, and define $F(S)$ to be the subset of $G(S)$ consisting of elements that do not lie in the image of $G(A)$ for any proper subset $A$ of $S$. Then there is a natural map from $F'(S)$ to $G(S)$ where we send, for each subset $A$ in $S$, $F(A) \subseteq G(A)$ to $G(S)$ under the functoriality map $G(A) \to G(S)$ This map is surjective because each element either lies in the image of some proper subset or doesn't, and we can induct in the first case. To see the map is injective, suppose that $x \in G(S)$ lies in the image of $y \in F(A) \subseteq G(A) $ and $z \in F(B)\subseteq G(B)$ for two subsets $A, B$ of $S$. If $A=B$ then we can apply a left inverse of the inclusion $A \to S$ to see that $y=z$ in $G(A)$ and so $(A,y)= (B,z)$ in $F'(S)$. If $B$ contains a nonempty proper subset which contains $A \cap B$, let $m$ be a left inverse of the inclusion $B \to S$ which sends $A \setminus B$ into this proper subset. Let $i_A$ and $i_B$ be the inclusions of $A $ and $B$ into $S$. Then by functoriality $$ G_{i_A} (y) = x = G_{i_b}(z)$$ and so $$z = G_m ( G_{i_b}(z)) = G_m( G_{i_A}(x)) = G_{m \circ i_A} (x) $$ but $m \circ i_A$ factors through the inclusion of $A \cap B$ into $B$, so $z$ lies in the image of $G(A \cap B)$ which contradicts the assumption that $z \in F(B)$. This second case only fails if $B \subseteq A$ or $B$ has one element. By symmetry, we are also good unless $A \subseteq B$ or $A$ has one element. Because we also handled the cases $A = B$, the remaining cases are when $B$ and $A$ both have one element and are equal. In this case we have an isomorphism between $A$ and $B$ and left inverses show that this isomorphism sends $y$ to $z$, but injectivity can fail. To fix injectivitiy, we move these elements from $F(1)$ into $F(0)$ and place them in $E$ (the set of evil elements). This gives the formula $$| G(n)| = \sum_{k=0}^n { n \choose k} |F(k) | $$ (or $|F(0)|-|E|$ if $n=0$) so if a sequence $a_n$ arises from any functor from finite sets to finite sets (and not just a monad) we must have $$ a_n = \sum_{k=0}^n { n \choose k} b_k $$ for some sequence of natural numbers $b_k$, with $a_0=b_0-e$. This gives the inequalities described by Valery Isaev. Conversely, for any functor $F$ from $(\operatorname{Finite Sets}, \operatorname{Isomorphisms})$ to $\operatorname{FinSet}$ and evil subset $E$, if $F(0)\setminus E \neq 0$ and $F(1)\neq 0$ we can extend $F'$ to a functor from $\operatorname{FinSet}$ to $\operatorname{FinSet}$, and in fact to a monad. To do this, fix $c \in F(0)\setminus E$ and thus in $F'(S)$ for all $s$, and fix $i \in F(1)$ and thus an injective natural transformation $i: S \to F'(S)$ for all $S$. To extend $F'$ to a functor, given a map $f: S_1 \to S_2$, a subset $A\subseteq S_1$, and an element $x \in F(A)$, send $(A,x)$ to $(f(A), f(x))$ if $f$ is injective on restriction to $A$ and to $(\emptyset, c)$ otherwise. Functoriality of this is easily checked. To extend $F'$ to a monad, our unit will be the map $i$ and our multiplication will send an element in $F'(F'(S))$ associated to a subset $A$ of $F'(S)$ and an element $x$ in $F(A)$ to, if $A$ is contained in the image of $i$, the element $(i^{-1}(A), i^{-1}(x)) \in F'(S)$, if $A$ has one element $y$ and $x=i$, the element $y \in F'(S)$, and $c$ in every other case. The compatibility with the left and right unit come from the first two cases. The associativity is not too hard to check. This shows a stronger functorial version of the claim that for any sequence $b_k$, as long as $b_0$ and $b_1$ are nonzero, we can find a monad with sequence $a_n$. Also Valery showed that if $b_1$ is zero, we must have $b_0=0$ or $1$ and all higher $b_i=0$. So the main case of interest is when $b_0=0$ and $b_1$ is not zero. In this case there may be more restrictions on the $b_i$s.<|endoftext|> TITLE: Proving automorphy of the Galois representations of number fields without considering the residual representation QUESTION [5 upvotes]: All the papers proving automorphy of the representations of Galois groups of number fields that I have come across seem to first reduce the representation modulo a prime, prove the automorphy of the residual representation (or make that an assumption) and then infer automorphy of the original representation using deformation-theoretic arguments. Is it possible to prove automorphy (in the number field context) directly, i.e. without proving the automorphy of the residual representation at any prime? REPLY [7 votes]: The canonical answer to that question is certainly the world of so called converse theorems, whose basic ideas go back to Hecke's remark that an holomorphic $L$-function satisfying a suitable functional equation should be automorphic. In the legendary paper Über die Bestimmung Dirichletscher Reihen durch Funktionalgleichungen, Math. Ann. 168 (1967) (legendary in particular because it leaves the modularity of elliptic curves over $\mathbb Q$ as an exercise for the interested reader), André Weil made this precise and deduced that the $L$-function of CM elliptic curves were automorphic. In the language of your question, this shows that Galois representation attached to the torsion points of a CM elliptic curve is automorphic without any consideration of the residual representation. If you want a conditional, modern incarnation of this principle, consider for instance the following theorem. Theorem: Let $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\longrightarrow\operatorname{GL}_{2}(\mathbb C)$ be a continuous representation whose $L$-function $L(\rho,s)$ is holomorphic. Then $\rho$ is automorphic. This theorem (due I believe to Andrew Booker and Muthu Krishnamurty) is proved by a generalization of Weil's converse theorem, and with no consideration of the residual representation. In theory, it applies in situations in which the residual techniques of your question are not at present applicable, namely when $\rho$ is even. On the other hand, I don't believe we have any way to check the hypothesis that $L(\rho,s)$ is holomorphic except by proving that $\rho$ is automorphic, so in that sense the theorem is only a partial success.<|endoftext|> TITLE: Does the sheaf $\mathcal{O}^*$ on a complex manifold have an acyclic cover? QUESTION [8 upvotes]: Let $X$ be a complex manifold and let $\mathcal{O}^*$ be the sheaf nonvanishing holomorphic functions on it. Does it have an acyclic cover? That is, a cover for which all open sets and all intersections are acyclic for this sheaf. According to this related MathSE question, the sheaf $\mathcal{O}$ does since any complex manifold can be covered by Stein manifolds. Therefore, the result would follow if the open sets and all intersections of this cover by Stein manifolds could be taken to be contractible, due of the long exact sequence in cohomology associated to the short exact sequence $0\to \mathbb{Z}\to \mathcal{O}\to \mathcal{O}^*\to 0\,.$ But I don't know if this can be done. REPLY [2 votes]: The answer is Yes for complex manifolds of dimension one. Indeed for any open subset $U\subset X$ the long exact cohomology sequence associated to the exponential sequence you mention yields the fragment $$\cdots\to H^i(U,\mathcal O_X)\to H^i(U,\mathcal O_X^*)\to H^{i+1}(U,\mathbb Z)\to \cdots$$ Now $U$, like any non-compact Riemann surface, is Stein and thus $H^i(U,\mathcal O_X)=0$ for $i\geq 1$. And $ H^{i+1}(U,\mathbb Z)=0$ for $i\geq 1$: for dimension reason if $i\geq 2$ and because $U$ is non-compact $i=1$. In conclusion $H^i(U,\mathcal O_X^*)=0$ ($i\geq 1$) for $U$ non-compact of dimension one, which of course shows that any covering is acyclic for $\mathcal O^*_X$. Remarks a) On a non-compact Riemann surface all holomorphic vector bundles, of whatever rank, are trivial ! This amazing result is Theorem 30.4, page 229, of Forster's wonderful book Lectures on Riemann Surfaces. b) The acyclicity result in the answer is completely false for a smooth algebraic curve $Y$ of dimension one over $\mathbb C$ of dimension one, for example for the algebraic curve underlying a Riemann surface. Indeed for any open (in the Zariski topology!) subset $V\subset Y $, the group $H^1(V, \mathcal O_{Y,\operatorname {alg}}^*)=\operatorname {Pic}_{alg}(V)$ is non denumerable. This means that there are more than denumerably many non-isomorphic algebraic line bundles on $V$ which are all holomorphically trivial! c) See also this answer.<|endoftext|> TITLE: When is the number of areas obtained by cutting a circle with $n$ chords a power of $2$? QUESTION [15 upvotes]: Also posted on the Math Stackexchange: When is the number of areas obtained by cutting a circle with $n$ chords a power of $2$? Introduction Recently, a friend told me about the following interesting fact: Place $n$ points on a circle and draw a line between every pair of points. Suppose that no three lines intersect at one point. Then the number of regions which are separated by the lines is equal to the sum of the first five numbers in the $n-1$st row of Pascal's triangle! See this image image (from Wikipedia). Here, $n$ is the number of points, $c$ is the number of lines and $r_G$ is the number of regions: Here is a great video by 3Blue1Brown on this subject: Circle Division Solution. The series is A000127 in the OEIS. Preliminary results The following is known (see again Wikipedia for instance): For $n$ points, the number of resulting regions is $$1+\binom n2+\binom n4 = \sum_{i=0}^4 \binom{n-1}i=\text{sum of first } 5 \text{ numbers in $n$th row of Pascal's triang.}=\frac{1}{24}n(n^3-6n^2+23n-18)+1.$$ In particular, for $n\in\{1,2,3,4,5,10\}$, the number of areas is a power of $2$. My question Is it true that, for any other $n$, the number of areas is not a power of two? Some attempts First off, we can simply check that for $n\in\{6,7,8,9\}$, the number of areas is not a power of two. So the question is equivalent to: Is it true that, for any $n\geq 11$, the number of areas is not a power of $2$? The following Proposition is easy to prove: Proposition. For $n> 5$, we have that $f(n)< 2^{n-1}$, where $f(n)$ denotes the number of regions. Proof. For $n>5$ we have $$f(n)=\sum_{i=0}^{n-1} \binom{n-1}i-\sum_{i=5}^{n-1}\binom{n-1}i = 2^{n-1}-\sum_{i=5}^{n-1}\binom{n-1}i<2^{n-1}.\square$$ However, this only proves that $f(n)\neq 2^{n-1}$ for any $n>6$. There could still be some $m\in\mathbb N$ with $m2$ we can apply the theorem to show that $f(n)=2^m$ has only finitely many solutions and they can be found effectively. So this at least reduces the problem to a finite search. REPLY [7 votes]: I can prove that there are only finitely many solutions as a consequence of Theorem 2 in The $p$-adic generalization of the Thue-Siegel-Roth theorem (Ridout, Mathematika 2010). Specializing this theorem slightly, it states Let $F(x,y)$ be a homogenous irreducible polynomial with integral coefficients of degree $n \geq 3$. Let $p$ be a prime and let $G_p(x,y)$ be the greatest power of $p$ dividing $F(x,y)$. Let $\kappa > 2$. Then there are only finitely many solutions to $$\frac{|F(x,y)|}{G_p(x,y)} < \max(|x|,|y|)^{n-\kappa}$$ with $GCD(x,y)=1$. Putting $y=1$, we deduce Let $f(x)$ be an irreducible polynomial with integral coefficients of degree $n \geq 3$. Let $p$ be a prime and let $g(x)$ be the greatest power of $p$ dividing $f(x)$. Let $\kappa > 2$. Then there are only finitely many solutions to $$\frac{|f(x)|}{g_p(x)} < |x|^{n-\kappa}.$$ It is quite probable that the version with $y=1$ is much easier, but I didn't find a source in the literature for the easier version. Now, let $$\binom{x}{0} + \binom{x}{1} + \binom{x}{2} + \binom{x}{3} + \binom{x}{4} = \frac{f(x)}{24}.$$ The polynomial $f(x)$ is $x^4-2 x^3+11 x^2+14 x+24$, which is easily checked to be irreducible. So there are only finitely many solutions to $$\frac{|f(x)|}{g_2(x)} < \max(|x|)^{1.9}.$$ But, if $f(x)$ is a power of $2$, then $\tfrac{|f(x)|}{g_2(x)} = 3$. So this can only happen finitely often.<|endoftext|> TITLE: Genus 2 3-manifolds bounding only $X^4$ with $b_2(X^4)$ big? QUESTION [9 upvotes]: The genus of a closed orientable 3-manifold $M^3$ is the minimum genus among all Heegaard splitting surfaces for $M$. Every such 3-manifold bounds a compact 4-manifold. Let $I(M)$ denote the minimum second Betti number $b_2$ amongst all such bounding $X^4$. Is there a sequence of genus 2 3-manifolds $M_n$ such that $I(M_n) \to \infty$. REPLY [2 votes]: I think it follows that a Heegaard genus 2 manifold will bound a topological 4-manifold with $b_2\leq 2$ from a theorem of Steve Boyer. Boyer, Steven, Simply-connected 4-manifolds with a given boundary, Trans. Am. Math. Soc. 298, 331-357 (1986). ZBL0615.57008. MR0857447 I'm guessing that you're interested in the smooth category though?<|endoftext|> TITLE: The homotopy category of the category of enriched categories QUESTION [5 upvotes]: We know that if $\mathcal C$ is a combinatorial monoidal model category such that all objects are cofibrant and the class of weak equivalences is stable under filtered colimits, then $\mathsf{Cat}_{\mathcal C}$, the category of all (small) $\mathcal C$-enriched categories, has the categorical model structure. My question is: Is the homotopy category of $\mathsf{Cat}_{\mathcal C}$, with respect to the categorical model structure, the category of all (small) $\operatorname{Ho}\mathcal C$-enriched categories, $\mathsf{Cat}_{\operatorname{Ho}\mathcal C}$? In other words, does every $\operatorname{Ho}\mathcal C$-enriched category arises from a $\mathcal C$-enriched category? (I am asking this question for the following reason. We know that $(\infty,1)$-categories are schematically categories enriched over the category of homotopy types, which is equivalent to $\operatorname{Ho}\mathsf{SSet}$. If the above question is not true for $\mathcal C=\mathsf{SSet}$, then there exists a $\operatorname{Ho}\mathsf{SSet}$-enriched category that does not arise from a simplicial category. This seems weird, since the theory of simplicial categories is a well-accepted model for $(\infty,1)$-categories. Therefore I expect the answer to the above question to be ''Yes'', at least in the case where $\mathcal C=\mathsf{SSet}$. But I am not able to give a proof, nor can I find any references to this question.) REPLY [5 votes]: Summarizing a discussion in the comments on Harry's answer, we can consider the case $\mathcal C=\mathrm{Gpd}$, with the canonical model structure: weak equivalences are just the equivalences of groupoids, cofibrations are functors injective on objects, and fibrations are isofibrations. A $\mathcal C$-enriched category with one object is then precisely a strictly monoidal groupoid. In contrast, a $\mathrm{Ho}(\mathcal C)$-enriched category with one object is essentially the same thing as an $A_3$-monoidal groupoid-we get a tensor product, which is certainly associative up to non-specified isomorphism, but there is no reason why the associators should satisfy the pentagon identity, and indeed they need not. For a counterexample, consider the $A_3$-monoidal groupoid $G$ freely generated by a single object $x$. For simplicity, let's also make it strictly unital. Then the objects of $G$ are given by rooted binary trees, corresponding to parenthesizations of finite strings of $x$'es. The morphisms are freely generated by the associator isomorphism between the two rooted binary trees of three leaves, together with the functoriality of the tensor product $\otimes:G\times G\to G$ which fuses two trees by adding a new root to their disjoint union. It is in fact possible to describe the morphisms in $G$ without reference to the tensor product: they are sequences of elementary moves, where an elementary move out of a binary tree $T$ is given by taking a node $N$ which is the left child of its parent $P$ and replacing $P$'s right child with $N$, $P$'s left child with $N$'s left child, $N$'s left child with its right child, and $N$'s right child with $P$'s right child. Under this description, one can see that the two different paths from $((xx)x)x$ to $x(x(xx))$ around the pentagon are not equal in $G$. (For comparison, the free strictly unital $A_4$-monoidal groupoid on $x$ has exactly one morphism between two binary trees whenever there exists any such sequence of elementary moves transforming one into the other.) Thus $G$ gives an $A_3$-monoidal groupoid which does not arise from an $A_4$-monoidal one, equivalently, not from a strictly monoidal one. EDIT: As Harry notes, this is not quite a counterexample to your question, as it's not clear that it's impossible to make a different choice of the associators that would extend to $A_4$. It seems unlikely, but also that it would be complicated to make an elementary argument against it. Here is a proposed easier example, coming from $\mathcal C=\mathrm{Cat}$ instead of $\mathrm{Gpd}$, again with the canonical model structure. We let $C$ be the free $A_3$-monoidal category generated by a co-pointed object. So as compared to $G$ from above, $C$ contains a map $p:x\to I$ contracting a leaf which freely generates $C$ from $G$ under $\otimes$ and naturality with respect to the elementary moves. I'll refer to any tensor product of $p$'s and identities as a projection. Call the canonical associators, given by elementary moves, $\alpha$; I claim there is no alternative choice $\beta$ of associators for an $A_3$-structure on $C$, so that $C$'s underlying $\mathrm{Ho}(\mathrm{Cat})$-enriched category arises from no monoidal category. To show that $C$ admits no $A_4$-monoidal structure, note that $\beta_{x,x,x}:(xx)x\to x(xx)$ is uniquely determined as it was in $G$-we've added no new morphisms between isomorphic trees. Furthermore, while morphisms between trees with different numbers of leaves cannot be uniquely written as strings of elementary moves and projections, the only relation on them is naturality of projections with respect to elementary moves. The induced equivalence relation respects lengths of strings of elementary moves and projections, so that any morphism of $C$ has a well defined length given by the length of any representing string of elementary moves and projections. Then to show for instance that $\beta_{xx,x,x}=\alpha_{xx,x,x}$, we can consider the equality $$(px)(xx)\circ \beta_{xx,x,x}=\beta_{x,x,x}\circ ((px)x)x:((xx)x)x\to x(xx).$$ This implies that $\beta_{xx,x,x}$ must be an elementary move, by computing lengths. And there is a unique elementary move $((xx)x)x\to (xx)(xx)$, namely $\alpha_{xx,x,x}$. Similarly one shows all the associators between four-leaf trees are the canonical ones, so that the pentagon cannot commute. If I haven't missed anything in this example, it seems as if its nerve should give rise to an $A_3$-but-not-$A_4$ space. I can't tell if it can be made to give such a groupoid, though.<|endoftext|> TITLE: Whether Krein-Milman property implies Radon-Nikodym property QUESTION [6 upvotes]: A Banach space is said to have Krein-Milman property (KMP in short) if every closed bounded convex set of it is a closed convex hull of its extreme points. Eg. Any reflexive space has KMP, $\ell_1$ has KMP. A Banach space is said to have Radon-Nikodym property (RNP in short) if every closed bounded convex set has slices of arbitrary small diameter. Eg. Any reflexive space has RNP, $\ell_1$ has RNP, a dual separable Banach space has RNP. RNP has many other characterisations in terms of geometrical and also analytical. It can be proved that that RNP implies KMP. Whether KMP implies RNP or not was not known for a long time. Is there any progress in recent past? REPLY [11 votes]: These two properties are equivalent for: Dual spaces (R. E. Huff, P. D. Morris, Proc. Amer. Math. Soc. 49 (1975), 104-108, C. Stegall, Trans. Amer. Math. Soc. 206 (1975), 213-223, C. Stegall, Trans. Amer. Math. Soc. 264 (1981), 507-519); Banach spaces isomorphic to their squares (W. Schachermayer, Studia Math. 81 (1985), 329-339). certain spaces with a shrinking basis (G. López, J. F. Mena, Studia Math. 118 (1996), 11-17). I do not think any significant progress has been obtained since.<|endoftext|> TITLE: rationality of weighted projective space QUESTION [5 upvotes]: A complex weighted projective is $\mathbb{P}(k_1, \cdots, k_{n+1})=Proj(\mathbb{C}[x_1, \cdots, x_{n+1}])$ with $x_i$ of degree $k_i$ (sometimes people ask for each $n$ of the weights being coprime). My first question is whether all weighted projective spaces are rational. The rationality of $\mathbb{P}(k_1, \cdots, k_{n+1})$ is equivalent to the functional field $\mathbb{C}_0(x_1, \cdots, x_{n+1})$ being free generated by $n$ elements. When $n=1$, it is clear that $\mathbb{C}_0(x_1, x_2)=\mathbb{C}(\frac{x_1^{k_2}}{x_2^{k_1}})$, hence $\mathbb{P}(k_1, k_2)$ is always a rational curve. However, I guess in higher dimension this could not hold. If the answer to the first question is no, then my second question is that whether some weaker rationality properties, like rational connectivity, stable rationality, etc hold for weighted projective space. Thank you in advance! REPLY [9 votes]: The $n$-dimensional weighted projective space $X = \mathbb{P}(k_1,\dots,k_{n+1})$ is a toric variety, hence is automatically rational since it contains an isomorphic copy of $\mathbb{G}_m^n$ as a dense open subvariety. Indeed, let $T$ be the torus $\{(t_1,\dots,t_{n+1}) \mid t_i \in \mathbb{G}_m\}/\{(t,\dots,t)\mid t\in \mathbb{G}_m \}$. Then $T$ acts on $X$ via $$(t_1,\dots,t_{n+1}) \cdot [x_1 : \dots :x_{n+1}] = \left[t_1^{k_1}x_1 :\dots :t_{n+1}^{k_{n+1}}x_{n+1} \right]$$ and the orbit of $[1: \dots : 1]$ is isomorphic to $T$, since its stabilizer is a product of cyclic groups.<|endoftext|> TITLE: Why does K-theory need schemes to be Noetherian? QUESTION [11 upvotes]: The definition of K-theory of a scheme $X$ is defined as $G_i(X):=K_i(\mathrm{Coh}(X))$ or $K_i(X):=K_i(\mathrm{Vec}(X))$. But usually the schemes are required to be (at least locally) Noetherian, and usually it is said that if it is not then the $G_i$'s are pretty bad. But for what reasons that we really need the condition of being noetherian? (If it is not, then $\mathrm{Coh}(X)$ is not abelian, but we only need it to be exact, which might be satisfied.) REPLY [21 votes]: You don't need the Noetherianness hypothesis to talk about K-theory. But the definition you propose in your question is not suited for the most general case. From a notion of K-theory we want at least the following properties K-theory of an affine scheme $\mathrm{Spec}\,R$ is given by the algebraic K-theory of projective $R$-modules in the sense of Quillen K-theory satisfies Zariski descent (as a spectrum) Since every scheme has a Zariski cover by affine schemes it is clear that there's at most one definition that satisfies the above two properties. At least when $X$ is quasi compact quasi separated this can be expressed more geometrically as the algebraic K-theory of perfect complexes. An introduction that does not require much background can be found in the classical paper by Thomason and Trobaugh, Higher Algebraic K-theory of schemes and of Derived categories (although if you want to invest some time learning some modern homotopy theory can only be beneficial -- many of the proofs in Thomason-Trobaugh can be simplified if you have more modern technology at your disposal). Similarly for G-theory the "correct" analogue for qcqs schemes is the algebraic K-theory spectrum of bounded pseudocoherent complexes (also known as "almost perfect" complexes in parts of the literature). Both notions of K-theory and G-theory recover Quillen's definition when working on Noetherian schemes (and for K-theory in a much greater generality -- whenever the scheme has an ample family of line bundles).<|endoftext|> TITLE: Constructing computable synthetic differential geometry? QUESTION [14 upvotes]: I'm a computer scientist, not a mathematician, so apologies if I've messed up a lot of things greatly. I've been reading about synthetic differential geometry, and trying to formalize it in Coq. While dealing with the axiomatic specification is quite pleasing, actually constructing a (computable / effective) model of this seems frightfully hard. The simplest model that I could find comes from Differential Geometry in Toposes: Ryszard Paweł Kostecki is $\mathbf{Set}^{\mathbb R- \mathbf{Alg}}$: That is, functors from $\mathbb R$ algebras to $\mathbf{ Set}$. This is quite painful to formalize within Coq, and at the end, I don't think what's left will be computable (since the reals are not computable) My questions are (in descending order of importance) How do I get a computable model of SDG, in the sense that, I should at the end of the whole process be abel to use a computable version of (say) the derivative operator within Coq. Is this possible? If yes, what model of synthetic differential geometry is this? Does restricting to the case of discrete differential geometry make life any easier for me? Is there a study of "synthetic discrete differential geometry"? EDIT: adding more details about what I'm looking for I know that one can impement differentiable programming languages by using implementations of automatic differentiation. There's a categorical interpretation to this, for example, see The simple essence of automatic differentiation. What I'm looking for is a way to perform computational differential geometry. So, not only do I want to be able to be able to calculate the value of $f'(x_0)$ at a given $x_0$, I want to be able to compute the differential of $f$ as a computable function. So, for example, I want there to exist an operator $d: (f : M \rightarrow N) \rightarrow (T_x M \rightarrow T_{f(x)}N)$. Ideally, I want this setup such that I can: prove things about the operator $d$ within the axiomatic system as laid out by SDG (in Coq). Create a computable model that satisfies those axioms (implement the axiomatic system in Coq) Finally, extract out runnable Haskell / OCaml code that allows computable access to things like the differential map $d$, such that when I feed it in $f(x) = x +2$, I should get $d f \equiv 1$ ($\equiv$ reasoning between equality of functions extentionally). I don't know if this is too much of an ask, or indeed, a coherent ask. The goal really for me is to have a verified, computable differential geometry (or at least, discrete differential geometry) library, with proofs that can be done easily, which is the whole point of SDG. REPLY [3 votes]: I still don't quite understand what OP wants, but let me just cite a few papers that I think might be relevant to such questions. First, there are a lot of literature that describe how to work with real numbers in a computationally meaningful way. To give a few examples: Andrej Bauer, Iztok Kavkler, A constructive theory of continuous domains suitable for implementation Ulrich Berger, From coinductive proofs to exact real arithmetic: theory and applications Helmut Schwichtenberg, Constructive analysis with witnesses Michal Konecný, Eike Neumann, Implementing evaluation strategies for continuous real functions Dirk Pattinson, Mina Mohammadian, Constructive Domains with Classical Witnesses I think that ideas presented in these papers can be used to implement a library in a proof assistant with the goal of extracting computable functions on real numbers. If you really want to work with SDG, I'd like to mention that there are other options. Differential λ-calculus was already mentioned in the comments. I also would like to mention tangent categories, which generalize both differential categories (which are a categorical model of differential λ-calculus and SDG). The question is at which level of abstraction you want to work. Differential categories is the simplest setup and tangent categories is the most general since they generalize both differential categories and SDG, but it is also simpler (in some sense) than SDG since it is more abstract. All of these frameworks are abstractions of the usual differential geometry (and also other settings in which the differential operator occurs). The more abstract framework, the more models it has, but it also means that it allows to prove less theorems in specific models. It seems that you do not really care about other models and wants to work formalize just ordinary differential geometry. If this is true, I'd say it does not make sense to start with a complicated framework such as SDG. Abstract framework does not really help if you care about program extraction since you still need to construct specific models in a proof assistant. They are useful if you want to reason abstractly about such structures. If you want to implement a library with a program extraction, I'd suggest to start with a direct implementation of real numbers as described in the papers I cited at the beginning (or any other similar paper). Then you can implement an abstract framework on top of that. You will need a concrete model of such a framework anyway if you want to get a concrete implementation of real functions.<|endoftext|> TITLE: Which cohomology classes are detected by tori? QUESTION [19 upvotes]: Given a space $X$, I am looking for a characterization of classes $\alpha \in H^n(X;\bf Q)$ such that there is a map $f\colon T^n \to X$ so that $f^{\ast} \alpha$ pairs non-trivially against the fundamental class of $T^n = (S^1)^n$. Let us denote the subset of such classes by $T^*(X) \subset H^*(X;\bf Q)$. Here are a few observations that might be helpful: 1) Since there is a degree 1 map $T^n \to S^n$, all classes that are detected by the image of the Hurewicz $\pi_n(X) \to H_n(X;\bf Q)$ are contained in $T_n(X)$. The examples $X = T^n$ and $X= $ a surface of genus $g > 1, n = 2$ show that $T^n(X)$ might contain more than these classes, but does not have to equal all of $H^n(X)$, in general. 2) There are obstructions coming from the algebraic structure of $H^{\ast}(X;\bf Q)$, for instance the above observation about the fundamental class of higher genus surfaces can be proven this way. 3) An application of Gromov's hyperbolization technique shows that every cohomology class can be detected by some aspherical manifold. The question asks when a torus works. REPLY [17 votes]: I'd prefer to work in homology (with coefficients $\mathbb{Q}$ everywhere). I'll say that a class $x\in H_k(X)$ is basically toral if there is a map $f\colon T^k\to X$ sending the fundamental class of $T^k$ to $x$. I'll say that $x$ is toral if it is a linear combination of basically toral classes, and I'll write $T_*(X)$ for the group of toral classes. This is clearly a subfunctor of $H_*(X)$, and $T^*(X)$ is the complement of the annihilator of $T_*(X)$. I'll say that $X$ is toral if $T_*(X)=H_*(X)$. It is easy to see that $T^n$ is toral and thus that $T_*(X)$ is the sum of the images $f_*(H_*(T^n))$ for all maps $f\colon T^n\to X$. This makes it clear that $T_*(X)$ is a subcoalgebra of $H_*(X)$, and also that $T_*(X\times Y)$ contains $T_*(X)\otimes T_*(Y)$. From this it follows that the class of toral spaces is closed under products. It is also easily seen to be closed under disjoint unions. Also, if $f\colon X\to Y$ and $X$ is toral and $f_*\colon H_*(X)\to H_*(Y)$ is surjective then $Y$ is toral. Using this we see that the class of toral spaces is closed under wedges and smash products. Note also that if $G$ is a connected Lie group and $T$ is a maximal torus in a maximal compact subgroup of $G$, then the inclusion $i\colon T\to G$ gives a surjection of rational homology groups. It follows that all connected Lie groups are toral, and the connectivity constraint is easily removed. Next, there is an $H_*$-epimorphism $T^n\to S^n$, so spheres are toral, so products of spheres are toral. Now let $X$ be a connected toral space. Using the James construction, we see that there is a family of maps $X^k\to\Omega\Sigma X$ that are jointly surjective in homology, so $\Omega\Sigma X$ is again toral. There is a standard map $\Omega\Sigma S^2\to\mathbb{C}P^\infty$ that gives an isomorphism in rational homology, so the space $\mathbb{C}P^\infty=BU(1)$ is toral. It follows that for any $n$ the space $BT^n=BU(1)^n$ is toral. By considering maximal tori again, we see that $BG$ is toral for any connnected Lie group $G$. In particular, the infinite Grassmannian $BU(n)=\text{Grass}_n(\mathbb{C}^\infty)$ is toral. Now suppose that $X$ is toral and that $Y$ is a homological $d$-skeleton of $X$, meaning that $Y$ is a $d$-skeleton with respect to some CW structure and that the map $i_*\colon H_*(Y)\to H_*(X)$ is injective. If $y\in H_k(Y)$ then wlog $k\leq d$ and we can express $i_*(y)$ as a linear combination of terms carried by maps $f_j\colon T^k\to X$. All these maps $f_j$ can be deformed so that they land in $Y$, and using this, we see that $y$ is toral. As $y$ was arbitrary we deduce that $Y$ is toral. In particular, we can take $X=\mathbb{C}P^\infty$ to see that the space $\mathbb{C}P^m=\text{Grass}_1(\mathbb{C}^{m+1})=U(m+1)/(U(1)\times U(m))$ is toral. I would guess that all homogeneous spaces $G/H$ (with $G$ a compact Lie group and $H$ a closed subgroup) are toral, but I do not immediately see a proof. This would include the finite Grassmannians $G_k(\mathbb{C}^m)$ for example. Toric varieties would be another interesting test case. Now let $X$ be a connected infinite loop space, so $X\simeq\Omega^\infty T$ for some $0$-connected spectrum $T$. By standard arguments in stable homotopy, we can choose a wedge of spheres $W$ and a map $f\colon W\to T$ that gives an isomorphism $\pi_*(W)\otimes\mathbb{Q}\to\pi_*(T)\otimes\mathbb{Q}$, or equivalently an isomorphism $H_*(W)\to H_*(T)$. This gives a map from the space $Y=\Omega^\infty W$ to $X$ that is surjective in rational homology. Now $Y$ is a filtered colimit of finite products of spaces of the form $QS^{2n+1}$ or $QS^{2n+2}$. There are $H_*$-epimorphisms $S^{2n+1}\to QS^{2n+1}$ and $\Omega\Sigma S^{2n+2}\to QS^{2n+2}$, and using these we see that $Y$ is toral, and thus that $X$ is toral. Again, the connectivity constraint is easily removed. For an interesting example in the opposite direction, I would like to consider the configuration space $F_n\mathbb{C}$ of $n$-tuples of distinct points in $\mathbb{C}$. There is a well-known calculation of the cohomology: it has generators $a_{pq}\in H^1$ for $1\leq p,q\leq n$ subject to $a_{pp}=0$ and $a_{pq}=a_{qp}$ and $a_{pq}^2=0$ and $a_{pq}a_{qr}+a_{qr}a_{rp}+a_{rp}a_{pq}=0$. The space can also be described as $B\Gamma_n$, where $\Gamma_n$ is the braid group on $n$ strings, so $[T^k,F_n\mathbb{C}]=\text{Hom}(\mathbb{Z}^k,\Gamma_n)/\text{conjugacy}$. From this one can probably work out $T_*(F_n\mathbb{C})$ and prove that it is not equal to $H_*(F_n\mathbb{C})$ when $n>2$, but I am not sure of the details. Other discrete groups such as mapping class groups, automorphisms of free groups and $GL_n(\mathbb{Z})$ may also be interesting test cases. The question mentions compact oriented surfaces, which can also be seen as classifying spaces of discrete groups. Along the same lines, we can describe $T^n\vee T^n$ as $B(\mathbb{Z}^n*\mathbb{Z}^n)$ (where $*$ denotes free product of groups). As the free product is highly non-commutative, there are not so many homomorphisms $\mathbb{Z}^m\to\mathbb{Z}^n*\mathbb{Z}^n$. Using this, I think we can show that the sum of the two top classes in $H_n(T^n\vee T^n)$ is not basically toral, so we do not get a good theory by restricting to the basically toral case.<|endoftext|> TITLE: Grauert's Contractibility Theorem QUESTION [5 upvotes]: I am interested in reading the proof of Grauert's Contractibility Theorem, asserting that an integral compact curve in a smooth compact surface (without the projectivity assumption - this is the case I am mostly interested) is exceptional iff it has negative self-intersection. The reference is this: Grauert, H.: Uber Modifikationen und exzeptionelle analytischen Mengen, Math. Ann. 146 (1962), 331-368. Unfortunately I do not read german; all the sources I tried only state the result without proof (e.g. Barth-Hulek-Peters-Van de Ven) or the prove the algebraic version. Does anyone know a reference where I can read the proof, or at least the main ideas or main steps? Thank you. REPLY [4 votes]: Fujiki has proved (in English!) the following generalization of Grauert Theorem in the analytic category. Let $X$ be a complex space, $A$ be an effective Cartier divisor on $X$ and $f:A \longrightarrow A'$ be a proper surjective morphism. Let $L =[A]$ be the line bundle corresponding to $A$. Assume the following conditions: _$L^*|_A$ is $f$-ample, _$R^1f_*(L^*|_A)^{\otimes m} = 0$ for every $m>0$. Then, there exists a birational morphism $g : X \longrightarrow X'$ such that $g|_A = f$. See Theorem 1 page 495 of Fujiki's paper In case $X$ is a compact surface and $A$ is a compact smooth (=integral) curve, the first hypothesis is equivalent to the self-intersection of the curve being negative by the normal bundle formula. The second hypothesis is always satisfied (provided that $L^*|_A$ is ample) if $A$ is a compact smooth curve by Serre duality and the fact the a negative line bundle on a smooth compact curve has no sections. The result above by Fujiki is the exact translation of the same result proved by M. Artin in the category of algebraic spaces. It seems however the proof in the analytic category needs some extra-arguments, as Fujiki relies on the theory of plusri-subharmonic functions.<|endoftext|> TITLE: Krull dimension of a local ring and completion QUESTION [10 upvotes]: Let $A$ be a local ring (not noetherian) of finite Krull dimension such that its maximal ideal $\mathfrak{m}$ is of finite type. Let $\hat{A}$ be its $\mathfrak{m}$-adic completion. Do we have that $\dim(A)=\dim(\hat{A})$? REPLY [3 votes]: There exist examples of commutative local rings of finite dimension whose completions have infinite dimension (all dimensions are Krull, of course). Take $A$ to be a complete valuation ring, whose valuation of non-discrete but of rank 1 (i.e., the value group can be embedded in $\mathbb{R}$, but is not cyclic), and consider the ring $A[X]$. Since $A$ has dimension 1, the dimension of $A[X]$ is at most 3 (since $\dim A[X]$ is bounded by $2\dim A+1$ in general). On the other hand, the completion of $A[X]$ in any ideal containing X is the ring of formal power series $A[\![X]\!]$ (since A itself is already complete). But in this case the dimension of $A[\![X]\!]$ is infinite --- see Krull dimension in power series rings'' by Arnold, Trans of the AMS, 1973. If byideal of finite type'' you mean an ideal generating the same topology as a finitely generated ideal (since here the maximal ideals themselves are not finitely generated), then this answers your question fully, and also shows that the inequality from the remark to the answer does not always hold.<|endoftext|> TITLE: Does the symmetric square L-function vanish at one? QUESTION [8 upvotes]: Take a cuspidal automorphic representation $\pi$ of $GL(3)$ over a number field. My question is quite straightforward and can be related to this one : Can $L(1, \pi, \mathrm{sym}^2)$ be zero? If yes, is there any extra assumption ensuring it cannot? I know that we have the splitting $L(s, \pi \times \tilde{\pi}) = L(s, \pi, \mathrm{sym}^2)L(s, \pi, \wedge^2 \pi)$. If we assume $\pi$ self-contragredient for instance, then the Rankin-Selberg convolution has a simple pole at $s=1$. Either $L(s, \pi, \mathrm{sym}^2)$ has a pole at $s=1$ (Gelbart-Jacquet lift case) and in that case it does not vanish ; or it has none and in that case the question rephrases as : can $L(s, \pi, \wedge^2 \pi)$ have a pole of order 2 at $s=1$? Any new insight or reference on this precise question is welcome. REPLY [10 votes]: For $GL(3)$, the exterior square $L$-function $L(s,\wedge^2\pi)$ is entire as it agrees with $L(s,\tilde\pi\otimes\omega)$, where $\omega$ is the central character of $\pi$. Therefore, $L(1,\mathrm{sym}^2\pi)=0$ would imply that $L(1,\pi\otimes\pi)=0$, contradicting a result of Shahidi (1980). The best known zero-free region for general Rankin-Selberg $L$-functions is due to Brumley (2012): see the Appendix here.<|endoftext|> TITLE: Example of a space X exhibiting the Landweber non-exactness of the additive formal group over the integers? QUESTION [6 upvotes]: Landweber exactness gives a criterion for when a complex oriented cohomology theory $E$ can be recovered from the formal group law over $E_{*}$ determined by the complex orientation. That is it gives a criterion for the following to hold (see for example here) $$E_{*}(X)\cong MU_{*}(X)\otimes_{MU_{*}}E_{*}$$ Since the additive formal group law over $\mathbb{Z}$ is NOT Landweber exact, when $E=H\mathbb{Z}$ this isomorphism cannot hold for all spaces $X$. My question is: what is an example of such a space? REPLY [8 votes]: Firstly, you ask for a space $X$. I will instead talk about finite spectra, but they become spaces if you suspend them enough times, so that does not really make a difference. There is a kind of tautological answer to your question as follows. By the nilpotence technology of Hopkins, Devinatz and Smith, for suitable sequences of natural numbers $i_k$ there are generalised Moore spectra $S/(v_0^{i_0},\dotsc,v_n^{i_n})$ such that $$ MU_*(S/(v_0^{i_0},\dotsc,v_n^{i_n})) = MU_*/(v_0^{i_0},\dotsc,v_n^{i_n}) $$ and there are cofibre sequences $$ \Sigma^{|v_n^{i_n}|} S/(v_0^{i_0},\dotsc,v_{n-1}^{i_{n-1}}) \xrightarrow{} S/(v_0^{i_0},\dotsc,v_{n-1}^{i_{n-1}}) \to S/(v_0^{i_0},\dotsc,v_n^{i_n}) $$ with the obvious effect in $MU$-homology. Any spectrum of the form $S/(v_0^{i_0},v_1^{i_1})$ will be an answer for your question. Examples for this $n=1$ case were already constructed by Adams before the nilpotence theory was available (and this was a big part of the motivation for the nilpotence programme). However, one might ask for a more elementary example. I think that one can proceed as follows. The Hopf map $\eta\colon S^1\to S^0$ has order $2$ and so extends to give a map $\eta'\colon S^1/2\to S^0$. Let $Q$ be the third suspension of the Spanier-Whitehead dual of the cofibre of $\eta'$. This has cells in dimensions $0$, $1$ and $3$. In mod $2$ homology, the bottom two cells are connected by $\text{Sq}^1$ and the top two are connected by $\text{Sq}^2$ so the cell diagram looks like a question mark and the complex is sometimes called the question mark complex. I think it works out that $MU_*Q=MU_*/(2,v_1)x\oplus MU_*y$ with $|x|=0$ and $|y|=3$. On the other hand, we have $H_*Q=\mathbb{Z}/2x\oplus\mathbb{Z}z$ with $x$ mapping to $x$ and $y$ mapping to $2z$, so the map $\mathbb{Z}\otimes_{MU_*}MU_*Q\to H_*Q$ is not surjective. However, the argument is a bit intricate.<|endoftext|> TITLE: Fundamental group under Gelfand duality QUESTION [11 upvotes]: Gelfand duality states that the functor of continuous functions $C(-)$ from compact Hausdorff topological to commutative $C^*$-algebras is an equivalence of categories. In other words, all topological properties of such a topological space $X$ are encoded in the algebraic properties (and of course the norm) of $C(X)$. I have the following related questions: How does one recognize that $X$ is simply connected from the algebraic properties of $C(X)$? How can the fundamental group $\pi_1(X)$ be defined in terms of the algebraic properties of $C(X)$? If $\tilde{X}$ is a universal cover of $X$, what is the algebraic relation between $C(\tilde{X})$ and $C(X)$? REPLY [2 votes]: It seems that this question has been investigated in the literature, where the fundamental group is approached through the notion of a regular covering. For instance, the following reference establishes an equivalence between the categories of regular coverings $\tilde{X} \to X$ and Galois extensions of $C(X)$, meaning that the algebraic analog of a universal cover is a universal Galois extension and the fundamental group is the group associated to that extension: Høst-Madsen, Anders, Separable algebras and covering spaces, Math. Scand. 87, No. 2, 211-239 (2000). ZBL1002.54010. MR1795745. More literature on this topic is cited therein.<|endoftext|> TITLE: Searching for a Thurston paper with egg / 3-manifold analogy? QUESTION [11 upvotes]: I remember coming across some article of Bill Thurston’s where he describes a 3-manifold (with boundary?) as being like an egg. In my recollection the interior of the egg, the shell, and even the shell after being cracked all had meaning. As you can imagine, Googling “like an egg three-manifold Thurston” is not very fruitful. If you happen to remember this article’s title, please point me in the right direction! If I had to guess, I would imagine that this article is more likely to be a research announcement or notes of some kind rather than a paper proving a major result. I guess this because the paper was somewhat “talk-y”. I hope that this paper really exists out there and I haven’t just imagined it! REPLY [15 votes]: I believe you're looking for page 211 of Hyperbolic Structures on 3-Manifolds I: Deformation of Acylindrical Manifolds (Annals of Math., 1986), which includes the following paragraph: A complete description of the three spaces $AH(M)$, $GH(M)$, and $QH(M)$ is certainly not rigorously known, but here is a conjectural image, of which certain features can be rigorously proved. Let us stick to the case that $M$ is a compact, acylindrical manifold. Then $H(M)$ is a hard-boiled egg. The egg complete with shell is $AH(M)$; it appears to be homeomorphic to a closed unit ball. $GH(M)$ is obtained by thoroughly cracking the egg shell on a convenient hard surface. Apparently no material is physically separated from the egg, but many cracks are developed - cracks are dense in the boundary - and at the same time, the material of the egg just inside the shell is weakened, so that neighborhood systems of points on the boundary become thinner. Finally, $QH(M)$ has uncountably many components, which are obtained by peeling off the shell and scattering the pieces all over. Each component is homeomorphic to some Teichmuller space - it is parametrized by Euclidean space of some even dimension. "Most" of the components have dimension zero, for they describe groups whose limit set is all of $S_\infty^2$. Here $H(M)$ is the set of complete hyperbolic manifolds $N$ with a homotopy equivalence $f:M\to N$, $AH(M)$ is $H(M)$ equipped with the 'algebraic topology', $GH(M)$ with the 'geometric topology', and $QH(M)$ with the 'quasi-isometric topology'.<|endoftext|> TITLE: Adjoints for radical and socle functors QUESTION [6 upvotes]: Let $R$ be a ring and $M$ be a $R$-module. Let $rad(M)$ be the radical of $M$, that is, the intersection of all maximal submodules of $M$. Moreover, let $soc(M)$ be the socle of $M$, that is, the sum of all simple submodules of $M$. We know that both rad and soc define covariant subfunctors of $Id:Mod_R\rightarrow Mod_R$. Do radical and socle functors admit left or right adjoints? Thanks in advance for answers. REPLY [8 votes]: In abelian groups: $$\text{soc}\left(\prod_{p\text{ prime}} \mathbb{Z}/p\mathbb{Z}\right) = \bigoplus_{p\text{ prime}} \mathbb{Z}/p\mathbb{Z}\not\cong \prod_{p\text{ prime}}\mathbb{Z}/p\mathbb{Z} = \prod_{p\text{ prime}} \text{soc}(\mathbb{Z}/p\mathbb{Z})$$ so the socle functor does not preserve limits and thus does not have a left adjoint. I also doubt that the radical functor preserves infinite products, but I don't have an example off the top of my head. Also, we have a coequalizer diagram $$\mathbb{Z}\rightrightarrows \mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$$ where the arrows on the left are the identity and the multiplication by $4$ map. We have $\text{rad}(\mathbb{Z}) = \text{soc}(\mathbb{Z}) = \{0\}$ and $\text{rad}(\mathbb{Z}/4\mathbb{Z}) = \text{soc}(\mathbb{Z}/4\mathbb{Z}) = \{0,2\}$. So taking radicals or socles gives $$\{0\}\rightrightarrows \{0\} \to \{0,2\}$$ which is not a coequalizer diagram. So neither functor preserves colimits, and neither has a right adjoint. REPLY [7 votes]: While it does not work for general rings, for Artin algebras one has that the left adjoint of the socle functor is the functor $M \rightarrow M/rad(M)$. I would think that for general rings that is the only choice in case a left adjoint exists.<|endoftext|> TITLE: What are the projective dimensions of big fraction fields? QUESTION [11 upvotes]: Let $A$ be an integral domain, $B$ is its fraction field. Can the projective dimension of the $A$-module $B$ be greater than $1$? This surely cannot happen if the spectrum of $A$ is countable (since then the presentation of $B$ as a countable direct limit of $A$-modules gives a length $1$ projective resolution for it). REPLY [5 votes]: Yes, this can happen. See for example https://projecteuclid.org/download/pdf_1/euclid.nmj/1118801622<|endoftext|> TITLE: Holomorphic extensions of a non-vanishing real-analytic function QUESTION [5 upvotes]: Let f(z) be a holomorphic function defined on an open neighborhood $R$ of the interval $I=[0,1]\subset \mathbb{R}$. Assume $f$ does not vanish on $I$. Then $g(x) = |f(x)|$ is a real-analytic function on $I$, and thus extends to a holomorphic function $g$ on some neighborhood of $R'\subset R$ of $I$. Now, $g$ does not necessarily extend to all of $R$. Take, for instance, $f(z) = z + i$, which is of course entire. Then $g(x) = |f(x)| = \sqrt{x^2+1}$ cannot be extended to any domain including $-i$, as then we hit the branch point of the square-root function at $0$. May we actually construct a region $R'$ where $g$ is guaranteed to be holomorphic, given $R$ and, say, a lower bound on $(\min_{x\in I} |f(x))/(\max_{x\in I} |f(x)|)$? If not, what other conditions could be helpful? Added remark: what if we have a lower bound on $\frac{\min_{z\in R} |f(z)|}{\max_{z\in R} |f(z)|}$? Note: the question came out of a conversation with F. Johannson. The added remark is also his. REPLY [4 votes]: To say something about the size of $R'$, you need a bound for the distance of the nearest zero to the real line. Such a bound can be obtained if your conditions on your function define a normal family in $R$. No reasonable condition involving only restriction of $f$ on $I$ will do this. However one can give bounds in terms of $K:=\min_I|f|/\max_R|f|$. Namely, there exists $\delta>0$ which depends on $K$ and $R$, such that $f$ has no zero in a $\delta$-neighborhood of $I$. To give an explicit estimate of this $\delta$ one needs to know the shape of $R$. To obtain an explicit estimate, you may argue as follows. Let $f$ be a function in the unit disk. $|f(z)|\leq 1$, and $|f(0)|\geq K$. Then by Cauchy, $|f'(z)|\leq 4, |z|\leq 1/2,$ so $$|f(z)-f(0)|\leq 4|z|,\quad |z|\leq 1/2,$$ so $$|f(z)|\geq K-4\delta\quad |z|\leq\delta<1/2.$$ taking $\delta TITLE: The convolution of comonads is a comonad QUESTION [8 upvotes]: $\def\Cat{\mathbf{Cat}}\def\Set{\mathbf{Set}}\def\A{\mathcal{A}}$I stumbled into the following statement: Let $\Cat(\Set,\Set)_s$ be the category of small functors[¹] $F : \Set \to\Set$ and let $F,G$ be two comonads on this category; then, the Day convolution $F * G$ is itself a comonad. I find this both interesting and difficult to prove: I can come up easily with a counit: $$ F * G \overset{\epsilon * \epsilon'}\Rightarrow 1 * 1 = 1 $$ but what about comultiplication? The only way to build a valid enough candidate for a comultiplication $\sigma_{F*G}$ was to consider a rather complicated cowedge $$i_{UV} : [U\times V, A]\times FU\times GV \to F*G(F*G(A))$$ and subsequently, the map induced between coends seems to be a natural transformation $$\sigma_{F*G} : F*G \Rightarrow (F*G)(F*G) $$ it seems however extremely painful to prove that this is coassociative and counital. Do you know this result, and a conceptual reason for it to be true? Or, in the worst case, a counterexample? [¹]: An endofunctor of $\Set$ is called small if it results as the left Kan extension of a functor $\A\to\Set$, where $\A\subseteq\Set$ is a small subcategory, along said inclusion. This restriction is needed in order for $\Cat(\Set,\Set)$ to (exist and to) form a locally small category. REPLY [8 votes]: This is an instance of a standard fact about duoidal categories. Proposition. Let $(\mathcal{V},\ast,J,\circ,I,\ldots)$ be a duoidal category (see e.g. Section 4.2 of Street's lecture notes), and let $(G,\delta,\varepsilon)$ and $(H,\delta,\varepsilon)$ be $\circ$-comonoids in $\mathcal{V}$. Then $G \ast H$ is a $\circ$-comonoid with comultiplication $\require{AMScd}$ \begin{CD} G\ast H @>\delta\ast\delta>> (G \circ G)\ast (H \circ H) @>\gamma>> (G \ast H) \circ (G \ast H) \end{CD} and counit \begin{CD} G\ast H @>\varepsilon\ast\varepsilon>> I \ast I @>\mu>> I. \end{CD} (Here $\gamma$ and $\mu$ are further parts of the duoidal structure on $\mathcal{V}$.) Proof. By the definition of a duoidal category, the functor $\ast \colon \mathcal{V} \times \mathcal{V} \longrightarrow \mathcal{V}$ is opmonoidal (a.k.a. oplax monoidal) with respect to the $\circ$-monoidal structures, and therefore preserves comonoids by a standard argument (which yields the above expressions of the comultiplication and counit). Example. The category of accessible endofunctors of $\mathbf{Set}$ is (normal) duoidal with $\circ$ given by composition and $\ast$ given by Day convolution, and so on. (See Section 8.1 of Garner & López Franco's paper 'Commutativity'.) Since the $\circ$-comonoids in this duoidal category are precisely the accessible comonads on $\mathbf{Set}$, the above Proposition implies that the Day convolution of two accessible comonads on $\mathbf{Set}$ is a comonad on $\mathbf{Set}$ in a canonical way.<|endoftext|> TITLE: Positive-dimensional Seiberg-Witten moduli spaces QUESTION [7 upvotes]: I am looking for examples of (symplectic or not) 4-dimensional manifolds $X$ that have positive dimensional Seiberg-Witten moduli spaces (and $b^{2+}>1$). Of course, the result/conjecture is that the resulting SW invariants should be zero (see page 261 and Corollary 13.22 of Salamon's book). REPLY [6 votes]: Exercise: The non-symplectic $K3\#n(S^1\times S^3)$ for $n>0$ with the spin-c structure induced from the canonical one on $K3$; it will have $n$-dimensional SW moduli. If you want symplectic examples: Assume for simplicity on a symplectic 4-manifold $(X,\omega)$ that $[\omega]\in H^2(X;\mathbb Z)$. A theorem of Donaldson says that for $k\in\mathbb N$ sufficiently large ($k\ge k_0$ where $k_0$ depends not only on the geometry of $X$ but also on $\omega$), $k[\omega]$ is Poincaré dual to an embedded symplectic surface $\Sigma\subset X$. We can build an $\omega$-compatible $J$ (hence many) for which $\Sigma$ is a $J$-holomorphic curve corresponding to the spin-c structure $s_k=s_\omega+k[\omega]$, where $s_\omega$ is the canonical spin-c structure. Then for $J$ generic the $\omega$-perturbed SW moduli space with respect to $s_k$ must be nonempty (via Taubes’ SW = Gr theorem), of even index $k(k[\omega]^2-K_\omega\cdot[\omega])>0$ where $K_\omega$ is the canonical class. Other examples are given in Jabuka's paper "Symplectic surfaces and generic J-holomorphic structures on 4-manifolds".<|endoftext|> TITLE: Why does a principal G-bundle with a discrete structure group G have a unique flat connection? QUESTION [6 upvotes]: I'm reading the Dijkgraaf–Witten paper Topological gauge theories and group cohomology (Comm. Math. Phys. 129 (1990) pp 393–429, doi:10.1007/BF02096988) and on page 395, 2nd paragraph they write Suppose we choose a discrete group $G$. Every principal $G$-bundle has a unique, flat connection, and corresponds to a homomorphism $\lambda : \pi_1(M) \rightarrow G$ There are two parts to this question: why must such a bundle have a flat connection that's unique? And what exactly is the significance of the second part of the statement (the part about the homomorphism)? REPLY [4 votes]: Tsemo's answer addresses the first part of your question. To answer the second, for $G$ discrete, a $G$-bundle $p:P\to M$ is in particular a covering space for $M$, and so every path in $M$ has a unique lift (which happens to be the horizontal lift for the flat connection discussed in Tsemo's answer). So a closed path $\gamma:[0,1]\to M$ based at $x\in M$ lifts to the path $\tilde\gamma:[0,1]\to P$ (starting at some fixed $q\in p^{-1}(x)$). Since $\tilde\gamma(0)$ and $\tilde\gamma(1)$ lie in the same fibre, they differ by an element of $G$, so we can write $\tilde\gamma(1) = \tilde\gamma(0)\cdot \Lambda(\gamma)$. In fact, homotopies on $M$ also lift to $P$, and so $\Lambda$ factors through $\pi_1(M)$ to give a map $\lambda:\pi_1(M)\to G$. It is not too hard to show that $\lambda$ is a homomorphism. $\lambda$ essentially defines $P$ (for $M$ connected): if $\tilde M$ denotes the universal cover of $M$, then $P \simeq \tilde{M}\times_\lambda G$. You can find a nice discussion of covering spaces in Hatcher's Algebraic Topology (particularly Section 1.3, Lifting Properties). The lifting arguments above apply more generally to principal bundles with flat connections (even if $G$ is not discrete): a discussion of flat bundles and holomorphy homorphisms is contained in Section 2.1.4 of Morita's "Geometry of Characteristic Classes".<|endoftext|> TITLE: Translation between formal geometry and rigid geometry QUESTION [8 upvotes]: I'm reading a paper that translates between formal geometry and rigid geometry. In particular, this paper begins with two rigid analytic spaces $A$ and $C$ (each coming from a scheme over $\mathbb{Z}_p$), both defined over $B$, and finds the formal completion of each, calling them $\mathfrak{A}$, $\mathfrak{C}$, and $\mathfrak{B}$ respectively. It then forms the fiber product $X = \mathfrak{A} \times_{\mathfrak{B}} \mathfrak{C}$, and proceeds to treat it as a rigid analytic space, including forming the fiber product of $X$ with another rigid analytic space over $C$. My question is, are the viewpoints of formal geometry and of rigid geometry essentially the same? The formal geometry doesn't seem to do anything else in the entire paper, except be the source for the map $C \to B$. But it's also used in the main reference for this section! Is there a subtle advantage that formal geometry affords us, or some difference in how the fiber product is formed, or is it a quirk of the original paper that this one carried over? Is there some way to add in the algebraic geometry over $\mathbb{Z}_p$ that the spaces we're talking about come from? REPLY [13 votes]: No, these are not the same thing. Formal schemes are to rigid-analytic spaces as $\mathbf{Z}_p$-schemes are to $\mathbf{Q}_p$-schemes. The book Lectures in Formal and Rigid Geometry by Bosch is an excellent and friendly reference on this subject - take a look especially at sections 7.4 and 8.3. In particular, let $K$ be a non-archimedean field (i.e. a field complete with respect to some $\mathbf{R}_{>0}$-valued multiplicative norm) and let $\mathscr{O}_K$ be its valuation ring. Then to any "reasonable" $\mathscr{O}_K$-formal scheme $\mathfrak{X}$, we can associate a rigid-analytic "generic fiber" $X = \mathfrak{X}_K$. (This is literally the generic fiber in the broader context of adic spaces, which subsume both formal schemes and rigid-analytic varieties). We say that a formal scheme $\mathfrak{X}$ with $X = \mathfrak{X}_K$ is a formal model of $X$. It is a deep theorem of Raynaud that formal models of (reasonable) rigid-analytic spaces always exist, and are unique up to the operation of "admissible formal blowing up" (more precisely, the category of reasonable rigid-analytic spaces over $K$ is equivalent to the localization of the category of reasonable formal schemes over $\mathscr{O}_K$ with respect to this operation). One warning: this "generic fiber" operation is not compatible with the usual one for schemes under analytification and formal completion. For example, consider the affine $\mathbf{Z}_p$-line $\mathrm{Spec}(\mathbf{Z}_p[T])$. Its generic fiber is the affine $\mathbf{Q}_p$-line $\mathrm{Spec}(\mathbf{Q}_p[T])$. The analytification is the rigid-analytic affine line, which includes all elements of $\mathbf{Q}_p$ as $\mathbf{Q}_p$-points. On the other hand, the formal completion of $\mathrm{Spec}(\mathbf{Z}_p[T])$ at $p$ is the formal unit ball $\mathrm{Spf}(\mathbf{Z}_p\{T\} := \varprojlim \mathbf{Z}/p^n[T])$. The generic fiber of this is the rigid-analytic unit ball, given by the max-spectrum of the ring $\mathbf{Q}_p\{T\} = \mathbf{Z}_p\{T\}[1/p]$. The $\mathbf{Q}_p$-points of this space are the elements of $\mathbf{Z}_p$. EDIT For completeness (i.e. in case my advisor is reading this), let me add a few details: Let $\varpi \in \mathscr{O}_K$ be a pseudo-uniformizer, i.e. a non-zero element with $|\varpi| < 1$. We define rings of restricted power series over $\mathscr{O}_K$ to be $$\mathscr{O}_K\{T_1, \ldots, T_n\} := \varprojlim_n (\mathscr{O}_K/\varpi^n)[T_1, \ldots, T_n] = \left\{\sum_\alpha a_\alpha T^\alpha \in \mathscr{O}_K[[T_1, \ldots, T_n]] \mid a_\alpha \rightarrow 0\right\} $$ Then we define $$ K\{T_1, \ldots, T_n\} := \mathscr{O}_K\{T_1, \ldots, T_n\}[1/\varpi] = \left\{ \sum_\alpha a_\alpha T^\alpha \in K[[T_1, \ldots, T_n]] \mid a_\alpha \rightarrow 0\right\} $$ The "formal closed unit ball" in $n$ variables over $\mathscr{O}_K$ is $\mathrm{Spf} \mathscr{O}_K\{T_1, \ldots, T_n\}$. Note that it is the formal completion of affine $n$-space over $\mathscr{O}_K$ at the special fiber $\{\varpi = 0\}$. A formal scheme $\mathfrak{X}$ over $\mathscr{O}_K$ is admissible if it is locally of the form $\mathrm{Spf} A$ where $A = \mathscr{O}_K\{T_1, \ldots, T_n\}/I$ with $I$ a finitely generated ideal, and $A$ has no $\varpi$-torsion. Essentially, this means that $\mathfrak{X}$ is locally the $\varpi$-adic completion of a flat finite-type $\mathscr{O}_K$-scheme. The rigid-analytic closed unit ball in $n$ variables over $K$ is $\mathrm{Sp} K\{T_1, \ldots, T_n\}$ (as a set, it consists of maximal ideals in this ring). Rigid-analytic spaces are constructed by gluing together spaces of the form $\mathrm{Sp} (K\{T_1, \ldots, T_n\}/I)$, where $I$ is any ideal (automatically finitely generated and closed). The generic fiber functor sends $\mathrm{Spf} (\mathscr{O}_K\{T_1, \ldots, T_n\}/I)$ to $\mathrm{Sp} (K\{T_1, \ldots, T_n\}/I)$. In particular, it sends the formal closed unit ball to the rigid-analytic closed unit ball. It extends to a functor from admissible formal schemes to rigid-analytic spaces. Raynaud's theorem applies once we add additional (very mild) compactness assumptions on both sides: the rigid spaces must be "quasiseparated and quasi-paracompact", and the formal schemes must be "quasi-paracompact". An admissible formal blowing up of an admissible formal scheme is a certain sort of blowup along a closed subset supported on the special fiber.<|endoftext|> TITLE: Existence of operator with certain properties QUESTION [5 upvotes]: I am curious to know the answer to the following question: Does there exist a continuous linear operator on some Banach space $X$ such that $\Vert T \Vert=1$, and $\sigma(T)\supset \{1\}$ is isolated in the spectrum of $T$ even though $\{1\}$ is not in the point spectrum? Or does an operator like that not exist? REPLY [8 votes]: Let $V$ be the Volterra operator, $(Vf)(t)=\int_0^t f(s) ds$, acting on the Hilbert space $L_2(0,1)$, and let us denote $A=(I+V)^{-1}$. Then $\|A\|=1$ and $\sigma(A)=\{1\}$ [Halmos, A Hilbert space problem book, 2nd ed. Problem 190], but $A\neq I$ because $V\neq 0$. $A$ has empty point spectrum because so has $V$.<|endoftext|> TITLE: About "strict" short exact sequences in quasi-abelian subcategory of a derived category QUESTION [5 upvotes]: I'm reading Bridgeland's Stability conditions on K3 surfaces. In Lemma 4.4 there appears a full quasi-abelian subcategory $\mathscr{A} \subset \mathscr{D}$ of a triangulated category $\mathscr{D} = \mathscr{D}(X)$ of a smooth variety $X$. Then he considers a "strict" short exact sequence $$0 \to A \to B \to C \to 0, \tag{$*$}$$ with $A, B, C \in \mathscr{A}$. 1. Question: How exactly does one define (short) exact sequences in a quasi-abelian category? The definition of abelian categories $\operatorname{Ker}(f_i) = \operatorname{Im}(f_{i+1})$ seems problematic, because I don't know how to define the image. In general it seems that $\operatorname{coker ker} f \neq\operatorname{ker coker} f$, see Wikipedia. 2. Question: What does "strict" mean in that context? Does it have anything to do with the first question, or does it just mean that $A \neq 0 \neq B$, i.e. those are non-trivial subobjects / quotients? After that Bridgeland goes on to argue that $f(B) = f(A) + f(C)$, where $f: K(\mathscr{D}) \to \mathbb{R}$ is an additive function. 3. Question: Why does a short exact sequence $(*)$ in $\mathscr{A}$ induce a triangle in $\mathscr{D}$? I know that this is true if $\mathscr{A}$ is abelian, and $\mathscr{D} = \mathscr{D}(\mathscr{A})$ is the derived category of $\mathscr{A}$, but I see no reason why this is true in our case. REPLY [2 votes]: I found the answer two 1. and 2. in Bridgeland's previous work Stability Conditions on Triangulated Categories: Let $\mathscr{A}$ be an additive category with kernels and cokernels. A morphism $f: A \to B$ is called strict, if the canonical map $\operatorname{coker ker} f \to \operatorname{ker coker} f$ is an isomorphism. $\mathscr{A}$ is called quasi-abelian if the pullback of every strict epi is a strict epi, and the pushout of every strict mono is a strict mono. Then a strict short exact sequence is a diagram $$ 0 \to A \xrightarrow{i} B \xrightarrow{j} C \to 0$$ in which $i$ is the kernel of $j$ and $j$ is the cokernel of $i$. In particular $i$ is mono and $j$ is epi, so $\operatorname{ker} i = 0$ and $\operatorname{coker} j = 0$. Hence \begin{align} \operatorname{ker coker} i = \operatorname{ker}j & = i = \operatorname{coker ker} i \\ \operatorname{coker ker} j = \operatorname{coker} i & = j = \operatorname{ker coker} j, \end{align} so both $i$ and $j$ are strict. In Lemma 4.3 Bridgeland then proceeds to prove that in exactly the situation which appears in the K3 surface paper, strict short exact sequences in $\mathscr{A}$ are in one-to-one correspondence to exact triangles in $\mathscr{D}$.<|endoftext|> TITLE: More recent introductory text on Differential Geometry similar to Kobayashi/Nomizu QUESTION [6 upvotes]: We began an introductory course on Differential Geometry this semester but the text we are using is Kobayashi/Nomizu, which I'm finding to be a little too advanced for an undergraduate introductory course in DG. There are also no graded homeworks, quizzes, or exams so a text with solved problems would be preferred. Textbook recommendations for introductory DG books is not a new question here, but I was specifically looking for books that follow a similar formalism as Kobayashi/Nomizu. REPLY [4 votes]: Two (bright new) interesting books by Jean Gallier and Jocelyn Quaintance: Differential Geometry and Lie Groups: A Computational Perspective. Differential Geometry and Lie Groups: A second course. A nice introduction: First Steps in Differential Geometry: Riemannian, Contact, Symplectic, by Andrew McInerney Differential Geometry: Curves - Surfaces - Manifolds, by Wolfgang Kühnel Applied to physics: Differential Geometry and Relativity Theory: An Introduction, by Richard L. Faber General Relativity Without Calculus: A Concise Introduction to the Geometry of Relativity, by Jose Natario Tensors, Differential Forms, and Variational Principles, by David Lovelock & Hanno Rund and many, many more... Update 1: After the comment by @BenMcKay, I found a set of (not as well-known as they should) lectures by Nomizu himself, published by the Mathematical Society of Japan in 1956. These are titled: Lie Groups and Differential Geometry. Just for your information, these notes are about 80 pages long, and has three chapters: Differential manifolds. Connections in fibre bundles. Linear connections. Another useful reference is: Analysis, manifolds and physics Part I: Basics. by Y. Choquet-Buhat, C. DeWitt-Morett and M. Dillard-Bleick. There is a Part II of this book, by the first two authors, focus in the applications.<|endoftext|> TITLE: Where should I search for computations of group cohomology rings of not-too-complicated finite groups? QUESTION [17 upvotes]: A computation I'm trying to make uses as input the cohomology rings of not-too-complicated finite groups in low degrees, and I'd like to determine where to search for preexisting computations. Specifically, I am interested in: Finite groups such as $D_{2n}$; $A_n$ and $S_n$ for $n\le 5$; binary dihedral/tetrahedral/etc. groups. In general, small and/or well-studied groups. The ring structures for their group cohomology with $\mathbb Z$ and $\mathbb Z/2$ coefficients, at least in degrees up to about 6. I don't mind computing these myself, but if they've already been computed it would be nice to know where to look, and these are the kinds of groups whose cohomology rings have presumably already been computed. But when I search for such computations, I can usually find cohomology groups, but not the ring structure. I suspect I'm looking in the wrong places. So, what are some good starting places to search for preexisting computations of the ring structure on group cohomology? Examples of what I might hope for: It would be wonderful if there were a database with this information, but this is a lot to hope for. I've had some luck finding computations in papers which use group cohomology on the way to some other result (e.g. computing bordism groups of $BG$), but there are surely plenty of applications I'm unaware of, so if you know of applications that require these kinds of computations as input, I'd be interested in hearing about them. REPLY [31 votes]: Simon King and David Green maintain a computer calculated computation of the mod p cohmology of many finite $p$-groups ('order at most 128, of all but 6 groups of order 243, and of some sporadic examples of order up to 1024') here. Simon King also maintains computations of the mod p cohomology of certain other finite groups here. There is also an excellent book 'Cohomology Rings of Finite Groups' by Carlson, Townsley, Valero-Elizondo and Zhang.<|endoftext|> TITLE: Multiplication in Thompson's Group F QUESTION [6 upvotes]: Does there exist an algorithmic way to multiply two elements of the Thompson Group F together? Specifically when looking at it from the perspective of pairs of binary trees. To multiply two elements together you need to add carets on the ends of the trees in order to compose them, but is there an algorithm to decide exactly how to add to the trees? If not, is anyone aware of a way to easily go from a pair of binary trees to a word in the generators of the group? Thank you for the help! REPLY [12 votes]: If you want to multiply a pair of trees $(T,T')$ and $(G,G')$, put $T'$ on top of $G$, that is, identify their roots and then if there is a vertex with two carets $a\to (b,c), a\to (x,y)$, identify the edges $(a,b), (a,x)$ and $(a,c), (a,y)$. That is if two pairs of children have the same parent, identify the pairs of children. After all these foldings are done, you get a finite binary tree $T''$, containing $T'$ and $G$ as rooted subtrees. Therefore $T''$ differs from $T'$ by a bunch of carets. Add the corresponding carets to $T$, get a tree $T'''$. The pair $(T''',T'')$ represents the same element as $(T, T')$. Similarly modify $G'$ and $G$ to obtain an equivalent pair of trees $(T'', G'')$. Then the product is $(T''', G'')$. An easy procedure to get from a pair of trees (i.e. a "diagram") to a word in generators is desribed in chapter 5 of my book "Combinatorial algebra: syntax and semantics".<|endoftext|> TITLE: What are the manifolds whose Curvature tensor has a globally vanishing $k$th order covariant derivative QUESTION [7 upvotes]: Let $(M,g)$ be a boundaryless Riemannian manifold whose curvature tensor have the property that there exists $k\geq 2$ such that $\nabla^k R\equiv0$. What is known about such Riemannian manfiolds ? Is there a classification ? I vaguely remember that I came (a long time ago) across a paper that claims that if $(M,g)$ is also known to be irreducible then we must have $\nabla R\equiv 0$. However, I cant find this paper anymore and I m starting to doubt myself so I decided to ask the question here. REPLY [4 votes]: In fact, the result is true for any complete Riemannian manifold as I remember. The result was proved by Katsumi Nomizu and Hideki Ozeki Here Here 2 and here 3 you may find many related interesting results.<|endoftext|> TITLE: Computation of \tau invariant QUESTION [5 upvotes]: I am trying to understand the following inequality, $$0 \leq \tau (K_{+}) - \tau(K_{-}) \leq 1$$ from the following paper by Livingston. \ https://arxiv.org/pdf/math/0311036.pdf . At page 737 , he argues, a negative crossing change converts $-T_{2,3}$ into the unknot.So $(K_{+} \# -T_{2,3} \#-K_{-})$ bounds a disks with 2 double point. Resolving the double point we get a genus 1 surface. I am trying to construct an explicit movie from $K_{+}$ to $T_{2,3} \# K_{-}$ with a genus one cobordism between them. I end up having a genus 2 cobordism instead of 1. Please help me with the movie move. REPLY [3 votes]: See Livingston's "Computations of the Ozsvath-Szabo knot concordance invariant" (https://arxiv.org/abs/math/0311036), Corollary 3. Or see my thesis for a picture of Livingston's cobordism (p. 19, http://lewark.de/lukas/PhDthesis-Lukas-Lewark.pdf).<|endoftext|> TITLE: Adding something to a book from an unpublished paper QUESTION [5 upvotes]: As many of the people that I am spamming in real life might at this point know, I am turning my coend note into a book. I would like to add a few pages taken from a (still) unpublished paper of mine (et al.), but I don't know exactly what is the policy here: considering that the material I want to add is not exactly original; it is considered folklore or at maximum it is just an alternative proof of a known fact, or it is just recomposing a puzzle of pieces scattered in the literature. my coauthor agrees to see that section turned into a chapter of my book. it is not clear to me/us whether, should I decide to take this path, this would compromise the possibility for the paper to be published because it contains non-inedited material. the paper has been submitted, has been rejected, and will probably be resubmitted as soon as we edit it accordingly to the referee comments. It will probably be at the same time I submit the book to the publishing house. the book will not suffer from me not adding these 4-5 pages; only, I believe that those two particular proofs are useful, in-topic, and particularly aesthetically clean. I'd be inclined to see who comes first, and add "the present section already appears in [X, §y]" for suitable choices of $X\in \{journal,book\}$ and $y\in \sum_{x : X} \{sections_x\}$. Considering all this, what's better to do? How free am I? Is it something I can (and should) discuss with the editor? Is this usually something regulated by the contract I signed? Thanks! REPLY [5 votes]: If the overlapping material is being reproduced verbatim (or nearly so), then technically, if you assign the copyright to the book publisher first, then you have to obtain the permission of the book publisher to include it in the paper (or vice versa, if you assign the copyright to the journal first). In practice, I would expect most publishers to be reasonable, but in principle, it isn't guaranteed that the publisher will grant this permission, so to be on the safe side, you may want to clarify this issue ahead of time. Some publishers will let you retain copyright, or at least some rights, over your work. This is the best solution if you can negotiate it. (In fact, it's a good thing in general, even when you don't have plans to publish the material in two different places.) Especially if it's just one chapter of a book, I would be optimistic that the book publisher will let you retain copyright over it. If the material is conceptually the same but you're going to substantially rewrite it so that there are no long passages of text that are exactly the same in both works, then you shouldn't need to worry about copyright. But I'm guessing that this is not the case.<|endoftext|> TITLE: Is there any theory why (for Bitcoin) the discrete logarithm problem is so hard to solve? QUESTION [31 upvotes]: Note I am an active member and contributor at the sister site https://bitcoin.stackexchange.com while studying Bitcoin and as a person who studied mathematics 10 years ago there is one thing I kept asking people and I am getting unsatisfying answers. Why is the discrete logarithm problem (in particular in the case of bitcoin) so hard? In Bitcoin we take a an elliptic curve as the zeros of $E: y^2 = x^3 + 7$ over $F_p$ with: p = FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFF FFFFFFFE FFFFFC2F which has a nice representation as $p = 2^{256} - 2^{32} - 2^9 - 2^8 - 2^7 - 2^6 - 2^4 - 1$ The we chose a generator point (with the standard encoding in hex): g = 04 79BE667E F9DCBBAC 55A06295 CE870B07 029BFCDB 2DCE28D9 59F2815B 16F81798 483ADA77 26A3C465 5DA4FBFC 0E1108A8 FD17B448 A6855419 9C47D08F FB10D4B8 This is used to select a cyclic subgroup $C$ of cofactor $h=1$ and order $n$ with n=115792089237316195423570985008687907852837564279074904382605163141518161494337 We already know there is a group homomorphism $f: F_n \rightarrow C$ with $x\mapsto g^x$ In particular by our construction we know that $f$ has to be bijective but apparently it is very hard to give an analytical construction of the inverse map. In Bitcoin / Computer science we chose $p$ and $g$ in such a way to make $n$ large enough that with pure brute force calculation we won't be able to solve the discrete logarithm. But why can't we construct an inverse map explicitly? Would anyone here please share some insights why the construction of an analytic closed form for this is so hard? REPLY [5 votes]: As commented by the two answers, Pollard Rho is the best known algorithm for discrete logarithms in a generic cyclic group (where no other special structure is used, and no such special structure, e.g., amenability to index calculus). The so called baby step giant step algorithm can also be used with essentially the same time complexity $O(\sqrt{n})$ where $|G|=n,$ as Pollard Rho. Unfortunately the bsgs needs memory of the same order as well, while Pollard Rho requires negligible memory. So, if $p$ has size $b$ bits, the time complexity for both Pollard Rho and bsgs is $O(2^{b/2}),$ and thus still exponential in input size $b.$ The bsgs is based on a very neat idea, see below: Input: $x=g^k,$ where $g$ is a generator of a multiplicative cyclic group of size $n$, say $\mathbb{Z}_p^\ast$ for simplicity. The goal is to recover $k,$ and $g$ is public as well as $p$ and the group operation. Let $m=\lceil \sqrt{n}~\rceil.$ Step 1. Precomputation: Form the list $$L=\{(j,g^{jm}):j=0,1,\ldots,m-1\}$$ and store it sorted on the second component (or you could use a hash table and a lookup to find an entry in step 2 below). Complexity: $O(\sqrt{n}\log n)$ time (with hash sorting time complexity would be $O(\sqrt{n})$ but generally additional memory is needed to control collisions in that case) and $O(\sqrt{n})$ memory. Think of the elements of $G$ in a $\lceil m\rceil \times \lceil m \rceil$ array (with some repeats at the end): $$ \begin{array}{cccccc} 1 & g & g^2 & g^3 &\cdots & g^{m-1} \\ g^m & g^{m+1} & g^{m+2} & g^{m+3} & \cdots & g^{2m-1} \\ \vdots & & & & \vdots\\ g^{m(m-1)} & g^{m(m-1)+1} & \cdots & g^{n-1} & 1 & \cdots\\ \end{array} $$ Step 2. Online Phase Note that the list $L$ has the entries in the first column sorted as integers. Now, form the elements $x,xg,\ldots, xg^i,\ldots,$ sequentially and lookup in $L$ until the element is found in $L$ (clearly this is guaranteed, as long as we continue until $xg^{m-1}$, since this operation spans two consecutive rows of the array starting at $x$ and ending at $x g^m$ which is below $x$ and one position to the left). When we find an element in $L,$ at the $i_0$th iteration of Step 2, we then know $$ xg^{i_0-1}=g^{j_0m} $$ where $i_0,j_0$ are known. Solving we get $x=g^{j_0 m-i_0+1}$ so $k=j_0 m-i_0+1\pmod n.$ Note that if you have a new $x,$ you can just repeat Step 2. One final comment which may be of interest. About 6 years ago there was quite a lot of progress in the DLP algorithms for composite order fields with special exponents (I am quoting a cryptography stack exchange question here below): A recent paper by Göloğlu, Granger, McGuire, and Zumbrägel: Solving a 6120-bit DLP on a Desktop Computer seems to "demonstrate a practical DLP break in the finite field of $2^{6120}$ elements, using just a single core-month". They credit a 2012 paper by Antoine Joux: Faster index calculus for the medium prime case. Application to 1175-bit and 1425-bit finite field for paving the way they explore. In 2013 Joux published A new index calculus algorithm with complexity $L(1/4+o(1))$ in very small characteristic, and very recently announced he "is able to compute discrete logarithms in $GF(2^{6168})=GF({(2^{257})}^{24})$ using less than 550 CPU.hours". This puts some pairing-based cryptographic schemes relying on the hardness of DLP in fields of characteristic 2 at risk, but not prime field based schemes, be it classical integer residue field DLP or elliptic curve DLP.<|endoftext|> TITLE: If $\Omega X \simeq \Omega Y$ then is $X \simeq Y$ for $X,Y$ simply connected? QUESTION [13 upvotes]: If $\Omega X \simeq \Omega Y$ then is $X \simeq Y$ for $X,Y$ simply connected? Assuming $X,Y$ are nice spaces like CW of course. Clearly this is true by Whitehead, but I am looking for a more enlightening proof. REPLY [37 votes]: To expand on my comment: As written (i.e. without requiring a map $f:X\to Y$), this is false in general. For an example, write $S^2$ as homogeneous space, $S^2=SU(2)/U(1)$. This exhibits $S^2$ as the homotopy fiber of a map $$ BU(1)\to BSU(2). $$ The space $BU(1)\times SU(2) \simeq \mathbb{C}P^\infty \times S^3$ is also the homotopy fiber of a map $BU(1)\to BSU(2)$, namely the constant map. $S^2$ and $\mathbb{C}P^\infty\times S^3$ are obviously not homotopy equivalent. But after looping once, both our maps $BU(1)\to BSU(2)$ turn into maps $U(1)\to SU(2)$, i.e. $S^1\to S^3$, and thus become homotopic. So $$ \Omega S^2 \simeq \Omega (\mathbb{C}P^\infty \times S^3). $$<|endoftext|> TITLE: Free models of finitely presented essentially algebraic theories in elementary toposes? QUESTION [8 upvotes]: The following result is well-known in folklore (I think), but I’ve been unable to find a reference in the literature: Let $T$ be a finitely presented essentially algebraic theory, and $\newcommand{\E}{\mathcal{E}}\E$ an elementary topos with NNO. Then there is an initial model of $T$ internal to $\E$. (By essentially algebraic I mean the notion also known as a cartesian theory or finite limit theory.) As with many folklore results, various generalisations of this (to settings weaker than elementary toposes, and giving a monadic left adjoint not just an initial model) should also hold, and are also I think fairly well-known or “clearly straightforward generalisations” to people sufficiently well-versed in the field — I’d be equally interested to hear a reference for any such generalisations. I’d also be interested if anyone can see a way to deduce this directly from some other result(s) in the literature. Proving this by hand isn’t terribly hard (essentially: take the standard construction of a free set model using syntax, and internalise it to an elementary topos), but there’s a fair bit of careful detail-checking to do there (especially when weakening the setting to less than a topos), and I haven’t managed to find a simpler way to deduce it from results in the literature. The closest results I’ve found are: Theorem 7.43 of Johnstone 1977 Topos Theory, due to Lesaffre, which is the special case of single-sorted algebraic theories. Ch.VI of Johnstone–Wraith 1978 Algebraic theories in toposes (in Indexed categories and their applications, LNM 661 1978). This gives some very relevant results, for a class of theories nearly as general as EAT’s (and sufficient for my interests), but doesn’t (as far as I can see) give free models or any result which immediately implies their existence, except in the special case of the theory of categories, which is rather easier than the general case. The material in §B2 of Johnstone’s Sketches of an Elephant, particularly Theorem 2.4.6, the indexed special adjoint functor theorem. One can deduce this result from that theorem, but the application requires (among other ingredients) construction of a separating family, which seems to me not much easier than just constructing free models by hand directly (but perhaps I’m overcomplicating something). REPLY [11 votes]: If you are willing to accept internal argument instead of purely categorical (external) one, a very good reference for this is Palmgren and Vickers' paper: " Partial Horn Logic and cartesian categories". They give a construction of the initial model for "partial horn theories" (which are equivalent to cartesian theories) which is constructive and predicative. They don't completely clarify what framework is needed for their proof, but looking at the paper it seems clear that it can be applied internally in any exact locally cartesian closed category with a natural number objects. I believe (and Steve Vickers seemed to think it was the case as well last time we spoke) their proof also applies within an "arithmetic universe" (a pretopos with parametrized list object) but that is not so easy to extract from the paper. In both case it applies in elementary toposes with NNO. This is explicitly claimed in the introduction of the paper : they mentioned it can be applied internally within the "predicative toposes" of Moerdijk and Palmgren, and this includes elementary toposes with NNO. For the record, I would be quite interested if someone could give a satisfying proof that this construction works in an arithmetic universe.<|endoftext|> TITLE: Non-example for Whitney (a) stratifications QUESTION [7 upvotes]: Given a $C^1$ stratification $\mathscr{S}$ of a $C^1$ manifold $M$, we write $N^\ast \mathscr{S}$ for the union of conormals to the strata. The stratification is said to be Whitney (a) if $N^\ast \mathscr{S}$ is closed. Or equivalently, for strata $X \subseteq \overline{Y}$ and points $x \in X$ and $y \in Y$, as $y \rightarrow x$ the tangent space $T_y Y$ become arbitrarily close to containing $T_x X$ (uniformly over compact subsets of $X$). What's a typical non-example of such a stratification not satisfying Whitney's conditions (a)? (A non-example for Whitney (b) stratification can be found in this question as well as a non-example for Whitney (a) of pairs of manifolds.) REPLY [3 votes]: I do not know of a simpler concrete example (as in the case of Whitney (b) condition) of a non-example for Whitney (a). But, a typical non Whitney (a) is as depicted in the picture. Observe that $X \subset \overline Y$, but $Y$ is not Whitney (a)-regular over $X$. Take for example a sequence of points $\{y_n\}$ converging to $x$ on the edge of the 'turn' of $Y$, then the limit of the tangent spaces at $y_i$'s does not contain the tangent space at $x$ of $X$.<|endoftext|> TITLE: Concrete examples of Freyd-Mitchell embedding QUESTION [5 upvotes]: I originally posted this on math.SE (https://math.stackexchange.com/questions/3438528/concrete-examples-of-freyd-mitchell-embedding) but since it's been a few days I figured I would crosspost it here. If this isn't the right level I'm happy to delete. By the Freyd-Mitchell Embedding Theorem, any Abelian category admits an exact embedding into the category of modules over some ring. I'm not (currently) hoping to learn a proof, but instead I want to know if there are specific cases of this, aside from the obvious ones, which we can work out explicitly. For example: is there a reasonably explicit way to describe the category of sheaves of Abelian groups on some space as a category of modules? Or the category of chain complexes of Abelian groups? Let's say the threshold for explicit is, at minimum, that we can cleanly describe the ring over which everything is a module. REPLY [11 votes]: For some abelian categories it is also very easy to describe such a ring quite explicitly if the category you start with is similar enough to a module category. Let's say you consider $\mathsf{Ch}(A\mathsf{-mod})$ and that $A$ is a $\mathbb{Q}$-algebra. Then $\mathsf{Ch}(A\mathsf{-mod})$ embeds as a full subcategory into $B-\mathsf{mod}$ where $$B:=A \otimes \mathbb{Z}\langle z,d\rangle / ([z,d]=d, d^2=0)$$ by mapping a chain complex $(X_\ast,\partial)\in\mathsf{Ch}(A\mathsf{-mod})$ to $\bigoplus_{n\in\mathbb{Z}} X_n$ and defining the $B$-action by $$\forall x\in X_n: z\cdot x := nx, d\cdot x := \partial(x).$$ In particular $z$ acts diagonalisable on this $B$-module, has spectrum $\mathbb{Z}$ and $X_n$ is exactly the eigenspace of $z$ for the eigenvalue $n$. It is easy to verify that every chain-map defines a $B$-linear map and vice versa so that this construction gives an equivalence between $\mathsf{Ch}(A\mathsf{-mod})$ and the full subcategory of $B\mathsf{-mod}$ consisting of those $B$-modules on which $z$ acts diagonalisable with spectrum $\mathbb{Z}$. One can remove the condition $\mathbb{Q}\subseteq A$ by using slightly more complicated algebras, like $$C:=A \otimes \mathbb{Z}\langle e_n, d | n\in\mathbb{Z}\rangle / (e_i e_j = \delta_{ij} e_i, d e_n = e_{n-1} d, d^2=0)$$ This also algebra also acts on $\bigoplus_n X_n$: $e_n$ is the projection onto the $n$-component and $d x_n =\partial(x_n)$ for all $x_n\in X_n$. In this way $\mathsf{Ch}(A\mathsf{-mod})$ gets identified with the full subcategory of $C\mathsf{-mod}$ of all those modules that satisfy $M=\sum_{n\in\mathbb{Z}} e_n M$. This second construction can be generalised: $C$ is a special case of a category algebra of an $\mathsf{Ab}$-enriched category $\mathsf{Ch}$: It has object set $\mathbb{Z}$ and $Hom(n,m)=\mathbb{Z}\partial$. The category of chain complexes of $A$-modules is nothing else than the functor category (of additive functors) $Fun_\mathbb{Z}(\mathsf{Ch},A\mathsf{mod})$. Generally: Every category that happens to be the functor category (of $k$-linear functors) $Fun_k(\mathsf{C},A\mathsf{-mod})$ from some $k$-linear category $\mathsf{C}$ into a module category can be embedded into $(A\otimes_k k[\mathsf{C}])-\mathsf{mod}$ in a similar spirit. Even more generally: If $\mathsf{A}$ is an abelian category which you already know how to embed into a module category, then you can adapt this construction to get an embedding of $Fun_k(\mathsf{C},\mathsf{A})$ into a module category as well. Even more generally yet: Any category can be embedded into a free $\mathbb{Z}$-linear category (using the same objects but the free abelian group generated by the original Hom-sets as new Hom-sets). Using this construction one can embed the category of all functors $Fun(\mathsf{C},\mathsf{A})$ into a module category. In particular you can do this with the category of open sets $\mathsf{Ouv}_X$ of a topological space $X$ and get an embedding of the category of $A\mathsf{-mod}$-valued sheafs on $X$ into $A \otimes \mathbb{Z}[\mathsf{Ouv}_X]\mathsf{-mod}$.<|endoftext|> TITLE: Is Stoch enriched in Met? QUESTION [9 upvotes]: Let $\mathsf{Stoch}$ denote the Kleisli category of the Giry monad. That is, $\mathsf{Stoch}$ is a category whose objects are measurable spaces and for which a morphism $f\in\mathsf{Stoch}(X,Y)$ is a map sending points in $X$ to probability measures on $Y$; see nLab: Giry monad for details. Let $\mathsf{Met}$ denote the symmetric monoidal category of metric spaces $(X,d)$ and short maps (functions $f\colon X\to Y$ satisfying $d(x_1,x_2)\geq d(fx_1,fx_2)$ for all $x_1,x_2\in X$). Question: Is there a way to enrich $\mathsf{Stoch}$ in $\mathsf{Met}$? In other words, I'm looking for a way to endow each hom-set $\mathsf{Stoch}(X,Y)$ with a metric space structure in such a way that the composition operation is short: $$ d(f_1,f_2)+d(g_1,g_2) \geq d(g_1\circ f_1, g_2\circ f_2) $$ The upshot would be that when working in $\mathsf{Stoch}$, diagrams would not simply either commute or not commute; they would commute up to some distance, and this distance would behave sensibly under "diagram chasing". REPLY [9 votes]: A way to enrich $\mathsf{Stoch}$ could be as follows. Let $X$ be a measurable space and let $PX$ be the space of probability measures over $X$. We can equip $PX$ with the total variational distance, which is given by the following equivalent forms. $$ d(p,q) \;=\; \sup_{A\subseteq X} |p(A) - q(A)| \;=\; \sup_{f:X\to[0,1]} \left| \int_X f\, dp - \int_X f\; dq \right| . $$ Above, and in what follows, the subsets $A\subseteq X$ and the functions $f:X\to\Bbb{R}$ are assumed measurable. Now first of all let $f:X \to Y$ be a measurable (deterministic) function. Denote by $f_*:PX\to PY$ the pushforward of measures (for category theorists, the action of the functor $P$ on the morphisms). We have \begin{align*} d( f_*p, f_*q ) \;&=\; \sup_{B\subseteq Y} |f_*p(B) - f_*q(B)| \\\ \;&=\; \sup_{B\subseteq Y} |p(f^{-1}(B)) - q(f^{-1}(B))| \\\ \;&\le\; \sup_{A\subseteq X} |p(A) - q(A)| \\\ \;&=\; d( p, q ) , \end{align*} which means that $f_*:PX\to PY$ is short for the metric just defined. This will be of use later. We can equip now the sets $\mathsf{Stoch}(X,Y)$ with the "product" or "$L^\infty$" metric construction induced by the one on $PY$, as follows. We use the following notation, a stochastic map $k:X\to PY$ evaluated at a point $x\in X$ gives the measure $k_x\in PY$. Now for stochastic maps $k,h:X\to PY$, we set $$ d(k,h) \;=\; \sup_{x\in X} d\big( k_x, h_y \big) . $$ This number is bounded by $1$. (Thanks to Martin Hairer for pointing this out, see the comments.) We now have to prove that the (Kleisli) composition of $\mathsf{Stoch}$ is short. Given $h:X \to PY$ and $l: Y \to PZ$, their Kleisli composition, which we denote by $lh$, is given by the Chapman-Kolmogorov formula. Explicitly, for every measurable $C \subseteq Z$, we have $$ lh_x(C) \;=\; \int_{PZ} q(C) \,d(l_*h_x)(q) \;=\; \int_{Y} l_y(C) \,dh_x(y) . $$ Now, to see that postcomposition is short, let $h,k:X \to PY$ and $l: Y \to PZ$. We have \begin{align*} d( lh, lk ) \;&=\; \sup_{x\in X} \sup_{C\subseteq Z} | lh_x(C) - lk_x(C) | \\\ \;&=\; \sup_{x\in X} \sup_{C\subseteq Z} \left| \int_{PZ} q(C) \,d(l_*h_x)(q) - \int_{PZ} q(C) \,d(l_*k_x)(q) \right| \\\ \;&\le\; \sup_{x\in X} d\big( l_*h_x , l_*k_x \big) \\\ \;&\le\; \sup_{x\in X} d( h_x , k_x ) \\\ \;&=\; d( h, k ) . \end{align*} (We used the fact that the evaluation map $PX\to \Bbb{R}$ given by $p\mapsto p(A)$ for a measurable set $A\subseteq X$ is measurable for the $\sigma$-algebra on $PX$ in the construction of the Giry monad.) To see that precomposition is short, let also $g: W \to PX$. Then \begin{align*} d( hg, kg ) \;&=\; \sup_{w\in W} \sup_{B\subseteq Y} | hg_w(B) - kg_w(B) | \\\ \;&=\; \sup_{w\in W} \sup_{B\subseteq Y} \left| \int_{X} h_x(B) \,dg_w(x) - \int_{X} k_x(B) \,dg_w(x) \right| \\\ \;&\le\; \sup_{w\in W} \int_{X} \sup_{B\subseteq Y} | h_x(B) - k_x(B) | \,d(g_w)(x) \\\ \;&=\; \sup_{w\in W} \int_{X} d( h_x, k_x ) \,d(g_w)(x) \\\ \;&\le\; d( h_x, k_x ) . \end{align*} Consider now the set $\mathsf{Stoch}(X,Y)\times\mathsf{Stoch}(Y,Z)$. We can equip it either with the metric $$ d \big( (h_1,l_1), (h_2,l_2) \big) \;=\; \max\{ d(h_1,h_2), d(l_1,l_2) \}, $$ which corresponds to the metric of the cartesian (categorical) product of $\mathsf{Met}$, or with the metric $$ d \big( (h_1,l_1), (h_2,l_2) \big) \;=\; d(h_1,h_2) + d(l_1,l_2) , $$ which is the metric of the closed monoidal product of $\mathsf{Met}$ as an (i.e. enhibiting a hom-tensor adjunction). For either choice, the map $\circ: \mathsf{Stoch}(X,Y)\times\mathsf{Stoch}(Y,Z) \to \mathsf{Stoch}(X,Z)$ given by composition is short. This gives the desired enrichment. By the way, if we want talk about diagrams "commuting up to $\varepsilon$", then by definition of the metric, $\varepsilon$ has to be a uniform bound. For example, for the diagram $\require{AMScd}$ \begin{CD} A @>f>> B\\ @V m V V @VV n V\\ C @>>g> D \end{CD} to commute up to $\varepsilon$, in the sense that $d(n\circ f,g\circ m)\lt\varepsilon$, we need that $$ \sup_{a\in A} d\big(n(f(a)),g(m(a))\big) \;\lt\;\varepsilon, $$ where the distance above is the one of $D$. For that to happen, $\varepsilon$ needs to be independent of $a$, which is quite a strong condition in practice (if $A$ is not compact). REPLY [3 votes]: A boring way is to set $d(f,g)=1$ if $f$ and $g$ are different. A slightly less boring way is to take $d(f,g)$ to be the supremum over $x$ of the probability that two independent variables with laws $f(x)$ and $g(x)$ differ. Different functions may then have a small distance if their image consists of distributions with atoms. REPLY [3 votes]: If you're willing to work with generalized metric spaces, then the answer is yes. $\mathsf{Stoch}$ is enriched over the category of convex spaces, which in turn is enriched over $V$-cat as there exists a closed functor from convex spaces to $V$-cat, where $V=([0,\infty], \ge)$ is the SMCC whose objects can be viewed as generalized metric spaces, and arrows viewed as short maps. This all follows from Meng's thesis, proposition 5. (Available at nLab) You can actually say a little more by noting that $\mathsf{Stoch}$ is enriched over the category of super convex spaces, which differs from convex spaces in that one considers countable sums of one, rather than finite sums, and requires the morphisms to preserve the countable sums. (The proof of enrichment over super convex spaces is essentially identical to that found in Meng.) The benefit of working with (coseparated) super convex spaces is that it is a $*$-autonomous category.<|endoftext|> TITLE: Universal example of Lie algebra QUESTION [11 upvotes]: In the recent IAS talk (available here: https://www.youtube.com/watch?v=LeaiPHAh0X0 - from 45:20) Lurie mentioned an (additive monoidal, with all colimits) category $\mathcal E$ together with a Lie algebra object $L_U$ in it such that for any other (additive monoidal, with all colimits) category $\mathcal C$ there is an identification between Lie algebra objects in $\mathcal C$ and functors $F{:}\mathcal E\to\mathcal C$ that preserve monoidal structure and colimits. It looks similar to a classical story of the "walking monoid" $\Delta$ and monoid objects in monoidal categories. I'm interested in some references on this category $\mathcal E$, is it a well known thing ? REPLY [15 votes]: For any operad $O$ there is a symmetric monoidal category $P(O)$ constructed as follow: the set of objects is $\mathbb{N}$ the tensor product is given by addition and the symmetry by the equality $m+n=n+m$ then there is a unique way to define morphisms in such a way that $$Hom(n,1)=O(n).$$ You can look at https://arxiv.org/abs/math/0005197 for a more concrete construction. This is nothing but the free (symmetric monoidal category with an $O$-algebra), or equivalently the free PROP on $O$. Then by construction $1\in P(O)$ is an $O$-algebra and if $S$ is any symmetric monoidal category, an $O$-algebra in $S$ is the same as a symmetric monoidal functor $P(O)\rightarrow S$. In a way this should be thought as a nice way of defining the notion of $O$ algebra in arbitrary symmetric monoidal categories. Formally $P$ is the left adjoint of the forgetful functor from PROPs to operads. Now if $O(n)$ is an abelian group or a vector space you can take the additive/idempotent completion of $P(O)$, then formally add all colimits if you like, or replace everything by $\infty$-stuffs etc.. and you'll get the free symmetric monoidal (adjective) category on $O$ where adjective is whatever world you choose to work in.<|endoftext|> TITLE: Adjoint Selmer groups and Deformation rings QUESTION [5 upvotes]: Let $\rho:\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow \operatorname{GL}_2(\bar{\mathbb{Z}}_p)$ be a $p$-adic Galois representation associated to a $p$-ordinary Hecke eigencuspform, let $\bar{\rho}:\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})\rightarrow \operatorname{GL}_2(\bar{\mathbb{F}}_p)$ denote the residual representation. The adjoint representation $\operatorname{Ad}\rho$ is the module of $2\times 2$ trace-zero matrices over $\bar{\mathbb{Q}}_p$ on which $\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ acts by conjugation. There is a relationship between the Selmer group over $\mathbb{Q}$ of the (characteristic zero) adjoint representation $\operatorname{Ad}\rho$ and a certain universal Galois deformation ring associated to $\bar{\rho}$. Namely, the dual of the adjoint Selmer group over $\mathbb{Q}$ is isomorphic to the module of $1$-forms on the deformation ring at the point corresponding to $\rho$. In the case when $\rho$ is unramified outside $p$, a proof of this statement is in Hida's book "Hilbert Modular forms and Iwasawa theory" (cf. Proposition 1.47). The relationship generalizes to totally real fields. The relationship here tells us that since the deformation ring comes from a Hecke algebra, then indeed the Selmer group of the Adjoint representation over $\mathbb{Q}$ is finite (cf. Proposition 1.53), and the same statement holds over a totally real field. This requires the hypothesis that $\rho$ is unramified outside $p$. A similar idea was exploited by Khare and Ramakrishna to show that if one starts with a residual Galois representation which is odd and absolutely irreducible then it may be lifted to a $p$-adic Galois representation for which the adjoint Selmer group is finite, this does not require the hypothesis that $\rho$ is unramified outside $p$. My question is the following, can one use similar lines of reasoning to say anything about the structure of the adjoint Selmer group over the cyclotomic $\mathbb{Z}_p$ extension of $\mathbb{Q}$? Has this been explored anywhere? REPLY [2 votes]: As far as I know, it is difficult to extract much information about the adjoint Selmer group over the cyclotomic $\mathbb{Z}_p$-extension. If the modular form $f$ corresponding to $\rho$ is ordinary at $p$, then one can deduce straightforwardly from the finiteness of the Selmer group over $\mathbb{Q}$ that the cyclotomic Selmer group is co-torsion; but it doesn't seem to be possible to get much more information than this -- in particular, I am not aware of any way to deduce the Iwasawa Main Conjecture for $Ad(\rho)$ by these methods. The main difficulty is that if you pick an integer $n$ and apply deformation theory to Hilbert modular form given by base-changing $f$ to the $n$-th layer $\mathbb{Q}_n$ in the cyclotomic $\mathbb{Z}_p$-extension, it tells you something about the size of $Sel(Ad(f) / \mathbb{Q}_n)$ as an abelian group, but not about its structure as a $\mathbb{Z}_p[\mathbb{Z}/p^n]$-module (essentially because the deformation theory doesn't "know" that the Hilbert form is a base-change from $\mathbb{Q}$). So you don't get information about characteristic ideals, etc. This is actually rather similar to the situation in the Iwasawa theory of the trivial representation (studying class groups in $\mathbb{Z}_p$-extensions, i.e. the original setting studied by Iwasawa himself). In this setting, the analogue of the results you quote is the analytic class number formula; and again, this only tells you the cardinality of the Selmer group over $\mathbb{Q}_n$, but not its Galois module structure. This isn't enough to deduce the Main Conjecture; but it does allow you to say that if you have proved "half" of the main conjecture, i.e. one inclusion of ideals, then you must in fact have equality. (So you can prove the Main Conjecture over $\mathbb{Q}_\infty$ using any 2 of the following 3 things: the cyclotomic unit Euler system, Eisenstein congruences a la Ribet/Mazur-Wiles, and the analytic class number formula.)<|endoftext|> TITLE: An intuitive explanation for group cohomology via cochains? QUESTION [7 upvotes]: I'm fairly new to topology, and so far I've understood cohomology via cochains. First we build an object called a cochain ($C^n$), then define a differential map that takes you from $C^n$ to $C^{n+1}$. Then all the types of cohomology groups I've encountered can be easily defined as: \begin{equation} H^n(M) = Z^n / B^n \end{equation} Where $Z^n$ is the n-cocyle and $B^n$ is the n-coboundary. The trouble I'm facing is in defining the map $\partial^n: C^n \rightarrow C^{n+1}$ In other cohomologies, the map has an intuitive explanation. For example, in deRham cohomology, the map just sends a differential form to a higher differential form, but due to Stokes' theorem we can interpret the nth form being the boundary of an n+1th form. In simplical homology, we define the map to be the boundary of the n-chain, so a line segment will have the two endpoints as its boundary etc. In group cohomology however, the map is defined as: $f:G^n \rightarrow A$ where A is a G module, and: \begin{equation} \partial^i f(g_0,\cdots,g_i) = g_o f(g_1,\cdots,g_i) + \sum_{j = 1}^i (-1)^j f(g_0,\cdots, g_{j-1} g_{j}, g_{j+1}, \cdots, g_i) + (-1)^{i+1} f(g_0,\cdots, g_{i-1}) \end{equation} Now I understand that you can prove the various "standard" properties of this map using this definition (like $\partial^{i+1} \partial^i f = 0 $ etc.), but what is the intuitive explanation for what's going on? Why is this map defined the way it is? How does this correspond to the boundary interpretation of other types of cohomologies? How can I "see" it being a boundary without explicitly computing $\partial^2 = 0$? And what happens when the groups are continuous? For lie groups can we apply deRham cohomology? What about cohomology on principal G-bundles? Any introductory resources on this subject would be much appreciated! REPLY [4 votes]: Both answers are wonderful, but I have to admit that there's still some gap between my understanding and why modern topologists think of $BG$ in that way. That being said, you can see why group cocycles/coboudaries are defined in that way by looking at a classical problem, at least for small $n$. I will lay out a classical problem, and explain how the cocycle/coboundary conditions show up naturally. For my own good, I hope someone will explain how topologists went from this classical problem to the construction of $BG$ (or bar constructions in general). Anyway, here we start. Let $K$ and $G$ be two finite groups. You can easily form their trivial product $K\times G$. This is another group, and it helps us to understand group theory a bit more because now we can investigate big groups by breaking them down to smaller ones. However, not every big group is a trivial product of smaller groups, so following the same thought we want to find more creative products of $K$ and $G$. One easy example all freshmen know is the semi-direct product, provided that $G$ acts on $K$. It has been successful, so people wondered all possible nontrivial products we can build from $K$ and $G$. This is the so-called group extension problem. More precisely, given two finite groups $K$ and $G$, we want to find all groups $E$, so that $E/K$ is isomorphic to $G$. It is hard to construct $E$ directly, so let's think in an opposite way. We imagine that we have found such extension $E$. Its group structure is still unknown, but as a set it must be $K\times G$. We will try to describe the group structure of $E$ by looking at the $K\times G$ side. This way has its pros and cons. Pros: it's easier to write things down concretely. Cons: the structure of $E$ might be highly twisted, so writing everything down in terms of $K\times G$ might make things messy. (Aside: if you have some experience with differential geometry, the way I'm using now is like drawing a map for $E$. $K\times G$, as a map, gives you a convenient coordinate system for you to write things down, but since $E$ is "curved", we need to develop a corresponding calculus on the map side.) To go between both side, we want to find a good map that sends $K\times G$ to $E$. From $K$ to $E$ is simple, since by assumption (the "imagination" given above) $K$ already sits in $E$. Unfortunately, for the other side, there's no good map from $G$ to $E$ yet. So lets just pick an arbitrary one $s: G\to E$, so that $s$ serves as a section. (Aside: I would appreciate if someone can draw a short exact sequence that will make things clearer.. I'm terrible at drawing with latex.) Note that we cannot expect $s$ to be a group homomorphism! In fact, if that's the case we are back to the semi-direct product, and vice versa (exercise). We want to create more creative products, so we really want some section $s$ that is highly distorted (from being a group homomorphism)! To measure how distorted it is, we introduce a set-theoretic function $f(g_1,g_2)=s(g_1)s(g_2)s(g_1g_2)^{-1}$. Note that $s$ is not distorted if and only if $f$ is trivial. We want $f$ to be as nontrivial as possible. Now, after the set-theoretic section $s$ has been chosen, we have a set-theoretic bijection (exercise) between our "space" $E$ and out "map" $K\times G$: $$ K\times G \to E: (k,g)\mapsto k\,s(g). $$ A fun exercise is to write down how this maps tells us the group structure of $E$. The result is: $$ (k_1,g_1)(k_2,g_2) = (k_1\, ^{s(g_1)}k_2\, f(g_1,g_2), g_1g_2),$$ where $^xk$ means $xkx^{-1}$, which lies in $E$ because $K$ is a normal subgroup of $E$ by assumption (the imagination made above). It should not be surprising that the distortion measure $f$ shows up in our map. Convince yourself! Another fun exercise is to translate the associativity on the "space" side to the "map" side: this simply gives you $$^{s(g_1)}f(g_2,g_3)\,f(g_1,g_2g_3)f(g_1g_2,g_3)^{-1}f(g_1,g_2)^{-1} = 1 \in K.$$ Cocycles If $K$ is abelian, then $f$ is a $2$-cocycle with the action of $G$ on $K$ given by conjugation. This gives a motivation of why cocycles are defined that way. (Aside: It should not be surprising why such horrible formula show up, since $s$ is chosen arbitrarily as a set-theoretic map!) (Aside: Exercise -- what does the other group structures of $E$ translate to? I don't know.. ) Conversely, given a left group action $G$ on $K$ and a $2$-cocycle $f\in Z^2(G,K)$, we can construct a space $E$ simply by reversing the process above. Note that in the space $Z^2(G,K)$ includes the action implicitly. Here is an explicit construction: set $E$ to be $K\times G$ as a set, and define the group operation by $$(k_1,g_1)(k_2,g_2) = (k_1\,^gk_2\,f(g_1,g_2),g_1g_2).$$ It is an easy exercise to write down the inverses and the identity element in this set. Conclusion of the first part: Fix a left group action $G$ on $K$. Any $2$-cocycle $f \in Z^2(G,K)$ gives an interesting product of $K$ and $G$! If $f$ is trivial, then the interesting product is the semi-direct product, which is less interesting. (Remark: to check that $K$ is central.) Coboundaries The next question is of course to find redundancy: do any two given $2$-cocycle $f$ and $h$ give "different" products? If not, how do we identify which two give the same product? Here, by different I mean that they are not the same. Of course they aren't the same set-theoretically, so I should be clearer: two products $E_f=(K\times G)_f$ and $E_h=(K\times G)_h$ are the same if there is a group isomorphism from one to another that respect all given structures. So let's suppose $E_f$ and $E_g$ are the same, i.e. there is a good isomorphism between them. The image of $(e,g)_f \in E_f$ in $E_h$ under this isomorphism must be in the form $(k_g,g)_h \in E_h$, for some $k_g \in K$ (exercise). Lets denote $k_g$ by $\phi(g)$ -- it is a function in $g$ -- it is a function $\phi:G\to K$! The last fun exercise is to show that $f/h$ is the coboundary of $phi$ provided that $K$ is abelian, and vice versa! This exercise is even more fun because you have to find a way to write down the image of $(k,g)_f$ in $E_h$. This gives a motivation of why coboudaries are defined that way. My questions How did modern topologist come up with the idea of bar construction from this kind of classical problems? It's kinda sad that nowadays it's so hard to bridge the classical themes to modern considerations.. at least for me! If we did not assume $K$ to be abelian, we still get something that could be call the nonabelian cocycles/coboudaries, but it is much harder to deal with. How are this done in modern language? And again can you connect the modern treatment of nonabelian cohomology with this classical picture?<|endoftext|> TITLE: Reconciling Sullivan's theorem with the hyperbolic structure of the Figure–8 knot complement QUESTION [5 upvotes]: I am interested in the 3-manifolds with hyperbolic structures from the physics (gravity) perspective. I encounter this paper https://arxiv.org/pdf/hep-th/9812206.pdf whose Eq. (9) mentions a theorem by Sullivan, which states that if a 3-manifold $M$ allows at least one hyperbolic structure, there is a 1-1 correspondence between hyperbolic structures on $M$ and conformal structures on $\partial M$. I want to apply this theorem to the figure-8 knot complement $M=S^3- K$, where $K$ denotes the figure-8 knot. The boundary of $M$ is $\partial M = T^2$. Hence, its conformal structure can be parametrized by a complex number $\tau$ (modulo $SL(2,Z)$). My question is does Sullivan's theorem mean that for any choice of $\tau$ there is a corresponding hyperbolic structure on $M$ (regardless of whether the volume of the $M$ is finite or not)? Thurston's note constructs a hyperbolic structure for the figure-8 knot complement $M=S^3- K$ by gluing two tetrahedra together. The $\tau$ parameter for the boundary conformal structure seems to be uniquely fixed (to be some third root of unit?) in this construction. And the hyperbolic volume of $M$ is finite. If the answer to my question is yes, does it mean that, if $\tau$ deviates from the values given by Thurston's construction, $M$ can still have a hyperbolic structure but the volume has to be infinite? If the answer to my question is no, then what exactly is the statement of Sullivan's theorem? REPLY [7 votes]: The statement is only true if you restrict to geometrically finite hyperbolic metrics (possibly of infinite volume) and ignore parabolic elements, which basically means that you ignore the boundary components of genus < 2. For a precise statement you may look at Section 3.1 of http://www.math.harvard.edu/~ctm/papers/home/text/papers/iter/iter.pdf (I don’t know what‘s the original source for Sullivan‘s theorem.) So in the case of the figure eight knot complement, there is only one hyperbolic structure. In this case, this already follows from Mostow-Prasad rigidity.<|endoftext|> TITLE: Imperfect Tate (cup product) pairing in Galois cohomology? QUESTION [7 upvotes]: Let $k$ be a field of characteristic 0 with a fixed algebraic closure $\bar k$ and absolute Galois group $\Gamma={\rm Gal}(\bar k/k)$. Let $M$ be a finite $\Gamma$-module, that is, a finite abelian group endowed with a continuous action of $\Gamma$. Write $$M^D={\rm Hom}(M,\bar k^\times).$$ We have a canonical pairing $$ M\times M^D\to \bar k^\times,\quad (m,d)\mapsto d(m),$$ which induces a cup product pairing $$ H^0(\Gamma, M)\times H^2(\Gamma, M^D)\to H^2(\Gamma,\bar k^\times)=:{\rm Br}(k), \quad (m, [d_{s,t}])\mapsto [d_{s,t}(m)],$$ where ${\rm Br}(k)$ denotes the Brauer group of $k$, and $(d_{s,t})\in Z^2(\Gamma,M^D)$. Note that $ H^0(\Gamma, M)=M^\Gamma$, the group of fixed points of $\Gamma$ in $M$. Let $x\in H^2(\Gamma, M^D)$. Using the cup product pairing, we obtain a homomorphism $$ x^0\colon M^\Gamma\to {\rm Br}(k),\quad m\mapsto m\,.\, x,$$ where $ m\,.\, x\in {\rm Br}(k)$ denotes the cup product of $m$ and $x$. If $k$ is a $p$-adic field, then by the Tate duality theorem (see Serre, "Galois cohomology", or Milne, "Arithmetic duality theorems", or Harari, "Cohomologie galoisienne et théorie du corps de classes") the cup product pairing is perfect, that is, the homomorphism $$ H^2(\Gamma,M^D)\to {\rm Hom}(M^\Gamma, {\rm Br}(k)), \quad x\mapsto x^0$$ is an isomorphism. It follows that $x^0=0$ if and only if $x=0$. Moreover, the assertion "$x^0=0$ if and only if $x=0$" is true also for any finitely generated $\Gamma$-module $M$. If $k=\bf R$, then using the Tate duality over $\bf R$ (see Milne, "Arithmetic duality theorems") one can show that again $x^0=0$ if and only if $x=0$ for any finitely generated $\Gamma$-module $M$. Question: What is an example of a field $k$ of characteristic 0, a finitely generated $\Gamma$-module $M$, and a cohomology class $x\in H^2(\Gamma, M^D)$ such that $x\neq 0$, but $x^0=0$? REPLY [2 votes]: Let us take $k=\mathbf{Q}$ and $M=\mu_3$. Then $H^0(\Gamma,M)=0$. On the other hand, $M^D\simeq \mathbf{Z}/3\mathbf{Z}$, and one can show that $H^2(\Gamma,M^D)\neq 0$. It follows that there exists an element $x\neq 0$ in $H^2(\Gamma,M^D)$, and of course we have $$x^0=0\in {\rm Hom}( H^0(\Gamma,M), {\rm Br}(k))$$ because $H^0(\Gamma,M)=0$. (We thank Ofer Gabber for this example.) Alternatively, there exists a field $k$ of characteristic 0 such that ${\rm Br}(k)=0$, but $k$ is not of dimension $\le 1$; see Serre, Galois Cohomology, II.3.1, Exercise 1. For such $k$, there exists a finite $\Gamma$-module $A$ such that $H^2(\Gamma,A)\neq 0$. Take $M=A^D$; then $A=M^D$. It follows that there exists an element $x\neq 0$ in $H^2(\Gamma,M^D)$, and of course we have $$x^0=0\in {\rm Hom}( H^0(\Gamma,M), {\rm Br}(k))$$ because ${\rm Br}(k)=0$. (We thank David Harari for this example.)<|endoftext|> TITLE: Resolutions of $\mathbb{Z}_{(p)}$ as $\mathbb{Z}$-module QUESTION [5 upvotes]: Are there any interesting canonical (maybe unbounded) projective resolutions of $\mathbb{Z}_{(p)}$ over $\mathbb{Z}$, for instance by tensoring together all the $\mathbb{Z}[x] \stackrel{qx-1}\to \mathbb{Z}[x]$ for all primes $q$ different from $p$? REPLY [6 votes]: Put $a_n=p^{n!}-1$. It is easy to see that $a_n$ divides $a_{n+1}$, so we can define $b_n=a_{n+1}/a_n\in\mathbb{N}$. Put $P=\bigoplus_n\mathbb{Z}$ and define $f\colon P\to P$ and $g\colon P\to\mathbb{Z}_{(p)}$ by $f(e_n)=b_ne_{n+1}-e_n$ and $g(e_n)=1/a_n$. Then $f$ is injective and $g\circ f=0$ and the induced map $\text{cok}(f)\to\mathbb{Z}_{(p)}$ is an isomorphism, so we have a two-stage projective resolution of $\mathbb{Z}_{(p)}$. The key point here is that ($m$ divides $a_n$ for some $n$) iff (some power of $p$ is one in $\mathbb{Z}/m$) iff ($p$ is a unit in $\mathbb{Z}/m$) iff $m$ is coprime with $p$. All this works even when $p$ is composite, provided that we interpret $\mathbb{Z}_{(p)}$ as the ring of rationals with denominator coprime to $p$. UPDATE: Here is another version with some advantages. Put $$ c_n=\prod_{i=1}^n(p^i-1) = |GL_n(\mathbb{Z}/p)|/p^{n(n-1)/2}, $$ and let $R$ be the subgroup of $\mathbb{Q}[x]$ with basis $\{x^n/c_n\;:\;n\in\mathbb{N}\}$ over $\mathbb{Z}$. Using the fact that $GL_n\times GL_m$ embeds in $GL_{n+m}$, we can see that $c_nc_m$ divides $c_{n+m}$,so $R$ is a subring of $\mathbb{Q}[x]$ containing $\mathbb{Z}[x]$. There is a surjective ring map $h_1\colon\mathbb{Q}[x]\to\mathbb{Q}$ given by $h_1(x)=1$, and this restricts to give a surjective ring map $h\colon R\to\mathbb{Z}_{(p)}$. One can check that $x-1$ is a regular element of $R$ that generates $\ker(h)$ as an ideal in $R$, so $R\xrightarrow{x-1}R\xrightarrow{h}\mathbb{Z}_{(p)}$ is a projective resolution of $\mathbb{Z}_{(p)}$ over $\mathbb{Z}$. This can be regarded as a Koszul resolution and so has a multiplicative structure.<|endoftext|> TITLE: Can the diameter be controled by the injectivity radius and the volume? QUESTION [11 upvotes]: Diameter bounded from above is usually needed in the finiteness theorem or other convergence theorems in Riemannian Geometry. Let $M^n$ be a closed manifold and {$g_i$} be a family of smooth Riemannian metrics on it with $Inj_{g_i}\geq \alpha>0$ and $Vol_{g_i}\leq \beta$. Are those two conditions enough to imply $Diam_{g_i}\leq \gamma$? If so, does the $\gamma$ depend only on $\alpha,\beta$ and $n$? REPLY [14 votes]: By Croke, Some isoperimetric inequalities and eigenvalue estimates, Proposition 14, on an $n$-dimensional Riemannian manifold with injectivity radius $\alpha$, a ball of radius $\alpha/2$ has volume at least $C\alpha^n/ (2n)^n$ for a constant $C$. Specifically, by their theorem 11, $C=2^{n-1}V_{n-1}^{n\phantom{1}} / V_{n\phantom{1}}^{n-1}$, where $V_n$ is the volume of the $n$-dimensional sphere. Let $x_1,\dots,x_m$ be a maximal collection of points such that the balls of radius $\alpha/2$ centered at $x_1,\dots,x_m$ do not overlap. Then clearly $$m\frac{ C\alpha^n }{ (2n)^n} \leq \beta.$$ By maximality, the balls of radius $\alpha$ centered at $x_1,\dots, x_m$ cover $M$. Hence because $M^n$ is connected, the graph with vertices $x_1,\dots,x_m$ and edges connecting $x_i$ to $x_j$ if the distance from $x_i$ to $x_j$ is at most $2\alpha$ is connected and thus has diameter at most $m-1$. Thus the diameter is at most $$\alpha + (m-1) 2\alpha+ \alpha = 2m\alpha \leq 2 \beta \frac{ (2n)^n}{ C \alpha^n} \alpha = \frac{ \beta}{\alpha^{n-1} } \frac{2^{n+1} n^n} {C} $$ as for the distance between $x$ and $y$, it takes a distance of $\alpha$ to get from $x$ to one of the $x_i$, then $(m-1)$ distances of $2\alpha$ to travel to an appropriate $x_j$, and then $\alpha$ to get from $x_j$ to $y$. This is sharp up to a constant depending on $n$, as demonstrated by a suitable long thin torus.<|endoftext|> TITLE: Unstable manifolds of a Morse function give a CW complex QUESTION [15 upvotes]: A coauthor of mine and I want to use the following innocent looking statement in a forthcoming paper: Statement. Let $M^{2n}$ be a compact manifold and let $f$ be a Morse function with critical points of even indices. Then for some choice of a metric $g$ on $M^{2n}$, the closure of each unstable manifold in $M^{2n}$ is a cycle of dimension equal to the index of the corresponding critical point. I thought naively that such a statement should be contained in some classical book (but the answers given below indicate that this might be not the case). Here is an idea of how to deduce the statement from the literature. Let's take the paper of Francois Laudenbach http://www.numdam.org/article/AST_1992__205__219_0.pdf and look into Remark 3. This remark claims something much stronger, namely that even without assumption on even indices the union of unstable manifolds give a structure of a CW complex on $M$ in case there exists a metric $g$ on $M$ such that the gradient flow satisfies the Morse-Smale condition and additionally the gradient vector field is Special Morse (i.e. looks like $\sum_i{\pm}x_i\frac{\partial}{\partial x_i}$). Unfortunately, it is not stated in this paper whether such a metric $g$ always exists (added: according to John and Alessia this is very simple) Question. Is there a reference or short proof for the above Statement? Or maybe one can say that a metric satisfying Morse-Smale condition and Special Morse condition always exists? Added. I would like to thank John, Pietro and Alesia for answers. I still hope that the exact Statement that I want might be from 20th century, not 21st. Indeed, suppose that all the indices are even, and $g$ is Morse-Smale. Then for each unstable cell $W$ the set $\bar W\setminus W$ has Hausdorf dimension at most $\dim W-2$. Should not this give a well-defined cycle in $M^{2n}$? Question 2 I don't quite understand what is Morse Homology, but should not the above Statement be a trivial part of this theory? (what about this preprint: https://arxiv.org/pdf/math/9905152.pdf ? looks relevant) REPLY [14 votes]: (1). Some experts tell me that Laudenbach's paper is incomplete and contains gaps. I will retract this for now. I do recall being told this, but I am not aware at this point in time where the gaps in his paper are, if any. (I do stand by my belief that a number papers in this area are incomplete.) (2). The result you seek can be deduced in the following papers by Lizhen Qin (disclaimer: he was my student): On moduli spaces and CW structures arising from Morse theory on Hilbert manifolds. J. Topol. Anal. 2 (2010), no. 4, 469–526 An application of topological equivalence to Morse theory. arXiv:1102.2838 In fact, what you ask can be deduced from the first of these papers which handles a metric that is flat near the critical points. The second paper shows that one can actually use any metric such that the function is Morse-Smale. Alternatively, there is a paper of Burghelea and Haller which also handles the case of a metric that is flat near the critical points. Lizhen tells me that the Burghelea-Haller paper is correct and one can deduce the desired result from their methods. (The reason I am ranting about this is that I believe this area to be a hotbed of papers containing gaps--with no disrespect to the authors of those papers intended.) REPLY [9 votes]: A generic perturbation of the metric makes the flow Morse Smale, (stable and unstable manifolds meet transversally). In this situation, the unstable manifold do form a CW-complex. The unstable manifold of a critical point $x$ of index $k$ is an embedded disk of dimension equal to the Morse index; its closure is made adding a union of unstable manifolds of strictly less index. What is also true, and less obvious, is that the unstable manifold $W_x$ of a critical point $x$ also admits a "cell map", that is a homeomorphism from the open disk of dimension $i(x)$ to $W_x$ that extend continuously to the closures (i.e. from the closed disk to the closure of $W_x$), which makes the collections of the unstable manifolds a true CW-complex. The first complete proof I think is in this paper: L. Qin, On moduli spaces and CW structures arising from Morse theory on Hilbert manifolds, J. Topol. Anal. 2 (2010), 469-526. REPLY [6 votes]: Let us show that there exists a metric for which stable and unstable manifolds of a given Morse function are transverse. By Kupka-Smale theorem, a Morse function $f$ on a manifold with a Riemannian metric $m$ can be perturbed to become Morse-Smale, by genericity. The perturbed function $g$ has identical level-set foliation up to diffeomorphism so it is conjugate by a diffeomorphism $\phi$, more precisely $g = u \circ f \circ \phi$, where $u$ is an increasing diffeomorphism on the real line. The stable and unstable manifolds of $u^{-1}\circ g$ for $m$ are transverse as composition by $u^{-1}$ preserves this property. Next we apply a global change of coordinate to both $u^{-1}\circ g$ and $m$ using $\phi^{-1}$. This will send $u^{-1}\circ g$ back to $f$ and $m$ to its pull-back by $\phi$. Also this change of coordinate will send the stable/unstable manifolds of $u^{-1}\circ g$ for $m$ to the ones of $f$ for the pull-back of $m$. In particular the new stable/unstable manifolds are transverse. So the pull-back metric satisfies the transversality condition. It remains to address the special Morse condition. It is not difficult to modify $m$ locally around critical points of $f$ so that the modified metric $m'$ is special Morse for $f$, using partitions of unity. We can then repeat the above construction with $m'$ instead of $m$. While changes of coordinates preserve the special Morse condition, the perturbation of $f$ might destroy it if it affects neighborhoods of critical points. Fortunately the proof of the genericity of Morse-Smale condition produces a perturbation that does not affect those neighborhoods, so the above construction will satisfy the special Morse condition as well.<|endoftext|> TITLE: Is every finite-order unimodular matrix conjugate to a $0,1,-1$ matrix? QUESTION [13 upvotes]: Problem. Given a matrix $A\in\mathrm{GL}(n,\mathbb{Z})$ such that $A^k=1$ for some $k\geq 1$, is there a matrix $g\in\mathrm{GL}(n,\mathbb{Z})$ such that $gAg^{-1}$ has only $0$, $1$, and $-1$ as possible entries? Edit: after the remark by Mark Sapir that it is a famous open problem (which for me was already sufficient as an answer), I changed my question into the following ones, so now maybe it is more suitable for staying on MO without being closed. What is known about this problem? Which other parts of mathematics is it connected to? REPLY [8 votes]: I found a proof here for $n=4$: Yang, Qingjie, Conjugacy classes of torsion in (\mathrm{GL}_N(\mathbb Z)), Electron. J. Linear Algebra 30, 478-493 (2015). ZBL1329.15063. MR3414308 See the discussion in the last paragraph on p. 482 for the case that the characteristic polynomial is irreducible, and Theorem 1.7 for the reducible case. On the other hand, I suppose it's possible that the number of conjugacy classes of finite-order elements in $GL_n(\mathbb{Z})$ could grow faster than the number of $0,\pm1$ matrices intersected with $GL_n(\mathbb{Z})$. One can get a lower bound on the number of conjugacy classes of finite-order elements in $GL_n(\mathbb{Z})$ by counting the number which are block-diagonalizable with irreducible blocks. This should correspond to a sum over decompositions of $n$ into $\varphi(m)$ by $|Cl(\mathbb{Z}[e^{2\pi i/m}])|$, a sum over class numbers, since one obtains a conjugacy class of element of $GL_{\varphi(m)}(\mathbb{Z})$ of order $m$ for every ideal class in $\mathbb{Z}[e^{2\pi i/m}]$. I have no intuition though for the growth of this function, especially since the class numbers of cyclotomic fields behave erratically. REPLY [3 votes]: For the record, the case $n=3$ of the problem can also be easily deduced by the lists presented in Tahara, On the finite subgroups of GL(3,Z)<|endoftext|> TITLE: Divisibility in the homotopy groups of spheres? QUESTION [8 upvotes]: Consider the $k$-sphere (I'm particularly interested in $k=3$). For each positive integer $n$ let $P_{n}$ be the set of primes that divide the order of $\pi_{i}S^{k}$ for some $i\leq n$. Does the cardinality of $P_{n}$ tend to infinity with $n$? REPLY [20 votes]: Yes. In fact $\bigcup_{n\geq 1} P_n $ is the set of all primes. Serre proved that, for each odd prime $p$, there is some very predictable $p$-torsion in $\pi_{k+(p-1)}(S^k)$, for example. There is a very nice theorem of McGibbon and Neisendorfer (proving, I think, a conjecture of Serre) that implies and vastly generalizes this. Theorem (McGibbon & Neisendorfer). Let $X$ be a simply-connected finite complex such that $\widetilde{H}_*(X; \mathbb{Z}/p) \neq 0$. Then $\pi_n X$ contains $p$-torsion for infinitely many values of $n$.<|endoftext|> TITLE: Nontrivial expansion in sumsets QUESTION [5 upvotes]: Let $A \subset \mathbb{Z}/p$, let $f$ be a function on $\mathbb{Z}/p$ and let $B:=\{f(a): a \in A\}$. Can we conclude that $|A+B|$ is large if $f$ is a sufficiently "nice" function? For instance say that $f(a)=a^2$. Then can we say that if $|A|=1000\sqrt{p}$ or even if $|A|=\frac{p}{\log^{*}(p)}$, then $|A+B| \ge p/100$? The idea should be that $A+B$ is only small if $A,B$ are related by some additive structure, and if $f$ is a sufficiently random function then it should kill any structure so that $A+B$ should be large. REPLY [6 votes]: Here is a partial "yes", to complement Sam Zbarsky's negative answer to the question. There are a bunch of different papers on this topic focusing on different functions $f$ and in different ranges. To simplify things, I will focus on $f(x)=x^2$, since you mentioned it. For $|A|< p^{5/8}$, a result of Pham, Vinh and de Zeeuw gives $|A+f(A)| \gg |A|^{6/5}$. For larger sets, I cannot find a reference from the top of my head, but I am quite confident that exponential sum techniques could prove that \begin{equation} \label{claim} |A+f(A)| \gg \min \left \{ \sqrt{p|A|}, \frac{|A|^2}{\sqrt{p}} \right \}. \end{equation} See for example Theorem 7 in this paper of Balog, Broughan and Shparlinksi, which implies this result for $f(x)=x^{-1}$. I am pretty sure that I have seen this result with $f(x)=x^2$ in the literature at some point, but cannot remember where right now. If this claim is correct then it is optimal, as you can construct a set $A$ of any given size with $|A+A^2| \ll \sqrt{|A|p}$.<|endoftext|> TITLE: Does Thompson's group F have Yu's property A? QUESTION [6 upvotes]: As far as I know, it is unknown whether Thompson's group F has Yu's Property A (that is, whether it is exact) or not. See for instance this MO question. The question is said to be open at several places in the litterature, but these references are several years old. I would like to know if someone is aware of a recent development. Thanks for your help ! For the definition of Property A, see G. Yu, Invent. math. 139 (2000), 201-240. REPLY [10 votes]: It is unknown. I (and some others) believe it is as hard as amenability. There were two approaches to this problem. One was discussed in Arzhantseva, Guba, Sapir, Metrics on diagram groups and uniform embeddings in a Hilbert space. Using that approach one would need to construct an embedding of $F$ into a Hilbert space with compression function $\gg \sqrt{n}$. That approach was killed in Gournay's The Liouville property and Hilbertian compression (Numdam) where it was proved that the compression function cannot exceed $\sqrt{n}$. Another approach from Dranishnikov and Sapir's On the dimension growth of groups uses the so-called dimension growth. We hoped that the dimension growth of $F$ is subexponential which would imply $A$ (see Ozawa, Metric spaces with subexponential asymptotic dimension growth). But it turned out to be not quite the case, although see this Corrigendum to “On the dimension growth of groups” [J. Algebra 347 (1) (2011) 23–39], by Dranishnikov and Sapir. The idea of using various dimension growth functions is still alive though. See, for example, Dranishnikov and Zarichnyi, Asymptotic dimension, decomposition complexity, and Haver's property C<|endoftext|> TITLE: Formal Schemes Methods: Applications QUESTION [5 upvotes]: Possibly this question is bit too broad but up to now I was not able to find a satisfying answer. Let $X$ be a locally Noetherian scheme and $X' \subset X$ be a closed subscheme of $X$ which is defined by an ideal $\mathcal{I} \subset \mathcal{O}_X$. Again one denotes $X_n = Spec\mathcal{O}_X/\mathcal{I}_{n+1}$ and obtains a chain of thinkening $X_• = (X_0 \to X_1 \to ...)$. Taking the colimit of $X_•$ we get a formal scheme $\hat{X}$ where $|\hat{X}| = |X'|$ and $\mathcal{O}_{\hat{X}} = \varprojlim_n \mathcal{O}_{X_n}$, also called the formal completion of $X$ wrt $X'$. My main reference is Doan Trung Cuong's excellent Minicourse. My question is simply where in algebraic geometry this concept of formal schemes used in a fruitful way. The only "big" theorem that I know based on this concept is the "Theorem on formal functions". Remark: By "fruitful" I mean that we can make usage of this theory as "new" toolbox in order to obtain new conclusions about schemes in "common" sense (note that formal schemes are ringed spaces and not schemes in usual sense). An excellent prototype of such interplay with the concept of formal schemes is again Theorem on formal functions, as we can deduce from it for example the Stein factorisation, a variant of Zariski main theorem. Again, unfortunatelly up to now during my research the TofF was the only "big" result that is provided by this formalism. Are there more such notable results of similar caliber? What is the philosophy of taking formal completions of usual schemes? Or what is the motivation, so which kind of "new" information about the scheme one intends to find out applying concept, which seems be not extractable without it? REPLY [2 votes]: Instead of trying to answer the question in full, let me give some further appearances of the notion of formal scheme. There are several situations when a consideration of a topology in certain rings makes the situation simpler, e.g. every complete regular noetherian algebra over a field is a ring of power series, therefore infinitesimal neighborhoods makes clear certain features of a geometric problem that in the usual context is hidden. This is in line with Zariski's philosophy of "algebraic holomorphic functions". For example, to define algebraic residues one needs to complete, and in the complete setting, the duality between local cohomology and derived completion plays a certain role. In certain circumstances you can restore non singularity by completing. Let $X$ be a projective variety over a field of characteristic 0. If you complete the ambient projective space along $X$ you obtain a formal scheme that is smooth (in an appropriate sense) over the base field and whose De Rham cohomology corresponds to the singular cohology of the underlying space of $X$ if the base field is $\mathbb{C}$. This is the philosophy of De Rham cohomology of algebraic varieties as developed by Hartshorne. See: Hartshorne, Robin: On the De Rham cohomology of algebraic varieties. Publ. Math. IHES No. 45 (1975), 5–99. There is a further trait, that is the use of formal schemes as algebraic models of rigid analytic varieties over a complete valued field. To begin, look at: Lütkebohmert, Werner: Formal-algebraic and rigid-analytic geometry. Math. Ann. 286 (1990), no. 1-3, 341–371. Formal schemes show up quite naturally in the context of Lefschetz theory, i.e. the reconstruction of properties of an algebraic variety by knowing properties of its hyperplane sections. See (especially chapters IV and V): Hartshorne, Robin: Ample subvarieties of algebraic varieties. Lecture Notes in Mathematics, Vol. 156 Springer-Verlag, Berlin-New York, 1970. This a sample of some different context where formal schemes arise. Sometimes a topology allows to tame a situation where non-finitely generated algebras arise. For some reason, there are few systematic development of the main properties of formal schemes but their uses are pervasive in algebraic geometry. Hope my pointers help.<|endoftext|> TITLE: References for literature from mathematicians who provided critiques and proposals concerning ethical aspects of mathematics research QUESTION [13 upvotes]: I would like to know sources, articles, books or other, that provide information on ethical aspects in the research of mathematics, I wondered what is the literature that this community knows about ethical issues and proposals that were proposed in the context of mathematical research. Question. What is the literature about ethical aspects in the context of the mathematical research? Many thanks. I'm asking about it as a reference request, thus only is required to add the reference of such remarks, articles or books. From your answer to this reference request I can to contrast with the information that I know. I think that my question is legitimate. I think that this site MathOverflow should be a safe site to ask good questions, and I think that my Question as reference request can be useful for more users. REPLY [3 votes]: Oliver Rosten published a paper On functional representations of the conformal algebra in the European Physical Journal in 2017. In the acknowledgements section he included the following in reference to the death by suicide of a friend of his, Francis Dolan, in 2011. I am firmly of the conviction that the psychological brutality of the post-doctoral system played a strong underlying role in Francis’ death. I would like to take this opportunity, should anyone be listening, to urge those within academia in roles of leadership to do far more to protect members of the community suffering from mental health problems, particularly during the most vulnerable stages of their careers. There was resistance to having these lines included in the published article with at least two journals refusing to publish them. See this article for more details.<|endoftext|> TITLE: Can random variables that almost surely solve equations be repaired to surely solve these equations? QUESTION [39 upvotes]: Let $(X_\alpha)_{\alpha \in A}$ be a family of boolean random variables $X_\alpha: \Omega \to \{0,1\}$ on a probability space $\Omega = (\Omega, {\mathcal F}, {\mathbf P})$. Let ${\mathcal S}$ be a family of boolean sentences that each involve finitely many of the $X_\alpha$. Suppose that each sentence $S \in {\mathcal S}$ is almost surely satisfied by the $(X_\alpha)_{\alpha \in A}$. Can one then "repair" the random variables by locating further random variables $(\tilde X_\alpha)_{\alpha \in A}$ with each $\tilde X_\alpha$ almost surely equal to $X_\alpha$, such that the $\tilde X_\alpha$ surely satisfy all the sentences $S \in {\mathcal S}$? If $|A| \leq \aleph_0$ (that is to say there are at most countably many random variables) then the task is easy, for then the set of sentences $S$ is also at most countable, and (because the countable union of null events is null) there is a single null event $N$ outside of which the $X_\alpha$ already surely satisfy all the sentences $S$. In particular there is a deterministic choice $X_\alpha^0 \in \{0,1\}$ of boolean inputs that satisfy all the sentences, and if one sets $\tilde X_\alpha$ to equal $X_\alpha$ outside of $N$ and $X_\alpha^0$ in $N$, we obtain the claim. If $|A| \leq \aleph_1$ (that is to say $A$ has at most the cardinality of the first uncountable ordinal) and $\Omega$ is complete, then a slight variant of the above argument also works. We may well order $A$ so that every element $\alpha$ has at most countably many predecessors. We then use transfinite induction to recursively select $\tilde X_\alpha$ almost surely equal to $X_\alpha$, with the property that for all (not just almost all) sample points $\omega \in \Omega$, the tuple $(\tilde X_\beta(\omega))_{\beta \leq \alpha}$ may be extended to a tuple $(x_\beta)_{\beta \in A}$ solving all the sentences $S \in {\mathcal S}$. Indeed, if such variables $\tilde X_\beta$ have already been constructed for all $\beta < \alpha$, then the random variable $X_\alpha$ will already have this property outside of a null set $N_\alpha$ (here we use the fact that the set of tuples in the metrisable space $\{0,1\}^{\{ \beta: \beta \leq \alpha\}}$ that can be extended is the continuous image of a compact set and is thus closed and measurable). For each $\omega \in N_\alpha$, there exists at least one choice of $\tilde X_\alpha(\omega)$ that will obey the required extension property, thanks to the compactness theorem; using the axiom of choice to arbitrarily define $\tilde X_\alpha$ on this null set, we obtain a $\tilde X_\alpha$ with the required properties (it is measurable because $\Omega$ is assumed complete), and then the entire tuple $(\tilde X_\alpha)_{\alpha \in A}$ will surely satisfy all the sentences $S \in {\mathcal S}$. [It may be possible to drop the completeness hypothesis here by appealing to a measurable selection theorem; I have not thought about this carefully.] Another illustrative case where the answer is affirmative is if $A$ is arbitrary and ${\mathcal S}$ is just the collection of equality sentences $X_\alpha = X_\beta$ for $\alpha,\beta \in A$. Thus we have $X_\alpha=X_\beta$ almost surely for each $\alpha,\beta$, and we wish to modify each $X_\alpha$ on a null set to create new random variables $\tilde X_\alpha$ such that $\tilde X_\alpha = \tilde X_\beta$. Note that for each $\omega \in \Omega$ it is not necessarily the case (even after deleting a null set) that all the $X_\alpha(\omega)$ are equal to each other (e.g., suppose $A=\Omega=[0,1]$ and $X_\alpha(\omega) = 1_{\alpha=\omega}$), but nevertheless the problem is easily solved in this case by arbitrarily selecting one element $\alpha_0$ of $A$ and defining $\tilde X_\alpha := X_{\alpha_0}$. However, I do not have a good intuition as to whether the answer to this question is affirmative in general, even if one assumes good properties on the probability space $\Omega$ (e.g., that it is a standard probability space). The appearance of the cardinal $\aleph_1$ hints that perhaps the answer is sensitive to undecidable axioms in set theory. (For my ultimate application I would eventually like to replace the boolean space $\{0,1\}$ with the interval $[0,1]$ or other Polish spaces, and the sentences $S$ with closed conditions involving finitely many or countably many of the variables at a time, but the Boolean case already seems nontrivial and captures much of the essence of the problem.) EDIT: The following "near-counterexample" may also be suggestive. Set $\Omega = [0,1]$, let $A = 2^{[0,1]}$ be the power set of $\Omega$, and let $\mathcal{S}$ be the set of sentences $X_\alpha = X_\beta$ where $\alpha,\beta \subset [0,1]$ differ by at most one point. If one sets $X_\alpha(\omega) := 1_{\omega \in \alpha}$, then one morally has that the $X_\alpha$ almost surely satisfy all the sentences in $S$, but that there is no way to repair the $X_\alpha$ to random variables $\tilde X_\alpha$ that surely satisfy the equations as this would force $\tilde X_{[0,1]} = \tilde X_\emptyset$ while $X_{[0,1]}=1$ and $X_\emptyset = 0$. However this is not actually a counterexample because most of the $X_\alpha$ are non-measurable. (Removed due to errors) REPLY [23 votes]: In Terry's answer, he shows that his original question reduces to the question of whether, given a $\sigma$-algebra $\mathcal F$ on some set $X$ and a measure $\mu$ on $(X,\mathcal F)$, there is a ``splitting'' of the quotient algebra $\mathcal F / \mathcal N$, where $\mathcal N$ denotes the ideal of $\mu$-null sets. In this context, a splitting is a Boolean homomorphism $\Phi: \mathcal F / \mathcal N \rightarrow \mathcal F$ such that $\Phi([A]) \in [A]$ for all $A \in \mathcal F$. (Some authors call this a lifting instead of a splitting.) When some such $\Phi$ exists, let us say that $(X,\mathcal F,\mu)$ has a splitting. I did some digging on this question this afternoon, and found two very good sources of information: David Fremlin's article in the Handbook of Boolean Algebras (available here) and a survey paper by Maxim Burke entitled "Liftings for noncomplete probability spaces" (available here). I'll summarize some of what I found below to supplement what Terry mentions in his answer. He mentions already that it is independent of ZFC whether $([0,1],\text{Borel},\text{Lebesgue})$ has a splitting: $\bullet$ (von Neumann, 1931) Assuming $\mathsf{CH}$, $([0,1],\text{Borel},\text{Lebesgue})$ has a splitting. $\bullet$ (Shelah, 1983) There is a forcing extension in which $([0,1],\text{Borel},\text{Lebesgue})$ has no splitting. Also mentioned already is the fact that if we expand the $\sigma$-algebra in question from the Borel sets to all Lebesgue-measurable sets, then the situation is more straightforward: $\bullet$ (Maharam, 1958) If $(X,\mu)$ is a complete probability space, then $(X,\mu\text{-measurable},\mu)$ has a splitting. Now on to some not-yet-mentioned results. First, it's worth pointing out that one can obtain splittings with nice extra properties. $\bullet$ (Ioenescu-Tulcea, 1967) Let $G$ be a locally compact group, and let $\mu$ denote its Haar measure. Then $(G,\mu\text{-measurable},\mu)$ has a translation-invariant splitting (which means $\Phi([A+c]) = \Phi([A])+c$ for every $\mu\text{-measurable}$ set $A$). Once again, shrinking our $\sigma$-algebra from all $\mu$-measurable sets to only the Borel sets causes problems. $\bullet$ (Johnson, 1980) There is no translation-invariant splitting for $([0,1],\text{Borel},\text{Lebesgue})$. Thus, interestingly, Shelah's consistency result becomes a theorem of $\mathsf{ZFC}$ if we insist on the splitting being translation-invariant (with respect to mod-$1$ addition). More generally: $\bullet$ (Talagrand, 1982) If $G$ is a compact Abelian group and $\mu$ is its Haar measure, then there is no translation-invariant splitting for $(G,\text{Borel},\mu)$. What stood out to me most in Fremlin and Burke's articles is how many questions seem to be wide open. Open question: Is it consistent that every probability space has a lifting? If yes, this would give a consistent positive answer to Terry's original question. Open question: Is it consistent with $2^{\aleph_0} > \aleph_2$ that $([0,1],\text{Borel},\text{Lebesgue})$ has a splitting? (Carlson showed that it is consistent to have $2^{\aleph_0} = \aleph_2$ and for $([0,1],\text{Borel},\text{Lebesgue})$ to have a splitting. Specifically, he showed that this holds whenever one adds precisely $\aleph_2$ Cohen reals to a model of $\mathsf{CH}$.) Open question: Does Martin's Axiom (or $\mathsf{PFA}$, or $\mathsf{MM}$) imply that $([0,1],\text{Borel},\text{Lebesgue})$ has a splitting? Open question: What of the same question in other well-known models of set theory (the random model, Sacks model, Laver model, etc.)?<|endoftext|> TITLE: Are all pointed maps of classifying spaces of abelian groups homotopy equivalent to homomorphisms? QUESTION [5 upvotes]: Let $G$ be a commutative topological group (e.g. $S^1$), and let $BG$ be its classifying space. Since $G$ is commutative, the space $BG$ is a group up to homotopy. It is well-known that we have a natural isomorphism $$\pi_0Map_*(BG,BG) \cong \pi_0Hom(G,G)$$ where $Map_*(BG,BG)$ is the space of pointed maps $BG \to BG$ with pointwise group structure. Clearly, we have a map $\pi_0Hom(BG,BG) \to \pi_0Map_*(BG,BG)$ induced by the map $$Hom(BG,BG) \to Map_*(BG,BG).$$ Is the map on path-components an isomorphism of groups? Even better, is the latter map a homotopy equivalence of spaces? REPLY [7 votes]: I don't think this is true, even for very simple examples. It is not completely evident that your question is well-defined (i.e., independent of the individual model that is chosen for $BG$), but let me try to describe a counterexample nonetheless. Let $\alpha \colon K({\bf Z},2) \to K({\bf Z},4)$ be the map of spaces classifying the square of a generator of $H^2(K({\bf Z},2);{\bf Z}) \cong {\bf Z}$, and let $\Omega \alpha \colon K({\bf Z},1) \to K({\bf Z},3)$ denote the corresponding $E_1$-map that arises from looping $\alpha$. Then the E_1-map $$K({\bf Z},1) \times K(Z,{\bf 3})\to K({\bf Z},1) \times K(Z,{\bf 3}),(x,y) \mapsto (*,\Omega\alpha(x))$$ is nullhomotopic as a map of spaces, but not as a E_1-map, since otherwise $\alpha$ would be null. Thus, for $BG = K({\bf Z},1) \times K({\bf Z},3)$, the map you wrote down fails to be injective on $\pi_0$.<|endoftext|> TITLE: Parallelizability of 3-manifolds QUESTION [7 upvotes]: Robert Bryant's answer here ( https://mathoverflow.net/a/149496/85500 ) states that any orientable 3-manifold is parallelizable. Previously I was under the impression that only closed (compact & without boundary) orientable 3-manifolds were necessarily parallelizable. I have also heard somewhere (unfortunately I don't remember the source) that most noncompact 3-manifolds are parallelizable, but even if this statement is correct in some sense, I have no idea whether this "most" was meant informally or if this statement can be interpreted in a technical sense. In either case, my knowledge of differential/algebraic topology is nowhere near my knowledge of local differential geometry so I do not know any proofs and would probably have difficulty understanding one anyways. Primary motivation for asking this question is to try to understand how restrictive is demanding spacetimes in general relativity to be parallelizable. If we want to have a well-posed initial value problem, spacetime must be of the form $\mathbb R\times\Sigma$, where $\Sigma$ is a 3-manifold, so it is parallelizable iff $\Sigma$ is parallelizable. It is also known that a spacetime admits spin structures if and only if it is parallelizable. So if one wishes to include fermions on spacetime, one wants a parallelizable spacetime. Question: So what is the proper statement about the parallelizability of orientable 3-manifolds, especially in regards to non-compact ones? REPLY [4 votes]: Suppose we have a non-compact $3$ dimensional manifold $M$ which is spinable. Fix a CW structure of the manifold, then being spinable is equivalent to the tangent bundle being trivial on the $2$-skeleton. There are two ways to proceed now. One can either use that every non-compact $n$-manifold is homotopy equivalent to a subset of its $n-1$-skeleton or one can use that $\pi_2(SO(3))$ vanishes and hence every trivialization on the two skeleton can be extended to the $3$-skeleton. So all that is left to show is that an orientable non-compact $3$-manifold is spinable. Since $H^2(M;\mathbb{Z}/2)\cong \text{Hom}(H_2(M;\mathbb{Z}/2),\mathbb{Z}/2)$ the second Stiefel-Whitney class is non-zero if and only if there exists an element $\alpha$ in $H_2(M;\mathbb{Z}/2)$ such that $w_2(\alpha)\neq 0$. Note that $\alpha$ can be represented by an embedded compact surface $S$ i.e a subsurface such that $\alpha=[S]$ (Represent $\alpha$ by an immersed subsurface and then resolve the singularities). Since $S$ is compact we know that it is contained in a compact $3$-dimensional submanifold $N$. Since $i^*(w_2(M))=w_2(N)=0$ we see that this implies that $w_2(M)([S])=0$ and hence $w_2(M)$ is zero i.e. $M$ is spinable.<|endoftext|> TITLE: Distribution of pairwise distances QUESTION [12 upvotes]: I am seeking results that describe the distribution of the set of Euclidean distances between pairs of $n$ points in a unit square in the plane. For example: All the distances could be short (a tight cluster), but not all the distances could be long (all long distances force some short distances). Are there results of this general form, constraining the possible shapes of the histogram of distances, for large $n$?                       Distribution of point-to-point distances in a unit square. My actual application requires some constraints on the lengths of geodesics between points sprinkled on a surface embedded in $\mathbb{R}^3$, a surface homeomorphic to a sphere. I'm hoping the planar case will shed light on geodesic constraints. REPLY [3 votes]: Let A be the measure on R^2 invariant wrt rotations about the origin with total weight at each radius equal to the frequency of that distance in your histogram. This measure is obtained by rotationally averaging the autocorrelation measure. The Fourier transform of the autocorrelation measure is nonnegative. Since the Fourier transform is equivariant under rotation, I believe this means that the Fourier transform of A must also be nonnegative. This yields some strong constraints on the histogram of distances, which are basically the entire theory behind LP bounds.<|endoftext|> TITLE: Infinite switching on lego duplo railway QUESTION [14 upvotes]: I bought a Lego Duplo railway for my child and it has different types of parts. There are ordinary ones (straight or curved, with two ends), and there are forks - they have three ends, with one highlighted (1, 2a, 2b). If the train comes at the fork through 2a or 2b, then it leaves from 1. And if it comes at the fork through 2a, then the next time it comes from 1, it will leave from 2a. In order to change the direction of the fork, it is necessary that the train will come on it from 2b. It is easy to build a railway road in which there are 2 such forks, and when the train starts, each of these forks will switch an infinite number of times. But after buying additional parts (forks and ordinary parts), my son cannot build the road so that all the forks switch. He asked me, but I also failed. Is it true that when starting a train on any road made up of these parts, no more than 2 forks will switch endlessly? REPLY [7 votes]: It is not possible to have more than two intersections switch endlessly. After following the train for some time we can assume that the train follows some cyclic sequence of states. Suppose the train visits intersections $1$, $2$, ..., $k$ which are distinct and then returns to $i\leq k$, i.e. this is the first time it visits an intersection twice after visiting $1$. There are two cases: either the train will continue in the loop $i,\ldots,k$, or it will go backwards to $1$ following the same intersections. The first case is clear. So assume the train visits $i, i-1, \ldots, 1$. Let us follow the train until it visits some intersection twice again. By relabeling we can assume that after $1$ it visits $j$. Again two cases, again assume we are in the second case. If $ji$ we have $1,2,3,\ldots,k,i,i-1,i-2,\ldots,1,j,j+1,\ldots,k,i,i-1,i-2,\ldots,1,$ and the sequence repeats. This is a contradiction. So there are only $2$ possibilities for the cyclic behavior of the train. Either it just follows a loop and no intersections switch at all, or it has two disjoint loops joined by a path.<|endoftext|> TITLE: Packing disks of infinitesimal diameter on a sphere: the asymptotics of the Tammes problem QUESTION [7 upvotes]: This is an elaboration on MO Question 212550: given $ 0 < 2a << 1 $, how many points can be placed on the unit sphere, subject to the constraint that any two of these points must be at Euclidean distance at least $2a$ ? Obviously, every point must be allocated at least the area of a disk with radius $a$, so the maximum number $M(a)$ cannot exceed $4/a^2$; however, Fejes Tóth famously proved that the allocated area must be at least that of the circumscribing hexagon, so that: $2 \sqrt 3 a^2 M(a) = 4\pi - S(a)$ with $S(a) > {2\over3}\pi a^2 $. Then, we have a 1955 (Edit: 1951) paper by Habicht & Van der Waerden offering a very contorted construction, and the estimate: $S(a) = O(a^{2/3}) $. They concluded with an oddly phrased comment, acknowledging their explicit value for the constant is far from minimal, while implying the order of magnitude is tight, i. e. their $O(a^{2/3}) $ should really be $\Theta(a^{2/3}) $. Then... that's all. Or at least, that's all I could google regarding my question. To wit: How much is proven, and how much more is plausible, of the behaviour of $S(a)\over a^{2/3}$ for inifinitesimal $ a $? If it is $o(1) $, what is its order of magnitude? If it is $\Theta(1) $, what are its upper & lower limits? Edit: I do not mean Google has no answer to this question; only that they lack relevance. J. Park pointed out in his answer that any explicit construction provides a lower bound; a plethora of such constructions result in $S(a) < C a^{2/3}$ for various $C$'s, some better, some worse than the original one by H.-v.d.W. What's wanting is a public claim of optimality. REPLY [7 votes]: The second-order behavior of $M(a)$ for $\mathbb{S}^2$ was investigated in van der Waerden's paper Punkte auf der Kugel shortly after the first paper by Habicht and van der Waerden was published (Lagerung von Punkten aut der Kugel appears to be the paper you are referring to, although it was published in 1951). There is only an improved lower bound for $M(a)$ given in the later paper (one needs also to renormalize quantities there to agree with those in this question). Edit: Note that any construction gives a lower bound on $M(a)$. It is difficult to give asymptotically optimal constructions to the first order (see Chapter 3.3 of Hamkin's thesis here for a review of lower bounds). Fejes Tóth's upper bound gives to first order a comparable bound (for $d=2a$): $$M(d)\leq 2\left[1-\frac{\pi/6}{\cot^{-1}\sqrt{3-d^2}}\right]^{-1}.$$ It is not clear whether there are matching bounds known to the second order.<|endoftext|> TITLE: Does the mean curvature flow naturally come with less applications than intrinsic curvature flows? QUESTION [10 upvotes]: I know studying the mean curvature flow is a very interesting area of research, I've fooled around with it a bit myself. But it honestly doesn't look like it has much applications within mathematics itself (at least when I try to search for "mean curvature flow applications", everything I find is phyics related or something like that), while, say, the Ricci flow, besides being interesting by itself (like the mean curvature flow), eventually provides a classification of all closed 3-manifolds (even in a lower level, you can use it to prove the sphere theorem or the uniformization theorem in dimension 2). My intuition tells me that the mean curvature flow, just like any other extrinsic curvature flow, can't ever be used to get topological results about our manifolds, for example. Is this correct? If not, I'd really like seeing some interesting applications like this for the mean curvature flow (or other extrinsic flows). I don't know if this is too elementary for mathoverflow but I decided to risk it since I didn't really find any questions like this on MSE. EDIT: To make things a little bit more objective, what I really wanna see are topological results obtained from mean curvature flow. Naturally I think results using mean curvature flow can only be as strong as results that relate topology and mean curvature, and as far as I know, those are kinda pretty weak. REPLY [9 votes]: Mu-Tao Wang (Math. Res. Lett. 2001) showed that any diffeomorphism $f:S^2\to S^2$ is isotopic to an isometry, which was originally shown by Smale (Proc. AMS 1959) Mao-Pei Tsui and Mu-Tao Wang (Comm. Pure Appl. Math. 2004) showed that if $f:S^n\to S^m$ is area-decreasing on two-dimensional submanifolds, then $f$ is null-homotopic. (Gromov had shown this in the weaker context that the two-dimensional area distortion factor is sufficiently close to 0. Larry Guth (Geom. Func. Anal. 2013) has counterexamples if "two" is replaced by "three".) Ivana Medos and Mu-Tao Wang (J Diff. Geom. 2011) showed that if $f:\mathbb{CP}^n\to\mathbb{CP}^n$ is a symplectomorphism such that $f$ and $f^{-1}$ have Lipschitz constants sufficiently close to one, then $f$ is symplectically isotopic to an isometry. (Gromov (Invent. Math. 1985) showed that in the case $n=2$ this is true without a condition on the Lipschitz factors.) The method in each case is to deform the graph of $f$ by the mean curvature flow and to show a long-time existence and convergence result. So it is mean curvature in codimension larger than one, as opposed to most research in MCF. As for mean curvature flow in codimension one, any smooth compact four-manifold which is homeomorphic to $S^4$ can be smoothly embedded in $\mathbb{R}^5$, and I think some people hope that a sufficiently good understanding of its mean curvature flow could prove the four-dimensional smooth Poincaré conjecture, roughly analogously to the Hamilton-Perelman proof of the three-dimensional Poincaré conjecture using Ricci flow. But it would probably be much more complicated, for analytic reasons.<|endoftext|> TITLE: Bubbling off of a pseudo holomorphic sphere on surface with cylindrical ends QUESTION [5 upvotes]: I need some clarification about the reason why we have a sphere bubbling off in the situation described by Seidel in The Symplectic Floer Homology of a Dehn Twist. I’ll try to summarize to the best of my abilities the situation I’m interested in: We are stretching the neck along a circle in $\Sigma$. Let us fix an identification of the tubular neighborhood of such circle with $[-1,1]\times S^1$. Let $R>0$, by $\Sigma^R$ I’ll denote the surface diffeomorphic to $\Sigma$ but with neck $[-R,R]\times S^1$. Assume that for $R_i\to \infty$, we have a sequence of $J^{R_i}$-holomorphic strips $\{u^i\}_i$, with $u^i\in \mathcal{M}^{R_i}(x_-,x_+)$ (for a definition see bottom of page 832 of the paper) with uniform bounded energy. Notice that we can’t have uniform bound on $|du^i|_{L^{\infty}}$, in particular we can find a sequence of points $z_i \in [-R_i,R_i]\times S^1$ such that $|du^i|$ has a maximum there, with value $C_i$, and $C_i\to \infty$. Let $\Sigma’=\Sigma \setminus [-1,1]\times S^1$, (the definition of $\Sigma’$ is independent of the length of the neck!) now assume that, for $i$ big enough, $$d(u^i(z_i),\Sigma’)\leq k < \infty$$ (I.e. no matter the length of the neck that is stretching to infinity, my Maximums stays within finite distance to some edge of the neck) From there the author immediately concludes that after some reparametrization, the $u^i$’s converged to a $J$-holomorphic sphere in the surface obtained from $\Sigma’$ by attaching a semi-infinite cylinder to each boundary component. I can show that, after reparametrization my curves converges to a non-constant $J$-hol map defined on $\Bbb C$, but I don’t see why this map should extend to the sphere. My understanding is that in order to use the removable of singularity theorem we need a compact image, but $\Sigma’$ With the cylindrical ends is non compact. Most likely there is some clever way to reparametrize this curves (i used the standard conformal map $z\mapsto z/C_i+z_i$ that centers $u^i$ at $0$ and normalize the norm of its differential at $0$), or to infer something about the image of the limit, but I’m unable to see it. Any hint is really appreciated, since this is bugging me. Thanks in advance! Update I agree with the given answer that monotonicity lemma could be a way to establish compactness of the image of the limiting function. What I don't understand are the following point: 1) Assuming that for small enough balls we can ensure that the topological boundary of our curve is outside such ball (as required by the monotonicity lemma) how do we deal with the fact that the target manifold is not compact? That's the key assumption in the lemma. As I wrote in the comments I though that we could just restrict our attention to some compact sub manifolds of $\Sigma' \cup \partial \Sigma' \times [0,\infty)$ but the lower bound provided by the lemma would depend on such a choice. Hence it's not clear to me how to obtain a uniform lower bound on the energy of the curve passing through a given ball. 2)If we work with a vertical almost complex structure on the neck, then I think the lower bound provided by the monotonicity lemma is the same in every sector $[a-n,a+n]\times \partial \Sigma'$ (since they all are isometric and with the same a.c.s.) In that case I can see how to get a uniform lower bound. But what worries me is that I don't see any problem in applying this reasoning to case #1, but as far as I understood, in that case we can't rule out non-constant $j$-hol planes that easily and we must use a different strategy (namely integrability of the a.c.s.) REPLY [4 votes]: It turns out that the Removable Singularities Theorem and the Monotonicity Lemma do not require compactness, but that the target manifold should have bounded geometry, such as in our case of inserting a cylindrical piece to the compact surface. See for example, Lemma 5.11 in "Compactness results in symplectic field theory" by Bourgeois-Eliashberg-Hofer-Wysocki-Zehnder and Proposition 2.71 of Abbas' book "An introduction to compactness results in symplectic field theory". Seidel brought to my attention Theorem 4.5.1 in Sikorav's paper "Some properties of holomorphic curves in almost complex manifolds", which only needs the (possibly noncompact) target to have bounded geometry and does not need the disc to have bounded image, for removing singularities. By assumption our points $u^i(z_i)$ don't run away along the semi-infinite ends attached to $\Sigma'$, i.e. our bubble points are stuck in $\Sigma'$ union a collar $\partial\Sigma'\times[0,k]$ and cannot escape along $\partial\Sigma'\times[0,\infty)$, otherwise we're in Case#1 (in the paper) which has the points running away along the neck of $\Sigma^{R_i}$. We now zoom in our around these bubble points (hence away from the boundary of the holomorphic strips), forming a rescaled sequence of holomorphic discs. Note that in both the case at hand (Case#2 in the paper) and Case#1 in the paper, our sequence of rescaled maps have domains being open disks centered around the bubble points. It can happen in Case#1 that these disks just shift along with the bubble points down the infinite neck region, which doesn't happen in Case#2. Aside: I hoped that the monotonicity lemma would prevent the sequence of rescaled holomorphic discs from "thinning" out and having unbounded image, that is, our maps must use up at least a certain amount of energy for every ball whose center it passes through. We couldn't a priori apply the usual monotonicity lemma (for compact target manifolds) to each compact set in a compact exhaustion of the noncompact surface, because the resulting (lower) energy bound may vary (hence decrease) with each set. But due to the cylindrical end there is a uniform lower bound. In any case, the a priori issue is that the (image of the) boundary of the discs might squeeze into the balls with which we are trying to compute the area bounds.<|endoftext|> TITLE: Motivation behind Fontaine's Theory QUESTION [5 upvotes]: I am reading Fontaine's theory of $p$-adic Galois representations. But I am not able find the motivation behind it. Please let me know some good reference where I can study the motivation behind Fontaine Theory. REPLY [6 votes]: Fontaine's program is the classification of $p$-adic representations of $\operatorname{Gal}(\bar{K}/K)$ where $K$ is a discrete valuation field of residual characteristic $p$. If by motivation, you meant why Fontaine himself wanted to do that, you could do worse than reading https://webusers.imj-prg.fr/~pierre.colmez/FW.pdf and/or https://www.math.u-psud.fr/~illusie/Illusie-Pisa5.pdf depending on whether your background is in arithmetic or algebraic geometry. If by motivation, you meant why we (for some value of we) might want to do this, well you could have a look at the Proceedings of the ICM of 2002* for instance, look up the lectures which mention Fontaine's theory and what they deal with (I found 5 different ones with topics ranging from special values of $L$-function to $K$-theory to Galois representations and automorphic forms). *I chose the year rather arbitrarily, I'm guessing that any other years since 1984, and certainly 2018, would do.<|endoftext|> TITLE: Non-nesting matchings and Catalan numbers QUESTION [5 upvotes]: It is well-known that the number of non-nesting perfect matchings on $2n$ points is given by the Catalan number $C_n$; see part (a) of the figure below. This is item e^5 in Stanley's list (http://www-math.mit.edu/~rstan/ec/catadd.pdf). [The following section has been edited to account for an initial mistake in the description.] Now I am interested in non-nesting arc diagrams on $n+1$ points, where no arc connects two neighboring points (two arcs may meet in the same point, one arc from above and one arc from below, so the arcs may not form a matching); see part (b) of the figure below. These diagrams are also counted by $C_n$, and it is easy to prove this. This must be a known fact, but the second type of arc diagrams is not in Stanley's book (maybe I overlooked it), so who has references for this type of arc diagrams with regards to Catalan numbers? REPLY [3 votes]: These are nonnesting partitions, which are usually defined as anti-chains in the root poset. Namely, the root poset (in $S_n$) is $\{ e_j-e_i : 1 \leq i < j \leq n \}$ with $e_k - e_j \leq e_{\ell} - e_i$ if $i \leq j < k \leq \ell$. Your arc diagrams turn into nonnesting partitions by $(i,j+1) \mapsto e_j-e_i$. Googling "nonnesting partition" will give you a ton of hits.<|endoftext|> TITLE: When is the Jacobson radical reflexive? QUESTION [6 upvotes]: Let algebras be Artin algebras. It is well known that a an algebra has global dimension at most one if and only if the Jacobson radical is projective. As reflexive is a natural generalisation of projective, one might ask when the Jacobson radical $J$ is reflexive. I noted that for Nakayama algebras one has that $J$ is reflexive if and only if the finitistic dimension is at most one. But in general such a statement would probably be too good to be true. Question: Is there an easy counterexample to $J$ being reflexive implies finitistic dimension at most 1? (the other direction is not true, see answer below) Reflexive means that the canonical evaluation map $f_M:M \rightarrow M^{**}$ is an isomorphism where for an algebra $A$: $M^{**}=Hom_A(Hom_A(M,A),A)$. Here $f_M(m)=g$ with $g(h)=h(m)$. REPLY [2 votes]: The algebra $K[x,y]/(x^2,y^2,xy)$ has finitistic dimension 0 but $J$ is semisimple and not reflexive. I have not yet found an example where $J$ is reflexive but the algebra has finitistic dimension larger than 1.<|endoftext|> TITLE: Origin of the noun "mathematician" QUESTION [12 upvotes]: I have read that Pythagoras's fraternity had two kinds of members, the 'acousmaticians', who were allowed to attend the lectures, and the 'mathematicians', who had been initiated. Is this the origin of the noun 'mathematician' ? REPLY [11 votes]: In classic Greek, μάθημα is a neuter noun, formed by a standard procedure from the root of the verb μανθάνω, to learn, and denotes in general the object of learning. Also standard is the derivation of the adjective μαθηματικός, "what concerns the object of learning". Plato refers this adjective to persons still in the generic sense of wishful to learn. Aristoteles uses the neuter plural, τὰ μαθηματικά, to denote theoretical studies, including mathematics, mathematical physics, astronomy. As far as I know, Aristoteles is also the first who uses μαθηματικός as a substantive, ὁ μαθηματικός, the mathematician, of course in the wider acceptation of "theoretical scientist" that this word had in all ancient science. In fact, throughout all medieval times, in this sense "Mathematici" especially denoted the followers of Aristotheles in natural sciences, (often as opposed to "Medici", the followers of Hippocrates, scientists with an empiric approach based on observations). The noun akousmatikos, that however we know from later authors, like Porphyry (born 230 AC) and Iamblichus (born 245AC) has a parallel formation: ἀκούω (to listen) > ἀκουσμα (oral teaching) > ἀκουσματικός (listener, follower of oral lectures). The distinction is between advanced students and not initiated, simple listeners, and also emanated the two branches after the scission of the Pythagorean school: one more interested in scientific aspects of Pythagorism, the other attached to the religious aspects of it.<|endoftext|> TITLE: Permutation groups generated by finitely many point stabilisers QUESTION [7 upvotes]: Assume that $G\leq\operatorname{Sym}(X)$ is a permutation group generated by all its point stabilisers, i.e. $G=\langle G_x \mid x\in X\rangle$. There is no cardinality restriction on $X$. Furthermore, assume that $G$ has finitely many orbits, and that $G$ is subdegree-finite, i.e. all point stabilisers have only finite orbits. Is $G$ then necessarily generated by finitely many point stabilisers? The question arises in the context of the permutation topology. When we endow $G$ with the permutation topology, $G$ is a totally disconnected locally compact group when all $G_x$ are compact, and I want to know if having finitely many orbits is then sufficient for $G$ to be compactly generated. REPLY [5 votes]: Let $A$ be an infinite abelian group of odd finite exponent. For example we could take $A$ to be the direct product of infinitely many copies of a cyclic group $C_n$, with $n>1$ odd. Let $\langle t \rangle$ be a cyclic group of order $2$, define $\phi:\langle t \rangle \to {\rm Aut}(A)$ by $\phi(t): a \mapsto a^{-1}\ (a \in A)$, and let $G = A \rtimes_\phi \langle t \rangle$ be the associated semidirect product (so $tat^{-1}= a^{-1}$ for all $a \in A$). Now consider the action of $G$ by left multiplication on the left cosetsof $\langle t \rangle$ in $G$. These have the form $a\langle t \rangle$ for $a \in A$, and the stabilizer in the action of this coset is the subgroup $a\langle t \rangle a^{-1}$ Now, for a subset $B$ of $A$, the subgroup of $G$ generated by the stabilizers of the cosets $\langle b \langle t \rangle$ is contained in the subgroup $\langle B,t \rangle$ of $G$ which is equal to $\langle B \rangle \langle t \rangle$ and has order $2\langle B \rangle$. So, if $B$ is finite then the subgroup generated by these stabilizers is also finite, and hence $G$ cannot be generated by finitely many stabilizers. In fact we need $|A|$ stabilizers to generate $G$.<|endoftext|> TITLE: How much of the Cantor-Schröder-Bernstein theorem is constructively recoverable if the injections have retractions and decidable images? QUESTION [7 upvotes]: This is cross-posted from MSE at the suggestion of a comment after receiving no answers over a few weeks. Suppose we have $f : A \to B$ and $g : B \to A$, as well as left inverses $f_r : B \to A$ of $f$ and $g_r : A \to B$ of $g$. Suppose further that $f$ and $g$ have decidable images; i.e., we have $\forall b : B.(\exists a : A. f(a) = b) \vee \neg(\exists a : A. f(a) = b)$, and similarly in the other direction (replace $\vee$ with $+$ in a proof-irrelevant setting). So this is the structure necessary to work with an expression like $f^{-1}(b)$; we can ask whether it is defined, and what its value is if so. My impression is that the inability to invert an arbitrary function is ordinarily the biggest obstacle in Cantor-Schröder-Bernstein, so I'm interested in knowing what difficulties are left if you skip over that issue by fiat. I've been playing with these assumptions in Coq, and so far I've managed to adapt the standard proof by König to show that there exist sequences of functions $F_n : A \to B$ and $G_n : B \to A$ which gradually approximate the classically defined bijections. Most importantly, their sequences of compositions $G_n \circ F_n$ and $F_n \circ G_n$ "converge pointwise to" the identities on $A$ and $B$ in the following weak sense: $$\forall a : A. \neg\neg\exists n : \mathbb N.\forall m \ge n.G_m(F_m(a)) = a$$ $$\forall b : B. \neg\neg\exists n : \mathbb N.\forall m \ge n.F_m(G_m(b)) = b$$ But this seems like a fairly weak result, and I'd like to know how much it may be possible to strengthen it! I did find the Myhill isomorphism theorem, but having looked at the proof, it seems like any adaptation of it to more general premises like the ones I've given might require the types in question to be subcountable or something like that. REPLY [4 votes]: Sorry that I come back to this question after 6 months, but I was looking for something else and duckduck brought me here. But if we go back even further, a short proof that a topos with NNO satisfies the Cantor-Schroeder-Bernstein if and only if it is boolean bubbled up on the CATEGORIES list in early 1994 (the following is from February '94): The other way around, the validity of CSB in a topos ALMOST implies that the topos is boolean. "Almost" means a slightly stronger version needs to be considered: (CSB*) if f:A->B and g:B->A are monics then there is an iso h:A->B such that for every x in A holds h(x) = f(x) or h(x) = g^{-1}(x) The usual set-theoretical proof of CSB (which goes through in boolean toposes) actually yields (CSB*). In return, (CSB*) implies booleanness. We assume that every topos has an object of natural numbers. PROPOSITION. A topos is boolean if and only if it satisfies (CSB*). PROOF of the "if" part: Let X be a subobject of Y. We use (CSB*) to construct its complement. Define A = YxN B = X+(YxN) (where N is the object of natural numbers). Let f:A-->B be the inclusion, and g:B-->A the unique arrow which maps X into the first copy of Y in A, while it takes the copies of Y contained in B into the second one, the third one and so on. (Formally, g is defined using the universal property of coproducts, and the structure of N.) It is clear that both f and g are monics. By (CSB*), there is a bijection h:A-->B with h(x)=f(x) or h(x)=g^{-1}(x). This means that the union of C = {z | h(z)=f(z)} and D = {z | h(z)=g^{-1}(z)} covers A. On the other hand, the definitions of f and g imply that C and D are disjoint. So they are complements in A. By pulling back C and D along the inclusion of Y iso Yx{0} in A, we get two complementary subobjects of Y. The claim is now that the intersection (pullback) X' of D and Yx{0} is isomorphic to X -- so that the intersection of C and Yx{0} yields the complement of X in Y. First of all, each element of D must surely be in the image of g. Thus, if is in D, then y must be in X. So X' is contained in Xx{0}. The other way around, since h(C) is contained in f(C), the intersection in B of X and h(C) must be empty. Therefore, X must be contained in h(D): indeed, h is an iso, and the direct images h(C) and h(D) must be complementary in B. But h(D) is contained in g^{-1}(D). So X is in g^{-1}(D), i.e. g(X) is in D. Hence, Xx{0}=g(X) is contained in X'.<|endoftext|> TITLE: Total length of a set with the same projections as a square QUESTION [17 upvotes]: Take some convex polygon $P$. I'm mostly asking about the unit square, but would also appreciate thoughts on general polygons. We want to take a family of line segments inside $P$ that have the same projections as $P$ (that is, any line intersecting $P$ intersects one of the segments). Let $L$ be the minimum total length of such segments. What is $L$? I have the following bounds: Upper bound: $L\le\sqrt{2}+\frac{\sqrt6}{2}\approx 2.639$ Take the following construction such that the three segments meet at angles of $2\pi/3$ (the gray dahsed diagonal is not one of the segments in the construction). The sum of the lengths is $\sqrt{2}+\frac{\sqrt6}{2}$. Lower bound: $L\ge 2$. The sum of the lengths of orthogonal projections onto the two diagonals is $2\sqrt{2}$. For any segment of length $a$ and angle $\theta$ to the diagonals, the sum of its projections is $a|\sin\theta|+a|\cos\theta|\le a\sqrt{2}$. Thus $L\sqrt{2}\ge 2\sqrt{2}$, hence the lower bound. I've tried improving the lower bound by using other combinations of projections, or splitting the square into subshapes and using different projections on different subshapes, but couldn't get anything better. REPLY [12 votes]: This problem has been heavily studied. For example, this paper Dumitrescu, Adrian, and Minghui Jiang. "The opaque square." In Proceedings of the 13th Annual Symposium on Computational Geometry, p. 529. ACM, 2014. ACM link. Preliminary PDF. includes an analog of your figure:                     Dumitrescu-Jiang Fig.1 (detail). and recites the history of the problem. One of their main results is: Theorem 2. The length of any interior finite segment barrier for a unit square is at least $2 + 10^{−5}$. This accords with @alesia's post.<|endoftext|> TITLE: 'Trapping' 3D regions with sheets of paper QUESTION [11 upvotes]: Given a square sheet of paper, how does one create a bag (a closed surface) with it such that the 3D region contained within this closed surface has maximum volume (operations allowed include wrinkling and sewing/gluing at the edges but not stretching or tearing)? The closed surface need not be smooth. This question might be related to the 'paper bag (or teabag) problem'. Further, by varying the shape of the paper sheet (to say a circular or elliptical sheet), one has a range of questions. Apart from solving each specific case, are there 'global' properties? For instance one could ask if these claims hold: "given any convex sheet, to produce a bag that can hold max volume, all sewing/gluing operations are done necessarily at the edges." "for any convex sheet, the bag formed with it holding maximum volume cannot be smooth when filled to capacity" REPLY [10 votes]: A section of Geometric Folding Algorithms: Linkages, Origami, Polyhedra reports on an exploration1 of all the convex polyhedra that can be folded from a unit square (by "gluing" the perimeter to itself without overlap or gaps). The polyhedra fall into six interconnected continua. We found that the largest volume was achieved by this folding to an irregular octahedron:                     Volume $\approx 0.056$. Joe Malkevitch posed the max-volume question (for squares and convex polyhedra). Unfortunately we gained no geometric insight into why that folding achieved the max volume (and our calculations were not exact). The max volume polyhedron lies on ring $A$ near 12 o'clock. 1 By myself and two students, Rebecca Alexander and Heather Dyson. REPLY [4 votes]: There is some recent fascinating relevant work in the physics-of-materials community: Paulsen, Joseph D., Vincent Démery, Christian D. Santangelo, Thomas P. Russell, Benny Davidovitch, and Narayanan Menon. "Optimal wrapping of liquid droplets with ultrathin sheets." Nature materials 14, no. 12 (2015): 1206. Journal link. "ultrathin sheets automatically achieve optimally efficient shapes that maximize the enclosed volume of liquid for a fixed area of sheet"           Concerning the OP's mention that "operations allowed include wrinkling," Paulsen et al. say, "our experiments and simulations, as well as previous work, indicate that the shape is wrinkled but has no folds."<|endoftext|> TITLE: Longest subgroup chain in $\mathrm{SL}_n(\mathbb{F}_p)$? QUESTION [10 upvotes]: What is the length $\ell$ of longest chain of subgroups $$\{e\} \lneq H_1 \lneq \dotsc \lneq H_\ell = G$$ in $\mathrm{SL}_n(\mathbb{F}_p)$? REPLY [15 votes]: We write $\ell(G)$ for the length of the longest subgroup chain of a group $G$. If $G$ is of Lie type, write $B$ for a Borel subgroup of $G$, and note that $\ell(B)$ is equal to the number of prime divisors of $|B|$ (since $B$ is solvable). Let me explain the situation in the literature as I understand it. Unfortunately it will only yield a partial answer to the question. We start with this paper.... Solomon, Ron; Turull, Alexandre, Chains of subgroups in groups of Lie type. III, J. Lond. Math. Soc., II. Ser. 44, No. 3, 437-444 (1991). ZBL0776.20007. ... whose main theorem is as follows: Theorem: For all primes p, there exist an integer $F$ such that if $G=G_r(p^m)$ is a quasisimple of Lie type and $m\geq F$, then $\ell(G)=\ell(B)+r$. Unfortunately, the requirement that $m\geq F$ means that the result does not apply here... However, one of the main theorems of this paper Solomon, Ron; Turull, Alexandre, Chains of subgroups in groups of Lie type. I, J. Algebra 132, No. 1, 174-184 (1990). ZBL0714.20011. implies the following Theorem: Suppose that $G_r(2^m)$ is a quasisimple group of Lie type. Then $\ell(G)=\ell(B)+r$. So that, at least, answers your question for $p=2$. For $p\neq 2$, the middle paper in the above series (which includes Seitz as an author) is the following: Seitz, Gary M.; Solomon, Ron; Turull, Alexandre, Chains of subgroups in groups of Lie type. II, J. Lond. Math. Soc., II. Ser. 42, No. 1, 93-100 (1990). ZBL0728.20018. Theorem B of that paper implies the following: Theorem: Suppose that $K$ is a positive real number, $m$ a positive real number and $G_r$ any group scheme of Lie type. Then there exists $p$ a prime such that $\ell(G)>\ell(B)+r+K$. Thus the bound for $p=2$ fails "as badly as possible" for odd $p$. I've scanned this paper, and the way they show this theorem for $SL_n(p^m)$ is by looking at the normalizer, $N$, of a torus $T$. Edit: I wrote earlier that one should consider the normalizer of a maximal split torus to see why the latter theorem works. In actual fact one should use the normalizer of non-split torus to get a longer chain. So, for instance, if $n$ is odd and if $p^m+1$ has many more prime divisors than $p^m-1$, then one might look $N$, the normalizer of the torus $(p^{2m}-1)^d$ where $2d+1=n$. One can see that $$\ell(N) = d\pi(p^{2m}-1) + \ell(S_d),$$ which, depending on $p$ and $n$, will get you a longer chain. To calculate precisely how long a chain you obtain you can use the fact that $\ell(S_n)=\lfloor{\frac{3n-1}{2}}\rfloor - b_n$, where $b_n$ is the number ones in the base $2$ expansion of $n$, by a theorem here: Cameron, Peter J.; Solomon, Ron; Turull, Alexandre, Chains of subgroups in symmetric groups, J. Algebra 127, No. 2, 340-352 (1989). ZBL0683.20004. Speculation: Let me conjecture what might be the actual answer to your question. It's easier to work inside $GL_n(p)$ rather than $SL_n(p)$. It seems to me that one can do even better than torus normalizers, by looking at Levi subgroups. Consider a Levi subgroup, $$L=GL_{d_1}(p^{i_1}) \times \cdots \times GL_{d_k}(p^{i_k}),$$ where $\sum d_j i_j=n$. By taking the Borel subgroup of this, one can obtain a subgroup chain in $L$ of length $$ f(L)= n-k + \sum\limits_{j=1}^k \left( d_i\pi(p^{i_j}-1) + \frac12d_j(d_j+1)\right).$$ This would then yield a subgroup chain in $G$ of length something like $g(L)=f(L)+k-1+\frac12\left(n^2-\sum_{i=1}^k d_i^2\right)$ (I think). Now $\ell(G)$ would be the maximum of all such values (so maximise over $L$, where we think of $L$ as a partition of the integer $n$).<|endoftext|> TITLE: Gadgets as primality tests QUESTION [5 upvotes]: From the literature, showed below, I know two gadgets that provide a way to know if a positive integer (a positive quantity of units) is composite or a prime number. I would like to know if in the literature or from your invention it is possible to show other different gadgets that provide us primality tests. Question. Do you know different gadgets or mechanisms from the literature that can be used as a primality test? Then, please add the references asnwering my question as a reference request and I try to read it from the literature. Are you able to provide a different gadget from your invention that provide (its idealization as a physical machine) a way to determine if a quantity (of something) is prime or composite? Many thanks. Feel free to provide draws of your machine as companion of your explanation of how and why works it as a primality test. References: [1] A. K. Dewdney, On the spaghetti computer and other analog gadgets for problem solving, Scientific American Volume 250  Issue 6 (June 1984), Computer Recreations p. 19-26. [2] Francisco Javier Díaz Aspe, Cómo detectar primos usando una cuerda con nudos, Miniaturas matemáticas de La Gaceta de la RSME, La Gaceta de la Real Sociedad Matemática Española, Núm. 1, Pág. 80 Vol. 22 (2019). REPLY [5 votes]: I interpret a "gadget" as a physical device that operates in an analog, rather than a digital way (to exclude a computer). The OP asks for "primality tests", but if I may broaden the question to include "prime number generators", there is a variety of such gadgets, collected at unusual and physical methods for finding prime numbers. The gadgets use effects from chemistry (Biochemical identification of prime numbers), biology ( A Biological Generator of Prime Numbers, and physics An optical Eratosthenes' sieve for large prime numbers. The latter would qualify as a primality test, I cite the abstract and show a figure from that paper: We report the first experimental demonstration of prime number sieve via linear optics. The prime numbers distribution is encoded in the intensity zeros of the far field produced by a spatial light modulator hologram, which comprises a set of diffraction gratings whose periods correspond to all prime numbers below 149. To overcome the limited far field illumination window and the discretization error introduced by the finite spatial resolution, we rely on additional diffraction gratings and sequential recordings of the far field. This strategy allows us to optically sieve all prime numbers below $149^2 = 22201$.<|endoftext|> TITLE: System of linear pde with non constant coefficients QUESTION [7 upvotes]: I'm a physicist and during my research work, I found a system of linear pde with non constant coefficients that I have to study, since I have totally no experience about systems of pde and I have even problems to find some references. I would be really great to have some advice about how to face the problem. What I would like to do is of course find a solution of this system but even proving the existence of the solution can be a first great step. So thank you in advance for any possible help and some good reference. Here in the link you can find the system: System of pde $$\begin{pmatrix} 0 & f_1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & f_1 \end{pmatrix}\partial_x \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} + \begin{pmatrix} f_2 & 0 & 0 \\ 0 & 0 & f_2 \\ 0 & 0 & 0 \end{pmatrix}\partial_y \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 0 \\ 0 & f_3 & 0 \\ f_3 & 0 & 0 \end{pmatrix}\partial_z \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix}= \begin{pmatrix} \omega_1 \\ \omega_2 \\ \omega_3\end{pmatrix}$$ Where we have three independent variables $(x,y,z)$ and three dependent variables $(N_1,N_2,N_3)$. The functions $\omega_i:R^3\rightarrow R$ $f_i: R^3\rightarrow R $, $N_i:R^3\rightarrow R $ are all smooth functions. Given that the matrix of coefficients are all singular I don't know how to proceed even about the classification of this system. Like I said I would be interested in a local solution or at least the proof of the existence of the solution. Any reference or help about this system will be really appreciated. *Edit, sorry but for some reason the editor was not allowing me to write the matrices that I need in a good way, so I put a link to the image of the system that I create with latex. REPLY [5 votes]: Recall that a first order system $A_{ij}^\mu \partial_\mu \phi^j + B_{ij} \phi^j = 0$ (the right hand-side could also be inhomogeneous) is symmetric hyperbolic when there exists at least one covector $p_\mu$ such that $A_{ij}^\mu p_\mu > 0$, the contraction is a positive definite symmetric matrix. If the coefficients $A_{ij}^\mu = A_{ij}^\mu(x)$ are $x$-dependent, then the positivity condition should be satisfied for all $x$. There are no special conditions imposed on $B_{ij}$, which could also be $x$-dependent. This is all standard. And the initial value problem for such systems is well posed, at least on those domains where the causal structure of formed by the cones of $p_\mu$'s satisfying the above positivity conditions is globally hyperbolic (the domain admits a foliation by Cauchy surfaces). See for instance the excellent account in Chapter 7 of [1]. Let me now introduce some non-standard terminology. I call a linear system $$A_{ij}^{\mu_1\cdots \mu_k} \partial_{\mu_1} \cdots \partial_{\mu_k} \phi^j + B(\phi,\partial\phi, \cdots \partial^{k-1}\phi) = 0 \tag{*}$$ higher order symmetric hyperbolic when it can be put in symmetric hyperbolic form by reduction to first order. Next, I call a linear system (using the notation $A_{ij}(\partial) = A_{ij}^{\mu_1\cdots \mu_k} \partial_{\mu_1} \cdots \partial_{\mu_k}$ and writing l.o.t for what I wrote as $B_{ij}(\cdots)$ above) $$A_{ij}(\partial) \phi^j + l.o.t = 0$$ generalized symmetric hyperbolic when there exists a complementary operator $C_k^k(\partial) = (C_k^i)^{\mu_1\cdots \mu_l} \partial_{\mu_1} \cdots \partial_{\mu_l}$ such that applying it to (*) gives a system $$ C_k^i(\partial) A_{ij}(\partial) \phi^j + l.o.t = 0$$ that is higher order symmetric hyperbolic. The point is that a generalized symmetric hyperbolic system has an equally well-posed initial value problem, with or without a source term, as a symmetric hyperbolic one. Unfortunately, I don't have a reference for this, except for the parallel discussion that I gave for generalized normally hyperbolic systems in the recent paper [2] (just replace "normally hyperbolic" by "symmetric hyperbolic" everywhere). Claim: Your PDE system is generalized symmetric hyperbolic. Let me first illustrate by a relevant example, what it means for a system to be higher order symmetric hyperbolic. Take the equation $$\delta_{ij} \partial_x \partial_y \partial_z N^j + l.o.t = 0 , \tag{**}$$ where $\delta_{ij}$ is just the Kronecker delta. By introducing the auxiliary variables $N_z^j = \partial_z N^j$ and $N_{yz}^j = \partial_y N_z^j$, it can be reduced to the first order system $$\begin{pmatrix} \delta_{ij}\partial_z & 0 & 0 \\ 0 & \delta_{ij}\partial_y & 0 \\ 0 & 0 & \delta_{ij}\partial_x \end{pmatrix} \begin{pmatrix} N^j \\ N_z^j \\ N_{yz}^j \end{pmatrix} + l.o.t = 0 ,$$ where the desired positivity is satisfied by any covector from the first octant, $p_x,p_y,p_z>0$. This shows that (**) is indeed higher order symmetric hyperbolic. Your PDE system has the special form $$\begin{pmatrix} f_2 \partial_y & f_1 \partial_x & 0 \\ 0 & f_3 \partial_z & f_2 \partial_y \\ f_3 \partial_z & 0 & f_1 \partial_x \end{pmatrix} \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} = \begin{pmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \end{pmatrix} .$$ As mentioned in Deane Yang's answer, this equation cannot be put into symmetric hyperbolic form. That is, there exists no matrix $C_k^i$ such that the principal symbol $C_k^i A_{ij}^\mu p_\mu$ satisfies both the symmetry and positivity conditions. You can find $C_k^i$ such that the symbol becomes symmetric, but not positive. At least one eigenvalue of the resulting $p$-dependent matrix will always remain negative. However, assuming that $f_1,f_2,f_3 \ne 0$ are nowhere vanishing, one can multiply this system by a second order matrix differential operator (which you can read off from the formula below) such that the system becomes $$\partial_x \partial_y \partial_z \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} + l.o.t = \frac{1}{2 f_1 f_2 f_3} \begin{pmatrix} f_1 f_3 \partial_x \partial_z & -f_1^2 \partial_x^2 & f_1 f_2 \partial_x \partial_y \\ f_2 f_3 \partial_y \partial_z & f_1 f_2 \partial_x \partial_y & -f_2^2 \partial_y^2 \\ -f_3^2 \partial_z^2 & f_1 f_3 \partial_x \partial_z & f_2 f_3 \partial_y \partial_z \end{pmatrix} \begin{pmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \end{pmatrix} .$$ From the example (**) it should now be obvious that this system is higher order symmetric hyperbolic, verifying the claim that your system is generalized symmetric hyperbolic. Therefore, it has a well-posed initial-value problem on any hyperplane whose conormal $p_\mu$ lies in the first octant. [1] Ringström, Hans, The Cauchy problem in general relativity., ESI Lectures in Mathematics and Physics. Zürich: European Mathematical Society (EMS) (ISBN 978-3-03719-053-1/pbk). xiii, 294 p. (2009). ZBL1169.83003. [2] See the discussion around Lemma 3 in García-Parrado, Alfonso and Khavkine, Igor, Conformal Killing Initial Data, Journal of Mathematical Physics 60 122502 (2019). [arXiv:1905.01231]<|endoftext|> TITLE: Almost complex structures on cotangent bundles of almost complex manifolds QUESTION [6 upvotes]: Let $M$ be a smooth manifold and $I:TM\to TM$ an integrable almost complex structure. Then, the cotangent bundle $T^*M$ admits a canonical complex structure, which can be built from holomorphic charts on $M$. If $I$ is not necessarily integrable, is there always an almost complex structure on $T^*M$ such that $T^*M\to M$ is holomorphic? Is it canonical? If not, can we construct it using some additional choice on $M$? REPLY [4 votes]: There is one due to Satô, "Almost analytic vector fields in almost complex manifolds". He constructs a $J'$ and $J''$ on $T^\ast M$ using a $J$ on $M$, such that $J''$ is always an almost complex structure but $J'$ is only an almost complex structure when $J$ is integrable (in which case $J'$ is also integrable), and $J''$ is a modified version of $J'$ using the nontrivial Nijenhuis tensor of $J$. This is elaborated on in Yano--Patterson's paper "Vertical and complete lifts from a manifold to its cotangent bundle" (which might be easier to digest). There is also a way using connections on $M$ (which generalizes Satô's work) due to Yano--Patterson, "Horizontal lifts from a manifold to its cotangent bundle". Then there is a unifying generalization of both works, see Bertrand's paper "Almost complex structures on the cotangent bundle" (https://arxiv.org/abs/math/0507068); here Theorem 3.1 asserts holomorphicity of the projection $\pi$.<|endoftext|> TITLE: Is there ever a Kunneth isomorphism just for powers? QUESTION [6 upvotes]: It's pretty rare for a multiplicative cohomology theory $E$ to have a Kunneth isomorphism $E^\ast(X \times Y) \cong E^\ast(X) \otimes_{E^\ast(pt)} E^\ast(Y)$ for all spaces $X,Y$. Are there any examples of cohomology theories $E$ which have a Kunneth isomorphism just for powers, but not for all products of spaces? That is, Question: Are there examples of multiplicative cohomology theories $E$ such that for all spaces $X$ and all $n \in \mathbb N$, the canonical map $$E^\ast(X) \otimes_{E^\ast(pt)} \dots \otimes_{E^\ast(pt)} E^\ast(X) \to E^\ast(X\times \dots \times X)$$ is an isomorphism (where the tensor product and cartesian product are taken $n$ times), and yet there are spaces $X,Y$ such that the canonical map $$E^\ast(X) \otimes_{E^\ast(pt)} E^\ast(Y) \to E^\ast(X \times Y)$$ is not an isomorphism? I'd be happy to understand this question in slightly different contexts, such as assuming $X,Y$ to be finite, or taking them to be spectra or pointed spaces rather than spaces, or assuming that $E$ has more structure. As alluded to at the start, the condition that $E$ doesn't have a Kunneth isomorphism in general doesn't rule out too many $E$'s -- I think just $H\mathbb F_p$, $K(n)$, $H \mathbb Q$, and things constructed from them. REPLY [20 votes]: The Künneth map is an isomorphism for trivial reasons if $X$ or $Y$ are empty. Otherwise, pick points $x_0\in X,y_0\in Y$; the idempotents $p_x:x\mapsto x,y\mapsto y_0, p_y:x\mapsto y_0,y\mapsto y$ induce the projections of $E^*(X\sqcup Y)\cong E^*(X)\oplus E^*(Y)$ onto its two summands. By naturality, the Künneth map $E^*(X\sqcup Y)\otimes_{E^*(pt)} E^*(X\sqcup Y)\to E^*((X\sqcup Y)\times (X\sqcup Y))$ preserves the decomposition of its source and target into the canonical four summands, and on each of these it is just the corresponding Künneth map. In particular, it is invertible if and only if the Künneth maps for $(X,X),(Y,Y),(X,Y),(Y,X)$ are all invertible. So your assumption on powers implies the Künneth map is always an isomorphism. (This is of course just the usual polarization identity which recovers a bilinear form from the corresponding quadratic form, lifted to spaces.)<|endoftext|> TITLE: Holomorphic deformation of complex structure on the real plane QUESTION [8 upvotes]: It is known that each complex structures on $\mathbb{R}^2$ is biholomorphic to either $\mathbb{C}$ or the open unit disk $\Delta$. One can continuously deform one complex structure to the other as is for example done in Winkelmann - Deformations of Riemann surfaces (page 3). My question is Can this deformation be taken to be holomorphic on the deformation parameter? That is, does there exist a non-trivial complex analytic family $M \to D$ where $D \subset \mathbb{C}$ is a small disk, the central fiber is biholomorphic to $\mathbb{C}$, and the generic fiber is biholomorphic to $\Delta$? Note that all the theorems that assure complex analytic triviality of deformations when $H^1(X,TX)$ vanishes, use the hypothesis that $X$ is compact. REPLY [7 votes]: The answer is no. This follows from the so-called $\lambda$-lemma of Sullivan, Mane Sad and Lyubich: Let $D$ be a disk, $C$ the complex plane and $A$ any set in C. Let $f:D\times A\to C$ be a function with the following properties: $\lambda\mapsto f(\lambda,z)$ is holomorphic for every $z\in A$. $z\mapsto f(\lambda,z)$ is injective for every $\lambda\in D$, $z\mapsto f(0,z)=z$ for every $z\in A$. Then for every $\lambda\in D$ the map $z\mapsto f(\lambda,z)$ is quasisymmetric (=quasiconformal if $A$ is open) Since the plane is not conformally equivalent to a disk, this implies that the answer to your question is negative. Ref. MR0732343 Mañé, R.; Sad, P.; Sullivan, D., On the dynamics of rational maps, Ann. Sci. École Norm. Sup. (4) 16 (1983), no. 2, 193–217.<|endoftext|> TITLE: A good reference to the Gauss result on the structure of the multiplicative group of a residue ring QUESTION [7 upvotes]: I need a good reference (desirably some textbook in Number Theory) to the following known result, attributed to Gauss in Wikipedia. Theorem (Gauss). Let $p$ be a prime number, $k\in\mathbb N$ and $\mathbb Z_{p^k}^\times$ be the multiplicative group of invertible elements of the residue ring $\mathbb Z_{p^k}:=\mathbb Z/p^k\mathbb Z$. If $p$ is odd, then the group $\mathbb Z_{p^k}^\times$ is cyclic. If $p=2$ and $k\ge 3$, then the element $-1+2^k\mathbb Z$ generates a $2$-element subgroup $C_2\subset\mathbb Z_{2^k}^\times$ and the element $5+2^k\mathbb Z$ generates a cyclic subgroup $C_{2^{k-2}}\subset \mathbb Z_{2^k}^\times$ of order $2^{k-2}$ such that $\mathbb Z_{p^k}^\times=C_2\oplus C_{2^{k-2}}$. Please help! REPLY [6 votes]: Another reference is Proposition 2.1.24, page 22, in Henri Cohen's "Number Theory – Volume I: Tools and Diophantine Equation".<|endoftext|> TITLE: What makes skew characters of the symmetric group special? QUESTION [7 upvotes]: For integer partitions $\mu\subset\lambda$ we can define the skew character $\chi^{\lambda/\mu}$ (for example?) via the Littlewood-Richardson rule. Many combinatorial gadgets and algorithms extend in a very natural way from the case of partitions to the skew case. I think it is fair to expect that characters corresponding to partitions are special, because they are irreducible. I would like to know whether there are (representation theoretic) properties of skew characters that make them special. REPLY [5 votes]: Here are a few references to expand on my comment about "convexity" properties of skew Schur functions $s_{\lambda/\mu}$. As the OP points out, it's maybe most natural for the question to assume that someone has handed you the decomposition of $s_{\lambda/\mu}$ into irreducibles, i.e., they've given you a list of LR coefficients. For a "convexity" conjecture on LR coefficients (which could be translated to say something about the list you've been given) see Conjecture 1 (due to Lam-Postnikov-Pylyavskyy) of https://arxiv.org/abs/math/0608134. Many special cases of that conjecture are known- see the references given in that paper. But another thing you can do is think of $s_{\lambda/\mu}$ as a polynomial (i.e., do the monomial expansion), and ask questions about its convexity properties. For instance, it is known that the nonzero coefficients of this polynomial are exactly the lattice points of some convex polytope (this is called the "saturated Newton polytope" (SNP) condition); see https://arxiv.org/abs/1703.02583. It is also conjectured that the "normalized" version of $s_{\lambda/\mu}$ is "Lorentzian" (another kind of convexity property) in https://arxiv.org/abs/1906.09633. These kind of convexity properties are likely enough to tell the character $s_{\lambda/\mu}$ apart from a "random" character of a symmetric group; but it is doubtful they would lead to a procedure for exactly recognizing $s_{\lambda/\mu}$. For instance, the Stanley symmetric functions (mentioned in a previous answer), also satisfy the SNP property (and indeed this is how the SNP property is deduced for skew Schur polynomials).<|endoftext|> TITLE: Holomorphic Weinstein Lagrangian neighborhood theorem QUESTION [7 upvotes]: The Weinstein Lagrangian neighborhood theorem says that if $(M,\omega)$ is a symplectic manifold and $L\subset M$ is a Lagrangian submanifold, then there are neighbourhoods $U$ of $L$ in $M$, and $U'$ of the zero-section in $T^*L$, and a symplectomorphism $U\to U'$ which restricts to the identity on $L$. Question: Is there a holomorphic version of this theorem for holomorphic-symplectic manifolds and complex-Lagrangian submanifolds? Presumably, we would need some additional conditions on the embedding since the tubular neighborhood theorem doesn't necessarily hold in the holomorphic setting. REPLY [9 votes]: A classic example that shows that WLNT doesn't always hold in the holomorphic category is an elliptically fibered $K3$ surface. A K3 surface $S$ is a compact complex symplectic manifold of complex dimension $2$. Any smooth curve $C\subset S$ is a Lagrangian submanifold. If the Darboux Theorem were true in the sense that a neighborhood of the curve were always symplectically biholomorphic with a neighborhood of the zero section of the cotangent bundle, then, for a nonsingular elliptic curve $C\subset S$, the neighborhood would be a product and so the $1$-parameter family of nearby deformations of the curve $C$ in $S$ would all be isomorphic to it, i.e., they would all have the same $j$-invariant. However, this is known not to be the case: When you have an elliptically fibered $K3$, the $j$-invariant of the elliptic fibers is not constant. If one doesn't need the ambient symplectic manifold to be compact, there is an easier example: Let $z$ be the standard complex coordinate on $\mathbb{C}$ and let $w$ be the standard complex coordinate on the upper half-plane $\mathbb{U}^+\subset\mathbb{C}$. Let $\mathbb{Z}^2$ act (freely) on $X = \mathbb{C}\times\mathbb{U}^+$ by $$ (m,n)\cdot(z,w) = (z + m + nw,\,w). $$ Let $M$ be the quotient of $\mathbb{C}\times\mathbb{U}^+$ by this $\mathbb{Z}^2$-action. This action preserves the symplectic form $\mathrm{d}z\wedge\mathrm{d}w$, so that $M$ is a symplectic surface, as well as the projection $w:M\to \mathbb{U}^+$, whose fiber in $M$ over $a\in \mathbb{U}^+$ is a torus $w^{-1}(a) \simeq \mathbb{C}/(\mathbb{Z}{+}\mathbb{Z}a)$. These tori are Lagrangian curves in $M$, but no neighborhood of such a torus is biholomorphic to the product of the torus and a disk, precisely because the tori near a given $w$-fiber (which are all $w$-fibers since $w$ would have to be constant on any compact, connected curve in $M$) are not biholomorphic to the given $w$-fiber. Maybe the result you want is to know that any symplectic form on a neighborhood of the zero section of the cotangent bundle that has the zero section as a Lagrangian submanifold is actually symplectomorphic to the canonical symplectic form on some neighborhood of the zero section of the cotangent bundle. This statement is true, and the standard homotopy argument proves it. (You don't need to use local coordinates for the proof, just the radial vector field on the cotangent bundle to construct the homotopy.) Added Remark: However, this result is particular for the zero section of the cotangent bundle of $L$. It is not in general true that, for any two holomorphic symplectic structures $\omega_1$ and $\omega_2$ on a complex manifold $M$ that have a common holomorphic Lagrangian submanifold $L\subset M$, there exist open $L$-neighborhoods $U_1$ and $U_2$ in $M$ such that there exists a biholomorphic mapping $\phi:U_1\to U_2$ fixing $L$ and satisfying $\phi^*\omega_2 = \omega_1$. For example, taking the above torus fibration $w: M\to \mathbb{U}^+$, for different nonvanishing holomorphic functions $f_1,f_2:\mathbb{U}^+\to\mathbb{C}$, the symplectic forms $\omega_i = f_i(w)\,\mathrm{d}z\wedge\mathrm{d}w$ for $i=1,2$ will not generally be locally holomorphically symplectomorphic near any $w$-fiber.<|endoftext|> TITLE: Rank matrices for type $D$ Bruhat order QUESTION [8 upvotes]: Roughly, this question asks how the Bruhat (strong) order in type $D$ can be understood like the Bruhat orders in types A and B=C. I'll review how types A and B work before asking my question. As a side note, I tried to ask this question earlier today, then deleted it with the intention of fixing some errors and reasking it, but I can't find the deleted question. Notation: I write $[n]$ for $\{1,2,\ldots, n \}$. Type A The Bruhat order of type $A_{n-1}$ is a partial order on the group of permutation of $[n]$. It can be described in the two following manners: ("Tableaux criterion") Put a partial order on the set of $k$ element subsets of $[n]$ as follows: For $I$, $J$ two such subsets, sort $I = \{ i_1 < i_2 < \cdots < i_k \}$ and $J = \{ j_1 < j_2 < \cdots < j_k \}$, and define $I \leq J$ if $i_1 \leq j_1$, $i_2 \leq j_2$, ..., $i_k \leq j_k$. Then, for permutations $u$ and $v$, we have $u \leq v$ if and only if $u[k] \leq v[k]$ for $1 \leq k \leq n-1$. ("Rank matrix criterion") We have $u \leq v$ if and only if, for all $1 \leq k, \ell \leq n-1$, we have $\#(u[k] \cap [\ell]) \leq \#(v[k] \cap [\ell])$. The numbers $r_{k \ell} = \#(u[k] \cap [\ell])$ are called the rank matrix of $u$. It is often useful to formally define $r_{0k}=r_{k0} = 0$ and $r_{kn}=r_{nk}=k$. With those boundary definitions, rank matrices are characterized by $0 \leq r_{(k+1)\ell} - r_{k \ell},\ r_{k(\ell+1)} - r_{k \ell} \leq 1$ and, if $r_{(k+1)\ell} = r_{k(\ell+1)} = r_{k\ell}+1$ then $r_{(k+1)(\ell+1)} = r_{k\ell}+2$. Type B Let $\sigma : [2n] \to [2n]$ be the involution $\sigma(k) = 2n+1-k$. Then the Coxeter group of type $B$ is the centralizer of $\sigma$ in $S_{2n}$. Bruhat order in type $B$ is the induced order from $S_{2n}$, and thus can be described as by either the Tableaux Criterion or the Rank Matrix Criterion. In either case, we may cut roughly in half the number of cases which need to be checked, because $u[k]$ determines $u[2n-k]$, so we only need check the conditions for $1 \leq k \leq n$. Type D The Coxeter group of type $D_n$ is the index two subgroup of $B_n$ consisting of permutations for which $\#(u [n] \cap [n]) \equiv n \bmod 2$. The Bruhat order is no longer induced from $B_n$. It seems to follow from other things I will say below that there is a quick definition of it as an induced order though, so that will be my first question: Let $\tau$ be the permutation $(n \ n+1)$ in $B_n$; this permutation is not in $D_n$ but normalizes $D_n$. Embed $D_n$ into $B_n \times B_n$ by $u \mapsto (u, \tau u \tau^{-1})$. Is the Bruhat order on $D_n$ simply the $B_n \times B_n$ Bruhat order restricted to the image? In any case, what I can find in sources is that something like the tableaux criterion holds. Namely, in any Coxeter group, there is a collection of subgroups called the maximal parabolics, and there are partial orders on the quotients by the maximal parabolics such that $u \leq v$ if and only if their cosets in by each maximal parabolic $P$ obey $u P \leq v P$. To make this sound more like the tableaux criterion, note that the maximal parabolics in type $D_n$ are the stabilizers of $[1]$, $[2]$, ..., $[n-2]$, $[n]$ and $[n]':=\tau([n])$. So we can identify cosets of maximal parabolics with $D_n$ orbits of these sets. So we can detect whether $u \leq v$ by comparing $uX$ and $vX$ for some poset relation, with $X$ in the list $[1]$, $[2]$, ..., $[n-2]$, $[n]$ and $[n]':=\tau([n])$. But I haven't found a source which spells out how to order the $D_n X$'s. What, explicitly, is the order on the $D_n X$'s? I tried to work this out myself, and I believe I got the following: Let $Y$ and $Z \in D_n X$ for $X$ as above. Then the poset relation is that we have both $Y \leq Z$ and $\tau(Y) \leq \tau(Z)$, in our order on subsets of $[2n]$. Is this right? Is there a source for this? It seems to me I can also encode this in a rank matrix style. Namely, for $X$ and $Y$ one of $[1]$, $[2]$, ..., $[n-2]$, $[n]$, $[n]'$, $[n+2]$, ..., $[2n-2]$, $[2n-1]$ and $u$ in the $D_n$ Coxeter group, let $r_{XY}(u) = \#(uX \cap Y)$. Is it true that $u \leq v$ if and only if $r_{XY}(u) \leq r_{XY}(v)$ for all such $X$, $Y$? Is there a source for this? Finally, one could ask for a simple local characterization of the $r_{XY}$'s, similar to the characterization I gave above for type $A$ rank matrices. Is such a characterization known? REPLY [2 votes]: Here are some more thoughts about how the Type D Bruhat order is more complicated than the Type A and Type B/C orders. These ideas might even suggest that giving a "rank matrix"-like description of the partial order is "impossible" in Type D. There is a certain property of posets called "clivage" (by Lascoux and Schützenberger) or "dissective" (by Reading). Lascoux and Schützenberger (cited below) showed that the Type A and Type B Bruhat orders are dissective. Meanwhile, Geck and Kim (cited below) showed that Type D Bruhat order is not dissective (they address the exceptional types as well). Also, it is known that a finite poset $P$ is dissective if and only if the MacNeille completion of $P$ is distributive (see Theorem 7 of the paper of Reading cited below). (As an aside, the MacNeille completion of the Bruhat order of Type A is the distributive lattice of Monotone Triangles (a.k.a. Alternating Sign Matrices) with componentwise order. See the paper of Brualdi and Schroeder cited below for more on this lattice.) So, Type D Bruhat order lacks a nice property ("dissective") which Type A and B have. But what does this have to do with the possibility of a "rank matrix"-like description of the partial order in Type D? Well, intuitively at least, if you have a partial order defined by componentwise order on some arrays of numbers satisfying certain inequalities, then to formally extend your partial order to be a lattice, what you'll end up doing is taking $\min$'s and $\max$'s of the entries until you get enough new arrays so that everyone has a meet and join. And if so, the result will be distributive because $\min$'s and $\max$'s distribute over one another. But we know Type D Bruhat order does not have a distributive lattice as its completion, so it "can't" have a partial order given by comparing arrays of numbers in this way. Brualdi, Richard A.; Schroeder, Michael W., Alternating sign matrices and their Bruhat order, Discrete Math. 340, No. 8, 1996-2019 (2017). ZBL1366.15024. Geck, Meinolf; Kim, Sungsoon, Bases for the Bruhat-Chevalley order on all finite Coxeter groups, J. Algebra 197, No. 1, 278-310 (1997). ZBL0977.20033. Lascoux, Alain; Schützenberger, Marcel-Paul, Lattices and bases of Coxeter groups, Electron. J. Comb. 3, No. 2, Research paper R27, 35 p. (1996); printed version J. Comb. 3, No. 2, 633-667 (1996). ZBL0885.05111.. Reading, Nathan, Order dimension, strong Bruhat order and lattice properties for posets, Order 19, No. 1, 73-100 (2002). ZBL1007.05097.<|endoftext|> TITLE: Sphere packing processes during biological development QUESTION [6 upvotes]: Within the context of mathematical biology, a sphere packing problem occurred to me. I must note that unlike the typical sphere packing problems, the variant I consider involves minimising the average distance between spheres rather than maximising the density of spheres. Furthermore, the spheres I consider are compressible rather than hard. I have attempted to solve this problem numerically for a finite number of spheres but I'd also like to learn more about exact solutions in the finite regime as well as the asymptotic regime. The mathematical problem is introduced in a self-contained manner but for those who are curious I decided to share some motivation below. Motivation: A number of excellent scientists including Alan Turing[2] and D'Arcy Wentworth Thompson[3] have turned to mathematical modelling in order to understand morphogenesis. Though I may be slightly biased due to my training in mathematics, it recently occurred to me that early organogenesis may be modelled as a sphere packing process. I think this might partially explain the highly symmetric forms of amphibian eggs. Consider a mass of spherical cells contained in a small region of space undergoing processes of division and specialisation in order to develop a specific geometrical structure. Now, let's suppose that vascularisation hasn't started so inter-cellular signalling via chemical signals is diffusion-constrained. Furthermore, let's make the reasonable assumption that the strength of a chemical signal diminishes as $\sim \frac{1}{r^2}$ where $r$ is the euclidean distance from the source cell. We may then conjecture that in order for growth to be coordinated the average pair-wise distance between cells must be minimised at this stage of development. Allow me to clarify the details of this mathematical problem. Finite sphere-packing with soft constraints: Let us suppose that there are $N$ compressible balls whose centres,$\{r_i\}_{i=1}^N$, are initialised by sampling uniformly from $[-l,l]^3$. Our challenge is to find $\{r^*_i\}_{i=1}^N \in \mathbb{R}^3$ that minimises: \begin{equation} U_{\text{net}} = \sum_{i=1}^N \sum_{j \neq i} U(r_i,r_j) \tag{1} \end{equation} where \begin{equation} U(r_i,r_j)=(\lVert r_i-r_j \rVert^2 - 2l^2)^2 \tag{2} \end{equation} is the form of each local potential function. Furthermore, my intuition for introducing the constant $2l^2$ is that this would allow cells to synchronise their chemical signalling activity. As pointed out below, I managed to solve this problem approximately but I am also interested in exact solutions that may be found analytically. Might the general solution for all $N \in \mathbb{N}^*$ be known to pure mathematicians? I am particularly curious about the geometry and topology of the optimal shape as $N$ increases. Does this geometry become unique as $N\to \infty$? Note: I must add that while the constants $l$ and $2l^2$ may be important in a numerical setting, I don't see how they affect the topology or the number of symmetries of the optimal shape in an analytical setting. To a pure mathematician these constants are probably irrelevant.  Approximate solutions: I managed to find approximate solutions in the case where $l=5$ and $4 \leq N \leq 20$ by solving this optimisation problem using gradient descent with the hyperopt library: https://github.com/hyperopt/hyperopt. For a sanity check I found that when $N=4$ I obtain a tetrahedron as expected. The python code I used is also available on Github: https://gist.github.com/AidanRocke/f48cea76eeaae5c25e11e9e47c9315c7 References: Weisstein, Eric W. "Sphere Packing." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/SpherePacking.html Turing, A. M. (1952). "The Chemical Basis of Morphogenesis" (PDF). Philosophical Transactions of the Royal Society of London B. Thompson, D. W., 1992. On Growth and Form. Dover reprint of 1942 2nd ed. (1st ed., 1917) Bergstra, J., Yamins, D., Cox, D. D. (2013) Making a Science of Model Search: Hyperparameter Optimization in Hundreds of Dimensions for Vision Architectures. To appear in Proc. of the 30th International Conference on Machine Learning (ICML 2013). REPLY [5 votes]: It is unlikely that the solution is known generally (for all $N$), but there are partial results known for similar kernels, or for these kernels with additional constraints. At first approximation your kernels are known as power-law attractive-repulsive kernels, meaning that potential energy between two points is decreasing at first as a function of the distance but then increasing for large distances. A closer look shows that the potentials you are interested in fit in the framework of a number of recent papers. When you take $l=1/\sqrt{2}$ for instance, the potential function $U(x)$ may be written as a function of distance $f(x)=f(|x|)$, with $f(x)=\frac{|x|^\alpha}{\alpha}-\frac{|x|^\beta}{\beta}$, and where $(\alpha,\beta)=(4,2)$ (the constant term does not play into the minimization problem, so it can be subtracted off from the potential function). These potentials are the main object of an increasing number of papers, and take the name of power-law mildly repulsive kernels, whenever $\alpha>\beta>2$ ([BCLR1, BCLR2, CFP]). In many of these papers the problem of minimizing over measures is considered instead of the discrete problem. Since the minimizing probability measures still are often supported on discrete sets, this still proves to be fruitful for understanding discrete energy minimizers. With $\beta=2$, the kernel in question is outside the mildy repulsive case, and this case appears more subtle (since many results that apply for $\beta>2$ cannot generally apply to $\beta=2$). A quantitative version of Theorem 1.4 in [LM] might explain why the regular simplex (tetrahedron) shows up as a minimizer for $ N=4$, but the result which appears there does not establish this. It only establishes that there exists a value of $\beta\geq\alpha$ for which the (repeated) regular simplex minimizes the discrete energy for all power-law potentials of this form and all $N$ divisible by $4$. Incidentally, if you add the constraint that all the minimizing configuration 'particles' lie on a sphere (as happens for the regular simplex) one can use a linear programming approach, as appears in our recent pre-print [BGMPV], to show optimality of the regular octahedron, regular simplex, and the icosahedron for similar potentials (those whose derivatives, as a function of the inner product $\langle x,y \rangle$, are positive up to some order) when $3,4,6$, or $12$ divides the number of particles $N$. The approach here is similar to that in [CK], relying on Hermite interpolation at certain nodes, but the continuous setting introduces an extra interpolation point (due to the probability measure constraint). [BCLR1] D. Balague, J. Carrillo, T. Laurent, and G. Raoul. Nonlocal interactions by repulsive–attractive potentials: radial ins/stability. Phys. D 260 (2013), 5–25. arXiv:1109.5258 MR3143991 [BCLR2] D. Balague, J. A. Carrillo, T. Laurent, and G. Raoul. Dimensionality of local minimizers of the interaction energy. Arch. Ration. Mech. Anal., 209 (2013) 1055–1088. [BGMPV] D. Bilyk, A. Glazyrin, R. Matzke, J. Park, and O. Vlasiuk. Optimal measures for p-frame energies on spheres. pre-print. arXiv:1908.00885 [CFP] J. Carrillo, A. Figalli, and F. S. Patacchini. Geometry of minimizers for the interaction energy with mildly repulsive potentials. Ann. Inst. H. Poincar´e Anal. Non Lin´eaire 34 (2017), 1299–1308. arXiv:1607.08660 MR3742525 [CK] H. Cohn and A. Kumar. Universally optimal distribution of points on spheres. J. Amer. Math. Soc. 20 (2007), 99–149. arXiv:math/0607446 MR2257398 [LM] T. Lim and R. J. McCann. Isodiametry, variance, and regular simplices from particle interactions. pre-print. arXiv:1907.13593<|endoftext|> TITLE: Almost all non-negative real numbers have only finitely many multiples lying in a measurable set with finite measure QUESTION [14 upvotes]: Let $A$ be Lebesgue measurable subset of $[0,\infty)$ such that Lebesgue measure of $A$ is positive i.e. $0<\lambda(A)<\infty$. Let $S$ be the set defined as follows: $$S:=\{t\in [0,\infty):nt\in A\text{ for infinitely many }n\in\Bbb N\}$$ What can we conclude about the measure of $S$? I can guess that $\lambda (S)=0$ for when $A$ is an open set but can't prove it. More particular case, when $A$ is open with finitely many components then I can conclude that $\lambda(S)=0$ REPLY [17 votes]: Let $f(t) = 1$ if $t \in A$ and $f(t) = 0$ otherwise. Suppose that $a > 0$. Then $$ \begin{aligned} \int_a^{2 a} \operatorname{card} \{n : n t \in A\} dt & = \int_a^\infty \biggl(\sum_{n = 1}^\infty f(n t) \biggr) dt \\ & = \sum_{n = 1}^\infty \int_a^{2 a} f(n t) dt \\ & = \sum_{n = 1}^\infty \frac{1}{n} \int_{n a}^{2 n a} f(s) ds \\ & = \int_a^\infty \biggl( \sum_{n = \lceil s/(2 a) \rceil}^{\lfloor s/a \rfloor} \frac{1}{n} \biggr) f(s) ds \\ & \leqslant \int_a^\infty \frac{\lfloor s/a \rfloor - \lceil s/(2 a) \rceil + 1}{\lceil s/(2 a) \rceil} \, f(s) ds \\ & \leqslant \int_a^\infty \frac{s/a - s/(2 a) + 1}{s/(2 a)} \, f(s) ds \\ & \leqslant \int_a^\infty 3 f(s) ds \leqslant 3 \lambda(A) < \infty . \end{aligned} $$ Therefore, $\operatorname{card}\{n : n t \in A\}$ is finite for almost all $t \in (a, 2 a)$. Since $a$ is arbitrary, we conclude that $\operatorname{card}\{n : n t \in A\}$ is finite for almost all $t > 0$, that is, $\lambda(S) = 0$.<|endoftext|> TITLE: What is the elementary proof of Weil's polynomial theorem of decomposition? QUESTION [5 upvotes]: André Weil sometimes glosses his Theorem of Decomposition in a simplified polynomial form: If $P(x,y)$ and $Q(x,y)$ are homogeneous polynomials algebraically prime to each other, with integer coefficients, and $x,y$ are integers prime to each other, then $P(x,y)$ and $Q(x,y)$ are ``almost" prime to each other, that is to say, their GCD admits a finite number of possible values.  (The Apprenticeship of a Mathematician (1992), p. 46) I think this must have an elementary proof which i am just not able to see.  Can someone help me? REPLY [11 votes]: If $P(x,y),Q(x,y)$ are relatively prime, then so are the one-variable polynomials $p(x)=P(x,1),q(x)=Q(x,1)$ (since we can homogenize any common factor of $p,q$ to a common factor of $P,Q$). It follows that in $\mathbb Q[x]$ there are two polynomials $a(x),b(x)$ such that $a(x)p(x)+b(x)q(x)=1$. Dehomogenizing and multiplying by a common denominator $M$ we get $$A(x,y)P(x,y)+B(x,y)Q(x,y)=My^k.$$ Hence every common factor of $P(x,y),Q(x,y)$ divides $My^k$. Similarly for some integer $N$ it must divide $Nx^l$. If $x,y$ are relatively prime, we get that $\gcd(P(x,y),Q(x,y))$ must divide $MN$ (by looking at exponent of each prime power separately).<|endoftext|> TITLE: conductor formula QUESTION [11 upvotes]: Let $\pi_p$ be an irreducible representation of $GL_2(\mathbb{Q}_p)$. Assume $\pi_p$ is ramified,hence it will have a positive conductor. Consider $sym^3(\pi_p)$ which is a representation of $GL_4(\mathbb{Q}_p)$. I want to know how to calculate the conductor of $sym^3(\pi_p)$. What is the relation between conductor of $\pi_p$ and conductor of $sym^3(\pi_p)$? Is there a way to calculate it? If so, then please suggest some good references. Thank you in advance. REPLY [10 votes]: The PhD thesis of Manami Roy (Univ. Oklahoma, 2019) gives a very detailed formula for the conductor of $\operatorname{Sym}^3(\pi_p)$, where $\pi_p$ is the representation of $\mathrm{GL}_2(\mathbf{Q}_p)$ coming from an elliptic curve. See Chapter 5 in particular. (More precisely, the $\operatorname{Sym}^3$ lifting from $\mathrm{GL}_2$ to $\mathrm{GL}_4$ factors through $\operatorname{GSp}_4$, and Roy computes the conductors of the resulting representations of $\operatorname{GSp}_4(\mathbf{Q}_p)$. But the lifting from $\operatorname{GSp}_4$-representations to $\mathrm{GL}_4$ preserves conductors.)<|endoftext|> TITLE: Can $L^1_{loc}$ be represented as colimit? QUESTION [5 upvotes]: Let $L^1_{loc}$ denote the set of all functions from $\mathbb{R}$ to itself which are locally integrable. For every infinite compact subset $K\subseteq \mathbb{R}$, let $L^1_{m_K}$ denote the space of Lebesgue measurable functions supported on $K$. Clearly the collection $\mathcal{K}$ of all such compact subsets of $\mathbb{R}$ form a poset wrt inclusion $i^{K_1}_{K_2}:K_1\hookrightarrow K_2$ if and only if $K_1\subseteq K_2$, for $K_i \in \mathcal{K}$. Therefore, we may define the colimit $$ \operatorname{colim}_{\mathcal{K}} L^1_{m_K}, $$ in Top. How are $\operatorname{colim}_{\mathcal{K}} L^1_{m_K}$ and $L^1_{loc}$ related? Note/Edit: Top is the category of topological spaces and continuous maps and LCS is the category of locally convex spaces and continuous linear maps. Related: $L^1_{\mu}$ as limit REPLY [7 votes]: Interpreting the question as Dmitri Pavlov I assume that $L^1_K$ is the space of $L^1$-functions with support in $K$ so that we have an inductive spectrum of Banach spaces (identifying almost everywhere equal functions). The colimit (=inductive limit) in the category of locally convex spaces (i.e., the union endowed with the finest locally convex topology making the inclusions from all $L^1_K$ continuous) is then a complete locally convex space (by a classical result of Dieudonne and Schwartz about strict countable inductive limits -- instead of all compact sets it is of course enough to consider a countable exhaustion). On the other hand it is dense in $L^1_{loc}$ (the projective limit of all $L^1_K$ with respect to the restrictions). Therefore, the inductive limit topology is strictly finer. (There are several other ways to see this. For example, a countable inductive limit of normed spaces $X_n$ such that $X_n\neq X_{n+1}$ is never metrizable.) The limit topology in the category TOP is even finer than that in LCS. If you consider uncountable colimits in LCS the situation is slightly different: As a Frechet spaces $L^1_{loc}$ is ultrabornological and hence the inductive limit of Banach spaces, namely of all Banach spaces generated by absolutely convex closed bounded sets (generated means that you take the linear span endowed with the Minkowski functional). You can describe these spaces as weighted $L^1$-spaces $\{f\in L^1: \int |f|wd\mu<\infty\}$ with suitable weights.<|endoftext|> TITLE: Strange formula for area of a convex polygon QUESTION [11 upvotes]: Consider a convex $n-$gon in $\mathbb{R}^2$ with sides contained in the lines $y=k_ix+b_i, 1\leq i\leq n.$ Then its area equals to $$ S=\frac{1}{2}\sum_{i=1}^{n} \frac{(b_{i+1}-b_i)^2}{k_{i+1}-k_i}. $$ Of course, it is not hard to prove this formula, but it still looks a bit mysterious to me. Question Has this expression appeared somewhere? It will be wonderful to see a pretty proof of it. Remark Another expression for $S$ can be found here, formula (17). It is a toy analog of a MHV scattering amplitude in N=4 SYM. REPLY [5 votes]: Fedor's derivation is very slick and elegant - surely the best way to guess the formula if you didn't know it already. It does however require additional justification to show that the oriented areas for all possible line positions will cancel appropriately to get the polygon area, especially if you are aiming to prove the formula for non-convex shapes as well (not asked by the OP but it does apply in this case as well.) If you want to avoid this and derive it from a standard area formula for polygons you can use Fedor's points of intersection $P_i$ and $P_{i+1}$ and plug them in to obtain: \begin{equation} \begin{aligned} &\frac{1}{2}\sum_{i=1}^n \det|P_iP_{i+1}|\\ &=\frac{1}{2}\sum_{i=1}^n\frac{(b_{i+1}-b_i)(k_{i+1}b_{i+2}-k_{i+2}b_{i+1})-(b_{i+2}-b_{i+1})(k_{i}b_{i+1}-k_{i+1}b_{i})} {(k_i-k_{i+1})(k_{i+1}-k_{i+2})}\\ &=\frac{1}{2}\sum_{i=1}^n\frac{b_{i+1}^2(k_i-k_{i+2})-b_{i}b_{i+1}(k_{i+1}-k_{i+2})-b_{i+1}b_{i+2}(k_i-k_{i+1})} {(k_i-k_{i+1})(k_{i+1}-k_{i+2})}\\ &=\frac{1}{2}\sum_{i=1}^n\frac{b_{i+1}^2(k_i-k_{i+1}+k_{i+1}-k_{i+2})} {(k_i-k_{i+1})(k_{i+1}-k_{i+2})}-\frac{1}{2}\sum_{i=1}^n\frac{b_{i}b_{i+1}}{(k_i-k_{i+1})}-\frac{1}{2}\sum_{i=1}^n\frac{b_{i+1}b_{i+2}}{(k_{i+1}-k_{i+2})}\\ &=\frac{1}{2}\sum_{i=1}^n\frac{b_{i+1}^2} {(k_i-k_{i+1})} +\frac{1}{2}\sum_{i=1}^n\frac{b_{i+1}^2} {(k_{i+1}-k_{i+2})} -\frac{1}{2}\sum_{i=1}^n\frac{b_{i}b_{i+1}}{(k_i-k_{i+1})} -\frac{1}{2}\sum_{i=1}^n\frac{b_{i+1}b_{i+2}}{(k_{i+1}-k_{i+2})}\\ &=\frac{1}{2}\sum_{i=1}^n\frac{b_{i+1}^2} {(k_i-k_{i+1})} +\frac{1}{2}\sum_{i=1}^n\frac{b_{i}^2} {(k_{i}-k_{i+1})} -\frac{1}{2}\sum_{i=1}^n\frac{b_{i}b_{i+1}}{(k_i-k_{i+1})} -\frac{1}{2}\sum_{i=1}^n\frac{b_{i}b_{i+1}}{(k_{i}-k_{i+1})}\\ &=\frac{1}{2}\sum_{i=1}^n\frac{b_{i+1}^2+b_{i}^2-2b_{i}b_{i+1}} {(k_i-k_{i+1})}\\ &=\frac{1}{2}\sum_{i=1}^n\frac{(b_{i+1}-b_{i})^2} {(k_i-k_{i+1})} \end{aligned} \end{equation} Update: Actually I just realised that Fedor's method looks to be the exact dual of the method behind the formula I used: "Sum the areas to a point between two points meeting at consecutive pairs of polygon sides" (The method I used) $\Updownarrow$ "point-line duality" "Sum the areas to a line between two lines joining consecutive pairs of polygon points" (Fedor's method) So this might be a good way to justify it.<|endoftext|> TITLE: How did Riemann prove that the moduli space of compact Riemann surfaces of genus $g>1$ has dimension $3g-3$? QUESTION [22 upvotes]: Consider the moduli space $M_g$ of compact Riemann surfaces (i.e., smooth complete algebraic curves over $\mathbb{C}$) of genus $g$ for some $g>1$. I'm interested in knowing how Riemann proved that $M_g$ has dimension $3g-3$. A modern proof involves deformation theory and Riemann-Roch theorem. In particular, one needs the notion of sheaf cohomology, which was not available at Riemann's time. How did Riemann prove that $M_g$ has dimension $3g-3$ for $g>1$? I would appreciate if someone could provide some reference to the original proof by Riemann. REPLY [31 votes]: Riemann combines what is called Riemann-Roch and Riemann-Hurwitz nowadays. He considers the dimension of the space of holomorphic maps of degree $d$ from the Riemann surface of genus $g$ to the sphere. He computes this dimension in two ways. By Riemann-Roch this dimension is $2d-g+1$, for a fixed Riemann surface. (Indeed, Riemann-Roch says that the dimension of the space of such functions with $d$ poles fixed is $d-g+1$ (when $d\geq 2g-1$ which we may assume), but these poles can be moved, so one has to add $d$ parameters). On the other hand, such a function has $2(d+g-1)$ critical points by Riemann-Hurwitz. Generically, the critical values are distinct, and can be arbitrarily assigned, and this gives the dimension of the set of all such maps on all Riemann surfaces of genus $g$. So the space of all Riemann surfaces of genus $g$ must be of dimension $$2(d+g-1)-(2d-g+1)=3g-3.$$ Riemann-Roch is proved in section 5 of Part I and and the dimension of the moduli space is counted in section 12 of Part I of the paper cited in F. Zaldivar's answer. Remark. Indeed, Riemann did not know about sheaves, cohomology and Serre duality. Neither he knew the general definition of a Riemann surface (which is due to Weyl). But one should take into account that all these notions were developed for the purpose to explain and digest what Riemann wrote in this paper. Remark 2. A pair $(S,f)$, where $S$ is a Riemann surface, and $f$ a meromorphic function from $S$ to the Riemann sphere is called "a Riemann surface spread over the sphere'' (Uberlagerungsflache). All such pairs can be constructed in the following way: choose critical values of $f$ and make some cuts between them so that the remaining region on the sphere is simply connected. Then take $d$ copies of this region (they are called sheets) stack them over the sphere, and paste them together along the cuts. You obtain a surface $S$ together with a map $f$, the "vertical" projection onto the sphere. Parameters are critical values. Riemann did not have any exact definition of "Riemann surface", he just explained this procedure of gluing as a visualization tool. For him, $S$ is a "class of algebraic curves $F(x,y)=0$ under birational equivalence". Until the work of Weyl, these pairs $(S,f)$ were called Riemann surfaces, and only Weyl defined exactly what $S$ is. Nowadays $S$ is called a Riemann surface, and a pair $(S,f)$ a "Riemann surface spread over the sphere".<|endoftext|> TITLE: Harmonic sums and elementary number theory QUESTION [22 upvotes]: Which natural number can be represented as a product of a sum of natural numbers and a sum of their inverses? I. e. does there exist for a natural $n$ a set of natural numbers $\{a_1, a_2,...a_m\}$ such that $n = (a_1 + a_2 + ...+a_m)(\frac{1}{a_1} + \frac{1}{a_2} + ... +\frac{1}{a_m})$? Call $n$ good if such a set exists, they do not have to be distinct. All $n = k^2$ are good, if n is good then $2n + 2$ is good as well, 10 and 11 are good, 2,3,5,6,7,8 are not. I'm too lazy to check any further especially if there is a solution out there, please point me if you know one. Does there exist a constant $C$ such that all $n \geq C$ are good? What if I let $a_i$s to be negative? Now mathoverflow gave me a link to another question Estimating the size of solutions of a diophantine equation which gives 14 as a good number (hard). Moreover there is a link to a very nice paper "An Unusual Cubic Representation Problem" by Andrew Bremner and Allan MacLeod which gives a whole bunch of solutions already for $m = 3$ on page 38 here http://ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf, for all their $N$ we get $2(N + 3)$ to be good. So if we take $C = 30$, then all even $N \geq C$ are good (needs a clear proof). Question There is still question about what happens for odd $N$s. And maybe it's worthy to ask if we can restrict $m$. I. e. What is the smallest $k$ such that any good $N$ can be represented as a product above with $k \geq m$? REPLY [29 votes]: Graham (1963) proved that any integer $n>77$ can be represented as $$ n = a_1 + \dots + a_m,$$ where $a_i$ are distinct positive integers with $$\frac{1}{a_1} + \dots + \frac{1}{a_m} = 1.$$ Hence, any integer $n>77$ is good. Btw, I've recently extended the result of Graham to sums of squares. UPDATE. As for the largest non-good (bad?) number, we know that $8$ is not good, while OEIS A028229 leaves just seven more candidates for the largest non-good number: $12, 13, 14, 15, 19, 21$, or $23$. Numbers $14, 15, 19, 21, 23$ are good with the corresponding $\{ a_i\}$ being $\{ 2, 3, 10\}$, $\{1, 2, 6\}$, $\{ 5, 8, 12, 15\}$, $\{ 8, 14, 15, 35\}$, and $\{76, 220, 285, 385\}$, respectively. Hence, there remains just 3 candidate for the largest non-good number: $8, 12$, or $13$. While the latter two candidates can be good only with $m=3$, MacLeod's approach (Sec. 6.2) here leads to elliptic curves of zero rank, thus implying that both $12$ and $13$ are not good. Therefore, $13$ is the largest non-good number. REPLY [19 votes]: A note on the observation "$n$ good implies $2n + 2$ good": First remark is that $n$ is good iff there are positive rational numbers $a_1, \dotsc, a_m$ such that $n = (a_1 + \dotsc + a_m)(1/a_1 + \dotsc + 1 / a_m)$. This is because one can multiply all $a_i$ by the lcm of their denominators. Second remark is that $n$ is good iff there are positive rational numbers such that $n = 1/a_1 + \dotsc + 1/a_m$ and $a_1 + \dotsc + a_m = 1$. This is because one can multiply all $a_i$ by the inverse of their sum. Finally, if $n = 1/a_1 + \dotsc + 1/a_m$ with $a_1 + \dotsc + a_m = 1$, then taking $a_i' = a_i / 2$ for $1\leq i \leq m$ and $a_{m + 1}' = 1/2$ gives us $2n + 2$. Looking at the proof by Graham cited above, I see that this observation is indeed halfway to the actual proof. Namely we need a second result that $n$ good implies $2n + 179$ good, to take care of the odd integers. Then the result follows from the fact that all integers from $78$ to $333$ are good. Since we don't require the $a_i$'s to be different, we may take $a_i' = a_i / 2$ for $1 \leq i \leq m$, $a_{m + 1}' = 1/3$, $a_{m + 2}' = 1/6$, and get $2n + 9$ good. This also suggests that we might improve significantly the bound of Graham. Following the comment of Max Alekseyev, this OEIS sequence says that all positive integers except $2, 3, 5, 6, 7, 8, 12, 13, 14, 15, 19, 21, 23$ are Egyptian, and hence good. Since our definition of "good" is weaker than "Egyptian", it doesn't a priori exclude the possibility that some of the above listed numbers are still good. Therefore it only remains to see whether the above numbers are good. The task is separated into $m = 2, 3, 4$. For $m = 2$, it is clear that only for $n = 4$, the equation $(x + y)(1/x + 1/y) = n$ has rational solution $(x, y)$. For $m = 3$, we are led to the equation $(x + y + z)(xy + yz + zx) = nxyz$. Assuming $n > 9$ and choosing the point $[x, y, z] = [-1, 1, 0]$ as the point at infinity, we get an elliptic curve. A Weierstrass equation is given by $Y^2 = X^3 + (n^2 - 6n - 3)X^2 + 16nX$, where $X = \frac{4n(x + y)}{x + y - (n - 1)z}$, $Y = \frac{4n(n - 1)(x - y)}{x + y - (n - 1)z}$. To get back $[x, y, z]$ from $X, Y$, we have the formula $[x, y, z] = [(n - 1)X + Y, (n - 1)X - Y, 2X - 8n]$. This curve has six torsion points, which correspond to useless solutions to our equation. Moreover, the Mordell-Weil group has rank $0$ for all the above $n$, except $n = 14, 15$. This proves that $n = 12, 13$ are not good. For $n = 14$, we have a solution $(3, 10, 15)$. For $n = 15$, we have a solution $(1, 3, 6)$. Thanks again to Max Alekseyev, $n = 19, 21, 23$ are all good, with solutions $(5,8,12,15), (8,14,15,35), (76,220,285,385)$, respectively. Conclusion: A positive integer is good if and only if it is not one of $2, 3, 5, 6, 7, 8, 12, 13$.<|endoftext|> TITLE: Eigenvectors of Kronecker Product QUESTION [8 upvotes]: Conjecture If $A$ and $B$ are two complex square matrices, then every eigenvector of $A\otimes B$ is of the form $x\otimes y$, where $x$ is an eigenvector of $A$ and $y$ is an eigenvector of $B$. Here, $A\otimes B$ denotes the Kronecker Product of two matrices. I would like to know if this conjecture is true. Motivation: I know that the following is true: Theorem Let $A$ and $B$ be two complex square matrices. If $\lambda$ is an eigenvalue of $A$ with corresponding eigenvector $x$ and $\mu$ is an eigenvector of $B$ with corresponding eigenvector $y$, then $\lambda\mu$ is an eigenvalue of $A\otimes B$ with corresponding eigenvector $x\otimes y$. Moreover, every eigenvalue of $A\otimes B$ arises as such a product. This is Theorem 4.2.12 in Horn and Johnson's "Topics in Matrix Analysis'', or Theorem 13.12 here. However, no statement is made that all eigenvectors arise in such fashion. Does the conjecture follow from this theorem? REPLY [8 votes]: Counterexample: the matrix $I \times I$ has eigenvectors that are not in product form, since every vector is an eigenvector of it and not every vector can be written in product form.<|endoftext|> TITLE: A new way of approaching the pole of the Riemann zeta function - and a new conjectured formula QUESTION [29 upvotes]: On the Wolfram page about the Euler-Mascheroni Constant $\gamma $, the following amazing limit is given without proof (referring to "personal communication"): $$\lim_{z\to\infty}\left[\zeta(\zeta(z))-2^z+\Bigl(\frac43\Bigr)^z\right]=\gamma -1$$ I have checked to thousands of decimals that likewise $$\lim_{z\to\infty}\left[\zeta\Bigl(\frac1{\zeta(z)}\Bigr)+2^z-\Bigl(\frac43\Bigr)^z\right]=\gamma $$ and started wondering what happens if we combine both, e.g. by fixing a weight $0\leqslant\lambda\leqslant1$ and looking at the behaviour of $$ f_\lambda(z):=\zeta \Bigl(\lambda{\zeta(z)+\frac{1-\lambda}{\zeta(z)}}\Bigr) $$ for again growing $z$ on the real axis. So the argument in the big parenthesis still tends to $1$, such that the pole at $\zeta(1)$ is approached in different ways depending on $\lambda$. And something cute happens: Numerically it looks like $$\boxed{\color{blue}{ \lim_{z\to\infty}\left[f_{\lambda}(z)- \frac1{2\lambda-1}\left(2^z- \Bigl(\frac43\Bigr)^z\right) \right]=\gamma -\frac{\lambda}{(2\lambda-1)^2}}}$$ for $\lambda\ne\frac12$ (even outside the interval $[0,1]$), while, completely off, $$ f_{1/2}(z)\sim 2\cdot4^z-4\cdot\Bigl(\frac83\Bigr)^z -2\cdot 2^z -\text {(5 other exp terms)}-4+\gamma. $$ I am wondering if such a formula has appeared before and how to prove it! REPLY [16 votes]: The first formula is trivial. $$f(s)= \frac1{s-1}+\gamma +O(s-1)$$ $$g(z)=1+2^{-z}+3^{-z}+4^{-z}+O(5^{-z})=1+2^{-z}(1+(3/2)^{-z}+(4/2)^{-z}+O(5/2)^{-z})$$ $$f(g(z)) = \frac1{2^{-z}(1+(3/2)^{-z}+(4/2)^{-z}+O(5/2)^{-z})}+\gamma+O(2^{-z})$$ $$=2^z \left(1-(3/2)^{-z}-(4/2)^{-z}+O(5/2)^{-z}+(O(3/2)^{-z})^2\right)+\gamma + O(2^{-z})$$ $$=2^z- (3/4)^{-z}-1+O(9/8)^{-z}+\gamma$$<|endoftext|> TITLE: The limit of a function with derivative at least $1_\mathbb{Q}$ QUESTION [8 upvotes]: Let $f:\mathbb{R}\to \mathbb{R}$ be differentiable, such that $f'(x) \ge 1_{\mathbb{Q}}(x)$. Is it true that $\lim_{x\to\infty}f(x) = \infty$? REPLY [6 votes]: The answer is "no", and here is how one can adapt Pietro Majer's construction from the other answer to work with the set of rational numbers. Step 1. We begin with the original construction of Pompeiu function. Let $q_n$ be an enumeration of rational numbers in $(0, 1)$, and define, as in the original construction, $$ u(x) = a + b \sum_{n = 1}^\infty \frac{\sqrt[3]{x - q_n}}{2^n} \, , $$ where $a$ and $b > 0$ are chosen in such a way that $u(0) = 0$ and $u(1) = 1$. Then $u$ is increasing and differentiable in an extended sense: $u'(x) \in (0, \infty]$ exists for every $x \in [0, 1]$, and $u'(x) = \infty$ for every rational $x \in (0, 1)$. Step 2. We now define $$ v(x) = u\biggl(\frac{1}{2} + \frac{x}{1 + 2 |x|}\biggr) . $$ Observe that $x \mapsto 1/2 + x / (1 + 2 |x|)$ is differtentiable with everywhere positive derivative, and it maps rational numbers to rational numbers. Thus, $v : \mathbb{R} \to (0, 1)$ is an increasing homeomorphism, $v$ is everywhere differentiable in the extended sense, $v'(x) \in (0, \infty]$ for every $x$, and $v'(x) = \infty$ whenever $x$ is rational. Step 3. We now find an increasing homeomorphism $\phi : (0, 1) \to (0, 1)$ which is differentiable with everywhere positive derivative, and which has the following property: for every $s \in \mathbb{Q}$ there is $x \in \mathbb{Q}$ such that $$x + \phi(v(x)) = x.$$ Such a function $\phi$ can be constructed recursively, as follows. We begin with $\phi_0(x) = s$. Let $r_n$ be an enumeration of all rational numbers. In step $n$ we suppose that we have already constructed a function $\phi_{n-1}$ with the following properties: $$|\phi_{n-1}' - 1| < 3^{-1} + \ldots + 3^{-(n-1)},$$ and there are rational numbers $x_1, \ldots, x_{n-1}$ such that $$x_j + \phi_{n-1}(v(x_j)) = r_j$$ for $j = 1, \ldots, n-1$. We now define $\phi_n$ to be an appropriate "tiny" modification of $\phi_{n-1}$. Since $\phi_{n-1}' > 0$, there is a unique number $\tilde{x}_n$ such that $$\tilde{x}_n + \phi_{n-1}(v(\tilde{x}_n)) = r_n.$$ We choose a non-negative, compactly supported, smooth function $\psi_n$ on $(0, 1)$ such that $\psi_n(v(x_j)) = 0$ for $j = 1, \ldots, n - 1$, $\psi_n(v(\tilde{x}_n)) > 0$, and $\|\psi_n'\|_\infty < 3^{-n}$. For every $\epsilon \in [0, 1)$ we have $$\phi_{n-1}' + \epsilon \psi_n' \ge 1 - (3^{-1} + \ldots + 3^{-(n-1)} + \epsilon 3^{-n}) > 0 ,$$ and hence there is a unique solution $x$ of $$x + \phi_{n-1}(v(x)) + \epsilon \psi_n(v(x)) = r_n .$$ This solution depends continuously on $\epsilon$, and since $\psi_n(v(\tilde{x}_n)) \ne 0$, it indeed changes with $\epsilon$. By the intermediate value property, there is an $\epsilon$ such that the corresponding solution $x$ is a rational number. For this $\epsilon$ and $x$, we set $$ \phi_n = \phi_{n-1} + \epsilon \psi_n $$ and $x_n = x$. This way, we have constructed $\phi_n$ with all the desired properties. By construction, $\psi_n$ converges in $C^1$ to an increasing homeomorphism $\psi : (0, 1) \to (0, 1)$ with the desired properties; namely, $|\psi' - 1| < 1/2$, and $x_n + \psi(v(x_n)) = r_n$. Step 4. We set $w(x) = \psi(v(x))$. Then $w : \mathbb{R} \to (0, 1)$ is differentiable everywhere in the extended sense, $w'(x) = \psi'(v(x)) v'(x) \in (0, \infty]$, $w'(x) = \infty$ for every rational $x$, and $$x_n + w(x_n) = r_n.$$ Now we follow Pietro Majer's argument: we set $g : (0, 1) \to \mathbb{R}$ to be the inverse function of $w$, $h(x) = g(x) + x$, and $f : \mathbb{R} \to (0, 1)$ to be the inverse function of $h$. Clearly, $g$, $h$ and $f$ are everywhere differentiable, with $g' \geqslant 0$, $h' \geqslant 1$ and $0 < f' \leqslant 1$, respectively. Observe that $h(w(x_n)) = g(w(x_n)) + w(x_n) = x_n + w(x_n) = r_n$. Since $w'(x_n) = \infty$ for every $n$, we have $g'(w(x_n)) = 0$, and hence $h'(w(x_n)) = 1$. Thus, $f'(r_n) = f'(h(w(x_n)) = 1 / h'(w(x_n)) = 1$. Since $r_n$ exhaust all rational numbers, $f'(x) = 1$ for every rational $x$.<|endoftext|> TITLE: How to prove local class field theory from global class field theory QUESTION [6 upvotes]: Historically (as I gather from Learning Class Field Theory: Local or Global First?), global class field theory was proved first, and then used to deduce local class field theory. But nowadays most treatments do the local theory first. Can someone give me a summary of how the global-to-local argument goes, or a reference to where the argument appears? I'm interested in seeing this as a toy model for how to think about the current state of Langlands, where it's still the case that many local statements (e.g. local Langlands for $GL_n$) are proved via global methods. I think these proofs start off by globalizing the local situation, getting a corresponding global object on the other side, and then extracting a local component. The issue is then showing that this construction doesn't depend on the globalization. Can you prove local class field theory along these lines? REPLY [5 votes]: That's the approach taken in Lang's Algebraic Number Theory Springer GTM 110. Lang develops global class field theory, and then in Chapter XI Section 4 he finishes "the proof of the complete splitting theorem and derives local class field theory, describing the effect of the Artin map on the local component $k_v^*$ for a fixed $v$."<|endoftext|> TITLE: Category of spaces/sheaves QUESTION [6 upvotes]: Consider the following category $\mathcal C$: An object of $\mathcal C$ is a pair $(X,\mathcal F)$ where $X$ is a space and $\mathcal F$ is a sheaf on $X$. A morphism $(X,\mathcal F)\to(Y,\mathcal G)$ is a map of spaces $f:X\to Y$ together with a map of sheaves $f^*\mathcal G\to\mathcal F$ (equivalently $\mathcal G\to f_*\mathcal F$). The words 'space' and 'sheaf' above can be taken in a number of senses: topological spaces and all sheaves, schemes and coherent sheaves, complex analytic spaces and coherent sheaves, etc. My question is somewhat general, and I will leave it up to the reader which context they wish to work in. Cohomology $H^*$ is a functor from $\mathcal C$ to the category of graded abelian groups, however we can say more. Namely, let $W$ denote the class of morphisms $(X,\mathcal F)\to(Y,\mathcal G)$ in $\mathcal C$ for which $\mathcal G\xrightarrow\sim f_*\mathcal F\xrightarrow\sim Rf_*\mathcal F$ are both isomorphisms. It follows by the Leray spectral sequence that $H^*$ sends morphisms in $W$ to isomorphisms. The functor $H^*:\mathcal C\to\operatorname{AbGrp}$ thus factors through the localization $\mathcal C\to\mathcal C[W^{-1}]$ (note though that we have not argued that this localization exists). The above suggests that cohomology makes sense not only for objects of $\mathcal C$ but also for what we might get by gluing together various objects of $\mathcal C$ along morphisms in $W$. For instance, take two objects $(X_1,\mathcal F_1),(X_2,\mathcal F_2)\in\mathcal C$. We could glue these together along a common open subset $(U_1,\mathcal F_1|_{U_1})\xrightarrow\sim(U_2,\mathcal F_2|_{U_2})$ to obtain another object of $\mathcal C$. On the other hand, we should also be able to, at least formally speaking, glue together along any morphism $(U_1,\mathcal F_1|_{U_1})\xrightarrow\sim(U_2,\mathcal F_2|_{U_2})$ in $W$ and obtain some sort of generalized object (though not an object of $\mathcal C$) to which it still makes sense to apply the functor $H^*$. I can imagine various ways of making the above discussion precise (i.e. making sense out of "objects of $\mathcal C$ glued together along morphisms in $W$"), however they all suffer from the following deficiency: there is no simple way to describe morphisms between two such objects. This is similar to how a topological groupoid presents a stack, but given two topological groupoids it is somewhat cumbersome to describe, purely in terms of topological groupoids, the space of morphisms between the associated stacks. The notion of a stack solves this issue (of describing morphism spaces) beautifully: a morphism of stacks (on a site $\mathcal D$) is just a natural transformation of functors $\mathcal D\to\operatorname{Groupoids}$. I can finally form my question: How can we form a category whose objects are "generalized objects" of $\mathcal C$ (i.e. gluings of objects of $\mathcal C$ along morphisms in $W$) and whose morphisms admit some sort of simple description? REPLY [5 votes]: Here is a construction which I think is at least close to what you're driving at. Let $\mathcal S$ be our category of spaces, and let $Shv: \mathcal S \to Cat$ be the pseudofunctor taking a space to its category of sheaves $Shv(X)$ and a taking a map $f$ to its pushforward $f_\ast$. Then by the Grothendieck construction there is a corresponding fibration $\mathcal C \to \mathcal S$. Moreover, this is the same $\mathcal C$ as in the question. That is, The category $\mathcal C$ is the result of applying the Grothendieck construction to the functor $Shv: \mathcal S \to Cat$. Let us assume that $\mathcal S$ is locally presentable; For each $X \in \mathcal S$, the category $Shv(X)$ is locally presentable; For each $f: X \to Y$, the functor $f_\ast: Shv(X) \to Shv(Y)$ has a left adjoint $f^\ast$; The functor $Shv : \mathcal S \to Cat$ preserves $\kappa$-filtered colimits for some $\kappa$. Then the fibration $\mathcal C \to \mathcal S$ is a presentable fibration, and by Thm 10.3 here, $\mathcal C$ is a locally presentable category. Now for locally presentable categories, there is a very nice theory of localization. Assuming that $W$ is closed under colimits in the arrow category and also under pushouts along arbitrary maps (satisfies a mild set-theoretic hypothesis), the localization $\mathcal C[W^{-1}]$ is a reflective subcategory of $\mathcal C$, consisting of the $W$-local objects. To connect to your setting, what I would do is take the $\infty$-categorical version of all of this (as in the paper of Gepner and Haugseng referenced above). So I would take $Shv(X)$ to be the $\infty$-category of sheaves on $X$ localized at the quasi-isomorphisms (i.e. the $\infty$-categorical enhancement of the derived category). From your examples, it sounds like neither $\mathcal S$ nor $Shv(X)$ is necessarily presentable -- you might have some finiteness conditions on them which prevent them from being cocomplete (every locally presentable category is cocomplete). But that's fine -- assuming that $\mathcal S$ and $Shv(X)$ have finite colimits, ($\infty$-categorical colimits in the latter case), you can just take the Ind-category of each to get something presentable. Having done this, we have a presentable $\infty$-category $\mathcal C$. I would take $\mathcal W$ to be the class of morphisms inverted by the cohomology functor. In order to apply the theory of localizations of presentable $\infty$-categories, $\mathcal W$ needs to be closed under colimits in the arrow category and pushout along arbitrary morphisms. I believe the colimits in $\mathcal C$ are going to be related to colimits in $\mathcal S$, so in order for this to work, you will probably need it to be the case that cohomology behaves well (i.e. satisfies Mayer-Vietoris) with respect to all pushouts in $\mathcal S$, which is probably not the case -- it probably only behaves well with respect to some kind of "homotopy pushouts". So probably $\mathcal S$ also needs to be modified by localizing it at some kind of "homotopy equivalences" to obtain an $\infty$-category $\mathcal S_L$ such that the $\infty$-categorical pushouts of $\mathcal S_L$ are computed via the "homotopy pushouts" of $\mathcal S_L$. So we'll end up with an $\infty$-category $\mathcal C_L$ by applying the $\infty$-categorical Grothendieck construction to the functor $Shv: \mathcal S_L \to Cat_\infty$. But once all of this is done, $\mathcal C_L[\mathcal W^{-1}]$ will be the full subcategory of $\mathcal C_L$ consisting of the $\mathcal W$-local objects. This may seems surprising, but really it's analogous to the fact that sheaves are a full subcategory of presheaves -- allowing more types of gluing makes your objects more local. Probably the trickiest part in general is constructing $\mathcal S_L$. When $\mathcal S$ is spaces, for example, $\mathcal S_L$ would be the $\infty$-category of spaces; when $\mathcal S$ is schemes, $\mathcal S_L$ might be motivic spaces, etc. Probably $\mathcal S_L$ can itself be constructed as some localization of sheaves of spaces on $\mathcal S$, as in motivic homotopy theory.<|endoftext|> TITLE: $h^{p,q} = h^{q,p}$ on complex smooth projective scheme QUESTION [5 upvotes]: I know that for compact Kähler manifolds $M$ there is an isomorphism: $$ H^p(M, \Omega_M^q) = H^q(M, \Omega_M^p) $$ where $\Omega_M$ is the sheaf of holomorphic $1$-forms. It is because $H^p(M, \Omega_M^q) = H^{p,q}_{\bar{\partial}}(M)=\mathcal{H}^{p,q}(M)$ the set of harmonic forms on $M$. We can then apply conjugation on $\mathcal{H}^{p,q}(M)$ to get $\mathcal{H}^{q,p}(M)$. So the Hodge numbers satisfy $h^{p,q} = h^{q,p}$. This means if I have a complex smooth projective scheme $X$ and take the sheaf of differentials $\Omega_X$, then we still have $h^{p,q} = h^{q,p}$ because analytification of $X$ is a compact Kähler manifold. I wonder if there is an algebraic argument for this fact. REPLY [9 votes]: I remember seeing such a proof in an article by Messing, who attributed it to Gabber. Let $X$ be a smooth projective variety of dimension $n$ over a field of characteristic $0$. Suppose that $p+q=i\le n$. Serre duality gives $h^{pq}= h^{n-p, n-q}$. Now put this together with algebraic proofs of hard Lefschetz for $\ell$-adic cohomology (Deligne); suitable comparison theorems, which implies HL for algebraic de Rham; the algebraic proof of degeneration of the Hodge to de Rham spectral sequence (Faltings, Deligne-Illusie). Then we get an isomorphism $$ H^{qp}\stackrel{L^{n-i}}{\cong} H^{n-i+q,n-i+p}=H^{n-p,n-q}$$ If you consider all the prerequisites, it's no easier than the analytic proof.<|endoftext|> TITLE: Generalized root systems and reflection groups QUESTION [10 upvotes]: Consider the following alternative definition of finite reflection group: Definition: A finite reflection group $\Gamma\subset\mathrm O(\Bbb R^d)$ is a finite group generated by orthogonal transformations $T\in\mathrm O(\Bbb R^d)$ with eigenvalues $\{-1^1,1^{d-1}\}$. (the exponents denote multiplicites) This definition suggests the following generalization: Definition: A finite $k$-reflection group $\Gamma\subset\mathrm O(\Bbb R^d)$ is a finite group generated by orthogonal transformations $T\in\mathrm O(\Bbb R^d)$ with eigenvalues $\{-1^k,1^{d-k}\}$. In other words: instead of inverting a 1-dimensional subspace, each generator inverts a $k$-dimensional subspace and leaves the orthogonal complement fixed. As there are root systems associated with finite reflection groups, one can defined analogous systems for $k$-reflection groups. The elements of these are not vectors, but $k$-dimensional subspaces which are invariant w.r.t. certain "generalized reflections". Question: Have such objects been studied before? Does there exist a classification? Some thoughts Let $\Gamma$ is a $k$-reflection group generated by "reflections" $T_U,U\in\mathcal U$, where $\mathcal U$ is the associated "generalized root system" that contains $k$-dimensional linear subspaces, and $T_U$ has eigenspace $U$ to eigenvalue $-1$. Then $\Gamma'$ generated by $T_{U^\bot}=-T_U,U\in\mathcal U$ is a $(d-k)$-reflection group. So, all $(d-1)$-reflection groups are already classified via the usual reflection groups. In particular, up to dimension three, all generalized reflection groups are classified in this way. The first interesting case are the 2-reflection groups in $\Bbb R^4$, which are probably related to complex reflection groups. REPLY [11 votes]: If we place no restrictions on $k$, then this is precisely the class of finite groups that are generated by involutions. In particular, if $G$ is any finite group of order $n$, then in the left regular representation of $G$ any involution acts as an $n\times n$ orthogonal matrix of order two and trace zero. Such a matrix must have eigenvalues $1$ and $-1$, each with multiplicity $n/2$, and is therefore a reflection across an $(n/2)$-dimensional subspace. Thus any group of order $n$ which is generated by involutions is an $(n/2)$-reflection group. This class of groups is fairly large. For example, it includes all non-abelian finite simple groups. For if $G$ is a finite simple group, then the elements of order two in $G$ must generate a normal subgroup of $G$. If $G$ is non-abelian, then $G$ has even order and hence at least one element of order two, and therefore $G$ is generated by its elements of order two.<|endoftext|> TITLE: Explicit short presentation of a 2-generated universal group? QUESTION [5 upvotes]: A result of Higman states that there exists a finitely-presented group $G$ in which all other finitely-presented groups embed - I'll call such a group universal. Every countable group embeds in a 2-generated group, so there are 2-generated universal groups. I was told that someone somewhere wrote down some explicit presentations of such 2-generated universal groups. Where can I find such presentations? Is the minimal number of relators for such a 2-generated universal group known? Lower bound? REPLY [3 votes]: As I wrote in my comment above, the OP is about two different classes of groups: 2-generated universal countable groups (these contain all countable groups and are not finitely presented) and universal finitely presented group (these contain all recursively presented groups and are finitely presented). The question about minimal number of relators makes sense for the second class only. The only known lower bound is 1 (1-related groups have solvable word problem and are not universal). Every universal finitely presented group has unsolvable word problem. I do not think there is an example of a universal group with few relators. The state of the art of few-relator groups with unsolvable word problem is in Collins paper of 1986: https://projecteuclid.org/download/pdf_1/euclid.ijm/1256044631. The smallest known presentation has 12 relators. See also Panov, D. , Petrunin, A. The telescopic construction: a microsurvey. Proceedings of the Gökova Geometry-Topology Conference 2013, 110–119, Gökova Geometry/Topology Conference (GGT), Gökova, 2014. for a different notion of universality of f.p. groups.<|endoftext|> TITLE: How can one define a kind of "determinant" on a reduced group $C^*$ algebra? QUESTION [7 upvotes]: Let $A$ be a unital $C^*$-algebra which is equipped with a faithful trace $T$. In particular we may consider $A=C^*_{\text{red}} (G)$ for some discrete group $G$. We consider the following differential equation on $A$: $$Z'=Z^2-Z.\tag{*}\label{star1}$$ (For $A=M_n(\mathbb{C})$ one can easily check that $$D'=D(T-n)\tag{**}\label{star2}$$ where $D$, $T$ are the standard determinant and trace respectively and $D'$ is the derivative of $D$ along solutions of \eqref{star1}. Note that $n$ in \eqref{star2} can be regarded as $\operatorname{trace}(I_n)$. We will modify this $n$ to $1$ in the case of normalized trace. In fact "determinant" is the unique analytic function on $M_n(\mathbb{C})$ satisfying the equation \eqref{star2} with initial condition $D(I_n)=1$.) We try to generalize this situation of matrix algebra to a $C^*$-algebra $A$ with a faithful normalized trace $T$. So we consider the following modified differential equation:$$D'=D(T-1)\tag{***}\label{star3}$$ where the unknown $D$ is a function on $A$ and $T$ is a normalized trace. Moreover $D'$ is the derivative of $D$ along solution of \eqref{star1}. What can be said about existence of a global solution $D$ for \eqref{star3} with initial condition $D(1)=1$? Does such a solution $D$ satisfy the multiplicativity condition $D(ab)=D(a)D(b)$? Is $D^{-1}(0)$ equal to the set of all non invertible elements? As a motivation for the later question, we note that the group of invertible elements is flow-invariant under system \eqref{star1} (see On differential equation $Z'=Z^2-Z$ on a $C^*$ algebra). If there are no some complete answers to the above questions for an arbitrary algebra with a faithful normalized trace, what would be the answer of those questions in the particular case $A=C^*_{\text{red}} (G)$? For which kind of groups the answer to the above questions are known? Remark: We conclude that "determinant" as a function on matrix algebra can be dynamically and uniquely extracted from "trace", at least in a neighborhood of the identity matrix since the identity matrix, as a singularity of \eqref{star1}, attracts all nearby orbits, as time goes to $-\infty$. This is somewhat compatible with the classical fact that "determinant" of a matrix $B$, as an invariant polynomial, can be generated by "trace" of powers of $B$, that is $\operatorname{trace}(B^k),\;k\in \mathbb{N}$. But this dynamical interpretation we provided, needs merely trace of power $1$ not higher powers. More precisely, if we denote by $\phi$ the flow of $(*)$, then for $B$ sufficiently close to identity matrix we have $$\operatorname{Det}(B)=\exp\left(\int_{-\infty}^0 (n-\operatorname{trace})(\phi_t(B))dt\right).$$ So knowing "trace" leads us to knowing "determinant". REPLY [5 votes]: This is a long comment, rather than an answer, so it perhaps fits here. Another way in which the trace determines the determinant, at least locally near the identity, is via the equation $$D(e^h)=e^{T(h)}.\tag{$\dagger$}\label{dagger}$$ I have tried to prove that the OP's condition $\eqref{star3}$ implies \eqref{dagger} but, although I still think this is true, I cannot come up with a proof (any ideas?). Regarding the existence of solutions for \eqref{dagger}, there is a clear obstruction: if $p$ is an idempotent element, then $e^{2\pi i p}=1$, so one would need $e^{2\pi i T(p)}=1$, which is equivalent to saying that $T(p)$ is an integer. Conversely, it is a consequence of Theorem (II.10) in my PhD thesis Rotation numbers for automorphisms of $C^*$ algebras that the existence of projections with non-integer trace is the only possible obstruction to the existence of a determinant. Regarding (reduced) group C*-algebras the question becomes very interesting. If the group $G$ has torsion, then the spectral projections of any nontrivial torsion element will have non-integer trace (for the standard normalized trace, of course), so no determinant function exists. On the other hand, if $G$ has no torsion one would hope to prove that all projections have integer trace, but doing so would solve the Kadison–Kaplansky conjecture which has been proved for many groups, including all hyperbolic groups, but I think the most general case is still open.<|endoftext|> TITLE: Elementary calculus estimate or not? QUESTION [5 upvotes]: Does there exist a constant $C>0$ such that for all $f \in H^3(\mathbb R)$ $$\int_{\mathbb R} \vert x f''(x) \vert^2 \ dx \le C \int_{\mathbb R} \vert f'''(x) \vert^2 + \vert x^3f(x) \vert^2 + \vert f(x) \vert^2 \ dx? $$ The question comes from the fact that it is very easy to see that $$\int_{\mathbb R} \vert x f'(x) \vert^2 \ dx \le C \int_{\mathbb R} \vert f''(x) \vert^2 + \vert x^2f(x) \vert^2 + \vert f(x) \vert^2 \ dx, $$ but the proof in that case (integrating by parts and using the Cauchy-Schwarz inequality) does not obviously carry over to the case of the first line. REPLY [18 votes]: This really belongs to MSE rather than to MO, but I'm too lazy to initiate the moving process, so I'll just answer. There may be more intelligent ways to do it, but you can also integrate by parts and get what you want, say, for smooth functions with compact support, after which you should carefully pass to the limit to extend it to the corresponding Sobolev class. Let's just solve a more general problem. For non-negative integer $a,b,c$ and an infinitely smooth compactly supported real-valued $f$, denote $$ I(a,b,c)=\int_{-\infty}^\infty x^af^{(b)}(x)f^{(c)}(x)dx. $$ Claim: If $r,R>0$ are integers and $b,c\le R$, $\frac ar+\frac{b+c-2R}{R}\le 0$, then $$|I(a,b,c)|\le C[I(0,0,0)+I(2r,0,0)+I(0,R,R)]$$ Indeed, there are only finitely many triples $(a,b,c)$ with the above property (admissible triples). Let $M$ be the maximum of $|I(a,b,c)|$ over the admissible triples. The integration by parts formula yields $$ |I(a,b,c)|\le a|I(a-1,b-1,c)|+|I(a,b-1,c+1)| $$ as long as $b>0$ and $c0$. If the extremal admissible triple is of the kind $(a,b,R)$, then we can use Cauchy-Schwarz to write $$ |I(a,b,R)|\le \delta I(2a,b,b)+\delta^{-1}I(0,R,R) $$ with any $\delta>0$ we want. Note that the triple $(2a,b,b)$ is still admissible if $(a,b,R)$ is. We'll choose $\delta=(2C_1)^{-1}$. Then, if $m$ is attained on a triple of this type, we have $$ M\le C_1m\le C_1[\delta M+\delta^{-1}I(0,R,R)]=\frac M2+2C_1^2I(0,R,R) $$ whence $M\le 4C_1^2I(0,R,R)$. Let us now consider extremal admissible triples of the kind $(a,0,c)$. If $a\le r$, then we can use Cauchy-Schwarz in the form $$ |I(a,0,c)|\le \delta I(0,c,c)+\delta^{-1}I(2a,0,0) $$ and, with the same choice of $\delta$ and same argument, get $$ M\le 4C_1^2I(2a,0,0)\le 4C_1^2[I(0,0,0)+I(2r,0,0)] $$ (the crucial point is that $(0,c,c)$ is admissible). If $a>r$, then we can use Cauchy-Schwarz in the form $$ |I(a,0,c)|\le \delta I(2(a-r),c,c)+\delta^{-1}I(2r,0,0) $$ and, with the same choice of $\delta$ and same argument, get $$ M\le 4C_1^2I(2r,0,0) $$ (the crucial point is that $(2(a-r),c,c)$ is now admissible). Either way, we get what we claimed. The inequality the OP asked about corresponds to the case $r=R=3$, $a=b=c=2$. REPLY [16 votes]: The inequality is true. One integration by parts and standard Cauchy-Schwarz gives $$\int |xf''|^2 \leq C\left(\int |f'''|^2 + x^4|f'|^2 + |f'|^2\right).$$ The second term can be handled by integrating by parts once and applying Cauchy-Schwarz: $$\int x^4|f'|^2 \leq \delta \int x^2|f''|^2 + C_{\delta}\left(\int(x^2+x^6)f^2\right),$$ and $x^2+x^6$ is smaller than a fixed constant times $(1+x^6)$. Finally, the third term in the first inequality can be handled using the Fourier transform and Parseval's identity: $$\int |f'|^2 = \int |\xi|^2|\hat{f}|^2 \leq \int(1+|\xi|^6)|\hat{f}|^2 = \int (|f|^2 + |f'''|^2).$$<|endoftext|> TITLE: Does midpoint-convex imply rationally convex? QUESTION [7 upvotes]: Say that a function $f$ defined on $\mathbb{Q}^n$ is midpoint convex if $f((x+y)/2) \le (f(x) + f(y))/2$. Say that it is rationally convex if, for $\lambda \in \mathbb{Q} \cap [0,1]$ and $\bar \lambda = 1-\lambda$, we have $f(\lambda x + \bar\lambda y) \le \lambda f(x) + \bar\lambda f(y)$. Clearly every rationally-convex function is midpoint-convex. Question: Is a midpoint-convex function on $\mathbb{Q}^n$ necessarily rationally-convex? Background: I'm looking for sufficient conditions for a function on $\mathbb{Q}^n$ to extend to a continuous function on $\mathbb{R}^n$. The following results are known and/or easy: Rationally-convex functions on $\mathbb{Q}^n$ are continuous; similarly real-convex functions on $\mathbb{R}^n$ are continuous. (I'm sticking to "proper" functions, with real values, for convenience.) Using the axiom of choice, there are rationally-convex functions on $\mathbb{R}$. (Take a $\mathbb{Q}$-basis of $\mathbb{R}^n$ and go from there.) Every rationally-convex function $f$ on $\mathbb{Q}^n$ extends continuously to a real-convex function. (Define the extension to be the $\liminf$ of the nearby rational values, and check the result is convex.) As a result, every rationally-convex function on $\mathbb{Q}^n$ extends to a continuous function on $\mathbb{R}^n$. Every measurable, midpoint-convex function on $\mathbb{R}^n$ is continuous. (Attributed by Wikipedia to Sierpinski, I have not read the reference. There are many prior questions on MathOverflow that are answered by this theorem.) These all seem like it should be implying midpoint-convex implies rationally-convex, but I couldn't quite close the loop. In the case I'm actually interested in, this is a side issue, since my function $f$ is additionally (rationally)-homogeneous, and homogeneous and midpoint-convex implies rationally-convex. But I was curious and could not find it in the literature. REPLY [11 votes]: Assume that $g$ defined on $\mathbb{Q}^n$ is midpoint convex. First we show that we can extend the midpoint inequality to arbitrary means: $g((x_1+\dots+x_m)/m)\leq (g(x_1)+\dots+g(x_m))/m$ for any $m\in\mathbb{Z}_{\geq1}$ We can easily prove this for $m=2^k$ by using midpoint convexity $k$ times. For general $m\leq 2^{i}$ you take $x_1,\dots,x_m$ plus $2^i-m$ copies of $x'=(x_1+\dots+x_m)/m$ This gives $$g(x')=g\left(\frac{(2^i-m)x'+x_1+\cdots+x_m}{2^i}\right) \le\frac{(2^i-m)g(x')+g(x_1)+\dots+g(x_m)}{2^i},$$ which implies $g(x')=g((x_1+\dots+x_m)/m) \leq (g(x_1)+\dots+g(x_m))/m$ for any $m\in\mathbb{Z}_{\geq1}$. Finally, taking $a$ copies of $x$ and $b$ copies of $y$ we obtain $$g(\frac{ax+by}{a+b})\leq \frac{ag(x)+bg(y)}{a+b}=\left[\frac{a}{a+b}\right]g(x)+\left[\frac{b}{a+b}\right]g(y)$$ for $a,b\in \mathbb{Z}_\geq0$ not both zero and hence that $g$ is rationally convex.<|endoftext|> TITLE: Conceptual insights and inspirations from experimental and computational mathematics QUESTION [18 upvotes]: I am interested in whether experiments on computers can help identifying new ideas or concepts in Mathematics. I am not talking about confirming particular conjectures up to certain numbers (for example on the Riemann hypothesis or Collatz conjecture). I wonder whether there are examples, where results found by computers have been used and understood by mathematicians, who then used this new insight to make real progress in their field? One example that I recently found is Casey Mann's results on the Heesch Problem called Heesch Numbers of Edge-Marked Polyforms. There, from some exhaustive computational calculation, he has a chapter talking about Interesting examples and observations. While I cannot evaluate the significance of these observations, it goes along the lines what I am searching for: Do you have examples and literature references with results (new concepts, ideas, insights), that have been inspired by computational search or experimentations? REPLY [4 votes]: I am quite sure that the paper Using the Logistic Map to Generate Scratching Sounds, with the amusing abstract This article presents a mathematical model for generating annoying scratching sounds. Such sounds are generated by frictional motion and have been attributed to the chaotic nature of the frequency spectrum thereby produced. The proposed model is based on the logistic map and is modified to have the stick-slip property of a frictional vibration. The resulting sound is similar to that generated by scratching a chalkboard or glass plate with the fingernails. used some sort of computer experiments to figure out what types of mathematical functions sound annoying. REPLY [2 votes]: There is this recent work "Adventures in Supersingularland" (Arxiv preprint: https://arxiv.org/pdf/1909.07779.pdf) in which several predictions are made about isogeny graphs of supersingular elliptic curves based on experimental data. I am not aware whether the claimed have been proven formally in the meanwhile (that is in the last two months), but I expect the investigation in the preprint to be fertile soil for further (more theoretical) research.<|endoftext|> TITLE: Isomorphism problem among Thompson's groups QUESTION [5 upvotes]: I am looking for a reference proving the following statement: For every $n,m \geq 2$, the groups $T_n$ and $T_m$ are isomorphic if and only if $n=m$. Here, $T_k$ denotes the variation of Thompson's group $T$ (compared to $F$, $T$ acts on the circle instead of the interval) where dyadic numbers are replaced with $k$-adic numbers. I do not focus only on the first appearance of this statement. If several proofs are available in the literature, I am also interested to know them. REPLY [2 votes]: I would like to add a small comment to Jim Belk's detailed and clear answer. There is a preprint Xiaobing Sheng's work describing quasi-isometric embeddings from $T_n$ to $T$ which also provides a partial answer along the lines of Liousse's paper mentioned in J. Belk's answer. I am not certain of the current publication status of Sheng's paper. The relevant proposition there is: Proposition 3.1.2. The order of a torsion element in $T_n$ is a divisor of $l(n − 1) + 1$ for some integer $l \geq 0$.<|endoftext|> TITLE: What are examples of (collections of) papers which "close" a field? QUESTION [78 upvotes]: There is sometimes talk of fields of mathematics being "closed", "ended", or "completed" by a paper or collection of papers. It seems as though this could happen in two ways: A total characterisation, where somehow "all of the information" about a field has been uncovered. A negative result, rendering the field somehow irrelevant. A possible example for 1 might be the classification of finite simple groups. Examples for 2 might be Goedel's theorem effectively halting Hilbert's programme, or results showing e.g. certain large cardinal axioms to be inconsistent undermining work which assumes it. What are some other examples of results "closing" a field? Are there examples of a small number of papers "completing" a field in the sense of 1 above? (Apologies for many scare quotes!) REPLY [3 votes]: Hilbert's famous work seemed to have killed Invariant Theory. At least that's what Gordan felt at that time. In the historical book by Dieudonn'e' and Carrell they say invariant theory, presumed dead, as coming back alive from the ashes like phoenix.<|endoftext|> TITLE: How to compute the cohomology of a local system? QUESTION [11 upvotes]: Suppose we have a reasonable topological space $X$ (i.e. a complex algebraic variety or a manifold) whose integral singular cohomology and fundamental group we understand well. Suppose that we are given a monodromy representation $\rho: \pi_1(X,x) \longrightarrow \text{GL}(V)$. How does one compute the cohomology $H^i(X,L)$ of the local system $L$ associated to $\rho$? REPLY [2 votes]: If you're looking for something very concrete in terms of things like explicit boundary map computations, you might be interested in looking at Section 31 of Steenrod's Topology of Fibre Bundles. As in the other answers, though, it sort of depends on what sort of information you already possess or can get at for your space.<|endoftext|> TITLE: Saturation of non-stationary ideal on $\omega_2$? QUESTION [6 upvotes]: It is known that $NS_{\omega_2}$ cannot be saturated (namely there cannot be $\aleph_3$ many stationary subsets of $\omega_2$ any two of which have non-stationary intersection). However, it may be the case when it is restricted to a stationary subset. It is also known that the stationary subset cannot be $\omega_2\cap cof(\omega)$. Can $NS_{\omega_2}\restriction cof(\omega_1)$ be saturated? or on some stationary subset of $\omega_2\cap cof(\omega_1)$? REPLY [3 votes]: I believe it is still open whether $\mathrm{NS}_{\omega_2} \restriction \mathrm{cof}(\omega_1)$ can be saturated. But it was known by early unpublished work of Woodin that $\mathrm{NS}_{\omega_2} \restriction S$ can be saturated, where $S$ is a stationary-costationary subset of $\mathrm{cof}(\omega_1)$. The complement provides a "safe space" to allow the iteration to work. Details are given by Foreman-Komjath (MR: 2151585) and by me. I have heard rumors that recent work related to higher analogues of forcing axioms might be used to tackle the problem. This approach would surely give a model with not-GCH, so we'd have an obvious next open question to tackle.<|endoftext|> TITLE: Bounding the spectral gap of a simple symmetric matrix QUESTION [8 upvotes]: I have a seemingly innocent linear algebra problem that I cannot solve, and which I hope that you would kindly offer some insight into. Here is the description: Let $\mathbf{a} = (a_1, a_2, \dots, a_d)^{T}$ be a positive probability vector, $i.e.$ $\Vert \mathbf{a}\Vert_1=1$ and $a_i > 0$ for all $i$. Let matrix $A$ be defined as follows: $$A = \textrm{diag}(\mathbf{a}) - \mathbf{a}\mathbf{a}^{T}$$ where $\textrm{diag}(\mathbf{a})$ means the diagonal matrix with the $i$th diagonal entry being $a_i$. It is straightforward to show that $\mathbf{1}_d$, the all-one vector of dimension $d$, is an eigenvector of $A$ of eigenvalue $0$. And Gershgorin circle theorem also shows that all $A$'s eigenvalues are greater or equal to $0$. My question is: What is the smallest eigenvalue of $A$ that is not zero? I carried out the calculation when $d = 3$ and realized that there may not be a simple analytic formula to it and hence a nice lower bound is also greatly appreciated. Thank you so much! REPLY [2 votes]: Here is an elementary bound. The second eigenvalue of $A$ satisfies $$\lambda_2(A)> \max_k\min_{i\ne k}a_ia_k=(\max_ka_k)(\min_ia_i).$$ To prove it, let $A_k$ be the principal submatrix obtained by deleting the $k$th row and column. By interlacing, we have $\lambda_2(A)>\lambda_1(A_k)$. Now apply Gershgorin to $A_k$: there exists an index $i\ne k$ such that $$\lambda_1(A_k)\ge a_i(1-a_i)-\sum_{j\ne i,k}a_ia_j=a_ia_k.$$ Improvement. Actually, one has $\lambda_2(A)\ge\min_ia_i$. To see this, write $A=D+B$ with $D={\rm diag}(a_1,\ldots,a_n)$ and $B=-aa^T$. By Weyl's inequality, we have $$\lambda_2(A)\ge\lambda_1(D)+\lambda_2(B)=\min_ia_i+0.$$ Remark that this bound is accurate, as if $a_1=\cdots=a_{n-1}=\min_ia_i$ and $a_n=1-(n-1)a_1$, then the spectrum of $A$ is given by $0$, $a_1$ (multiplicity $n-2$) and $na_1a_n$. Hence $\lambda_2=a_1$.<|endoftext|> TITLE: Most general definition of differentiation QUESTION [23 upvotes]: There are various differentiations/derivatives. For example, Exterior derivative $df$ of a smooth function $f:M\to \mathbb{R}$ Differentiation $Tf:TM\to TN$ of a smooth function between manifolds $f:M\to N$ Radon-Nikodym derivative $\frac{d\nu}{d\mu}$ of a $\sigma$-finite measure $\nu$ Fréchet derivative $Df$ of a function between Banach spaces $f:V\to W$ What is the most general definition of differentiation or derivative? REPLY [2 votes]: Other answers possibly did not mention multiplicative differentiation (inverse of multiplicative integral) and discrete differentiation (finding of finite difference, both backward and forward). There is also discrete multiplicative differentiation. I do not think there is any generalization that includes these ones.<|endoftext|> TITLE: Complexity of graph 3 coloring and counting algorithm QUESTION [5 upvotes]: 3-coloring a graph $G$ is equivalent to partitioning the vertices of $G$ in three independent sets. The smallest independent set $A$ is at most $n/3$ where $n$ is the order of $G$. We have $G \setminus A$ is bipartite graph, and the bipartitions can be found in polynomial in $n$ time. This gives graph coloring and counting of 3 colorings algorithm: For one coloring: Enumerate independent sets A up to size n/3 If G \ A is bipartite report 3 colorable and stop For counting colorings: set cols:=0 Enumerate independent sets A up to size n/3 If G \ A is bipartite set cols := cols + number_of_2_colorings of G \ A # G \ A might not be connected Q1 What is the complexity of these algorithms? Crude upper bound: All subsets (not necessary independent) of size at most $n/3$ experimentally is about $1.883^n$ Q2 What is state of the art of counting 3 colorings? REPLY [2 votes]: For finding colorings, let $A=K_4 \cup \bar{K^m}$, which is a $K_4$ and $m$ disjoint vertices. The algorithm keeps enumerating independent sets without success, as $A$ is not 3-colorable. For counting colorings, the complement of a complete graph takes the longest time. Both have complexity of the order $1.883^n c$ where $c$ is some constant.<|endoftext|> TITLE: Is property FA of Serre known for $SL_2(\mathbb{Z}[i])$ and $SL_2(\mathbb{Z}[\zeta_3])$ QUESTION [10 upvotes]: In Serre's book Trees [Se, p. 68] it says: 3) For $SL_2$ the situation is different. It is clear that $SL_2(\mathbf Z )$ does not have property (FA). It is the same with $SL_2(A)$ when $A$ is the ring of integers of an imaginary quadratic field not isomorphic to $\mathbf{Q}(\sqrt{- 1})$ or $\mathbf{Q}(\sqrt{-3})$, since such a group has a quotient isomorphic to $\mathbf{Z}$ (cf. [20], th. 9, p. 519). On the other hand, if $K$ is an algebraic number field not isomorphic to $\mathbf{Q}$ nor to an imaginary quadratic field, one can show that every arithmetic subgroup of $SL_2(K)$ has property (FA); this applies notably to the groups $SL_2(\mathbf Z[\sqrt{D}])$ and their subgroups of finite index ($D$ a square-free integer $> 1$). Question: Is it known whether $SL_2$ of the rings of integers of $\mathbf{Q}(\sqrt{- 1})$ and $\mathbf{Q}(\sqrt{- 3})$ has property FA? [Se]: Serre, Jean-Pierre, Trees. Transl. from the French by John Stillwell, Berlin-Heidelberg-New York: Springer-Verlag. IX, 142 p. DM 48.00; {$} 28.40 (1980). ZBL0548.20018. REPLY [10 votes]: Here is an answer for all Bianchi groups $SL(2, O_d)$: Such a group admits a nontrivial graph of groups decomposition (equivalently, does not have the property FA) unless $d=3$, in the latter case, it does not split, i.e. has the Property FA. For details, see: C. Frohman and B. Fine, Some Amalgam Structures for Bianchi Groups, Proceedings of the American Mathematical Society, Vol. 102, No. 2 (1988), pp. 221-229.<|endoftext|> TITLE: Study topology from existence of multiple smooth structures? QUESTION [16 upvotes]: There are classical existence results of a smooth structure on a topological manifold, and many results on the existence of multiple (i.e. exotic) smooth structures. Some utilize Freedman's theorem in dimension 4, and there's Milnor's twisted spheres in dimension 7, etc.. Morse theory and Riemannian geometry, for example, help to probe topology and only need a smooth structure. I questioned why we would want to know whether there is more than one smooth structure, and it leads to a less subjective question: Are there ways to study the underlying topological manifold from the existence of multiple smooth structures? Here is a more refined question: Can different smooth structures give you different "bounds" on some topological quantities (maybe minimal genus of surfaces, "optimal" presentations of the fundamental group, etc.), or where one smooth structure gives you information while the other smooth structure can't give you any? REPLY [13 votes]: Since this question might have many answers, I can propose one possible answer (I hope this is really an answer to the question.) So, suppose we have a smooth $4$-dimensional manifold. We want to triangulate it smoothly so that the number of vertices is the minimal possible. Then this minimal number of vertices might well depend on the choice of a smooth structure. Indeed, if we take a smooth compact $4$-fold that has infinite number of smooth structures, then for any $N$ there will be only finite number of such structures for which there exists a triangulation with at most $N$ vertices. It is well known that for $\mathbb C \mathbb P^2$ with the standard smooth structure the minimal number is $9$ . Such a triangulation was invented by Kühnel (and some other people, see references in the article below). If by any chance $\mathbb C \mathbb P^2$ has an infinite number of smooth structures, there will be those for which the minimal number of vertices is huge (the number is always $\ge 9$). By the way, there is also a triangulation of a K3 surface (also constructed by Kühnel with a collaborator) with 16 vertices. This result uses Freedman's theorem. As far as I understand, it is not yet known whether the corresponding K3 surface is diffeomorphic to the standard one, see https://www.sciencedirect.com/science/article/pii/S0040938399000828<|endoftext|> TITLE: Do infinite products commute with trivial cofibrations, for simplicial sets? QUESTION [5 upvotes]: I'm reading Voevodsky and Morel's book '$\mathbb{A}^1$-homotopy theory of schemes'. In Remark 3.1.15, it says that for any simplicial fibrant sheaf $F$ and open sets $U\subseteq V$, $F(V)\to F(U)$ is a fibration. Prove by definition. We have a bifunctor $$\begin{array}{ccccc}sSet&\times&Shv(Sm/k)&\to&sShv(Sm/k)\\(S&,&F)&\mapsto&S\times F\end{array},$$ where $(S\times F)(X)_n=S_n\times F(X)$. Consider the coequalizer $$\Lambda^n_k\times U\rightrightarrows\Lambda^n_k\times V\coprod\triangle^n\times U\to C.$$ Then there is a map $i:C\to \triangle^n\times V$ and the question is reduced to the RLP of $F$ w.r.t $i$. So I want to prove $i$ is a trivial cofibration. It's obviously a cofibration but I'm stuck at proving it's a weak equivalence. It suffices to prove that the functor $-\times F:sSet\to sShv(Sm/k)$ is a left Quillen functor since we could then use the pushout diagram of $C$. So we are going to prove that trivial cofibrations commute with infinite products, by passing to stalks... I think we have to prove that geometric realization functor commutes with infinite products, as least up to a weak equvalence. Is this true? Thanks a lot! REPLY [9 votes]: To answer the title question, here's an example which I think I learned from some standard reference but I forget which. Let $N$ be the following graph, considered as a 1-skeletal simplicial set: the vertices are the natural numbers $\mathbb N$, and there is an edge from $n$ to $n+1$ for each $n \in \mathbb N$; there are no other nondegenerate cells. Let $i: \Delta[0] \to N$ be the inclusion of your favorite point. Then clearly $N$ is weakly contractible, so that $i$ is an anodyne extension (i.e. an acyclic cofibration in the usual Kan-Quillen model structure). Then consider The power $N^\omega$, i.e. the product of $N$ with itself $\omega$ many times; The inclusion $i^\omega: \Delta[0] = \Delta[0]^\omega \to N^\omega$. The thing to notice is that $\pi_0(N^\omega)$ is nontrivial (I think it's the cardinality of the continuum): a point in $N^\omega$ is a sequence of natural numbers, and two sequences are connected by a path only if their sequence of differences is bounded! The upshot is that $i^\omega$ is not a weak equivalence, even though $i$ is an anodyne extension. So anodyne extensions are not closed under products. We can play a similar game by taking $j: N \to \bar N$ to be any fibrant replacement of $N$: again $j^\omega$ can't be a weak equivalence. This shows that even the class of anodyne extensions into Kan complexes is not stable under products. Also, instead of $N$, we can use its subdivision $sd(N)$, which has a point for each nonnegative half-integer, and 1-cells from $n+\frac 1 2$ to $n$ and $n+1$ for each $n \in \mathbb N$. The same conclusion applies: your favorite anodyne extension $\Delta[0] \to sd(N)$ fails to be preserved by powers. This is illustrative because $sd(N)$ has some nice properties that $N$ doesn't have -- it's the nerve of a poset (in particular it's the nerve of a category (in particular it's a quasicategory)). Thus we see that the power of an anodyne extension may fail to be an anodyne extension, even when it's the nerve of a map of posets (in particular even when it's the nerve of a functor (in particular when it's a map of quasicategories)).<|endoftext|> TITLE: Which group do two generic $2\times 2$ triangular matrices generate? QUESTION [8 upvotes]: Let $A,B$ be two generic (in particular invertible) $2\times 2$ upper-triangular complex matrices. They generate a countable group $G$, the commutator subgroup of $G$ is abelian. Are there other relations in $G$? How is this group called? REPLY [16 votes]: I think it is a old result of Magnus that the free metabelian group on $d$ generators can be embedded as group $T_2$ of $2\times 2$ matrices over $\mathbf{C}$. It is easy to deduce that every generic $d$-tuple in $T_2$ freely generates such a free metabelian group $\Gamma_d$. Here generic means: not in some countable union of proper Zariski closed subsets of $T_2^d$; hence it's generic both in the topological and the measured sense. Indeed, let $w(x_1,\dots,x_d)$ be a nontrivial element $\Gamma_d$. Let $F_w$ be the set of $d$-tuples $g\in T_2^d$ such that $w(g)=I_2$. Then $F_w$ is a proper Zariski closed subset. Hence any $d$-tuple outside $\bigcup_{w\in\Gamma_d-\{1\}}F_w$ works. (A group is metabelian if its derived subgroup is abelian, or equivalently if and only if any two commutators commute. So it doesn't mean a free metabelian group satisfies no other relation, but it indeed means that all its relations can be derived from this one.)<|endoftext|> TITLE: Compact manifold $X$ having fixed-point property but $X\times X$ does not QUESTION [16 upvotes]: A manifold $X$ has the fixed-point property if for every continuous map $f:X→X$ there is $x∈X$ with $f(x)=x$. Examples of such spaces are disks and the real projective plane $\mathbb{RP}^2$. Question: If a compact manifold $X$ has the fixed-point property, does $X\times X$ necessarily have the fixed-point property? Known: False for $X$ a polyhedron. See An example in the fixed point theory of polyhedra. Bull. Amer. Math. Soc. 73(1967), 922-924. False for various other spaces. See MO question 283930. True for compact surfaces: the only such surface with the fixed-point property is $\mathbb{RP}^2$, and $\mathbb{RP}^2\times \mathbb{RP}^2$ also has the fixed-point property, by the Lefschetz fixed point lemma. REPLY [8 votes]: Not a full answer, but the answer seems to not be known. The question is open for closed manifolds (see the mathscinet review of Kwasik and Sun's paper, MR3584128). The answer is yes for other manifolds like $\mathbb{CP}^2$, but no for the symmetric product of closed manifolds according to the same paper (Theorem 2). One can take $X=\mathbb{RP}^4\#\mathbb{RP}^4\#\mathbb{RP}^4$ which has the fpp, but $X(2)=(X\times X)/\mathbb{Z}_2$ does not.<|endoftext|> TITLE: Lifting a splitting of an Abelian variety to characteristic 0 QUESTION [11 upvotes]: Let $R$ be the ring of integers in a (complete) algebraic closure of $\mathbb Q_p$ with maximal ideal $\mathfrak p$. Suppose I have an Abelian surface $\mathcal A/R$ such that over every $R/\mathfrak p^n$, there exist elliptic curves $E_n, E_n'$ over $R$ with $\mathcal A$ isogenous to $E_n\times E_n'$ over $R/\mathfrak p^n$. Does this imply that $\mathcal A$ splits over $R$? Note that I do not require any compatibility conditions between the elliptic curves or the isogenies over varying $n$. REPLY [6 votes]: $\newcommand{\cA}{\mathcal{A}}\newcommand{\cB}{\mathcal{B}}\newcommand{\bZ}{\mathbb{Z}}$No, that does not imply that $\cA$ splits over $R$. In fact, if $\cA_1=\cA\times_R R/p$ is isogenous to a product of elliptic curves then all $\cA_n=\cA\times_R R/p^n$ are isogenous to products of elliptic curves: Lemma. Let $S'\twoheadrightarrow S$ is a surjection of $\bZ/p^n$-algebras with nilpotent kernel. If $\cA,\cB$ are abelian schemes over $S'$ then for any morphism $f:\cA\times_{S'}S\to\cB\times_{S'}S$ of abelian schemes over $S$ there exsists a morphism $g:\cA\to\cB$ over $S'$ reducing to $p^Nf$ for a certain $N$. Proof. The analogous statement for $p$-divisible groups is a part of Drinfeld's rigidity lemma for quasi-isogenies and the statement for abelian schemes follows from Serre-Tate theorem(Lemma 1.1.3 and Theorem 1.2.1 in Katz's 'Serre-Tate local moduli'). In fact, this particular consequence can be also derived by a more direct argument: we can define the desired $g$ on the level of functors of points. There exists $N$ such that for any $S'$-algebra $T$ the kernel of the surjection $\cB(T)\to \cB_{S}(T\otimes_{S'}S)$ is annihilated by $p^N$. For a section $x\in\cA(T)$ define then $g(x)$ as $p^N\widetilde{f(\overline{x})}$ where $\widetilde{\cdot}$ denotes an arbitrary lifting along the surjection $\cB(T)\to\cB_S(T\otimes_{S'}S)$, the result does not depend on the choice. $\square$ Applying the lemma to the thickening $R/p^n\to R/p$ and the abelian schemes $\cA_n$ and $\widetilde{E}_1\times\widetilde{E'_1}$ where $\widetilde{E}_1,\widetilde{E'_1}$ are arbitrary lifts of the elliptic curves $E_1,E_1'$ gives the claim. The key here is, of course, that we are not bounding the degree of a splitting isogeny. So, it remains to give an example of a simple abelian surface $\cA$ over $R$ with reduction isogenous to a product of elliptic curves. Here is one way to do this: pick a quaternion algebra $D$ over $\mathbb{Q}$ that is non-split at $p$ but is split over $\mathbb{R}$ and an abelian surface $\cA/R$ with $End_R(\cA)\otimes_{\bZ}\mathbb{Q}\simeq D$ (abelian surfaces with an action of an order in $D$ are parametrized by a Shimura curve and we just need to choose a general enough point on it so that the endomorphism algebra is not larger than $D$). Because of the assumption on $p$ the reduction of $\cA$ modulo $p$ is supersingular and is isogenous to a product of two supersingular elliptic curves(see e.g. Proposition 2 in Chapter III of Boutout-Carayol's 'Uniformisation p-adique des courbes de Shimura').<|endoftext|> TITLE: Partitioning $\beta \mathbb{Z} \setminus \mathbb{Z}$ QUESTION [7 upvotes]: Take the integers $\mathbb{Z}$ and the addition \begin{align*} +: \mathbb{Z} \times \mathbb{Z} &\to \mathbb{Z} \\ (a,b) &\mapsto a+b. \end{align*} Using the Stone-Čech compactification $\beta\mathbb{Z}$ in two steps, $A$ can be "continuously" extended to \begin{equation*} +: \beta\mathbb{Z} \times \mathbb{Z} \to \beta\mathbb{Z} \end{equation*} and then to \begin{equation*} +: \beta\mathbb{Z} \times \beta\mathbb{Z} \to \beta\mathbb{Z}. \end{equation*} The resulting operation $+$ is continuous in its right argument, and, when the right argument is an integer, it is continuous on its left argument. Call $\mathbb{Z}^* := \beta\mathbb{Z} \setminus \mathbb{Z}$. For any $p \in \mathbb{Z}^*$, we have the right ideal $I_p := p + \beta\mathbb{Z}$. Is it possible to partition $\mathbb{Z}^*$ in two (non trivial) pieces $A$ and $B$ such that \begin{equation*} p \in A \text{ and } q \in B \Rightarrow I_p \cap I_q = \emptyset? \end{equation*} In other words, if I generate an equivalence relation based on $p \sim q$ whenever $I_p \cap I_q \neq \emptyset$, will this equivalence relation partition $\mathbb{Z}^*$ in more then one piece? REPLY [6 votes]: Yes, this is possible. First of all, let me suggest a way of thinking about the $+$ operation. If you're familiar with the idea of taking limits along an ultrafilter, then given $p,q \in \mathbb Z^*$, you can think of $p+q$ as $$p+q = q\text{-lim}_{n \in \mathbb Z} \ (p+n).$$ I tend to look at this dynamically: we have a map (the shift map) that sends an ultrafilter $p$ to $p+1$, and $p+q$ just represents one limit point of the set $\{p+n \,:\, n \in \mathbb Z\}$, the (full) orbit of $p$ under the shift map. Also, this makes it clear that $I_p = p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}}$ for any $p \in \mathbb Z^*$. To answer your question, one partition of $\mathbb Z^*$ that does what you want is the partition into the two clopen sets $\mathbb N^*$ and $(-\mathbb N)^*$ (the ultrafilters concentrating on the positive and negative integers, respectively). If $p$ concentrates on the positive integers, then so does $p+n$ for any $n$. In other words, $p \in \mathbb N^*$ implies $p+n \in \mathbb N^*$ for all $n$. This means that $p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} \subseteq \mathbb N^*$. Similarly, if $p \in (-\mathbb N)^*$ then $p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} \subseteq (-\mathbb N)^*$. No other partition of $\mathbb Z^*$ into clopen sets has the property you want. (I'll sketch a proof of this below.) But there are other, more interesting partitions of $\mathbb Z^*$ that do what you want. Recall that $X \subseteq \mathbb Z^*$ is a weak $P$-set if $X \cap \overline{D} = \emptyset$ for any countable $D \subseteq \mathbb Z^* \setminus X$. In this paper, Jonathan Verner and I show that there is a minimal right ideal $R$ of $\mathbb Z^*$ that is also a weak $P$-set. This means that if $p \in R$, then $p + \beta \mathbb Z = R$ (because it's a minimal right ideal), and if $p \notin R$ then $p+\beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} \subseteq \mathbb Z^* \setminus R$ (because $R$ and its complement are both shift-map-invariant, and because $R$ is a weak $P$-set). Therefore the partition of $\mathbb Z^*$ into $R$ and $\mathbb Z^* \setminus R$ has the property you want. In fact, the proof in my paper with Jonathan shows that there are $2^{\mathfrak{c}}$ distinct right ideals of this kind; because each minimal right ideal that is a weak $P$-set constitutes an equivalence class, it follows that the equivalence relation you describe has the maximum possible number of equivalence classes, namely $2^{\mathfrak{c}}$. (Suppose $\mathcal P$ is a partition of $\mathbb Z^*$ into clopen sets other than the one described above. Some $A^* \in \mathcal P$ has the property that either (1) $\mathbb N^* \cap A^* \neq \emptyset$ and $\mathbb N^* \setminus A^* \neq \emptyset$, or else (2) $(-\mathbb N^*) \cap A^* \neq \emptyset$ and $(-\mathbb N^*) \setminus A^* \neq \emptyset$. Without loss of generality, let us say $\mathbb N^* \cap A^* \neq \emptyset$ and $\mathbb N^* \setminus A^* \neq \emptyset$. This means that the set $E$ of $n \in \mathbb N$ such that $n \in A$ but $n+1 \notin A$ must be infinite. If $p$ is an ultrafilter that concentrates on $E$, then we have $p \in A^*$ while $p+1 \notin A^*$. Thus $p$ and $p+1$ are in different partition elements, but clearly $p + \beta \mathbb Z = \overline{\{p+n \,:\, n \in \mathbb Z\}} = \overline{\{(p+1)+n \,:\, n \in \mathbb Z\}} = (p+1)+\mathbb Z^*$. Therefore this partition does not have the property you want.)<|endoftext|> TITLE: On the closed-form of $\int_0^1\int_0^1\int_0^1\frac{dxdydz}{1-\frac{z}{3}(x+\sqrt{xy}+y)}$ QUESTION [7 upvotes]: I would like to know if it is possible to calculate in closed-form, or well what work can be done about it, the definite integral $$\int_0^1\int_0^1\int_0^1\frac{3dxdydz}{3-z(x+\sqrt{xy}+y)},\tag{1}$$ where I was inspired in a well-known integral representation for the Apéry constant involving the volume $x\cdot y\cdot z$ in the denominator, and in the formula for the volume of a square frustum of basis $\sqrt{x}$ and $\sqrt{y}$ and height $z$, as reference for all I add the Wikipedia Heronian mean. It is easy to check the integration of the logartihm $$\int_0^1 \frac{1}{3-z(x+\sqrt{xy}+y)}dz=\frac{\log 3-\log(3-x-\sqrt{xy}-y)}{x+\sqrt{xy}+y}$$ but the computations using a CAS (and standard time of computation with my computer), that I tried after this step, seems to me very tedious to evaluate. I suspect that now an important key to evaluate it should be to exploit symmetry or a suitable change of variable. Question. I would like to know if it is possible to evaluate in closed-form (in terms of well-known constants and/or particular values of special functions) previous definite integral $(1)$. If isn't feasible to get the closed-form that I evoke explain why or add what work can be done. Many thanks. If this integral is in the literature feel free to refer the literature in your answer or comment and I can to read the result from the literature. Edit: (see comments, please). I don't know if from this step one can to get the integral in closed-form. Feel free to do more feedback, many thanks. Using previous hints in comments I can to write that it is possible to reduce the integral over $xy$ to one-dimensional integral by using polar coordinates in $xy$-plane $$\int_0^1\int_0^1\frac{dxdy}{1-\frac{z}{3}(x+\sqrt{xy}+y)}=2\int_0^{\frac{\pi}{4}}\int_0^{\sec \theta}\frac{drd\theta}{\frac{3}{r}-z(\cos\theta+\sqrt{\cos\theta\sin\theta}+\sin\theta)},$$ where the inner integral is equals to $$\int_0^{\sec\theta}\frac{dr}{\frac{3}{r}-z(\cos\theta+\sqrt{\cos\theta\sin\theta}+\sin\theta)}=\frac{-A\sec\theta-3\log(3-A\sec\theta)+3\log 3}{A^2},$$ being $A=z(\cos\theta+\sqrt{\cos\theta\sin\theta}+\sin\theta)$. Since the integral seems very difficult I am going to accept an answer showing what work can be done (see the Question) as soon as expires the bounty. REPLY [12 votes]: Let me expand the integrand in powers of $z$ and integrate over $z$, $$I=\int_0^1 dx\int_0^1 dy\int_0^1 dz\;\frac{3}{3-z(x+\sqrt{xy}+y)}$$ $$\qquad\qquad=\sum_{n=0}^\infty\,\frac{3^{-n}}{n+1} \int_0^1 dx\int_0^1 dy\;(x+\sqrt{xy}+y)^n.$$ The integral over $x$ and $y$ is an element $c_{n}\in\mathbb{Q}$, $$I=\sum_{n=0}^\infty\,\frac{3^{-n}}{n+1}c_n,\;\;c_n=\int_0^1 dx\int_0^1 dy\;(x+\sqrt{xy}+y)^n,$$ $$\{c_0,c_1,c_2,\ldots\}=\{1,\tfrac{13}{9},\tfrac{149}{60},\tfrac{1667}{350},\tfrac{18623}{1890},\tfrac{69667}{3234},\ldots\}.$$ I have not yet succeeded in finding a closed-form expression for $c_n$ valid for all integer $n$, but that seems doable. At least we would then have the desired integral as a sum over rational coefficients. With FusRoDah's input for $c_n$, and carrying out the sum over $n$, I find $$I=\sum_{a,b=0}^\infty\frac{12 \Gamma (a+b+1) \left(3^{-a-\frac{b}{2}} B_{\frac{1}{3}}\left(\frac{b}{2}+1,-a-b\right)-B_{\frac{1}{3}}(a+b+1,-a-b)\right)}{(2 a+b) (2 a+b+2) \Gamma (a+1) \Gamma (b+1)},$$ (with $B$ the incomplete beta function).<|endoftext|> TITLE: Operator norm of square root of matrix vs original QUESTION [7 upvotes]: If I have a nonsymmetric matrix whose operator norm is $\leq 1$ and square root it, does its operator norm remain below $1$? More formally, I want to know whether there is always at least one square root for which this is the case even if it isn't true for all of them and under reasonable assumptions that guarantee a square root exists. Specifically, suppose I have a square nonsymmetric matrix $A$ with $\|A\|_2 \leq 1$ where $\|\cdot\|_2$ denotes the operator norm (max singular value). Does there always exist a square matrix $B$ such that $B^2=A$ and $\|B\|_2 \leq 1$? I am willing to assume $A$ is diagonalizable, although not unitarily diagonalizable. I am also willing to assume $A$ is of the form $A' + \Delta$ for some small random perturbation $\Delta$ and matrix $A'$. Without at least one of these assumptions, the square root might not exist at all. I would be interested in a proof / counterexample under any nonempty subset of these assumptions. For example, a counterexample showing that the statement can be false when we assume $A$ is diagonalizable would be useful to me. REPLY [16 votes]: This is not true. For example, if $$ A = \frac{2}{1 + \sqrt{5}}\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} $$ then $\|A\| = 1$. Furthermore, $A$ has exactly $2$ square roots, which are $$ B_{\pm} = \frac{\pm\sqrt{2}}{\sqrt{1 + \sqrt{5}}}\begin{bmatrix} 1 & 1/2 \\ 0 & 1\end{bmatrix}. $$ It is straightforward to check that $\|B_{\pm}\| \approx 1.0068841364 > 1$. The matrix $A$ here is not diagonalizable, but if we add $\varepsilon > 0$ to its bottom-right entry then it becomes diagonalizable. It will then have four square roots, two of which are continuous in $\varepsilon$ (and thus still have norm $> 1$ when $\varepsilon$ is small), and the other two of which have a huge top-right entry (and thus have norm $> 1$ when $\varepsilon$ is small).<|endoftext|> TITLE: Is there an almost strongly zero-dimensional space which is not strongly zero-dimensional QUESTION [6 upvotes]: A Tychonoff space $X$ is called strongly zero-dimensional if each functionally closed subset $F$ of $X$ is a $C$-set, which means that $F$ is the intersection of a sequences of clopen sets in $X$. A Tychonoff space $X$ is called almost strongly zero-dimensional if each functionally closed subset of $X$ is the union of a sequence of $C$-sets. Question. Does there exists a (metrizable separable) Tychonoff space which is almost strongly zero-dimensional but not strongly zero-dimensional? This problem was posed on 30.11.2019 by Olena Karlova (from Chernivtsi) on page 35 of Volume 3 of the Lviv Scottish Book. Prize. A portrait of a mathematician who will solve this problem :) REPLY [2 votes]: Not an answer, but this may be helpful: Theorem 1. If $X$ is a Lindelöf Tychonoff almost strongly zero-dimensional space, then the following are equivalent: (i) $X$ is strongly zero-dimensional; (ii) $X$ is almost zero-dimensional, that is, $X$ has a neighborhood basis of C-sets. Proof. (i)$\Rightarrow$(ii) is trivial, and the converse follows from Theorem 4.3 in this paper (we assume separable metrizable there, but Lindelöf should be enough). $\square$ In light of Taras Banakh's comment above, for separable metrizable spaces I believe the question is: If $X$ is separable metrizable and $f:X\to Y$ is a continuous bijection onto a zero-dimensional space $Y$ which maps open sets to $G_\delta$-sets, then is $X$ almost zero-dimensional? Theorem 2. Every almost strongly zero-dimensional homogeneous Polish space $X$ is (strongly) zero-dimensional. Proof. If $U$ is any open subset of $X$, then $U$ is a $\sigma$C-set, so by the Baire property there is a C-set $F\subseteq U$ which contains a non-empty open set. Continuing this process we construct C-sets $F_n$ such that $F_{n+1}\subseteq F^\mathrm{o}_n$ and $\text{diam}(F_n)\leq 1/n$ in a complete metric. Then there exists $x\in \bigcap F_n$, and $x$ has a neighborhood basis of C-sets. By homogeneity, $X$ is almost zero-dimensional, so by Theorem 1 $X$ is strongly zero-dimensional. $\square$ More generally it is true that each almost strongly zero-dimensional Polish space is zero-dimensional at a dense $G_\delta$-set of points.<|endoftext|> TITLE: Polygonal paths and polygons with prescribed set of vertices QUESTION [7 upvotes]: Let $A$ be a finite set of points in the plane. How can we determine if there is a simple open polygonal path (i.e. without intersections), whose vertices are exactly $A$, with no straight angles between adjacent sides? Since in mathoverflow.net/q/226469/4312 question is only about cycles, the stronger question also remains. Namely - how to determine if there exists a polygon with non-intersecting sides and without straight angles, whose vertices are exacly $A$ ? Particular interesting case is when $A$ is a set of points $(x;y)$ with $0\leq x+y \leq 2n$ , where $x,y$ are nonnegative integers. Hypothesis : no for path for $n=1$ and $n=2$. REPLY [5 votes]: The following polygon is a counterexample to the hypothesis for $n=4$. Namely we consider the set of integer points $(x,y)|0\le x+y\le 8$. The picture is on a square-lined paper, where the size of one square is $\frac{1}{2}\times \frac{1}{2}$ The picture was constructed in collaboration with Svetlana Ermakova. So my guess that it will be possible to do the same thing for all $n\ge 4$<|endoftext|> TITLE: Can I assign the term "is eigenvector" and "is eigenmatrix" of matrix $P$ in my specific (infinite-size) case? QUESTION [5 upvotes]: Remark: I asked this in MSE, the question got views and votes but seemingly no one had an answer so far. Background: I'm rereading a couple of my exploratory (surely not research-level) math-essays and want to fix some possible wrong or misleading expressions. I've used the following notions/expressions a couple of years already but I would like to confirm that I can really use it in revisions of my web-essays. Consider the (upper triangular) infinite "Pascal"/"binomial"-matrix $P$ with top-left element as $$\small \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ . & 1 & 2 & 3 & 4 & 5 \\ . & . & 1 & 3 & 6 & 10 \\ . & . & . & 1 & 4 & 10 \\ . & . & . & . & 1 & 5 \\ . & . & . & . & . & 1 \end{bmatrix}.$$ Rightmultiplying it with the columnvector $E_1 = [1,1/1!,1/2!,1/3!, \dotsc]$ gives $$ P \cdot E_1 = e \cdot E_1 $$ which has the form of an eigenvector equation as known from the cases with matrices of finite size. However, using $P$ and $E_1$ truncated to finite size $P^\star$ and $E_1^\star$ this would never be correct since $P^\star$ has no diagonalization. Back to infinite size: in general with some columnvector $E(x)=[1, x^1, x^2/2!, x^3/3!, \dotsc]$ we have $$ P \cdot E(x) = e^x \cdot E(x) $$ thus for each $x$ we have that $P$ has $E(x)$ as eigenvector to eigenvalue $e^x$. Now what I'm discussing in a couple of essays are a second type of infinite matrices, namely the concatenation of vectors $E_n=E(n)$ to a matrix $$EZ=[E_0,E_1,E_2,E_3,\dotsc]$$ and following the example I could write $$ P \cdot EZ = EZ \cdot {^dV}(e) \\\qquad \qquad \qquad \text{ where ${^dV}(e) = \operatorname{diagonal}([1,e,e^2,e^3...])$} $$ which has again the form of an eigenmatrix-decomposition (or "diagonalization"). I always tended to say, that "$EZ$ is an eigenmatrix of $P$" (or is matrix-of-eigenvectors), or that "$P$ of infinite size has a diagonalization" and used this at several places in my manuscripts. But because for the case of finite size $P^\star$ has no diagonalization (it has only a Jordan-form), I feel it might be too sloppy to formulate this as an Eigenmatrix-relation or even as "diagonalization of $P$" (the latter is even more problematic since the matrix $EZ$ has no inverse/reciprocal and we cannot write $P=EZ \cdot {^dV}(e) \cdot EZ^{-1}$). Q: How could I correctly express that relation, even in a informal context? Can I still apply the terms "matrix of eigenvectors", "… of eigenvalues" and "diagonalization"? Update, added. One argument which is possibly against the use of the concept of diagonalization here, is perhaps that of the existence of a Jordan-decomposition for the finite-size case $P^\star$. The top-left $6 \times 6$ -truncation of that (finite-size) Jordan-decomposition $P^\star = S^\star \cdot J^\star \cdot S^{\star -1}$ shows known matrices $S^\star$ (from Stirlingnumbers $1$st kind, left hand, factorially scaled) , the simple matrix $J^\star$ (in the middle) and $S^{\star-1}$ (from Stirlingnumbers $2$st kind, right hand, factorially scaled) (or in a non-canonical rescaled version, but Stirlingnumbers nicer recognizable): I don't know, whether it is more appropriate to apply the generalization to the case of infinite-size for the Jordan-decomposition, but if we do this, than we had a parallel between "diagonalization" and "Jordan-decomposition" which likely points to some incompatibility here with respect to the "diagonalization"-concept for the case of infinite size. P.s.: don't know the best tagging for MO. Please feel free and improve if you think fit. REPLY [2 votes]: I would suggest to call $E(x)=(1, x^1, x^2/2!, x^3/3!, \cdots)$ a fixed point of the linear map $E\mapsto M_\lambda\cdot E$, with $M_\lambda=\lambda P$ and scale factor $\lambda=e^{-x}$. In this way you can avoid the words eigenvector and eigenvalue, which mean something different in this context.<|endoftext|> TITLE: Field of definition for general type surfaces QUESTION [6 upvotes]: In the survey paper https://arxiv.org/abs/1004.2583 of Bauer-Catanese-Pignatelli, they mention a question of Mumford: Can a computer classify all surfaces of general type with $p_g=0$? I've been playing a bit with the Craighero-Gattazzo (CG) surface (a particular surface of this kind) using various computer algebra systems, and my life has been made difficult by the fact that the standard equations for this surface are defined over a cubic extension of $\mathbb{Q}$ rather than over $\mathbb{Q}$ itself (it seems to make computations run for longer and various algorithms are not implemented). This made me worry about Mumford's question: because $\bar{\mathbb{Q}}$ is only countable, a generic complex surface in this moduli space will only be defined over some transcendental extension of $\mathbb{Q}$, which presumably makes Groebner basis calculations even less tractable. So my question is: In the moduli space of general type surfaces, is anything known about the existence or density of surfaces defined over $\mathbb{Q}$ or $\bar{\mathbb{Q}}$? Should I be able to perturb the pluricanonical ring of the CG surface and find a "nearby" surface defined over $\mathbb{Q}$? Should every component of moduli space contain a surface defined over $\bar{\mathbb{Q}}$? If this question is too general, I would be happy to know the answer to the following more concrete question. The CG surface has an explicit birational model as a quintic in $\mathbb{P}^3$ with four simple elliptic singularities. The standard model is defined over $\mathbb{Q}[r]/(r^3+r^2-1)$. Is it known that it is necessary to work over this cubic extension, or could there be a similar quintic defined over $\mathbb{Q}$ with the requisite properties, i.e. whose minimal resolution is biholomorphic to (or at least deformation equivalent to) the CG surface? REPLY [3 votes]: Firstly, since you are interested in the field of definition of the surfaces, you should work with the moduli stack, rather than the coarse moduli space. A $k$-rational point on the moduli stack corresponds to a surface defined over $k$, but the same is not true for the coarse moduli space (this is the issue that ``Field of definition $\neq$ Field of moduli'' in general for coarse moduli spaces; though this issue only occurs for surfaces with non-trivial automorphism group). So let $\mathcal{X}$ be moduli stack of interest. The functor of points is as follows: for a $\mathbb{Q}$-scheme $S$, we have that $\mathcal{X}(S)$ is the set of all smooth proper morphisms $Y \to S$ of relative dimensional two with connected fibres such that the relative canonical bundle $\omega_{Y/S}$ is relatively ample. You seem to be also interested in singular surfaces, so you should adapt the definition as required; but this will correspond to some compactification of this stack. As you are interested in density of rational points, one can just work with the open subset corresponding to smooth surfaces. Moreover, if you want a stack of finite type, you should restrict the Hodge numbers or Hilbert polynomial, as you have done in the question. Now note that the definition makes sense for any $\mathbb{Q}$-scheme $S$, so this means that $\mathcal{X}$ is defined over $\mathbb{Q}$. In fact, you can easily modify the functor of points definition for an arbitrary scheme $S$, which means that one can define a moduli stack over $\mathbb{Z}$, just as Deligne-Mumford defined the moduli stack over curves of given genus over $\mathbb{Z}$. Next, in general, unless there is a good reason otherwise, one expects that moduli spaces are of general type. Therefore conjectures of Lang-Vojta predict that the $\mathbb{Q}$-rational points are not Zariski dense. Hence in general there is no reason to expect to be able to approximate a surface over $\mathbb{C}$ by a surface over $\mathbb{Q}$. As for $\bar{\mathbb{Q}}$-points, as explained in the comments, for a non-empty finite type $\bar{\mathbb{Q}}$ scheme, the $\bar{\mathbb{Q}}$-points are Zariski dense.<|endoftext|> TITLE: Donaldson Invariants in 2 dimensions QUESTION [5 upvotes]: I am trying to understand the correspondence between Donaldson invariants and different correlation functions in certain topological quantum field theories. To be exact, among others I am reading Witten's "Supersymmetric Yang-Mills Theory on a Four-Manifold". However, topic seems to be a difficult one for me, and so I wonder, whether there are some (hopefully easier) examples of the same concept. Namely, are there some analogs of Donaldson Invariants in 2 dimensions which can be alternatively computed from the side of physics? REPLY [2 votes]: It depends what you mean by the Donaldson invariants. The smooth structure on a two manifold is unique, so there aren't any analogues of the Donaldson invariant in that sense. But there are 2d physics analogues of the Donaldson invariants, in that there are two-dimensional topologically twisted quantum field theories some of whose correlation functions reduce to intersection numbers of moduli spaces. The most direct analogy is twisted two-dimensional gauge theory, which computes some of the cohomology of the moduli space of bundles on a Riemann surface. A more famous example is the A-type topologically twisted string theory, which gives Gromov-Witten invariants. An old reference that sketches out these analogies is Witten's Introduction to Cohomological Field Theories. Also worth time if you have enough of it is Cordes, Moore, & Ramgoolam's Lectures on 2D Yang-Mills Theory, Equivariant Cohomology and Topological Field Theories.<|endoftext|> TITLE: Formalizing Entropy Compression (as used to constructify the Lovász Local Lemma) QUESTION [5 upvotes]: In 2009, Moser published a breakthrough paper constructifying the Lovász Local Lemma (LLL). His talk at STOC was described in a blog post by Fortnow that proves a slightly weakened result using Kolmogorov complexity. This led to a great write-up by Terry Tao that coined the term entropy compression, and this method has since been extended to various combinatorial problems. Both Fortnow and Tao describe how Moser's algorithm takes a (Kolmogorov or statistically) random bit string $r$ and produces a final assignment $x$ and a history record $h$ (essentially the witness tree of Moser's paper). Since $r$ can be reconstructed exactly from $x$ and $h$, they claim that, in expectation or with high probability, $|x| + |h| \geq |r|$, because you can't compress a random string beyond its Shannon entropy or compress a Kolmogorov random string at all. From this sort of inequality, you can deduce that Moser's algorithm runs efficiently. I found both of these explanations quite elegant at first but am now struggling to see their correctness as stated (though I in no way doubt the correctness of the proof in Moser's paper). My problem with Tao's entropy argument is that the source coding theorem applies to compression protocols, but Moser's algorithm doesn't seem to fall into this category exactly. It's not as if each string $r$ (or each $r$ of some fixed size) corresponds to an $x,h$ pair; for some $r$, the algorithm will terminate after seeing just a prefix of the bits, and only this prefix will be in correspondence with the final $x,h$ pair. With the Kolmogorov complexity argument, Fortnow starts by fixing a Kolmogorov random $r$ of proper length, but this length depends on the run time of the algorithm, which in turn depends on the selected string. I can't work my way out of this circularity. Any insight here? Specifically, I would be very interested in a formal description of an entropy compression principle guaranteeing that, for an algorithm $\mathcal{A}$ similar to Moser's, $$|\mathcal{A}(r)| \geq |r|,$$ in expectation or with high probability, where $r$ denotes the random bit string used by the algorithm if it is provided with fair coin flips until termination (that is, we are not sampling $r$ uniformly from all strings of a fixed size, but rather from the random process determined by the algorithm when provided with a fair coin). REPLY [4 votes]: In my formulation of the argument, the length of the string $r$ (which I call $R$) is fixed: one does indeed only read a prefix of this string, but the remaining unread bits of the string are saved as part of the output (and referred to in my writeup as $R'$). The precise length of $r$ is not terribly relevant (it cancels out on both sides of the entropy inequality) so long as it is longer than any possible number of bits read from it through the algorithm; in my writeup I choose the fixed length of $r$ to be $Mk$ (in my notation). Fortnow's argument is similar; the random string $r$ (which he calls $x$) has a fixed length (which in his notation is $n+sk$), and any bits not read by the "Fix" algorithm are appended to the output string.<|endoftext|> TITLE: Cohomology theory with only one Adams operation? QUESTION [7 upvotes]: Let $E$ be a multiplicative cohomology theory. Fix a prime p. Call a ring map $\psi^{p}:E\rightarrow E$ an Adams operation if it lifts the Frobenius map $E/p\rightarrow E/p$. It is of course well-known that $K$-theory has Adams operations. (If it weren't for $K$-theory, these operations would have a different name.) In fact, it has an Adams operation for every prime $p$. My question is: are there examples of multiplicative cohomology theories which have an Adams operation $\psi^{p}$ for only one prime $p$? By "only one prime" I don't mean that I require that operations at other primes necessarily don't exist, just that they don't necessarily exist. In other words, their (non)existence is much less obvious/clear/explicit than that of $\psi^{p}$. Motivation: This would turn $E^{*}(X)$ into a $\delta$-ring, which is something some people like to study. REPLY [5 votes]: Adams operations exist in quite wide generality. For any even periodic ring spectra $E$ and $F$, we have associated formal groups $G_E$ and $G_F$ over base schemes $S_E$ and $S_F$. There is a moduli scheme $\text{Hom}(G_E,G_F)$ parametrising pairs $(f,\widetilde{f})$ consisting of a map $f\colon S_E\to S_F$ and a homomorphism $\widetilde{f}\:G_E\to f^*G_F$. This contains an open subscheme $\text{Iso}(G_E,G_F)$ consisting of pairs where $\widetilde{f}$ is an isomorphism. There are natural comparison maps $\text{spec}(\text{Ind}(E_0\Omega^\infty F))\to\text{Hom}(G_E,G_F)$ and $\text{spec}(E_0F)\to\text{Iso}(G_E,G_F)$, both of which are isomorphisms when $E$ and $F$ are Landweber exact. By considering the case $(f,\widetilde{f})=(\text{id},k.\text{id})$ we see that $\psi^k$ exists as a ring automorphism of $E$ when $k$ is invertible in $\pi_0(E)$, and as a ring endomorphisms of $\Omega^\infty E$ when $k$ is not invertible. In other words, in the first case $\psi^k$ is an additive and multiplicative stable operation, and in the second case it is an additive and multiplicative unstable operation. This remains true in many cases when $E$ is not Landweber exact, by various less systematic arguments. It will always be easier to produce $\psi^k$ in cases where $k$ is invertible.<|endoftext|> TITLE: The density of the set of non-pathological primes QUESTION [7 upvotes]: An prime number $p$ is called pathological if there exists a prime number $q\ne p$ such that for every $n\in\mathbb N$ the number $2^n-1$ is divisible by $p$ if and only if $2^n-1$ is divisible by $q$. According to the comment of Gerhard Paseman and @YCor to this problem, pathological prime numbers exist and the smallest one is 23, the next is 53, then 89, 157, etc. Problem 1. How large is the set of non-pathological primes? Is a version of the Dirichlet density theorem true for non-pathological prime numbers: Problem 2. Is it true that for every natural number $a$ and any (square-free) number $b$, which is relatively prime with $a$, the arithmetic progression $a+b\mathbb N=\{a+bn:n\in\mathbb N\}$ contains a non-pathological odd prime number? Added in Edit. By Bang's Theorem, for any $n\ge 2$, the number $2^n-1$ has a non-pathologic prime divisor. Is this information sufficient for answering Problem 2 in affirmative? REPLY [2 votes]: One can phrase this is terms of divisors of (values when evaluated at $x=2$ of) cyclotomic polynomials $P_n(x)$. A pathological prime $p$ is a primitive prime divisor of $P_n(2)$ (so $p$ does not divide any $P_m(2)$ for $m$ smaller than $n$) such that there is $q$ different from $p$ which is also a primitive prime divisor of $P_n(2)$. Given the sparsity of Mersenne primes, one can find many pathological primes as divisors of Mersenne non-primes but with the exponent of this Mersenne number also being prime. So barring exceptional behaviour of the sequence of Mersenne primes, we have a potential source of infinitely many pathological prime numbers: just look at almost any Mersenne number with prime exponent ($P_n(2)$ is a Mersenne number when $n$ is prime). When $n$ is not prime, $P_n(2)$ may yield some more pathological primes, unless it is a nontrivial power of a prime. (For $n$ bigger than 6, there will be at least one primitive prime divisor; Zsigmondy and Bang showed this, and Granville had something to add more generally about the parity of the exponent of such primes.) However, there are some $n$ for which $P_n(2)$ is prime, and these provide nonpathological primes. Further, these primes are 1 mod n, and so knowing which $n$ these are give a start on approaching Problems 1 and 2. Looking at Cunningham tables suggests that these primes are frequent in occurrence and possibly also infinite in number, regardless of the infinitude of Mersenne primes (which are themselves nonpathological). Update 2019.12.12: After rereading notes of G. Jameson on cyclotomic polynomials (see comments below for an indirect reference), we can be more clear on the presence of nonpathologic primes. Although one can look at P_n(a) for values of a other than 2, I stick with Taras's choice of 2. Further, for n less than 7, one can study those numbers individually, so I will speak of n bigger than 6. Prime divisors of P_n(2) are bigger than n, except in one case: if n has a special form mq^k where q is the largest prime divisor of n and q is a primitive divisor of P_m(2), then q can also divide P_n(2). So there are two possibilities for p not to be a pathological prime: if n is not special and P_n(2) has one prime divisor (and I would expect all but finitely many to not be proper prime powers, but this is unproven), or if n is special and P_n(2) is qp^k (less is known about this, but k bigger than 1 would surprise me). I don't know how to apply this to Will Sawin's comment. This form suggests to me that maybe the upper bound on the number of non pathological primes below X could actually be log of his suggested upper bound. If indeed there are infinitely many, then each will be 1 mod n for some n, and when n is sufficiently large may fall into enough of the arithmetic progressions of part 2. End Update 2019.12.12. Gerhard "Tables Nice Place To Frequent" Paseman, 2019.12.09.<|endoftext|> TITLE: Is every prime the largest prime factor in some prime gap? QUESTION [19 upvotes]: Definition: In the gap between any two consecutive odd primes we have one or more composite numbers. One of these composite number will have a prime factor which is greater than that of any other number in the gap. E.g. $43$ is the largest prime in the gap between the consecutive primes $83$ and $89$. I am interested in the largest prime factor in the gap between two consecutive primes. Claim: Every prime is the largest prime factor in some prime gap. I am looking for a proof or disproof. Update, 21 Dec 2019: Conjecture verified for $p \le 10^{10}.$ Note: This question was posted in MSE and got many upvotes but no answer hence posting in MO. REPLY [3 votes]: Extra long comment sharing data as requested by @GerhardPaseman Here is the data for all primes $\le 16290041$. The way to read this data is as follows: Taking the first row as an example, of the $1048575$ primes $\le 16290041$, there are exactly $774792$ primes $p$ such that the integer $2p$ causes the largest prime factor in the prime gap containing it to be $p$, i.e. nearly $74\%$ of the times. And so on for the rest of the rows. This nearly $26\%$ of the times we need a multiple $k > 2$.<|endoftext|> TITLE: Algebraic condition that distinguishes embedded from immersed lie subgroups QUESTION [7 upvotes]: Let $L$ be a finite dimensional Lie algebra over $\mathbb{R}$, and $K$ a subalgebra of $L$. Then, by Lie's correspondence theorems, there exists a unique (up to isomorphisms) simply connected Lie group $G$ having $L$ as a Lie algebra. There also exists a unique connected Lie subgroup $H$ of $G$ having $K$ as its Lie algebra. Question: What algebraic condition on the pair $(L, K)$ determines if $H$ will be an embedded submanifold or not? REPLY [3 votes]: See the closed-subgroup theorem, in particular the conditions for being closed.<|endoftext|> TITLE: Subspaces of $\ell_p$ ($1 TITLE: A variant of the Moore-Mrowka problem QUESTION [10 upvotes]: A space $X$ is said to be sequential if whenever $A \subset X$ is not closed then $A$ contains a sequence converging to a point outside of $A$. A space $X$ is said to have countable tightness if for every non-closed set $A \subset X$ and every point $x \in \overline{A} \setminus A$ there is a countable subset $C$ of $A$ such that $x \in \overline{C}$. Clearly, every sequential space has countable tightness, and not every space of countable tightness is sequential. The Moore-Mrowka problem asks whether every compact Hausdorff space of countable tightness is sequential. This problem is known to independent of ZFC. Balogh proved that every compact space of countable tightness is sequential under PFA and the one-point compactification of Ostaszewski's example under $\Diamond$ of a locally compact countably compact hereditarily separable perfectly normal non-compact space, is an example of a compact space of countable tightness which is not sequential. Balogh, Zoltán, On compact Hausdorff spaces of countable tightness, Proc. Am. Math. Soc. 105, No. 3, 755-764 (1989). ZBL0687.54006. Even though it is not sequential, the one-point compactification of Ostaszewski space enjoys another natural convergence-type property known as pseudoradiality. A space is pseudoradial if whenever $A \subset X$ is not closed then $A$ contains a transfinite sequence (that is, a well-ordered net) converging outside of $A$. $MA_{\omega_1}$ negates the existence of Ostaszewski space for two reasons: 1) because it implies that every countably compact perfectly normal space is compact (Weiss) or 2) because it implies that every locally compact hereditarily separable space is hereditarily Lindelof (Szentmiklossy). However, $MA_{\omega_1}$ is not strong enough to imply a positive solution to the Moore-Mrowka problem (see Balogh's paper). QUESTION: Assume $MA_{\omega_1}$. Is it true then that every compact pseudoradial space of countable tightness is sequential? NOTE: 1) A byproduct of Szentmiklossy's result is that the Moore-Mrowka problem has a positive answer for hereditarily separable spaces under $MA_{\omega_1}$ (every hereditarily Lindelof space has points $G_\delta$ and every compact space with points $G_\delta$ is first-countable, and hence, sequential). 2) A Hausdorff (non-compact) pseudoradial non-sequential space of countable tightness was constructed in ZFC by Simon and Tironi. Simon, Petr; Tironi, Gino, Two examples of pseudo-radial spaces, Commentat. Math. Univ. Carol. 27, 155-161 (1986). ZBL0596.54005. REPLY [5 votes]: This question was answered in the negative by Alan Dow and Istvan Juhász in a recent preprint. On the cardinality of separable pseudoradial spaces.<|endoftext|> TITLE: Smallest ordinal modelling $\aleph_1$? QUESTION [5 upvotes]: Let $X_1$ be the class of all ordinals $\alpha$ such that there exists a transitive model $M$ of ZF(C) such that $M$ thinks that $\alpha$ is $\aleph_1$. Every class of ordinals has a minimum element (because ordinals are well-ordered), so let $\alpha_1$ be the smallest ordinal in $X_1$. So, I have 2 questions: Is the class $X_1$ actually well-defined? If it is well-defined, what can we determine about the actual value of $\alpha_1$? We could also ask the same question for $\aleph_\beta$ (for any ordinal $\beta$) instead of $\aleph_1$, which I'd also be interested in. My thoughts: There are two points I think to verify that $X_1$ is well-defined. The first is that "is a transitive model of ZF(C)" is definable in the language of ZF(C), and the second is that "$M$ thinks $\alpha$ is $\aleph_1" needs to be definable. Being transitive is definable, so really, the first part reduces to whether "being a model of ZF(C)" is definable. I'm a little unsure of whether or not it is, because ZFC is not finitely axiomatizable, but NBG is, which makes me think the idea behind this question might still be salvageable even if it's not definable in ZF(C). As for that "$M$ thinks $\alpha$ is $\aleph_1$" is definable, it seems very true, because countability is definable, and then "every ordinal less than $\alpha$ is countable, but $\alpha$ is not countable", which should definitely be definable. So, that means, that part 2 makes sense to talk about. We know that (at least for small enough $\beta$), $\alpha_\beta$ must be countable, because countable models of ZF(C) exist (by Lowenheim Skolem), which is already interesting. I suspect they are all fairly big countable ordinals, almost certainly larger than $\omega_1^{CK}$. REPLY [3 votes]: About well-definedness of $X_1$, "being a model of ZFC" is definable since ZFC is a recursive theory, so we could construct some $\Sigma_1^0$ predicate $\textrm{isZFCAxiom}(e)$ for $e$ a Godel-coding of a formula in the language of ZFC. Then I believe we can formalize "$M$ is a model of ZFC" by $\forall(e\in\mathbb N)(\textrm{isZFCAxiom}(e)\rightarrow M\vDash e)$ using some formalization of $\vDash$ for Godel-codes, such as the one in this set of notes "Models of Set Theory I" by Koepke. Bounding $\alpha_1$: The following paragraph is an argument I heard by personal communication. To avoid considering arbitrary transitive models, for any transitive $M$ where $(M,\in)\vDash\textrm{ZF}$ we have $L^M\subseteq M$ and $(L^M,\in)\vDash\textrm{ZF}$. These models $L^M$ must be of the form $L_\xi$ (or $(L_\xi)^V$) for some ordinal $\xi$, and we have $\aleph_1^{(L^M)}\leq\aleph_1^M$, so without loss of minimality we can just consider the ordinals $\aleph_1^{L_\xi}$ where $(L_\xi,\in)\vDash\textrm{ZF}$. Also, to ensure we're bounding $\alpha_1$ and not just the $\aleph_1$ of the minimal model, we can show that $\gamma<\delta$ implies $\aleph_1^{L_\gamma}\leq\aleph_1^{L_\delta}$, since any bijection $\omega\rightarrow \aleph_1^{L_\delta}$ that's a member of $L_\gamma$ must also be a member of $L_\delta$. So the $\aleph_1$ of the minimal model of ZF is indeed $\alpha_1$. From here on let $\xi$ be least such that $L_\xi\vDash\textrm{ZF}$. Each $L_\xi$ modeling ZF is admissible, so we can apply Theorem 10.2 of Arai's "A sneak preview of proof theory of ordinals", stating admissible $L_\xi$ satisfying "$\alpha$ is a cardinal $>\rho$" for some $\rho<\xi$ must have $L_\xi\cap\mathcal P(\omega)=L_\alpha\cap\mathcal P(\rho)$. Using some results from §4 of Marek and Srebrny's thesis "Gaps in the constructible universe" we obtain some weak lower bounds on $\alpha$: it's larger than the ordinal of ramified analysis $\beta_0$, and larger than the ordinal $\xi$ starting a gap of length $\beta^\beta$ of corollary 4.12. For the bonus question on $\aleph_n^M$, similar results apply when $\rho>\omega$, using the analogous notions of $\rho$-gaps (i.e. gaps in $L_\bullet\cap\mathcal P(\rho)$) introduced in §7. Strengthening the bounds: Since we know the gap starts at $\alpha_1$ and ends at $\xi$, we can strengthen this bound by showing that the gap is long, e.g. contains many admissibles: let $\sigma_0$ be the $\Pi_3$ formula expressing "the universe is an admissible set" from Richter and Aczel's "Inductive Definitions and Reflecting Properties of Admissible Ordinals". Working in $L_\xi$, ZF proves $\forall\rho((\rho\textrm{ is a cardinal }>\omega)\rightarrow(\sigma_0)^{L_\rho})$, we assumed ZF is sound by assuming ZF has a transitive model, so $L_\rho$ satisfies this sentence. Since each relativization $(\sigma_1)^{L_\rho}$ is $\Delta_0$, each one is absolute and therefore true in $V$, so each $L_\rho$-cardinal, while not really a cardinal, is really admissible. So we get many admissibles (e.g. $\aleph_2^{L_\xi}$,  $\aleph_\omega^{L_\xi}$,  $\aleph_{\omega_1^{L_\xi}+\varepsilon_0^{234}}^{L_\xi}$) between the start and end points of the gap. We can repeat this section but based on stronger properties of $L_\rho$-cardinals to improve this further. For instance Barwise's Admissible Sets and Structures has an exercise to show each $L_\rho$-cardinal $>\omega$ must be $\rho$-stable for admissible $\rho$, from which we show each aforementioned $L_\rho$-cardinal in our gap is inaccessibly-stable. (However we need to be careful about showing these properties are absolute w.r.t. $V$)<|endoftext|> TITLE: Indecomposable objects in bounded derived category of $\mathbb C[x]/x^2$-mod QUESTION [6 upvotes]: We know for any principal ideal domain, objects in the bounded derived category are all formal hence we can classify those objects with finitely generated cohomology using structure theorem for finitely generated modules. Now consider the bounded derived category of $\mathbb C[x]/x^2$-modules, how to classify indecomposable objects with finitely generated cohomology in this category? Examples include $\mathbb C[x]/x^2 \overset{x}{\rightarrow} \mathbb C[x]/x^2 \overset{x}{\rightarrow}... \overset{x}{\rightarrow} \mathbb C[x]/x^2 $ and $\mathbb C$. Note there exists non-formal object. REPLY [4 votes]: Up to shifts, every indecomposable object is of one of the forms described in the question. I don't know an explicit reference, but here's a sketch of a proof. By induction on the length, it's not hard to prove that every bounded complex of finite rank free modules, such that the image of each differential is contained in the radical of its codomain, is a direct sum of complexes of the form $\mathbb C[x]/x^2 \overset{x}{\rightarrow} \mathbb C[x]/x^2 \overset{x}{\rightarrow}... \overset{x}{\rightarrow} \mathbb C[x]/x^2 $. Now consider a minimal projective resolution of an object of the bounded derived category, truncated to the left of its homology. By the way, this is an example of a "derived discrete" algebra, and there's a fair amount of literature about these.<|endoftext|> TITLE: Completed tensor product is exact QUESTION [9 upvotes]: In the beginning of the 7th chapter of the book "Spectral theory and analytic geometry over non-Archimedean fields" by Vladimir Berkovich one can find the phrase "...tensor product functor is exact on the category of Banach spaces...". He gave no clue how to prove it, but it is known that the same fact is not true for Archimedean Banach spaces. Is the statement correct, and how can it be proved? UPD I know that tensor product of complex Banach spaces is not left exact. I'm interested in the proof (or counterexample) for non-Archimedean Banach spaces. REPLY [3 votes]: To make things precise, let me add the end of the quoted sentence: "with admissible linear operators as morphisms". Moreover, I believe that Berkovich refers here to tensor products over a fixed base field. In this case, the exactness result you are looking for may be found in Gruson's paper "Théorie de Fredholm $p$-adique" (Bulletin de la Société Mathématique de France, Tome 94 (1966), p. 67-95, http://www.numdam.org/item/BSMF_1966__94__67_0/). It is part of Theorem 1 on page 79. To be sure that it means what you want, you will have to unravel a few definitions and you should in particular have a look at the first section of the paper for the definition of the category (enc) and strict morphisms (which are Berkovich's admissible morphisms).<|endoftext|> TITLE: Examples of proofs using induction or recursion on a big recursive ordinal QUESTION [10 upvotes]: There are many proofs use induction or recursion on $\omega$, or on an arbitary (may be uncountable) ordinal. Are there some good examples of proofs which use a big but computable ordinal? The original proof of Ramsey theorem and Hales-Jewett theorem use induction on $\omega^2$, but the using is not essential, because Erdos and Shelah have given better bounds by using induction just on $\omega$. And further more $\omega^2$ shouldn't be considered big. A typical use of big ordinal induction is proving the consistence of axiom systems, for example, using $\varepsilon_0$-induction to prove the consistence of PA. This is one kind of examples. The existence of Goodstein function uses the induction on $\varepsilon_0$, and I think it's just a directly explaining of how do recursion on ordinal works. Are there more examples? REPLY [4 votes]: This is an expanded version of my comment. There are examples from wqo (well quasi-order) theory, if you accept that induction on a wqo is "induction on an ordinal" (specifically, the ordinal of the tree of "finite bad sequences" of the wqo). Kruskal's tree theorem can be proved by induction on a certain wqo whose ordinal is bigger than $\Gamma_0$, which is much bigger than the ordinals you mentioned. This is explained in detail in "What's so special about Kruskal's theorem and the ordinal $\Gamma_0$? A survey of some results in proof theory," by Jean H. Gallier, Ann. Pure Appl. Logic 53 (1991), 199–260. Related to this is Friedman's extension of Kruskal's theorem; let's call it EKT. In "The metamathematics of the graph minor theorem" by Friedman, Robertson, and Seymour, it is explained that EKT is equivalent (over RCA$_0$) to a weak version of the Graph Minor Theorem that they call the "bounded Graph Minor Theorem," i.e., the Graph Minor Theorem restricted to graphs of bounded tree-width. The relevant ordinal here is $\alpha_n$, the ordinal of the wqo of graphs of tree-width at most $n$, partially ordered by minor inclusion. I believe that the exact ordinal corresponding to the full Graph Minor Theorem is still unknown, but is conjectured to exceed $\lim_n \alpha_n$, which is the proof-theoretic ordinal of $\Pi_1^1$-CA$_0$.<|endoftext|> TITLE: Can the category of partial orders be fully embedded in the category of linear orders? QUESTION [5 upvotes]: Q(1): Can the category of partial orders be fully embedded in the category of linear orders? Vopěnka's principle, or VP, is a very intriguing axiom with many equivalent forms and consequences spanning universal logic (in the Barwise sense), large cardinals, model theory, and category theory. VP for $C$ for a (large locally presentable) category $C$ means that there is no large discrete full subcategory of $C$. It is widely known that VP for $C$ for most "sufficiently expressive" categories $C$ is simply equivalent to all of VP (see Joel David Hamkin's answer to this question). The notion of "sufficiently expressive" can be formalized as follows: if $A$ fully embeds into $B$ and VP for $B$ holds, then any large discrete full subcategory of $A$ necessarily produces large discrete full subcategory of $B$. So, perhaps a suitable notion of "sufficiently expressive" is simply a full embedding. For example, the categories of structures of a given language can be fully embedded in the category of graphs, showing that, for example, VP for graphs is the same as VP for structures. This leads me to the question above; an affirmative answer (which I believe is impossible) would also definitively show that VP for linear orders is indeed equivalent to VP itself. However, I have strong doubts about this. Perhaps it needs to be weakened? Q(2): Is the category of linear orders a reflective subcategory of the category of partial orders? It isn't immediately obvious that this even relates to VP anymore. However, as shown in Adamek-Rosicky's Locally Presentable and Accessible Categories the following is equivalent to WVP (Weak Vopenka's principle, weakened form): For $C$ a locally presentable category, every full subcategory $D↪C$ which is closed under limits is a reflective subcategory. Now it becomes clear! The category of partial orders is certainly locally presentable, and the category of linear orders is closed under limits. Under the Weak Vopenka's principle therefore, the answer to Q(2) is affirmative! But is using such a powerful axiom unnecessary? The original hypothesis also implies Q(2); if strong large cardinal axioms are indeed necessary for Q(2), that doesn't bode very well for our friend Q(1). This leads me to Q(3), where we investigate the consistency strengths of Q(1) and Q(2): Q(3): Is it consistent that Q(2) fails? If so, what is the consistency strength of Q(2)? What is the consistency strength of Q(1)? And finally, of course, the last question that motivated this whole thing in the first place: Q(4): Is VP for linear orders equivalent to VP? Adámek, Jiří; Rosický, Jiří, Locally presentable and accessible categories, London Mathematical Society Lecture Note Series. 189. Cambridge: Cambridge University Press. xiv, 316 p. (1994). ZBL0795.18007. REPLY [9 votes]: Expanding on Jeremy's comments, Q1) No, the category of partial orders does not admit a faithful functor $F$ into the category of linear orders. The poset $\{l,r\}$ of two incomparable elements has a nonidentity automorphism $n$. If the global elements $l$, $r$ satisfy $F(l) < F(r)$ or $F(r) < F(l)$, then $F(n)$ is not monotone, so $F(l) = F(r)$ and $F$ is not faithful. Q2) Neither the category of linear orders with $\leq$- or $<$ preserving functions is closed under products, and therefore cannot be a reflective subcategory. For $\mathbf{Lin}_\leq$, by considering global elements, we must have that $2 \times 2 \cong 4$, and then the function $(\pi_2, \pi_1)$ isn't monotone. For $\mathbf{Lin}_<$, there is no terminal object.<|endoftext|> TITLE: Contraction of a family of loops simultaneously QUESTION [5 upvotes]: Let $LX$ denote the free loop space on $X$. We have an evaluation map $ev\colon LX\to X$ and we have an inclusion $X\hookrightarrow LX$ (where $x\in X$ is mapped to the constant loop at $x$). Suppose that $A\subset X$ is a cofibration, where $X$ is simply connected (I wouldn't mind to assume that $A$ is also simply connected if that would simplify anything). Let $\varphi\colon A\to LX$ be a map that makes the following diagram commutes: In other words: $\varphi$ sends $a\in A$ to a loop in $X$ that is based in $a$. My question: Is it possible to contract all loops simultaneously but in a "basepoint preserving" way (see below)? I.e., is is possible to find a homotopy $\psi_t\colon A\to LX$ where $\psi_0=\varphi$ and $\psi_1$ is the inclusion $A\hookrightarrow LX$, where commutes for every $t\in [0,1]$? REPLY [10 votes]: Surely not. Let $S^1 \to LS^2$ be adjoint to the map $c: S^1 \times S^1 \to S^2$ which collapses $S^1\vee S^1$ to a point. The latter has degree one. Let $p\in S^1$ be any point but the basepoint. Take $A\to X$ to be the composite map $$ S^1 \times p \subset S^1\times S^1 \overset{c}\to S^2 $$ (this is an inclusion). If a homotopy of the kind you are requesting existed, then we could conclude, by taking adjoints, that $c: S^1 \times S^1 \to S^2$ is homotopic to a map that factors through $S^1 \times p$. This gives a contradiction, since such a map has degree zero.<|endoftext|> TITLE: Computing K-theory for cellular vector bundles QUESTION [9 upvotes]: One of the most computationally convenient properties of singular cohomology $X \mapsto H^\bullet(X;\mathbb{Z})$ is the fact that one can extract it from a good cover $\{U_i\}$ of $X$ via Cech cohomology, thus bypassing any scary spectral sequences. Is there a similar algorithmic shortcut available for other cohomology theories? I'm mostly asking about the possibility of computing the complex $K$-theory of a finite simplicial complex $X$ in the presence of a good cover (without using the Atiyah-Hirzebruch SS). There has been at least one old question about this with no answers in sight, which doesn't sound too promising. Perhaps it will help if I propose a candidate definition for vector bundles on $X$: Def: A vector bundle $V$ (of rank $n$) over a simplicial complex $X$ is a functor from the poset of simplices in $X$ to the group $\text{GL}_n(\mathbb{C})$. In other words, each face relation $\sigma < \tau$ is assigned an invertible complex $n \times n$ matrix $V_{\sigma<\tau}$ so that (1) the matrix $V_{\sigma < \sigma}$ is the identity for every simplex $\sigma$, and (2) the relation $V_{\tau < \gamma} V_{\sigma < \tau} = V_{\sigma < \gamma}$ holds across any triple $\sigma < \tau < \gamma$ of simplices. Let's say I don't care about honest vector bundles on the geometric realization of $X$, and focus exclusively on these discrete guys which are constructible with respect to the fixed simplicial decomposition. Is there a non-spectral-sequencey way to compute the obvious notion of $K_\bullet(X)$ for the collection of all such vector bundles? Maybe we can at least compute Chern classes? I'm happy to assume that $X$ is a triangulated smooth manifold if that helps in any way. REPLY [2 votes]: The fact that singular cohomology agrees with Cech cohomology of a good cover $\mathcal{U}$ is a sort of generalised Mayer-Vietoris principle, as explained in Section II.8 and III.15 of Bott and Tu's "Differential Forms in Algebraic Topology". You get a double complex $C^*(\mathcal{U},S^*)$ of Cech-singular cochains, and the resulting Mayer-Vietoris sequence degenerates in two ways to give the result. To generalise this to other cohomology theories, you'd need to use the spectral sequence of the pre-simplicial space $$ X \leftarrow \bigsqcup_{i_0} U_{i_0} \stackrel{\leftarrow}{\leftarrow}\bigsqcup_{i_0 TITLE: Is there a solution of the Yamabe problem using Ricci flow? QUESTION [13 upvotes]: Someone told me that it is possible to solve the Yamabe problem using Ricci flow. The proof I know of is the one originally proposed by Yamabe and then completed by Trudinger, Aubin and Schoen (in particular, the final step by Schoen made use of the positive mass theorem which had earlier been proved by Schoen and Yau). If the Ricci flow proof exists, could someone point me to a reference? Edit: When I say the 'Ricci flow' proof, I actually mean the Yamabe flow, since the two coincide on surfaces. REPLY [11 votes]: The references provided by Carlo Beenakker solve the problem only in special cases. The article "Global existence and convergence of Yamabe flow" by R. Ye assume -- for example -- conformal flatness. The problem was solved completely within the last years. Important progress in dimension 3,4 and 5 was obtained by Schwetlick and Struwe, Convergence of the Yamabe flow for "large'' energies. J. Reine Angew. Math. 562 (2003), 59–100. and then later, the general statement was proven by Simon Brendle in Brendle, Convergence of the Yamabe flow in dimension 6 and higher. Invent. Math. 170 (2007), no. 3, 541–576. However, Brendle's proof only solves all cases, if one assumes the general version of the positive mass theorem. This general version was the subject of several preprints recently. Besides several articles by J. Lohkamp (see arXiv), there is also a preprint by Schoen and Yau https://arxiv.org/abs/1704.05490. To my knowledge no one of these preprints has been published so far.<|endoftext|> TITLE: Closed Form for $\sum\limits_{n=-2a}^\infty(n+a){2a\choose-n}^4,~a\not\in\mathbb Z$ QUESTION [11 upvotes]: Do either $~S_4^+(a)~=~\displaystyle\sum_{n=0}^\infty(n+a){2a\choose n}^4~$ or $~S_4^-(a)~=~\displaystyle\sum_{n=-2a}^\infty(n+a){2a\choose-n}^4~$ possess a meaningful closed form expression1 in terms of the general parameter $a\not\in\mathbb Z$ ? Ramanujan provided the following result : $~S_4^-\Big(-\tfrac18\Big)~=~\dfrac1{\bigg[\Big(-\tfrac14\Big){\large!}\bigg]^2~\sqrt{8\pi}}~,~$ which would tentatively point to a possible closed form expression in terms of $~(2a)!~$ and/or $~(4a)!~$ For somewhat similar expressions with lesser values of the exponent, we have Dixon's identity : $$\sum_{n=0}^\infty(-1)^n{2a\choose n}^3 ~=~ \sum_{n=-2a}^\infty(-1)^n{2a\choose-n}^3 ~=~ \cos(a\pi)~{3a\choose a,a},$$ and Vandermonde's identity : $$\sum_{n=0}^\infty(-1)^n{2a\choose n}^2 ~=~ \sum_{n=-2a}^\infty(-1)^n{2a\choose-n}^2 ~=~ \cos(a\pi)~{2a\choose a},$$ $$\sum_{n=0}^\infty{a\choose n}^2 ~=~ \sum_{n=-a}^\infty{a\choose-n}^2 ~=~ {2a\choose a},$$ as well as the binomial theorem : $$\sum_{n=0}^\infty{a\choose n}^1x^n ~=~ (1+x)^a,\qquad\sum_{n=-a}^\infty{a\choose-n}^1x^n ~=~ \Big(1+\tfrac1x\Big)^a.$$ This question has already been asked on Mathematics Stack Exchange, where, despite having been posted a considerable amount of time ago, and also put up for a bounty, it nonetheless still failed to receive a satisfying answer. 1 As opposed to, say, a mere rewriting in terms of $($generalized$)$ hypergeometric functions, which, as already pointed out in the comment section to the original post, constitutes nothing more than a formal exercise in mathematical taxonomy. REPLY [12 votes]: Some experimentation suggests that the second series is given by $$ S_4^-(a) = \sum_{n=-2a}^\infty (n+a) \binom{2a}{-n}^4 = \frac{1}{4\cos(2\pi a) \,\Gamma(2a+1)^2\,\Gamma(-4a)}, $$ which agrees with Ramanujan's at $a = -1/8$, but I have no proof...<|endoftext|> TITLE: Can the automorphism group vary too much in families of complex projective varieties? QUESTION [5 upvotes]: In a family of smooth projective curves over a reduced complex scheme of finite type the list of isomorphism classes of automorphism groups of the fibers is finite. This follows from the Hurwitz's bound and the constancy of the genus on each connected component of the base. Can one prove a similar result for families of higher-dimensional varieties? REPLY [6 votes]: In the case of a family of minimal smooth projective varieties of general type, we can repeat exactly the same argument as in the case of curves of genus $\geq 2$. In fact, by a result of Hacon, McKernan and Xu [HMcKX13] we know that, if $X$ is any such a variety, we have $$|\mathrm{Aut}(X)| \leq c \cdot \mathrm{vol}(X, \, K_X), $$ where $c$ is a constant that only depends on $d:=\mathrm{dim}(X)$. Now, it sufficies to observe that the canonical volume $$\mathrm{vol}(X, \, K_X):=\limsup_{m \to + \infty} \frac{d !\; h^0(X, \, mK_X)}{m^d}$$ is deformation invariant, by the celebrated invariance of plurigenera in smooth families proved by Siu [S98]. References. [HMcKX13] Hacon, Christopher D.; McKernan, James; Xu, Chenyang, On the birational automorphisms of varieties of general type, Ann. Math. (2) 177, No. 3, 1077-1111 (2013). ZBL1281.14036. [S98] Siu, Yum-Tong, Invariance of plurigenera, Invent. Math. 134, No. 3, 661-673 (1998). ZBL0955.32017.<|endoftext|> TITLE: prime dividing no term in a sequence QUESTION [7 upvotes]: Let $q_1,\cdots,q_m$ be distincts elements of $\mathbb N\setminus\{0,1\}$. Does there exist non zero integers $A_1,\cdots,A_m$ such that infinitely many primes divide no elements of the sequence $\left(A_1q^n_1+\cdots+A_mq_m^n\right)_{n\in\mathbb N}$? Any aid or answer would be welcome. Thanks in advance. REPLY [2 votes]: You can prove this relatively easily for $m=2$. Chose $A_1, A_2$ so that there exist infinitely many primes with $q_1q_2$ a quadratic residue and $-A_1A_2$ a non-residue mod $p$. This can be done using quadratic reciprocity and gives the primes lying in an arithmetic progression. (For example if $p\equiv -1 \bmod 24$ then 2 is a residue and -3 is a non residue mod $p$.) Then $p\mid(A_1q_1^n+A_2q_2^n)$ means that $A_1q_1^n+A_2q_2^n\equiv0 \bmod p$ or $A_1^2q_1^{2n}+A_1A_2{(q_1q_2)}^n\equiv0 \bmod p$ which cannot be true if $-A_1A_2$ is a non residue and $q_1q_2$ is a residue. Hence all such $p$ cannot divide $(A_1q_1^n+A_2q_2^n)$.<|endoftext|> TITLE: Classification of $\operatorname{Rep} D(G)$ QUESTION [8 upvotes]: Let $G$ be a finite group and $D(G)$ its quantum double. Its finite dimensional complex representations are classified in this Dijkgraaf et al. Quasi-Quantum Groups Related To Orbifold Models. However, in the paper, the authors claimed the examples obtained by natural inductions are complete, but without a proof. My impression is that the representation theory for a Hopf algebra is not completely known yet in general. How would one classify the representations of $D(G)$? My understanding is still at the level of representation of a finite group $G$. In that case, the classification can be done due to $\mathbb{C}[G]$ is semi-simple $\lvert\operatorname{Irrep}(G)\rvert =\lvert\operatorname{Conj}(G)\rvert$. Are there similar statements for $D(G)$ (better with proofs)? Pointers towards anything relevant will be appreciated. Thank you! EDIT: While there's only one chosen answer, the others might be valuable for you. Here is a table of answers below so far. Konstantinos Kanakoglou pointed out several papers that directly answered my question. I wrote up a note spelling out Konstantinos Kanakoglou's wonderful answer. I am willing to discuss the detail of the proof. zibadawa timmy's higher categorical view point. REPLY [3 votes]: There's a higher way to come at this. I'll be a little light on the rigorous details here, but everything I mention can be found in the book "Tensor Categories" by Etingof, Gelaki, Nikshych, and Ostrik. The book's a very good starting point for moving from the Hopf algebras perspective to the tensor category perspective, which is where a lot of current research is being done. If one considers a semisimple Hopf algebra $H$, and take $\mathcal{C}=\operatorname{Rep}(H)$ to be the category of finite dimensional left (or right) modules of $H$, then there is a braided tensor equivalence $\operatorname{Rep}(D(H))\cong \mathcal{Z}(\operatorname{Rep}(H))$, where $\mathcal{Z}(\mathcal{C})$ denotes the categorical center of the category $\mathcal{C}$. This center construction works not just for the particular choice here, but any tensor (aka monoidal) category with sufficiently similar properties. The objects of the center are the pairs $(V,\gamma_V)$ where $V$ is an object of $\mathcal{C}$ and $\gamma_V$ is a natural family of isomorphisms called a "half-braiding" (because they piece together into a braiding on the entire category). In the case of $H=\mathbb{C}G$ with $G$ a finite group, we can go one step better. There is a Morita-equivalence between $\operatorname{Rep}(G)$ and $\text{Vec}_G$, where the latter is the space of $G$-graded finite-dimensional vector spaces (over $\mathbb{C}$). This is equivalent to saying these categories have the same centers, up to braided tensor equivalence, so we could just as well compute $\mathcal{Z}(\text{Vec}_G)$ instead. Once you actually write down what the half-braiding conditions are, this center becomes very easy to determine: it's $\text{Vec}_G^G$ (sometimes denoted ${}^G_G\mathcal{M}$, or some variation thereof depending on the use of left/right (co)actions), the category of finite dimensional $G$-graded, $G$-equivariant vector spaces. At this point it's easy to decide the isomorphism classes of the irreducibles, and you find that they are parameterized by pairs $(g,\chi)$ where $g$ is an element in a complete set of representatives of the conjugacy classes of $G$, and $\chi$ is an element in a complete set of representatives for the irreducible representations (or characters) of $C_G(g)$. So the isomorphism type of the module depends only on the conjugacy class of $g$ and the isomorphism class of $\chi$. When you understand the objects of $\text{Vec}_G^G$ it becomes readily apparent that the irreducible objects are just induced representations from $C_G(g)$ to $G$, but where the implicit grading of this induction via cosets of $C_G(g)$ is relevant to deciding the full action of $D(G)$. And if you want to go even further than that, you can change the associativity morphism of $\text{Vec}_G$ via a normalized 3-cocycle $\omega$ to obtain the category $\text{Vec}_G^\omega$, and then we have $\mathcal{Z}(\text{Vec}_G^\omega)\cong\operatorname{Rep}(D^\omega(G))$, where $D^\omega(G)$ is the twisted Drinfeld double, and is in general a quasi-Hopf algebra and not a Hopf algebra. These objects are also quickly described in the paper you mention. The description of the irreducibles is similar, except now we're using irreducible projective representations for particular 2-cocycles of $C_G(g)$ obtained from $\omega$. This category, as a braided tensor category, will only depend on the cohomology class of $\omega$, while $D^\omega(G)$ can have wildly different structures even for representatives of the same cohomology class. Since those structures are also quite nightmarish to deal with directly for any non-trivial 3-cocycle, most people end up gravitating towards dealing with them through their representation categories, instead.<|endoftext|> TITLE: Boardman's thesis or mimeographed notes QUESTION [22 upvotes]: I would like to know if there is some online source where Boardman's 1964 thesis is available or his Warwick mimeographed notes. This is because by what I've heard Boardman's construction has a more modern or categorical feel to it than Adams's account in his blue book and I would love to look at Boardman's construction with modern eyes, taking into account all we've learned about the $\infty$-category of spectra in the last years. REPLY [12 votes]: Boardman's thesis was (re)published a year later as three separate booklets, and a PDF scan of all three booklets is available on my scans page: J. M. Boardman. Stable homotopy theory. University of Warwick, Coventry. Chapters I–IV: 42 pages, November 1965. Chapter V: Duality and Thom spectra, 66 pages, January 1966. Chapter VI: Unoriented bordism and cobordism, 71 pages, July 1966.<|endoftext|> TITLE: Multivariate Lagrange inversion with zero powers QUESTION [6 upvotes]: (Also asked on MSE) The multivariate Lagrange inversion formula, which I found in a couple of papers (such as this and this), is as follows. If $f_i=t_ig_i(f)$, $1\le i\le k$, then $$ [t^n]h(f(t))=\frac{1}{n_1n_2\cdots n_k}[x^{n-1}]\sum_T \frac{\partial (h,g_1^{n_1},...,g_k^{n_k})}{\partial T},$$ where $t^n=t_1^{n_1}\cdots t_k^{n_k}$ and the derivative is taken with respect to some trees (as discussed in those papers). Not one of the papers in question has addressed the question of how this formula is to be used when some of the powers are zero, $n_j=0$, something that does not happen in the one variabe case (due to the assumption that $g(0)=0$) but can happen in the multivariable one. REPLY [2 votes]: The formula cannot be applied for multinomial terms where an $n_i$ is $0$. For $n=0$, Lagrange inversion formula and Lagrange-Bürmann formula have particular formulas. You have to use another multivariate Lagrange inversion formula. See Rosenkranz, M: Lagrange Inversion. Diploma thesis, RISC Linz 1997: page 40: "It turns out that the inversion formulas in their second form can be generalized to the multivariate case in a very natural way. (For the first form of the inversion formulas, the multivariate generalizations are very complicated.)" See theorem 42 at page 38, corollary 43 at page 39, and theorem 47 at page 41. Corollary 43 (univariate case) and theorem 47 (multivariate case) present formulas for the general series coefficients without the factor $\frac{1}{\boldsymbol{n}}$.<|endoftext|> TITLE: Multisections of the universal curve QUESTION [7 upvotes]: Fix some $g \geq 2$. Let $\mathcal{M}_g$ be the moduli space of smooth genus $g$ curves over $\mathbb{C}$. For some $d \geq 1$, let $X_{g,d} \rightarrow \mathcal{M}_g$ be the family whose fiber over $S \in \mathcal{M}_g$ is the $d^{\text{th}}$ symmetric power of $S$. I'm aware that due to the presence of curves with automorphisms this doesn't (strictly speaking) exist, but let's ignore that point (e.g. by adding a full level structure to $\mathcal{M}_g$ to rigidify things). Question 1: Does there exist some $d \geq 1$ such that $X_{g,d} \rightarrow \mathcal{M}_g$ has a section? I expect something like Weierstrass points will work here, but I don't know how they vary in families. Question 2: Assuming that Question 1 has a positive answer, what I'm really interested in is the following. Does there exit some $d,e \geq 1$ such that there exist sections $\sigma\colon \mathcal{M}_g \rightarrow X_{g,d}$ and $\sigma'\colon \mathcal{M}_g \rightarrow X_{g,e}$ such that for all $S \in \mathcal{M}_g$, the $d$ points making up $\sigma(S)$ are disjoint from the $e$ points making up $\sigma'(S)$? Here I don't have a candidate for the two disjoint multisections. REPLY [3 votes]: I would like to give an answer to the simple Question 1. The answer relies on the following lemma. Lemma. For any fixed $(r,g,n)$ there exists a hypersurface $V\subset \mathbb CP^n$ that doesn't contain any smooth genus $g$ curve of degree $\le r$. I think, this lemma can be proven by simple dimension count. Now, to solve the question, find $n$, such that (the universal curve) $M_{g,1}$ can be embedded in $\mathbb CP^n$. Let $r$ be the degree of all the genus $g$ curves in the embedding. Take, a hypersurface $V$ from the Lemma and let $d$ be its degree. Now, take and the intersection of $V$ with each curve. This will give you the desired section. (As for Question 2, I have no clue).<|endoftext|> TITLE: Conductor of Principal series representation QUESTION [5 upvotes]: Let $\mathbb{F}$ be a local field and let $\pi$ be a principal series representation of $GL_2(\mathbb{F})$ that is $\pi=Ind_B^{GL_2}(\chi_1\otimes\chi_2)$ for two characters $\chi_1$ and $\chi_2$ of the maximal torus. Then I came to know from the book "Automorphic Forms on Adele group" by Gelbert that due to Casselman , it follows that conductor of $\pi=$ (conductor of $\chi_1$)$\times$(xonductor of $\chi_2$). I want to know if there is similar result for $GL_n(\mathbb{F})$ for $n>2$. That is if $\pi=Ind_B^{GL_n}(\chi_1\otimes\chi_2\otimes\dots\otimes\chi_n)$ is a principal series representation of $GL_n(\mathbb{F})$, then is it true that conductor of $\pi= \prod$(conductor of $\chi_i$)? Please refer some good paper for this. Thank you in advance. REPLY [6 votes]: Here is the story as I know it. The more general setup is for induced representations of Langlands type. These are certain generic representations of $\mathrm{GL}_n(F)$, where $F$ is a nonarchimedean local field. The construction of such representations is the following. Suppose that $n = n_1 + \cdots + n_r$ with $n_1,\ldots,n_r \geq 1$. Let $\sigma_j$ be a discrete series representation of $\mathrm{GL}_{n_j}(F)$ (necessarily unitary and tempered), from which we can construct a (no longer necessarily unitary or tempered) generic irreducible admissible smooth representation $\pi_j = \sigma_j \otimes \left|\det\right|^{t_j}$ of $\mathrm{GL}_{n_j}(F)$. We form the representation $\pi_1 \otimes \cdots \otimes \pi_r$ of $\mathrm{GL}_{n_1}(F) \times \cdots \times \mathrm{GL}_{n_r}(F)$, which we view as a Levi subgroup in $\mathrm{GL}_n(F)$. Then by normalised parabolic induction, we can induce this to a representation $\pi = \pi_1 \boxplus \cdots \boxplus \pi_r$ of $\mathrm{GL}_n(F)$. This is known as an induced representation of Whittaker type (in particular, it has a Whittaker model); we say that $\pi$ is the isobaric sum of $\pi_1,\ldots,\pi_r$. If $\Re(t_1) \geq \cdots \geq \Re(t_r)$, then this is said to be an induced representation of Langlands type (which is isomorphic to its Whittaker model). These may not be irreducible, but the irreducible ones are generic irreducible admissible smooth representations, and conversely every generic irreducible admissible smooth representation is isomorphic to an induced representation of Langlands type. The case you are interested in is $n_1 = \cdots = n_r = 1$, so that each $\pi_j$ is simply a character. The theorem you are interested in is the following. The conductor exponent $c(\pi)$ of an induced representation of Langlands type $\pi = \pi_1 \boxplus \cdots \boxplus \pi_r$ is equal to $c(\pi_1) + \cdots + c(\pi_r)$. Here the conductor exponent is as in my answer here: On the consistency of the definition of the conductor for automorphic forms In "Conducteur des représentations du groupe linéaire" by Jacquet, Piatetski-Shapiro, and Shalika, this conductor exponent is defined (see in particular Section 5). In "Rankin-Selberg Convolutions" by the same authors, further properties are shown. In particular, a combination of Theorem 3.1, Proposition 8.4, and Proposition 9.4 shows that the epsilon factor $\varepsilon(s,\pi,\psi)$ satisfies $$\varepsilon(s,\pi,\psi) = \prod_{j = 1}^{r} \varepsilon(s,\pi_j,\psi),$$ where $\psi$ is an additive character of $F$. (More precisely, they show the same multiplicativity properties hold for $L(s,\pi)$ and $\gamma(s,\pi,\psi)$, from which the result for $\varepsilon(s,\pi,\psi)$ follows.) Finally, Théorème 5 of the first paper states that $$\varepsilon(s,\pi,\psi) = \varepsilon\left(\frac{1}{2},\pi,\psi\right) q^{-c(\pi)\left(s - \frac{1}{2}\right)}$$ (at least for unramified $\psi$), at which point the result follows.<|endoftext|> TITLE: Very large axiom of choice QUESTION [6 upvotes]: let me say that I am not a set theorist, but I have to settle up some things in category theory and I need your help. What I'd like to do is, in some way, use axiom of choice for proper classes. I say that axiom of choice holds for a set $X$ if there exist a function $f:X \to \bigcup X$ such that $f(x) \in x$. My attempt is the following: Assume the Tarski Grothendieck axiom, i.e. every set is contained in a Grothendieck Universe. Fix a Universe $U$ and an enlarged universe $U^+$ that contains $U$. Now note that every subset of $U$, i.e. a $U$ proper class, is contained in $U^+$ by the subset axiom. Now I read from wikipedia that "axiom of choice holds" in Tarski Grothendieck framework. My question is: Does the axiom of choice holds in TG for every element of some universe? This would mean that for every subset of $U$, axiom of choice holds, being an element of $U^+$. To be honest, I would be ok with the following: Does the axiom of choice holds for a Universe? Can you provide a reference? REPLY [8 votes]: If you have ZFC in the ambient theory, including the axiom of choice, then indeed the axiom of choice holds in every Grothendieck-Zermelo universe (also sometimes known as Grothendieck universes). A Grothendieck-Zermelo universe is a rank-initial segment $V_\kappa$ of the cumulative hierarchy, where $\kappa$ is an inaccessible cardinal. And every set in $V_\kappa$ has a well-ordering in $V$, by the axiom of choice in the ambient theory, and this order must also be in $V_\kappa$, since $V_\kappa$ is closed under subsets of its elements. Basic lesson: if the axiom of choice holds in the background theory, then it also holds in every Grothendieck-Zermelo universe. Meanwhile, it is not a consequence of the Universe axiom over ZF that AC must hold, since one can easily use forcing to make AC false, while preserving the truth of the universe axiom. So without AC in the ambient theory, you cannot conclude that AC holds in every Grothendieck-Zermelo universe.<|endoftext|> TITLE: Are unitarily equivalent permutation matrices permutation similar? QUESTION [17 upvotes]: Two matrices $A, B \in \mathbb R^{n \times n}$ are called unitarily equivalent if there exists an unitary matrix $U \in \mathbb C^{n \times n}$ such that $A = U B U^{\ast}$. If in addition $U$ is a permutation matrix, then we call $A$ and $B$ permutation similar. Suppose that $A$ and $B$ are unitarily equivalent permutation matrices. Are they permutation similar too? REPLY [17 votes]: Here's a direct proof that if permutation matrices $A,B$ (of size $n\ge 0$) have the same characteristic polynomial (or equivalently are linearly equivalent, since these are diagonalizable) then they are conjugate as permutations, i.e., have the same cycle decomposition. If $A,B$ have some cycle of common size, then we can "remove" it (which divides the characteristic polynomial by the same factor while reducing the matrix size). So we can suppose there are no cycles of common length in $A$ and $B$. In this case, we have to prove that $n=0$. Consider a cycle of maximal length $m$ occurring in either $A$ or $B$, say in $A$. Then $\xi=\exp(2i\pi/m)$ is a root of the characteristic polynomial of $A$, and hence of $B$. But since $B$ has only cycles of length $ TITLE: About contractibility of certain categories QUESTION [13 upvotes]: Let $\mathcal{C}$ be an ordinary 1-category and suppose that there exists some object $X \in \mathcal{C}$ such that the following conditions are satisfied, (1) For every $C \in \mathcal{C}$ we have $\operatorname{Hom}(X,C)\neq \emptyset$ , $\operatorname{Hom}(C,X)\neq \emptyset$. (2) $\operatorname{Hom}(X,X)=*$. It is easy to see that its 1-groupoidification must be categorically equivalent to the point. I would like to know if these 2 conditions already imply contractibility of the nerve of $\mathcal{C}$. REPLY [9 votes]: A counterexample is Connes’ cyclic category $\Lambda$: objects are $\langle n\rangle$ for all $n\in\mathbb N$; arrows $\langle n\rangle \to \langle m\rangle$ are homotopy classes of monotone, degree-$1$ maps $(S^1,\mu_{n+1})\to (S^1, \mu_{m+1}$) between circles with marked roots of unity. Its classifying space is known to be $\mathrm{B} S^1$ but it has an object $\langle 0\rangle$ satisfying your conditions because $\Lambda(\langle n\rangle, \langle 0\rangle ) \cong \Lambda (\langle 0\rangle, \langle n\rangle ) \cong \{0,\dots , n\}$ for all $n\in\mathbb N$<|endoftext|> TITLE: Wikipedia article on forbidden graph substructures QUESTION [8 upvotes]: I apologies if this is too trivial a question or if I am over complicating anything here. But I was hoping for some clarification in an article I was reading about forbidden graph substructures on Wikipedia. To explain, for any class of graphs $C$ let $\neg C$ be the class of all graphs not in $C$. Now given any pre-order $\leq$ on the class of all graphs we say that a class $F$ obstructs $C$ under $\leq$ iff for every graph $G$ we have the equivalence $G\not\in C\iff \text{There exists }H\leq G\text{ isomorphic to a graph in }F$. For example if we write that $H\leq G$ when a graph $H$ is a minor of a graph $G$ then $\leq$ is a pre-order and by Wagner's theorem if $C$ is the class of planar graphs then $\{K_5,K_{3,3}\}$ obstructs $C$ under $\leq$. With all that said, isn't it true that there exists an inclusion minimal obstruction class for any class of graphs $C$ under any pre-order $\leq$ if and only if $(1)$ membership in $C$ is closed under isomorphisms, $(2)$ $C$ is an ideal of $\leq$ and $(3)$ every chain in the pre-order $P=(\neg C,\leq)$ has some lower bound. Since if $(1)$ and $(2)$ are true then defining $\preceq$ to be $(\leq/\cong)$ we see $[A]_{\cong }\preceq [B]_{\cong}\iff A\preceq B$ and for any class $F$ that $(F/\cong)=\{[G]_{\cong}:G\in F\}$ is cofinal in $P'=(\neg C/\cong,\succeq)$ if and only if $F$ obstructs $C$ under $\leq$. Thus when every chain in $P$ has a lower bound this means there exists an inclusion minimal class $F$ cofinal in $P'$ so that every complete set of representatives $T$ for the graph isomorphism classes in $F$ must be an inclusion minimal obstruction class of $C$ under $\leq$. In fact if at least one non-proper class (a set) obstructs $C$ under $\leq$ then the minimum cardinality of those sets obstructing $C$ (note the existence of a non-proper class obstructing $C$ may be guaranteed or even independent of the axioms we are using for example if $\leq$ is the induced subgraph pre-order then the existence of such a set is equivalent to Vopenka's principle which is currently not known to be inconsistent with ZFC) under $\leq$ is exactly $\text{cf}(P')$ (the cofinality of the pre-order $P'$). While note in particular for all $G\in T$ that $G\not\in C$ and $\forall HH_4>H_5>\ldots$. An earlier version of the question spoke of unique minimal obstruction classes, which I think is an interesting question. It is easy to find failures of uniqueness for minimal obstruction classes by considering graphs that are bi-embeddable but not isomorphic. In particular, given a choice of $\leq$, let $\equiv$ denote the induced equivalence relation. Example. Let $\leq$ denote induced subgraph. Let $E$ and $F$ be graphs such that $E\equiv F$ but $E\neq F$. (For example, let $E$ be the disjoint union of the finite complete graph $K_n$ for all $n$, and let $F$ be the disjoint union of $K_n$ for even $n$.) Then $\mathcal{C}:=\text{Forb}_{\leq}(E)=\text{Forb}_{\leq}(F)$, but $\{E\}$ and $\{F\}$ are distinct minimal obstruction classes for $\mathcal{C}$. This issue doesn't arise in the case of forbidding finite graphs under subgraph or induced subgraph, since $\equiv$ is the same as isomorphism for finite graphs. At any rate, by focusing on $\equiv$-invariant $\leq$-obstruction classes, we obtain the following characterization. Let $\leq^*$ denote the partial order obtained by the quotient of $\leq$ by $\equiv$. Proposition 2. Let $\mathcal{C}$ be a $\leq$-closed class of graphs. Then the following are equivalent: $(i)$ $\mathcal{C}$ has a unique minimal $\equiv$-invariant $\leq$-obstruction class. $(ii)$ $\mathcal{C}$ has a minimal $\equiv$-invariant $\leq$-obstruction class. $(iii)$ $\mathcal{C}=\text{Forb}_\leq(\mathcal{F})$ for some $\equiv$-invariant $\leq^*$-antichain $\mathcal{F}$. $(iv)$ $\mathcal{C}=\text{Forb}_\leq(\mathcal{E})$ for some $\equiv$-invariant $\leq^*$-well-founded class $\mathcal{E}$. Proof. $(i)\Rightarrow (ii)\Rightarrow (iii)\Rightarrow(iv)$ are straightforward. $(iv)\Rightarrow(i)$. Assume $\mathcal{C}=\text{Forb}_{\leq}(\mathcal{E})$ for some $\leq^*$-well-founded $\mathcal{E}$. Let $\mathcal{H}$ be the class of graphs $H$ such that $H\not\in\mathcal{C}$, but if $H' TITLE: Geometric interpretation of the exceptional isomorphism $PSp(4,3)=PSU(4,2^2)$ QUESTION [15 upvotes]: It is well-known that there is an isomorphism between $PSp(4,3)$ (the symplectic group of dimension $4$ over $\mathbb F_3$) and $PSU(4,2^2)$ (the unitary group defined by $4\times4$ unitary matrices over $\mathbb F_4$). Question: Can the exceptional isomorphism be interpreted in (finite) geometry? Known examples of geometric interpretation: The isomorphism $PSL(3,2)=PSL(2,7)$ is presented here. The isomorphism $S_6=PSp(4,2)$ is explained in John Baez's article. The isomorphism $A_6=PSL(2,9)$ can be explained by the Taylor graph of $PSL(2,9)$ acting on 10 points (i.e. the projective line of $\mathbb F_9$): this is a distance-regular graph on 20 vertices with intersection array $\{9,4,1;1,4,9\}$. Observing that this is also the intersection array of $J(6,3)$ and Johnson graphs are unique, it lets $S_6$ acts on the graph. The isomorphism $PSU(3,3)=G_2(2)'$ can be explained by the unitary nonisotropics graph of $PΓU(3,q)$: for $q=3$, the intersection array is $\{6,4,4;1,1,3\}$, which is exactly that of a generalized hexagon of order 2. Furthermore, $GH(2,2)$ is unique up to duality, so it's also the Cayley generalized hexagon where $G_2(2)$ acts. REPLY [13 votes]: I believe that a finite geometric proof is given by Jean Dieudonné here: Dieudonné, Jean, Les isomorphismes exceptionnels entre les groupes classiques finis, Can. J. Math. 6, 305-315 (1954). ZBL0055.01904. If you go to Section 7 of the given paper you will see the proof. Dieudonné displays the required isomorphism using a 1-1 correspondence between the 40 lines in $(\mathbb{F}_3)^4$, and the non-isotropic lines in $(\mathbb{F}_4)^4$ (non-isotropic with respect to a non-degenerate hermitian form). A google-books version is here.<|endoftext|> TITLE: Invariants for $SO_n \backslash \mathfrak{gl}_n / SO_n$ QUESTION [10 upvotes]: Is there a nice theorem about the algebra of invariants $\mathbb{C}[\mathfrak{gl}_n]^{SO_n \times SO_n}$, where the action is by left and right multiplication? I'm hoping for something along the lines of the Chevalley restriction theorem, with a list of generators for the algebra being a nice bonus if possible (the subalgebra should have dimension $n$; I'm hoping it's free with $n$ generators). For example, for $n = 2$, the algebra of invariants is generated by the determinant and the sum of squares of components. My slightly-educated guess is that the theorem should be approximately: $\mathbb{C}[\mathfrak{gl}_n]^{O_n \times O_n} \simeq \mathbb{C}[\mathfrak{h}]^{W'}$, where $W' = W \rtimes C_2^n$, $C_2$ is the cyclic group of order 2, the action of $W = S_n$ on $C_2^n$ comes from permutation, and the action of $C_2^n$ on $\mathfrak{h}$ is by negation of a coordinate. Generators for the latter algebra would be symmetric polynomials on the squares of coordinates in $\mathfrak{h}$, with degrees $2, 4, 6, ... 2n$. Then $\mathbb{C}[\mathfrak{gl}_n]^{SO_n \times SO_n} = \mathbb{C}[\mathfrak{gl}_n]^{O_n \times O_n}[\text{det}]$; generators would be the same, except the generator of highest degree (degree $2n$) would be replaced with the determinant. REPLY [6 votes]: Let $X$ denote the $n \times n$ complex matrices and let $D$ be the diagonal $n \times n$ complex matrices. We first claim that a generic matrix in $X$ factors as $U\Sigma V$ with $U$ and $V \in SO(n)$ and $\Sigma \in D$. Proof: Let $\mu$ be the multication map $SO(n) \times D \times SO(n) \to X$. Then it is not bad to compute that $D\mu$ is an isomorphism $TSO(n) \oplus TD \oplus TSO(n) \to TX$ at $(\mathrm{Id}, \mathrm{diag}(\sigma_1, \sigma_2,\ldots,\sigma_n),\ \mathrm{Id})$. So the image contains an analytically open set around such a $\mathrm{diag}(\sigma_1, \sigma_2,\ldots,\sigma_n)$; the image is also constructible by Chevalley's theorem, so the image contains a Zarsiki open set around such a $\mathrm{diag}(\sigma_1, \sigma_2,\ldots,\sigma_n)$. So, an $SO(n) \times SO(n)$ invariant function is determined by its restriction to $D$. Its restriction to $D$ must be invariant for the subgroup of $SO(n) \times SO(n)$ which takes $D$ to itself, which can be described as the Coxeter group of type $D_n = S_n \ltimes \{ \pm 1 \}^{n-1}$: We can permute the $\sigma$ in any order, and can multiply and even number of them by $-1$. The invariants for this action are known to be generated by the polynomials $e_1(\sigma_1^2, \ldots, \sigma_n^2)$, $e_2(\sigma_1^2, \ldots, \sigma_n^2)$, ..., $e_{n-1}(\sigma_1^2, \ldots, \sigma_n^2)$, $\sigma_1 \sigma_2 \cdots \sigma_{n}$. Since you have already given explicit formulas for invariants of $n \times n$ matrices which restrict to these on the diagonal matrices, your formulas are generators as well. $\square$ When we see the Coxeter group $S_n \ltimes \{ \pm 1 \}^{n-1}$, we should expect its partner the Lie Group $SO(2n)$ to be close by. Indeed, consider $SO(2n)$ acting by conjugation on $(2n) \times (2n)$ skew symmetric matrices. The invariants for that action are known by the Chevalley restriction theorem. A generic skew symmetric matrix can be conjugated to a matrix of the form $\left[ \begin{smallmatrix} 0 & M \\ -M^T & 0 \end{smallmatrix} \right]$, and your description is the $SO(2n)$ invariants restricted to matrices of this block form.<|endoftext|> TITLE: Any real contribution of functional analysis to quantum theory as a branch of physics? QUESTION [27 upvotes]: In the last paragraph of this last paper of Klaas Landsman, you can read: Finally, let me note that this was a winner's (or "whig") history, full of hero-worship: following in the footsteps of Hilbert, von Neumann established the link between quantum theory and functional analysis that has lasted. Moreover, partly through von Neumann's own contributions (which are on a par with those of Bohr, Einstein, and Schrodinger), the precision that functional analysis has brought to quantum theory has greatly benefited the foundational debate. However, it is simultaneously a loser's history: starting with Dirac and continuing with Feynman, until the present day physicists have managed to bring quantum theory forward in utter (and, in my view, arrogant) disregard for the relevant mathematical literature. As such, functional analysis has so far failed to make any real contribution to quantum theory as a branch of physics (as opposed to mathematics), and in this respect its role seems to have been limited to something like classical music or other parts of human culture that adorn life but do not change the economy or save the planet. On the other hand, like General Relativity, perhaps the intellectual development reviewed in this paper is one of those human achievements that make the planet worth saving. To balance this interesting debate, if there actually exists real reasons to disagree with above bolded sentence of Klaas Landsman, let me ask the following: What are the real contributions of functional analysis to quantum theory as a branch of physics? Here "real" should be understood in the sense underlying the above paragraph. This question was asked on physics.stackexchange and on PhysicsOverflow. REPLY [7 votes]: The Birman-Schwinger principle bounds the number of eigenvalues of a Schrödinger operator below certain level (in terms of an integral operator involving the potential and the resolvent of the Laplacian). This has been used in many "real physics" articles, e. g. "Bound for the Kinetic Energy of Fermions Which Proves the Stability of Matter" by Lieb and Thirring. On might argue that Birman-Schwinger principle does not constitute a very deep application of functional analysis, which is probably true, but nevertheless even its general rigorous formulation (and, of course, proof) requires some FA.<|endoftext|> TITLE: A "polar dual" for projective varieties? QUESTION [8 upvotes]: Given a projective variety $X$ (over $\mathbb{C}$, say) with an affine paving $X=\sqcup_i C_i$, one can construct a poset $P_X$ on the set of cells $\{C_i\}$ by saying $C_i \leq C_j$ whenever $C_i \subseteq \overline{C_j}$. For example, doing this for Schubert cells in the flag variety gives the Bruhat order. The dual $P^*$ of a poset $P$ is obtained by "flipping $P$ upside down": $x \leq_{P^*} y$ if and only if $y \leq_P x$. Is there a natural construction of a "polar dual" variety $X^*$, along with an affine paving, such that $P_{X^*}\cong P_X^*$? Ideally $X^{**}$ should be "the same" stratified variety as $X$. I am willing to assume any niceness conditions on the decomposition $X=\sqcup_i C_i$ that you like. The reason I use the name "polar dual" is that if one views a convex polytope $Q$ as the union of its open faces, and defines the poset $P_Q$ using the closure relation as above, then $P_{Q^{\circ}}=P_Q^*$, where $Q^{\circ}$ is the polar dual polytope. REPLY [9 votes]: The answer to your question is no. If you have an irrdeucible projective variety $X$ of dimension $d$ with an affine paving, then for all $i < d/2$, the number of $i$-cells is less than or equal to the number of $(d-i)$-cells. This is the main idea of this paper by Bjorner and Ekedahl: https://arxiv.org/pdf/math/0508022.pdf The proof is as follows: the Hard Lefschetz Theorem for intersection cohomology tells you that multiplication by the $(d-2i)$th power of the ample class gives an isomorphism from $I\!H^{2i}(X)$ to $I\!H^{2(d-i)}(X)$. The existence of an affine paving implies that cohomology sits inside of intersection cohomology, so the above isomorphism restricts to an injection from $H^{2i}(X)$ to $H^{2(d-i)}(X)$. But the dimension of $H^{2i}(X)$ is equal to the number of $i$-cells, so we win. The upshot is that you can only hope to find a polar dual variety in the sense that you describe if the Poincare polynomial is palindromic, and this does not always happen. For example, let $V$ be a generic 3-dimensional subspace of $\mathbb{C}^4$, and let $X$ be its closure inside of $(\mathbb{P}^1)^4$. Then $X$ admits an affine paving in which the cells are indexed by looking at what subset of coordinates is equal to infinity. There is one 0-cell (all coordinates equal to infinity), four 1-cells (three coordinates equal to infinity), six 2-cells (two coordinates equal to infinity), and one 3-cell (no coordinates equal to infinity). (It's easy to convince yourself that it is not possible for exactly one coordinate to equal infinity.) Since 4 is not equal to 6, $X$ cannot have a polar dual.<|endoftext|> TITLE: Simple conjecture about rational orthogonal matrices and lattices QUESTION [24 upvotes]: The following conjecture grew out of thinking about topological phases of matter. Despite being very elementary to state, it has evaded proof both by me and by everyone I've asked so far. The conjecture is: Let $R$ be an $N \times N$ rational orthogonal matrix. Define a sublattice $\Lambda \subseteq \mathbb{Z}^N$ by $$ \Lambda = \{ v \in \mathbb{Z}^N : Rv \in \mathbb{Z}^N \} = \mathbb{Z}^N \cap R^{-1} \mathbb{Z}^N $$ Then if $N$ is not a multiple of 4, $\Lambda$ contains a vector of odd length-squared. Note that by this question, such matrices $R$ are in plentiful supply, so the statement is far from vacuous. To motivate this conjecture, we can first look at several examples where it is easy to prove. $N = 1$: Here it is trivially true. The only possible $R$ are $(\pm 1)$, so $\Lambda = \mathbb{Z}$. $N = 2$: In this case, $R$ takes the form $$ R = \frac{1}{c} \begin{pmatrix}a & \mp b \\ b & \pm a\end{pmatrix} $$ with $a^2 + b^2 = c^2$ a primitive Pythagorean triple, which in particular means that $c$ must be odd. Now it can be shown that $\Lambda$ admits a basis in which the inner product has matrix $$ \begin{pmatrix}c & 0 \\ 0 & c\end{pmatrix} $$ So $\Lambda$ once again contains vectors of odd length-squared. $N = 3$: I haven't found an elegant proof for this case, but did manage to reduce it to checking a finite list of cases by computer. Sparing the messy details, the conjecture turns out to be true. $N = 4$: Here something new happens! Consider the matrix $$ R = \frac{1}{2} \begin{pmatrix}+1 & -1 & -1 & -1 \\ -1 & +1 & -1 & -1 \\ -1 & -1 & +1 & -1 \\ -1 & -1 & -1 & +1\end{pmatrix} $$ Then a basis for $\Lambda$ consists of $(1,-1,0,0)$, $(0,1,-1,0)$, $(0,0,1,-1)$, $(0,0,1,1)$, which all have even length-squared. This 'counterexample' is the origin of the requirement $4 \nmid N$ in the statement of the conjecture. $N = 5$: The same strategy as for $N = 3$ works. However, it can't be pushed any further. To gain further confidence, we can turn to Monte-Carlo experiments. For each matrix size $N = 1, \dots, 16$, the following table shows the result of generating 1 million random rational orthogonal matrices $R$, and counting the number of them for which $\Lambda$ is an even lattice: $$ \begin{array}{r|cccccccccccccccc} N & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \#R & 0 & 0 & 0 & 95444 & 0 & 0 & 0 & 3299 & 0 & 0 & 0 & 136 & 0 & 0 & 0 & 5 \end{array} $$ Obviously, I haven't specified exactly what kind of "random" matrix generator is being used here. However, the important point is this: whatever the distribution, it has yielded many examples for each of $N = 4,8,12,16$. This suggests that, if there were examples to be found for $4 \nmid N$, we would have found some, which we haven't. What's more, this pattern is exactly what one expects from considerations of the physics of topological phases of fermions. Together with the above checks, this makes a compelling case in favour of the conjecture. So, assuming the conjecture to be true, my question is: why? If it can be related to any known result, then it would be nice simply to have a reference to it. And if not, it would be great to know of a proof all the same, in the hope that some of the ingredients may shed light on the physics, or vice-versa. REPLY [13 votes]: Proof Let $R$ be any matrix. We have the obvious exact sequence $$ 0 \longrightarrow\mathbb{R}^N \xrightarrow[\left(\begin{matrix} I \\ R \end{matrix}\right)]{} \mathbb{R}^N \oplus \mathbb{R}^N \xrightarrow[\left(\begin{matrix} I & -R^{-1} \end{matrix}\right)]{} \mathbb{R}^N \longrightarrow 0 $$ This contains as a subsequence $$ 0 \longrightarrow\mathbb{Z}^N \cap R^{-1} \mathbb{Z}^N \longrightarrow \mathbb{Z}^N \oplus \mathbb{Z}^N \longrightarrow \mathbb{Z}^N + R^{-1} \mathbb{Z}^N \longrightarrow 0 $$ By definition, $\mathbb{Z}^N \cap R^{-1} \mathbb{Z}^N = \Lambda$. Taking the dual of this equation, $\mathbb{Z}^N + R^T \mathbb{Z}^N = \Lambda^\star$. At this point, we need to invoke the fact that $R$ is orthogonal, so that $R^T = R^{-1}$. Then taking the quotient of the two sequences yields $$ 0 \longrightarrow \frac{\mathbb{R}^N}{\Lambda} \longrightarrow \frac{\mathbb{R}^N}{\mathbb{Z}^N} \oplus \frac{\mathbb{R}^N}{\mathbb{Z}^N} \longrightarrow \frac{\mathbb{R}^N}{\Lambda^\star} \longrightarrow 0 $$ Suppose that every vector of $\Lambda$ has even length-squared. Then $(1, \dots, 1)$ has even inner product with every vector in $\Lambda$, so $\tfrac{1}{2} (1, \dots, 1)^T \in \Lambda^\star$. This tells us that $\tfrac{1}{2}(1, \dots, 1) \oplus 0$, viewed as an element of the middle group above, maps to zero. By exactness, it must therefore be the image of some $v \in \mathbb{R}^N$. So we have $$ \begin{align} v &= \tfrac{1}{2}(1, \dots, 1) \mod \mathbb{Z}^N \\ Rv &= 0 \mod \mathbb{Z}^N \end{align} $$ Comparing length-squareds, $$ \underbrace{v^2 \vphantom{)^2}}_{\frac{N}{4} \text{ mod } 2} = \; \underbrace{(Rv)^2}_{\vphantom{\frac{N}{4}} 0 \text{ mod } 1} $$ we immediately read off that $N$ is a multiple of $4$. A connection A closely related problem is considered here. For an $N \times N$ rational orthogonal matrix $R$, and a sublattice $L \subseteq \mathbb{Z}^N$, define the coincidence index $$ \Sigma_L(R) := [ L : L \cap R L ] $$ It can be shown that $$ \frac{\Sigma_{\mathbb{Z}^N}(R)}{\Sigma_{D_N}(R)} \in \{1, 2\} $$ where $D_N$ is understood to mean the sublattice of vectors with even component sum. The original conjecture is equivalent to saying that $$ 4 \nmid N \; \implies \; \Sigma_{\mathbb{Z}^N}(R) = \Sigma_{D_N}(R) $$ The special case $N = 3$ is a known result, stated in the paper as Fact 3.<|endoftext|> TITLE: Sidon sets of $\mathbb{Z}/p\mathbb{Z}$ QUESTION [9 upvotes]: A set $S \subseteq \mathbb{Z}/p\mathbb{Z}$ is called a Sidon set if given $a, b, c, d \in S$ and $a+ b = c+ d$, then $\{a, b\} = \{c,d\}$. I was interested in knowing about the largest possible Sidon set $\mathbb{Z}/p\mathbb{Z}$ has (Here $p$ is prime). What are known about the largest possible Sidon sets contained in $\mathbb{Z}/p\mathbb{Z}$? Are there any explicit examples or constructive proof for `large' Sidon sets? Any information is appreciated. REPLY [9 votes]: I am not sure of your level of knowledge of Sidon sets, but a good reference is O' Bryant's survey (see section 4.3 in particular) which contains several constructions in the integers. If you are already aware of the three constructions then as far as I know, we do not know much more. We do know if you replace prime with a number of the form $n = p^2 - 1$, then the largest Sidon set of $\mathbb{Z}/n\mathbb{Z}$ has size $(1 + o(1))\sqrt{n}$, thanks to Bose and Chowla (see the cited survey for details). We can always make a large Sidon subset of $\mathbb{Z}/p\mathbb{Z}$ from one in the integers as follows: Given $N < p/2$ and a Sidon Set $$A \subset\{1 , \ldots , N\},$$ embed $A$ into $\mathbb{Z}/p\mathbb{Z}$ via the projection map. With the constructions in the aforementioned survey this gives a Sidon subset of $\mathbb{Z}/p\mathbb{Z}$ of size $(1 + o(1))\sqrt{p/2}$. On the other hand, there is an easy counting that gives a Sidon subset of $\mathbb{Z}/p\mathbb{Z}$ has size at most $(1 + o(1))\sqrt{p}$. Thus the largest Sidon subset, $A$, of $\mathbb{Z}/p\mathbb{Z}$ has size $$(1 + o(1))\sqrt{p/2} \leq |A| \leq (1 + o(1))\sqrt{p}.$$ Closing the gap in the constant is an interesting question.<|endoftext|> TITLE: Why was the factor $\frac12$ introduced in the Riemann $\xi$ function? QUESTION [8 upvotes]: The factor $\frac12$ in the Riemann $\xi$ function: $$\xi(s)=\frac12 s(s-1)\,\pi^{-s/2}\,\Gamma(s/2)\,\zeta(s)$$ was introduced by Riemann, however appears to be redundant. Once he had arrived at: $$\pi^{-s/2}\,\Gamma(s/2)\,\zeta(s)$$ the remaining poles at $s=0,1$ could have been removed by the simpler factor $s\,(s-1)$. This would have also made the function entire and retains the reflection formula $\xi(s) = \xi(1-s)$. A quite plausible explanation as to why the additional factor $\frac12$ was introduced, I found here. It boils down to the following trick: $$\frac12s(s-1)\,\Gamma\left(\frac{s}{2}\right)=2\,\Gamma\left(\frac{s+4}{2}\right)-3\,\Gamma\left(\frac{s+2}{2}\right)$$ that serves as the first step towards deriving the well known Fourier integral expression for $\xi(\frac12+it)$. Splitting the LHS into $\Gamma$'s with constant weights, only works when the factor $\frac12$ is introduced and apparently this was a trick known to Riemann. However, this is not the approach towards the Fourier integral that is attributed to Riemann in the well-known books about the Zeta-function of Titchmarsh (1986-edition p254/255) and Edwards (1974-edition p16/17, p41). These predominantly apply integration by parts to an integral representation of $\xi(s)$ and use a special relation involving $\psi(x)$ and $\psi'(x)$. But this approach doesn't seem to require the factor $\frac12$. The only reference about the factor $\frac12$ that I have ever come across was in a footnote in a book or paper that (paraphrased) said: 'the factor $\frac12$ was introduced by Riemann in his 1859 paper and has since then stuck' (unfortunately I can't recall the precise source of this quote...). Q1: Is there any reference to literature about why the factor $\frac12$ was introduced? Q2: Does it actually matter that the factor $\frac12$ got 'stuck' in $\xi(s)$? It is obviously just a factor, but I could imagine that such a redundant factor would make certain formula less 'beautiful', e.g without the factor, $\xi(0=\xi(1)=1$ and the Hadamard product would simply be: $\displaystyle \xi(s)=\prod_{\rho} \left(1-\frac{s}{\rho}\right)$. REPLY [2 votes]: This factor of $\frac12$ is ultimately due to the fact that the usual integral representation is written as a Mellin transform, that is an integral over $\mathbb{R}^\times_{>0}$, whereas the "correct" integral is over $\mathbb{R}^\times=\mathrm{GL}_1(\mathbb{R})$.<|endoftext|> TITLE: Geodesic sphere in the octonion projective plane QUESTION [8 upvotes]: I am considering Laplacian eigenvalues of a geodesic sphere in the octonion projective plane and I would like to know the metric on a geodesic sphere. Does the metric on a geodesic sphere in the octonion projective plane is a canonical variation associated with the Riemannian submersion $\pi : \mathbb{S}^{15} \rightarrow \mathbb{O}P^1 $ from the octonion Hopf fibration? Loosely speaking, canonical variation on a metric associated with the Riemannian submersion with totally geodesic fibers is a variation on the given metric only in the fiber. More precise definition can be found in p. 191, Section 5, Berard-Bergery and Bourguignon - Laplacians and Riemannian submersions with totally geodesic fibres [Illinois Journ of Math, 1982] (MSN). REPLY [6 votes]: Yes; every geodesic sphere $S_r$, $0 TITLE: Number of fixed points of an involution QUESTION [19 upvotes]: Let $X$ be a positive-dimensional, smooth, connected projective variety (say over $\Bbb{C}$), and let $\sigma $ be an involution of $X$ with a finite number of fixed points; then this number is even. The proof I have is somewhat artificial: blow up the fixed points, take the quotient variety, observe that the image of the exceptional divisor in the quotient is divisible by 2 in the Picard group, and compute its self-intersection. Does someone know a more natural proof? REPLY [9 votes]: Indeed, the number of fixed points is divisible by $2^{\dim X}$. This is actually an old result of Conner and Floyd, Periodic maps which preserve a complex structure, Bull. Amer. Math. Soc. 70, no. 4 (1964), 574-579. As @SashaP mentions, Atiyah and Bott observed that it is also a consequence of the holomorphic Lefschetz formula (Notes on the Lefschetz fixed point theorem for elliptic complexes, Matematika 10, no. 4 (1966), 101-139).<|endoftext|> TITLE: Why does the Steenrod algebra act faithfully on $H^\ast(BC_p)$? QUESTION [5 upvotes]: Define the Steenrod algebra $A^\ast$ to be the algebra of all stable mod $p$ cohomology operations. Without actually computing $A^\ast$, is it possible to see that $A^\ast$ acts faithfully on $H^\ast(BC_p; \mathbb F_p)$? My question is closely related to this one. Here is an attempt at an argument. (EDIT: As Tyler's comment shows, this argument doesn't work! I'll leave it up, though, as an example of the kind of thing I might hope could be true) Let $\phi: H \to \Sigma^r H$ be a nonzero stable cohomology operation (where $H = H\mathbb F_p$ is the Eilenberg-MacLane spectrum). Then $$\phi \wedge 1 : H \wedge H \to \Sigma^r H \wedge H$$ is a nonzero $H$-module map, and so is nonzero on homotopy. We have $H = \varinjlim_n \Sigma^{-n} K(\mathbb F_p, n)$ . It follows that $$\phi \wedge 1: H \wedge \Sigma^{-n} K(\mathbb F_p, n) \to \Sigma^r H \wedge \Sigma^{-n} K(\mathbb F_p, n)$$ is nonzero on homotopy for some $n$. I think it's the case that $H_\ast(K(\mathbb F_p, n))$ is generated under Pontryagin product by $H_\ast(\Sigma K(\mathbb F_p, n-1))$ -- but I'm not sure if this is true, much less whether there is a non-computational reason for it. This ought to allow us to induct downwards to show that $$\phi \wedge 1: H \wedge K(\mathbb F_p, 1)^N \to \Sigma^r H \wedge K(\mathbb F_p, 1)^N$$ is nonzero on homotopy for some $N$, which is almost the desired conclusion. REPLY [11 votes]: As is commented by @Connor Malin, the action of the Steenrod algebra on $H^*B\mathbb{Z}/p$ is not faithful. Consider the case $p=2$. $Sq^3Sq^1$ acts trivially on $H^*(B \mathbb{Z}/2)$, since $Sq^{2n+1}x^{2m}=0$ by the Cartan formula. As a matter of fact, the computation of the Hopf ring structure of $H_*K(\mathbb{Z}/p,*)$ shows that for no finite $n$, the action of the Steenrod algebra on $H^*((B\mathbb{Z}/p)^n)$ can be faithful. One can see this more easily using the last section of T.Kashiwabara Hopf Rings and Unstable Operations, JPAA 94 (1994) 183-193, or even from classical computations of Steenrod algebra.<|endoftext|> TITLE: Is there a topological group with the small index property that does not have automatic continuity? QUESTION [10 upvotes]: Here are the exact definitions of the terms: Let $G$ be a topological group. Then $G$ has the small index property if every subgroup of countable (including finite) index is open in $G$. Furthermore, $G$ has automatic continuity if every group homomorphism from $G$ to any separable topological group is continuous. It is well-known that $G$ having the small index property is equivalent to the fact that every homomorphism from $G$ to the symmetric group $S_{\infty}$ on a countably infinite set $X$ is continuous, where $S_{\infty}$ is given the usual pointwise topology with sub-basis $\{f\in S_{\infty}: f(x)=y\}$ over all points $x,y \in X$. Hence automatic continuity of $G$ implies the small index property. Is an example known that shows the converse is false? REPLY [10 votes]: Here is a counterexample. Consider $\mathbb{R}$ with addition. We define a topology on this group by giving the cosets of all countable index subgroups as a sub-basis. This subbasis is actually a basis as the intersection of finitely many subgroups of countable index is another subgroup of countable index. Moreover as $\mathbb{R}$ is abelian we can see this is a group topology, as if G is a countable index subgroup then $(G+x) + (G+y) \subseteq (G+x+y)$ and $-(G+x)= (G-x)$ for all $x,y \in \mathbb{R}$. This topolgical group has the small index property by the definition of the topology. To see it doesn't have automatic continuity, consider the idenity map from the reals with this topology the the reals with the standard topology. This is a homomorphism (even an isomorphism), but the preimage of $(-1, 1)$ is itself. This is not open in our new topology as all open sets in our new topology contain a translation of a non-trivial group, and are thus unbounded.<|endoftext|> TITLE: Does the basis graph of a matroid determine it? QUESTION [13 upvotes]: Let $M$ be a matroid with set of basis $\mathcal{B}$. The basis graph of $M$ is a graph with set of vertices $\mathcal{B}$ and edges $(B,B')$ always that $B$ and $B'$ differ (as sets) by exactly one element. It is easy to see that this graph can be thought as the $1$-skeleton of the basis polytope of $M$. My question is if it is indeed true that the isomorphism class of $M$ is determined by its basis graph. To avoid trivialities, we ask $M$ to be loopless and bridgeless (it is $M^*$ is also loopless), because otherwise there are simple counterexamples. My intuition says the answer is no, but I couldn't find a counterexample so far. EDIT: As pointed out in the answers and comments, since duality preserves the basis graph, it is reasonable to ask if there is a pair of indecomposable non isomorphic non dual matroids (loopless and bridgeless) such that they induce isomorphic basis graphs. REPLY [12 votes]: I think this question is answered in the paper A Graphical Representation of Matroids by C. A. Holzmann, P. G. Norton, and M. D. Tobey. From the abstract: "A base graph of a matroid is the graph whose points are the bases of the matroid. Two bases are adjacent if they differ by exactly one element. A definition of equivalence of matroids is given and it is shown that two matroids are equivalent if and only if their base graphs are isomorphic. In particular, if M and $M_1 $ are nonseparable matroids with isomorphic base graphs, then M is isomorphic to either $M_1 $ or its dual..." The notion of equivalence is $M = \sum_i M_i$ and $M' = \sum_i M'_i$ are equivalent if and only if $M'_i$ is isomorphic to $M_i$ or its dual for each $i$ (and some reordering if needed).<|endoftext|> TITLE: Is there something "Koszul dual" to formal groups? QUESTION [8 upvotes]: The Lie operad is Koszul dual to the commutative operad. In some sense, the data of a formal group is an "elaboration" of the data of a Lie algebra. Is there some corresponding "elaboration" of the data of an $E_\infty$-algebra which "corresponds" to the notion of a formal group under Koszul duality? Formal groups / formal group laws are not given as the algebras for an operad, I don't expect their "Koszul dual" to be either, so I'm not even sure what Koszul duality should mean here. So strictly speaking, this question doesn't even make sense. Nonetheless, Koszul duality seems to be a phenomenon which has many manifestations, so perhaps there's some adaptation of the notion for which my question at least makes sense. REPLY [3 votes]: I'm not sure what you mean by "elaboration" since the category of formal groups is equivalent to that of Lie algebras. There is however a way Koszul duality can enter this story (I'm not an expert so take this with a pinch of salt). Formal groups are (by definition) the same as conilpotent cocommutative bialgebras (which are thus automatically Hopf), and the equivalence with Lie algebras comes from Cartier-Milnor-Moore theorem. I feel like what you're asking for boils down to reproving this theorem using Koszul duality. Namely, I believe if you apply Koszul duality between coalgebras and algebras to the coproduct of a conilpotent cocommutative coalgebra, you'll get something like an algebra in $E_\infty$ algebra, which should just be the same as an $E_\infty$-algebra (by the "stable" version of Dunn thoerem). Applying now the duality between Com and Lie you'll get a dg-Lie algebra. So long story short I guess I'm trying to say Lie algebra and formal groups are both Kozsul dual to $E_\infty$-algebras and that the composition of these equivalences one way is the enveloping algebra, and composition the other way is taking the primitives.<|endoftext|> TITLE: Is the de Rham complex in characteristic $p$ a CDGA? QUESTION [10 upvotes]: In the paper by Bhatt and Scholze on prismatic cohomology (https://arxiv.org/pdf/1905.08229.pdf), it is stated that the de Rham comparison theorem for prismatic cohomology can be lifted to an equivalence of cdga's. This confuses me, because I don't see why the de Rham complex of a smooth and proper scheme over a field k of characteristic p which is not affine is a cdga. It seem to me that it might be an $E_\infty$ algebra which is not strictly commutative. So i'm asking: It it true that for a smooth and proper scheme X over a field k of characteristic $p$, the global sections of the de Rham complex $R\Gamma(X,DR_X)$, which is a priori an $E_\infty$ algebra in the $\infty$-category of $k$-modules, arises from a commutative differential graded algebra $k$-vector spaces? Note that this imposes many restrictions on the complex, namely the vanishing of most of Steenrod reduced powers on its cohomology. If it is a cdga, why is it the case?, or, alternatively, if it is not the case, what do I miss in the prismatic version of de Rham comparison stated in this paper? REPLY [8 votes]: For a smooth and proper scheme $X$ over a field $k$ of characteristic $p$, the $E_\infty$-algebra $R\Gamma(X, DR_{X/k})$ over $k$ is not represented by a $k$-cdga. Indeed, if it were, then most of the Steenrod operations on its cohomology groups would have to vanish, which is simply not true. You get a counterexample as in the first Bhatt-Morrow-Scholze (BMS1) paper by approximating $B(\mathbf{Z}/p)$ by smooth proper $k$-schemes as the Steenrod operations are nonzero on $H^*_{DR}(B(\mathbf{Z}/p))$ (it is just the singular cohomology of $B(\mathbf{Z}/p)$ with $k$-coefficients). I suspect the phrase "it can be upgraded naturally to an isomorphism of commutative differential graded algebras" in Theorem 1.8 (3) of the quoted paper really should be interpreted at the level of sheaves. For example, the proof of 15.4 in the same paper implicitly appears to suggest the following: Theorem: Fix a prism $(A,I)$ and a smooth formal scheme $X/(A/I)$. There is a natural identification $$\varphi_A^* \Delta_{X/A} \simeq L\eta_I \Delta_{X/A}$$ of sheaves of $E_\infty$-$A$-algebras. By the Bockstein lemma (6.12 in BMS1), the sheaf of $E_\infty$-$A/I$-algebras $(\varphi_A^* \Delta_{X/A}) \otimes_A A/I$ is represented by the sheaf of cdgas $H^*(\Delta_{X/A}/I)$, which is naturally identified with the de Rham complex $\Omega^*_{X/(A/I)}$ as a sheaf of cdgas.<|endoftext|> TITLE: Two conjectural series for $\pi$ involving the central trinomial coefficients QUESTION [6 upvotes]: For each $n=0,1,2,\ldots$, the central trinomial coefficient $T_n$ is defined as the coefficient of $x^n$ in the expansion of $(x^2+x+1)^n$. It is easy to see that $T_n=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}\binom{2k}k$. On December 7, 2019, I conjectured that $$\sum_{k=1}^\infty\frac{(105k-44)T_{k-1}}{k^2\binom{2k}k^23^{k-1}}=\frac{5\pi}{\sqrt3}+6\log3\tag{1}$$ and $$\sum_{k=2}^\infty\frac{(5k-2)T_{k-1}}{k^2\binom{2k}k^2(k-1)3^{k-1}}=\frac{21-2\sqrt3\,\pi-9\log3}{12}.\tag{2}$$ As the two series converge very fast, it is easy to check (1) and (2) numerically. The two identities and related congruences appear in Section 10 of my recent preprint New series for powers of $\pi$ and related congruences. I'm unable to find proofs of $(1)$ and $(2)$. So, here I ask the following question. Question. How to prove the conjectural identities $(1)$ and $(2)$? Your comments are welcome! REPLY [4 votes]: Not an answer, but a reduction to a definite integral. First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$ and thus $T_k = [z^k]\ {\cal T}(z)$. Notice that $${\cal T}'(z) = \frac{1+3z}{(1+z)(1-3z)}{\cal T}(z).$$ Second, by the property of the beta function, we have $$\frac{1}{k\binom{2k}k} = B(k+1,k) = \int_0^1 x^k(1-x)^{k-1}\,{\rm d}x = \frac12 \int_0^1 x^{k-1}(1-x)^{k-1}\,{\rm d}x.$$ It follows that $$I_1:=\sum_{k\geq 1} \frac{T_{k-1}}{k^2\binom{2k}k^2 3^{k-1}} = \frac14 \int_0^1\int_0^1 {\cal T}\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y$$ and $$I_2 := \sum_{k\geq 1} \frac{(k-1) T_{k-1}}{k^2\binom{2k}k^2 3^{k-1}} = \frac14 \int_0^1\int_0^1\, \frac{x(1-x)y(1-y)}3\, {\cal T}'\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y.$$ Since $105k-44 = 105(k-1)+61$, the first sum in question equals $105 I_2 + 61 I_1$. The second sum in question can be reduced to a definite integral similarly.<|endoftext|> TITLE: Do matrices of the form $A^\ast A$ where $A$ has entries in $R\subset\Bbb C$ account for all positive semidefinite matrices with entries in $R$? QUESTION [10 upvotes]: Let $R$ be a subring of $\Bbb C$ closed under complex conjugation and let $P$ be an $n\times n$ positive semidefinite matrix with entries in $R$. I'm curious if it is always possible to factor $P$ as $P=A^\ast A$ where $A$ is $m\times n$ with entries in $R$. If $P$ is $1\times 1$ and $R=\Bbb Z$, then Lagrange's four-square theorem says $P=A^\ast A$ where $A$ is $4\times 1$. Is there anything known beyond this observation? REPLY [5 votes]: $\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}$I did some searching and thinking, and this is what I have come up with. Disclaimer: I couldn't actually find full copies of the papers I'm going to cite by Mordell and Ko, so this is based on reading summaries of them in other papers. The case $R = \ZZ$ was considered by Mordell in "A new Waring’s problem with squares of linear forms", Quarterly Journal of Mathematics 1 (1930), 276–288 and "On the representation of a binary quadratic form as a sum of squares of linear forms", Mathematische Zeitschrift 35 (1932), 1–15. To see the connection, note that, if the $B = A^T A$ and the columns of $A$ are $a_1$, $a_2$, ..., $a_m$, then $x^T B x = \sum \langle a_i, x_i \rangle^2$. So Mordell thought about this problem as expressing a given quadratic form as a sum of squares of linear forms. Mordell showed that this is always possible for quadratic forms in $2$ variables, and that $5$ squares suffice. Without the bound $5$, this appeared on the 1995 IMO shortlist. (solution) Chao Ko showed in "On the representation of a quadratic form as a sum of squares of linear forms" The Quarterly Journal of Mathematics, Volume os-8, Issue 1, (1937), Pages 81–98 that, for $3$, $4$ or $5$ variables, $6$, $7$ or $8$ squares suffice. However, it was already noted by Mordell that the $E_6$ Cartan matrix (below) is not a sum of any number of squares! $$E_6 = \left[ \begin{smallmatrix} 2&-1&0&0&0 &0\\ -1&2&-1&0&0 &0 \\ 0&-1&2&-1&0 &-1 \\ 0&0&-1&2&-1 &0 \\ 0&0&0&-1&2 &0 \\ 0&0&-1&0&0 &2 \\ \end{smallmatrix} \right].$$ To see that this can be proved by a finite search, note that $\mathrm{Tr}(B) = \sum A_{ij}^2$, so in this case we would have $\sum A_{ij}^2 = 12$ and so the matrix $A$ can have at most $12$ nonzero entries. If you think a bit harder, it isn't too hard to get this down to a number of cases that can be done by a pencil and paper search; Hint: First show that, after reordering columns and switching their signs, we can assume that the first five rows of $A$ are $(e_2-e_1, e_3-e_2, e_4-e_3, e_5-e_4, e_6-e_5)$. Since then, the main people thinking about this seem to be Myung-Hwan Kim and Byeong-Kweon Oh. They've produced a series of papers on this sort of question over the last 20 years. Their paper "Bounds for quadratic Waring's problem", Acta Arith. 104 (2002), no. 2, 155–164 is pretty readable and seemed like a good starting point. I couldn't find many papers on the Hermitian variant, but I did find "On a Waring's problem for integral quadratic and hermitian forms" which also has many good references in it. Based on this, I would suspect that there is no simple answer for $\ZZ$, and thus no simple answer for subrings of $\CC$ more complicated than $\ZZ$. What that leaves are the subfields of $\CC$, so I'll turn to those next. If $K$ is a field of characteristic not $2$, and $B$ is a symmetric matrix with entries in $K$, then there is an invertible matrix $S$ such that $SBS^T$ is diagonal (see, for example, here.) I claim that a diagonal matrix $D$ is a sum of squares if and only if each diagonal entry of $D$ is a sum of squares. If each entry of $D$ is a sum of squares, the claim is clear; conversely, if $D = A^T A$ then $D_{jj} = \sum_i A_{ij}^2$. So we are reduced to the question of which elements of $K$ are sums of squares. Similarly, let $L$ be a field of characteristic not $2$ and let $\sigma : L \to L$ be an involution. For a matrix $A$, define $A^{\dagger}$ to be the matrix with $A^{\dagger}_{ij}= \sigma(A_{ji})$. A very similar argument shows that, if $B$ is a matrix with $B^{\dagger} = B$, then there is an invertible matrix $S$ such that $SBS^T$ is diagonal. So, as above, we are reduced to the case of diagonal matrices. Let $K$ be the fixed field of $L$, so a diagonal matrix obeys $D^{\dagger} = D$ if and only if its entries are in $K$. An argument as above shows that a diagonal matrix is of the form $A^{\dagger} A$ if and only if its diagonal entries are sums of norms from $L$ to $K$. The extension $L/K$ is quadratic; if $L = K(\sqrt{-1})$ then we can simply say that a diagonal matrix is of the form $A^{\dagger} A$ if and only if each entry is a sum of squares. Now, let $L$ be a subfield of $\CC$ and let $\sigma$ be complex conjugation, so $K = L \cap \RR$ and $L = K(\sqrt{-1})$. So the above argument reduces the question of which Hermitian matrices are of the form $A^{\dagger} A$ to the question of which elements of $K$ are sums of squares. In a number field $K$, an element is a sum of squares if and only if it is nonnegative for every embedding $K \to \RR$. I think I could also say something for finitely generated subfields of $\CC$, but this seems like it might be pretty far from the OP's interest.<|endoftext|> TITLE: $p$-torsion of class groups QUESTION [7 upvotes]: Let $p$ be a fixed odd prime and $\ell$ be another prime such that $\ell \equiv 1 \pmod{p}$. Consider the number field $\mathbb{Q}(\zeta_p)$ and its extension $\mathbb{Q}(\zeta_p, \zeta_\ell)$. Note that $\ell$ splits completely in $\mathbb{Q}(\zeta_p)$. There is a unique degree $p$-extension of $\mathbb{Q}(\zeta_p)$ contained in $\mathbb{Q}(\zeta_p, \zeta_\ell)$, call this extension $K$. Using the ambiguous class number formula it is possible to show that the $p$-torsion of the class group of $K$ is non-trivial. How does the $p$-torsion of the class group of $K$ depend on $\ell$? What happens as we vary over all $\ell$? Is it possible that the $p$-torsion of the class group of $K$ becomes arbitrarily large or is there an upper bound (depending on $\ell$ maybe)? REPLY [2 votes]: I do not think that it is possible to even bound the $l$-rank of ${\mathbb Q}(\zeta_p)$ in any simple way. Experiments with pari in the case $p = 5$ yield the following results for the $l$-rank $r$ of $K$: $r = 1$ for $l = 11$, $41$, $71$, $191$, $241$, $271$, $\ldots$ $r =2$ for $l = 31$, $131$, $151$, $251$, $\ldots$ $r = 3$ for $l = 101$, $181$, $281$, $\ldots$ $r = 4$ for $l = 211$, $401$, $\ldots$ If $l$ does not divide the class number of ${\mathbb Q}(\zeta_p)$, then the ambiguous class group has rank $\le p-1$, which implies that the $l$-rank is at most $(l-1)(p-1)$ (this is due to Inaba; see Prop. 2.4 in Chapter 2 here.<|endoftext|> TITLE: Flat metric on compact surface minus a point QUESTION [8 upvotes]: Let $T^2$ be a compact smooth surface and let $p\in T^2$. Suppose that $T^2$ admits a symmetric $\left(0,2\right)$-tensor which is a flat Riemannian metric restricted to $T^2-\{p\}$. Is it true that $\chi(T)=0$? REPLY [18 votes]: Here are explicit examples when $M$ is compact, connected, and $\chi(M)\le0$. Orientable case: Let $M$ be the 1-point compactification of the hyperelliptic Riemann surface defined in the affine plane $\mathbb{C}^2$ by $$ y^2 = x^{2g+1}-1. $$ This is a smooth Riemann surface of genus $g\ge1$ and hence $\chi(M) = 2-2g$. The holomorphic $1$-form $$ \omega = \frac{dx}{y} = \frac{dx}{\sqrt{x^{2g+1}-1}} $$ has only one zero (at the point $p$ where $x$ and $y$ have poles of order $2$ and $2g{+}1$ respectively). Consequently, the smooth $(0,2)$-tensor $$ h = \omega\circ\overline{\omega}, $$ which vanishes only at $p$ (and vanishes there to order $2g{-}1$), defines a flat metric on $M\setminus\{p\}$. Non-orientable case: Consider the Riemann surface $\tilde M$ that is the $2$-point compactification of the complex affine curve without real points $$ x^{2g+2} + y^2 + 1 = 0, $$ where $g\ge 1$. This is a hyperelliptic Riemann surface of genus $g$ and hence $\chi(\tilde M) = 2-2g$. The holomorphic $1$-form $$ \omega = \frac{\mathrm{d}x}{y} $$ has two zeroes, one at each of the two points where $x$ and $y$ have poles (of order $1$ and $g{+}1$, respectively), and hence $\omega$ has a zero of order $g{-}1$ at each of these points. The antiholomorphic involution $C(x,y)=(\bar x,\bar y)$ has no fixed points and pulls $\omega$ back to $\overline{\omega}$. Hence, the smooth quadratic form $\omega\circ\overline{\omega}$ is invariant under $C$ and thus descends to the quotient $M$ consisting of the pairs $\{q,C(q)\}$ for $q\in \tilde M$. This $(0,2)$-form on $M$ vanishes at the point $\{p,C(p)\}\in M$ where $p\in \tilde M$ is (either) pole of $x$ and nowhere else. Away from the point where it vanishes, it defines a flat metric on $M$. Meanwhile, $M$ is a compact nonorientable surface of Euler characteristic $\chi(M) = 1-g\le 0$. By the classification of surfaces, the above two cases cover all of the compact, connected surfaces $M$ with $\chi(M)\le 0$. Nonexistence when $\chi(M)>0$: In this case, there cannot be a smooth $(0,2)$-form $h$ on $M$ that vanishes only at one point and elsewhere defines a flat metric. This will follow from the Gauss-Bonnet Theorem. First, a local fact: If $h$ is a smooth $(0,2)$-form on a (not necessarily compact) surface $M$ that degenerates at a single point $p$ but defines a (positive definite) metric with Gauss curvature $0$ everywhere else, then one can show that $p$ has an open neighborhood $B\subset M$ on which there exists a complex coordinate chart $\zeta:B\to\mathbb{C}$ that is smooth except possibly at $p$, satisfies $\zeta(p)=0$, and satisfies $h = L^2|\zeta|^{2(L-1)}|\mathrm{d}\zeta|^2$ on $B\setminus\{p\}$ for some constant $L>0$. (This uses the fact that $h$ near $p$ can be bounded above by an actual smooth metric.) Now, suppost that $M$ is compact and let $D\subset B$ be the disk on which $|\zeta|<\epsilon$ for some small $\epsilon>0$. Apply the Gauss-Bonnet Theorem to the compact surface $C = M\setminus D$, which has the circle $|\zeta|=\epsilon$ as boundary and satisfies $\chi(C) = \chi(M)-1$. Since the Gauss curvature of $h$ on $C$ is zero, Gauss-Bonnet yields $$ -2\pi L = \int_{\partial C} \kappa_g\,\mathrm{ds} = 2\pi\chi(C) = 2\pi(\chi(M)-1), $$ so $\chi(M) = 1-L < 1$. Thus, $\chi(M)\le 0$ and $L$ is positive integer. (This works even when the surface $C$ is non-orientable, using the 'unoriented' version of Gauss-Bonnet. Alternatively, one could pass to the orientation double cover if necessary and argue there. The result is the same.) Smoothing conical flat metrics: It is known that, if $M$ is a compact, connected surface and $L_1,\ldots,L_m$ are positive real numbers such that $L_1+\cdots+L_m = m-\chi(M)$, then there exists a conformal structure on $M$ and $m$ points $p_1,\ldots,p_m$ and a flat Riemannian metric $h$ on $M\setminus\{p_1,\ldots,p_m\}$ such that each $p_i$ has a $p_i$-centered conformal coordinate chart $\zeta_i:B_i\to\mathbb{C}$ with the property that $h = (L_i)^2|\zeta_i|^{2(L_i-1)}|\mathrm{d}\zeta_i|^2$ on $B_i\setminus\{p_i\}$. This $h$ will not extend smoothly in the $\zeta_i$ coordinate to $\zeta_i=0$ (i.e., $p_i$) unless $L$ is a (positive) integer. However, defining a new (non-conformal) coordinate $z_i = |\zeta_i|^{(L_i-2)/2}\zeta_i$ on $B_i$, one finds that $$ (L_i)^2|\zeta_i|^{2(L_i-1)}|\mathrm{d}\zeta_i|^2 = (L_i)^2 |z_i|^2|\mathrm{d}z_i|^2 + \bigl(1-(L_i)^2/4)\bigr)\bigl(\mathrm{d}(|z_i|^2)\bigr)^2. $$ Thus, in the $z_i$ coordinate, $h$ is smooth at $z_i=0$ (i.e., $p_i$). Using the $z_i$ to build a smooth (in fact, real-analytic) atlas on $M$ (keeping the conformal atlas at all the points of $M$ other than the $p_i$) yields a new smooth structure on $M$ in which $h$ extends real-analytically to all of $M$ and vanishes to order 2 at each of the $p_i$.<|endoftext|> TITLE: For a spherical pair $(G, H)$, which $G$-representations appear in $k[G/H]$? QUESTION [10 upvotes]: Let $G$ be a reductive algebraic group (over some alg. closed field $k$ of char 0), and $H$ a subgroup such that $(G, H)$ is spherical (i.e., the Borel $B$ of $G$ has an open orbit on $G/H$). Then $k[G/H]$ is multiplicity-free as a $G$-module. I'm looking for a nice description of the irreducible $G$-representations it contains. More precisely, let $\Lambda(G)$ be the lattice of integral weights for $G$ and $\Lambda_+(G)$ be the cone of dominant integral weights. For each $\lambda \in \Lambda_+(G)$ we have an irreducible representation $V_\lambda$. Let me write $\Lambda_+(G, H)$ for the set of $\lambda \in \Lambda_+(G)$ such that $V_\lambda$ appears in $k[G/H]$ (equivalently, such that $(V_\lambda)^H \ne 0$). The set $\Lambda_+(G, H)$ is a sub-monoid of $\Lambda_+(G)$ (see this question). Playing around with some examples suggests the following: Claim: $\Lambda_+(G, H)$ is a lattice cone in $\Lambda(G)$, and its $\mathbb{Z}$-span is exactly the weights trivial on the $B$-stabiliser of a point in the open $B$-orbit on $G/H$. Is this true? If so, is there a good reference that is easily digestible by non-experts like myself? (Brion's paper "Spherical varieties" in the 1994 ICM proceedings, page 755, has a very general assertion about the sections of any $G$-equivariant line bundle on any spherical $G$-variety, which seems to be closely related; but I couldn't make head or tail of the proof, and surely there must be an easier approach for homogenous spherical varieties $G/H$ and trivial line bundles.) REPLY [4 votes]: Let $X_0=B/B_{x_0}=Bx_0$ be the open $B$-orbit in $X=G/H$. Then every character of $B$ which is trivial on $B_{x_0}$ yields a $B$-semiinvariant $f_\lambda$ on $X_0$. In general, it does not extend to a regular function on $X$. If it does, this means that $V_\lambda$ occurs in $k[X]$. So your second problems asks whether every $B$-semiinvariant on $X_0$ is a quotient of two regular $B$-semiinvariants on $X$. This is true if $X$ is quasiaffine (e.g., if $H$ is reductive) but false in general. Take, e.g., $G=SL(3,k)$ and $H=ker(\chi)$ where $\chi:B\to G_m$ is a character which is neither dominant nor antidominant (e.g. $\chi=a_{22}$). Then $V_\chi^H=0$ for all $\lambda$. I don't know what you mean by lattice cone but if it means a monoid which is the intersection of a convex cone by a lattice then your first claim is true. This is a consequence of the normality of $X$. More precisely, let $\overline\Lambda_+$ be the intersection of the convex cone and the lattice generated by $\Lambda_+(G,H)$. It is called a saturation. An element $\lambda$ is in $\overline \Lambda_+$ if there is $m\ge1$ such that $m\lambda\in\Lambda_+(G,H)$. Now, since $\lambda$ is in the lattice generated by $\Lambda_+$ it corresponds to a rational $B$-semiinvariant $f_\lambda$ on $X$. The power $f_\lambda^m$ is regular. So $f_\lambda$ is regular and $\lambda\in\Lambda_+$.<|endoftext|> TITLE: Dual of $End_A(M)$ QUESTION [5 upvotes]: Let $A$ be a finitely generated $\mathbb C$-algebra and an integral domain. Assume also $A$ is Gorenstein. Let $M$ be a finitely generated torsion-free $A$ module. Is it true that $Hom_A(End_A(M), A)\cong End_A(M)$? REPLY [2 votes]: As Mohan mentioned in the comments, this is false if one does not assume $M$ is reflexive, but $M$ being reflexive is still not good enough. I'll comment on the local case, and remark that this can easily be extended to more generality, including, for instance, the standard graded case. Proposition: Let $(A,\mathfrak{m},k)$ be a Gorenstein local ring with $\dim A \le 1$. If $\operatorname{Hom}_A(\operatorname{End}_A(\mathfrak{m}),A) \cong \operatorname{End}_A(\mathfrak{m})$, then $\mu_A(\mathfrak{m}) \le 2$. Proof: Let $(-)^*=\operatorname{Hom}_A(-,A)$ denote the $A$-dual. Then, as $A$-modules, we have $\operatorname{End}_A(\mathfrak{m}) \cong \mathfrak{m}^*$ (see, for example, this answer). Furthermore, since $\dim A \le 1$, $\mathfrak{m}$ is maximal Cohen-Macaulay, and thus reflexive, since $A$ is Gorenstein. In particular, we have $\operatorname{End}_A(\mathfrak{m})^* \cong \mathfrak{m}^{**} \cong \mathfrak{m}$. We claim that if $\mathfrak{m} \cong \mathfrak{m}^*$, then $\mu_A(\mathfrak{m}) \le 2$. To see this, we have the natural exact sequence $0 \to \mathfrak{m} \to A \to k \to 0$. Applying $(-)^*$ to this sequence, we get an exact sequence of the form $$ 0 \to k^* \to A^* \to \mathfrak{m}^* \to \operatorname{Ext}^1_A(k,A) \to 0.$$ Either $\dim A=0$ or $\dim A=1$. If $\dim A=0$, then, as $A$ is Gorenstein, $\operatorname{Ext}^1_A(k,A)=0$, so so we have a surjection $A \to \mathfrak{m}$, implying $\mu_A(\mathfrak{m}) \le 1$. In instead, $\dim A=1$, then $k^*=0$ and $\operatorname{Ext}^1_A(k,A) \cong k$; of course $A^* \cong A$. In particular, $\mu_A(\mathfrak{m}) \le \mu_A(A)+\mu_A(k)=2$. As a remark, note that in the dimension $1$ case we actually have the stronger claim that the Hilbert-Samuel multiplicity $e(A)$ of $A$ is at most $2$; I'll leave this as an exercise. To point out, some work has been done on understanding for which local rings every maximal Cohen-Macaulay module is self dual, rather than endomorphism modules; see this paper which also contains part of the above Proposition, though the argument is slightly different and a bit more general. This condition is extremely restrictive (for instance it forces $A$ to be a hypersurface and it is conjectured that $e(A) \le 2$ without the $\dim A \le 1$ hypothesis), and I would expect the same of the condition that $\operatorname{End}_A(M)$ be self dual for even every maximal Cohen-Macaulay $M$.<|endoftext|> TITLE: When does "sufficient genericity" actually suffice? QUESTION [6 upvotes]: Fix a forcing notion $\mathbb{P}$. Say that a formula $\varphi(x)$ with parameters is $\mathbb{P}$-enforceable if there is some countable set $\mathcal{D}$ of dense sets in $\mathbb{P}$ such that for every $\mathcal{D}$-generic filter $G\subseteq\mathbb{P}$ there is some $p\in G$ such that for every $\mathcal{D}$-generic filter $H\subseteq \mathbb{P}$ with $p\in H$ we have $$\varphi(G)\iff\varphi(H).$$ Note that this all takes place within $V$ itself, and since $\mathcal{D}$ is required to be countable enforceability isn't trivial. In general, we don't always have that $\varphi$ is $\mathbb{P}$-enforceable$^*$ - however, for reasonably natural $\varphi$ and $\mathbb{P}$ it does tend to be the case that $\varphi$ is $\mathbb{P}$-enforceable. I'm interested in the particular case when $\mathbb{P}$ is Cohen forcing $\operatorname{Fin}(\omega, 2)$ and $\varphi$ is projective with only real parameters. I've heard it stated repeatedly that the usual suspects guarantee tameness, but I've never seen a citation for this and the proof isn't immediate to me. So my question is: Is it actually the case that, assuming reasonable large cardinals, every projective formula $\varphi$ with only real parameters is $\mathbb{P}$-enforceable? If so, what's a citation for this fact? $^*$Here's one counterexample - indeed, where $\varphi$ is parameter-free and projective. Assume $V=L$, take $\mathbb{P}$ to be the usual Cohen forcing, let $a$ be a projective bijection from $\mathbb{R}$ to $\omega_1$, and let $b$ be a bijection between the set of countable sets of dense subsets of $\mathbb{P}$ and $\omega_1$. We can define by recursion a projective injection $f:\alpha\rightarrow\mathbb{R}$ such that $f(\alpha)$ is $b^{-1}(\beta)$-generic for all $\beta\le\alpha$. Now let $\varphi(x)$ be the projective formula "$x\in \operatorname{ran}(f)$ and $f^{-1}(x)$ is a limit ordinal." Of course, this sort of nonsense relies on wild projective sets, so large cardinals rule it out. REPLY [7 votes]: Assume $\text{AD}^{L(\mathbb R)}$. Let $A$ be projective. Since $A$ is ${}^\infty$Borel in $L(\mathbb R)$, there is a set of ordinals $S$ and a formula $\psi$ such that $\omega_1$ is strongly inaccessible in $L[S]$ and for all reals $x$, $x\in A$ if and only if $L[S,x]\vDash \psi(S,x).$ Let $\mathcal D$ be the set of dense subsets of Cohen forcing that belong to $L[S]$. Suppose $g$ is $\mathcal D$ -generic. Let $p\in g$ decide whether $L[S,g]\vDash \psi(S,g).$ Then for any $\mathcal D$-generic $h$ containing $p$, $L[S,h]\vDash \psi(S,h)$ if and only if $L[S,g]\vDash \psi(S,g).$ Hence $h\in A$ if and only if $g\in A$. Thus $A$ is enforceable. EDIT: One can prove in ZFC that all ${\bf \Pi}^1_1$ sets are enforceable. Suppose $A$ is ${\bf \Pi}^1_1$. Let $T$ be a tree on $\omega\times \omega$ such that $x\in A$ if and only if $T_x$ is wellfounded. Let $M$ be a countable transitive model of enough set theory containing $T$. Suppose $g$ is Cohen generic over $M$ and $g\notin A$. Then $T_g$ is illfounded, and hence $T_g$ is illfounded in $M[g]$: otherwise $T_g$ would be ranked in $M[g]$, and hence wellfounded in $V$, a contradiction. It follows that some $p\in g$ forces relative to $M$ that $T_g$ is illfounded in $M[g]$, so every $M$-generic Cohen real $h$ containing $p$ is such that $h\notin A$. Similarly, if $g\in A$, one can find a $p\in g$ that enforces this for $M$-generics. Thus $A$ is enforceable (taking $\mathcal D$ equal to the set of dense sets in $M$). EDIT 2: I think I had too much coffee yesterday. A set $A$ is enforceable (for Cohen forcing) if and only if $A$ has the Baire property. The reason is that the comeager filter on $\omega^\omega$ is generated by sets of the form $\{g\in \omega^\omega : g\text{ is }\mathcal D\text{-generic}\}$ for $\mathcal D$ a countable collection of dense subsets of Cohen forcing. Thus $A$ is enforceable if and only if $A$ and $\omega^\omega - A$ are open on a comeager set, which is easily equivalent to $A$ having the Baire property. This subsumes all the other things I said.<|endoftext|> TITLE: Average of the maximum matrix element over the Haar measure QUESTION [8 upvotes]: Let $U$ be a $d\times d$ unitary matrix, and $U_{i,j}$ be its matrix elements. I am interested in the following quantity $$\int dU \max_j |U_{1,j}|^2 \ , $$ where $dU$ is the uniform Haar measure over ${\rm SU}(d)$. Please let me know if you have any idea for calculating this integral for general $d$. REPLY [3 votes]: $\newcommand{\C}{\mathbb C}$$\newcommand{\R}{\mathbb R}$Any linear isometry of $\C^d$ is a unitary transformation. Therefore, the distribution of the random vector $V:=(X_1,Y_1,\dots,X_d,Y_d)$ is uniform on the unit sphere in $\R^{2d}$, where $X_j:=\Re U_{1,j}$ and $Y_j:=\Im U_{1,j}$. So, $V$ equals $W:=(W_1,\dots,W_{2d})$ in distribution, where $$W_j:=\frac{Z_j}{\sqrt{Z_1^2+\dots+Z_{2d}^2}} $$ and $Z_1,\dots,Z_{2d}$ are iid standard normal random variables (r.v.'s). So, the random vector $(|U_{1,1}|^2 ,\dots,|U_{1,d}|^2)$ equals $R:=(R_1,\dots,R_d)$ in distribution, where $$R_j:=\frac{T_j}{T_1+\dots+T_d} $$ and $T_k:=Z_{2k-1}^2+Z_{2k}^2$, so that $T_1,\dots,T_d$ are iid standard exponential r.v.'s. So, $\max_j|U_{1,j}|^2$ equals $\max_j R_j$ in distribution. The distribution and, in particular, the expectation of $\max_j R_j$ were found a long time ago; see e.g. Irwin (1955) and historical references therein going back to as far as 1897. In particular, according to formula (15) in Cochran's paper, $$E\max_j|U_{1,j}|^2=E\max_j R_j=\frac1d\,\Big(1+\frac12+\dots+\frac1d\Big). $$<|endoftext|> TITLE: Comonoids in the category of monoids QUESTION [6 upvotes]: Let us give the category of monoids $\mathbf{Mon}$ a monoidal structure with $\otimes = \sqcup$ (coproduct). How can we classify $\mathbf{CoMon}(\mathbf{Mon})$, the category of comonoids of monoids? This category has (at least) two different descriptions, namely it is also the category of cocontinuous functors $\mathbf{Mon} \to \mathbf{Mon}$ (given by tensoring with a comonoid), but also the opposite of the category of continuous functors $\mathbf{Mon} \to \mathbf{Mon}$ (the functors represented by comonoids). From the description of cocontinous functors $\mathbf{Mon} \to \mathbf{Mon}$ we also get a structure of a cocomplete monoidal category (with $\otimes=\circ$ being cocontinous in each variable) with a zero object. So far I have found three basic comonoids: $\langle x \rangle$ with $\nabla(x) = x_1 x_2$, $\langle x^{\pm 1} \rangle$ with $\nabla(x)=x_1 x_2$, and $\langle x \rangle$ with $\nabla(x)=x_2 x_1$. The continuous functors $\mathbf{Mon} \to \mathbf{Mon}$ represented by them are the identity functor, the group of units functor (which factors over $\mathbf{Grp}$), and the opposite monoid functor. Their left adjoints are the identity $\mathbf{1}$, the group completion $K$ and the opposite monoid functor $D$. Hence, every coproduct of them is also an example. Apart from the unit $\eta : \mathbf{1} \to K$, I think that all other morphisms between them are zero, and it seems that we have $D \circ K \cong K \circ D \cong K \circ K \cong K$. In case it turns out that this monoidal category is too complicated, what about the category of monoids in it, i.e. the category of Tall-Wraith monoids in $\mathbf{Mon}$? Can we classify them? For $\mathbf{Grp}$ it is well-known by a result of Freyd. REPLY [4 votes]: George Bergman characterizes representable endofunctors of monoids in 10.6 of https://math.berkeley.edu/~gbergman/245/3.2.pdf by classifying the comonoids in the category of monoids.<|endoftext|> TITLE: Can homotopy colimits recover cohomology sheaves? QUESTION [5 upvotes]: The question is basically the one outlined in the title. Let $\mathcal{T}$ be a triangulated category containing infinite direct sums (e.g. $D_{qc}(X)$ for some separated, finite type over a field $k$, scheme $X$) and consider the subcategory $\mathcal{E}$ generated by an object $E$ of $\mathcal{T}$. Here by subcategory generated I mean the smallest thick full triangulated subcategory containing all direct sums (and thus homotopy colimits). Is it true that $\mathcal{E}$ contains all the cohomology sheaves of $E$? Does it contain only some of them? It is clear that if I were to consider the subcategory generated in $\mathcal{T}^c$ (hence admitting only finite direct sums) this would necessarily be true. Indeed, considering the geometric example $\mathcal{T} = D_{qc}(X)$, then the cohomology sheaves of a perfect complex are not necessarily perfect. Thanks. REPLY [7 votes]: No. Let $j:\mathbb{A}^2_k\smallsetminus\{0\}\to \mathbb{A}^2_k$ be the canonical open embedding. Then the derived pushforward $Rj_*$ is fully faithful and colimit-preserving. In particular, the subcategory of $D_{qc}(\mathbb{A}^2)$ generated under colimits by $\mathscr{A}:=Rj_*\mathcal{O}$ is contained in this subcategory (in fact it coincides with the category of $\mathscr{A}$-modules, which is a subcategory because $\mathscr{A}$ is an idempotent algebra). However $H_0\mathscr{A}=k[x,y]$ is not.<|endoftext|> TITLE: Pythagorean theorems for other distances QUESTION [5 upvotes]: Question The usual projection in $\mathbb{R}^n$ on a subspace can be defined as the point that minimizes the squared distance to the subspace. I'll call the Pythagorean theorem the easy fact that, given a point $x$, its projection $Px$ and another point $y$ in the subspace, $$ |x-y|^2 = |x - Px|^2 + |Px-y|^2 $$ An amazing fact to me is that, in a easy result (due to, I believe, Csiszar), the Kullback-Leibler divergence also obeys this Pythagorean theorem. The question is, is this an example of a more general phenomenon? Are there other interesting 'distance' functions that obey the Pythagorean theorem? Note that the KL divergence does not obey the triangle inequality. Besides the convergence of the alternating projection algorithm (see below), does this lead to any other interesting mathematics? Background I came to this question studying the "RAS" algorithm in economics. This algorithm takes a matrix and gives a matrix with prescribed row and column sums by alternatingly scaling the rows and columns to have the prescribed sums. Csiszar [1] showed that this algorithm is exactly the alternating projection algorithm in disguise, except replacing the $L^2$ projection with the KL distance (Csiszar calls these I-projections). The proof of convergence is almost identical to the usual one, making use of Pinsker's inequality. This beautiful fact led me to the question above. [1] Csiszár, Imre. "I-divergence geometry of probability distributions and minimization problems." The Annals of Probability (1975): 146-158. https://www.jstor.org/stable/2959270?seq=1 REPLY [5 votes]: The so-called "Bregman divergences", which include both $\ell_2$ distance and KL divergence among others, obey the Pythagorean relation. For a nice summary, see Banerjee et al, "Clustering with Bregman Divergences", JMLR 2005, in particular p.1741.<|endoftext|> TITLE: Do spaces admit a weak cogenerating set? QUESTION [11 upvotes]: Let $\mathcal C$ be a category. Say that a class of objects $\mathcal S \subseteq \mathcal C$ is weakly cogenerating if the functors $Hom_{\mathcal C}(-,S)$ are jointly conservative, for $S \in \mathcal S$. That is, a map $X \to Y$ in $\mathcal C$ is an isomorphism if and only if it induces bijections $Hom_C(Y,S) \to Hom_C(X,S)$ for every $S \in \mathcal S$. Of course, every category $C$ admits a weakly cogenerating class -- namely, take $\mathcal S = \mathcal C$. But it's frequently important to have a cogenerating set -- i.e. to require that $\mathcal S$ is small. Question: Does the homotopy category (of spaces) admit a weak cogenerating set? It's clear that the homotopy category of simply-connected spaces admits a weak cogenerating set -- we can take $\mathcal S = \{K(\mathbb Z, n) \mid n \geq 2\}$ or alternatively $\mathcal S = \{K(k,n) \mid n \geq 2, k \in \{\mathbb Q, \mathbb F_p\}\}$ in this case by the cohomology Whitehead theorem. But I'm pessimistic about the chances of doing something similar with arbitrary spaces. Relatedly, I wonder whether the category of groups admits a weak cogenerating set. I also wonder whether the class of truncated spaces -- those spaces $S$ for which $\pi_k(S) = 0$ for $k$ sufficiently large -- is a cogenerating class for the homotopy category. What about the class of Eilenberg-MacLane spaces? REPLY [6 votes]: For any infinite set $X$ let $S_X$ be the group of bijections $\sigma \colon X\to X$ such that $\{x : \sigma(x)\neq x\}$ is finite. This still has signature homomorphism, and the alternating subgroup $A_X$ is simple, and has the same cardinality as $X$. Now let $\mathcal{G}$ be a set of groups, and put $\kappa = \max \{|G|:G\in\mathcal{G}\}$. Then $\text{Hom}(A_X,G)$ will be a singleton for all $G\in\mathcal{G}$ and $X$ with $|X|>\kappa$ (because the kernel of any homomorphism is nontrivial by cardinality, and so is the whole of $A_X$ by simplicity). So $\mathcal{G}$ is not a weak cogenerating set. It doesn't seem to be straightforward to deduce the corresponding result for the homotopy category. EDIT To summarize the discussion in the comments, we can indeed deduce the corresponding result for the homotopy category with a little more work. Choose an acyclic simple group $G$ bigger than the fundamental group of any space in $\mathcal S$. Then any map $f: BG \to S$ for $S \in \mathcal S$ is trivial on $\pi_1$ by simplicity, so it lifts to the universal cover $\tau_{\geq 2} S$. By acyclicity, the composite map $BG \to \tau_{\geq 2} S \to K(\pi_2(S),2)$ is trivial so $f$ lifts through the 2-connected cover $\tau_{\geq 3} S$. Continue in this manner, lifting through the Whitehead tower to see that $f$ is nullhomotopic. Thus $\mathcal S$ does not distinguish $BG$ from a point, and is not weakly cogenerating.<|endoftext|> TITLE: What is the outer automorphism group of $\operatorname{SL}(2,\mathbb{F}_q)$? QUESTION [8 upvotes]: I'm looking for a reference for a description of the outer automorphism groups of $\operatorname{SL}(2,\mathbb{F}_q)$ for $q = p^n$. I'm sure such a thing must exist somewhere, but I'm having trouble locating a reference. REPLY [5 votes]: "Each automorphism $\sigma$ of $G$ can be written $\sigma = g f d i$, with $i$, $d$, $f$, and $g$ being inner, diagonal, field, and graph automorphisms, respectively" (Steinberg - Automorphisms of finite linear groups, 3.2). Here, as best as I can tell, $G$ is $\operatorname{PSL}(2, \mathbb F_q)$ (not $\operatorname{PGL}(2, \mathbb F_q)$); the definition of $G$ relies on a set that looks like $\mathfrak B$, whose definition I cannot find. A diagonal automorphism is one that arises by conjugation in the diagonal subgroup of $\operatorname{PGL}(2, \mathbb F_q)$, not just of $\operatorname{PSL}(2, \mathbb F_q)$; the only non-inner automorphism that we get this way is conjugation by $\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}$, where $\epsilon$ is a non-square in $\mathbb F_q^\times$. If an automorphism $\sigma$ of $\operatorname{SL}(2, \mathbb F_q)$ induces the identity on $\operatorname{PSL}(2, \mathbb F_q)$, then $g \mapsto g^{-1}\sigma(g)$ maps $\operatorname{SL}(2, \mathbb F_q) \to \{\pm1\}$. However, $\operatorname{SL}(2, \mathbb F_q)$ is generated by its unipotent elements, hence admits no non-trivial homomorphisms to a 2-torsion group (assuming $p \ne 2$). (EDIT: If $p = 2$, then $\operatorname{SL}(2, \mathbb F_q) = \operatorname{PSL}(2, \mathbb F_q)$, so there is no ambiguity.) EDIT: I somehow misread the question as asking about the full automorphism group of $\operatorname{SL}(2, \mathbb F_q)$, not just the outer automorphism group; and it's always a good time to break out a result of Steinberg. As @RichardLyons and @YCor point out, since there are no diagram automorphisms in type $\mathsf A_1$, the outer automorphism group (of $\operatorname{SL}(2, \mathbb F_q)$, which, we argued above, is the same as that of $\operatorname{PSL}(2, \mathbb F_q)$) is $\langle\operatorname{Fr}\rangle \times \langle\operatorname{Int}\begin{pmatrix} \epsilon & 0 \\ 0 & 1 \end{pmatrix}\rangle \cong \operatorname C_n \times \operatorname C_2$, generated by the Frobenius and an appropriate conjugation in $\operatorname{PGL}(2, \mathbb F_q)$, when $p \ne 2$. If $p = 2$, then, again as @YCor points out, there is no extra conjugation coming from $\operatorname{PGL}(2, \mathbb F_q)$ (since every element of $\mathbb F_q$ is a square), so we get that the outer automorphism group is just $\langle\operatorname{Fr}\rangle \cong \operatorname C_n$.<|endoftext|> TITLE: A mysterious connection between primes and $\pi$ QUESTION [53 upvotes]: The Prime Number Theorem relates primes to the important constant $e$. Here I report my following surprising discovery which relates primes to $\pi$. Conjecture (December 15, 2019). Let $s(n)$ be the sum of all primes $p\le n$ with $p\equiv1\pmod4$, and let $s_*(n)$ be the sum of those $x_py_p$ with $p\le n$, where $p$ is a prime congruent to $1$ modulo $4$, and $p=x_p^2+y_p^2$ with $x_p,y_p\in\{1,2,3,\ldots\}$ and $x_p\le y_p$. Then $$\lim_{n\to+\infty}\frac{s(n)}{s_*(n)} = \pi.$$ Recall that a classical theorem of Euler (conjectured by Fermat) states that any prime $p\equiv1\pmod4$ can be written uniquely as $x^2 + y^2$ with $x,y\in\{1,2,3,\ldots\}$ and $x\le y$. Since $x^2 + y^2 \ge 2xy$ for any real numbers $x$ and $y$, we have $s(n) \ge 2s_*(n)$ for all $n=1,2,3,\ldots$. I have created the sequence $(s_*(n))_{n>0}$ for OEIS (cf. http://oeis.org/A330487). Via computation I found that $$s(10^{10}) = 1110397615780409147,\ \ s_*(10^{10}) = 353452066546904620, $$ and $$ 3.14157907 < \frac{s(10^{10})}{s_*(10^{10})} < 3.14157908. $$ This looks an evidence to support the conjecture. QUESTION. Is my above conjecture true? If true, how to prove it? Any further check of the conjecture is welcome! REPLY [73 votes]: Here is a proof of the conjecture. We shall use Hecke's theorem that the angles of the lattice points $(x_p,y_p)$ are asymptotically equidistributed in $[\pi/4,\pi/2]$, cf. this MO post. Let $t_p\in[\pi/4,\pi/2]$ be the angle of the lattice point $(x_p,y_p)$. Let us divide the interval $[\pi/4,\pi/2]$ into $R$ subintervals of equal length, where $R$ is large but fixed. For $r\in\{1,\dotsc,R\}$, the $r$-th subinterval is $$I_r:=[u_{r-1},u_r]\qquad\text{with}\qquad u_r:=\frac{\pi}{4}\left(1+\frac{r}{R}\right).$$ Observe that $$\frac{\sin(2u_r)}{2}\sum_{\substack{p\leq n\\t_p\in I_r}}p\leq \sum_{\substack{p\leq n\\t_p\in I_r}}x_p y_p\leq \frac{\sin(2u_{r-1})}{2}\sum_{\substack{p\leq n\\t_p\in I_r}}p.$$ By the quoted equidistribution theorem, $$\sum_{\substack{p\leq n\\t_p\in I_r}}p\sim\frac{s(n)}{R}\qquad\text{as}\qquad n\to\infty,$$ and so we infer that $$\frac{1}{R}\sum_{r=1}^R\frac{\sin(2u_r)}{2}\leq \liminf_{n\to\infty}\frac{s_\ast(n)}{s(n)}\leq \limsup_{n\to\infty}\frac{s_\ast(n)}{s(n)}\leq \frac{1}{R}\sum_{r=1}^R\frac{\sin(2u_{r-1})}{2}.$$ By letting $R\to\infty$, both sides tend to $$\frac{4}{\pi}\int_{\pi/4}^{\pi/2}\frac{\sin(2u)}{2}\,du=\frac{1}{\pi},$$ whence $$\lim_{n\to\infty}\frac{s_\ast(n)}{s(n)}=\frac{1}{\pi}.$$<|endoftext|> TITLE: Algebraic properties of graph of chess pieces QUESTION [11 upvotes]: For the purpose of this question, a chess piece is the King, Queen, Rook, Bishop or Knight of the game of chess. To a chess piece is attached a graph which represents the legal moves it can make on an empty chess board. Except for the bishop, the graph attached to a chess piece has 64 vertices, while the bishop's graph has 32 vertices. These graphs satisfy a number of interesting mathematical properties from the point of view of graph theory. Only the rook's graph is regular, and the rook's graph is even strongly regular (a fact well-known to chess player with no knowledge of mathematics). The knight's graph is bipartite. The king's, rook's, queen's and bishop's graph are hamiltonian (and after Euler, it is very well-known though to me non-obvious a priori that the knight's graphs is hamiltonian). Less obvious properties have of course been devised by graph theorists. Recently, I asked myself (with no other purpose than leisure) if these graphs were known to satisfy interesting algebraic properties. A quick computation based on its strong regularity or on the observation that it is the cartesian product of two copies of the complete graph $K_8$ show that spectrum of the rook's graph is $\{14^{(1)},6^{(14)},-2^{(49)}\}$ (in particular, it is an integral graph). Are there any other remarkable algebraic properties of the graphs of chess pieces, either of their spectra or of their group of automorphisms? REPLY [2 votes]: Wikipedia says the chromatic polynomial is "studied in algebraic graph theory". So here's an answer involving the chromatic polynomial of the rook's graph. The rook's graph is $K_8 \Box K_8$, i.e., the Cartesian product of $K_8$ and $K_8$. The chromatic polynomial $P(K_8 \Box K_8;n)$ of the rook's graph is the number of $8 \times 8$ generalized Latin squares, where we can use $n$ symbols (without repeated symbols in any row or column). In particular, evaluating it at its chromatic number... $$P(K_8 \Box K_8;8) = \text{number of Latin squares of order $8$}.$$<|endoftext|> TITLE: A semicartesian monoidal category with diagonals is cartesian: proof? QUESTION [5 upvotes]: The nLab states that a semicartesian monoidal category equipped with natural transformations $\delta_x : x \to x \otimes x$ such that $\pi_1 \circ \delta_x = 1_x$ and $\pi_2 \circ \delta_x = 1_x$ (where $\pi_1$ and $\pi_2$ are the definable projections), is automatically cartesian. I'm looking for a proof of this statement. I managed to show that given $f_1 : a \to b_1$ and $f_2 : a \to b_2$, we can construct a map $(f_1, f_2) : a \to b_1 \otimes b_2$ as $(f_1, f_2) = (f_1 \otimes f_2) \circ \delta_a$, and this map projects appropriately: \begin{equation} \pi_i \circ (f_1 \otimes f_2) \circ \delta_a = f_i \circ \pi_i \circ \delta_a = f_i. \end{equation} Now I'm trying to show that $(f_1, f_2)$ is the unique map with this property. However, I seem to require that $(\pi_1 \otimes \pi_2) \circ \delta_{x \otimes y} = 1_{x \otimes y}$. Then if there is another function $h : a \to b_1 \otimes b_2$ such that $\pi_i \circ h = f_i$, we have \begin{equation} h = (\pi_1 \otimes \pi_2) \circ \delta_{b_1 \otimes b_2} \circ h = (\pi_1 \otimes \pi_2) \circ (h \otimes h) \circ \delta_a = (f_1 \otimes f_2) \circ \delta_a = (f_1, f_2). \end{equation} Is it correct that we need to postulate this "$\eta$-equality"? I have an intuition that it may somehow follow from the unit laws (since it already holds if either $x$ or $y$ is the terminal object), but I don't know how to prove that. REPLY [3 votes]: As pointed out by Dylan Wilson, the nLab postulates that $\delta$ be a monoidal natural transformation. This requires that the functor $Gx = x \otimes x$ be a (lax/oplax) monoidal functor. Let us assume that the category is symmetric monoidal, then $G$ is strong (in the sense of non-lax) monoidal. Then we have an isomorphism \begin{equation} \iota : G(x \otimes y) = (x \otimes y) \otimes (x \otimes y) \cong (x \otimes x) \otimes (y \otimes y) = Gx \otimes Gy \end{equation} Of course the symmetric structure provides multiple such morphisms, but naturality w.r.t. $x \to \top$ and $y \to \top$ implies that $\pi_i \circ \iota = G\pi_i = \pi_i \otimes \pi_i$, which reveals that $\iota$ swaps the middle two components, and hence $\pi_1 \otimes \pi_2 = (\pi_1 \otimes \pi_2) \circ \iota$. Then we have \begin{align*} (\pi_1 \otimes \pi_2) \circ \delta_{x \otimes y} &= (\pi_1 \otimes \pi_2) \circ \iota \circ \delta_{x \otimes y} \\ &= (\pi_1 \otimes \pi_2) \circ (\delta_x \otimes \delta_y) \\ &= (1 \otimes 1) = 1, \end{align*} proving the equation that was missing in the question.<|endoftext|> TITLE: Representations of $\text{SL}_n(q)$ of degree $q^{O(n)}$ QUESTION [10 upvotes]: Every representation of $A_n$ of degree $n^{O(1)}$ is contained in a $O(1)$ tensor power of the defining permutation representation. Is there an analogous result for classical groups of Lie type, say $G = \text{SL}_n(q)$ ($q$ bounded, $n$ large)? The defining action of $G$ on $F_q^n$ gives rise to a complex (permutation) representation $V$ of degree $q^n-1$, and the dual action likewise a representation $W$ also of degree $q^n-1$. Is it true that every representation of degree $q^{O(n)}$ is contained in $V^{\otimes O(1)} \otimes W^{\otimes O(1)}$? REPLY [5 votes]: A complete answer to my question is available from recent work on "character level" of Guralnick, Larsen, and Tiep: see [1] for linear and unitary groups and [2] for symplectic and orthogonal groups. First of all, a correction to my question. There is no need to include the representation $W$, because it is isomorphic to $V$. Indeed, both have character $g\mapsto q^{\dim \ker (g-1)}$ (the permutation representations are not permutation-isomorphic, however). GLT actually define (in the connected cases, $\mathrm{SL}_n(q)$, $\mathrm{SU}_n(q)$, $\mathrm{Sp}_n(q)$, $\Omega_n^{(\pm)}(q)$, for simplicity) the level of an irreducible character $\chi$ to be the smallest $\ell$ such that $\chi$ is contained in $\psi^\ell$, where $\psi$ is the permutation character $\tau : g \mapsto q^{\dim \ker(g-1)}$ in the linear and orthogonal cases, the reducible Weil character $\zeta : g \mapsto (-1)^n (-q)^{\dim \ker(g-1)}$ in the unitary case (note that $\zeta^2 = (q^2)^{\dim \ker (g-1)}$ is the permutation character in this case), $\tau + \zeta$ in the symplectic even-characteristic case, $\omega + \omega^*$ in the symplectic odd-characteristic case, where $\omega$ and $\omega^*$ are the reducible Weil characters of $\mathrm{Sp}_n(q)$ (such that $\{\tau, \zeta\} = \{\omega^2, \omega\omega^*\}$). For this notion of level, GLT prove exactly what I asked for. (Closer to the spirit of my question would be, say, "permutation level" $\ell_\text{perm}$, which would be as above with just $\psi = \tau$ in all cases. I don't know whether this is equivalent, i.e., whether $\zeta,\omega,\omega^*$ are contained in $\tau^{O(1)}$.) [1] Guralnick, Robert M.; Larsen, Michael; Tiep, Pham Huu, Character levels and character bounds, ZBL07158138. [2] Guralnick, Robert M.; Larsen, Michael; Tiep, Pham Huu, Character levels and character bounds. II.<|endoftext|> TITLE: Is there a Kähler manifold with no anti-holomophic involution? QUESTION [10 upvotes]: That is, is there a Kähler manifold $X$ on which there is no map $$ \tau:X\to X $$ such that $$ d\tau\circ I=-I\circ d\tau $$ and $$ \tau\circ \tau=\mathrm{Id}_X? $$ REPLY [8 votes]: There even exists a (compact) Kähler manifold with no anti-holomorphic self-diffeomorphism at all. Namely a suitable elliptic curve. Indeed, consider $\mathbf{C}/\Lambda$, for some lattice $\Lambda$. Let $f$ be an anti-holomorphic self-diffeomorphism. Up to compose with a translation, we can suppose that $f(0)=0$. Let $F$ be the unique lift $\mathbf{C}\to\mathbf{C}$ mapping $0$ to $0$. Then $F$ is a self-covering, and hence is a diffeomorphism, and commutes with $\Lambda$-translations. Write $G(z)=\overline{F(z)}$. Then $G$ is a holomorphic self-diffeomorphism of $\mathbf{C}$ and fixes $0$, hence $G(z)=\bar{q}z$ for some $q\in\mathbf{C}^*$. Since $G$ maps $\Lambda$ to $\overline{\Lambda}$ which has the same covolume, we have $|q|=1$. We have $F(z)=q\bar{z}$ for all $z$, and $F(\Lambda)=\Lambda$. So $\Lambda=q\overline{\Lambda}$. That is, $\Lambda$ is preserved by some reflection $z\mapsto q\bar{z}$. Next is it enough to see that "most" lattices are not preserved by any reflection. Indeed, we can choose coordinates so that $\Lambda$ has the basis $(1,z)$ with $z$ belonging to the usual "strip" ($|z|\ge 1$, $\frac12<\mathrm{Re}(z)\le \frac12$, $\mathrm{Im}(z)>0$). Then such a lattice is preserved by a reflection if and only if $\mathrm{Re}(z)\in\{0,1/2\}$. (The latter point was already observed in the previous answer using another language, but the easy argument could easily be explained.) (Indeed we see in these cases that there exists a anti-holomorphic self-diffeomorphism if and only if there is one that is in addition involutive, and furthermore preserves the standard flat Kähler structure.)<|endoftext|> TITLE: Fundamental group of punctured simply connected subset of $\mathbb{R}^2$ QUESTION [18 upvotes]: (This question is originally from Math.SE where it was suggested that I ask the question here) Let $S$ be a simply connected subset of $\mathbb{R}^2$ and let $x$ be an interior point of $S$, meaning that $B_r(x)\subseteq S$ for some $r>0$. Is it necessarily the case that $\pi_1(S\setminus\{x\})\cong\mathbb{Z}$? This is true if $S$ is open by the Riemann mapping theorem. Moishe Kohan's answer on Math.SE shows that the result holds if $S$ is compact. Let $B=B_r(x)$ and let $G=\pi_1(S\setminus\{x\})$. I can show that $G^{ab}\cong\mathbb{Z}$. Consider the commutative diagram of topological spaces $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c}B\setminus\{x\}&\ra{}&S\setminus\{x\}&\ra{}&\mathbb{R}^2\setminus\{x\}\\\da{}&&\da{}\\B&\ra{}&S\end{array} $$ Applying the fundamental group functor gives a commutative diagram of groups $$ \begin{array}{c}\mathbb{Z}&\ra{}&G&\ra{}&\mathbb{Z}\\\da{}&&\da{}\\1&\ra{}&1\end{array} $$ where the composition of the top two maps is the identity homomorphism on $\mathbb{Z}$. By the Seifert-van Kampen theorem, the square is a pushout diagram of groups, meaning that the normal closure of the image of $\mathbb{Z}\to G$ is all of $G$. In particular, if $A$ is an abelian group then any two homomorphisms $G\to A$ that agree on the image of $\mathbb{Z}\to G$ must agree on all of $G$. Another way to put this is that if we have two homomorphisms $G\to A$ such that the two compositions $\mathbb{Z}\to G\to A$ are equal then the two homomorphisms $G\to A$ are equal. I claim that the map $G\to\mathbb{Z}$ is an abelianization map. To see this, let $A$ be an abelian group and let $G\to A$ be a homomorphism. Now recall that the composition $\mathbb{Z}\to G\to\mathbb{Z}$ is the identity. Then the composition $\mathbb{Z}\to G\to\mathbb{Z}\to G\to A$ agrees with the composition $\mathbb{Z}\to G\to A$. By the remark at the end of the previous paragraph, this means that the composition $G\to\mathbb{Z}\to G\to A$ agrees with the map $G\to A$. In other words, the composition $\mathbb{Z}\to G\to A$ makes the abelianization diagram commute. To show uniqueness, let $\mathbb{Z}\to A$ be a map making the abelianization diagram commute. Then the composition $G\to\mathbb{Z}\to A$ agrees with the map $G\to A$. Then the composition $\mathbb{Z}\to G\to\mathbb{Z}\to A$ agrees with the composition $\mathbb{Z}\to G\to A$. Since the composition $\mathbb{Z}\to G\to\mathbb{Z}$ is the identity, this shows that the map $\mathbb{Z}\to A$ is given by the composition $\mathbb{Z}\to G\to A$. This shows that the map $G\to\mathbb{Z}$ is an abelianization map. REPLY [6 votes]: Here is a proof that $\pi_1(S\setminus\{x\})\cong\mathbb Z$. It is along similar lines to Jeremy Brazas' answer, and also makes use of Lemma 13 from the paper The fundamental groups of subsets of closed surfaces inject into their first shape groups. Lemma 13: Let any set $A\subseteq\mathbb R^2$, and map $\alpha\colon S^1\to A$ be given. Let $U$ be the unbounded connected component of $\mathbb R^2\setminus{\rm Im}(\alpha)$. If $\alpha\colon S^1\to A$ is null-homotopic, then so is $\alpha\colon S^1\to A\setminus U$. As we know, the embedding $B_r(x)\setminus\{x\}\to\mathbb R^2\setminus\{x\}$ gives an isomorphism of fundamental groups, which are isomorphic to $\mathbb Z$. Hence, the embeddings $B_r(x)\setminus\{x\}\to S\setminus\{x\}\to\mathbb R^2\setminus\{x\}$ gives homomorphisms $$ \mathbb Z\stackrel{f}{\to}\pi_1(S\setminus\{x\})\stackrel{g}{\to}\mathbb Z $$ with $gf=\iota$. To show that $f,g$ are isomorphisms, it is enough to show that $g$ has trivial kernel. So, suppose that $g([\gamma])=0$ for some closed curve $\gamma\colon S^1\to S\setminus\{x\}$. This means that $\gamma$ is null-homotopic in $\mathbb R^2\setminus\{x\}$. Letting $U$ be the unbounded connected component of $\mathbb R^2\setminus{\rm Im}(\gamma)$, Lemma 13 states that $\gamma$ is null-homotopic in $\mathbb R^2\setminus(U\cup\{x\})$. If we can show that $\mathbb R^2\setminus U\subseteq S$ then we are done. It is enough to show that each bounded connected component $V$ of $\mathbb R^2\setminus{\rm Im}(\gamma)$ is contained in $S$. From the proof of Lemma 13 in the linked paper, $V$ is simply connected and, by the Riemann mapping theorem, there is a conformal $h\colon\mathbb D\to V$ with boundary map $h\vert_{\partial\mathbb D}$ mapping to $\partial V\subseteq {\rm Im}(\gamma)\subseteq S$. By shrinking this down to a circle in $\mathbb D$, we see that it has winding number $1$ about every point in $V$, so cannot be null-homotopic in $S$ unless $V\subseteq S$. [I was hoping to complete this proof only using the statement of Lemma 13, but it is a bit trickier than expected, so had to refer to the proof of the lemma from the linked paper].<|endoftext|> TITLE: Global dimension 2 Nakayama algebras , 321-avoiding permutations and Fibonacci combinatorics QUESTION [5 upvotes]: A $n$-LNakayama algebra (for $n \geq 2$) is a list $[c_0,c_1,...,c_{n-1}]$ with $c_{n-1}=1$, $c_i \geq 2$ for $i \neq n-1$ and $c_i -1 \leq c_{i+1}$ for all $i$. They are in bijection with Dyck paths, where one associates to such a list simply the Dyck path with area sequence $[c_0,c_1,...,c_{n-1}]$ , see page 6 of https://arxiv.org/pdf/1811.05846.pdf. Thus $n$-LNakayama algebras are enumerated by the Catalan numbers $C_{n-1}$. Call an $n$-LNakayama algebra strong in case for all $i \neq n-1$ we have that $c_{i+1}=c_i -1$ or ($c_{i+1}>c_i -1$ and $c_{i+c_i}=c_{i+1}-c_i +1$) (Those are exactly the $n$-LNakayama algebras with global dimension at most 2). Strong $n$-LNakayama algebras are enumerated by $2^{n-2}$ and thus in bijection with integer compositions. (although Im not sure what the best bijection might be) The coarea sequence $[d_0,d_1,...,d_{n-1}]$ is the area sequence of the "opposite" Dyck path. Formally : $d_i = min \{ k | k \geq c_{i-k} \}$. Now define a statistic $f(A)$ on the strong $n$-LNakayama algebras $A$: $f(A):= \{ i \in \{0,...,n-1 \} | c_i < d_{i+c_i -1}$ and $d_{i+c_i-1}-c_i=d_{i-1} \}$ (those are the number of indecomposable projective modules with injective dimension one. One can show that this is also the number of indecomposable injective modules with projective dimension one, since Nakayama algebras are QF-3 algebras). I can prove that the strong $n$-LNakayama algebras with $f(A)=0$ are enumerated by the Fibonacci numbers (note that Fibonacci numbers are enumerated by integer compositions without parts equal to one, see https://oeis.org/A000045). Experimenting with the computer and findstat suggests that much more is true. Guess 1: Via the Billey-Jockusch-Stanley bijection $g$ from Dyck paths to 321-avoiding permutations we have that $f(A)= p(g(A))$, where for a permutation $\pi$ we have that $p(\pi)$ is the number of fixed points of $\pi$. See http://www.findstat.org/StatisticsDatabase/St001008 . Guess 2: There is a bijection $h$ from strong $n$-LNakayama algebras to integer compositions such that $f(A)= t(h(A))$, where $t(U)$ for an integer composition $U$ counts the number of parts equal to one. I arrived at guess 2 since the strong $n$-LNakayama algebras with $f(A)=1$ seem to be counted by https://oeis.org/A105423 (integer compositions with exactly one part equal to 1) and those with $f(A)=2$ are counted by https://oeis.org/A105423 (integer compositions with exactly two parts equal to one) (and of course the $f(A)=0$ case counted by the Fibonacci numbers as mentioned before). For guess 2 there is a cyclic analogue for CNakayama algebras (Nakayama algebras with a cyclic quiver). But for simplicity I do not state this for now (since they do not have such a nice combinatorial model as Dyck paths). For the cyclic case the algebras with $f(A)=0$ where counted by the cyclic analogue of the Fibonacci numbers : https://oeis.org/A032190. And those with $f(A)=1$ were counted surprisingly again by the usual Fibonacci numbers. Guess 1 is better tested I guess since it comes from findstat, while guess 2 is based just on small values of $f(A)$ (namely $f(A) \leq 2$) and oeis. Here is the statistic f for the 4-LNakayama algebras: [ [ [ 2, 2, 2, 1 ], 0 ] [ [ 3, 2, 2, 1 ], 1 ] [ [ 2, 3, 2, 1 ], 1 ] [ [ 3, 3, 2, 1 ], 0 ] [ [ 4, 3, 2, 1 ], 3 ] ] Here is the statistic f for the strong 5-LNakayama algebras: [ [ 2, 3, 2, 2, 1 ], 0 ] [ [ 4, 3, 2, 2, 1 ], 2 ] [ [ 3, 2, 3, 2, 1 ], 2 ] [ [ 4, 3, 3, 2, 1 ], 1 ] [ [ 2, 4, 3, 2, 1 ], 2 ] [ [ 3, 4, 3, 2, 1 ], 1 ] [ [ 4, 4, 3, 2, 1 ], 0 ] [ [ 5, 4, 3, 2, 1 ], 4 ] edit: Here is the statistic on strong $n$-LNakayama algebras for $n \leq 6$ where they are displayed as permutations (via the Billey-Jockusch-Stanley bijection). Has anyone seens those $2^{n-2}$ or can recognize them? n=2: [1] => 1 n=3: [2, 1] => 0 [1, 2] => 2 n=4: [1, 3, 2] => 1 [2, 1, 3] => 1 [3, 1, 2] => 0 [1, 2, 3] => 3 n=5: [2, 1, 4, 3] => 0 [1, 2, 4, 3] => 2 [1, 3, 2, 4] => 2 [1, 4, 2, 3] => 1 [2, 1, 3, 4] => 2 [3, 1, 2, 4] => 1 [4, 1, 2, 3] => 0 [1, 2, 3, 4] => 4 n=6: [2, 1, 4, 3, 5] => 1 [2, 1, 3, 5, 4] => 1 [2, 1, 5, 3, 4] => 0 [2, 1, 3, 4, 5] => 3 [1, 3, 2, 5, 4] => 1 [1, 3, 2, 4, 5] => 3 [3, 1, 2, 5, 4] => 0 [3, 1, 2, 4, 5] => 2 [1, 2, 4, 3, 5] => 3 [1, 4, 2, 3, 5] => 2 [4, 1, 2, 3, 5] => 1 [1, 2, 3, 5, 4] => 3 [1, 2, 5, 3, 4] => 2 [1, 5, 2, 3, 4] => 1 [5, 1, 2, 3, 4] => 0 [1, 2, 3, 4, 5] => 5 REPLY [4 votes]: It looks like the permutations are of the form $\pi_1 \oplus \pi_2 \oplus \cdots \oplus \pi_k$ (see https://en.wikipedia.org/wiki/Skew_and_direct_sums_of_permutations) where the $\pi_i$ are long-cycles of the form $(1,\ell,\ell-1,\ldots,2)$. These evidently correspond to compositions, and in the correspondence the number of fixed points is evidently the number of parts of size one (as you desired). They seem like a kind of variant of layered permutations.<|endoftext|> TITLE: What is the smallest rank for a noncommutative fusion ring? QUESTION [5 upvotes]: A fusion ring $\mathcal{F}$ (of rank $r$) is given by a finite set $B = \{b_1,b_2, \dots, b_r \}$ such that $b_i b_j = \sum_k n_{i,j}^k b_k$ with $n_{i,j}^k \in \mathbb{Z}_{\ge 0}$, satisfying axioms slightly augmenting the group axioms (see the details here). The fusion ring $\mathcal{F}$ is called noncommutative if $\exists i,j$ with $b_ib_j\neq b_jb_i$. Question: What is the smallest rank for a noncommutative fusion ring? We already know that this smallest rank is at most $6$ because the Grothendieck ring of the Haagerup fusion category $H_6$ is noncommutative and of rank $6$. Here are its fusion rules (coming from this paper): $$\begin{smallmatrix}1&0&0&0&0&0 \\\ 0&1&0&0&0&0 \\\ 0&0&1&0&0&0 \\\ 0&0&0&1&0&0 \\\ 0&0&0&0&1&0 \\\ 0&0&0&0&0&1\end{smallmatrix} , \ \begin{smallmatrix}0&1&0&0&0&0 \\\ 0&0&1&0&0&0 \\\ 1&0&0&0&0&0 \\\ 0&0&0&0&1&0 \\\ 0&0&0&0&0&1 \\\ 0&0&0&1&0&0\end{smallmatrix} , \ \begin{smallmatrix}0&0&1&0&0&0 \\\ 1&0&0&0&0&0 \\\ 0&1&0&0&0&0 \\\ 0&0&0&0&0&1 \\\ 0&0&0&1&0&0 \\\ 0&0&0&0&1&0\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&1&0&0 \\\ 0&0&0&0&0&1 \\\ 0&0&0&0&1&0 \\\ 1&0&0&1&1&1 \\\ 0&0&1&1&1&1 \\\ 0&1&0&1&1&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&1&0 \\\ 0&0&0&1&0&0 \\\ 0&0&0&0&0&1 \\\ 0&1&0&1&1&1 \\\ 1&0&0&1&1&1 \\\ 0&0&1&1&1&1\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&0&1 \\\ 0&0&0&0&1&0 \\\ 0&0&0&1&0&0 \\\ 0&0&1&1&1&1 \\\ 0&1&0&1&1&1 \\\ 1&0&0&1&1&1\end{smallmatrix}$$ We also know that this smallest rank is at least $5$, because $\mathbb{C}B$ admits a structure of finite dimensional von Neumann algebra of the form $\mathbb{C} \oplus A$ (where the first component is generated by $\sum_id(b_i)b_i$). By noncommutativity, if $A$ has the smallest possible dimension then $A = M_2(\mathbb{C})$. Then a noncommutative fusion ring is of rank at least $5$. So the question reformulates as follows: Reformulated question: Is there a noncommutative fusion ring of rank $5$? Investigation at rank $5$ First note that $ n_{j^*,i^*}^{k^*} = n_{i,j}^k $ because $b_{j^*}b_{i^*} = (b_i b_j)^* = \sum_k n_{i,j}^k b_{k^*}$, so if $i^* = i$ for all $i$ then the fusion ring is commutative. Thus there is $i$ such that $i^* \neq i$. We can assume that $2^* = 3$. Then there are two cases: (1) $4^* = 5$, (2) $4^*=4$ (and so $5^* = 5$). Note that (1) implies commutativity (see Appendix below). So we can assume (2), and then by Frobenius reciprocity, the fusion rules must be as follows (with $16$ parameters): $$\begin{smallmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{smallmatrix}, \ \begin{smallmatrix}0&1&0&0&0\\0&a_1&a_{11}&a_6&a_{10}\\1&a_1&a_1&a_2&a_3\\0&a_4&a_6&a_7&a_8\\0&a_5&a_{10}&a_9&a_{12}\end{smallmatrix} , \ \begin{smallmatrix}0&0&1&0&0\\1&a_1&a_1&a_4&a_5\\0&a_{11}&a_1&a_6&a_{10}\\0&a_6&a_2&a_7&a_9\\0&a_{10}&a_3&a_8&a_{12}\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&1&0\\0&a_2&a_6&a_7&a_9\\0&a_6&a_4&a_7&a_8\\1&a_7&a_7&a_{13}&a_{15}\\0&a_9&a_8&a_{15}&a_{16}\end{smallmatrix} , \ \begin{smallmatrix}0&0&0&0&1\\0&a_3&a_{10}&a_8&a_{12}\\0&a_{10}&a_5&a_9&a_{12}\\0&a_8&a_9&a_{15}&a_{16}\\1&a_{12}&a_{12}&a_{16}&a_{14}\end{smallmatrix}$$ such that $a_i \in \mathbb{Z}_{\ge 0}$, and $\sum_s n_{i,j}^sn_{s,k}^t = \sum_s n_{j,k}^sn_{i,s}^t$ (associativity). Note that by brute-force computation, there is no noncommutative example of multiplicity at most three in this case. Appendix: the proof that (1) implies commutativity. $$b_2 b_3 = b_1 + n_{2,3}^2b_2 + n_{2,3}^3b_3 + n_{2,3}^4b_4 + n_{2,3}^5b_5,$$ but $(b_2b_3)^* = b_3^* b_2^* = b_2b_3$, so $n_{2,3}^2 = n_{2,3}^3$ and $n_{2,3}^4=n_{2,3}^5$. It follows that $$b_2 b_3 = b_1 + n_{2,3}^2(b_2 + b_3) + n_{2,3}^4(b_4 + b_5).$$ Idem, $b_3 b_2 = b_1 + n_{3,2}^3(b_2 + b_3) + n_{3,2}^4(b_4 + b_5)$. By Frobenius reciprocity, $n_{2,3}^2 = n_{3,2}^3$, but $d(b_2b_3) = d(b_3b_2)$, so $n_{2,3}^4d(b_4 + b_5) = n_{3,2}^4d(b_4 + b_5)$. Then $n_{2,3}^4 = n_{3,2}^4$ and $b_2b_3 = b_3b_2$. Idem $b_4b_5 = b_5b_4$. Next by Frobenius reciprocity: $n_{2,4}^2 = n_{3,2}^4 = n_{2,3}^4 = n_{4,2}^2$ (the second equality comes from $b_2b_3 = b_3b_2$), $n_{2,4}^3 = n_{3,3}^4 = n_{4,2}^3$, $n_{2,4}^4 = n_{4,5}^2 = n_{5,4}^2 = n_{4,2}^4$ (the second equality comes from $b_4b_5 = b_5b_4$), $n_{2,4}^5 = n_{5,5}^2 = n_{4,2}^5$. It follows that $b_2b_4=b_4b_2$. Idem $b_2b_5=b_5b_2$, $b_3b_4=b_4b_3$ and $b_3b_5=b_5b_3$. Conclusion: the fusion ring is commutative in this case. REPLY [5 votes]: I think that a noncommutative fusion ring of rank 5 does not exist. Namely, let $a$ and $b$ be the formal codegrees (see https://arxiv.org/pdf/0810.3242.pdf) of such ring. Then $a$ and $b$ are positive (EDIT: and rational, see the explanation by Noah) integers satisfying $\frac1a+\frac2b=1$ (see Proposition 2.10 in https://arxiv.org/pdf/1309.4822.pdf). It is easy to see that this equation has no solutions with $a\ge 5$. This is impossible as the Frobenius-Perron dimension should be $\ge 5$.<|endoftext|> TITLE: Regular morphisms and formal power series QUESTION [6 upvotes]: Let $A$ be a local noetherian ring. When (besides when $A$ is excellent) do we have that $\operatorname{Spec}(A[[t]])\rightarrow \operatorname{Spec}(A[t])$ is regular? REPLY [9 votes]: As you probably are aware, a sufficient condition is that $A$ is a $G$-ring [Matsumura, Thm. 79]. On the other hand, the following result says that one must put some strong assumptions on $A$ to have $A[t] \to A[[t]]$ be regular. Theorem [Sharp, Thm. 2.9]. There exist a local domain $B$ of dimension either two or three such that setting $L = \operatorname{Frac}(B)$, the generic fiber $B[[t_1,\ldots,t_n]] \otimes_B L$ of the homomorphism $B \to B[[t_1,\ldots,t_n]]$ is not Cohen–Macaulay for every integer $n > 0$. The example is based on Ferrand and Raynaud's construction of a local domain $A$ such that the zero ideal in the completion $\hat{A}$ has an embedded prime [Ferrand–Raynaud, Prop. 3.3]. This ring $A$ therefore has non-Cohen–Macaulay formal fibers. Sharp's construction also gives a ring $B$ that also has non-Cohen–Macaulay formal fibers by [Ooishi, Cor. 36]. We can now consider the composition $$B \longrightarrow B[t] \longrightarrow B[[t]],$$ where $B$ is as in Sharp's theorem. The first map is a regular homomorphism. If the second map were also regular, then the composition would be regular. But Sharp's theorem says that this cannot be the case.<|endoftext|> TITLE: Is the Wikipedia depiction of the ergosphere of a Kerr black hole a Cassini oval? QUESTION [5 upvotes]: Disclaimer: this a cross post from MSE, where this question was asked on November 4th 2019 and has so far received no upvote, no comment and no answer whatsoever. Glancing at https://en.wikipedia.org/wiki/Rotating_black_hole (current revision) I thought that the frontier of the ergosphere appearing in the picture at the beginning of the considered article looks very much like a Cassini oval. Is it actually one? If yes, how can we prove it? REPLY [8 votes]: It's not a Cassini oval. To see this, recall that we're talking about the outer static limit of the black hole, whose (Boyer-Lindquist) $r$ coordinate in function of $\theta$ is given by $r = M + \sqrt{M^2 - a^2\,\cos^2\theta}$ (where, as usual, $M$ is the black hole mass and $a$ its angular momentum per unit mass). However, this $r$ coordinate is not what is being plotted in a slice, because it does not reduce to the Euclidean distance-to-the-center coordinate when $M\to 0$, but rather the $\sqrt{r^2 + a^2\,\sin^2\theta}$ coordinate, let's call it $R$, which does (this $R$ is not to be confused with $\rho := \sqrt{r^2 + a^2\,\cos^2\theta}$ which features prominently in the expressions of the Kerr metric). Now eliminating $r$ from the algebraic equations $(r-M)^2 = M^2 - a^2\,\cos^2\theta$ and $R^2 = r^2 + a^2\,\sin^2\theta$ gives the implicit polar equation $$ R^4 + 2(a^2\cos(2\theta) - 2M^2)R^2 + a^2(a^2\cos^2(2\theta) + 4M^2\sin^2(\theta)) = 0 $$ In contrast, a Cassini oval, if I believe Wikipedia, has the implicit polar equation $$ R^4 - 2u^2\cos(2\theta)\, R^2 + (u^4-v^4) = 0 $$ where $u,v$ (called $a,b$ in Wikipedia) are real parameters. Clearly the curve cannot be put into this form.<|endoftext|> TITLE: Questions involving primes $p\equiv1\pmod4$ QUESTION [6 upvotes]: As claimed by Fermat and proved by Euler, any prime $p\equiv1\pmod4$ can be written uniquely as $s_p^2+t_p^2$ with $s_p,t_p\in\{1,2,3,\ldots\}$, $2\nmid s_p$ and $2\mid t_p$. For any positive integer $n$, let us define $$S(n):=\sum_{p\le n\atop p\equiv1\pmod4}s_p \ \ \ \text{and}\ \ \ T(n)=\sum_{p\le n\atop p\equiv1\pmod4}t_p.$$ Via computation, I have found that $$S(10^9)=334976550299,\ \ T(10^9)=334979004134,\ \ \frac{S(10^9)}{T(10^9)}\approx 0.99999267.$$ This leads me to pose the following conjecture. Conjecture. We have $$\lim_{n\to+\infty}\frac{S(n)}{T(n)}=1.$$ QUESTION 1. Is the conjecture true? If true, how to prove it? I have another question. QUESTION 2. Is there a positive contant $c$ such that $$\lim_{n\to+\infty}\frac{\sum_{p\le n\atop p\equiv1\pmod4}s_t/t_p}{\sum_{p\le n\atop p\equiv1\pmod4}t_p/s_p}=c$$ holds? Concerning this question, I conjecture that $c$ exists and its value is probably $1$. I have found that $$\frac{\sum_{p\le 10^{11}\atop p\equiv1\pmod4}s_t/t_p}{\sum_{p\le 10^{11}\atop p\equiv1\pmod4}t_p/s_p}\approx 0.896.$$ Your comments are welcome! REPLY [2 votes]: For $p\equiv1\bmod4$, we write $$p=s_p^2+t_p^2=\sigma_p\overline{\sigma}_p,\ \ \ \sigma_p\in\mathbf{Z}[i].$$ Note that $\sigma_p$ and $2$ are coprime in $\mathbf{Z}[i].$ For each positive integer $k$, we would like to evaluate $$ S_k(x):=\sideset{}{'}\sum_{\substack{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x,\\ \sigma_p+\overline{\sigma}_p\equiv0\bmod4}}(\Re\sigma_p)^k =\sideset{}{'}\sum_{\substack{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x,\\ \sigma_p+\overline{\sigma}_p\equiv0\bmod4}}(\sqrt{p}\cos\theta_p)^k, $$ where $'$ yields the restriction that $\arg(\sigma_p)\in(0,\pi/2).$ On the other hand, $$\sigma_p+\overline{\sigma}_p\equiv0\bmod4\Leftrightarrow \sigma_p^2+p\equiv0\bmod4\Leftrightarrow \sigma_p^2+1\equiv0\bmod4.$$ The last congruence is viewed $\mathbf{Z}[i]$, so that $$S_k(x)=\frac{1}{4}\sideset{}{'}\sum_{\substack{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x,\\ \sigma_p^2+1\equiv0\bmod4}}(\sqrt{p}\cos\theta_p)^k.$$ Introducing multiplicative characters in $(\mathbf{Z}[i]/4\mathbf{Z}[i])^\times$, we may write $$ S_k(x)=\frac{1}{4}\frac{1}{\Phi(4)}\sum_{\chi\in\widehat{(\mathbf{Z}[i]/4\mathbf{Z}[i])^\times}}\sum_{z\in\mathbf{Z}[i],z^2+1\equiv0\bmod4}\overline{\chi}(z)\sideset{}{'}\sum_{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x}\chi(\sigma_p)(\sqrt{p}\cos\theta_p)^k,$$ where $\Phi(4)=|(\mathbf{Z}[i]/4\mathbf{Z}[i])^\times|=8.$ For each $\chi$, the sum $$ \sideset{}{'}\sum_{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x}\chi(\sigma_p)(\cos\theta_p)^k, $$ can be evaluated via Hecke's argument since \begin{align*} \chi(\sigma_p)e^{\ell i\theta_p}=\chi(\sigma_p)\Big(\frac{\sigma_p}{|\sigma_p|}\Big)^\ell \end{align*} gives a Hecke Grossencharacter at evaluated at $\sigma_p.$ Similarly, we can also consider \begin{align*} T_k(x):=\sideset{}{'}\sum_{\substack{\sigma_p\in\mathbf{Z}[i],\sigma_p\overline{\sigma}_p\leqslant x\\ \sigma_p-\overline{\sigma}_p\equiv0\bmod4}}(\Im\sigma_p)^k, \end{align*} which is related to the congruence $z^2-1\equiv0\bmod4.$ After a collection of serious arguments, we find the limit is equal to the ratio $|\mathcal{A}|/|\mathcal{B}|$ with $$\mathcal{A}=\{z\bmod4:z^2+1\equiv0\bmod4\},\ \ \ \mathcal{B}=\{z\bmod4:z^2-1\equiv0\bmod4\}.$$ In fact, \begin{align*} \mathcal{A}=\{i,3i,2+i,2+3i\bmod4\},\ \ \ \mathcal{B}=\{1,1+i,3,3+2i\bmod4\}. \end{align*} Hence $|\mathcal{A}|/|\mathcal{B}|=1,$ which proves the first conjecture.<|endoftext|> TITLE: Tamagawa numbers QUESTION [9 upvotes]: Let $K$ be a finite extension of $\mathbb{Q}_p$ with absolute Galois group $G_K$. Let $A$ be an abelian variety defined over $K$. The (geometric) Tamagawa number is defined as the order of the quotient $$c(A/K)=A(K)/A_0(K)$$ where $A_0(K)$ denotes the $K$-points of $A$ that are sent to the identity component of the special fiber of the Néron model of $A$ via the natural reduction map. On the other hand, following Fontaine and Perrin-Riou (in "Cohomologie galoisienne et valeurs de fonctions L", §4), starting with the $p$-adic Tate module $T_p=T_p(A)$, one may define the ($p$-adic) Tamagawa number ${\rm Tam}_\omega(T_p)$. The latter depends on a choice of a generator $\omega\in\det(t_{V_p})^*$, where $V_p=T_p\otimes\mathbb Q_p$ and $t_{V_p}=(B_{\rm dR}/B_{\rm dR}^+\otimes V_p)^{G_K}$. As suggested by the vocabulary, can one expect a direct relation between $c(A/K)$ and ${\rm Tam}_\omega(T_p)$ to hold? Is there a proof of this? The number ${\rm Tam}_\omega(T_p)$ is a power of $p$, so one might expect it to be comparable with the $p$-part of $c(A/K)$. However, one would need to have an almost cannonical choice for $\omega$. REPLY [10 votes]: Denote by $\Phi$ the quotient of $\mathcal A^\vee$, the special fiber of the smooth (but not necessarily proper) model of the dual abelian variety $A^\vee$, by the connected component of $0$ of $\mathcal A^\vee$. Then it is indeed true that the Tamagawa number $c_p$ as defined in $L$-functions and Tamagawa number of motives (Bloch, Kato) or in Autour des conjectures de Bloch-Kato (Fontaine, Perrin-Riou) is then equal up to a $p$-adic unit to $|\Phi[p^\infty]|$, but this is not quite obvious. The main step of the proofs are as follows: First, relate $\Phi$ to the formal group of $A^\vee$, and then through the $\log$ map to $\operatorname{Lie}_{\mathbb Z_p}A^{\vee}$ (which is also the tangent space of the formal group with values in $\mathbb Z_p$). Then notice that this induces isomorphisms between the determinant of $\operatorname{Lie}_{\mathbb Z_p}A^{\vee}$ and the determinant of the first Bloch-Kato cohomology group $H^1_f(\mathbb Q_p,T_pA)$ of the $p$-adic Tate module of $A$. This isomorphism involves finite corrective terms which come from the fact that the formal group is a subgroup with finite index of $H^1_f(\mathbb Q_p,T_pA)$ (and similarly for the image of the $\log$ with respect to $\operatorname{Lie}_{\mathbb Z_p}A^{\vee}$). Now the Tamagawa number $c_p$ arises as an alternate product of these corrective terms and of the Euler factor at $p$ (because the exponential map of Bloch-Kato coincide in this case with $\log$). Finally, relate this alternate product to $\Phi$. This involves the classical description of $\Phi$ in terms of the Neron model as well as a study of the reduction of $A^{\vee}$. Personally, I don't find any of these steps particularly easy, and the last one certainly isn't. I don't have a good reference to mention that would cover all of them. A good but hard reference for the hardest part is SGA7, Exposé IX. One thing that I would mention is that, as you say, the computations of the proof above involve in several steps the choice of a $\mathbb Z_p$-basis, but as this plays a role both in the normalization of maps with target $\operatorname{Lie}A^\vee$ as well as in the definition of $c_p$, the resulting computation does not depend on the choice of the basis (up to a $p$-adic unit, which plays no role in the Tamagawa number conjectures of Bloch-Kato).<|endoftext|> TITLE: What are the algebras for the ultrafilter monad on topological spaces? QUESTION [16 upvotes]: Motivation: Let $(X,\tau)$ be a topological space. Then the set $\beta X$ of ultrafilters on $X$ admits a natural topology (cf. Example 5.14 in Adámek and Sousa - D-ultrafilters and their monads), giving rise to a functor $\beta: \operatorname{Top} \to \operatorname{Top}$ which admits the structure of a monad. It turns out that the algebras for this monad, which I'll call "$\beta$-spaces", admit the following description (which one can alternatively take as a definition). Definition: A $\beta$-space consists of a topological space $(X,\tau)$ equipped with an additional topology $\tau^\xi$ on $X$ such that $(X, \tau^\xi)$ is compact Hausdorff; The topology $\tau^\xi$ refines the topology $\tau$; and For every $x \in X$ and every $\tau$-open neighborhood $U$ of $x$, there exists a $\tau$-open neighborhood $V$ of $x$ such that the $\tau^\xi$-closure of $V$ is contained in $U$. Notes: From (1) and (2) it follows that $(X,\tau)$ is compact. So if $(X,\tau)$ is additionally Hausdorff, then it admits a unique $\beta$-space structure, namely the one with $\tau^\xi = \tau$ (since continuous bijections of compact Hausdorff spaces are homeomorphisms). $(X,\tau)$ need not be Hausdorff—e.g., if $\tau$ is the indiscrete topology, then the topology $\tau^\xi$ can be an arbitrary compact Hausdorff topology. The compact Hausdorff topology $\tau^\xi$ traces back to Manes' theorem, which says that the algebras for the ultrafilter monad on $\operatorname{Set}$ rather than $\operatorname{Top}$ are precisely the compact Hausdorff spaces. Questions: Are there additional restrictions on the topology $(X,\tau)$ such that it admits a refinement $\tau^\xi$ satisfying (1), (2), (3) (i.e. constituting a $\beta$-space), beyond the fact, as noted, that $X$ must be compact? Do $\beta$-spaces already have some other name? Or at least, is condition (3) above, relating a topology $\tau$ to a refinement $\tau^\xi$, something which has a name? REPLY [2 votes]: $\DeclareMathOperator\cp{cp}$We will derive some additional necessary conditions from the following Observation: Let $\tau$ be a topology on $X$ and $\tau'$ a topology refining $\tau$. Suppose that $(X,\tau')$ is compact. Then any $\tau'$-closed set is $\tau$-compact. Indeed, it is compact in $\tau'$ because it is closed in a compact, and so it is compact also in $\tau$ because the identity $\tau' \to \tau$ is continuous. Consequences: Let $(X,\tau)$ be a topological space admitting a $\beta$-structure $\tau^\xi$. Then: $(X,\tau)$ is compact (as noted in the question). $(X,\tau)$ is locally compact (in the sense that for every $x \in X$ there is a local base of compact neighborhoods). This follows from condition (3) on a $\beta$-space and the Observation. $(X,\tau)$ is "c-separated": For every disjoint $C,D \subseteq X$ which are either closed or singletons, there exist compact $K,L \subseteq X$ such that $C \cap K = \emptyset$, $D \cap L = \emptyset$, and $K \cup L = X$. This follows from the fact that $(X,\tau^\xi)$ is Hausdorff, regular, and normal and the Observation. $(X,\tau)$ is "c-completely separated": Let $C,D \subseteq X$ be disjoint and either closed or singletons. Then there exists a (not necessarily continuous) function $f: X \to [0,1]$ such that $f^{-1}(0) = C$, $f^{-1}(1) = D$, and $f^{-1}([a,b])$ is compact for every $a \leq b$. This follows from the fact that $(X,\tau^\xi)$ has the corresponding separation property and the Observation. Note also that if the collection of sets with compact complement forms a topology, this this topology is the unique $\beta$-structure on $(X,\tau)$. But this is not necessarily the case.<|endoftext|> TITLE: About a generalization of the Borsuk-Ulam theorem QUESTION [7 upvotes]: I've come upon this MO post, and I was wondering if there is a way to generalize what is said in the answer at the very bottom. Specifically, we have that given continuous maps $f: \mathbb{S}^n \to \mathbb{S}^n$, and $g: \mathbb{S}^n \to \mathbb{R}^n$, if $f$ is an involution, then there is a point $x \in \mathbb{S}^n$ such that $g(x) = g(f(x))$, i.e. the value at $x$ is fixed under the transformation of the sphere by $f$. We know from the answer of the post that for a general $f$, there need not be such a fixed point. But now what if instead $f$ is such that there exists a $k \in \mathbb{N}$ where $f^k$ is the identity map, do we still get such a fixed point? Furthermore, do we know about other spaces besides the sphere that have this property when $f$ is an involution? REPLY [3 votes]: A first observation is that a map $f:S^n\to S^n$ with $f^k=\operatorname{Id}_{S^n}$ generates a topological action of the cyclic group $\mathbb{Z}/k$ on $S^n$. So we are talking about generalising from $\mathbb{Z}/2$ actions to $\mathbb{Z}/k$ actions. There are many generalisations of the Borsuk-Ulam theorem of this kind in the literature. Most of them concern estimates of the dimension of the set $$ A(g)=\{x\in S^n \mid g(x)=g(f^i(x))\mbox{ for }i=1,\ldots , k\}, $$ for various flavours of map $g:S^n\to M^m$. Interpreted this way, the answer to your first question appears to be negative in general, by a theorem of Munkholm in Munkholm, H. J., On the Borsuk-Ulam theorem for (Z_{p^ a}) actions on (S^{2n-1}) and maps (S^{2n-1} \to R^ m), Osaka J. Math. 7, 451-456 (1970). ZBL0211.55701. In particular, let $k=p^a\neq 9$ be an odd prime power with $a>1$, and let $\omega=\operatorname{exp}(2\pi i/p^a)$ be a primitive $p^a$-th root of unity. Let $f:S^{2n-1}\to S^{2n-1}$ be given in complex coordinates by $$ f(z_1,\ldots , z_n)=(\omega z_1,\ldots ,\omega z_n). $$ Then Munkholm's "mod $p^a$ Borsuk-Ulam anti-theorem" seems to imply that there exists a continuous map $g:S^{2n-1}\to \mathbb{R}^{2n-1}$ with $A(g)=\emptyset$. For a summary of positive results in the literature, you could do worse than Chapter 1 of Yuri Turygin's thesis at the Unviersity of Florida.<|endoftext|> TITLE: Approximation of homeomorphism by diffeomorphism QUESTION [16 upvotes]: Let $M$ be a smooth closed manifold. Let $f\colon M\to M$ be a homeomorphism. Does there exist a sequence of diffeomorphisms $f_i\colon M\to M$ which conveges to $f$ uniformly, i.e. in $C^0$-topology: $$\sup_{x\in M}dist(f(x),f_i(x))\to 0,\, \sup_{x\in M}dist(f^{-1}(x),f^{-1}_i(x))\to 0 \mbox{ as } i\to\infty,$$ where the distance $dist$ is taken with respect to a Riemannian metric on $M$? A reference would be helpful. REPLY [26 votes]: No. The space of homeomorphisms of a compact manifold is locally contractible: A. V. Černavskiı̆. Local contractibility of the group of homeomorphisms of a manifold. Mat. Sb. (N.S.), 79 (121):307–356, 1969. So if there were such a sequence then for large enough $i$ the diffeomorphism $f_i$ would be topologically isotopic to $f$. But there are homomorphisms which are not isotopic to diffeomorphisms. For example, let $\Sigma$ be a smooth homotopy $d$-sphere which does not have order 2 in the group $\Theta_d$ of such (e.g. Milnor's exotic 7-sphere). As $\Sigma$ is homeomorphic to $S^d$ (by the topological Poincare conjecture) it admits an orientation-reversing homeomorphism $f : \Sigma \to \Sigma$. But this $f$ cannot even be homotopic to a diffeomorphism, for if it were it would mean that $[\Sigma] = - [ \Sigma] \in \Theta_d$.<|endoftext|> TITLE: Centralizer of a cyclic subgroup within the group algebra $\mathbb{C} S_N$ of the symmetric group QUESTION [6 upvotes]: Let us take the group algebra $\mathbb{C} S_N$ and the subgroup $H=Z_N$ generated by the element $\sigma=(123\dots N)$, which is a cyclic shift. What is the structure of the centralizer of $H$ within $\mathbb{C} S_N$? If we are looking at the Lie algebra $\mathcal{L}(\mathbb{C} S_N)$ obtained from the usual commutator of two elements, then what is the centralizer of the commutative sub-algebra $\mathbb{C} H$? My motivation for this comes from physics, where the special ordering $1,2,3,\dots N$ has a concrete physical meaning, for example actual atoms are placed next to each other. In this case those group algebra elements that commute with $\sigma$ are called ``translationally invariant'' permutation operators, under periodic boundary conditions. Ultimately I would like to know what is the maximal number of mutually commuting operators within $\mathbb{C} S_N$ that commute with $\mathbb{C} H$, and how to find explicit bases for them. I wanted to approach this through the structure of the Lie algebra $\mathcal{L}(\mathbb{C} S_N)$, decomposition into simple Lie algebras, etc. I understand that the irreducible subspaces in $\mathbb{C} S_N$ correspond to the Young symmetrizers. But it is not so clear what to do from here, how to add the action of the concrete element $\sigma$ into this. REPLY [2 votes]: $\newcommand{\IC}{\mathbb{C}}$ Let $V_\lambda$ be the irreducible $S_N$-module (Specht module). The representations $\rho_\lambda: \IC S_N \to \operatorname{End}_\IC(V_\lambda)$ give us an isomorphism of algebras $\IC[S_N] \to \prod_{\lambda \vdash N} \operatorname{End}_\IC(V_\lambda)$. Since we work with $\IC$, the endomorphisms of finite order are always diagonalisable. In particular we can decompose $V_\lambda = \bigoplus_{k=0}^{N-1} V_{\lambda,k}$ such that $\sigma$ acts as multiplication by $\exp(\frac{2\pi i}{N}k)$ on $V_{\lambda,k}$. The centraliser of $\sigma$ is therefore exactly equal to $\prod_{\lambda\vdash N, 0\leq k TITLE: Expressing the Riemann Zeta function in terms of GCD and LCM QUESTION [27 upvotes]: Is the following claim true: Let $\zeta(s)$ be the Riemann zeta function. I observed that as for large $n$, as $s$ increased, $$ \frac{1}{n}\sum_{k = 1}^n\sum_{i = 1}^{k} \bigg(\frac{\gcd(k,i)}{\text{lcm}(k,i)}\bigg)^s \approx \zeta(s+1) $$ or equivalently $$ \frac{1}{n}\sum_{k = 1}^n\sum_{i = 1}^{k} \bigg(\frac{\gcd(k,i)^2}{ki}\bigg)^s \approx \zeta(s+1) $$ A few values of $s$, LHS and the RHS are given below $$(3,1.221,1.202)$$ $$(4,1.084,1.0823)$$ $$(5,1.0372,1.0369)$$ $$(6,1.01737,1.01734)$$ $$(7,1.00835,1.00834)$$ $$(9,1.00494,1.00494)$$ $$(19,1.0000009539,1.0000009539)$$ Note: This question was posted in MSE. It but did not have the right answer. REPLY [2 votes]: Your summand is symmetric with respect to $k$ and $i$: $$f(n,s) = \frac{1}{n}\sum_{k = 1}^n \sum_{i = 1}^{k} \bigg(\frac{\gcd(k,i)}{\operatorname{lcm}(k,i)}\bigg)^s$$ We can sum along skew diagonals to evaluate the sum. That is, we can convert $(k,i)$ to polar form $(\sqrt{k^2 + i^2}, tan^{-1}\frac{i}{k})$. By symmetry of the summand, the rays from the origin have the same value. That is, when $\theta = tan^{-1}\frac{i}{k}$, $i = k \tan\theta$. We can vary $\theta$ from $[0, \frac{\pi}{4}]$. The $\gcd(j,j\tan \theta)$ is independent of $j$ but depends on $\theta$, we can therefore use $n$: $$\frac{1}{n}\int_{0}^{\frac{\pi}{4}} \sum_{j=0}^{n} \left[\frac{\gcd(j,j\tan\theta)}{\operatorname{lcm}(j,j\tan\theta)}\right]^{s} d\theta = \int_{0}^{\frac{\pi}{4}} \left[\frac{\gcd(n,n\tan\theta)}{\operatorname{lcm}(n,n\tan\theta)}\right]^{s} d\theta = \sum_{k=0}^{n} \frac{1}{k^{s+1}} \rightarrow \zeta(s+1)$$ The polar integral(The integral is not continuous on $\theta$ but over the irrational($\tan\theta = \frac{i}{k}$ implies $\theta$ is irrational) values that correspond to the rays) to the sum is calculated easily enough because the values along each ray is constant w.r.t to $j$. Every ray corresponds to one of the values of $\frac{1}{k^s}$ but they have a weight of $\frac{1}{k}$. E.g., the ray that accumulates $1$ has $\theta = \tan^{-1}(1) = \frac{\pi}{4}$, for $\frac{1}{2^s}$ it is $\theta = \tan^{-1}(2)$ but it has $\frac{1}{2}$ the density of $1$. Similarly for all the others. If you are having trouble following this, simply look at $\frac{\operatorname{lcm}(k,i)}{\gcd(k,i)}$ in "polar" form, I'll make it easy for you(the text format obscures the patterns but they are there): \begin{matrix} \color{green}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \\ \color{blue}2 & \color{green}1 & 6 & 2 & 10 & 3 & 14 & 4 & 18 & 5 & \\ \color{red}3 & 6 & \color{green}1 & 12 & 15 & 2 & 21 & 24 & 3 & 30 & \\ 4 & \color{blue}2 & 12 & \color{green}1 & 20 & 6 & 28 & 2 & 36 & 10 & \\ 5 & 10 & 15 & 20 & \color{green}1 & 30 & 35 & 40 & 45 & 2 & \\ 6 & \color{red}3 & \color{blue}2 & 6 & 30 & \color{green}1 & 42 & 12 & 6 & 15 & \\ 7 & 14 & 21 & 28 & 35 & 42 & \color{green}1 & 56 & 63 & 70 & \\ 8 & 4 & 24 & \color{blue}2 & 40 & 12 & 56 & \color{green}1 & 72 & 20 & \\ 9 & 18 & \color{red}3 & 36 & 45 & 6 & 63 & 72 & \color{green}1 & 90 & \\ 10 & 5 & 30 & 10 & \color{blue}2 & 15 & 70 & 20 & 90 & \color{green}1 & \\ \end{matrix} If you look you can see $k$th ray which has the constant value $\frac{\color{green}1}{k^s}$(displayed as just $k$ in the table) but they repeat at a rate of $\frac{\color{green}1}{k}$ along the ray. Alternatively, if write the table in polar coordinates(we are rotating coordinate space 45 degree's) we get \begin{matrix} \color{green}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ \color{green}1 & \color{blue}2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ \color{green}1 & 0 & \color{red}3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ \color{green}1 & \color{blue}2 & 0 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & \\ \color{green}1 & 0 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0 & \\ \color{green}1 & \color{blue}2 & \color{red}3 & 0 & 0 & 6 & 0 & 0 & 0 & 0 & \\ \color{green}1 & 0 & 0 & 0 & 0 & 0 & 7 & 0 & 0 & 0 & \\ \color{green}1 & \color{blue}2 & 0 & 4 & 0 & 0 & 0 & 8 & 0 & 0 & \\ \color{green}1 & 0 & \color{red}3 & 0 & 0 & 0 & 0 & 0 & 9 & 0 & \\ \color{green}1 & \color{blue}2 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & \color{green}10 & \\ \end{matrix} Where now the $k$th column is the "$k$th" ray. I.e., the first column in the above table corresponds to the diagonals/rays in the table above it.<|endoftext|> TITLE: Equivariant cohomology of $\text{Diff}S^1/ S^1$ and Virasoro QUESTION [8 upvotes]: Consider $$\mathcal{M}\ =\ \text{Diff}S^1/S^1$$ which is a contractible complex manifold with an action of $\text{Diff}S^1$ by translations. It is claimed in page 358 of [1] that $\mathcal{M}$ has single-dimensional equivariant cohomology $H^2_{\text{Diff}S^1}(\mathcal{M})$, which is related to the central charge in the Virasoro algebra. Question: is $H^*_{\text{Diff}S^1}(\mathcal{M})$ known in general, and what relation does it have to the Virasoro algebra? I think the first part might be easy because (I think) the stack $\mathcal{M}/\text{Diff}S^1$ is homotopic to $B\text{Diff}S^1$, which in turn is homotopic to $B \text{U}(1)$ which has cohomology $\mathbf{Z}[x]$. [1] M. Bowick, S. Rajeev: The holomorphic geometry of closed bosonic string theory and $\text{Diff}S^1/S^1$. Edit: I should maybe explain what [1] says. It says that the two-form $\text{Ric}$ (Ricci curvature) satisfies $$\text{Ric}(L_m,L_n)\ =\ \left(\frac{-26}{12}m^3+\frac{1}{6}m\right) \delta_{-m,n}$$ where $L_n$ are the vector fields on $\mathcal{M}$ generating the action of $\text{Diff}S^1$. It then goes on to make the relationship between the generator of $H^2$ and the central charge more precise (I've not read the paper closely but I don't think it contains an answer to my question). REPLY [5 votes]: For any group $G$ and subgroup $K$, there is an isomorphism $$ H^*_G(G/K)= H^*_K(pt) $$ So the cohomology in question is $H^*_{S^1}(pt)=\mathbb Z[x]$, and it does not seem to be very related to the Virasoro algebra. (Here, the generator $x\in H^2_{S^1}(pt)$ is in degree two.) If I remember correctly, the Lie algebra cohomology of the Witt algebra (with coefficients in the trivial module $\mathbb C$) is a polynomial algebra generated by a class in degree $2$ tensor an exterior algebra generated by a class in degree $3$. From this, I think that one can deduce that the Lie algebra cohomology of the Virasoro algebra (= the universal central extension of the Witt algebra) is an exterior algebra generated by a class in degree $3$.<|endoftext|> TITLE: Fundamental Theorem of Algebra, via algebra QUESTION [6 upvotes]: I know there are already a couple of questions on this on the site, but I haven't seen an answer to this particular form... We know, from the Fundamental Theorem of Algebra, that the complex algebraic numbers contain a unique maximal ordered subfield, namely the real algebraic numbers, and the complex algebraic numbers are obtained by adjoining a square root of $-1$ to the real algebraic numbers. This is a purely algebraic statement (I think!) and one could reasonably ask for a purely algebraic proof. Is there such a proof? Or is there a barrier to such a proof (such as a model of ZF where the statement is false)? I'm aware of the standard (purely algebraic, or at least I'm happy to call it so) Zorn's Lemma proof that the complex algebraic numbers exist, since any field $K$ has an algebraic closure and this is unique up to isomorphism. However the 'up to isomorphism' here includes a great many isomorphisms of the complex algebraic numbers that fix the rationals. I think a more or less equivalent question to what I am asking is: from this construction, is there a good algebraic way to pick out the privileged automorphism $i\to-i$ and put an order on the subfield this automorphism fixes? There do exist proofs of the usual Fundamental Theorem of Algebra which use a `minimal' amount of analysis, such as https://arxiv.org/abs/1504.05609 (due to Piotr Błaszczyk) which doesn't really use more analysis than required to define the reals. However even this is more than (I think!) I am asking for. REPLY [10 votes]: Despite the extensive discussion in the comments, I'm not sure I completely understand the question, but I think the following fact may settle the question in the negative. Work of Läuchli implies that there is a model of ZF in which $\mathbb Q$ has an algebraic closure with no real-closed subfield. For more details, see for example Wilfrid Hodges, "Läuchli's algebraic closure of $Q$," Math. Proc. Cambridge Philos. Soc. 79 (1976), 289–297. I think your desired conclusion requires every algebraic closure of $\mathbb Q$ to have a real-closed subfield. If you agree, then the above result shows that you at least need some weak form of AC, which arguably is "not algebraic." But this depends on what you mean by "algebraic"; maybe a weak form of choice doesn't automatically disqualify a proof from being "algebraic"?<|endoftext|> TITLE: Do there exist acyclic simple groups of arbitrarily large cardinality? QUESTION [12 upvotes]: Recall that a group $G$ is acyclic if its group homology vanishes: $H_\ast(G; \mathbb Z) = 0$. Equivalently, $G$ is acyclic iff the space $BG$ is acyclic, i.e. $\tilde H_\ast(BG;\mathbb Z) = 0$. In order to tie up a loose end over at this question, I wonder Questions: Do there exists arbitrarily large simple acyclic groups? More generally, do there exist arbitrarily large simple groups $G$ such that there exists an acyclic space $X(G)$ with $\pi_1(X(G)) = G$? Do there exist arbitrarily large simple groups $G$ with $H_2(G; \mathbb Z) = 0$ -- or equivalently (I think) for which there are no nontrivial central extensions? Heck, what is one example of a simple nonabelian group $G$ with $H_2(G;\mathbb Z) = 0$? (2) is all I really need, for which (3) will suffice (see below); (1) is just a natural strengthening. Notes: There is a proper class of simple groups; e.g. the alternating group on any set is simple (though not acyclic). There are also acyclic spaces with arbitrarily large fundamental group, cf. Kan-Thurston, but the constructions I've seen don't produce spaces with simple fundamental group. In the comments at the above-linked question, Tom Goodwillie points out that a positive answer to (3) implies a positive answer to (2) by taking $X(G)$ to be the fiber of $BG \to BG^+$. I've included the "model theory" and "logic" tags mostly because I suspect maybe the people who know the most about very large simple groups might just be logicians. But if these tags seem inappropriate, I wouldn't object too strongly to removing them. REPLY [6 votes]: Here are some explicit examples. Let $\alpha$ be a cardinal $\ge\aleph_1$ and $X$ a set of cardinal $\alpha$ (we can choose $X=\alpha$). Let $G_\alpha=S_\alpha/D_\alpha$, where $S_\alpha$ (resp.\ $D_\alpha$) is the group of permutations of $\alpha$ whose support has cardinal $\le\aleph_1$ (resp. $<\aleph_1$). This is a simple group (particular case of a result of Baer). Claim: $G_\alpha$ is acyclic. Indeed, in a paper of P. de la Harpe and D. McDuff (CMH 1983), one has the definition (given below) of a "flabby" group, with the lemma, attributed to Wagoner "every flabby group is acyclic". I claim: $G_\alpha$ is flabby for every $\alpha\ge\aleph_2$. Hence this is a simple acyclic group (of cardinal $\ge\alpha$, namely the same as the set of subsets of $\alpha$ of cardinal $\le\aleph_1$). I start with the definition: $G$ is flabby if there exist homomorphisms: $\sqcup:G\times G\to G$ ("concatenation") and $\tau:G\to G$ ("countable repetition") satisfying: for every finite subset $F\subset G$, there exist $u,v,w\in G$ such that $g\sqcup 1=ugu^{-1}$ and $1\sqcup g=vgv^{-1}$, and $g\sqcup \tau(g)=w\tau(g)w^{-1}$ for every $g\in F$. Indeed, let $s$ be a bijection $X\to X\times\omega$; think of $X\times\{n\}$ as the $n$-th copy of $\alpha$. Define $g\sqcup h$ as "$g$ on the $0$-th copy, $h$ on the $1$-st copy, and identity on other copies, and $\tau(g)$ as "$g$ on each copy". Note that $\tau$ is well-defined (if we were modding out the finitely supported subgroup, this would fail). Now fix $F$ finite ($F$ of cardinal $<\alpha$ would also work); the union $X_F$ of supports of all $g\in F$ has cardinal $\alpha$. Extend the inclusion $X_F\to X_F\times\{0\}$ to a bijection $U:X\to X\times\omega$ and define $u=s^{-1}\circ U$. Then it satisfies the required equality. The other two conjugacy are obtained similarly. Notes: let $S(\alpha,\beta)$ be the group of permutations of $\alpha$ with support of cardinal $<\beta$ (it is understood that $\beta$ is infinite or $1$). Noyte that $G_\alpha=S(\alpha,\aleph_2)/S(\alpha,\aleph_1)$. The argument works without change to prove that for all cardinals $\alpha,\beta,\gamma$, the group $S(\alpha,\beta)/S(\alpha,\gamma)$ is flabby, acyclic if $\beta\le\alpha$ and $\gamma$ has uncountable cofinality. Probably the conclusion that it is acyclic holds for $\beta=\alpha^+$ (for $\gamma=1$ this is done in Harpe-McDuff).<|endoftext|> TITLE: Is the tensor product of compactly generated Hausdorff abelian groups again Hausdorff? QUESTION [5 upvotes]: Consider the tensor product $G \otimes_{\mathbb{Z}} H$ of two abelian groups $G$ and $H$. If $G$ and $H$ are topological groups, we can give $G \otimes_{\mathbb{Z}} H$ a topology as follows. For any $k$, consider the function. \begin{align} \phi_k: G^k \times H^k = G \times \dots \times G \times H \times \dots \times H &\to G \otimes_{\mathbb{Z}} H, \\ (g_1, \dots, g_k, h_1, \dots, h_k) &\mapsto \sum_{i=1}^k g_i \otimes h_i. \end{align} We topologize $G \otimes_{\mathbb{Z}} H$ using the finest topology making all functions $\phi_k$ continuous. It follows that $G \otimes_{\mathbb{Z}} H$ has the universal property with respect to all continuous bilinear maps $G \times H \to G'$ into a topological group $G'$. The question: If we assume that $G$ and $H$ are compactly generated Hausdorff topological spaces, is $G \otimes_{\mathbb{Z}} H$ again compactly generated Hausdorff? (It is automatically compactly generated, being a quotient of the compactly generated space $\bigsqcup_{k=1}^\infty G^k \times H^k$. The problem is Hausdorffness.) If the above is not true, we could replace $G \otimes_{\mathbb{Z}} H$ by its Hausdorff quotient. Is it true that the resulting functor $(G,H)\mapsto {\rm Haus}(G \otimes_{\mathbb{Z}} H)$ is universal with respect to bilinear maps $G \times H \to G'$ into a compactly generated Hausdorff topological abelian group $G'$? In particular, is this functor naturally associative? REPLY [3 votes]: This is not true. Our example will be $G = H = S^1$ with its standard topology. Let $\mu \subset S^1$ be the subgroup of roots of unity. The group $S^1$ is divisible and $\mu$ consists entirely of torsion, and so $$ \mu \otimes S^1 = 0. $$ This lets us conclude that the natural map of groups $$ S^1 \otimes S^1 \to (S^1/\mu) \otimes (S^1/\mu) $$ is an isomorphism. (Here $S^1/\mu$ is the group theoretic quotient, not the set-theoretic one.) When we bring topologies in, the quotient group $S^1/\mu$ has a natural quotient topology: this topology is the indiscrete topology because $\mu$ is dense in $S^1$. Similarly, the quotient map $S^1 \times S^1 \to (S^1/\mu) \times (S^1/\mu)$ makes the target indiscrete, and so in this case the product of the quotient topologies is the quotient of the product topologies. The universal property of the topologized tensor product is that a topological group homomorphism $S^1 \otimes S^1 \to G$ is the same as a continuous map $S^1 \times S^1 \to G$ that is a group homomorphism in each variable; this is the same as a continuous map $S^1/\mu \times S^1/\mu \to G$ that is a group homomorphism in each variable and which pulls back to a continuous map on $S^1 \times S^1$; because of the above quotient topology property, that's the same as a map of topological groups $(S^1/\mu) \otimes (S^1/\mu) \to G$. So $S^1 \otimes S^1$ coincides with $(S^1/\mu) \otimes (S^1/\mu)$ as a topological group. We now have to show that $S^1/\mu \otimes S^1/\mu$ is not Hausdorff. It's not immediate that it has the indiscrete topology and so we need to be a little careful, but the map $x \mapsto x \otimes \pi$ does determine a continuous 1-to-1 map from $S^1/\mu$ into the tensor product, and so the target can't possibly be Hausdorff.<|endoftext|> TITLE: Riemann-Hurwitz for real maps QUESTION [6 upvotes]: Let $S$ be a (compact, connected) Riemann surface of genus $g$ and $f: S\to \mathbb CP^1$ be a degree $d$ meromorphic function. Then the Riemann-Hurwitz formula tells us that the number of ramifications of $f$ counted with multiplicity is $2(d-1)+2g-2$. Suppose we consider instead a map $\varphi: S\to \mathbb CP^1$ of degree $d$ that is smooth, but not necessarily holomorphic. It will have singularities, like folds, etc. Question. Can one express the number $2(d-1)+2g-2$ as a sum of contributions, involving various types of singularities of the map $\varphi$ (say, if $\varphi$ is analytic)? REPLY [4 votes]: This is not possible, just using the singularities and the homological degree of the map. Let $f_1$ be a generic degree $3$ holomorphic map $\mathbb P^1 \to \mathbb P^1$. It has four quadratic singularities. Let $E$ be an elliptic curve, and $f_2$ a degree $2$ map $E \to \mathbb P^1$, with four simple ramification points singularities. Let $f_1'$ and $f_2'$ be $f_1$ and $f_2$ with the orientations of the source reversed. Then we can glue $f_1$ to $f_1'$, and $f_2$ to $f_2'$, by cutting out a disc from each component, lying over the same disc of $\mathbb P^1$, and gluing the holes together, introducing orientation-reversing fold singularities. These glued maps are from Riemann surfaces of genus $0$ and $2$ respectively, but have the same singularities, being four positive-orientation simple ramification points, four negative-orientation simple ramification points, and a loop of orientation-reversing folds, and the same degree, $0$.<|endoftext|> TITLE: Is the "homotopy category" functor well-defined? QUESTION [5 upvotes]: $$\def\Cat{\mathbf{Cat}} \def\qCat{\mathbf{qCat}} \def\Catinf{{\mathcal{C}at_\infty}} \def\Catone{{\mathcal{C}at_1}} \def\cC{\mathcal{C}}$$ Let $\Cat$ be the 1-category of small categories, $\qCat$ be the 1-category of small quasi-categories, $\Catinf$ be the $\infty$-category of small $\infty$-categories, and $\Catone$ be the $\infty$-category of small $1$-categories. A typical construction of the "homotopy category" operation on $\infty$-categories looks starts off like the following construction on $\qCat$: Take the objects of $\mathrm{h}(X)$ to be $X_0$ Define the hom-sets to be ... One then shows that this defines a functor $\qCat \to \Cat$ that preserves equivalences, and thus induces a functor $\Catinf \to \Catone$. However, this does not obviously induce a $\Cat$-valued functor, since the choice of objects was not made in an equivalence-preserving way. So, I ask Question: Is there a "homotopy category" functor $\Catinf \to \Cat$? To better emphasize the problem, consider another approach. the inclusion $\Catone \to \Catinf$ is an accessible, limit preserving functor between presentable $\infty$-categories, and so it has a left adjoint which is the aforementioned functor. However, $\Cat \to \Catinf$ does not preserve limits, so this method does not work to produce a functor $\Catinf \to \Cat$. REPLY [9 votes]: So by this standard, I don't think that the identity functor $Cat_1 \to Cat_1$ satisfies your desired constraints. But perhaps I should elaborate. What you are describing might be the following. If $W$ is the class of equivalences in $Cat_\infty$, then you have shown that "taking the homotopy category" does not define a functor from $W^{-1} Cat_\infty$, the 1-category obtained by inverting equivalences, to the $1$-category $Cat_1$ of categories and functors: if it did, then it would take equivalences of quasicategories to isomorphisms of categories. The category $Cat_1$ embeds in $Cat_\infty$; a functor becomes an equivalence in $Cat_\infty$ if and only if it was an equivalence in $Cat_1$; the "homotopy category" functor is basically the identity on $Cat_1$; and we also have the result that the identity doesn't extend to a functor from the 1-category $V^{-1} Cat_1$, obtained by inverting equivalences, to the 1-category $Cat_1$. In a sense, though, these 1-category localizations are ignoring higher structure. Namely, $V^{-1} Cat_1$ is the 1-category of categories and isomorphism classes of functors; similarly $W^{-1} Cat_\infty$ is the 1-category of quasicategories and equivalence classes of functors. We do have a functor $W^{-1} Cat_\infty \to V^{-1} Cat_1$: equivalent quasicategories have equivalent homotopy categories. This is ... possibly? ... fixable. If we have access to some kind of global choice operator for proper classes, then we can construct a functor $sk: Cat_1 \to Cat_1$ that takes a category $C$ and sends it to a skeleton $sk(C) \subset C$, and construct a natural equivalence from $sk$ to the identity functor. (To do this, we need to choose one representative for every isomorphism class of object in every category $C$ simultaneously.) If we do that, then maybe we can compose the "homotopy category" functor with $sk$ and obtain a functor that does factor through $W^{-1} Cat_\infty$, because $sk$ factors through $V^{-1} Cat_1$. But I'm concerned that this is trying to provide a section of $Cat_1 \to V^{-1} Cat_1$ and I don't think that exists. However, there are reasons why one might object to this. The idea that every category should be replaced by its skeleton, so that equivalent categories are isomorphic, requires hard set-theoretic theorems or is impossible. It is also, in some sense, solving a fake problem. We already have a well-established foundation for how to deal with equivalences of categories. Except in special cases, we are often not as interested in isomorphisms of categories. $Cat_1$, as a $1$-category, doesn't necessarily reflect our interests. I would instead contend that the "homotopy" category functor $Cat_\infty \to Cat_1$ should at least reflect the 2-categorical nature of $Cat_1$. It takes quasicategories to categories and functors to functors, but it takes natural equivalences of functors to natural isomorphisms of functors. This means that equivalent quasicategories have equivalent homotopy categories, rather than isomorphic ones, and it seems likely that we shouldn't expect more.<|endoftext|> TITLE: Derived invariant for Gorenstein algebras? QUESTION [6 upvotes]: Let $A$ be a finite dimensional algebra with simple modules $S_i$ and projective indecomposable modules $P_l$ and global dimension $g< \infty$ (and $n$ is the number of simple modules). I noted that the number $\sum\limits_{l=1}^{n}{\sum\limits_{i=0}^{g}{(-1)^i Ext_A^i(S_l,P_l)}}$ is a derived invariant of the algebra. In case I made no mistake, the proof goes by showing that this is the negative trace of the transpose of the inverse of the Coxeter transformation of the algbra, which is a derived invariant. One can also express the other coefficients of the Coxeter polynomial like this. Did this appear somewhere before? Question: Is this still true when $A$ is just assumed to be Gorenstein? (that is the injective dimension of $A$ is finite on both sides). Restricted to selfinjective algebras this would be implies by a positive answer to the question whether the Nakayama permutation is a derived invariant, which I forgot whether it is true. REPLY [5 votes]: Doesn't basically the same proof work? Let $A$ be a Gorenstein finite dimensional algebra over an algebraically closed field. Let $K^b(P_A)$ be the subcategory of the bounded derived category $D^b(\text{mod-}A)$ consisting of perfect objects (i.e., bounded complexes of finitely generated projectives). Since $A$ is Gorenstein, this is also the subcategory consisting of objects isomorphic to bounded complexes of injectives. The inverse derived Nakayama functor $\mathbf{R}\text{Hom}_A(DA,-)$ induces a self-equivalence of $K^b(P_A)$, and hence induces an endomorphism of the Grothendieck group $K_0\left(K^b(P_A)\right)$, which has a basis given by the classes of the indecomposable injectives, and its matrix with respect to this basis has entries $\sum_i(-1)^i\dim\text{Ext}^i(S_m,P_l)$, the coefficient of $[I_m]$ when $[P_l]$ is written in terms of the basis of indecomposable injectives. So the trace of this endomorphism is precisely the quantity that you are considering. This is derived invariant, since $K^b(P_A)$ and the inverse derived Nakayama functor are derived invariant.<|endoftext|> TITLE: Vafa-Witten invariants for mathematicians QUESTION [10 upvotes]: As Richard Thomas has written (we paraphrase just slightly), mathematical physicists Vafa and Witten introduced new "invariants" of four-dimensional spaces in a paper: A Strong Coupling Test of S-Duality (1994), Cumrun Vafa, Edward Witten https://arxiv.org/abs/hep-th/9408074 Nucl.Phys.B431:3-77,1994 These invariants "count" solutions of a certain equation (the N=4 supersymmetric Yang-Mills equations) over the four dimensional space, and should tell us (both physicists and mathematicians) something about the space. There is one for every integer charge of the Yang-Mills field. Motivated by a generalization of electromagnetic duality in string theory, Vafa and Witten predicted that on a fixed space, one could put all these invariants together in a generating series (a Taylor series or Fourier series, with coefficients the Vafa-Witten invariants) and get a very special function called a "modular form". In particular, the invariants should have hidden symmetries that mean that only a finite number of them determine all the rest. If I was told correctly, until now mathematicians have been unable to make sense of how this Vafa-Witten "counting" should be done without getting infinity. question 1: So what are Vafa-Witten invariants meant for mathematicians in your research fields or subfields? (questionable since the mathematical "counting" so far involves getting infinity.) question 2: Are there similar invariants describing "topologically twisted maximally supersymmetric 5d Yang-Mills theory"? REPLY [2 votes]: I'm not trying to give an answer, I just want to make a few remarks. question 2: Are there similar invariants describing "topologically twisted maximally supersymmetric 5d Yang-Mills theory"? No. A crucial aspect of the Vafa-Witten computation is supersymmetry. For supersymmetry to exist, a globally covariant section of the spinor bundle (the quare root of the canonical bundle) is needed; in other words, the manifold should be a local model of some even-dimensional Calabi-Yau manifold. Five dimensional supersymmetric theories exist. But always embedded on higher dimensional spaces. question 1: So what are Vafa-Witten invariants meant for mathematicians in your research fields or subfields? (questionable since the mathematical "counting" so far involves getting infinity.) You can think on the Vafa-Witten theory as a lower dimensional analog of the Donaldson-Thomas theory. Indeed, the cases explicitly analized in his famous paper were choosen because on those cases, the Vafa-Witten partition function localizes to the Euler characteristic of the moduli space of instantons for the relevant twisted version of $N=4$ super Yang-Mill theory. In mathematical terms that means that aforementioned invariants provide an enummerative theory of ideal sheaves on a given 4-manifold with spin structure. For an explicit computation of the Vafa-Witten invariants as "lower dimensional analogue" of the DT ones see: Crystals and intersecting branes.<|endoftext|> TITLE: On the existence of a family of countably additive extensions of Lebesgue measure QUESTION [14 upvotes]: Let $m$ be Lebesgue measure on $\mathbb R$, and let $m_i$ and $m_o$ be the inner and outer measures respectively. Is it the case that for all $A \subset \mathbb R$ and all $x \in [m_i(A), m_o(A)]$ there exists a countably additive extension $m^+$ of $m$ to the powerset of $\mathbb R$ such that $m^+(A)=x$? A result of Solovay's says that the existence of a measurable cardinal is equiconsistent with the weaker assertion that There exists a countably additive extension of $m$ to the powerset of $\mathbb R$. So an affirmative answer to my question will need to assume the existence of a measurable cardinal. Is anything more needed? I'll note that if we allow $m^+$ to be merely finitely additive, then the answer to the question is affirmative without any large cardinal assumptions. REPLY [10 votes]: Suppose $\kappa$ is the least real-valued measurable cardinal and $\nu:\mathcal{P}(\kappa) \to [0, 1]$ is a witnessing $\kappa$-additive probability measure. Gitik and Shelah showed that the Maharam type of the measure algebra $\mathbb{M}$ of $\nu$ is $\geq \kappa^+$ (See Theorem 2.6 in Gitik-Shelah, Forcing with ideals and simple forcing notions, Israel J. Math. 68 (1989), 129-160 or Theorem 3G in Fremlin's survey on real-valued measurable cardinals here). In particular, forcing with $\mathbb{M}$ adds $\geq \kappa^+$ random reals. Let $G$ be an $\mathbb{M}$-generic filter over $V$. Then $V[G]$ has a set $X \subseteq [0, 1]$ of size $\kappa$ all of whose Lebesgue null subsets are countable. In $V[G]$, let $N$ be the transitive collapse of the well founded $G$-ultrapower of $V$ and suppose $j: V \to N \subseteq V[G]$ is the generic ultrapower embedding with critical point $\kappa$ (See Foreman's chapter in Handbook of Set Theory for background on generic utrapowers). Since the null ideal of $\nu$ is countably saturated, $N$ contains every $\kappa$-sequence of ordinals from $V[G]$ (Proposition 2.14 in Foreman's chapter in Handbook of Set Theory). In particular, $X \in N$. Since $N$, $V[G]$ have the same reals, $N \models$ "$X$ is Lebesgue non null and every Lebesgue null subset of $X$ is countable". Therefore $N \models$ "There is a Lebesgue non null set of cardinality $\aleph_1$". By elementarity of $j$, $V \models$ "There is a Lebesgue non null set of cardinality $\aleph_1$". For a forcing-free proof of this, see Proposition 6F in Fremlin's survey linked above. So we can fix $A \subseteq [0, 1]$ of positive Lebesgue outer measure such that $|A| = \aleph_1$. Let $\mu:\mathcal{P}([0, 1]) \to [0, 1]$ be a total extension of Lebesgue measure. Since $\kappa$ is the least real-valued measurable cardinal, the additivity of the null ideal of $\mu$ is at least $\kappa$. Now $\kappa$ is weakly inaccessible (and much more) and so $\kappa > \aleph_1$. It follows that $\mu(A) = 0$. Hence no total extension of Lebesgue measure can assign positive measure to $A$.<|endoftext|> TITLE: Arithmetic progressions in stopping time of Collatz sequences QUESTION [15 upvotes]: Inspired by the question here, we did a few more simulations of numbers of some specific forms and noticed a pattern. We consider the original $3n+1$ transform where we divide by $2$ if it's even and multiply $3$ and add $1$ if it's odd. The stopping times of the sequences are defined as the number of iterations it took to reach $1$ (which is believed to be finite for all natural numbers in this case). We look at numbers of the form $2^n+1$ (as in the link above) and note that the stopping times create the following sequence: $7, 5, 19, 12, 26, 27, 121, 122, 35, 36, 156, 113, 52, 53, 98, 99, 100, 101, 102, 72, 166, 167, 168, 169, 170, 171, 247, 173, 187, 188, 251, 252, 178, 179, 317, 243, 195, 196, 153, 154, 155, 156, 400, 326, 495, 496, 161, 162, 331, 332, 408, 471, 410, 411, 337, 338, 339, 340, 553, 479, 480, 481, 482, 483, 559, 560, 561, 562, 563, 564, 565, 566, 567, 568, 569, 570, 571, 572, 573, 574, 575, 576, 626, 578, 628, 629, 630, 631, 583, 584, 634, 635, 636, 637, 894, 895, 640, 641, 898, 643, 644, 645, 902, 903, 648, 649, 769, 907, 652, 653, 654, 655, 656, 657, 658, 915, 916, 917, 918, 919, 920, 921, 914, 923, 916, 917, 918, 919, 920, 921, 930, 923, 994, 995, 996, 997, 998, 999, 938, 1001, 1002, 1003, 1004, 1005, 1006, 1007, 1145, 1009, 1147, 1148, 1012, 1013, 1151, 1152, 1016, 1017, 1018, 1019, 1020, 1021, 1022, 1023, 1528, 1529, 1530, 1531, 1532, 1533, 1534, 1535, 1169, 1537, 1445, 1446, 1447, 1448, 1542, 1543, 1544, 1545, 1546, 1547, 1548, 1549, 1550, 1551, 1552, 1553, 1554, 1555, 1556, 1557, 1527, 1528, 1560, 1561, 1562, 1563, 1533,\ \ldots$ The curious fact is that there seems to exist arithmetical sequences with common difference $1$ and of various lengths existing in the sequence. In this sequence shown, the largest such sequence starts at $559$ and ends at $576$ with a length of $18$. Next we looked at numbers of the form $3^n+1$ and noted the following sequence of stopping times: $2, 6, 18, 110, 21, 95, 32, 75, 74, 42, 134, 133, 132, 131, 143, 204, 128, 189, 139, 94, 93, 260, 427, 90, 257, 393, 330, 254, 253, 389, 388, 387, 461, 460, 459, 458, 457, 456, 455, 454, 453, 452, 500, 499, 449, 497, 496, 751, 494, 493, 492, 747, 490, 745, 488, 487, 486, 741, 740, 739, 738, 737, 728, 727, 726, 725, 794, 793, 792, 791, 790, 789, 788, 787, 923, 785, 921, 783, 782, 781, 780, 1283, 1282, 1281, 1280, 1279, 1185, 1184, 1276, 1275, 1274, 1273, 1272, 1271, 1270, 1269, 1237, 1267, 1266, 1234, 1264, 1263, 1231, 1230, 1229, 1228, 1227, 1226, 1225, 1255, 1422, 1222, 1420, 1419, 1418, 1417, 1248, 1216, 1246, 1214, 1213, 1212, 1211, 1409, 1408, 1407, 1406, 1343, 1342, 1403, 1733, 1339, 1338, 1337, 1336, 1335, 1727, 1333, 1681, 1724, 1723, 1678, 1677, 1720, 1675, 1718, 1717, 1716, 1715, 1714, 1713, 1712, 1711, 1710, 1709, 1708, 1707, 1706, 1705, 1704, 1703, 1702, 1701, 1700, 1699, 1698, 1697, 1696, 1695, 1694, 1693, 1692, 1691, 1690, 1707, 1688, 1687, 1686, 1703, 1702, 1701, 1700, 1681, 1680, 1679, 1678, 1695, 1676, 1675, 1692, 1691, 1690, 1689, 2585, 1687, 1686, 2334, 2333, 2332,\ \ldots$ Here also, we notice that there exist arithmetic sequences with common difference $-1$ and again of various lengths in the sequence. The largest such sequence here starts at $1718$ till $1690$ with a length of $29$. We did these simulations a few more times with more and more numbers and sequences and essentially noted the following: So we tried combining the two observations and guessed that numbers of the form $6^n+1$ should create sequences would have arithmetic sequences with common difference $0$. And yes it does so. The below sequence shows it: $16, 21, 26, 101, 83, 83, 145, 145, 220, 158, 145, 207, 114, 114, 450, 114, 357, 357, 282, 419, 419, 494, 494, 494, 494, 494, 494, 494, 543, 494, 543, 799, 799, 543, 543, 799, 543, 543, 799, 799, 791, 791, 791, 791, 861, 861, 861, 861, 998, 998, 998, 861, 861, 861, 1365, 1365, 998, 1272, 1365, 1365, 1365, 1365, 1365, 1365, 1365, 1365, 1365, 1365, 1334, 1334, 1334, 1334, 1365, 1533, 1533, 1533, 1533, 1365, 1365, 1334, 1533, 1334, 1533, 1533, 1471, 1471, 1864, 1471, 1471, 1471, 1864, 1471, 1864, 1864, 1820, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1864, 1882, 1882, 1882, 1864, 1864, 1864, 1864, 1864, 1882, 1882, 2779, 1882, 2531, 2531, 2779, 1882, 1882, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2805, 2805, 2805, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 2779, 3203, 2779, 2779, 3203, 2779, 2779, 3203, 3203, 3203, 3932, 3932, 3932, 3932, 3932, 3203, 3932, 3932, 3203, 3203, 3203, 3203, 4586,\ \ldots$ Also: The final thing that we observed (sorry for not having a table ready, I will edit it as soon as possible) is that any number of the form $(2^a3^b)+1$ had stopping time sequences with the existence of arithmetic progressions with common difference $a-b$. Claim: Any number of the form $(2^a3^b)+1$ has stopping time sequences with the existence of arithmetic progressions with common difference $a-b$. We know this is true, but a proof eludes us. Following everything we found, we made a conjecture : Strong Conjecture : If the Collatz conjecture is true then the sequence of stopping times of the Collatz sequence for numbers of the form $(2^a3^b)^n +1$ has arbitrarily long arithmetic sequences and with common difference $a-b$. We were not successful in proving or disproving the result, neither did we find any literature on this. Any input on the topic would be highly appreciated. Edit: I added a claim separately to highlight a part I just mentioned in passing. We believe that there may not be arbitrarily long progressions because clearly we were missing a lot of values in the length sequences. But they did seem to be unbounded (no upper limit on the length of the progressions). So we weakened the conjecture a bit: Weak Conjecture : If the Collatz conjecture is true then the sequence of stopping times of the Collatz sequence for numbers of the form $(2^a3^b)^n +1$ has arithmetic sequences with no length bound and with common difference $a-b$. Evidence for the weakening: For numbers of the form $6^n+1$, the sequence of length of progressions (what progression lengths are there in the stopping time sequences) looked like this: For the first $1000$ terms: $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 27, 29, 38, 48, 51, 52$ For the first $2000$ terms: $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 27, 28, 29, 36, 38, 45, 46, 48, 51, 52, 59, 110$ For the first $10000$ terms: $1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 31, 36, 38, 41, 45, 46, 48, 50, 51, 52, 54, 55, 56, 58, 59, 60, 61, 65, 71, 72, 73, 74, 78, 82, 83, 86, 92, 99, 101, 102, 107, 109, 110, 111, 116, 126, 130, 132, 133, 160, 161, 163, 172, 180, 182, 184, 185, 186, 189, 220, 248, 271, 337$ We are clearly missing $12$ and $24$. Probably $30$ is also missing. We are not sure if these are actually missing, but appears to be so. If indeed they are missing, then the strong version is false. If not i.e. if the values will be available as the number of terms increase, then it's true. But as of this moment, both the conjectures are open (possibly). So the main questions to move forward will be to the answer the following: Why are their arithmetic sequences in the stopping times of these specific numbers? Can we prove the claim? Are the conjectures true? (I will be editing the question with more data and possibly some graphs as soon as I have some more time on hand. Till then, please let us know of your thoughts. Thank you). REPLY [3 votes]: Here an attempt to try to explain the very interesting observations. I believe it does not have much to do with powers but with the fact that consecutive integers often converge to $1$ with the same number of $/2$ and $*3+1$ steps in a different order. For example if you start with $8n+4$ and $8n+5$ you end up with $3n+2$ in both cases with three $/2$ and one $*3+1$ step. So any consecutive numbers equal to $4$ and $5$ modulo 8 end will use the same number of $/2$ and $*3+1$ steps to converge to 1 (if they converge at all). You can find more similar rules for higher powers of two. For example \begin{eqnarray} 4,5 \text{ mod } 8\\ 2,3 \text{ mod } 16\\ 5,6 \text{ and } 22,23 \text{ mod } 32\\ 14,15 \text{ and } 45,46 \text{ mod } 64\\ 29,30 \text{ and } 49,50 \text{ and } 94,95 \text{ mod } 128\\ 62,63 \text{ and } 99,100 \text{ and } 145,146 \text{ and } 189,190 \text{ mod } 256 \end{eqnarray} The number of pairs that converge to the same value $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ \hline \text{number of new "converging" pairs mod } 2^n & 1 & 1 & 2 & 2 & 3 & 4 & 6 & 11 & 19 & 34 & 60 \\ \hline \end{array} $$ If we now compute the probability that two consecutive numbers converge with the same number of steps we and up with more than $1/3$. If we start with $3^n+1$ then there is the chance it will the need the same number of iterations as $3^n$. The number $3^n$ will after one step be at $3^{n+1}+1$. Hence the chances are that $3^{n+1}+1$ needs one step less than $3^n+1$.<|endoftext|> TITLE: Left-right non-bimodule examples QUESTION [8 upvotes]: Let $A$ be a unital algebra, defined over the complex numbers. Any bimodule $M$ over $A$ must, by definition, be a left, and right, module satisfing $$ a.(m.b) = (a.m).b, ~~~~~~~ \textrm{ for all } a,b \in A, ~ m \in M $$ What is a "natural" or "well-motivated" example of a an object which is both a left and right module, but does not satisfy the above condition? REPLY [3 votes]: Suppose $A$ is a non-commutative Hopf algebra. Then you can use $M=A$ with the left adjoint and right regular actions: $$ a\cdot m = \sum_{(a)}a_{(1)}mS(a_{(2)}), \ m\cdot b = mb . $$ In particular, you can use the group algebra of a a group $G$ so that $$ g\cdot m = gmg^{-1}, \ m\cdot h = mh, \ g,h,m\in G. $$<|endoftext|> TITLE: Counting nilpotent self-maps of $\{1,\dots,n\}$ with image of a given cardinal QUESTION [7 upvotes]: Let $\mathcal{C}_n$ be the monoid of self-maps $\alpha$ of $\{1\dots,n\}$ that are order-preserving ($\forall x,y$, $x\le y$ $\Rightarrow$ $\alpha(x)\le\alpha(y)$ and decreasing ($\forall x$, $\alpha(x)\le x$). Let N($\mathcal{C_n})$ be the set of nilpotent elements of $\mathcal{C}_n$, i.e., those $\alpha\in\mathcal{C}_n$ such that some power of $\alpha$ equals the absorbing element (the constant function $1$). As mentioned in a comment, this means that $\alpha(x)1$. Given $r$, what is the cardinality of the subset $$X_{n,r}=\{\alpha\in N(\mathcal{C_n}): \lvert \mathrm{im}(\alpha)\rvert=r\}?$$ REPLY [5 votes]: The answer is ${n-2\choose r-1}{n-1\choose r-1}-{n-2\choose r-2}{n-1\choose r}=\frac1{n-r}{n-2\choose r-1}{n-1\choose r}$. Let $\{1=a_0 TITLE: abelianization and homotopy QUESTION [10 upvotes]: I deleted by previous questions, seems they are too vague. Let me try to ask a more precise question. Let $f:G\rightarrow K$ a morphism of simplicial groups such that $f$ is a weak homotopy equivalence of underlying simplicial sets. We will make to assumptions: 1) for each natural number $i$, $G_{i}$ is a free group. 2) for each natural number $i$, $K_{i}$ is a subgroup of a product of free groups, I mean that $K_{i}\subset \prod_{j\in J} F_{j}$ where each $F_{j}$ is free. Now lets take the corresponding abelianization, we get $f^{ab}:G^{ab}\rightarrow K^{ab}$. Is $f^{ab}$ a weak homotopy equivalence of underlying simplicial sets? REPLY [7 votes]: The map is not necessarily a weak equivalence, even if $K_i$ is actually the whole product of free groups rather than just a subgroup. Let $K$ be the constant simplicial set which is $\Bbb Z^2$ in each degree. The map $K \to K_{ab}$ is an isomorphism, and $K = K_{ab}$ has no higher homotopy groups. Let $G$ be a "cofibrant replacement" of $K$. This is work to describe explicitly, but there exists a version of $G$ which, in degree $n$, is free on $(n+2)$ generators. Namely, we let $G_n$ be the free group on generators $x, r_1, \dots, r_n, y$, subject to relations: all the face & degeneracy maps take $x$ to $x$ and $y$ to $y$, the generator $r_1$ in $G_1$ satisfies $d_0(r_1) = [x,y]$ and $d_1(r_1) = 1$, and the elements $r_i$ in $G_n$ are the images of $r_1$ under the $i$ different degeneracy maps $G_1 \to G_n$. This forces the rest of the face and degeneracy maps by multiplicativity. There is a map of simplicial groups $G \to K$ sending $x$ to $(1,0)$, $y$ to $(0,1)$, and $r_i$ to $0$. One can check that this is an equivalence. (It roughly corresponds to the presentation of a torus, which is a $K(\Bbb Z^2,1)$, using two 1-dimensional cells and a 2-dimensional cell.) However, in $G_{ab}$ the first homotopy group is nonzero: the element $r_1$ becomes a cycle because its boundary $[x,y]$ became trivial, but it's not in the image of the boundary map. All of this is an instance of an insight of Quillen's. If you take a simplicial group $K$, replace it by a weak equivalence $G \to K$ from a simplicial group $G$ that is levelwise free, and form $G_{ab}$, then the homotopy group $\pi_k G_{ab}$ is the homology group $H_{k+1}(BK;\Bbb Z)$. In this respect, group homology is a kind of derived abelianization procedure. Because, in your question, the groups $K_i$ are subgroups of products of free groups and may have nontrivial higher group homology, we can't use them as replacements for free groups when forming these resolutions.<|endoftext|> TITLE: Mathematical words outside of mathematics QUESTION [13 upvotes]: We've all heard expressions like "We need to factor this into the equation," where mathematical words have broader meanings than strictly mathematical. I'd like to develop a collection of such usages. Of course, there's "grows exponentially" for just about any rapid growth, and "the rate has accelerated" not really meaning the third derivative. Also, there's Friedman's cool usage of "orthogonal" at the Supreme Court (https://www.librarything.com/topic/193156). Here's my two-part question: is there already list of such usages, somewhere? Can you contribute some examples? REPLY [5 votes]: I've often wondered whether whoever created Delta Airlines' slogan "Delta means change" had some mathematical training . . . REPLY [2 votes]: French has an expression "c'est epsilon" meaning something like "it's negligible". French wiktionary says this sense is "par extension" from the mathematical usage. There's also lambda "(Éthologie) Un membre considéré comme le plus ordinaire d’un ensemble", which might have a mathematical origin.<|endoftext|> TITLE: Aspherical manifold with superperfect fundamental group and non-trivial center? QUESTION [7 upvotes]: I am interested in knowing if there is a closed, (smooth) aspherical manifold $M$ (hyperbolic would be best) with superperfect fundamental group (that is to say, with $H_1(\pi_1(M);\mathbb{Z}) = H_2(\pi_1(M);\mathbb{Z}) = 0$; note $H_1(\pi_1(M);\mathbb{Z}) = 0$ is equivalent to perfect) and non-trivial center ($\mathbb{Z}_2$ would be best, but any f.g. Abelian group will do). Also, assuming there is a manifold that fits the criteria, I would likely need a handlebody decomposition for the manifold, assuming the "standard handlebody procedure" for producing a closed, smooth manifold from a prescribed finite presentation of a/the fundamental group does not yield the smooth manifold in question (e.g., the "standard manifold" is not aspherical). I found many hyperbolic 3-manifolds with superperfect fundamental group using SnapPy, but SnapPy evidently doesn't have a center "method" for the fundamental group method/class attached to 3-manifolds. Sage/GAP/MAGMA also appear not to be able to compute the center for an infinite finitely-presented fundamental group. Thanks much in advance. I realize this is kind-of "shooting for the moon/stars". REPLY [3 votes]: Per Lee's request, I've turned the discussion into an answer. Suppose that $M$ is a closed connected oriented hyperbolic $n$ manifold, for $n$ at least two. Then the fundamental group contains no torsion elements; also its center is trivial. Both claims are exercises from the classification of isometries of hyperbolic $n$-space. Thus there are no hyperbolic homology three-spheres with the property you desire. However, there are many three-manifolds that are (integral) homology spheres (and so $\pi_1$ is super-perfect) and where $\pi_1$ has non-trivial abelian centre. These are found among the Seifert fibered spaces. The most famous of these is the Poincare homology sphere, but this example is ruled out by your requirement that $M$ be aspherical. More specifically, you should consider the "aspherical Brieskorn homology spheres" $\Sigma(p, q, r)$. These are described at the Wikipedia page linked to immediately above, which also briefly sketches a presentation of their fundamental group. If you want Snappy to compute things for you, then consult Figure 1 of this paper (say) for a surgery description of $\Sigma(p, q, r)$. This paper also gives a short but useful discussion of the fundamental group.<|endoftext|> TITLE: Free linear group actions on spheres with "strong" angle preservation QUESTION [6 upvotes]: Suppose $G$ is a finite group and that $\rho: G\rightarrow O(d)$ is a faithful orthogonal representation, with action on $\mathbb{R}^d$ denoted $\cdot$. Let's say that $\rho$ is "strongly" angle preserving if for each $g\in G$ one has \begin{equation}\langle g\cdot v, v\rangle = \langle g\cdot w, w\rangle \end{equation} for all $v, w$ on the (representation) sphere $S^{d-1}$. (There must be a name for such an action, but let's stick with this name and let someone edit it as desired). Examples of such representations would be all the finite subgroups of the classical algebras $\mathbb{R},\mathbb{C},$ and $\mathbb{H}$ under their standard embeddings in $O(1)$, $U(1)$, and $Sp(1)$. Question: Can anyone point me towards literature on any other groups admitting such a representation? These representations give free linear actions on spheres (since $g\cdot v=v$ forces $\langle g\cdot w, w\rangle =1$ for all $w\in S^{d-1}$), and these are classified, but it seems something stronger is required of the representation. REPLY [3 votes]: I believe, the examples you gave are essentially the only ones. Indeed, let $\rho\colon G\to O(d)$ be a faithful irreducible “strongly angle-preserving” representation. Claim. The image of $\mathbb{R}G$ under (the natural linear extension of) $\rho$ is a division algebra. Proof. By linearity of the scalar product, for each $x\in\mathbb RG$ we have $$ \langle \rho(x)v,v\rangle = \langle \rho(x)w,w\rangle,\quad v,w\in S^{d-1}. $$ Let $A=\sum_{g\in G} \alpha_g \rho_g\in M_d(\mathbb R)$ be an element with nontrivial kernel and let $v\in\ker A = \ker A^*A$. We then have $$ 0 =\langle A^*Av,v\rangle=\langle A^*Aw,w\rangle, $$ whence $Aw=0$ for all $w\in S^{d-1}$, so $A=0$. This proves the claim. Therefore, by Frobenius theorem on division algebras, the image $D$ of $\mathbb R G$ under $\rho$ is isomorphic to $\mathbb R$, $\mathbb C$ or $\mathbb H$, and we therefore can identify $G$ with its image in this division algebra. The representation induces a natural scalar product on $D$ by $$ \langle x,y\rangle:= \frac{1}{d} \mathrm{Tr}(y^*x) = \frac{1}{d} \sum_{i=1}^d\langle y^*xv_i,v_i\rangle = \langle xv_1,yv_1\rangle = \langle xw,yw\rangle,\,w\in S^{d-1} $$ and as $\rho$ is (in view of irreducibility) generated by a vector $w\in S^{d-1}$, it is naturally isomorphic to the multiplication representation of $D$ on itself (with the scalar product from above and the isomorphism sending $1\in D$ to $w$).<|endoftext|> TITLE: Groups containing each other as finite index subgroups QUESTION [16 upvotes]: Suppose that $G, H$ are finitely generated groups such that $H$ is isomorphic to a finite index subgroup of $G$ and vice versa. Does it follow that $G$ is isomorphic to $H$? I am sure that the answer is negative but cannot find an example. I am mostly interested in the case of finitely presented groups. The assumption of finite index is, of course, necessary, otherwise one can take any two nonabelian free groups of finite rank. Here is what I know: Given a pair of groups $G, H$ as above, there is, of course, a sequence of isomorphic proper subgroups of finite index $$ ... G_n\lneq G_{n-1}\lneq ...\lneq G_1\lneq G $$ One can rule out the existence of such a sequence when $G$ is nonelementary hyperbolic, but this does not say much. PS. Noam's example makes me feel rather silly since I have seen such groups in vivo: The affine Coxeter groups $\tilde{B}_n, \tilde{C}_n$, $n\ge 3$. REPLY [34 votes]: Simple counterexample: $G$ is the square of an infinite dihedral group, consisting of symmetries of the ${\bf Z}^2$ lattice of the form $(x,y) \mapsto (\pm x + a, \pm y + b)$ with $a,b \in \bf Z$; and $H$ is the index-$2$ subgroup where $a \equiv b \bmod 2$. Then $H$ has index-$2$ subgroup consisting of the symmetries $(x,y) \mapsto (\pm x + 2a, \pm y + 2b)$, and this subgroup is isomorphic with $G$, but $G \not \cong H$. REPLY [16 votes]: The answer is "no". It is proved in https://arxiv.org/abs/0711.1014 for R. Thompson group $F$. That group contains a finite index subgroup $H$ which is not isomorphic to $F$ but contains a copy of $F$ as a subgroup of finite index.<|endoftext|> TITLE: Elementary topology of surfaces QUESTION [13 upvotes]: Let $S$ be a compact connected orientable bordered surface of genus $g$ with $n$ holes (a hole is a component of the border homeomorphic to a circle). Consider a cell decomposition (the closure of each cell is a closed disk of the same dimension as the cell) with $f$ faces, $e_i$ interior edges, $e_b$ boundary edges, $v_i$ interior vertices and $v_b$ boundary vertices. Is there a function $F$ such that $g=F(f,e_i,e_b,v_i,v_b)$? If the answer is positive what is this function $F$? The Euler characteristic gives me $2g+n$, and I want to recover $g$ and $n$ separately from $f,e_i,e_b,v_i,v_b$. A negative answer would be an example of two non-homeomorphic surfaces with cell decompositions for which the five numbers $f,e_i,e_b,v_i,v_b$ are the same. REPLY [12 votes]: The answer is negative. First, take two hexagons $A_1 B_1 C_1 D_1 E_1 F_1$ and $A_2 B_2 C_2 D_2 E_2 F_2$ and identify vertices $(A_1,A_2), (B_1,B_2)$, as well as glue the pairs of edges $(F_1A_1,F_2A_2), (B_1C_1, B_2C_2)$. Now add edges $D_1D_2, E_1E_2$ and a $2$-cell $D_1D_2E_2E_1$. You obtain a cell decomposition with $g=0, n=3$ and $(f,e_i,e_b,v_i,v_b)=(3,4,8,0,8)$. On the other hand you can cut out a small octagon $ABCDEFGH$ from a torus and draw two non intersecting edges $AD,EH$ along longitudes, as well as two edges $CF, BG$ along meridians. This produces three faces and gives $g=1, n=1$ with the same vector $(f,e_i,e_b,v_i,v_b)=(3,4,8,0,8)$. If we are given the more refined data of all incidence relations between cells then there are multiple ways to count the number of boundary components. If one doesn't wish to chase the cycles formed by boundary edges, one can take a more algebraic approach. For example, if the matrix $M\in \mathbb R^{v_b\times e_b}$ records incidences between boundary vertices and boundary edges (each entry being $1$ when two cells are incident and $0$ otherwise) then the matrix $MM^{T}$ is equal to $2I+A$, where $I$ is the identity matrix and $A$ is the adjacency graph of the boundary graph. Since the number of connected components of a graph is given by the dimension of the nullspace of its Laplacian matrix, this means that the number of boundary components of our original complex is given by $\dim \text{null} (2I-A)=\dim \text{null} (4I-MM^{T})$.<|endoftext|> TITLE: Quick reference for general Weyl's inequality in number theory QUESTION [7 upvotes]: I would like a reference for the result here. Having that $t$ there makes me happy. I would prefer not to have to, in my paper, run through and (not trivially but not too greatly) alter the proof of the standard Weyl inequality to get the result with the $t$, but rather just cite a book or paper. Feel free to just comment so I can delete the question (and edit the Wikipedia page). REPLY [11 votes]: This form of Weyl's inequality is due to Ivan Matveevich Vinogradov and the relevant reference is the 1927 paper [3]. Precisely, Lemma III at pages 568-569 states the following equivalent form: if $$ S=\sum_{x=N+1}^{N+P} e^{2\pi i f(x)},\quad f(x)=\lambda x^n+\ldots+\lambda_n,\label{WS}\tag{WS} $$ and $$ \left|\lambda -\frac{a}{q}\right|<\frac{\tau}{q^2},\quad (a,q)=1, \quad 0 q$ we can say that this asymptotic estimate holds irrespectively of any upper bound on the value of $\tau$: of course, in such condition it loses its usefulness, since it is far worser than the trivial estimate. A drawback of formula \eqref{WI} was noted by Vinogradov ([5], pp. 185-186, or [6], p. 6): the estimate become quickly less accurate as $n$ increases, since its left side is (far, as he says) larger than $P^{1-\sigma}$, and this term tends rapidly to $P$. Addendum: incidentally I recently noted the work [A1]. The Author, while proving a refinement of \eqref{WI} valid for polynomials $f(x)$ for which the coefficient of the $(n-1)$th power is $0$, acknowledges the work of Vinogradov on this formula without citing [3] ([A1] p. 1) and cites Vaughan's monograph as a reference for a proof ([A2] §2.1, lemma 2.4, pp. 11-12). This monograph can thus be used as a modern reference on Vinogradov's form of Weyl's inequality for the English reader. Addendum Bibliography [A1] Allakov, Ismail A., On an estimate by Weyl and Vinogradov, Sibirskiĭ Matematicheskiĭ Zhurnal 43, No. 1, 9-13 (2002); translation in Siberian Mathematical Journal 43, No. 1, 1-4 (2002), MR1888113 ZBL1008.11031. [A2] Vaughan, Robert C., The Hardy-Littlewood method, Cambridge Tracts in Mathematics, 125. Cambridge: Cambridge University Press. pp. vii+232 (1997), ISBN: 0-521-57347-5, MR1435742 ZBL0868.11046. Bibliography [1] Chandrasekharan, Komaravolu, Arithmetical functions, Die Grundlehren der Mathematischen Wissenschaften in Einzeldarstellungen. 167. Berlin-Heidelberg-New York: Springer-Verlag. XI, 231 p. (1970), MR0277490, ZBL0217.31602. [2] Landau, Edmund, Vorlesungen über Zahlentheorie. I: Aus der elementaren und additiven Zahlentheorie. II: Aus der analytischen und geometrischen Zahlentheorie. III: Aus der algebraischen Zahlentheorie und über die Fermatsche Vermutung, Leipzig, S. Hirzel. I: xii, 360 S. II: viii, 308 S. III: viii, 342 S. (1927). JFM 53.0123.17. [3] Vinogradov, Ivan Matveevich, "Démonstration analytique d'un théorème sur la distribution des parties fractionnaires d'un polynôme entier", Bulletin de l’Académie des Sciences de l’Union des Républiques Soviétiques Socialistes, (6) 21, 567-578 (1927), JFM 53.0160.02. [4] Vinogradov, Ivan Matveevich,The method of trigonometrical sums in the theory of numbers. Translated, revised and annotated by K. F. Roth and Anne Davenport, New York: Interscience Publishers Inc. X, 180 p. (1954), MR0062183, ZBL0055.27504. [5] Vinogradov, Ivan Matveevich, Selected works. Prepared by the Steklov Mathematical Institute of the Academy of Sciences of the USSR on the occasion of his ninetieth birthday. Ed. by L. D. Faddeev, R. V. Gamkrelidze, A. A. Karatsuba, K. K. Mardzhanishvili and E. F. Mishchenko, Berlin-Heidelberg-New York: Springer-Verlag pp. viii+401 (1985), ISBN: 3-540-12788-7, MR0807530, ZBL0577.01049. [6] Vinogradov, Ivan Matveevich; Karatsuba, Anatoliĭ Alekseevich, "The method of trigonometric sums in number theory", Proceedings of the Steklov Institute of Mathematics 168, 3-30 (1986), MR0755892, ZBL0603.10037. [7] Weyl, Hermann, "Über die Gleichverteilung von Zahlen mod. Eins", Mathematische Annalen 77, 313-352 (1916). ZBL46.0278.06. [8] Weyl, Hermann, "Zur Abschätzung von $\zeta(1+ti)$", Mathematische Zeitschrift 10, 88-101 (1921). ZBL48.0346.01.<|endoftext|> TITLE: Do 1-additive maps admit tensor products? QUESTION [5 upvotes]: Let $\mathcal{F}$ be a set algebra (or a Boolean algebra). Following Kalton, let me call a function $f\colon \mathcal{F}\to \mathbb R$ $\delta$-additive ($\delta \geqslant 0$), whenever $f(\varnothing) = 0$ and $$| f(A) + f(B) - f(A\cup B) | \leqslant \delta$$ as long as $A\cap B=\varnothing$ for $A,B\in \mathcal{F}$. Surely 0-additive functions are nothing but finitely additive signed measures. I am interested in the notion of a tensor product that would be analogous to a product measure but actually only in a very simplistic setting. Let $X$ and $Y$ be finite sets and suppose that $f\colon \wp(X)\to \mathbb R, g\colon \wp(Y)\to\mathbb R$ are 1-additive functions. Is there a function $h\colon \wp(X\times Y)\to \mathbb{R}$ such that $h$ is 1-additive, $h(A\times B) = f(A)\cdot g(B)\quad (A\subset X, B\subset Y)$. The problem is, I think, non-trivial as we need some sort of a canonical decomposition of any given set into a union of rectangles, which is highly non-unique. On the other hand, working only with singletons (trivial rectangles) is not good enough to retrieve the tensorial property of $h$. REPLY [3 votes]: As you guessed in the comments, it does not exist. The idea is the following: you are searching for some values that have small distance from some given values (the one on rectangles). If for example a new value $v$ must be close to some given values $v_1,v_2$, then necessarily $v_1, v_2$ must be close. Conditions of Delta additivity in each variable is not enough to guarantee that $v_1$ and $v_2$, will be always close. If you want to find a condition, I guess you should do something like you do in elimination theory with equations, but in this "distance" fashion. An interesting question so would be, mimicking elimination style: if such inequalities are satisfied, does there exist a solution? Without such conditions there is a counterexample. Take both sets to have 2 elements that we call $x,y$. In the first one: $x$ and $y$ have measure 0; $\{x,y\}$ has measure 1. In the second one: $x$ and $y$ have measure $r$; $\{x,y\}$ has measure $2r$. Emptyset has zero measure in both. Note that singletons in the product has measure zero. Now take the L-shaped set $L=\{(x,x), (x,y), (y,x) \}$ and suppose it has measure $A$. If we add the last brick to get the rectangle, we have $$ | A + 0- 2r| \le 1$$ If we take out the the brick $(y,x)$ the rectangle we are left with has projection $x$ on the first set, thus it has measure zero. On balance we get $$ |A -0-0| \le 1$$ In contradiction with the previous one for big $r$. Also, note that this yield that in general it does not exist a delta additive tensor product for any fixed $\delta$, and that even if one of them is a measure the tensor could not exist.<|endoftext|> TITLE: Is every locally free module of rank $1$ over a commutative ring concretely invertible? QUESTION [11 upvotes]: Since this subject is full of misunderstandings (see here, here, here, and here) let us fix a precise terminology. Let $A$ be a commutative ring and $P$ an $A$-module. I) We'll say that $P$ is a locally free module of rank one or is invertible if $P$ is finitely generated, projective and of rank one in the sense that for every prime ideal $\mathfrak p$ of $A$ the localized $A_\mathfrak p$- module $P_\mathfrak p$ (which is free by projectiveness) is of dimension $1$. These modules correspond bijectively, by a well known result of Serre in FAC, to locally free sheaves $\tilde P$ of rank $1$ on $\operatorname {Spec}A$, also known as invertible sheaves. This is one motivation for the above terminology. Another justification for the terminology "invertible" is that these modules are exactly those for which the canonical evaluation map $ P^*\otimes_AP\to A$ is an isomorphism. II) If $B\supset A$ is an overring of $A$ and $P\subset B$ is an $A$-module, we'll say that it is concretely invertible with respect to $B$ if $P.(A:P)_B=A$. [As is standard $(A:P)_B$ denotes the set of elements $b\in B$ such that $bP\subset A$] Lam proves in his Lectures on Modules and Rings, that these concretely invertible modules are invertible. What about the converse? Question: Is an invertible $A$-module $P$ isomorphic as an $A$-module to a concretely invertible module $P'\subset B$ with respect to a suitable overring $B\supset A$? Remarks a) Denote by $\operatorname {Quot} A$ the total quotient ring of $A$ obtained by inverting the regular (=not zero-divisors) of $A$, so that $A\hookrightarrow \operatorname {Quot} A$ is injective. Then a submodule $P\subset \operatorname {Quot}A$ is invertible if and only if it is concretely invertible. b) The answer is "yes" if $A$ is an integral domain: we can take $P'$ sitting inside $B=\operatorname {Frac}A$. c) The answer is "yes" if $A$ is semi-local, since then $P$ is free of rank $1$: see here. d) The answer is "yes" if $A$ is noetherian: in Eisenbud's Commutative Algebra, page 253, Theorem 11.6 b., it is proven that every invertible module $P$ over a noetherian ring $A$ is isomorphic to a concretely invertible submodule $P'\subset \operatorname {Quot} A$ of its total quotient ring. e) Whatever the answer to the question is, it is definitely not true that we can always find the required $P'$ inside the total quotient ring $B=\operatorname {Quot} A$. Lam gives a counter-example in his Lectures on Modules and Rings, Example (2.22)(A), page 37. REPLY [9 votes]: The answer is yes. Recall that given an invertible $A$-module $P$ and $n \in \mathbf{Z}$ there is an invertible $A$-module $P^{\otimes n}$ such that $P^{\otimes 0} = A$, $P^{\otimes 1} = P$, and $P^{\otimes n} \otimes_A P^{\otimes m} = P^{\otimes n + m}$. Set $B = \bigoplus_{n \in \mathbf{Z}} P^{\otimes n}$; this is a commutative $\mathbf{Z}$-graded $A$-algebra (details omitted). Then $P \subset B$ in degree $1$ and $(P : A)_B = P^{\otimes -1}$ sitting in degree $-1$ and we have $P \cdot P^{\otimes -1} = A$. Remark. The spectrum of $B$ is the $\mathbf{G}_m$-torsor over $\text{Spec}(A)$ corresponding to $P$.<|endoftext|> TITLE: When do elements in the braid group $B_n$ commute? QUESTION [8 upvotes]: I have been looking around for an answer to this question, but I have not been able to find anything. My question is: Is it known how to tell whether two elements $b_1, b_2 \in B_n$ commute? EDIT: Is there perhaps a nicer/cleaner rule for $B_3$ and $B_4$? REPLY [4 votes]: You are asking: "Suppose that a and b are braids, with inverses A and B. Is the element abAB trivial?" This is a special case of the word problem for the braid group, which has many solutions. Perhaps Artin's 1947 paper (Theorems 17 and 18) is the first. Many of the papers mentioned by other answers in the thread also "quickly" give solutions to the word problem (eg for linear groups over $\mathbb{Z}$, for groups acting on free groups, etc).<|endoftext|> TITLE: Why can Euler systems constructed from algebraic cycles only be anticyclotomic? QUESTION [8 upvotes]: In a footnote to the 2018 Zerbes-Loeffler lecture notes from the Arizona Winter School, it's stated that Euler systems constructed from algebraic cycles "cannot give a full Euler system, only an anticyclotomic one." Why is this the case? It is not obvious to me. REPLY [13 votes]: Let me explain a bit more what that footnote was supposed to mean. As I'm sure you know, an Euler system for a Galois representation $V$ over a number field $K$ consists of a bunch of classes in $H^1(L, V)$, as $L$ varies over a suitable class of abelian extensions of $K$. If we're willing to temporarily forget about integrality, we can project to eigenspaces for the action of $Gal(L/K)$ and think of an Euler system equivalently as a collection of classes $z_\tau \in H^1(K, V(\tau))$, as $\tau$ varies over some collection of finite-order characters of the Galois group $G_K$ -- either all finite-order characters (a full Euler system) or only anticyclotomic ones, assuming $K$ is CM (an anticyclotomic Euler system). Now, if your classes $z_\tau$ come from something geometric, they will have a special property: they will lie in the Bloch--Kato "$H^1_{\mathrm{g}}$" subspace. Up to a minor correction coming from local Euler factors, which will be trivial for almost all $\tau$ and thus can be ignored [*], this is the same as the Bloch--Kato "$H^1_{\mathrm{f}}$" subspace. Now, the Bloch--Kato conjecture predicts that the dimension of $H^1_{\mathrm{f}}(K, V(\tau))$ is the order of vanishing of $L(V^* \otimes \tau)$ at $s = 1$. So, the moral of that is the following: for an Euler system coming from geometry to exist, you need the L-values of all of these character twists of $V$ to vanish. There are basically two possible mechanisms for forcing lots of L-values to vanish in a systematic way. One is poles of Archimedean $\Gamma$-factors (which is what gives the "trivial" zeroes of $\zeta(s)$ at negative even integers); the other is from sign considerations when $s = 1$ is the central value. (These are mutually exclusive, because Archimedean $\Gamma$-factors can only force vanishing when $s = 1$ is not the central value). However, "sign-related" vanishing is somehow rather fragile: it only applies when $W = Ind_K^{\mathbb{Q}} V$ is Tate self-dual ($W = W^*(1)$). If you want to twist $V$ by a character $\tau$ while preserving this self-duality, you need $\tau$ to be self-dual in a suitable sense -- and this ends up forcing you to consider only anticyclotomic characters. If your geometric classes come from algebraic cycles (i.e. from $K_0$ of algebraic varieties), then they correspond to an $L$-value at the centre of the functional equation (motivic weight $-1$) so the only possibility is sign-induced vanishing. This is why "algebraic cycle" Euler systems, like Heegner points and the more recent constructions of Cornut and of Jetchev, are always anticyclotomic. To get a full (non-anticyclotomic) Euler system, you have to use geometric classes coming from K-theory in positive degrees in a setting where there is "$\Gamma$-factor" vanishing, as in the case of cyclotomic units and Kato's Euler system. [*] Historical remark: this gap between $H^1_{\mathrm{g}}$ and $H^1_{\mathrm{f}}$ is precisely where Flach's cohomology classes for the symmetric square of an elliptic curve live; and the fact that this gap closes up again when you twist by a character is why it is so hard to extend Flach's construction to a full Euler system.<|endoftext|> TITLE: Are Hölder manifolds a thing? QUESTION [17 upvotes]: We know topological manifolds and we know Lipschitz manifolds. It seems that "Hölder manifolds" should be somewhere in between but not much seems published about them. In the context of this question, a Hölder manifold is a topological manifold equipped with an atlas whose transition functions are locally Hölder continuous (that is, belong to the Hölder class $C^\alpha$ for some fixed $\alpha \in (0,1)$. The definition above just mimics the definition of Lipschitz manifolds. What is known about Hölder manifolds, and is there any particular reason less has been published about them than other classes of manifolds? REPLY [10 votes]: I don't think this has been studied but I would note two things. First, Sullivan proved that in dimensions $n\ne 4$ any topological manifold $M^n$ has a Lipschitz structure and any two such structures are Lipschitz equivalent. Therefore the same is true for Hölder structures. So your question is only interesting in dimension 4. In this case AFAIK nothing is known but I would guess that the same is true. Note that this would be different from the Lipschitz case where Sullivan and Donaldson showed that there are 4-manifolds without Lipschitz structures and there are homeomorphic 4-manifolds with nonequivalent Lipschitz structures. But I don't believe this can happen with Hölder structures because it seems to me that it should be possible to make Casson handles Hölder so Freedman's theory should work in the Hölder category for example. This is pure speculation though and one would need to look at the details of the construction and see if it works.<|endoftext|> TITLE: Elliptic operator with finite spectrum? QUESTION [6 upvotes]: Is it possible for a (non-symmetric) elliptic differential operator to have finite spectrum. If so, is there an explicit example? REPLY [2 votes]: This is more a long comment than an answer since some details should be checked. The answer shoould be not for (say) a second order differential operators with continuous top order coefficients and bounded first and zero order ones in bounded domain (with some reularity on the boundary), under say Dirichlet bc. A very nice theorem in the second volume of Dunford Schartz (Theorem 29 pag 1115 and subsequent corollaries) implies that if the resolvent is of trace class and decays like $1/|\lambda|$ on two lines from the origin such that the smallest angle between them can be arbitrarily small, then the linear span of the generalized eigenfunctions is dense in $L^2$. In our situation the domain of the operator is $H^{2}\cap H^1_0$ and the resolvent to the power $k$ is trace class if $k>d/2$ ($d$ is the dimension). The rays above from the origin exist since the operator generates an analytic semigroup of angle $\pi/2$. I admit that many details are left...but it should work.<|endoftext|> TITLE: Correspondences of $\infty$-categories QUESTION [10 upvotes]: In Higher topos theory and Higher algebra Lurie defines (see section 2.3.1 of HTT) a correspondence between two $\infty$-categories $C$ and $D$ as being an $\infty$-category $\mathcal{M}$ over $\Delta[1]$ whose fiber over $0$ and $1$ are respectively $C$ and $D$. Of course this is supposed to be equivalent to presheaves over $C^{op} \times D$, or to left adjoint functor $\widehat{D} \rightarrow \widehat{C}$. This is vaguely claimed as remark 2.3.1.4 in Higher topos theory, but I do not know if the details of this correspondence have been developped somewhere... Is there a reference for this equivalence ? It is possible to deduce it without too much work from other results in the literature ? I need this results for a technical lemma, and so far I don't see how to prove without some hard work on simplicial sets taking several pages... but I'm hopping I'm missing a nicer argument. Remark: it is easy using Lurie's work to attach to $\mathcal{M} \rightarrow \Delta[1]$ a 'profunctor' $C^{op} \times D \rightarrow$ Space, In fact using the twisted arrow category construction (see section 5.2.1 of Higher algebra) one can even give a very explicit right fibration $T \rightarrow C^{op} \times D$ representing this profunctor, but going the other way seem harder... REPLY [12 votes]: Your question is answered by the following result, for which I will give a few references. Theorem. For each pair of simplicial sets $A$ and $B$, there is a functor $$a_{A,B}^* \colon \mathbf{Cyl}(A,B) \longrightarrow \mathbf{sSet}/(A^\mathrm{op} \times B)$$ which is both a left and a right Quillen equivalence between the Joyal model structure on $\mathbf{Cyl}(A,B)$ and the covariant model structure on $\mathbf{sSet}/(A^\mathrm{op} \times B)$. Definitions. (1) The category $\mathbf{Cyl}(A,B)$ of cylinders from $A$ to $B$ (this is Joyal's terminology) is the fibre over the pair $(A,B)$ of the functor $$\mathbf{sSet}/\Delta[1] \longrightarrow \mathbf{sSet} \times \mathbf{sSet}$$ that sends a map $X \to \Delta[1]$ to its fibres over $0$ and $1$. One can show that the category $\mathbf{Cyl}(A,B)$ admits a model structure (called the "Joyal model structure" above) created from Joyal's model structure for quasi-categories by the forgetful functor $$\mathbf{Cyl}(A,B) \longrightarrow \mathbf{sSet}/\Delta[1] \longrightarrow \mathbf{sSet}.$$ If $A$ and $B$ are quasi-categories, then the fibrant objects of this model structure are precisely the correspondences from $A$ to $B$ as you have defined them. (2) The functor $a_{A,B}^*$ sends a cylinder $X \in \mathbf{Cyl}(A,B)$ to the pullback $\require{AMScd}$ \begin{CD} a_{A,B}^*(X) @>>> \mathrm{Tw}(X)\\ @V V V @VV V\\ A^\mathrm{op} \times B @>>i^\mathrm{op}\times j> X^\mathrm{op} \times X \end{CD} where $\mathrm{Tw}(X)$ denotes the twisted arrow of $X$ (whose $n$-simplices are maps $\Delta[n]^\mathrm{op} \star \Delta[n] \to X$), and where $i \colon A \to X$ and $j \colon B \to X$ denote the inclusions of the fibres of the structure map $X \to \Delta[1]$. Proofs. In the case where $A$ and $B$ are quasi-categories, the theorem above is proved by Danny Stevenson in his recent preprint (see Theorem C therein) Danny Stevenson. Model structures for correspondences and bifibrations. arXiv:1807.08226. (Note that Stevenson uses the term "correspondence" for what I have called "cylinder".) As I shall explain below, the general case of the theorem can be proved by a combination of results from my recent preprint Alexander Campbell. Joyal's cylinder conjecture. arXiv:1911.02631. and from Cisinski's recent book Denis-Charles Cisinski. Higher categories and homotopical algebra, vol. 180 of Cambridge studies in advanced mathematics. Cambridge University Press, 2019. First, note that the category $\mathbf{Cyl}(A,B)$ is equivalent to the category $\mathbf{ssSet}/(A^\mathrm{op} \boxtimes B)$ of bisimplicial sets over the exterior product of $A^\mathrm{op}$ and $B$. It follows from Remark 3.23 and Theorem 5.4 of my preprint that, under this equivalence, the Joyal model structure on $\mathbf{Cyl}(A,B)$ corresponds to the bicovariant model structure on $\mathbf{ssSet}/(A^\mathrm{op} \boxtimes B)$ defined in Section 5.5 of Cisinski's book. Now, it is not difficult to see that, under this equivalence, the functor $a_{A,B}^*$ defined above corresponds to the functor $$\delta_{A^\mathrm{op},B}^* \colon \mathbf{ssSet}/(A^\mathrm{op} \boxtimes B) \longrightarrow \mathbf{sSet}/(A^\mathrm{op} \times B)$$ which sends a bisimplicial set over $A^\mathrm{op} \boxtimes B$ to its diagonal (note that the diagonal of $A^\mathrm{op} \boxtimes B$ is $A^\mathrm{op} \times B$). Hence the theorem follows from Theorem 5.5.24 in Cisinski's book, where he proves that the functor $\delta_{A^\mathrm{op},B}^*$ is both a left and a right Quillen equivalence between the bicovariant model structure on $\mathbf{ssSet}/(A^\mathrm{op} \boxtimes B)$ and the covariant model structure on $\mathbf{sSet}/(A^\mathrm{op} \times B)$.<|endoftext|> TITLE: Cyclotomic type criterion for unimodular matrices QUESTION [6 upvotes]: Here unimodular usually means $A \in GL(n,\mathbb{Z})$ (but if you like you can also assume $A \in SL(n, \mathbb{Z})$. In an article I read the following (the problem comes from representation theory of quiver algebras): A unimodular $n \times n$ matrix $A$ has all eigenvalues on the unit circle (this is called cyclotomic) if and only if we have $|Tr(A^k)| \leq n$ for all $k=0,...,n$. (maybe there is a typo in the article and it is meant $|Tr(A)^k| \leq n$ instead?) Usually Im wrong when I find something that looks wrong in an article and since I got noone to ask this in the next weeks I thought about asking here (I can delete the thread in case I made a stupid thinking error,so please do not answer in case I just have a thinking error. ). I think I found a counterexample with GAP (I post the GAP input/output so you can check it if you like): Let $A$ be the following unimodular matrix (this is the coxeter matrix of the Nakayama algebra with Kupisch series [5,5,5,5,5,5,4,3,2,1]): [ [ 0, 0, 0, 0, 0, 0, 1, 0, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 1, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 1, 0 ], [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1 ], [ -1, 0, 0, 0, 0, 0, -1, -1, -1, -1 ], [ 0, -1, 0, 0, 0, 0, 1, 0, 0, 0 ], [ 0, 0, -1, 0, 0, 0, 0, 1, 0, 0 ], [ 0, 0, 0, -1, 0, 0, 0, 0, 1, 0 ], [ 0, 0, 0, 0, -1, 0, 0, 0, 0, 1 ], [ 0, 0, 0, 0, 0, -1, -1, -1, -1, -1 ] ] Then we have W:=[];for i in [0..10] do Append(W,[AbsoluteValue(Trace(g)^i)]);od;W; [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ] and W:=[];;for i in [0..10] do Append(W,[AbsoluteValue(Trace(g^i))]);od;W; [ 10, 1, 1, 2, 1, 4, 4, 6, 1, 2, 6 ] . So by the above theorem, $A$ should be cyclotomic. But GAP gives me for the characteristic polynomial x_1^10+x_1^9-x_1^7-x_1^6-x_1^5-x_1^4-x_1^3+x_1+1 and wolfram alpha says it has a zero 1.17628... which does not lie on the unit circle: https://www.wolframalpha.com/input/?i=x_1%5E10%2Bx_1%5E9-x_1%5E7-x_1%5E6-x_1%5E5-x_1%5E4-x_1%5E3%2Bx_1%2B1 Question: In case there is really something wrong, can the criterion be fixed by using $|Tr(A^k)| \leq n$ for $i=0,1,...,f(n)$ instead (so we replace $n$ by a polynomial function $f(n)$. Maybe $f(n)=n^2$ already works (it does in this example)). It might be interesting to see what a "minimal" f(n) might look like (even if when it is not polynomial). For example how does $f(n)$ start for $n=2,3,...$ (in case it exists). Of course one might also think about this question for various types of matrices, such as $A \in SL(n,K)$ or $GL(n,K)$ , where $K$ is $\mathbb{Z}$, $\mathbb{R}$ or $\mathbb{C}$. edit: Fun questions: 1) What is the number of cyclotomic 0-1 matrices with determinant 1? It seems to start with 1, 3, 27, 756, see https://oeis.org/A185149. 2) What is the number of cyclotomic 0-1 matrices with determinant 1 or -1? It seems to start with 1, 4, 48, 1536, see https://oeis.org/A011266. Sadly my computer seems to be willing to calculate this only until n=4... REPLY [4 votes]: This doesn't have much to do with matrices, I think. Since being cyclotomic and the trace are conjugacy invariants, we may as well assume that $A$ is in Jordan form, say $A=J+N$ with $J$ diagonal and $N$ upper nilpotent. Then WLOG, we may take $A=J$, since eigenvalues and traces of $J^k$ and $(J+N)^k$ are the same. Now one direction is easy. If $A$ is cyclotomic with eigenvalues $e_1,\ldots,e_n$, then $$ |\text{Trace}(A^k)| = |e_1^k+\cdots+e_n^k|\le |e_1|^k+\cdots+|e_n|^k=n. $$ For the other direction, let $e_1,\ldots,e_n$ be the eigenvalues of $A$, or for that matter, an arbitrary Galois invariant set of algebraic integers. If we assume that $$ |e_1^k+\cdots+e_n^k|\le C\quad\text{for $k\ge1$,} $$ then by the usual formulas relating elementary symmetric polynomials to elementary power polynomials, the elementary symmetric polys of $e_1,\ldots,e_n$ are bounded by a function of $C$ and $n$, so there are only finitely many possibilities for $e_1,\ldots,e_n$. But any set of powers $e_1^k,\ldots,e_n^k$ has the same property, so for each $i$, the sequence $(e_i^k)_{k\ge1}$ is a finite set, and hence every $e_i$ is a root of unity. (This is all more-or-less Kronecker's theorem, and can undoubtedly be proven more slickly.)<|endoftext|> TITLE: Example of a weak basic localizer which is not a basic localizer? QUESTION [6 upvotes]: In Grothendieck's homotopy theory, the category $Cat$ of small categories is used to model spaces, or some localization thereof. Grothendieck gives two sets of axioms which the relevant weak equivalences $\mathcal W \subseteq Mor(Cat)$ might be required to satisfy. They are as follows: Definition 1: $\mathcal W \subseteq Mor(Cat)$ is a weak basic localizer if $\mathcal W$ is weakly saturated; For every $C \in Cat$ with a terminal object, the functor $C \to 1$ is in $\mathcal W$; If $u: A \to B$ is a functor and for every $b \in B$, the induced functor of slice categories $u/b: A/b \to B/b$ is in $\mathcal W$, then $u \in \mathcal W$. Definition 2: $\mathcal W \subseteq Mor(Cat)$ is a basic localizer if it is a weak basic localizer and in addition satisfies 3'. If $A \xrightarrow u B \xrightarrow v C$ are functors and for every $c \in C$ the induced functor of slice categories $u/c: A/c \to B/c$ is in $\mathcal W$, then $u \in \mathcal W$. Notes: In condition (1), "weakly saturated" means that $\mathcal W$ contains the identity functors, is closed under 2/3, and has the property that if an idempotent $e$ is in $\mathcal W$, then so are the maps splitting the idempotent. In all the important examples, $\mathcal W$ is actually strongly saturated in the sense that it consists of exactly those functors inverted upon passage to the homotopy category $Cat \to Cat[\mathcal W^{-1}]$. Condition (3) is the case of Condition (3') where $v$ is the identity functor. There are many examples of basic localizers. The minimal one is the class $\mathcal W$ of weak homotopy equivalences (i.e. functors inducing a weak homotopy equivalence of geometric realizations), and they also include cohomological localizations, a localization coming from the plus construction, etc. For much of the theory of test categories, one only needs a weak basic localizer, but e.g. to construct the Cisinski model structure on a presheaf category one needs a basic localizer. Questions: What is an example of a weak basic localizer which is not a basic localizer? What is an example of a (weak) basic localizer for which $\mathcal W$ is not strongly saturated? REPLY [6 votes]: As for basic localizers: they are all strongly saturated (Prop. 4.2.4 in Astérisque 308). I do not know any weak basic localizer which is not a basic localizer, but it is quite likely that there are such things. Here are some suggestions to determine where to look. If $W$ is a basic localizer, we may consider the class $A$ of small categories $C$ such that the map to the final category $C\to e$ is in $W$. There is then: the smallest basic localizer $W'$ containing maps of the form $C\to e$ for $C$ in $A$. the smallest weak basic localizer $W''$ containing maps of the form $C\to e$ for $C$ in $A$. Here are several useful observations: Saying that $W$ is accessible is equivalent to the property that there is a cofibrantly generated model structure with the elements of $W$ as weak equivalences. Furthermore, accessible basic localizers correspond to accessible left Bousfield localizations of the homotopy theory of CW-complexes. This is what Thm. 4.2.15 in Astérisque 308 means. If $W$ is accessible, then, with the notations introduced above, $W'=W''$ and $W'$ is accessible as well. Furthermore, the model structures associated to $W'$ (on the category of small categories or on the category of presheaves on any local test category) are proper, and the equality $W=W'$ holds if and only if the model category structures associated to $W$ are proper; i.e. the proper left Bousfield localizations of the homotopy theory of spaces are exactly the nullifications. This is Theorem 6.1.11 and Corollary 6.1.17 in Astérisque 308. In particular, the smallest weak basic localizer is equal to the smallest basic localizer and thus corresponds to classical homotopy theory; see Thm. 6.1.18 in Astérisque 308. There is another approach to this which I find enlightening. Given any bicomplete $\infty$-category $C$ (or Grothendieck derivator $C$) there is an associated basic localizer $W_C$: the class of functors $f:X\to Y$ such that the induced functor $$f^*:Fun(Y,C)\to Fun(X,C)$$ is fully faithful on constant functors. There is also an associated weak basic localizer $W'_C$: the class of functors $f:X\to Y$ such that the induced functor $$f^*:Fun_{lc}(Y,C)\to Fun_{lc}(X,C)$$ is an equivalence of categories, where $Fun_{lc}(Y,C)$ is the full subcategory of $Fun(Y,C)$ which consists of locally constant functors (i.e. functors $Y\to C$ sending all maps in $Y$ to invertible maps in $C$). This is proven in Prop. 1.12 and 1.16 in my paper Locally constant functors. In the case where $C$ is a proper left Bousfield localization of the Thomason model structure on the category of categories, we can see by descent that $Fun_{lc}(Y,C)$ the induced homotopy theory of $Cat/Y$ (i.e. is equivalent to an appropriate homotopy theory of spaces over $Y$) and thus that $W_C=W'_C$. But in general, there is no reason that $W_C$ and $W'_C$ will agree. All this suggests that if we choose $C$ such that $W_C$ is not proper, there is a good chance that it will not agree with $W'_C$, which would be good starting point towards proving that $W'_C$ is not a basic localizer. I would first try with $C$ the homotopy theory of complexes of vectors spaces over $\mathbf Q$ (the fact that $W_C$ is not proper is Prop. 9.4.6 in Astérisque 308; more generally, Scholie 9.4.7 in loc. cit. is also a fun way to see that the properties of the +-construction show that any Bousfield localization by singular homology with any coefficients is not proper).<|endoftext|> TITLE: What are known properties of matrices where off-diagonal elements are 1? QUESTION [7 upvotes]: Consider a matrix where the diagonal entries are anything but the off-diagonal entries are all one. I was able to find a formula for the determinant of this matrix, but what are other known properties? Does this matrix have a name? In particular is there a formula for its inverse? REPLY [6 votes]: Such a matriix has the form $J +D,$ where $D$ is a diagonal matrix, and $J$ is a square matrix with all entries $1$. One small remark is that if $D$ has two of its diagonal entries equal to $\lambda$, then $\lambda$ is also an eigenvalue of $J+D$. This is because the $\lambda$-eigenspace of $D$ is at least two-dimensional and the $0$-eigenspace of $J$ has codimension one.<|endoftext|> TITLE: The position of a prime between the two neighboring primes QUESTION [5 upvotes]: Let $$ p_0:=2\, \ldots\, p_{n-1}\,\,p_n\,\,p_{n+1}\,\ldots $$ be the increasing sequence of all primes. How often (three questions): $\,p_n > \frac{p_{n-1}+p_{n+1}}2 $ $\,\sqrt{p_{n-1}\cdot p_{n+1}}\, <\, p_n\, \le\, \frac{p_{n-1}+p_{n+1}}2 $ $\, p_n < \sqrt{p_{n-1}\cdot p_{n+1}} $ ? REPLY [5 votes]: First of all, I wouldn't expect anything to be provable, but I expect you can get a good guess for the first question by using Granville's modification of Cramér's random model for prime distributions. For the second question, assume that $p_{n+1}-p_{n-1}<\sqrt{n}/10$ (this is conjectured true for all sufficiently large $n$, but even just the prime number theorem says this is true of all but $O(\sqrt{N}\log^2 N)$ primes less than $N$). Then $$ \frac{p_{n-1}+p_{n+1}}{2}-\sqrt{p_{n-1}p_{n+1}}<0.5 $$ so the question is asking how often $p_{n-1},p_n,p_{n+1}$ are in an arithmetic progression. The difference would have to be a multiple of 6 for all three numbers to be prime. You can get a good guess by summing over all possible multiples of six and using the Hardy-Littlewood conjecture.<|endoftext|> TITLE: Distinguish smooth affine algebraic group by its underlying variety QUESTION [11 upvotes]: If $G_1, G_2$ are two connected affine smooth algebraic groups over an algebraically closed field $k$, and $G_1 \cong G_2$ as algebraic varieties, must they be isomorphic as algebraic groups? Although connected unipotent groups of same dimension all have same underlying varieties (i.e the affine space), I suspect that with suitable reductive assumptions (e.g one of them is reductive + other conditions), $G_1$ and $G_2$ must be isomorphic as algebraic groups. There is a classification of connected reductive groups, but do the topological invariants (e.g fundamental group + dimension + cohomology) determine the group? What if $k$ is only assumed to be a perfect field? Example: the dimension $1$ case and semisimple rank $1$ case are easy by classification. For instance, $SL_2$ and $PGL_2$ are different because they have different fundamental groups. Edit: another way of saying the question is: how to characterize those $G$ on which there exists only one / finitely many smooth (reductive) algebraic group structures? If one is reductive, must the other be? REPLY [23 votes]: $G_1:=GL(2)(\cong SL(2)\rtimes \mathbb G_m)$ is isomorphic as a variety to $G_2:=SL(2) \times \mathbb G_m$ via the map $$ A\mapsto \Big(\big(\begin{smallmatrix}\det(A)^{-1} & 0 \\ 0 & 1 \end{smallmatrix}\big)A , \det(A)\Big). $$ We will show that $G_1$ and $G_2$ (which are both reductive) are not isomorphic as algebraic groups over fields of characteristic 0. This is not immediate, as for $k$ a perfect field of characteristic 2 there is an isomorphism of abstract groups $\psi: GL(2)(k) \to (SL(2) \times \mathbb G_m)(k)$ given by $\psi(A) = (\det(A)^{-\frac{1}2}A, \sqrt{\det(A)})$. So suppose $\varphi: G_1 \to G_2$ is an isomorphism. First proof: Note that $Z(G_1)=\mathbb G_m$ and $Z(G_2)=C_2 \times \mathbb G_m$. Then $\varphi$ induces an isomorphism $Z(G_1) \cong Z(G_2)$, which is absurd since over a field with characteristic different from 2 the equation $x^2=1$ has 2 solutions in $\mathbb G_m$ and 4 solutions in $C_2 \times \mathbb G_m$. Second proof: As $\mathbb G_m$ is diagonalizable and $\varphi^{-1}$ preserves this property, so is $\varphi^{-1}(\mathbb G_m)$ and hence is contained in a maximal torus, which are all conjugate. So $\psi:= \varphi^{-1}|_{\mathbb G_m}$ gives rise to a so-called cocharacter $\psi \in Hom(\mathbb G_m, T)$ for $T$ the maximal torus of diagonal matrices of $GL_2$. By standard facts about cocharacters, $\psi$ has the form $z\to (z^a, z^b)$ for some integers $a, b$. Since the domain of $\psi$ is central in $G_2$, so must be $\text{im } \psi$, which forces $a=b$. Since $\psi$ is injective on $\bar{k}$-rational points, we have $a=\pm 1$. Then $\psi(-1)=(-1, -1)$. On the other hand, $\varphi^{-1}$ affords a 2-dimensional algebraic representation of $SL_2$, so $\varphi^{-1}(SL_2)=SL_2$. This finally gives a contradiction as the images of $\varphi^{-1} (SL_2)$ and $\varphi^{-1}(\mathbb G_m)$ are not disjoint. REPLY [3 votes]: Let's consider the last version of the question over $\mathbb C$. We will prove two facts: I, II. I. If two complex algebraic groups $G_1$ and $G_2$ are diffeomorphic, then they are either both reductive or both non reductive. II. For any reductive complex algebraic group there exist at most finite number of diffeomorphic to it complex algebraic reductive groups. Obvious remark. Of course, two complex Lie groups that the same as algebraic varieties are diffeomorphic. Proof of I. This follows from the following claim. Claim. A complex algebraic $n$-dimensional group is reductive if and only if it is homotopy equivalent to an (orientable) real compact manifold of dimension $n$. Clearly, this shows that a non-reductive group can not be even diffeomorphic to a reductive one. Let's see how to prove this claim. We will use two alternative characterisations of reductive groups. First, a complex algebraic group is reductive if and only if it has a compact real form, see for example here: http://www.math.uchicago.edu/~mbergeron/ComplexReductive.pdf If follows from this, that any complex reductive group is homotopy equivalent to a compact manifold of half dimension, its real compact form. See, for example, "Other characterisations" here: https://en.wikipedia.org/wiki/Reductive_group . For example $(\mathbb C^*)^n$ is homotopy equivalent to $(S^1)^n$. In particular $H_n(G,\mathbb Z)$ of a complex reductive group $G$ of dimension $n$ is $\mathbb Z$. Second. The mid-dimensional homology group vanishes for non-reductive groups, even more, non-reductive groups are homotopy equivalent to compact manifolds of dimensions less than the half. To prove this one uses the following definition of reductive groups: A non-reductive group has a nontrivial normal unipotent subgroup. See here https://en.wikipedia.org/wiki/Reductive_group. Now, if we quotient a group by a normal unipotent subgroup, the homotopy type doesn't change. So we can quotient until we get a reductive group. Proof of II. If is easy to see from the proof of I, that if $G_1$ is diffeomorphic to $G_2$, then their (half-dimensional) compact forms are homotopy equivalent. So we just need to know that in each dimension there exists only finite number of connected compact Lie groups. This is indeed well known, see for example the references in the following mathoverflow answer: Classification of (compact) Lie groups<|endoftext|> TITLE: Do there exist three pairwise independent random variables, such that their sum is zero? QUESTION [6 upvotes]: Do there exist such three non-constant pairwise independent random variables $X, Y, Z$ such that $X + Y + Z = 0$? I managed only to prove the following two facts: If such $X, Y, Z$ exist, they are not independent. Proof: If they are, then $X$ and $-X = Y + Z$ are also independent, which is impossible. If such $X, Y, Z$ exist, then at least two of them do not have finite second moment. Proof: $\DeclareMathOperator\Var{Var}$Suppose, they all have finite second moments. Then $\Var(X) + \Var(Y) + \Var(Z) = 0$, which implies that all $X$, $Y$ and $Z$ are constants. Now suppose that without the loss of generality $Y$ and $Z$ have finite second moment. Then $\Var(X) = \Var(-Y-Z) = \Var(Y) + \Var(Z) \leq \infty$ and we return to the previous case. However, those facts are clearly insufficient to solve this problem. REPLY [10 votes]: Replace $Z$ by $-Z$, so that $Z=X+Y$. Let $f_X$ and $f_Y$ the characteristic functions of $X$ and $Y$, so that $f_X(s)=Ee^{isX}$ for real $s$. Suppose the pairwise independence. Then for all real $s$ and $u$ $$f_X(u)f_Y(u)f_X(s)=f_Z(u)f_X(s)=Ee^{iuZ+isX} \\ =Ee^{i(u+s)X+iuY}=f_X(u+s)f_Y(u). \tag{1} $$ Therefore and because $f_Y$ is continuous with $f_Y(0)=1\ne0$, we have $$f_X(u+s)=f_X(u)f_X(s) \tag{2} $$ for all real $u$ close enough to $0$ and all real $s$. Note that (2) (together with the conditions that $f_X$ is continuous with $f_X(0)=1\ne0$) implies that $f_X$ is nowhere $0$. Similarly, $f_Y$ is nowhere $0$. So, (2) actually holds for all real $u$ and $s$. So, $f_X(s)=e^{isa}$ for some real $a$ and all real $s$. So, $X$ is a constant almost surely. Similarly, $Y$ is a constant almost surely.<|endoftext|> TITLE: Isomorphic Jacobian Varieties Just Like Abelian Varieties — Torelli's Theorem QUESTION [9 upvotes]: Torelli's theorem states: Let $R$, $R'$ be compact Riemann surfaces of genus $g$, $J(R)$, $J(R')$ their Jacobian varieties, $\Theta$, $\Theta'$ their respective theta divisors. The Riemann surfaces $R$ and $R'$ are isomorphic if and only if $(J(R), \Theta)$ and $(J(R'), \Theta')$ are isomorphic as principally polarized Abelian varieties. In this theorem, $J(R)$ and $J(R')$ are required to be isomorphic not only as Abelian varieties but also as principally polarized Abelian varieties. It turns out that the condition for $J(R)$ and $J(R')$ to be isomorphic as Abelian varieties alone need not imply that $R$ and $R'$ are isomorphic. Where can I find an example that shows that $J(R)$ and $J(R')$ being isomorphic just as Abelian varieties, does not imply that $R$ and $R'$ are isomorphic? REPLY [6 votes]: Consider the case of curves of genus $2$. If $\mathrm{A}$ is an abelian surface and $\mathrm{C}$ a smooth curve in $\mathrm{A}$ of genus $2$, then $\mathrm{A}\simeq\mathrm{J}(\mathrm{C})$ and $\mathrm{C}$ is the theta divisor of $\mathrm{J}(\mathrm{C})$. The special case $\mathrm{A}=\mathrm{E}\times\mathrm{E}$ (where $\mathrm{E}$ is an elliptic curve) was studied in this paper by Hayashida. It is known that for a given abelian variety $\mathrm{A}$ there are only finitely many curves with Jacobian $\mathrm{A}$ (see this paper by Narasimhan and Nori; for surfaces this was proven much earlier by Hayashida and Nishi). The first paper I mentioned gives in fact formulae (depending on the nature of $\mathrm{End(E)}$) for the number of curves $\mathrm{C}$ with Jacobian $\mathrm{A}=\mathrm{E}\times\mathrm{E}$. More explicit examples (in the sense that the equations for the curves $\mathrm{C}$ can be written down) were constructed by Howe.<|endoftext|> TITLE: Lower bound for the order of a simple group with a given class number QUESTION [8 upvotes]: Every simple group below are assumed non-abelian. Let us call the class number $k(G)$ of a finite group $G$ the number of its conjugacy classes (also, the number of its irreducible complex representations, up to equivalence). Here is the computation of the order and the class number of the first finite simple groups: gap> it:=SimpleGroupsIterator(10,1000); gap> for G in it do Print([G,Order(G),NrConjugacyClasses(G)]); od; [ A5, 60, 5 ][ PSL(2,7), 168, 6 ][ A6, 360, 7 ][ PSL(2,8), 504, 9 ][ PSL(2,11), 660, 8 ] The points of the following picture corresponds to the $161$ finite simple groups of order less than $10^8$, where $x$-axis is the class number and $y$-axis the order. We observe that these points are bounded below by the curve interpolating the points for the groups $\mathrm{PSL}(2,2^n)$. We checked that $k(\mathrm{PSL}(2,2^n)) = 2^n+1$, for $n \le 50$, so that the equality should be known true in general. Now $$|\mathrm{PSL(2,2^n)}| = 2^n (2^{2n}-1) = (r-1)((r-1)^2-1)) = (r-2)(r-1)r,$$ with $r = 2^n+1$, which leads to: Question: Let $G$ be a finite simple group of class number $r$. Is it true that $|G| \ge (r-2)(r-1)r$? If so, do you expect the existence of a proof without CFSG? (because motivation: extend such a result to the simple integral fusion rings) gap> for r in [5..15] do Print([r,(r-2)*(r-1)*r]); od; [ 5, 60 ][ 6, 120 ][ 7, 210 ][ 8, 336 ][ 9, 504 ][ 10, 720 ][ 11, 990 ][ 12, 1320 ][ 13, 1716 ][ 14, 2184 ][ 15, 2730 ] Bonus We checked that $k(\mathrm{PSL}(2,3^n)) = (5 + 3^n) / 2$ for $n \le 40$ (see A289521), moreover, $$|\mathrm{PSL}(2,3^n)| = \frac{1}{2}3^n (3^{2n}-1) = \frac{1}{2}(2r-5)((2r-5)^2-1)) = \frac{1}{2}(2r-6)(2r-5)(2r-4),$$ with $r = \frac{1}{2}(5 + 3^n)$. These groups seems to be in the main curve in the middle. Then: Augmented question: Let $G$ be a finite simple group of class number $r$. Is it true that $$|G| \ge (r-2)(r-1)r,$$ and that the equality holds iff $G = \mathrm{PSL}(2,2^n)$, otherwise that $G=\mathrm{PSU}(3,3)$ xor $$|G| \ge \frac{1}{2}(2r-6)(2r-5)(2r-4),$$ and that the equality holds iff $G = \mathrm{PSL}(2,q)$ with $q$ a odd prime power? Remark: It is checked by GAP for $|G| < 15000000$. gap> for r in [5..15] do Print([r,(2*r-6)*(2*r-5)*(2*r-4)/2]); od; [ 5, 60 ][ 6, 168 ][ 7, 360 ][ 8, 660 ][ 9, 1092 ][ 10, 1680 ][ 11, 2448 ][ 12, 3420 ][ 13, 4620 ][ 14, 6072 ] [ 15, 7800 ] REPLY [7 votes]: Apart from finitely many examples, such results would follow from the work of Liebeck and Shalev (Proc. London Math. Soc. 2005). In Corollary 5.2 of that paper they show that for some constant $c >0$ one has $$ |G| \ge c k(G)^4, $$ except for the finite simple groups in the families $L_2(q)$, $L_3(q)$ and $U_3(q)$. Naturally in the remaining families one can compute $k(G)$ directly. REPLY [5 votes]: It has been proved by J. Fulman and R. Guralnick that is $G$ is ``almost simple", (and $F(G) = 1$), then $k(G) < |G|^{0.41}.$ This yields $|G| > k(G)^{2.436}.$ This result needs the classification of finite simple groups. I do not know how much it can be improved by restricting to simple groups. I would not expect your question to be answered without the classification of finite simple groups. It would seem to be necessary, for example, to at least know that there are only finitely many sporadic simple groups.<|endoftext|> TITLE: Cubic polynomials over finite fields whose roots are quadratic residues or non-residues QUESTION [11 upvotes]: For a cubic polynomial $f(x)=x^3+x^2+\frac{1}{4}x+c$ over $\mathbb{F}_q$, where $q$ is a odd prime power, I find that for a lot of $q$, there does not exist $c\in\mathbb{F}_q$ such that $f$ has three distinct roots in $\mathbb{F}_q$, one of which is a quadratic residue and the other two are non-residues. I have not found any counter examples, so my question is, does it hold for any $q$? If so, what forms of cubic polynomials have such property? REPLY [4 votes]: I have an indirect proof using a result from Leonard's paper in 1969 "On Factoring Quartics (mod $p$)": Lemma. Let $f(x)=x^4+a_2x^2+a_1x+a_0$ be a quartic polynomial over $\mathbb{F}_q$ having four distinct roots in its splitting field, where $q$ is an odd prime power and $a_1\ne 0$, then $$g(x)=x^3+8a_2x^2+(16a_2^2-64a_0)x-64a_1^2$$has one root being a non-zero quadratic residue in $\mathbb{F}_q$, and the other two roots being quadratic non-residues in $\mathbb{F}_q$, if and only if $f(x)=h_1(x)h_2(x)$, where $h_1,h_2$ are irreducible quadratics over $\mathbb{F}_q$. Assume that among the roots of $g(x)=x^3+x^2+\frac{1}{4}x+c$, there are one non-zero quadratic residue and two non-residues, then $-c$ must be a square in $\mathbb{F}_q$. It is easy to see that $g(x)$ is a resolvent cubic polynomial of $f(x)=x^4+\frac{1}{8}x^2+\frac{1}{8}\sqrt{-c}x$. This contradicts the above lemma. Furthermore, for $f(x)=x^4+a_2x^2+a_1x+a_0$ , let $a_0=0$, then $b=\frac{a^2}{4}$ for its solvent cubics $g(x)=x^3+ax^2+bx+c$. Thus $b=\frac{a^2}{4}$ is a sufficient condition for the property.<|endoftext|> TITLE: Small automorphism groups of groups QUESTION [10 upvotes]: I do not know much about group theory, so sorry in case this question is not for MO. For a finite group $G$, denote by $f(G)$ the number of elements of the automorphism group of $G$. Question: For which $n$ is there a unique group with $n$ elements such that $f(G)$ is minimal among all groups with n elements? What is the sequence $a_n$ giving the minimal $f(G)$ for a given $n$. For $n \leq 63$ (except $n=32$ which has too many groups for my computer it seems and I was not able to finish the calculation with GAP. Ill try it after the holidays with a better computer) it was true that there is a unique such group namely $G=Z/Zn$, so $a_n$ is the Euler totient function $\phi(n)$ for those values. This suggests the following question: Question: Do we have $f(G) \geq \phi(n)$ in case $G$ has $n$ elements? REPLY [9 votes]: The following conjecture is problem 15.43 in The Kourovka Notebook, attributed to M. Deaconescu : Let $G$ be a finite group of order $n$. a) Is it true that $|Aut(G)|\geq \varphi (n)$ where $\varphi$ is the Euler function? b) Is it true that $G$ is cyclic if $|Aut(G)|=\varphi (n)$? The problem is now in the solved section. In fact J. N. Bray and R. A. Wilson in the paper "On The Orders of Automorphism Groups of Finite Groups" show that $\frac{|Aut(G)|}{\varphi(|G|)}$ can be arbitrarily close to zero. And the same holds if we add the extra restrictions of $G$ being perfect or soluble as shown in "On The Orders of Automorphism Groups of Finite Groups II" The failure of the conjecture in this strong form (ratio can be arbitrarily close to zero) was shown to hold even when restricting to $p$-groups $G$ by J. González-Sánchez and A. Jaikin-Zapirain in "Finite p-groups with small automorphism group".<|endoftext|> TITLE: Question about functional derivatives QUESTION [14 upvotes]: This page on Wikipedia defines the so-called functional derivative as follows: "Given a manifold $M$ representing (continuous/smooth) functions $\rho$ (with certain boundary conditions, etc.) and a functional $F: M \to \mathbb{C}$, the functional derivative of $F[\rho]$, denoted by $\delta F/\delta \rho$ is defined by: $$\int \frac{\delta F}{\delta \rho}(x) \phi(x) dx = \lim_{\epsilon \to 0}\frac{F[\rho+\epsilon \phi]-F[\rho]}{\epsilon} = \bigg{[}\frac{d}{d\epsilon}F[\rho+\epsilon \phi]\bigg{]}_{\epsilon = 0} \label{1}\tag{1}$$ where $\rho$ is an arbitrary function. In other words, $$ \phi \mapsto \bigg{[}\frac{d}{d\epsilon}F[\rho+\epsilon \phi]\bigg{]}_{\epsilon = 0} \label{2}\tag{2}$$ is a linear functional, so by Riesz-Markov-Kakutani Representation Theorem, this functional is given by integration against some measure. Then $\delta F/\delta \rho$ is defined to be the Radon-Nikodym derivative of this measure." Now, regarding this definition, I have two questions: I understand that the linear functional \eqref{2} is nothing but the Gâteaux derivative of $F$ (if it exists). Now, as far as I know, the Riesz-Markov-Kakutani Representation Theorem is related to positive linear functionals, not just arbitrary linear functionals and I see no reason why the Gâteaux derivative \eqref{2} should be (always) positive. Does it mean that the functional derivative of $F$ exists if it is Gâteaux differentiable and its Gâteaux derivative is positive? If this is the case, this seems to imply that existence of Gâteaux derivative does not imply existence of functional differentiable but the converse is holds. Is the limit in \eqref{1} uniform, i.e., does it depend on the choice of $\phi$? I assume it does not because functional derivatives are usually referred to as Frechet derivatives and the latter is some sort of uniform Gâteaux derivative. Is this correct? REPLY [15 votes]: Premise (a long one): before answering your questions, I must say that, if your are searching mathematically rigorous informations, you should not rely on Wikipedia entry "Functional derivative" in its current status, since it is seriously flawed due to an "edit war" between me and another contributor (or perhaps it would be better to say between him and all other contributors, as you can notice having a look at the talk page of the entry). Due to this, the entry is written more from a theoretical physicist's point of view than from a contemporary mathematical perspective, and its content adheres strictly and tacitly to the hypotheses assumed (even implicitly) by Vito Volterra ([6], §II.1.26-II.1.28, pp. 22-24). Namely Volterra implicitly assumes that the functional $F$ is of integral type, i.e. similar to the functionals encountered in the classical calculus of variation. However, in general functional analysis, this is not always true: for example, the following functional, defined defined on ${C}^1(\Omega)$, $\Omega\subseteq\Bbb R^n$ $$ F[\rho]=\sum_{i=1}^n\frac{\partial\rho}{\partial x_i}(0)=\langle\vec{\mathbf{1}},\nabla \rho(0)\rangle \neq\int\limits_{\Omega}\!\rho(x)\,\mathrm{d}\mu_x,\label{nif}\tag{NIF} $$ cannot be expressed in the form of an integral respect to any given measure, as it is well known from the theory of distributions. In general, the functional derivative cannot always be represented as the left side term of \eqref{1} since it may not be defined and, even if it happens to be so, it can be different from the central and right side ones (which represent however the true definition of functional derivative), unless it is interpreted as a distribution or as another kind of generalized function by abuse of notation. However, there is a deeper issue, described in the following point. Volterra explicitly assumes that the variation of $F$ i.e. the quantity $$ \Delta F[\rho]=F[\rho+\delta\rho]-F[\rho]=F[\rho+\varepsilon\phi]-F[\rho] $$ is linear respect to the increment $\delta\rho=\varepsilon \phi$ apart from a remainder behaving as $o(\varepsilon)$ as $\varepsilon\to 0$. Now , while the requirement on asymptotic behavior is basically equivalent to the existence of the limit \eqref{1}, the linearity hypothesis is not always satified ([3], §3.1-3.3, pp. 35-40, and [4] §2.1 p. 15, §3.1-3.3 pp. 30-33). For example, the following functional defined on ${C}^1(\Bbb R^n)$ by using a function $\rho_o\in C^1(\Bbb R^n)$ such that $\rho_0\not\equiv 0$, $$ F[\rho] = \int\limits_{G} \frac{|\nabla(\rho(x)-\rho_o(x))|^2}{\rho(x)-\rho_o(x)}\exp\left(-\frac{|\nabla(\rho(x)-\rho_o(x))|^4}{|\rho(x)-\rho_o(x)|^2}\right) \mathrm{d}x, \quad G\Subset\Bbb R^n \label{nlf}\tag{NLF} $$ (the specification of the precise form of the integrand on the zero set of $\rho-\rho_0$, as well as on intersection between this set and the zero set of its gradient, would require a little more care, but this is only a technical detail and adds nothing to the answer) has a functional derivative which is not linear at the point $\rho_o$. Indeed, given $\phi\in C^1(\Bbb R^n)$ such that $\phi\neq 0$ in $G$, $$ F[\rho_o+\varepsilon \phi] = \varepsilon\int\limits_{G} \frac{|\nabla \phi(x)|^2}{\phi(x)}\exp\left(-\varepsilon^2\frac{|\nabla\phi(x)|^4}{|\phi(x)|^2}\right) \mathrm{d}x, $$ thus $$ \bigg{[}\frac{\mathrm{d}}{\mathrm{d}\varepsilon}F[\rho+\varepsilon \phi]\bigg{]}_{\varepsilon = 0} = \int\limits_{G} \frac{|\nabla \phi(x)|^2}{\phi(x)} \mathrm{d}x $$ Furthermore, while Volterra developed his functional calculus having in mind Banach spaces of continuous functions with respect to the uniform norm (even if the concept of a Banach spaces was not yet defined at the time), theoretical physicists apply it to far more general contexts, in general without any formal justification. Said that, I can proceed and answer to your questions. I understand that the linear functional \eqref{2} is nothing but the Gâteaux derivative of $F$ (if it exists). Now, as far as I know, the Riesz-Markov-Kakutani Representation Theorem is related to positive linear functionals, not just arbitrary linear functionals and I see no reason why the Gâteaux derivative \eqref{2} should be (always) positive. Does it mean that the functional derivative of $F$ exists if it is Gâteaux differentiable and its Gâteaux derivative is positive? If this is the case, this seems to imply that existence of Gâteaux derivative does not imply existence of functional differentiable but the converse is holds. As is, that statement in the entry is not correct without assuming something on where the functional $F$ is defined and thus on its structure. You have correctly noticed one of the basic issues: the functional derivative of $F$ is assumed to be a Gâteaux derivative, but this does not implies its positivity, and moreover it does not need to be representable as a measure, as example \eqref{nif} above shows. For example it can be thought as a distribution, as shown in this answer. Volterra derives the integral representation for the functional derivative on the left side of \eqref{1} under precise hypothesis ([6] §II.1.27 pp. 23-24 and reference [5] §2, pp. 99-102 cited therein), having in mind applications to the classical calculus of variation: under different hypotheses, this may not be true. Is the limit in \eqref{1} uniform, i.e., does it depend on the choice of $\phi$? I assume it does not because functional derivatives are usually referred to as Frechet derivatives and the latter is some sort of uniform Gâteaux derivative. Is this correct? The limit depends on the structure of $\phi$, not only on its "size" (i.e. its norm when $M$ is a Banach space): this is probably the core difference between Gâteaux and Fréchet derivatives of functionals, with the former one playing the infinite dimensional analogue of the directional derivative ([1] §1.1 p. 12 and [2] §1.B p. 11). When $M$ is Banach, the statement is clear since $\phi$ enters the definition, equivalent to \eqref{2}, of Fréchet derivative only with its norm, and this implies that any $\phi$ with the same norm does the job: for more general topological vector spaces, things are more complex, but you can have a look at references [4], §3.2-3.2 pp. 30-32 for Gâteaux derivatives and to [2] §1.B p. 11 for Fréchet derivatives (see however [1] remark 1.2 pp. 11-12, on the definition of Fréchet derivatives in locally convex spaces and the issues involved in defining higher order derivatives). Bibliographical note Vainberg ([3], [4]) explicitly says that the functional derivative can be a nonlinear functional of the increment: however, he calls it Gâteaux differential, reserving the name "derivative" for the cases where it is a linear functional, and this nomenclature seems to be non standard. All other authors deal extensively only with functionals having linear functional derivatives, sometimes not even mentioning the possibility of the existence of functionals like \eqref{nlf}. Bibliography [1] Ambrosetti, Antonio; Prodi, Giovanni, A primer of nonlinear analysis, Cambridge Studies in Advanced Mathematics, 34. Cambridge: Cambridge University Press, pp. viii+171 (1993), ISBN: 0-521-37390-5, MR1225101, ZBL0781.47046. [2] Schwartz, Jacob T., Nonlinear functional analysis, Notes by H. Fattorini, R. Nirenberg and H. Porta. With an additional chapter by Hermann Karcher. (Notes in Mathematics and its Applications.) New York-London-Paris: Gordon and Breach Science Publishers, pp. VII+236 (1969), MR0433481, ZBL0203.14501. [3] Vaĭnberg, Mikhail Mordukhovich, Variational methods for the study of nonlinear operators. With a chapter on Newton’s method by L.V. Kantorovich and G.P. Akilov, translated and supplemented by Amiel Feinstein, Holden-Day Series in Mathematical Physics. San Francisco-London- Amsterdam: Holden-Day, Inc. pp. x+323 (1964), MR0176364, ZBL0122.35501. [4] Vaĭnberg, Mikhail Mordukhovich, Variational method and method of monotone operators in the theory of nonlinear equations. Translated from Russian by A. Libin. Translation edited by D. Louvish, A Halsted Press Book. New York-Toronto: John Wiley & Sons; Jerusalem-London: Israel Program for Scientific Translations, pp. xi+356 (1973), MR0467428, ZBL0279.47022. [5] Volterra, Vito, "Sulle funzioni che dipendono da altre funzioni [On functions which depend on other functions]" (in Italian), Atti della Reale Accademia dei Lincei, Rendiconti (4) III, No. 2, 97-105, 141-146, 153-158 (1887), JFM19.0408.01. [6] Volterra, Vito, Theory of functionals and of integral and integro-differential equations. Dover edition with a preface by Griffith C. Evans, a biography of Vito Volterra and a bibliography of his published works by Sir Edmund Whittaker. Unabridged republ. of the first English transl, New York: Dover Publications, Inc. pp. 39+XVI+226 (1959), MR0100765, ZBL0086.10402.<|endoftext|> TITLE: Learning roadmap for Foundations of Mathematics (for the working mathematician) QUESTION [39 upvotes]: (At the risk of being vapulated and downvoted, I'll ask this here.) Suppose you work in a field that has nothing to do with the foundations of mathematics, but thanks to MO, you are becoming more and more interested in topics like axiomatic set theory, the different logical systems (intuitionistic, classical, finitist), category theory, type theory, etc. Thus, you want to understand all of this, and you can spend some time learning from books and articles, without any hurry. But you don't want to actually do research is this field. This is the case for me. Question: Is there any organized way to achive this? Any book recomendatons to achive this goal? At my university there is no one working in this area, hence I have no one to ask this question directly. My background is naive set theory, naive category theory and some basic logic. Let me explain with an example where I want to get. Suppose you saw the recent Zizek/Peterson debate, and you have read some of their books, you understood their ideas and you have an opinion. But you can't, and don't want, to sit in front of public and debate with either of them. (Of course, you would like to sit and chat with Zizek for hours :)) Thank you very much. REPLY [2 votes]: Paul Taylor's book should definitely be on this big list! It's not the book I would start with, but the sooner you open it, the better. Paul Taylor, Practical Foundations Of Mathematics.<|endoftext|> TITLE: Supremum of $ a_n = a_{n-1}^3 - a_{n-2} $ QUESTION [13 upvotes]: Let $a_1=0$ and let $ - \ln(2) < a_2 < \ln(2) $ Define $$ a_n = a_{n-1}^3 - a_{n-2} $$ Then $$ \sup_{n>2} a_n = a_2 $$ And $$ \inf_{n>2} a_n = - a_2 $$ How to prove that ? REPLY [6 votes]: I encountered a similar problem with the iteration $u_n=\sqrt{u_{n-1}^2+1}-u_{n-2}$, where there is fundamentally a $9$-cycle in the shape of a maple leaf (replace the +1 by 0 to see this). I asked an expert 20 years ago who told me that using the KAM (Kolmogorov-Arnold-Moser) theorem, one could prove that the "curve" drawn above by R. Israel is not a curve at all, but a very thin strip of width 0.07 or something. Don't ask me for the proof, I have no idea.<|endoftext|> TITLE: Equivariant cohomology algebra of toric variety QUESTION [5 upvotes]: Let $X$ be a complex projective and smooth toric variety of complex dimension $n$. It is acted by the real torus $T=(S^1)^n$. Is it true that the $T$-equivariant cohomology $H^*_T(X,\mathbb{Z})$ of $X$ is isomorphic as a graded algebra to $$H^*(X,\mathbb{Z})\otimes H^*_T(pt,\mathbb{Z})\simeq H^*(X,\mathbb{Z})\otimes \mathbb{Z}[x_1,\dots,x_n],$$ where $x_i$ have degree 2? (to prove such an isomoprhism with coefficients in a field would be fine too.) Remark. I think I have a proof of such an isomorphism (with $\mathbb{Q}$-coefficients) in the category of graded $H_T^*(pt, \mathbb{Q})$-modules rather than graded algebras (which seems to be well known). REPLY [4 votes]: As per Mark Grant's suggestion, here are some additional details for my comment. We may write $\Bbb P^1$ with its circle action as the adjunction space $$* \cup_{S^1 \times 0} S^1 \times [0,1] \cup_{S^1 \times 1} *,$$ with the circle acting in the obvious way on $S^1 \times [0,1]$. Passing to the Borel construction, we find that $$\Bbb P^1 \times_{S^1} ES^1 = BS^1 \cup_{ES^1 \times 0} ES^1 \times [0,1] \cup_{ES^1 \times 1} BS^1.$$ The picture is that we are connecting two copies of $BS^1$ by a contractible bit. To make this precise, collapse the closed contractible subspace $ES^1 \times 1/2$ to a point. As this is contractible, the projection map is a homotopy equivalence; thus $\Bbb P_{hS^1}$ is homotopy equivalent to the wedge of two mapping cones $$\text{Cone}(ES^1 \to BS^1) \vee \text{Cone}(ES^1 \to BS^1).$$ Because $ES^1$ is contractible, each of these cones are homotopy equivalent to $BS^1$ itself. Thus we see that $$\Bbb{CP}^\infty \vee \Bbb{CP}^\infty \simeq \Bbb P^1_{hS^1};$$ further the projection map $\Bbb P^1_{hS^1} \to \Bbb{CP}^\infty$ sends the two wedge summands identically onto $\Bbb{CP}^\infty$. Thus $$H^*_{S^1}(\Bbb P^1;\Bbb Z) \cong \Bbb Z[x,y]/(xy),$$ with action of $u \in H^2_{S^1}(pt)$ given by $u \cdot x = x^2, u \cdot y = y^2.$ This is isomorphic as a graded module to $$H^*(\Bbb P^1;\Bbb Z) \otimes H^*_{S^1}(pt;\Bbb Z),$$ but not as an algebra: the equivariant cohomology has no nilpotent elements, whereas the tensor-product algebra does, given by the generator of $H^2(\Bbb P^1) \otimes H^0_{S^1}(pt)$.<|endoftext|> TITLE: Are the L-functions of a normalized newform and the corresponding cuspidal representation equal? QUESTION [9 upvotes]: Let $f \in S_k(\Gamma_0(N))$ be a normalized newform with Fourier expansion $$f(z) = \sum\limits_{n=1}^{\infty} a_n e^{2\pi i z n}$$ and $a_1 = 1$. Then $f$ is an eigenfunction of all Hecke operators $T_p$, not just those with $(p,N) = 1$, and for the normalized L-function $$L^{S_{\infty}}(f,s) = \sum\limits_{n=1}^{\infty} \frac{n^{(1-k)/2}a_n}{n^s}$$ one can add in an archimedean factor to make a completed L-function $L(f,s)$ which is an Euler product $$L(f,s) = \prod\limits_{p \leq \infty} L_p(f,s)$$ and, for a suitable "contragredient" $g \in S_k(\Gamma_0(N))$ and epsilon root number $\epsilon(f,s)$, satisfies the functional equation $$L(f,s) = \epsilon(f,s) L(g,1-s).$$ See for example Kowalski's article "Classical Automorphic Forms" in the book "An Introduction to the Langlands Program." Now, $f$ can be identified in various ways with a cuspidal automorphic form on $\operatorname{GL}_2(\mathbb Q) \backslash \operatorname{GL}_2(\mathbb A_{\mathbb Q})$. There is a cuspidal automorphic representation $\pi = \otimes \pi_p$, unique up to isomorphism, which contains $f$. The normalized newform $f$ is conversely uniquely determined by $\pi$. This is one way to state the "Multiplicity One" theorem. Also, $\pi$ itself has an L-function $L(\pi,s) = \prod\limits_{p \leq \infty} L(\pi_p,s)$ satisfying its own functional equation $L(\pi,s) = \epsilon(\pi,s) L(\tilde{\pi},1-s)$. At the primes $p$ not dividing $N$, the representation $\pi_p$ is spherical, and the L-function $L(\pi_p,s)$ arises from the local spherical Hecke algebra of $\operatorname{GL}_2(\mathbb Q_p)$. Here things can be normalized so that $$L(\pi_p,s) = L_p(f,s). \tag{1} $$ What about the primes dividing $N$? Here $L_p(f,s)$ arises from a Hecke operator $T_p$, but the representation $\pi_p$ could be supercuspidal, or it could have an Iwahori-fixed vector but be non-spherical. With the proper normalizations, can we have equation (1) holding for all primes $p$ (and consequently $L(\pi,s) = L(f,s)$)? Note that when $\pi_p$ is supercuspidal or is induced from a ramified character, $L(\pi_p,s) = 1$. I expect it's possible to normalize things so that the local factors of $L(f,s)$ and $L(\pi,s)$ agree at all places, but I have never seen any reference which does this. REPLY [10 votes]: $L(\pi,s)$ agrees with $L(f,s)$ if $f\in\pi$ is a newform, and this is even true for $\mathrm{GL}_n$. Of course, things are complicated by the fact that there are many ways to define $L(\pi,s)$ and $L(f,s)$. I found the following papers very useful to check various consistencies: Schmidt and Kondo-Yasuda. Here is a summary based on Kondo-Yasuda. Let $G:=\mathrm{GL}_n$. Let $(\pi,V_\pi)$ be a cuspidal automorphic representation of $G$ over $\mathbb{Q}$ of unitary central character $\omega$. Let $K_p(c)$ denote the subgroup of elements of $G(\mathbb{Z}_p)$ whose bottom row is congruent to $\begin{pmatrix}0 & \cdots & 0 & 1\end{pmatrix}$ modulo $c$. Let $K(c):=\prod_p K_p(c)$. The conductor $c_\pi$ is the smallest $c$ such that $V_\pi^{K(c)}\neq\{0\}$. An element of $V_\pi^{K(c_\pi)}$ is called a global newform: it is an eigenfunction of the convolution by any element of $C_c(K_p(c_\pi)\backslash G(\mathbb{Q}_p)/K_p(c_\pi))$. Let $c_\omega$ be the conductor of $\omega$, and consider the corresponding primitive Dirichlet character modulo $c_\omega$: $$\chi_\omega(m):=\omega(1,\underbrace{1,\dotsc,1}_{v\mid c_\omega},\underbrace{m,m,\dotsc}_{v\nmid c_\omega})= \omega(1,\underbrace{m^{-1},\dotsc,m^{-1}}_{v\mid c_\omega},\underbrace{1,1,\dotsc}_{v\nmid c_\omega}),$$ for $m>0$ and $(m,c_\omega)=1$. Note that $c_\omega\mid c_\pi$. Let $\chi_\pi$ denote the Dirichlet character modulo $c_\pi$ induced by $\chi_\omega$. For $k\in\{1,\dotsc,n-1\}$, we define the $k$-th Hecke operator at $p$ as the characteristic function of $$H_k(p):=K_p(c_\pi)\,\mathrm{diag}(\underbrace{p,\dotsc,p}_\text{$k$ entries},\underbrace{1,\dotsc, 1}_\text{$n-k$ entries})\,K_p(c_\pi).$$ Accordingly, let $\lambda_{\pi,k}(p)$ be the $k$-th normalized Hecke eigenvalue at $p$: $$\int_{H_k(p)}f(xg)\,dg=\lambda_{\pi,k}(p)p^{\frac{k(n-k)}{2}}\mathrm{vol}(K_p)f(x),\qquad f\in V_\pi^{K(c_\pi)}.$$ Further, let $\lambda_{\pi,0}(p):=1$ and $\lambda_{\pi,n}(p):=\chi_\pi(p)$. Theorem (Tamagawa 1963, Satake 1963, Kondo-Yasuda 2010). $$L(\pi,s)=\prod_p\left(\sum_{k=0}^n(-1)^k\lambda_{\pi,k}(p)p^{-ks}\right)^{-1}.$$<|endoftext|> TITLE: The direct product of the geometric fundamental group and the absolute Galois group QUESTION [8 upvotes]: Given a geometrically connected variety $X$ over $\mathbb{Q}$ we have a short exact sequence $$ 1\to \pi_1(X_{\overline{\mathbb{Q}}})\to \pi_1(X)\to Gal(\overline{\mathbb{Q}}/\mathbb{Q})\to 1. $$ A rational point on $X$ provides a splitting of this sequence i.e. it exhibits $\pi_1(X)$ as a semidirect product of $\pi_1(X_{\overline{\mathbb{Q}}})$ and the absolute Galois group of $\mathbb{Q}$. Does there exist a smooth and proper variety (to avoid trivialities with an infinite geometric fundamental group) such that $\pi_1(X)$ is actually the direct product of these two groups? Is there an example where such a splitting comes from a rational point? REPLY [4 votes]: If $X$ is an abelian variety, then it is not a direct product. For example, because the absolute galois group cannot act trivially on the torsion points (e.g. by Mordell-Weil). It follows that it is not a direct product for any $X$ with a non-trivial Albanese.<|endoftext|> TITLE: Examples of non-cubulated hyperbolic groups QUESTION [8 upvotes]: What is known regarding which hyperbolic groups are cubulated? I take it the usual definition of cubulated is acting properly and cocompactly on a CAT(0)-cube complex. My impression is that not all of them are, but I didn't manage to find references with a counterexample. Are there known ways to create non-cubulated hyperbolic groups? Are there famous examples of non-cubulated groups? REPLY [6 votes]: As @AGenevois says in his answer, the standard examples of non-cubulated hyperbolic groups are those with Kazhdan's property (T), such as quaternionic hyperbolic lattices. Complex hyperbolic lattices (in dimension >2) provide a more delicate class of examples. On the one hand, they are not cubulable, by a theorem of Delzant--Py. On the other hand, they are known to have the Haagerup property (I think this is proved in the book by Bekka--de la Harpe--Valette), so they also don't have property (T). As far as I know, they are the only class of examples of hyperbolic groups known to be Haagerup but non-cubulable. I'd be interested to hear of others.<|endoftext|> TITLE: $GL(\infty)$ group action through the boson-fermion correspondence QUESTION [6 upvotes]: Every point of the Sato Grassmannian can be used to generate a tau function of the KP hierarchy. In addition, the Sato Grassmannian can be seen as a subset of the "second quantized fermion Fock space" $\mathcal{F}$ of the Clifford algebra. The infinite dimensional group $GL(\infty)$ has a representation on the Fock space $\mathcal{F}$. An element $A_{ij} \in GL(\infty)$ describes an operator, $$U = \exp(\sum_{ij}A_{ij}\psi_i \psi_j^*)$$ that acts transitively on the space of tau functions. The boson-fermion correspondence gives a description of this picture in terms of the Heisenberg algebra and it's Fock space. The precise isomorphism is given by, $$\sigma: \mathcal{F}^{f} \to \mathcal{F}^{b}, \quad \sigma(:\psi(z)\psi^*(z):)\sigma^{-1}=\alpha(z)$$ where $\alpha(z)=\sum \alpha_n z^{-n-1}$ is the Heisenberg field and the superscripts in the domain and range denote the fermionic and bosonic Fock space respectively. $::$ denotes normal ordering of the modes. Question Can $U$ be written in terms of modes of the Heisenberg algebra $\alpha_n$ using the boson-fermion correspondence? What does this look like? REPLY [2 votes]: The full question is beyond my understanding, but let me suggest a substitution to your operator that can in some cases simplify handing the infinite matrices. If you introduce a new index $k$, set $k=i+j$ and consider the sequence $s_k = \sum_{i\in [0,k]}A_{i(k-i)}\psi_i\psi^*_{k-i}$, then each element $s_k$ is a finite partial sum since $k$ is finite and you should be able to decompose $U = \exp(\sum_k s_k)$, which can be easier to manipulate, because you'd only have one dimension where indices go to infinity, and evaluating the partial sums $s_k$ can often utilize symmetries inherent in the matrices.<|endoftext|> TITLE: Prove $\frac{\text{Area}_1}{c_1^2}+\frac{\text{Area}_2}{c_2^2}\neq \frac{\text{Area}_3}{c_3^2}$ for all primitive Pythagorean triples QUESTION [16 upvotes]: A while ago I asked this question on MSE here. After placing a bounty it got quite a bit of attention but unfortunately it has yet to be resolved. After getting some advice from MO Meta I have decided to post the question here (note this is the same as area just multiplied by two and $a_n,b_n > 0$). Does, $$\frac{a_1b_1}{c_1^2}+\frac{a_2b_2}{c_2^2} = \frac{a_3b_3}{c^2_3}$$ for any three different primitive Pythagorean triples $(a_n,b_n,c_n)$? My personal belief, for what it's worth, is that this does not occur and I am actively trying to disprove it. However, I would appreciate a counter example just as much. Some results so far: User @mathlove on MSE has found the following necessary condition, The following is a necessary condition for $c_i.$ It is necessary that for every prime $p$, $$ \nu_p(c_1) \leq \nu_p(c_2)+\nu_p(c_3)$$ $$\nu_p(c_2)\le \nu_p(c_3)+\nu_p(c_1)$$ $$\nu_p(c_3)\le \nu_p(c_1)+\nu_p(c_2)$$ where $\nu_p(c_i)$ is the exponent of $p$ in the prime factorization of $c_i$. (Proof can be found here) In order to search for these values I created an exhaustive algorithm to search for these triples with help from here. This is what I've found, For $c^2 < 10^{14}$ $$\frac{a_1b_1}{c_1^2}+\frac{a_2b_2}{c_2^2} \neq \frac{a_3b_3}{c^2_3}$$ and, $$\frac{1}{c_1^2}+\frac{1}{c_2^2} \neq \frac{1}{c^2_3}$$ Note that the trouble finding these triples appears to come from dividing by the square of the hypotenuse as there are many solutions to $a_1b_1 + a_2b_2 = a_3b_3$. It seemed at first that these solutions may just be extremely unlikely to occur (ratios lining up perfectly) hence why nothing was found but it looks like there is a little more to this. A bug in my initial code made me accidentally search for solutions to this, $$\frac{a_1b_1}{c_1}+\frac{a_2b_2}{c_2} = \frac{a_3b_3}{c_3}$$ Which yielded these very interesting values for $c < 10^7$, $$\frac{3*4}{5} + \frac{20*21}{29} = \frac{17*144}{145}$$ $$\frac{20*21}{29} + \frac{119*120}{169} = \frac{99*4900}{4901}$$ $$\frac{119*120}{169} + \frac{696*697}{985} = \frac{577*166464}{166465}$$ $$\frac{696*697}{985} + \frac{4059* 4060}{5741} = \frac{3363*5654884}{5654885}$$ This pattern has a well defined structure to it. Notice the recursive nature where one of the terms always comes from the sum of the previous. Additionally the LHS numerators are both one apart and on the RHS the $b_3$ and $c_3$ are also a distance one from each other. Background and motivation A resolution one way or another to the original question could help to resolve a couple of (presumably not so important) but pesky open problems in number theory. I am preparing a website that I will link to eventually that will give the full background. However, it is too lengthy for this post and in accordance with advice from META MO I will omit it to keep this as brief as possible. Additionally I am not a research level mathematician so please forgive any unintended ignorance when responding to comments. Edit for bounty Joe Silverman and Constantin-Nicolae Beli have already given some good insight into the problem, I am putting a bounty on this with the hope that it will get more attention. I don't have much reputation so doing anything, even just commenting would go a long way for me. Looking at the problem as it stands I see one main problem and a subproblem that may help answer the main problem. Main problem Prove the original statement or find a counter example. Subproblem Is the solution set mentioned by Constantin-Nicolae Beli the only solutions to $\frac{a_1b_1}{c_1}+\frac{a_2b_2}{c_2} = \frac{a_3b_3}{c_3}$ and would that also be the same solution set for when the hypotenuse is squared? Important update Going back and looking at the background of where this comes from. I found that what I am asking for is equivalent to this, $$\frac{\left(c_{1}-x_{1}\right)\left(c_{1}+x_{1}\right)}{c_{1}^2}+\frac{\left(c_{2}-x_{2}\right)\left(c_{2}+x_{2}\right)}{c_{2}^2}=\frac{\left(c_{3}-x_{3}\right)\left(c_{3}+x_{3}\right)}{c_{3}^2}$$ Where $c_n$ is the associated hypotenuse of the primitive triple and $x_n$ is an integer solution to the circle, $$x^2+y^2=2c^2$$ I wanted to mention this connection as it is related to the solution set mentioned by Constantin-Nicolae Beli. REPLY [4 votes]: Not really an answer, but a suggestion. Did you try to solve the "wrong" problem, where at the denominators you have $c_i$, not $c_i^2$? It seems quite interesting. If we restrict ourselves to solutions of the type you found, then a Pythagorean triple with $b=a+1$ satisfies $a^2+(a+1)^2=c^2$, which writes as $(2a+1)^2-2c^2=-1$. The fundamental unit of ${\mathbb Z}[\sqrt 2]$ is $1+\sqrt 2$, of norm $-1$. Hence if $(x_n,y_n)$ are the solutions of the Pell equation $x^2-2y^2=\pm 1$, with $x_n+y_n\sqrt 2=(1+\sqrt 2)^n$, then $x_n^2-2y_n^2=(-1)^n$. So we have Pythagorean triplets $(a,a+1,c)$ if we take $2a+1=x_n$ and $c=y_n$, with $n$ odd. Such a triplet looks like $(\frac{x_n-1}2,\frac{x_n+1}2,y_n)$. If we want a triplet with $c=b+1$ then we have the equation $a^2+b^2=(b+1)^2$ and we get $(a,b,c)=(a,\frac{a^2-1}2,\frac{a^2+1}2)$, with $a$ odd. Then one can verify that $(a_1,b_1,c_1)=(\frac{x_{2k-1}-1}2,\frac{x_{2k-1}+1}2,y_{2k-1})$, $(a_2,b_2,c_2)=(\frac{x_{2k+1}-1}2,\frac{x_{2k+1}+1}2,y_{2k+1})$ and $(a_3,b_3,c_3)=(x_{2k},\frac{x_{2k}^2-1}2,\frac{x_{2k}^2+1}2)$, with $k\geq 2$, satisfy the equation $\frac{a_1b_1}{c_1}+\frac{a_2b_2}{c_2}=\frac{a_3b_3}{c_3}$. This checks straightforwardly if we write $x_n=\frac{\alpha^n+\beta^n}2$ and $y_n=\frac{\alpha^n-\beta^n}{2\sqrt 2}$, where $\alpha =1+\sqrt 2$ and $\beta =1-\sqrt 2=-\alpha^{-1}$. A legitimate and nice problem you may think of is whether these are the only solutions or not. It also has some geometric meaning since $\frac{ab}c$ is the height perpendicular on $c$ in the triangle.<|endoftext|> TITLE: Random walks: How many times does the largest component change? QUESTION [6 upvotes]: My understanding is that for an unbiased random walk (starting at the origin) on $\mathbb R$ with $N$ steps that the expected number of sign changes is $O(\sqrt N)$. For a biased walk I believe the expected number is $O(1)$. In either case it is $O(\sqrt N)$. Now suppose we have a random walk $(X_n,Y_n)$ on $\mathbb R^2$ and are interested in the number of times the largest component changes. Formally define $M(n) = 1$ if $X_n \ge Y_n$ and $M(n) = 2$ if $X_n < Y_n$. We want to bound $\mathbb E|\{n \le N: M(n) \ne M(n+1)\}|$. To get a bound we only need to consider the random walk $X_n-Y_n$ on $\mathbb R$. By the previous result the sign changes $O(\sqrt N)$ times on expectation. But sign changes correspond to the largest component changing and we're done. For dimensions $d>2$ this trick no longer works. In that case is anything known about the expected number of times the largest component changes? Are there any known order bounds depending on $d$ and $N$? I suspect $O(\log(d) \sqrt N)$ bounds might be possible. I am trying to find answers online but I can't seem to even find a reference for the $O(1)$ result I mentioned. REPLY [4 votes]: For a large class of unbiased random walks, the expected number of switches is of order $\Theta (\sqrt{N \log d})$. It is closely related to the growth of regret in online learning, see, e.g., [1]; it is fine if the increments are biased as long as all components have the same mean. The exact answer depends on the step distribution of the random walk. For concreteness, suppose $X_n=(X_n(i) : 1 \le i \le d)$ is a random walk in $\mathbb R^d$ with $d$ independent components, and each component has $\pm 1$ increments of the same mean $\mu \in (-1,1)$. Denote $M_n:= \max_{j\le d} X_n(j)$. Let $J_n$ denote the index of a maximal component at time $n$. (Precisely, let $J_0=1$. Given an integer $n \ge 1$, suppose that $J_{n-1}$ has already been defined. If $X_{n}(J_{n-1})=M_n$ then take $J_n:=J_{n-1}$; otherwise, set $J_n$ to be the minimal $j$ such that $X_{n}(j)=M_{n}$.) Finally, let $S_n:=\sum_{k=1}^n {\mathbf 1}_{J_k \ne J_{k-1}}$ be the number of times the maximal component switches by time $n$. Observe that for $n \ge 1$, $$ M_{n+1}-M_n =X_{n+1}(J_n)-X_n(J_n)+ 2 \cdot{\mathbf 1}_{J_{n+1} \ne J_n} \, . $$ Therefore $M_n-n\mu-2S_n$ is a martingale for $n \ge 0$, so for all $N>0$, $${\mathbb E} M_N-N\mu- 2 {\mathbb E}S_N=0 \,. \quad (*)$$ The multivariate central limit theorem, and the standard asymptotics for the maximum of $d$ Gaussians (see, e.g., Solution 18.7, page 348 in [2]), imply that as $N \to \infty$, $${\mathbb E}\Bigl[\frac{M_N-N\mu}{\sigma\sqrt{N}}\Bigr] \to \sqrt{2\log d} \, , $$ where $\sigma^2=1-\mu^2$ is the variance of the increments. By (*), as $N \to \infty$, $${\mathbb E}\Bigl[\frac{S_N}{\sigma\sqrt{N}}\Bigr] \to \sqrt{(\log d)/2} \, . $$ The analysis above can be extended to the case where each independent component has increments of the same mean $\mu$ and variance bounded above and below by positive constants. (In that case $M_n-n\mu-cS_n$ will be a super- or sub-martingale depending on the value of $c>0$.) If the increments of different components have different means, then one can restrict attention just to those components where the increments have a maximal mean. [1] Towards Optimal Algorithms for Prediction with Expert Advice. Nick Gravin, Yuval Peres, and Balasubramanian Sivan. Proceedings of the Twenty-Seventh Annual ACM-SIAM Symposium on Discrete Algorithms, 2016, 528-547 [2] Karlin, Anna R., and Yuval Peres. Game theory, alive. Vol. 101. American Mathematical Soc., 2017.<|endoftext|> TITLE: Functional approach vs jet approach to Lagrangian field theory QUESTION [19 upvotes]: Context: I am a PhD student in theoretical physics with higher-than-average education on differential geometry. I am trying to understand Lagrangian and Hamiltonian field theories and related concepts like Noether's theorem etc. in a mathematically rigorous way since the standard physics literature is sorely lacking for someone who values precision and generality, in my opinion. I am currently studying various text by Anderson, Olver, Krupka, Sardanashvili etc. on the variational bicomplex and on the formulation of Lagrangian systems on jet bundles. I do not rule the formalism yet, but made significant steps towards understanding. On the other hand, most physics literature employs the functional formalism, where rather than calculus on variations taking place on finite dimensional jet bundles (or the "mildly infinite dimensional" $\infty$-jet bundle), it takes place on the suitably chosen (and usually not actually explicitly chosen) infinite dimensional space of smooth sections (of the given configuration bundle). Even relatively precise physics authors like Wald, DeWitt or Witten (lots of 'W's here) seems to prefer this approach (I am referring to various papers on the so-called "covariant phase space formulation", which is a functional and infinite dimensional but manifestly "covariant" approach to Hamiltonian dynamics, which also seems to be a focus of DeWitts "The Global Approach to Quantum Field Theory", which is a book I'd like to read through but I find it impenetrable yet). I find it difficult to arrive at a common ground between the functional formalism and the jet-based formalism. I also do not know if the functional approach had been developed to any modern standard of mathematical rigour, or the variational bicomplex-based approach has been developed precisely to avoid the usual infinite dimensional troubles. Example: Here is an image from Anderson's "The Variational Bicomplex", which shows the so-called augmented variational bicomplex. Here $I$ is the so-called interior Euler operator, which seems to be a substitute for integration by parts in he functional approach. Later on, Anderson proves that the vertical columns are locally exact, and the augmented horizontal rows (sorry for picture linking, xypic doesn't seem to be working here, don't know how to draw complices) are locally exact as well. In fact for the homotopy operator $\mathcal H^1:\mathcal F^1\rightarrow\Omega^{n,0}$ that reconstructs Lagrangians from "source forms" (equations of motion) he gives (for source form $\Delta=P_a[x,y]\theta^a\wedge\mathrm d^nx$) $$ \mathcal H^1(\Delta)=\int_0^1 P_a[x,tu]u^a\mathrm dt\ \mathrm d^nx. $$ On the other hand, if we use the functional formalism in an unrigorous manner, the functional derivative $$ S\mapsto\frac{\delta S[\phi]}{\delta \phi^a(x)} $$ behaves like the infinite dimensional analogue of the ordinary partial derivative, so using the local form of the homotopy operator for the de Rham complex (which for the lowest degree is $f:=H(\omega)=\int_0^1\omega_\mu(tx)x^\mu\mathrm dt$) and extending it "functionally", one can arrive at the fact that if an "equation of motion" $E_a(x)[\phi]$ satisfies $\frac{\delta E_a(x)}{\delta\phi^b(y)}-\frac{\delta E_b(y)}{\delta\phi^a(x)}=0$, then $E_a(x)[\phi]$ will be the functional derivative of the action functional $$ S[\phi]=\int_0^1\mathrm dt\int\mathrm d^nx\ E_a(x)[t\phi]\phi^a(x). $$ I have (re)discovered this formula on my own by simply abusing the finite dimensional analogy and was actually surprised that this works, but it does agree (up to evaluation on a secton and integration) with the homotopy formula given in Anderson. This makes me think that the "variation" $\delta$ can be considered to be a kind of exterior derivative on the formal infinite dimensional space $\mathcal F$ of all (suitable) field configurations, and the Lagrangian inverse problem can be stated in terms of the de Rham cohomology of this infinite dimensional field space. This approach however fails to take into account boundary terms, since it works only if integration by parts can be performed with reckless abandon and all resulting boundary terms can be thrown away. This can be also seen that if we consider the variational bicomplex above, the $\delta$ variation in the functional formalism corresponds to the $\mathrm d_V$ vertical differential, but in the augmented horizontal complex, the $\delta_V=I\circ\mathrm d_V$ appears, which has the effect of performing integrations by parts, and the first variation formula is actually $$ \mathrm d_V L=E(L)-\mathrm d_H\Theta, $$ where the boundary term appears explicitly in the form of the horizontally exact term. The functional formalism on the other hand requires integrals everywhere and boundary terms to be thrown aside for $\delta$ to behave as an exterior derivative. Moreover, integrals of different dimensionalities (eg. integrals over spacetime and integrals over a hypersurface etc.) tend to appear sometimes in the functional formalism, which can only be treated using the same concept of functional derivative if various delta functions are introduced, which makes me think that de Rham currents (I am mostly unfamiliar with this area of mathematics) are also involved here. Question: I would like to ask for references to papers/and or textbooks that develop the functional formalism in a general and mathematically precise manner (if any such exist) and also (hopefully) that compare meaningfully the functional formalism to the jet-based formalism. REPLY [5 votes]: This is meant as a long comment to the very good answer by Pedro Riberio. There is a nice analog of the variational bicomplex in the functional framework. Namely, the space of differential forms on $M \times \Gamma(E)$ (where $E$ is a fiber bundle over $M$) comes with a natural bigrading $\Omega^{p, q}(M \times \Gamma(E))$ induced by the product structure, i.e. the dual of the decomposition $T_{m, \phi} (M \times \Gamma(E)) = T_m M \times T_\phi \Gamma(E)$ of the tangent space. Moreover, the jet map $$ j^k: M \times \Gamma(E) \to J^k E, \qquad (m, \phi) \mapsto j^k_m \phi $$ yields a morphism from the variational bicomplex $\Omega^{p, q}(J^k E)$ to the bicomplex $\Omega^{p, q}(M \times \Gamma(E))$ with the exterior differential. Personally, I find the functional bicomplex easier to understand than the variational one; and as remarked by Pedro the functional framework is more flexible as it also handles non-local Lagrangians. On the other hand, the jet bundle approach has advantages for simulation, because you stay in the finite-dimensional setting which makes it easier to discretize while preserving the (symplectic) geometry.<|endoftext|> TITLE: The (possibly negative) integers arising from sum of elements in any column of the character table QUESTION [8 upvotes]: Let $G$ be a finite group and $A$ be the character table for the irreducible complex representations. The sum of elements in any row of the character table is a positive integer as its equals to the multiplicity of irreducible representation corresponding to that row inside $V$, where $V$ is the representation on the group algebra induced by the conjugacy action of $G$. But how to understand the sum of elements in any column of the character table in terms of representation theory? It's an integer by Galois theory but might not be positive (e.g $G=M_{11}$). In the case of Weyl groups, is there a simple formula? At least, can we determine its sign from $G$? REPLY [10 votes]: If $G$ is a finite group whose complex irreducible representations are all realizable over $\mathbb{R}$, (eg $G = S_{n}$), then for any $x \in G$, the sum of the entries of the column corresponding to (the conjugacy class of ) $x$ is precisely the number of square roots of $x$ in $G$, so is, in particular, a non-negative integer (and is strictly positive if $x$ has odd order).<|endoftext|> TITLE: Uncountability of the real numbers from LLPO without countable choice QUESTION [6 upvotes]: Does there exist a proof of the uncountability of the real numbers that uses analytic LLPO (the statement that any real number $x$ satisfies either $x \leq 0$ or $x \geq 0$) but avoids Excluded Middle and the Axiom of Choice (including dependent and countable choice)? You may use Unique Choice. I find it difficult to apply LLPO because of the overlapping cases and the unavailability of Countable Choice. REPLY [2 votes]: This is not an answer but rather an alternative proof of @FrankaWaaldijk's observation that explicit Cauchy sequences are uncountable. The only difference is that we use explicitly given moduli of convergence, rather than a fixed rate of convergence, and that we explicitly impose an equality relation (which I think is implied in @FrankaWaaldijk's answer). We work in Bishop-style constructive mathematics but without countable choice. In particular, a set is a collection with an imposed equality relation, and a function is an operation that respects the imposed equalities on the domain and the codomain. Definition: An (explicit) Cauchy sequence is a sequence of rational number $q : \mathbb{N} \to \mathbb{Q}$ together with a strictly increasing function $\mu : \mathbb{N} \to \mathbb{N}$, called modulus, such that $\forall k, m, n \in \mathbb{N} \,.\, |q_{\mu(k) + m} - q_{\mu(k) + n}| < 2^{-k}$. Two Cauchy sequences $(q, \mu)$ and $(q', \mu')$ are considered equal when $|q_{\mu(i)} - q'_{\mu'(j)}| \leq 2^{-i-j}$ for all $i, j \in \mathbb{N}$. Lemma: There is an operation which takes as input an interval $[\ell, r]$ with rational endpoints and a Cauchy sequence $(q, \mu)$, and outputs an interval $[\ell', r']$ such that $|r' - \ell'| = |r - \ell|/3$ and $[\ell', r'] \subseteq [\ell, r]$, and at most finitely many terms of $q$ are inside $[\ell', r']$. Proof. We describe the procedure for computing $r'$ and $\ell'$. Let $k$ be the least number such that $2^{-k} < |r - \ell|/12$. Now, if $q_{\mu(k)} > (\ell + r)/ 2$ then let $[\ell', r'] = [\ell, (2 \ell + r)/3]$, otherwise let $[\ell', r'] = [(\ell + 2 r)/3, r]$. In the first case, we have for all $n$ $$ q_{\mu(k)+n} > q_{\mu(k)} - 2^{-k} > (\ell + r)/2 - |r - \ell|/12 > (\ell + 2 r)/3. $$ Thus, at most the first $\mu(k)$ terms of $q$ are contained in $[\ell', r']$. The other case is treated similarly. $\Box$ Theorem: The set of explicit Cauchy sequences is uncountable. Proof. Let $C$ be the set of all rational Cauchy sequences and $a : \mathbb{N} \to C$ a sequence of Cauchy sequences. We construct a sequence $s$ which avoids every sequence enumerated by $a$. First, define a sequence of nested intervals $[\ell_{-1},r_{-1}] \supseteq [\ell_0, r_0] \supseteq \cdots$ by taking $[\ell_{-1}, r_{-1}] = [0, 1]$ and letting $[\ell_{k+1}, r_{k+1}]$ be the interval obtained from the Lemma applied to $[\ell_k, r_k]$ and $a(k)$. Now define the sequence $s : \mathbb{N} \to \mathbb{Q}$ by $s_k = (\ell_k + r_k)/2$. By the Lemma, $s_k$ differs from $a(k)_k$ and therefore $s$ is not enumerated by $a$. $\Box$ Discussion Authors often require a fixed modulus of convergence for all sequences. For example, Bishop requires $|q_m - q_n| < 1/m + 1/n$ (if memory serves me right), and another popular choice is $|q_m - q_n| \leq 2^{-min(m, n)}$. Which particular modulus we prefer, if any, is not too important because we can always find a subsequence of a given sequence that has any desired modulus: Lemma: Suppose $(q, \mu)$ is a Cauchy sequence and $\nu : \mathbb{N} \to \mathbb{N}$ a strictly increasing function. Then there is a reindexing function $\iota : \mathbb{N} \to \mathbb{N}$ such that $(q, \mu)$ is equal to $((q \circ \iota), \nu)$. Proof. Exercise. $\Box$ Of course, one would want to show the uncountability of the set of Cauchy sequences without explicit moduli. That is, suppose we say that $q : \mathbb{N} \to \mathbb{Q}$ is standard Cauchy when $$\forall k \in \mathbb{N} . \exists \mu \in \mathbb{N} . \forall m, n \in \mathbb{N} \,.\, |q_{\mu+m} - q_{\mu+n}| < 2^{-\mu}. \tag{1}$$ Can we show uncountability of standard Cauchy sequences? Well, one seemingly needs to choose a suitable $\mu$ at each step of diagonalization. However, as a mitigating factor I want to point out that (1) is not a workable definition of Cauchy sequences, unless we do have countable choice, because we cannot even show that a Cauchy sequence of Cauchy sequences converges. @jkabrg asked whether the explicit Cauchy sequences (with the given equality) form a complete field. I did not verify all the details, but I think the answer is positive so long as one sticks to the principle that without countable choice $\forall\exists$ statements should be replaced by moduli. That is, an explicit Cauchy sequence of explicit Cauchy sequences converges to an explicit Cauchy sequence. I personally am not a big fan of using setoids, or sets with imposed equality relations. The reason is as follows. Suppose we want to work in a "naturally occurring" constructive mathematical environment $\mathcal{E}$ (a topos, a higher-order fibration, a realizability model, a type theory). Then we do not get to pick equality relations because $\mathcal{E}$ already has a built in notion of equality. If $\mathcal{E}$ also has quotients then we may get the desired equality on a set by quotienting it. Setoids and Bishop's sets-with-equality are not even object of $\mathcal{E}$, but rather objects of an exact completion of $\mathcal{E}_{\mathrm{ex}}$. They live in the wrong place! To put it differently, even though Bishop succeeds in being agnostic about excluded middle, he fails to be ecumenical about equality because he commits to a certain kind of semantic models known as exact completions. These include realizability models but exclude sheaves on a space.<|endoftext|> TITLE: A definable gender problem related to Sacks reals QUESTION [10 upvotes]: Let $L$ be the ground model, and $a\in2^\omega$ be a Sacks-generic real over $L$. Note that any real $x\in S=(2^\omega\cap L[a])\setminus L$ is still Sacks-generic over $L$. Now assume that $\mathsf E$ is an OD (ordinal-definable) equivalence relation in $L[a]$ on the set $S$, with exactly two equivalence classes, say $M$ and $F$. (Two genders of the Sacks reals.) Are $M$ and $F$ necessarily OD themselves? My idea of a counterexample is as follows. Let $F$ be the set of all continuous 1-1 maps $2^\omega$ onto $2^\omega$, coded in $L$. Then $F$ is a group under the superposition. Moreover if $x,y\in S$ then it is known that $y=f(x)$ for some $f\in F$. Now if $H\subseteq F$ is a subgroup coded in $L$ then the relation: $x \mathrel{\mathsf E_H} y$ iff $y=f(x)$ for some $f\in H$ is an OD equivalence on $S$, and there is no immediate idea as how to OD-define the $\mathrel{\mathsf E_H}$-class of $a$ (w/o a reference to $a$). Now the goal is to define $H$ such that $\mathrel{\mathsf E_H}$ has exactly two equivalence classes on $S$ in $L[a]$. The principal non-commutativity of $F$ looks to be a huge obstacle though. [Added Feb 4 at 5:50] And finally it is established, there are two distinct but OD-indiscernible populations of Sacks-generic reals over $L$, arXiv . In fact this is an unpublished result of Solovay dated back to 2002. A similar result holds for $E_0$-large forcing, via a known ``canonization'' theorem, but the problem is open for other popular forcing notions. REPLY [6 votes]: The question is now answered in this recent paper, where it is shown that in the extension $L(s)$ of the constructible universe $L$ by a single Sacks real $s$, there is a definable pair $\{a,b\}$ of objects (where $a$ and $b$ are subsets of real numbers) such that neither $a$ nor $b$ is ordinal definable. This shows that $a$ and $b$ are indiscernible in $L(s)$ in the following strong sense: for any formula of set theory $\phi(x,y,\alpha)$, where $\alpha$ is an ordinal parameter, $\phi(a,b,\alpha) \leftrightarrow\phi(b,a,\alpha)$ holds in $L(s)$.<|endoftext|> TITLE: Is there a reasonable notion of spectral theorem on a pre-Hilbert space? QUESTION [5 upvotes]: I'm trying to understand how bad things could possibly get without Cauchy completeness as a criterion for Hilbert spaces in quantum mechanics. Obviously, doing calculus on a pre-Hilbert space would be complicated but could there still be some reasonable version of the spectral theorem? Why or why not? Some elaboration and perhaps even some references would be appreciated. REPLY [8 votes]: Here is a simple example that shows that the idea of spectral theory on pre-Hilbert spaces in the sense you are asking is hopeless. Consider the pre-Hilbert space consisting of the restrictions of all complex polynomials to $[0,1]$, as a dense subspace of $L^2[0,1]$. Then let $A$ be the operator of multiplication by $x$. The spectral projections of this operator are characteristic functions; none of them except $0$ and $1$ are polynomials.<|endoftext|> TITLE: Is the result of Schmidt conditional to RH QUESTION [5 upvotes]: From this page: https://en.wikipedia.org/wiki/Chebyshev_function#Asymptotics_and_bounds A theorem due to Erhard Schmidt states that, for some explicit positive constant $K$, there are infinitely many natural numbers $x$ such that $$ψ(x)>x+K√x$$ Since the original paper is in German. I read this paper: https://projecteuclid.org/download/pdf_1/euclid.acta/1485887467 My question is: Is the result of Schmidt conditional to Riemann Hypothesis. REPLY [8 votes]: It is elementary, take $K=1/30 < 1/|\rho_0|$ where $\rho_0\approx 1/2+i14.13$ is the first zero, if $\psi(x)-x\le Kx^{1/2}$ for $x$ large enough, then $x-\psi(x)+K x^{1/2}+C\ge 0$ for all $x$, where $C$ is a suitable real constant. Let $$F(s):=\int_1^\infty (x-\psi(x)+Kx^{1/2}+C) x^{-s-1}dx= \frac1{s-1}+\frac{\zeta'(s)}{s\zeta(s)}+\frac{K}{s-1/2}+\frac{C}{s}.$$ By the non-negativity of the integrand, it has a singularity at its abscissa of convergence $\sigma$. But the RHS is analytic on $(1/2,\infty)$, thus $\sigma=1/2$. And by the non-negativity again we have, as $\Re(s) \to 1/2$, $$|F(s)|\le F(\Re(s))\sim \frac{K}{\Re(s)-1/2}.$$ This contradicts that $$F(s)\sim\frac{\zeta'(s)}{s\zeta(s)}\sim\frac{1/\rho_0}{s-\rho_0}\qquad\text{as $s\to \rho_0$}.$$<|endoftext|> TITLE: Limit of alternated row and column normalizations QUESTION [6 upvotes]: Let $E_0$ be a matrix with non-negative entries. Given $E_n$, we apply the following two operations in sequence to produce $E_{n+1}$. A. Divide every entry by the sum of all entries in its column (to make the matrix column-stochastic). B. Divide every entry by the sum of all entries in its row (to make the matrix row-stochastic). For example: $E_0=\begin{pmatrix} \frac{2}{5} & \frac{1}{5} & \frac{2}{5} & 0 & 0\\ \frac{1}{5} & 0 & \frac{7}{10} & \frac{1}{10} & 0\\ 0 & 0 & 0 & \frac{3}{10} & \frac{7}{10} \end{pmatrix}\overset{A}{\rightarrow}\begin{pmatrix} \frac{2}{3} & 1 & \frac{4}{11} & 0 & 0\\ \frac{1}{3} & 0 & \frac{7}{11} & \frac{1}{4} & 0\\ 0 & 0 & 0 & \frac{3}{4} & 1 \end{pmatrix}\overset{B}{\rightarrow}\begin{pmatrix} \frac{22}{67} & \frac{33}{67} & \frac{12}{67} & 0 & 0\\ \frac{44}{161} & 0 & \frac{12}{23} & \frac{33}{161} & 0\\ 0 & 0 & 0 & \frac{3}{7} & \frac{4}{7} \end{pmatrix}=E_1$ What is the limit of $E_n$ as $n \to \infty$? Additional remarks. In my problem, the matrix has $c\in \{1,2,\dots,5\}$ rows and $r=5$ columns (note that the two letters are reversed, but in the original context of this problem these letters $r$ and $c$ do not actually stand for rows and columns). So $E_0$ can be $1\times 5$, $2\times 5$, ... or $5\times 5$. We denote with $(e_n)_{ij}$ the entries of $E_{n}$; hence $(e_n)_{ij}\in[0;1]$ and $\forall i \sum_{j=1}^{r}(e_n)_{ij}=1$ for $n>0$. I managed to express $(e_{n+1})_{ij}$ as a function of $(e_{n})_{ij}$ : $$(e_{n+1})_{ij}=\frac{\frac{(e_{n})_{ij}}{\sum_{k=1}^{c}(e_n)_{kj}}}{\sum_{l=1}^{r}\frac{(e_n)_{il}}{\sum_{k=1}^{c}(e_n)_{kl}}}$$ What I can't seem to find now is an expression $(e_{n})_{ij}$ as a function of $(e_{0})_{ij}$, to be able to calculate $\underset{n \to +\infty }{lim}(e_n)_{ij}$ I wrote code to compute this iteration; when I ran it with the previous example $E_0$, I found out that: $E_0=\begin{pmatrix} \frac{2}{5} & \frac{1}{5} & \frac{2}{5} & 0 & 0\\ \frac{1}{5} & 0 & \frac{7}{10} & \frac{1}{10} & 0\\ 0 & 0 & 0 & \frac{3}{10} & \frac{7}{10} \end{pmatrix}\overset{n \rightarrow+\infty}{\rightarrow}E_n=\begin{pmatrix} \frac{7}{25} & \frac{3}{5} & \frac{3}{25} & 0 & 0\\ \frac{8}{25} & 0 & \frac{12}{25} & \frac{1}{5} & 0\\ 0 & 0 & 0 & \frac{2}{5} & \frac{3}{5} \end{pmatrix}$ Not only do the row sums equal $1$, but the column sums equal $\frac{3}{5}$: it seems that in this process column sums converge to $\frac{c}{r}$. I'm not a mathematician so I was looking for a simple inductive proof. I tried to express $E_2$ (and so on) as a function of $E_0$, but it quickly gets overwhelming, starting from $E_2$... REPLY [5 votes]: When $E_0$ is square (i.e., $r = c$) this procedure is called Sinkhorn iteration or the Sinkhorn-Knopp algorithm (see this Wikipedia page). You can find a wealth of results by Googling those terms, the most well-known of which is that if $E_0$ has strictly positive entries (and again, is square) then the limit of $E_n$ indeed exists and is doubly stochastic.<|endoftext|> TITLE: Software and ideas for workshops and conferences with long-distance participants QUESTION [26 upvotes]: Conferences and workshops are often great - getting together and being together is an ideal setting for doing research and learning things. However, there are various reasons to encourage the possibility of having conferences and workshop through "cyber space". The cost of travel, in terms of time, money, $CO_2$, is large. Q1: My first question is about software that makes it possible to run lectures with long-distance participation. What are the features and what are additional features we would like to see. Q2 My second question is about what could be good ideas for running a long-distance workshops or conferences. (This question is mainly experience-based rather than opinion-based.) Hybrid setting with conventional conference that allows some participation from cyber space? Completely cyber conference where each participant stays in his office/study.? I have some experience with the TCS+ seminars which are completely in cyberspace and participants from the audience can ask questions. They work rather well. (But only a few participants per talk.) I am not familiar of similar things in mathematics. Very related question: Tools for long-distance collaboration (This is about tools for long distance collaboration, in general. Not specific to seminars/workshops/conferences and also asked nine years ago.) REPLY [7 votes]: The Northeastern Combinatorics Network has been running a Virtual Combinatorics Colloquium since March 2018 using Zoom and encouraging local viewing parties. Not exactly a conference, but perhaps a relevant existing program. https://sites.google.com/view/northeastcombinatoricsnetwork/virtual-combinatorics-colloquium<|endoftext|> TITLE: Is there a Poincare residue in characteristic $p$? QUESTION [9 upvotes]: The Poincare residue I mean is there one here: https://en.wikipedia.org/wiki/Poincar%C3%A9_residue Basically, I would like a nice way to use a meromorphic $n$-form on $\mathbf{P}^n_{\mathbf{F}_p}$ to get an $(n-1)$-form on the hypersurface given by the pole. I suspect that one exists, but phrased in some fancy language. I'm not sure what to look for. A reference (and some decoding of it) would be very welcome. REPLY [2 votes]: I've been thinking about this, and I want to record a few thoughts. Let $k$ be a field of characteristic $p$, let $X$ be a smooth $n$-dimensional variety, let $D$ be a Cartier divisor and let $U = X \setminus D$. We cannot hope to have a natural map from $H_{DR}^n(U)$ to $H^{n-1}_{DR}(D)$ which looks anything like the residue map (also known as the Gysin map). Take $p$ odd. Take $X$ to be the affine plane with coordinates $(x,y)$, and let $D$ be $\{ y=0 \}$. Let $\alpha$ be the $2$-form $x^{2p-1} y^{-p-1} dx \wedge dy$ and consider the automorphism $\phi(x,y) = (x+y,y)$ of $X$. This preserves the divisor $D$ and acts trivially on $D$, so $\alpha$ and $\phi^{\ast}(\alpha)$ should have the same residue. In other words, $\phi^{\ast} \alpha - \alpha$ should have residue $0$. Now, $\phi^{\ast} \alpha - \alpha = \sum_{j=0}^{2p-2} \binom{2p-1}{j} x^j y^{p-2-j} dx \wedge dy$. If we compute residues naively, the residue of $\sum_{j=0}^{2p-2} \binom{2p-1}{j} x^j y^{p-2-j} dx \wedge dy$ should be $\binom{2p-1}{p-1} x^{p-1} dx$. Also, $\binom{2p-1}{p-1} \equiv 2 \neq 0 \bmod p$ by Lucas' theorem. But $x^{p-1} dx$ is not exact in characteristic $p$. So working naively can't give us a residue which is well defined in $H^{\ast}_{dR}$. Moreover, it doesn't make sense to fix this by defining $\binom{2p-j}{p-j} x^j y^{p-2-j} dx dy$ to have a nonzero residue for other values of $j$, because $x^j y^{p-2-j} dx dy$ is exact for all $0 \leq j \leq 2p-2$ except $p-1$. There is a very deep thing to do. We can lift $X$, $D$ and $U$ up to flat schemes over some dvr of mixed characteristic (for example, if $k = \mathbb{F}_p$, we could take $p$-adic lifts) and take the de Rham cohomology of these lifts. There is a ton of very hard literature on this sort of idea, starting with the research of Monsky and Washnitzer. Indeed, there is a Gysin sequence in Monsky-Washnitzer cohomology: See Monsky, P., Formal cohomology. II: The cohomology sequence of a pair, Ann. Math. (2) 88, 218-238 (1968). ZBL0162.52601. I don't feel confident to summarize this paper. I went looking for something more elementary to do and came up with an interesting idea: Although $x^{p-1} dx$ isn't exact, it is in a sense "almost exact". The exact forms are the kernel of the Cartier operator, and $x^{p-1} dx$ is in the kernel of the square of the Cartier operator. Define $EH^n$ to be $n$-forms modulo forms which are killed by some power of the Cartier operator. (This is a definition for top dimensional forms only; see my recent question for what I think the more general definition should be.) I think I can build a Gysin map $EH^n(X) \to EH^{n-1}(D)$. But I'm going to wait a bit to see if someone answers my other question before I write more. Okay, let me spell out this idea in a bit more detail. First of all, let's recall how residue works when $\omega$ only has a simple pole along $D$. First, choose an open set $X'$ on which $D$ is principal, with generator $t$, and on which there is a vector field $\vec{v}$ with $\langle \vec{v}, dt \rangle = 1$. Set $U' = X' \cap U$ and $D' = X \cap D$. If $\omega$ has only a simple pole on $D'$, then $t \omega$ extends to $X'$. Contracting $t \omega$ against $\vec{v}$ gives an $(n-1)$-form, which we can then restrict to $D'$. The final result is independent of the choices of $t$ and $\vec{v}$, and is the residue of $\omega$ to $D'$. We can cover $X$ by open sets $X'$ as above and compute the residue on each such set, and since the result is independent of our choices, we get a well defined residue on $D$. Nothing here uses characteristic $0$ (and we even get a specific differential form for our residue, not a cohomology class.) Now, suppose that $\omega$ has a pole of order $N$, and let $\mathcal{C}$ be the Cartier operator. Then $\mathcal{C}(\omega)$ has a pole of order at most $1+(N-1)/p$. Applying the Cartier operator $k$ times for $k$ large enough that $p^k \geq N$, we get a differential form with a pole of order $\leq 1$. We can take the residue $\mathrm{Res}(\mathcal{C}^k \omega)$ of that form. But then we should apply the "inverse Cartier operator" $k$-times to this residue. The Cartier operator from top dimensional forms to top dimensional forms is surjective, but has a kernel, so what this really means is to find some $n-1$ form $\alpha$ on $D$ with $\mathcal{C}^k(\alpha) = \mathrm{Res}(\mathcal{C}^k \omega)$. So $\alpha$ is only defined modulo the kernel of $\mathcal{C}^k$. In other words, this residue is a class in $EH^{n-1}(D)$ in the sense I describe above. This is a map $\Omega^n(U) \to EH^{n-1}(D)$. It is also not hard to show that this map passes down to a map $EH^n(U) \to EH^{n-1}(D)$. I don't know if this is helpful, but I think it is the best you can do.<|endoftext|> TITLE: Rational surgery and attaching $2$-handles QUESTION [5 upvotes]: It is well-known fact that integral Dehn surgeries on $3$-sphere $S^3$ are viewed as the result on the boundary of attaching $2$-handles $B^2 \times B^2$ to the $4$-ball $B^4$. Is there an analogue of rational surgeries relating handle attachment of rational framing? If not, which problem occurs? REPLY [10 votes]: To attach a 4-dimensional 2-handle to the 4-ball, one requires an attaching region in $S^3=\partial B^4$ and a map from the attaching region of the handle (which has a natural parametrization as $S^1\times D^2\subset \partial(D^2\times D^2)$) to the attaching region in $S^3$. The attaching region in $S^3$ is determined by specifying a knot $K\subset S^3$ (and then convention dictates that the attaching region $\nu(K)\cong S^1\times D^2$ is parametrized by identifying the Seifert longitude $\lambda$ for $K$ with $S^1\times\{pt\}$). Thus the handle may be attached via any orientation reversing homeomorphism from the $S^1\times D^2$ in the boundary of the handle to the $S^1\times D^2$ neighborhood of $K$. There are only an integers worth of such maps up to isotopy (see eg Rolfsen Knots and Links 2D4 and 2E5); in particular $S^1\times \{pt\}$ has to be mapped to $\lambda+n\mu$ and $\{pt\}\times \partial D^2$ has to be mapped to $\mu$, where $\mu$ denotes a meridian of $K$. The resulting boundary after the handle attachment should be thought of as (the bits of the boundary of the handle that didn't get stuck to anything)$\cup$(the bits of the boundary of $S^3$ that didn't get something stuck to them) . That's $(D^2\times S^1) \cup (S^3\smallsetminus\mathring{\nu(K))}$, so the boundary is some Dehn surgery on $K$. And we can see which; we had to send $\partial D^2\times \{pt\}$ to $\lambda+n\mu$, so the only surgeries we can obtain are integral. REPLY [5 votes]: As Lisa points out, 2-handle attachments correspond exactly to integral surgeries. However, a general Dehn surgery corresponds to a sequence of integral surgeries, and hence to multiple 2-handle attachements. This is a bit hard to do without pictures, so I'll just refer you to Section 5.3 in Gompf and Stipsicz's 4-manifolds and Kirby calculus, where they explain how to use slam dunks (Figure 5.30) to convert a rational surgery into a sequence of integral surgeries. (Well, technically they do it for lens spaces, in Exercise 5.39, but the idea is completely general.)<|endoftext|> TITLE: Topology/geometry of $O(2n)/U(n)$ QUESTION [6 upvotes]: I am interested in what is known about the topology (diffeomorphism type) or geometry (is it complex? non-complex? symplectic?) of either the compact space of orthonormal complex structures on $\mathbb{R}^{2n}$, $O(2n)/U(n)$, or the full space of complex structures, $Gl(2n,\mathbb{R})/Gl(n,\mathbb{C})$ (which retracts onto the compact one). I know that the algebraic topology and CW-structure of the former was worked out by Yokota. I think I stumbled upon a $Gl(2n,\mathbb{R})$-invariant (integrable) complex structure on the non-compact quotient, and I feel it already is somewhere in the literature (hence the tag), but a brief search didn't turn out anything. Maybe these manifolds have a specific name and/or are studied extensively in a specific subfield? Any pointers will be greatly appreciated. REPLY [13 votes]: Well, $\mathrm{SO}(2n)/\mathrm{U}(n)$ is a well-known Hermitian symmetric space, DIII. In particular, it is a compact complex manifold. See, for example, Helgason's treatment in his Differential Geometry, Lie Groups, and Symmetric Spaces. REPLY [9 votes]: You may be remembering papers by Vogan (1987, p. 262; 2008, prop. 6.9). There he describes: (a) $\mathrm{GL}(2n,\mathbf R)/\mathrm{GL}(n,\mathbf C)\cong\{\!$complex structures on $\mathbf R^{2n}\}$: an elliptic coadjoint orbit of $\mathrm{GL}(2n,\mathbf R)$, hence symplectic and pseudo-kähler — with signature $\left({\frac12}(n^2-n),{\frac12}(n^2+n)\right)$ (over $\mathbf C$). (b) $\mathrm O(2n)/\mathrm U(n)\cong\{\!$orthogonal complex structures on $\mathbf R^{2n}\}=$ the $\mathrm O(2n)$-orbit of $\bigl(\begin{smallmatrix}0&-I\\I&\phantom{-}0\end{smallmatrix}\bigr)$ in (a): a complex submanifold with signature $\left({\frac12}(n^2-n),0\right)$, hence symplectic and a (Kähler) coadjoint orbit of $\mathrm O(2n)$.<|endoftext|> TITLE: Is there a relation between type (maximum linearization) of a computable WQO and the ordinal strength of a theory needed to prove it? QUESTION [10 upvotes]: Background: Given a well partial order $X$ (more commonly studied with antisymmetry dropped as well-quasi-orders, but I'm going to say well partial order to make this definition simpler, obviously the two theories are essentially the same), De Jongh and Parikh showed there's always a largest ordinal that can be realized as the order type of a linearization of $X$. There are also several other equivalent characterizations of this quantity. Following De Jongh and Parikh I'll denote this $o(X)$, and following Kriz and Thomas I'll call it the type of $X$. Consider the WPO of finite plane trees (ordered trees) ordered under topological, order-preserving embedding; the statement that this is a WPO is Kruskal's tree theorem. Let's call this $T$. Fact 1: The type of $T$ is the small Veblen ordinal. Upper bound due to Schmidt, lower bound due (as far as I'm aware) to Jervell (though he didn't state it in that form). Fact 2: Rathjen and Weiermann -- using Schmidt's techniques -- showed that, in a certain sense, the proof-theoretic ordinal of Kruskal's tree theorem is the small Veblen ordinal. Note that this seems to be a somewhat unusual sense... the sense seems to be something like, for a theory to prove Kruskal's tree theorem, it has to have proof-theoretic ordinal (in the usual sense) of at least the small Veblen ordinal? Apologies, I'm not much of a logician, I'm having some trouble reading the paper... (maybe this is actually a common thing in logic, I wouldn't be familiar). Question: In general, given a Turing machine $M$ computing a countable WPO $X$ -- probably satisfying additional nice conditions but I'll get to that -- is there a relation between $o(X)$, and the proof-theoretic ordinal in the above sense -- to the extent that this is a well-defined thing, which I'm not clear on -- of the statement "$M$ computes a WPO"? (It may be worth noting here that Montalbán showed that the type of a computable WPO is always a computable ordinal.) As in, are they equal? Or at least, is there an inequality in one direction? I've wondered about this some time but am bringing it up now but am not really qualified to answer it; I bring it up now in particular because discussion on this answer made me realize other people may be wondering about it too and I thought it would be good to have one place where the question is properly stated. (Although that one talks about $\Gamma_0$, rather than the small Veblen ordinal, as being an ordinal related to Kruskal's tree theorem? I'm not sure what's going on here; it's not out of nowhere, they reference a paper explaining this, but I'm still not clear what's going on there. Nevermind, see Sylvain's comment about this.) Two additional notes: Like I said above, we presumably need additional conditions for this to work. These would likely be expressible as conditions on $o(X)$. Like, obviously $o(X)$ should be a power of $\omega$; or really I'd expect it would need to be of the form $\omega^{\omega^\alpha}$. Possibly it'd need to be an epsilon number, or something further along those lines. Possibly the whole question is dependent on some sort of implicit base theory, and how strong this sort of condition needs to be would depend on that. One direction of this maybe seems like it should be easy, but I don't think it actually is. Specifically, it seems like the type $\le$ proof-theory direction ought to be easy, because if a theory can prove that $X$ is a WPO, it ought to be able to prove that $o(X)$ is a well order, right? Since any linearization of a WPO is a well-order. Except, I think this doesn't actually work, because while computable WPOs have computable types, there isn't (according to the paper on this linked above) any computable way to realize this mapping of computable WPOs to their types? (Do I have that right?) Too bad -- if even just that direction were true, that alone would be quite significant... Anyway this may be an unreasonably difficult question but like I said, thought it would be good to have one place on the internet where this question is explicitly being asked, since I'm not aware of it appearing in the literature already! But, if these were indeed the same thing, or if there were even an inequality, it would be quite something, because it would mean that all the work on computing one could be ported over to the other. (I'd like to be able to port over proof-theoretic computations to computations of types, but I imagine many other people would like the reverse.) Thanks all! REPLY [5 votes]: Let me show that for extensions $T\supseteq\mathsf{ACA}_0$ the usual proof-theoretic ordinal $|T|_{WO}$ coincide with $|T|_{WPO}$ that is the suprema of $\mathsf{o}(X)$, for recursive wpo $X$, for which $T$ proves that $X$ is a wpo. Here $|T|_{WO}$ is the suprema of order types $\mathsf{ot}(X)$ of recursive well-orderings $X$ for which $T$ proves that they are well-orderings. Since any well-ordering $X$ is a wpo and $\mathsf{o}(X)=\mathsf{ot}(X)$ (and this is provable in $\mathsf{ACA}_0$), we have $|T|_{WO}\le |T|_{WPO}$. To establish that $|T|_{WO}\ge |T|_{WPO}$ we consider the rank $|T|_{WF}$ that is the suprema of ranks $\mathsf{rk}(X)$ of well-founded recursive binary relations for which $T$ proves well-foundedness. Since in $\mathsf{ACA}_0$ it is possible to prove that any well-founded relation is embeddable into a well-ordering (and for recursive well-founded relation the corresponding well-ordering is recursive as well), we have $|T|_{WF}=|T|_{WO}$. Thus we Any anti-chain $Q$ in a partial order $X$ determines a downward-closed set $E(Q)=\{x\in X\mid \forall q\in Q(q\not<_X x)\}$. This gives us the partial order $A^{\mathsf{fin}}(X)$ on all finite anti-chains in $X$: $$Q\le_{A^{\mathsf{fin}}(X)} P \iff E(Q)\subseteq E(P).$$ The function $E$ is an embedding of $A^{\mathsf{fin}}(X)$ into the inclusion order $C(X)$ on all downward-closed subsets of $X$. It is straightforward to prove that the following are equivalent: $X$ is a wpo $C(X)$ is well-founded $A^{\mathsf{fin}}(X)$ is well-founded. And it is easy to see if $X$ is a wpo, then $E$ is an isomorphism of $A^{\mathsf{fin}}(X)$ and $C(X)$. The proofs of both the facts here require only Ramsey theorem for pairs and could be proved in $\mathsf{ACA}_0$. Fairly obviously, any linearization $L$ of $X$ is embeddable into $C(X)$ by the map $x\longmapsto \{y\in X\mid y<_L x\}$. Thus for wpo's $X$ we have $\mathsf{o}(X)\le \mathsf{rk}(C(X))=\mathsf{rk}(A^{\mathsf{fin}})$. Notice that for recursive orders $X$ the order $A^{\mathsf{fin}}(X)$ is recursive as well. Thus, $|T|_{WPO}\le |T|_{WF}$.<|endoftext|> TITLE: Universal cover or Bass-Serre tree: difference between definitions given by Bass and Serre QUESTION [5 upvotes]: Let $(\mathbb G,\Gamma)$ be a graph of groups. A $G$-path from $u_0$ to $u$ is $$g_0e_1g_1\cdots e_{n}g_n,$$ where $e_1\cdots e_{n}$ is a walk in $\Gamma$ from $u_0$ to $u$ and each $g_i\in G_{s(e_i)}$. We denote the set of all $\mathbb G$-paths from $u_0$ to $u$ by $\pi[u_0,u]$. Let $\mathbb F(\mathbb G,\Gamma)$ be the group generated by the vertex groups and the elements $e$ of edge $\Gamma$, subject to the relations $$\{s_e(g)e=et_e(g)\mid g\in G_e,e\in E(\Gamma)\},\{e^{-1}=\bar{e}\mid e\in E(\Gamma)\}$$ Recall that the fundamental group of $(\mathbb G,\Gamma)$ on the base point $u_0$ is the set of all elements $g_0e_1g_1\cdots e_{n}g_n$ in $\mathbb F(\mathbb G,\Gamma)$, where $e_1\cdots e_{n}$ is a closed walk from $u_0$ to $u_0$. In other words, the fundamental group of $(\mathbb G,\Gamma)$ is $\pi[u_0,u_0]$. My question is regarding the universal cover (Bass-Serre tree) of $(\mathbb G,\Gamma)$. In the book "Trees'' by Serre page 51, the universal cover is defined in the following way: The vertices are the left cosets of the vertex groups in $\pi[u_0,u_0]$, i.e. $$\bigcup_{u\in V(\Gamma)} \pi[u_0,u_0]G_u$$ However in a paper by Bass "Covering theory for graphs of groups ", he defined in the following way: The vertices are $$\bigcup_{u\in V(\Gamma)} \pi[u_0,u]G_u$$ My question is that Are they $\pi[u_0,u_0]$-isomorphic?or they are completely different? If yes, I really wonder to know the $\pi[u_0,u_0]$-isomorphism map. REPLY [4 votes]: To address a small point: the sets $\pi[u_0,u]$ should really be considered as their images in $\mathbb F(\mathbb G,\Gamma)$. This has the effect of doing some algebraic version of passing to homotopy classes of paths rel endpoints. For example, if $e$ is an edge of $\Gamma$, then $e\bar e e$ and $e$ should be considered as equivalent. This is already what happens when constructing the universal cover of a graph, say. Modulo agreement on that point, my claim is that the definitions of the vertex sets of the universal cover of $(\mathbb G,\Gamma)$ are equivalent. Here is how to see it. Fix a choice of spanning tree $T$ in $\Gamma$. For every vertex $u$ of $\Gamma$, there is a unique path $\sigma_u$ from $u_0$ to $u$ that stays entirely within $T$. We can think of $\sigma_u$ as an element of $\pi[u_0,u]$ by sending $\sigma_u$ to (the equivalence class of) the ordered sequence of edges it traverses. We get maps $\pi[u_0,u_0] \to \pi[u_0,u]$ and $\pi[u_0,u]\to \pi[u_0,u_0]$. The former sends a class $\tau \in \pi[u_0,u_0]$ to the class of the concatenation $\tau\sigma_u$. The latter sends $\rho \in \pi[u_0,u]$ to the class of the concatenation $\rho\bar\sigma_u$. Since the classes $\tau$ and $\tau\sigma_u\bar\sigma_u$ are equal, it’s easy to see that these maps are inverse bijections.<|endoftext|> TITLE: So what exactly are perverse sheaves anyway? QUESTION [19 upvotes]: Is there a way to define perverse sheaves categorically/geometrically? Definitions like the following from lectures by Sophie Morel: The category of perverse sheaves on $X$ is $\mathrm{Perv}(X,F):=D^{\leq 0} \cap D^{\geq 0}$. We write $^p\mathrm{H}^k \colon D_c^b(X,F) \to \mathrm{Perv}(X,F)$ for the cohomology functors given by the $\mathrm{t}$-structure. are well and good, but to me, it feels like they come with too many "luggages", and therefore make me feel like I don't have as good of an intuition of derived categories as I should. REPLY [10 votes]: After some searching I've found the notes An illustrated guide to perverse sheaves, by Geordie Williamson, which is a beautifully illustrated (and sort of topologically oriented) introduction to perverse sheaves.<|endoftext|> TITLE: Has gnu(2048) been found? QUESTION [42 upvotes]: The gnu (or Group NUmber) function describes how many groups there are of a given order. The number of groups of each order are known up to 2047, see https://www.math.auckland.ac.nz/~obrien/research/gnu.pdf Has any progress been made on the number of groups of order 2048? This case is particularly difficult due to 2048 being a large power of 2. It is known that the number of groups of order 2048 of nilpotency class 2 is 1,774,274,116,992,170 (according to the above link), and apparently the full group number is expected to agree with this number in the first three digits. REPLY [49 votes]: No, it is unknown, and I don't think we will find it anytime soon. For the state of the art, see our 2017 paper "Constructing groups of ‘small’ order: Recent results and open problems" DOI (here is a PDF). I collected the known data on a little website for easier browsing. And am working as I type this on packaging it up for GAP.<|endoftext|> TITLE: For regular tetrahedron $ABCD$ with center $O$, and $\overrightarrow{NO}=-3\overrightarrow{MO}$, is $NA+NB+NC+ND\geq MA+MB+MC+MD$? QUESTION [7 upvotes]: Let $ABCD$ be a regular tetrahedron with center $O.$ Consider two points $M,N,$ such that $\overrightarrow{NO}=-3\overrightarrow{MO}.$ Prove or disprove that $$NA+NB+NC+ND\geq MA+MB+MC+MD$$ I tried to use CS in the Euclidean space $E_3$, but it does not help, because the minoration is too wide. Note: I also posted this on the Mathematics Stack Exchange, but not much progress has been made on this question. This is why I thought that posting here too would be all right (this problem is open in the sense that its proposer doesn't have a proof, so I guess it is fit the for this forum). EDIT: The bounty expired, so this may be reopened. REPLY [8 votes]: Following suggestions on Stack Exchange, we use the homothety with respect to $O$ and factor $-3$. If $X$ is a point, then $X'$ denotes the point for which $\overrightarrow{XO}= -3 \overrightarrow{X'O}$. So $M=N'$ and $A'$ is the midpoint of the face opposite to $A$. One has $XY=3X'Y'$ and the desired inequality becomes $3MA'+3MB'+3MC'+3MD'\geq MA+MB+MC+MD$. It thus suffices to prove inequalities of the type $MB'+MC'+MD'\geq MA$. Thus put $f(X)=XB'+XC'+XD'-XA$. We have to show that $f(M)\geq0$. Let us fix our tetrahedron as follows. $A=\left(0,0,\sqrt{\frac{2}{3}}-\frac{1}{2 \sqrt{6}}\right)$, $B= \left(-\frac{1}{2 \sqrt{3}},-\frac{1}{2},-\frac{1}{2 \sqrt{6}}\right)$, $C=\left(-\frac{1}{2 \sqrt{3}},\frac{1}{2},-\frac{1}{2 \sqrt{6}}\right)$, $D=\left(-\frac{1}{2 \sqrt{3}},\frac{1}{2},-\frac{1}{2 \sqrt{6}}\right)$. Put $f_1(X)=XB'+XC'$, $f_2(X)=XD'-XA$, so that $f(X)=f_1(X)+f_2(X)$. Notice that $f(O)=0$ and notice that $f(M)$ is positive if $M$ is far away from $O$. Now take $M$ so that $f(M)$ is an absolute minimum. Note that $f$ is positive at $A$, $B'$, $C'$, $D'$, so that $M$ is none of those points. We have $f_2(M)<0$ and the gradients of $f_1$, $f_2$ cancel each other at $M$. That means that the level sets through $M$ of $f_1$ and $f_2$ touch at $M$. (There is no point where both gradients vanish.) The level set through $M$ of $f_1$ is an ellipsoid, with $f_1$ smaller inside, and the level set through $M$ of $f_2$ is the lower sheet of a hyperboloid of two sheets, with $f_2$ more negative inside the sheet. As each level set is the curved boundary of a convex region, there is no other point where the two level sets touch. The level sets are symmetric with respect to the plane $ \left\{X\mid XB'=XC' \right\}$, so $M$ must lie on that plane. We have shown that $MB'=MC'$. Similarly $MC'=MD'$ and $M$ must be of the form $(0,0,x)$, so that $f(M)=\frac{\sqrt{72 x^2-4 \sqrt{6} x+3}-\sqrt{8 x^2-4 \sqrt{6} x+3}}{2 \sqrt{2}}\geq0$.<|endoftext|> TITLE: Can Yoneda lemma for smooth projective varieties only use curves? QUESTION [17 upvotes]: Let $X,Y$ be two projective smooth varieties over an algebraically closed field $k$. If we have natural isomorphism $\operatorname{Hom}(C,X) \cong \operatorname{Hom}(C,Y)$ as sets for every smooth projective curve $C$, then do we know $X \cong Y$? In other words, $Hom(-,Y)$ and $Hom(-,X)$ are two isomorphic functors when restricted to the subcategory of curves. If not, for a fixed $X=X_0$, can we classify all such $Y$ ? Are they all birational? When is there a unique $Y$ (namely $X_0$)? What if we require all curves $C$ (not necessarily smooth or projective)? REPLY [16 votes]: This is a twice updated version: Thanks to @PiotrPstrągowski for pointing out some issues that require more care. Some thoughts were in response to the original wording of the question. I left them here, because they might be interesting to some. As @AndréHenriques points out this still doesn't work -- at least compared to the updated question. My initial reaction was to delete the answer, but perhaps it is still interesting, so I will leave it here. The main issue here is to decide what you mean by "functorial isomorphism"? If, for instance, you mean that it is induced by a morphism $\phi:X\to Y$, then sure, it is true: By your condition $\phi$ would have to be surjective and have only zero dimensional fibers, so by ZMT an isomorphism. On the other hand, if we're not asking for such a strong funtoriality, only for some "natural" way to have the two sets be bijective, then one can do this. As @AndréHenriques points out this does not work if we allow constant maps between curves. For whatever it is worth it still seems to work if we restrict to dominant maps... Choose an $n\in\mathbb N$, $n\gg 0$ and choose two disjoint sets of $n$-points in general position in $\mathbb P^2$, say $p_i$ and $q_i$, such that there is no automorphism of $\mathbb P^2$ that takes one set to the other. Let $X$ and $Y$ be $\mathbb P^2$ blown-up along these two sets of points respectively (say $X$ along the $p_i$'s and $Y$ along the $q_i$'s). For any $C\to X$ or $C\to Y$ that when composed with the blow-down to $\mathbb P^2$, it does not map into any $p_i$'s or $q_i$'s, there is a well-defined proper transform on the other one (between $X$ and $Y$), so make these correspond to each other. For each $p_i$ and $q_i$ let the corresponding exceptional curves be $E_i\subset X$ and $F_i\subset Y$ respectively and choose a point on each: $P_i\in E_i$ and $Q_i\in F_i$. In addition, choose an isomorphism $\phi_i:E_i\to F_i$ such that $\phi_i(P_i)=Q_i$. Now let $\alpha:C\to E_i\subset X$ be a morphism. If $\alpha(C)=\{P_i\}$ then make it correspond to the "same" map $\alpha':C\to p_i\in Y$, otherwise make it correspond to the map $\alpha''=\phi_i\circ\alpha:C\to F_i\subset Y$. Finally, let $\alpha:C\to q_i\in X$ be a morphism and make it correspond to the "same" map $\alpha':C\to _i\in Y$. This, with the proper transforms above, gives a bijection $\operatorname{Hom}(C,X)\leftrightarrow \operatorname{Hom}(C,Y)$ which is functorial in $C$. Note that this is for the case when $C$ is smooth or at least irreducible. If we allow singular reducible curves, then this does not work, because one could consider maps from two intersecting lines, one mapping to an exceptional curve, say $E_i$, and the other to the proper transform on $X$ of a general line through $P_i\in\mathbb P^2$. If the bijection between the $\operatorname{Hom}$'s is functorial, then it would have to map the proper transform to its proper transform on $Y$ which will not intersect the image of the other line by the morphism that is assigned to it by the above process. As @PiotrPstrągowski points out, the assumptions imply that there is a bijection between the points of $X$ and the points of $Y$. The above example shows that as long as $C$ is smooth, this bijection does not have to be a morphism, and accordingly $X$ and $Y$ are not necessarily isomorphic. However, it does leave the possibility open that $X$ and $Y$ may be necessarily birational, and Allowing $C$ to be singular may force $X$ and $Y$ to be isomorphic. Remark: I don't think allowing non-projective curves would make a difference as long as $X$ and $Y$ remain projective.<|endoftext|> TITLE: How many steps are required for double transitivity? QUESTION [11 upvotes]: Let $A$ be a set of generators of $S_n$, or of a doubly transitive subgroup of $S_n$. Assume $e\in A$, $A=A^{-1}$. What is the least $k$ such that $A^k$ is doubly transitive as a set? That is, what is the least $k$ such that there is a pair $x = (i,j)$, $i,j\in \{1,\dotsc,n\}$, $i\ne j$, for which $A^k x$ is the set of all pairs of distinct elements of $\{1,2,\dotsc, n\}$? The bound $k = O(n^2)$ is very easy. Can we prove $k = O(n \log n)$? $k = O(n)$? As a starting exercise, can we at least prove $k = O(n^{3/2})$? Alternatively, can one construct a counterexample to $k=O(n)$? (Note the classical example $A = \{(1 2), (1 2 \dotsc n)\}$ is not a counterexample.) REPLY [9 votes]: It seems that this is a lower bound of $\Omega(n^2)$. Take an $n$ and an $a=\Theta( n) $ coprime with $n$ (with $a TITLE: Updates to Stanley's 1999 survey of positivity problems in algebraic combinatorics? QUESTION [34 upvotes]: [I am a co-moderator of the recently started Open Problems in Algebraic Combinatorics blog and as a result starting doing some searching for existing surveys of open problems in algebraic combinatorics.] In 1999 R. Stanley wrote a very nice survey on open problems in algebraic combinatorics, with a specific focus on positivity, called "Positivity problems and conjectures in algebraic combinatorics", available online here. It includes 25 specific open problems, as well as a lot of discussion/context. Question: 20 years later, which problems from Stanley's list have been resolved? On his website he has a page with updates from 2004, but still, even 2004 was 15 years ago. REPLY [37 votes]: I'm posting a community wiki answer to compile all the known information about the status of all the problems. Problem 1 (The Generalized Lower Bound Theorem/g-theorem for Gorenstein* complexes): In December 2018, Adiprasito posted a preprint (see also this summary) announcing a proof of the g-theorem for homology spheres that are also homology manifolds, which are the same thing as Gorenstein* complexes. Problem 2 (The GLBT for toric h-vectors of polytopes/Gorenstein* lattices): Karu established the GLBT for toric h-vectors of arbitrary convex polytopes. It would seem the extension to Gorenstein* lattices remains open though, and is discussed in this paper of Billera and Nevo. Problems 3 and 3' (Kalai's $3^d$ conjecture for $f$-vectors of centrally symmetric polytopes): It would seem this is still open (see Wikipedia). Some stronger versions of the conjecture were disproved by Sanyal-Werner-Ziegler. The latest discussion I can find is in Freij-Henze-Schmitt-Ziegler. Problem 4 (The Charney-Davis conjecture on $h$-vectors of flag spheres): This seems wide-open still. In dimensions $\leq 3$ it has been proved by Davis and Okun. The most significant progress for arbitrary dimensions is work of Gal in which he defines the $\gamma$-vector of a flag complex as the "right" analog of the $g$-vector for flag complexes, and conjectures that the $\gamma$-vector of a flag generalized homology sphere is nonnegative. This would in particular imply the Charney-Davis conjecture (which is essentially the statement that a particular coefficient in the $\gamma$-vector is nonnegative). See this nice survey of Zheng. Problem 5 (A generalization of the decomposition of acyclic complexes to "$k$-fold" acyclic complexes): This conjecture of Stanley was resolved in the negative by Doolittle and Goeckner. Problem 6 (Are Cohen-Macaulay complexes partitionable?): This conjecture of Stanley and Garsia was resolved in the negative by Duval–Goeckner–Klivans–Martin. Problem 7 (Positivity of the cd-index of a Gorenstein* poset): This was proved by Karu. Problem 8 (Positivity of cubical $h$-vectors of Cohen-Macaulay cubical complexes): I believe that this is still open. The one significant result I can find is that Athanasiadis has proved that for a Cohen-Macaulay cubical complex of dimension $d$ (or more generally a Cohen-Macaulay cubical poset) we have $h^{(c)}_{d-1}\geq 0$ (that $h^{(c)}_{d}\geq 0$ is easy). Problem 9 (Combinatorial interpretation of plethysm coefficients): Still wide-open in general but solved in many special cases. The maximal and minimal constituents of an arbitrary plethysm were found by Paget and Wildon; this article has a survey of the main results known in 2016. Some relations between plethysm coefficients, generalizing results of Brion, Bruns–Conca–Varbaro, Ikenmeyer, and Paget–Wildon, are in this preprint. Foulkes' Conjecture is the special case that $h_m \circ h_n - h_n \circ h_m$ is Schur positive when $m \ge n$: it is known when $m \le 5$ (Cheung–Ikenmeyer–Mkrtchyan) and when $m$ is large compared to $n$ (Brion). Problem 10 (Combinatorial interpretation of Kroenecker coefficients): Still wide-open in general. But some special cases are known, such as when some of the partitions are hooks or two-rowed shapes (see Blasiak and Liu and its references). Problem 11 (Combinatorial interpretation of Schubert polynomial structure constants): Still wide-open in general. But some special cases are known, such as some cases of "Schur times Schubert" (see Mészáros–Panova–Postnikov). Problem 12 (Combinatorial interpretation of row sums of the character table of the symmetric group): Still open in general. Some special cases are discussed by Baker and Early in "Character Polynomials and Row Sums of the Symmetric Group", see here, and Sundaram in "The conjugacy action of $S_n$ and modules induced from centralisers", see here. Problem 13 (The Macdonald positivity conjecture): This was resolved by Haiman, using advanced machinery from algebraic geometry like the Hilbert scheme of points. A combinatorial interpretation of the $(q,t)$-Kostka polynomials remains elusive in general, but there are some partial results (see e.g. this paper of Assaf). Problem 14 (LLT polynomials- combinatorial proof of symmetry, and Schur positivity): The Schur positivity of LLT polynomials was proved by Grojnowski and Haiman in an unpublished manuscript from 2006. I imagine a combinatorial proof of the symmetry of these polynomials remains open. Problem 15 (Positivity of the coefficients of Kazhdan-Lusztig polynomials for arbitrary Coxeter groups): This was resolved by the work of Elias and Williamson on Soergel bimodules. Problem 16 (Combinatorial interpretation of the coefficients of Kazhdan-Lusztig polynomials for Weyl groups/affine Weyl groups): ??? Problems 17, 17', 18 (Total positivity and Schur positivity of monomial immanants): These are apparently open (for instance, Stanley notes that an affirmative answer to Problem 17 would imply one for Problem 21), but some special cases are addressed in the work of Clearman–Shelton–Skandera. Problem 19 (Positivity/symmetry/unimodality of monomial characters of Hecke algebra evaluated on Kazhdan-Lusztig basis elements): According to Clearman–Hyatt–Shelton–Skandera, this is still open (or at least was in 2016). Problems 20 and 20' (The Stanley-Neggers conjecture about real rootedness of poset descent polynomials): Counterexamples to these conjectures were first found by Brändén and Stembridge. (As explained in the survey by Stanley, Problem 20' about chain polynomials is equivalent to Problem 20 and hence also has a negative answer.) Problem 21 (The Stanley-Stembridge conjecture about e-positivity of chromatic symmetric functions of (3+1)-free posets): This is currently a hot topic, and still open, although many special cases are known as documented here. The most significant advances on the problem are a result of Guay-Paquet which reduces the conjecture to the case of (3+1)- and (2+2)-free posets, i.e., unit interval orders; as well as work of Shareshian and Wachs, Brosnan and Chow, and Guay-Paquet, which connects the conjecture to the cohomology of Hessenberg varieties. Problems 22 (Gasharov's conjecture about s-positivity of chromatic symmetric functions of claw-free graphs): ??? Problem 23 (Real-rootedness of stable set polynomials of claw-free graphs): This was proved by Chudnovsky and Seymour. Problem 24 (The Monotone Column Permanent Conjecture): Solved by Brändén–Haglund–Visontai–Wagner using the theory of real stable polynomials. Problem 25 (Unimodality/log concavity of (a) coefficients of characteristic polynomial of graph/matroid, (b) number of size $i$ independent sets of graph/matroid, (c) rank sizes of a geometric lattice): For graphs, (the absolute value of) the coefficients of the chromatic polynomial were shown to be log-concave by Huh; this was extended to realizable matroids by Huh–Katz; and then to all matroids by Adiprasito–Huh–Katz. The result for the characteristic polynomial actually implies the result for independent sets, as observed by Lenz using a result of Brylawski. The unimodality of the rank sizes of a geometric lattice (i.e., the so-called "Whitney numbers of the second kind") is apparently a harder question and remains open: see section 5.10 of this survey of Baker. But ``half'' of the conjecture has been proved (i.e., that the rank sizes increase up to halfway), in the case of realizable matroids by Huh–Wang, and in the case of arbitrary matroids in this preprint of Braden–Huh–Matherne–Proudfoot–Wang.<|endoftext|> TITLE: Universal property of the cocomplete category of models of a limit sketch QUESTION [5 upvotes]: Let $\mathscr{S}$ be a limit sketch in a small category $\mathcal{E}$, i.e. just a collection of cones in $\mathcal{E}$. Then its category $\mathbf{Mod}(\mathscr{S})$ of models (i.e. functors $\mathcal{E} \to \mathbf{Set}$ which send the cones in $\mathscr{S}$ to limit cones) is cocomplete, in fact locally presentable. It enjoys the following universal property in the $2$-category of cocomplete categories: If $\mathcal{C}$ is any cocomplete category, then $$\mathrm{Hom}_c(\mathbf{Mod}(\mathscr{S}),\mathcal{C}) \simeq \mathbf{Mod}_{\mathcal{C}}(\mathscr{S}^{\mathrm{op}}).$$ Here, $\mathrm{Hom}_c$ denotes the category of cocontinuous functors, $\mathscr{S}^{\mathrm{op}}$ denotes the dual colimit sketch, and $\mathbf{Mod}_{\mathcal{C}}$ refers to $\mathcal{C}$-valued models. A typical example of this universal property is $\mathrm{Hom}_c(\mathbf{Grp},\mathcal{C}) \simeq \mathbf{CoGrp}(\mathcal{C})$, that is, $\mathbf{Grp}$ is the universal example of a cocomplete category with an internal cogroup object. The proof is a combination of two well-known results, so I assume that this universal property is also well-known. Can someone confirm this and point me to literature which I can cite for this result? Maybe it follows from some of the theorems in Kelly's book on enriched categories, chapter 6. But I am not sure. Here is a sketch of the two-step proof I was thinking of. Let $G : \mathbf{Mod}(\mathscr{S}) \to [\mathcal{E},\mathbf{Set}] $ be the inclusion and $F : [\mathcal{E},\mathbf{Set}] \to \mathbf{Mod}(\mathscr{S})$ its left adjoint. Then by a "tensor-less" variant of Prop. 2.3.6. in Tensor functors between categories of quasi-coherent sheaves (as you see, this is not quite the reference I need!) the category $\mathrm{Hom}_c(\mathbf{Mod}(\mathscr{S}),\mathcal{C})$ is equivalent to the category of those cocontinuous functors $Q^* : [\mathcal{E},\mathbf{Set}] \to \mathcal{C}$ such that $Q^* \to Q^* \circ G \circ F$ is an isomorphism; this is true iff the right adjoint $Q_* : \mathcal{C} \to [\mathcal{E},\mathbf{Set}]$ factors over $\mathbf{Mod}(\mathscr{S})$. By the well-known universal property of $[\mathcal{E},\mathbf{Set}]$ as the free cocompletion of $\mathcal{E}^{\mathrm{op}}$ (for instance, Section 4.4 in Kelly's book), $Q^*$ corresponds to a functor $P : \mathcal{E}^{\mathrm{op}} \to \mathcal{C}$ via $Q^* = P \otimes_{\mathcal{E}} -$, and hence $Q_* = \mathrm{Hom}(P(-),-)$. By the Yoneda Lemma, the requirement that $Q_*$ factors over $\mathbf{Mod}(\mathscr{S})$ exactly means that $P$ is a model of $ \mathscr{S}^{\mathrm{op}}$. I have found a reference: Theorem 2.2.4 in this paper (actually, I knew this paper, but I somehow forgot that the theorem is in there). But the authors also explicitly state that this theorem is well-known. I would be happy with a more "classical" or "original" reference. REPLY [2 votes]: Probably the earliest reference is Theorem 2.5 in A. Pultr, The right adjoints into the categories of relational systems, Reports of the Midwest Category Seminar IV. Springer, Berlin, Heidelberg, 1970 Pultr's "relational theories" are exactly small realized limit sketches. Remark: I have recently generalized the result to large limit sketches (arXiv:2106.11115).<|endoftext|> TITLE: Relation between "triangulated bordism", MO, and $H\mathbb{F}_2$ QUESTION [5 upvotes]: The unoriented bordism theory $MO$ has a map to $H\mathbb{F}_2$ which is easily described for a space $X$ by pushing forward the fundamental class of a singular manifold to $H_*(X)$. Since $MO$ and $H\mathbb{F}_2$ both factor through chain complexes, it is tempting to ask if this can be realized as a map of chain complexes. I believe this is unlikely because although all smooth manifolds admit a triangulation, this obviously cannot be done naturally. However, this makes me wonder about the following: For a space $X$, let $D_n(X)$ denote the set of pairs $(M,f)$ where $M$ is a smooth manifold with a specified triangulation and $f$ is a map from $M$ into $X$. This is a chain complex by letting the boundary map take a manifold to its boundary. Denote the nth homology of this chain complex by $MT_n(X)$ (meant to stand for triangulated bordism). Alternatively, I believe we can describe $D_n(X)$ as tuples $(M,f,\sigma)$ where $(M,f)$ is a singular manifold and $\sigma$ is a cycle lifting the fundamental class $[M]$ subject to a few conditions. Recall that one can describe $H\mathbb{F}_2(X)$ as the homology of the chain complex $C(X)$ where $C_n(X)$ is the set of pairs $(S,f)$ where $S$ is a simplicial complex that is the union of its n-simplices and $f$ is a map from $S$ into $X$. The boundary operator takes $S$ to the union of n-1 simplices that are incident to an odd number of n-simplices. There is a chain map $i:D(X) \rightarrow C(X)$ given by forgetting the fact that the domain of $f$ is a manifold. As well, we may also forget the triangulation to get a map $j$ from $D(X)$ to the chain complex of singular manifolds. Since every smooth manifold has a triangulation, $j$ (hence $j_*$) is surjective. This implies $i_*$ is surjective since it factors through $j_*$ and the map $MO \rightarrow H\mathbb{F}_2$. (1) Is $MT$ a homology theory? (2) Supposing (1), is $MT$ a wedge of $H\mathbb{F}_2$? (3) Supposing (1), is $MT$ a pullback of $H\mathbb{F}_2$ and $MO$? REPLY [6 votes]: There is an implicit assumption in your question, namely that one can define a chain complex calculating the functor $X\mapsto MO_*(X)$ which is based on maps from unoriented manifolds into $X$. As far as I know, this is not known to be the case. Though, as you point out, since $MO$ is a wedge sum of shifts of copies of $H\mathbb F_2$, it's at least remotely possible that such a complex exists. It would have to be very special, though, since most (all?) other bordism theories are not wedge sums of Eilenberg--MacLane spectra. The following obvious thing to try definitely does not work: take $C^{MO}_*(X):=$ the free $\mathbb F_2$ vector space on all isomorphism classes of pairs $(M,f)$ where $M$ is a compact unoriented manifold with boundary and $f:M\to X$ (with the obvious boundary map $\partial(M,f):=(\partial M,f|_{\partial M})$). The homology $H^{MO}_*(X)$ is not $MO_*(X)$, even for $X=\operatorname{pt}$. Indeed, $MO_1(\operatorname{pt})=0$ (every $1$-manifold is null cobordant), whereas there is a nontrivial cycle in $C^{MO}_1(\operatorname{pt})$ given by the interval $[0,1]$ equipped with the constant map to $\operatorname{pt}$. This cycle cannot be a boundary, since the only boundaries in $C^{MO}_*$ are maps from closed manifolds. There are, of course, a few variations one could try on the above construction, but there are also what seem to me to be a number of fundamental difficulties. Suppose we start by taking as generators pairs $(M,f:M\to X)$ with $M$ closed (these having zero boundary). To get a bordism theory as the homology, clearly we want to also add in $M$ with boundary, so that cobordant manifolds are homologous. But this creates many new cycles! Namely, we could have some formal sum $c$ of manifolds with boundary such that the terms comprising $\partial c$ all cancel (like $[0,1]$ in the example above). Moreover, there may be many different ways of realizing this cancelling geometrically by gluing together the given manifolds comprising $c$ to create a closed manifold. Note that there's no good reason that all of these resulting manifolds to be cobordant! This is somewhat unnerving, but things are about to get worse. If we forge ahead and add in isomorphism classes of pairs $(M,f:\to X)$ where $M$ is allowed to have corners, then given a cycle in this chain complex, it's no longer even clear whether it makes sense to glue up the resulting manifolds in a consistent way (note that I have written isomorphism classes here---this is very important: without these words the generators wouldn't even form a set). This issue can be remedied by fixing a set of manifolds with corners and consistent identifications of their boundary strata with other manifolds with corners on the same list. For example, this is exactly what we do to define singular homology! (take the manifolds with corners to be $\Delta^n$). I had to do something like this in section 8.7.1 of this paper, to calculate singular homology. But this seems likely to just give singular homology no matter how you do it, not any bordism theory.<|endoftext|> TITLE: Does every $C_4$-free bipartite graph lies in some finite projective plane? QUESTION [10 upvotes]: A projective plane $Π$ is a 3-tuple $(P,L,I)$ where $P$ and $L$ are sets, and $I$ is a relation between $P$ and $L$, such that: For every two elements $p_1$, $p_2\in P$, there exists a unique element $l \in L$ such that $p_1 I l$ and $p_2 I l$. For every two elements $l_1$, $l_2\in P$, there exists a unique element $p \in L$ such that $p I l_1$ and $p I l_2$. A finite projective plane is a projective plane where $P$ and $L$ are finite. Identify projective planes with the bipartite graph with two parts $P$ & $L$ where $p\in P$ is connected to $l \in L$ iff $pIl$. Such graphs do not have $C_4$ as a subgraph: Suppose there's a $C_4$ subgraph formed by the vertices $p_1,l_1,p_2,l_2$ where $p_1$ and $p_2$ are both connected to $l_1$ and $l_2$. Then the element $l$ where $p_1 I l$ and $p_2 I l$ is not unique, thus violating the rules. Their induced subgraphs has no $C_4$s, either. Q: Is the converse true, i.e. if $G$ is a $C_4$-free bipartite graph, is there a projective plane $Π$ where $G$ is an induced subgraph of $Π$? One cannot expect the conjecture above holding for finite-field planes: Let $Γ$ be the Desagures graph. Let $e$ be an edge of $Γ$(The Desagures graph is edge-transitive, so all edges are the same) . The Desagures theorem states, if the configuration $Γ＼e$ can be found in a finite-field plane, then the edge $e$ is also present there. So $Γ＼e$ is not an induced subgraph of any finite-field plane. Known: All graphs with 13 vertices or less are subgraphs of finite-field projective planes, as checked by computer. By the Lefschetz principle, if $G$ is a subgraph of the incidence graph of $\mathbb{RP}^2$, then $G$ is the subgraph of an incidence graph of some finite-field plane. So trees satisfy the conjecture. REPLY [12 votes]: This is an open problem posed by Erdos in "Some old and new problems in various branches of combinatorics" (see section 6). There hasn't been any substantial progress since then. After posing the question Erdos writes "I have no idea how to attack this problem", and that seems to be state of the art.<|endoftext|> TITLE: Research work on $ax^n-by^m=1$ QUESTION [5 upvotes]: I am looking for results on the equation $$ax^n-by^m=1 \tag 1 $$ where $\gcd(m,n)=1$ and $a,b,n,m$ are constants. I found literature for $ax^n-by^n=1$ (R. A. Mollin, D. T. Walker) but couldn't find anything on equation $(1)$. The closest I got is Pillai's conjecture which is a generalization of Catalan's conjecture. Does $ax^n-by^m=1$ has infinite solution? If it does how do we find them? Please provide related literature/reference if possible. REPLY [8 votes]: A lot is known about such diophantine equations of the form $F(x,y)=0$ for a polynomial $F$, including results about when the number of solutons can be infinite (this happens in very rare cases that don't include the ones you are interested in) as well as efficient methods to find the solutions. I would recommend looking at the paper "The Diophantine Equation $f(x)=g(y)$" by Y. F. Bilu and R. F. Tichy. It has an extensive list of references that cover most classical results on such equations.<|endoftext|> TITLE: Morita-invertible C*-algebras QUESTION [6 upvotes]: I am familiar with the Morita theory of rings, and the hermitian Morita theory of rings with involution, and I am trying to understand some parallels and differences with the Morita theory of C*-algebras. In the algebraic version, we are interested in the monoid structure of the Morita equivalence classes of $R$-algebras (where R is a commutative ring), given by the tensor product over $R$. In particular, the invertible elements of this monoid are given by the Azumaya algebras over $R$, and they form the Brauer group of $R$. Do similar phenomena occur for C*-algebras? For von Neumann algebras, I asked a question on Math.SE, and someone commented that the theory is basically empty: since factors of type I are Morita-trivial, and a tensor product of a factor of type II or III with another factor is again of type II or III, in the end the only way to be Morita-invertible is to be Morita-trivial. Do C*-algebras offer more theory? I understand that in that context one has to be a little more careful: the Morita-equivalence I care about is the strong Morita equivalence (as defined by imprimitivity bimodules). Also, talking about tensor products can be awkward, so maybe I should restrict to nuclear C*-algebras (but if there are results of Morita-invertibility for well-chosen tensor products on non-nuclear algebras, I am also interested in hearing about it). Clearly, one has to restrict to unital and central algebras, so in the end my question is the following: If $A$ is a central unital C*-algebra (maybe nuclear), and there exists $B$ such that $A\otimes B$ is strongly Morita-equivalent to $\mathbb{C}$ (for some tensor product if $A$ is not nuclear), does it follow that $A$ itself is strongly Morita-equivalent to $\mathbb{C}$? I am also interested in similar results for real C*-algebras (maybe even more so). REPLY [7 votes]: I know my answer is coming a bit late, but the answer to your question is: yes. If $A$ is a $C^\ast$-algebra, and there exists a $C^\ast$-algebra $B$ such that $A\otimes_\alpha B$ is strongly Morita equivalent to $\mathbb C$ for some $C^\ast$-tensor product $\otimes_\alpha$, then $A\cong \mathcal K(H)$ is the compact operators on some Hilbert space $H$, and in particular $A$ is strongly Morita equivalent to $\mathbb C$. If $A$ is unital (as in the question), then $H$ is finite dimensional, so $A$ is a matrix algebra. It is well-known that strong Morita equivalence preserves: (1) being a Type I $C^\ast$-algebra (also called GCR); and (2) the spectrum. Hence $A\otimes_\alpha B$ is of Type I and its spectrum is a point. In particular $A\otimes_\alpha B$ is simple so $\otimes_\alpha$ is the minimal tensor product (since the $A\otimes_{\min{}} B$ is a quotient of $A\otimes_{\alpha} B$). By Theorem 2 in Tomiyama's paper "Applications of Fubini type theorem to the tensor products of C∗-algebras. Tohoku Math. J. (2) 19 (1967), 213–226." it follows that $A$ and $B$ are Type I. By Theorem B.45 in Raeburn and William's book on Morita equivalence, it follows that the spectrum of $A$ is a singleton. As Naimark's problem* is true amongst Type I $C^\ast$-algebras it follows that $A \cong \mathcal K(H)$ for some Hilbert space $H$. *Naimark's problem: if $A$ is a $C^\ast$-algebra for which the spectrum is a singleton, is $A$ isomorphic to the compact operators on some Hilbert space?<|endoftext|> TITLE: Mathematicians with both “very abstract” and “very applied” achievements QUESTION [61 upvotes]: Gödel had a cosmological model. Hamel, primarily a mechanician, gave any vector space a basis. Plücker, best known for line geometry, spent years on magnetism. What other mathematicians had so distant interests that one wouldn’t guess one from the other? (Best if the two interests are not endpoints of a continuum, as may have been the case of past universalists like Euler or Cauchy. For this reason, maybe best restrict to post-1850 or so?) The point of asking is that it seems not so rare, but you don’t normally learn it other than by chance. Edit: Now CW, works best with “one mathematician per answer” (and details of actual achievement, e.g. “war work on radar” may have been creative for some but maybe not all who did it).$\,\!$ REPLY [2 votes]: The big bird Yuri Manin Manin is known for his work in algebraic geometry. He is also father of quantum computing together with Richard Feynmann.<|endoftext|> TITLE: Circular sequences continuous? QUESTION [6 upvotes]: I noticed something interesting when playing around with Mathematica. Consider the sum $$x(N)= \frac{1}{N^2} \sum_{i=1}^{N-1} \frac{1}{1-\cos(2\pi i/N)}$$ this sequence will converge to $1/6$ as $N$ tends to infinity. Now, if I look at $$y(N)= \frac{1}{N^2} \sum_{i=1}^{N-1} \frac{1}{1-\cos((2\pi i+\pi)/N)}$$ this sequence will no longer converge to $1/6$, yet we get, according to mathematica, something close to like $0.297358.$ Now, I got curious and tried to see whether the limiting expression is continuous in its argument, i.e. I wanted to start with something that for $N$ small is more like $y(N)$ and whose summands tend pointwise to the summands of $x(N)$ as $N$ tends to infinity $$z_{\alpha}(N)= \frac{1}{N^2} \sum_{i=1}^{N-1} \frac{1}{1-\cos((2\pi i+\pi/N^{\alpha})/N)}.$$ From Mathematica I seem to get that if $\alpha=1$ then $z_{1}(N)$ tends to $1/6$. The same is true if $\alpha=1/4$, for example. Then, I tried something like $\alpha=1/100$ and it was no longer clear whether it would converge. Therefore, my question is: Does $z_{\alpha}(N)$ tends to $1/6$ for all $\alpha>0$ or is there some threshold at which it stops converging? Is the convergence with some explicit rate? Please let me know if you have any questions. REPLY [3 votes]: Claim: The sequence $z_{\alpha}(N)$ tends to $1/6$ for all $\alpha>0$, and the error is $O(N^{-\min\{1,\alpha\}})$. Let's assume $\alpha \le 1$. For fixed $\alpha$ and $1 \le kk^2/2$, we deduce that $$w_{\alpha}(N)= \frac{1}{2\pi^2} \sum_{k=1}^{N/7} \Bigl(\frac{1}{k^2}+O(N^{-\alpha}/k^3)+O(N^{-2}) \Bigr) = \frac{1}{2\pi^2}\Bigl(\pi^2/6+O(N^{-\alpha})\Bigr)\,. $$ This proves (*). The same method shows that $$y(N):= \frac{1}{N^2} \sum_{k=1}^{N-1} \frac{1}{1-\cos((2\pi k+\pi)/N)}$$ converges to $$ \frac{2}{\pi^2} \Bigl(\sum_{k \ge 1} \frac{1}{(2k+1)^2}+ \sum_{k \ge 1}\frac{1}{(2k-1)^2} \Bigr)=\frac{2}{\pi^2}\Bigl(2\frac{\pi^2}{8}-1\Bigr)=1/2-2/\pi^2=0.2973576327153...$$<|endoftext|> TITLE: A variant on characteristic $p$ de Rham cohomology QUESTION [8 upvotes]: I was thinking about de Rham cohomology in characteristic $p$, and in particular the recent question about Poincare residues, and I came up with the following construction. Let $k$ be a perfect field of characteristic $p$ and let $A$ be a regular $k$-algebra. Let $\Omega^j$ be the Kahler $j$-forms, let $Z^j$ be the closed $j$-forms, $B^j$ the exact $j$-forms and $H^j = Z^j/B^j$. The inverse Cartier operator is the unique isomorphism $C^{-1} : \Omega^j \to H^j$ satisfying $$C^{-1}(\alpha \wedge \beta) = C^{-1}(\alpha) \wedge C^{-1}(\beta) \quad C^{-1}(f) = f^p \quad C^{-1}(df) = f^{p-1} df$$ for $f \in A$. (It is easy to see that there is at most one such map, a nice exercise to see that it is well defined, and not at all clear that it is an isomorphism.) The inverse operator is an isomorphism $Z^j/B^j \to \Omega^j$, which we can also consider as a surjection $Z^j \to \Omega^j$. By abuse of notation, I'll write $C$ for the surjection $Z^j \to \Omega^j$ as well. We thus have two maps $Z^j \to \Omega^j$: The surjection $C$, and the obvious inclusion. Define a differential form $\alpha \in \Omega^j$ to be forever closed if, for all $i$, we have $C^i(\alpha) \in Z^j$. Note that we must have $C^{i-1}(\alpha) \in Z^j$ for it to make sense to define $C^i(\alpha)$, so this condition spells out as "we impose that $\alpha$ is closed, and therefore $C(\alpha)$ is defined, and we impose that $C(\alpha)$ is closed, and therefore $C^2(\alpha)$ is defined, etcetera." Define a forever closed form $\alpha$ to be "eventually exact" if $C^k(\alpha)$ is $0$ for $k$ sufficiently large. Note that exact forms are eventually exact, since the exact forms are the kernel of $C$. Define the eventual cohomology, $EH^j$, to be the forever closed forms modulo the eventually exact forms. It looks like $EH^{\bullet}$ is always finite dimensional, and forms a graded ring. It does not appear that the dimension of $EH^j$ gives topological betti numbers -- it appears to give something like the multiplicity of the highest weight part of the cohomology. Is this some object people have studied before? REPLY [6 votes]: To be able to compute the iterates of the Cartier operator it is convenient to understand how $C$ interacts with the de Rham differential: Cartier isomorphism induces an isomorphism of complexes $$(\Omega^{*}_A,d_{dR})\simeq (H^{*}(\Omega^{\bullet}_A),\beta)$$ where $\beta$ is the Bockstein differential provided by the distinguished triangle $$\Omega^{\bullet}_A\to R\Gamma_{cris}(A/W_2(k))\to \Omega^{\bullet}_A$$ It shows that for a closed form $\alpha$ the image $C(\alpha)$ is closed iff the class $[\alpha]$ is annihilated by the Bockstein homomorphism which in turn is equivalent to the liftability of $\alpha$ to class in $H^i_{cris}(A/W_2(k))$. Passing to cohomology in the above isomorphism, composing it with the Cartier isomorphism and iterating this procedure $(i-1)$ times we get an isomorphism $$(\Omega^{*}_A,d_{dR})\simeq (E_i^{(1-i)*,*},\beta_i)$$ of the de Rham complex with the complex appearing on the $i$-th page of the Bockstein spectral sequence associated to the crystalline cohomology complex. These facts can be seen easily from the following description of the Cartier isomorphism: choose a lift $\tilde{A}$ of $A$ to a complete formally smooth algebra over $W(k)$ equipped with a lift $\widetilde{Fr}$ of the Frobenius endomorphism of $A$(the existence of such lift follows from the vanishing of the relevant obstruction groups which is implied by smoothness of $A$ over $k$). The Cartier operator applied to a form $\omega\in \Omega^i_A$ is then given by $C(\omega)=\overline{\frac{\widetilde{Fr}^*(\tilde{\omega})}{p^i}}$ where $\tilde{\omega}$ is any lift of $\omega$ to a form on $\tilde{A}$ and $\overline{\cdot}$ denotes the reduction. By tracing through the construction of the Bockstein differentials we get the following Lemma. For a closed form $\alpha$ the $i$-th iteration of the Cartier operator is defined and gives a closed form if and only if $[\alpha]\in H^j(\Omega_A^{\bullet})$ lifts to a class $\widetilde{[\alpha]}$ in the crystalline cohomology of $A$ over $W_{i+1}(k)$. The $(i+1)$-th iteration of the Cartier operator is zero if and only if the class $p^i\widetilde{[\alpha]}\in H^j_{cris}(A/W_{i+1}(k))$ vanishes. Combining these conditions for all $i$ we get that a form is forever closed iff its class is in the image of the map $H^j_{cris}(A/W(k))\to H^i_{dR}(A/k)$ and it is eventually exact iff the class is in the image of $H^j_{cris}(A/W(k))[p^{\infty}]\to H^i_{dR}(A/k)$. Crystalline cohomology of $A$ coincides with the cohomology of the $p$-adically completed de Rham complex of any lift of $A$ to $W(k)$ and for the purposes of computing the above invariants we can replace $H^j_{cris}(A/W(k))$ by the (non-complete) de Rham cohomology $H^j_{dR}(\widetilde{A}/W(k))$ where $\widetilde{A}$ is any lift. The quotient $H^j_{dR}(\widetilde{A})/H^j_{dR}(\widetilde{A}){[p^{\infty}]}$ is a $W(k)$-lattice in the finite-dimensional vector space $H^j_{dR}(\widetilde{A}[1/p]/W(k)[1/p])$(it is finite-dimensional e.g. by comparison with singular cohomology). It indeed seems to follow that $EH^j$ is a finite-dimensional vector space over $k$ with dimension at most the $j$-th rational Betti number of any lift of $A$.<|endoftext|> TITLE: Type theory - category theory correspondence QUESTION [8 upvotes]: As explained here, simply typed lambda calculus can be viewed as a syntactic language for category theory. My question is, can the following modification make it equally well a formal syntactic language for 2-category theory? It goes something like this: we have a universe $U$. For each $X : U$, we make $X$ into the universe of a simply typed lambda calculus by introducing the necessary type formation rules. Maps $a \rightarrow b$ for $a, b : X$ are like maps of objects, under the usual correspondence. In this setup, I would expect maps $X \rightarrow Y$ for $X, Y : U$ to correspond to functors in some way, and I would expect some kind of correspondence between type theories of this kind and $2$-categories. More generally, I expect that introducing a higher universe is like passing from $n$-categories to $n+1$-categories. For a similar idea, applied instead to homotopy type theory, we might expect to get a model for $(\infty, 1)$-categories. Has anyone pursued an approach like this? REPLY [5 votes]: Some references to some approachesthat exist in the literature have been given in Andrej's answer and Ivan's comment. A couple others are Licata-Harper's work on two-dimensional directed type theory and Licata-Riley-Shulman's work on fibrational frameworks (specifically the "mode theory" is a 2-dimensional type theory).<|endoftext|> TITLE: What is known about the cohomological dimension of algebraic number fields? QUESTION [8 upvotes]: What is the cohomological dimension of algebraic number fields like $\Bbb{Q}$, $\Bbb{Q}[i]$, $\Bbb{Q}[\sqrt{3}]$ or similar? I'm interested in computing the cohomological dimension of $\Bbb{A}^1_k$ with $k$ a finite extension of $\Bbb{Q}$. REPLY [22 votes]: By definition, an algebraic number field is a finite extension of the field of rational numbers $\Bbb Q$. An algebraic number field $K$ is called totally imaginary if it has no embeddings into $\Bbb R$. For example, the field $\Bbb Q(i)$ is totally imaginary, while $\Bbb Q$ and $\Bbb Q(\sqrt{3})$ are not. We fix an algebraic closure $\overline K$ of $K$, and we write $\Gamma_K={\rm Gal}(\overline K/K)$ for the absolute Galois group of $K$. Let $p$ be a prime number. By definition, the $p$-cohomological dimension ${\rm cd}_p(K)$ is the supremum of the degrees of nonzero cohomology over all finite $\Gamma_K$-modules $M$ whose cardinality is a power of $p$. The cohomological dimension ${\rm cd}(K)$ is the supremum of ${\rm cd}_p(K)$ over all prime numbers $p$. Theorem 1 (well-known) The cohomological dimension ${\rm cd}(K)$ of an algebraic number field $K$ is $2$ if $K$ is totally imaginary, and $\infty$ otherwise. Following a suggestion of David Loeffler, I state and prove two other theorems, from which Theorem 1 follows. Theorem 2. The 2-cohomological dimension ${\rm cd}_2(K)$ of an algebraic number field $K$ is $2$ if $K$ is totally imaginary, and $\infty$ otherwise. Proof. Assume that $K$ is not totally imaginary. Let $\Omega_{\Bbb R}$ denote the set of real embeddings of $K$; then $\Omega_{\Bbb R}$ is nonempty. We take $M=\mu_2$. For any odd natural number $n\ge 3$ we have $$ H^{n}(K,\mu_2)\cong\bigoplus_{v\in\Omega_{\Bbb R}} H^n(K_v,\mu_2)\cong \bigoplus_{v\in\Omega_{\Bbb R}} H^1(\Gamma_{\Bbb R},\mu_2) \cong\bigoplus_{v\in\Omega_{\Bbb R}}\Bbb Z/2\Bbb Z\neq 0.$$ (For the first isomorphism, see J. S. Milne, Arithmetic Duality Theorems, Theorem I.4.10(c). For the second isomorphism, see Atiyah and Wall, Cohomology of Groups, IV.8, Theorem 5, in: Cassels and Fröhlich (eds.), Algebraic Number Theory.) Thus ${\rm cd}_2(K)=\infty$, as required. Now assume that $K$ is totally imaginary. Then for any $n\ge 3$ and any finite $\Gamma_K$-module $M$ (in particular, for any $\Gamma_K$-module $M$ whose cardinality is a power of 2) we have $$H^n(K,M)\cong\bigoplus_{v\in\Omega_{\Bbb R}}H^n(K_v,M)=0,$$ because $\Omega_{\Bbb R}=\varnothing$. (For the isomorphism, again see Milne, Theorem I.4.10(c).) Thus ${\rm cd}_2(K)\le 2$. Let $\Omega_f$ denote the set of finite places of $K$. We take $M=\mu_2$. We have $$H^2(K, \mu_2)\cong\left\{(a_v)\in\bigoplus_{v\in\Omega_f}\Bbb Z/2\Bbb Z\ \mid\ \sum_{v\in \Omega_f} a_v=0\right\}, $$ and hence, $H^2(K,\mu_2)\ne 0$. Thus ${\rm cd}_2(K)=2$, as required. Theorem 3. For any odd prime $p$, the $p$-cohomological dimension ${\rm cd}_p(K)$ of any algebraic number field $K$ equals $2$. Proof. For any $n\ge 3$ and any finite $\Gamma_K$-module $M$ whose cardinality is a power of $p$, we have $$H^n(K,M)\cong\bigoplus_{v\in\Omega_{\Bbb R}}H^n(K_v,M)\cong \bigoplus_{v\in\Omega_{\Bbb R}} H^1(\Gamma_{\Bbb R},M)=0,$$ where $H^1(\Gamma_{\Bbb R},M)=0$ because the group $\Gamma_{\Bbb R}$ is of order 2 while $M$ is of odd cardinality (this is an easy exercise; see also Atiyah and Wall, IV.6, Corollary 1 of Proposition 8). Thus ${\rm cd}_p(K)\le 2$. Let $\Omega_f$ denote the set of finite places of $K$. We take $M=\mu_p$. We have $$H^2(K, \mu_p)\cong\left\{(a_v)\in\bigoplus_{v\in\Omega_f}\Bbb Z/p\Bbb Z\ \mid\ \sum_{v\in \Omega_f} a_v=0\right\}, $$ and hence, $H^2(K,\mu_p)\ne 0$. Thus ${\rm cd}_p(K)=2$, as required.<|endoftext|> TITLE: Conjugacy classes of the mapping class group QUESTION [12 upvotes]: I am not sure if this is a well known problem, but I was not able to find anything online that answered my question. Is it known how to tell whether two elements of the mapping class group of a surface are conjugate? REPLY [16 votes]: An exponential-time solution to the conjugacy problem in the mapping class group was given by Jing Tao, in: Tao, Jing(1-OK) Linearly bounded conjugator property for mapping class groups. (English summary) Geom. Funct. Anal. 23 (2013), no. 1, 415–466. Tao's main contribution is to prove that two conjugate periodic mapping classes have a conjugator of linear length; the pseudo-Anosov case had already been done by Mazur--Minsky. To actually determine conjugacy in practice, I think the state of the art in this area is the recent papers of Bell--Webb, who show how to compute distance in the curve graph and Nielsen--Thurston type in practice. Bell has even implemented some of their algorithms (see here). See also the parallel results of Birman--Margalit--Menasco. Added (4 March 2020): Mark Bell pointed out to me that his software can indeed effectively solve the conjugacy problem in specific examples. Flipper decides conjugacy of pseudo-Anosovs, while Curver decides conjugacy of periodic automorphisms.<|endoftext|> TITLE: Freedom problem in hyperbolic groups QUESTION [8 upvotes]: I will start with a general algorithmic question: Question. (A faithfulness decision problem.) Suppose that $G, H$ are finitely-presented groups with decidable word problem. Is injectivity decidable for homomorphisms $f: G\to H$ (defined by the images of generators of $G$)? [A very minor side issue: I am not even sure if this decision problem has an established name. For some reason, Max Dehn did not consider it.] The non-injectivity problem is trivially semidecidable. More specifically, I am interested in decidability of faithfulness for homomorphisms from (finitely generated, nonabelian) free groups to (nonelementary) hyperbolic groups $H$. The decision problem then becomes: Freedom problem. Do given elements $g_1,...,g_n\in H$ generate a rank $n\ge 2$ free subgroup? Things that I know: If $H$ is locally quasiconvex (or, more generally, all finite rank free subgroups of $H$ are undistorted), then the freedom problem is decidable. The same applies to the injectivity problem for homomorphisms from general finitely-presented groups $G\to H$ to locally quasiconvex hyperbolic groups. The injectivity problem is decidable if $H$ is the fundamental group of a compact hyperbolic 3-manifold. It is natural to expect that the decidability of the freedom problem in hyperbolic groups $H$ is closely related to distortion of free subgroups in $H$. One can conjecture that if every f.g. free subgroup in $H$ is recursively distorted then the freedom problem is decidable. One can also conjecture that hardness of the freedom problem correlates to the degree of distortion of free subgroups. (For instance, for "hyperbolic hydra groups" of Brady-Dison-Riley, the distortions of some free subgroups are given by Ackerman functions.) I do not know how to approach any of these conjectures. So, my question is if anything else is known about these problems. One more thing: There is a "dual" problem about surjectivity of group homomorphisms. I am not asking about this one. (It is more related to the subgroup membership problem and more is known about it.) REPLY [4 votes]: As regards the first question (with $H$ not hyperbolic), it has a negative answer even with $G$ free: This question has an answer saying that there exists a f.p. $H$ group with solvable word problem in which injectivity of homomorphisms $\mathbf{Z}\to H$ is undecidable. (Hence injectivity of homomorphisms $F_n\to H\ast F_{n-1}$ is undecidable too.) Nevertheless for $H$ hyperbolic, it is decidable whether an element has infinite order (because the conjugacy problem is solvable and there are finitely many finite order conjugacy classes, so it is enough to test conjugacy with each representative).<|endoftext|> TITLE: How to obtain the rational solution of a linear system efficiently? QUESTION [8 upvotes]: Suppose that I have a linear system $AX=b$ with $A\in\mathbb{Z}^{n\times m}$ and $b\in\mathbb{Z}^{n}$. Assume that $AX=b$ has exactly one rational solution. Then how can I obtain this solution efficiently? I know LU decomposition is a choice for this purpose. But it seems that the denominators may expand in the process of computing a LU decomposition, which I hope to avoid if the rational solution has small denominators. If I already have a numerical solution of $AX=b$, does it help? REPLY [4 votes]: This is standard and can be done using $p$-adic lifting, see for example the following article: J.D. Dixon, Exact solution of linear equations using $P$-adic expansions, Numer. Math. 40 (1982) pp. 137–141, doi:10.1007/BF01459082. This also exploits that there is a unique solution.<|endoftext|> TITLE: Division of space by hyper-planes QUESTION [6 upvotes]: It is a well known and lovely result that the maximum number of regions that $\mathbb R^{k}$ (with $k$ positive) can be divided into by $n$ hyperplanes is given by $$1+n+\binom{n}{2}+\cdots+\binom{n}{k}. $$ and occurs when they are in general position. It is clear that the minimum with distinct hyperplanes is $n+1$ (when they are parallel ) and the next smallest is $2n$ when all but one are parallel and also when each pair has the same (nonempty ) intersection. For which $n,k,m$ is it possible to divide $\mathbb R^{k}$ into exactly $m$ regions using $n$ distinct hyperplanes? Fir purposes of induction it might be better to drop the requirements that the hyperplanes be distinct. REPLY [2 votes]: The motivation of the question was the observation that the least number of lines in the plane that give five regions is $n=4$ parallel lines. Surprisingly, as long as $m \gt 5$ and $m \leq \binom{n}2+n+1$, the maximum number of regions possible using $n$ lines, there is a division of the plane into $m$ regions using no more than $n$ lines. I suspected this based on this easy observation (and some computation): If the plane is divides by $t$ parallel classes of lines with sizes $a_1,\cdots,a_t$, no three lines concurrent, then the plane is split into $\frac{n^2-s}2+n+1$ regions where $n=\sum a_i$ and $s=\sum a_i^2.$ The nice construction by Manturuv does capture some small numbers of regions which this does not, but those attained are enough to establish the result claimed above up to $n=50$ except for $m=5$ and $m=17.$ Naively running over all partitions of $n$ becomes time consuming. So for $k=2$ then set of attainable number of regions is the union of intervals. Once $n-r \leq \binom{r}2$ we have an interval starting at $$(n-r)(r+1)+\binom{r}{2}-(n-r)=nr-\binom{r+1}2$$ with the previous interval ending at $$(n-r+1)r+\binom{r-1}2=nr-\binom{r+1}2+1.$$ Since this happens near $r=\sqrt{2n},$ we can certainly attain any number of regions $n\sqrt{2n} \leq m \leq n\frac{n+1}2+1$ using exactly $n$ lines.<|endoftext|> TITLE: The $r$-dimensional volume of the Minkowski sum of $n$ ($n\geq r$) line sets QUESTION [11 upvotes]: Let $n$ line sets be $\mathcal{S}_i=\{a\mathbf{h}_i:0 \le a \le 1\}$, for $1 \le i \le n$, where $\{\mathbf{h}_1,\cdots,\mathbf{h}_n\}$ is a vector group of rank $r$ in the $r$-dimensional Euclidean space. Define the Minkowski sum of two sets as $\mathcal{S}_1+\mathcal{S}_2=\{\mathbf{s}_1+\mathbf{s}_2:\mathbf{s}_1\in\mathcal{S}_1,~\mathbf{s}_2\in\mathcal{S}_2\}$. How to compute the $r$-dimensional volume of $\mathcal{S}_1+\cdots+\mathcal{S}_n$? REPLY [13 votes]: Let $M$ be the matrix whose rows are the vectors $\boldsymbol{h_i}$. Then the $r$-dimensional volume of $\mathcal{S}=\mathcal{S}_1+\cdots+\mathcal{S}_n$ is equal to the sum of the absolute values of the $r\times r$ minors of $M$. I don't know who originally showed this, but one can show that $\mathcal{S}$ can be tiled with $r$-dimensional parallelotopes whose volumes are the $r\times r$ minors. This follows for instance from the proof of Lemma 2.1 here. REPLY [13 votes]: The keyword you are looking for is "zonotope", which is defined to be the Minkowski sum of line segments. An early reference for zonotope is: P. McMullen, “On zonotopes”, Transactions of the American Mathematical Society, Vol. 159, 1971. Following your notation, the $r$-dimensional volume of the zonotope $\mathcal{S}_{1} + ... + \mathcal{S}_{n}$ is equal to $$\displaystyle\sum_{1\leq i_{1} < i_{2} ... < i_{r}\leq n} \big\vert{\rm{det}}\left(\mathbf{h}_{i_{1}},\mathbf{h}_{i_{2}},...,\mathbf{h}_{i_{r}}\right)\big\vert.$$ For reference, see eqn. (57) in "Combinatorial Properties of Associated Zonotopes" by G.C. Shephard, Canadian Journal of Mathematics, 1974. In that paper, there is an extra factor $2^{r}$ in front of the above expression since the line segments there are defined by $\{a\mathbf{h}_{i} : -1\leq a \leq 1\}$ instead of the OP's convention: $0\leq a \leq 1$. In the very end of this paper, Shephard credits McMullen for drawing attention to this formula. The same formula also appears as Exercise 7.19 in G.M. Ziegler, Lectures on Polytopes, Vol. 152, Springer, 2012; screenshot below:<|endoftext|> TITLE: Are primes linearly separable? QUESTION [16 upvotes]: Let $X_1,\cdots,X_n$ be finite subsets of some set $Z$. Then the symmetric difference metric space: $$d(X_i,X_j) = \sqrt{ |X_i|+|X_j|-2|X_i\cap X_j|}$$ can be embedded in Euclidean space. The value $|X_i \cap X_j| = \langle \phi(X_i),\phi(X_j) \rangle$ is equal to the dot product of the embeddings $\phi(X_i),\phi(X_j)$. Taking $X_i = \{ i/k | 1 \le k \le i\}$ for a natural number $i \ge 2$, then one observes that $X_i \cap X_j = X_{\gcd(i,j)}$ hence $\langle \phi(X_i),\phi(X_j) \rangle =|X_i \cap X_j| = |X_{\gcd(i,j)}|=\gcd(i,j)$ and the gcd-matrix $\gcd(i,j)$ is a matrix which consists of pairwise dot-products, hence a Gram-Matrix, which is positive definite. Thus we might want to "define" a kernel $K(i,j) := \gcd(i,j)$ and apply the kernel trick in Support Vector Machines to decide if the "training set" $2 \le i \le n, (i,\pm 1),$ where $+1$ if $i$ is prime, $-1$ otherwise, is linear separable or not. The "kernel trick" in this context is just a method to do the svm computation without explicitly constructing the embedding $\phi$, to determine the linear separability of the training set. For future values, this might or not be useless. In fact the "kernel trick" is not needed technically since we can use the standard dot product after embedding the numbers (but for this one would have to explicitly construct the embedding $\phi$). For the definition of the embedding $\phi$, see: Is this function embeddable in Euclidean space? I have implemented this idea in scikit learn and it seems that primes are in this sense linear separable. However, I do not understand why this is so. Is there any theoretical explanation for this? Thanks for your help. import math import math,csv import numpy as np import pandas as pd import csv,math,collections import math,csv, random import numpy as np import re from collections import Counter from sklearn import svm, metrics from functools import partial from sklearn.model_selection import cross_val_score from sklearn.decomposition import KernelPCA from sklearn.neighbors import NearestNeighbors from sklearn.metrics import confusion_matrix import random,nltk,sys from fractions import gcd def is_prime(n): if int(math.sqrt(n))**2 == n: return(False) for i in range(2,int(math.ceil(math.sqrt(n)))): if n%i==0: return(False) return(True) X = range(2,50) y = [ is_prime(x)*1 for x in X] def d1(a,b): return np.sqrt(a+b-2*gcd(a,b)) def d2(a,b): return np.sqrt(np.abs(a-b)) dd = d1 def simLeinster(x,y,dist=dd): return np.exp(-dist(x,y)) def psdKern(x,y,dist=dd): """This is a psd-Kernel if dist can be embedded in Euclidean Space """ t = 1.0/2.0*( dist(x,1)**2+dist(y,1)**2-dist(x,y)**2) return( t ) def proxy_kernel(X,Y,K): gram_matrix = np.zeros((X.shape[0], Y.shape[0]),dtype=np.float64) for i, x in enumerate(X): for j, y in enumerate(Y): gram_matrix[i, j] = K(x[0], y[0]) return gram_matrix XX_train = np.array(X).reshape(-1,1) Y = np.array(y) kernel = partial(proxy_kernel, K=gcd) Gram = (kernel(XX_train,XX_train)) clf = svm.SVC(kernel=kernel) print(Gram.shape) import scipy.linalg print(Gram) print(scipy.linalg.eigvalsh(Gram)) scores = cross_val_score(clf, XX_train, Y, cv=5,scoring="accuracy") print(scores) #The mean score and the 95% confidence interval of the score estimate are hence given by: print("Score: %0.2f (+/- %0.2f)" % (scores.mean(), scores.std() * 2)) clf = clf.fit(XX_train,Y) y_pred = clf.predict(XX_train) print( sum(Y==y_pred)/len(Y) ) print(confusion_matrix(Y,y_pred)) print(y_pred) Related: Is this function embeddable in Euclidean space? Trigonometry / Euclidean Geometry for natural numbers? Edit: The linear separability of primes in this case might be stated as this conjecture: Let $x_i=i+1, 1\le i \le n$, $y_i = +1,$ if $x_i$ is prime, $-1$ otherwise. Then (?) there exists $b,c_i \in \mathbb{R}, 1 \le i \le n$ such that for all $j=1,\ldots,n$: $$y_j = \text{ sgn }\left(\left[\sum_{i=1}^n c_iy_i\gcd( x_i, x_j)\right] - b\right)$$ After waiting 2 days also posted on MSE: https://math.stackexchange.com/questions/3497098/can-gcd-separate-primes-from-composite-numbers REPLY [12 votes]: The purely number-theoretic problem can be stated as: For every $n$, are there reals $b$ and $c_i$ such that $$j\text{ is prime iff }\sum_{i=2}^n c_i \gcd(i,j)>b\ ?$$ This is the same as the version above, after removing the $x$'s and $y$'s, shifting the indices by 1, and incorporating a factor of $y_i$ into the $c_i$. For example, for $n=6$, this is asking for reals $b$ and $c_i$ such that $$ \begin{pmatrix} 2 & 1 & 2 & 1 & 2 \\ 1 & 3 & 1 & 1 & 3 \\ 2 & 1 & 4 & 1 & 2 \\ 1 & 1 & 1 & 5 & 1 \\ 2 & 3 & 2 & 1 & 6 \\ \end{pmatrix} \begin{pmatrix} c_2 \\ c_3 \\ c_4 \\ c_5 \\ c_6 \end{pmatrix} \text{ is of the form } \begin{pmatrix} >b \\ >b \\ b \\ TITLE: Drinfeld center of a braided category QUESTION [6 upvotes]: Suppose I have a braided monoidal category $\mathcal{C}$. I specifically am interested in the case where $\mathcal{C}$ is the category of finite-dimensional modules of a quantum group, say $\mathcal{U}_q(\mathfrak{sl}_2)$ (or a variant of it.) The braiding $c_{-,-}$ embeds $\mathcal{C}$ into its Drinfeld center $\mathcal{Z}(\mathcal{C})$ via $$ V \mapsto (V, c_{V,-} ). $$ Does this give the entire Drinfeld center? If not, is it easy to see what parts of $\mathcal{Z}(\mathcal{C})$ it misses, at least in this case? Is there a reference that discusses this? I think this should be related to a theorem of the form $D(D(H)) \cong D(H)$ (where $D(H)$ is the Drinfeld double of a Hopf algebra) but I don't recall a reference for that result either. REPLY [9 votes]: No, the functor $\mathcal C \to \mathcal Z(\mathcal C)$ is not essentially surjective in general. For example, in the case you have in mind, $\mathcal C = Rep_q(G)$ (say $G$ a semisimple algebraic group), the Drinfeld center can be identified with the category $HC_q := \mathcal O^{RE}_q(G)-mod_{Rep_q(G)}$ of modules for the so-called reflection equation algebra $\mathcal O_q^{RE}(G)$ internal to $Rep_q(G)$. The image of $Rep_q(G)$ in thus identified with those modules on which the REA acts trivially (i.e. via the augmentation $\varepsilon: \mathcal O_q^{RE}(G) \to \mathbb C$). Note that this holds even when $q=1$ (and so $Rep(G)$ is symmetric monoidal). Then $HC_{q=1}$ is the same thing as $Coh(G/G)$, the category of $G$-equivariant coherent sheaves on $G$. The image of $Rep(G)$ consists of coherent sheaves supported on the identity element of $G$. This example also makes sense for a finite group. Note also that the Drinfeld center may be non-symmetrically braided even when $\mathcal C$ is symmetric.<|endoftext|> TITLE: Aronszajn Trees when AC fails QUESTION [7 upvotes]: This question may be easy and indicative of my ignorance about the failure of the axiom of choice. If so, I apologize. Below assume $\mathsf{DC}$ but not $\mathsf{AC}$. Suppose we have a partial order $A = (A, \leq)$ satisfying the following: $A$ is uncountable in the sense that there is no surjection of $\omega$ onto it. $\leq$ is a transitive partial order so that for any $a \in A$ the set of $b \leq a$ is linearly ordered $\leq$ is well founded: there is no infinite $ \leq$-descending sequence There is no uncountable $\leq$-linearly ordered subset of $A$ For each $a \in A$, the set of $b\in A$ for which there is an order isomorphism between $\{c \; | \; c < b\}$ and $\{ c\; | \; c < a\}$ is countable. Of course if choice holds then we call such an object an Aronszajn tree. What I want to know is whether, under simply $\mathsf{DC}$, this is still acceptable. Specifically, do the above conditions guarantee the existence of a rank function into $\omega_1$ so that we can make sense of "the $\alpha^{\rm th}$ level of $T$" ? More to the point is the following. It's a well known theorem of Solovay that under $\mathsf{AD}$ $\omega_1$ is measurable and hence there are no Aronszajn trees under $\mathsf{AD}$. Does this suffice to rule out the existence of the partial order $A$ described above? REPLY [7 votes]: Is it acceptable? Sure. In some sense, it is an Aronszajn tree. The condition of being well-founded, which in the presence of $\sf DC$ is the same as saying there are no decreasing sequences, is equivalent to having a rank function. So much is true in $\sf ZF$. So you can make sense of this tree having height $\omega_1$. Even more, if you just combine 2 and 3 by saying that the predecessors of each node is a well-ordered set, then you don't even need $\sf DC$ to make sense of this $A$ being a tree of height $\omega_1$. But here comes the thing that will haunt your nightmares.1 Not everything is well-ordered in the universe. Start with a model of $\sf AD$ which has enough structure (e.g. $L(\Bbb R)$ inside some model of $\sf ZFC$). Now force (or rather, take a symmetric extension) over this model to add some sets well above $\Theta$, so you do not add any set of rank $<\Theta$, so that the new sets form a copy of an Aronszajn tree in some sense. If you do it well, every set of ordinals will be given by some "small" part of this tree and will not interfere with $\sf AD$, which still holds as you did not add reals or sets of reals. And $\sf DC$ will hold because if you chose your weapons appropriately the forcing is at least $\sigma$-closed. If you want to think about this in a more "hands-on" kind of way, force an Aronszajn tree, use it to force your new sets up beyond the clouds of $\Theta$, and then look at $L(\Bbb R,T^\omega)$ where $T$ is this copy of the tree. And again, if you were wise enough, you will get the promised model. The key point here is that: Not everything is a set of ordinals. Not everything is below $\Theta$. 1. Or not, I'm not a witch.<|endoftext|> TITLE: Fractional powers of an operator QUESTION [6 upvotes]: What is the large class of operators for which one can define fractional powers? For example, we can consider an operator $A: D(A) \subset X \rightarrow X$, generator of an analytic semigroup on a Banach space $X$. Can we define the powers $(-A)^\alpha$ for $\alpha>0$ without additional assumptions? I found some references with some restrictions on the spectrum of $A$ or on the associated semigroup. I'm wondering if there is a general definition without further assumption. Any reference would be helpful. REPLY [7 votes]: This question was considered by Functional Analysis specialists around 1960 or earlier. In the case of Banach Algebras $C(X)$ (for Hausdorff compact $X$) this gets reduced often to studying the auto-homeomorphisms of $X$, and it is extra interesting when the compact space is nice. In this context, in 1961, I have rediscovered that orientation preserving homeomorphisms of $\,X:=[0;1]\,$ admit square root (and quite a bit more but I didn't know at the time that the group of homeomorphisms was described already in full by someone else, sometimes earlier). This means that there are root squares (and similar) of the respective operators. I also constructed an orientation preserving homeomorphism of $\,\Bbb S^1\,$ which does not admit a square root hence the respective operator didn't either. The later result was presented in a monograph by Marek Kuczma, on functional equations, (which made me feel good, especially that I was a student at the time).<|endoftext|> TITLE: Can the dual of a finitely-accessible category be accessible? QUESTION [8 upvotes]: What is an example of an accessible category $\mathcal C$ which is not essentially small, such that $\mathcal C^{op}$ is finitely-accessible? More generally, what is an example of an accessible category $\mathcal C$ (not essentially small) such that one of the following related conditions holds? $\mathcal C^{op}$ continuous (i.e. $\mathcal C$ has cofiltered limits and the colimit functor $Ind(\mathcal C^{op}) \to \mathcal C^{op}$ has a left adjoint); $\mathcal C^{op}$ is precontinuous (i.e. $\mathcal C$ has colimits and cofiltered limits and the colimit functor $Ind(\mathcal C^{op}) \to \mathcal C^{op}$ preserves limits); $\mathcal C$ has finite colimits and cofiltered limits, and they commute; $\mathcal C^{op}$ is finitely accessible. The closest thing to an example I can think of is the category $Hilb$ of Hilbert spaces and contractive maps, which is a self-dual $\aleph_1$-accessible category. But I don't believe that finite limits commute with filtered colimits in $Hilb$. REPLY [3 votes]: Any locally presentable category where epimorphisms are stable under $\lambda$-codirected limits is equivalent to a complete lattice (see http://www.tac.mta.ca/tac/volumes/33/10/33-10.pdf, 3.10).<|endoftext|> TITLE: Why C*-algebras is not as popular as other areas of pure mathematics? QUESTION [7 upvotes]: I am applying for graduate school in pure mathematics and I recently got very interested in C*-algebra. I am definitely wrong but I get the feeling that C*-algebras is not as popular as other areas of pure mathematics like number theory, analysis, algebraic geometry, etc. It also seems that most top ranked universities like MIT, Harvard, Stanford, Princeton, etc do not have any active research group in C*-algebras. If my observations is right, then what is the reason? Is it because C*-algebra is harder than other areas of pure mathematics or is it because it is still a young area of pure mathematics? Given that I am interested in C*-algebras and most top ranked universities are not active in this area, where can I apply? Also, will doing graduate work in C*-algebras instead other more popular areas of pure mathematics have a negative effect on my academic career? REPLY [23 votes]: One way to tell how active a field is is by looking at what's appearing on the arXiv in that area. I think that will show you that operator algebra is a robust subject with a lot of activity. In the comments, MaoWao points out that UC Berkeley and UCLA have very strong operator algebra groups, and Ulrich Pennig mentions groups in Münster, Göttingen, Erlangen, and Glasgow as places with substantial groups. Copenhagen is another good example. On the other hand, the OP's observation that most top schools don't have an operator algebra group is quite correct. I would think that simply has to do with the size of the field --- there aren't enough C*-algebraists to populate that many departments. The two Fields Medals in the subject (to Connes and Jones) show that people in other areas do respect the field, I think. The question is partly about career advice. All I can do there is report my impression that C*-algebraists don't seem to have more trouble finding employment than mathematicians of equal ability in other areas. You ask which schools you should apply to --- in the comments MaoWao gave this site, which I was not aware of, which lists operator algebraists worldwide. It looks pretty complete to me. However, my last comment is that as a prospective graduate student, you are at a very early stage to be settling on a specialty. Not saying you shouldn't, but I think most of us would recommend putting your main effort into getting a broad education during the first two years of grad school. Related MathOverflow questions: What are the applications of operator algebras to other areas? States in C*-algebras and their origin in physics? Quantum functional analysis applications of C*-algebras in the field of PDEs<|endoftext|> TITLE: Help with definition of Liouville measure QUESTION [7 upvotes]: $\require{AMScd}$For a Riemannian manifold $M$, I have read authors talking about a 'Liouville measure' on the unit tangent bundle $\operatorname{T}^1(M)$ and then proceed to claim/prove that it is invariant under the geodesic flow. See for example Mautner's $1957$ paper 'Geodesic Flows on Symmetric Riemann Spaces.' This measure is loosely defined as 'the measure on $\operatorname{T}^1(M)$ locally defined by $\omega \wedge \theta$ where $\omega$ is the volume form on $M$ associated to the metric and $\theta$ is an invariant volume form on the sphere' (again, see equation ($17$) of Mautner's paper). I would like a precise definition/construction of such a measure on a general symmetric space. I have seen the example of the upper half-plane where the unit tangent bundle can be identified with a group and one uses the Haar measure. Setup: Assume $(G,\mathfrak{g})$ is a non-compact, real, semi-simple, Lie group with finite center. Fix a Cartan involution $\sigma: G \to G, d\sigma: \mathfrak{g} \to \mathfrak{g}$ which then induces the decomposition $\mathfrak{g}= \mathfrak{k}\oplus\mathfrak{p}$ into the $\pm 1$ eigenspaces respectively. Moreover, if $K = G_{\sigma}$ (the fixed point set of $\sigma$), then $\operatorname{Lie}(K)= \mathfrak{k}$ and $K$ is compact since we assumed $G$ to have finite center. One can also check that $\mathfrak{k}$ and $\mathfrak{p}$ are invariant under $\operatorname{Ad}(K)$. By giving $\mathfrak{p}$ an $\operatorname{Ad}(K)$-invariant inner product, one can induce a $G$-invariant Riemannian metric on the tangent bundle of $G/K$. Additionally, $G/K$ has Riemannian geodesics of form $t\mapsto g\exp(tX)K$, with $X\in \mathfrak{p}$. Perhaps the example to keep in mind is $\operatorname{SL}_n(\mathbb{R}) \simeq \exp(\mathfrak{p}) \times \operatorname{SO}(n)$ where $\mathfrak{p}$ are symmetric, trace-zero matrices. What I want: I would like to define a non-vanishing top form (Liouville measure) on $\operatorname{T}^1(G/K)$ which is 'locally of the type $\omega \wedge \theta$', where $\omega$ is the Riemannian volume form and $\theta$ is a non-vanishing top form on the sphere. Note that $l_x$, left multiplication by $x\in G$, is an isometry on $G/K$. And, of course, I would like the Liouville form to be invariant under the maps $dl_x$ so as to pass to the interesting class spaces $\operatorname{T}^1(\Gamma\backslash(G/K)) \simeq \Gamma\backslash \operatorname{T}^1(G/K)$, where $\Gamma$ is taken to be a freely acting lattice. Lastly, I would like to know if there is some sort of uniqueness statement that can be formulated. An attempt: A naive attempt would be to say that $\operatorname{T}^1(G/K) \simeq G/K \times S^d$ is a trivial sphere-bundle since $G/K\simeq \mathfrak{p}$ is contractible. But I don't know how the action of $G$ would translate to $G/K \times S^d$. Another attempt would be to use the commutative diagram $$ \begin{CD} G\times \mathfrak{p}_1 @>>> G \times_{K} \mathfrak{p}_1 \\ @V{\pi_1}VV @VV{\overline{q\circ\pi_1}}V \\ G @>{q}>> G/K \end{CD}$$ Here $\mathfrak{p}_1$ are the vectors in $\mathfrak{p}$ of norm $1$, $K$ acts on $G\times \mathfrak{p}_1$ on the right by $$(g,Y)\cdot k := (gk,\operatorname{Ad}(k^{-1})Y),$$ and $G \times_{K} \mathfrak{p}_1$ is shorthand for the quotient by this action. Note that $G \times_{K} \mathfrak{p}_1$ can be identified with $\operatorname{T}^1(G/K)$ by using the map $$(g,Y) \mapsto (g\exp(tY)K)'_{t=0}.$$ Under this identification, the action of $dl_x: \operatorname{T}^1(G/K) \to \operatorname{T}^1(G/K)$ for $x\in G$ is intertwined with the action $x\cdot(g,Y)K := (xg,Y)K$ on the first coordinate. Thus it perhaps makes sense to try and construct a form $\alpha$ (of appropriate dimension) on $G\times \mathfrak{p}_1$ which is invariant (under both the left action of $G$ on the first component and the right action of $K$ specified above) and is also $K$-horizontal, that is to say that the interior product $\iota_X \alpha =0$ whenever $X$ is a vector field induced by the action of $K$. I guess such a form would induce a top form on $G\times_K \mathfrak{p}_1$ with the required properties? Any help/references would be appreciated. REPLY [5 votes]: The construction doesn’t really simplify on symmetric spaces. On $TM\cong T^*M$ (using the metric) consider the canonical 1-form $\alpha=“\langle p,dq\rangle”$ and symplectic form $d\alpha$ and hamiltonian vector field $\xi$ of $H=\frac12\|p\|^2$: $\mathrm i_\xi d\alpha=-dH$. Then $\alpha$ and $\xi$ restrict to a contact structure and its Reeb vector field on the level $T^1M$ of $H$: $$ \mathrm i_\xi d\alpha = 0, \qquad\quad \mathrm i_\xi \alpha=1. $$ Moreover the geodesic flow is the Reeb flow. (For this Besse cites Weinstein (1974) who cites Berger (1965) who doesn’t cite Reeb (1950).) Now $\mathrm L_\xi\alpha=\mathrm i_\xi d\alpha + d\mathrm i_\xi \alpha=0$, so that flow preserves $\alpha$ and hence the volume form $$ \alpha\wedge(d\alpha)^{\dim M-1} $$ of which Besse also gives a base $\times$ fiber description. Finally, any diffeo $g$ of $M$ lifts to a diffeo of $T^*M$ characterized by $\langle g(p),Dg(q)(\delta q)\rangle$ $=$ $\langle p,\delta q\rangle$, which “by construction” preserves $\alpha$. When $g$ is an isometry, it also preserves $T^1M$ $\subset$ $TM$ $\cong$ $T^*M$ and hence everything in sight. Added: This “Lie” view of geodesics as produced by a contact flow (1896, pp. 96-102) works directly on $\Gamma{\small\backslash} G/K$; it may not have been that of Mautner, Gelfand-Fomin, or Hopf — they seem closer to the (of course equivalent) idea of putting a canonical (“Sasaki”) metric on $TM$ and $T^1M$ and using the resulting volume form, as in e.g. Paternain (1999, 1.17) or Berger (2003, pp. 195, 359, 472).<|endoftext|> TITLE: Homotopicity of two certain sections of frame bundle of $GL(n,\mathbb{R})$ QUESTION [7 upvotes]: Edit: According to comment of Prof. GoodWillie we revise the question. Put $M=GL(n,\mathbb{R})$. We identify $M_n(\mathbb{R})$ with $\mathbb{R}^{n^2}$: The identification is based on the lexicographic order on the index $i,j$ in $(a_{ij})$. For example $$ \begin{pmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}$$ is identified with $$(a_{11}, a_{12},a_{21}, a_{22})$$ So $M$ being an open subset of $\mathbb{R}^{n^2}$ has trivial tangent bundle and there is an obvious description for a tangent vector at a point $A\in M$. The mapping $A\mapsto A\otimes A$ defines a section of fram bundle of $M$.Because each row of $A\otimes A$ is counted as a $n×n$ matrix via the above identification. So $A\otimes A$ actualy determines $n^2$ independent vectors in the tangent space to $M$ at $A$. In a similar manner $A\otimes A^{tr}$ is another section of the frame bundle of manifold $M$ where $tr$ is transpose operator. Are the above two sections $A\otimes A$ and $A\otimes A^{tr}$ homotopic sections of frame bundle of $M$? REPLY [3 votes]: The two sections are not homotopic for large $n$. Here's why. First, as you note, the tangent bundle of $GL_{n}$ is trivial, which means its frame bundle is as well. Next, as noted in the comments, we might as well replace $GL_{n}$ with $O(n)$, which is nicer because it is easier to typeset, and we can swap the transpose operator for the inverse operator. The two sections you define are therefore maps $s,t:O(n)\rightarrow O(n)\times O(n^{2})$ given by $A\mapsto (A, A\otimes A)$ and $A\mapsto (A, A\otimes A^{-1})$. So to check homotopicity it's enough to check homotopicity of the maps $O(n)\rightarrow O(n^{2})$ that you get from the sections after projecting on to the second factor. Let's check what the maps do on $\pi_{3}$. First note that $\pi_{3}O(n)=\mathbb{Z}$ for all $n>5$. Next, let's factor our maps $s=S\circ\Delta$, $t=T\circ\Delta$ where $\Delta:O(n)\rightarrow O(n)\times O(n)$ is the diagonal, $S(A,B)=A\otimes B$, and $T(A,B)=A\otimes B^{-1}$. As Tom Goodwillie noted in the comments, the inverse map acts as $-1$ on all homotopy groups. So the maps induced by $S$ and $T$ are two $1\times2$ integer matrices, and the second column of $T$ is the negative of the second column of $S$; i.e. $S=[a\ \ b]$ and $T=[a\ \ -b]$. If $s=t$ (on homotopy) then $a+b=a-b$ so $S=[a\ \ 0]$. But that clearly can't be, for symmetry reasons.<|endoftext|> TITLE: Can local flip moves connect dimer matchings on 'quadrangulated' planar bipartite graphs? (perfect matching reconfiguration problem) QUESTION [5 upvotes]: I'm interested in the structure of dimer matchings on planar graphs with a bipartite structure. In particular, I'm interested in whether any two perfect matchings can be connected, i.e. transformed into each other, by 'flip moves', which replace a dimer pair on opposite edges of a 4-cycle with the other possible pair. This problem was dubbed the "Perfect Matching Reconfiguration Problem" by the paper by Bonamy et al (where the picture is from). The linked paper studies this problem for many different types of graphs, but not for the general class of planar bipartite graphs. Similar constructions for 'quadriculated graphs' in the plane that consist of squares meeting in an array in the plane have been in the affirmative by Thurston, in "Conway's Tiling Groups", and given necessary and sufficient conditions for other flat 2D surfaces like the torus, Mobius band Klein bottle in work by Saldanha et al, in "Spaces of domino tilings". And, similar statements showed some structure for the case of a 3D grid in work by Freire, Saldanha, et al, where they consider an additional 'trit' move. My question is, for the case of planar graphs, when can we say that the space of tilings is connected under face flip moves? This is obviously false for the cycle graph of even length, since there are no 4-cycles at all. But, the case I'm specifically interested in is for the case, similar to the picture and references above, where the planar bipartite graph is generated by 2-cells which are 4-cycles. This is more general than the case of 'quadriculated' regions answered previously, but I can't seem to find if this is shown by anyone. EDIT: I want to consider the case where the cellulated manifold is simply connected, since there are simple counterexamples if this condition isn't met, as indicated in Brendan's answer. EDIT: Brendan has shown a counterexample for certain configurations of quadrangulations. EDIT: See Lev's proof below, which verifies the conjecture for a large class of quadrangulations, where each vertex is attached to at least 4 facets. REPLY [2 votes]: I think I've come up with a condition for flip-connectivity for which Lev's elegant answer is a special case and that avoid Brendan's counterexamples. We'll also see that it supports Lev's intuition that a limited number of vertices with degree<4 are okay in some cases. The condition is the non-existence of certain configurations of cycles on the graph's interior and the faces surrounding them, which we'll call "daisy chains". A daisy chain $\mathcal{D}$ is a loop in the graph G (with possibly repeating vertices) that satisfies two conditions. The loop of $\mathcal{D}$ has an interior which is a set of polygons that share a vertex or are connected by line segments that the loop traverses along. The loop must always be on the boundary of these polygons/line segments. Every two edges adjacent on the loop whose vertex colors in sequence are "white $\rightarrow$ black $\rightarrow$ white" are part of a facet on the exterior of $\mathcal{D}$ (we could just as easily switch "black $\leftrightarrow$ white" in this part). We'll refer to this as the "outer facet condition" An example of a daisy chain looks like Here, the loop of vertices is colored red, the polygons and connecting lines are drawn in blue, the facets on the outside of the loop are drawn in yellow, and the dashed blue lines indicate either more polygons or a closure of of the polygon, all for which the loop goes around. Note that we may allow the outer facets to share vertices or edges. Given this definition, our main claim is Main Claim: Let $G$ be a bipartite quadrangulated graph. Then $G$ is flip-connected if $G$ has no daisy chains where all vertices in the loop are internal. First before proving the claim, we'll show a stronger version of Lev's proposition, i.e. that flip-connectedness is true if every internal vertex has degree$\ge$4 holds if we assume this claim. Proof of Lev's proposition assuming the main claim Assuming the main claim, we'll show a stronger claim than Lev's. Namely if for every loop of internal vertices, the average degree of the vertex set $V_{in} := \{v | v \in \text{interior of loop OR } v \text{ is black and on the boundary of the loop} \}$ is at least 4, then no daisy chain of internal vertices can exist. This will be analogous to Brendan's observation in the comment section that the average degree of all vertices in a planar quadrangulated graph must be less than 4. But we'll need to write it out in more detail for this specfic case. Define the set of white vertices contained on $\mathcal{D}'s$ boundary loop as $V^{w,\partial}$. Let $|V|, E, F$ be the number of vertices, edges, faces of subgraph $S \subset G$ completely contained in the loop and inside the polygon. Then we'll have Euler's formula saying that $|V|-E+F=1$. First, we'll have that $|V| = |V_{in}| + |V^{w,\partial}|$. To get expressions for $E$, the outer facet condition for the daisy chain is critical, since it ensures that the every edge of any vertex in $V_{in}$ is contained in the subgraph, i.e. that the degree of any $v \in V_{in}$ is the same when considered as a part of either the subgraph $S$ or the full graph $G$. We denote $\bar{d_{in}}$ as the average degree of vertices in $V_{in}$. And, we let $\bar{d_{in}}$ We'll note that there exists a positive integer $A \ge 0$ such that $\bar{d_{in}} |V_{in}| = 2E - 2 |V^{w,\partial}| - A$. This is because $\bar{d_{in}} |V_{in}|$ double counts every edge between vertices in $V_{in}$, but undercounts $2E$ by the sums of degrees of vertices in $V^{w,\partial}$ (where the 'degree' here means as a vertex in $S$). But every vertex in $V^{w,\partial}$ has at least two edges in $S$, which shows this formula. Now, let's get one last expression for $E$. Since the graph is quadrangulated, we'll have that $4F = 2E - 2 |V^{w,\partial}| - A - B$, where $B \ge 0$ is an integer. This is because $4F$ gives a double count of all edges on the interior of S, but again undercounts because of the edges boundary loop and connectors. In fact, one can see that it undercounts by at least the same edges that $\bar{d_{in}} |V_{in}|$ does, which gives the "$-2 |V^{w,\partial}| - A$" terms. The extra positive number $B$ is because there might be vertices on the 'connecting part' between the polygons that don't get counted in $4F$ but do get counted in $\bar{d_{in}} |V_{in}|$ Putting these together, we get that $E = 2F + |V^{w,\partial}| + \frac{1}{2}(A + B)$, that $F = \frac{1}{4}(\bar{d_{in}} |V_{in}| - B)$, and that: $$V - E + F = (|V_{in}| + |V^{w,\partial}|) - (2F + |V^{w,\partial}| + \frac{1}{2}(A + B)) + F$$ So, we'll get that $$|V_{in}| - F - \frac{1}{2}(A + B) = |V_{in}| - \frac{1}{4}(\bar{d_{in}} |V_{in}| - B) - \frac{1}{2}(A + B)= 1$$ This gives us that $$(1-\frac{\bar{d_{in}}}{4}) |V_{in}| = 1 + \frac{A}{2} + \frac{B}{4} > 0$$ So, this is impossible if $\bar{d_{in}} \ge 4$, which implies the proposition of Lev. Now it's time to prove the Main Claim. Proof of Main Claim The main idea is that these daisy chains end up being obstructions to being able to perform Lev's "global move" (see his answer for the definition) via local moves. I.e., if the global move is always possible to do from local moves, then the graph is flip-connected. However, if there is a loop in the graph such that the global move is impossible, we'll show that there must be a daisy chain in the graph. Suppose that there is a perfect matching $\mathcal{P}$ of $G$ such that for some cycle $C$ in the graph with every alternate edge being in the perfect matching, the global move for $C$ is not possible (this cycle can't repeat vertices and is NOT the loop in the daisy chain). Without loss of generality, we'll focus our attention to the subgraph completely contained in $G$, so that the cycle $C$ is the boundary of $G$. From this, we can consider all matchings connected to $\mathcal{P}$ via flip moves. For all of these matchings, there must be a cycle $C'$ for which the global move is impossible, otherwise the graph would be flip-connected. We'll call such a cycle $C'$ irreducible. Furthermore, we will choose the matching containing the "minimal" such irreducible cycle $C'$, defined as follows. $C'$ will be minimal if It contains the minimum number, $N$, of total vertices on the boundary of $C'$ and on the interior of $C'$ Given the minimal $N$, $C'$ has a minimal number of strictly interior points for an irreducible cycle (i.e. maximal number of points on the boundary loop, $C'$). So, again we can WLOG choose the graph $G$ with matching on it's boundary that is minimal and irreducible boundary cycle. First, we claim that there must be internal vertices inside $C$. Suppose there are no internal vertices. We note that it's always possible to do at least one flip moves, since there are more edges in the matching than faces, so at least one face has two matched dimers on it. Performing a flip move on this face will split the outer cycle into two smaller cycles, contradicting the minimality of $C$. Induction will actually show that such an "outerplanar" graph is flip connected. (A statement for more general outerplanar graphs is in Bonamy et al). So, let $w_0$ be an internal vertex that shares an edge with a boundary vertex $v$, and let's say that it is colored white. Note that $w_0$ must be matched with a black internal vertex $b_1$. Now, consider all (white) vertices $w_{1,i}$ indexed by $i$, that are connected to $b_1$. We'll have that the minimality and irreducibility of $C$ will guarantee that none of the $w_{1,i}$ are on the boundary. If $w_{1,I}$ were on the boundary, then tracing the path $C'' := v \rightarrow w_0 \rightarrow b_1 \rightarrow w_{1,i} \rightarrow ...\text{boundary vertices}... \rightarrow v$ would create a cycle $C''$ with more boundary points or fewer internal vertices than $C$. So, since $C$ was minimal and irredicuble, the global move on $C''$ would be possible by local moves inside $C''$. And then the path in the other direction from the boundary point $w_{1,i}$ to $v$ would be a cycle of matches which also admits a global move, since it's not irreducible. This whole process is the global move on $C$. So since $C$ was irreducible and minimal, $w_{1,i}$ isn't on the boundary. So, since the $w_{1,i}$ aren't on the boundary, if they are distinct from $w_0$, they have new vertices $b_{2,i}$ that they are matched with. Then consider all the white vertices $w_{2,I}$ (labeled by some appropriate multi-index $I$) who neighbor the $b_{2,i}$. The $w_{2,I}$ can't be on the boundary by the same argument as for the $w_{1,i}$. Iterating this process, we can consider the perfect matching partners $b_{n,I}$ of the new vertices in $w_{n-1,I}$ and then considering all the white neighbors of the $b_{n,I}$. The $w_{n-1,I}$ can never be on the boundary, again since $C$ is minimal and irreducible. This process must terminate and eventually give no new vertices since $G$ is finite. When the process terminates, there will be a set of polygons and paths of links connecting them whose boundary consists of the vertices $w_{n,I}, b_{n,I}$ chosen, like the blue part of the figure above. (It is not necessary that all the vertices inside the polygons have been reached by the process, just that the boundary consists of ones that have been reached.) Since all these vertices are on the interior, each of them must be part of a 2-cell that is on the exterior of the polygons and connecting paths. But, any new edges coming from these external facets must be connected to the white vertices, since all the neighbors of black vertices have been chosen in the process. This means that on the boundary loop of the polygons and paths, every adjacent edges along the loop that have colors "white $\rightarrow$ black $\rightarrow$ white" must be part of a common facet. These outer facets give our daisy chain, so we've completed out proof.<|endoftext|> TITLE: Building a hypersurface from hyperplane sections QUESTION [7 upvotes]: Let $P_0,\ldots,P_n$ denote the coordinate hyperplanes in $\mathbb P^n$, and suppose that in each $P_i$ I have a degree $d$ hypersurface $V_i$. I am trying to understand what is the obstruction to the existence of a degree $d$ hypersurface $V \subset \mathbb P^n$ such that $V \cap P_i = V_i$. There are some obvious obstructions: for example, my prescribed hyperplane sections $V_i$ must agree on intersections so that $V_i \cap P_j = V_j \cap P_i$. I don't think this alone is quite sufficient. In the case $n=3$ for example, let $p = P_1 \cap P_2 \cap P_3$ and suppose that $V_1,V_2,V_3$ all go through this point. Then $T_p V_1 + T_p V_2 + T_p V_3$ had better have dimension $2$, not $3$. I am sure there is some exact sequence I'm missing - what is it? (note: of course $V$ does not have to be unique, at least if $d \geq n+1$. You can add monomials containing all of the variables to the defining equation, and it doesn't change the intersection of the resulting surface with the coordinate hyperplanes) REPLY [7 votes]: The condition $V_i \cap P_j = V_j \cap P_i$ (if understood properly) is the only obstruction. To see this denote by $Z$ the union of $P_i$. Then we have an exact sequnce $$ 0 \to \mathcal{O}_Z \to \bigoplus \mathcal{O}_{P_i} \to \bigoplus \mathcal{O}_{P_i \cap P_j} \to \bigoplus \mathcal{O}_{P_i \cap P_j \cap P_k} \to \dots $$ (such an exact sequence exists for any union of transverse hypersurfaces and can be proved by induction on the number of components). Tensoring it by $\mathcal{O}(d)$ one deduces an isomorphism $$ H^0(Z,\mathcal{O}_Z(d)) = \mathrm{Ker}\Big(\bigoplus H^0(P_i,\mathcal{O}_{P_i}(d)) \to \bigoplus H^0(P_i \cap P_j, \mathcal{O}_{P_i \cap P_j}(d)) \Big). $$ Thus, to give a section of $\mathcal{O}_Z(d)$ one needs to give a collection of sections of $\mathcal{O}_{P_i}(d)$ that agree on pairwise intersections (this is the right way to state the condition). On the other hand, $Z$ is a hypersurface of degree $n+1$, hence there is an exact sequence $$ 0 \to \mathcal{O}(d-n-1) \to \mathcal{O}(d) \to \mathcal{O}_Z(d) \to 0 $$ on $\mathbb{P}^n$, and since $H^1(\mathbb{P}^n,\mathcal{O}(d-n-1)) = 0$ (I assume that $n \ge 2$), it follows that the morphism $$ H^0(\mathbb{P}^n, \mathcal{O}(d)) \to H^0(Z, \mathcal{O}_Z(d)) $$ is surjective, hence any such section lifts to an equation of a hypersurface in $\mathbb{P}^n$.<|endoftext|> TITLE: Calculation of $H^{10}(K(\mathbb{Z}, 3); \mathbb{Z})$ QUESTION [15 upvotes]: I was trying to calculate $H^q(K(\mathbb{Z}, 3); \mathbb{Z})$ for some $q$ with the Serre spectral sequence associated to the fibration $K(\mathbb{Z}, 2) \to PK(\mathbb{Z}, 3) \simeq * \to K(\mathbb{Z}, 3)$. I obtained that: $$ H^q(K(\mathbb{Z}, 3)) = \mathbb{Z}, 0, 0, \mathbb{Z}x, 0, 0, \mathbb{Z}_2x^2, 0, \mathbb{Z}_3y, \mathbb{Z}_2x^3, 0, \mathbb{Z}_3xy, \mathbb{Z}_2x^4\oplus\mathbb{Z}_5w. $$ But in Hatcher's book on spectral sequences, chapter 1, he claims that $$ H^q(K(\mathbb{Z}, 3)) = \mathbb{Z}, 0, 0, \mathbb{Z}x, 0, 0, \mathbb{Z}_2x^2, 0, \mathbb{Z}_3y, \mathbb{Z}_2x^3, \mathbb{Z}_2z, \mathbb{Z}_3xy, \mathbb{Z}_2x^4\oplus\mathbb{Z}_5w. $$ The only difference is in $H^{10}(K(\mathbb{Z}, 3))$, that for me is $0$, while for Hatcher is $\mathbb{Z}_2z$. I cannot understand why this happens. My reasoning is: $H^{10}(K(\mathbb{Z}, 3))$ is in position $(10, 0)$ and it can be reached by groups in position $(9 - r, r)$. These are non trivial only if $r$ is even and $9-r = 0, 3, 6$. The only possibility is then $r = 6$, i.e., $$ (3, 6) = H^3(K(\mathbb{Z}, 3); H^6(K(\mathbb{Z}, 2))) = H^3(K(\mathbb{Z}, 3); \mathbb{Z}n^3) = \mathbb{Z}n^3x. $$ But my claim is that $(10, 0)$ could not be reached neither by $(3, 6)$ because this dies turning $E_2$. In fact $d_2(n) = x$, so $d_2(n^3x) = 3n^2x^2 = n^2x^2$ and so $d_2: (3, 6) \to (6, 4) = \mathbb{Z}_2n^2x^2$ is an isomorphism (edit: the error is here, it is not an isomorphism, but it has kernel $2\mathbb{Z}n^3x$). Then $(10, 0)$ would survive at $\infty$, which is not possible. What's wrong with this? REPLY [20 votes]: I do not like naming a cohomology class $n$ because that deserves to be the name of an integer. I will use the name Hatcher does and call the generator of $H^2(K(\Bbb Z, 2); \Bbb Z)$ by the name "$a$". The map $d_3: E_3^{0, 8} \to E_3^{3,6}$ sends $d_3(a^4) = 4a^3 x$ by the Leibniz rule. The map $d_3: E_3^{3,6} \to E_3^{6,4} \cong \Bbb Z_2 a^3 x^2$ sends $a^3x$ to $a^3x^2$; that is, this map is reduction mod 2 in this basis. (You already calculated that this must be true earlier in the computation.) The homology group of $$\Bbb Z \xrightarrow{4} \Bbb Z \xrightarrow{\pmod 2} \Bbb Z_2$$ in the middle is $\Bbb Z_2$. Thus $$E_4^{3,6} = E_7^{3,6} = \Bbb Z_2\langle 2a^3 x\rangle,$$ where the angle brackets indicate that the nonzero class came from the element $2a^3 x$ on the $E_3$ page. Thus there is indeed something left at $E_7$ in this position, so that the differential $d_7: E_7^{3,6} \to E_7^{10,0}$ must be an isomorphism. It seems what you missed is that $d_3: \Bbb Z = E_3^{0,8} \to E_3^{3,6} = \Bbb Z$ is $4$, not $2$.<|endoftext|> TITLE: Homotopy extension of $E_{\infty}$-spaces QUESTION [10 upvotes]: Suppose that $X$ is a connected $E_{\infty}$-space, naturally $\Omega X$ is also an $E_{\infty}$-space. Can we classify all $E_{\infty}$-extensions of $X$ by $\Omega X$ (up to homotopy). I mean the following: we would like to classify of homotopy fiber sequences $A\rightarrow B\rightarrow C$ where $A\sim \Omega X$ and $C\sim X$ as $E_{\infty}$-spaces and $B\rightarrow C$ , $A\rightarrow B$ are maps of $E_{\infty}$-spaces in particular we assume that $B$ is an $E_{\infty}$-space. (space could mean a simplicial set or a topological space) There are two obvious extensions of $E_{\infty}$-space the first one is when $B\sim X\times \Omega X$ and the second one is when $B=PX$ the path space. Is there others ? My first guess was that the set of extensions is a group and more precisely the set of homotopy classes of maps of spectra $[X,X]$ where $X$ is seen as a connective spectra. Maybe I'm wrong but the first extension corresponds to the homotopy class of the trivial map $X\rightarrow \ast\rightarrow X$ and the second extension corresponds to the identity map $id: X\rightarrow X$. But there should be other extension in bijective correspondence with $[X,X]$. Is it correct ? REPLY [19 votes]: This is probably belaboring the obvious, but just take seriously the equivalence between grouplike $E_{\infty}$ spaces and connective spectra. See for example Equivalence between $E_\infty$-spaces and connective spectra The asumption that X is connected means that the associated spectrum is connected and not just connective. So we may as well just ignore $E_{\infty}$ spaces (much as I hate to do that!) and take $X$ to be a connected spectrum. One is asking for all fiber sequences $\Sigma^{-1}X \to Y \to X$ of spectra or equivalently all fiber sequences $Y \to X \to X$. That is, Y is equivalent to the fiber of a map $X\to X$. A quick triangulated category type argument shows that equivalence classes of such fibers correspond bijectively to maps $X\to X$, that is to elements of $[X,X]$.<|endoftext|> TITLE: Why do stochastic integrals depend on the choice of partitioning points? QUESTION [16 upvotes]: When we integrate a function, we must make some choice about how we approximate it before we take the limit. In principle, we can choose $\tau_i$ to be any value between $t_{i-1}$ and $t_i$. But for an ordinary Riemann integral our choice doesn't matter since for any value of the intermediate point $\tau \equiv \frac{\tau_i}{t_i-t_{i-1}}$, we find the same value in the limit of vanishing box sizes. For stochastic integrals, however, this is no longer the case. For example, for the Itô integral, we choose $\tau =0$, while for the Stratonovich integral we choose $\tau = 0.5$. I'm wondering what feature of stochastic integrals leads to their dependence on the choice of $\tau$? (Since I'm a physicist by trade, a somewhat intutive argument would be great.) REPLY [7 votes]: To complement the excellent answer by Ofer Zeitouni, let me offer a functional analysis perspective. We want to define an integral of the following form: $\int F(W_t)dW_t=\int F(W_t)W'_tdt$, say, for a nice $F$. We can ask, generally, when is the integral $\int G(t)H(t)dt$ naturally defined? An obvious answer is: whenever $G$ belongs to some function space and $H$ belongs to the dual of that space. Then, in particular, any "reasonable" approximation scheme $\int G_n(t)H_n(t)dt$, where $G_n,H_n$ approximate $G,H$ in their corresponding spaces, will produce the same result. Which function spaces are we talking about? Well, note that $W'_t$ only makes sense as a distribution, and $F(W_t)$ has the same regularity as $W_t$, that it, it is not smooth. Therefore, "soft" tools like Schwarz spaces will not do. A natural scale is then that of Sobolev spaces; a function $f$ is in $H^s$ if $(1+|\xi|^2)^\frac{s}{2}\hat{f}(\xi)$ is square integrable; to a very rough first approximation this means that $f$ is (almost) s-Holder continuous. It is then clear that the dual of $H^s$ is $H^{-s}$, and that differentiation takes away $1$ from $s$. This implies that the integral $\int F(W_t)W'_t dt$ would be naturally defined if we had $W_t\in H^s$ for some $s\geq\frac12$. But by Wiener's construction, we control the Fourier coefficients very well: we know that $\hat{W}(n)=\text{sgn}(n)n^{-1}\zeta_{|n|}$, where $\zeta_n$ are i. i. d. Gaussians, and so, we are essentially asking for which $s$ does the series $\sum_n \zeta^2_n (1+n^2)^sn^{-2}$ converge. The answer is provided by Kolmogorov's three series theorem: it converges (almost surely) if and only if the series of expectations and of variances converge, which happens if and only if $s<\frac12$. So, the condition we are after fails just barely. This explains why we cannot define the integral $\int F(W_t)dW_t$ "pathwise" just by applying Riemann, Lebesgue or whatever integration to each realization. But also the fact that the required condition is just barely missed indicates why something like Ito integration, exploiting randomness additionally, has a chance to work.<|endoftext|> TITLE: Hahn's approach to Hilbert's 17th problem? QUESTION [9 upvotes]: The Wikipedia article on Hahn Series mentions mentioned that these were studied by Hahn "in his approach to Hilbert's seventeenth problem". Is this correct? If so, what was this approach, and where can I read about it? I have read most of Hahn's paper über die nichtarchimedischen Größensysteme, where Hahn series were introduced, but I have not seen Hilbert's 17th problem on positive polynomials being mentioned there. I have also skimmed his list of publications and not found anything else that looks relevant. REPLY [4 votes]: As you suspect, Hahn does not discuss Hilbert's 17th problem in the paper you cite, and, like you, I am not aware of any of Hahn's publications that applies his work on non-Archimedean ordered systems to Hilbert's 17th problem. However, in the paper you mention Hahn does discuss Hilbert's (arithmetic) completeness condition, which is discussed by Hilbert in the second problem of his famous list of problems. There, however, Hahn is not concerned with the consistency of the real number system, which is the subject of the second problem, but rather with generalizing Hilbert's completeness condition so as to be applicable to non-Archimedean as well as Archimedean ordered Abelian groups and ordered fields.<|endoftext|> TITLE: When is Thom isomorphism a ring map? QUESTION [9 upvotes]: Let $R$ be an $E_{\infty}$-ring spectrum and $B$ be an $E_\infty$-space. Suppose we have an $E_\infty$-map $$ f: B \to BGL_1(S^0)$$ such that the composite $$f_R: B \to BGL_1(S^0) \to BGL_1(R) $$ is null, then a choice of null-homotopy produces Thom isomorphism which is a weak-equivalence $$u: Mf \wedge R \simeq B_+ \wedge R,$$ where $Mf$ is the Thom spectrum associated to $f$. Note that, both sides of the Thom isomorphism are ring spectra. Q: I wonder when $u$ is a ring-map? A possible guess is that if $f_R$ is homotoped to null via infinite-loop space maps, then maybe $u$ is a ring map. I am not quite sure if this is true or how to see this. REPLY [9 votes]: One comment is that you have to be careful, because to be an $E_{\infty}$-map is not a property but rather additional structure. You are exactly right in that what is needed is a null-homotopy of $B \rightarrow BGL_{1}(R)$ as a map of $E_{\infty}$-spaces. This is Proposition 3.16 in Omar and Toby's "A simple universal property of Thom ring spectra" when you set $A = R$.<|endoftext|> TITLE: Implications of Geometrization conjecture for fundamental group QUESTION [7 upvotes]: Hempel proved that Haken manifolds have residually finite fundamental groups. With the Geometrization conjecture, this now holds for any compact and orientable 3-manifold. How exactly does the Geometrization conjecture imply that the only non-Haken compact orientable irreducible 3-manifolds are compact hyperbolic manifolds with no cusps? Thanks a lot. REPLY [6 votes]: There are a few ways to look at this, but let me give a synopsis of Peter Scott's discussion in Section 6 of his 8 geometries paper, because it seems to directly address the question: Scott, Peter, The geometries of 3-manifolds, Bull. Lond. Math. Soc. 15, 401-487 (1983). ZBL0561.57001. (Note: there is also an errata for this reference: http://www.math.lsa.umich.edu/~pscott/errata8geoms.pdf. The corrections mentioned there do not affect this summary. ) Let's set up some standard 3-manifold terminology (anything left out is clearly defined in the reference). A compact orientable 3-manifold is irreducible if is either $S^1 \times S^2$ or every embedded $S^2$ bounds a 3-ball. From now on, assume $M$ is a compact, orientable, irreducible 3-manifold (unless otherwise specified). A surface $S$ (smoothly) embedded in $M$ is incompressible if every embedded curve $\gamma$ in $S$ which bounds a disk in $M$ also bounds a disk in $S$. A compact, orientable, irreducible 3-manifold is toroidal if it contains an incompressible torus. The most natural thing to say is that the affirmative solution to the Geometrization Conjecture implies that a compact, irreducible 3-manifold is either toroidal or is homeomorphic to a quotient of the form $X/\Gamma$ where $X$ is one of the following geometric spaces: $S^3,S^1 \times \mathbb{R},E^3,Nil,Sol, H^2 \times \mathbb{R}, \widetilde{PSL(2,\mathbb{R})}$ or $H^3$ and $\Gamma \subset Isom^+(X)$ acting properly and discontinuously. If a space admits a Sol geometry it is Haken. In fact, it is both toroidal and has positive first Betti number, (see [Scott, Theorem 4.17]). The remaining geometries are either $H^3$ or Seifert fibered. Of course, some Seifert fibered manifolds like $T^3 \cong S^1 \times S^1 \times S^1$ are both toroidal and Seifert fibered and there are other minor pathologies: for example, $RP^3 \# RP^3$ is Seifert fibered and reducible and in the wake of Geometrization manifolds with $S^3$ geometry are exactly those with finite fundamental group. Happily, Scott gives a clean statement of what you want in the conjecture on page 484 (of course the affirmative solution to the Geometrization Conjecture implies this conjecture is now known to be true): Conjecture (now theorem): Let $M$ be a closed, irreducible, non-Haken 3-manifold with infinite fundamental group. Then $M$ is either a Seifert fibered space or admits a hyperbolic structure. To connect this statement to the comments mentioned above, if we assume $M$ is Seifert fibered, non-Haken, irreducible, and has infinite fundamental group, then it is small Seifert fibered, which implies $M$ that the base orbifold of $M$ is $S^2(a_1,a_2,a_3)$ with $1/a_1 + 1/a_2 + 1/a_3 \leq 1$.<|endoftext|> TITLE: On the sum $\sum_{k=b}^{a-1} \binom{2k+1+n}{2n+1}\binom{2a-1}{a+k}$ QUESTION [6 upvotes]: Let $n$ be a non-negative integer. Does there always exist a polynomial $P_n(a,b)$ such that for all integers $a > b \geq n/2$ we have $$ \sum_{k=b}^{a-1} \binom{2k+1+n}{2n+1}\binom{2a-1}{a+k} = \binom{2a-1}{a+b} P_n(a,b)\quad ? $$ For small values of $n$ this is easily verified using Gosper's algorithm, for example $$P_0(a,b) = a+b,\qquad P_1(a,b) = \tfrac{2}{3}(a+b)(b^2 +a -1), $$ but I am struggling to prove the general case. Any suggestions or literature references on this problem? Here are some details on Gosper's algorithm. Let us denote the summand by $t(k) = \binom{2k+1+n}{2n+1}\binom{2a-1}{a+k}$ and let $$p(k) = (k-\tfrac12n+1)_n(k-\tfrac12n+\tfrac12)_{n+1}, \quad q(k) = a-k-1, \quad r(k)=a+k$$ where $(x)_n = x(x+1)\cdots(x+n-1)$. Then we have $$ \frac{t(k+1)}{t(k)} = \frac{q(k)}{r(k+1)}\frac{p(k+1)}{p(k)}.$$ According to Gosper, $T(k+1)-T(k)=t(k)$ has a hypergeometric term solution $T(k)$ iff there exists a polynomial $s(k)$ of degree $2n$ (with coefficients that depend on $a$) that solves \begin{equation} \tag{1} p(k) = q(k) s(k+1) - r(k)s(k). \end{equation} If such a solution exists, then $$T(k) = \frac{r(k)}{p(k)} t(k)s(k)\quad \text{and}\quad P_n(a,b) = \frac{T(a)-T(b)}{\binom{2a-1}{a+b}} = -\frac{2^{2n+1}}{(2n+1)!}(a+b)\,s(b)$$ fulfilling my request. The linear mapping $s(k) \mapsto q(k) s(k+1) - r(k)s(k)$ from polynomials of degree $2n$ to those of rank $2n+1$ is easily seen to be injective, but it seems difficult to show that $p(k)$ lies in its image for general $n$. Update: I have included a proof in an answer below, but I'd still be interested in literature references. REPLY [2 votes]: Actually, checking whether $p(k)$ lies in the image of $s(k) \mapsto q(k) s(k+1) - r(k)s(k)$ turns out to be not too difficult after all. We need to determine a single linear condition that spans the cokernel of the linear mapping. A convenient way to do this is by turning $q(k) s(k+1) - r(k)s(k)$ into a differential operator and seeking a (formal) power series $V(x)$ solving \begin{align} 0 &= \left[s(1+\partial_x)q(\partial_x) - s(\partial_x)r(\partial_x) \right] V(x)\big|_{x=0} \\ &=\left[e^{-x}s(\partial_x)e^{x}q(\partial_x) - s(\partial_x)r(\partial_x) \right] V(x)\big|_{x=0} \\ &=s(\partial_x)\left[e^{x}q(\partial_x) - r(\partial_x) \right] V(x)\big|_{x=0}. \end{align} Hence, a solution to \begin{equation} 0 = \left[e^{x}q(\partial_x) - r(\partial_x) \right] V(x) = (e^x(a-1)-a)V(x)-(e^x+1)V'(x) \end{equation} will do for any $n\geq 0$ and any polynomial $s(k)$. One easily finds \begin{equation} V(x) = e^{-ax} (1+e^{x})^{2a-1} = e^{-x/2} (2\cosh(x/2))^{2a-1}. \end{equation} It remains to show that $p(\partial_x) V(x)|_{x=0}=0$. This follows from \begin{equation} p(\partial_x) V(x)\big|_{x=0} = p(\partial_x-\tfrac{1}{2}) e^{x/2}V(x)\big|_{x=0} = p(\partial_x-\tfrac{1}{2}) (2\cosh(x/2))^{2a-1}\big|_{x=0}, \end{equation} which vanishes because $p(\partial_x-1/2)$ is odd in $\partial_x$, while $(2\cosh(x/2))^{2a-1}$ is an even power series in $x$. This shows that the recursion equation (1) has a unique polynomial solution $s(k)$. Moreover the coefficients of $s(k)$ are seen to be polynomials in $a$, since the linear mapping $s(k) \mapsto q(k) s(k+1) - r(k)s(k)$ is essentially upper-triangular (in the monomial basis) with a diagonal that is independent of $a$. QED. In fact this demonstrates an equivalence: Let $p(k)$ be a polynomial. Then $\sum_{k=b}^{a-1} p(k)\binom{2a-1}{a+k}$ is given by a hypergeometric term in $b$ if and only if $p(k-1/2)$ is odd in $k$. In that case there exists a polynomial $P(a,b)$ such that \begin{equation}\sum_{k=b}^{a-1} p(k)\binom{2a-1}{a+k} = \binom{2a-1}{a+b} P(a,b).\qquad (a>b) \end{equation}<|endoftext|> TITLE: $\sum_{k =1, k \neq j}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)=?$ QUESTION [7 upvotes]: It is well-known that one can evaluate the sum $$\sum_{k =1}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)=\frac{N^2-1}{3}.$$ The answer to this problem can be found here click here. I am now interested in the more difficult problem of evaluating for some $j \in \{1,...,N-1\}$ the sum (we do not sum over $j$) $$\sum_{k =1, k \neq j}^{N-1} \csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)=?$$ Does this one still allow for an explicit answer? REPLY [6 votes]: Start from the well known formula \begin{equation} 2^{N-1} \prod _{k=1}^N \left[\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)\right]=\cos N(x-y)-\cos N(x+y), \end{equation} take logarithmic derivative \begin{equation} \sum _{k=1}^N \frac{\sin(x-y)}{\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)}=\frac{N\sin N(x-y)}{\cos N (x-y)-\cos N(x+y)}, \end{equation} rewrite it as \begin{equation} N+\sum _{k=1}^N \frac{\cos(x-y)+\cos \left(x+y+\frac{2\pi k}{N}\right)}{\cos (x-y)-\cos \left(x+y+\frac{2\pi k}{N}\right)}=\frac{2N\sin N(x-y)}{\cos N (x-y)-\cos N(x+y)}\cot(x-y), \end{equation} and simplify to get \begin{equation} \sum_{k=1}^N\cot\left(x-\frac{\pi k}{N}\right)\cot\left(y-\frac{\pi k}{N}\right)=\frac{2N\cot(x-y)\sin N(x-y)}{\cos N(x-y)-\cos N(x+y)}-N. \end{equation} Then take derivatives wrt $x$ and $y$ $$ \sum_{k=1}^N\csc^2\left(x-\frac{\pi k}{N}\right)\csc^2\left(y-\frac{\pi k}{N}\right)=\frac{N}{\sin ^2(x-y)} \left(\frac{N}{\sin ^2Nx}+\frac{N}{\sin ^2Ny}-\frac{2\sin N(x-y)}{\sin Nx\sin Ny}\,\cot (x-y) \right). $$ REPLY [4 votes]: Yes, it may be simplified. The text below is not probably the shortest way, but it explains how to calculate many similar sums. We start with $$\sin^2 x=-\frac14(e^{ix}-e^{-ix})^2=-\frac14e^{-2ix}(e^{2ix}-1)^2.$$ So, denoting $\omega_k=e^{2\pi i k/N}$ for $k=0,\ldots,N-1$ we get $$S:=\sum_{1\leqslant k\leqslant N-1,k\ne j}\csc^2\left(\pi \frac{k}{N} \right)\csc^2\left(\pi \frac{j-k}{N} \right)= 16\sum_{k\ne 0,j}\frac{\omega_k \omega_{k-j}}{(\omega_k-1)^2(\omega_{k-j}-1)^2}=\\ 16\omega_j \sum_{k\ne 0,j}\frac{\omega_k^2}{(\omega_k-1)^2(\omega_{k}-\omega_j)^2}.$$ Consider the rational function $f(x):=\frac{x^2}{(x-1)^2(x-a)^2}$ (where $a=\omega_j$). We have $$f(x)= \frac{a^2}{(a - 1)^2 (x - a)^2} + \frac{2 a}{(a - 1)^3 (x - 1)} - \frac{2 a}{(a - 1)^3 (x - a)} + \frac1{(a - 1)^2 (x - 1)^2}.$$ Each of four summands may be easily summed up over $x\in \{\omega_0,\ldots,\omega_{N-1}\}\setminus \{1,a\}$. Indeed, we have $$ \sum_{k} \frac 1{t-\omega_k}=\frac{(t^n-1)'}{(t^n-1)}=\frac{nt^{n-1}}{t^n-1},\,\,\, \sum_{k} \frac 1{(t-\omega_k)^2}= -\left(\frac{nt^{n-1}}{t^n-1}\right)'=n\frac{t^{n-2}(t^n+n-1)}{(t^n-1)^2}. $$ If we want to substitute $t=\omega_j$, we get something like $$ \sum_{k:k\ne j} \frac1{\omega_j-\omega_k}=\left(\frac{nt^{n-1}}{t^n-1}-\frac1{t-\omega_j}\right)_{t=\omega_j}=\left(\frac{(n-1)t^{n}-n\omega_j t^{n-1}+1}{(t^n-1)(t-\omega_j)}\right)_{t=\omega_j}=(n-1)\omega_j^{-1} $$ by L'Hôpital, analogously for the squares.<|endoftext|> TITLE: Gentzen's result on PA QUESTION [8 upvotes]: The Wikipedia states that Gentzen proved that "in modern terms, the proof-theoretic ordinal of PA is $\varepsilon_0$." Further down, the article defines what the "proof theoretic ordinal" of a theory means. However, I'm not sure what this means regarding PA, since PA can only make finitary statements. Let me elaborate. Define some encoding of the ordinals ${}<\varepsilon_0$ as natural numbers. This encoding allows us to express statements involving ordinals ${}<\varepsilon_0$ in PA. Then, allegedly PA cannot prove transfinite induction using this encoding. But what does this mean exactly? One option would be using sets: "Suppose $S$ is a set of ordinals such that, for every $\beta<\varepsilon_0$, whenever $\alpha\in S$ for every $\alpha<\beta$, we also have $\beta\in S$. Then $\beta\in S$ for all $\beta<\varepsilon_0$." But this is an infinitary statement, as far as I understand, so it cannot be stated in PA. Another option is to have a schema of infinitely many statements, one for each possible formula $\varphi$ (just like the "induction schema" contains infinitely many axioms, one for each possible formula): "Suppose that for every $\beta<\varepsilon_0$, whenever $\varphi(\alpha)$ holds for every $\alpha<\beta$, we also have $\varphi(\beta)$. Then $\varphi(\beta)$ holds for every $\beta<\varepsilon_0$." So what is it exactly that Gentzen proved? Presumably PA can prove the above statement for some formulas $\varphi$. So did Gentzen find some specific $\varphi$ for which PA cannot prove the above statement? Or what? REPLY [10 votes]: Yes, Gentzen found a single "induction instance" which is PA-unprovable: more-or-less $\varphi(\alpha)$ = "Every sentence with a proof of cut-rank $\alpha$ has a cut-free proof." Now, this $\varphi$ is a $\Pi^0_2$ formula. If memory serves, this is suboptimal: with some coding work this can be improved from a $\Pi^0_2$ formula to a $\Sigma^0_1$ formula. The basic idea is to assign in a primitive recursive way to each ordinal $\alpha<\epsilon_0$ a sentence $p_\alpha$ and a candidate proof $s_\alpha$ of cut rank $<\alpha$ such that each pair occurs cofinally often, and then look at the formula $\psi(\alpha)$ = "Either $s_\alpha$ is not a proof of $p_\alpha$ or there is a cut free proof of $p_\alpha$." And I think even that's suboptimal - that we can get to the level of $\Delta_0$ - but I'm not sure.<|endoftext|> TITLE: On Exercise 2.5.10 in Ram. M. Murty's book, "Problems in Analytic Number Theory." QUESTION [5 upvotes]: I have just been told about this result, available as Exercise 2.5.10 in Ram. M. Murty's book, "Problems in Analytic Number Theory (2nd edition)". It says: Let $\alpha>0$. Suppose $a_n \ll n^{\alpha}$ and $A(x) \ll x ^{\delta} $ for some $\delta<1$, where $A(x) = \sum_{n\leq x} a_n$. Define $b_n = \sum_{d|n} a_d$. Then one has $$\sum_{n\leq x} b_n = cx + O\Big(x^{(1-\delta)(1+\alpha)/(2-\delta)}\Big)$$ for some constant $c$. Does anyone know who first came up with this result, or maybe it's just too straightforward to be attributed to anyone ? The reason why i'm particularly interested in knowing the originator of this result is that, Murty's proof (which is on pages 262-263 of the aforementioned book) doesn't look quite right to me (but of course, i may be mistaken). REPLY [3 votes]: The computation of the powers is wrong and the result stated in the book is incorrect and it should be $cx+ O(x^{(1+\delta-\delta^2)/(2-\delta)})+O(x^{\frac{(1-\delta)(1+\alpha)}{2-\delta}})$ If $y=x^{(1-\delta)/(2-\delta)}$, $y^{\delta-1}=x^{-(1-\delta)^2/(2-\delta)}$, so $xy^{\delta-1}=x^{(1+\delta-\delta^2)/(2-\delta)}$ and it is not true that $\frac{1+\delta-\delta^2}{2-\delta} \le \frac{(1-\delta)(1+\alpha)}{2-\delta}$ in general, only for $\alpha \ge \delta + \frac{\delta}{1-\delta}$ Note that as stated, the result doesn't make sense because one can always increase $\delta$ in the hypothesis, while keeping $\alpha$ fixed, so in particular if $A(x) << x^{\delta}$ for a given $\delta < 1$, then $A(x) <<_{\epsilon} x^{1-\epsilon}$ for arbitrary $\epsilon >0$, hence we would get that $B(x) =cx + O_\epsilon(x^{\epsilon})$ under very general conditions and I am sure lots of counterxamples to that can be found<|endoftext|> TITLE: Diagonal of a diagram of codescent objects QUESTION [5 upvotes]: Given the following diagram in a $2$-category, in which squares of the same "type" commute, where each column and each row is a strong codescent diagram (Edit: it should be reflexive as well), is then the diagonal a codescent diagram as well? Actually, for each row and each column we also have six $2$-isomorphisms (for example, $\xi^h_i : x^h_i d^h_i \to x^h_i c^h_i$ for the $i$th row), which I didn't write down here. This question is motivated by Mike Shulman's comment here. In fact, this statement would probably be an important step to construct bicategorical pushouts of symmetric pseudomonoids. I tried to find a proof, but basically get lost because of this huge amount of data. Notice that this statement (if it is true) is a generalization of the corresponding $1$-dimensional statement about reflexive coequalizers (Sketches of an elephant, Lemma A.1.2.11). I would be very happy for a reference to the literature where this is proven. Instead of writing down a proof on my own, which will be probably too long anyway, I would like to cite and then use this result in a paper. REPLY [7 votes]: As I wrote in a comment above, for this result to hold, each row and column of your diagram must be a reflexive codescent diagram. I do not know of any place in the literature where this result is explicitly stated, but, as I will explain below, it follows without difficulty from a result of Steve Lack. (For simplicity, let me deal only with strict reflexive codescent objects. Since these are flexible colimits, one can deduce the fully weak bicategorical version of this result from the strict version by standard arguments.) Definition. Let $\Delta_{\leq 2}$ denote the full subcategory of the simplex category $\Delta$ containing the objects $[0]$, $[1]$, and $[2]$, and let $W : \Delta_{\leq 2} \longrightarrow \mathbf{Cat}$ denote the composite of the full inclusion $\Delta_{\leq 2} \longrightarrow \mathbf{Cat}$ with the groupoid reflection functor $\mathbf{Cat} \longrightarrow \mathbf{Cat}$. For any $2$-category $\mathcal{K}$, the reflexive codescent object of a functor $X : \Delta_{\leq 2}^\mathrm{op} \longrightarrow \mathcal{K}$ is the colimit $W \ast X$ of $X$ weighted by $W$. We will deduce the "diagonal lemma" of your question for reflexive codescent objects from the fact that reflexive codescent objects are sifted colimits (in the $\mathbf{Cat}$-enriched sense), i.e. that the functor $$W \ast (-) : [\Delta_{\leq 2}^\mathrm{op},\mathbf{Cat}] \longrightarrow \mathbf{Cat}$$ preserves finite products. This fact is due to Steve Lack -- see Proposition 4.3 of Lack, Stephen. Codescent objects and coherence. J. Pure Appl. Algebra 175 (2002), no. 1-3, 223--241. doi and Proposition 4 of Bourke, John. A colimit decomposition for homotopy algebras in Cat. Appl. Categ. Structures 22 (2014), no. 1, 13--28. doi Thanks to the "Fubini theorem" for iterated weighted colimits, we may state the diagonal lemma for reflexive codescent objects in the following form. Lemma (diagonal lemma for reflexive codescent objects). Let $\mathcal{K}$ be a $2$-category and let $X \colon \Delta_{\leq 2}^\mathrm{op} \times \Delta_{\leq 2}^\mathrm{op} \longrightarrow \mathcal{K}$ be a functor. Then we have an isomorphism of weighted colimits in $\mathcal{K}$ $$W \ast (X \circ \delta) \cong (W \times W) \ast X,$$ either side existing if the other does. (Here $\delta$ denotes the diagonal functor $\Delta_{\leq 2}^\mathrm{op} \longrightarrow \Delta_{\leq 2}^\mathrm{op} \times \Delta_{\leq 2}^\mathrm{op}$). Remark. It is also worth displaying the isomorphism of this lemma in coend form: $$\int^{[k]} W^k \times X_{k,k} \cong \int^{[n],[m]} W^n \times W^m \times X_{n,m}.$$ Proof of lemma. The preservation of binary products of representables by the functor $W \ast (-) : [\Delta_{\leq 2}^\mathrm{op},\mathbf{Cat}] \longrightarrow \mathbf{Cat}$ implies, via the weighted colimit formula for left Kan extensions, that the functor $W \times W : \Delta_{\leq 2} \times \Delta_{\leq 2} \longrightarrow \mathbf{Cat}$ is the left Kan extension of $W \colon \Delta_{\leq 2} \longrightarrow \mathbf{Cat}$ along the diagonal functor $\delta \colon \Delta_{\leq 2} \longrightarrow \Delta_{\leq 2} \times \Delta_{\leq 2}$. Hence the lemma follows from Theorem 4.38 of Kelly's Basic concepts of enriched category theory. $\Box$ It is worth mentioning that, when working bicategorically (i.e. "up to equivalence"), the codescent object of a (pseudo)functor $X : \Delta_{\leq 2}^\mathrm{op} \longrightarrow \mathcal{K}$ is simply its bicolimit. Hence the bicategorical version of the diagonal lemma for reflexive codescent objects -- which follows from the strict version by standard arguments -- is simply the statement that the diagonal functor $\Delta_{\leq 2}^\mathrm{op} \longrightarrow \Delta_{\leq 2}^\mathrm{op} \times \Delta_{\leq 2}^\mathrm{op}$ is 2-final.<|endoftext|> TITLE: local UFD with dimension less than or equal 3 is catenary QUESTION [6 upvotes]: Let $R$ be a commutative ring with identity. Then $R$ is $\textit{catenary}$ if for each pair of prime ideal $p \subsetneq q$, all maximal chains of prime ideals $p = p_0 \subsetneq p_1 \subsetneq \dots \subsetneq p_n = q$ have the same length. In some (informal) texts the author conclude that (without further explanation) the ring is catenary since it is a local UFD and its dimension is less than or equal 3. Is this considered trivial? thank you. REPLY [8 votes]: This result appears as Proposition II.3 in Hamet Seydi, Anneaux henséliens et conditions de chaînes. III, Bull. Soc. Math. France 98 (1970), 329–336. Numdam: BSMF_1970__98__329_0. DOI: 10.24033/bsmf.1706. MR: 276222. Namely, Seydi proves that "every (Noetherian) UFD of dimension three is catenary." The Noetherian assumption is necessary due to a counterexample of Fujita. There is no proof provided, but I think that Seydi is pointing out that this result is a consequence of the proofs of previous results, namely, Théorème II.2 and Corollaire II.2.4. We give a version of Seydi's proof here, by showing that Noetherian UFD's of dimension at most three are catenary. Proof. Let $A$ be a Noetherian UFD of dimension at most three. Since the property of being catenary can be checked after localizing at every maximal ideal, it suffices to consider the case when $A$ is local. We recall that Ratliff's criterion [Matsumura, Theorem 31.4] says that a Noetherian local domain $B$ is catenary if and only if $$\operatorname{ht} \mathfrak{p} + \dim(B/\mathfrak{p}) = \dim B\tag{$*$}\label{eq:ratliff}$$ for every prime ideal $\mathfrak{p} \subseteq B$. We also recall that Noetherian domains of dimension $\le 2$ are catenary [Matsumura, Corollary 2 to Theorem 31.7], and so it suffices to consider the case when $\dim A = 3$. Consider a prime ideal $\mathfrak{p} \subseteq A$. If $\operatorname{ht} \mathfrak{p} = 0$, then $\mathfrak{p} = 0$, in which case \eqref{eq:ratliff} trivially holds for $B$ replaced by $A$. Otherwise, suppose that $\operatorname{ht} \mathfrak{p} \ge 1$. Then, there exists a prime ideal $\mathfrak{q} \subseteq \mathfrak{p}$ such that $\operatorname{ht} \mathfrak{q} = 1$ and such that $\operatorname{ht}(\mathfrak{p} \cdot A/\mathfrak{q}) + 1 = \operatorname{ht}\mathfrak{p}$. Since $A$ is a local UFD, the ideal $\mathfrak{q}$ is principal, and we have $\dim(A/\mathfrak{q}) = 2$. We then have $$\operatorname{ht}(\mathfrak{p} \cdot A/\mathfrak{q}) + \dim(A/\mathfrak{p}) = \dim(A/\mathfrak{q}) = 2$$ by Ratliff's criterion, since $A/\mathfrak{q}$ is a Noetherian domain of dimension $2$. But $\operatorname{ht}(\mathfrak{p}\cdot A/\mathfrak{q}) + 1 = \operatorname{ht}\mathfrak{p}$, and hence by adding $1$ to both sides of the equation above, we obtain $$\operatorname{ht}\mathfrak{p} + \dim(A/\mathfrak{p}) = \dim(A/\mathfrak{q}) + 1 = 3 = \dim A.\tag*{$\blacksquare$}$$<|endoftext|> TITLE: Fredholm elements of a Lie algebra QUESTION [5 upvotes]: An element $a$ of a Lie algebra $L$ is called a Fredholm element if the adjoint operator $\mathrm{ad}_a:L \to L$ is a Fredholm linear map. That is: its kernel is a finite-dimensional space and its range $\mathrm{ad}_a(L)$ is a finite-codimensional subspace. Is there an infinite-dimensional Lie algebra with at least one Fredholm element? Is there a Lie algebra $L$ whose only non-Fredholm element is $0$? REPLY [4 votes]: The infinite-dimensional Lie algebra with basis $(e_n)_{n\in\mathbf{N}}$ and brackets $[e_i,e_j]=(i-j)e_{i+j}$, over a field of characteristic zero, satisfies the required condition: every nonzero element is Fredholm. First let me mention it's immediate that $\mathrm{ad}_{e_i}$ is Fredholm for every $i$. Denote by $V_k$ the subspace generated by $(e_i)_{i\le k}$. Let $f$ be a nonzero element, say $f=\sum_{i=0}^ka_ie_i$ with $a_k\neq 0$ and $T=\mathrm{ad}_f$ (let's call $k$ the degree of $f$). Then the kernel of $T$ is contained in $V_k$ (indeed, if $g\notin V_k$, say of degree $\ell$, then $[f,g]$ has degree exactly $k+\ell$). Next computing $[f,e_i]$ for each $i\ge k+1$, we see by induction that $V_i/V_k$ is contained in the image of $p_k\circ T$, where $p_k$ is the projection $L\to L/V_k$. Hence $p_k\circ T$ is surjective. Hence $\mathrm{Im}(T)+V_k=L$, thus $\mathrm{Im}(T)$ is finite-codimensional.<|endoftext|> TITLE: Minimize spectral radius with orthogonal matrix QUESTION [7 upvotes]: Let $A$ be a real square and invertible matrix. I would like to find $$ s(A) = \min_U \rho(U A), $$ Where $U$ is orthogoal, i.e. $U U^T = I$ and $\rho(A)$ is the spectral radius, i.e. the largest eigenvalue of $A$ in absolute values. I am interested in a numerical solution to determine $s(A)$. REPLY [9 votes]: For an $n\times n$ matrix, the answer is $$|\det A|^{\frac1n}.$$ Explanation: on the one hand, $\rho(UA)\ge|\det (UA)|^{\frac1n}=|\det A|^{\frac1n}$. On the other hand, singular value decomposition gives $A=PDQ$ where $P,Q$ are orthogonal and $D>0$ is diagonal. Then $\rho(UA)=\rho(QUPD)$ and $$\min_U\rho(UA)=\min_V\rho(VD)$$ where $V$ runs over the orthogonal group. Take for $V$ the permutation matrix associated with the cycle $(12\ldots n)$. We have $$VD=\begin{pmatrix} 0 & \cdots & \cdots & 0 & s_1 \\ s_2 & \ddots & & & 0 \\ 0 & \ddots & \ddots & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \cdots & 0 & s_n & 0 \end{pmatrix},$$ whose spectrum is made of the $n$-roots of $s_1\cdots s_n=\det D=|\det A|$. Remark that the answer is valid for an arbitrary matrix, singular or not. It is valid for complex matrices too, provided one takes the minimum over unitary matrices $U$.<|endoftext|> TITLE: Sign in May’s General algebraic approach to Steenrod operations QUESTION [5 upvotes]: In the first section of J. P. May’s General algebraic approach to Steenrod operations, May defines for $\pi\subseteq\Sigma_r$ an integer $q\in\mathbb{Z}$ and a commutative ring $\Lambda$, the $\Lambda\pi$-module $\Lambda(q)=\Lambda$ with sign action $\sigma\lambda = (-1)^{qs(\sigma)}\cdot \lambda$ where $(-1)^{s(\sigma)}$ is the sign of $\sigma$. For a $\Lambda$-chain complex $K$ we consider $K^{\otimes r}(q)= K^{\otimes r}\otimes \Lambda(q)$ with the diagonal action. I assume that this translates to the explicit sign rule for the transposition $\sigma_{i,i+1}$ in $K^{\otimes r}(q)$ $$\sigma_{i,i+1}\cdot (a_1\otimes\dotsb\otimes a_r) = (-1)^{q+|a_i|\cdot |a_{i+1}|}\cdot (a_1\otimes\dotsb \otimes a_{i+1}\otimes a_i\otimes\dotsb\otimes a_r).$$ Am I correct with this? He later considers cycles $a,b\in K_q$ and $c\in K_{q+1}$ with $dc=a-b$. Now let $I$ be the cellular chain complex of the intervall, i.e. $I_1=\Lambda\langle e\rangle$ and $I_0=\Lambda \langle e_0,e_1\rangle$, and $de=e_1-e_0$. We consider the chain map of degree $q$ $$f:I\to K, e\mapsto (-1)^qc, e_1\mapsto a, e_2\mapsto b.$$ This satisfies $dfe=(-1)^qfde$ and hence is a chain map. May now claims that $f^{\otimes r}:I^{\otimes r}\to K^{\otimes r}(q)$ is $R\pi$-equivariant. I don’t see why: We have $$f^{\otimes r}(\sigma_{i,i+1}\cdot (a_1\otimes\dotsb\otimes a_r)) = (-1)^{|a_i|\cdot |a_{i+1}|} f(a_1)\otimes\dotsb\otimes f(a_{i+1})\otimes f(a_i)\otimes\dotsb\otimes f(a_r),$$ whereas on the other side, we have $$\sigma_{i,i+1}\cdot f^{\otimes r}(a_1\otimes\dotsb\otimes a_r)=(-1)^{q+(|a_i|+q)\cdot(|a_{i+1}|+q)}\cdot f(a_1)\otimes\dotsb\otimes f(a_{i+1})\otimes f(a_i)\otimes\dotsb\otimes f(a_r).$$ There are equal iff $q\cdot (|a_i|+|a_{i+1}|)$ is even, but this is not necessarily the case. Concretely, if $q=1$ and $\sigma=\sigma_{1,2}$, then $$f^{\otimes 2}(\sigma\cdot e\otimes e_1) = f^{\otimes 2}(e_1\otimes e)=a\otimes (-c)\ne a\otimes c=\sigma\cdot (-c)\otimes a=\sigma\cdot f^{\otimes 2}(e\otimes e_1).$$ What am I missing? REPLY [9 votes]: @FKranhold You mean I got it right? You had me fooled. I should apologize for leaving that detail to the reader, but let me give two excuses. First, one does not actually need that detail to prove Lemma 1.1(iv). It just answers an obvious question the reader might have about the proof. Second, that paper was from the good old days when things went fast. There was a conference, March 30 to April 4, 1970, for Steenrod's 60th birthday. Its Proceedings were submitted June 15, 1970. No time to polish the writing. I remember Steenrod telling me that mine was the only paper in the Proceedings that he understood. He died the next year.<|endoftext|> TITLE: Group cohomology of modular representations for finite groups of Lie type QUESTION [9 upvotes]: $GL_n(\mathbb F_q)$ naturally acts on the vector space $V=\mathbb F_q^n$. As $GL_n(\mathbb F_q)$ is a finite group, the cohomology group $H^i(GL_n(\mathbb F_q),V)$ are all finite abelian groups. Can we compute those cohomology groups explicitly? This is a baby example, and for odd $p$ one can use the trick in Cohomology of SL(2,R) with coefficients given by linear action. In general, Let $G=\mathbb G(\mathbb F_q)$ where $\mathbb G$ is a connected reductive group over $\mathbb F_q$ (or more generally a finite group of Lie type), $V$ be an irreducible algebraic representation of $\mathbb G$ defined over $\mathbb F_{q^n}$ (or more generally any irreducible equal characteristic modular representation), can we compute $H^i(G,V)$ explicitly or at least give some bounds? REPLY [3 votes]: As Derek Holt comments, cohomology has complications even for fimite general linear groups. Probably you are using the term "reductive" too casually and should replace it by "simple" or perhaps "semisimple" to get a finite group of Lie type: for example, an algebraic torus (direct product of copies of the multiplicative group of the field involved) is reductive but does not lead to a group of Lie type. Here "simple" refers to a connected algebraic group with no proper normal connected subgroups except the trivial group. For example, SL$_2$ is simple in this sense and the associated finite group of Lie type has tricky cohomology to compute: see the old paper by Jon Carlson, which has not been improved on much, here . See also the many papers of Cline-Parshall-Scott, which lead to some bounds on dimension of cohomology and some other approaches to computation. A summary and references are given in my LMS Lecture Notes 326 (2006) here .<|endoftext|> TITLE: Representable diagonal map $\Delta: \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ for DM-Stacks/algebraic spaces QUESTION [8 upvotes]: Following the standard definitions of a algebraic space or Deligne–Mumford stack one imposed condition is that the diagonal morphism $\Delta: \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ has to be representable. The latter means that if $X,Y$ are schemes, $h_X,h_Y$ their Yoneda representations and we have natural maps $h_X \to \mathcal{X}, h_Y \to \mathcal{X}$ then the fiber product $h_X \times_{\mathcal{X}} h_Y$ has to be representable in "usual" sense by a scheme. That's a clear formulated technical condition that one can start to verify. Question: What is the intuition/philosophy one should have in mind associating with this condition on representability of the diagonal $\Delta: \mathcal{X} \to \mathcal{X} \times \mathcal{X}$? What makes this condition so "interesting" in order to study stacks? What is the wittness of this condition? Recall then we talking about schemes a scheme $S$ is called separated if the diagonal map $S→S×S$ is a closed immersion. It allows to "transfer" in certain way a "category theoretical Hausdorff axiom" the algebraic geometry, since the classical one fails for Zariski topology. For schemes the property "be separated" is is often used in following sence: Let $f: X\to Y, g:Y \to Z$ morphisms, composition $g \circ f$ has certain property P well behaving under pullbacks and $g$ separated. Then $f$ has also P. Is the the motivation for representability of $\Delta: \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ of similar manner? That is my question is which intuitive property is "transfered" by imposing the representability for the diagonal map to the world of stack? REPLY [6 votes]: The basic issue is the following Theorem. If the diagonal is representable (by schemes, algebraic spaces, etc.), then any morphism $S\rightarrow \mathcal{X}$ with $S$ a scheme, algebraic space, etc. is representable (by schemes, algebraic spaces, etc.). For a proof, see MathOverflow "Diagonal is representable then any morphism is representable". Note, for instance that if the diagonal is affine, then on a scheme the intersection of two open affines is open (which is a kind of weak separability that is often useful in cohomological situations.) A representability condition on the diagonal gives some information on the wildness of the (possible) non-separability of the stack/space considered. Being representable by a closed embedding is as good as it gets (i.e. the Hausdorff condition) but as long as you are willing to consider patchings with bad properties, you need some degree of control of its behavior. So some condition on the diagonal tames this. You might weaken from closed embeddings to affine schemes or to just schemes (this would make the patching scheme-like) or even to algebraic spaces (then the patching would look schematic form the etale toposic point of view). Beyond that, the rough idea is that very complicated behaviors might arise and one should avoid it. Thus the condition.<|endoftext|> TITLE: Corollaries of the halo conjecture that do not involve the eigencurve QUESTION [6 upvotes]: In the theory of p-adic modular forms there is a certain construction called the Coleman-Mazur eigencurve. The spectral halo conjecture roughly states that if you remove a closed subdisc of the weight space, the eigencurve is an infinite disjoint union of finite flat covers of what remains of the weight space. For a person who does not intrinsically care about the eigencurve, what interest does the halo conjecture pose? What corollaries does it have that do not directly involve the eigencurve? REPLY [3 votes]: There was just a paper posted by Newton and Thorne (https://arxiv.org/abs/1912.11261) on automorphy of symmetric powers of eigenforms that might be something you care about. See also a recent blog post here: https://www.galoisrepresentations.com/2019/12/30/new-results-in-modularity-christmas-update-ii/. Of course, you might also not care about that thing. Perhaps someone can answer you more usefully if you describe more carefully what you actually care about?<|endoftext|> TITLE: A binary operation on vector bundles that adds Chern classes? QUESTION [11 upvotes]: Let $E$ and $F$ be two complex vector bundles over a space $X$. There's a fairly well-known binary operation called the direct sum, written $E\oplus F$, which has the property that its first Chern class is the sum of the Chern classes of the constituents: $c_{1}(E\oplus F)=c_{1}E+c_{1}F$. My question is: For $k\geq 1$, are there binary operations $B^{k}(E,F)$ with the property that $c_{k}(B^{k}(E,F))=c_{k}E+c_{k}F$? I am happy to allow the binary operation to take values in virtual bundles if need be. Motivation: These operations would of course be pretty useful for contstructing (virtual) vector bundles with desired Chern classes. REPLY [5 votes]: Such an operation with values in bundles does not exist for $k = 4$ and the base space $\mathbb{HP}^2$. For virtual vector bundles, it depends how you extend the definition of the Chern classes; for the natural extension such that the total Chern class is multiplicative under sums, it also doesn't exist (in fact, this extension is used in the proof). Pullback along the evident inclusion $\mathbb{HP}^n\hookrightarrow \mathbb{HP}^\infty\cong BSU(2)$ gives a ring homomorphism $R(SU(2))\to K^0(\mathbb{HP}^n)$ preserving the augmentation ideals $I$. Since $\mathbb{HP}^n$ is covered by $n+1$ contractible sets (the standard coordinate charts), we have $I^3_{K^0(\mathbb{HP}^n)} = 0$, and therefore an induced morphism $\phi_n:R(SU(2))/I_{R(SU(2))}^{n+1}\to K^0(\mathbb{HP}^n)$. Now the representation ring $R(SU(2))$ is additively generated by the representations $[V_n]$, of dimension $n+1$, subject to the relations $[V_m]\otimes [V_n] = [V_{|m-n|}] + [V_{|m-n|+2}] + \dots + [V_{m+n}]$. Setting $v = [V_1]-2$, it follows easily that $I_{R(SU(2))}^n/I_{R(SU(2))}^{n+1}$ is generated by $v^n$; multiplicativity of the Atiyah-Hirzebruch spectral sequence then shows that $\phi_n$ is an isomorphism. (This calculation is a special case of the Atiyah-Segal completion theorem.) Now $I_{R(SU(2))}/I_{R(SU(2))}^3$ is generated by the classes of $V_1-2$ and $V_2-3$, and their images $E_1-2$ and $E_2-3$ define virtual vector bundles over $\mathbb{HP}^2$ whose classes generate $\widetilde K^0(\mathbb{HP}^2)$. Let's calculate their total Chern classes: $E_1$ is just the canonical $2$-dimensional bundle over $\mathbb{HP}^2$, so its total Chern class is $c(E_1) = 1 + x$ where $x\in H^4(\mathbb{HP}^2;\mathbb Z)$ is a generator. Since $E_2$ is $3$-dimensional, only its second Chern class can be nonzero, and one easily calculates $c(E_2) = 1 + c_2(E_2) = 1 + 4x$: we have $E_2\oplus \mathbb C = E_1^{\otimes 2}$, so pulling back along the map $\iota:\mathbb{CP}^2\to \mathbb{HP}^2$, noting that $\iota^*E_{1} = L\oplus L^{-1}$ splits as a sum of the canonical line bundle and its dual and setting $u = c_1(L)$ we have \begin{align*} \iota^*c(E) &= (1+u)(1-u) = 1-u^2\\ \implies \iota^*x &= -u^2\implies \iota^*|_{H^4}\text{ is injective}\\[12pt] \iota^*E_2 &= (\iota^*E_1)^{\otimes 2} - 1 \\ &= (L\oplus L^{-1})^{\otimes 2} - 1 \\ &= L^{\otimes 2} + L^{\otimes -2} + 1\\ \iota^*c(E_2) &= (1 + 2u)(1 - 2u)\\ &= 1 - 4u^2 \end{align*} The total Chern class can be extended to a natural transformation $K^0(X) \to \bigoplus H^{2*}(X;\mathbb Z)$ by $[E] - [F]\mapsto c(E)c(F)^{-1}$. We immediately obtain \begin{align*} c(k\cdot E_1 + l\cdot E_2) &= (1+x)^k(1+4x)^l \\ &= 1 + (k + 4l)x + \left(\frac{k(k-1)}{2} + 4kl + 8l(l-1)\right)x^2\\ &= 1 + (k+4l)x + \left(\frac{(k+4l)(k+4l-1)}{2} - 6l\right)x^2 \end{align*} In particular, no cohomology class of the form $(3n+2)x^2$ with $n\in\mathbb Z$ can be the fourth Chern class of a (virtual) bundle over $X$, so there is no (virtual) bundle $F$ with $c_4(F) = c_4(E_1^{\oplus 2}) + c_4(E_1^{\oplus 2})$.<|endoftext|> TITLE: Absolutely continuous coupling of probability measures QUESTION [5 upvotes]: I have a Borel probability measure $\pi$ on $\mathbb{R}^{n+1}$ such that $\pi_1=\mu_1, \ldots, \pi_{n+1}=\mu_{n+1}$ for some fixed Borel probability measures $\mu_1, \ldots, \mu_{n+1}$ (where each $\mu_i$ is absolutely continuous with respect to Lebesgue measure). I want to construct a sequence of probability measures $\pi^{(k)}$ such that $\pi^{(k)}$ converges weakly to $\pi$ and has the same marginals as $\pi,$ that is, $\pi^{(k)}_j=\mu_j$ for $j=2,\ldots, n+1.$ Now of course, I can take $\pi^{(k)}=\pi.$ But, I am trying to prove some inequality where I need $\pi^{(k)}$ to be absolutely continuous with respect to Lebesgue measure (on $\mathbb{R}^n$). Now I have a twofold goal: I can make $\pi^{(k)}$ absolutely continuous but then in the process I end up messing with the marginals. Can someone point it out to me if it is possible at all to cook up such a sequence $\pi^{(k)}$ which is absolutely continuous with respect to Leb, and has the fixed marginals and which weakly converges to $\pi.$ REPLY [4 votes]: Let me formulate and prove it in greater generality (which actually makes your question easier). Let $X$ be a metric space, and $\mu$ be a probability measure on $X\times X$ (for simplicity I consider the product of two copies of $X$ only; the general case is precisely the same). You want to obtain a sequence of measures $\nu_n$ on $X\times X$ which (1) have the same marginals $\mu_1,\mu_2$ as $\mu$; (2) are absolutely continuous with respect to the product measure $\mu_1\times\mu_2$; (3) weakly converge to $\mu$. For each $n$ take a countable partition of $X$ into measurable sets $X^i$ with diameter $\le 1/n$ (presuming the space $X$ is such that partitions like this exist for any $n$) and denote by $\mu_\epsilon^i$ the normalized restriction of the measure $\mu_\epsilon$ to $X^i$. Then put $$ \nu_n = \sum_{i,j} \mu(X^i\times X^j) \mu_1^i\times \mu_2^j \;. $$<|endoftext|> TITLE: Usage of étale cohomology in algebraic geometry QUESTION [16 upvotes]: I'm a student interested in arithmetic geometry, and this implies I use étale cohomology a lot. Regarding its definition, étale cohomology is a purely algebro-geometric object. However, almost every material I found on étale cohomology focus on its number-theoretic applications, such as the Weil conjectures and Galois representations. So, this is my question: Are there some applications of étale cohomology on pure algebro-geometric problems? Here, "pure algebro-geometric problems" means some problems of algebraic geometry without number-theoretic flavors, such as birational geometry, the minimal model program, classifying algebraic varieties (curves, surfaces, etc..), especially over algebraically closed fields. Since étale cohomology coincides with singular cohomology over $\mathbb{C}$, there must exist such problems over the complex numbers. Hence, I am looking for applications of étale cohomology which are also useful over algebraically closed fields which are not $\mathbb{C}$. REPLY [5 votes]: The Betti numbers of many (complex) moduli spaces have been computed by counting points over finite fields, using the Weil conjectures, as proved by Deligne, and comparison theorems for étale and singular cohomology. The first example of which I am aware is Lothar Göttsche's calculation of the Betti numbers of Hilbert schemes of points on a smooth projective surface.<|endoftext|> TITLE: Absolutely irreducible representations of affine group schemes of finite type over a field QUESTION [6 upvotes]: Let $G$ be an affine group scheme of finite type over an algebraically closed field $k$. Suppose that $V$ is a finite dimensional representation of $G$. For every $k$-algebra $A$ we have the base change representation $V_A = V\otimes_kA$ of the group $G_A = \mathrm{Spec}\, A\times_{\mathrm{Spec}\,k}G$. Definition 1. $V$ is absolutely irreducible if for every $k$-algebra $A$ there are no nontrivial subspaces $W\nsubseteq V$ such that $W_A = A\otimes_kW$ is $G_A$-stable in $V_A$. Definition 2. $V$ is absolutely irreducible if for every field extension $L$ of $k$ the representation $V_L$ of the group $G_L$ is irreducible. Two definitions above capture the same notion. Indeed, if $W_A$ is $G_A$-stable for some $k$-algebra $A$, then pick a nonzero morphism $A\rightarrow L$ into a field $L$ over $k$. Then $W_L$ is $G_L$-stable. So if there are no stable subspaces in the second sense, there are also no stable subspaces in the first. The other implication is trivial. Fact. Suppose that $G$ is reduced. If $V$ is irreducible, then $V$ is absolutely irreducible. Sketch of proof. Since $G$ is reduced and $k$ is algebraically closed, we deduce that $V$ is an irreducible representation of the abstract group $G(k)$. Suppose that the dimension of $V$ is $n$ and consider a morphism $\rho:G(k)\rightarrow \mathbb{M}_n(k)$ inducing action of $G(k)$ on $V$. Extend $\rho$ to a morphism of $k$-algebras $\tilde{\rho}:k[G(k)]\rightarrow \mathbb{M}_n(k)$. By Jacobson's density theorem we derive that $\tilde{\rho}$ is surjective. This property is stable under base change. This implies that $V_L$ is irreducible over $G_L(L)$ and hence $V_L$ is irreducible over $G_L$. Now my question: Suppose that $G$ is nonreduced. If $V$ is irreducible, then is $V$ absolutely irreducible? REPLY [2 votes]: The classical viewpoint is captured well in the 1962 book Representation Theory of Finite Groups and Associative Algebras by Curtis and Reiner (Wiley), Corollary 29.15. This of course doesn't directly answer the question about non-reduced representations of affine group schemes. But I suspect this generalizes. See for example II, $\S2$ of the 1970 book by Demazure and Gabriel, Groupes Algebriques (Masson, Paris). This book is written in the language of affine group schemes. (By the way, my comment on the question here was being edited when I was interrupted by a phone call and didn't finish the editing.)<|endoftext|> TITLE: $B_{\mathrm{dR}}=B_{\mathrm{cris}}+{B_{\mathrm{dR}}^+}$? QUESTION [6 upvotes]: $B_{\mathrm{cris}}\subseteq B_{\mathrm{dR}}$ and $B_{\mathrm{dR}}^+$ are well-known period rings in $p$-adic Hodge. I know $B_{\mathrm{dR}}=B_{\mathrm{dR}}^+[\frac{1}{t}]$ and $\frac{1}{t}\in B_{\mathrm{cris}}$ where $t$ is Fontaine's $2\pi i$. I want to ask if the natural map $B_{\mathrm{cris}}\rightarrow \frac{B_{\mathrm{dR}}}{B_{\mathrm{dR}}^+}$ induced by the inclusion is surjective. This is to say if $B_{\mathrm{dR}}=B_{\mathrm{cris}}+{B_{\mathrm{dR}}^+}$. For example, if $x\in B_{\mathrm{dR}}^+$, then $\frac{x}{t}\in B_{\mathrm{cris}}+{B_{\mathrm{dR}}^+}$? Thanks! REPLY [9 votes]: Much more is true: the subring $B_{\mathrm{cris}}^{\varphi = 1}$ surjects onto $B_{\mathrm{dR}} / B_{\mathrm{dR}}^+$, so there is an exact sequence $$ 0 \to \mathbf{Q}_p \to B_{\mathrm{cris}}^{\varphi = 1} \to B_{\mathrm{dR}} / B_{\mathrm{dR}}^+ \to 0.$$ This is the Bloch--Kato fundamental exact sequence which is used to construct the Bloch--Kato exponential map; you can read all about it in the paper by Bloch and Kato in the Grothendieck Festschrift.<|endoftext|> TITLE: Limits (growth rates) of power series coefficients QUESTION [6 upvotes]: Take two positive integers $m$ and $n$ and consider the rational function $$G_{m,n}(x,t)=\frac{d}{dx}\left(\frac1{(1-x^m)(1-tx^n)}\right)$$ and the corresponding Taylor expansion as $$G_{m,n}(x,t)=u_0(t)+u_1(t)x+u_2(t)x^2+\cdots$$ where $u_j(t)$ itself is expanded as a polynomial in $t$ (note: coefficients in each $u_j(t)$ are all equal). Now, read-off the coefficients of $G_{m,n}(x,t)$ in the exact order they appear and be listed (including multiplicities) as $\beta_{\ell}(m,n)$. QUESTION. Is this limit true? $$\lim_{\ell\rightarrow\infty}\frac{\beta_{\ell}^2(m,n)}{\ell}=2mn.$$ EXAMPLE 1. If $m=n=1$ then $$G_{1,1}(x,t)=1+t+(2+2t+2t^2)x+(3+3t+3t^2+3t^3)x^2+\cdots$$ and hence $\beta_{\ell}(1,1)$ starts with (keep in mind: $\beta_1=1, \beta_2=1, \beta_3=2$, etc) $$1,1,2,2,2,3,3,3,3,\dots.$$ Hence $$\beta_{\ell}(1,1)=\left\lfloor\frac{\sqrt{8\ell+1}-1}2\right\rfloor \qquad \Longrightarrow \qquad \lim_{\ell\rightarrow\infty}\frac{\beta^2_{\ell}(1,1)}{\ell}=2.$$ EXAMPLE 2. If $m=1$ and $n=2$ then $$G_{1,2}(x,t)=1+(2+2t)x+(3+3t)x^2+(4+4t+4t^2)x^3+\cdots$$ and hence $\beta_{\ell}(1,2)$ starts with (keep in mind: $\beta_1=1, \beta_2=2, \beta_3=2$, etc), $$1,2,2,3,3,4,4,4,\dots.$$ $$\beta_{\ell}(1,2)=\left\lfloor\sqrt{4\ell+1}-1\right\rfloor \qquad \Longrightarrow \qquad \lim_{\ell\rightarrow\infty}\frac{\beta^2_{\ell}(1,2)}{\ell}=4.$$ REPLY [7 votes]: Yes, the conjectured limit is true. Let $d=\gcd (m,n)$ and $m=m_1d, n=n_1d$. Suppose $a_k$ denotes the number of solutions to $k=m_1r+n_1s$ with $r,s\geq 0$, so that $$a_0+a_1x+a_2x^2+\cdots =\frac{1}{(1-x^{m_1})(1-x^{n_1})}.$$ We have $\beta_{\ell}(m,n)=Nd$ if and only if $\sum_{i=0}^{N-1}a_i< \ell\le \sum_{i=0}^{N}a_i$. Notice that $a_i\in \{0,1\}$ for $0\le i\le m_1n_1-1$, and the rest satisfy $a_{mn+i}=1+a_i$.So we have $$\sum_{i=0}^{N-1}a_i\geq \sum_{i=0}^{m_1n_1\lfloor \frac{N-1}{m_1n_1}\rfloor}a_i\geq m_1n_1\binom{\lfloor \frac{N-1}{m_1n_1}\rfloor}{2}=\frac{N^2}{2m_1n_1}+O(N)$$ and similarly $$\sum_{i=0}^{N}a_i\le \sum_{i=0}^{m_1n_1\lceil \frac{N}{m_1n_1}\rceil}a_i\le m_1n_1\binom{1+\lceil \frac{N}{m_1n_1}\rceil}{2}=\frac{N^2}{2m_1n_1}+O(N).$$ So we have $$\lim_{\ell\to \infty} \frac{\ell}{\beta_{\ell}^2(m,n)}=\lim_{N\to \infty}\frac{\frac{N^2}{2m_1n_1}}{N^2d^2}=\frac{1}{2mn}$$ as desired.<|endoftext|> TITLE: A plausible hyperbolic link QUESTION [9 upvotes]: This link is hyperbolic according to SnapPy's computation. There is an obvious non-boundary parallel annulus spanned by two components at the very top in the diagram. If this annulus is essential, then the link is not hyperbolic. Is this annulus inessential in the link exterior? If yes, how can one see it? REPLY [8 votes]: As a complement (sorry) to Josh Howie's answer, here is my Snappy session. In[1]: M = Manifold() Starting the link editor. Select Tools->Send to SnapPy to load the link complement. New triangulation received from PLink! In[2]: M.volume() Out[2]: 7.327724753 In[3]: M.solution_type() Out[3]: 'contains degenerate tetrahedra' Snappy is "guessing" that the manifold is not hyperbolic, but does have a hyperbolic piece in its JSJ decomposition. (There is a lot of accuracy in the volume computation!) As Josh Howie points out, you can actually use Snappy (running under Sage) to prove that the manifold is not hyperbolic, and that that piece is a Borromean rings complement. (The .identify() method is your friend here.) Added later: So I got snappy running under sage. Here is that session. sage: %gui tk sage: M = snappy.Manifold() Starting the link editor. Select Tools->Send to SnapPy to load the link complement. New triangulation received from PLink! sage: M.volume() 7.32772475341777 sage: M.solution_type() 'contains degenerate tetrahedra' sage: M.splitting_surfaces() [Orientable two-sided with euler = 0] sage: M.split(0) [unnamed link.a(0,0)(0,0)(0,0), unnamed link.b(0,0)(0,0)(0,0)] sage: A, B = M.split(0) sage: A.identify() [t12067(0,0)(0,0)(0,0), 6^3_2(0,0)(0,0)(0,0), L6a4(0,0)(0,0)(0,0), ooct02_00005(0,0)(0,0)(0,0)] sage: B.identify() [6^3_3(0,0)(0,0)(0,0)] sage: B.volume() -1.45994327738208e-14 sage: A.volume() 7.32772475341775 and we are done. Snappy identified the two pieces (after splitting) as link complements. Thus the splitting torus is incompressible. So the manifold A (which is the Borromean rings complement) is a geometric piece of the JSJ decomposition. The other piece, a three-times punctured sphere crossed with the circle, is Seifert fibered and so is the other piece of the JSJ decomposition.<|endoftext|> TITLE: Hom spaces in (∞, 1)-categories QUESTION [7 upvotes]: In ordinary category theory it is a well-known and important fact that the $\hom$ bifunctor into $\text{Set}$ preserves limits. I am unable to find a reference for the corresponding fact in infinity category theory nor am I able to write down a proof myself. I am looking for a statement of the following form: Let $\mathcal{C}, \mathcal{D}$ be $(\infty, 1)$-categories (considered as a weak Kan complex preferably) and let $F: \mathcal{D} \to \mathcal{C}$ be a diagram. Then, for any object $X \in \mathcal{C}$, we have a natural equivalence between $\lim \hom(X, F(D))$ and $\hom (X, \lim F(D))$ where each $\hom$ is considered as a weak Kan complex. I would presume that HTT would have a statement to this effect, but I am unable to find one. REPLY [11 votes]: This is explained in the opening paragraph of HTT.5.5.2; it’s a combination of the fact that both the Yoneda embedding and evaluation in functor categories preserve limits.<|endoftext|> TITLE: When does a graph have a circular orientation? Or equivalently can anyone help me characterize this particular class of $3$-colorable perfect graphs? QUESTION [10 upvotes]: Call an oriented digraph $D=(V,A)$ circular when for all $\small x,y,z\in V$ if $(x,y)\in A$ and $(y,z)\in A$ then $(z,x)\in A$ or equivalently if $D$ is any oriented digraph whose arc set is a circular relation. With that said, when does an undirected graph have a circular orientation? I can prove these graphs are perfect i.e. graphs with circular orientations have identical clique and chromatic numbers for all their induced subgraphs, also I can show they are $3$-colorable and there exists a family of forbidden induced subgraphs which characterizes them. Its also easy to see these graphs look similar in their definition to comparability graphs i.e. graphs which have an orientation $D=(V,A)$ such that for all $x,y,z\in V$ if $(x,y)\in A$ and $(y,z)\in A$ then $(x,z)\in A$ (this last arc in the definition of circular orientations is flipped) also like graphs with circular orientations, these (graphs with transitive orientations) are similarly perfect graphs. Now Gallai proved the countable set $S$ below, of forbidden induced subgraph isomorphism types characterized every comparability graph: $$\small S=\{(G_k)^{\complement}:1\leq k\leq 8\}\cup\{B_1^{\complement},B_2^{\complement}\}\cup\bigcup_{n=2}^{\infty}\{C_{2n+1},J_n,J'_{n+1},J''_n,(K_n)^{\complement},(C_{n+4})^{\complement},(L_{n-1})^{\complement},(L'_{n-1})^{\complement}\}$$ Where the indexed types $G,B,K,L,C,J,J',J''$ are each defined diagrammatically as follows: Thus surely characterizing those graphs with circular orientations by a family of forbidden induced subgraphs can not be harder then this? I mean they have bounded clique/chromatic number so I'd expect them not to be that complicated, not like the above graph types. Perhaps even looking for a forbidden induced subgraph characterization is overkill, though there must be some simple way to characterize these graphs, if fifty years ago Tibor Gallai managed to do this for the class of far more complicated, comparability graphs. Right? If so I'd appreciate any help. REPLY [9 votes]: The class is exactly the class of bipartite graphs ∪ complete 3-partite graphs. Such a graph must be paw-free. To check this, note that the orientation of the triangle in the paw must be cyclic, and check the two cases where the remaining edge points to or away from the triangle. According to ISGCI, a graph both paw-free and perfect is a bipartite graph or a complete multipartite graph. As such graphs have chromatic number≤3, they are bipartite graphs or complete 3-partite graphs. And it's easy to check that these two types of graphs are circularly orientable. EDIT: There's a proof that a 3-partite circularly orientable graph is complete. Proof. It's easy to check the case where the graph have 3 vertices. Suppose $H$ is the smallest counterexample. As $H$ is 3-partite, call the three color classes $A$, $B$ and $C$. Without loss of generality, assume $A$ has more that 2 vertices and $a$ is a vertex of $A$, and every vertex of $A\backslash a$ is connected to every vertex of $B$, and that holds for $B$ to $C$ and $C$ to $A\backslash a$. Since the graph is connected, $a$ is connected either to some vertex in $B$ or some vertex in $C$. If $a$ is connected to some vertex in $C$, then by circularity, every vertex in $A\backslash a$ is connected to $a$, which is a contradiction. So $a$ is connected to some vertex in $B$. By circulatity, every vertex in $C$ is connected to $a$, and by the same reasoning, $a$ is connected to every vertex in $B$. Thus $H$ is complete, and is not a counterexample. So there are no counterexamples, which proves the claim.<|endoftext|> TITLE: When is an $\infty$-categorical localization left exact? QUESTION [7 upvotes]: Let $L: \mathcal C^\to_\leftarrow L\mathcal C : i$ be an adjunction with $i$ fully faithful. In ordinary category theory, $L$ is left exact iff the class of $L$-local morphisms is stable under base change [1]. This appears to be true $\infty$-categorically, as well. Is there a proof in the literature? But $\infty$-categorically, this is no longer true: Example: Let $\mathcal C$ be the $\infty$-category of spaces and $L\mathcal C$ the full subcategory of $n$-truncated spaces for some fixed $n$, so that $L$ is the $n$-truncation functor and the $L$-local morphisms are those with $(n+1)$-connected fibers. Then $L$ is not left exact (failing to preserve, for example, the pullback square $K(\mathbb Z, n) \rightrightarrows \ast, \ast \rightrightarrows K(\mathbb Z, n+1)$), but the $L$-local morphisms are stable under base change. Question: Let $L: \mathcal C^\to_\leftarrow L\mathcal C: i$ be an adjunction of finitely-complete $\infty$-categories with $i$ fully faithful. Let $\mathcal W = L^{-1}(\{\textrm{isos}\}) \subseteq \textrm{Mor} \mathcal C$ be the class of $L$-local morphisms. What are necessary and sufficient closure conditions on $\mathcal W$ ensuring that $L$ is left exact? I'm happy to assume that $\mathcal C$ is presentable, or even an $\infty$-topos, and that $\mathcal W$ is of small generation. [1] Here we assume that $\mathcal C$ is finitely complete. A morphism $f$ is said to be $L$-local if $L(f)$ is an isomorphism, and a class of morphisms $\mathcal W$ is stable under base change if $f \in \mathcal W$ implies $f' \in \mathcal W$ where $f'$ is any pullback of $f$ along an arbitrary morphism. REPLY [3 votes]: Unless I misunderstand the statement, this is precisely proposition 6.2.1.1 in Higher Topos Theory.<|endoftext|> TITLE: Is ETCS well-founded? QUESTION [5 upvotes]: I can't find a statement about the axiom of regularity anywhere in treatments of ETCS. Perhaps this is due to the unfortunate clash of terminology with 'foundations'. REPLY [11 votes]: As Noah suggests, the axiom of regularity doesn't make sense in the language of ETCS. What we can say is: To construct a model of ETCS+R from a model of ZFC, one doesn't need regularity. From a model of ETCS+R, one can construct a model of ZFC with regularity, and also separately models of ZFC with regularity replaced by various anti-foundation axioms. The composite ETCS+R -> ZFC -> ETCS+R is the identity, as are the ill-founded versions. And the composite ZFC -> ETCS+R -> ZFC is also the identity, as are ill-founded versions whose anti-foundation axiom is sufficiently strong to characterize possible set-membership diagrams structurally. Among other places, more details can be found in this paper of mine.<|endoftext|> TITLE: Künneth theorem for G-space QUESTION [5 upvotes]: Let $X$ and $Y$ be a right $G$-space and a left $G$ space, respectively, where $G=H \rtimes K$, $H$ a finite group and $K$ a compact Lie group. Moreover, suppose that the $G$-action on $X$ is free. Denote $H_*(-)$ the singular homology with coefficient in a field. Is $H_*(X \times_G Y )$ isomorphic to $H_*(X) \otimes_{H_*(G)} H_*(Y)$? Is it possible to prove it without using spectral sequences? REPLY [13 votes]: This is very untrue. Taking $X=EG$, a contractible space with a free $G$-action, and $Y$ to be a point we have $H_*(X\times_G Y)=H_*(BG)$, which may be non-trivial in arbitrarily high degrees, while $H_*(X)\otimes_{H_*(G)} H_*(Y)$ is the homology of a point.<|endoftext|> TITLE: Why are homeomorphism groups important? QUESTION [10 upvotes]: For a compact metric space $X$ let $\mathcal H(X)$ denote the set of homeomorphisms in the compact-open topology (also generated by sup metric). It is known that $\mathcal H(X)$ is a Polish topological group under the composition operation. Many of the classical results in this area concern the homeomorphism type of $\mathcal H(X)$, for instance: (0) $\mathcal H(2^\omega)\simeq \mathbb R \setminus \mathbb Q$; (1) $\mathcal H_\partial([0,1])\simeq \ell^2$ (here $\mathcal H_\partial$ is the space of homeomorphisms which fix the boundary pointwise); (2) $\mathcal H_\partial([0,1] ^2)\simeq \ell^2$; (3) $\mathcal H_\partial([0,1]^n)$ is a mystery for $3\leq n<\omega$; (4) $\mathcal H([0,1]^\omega)\simeq \ell^2$; (5) $\mathcal H(\text{Sierpinski carpet})$ and $\mathcal H(\text{Menger curve})$ have dimension $1$. It is conjectured that these homeomorphism groups and others are homeomorphic to the $\omega$-power of $\{x\in \ell^2:x_i\notin \mathbb Q\text{ for all }i<\omega\}$; (6) $\mathcal H(\text{Pseudo-arc})$ contains no continuum. It is unknown whether this group is zero-dimensional, connected, or something in-between. (7) It is unknown if there is a compact space $X$ with $1<\dim(\mathcal H(X))<\infty$. Question 1. What are some other major results and questions along these lines? Question 2. I suppose the dimension/connectedness of $\mathcal H(X)$ says something about generalized homotopy between homeomorphisms. But specifically why is knowing the homeomorphism type of $\mathcal H(X)$ important/useful? Also are there applications and/or interpretations of $\mathcal H(X)$ which lead to a greater appreciation? REPLY [3 votes]: Of course the automorphism group of any mathematical object is interesting if the object itself is interesting. But that doesn't single out the homeomorphism group . Let me give some examples. Let $X$ be a space and $H=H(X)$ the homeomorphism group. The homeomorphism group is a space. Knowing its homotopy type is much weaker than knowing its homeomorphism type, but still can give interesting results. Let me give two examples. Fiber bundles over the sphere $S^n$ with fiber $X$ are in 1 to 1 correspondence with elements of $\pi_{n-1}(H)$. To classify the fiber bundles of any reasonable space $Y$, you need to understand all homotopy classes $[Y,BH]$, where $BH$ is the classifying space of $H$. Sometimes it is possible to partially compute these sets. This allows you to construct new bundles, with interesting properties. Here is another nice theorem of Reeb, which depends on the fact that the group $H_\partial(D^n)$ is connected. Let $f:M\rightarrow \mathbb{R}$ be a smooth function on a smooth compact manifold $M$. Suppose that $f$ has only two critical points. Then $M$ is homeomorphic to a sphere. This theorem is false if we want to construct a diffeomorphism. Indeed, the group of diffeomorphisms $\mathrm{Diff}_{\partial}(D^n)$ is not always connected. Milnor famously found the first examples of manifolds that are homeomorphic but not diffeomorphic.<|endoftext|> TITLE: Why are we interested in permutahedra, associahedra, cyclohedra, ...? QUESTION [29 upvotes]: The following families of polytopes have received a lot of attention: permutahedra, associahedra, cyclohedra, ... My question is simple: Why? As I understand, at least the latter two were initially constructed by their face lattice representing certain combinatorial objects (e.g. ways to insert parentheses into a string). So I assumed that representing these structures as a face lattice was of some use. But then people got interested in realizing these objects geometrically, and it turns out that, e.g. the associahedron can be realized in many ways. Was this surprising? Is there something to be learned from that fact? On the other hand, for the permutahedron the realization came probably first, so is there anything deep to learn from its combinatorial structure? Further, there seem to exist connections to algebra, e.g. homotopy theory. I cannot wrap my head around these connections. For me, these polytopes are just further examples of polytopes, nothing else. So what's up? Do they have some extremal properties? Are they especially symmetric (i.e. are they interesting for their symmetries)? Does the geometric point of view make apparent some hidden combinatorial properties of the underlying structures (e.g. the cyclohedron is said to be "useful in studying knot invariants")? What justifies this interest? REPLY [3 votes]: If I were an orthodox Bourbakist, I would say the polytopes per se are of no interest, but I'm not. The geometric enumerative combinatorics of the faces of the permuto/permutahedra and the associahedra respectively reflect the algebra of multiplicative and compositional inversion of formal power series, and I find that alone fascinating. But, how useful are the geometric realizations of the combinatorics of these relations? First consider the algebraic equation of a circle and the geometric rep. Through which rep. is it easier to discern the possible number of intersections of two circles of different radii, not given their locations in a plane? Now consider two particles traveling on two circles, one on each. The geometry of the intersections tells us a lot about the possible trajectories of collisions of the particles and would guide us in developing a program to calculate and depict them. Ramp up your knowledge of algebra, analysis, geometry, and physics, and now you are prepared to understand "Combinatorics and Topology of Kawai–Lewellen–Tye Relations" by Sebastian Mizera. The abstract: We revisit the relations between open and closed string scattering amplitudes discovered by Kawai, Lewellen, and Tye (KLT). We show that they emerge from the underlying algebro-topological identities known as the twisted period relations. In order to do so, we formulate tree-level string theory amplitudes in the language of twisted de Rham theory. There, open string amplitudes are understood as pairings between twisted cycles and cocycles. Similarly, closed string amplitudes are given as a pairing between two twisted cocycles. Finally, objects relating the two types of string amplitudes are the $\alpha_0$-corrected bi-adjoint scalar amplitudes recently defined by the author 1. We show that they naturally arise as intersection numbers of twisted cycles. In this work we focus on the combinatorial and topological description of twisted cycles relevant for string theory amplitudes. In this setting, each twisted cycle is a polytope, known in combinatorics as the associahedron, together with an additional structure encoding monodromy properties of string integrals. In fact, this additional structure is given by higher-dimensional generalizations of the Pochhammer contour. An open string amplitude is then computed as an integral of a logarithmic form over an associahedron. We show that the inverse of the KLT kernel can be calculated from the knowledge of how pairs of associahedra intersect one another in the moduli space. In the field theory limit, contributions from these intersections localize to vertices of the associahedra, giving rise to the bi-adjoint scalar partial amplitudes. (For more on associahedra and scattering amplitudes in quantum physics, just google the terms. See also this MO-Q and the attendant comments and links.) Conceivably, the permutohedra might prove useful since they can be deformed into associahedra. Analogously, conic sections other than the circle are important in understanding gravity and collisions of asteroids with planets as well as SL2 transformations. Good to have diverse reps shedding different light on mathematical and physical relations, in general. For connections to extremal properties, see the Aquiar and Ardila ref in https://oeis.org/A133314 for permutahedra (also for multiplicative inversion of power series, surjective mappings, the inversion of modified Pascal matrices and other matrices derived from infinitesimal generators, lattice paths, the formalism of Appell polynomials related to Weyl algebras, quantum field theory) and https://oeis.org/A133437 for the associahedra (compositional inversion, quantum physics, lattice paths, and a slew of connections to other diverse geometric / combinatorial structures).<|endoftext|> TITLE: Is there a version of algebraic de Rham cohomology that can be used to calculate torsion classes? QUESTION [13 upvotes]: Much work has gone into the construction of cohomology theories which are defined on algebraic varieties (étale, crystalline, etc.) and comparison isomorphisms between them. Say $X$ is an algebraic variety over $\mathbb Z$. I am interested in computing $H^*_{\rm sing}(X_{\mathbb C}, \mathbb{Z}_p)$ using a variant of the de Rham complex. However, the obvious integral version of de Rham cohomology, gives the completely wrong answer even for $X = \mathbb A^1$. This is surprising to me, because (if I am reading the literature correctly, as a non-expert) the de Rham complex of $X_{\mathbb Z_p}$ computes the crystalline cohomology of $X_{\mathbb F_p}$. And crystalline cohomology is often claimed to be the "correct" $p$-adic replacement for the etale cohomology groups $H^*_{et}(X_{\overline{\mathbb F_p}}, \mathbb Z_l)$. However, it seems that crystalline cohomology/de Rham cohomology groups of affine space are "wrong", in the moral sense of not lining up with what I would naively expect. Is there a correction/explanation for this discrepancy? Is there a variant of de Rham/crystalline cohomology which computes the "right" cohomology for $\mathbb A^1$? Morally, is there a reason why we would want integral crystalline cohomology to carry so much extraneous torsion? Is there something I'm missing? REPLY [9 votes]: You should read the introduction to Bhargav Bhatt's lecture notes on prismatic cohomology: available here. This is a new cohomology theory introduced by Bhatt-Scholze (closely related to prior work by Bhatt-Morrow-Scholze) for explaining these torsion phenomena. The presence of extra torsion in de Rham (or crystalline etc.) cohomology is unavoidable, as SashaP's answer demonstrates. These cohomology theories are only supposed to be reasonable analogues of singular cohomology (i.e. Weil cohomology theories) when you use torsion free coefficients. However, we can explain where the extra torsion "comes from" (relative to the "true" etale/singular cohomology with e.g. $\mathbf F_p$ coefficients) with a certain long exact sequence. Let $X$ be a reasonable variety over (e.g.) $\mathbf Z$. Then we can consider its etale cohomology $H^i(X_{{\overline{\mathbf{Q}}}}, \mathbf Z_p)$. This agrees as a module with the singular cohomology of $X_{\mathbf{C}}$, so it's the "right answer" (in particular it's $\mathbf A_1$-invariant). On the other hand, we can consider the algebraic de Rham cohomology of $X_{\mathbf Z_p}$ (this is essentially the algebraic de Rham cohomology of $X$ tensored with $\mathbf Z_p$), $H^i_{\mathrm{dR}}(X/\mathbf Z_p)$. This agrees with the crystalline cohomology of the special fiber - in particular it only depends on the special fiber, so we shouldn't expect it to capture the same information as the etale/singular cohomology of the generic fiber. These two $\mathbf Z_p$-modules have the same rank by Fontaine's comparison theorem (at least in the proper case, but if you use some sort of log theory you can extend to the open case), but in general the de Rham cohomology has extra $p$-torsion. To explain this, Bhatt-Scholze construct a third cohomology theory for schemes over $\mathbf Z_p$ (it's essentially a variant of the construction of crystalline cohomology), valued in modules over the ring $\mathbf Z_p[[u]]$. Let's call this $\mathcal{H}(X)$. Then: $\mathcal{H}(X)/u$ is isomorphic to de Rham cohomology of $X_{\mathbf{Z}_p}$ (equivalently, it's isomorphic to the crystalline cohomology of $X_{\mathbf{F}_p}$). $\mathcal{H}(X)[1/u]$ is non-canonically isomorphic to the etale cohomology with $\mathbf Z_p$ coefficients of $X_{{\overline{\mathbf Q}}}$ (equivalently to singular cohomology of $X_{\mathbf C}$), tensored with $\mathbf Z_p((u))$. $\mathcal{H}(X)$ (at least after some scalar extension) can in principle be computed using a "q-de Rham complex" (I'm sweeping some homological algebra consideration, about e.g. exactness of modding out by u, under the rug - see the Bhatt-Scholze paper for precise statements. You really need to use derived categories everywhere.)<|endoftext|> TITLE: Isomorphism of $\mathbb{Z}\ltimes_A \mathbb{Z}^m$ and $\mathbb{Z}\ltimes_B \mathbb{Z}^m$ QUESTION [20 upvotes]: here it's a question that I've posted in MSE but unfortunately got no answers: Let $A$ and $B$ be matrices of finite order with integer coefficients. Let $n\in\mathbb{N}$ and let $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^n$ be the semidirect product, where the action is $\varphi(n)\cdot (m_1,\ldots,m_n)=A^n (m_1,\ldots,m_n)$, and similarly with $B$. It is easy to construct an isomorphism between $G_A$ and $G_B$ if $A$ is conjugate in $\mathrm{GL}(n,\mathbb{Z})$ to $B$ or $B^{-1}$. But, this is also a necessary condition? I mean, does $G_A\cong G_B$ implies $A\cong B$ or $A\cong B^{-1}$ in $\mathrm{GL}(n,\mathbb{Z})$ or is there a counterexample? I've seen at this MSE question that it is true if $A$ and $B$ are hyperbolic, i.e none of their eigenvalues have module 1, but it isn't the case. Thank you very much! REPLY [3 votes]: This is a complement to Johannes Hahn's answer. Corrigendum. In the previous version of this answer, I have made an erroneous claim, allowing $\omega$, the order of $A$ and $B$, to be any positive number. The claim below is valid only if $$\omega \in \{1, 2, 3, 4, 6 \},$$ which is sufficient to address OP's subsequent examples. Following Johannes Hahn's approach, we can prove the following: Claim. Assume that $G_A$ and $G_B$ are isomorphic. Then $\begin{pmatrix} 1 & 0 \\ 0 & A \end{pmatrix}$ is a conjugate of $\begin{pmatrix} 1 & 0 \\ 0 & B \end{pmatrix}$ or $\begin{pmatrix} 1 & 0 \\ 0 & B^{-1} \end{pmatrix}$ in $\text{GL}_{n + 1}(\mathbb{Z})$. In particular $A$ is a conjugate of $B$ or $B^{-1}$ in $\text{GL}_{n}(\mathbb{Q})$. Proof. Let $K_A$ be the centraliser of the derived subgroup $G_A' = [G_A, G_A]$ of $G_A$. It is clearly a characteristic subgroup of $G_A$. Let $C_A$ be the infinite cyclic subgroup of $G_A$ generated by $a \Doteq (1, (0, \dots, 0))$. The conjugation by $a$, or equivalently, the multiplication by $\begin{pmatrix} 1 & 0 \\ 0 & A \end{pmatrix}$ induces a structure of $\mathbb{Z}[C_A]$-module on $K_A$. This structure is almost invariant under isomorphism in the following sense: if $\phi:G_A \rightarrow G_B$ is a group isomorphism, and if we identify $C_A$ with $C_B$ via $a \mapsto b = (1, (0, \dots, 0)) \in G_B$ then $K_A$ is isomorphic to $K_B$ or to $K_{B^{-1}}$ as a $\mathbb{Z}[C]$-module with $C = C_A \simeq C_B$, depending on whether $\phi(a) = bk$ or $b^{-1}k$ for some $k \in K_B$. This is so because conjugation by $bz$ induces a group action on $K_B$ which is independent of $k$. Now the claimed result immediately follows. Thus the pair of modules $\{K_A, K_{A^{-1}}\}$ is a group isomorphism invariant of $G_A$ It turns out to be useful for this example and this one. Addendum. Here are some details on the module $K_A$. An element of $\mathbb{Z}[C_A]$ is a Laurent polynomial with coefficients in $\mathbb{Z}$ of the form $P(a) = \sum_{i = 0}^d c_i a^{e_i}$ where $e_i \in \mathbb{Z}$ for every $i$. The structure of $\mathbb{Z}[C_A]$-module of $K_A$ is defined in the following way: $$P(a) \cdot k = (a^{e_0}k^{c_0}a^{-e_0}) \cdots (a^{e_d}k^{c_d}a^{-e_d})$$ for $k \in K_A$. Assume now that there is an isomorphism $\phi: G_A \rightarrow G_B$. As $\phi$ is surjective and $\phi(K_A) = K_B$, there is $f \in \mathbb{Z}$ coprime with $\omega$ and $z \in \mathbb{Z}^n \triangleleft G_B$, such that $\phi(a) = b^f z$. Since $\omega \in \{1, 2, 3, 4, 6\}$, we infer that $\phi(a) = b^{\epsilon}k'$ for some $\epsilon \in \{\pm 1\}$ and some $k' \in K_B$. Thus $\phi(a^e) = b^{\epsilon e}k''$ where $k'' \in K_B $ depends on $e$, $k$ and $\epsilon$. The image of $P(a) \cdot k$ by $\phi$, after substituting $\phi(a^{e_i})$ with $b^{\epsilon e_i}k_i''$, and after simplification ($K_B$ is Abelian), results in $$(b^{\epsilon e_0}\phi(k)^{c_0}b^{- \epsilon e_0}) \cdots (b^{\epsilon e_d}\phi(k)^{c_d}b^{- \epsilon e_d}) = P(b^{\epsilon}) \cdot \phi(k).$$ Therefore $\phi$ induces an isomorphism of $\mathbb{Z}[C]$-module if $\epsilon = 1$, where $C = C_A \simeq C_B$. Let $e_0 \Doteq (\omega, (0, \dots, 0))$. Let $(e_1, \dots, e_n)$ denote the canonical basis of $\mathbb{Z}^n$ and let $C \in \text{GL}_{n + 1}(\mathbb{Z})$ be the matrix of $\phi$ with respect to $(e_0, e_1, \dots, e_n)$. If $\epsilon = 1$, then the following identity $\phi(a \cdot k) = b \cdot \phi(k)$ holds true and translates into $$C \begin{pmatrix} 1 & 0 \\ 0 & A \end{pmatrix} k = \begin{pmatrix} 1 & 0 \\ 0 & B \end{pmatrix}C k$$ simply because of the way we defined the action of $a$ and $b$ on $K_A$ and $K_B$ respectively. The claimed result on the matrix conjugation follows.<|endoftext|> TITLE: Does any real projective plane incidence theorem follow from axioms? QUESTION [6 upvotes]: Is it known whether any projective geometry statement that holds true in the real projective plane (equivalently, can be deduced from Hilbert axioms) follows from the standard projective axiomatics? By projective geometry statement I mean any statement it terms of incidence of points and lines, that is a statement of the form: Let $I\subset \mathbb{N}\times\mathbb{N}$ be a finite subset and $(i_0,j_0)\in\mathbb{N}\times\mathbb{N}$. For all sequences $(p_i)$ of points and sequences $(l_j)$ of lines, if $p_i\in l_j$, for all $(i,j)\in I$ then $p_{i_0}\in l_{j_0}$. By standard projective axiomatics I mean four incidence axioms plus Desargues, Pappus and Fano. REPLY [3 votes]: This is false. I don't see how to get a counterexample from Andreas Blass's comment (the only uses for the existence of $\sqrt2$ which are obvious to me require a more flexible notion of incidence statement), so I am posting this as an answer although it is probably more complicated than necessary. For fixed prime $p$, one can take a point and line configuration corresponding to the rank 3 Dowling geometry of $\mathbb{Z}/p$, in which a certain equality must occur unless the underlying field has a primitive p-th root of unity. The presence of a root of unity of prime order $p \ge 3$ is not ruled out by these axioms, which can therefore never prove the equality, although it holds over $\mathbb{R}$. Explicitly, what you do is the following: take 3 points and the 3 lines connecting them (think of this as a triangle). On each of these three lines, place $p$ points, labelled $0,1,\ldots,p-1$. Label the lines by $a,b,c$, and denote by $i_a, i_b, i_c$ the point labelled $i$ on the corresponding line. Define a new line on which $i_a, j_b, k_c$ are present whenever $i+j+k = 0 \mod p$. Any realization of this point and line arrangement over a field, in which the corners of the triangle are all distinct, corresponds to a representation of $\mathbb{Z}/p$ in the multiplicative group of the field. Hence if $p\ge 3$, either some two corners coincide or the representation is trivial, so all points $\{i_a\}_i$ are identical, as are all points $\{i_b\}_i$ and all points $\{i_c\}_i$. The many lines connecting various triples of points also imply that if two corners of the triangle coincide, so do all points $i_a, j_b, k_c$ for all $i,j,k$. Hence in $\mathbb{R}$, one has that $1_a$ lies on the line through $0_a, 0_b, 0_c$. This incidence statement is not true over $\mathbb{C}$, hence it is not implied by the axioms given.<|endoftext|> TITLE: The generalized word problem on groups QUESTION [5 upvotes]: Given some group $G$ that is generated by $a$ and $b$, each of which has infinite order, and some free subgroup $N$ generated by $a^k$ and $b^k$, is there any algorithm that tells me if some $x \in G$ is also in $N$? If not generally, what about for $k=2,3,4$? REPLY [10 votes]: The answer is "no". Take the free group of rank 2 $F=$. Take an infinite recursively enumerable non-recursive set $X=\{u_1,u_2,...\}$, a subset of a recursive set $X'$ of words in $a,b$ satisfying the small cancelation condition $C'(1/12)$ and containing neither $a^2$ nor $b^2$. Take another infinite set of words $Y$ in $a^2,b^2$ satisfying $C'(1/12)$, $Y=\{v_1,v_2,...\}$. Consider the factor group $G$ of $F$ by relations $u_1=v_1,u_2=v_2,...$. Then the subgroup $H$ of $G$ generated by the images of squares $a^2,b^2$ is free and the membership problem for $H$ is undecidable. The group $G$ is infinitely presented, the question does not ask for a finitely presented example, which is more difficult and I do not know the answer in that case. If rank can be assumed bigger than 2, the question might be manageable. In the above construction one can replace $a^2, b^2$ by $a^m,b^n$ for any $m,n>1$.<|endoftext|> TITLE: Incorrect information in an old article about the Kervaire invariant QUESTION [35 upvotes]: In the Soviet times there was a famous Encyclopedia of Mathematics. I think it is still familiar to every Russian mathematician maybe except very young ones, and yours truly is in possession of all 5 volumes. Browsing it recently (with no real purpose) I came across a certain peculiarity. In the article "Kervaire invariant" by M.A.Shtan'ko there was a claim that the Kervaire invariant is nontrivial in all dimensions $2^k-2$ for $2\le k\le 7$ - yes, including 126. Which, for what I know, is still an open problem Kervaire invariant: Why dimension 126 especially difficult? . In this article, the credit for the $k=6$ and $k=7$ cases (lumped together) was given to M. Barratt, M. Mahowald, and A. Milgram, but with no actual reference. To be fair, the absence of references is understandable (in the original article, that is) because it was written in 1978 while a complete proof for dimension 62 was only published six years later https://web.math.rochester.edu/people/faculty/doug/otherpapers/barjoma.pdf It is possible that back in 1978 the result was just announced. But, what happened to the 126? And to Milgram? The simplest possible explanation is that a proof for 126 was also announced but later retracted. However this is by no means the only possibility, so I am curious what really happened. Besides, those MO folks who know more about the subject then myself might wonder what the attempted proof was like. After a bit of search I found a reference which may be relevant (hopefully). In "Some remarks on the Kervaire invariant problem from the homotopy point of view" by M.E.Mahowald (1971) there is Theorem 8 attributed to Milgram and after it the following Remark: "It can be shown that $\theta_4^2=0$ and thus Milgram's theorem implies $\theta_6$ exists". If I get it right this indeed means a nontrivial Kervaire invariant in dimension 126, so probably there is a mistake somewhere in this argument. (But even if it is so, damned if I have a clue who has made it: Milgram, Mahowald, or somebody else.) I have to admit that a few things about this story look suspicious. To begin with, A. Milgram died in 1961 so it should probably be R. J. Milgram if any. In the introduction to "The Kervaire invariant of extended power manifolds " J. Jones stated explicitly that the 62 case is solved by Barratt and Mahowald but not published yet while in the higher dimensions the problem is open, in contradiction to what Shtan'ko wrote the same year. In a couple of papers between 1978 and 1981 I spotted references like [Barratt M. G., Mahowald M., The Arf invariant in dimension 62, to appear] but no traces whatsoever of 126 and Milgram. (Besides this article of Mahowald from almost a decade before.) I am at a loss what to make of all this. It would be nice if someone can set it straight - at the very least, I want to know if Shtan'ko made it up. By the way, an English translation of this Encyclopedia article can be found here https://www.encyclopediaofmath.org/index.php/Kervaire_invariant Only, the year is written 1989 instead of 1978 (a second edition, apparently). REPLY [7 votes]: I'll try to give more precise detail soon, but here's my understanding of this history. The 'proof' Peter May mentions is the 'standard mistake' in the subject. Consider elements in a spectral sequence coming from the homotopy exact couple of a tower. If you have a geometric construction that the boundary of $x$ is $y$, and you can show that $y$ is itself null-homotopic, it is tempting to think you have shown that $x$ survives to a non-zero class. However, $x$ is one of the reasons that $y$ is null-homotopic, so you haven't really observed anything about $x$ from knowing only that $y$ is null-homotopic. What you need is that $y$ was already null-homotopic before $x$ got there to kill it. In other words, you need that $y$ is null-homotopic in an appropriately high stage of the tower, not just in the $0^{th}$ term. For an example related to this case, see pp. 38-39 of http://www.rrb.wayne.edu/papers/fin_conj_handout.pdf The mistake Milgram made in "Symmetries and operations" was just a miscalculation in the $\Sigma_4$ extended power of the $30$-sphere, or perhaps of $S^{30} \cup_2 e^{31}$. It did seem like this would give $\theta_6$ in the $126$-stem, given what was then known about $\theta_4$, until the mistake was noticed, and apparently Shtan'ko wrote his report during this burst of enthusiasm. Shtan'ko's 'A. Milgram' was just a mistake. Surely he meant R. J. Milgram'.<|endoftext|> TITLE: Relationship between $q$-Weyl dimension formula and $q$-analog of weight multiplicity? QUESTION [7 upvotes]: $\DeclareMathOperator\dim{dim}$For a dominant (integral) weight $\lambda$ and any (integral) weight $\mu$ of a simple Lie algebra $\mathfrak{g}$, Lusztig's $q$-analog of weight multiplicty $K_{\lambda,\mu}(q)$ is a $q$-analog of the dimension $\dim V^{\lambda}_{\mu}$ of the $\mu$-weight space $V^{\lambda}_{\mu}$ of the highest weight $\mathfrak{g}$-irrep $V^{\lambda}$ with highest weight $\lambda$. There are a few ways to define it, but probably the simplest is to think of it as the Hilbert series associated to a filtration on $V^{\lambda}_{\mu}$ induced by the action of any regular nilpotent element of $\mathfrak{g}$. See for instance Lusztig's original paper "Singularities, character formulas, and a $q$-analog of weight multiplicities" and Joseph, Letzter and Zelikson - "On the Brylinski–Kostant filtration". In Type A, the $K_{\lambda,\mu}(q)$ are called Kostka–Foulkes polynomials and sometimes this name is used in other types too. (Other ways to define $K_{\lambda,\mu}(q)$: via a $q$-analog of Kostant's partition function; in terms of intersection cohomology of Schubert varieties; as certain affine Kazhdan–Lusztig polynomials.) On the other hand, let us define the $q$-dimension of $V^{\lambda}$ to be $\dim_q(V^{\lambda}) \mathrel{:=} \sum_{\mu} (\dim V^{\lambda}_{\mu}) q^{\langle \mu, \rho^{\vee}\rangle}$, where $\rho^{\vee}$ is the dual of the Weyl vector $\rho$, which is the sum of the fundamental weights. I believe the Weyl character formula tells us that $\dim_q(V^{\lambda}) = \prod_{\alpha \in \Phi^+} \frac{[\langle\lambda+\rho,\alpha\rangle]_q}{[\langle\lambda,\alpha\rangle]_q}$, where $\Phi^+$ are the positive roots of $\mathfrak{g}$ and $[m]_q \mathrel{:=} \frac{(q^{1/2}-q^{-1/2})^m}{(q^{1/2}-q^{-1/2})}$. Maybe I'm slightly off here, but something more-or-less like this should be true, and I think the notion of $q$-Weyl dimension formula is anyways an established thing. Question: what is the relationship between the $K_{\lambda,\mu}(q)$ and $\dim_q(V^{\lambda})$? In particular, is there some way to write $\dim_q(V^{\lambda}) = \sum_{\mu} c_{\lambda,\mu}(q) K_{\lambda,\mu}(q)$ for some "simple" coefficients $c_{\lambda,\mu}(q)$? Note that in Type A we have $s_{\lambda}(\mathbf{x}) = \sum_{\text{dominant $\mu$}} K_{\lambda,\mu}(q) P_{\lambda}(\mathbf{x})$, where $s_{\lambda}(\mathbf{x})$ is the Schur polynomial and $P_{\lambda}(\mathbf{x})$ is the Hall–Littlewood polynomial. And the $q$-Weyl dimension is essentially a principal specialization of $s_{\lambda}(\mathbf{x})$, and I believe the corresponding specialization of $P_{\lambda}(\mathbf{x})$ should give something like a $q$-binomial although I didn't fully work it out. My main conceptual difficulty in linking $K_{\lambda,\mu}(q)$ and $\dim_q(V^{\lambda})$ is that, for $\dim_q(V^{\lambda})$, each weight space contributes a single power of $q$, while for the $K_{\lambda,\mu}(q)$ the weight spaces contribute many different powers of $q$. But I still suspect something like what I'm asking for should be true and probably well-known, although I'm having trouble googling for the answer. EDIT: I just noticed that in a comment to this MathOverflow answer of Jim Humphreys, Victor Protsak says "There is an interesting $q$-analogue of the character of a finite-dimensional module that involves the notion of $q$-multiplicity of weight due to Lusztig, and it specializes into a natural $q$-analogue of the Weyl dimension formula." An expansion of Victor's comment would likely answer my question. REPLY [4 votes]: I think that the best reference for what I mention here is Stembridge's notes "Kostka-Foulkes Polynomials of General Type". The picture you describe for type A actually generalizes nicely to all types. Let's denote by $\Lambda$ the weight lattice, by $\Lambda^{+}$ the dominant weights, $W$ the Weyl group, and by $\chi_{\lambda}=\sum_{\mu} \dim(V^{\lambda}_{\mu})e^{\mu}$ the Weyl characters that live in the ring $\mathbb Z[\Lambda]^W$. There are Hall-Littlewood polynomials $P(\lambda,q)$ defined in the notes above, that satisfy the important identity $$\chi_{\lambda}=\sum_{\mu\in \Lambda^+}K_{\lambda, \mu}(q)P(\mu,q) \tag{1}$$ In Lusztig's paper that you refer to, he proves this for one of the definitions of the $K_{\lambda,\mu}$ (Kazhdan-Lusztig for affine Weyl group) and states that conjecturally this is the same if one defines them by taking the q-analog of the Kostant's partition function (conjecture 9.6). This conjecture was proved by Kato in "Spherical Functions and a q-Analogue of Kostant's Weight Multiplicity Formula" using methods from spherical analysis on p-adic groups. A more elementary proof was given by R. K. Brylinski in "Characters and the q-analog of weight multiplicity". In your question you want to take the principal specialization of $(1)$, so that $\chi_{\lambda}$ becomes $\operatorname{dim}_q(V^{\lambda})$. As a result, $c_{\lambda,\mu}(q)$ becomes the principal specialization of the Hall-Littlewood polynomial $P(\mu,q)$. Just like in type A, this is always a nice q-product. In fact, this was one of the motivating examples behind the evaluation conjecture for Macdonald Polynomials (See conjecture 12.10 in Macdonald's "Orthogonal polynomials associated with root systems", and the following discussion in (v) where it is explained that the principal specialization is known for the Hall-Littlewood case). The general principal specialization of Macdonald polynomials was proven later by Cherednik in "Macdonald's Evaluation Conjectures and Difference Fourier Transform".<|endoftext|> TITLE: W H Lin's thesis and Hopf subalgebras of the Steenrod algebra QUESTION [11 upvotes]: If $B$ is a subalgebra of $A$, you can ask whether the $B$-module structure on $B$ can be extended to give an $A$-module structure on $B$. W H Lin, in his 1973 PhD thesis at Northwestern, showed that the only Hopf subalgebras of the mod 2 Steenrod algebra for which this can be done are the algebras $A(n)$ — this is the algebra generated by $\text{Sq}^{2^i}$ for $i\leq n$. Are there any electronic copies of the thesis, or at least this particular proof, available? REPLY [11 votes]: Using @CarloBeenakker's answer, our librarian found an electronic version, produced from the microfilm copy of the original: https://search.proquest.com/docview/302701183 (full text may require accessing through a university library or similar). At a quick glimpse, it looks complete.<|endoftext|> TITLE: Monoidal categories from the projective modules of a ring QUESTION [6 upvotes]: Let $R$ be a not necessarily commutative ring, and denote by $_R\mathrm{lp}_R$ the category of $R$-bimodules, which are finitely generated projective as left modules, with morphism $R$-bimodule maps, denoted by $_R\mathrm{Hom}_R(M,N)$, for any two objects $M,N$. i) Is $_R\mathrm{lp}_R$ a monoidal category? In other words is $M \otimes_R N$ again projective as a left $R$-module? ii) Denote by $_R\mathrm{Hom}(M,R)$ the left $R$-module maps from $M$ to $R$. Is $M^* := _R\mathrm{Hom}(M,R)$, endowed with its usual bimodule structure, projective as a left $R$-module? iii) If $M^*$ is projective, then is it a dual for $M^*$, that is, is $_R\mathrm{lp}_R$ a rigid monoidal category? REPLY [10 votes]: Yes to (i). It's easier to think about if you weaken/generalize. Suppose that $M$ is an $(R,S)$-bimodule and $N$ is a left $S$-module. If $N$ is projective then the $R$-module $M\otimes_SN$ is a summand of a direct sum of copies of $M\otimes_SS\cong M$, so that it is a projective module if $M$ is. No to (ii). $_RHom(M,R)$ gets its left module structure module from the right module structure of $M$. You can have an example where $R$ is a domain and where $M$ as a right module is a torsion module and $_RHom(M,R)$ as a left module is a torsion module. For example, let $R=k[X]$, $k$ a field, take $M$ to be $R$ with the usual left module structure but right module structure given by $f(X)g(X)=f(X)g(0)$. (Bad notation, but I think you get the idea.)<|endoftext|> TITLE: Numerical method for simultaneous computation of eigenvalues of a family of commuting matrices QUESTION [6 upvotes]: I have a problem where I have $n$ commuting matrices $M_1,\dots,M_n$. It is a well-known fact that commuting matrices are simultaneously diagonalizable/triangularizable. I need to find the eigenvalues of these matrices, but I need to know the eigenvalues grouped up by the common eigenspaces. In exact arithmetic, this would be as easy as Schur-factorizing $M_1 = UT_1U^*$, and then computing $T_i = U^*M_iU$. However, in floating-point arithmetic, it is my understanding that U may be computed inaccurately when the eigenspaces are poorly conditioned (i.e. when the eigenvalues are clustered). Is there a stable numerical method for performing this computation? REPLY [4 votes]: For small Hermitian (or real symmetric) matrices, yes, but really this is a hard problem not fully solved. See [1,2] for algorithms. The Cardoso paper [2] looks at the non-commuting case, but in the commuting case should minimize the off diagonal errors with respect to the Frobeneius norm. I don't know about about getting matrices simultaneously into upper triangular form. I would look at papers that cite these two papers. [1] Bunse-Gerstner, Angelika, Ralph Byers, and Volker Mehrmann. "Numerical methods for simultaneous diagonalization." SIAM journal on matrix analysis and applications 14.4 (1993): 927-949. [2] Cardoso, Jean-François, and Antoine Souloumiac. "Jacobi angles for simultaneous diagonalization." SIAM journal on matrix analysis and applications 17.1 (1996): 161-164.<|endoftext|> TITLE: Properness of reductive group actions on smooth varieties QUESTION [6 upvotes]: Suppose that $G$ is a reductive algebraic group acting on a smooth variety $X$, and that the action has finite stabilizers. When is the action of $G$ on $X$ proper? What is an example where the action is not proper? I am aware of a similar statement which is Proposition 0.8 in Mumford's GIT, which says that the action is proper if the geometric quotient $\phi : X \to X / G$ exists and $\phi$ is affine. I would like to know in what situations the action can be assumed proper without this assumption. REPLY [7 votes]: Actions of reductive groups with finite stabilizers on quasi-projective varieties are often not proper. The simplest example I know is given by he action of $\mathrm{PGL}_2$ on the projective space $\mathbb P^4$ of effective divisors of degree $4$ on $\mathbb P^1$. Consider the open subset $X \subseteq \mathbb P^4$ of points whose stabilizer is finite. These are of two types. (1) four distinct points in $\mathbb P^1$, and (2) three distinct points, one of them double. The stabilizer of a point of $X$ of type (1) is well-known to be of type $\mathbb Z/2 \times \mathbb Z/2$, while one of type (2) has $\mathbb Z/2$ as a stabilizer. Thus the generic stabilizer is larger than the stabilizer at a special point: this means that the inertia can not be finite, and the action can not be proper.<|endoftext|> TITLE: Is the representation of $GL_n(\mathcal{O})$ in functions on Grassmannian multiplicity free? QUESTION [9 upvotes]: Let $\mathbb{F}$ be a local non-Archimedean field. Let $\mathcal{O}\subset \mathbb{F}$ be its ring of integers. Let $GL_n(\mathcal{O})$ be the (compact) group of $n\times n$ invertible matrices with entries in $\mathcal{O}$ such the inverse matrix also has entries in $\mathcal{O}$. Let $Gr_{i,n}$ be the Grassmannnian of linear $i$-dimensional subspaces in $\mathbb{F}^n$. Is it true that the natural representation of $GL_n(\mathcal{O})$ in the space of locally constant functions on $Gr_{i,n}$ is multiplicity free? In other words is it true that any irreducible representation of $GL_n(\mathcal{O})$ appears with multiplicity at most 1? A reference would be helpful. Remark. Archimedean analogue of the above statement, i.e. $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$, is true; here $GL_n(\mathcal{O})$ should be replaced by $O(n)$ or $U(n)$ respectively. (That follows from the fact that the Grassmannian is a symmetric space.) REPLY [5 votes]: This may be redundant as a complete answer with references has already been posted. Just in case this is still useful: I think multiplicity one can be proved by the usual Gefland trick: we need to check that endomorphisms of the induced representation is a commutative ring, this follows once we check that the identity automorphism of that algebra is an anti-involution. The endomorphism algebra coincides with the Hecke algebra ${\mathbb{C}}[P(O)\backslash G(O)/P(O)]$ where $P$ is the group of block upper triangular matrices with blocks of sizes $i, \, n-i$. This can also be realized as the space of $G(O)$-invariant functions on pairs of rank $i$ summands in $O^n$. Switching the two elements in the pair induces an anti-involution. Looking at relative positions of two direct summands in $O^n$ one sees that the relative position of a pair $(N,M)$ is the same as the relative position of $(M,N)$, which yields the statement.<|endoftext|> TITLE: Polynomial approximation (in $L^1$ norm) at minimal cost QUESTION [5 upvotes]: Define the cost of a polynomial $\sum_{i=0}^N a_i x^n$ to be $\sum_{i=0}^N |a_i|$. Let $g:[0,1]\to \mathbb{R}$ be a function to be approximated — say, $g(x)=0$ if $0\leq x < e^{-1}$, $g(x)= 1/x$ if $e^{-1}\leq x\leq 1$. (That function comes up in practical contexts.) We are interested in polynomials $P_+$, $P_-$ such that $P_-(x)\leq g(x)\leq P_+(x)$. We define the tightness $\epsilon(P)$ of $P$ to be $\epsilon(P) = \epsilon = \int_0^1 |P(x)-g(x)| dx$. For given $\epsilon>0$ and $N$ what are the polynomials $P_+$, $P_-$ of tightness $\leq \epsilon$ and minimal cost? What are those minimal costs $c_+(\epsilon,N)$, $c_-(\epsilon,N)$? What if we allow the degree $N$ to be arbitrary? In other words, what are $c_-(\epsilon) = \inf_N c_-(\epsilon,N)$ and $c_+(\epsilon) = \inf_N c_+(\epsilon,N)$? Is there a simple way to see the right order of magnitude of $c_+(\epsilon)$ and $c_-(\epsilon)$? As an alternative, we could allow $P_+(x)$, $P_-(x)$ to be a linear combination of fractional powers $x^r$, $r\geq 1$. REPLY [3 votes]: And why is that terribly costly? @AlexandreEremenko will, probably, have a simpler explanation, but this is how I see it if no pen and paper is allowed. Let $\varphi$ be the conformal mapping of the horizontal strip $\{|\Im z|<1\}$ to the unit disk such that $\varphi(0)=0$ and the real axis is mapped to $[-1,1]$ so that $-\infty$ goes to $-1$ and $+\infty$ goes to $1$. Suppose $f$ is an analytic function in the unit disk bounded by $M$ and such that $\int_{[-1,1]}|f-H|\le\varepsilon$ where $H(x)=0$ on $(-\infty,0)$ and $H(x)=1$ on $[0,+\infty)$. Consider $F=(f\circ \varphi)\cdot\varphi'$. We have $F$ analytic and dominated by $M|\varphi'|$ in the strip and $\int_{-\infty}^\infty |F-H\varphi'|\le\varepsilon$, whence for every $y>0$, we must have $\left|\int_{-\infty}^\infty F(x)e^{iyx}dx-\int_{0}^\infty \varphi'(x)e^{iyx}dx\right|\le\varepsilon$. Shifting the contour in the first integral to $i+\mathbb R$ and taking into account that $\int_{-\infty}^{\infty}|\varphi'(i+x)|dx=\pi$ (the top line is mapped to the top half-circle), we see that the first integral is at most $\pi Me^{-y}$. On the other hand, the second one is about $i\frac{\phi'(0)}{y}$ for decent size $y$ (shift the contour to the vertical interval $[0,i]$ followed by $i+[1,\infty)$ and use the trivial version of the "Laplace formula"). Thus, we get $$ \frac cy\le \varepsilon+\pi Me^{-y} $$ Now choose $y=\frac c{2\varepsilon}$, say. Is there another, possibly simpler example? Yes, if you want just an $L^1$ approximation but adjusting it to your requirement that the polynomial should stay on one side of the function is a bit cumbersome exercise, so I'll not bother going into it.<|endoftext|> TITLE: Non real eigenvalues for elliptic equations QUESTION [20 upvotes]: I am looking for an example of a pure second order uniformly elliptic operator $L=\sum_{i,j=1}^da_{ij}(x)D_{ij}$ in a bounded domain $\Omega$ (with Dirichlet boundary conditions, for example) having a non-real eigenvalue in $L^2(\Omega)$. Under the above assumptions, the operator $L$ has a discrete spectrum and the principal eigenvalue is real; however it is non-symmetric. Note that if $a_{ij}=a$ (in particular if $d=1$), then every eigenvalue is real since the measure $d\mu=a^{-1}(x)\, dx$ symmetrizes $L$. Does anybody know such an example? REPLY [18 votes]: Here is a construction. It elaborates from perturbation analysis of eigenvalues. However it starts from the situation of a non-simple eigenvalue. So, let me start with the standard self-adjoint $L_0=-\Delta$. I assume that the Dirichlet problem admits an eigenvalue $\lambda_0$ of multiplicity $2$ exactly. I denote $(u,v)$ an orthonormal basis of the eigenspace. A well-known example is $\lambda_0=5$ for the domain $K=(0,\pi)\times(0,\pi)$ and $$u(x)=\frac2\pi\,\sin x_1\sin2x_2,\quad v=\frac2\pi\,\sin2x_1\sin x_2.$$ Let $M=\sum a_{ij}D_{ij}$ be given with bounded (smooth) functions $a_{ij}$, and form $L_\epsilon=L+\epsilon M$. Perturbation analysis tells us that $L_\epsilon$ admits a stable plane $\Pi_\epsilon$, which depends smoothly upon $|\epsilon|<\!<1$, and $P_0={\rm vec}(u,v)$. In addition, the restriction of $L_\epsilon$ over $\Pi_\epsilon$ is similar to a matrix $C_\epsilon$ such that on the one hand $C_0=\lambda_0I_2$, and on the other hand $$\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}C_\epsilon=\begin{pmatrix} \langle Mu,u\rangle & \langle Mu,v\rangle \\ \langle Mv,u\rangle & \langle Mv,v\rangle \end{pmatrix}=:X_0.$$ Suppose now that $M$ has been chosen so that $$(\dagger)\qquad ( \langle Mu,u\rangle - \langle Mv,v\rangle )^2+4 \langle Mu,v\rangle \langle Mv,u\rangle <0. $$ Then its eigenvalues are complex conjugated. Therefore, for $\epsilon\ne0$ small enough, $C_\epsilon$ will have non-real eigenvalues. Since these are eigenvalues of the Dirichlet problem for $L_\epsilon$, this provides an example. There remains to find coefficients $a_{ij}$ satisfying ($\dagger$). Let me just choose in my example (on the square $K$) $M=aD_{12}$. The functions $uD_{12}u-vD_{12}v$, $uD_{12}v$ and $vD_{12}u$ are linearly independent. Thus there exists a function $a$ such that $$\int_Ka(uD_{12}u-vD_{12}v)\,dx=0,\qquad\int_KauD_{12}v\,dx=1,\qquad\int_KavD_{12}u\,dx=-1.$$ This ends the construction. Edit. It is known that one-dimensional elliptic second-order operators have a real spectrum. Actually the eigenvalues are simple, therefore the construction described above cannot be implemented. Redit. In the construction above, the eigenvalues depend smoothly upon $\epsilon$. There is no branching phenomenon. This happens because the unperturbed operator $L_0$ is self-adjoint, thus diagonalisable.<|endoftext|> TITLE: "Robinson arithmetic" for (some) levels of $L$? QUESTION [7 upvotes]: I'll write "$\mathcal{L}_\alpha$" for the fragment $\mathcal{L}_{\infty,\omega}\cap L_\alpha$. Say that a countable admissible $\alpha$ is Robinsonian if there is some sentence $\varphi\in\mathcal{L}_\alpha$ such that $L_\alpha\models\varphi$ and there is no $T\subseteq\mathcal{L}_\alpha$ which is consistent, complete with respect to $\mathcal{L}_\alpha$, and $\Delta_1$ over $L_\alpha$. Intuitively, such a $\varphi$ is the "$L_\alpha$-analogue" of Robinson arithmetic. By Barwise completeness, if $\alpha$ is a limit of admissibles then the set of satisfiable $\mathcal{L}_\alpha$-sentences is $\Delta_1$ over $L_\alpha$. Hence via a Henkinization argument we have that limits of admissibles are not Robinsonian. On the other hand, $\omega$ is clearly Robinsonian and it's not hard to show that $\omega_1^{CK}$ is Robinsonian as well. My question is: What are the Robinsonian ordinals? I'd love it if the answer were exactly the successor admissibles, but I suspect it isn't; the stumbling point seems to be the non-Gandy ordinals (at a glance I think we do get that every successor admissible of a Gandy ordinal is Robinsonian by generalizing the argument for $\omega_1^{CK}$, but I haven't checked the details). Note that it's not hard to show that for every $\alpha$ which is either admissible or a limit of admissibles, an analogue of Godel's first incompleteness theorem does hold: there is a $\Sigma_1$-over-$L_\alpha$ theory $T\subseteq\mathcal{L}_\alpha$ such that $L_\alpha\models T$ but $T$ has no $\Delta_1$-over-$L_\alpha$ consistent completion with respect to $\mathcal{L}_\alpha$. Moreover, there is a single $\Sigma_1$ formula which describes such a $T$ in every $L_\alpha$ with $\alpha$ pre-admissible. So it's plausible that there are lots of Robinsonian ordinals. REPLY [2 votes]: EDIT: to my chagrin, the notion of "$n$-admissibility" is not what I thought it was! What I wanted was $\Sigma_n$-admissibility. You can find the definition of $n$-admissibles here; they are vastly smaller than their $\Sigma_n$ counterparts, and indeed for each $n$ the least $n$-admissible is less than the least $\Sigma_2$-admissible. Now $n$-admissibility is a rare notion these days and I've seen "$n$-admissible" used for "$\Sigma_n$-admissible" before, but given the relevance of older papers to this topic it's probably a good idea for me to not butcher this distinction. Embarrassingly, I think I was overthinking this: I believe that the Robinsonian admissibles are exactly the successor admissibles. The idea is to lift the following argument for the essential undecidability of $Q$ in the FOL-context to $\mathcal{L}_\alpha$: "If $T\supseteq Q$ is recursive then there is some $\psi$ such that $\psi^N\cap\mathbb{N}=T$ for all $N\models Q$, and if $T\supseteq Q$ is complete and consistent there is some $M\models T$; putting this together we get an $M\models Q$ with $Th(M)$ the standard part of a parameter-freely-definable set in $M$, contradicting (a version of) Tarski's theorem." So suppose $\alpha$ is the next admissible above some admissible $\beta$ ... Below, by "definable$_\eta$" I mean "definable by a parameter-free $\mathcal{L}_\eta$-formula," and "$Th_\eta(K)$" is the parameter-free $\mathcal{L}_\eta$-theory of $K$ - thought of as a subset of $L_\eta$. Note that it does make sense to ask whether a structure satisfies an $\mathcal{L}_\eta$-sentence even when that structure is not in $L_\eta$: $\mathcal{L}_\eta$ is just a sublogic of $\mathcal{L}_{\infty,\omega}$. Also, I'll conflate transitive sets with the corresponding $\{\in\}$-structures and conflate $\mathcal{L}_\alpha$-formulas with sets in $L_\alpha$ in some appropriate way. First, define by recursion a formula $\sigma_s$ assigned to each set $s$ as follows: $$\sigma_s(x): \forall y(y\in x\leftrightarrow\bigvee_{t\in s}\sigma_t(s)).$$ Intuitively, $\sigma_s$ defines $s$ in a parameter-free way. For $s$ a set, let $\theta_s$ be the sentence $\bigwedge_{t\in s\cup\{s\}}\exists!y(\sigma_t(y))$. The point of all this is that if $M\models$ Extensionality + $\theta_s$, then there is a unique embedding of $tc(\{s\})$ as an initial segment of $M$. Now consider the $\mathcal{L}_\alpha$-sentence $(*)$ = "KP + Inf + V=L + $\theta_\beta$." I claim that $(*)$ witnesses the Robinsonian-ness of $L_\alpha$. We observe the following: for every $M\models(*)$ there is a unique end-embedding $l_M: L_\alpha\subseteq_{end}M$, and every element of $im(l_M)$ is definable$_\alpha$ in $M$. The second half of this is trivial given the first half, and the first half combines the initial segment observation from the previous section with the fact that the well-founded part of an admissible set is admissible. That last bit is what I was missing when I was worrying about non-Gandy-ness. I think it's worth elaborating on: First, note that it fails for $\Sigma_2$-admissibility, since by the Gandy Basis Theorem there is a model of $KP2$ with wellfounded part having height $\omega_1^{CK}$. The reason it works for ($\Sigma_1$-)admissibility is the upwards absoluteness of $\Sigma_1$ formulas. Let $M\models KP$ and $N$ be the wellfounded part of $M$. Let $a,\varphi$ be a $\Sigma_1$-Replacement instance in $N$: that is, $\varphi$ is $\Sigma_1$ and for each $b\in a$ there is exactly one $c\in N$ such that $N\models\varphi(b,c)$. Then in $M$ we can apply absoluteness to argue that $a,\hat{\varphi}$ is also a $\Sigma_1$-Replacement instance with the same solution class, where $\hat{\varphi}(x,y)$ is the formula "$\varphi(x,y)$ and no $z$ of rank $ TITLE: Naive point count underestimates the number of mod $p$ points of an elliptic curve for infinitely many primes QUESTION [13 upvotes]: Let $E$ be an elliptic curve over $\mathbb{Z}[1/N]$ where $N$ is some non-zero integer. Can one show that that the integer $n_p-p-1$ (where $n_p$ is the number of points of $E$ mod $p$) is positive for infinitely many primes $p$ (not dividing $N$) without invoking the modularity theorem or the Sato-Tate conjecture? REPLY [4 votes]: I don't think that anyone has any idea how to prove such a theorem without also being able to prove that $E$ is (potentially) modular. For example you can ask the same question for curves of higher genus. For curves of genus $2$, this problem appears to have been open before 2018, and for curves of higher genus it is still open, see the introduction to http://www.math.uchicago.edu/~fcale/papers/CDM.pdf<|endoftext|> TITLE: Phase transition in matrix QUESTION [6 upvotes]: Playing around with Matlab I noticed something very peculiar: Take the symmetric matrix $A \in \mathbb R^{n \times n}$ defined by $$A_{ij}= i \delta_{ij} - \frac{\varepsilon}{\sqrt{i}\sqrt{j}}\,.$$ Here $\delta_{ij}$ is the Kronecker delta. We first note that this matrix is not diagonally dominant if $n$ is large enough. This is because $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \vert A_{i,1}\vert=\infty >\vert A_{1,1} \vert.$ It is obvious that we require $\varepsilon<1$ in order for $A$ to be positive definite, since otherwise $A_{1,1}\le 0.$ However, I noticed that for let's say $\varepsilon=0.1$ one can make the dimension as large as one wants and the matrix remains positive definite. Question: How can one show that $A$ is positive definite independent of the dimension if $\varepsilon$ is sufficiently small but fixed ? REPLY [7 votes]: The claim is true with $\epsilon=\frac6{\pi^2}\,$. To see this, remark that by changing variable $x_i=y_i\sqrt i\,$, this is equivalent to proving that $$\epsilon\left(\left(\frac1{ij}\right)\right)_{1\le i,j}\le I_\infty.$$ The first (infinite) matrix is $V\otimes V$ with $V=(1,\frac12\,,\ldots,\frac1n\,,\ldots)$. It is symmetric, rank-one, with eigenvalues $0$ (infinitely many times) and ${\rm Tr}(V\otimes V)=\frac{\pi^2}6$ (simple).<|endoftext|> TITLE: When is $\mathcal{D}(\mathcal{F}):\mathcal{D}(\mathcal{A})\to \mathcal{D}(\mathcal{B})$ fully faithful? QUESTION [5 upvotes]: Let $\mathcal{A}$ and $\mathcal{B}$ be two abelian categories and let $\mathcal{F}:\mathcal{A}\to \mathcal{B}$ be an additive functor. Assume that $\mathcal{F}$ is exact and let $\mathcal{D}(\mathcal{F}):\mathcal{D}(\mathcal{A})\to \mathcal{D}(\mathcal{B})$ denotes its (right and left) derived functor. Under which assumptions on $\mathcal{F}$ is $\mathcal{D}(\mathcal{F})$ fully faithful? (Surely, $\mathcal{F}$ would be fully faithful, but is it enough?) Many thanks! REPLY [3 votes]: I suspect that it is fully faithful if and only if $\mathcal F$ preserves $Ext^n$-s between objects. The "only if" part is clearly necessary (see below). I am slightly hesitant about the "if" part, especially for the unbounded categories. On the other hand, you can write many explicit sufficient conditions for that for bounded or bounded on one side categories. It is slightly harder for unbounded categories but I can think of the following sufficient condition: every object of $A$ has an injective envelope in $B$, which is also in $A$, and $B$ has finite injective dimension... Having $\mathcal F$ fully faithful is not enough. Let $B=mod({\mathbb C}[X])$ and $A=mod({\mathbb C}[X]/(X))$ or $A=mod({\mathbb C}[X]/(X^2))$. In both cases, the hom-s in the derived categories are different.<|endoftext|> TITLE: Efficient techniques for computing singular locus QUESTION [5 upvotes]: Let $X = \operatorname{Spec} k[x_1,\ldots,x_n]/I$ be an affine variety of dimension $d$. By definition the singular locus of $X$ is the locus of points of $X$ where the rank of the Jacobian matrix is $< n - d$. In theory the singular locus is easy to compute: its ideal is generated by $I$ and all the $(n - d) \times (n - d)$ minors of the Jacobian. In practice, computing the singular locus can be infeasible. For instance, suppose $X \subset \mathbf{A}^{12}$ is a dimension $4$ variety defined by $18$ equations. (Let us assume this is a minimal number of equations for $X$.) The the Jacobian is an $18 \times 12$ matrix, and one must compute all the $8 \times 8$ minors of this matrix, of which there are over $21$ million! Even a computer will struggle greatly with this task. Are there any known algorithms or techniques for getting a handle on the singularities of $X$ when faced with such a computationally difficult situation? REPLY [4 votes]: There's a package in the Macaulay2 build tree and the latest version of M2, FastLinAlg, which can affirmatively check that a variety is nonsingular, or for instance Rn (or R1, if you are trying to verify it is normal). This just computes some of the minors (in a somewhat smart way). This works surprisingly well in the examples I've tried. Frankly, I'd love to try it on your example. If you email me, I can send you an in-development version of the package which probably has some improvements over that version. If you want to verify a particular point is nonsingular of course, just plug in the point into that Jacobian matrix and compute the rank. In characteristic 2, there are some other tricks that might be reasonable in speed related to Kunz' criterion for regularity. Other things I've tried related to this question all seem to boil down to the same determinants unfortunately.<|endoftext|> TITLE: presentability rank of categories of coalgebras QUESTION [14 upvotes]: The following theorem is relatively classical: Theorem: Given an accessible endofunctor, (co)pointed endofunctor or (co)monad $T$ on a locally presentable category $C$, then the category of $T$-(co)algebra is also locally presentable. The proof goes as follows: in each case the category of (co)algebra can be written as a certain weighted bilimits in the category of accessible categories and accessible functors hence it is an accessible category. Moreover it is well know that categories of algebras are complete and categories of co-algebras are co-complete (in both case either limits or colimits are created by the forgetfull functor) so in both case they are locally presentable categories. Unfortunately the argument above give very little control on the presentability rank of the category of (co)algebras. And this what this question is about: can we give a good bound on the presentability rank of the categories of (co)algebras ? In the special case of algebra on a monad it is easy to see explicitly that if $C$ is locally $\lambda$-presentable and $T$ is $\lambda$-accessible then the category of $T$ algebras is locally $\lambda$-presentable, by showing that the free algebra on $\kappa$-presentable objects form a dense subcategory of $\kappa$-presentable objects. This done for example in Bird's Phd thesis (and probably in other places as well). I convinced myself that the following was true: Conjecture: Given $\kappa$ an uncountable regular cardinal. If in the theorem above $C$ is locally $\kappa$-presentable and $T$ is $\kappa$-accessible then the category of $T$-(co)-algebras is locally $\kappa$-presentable. Assuming it is correct, I like would to know if it was proved somewhere, or if some other result of this kind is known (or if on the contrary counter-example where known) or not. I'm stating both the case of algebras and coalgebras, but I am considerably more interested by the case of coalgebras. REPLY [8 votes]: The case of algebras for a monad is discussed explicitly in Gregory Bird's thesis (see theorem 6.9). The case of the categories of algebras for an endofunctor or pointed endofunctor can be deduced from the fact that if $F$ is a (pointed) endofunctor on $C$, then $F$-Alg $\rightarrow C$ obviously satisfies the condition of Beck's monadicity theorem, and the induced monad preserves $\lambda$-filtered colimits if $F$ does. All this works for any regular $\lambda$, even $\omega$. For the case of coalgebras, Jiří Rosický pointed out the key references to me by email: The following theorem is due to Adámek and Porst in On tree coalgebras and coalgebra presentations as their Theorem 4.2. We fix $\lambda$ an uncountable regular cardinal. Theorem: Let $A$ be a $\lambda$-accessible category that admits colimits of $\omega$-chains, and let $F: A \rightarrow A$ be a $\lambda$-accessible endofunctor. Then: The category of $F$-coalgebra is $\lambda$-accessible. A $F$-coalgebra is $\lambda$-presentable if and only if its underlying object is $\lambda$-presentable in $A$. Corollary: If $A$ is a locally $\lambda$-presentable category and $F$ is a $\lambda$-accessible endofunctor on $A$ then the category of $F$-coalgebra is locally $\lambda$-presentable. The corollary follows immediately: as $A$ is cocomplete it has colimits of $\omega$-chains, and the forgetful functor $F$-coalg $\rightarrow A$ create colimits, so $F$-coalg is $\lambda$-accessible and cocomplete, hence $\lambda$-presentable. We can immediately deduce that: Theorem: If $F$ is a $\lambda$-accessible copointed endofunctor or comonad on a locally $\lambda$-presentable $A$, then: The category of $F$-coalgebras is locally $\lambda$-presentable. An $F$-coalgebra is $\lambda$-presentable if and only if its underlying object is $\lambda$-presentable. Indeed, this can be deduced from the corollary above using that (for $\lambda$ an uncountable cardinal) the category of $\lambda$-presentable categories and left adjoint functors between them preserving $\lambda$-presentable objects is closed under $\lambda$-small cat weighted pseudo-limits. The category of $M$-coalgebras for a copointed endofunctor $M$ can be constructed as a full subcategory of the category of $M_0$-coalgebra where $M_0$ is the underlying endofunctor of $M$ as the equifier of $Id,v:U \rightrightarrows U$ where $U:M_0\text{-Coalg} \rightarrow C$ is the forgetful functor, and $v$ is the natural transformation which on each $M_0$-coalgebra $X$ is the composite $X \rightarrow M(X) \rightarrow X$. When $M$ is a comonad this is a bit more complicated as we would like to take the equifier of the two natural transformation $X \rightrightarrows M_0^2(X)= M_0(M_0(X))$ corresponding to the two side of the usual square, but as $F^S$ is not a left adjoint functor we cannot directly conclude using 2-limits of diagrams of left adjoint functors. Instead we consider the category: $$E=\{X \in C, v_1,v_2:X \rightrightarrows M_0^2(X) \}$$ which is the category of coalgebra for the endofunctor: $$ X \mapsto M_0^2(X) \times M_O(X)^2$$ which is indeed $\lambda$-accessible, so $E$ is locally $\lambda$-presentable and its $\lambda$-presentable objects are these whose underlying object $X$ is $\lambda$-presentable. One has a natural functor $M_0$-coalg to $E$ which sends each $M_0$-algebra to the pair of maps $X \rightrightarrows M_0^2 $ corresponding to square defining $M$-algebras and another functor from $M_0^2$-Coalg to $E$ that sends each $f:X \rightarrow M^2_0(X)$ to $(X,f,f)$. taking the (pseudo)pullback of these two functors give us exactly the category of $M_0$-coalgebras compatible with the comultiplication of $M$. Both these functors clearly preserve all colimits and $\lambda$-presentable objects, so by the results mentioned above, this category is locally $\lambda$-presentable. Combining this with the case of copointed endofunctors we obtain the result. I've included this material with a bit more details and other related results in appendix A of this paper. Regarding relaxing the assumption that $\lambda$ is uncountable, Adámek and Porst show in their paper that the endofunctor: $$ \mathcal{P}_f(X) = \{ F \subset X | F \text{ is finite} \} $$ as an endofunctor of the category of sets (with the direct image functoriality) is a counter example to the first theorem in the case $\lambda=\omega$. That is the category of $\mathcal{P}_f$ coalgebra is not finitely accessible. For the case of comonads, there seems to be a counter-example in the comments of the question.<|endoftext|> TITLE: Does $\mathbb{Q}$ embed into a finitely generated solvable group? QUESTION [15 upvotes]: Does $\mathbb{Q}$ embed into a finitely generated solvable group? I've checked that $\mathbb{Q}$ is not a subgroup of any finitely generated metabelian group. I don't know how to show this (or whether it is true) for solvable groups of higher step. REPLY [28 votes]: Yes, it's due to Ph. Hall. It embeds into a f.g. 3-step solvable group. Let $s:\mathbf{Z}\to\mathbf{Q}^*$ be a map (thought as an bi-infinite word) such that every finite sequence of nonzero rational numbers occurs as subword. Define two automorphisms $u,v$ of $\mathbf{Q}^{(\mathbf{Z})}$ (vector space over $\mathbf{Q}$ with basis $(e_n)_{n\in\mathbf{Z}}$) as follows: $u$ is the shift $e_n\mapsto e_{n+1}$, and $v$ is the diagonal map $e_n\mapsto s(n)e_n$. Since $u^nvu^{-n}$ is also diagonal for every $n$ it commutes with $v$, and hence the pair $\langle u,v\rangle$ generates a quotient of the wreath product $\mathbf{Z}\wr\mathbf{Z}$ (actually a copy of it). Moreover, as $\mathbf{Z}[\langle u,v\rangle$]-module, $\mathbf{Q}^{(\mathbf{Z})}$ is readily seen to be simple. Hence $\langle u,v\rangle\ltimes \mathbf{Q}^{(\mathbf{Z})}$ is generated by 3 elements, contains a copy of $\mathbf{Q}$ and is 3-step solvable. (Reference of roughly the same construction: Ph. Hall, On the finiteness of certain soluble groups, Proc. London Math. Soc. (3) 9 (1959).) It was also proved by Neumann-Neumann that every countable $k$-step solvable group embeds into a finitely generated $(k+2)$-step solvable group (reference: B. H. Neumann, H. Neumann. Embedding Theorems for Groups, J. London Math. Soc. 34 (1959), 465-479). Edit: Ph. Hall earlier (Finiteness conditions for soluble groups, Proc. London Math. Soc. (3) 4 (1954)) checked that there exists a 3-step solvable finitely generated group, whose center is free abelian of infinite rank (such a group can be viewed as group of upper triangular $3\times 3$ matrices over $\mathbf{Z}[t^{\pm 1}]$). It is straightforward that every abelian group can be embedded into (the center of) a quotient of the latter group (I guess he was aware of this, and I think I remember that Neumann-Neumann claim to generalize Hall's result but don't have their paper right now).<|endoftext|> TITLE: Characteristic polynomial of an $8 \times 8$ symmetric matrix with indeterminate entries related to octonionic multiplication QUESTION [12 upvotes]: I consider $1,i,j,k,l,m,n,o$ the standard basis of the (complexified if you like) octonions ($\mathbb{O}$ for the octonions). Let $a = x_1.1 +\ldots + x_8.o$, $b = x_9.1+ \ldots + x_{16}.o$ and $c = x_{17}.1+ \ldots +x_{24}.o$, where $x_1, \ldots, x_{24}$ are indeterminates over the base field (say $\mathbb{C}$). I denote by $L_a$ the $8 \times 8$ matrix which represents the left multiplication by $a$ in $\mathbb{O} \simeq \mathbb{C}^8$ and $R_a$ the $8 \times 8$ matrix which represents the right multiplication by $a$. Similar notations for $b$ and $c$. I would like to compute the characteristic polynomial of the symmetric matrix: $$ S = \dfrac{1}{2}(R_a L_b L_c + {}^{t}(R_a L_b L_c)),$$ where ${}^{t} X$ is the transpose of $X$. I have tried with Macaulay2 and this computation seems to be far beyond the possibilities my machine (which is supposed to be a quite powerful portable workstation). A simple reforrmulation of the eigenvalue problem in a well-chosen basis (namely let $\mathbb{H}$ be the quaternionic subalgebra generated by $b$ and $c$, split $\mathbb{O}$ as $\mathbb{H} \bigoplus \mathbb{H}.e$, where $e$ is orthogonal to $\mathbb{H}$ and take a basis adapted to this decomposition) shows that: $$ (T - \mathrm{Re}((bc)\overline{a}))^4 \ \textrm{divides} \ det(S- T.id),$$ where $\mathrm{Re}(z)$ is the real part of $z \in \mathbb{O}$. I put $f(T) = \dfrac{det(S- T.id)}{(T - \mathrm{Re}((bc)\overline{a})^4}$. A vast number of computations over finite fields and specializing the $x_i$ to random values suggests that $f(T)$ is in fact a square, say $f(T) = g(T)^2$, where $g$ is a quadratic polynomial in $T$. I would like to get a closed expression of $g(T)$. May it be a clean formula involving $a,b$ and $c$ or a dirty "in coordinates" polynomial. I would really appreciate any suggestion. I would also be interested in a theoretical argument which shows that $f(T)$ is indeed a square. Thanks a lot! REPLY [10 votes]: Let $a,b,c\in\mathbb{O}$ be octonions and consider the linear map $L:\mathbb{O}\to\mathbb{O}$ defined by $$ L(x) = (b(cx))a = R_aL_bL_c(x). $$ One desires a formula for the characteristic polynomial of $S$, the symmetric part of $L$, i.e., $$ S(x) = \tfrac12\bigl(R_aL_bL_c + {}^t(R_aL_bL_c) \bigr). $$ (I note that the OP seems to have inadvertently omitted the factor of $\tfrac12$ in the formula for $S$; this becomes apparent when one compares the claimed formula for the characteristic polynomial of $S$ when $a = b = c = 1$.) This is equivalent to knowing the symmetric functions of the eigenvalues of the quadratic form $$ Q(x) = L(x)\cdot x = (b(cx))a\cdot x $$ relative to the quadratic form $Q_0(x) = x\cdot x$. First, note that, if any of $a$, $b$ or $c$ vanishes, then, of course, $L$ and $Q$ vanish identically, and all of the eigenvalues of $S$ are equal to zero. Thus, we can assume that none of $a$, $b$, or $c$ vanishes. Then, dividing by $|abc|\not=0$, we can assume that $|a| = |b| = |c| = 1$. In this case, since $L$, being a product of elements of $\mathrm{SO}(8)$, belongs to $\mathrm{SO}(8)$, it follows that $L$ is conjugate in $\mathrm{SO}(8)$ to an element of the maximal torus ${\mathrm{SO}(2)}^4$, i.e., a blocked diagonal matrix where the diagonal elements are the $2$-by-$2$ rotation matrices $R(\theta_i)$ for $1\le i\le 4$. The matrix $S$ (the symmetric part of $L$) is then diagonal with double eigenvalues $\cos(\theta_i)$, and hence the characteristic polynomial of $S$ is $$ p(t) = (t-\cos\theta_1)^2(t-\cos\theta_2)^2(t-\cos\theta_3)^2(t-\cos\theta_4)^2. $$ When $p(t)= t^8 + r_1\,t^7 + r_2\,t^6 + \cdots + r_8$, we have $p(t) = q(t)^2$ where $$ q(t) = t^4 + \tfrac12\,r_1\,t^3 + \tfrac18(4r_2-{r_1}^2)\,t^2 + \cdots $$ (I leave it to the interested reader to work out the formulae for the $t$ and constant coefficients of $q$ as polynomials in $r_1,r_2,r_3,r_4$). Now, by a theorem of Dickson, $b$ and $c$ lie in a quaternion subalgebra $\mathbb{A}\subset\mathbb{O}$. Also, we can write $a = \cos\theta\,a_0 + \sin\theta\, u$ where $a_0\in\mathbb{A}$ is a unit vector, $u\in\mathbb{A}^\perp$ is a unit octonion, and $0\le \theta\le \tfrac12\pi$. [This expression for $a$ will be unique relative to $\mathbb{A}$ if $0<\theta<\tfrac12\pi$.] In what follows, it will be useful to remember that elements of $\mathbb{O}$ satisfy $xy\cdot z = x\cdot z\bar y = y\cdot \bar x z$. Recalling that $\mathrm{Re}(x) = x\cdot\mathbf{1}$ (where $\mathbf{1}\in\mathbb{O}$ is the multiplicative unit), these identities imply that $\mathrm{Re}(xy) = \mathrm{Re}(yx)$ and that $\mathrm{Re}\bigl(a(bc)\bigr) = \mathrm{Re}\bigl((ab)c\bigr)$, so that $\mathrm{Re}(abc)$ is unambiguous, even though $\mathbb{O}$ is not associative. We also have $$ \mathrm{Re}(abc) = \mathrm{Re}(bca) = \mathrm{Re}(cab) = \mathrm{Re}(\bar c\,\bar b\,\bar a), $$ but note that $\mathrm{Re}(abc)\not=\mathrm{Re}(bac)$ in general. While $\mathbb{O} = \mathbb{A}\oplus \mathbb{A} u$ (note the orthogonal direct sum) is not associative, we have the product formula of Cayley and Dickson: $$ (a+b\,u)(c+d\,u) = \bigl( ac - \bar d b\bigr) + (da + b\bar c)\,u $$ for all $a,b,c,d\in\mathbb{A}$. Writing $x\in\mathbb{O}$ as $x = x_0 + x_1\,u$ where $x_i\in\mathbb{A}$ and using the Cayley-Dickson formula several times, we have $$ \begin{aligned} Q(x) &= (b(cx))a\cdot x = b(cx)\cdot x\bar a \\ &= b(c(x_0+x_1\,u))\cdot (x_0+x_1\,u)(\cos\theta\,\overline{a_0} -\sin\theta\,u)\\ &= \bigl(bcx_0 + (x_1cb)\,u\bigr)\cdot\bigl((\cos\theta\,x_0\overline{a_0}+\sin\theta\,x_1) + (\cos\theta\,x_1a_0-\sin\theta\,x_0)\,u\bigr)\\ &=\cos\theta\,bcx_0\cdot x_0\overline{a_0} + \sin\theta(bcx_0\cdot x_1-x_1cb\cdot x_0) + \cos\theta\,x_1cb\cdot x_1a_0\\ &= \cos\theta\,bcx_0\cdot x_0\overline{a_0} + \sin\theta(bcx_0-x_0\bar b\bar c)\cdot x_1 + \cos\theta\,x_1\cdot x_1a_0\bar b\bar c\\ &= \cos\theta\,bcx_0a_0\cdot x_0 + \sin\theta(bcx_0-x_0\bar b\bar c)\cdot x_1 + x_1{\cdot}x_1\,\mathrm{Re}(\cos\theta\,a_0\bar b\bar c) \end{aligned} $$ (Note that I have used $x_1\cdot x_1a_0\bar b\bar c = \overline{x_1}x_1\cdot a_0\bar b\bar c = |x_1|^2 \mathrm{Re}(a_0\bar b\bar c)$. This expression can be further simplified. Since $b$ and $c$ are unit vectors in $\mathbb{A}$, we can write $bc = w$, which implies that $b = w\bar c$ and hence that $\bar b\bar c = c\bar w\bar c$. Making the substitution $x_0 = y_0\bar c$ and $x_1 = y_1\bar c$ then yields $Q_0(x) = |x_0|^2+|x_1|^2 = |y_0|^2+|y_1|^2$ while setting $v = \bar c a_0 c$ yields $$ \begin{aligned} Q(x) &= \cos\theta\,wy_0{\bar c}a_0 \cdot y_0\bar c + \sin\theta\,(wy_0-y_0\bar w)\cdot y_1 + |y_1|^2\,\mathrm{Re}(\cos\theta\,a_0\bar b\bar c)\\ & = \cos\theta\,wy_0v \cdot y_0 + \sin\theta\,(wy_0-y_0\bar w)\cdot y_1 + |y_1|^2\,\mathrm{Re}(\cos\theta\,v\bar w) \end{aligned} $$ Using this reduced form, it is straightforward to compute that the characteristic polynomial of $S$ is $$ \bigl(t-\mathrm{Re}(\cos\theta\,v\bar w)\bigr)^4 \bigl(t^2-2\cos\theta\,\mathrm{Re}(v)\mathrm{Re}(w)\,t + \cos^2\theta\,\mathrm{Re}(v)^2w\bar w - |\mathrm{Im}(w)|^2\bigr)^2, $$ which can also be written as $$ \bigl(t-\mathrm{Re}(\cos\theta\,v\bar w)\bigr)^4 \bigl((t-\cos\theta\,\mathrm{Re}(v)\mathrm{Re}(w))^2 - |\mathrm{Im}(w)|^2(1-\cos^2\theta\,\mathrm{Re}(v)^2)\bigr)^2, $$ so that its roots are $t = \mathrm{Re}(\cos\theta\,v\bar w)$ with multiplicity $4$ and $$ t = \cos\theta\,\mathrm{Re}(v)\mathrm{Re}(w) \pm |\mathrm{Im}(w)|\bigl(1-\cos^2\theta\,\mathrm{Re}(v)^2)^{1/2}, $$ each with multiplicity $2$. Finally, tracing back through the definitions and normalizations, we see that the characteristic polynomial of $S$ can be written in the form $$ \bigl(t-\mathrm{Re}(a\bar b\bar c)\bigr)^4 \bigl(\bigl(t-\mathrm{Re}(a)\,\mathrm{Re}(bc)\bigr)^2-\mathrm{Im}(a)^2\,\mathrm{Im}(bc)^2\bigr)^2. $$ Note on a computation: Since the OP asked, here is a bit of detail of the computation of the characteristic polynomial. We can choose a basis of $\mathbb{A}\simeq\mathbb{H}$ as $(\bf{1},\bf{i},\bf{j},\bf{k})$ in such a way that $w = w_0\,{\bf{1}}+w_1\,\bf{i}$ while $v = v_0\,{\bf{1}}+v_1\,{\bf{i}}+v_2\,{\bf{j}}+v_3\,{\bf{k}}$. Now write out $y_0$ in the orthonormal basis $(\bf{1},\bf{i},\bf{j},\bf{k})$ and $y_1$ in the orthonormal basis $(-\bf{i},\bf{1},\bf{j},\bf{k})$. Set $\lambda_1 = c(v_{{0}}w_{{0}}{-}v_{{1}}w_{{1}})$ and $\lambda_2=c(v_{{0}}w_{{0}}{+}v_{{1}}w_{{1}})$ where $c = \cos\theta$ and $s=\sin\theta$ (to make the matrix below more readable). Then the matrix of the quadratic form $Q$ in this basis is $$ \left( \begin{array}{cccccccc} \lambda_1&0&-cw_{{1}}v_{{3}}&cw_{{1}}v_{{2}}&sw_{{1}}&0&0&0\\ 0&\lambda_1&-cw_{{1}}v_{{2}}&-cw_{{1}}v_{{3}}&0&sw_{{1}}&0&0\\ -cw_{{1}}v_{{3}}&-cw_{{1}}v_{{2}}&\lambda_2&0&0&0&0&0\\ cw_{{1}}v_{{2}}&-cw_{{1}}v_{{3}}&0&\lambda_2&0&0&0&0\\ sw_{{1}}&0&0&0&\lambda_2&0&0&0\\ 0&sw_{{1}}&0&0&0&\lambda_2&0&0\\ 0&0&0&0&0&0&\lambda_2&0\\ 0&0&0&0&0&0&0&\lambda_2 \end{array} \right) $$ Now tell MAPLE to compute the characteristic polynomial of this matrix and tell it to factor the result. (I confess that it's a bit surprising that $\lambda_2$ turns out to be a root of multiplicity $4$ instead of just $2$. I don't have a 'theoretical' understanding of this. Looking at the matrix, you can see that by choosing the basis of $\mathbb{A}$ a bit more carefully, you could arrange that $v_2=0$, and that makes it clear that the characteristic polynomial will be a square, but we knew that already.)<|endoftext|> TITLE: Properties of Mod $\ell^m$ Galois representation associated to modular form QUESTION [5 upvotes]: (Sorry for my poor english..) Let $F(z)\in S_{2k}(SL_2(\mathbb{Z})$) be a newform and $\ell$ be a prime larger than $3$. Let $K$ be a some number field and $v$ be a prime of $K$ over $\ell$. Let $K_v$ be a $v$-adic completion of $K$ and $O_{K,v}$ be a ring of integers of $K_v$. Let $v=(\pi)$. I already know that Serre and Deligne proved that there exists a Galois representation $\rho_{F,v}$ such that \begin{equation} \rho_{F,v} : Gal(\overline{\mathbb{Q}}/\mathbb{Q})\to GL_2(\mathcal{O}_{K,v}) \end{equation} with for primes $p\neq \ell$, \begin{equation} \text{Tr}(\text{Frob}_p)=\lambda_p ,\quad \det(\text{Frob}_p)=p^{2k-1} \end{equation} where $\text{Frob}_p$ is a Frobenius element. Let $\rho_{F,v}^{m}$ be a reduction of $\rho_{F,v}$ modulo $(\pi^{m})$. In other words, \begin{equation} \rho_{F,v}^{m} : Gal(\overline{\mathbb{Q}}/\mathbb{Q})\to GL_2(\mathcal{O}_{K,v}/(\pi^m)) \end{equation} with for primes $p\neq \ell$, \begin{equation} \text{Tr}(\text{Frob}_p)\equiv \lambda_p, \quad \det(\text{Frob}_p)\equiv p^{2k-1} \pmod{(\pi^m)}. \end{equation} Does there exist $N\in \mathbb{N}$ such that for $p_1\equiv p_2 \pmod{N}$ implies that $\rho_{F,v}^m(\text{Frob}_{p_1})\equiv \rho_{F,v}^m(\text{Frob}_{p_2})$? REPLY [4 votes]: Write $L$ for the finite Galois extension of $\mathbb Q$ with Galois group $G_{\mathbb Q}/\operatorname{Ker}\rho_{F,v}^m$. Then $\rho_{F,v}^m(\operatorname{Frob}_p)$ is the identity in $\operatorname{GL}_2(\mathcal O_{K,v}/\pi^m)$ if and only if $\operatorname{Frob}_p$ is the identity in $\operatorname{Gal}(L/\mathbb Q)$ if and only if $p$ splits completely in $L/\mathbb Q$. If an $N$ as in your question exists, then choosing it large enough so as to eliminate at most finitely many exceptions, we can arrange that there exists $a$ such that $p$ splits completely in $L/\mathbb Q$ if $p\equiv a$ modulo $N$. This implies that the extension $L/\mathbb Q$ is abelian, or equivalently that $\rho_{F,v}^m$ has abelian image. For $m$ large enough, this is never true for $\rho_{F,v}^m$. So the $N$ in your question does not exist. The statement that $L/\mathbb Q$ must be abelian if $p\equiv a$ modulo $N$ implies that $p$ splits completely in $L/\mathbb Q$ is closely related to classical questions in class field theory, but is not quite stated in the usual form (for instance, it is much easier to prove that $L/\mathbb Q$ must be abelian if $p\equiv a$ modulo $N$ is equivalent to the statement that $p$ splits completely in $L/\mathbb Q$). Nevertheless, a complete proof can be found for instance at the following MO answer. Why do congruence conditions not suffice to determine which primes split in non-abelian extensions?<|endoftext|> TITLE: Domain of $(-\Delta)^\alpha$, $\alpha \in (0,1)$ QUESTION [5 upvotes]: Is there any simple way to characterize explicitly the domain of fractional powers for a given operator? For example, the domain of Dirichlet Laplacian on a bounded nice domain $\Omega \subset \mathbb{R}^n$: $$(-\Delta)^\alpha, \quad \alpha \in (0,1).$$ I know that there is a characterization using interpolation theory, so one can find $H^{2\alpha}(\Omega)$ or $H^{2\alpha}_0(\Omega)$. But I'm asking if there is a systematic and simple way to do that, so we can use it for more complicated operators, for example matrix valued operators. Any reference which gives the proof of the above characterization or a related topic will be very helpful. REPLY [2 votes]: The problem can be, in general, very hard. A famous example is the Kato's conjecture (Wikipedia link) concerning the domain of the square root of certain elliptic operators, which was solved only in 2001, almost a half-century after Kato posed the question.<|endoftext|> TITLE: Alexandrov's rigidity in higher dimensions QUESTION [7 upvotes]: If $\Phi_1,\Phi_2$ are convex polyhedra in $\mathbb{R}^3$ such that the sets of outer normals to facets coincide, but $\Phi_1$ is not a translate of $\Phi_2$, then there exist two corresponding facets $F_1,F_2$ (with the same outer normal) such that one of them is a translate of a proper subset of another. This is A. D. Alexandrov's theorem, which generalizes the theorem of H. Minkowski which assumes that the areas of corresponding facets are always equal. If I remember well, then in dimensions greater than 3 this is no longer true (while Minkowski theorem holds true in any dimension.) The request is a reference to counterexamples. REPLY [3 votes]: Here is a counterexample in dimension four. Consider positive numbers $x_1,x_2,x_3,x_4\in\Bbb R$ with $x_1 TITLE: When is there a unique perfect group of order $n$? QUESTION [9 upvotes]: For which $n$ is there a unique perfect group of order $n$? Are there infinitely many such $n$? Some guesses for infinite sequences of such $n$: $|\mathrm{PSL}(2,p)|$, $|\mathrm{SL}(2,p)|$, $|A_m|$, $|S_m|$. If you replace "perfect" with "simple", a complete understanding follows from the Classification. Apart from a few small coincidences and one well-understood infinite family, there are no coincidences. See this previous question for instance. But the picture for perfect groups is less clear to me. A complete list of the orders of perfect groups up to $10^6$ is contained in Section 5.4 of the book by Holt and Plesken (Holt, Derek F.; Plesken, Wilhelm, Perfect groups. (1989). ZBL0691.20001.). There seem to be many $n$ for which there is a unique perfect group of order $n$, but there are also many $n$ (e.g. $2^{14} \cdot 60$) for which the number of perfect groups is very large. There are precisely five $n < 10^6$ for which there are both simple and non-simple perfect $G$ of order $n$: 20160: $A_8$, $L_3(4)$, $2.(A_5 \times L_3(2))$ 181440: $A_9$, $A_6 \times L_2(8)$, $L_3(2) \times 3.A_6$ 262080: $L_2(64)$, $2^2.(A_5 \times L_2(13))$ 443520: $M_{22}$, $2^2.(L_3(2)\times L_2(11))$ 604800: $J_2$, $A_5^2 \times L_3(2)$, $2^2.(A_5 \times A_7)$ REPLY [13 votes]: For the case $n = p(p-1)(p+1)$ for $p >3$ a prime, a theorem of M. Herzog (to be found in the article "Finite groups with a large cyclic Sylow subgroup" (MSN) in the conference proceedings "Finite Simple Groups" (Oxford, editors M. Powell and G. Higman, published 1971 by Academic Press (MSN))) answers your question postively. The Theorem of Herzog uses a 1958 Theorem of Brauer and Reynolds ("On a problem of E. Artin", Annals of Mathematics, 68 (MSN)) which probably suffices to cover the cases you are interested in. Note that if $n = p(p-1)(p+1)$ with $p >3$ a prime, and $G$ is a perfect group of order $n$, then $G$ is not $p$-solvable (for if $G$ is $p$-solvable of order $n$, then $G$ either has a factor group of order $p$ or a non-trivial cyclic factor group of order dividing $p-1$)-(later edit: more generally, if $G$ is any $p$-solvable group with a Sylow $p$-subgroup of order $p$, then $G$ is not perfect). The Theorem of Herzog/Brauer–Reynolds proves that a finite group $G$ of order divisible by $p$, but of order less than $p^{3}$ (for a prime $p >3$) which is not $p$-solvable is one of the following groups: $\operatorname{PSL}(2,p)$, $\operatorname{PSL}(2,p-1)$ (for $p >5$ a Fermat prime), $\operatorname{SL}(2,p)$, $\operatorname{PGL}(2,p)$, $\operatorname{PSL}(2,p) \times \mathbb{Z}/2\mathbb{Z}$. The only perfect group of order $n$ on the list is ${\rm SL}(2.p)$ and the only perfect group of order $\frac{n}{2}$ is ${\rm PSL}(2,p)$, showing that ${\rm SL}(2,p)$ and ${\rm PSL}(2,p)$ ( for $p >3$ a prime) are the unique perfect groups of their orders.<|endoftext|> TITLE: Geometric models for 2-gerbes QUESTION [12 upvotes]: One can think of a complex line bundle as a geometric model for an integral cohomology class of degree 2. Similarly, a locally-trivial bundle of $C^*$-algebras with fiber B(H) (the $C^*$-algebra of bounded operators on an infinite-dimensional Hilbert space) can be thought of as a geometric model for an integral cohomology class of degree 3. One can say that such a bundle is a 1-gerbe, while a complex line bundle is a 0-gerbe. Are there similarly nice models for integral cohomology classes of degree 4 (that is, for 2-gerbes)? REPLY [2 votes]: Bundle $n$-gerbes seem to be what you are looking for. Bundle gerbes can be defined w.r.t. an open cover (then it is what John Greenwood wrote), but don't have to. Instead of a cover, any surjective submersion is ok. $k$-fold intersections are then replaced by $k$-fold fibre products. A nice example are basic bundle gerbes over compact Lie groups. For the groups $SU(n)$ and $Sp(n)$, one can construct them using an open cover of the Lie group, with a canonical line bundle over the double intersections. For the other simply-connected Lie groups, one needs to "resolve" the open sets by principal bundles over them, related to certain stabilizer subgroups. The union of their total space then gives a surjective submersion mapping to $G$, over whose 2-fold fibre product then again a canonical line bundle can be defined. A further example are so-called lifting bundle gerbes, which are constructed from the problem of lifting the structure group of a principal bundle along a central extension. In this case, the surjective submersion is the bundle projection. For bundle 2-gerbes, there is the Chern-Simons bundle 2-gerbe. Its surjective submersions is the bundle projection of a principal $G$-bundle. Over its 2-fold fibre product, it has a bundle gerbe, over its three-fold fibre product, a bundle gerbe isomorphism, over its 4-fold fibre product, a 2-isomorphism, and over its 5-fold fibre product, a condition. The Chern-Simons bundle 2-gerbe represents a class in $H^4(M,\mathbb{Z})$, the level of a Chern-Simons theory over $M$. It also carries a canonical connection, whose 3-holonomy is the value of the Chern-Simons theory on closed oriented 3-manifolds. This works for arbitrary Lie groups $G$, with no need to assume simply-connectedness.<|endoftext|> TITLE: Second moment estimates for $\zeta(s)$: different methods? QUESTION [11 upvotes]: What are some different ways to achieve the bound $$\int_0^T \left|\zeta\left(\frac{1}{2} + i t\right)\right|^2 dt = T \log \frac{T}{2 \pi} + (2 \gamma - 1) T + E(T)$$ with an error term $E(T) = O(T^{\alpha+\epsilon})$ with $\alpha=1/2$ or $\alpha=1/3$? Note I am not asking for the best $\alpha$; I am asking for different ways - the simpler the better - with the intention of choosing one of them to make explicit. Methods I know: Proofs based on Atkinson's 1949 formula. This formula yields $\alpha=1/2$ automatically and $\alpha=1/3$ after some smoothing and some work. The problem is that there is an asymptotically tiny error term that seems hard to make explicit without applying the saddle-point method explicitly (twice) - something that I know to be time-consuming and sometimes painful. Balasubramanian (1978). Again, this doesn't look simple. Ingham/Atkinson (1939). Short and relatively simple. Yields $E(T) = O(T^{1/2} \log^2 T)$. I'm also curious about the same question (in the same way) for $$\int_0^T \left|\zeta\left(\sigma + i t\right)\right|^2 dt,$$ where $0\leq \sigma <1$. REPLY [2 votes]: Theorem 7.2(A) of Titchmarsh's 'The Theory of the Riemann zeta function' uses a simplified form of the special case $x=t$ of the approximate functional equation of the Riemann zeta function which holds for $1/2<\sigma\leq 1$, $$ \zeta(s)=\sum_{n1/2$, Theorem 7.2 of the same book gives $$ \int_1^T |\zeta(\sigma+it)|^2 \ dt \sim \zeta(2\sigma) T. $$<|endoftext|> TITLE: Necessary use of large cardinals in mathematics QUESTION [12 upvotes]: There are some statements, whose consistency (or the consistency of their negation) require the existence of large cardinals (in the sense that if the statement (or its negation) is consistent, then it is consistent that there are some large cardinals). I know many of such examples inside set theory. Question. What are examples of statements in mathematics (other than set theory), where it is known we need large cardinals to prove their consistency (or the consistency of their negation)? For each example, giving what kind of large cardinals are sufficient and what kind of large cardinals are necessary is appreciated. Please also give references. REPLY [19 votes]: The dual of an abelian group $A$ is defined to be the group $\text{Hom}(A,\mathbb Z)$ of homomorphisms to the infinite cyclic group. As usual with such dualities, there's a canonical homomorphism from $A$ to its double dual $$ A\to A^{**}:a\mapsto(h\mapsto h(a)). $$ If this is an isomorphism, $A$ is said to be reflexive. Question: Are all free abelian groups reflexive? Answer: Yes if and only if there are no measurable cardinals.<|endoftext|> TITLE: Existence of a finite group with a given decomposition for a tensor square of one irreducible complex representation QUESTION [8 upvotes]: In this post, irrep and dim mean "irreducible complex representation" and "dimension", respectively. It would be helpful (in a problem of monoidal category) to find a finite group $G$ with (at least) two irreps of dim $5$ (denoted $5_1$ and $5_2$) and (at least) two irreps of dim $7$ (denoted $7_1$ and $7_2$) with $$ 5_1 \otimes 5_1 \simeq 1 \oplus 5_1 \oplus 5_2 \oplus 7_1 \oplus 7_2$$Question: Is there such a finite group $G$? Remark: such group should be of order a multiple of $35$ and should admit irreps of dims $5$ and $7$, which is helpful to rule out the following cases with GAP on a laptop: simple, of order less than $10^6$, perfect, of order less than $15120$, general, of order less than $2240$. Let $a_n$ be the smallest order of a group with an irrep of dim $n$ (oeis.org/A220470): $1, 6, 12, 20, 55, 42, 56, 72, 144, 110, 253, 156, 351, 336, 240, 272,\dots$ In particular, $a_5=55$ (given by $C_{11} : C_5$) and $a_7=56$ (given by $C_2^3 : C_7$). Now, I don't even know if there exists a group $G$ with $|G|<55 \times 56=3080$, and with irreps of dims $5$ and $7$. REPLY [11 votes]: I think that there is indeed no such finite group $G$, whether simple or otherwise. Note first that the representation $5_{1}$ can be assumed to be faithful ( for if $K$ is its kernel, then the group $G/K$ has the same property), so from now on, we assume it faithful. Note next that $Z(G) = 1$, since if $5_{1}$ lies over a linear character $\lambda$ of $Z(G)$, then we must have $\lambda^{2} =1 $ since the trivial character occurs in $5_{1} \otimes 5_{1}$. However $\lambda = \lambda^{2}$ since $5_{1}$ also occcurs in $5_{1} \otimes 5_{1}$. Hence $\lambda$ is trivial. Now $ N = O_{5^{\prime}}(G)$ is Abelian by Clifford's Theorem. If $N$ is non-trivial, then it is also non-central, sincee $Z(G) = 1$, and it follwws from Clifford's Theorem that the representation $5_{1}$ is (up to equivalence) monomial. Then $G$ has an Abelian normal subgroup $A$ such that $G/A$ is isomorphic to a subgroup of $S_{5}$. But in that case, $G$ has an Abelian normal Sylow $7$-subgroup, and $G$ has no irreducible character of degree $7$ (by a theorem of Ito, the degree of a complex irreducible character of $G$ divides $[G:A]$ whenever $A \lhd G$ is Abelian). Hence it follows that $N = 1$. More generally, this argument shows that the representation $5_{1}$ is primitive, ie not (equivalent to one) induced from any proper subgroup of $G$. There are several ways to finish from here. One is to invoke Brauer's classification of the finite primitive subgroups of ${\rm GL}(5, \mathbb{C})$ and note that none of these has tthe order of $G/Z(G)$ divisible by $7$. Another is to note that if $O_{5}(G)$ is non-trivial, then it is irreducibly represented by $5_{1}$, in which case $Z(G)$ has order divisible by $5$, a contradiction. Now we are reduced to the case $F(G) = 1$ and we continue until we see that $M = F^{\ast}(G)$ is a finite simple subgroup of ${\rm GL}(5,\mathbb{C})$ of orer divisible by $35$. But by a theorem of Feit, if $G$ is a finite simple subgroup (of order divisible by the prime $p$) of ${\rm GL}(p-2,\mathbb{C})$ for some prime $p$, then $p$ is a Fermat prime and $G = {\rm SL}(2,p-1)$ ( we may apply this with $p = 7$, so obtain a contradiction).<|endoftext|> TITLE: Two equivalent irreducible representations given by integer matrices QUESTION [12 upvotes]: Let $G$ be a finite group, and $\rho_1, \rho_2: G\to GL_n(\mathbb C)$ be two representations. Suppose that $\rho_1$ and $\rho_2$ are equivalent (i.e. conjugate over $\mathbb C$), and suppose that both groups $\rho_1(G)$, $ \rho_2(G)$ belong to $GL(n,\mathbb Z)$. Is it true that these two groups are conjugate in $GL(n,\mathbb Z)$? If not, is this at least true in the case when $G$ is a symmetric group $S_n$ and the representation $\rho$ is irreducible? The motivation for this question is the following: I know that all complex irreducible representations of $S_n$ can be defined over integers. I wonder whether there is somehow a canonical choice. REPLY [5 votes]: An instructive example (for general $G$, not for symmetrc groups) is provided by the case that $G$ is a dihedral group with eight elements. Then $G$ has a unique complex irreducible character $\chi$ which may be expressed as ${\rm Ind}_{U}^{G}(\lambda) $ and ${\rm Ind}_{V}^{G}(\mu)$, where $U$ and $V$ are the two Klein $4$-subgroups of $G$, and $\lambda, \mu$ are non-trivial linear characters of $U,V$ respectively. These representations exhibit $G$ as an absolutely irreducible subgroup of ${\rm GL}(2, \mathbb{Q})$ with all matrix entries in $\mathbb{Z}$ (even in $\{0,1,-1\}$). The two given representations are equivalent over $\mathbb{C}$, but they are not equivalent as integral representations. One way to explain this is via J.A. Green's theory of vertices and sources : both these integral representations are indecomposable on reduction ( mod $2$). One of the reductions has vertex $U$ and one has vertex $V$. Since $U \lhd G$ and $V \lhd G$ , but $U \neq V$, we see that $U$ and $V$ are not $G$-conjugate. Since (by Green's theory) the vertex of an indecomposable module is unique up to conjugacy, these two modules are not isomorphic on reduction (mod $2$), so they are certainly not isomorphic as $\mathbb{Z}G$-modules<|endoftext|> TITLE: Topological mapping class groups of 4-manifolds QUESTION [8 upvotes]: It is a classical result of Quinn that for a simply-connected closed $4$-manifold $X$ the isometries of its intersection form are in one-to-one correspondence with $\pi_0 \text{Homeo}(X)$. (Isotopy of 4-manifolds, 1986) Let $X$ have some simple fundamental group, say $\mathbb{Z}_2$, and let $h\colon X \to X$ be a diffeomorphism which acts trivially on $H_2(X;\mathbb{Z})$. Is $h$ isotopic to the identity? (through $\text{Homeo}(X)$) Edit: let $\tilde{X}$ denote the universal cover of $X$, and let $s \colon \tilde{X} \to \tilde{X}$ be the covering involution so that $\tilde{X}/s \cong X$. One can take a lift $\tilde{h} \colon \tilde{X} \to \tilde{X}$ of $h$ (there are two, but take any of them.) Another reasonable assumption on $h$ is that we want $\tilde{h}$ to be isotopic to either $\text{id}$ or to $s$. Edit: The second question has been removed. REPLY [9 votes]: Let $X = (S^2\times S^2)/\mathbb{Z}_2$ where the $\mathbb{Z}_2$ action is generated by $(x, y) \mapsto (-x, -y)$. Note that $H_2(X; \mathbb{Z}) \cong \mathbb{Z}_2$, so every diffeomorphism acts trivially. Consider the diffeomorphism $f : S^2\times S^2 \to S^2\times S^2$ given by $(x, y) \mapsto (x, -y)$. This descends to a diffeomorphism $g : (S^2\times S^2)/\mathbb{Z}_2 \to (S^2\times S^2)/\mathbb{Z}_2$. Letting $\pi : S^2\times S^2 \to (S^2\times S^2)/\mathbb{Z}_2$ be the universal covering map, we have a commutative diagram $$\require{AMScd} \begin{CD} S^2\times S^2 @>{f}>> S^2\times S^2\\ @V{\pi}VV @VV{\pi}V \\ (S^2\times S^2)/\mathbb{Z}_2 @>{g}>> (S^2\times S^2)/\mathbb{Z}_2 \end{CD}$$ Taking $\pi_2$ of this diagram, we get a commutative diagram of abelian groups $$\require{AMScd} \begin{CD} \mathbb{Z}\oplus\mathbb{Z} @>{f_*}>> \mathbb{Z}\oplus\mathbb{Z}\\ @V{\pi_*}VV @VV{\pi_*}V \\ \mathbb{Z}\oplus\mathbb{Z} @>{g_*}>> \mathbb{Z}\oplus\mathbb{Z} \end{CD}$$ Note that $\pi_* = \operatorname{id}$ as $\pi$ is a covering map, but $f_*$ is given by $(a, b) \mapsto (a, -b)$. By commutativity, the same is true of $g_*$. In particular, $g_* \neq \operatorname{id}$ and therefore $g$ is not homotopic to the identity map. Alternatively, note that $(S^2\times S^2)/\mathbb{Z}$ is orientable. As $\pi\circ f = g\circ\pi$, the maps $f$ and $g$ have the same degree, and it is easy to see that $f$ has degree $-1$. Again we see that $g$ is not homotopic to the identity map. Geometrically, $X = \operatorname{Gr}(2, 4)$, the Grassmannian of unoriented two-planes in $\mathbb{R}^4$, and $S^2\times S^2 = \operatorname{Gr}^+(2, 4)$ the corresponding oriented Grassmannian. The diffeomorphisms $f$, $g$ are the maps given by $P \mapsto P^{\perp}$.<|endoftext|> TITLE: Could a motivic spectrum have a "zeta function"? QUESTION [6 upvotes]: I'm currently learning about zeta functions, so I apologize in advance if this is riddled with nonsense. Suppose you have a sequence $E=(E_0,E_1,...)$ of motivic spaces along with structure maps $s_i:\mathbb{P}^1\wedge E_i\to E_{i+1}$ that induce weak equivalences $\Sigma_{\mathbb{P}^1}E_i\simeq E_{i+1}$, so that $E$ is a motivic spectrum. Is there a sequence of zeta functions $\zeta_E=(\zeta_{E_0},\zeta_{E_1},...)$ that respects the structure maps? There are several pieces of this question that I'm not sure about: If $E_i$ is a projective variety over $\mathbb{F}_q$, then we can assign a local zeta function $\zeta_{E_i}$ to it. Is it possible to do the same for a more general motivic space? If $X$ and $Y$ are motivic spaces that have zeta functions $\zeta_X$ and $\zeta_Y$, is there a zeta function $\zeta_{X\wedge Y}$ for their smash product? If so, is there any way to express $\zeta_{X\wedge Y}$ in terms of $\zeta_X$ and $\zeta_Y$? If $X$ and $Y$ are motivic spaces that have zeta functions $\zeta_X$ and $\zeta_Y$, and if $f:X\to Y$ is a sufficiently nice morphism, is there a way to make sense of $f_*\zeta_X$ (for example, would this be $\zeta_{f(X)}$)? If so, is there a way to relate $f_*\zeta_X$ and $\zeta_Y$? REPLY [9 votes]: Etale cohomology factors through the stable $\mathbb A^1$ homotopy type. So you should simply consider your local zeta factor in terms of traces of frobenius on etale cohomology. Since smashing with $\mathbb P^1$ corresponds to tensoring with $\mathbb Q_l(q)$ the eigenvalues that give you the local factor will be multiplied by $q$ each time. This means that the local factor will undergo the transformation $f(t) \mapsto f(qt)$. In terms of the global zeta function, this should correspond to translating by $1$.<|endoftext|> TITLE: Quadratic reciprocity for three primes? QUESTION [8 upvotes]: The quadratic reciprocity law states that for $p_1\ne p_2$ prime, the product $\left(\frac{p_1}{p_2}\right)\left(\frac{p_2}{p_1}\right)$ takes values $1$ or $-1$ depending on whether $p_1$ and $p_2$ satisfy some set of restrictions mod $4$. Is there a "quadratic reciprocity law for three primes"? I suspect that the answer is negative. Is it true that for any integers $M>0$, $\varepsilon\in\{-1,1\}$, and $r_1,r_2,r_3$ coprime with $M$, there exist primes $p_1\equiv r_1\pmod M$, $p_2\equiv r_2\pmod M$, and $p_3\equiv r_3\pmod M$ such that $$ \left(\frac{p_1}{p_2}\right)\left(\frac{p_2}{p_3}\right)\left(\frac{p_3}{p_1}\right) = \varepsilon ? $$ What about, say, five primes? REPLY [12 votes]: Turns out the details are easy so I worked them out myself :) The statement in yellow is true. Let me assume $4\mid M$. Pick $p_2,p_3$ arbitrary satisfying the congruence modulo $M$ (they exist by Dirichlet). Take any $p_1$ which is congruent to $r_1\pmod M$, congruent to $p_2^{-1}\pmod{p_3}$, and such that $$\left(\frac{p_1}{p_2}\right)=\varepsilon\cdot(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}$$ (which exists by Dirichlet, CRT, and existence of (non)residues modulo $p_2$.) We have $$\left(\frac{p_2}{p_3}\right)\left(\frac{p_3}{p_1}\right)=\left(\frac{p_2}{p_3}\right)\left(\frac{p_1}{p_3}\right)\cdot(-1)^{\frac{p_1-1}{2}\frac{p_3-1}{2}}=\left(\frac{p_1p_2}{p_3}\right)\cdot(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}=\left(\frac{1}{p_3}\right)\cdot(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}=(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}$$ (second inequality follows since $p_1\equiv r_1,p_3\equiv r_3\pmod 4$), so multiplying by $\left(\frac{p_1}{p_2}\right)$ leaves $\varepsilon$.<|endoftext|> TITLE: moving from sphere spectrum to finite spectrum QUESTION [6 upvotes]: I am reading Hatcher's treatment of the Adam's spectral sequence. http://pi.math.cornell.edu/~hatcher/SSAT/SSch2.pdf On page 20, he states "Thus for each $i$ the groups $\pi_i(Z^k)$ are zero for all sufficiently large $k$. The same is true for the groups $\pi^Y_i (Z^k)$ when $Y$ is a finite spectrum, since a map $\Sigma^i Y \to Z^k$ can be homotoped to a constant map one cell at a time if all the groups $\pi_j (Z^k)$ vanish for $j$ less than or equal to the largest dimension of the cells of $\Sigma^i Y$.'' Context: $Z^k$ is a CW spectrum of finite type. Also, $\pi^Y_n(Z) = [ \Sigma^n Y, Z ] = colim_k [ \Sigma^{n} Y_k, Z_k]$, where the $Y_k$ and $Z_k$ are spaces. I don't really understand the details of his statement. So my questions are 1) Can someone explain the details of this more? 2) (I'd prefer) Can we say instead that this holds because we can write $Y$ as a finite limit of sphere spectra, and finite limits and filtered colimits commute? And if so, how does one write $Y$ as a finite limit of sphere spectra?... Does this work exactly the same as if we had spaces? REPLY [8 votes]: Let $\{Y,Z\}$ denote the homotopy group of maps between a spectrum $Y$ and a spectrum $Z$. We are assuming that one has a sequence of spectra $Z^1, Z^2, \dots$ such that for any fixed $i$, $\{S^i,Z^k\} = 0$ for $k>>0$, and we want to show that, if $Y$ is a finite spectrum, then $\{Y,Z^k\} = 0$ for $k>>0$. An easy way to see this is by induction on the number of cells of $Y$, with inductive step as follows: if $Y$ is obtained from $X$ by attaching an $n$--cell, then, by basic homotopy theory, the cofibration sequence $X \rightarrow Y \rightarrow S^n$ induces $\{S^n,Z^k\} \rightarrow \{Y,Z^k\} \rightarrow \{X,Z^k\}$ exact in the middle. By assumption and inductive hypothesis, the first and last of these are 0 for $k>>0$, and thus the so is the middle term. [This is clearer than Hatcher, who is sort of saying the same thing much more awkwardly.]<|endoftext|> TITLE: Generalized Schoenflies - formalizing step in proof? QUESTION [7 upvotes]: [Sorry if the level here is wrong, I asked this on math.SE, but even with a bounty, it got no attention.] I am currently reading Hatcher's 3-Manifolds notes, the part proving Alexander's theorem, which is a specific case of the generalized Schoenflies theorem: Every smoothly embedded $S^2\subset \mathbb{R}^3$ bounds a smooth 3-ball. The proof seems to rely on intuition for these low-dimensional arguments, which I find disconcerting, because I have not yet developed that intuition, so I am trying to give actual formal proofs for the statements in Hatcher's proof. The proof begins with a generic smoothly embedded closed surface $S\subset\mathbb{R}^3$. I have been able to prove that I can isotope $S$ so that projection on the last coordinate $\pi:\mathbb{R}^3\rightarrow\mathbb{R}$ is a Morse function on $S$. Hatcher then argues that if $t$ is a regular value for $\pi$, then $\pi^{-1}(t)\cap S$ is a finite collection of circles. The proof continues by taking an innermost circle $C\subset \pi^{-1}(t)\cap S$, which by 2-dimensional Schoenflies bounds a disk $D$, and $D\cap S=\partial D=C$. Hatcher then uses surgery to cut away a neighborhood of $C$ in $S$, and cap the cuts with two disks. This last part is what I want to formalize. It seems we are finding a small-enough tubular neighborhood $C\times(-\epsilon,\epsilon)\subset S$, and then removing that, leaving $S_-=C\times\{-\epsilon\}$ and $S_+=C\times\{\epsilon\}$. Again by 2-dimensional Schoenflies, these bound disks $D_-$ and $D_+$. What I don't get is: why are $S_-$ and $S_+$ still innermost? Or, put differently, why is $D_-\cap S=S-$ (and similarly for $S_+$)? Intuitively, this seems obvious, and it seems like some sort of "continuity" argument would work, but I cannot figure out how to make this formal. I tried proving that in fact all the disks, "stacked" together for the tubular neighborhood, gave a smooth $D\times [-\epsilon,\epsilon]$, but again I find it hard to make topological arguments when one step of the construction is "apply Schoenflies to get a disk". In particular, I can't prove the projection of this "solid neighborhood" to $D$ is continuous. Does anyone know how to formalize this? Or, even better, a reference where this type of surgery is discussed? I checked a few places, but only found surgery on a single manifold, not the type discussed here, where we're surgering an embedded submanifold in some ambient manifold. REPLY [2 votes]: You have a collection of circles which move smoothly (hence homotopically, see below) with the height (due to regularity and implicit function theorem). Let us define what means "innermost": It means that if you take a point $x$ from the inner circle then every outer circle will have winding number 1 or -1 around $x$. Or if you take a point $x$ from an outer circle then the inner circle will have winding number 0 around $x$. Now simply use the fact that the winding number is invariant under homotopies, that is, if $H\colon[a,b]\times[0,1]\to\mathbb R^2$ and $x\colon[0,1]\to\mathbb R^2$ are continuous with $H(t,0)=H(t,1)$ and such that $x(t)\notin H(\{t\}\times[0,1])$ for all $t\in[a,b]$ then the winding number $n(H(t,\cdot),x(t))$ is independent of $t\in[a,b]$.<|endoftext|> TITLE: Properties of convolutions QUESTION [7 upvotes]: Consider the function $$f_{n}(x)=e^{-x^2}x^n.$$ and the function $$h_p(x):=e^{-\vert x \vert^p}.$$ My goal is to analyze $$ F_p(y):=\frac{(f_2*h_p)(y)}{(f_0*h_p)(y)}- \left(\frac{(f_1*h_p)(y) }{(f_0*h_p)(y)}\right)^2$$ Question: Can we show that $F_p$ has a global maximum at zero for $p>2$ and a global minimum at zero for $p<2$? Here are graphs from Mathematica for $F_1$, with a unique minimum $F_4$, with a unique maximum $F_{2+10^{-4}}$, i.e. with an exponent slightly above $2$ $F_{2-10^{-4}}$, i.e. with an exponent slightly below $2$ Further observations: Iosif Pinelis showed that $F_2(y)$ is constant, see this earlier question of mine question I asked some days ago. All functions $F_p$ are positive by the Cauchy-Schwarz inequality. The functions $F_p$ are not log-convex or log-concave in general. Finally, I computed the first derivative for convolving with $e^{-\vert x \vert^p}$ at $p=2$ by numerically differentiating If you have any other conjectures you would like to me to verify, I am happy to do so. Just leave a comment! REPLY [7 votes]: Global minimum at $0$ for $p<2$ is trivial in the sense that you can see it without writing a single equation or inequality (i.e., a non-trivial formula with two sides and a sign in between comparing them). Observation 1 $e^{-|x|^p}$ is some weighted average of $e^{-ax^2}$ with positive $a$. Observation 2 The thingy you are interested in is just the variance of $x$ with respect to the probability measure $\mu_y$ whose density is proportional to $e^{-x^2}e^{-|x+y|^p}$. By observation 1, this measure is a mixture of the probability measures $\mu_{a,y}$ with densities proportional to $e^{-x^2}e^{-a(x+y)^2}$. The weight of $\mu_{a,y}$ in that mixture is proportional to something independent of $y$ times $e^{-\frac a{a+1}y^2}$, i.e., when we move $y$ away from the origin, the measures $\mu_{a,y}$ with lower $a$ gain more weight in the composition. Observation 3. The variance of $x$ with respect to $\mu_{a,y}$ is independent of $y$ and decreases in $a$. To be exact, it is just inversely proportional to $1+a$. Observation 4. The variance in the mixture is at least the mixture of the variances, which is minimized at $0$ by the independence of individual variances of $y$, observation 3, and the last sentence of observation 2. Also at $y=0$ we have equality because all means are at $0$ by symmetry. The end. I wish I could come up with an equally simple argument for $p>2$, but, alas, I don't have one at the moment. Edit OK, I guess I finally figured it out. We'll prove the following statement. Let $\varphi$ be an even convex function such that $\varphi''$ increases on $[0,+\infty)$. Let $p_y(x)$ be the probability density proportional to $e^{yx}e^{-\varphi(x)}$. Then the variance of $x$ with respect to the corresponding probability measure is a non-increasing function of $y$ for $y>0$. Indeed, let's differentiate in $y$. We have $p_{y+\delta y}(x)$ proportional to $p_y(x)(1+x\delta y)$ (the linearization of $e^{x\delta y}$), but if we leave it at that, the mass will change to $1+\delta y\int p_y(x)xdx=1+c\delta y$ where $c$ is the expectation of $x$ with respect to $p_y$, so we need to compensate by dividing by that factor, which will result in the linearization $$ p_{y+\delta y}(x)=(1+(x-c)\delta y)p_y(x) $$ Taking the linearization of the variance of $x-c$ now (which is the same as the variance of $x$ but easier to compute), we see that what we need to show is that $$ \int (x-c)^3p_y(x)dx\le 0. $$ We will show even that $\int_{x:|x-c|>a}(x-c)p_y(x)\le 0$ for all $a\ge 0$. The equality obviously holds for $a=0$ (the definition of $c$) and for $a=+\infty$. The derivative in $a$ is just $a(p_y(c-a)-p_y(c+a))$. I claim now that it can change sign only once for $a>0$ and that change is from $-$ to $+$. That is equivalent to saying that $\Phi(a)=\varphi_y(c-a)-\varphi_y(c+a)$, where $\varphi_y(x)=\varphi(x)-yx$, can change sign only once for $a>0$ and that change is from $+$ to $-$. We clearly have $c>0$ for $y>0$. Therefore, the point $c-a$ is always closer to the origin than $c+a$ for $a>0$ whence, by the assumed property of the second derivative of $\varphi$ (the linear term coming from $yx$ can do nothing with the second derivative), $\Phi''<0$. Thus, if we start from $\Phi(0)=0$ in the positive direction, we change sign once from $+$ to $-$ as promised. If we had started in the negative direction, we would never be able to change sign at all and the integral we are interested in would be monotone all the way, which is ridiculous, because its values at $0$ and $+\infty$ are both $0$, so that case is impossible. That takes care of $p>2$. If the second derivative of $\varphi$ were decreasing, then all inequalities would be reversed, so we can cover $p<2$ by this method as well and even get unimodality you see on the pictures, not just the global minimum at $0$. I hope I haven't made any stupid mistake in the computations but, since it is nearly midnight here now, you'd better check them carefully :-)<|endoftext|> TITLE: Can $H_{\omega_1}$ and $H_{\omega_2}$ be in bi-interpretation synonymy? QUESTION [5 upvotes]: This question concerns the possibility of the bi-interpretation synonymy of the structure $\langle H_{\omega_1},\in\rangle$, consisting of the hereditarily countable sets, and the structure $\langle H_{\omega_2},\in\rangle$, consisting of sets of hereditary size at most $\aleph_1$. These are both models of Zermelo-Fraenkel set theory $\text{ZFC}^-$, without the power set axiom. The structure $\langle H_{\omega_1},\in\rangle$ is of course a definable submodel of $\langle H_{\omega_2},\in\rangle$, which provides one direction of interpretation. Depending on the set-theoretic background, it is also possible that there is a converse interpretation in the other direction. Indeed, in my recent paper with Afredo Roque Freire, Joel David Hamkins and Alfredo Roque Freire, Bi-interpretation in weak set theories, blog post, arXiv:2001.05262 we prove that it is relatively consistent with ZFC that the structures $\langle H_{\omega_1},\in\rangle$ and $\langle H_{\omega_2},\in\rangle$ are bi-interpretable (see theorem 17). This is true, specifically, in the Solovay-Tennenbaum model, obtained by c.c.c. forcing over $L$ to achieve $\text{MA}+\neg\text{CH}$. What is needed is (i) $H_{\omega_1}$ has a definable almost disjoint $\omega_1$-sequence of reals; and (ii) every subset $A\subseteq\omega_1$ is coded by a real via almost-disjoint coding with respect to that sequence. The basic idea is that objects in $H_{\omega_2}$ are coded by a well-founded relation on $\omega_1$, which is in turn coded by a real, and so in $H_{\omega_1}$ we can define the class $U$ of reals coding a set in this manner and an equivalence relation on those reals $x\equiv y$ for when they code the same set, and a relation $\bar\in$ on those reals, so that $\langle H_{\omega_2},\in\rangle$ is isomorphic to the quotient structure $\langle U,\bar\in\rangle/\equiv$. The argument seems to use the equivalence relation in a fundamental manner, and the question I have about this here is whether one can omit the need for the equivalence relation. This ultimately amounts to the following, which is question 18 in the paper: Question. Is it relatively consistent with ZFC that there is a binary relation $\bar\in$ that is definable in $H_{\omega_1}$ such that $$\langle H_{\omega_1},\bar\in\rangle\cong \langle H_{\omega_2},\in\rangle?$$ This is what it would mean for these structures to form a bi-interpretation synonymy. For a positive answer, it would be enough to show the consistency with ZFC of the existence of a definable global well-order in $H_{\omega_1}$, together with the almost-disjoint coding of hypothesis (ii) above. Is that possible? Apart from $H_{\omega_1}$ and $H_{\omega_2}$ specifically, a related question we have is whether one can prove any instance of interpretation in a model of $\text{ZFC}^-$ that requires the quotient by an equivalence relation. Question. Is there a structure that is interpretable in a model of $\text{ZFC}^-$, but only by means of a nontrivial equivalence relation? This is question 9 in the paper. REPLY [7 votes]: A theorem of Harrington (Theorem B of his paper "Long projective wellorders") says $\text{MA} + \neg\text{CH}$ is consistent with a projective wellorder of the reals, hence a wellorder of $H_{\omega_1}$ definable over $H_{\omega_1}$. Since $\text{MA}_{\omega_1}$ implies a strong form of almost disjoint coding (i.e., any almost disjoint family $\langle A_\alpha : \alpha < \omega_1\rangle$ almost-disjoint-codes any subset of $\omega_1$ relative to some real), this gives an answer to your first question. For the second question, assuming $\text{AD}^{L(\mathbb R)}$ and $\delta^1_2 = \omega_2$, there is a prewellorder of $\omega^\omega$ with rank $\omega_2$ definable over $H_{\omega_1}$. On the other hand, there is no wellorder of a subset of $H_{\omega_1}$ in ordertype $\omega_2$ definable over $H_{\omega_1}$, because, I believe, there is no injection from $\omega_2$ into $H_{\omega_1}$ in $L(\mathbb R)$. (I am looking for a reference for this, but I am pretty sure it is true; maybe a ZF/AD expert can help me.) So $(\omega_2,<)$ is interpretable in $(H_{\omega_1},\in)$ via the prewellorder, but not via a trivial equivalence relation. UPDATE: Here is a proof that $\omega_2$ does not inject into $H_{\omega_1}$ under AD + $V = L(\mathbb R)$. In fact, we use the weaker hypothesis of $\text{AD}^+ + V = L(P(\mathbb R))$. We cite the following theorems: Theorem 1 (Woodin, $\text{AD}^+ + V= L(P(\mathbb R))$). For any set $X$, either $X$ can be wellordered or there is an injective function from $\mathbb R$ to $X$. This is Theorem 1.4 of Caicedo-Ketchersid's A trichotomy theorem in natural models of $\text{AD}^+$. I think the $\text{AD}^+ + V= L(P(\mathbb R))$ version is due to Woodin, but it's not clear from the paper. Theorem 2 (AD) There is no prewellorder of $\mathbb R$ whose proper initial segments are countable. This follows from the Kuratowski-Ulam Theorem since every set of reals has the Baire property. See Moschovakis's Descriptive Set Theory Exercise 5A.10 for a more general result. Using these we obtain the following corollary: Corollary ($\text{AD}^+ + V= L(P(\mathbb R)$). Suppose $X$ is a set and there is no injection from $\omega_1$ to $X$. Then there is no injection from $\omega_1$ to the set $P_{\aleph_1}(X)$ of all countable subsets of $X$. Proof. Suppose towards a contradiction that $f : \omega_1\to P_{\aleph_1}(X)$ is an injection. It follows that $A = \bigcup_{\alpha < \omega_1} f(\alpha)$ is an uncountable subset of $X$. Hence $A$ is not wellorderable. Moreover, there is a prewellorder of $A$ all of whose initial segments are countable: let $\varphi(x) = \min \{\alpha : x\in f(\alpha)\}$, and set $x \leq y$ if $\varphi(x)\leq \varphi(y)$. Since $A$ is not wellorderable, Theorem 1 yields an injection $g : \mathbb R \to A$. We define a prewellorder of $\mathbb R$ by setting $u \prec w$ if $g(u) < g(w)$. Its initial segments are countable since $g$ is injective, and this contradicts Theorem 2. THis proves the corollary. The corollary allows us to build a rank hierarchy for $H_{\omega_1}$. Define $R_0 = \emptyset$, $R_{\alpha+1} = P_{\aleph_1}(R_\alpha)$, and for limit ordinals $\gamma$, $R_\gamma = \bigcup_{\alpha < \gamma} R_\alpha$. There are now two key observations: (1) A trivial induction using the corollary shows that for any ordinal $\alpha < \omega_1$, there is no injection from $\omega_1$ to $R_\alpha$. (2) $R_{\omega_1} = H_{\omega_1}$. On the one hand, clearly every element of $R_{\omega_1}$ is hereditarily countable so $R_{\omega_1}\subseteq H_{\omega_1}$. On the other hand, $P_{\aleph_1}(R_{\omega_1}) = R_{\omega_1}$, and so by $\in$-induction one can prove $H_{\omega_1}\subseteq R_{\omega_1}$. Now suppose towards a contradiction that $f : \omega_2\to H_{\omega_1}$ is an injection. Since $\omega_2$ is not the union of $\omega_1$ countable sets, we can find an ordinal $\alpha < \omega_1$, such that $f[A]\subseteq R_\alpha$ for an uncountable set $A\subseteq \omega_2$. But this obviously yields an injection from $\omega_1$ to $H_\alpha$, contradicting the previous paragraph.<|endoftext|> TITLE: Is $\prod_{X : \mathcal{U}} (X \to X) \cong 1$ consistent with type theory? QUESTION [7 upvotes]: Assume we work in some minimalistic version of Martin-Löf type theory. Does it break consistency to postulate that the function that selects the identity function has an inverse? $$\prod_{X : \mathcal{U}} (X \to X) \cong 1.$$ From "Parametricity, automorphisms of the universe, and excluded middle", I understand that if it could be proved, excluded middle would not be consistent. More generally, can we assume this form of Yoneda reduction for $F : \mathcal{U} \to \mathcal{U}$ without breaking consistency? $$ \left(\prod_{X : \mathcal{U}} (A \to X) \to FX\right) \cong FA, $$ Or this form of coYoneda? $$ \left(\sum_{X : \mathcal{U}} (X \to A) \times FX \right) \cong FA, $$ The same question, but with paths instead of functions, points me to work by Rijke and later by Escardó relating the J-elimination rule and the Yoneda lemma. REPLY [8 votes]: $\prod_{X : \mathcal{U}} (X \to X) \cong 1$ is consistent. It follows from parametricity and function extensionality. Usual parametric models also support function extensionality. The simplest one which suffices would be the Fam model where every closed type is a set together with a family of sets over it, functions are predicate-preserving functions and the universe is the set of sets together with the family which maps each $A : \mathsf{Set}$ to $A \to \mathsf{Set}$. Here's a reference for a model which also works for our purpose but which is more complicated than necessary. $(\prod_{X : \mathcal{U}} (A \to X) \to FX) \cong FA$ is provably false in plain MLTT. Let $F\,X := X \to \bot$ and $A := \bot$. Now the statement simplifies to $(\prod_{X : \mathcal{U}} X \to \bot) \cong \top$, which is evidently false. Same for coYoneda. Pick $A := \top$ and $F$ as before, and coYoneda simplifies to $(\sum_{X : \mathcal{U}} X \to \bot) \cong \bot$. The problem is that $F$ is just an $\mathcal{U}\to \mathcal{U}$ function, and not a functor with respect to functions-as-morphisms in $\mathcal{U}$.<|endoftext|> TITLE: Different definitions of formality for groups QUESTION [8 upvotes]: Let $X$ be a space with fundamental group $G$. Recall that the de Rham fundamental group of $X$ is the inverse limit of the Malcev completions of the nilpotent truncations of $G$. This has a Lie algebra, which I will denote by $\mathfrak{g}$. The Lie algebra $\mathfrak{g}$ has a natural filtration, and thus has an associated graded Lie algebra $\text{gr}\ \mathfrak{g}$. In the literature, I have seen two different definitions of what it means for $G$ to be formal: The Lie algbera $\mathfrak{g}$ is isomorphic to its associated graded $\text{gr}\ \mathfrak{g}$. The Lie algebra $\mathfrak{g}$ has a presentation with only quadratic relations. There is also a notion of $X$ being formal: The minimal model of $X$ is quasi-isomorphic to the cohomology algebra (with the trivial differential). I assume that 3 implies 1 and 2 (though presumably it is stronger). Question: What is the precise relationship between the above notions? Is there a good place to read proofs of whatever implications exist? REPLY [4 votes]: There is a notion of $q$-equivalence between cdgas: it is a chain of morphisms each being isomorphism on cohomology in degrees $\leq q$ and monomorphism in degree $q+1$. Now, we call something $q$-formal if it is $q$-equivalent to its cohomology. Obviously, a space $X$ is $1$-formal if and only if $K(\pi_1(X), 1)$ is $1$-formal because killing homotopy groups map $X \to K(\pi_1(X), 1)$ is evidently an 1-equivalence. Define holonomy algebra $hol(X)$ of a cdga as quotient of free Lie algebra on $(H^1)^*$ by ideal spanned by image of comultiplication $\mu^*: (H^2)^* \to (H^1)^* \wedge (H^1)^*$. It's obviously a graded quadratic Lie algebra. Notice that 1-equivalences give isomorphisms on holonomy Lie algebras. So, for spaces, it depends only on fundamental group, because if you have $G \cong \pi_1(X) \cong \pi_1(Y)$, then both map into $K(G, 1)$ with maps being $1$-equivalences. Malcev completion of a group $\mathfrak m (G)$ is defined as Lie algebra of primitive elements in $\lim_{\leftarrow} \Bbb QG/IG^n$ where $IG$ is augmentation ideal. There is a homomorphism from Magnus algebra of a group $L(G) := \bigoplus \gamma_kG/\gamma_{k+1}G \otimes \Bbb Q$ to Malcev algebra which becomes isomorphism after taking associated graded quotients, as proven by Quillen in paper On the associated graded ring of a group ring, 1968. We have a theorem due to Sullivan. $\square$ Space $X$ is $1$-formal if and only if degree completion of holonomy Lie algebra $hol(X)$ is filtered isomorphic to $\mathfrak m (\pi_1(X))$. $\square$ Good reference is two books by Felix, Halperin, Thomas, called (unsurprisingly) Rational homotopy theory I/II. 1-formality is very clearly explained in the beginning of second tome. Now, to your question. If a space is formal, it is indeed $q$-formal for all $q$, in particular, 1, so your (3) implies both (2) and (1). YCor mentioned in comments that (1) is strictly weaker as witnessed by integral Heisenberg group. More generally, every two-step nilpotent Lie algebra is isomorphic to its associated lower central quotient, but only virtually abelian nilpotent groups are 1-formal. (If some commutator $[a, b]$ is nontrivial and there are no relations in weight 3, then $[a, [a, b]]$ will be nontrivial). Let's prove that (2) implies (3). Suppose $\mathfrak m(G)$ is a completion of some quadratic algebra $E = FreeLie(V)/(R), R \subset \wedge^2 V$. Now, going back to Quillen, we have a homomorphism $L(G) \to \mathfrak m(G)$ which becomes isomorphism on associated graded (in particular, it's injective). As Malcev algebra was graded from the beginning, $L(G)$ inherits this grading, and the only thing it can be is $E$. From Sullivan's Infinitesimal computations in topology we know that kernel of Lie bracket induced by group commutator on $G_{ab} \wedge G_{ab}$ rationally isomorphic to image of comultiplication $\mu^*$, so $E \cong hol(G)$ q.e.d.<|endoftext|> TITLE: Is the $\mathbb A^1$ homotopy type of a punctured curve independent of the choice of puncture? QUESTION [8 upvotes]: Let $C$ be a smooth connected algebraic curve over $\mathbb C$. Assume that $C$ admits no automorphisms. Let $p_1, p_2 \in C$ be two distinct points. Question: Are the $\mathbb A^1$ homotopy types of $C - p_1$ and $C- p_2$ the same? I think that the answer is "no", but I do not know how to construct an invariant that distinguishes the two. Of course, the ordinary homotopy types of $C-p_1$ and $C-p_2$ are the same. I also think that the stable $\mathbb A^1$ homotopy types are the same because of the purity cofiber sequence $$C-p_{i} \to C \to (\mathbb A^1, \mathbb A^1 - 0).$$ Is there a way of seeing that these two are not $\mathbb A^1$ homotopic, or of constructing an equivalence between them? REPLY [6 votes]: If I understand it correctly, this follows from the results of Severitt's master's thesis. (I am not an expert on this, so it is possible I misread one of the statements.) Indeed, Lemma 9.1.1 shows that smooth projective curves of genus $> 0$ are $\mathbf A^1$-rigid. But the proof actually shows the following: Lemma. Let $X$ be a smooth variety such that every map $\mathbf A^1 \to X$ is constant. Then $X$ is $\mathbf A^1$-rigid. $\square$ Now if $X = C - p$ for some smooth projective curve $C$ with $\operatorname{Aut}(C) = 1$, then in particular $g(C) > 0$ so every map $\mathbf A^1 \to C$ (or $\mathbf A^1 \to C \setminus p$) is constant. Thus, the lemma implies $X$ is $\mathbf A^1$-rigid. Then Theorem 6.3.7 implies the result: if $C - p$ and $C - q$ are (weakly) $\mathbf A^1$-homotopy equivalent, then the theorem gives $C - p \cong C - q$, which implies that there is a nontrivial automorphism of $C$ taking $p$ to $q$. $\square$<|endoftext|> TITLE: Why do we study complex orientable cohomology theories QUESTION [7 upvotes]: It seems that much of the literature in stable homotopy theory seems to study complex orientable cohomology theories. What is the reason of restricting to this class of multiplicative cohomology theories? Is it simply that they are more computable? Is there a good a priori reason that this is an important class of cohomology theories to study? REPLY [22 votes]: There is a sort of a priori reason why one would consider the cohomology theory $MU$, without first knowing of its connection to manifold geometry, to formal groups, … . Since complex-oriented cohomology theories are those cohomology theories with a ring map from $MU$, perhaps having a sufficiently strong interest in $MU$ would suffice to then motivate them. $\newcommand{\co}{\colon\thinspace}\newcommand{\F}{\mathbb F}\newcommand{\Z}{\mathbb Z}\newcommand{\A}{\mathcal A}\newcommand{\from}{\leftarrow}$ There are a variety of constructions in homotopy theory that allow one to delete a class, where "class" can have different meanings and "delete" different levels of finesse. For instance, if you have a map $f\co S^n \to X$ which on homology sends the fundamental class $\iota_n \in H_n S^n$ to some class $f_* \iota_n \in H_n X$, then the mapping cone $C(f)$ has the property that $f_* \iota_n$ pushes forward to $0$ in $H_n C(f)$. In fact, if $f_* \iota_n$ is not torsion, then $H_* C(f)$ will be exactly $(H_* X) / f_* \iota_n$, as one can see by writing out the homology long exact sequence. By the same token, if $f_* \iota_n$ is $m$–torsion, the same long exact sequence will produce a fresh class in $H_{n+1} C(f)$, due to the death of $m \iota_n$ under $f_*$. This is a general truism in homotopy theory: if a class has been deleted "twice", then you get a fresh class one degree higher. For a slight twist on this same idea, consider the mod–$p$ cohomology of a space, which is related to its homology by the universal coefficient sequence $$ 0 \to \operatorname{Ext}(H_{n-1}(X; \Z), \F_p) \to H^n(X; \F_p) \to \operatorname{Hom}(H_n(X; \Z), \F_p) \to 0. $$ A torsion-free class in $H_n(X; \mathbb Z)$ contributes a class only to $H^n(X; \F_p)$, but a $p$–power–torsion class contributes classes both to $H^n(X; \F_p)$ and $H^{n+1}(X; \F_p)$. There is actually a process by which one can marry these pairs of classes and reconstruct integral cohomology: the Bockstein spectral sequence has signature $$H^*(X; \F_p) \otimes \F_p[b] \Rightarrow H^*(X; \Z_p),$$ it converges for connected $X$ of finite type, and its differentials perform exactly these marriages. Its construction relies on the facts that $\Z_p$ is $p$–complete and that it participates in the short exact sequence $$0 \to \Z_p \xrightarrow{p \cdot -} \Z_p \to \F_p \to 0,$$ and for us it suffices to leave its explanation at that. The first differential in the Bockstein spectral sequence takes the form of a stable, additive homomorphism $$\beta\co H^n(X; \F_p) \to H^{n+1}(X; \F_p).$$ In fact, this map is natural in $X$, and such natural transformations in general are called stable cohomology operations. The Steenrod algebra is the collection of all stable mod–$p$ cohomology operations, given by $$\A^* := [H\F_p, H\F_p]_* = H^*(H\F_p; \F_p).$$ This associative algebra is calculable from first principles [1]: $$\A^* = \F_p\langle \beta, P^n \mid j \ge 1 \rangle \, / \, (\text{various relations}),$$ where the angle brackets indicate that these generators do not commute; where $P^n$ is the "$p$th power operation", which sends a cohomology class of degree $2n$ to its $p$th power [2]; and where $\beta$ is the same Bockstein operation as above. Taking this calculation for granted, one can then make a further calculation of a cousin of these operations: $$H^*(H\Z_p; \F_p) = [H\Z_p, H\F_p]_*.$$ Starting with same the quotient sequence $$H\Z_p \xrightarrow{p} H\Z_p \to H\F_p,$$ and applying $[-, H\F_p]_*$, the multiplication-by-$p$ map induces zero on cohomology, and hence the going-around map participates in a short-exact sequence of $\A$–modules $$0 \from [H\Z_p, H\F_p]_* \from \A^* \from [H\Z_p, H\F_p]_{*+1} \from 0.$$ The first map is onto, hence our $\A^*$–module of interest is cyclic; the second map presents it as a submodule of $\A^*$ generated by a class in degree $1$; and, since $\beta$ is the only operation in degree $1$, we learn $$[H\Z_p, H\F_p]_* = \A^* / (\beta \cdot \A^*).$$ In this sense, $\beta$ witnesses the double-quotient of $H\Z_p$ by $p$, in that the quotient appears on the left- and on the right-hand sides of $[H\F_p, H\F_p] = [H\Z_p / p, H\Z_p / p]$. By removing the quotient from the left and instead studying $[H\Z_p, H\Z_p/p] = [H\Z_p, H\F_p]$, we avoid killing $p$ twice, and $\beta$ disappears. However, longer monomials in which $\beta$ appears in the middle still survive, and they contribute other odd-degree cohomology operations. One might wonder whether every such odd-degree cohomology operation belongs to some spectrum $X$ with a non-nilpotent endomorphism $v\co \Sigma^n X \to X$ whose quotient recovers $H\F_p$ and which admits an associated Bockstein spectral sequence. It turns out that a version of this is true: $[MU, H\F_p]_*$ is (almost [3]) the quotient of $\A^*$ with $\beta$ fully deleted. In this sense, $MU$ is the "maximally unquotiented" version of $H\F_p$ within even-concentrated ring spectra. Even more than this, there are multiple theorems along these lines, coming at the same problem from different angles. The study of the mod–$p$ cohomology of $MU$ (and its various features, including its relation to that of $H\F_p$) is originally due to Milnor. The homotopy of $MU$ is given by $$\pi_* MU = \Z[x_1, x_2, \ldots],$$ and the generators $x_n$ (again, almost [3]) iteratively play the role of $v$ in the above fantasy resolution of $H\F_p$. Starting with the $p$–local sphere, one can iteratively remove its odd-degree homotopy while retaining even-concentrated homology. Priddy showed that this ultimately leads to a spectrum called $BP$, which is an indivisible $p$–local summand of $MU$. Starting with the $p$–local sphere, one can iteratively remove its odd-degree homotopy by $A_\infty$–algebra maps. This leads to a sequence of spectra $X(n)_{(p)}$ with $X(\infty)_{(p)} = MU_{(p)}$. The original study of this sequence of spectra is due to Ravenel, then taken up by Hopkins, Devinatz, and Smith, and this perspective in terms of iterated quotients is due to Beardsley. Each of these objects begins with some desirable properties: the sphere spectrum has pleasant homology (but very knotty homotopy), and the Eilenberg–Mac Lane spectrum has pleasant homotopy (but knotty co/homology). By trying to correct the unpleasant part, one keeps ending up at (a chunk of) $MU$, which has even-concentrated homotopy, even-concentrated homology, even-concentrated co/operations, … . All I mean to point out by this is that $MU$ is an extremely natural object to bump into, especially if one has a preference for even-concentration, whether due to a preference for commutativity over graded-commutativity or due to an aversion toward unnecessarily killing classes twice. –––––––––– [1] - Here I'm restricting to odd primes, but you can say all these same words with slightly different formulas at $p = 2$. [2] - $P^n$ does other, more mysterious things in degrees other than $2n$. [3] - The precise statement is that the $p$–localization $MU_{(p)}$ splits as a sum of shifts of copies of a ring spectrum $BP$. This new spectrum has homotopy given by $$\pi_* BP = \Z_{(p)}[v_1, v_2, \ldots],$$ $[BP, H\F_p]$ is the submodule of $\A^*$ where $\beta$ is totally deleted, and these $v_j$s are the desired self-maps $v$. (In terms of $\pi_* MU$, $v_j$ is equivalent to $x_{p^j}$ modulo decomposables.) Although $BP$ has all these nice properties, it depends on the prime $p$, and $MU$ is to be thought of as the best integral object capturing all of them at once.<|endoftext|> TITLE: Lifting $G$-invariants from characteristic $p\gg 0$ to characteristic 0 for a reductive algebraic group $G$ QUESTION [10 upvotes]: Let $S\subset \mathbb{C}$ be a finitely generated ring, let $R$ be a finitely generated commutative ring over $S$. Let $G$ be a linear algebraic group over $S$, such that $G_{\mathbb{C}}$ is reductive. Suppose that Spec$(R)$ is equipped with a $G$-action over $S$. In this setting, I hope that the following statement holds. For any large enough prime $p$, given a base change $S\to \bf{k}$ to an algebraically closed field of characteristic $p,$ then $G_{\bf{k}}$-invariants of $R_{\bf{k}}$ are generated by the image of $G$-invariants of $R.$ It feels that the above statement is either explicitly well-known, or at least should follow from a well-known result. Any suggestions or references will be greatly appreciated. REPLY [5 votes]: We offer two facts and a Theorem. Let $S$ be a commutative noetherian ring containing $\mathbb Z$ and let $G=G_S$ be reductive over $S$ in the sense of SGA3. That is, $G$ is smooth over $S$ with geometric fibers that are connected reductive. Let $R$ be a finitely generated commutative $S$-algebra. Suppose that $\mathrm{Spec}(R)$ is equipped with a right $G$-action over $S$. Fact 1 Let us be given a base change $S\to \bf k$ to a field of positive characteristic $p$. For every $x\in (R\otimes_S {\bf k})^G$ there is an $r\geq1$ so that $x^{p^r}$ lies in the $\bf k$-span of the image of $R^G$. Fact 2 For almost all primes $p$ the map $R^G\to (R/pR)^G$ is surjective. Remark. We do not need to distinguish between $(R/pR)^G$ and $(R/pR)^{G_{S/pS}}$. Base change of the group is unnecessary when computing invariants or cohomology. (See 1.13. Restriction in our survey Reductivity properties over an affine base .) Theorem Assume further that $R$ is flat over $S$ and that $S$ is of finite global homological dimension. Then there is an integer $n\geq1$ so that if $S[1/n]\to \bf k$ is a base change to a field, then the map $R^G\otimes_S{\bf k}\to (R\otimes_S{\bf k})^{G_{\bf k}}$ is surjective. To prove Fact 1, let $D$ be the image of $S$ in $\bf k$. As $D\to \bf k$ is flat, we have $(R\otimes_S {\bf k})^G=(R\otimes_S D)^G\otimes_D {\bf k}$, so it suffices to show that for every $x\in (R\otimes_S D)^G$ there is an $r\geq1$ so that $x^{p^r}$ lies in the the image of $R^G$. Now $G$ is power reductive, so we may apply Proposition 41 of our paper Power Reductivity over an Arbitrary Base . See also my survey Reductivity properties over an affine base . To prove Fact 2, recall from Theorem 10.5 of Good Grosshans filtration in a family that $H^1(G,R)$ is a finitely generated module over $R^G$. This module is also $\mathbb Z$-torsion. To see this, first take an fppf base change to reduce to the case that $G$ is split over $S$ (see SGA3). Then $G_{\mathbb Q}$ makes sense and $H^1(G,R)\otimes_{\mathbb Z}{\mathbb Q}=H^1(G_{\mathbb Q},R\otimes_{\mathbb Z}{\mathbb Q})=0$. Choose $n\geq1$ so that $n$ annihilates the generators of $H^1(G,R)$. Choose $m\geq1$ so that $m$ annihilates the generators of the $\mathbb Z$-torsion ideal of $R$. Now if $p$ does not divide $mn$, then $\partial$ vanishes in the exact sequence $0\to R^G\stackrel{\times p}\to R^G\to (R/pR)^G\stackrel\partial\to H^1(G,R)$. Now we turn to the proof of the Theorem. If $G$ is split over $S$ then we may apply Remark 31 and Theorem 33 of Power Reductivity over an Arbitrary Base to obtain $n$ so that $H^i(G,R[1/n])$ vanishes for $i\geq1$. If $G$ is not yet split the same result is true. Indeed we may by SGA3 do an fppf base change $S\to T$ so that $G_T$ is split over $T$. Then we may arrange that $H^i(G,R[1/n])\otimes_ST=H^i(G_T,R[1/n]\otimes_ST)$ vanishes for $i\geq1$. This implies that $H^i(G,R[1/n])$ vanishes for $i\geq1$. Having chosen $n$ this way we now claim that for every $S$-module $N$ with trivial $G$ action, the $H^i(G,R[1/n]\otimes_SN)$ also vanish for $i\geq1$. This is clear if $N$ is free and then it follows by induction on the projective dimension of the $S$-module $N$. (If $0\to N'\to F \to N\to0$ is exact, with $F$ free, consider the long exact sequence for $G$-cohomology associated with the exact sequence $0\to R[1/n]\otimes_SN'\to R[1/n]\otimes_SF \to R[1/n]\otimes_SN\to0$.) Now let us be given a base change $S[1/n]\to \bf k$ to a field. Let $N$ be the kernel of $S[1/n]\to \bf k$ and let $D$ be its image. Note that $0\to R\otimes_SN\to R\otimes_SS[1/n] \to R\otimes_SD\to0$ is exact and that $R\otimes_SN=R[1/n]\otimes_SN$. As $H^1(G,R[1/n]\otimes_SN)=0$ we have a surjection $(R\otimes_SS[1/n])^G\to (R\otimes_SD)^G$. As $D\to \bf k$ is flat, $(R\otimes_S{\bf k})^{G_{\bf k}}=(R\otimes_SD)^G\otimes_D{\bf k}$. We see that $(R\otimes_SS[1/n])^G\otimes_S{\bf k}$ maps onto $(R\otimes_S{\bf k})^{G_{\bf k}}$. But $(R\otimes_SS[1/n])^G\otimes_S{\bf k}$ equals $R^G\otimes_SS[1/n]\otimes_S{\bf k}=R^G\otimes_S{\bf k}$. The result follows.<|endoftext|> TITLE: Do topos-valued sheaves form a topos? QUESTION [14 upvotes]: Let $\bf C$ be a category, $\mathcal S$ an (elementary) topos. If $\mathcal S$ is a presheaf category over $\bf D$, then it's easy to see $[\mathbf C^{\rm op},\, \mathcal{S}] \cong [(\mathbf C \times \mathbf D)^{\rm op},\, \mathcal{Sets}]$ is still a topos. In more general situations I struggle to see an easy reason for it to be true as well. Thus: When is the category $[\mathbf C^{\rm op},\, \mathcal S]$ of contravariant functors from $\bf C$ to $\mathcal S$ a topos? REPLY [3 votes]: Rather than a full-fledged answer, this is a sketch of a plan: As you yourself pointed out, if $\mathcal S=[\bf D^{\rm op},\, \mathcal \cal{Sets}] $, the result is trivial. Now, suppose you consider an arbitrary elementary topos $\mathcal S$. The steps would be: Use the representation theorem of Joyal-Tierney to embed your target topos as the equivariant sheaves over a localic topos, see here. $E:\mathcal S \to Set^{\mathcal{L}^{op}}$ Now each map $[\bf C^{\rm op},\, \mathcal S]$ composes with the embedding $E$ and thus lands into a localic topos. The last step would be to see how $[\bf C^{\rm op},\, E(\mathcal S)]$ sits inside $[\bf C^{\rm op},\, Set^{\mathcal{L}^{op}}]$. Conjecture: under some some conditions (to be determined, but see this post When is a reflective subcategory of a topos a topos? ) there is a reflection which is sufficiently exact to ensure that it is indeed a subtopos of $[(\bf C \times \bf\mathcal{L})^{\rm op},\, \cal{Sets}]$<|endoftext|> TITLE: Optimization algorithm sought QUESTION [5 upvotes]: Suppose I have $N$ pairs of positive numbers $(a_1, b_1), (a_2, b_2), \dotsc, (a_N, b_N).$ and I want to find a subset of $M$ of them maximizing $$ \frac{\sum_{j=1}^M a_{i_j}}{\sum_{j=1}^M b_{i_j}}. $$ Can this be done in polynomial time? REPLY [2 votes]: The paper "Dropping Lowest Grades" by Daniel M. Kane and Jonathan M. Kane addresses this question in the context of dropping $r$ quiz grades from a collection of weighted grades. The solution described there is fundamental the same as that in David Eppstein's answer. However, there is also a discussion of a practical implementation that may be even faster in practice.<|endoftext|> TITLE: Reference for computation of $K_8(\mathbb{Z})$ QUESTION [7 upvotes]: In an Oberwolfach report from 2016 [1, page 2] it is said that $K_8(\mathbb{Z})$ has recently been computed. Does anyone know a reference for the computation? [1] https://orbilu.uni.lu/bitstream/10993/29499/1/preliminary_OWR_2016_52.pdf REPLY [9 votes]: There are two preprints available: https://arxiv.org/abs/1910.11598 https://www.utsc.utoronto.ca/people/kupers/wp-content/uploads/sites/50/2021/01/k8zshorter.pdf<|endoftext|> TITLE: Relationship between "infinitely unequal" and "eventually different" QUESTION [5 upvotes]: Suppose that $\kappa$ has the property that for every family $A\subseteq\omega^{\omega}$, if $|A|<\kappa$, then there exists some $g\in\omega^{\omega}$ such that for any $f\in A, \exists^{\infty}n\;f(n)\neq g(n)$. Does it then follow that for any family $A\subseteq\omega^{\omega}$ of size $<\kappa$, there exists a $g\in\omega^{\omega}$ such that for any $f\in A, \forall^{\infty}n\;f(n)\neq g(n)$? REPLY [5 votes]: It does not follow. In fact, it is not true in the Cohen model. Let $X$ be a set of $\aleph_2$ mutually generic Cohen reals over $V$. (Recall: ``mutually generic'' means that if $x \in X$ and $Y \subseteq X$, then $x \in V[Y]$ if and only if $x \in Y$, and otherwise $x$ is Cohen-generic over $V[Y]$.) In $V[X]$, if $A \subseteq \omega^\omega$ with $|A| < \aleph_2$, then (because $A$ is an object that is hereditarily of size $\leq\!\aleph_1$) there is some $X_0 \subseteq X$ with $|X_0| = \aleph_1$ such that $A \subseteq V[X_0]$. If $g \in X \setminus X_0$, then $g$ is Cohen-generic over $V[X_0]$, and hence over any $f \in A$. In particular, for any $f \in A$, $\exists^\infty n f(n) \neq g(n)$. On the other hand, let $A \subseteq X$ with $|A| = \aleph_1$. If $g \in \omega^\omega$, then (because $g$ is a hereditarily countable object) there is some countable $A_0 \subseteq A$ such that $g \in V[A_0]$. If $f \in A \setminus A_0$, then it is Cohen-generic over $V[A_0]$, and hence $\exists^\infty n f(n) = g(n)$.<|endoftext|> TITLE: Intuition and analogue of Wraith axiom from synthetic differential geometry QUESTION [6 upvotes]: In synthetic differential geometry, an object $M$ verifies the Wraith axiom if for all functions $\tau:D\times D\to M$ which are constant on the axes $\tau(d,0)=\tau(0,d)=\tau(0,0)$ for all $d\in D$, there's a unique factorization through the multiplication map, i.e there's a unique function $t:D\to M$ such that $t(d_1\cdot d_2)=\tau(d_1,d_2)$. What is the geometric/physical intuition behind this axiom? What is the analogue in the category of smooth manifolds? Update. Following the answers I think I should add some motivation. Lavendhomme's book defines the commutator of vector fields as $\tau(d_1,d_2)=Y_{-d_2}X_{-d_1}Y_{d_2}X_{d_1}$. Because this is constant on axes, it factors through the multiplication map to give a vector field $t$ charaterized by $t(d_1d_2)=\tau(d_1,d_2)$. I understand $t$ is desirable since it is a vector field, but I don't know how to geometrically interpret its characterizing property. For instance, why not consider the vector field $\tau(d_1,d_1)$ given by precomposing the diagonal? This motivated my question. REPLY [5 votes]: Axiom W is about the behaviour of the second tangent bundle - it ensures that the vertical bundle of the tangent bundle, $V(M) \subseteq T\circ T(M)$, where $V(M) = T(p)^{-1}(0)$, decomposes as the pullback of the projection $p_M: T(M) \to M$ along itself. The map $[\bullet, M]:[D,M] \to [D \times D, M]$ would be written fiberwise as $\ell(v) = \frac{d}{dt}_{t = 0} (vt)$. If you look at Robin Cockett and Geoff Cruttwell's first paper on tangent categories, you can see they spend a fair bit of time talking about the universality of the vertical lift and its relationship with the Lie bracket. You may also want to look at the definition of the core of a double vector bundle - axiom W can also be read as saying the core of the second tangent bundle of M is $TM$. Edit: If you look at the answer here, you can see how the lift from the fibred product $TM \times_M TM \to T^2(M)$ is written. Using infinitesimals, you would write $\gamma,\beta:D\to M, \gamma(0)=\beta(0)$ is sent to the map $D\times D \to M$ which is given by $d_1,d_2 \mapsto \gamma(d_1) + \beta(d_1d_2)$. The condition that this be the kernel of $T(p)$ is provable from property W (and a good exercise). The square root business you mentioned looks to be how you would show property $W$ holds in the category of smooth manifolds (when rewritten as the universality of the vertical lift from the tangent category axioms).<|endoftext|> TITLE: Categorical description of log as approximate rig homomorphism? QUESTION [16 upvotes]: Summary The base-$\beta$ logarithm gives an isomorphism of topological spaces $$ \log_\beta\colon\mathbb{R}_{\geq0}\xrightarrow{\cong}[-\infty,\infty). $$ This continuous map preserves some algebraic structure, e.g. sending multiplication to addition. As is well-known in tropical geometry, it also approximately sends the binary operation addition to the binary operation $\max$, $$\log_\beta(a+b)\approx\max\left(\log_\beta(a),\log_\beta(b)\right),$$ and this approximation gets better as the base $\beta\in(1,\infty)$ increases. Question: Is there a category-theoretic way to tell this story? Background A commutative rig ("ring without negatives") is a tuple $(S, 0, +, 1, *)$, where $(S,0,+)$ and $(S,1,*)$ are both commutative monoids, and where multiplication distributes over addition, $a*(b+c)=(a*b)+(a*c)$. Both $$ \big(\mathbb{R}_{\geq0},0,+,1,*\big)\quad\text{and}\quad\big([-\infty,\infty), -\infty, \max, 0, +\big) $$ are rigs; in fact, they are topological rigs. That is, each of the sets $\mathbb{R}_{\geq0}$ and $[-\infty,\infty)$ has a well-known topology (e.g. in $[-\infty,\infty)$, the open sets containing $-\infty$ are generated under union by those of the form $[-\infty,x)$ for $x\in \mathbb{R}$), and in each case the two binary operations, namely $+,*$ and $\max,+$, are continuous with respect to that topology. As mentioned above, the $\log_\beta$ function is an isomorphism of topological spaces, but it is not a homomorphism of rigs. It preserves almost all of the structure ($0\mapsto-\infty, 1\mapsto 0, *\mapsto +$), the one exception being that the addition operation is not preserved. However, we do have simple bounds on the difference between $\log_\beta(a+b)$ and $\max(\log_\beta a,\log_\beta b)$ for all $a,b\in\mathbb{R}_{\geq0}$. One checks easily (using the two exhaustive cases $a\leq b$ and $b\leq a$) that $$ 0\leq \log_\beta(a+b)-\max\big(\log_\beta(a),~\log_\beta(b)\big)\leq\log_\beta 2. $$ Thus as $\beta$ gets bigger—one might think of $\beta$ physically as "coldness", so "as the system gets colder"—the approximation gets better: $\log_\beta$ becomes closer and closer to an isomorphism of topological rigs. Motivation Tropical geometry is widely studied field, and I imagine that nothing in this section is particularly novel in that context. But let me just say what motivates me. Having an almost-isomorphism (asking precisely what that should mean is the goal of this post) between these two topological rigs is interesting because, for one thing, $\max,+$ matrix multiplication is much easier for humans and faster to compute than $+,*$ matrix multiplication. Moreover, lots of ideas port over. For example, one can discuss what I might call "log-stochastic matrices" where the max of each column is 0. These form a category that is similar in certain ways to the usual category Stoch of finite sets and stochastic maps, but of course having key differences. Working a bit in this category, I could imagine it being relevant in behavioral economics, where there may be multiple "best choices" in a given situation (multiple 0's in a given column), and calculations need to be easy (again max, + being easier than +,*). Categorical issues My goal is to be able to say, category-theoretically, both that $\log_\beta$ is an approximate rig map for each $\beta\in(1,\infty)$, and also that these approximations get better as $\beta$ increases. To tell this story categorically, one needs to define what "better approximation" means. One attempt could be to give something like a distance between $\log(a+b)$ and $\max(\log a, \log b)$, and for this we would need something like a metric. The usual category of metric spaces has short maps (distance non-decreasing functions) as its morphisms. This doesn't work too well for us, because $(\mathbb{R},0,+,1,*)$ is not an internal rig in that category; in particular, the multiplication operation $*\colon\mathbb{R}_{\geq0}\times\mathbb{R}_{\geq0}\to\mathbb{R}_{\geq 0}$ is not a short map. In fact, it is not even Lipschitz. But once one moves to the category of topological spaces or sets—where our two structures do become internal rigs—how will one measure the distances between their distance addition operations? The goal is not to necessarily stick with the viewpoint of the previous paragraph, but to come up with a categorical viewpoint in which the successive approximations story is most at home. REPLY [2 votes]: Summary While it could certainly be prettied up, it seems that the basic idea is actually fairly straightforward: use each $\log_\beta$ to transport the rig structure from $\mathbb{R}_{\geq0}$ to the space $[-\infty,\infty)$ and use the metric structure there. Transporting the rig structure For any $\beta$, consider the operation $P_\beta\colon[-\infty,\infty)\times[-\infty,\infty)\to[-\infty,\infty)$ given by $$P_\beta(a,b):=\log_\beta\left(\beta^a+\beta^b\right).$$ This is obtained by transporting $+$ along the $\log_\beta$ bijection. As such, it is easy to see that $P_\beta$ is unital with respect to $-\infty$ and associative, and that $+$ distributes over $P_\beta$. We can also see this directly, e.g.: \begin{align*} a+P_\beta(b,c)&= \log_\beta(\beta^a)+\log_\beta(\beta^b+\beta^c)\\&= \log_\beta(\beta^{a+b}+\beta^{a+c})\\&= P_\beta(a+b,a+c). \end{align*} The limit of these operations $P_\beta$ as $\beta$ increases is $$ \lim_{\beta\to\infty}P_\beta(a,b)= \lim_{\beta\to\infty}\log_\beta\left(\beta^a+\beta^b\right)= \max(a,b). $$ As an aside, a similar phenomenon occurs for the family of $\ell_p$ norms, and there seems to be a relationship. Namely, the variables in the exponential terms have switched roles ($\beta\leftrightarrow p$ and $\ell_p\leftrightarrow P_\beta$). Writing $\ell$ for $\ell_p$ and $P$ for $P_\beta$, one sees the resemblance immediately: $$ \ell(a,b)^p=a^p+b^p \qquad\text{vs.}\qquad \beta^{P(a,b)} = \beta^a + \beta^b. $$ We'll save this curiosity for another time. A family of metric rigs Going back to the main story, for any $\beta\in(1,\infty)$ we have a rig $$ R_\beta:=\big([-\infty,\infty),-\infty,~P_\beta~,0,+\big). $$ Each of these is of course isomorphic to the original rig $(\mathbb{R}_{\geq0},0,+,1,*)$, but the point is to show that these are rig objects in the category of (Lawvere) metric spaces. That is, we need to show that both operations are short with respect to the usual metric on $[-\infty,\infty)$. Checking shortness for the "multiplication" operation $+$, we choose $a,b_1,b_2$ and see immediately that we have $d(a+b_1,a+b_2)=d(b_1,b_2)$ and hence the necessary inequality holds $$d(a+b_1,a+b_2)\leq d(b_1,b_2).$$ Checking shortness for the "addition"operation $P_\beta$ is slightly more involved. It uses the fact that for nonnegative reals $0\leq x$ and $0 TITLE: An isogeny between Jacobians of hyperelliptic curves QUESTION [12 upvotes]: Let $\mathbf{F}_q$ be a finite field of odd characteristic. Let $X_t$ be the hyperelliptic curve over $\mathbf{F}_{q^2}(t)$ with affine equation $$y^2 = \left((x^{(q+1)/2}-(x-1)^{(q+1)/2})^2 - t\right) \left((x^{(q+1)/2}+(x-1)^{(q+1)/2})^2 - t\right).$$ This family of hyperelliptic curves is smooth of genus $q-1$ away from $t=0,1,\infty$. For theoretical reasons I expect the following claim to be true: The Jacobians of $X_t$ and $X_{1-t}$ are isogenous. Equivalently, there is a nontrivial correspondence between $X_t$ and $X_{1-t}$. It is easy to verify (with MAGMA for instance) that the zeta functions of $X_t$ and $X_{1-t}$ agree for small values of $q$ and specific values of $t$, to the point that I'm quite convinced of the claim. To prove it, one could try to construct the correspondence explicitly. There is some literature about pairs of hyperelliptic curves with isogenous Jacobians (for instance this paper of Mestre), but it seems to involve finding a congruence between bivariate polynomials, which comes out of nowhere. Short of constructing the actual correspondence, is there an algorithm for deciding whether two hyperelliptic curves admit an isogeny of given degree between their Jacobians? REPLY [4 votes]: In the most easy case $q=3$, the curve $X_t$ is bielliptic (the bielliptic involution given by $x\mapsto 1-x$), and the Jacobian of $X_t$ is then $(2,2)$-isogenous to the product $E_{1,t}\times E_{2,t}$, where $E_{1,t}$ and $E_{2,t}$ are the elliptic curves $$E_{1,t}: y^2=x(x^2 - (t+1)x + (t- 1)^2)$$ and $$E_{2,t}: y^2=x(x^2 + x - (t^3 - 1))$$ Now, $E_{1,t}$ is 2-isogenous to $$E_{1,t}': y^2= x(x+1)(x-(t-1))$$ and one has that $$E_{1,t}'\cong E_{1,1-t}$$ by sending $x\mapsto 1-x$ and then twisting by $-1$, which is an isomorphic curve over $\mathbb{F}_9$ (this is the only place where we use we are over $\mathbb{F}_9$ and not over $\mathbb{F}_3$). while $E_{2,t}$ is 2-isogenous to $$E_{2,t}': y^2 = x(x^2 + x + t^3)$$ which in turn is clearly equal to $E_{2,1-t}$.<|endoftext|> TITLE: Proof of Prop 1.1 in Wiles' "Modular elliptic curves and Fermat's last theorem" QUESTION [8 upvotes]: On p. 459 of "Modular elliptic curves and Fermat's last theorem", proof of Prop. 1.1, where it says "Since $H^2(G,\mu_{p^r}) \rightarrow H^2(G,\mu_{p^s})$ is injective for $r \leq s$...", is there any chance that that's a typo and he really means to say $H^1$ both times rather than $H^2$? He uses it to conclude that the tensor product of one $H^1$ with a certain module is isomorphic to another $H^1$, so if the hypothesis really is with an $H^2$ both times then that presumably must involve exact sequences in some way. I'd appreciate any help in understanding the argument. REPLY [23 votes]: Well, although there is a typo (Wiles forgot to close his parenthesis, and wrote $H^2(G,\mu_{p^r}\to H^2(G,\mu_{p^s})$ in his proof), his claim is correct. Let, as ibid. $F$ be the finite extension of $\mathbb{Q}_p$ fixed by $G$, so that your arrow can be written $H^2(F,\mu_{p^r})\to H^2(F,\mu_{p^s})$. Consider the Kummer sequence (for any $n\geq 1$) $$ 1\longrightarrow \mu_{p^n}\longrightarrow \overline{F}^\times\overset{(\cdot)^{ p^n}}{\longrightarrow}\overline{F}^\times\longrightarrow 1. $$ Taking Galois cohomology and using Hilbert '90, which tells you $H^1(F,\overline{F}^\times)=1$, you find $$ 1\longrightarrow H^2(F,\mu_{p^n})\longrightarrow H^2(F,\overline{F}^\times)\overset{\cdot p^n}{\longrightarrow }H^2(F,\overline{F}^\times). $$ Now, local class field theory tells you that the Brauer group $H^2(F,\overline{F}^\times)$ is isomorphic to $\mathbb{Q}/\mathbb{Z}$, so the above sequence identifies $H^2(F,\mu_{p^n})$ with the group $\mathbb{Z}/p^n$, seen as the kernel of multiplication by $p^n$ on $\mathbb{Q}/\mathbb{Z}$. If now you apply this for $r\leq s$, you see that the arrow you were firstly interested in is the injection $\mathbb{Z}/p^r\hookrightarrow \mathbb{Z}/p^s$ or, if you prefer, the injection $$ \Big(\ker(\cdot p^r)\hookrightarrow \ker(\cdot p^s)\Big)\subseteq \mathbb{Q}/\mathbb{Z}. $$ Added Jan 24th Concerning the second exact sequence (that involving $H^1$), there, there is a typo! What Wiles wanted to write was that the natural map $$\tag{1} H^1(G,\mu_{p^n})\otimes M\longrightarrow H^1(G,M(1)) $$ is an isomorphism, but in his paper he made (again!) a mistake with parenthesis and ibid he replaces the first term by $H^1(G,\mu_{p^n}\otimes M)$. You can see (1) in the quoted paper by Diamond and, moreover, it is (1) which shows up later in Wiles' proof: indeed, he deduces from (1) something about another sequence where $H^1(G,\mu_{p^n})\otimes M$ is replaced, by Kummer theory, by $((F^\times/(F^\times)^{p^n})\otimes M$, which he could not do had he only some result about $H^1(G,\mu_{p^n}\otimes M)$. To prove (1), observe that Wiles has managed to make the action of $G$ on $M$ trivial, and $n$ is such that $p^nM=0$. So, $M$ is a finite abelian $p$-group of exponent bounded by $n$, and isomorphic (as $G$-module!) to a finite number of copies of $\mathbb{Z}/p^a\mathbb{Z}$ for $a\leq n$. It thus suffices to prove (1) assuming $M=\mathbb{Z}/p^a$ for $a\leq n$. Consider, for any $s\geq 0$, the exact sequence $$ 0\longrightarrow \mu_{p^s}{\longrightarrow}\mu_{p^{s+a}}\longrightarrow \mu_{p^a}\longrightarrow 0. $$ By the first result on $H^2$ it gives rise to a surjective map $$ H^1(G, \mu_{p^s}){\longrightarrow}H^1(G,\mu_{p^{s+a}})\longrightarrow H^1(G,\mu_{p^a})\longrightarrow 0 $$ and, by taking projective limit over $s$, $$ H^1(G, \mathbb{Z}_p(1))\overset{p^a}{\longrightarrow}H^1(G,\mathbb{Z}_p(1)\longrightarrow H^1(G,\mu_{p^a})\longrightarrow 0 $$ which means $H^1(G, \mathbb{Z}_p(1))/p^aH^1(G, \mathbb{Z}_p(1))\cong H^1(G,\mu_{p^a})$. But this can be rewritten as $$ H^1(G, \mathbb{Z}_p(1))\otimes\mathbb{Z}/p^a\cong H^1(G, \mathbb{Z}/p^a(1)). $$ This is (1) for the module $\mathbb{Z}/p^a$ and finished the proof.<|endoftext|> TITLE: (Co)tensoring of enriched slice categories QUESTION [6 upvotes]: In an answer to this question: Enriched slice categories, a description of the enrichment of the slice category in an enriched category is given. I'm interested in going a bit further. If we assume that the original enriched category, say $\mathcal{C}$, is $\mathcal{V}$-enriched but also $\mathcal{V}$-tensored and cotensored, and that $\mathcal{V}$ is semi-Cartesian (i.e. the monoidal unit is terminal) how can we describe the tensoring and cotensoring of a slice $\mathcal{C}_{/X}$ for $X\in\mathcal{C}$? It seems to me that the tensoring should probably be some kind of weighted colimit, which I believe would mean we just compute it on the "underlying object." My immediate guess about what this would look like is the following: given an object $f:A\to X\in\mathcal{C}_{/X}$, the tensoring with $K\in\mathcal{V}$ is $$f\otimes K:A\otimes K\overset{A\otimes\varepsilon_K}\longrightarrow A\otimes 1_{\mathcal{V}}\cong A\overset{f}\to X$$ But first of all, I think this is probably wrong, and second of all, I'm wondering if there is an explicit description for the cotensoring as well, such that we have the usual isomorphisms $$\mathit{Map}_{\mathcal{C}_{/X}}(f\otimes K, g)\cong \mathit{Hom}_\mathcal{V}(K,Map_{\mathcal{C}_{/X}}(f,g))\cong \mathit{Map}_{\mathcal{C}_{/X}}(f, g^K)$$ I happen to be only interested in the case that $\mathcal{V}=sSet$, but perhaps there's a more general answer. I'm not even sure we need the condition that $\mathcal{V}$ is semi-Cartesian honestly, since it seems like the statement about tensors being colimits, and the forgetful functor preserving colimits, wouldn't require it. REPLY [3 votes]: Your guess for the copower (née tensor) is correct. You can check it by giving it the right universal property with respect to the enriched homs as described in the question you linked. Similarly, for the power (née cotensor) you can check that $g^K$ for $g:B\to X$ can be defined by the pullback of the induced map $B^K \to X^K$ along $X \cong X^{1_V} \to X^K$. However, I don't know offhand how to generalize away from the semicartesian case. Possibly the right point of view in that case is that the slice $C/X$ is not enriched over $V$ itself but over $V/I$; I think that it can be given powers and copowers for that enrichment.<|endoftext|> TITLE: Singularities of power series QUESTION [12 upvotes]: The power series $\sum_{n=1}^\infty \ln(n)z^n$ has radius of convergence $1$ and $z=1$ is a singular point. Is $z=1$ an isolated singularity? If yes, what kind of isolated singularity? I am only able to deduce that $z=1$ cannot be a pole. Such type of questions appear naturally when one tries to relate the singularities of the power series and those of the Dirichlet series associated to the same sequence. REPLY [18 votes]: Let $f(z)$ be your function. Then $g(z)=f(z)(1-z)$ is equal to $$ (1-z)\sum_n \ln(n)z^n=\sum_{n\geq 2} (\ln(n)-\ln(n-1))z^n $$ Now, $\ln(n)-\ln(n-1)=\frac{1}{n}+O\left(\frac{1}{n^2}\right)$, which gives us $$ g(z)=\sum_{n≥1} \left(\frac{1}{n}+g_n\right)z^n, $$ where $g_n=\ln(n)-\ln(n-1)-\frac{1}{n}=O\left(\frac{1}{n^2}\right)$ for $n>1$ and $g_1=-1$, so that $$ f(z)=\frac{\ln(1-z)}{z-1}+\frac{h(z)}{z-1}. $$ Here $h(z)$ is holomorphic in the unit disc and continuous on its boundary. Hope this answers your question on the type of singularity in $z=1$. Edit: Let me also present a slightly less elementary way to study properties of this series, based on my favorite method of lots of contour integration. We will use the derivative of Riemann zeta-function, so this is more in the spirit of the question. Let $x\in \mathbb C$ be a number with positive real part. Using the formula $$ e^{-nx}=\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} \Gamma(s)(nx)^{-s}ds, $$ we obtain $$ f(e^{-x})=\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty}\Gamma(s)\zeta'(s)x^{-s}ds. $$ From this we easily get $$ f(e^{-x})=\mathrm{Res}_{s=1}\,\Gamma(s)\zeta'(s)x^{-s}+\sum_{n\geq 0}\mathrm{Res}_{s=-n}\,\Gamma(s)\zeta'(s)x^{-s}. $$ The first summand is actually a bit different from all the other, because we get double pole. From expansions $$ \Gamma(s)=1-\gamma(s-1)+O((s-1)^2), \zeta'(s)=\frac{1}{(s-1)^2}+O(1) $$ and $$ x^{-s}=\frac{1}{x}-\frac{(s-1)\ln x}{x}+O((s-1)^2) $$ (here $\gamma$ is the Euler-Mascheroni constant) we get $$ \mathrm{Res}_{s=1}\,\Gamma(s)\zeta'(s)x^{-s}=-\frac{\ln x+\gamma}{x}, $$ which corresponds to the first part of my answer and also gives $h(1)=-\gamma$. The rest is way easier to compute and we obtain $$ f(e^{-x})=-\frac{\ln x+\gamma}{x}+\sum_{n\geq 0}\frac{(-1)^n\zeta'(-n)x^n}{n!}. $$ Now, from this answer about derivative of zeta we see that this series has a nonzero radius of convergence (namely, $2\pi$) and we can even see singularities at $x=2\pi i n$ for $n\in \mathbb Z$, which is of course what one should expect because of singularity of $f$ at $z=1$.<|endoftext|> TITLE: Forcing square introduces diamond QUESTION [5 upvotes]: Let $\mathbb S_\kappa$ be the standard forcing for $\square_\kappa$ by initial segments. This is $(\kappa+1)$-strategically closed. Observation: Let $T \subseteq \kappa^+$ be stationary. If $T$ concentrates on $\mathrm{cof}({<}\kappa)$,then $\mathbb S_\kappa$ forces $\diamondsuit(T)$. If $T$ concentrates on $\mathrm{cof}(\kappa)$ and $T$ is approachable, then $\mathbb S_\kappa$ forces $\diamondsuit(T)$. The proof is like the argument that adding a Cohen subset of a regular cardinal $\kappa$ forces $\diamondsuit_\kappa$. The coding goes by fixing bijections $f_\alpha : \alpha \to \kappa$ for $\alpha <\kappa^+$ and taking the diamond sequence to be $a_\alpha = \{ \beta<\alpha: \alpha+f_\alpha(\beta)+1 \in C_{\alpha+\kappa} \}$. It works because of the freedom we have in choosing $C_{\alpha+\kappa}$. In the $\mathrm{cof}(\kappa)$-concentration case, we use an elementary submodel argument plus approachability so that we can work with a chain of conditions of length only $\kappa+1$. Question 1: This seems like it might have been observed before. Does it appear in a paper? Question 2: Can we eliminate the approachability assumption? REPLY [5 votes]: Regarding Question 2, the assumption of approachability is necessary (or, more precisely, the assumption that $T$ has a stationary subset that is approachable), at least if we have $2^\kappa = \kappa^+$ in the ground model. The reason is that, if $T$ does not have a stationary subset that is approachable, then it becomes non-stationary after forcing with $\mathbb{S}_\kappa$, so $\diamondsuit(T)$ necessarily fails. To see this, note that if $2^\kappa = \kappa^+$, then we can fix in $V$ an enumeration $\vec{a} = \langle a_\alpha \mid \alpha < \kappa^+ \rangle$ of all bounded subsets of $\kappa^+$. Then a set $S \subseteq \kappa^+$ is in $I[\kappa^+]$ if and only if there is a club $C \subseteq \kappa^+$ such that $\gamma$ is approachable with respect to $\vec{a}$ for every $\gamma \in S \cap C$. This is because, if $\vec{b} = \langle b_\alpha \mid \alpha < \kappa^+ \rangle$ is any other sequence of bounded subsets of $\kappa^+$, then there is a club of $\delta < \kappa^+$ such that all entries in $\langle b_\alpha \mid \alpha < \delta \rangle$ appear in $\langle a_\alpha \mid \alpha < \delta \rangle$. But now the assumption that $T$ does not have a stationary subset in $I[\kappa^+]$ means that there is in fact a club $C \subseteq \kappa^+$ such that, for every $\gamma \in T \cap C$, $\gamma$ is not approachable with respect to $\vec{a}$. Forcing with $\mathbb{S}_\kappa$ does not add any new bounded subsets to $\kappa^+$, so if $G$ is generic, then $\vec{a}$ remains an enumeration of all bounded subsets of $\kappa^+$ in $V[G]$ and every ordinal $\gamma < \kappa^+$ is approachable with respect to $\vec{a}$ in $V[G]$ if and only if it is approachable with respect to $\vec{a}$ in $V$. However, $\square_\kappa$ holds in $V[G]$, so also $\mathrm{AP}_\kappa$ holds. Then, in $V[G]$, $T \cap C \in I[\kappa^+]$, but every ordinal in this set is not approachable with respect to $\vec{a}$ so, there must be a club $D \subseteq \kappa^+$ disjoint from $T \cap C$. Then $C \cap D$ is disjoint from $T$, so $T$ is non-stationary. Regarding Question 1, I made a similar but slightly different observation in my paper "Aronszajn trees, square principles, and stationary reflection" (see Lemma 3.11). I was forcing $\square(\lambda)$ rather than $\square_\kappa$, which slightly simplifies things. I just state the lemma for $\diamondsuit(\lambda)$, but the same argument works for $\diamondsuit(T)$ for any stationary $T \subseteq \lambda$. Approachability is not needed there, since the forcing notion for $\square(\lambda)$ is fully $\lambda$-strategically closed. I'm sure this had been recognized before, but I haven't seen it elsewhere in print.<|endoftext|> TITLE: Commutator subgroup of the absolute Galois group - a closed subgroup QUESTION [8 upvotes]: Let $K$ be a finite extension of $\mathbb{Q}$. Is it possible that the commutator subgroup of the absolute Galois group of $K$ (considered as an abstract group) is a closed subgroup? This property holds for the absolute Galois groups of $p$-adic fields because they are topologically finitely generated. REPLY [8 votes]: No, the abstract commutator subgroup $[G_K,G_K]$ of the absolute Galois group $G_K$ of a number field $K$ is never closed: Write $[G,G]$ for the commutator subgroup of $G$ as an abstract group, and $c(G)$ for the commutator width of $G$, i.e. the minimal $n$ such that every element in $[G,G]$ is a product of at most $n$ commutators. From the Baire category theorem one gets that $(*)$ the commutator subgroup $[G,G]$ of a profinite group $G$ is closed if and only if $c(G)$ is finite. Moreover, for a profinite group $G$ one clearly has $(**)$ $c(G) = \sup_{N\lhd_c G} c(G/N)$ i.e. the commutator width is the supremum of the commutator width of the (continuous) finite quotients. So in order to show that $c(G_K)=\infty$ (and hence $[G_K,G_K]$ is not closed) it would suffice to exhibit a family of finite quotients of $G_K$ of unbounded commutator width. One such family is the family of finite $p$-groups, for a fixed prime $p$: (1) By a celebrated theorem of Shafarevich [1], every such group (in fact every finite solvable group) is the Galois group of a finite Galois extension of $K$. (2) By a result of Roman'kov (Theorem 2 in [2]), there exists a finitely generated pro-$p$ group $F$ whose (abstract) second commutator group $F^{(2)}$ is not closed. By $(*)$ that means that the commutator subgroup $F'=[F,F]$ has $c(F')=\infty$. Therefore, by $(**)$, the family of finite quotients of $F'$, all of which are $p$-groups, has unbounded commutator width. EDIT: For a direct and short proof of this see the comment of YCor below. [1] https://en.wikipedia.org/wiki/Shafarevich%27s_theorem_on_solvable_Galois_groups [2] https://link.springer.com/content/pdf/10.1007/BF01987820.pdf<|endoftext|> TITLE: How to determine a highest weight corresponding to a parabolic subgroup? QUESTION [5 upvotes]: Let $G$ be a simply connected, semisimple algebraic group over $\mathbb C$ with maximal torus $T$ and Borel subgroup $B$ containing $T$. If $(V,\pi)$ is an irreducible representation of $G$, then $(V,d\pi)$ is an irreducible representation of the Lie algebra $\mathfrak g$ which has a unique highest weight $\lambda \in \mathfrak t^{\ast}$. I have read that if we identify the one-dimensional weight space $V_{\lambda} \subset V$ with a point in projective space $\mathbb P(V)$, then under the action $$ G \xrightarrow{\pi} \operatorname{GL}(V) \rightarrow \operatorname{Aut}(\mathbb P(V))$$ the stabilizer of $V_{\lambda}$ is a parabolic subgroup of $G$, and every parabolic subgroup of $G$ arises this way. How does one go in the opposite direction? If $P$ is a (let's say maximal) parabolic standard subgroup of $G$, how does one find a dominant integral weight whose corresponding irreducible representation determines $P$ in the above sense? Can the highest weight occur as the highest root of $T$ in the unipotent radical of $P$? REPLY [3 votes]: This is done in e.g. Baston-Eastwood (1989, pp. 40, 55): with $P$ characterized as usual by a subset $\mathcal S_{\mathfrak p}$ of the simple roots (for a maximal parabolic, $\mathcal S_{\mathfrak p}$ is all but one simple root), take $\lambda=$ sum of the fundamental weights $\varpi_i$ corresponding to simple roots $\alpha_i$ not in $\mathcal S_{\mathfrak p}$. Then $G/P$ is the coadjoint orbit of $\lambda$ under the compact real form, and $V$ is its geometric quantization. $\lambda$ can be the highest root (giving $V=$ adjoint representation) but then $P$ is usually not maximal.<|endoftext|> TITLE: Question about a lesser-known "class number formula" of Gauss QUESTION [12 upvotes]: My question refers to article 301 of section 5 of Gauss's D. A. - there Gauss gives an asymptotic formula for the mean number of classes of forms with positive discriminant ($D>0$): $$h(D) = \frac{4}{\pi^2}\log (D) + \delta$$ where $\delta$ is the following constant: $$\delta = \frac{8}{\pi^2}C+\frac{48}{\pi^4}\sum_{n=2}^{\infty}\frac{\log (n)}{n^2} - \frac{2 \log 2}{3\pi^2}$$ and $C$ is the Euler-Mascheroni constant. This formula is noteworthy because Gauss also evaluates the error term $\delta$ in this class formula. I tried to search on the web for references to this analytic result of Gauss but all I found is articles about his asymptotic class number formula for forms with negative discriminant ($D<0$), which Gauss gives in the next article of D. A. (article 302). My question is mainly about getting a reference for this result of Gauss. If such a reference doesn't exist, I'll be glad to hear an expert opinion on the historic significance of this formula (Gauss did publish this formula; thereby it's plausible that it did have influence). REPLY [3 votes]: I think the only detailed discussion of this particular result of Gauss is due to Dirichlet, and appears in his 1838 publication "on the use of infinite series in the theory of numbers", which is available on internet archive on p.376-391 of G. Lejeune Dirichlet's Werke. In fact, this publication of Dirichlet is discussed in a recent (2018) biography of Dirichlet (by Uta C. Merzbach), which cites the pages in Dirichlet's publication on Gauss's formula from D.A article 301: Finally, Dirichlet observed that by the same kind of analysis he could find the formulas presented in article 301 of Gauss's beautiful work: Suppose, for example, that it's a question of obtaining the mean number of genera for the determinant $-n$, a number which we shall denote by $F(n)$. If one compares art. 231 (of the D.A) where all the complete characters assignable a priori are enumerated, with art. 261 and 287, where the illustrius author showed that only half of these characters correspond to really existing genera, one could easily find (...) five equations (...) which (on appropriate summing and substituting), result in the asymptotic formula of the mean value of the number of genera for a determinant $-n$ as $$\frac{4}{\pi^2}(\mathbb{log (n)}+\frac{12C'}{\pi^2}+2C-\frac{1}{6}\mathbb{log (2)})$$ which coincides with the result of M. Gauss. Looking at Dirichlet's relatively long reconstruction of Gauss's result, i gained the impression that this is one of those results where truly rigorous analytic methods were needed in it's derivation (this is also evident from the title of Dirichlet's memoir). I think this makes clear that, despite not giving a proof for this result (not even in his Nachlass), Gauss was fully aware of some equivalent form of Dirichlet's L-series technique; Gauss's formula is too precise, and therefore it cannot be assumed that he conjectured it on empirical basis. This might seem to contradict Gauss's own statement from art. 301, according to which he discovered this result after a long tables study - but i guess the "tables study" needs to be interpreted as a "semi-empirical derivation".<|endoftext|> TITLE: Existence of fibered surfaces in arbitrary 4-manifolds? QUESTION [9 upvotes]: It is apparently a result of F. González-Acuña that all closed orientable 3-manifolds contain a fibered knot. (I am not sure exactly where to find a published proof of this result and as an aside I would be interested in hearing about any proofs/references that anyone knows.) I am wondering to what extent this is true in higher dimensions and specifically in dimension 4. In particular, is there always a fibered 2-sphere in a closed orientable 4-manifold? If not, what about arbitrary genus surfaces? REPLY [8 votes]: A fibered codimension $2$ link in a manifold $M$ can be identified with an open book decomposition of $M$ (the object that is described at the beginning of Bruno Martelli's answer). The existence of open books was studied in Quinn's paper: Open book decompositions, and the bordism of automorphisms. In particular, any odd-dimensional (closed oriented) manifold admits the structure of an open book. In even dimensions there are obstructions. For example, the only surface admitting an open book is $S^2$. A general obstruction is coming from the signature as Bruno Martelli was pointing out. Given an open book decomposition of $M$, it is relatively easy to construct an orientation reversing diffeomorphisms on the neighborhood of the codimension $2$ manifold and on its complement. Thus the signature has to vanish. In the above paper, Quinn identifies another (more complicated) algebraic obstruction to the existence of open books in even dimensions larger than $4$ and shows on the other hand that these are the only obstructions. In dimension $4$ the question about the existence of open book structures seems to be unsolved (apart from that the signature has to vanish). (Quinn's paper uses the $s$-cobordism theorem and thus the argument is probably not working in dimension $4$.)<|endoftext|> TITLE: Is there a ring for which the reducibility of a polynomial is undecidable? QUESTION [13 upvotes]: Let $R$ be a ring such that all of its elements have a finite number of divisors, ie $\forall r\in R\, |\{x\in R: x|r\}|<\infty$. Then we can decide whether a polynomial in $R[t]$ is reducible through Kronecker's method. Even in the ring $\mathbb{Q}[x,\sqrt{x},\sqrt[4]{x},...]$, where $x$ has an infinite number of divisors, it is easy to list any polynomial's possible factors, and reducibility is decidable. Is there a ring for which the reducibility of a polynomial is undecidable? Is there a countable ring with computable ring operations and decidable equality for which the reducibility of a polynomial is undecidable? REPLY [23 votes]: Yes, but the answer is a bit unsatisfying. This answer is a summary of the very nice paper Computable Fields and Galois Theory, Russel Miller, Notices of the AMS, 2008. First of all, if one could not even compute with the elements of the ring $R$ at all, it would be unclear what it would mean for factorization to be computable. The usual solution is to talk about "computable rings", meaning a countable (or finite) ring $R$ where the elements are indexed by integers and the operations of addition and multiplication are given by computable functions. The following is an example of a computable field in which factorization is not decidable: Let $p_n$ be the $n$-th prime and let $T_n$ be the $n$-th Turing machine and let $K = \mathbb{Q}(\sqrt{p_n} \ : \ T_n \ \mbox{halts})$. Given any element $\theta$ of $K$, there is a finite expression which witnesses that $\theta$ is in $K$: namely, the algebraic expression for $\theta$ in terms of finitely many $p_n$'s and a transcript of the ruunning of the corresponding Turing machines. This can be made into a proof that $K$ is computable, in particular (in response to Matt F.'s comment) there is no issue with testing equality. However, $x^2 - p_n$ is reducible iff $T_n$ halts, so we cannot test reducibility. The fields one normally meets in life do not have this issue. Reducibility is computable over $\mathbb{Q}$, $\mathbb{R} \cap \mathbb{Q}^{\mathrm{alg}}$ and over finite fields. If reducibility is computable over $k$, then it is computable over $k(t)$ and over $k[t]/f(t)$ for $f$ irreducible. That already covers the most obvious fields you know. Searching for "computable fields" turns up a fair bit of recent research.<|endoftext|> TITLE: Associativity may fail by little? QUESTION [14 upvotes]: It is a well-known result on group theory that if a group has many pairs of commuting elements then it is abelian. This motivated the following pseudo-conjecture. If a (possibly infinite) set $S$ with a binary operation $\cdot$ is such that for many triples $a,b,c\in S$ it holds $(a\cdot b)\cdot c = a \cdot (b\cdot c)$, then $(S,\cdot)$ is a semigroup. Exploring a little bit on this, a colleague told me he has read somewhere (he can't remember where) that the statement above is false. If $S$ is a finite set, then there exists a binary operation $\star$ that satisfies $(a\star b)\star c = a\star(b\star c)$ for all but just one ordered triple $(a,b,c)\in S\times S\times S$. Such a result implies that an algorithm that checks if a certain operation is associative must check indeed all triples of elements, which is kind of funny and rather unintuitive (to me, at least). We couldn't find a proof of this last result, although playing with small sets it seems to be true. My questions are the following: 1) Does anybody here know a reference for this result? 2) Is there a constructive proof for such an example? REPLY [15 votes]: The result you quoted appears in this reference: G. Szasz, Die Unabhängigkeit der Assoziativitätsbedingungen, Acta. Sci. Math. Szeged 15 (1953), 20-28. The Szasz theorem requires that the set $S$ have at least four elements, though it is also true for sets of size $3$. Szasz' proof is constructive and goes as follows. Assuming $a$, $u$, $v$ and $w$ are distinct members of $S$, define $a\cdot a = u$, $a\cdot u = v$ and $x\cdot y = w$, for any $(x,y)$ other than $(a,a)$ and $(a,u)$. Then a case by case verification shows that $(a\cdot a)\cdot a = w\neq v = a\cdot(a\cdot a)$ but that, for every other triple $(x,y,z)\in S^3$, we have $(x\cdot y)\cdot z = x\cdot(y\cdot z)$. Direct enumeration shows that there are (up to isomorphism) $10$ magmas of order $3$ with exactly one non-associative triple. (There are $124$ of order $4$.)<|endoftext|> TITLE: A special type of generating function for Fibonacci QUESTION [14 upvotes]: Notation. Let $[x^n]G(x)$ be the coefficient of $x^n$ in the Taylor series of $G(x)$. Consider the sequence of central binomial coefficients $\binom{2n}n$. Then there two ways to recover them: $$\binom{2n}n=[x^n]\left(\frac1{\sqrt{1-4x}}\right) \tag{1-1}$$ and $$\binom{2n}n=[x^n]\left((1+x)^2\right)^n. \tag{1-2}$$ This time, take the Fibonacci numbers $F_n$ then, similar to (1-1), we have $$F_n=[x^n]\left(\frac1{1-x-x^2}\right). \tag{2-1}$$ I would like to ask: QUESTION. Does there exist a function $F(x)$, similar to (1-2), such that $$F_n=[x^n]\left(F(x)\right)^n? \tag{2-2}$$ REPLY [5 votes]: The power series $F(x)$ is closely related to the series of the "exponential reversion of Fibonacci numbers" $$R(x)=\sum_{n\ge1}r_n\frac{x^n}{n!}$$ (the $r_n$ are A258943, quoted in a comment). In fact it appears that, again in the notation of Henri Cohen, $$a_{n+1}=nr_n,$$ equivalently $$F'(x)=xR'(x).$$ So if the Fibonacci numbers are encapsulated by $$x=\sum_{n\ge1}F_n\frac{y^n}{n!}=y+\frac{y^2}{2!}+2\frac{y^3}{3!}+3\frac{y^4}{4!}+5\frac{y^5}{5!}+\cdots,$$ the reverse series of this is $$ y=R(x)=\sum_{n\ge1}r_n\frac{x^n}{n!}=x-\frac{x^2}{2!}+\frac{x^3}{3!}+\color{red}{2}\frac{x^4}{4!}-\color{red}{25}\frac{x^5}{5!}+-\cdots,$$ while $$\begin{align}F(x)=1+\sum_{n\ge1}a_n {x^n} &=1 + x+\frac{x^2}{ 2!}-2\frac{x^3}{ 3!}+3\frac{x^4}{4!}+8\frac{x^5}{5!}-125\frac{x^6}{6!}+-\cdots\\ &=1 + x+\frac{x^2}{ 2!}-2\frac{x^3}{ 3!}+3\frac{x^4}{4!}+4\cdot\color{red}{2}\frac{x^5}{5!}-5\cdot\color{red}{25}\frac{x^6}{6!}+-\cdots\end{align}.$$ Possibly this relationship is not even specific to the Fibonacci numbers. EDIT: It looks like the sequence $\{a_n\}$ has finally yielded its secrets. Given the conjectured "smoothness" of these coefficients (i.e. all prime factors are relatively small) as mentioned in Henri Cohen's answer, I have looked again into the factors and those quadratic sequences mentioned in the comments, and fortunately there are enough primes in them, such that finally I was able to find the pattern! We have for the sequence $\{r_n\}$ $${ r_n=\begin{cases} {(-1)^k} \prod\limits_{j=1}^k(n^2-5nj+5j^2) \quad\text{for }\ n=2k-1, \\ \\ {(-1)^kk\cdot} \prod\limits_{j=1}^{k-1}(n^2-5nj+5j^2) \quad\text{for }\ n=2k. \end{cases}}$$ Once found, it should not be hard to prove that rigorously. As pointed out by Agno in a comment, we can reduce to linear factors and write the product in terms of the Gamma function simply as $$r_n= {\sqrt5^{ \,n-1 }\frac { \Gamma \left( \frac{5-\sqrt {5}}{10}n \right)}{ \Gamma \left( 1-\frac{5+\sqrt {5}}{10}n \right) }}.$$More generally, if we start with a Lucas sequence $$f_0=0,\ f_1=1,\ f_n=pf_{n-2}+qf_{n-1}\quad(n\ge2),$$ the reversed series has $$\boxed{r_n= {\sqrt{4p+q^2}^{ \,n-1 }\frac { \Gamma \left[\dfrac n2 \Bigl(1-\dfrac{q}{\sqrt{4p+q^2}} \Bigr)\right]}{ \Gamma \left[1-\dfrac n2 \Bigl(1+\dfrac{q}{\sqrt{4p+q^2}} \Bigr)\right]}}}.$$ Note that whenever the argument in the denominator is a negative integer, the coefficient $r_n$ vanishes, e.g. this happens when $p=3,q=2$ for all $n\equiv0\pmod4$. As far as the sequence of the signs, it is after all quite regular and is in fact self-similar (that is, unless $\sqrt{4p+q^2}$ is rational). This self-similar behaviour of the signs can be seen by virtue of the (negative) argument of the Gamma function in the denominator, knowing that $\Gamma$ changes signs at each negative integer and the multiples of $\sqrt{4p+q^2}$ occurring in the argument do the rest. (Think e.g. of the self similarity features of the Wythoff sequence.) For the reversion of the original Fibonacci sequence, I have displayed here the signs of the first $1500$ even and then the first $1500$ odd coefficients and found that their quasi periodicity comes out nicely when putting exactly $76$ in each row (writing "o" instead of "$-$" for better visibility). The "longest pairings" are colored: all patterns "$++--++--++$" in yellow and all patterns "$--++--++--$" in blue.<|endoftext|> TITLE: What (or how) are the new spaces of derived algebraic geometry? QUESTION [26 upvotes]: I am a beginner in derived algebraic geometry and I am trying to develop some visual and geometrical intuition about derived schemes (and stacks), or more precisely about the new geometrical phenomena that they introduce. It seems to me that (broadly speaking) the new spaces that derived geometry gives rise to are: (Possibly higher dimensional) Loop spaces. They arise as self-intersections: e.g. see comments of J.Pridham and the answer of DamienC (1) below. Derived infinitesimal disks/Formal neighbourhoods. Originated by nilpotent extensions. See for example the definition 1.1 in Vezzosi - A note on the cotangent complex in derived algebraic geometry. QUESTION: What else? (See also an answer of DamienC below (2)). I think that while 1 and 2 are already present in derived schemes other phenomena require derived stacks. I would like to see more examples that have some geometrical interpretation. There are cases of derived stacks for example in Toen - Higher and derived stacks: A global overview. Such examples include the derived stack of rank $n$ local systems over some topological space (and the derived moduli stack of vector bundles), derived linear stacks, and the derived stack of perfect complexes. However I am unable to obtain a geometrical meaning for this examples. EDIT: What is the geometrical interpretation of the higher homology groups in (for example) the derived stack of vector bundles over a projective variety $\mathbb{R} \underline{{Vect}_{n}}(X)$? According to this paper from Toen-Vezzosi some of the motivation for this derived stack comes from the will to build a smooth moduli space (unlike the underived case). When $X=S$, a smooth projective surface, they claim that the tangent space at a point $E$ is: $T_{E} \mathbb{R} \underline{\operatorname{Vect}}_{n}(S) \simeq-H^{2}(S, \underline{E n d}(E))+H^{1}(S, \underline{E n d}(E))-H^{0}(S, \underline{E n d}(E))$. However here the $H^{2}$ term (which is the derived part) seems to come from the fact of $S$ is $2$-dimensional and not from any singularity or self-intersection (which seems strange to me) If you look at the example of (2), which is quite similar (I think it is the derived stack of local systems), the $H^{2}$ term appears when you take into account the self intersection of $0$ in $\mathbb{A}^{1}$, (i.e. a truly derived structure). What I am misunderstanding here? REPLY [4 votes]: $\DeclareMathOperator\Map{Map}\DeclareMathOperator\ad{ad}$This is an attempt to answer the third question: what else? Let $X$ be a compact space and let $G$ be an affine algebraic group. One can contemplate the following (underived) higher stacks: $BG$: the classifying stack of $G$-torsors. $X_B$: the constant stack associated to $X$. One can consider the higher underived mapping stack $\Map(X_B,BG)$, which is nothing but the ordinary (ie non-derived) stack of $G$-local systems on $X$. Its tangent complex has amplitude $[-1,0]$: in degree $-1$, at a $k$-point $P$ ($P$ is a $G$-local system), its cohomology is $H^0(X,\ad(P))$, where $\ad(P)$ is the linear local system associated with $P$ and the adjoint $G$-representation $\mathfrak{g}$: $\ad(P)=P\times_G\mathfrak{g}$. in degree $0$, at a $k$-point $P$, its cohomology is $H^1(X,\ad(P))$. The infinitesimal theory $\Map(X_B,BG)$ doesn't capture anything about higher cohomology groups $H^{*\geq 2}(X,\ad(P))$. If you're looking at the derived mapping stack $\mathbb{R}{\Map}(X_B,BG)$ instead, then its tangent complex at a $k$-point $P$ is the full de Rham cohomology $H^{*+1}(X,\ad(P))$. Why is this so? The point is that the underived stack $\Map(X_B,BG)$ doesn't see anything else than the fundamental groupoid of $X$: this is because $BG$ is a $1$-truncated Artin stack. But if we allow families of $G$-local systems parametrized by geometric objects intrinsically carrying homotopical information (affine derived schemes), then we get back the missing information. For instance, it's a good exercise to check that if $Y$ is the derived self-intersection of $0$ in the affine line that was mentionned in previous answers (i.e. $Y=\operatorname{Spec}(k[\tau])$, $\operatorname{deg}(\tau)=-1$), then a $Y$-point in $\mathbb{R}{\Map}(X_B,BG)$ is the datum of a $k$-point $P$ and a class in $H^2(X,\ad(P))$.<|endoftext|> TITLE: Classifying space for Thompson's group F? QUESTION [12 upvotes]: Let $\mathcal C$ be the free monoidal category generated by an object $X$, and a morphism $X \otimes X \to X$. This category contains exactly two connected components: that of the monoidal unit $1\in \mathcal C$, and that of $X\in \mathcal C$. (In general, two object $A$ and $B$ of a category are said to be in the same connected component if they are related by a zig-zag of arrows $$A\to Y_1\leftarrow Y_2\to Y_3\leftarrow Y_4\to Y_5\leftarrow\ldots \to B.$$ In the case of $\mathcal C$, any two non-unit objects are related by a single morphism.) Let $\mathcal C'\subset \mathcal C$ be the connected component of $X$ and let $|\mathcal C'|$ denote the geometric realisation (of the simplicial nerve) of $\mathcal C'$. The article Marcelo Fiore, Tom Leinster, An abstract characterization of Thompson's group $F$, Semigroup Forum 80 (2010), 325-340, doi:10.1007/s00233-010-9209-2, arXiv:math/0508617. proves that $\pi_1(|\mathcal C'|)$ is isomorphic to Thompson's group $F$. Question: Is $|\mathcal C'|$ a classifying space for Thompson's group $F$? REPLY [21 votes]: What you describe is the so called Squier complex of the semigroup presentation $\langle x \mid x^2=x\rangle$ (you did not describe the 2-cells, but it is straightforward). The fact that its fundamental group is $F$ was proved by Guba and myself in 1997, "Diagram groups", Memoirs of the AMS, November, 1997 (link). Farley in "Finiteness and CAT(0) properties of diagram groups", Topology 42 (2003), no. 5, 1065–1082 doi:10.1016/S0040-9383(02)00029-0, author pdf proved that its universal cover is a $CAT(0)$ cube complex. So indeed the Squier complex is a classifying space for $F$. The proof of all these from the category theory point of view can be found in Guba, V. S.; Sapir, M. V. "Diagram groups and directed 2-complexes: homotopy and homology. " J. Pure Appl. Algebra 205 (2006), no. 1, 1–47.<|endoftext|> TITLE: Sign in Dirichlet's approximation theorem QUESTION [6 upvotes]: Fix $\alpha \in \mathbf{R}$. The classical Dirichlet's approximation theorem states there exist infinitely many rationals $p/q$ such that $$ \left|\alpha-\frac{p}{q}\right|<\frac{1}{q^2}. $$ Question. Fix $\alpha \in \mathbf{R}$. Is it true that there exist infinitely many rationals $p/q$ such that $$ 0\le \alpha- \frac{p}{q}\ll\frac{1}{q^2}\,\,? $$ REPLY [10 votes]: Yes, this follows from considering the continued fraction of $\alpha$. If $p_n/q_n$ is the $n$th convergent to $\alpha$ and $n$ is odd then $$ 0\leq \alpha - \frac{p_n}{q_n} \leq \frac{1}{q_n^2}.$$<|endoftext|> TITLE: When is a homotopy pushout contractible? QUESTION [16 upvotes]: Let $B \leftarrow A \to C$ be a span of spaces, and consider the homotopy pushout $B \cup_A C$. Question: When is $B \cup_A C$ contractible? This is a pretty open-ended question. I'm interested in necessary conditions or sufficient conditions or interesting examples or special cases, etc. Some additional conditions I'd be happy to take as blanket assumptions: I don't think too much is lost if we assume that $A,B,C$ are connected. I'm happy to assume that $A \to B$ is 1-connected (conventions vary about what this means; I mean that the homotopy fiber of $A \to B$ is 1-connected, i.e. that $A \to B$ is $\pi_1$-surjective). For somewhat obscure reasons, I'm particularly interested in the case where $A \to C$ is the projection $A_0 \times C \to C$ out of a binary product. For similarly obscure reasons, I'm particularly interested in the case where $C$ is a loopspace, or even where $C = \Omega \Sigma C_0$ is a free loopspace. Observation: By pasting on a few more pushout squares, we obtain the identity $\Sigma(B \vee C) = \Sigma A$. I'm not sure what to make of this, though. REPLY [10 votes]: Let me add some additional remarks on the enumeration question: How many such spaces $A$ are there sitting over $B\times C$ such that the homotopy pushout $$ B \leftarrow A \to C $$ is contractible? If $B\cup_A C$ is contractible, then the sum of the maps $\Sigma A \to \Sigma B$ and $\Sigma A \to \Sigma C$ gives a homology isomorphism $$ \Sigma A \to \Sigma B \vee \Sigma C\, . $$ If $A$ is connected, then we conclude that the map is also a weak homotopy equivalence, so $\Sigma A$ is a wedge of $\Sigma B$ and $\Sigma C$ in this case. The converse to this statement is also true. Hence, we are reduced to classifying those spaces $A$ whose suspension splits as a wedge of $\Sigma B$ and $\Sigma C$. Hence, this problem can be formulated as a kind desuspension problem. There is a classifying space ${\cal D}(B,C)$ which is the realization of (the nerve of) a category whose objects are spaces $A$ with structure map $A \to B\times C$ such that $\text{hocolim}(B \leftarrow A \to C)$ is contractible. A morphism of this category is a weak homotopy equivalence of spaces over $B\times C$. Classification Result: Assume $A,B,C$ are $1$-connected. There is a function $$ \pi_0({\cal D}(B,C)) \to \{B\vee C, B\wedge C\} $$ (the target is the abelian group of homotopy classes of stable maps) which is a bijection in a metastable range $$ \max(b,c) \le 3\min(r,s)-1\, , $$ where $b$ is the homotopy dimension of $B$ (the CW complex of minimal dimension having the homotopy type of $B$), $c$ is the homotopy dimension of $C$, $r$ is the connectivity of $B$ and $s$ is the connectivity of $C$. Here I have given $B$ and $C$ basepoints. The function in (4) is easily described as the stable homotopy class of the weak map $$ \Sigma B\vee \Sigma C \overset{\simeq}\leftarrow \Sigma A \to \Sigma (B \wedge C) $$ where the right arrow is given by suspending the composition $A \to B\times C \to B\wedge C$. If we want to work beyond the metastable range, the homotopy type of ${\cal D}(B,C)$ can be determined "up to extensions" via the coefficients of the identity functor in homotopy functor calculus. See the paper Klein, John R.; Peter, John W. Fake wedges. Trans. Amer. Math. Soc. 366 (2014), no. 7, 3771–3786 which extends my earlier paper: John R. Klein, Structure theorems for homotopy pushouts. I. Contractible pushouts, Math. Proc. Cambridge Philos. Soc. 123 (1998), no. 2, 301–324.<|endoftext|> TITLE: Do these surfaces intersect? QUESTION [9 upvotes]: For any real numbers $a_{1},a_{2},\cdots a_{6}$ and $b_{1},b_{2},\cdots b_{6}$ with $\sum_{i=1}^{6}a_{i}^{2}=1$ and $\sum_{i=1}^{6}b_{i}^{2}=1$, does the equation $$ x_{1}^{2}x_{2}^{2}x_{3}^{2}x_{4}^{2}\left(\sum_{i=1}^{6}a_{i}x_{i}\right)^{2}\left(\sum_{i=1}^{6}b_{i}x_{i}\right)^{2}=1 $$ always have a solution $x_{1},x_{2},\cdots x_{6}$ in $\mathbb{R}$ with $\sum_{i=1}^{6}x_{i}^{2}=6$? Thanks. REPLY [5 votes]: The answer is "yes" though the argument is rather ad hoc and doesn't generalize to vectors in more general positions. We have $6$ unit vectors $v_j$ out of which the first $4$ are pairwise orthogonal and want to show that there exists a vector of length $\sqrt 6$ such that $\prod_{j=1}^6 |\langle x,v_j\rangle|\ge 1$ (to get below $1$ is trivial). Consider all sums $y=\sum_{j=1}^6\varepsilon_j v_j$ where $\varepsilon_j=\pm 1$ and choose the one with the largest length. Replacing some $v_j$ with $-v_j$, if necessary, we can assume WLOG that it is $y=\sum_j v_j$. Comparing $y$ with $y-2v_j$ (one sign flip), we see that $\langle y,v_j\rangle\ge 1$ for all $j$. Unfortunately, $y$ is a bit long, but it cannot get the length greater than $4$ (the $4$ pairwise orthogonal vectors produce length $2$) and we have $$ \|y\|^2=\sum_j \langle y,v_j\rangle=:\sum_j (1+u_j), \quad 0\le u_j\le 3 $$ Reducing the length to $\sqrt 6$ means that we have to multiply $y$ by $\left(1+\frac 16\sum_j u_j\right)^{-1/2}$, so it suffices to show that $$ \prod_j(1+u_j)\ge \left(1+\frac 16\sum_j u_j\right)^3 $$ i.e. $$ \prod_j(1+u_j)^{1/3}\ge 1+\frac 16\sum_j u_j. $$ However, on $[0,3]$, we have $(1+u)^{1/3}\ge 1+\frac u6$ (the LHS is concave, so it is enough to check the endpoints) and Bernoulli finishes the story.<|endoftext|> TITLE: Fubini's theorem on arbitrary foliations QUESTION [6 upvotes]: In what follows $ \mathbb{R}^{n+m} = \{(x,y): x \in \mathbb{R}^n, \ y \in \mathbb{R}^m \} \ .$ Suppose $G: U \to V $ is a $C^1$-diffeomorphism from an open subset of a manifold to an open subset of $\mathbb{R}^{n+m}$. We write $$ (\xi,\eta)= G^{-1}(x,y) \ . $$ This is a local parameterization of $U$. We will think of $$ U_\eta: = G^{-1}(\mathbb{R}^n \times \{y\} \cap V) $$ as the "horizontal" fibers, which are $n$-dimensional $C^1$-manifolds. Similarly, we refer to $$ U_\xi = G^{-1}( \{x\} \times \mathbb{R}^m \cap V) $$ as the ``vertical" fibers, which are $m$-dimensional $C^1$-manifolds. Question: Is there a version of the Fubini's theorem that equates the integral over $U$ to a double integral over fibers? I am wondering if such a result exists in any textbook. In any other publication? It must be, because it is so natural to desire an integration over the original foliations. This is a paraphrase of the question asked previously here. I came up with the following: Let$ DG_{|U_\xi} (\xi,\eta)$ be the derivative, at point $(\xi,\eta)$, of the map $G_{|U_\xi}$, i.e. the restriction of $G$ to $U_\xi$. This restriction is from an $m$-dimensional $C^1$-manifold to a subset of $\mathbb{R}^m$.} So its derivative and its Jacobian are defined. Suppose for some fixed $\eta_0$, the union of vertical fibers along $U_{\eta_0}$ covers the set $U$. Lemma Under the assumptions above, for any integrable function $f: U \to \mathbb{R} $, $$ \int_U f = \int_{U_{\eta_0}}\left(\int_{U_\xi} f(\xi,\eta) \frac{|\det DG_{U_\xi} (\xi,\eta)| \cdot |\det DG_{U_{\eta_0}} (\xi,\eta_0)|}{|\det DG(\xi,\eta)|} \ d\mathcal{H}^m(\eta)\right) d\mathcal{H}^n(\xi) \cdot $$ That is saying that to integrate over $U$ simply integrate along the fibers and then sum over the "base", which is what Fubini's theorem says in the standard orthogonal coordinates of the Euclidean space. The correcting factor is can be memorized as $$ \frac{\text{ Jacobian along fibers} \times \text {Jacobian along base}}{\text{full Jacobian}} \ .$$ Corollary: Integration in spherical coordinates. There, the fibers are orthogonal to the base, and so, the full Jacobian also factors, cancelling the numerator. Therefore, the correcting factor disappears (=1), and we get the familiar $$ \int_U f \ d \mathcal{L}^n = \int_0^\infty \left( \int_{\mathbb{S}^{n-1}(r)\cap U} f \ d \sigma \right) dr \ . $$ Example: Let $P$ be a parallelepiped in the plane, resulting from tilting a rectangle so that the acute angle between its edges $A$ and $B$ is $\theta$. Then, foliating $P$ by parallel copies of the edges $B$, indexed $B_x$, leads to $$ \int_P f \ d \mathcal{L}^2 = \int_A \left( \int_{B_x} f \sin(\theta) \ d \mathcal{H}^1 \right) dx \ = \sin(\theta) \int_A \left( \int_{B_x} f \ d \mathcal{H}^1 \right) dx \ . $$ Again my questions are: 1) Is there some alternative formula known out there? 2) Are there references giving specifically this? (I am not interested in "it can follow"s and "it can be done"s!) REPLY [7 votes]: A nice version of the manifold version of Fubini's theorem is in Differentialgeometrie und Fasserbündel by Rolf Sulanke and Peter Wintgen: Let $\phi \in C^1(M,N)$, where $M,N$ are smooth manifolds of dimensions $m,n$, respectively, with $m \ge n$. Let $\omega \in \Omega^{m-n}(M)$ and $\eta \in \Omega^n(N)$, and let $f : M \to {\bf R}$ be measurable (meaning, its superposition with any map is Lebesgue measurable). Assume the set of critical values of $\phi$ has measure zero in $N$ (again, this means the image under any map has Lebesgue measure zero). If the $m$-form $f\omega\wedge\phi^*\eta$ is integrable on $M$, then for almost all $x \in N$ the integral: $$ \int\limits_{\phi^{-1}(x)} f\omega $$ is well-defined and, moreover, when treated as a function of $x$ and multiplied by $\eta$, it is integrable on $N$ and: $$ \int\limits_M f\omega\wedge\phi^*\eta = \int\limits_N \bigg( \int\limits_{\phi^{-1}(x)} f\omega \bigg)\, \eta $$<|endoftext|> TITLE: Scaling in Mehta's integral QUESTION [9 upvotes]: The following expression is known as Mehta's integral and deeply connected to random matrix theory: $$\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \prod_{i=1}^n e^{-t_i^2/2} \prod_{1 \le i < j \le n} |t_i - t_j |^{2 \gamma} dt_1 \cdots dt_n =\prod_{j=1}^n\frac{\Gamma(1+j\gamma)}{\Gamma(1+\gamma)}.$$ An interesting question is what happens if one assumes $\gamma$ to be a function of $n.$ For example by choosing $\gamma=1/n$ one finds that as $n$ tends to infinity, the value of the integral tends to zero whereas for $\gamma=1/n^2$ the value of the integral approaches a positive constant value as $n$ tends to infinity. These properties one can deduce from the asymptotics of the product of gamma functions. I would like to ask: It is not too surprising that for some suitable scaling $\gamma=1/n^{\alpha}$ one approaches a constant value, as $\vert t_i-t_j \vert^{1/n} \xrightarrow 1$ for fixed $t_i,t_j$ and $$\frac{1}{(2\pi)^{n/2}}\int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \prod_{i=1}^n e^{-t_i^2/2} dt_1 \cdots dt_n =1.$$ Can one also conclude these two properties from the integral directly without evaluating it? REPLY [10 votes]: Yes, this follows by the de la Vallée-Poussin necessary and sufficient condition for the uniform integrability. Indeed, suppose that \begin{equation} \gamma n^2\to a \end{equation} (as $n\to\infty$) for some real $a\ge0$. Your integral is $$E\prod_{1\le i TITLE: Prove category of constructible sheaves is abelian QUESTION [6 upvotes]: Let $X$ be a nice enough topological space, perhaps a complex algebraic variety with its analytic topology. I'm hoping someone could help me prove that the category $\text{Constr}(X)$ of constructible sheaves on $X$ is abelian. I think one could use that subsheaves and quotients of local systems are local systems. However, conceptually what is confusing me is that given two constructible sheaves $\mathcal{F}, \mathcal{G}$, they might be constructible with respect to completely different Whitney stratifications of X, right? So given a morphism $\mathcal{F} \to \mathcal{G}$ of constructible sheaves, can you always "refine" the two Whitney stratifications to get one such that the morphism restricts to a morphism of local systems? Or should I be thinking along another lines? REPLY [12 votes]: Verdier has proved (for multiple classes of spaces, including in particular complex varieties) that for any finite set of analytic subsets, there exists a Whitney stratification for which all these analytic subsets are unions of strata. In particular, given two constructible sheaves, and a stratification on which they are lisse, there exists a Whitney stratification that refines both of them. This suffices to check that kernels and cokernels of maps of constructible sheaves are constructible, as well as direct sums of constructible sheaves.<|endoftext|> TITLE: The diamond principle for functors QUESTION [7 upvotes]: Let $F:\mathbf{Comp}\to\mathbf{Set}$ be a continuous functor from the category of compact Hausdorff spaces to the category of sets such that $|Fn|\le\mathfrak c$ for any finite ordinal $n$. The continuity of $F$ means that $F$ preserves limits of inverse spectra. Typical examples of such a functor $F$ are the functors of countable power, of the hyperspace, or of spaces of probability measures. I am interested in validity of the following functorial version of the Jensen Diamond principle: $\diamondsuit_F$: there exists a transfinite sequence $\langle \mu_\alpha:\alpha\in\omega_1\rangle$ such that $\bullet$ $\mu_\alpha\in F(2^{\alpha})$ for every $\alpha\in\omega_1$; $\bullet$ for every $\mu\in F(2^{\omega_1})$ the set $\{\alpha\in\omega_1:\mu_\alpha=F\pi_\alpha(\mu)\}$ is stationary in $\omega_1$. Here $\pi_\alpha:2^{\omega_1}\to 2^\alpha$, $\pi_\alpha:f\mapsto f{\restriction}\alpha$, is the projection onto the $\alpha$th face of $2^{\omega_1}$. Observation. The classical Jensen Diamond Principle is just $\diamondsuit_{Id}$ for the identity functor $Id$. It is equivalent to the Principle $\diamondsuit_{Id^\omega}$ for the functor $Id^\omega$ of countable power. Using the Parovichenko Theorem, it can be shown that the Jensen Diamond Principle is equivalent to $\diamondsuit_{\exp}$ for the functor $\exp$ of hyperspace. I am interested in $\diamondsuit_P$ for the functor $P$ of probability measures. Problem. Does $\diamondsuit_P$ follow from the Jensen diamond principle? Or it is a stronger statement (still holding in the Constructible Universe)? REPLY [3 votes]: The answer to this question is affirmative and follows from a more general Theorem. Let $X_{\omega_1}$ be the limit of a continuous well-ordered spectrum $\langle X_\alpha,p_\alpha^\beta:\alpha\le \beta<\omega_1\rangle$ in the category of sets such that each set $X_\alpha$, $\alpha<\omega_1$ has cardinality $\le\mathfrak c$. The Jensen Diamond Principle $\diamond$ implies that there exists a transfinite sequence $(\mu_\alpha)_{\alpha\in\omega_1}\in\prod_{\alpha\in\omega_1}X_\alpha$ such that for any $\mu\in X_{\omega_1}$ the set $\{\alpha\in\omega_1:\mu_\alpha=p^{\omega_1}_\alpha(\mu)\}$ is stationary in $\omega_1$. Proof. Let $C=2^\omega$ be the Cantor cube. For ordinals $\alpha\le\beta\le\omega_1$ let $\pi_\alpha^\beta:C^{\beta}\to C^\alpha$, $\pi_\alpha:x\mapsto x{\restriction}\alpha$, the projection onto the $\alpha$th face of the cube $C^\beta$. It is well-known that $\diamondsuit$ implies the existence of a transfinite sequence $(x_\alpha)_{\alpha\in\omega_1}\in C^{\omega_1}$ such that for any $x\in C^{\omega_1}$ the set $\{\alpha\in\omega_1:x_\alpha=\pi^{\omega_1}_\alpha(x)\}$ is stationary in $\omega_1$. Taking into acount that the spectrum $\langle X_\alpha,p_\alpha^\beta:\alpha\le \beta<\omega_1\rangle$ is continuous and consists of sets of cardinality $\le\mathfrak c=|C|$, it is possible to construct inductively a transfinite sequence of injective maps $(f_\alpha:X_\alpha\to C^\alpha)_{\alpha\le\omega_1}$ such that for any $\alpha<\beta\le\omega_1$ we have the equality $f_\alpha\circ p_\alpha^\beta=\pi^\beta_\alpha\circ f_\beta$. Let $\Omega=\{\alpha\in\omega_1:x_\alpha\in f_\alpha(X_\alpha)\}$. Let $(\mu_\alpha)_{\alpha\in\omega_1}\in\prod_{\alpha\in\omega_1}X_\alpha$ be any transfinite sequence such that $f_\alpha(\mu_\alpha)=x_\alpha$ for any $\alpha\in\Omega$. We claim that this sequence has the required property. Given any $\mu\in X_{\omega_1}$ consider the element $x=f_{\omega_1}(\mu)\in C^{\omega_1}$. The choice of the transfinite sequence $(x_\alpha)_{\alpha\in\omega_1}$ ensures that the set $S=\{\alpha\in\omega_1:x_\alpha=x{\restriction}\alpha\}$ is stationary in $\omega_1$. For every $\alpha\in S$ we have $$x_\alpha=x{\restriction}\alpha=\pi_\alpha^{\omega_1}(x)=\pi_\alpha^{\omega_1}\circ f_{\omega_1}(\mu)=f_\alpha\circ p_\alpha^{\omega_1}(\mu)\in f_\alpha(X_\alpha)$$ and hence $\alpha\in\Omega$ and $f_\alpha(\mu_\alpha)=x_\alpha=f_\alpha\circ p_\alpha^{\omega_1}(\mu)$. Now the injectivity of the map $f_\alpha$ ensures that $\mu_\alpha=p_\alpha^{\omega_1}(\mu)$. Therefore, the set $\{\alpha\in\omega_1: \mu_\alpha=p_\alpha^{\omega_1}(\mu)\}\supset S$ is stationary in $\omega_1$.<|endoftext|> TITLE: Multisignature and homeomorphism type QUESTION [5 upvotes]: In classical surgery theory, there is a map $$L_{n+1}(\pi_1M)\to S(M^n)$$ Element in $L_{n+1}(\pi_1M)$ is realized as surgery obstruction of a surgery problem to $M\times I$ with one boundary piece the identity map and the other a homotopy equivalence (Wall's realization). The map is defined by sending the element in $L$ to the boundary piece which is a homotopy equivalence (a structure on $M$).It is NOT clear to me if the domain manifold of the structure is homeomorphic to $M$. In Homology spheres and fundamental group (when $n+1$ is even) Danny Ruberman commented that "The effect of the action is to change the multisignature, and hence it changes the homeomorphism type of M" What's the correct reference if i want to understand some details about the effect of Wall's realization on multisignature? REPLY [2 votes]: There is perhaps some confusion over the terminology. Wall (chapter 13A) uses the term multisignature to denote a collection of invariants of certain Hermitian forms over group rings, giving rise to a function from $L_{2k}(\pi) \to \mathbb{Z}^n$. In that chapter, he interprets the multisignature in terms of equivariant signatures. Such signatures occur in many places in geometric topology, for instance as the Tristram-Levine signatures and Casson-Gordon invariants of knots. One main use of the multisignature comes in the study of odd-dimensional manifolds. Given such a manifold $Y^{2k-1}$ and a finite regular covering with covering group $G$, classified by a homomorphism $\phi:\pi_(Y) \to G$, then one shows that some multiple (disjoint union) $N\cdot(Y,\phi)$ extends to a $2k$-manifold $(X,\Phi)$. Then (suitably normalized) the equivariant signature of $(X,\Phi)$ is an invariant $\rho(Y,\phi)$. This can be packaged in various ways, see for instance chapter 14E in Wall, where for a manifold with finite fundamental group $G$, one obtains an invariant $\rho$. Technically one has to choose an identification of $\pi_1(Y)$ with G, and $\rho$ depends on this identification. A priori, the fact that one gets an invariant depends on the Atiyah-Singer index theorem, and so is really an invariant up to diffeomorphism. But a bordism argument (Wall 14B) shows that it is actually a homeomorphism invariant. Putting these notions together, the way that the invariant $\rho$ is defined tells you that if one acts on (the identity map from $Y$ to itself) by an element $A \in L_{2k}(\pi_1(Y))$ then the invariant $\rho$ of the manifold $Y'$ at the other end of the resulting normal cobordism satisfies $\rho(Y') - \rho(Y) = $ multisignature of $A$. If this multisignature is non-trivial, then there is no homeomorphism of $Y$ and $Y'$ that respects the identification of their fundamental groups given by the normal cobordism. If you want to argue that they are simply not homeomorphic you need to look a little more closely.<|endoftext|> TITLE: Smoothness of family of distributions QUESTION [6 upvotes]: Let $X$ be a compact manifold. Denote by $\mathscr{D}^\prime(X \times X)$ the space of tempered distributions on the cartesian product $X \times X$. Given two test functions $\varphi, \psi \in \mathscr{D}(X)$, an element $T \in \mathscr{D}^\prime(X \times X)$ can be evaluated at the function $\varphi \otimes \psi$ on $X \times X$ defined by $(\varphi \otimes \psi)(x, y) := \varphi(x) \psi(y)$. Suppose now that $T_\lambda \in \mathscr{D}^\prime(X \times X)$, $\lambda \in\mathbb{R}$, is a family of distributions such that $$\lambda \longmapsto T_\lambda[\varphi \otimes \psi]$$ is a smooth function from $\mathbb{R}$ to $\mathbb{R}$ for any two $\varphi, \psi \in \mathscr{D}(X)$. Q: Does it follow that also the function $$ \lambda \longmapsto T_\lambda[\Phi]$$ is smooth for every $\Phi \in \mathscr{D}(X \times X)$? REPLY [3 votes]: Here is a proof using convenient analysis: By the kernel theorem $\mathscr{D}^\prime(X \times X) = L(\mathscr{D}(X),\mathscr{D}^\prime(X))$ and by The Convenient setting of Global Analysis, 5.18 (which is just the uniform boundedness principle) and 2.14 we have \begin{align*} &\lambda\mapsto T_\lambda(\Phi) \in \mathbb R \text{ is }C^\infty \quad\forall \Phi \in \mathscr{D}(X \times X) \\ \iff &\lambda\mapsto T_\lambda \in \mathscr{D}^\prime(X \times X) \text{ is }C^\infty \quad &\text{by 2.14} \\ \iff &\lambda\mapsto T_\lambda \in L(\mathscr{D}(X),\mathscr{D}^\prime(X)) \text{ is }C^\infty \\ \iff & \lambda\mapsto T_\lambda(\varphi) \in \mathscr{D}^\prime(X) \text{ is }C^\infty\quad \forall \varphi\in \mathscr{D}(X) &\text{by 5.18} \\ \iff & \lambda\mapsto T_\lambda(\varphi)(\psi) \in \mathbb R \text{ is }C^\infty \quad\forall \varphi,\psi\in \mathscr{D}(X) \quad &\text{by 2.14} \end{align*} Up to Frechet spaces convenient smoothness equals each other reasonable notion, but beyond it differs. A short description of convenient analysis can be found in Wikipedia.<|endoftext|> TITLE: Self-embeddings of uncountable total orders, 2 QUESTION [6 upvotes]: Let $S = (\Omega,\leq)$ be an uncountable dense total order, such that for all positive integers $m$ and all finite ordered sequences $a_1 < a_2 < \ldots < a_m$ and $b_1 < b_2 < \ldots < b_m$, we have an order automorphism of $S$ mapping the former sequence to the latter. Does there exist a proper subset $\Omega' \subset \Omega$ such that the induced total order $S'$ is order isomorphic to $S$ ? If so, can $\Omega'$ be chosen to be an interval ? (I guess the answer will follow from some general theorem on total orders, and I would love to know what that is.) Thanks ! REPLY [8 votes]: For the first question, the answer is YES, even if we just assume that $S$ is not rigid. Indeed, fix $f\in\mathrm{Aut}(S)$ and $a\in S$ such that $a TITLE: A non-recursive, explicit formula for the Fabius function QUESTION [8 upvotes]: The Fabius function $F\colon\mathbb R\to[-1,1]$ may be defined as the unique solution of the functional integral equation $F(x)=\int_0^{2x}F(t)\,dt$ for all real $x$ such that $F(1)=1$. The recent MO post provided a link to the MathSE question, asking to confirm a conjectured "non-recursive, self-contained formula for the Fabius function". The MO post has been overall negatively received and may get closed. I think the mentioned MathSE question may be of interest. On this page, whereas the mentioned conjectured formula will not be confirmed so far, a simpler non-recursive, explicit formula for the Fabius function will be offered, which is expressed in terms similar to, but simpler than, those in the conjectured formula. So, a question yet remains: Can one use the simpler formula below to confirm the conjecture on MathSE? Or maybe one could do that otherwise? REPLY [8 votes]: As noted in the linked Wikipedia article on the Fabius function, on the interval $I:=[0,1]$ the Fabius function coincides with the cumulative distribution function (cdf) of $$\sum_{j=1}^\infty 2^{-j}U_j,$$ where the $U_j$'s are independent random variables uniformly distributed on $I$. So, for each $x\in I$ $$F(x)=\lim_{n\to\infty} F_n(x),\tag{1}$$ where $F_n$ is the cdf of $\sum_{j=1}^n 2^{-j}U_j$. Next (see e.g. formula (2.2)), for any real $x$ $\newcommand\vp{\varepsilon}$ \begin{equation} F_n(x)=\text{vol}_n(I^n\cap H_{n;c^{(n)},x}) =\frac1{n!\prod_1^n c_i}\,\sum_{\vp\in\{0,1\}^n}(-1)^{|\vp|}\, \big(x-c^{(n)}\cdot\vp\big)_+^n, \end{equation} where $\text{vol}_n$ is the Lebesgue measure on $\mathbb R^n$, $H_{n;b,x}:=\{v\in\mathbb R^n\colon b\cdot v\le x\}$, $c^{(n)}:=(c_1,\dots,c_n)$, $c_j:=2^{-j}$, $|\vp|:=\vp_1+\dots+\vp_n$, $\cdot$ denotes the dot product, and $t_+^n:=\max(0,t)^n$. So, for $x\in I$ \begin{equation} F_n(x)=\frac{2^{n(n+1)/2}}{n!}\,\sum_{y\in D_{n,x}}(-1)^{s(y)}\, \big(x-y\big)^n, \tag{2} \end{equation} where $D_{n,x}$ is the set of all dyadic numbers in $[0,x]$ of the form $m2^{-n}$ for integers $m$, and $s(y)$ is the sum of the binary digits of $y$. Formulas (1) and (2) provide the answer to the question. Some of the differences between (1)--(2) and the formula conjectured in the linked MathSE post are as follows: In the MathSE post, the main conjectured formula for $F(x)$ is stated only for dyadic numbers $x$, and then extended to all values of $x$ by the continuity of $F$. The expression in that post for $F(x)$ for dyadic $x$ contains a double summation and a number of $q$-Pochhammer symbols (I have had no experience with those symbols).<|endoftext|> TITLE: Gluing filtered object from associated graded pieces QUESTION [8 upvotes]: So, I believe the following result is correct but do not know the exact reference (and not sure to what extent what I'm saying is true). If anyone could give a reference for this it would be great. 1) Consider three objects in some abelian category, $F_1, F_2, F_3$. Then objects with filtrations $F^1 \subset F^2 \subset F^3$ such that $F^1 = F_1, F^2/F^1 = F_2, F^3/F^2 = F_3$ are classified by triples $(\alpha, \beta, \gamma)$ of extension elements, $\alpha \in Ext^1(F_2, F_1), \beta \in Ext^1(F_3, F_2), \gamma \in Ext^1(F_3, F_1)$, such that the Yoneda product $\alpha \beta \in Ext^2(F_3, F_1)$ vanishes. 2) Similar result should hold for any amount of objects at least for the case of linear category in characteristic zero - the ways of gluing objects $F_1, ..., F_n$ into a filtered object $F^1 \subset F^2 \subset ... \subset F^n$ such that $F^k / F^{k-1} = F_k$ should be classified by Maurer-Cartan elements in the algebra $\bigoplus_{j>i}RHom^{\bullet}(F_j, F_i)$ (considered as either dgla or L-infinity algebra). I also know that this type of questions frequently appear in the theory of mixed hodge structures (but unable to find any direct reference, too). Edit: added forgotten $\gamma$ REPLY [2 votes]: Here is how I would think about this: $\alpha \in Ext^1(F_2,F_1)$ corresponds to a short exact sequence $$0 \rightarrow F_1 \xrightarrow{i} F(1,2) \xrightarrow{p} F_2 \rightarrow 0.$$ Similarly $\beta \in Ext^1(F_3,F_2)$ corresponds to a short exact sequence $$0 \rightarrow F_2 \rightarrow F(2,3) \rightarrow F_3 \rightarrow 0.$$ The first short exact sequence induces a long exact sequence including $$ Hom(F_3,F_2) \xrightarrow{\alpha \circ} Ext^1(F_3,F_1) \xrightarrow{i_*} Ext^1(F_3,F(1,2)) \xrightarrow{p_*} Ext^1(F_3,F_2) \xrightarrow{\alpha \circ} Ext^2(F_3,F_1),$$ and from this one sees that an $F(1,2,3) \in Ext^1(F_3,F(1,2))$ exists such that $F(1,2,3)/F_1 = F(2,3)$ if and only if $\alpha \circ \beta = 0 \in Ext^2(F_3,F_1)$. Furthermore, choices correspond to the image of $i_*$. Perhaps this is the classification you desire. I don't know of a reference, but the argument is just using basic triangulated category/homological algebra techniques. Understanding filtered objects with 4 or more composition factors leads one quickly to Massey products. Ambiguities tend to get out of hand unless one has something special going on in the case in hand.<|endoftext|> TITLE: Models of ZF intermediate between a model of ZFC and a generic extension QUESTION [10 upvotes]: Let $M$ be a countable model of $ZFC$ and $M[G]$ be a (set) generic extension of $M$. Suppose $N$ is a countable model of $ZF$ with $$M\subseteq N \subseteq M[G]$$ and that $N=M(x)$ for some $x\in N$; i.e., it is the smallest inner model of $M[G]$ which contains $x$ and $M$. Is $N$ a symmetric extension of $M$? REPLY [10 votes]: Yes, if $N=M(x)$ (taking the modern notation over Grigorieff's $M[x]$), then it is a symmetric extension. This is a very recent result of Toshimichi Usuba (see this and that). However, by the Bristol model construction, this symmetric extension need not be obtained by the forcing that was used to get $M[G]$.<|endoftext|> TITLE: How to add two numbers from a group theoretic perspective? QUESTION [18 upvotes]: It is known that adding two numbers and looking at the carrying operation has a link with cocycles in group theory. (https://www.jstor.org/stable/3072368?origin=crossref) When we add two numbers by elementary addition, we choose a basis $b$ for example $b=2$ which corresponds to the cyclic group $C_2$. Suppose we have words $w_1,w_2$ of (possibly different) lengths from this group $C_2$, how do we add them to get a new word $w$ in elementary addition? For example: $2=10_2=w_1$, $3=11_2=w_2$. Consider these as $w_1$ and $w_2$. Adding these numbers we get $5=101_2$ so the new word $w=101$. (1) But how exactly is the process of adding these two "words" from $C_2 = \{0,1\}$ in group theoretic means? (2) Is this "elementary addition" also possible for example for a non-cylcic group such as the Klein Four group? (3) We also assign a number to such a word (b-adic expansion). Is this assignment also possible for the Klein Four group? Thanks for your help. Edit: In view of the plot given below, I decided to put the tag "fractals" to this question. REPLY [4 votes]: I think I found a way how to mimic the elementary addition for arbitrary finite groups $G$: Let $G$ be a finite group, $S \subset G$ a generating set, $|g|:=|g|_S=$ word-length with respect to $S$. Let $\phi(g,h)=|g|+|h|-|gh| \ge 0$ be the "defect-function" of $S$. The set $\mathbb{Z}\times G$ builds a group for the following operation: $$(a,g) \oplus (b,h) = (a+b+\phi(g,h),gh)$$ On $\mathbb{N}\times G$ is the "norm": $|(a,g)| := |a|+|g|$ additive, which means that $|a \oplus b| = |a|+|b|$. Define the multiplication with $n \in \mathbb{N_0}$ to be: $$ n \cdot a := a \oplus a \oplus \cdots \oplus a$$ (if $n=0$ then $n \cdot a := (0,1) \in \mathbb{Z} \times G$). A word $w := w_{n-1} w_{n-2} \cdots w_0$ is mapped to an element of $\mathbb{Z} \times G$ as follows: $$\zeta(w) := \oplus_{i=0}^{n-1} (m^i \cdot (0,w_i))$$ where $m := \min_{g,h\in G, \phi(g,h) \neq 0} \phi(g,h)$. We let $|w|:=|\zeta(w)|$ and $w_1 \oplus w_2:=\zeta(w_1)\oplus \zeta(w_2)$ Then we have $|w_1 \oplus w_2| = |w_1|+|w_2|$. For instance for the Klein four group $\{0,a,b,c=a+b\}$ generated by $S:=\{a,b\}$, we get sorting the words $w$ by their word-length: $$0,a,b,c,a0,aa,ab,ac,b0,ba,bb,bc,c0,ca,cb,cc,a00,a0a,a0b,a0c$$ corresponding to the following $\mathbb{Z}\times K_4$ elements $\zeta(w)$: $$(0,0),(0,a),(0,b),(0,c),(2,0),(2,a),(2,b),(2,c),(2,0),(2,a),(2,b),(2,c),(4,0),(4,a),(4,b),(4,c),(4,0),(4,a),(4,b),(4,c)$$ corresponding to the the following "norms" of words $|w| = |\zeta(w)|$: $$0,1,1,2,2,3,3,4,2,3,3,4,4,5,5,6,4,5,5,6$$ It would be interesting to see what sequence one gets for the smallest non-abelian group $S_3$. If someone likes to write a computer program to compute this, that would be great. Related questions: How is this group theoretic construct called? Edit: Here is some python code for the cyclic groups and an example for $b=3$: def add(a,b,n=2): x,y = a c,d = b return(x+c+(y%n+d%n-(y+d)%n),(y+d)%n) def sumadd(l,n=2): x = (0,0) for y in l: x = add(x,y,n=n) return(x) def norm(a): return(abs(a[0])+abs(a[1])) def mult(x,a,n=2): return(sumadd([a for i in range(x)],n=n)) def zeta(w,n=2): return sumadd([mult(n**(len(w)-1-i),(0,w[i]),n=n) for i in range(len(w))],n=n) def digits(n, b): if n == 0: return [0] digits = [] while n: digits.append(int(n % b)) n //= b return digits[::-1] b = 3 for m in range(1,20): w = digits(m,b) print(m, norm(zeta(w,n=b))) (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6) (7, 7) (8, 8) (9, 9) (10, 10) (11, 11) (12, 12) (13, 13) (14, 14) (15, 15) (16, 16) (17, 17) (18, 18) (19, 19) Update Here is some Python Code, to do the computations for the Klein Four group: K4_elements = {'0':0,"a":1,"b":2,"c":3} K4_group_table = [ ["0","a","b","c"], ["a","0","c","b"], ["b","c","0","a"], ["c","b","a","0"] ] K4_lengths = {"0":0,"a":1,"b":1,"c":2} def K4_add(g,h): i = K4_elements[g] j = K4_elements[h] return(K4_group_table[i][j]) def K4_phi(g,h): return(K4_lengths[g]+K4_lengths[h]-K4_lengths[K4_add(g,h)]) def add_ZxK4(a,b): a0,a1=a b0,b1=b return((a0+b0+K4_phi(a1,b1),K4_add(a1,b1))) def sumadd_ZxK4(l): x = (0,"0") for y in l: x = add_ZxK4(x,y) return(x) def norm_ZxK4(a): return(abs(a[0])+K4_lengths[a[1]]) def mult_ZxK4(x,a): return(sumadd_ZxK4([a for i in range(x)])) def zeta_ZxK4(w): m = min([K4_phi(g,h) for g in K4_elements.keys() for h in K4_elements.keys() if K4_phi(g,h)!=0]) return sumadd_ZxK4([mult_ZxK4(m**(len(w)-1-i),(0,w[i])) for i in range(len(w))]) def operate_ZxK4(h,a): return(add_ZxK4((0,h),a)) from itertools import product K4 = ['0',"a","b","c"] words = [] words.extend(list(product(K4,K4,K4))) for word in words: print(".".join(word), zeta_ZxK4(word),norm_ZxK4(zeta_ZxK4(word))) 0.0.0 (0, '0') 0 0.0.a (0, 'a') 1 0.0.b (0, 'b') 1 0.0.c (0, 'c') 2 0.a.0 (2, '0') 2 0.a.a (2, 'a') 3 0.a.b (2, 'b') 3 0.a.c (2, 'c') 4 0.b.0 (2, '0') 2 0.b.a (2, 'a') 3 0.b.b (2, 'b') 3 0.b.c (2, 'c') 4 0.c.0 (4, '0') 4 0.c.a (4, 'a') 5 0.c.b (4, 'b') 5 0.c.c (4, 'c') 6 a.0.0 (4, '0') 4 a.0.a (4, 'a') 5 a.0.b (4, 'b') 5 a.0.c (4, 'c') 6 a.a.0 (6, '0') 6 a.a.a (6, 'a') 7 a.a.b (6, 'b') 7 a.a.c (6, 'c') 8 a.b.0 (6, '0') 6 a.b.a (6, 'a') 7 a.b.b (6, 'b') 7 a.b.c (6, 'c') 8 a.c.0 (8, '0') 8 a.c.a (8, 'a') 9 a.c.b (8, 'b') 9 a.c.c (8, 'c') 10 b.0.0 (4, '0') 4 b.0.a (4, 'a') 5 b.0.b (4, 'b') 5 b.0.c (4, 'c') 6 b.a.0 (6, '0') 6 b.a.a (6, 'a') 7 b.a.b (6, 'b') 7 b.a.c (6, 'c') 8 b.b.0 (6, '0') 6 b.b.a (6, 'a') 7 b.b.b (6, 'b') 7 b.b.c (6, 'c') 8 b.c.0 (8, '0') 8 b.c.a (8, 'a') 9 b.c.b (8, 'b') 9 b.c.c (8, 'c') 10 c.0.0 (8, '0') 8 c.0.a (8, 'a') 9 c.0.b (8, 'b') 9 c.0.c (8, 'c') 10 c.a.0 (10, '0') 10 c.a.a (10, 'a') 11 c.a.b (10, 'b') 11 c.a.c (10, 'c') 12 c.b.0 (10, '0') 10 c.b.a (10, 'a') 11 c.b.b (10, 'b') 11 c.b.c (10, 'c') 12 c.c.0 (12, '0') 12 c.c.a (12, 'a') 13 c.c.b (12, 'b') 13 c.c.c (12, 'c') 14 Plotting this sequence one recognizes a fractal structure:<|endoftext|> TITLE: Simple examples of colimits of affine schemes (evaluated in the presheaf category) which are not affine schemes QUESTION [9 upvotes]: Notation and Setting: let $\operatorname{Aff}$ denote the category of affine schemes whose objects are covariant representable functors $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ and $\operatorname{Spec}:\operatorname{Ring^{op}}\rightarrow\operatorname{Func(Ring, Set)} $ be the contravariant Yoneda embedding of $\operatorname{Ring^{op}}$ in its category of presheaves so that $\operatorname{Aff}\simeq\operatorname{Ring^{op}}$. In addition, let $\mathcal{O}:\operatorname{Func(Ring, Set)}\rightarrow\operatorname{Ring^{op}}$ be the functor that sends a functor $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ to the ring of maps $\operatorname{X}\rightarrow \mathbb{A}^1$ (where $\mathbb{A}^1$ is the forgetful functor) so that $\operatorname{Spec}$ and $\mathcal{O}$ are inverse of one another. Let $\widehat{\operatorname{Aff}}$ be the indization of $\operatorname{Aff}$, i.e. the category whose objects are functors $\operatorname{X}:\operatorname{Ring}\rightarrow\operatorname{Set}$ that are small filtered colimits of affine schemes. My question: I am looking for (simple) examples of functors which are objects of $\widehat{\operatorname{Aff}}$ but are not affine schemes. I am particularly interested in examples of the following form: Let $\operatorname{X}$ be an affine scheme, $I\subseteq\mathcal{O}_{X}$ an ideal and consider the following diagram in $\operatorname{Func(Ring, Set)}$ over $\mathbb{Z_{\geq0}}$ $0=\operatorname{Spec(\mathcal{O}_{X}/I^{0})}\hookrightarrow\ldots\hookrightarrow\operatorname{Spec(\mathcal{O}_{X}/I^{n-1})}\hookrightarrow\operatorname{Spec(\mathcal{O}_{X}/I^{n})}\hookrightarrow\ldots$ Since $\operatorname{Func(Ring, Set)}$ admits small colimits, $\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathcal{O}_{X}/I^{n})}$ exists. Thus I am looking for examples of affine schemes $\operatorname{X}$ and ideals $I\subseteq\mathcal{O}_{X}$ for which $(\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathcal{O}_{X}/I^{n})})\neq \mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}(\operatorname{Spec(\mathcal{O}_{X}/I^{n})})$ The only example that I could find so far was that of $\operatorname{Spec(\mathbb{Z}[x])}$ and the ideal $(x)$ which give the functor $\operatorname{Nil}\simeq\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathbb{Z}[x]/(x)^{n})}\neq(\mathrm{colimit}_{n\in\mathbb{Z}_{\geq0}}\operatorname{Spec(\mathbb{Z}[x]/(x)^{n})})\simeq\operatorname{Spec}(\mathbb{Z}[\![ x ]\!])$. To see that you can either show that $\operatorname{Nil}$ is not representable1 or check, for example, that $\operatorname{Spec}(\mathbb{Z}[\![ x ]\!])(\mathbb{Z})\neq\operatorname{Nil}(\mathbb{Z})$. REPLY [10 votes]: A basic standard example is the colimit of $\mathbb{A}^0 \to \mathbb{A}^1 \to \mathbb{A}^2 \to \cdots$ with transition maps $x \mapsto (x,0)$. The $R$-valued points are finite sequences in $R$. This functor is not representable. More generally, let $A$ be a commutative ring with a sequence of ideals $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$. (In the mentioned example, $A = \mathbf{Z}[x_0,x_1,\dotsc]$ and $I_n = \langle x_n,x_{n+1},\dotsc\rangle$.) Then the colimit $X$ of $\mathrm{Spec}(A/I_0) \to \mathrm{Spec}(A/I_1) \to \cdots$ is the subfunctor of $\mathrm{Spec}(A)$ whose $R$-valued points are those $A \to R$ whose kernel contains some $I_n$. We have $\mathcal{O}(X) = \lim_n A/I_n =: \widehat{A}$. Then $X$ is representable aka affine iff the canonical morphism $X \to \mathrm{Spec}(\widehat{A})$ is an isomorphism. It is injective anyway, and it is surjective on $R$-valued points iff every homomorphism $\widehat{A} \to R$ factors through some projection $\widehat{A} \to A/I_n$. So $X$ is representable iff the identity $\widehat{A} \to \widehat{A}$ factors through some projection $\widehat{A} \to A/I_n$. But this clearly implies $I_n = I_{n+1} = \dotsc$ and the sequence is stationary. Conversely this means that for every non-stationary sequence $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$ the functor $X$ is not representable.<|endoftext|> TITLE: Direct proof that Chern-Weil theory yields integral classes QUESTION [17 upvotes]: Suppose $E$ is a complex vector bundle of rank $n$ on a compact oriented manifold (both assumed smooth). Let $h$ be a Hermitian metric on $E$, and let $A$ be a Hermitian connection on $E$ and $F_A$ its curvature form, an element of $\Omega^2(M,\text{End }E)$. Chern-Weil theory produces a closed even differential form $c(A) = \det(1 + \frac i{2\pi}F_A) = c_0(A)+c_1(A) + \cdots + c_n(A)$. These classes have the property that for all compact oriented submanifolds $\Sigma\subset M$ of dimension $2k$, the expression $\int_\Sigma c(A)$ is an integer. The usual way to prove this seems to be to first show that the de Rham cohomology class of $c(A)$ is independent of the connection $A$ and functorial under pullbacks. We then check that it has the desired integrality property for the tautological bundles on Grassmannians by direct computation (which says this class lies in singular cohomology with $\mathbb Z$ coefficients) and then use the fact that every bundle on $M$ is a pullback of the tautological bundle from a sufficiently large Grassmannian. Alternatively, we show that the above definition satisfies the axiomatic characterization of Chern classes and then we identify it with the image in $H^\bullet(-,\mathbb C)$ of Chern classes constructed in $H^\bullet(-,\mathbb Z)$ by using a computation of the cohomology ring of infinite Grassmannians. Is there a direct proof that $\int_\Sigma$ c(A) is an integer for all compact oriented submanifolds $\Sigma\subset M$? By a direct proof, I mean something which only uses only local reasoning about curvature and connections and doesn't have to pass through much algebraic topology. For instance for the case $n=1$ (i.e., a line bundle) and taking $\Sigma = M$ to be a compact oriented surface, we note by elementary local computations (using Stokes' theorem) that for any disc $D$ inside $\Sigma$ with oriented boundary circle $\gamma$, the number $e^{\pm\int_D F_A}\in U(1)$ measures the holonomy of the parallel transport along the loop $\gamma$. From this, it easily follows that we can extend the above statement to the case when $D$ is a more general region with smooth oriented boundary $\gamma$. As $\Sigma$ has no boundary, applying this to $D = \Sigma$ gives $\int_\Sigma F_A\in \frac{2\pi}i\mathbb Z$ as desired. (This proof is similar to the moral proof of Stokes' theorem given by checking it at the infinitesimal level -- essentially the definition of the exterior derivative -- and then triangulating the domain of integration and noting that boundary terms with opposite orientations cancel.) The $c_1$ case seems to be simple because of $U(1)$ being abelian. REPLY [10 votes]: Yes, the Chern–Weil homomorphism lifts to differential cohomology, which guarantees that periods are integral. See the original paper by Cheeger and Simons, or the paper by Hopkins and Singer. The (modernized) construction of such a refinement relies on the computation of the de Rham complex of the stack B_∇(G) of principal G-bundles with connection and their isomorphisms, which is isomorphic (and not just quasi-isomorphic) to the algebra of G-invariant polynomials on the Lie algebra of G. This is proved by Freed and Hopkins, and their argument is a local argument that uses computations with differential forms. There is a bit of linear algebra and invariant theory at the very end. This argument is essentially an abstract higher-dimensional version of the n=1 argument above with the Stokes theorem.<|endoftext|> TITLE: Regularity of conformal maps QUESTION [14 upvotes]: In order to define what it means for a map $f \colon \Omega \subseteq \mathbb R^n \to \mathbb R^n$ to be conformal, it is sufficient to require that $f$ is everywhere differentiable. Does conformality automatically implies that $f$ is $\mathcal C^1$ (hence real-analytic, see below)? By complex analysis, we know the answer is positive when $n=2$. In higher dimensions, Liouville's theorem characterizes conformal maps as Möbius transformations, but it is stated for $f \in W^{1,n}$ in Wikipedia. Is it known whether it also holds when $f$ is assumed everywhere differentiable? REPLY [7 votes]: Let $n\geq 3.$ Let $\Omega$ be an open connected subset of $\mathbb R^n,$ and let $f:\Omega\to\mathbb R^n$ be a function having a pointwise derivative $Df(x)$ everywhere satisfying $(Df)^T(Df)=g(x)I$ with $g(x)>0.$ Then $f$ is continuously differentiable. By the inverse function theorem, $f$ is a local homeomorphism. By shrinking $\Omega$ we can assume $f$ maps $\Omega$ homeomorphically to $f(\Omega),$ and that $\Omega$ and $f(\Omega)$ are bounded. Note $$\|Df\|^2=c_ng(x)=c_n\det(g(x)I)^{1/n}=c_n|\det Df(x)|^{2/n}$$ where $\|\cdot\|$ is Frobenius norm, and $c_n$ is a constant depending on $n.$ By the change of variables formula (proof sketch), $$\int_\Omega |\det Df(x)|\;dx=\mu(f(\Omega))<\infty.$$ So $f\in W^{1,n}(\Omega,\mathbb R^n)$ and you can use the $W^{1,n}$ result you mentioned.<|endoftext|> TITLE: Functional equation of twisted triple product L-function QUESTION [6 upvotes]: Let $\mathbb{E}=E_1\times E_2\times E_3$ denote the product of three elliptic curves over $\mathbb{Q}$ of prime level $p$ and consider the $p$-adic Galois representation $$V_p(\mathbb{E})=H^1_{et}(E_{1/\bar{\mathbb{Q}}}, \mathbb{Q}_p)\otimes H^1_{et}(E_{2/\bar{\mathbb{Q}}}, \mathbb{Q}_p)\otimes H^1_{et}(E_{3/\bar{\mathbb{Q}}}, \mathbb{Q}_p).$$ Denote by $L(\mathbb{E}, s)=L(V_p(\mathbb{E}),s)$ the associated triple product $L$-function. It has a functional equation centered at $s=2$ with global sign equal to $a_p(E_1)a_p(E_2)a_p(E_2)\in \{ \pm 1 \}$ (cf. Gross-Kudla '92). Here, $a_p(E_i)$ denotes the $p$-th Fourier coefficient of the weight 2 normalized newform of level $\Gamma_0(p)$ associated to $E_i$ by modularity. Let $\chi$ be a Dirichlet character modulo $p$ and denote by $L(\mathbb{E}\otimes \chi, s)$ the $L$-function attached to the Galois representation $V_p(\mathbb{E})\otimes \chi$. My question is: what is the functional equation of $L(\mathbb{E}\otimes \chi, s)$ and what is its global sign? Thank you in advance for any help. REPLY [5 votes]: There is a functional equation for $L(\mathbb{E} \otimes \chi, s)$, but it relates $L(\mathbb{E} \otimes \chi, s)$ to $L(\mathbb{E} \otimes \bar\chi, 4-s)$. If $\chi$ is not trivial or quadratic, then $L(\mathbb{E} \otimes \chi, s)$ and $L(\mathbb{E} \otimes \bar\chi, s)$ are different functions, so you cannot use this to deduce anything in particular about vanishing at the central value. The Langlands $\varepsilon$-factor is still defined, and it is a complex number of absolute value 1, but it isn't $\pm 1$, so you can't reasonably call it a "sign"; at a guess it is probably something like $\tau(\chi)^4 / p^{2n}$, where $p^n$ is the conductor of $\chi$ and $\tau(\chi)$ is the Gauss sum. (You can try to cheat by considering the function $M(s) = L(E \otimes \chi, s) L(E \otimes \bar\chi, s)$, which does satisfy a functional equation relating $M(s)$ and $M(4-s)$, but unfortunately the order of vanishing of $M(s)$ at $s = 2$ is automatically even anyway, since $L(E \otimes \bar\chi, s) = \overline{L(E \otimes \chi, \overline{s})}$, so both factors have the same order of vanishing. So although $M(s)$ does have a functional equation, you can't get any nontrivial vanishing information out of it.) The moral here is that the whole story of L-values vanishing because of "sign" phenomena is specific to self-dual settings.<|endoftext|> TITLE: Existence and uniqueness of Haar measure on compacta; a cohomological approach QUESTION [44 upvotes]: I am trying to use a modification of group cohomology to prove the existence and uniqueness of Haar measure on a compact Hausdorff group. I think the best way of introducing the idea I am pursuing is via analogy. Let $G$ be a finite group and let $A = \mathbb{C}[G]$ be the group algebra. For each $G$-set $X$, there is an $A$-module $[X, \mathbb{C}]$ (functions from $X$ to $\mathbb{C}$). For $g \in G$, $f^g$ sends $x$ to $f(xg)$. We may view $\mathbb{C}$ as a $\mathbb{C}[G]$-module where $g z = z$ for each $z \in \mathbb{C}$. $\mathbb{C}$ embeds into $[G, \mathbb{C}]$ as constant functions. We may then consider the exact sequence $$0 \rightarrow \mathbb{C} \rightarrow [G, \mathbb{C}] \rightarrow [G, \mathbb{C}]/ \mathbb{C} \rightarrow 0$$ We know that this sequence splits because of the symmetrization map $[G, \mathbb{C}] \rightarrow \mathbb{C}$ sending $f \in [G, \mathbb{C}]$ to $\frac{1}{|G|} \sum_{g \in G} f(g)$. For the question at hand, let $G$ be a compact hausdorff group. Let $A = \mathbb{C}[G]$. This is a $\mathbb{C}$-algebra. Let $C$ be the category of topological $\mathbb{C}[G]$-modules (topological abelian groups $A$ with a map of abelian groups $\mathbb{C}[G] \rightarrow \text{End}_{\text{TopAb}}(A, A)$. We might wish to tweak this later on to get something abelian). For each topological $G$-set $X$, $[X, \mathbb{C}]_{\text{Top}}$ (continuous functions from $X$ to $\mathbb{C}$) is such a $\mathbb{C}[G]$-module (it is also a $C^*$-algebra). We may view $\mathbb{C}$ as a $\mathbb{C}[G]$-module where $gz = z $ for each $z \in \mathbb{C}$ and each $g \in G$. $\mathbb{C}$ embeds into $[G, \mathbb{C}]_{\text{Top}}$ as constant functions. We may then consider the exact sequence $$0 \rightarrow \mathbb{C} \rightarrow [G, \mathbb{C}]_{\text{Top}} \rightarrow [G, \mathbb{C}]_{\text{Top}}/ \mathbb{C} \rightarrow 0$$ We know that this sequence splits because of Haar measure $[G, \mathbb{C}]_{\text{Top}} \rightarrow \mathbb{C}$ sending $f \in [G, \mathbb{C}]_{\text{Top}}$ to $\int_{G} f d \mu$. On the other hand, we may wish to show that the sequence splits another way, from which it would follow that a unique Haar measure exists. For instance, what is $\text{Ext}^1_{\mathbb{C}[G]}([G, \mathbb{C}]/\mathbb{C}, \mathbb{C})$? We may need to tweak the category $C$ to make sense of this. But if we succeed, and this cohomology group vanishes, then the sequence will split by the usual characterization of $\text{Ext}^1$ as classifying extensions. Does anyone see a say to show that $\text{Ext}^1$ vanishes? No doubt we must use somewhere that $G$ is compact hausdorff (or at least locally compact hausdorff), since the theorem does not hold otherwise. Recall that there is an equivalence of categories between $C^*$-algebras and compact hausdorff topological spaces, Gelfand duality. The $C^*$-algebra structure of $[G, \mathbb{C}]_{\text{Top}}$ is one way that the compact hausdorff property of $G$ could show up in the proof (there is an equivalence of categories between compact hausdorff topological spaces with a continuous left $G$-action on the one hand, and $C^*$-algebras with a right $G$-action on the other; the second is equivalent to $C^*$-algebras which are $\mathbb{C}[G]$-modules in a compatible way). REPLY [26 votes]: Fix a compact group $G$ and consider its category of Banach representations: the objects are (complex) Banach spaces $X$ endowed with a $G$-action by automorphims (not necessarily isometries) such that the action maps $G\times X\to X$ are jointly continuous and the morphisms are bounded maps $X\to Y$ commuting with the $G$-actions. Denote the trivial representation by $\bf 1$. It is an injective object in this category, that it for every $0 \to X \to Y$ every $f: X\to \bf 1$ can be extended to $Y\to \bf 1$. This fact could be seen as an equivariant Hahn-Banach theorem. Let me explain a way to see this. By the classical Hahn-Banach theorem, $f$ extends to a functional $\bar{f}\in Y^*$, but it might not be invariant. Consider the subset of all such extensions in $Y^*$ and consider further the collection of all $G$-invariant compact convex subsets of it. This collection is not empty, as it contains the convex hull of the $G$-orbit of $\bar{f}$. Use compactness and Zorn's lemma to find a minimal element $C$ in this collection, with respect to inclusion. $C$ must be a singleton, otherwise it would contain a point $h$ which is not extremal and the convex hull of the $G$-orbit of $h$ would be a proper $G$-invariant comapct convex subset of $C$, contradicting its minimality, as it contains no extreme point of $C$. It follows that $C=\{h\}$ for some $G$-invariant functional, that is a morhism $Y\to \bf 1$ extending $f$. In particular, the short exact sequence $0\to {\bf 1}\to C(G) \to C(G)/{\bf 1}\to 0$ splits, by taking $X=\bf 1$ and $Y=C(G)$. The morphism $C(G)\to \bf 1$ is the Haar integral. Remark 1: I wouldn't regard the above as a categorical or cohomological approach for proving the existence of the Haar measure. Whatever way you prove it, you prove a splitting theorem. Remark 2: The fact that $\bf 1$ is injective in the category of Banach representations characterizes compact groups among all locally compact second countable ones. Remark 3: Take care with standard categorical interpretations of injectivity and alike, as categories of Banach representations are not abelian. Let me justify Remark 2. See here that every locally compact second countable group has a proper isometric affine action on some reflexive Banach space $V$, given by $v \mapsto \rho(g)(v)+c(g)$ where $\rho$ is an isometric representation of $G$ on $V$ and $c:G\to V$ is a 1-cocycle. Assuming $G$ is non-compact, this action has no fixed point. Consider the space $V\oplus \mathbb{C}$ and endow it with the linear representation $\pi$ given by $\pi(g)(v,t)=(\rho(g)(v)+tc(g),t)$. Note that the projection $(v,t)\mapsto t$ is $G$-invariant and does not split. Let $Y=(V\oplus \mathbb{C})^*$ endowed with the contragredient representation. Then we have a morphism ${\bf 1} \to Y$ which does not split. Alternatively, for a non amenable group we can take $Y$ to be the space of continuous functions over a compact $G$-space having no invariant measure and for a non property (T) group we can take the dual space of a linearization, as above, of an isometric affine action on a Hilbert space having no fixed point, and conclude by the fact that every non-compact group is either amenable or has (T).<|endoftext|> TITLE: Is a representation of $\operatorname{SL}_n$ defined over a field $k$ if its image is contained in $\operatorname{GL}_n(k)$? QUESTION [14 upvotes]: Let $k$ be a subfield of $\mathbb{C}$ and let $f\colon \operatorname{SL}_n(\mathbb{C}) \rightarrow \operatorname{GL}_m(\mathbb{C})$ be an algebraic homomorphism such that $f(\operatorname{SL}_n(k)) \subset \operatorname{GL}_m(k)$. Question: must $f$ be defined over $k$? In other words, are the matrix entries of $f(x)$ polynomials in the entries of $x \in \operatorname{SL}_n(\mathbb{C})$ whose coefficients lie in $k$? I'd also be interested in the question with $\operatorname{SL}_n$ replaced by other semisimple groups, but the above is the most important special case for what I am doing. REPLY [15 votes]: Yes, and in fact something much more general is true Let $X$ and $Y$ be affine varieties defined over a field $k$. If the $k$-points of $X$ are Zariski dense, $X$ is reduced, and $f: X_{\mathbb C} \to Y_{\mathbb C}$ sends $X(k)$ to $Y(k)$, then $f$ is defined over $k$. This was inspired by comments of Martin Brandenburg, Jef L, Andy Putman, and Piotr Achinger. Proof: First note that by embedding $Y$ in affine space, we may assume $Y = \mathbb A^n$. Second note that by viewing a map to $\mathbb A^n$ as an $n$-tuple of maps to $\mathbb A^1$, we may assume $Y =\mathbb A^1$. Thus $f$ is a polynomial function in $\mathbb C[X]$, and we want to check it lies in $k[X]$. Because $\mathbb C[x]= k[X] \otimes_k \mathbb C$, we may write $f = \sum_{i=1}^n \alpha_i f_i$ where $\alpha_i \in \mathbb C$ and $f_i \in k[X]$. Without loss of generality, we may assume that $\alpha_1=1$ and that the $\alpha_i$ are $k$-linearly independent. (Add $1$ to the list of $\alpha_i$s, then for any linear relation, use that relation to remove whichever $\alpha_i$ is not $1$ and adjust the $f_i$s appropriately.) Now for $x \in X(k)$, we have $f(x) = \sum_{i=1}^n \alpha_i f_i(x)$. Because the $\alpha_i$ are $k$-linearly independent, and $f(x)\in k$, this implies $f_i(x)=0$ for $i>1$. Then because $X(k)$ is Zariski dense, this implies $f_i=0$ for $i>1$, so $f=f_1 \in k[X]$. QED The Zariski density can be checked for $SL_n$ using the birational map to affine space obtained by forgetting one entry, and for other semisimple groups using the Bruhat decomposition as suggested by Mikhail Borovoi.<|endoftext|> TITLE: multiplicative structure of Ext QUESTION [6 upvotes]: Basically, I am trying to compute something with the Adams spectral sequence (as a toy example). The $E^2$ page reduced to computing $Ext^{s,t}_{\Gamma} (\mathbb{F}_2, \mathbb{F}_2)$, where $\Gamma = \mathbb{F}_2 x/ x^2$ and $deg(x) =1$. Keeping very close track of degree, I found that $Ext^{s,t}_{\Gamma} (\mathbb{F}_2, \mathbb{F}_2)$ is only nonzero in degree $s=t$. In this case, we have $\mathbb{F}_2 \{ \phi_t \} $ where $\phi_t \in Hom^t_{\mathbb{F}_2 x/x^2} (\Sigma^t \mathbb{F}_2 x/x^2, \mathbb{F}_2)$ sends $1$ to $1$ and $x$ to $0$. (the superscript $t$ denotes morphisms that lower degree by $t$.) So I have come to the point that I need to understand the multiplicative structure of $Ext$. I know I am supposed to get that $\phi_1^2 = \phi_2$ (because I know what the spectral sequence is supposed to converge to--the 0th column should have a copy of $\mathbb{F}_2$ in each degree, but all of it generated by $\phi_1$ ). However, I am not really sure how to make sense of $\phi_1^2$. So my questions are 1) how to make sense of $\phi_1^2$? or have I confused things somewhere along the way 2) in order to work with the multiplicative structure in the second page of Adams spectral sequence, I need to understand the multiplicative structure of $Ext$. I would like to work directly with objects in $Ext$, but how does one multiply? and is it easier to do it this way or does one typically use the isomorphism between objects of $Ext$ and extensions, multiply the extensions (which is simple), and then translate back? But the isomorphism for higher Ext does not seem easy to work with. Note: I specifically am trying to do with without the cobar complex. Edit: I should add that the resolution I took was $\ldots \Sigma^2 \mathbb{F}_2 x/x^2 \xrightarrow{d} \Sigma \mathbb{F}_2 x/x^2 \xrightarrow{d} \mathbb{F}_2 x/x^2 \to \mathbb{F}_2$, where $d$ sends $\Sigma^{t} 1$ to $\Sigma^{t-1} x$ and everything else to $0$. I'm adding the $\Sigma$ to help keep track of degree. REPLY [3 votes]: The composition product $Ext(N,P) \otimes Ext(M,N) \to Ext(M,P)$ can be computed as follows. Given cocycles $x : N_{s_1} \to \Sigma^{t_1} P$ and $y : M_{s_2} \to \Sigma^{t_2} N$, let $\{ y_s : M_{s+s_2} \to \Sigma^{t_2} N_s\}$ be a chain map lifting $y$. Then $xy$ is represented by the cocycle $x y_{s_1} : M_{s_1+s_2} \to \Sigma^{t_2} N_{s_1} \to \Sigma^{t_1+t_2} P $. The point is that one needs (a finite bit of) the chain map lifting $y$, but only the cocycle $x$. Saves effort. (The $M_s$ and $N_s$ are the terms in projective resolutions of $M$ and $N$, of course.)<|endoftext|> TITLE: Constructible étale sheaves on X are étale algebraic spaces over X QUESTION [5 upvotes]: I saw the following statement in a paper of Bhatt-Mathew: Let $X$ be a quasicompact quasiseparated scheme. Then there is an equivalence of categories between constructible étale sheaves (of sets) on $X$ and algebraic spaces $\mathcal{Y}\to X$ over $X$ whose structure map is étale. I'm familiar with the proof that lcc étale sheaves of sets $F$ are equivalent to finite étale schemes $Y\to X$, which works by showing that constant finite sheaves of sets on $X$ correspond to trivial finite étale surjections over $X$ then using a descent argument (namely, if $F$ is the constant étale sheaf with value $S$ on $X$, it is representable by the étale $X$-scheme $X\times S$. Then the locally constant constructible case follows by unwinding the sheaf data to glue the $U_i \times S_i$ where the $U_i$ form a cover of $X$ such that the restriction of $F$ to $U_i$ is the constant sheaf valued in the set $S_i$. For the case of an algebraic space $\mathcal{Y}\to X$ étale over $X$, I don't understand why there is a stratification $\{X_i\}$ of $X$ such that the pullback $Y_i:=\mathcal{Y}\times_X X_i \to X_i$ to each stratum is finite étale, and conversely, given a sheaf that is constructible with respect to a particular stratification $\{X_i\}$ of $X$, I don't see why this helps us build an algebraic space from the corresponding family of finite étale schemes $Y_i \to X_i$. REPLY [5 votes]: In case anyone wants to know a reference, I found it now: SGA4, Exp. IX, Prop. 2.7 Statement (Translated): Proposition 2.7 Let $X$ be a quasicompact and quasiseparated scheme, and let $F$ be a sheaf of sets. For $F$ to be constructible, it is necessary and sufficient that it be isomorphic to the coequalizer of a pair of morphisms $H\rightrightarrows G$, where $H$ and $G$ are sheaves of sets representable by by étale schemes of finite presentation over $X$.<|endoftext|> TITLE: Are open $\mathbb{G}_m$-invariant subschemes of an affine scheme precisely the homogeneous radical ideals of its coordinate ring? QUESTION [5 upvotes]: This question is certainly not research level and in fact quite elementary which is why I asked it on math.stackexchange before: math.stackexchange. However it doesn't seem to get much attention there and I thought I would try it here. My questions are regarding actions of a group scheme $G$ on a scheme $X$. I'm fine with assuming $G$ affine. $\newcommand{\IG}{\mathbb{G}} \newcommand{\pmo}{{\pm 1}} \newcommand{\IZ}{\mathbb{Z}} \newcommand{\Spec}{\mathrm{Spec}} \newcommand{\tensor}{\otimes} \newcommand{\into}{\hookrightarrow} \newcommand{\iso}{\cong} \newcommand{\onto}{\twoheadrightarrow} \newcommand{\sheaf}{\mathcal} \newcommand{\inv}{{-1}}$ Originally, I was thinking about actions of the multiplicative group $\IG_m = \Spec(\IZ[T^\pmo])$ on affine schemes $X = \Spec(R)$. The category of affine schemes with a $\IG_m$-action is equivalent to the category of $\IZ$-graded rings as follows: To a $\IZ$-graded ring $R$ associate the action $\IG_m \times \Spec(R) \to \Spec(R)$ which is given by the map of rings $R \to \IZ[T^\pmo] \tensor R$, $f = \sum f_d \mapsto \sum f_d \tensor T^d$, where $f_d$ are the homogenous components of $f$. Conversely, if the action $\IG_m \times \Spec(R) \to \Spec(R)$ is given, call $f_d \in R$ homogenous of degree $d$ if it is sent to a homogenous element of rank $d$ by the morphism $R \to \IZ[T^\pmo] \tensor R \cong R[T^\pmo]$, where we regard the latter ring as graded by the powers of $T$. If $Y \subseteq X$ is a closed subscheme, I call $Y$ invariant under the action if the morphism $\IG_m \times Y \into \IG_m \times X \to X$ factors over $Y \into X$. Proposition: The correspondence $$\{\text{Closed subschemes of }X\} \iso \{\text{Ideals of }R\}$$ restricts to a bijective correspondence $$\{\IG_m\text{-invariant closed subschemes of }X\} \iso \{\text{Homogeneous ideals of }R\}.$$ For example if $Y$ is $\IG_m$-invariant, then by definition the given action restricts to an action of $\IG_m$ on $Y$ and the inclusion $Y \into X$ becomes a morphism of affine schemes with a $\IG_m$-action. Hence the surjective map $R=\sheaf{O}(X) \onto \sheaf{O}(Y)$ is a morphism of graded rings. But its kernel is $I$ which is, hence, homogeneous. I was trying to prove the corresponding result for open subschemes: Conjecture 1: The correspondence $$\{\text{Open subschemes of }X\} \iso \{\text{Radical ideals of }R\}$$ restricts to a bijective correspondence $$\{\IG_m\text{-invariant open subschemes of }X\} \iso \{\text{Homogeneous radical ideals of }R\}.$$ It is easy to see that a homogeneous radical ideal defines an invariant open subscheme. First note that the union of invariant open subschemes is again invariant. If the open subschemes $U_j$ are defined by $I_j$, then $\bigcup U_j$ is defined by $\sqrt{\sum I_j}$. Hence it suffices to consider $U = D(f)$ for $f$ a homogeneous element. But then it is easy to see that $U$ is invariant. In fact, the restricted action $\IG_m \times U \to U$ makes $U = \Spec(R[f^\inv])$ into an affine schemes with a $\IG_m$-action. The associated grading on $R[f^\inv]$ is the natural grading that you would expect on a localisation at a homogeneous element. However I cannot manage to prove the converse, i.e. if $I$ is a radical ideal of $R$ such that the open subscheme $X_I \into X$ defined by $I$ is $\IG_m$-invariant then $I$ is homogeneous. By the proposition it would suffice to prove Conjecture 2: If $G$ is an (affine) group scheme acting on a scheme $X$ then the bijection $$\{\text{Open subschemes of }X\} \iso \{\text{Reduced closed subschemes of }X\}$$ given by “reduced closed complement” and “open complement” restricts to a bijection $$\{G\text{-invariant open subschemes of }X\} \iso \{G\text{-invariant reduced closed subschemes of }X\}.$$ This certainly sounds reasonable if one thinks for example of an action of a topological group on a topological space, where the complement of an invariant subset is clearly invariant again. However I cannot prove it in the context of schemes and I'm not quite sure that it is correct. If $X = \Spec(R)$ is reduced and of finite type over an algebraically closed field and $G = \IG_m$ (so that $\IG_m \times X$ is again reduced and of finite type), then it suffices to consider $k$-valued points, so that the reduced complement of an invariant open subscheme is indeed invariant again. If $G = \IG_m$ and $X$ is affine then this conjecture is equivalent to the first one. My conjecture is also equivalent to the claim that if $U \subseteq X$ is an open $\IG_m$-invariant subscheme then $U$ is a union of subschemes of the form $D(f) \subseteq X$ where $f$ in $R$ is a homogeneous element. In particular, this would imply (a special case of) the following conjecture: Conjecture 3: If $G$ is an (affine) group scheme acting on a scheme $X$ and if $U \subseteq X$ is an open $G$-invariant subscheme then $U$ is a union of $G$-invariant affine open subschemes. Unfortunately I really don't have a good intuition for actions of groups schemes and I would be glad about some clarification. REPLY [2 votes]: Your conjecture 2 is false if you don't assume that $G$ is reduced (in positive characteristic there are affine group schemes that are not reduced). As to conjecture 3, it is hopelessly wrong (think of the action of $\mathrm{GL}_n$ on $\mathbb P^{n-1}$). The only non-trivial case I know is Sumihiro's theorem: a normal algebraic variety with an action of a torus can be covered by invariant affine open subsets. Being normal is essential: consider the standard action of $\mathbb{G}_{\mathrm m}$ on $\mathbb P^1$, and glue together the origin and the point at infinity.<|endoftext|> TITLE: Packing circles with radii 1, 2, 3, ..., n in a rectangle QUESTION [5 upvotes]: For each positive integer n, let $a_n$ be the area of the smallest rectangle whose area is a whole number, and inside which it is possible to pack all n circles of radii 1, 2, 3, ..., n respectively (with no overlaps). Is it possible to determine $a_n$ precisely? For example $a_{12}$ is at most 2466 (https://puzzling.stackexchange.com/questions/92949/my-mothers-dish-collection), and can perhaps be proved to be precisely that. REPLY [14 votes]: Here's a better solution for $n=12$, with area approximately 2496: Even better, with area approximately 2463: Here's @MattF's suggestion, but it's worse in both dimensions: @GerhardPaseman, if I consider only circles 6 through 12, this is the best solution I have found:<|endoftext|> TITLE: A trapping set with finite measure QUESTION [7 upvotes]: Does there exist a measurable subset $T$ of $[0, \infty)$ with finite measure and some $\epsilon > 0$ such that for every $r$ with $0 < r < \epsilon$, $nr$ is in $T$ for infinitely many positive integers $n$? Note: The integers $n$ such that $nr$ lie in $T$ can depend on $r$. REPLY [10 votes]: No. Denote $T_k=T\cap [k,k+1)$. Then $\sum |T_k|<\infty$ (where $|X|$ stands for the measure of $X\subset \mathbb{R}$). Choose a segment $[a,b]\subset (0,\epsilon)$. Note that if $r\in [a,b]$ and $nr\in T_k$, then $na\leqslant nr< k+1$ and $nb\geqslant nr\geqslant k$, thus $n\in [k/b,(k+1)/a]$. The union of $n^{-1}T_k$ over all positive integers $n\in [k/b,(k+1)/a]$ has measure at most $$|T_k|\cdot \sum_{n\in [k/b,(k+1)/a]} n^{-1}\leqslant C|T_k|,$$ where $C$ depends only on $a$ and $b$. Now choose $k_0$ so that $C\sum_{k>k_0} |T_k|k_0$ and positive integer $n$.<|endoftext|> TITLE: Example of nef divisor with Iitaka dimension 0 QUESTION [6 upvotes]: Does anyone know examples of a numerically non trivial nef divisor with Iitaka dimension 0? (Unfortunately, this question might be trivial as right now I can't think of any effective divisors with Iitaka dim 0 other than exceptional divisors) REPLY [4 votes]: This is not exactly what was asked, but I think it is close to the spirit of the question: Mumford's example of a non-ample divisor that is positive on every curve gives an example of a numerically non-trivial nef divisor with Iitaka dimension $-1$. Here is a sketch, you can find a complete proof in many places, or fill in the gaps yourself: Let $B$ be a curve of genus at least $2$. Then there exists $\mathscr E$, a locally free sheaf on $B$ of rank $2$ and degree $0$ such that $\operatorname{Sym}^m(\mathscr E)$ is stable for all $m>0$. (Proof: HW). Let $X=\mathbb P(\mathscr E)$ and $D$ the divisor corresponding to $\mathscr O_{\mathbb P(\mathscr E)}(1)$. Then $D\cdot C>0$ for any effective curve $C\subseteq X$. (Proof: HW. Hint: use the stability assumption). So, $D$ is nef, in fact as nef as it can be $\overset{..}\smile$. However, $D$ is not ample, because $D^2=0$. (Proof: $\deg\mathscr E=0$). This is the usual reason this example is mentioned, that is, that being positive on curves is not enough for being ample. But it actually gives an example of a numerically non-trivial nef divisor that does not have any effective representative, because if $D$ was represented by an effective divisor, then it would be positive on it, which would contradict that $D^2=0$. (And, in fact in that case, $D$ would be ample. In other words, any non-ample divisor on a surface that is positive on every curve must have an empty linear system).<|endoftext|> TITLE: Constant Martin kernel and amenability QUESTION [8 upvotes]: Consider a finitely supported random walk on a discrete group G such that the support generates $G$ as a semigroup. The Martin kernels are then non-negative functions on the product $G \times M$ where $M$ denotes the Martin boundary of the random walk. Does anyone know an example with $G$ amenable such that there does not exist a point m in the Martin boundary for which the function $K( . , m)$ is constant $1$? And does anyone know an example of a non-amenable group for which there is an element $m$ in the Martin boundary such that the corresponding Martin kernel $K( . ,m)$ is constant $1$? REPLY [3 votes]: This is a partial answer to your second question. If the function 1 is minimal, then every bounded harmonic function is constant. To reformulate, if there exists a point $m$ in the minimal Martin boundary such that $K(\cdot, m)$ is constant 1,then the Poisson boundary is trivial, which cannot happen if the group is non-amenable. So you have to look at non-minimal points. EDIT (taking into account the comments of R W) However, for finitely supported measures, in several classes of groups (such as hyperbolic groups and relatively hyperbolic groups with respect to virtually abelian subgroups), the Martin boundary is minimal. So in particular, for such examples of groups, the answer is no. You thus have to look for groups such that the minimal boundary is not the whole Martin boundary (such as the examples given by R W).<|endoftext|> TITLE: The multipartition of the dual of a $C_m \wr S_n$ module QUESTION [5 upvotes]: I know that the irreducible modules of $C_m \wr S_n$ over $\mathbb{C}$ are parametrised by m-multipartitions. The parts of the multipartition are indexed by the elements of $C_m$. My question now is: If $V$ is an irreducible module over $\mathbb{C}$ with multipartition $\underline{\lambda} = (\lambda_0, \lambda_1, ..., \lambda_{m-1})$, what is the multipartition of $V^*$? I feel like it should be a permutation of $\underline{\lambda}$ corresponding to the map $i \mapsto m-i$, but I cannot find that statement anywhere. It should be fairly easy to prove but I have not managed to do so yet. If the proof turns out to be too complicated, any source would suffice. Thanks in advance! REPLY [3 votes]: I assume that by a multipartition you mean here that $\lambda_i$ is a partition of $k_i$ and $\sum_i k_i = n$. In this case the correspondence you described is indeed the duality correspondence. I believe the easiest way to see this is via Clifford Theory: Consider the short exact sequence $$1\to C_m^n\to G\to S_n\to 1$$ where $G=C_m\wr S_n$. Let $\zeta$ be an $m$-th root of unity and let $g$ be a generator of $C_m$. Every irreducible representation of $C_m^n$ is of the form $$\rho_{t_1,\ldots,t_n}\big((g^{a_1},\ldots,g^{a_n})\big)=\zeta^{\sum_i t_ia_i}$$ where $t_i\in \{0,\ldots, m-1\}$. Every irreducible representation is $S_n$-conjugate to a unique representation in which $t_1\leq t_2\ldots \leq t_{n-1}$. Fix such a tuple. For $i\in \{0,\ldots, m-1\}$ write $k_i:= |\{j| t_j=i\}|$. The stabilizer of $\rho_{t_1,\ldots, t_n}$ in $S_n$ will then be isomorphic to $S_{k_0}\times\cdots\times S_{k_{m-1}}$. If $(\lambda_i,\ldots, \lambda_m)$ is a multipartition in the sense that $\lambda_i$ is a partition of $k_i$, then the corresponding tensor product of Specht modules $\mathbb{S}_{\lambda_0}\otimes\cdots\otimes \mathbb{S}_{\lambda_{m-1}}$ is an irreducible representation of $S_{k_0}\times\cdots\times S_{k_{m-1}}$. By letting $C_m^n$ act via the character $\rho_{t_1\ldots,t_n}$ on this vector space you get a representation $V$ of $$H:=C_m^n\ltimes (S_{k_1}\times\cdots\times S_{k_m}).$$ By taking the induced representation $\text{Ind}^G_H V$ you get a representation of $G$. Clifford Theory asserts that this is an irreducible representation of $G$, and that you get all the irreducible representations of $G$ this way. Now for the duality: it holds that $$(\text{Ind}^G_H V)^* \cong \text{Ind}^G_H (V^*).$$ Since all the irreducible representations of $S_{k_0}\times\cdots\times S_{k_{m-1}}$ are self dual, you are left with the same representation of this group. However, $C_m^n$ acts now via the dual character of $\rho_{t_1,\ldots, t_n}$, which is $\rho_{m-t_1,\ldots, m-t_n}$. When you calculate the numbers $k'_i$ which correspond to this representation you will get now that $k'_i=k_{m-i}$. So the multipartition which corresponds to the dual representation is indeed $(\lambda_0,\lambda_{m-1},\ldots, \lambda_1)$<|endoftext|> TITLE: Ultrafilter on the ordinal $\omega^\omega$ QUESTION [7 upvotes]: For any ultrafilter $\mathcal{U}$ on $\omega$ and any finite $k$ we can construct tensor power $\mathcal{U}^{\otimes k}$ which is ultrafilter on $\omega^k$. Does there exist some natural extension of this construction for the ordinal $\omega^\omega$? Edit: my suggestion: $$ \mathcal{U}^{\otimes\omega}=\{B\subset\omega^\omega~|~\{k<\omega~|~B\cap\omega^k\in\mathcal{U}^{\otimes k}\}\in\mathcal{U}\} $$ But is it good idea? REPLY [11 votes]: The relevant general construction is the sum of a family $\{\mathcal V_i:i\in I\}$ of an indexed family of ultrafilters, with respect to an ultrafilter $\mathcal U$ on the index set $I$. If $\mathcal V_i$ is an ultrafilter on $X_i$, then the sum is the ultrafilter $\mathcal W$ on the disjoint union $\bigsqcup_{i\in I}X_i$ defined by $$ \mathcal W=\{A:\{i\in I:A\cap X_i\in\mathcal V_i\}\in\mathcal U\}. $$ In your situation, taking $\mathcal V_i$ to be $\mathcal U^{\otimes i}$, you get a sum ultrafilter on $\bigsqcup_{i\in\omega}\omega^i$, which can be identified with the ordinal $\omega^\omega$ to produce the ultrafilter $\mathcal U^{\otimes\omega}$ in the question.<|endoftext|> TITLE: What is the largest possible probability that a random matrix over $\mathbb{F}_2$ is non-singular? QUESTION [7 upvotes]: Suppose $A(p, n)=(a_{ij}(p))_{i, j \leq n}$ is an $n\times n$ random matrix over $\mathbb{F_2}$, with all its entries being i.i.d. and such that $P(a_{ij}(p) = 1) = p$, where $p$ is some real number from $[0; 1]$. What is the largest possible probability, that $A(p, n)$ is non-singular and with what $p$ is it reached? Note, that $A(p, n)$ is non-singular iff $\det(A(p, n)) = 1$. Solution for $n=1$: $\det(A(p, 1)) = 1$ with probability $p$. The maximum of $\det(A(p, 1))$ is $1$ and it is reached with $p = 1$. Solution for $n = 2$: $\det(A(p, 2)) = 1$ with probability $2p^2(1 - p^2)$. The maximum of $P(\det(A(p, 2))=1)$ is $\frac{1}{2}$ and it is reached with $p = \frac{1}{\sqrt{2}}$. However, I would like to know some sort of general formula (or at least asymptotics). After I failed to solve this problem using determinants, I tried to prove this using the fact that a square matrix is non-singular iff its rows are linearly dependent. As there exists only one non-zero element in $\mathbb{F_2}$, we can write linear dependence of the vector system $\{v_i\}_{i \leq n}$ in $\mathbb{F}_2^n$ as $\forall S \subset \{1, ... , n\}$ such that $S \neq \emptyset$ we have $\sum_{i \in S} v_i \neq \overline{0}$. I know the probability that a given set of vectors with i.i.d. random entries Bernoulli distributed with parameter $p$ $\{v_i\}_{i \leq k}$ over $\mathbb{F}_2^n$ satisfy $\sum_{i = 1}^k v_i \neq \overline{0}$ is $(1 - \frac{p((1 - 2p)^k - 1)}{1 - 2p})$. However, I do not know how to proceed further in this direction. This question on MSE REPLY [8 votes]: For $n\rightarrow\infty$ the probability ${\cal P}_\infty$ that $A(p,n)$ is nonsingular becomes independent of $p\in(0,1)$, given by $${\cal P}_\infty=\prod_{i=1}^\infty(1-2^{-i})=0.2887880951$$ See theorem 3.2 in Properties of random matrices and applications (2007). For finite $n$, there is a result that could be instructive, which is the expectation value $E(p,n)$ of the number of linear dependencies among the rows of $A(p,n)$. This is given by theorem 4.5, $$E(p,n)=2^{-n}\sum _{j=1}^n \binom{n}{j} \left(1+(1-2 p)^j\right)^n.$$ A plot of $E(p,n)$ as a function of $p$ becomes flatter and flatter with increasing $n$ (see below for $n$ up to 20), consistent with the understanding that for large $n$ the probability that the matrix is singular no longer depends on $p$.<|endoftext|> TITLE: Can I build infinitely many polytopes from only finitely many prescribed facets? QUESTION [16 upvotes]: Given a finite set of convex $d$-dimensional polytopes $\mathcal P$, for some $d\ge 2$. Question: Is it true that there are only finitely many different convex $(d+1)$-dimensional polytopes whose facets are solely (scaled and rotated versions of) polytopes in $\mathcal P$? Some clarifications A face of a polytope is the intersection of the polytope with a touching hyperplane, so subdividing facets does not count here. In general, two $(d+1)$-polytopes shall be considered as different if they differ not just in scale and orientation. By the usual rigidity arguments, given the shape of facets and their connections, the metric of the polytope is uniquely determined. Hence, if we can built different polytopes, they will be combinatorially different as well. Example There are only finitely many polyhedra that can be built from any finite set of regular polygons, but as far as I know, this result is by enumeration (see, e.g. Johnson solids). REPLY [6 votes]: Not an answer, but an "almost counterexample" for $d=2$ that came to my mind when I saw that question. Starting with a regular dodecahedron, we can successively insert circular "belts" of hexagons, in a way that the whole polyhedron is convex at each stage. The illustrations should give an idea. The shapes of the hexagonal facets can come very close to each other (in the appropriate sense), though it is clear that there can only be so many of a precise shape.<|endoftext|> TITLE: What is the convex hull of the quaternionic symmetries of the 3 dimensional cube? QUESTION [10 upvotes]: It is well known that there are exactly five 3-dimensional regular convex polyhedra, known as the Platonic solids. In 1852 the Swiss mathematician Ludwig Schlafli found that there are exactly six regular convex 4-polytopes (the generalization of polyhedra to 4 dimensions) and that, for dimensions 5 and above, there are only three! The six regular 4-polytopes are: NAME VERTEXES EDGES FACES CELLS Hypertetrahedron 5 10 10 5 Hypercube 16 32 24 8 Hyperoctahedron 8 24 32 16 24-cell 24 96 96 24 Hyperdodecahedron 600 1200 720 120 Hypericosahedron 120 720 1200 600 The easiest ones to be described are the first two: a model for the hypertetrahedron may be obtained as the convex hull of the canonical basis in $\mathbb R^5$ (hence a 4-dimensional object), while a model for the hypercube is the Cartesian product $[0, 1]\times[0, 1]\times[0, 1]\times[0, 1]$. As in the case of 3 dimensions, the dual of a regular 4-polytope is also a regular 4-polytope and it turns out that the six regular 4-polytopes found by Schlafli are related to each other via duality as follows. Hypetetrahedron <-> Itself 24-cell <-> Itself Hypercube <-> Hyperoctahedron Hyperdodecahedron <-> Hypericosahedron This means that one needs only describe the 24-cell and the hypericosahedron for all of them to be known. In other words: Hypertetrahedron = convex hull of the canonical basis in 5 dimensions Hypercube = [0,1]x[0,1]x[0,1]x[0,1] Hyperoctahedron = dual of the hypercube Hyperdodecahedron = dual of the hypericosahedron 24-cell ??? Hypericosahedron ??? The description of the last two 4-polytopes above may be obtained by considering the quaternions $\mathbb H$. Viewing $\mathbb R^3$ within $\mathbb H$ via the map $$(x,y,z)\mapsto xi+yj+zk, $$ it is well known that every quaternion $q$, with $\Vert q\Vert=1$, gives a rotation $R_q$ on $\mathbb R^3$ via the formula $$ R_q(v) = qvq^{-1}, \quad \forall v \in \mathbb R^3. $$ In fact the correspondence $q\mapsto R_q$ is a two-fold covering of $SO(3)$ by the unit sphere in $\mathbb H$. Letting $P_{20}$ be the icosahedron in $\mathbb R^3$, consider the quaternionic symmetries of $P_{20}$, which I will write as $\mathbb {HS}(P_{20})$, defined to be the set of all unit quaternions $q$ such that $R_q$ leaves $P_{20}$ invariant. In symbols $$ \mathbb {HS}(P_{20}) =\{q\in \mathbb H: \Vert q\Vert=1,\ R_q(P_{20})=P_{20}\}. $$ Well, the convex hull of $\mathbb {HS}(P_{20})$ in $\mathbb R^4$ turns out to be a model for the hypericosahedron! Since the symmetries of a regular polyhedron are the same as the symmetries of its dual, it is clear that the symmetries of $P_{12}$, the dodecahedron, gives nothing new: the convex hull of $\mathbb {HS}(P_{12})$ is just another model for the hypericosahedron. Passing to the (self dual) tetrahedron, call it $P_4$, the convex hull of $\mathbb {HS}(P_{4})$ gives a model for the remaining 4-polytope, namely the 24-cell, completing the description of the six Schlafli's 4-polytopes. Question: What is the convex hull of the quaternionic symmetries of the 3 dimensional cube? If I am not mistaken, this 4-polytope has 48 vertexes and 144 edges, so it is not in Schlafli's list and hence cannot be regular. EDIT: Yes I was mistaken about the number of edges which is in fact 336 according to M. Winter's answer below! REPLY [14 votes]: This is the disphenoidal 288-cell, which is the dual of the bitruncated 24-cell. This is also mentioned in the "Geometry" section of the Wikipedia article on the 288-cell. It has 48 vertices, and 336 edges. However, 144 of these are of the shortest length, and I suppose you have counted these. The symmetry group of the tetrahedron is "the half" of the symmetry group of the cube (as the tetrahedron is the 3-dimensional demicube). You already know that the tetrahedral symmetries give you the 24-cell. In the same way, the 24-cell is "the half" of the disphenoidal 288-cell: the latter is the convex hull of the union of a 24-cell and its dual (which is a 24-cell as well, but differently oriented).<|endoftext|> TITLE: Does every f.g. group have a minimal presentation? QUESTION [12 upvotes]: Call a group presentation $\langle X \,\|\,R \rangle$ minimal if no relator from $R$ is a consequence of the remaining relators, i.e., no $r \in R$ belongs to the normal closure of $R\setminus \{r\}$ in the free group $F(X)$. Question: Does every finitely generated group have a minimal presentation (with $X$ finite)? Remark: evidently every finite presentation $\langle X \,\|\,R \rangle$ has a minimal sub-presentation $\langle X \,\|\,R' \rangle$ (where $R' \subseteq R$ has the same normal closure in $F(X)$ as $R$), so the question really concerns infinitely presented groups. It is possible to construct infinite presentations which do not have minimal sub-presentations. Indeed, let $F=F(a,b)$ be the free group on $\{a,b\}$. One can choose "sufficiently independent" elements $w_1,w_2,\dots \in F$ so that the presentation $$\langle a,b \,\|\, w_i^2,w^2_{i+1}w_i, ~i \in \mathbb{N} \rangle$$ has no minimal sub-presentation. Here $R$ consists of the elements $w_1^2,w_2^2,\dots$, and $w_2^2w_1,w_3^2w_2,\dots$. However, the above group also has the presentation $\langle a,b \,\|\,w_1,w_2,\dots \rangle$, which may be minimal. REPLY [9 votes]: The answer is no. In order to see this, you may combine Theorem 3.9 and Remark 5.3 of [1]. A counter-example is given by the nilpotent-by-Abelian group $B$ of Equation (3.2). Further examples are provided by Remark 5.15. [1] R. Bieri, Y. de Cornulier, L. Guyot and R. Strebel, "Infinite presentability of groups and condensation", 2014.<|endoftext|> TITLE: Irreducible global Galois representation with weights 0, 1, 3? QUESTION [7 upvotes]: Fix a prime number $p$. Can there exist a continuous irreducible representation $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \mathrm{GL}_3(\mathbb{Q}_p)$ that is unramified at almost all primes, is de Rham at $p$ and whose Hodge-Tate weights at $p$ are 0, 1 and 3? REPLY [6 votes]: Here are two arguments for why such a representation $\rho$ cannot exist. Automorphic argument: Fontaine and Mazur have conjectured that any irreducible $n$-dimensional geometric representation $\rho$ of $Gal(\overline{\mathbf{Q}} / \mathbf{Q})$ comes from a cuspidal automorphic representation $\pi$ of $GL_n(\mathbf{A}_{\mathbf{Q}})$, and that "local-global compatibility at $\infty$" should hold, which amounts to saying that the Archimedean component $\pi_\infty$ should be determined by the Hodge–Tate weights of $\rho$ – up to a certain explicit shift, the multiset of Hodge–Tate weights of $\rho$ is the Harish-Chandra parameter of $\pi_\infty$. However, the possibilities for the representations $\pi_\infty$ which can show up as Archimedean components of automorphic representations are pretty restricted, so $(0, 1, 3)$ isn't possible. (This is essentially the argument sketched in David Hansen's comment from 2010 that you linked to.) Motivic argument: Fontaine and Mazur also made a (separate) conjecture that any such $\rho$ is the $p$-adic realisation of a pure motive over $\mathbf{Q}$. By the comparison isomorphism of Faltings–Tsuji relating étale and de Rham cohomology, this implies that the Hodge–Tate weights of $\rho$ give the graded pieces of a pure Hodge structure. Since a pure Hodge structure has a weight $w$ and an action of complex conjugation which switches the $(p, w-p)$ and $(w-p, p)$ parts, this means the set of weights must be symmetric around $w/2$.<|endoftext|> TITLE: A zeta function for the Klein Four group? QUESTION [5 upvotes]: Let $G$ be a finite group, $S \subset G$ a generating set, $|g|:=|g|_S=$ word-length with respect to $S$. Let $\phi(g,h)=|g|+|h|-|gh| \ge 0$ be the "defect-function" of $S$. The set $\mathbb{Z}\times G$ builds a group for the following operation: $$(a,g) \oplus (b,h) = (a+b+\phi(g,h),gh)$$ On $\mathbb{N}\times G$ is the "norm": $|(a,g)| := |a|+|g|$ additive, which means that $|a \oplus b| = |a|+|b|$. Define the multiplication with $n \in \mathbb{N_0}$ to be: $$ n \cdot a := a \oplus a \oplus \cdots \oplus a$$ (if $n=0$ then $n \cdot a := (0,1) \in \mathbb{Z} \times G$). A word $w := w_{n-1} w_{n-2} \cdots w_0$ is mapped to an element of $\mathbb{Z} \times G$ as follows: $$\zeta(w) := \oplus_{i=0}^{n-1} (m^i \cdot (0,w_i))$$ where $m := \min_{g,h\in G, \phi(g,h) \neq 0} \phi(g,h)$. We let $|w|:=|\zeta(w)|$ and $w_1 \oplus w_2:=\zeta(w_1)\oplus \zeta(w_2)$ Then we have $|w_1 \oplus w_2| = |w_1|+|w_2|$. For instance for the Klein four group $\{0,a,b,c=a+b\}$ generated by $S:=\{a,b\}$, we get sorting the words $w$ by their word-length: $$0,a,b,c,a0,aa,ab,ac,b0,ba,bb,bc,c0,ca,cb,cc,a00,a0a,a0b,a0c$$ corresponding to the following $\mathbb{Z}\times K_4$ elements $\zeta(w)$: $$(0,0),(0,a),(0,b),(0,c),(2,0),(2,a),(2,b),(2,c),(2,0),(2,a),(2,b),(2,c),(4,0),(4,a),(4,b),(4,c),(4,0),(4,a),(4,b),(4,c)$$ corresponding to the the following "norms" of words $|w| = |\zeta(w)|$: $$0,1,1,2,2,3,3,4,2,3,3,4,4,5,5,6,4,5,5,6$$ Let $a_n, n\ge 0$ be the sequence of numbers generated by the Klein four group. 1) Is $$\sum_{n=1}^\infty \frac{1}{a_n^s} = \sum_{n=1}^\infty \frac{n+1}{n^s} = \zeta(s-1) + \zeta(s)$$ where $\zeta$ denotes the Riemann zeta function? I have checked this with SAGE math up to a certain degree and it seems plausible, however I have no idea how to prove it. 2) Is every $a_n$ the product of primes $p=a_k$ for some $k\le n$? 3) Let $\pi_{K_4}(n) = |\{ k : \text{$a_k$ is prime, $k \le n$}\}|$ be the prime counting function of the sequence. What is the approximate relationship to the usual prime counting function $\pi(n)$? REPLY [2 votes]: Consider a word $w=w_0w_1\cdots w_{n-1}$ with each $w_i\in\{0,a,b,c\}$. Following your notation, $$\lvert w\rvert=\lvert\zeta(w)\rvert=\left\lvert\bigoplus_{i=0}^\infty m^i\cdot(0,w_i)\right\rvert=\sum_{i=0}^{n-1}\lvert 2^i\cdot(0,w_i)\rvert=\sum_{i=0}^{n-1} 2^i\lvert w_i\rvert$$ where $\lvert0\rvert=0$, $\lvert a\rvert=1$, $\lvert b\rvert=1$, $\lvert c\rvert=2$. We now consider the generating function $$F_n(x)=\sum_wx^{\lvert w\rvert}$$ where the sum is over all words $w=w_0w_1\cdots w_{n-1}$. Then \begin{align*} F_n(x)&=\sum_{w_0}\sum_{w_1}\cdots\sum_{w_{n-1}}x^{|w_0|+2|w_1|+\cdots+2^{n-1}|w_{n-1}|}\\ &=\left(\sum_{w_0}x^{|w_0|}\right)\left(\sum_{w_1}x^{2|w_1|}\right)\cdots\left(\sum_{w_{n-1}}x^{2^{n-1}|w_{n-1}|}\right)\\ &=(1+2x+x^2)(1+2x^2+x^4)(1+2x^4+x^8)\cdots(1+2x^{2^{n-1}}+x^{2^n})\\ &=(1+x)^2(1+x^2)^2(1+x^4)^2\cdots(1+x^{2^{n-1}})^2\\ &=(1+x+x^2+\cdots+x^{2^n-1})^2\\ &=1+2x+\cdots+(2^n-1)x^{2^n-2}+2^nx^{2^n-1}+(2^n-1)x^{2^n}+\cdots+x^{2^{n+1}-2}. \end{align*} What this shows is that among the first $4^n$ terms of the sequence: \begin{align*} &0\text{ appears }1\text{ times},\\ &1\text{ appears }2\text{ times},\\ &2\text{ appears }3\text{ times},\\ &\cdots\\ &2^n-2\text{ appears }2^n-1\text{ times},\\ &2^n-1\text{ appears }2^n\text{ times},\\ &2^n\text{ appears }2^n-1\text{ times},\\ &\cdots\\ &2^{n+1}-4\text{ appears }3\text{ times},\\ &2^{n+1}-3\text{ appears }2\text{ times},\\ &2^{n+1}-2\text{ appears }1\text{ times}.\\ \end{align*} This resolves question 1. Since this seems to be purely a combinatorics question, I would not expect the second and third questions to have particularly interesting answers. I will remark that $$\pi_{K_4}(4^n)=\sum_{p\leq2^{n+1}-2}(2^n-\lvert p-2^n+1\rvert)p$$ so you could use some analytic number theory to get asymptotics for $\pi_{K_4}(n)$ if you wanted.<|endoftext|> TITLE: How to properly verify that $E\times E'$ has no non-trivial effective divisors with Kodaira dimension zero QUESTION [5 upvotes]: Let $E$ and $E'$ be elliptic curves over $\mathbb{C}$. I am pretty confident that the only effective divisor $D\subset E\times E'$ with Kodaira dimension zero is the trivial divisor. How to prove this in a simple manner? REPLY [7 votes]: If $D$ has Iitaka dimension zero, then $\dim H^0 ( n E) =1$ for all $n$, because if it were any larger than $\dim H^0(k ne) \geq k+1$. If we have an automorphism $\sigma$ with $[ n \sigma (D)] = [n D]$ in the divisor class group, we would have two linearly independent sections unless in fact $\sigma(D)= D$. For $\sigma$ translation by an $n$-torsion point, for any $D$, $\sigma(D)- D$ is $n$-torsion, and so $[n \sigma(D) ] =[nD]$. (To see this, one can use the exponential exact sequence $H^1( A, \mathcal O_X) \to H^1(A, \mathcal O_X^\times) \to H^2(A, \mathbb Z)$. Translation acts trivially on the first and last terms, meaning that $\sigma (\sigma(D)-D) =\sigma(D)-D$ and thus $\sigma^n=1$ implies $\sigma(D)- D$ is $n$-torsion.) So to have Kodaira dimension $0$, $D$ would have to be invariant under translation by $n$-torsion for all $n$. Thus, it must be either empty or Zariski dense, and therefore it is empty.<|endoftext|> TITLE: functors $\text{Vect} \to \text{Vect}$ that preserve filtered and sifted colimits QUESTION [9 upvotes]: I'm considering various functors from the category $\text{Vect}$ of real vector spaces to itself, and would like to know that they preserve filtered colimits and possibly even sifted colimits. The functors I'm interested in send $V$ to the power $V^k$, the tensor power $V^{\otimes k}$, the exterior power $\Lambda^k(V)$, and the free vector space $F(V)$ on the underlying set of $V$. I have proved by hand that some of these preserve filtered colimits or sifted colimits, but I'm looking for references and/or conceptual arguments. Of course, I'd also like to know if some of these don't preserve filtered and/or sifted colimits. (I doubt it matters that I'm working over the reals.) REPLY [9 votes]: Maybe the following tools can help. $G: \mathcal{C}_1 \times \cdots\times \mathcal{C}_k\to \mathcal{D}$ preserves sifted colimits separately in each variable if and only if it preserves sifted colimits. Indeed, given a diagram $p: K \to \mathcal{C}_1\times\cdots \times \mathcal{C}_k$ with $K$ sifted, observe that $p= (q_i)\circ \delta$ where $(q_i): \prod_{i=1}^kK \to \prod_{i=1}^k\mathcal{C}_i$ is the product of $q_i=\mathrm{proj}_i\circ p$ and $\delta$ is the diagonal. Since $K$ is sifted, $\delta$ is final, and the colimit may be computed over $K^{\times k}$ instead of over $K$. But then we can apply the hypothesis $k$ times to see that this colimit is preserves by $G$. The diagonal map $\mathcal{C} \to \mathcal{C}^{\times j}$ preserves all colimits. The forgetful functor $\mathbf{Vect} \to \mathbf{Set}$ preserves all sifted colimits (since it preserves filtered colimits and reflexive coequalizers by inspection). The free vector space functor preserves all colimits, being a left adjoint. Combining (4) and (3) gives the "F(V)" example you wanted, so let's move on to the others. Notice that the tensor power functor $V \mapsto V^{\otimes n}$ is a composite of the diagonal and the tensor product functor, which preserves all colimits separately in each variable, so it preserves sifted colimits. Moreover, there is an action of $\Sigma_n$ on $V^{\otimes n}$ so this functor refines to $\mathbf{Vect} \to \mathbf{Fun}(\mathrm{B}\Sigma_n, \mathbf{Vect})$. This also preserves sifted colimits since colimits in functor categories are computed pointwise. It's also true that the product $V \mapsto V^{\times n}$, as a functor to vector spaces with a $\Sigma_n$ action, preserves sifted colimits, since the cartesian product of sets preserves all colimits of sets in each variable, and sifted colimits of vector spaces may as well be computed on the underlying set, so the same reasoning applies (even though the cartesian product functor does not preserve all colimits of vector spaces separately in each variable). So now, given a functor $\mathbf{Fun}(\mathrm{B}\Sigma_n, \mathbf{Vect})\to \mathbf{Vect}$ that preserves sifted colimits, we can apply it to the tensor power functor to get new functors. For example, we can take $\Sigma_n$-coinvariants (i.e. symmetric powers). We can tensor with the sign representation and then take $\Sigma_n$-coinvariants (which gives the exterior power). Any Schur functor works too, since we're just tensoring over $\mathbb{C}[\Sigma_n]$ (or $\mathbb{R}[\Sigma_n]$ over the reals) with the corresponding representation, and that preserves colimits.<|endoftext|> TITLE: Quirky, non-rigorous, yet inspiring, literature in mathematics QUESTION [18 upvotes]: In contrast with such lucid, pedagogical, inspiring books such as Visualizing Complex Analysis by Needham and Introduction to Applied Mathematics by Strang, I've had the pleasure of coming across the non-rigorous, thought-provoking/stimulating, somewhat quirky works of Heaviside on operational calculus and divergent series; of Ramanujan on series, in particular, his use of his master theorem/formula (as explicated by Hardy); and the relatively unknown posthumous notes of Bernard Friedman on distributions and symbolic/operational calculus Lectures on Applications-Oriented Mathematics. These works just blindside you. You think, "What the hey?" and slowly they grow on you and you start to understand them after further research using other texts, translating the terminology/concepts, and working out details. You're left with a deeper understanding and appreciation of the originality and applicability of the work--much like Hardy professed the day after he received Ramanujan's letter, no doubt. (Friedman's Applied Mathematics, in contrast, is of a very different nature and an immediately enlightening intro to its topics.) Any similar experiences with other mathematical works? (This question is not research-level per se and may be more appropriate for MSE, but certainly the works cited have inspired and continue to inspire much advanced research into the related topics, and the question falls in a similar category to MO-Q1 and the MO-Qs that pop up in the Related section of that question). REPLY [3 votes]: I enjoyed reading the book 'Quantization, Classical and Quantum Field Theory and Theta Functions' by Andrej Tyurin very much. It is certainly not rigorous but it was very inspiring for me.<|endoftext|> TITLE: Reference request: proof of Ramanujan's Cos/Cosh Identity QUESTION [10 upvotes]: The Ramanujan Cos/Cosh Identity, as stated here, is $$\left[1+2\sum_{n=1}^{\infty}\frac{\cos n\theta}{\cosh n\pi}\right]^{-2}+ \left[1+2\sum_{n=1}^{\infty}\frac{\cosh n\theta}{\cosh n\pi}\right]^{-2}= \frac{2\Gamma^4\left(\frac34\right)}{\pi}.$$ I am looking for a proof, preferably from a reputable source. I hoped I would find something in Ramanujan's Notebooks, but have so far found no mention of it. REPLY [24 votes]: I expand my comment into an answer. The key here is the Fourier series for the elliptic function $\operatorname {dn} (u, k) $ given as $$\operatorname {dn} (u, k) =\frac{\pi} {2K}\left(1+4\sum_{n=1}^{\infty} \frac{q^n} {1+q^{2n}}\cos\left(\frac{n\pi u} {K} \right) \right) $$ where $K$ is the complete elliptic integral of first kind corresponding to modulus $k$ and $q=\exp(-\pi K'/K) $ is the nome corresponding to modulus $k$. Let $\theta=\pi u/K$ then we have $$\operatorname {dn} \left(\frac{K\theta} {\pi}, k\right) =\frac{\pi} {2K}\left(1+2\sum_{n=1}^{\infty}\frac{\cos n\theta} {\cosh (n\pi K'/K)} \right) $$ Putting $K'=K$ so that $k=1/\sqrt{2}$ and $K=\dfrac{\Gamma^2(1/4)}{4\sqrt{\pi}}$ we get $$1+2\sum_{n=1}^{\infty}\frac{\cos n\theta} {\cosh n\pi} =\frac{2K}{\pi}\operatorname {dn} \left(\frac{K\theta} {\pi}, \frac{1}{\sqrt{2}}\right)$$ Let the above expression be denote by $A$ and the expression obtained from it by replacing $\theta$ with $i\theta$ be denoted by $B$. Then we have to show that $$\frac{1}{A^2}+\frac{1}{B^2}=\frac{2\Gamma^{4}(3/4)}{\pi}$$ Note that we have $$\Gamma(1/4)\Gamma (3/4)=\sqrt{2}\pi$$ and therefore $$\frac{2K} {\pi} =\frac{\sqrt{\pi}}{\Gamma ^2(3/4)}$$ and we have then $$A^{-2}+B^{-2}=\frac{\Gamma ^{4}(3/4)}{\pi}\left(\operatorname {dn} ^{-2}\left(\frac{K\theta}{\pi}\right)+\operatorname {dn} ^{-2}\left(\frac{iK\theta}{\pi}\right)\right)$$ The expression in parentheses is easily seen to be $2$ if we note that $$\operatorname {dn} (iu, k) =\frac{\operatorname {dn} (u, k')} {\operatorname {cn} (u, k')} $$ and here $k=k'=1/\sqrt{2}$.<|endoftext|> TITLE: Is the lexicographic ordering on the unit square perfectly normal? QUESTION [6 upvotes]: It's known that order topology is completely normal, so the lexicographic ordering on the unit square is also completely normal. It's also known that the lexicographic ordering on the unit square is not metrizable. I am interested in whether it is perfectly normal. (A space is perfectly normal if for any two disjoint nonempty closed subsets, there is a continuous function $f$ to $[0,1]$ that "precisely separates" the two sets, meaning that the two closed sets are $f^{-1}(0)$ and $f^{-1}(1)$.) And how do we prove it? REPLY [3 votes]: For a compact Hausdorff $X$ it is equivalent that $X$ is hereditarily Lindelöf or that $X$ is perfectly normal. Sketch: $X$ is hereditarily Lindelöf implies that every open set is an $F_\sigma$ (as $X$ is regular) so dually every closed set is a $G_\delta$. OTOH if $X$ is perfectly normal, every open set is an $F_\sigma$ so $\sigma$-compact and Lindelöf, making all open sets Lindelöf and $X$ hereditarily Lindelöf again. $[0,1]^2$ in the lexicographic order topology has a discrete subset $[0,1]\times \{\frac12\}$ so is definitely not hereditarily Lindelöf, so not perfectly normal either.<|endoftext|> TITLE: Applications of quantum representations of the mapping class group to quantum computers QUESTION [5 upvotes]: Quantum representations of the mapping class group of a surface are certain representations constructed from the data of a TQFT and described, for example, in and 1 and 2. The following sources 3 and 4 imply that there are applications of quantum representations of the mapping class group to quantum computing. What are some of these applications? Which concrete mathematical questions (e.g. from low dimensional topology or quantum algebra) have applications/are relevant to quantum computing? REPLY [3 votes]: The context is topological quantum computation, where quantum information is stored nonlocally in a physical system, so that it is protected from decoherence by local sources of noise. The nonlocal degree of freedom is a socalled non-Abelian anyon, a particle-like excitation which is described by a (2+1)-dimensional topological quantum field theory. Arxiv:1705.06206 provides a survey of the mathematics of topological quantum computing, with a list of conjectures and open problems. I should add, as a physicist, that I am uncertain whether any of these mathematical questions are relevant in the quest to actually build and operate a quantum computer. The key challenge there is to identify a physical system that has these exotic particles. The fractional quantum Hall effect was a primary candidate for several decades, but this system has now been largely abandoned in favor of superconducting systems, where the energy gaps can be much larger (allowing for operation in a realistic temperature range). Microsoft is heavily invested in the design of a topological quantum computer using anyons in the Ising universality class (Majorana fermions). These are "trivial" from a mathematical perspective, since they only implement a Clifford algebra and do not provide access to the full unitary group. There are no realistic options for Fibonacci anyons (which would cover all unitary operations).<|endoftext|> TITLE: $KO_*$ groups of $\mathbb{R}P^\infty$, "Snaiths" theorem for $KO$ QUESTION [5 upvotes]: I posted this question some days ago at math.stackexchange, but didn't receive an answer. I have two questions: I wonder whether anyone has taken the time to compute $KO_*(\mathbb{R}P^\infty)$? The standard tools to compute these Groups in the complex case rest on the requirement for the cohomology theories $E$ to be complex orientable. Naturally, I looked up whether something like real orientable cohomology theories exist in the literature but found out that $KO$ is not real-oriented. Anyway, there is a way to "circumvent" Snaiths theorem for the spectrum $K$ if one is only interestd in the algebra of cooperations, in the sense that one can show that $$K_*(\mathbb{C}P^\infty) \xrightarrow{i_*} K_*K$$ is an injection of rings, where $i$ is induced from the inclusion $\mathbb{C}P^\infty \simeq BU(1) \hookrightarrow BU$. In fact, one only needs to invert the Bott element $\beta$ to turn it into an isomorphism, so it is a localization. This can be concluded from Robert M. Switzer. Algebraic topology—homotopy and homology. Classics in Mathematics. Springer-Verlag, Berlin, 2002. Reprint of the 1975 original [Springer, New York; MR0385836 (52 #6695)]. 17.33, which states $K_*K$ is generated over $\mathbb{Z}[u,u^{-1},v^{-1}]$ by the polynomials $\{p_1,p_2,\ldots\}$. By a process reminiscent of J. F. Adams. Stable homotopy and generalised homology. University of Chicago Press, Chicago, Ill.-London, 1974. Chicago Lectures in Mathematics. p. 44 we can describe the relations of the generators of $\beta_i$ of $K_*(\mathbb{C}P^\infty) = K_* \{\beta_0 , \beta_1 , \ldots \}$ such that $$\beta_1\beta_n = n \beta_n +(n+1)\beta_{n+1}$$ and by setting $$\binom{x}{i} = \frac{x(x-1)\cdots (x-(i-1))}{i!} \in \mathbb{Q}[x]$$ with $x:=\beta_1$ one can see that $K_*(\mathbb{C}P^\infty)\otimes \mathbb{Q}$ is the polynomial algebra $K_* \otimes \mathbb{Q}[x]$ over $K_*\otimes \mathbb{Q} = \mathbb{Q} [t,t^{-1}]$ and $K_*(\mathbb{C}P^\infty)$ can be identified with the subalgebra of $K_* \otimes \mathbb{Q}[x]$ generated by $\binom{x}{i}$ for $i=0,1,2, \ldots$, where we set $\binom{x}{0}=1$. While snaiths theorem works on the spectrum level and the aforementioned result follows, I wonder whether a similar result holds in the real case, i.e. $$ KO_*(\mathbb{R}P^\infty)[\alpha^{-1}] \cong KO_*KO$$ for some element $\alpha \in KO_*(\mathbb{R}P^\infty)$? REPLY [15 votes]: There are many ways to do this. One elementary approach is to use the Adams spectral sequence $Ext_A(H^*ko \wedge RP^\infty, F_2) \cong Ext_{A(1)}(H^*RP^\infty,F_2) \Rightarrow ko_*(RP^\infty) $ and invert the Bott map. An $A(1)$ resolution giving the answer (since $E_2 = E_\infty$) can be seen at the bottom of the page http://www.rrb.wayne.edu/art/index.html We see the groups, starting with $KO_0$, are $0, Z/2, Z/2, Z/2^\infty, 0, 0, 0, Z/2^\infty$ and repeat by Bott periodicity, with the evident action of the coefficients $KO_*$. (This action follows from the homological algebra of the $Ext$ calculation.)<|endoftext|> TITLE: Physical interpretation of the Manifold Hypothesis QUESTION [15 upvotes]: Motivation: Most dimensionality reduction algorithms assume that the input data are sampled from a manifold $\mathcal{M}$ whose intrinsic dimension $d$ is much smaller than the ambient dimension $D$. In machine learning and applied mathematics circles this is typically known as the manifold hypothesis. Empirically, this is observed to be true for many kinds of data including text data and natural images. In fact, Carlsson et al. found that the high-dimensional space of natural images has a two-dimensional embedding that is homeomorphic to the Klein bottle [1]. Question: Might there be a sensible physical interpretation of this phenomenon? My current intuition, from the perspective of dynamical systems, is that if we can collect large amounts of data for a particular process then this process must be stable. Might there be a reason why stable physical processes would tend to have low-dimensional phase spaces? Might there be a better physical perspective for interpreting this phenomenon? References: Carlsson, G., Ishkhanov, T., de Silva, V. et al. On the Local Behavior of Spaces of Natural Images. Int J Comput Vis 76, 1–12. 2008. Charles Fefferman, Sanjoy Mitter, and Hariharan Narayanan. TESTING THE MANIFOLD HYPOTHESIS. JOURNAL OF THE AMERICAN MATHEMATICAL SOCIETY Volume 29, Number 4, October 2016, Pages 983–1049. 2016. Bastian Rieck, Markus Banagl, Filip Sadlo, Heike Leitte. Persistent Intersection Homology for the Analysis of Discrete Data. Arxiv. 2019. D. Chigirev and W. Bialek, Optimal manifold representation of data : an information theoretic approach, in Advances in Neural Information Processing Systems 16 161–168, MIT Press, Cambridge MA. 2004. T. Roweis and L. K. Saul. Nonlinear dimensionality reduction by locally linear embedding. Science, 290(5500):2323–2326, 2000. Henry W. Lin, Max Tegmark, and David Rolnick. Why does deep and cheap learning work so well? Arxiv. 2017. REPLY [5 votes]: Q: Might there be a reason why stable physical processes would tend to have low-dimensional phase spaces. Yes. One reason is physical processes have dissipation. E.g., turbulence is "known" to be chaotic dynamics on a low dimensional manifold (i.e., strange attractor) in the infinite dimensional phase space (of $L^2$ velocity fields). Even its dimension can be estimated. See for example: Doering, Charles R., and John D. Gibbon. "Note on the Constantin-Foias-Temam attractor dimension estimate for two-dimensional turbulence." Physica D: Nonlinear Phenomena 48.2-3 (1991): 471-480. .<|endoftext|> TITLE: Global obstructions for being a quotient of a rank $d$ vector bundle QUESTION [9 upvotes]: In this recent question (which now has an answer), Richard Thomas asked whether any projective $k$-scheme $X$ of (local) embedding dimension $d(X)$ can be embedded in a smooth $k$-scheme of dimension $d(X)$. If $i \colon X \hookrightarrow Y$ is such an embedding, then in particular we get a surjection $i^*\Omega_Y \twoheadrightarrow \Omega_X$. My (so far unsuccessful) strategy was to obstruct such a surjection from existing. For a coherent sheaf $\mathscr F$, write $d_x(\mathscr F) = \dim_{\kappa(x)} \mathscr F_x \otimes_{\mathcal O_{X,x}} \kappa(x)$ and $$d(\mathscr F) = \max \left\{d_x(\mathscr F)\ |\ x \in X\right\}.$$ Question. If $X$ is a quasi-projective $k$-scheme, and $\mathscr F$ a coherent sheaf, does there exist a surjection $\mathscr E \twoheadrightarrow \mathscr F$ from a locally free sheaf of rank $d(\mathscr F)$? Already if $X = \mathbf A^n$ this seems false to me; for example there should exist finite modules $M$ with $d(M) = 2$ that cannot be generated by $2$ elements (here I am using the Quillen–Suslin theorem that a finite projective module on $\mathbf A^n$ is free). But I don't know so many ways to prove that something is not generated by $2$ elements, except for a local obstruction $d_x(\mathscr F) > 2$. I think it should be possible to give a negative answer to Thomas's question along these lines, by exhibiting a finite flat cover $\pi \colon X \to \mathbf A^n$ such that $\pi_*\Omega_X$ does not admit a surjection from a vector bundle of rank $\deg(\pi) \cdot d(\Omega_X)$. A great answer would incorporate something like this, but I would already be very happy with some global obstruction to surjecting from a vector bundle of a given rank. REPLY [12 votes]: Let me explain a simple example. Let $C \subset \mathbb{P}^3$ be a twisted cubic curve. It is a locally complete intersection of codimension 2, hence its ideal $I_C$ is locally generated by two sections. Let me show that there are no surjections $E \twoheadrightarrow I_C$ from a locally free sheaf $E$ of rank 2. Indeed, assume such a surjection exists. Its kernel is a reflexive sheaf of rank 1, hence is a line bundle, so we have an exact sequence $$ 0 \to L \to E \to I_C \to 0. $$ Restricting to $C$ we obtain an exact sequence $$ 0 \to \det N^* \to L\vert_C \to E\vert_C \to N^* \to 0, $$ where $N^*$ is the conormal bundle. But $N^*$ is locally free of rank 2, hence the surjection $E\vert_C \to N^*$ is an isomorphism, hence the middle arrow is zero, hence $$ \det N^* \cong L\vert_C. $$ But the adjunction fromula shows that $\det N^* \cong \mathcal{O}_C(-10)$, and this line bundle does not restrict from $\mathbb{P}^3$ (because 10 is not divisible by 3). This contradiction proves that no surjection from $E$ as above exists. Of course, the same argument works for many other lci of codimension 2.<|endoftext|> TITLE: Plane partitions with equal margins QUESTION [16 upvotes]: A plane partition of $n$ is an table of integers $A=(a_{ij})$ which add up to $n$ and non-increase in rows and columns. For example, $$A= \begin{matrix} 331 \\ 32 \ \ \\ 11 \ \ \end{matrix} $$ is a plane partition of $(3+3+1)+(3+2)+(1+1)=14$. One can view plane partitions as an arrangement of cubes stacked in the corner, see more on Wikipedia. Define margins to be 1-dim projections of cubes on all coordinate axis. These margins are triples $(\lambda,\mu,\nu)$ of partitions of $n$. For example, for $A$ as above we have $\lambda = (7,5,2)$, $\mu=(7,6,1)$ and $\nu=(7,4,3)$. Question: What is the smallest $n$ for which there exist two different plane partitions of $n$ with the same margins $(\lambda,\mu,\nu)$? Note: I know there are two different plane partitions of $2100$ with equal margins. This is a consequence of $p_2(n) TITLE: Convolution algebra associated to a finite dimensional algebra QUESTION [5 upvotes]: Given a finite dimensional $k$-algebra $A$ (we can assume it is given by a connected quiver with relations). One can form its trivial extension $T(A)$ (see for example https://math.stackexchange.com/questions/229412/trivial-extension-of-an-algebra ), which is a Frobenius algebra and thus it has a coalgebra structure, see for example the book by Kock on Frobenius algebras and 2D topological quantum field theories. Now given any $k$-coalgebra $C$ and $k$-algebra $A$, we can form the convolution algebra $W_{C,A}={\rm Hom}_k(C,A)$. This gives rise to (at least) two new algebras from a given algebra $A$, namely $W_{T(A),A}$ and $W_{T(A),T(A)}$ using that $T(A)$ has a coalgebra structure as a Frobenius algebra. I have no real experience with this topic, so sorry in case my questions are stupid. Question 1: Have either of the algebras $W_{T(A),A}$ or $W_{T(A),T(A)}$ been considered/described before? Are they quiver algebras and if so, can their quivers be described? Question 2: What do those two algebras look like in the concrete example where $A=kQ$ is a path algebra (maybe of Dynkin type)? Question 3: In case those questions are non-trivial, can those algebras be calculated by a computer algebra system like GAP/QPA? REPLY [3 votes]: Question 3: The algebra constructions $W_{C,A}$ or even $W_{T(A),A}$ are not available in QPA. There are no structures made for co-algebras in QPA.<|endoftext|> TITLE: Approximations to $\pi$ QUESTION [12 upvotes]: Is there a way to efficiently solve the following problem besides brute-force calculation? Fix $n\in\mathbb{N}$ (say $n=100$). Find the integers $p,q,r,s$ with $0\leq p,q,r,s\leq n$ such that $$\pm\frac{p}{q}\pm\sqrt{\frac{r}{s}}$$ most closely approximates $\pi$. Some cases can be handled by finding the continued fraction expansion of $(\pi-\frac{p}{q})^2$ for various $p,q$. Playing with this method I found the approximation $4-\sqrt{\frac{14}{19}}$, which is really quite good, but may not be best for $n=20$. Note that the solution will be unique (for a given $n$). REPLY [6 votes]: This is not exactly an answer to the stated question, but it's too long for a comment. Rather than the form given in the question, one could represent a number in the form $\frac{a + b \sqrt{d}}{c}$, where $d$ is a squarefree positive integer, and this form lends itself to finding good approximations to $\pi$ using lattice reduction. Fix a bound $n$. For each squarefree $d$ with $1 \leq d \leq n$, choose a constant $X \approx n^{2} \sqrt{d}$ and create the lattice in $\mathbb{R}^{4}$ spanned by $$ v_{1} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ X \pi \end{bmatrix}, v_{2} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ -X \end{bmatrix}, v_{3} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ -X \sqrt{d} \end{bmatrix}. $$ A short vector in this lattice with respect to the $\ell_{2}$ norm is a linear combination $a v_{1} + bv_{2} + cv_{3}$ and because the fourth coordinate of these vectors are so large, this forces $\frac{a+b \sqrt{d}}{c}$ to be a close approximation to $\pi$. Finding the shortest vector in a lattice is a hard problem, even in small dimensional lattices. However, one can get within a constant multiple of the true minimum using the LLL-algorithm. With this, the above algorithm would run in time $O(n \log^{3} n)$ and find "some good solutions", but isn't guaranteed to find the optimal representation (even in this modified form). I ran this with $n = 10^{6}$ and $X = n^{2} \sqrt{d} \log(n)$ and obtained (after about a minute and a half) $$ \pi \approx \frac{-327031 + 7075 \sqrt{224270}}{962406}. $$ The approximation differs from the truth by about $8 \cdot 10^{-22}$.<|endoftext|> TITLE: What happens to eigenvalues when edges are removed? QUESTION [9 upvotes]: I am stuck at the following : Let $G$ be a graph and $A$ is its adjacency matrix. Let the eigenvalues of $A$ be $\lambda_1\le \lambda_2\leq \cdots \leq \lambda_n$. If we remove some edges from the graph $G$ and form the graph $H$ keeping the number of vertices same, is there any result how the smallest eigenvalue of $H$ is related to the smallest eigenvalue of $G$? I know Cauchy Interlacing Theorem which gives the relation between eigenvalues of a graph and its induced subgraph when some vertices are removed. I want to know what happens when edges are removed keeping the number of vertices same. Can someone help please? The question stands as: Let $G$ be a graph and $A_G$ is its adjacency matrix. Let the eigenvalues of $A_G$ be $\lambda_1\le \lambda_2\leq \cdots \leq \lambda_n$. Let $H$ be a subgraph of $G$ which has $n$ vertices as $G$ but some edges have been removed from $G$ to form $H$.$A_H$ is its adjacency matrix. Let the eigenvalues of $A_H$ be $\mu_1\le \mu_2\leq \cdots \leq \mu_n$. Is $\mu_1\ge \lambda_1$ or $\mu_1\le \lambda_1$? If someone can give any reference like book or paper , I will be grateful. REPLY [15 votes]: The smallest eigenvalue can go up or down when an edge is removed. For "down": $G=K_n$ for $n\ge 3$. For "up": Take $K_n$ for $n\ge 1$ and append a new vertex attached to a single vertex of the original $n$ vertices. Now removing the new edge makes the smallest eigenvalue go up. Both of these follow from the fact that the smallest eigenvalue of a connected graph with $n\ge 2$ vertices is $\le -1$ with equality iff the graph is complete. It has something to do with whether the two corresponding eigenvector entries have the same or opposite sign, but I don't know if that relationship can be made precise. REPLY [3 votes]: Here's a statement from the book "Spectra of Graphs" by Brouwer and Haemers concerning the largest eigenvalue of the adjacency matrix. It implies that $\lambda_n \geq \mu_n$. Proposition 3.1.1 The value of the largest eigenvalue of a graph does not increase when vertices or edges are removed.<|endoftext|> TITLE: Presentation of $H^2(\overline{M}_{0,n},\mathbb{Z})$ as an $S_n$-module? QUESTION [6 upvotes]: Let $\overline{M}_{0,n}$ be the moduli space of genus zero curves with $n$ marked points. Let $I=\{\{S,S^c\}|S\subset\{1,\dots,n\},|S|\geq2, |S^c|\geq2\}$ be the set of partitions of $\{1,\dots n\}$ into two subsets, each has at least two elements. Keel shown that the cohomology group $H^2(\overline{M}_{0,n},\mathbb{Z})$ is generated by boundary divisor classes $\delta_{\{S,S^c\}}$, where $\{S,S^c\}\in I$. (Here $\delta_S$ is the divisor in $\overline{M}_{0,n}$, consisting of curves with two components, marked by $S$ and $S^c$, and their further degenerations.) Thus we have a presentation $$\mathrm0\to K\to\bigoplus_{\{S,S^c\}\in I}\mathbb{Z}\cdot\delta_{\{S,S^c\}}\to H^2(\overline{M}_{0,n},\mathbb{Z})\to 0.$$ Keel shown that the kernel $K$ is generated by equations $$\sum_{i,j\in S;k,l\notin S}\delta_{\{S,S^c\}}=\sum_{i,k\in S;j,l\notin S}\delta_{\{S,S^c\}}=\sum_{i,l\in S;k,j\notin S}\delta_{\{S,S^c\}},$$ for any four distinct elements $\{i,j,k,l\}\subset\{1,\dots,n\}$. These give $2\cdot{n\choose 4}$ such equations. But the rank of $K$ is only $\frac{n(n-3)}{2}$. ($\#I=2^{n-1}-n-1$, $\mathrm{rank}H^2(\overline{M}_{o,n},\mathbb{Z})=2^{n-1}-{n\choose2}-1$), so these relations are very much dependent.. The question is, would there be a good presentation of $K$, my goal is to calculate the group cohomology $$H^1(S_n,H^2(\overline{M}_{0,n},\mathbb{Z}))?$$ REPLY [2 votes]: This is a bit too long for a comment. If you take the relation $$ \sum_{i,j\in S;k,l\notin S}\delta_{\{S,S^c\}}=\sum_{i,k\in S;j,l\notin S}\delta_{\{S,S^c\}} $$ and add to both sides $\sum\limits_{i,j,k\in S;l\notin S}\delta_{\{S,S^c\}}$, you get $$ \sum_{i,j\in S;l\notin S}\delta_{\{S,S^c\}}=\sum_{i,k\in S;l\notin S}\delta_{\{S,S^c\}}. $$ Of course, this is totally reversible, so you can take these elements instead. (This essentially recovers the presentation of cohomology of $\overline{M}_{0,n}$ when this space is constructed as a De Concini-Procesi wonderful model of the Coxeter hyperplane arrangement of type A.) Now we note that for fixed $i,l$, it is enough to choose $n-3$ equations among the $(n-2)(n-3)/2$ equations indexed by different choices of $j,k$ to force all of them to hold, and the symmetric group action on those seems rather straightforward; I suspect that this would be quite useful for your purposes. [Things would be even simpler if you could fix $l$ once and forever (that is, restrict to $S_{n-1}$ inside $S_n$), but I suspect you do not want to do that.]<|endoftext|> TITLE: An analogue of the exponential function by replacing infinite series with improper integral QUESTION [16 upvotes]: For every positive real number $x$ we define $$E(x)= \int_0^{\infty} x^t/t!\,\mathrm dt$$ where $t!=\Gamma(t+1)$. This is motivated by classical exponential function. Is this function well defined (the problem of convergence)? Is there a real analytic extention of $E$ to all real numbers? What about a holomorphic extention to complex numbers? How can we compare $E(x+y)$ with $E(x)$ and $E(y)$? What kind of differential equation can be satisfied by $E$? Is $E$ one to one on real numbers?What can be said about its possible inverse? REPLY [31 votes]: This is particular case of a classic integral studied by Ramanujan. See Chapter 11 in Hardy's book, "Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work", where it is shown that $$ \int_{-\xi}^\infty\frac{x^t}{\Gamma(1+t)}\,dt+\int_0^\infty t^{\xi-1}e^{-xt}\left(\cos\pi \xi-\frac{\sin\pi \xi}{\pi}\ln t\right)\frac{dt}{\pi^2+\ln^2t}=e^x, \quad (x\ge 0, \xi\ge 0). $$ From this it follows that your integral can be represented as an integral of elementary functions as follows $$ \int_0^\infty\frac{x^t}{\Gamma(1+t)}\,dt=e^x-\int_0^\infty \frac{e^{-xt}\,dt}{t(\pi^2+\ln^2t)},\quad (x\ge 0). $$ Also, see Fransen-Robinson constant $$ C=\int_0^\infty\frac{dt}{\Gamma(t)}= 2.8077702420285... $$ REPLY [19 votes]: (Some obvious properties of $E$; too long for a comment, though). The holomorphic extension of $E$ to $\mathbb{C} \setminus (-\infty, 0]$ (in fact, to the entire Riemann surface of the complex logarithm) is given by $$E(x) = \int_0^\infty \frac{\exp(t \log x)}{\Gamma(t+1)}\, dt,$$ where $\log$ denotes the principal branch of the complex logarithm. This follows from a standard application of Morera's theorem, involving Fubini's theorem, the estimate $$|\exp(t \log x)| = \exp(t \log |x|) = |x|^t \le a^t + b^t$$ when $a \le |x| \le b$, and integrability of $(a^t + b^t) / \Gamma(t + 1)$ over $(0, \infty)$. In particular, for $x > 0$ we have $\log(-x + 0 i) = \log x + i \pi$, and hence $$ E(-x+0i) = \int_0^\infty e^{i \pi t} \frac{x^t}{\Gamma(t+1)} \, dt $$ is not real-valued in any neighbourhood of $0$. Thus, there is no real-analytic extension of $E$ to $(-\epsilon, \infty)$ (as already follows from Carlo Beenakker's comment). By dominated convergence theorem, the integral can be differentiated under the integral sign, so $$E^{(n)}(x) = \int_0^\infty \frac{x^{t - n}}{\Gamma(t+1 - n)}\, dt = \int_{-n}^\infty \frac{x^t}{\Gamma(t+1)}\, dt .$$ This does not seem to lead to any interesting differential equation. Since $E'(x) > 0$, clearly $E$ is increasing on $(0, \infty)$, with $E(0) = 0$ and $E(\infty) = \infty$. One can easily find the Laplace transform of $E(x)$: when $\operatorname{Re} \xi > 1$, we have $$ \int_0^\infty e^{-\xi x} E(x) dx = \int_0^\infty \frac{1}{\xi^{t + 1}} \, dt = \frac{1}{\xi \log \xi} . $$ (EDIT: This was meant to be an extended comment only, but since it has received a number of upvotes, let me add further remarks, inspired by Nemo's answer.) The Laplace transform $\mathcal{L} E$ of $E$ has a simple pole at $\xi = 1$ with residue $1$, and a branch cut along $(-\infty, 0]$. Since it decays (barely) sufficiently fast at infinity, one can (carefully) write the usual inversion formula and then deform the contour of integration to the Hankel contour to find that $$ E(x) = e^x - \frac{1}{\pi} \int_0^\infty e^{-t x} \operatorname{Im} (\mathcal{L} E(-t + 0i)) dt .$$ This leads to the formula given in Nemo's answer: since $$\mathcal{L} E(-t + 0i) = -\frac{1}{t \log(-t + 0 i)} = -\frac{1}{t (\log t + i \pi)} \, ,$$ we obtain $$ E(x) = e^x - \int_0^\infty \frac{e^{-t x}}{t (\pi^2 + \log^2 t)} \, dt .$$ As a consequence, $e^x - E(x)$ is completely monotone, and $$ E(x) = e^x - \frac{1 + o(1)}{\log x} $$ as $x \to \infty$. Further terms can be obtained in a similar way. The function $E(x)$ itself is the Mellin transform of $1 / \Gamma(t + 1)$. Thus, $1 / \Gamma(t + 1)$ can be written as the inverse Mellin transform: $$ \frac{1}{\Gamma(t + 1)} = \frac{1}{2 \pi i} \int_{c + i \mathbb{R}} t^{-1 - x} E(x) dx , $$ or, equivalently, $$ \frac{1}{\Gamma(t)} = \frac{1}{2 \pi i} \int_{c + i \mathbb{R}} t^{-x} E(x) dx . $$ The definition of $E(x)$ looks a little bit like Mellin–Barnes integral, but the contour is wrong. Finally, the (fractional) integral of $E(x)$ of order $\alpha$ is given by $$ I_\alpha E(x) = \frac{1}{\Gamma(\alpha)} \int_0^x E(t) (x - t)^{\alpha - 1} dt = \int_0^\infty \frac{x^{t + \alpha}}{\Gamma(t + 1 + \alpha)} \, dt , $$ and so $$ I_\alpha E(x) = \int_\alpha^\infty \frac{x^t}{\Gamma(t + 1)} \, dt . $$ This agrees with the expression for the derivatives of $E$ (which correspond to negative integer $\alpha$).<|endoftext|> TITLE: Conjectures and open problems in representation theory QUESTION [6 upvotes]: Are there very famous open problems or conjectures in representation theory, or in enumerative geometry, like the volume conjecture in topology? REPLY [6 votes]: The Clemens conjecture in enumerative geometry: a general quintic threefold has only finitely many rational curves in each positive degree. REPLY [6 votes]: There are many open, and seemingly deep, conjectures in modular representation theory (or block theory) in connection with enumerating representation-theoretic invariants: a start of a list might be : Brauer's $k(B)$-problem, the Alperin-McKay Conjecture, the Alperin Weight Conjecture, Dade's conjectures, Isaacs-Navarro conjecture. Gabriel Navarro has several recent survey papers discussing these and other conjectures. In a different part of (modular) representation theory, with perhaps a more geometric flavor, there are problems such as the Lusztig Conjecture (now known to be false in its original formulation), and work of Geordie Williamson. As noted by Julian Kuelshammer, Representation Theory is a vast subject, and it might be helpful to point out which specific areas you are most interested in (I only mention two facets of the subject which are most familiar to me).<|endoftext|> TITLE: Looking for a "cute" justification for a Catalan-type generating function QUESTION [8 upvotes]: The Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$ have the generating function $$c(x)=\frac{1-\sqrt{1-4x}}{2x}.$$ Let $a\in\mathbb{R}^+$. It seems that the following holds true $$\frac{c(x)^a}{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{a+2n}nx^n.$$ QUESTION. Why? REPLY [3 votes]: Let $C_a(x)=\frac{c(x)^a}{\sqrt{1-4x}}$ and $B_a(x) =\sum_{n=0}^{\infty}\binom{a+2n}nx^n.$ The identity $c(x)=1+xc(x)^2$ implies $C_{a+1}(x)= C_{a}(x)+x C_{a+2}(x).$ The recursion for the binomial coefficients implies $B_{a+1}(x)= B_{a}(x)+x B_{a+2}(x)$. If we show that $B_a(x)=C_a(x)$ holds for $a=1$ then it holds for all positive integers. This follows from $B_1(x)=\frac{1}{2} \sum_{n=0}^{\infty}\binom{2+2n}{n+1}x^n= \frac{1}{2x}(B_0(x)-1)=C_1(x).$<|endoftext|> TITLE: Algorithm for computing external angles for the Mandelbrot set QUESTION [12 upvotes]: Let $M$ be the Mandelbrot set: there exists a unique series $$ \psi(z) := z + \sum_{m=0}^{+\infty} b_m z^{-m} = z - \frac{1}{2} + \frac{1}{8} z^{-1} - \frac{1}{4} z^{-2} + \cdots $$ which defines a conformal bijection between the complement $\{z\in\mathbb{C} : |z|>1\}$ of the closed unit disk in $\mathbb{C}$ and the complement of the Mandelbrot set. This function $\psi$ is sometimes known as the “Jungreis function”, see the answers to this question for more. The argument $\arg(\psi^{-1}(w))$ for a point $w\not\in M$ is called the external angle of $w$. There exists an easy way to “almost” compute $\arg(\psi^{-1}(w))$, or $\psi^{-1}(w)$ itself for that matter: indeed, if we let $p_0(w) := w$ and $p_{i+1}(w) := p_i(w)^2 + w$, then $$ p_n(\psi(z)) = z^{2^n} + o(1) $$ as $z\to\infty$, so $\psi^{-1}(w)$ can be “almost” computed as the limit of the $(2^n)$-th root of $p_n(w)$ as $n\to+\infty$. The reason for the “almost” is that, while this indeed allows for computation of the modulus $|\psi^{-1}(w)|$, it leads to an indetermination between $2^n$ values on the argument. The formula $$ \psi^{-1}(w) = w \mskip3mu \prod_{n=1}^{+\infty} \left(1 + \frac{w}{p_{n-1}(w)^2}\right)^{1/2^n} $$ is no better (it also requires computing $(2^n)$-th roots and one cannot simply take the principal determination; my understanding is that one needs to find a determination of $\Big(1 + \frac{w}{p_{n-1}(w)^2}\Big)^{1/2^n}$ that is continuous outside of $M$, which seems computationally intractable). So, is there a way to lift this square root indeterminacy and compute external arguments for arbitrary $w\not\in M$? Is there an algorithm that does this in a reasonably efficient way (which excludes, e.g., trying to trace external rays outwards towards infinity)? I was unable to find anything relevant in the literature. There is a 1986 paper by Douady titled “Algorithms for computing angles in the Mandelbrot set” which seems promising, but it seems to concerns the computation for points of $M$, not points outside $M$. This web page about the Mandel program actually discusses the issue (in the section called “Computation of the external argument”), but the description is vague (e.g., where it speaks of a “modified” function $\arg(z/(z-c))$), and the conclusion that “the discontinuities are moved closer to the Mandelbrot set” is not too promising. REPLY [5 votes]: I have implemented some algorithms based on Kawahira's paper, which as presented goes $\theta \to c \not\in M$, but can be adapted to go $c \to \theta$. $\theta$ is conveniently expressed in turns as a binary expansion. When tracing inwards, one peels off the most-significant bit (aka angle doubling) each time the ray crosses a dwell band (integer part of normalized iteration count increases by 1). The trick when tracing outwards is to prepend bits when crossing dwell bands, depending if the outer cell was entered from its left or right inner cell. A picture may make it clearer: The exterior grid is generated from the fractional part of the smoothed iteration count, and the argument of the final (first to escape) $z$ iterate. One can see that approaching the $\frac{1}{2}$ bond point, the cells alternate left/right corresponding to the binary expansion $.(01)$ or $.(10)$. The argument of the first iterate to escape (typically floating point), within the cell of the starting point, can be used to get a few more least significant bits for the accumulated angle, but the prefix is found by accumulating bits one by one when tracing the ray outwards. Source code: m_d_exray_in.c $\theta \to c$, machine double precision m_r_exray_in.c $\theta \to c$ in arbitrary (dynamically changed as necessary) precision m_d_exray_out.c $c \to \theta$, machine double precision m_r_exray_out.c $c \to \theta$, arbitrary (but fixed, dynamic is possible but still TODO) precision m-exray-in.c command line driver program showing usage of the library functions m-exray-out.c command line driver program showing usage of the library functions git clone https://code.mathr.co.uk/mandelbrot-numerics.git However, tracing external rays to/from dwell $n$ takes $O(n^2)$ time (even ignoring that higher $n$ needs higher precision which costs more), which may make it too slow in practice. I am also looking for faster methods since some years, but I haven't found any yet. See my related question: fast algorithms for external angle computations<|endoftext|> TITLE: The "contrary" of an isomorphism QUESTION [6 upvotes]: Roughly, my question is: is there a standard name for functions which one might characterise as the "contrary" of an isomorphism? Here is a more precise version of my question. Working model-theoretically, consider the following definition. Definition. Let $\mathscr{L}$ be a relational signature, and let $\mathcal{A}, \mathcal{B}$ be $\mathscr{L}$-structures. An $\mathscr{L}$-anti-embedding $\pi : A \longrightarrow B$ is any injection such that: $(a_1, \ldots, a_n) \in R^\mathcal{A}$ iff $(\pi(a_1), \ldots, \pi(a_n)) \notin R^\mathcal{B}$, for each $n$-place predicate $R \in \mathscr{L}$ and all $a_1, \ldots, a_n \in A$ A bijective $\mathscr{L}$-anti-embedding is an $\mathscr{L}$-anti-isomorphism. An $\mathscr{L}$-anti-isomorphism from a structure to itself is an $\mathscr{L}$-anti-automorphism. My question is: are there standard names for the kinds of functions I just defined? Indeed, are they discussed anywhere? I have only encountered one such function "in the wild". I was reading Thomas Forster on Church-Oswald set theory; he called his $\{\in\}$-anti-automorphism an "antimorphism". (NB that, in the Church-Oswald setting, self-membered sets are perfectly ok.) Update: Googling the phrase "anti-isomorphism" teaches me that, in group theory, an anti-isomorphism is standardly defined as a bijection $\pi : \mathcal{G} \longrightarrow \mathcal{H}$ such that $\pi(x \cdot^\mathcal{G} y) = \pi(y) \cdot^\mathcal{H} \pi(x)$. That's obviously a totally different idea. So I should use a different name! If there is no standard name, I welcome suggestions! REPLY [4 votes]: Tim has drawn my attention to this (Thank you, Tim!) Perhaps i should provide what Old Lags do in this kind of setting, namely provide some ancient history. The context is Quine's NF. Let us say the dual of a formula of the language of set theory is the result of replacing all occurrences of $\in$ by `$\not\in$'. It's routine to show that the dual of any axiom of NF is a theorem of NF. Thus if we take $<$M,$\in>$ a model of NF and consider M equipped with the complement (in MxM) of the membership relation we get another model of NF! Are these two structures elementarily equivalent? Not reliably! Might they be isomorphic? If they are, then the isomorphism is an antimorphism. Of course - since this is NF - there is the possibility that (not only might there be an antimorphism but) the antimorphism is a set of the model! It's unknown if this can happen. The best i can do is show that every model of NF is elementarily equivalent (with respect to stratifiable formul{\ae}) to one with two (set) permutations $\sigma$ and $\tau$ satisfying both $$(\forall x y)(x \in y \leftrightarrow \sigma(x) \not\in \tau(y))$$ and $$(\forall x y)(x \in y \leftrightarrow \tau(x) \not\in \sigma(y))$$ Arranging for $\sigma = \tau$ is beyond me at the moment. The word `antimorphism' is a coinage of your humble correspondent, tho' i can't give you chapter and verse. Sorry if some readers find this perhaps off-piste, but this is the context of Tim's question, and might be helpful. Thomas Forster www.dpmms.cam.ac.uk/~tf<|endoftext|> TITLE: Index of Dirac operator and Chern character of symmetric product twisting bundle QUESTION [10 upvotes]: I am having trouble understanding a couple of lines of computation from Theorem 13.30 in Besse's Einstein Manifolds text We are twisting the spinor bundle (on Einstein 4-manifold) $\Sigma$ with an auxiliary bundle $S^3\Sigma^-$. We form the Dirac operator $\mathscr{D}^D$ formed by twisting the Levi-Civita connection on $\Sigma$ with a copy $D$ acting on $S^3\Sigma^-$ (and composing with the Clifford action acting trivially on $S^3\Sigma$). Besse evaluates the index of this operator using the APS theorem as an integral over Chern and Pontrjagin classes. I understand that the $(1-\frac{1}{24}p_1)$ terms comes from the $\widehat{A}$-genus of the manifold, and further the signature theorem, $\tau=\frac13 \int_M p_1(M) $. However, I do not understand the evaluation of the Chern character of the twisting bundle as $(4-10c_2)$ and the subsequent evaluation in terms of Euler characteristic and signature. Any insight would be greatly appreciated. Edit: I am still not sure about the final calculation relating the index to the Euler character and signature, however, working backwards it seems we require $c_2(\Sigma^-)=\frac12e(M)-\frac14p_1(M)$ (or perhaps something a bit different if there are lower degree terms which could multiply with with the $\frac{1}{24}p_1$ term from the $\widehat{A}-$genus), where $e(M)$ is the Euler class of $M$. Edit#2: Solved by Michael Albanese. REPLY [9 votes]: Your first question can be answered by using the splitting principle. If $V \to X$ is a complex vector bundle of rank two, then $c_1(S^3V) = 6c_1(V)$ and $c_2(S^3V) = 11c_1(V)^2 + 10c_2(V)$. Proof: By the splitting principle, there is a map $p : Y \to X$ such that $p^*$ is injective on integral cohomology and $p^*V \cong L_1\oplus L_2$, so $p^*(S^3V) \cong S^3(p^*V) \cong S^3(L_1\oplus L_2)$. In general, we have $S^n(E_1\oplus E_2) \cong \bigoplus_{i+j=n} S^i(E_1)\otimes S^j(E_2)$, so \begin{align*} &\, S^3(L_1\oplus L_2)\\ \cong&\, S^3(L_1)\otimes S^0(L_2)\oplus S^2(L_1)\otimes S^1(L_2)\oplus S^1(L_1)\otimes S^2(L_2)\oplus S^0(L_1)\otimes S^3(L_2)\\ \cong&\, L_1^3\oplus L_1^2\otimes L_2\oplus L_1\otimes L_2^2\oplus L_2^3. \end{align*} It follows that $c_1(S^3(L_1\oplus L_2)) = 6c_1(L_1) + 6c_1(L_2) = 6c_1(L_1\oplus L_2)$. So $$p^*c_1(S^3V) = c_1(p^*S^3V) = c_1(S^3(L_1\oplus L_2)) = 6c_1(L_1\oplus L_2) = 6c_1(p^*V) = p^*(6c_1(V)).$$ By the injectivity of $p^*$, we have $c_1(S^3V) = 6c_1(V)$. Similarly, one can compute that \begin{align*} c_2(S^3(L_1\oplus L_2)) &= 11c_1(L_1)^2 + 11c_1(L_2)^2 + 32c_1(L_1)c_1(L_2)\\ &= 11(c_1(L_1)+c_1(L_2))^2 + 10c_1(L_1)c_1(L_2)\\ &= 11c_1(L_1\oplus L_2)^2 + 10c_2(L_1\oplus L_2) \end{align*} and hence $c_2(S^3V) = 11c_1(V)^2 + 10c_2(V)$. $\quad\square$ In this case, $\Sigma^-$ is an $SU(2)$ bundle and hence $c_1(\Sigma^-) = 0$. So we see that $c_1(S^3\Sigma^-) = 0$ and $c_2(S^3\Sigma^-) = 10c_2(\Sigma^-)$. Therefore $$\operatorname{ch}(S^3\Sigma^-) = \operatorname{rank}(S^3\Sigma^-) + c_1(S^3\Sigma^-) + \frac{1}{2}(c_1(S^3\Sigma^-)^2 - 2c_2(S^3\Sigma^-)) = 4 - 10c_2(\Sigma^-).$$ To find $c_2(\Sigma^-)$, or $c_2(\Sigma^+)$, we can proceed as follows. As $\Sigma^{\pm}$ is an $SU(2)$-bundle which is a lift of the $SO(3)$-bundle $\Lambda^{\pm}$, there is a relationship between $c_2(\Sigma^{\pm})$ and $p_1(\Lambda^{\pm})$, namely $p_1(\Lambda^{\pm}) = -4c_2(\Sigma^{\pm})$; see Appendix E of Instantons and Four-Manifolds by Freed and Uhlenbeck, also this question. So now we just need to know $p_1(\Lambda^{\pm})$, but this is given by $\pm 2e(M) + p_1(M)$; see Chapter $6$, Proposition $5.4$ of Metric Structures in Differential Geometry by Walschap. Therefore $$c_2(\Sigma^{\pm}) = \mp\frac{1}{2}e(M) - \frac{1}{4}p_1(M).$$ Note, as $M$ is assumed to be spin, its signature is a multiple of $16$ by Rohklin's Theorem, so $\frac{1}{4}p_1(M)$ is an integral class. As the signature of $M$ is even, so is the Euler characteristic, and hence $\frac{1}{2}e(M)$ is also an integral class. Finally, as $c_2(\Sigma^-) = \frac{1}{2}e(M) - \frac{1}{4}p_1(M)$, we see that \begin{align*} \int_M(10c_2(\Sigma^-) - 4)\left(1 - \frac{1}{24}p_1(M)\right) &= \int_M 10c_2(\Sigma^-) + \frac{1}{6}p_1(M)\\ &= \int_M 5e(M) - \frac{5}{2}p_1(M) + \frac{1}{6}p_1(M)\\ &= \int_M 5e(M) - \frac{7}{3}p_1(M)\\ &= 5\chi(M) - 7\tau(M) \end{align*} as claimed.<|endoftext|> TITLE: Existence of a multiplication bifunctor for the category of groups QUESTION [5 upvotes]: For $\mathsf{Grp}$ the category of groups, a bifunctor $M: \mathsf{Grp} \times \mathsf{Grp}\to \mathsf{Grp}$ is a multiplication bifunctor if: $M(C_n,C_m) \simeq C_{nm}$, $M(C_1,G) \simeq M(G,C_1) \simeq G$, for every group $G$ and every $n,m>0$, with $C_n$ the cyclic group of $n$ elements. Question: Is there a multiplication bifunctor for the category of groups? (or for the subcategory of countable groups, or of finite groups) Stronger question: Is there a multiplication bifunctor providing a monoidal structure? This post is a multiplicative analogous of that additive one. REPLY [8 votes]: No. $C_1$ is a retract of $C_2$, so $M(C_2,C_1)\simeq C_2$ would have to be a retract of $M(C_2,C_2)\simeq C_4$, which it isn't.<|endoftext|> TITLE: On the relation between categorification and chromatic redshift QUESTION [11 upvotes]: In the introduction to the paper Higher traces, noncommutative motives, and the categorified Chern character, Hoyois, Scherotzke and Sibilla write the following. An important insight emerging from topology is that climbing up the chromatic ladder is related to studying invariants of spaces that are higher-categorical in nature. The only example that I am aware of is the theory of categorified vector bundles, and its relation with chromatic-level-$2$ cohomology theories, including elliptic cohomology, and the algebraic K-theory spectrum of complex K-theory. See for instance this nLab page and the references therein. Question. Are there other examples of this phenomenon, and is there a conceptual reason why categorification should be related to chromatic redshifting? REPLY [6 votes]: The paper https://arxiv.org/abs/1312.5699 by Torlief Veen involves arbitrarily large chromatic redshifts via higher topological Hochschild homology. I know of no other examples like that.<|endoftext|> TITLE: How are number theory and C*-algebras connected? QUESTION [9 upvotes]: I came across this research profile where under Research Overview, it states that These days C*-algebra theory is a very active area of mathematical research in its own right, and enjoys deep connections with other areas of mathematics such as symbolic dynamics, ergodic theory, group theory and even number theory. My question is how C*-algebras and number theory are connected and if this is an active area of research? REPLY [6 votes]: I think the most prominent example of this is Connes' work on the Riemann hypothesis from a C*-algebra perspective.<|endoftext|> TITLE: Horizontal categorification: Two questions QUESTION [9 upvotes]: According to the nlab, horizontal categorification is a process in which a concept is realized to be equivalent to a certain type of category with a single object, and then this concept is generalized to the same type of categories with an arbitrary number of objects. The prototypical example is the concept of a group, which horizontally categorifies to the concept of a groupoid. It is well-known that in certain parts of algebraic topology groupoids are much more convenient than groups. Similarly, monoids categorify to categories, rings to linear categories, etc. I have two questions about this. Let C be a concept which horizontally categorifies to a concept D. 1) In many examples, C and D have been known and developed independently from each other. Or at least, D was not introduced as the horizontal categorification of C, but rather this connection was realized afterwards. For example, I am pretty sure that k-linear categories were not introduced as a horizontal categorificiation of k-algebras; instead they were introduced because of the abundance of examples of (large) k-linear categories which appear in everyday mathematics. Although representation theory seems to be in a current progress of a generalization from k-algebras to small k-linear categories, the concept of a k-linear category was already there before that. Are there examples where D was developed with the purpose to categorify C, say in order to solve some problems which deal with C but which are not solvable with C alone? Perhaps C*-categories (categorifying C*-algebras) could provide such an example, but I am not familiar with the history of this concept. And maybe there are other examples as well? 2) In all the examples I know of, it is trivial that C has a horizontal categorification and that it is D. Are there any more interesting examples where, when you look at C, it is not even clear how to interpret C as a type of category with one object? I would like to see examples where the connection between C and D is deep and surprising. These examples should illustrate why horizontal categorification is an important and useful concept in practice. I could also ask a more provocative question: if any mathematician working outside of category theory reads the nlab article in its current form, why should he/she even care, since, after all, both concepts C and D have been there already without the categorification process? REPLY [2 votes]: Often when a horizontal categorification is introduced, the author has intended examples, which the author gives as the motivation. It thus makes it a little tricky to determine when a structure has been motivated explicitly by horizontal categorification. However, here are a few examples where (1) the concept had not previously appeared in the literature, and (2) the authors make it clear horizontal categorification was intended (though this precise language may not appear). In Street's Elementary cosmoi I, the concept of extension system is introduced, which is explicitly intended to be to closed categories what bicategories are to monoidal categories. Intuitively, the analogue of closed structure is proved by right-extensions. Two more examples are the multi- and poly-bicategories of Cockett–Koslowski–Seely's Morphisms and modules for poly-bicategories, which are horizontal categorifications of multicategories and polycategories respectively. My favourite example of a surprising horizontal categorification is that of symmetric monoidal categories. A symmetric bicategory (not to be confused with a symmetric monoidal bicategory) is a bicategory with an involution. Explicitly, a bicategory $\mathcal K$ is symmetric when equipped with a biequivalence $\mathcal K \simeq \mathcal K^{\mathrm{op}}$ that is a bijection on objects. A symmetric monoidal category is precisely a one-object symmetric bicategory for which the involution is the identity. This is interesting in two ways: the generalisation from symmetry to involution is a highly non-obvious step; and one is forced to make a generalisation to the original structure in order to horizontally categorify. There is an analogous concept of closed symmetric bicategory, which is a symmetric bicategory with left- and right-extensions. (Closed) symmetric bicategories are defined and studied in May–Sigurdsson's Parametrized Homotopy Theory. I think these examples give some indication that studying horizontal categorifications in their own right can lead to interesting structures that may then be motivated by concrete examples.<|endoftext|> TITLE: Are all maps $\mathbb{R}^2 \to \mathbb{R}^2$ with fixed singular values affine? QUESTION [12 upvotes]: Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a smooth map whose differential has fixed distinct singular values $0<\sigma_1<\sigma_2$ and an everywhere positive determinant (which is the product $\sigma_1\sigma_2$). Must $f$ be affine? My assumption is equivalent to $df_x \in \text{SO}(2) \cdot \text{diag}(\sigma_1,\sigma_2) \cdot \text{SO}(2)$ for every $x \in \mathbb{R}^2$. If we were only allowing a copy of $\text{SO}(2)$ from one of the sides of $ \text{diag}(\sigma_1,\sigma_2)$, then the answer would be positive. (This reduces to the case of isometries). Similarly, if we had $\sigma_1=\sigma_2$, the answer would also be positive. REPLY [19 votes]: Answer modified on 1 February 2020: It's not true 'locally' in the sense that non-affine $f$'s satisfying this system of PDE can be constructed on some open sets in $\mathbb{R}^2$. This first order, determined PDE system is hyperbolic, so there are many local solutions. However, it turns out (see below) that all $C^3$ solutions with domain equal to $\mathbb{R}^2$ are affine. (The proof I give below does not work for solutions of lower regularity.) Let $D\subset\mathbb{R}^2$ be a $1$-connected open domain on which there exists a $C^3$ mapping $f:D\to\mathbb{R}^2$ whose differential $\mathrm{d}f$ has constant, distinct singular values $0<\sigma_1<\sigma_2$. Because $D$ is simply connected, one can choose an orthonormal frame field $E_1,E_2$ on $D$ such that, at each point $p\in D$, the image vectors $F_i(p) = \mathrm{d}f\bigl(E_i(p)\bigr)$ are orthogonal and satisfy $|F_i(p)|=\sigma_i$. Let $\omega = (\omega_1,\omega_2)$ be the dual coframing on $D$, which is of regularity type $C^2$. The $1$-forms $\eta_i = \sigma_i\,\omega_i$ for $i=1,2$ have the property that $(\eta_1)^2+(\eta_2)^2$, being the $f$-pullback of the flat metric on $\mathbb{R}^2$, must also be a flat metric. Let $\omega_{12}$ be the connection $1$-form associated to the coframing $\omega$, i.e., it satisfies the structure equations $$ \mathrm{d}\omega_1 = -\omega_{12}\wedge\omega_2 \qquad\text{and}\qquad \mathrm{d}\omega_2 = \omega_{12}\wedge\omega_1\,.\tag1 $$ Write $\omega_{12} = -\kappa_1\,\omega_1 + \kappa_2\,\omega_2$. The function $\kappa_i$ is the curvature of the $E_i$-integral curve. Since $\omega_{12}$ is $C^1$, so are the functions $\kappa_i$. A straightforward computation shows that the $1$-form $\eta_{12}$ that satisfies the corresponding structure equations $$ \mathrm{d}\eta_1 = -\eta_{12}\wedge\eta_2 \qquad\text{and}\qquad \mathrm{d}\eta_2 = \eta_{12}\wedge\eta_1\,.\tag2 $$ is given by $$ \eta_{12} = -(\sigma_1/\sigma_2)\,\kappa_1\omega_1 + (\sigma_2/\sigma_1)\,\kappa_2\omega_2\,. $$ Since $\sigma_1\not=\sigma_2$, the conditions $\mathrm{d}\omega_{12} = \mathrm{d}\eta_{12}=0$ (which hold because the domain metric and the $f$-pullback of the range metric are both flat) are equivalent to $$ 0 = \mathrm{d}(\kappa_i\,\omega_i) = \bigl(\mathrm{d}\kappa_i - {\kappa_i}^2\,\omega_{3-i}\bigr)\wedge\omega_i\,\qquad i = 1,2.\tag3 $$ Proposition: If $D = \mathbb{R}^2$, then $\kappa_1 \equiv \kappa_2 \equiv 0$, and $f$ is an affine map. Proof: Suppose that, say, $\kappa_1$ be nonzero at some point $p\in\mathbb{R}^2$ and consider the value of $\kappa_1$ along the $E_2$ integral curve through $p$, which, since $E_2$ has unit length, is necessarily complete. Let $p(s)$ be the flow of $E_2$ by time $s$ starting at $p = p(0)$. Then (3) implies that the function $\lambda(s) = \kappa_1\bigl(p(s)\bigr)$ satisfies $\lambda'(s) = \lambda(s)^2$. Consequently, $$ \kappa_1\bigl(p(s)\bigr) = \frac{\kappa_1\bigl(p(0)\bigr)}{1-\kappa_1\bigl(p(0)\bigr)s}. $$ Hence $\kappa_1$ cannot be continuous along this integral curve, which is a contradiction. Thus, $\kappa_1$ and, similarly, $\kappa_2$ must vanish identically when $D = \mathbb{R}^2$. In particular, $\mathrm{d}\omega_i = 0$, from which one easily concludes that $f$ is affine. QED More interesting, locally, is what happens near a point where $\kappa_1\kappa_2\not=0$. (There is a similar analysis when one of $\kappa_i$ vanishes identically that can safely be left to the reader, but see the note at the end.) One might as well assume that $\kappa_1\kappa_2$ is nowhere vanishing on $D$. Then one can write $$ \kappa_1\,\omega_1 = \mathrm{d}u \qquad\text{and}\qquad \kappa_2\,\omega_2 = \mathrm{d}v $$ for some $C^2$ functions $u$ and $v$ on $D$, uniquely defined up to additive constants. Writing $\omega_1 = p\,\mathrm{d}u$ and $\omega_2 = q\,\mathrm{d}v$ for some non-vanishing functions $p$ and $q$, one finds that the structure equations (1), with $\omega_{12} = -\mathrm{d}u + \mathrm{d}v$, yield the equations $$ p_v = - q \qquad\text{and}\qquad q_u = -p. $$ In particular, note that $p_v$ is $C^1$ and $p_{uv}-p = 0$. Conversely, if $p$ be any nonvanishing $C^2$ function on a domain $D'$ in the $uv$-plane that satisfies the hyperbolic equation $p_{uv}-p=0$ and is such that $p_v$ is also nonvanishing on $D'$, then the $1$-forms $$ \omega_1 = p\,\mathrm{d}u,\quad \omega_2 = -p_v\,\mathrm{d}v,\qquad \omega_{12} = -\mathrm{d}u+\mathrm{d}v\tag4 $$ satisfy the structure equations of a flat metric, and so do $$ \eta_1 = \sigma_1\,p\,\mathrm{d}u,\quad \eta_2 = -\sigma_2\,p_v\,\mathrm{d}v,\qquad \eta_{12} = -(\sigma_1/\sigma_2)\,\mathrm{d}u+(\sigma_2/\sigma_1)\,\mathrm{d}v.\tag5 $$ Indeed, one now sees that the $1$-forms $$ \begin{aligned} \alpha_1 &= \cos(u{-}v)\,p\,\mathrm{d}u +\sin(u{-}v)\,p_v\,\mathrm{d}v\\ \alpha_2 &= \sin(u{-}v)\,p\,\mathrm{d}u -\cos(u{-}v)\,p_v\,\mathrm{d}v \end{aligned} $$ are closed, and therefore can be written in the form $\alpha_i = \mathrm{d}x_i$ for some $C^3$ functions $x_i$ on $D'$. $$ (\mathrm{d}x_1)^2 + (\mathrm{d}x_2)^2 = (\alpha_1)^2 + (\alpha_2)^2 = (\omega_1)^2 + (\omega_2)^2 $$ and, hence, they define a $C^3$ submersion $x = (x_1,x_2):D'\to\mathbb{R}^2$ that pulls back the standard flat metric on $\mathbb{R}^2$ to the metric $(\omega_1)^2 + (\omega_2)^2$ on $D'$. Similarly, setting $\rho = \sigma_2/\sigma_1$ and $$ \begin{aligned} \beta_1 &= \cos(u/\rho{-}\rho v)\,\sigma_1\,p\,\mathrm{d}u +\sin(u/\rho{-}\rho v)\,\sigma_2\,p_v\,\mathrm{d}v\\ \beta_2 &= \sin(u/\rho{-}\rho v)\,\sigma_1\,p\,\mathrm{d}u -\cos(u/\rho{-}\rho v)\,\sigma_2\,p_v\,\mathrm{d}v, \end{aligned} $$ one finds that $\mathrm{d}\beta_i = 0$ and hence there exist $C^3$ functions $y_i$ on $D'$ such that $\beta_i = \mathrm{d}y_i$. Set $y = (y_1,y_2)$. Restricting to a subdomain $D''\subset D'$ on which $x$ is $1$-to-$1$ onto its image $D = x(D'')$ yields a domain on which $x^{-1}:D\to D''$ is a $C^3$ diffeomorphism. Now set $f = y\circ x^{-1}:D\to\mathbb{R}^2$, and one has a $C^3$ solution of the original PDE system. This completely determines the structure of the 'generic' local $C^3$ solutions. The case when one of the $\kappa_i$, say, $\kappa_1$, vanishes identically (so that the corresponding integral curves are straight lines) and the other is nonvanishing can easily be reduced to the normal form $$ \omega_1 = \mathrm{d}u,\qquad \omega_2 = \bigl(p(v)-u\bigr)\,\mathrm{d}v,\qquad \omega_{12} = \mathrm{d}v\tag6 $$ where, now, $p$ is a $C^2$ function of $v$, and the rest of the analysis goes through essentially unchanged. REPLY [7 votes]: I would like to propose a simple local example: Consider the map in polar coordinates, $\mathbb C\to \mathbb C$ that takes a complex number $z=e^{2\pi i \theta}r$ to $e^{(\sigma_1/\sigma_2)\cdot 2\pi i \theta}r\sigma_2$. (apologies for the previous wrong example, I confused in it singular values with eigenvalues...)<|endoftext|> TITLE: Real rootedness of a polynomial QUESTION [21 upvotes]: Let's consider $m$ and $n$ arbitrary positive integers, with $m\leq n$, and the polynomial given by: $$ P_{m,n}(t) := \sum_{j=0}^m \binom{m}{j}\binom{n}{j} t^j$$ I've found with Sage that for every $1\leq m \leq n \leq 80$ this polynomial has the property that all of its roots are real (negative, of course). It seems these roots are not nice at all. For example for $m=3$ and $n=10$, one has $$P(t) = 120t^3 + 135 t^2 + 30t+1$$ and the roots are: $$ t_1 = -0.8387989...$$ $$ t_2 = -0.2457792...$$ $$ t_3 = -0.0404217...$$ Is it true that all roots of $P_{m,n}(t)$ are real? REPLY [24 votes]: If you have two polynomials $f(x)=a_0+a_1x+\cdots a_mx^m$ and $g(x)=b_0+b_1x+\cdots+b_nx^n$, such that the roots of $f$ are all real, and the roots of $g$ are all real and of the same sign, then the Hadamard product $$f\circ g(x)=a_0b_0+a_1b_1x+a_2b_2x^2+\cdots$$ has all roots real. This was proven originally in E. Malo, Note sur les équations algébriques dont toutes les racines sont réelles, Journal de Mathématiques Spéciales, (4), vol. 4 (1895) One can make stronger statements, such as the result by Schur that says that $\sum k!a_kb_k x^k$ will only have real roots, under the same conditions. Schur's theorem combined with the fact that $\{1/k!\}_{k\geq 0}$ is a Polya frequency sequence, implies Malo's theorem. I'm not sure what the best reference to learn the theory of real rooted polynomials, and the associated operations that preserve real rootedness is. One textbook I know that discusses some of these classical results is Marden's "Geometry of Polynomials". REPLY [16 votes]: According to the representation for Jacobi polynomials https://en.wikipedia.org/wiki/Jacobi_polynomials#Alternate_expression_for_real_argument $$ P^{(0,n-m)}_m(x)=\sum_{j=0}^m \binom{m}{j}\binom{n}{j}\left(\frac{x-1}{2}\right)^{j}\left(\frac{x+1}{2}\right)^{m-j} $$ OPs polynomial equals $$ P_{m,n}(t)=(1-t)^mP^{(0,n-m)}_m\left(\frac{1+t}{1-t}\right). $$ Since zeroes of Jacobi polynomials are real valued, all roots of the polynomial $P_{m,n}(t)$ are also real valued (see D.Dominici, S.J.Johnston, K.Jordaan, Real zeros of hypergeometric polynomials).<|endoftext|> TITLE: Theory of integration for functions from $\mathbb{Z}_{p}$ to $\mathbb{Z}_{q}$ for distinct primes $p,q$ QUESTION [6 upvotes]: Let $p$ and $q$ be prime numbers. When $p=q$, Mahler's Theorem gives a complete description of $C\left(\mathbb{Z}_{p};\mathbb{Z}_{p}\right)$, the space of continuous functions from $\mathbb{Z}_{p}$ to $\mathbb{Z}_{p}$. I'm wondering (possibly in vain) if there might be a comparable classification of $C\left(\mathbb{Z}_{p};\mathbb{Z}_{q}\right)$ when $p$ and $q$ are distinct. I ask only because I've been doing $p$-adic harmonic analysis, but have found myself having to brave the wilds of $L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)$, the space of all $f:\mathbb{Z}_{p}\rightarrow\mathbb{C}_{q}$ so that:$$\sup_{\mathfrak{z}\in\mathbb{Z}_{p}}\left|f\left(\mathfrak{z}\right)\right|_{q}<\infty$$ Pontryagin duality lets me do Fourier analysis on $L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}\right)$; for $p=q$, on the other hand, I can use things like the volkenborn integral, or the amice transform / mazur-mellin transform—$p$-adic distributions, in general. The problem is, without a structure theorem like Mahler's for the $p\neq q$ case, though I can define “integration” on $L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)$ by elements of its dual space (continuous functionals $\varphi:L^{\infty}\left(\mathbb{Z}_{p};\mathbb{C}_{q}\right)\rightarrow\mathbb{C}_{q})$, I don't see a way to do useful computations for the specific, non-abstract functions that I'm trying to fourier analyze. So, I guess what I'm really asking is: how do you take the "integral" or "fourier transform" of such a function? Any thoughts? Reference recommendations? Etc.? REPLY [4 votes]: ${\mathbb Z}_p$ and ${\mathbb Z}_q$ are homeomorphic; hence so are $C({\mathbb Z}_p,{\mathbb Z}_p)$ and $C({\mathbb Z}_p,{\mathbb Z}_q)$.<|endoftext|> TITLE: Classification of the functors on the category of cyclic groups QUESTION [8 upvotes]: Let $\mathsf{Grp}$ be the category of groups and let $\mathsf{Cyc}$ be the subcategory of cyclic groups. As seen in the posts here and there (and their answers), a functor $F: \mathsf{Cyc} \to \mathsf{Cyc}$ is a very structured/restrictive notion, we are then lead to wonder whether there exists such a functor which is non-equivalent to the identity or the trivial functor, or if there is a such functor with $F(C_1) \not \simeq C_1$. As pointed out by Martin Brandenburg and Jeremy Rickard, $C_1$ is a retract of $F(C_1)$, so that $F(C_1)$ must be a retract of $F^2(C_1)$, and more generally, $F^n(C_1)$ is a retract of $F^{n+1}(C_1)$, which means that $F^{n+1}(C_1)$ is isomorphic to a semidirect product $F^n(C_1) \ltimes N_n$; now $F^{n+1}(C_1)$ is a cyclic group, so the semidirect product is in fact a direct product and moreover $gcd(|F^n(C_1)|,|N_n|) = 1$. Question: What are the functors on the categroy of cyclic groups? Remark: $Aut(-)$ is not such a functor because $Aut(C_8) \simeq C_2 \times C_2$ (and $Aut^2(C_8) \simeq S_3$). In his answer, Neil Strickland provides examples of functors $F$ with $F(C_1) \not \simeq C_1$ and with $F^2(C_1) \not \simeq F(C_1)$, but with $F^3(C_1) \simeq F^2(C_1)$. Bonus question: Is there a functor $F: \mathsf{Cyc} \to \mathsf{Cyc}$ such that $F^{n+1}(C_1) \not \simeq F^n(C_1)$ for all $n$? Remark: If so, the sequence $(F^n(C_1))_n$ cannot be periodic (for $n$ large enough), because then (as shown above) $F^{n+1}(C_1) \simeq F^{n}(C_1) \times N_n$ with $|N_n|>1$ for all $n$. REPLY [9 votes]: I'll use additive notation, and I'll assume that you are only considering finite cyclic groups. Let $\mathbf{Cyc}_p$ be the category of cyclic $p$-groups. Given $i,j\geq 0$ we can define $Q(p;i,j)\colon\mathbf{Cyc}\to\mathbf{Cyc}_p$ by $Q(p;i,j)(A)=\{a\in p^iA\colon p^ja=0\}$. We can also define a constant functor $C(p;i)\colon\mathbf{Cyc}\to\mathbf{Cyc}_p$ by $C(p;i)(A)=\mathbb{Z}/p^i$. Now suppose we have a collection of functors $F_p$, one for each prime $p$, each of the form $Q(p;i,j)$ or $C(p;i)$, and that only finitely many of the functors $F_p$ are constant. Then the group $F(A)=\prod_pF_p(A)$ is cyclic for all $A$, so we get a functor $F\colon\mathbf{Cyc}\to\mathbf{Cyc}$. I don't know if that gives all possible functors, but it certainly gives a reasonably rich supply of them. As a very specific example, the functor $F(A)=(A/2A)\times(\mathbb{Z}/3)$ is non-constant with $F(0)\neq 0$. UPDATE: Here's a more exotic example. If $X$ is a based set of size $1$ or $3$, there is a unique group structure for which the basepoint is the identity. If $A$ is cyclic of order $1$ or $7$ then we can impose an equivalence relation with $a\sim a^2\sim a^4$ for all $a$, and then $A/\sim$ has size $1$ or $3$ with basepoint $0$ and so has a group structure. This construction gives a functor on the category of groups of order $1$ or $7$, and we can compose with $A\mapsto A/7$ to get a functor $\mathbf{Cyc}\to\mathbf{Cyc}_3$. I am sure that there are many variations on this theme.<|endoftext|> TITLE: On the iterated automorphism groups of the cyclic groups QUESTION [12 upvotes]: Let $C_n$ be the cyclic group of order $n$. Its automorphism group $Aut(C_n)$ is a group of order $\varphi(n)$ isomorphic to $(\mathbb{Z}/n\mathbb{Z})^{\times}$ the multiplicative group of integer modulo $n$. This last group is abelian but not always cyclic, the first non-cyclic being $Aut(C_8) \simeq C_2 \times C_2$. Moreover the iterated automorphism groups $Aut^m(C_n)$ are not always abelian, as $Aut^2(C_8) \simeq S_3$. By watching the table here for the group structure of $(\mathbb{Z}/n\mathbb{Z})^{\times}$, we cannot expect an easy classification for the set of groups $Aut(C_n)$, but (Q1) what about the following set? $$\{Aut^m(C_n) \ | \ n \ge 1, m \ge 0 \}$$ If it is also non-easy, (Q2) what about the set of groups $Aut^m(C_n)$ which are isomorphic to $Aut^{m+1}(C_n)$? For $n \le 15$, it is exactly $\{C_1,S_3,D_8 \}$ as shown by the following table: $$\scriptsize{ \begin{array}{c|c} G &C_1&C_2&C_3&C_4&C_5&C_6&C_7&C_8&C_9&C_{10}&C_{11}&C_{12}&C_{13}&C_{14}&C_{15} \newline \hline Aut(G) &-&C_1&C_2&C_2&C_4&C_2&C_6&C_2^2&C_6&C_4&C_{10}&C_2^2&C_{12}&C_6&C_2 \times C_4 \newline \hline Aut^2(G) &-&-&C_1&C_1&C_2&C_1&C_2&S_3&C_2&C_2&C_4&S_3&C_2^2&C_2&D_8 \newline \hline Aut^3(G) &-&-&-&-&C_1&-&C_1&-&C_1&C_1&C_2&-&S_3&C_1&- \newline \hline Aut^4(G) &- &- &-&-&-&-&-&-&-&-&C_1&-&-&-&- \end{array} }$$ Let $\alpha(n)$ be the smallest $m \ge 0$ such that $Aut^m(C_n) \simeq Aut^{m+1}(C_n)$. Then: $$\scriptsize{ \begin{array}{c|c} n &1&2&3&4&5&6&7&8&9&10&11&12&13&14&15 \newline \hline \alpha(n) &0&1&2&2&3&2&3&2&3&3&4&2&2&3&2 \end{array} }$$ Surprisingly, for $n<15$ (but not for $n=15$), $\alpha(n)$ is exactly the number of iterations of the Carmichael lambda function $\lambda$ needed to reach $1$ from $n$ (OEIS A185816). (Q3) Why? A specific OEIS sequence has just been created (A331921) providing the value of $\alpha(n)$ for $n<32$; in this case, $\alpha(n) \le 5$ and $\{\mathrm{Aut}^5(C_n) \ | \ n < 32 \} = \{C_1,S_3,D_8,D_{12},\mathrm{PGL}(2,7) \}$. A usual laptop cannot compute $\alpha(n)$ for $n \ge 32$ (you are welcome to contribute to this sequence), we just know that $\alpha(32) \ge 6$. Here is the sequence $|Aut^n(C_{32})|$ for $n \le 6$: $$\scriptsize{ \begin{array}{c|c} n &0&1&2&3&4&5&6&7&8 \newline \hline |Aut^n(C_{32})| & 2^5&2^4&2^4&2^6&2^{7}3&2^{9}3&2^{11}3&?&? \newline \hline \text{IdGroup}(Aut^n(C_{32})) & [32,1]&[16,5]&[16,11]&[64,138]&[384,17948]&[1536,?]&[6144,?]&[?,?]&[?,?] \end{array} }$$ We can also consider the sequence of $n$ such that $\exists m \ge 0$ with $Aut^m(C_n) \simeq C_1$: $1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 14, 18, 19, 22, 23, 27, 38, 46, 47, 54, 81, \dots$ (A117729). It seems to be a (strict) subsequence of A179401, so (Q4) why $Aut^m(C_n) \simeq C_1$ $\Rightarrow$ $\varphi^2(n) = \lambda^2(n)$? It is an open problem whether the sequence $Aut^n(G)$ stabilizes (see here). For a given finite group $G$ there are three possibilities: (1): $\exists m \ge 0 $ such that $Aut^{m+1}(G) \simeq Aut^{m}(G)$, (2): not (1) and $\exists m \ge 0$ such that $\exists r>0$ with $Aut^{m+r}(G) \simeq Aut^{m}(G)$, (3): not (1) and not (2), i.e., $\forall m \ge 0$ and $\forall r>0$ then $Aut^{m+r}(G) \not \simeq Aut^{m}(G)$. The case (1) means that the sequence $(Aut^m(G))_m$ is constant for $m$ large enough, (2) means that it is periodic non-constant for $m$ large enough, and (3) means never periodic. The existence of finite groups in cases (2) or (3) is open. (Q5) Can we rule out the cases (2) and (3) for the cyclic groups? If not, let redefine $\alpha(n)$ as the smallest $m \ge 0$ such that $\exists r>0$ with $Aut^{m}(C_n) \simeq Aut^{m+r}(C_n)$ if $C_n$ is in cases (1) or (2), and $\infty$ in the case (3). REPLY [6 votes]: It seems to me that the structure of ${\rm Aut}^{m}(C_{n})$ also depends heavily on the prime factorization of $n$, and I don't really see any reason to expect the answer to Q1 to be any more tractable than determining the structure of ${\rm Aut}(C_{n})$. For example (just to illustrate) , if we choose a prime $p$, greater than $3$, and then we take a pair of primes $q_{1}$ and $q_{2}$ so that each $q_{i}-1$ is divisible by $p$, but not by $p^{2}$, and we set $n = q_{1}q_{2}$, then ${\rm Aut}^{2}(C_{n})$ has a direct factor isomorphic to ${\rm GL}(2,p)$, and is not solvable. Furthermore the involvement of ${\rm PSL}(2,p)$ in ${\rm Aut}^{m}(C_{n})$ persists for $m >2$.<|endoftext|> TITLE: Are exterior algebras intrinsically formal as associative dg algebras? QUESTION [7 upvotes]: (Cross-posted from mathematics stackexchange.) Fix a finite dimensional vector space $V$ over a field of characteristic zero, and let $R=Sym(V[1])$ be the free graded commutative algebra generated by $V$ in cohomological degree $-1$, but thought of as a formal associative dg algebra (we forget that it is commutative). Then it is easy to see that (edit: any $A_{\infty}$-deformation of) $R$ is almost formal, meaning that in a minimal $A_{\infty}$-model, the higher products eventually vanish. Indeed, $R$ is concentrated in non-positive cohomological degrees, so all products of an $A_{\infty}$-structure $\mu_{n} : R^{\otimes n} \rightarrow R[2-n]$ have to vanish for $n>>0$. Question 1: Is $R$ intrinsically formal, so that for any deformation, $\mu_{n}$ can be assumed to be zero for $n>2$? Question 2: What if $R=Sym(V[-1])$, so that $R$ is concentrated cohomologically in non-negative degrees? (I'm thinking of the cohomology ring of a torus, but as an associative algebra, rather than a commutative algebra.) Can there be non-trivial $A_{\infty}$-structures on $R$, or is $R$ intrinsically formal? REPLY [8 votes]: Here are examples of nontrivial $A_\infty$-structures extending the product on $\operatorname{Sym}(\mathbb R^3[-1])$ and $\operatorname{Sym}(\mathbb R^5[-1])$, respectively. Below the fold, I have kept my original answer, which got the main ideas right but almost all degrees wrong. Fix coordinates $\xi_1,\xi_2,\xi_3$ on $\mathbb R^3[-1]$ and set $$ \mu_3(f_1,f_2,f_3) = \xi_2\xi_3 (\partial_{\xi_1}f_1)(\partial_{\xi_1}f_2)(\partial_{\xi_1}f_3)\ . $$ This defines a $A_\infty$-structure: Every term in $[\mu_3,\mu_3](f_1,\dots,f_5) = \sum \mu_3(\dots,\mu_3(\dots),\dots)$ contains $\xi_1^2 = 0$ and hence vanishes; similarly, $[\mu_3,\mu_2](f_1,\dots,f_4)$ vanishes if one of the $f_i$ is a multiple of $\xi_2$ or $\xi_3$, and it also vanishes if one of them is the identity since $[-,-]$ preserves normalized Hochschild cochains. Thus the only expression we have to check is \begin{align*} [\mu_2,\mu_3](\xi_1,\xi_1,\xi_1,\xi_1) &= \color{red}{-}\xi_1\mu_3(\xi_1,\xi_1,\xi_1) -\mu_3(\xi_1^2,\xi_1,\xi_1) \\ &\ + \mu_3(\xi_1,\xi_1^2,\xi_1) - \mu_3(\xi_1,\xi_1,\xi_1^2) \\ &\ + \mu_3(\xi_1,\xi_1,\xi_1)\xi_1\\ &= -\xi_1\xi_2\xi_3 + \xi_2\xi_3\xi_1 = 0 \end{align*} (The red sign comes from the Koszul sign rule.) This $A_\infty$-algebra has a nontrivial Massey product $0\notin \langle\xi_1,\xi_1,\xi_1\rangle = \xi_2\xi_3 + \xi_1A + A\xi_1$ and hence is not formal. Essentially the same argument applies to the $A_\infty$-structure on $\mathbb R^5[-1]$ defined by the only nontrivial bracket $$ \mu_3(f_1,f_2,f_3) = \xi_2\xi_3\xi_4\xi_5 (\partial_{\xi_1}f_1)(\partial_{\xi_1}f_2)(\partial_{\xi_1}f_3)\ . $$ Recall that we can associate to any (cohomologically) graded algebra $(A,\mu_A)$ the Hochschild cochains $$ C^n(A) = \prod_{p+q = n}\operatorname{Hom}^p(A^{\otimes q},A) $$ which come equipped with a differential defined by precomposing with the multiplication of cyclic neighbours in the tensor product. As the totalization of a double complex, it also comes with a canonical filtration, namely $C^n_{\ge m}(A) = \prod_{{\substack{p+q=n\\q\ge m}}} \operatorname{Hom}^p(A^{\otimes q},A)$. The skew-symmetrization of the pre-Lie structure $f\circ g(c_1,\dots,x_{m+n-1}) = \sum_{i=1}^m \pm f(x_1,\dots,g(x_i,\dots,x_{i+n-1}),\dots,x_{m+n-1})$ defines a filtered shifted dgla structure on $C^*(A)$, and $A_\infty$-structures with trivial differential $(A,0,\mu_A,\mu_3,\dots)$ correspond to Maurer-Cartan elements $(\mu_3,\dots)$ in $C^2_{\ge 3}(A)$. Now consider any such element $\mu$; its component of lowest filtration defines a degree $2$ cohomology class in the associated graded, since $\mathrm d\mu = -\frac{1}{2}[\mu,\mu]$ lies in higher filtration (as $[C^p_{\ge m}(A),C^q_{\ge n}(A)]\subset C^{p+q-1}_{\ge m+n-1}(A)$). If this cohomology class is $0$, i.e. $\mu = \mathrm d\eta$ up to terms of higher filtration, we can act by the gauge symmetry $\mu\mapsto e^{-\eta}\circ\mu = \mu - \mathrm d\eta\, + (\text{higher filtration})$ to obtain an equivalent Maurer-Cartan element which lies in higher filtration. Since the Hochschild cochains are complete with respect to the canonical filtration, the limit of this process is well-defined, i.e. if all obstructions vanish we obtain that our Maurer-Cartan element is gauge equivalent to $0$. If $A = \operatorname{Sym}(V)$ is the commutative algebra on a graded vector space $V$, we can compute $C^*(A)$ with its filtration explicitly: We have $C^*(A) \simeq \operatorname{Ext}_{A\otimes A^{op}}(A,A)$, and since the multiplication map $A\otimes A^{op}\to A$ is $\operatorname{Sym}$ of the codiagonal $V\oplus V\to V$, we obtain a Koszul resolution $A\simeq (\operatorname{Sym}(V\oplus V\oplus V[1]),\mathrm d)$. Explicitly, we choose a basis $x_i$ of $V$; then the resolution is generated by elements $x_i,x_i'$ in degree $|x_i|$ and $y_i$ in degree $|x_i|-1$, with $\mathrm dy_i = x_i' - x_i''$. (This is where we use characteristic $0$.) It follows that $C^*(\operatorname{Sym}(V))\simeq \widehat{\operatorname{Sym}}(V\oplus V^*[-1])$, where the filtration is given by the degree in the $V[-1]$-variables, and the hat indicates that we take the completion with respect to this filtration. Now let us specialize to the two cases in your question, namely $V$ is concentrated in degree $-1$ or $+1$. In the second case, $C^*(A)$ is concentrated in nonnegative degrees, and $HH^2(A) := H^2(C^*(A))\cong \operatorname{Sym}^* V^*\otimes \Lambda^2 V$, with the relevant part in $\operatorname{Sym}$-filtration $\ge 3$. In the first case, $V\oplus V^*[-1]$ lives in degrees $-1$ and $2$, and the deformation group is given by $HH^2_{\ge 3}(A) = \prod_{p\ge 3} \Lambda^{2p-2}V\otimes \operatorname{Sym}^p V^*$ and thus also does not vanish. Indeed, Kontsevich's formality theorem shows that Maurer-Cartan elements of $\hbar C^*_{\ge 3}(A)[[\hbar]]$ are in bijection with those of its cohomology, i.e. an infinitesimal deformation can be extended to a formal deformation iff it satisfies the Maurer-Cartan equation in in the cohomology. In some cases, the formal infinite series involve only finitely many terms and therefore give rise to actual deformations, as is the case in the two examples above.<|endoftext|> TITLE: Does minimal degree $n$ imply a $K_n$ minor QUESTION [10 upvotes]: Is it true that any finite graph has a $K_n$ minor, where $n$ is a minimal vertex degree? REPLY [22 votes]: More generally, it is a classic result (independently due to Kostochka and Thomason) that minimum degree $(\alpha+o(1))n \sqrt{\log n}$ suffices to force a $K_n$ minor, where $\alpha$ is an explicit constant. Conversely, there are random graphs with minimum degree $\Omega(n\sqrt{\log n})$ that do not contain a $K_n$ minor. See here to access the paper by Thomason. Update. Alon, Krivelevich, and Sudakov have recently given a new short proof of the Kostochka-Thomason result.<|endoftext|> TITLE: Extent of “unscientific”, and of wrong, papers in research mathematics QUESTION [97 upvotes]: This question is cross-posted from academia.stackexchange.com where it got closed with the advice of posting it on MO. Kevin Buzzard's slides (PDF version) at a recent conference have really unsettled me. In it, he mentions several examples in what one would imagine as very rigorous areas (e.g., algebraic geometry) where the top journals like Annals and Inventiones have published and never retracted papers which are now known to be wrong. He also mentions papers relying on unpublished results taken on trust that those who announced them indeed have a proof. He writes about his own work: [...] maybe some of my work in the p-adic Langlands philosophy relies on stuff that is wrong. Or maybe, perhaps less drastically, on stuff which is actually correct, but for which humanity does not actually have a complete proof. If our research is not reproducible, is it science? If my work in pure mathematics is neither useful nor 100 percent guaranteed to be correct, it is surely a waste of time. He says that as a result, he switched to formalizing proofs completely, with e.g. Lean, which guarantees correctness, and thus reusability forever. Just how widespread is the issue? Are most areas safe, or contaminated? For example, is there some way to track the not-retracted-but-wrong papers? The answer I accepted on academia.stackexchange before the closure gives a useful general purpose method, but I'd really appreciate more detailed area-specific answers. For example, what fraction of your own papers do you expect to rely on a statement "for which humanity does not actually have a complete proof" ? REPLY [44 votes]: As Kevin Buzzard himself admits in his answer, he somewhat exaggerated his point for effect. However, I'd submit that if you were unsettled by his talk, then that's a good thing. I don't think that the proper reaction is to look for reassurance that mathematics really is fine, or that the problems are restricted to some easily quarantined corner. Rather, I think the proper reaction is to strive for a more accurate view of the true state of the mathematical literature, and refuse to settle for comforting myths that aren't based on reality. Some of the literature is rock-solid and can stand on its own, much more of it is rock-solid provided you have access to the relevant experts, and some of it is gappy but we don't really care. On the other hand, some small percentage of it is gappy or wrong and we do care, but social norms within the mathematical community have caused us to downplay the problems. This last category is important. It is a small percentage, but from a scholarly point of view it is a serious problem, and we should all be aware of it and willing to acknowledge it. If, every time someone brings it up, we try to get them to shut up by repeating some "propaganda" that makes us feel good about mathematics, then we are not solving the problem but perpetuating it. Some related concerns were raised over 25 years ago by Jaffe and Quinn in their article on Theoretical Mathematics. This generated considerable discussion at the time. Let me quote the first paragraph of Atiyah's response. I find myself agreeing with much of the detail of the Jaffe–Quinn argument, especially the importance of distinguishing between results based on rigorous proofs and those which have a heuristic basis. Overall, however, I rebel against their general tone and attitude which appears too authoritarian. My takeaway from this is that Jaffe and Quinn made many valid points, but because this is a sensitive subject, dealing with societal norms, we have to be very careful how we approach it. Given the way that the mathematical community currently works, saying that someone's work has gaps and/or mistakes is often taken to be a personal insult. I think that if, as a community, we were more honest about the fact that proofs are not simply right or wrong, complete or incomplete, but that there is a continuum between the two extremes, then we might be able to patch problems that arise more efficiently, because we wouldn't have to walk on eggshells.<|endoftext|> TITLE: Vanishing of Hochschild homology of a category QUESTION [7 upvotes]: Let $A$ be a dg- or $A_{\infty}$-category (with $\mathbb{Z}$-graded Hom sets, over a field of characteristic $0$). Let $HH_*(A)$ be the Hochschild homology of $A$. Suppose that $HH_n(A)=0$ for all $n \in \mathbb{Z}$. Does this imply that $A$ is the zero category? If not, then what assumptions can I add to $A$ to make this true (e.g. smoothness, properness, ect)? Remark: I have heard the heuristic that Hochschild homology can be viewed as "differential forms" on the "spectrum" of the non-commutative category $A$ (this heuristic is presumably motivated by the Hochschild-Kostant-Rosenberg theorem). Hence it's natural to expect that $A$ vanishes if it admits no non-zero differential forms. Note also that if $A$ is a (possibly non-commutative) $k$-algebra, then one can check that $HH_0(A) =0$ iff $A=0$. REPLY [13 votes]: This precise question was phrased as the vanishing conjecture in Hochschild homology and semiorthogonal decompositions. But we now know that there exist so called (quasi)phantom categories, which give counterexamples. These are categories with vanishing Hochschild homology, and vanishing (resp. torsion) Grothendieck group. As they are admissible subcategories of derived categories of smooth projective varieties, they have all nice properties you could want for their dg enhancements. An overview of some constructions: Gorchinskiy, Sergey; Orlov, Dmitri, Geometric phantom categories, Publ. Math., Inst. Hautes Étud. Sci. 117, 329-349 (2013). ZBL1285.14018. Böhning, Christian; Graf von Bothmer, Hans-Christian; Katzarkov, Ludmil; Sosna, Pawel, Determinantal Barlow surfaces and phantom categories, J. Eur. Math. Soc. (JEMS) 17, No. 7, 1569-1592 (2015). ZBL1323.14014. Galkin, Sergey; Katzarkov, Ludmil; Mellit, Anton; Shinder, Evgeny, Derived categories of Keum’s fake projective planes, Adv. Math. 278, 238-253 (2015). ZBL1327.14081. Galkin, Sergey; Shinder, Evgeny, Exceptional collections of line bundles on the Beauville surface, Adv. Math. 244, 1033-1050 (2013). ZBL1408.14068. Böhning, Christian; Graf von Bothmer, Hans-Christian; Sosna, Pawel, On the derived category of the classical Godeaux surface, Adv. Math. 243, 203-231 (2013). ZBL1299.14015. Alexeev, Valery; Orlov, Dmitri, Derived categories of Burniat surfaces and exceptional collections, Math. Ann. 357, No. 2, 743-759 (2013). ZBL1282.14030. What is interesting is that Hochschild cohomology can detect their non-vanishing, and all kinds of interesting behavior regarding deformation theory arises, see Kuznetsov, Alexander, Height of exceptional collections and Hochschild cohomology of quasiphantom categories, J. Reine Angew. Math. 708, 213-243 (2015). ZBL1331.14024.<|endoftext|> TITLE: The isometry group of a product of two Riemannian manifolds QUESTION [5 upvotes]: Under what conditions is the isometry group of a product of two Riemannian manifolds the product of the isometry groups of each one of the components? One counterexample is a product of two isometric manifolds, since the isometry contains the involution which cannot be a product. I was wondering if there are some general criterion. REPLY [2 votes]: You may find this answer of mine of some interest. There, I explain a special case when one can prove that $I(M \times N) \cong I(M) \times I(N)$ with simple differential geometric arguments involving sectional curvatures. And then (in the edit) I sketch a proof of a much more general result using some holonomy techniques taken from Kobayashi & Nomizu. In a nutshell, this result says that if you have a product of some irreducible Riemannian manifolds and at most one flat Riemannian manifold, then the isometry group of the product is the product of the isometry groups of the factors plus permutations of isometric factors (no compactness or even completeness assumption required). One can also try to drop the irreducibility assumption, but then to say something interesting about the isometry group and how it relates to those of the factors, one needs to add the assumptions of completeness and simply connectedness (at least I don't see how to proceed otherwise). Here is the resulting statement: Proposition. Let $M = M_1 \times \ldots \times M_k$ be a Riemannian product of complete simply connected Riemannian manifolds. We have an obvious embedding of Lie groups $i \colon I(M_1) \times \ldots \times I(M_k) \hookrightarrow I(M)$. The following are equivalent: $i$ is a local isomorphism, i. e. it gives an isomorphism on the identity components of the above groups: $I^0(M_1) \times \ldots \times I^0(M_k) \simeq I^0(M)$; $\dim I(M) = \dim I(M_1) + \cdots + \dim I(M_k)$; At most one of $M_i$'s has a nontrivial Euclidean de Rham factor. Moreover, if these conditions are satisfied, then $i$ is an isomorphism if and only if there does not exist $i \ne j$ such that $M_i$ and $M_j$ share isometric de Rham factors. (If all $M_i$'s are irreducible, it means that no two of them are isometric.) In order to prove this, just de Rham decompose each factor, stack all Euclidean subfactors together, and apply the second proposition from my answer linked above.<|endoftext|> TITLE: Using irreducible characters of the orthogonal group as basis for homogeneous symmetric polynomials QUESTION [10 upvotes]: The irreducible characters of the orthogonal group $O(2N)$ are given by $$ o_\lambda(x_1,x_1^{-1},...x_N,x_N^{-1})=\frac{\det(x_j^{N+\lambda_i-i}+x_j^{-(N+\lambda_i-i)})}{\det(x_j^{N-i}+x_j^{-(N-i)})}$$ I was playing with them as basis for the space of homogeneous symmetric polynomials. I wanted to write the function $$p_3=\sum_{i=1}^N (x_i^3+x_i^{-3})$$ as a linear combination of $o_\lambda$'s. I started with $N=2$ and in that case I found that $$ p_3=o_{(3)}-2o_{(2,1)}+o_{(1)}.$$ However, I then tried with $N=4$ and in that case I found that $$ p_3=o_{(3)}-o_{(2,1)}+o_{(1,1,1)}.$$ I was expecting the coefficients to be the same, i.e. to be independent of $N$. (Independence of $N$ indeed holds when using characters of the unitary group, in which case the coefficients are the characters of the permutation group.) Have I made some mistake or are the coefficients indeed dependent on $N$? REPLY [3 votes]: (I have looked more carefully at this theory of "universal characters" mentioned by Stanley and am updating this answer according to what I learned. All of this was contained in the answer by Stanley, I am just unpacking it for the sake of those like me and the OP who may be confused.) The following material is in the paper Young-diagrammatic methods for the representation theory of the classical groups of type $B_n$, $C_n$, $D_n$ by Koike and Terada, also in works by King (such as Modification Rules and Products of Irreducible Representations of the Unitary, Orthogonal, and Symplectic Groups, Journal of Mathematical Physics 12, 1588, 1971) and a very readable 1938 book by Murnaghan (The Theory of Group Representations). $\DeclareMathOperator\Or{O}\DeclareMathOperator\U{U}$So functions $o_\lambda$ correspond to irreducible characters of $\Or(2N)$ only when $\ell(\lambda)\le N$. The set of such actual characters forms a basis for the space of symmetric functions on variables $\{x_1,x_1^{-1},\dotsc,x_N,x_N^{-1}\}$. However, the coefficients in the expansion of power sums in general depend on $N$. On the other hand, partitions with $\ell(\lambda)>N$ do not define irreducible representations, so $o_\lambda$ is not an actual character. In that case they are replaced by another symmetric function, $o_\lambda\to o_{\widetilde{\lambda}}$ with a modified partition $\widetilde{\lambda}$. When these functions are used, the large-$N$ expansions continue to hold for small $N$. In that sense they are 'universal'. When a Schur function $s_\mu$ is decomposed in terms of $o_\lambda$, which corresponds to the branching rule of $\U(2N)\supset \Or(2N)$, one has $\ell(\lambda)\le \ell(\mu)$, so if $\ell(\mu)\le N$ only actual characters appear. This is not true if $N<\ell(\mu)\le 2N$, and then universal characters must also be used. This is not discussed in the classic book "Representation theory" by Fulton and Harris, for example, where only the case $\ell(\mu)\le N$ appears. The modified partition $\widetilde{\lambda}$ is defined as follows. For $O(2N)$, let $m=2\ell(\lambda)-2N$. Then remove from the Young diagram of $\lambda$ a total of $m$ adjacent boxes, starting from the bottom of the first column and keeping always at the boundary of the diagram. In this way the changes to be implemented are, in sequence, $\lambda'_1\to\lambda'_2-1$, then $\lambda'_2\to\lambda'_3-1$, etc. until the procedures stops at some column $c$. If $m$ is too large and there are not enough boxes to accommodate this procedure, or if the remaining diagram is not a partition, then $o_\lambda=0$; otherwise $o_\lambda=(-1)^{c-1}o_{\widetilde{\lambda}}$. For example, the universal decomposition for $p_4$ is $$p_4=o_4-o_{31}+o_{211}-o_{1111}+o_0.$$ If $\lambda=(1,1,1,1)$ and $N=6$, we must remove $8-6=2$ boxes, in which case we get $c=1$ and $\widetilde{\lambda}=(1,1)$. Hence, for $O(6)$ we have $o_{1,1,1,1}=o_{1,1}$ and the universal relation reduces to $p_4=o_4-o_{31}+o_{211}-o_{11}+o_0$. When $N=4$, we must remove $8-4=4$ boxes from $\lambda=(1,1,1,1)$. We end up with no boxes at all, so $o_{1,1,1,1}=o_\emptyset$ for $O(4)$. We must remove $6-4=2$ boxes from $\lambda=(2,1,1)$, arriving at $o_{2,1,1}=o_2$. The relation reduces to $p_4=o_4-o_{31}+o_{2}$. Finally, take $N=2$. We cannot remove $8-2=6$ boxes from $(1,1,1,1)$, so $o_{1,1,1,1}=0$ for this group; when we remove $4-2=2$ boxes from $(3,1)$, the result is not the diagram of a partition, so $o_{3,1}=0$ for this group; removing $6-2=4$ boxes from $(2,1,1)$ leads to the empty partition, and the removing procedure ends in the second column so $c=2$, hence $o_{2,1,1}=-o_\emptyset$ for this group. Thereby the relation reduces to $p_4=o_4$. Unfortunately, the modification rule is incorrectly stated in the recent "The Random Matrix Theory of the Classical Compact Groups" (Cambridge University Press, 2019), by Elizabeth Meckes.<|endoftext|> TITLE: 3 term van der Waerden with large step size QUESTION [7 upvotes]: Let $P(n)$ be the statement "any $n$ coloring of $\mathbb{N}$ contains a monochromatic progression $a, a+d, a+2d$ such that $d>a$". For which $n$ is $P(n)$ true? It's easy to see that $P(2)$ is true by a simple modification of the color focusing argument that is used in the traditional proof of van der Waerden's theorem. However, this argument does not seem to generalize to more colors, or at least not very easily. It's also easy to see that a similar statement is not true for progressions of length $4$, even in the $2$ color case: just color $[2^n,2^{n+1}-1]$ red if $n$ is even and blue if $n$ is odd. If $2^n 2$. To see this, it suffices to show that $P(3)$ is false, because if we start with a $3$-coloring that witnesses the failure of $P(3)$, then we can always add a few more colors, say by using each of the new colors on just one or two numbers each, to obtain a witness to the failure of $P(n)$ for larger $n$. Here is a coloring that shows $P(3)$ is false, inspired by your example for $2$ colors and length-$4$ progressions. Let $F_n$ denote the $n^{th}$ term of the Fibonacci sequence (with $F_1 = 1$ and $F_2 = 2$), and then color $m$ red if $m \in [F_{3k},F_{3k+1})$ for some $k$, color $m$ green if $m \in [F_{3k+1},F_{3k+2})$ for some $k$, color $m$ blue if $m \in [F_{3k+2},F_{3k+3})$ for some $k$. To see that this coloring really is a counterexample, let's suppose that $a$, $a+d$, and $a+2d$ all have the same color, and that $a < d$. Notice that this implies $$\frac{3}{2} < \frac{a+2d}{a+d} < 2.$$ This does not immediately give us a contradiction, but it does tell us that $a+d$ and $a+2d$ must both lie in a single interval of the form $[F_n,F_{n+1})$. (This is because $2(F_n-1) < F_{n+3}$ for all $n$. This isn't too hard to prove -- for large $n$ it follows from the fact that $F_{n+3}/F_n \approx \varphi^3 \approx 4.2$.) Once we know that $F_n \leq a+d < a+2d < F_{n+1}$, we get that $$a = 2(a+d) - (a+2d) > 2F_n - F_{n+1} = 2F_n - (F_n+F_{n-1}) = F_n - F_{n-1} = F_{n-2}.$$ This tells us that if $a$ is to have the same color as $a+d$ and $a+2d$, then we must also have $a \in [F_n,F_{n+1})$. But now this is absurd: we have $\frac{a+2d}{a} > 3$ while $\frac{F_{n+1}}{F_n} \leq 2$ for all $n$.<|endoftext|> TITLE: Does the homotopy category of spaces admit a weak generating set? QUESTION [11 upvotes]: As a follow-up to this question, let $\mathcal C$ be a category and $\mathcal S \subseteq \mathcal C$ a class of objects. Say that $\mathcal S$ is weakly generating if the functors $Hom_{\mathcal C}(S,-)$ are jointly conservative, for $S \in \mathcal S$. That is, a map $X \to Y$ in $\mathcal C$ is an isomorphism if and only if it induces a bijection $Hom_{\mathcal C}(S,X) \to Hom_{\mathcal C}(S,Y)$ for each $S \in \mathcal S$. Question 1: Does the homotopy category of spaces admit a small generating set? (For example, as Simon Henry asks, do finite CW complexes work? How about the spheres?) Of course, by Whitehead's theorem, the homotopy category of pointed connected spaces admits a generating set given by the spheres. But I'm not sure about unpointed spaces. Note that the singleton set comprising the contractible space $\ast$ is a generator in the $\infty$-category of spaces, since $X \to Y$ is an equivalence if and only if $Map(\ast, X) \to Map(\ast,Y)$ is an equivalence. But passage to the the homotopy category discards the higher homotopy of the mapping spaces. Question 2: More generally, if $\mathcal C$ is an accessible $\infty$-category, then does the homotopy category $h\mathcal C$ admit a small generating set? What if we assume that $\mathcal C$ is presentable? Again, if $\mathcal C$ is $\kappa$-accessible, then the class $\mathcal C_\kappa$ of $\kappa$-compact objects forms a generating set in $\mathcal C$, but it's not clear if it forms a generating set in $h\mathcal C$. In fact, I think that Question 2 (in the "presentable" case) is equivalent to Question 1: if the answer to Question 1 is affirmative, so that $\mathcal S$ is a generating set for the homotopy category of spaces and $\mathcal T$ is a generating set for $\mathcal C$, then the set of spaces $S \ast T$ for $S \in \mathcal S, T \in \mathcal T$ forms a generating set for $h\mathcal C$. Here $\ast$ denotes copowering. One result in this direction is Rosicky's Theorem, which says (in model-independent language) that if $\mathcal C$ is a presentable $\infty$-category, then the canonical functor $h\mathcal C \to Ind_\kappa(h\mathcal C_\kappa)$ is essentially surjective and full for some $\kappa$. For my purposes, it would suffice to know that this functor is conservative for some $\kappa$. REPLY [13 votes]: This paper by Kevin Carlson and Dan Christensen says that the answer to question one is no: No set of spaces detects isomorphisms in the homotopy category, arXiv:1910.04141.<|endoftext|> TITLE: Undetermined Banach-Mazur games in ZF? QUESTION [22 upvotes]: This question was previously asked and bountied on MSE, with no response. This MO question is related, but is also unanswered and the comments do not appear to address this question. Given a topological space $\mathcal{X}=(X,\tau)$, the Banach-Mazur game on $\mathcal{X}$ is the (two-player, perfect information, length-$\omega$) game played as follows: Players $1$ and $2$ alternately play decreasing nonempty open sets $A_1\supseteq B_1\supseteq A_2\supseteq B_2\supseteq ...$. Player $1$ wins iff $\bigcap_{i\in\mathbb{N}} A_i=\emptyset$. ZFC implies that there is a subspace of $\mathbb{R}$ with the usual topology whose Banach-Mazur game is undetermined; on the other hand, it's consistent with ZF+DC (and indeed adds no consistency strength!) that no subspace of $\mathbb{R}$ does this ("every set of reals has the Baire property"). However, when we leave $\mathbb{R}$ things get much weirder. My question is: Does ZF alone prove that there is some space $\mathcal{X}$ whose Banach-Mazur game is undetermined? Controlling the behavior of all possible topological spaces in a model of ZF is extremely hard for me, and I suspect the answer to the question is in fact yes. In fact, I recall seeing a pretty simple proof of this; however, I can't track it down or whip up a ZF-construction on my own (specifically, everything I try ultimately winds up being a recursive construction killed by having too many requirements to meet in the given number of steps). REPLY [10 votes]: This is only a partial answer. ZF + DC + 'every Banach-Mazur game is determined' is inconsistent. Let $X$ be the set of all functions of the form $f: \alpha \rightarrow \{0,1\}$, with $\alpha$ an ordinal such that for any ordinals $\beta < \gamma$ with $\omega \cdot \gamma + \omega \leq \alpha$, the sets $(n \in \mathbb{N} : f(\omega \cdot \beta + n) = 1)$ and $(n \in \mathbb{N} : f(\omega \cdot \gamma + n) = 1)$ are distinct. Let $\tau$ be the topology on $X$ generated by sets of the form $U_f = (g \in X : g \supseteq f)$, and let $\mathcal{X} = (X,\tau)$. Claim: Player $1$ does not have a winning strategy for the Banach-Mazur game on $\mathcal{X}$. Proof of claim: For any strategy $S$ for player $1$. Let $T$ be the tree of all initial segments of plays against $S$ such that each play by player $2$ is of the form $U_f$ for some $f \in X$ of length $\omega \cdot \alpha$ for some $\alpha$. If this tree fails to be pruned, then player $1$ has in some play of the game played a set $V$ such that for any $U_f \subseteq V$, $f$ enumerates every element of $2^{\mathbb{N}}$ extending $\sigma$ for some $\sigma \in 2^{<\omega}$. This would imply that $2^{\mathbb{N}}$ can be well-ordered, allowing us to construct a Bernstein set, contradicting our assumptions. Hence this must be a pruned tree of height $\omega$, so by dependent choice it has a path. Let $g$ be the union of all $f$ such that $U_f$ is on that path somewhere. By construction, $g \in X$, so we have that the strategy where player $2$ blindly plays the moves in this path wins against $S$. So, it must be the case that player $2$ has a winning strategy $S$. For any $f \in X$ with length $\omega \cdot \alpha$ for some $\alpha$, let $T_f$ be the strategy for player $1$ that plays $U_f$ on player $1$'s zeroth move and on player $1$'s $n+1$st move, if player $2$ played $V$, then player $1$ plays $V \cap \bigcup_{\sigma \in 2^n} U_{f\frown \sigma\frown 0}$, if this is non-empty, and otherwise player $1$ plays $V \cap \bigcup_{\sigma \in 2^n} U_{f\frown \sigma\frown 1}$. (Note that since the union of these two is $V$, one of these must be non-empty.) Since $S$ is a winning strategy, for any $f\in X$ of length $\omega \cdot \alpha$ for some $\alpha$, the play of $T_f$ against $S$ must result in a nonempty intersection. By construction, for any $g$ and $h$ in that intersection, $g(\omega \cdot \alpha + n) = h(\omega \cdot \alpha +n)$ for all $n<\omega$, and the set $(n \in \mathbb{N} : h(\omega \cdot \alpha + n) = 1)$ must be distinct from $(n \in \mathbb{N} : f(\omega \cdot \beta + n) = 1)$ for any $\beta < \alpha$. Therefore $S$ gives us a uniform procedure for picking a real not on a given well-ordered list of reals. By iterating this gives us a well-ordering of the reals. Therefore we can construct a Bernstein set from $S$ and we have that the Banach-Mazur game on that set is not determined, which contradicts our assumption. Therefore ZF + DC proves that there is an undetermined Banach-Mazur game. Right now I don't see how to use a failure of DC to build an undetermined game.<|endoftext|> TITLE: Do there exist any variational principles on the space of braids (or knots)? QUESTION [5 upvotes]: This is very speculative question and I do not know where to start looking up the literature, or if what I am looking for is even mathematically possible/meaningful. Q: I am interested in finding out whether any variational or action principles have been derived anywhere in the physics or mathematical literature (e.g. in study of geometry of configuration spaces etc.) on the space of braids (or knots). Explicitly, some example where a Lagrangian is a function of a braid representation, and it is minimized/extremized under some constraints. The resulting equation would be differential equation whose solution gives a trajectory in the space of braids. REPLY [4 votes]: O'Hara introduced knot energies, and a Möbius invariant case was studied by Freedman-He-Wang. For prime knots, Zheng-Xu He subsequently showed that there exists a smooth minimizer (up to Möbius transformation), and now it is known that critical points are analytic. For braids, one can look at various variational principles by looking at metrics on Teichmuller space of the punctured sphere. The Teichmuller metric is a Finsler metric, with unique minimizing geodesics for pseudo-Anosov braids. The Weil-Petersson metric is a Riemannian metric which is incomplete, but for which pseudo-Anosovs also have unique minimizing representatives. The minimizing geodesic then gives a canonical braid representative up to conjugacy (by choosing the representatives to have fixed center of mass and constant moment). One can also find unique representatives for braids up to choosing basepoint surfaces (e.g. $n$ points lying on the $x$-axis in $\mathbb{R}^2$). For the 3-punctured sphere, the Teichmuller metric is the hyperbolic plane, and one can easily compute the geodesic. One can also look at solutions to the planar $n$-body problem to get braid realizations. Richard Montgomery first proved the existence of solutions to the 3-body problem.<|endoftext|> TITLE: Conjectured primality test for specific class of $N=4kp^n+1$ QUESTION [9 upvotes]: Can you provide a proof or counterexample for the following claim? Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ . Let $N= 4kp^{n}+1 $ where $k$ is a positive natural number , $ 4k<2^n$ , $p$ is a prime number and $n\ge3$ . Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{N}\right)=-1$ and $\left(\frac{a+2}{N}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol. Let $S_i=P_p(S_{i-1})$ with $S_0$ equal to the modular $P_{kp^2}(a)\phantom{5} \text{mod} \phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} \equiv 0 \pmod{N}$ . You can run this test here. I have verified this claim for $k \in [1,500]$ with $p \leq 97$ and $n \in [3,50]$ . Further generalization of the claim A Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ . Let $N= 2kp^{n} + 1 $ where $k$ is a positive natural number , $ 2k<2^n$ , $p$ is a prime number and $n\ge3$ . Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{N}\right)=-1$ and $\left(\frac{a+2}{N}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol. Let $S_i=P_p(S_{i-1})$ with $S_0$ equal to the modular $P_{kp^2}(a)\phantom{5} \text{mod} \phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} \equiv -2 \pmod{N}$ . You can run this test here. B Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ . Let $N= 2kp^{n} - 1 $ where $k$ is a positive natural number , $ 2k<2^n$ , $p$ is a prime number and $n\ge3$ . Let $a$ be a natural number greater than two such that $\left(\frac{a-2}{N}\right)=1$ and $\left(\frac{a+2}{N}\right)=-1$ where $\left(\frac{}{}\right)$ denotes Jacobi symbol. Let $S_i=P_p(S_{i-1})$ with $S_0$ equal to the modular $P_{kp^2}(a)\phantom{5} \text{mod} \phantom{5} N$. Then $N$ is prime if and only if $S_{n-2} \equiv -2 \pmod{N}$ . You can run this test here. REPLY [7 votes]: The "if and only if" statement fails for $$[p,n,k,a] \in \{ [3, 4, 1, 100], [3, 4, 1, 225], [3, 6, 13, 2901] \}$$ and many others. In these cases, $N$ is not prime, but the congruence $S_{n-2}\equiv 0\pmod{N}$ still holds.<|endoftext|> TITLE: Notation for "the" left adjoint functor QUESTION [13 upvotes]: As far as I know, there is no "official" notation for the left adjoint of a functor $F : \mathcal{C} \to \mathcal{D}$ if it exists. I have seen the notation $F^*$ sometimes, but this looks only nice when $F$ is already written as $F_*$, which is not practical. (This notation is then motivated by direct and inverse image functors. And it seems to be quite common for adjunctions between preorders aka Galois connections.) Similarly, I have seen the notation $F_*$ for the right adjoint of $F$, which only looks nice when $F$ is already written as $F^*$. I have also seen the notation $F^{\dagger}$ for the right adjoint, which looks nice, but then how would you denote the left adjoint if it exists? Perhaps ${}^{\dagger} F$? I don't want to start a debate here what is a good notation or not, since this is subjective anyway and is not suited for mathoverflow. I would like to know: Are there any textbooks, influential papers or monographs on category theory which have introduced a notation for the left adjoint of $F$? Is there any notation which has been used by multiple authors? Just to avoid any misunderstanding: Of course there is the official notation $F \dashv G$ when $F$ is left adjoint to $G$, but $\dashv$ is a relation symbol. I am interested in a function symbol (which makes sense since left and right adjoints are unique up to canonical isomorphism if they exist). REPLY [4 votes]: P. Gabriel uses $F_\lambda$ for the left adjoint and $F_\rho$ for the right adjoint. See for instance p. 344 of the article "Covering spaces in representation theory" by K. Bongartz and P. Gabriel (Invent. Math. 65, 1982) (EuDML).<|endoftext|> TITLE: Automorphism groups of odd order QUESTION [11 upvotes]: This is inspired by this question. Is there a description of finite groups without automorphisms of order $2$? REPLY [4 votes]: New version (existence hinted in previous version): If $G$ is a non-trivial finite (solvable) group of odd order with $\Phi(G) = 1$, then $G$ has an automorphism of order $2$. It is well-known and easy to check that $F = F(G)$ is a product of minimal normal subgroups of $G$, each an elementary Abelian $p_{i}$-group for some prime $p_{i}$. Also, $F$ is well-known to be complemented in $G$ in this case (I give a proof for completeness: Choose a proper subgroup $H$ of $G$ minimal subject to $G = FH$ (such exists because $1 \neq F \not \leq \Phi(G)$). Then $(H \cap F) \leq \Phi(H)$ by minimality of $H$. Also $F \cap H$ is normal in $\langle H,F \rangle = G$, since $F$ is Abelian and $F \lhd G$. If $F \cap H \neq 1$, then there is a maximal subgroup $M$ of $G$ with $G = (F \cap H)M$ since $\Phi(G) = 1$. Then $H = (F \cap H)(M \cap H)$ by Dedekind's modular law. But then $H = H \cap M \leq M$ since $F \cap H \leq \Phi(H)$. But then $G = (F \cap H)M \leq M$, contrary to the fact that $M$ is maximal). Now $G = FH$ for some subgroup $H$ of ${\rm Aut}(F)$, and the product is semidirect. Thus $G$ is isomorphic to a subgroup of the holomorph $ X = F{\rm Aut}(F)$ (the semidirect product of $F$ with its automorphism group). Here, we have $G \cong F{\rm Aut}_{G}(F)$, where ${\rm Aut}_{G}(F)$ is the subgroup of ${\rm Aut}(F)$ induced by the conjugation action of $G$ on $F$. Now let $t$ be the central element of ${\rm Aut}(F)$ of order $2$ which inverts $F$ elementwise (note that $t$ is indeed central in ${\rm Aut}(F)$, because $\alpha(f)^{-1} = \alpha(f^{-1})$ for every $\alpha \in {\rm Aut}(F)$). Then $F\langle t \rangle$ normalizes every subgroup of $X$ containing $F$, so normalizes $F{\rm Aut}_{G}(F) \cong G$. Now $|(F{\rm Aut}_{G}(F))(F \langle t\rangle)| = 2|F{\rm Aut}_{G}(F)|$, so that $t$ induces an automorphism of order $2$ of $F{\rm Aut}_{G}(F) \cong G$ (recall that $t$ already inverts $F$ elementwise). Note that $F{\rm Aut}_{G}(F)$ is of index $2$ in $(F{\rm Aut}_{G}(F))(F \langle t\rangle)$, so is normal in the latter group.<|endoftext|> TITLE: Almost transferred model structures QUESTION [6 upvotes]: Let $F : \mathcal{C} \leftrightarrows \mathcal{D} : U$ be a Quillen adjunction between cofibrantly generated model categories. The model structure on $\mathcal{D}$ is called transferred if $U$ preserves and reflects weak equivalences and fibrations. It follows that cofibrations of $\mathcal{D}$ are generated by $F(\mathrm{I})$, where $\mathrm{I}$ is a set of generating cofibrations of $\mathcal{C}$. Now, assume that cofibrations of $\mathcal{D}$ are generated by $F(\mathrm{I})$ and that $U$ reflects fibrant objects (or even all fibrations). Is the model structure on $\mathcal{D}$ necessarily transferred in this case? If these conditions hold and the transferred model structure exists, then it necessarily coincides with the given one. Thus, the question can be reformulated as follows. Is there a model category $\mathcal{D}$ satisfying conditions given above such that the transferred model structure on $\mathcal{D}$ does not exist? REPLY [5 votes]: This is too long for a comment, so I'm putting it here. I can think of three places to look for such an example. First, you could check out Example 2.8 in this paper I wrote with Michael Batanin: Bousfield Localization and Eilenberg-Moore Categories, arXiv:1606.01537. In that example, we proved that the transferred model structure doesn't exist, so you could check if there is a different model structure on the category of operads valued in Ch($\mathbb{F}_2)$ that has the property you want. Second, it is well-known that the transferred model structure on commutative differential graded algebras does not exist in characteristic $p>0$. So, CDGA($\mathbb{F}_p)$ doesn't have a transferred model structure. It does have a model structure discovered by Donald Stanley: Determining Closed Model Category Structures, (1998) (link), and you could check if Stanley's model structure has the property you want. Seems plausible. Third, you could try to construct such an example using the techniques of Reid Barton at this answer: Counter-example to the existence of left Bousfield localization of combinatorial model category You'd want to pick $C$ and $D$ to be very small, like Reid did. I would try to work this out, but I just don't have time (and probably won't all semester).<|endoftext|> TITLE: Cyclic action on Kreweras walks QUESTION [22 upvotes]: A Kreweras walk of length $3n$ is a word consisting of $n$ $A$'s, $n$ $B$'s, and $n$ $C$'s such that in any prefix there are at least as many $A$'s as $B$'s, and at least as many $A$'s as $C$'s. For example, with $n = 3$, one Kreweras walk is: $w = AABBCACCB$. These are the same as walks in $\mathbb{Z}^2$ from the origin to itself consisting of steps $(1,1)$, $(-1,0)$, and $(0,-1)$ which always remain in the nonnegative orthant (treat $A$'s as $(1,1)$ steps, $B$'s as $(-1,0)$ steps, and $C$'s as $(0,-1)$ steps). Kreweras in 1965 proved that the number of Kreweras walks is $\frac{ 4^n(3n)!}{(n+1)!(2n+1)!}$ (OEIS sequence A006335). Many years later, in the 2000's, the Kreweras walks became a motivating/foundational example in the theory of "walks with small steps in the quarter plane" as developed by Mireille Bousquet-Mélou and her school. They also are related to decorated planar maps and in particular are a key ingredient in recent breakthrough work relating random planar maps to Liouville quantum gravity. I discovered a very interesting cyclic action on Kreweras walks, which apparently had not been noticed previously. Let me refer to this action as rotation. To perform rotation on a Kreweras walk $w$, first we literally rotate the word $w =(w_1,w_2,...,w_{3n})$ to $w' = (w_2,w_3,...,w_{3n},w_1)$. With the above example of $w$, we get $w' = ABBCACCBA$. This is, however, no longer a valid Kreweras walk. So there will be a smallest index $i$ such that $(w'_1,...,w'_i)$ has either more $B$'s than $A$'s, or more $C$'s than $A$'s. Then we create another word $w''$ by swapping $w'_i$ and $w'_{3n}$ (which is always an $A$). For example, with the previous example, we have $i = 3$ (corresponding to the second $B$ in the word), and we get $w'' = ABACACCBB$. It's not hard to see that the result is a Kreweras walk, which we call the rotation of the initial Kreweras walk. The sequence of iterated rotations of our initial $w = AABBCACCB$ example looks like $$ 00 \; AABBCACCB \\ 01 \; ABACACCBB \\ 02 \; AACACCBBB \\ 03 \; ACACABBBC \\ 04 \; AACABBBCC \\ 05 \; ACABBACCB \\ 06 \; AABBACCBC \\ 07 \; ABAACCBCB \\ 08 \; AAACCBCBB \\ 09 \; AACCBABBC \\ ...$$ In particular, $3n = 9$ applications of rotation yields the Kreweras walk which is the same as our initial $w$ except that the $B$'s and $C$'s have swapped places. If we did another $9$ applications we would get back to our initial $w$. Conjecture: For a Kreweras walk of length $3n$, $3n$ applications of rotation always yields the Kreweras walk which is the same as the initial walk except with the $B$'s and $C$'s swapped (so $6n$ applications of rotation is the identity). (So my question is obviously: is my conjecture right?) I've thought about this conjecture a fair amount with little concrete progress. I've done a fair amount of computational verification of this conjecture: for all $n \leq 6$, and for many thousands more walks with various $n \leq 30$. Where this action comes from: The Kreweras walks of length $3n$ are in obvious bijection with the linear extensions of a poset $P$, namely, $P=[n] \times V$, the direct product of the chain on $n$ elements and the 3-element ``$V$'' poset with relations $A < B$, $A < C$. I became aware of this poset thanks to this MO answer of Ira Gessel to a previous question of mine, which cited this paper of Kreweras and Niederhausen in which the authors prove not just a product formula for the number of linear extension of $P$, but for the entire order polynomial of $P$. The rotation of Kreweras walks as I just defined it is nothing other than the famous (Schützenberger) promotion operation on linear extension of a poset (see this survey of Stanley's for background on promotion). There are few non-trivial classes of posets for which the behavior of promotion is understood (see section 4 of that survey of Stanley's), so it is very interesting to discover a new example. In particular, all known examples are connected to tableaux and symmetric functions, etc.; whereas this Kreweras walks example has quite a different flavor. Some thoughts: The analogous rotation action on words with only $A$'s and $B$'s (i.e., Dyck words) is well-studied; as explained in section 8 of this survey of Sagan's on the cyclic sieving phenomenon, it corresponds to promotion on $[2]\times[n]$, and in turn to rotation of noncrossing matchings of $[2n]$. There is a way to view a Kreweras walk as a pair of noncrossing partial matchings on $[3n]$ (basically we form the matching corresponding to the $A$'s and $C$'s, and to the $A$'s and $B$'s). But this visualization does not seem to illuminate anything about the rotation action (in particular, when we rotate a walk, one of the noncrossing partial matchings simply rotates, but something complicated happens to the other one). As mentioned earlier, there is a bijection due to Bernardi between Kreweras walks and decorated cubic maps, but I am not able to see any simple way that this bijection interacts with rotation. On a positive note, it seems useful to write the $3n$ rotations of a Kreweras walk in a cylindrical array where we indent by one each row, as follows: $$ \begin{array} \, A & A & B & b & C & A & C & C & B \\ & A & b & A & C & A & C & C & B & B \\ & & A & A & C & A & C & c & B & B & B \\ & & & A & c & A & C & A & B & B & B & C \\ & & & & A & A & C & A & B & B & b & C & C \\ & & & & & A & c & A & B & B & A & C & C & B \\ & & & & & & A & A & B & b & A & C & C & B & C \\ & & & & & & & A & b & A & A & C & C & B & C & B \\ & & & & & & & & A & A & A & C & C & B & c & B & B \\ & & & & & & & & & A & A & C & C & B & A & B & B & C \end{array} $$ In each row I've made lowercase the $B$ or $C$ that the initial $A$ swaps with. We can extract from this array a permutation which records the columns in which these matches occur (where we cylindrically identify column $3n+i$ with column $i$). In this example, the permutations we get is $p = [4,3,8,5,11,7,10,9,15] = [4,3,8,5,2,7,1,9,6]$. The fact that this list of columns is actually a permutation (which I don't know how to show) is equivalent to the assertion that the position of the $A$'s after $3n$ rotations is the same as in the initial Kreweras walk. Furthermore, this permutation $p$ determines position of $A$'s. Namely, the $A$'s are exactly the $p(i)$ for which $p(i) < i$. In our example, these are $2$, $1$, and $6$, corresponding to $i = 5,7,9$. Also, you can see how the $3n$ rotations "permute" the position of the $A$'s from $p$ as well. To do that, write down a new permutation $q$ from $p$: $q$ is the product of transpositions $q = (3n, p(3n)) \cdots (2, p(2)) \cdot (1, p(1))$. Then $q$ exactly tells us how the $A$'s are permuted. In our example, as we process the transpositions of $q = (9,6)(8,9)(7,1)(6,7)(5,2)(4,5)(3,8)(2,3)(1,4)$ right-to-left on the positions $\{1,2,6\}$ of $A$'s we see $1 \to 4 \to 5 \to 2$; $2 \to 3 \to 8 \to 9 \to 6$; and $6 \to 7 \to 1$. Note that the $A$'s end up changing places, and that they each are involved in a different number of swaps. Another thing worth noting is that the permutation $p$ does not determine the Kreweras walk (even after accounting for the $B \leftrightarrow C$ symmetry). In spite of these observations, the lack of any connection to algebra (e.g., the representation theory of Lie algebras), and the lack of any good "model" for these words, makes it really hard to reason about how they behave under rotation. EDIT: Let me add one example which may indicate some subtlety. Let's define a $k$-letter Kreweras word of length $kn$ to be a word consisting of $n$ A's, $n$ B's, $n$ C's, $n$ D's, etc. for $k$ different letters such that in any prefix there are at least as many $A$'s as $B$'s, at least as many $A$'s as $C$'s, at least as many $A$'s as $D$'s, etc. So $3$-letter Kreweras words are the Kreweras walks discussed above, and $2$-letter Kreweras words are the Dyck words. We can define rotation for $k$-letter Kreweras words in exactly the same manner: literally rotate the word, find the first place the inequalities are violated, swap this place with the final $A$ to obtain a valid word (and this corresponds to promotion on a certain poset). For the case $k=2$, note that $kn$ applications of rotation to a $k$-letter Kreweras word of length $kn$ results in a word with the $A$'s in the same position (because this is just rotation of noncrossing matchings). For the case $k=3$, apparently $kn$ applications of rotation results in a word with the $A$'s in the same position (because apparently the $B$'s and $C$'s switch). But for $k > 3$, it is not true necessarily that $kn$ applications of rotation results in a word with the $A$'s in the same position. For instance, with $k=4$ and $n=3$, starting from the word $w=AADACCDCBDBB$, 12 rotations gives us: $$ 00 \; AADACCDCBDBB \\ 01 \; ADACCDABDBBC \\ 02 \; AACCDABDBBCD \\ 03 \; ACADABDBBCDC \\ 04 \; AADABDBBCDCC \\ 05 \; ADABDBACDCCB \\ 06 \; AABDBACDCCBD \\ 07 \; ABDAACDCCBDB \\ 08 \; ADAACDCCBDBB \\ 09 \; AAACDCABDBBD \\ 10 \; AACDCABDBBDC \\ 11 \; ACDAABDBBDCC \\ 12 \; ADAABDBBDCCC $$ where the $A$'s do not end up in the same positions they started in. So something kind of subtle has to be happening in the case $k=3$ to explain why they do. REPLY [6 votes]: Martin Rubey and I solved my conjecture. The basic idea of the proof is as follows. First to a Kreweras word $w$ we associate what we call its bump diagram, which is just the union of the two noncrossing partial matchings of $\{1,2,...,3n\}$ associated to $w$ (the one for the A's and B's and the one for the A's and C's), drawn as a graph in the obvious way. For example, with $w=AABBCACCB$ its bump diagram is We also think of this diagram as a set of ordered pairs ('arcs'); in this example that set is $$\{ (1,4),(1,8),(2,3),(2,5),(6,7),(6,9)\} $$ We extract a permutation $\sigma_w$ of $\{1,2,...,3n\}$ from the bump diagram as follows. For $i=1,2,...,3n$, we take a trip from position $i$. We start traveling from position $i$ along the unique arc ending at $i$ (if $w_i=B$ or $C$), or the "shorter arc" beginning at $i$ (if $w_i=A$), and we continue until we reach a "crossing" of arcs. When we hit a crossing of arcs $(i,k)$ and $(j,\ell)$ with $i \leq j < k < \ell$ (note that we allow the case $i=j$), we follow the following "rules of the road": if we were heading towards $i$, then we turn right and head towards $\ell$; if we were heading towards $\ell$, then we turn left and head towards $i$; otherwise we continue straight to where we were heading. When we finish our trip at position $j$ then we define $\sigma_w(i) := j$. For example, to compute $\sigma_w(3)$: we start traveling along the arc $(2,3)$ heading towards $2$; then we come to the crossing of $(2,3)$ and $(2,5)$ and we turn right, heading towards $5$; then we come to crossing of $(1,4)$ and $(2,5)$ and we turn left, heading towards $1$; then we come to the crossing of $(1,4)$ and $(1,8)$ and we turn right, heading towards $8$; then we come to the crossing of $(1,8)$ and $(6,9)$, but we just continue straight to $8$; and so we finish our trip at $8$. So $\sigma_w(3)=8$. Or to compute $\sigma_w(7)$: we start traveling along the arc $(6,7)$ heading towards $6$; then we come to the crossing of $(6,7)$ and $(6,9)$ and we turn right, heading towards $9$; then we come to the crossing of $(1,8)$ and $(6,9)$ and we turn left, heading towards $1$; and then we come to the crossing of $(1,4)$ and $(1,8)$, but we just continue straight to $1$; and so we finish our trip at $1$. So $\sigma_w(7)=1$. We could compute the whole permutation is $\sigma_w = [4,3,8,5,2,7,1,9,6]$. You might notice that this example $w$ is the same as the original post and that this permutation $\sigma_w$ is the same as the "permutation" $p$ defined in terms of the cylindrical rotation array. Indeed, this always happens (that the trip permutation is the same as the permutation from the cylindrical rotation array). It follows from the main lemma behind the overall proof, which is Lemma. If $w'$ is the rotation of $w$, then $\sigma_{w'} = c^{-1} \sigma_w c$ where $c= (1,2,3,...,3n)$ is the "long cycle." As a remark, these trip permutations come from the theory of plabic graph (cf. Section 13 of Postnikov's paper https://arxiv.org/abs/math/0609764). Since $\sigma_w$ does not completely determine $w$, to finish the proof we need to keep track of a little more data. For that purpose, we define $\varepsilon_w=(\varepsilon_w(1),...,\varepsilon_w(3n))$, a sequence of $3n$ letters which are B's or C's, defined by $$ \varepsilon_w(i) := \begin{cases} w_{\sigma(i)} &\textrm{if $w_{\sigma(i)}\neq A$} \\ w_{\sigma(\sigma(i))} &\textrm{if $w_{\sigma(i)}= A$}. \end{cases} $$ Similarly to the previous lemma, we can show Lemma. If $w'$ is the rotation of $w$, then $\varepsilon_{w'} = (\varepsilon_w(2),\varepsilon_w(3),...,\varepsilon_w(3n),-\varepsilon_w(1))$ with the convention that $-B=C$ and vice-versa. The above lemmas easily imply that the $3n$th rotation of $w$ is obtained from $w$ by swapping B's and C's. Martin and I will post a preprint to the arXiv with all the details soon. EDIT: The paper is now on the arXiv at https://arxiv.org/abs/2005.14031.<|endoftext|> TITLE: A classification of $G_{\delta\sigma}$ zero-dimensional spaces? QUESTION [6 upvotes]: Among separable metrizable spaces: Cantor set is the unique compact zero-dimensional space without isolated points. $\mathbb Q$ is the unique countable space without isolated points $\mathbb R \setminus \mathbb Q$ is the unique zero-dimensional, $G_\delta$-space with no compact neighborhood. $\mathbb Q ^\omega$ is the unique zero- dimensional, first category $F_{\sigma\delta}$-space with the property that no nonempty clopen subset is a $G_{\delta\sigma}$-space. Question. Is there a simple classification of zero-dimensional $G_{\delta\sigma}$-spaces which have no compact neighborhoods? The simplest examples would be $\mathbb Q$, $\mathbb R\setminus \mathbb Q$, and $\mathbb Q\times (\mathbb R\setminus \mathbb Q)$. Are there many others? Is there a nice characterization of $\mathbb Q\times (\mathbb R\setminus \mathbb Q)$? REPLY [5 votes]: This paper by Van Mill from 1981 gives a characterisation of $\Bbb Q \times \Bbb P$ (where $\Bbb P$ is a common notation for the irrationals) in Thm 5.3: If $X$ is separable metrisable and zero-dimensional, $\sigma$-complete and nowhere complete and nowhere $\sigma$-compact then $X \simeq \Bbb Q \times \Bbb P$. Where by complete I mean topologically complete (i.e. in this context: completely metrisable) and a nowhere-$P$ space is one where no non-empty open subset has property $P$, so $\Bbb P$ and $\Bbb Q$ are nowhere locally compact, e.g.) I think $\Bbb Q \times C$ (with $C$ the Cantor set) is another example for your list. A lot of information can be found in van Engelen's PhD-thesis from 1985: homogeneous zero-dimensional absolute Borel sets, where he shows there are are $\omega_1$ many homeomorphism types of subsets of $C$ that are both $F_{\sigma \delta}$ and $G_{\delta\sigma}$, also separately written up here. In his thesis he also gave the first characterisation of $\Bbb Q^\omega$ (now relegated to the appendix of it). Look up Van Engelen's work from around that time for related results.<|endoftext|> TITLE: Differential forms on standard simplices via Whitney extension vs diffeological structure QUESTION [7 upvotes]: The standard simplices $\Delta^n \subset \{\mathbf{x}\in\mathbb{R}^{n+1}\mid x_0 + \ldots + x_n =1 \} =: \mathbb{A}^n$ carry two natural sorts of smooth differential forms: Those differential forms on the interior of $\Delta^n$ that extend smoothly to a neighbourhood of $\Delta^n$ in $\mathbb{A}^n$ (this definition is used for instance by Dupont in his book Curvature and Characteristic Classes). This is implicitly using Whitney's extension theorem, I think. Consider $\Delta^n$ as a diffeological subspace of $\mathbb{A}^n$, where a function $\phi\colon \mathbb{R}^k \to \Delta^n$ is a plot if and only if the composite $\mathbb{R}^k \to \Delta^n \hookrightarrow \mathbb{A}^n$ is smooth. A differential form $\omega$ on $\Delta^n$ then consists of the data of a differential form $\phi^*\omega$ for each plot $\phi$ with a compatibility condition when a plot factors through another via a smooth map between Euclidean spaces. The first of these is more of a "maps out" viewpoint, and probably corresponds to a natural smooth space structure defined via smooth real-valued functions. The second is a "maps in" viewpoint. Note that the $D$-topology arising from the diffeology in 2. above is the standard topology on the simplex. We then get a cochain complex of differential forms of each type, as exterior differentiation can be defined in the more-or-less obvious way in each case. My question is: how do these relate? Is one a subcomplex of the other? Or are they quasi-isomorphic, via a third cochain complex? The motivation is that Dupont's simplicial differential forms on semisimplicial manifolds $X_\bullet$ look like they should be differential forms on the fat geometric realisation considered as a diffeological space, since a simplicial differential form is more or less descent data for the sheaf of differential forms and the 'cover' $\coprod_{n\geq 0} \Delta^n\times X_n \to ||X_\bullet||$, assuming the first definition above. However, if his differential forms on $\Delta^n$ (or more precisely on $\Delta^n\times X_n$) aren't diffeological differential forms, they don't give a form on $||X_\bullet||$ as a diffeological space. I guess all we really need is a map of cochain complexes from the first to the second given above. REPLY [3 votes]: The two chain complexes are isomorphic for any $n≥0$. Fix some $n≥0$, $k≥0$ and consider $k$-forms on the $n$-simplex. I will use the notations $Ω_e^k$ and $Ω_d^k$ for forms of type 1 and 2 respectively. I also use the notation $Δ_d^n$ for the diffeological $n$-simplex so that $\def\Hom{\mathop{\rm Hom}} \def\R{{\bf R}} \def\d{\,{\rm d}} Ω_d^k=\Hom(Δ_d^n,Ω^k)$. First, we have a canonical map $ι\colon Ω_e^k→Ω_d^k$ that sends $ω∈Ω_e^n$ to the map $Δ_d^n→Ω^k$ that sends a smooth map $φ\colon S→Δ_d^n$ to the $k$-form $φ^*ω∈Ω^k(S)$. Secondly, the map $ι$ is injective, which follows from the following two observations. By the Yoneda lemma the restriction of $ι(ω)\colon Δ_d^n→Ω^k$ to the sheaf represented by the open interior of $Δ_d^n$ equals the restriction of $ω$ to the open interior of $Δ^n$. Furthermore, any two forms in $Ω_e^k$ whose restrictions to the open interior of $Δ^n$ coincide must be equal (by continuity). Thirdly, the map $ι$ is surjective. This can be shown using a two-step construction: in the first step we construct a certain element $ω$ of $Ω_e^k$ starting from some given element $ψ$ of $Ω_d^k$ and in the second step we show that $ι(ω)=ψ$. First, let's briefly examine the simplest nontrivial case $d=n=1$. We have a form $ψ∈Ω_d^1(Δ_d^1)=Ω_d^1([0,1])$, and we already know its restriction $f(x) \d x$ to the open interval $(0,1)$. In particular, $f$ is a smooth function on $(0,1)$. Pulling back $ψ$ along the plot $\R→\R$ ($x↦x^2$) yields some 1-form $g(x) \d x$, where for any $x≠0$ we have $g(x)=2xf(x^2)$ and $g$ is a smooth function on $(-1,1)$. Since $g$ is odd, we have $g(0)=0$, so the even function $h$ given by $h(x)=g(x)/(2x)$ is smooth on $(-1,1)$ and we have $h(x)=f(x^2)$ for all $x≠0$. Set $f(0)=h(0)$. We claim $f$ is smooth on $(-1,1)$. Indeed, $h'(x)=2xf'(x^2)$ for $x≠0$, and since $h'$ is odd, we have $h'(0)=0$ and $h'(x)/(2x)$ is a smooth function on $(-1,1)$ whose value at $x=0$ is $f'(0)$. We repeat this step by induction, proving $f$ has derivatives of all orders at 0. The general case is nothing else than a multivariable paremetrized version of the above argument. More precisely, we argue as follows. For the first step, suppose we are given some $ψ∈Ω_d^k$. Given some point $x∈Δ^n$, we would like to define $ω(x)$. Denote by $d≥0$ the codimension of the stratum of $Δ^n$ that contains $x$. Parametrize the simplex $Δ^n$ using $n$ coordinates in such a way that $x_1=⋯=x_d=0$ and the other $n-d$ coordinates of $x$ are strictly positive, so that the points $y$ of the stratum containing $x$ are defined by the relations $y_i≥0$ for $1≤i≤d$. Decompose $$ω=∑_I g_I ∏_{i∈I} \d x_i$$ into its individual components with respect to this coordinate system. Pick one such component $g_I ∏_{i∈I} \d x_i$; we would like to define $g_I$ as a smooth function on some open neighborhood $U$ of the given stratum of $Δ^n$, which is parametrized by the remaining $n-d$ coordinates $x_{d+1}$, …, $x_n$. Consider the smooth map $U→Δ^n$ that (in the coordinates introduced above) sends $x_i↦x_i^2$ for $1≤i≤d$ and $x_i↦x_i$ for $i>d$. This defines a morphism $τ\colon U→Δ_d^n$, so we have a form $τ^*ψ∈Ω^k(U)$. Now take the coefficient $h$ of $τ^*ψ$ before $∏_{i∈I} dx_i$. Take the partial derivative of $h$ with respect to all coordinates $x_i$ such that $i∈I$ and $i≤d$. Divide the resulting function by $2^{\#(I∩\{1,…,d\})}$. This is the function $g_I$. This argument also proves that the resulting $k$-form $ω$ is smooth. For the second step, we have to show that $ι(ω)=ψ$. Reusing the notation of the previous paragraph, consider some arbitrary plot $φ\colon S→Δ_d^n$ such that $φ(s)=x$ for some $s∈S$. The first $d$ coordinates of $φ$ must be nonnegative in a neighborhood $U$ of $s$. Since $φ$ is differentiable, the first derivatives of these $d$ coordinates must vanish at the point $s$. Thus, taking the square root of each of the first $d$ coordinates produces a smooth map $ε\colon U→\R^n$ such that $φ=τε$, where the map $τ$ was defined in the previous paragraph. Now $φ^*ι(ω) = ε^* τ^* ι(ω) = ε^* τ^*ψ = φ^*ψ$, where $τ^*ι(ω)=τ^*ψ$ by definition of $ω$.<|endoftext|> TITLE: Empty interior of union of cosets? QUESTION [5 upvotes]: The following question arises from trying to understand Lemma 1.3(ii) of arXiv:math/0405063. I believe a particular case of the proof (and in fact I think the proof is essentially equivalent to this claim) is: Let $G$ be a locally compact group. Let $C,D$ be cosets (not assumed open, closed etc.) each of which has empty interior. Then $C\cup D$ also has empty interior. This is not try in general topology, of course: let $C,D$ be the rational, respectively, irrationals, in $\mathbb R$. However, I cannot decide if being a coset rules out this sort of example. Is the claim true, and if so, what is a proof? REPLY [4 votes]: This is false. Take the (compact abelian) group $G=(\mathbf{Z}/2\mathbf{Z})^\mathbf{N}$ and let $H$ be a dense subgroup of index 2 (there are many, since $G$ has only countably many closed subgroups of index 2 but has $2^c$ subgroups of index $2$, and clearly a subgroup of index 2 is either closed or dense). Then $G=H\cup (G\smallsetminus H)$ and both $H$ and its coset $G\smallsetminus H$ have empty interior.<|endoftext|> TITLE: Scattering theory for Coulomb potential QUESTION [5 upvotes]: Both physical and mathematical theories of quantum scattering seem to be well developed in the case when the potential (or a more general perturbation of the Laplacian) decays fast enough at infinity and satsifies some regularity assumptions. For example under such rather general assumptions existence of $S$-matrix is proved in Ch. XIV of Hormander's "The Analysis of Linear Partial Differential Operators II". However the Coulomb potential does not satisfy these assumptions of fast decay at infinity. Moreover my impression is that this is not just a technical issue, but the whole theory should look different for the Coulomb potential. I am wondering if there is any notion of $S$-matrix for the Coulomb potential and how it looks like. A reference would be helpful. Apologies if the question is not advanced enough- I am not an expert. REPLY [5 votes]: The $1/r$ Coulomb potential needs to be regularized, typically this is done by studying the Yukawa potential $e^{-\alpha r}/r$ and taking the limit $\alpha\rightarrow 0$ at the end. A recent critical examination of this procedure can be found in Regularization of the Coulomb scattering problem (2004).<|endoftext|> TITLE: Maximizing $\iiint|(x-z)\times(y-z)|d\mu d\mu d\mu$ over probability measures on the unit circle QUESTION [5 upvotes]: What probability measure(s) maximize the quantity $\iiint_{\mathbb{S}^1}|(x-z)\times(y-z)|d\mu(x)d\mu(y)d\mu(z)$? The answer appears to be uniform measure, since informally it appears better to have more triangles in the support of $\mu$ which the function $|(x-z)\times(y-z)|$ computes the area of. Is there an argument that shows this is true generally for $\mathbb{S}^{d-1}$? Edit: Thanks to @fedja for the answer to the main question and the request for clarification. While it could be possible to replace the area of triangles with volumes of simplices (and so on), what was meant by generalization was most immediately the question of whether uniform measure maximizes $$\iiint_{\mathbb{S}^2}|(x-z)\times(y-z)|d\mu(x)d\mu(y)d\mu(z).$$ REPLY [6 votes]: Yes, it is true for the circle (though the reason is not quite the one you suggested). We shall consider the discrete version of the problem, which is to put some odd number $n\ge 3$ of points (think of them as of being assigned the probability of $1/n$ each) on the circle so that the sum of triangle areas is maximized. Then an optimal configuration exists by compactness. We just want to show that it should be an equispaced distribution. Since you can obtain any measure on the circle as a weak limit of such discrete measures (and the integrand is continuous, so a small displacement of the measure does not change the integral much), you'll get the required optimality of the uniform measure as a limiting statement. Take any of the points $e$ and enumerate all other points by the numbers $1,\dots,n-1$ clockwise starting from $e$ (so the whole system is $e,e_1,\dots,e_{n-1}$. Fix all points except $e$ and think a bit of how the total double area depends on the position of $e$. You'll realize quite soon that it is just some constant plus $v_e\times e$ where $$v_e=\sum_{1\le k< m \le n-1}(e_m-e_k).$$ If you now consider for fixed $\ell<\frac n2$ all vectors $e_m-e_k$ with [$(\ell\le k< m\le n-\ell$) and ($k=\ell$ or $m=n-\ell$)], you'll see that $v_e=\sum_{\ell<\frac n2}A_\ell(e_{n-\ell}-e_\ell)$ with some $A_\ell>0$. You will be able to increase the sum of (even oriented) areas if $e$ is not perpendicular to $v_e$, so in the optimal configuration we must always have $\langle e,v_e\rangle=0$ for every choice of $e$. Now recall the famous "drive around the circle without running out of gas" problem (more precisely, its particular case when all gas stations have the same amount of fuel sufficient for driving $\frac 1n$ of the entire circle). The conclusion of it is that you can choose $e$ so that the arc distance to $e_k$ is at most $\frac k n$ of the whole circle for every $k=1,\dots,n-1$, which makes all scalar products $\langle e,e_{n-\ell}-e_\ell\rangle$ with $\ell<\frac n2$ non-positive with the only chance of the equality $\langle e,v_e\rangle=0$ when the arc distance from $e$ to $e_k$ is exactly $\frac kn$ for all $k=1,\dots,n-1$. As to higher dimensions, can you, please, specify first what you mean by "this"? (there are several ways to try to generalize the setup, so the word seems rather ambiguous to me :-) )<|endoftext|> TITLE: Is it possible to multiply two series to get as a result all composite numbers? QUESTION [6 upvotes]: I was toying with the following problem: Is it possible to find two infinite integer sequences $(a_n), (b_n)>0$ such that $\sum_{n=1}^{\infty}\frac{1}{(a_n)^s}\cdot \sum_{n=1}^{\infty}\frac{1}{(b_n)^s}=\sum_{n=1}^{\infty}\frac{1}{(c_n)^s}$ for every $s>1$? Here $c_n$ denotes the $n$-th composite number. I can show that without loss of generality, $1\in a_n$ and for every $x$ with $\Omega(x)=2, x\in (b_n)$ but this did not help much. Can someone provide an answer to this problem? REPLY [6 votes]: This is the answer of Greg Martin, with the correction of Mark Sapir, and details added. Write $\Omega(n)$ for the number of prime factors of $n$ (counted with multiplicity), and $\Omega_{\operatorname{odd}}(n)$ for the number of odd prime factors of $n$ (counted with multiplicity), so $\Omega(n) = \Omega_{\operatorname{odd}}(n) + v_2(n)$ (where $v_2$ is the valuation at $2$). Example. Let $A = \{1,2,4,8,\ldots\} = 2^{\mathbf Z_{\geq 0}}$, and let $$B = \{n\ |\ \Omega(n) = 2\} \cup \{n \text{ odd}\ |\ \Omega(n) \geq 3\}.$$ For $n \in \mathbf Z_{>0}$, the number of representations $n = a \cdot b$ with $a \in A$ and $b \in B$ is $1$ if $n$ is composite, and $0$ otherwise. Proof. Given $n \in \mathbf{Z}_{>0}$ composite (i.e. $\Omega(n) \geq 2$), define $k \in \mathbf Z_{\geq 0}$ as follows: If $\Omega_{\operatorname{odd}}(n) \geq 2$, set $k = v_2(n)$. If $\Omega_{\operatorname{odd}}(n) = 1$, set $k = v_2(n) - 1$. If $\Omega_{\operatorname{odd}}(n) = 0$, set $k = v_2(n) - 2$. In cases 2 and 3, note that $k \geq 0$ since $\Omega(n) \geq 2$. Then set $a = 2^k$ and $b = \tfrac{n}{a}$. Then $n = a \cdot b$, and clearly $a \in A$. We also have $b \in B$: In case 1 above, $b$ is odd with $\Omega(b) \geq 2$; In case 2 above, $b$ is even with $\Omega(b) = 2$; In case 3 above, $b = 4$. This shows existence of the desired decomposition. For uniqueness, assume $n = a \cdot b$ with $a \in A$ and $b \in B$. Let $m = v_2(n)$. Then $\Omega_{\operatorname{odd}}(b) = \Omega_{\operatorname{odd}}(n)$, so If $\Omega_{\operatorname{odd}}(n) \geq 2$, then $\Omega_{\operatorname{odd}}(b) \geq 2$, which by definition of $B$ forces $b$ odd, hence $a = 2^m$. If $\Omega_{\operatorname{odd}}(n) = 1$, then $\Omega_{\operatorname{odd}}(b) = 1$, which by definition of $B$ forces $b$ even and $\Omega(b) = 2$, hence $a = 2^{m-1}$. If $\Omega_{\operatorname{odd}}(n) = 0$, then $\Omega_{\operatorname{odd}}(b) = 0$, which by definition of $B$ forces $b = 4$, hence $a = 2^{m-2}$. This shows that $(a,b)$ must be as constructed above. Finally, since all elements of $B$ are composite, any integer of the form $n = a \cdot b$ with $a \in A$ and $b \in B$ is composite. $\square$<|endoftext|> TITLE: Is the complement of a zero-dimensional subset of the plane path-connected? QUESTION [10 upvotes]: Let $X$ be a zero-dimensional subset of the plane $\mathbb R ^2$. Is $\mathbb R ^2\setminus X$ necessarily path-connected? I feel the answer must be yes but I need a reference. If it helps, assume $X$ is nowhere dense. REPLY [14 votes]: If the zero-dimensional set $X$ is not closed, then the answer is "no". To construct a suitable example, take any open bounded neighborhood $U\subset\mathbb R^2$ of zero, whose boundary $\partial U$ does not contain a topological copy of $[0,1]$. For example, for $U$ we can take a bounded connected component of the complement of the union of two suitable pseudoarcs in the plane. Then the set $$X=\mathbb R^2\setminus\{\vec a+\tfrac1n\partial U:\vec a\in\mathbb Q^2,\;n\in\mathbb N\}$$ will have the desired property: it is zero-dimensional and its complement $\mathbb R^2\setminus X$ does not contain a copy of $[0,1]$ (by the Baire Theorem) and hence is not path-connected.<|endoftext|> TITLE: A new cardinal characteristic (related to partitions)? QUESTION [11 upvotes]: In this post I will discuss some cardinal characteristic of the continuum, related to partitions of $\omega$ and would like to know if it is equal to some known cardinal characteristic. By a partition of $\omega$ I understand a cover of $\omega$ by pairwise disjoint nonempty subsets. A partition $\mathcal P$ is called finitary if $\sup_{P\in\mathcal P}|P|$ is finite. A family $\mathfrak P$ of partitions of $\omega$ is called directed if for any two partitions $\mathcal A,\mathcal B\in\mathfrak P$ there exists a partition $\mathcal C\in\mathfrak P$ such that each set $S\in\mathcal A\cup\mathcal B$ is contained in some set $C\in\mathcal C$. Let $\mathfrak P$ is a family of partitions of $\omega$. An infinite subset $D\subset\omega$ is called $\mathfrak P$-discrete if for any partition $\mathcal P\in\mathfrak P$ there exists a finite set $F\subset D$ such that for any $P\in\mathcal P$ the intersection $P\cap (D\setminus F)$ contains at most one point. Let $\kappa$ be the smallest cardinality of a directed family $\mathfrak P$ of finitary partitions of $\omega$ admitting no infinite $\mathfrak P$-discrete set $D\subset\omega$. It can be shown that $\mathfrak b\le\kappa\le\mathfrak c$ (the upper bound follows from the observation that any maximal directed family of finitary partitions has no infinite discrete set, see Proposition 6.5 in this preprint). Problem 1. Is $\kappa$ equal to some known cardinal characteristic of the continuum? Problem 2. Is $\kappa=\mathfrak c$ in ZFC? Problem 3. Find lower and upper bounds on $\kappa$ (which are better than $\mathfrak b\le\kappa\le\mathfrak c$). Added in Edit. The lower bound $\sup_{U\in\beta\omega}\pi(U)\le\kappa$, suggested by Todd Eisworth can be improved to $\mathfrak s\le \kappa$. One can also prove that $\max\{\mathfrak b,\mathfrak s,\}\le\mathfrak j\le\kappa\le\mathrm{non}(\mathcal M)$ and hence $\kappa$ is not equal to $\mathfrak c$. The cardinal $\mathfrak j$ is discussed in this MO-post. REPLY [7 votes]: This is not an answer, but hopefully it's a helpful observation: (1) If $U$ is an ultrafilter on $\omega$ and $\mathcal{P}$ is a finitary partition of $\omega$, then there is $A\in U$ such that $A\cap P$ contains at most one element for each $P\in\mathcal{P}$. (As if each piece of the partition has cardinality at most $n$, then there is a $k\leq n$ such that the union of pieces with size exactly $k$ is in $U$. Now split this union up into $k$ pieces in the obvious way, and one of these is in $U$.) (2) Given an ultrafilter $U$, let $\tau(U)$ be the least cardinal $\tau$ such that some subfamily of $U$ of cardinality $\tau$ fails to have an infinite pseudo-intersection. (We do not require the pseudo-intersection to be in $U$, so $\aleph_1\leq\tau(U)\leq\mathfrak{c}$.) Observation: If $U$ is an ultrafilter on $\omega$, then $\tau(U)\leq\kappa$. Proof. Given a family $\mathfrak{P}$ of finitary partitions of $\omega$ (directed or not), we fix for each $P\in\mathfrak{P}$ a set $A_P\in U$ meeting each element of $P$ in at most one point. If $|\mathfrak{P}|<\tau(U)$ then we can find an infinite pseudo-intersection $X$ for the collection $\{A_P:P\in\mathfrak{P}\}$, and $X$ is $\mathfrak{P}$-discrete.$_\square$ I don't know anything about the cardinals $\tau(U)$. I note that at one point Blass and Shelah claimed to have model containing both simple $P_{\aleph_1}$ and simple $P_{\aleph_2}$ points, but Alan Dow discovered an error in the paper, and I'm not sure if it has ever been repaired. (The existence of a simple $P_{\aleph_1}$-point implies $\mathfrak{b}=\mathfrak{u}=\aleph_1$, while the simple $P_{\aleph_2}$ point is an ultrafilter $U$ with $\tau(U)=\aleph_2$. In such a model, $\kappa$ would be strictly greater than $\mathfrak{b}$.) Clearly this is all tied up with the topology of $\beta\omega$, so I suspect much more is known by the experts.<|endoftext|> TITLE: Cutting the unit square into pieces with rational length sides QUESTION [5 upvotes]: The following questions seem related to the still open question whether there is a point(s) whose distances from the 4 corners of a unit square are all rational. To cut a unit square into n (a finite number) triangles with all sides of rational length. For which values of n can it be done if at all? To cut a unit square into n right triangles with all sides of rational length. For which values of n can it be done, if at all? Remark: If one can find a finite set of 'Pythagorean rectangles' (rectangles whose sides and diagonal are all integers) that together tile some square (of integer side), that would answer this question. 3.To cut a unit square into n isosceles triangles with all sides of rational length. For which values of n can it be done, if at all? Now, one can add the requirement of mutual non-congruence of all pieces to all these questions. Further, one can demand rationality of area of pieces or replace the unit square with other shapes (including asking for a triangulation of the entire plane into mutually non-congruent triangles all with finite length rational length sides)... Note: From what has been shown by Yaakov Baruch in the discussion below, cutting the unit square into mutually non-congruent rational sided-right triangles can be done for all n>=4. Indeed, he has shown n=4 explicitly; for higher n, one can go from m non-congruent pieces to m+1 pieces by recursively cutting any of the m right triangular pieces n by joining its right angle to the hypotenuse to cut it into two smaller and mutually similar but non-congruent pieces. That basically settles questions 1 and 2 - the non-congruent pieces case. However, if we need all pieces to be non-congruent and non-similar, the n=4 answer has no obvious generalization to higher n. References: 1. https://nandacumar.blogspot.com/2016/06/non-congruent-tiling-ongoing-story.html?m=1 On dissecting a triangle into another triangle REPLY [5 votes]: @YaakovBaruch's beautiful construction for $n=4$:           Edit by Yaakov Baruch. A partition with 8 isoceles triangles (I'm sure not minimal):<|endoftext|> TITLE: Which plane curves can be harmonically parametrized? QUESTION [5 upvotes]: In this question, a “(closed oriented plane) curve” $\Gamma$ will mean a continuous map $f \colon \mathbb{U} \to \mathbb{C}$ where $\mathbb{U} := \{z\in\mathbb{C} : |z|=1\}$ is the unit circle, modulo right-composition by orientation-preserving homeomorphisms $\varphi\colon\mathbb{U}\to\mathbb{U}$. Any such $f$ (i.e. any of the $f\circ\varphi$ defining the curve) will be called a “parametrization” of the curve $\Gamma$. I'm willing to add any reasonable regularity conditions on $\Gamma$ if they help in answering the question, e.g., piecewise $C^1$ or even $C^\infty$. Say that a curve $\Gamma$ is harmonically parametrizable iff it admits a parametrization $f \colon \mathbb{U} \to \mathbb{C}$ (see above) such that the Fourier coefficients $(c_k(f))_{k\in\mathbb{Z}}$ of $f$ are zero for all $k<0$: this then allows us to see $f$ as the restriction to $\mathbb{U} = \partial\Delta$ of a continuous function $F$ on the closed unit disk $\overline{\Delta}$ that is holomorphic on the open unit disk $\Delta$ (namely, $F$ is the Poisson integral of $f$, or equivalently $F(z) = \sum_{k=0}^{+\infty} c_k z^k$; see also this question). More generally, if $k_0\in\mathbb{Z}$ say that $\Gamma$ is $k_0$-harmonically parametrizable iff it admits a parametrization $f \colon \mathbb{U} \to \mathbb{C}$ that $c_k(f)=0$ for all $k0$, being $k_0$-harmonically parametrizable seems to say something about the moments of $\Gamma$ for the harmonic measure vanishing, and this is probably less interesting.) PS: This question seems related to a kind of converse to the one I'm asking (I want to know if there exists a parametrization change which becomes harmonic, that other question is about what happens upon such a change). REPLY [2 votes]: First of all, this is a purely topological problem. Let $\gamma:U\to C$ be a curve. Your question is when this curve can be reparameterized so that the new parameterization is by boundary values of an analytic function in the unit disk. It is necessary that $\gamma$ extends to a topologically holomorphic map of the unit disk to $C$ (topologically holomorphic map, a. k. a. polymersion, is a map which is topologically equivalent to $z\mapsto z^n$ near every point. This condition is also sufficient. Indeed, once we have a topologically holomorphic map, we can pull back the complex analytic structure to the disk, and then use the Uniformization theorem. So for every topologically holomorphic map $f$ there is a homeomorphism $\phi$ of the disk such that $f\circ\phi$ is holomorphic. Second, this topological problem is solved in the paper (under some smoothness conditions on the curve): C. Curley and D. Wolitzer, Btranched immersions of surfaces, Michigan Math. J. 33 (1986) 131-144. Unfortunately the answer is somewhat complicated and I do not reproduce it here.<|endoftext|> TITLE: RSK correspondence QUESTION [8 upvotes]: Up to now, what are the difference ways we know to define RSK correspondence? I already know: By insertion and recording tableau. Ball construction or Viennot's geometric construction. Growth diagram proposed by Sergey Fomin. Do you know other models? REPLY [4 votes]: Here, slightly edited, is the first paragraph of Steinberg's paper, An occurrence of the Robinson–Schensted correspondence. Let $V$ be an $n$-dimensional vector space over an infinite field, $\mathscr F$ the flag manifold of $V$, $u$ a unipotent transformation of $V$, and $\lambda$ the type of $u$, a partition of $n$ whose parts are the sizes of the Jordan blocks for $u$. … The components of $\mathscr F_u$, the variety of flags fixed by $u$, correspond naturally to the standard tableaux of shape $\lambda$. The purpose of this note is to show that the "relative position" of any two components of $\mathscr F_u$ (in general an element of the Weyl group, in the present case an element of $S_n$) is given, in terms of the corresponding tableaux, by the Robinson–Schensted correspondence.<|endoftext|> TITLE: Asymptotic behavior of a certain sum of ratios of consecutives primes QUESTION [10 upvotes]: I am looking for the asymptotic growth of the following sum $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}$$ where $p_k$ stands for the prime of index $k$. Manual computations show, for small values of n, a behavior quite similar to that of the sum over naturals $$\sum_{k=0}^{n-1}(2k+1)=n^2$$ But more accurate simulations with Python suggest that $\sum_{k=1}^{n}\frac{p_{k+1}+\,p_k}{p_{k+1}-\,p_k}$ ~ $\frac{2}{e}\,n^2\log\log n$ Is there anybody who can confirm this asymptotic behavior and, if it is correct, give a sketch of a proof? REPLY [10 votes]: It is elementary to prove that the sum grows at least as fast as $n^2$, and at most as fast as $n^2\log n$. The precise asymptotic behavior depends on the distribution of prime gaps $p_{k+1}-p_k$, on which we only have conjectures (see also my Added section below). It is clear that $$\#\{k\leq n: p_{k+1}-p_k>\log n\}<\frac{p_{n+1}}{\log n},$$ hence the contribution of $p_{k+1}-p_k>\log n$ is $$\sum_{\substack{1\leq k\leq n\\p_{k+1}-p_k>\log n}}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}<\frac{p_{n+1}}{\log n}\cdot\frac{p_{n+1}+p_n}{\log n}=O(n^2).$$ Now, for a fixed even $h\leq\log n$, Hardy and Littlewood conjectured that $$\#\{k\leq n: p_{k+1}-p_k=h\}\sim\frac{n}{\log n}\cdot 2C_2\cdot D_h,\tag{$\ast$}$$ where $$C_2:=\prod_{p>2}\left(1-\frac{1}{(p-1)^2}\right)=0.66016\dots\qquad\text{and}\qquad D_h:=\prod_{\substack{p|h\\{p>2}}}\frac{p-1}{p-2}.$$ If we believe in this, then integration by parts gives that the contribution of $p_{k+1}-p_k=h$ is asymptotically $n^2 C_2 D_h/h$. Based on this heuristic, it is reasonable to conjecture that $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C_2\, n^2\sum_{h\leq\log n}\frac{D_h}{h}.$$ It is straightforward that the Dirichlet series of $D_h$ factors as $$\sum_{h=1}^\infty\frac{D_h}{h^s}=\zeta(s)F(s),$$ where $F(s)$ is an explicit Euler product converging uniformly in $\Re(s)>3/4$, say. Therefore, heuristically, $$\sum_{k=1}^{n}\frac{p_{k+1}+p_k}{p_{k+1}-p_k}\sim C\, n^2\log\log n,$$ where $C:=C_2F(1)$. Most likely, the constant $C$ is not equal to $2/e$ as suggested by the original post, but I have not checked this. Added. It think that the known upper bounds for the left hand side of $(\ast)$ allow one to show, unconditionally, that the sum in question is $O(n^2\log\log n)$. As Lucia kindly pointed out, a result of Gallagher's implies that $C=1$.<|endoftext|> TITLE: Importance of textbooks in health of a sub-discipline QUESTION [18 upvotes]: I am interested in published articles, and also more informal writing (blog posts, talk slides etc.) which discuss the importance of textbooks (where this word encompasses research monographs etc.) in the long-term health of a sub-discipline in Mathematics. Motivation: I have been thinking of late about how large Mathematics is getting (compared to, say, 50-60 years ago) with many more mathematicians and many more published papers. It also seems to me that at least some areas are becoming increasingly technical. Furthermore, it can be very hard to follow a field by just reading the original papers: there can be false steps, or incomplete results, which only reach final form after some attempts. Often the pressures of space mean that motivation, or background material, is omitted in articles. A good textbook can solve all of these problems. It seems to me that especially graduate students, or more established mathematicians seeking to move field, or use results of a different field to their own, face these sorts of problems in the extreme. By contrast, people working in the field probably carry around a lot of the "missing content" in their heads. This then makes me wonder if the lack of textbooks might lead to an ever increasing barrier to entry, and perhaps to sub-disciplines dying out as younger/newer mathematicians do not take up the study. Hence my question of whether these thoughts have been stated in a longer, more thought out way before. REPLY [10 votes]: The following is purely my opinion, but the specific question that you posed seems to invite such answers. Strictly speaking, I'd say that the answer to your question is "Yes", a lack of textbooks in a new area is a barrier to entry. However, my experience over the past 30+ years is that areas don't stay "new" for long, and as things become more solidified, people write introductory (graduate level) textbooks. This may be to promote their vision, or it may simply be because they find the subject beautiful and want to share that beauty with others. To take an older example, Grothendieck's revolution in algebraic geometry created a large barrier for entry, but Hartshorne's book appeared when I was in graduate school, and it provided a way in. Was it perfect? No. Have other books, possibly better introductions, appeared since. Sure. But it was there, and I think it's fair to say that it helped train a generation (or more) of algebraic geoemters and those in allied fields. (I'm in the latter group.) So "yes", lack of graduate level texts in a area is a barrier to entry. But is it a long-term problem. I'd suggest that the answer is "No", because a new area of mathematics that's thriving tends to acquire such textbooks.<|endoftext|> TITLE: Logical completeness of Hilbert system of axioms QUESTION [10 upvotes]: This is really a question about references. The entry in Russian Wikipedia about Hilbert's axioms states, in particular, that completeness of Hilbert's system was proven by Tarski in 1951. The reference is to the Russian encyclopedia of elementary mathematics, to which I don't have access, and I somehow am not able to find any references to this statement in the literature. I have two questions: The continuity axioms are not first-order-logic statements. What does then completeness mean in this situation? Does anybody know a reference to the original proof by Tarski or any other proof of this statement, for that matter? REPLY [13 votes]: The original is Alfred Tarski's book "The completeness of elementary algebra and geometry", which was due to appear in 1940 but never made it into print because of the outbreak of WW2. An edition appeared after all in 1967 (Institut Blaise Pascal, Paris), but is not easy to come by. Essentially the same argument is presented in Tarski's 1948/51 "A decision method for elementary algebra and geometry", available here. Tarski uses an axiomatic setup of elementary geometry different (but equivalent) to Hilbert's, using only points (not lines or planes) as primitive terms, and two relation symbols, $B$ and $D$ (ternary and quaternary, respectively). $Bxyz$ signifies that the point $y$ is on the "line" $xz$ between the points $x$ and $z$ ("betweenness"), while $Dxyzw$ means that the distance between $x$ and $y$ equals that between $z$ and $w$ (the "equidistance" relation). Here, $x$, $y$, $z$ (and $w$) are allowed to coincide. Full continuity is the second order axiom $\forall_{X}\forall_{Y}((\exists_{a}\forall_{x\in X}\forall_{y\in Y}Baxy)\rightarrow(\exists_{b}\forall_{x\in X}\forall_{y\in Y}Bxby))$, where $X$ and $Y$ are variables ranging over sets of points. First order continuity is weaker, and is expressed as an axiom schema where $X$ and $Y$ are given as $\lbrace x\mid \phi(x)\rbrace$ and $\lbrace y\mid \psi(y)\rbrace$, respectively, for arbitrary first order formulas $\phi(x)=\phi(x,p_{1},\cdots,p_{n})$ and $\psi(y)=\psi(y,p_{1},\cdots,p_{n})$ that are allowed to contain parameters $p_{1},\cdots,p_{n}$. A special case of first order continuity is the Circle Axiom, by which a line that contains an interior point of a circle (in the same plane) must meet that circle. Completeness is the statement that any model $\mathfrak A$ of the axioms of elementary $n$-dimensional geometry without continuity is isomorphic to $K^{n}$ (with the obvious interpretations for the $B$ and $D$ relations) for a Pythagorean ordered field $K$ (that is $K\models\forall_{a}\forall_{b}\exists_{c}(a^{2}+b^{2}=c^{2})$), uniquely determined by $\mathfrak A$ up to isomorphism. Under full continuity, $K$ must be $\mathbb{R}$, under first order continuity $K$ must be real closed, and for the Circle Axiom $K$ merely needs to be Euclidean (i.e., $K\models\forall_{a}\exists_{b}(a=b^{2}\vee -a=b^{2})$). This is the content of the Representation Theorem, Th. I, (16.15) in "Metamathematische Methoden in der Geometrie" by W. Schwabhäuser, W. Szmielew and A. Tarski, Springer Hochschultext, 1983, an excellent reference for the metamathematics of elementary geometry (in German). Edit: Let me add a few comments. The statement above that Tarki's setup is equivalent to Hilbert's is rather imprecise, as noted by Matt F. and others. Tarski works in first order logic, while a formalization of Hilbert's system is at least unclear. (Still, the axioms in Hilbert's axiom groups I-IV can be derived from Tarski's axioms, as shown in the Schwabhäuser, Szmielew, Tarski text). For the same reason, it is not clear what it would mean for Hilbert's system to be complete (in the modern sense), and I do not claim that "completeness" of Hilbert's system follows from that of Tarski. Hilbert includes a "completeness axiom", to the effect that his "model" of the axioms in groups I-V (where V is archimedeanity) cannot be extended to a "model" with a larger universe. To add to the confusion, my use of the word Completeness above (in the body of the answer, in reference to the Representation Theorem) was also unfortunate. Tarski has shown that the first order theory of real closed fields is complete (in the modern sense), and that, as a result, the same goes for the theory of $n$-dimensional elementary geometry (based on Tarski's axioms, with the first order continuity axiom schema included).<|endoftext|> TITLE: Sparse cofinal families QUESTION [7 upvotes]: Let $\kappa$ be an infinite cardinal and as usual denote by $[\kappa]^\mu$ the set of all subsets of $\kappa$ having cardinality $\mu$: Call a family $\mathcal{F} \subset [\kappa]^\omega$ sparse if for all $\mathcal{G} \in [\mathcal{F}]^{\omega_1}$ the set $\bigcup \mathcal{G}$ is uncountable. This is a pretty natural notion that has been rediscovered by various authors and is the combinatorial core of various seemingly unrelated problems in topology, analysis and algebra (see the list of references at the end). Getting a sparse family isn't that much of a deal. The problem is often getting a big one, or more precisely, a cofinal one, with respect to containment. It's easy to see that there is a sparse cofinal family in $([\aleph_n]^\omega, \subseteq)$), for every $n<\omega$. This is a simple consequence of the fact that $cf([\aleph_n]^\omega, \subseteq)=\aleph_n$. QUESTION: Is there (in ZFC) a cardinal $\kappa$ such that $cf([\kappa]^\omega, \subseteq) > \kappa$ and there is a sparse cofinal family on $[\kappa]^\omega$? This cardinal, if it exists, must be larger than $\aleph_\omega$, because the existence of a sparse cofinal family of countable subsets of $\aleph_\omega$ can be proved to be independent of ZFC, modulo very large cardinals. Indeed (see Todorcevic's book or Blass's article for the proofs): If the Chang's Conjecture variant $(\aleph_{\omega+1}, \aleph_\omega) \twoheadrightarrow (\aleph_1, \aleph_0)$ holds then there is no sparse cofinal family of countable subsets of $\aleph_\omega$. The consistency of this Chang's Conjecture variant has been proven from (slightly less than) a 2-huge cardinal by Levinski, Magidor and Shelah. If $\square_{\aleph_\omega}+cf([\aleph_\omega]^\omega, \subseteq)=\aleph_{\omega+1}$ holds then there is a sparse cofinal family of subsets of $\aleph_\omega$. REFERENCES: Blass, Andreas, On the divisible parts of quotient groups, Göbel, Rüdiger (ed.) et al., Abelian group theory and related topics. Conference, August 1-7, 1993, Oberwolfach, Germany. Providence, RI: American Mathematical Society. Contemp. Math. 171, 37-50 (1994). ZBL0823.20057. Kojman, Menachem; Milovich, David; Spadaro, Santi, Noetherian type in topological products, Isr. J. Math. 202, 195-225 (2014). ZBL1302.54008. Peter Nyikos, Generalized Kurepa and MAD families and Topology, preprint. Spadaro, Santi, On two topological cardinal invariants of an order-theoretic flavour, Ann. Pure Appl. Logic 163, No. 12, 1865-1871 (2012). ZBL1269.03047. Todorcevic, Stevo, Walks on ordinals and their characteristics, Progress in Mathematics 263. Basel: Birkhäuser (ISBN 978-3-7643-8528-6/hbk). vi, 324 p. (2007). ZBL1148.03004. REPLY [4 votes]: Your problem is related to the generalized Chang conjectures of the form $(\kappa^+,\kappa) \twoheadrightarrow (\aleph_1,\aleph_0)$. Observation: If $(\kappa^+,\kappa) \twoheadrightarrow (\aleph_1,\aleph_0)$ holds, and if $\mathrm{cf}([\kappa]^{\omega},\subseteq) > \kappa$, then $([\kappa]^{\omega},\subseteq)$ has no sparse cofinal family. So if we'd like to answer your question in the negative (or restrict our search for a positive answer), we should ask when it's possible to have lots of instances of $(\kappa^+,\kappa) \twoheadrightarrow (\aleph_1,\aleph_0)$ holding simultaneously. A paper relevant to this question is "Global Chang's Conjecture and singular cardinals" by Monroe Eskew and Yair Hayut (available here). In it, they articulate the following conjecture: Singular Global Chang's Conjecture: For all infinite $\mu < \kappa$ of the same cofinality, $(\kappa^+,\kappa) \twoheadrightarrow (\mu^+,\mu)$. The word "conjecture" is used here to allude to Chang's Conjecture. No one is suggesting the SGCC as a potential theorem of $\mathsf{ZFC}$, and in fact it's not hard to show that the negation of the SGCC is consistent with $\mathsf{ZFC}$. The suggestion is that the SGCC might be consistent with $\mathsf{ZFC}$, relative to large cardinals. My point is that if this conjecture is consistent, then the answer to your question is no. Based on the observation above, If this conjecture holds, then there are no cardinals $\kappa$ with the property described in your question. In fact, SGCC implies that the answer to your question is no even when $\omega$ is replaced by an arbitrary infinite cardinal $\mu$. So the conjecture gives a conditional answer to your question. In the same paper (Theorem 32), the authors show that a fragment of this conjecture is consistent. This fragment is enough to show, using the observation above, that If $\alpha < \omega_1$, then it is consistent (relative to two supercompact cardinals) that the answer to your question is no for every $\kappa < \aleph_\alpha$. That is, it is consistent that if $\kappa < \aleph_\alpha$ and $\mathrm{cf}([\kappa]^{\omega},\subseteq) > \kappa$, then $[\kappa]^\omega$ has no sparse cofinal family. A proof of the observation above: Recall that $(\kappa^+,\kappa) \twoheadrightarrow (\mu^+,\mu)$ is an abbreviation for the following statement: $\bullet \ \ $ For every model $M$ for a countable language $\mathcal L$ that contains a predicate $A \subseteq M$, if $|M| = \kappa^+$ and $|A| = \kappa$ then there is an elementary submodel $M'$ of $M$ such that $|M'| = \mu^+$ and $|M' \cap A| = \mu$. To prove the observation above, suppose $(\kappa^+,\kappa) \twoheadrightarrow (\aleph_1,\aleph_0)$, and suppose that $\mathrm{cf}([\kappa]^{\omega},\subseteq) > \kappa$. Let $\mathcal C$ be any cofinal family in $[\kappa]^\omega$. Using the Lowenheim-Skolem theorem in the usual way, let $M$ be an elementary submodel of $H(\theta)$ (for some sufficiently big regular cardinal $\theta$) such that $\mathcal C \in M$, $\kappa \subseteq M$, and $|M| = |M \cap \mathcal C| = \kappa^+$. Let $\phi$ be a bijection $M \rightarrow \mathcal C \cap M$. (Note that this $\phi$ cannot be a member of $M$.) Consider the language consisting of a single relation symbol and a single function, and consider the model $$(M,\in,\phi,\kappa)$$ in this language. Applying our hypothesis $(\kappa^+,\kappa) \twoheadrightarrow (\aleph_1,\aleph_0)$, there is some $M' \subset M$ such that $(M',\in,\phi,\kappa \cap M')$ is an elementary substructure of $(M,\in,\phi,\kappa)$ with $|M'| = \aleph_1$ and $|M' \cap \kappa| = \aleph_0$. Let $\mathcal C' = M' \cap \mathcal C$. The restriction of $\phi$ to $M'$ is (by elementarity) a bijection $M' \rightarrow \mathcal C'$. Therefore $|\mathcal C'| = |M'| = \aleph_1$. If $X \in \mathcal C'$, then $X$ is a countable set in $M'$. Because $M'$ is a model of (enough of) $\mathsf{ZFC}$, every countable set in $M'$ is a subset of $M'$. So $X \subseteq M'$. But $X \subseteq \kappa$ as well, so every $X \in \mathcal C'$ is a subset of $M' \cap \kappa$. Therefore $\bigcup \mathcal C' = M' \cap \kappa$. Because $|M' \cap \kappa| = \aleph_0$, this means that $\mathcal C'$ is an uncountable subset of $\mathcal C$ whose union is countable. So $\mathcal C$ is not sparse. Because $\mathcal C$ was an arbitrary cofinal family in $[\kappa]^\omega$, this shows that $[\kappa]^\omega$ does not admit a sparse cofinal family.<|endoftext|> TITLE: Bounds on the lowest dimension of a faithful representation of a finite group QUESTION [12 upvotes]: Let $G$ be a finite group. Can one always construct a faithful representation of $G$ over $\mathbb C$ of dimension $\le \sqrt{|G|}$? What would be the order of the minimal such dimension for a "random" group? Are the worst groups (with the largest minimal faithful dimensions) special in some way? REPLY [10 votes]: I'm writing a paper where I prove that all counterexamples are related to Derek Holt's example: if G is a finite group then either G has a faithful representation over $\mathbb{C}$ of dimension $\leq\sqrt{|G|}$ or $G$ is a $2$-group with center is elementary abelian of order 8 and all irreducible characters of $G$ whose kernel does not contain $Z(G)$ vanish on $G-Z(G)$. For any of these $2$-groups, the minimal dimension of a faithful representation is $\frac{3}{\sqrt{8}}\sqrt{|G|}$. I also prove that this minimal dimension is equal to $\sqrt{|G|}$ if and only if $G$ is a $2$-group with center elementary abelian of order either $4$ or $16$ and all irreducible characters of $G$ whose kernel does not contain $Z(G)$ vanish on $G-Z(G)$. I'd be happy to share my preprint. It would help me to know the motivation for this question.<|endoftext|> TITLE: Lengths of closed geodesics on a flat vs hyperbolic punctured torus QUESTION [5 upvotes]: Let $T$ be a torus (oriented closed surface of genus 1), $p\in T$, and $T^* := T - \{p\}$. Let $\mu$ denote a flat structure on $T$. This can be obtained for example by choosing a uniformization $p_f: T\cong\mathbb{R}^2/\Lambda$ for some lattice $\Lambda$ and descending the standard flat structure on $\mathbb{R}^2$ down to $T$. The flat structure on $T$ restricts to one on $T^*$. Moreover, if we fix an identification $\mathbb{C}\cong\mathbb{R}^2$, this flat structure gives a complex structure on $T^*$. This complex structure yields a holomorphic universal covering $p_h : \mathbb{H}\rightarrow T^*$, such that the deck transformation group consist of hyperbolic isometries of $\mathbb{H}$, and hence this complex structure determines a hyperbolic structure on $T^*$. By a paper of Gutkin and Judge (Affine mappings of translation surfaces), it seems to follow from their Theorem 6.5 that the number of (simple) closed geodesics on a flat (punctured) torus (up to translation) of length bounded by $L$ is quadratic in $L$. If the torus is the unit square torus, the leading term turns out to be $\frac{3}{\pi}L^2$. On the other hand, by a result of McShane and Rivin (A norm on homology of surfaces and counting simple geodesics, as pointed out to me in a previous question), there is a natural bijection between the simple closed geodesics in the flat torus up to translation with those in the corresponding hyperbolic torus (both sets are bijective onto the set of primitive homology classes under the natural map), and that the number of hyperbolic simple closed geodesics of length bounded by L is also quadratic in L. This suggests the flat geodesic representatives of primitive homology classes might have length which is universally proportional to that of the hyperbolic geodesic representative. Could this be true? If not, what is known about the relationship between these lengths under the natural bijection described above? References would be appreciated! REPLY [4 votes]: Yes, this is true. It is shown (either in the paper you cite or the other McShane-Rivin paper) that the length of a simple closed geodesic is quasi-the-same as the combinatorial length ($m+n$) (this is easy, because a simple geodesic stays away from the cusp), and that, in turn, is easily seen to be quasi-the-same as the Euclidean length.<|endoftext|> TITLE: Vector bundles on $\mathbb{A}^n / G$ QUESTION [23 upvotes]: Let $G$ be a finite group acting linearly on $\mathbb{A}^n$. Do we expect algebraic vector bundles on $X := \mathbb{A}^n/G$ to be trivial? Here by the quotient I mean the singular scheme, not the stack quotient. What is known: (1) For finite abelian groups $G$, $X$ is an affine toric variety, and vector bundles on $X$ are trivial by a Theorem of Gubeladze: https://iopscience.iop.org/article/10.1070/SM1989v063n01ABEH003266. In particular, when $G$ is the trivial group, triviality of vector bundles on $\mathbb{A}^n$ is an older result by Quillen-Suslin. (2) Line bundles on $X$ are trivial. This can be shown by lifting a line bundle on $X$ to a $G$-line bundle on $\mathbb{A}^n$; then $\mathrm{Pic}^G(\mathbb{A}^n)$ is the group of characters of $G$, and since the linear $G$-action has a fixed point $0$, this character will be trivial, hence coming from a trivial line bundle on $X$. (2') For vector bundles of higher rank the argument in (2) does not work. This has to do in particular with $G$-actions on $\mathbb{A}^N$ not being linearizable in general: https://link.springer.com/chapter/10.1007%2F978-94-015-8555-2_3 (3) The Grothendieck group of vector bundles is $\mathrm{K}_0(X) = \mathbb{Z}$. We prove it in https://arxiv.org/pdf/1809.10919.pdf, Prop. 2.1 indirectly, using comparison with cdh topology of differential forms. (3') By homotopy invariance of K-groups in the smooth case, $\mathrm{K}^G_0(\mathbb{A}^n) \simeq \mathrm{K}^G_0(\mathrm{Spec}(k))$ which is the Grothendieck ring of $G$-representations; however this does not seem to help. Is there any more evidence for/against the triviality of vector bundles on $X$? Is this question mentioned anywhere in the literature? REPLY [9 votes]: In the paper Affine varieties dominated by $\mathbf{C}^2$ Gurjar considers a slightly more general situation, namely an affine normal variety $\mathrm{X}$ with a proper surjective morphism $\mathbf{A}^2\rightarrow\mathrm{X}$. He shows that every line bundle on $\mathrm{X}$ is trivial; together with a result of Anderson (every vector bundle on $\mathrm{X}$ is the direct sum of a trivial bundle and a line bundle) this shows in particular that every vector bundle on $\mathbf{A}^2/\mathrm{G}$ is trivial.<|endoftext|> TITLE: Derivators and fibred $\infty$-categories QUESTION [11 upvotes]: In his Cohomological methods in intersection theory, Cisinski writes: "[...] note however that, by Balzin’s work [Bal19, Theorem 2], it is clear that one can go back and forth between the language of fibred $\infty$-categories and the one of algebraic derivators." By Bal19 he intends Balzin's Reedy model structures in families. Let me quote the abovementioned Theorem 2. Thm. 2 in Bal19. Let $\mathsf{E} \to \mathsf{R}$ be a left model Reedy fibration. Then the induced $\infty$-functor $\mathsf{L}\text{Sect}(\mathsf{R}, \mathsf{E}) \to \text{Sect}(\mathsf{R}, \mathsf{LE})$ is an equivalence of quasicategories. Even though I sense a connection between derivators, fibred $\infty$-categories and this statement, I cannot make it precise. Could someone help me spelling out what Cisinski means precisely? REPLY [4 votes]: I am no Denis-Charles but given the other paper you quoted let me think of a sketch, perhaps you will be able to make the right out of it. Let $\mathcal E \to \mathcal C$ be a Quillen presheaf (model categorical fibres, Quillen pairs as transition functors) where each $\mathcal E(c)$ is stable. This would imply, if we had the cofibrant generation in fibres for example, that for any functor $F: \mathcal D \to \mathcal C$, the category of sections $Sect(\mathcal D,F^*\mathcal E)$ is a stable model category. You can then introduce a $2$-functor $\mathbb D: (Cat/ \mathcal C)^{op} \to CAT$ that sends $F: \mathcal D \to \mathcal C$ to the homotopy category $Ho (Sect(\mathcal D,F^*\mathcal E))$ and study its properties. Something like that could be called a relative (pre)derivator (maybe it already is, forgive my ignorance, then) valued in triangulated categories. We can also start with the infinity-localisation $L \mathcal E \to \mathcal C$ and study its infinity-categorical sections $Sect(\mathcal D,F^* L \mathcal E)$ for functors $F: \mathcal D \to \mathcal C$. Since a localisation of a stable model category is a stable infinity-category (a statement true even without cofibrant generation due to Proposition 3 mentioned below), $L \mathcal E \to \mathcal C$ is a bicartesian fibration with stable fibres and using Section 5 of HTT (co/limits of sections computed fibrewise) we can verify that each $Sect(\mathcal D,F^* L \mathcal E)$ is stable. Taking their homotopy categories would again yield another $2$-functor $\mathbb D':(Cat/ \mathcal C)^{op} \to CAT$ valued in triangulated categories. Proposition 3 of the Reedy paper means however that $Ho (Sect(\mathcal D,F^*\mathcal E)) \cong Ho (Sect(\mathcal D,F^* L \mathcal E))$ where the latter is the homotopy category of an infinity-category and the former is that of a model category. In other words, the relative derivators $\mathbb D$ and $\mathbb D'$ are equivalent; the equivalence comes from the fairly canonical map $L Sect(\mathcal D,F^*\mathcal E)) \to Sect(\mathcal D,F^* L \mathcal E)$ (universality of localisation is used here) so it should be trackable along base changes. Thus you can study your relative derivators any way you want, via the model-categorical presentation or the infinity-categorical one. I suppose you get an algebraic derivator if you replace $\mathcal C$ by $Sch$ (no restriction on the size of $\mathcal C$ was made above) and consider something like $QCoh \to Sch$ for the Quillen presheaf, but you will have to take it from here to see if that fits. Maybe I should add something about such relative derivators to the Reedy paper if life permits.<|endoftext|> TITLE: Non-constant holomorphic map onto a smooth curve QUESTION [5 upvotes]: Let $\Gamma$ be a smooth projective curve in $\mathbb{P}^2$ and let $U$ be an open neighborhood of $\Gamma$. Denote by $\Gamma_1,\Gamma_2,\ldots,\Gamma_n$ a finite collection of smooth curves intersecting $\Gamma$ transversally. My question is: Does there exist a non-constant holomorphic function $\pi: U \setminus \bigcup\limits_{i=1}^n \Gamma_i \to \Gamma \setminus \bigcup\limits_{i=1}^n \Gamma_i$ ? The situation as in the following picture. I can build a smooth map $\pi$ easily by partition of unity. Intuitively, $\pi$ seems like the tubular neighborhood projection of $\Gamma$ ? But unfortunately, a holomorphic tubular neighborhood does not exist in general as discussed in Is there any holomorphic version of the tubular neighborhood theorem?. And I realized that even the existence of a non-constant holomorphic map $\pi : U \to \Gamma$ (without removing any curve) is not obvious to me. Any comment is welcome. Thanks in advance! REPLY [4 votes]: Let me show that such a map usually doesn't exist (even if we don't remove any additional curves). Consider the case when $\Gamma$ is a curve of degree $\ge 4$. Then one can slightly perturb $\Gamma$ in $\mathbb CP^2$ to a curve $\Gamma'$ that is not isomorphic to $\Gamma$ (so that it stays in a small neighbourhood of $\Gamma$). If the map $\pi$ would exists, we can restrict it to $\Gamma'$, and it will be extendable to the whole $\Gamma'$. So we get a holomorphic (non constant) map $\pi: \Gamma'\to \Gamma$, which is impossible, since $\Gamma$ and $\Gamma'$ are not isomorphic (and by Riemann-Hurwitz for two curves of the same genus $\ge 2$ a non-constant holomorphic map $\Gamma\to \Gamma'$ exits only if the curves are isomorphic). This argument can be easily generalised to curves $\Gamma$ of degree $3$, i.e. cubics. The case of conics and lines should be analysed separately (and for lines $\Gamma$ such a map sometimes exists, for example, when all $\Gamma_i$'s are lines transverse to the line $\Gamma$ and intersecting at one point).<|endoftext|> TITLE: Is $\mathbb{F}_{p}(t)^{h}$ an elementary substructure of/existentially closed in $\mathbb{F}_{p}((t))$? QUESTION [7 upvotes]: It is a well-known fact that the Henselization of the function field $\mathbb{F}_{p}(t)$ in regard to the $t$-adic valuation is $\mathbb{F}_{p}(t)^{alg} \cap \mathbb{F}_{p}((t))$, so of course $\mathbb{F}_{p}(t)^{h}$ embeds into $\mathbb{F}_{p}((t))$, but is it known whether this embedding is elementary in the language of rings $\mathcal{L}_{\text{Ring}}$ or respectively the language of valued fields $\mathcal{L}_{\text{Ring}, \ \mid}$ (this does not matter as the valuation is uniformly definable in henselian valued fields with finite residue field). I would assume that, if this is true, it is not known because we do not know whether $\mathbb{F}_{p}((t))$ is decidable, but do we at least know whether $\mathbb{F}_{p}(t)^{h}$ is existentially closed in $\mathbb{F}_{p}((t))$? This at least should be guaranteed for the perfect hull by Ax-Kochen-Ershov for tame fields. REPLY [3 votes]: There is apparently not a very short answer to this. It is, I think, really not known whether this extension is elementary and it would be very surprising if it was known, since we do not even know if the theory of $\mathbb{F}_{p}((t))$ is decidable. But there is the following result by Franz-Viktor Kuhlmann in his paper "The Algebra and Model Theory of Tame Valued Fields" (https://arxiv.org/pdf/1304.0194.pdf, Theorem 5.9) Let $(K,v)$ be a henselian field. Assume that $(L/K,v)$ is a separable subextension of $(K^{c}/K,v)$. Then $(K,v)$ is existentially closed in $(L,v)$. In particular, every henselian inseparable defectless field is existentially closed in its completion. Here $K^{c}$ stands for the completion in regard to $v$. Now we take a look at $\mathbb{F}_{p}(t)^{h}$, it is obviously henselian. And from that it obviously follows that any henselian valued field is existentially closed in its completion. Of course this does not give a lot of insight on the problem, so let me try and elaborate a little: The proof of Theorem 5.9 makes use of the fact that it is enough to be existentially closed in every finitely generated subfield to be existentially closed. So we take such a subfield $(F,v)$ and go over to its Henselization, which still lies in the completion (because the completion is Henselian). This is of course a subextension hence it is separable. We can see that indeed $(F,v)^{h}=(K(x_{1}, \ldots, x_{n}),v)^{h}$. Then it boils down to show that if we set $K_{i}:=(K(x_{1}, \ldots, x_{i}), v)^{h}$, then $K_{i} \leq_{1} K_{i+1}$, which we can show by realizing that $x$ is the limit of a Cauchy sequence and further results in the linked paper. This is of course a very rough sketch. It is better to read the paper to grasp the proof in detail.<|endoftext|> TITLE: Yet another real-rooted polynomial QUESTION [12 upvotes]: In this entry I asked for the real-rootedness of a polynomial, and two very interesting answers were given: one using Malo's theorem and the other a clever rewriting of the expression using Jacobi Polynomials which are known to be real-rooted. Now, in the middle of a reasoning I had to somehow prove that the following polynomials are real-rooted: $$Q_{m,n}(t) = \sum_{j=0}^m \binom{m}{j}\binom{n}{j} \binom{t-j}{m+n+1}$$ defined for $m\leq n$, integers. Observe that where in the last question one had $t^j$, here one has a $\binom{t-j}{m+n+1}$. Of course, it has been verified numerically for the first small cases and it is true and, in fact, it seems to be true that all the roots are $\leq m+n$. It's possible to prove that many of the roots are integers, which suggests one may "factor out" something like $\binom{t-m}{n+1}$ out of the expression. The problem is that the remaining factor does not seem to be friendly at all. This leads me to the following questions: 1) Any ideas to prove the real-rootedness of this thing? 2) (Much wider and maybe not useful for this concrete problem but interesting on its own) There's a theorem of Schur (see here Theorem 1) that gives a sufficient condition for a polynomial to be real rooted. I want to know if there's any kind of generalization in the case one changes the basis $\{1,x,x^2,\ldots\}$ for example with $\{\binom{x}{0},\binom{x}{1},\binom{x}{2},\ldots\}$. REPLY [13 votes]: First, we write the polynomial in hypergeometric form $$ Q_{m,n}(t)=\frac{(-1)^{m+n+1} \Gamma (m+n-t+1) }{\Gamma (-t) \Gamma (m+n+2)}\, _3F_2\left({-m,-n,m+n-t+1\atop 1,-t};1\right). $$ Applying Thomae's transformation we get (see eq. 1.3 in W. N. Bailey, Contiguous Hypergeometric Functions of the Type 3F2(1)): $$ Q_{m,n}(t)=\frac{(-1)^{n+1} m! \Gamma (m+n-t+1) }{\Gamma (m+n+2) \Gamma (m-t)}\, _3F_2\left({-m,n+1,-m-n+t\atop 1,-m};1\right). $$ Here the polynomial is understood as $\displaystyle{\lim_{\epsilon\to 0}{}_3F_2\left({-m,n+1,-m-n+t\atop 1,-m+\epsilon};1\right) }$, so $-m$ in the numerator and denominator of the hypergeometric function can not be cancelled. The hypergeometric polynomial in the last expression can be expressed in terms of Hahn polynomials, which are discrete orthogonal polynomials (see M. Ismail, Classical and Quantum Orthogonal Polynomials in One Variable): Thus $$ Q_{m,n}(t)=(-1)^{n+1}\frac{ (m-t)_{n+1} }{(m+1)_{n+1}}\,Q_m(m+n-t;0,n-m,m). $$ It is known that zeroes of Hahn polynomials are real and simple. In particular, one can see that some roots are indeed integers due to the factor $(m-t)_{n+1} $, in agreement with OP's observation.<|endoftext|> TITLE: Better trigonometrical inequalities for $\zeta(s)$? QUESTION [8 upvotes]: The inequality $$3 + 4 \cos \theta + \cos 2 \theta \geq 0$$ plays a key role in the proof of the classical zero-free region of the Riemann zeta function. Are there other inequalities of the form $$\sum_{i=0}^k a_i \cos b_i \theta \geq 0,\;\;\;\;\;a_\geq 0$$ such that $a_{i_0} = \sum_{i\ne i_0} a_i$ for some $0\leq i_0\leq k$ and $a_{0} < \frac{3}{4} a_{i_0}$, $b_0=0$? REPLY [17 votes]: Assuming the $b_i$ are all distinct (or at least non-zero for $i \neq 0$), this is not possible. (Otherwise there are trivial examples, e.g. $1 + 2 \cos(0 \theta)+ \cos(0 \theta) \geq 0$ or $1 + 4 \cos \theta + \cos(2\theta) + 2 \cos(0 \theta) \geq 0$.) Suppose that $\sum_{i=0}^k a_i \cos b_i \theta \geq 0$. Since $a_{i_0} = \sum_{i \neq i_0} a_i$, this implies that whenever $\cos b_{i_0} \theta = -1$, one must have $\cos b_i \theta = +1$ for all other $i$. In particular, the other $b_i$ must be integer multiples of $2b_{i_0}$. We now have $$ a_0 + a_{i_0} \cos b_{i_0} \theta + \sum_{i \neq 0, i_0} a_i \cos b_i\theta \geq 0$$ with the $b_i$ in the sum nonzero integer multiples of $2b_{i_0}$. Performing a Taylor expansion around $\theta = \pi / b_{i_0}$ to second order, we conclude that $$ - a_{i_0} \frac{b_{i_0}^2}{2} + \sum_{i \neq 0, i_0} a_i \frac{b_i^2}{2} \geq 0$$ and hence (since $b_i^2 \geq 4 b_{i_0}^2$ and $b_{i_0} \neq 0$) $$ \sum_{i \neq 0, i_0} a_i \leq \frac{1}{4} a_{i_0}$$ or equivalently $$ a_0 \geq \frac{3}{4} a_{i_0}.$$ Thus one cannot have $a_0 < \frac{3}{4} a_{i_0}$. This argument also shows that up to rescaling and other trivial rearrangements, Mertens' inequality $3 + 4 \cos(\theta)+\cos(2\theta) \geq 0$ is the unique inequality that attains $a_0 = \frac{3}{4} a_{i_0}$. At a more metamathematical level, if there were a variant of Mertens' trigonometric inequality that gave superior numerical results towards the classical zero free region, I would imagine that this would already have been noticed by now. :-)<|endoftext|> TITLE: Examples of incorrect arguments being fertilizer for good mathematics? QUESTION [24 upvotes]: Sometimes (perhaps often?) vague or even outright incorrect arguments can sometimes be fruitful and eventually lead to important new ideas and correct arguments. I'm looking for explicit examples of this phenomenon in mathematics. Of course, most proof ideas start out vague and eventually crystallize. So I think the more incorrect/vague the original argument or idea, and the more important the final fruit, the better, as long as there is still a pretty direct connection from the vague idea to the final fruit. Note: Many "paradoxes" sort-of are like this, but I think aren't what I'm looking for. (William Byers's book "How Mathematicians Think" has several examples and lots of discussion of the important role of paradox in mathematical research.) For example, the relationship between Russell's paradox, Godel's Incompleteness Theorem, and the undecidability of the halting problem (Church; Turing). But I think, unless the paradox has some other aspects of the vague-idea-as-fertilizer phenomenon, that I'm not looking for examples of paradoxes, though I am willing to be convinced otherwise. Edit: It's been suggested that this a duplicate of this other question, but I really think it is not. I am more interested in examples of outright incorrect (or nearly so) original statements that nonetheless lead to fruitful mathematics, whereas the other question seems to essentially be asking about ideas that start out intuitive, non-rigorous, or ill-defined and then are turned into rigorous arguments but along the same intuitive lines. (And, as I said above, I think I agree with one of the answers there that that is simply much of mathematics.) By comparing the answers to the other question to the three great answers already on this question (knot theory rising because Kelvin thought atoms were knotted strings; Lame's erroneous proof of FLT leading Kummer to develop algebraic integers; Lebesgue's incorrect proof that projections of Borel sets are Borel leading to Suslin's development of analytic sets), one can get a sense of the difference. REPLY [5 votes]: Obviously König's theorem should appear on this page. König suggested a proof by which the real numbers cannot be well-ordered. Unfortunately, he misunderstood some of the work he relied on, and thence we have this wonderful theorem known as König's theorem or Zermelo–König's theorem: If $I$ is any set, and for each $i\in I$, $|A_i|<|B_i|$, then $\left|\bigcup_{i\in I}A_i\right|<\left|\prod_{i\in I}B_i\right|$.<|endoftext|> TITLE: Acyclic group and finite CW-complex QUESTION [13 upvotes]: Is there a nontrivial example of an acyclic group $G$ such that its corresponding Eilenberg space $K(G,1)$ is homotopy equivalent to a finite CW-complex ? REPLY [17 votes]: The Higman group with presentation $$\langle{a,b,c,d}\mid{aba^{-1}b^{-2}},~bcb^{-1}c^{-2},~cdc^{-1}d^{-2},~ dad^{-1}a^{-2}\rangle$$ is perfect, and the 2-complex associated to this presentation has Euler characteristic 0. Hence this complex is acyclic. It is in fact aspherical, but it may be simpler to observe that Higman's group is also an iterated generalized free product with amalgamations $(A*_{\langle{b}\rangle}B)*_{F(a,c)}(C*_{\langle{c}\rangle}D)$, where $A,B,C$ and $D$ are copies of the Baumslag-Solitar group $BS(1,2)$, generated by $\{a,b\}$, $\{b,c\}$, $\{c,d\}$ and $\{d,a\}$, respectively. We may assemble a 2-dimensional Eilenberg-Mac Lane complex for the Higman group in a similar way.<|endoftext|> TITLE: A question on simple $P_{\aleph_2}$-points QUESTION [6 upvotes]: This question is motivated by discussion surrounding this MO question. An ultrafilter $U$ on $\omega$ is a simple $P_{\aleph_2}$-point if it is generated by a sequence $\langle X_\alpha:\alpha<\omega_2\rangle$ such that $$\alpha<\beta\Longrightarrow |X_\beta\setminus X_\alpha|<\aleph_0,$$ that is, generated by an $\omega_2$ sequence of subsets of $\omega$ that is decreasing modulo the ideal of finite sets. Question Is the existence of a simple $P_{\aleph_2}$-point consistent with $\mathfrak{b}=\aleph_1$? REPLY [3 votes]: The answer is "yes". Alan Dow pointed out in correspondence that the construction of Blass and Shelah referenced below can be modified to yield such a model. The particular model arises by using finite support to add an $\omega_1$-sequence of Cohen reals, followed by a finite support iteration adding the generating set for the $P_{\aleph_2}$-point using a variant of Mathias forcing. The salient point of the construction is that the Mathias-type forcing (really Mathias forcing with respect to a carefully chosen ultrafilter at each stage) does not add a real dominating the Cohen reals, and so $\mathfrak{b}=\aleph_1$ in the extension. Referring to the discussion on the question of Banakh that motivated our question, this shows there is a model in which Banakh's cardinal $\kappa$ is strictly greater than $\mathfrak{b}$, as the $P_{\aleph_2}$-point is an ultrafilter $U$ with $\tau(U)=\aleph_2$. Blass, Andreas; Shelah, Saharon, Ultrafilters with small generating sets, Isr. J. Math. 65, No. 3, 259-271 (1989). ZBL0681.03033.<|endoftext|> TITLE: The abc-conjecture as an inequality for inner-products? QUESTION [21 upvotes]: The abc-conjecture is: For every $\epsilon > 0$ there exists $K_{\epsilon}$ such that for all natural numbers $a \neq b$ we have: $$ \frac{a+b}{\gcd(a,b)}\,\ <\,\ K_{\epsilon}\cdot \text{rad}\left(\frac{ab(a+b)}{\gcd(a,b)^3}\right)^{1+\epsilon} $$ I have two questions after doing some experiments with SAGEMATH: 1) Is the matrix $$L_n = \left( \frac{\gcd(a,b)}{a+b}\right)_{1\le a,b \le n}$$ positive definite? 2) Is the matrix: $$ R_n = \left( \frac{1}{\text{rad}\left(\frac{ab(a+b)}{\gcd(a,b)^3}\right)} \right)_{1\le a,b \le n} $$ positive definite? If both of the questions can be answered with yes, then we would have "mappings" $$\psi ,\phi: \mathbb{N} \rightarrow \mathbb{R}^n$$ and the abc-conjecture might be stated as an inequality in the inner-product of these mappings: $$\left< \psi(a),\psi(b) \right>^{1+\epsilon} < K_{\epsilon} \left < \phi(a), \phi(b) \right >$$ which I think would be very interesting. Edit: I realized that it is better to ask the following question: Is $$R^{(\epsilon)}_n := (\frac{2^{\epsilon}}{\text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^{1+\epsilon}})_{1\le a,b\le n}$$ positive definite for all $\epsilon \ge 0$? If "yes", then we would have: For all $\epsilon \ge 1$ and all $a \neq b$ the following are equivalent: $$1) d_R^{(\epsilon)}(a,b) = \sqrt{1-\frac{2^{1+\epsilon}}{\text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^{1+\epsilon}}}>d_L(a,b) = \sqrt{1-2\frac{\gcd(a,b)}{a+b}}$$ $$2) \left < \psi^{(\epsilon)}_R(a),\psi^{(\epsilon)}_R(b) \right > < \left < \psi_L(a),\psi_L(b) \right >$$ 3) The abc conjecture for $\epsilon \ge 1$ with $K_{\epsilon} = \frac{1}{2^{\epsilon}}$ Related question Two questions around the $abc$-conjecture Also the metrics $d_R^{(\epsilon)},d_L$ would be embedded in Euclidean space. Yet another edit: It seems that $$\frac{\phi(n)}{n} = \sum_{d|n} \frac{\mu(d)}{\text{rad}(d)}$$ wher $\mu, \phi$ are the Moebius function and the Euler totient function. From this it would follow using Moebius inversion, that : $$\frac{1}{\text{ rad}(n)} = \sum_{d|n} \frac{\mu(d)\phi(d)}{d}$$ which could (I am not sure about that) be helpful for question 2). Edit with proof that $k(a,b)$ is a kernel: Let $$k(a,b) := \frac{1}{\frac{ab(a+b)}{\gcd(a,b)^3}} = \frac{\gcd(a,b)^3}{ab(a+b)} = \frac{\gcd(a,b)^2}{ab} \cdot \frac{\gcd(a,b)}{a+b} = k_1(a,b) \cdot k_2(a,b)$$ It is known that: $$\int_0^1 \psi(at)\psi(bt) dt = \frac{1}{12} \frac{(a,b)^2}{ab} = \frac{1}{12} k_1(a,b).$$ Where $\psi(t) = t - \lfloor t \rfloor - \frac{1}{2}$ is the sawtooth function. Hence $k_1(a,b)$ is a kernel. On the other hand, it is known for example by the answer of @DenisSerre, that $k_2(a,b)$ is also a kernel. Hence the product $k(a,b) = k_1(a,b) \cdot k_2(a,b)$ is also a kernel. Update: I found this paper online which is interesting (Set there: $X_a = \{ a/k | 1 \le k \le a \}$ then: $|X_a \cap X_b| = |X_{\gcd(a,b)}| = \gcd(a,b)$ ) and may be of use for the questions above: https://www.researchgate.net/publication/326212690_On_the_positive_semi-definite_property_of_similarity_matrices Setting in the paper above $A_i = \{ i/k | 1 \le k \le i \}$ we see that $|A_i \cap A_j| = |A_{\gcd(i,j)}| = \gcd(i,j)$ and $|A_i|=i$. Since in the paper it is proved that: 1) The Sorgenfrei similarity $\frac{|A_i \cap A_j|^2}{|A_i||A_j|}$ is a (positive definite $\ge0$, symmetric) kernel, we have another proof, that $\frac{\gcd(a,b)^2}{ab}$ is a kernel. 2) The Gleason similarity $\frac{2|A_i \cap A_j|}{|A_i|+|A_j|}$ is a (positive definite $\ge0$, symmetric) kernel, we have another proof, that $\frac{\gcd(a,b)}{a+b}$ is a kernel. Using the product of these kernels, we get the new kernel $\frac{\gcd(a,b)^3}{ab(a+b)}$. REPLY [15 votes]: The matrix $L_n$ is positive definite. Proof. The matrix $G_n$ with entries ${\rm gcd}(a,b)$ is positive definite because of $G=D^T\Phi D$ where $\Phi={\rm diag}(\phi(1),\ldots,\phi(n))$ ($\phi$ the Euler's totient function) and $d_{ij}=1$ if $i|j$ and $0$ otherwise. Then the matrix $H_n$ with entries $\frac1{a+b}$ is positive definite because $$h_{ij}=\int_0^1 x^{i+j-1}dx$$ and the matrix with entries $x^{i+j-1}$ is positive semi-definite for $x>0$. Finally $L_n=G_n\circ H_n$ (Hadamard product) is positive definite.<|endoftext|> TITLE: Around the diophantine equation $\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{2a+3b}=\text{odd integer}$, over positive integers QUESTION [7 upvotes]: I am interested to know if a similar theorem that shows this answer of the post Estimating the size of solutions of a diophantine equation (this MathOverflow, January 5th 2016) is feasible for a different diophantine problem. Conjecture. Let $a,b$ and $c$ be integers greater or equal than $1$. Then $$\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{2a+3b}$$ can never be an odd (positive) integer. Question. I would like to know if the reasonings in the answers of the linked post also works for the diophantine equation in previous conjecture. Is it possible to prove previous conjecture, or is there any counterexample? Many thanks. I don't know if this equation is in the literature (as reference I've added the mentioned post and [1]) as a special case of the expression $$\frac{\lambda a}{\beta b+\eta c}+\frac{\lambda b}{\beta c+\eta a}+\frac{\lambda c}{\beta a+\eta b}=N$$ with $N$ a non-zero integer, and for given integers $1\leq \lambda\leq \beta\leq \eta$. Our case, is similar than the linked post with $(\lambda,\beta,\eta)=(1,2,3)$. Our curve can be written (I did the calculation using Wolfram Alpha online calculator) $$6 a^3 + 9 a^2 b + 4 a b^2 + 6 b^3 + 4 a^2 c + 18 a b c + 9 b^2 c + 9 a c^2 + 4 b c^2 + 6 c^3$$ $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad-n((2 a + 3 b) (3 a + 2 c) (2 b + 3 c))=0$$ from $\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{2a+3b}=n$. I think that this is a good companion of previous post, but if it is in the literature feel free to comment or provide your answer as a reference request and I search and read the statement about previous Conjecture from the literature. References: [1] Andrew Bremner and Allan MacLeod, An Unusual Cubic Representation Problem, Annales Mathematicae et Informaticae Volume 43 (2014), pp. 29-41. REPLY [7 votes]: The problem of this question is qualitatively and quantitatively different in some ways from that considered by Andrew Bremner and myself. If we take the cubic, with $N$ as a fixed constant, it is possible to show that the related elliptic curve is \begin{equation*} E_N:G^2=H^3+((35N+18)H+4(1260N+1441))^2 \end{equation*} with the formulae linking $(H,G)$ to solutions $a,b,c$ being lengthy, but straightforward to find. The discriminant is \begin{equation*} \Delta=2^{12}5^27^2(5N-3)(1260N+1441)^3(7N^2+15N+9) \end{equation*} so that $E_N$ has two components for all $N\ge1$. This curve is in the standard form for one with $\mathbb{Z}3$ torsion, with points of order $3$ when $H=0$. The algebra reducing the original cubic to the elliptic curve showed that there are other rational points. I found \begin{equation*} H=\frac{288(18N+31)(36N+43)}{361} \end{equation*} giving \begin{equation*} G=\pm \frac{4(468N+635)(79056N^2+186120N+115297)}{6859} \end{equation*} which just lead to trivial solutions. What I then found was the very surprising fact that this point is double $(-144,\pm 2660)$ for all $N$. The positive $G$ value gives the parametric solution \begin{equation*} a=12(34992N^2+78120N+40499) \end{equation*} \begin{equation*} b=-279936N^2-1271340N-1096057 \end{equation*} \begin{equation*} c=18(10368N^2+54180N+52811) \end{equation*} The elliptic curve has positive rank and simple numerical experiments show rank $3$ is fairly common.<|endoftext|> TITLE: A minimality problem for a class of Banach spaces QUESTION [6 upvotes]: The following question is related to the previous question Minimality properties of James' space; I post it as a new question since the system does not allow me to add a comment. Question Consider the following class of non-Hilbertian spaces: $X_{p,2}=(\sum_{n=1}^\infty \oplus\ell^p_n)_2$, $1\le p\le \infty$, $p\neq 2$. Is it true that the only infinite dimensional Banach space that is isomorphically embedded into anyone of them is the Hilbert space? Notice that all these spaces are subspaces of the space $\mathcal{J}$. REPLY [2 votes]: I think such a space has type 2 and cotype 2 so by Kwapien's theorem it is isomorphic to a Hilbert space.<|endoftext|> TITLE: What are Harish-Chandra bimodules used for? QUESTION [12 upvotes]: There are many recent papers on classification of Harish-Chandra bimodules for rational Cherednik algebras and, more generally, non-commutative algebras which are quantizations of symplectic singularities (Losev). What is the meaning of Harish-Chandra bimodules in terms of representation theory of the underlying algebra/its category O? Are Harish-Chandra bimodules related to the classical notion of Harish-Chandra modules? REPLY [7 votes]: Here is an answer from a mathematician who prefers me to post it here myself: Harish-Chandra bimodules make sense in a very wide context. Take two filtered algebras A, A' that quantize the same commutative algebra $C$, and fix isomorphisms ${\rm gr} A \to C$, ${\rm gr} A^{'} \to C$. Then one can make sense of the definition of a $HC (A, A^{'})$-bimodule. These are (A, A')-bimodules, say B, that admit a filtration such that \gr B is a finitely generated C-module, meaning that the left and right actions on C coincide. It is not hard to see that if A, A' are $U(g)$ for a simple Lie algebra g, this coincides with the notion of HC bimodule that I alluded above. In the context of symplectic singularities, note that you need to have a Hamiltonian $\mathbb C^*$-action to define the category O. Such an action does not always exist (e.g. for Kleinian singularities outside of type A). In this sense, HC bimodules are a substitute for the category O. See for example Ginzburg https://arxiv.org/pdf/0807.0339.pdf When you do have categories O, HC bimodules give, via tensor product, functors between categories O for different quantization parameters. For example, projective functors in Lie theory are a special case of tensoring with a HC $U(g)$-bimodule. In this sense, HC bimodules also generalize the notion of projective functors. Translation functors for Cherednik algebras are a special case of this. I must warn, however, that tensoring with a HC bimodule is in general a very bad functor -- it can kill many things and it is not exact. Nevertheless, these functors were used by Losev to construct derived equivalences between categories O for Cherednik algebras https://arxiv.org/pdf/1406.7502.pdf Also, Harish-Chandra bimodules are much more sensitive to the quantization parameter than the category O is. Category O always has the same number of simples = number of fixed points under Hamiltonian torus action. This is far from being true for HC bimodules. For example, for type A Cherednik algebras the quantization parameter is a complex number $c$ (I apologize if I am overexplaining, I don't know how familiar you are with these). If $c$ is not a rational number with denominator $1 < d \leq n$ ($n =$ rank of symmetric group) then the category O is semisimple and equivalent to reps of $S_n$. This is not true for the category of HC $H_{c}$-bimodules. For these parameters, the category is still semisimple, but it is only equivalent to reps of $S_n$ when c is an integer. Otherwise, it is equivalent to Vec. In this sense, HC bimodules detect how integral the parameter is. See https://arxiv.org/pdf/1409.5465.pdf Theorem 1.1 for the case of rational Cherednik algebras (the subgroup $W_{c}$ essentially detects how far c is from being integral). This was generalized by Losev to symplectic singularities in https://arxiv.org/pdf/1810.07625.pdf One more thing, the simplest example of a HC $A$-bimodule is the regular bimodule. So one can use HC bimodules to answer questions about, for example, ideals in $A$ (usually these techniques come from constructing restriction functors for HC bimodules, similar to the Bezrukavnikov-Etingof functors for category O and applying them to the regular bimodule). This was used by Losev for Cherednik algebras in https://arxiv.org/pdf/1001.0239.pdf (see Thms 1.3.1 and 5.8.1) and for finite W-algebras in https://arxiv.org/pdf/0807.1023.pdf Finally, in the context of symplectic resolutions it is believed that HC bimodules should categorify the homology of the generalized Steinberg variety. This is of course not true in general (even for Cherednik algebras for the reasons above -- for some parameters there are simply not enough irreducibles) but it should be true for integral parameters, for an appropriate notion of integral. See Braden-Proudfoot-Webster, https://arxiv.org/pdf/1208.3863.pdf Proposition 6.16 (later in that paper they show that wall-crossing functors are always tensoring with an appropriate HC bimodule, Proposition 6.23)<|endoftext|> TITLE: Does $Π^1_{2n+1} = (Σ^1_{2n+1})^{M_{2n-1}}$? QUESTION [10 upvotes]: Every $Π^1_1$ formula $φ$ without free second order variables can be converted into a $Σ^1_1$ $ψ$ such that $φ ⇔ ψ^\mathrm{HYP}$, and vice versa. ($\mathrm{HYP}$ is the hyperarithmetical universe, which can be identified with $L_{ω_1^\mathrm{CK}}$.) Essentially, arbitrary $Π^1_1$ statement $⇔$ well-foundedness of some recursive relation '$≺$' $⇔$ existence of a hyperarithmetical set iterating the Turing jump along '$≺$' $⇔$ arbitrary $(Σ^1_1)^\mathrm{HYP}$ statement. My question is whether, assuming projective determinacy, an analogous correspondence holds for $Π^1_{2n+1}$; and I conjecture $Π^1_{2n+1} = (Σ^1_{2n+1})^{M_{2n-1}}$. $M_n$ is the minimal iterable inner model with $n$ Woodin cardinals. $M_0$ is the constructible universe $L$, and $M_{-1}$ (not an inner model) would be $L_{ω_1^{\mathrm{CK}}}$. The reals in $M_n$ are precisely those that are $Δ^1_{n+2}$ in a countable ordinal. For even $n$, $M_n$ is $Σ^1_{n+2}$ correct, but this is not the case for odd $n$. A positive answer to the question should enhance our understanding of $Π^1_{2n+1}$ prewellordering and uniformization. True $Σ^1_{2n}$ statements can be 'graded' (assuming projective determinacy) by the complexity of the least witness. For example, consider a theory $T$ such as ZFC + "there is a supercompact cardinal", and assume that $T$ has sufficiently sound models. At the $Σ^1_2$ level, we can assign $T$ an ordinal based on the least height of a transitive model of $T$; furthermore, there is a $Δ^1_2$ example of such a model. This generalizes to $Σ^1_{2n}$ and models of $T$ that are closed under $M_{2n-3}^\#$. But what kind of witnesses do we have for $Π^1_{2n+1}$ statements? The first paragraph gives an answer for $Π^1_1$, and we can generalize it as follows. Assuming $M_{2n+1}^\#$ exists, a $Π^1_{2n+3}$ statement $T$ is true iff there is an iterable model $M$ of ZFC + "$2n+1$ Woodin cardinals" such that $M^{\mathrm{Coll}(ω,δ)}⊨T$ where $δ$ is the least Woodin cardinal in $M$. (The use of ZFC in $M$ is essentially arbitrary; also, genericity iterations allow Woodin cardinals to 'absorb' real quantifiers.) Presumably, such an $M$ can be chosen to be $Δ^1_{2n+3}$ and its existence is $(Σ^1_{2n+3})^{M_{2n+1}}$ (but how?) (Also, the least complexity of such $M$ should be connected to $Π^1_{2n+3}$ prewellordering, but the exact connection is unclear to me as the prewellordering is about sets of reals.) REPLY [10 votes]: Your conjecture is true. We assume PD throughout. The proof I see requires citing a number of facts from inner model theory and descriptive set theory. First, it uses Woodin's theorem characterizing the reals of $M_{2n-1}$ as the set $Q_{2n+1}$ of points in Baire space that are $\Delta^1_{2n+1}$ definable from a countable ordinal. In symbols: Theorem (Woodin). $\omega^\omega\cap M_{2n-1} = Q_{2n+1}$. It uses a correctness theorem for the odd levels: Theorem $M_{2n-1}$ is $\Pi^1_{2n}$-correct. A set $A\subseteq \omega^\omega$ is $\Pi^1_{2n+1}$-bounded if $\Pi^1_{2n+1} = \exists^{A} \Pi^1_{2n+1}$. We need that $Q_{2n+1}$ is $\Pi^1_{2n+1}$-bounded. In fact, something stronger is true (see Kechris-Martin-Solovay's "Introduction to $Q$-theory"): Theorem (Kechris-Martin-Solovay). $Q_{2n+1}$ is the largest $\Pi^1_{2n+1}$-bounded subset of Baire space. We need Moschovakis's "Spector-Gandy theorem for the odd levels" (Moschovakis, Descriptive Set Theory, 6E.7): Theorem (Moschovakis) $\Pi^1_{2n+1}\cap\omega^\omega = \exists^{\Delta^1_{2n+1}\cap \omega^\omega}\Pi^1_{2n}\cap \omega^\omega$. Moschovakis's theorem will actually be applied to $Q_{2n+1}$ using the $Q$-Theory Reflection Theorem: Theorem (Kechris-Martin-Solovay) If $A\subseteq \omega^\omega$ is $\Pi^1_{2n+1}$, then $\exists x\in \Delta^1_{2n+1}\ A(x)$ if and only if $\exists x\in Q_{2n+1}\ A(x)$. Given these facts, the calculation becomes a straightforward pointclass calculation. By definition, $\Sigma^1_{2n+1} = \exists^{\omega^\omega}\Pi^1_{2n}$, so $(\Sigma^1_{2n+1})^{M_{2n-1}} = \exists^{\omega^\omega\cap M_{2n-1}}(\Pi^1_{2n})^{M_{2n-1}}$. Woodin's theorem characterizing $\Pi^1_{2n+1}$ along with the $\Pi^1_{2n}$-correctness of $M_{2n-1}$ imply $\exists^{\omega^\omega\cap M_{2n-1}}(\Pi^1_{2n})^{M_{2n-1}}\cap\omega^\omega = \exists^{Q_{2n+1}}\Pi^1_{2n} \cap\omega^\omega$. Moschovakis's Spector-Gandy Theorem along with the $Q$-Theory Reflection Theorem yields that $\exists^{Q_{2n+1}}\Pi^1_{2n}\cap \omega^\omega = \exists^{\Delta^1_{2n+1}\cap\omega^\omega}\Pi^1_{2n}\cap \omega^\omega = \Pi^1_{2n+1}\cap \omega^\omega$. Stringing together a bunch of pointclass identities, one can conclude that $(\Sigma^1_{2n+1})^{M_{2n-1}}\mathrel{\cap} \omega^\omega = \Pi^1_{2n+1} \cap \omega^\omega$. You might also want to look at Theorem 4.12 of John Steel's paper Projectively Well-Ordered Inner Models. I think you can use the proof to get a more inner model theoretic proof of your conjecture, but Steel's result is closely related and of independent interest.<|endoftext|> TITLE: Are evil properties really evil QUESTION [15 upvotes]: I have this question for a moment now, so I think it is time that I sort it out. I got into category theory and homotopy type theory at the same time, and so I have always read and been told that one should be careful about "evil" properties, i.e. the ones that are not invariant under equivalence of categories. So in order to try and disambiguate the terminology, what I will refer to as "categories" are up to equivalence, and I will call a precategory the objects up to isomorphism. In other words, categories are objects of the category $\operatorname{Ho}(\operatorname{Cat})$, for the folk model structure of $\operatorname{Cat}$, and the precategories are the objects of $\operatorname{Cat}$. It is my understanding that category theory is in fact about categories and not precategories, and it makes perfect sense, since the point of category theory is to give the proper meaning to isomorphisms. The intuition that I always keep in mind is that it should not matter if I see the same object many times as long as I know that it is the same. In order to really talk about these categories, people have started looking at the appropriate language, which is a language in which evil property do not make sense, or are not expressible. This is a way to get to the theory of categories, or their internal language (I am not sure how to refer to this notion exactly). However, there are many constructions that are used on precategories, the two main ones that I have in mind are the Reedy categories and the contextual categories (aka C-systems), and some results are proved by making heavy use of these constructions. My understanding of the situation The precatgories can be thought of as some sort of a "presentation" of a given category. It is of course not unique since many non-isomorphic can be equivalent, but that's fine, I am used to that in groups or vector spaces : A group can have different presentations. Now the using an "evil" notion would amount to working in a given presentation of a group, so whatever non-evil property you prove for a Reedy category is in fact a property of the category presented by my precategory which is Reedy. In group theory would be analogue to a theorem of the form "If a group has a finite presentation, then it satisfies $P$" And then I can freely state that $\mathbb{Z}$ satisfies $P$, since it is finitely presented as $\langle x\rangle$ , although I might as well give the (different) presentation $\langle x_1,x_2,\ldots | x_1=x_2, x_1 =x_3,\ldots\rangle$, which is not finite. It is sort of picking a representative in an equivalence class to prove a property which goes through the quotient. This is something we do a lot, and it gives a perfectly valid proof. So here I can very easily transform the property "being a Reedy category" to "being equivalent to a Reedy category", and I have changed my evil property into a non-evil one. Every non-evil property I can prove for Reedy categories is in fact provable for all the precategory that are categorically equivalent to a Reedy category Possible limitations There are two main "problems" that I can see using evil notions in order to prove non-evil ones Firstly the construction that I get may not be "canonical" or "natural" (again I am not sure what the proper terminology should be here). If I take the example of a Reedy model structure, there might be many equivalent precategory that present the same category and that are Reedy. The Reedy model structure I get using one might not agree with the Reedy model structure I get using the other - even up to the equivalence between them. This is in my intuition much like the fact that every finite dimensional vector space is isomorphic to its dual, but there is necessarily a choice of a basis, leading to really different isomorphisms. Here a choice of a basis is a given presentation. This also makes sense with my previous correction of Reedy to be non-evil : the property "being equivalent to a Reedy precategory" gives an equivalence of categories, and then a specific choice of a presentation, and there might be many ways to "be equivalent to a Reedy precategory", leading to different constructions for the model structure. But again, constructing something non-canonical is not that big of a deal, if we carry out the constructions everywhere. After it is true that all finite dimensional vector space is equivalent to its dual even if there is no canonical way to do it - Why not imagine properties in non-evil properties in categories that are true, but never "non-evily" true (i.e. never true in a canonical way)? The proof that we get is not in the language of categories (and I think this is basically saying the same thing). It is a proof in the language of precategories. But again it sounds fine to me, with finite dimensional vector spaces we usually say "take a basis", which is not expressible in their internal language. It does not make the fact true, it just makes it non-canonical. My Question So firstly, I would like to know if my intuition is correct. If so, the "evil" properties do not seem too evil to me, and we can freely use them as long as we don't claim canonicity. I have seen examples of this in different areas of mathematics, so why not in category theory? If my intuition is incorrect, am I missing something that would make these concepts truly evil? Then a related question that I have (assuming my understanding is correct) is are there any known examples of a property that is always true but never canonically in category theory? What I mean by that is a theorem that we can prove about categories but using precategories, and that is not provable in the internal language of categories? If we know for some reason that no such thing exists, then I would be convinced to avoid evil properties, as they are unnecessary, but otherwise how to know that we are not missing something by avoiding them? REPLY [14 votes]: Before I try to answer the question itself, let me make a few preliminary remarks. Firstly, a minor point: it's not really correct to say that categories are the objects of Ho(Cat), if by Ho(Cat) you mean the homotopy 1-category of Cat; the correct thing to say is that categories are the objects of the 2-category Cat. The point is that in Ho(Cat), whose morphisms are isomorphism classes of functors, you lose the ability to distinguish between multiple isomorphisms between two isomorphic functors. Secondly, it's perfectly correct, as you say, that there are some operations on things sometimes called "categories" that require treating them up to isomorphism rather than equivalence. But instead of your "precategories", I will call those things "strict categories", to match the nLab and the HoTT Book (of which the latter got the terminology from the former). So Reedy categories and contextual categories are best thought of as strict categories, rather than categories proper. Thirdly, I would approach the problem of defining a "Reedy non-strict category" somewhat differently. It's true that a given category could be equivalent to a Reedy strict category in more than one way --- but even more than that, it's already true that a given strict category can be Reedy in more than one way! That is, being Reedy is structure on a strict category, not a property of it, and different choices of Reedy structure on the domain can yield different Reedy model structures, even before we start to worry about strictness of categories. So similarly, being Reedy must also be structure of a non-strict category, and once we realize that it is quite natural to say that a Reedy structure on a non-strict category $C$ consists of a specified strict category $C'$, a specified equivalence $C\simeq C'$, and a Reedy structure on $C'$. With this definition, a Reedy structure on a non-strict category induces a Reedy model structure in a canonical and equivalence-invariant way. (With that said, there seems relatively little reason to do such a thing, since most naturally-arising Reedy categories are in fact strict. There's nothing wrong with using strict categories, as long as we don't assume that every category is strict.) Now, your question seems to be roughly "what is wrong with non-equivalence-invariant constructions on categories?" As you say, this is really the same question as "what is wrong with non-isomorphism-invariant constructions on vector spaces?". The first answer is that in practice we don't keep track of "which equivalent/isomorphic copy" of an object we are using. For instance, the tangent space of a manifold at a point can be defined in many ways, yielding isomorphic but non-equal vector spaces. If we have a construction on vector spaces that is not isomorphism-invariant, then when someone wants to apply that construction to the tangent space of a manifold, we have to ask "which tangent space?" Or, even worse, if we have a construction on groups that is not isomorphism-invariant and someone wants to apply that construction to the two-element group (or even the one-element group!) we have to ask "which one?" The experience of abstract/structural mathematics over the past century strongly suggests that such constructions should not be considered part of linear algebra or group theory. (You seem to believe that such constructions are used in some areas of mathematics, but I would dispute that; can you give examples of the sort of thing you are thinking of?) Essentially the same is true for categories and equivalences. The second answer, which to my mind is even more powerful, is that only invariant constructions can be generalized to "variable" objects. Consider for instance the notion of vector bundle over a space. Even though we can choose bases for each individual fiber, if the bundle is nontrivial it may not be possible to choose a family of global sections that form a basis at every point simultaneously. Thus, a construction on vector spaces which depends on the choice of basis cannot be performed on vector bundles. Another way of thinking about this is that a vector bundle can be represented by a functor into the groupoid of vector spaces, so any construction on vector spaces that is to be "composed" with such a functor must itself be functorial on the groupoid of vector spaces, i.e. respect isomorphism. Exactly the same thing is true for categories, although the relevant kind of "variation" also has to be categorified. There are "categorified spaces" and "bundles of categories" over them such that although each individual "fiber" can be presented by some strict category, the entire bundle cannot be simultaneously so presented. Thus, a construction on categories that doesn't respect equivalence, or equivalently that depends on a choice of strict presentation, cannot be performed on such "bundles of categories".<|endoftext|> TITLE: Is the product of commuting ultrafilters an ultrafilter? QUESTION [9 upvotes]: If $U$ is a filter on $X$ and $W$ is a filter on $Y$, their product is the filter $U\times W$ on $X\times Y$ generated by rectangles $A\times B$ where $A\in U$ and $B\in W$. In certain circumstances (e.g., when $U$ is $|W|$-complete), the product of two ultrafilters $U$ and $W$ is again an ultrafilter. In this situation, $U$ and $W$ must commute in the following sense: for any binary relation $R$, we have $\forall^U x\ \forall^W y\ (x\mathrel{R} y)$ if and only if $\forall^W y\ \forall^U x\ (x\mathrel{R} y)$. (Here, for $P$ a unary predicate, we write $\forall^U x\ P(x)$ to mean that $\{x\in X : P(x)\}\in U$.) The question is whether the converse holds. Question: If two ultrafilters $U$ and $W$ commute, must $U\times W$ be an ultrafilter? Conceivably, a positive answer is provable in ZFC or assuming GCH. The question has a vacuous positive answer under the assumption that there are no measurable cardinals since this implies all commuting ultrafilters are principal. Put another way, constructing a counterexample would require the use of large cardinals. Background: Ultrafilters $U$ and $W$ such that $U\times W$ is ultra were studied by Blass in his thesis. Blass showed that this is equivalent to the statement that $U$ is complete modulo $W$ in the sense that for any sequence $\langle A_i : i\in I\rangle\subseteq U$ defined on a $W$-large set $I$, $\bigcap_{i\in J} A_i\in U$ for some $W$-large set $J$. In particular, despite all appearances, the relation $U$ is complete modulo $W$ is symmetric in $U$ and $W$. Commuting ultrafilters are related to Kunen's Commuting Ultrapowers Lemma, which says that if $U$ is $|W|$-complete, then $j_U(j_W) = j_W\restriction \text{Ult}(V,U)$ (which is pretty clear) and $j_W(j_U) = j_U\restriction \text{Ult}(V,W)$ (which is nontrivial). Here $j_U : V\to \text{Ult}(V,U)$ denotes the (transitive collapse of the) ultrapower embedding associated to $U$ and $j_U(j_W) = \bigcup_{x\in V} j_U(j_W\restriction x)$. It is a not-so-easy exercise to see that for countably complete ultrafilters $U$ and $W$ commute if and only if $j_U(j_W) = j_W\restriction \text{Ult}(V,U)$. In particular, despite all appearances, the relation $j_U(j_W) = j_W\restriction \text{Ult}(V,U)$ is symmetric in $U$ and $W$. REPLY [3 votes]: The following may give a hint: Suppose $U$ is a uniform ultrafilter on $\omega$ and $W$ is an ultrafilter on $\kappa$ such that $W$ commutes with $U$, then $W$ is countably complete. Otherwise, there exist $\langle A_i \in W: i\in \omega\rangle$ decreasing such that $\bigcap_{i\in \omega} A_i =\emptyset$. Let $R\subset \omega\times \kappa$ be such that $i R \gamma$ iff $\gamma\in A_i$. It can't be the case that $\forall^W \gamma \forall^U i \ iR\gamma$. Since fix some such $\gamma\in \kappa$, we have $\gamma\in \bigcap_{i\in \omega} A_i$ which is impossible. By commuting, it can only be that $\neg\forall^U i \forall^W \gamma \ iR\gamma$, hence $\forall^U i \forall^W \gamma \ \neg iR\gamma$. But this is bogus too since fix some such $i\in \omega$, we have $D\in W$ such that for all $\gamma\in D$, $\gamma\not\in A_i$, contradicting with the fact that $A_i\in W$.<|endoftext|> TITLE: Parity of shuffle permutations QUESTION [5 upvotes]: A $(p,q)$-shuffle is a permutation $\sigma$ of the set $\{1, \dots, p, p+1, \dots,p+q\}$ such that $$\sigma(1)<\dots<\sigma(p)$$ and $$\sigma(p+1) < \dots<\sigma(p+q)\,.$$ It is known that the number of $(p,q)$-shuffles is ${p+q \choose p}$. Looking at the case $p+q = 6$ and considering the parity of shuffles, I was surprised to find that $3$ of the ${6 \choose 1} = 6$ $(1,5)$-shuffles have even parity $9$ of the ${6 \choose 2} = 15$ $(2,4)$-shuffles have even parity $10$ of the ${6 \choose 3} = 20$ $(3,3)$-shuffles have even parity $9$ of the ${6 \choose 4} = 15$ $(4,2)$-shuffles have even parity $3$ of the ${6 \choose 5} = 6$ $(5,1)$-shuffles have even parity It's not so difficult to believe that the number of even $(p,q)$-shuffles should equal the number of even $(q,p)$-shuffles. But I was very surprised to see that there are so many more even than odd $(2,4)$-shuffles. Are any closed-form solutions known for the number of even $(p,q)$-shuffles? If not, is anything known about the asymptotic behaviour? REPLY [3 votes]: As Philippe suggests in the comments, it is well know that $$\sum_{\sigma \text{ is a $(a,b)$-shuffle}} q^{\ell(\sigma)}={a+b \choose b}_q$$ where the $q$-binomial coefficient (or Gaussian binomial coefficient) on the right side is defined by $${n \choose k}_q=\frac{[n]_q! }{[k]_q! [n-k]_q!}$$ where $[n]_q!=[1]_q [2]_q \cdots [n]_q$ and $[n]_q=1+q+\cdots +q^{n-1}$. Thus the difference between the number of even and odd shuffles is obtained by specializing at $q=-1$. This polynomial is palindromic, so, depending on the parity, one either gets 0 or the middle (and largest) coefficient of the polynomial ${a+b \choose b}_q$. Precise asymptotics for this coefficient (and others) are given in Theorem 1 of this paper by Melczer, Panova, and Pemantle.<|endoftext|> TITLE: Kirby calculus on Mazur manifolds QUESTION [10 upvotes]: I have questions about Akbulut and Kirby's paper Mazur manifolds. I couldn't figure out the following equality passages: Any help will be appreciated. REPLY [9 votes]: Here is a proof with pictures. Observe that blowing up and blowing own doesn't change the 3 manifold,i.e, the boundary. So all these above pictures have same boundary. As a 3 manifolds all of these are isomorphic.<|endoftext|> TITLE: Factorization and vertex algebra cohomology QUESTION [12 upvotes]: A chiral algebra on a smooth curve $X$, in the sense of Beilinson-Drinfeld, is a right $D_{X}$-module with a chiral bracket, which is a map $\mathcal{V}^{\boxtimes 2}(\infty\Delta)\rightarrow \Delta_{*}\mathcal{V}$ satisfying certain conditions making it look like a Lie bracket. In fact $\mathcal{V}$ is a Lie algebra with respect to a certain pseudo-tensor structure on $D_{X}$-modules. A pseudo-tensor structure is what one obtains when one forgets the tensor product but remembers the associated multilinear maps. Multi-linear maps from a collection $\{\mathcal{V}_{i\in I}\}$ to $\mathcal{W}$ are defined to be $\mathrm{Hom}_{D_{X^{I}}}(\boxtimes_{i\in I}\mathcal{V}_{i}(\infty\Delta_{\mathrm{big}}),\Delta_{*}\mathcal{W})$. One thus obtains a complex of vector spaces $C^{*}_{CE}(\mathcal{V})$ via the appropriate construction of Chevalley-Eilenberg cochains, which makes sense in a pseudo-tensor category. Notably this formalism does not produce CE chains. If $X$ is taken to a be a formal disc $D$, then we obtain a vertex algebra. I believe in this case the above construction produces the vertex algebra cohomology (with coefficients in the adjoint module) studied in the work of Bakalov, de Sole, Heluani and Kac (arXiv link). Is this true? If not what is the precise relation? Now if $\mathcal{V}$ is a chiral algebra, then BD construct its factorization homology. This is constructed as de Rham homology of the associated $D$-module on the Ran space of $X$. In chiral terms I believe it is gotten by taking (the colimit over finite sets $I$) of the Chevalley Cousins complexes on $X^{I}$, cf. these notes by Rozenblyum (on Gaitsgory's webpage). This can also be performed on a formal disc $D$*, in which case one obtains what one might call Vertex Homology. Has this been studied anywhere? Is there a relation with the work of BdSHK cited above, in particular does the BdSHK complex act on it? For example, if I'm not mistaken we compute $H^{\mathrm{vtx}}_{0}(V)=V/\operatorname{Res}(V)$, where $\operatorname{Res}(V)$ is the image of all the residues $\operatorname{res}_{z=0}(v(z))$ of the fields, which has a very similar flavour to the complex of BdSH. *Edit, in accordance with David Ben-Zvi's comments it seems that vertex algebra homology as defined in this question would be better defined over a punctured disc $D^{*}$. REPLY [8 votes]: If $X$ is taken to a be a formal disc $D$, then we obtain a vertex algebra. I believe in this case the above construction produces the vertex algebra cohomology (with coefficients in the adjoint module) studied in the work of Bakalov, de Sole, Heluani and Kac. Is this true? If not what is the precise relation? This is not exactly so. If you take $X=\mathbb{A}^1$ and your D-modules are translation equivariant and you consider the translation equivariant CE complex you will get the same complex that we studied. Chiral algebras on the disc are not the same thing as vertex algebras. If you consider chiral algebras equivariant with respect to the action of the group $Aut_\mathcal{O}$ of changes of coordinates then they are the same as quasi-conformal vertex algebras (notation of Frenkel Ben-Zvi). The $Aut_\mathcal{O}$-equivariant complex as you describe on the punctured disk will be equivalent to ours in the quasi-conformal setting. This can also be performed on a formal disc $D^*$, in which case one obtains what one might call Vertex Homology. Has this been studied anywhere? Is there a relation with the work of BdSHK cited above, in particular does the BdSHK complex act on it? Again if you want something that should be called "vertex homology" then you should consider either $Aut_{\mathcal O}$ equivariant objects on the disk or translation equivariant objects on the line. There are a few statements in this answer that are plain wrong cause I had in mind translation invariants sections on the disk. I'll strike them through. Factorization homology of a unital chiral algebra vanishes on an open proper subset of a curve. Factorization homology of the restriction of a chiral algebra to an affine proper subset of a curve vanishes. This is the content of Lemma 4.3.4 in Beilinson and Drinfeld's book. I also believe that David's comments about Factorization homology on the disk are wrong. There's a big difference between the topological setting and the algebraic setting and between $E_2$ algebras and the chiral/factorization algebras of Beilinson and Drinfeld. In particular, the chiral homology of the disk with coefficient on any unital chiral algebra vanishes. This is because chiral homology is not just the CE complex of a Lie algebra on the Ran space, but also it's de Rham complex. In particular as you point out the degree zero homology of this complex is $V/V_{(-1)}V = 0$ and this is because we have the vacuum vector to play around. Therefore the image of $\mathbb{1}$ in chiral homology vanishes and then by the last comment before the mentioned Lemma 4.3.4 the whole complex vanishes. Chiral homology of a curve is really a global object and you cannot detect any of it by looking at formal disks (see however 2.4.12 in BD in agreement with David's comments). The complex studied in BdSHK is purely a local object, or rather a translation equivariant complex on the line. I am not aware of any direct connection between them. There are however a couple of instances where you can compute the global chiral homology by local considerations: the case of $X=\mathbb{P}^1$ and the case of an elliptic curve. Both cases because you have global coordinates. In degree $0$ this is the work of Zhu way before chiral homology was defined. In the elliptic curve case we have a preprint with J. van Ekeren describing the nodal curve limit of the first chiral homology group and we are supposed to post a continuation describing the general elliptic curve soon. Edit: To reflect David's comments. Indeed by $U\subset X$ open they really mean an open affine, so that the example of a disconnected curve is not allowed. The point needed is the vanishing of the de Rham cohomology. They use this lemma heavily in proving the quasi-isomorphism of the complex with supports in 4.4.2--4.4.3. Commutative vertex algebras are not a counter-example to this statement as the functor of coinvariants on the disk is canonically zero as long as your algebras are unital. Here the very naive notion of coinvariants is clear, the Cousin complex on $D^2$ is already enough to compute the homology in degree $0$ and this group is $V/V_{(-1)}V$ (think of homologies with support at the origin). As for the $E_2$ algebras business: here is where things get sloppy. As far as I know the folklore is that $E_2$ algebras are essentially the same as topological vertex algebras. With this I have no objection, however notice that the latter is a vertex algebra together with a BRST differential, and that this equivalence is inherently derived. Since the BRST cohomology of a topological vertex algebra is typically stupid (it lies in conformal weight zero and is a commutative algebra) then it's no wonder that the chiral homology viewed as a complex with this extra BRST differential is silly. However the complex without taking this differential into account is still very interesting, you capture irreducible modules and Jacobi elliptic functions on genus $1$ (at least in rational cases) and so forth. The second point is more subtle: I have seen in many places the analogy drawn between Factorization homology in the topological world and chiral homology in the algebraic setting. I may be wrong and very much outdated in the literature that I can cope with, but I have not seen anything written to prove an actual dictionary. There is no formal way of constructing a factorization algebra (in the topological meaning) coming from a factorization algebra in the algebraic world. Folklore expected that if one takes the complex that computes chiral homology on the Ran space of $X$ á la BD and one takes an analitification of this and then sections with compact support, this is something that looks like a Factorization algebra in the topological setting. As far as I know the problem is that the process of analytification (or tensoring with the analytic de Rham complex) does not commute with tensor products and this breaks down being a Lie/Commutative algebra. I don't claim that having a topological intuition is wrong, just that at least when trying to translate that intuition into an actual theorem in BD's setting the situation is not so transparent even in the simplest of cases (for example the zeoro-th Hochshild Homology of higher Zhu algebras equals zero-th chiral homology of vertex algebras, this is a theorem that requires a lot of theory and a bunch of further assumptions that have no topological counterparts).<|endoftext|> TITLE: Weyl map for $SU(n)$ QUESTION [9 upvotes]: Let $G= SU(n)$ and let $\mathbb{T}$ be the maximal torus in $G$ given by diagonal matrices. We have $$ H^*(G,\mathbb{Q}) \cong \Lambda_{\mathbb{Q}}[x_3, x_5, \dots, x_{2n-1}] \ . $$ Now consider the Weyl map $$ p \colon G/\mathbb{T} \times \mathbb{T} \to G \quad , \quad ([g],z) \mapsto gzg^{-1}\ . $$ The induced map in rational cohomology $$ p^* \colon H^*(G,\mathbb{Q}) \to H^*(G/\mathbb{T}, \mathbb{Q}) \otimes H^*(\mathbb{T},\mathbb{Q}) $$ is injective. In fact, if we restrict the codomain to fixed points with respect to a certain action of the Weyl group $W$ it becomes an isomorphism (see for example Reeder, On the Cohomology of Compact Lie Groups). The cohomology ring $H^*(\mathbb{T},\mathbb{Q})$ is isomorphic to another exterior algebra and there are also explicit descriptions of $H^*(G/\mathbb{T},\mathbb{Q})$ (see Reeder's paper again for a reference). Is there a formula describing $p^*(x_{2i+1})$ in terms of any set of natural generators for the codomain? REPLY [7 votes]: First, let me fix generators for $H^*(SU(n))$ and $H^*(SU(n)/\mathbb T)$: For the first, consider the vector bundle on $\Sigma SU(n)$ with clutching map $\operatorname{id}_{SU(n)}$, i.e. with classifying map $f_n\colon\Sigma SU(n)\simeq \Sigma\Omega BSU(n)\to BSU(n)$, and let $\Sigma x_{2i-1} = f^*c_i$. For the second, let $\pi_k\colon \mathbb T\subset U(1)^k\to U(1)$ be the projection to the $k$-th coordinate and $L_k = SU(n)\times_{T,\pi_k} \mathbb C$ be the associated line bundle, and set $y_k = c_1(L_k)$; since $\bigoplus_{k=1}^n \pi_k$ is the restriction of the defining representation of $SU(n)$ to $\mathbb T$, the sum $E := \bigoplus_{k=1}^n L_k$ is trivial, so that all symmetric polynomials in the $y_k$ vanish since they can be expressed via Chern classes of $E$. The induced map $Q[y_1,\dots,y_n]/Q[y_1,\dots,y_n]^{S_n}\to H^*(SU(n)/\mathbb T)$ is an isomorphism; a basis for this cokernel is given by the monomials $y_1^{\alpha_1}\dots y_n^{\alpha_n}$ with $0\le \alpha_k < k$. Now think of $p:G/\mathbb T\times\mathbb T\to SU(n)$ as an automorphism $\phi$ of $E\boxtimes\underline{\mathbb C}$; it splits as $\phi = \phi_1\oplus\dots\oplus \phi_n$, where $\phi_k = \operatorname{id}_{L_k}\boxtimes \pi_k$ is the automorphism of $L_k\boxtimes \underline{\mathbb C}$ obtained as the external tensor product of the identity of $L_k$ with the projection $\pi_k:\mathbb T\subset U(1)^k\to U(1)$. Consider the clutching construction $(E\boxtimes\underline{\mathbb C})^\phi\to G/\mathbb T\times\mathbb T\times S^1$ which is obtained by gluing the two ends of $E\boxtimes\underline{\mathbb C}$ together using $\phi$. Denote by $z_k = \pi_k^*([U(1)]), u = [S^1]$ the obvious cohomology classes in degree $1$, and itdentify them and the $y_k$ with their image in the product. An easy argument shows that $c_1(\mathbb C^{\pi_k}) = uz_k\in H^2(\mathbb T\times S^1)$. By naturality, we have \begin{align*} (E\boxtimes\underline{\mathbb C})^\phi &\cong \bigoplus_{k=1}^n(L_k\boxtimes\underline{\mathbb C})^{\phi_k}\\ c_1\big((L_k\boxtimes\underline{\mathbb C})^{\phi_k}\big) &= c_1(L_k) + c_1(\underline{\mathbb C}^{\pi_k}) = y_k + uz_k\\ c\big((E\boxtimes\underline{\mathbb C})^\phi\big) &= \prod_{k=1}^n c\big((L_k\boxtimes\underline{\mathbb C})^{\phi_k}\big)\\ &= \prod_{k=1}^n (1 + y_k + uz_k)\\ &= \underbrace{c(E)}_{= 0} + u\sum_{k=1}^n z_k \sum_{S\subset \{1,\dots,n\}\smallsetminus\{k\}}\prod_{l\in S} y_l \end{align*} It is no surprise that this expression vanishes for $u=0$ since we can use the chosen trivialization of $E\boxtimes \underline{\mathbb C}$ to descend $(E\boxtimes \underline{\mathbb C})^\phi$ to $\Sigma(G/\mathbb T\times \mathbb T)$, and the resulting vector bundle has classifying map $f_n\circ \Sigma p$. Chasing through the definitions, we see that $$ p^* x_{2i-1} = \sum_{k=1}^n\Big[\sum_{\substack{S\subset \{1,\dots,n\}\smallsetminus\{k\}\\|S| = i - 1}}\prod_{l\in S} y_l\Big]\otimes z_k\ . $$<|endoftext|> TITLE: Show that this process is not a martingale QUESTION [9 upvotes]: I am cross-posting this question from MSE since I did not received any answer, furthermore I tried asking some professors in my university but still we could not find an answer. The most surprising thing is that this exercise was taken from an introductory text on Stochastic Analysis. (Introduction to Stochastic Integration by Hui-Hsiung Kuo) Assume we have the following stochastic process: $$X_t=\int_0^t e^{B(s)^2}dB(s)\, ,0\leq t \leq 1$$ where $(B)_{t\geq 0}$ is a Brownian Motion. I have to show that $X_t$ is not a martingale. I know that if $t< \frac 1 4$ then $\int_0^t \mathbb E(e^{2B(s)^2})ds < \infty $ and then the process is a martingale, this makes me think that $X_t$ is actually a local martingale, but I don't see how to prove that it's not a proper martingale. Thanks in advance. REPLY [7 votes]: Here's an approach that comes from Li, Xue-Mei, Strict local martingales: examples, Stat. Probab. Lett. 129, 65-68 (2017). ZBL1386.60159, https://arxiv.org/abs/1609.00935. Indeed, she mentions this very example after Corollary 4 (bottom of page 4 in the arXiv version). Lemma. Suppose $X_t$ is a continuous martingale, and let $\langle X \rangle_t$ be its quadratic variation. Then for every $0 < \alpha < 1$ and every $t$, we have $E[\langle X \rangle_t^{\alpha/2}] < \infty$. Proof. Let $M_t = \sup_{0 \le s \le t} |X_s|$. By Doob's maximal inequality, for any $x > 0$ we have $$P(M_t \ge x) \le \frac{1}{x} E|X_t|.$$ So $$\begin{align*} E[M_t^\alpha] &= \int_0^\infty P(M_t^\alpha \ge x)\,dx \\&\le 1 + \int_1^\infty P(M_t^\alpha \ge x)\,dx \\& \le 1 + E|X_t| \cdot \int_1^\infty x^{-1/\alpha}\,dx \\&< \infty.\end{align*}$$ Then by the Burkholder-Davis-Gundy inequality we have$$E[\langle X \rangle_t^{\alpha/2}] \le C_\alpha E[M_t^\alpha] < \infty$$ as desired. $\Box$ Now for the process at hand, we have $\langle X \rangle_t = \int_0^t e^{2 B_s^2}\,ds$. By Hölder's inequality or Jensen's we have $\langle X \rangle_t^\alpha \ge t^{\alpha-1} \int_0^t e^{2 \alpha B_s^2}\,ds$, so by Fubini/Tonelli $E[ \langle X \rangle_t^\alpha] \ge t^{\alpha - 1} \int_0^t E[e^{2 \alpha B_s^2}]\,ds$. The integrand is infinite for all $s > 1/4$, so by the lemma $X_t$ cannot be a martingale for any $t > 1/4$. I feel like perhaps there should be a more direct way to relate $E[\langle X\rangle_t^{\alpha/2}]$ to $E[X_t]$, perhaps by some clever application of Itô's formula and Hölder's inequality, but I don't quite see how. REPLY [4 votes]: Let $B_t:=B(t)$. By the Itô formula $$f(B_1)-f(B_0)=\int_0^1 f'(B_t)\,dB_t+\frac12\,\int_0^1 f''(B_t)\,dt$$ with $f(b):=\int_0^b e^{a^2}da$ (with $\int_0^b:=-\int_b^0$ for $b<0$), we have \begin{equation*} X_1=\int_0^1 f'(B_t)\,dB_t =f(B_1)-\frac12\,\int_0^1 f''(B_t)\,dt =f(B_1)-\int_0^1 B_t e^{B_t^2}\,dt. \tag{1} \end{equation*} From here, it is not hard to see that $EX_1$ does not exist. Indeed, consider the event \begin{equation*} A:=\{B_u\in[b,b+1/b],B_1-B_u<-2/b,M_u>-b\}, \end{equation*} where $b\to\infty$, \begin{equation*} u:=1-1/b^2, \end{equation*} \begin{equation*} M_u:=\min_{t\in[u,1]}(B_t-B_u). \end{equation*} Note that on the event $A$ we have $B_u\ge b$, $B_10$ for all $t\in[u,1]$. Therefore and because (by the l'Hospital rule) $f(b)\sim e^{b^2}/(2b)$, it follows from (1) that on $A$ \begin{align*} X_1&\le\frac{e^{(b-1/b)^2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\ &=\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u B_t e^{B_t^2}\,dt \\ &\le\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u (\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}\,dt, \end{align*} where $B_t^u:=B_t-\frac tu\,B_u$; for the latter, displayed inequality, we recall that $B_u\ge b$ on $A$ and use the fact that $se^{s^2}$ is increasing in real $s$. The Brownian bridge $(B_t^u)_{t\in[0,u]}$ is a zero-mean (Gaussian) process independent of $(B_u,B_1-B_u,M_u)$; so, $(B_t^u)_{t\in[0,u]}$ is independent of the event $A$. Introduce also the event $$C_x:=\{\max_{t\in[0,u]}|B_t^u|\le x\}$$ for real $x>0$, which allows us to use the Fubini theorem to get \begin{align*} EX_1\,1_{A\cap C_x}\le\Big(&\frac{e^{b^2-2}}{(2+o(1))b}P(C_x) \\ &-\int_0^u E(\tfrac tu\,b+B_t^u) \exp\{(\tfrac tu\,b+B_t^u)^2\}1_{C_x}\,dt\Big)\,P(A). \tag{2} \end{align*} Because $g_a(b):=(a+b) e^{(a+b)^2}+(a-b)e^{(a-b)^2}$ is convex and even in real $b$ for each real $a\ge0$, we have $g_a(b)\ge g_a(0)=2ae^{a^2}$. Therefore and because the distribution of the Brownian bridge $(B_t^u)_{t\in[0,u]}$ is symmetric, \begin{align*} E(a+B_t^u) \exp\{(a+B_t^u)^2\}1_{C_x} =\tfrac12\,Eg_a(B_t^u)\,1_{C_x} \ge ae^{a^2}\,P(C_x) \end{align*} for $a\ge0$. So, by (2), \begin{align*} EX_1\,1_{A\cap C_x}&\le\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\int_0^u \tfrac tu\,b \exp\{(\tfrac tu\,b)^2\}\,dt\Big)\,P(C_x)P(A) \\ &=\Big(\frac{e^{b^2-2}}{(2+o(1))b}-\frac u{2b} \, (e^{b^2}-1)\Big)\,P(C_x)P(A) \\ &=-e^{b^2(1+o(1))}\,P(C_x)P(A). \end{align*} On the other hand, letting now $x\to\infty$, we have $P(C_x)\to1$ and, still with $b\to\infty$, \begin{align*} P(A)&=P(B_u\in[b,b+1/b])P(B_1-B_u<-2/b,M_u>-b) \\ &\ge P(B_u\in[b,b+1/b])[P(B_1-B_u<-2/b)-P(M_u\le-b)] \\ &=e^{-b^2/(2+o(1))}[P(B_1<-2)-o(1)]=e^{-b^2/(2+o(1))}. \end{align*} Thus, \begin{align*} EX_1\,1_{A\cap C_x}&\le-e^{b^2(1+o(1))}\,(1-o(1))\,e^{-b^2/(2+o(1))}=-e^{b^2/(2+o(1))}\to-\infty, \end{align*} which shows that indeed $EX_1$ does not exist.<|endoftext|> TITLE: Stability of fractional Sobolev spaces under diffeomorphisms QUESTION [5 upvotes]: Let $H^s_p(\mathbb{R}^n)$ be the fractional Sobolev space of fractional order $s\in \mathbb{R}$, for $1 TITLE: Classification of $\operatorname{Rep}D(H)$ QUESTION [5 upvotes]: Question Let $H$ be a finite dimensional complex Hopf algebra and $D(H)$ its quantum double. Can we classify the simple objects in $\operatorname{Rep}D(H)$ if the representations of $H$ are well-understood? My impression is that this is not yet completely developed. However, the more examples the better! Would you mind sharing papers/resources that deal with the representations of quantum doubles, even those that deal with specific examples? Example: finite group algebra For example, if $H$ is the complex group algebra of some finite group $G$, both $\operatorname{Rep}(G)$ and $\operatorname{Rep}D(G)$ are well-enough understood (to my standard). See my previous questions about this Representations of $D(G)$ as an object in the center of $\operatorname{Rep}(G)$ Classification of $\operatorname{Rep} D(G)$ Interestingly, irreducible representations of $D(G)$ do not restrict to irreducible ones as $G$-reps. Viewing them as objects of $Z(\operatorname{Rep}G)$ reveals hidden structures among irreps of $G$: nontrivial half-braidings arise naturally! Example: Taft algebra As another example, the representations of the double of Taft algebras are examined here by Chen. REPLY [2 votes]: To any skew-pairing $\lambda:U\otimes H\rightarrow k$, one can associate a hopf algebra $D(U, H)$ (built on $U\otimes H$) which is called the generalized quantum double of $U$ and $H$. (If $H$ is finite dimensional, $U=H^{*cop}$ and $\lambda$ is the usual evaluation map, then, this corresponds to the usual quantum double $D(H)$ introduced by Drinfeld). In the article On the irreducible representations of generalized quantum doubles, the authors describe the irreducible representations of generalized quantum double hopf algebras $D_{\lambda_f}(U,H)$, where $\lambda_f:U\otimes H\rightarrow k$ is a skew-pairing induced by a surjective hopf algebra map $f:U\rightarrow H^{*cop}$. One of the main results of the paper (as i understand it) is theorem 1.1 (p.2), which describes the simple objects of $\operatorname{Rep}\big(D_{\lambda_f}(U,H)\big)$, in a way which generalizes the corresponding description of the simple objects of $\operatorname{Rep}D(G)$ (discussed in the questions linked to the OP). Then the paper proceeds in a classification of the generalized quantum double, simple modules, for the case where both $U,H$ are semisimple hopf algebras and there is a surjective map $U\rightarrow H^{*cop}$. Furthermore: In The Representations of Quantum Double of Dihedral Groups, the finite dimensional, indecomposable, left $D(kD_n)$-modules are classified, where $D_n$ is the dihedral group of order $2n$ and $k$ is an algebraically closed field, of odd characteristic $p\ |\ 2n$.<|endoftext|> TITLE: fundamental groups of complements to countable subsets of the plane QUESTION [14 upvotes]: This question is a follow-up of this MSE post and a comment by Henno Brandsma: Question 1. Let $S$ be the set of isomorphism classes of fundamental groups $\pi_1(E^2 - C)$, where $C$ ranges over all countably infinite subsets of the Euclidean plane $E^2$. What is the cardinality of $S$? All what I can say is that $S$ contains at least two non-isomorphic groups: One is the free group $F_\omega$ of countably infinite rank, the fundamental group of the complement to a closed discrete countably infinite subset of $E^2$, does not matter which one, say, $C={\mathbb Z}\subset {\mathbb R}\subset {\mathbb R}^2$. The other is $G_{{\mathbb Q}^2}=\pi_1(E^2-C)$, where $C$ is a dense countable subset of $E^2$, again, does not matter which one, for instance, $C={\mathbb Q}^2$. This group is not free since it contains, for instance, the fundamental group of the Hawaiian Earrings. (Actually, it contains $\pi_1$ of every nowhere dense planar Peano continuum.) The natural expectation is that $S$ is uncountable (more precisely, has the cardinality of continuum: It is clear that the cardinality of $S$ cannot be higher than that). Edit. Following Yves' suggestion: Question 2. Let ${\mathbb H}$ denote the Hawaiian Earrings. Is the fundamental group $\pi_1({\mathbb H})$ essentially freely indecomposable? (Here a group $G$ is essentially freely indecomposable if in every free product decomposition $G\cong G_1\star G_2$, one of the free factors $G_1, G_2$ is free of finite rank.) One can also ask for the weaker property of $G=\pi_1({\mathbb H})$, namely, that $G$ does not admit free product decompositions $G\cong G_1\star G_2$ with two uncountable factors. The only relevant result I could find in the literature is the theorem (due to Higman) which (according to "The combinatorial structure of the Hawaiian earring group" by Cannon and Conner) implies that that every freely indecomposable free factor of $G=\pi_1({\mathbb H})$ is either trivial or infinite cyclic. Maybe Higman's methods prove more, but his paper ("Unrestricted Free Products, and Varieties of Topological Groups", Journal of LMS, 1952) is behind the paywall. If Q2 (even in the weaker form) has positive answer, then in Q1 at least one can say that $S$ is infinite. REPLY [4 votes]: Thanks to the comments, my original cardinality bound $\aleph_1\leq |S|\leq \mathfrak{c}$ has been refined to the equality $|S|=\mathfrak{c}$ that I originally suspected. For Question 1: $S$ has the cardinality of the continuum. It's clear that $|S|\leq \mathfrak{c}$. Below, I'll argue that $|S|$ is at least the cardinality of the set of homeomorphism types of closed nowhere dense subsets of $[0,1]$. Since this set has cardinality $\mathfrak{c}$ (using Pierre PC's comment below), we have $|S|\geq \mathfrak{c}$. For the lower bound, I'll use a construction from this paper. Consider any infinite closed nowhere dense subset $A\subseteq[0,1]$ containing $\{0,1\}$. Let $\mathcal{I}(A)$ denote the ordered set of components of $[0,1]\backslash A$. For each $I=(a,b)\in\mathcal{I}(A)$, let $$C_I=\left\{(x,y)\in\mathbb{R}^2\mid y\geq 0,\left(x-\frac{a+b}{2}\right)^2+y^2=\left(\frac{b-a}{2}\right)^2\right\}$$ be the semicircle whose boundary is $\{(a,0),(b,0)\}$. Let $$\mathbb{W}_{A}=([0,1]\times\{0\})\cup \bigcup_{I\in\mathcal{I}(A)}C_I$$ with basepoint $(0,0)$. Here is an example where $\mathcal{I}(A)$ has the order type of the linear order sum $\omega^{\ast}+\omega+1+\omega^{\ast}$ where $\ast$ denotes reverse order. First, notice that $\mathbb{W}_A$ is a one-dimensional Peano continuum (connected, locally path-connected, compact metric space). By picking a single point in the interior of each simple closed curve $C_I\cup (\overline{I}\times \{0\})$, we can see that $\mathbb{W}_A$ is homotopy equivalent to $E^2\backslash C$ for some countably infinite set $C$. The fundamental groups $\pi_1(\mathbb{W}_A)$ ranging over all such $A$ realize continuum-many non-isomorphic groups. Here are the heavy hitting theorems that get the job done. Eda's homotopy classification of 1-dimensional Peano continuua: two one-dimensional Peano continua are homotopy equivalent if and only if they have isomorphic fundamental groups. I would just like to pause and emphasize how incredible and powerful this result is. When you first hear about it and realize the kind of groups and spaces it applies to, you might get the impression that you are being scammed. Let $\mathbf{w}(X)$ denote the subspace of $X$ consisting of the points at which $X$ is not semilocally simply connected, i.e. the 1-wild set of $X$. For general spaces, the homotopy type of $\mathbf{w}(X)$ is a homotopy invariant of $X$, but monodromy actions in one-dimensional spaces have discrete graphs (see 9.13 of this paper of mine with H. Fischer). Among spaces whose monodromy actions have discrete graphs, the homeomorphism type of $\mathbf{w}(X)$ becomes a homotopy invariant of $X$ (see 9.15 of the same paper). Thus, among 1-dimensional spaces, $\mathbf{w}(X)$ is a homotopy invariant. This result is kind-of embedded in Eda's work leading up to 1. but the core idea behind what is really going on is fleshed out in Section 9 of the linked paper. By combining 1. and 2. we have: Corollary: If one-dimensional Peano continua $X$ and $Y$ have isomorphic fundamental groups, then $\mathbf{w}(X)\cong \mathbf{w}(Y)$. A direct consequence is that the Hawaiian earring group and the free product of the Hawaiian earring group with itself are not isomorphic because two copies of $\mathbb{H}$ adjoined by an arc has two 1-wild points. Returning back to the spaces $\mathbb{W}_A$, notice that $\mathbf{w}(\mathbb{W}_A)$ is homeomorphic to the Cantor Bendixsion derivative of $A$, i.e. the subspace of non-isolated points of $A$. Every closed nowhere dense set $B\subseteq [0,1]$ is the Cantor Bendixsion derivative of some other $A$. Hence, the 1-wild sets of the $\mathbb{W}_A$ realize all closed nowhere dense subsets of $B\subseteq [0,1]$. By the corollary, each homeomorphism class of a nowhere dense closed subset of $[0,1]$, gives a unique isomorphism class of fundamental group $\pi_1(\mathbb{W}_A)$ and hence a unique isomorphism class of a fundamental group $\pi_1(E^2\backslash C)$ for some countably infinite set $C$. In the comments below, Pierre PC gives a construction of $\mathfrak{c}$-many non-homeomorphic closed nowhere dense subsets of $[0,1]$ (one can confirm this by analyzing the Cantor Bendixson derivatives of the neighborhoods of the described "super limit points"). Hence, $|S|=\mathfrak{c}$.<|endoftext|> TITLE: Nontrivially fillable gaps in published proofs of major theorems QUESTION [80 upvotes]: Prelude: In 1998, Robert Solovay wrote an email to John Nash to communicate an error that he detected in the proof of the Nash embedding theorem, as presented in Nash's well-known paper "The Imbedding Problem for Riemannian Manifolds" (Annals of Math, 1956), and to offer a nontrivial fix for the problem, as detailed in this erratum note prepared by John Nash. This topic is also discussed in this MO question. Of course, any mathematician who has been around long enough knows of many published proofs with significant gaps, some provably irreparable, and some perhaps authored by himself or herself. What makes the above situation striking--and discomforting to many of us--is the combination of the following three factors: (1) The theorem whose proof is found faulty is a major result that was published in 1950 or after, in a readily accessible source to experts in the field . (I chose the 1950 lower bound as a way of focusing on the somewhat recent past). (2) The gap detected is filled with a nontrivial fix that is publicly available and consented to by experts in the field (so we are not talking about gaps easily filled, or about gaps alleged by pseudomathematicians, or about false publicly accepted theorems, as discussed in this MO question). (3) There is an interlude of 30 years or more between the publication of the proof and the detection of the gap (I chose 30 years since it is approximately the age difference between consecutive generations, even though the interlude is 42 years in the case of the Nash embedding theorem). Question to fellow mathematicians: what is the most dramatic instance you know of where all of the three above factors are present? REPLY [3 votes]: Not too long to wait to find the mistake, but the following might be interesting: In 1905, Henri Lebesgue claimed to have proved that if B is a subset of $\mathbb{R}^2$ with the Borel property, then its projection onto a line is a Borel subset of the line. This is false. In 1912 Souslin and Luzin defined an analytic set as the projection of a Borel set. All Borel sets are analytic, but, contrary to Lebesgue's claim, the converse is false.<|endoftext|> TITLE: Compact complex surface that admits a Kodaira fibration is Kahler QUESTION [5 upvotes]: A Kodaira fibration is a compact complex surface X endowed with a holomorphic submersion onto a Riemann surface $\pi: X\to\Sigma$ which has connected fibers and is not isotrivial. Is there an easy way to see why a compact complex surface that admits a Kodaira fibration is Kahler? I know for a complex compact surface is Kahler if and only if its first Betti number is even. I wonder it's possible to deduce that a compact complex surface that admits a Kodaira fibration has even Betti number? REPLY [2 votes]: Let $f \colon S \longrightarrow B$ be a Kodaira fibration, and let $F$ be a general fibre. Then by [Kas68, Thm. 1.1] we have $g(B) \geq 2$ and $g(F) \geq 3$. In particular, $S$ contains no rational or elliptic curves: in fact, such curves cannot neither dominate the base (because $g(B) \geq 2$) nor be contained in fibres (because the fibration is by assumption smooth). So every Kodaira fibred surface $S$ is minimal and, by the superadditivity of the Kodaira dimension, it is of general type. In particular, it is not only Kähler but actually algebraic (i.e., projective). References. [Kas68] A. Kas: On deformations of a certain type of irregular algebraic surface, American J. Math. 90, 789-804 (1968). ZBL0202.51702.<|endoftext|> TITLE: Consequences of existence of a certain function from $\omega_1$ to $\omega_1$ QUESTION [14 upvotes]: In his book [1], Paul Larson remarks (Remark 1.1.22) that in L there is a function $h:\omega_1\rightarrow\omega_1$ such that for any countable elementary submodel $X$ of $V_\gamma$ (where $\gamma$ is the first strong limit cardinal), we have the order-type of $X\cap Ord$ is strictly less than $h(X\cap\omega_1)$. What is known about the relationship between the existence of such a function and large cardinal phenomena? Larson remarks that the existence of such a function is consistent with many large cardinals, and later in the book sketches a result of Velickovic showing that no such function exists in the presence of a precipitous ideal on $\omega_1$. What more is known? Happy to also learn results concerning other related functions, or be pointed to associated references. [1] Larson, Paul B., The stationary tower. Notes on a course by W. Hugh Woodin, University Lecture Series 32. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3604-8/pbk). x, 132 p. (2004). ZBL1072.03031. REPLY [6 votes]: Not sure if this is what you're interested in, but let me take up the question of what large cardinal axioms are known to be consistent with the existence of $h$. The answer is all large cardinal axioms known to have a canonical inner model. Beyond that, the consistency question remains open: indeed, it remains open whether large cardinal axioms beyond the reach of inner model theory could outright imply the existence of a precipitous ideal on $\omega_1$, and thereby refute the existence of $h$. It seems plausible, however, that $h$ in fact exists in the canonical inner models yet to be discovered. This is because one can construct a function that is almost exactly like $h$ (see the fourth-to-last paragraph below) using a general condensation principle due to Woodin called Strong Condensation, which likely holds in all canonical inner models. If $\kappa$ is a cardinal, Strong Condensation at $\kappa$ states that there is a surjective function $f : \kappa\to H(\kappa)$ of such that for all $M\prec (H(\kappa),f)$, letting $(H_M,f_M)$ be the transitive collapse of $(M,f)$, $f_M = {f}\restriction H_M$. All the known canonical inner models satisfy Strong Condensation at their least strong limit cardinal. (Strong condensation cannot hold at any cardinals past the first Ramsey cardinal.) Moreover by a theorem of Woodin, under $\text{AD}^+ + V = L(P(\mathbb R))$, for a Turing cone of reals $x$, $\text{HOD}_x$ satisfies Strong Condensation at its least strong limit cardinal. This heuristically argues that Strong Condensation should hold at the least strong limit cardinal in canonical inner models satisfying arbitrarily strong large cardinal axioms, assuming that such models exist. The reason is that the pattern observed in inner model theory to date suggests that these models should locally resemble the $\text{HOD}$s of determinacy models. Here is the approximation to the existence of $h$ one gets assuming Strong Condensation at the least strong limit cardinal $\gamma$: There is a set $a\subseteq \omega_1$ and a function $g:\omega_1\to\omega_1$ such that for any $N\prec V_\gamma$ with $a\in N$, $g(N\cap \omega_1) > \text{ot}(N\cap \gamma)$. The rest of this answer consists of a proof of this fact. Fix $f : \gamma\to H(\gamma)$ witnessing Strong Condensation at $\gamma$. The first step of the proof is cosmetic. One uses a theorem of Woodin which states that $f$ is definable over $H(\gamma)$ from the parameter $f \restriction \omega_1$. Let $a\subseteq \omega_1$ code $f\restriction \omega_1$. (It is an easy exercise to show that $f\restriction \omega_1\in H(\omega_2)$.) Then every $N\prec V_\gamma$ with $a\in N$ has the property that $N\cap H(\gamma)\prec (H(\gamma),f)$. For $\alpha < \gamma$, let $P_\alpha = f[\alpha]$. Note that the $P_\alpha$ are increasing with union $H(\gamma)$, and if $M\prec (H(\gamma),f)$ and $\text{ot}(M\cap \gamma) = \alpha$, then the transitive collapse of $M$ is equal to $P_\alpha$. The structures $P_\alpha$ will play the role of the $L_\alpha$ hierarchy in Larson's proof. For every $\xi < \omega_1$, let $g(\xi)$ be the least ordinal $\alpha$ such that there is a surjection from $\omega$ to $\xi$ in $P_\alpha$. Suppose $N\prec V_\gamma$ and $a\in N$. We will show that $g(N\cap \omega_1) > \text{ot}(N\cap \gamma)$. Let $M = N\cap H(\gamma)$, so $M\prec (H(\gamma),<)$. Clearly it suffices to show that $g(M\cap \omega_1) > \text{ot}(M\cap \gamma)$. Let $H_M$ be the transitive collapse of $M$. Then $M\cap \omega_1 = \omega_1^{H_M}$ and letting $\beta = \text{ot}(M\cap \gamma)$, $H_M = P_\beta$. Assume towards a contradiction that $\beta \geq g(\omega_1^{H_M})$. By the definition of $g$, there is a surjection from $\omega$ to $\omega_1^{H_M}$ in $P_\beta$. But since $P_\beta = H_M$, this contradicts that $\omega_1^{H_M}$ is uncountable in $H_M$.<|endoftext|> TITLE: Cobordism and Kirby calculus QUESTION [12 upvotes]: It may be a simple question but I wonder to ask: Is it possible to draw a homology cobordism between $3$-manifolds by using the techniques of Kirby calculus? At least, for instance, Brieskorn spheres? REPLY [13 votes]: As Golla pointed out that since every smooth $4$-manifold has a handle decomposition, you can draw a Kirby diagram. See the following pretty nice picture from Akbulut's lecture notes (now it is a published book titled $4$-manifolds). Golla listed that lots of Brieskorn spheres which are known to be bound integral or rational homology balls, i.e., they are all integral or rational homology cobordant to $S^3$. Following the technique of Akbulut and Larson, I also recently found new Brieskorn spheres bounding rational homology balls: $\Sigma (2,4n+3,12n+7)$ and $\Sigma(3,3n+2,12n+7)$, see the preprint. It is interesting to note that the following Brieskorn spheres bound rational homology balls but not integral homology balls (these $4$-manifolds must contain $3$-handle(s)): $\Sigma(2,3,7)$, $\Sigma(2,3,19)$, $\Sigma(2,4n+1,12n+5)$ and $\Sigma(3,3n+1,12n+5)$ for odd $n$, $\Sigma(2,4n+3,12n+7)$ and $\Sigma(3,3n+2,12n+7)$ for even $n$. On the other hand, we know that every closed oriented $3$-manifold is cobordant to $S^3$ due to the celebrated theorem of Lickorish and Wallace. In the following picture, you can see the explicit cobordism from $\Sigma(2,3,13)$ to $\Sigma(2,3,7)$ which is constituted by adding the red $(-1)$-framed $2$-handle. (The knot pictures are from KnotInfo). Here, blow down the red one to get the right-hand side. (Of course, they are not integral homology cobordant.) Note that the Brieskorn spheres $\Sigma(2,3,6n+1)$ are obtained by $(+1)$-surgery on the twist knots $(2n+2)_1$, see for example Saveliev's book pg. 49-50. Here, $6_1$ is the stevedore knot (the left knot in the figure) and $4_1$ is the figure-eight knot (the right one). REPLY [8 votes]: There are many examples of the sort, in effect. As far as I know, Akbulut and Kirby (Mazur manifolds, Michigan Math. J. 26 (1979)) proved that $\Sigma(2,5,7)$, $\Sigma(3,4,5)$, and $\Sigma(2,3,13)$ bound contractible 4-manifolds; their work was then extended by Casson and Harer (Some homology lens spaces which bound rational homology balls, Pacific Math. J. 96 (1981)). Stern, Fintushel-Stern, and Fickle have more examples. I'm sure that the Akbulut-Kirby (and some of the Casson-Harer) examples were done by Kirby calculus. Additionally, there are also examples of Brieskorn spheres bounding rational homology 4-balls (but not integral ones, because they have non-zero Rokhlin invariant). The first example was $\Sigma(2,3,7)$, and the rational ball was produced by Fintushel and Stern (A $\mu$-invariant one homology 3-sphere that bounds an orientable rational ball, in Four-manifold theory (Durham, NH, 1982) (1984)) by explicit handle moves. This has been extended further; the latest news I have are from a paper of Akbulut and Larson (Brieskorn spheres bounding rational balls, Proc. Amer. Math. Soc. 146 (2018)), where they provide two infinite families of examples: $\Sigma(2,4n+1,12n+5)$ and $\Sigma(3,3n+1,12n+5)$, as well as $\Sigma(2,3,19)$. Akbulut's book 4-manifolds (Oxford University Press) contains a wealth of examples along these lines (not many more with Brieskorn spheres, I should think). Finally, I am not aware of (but would be interested in seeing) explicit, non-trivial examples of (rational or integral) homology cobordisms between Brieskorn spheres.<|endoftext|> TITLE: Homomorphism induced by the second exterior power of a linear map QUESTION [9 upvotes]: Consider the map from $M(n, \mathbb Z) \rightarrow M(\binom{n}{2}, \mathbb Z)$ taking a matrix A to its second compound, i.e, $\bigwedge^2 A$. Restricting this map to the invertible matrices we get a homomorphism of groups from $\mathrm{GL}(n, \mathbb Z)$ to $\mathrm{GL}(\binom{n}{2}, \mathbb Z)$. How can we determine if a given matrix $B \in \mathrm{GL}(\binom{n}{2}, \mathbb Z)$ is contained in the image of this map? REPLY [11 votes]: First, let us discuss the same question over an algebraically closed field (e.g., over $\overline{\mathbb{Q}}$). Let $V$ be a vector space of dimension $n$. The question is to understand the image of the homomorphism $$ \lambda \colon \mathrm{GL}(V) \to \mathrm{GL}(\wedge^2V). $$ Note that $\mathrm{GL}(\wedge^2V)$ acts naturally on the projective space $\mathbb{P}(\wedge^2V)$, which contains as a subvariety the Grassmannian $$ \mathrm{Gr}(2,V) \subset \mathbb{P}(\wedge^2V). $$ Clearly, it is preserved by the action of $\mathrm{GL}(V)$. The converse is also true for $n > 4$, i.e., if $g \in \mathrm{GL}(\wedge^2V)$ is such that $$ g(\mathrm{Gr}(2,V)) \subset \mathrm{Gr}(2,V), $$ then $g \in \mathrm{Im}(\lambda)$. This follows immediately from the isomorphism $$ \mathrm{Aut}(\mathrm{Gr}(2,V)) \cong \mathrm{PGL}(V). $$ Over $\mathbb{Z}$, I guess, the equality $$ \det(\wedge^2g) = \det(g)^{n-1} $$ gives an extra constraint; so besides preserving the Grassmannian one should impose the condition that the determinant is $(n-1)$-st power.<|endoftext|> TITLE: Hard: One more generator needed for a Z/6 elliptic curve QUESTION [8 upvotes]: We are searching for rank 8 elliptic curves with the torsion subgroup $\mathbb{Z}/6$ using newly discovered families similar to Kihara's as described in A. Dujella, J.C. Peral, P. Tadić, Elliptic curves with torsion group $\mathbb{Z}/6\mathbb{Z}$, Glas. Mat. Ser. III 51 (2016), 321-333 doi:10.3336/gm.51.2.03, 1503.03667 and came across a curve [1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988] Both Magma Calculator and mwrank return $7$ generators for this curve: SetClassGroupBounds("GRH"); E:=EllipticCurve([1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988]); MordellWeilShaInformation(E); Both Magma and mwrank return $8$ for the upper bound on rank: E:=EllipticCurve([1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988]); TwoPowerIsogenyDescentRankBound(E); 8 [ 4, 4, 4, 4, 4 ] [ 6, 6, 6, 6, 6 ] mwrank -v0 -p200 -s [1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988] Version compiled on Oct 29 2018 at 22:35:09 by GCC 7.3.0 using NTL bigints and NTL real and complex multiprecision floating point Enter curve: [1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988] Curve [1,0,1,-728177856117250596635013323700992100749546784263413,7725511368374502384905062271934799362136250437256099440874528514783779254988] : selmer-rank = 9 upper bound on rank = 8 Considering parity, there should be one more generator on the curve. Is there a way to find it? We would greatly appreciate any hint leading to the discovery of the extra generator. A bounty of $100$ has been offered for obtaining it. Also, if you can compute an extra generator, your name will be published at the bottom of the page here: https://web.math.pmf.unizg.hr/~duje/tors/z6.html REPLY [17 votes]: A set of points that generate $E(\mathbb{Q})$ modulo torsion is given by (1955516573881233507049678279 : -86467145649172260650105545143411861089140 : 1), (49225691888888099223656060329/10201 : 67749663895993353685065159554645568700902610/1030301 : 1), (61339810590192565389735634 : -440289331793622522908840423931186017125 : 1), (301884243790342804873202050999/1681 : 164095919303197903219089875947912899634054060/68921 : 1), (12495717670305680867142229 : -24031745881863415519418908823242701040 : 1), (48812081421189741670987918753619270029/14228919471376 : -3895612939954697213016286372117889003488190324193605593985/53673248632044722624 : 1), (5561842419887590167868100830494509281/162696869449 : 9905381606012663087305509196041719017978015930195439090/65624921170340293 : 1), (-24644413733187137559835573003063695698428162289232517969749039/810893447144357785058346728220801409 : 30847724470076383865716266151756242512110696731502256770076024073253839003102120576612459770/730206486187013450403786627354716551758061149557632577 : 1) My guess is that you were missing the final one. We can find the final point by applying $4$-descent to the $2$-covering $C_2$ given by $$y^2 = 3600489235862039958255625x^4 - 26108156374576368607091450x^3 + 135553629286468859778411799x^2 + 184563701310722380421312754x + 49111306298020667812024521$$ of $E$, which is one of the 511 $2$-coverings returned by running the command TwoDescent(E) magma. Running the command FourDescent(C2) returns 256 $4$-coverings of $E$. One of them is the curve $C_4$ defined by the intersection of the quadrics $$46500x^2 + 74693xy + 54170xz + 647076xw - 121026y^2 - 196538yz + 862965yw - 212375z^2 - 238791zw + 333744w^2$$ and $$722768x^2 - 2936122xy + 3336517xz - 2782182xw - 2731148y^2 - 13360024yz - 950117yw - 4385375z^2 + 2688700zw - 199207w^2$$ in $\mathbb{P}^3(\mathbb{Q})$. Running PointsQI(C4,2^11) returns a single point $Q = (2834:53:2444:376)$. We can map $Q$ to a rational point on $E$ using the map given as the second return value of the command AssociatedEllipticCurve(C4:E:=E), and the point we get is the final point above.<|endoftext|> TITLE: Could groups be used instead of sets as a foundation of mathematics? QUESTION [28 upvotes]: Sets are the only fundamental objects in the theory $\sf ZFC$. But we can use $\sf ZFC$ as a foundation for all of mathematics by encoding the various other objects we care about in terms of sets. The idea is that every statement that mathematicians care about is equivalent to some question about sets. An example of such an encoding is Kuratowski's definition of ordered pair, $(a,b) = \{\{a\},\{a,b\}\}$, which can then be used to define the cartesian product, functions, and so on. I'm wondering how arbitrary the choice was to use sets as a foundation. Of course there are alternative foundations that don't use sets, but as far as I know all these foundations are still based on things that are quite similar to sets (for example $\sf HoTT$ uses $\infty$-groupoids, but still contains sets as a special case of these). My suspicion is that we could instead pick almost any kind of mathematical structure to use as a foundation instead of sets and that no matter what we chose it would be possible to encode all of mathematics in terms of statements about those structures. (Of course I will add the caveat that there has to be a proper class of whichever structure we choose, up to isomorphism. I'm thinking of things like groups, topological spaces, Lie algebras, etc. Any theory about a mere set of structures will be proved consistent by $\sf ZFC$ and hence be weaker than it.) For concreteness I'll take groups as an example of a structure very different from sets. Can every mathematical statement be encoded as a statement about groups? Since we accept that it is possible to encode every mathematical statement as a statement about sets, it would suffice to show that set theory can be encoded in terms of groups. I've attempted a formalization of this below, but I would also be interested in any other approaches to the question. We'll define a theory of groups, and then ask if the theory of sets (and hence everything else) can be interpreted in it. Since groups have no obvious equivalent of $\sf ZFC$'s membership relation we'll instead work in terms of groups and their homomorphisms, defining a theory of the category of groups analogous to $\sf{ETCS+R}$ for sets. The Elementary Theory of the Category of Sets, with Replacement is a theory of sets and functions which is itself biinterpretable with $\sf ZFC$. We'll define our theory of groups by means of an interpretation in $\sf{ETCS+R}$. It will use the same language as $\sf{ETCS+R}$, but we'll interpret the objects to be groups and the morphisms to be group homomorphisms. Say the theorems of our theory are precisely the statements in this language whose translations under this interpretation are provable in $\sf{ETCS+R}$. This theory is then recursively axiomatizable by Craig's Theorem. Naturally we'll call this new theory '$\sf{ETCG+R}$'. The theory $\sf{ETCS+R}$ is biinterpretable with $\sf ZFC$, showing that any mathematics encodable in one is encodable in the other. Question: Is $\sf{ETCG+R}$ biinterpretable with $\sf ZFC$? If not, is $\sf ZFC$ at least interpretable in $\sf{ETCG+R}$? If not, are they at least equiconsistent? REPLY [31 votes]: The answer is yes, in fact one has a lot better than bi-interpretability, as shown by the corollary at the end. It follows by mixing the comments by Martin Brandenburg and mine (and a few additional details I found on MO). The key observation is the following: Theorem: The category of co-group objects in the category of groups is equivalent to the category of sets. (According to the nLab, this is due to Kan, from the paper "On monoids and their dual" Bol. Soc. Mat. Mexicana (2) 3 (1958), pp. 52-61, MR0111035) Co-groups are easily defined in purely categorical terms (see Edit 2 below). The equivalence of the theorem is given by free groups as follows: if $X$ is a set and $F_X$ is the free group on X then Hom$(F_X,H)=H^X$ is a group, functorially in H, hence $F_X$ has a cogroup object structure. As functions between sets induce re-indexing functions: $H^X \rightarrow H^Y$ that are indeed group morphisms, morphisms between sets indeed are cogroup morphisms. Explicitly, $\mu:F_X \rightarrow F_X * F_X$ is the map that sends each generator $e_x$ to $e_x^L * e_x^R$, and $i$ is the map that sends each generators to its inverse. An easy calculation shows that the generators are the only elements such that $\mu(y)=y^L*y^R$ and hence that any cogroup morphism comes from a function between sets. So the only co-group morphisms are the ones sending generators to generators. And with a bit more work, as nicely explained on this other MO answer, one can check that any cogroup object is of this form. Now, as all this is a theorem of $\sf{ETCS}$, it is a theorem of $\sf{ETCG}$ that all the axioms (and theorems) of $\sf{ETCS}$ are satisfied by the category of cogroup objects in any model of $\sf{ETCG}$, which gives you the desired bi-interpretability between $\sf{ETCS}$ and $\sf{ETCG}$. Adding supplementary axioms to $\sf{ETCS}$ (like R) does not change anything. In fact, one has more than bi-interpretability: the two theories are equivalent in the sense that there is an equivalence between their models. But one has a lot better: Corollary: Given $T$ a model of $\sf{ETCS}$, then $Grp(T)$ is a model of $\sf{ETCG}$. Given $A$ a model of $\sf{ETCG}$, then $CoGrp(A)$ is a model of $\sf{ETCS}$. Moreover these two constructions are inverse to each other up to equivalence of categories. Edit: this an answer to a question of Matt F. in the comment to give explicit example of how axioms and theorems of $\sf{ECTS}$ translate into $\sf{ECTG}$. So in $\sf{ECTS}$ there is a theorem (maybe an axioms) that given a monomorphism $S \rightarrow T$ there exists an object $R$ such that $T \simeq S \coprod R$. In $\sf{ECTG}$ this can be translated as: given $T$ a cogroup object and $S \rightarrow T$ a cogroup monomorphism* then there exists a co-group $R$ such that $T \simeq S * R$ as co-groups**. *: It is also a theorem of $\sf{ECTG}$ that a map between cogroup is a monomorphism of cogroup if and only if the underlying map of objects is a monomorphisms. Indeed that is something you can prove for the category of groups in $\sf{ECTS}$ so it holds in $\sf{ECTG}$ by definition. ** : We can prove in $\sf{ECTG}$ (either directly because this actually holds in any category, or proving it for group in $\sf{ECTS}$) that the coproduct of two co-group objects has a canonical co-group structure which makes it the coproduct in the category of co-groups. Edit 2: To clarify that the category of cogroup is defined purely in the categorical language: The coproduct in group is the free product $G * G$ and is definable by its usual universal property. A cogroup is then an object (here a group) equipped with a map $\mu: G \rightarrow G * G$ which is co-associative, that is $\mu \circ (\mu * Id_G) = \mu \circ (Id_G * \mu)$, and counital (the co-unit has to be the unique map $G \rightarrow 1$), that is $(Id_G,0) \circ \mu = Id_G$ and $(0,Id_G) \circ \mu = Id_G$, where $(f,g)$ denotes the map $G * G \rightarrow G$ which is $f$ on the first component and $g$ on the other component, as well as an inverse map $i:G \rightarrow G$ such that $(Id_G ,i ) \circ \mu = 0 $. Morphisms of co-groups are the map $f:G \rightarrow H$ that are compatible with all these structures, so mostly such that $ (f * f) \circ \mu_H = \mu_G \circ f $. If you have doubt related to the "choice" of the object $G * G$ (which is only defined up to unique isomorphisms) a way to lift them is to define "a co-group object" as a triple of object $G,G *G,G * G *G$ with appropriate map between them satisfying a bunch of confition (includings the universal property) and morphisms of co-group as triple of maps satisfying all the expected conditions. This gives an equivalent category.<|endoftext|> TITLE: Can filtered colimits be computed in the homotopy category? QUESTION [10 upvotes]: For $\mathcal{S}$ the $(\infty,1)$-category of spaces its homotopy category $h\mathcal{S}$ does not have pushouts or pullbacks. Even if it does, they won't always agree with the (homotopy) pushouts or pullbacks in $\mathcal{S}$. Generally, filtered colimits are much better behaved than general ones. For example filtered colimits in the (Quillen) model category $\mathit{sSet}$ compute homotopy colimits in $\mathcal{S}$. In light of this I was wondering the following: Does $h\mathcal{S}$ have filtered colimits and, if so, does the functor $\mathcal{S} \to h\mathcal{S}$ preserve them? More generally one could ask the same for any presentable $(\infty,1)$-category $\mathcal{C}$; I particularly care about the case of the derived category $\mathcal{D}(R)$ of a ring $R$. REPLY [5 votes]: As has already been said, the homotopy category does not admit filtered colimits in general, but it’s much worse than that. Even colimits in an $\infty$-category which don’t give rise to colimits in the homotopy category sometimes do give rise to weak colimits. (A weak colimit cocone gives the existence, but not the uniqueness, of the factorizations a colimit cocone gives.) This is the case, for instance, with sequential colimits, as well as those along any free category, at least in spaces. So one might ask whether at least every filtered colimit in $\mathcal S$ gives rise to a weak filtered colimit in $h\mathcal S$. Alas, this is still not true. In our paper kindly referenced by Tim, Christensen and I give an $\aleph_1$-indexed sequence of spaces whose homotopy colimit is not a weak colimit in the homotopy category, namely the sequence mapping a countable ordinal $\alpha$ to the wedge of $\alpha$ circles. The homotopy colimit is a wedge of $\aleph_1$ circles, and the problem is that a map out of that just requires too much coherence to be constructed out of a cocone over countable wedges in $h\mathcal S$. So there is not much hope for filtered colimits in $h\mathcal S$. I expect the same counterexample would work, though I have no idea how to make the argument, in higher homotopy categories $h_n\mathcal S$. Regarding “minimal” or “distinguished” weak colimits, the general idea is that you want some weak colimits which are distinguished up to at least non-unique isomorphism, as occurs for cones in triangulated categories. Since “homotopy colimits” of sequences in triangulated categories with countable coproduct a are constructed out of those coproduct together with cones, they are also distinguished in this sense. It is possible to get at the idea of minimal weak colimit of at least a filtered diagram in a category which may not be triangulated, but which has some set of objects detecting isomorphisms, by asking that $Hom(S,\mathrm{wcolim} D_i)\cong \mathrm{colim} Hom(S, D_i)$ for every $S$ in your isomorphism-detecting set. Such weak colimits are then indeed determined up to isomorphism, and they’re also nice because they see the objects $S$ as compact. However, this is not to say that such distinguished weak colimits are common! Our diagram from above actually admits no weak colimit which views even $S^1$ as compact in this way. (Though note that some weak colimit always exists-homotopy pushouts give weak pushouts, coproducts exist, and then the usual construction applies.) If your category actually has a set of compact generators in a model, as for $D(R)$, then a distinguished weak colimit must come from a homotopy colimit. Franke gives an argument, cited in our paper, that on these grounds distinguished weak colimits of uncountable chains should essentially never exist in $D(R)$. The problem is that there’s a spectral sequence converging to homs out of a homotopy colimit indexed by $J$ whose $E_2$ page involves the derived functors $R^n\mathrm{lim}^J$ for all $n$. These derived functors were shown by Osofsky to be non vanishing up through $n$ when $J=\aleph_n$, and the homotopy colimit is a weak colimit only if the spectral sequence collapses, so this probably shouldn’t happen. However Franke doesn’t give an argument that there couldn’t in principle be enough unlikely differentials to produce the collapse. Christensen and I tried for a while to work with the analogous spectral sequence for spaces, but it seemed to require a proficiency with calculating higher derived limits unsupported by the literature-Osofsky gives a special example, and for all I can tell nobody else has ever calculated a derived limit over $\aleph_n$. So our approach turns out to be entirely different and doesn’t immediately apply to the stable case. Thus I think it’s unknown, though highly doubtful, whether $D(R)$ admits minimal filtered colimits in general.<|endoftext|> TITLE: When homology isomorphism implies homotopy isomorphism QUESTION [8 upvotes]: Let's suppose that $f:X\rightarrow X$ is a continuous map such that $H_{\ast}(f): H_{\ast}(X)\rightarrow H_{\ast}(X)$ is a homology isomorphism (with integral coefficients) $X$ is a finite connected CW-complex. $\pi_{1}(f): \pi_{1}(X)\rightarrow \pi_{1}(X)$ is an isomorphism of fundamental groups. $\pi_{1}(X)$ is a finitely presented group. $\pi_{n}(f)=0$ for $n>1$. the homotopy colimit $$hocolim(X\rightarrow_{f} X\rightarrow_{f} X\dots)$$ is homotopy equivalent to a finite CW-complex. Does it imply that $f$ has to be a weak homotopy equivalence ? My guess is that the answer should be no but I don't have a counterexample. REPLY [15 votes]: Here's a counterexample. Set $X'=S^1\vee S^2$. Consider the following map $F':S^2\vee S^2\vee S^2\rightarrow X'$: It maps the first $S^2$ summand to the $S^2$ summand of $X'$ via a map that represents $2\in\pi_{2}S^2$; it maps the second summand once around the $S^1$ factor of $X'$, and maps the third $S^2$ summand to the $S^2$ summand in $X'$ by a map that represents $-1\in\pi_{2}S^2$. Let $F$ be the composition of $F'$ with the map $S^2\rightarrow S^2\vee S^2\vee S^2$ which collapses 2 different latitudinal circles. Form $X$ by attaching a 3-cell to $X'$ by the map $F$. Note that $\pi_{1}X=\pi_{1}S^1$, and the inclusion $S^1\hookrightarrow X$ is a homology isomorphism. The map $f:X\rightarrow X$ which collapses $X$ to its $S^1$ summand satisfies all the requirements. In this case the hocolim in requirement 6 is $\simeq S^1$.<|endoftext|> TITLE: How to check whether a given matrix is in the image of a representation? QUESTION [6 upvotes]: Let $G$ be a compact simple Lie group, and let $\rho$ be a (faithful, unitary) irreducible representation thereof of $\mathbb K$-dimension $n$, where $\mathbb K=\mathbb C/\mathbb R/\mathbb H$ if $R$ is real/complex/pseudo-real, respectively. It then follows that there is a subgroup of $SU(n)/SO(n)/Sp(n)$, respectively, isomorphic to $G$. One can think of $\rho$ as a map from $G$ to this subgroup. How can I check whether a given matrix $M\in SU(n)/SO(n)/Sp(n)$ is in the image of $\rho$? In other words, given one such matrix $M$, how can I decide whether there exists some $g\in G$ such that $\rho(g)=M$? For the sake of concreteness, say $G=G_2$ is the first exceptional simple group, and let $\rho$ be the representation with highest weight $2\omega_2$ (which is real and $27$-dimensional). This means that for any $g\in G_2$, $\rho(g)$ is a $27$-dimensional orthogonal matrix. If I take some arbitrary $27$-dimensional orthogonal matrix $M$, how can I check whether it can be written as $M=\rho(g)$ for some $g\in G_2$? Note: I am particularly interested in the case where $M$ is diagonal, but I'd be interested in hearing about the general case as well. In the diagonal case, where everything is abelian, and one can essentially focus on a Cartan subalgebra, I assume one can be quite explicit about the image of $\rho$. In the general case, I wouldn't be surprised if one has to work harder. REPLY [4 votes]: Suppose that $G$ is a compact Lie group (or, more generally, an algebraic group over some field $F$), ${\mathfrak g}$ is its Lie algebra. You are given a linear representation $\rho: G\to GL(n, F)$. From this, compute the representation of the Lie algebra $\rho': {\mathfrak g}\to End(F^n)$. In fact, the representation $\rho$ is frequently given in terms of the highest weight of $\rho'$. Now, you check section 4.5 "Computing defining polynomials of an algebraic group in terms of its Lie algebra" in W. de Graaf, "Computing with Linear Algebraic Groups", CRC, 2017. He describes an algorithm for computing defining polynomial equations $p_i, i=1,...,N$, for $\rho(G)$ in terms of $\rho'({\mathfrak g})$ (more precisely, in terms of a basis for this subalgebra in $End(F^n)$). He even has software (check his webpage https://www.science.unitn.it/~degraaf/) for practical computations of this type. Lastly, to verify if the given matrix $A\in GL(n, F)$ belongs to $\rho(G)$, evaluate the polynomials $p_i$ on $A$.<|endoftext|> TITLE: 3-balls with the same boundary in $S^4$ differ up to diffeomorphism QUESTION [6 upvotes]: I am looking at this recent paper by Budney and Gabai and I am confused by a certain sentence in it. Theorem 4.7 states that if $\Delta_1$ and $\Delta_2$ are two 3-balls smoothly embedded in $S^4$ that are identical near there boundary, then there is a diffeomorphism $\Phi: S^4 \to S^4$ that is the identity on the neighborhood of the boundary of these two balls where they agree and $\Phi(\Delta_1) = \Delta_2$. The proof of this theorem is one sentence long, and this is the sentence where I am seeking some clarification. The sentence cites two facts: (1) Regular neighborhoods are unique (2) $\operatorname{Diff}_0(S^2)$ is connected and two papers: (1) J. Cerf, Topologie de certains espaces de plongements, Bull. Soc. Math. France, 89 (1961), 227–380. (2) R. Palais, Extending diffeomorphisms, Proc. Amer. Math. Soc. 11 (1960), 274–277. I am not sure exactly what parts of these two references are being used. I believe for the Palais paper, maybe the authors are using Corollary 2 to say that there is some diffeomorphism of $S^4$ taking $\Delta_1$ to $\Delta_2$. I am not sure what is in the Cerf paper (it is written in french and over 100 pages so that has kept me away). However, maybe this is the standard reference for the fact that every diffeomorphism of $S^3$ extends over $B^4$? I'm also not exactly sure what the statement "regular neighborhoods are unique" means. I suppose it means that any two regular neighborhoods for a smooth submanifold differ up to homotopy rel the submanifold. I would love it if someone could tell me how to fit these pieces together and understand the proof. REPLY [5 votes]: The cited (early) work by Cerf proves that, given a submanifold Y in a manifold X, the obvious map Diff(X)->Emb(Y,X) is a locally trivial fibration. I guess that Budney and Gabai mean the following. By Palais, all embeddings D^3->S^4 are isotopic. Hence, for i=0, 1, the complement C_i of a small open tubular neighborhood U_i of Delta_i in S^4 is diffeomorphic with the compact 4-ball B^4. One has two disjoint embeddings phi_i, psi_i of B^3 in the boundary S^3 of B^4=C_i, one orientation-preserving, the other orientation-reversing, representing the two sides of Delta_i. It remains to extend the diffeomorphism between C_0 and C_1 through U_0 and U_1. This amounts to find a diffeomorphism f:B^4->B^4 such that f o phi_0=phi_1 and f o psi_0=psi_1. The papers by Palais and Cerf precisely give this. The connexity of Diff_+(S^2), and the unicity of the tubular neighborhood up to isotopy, serve to arrange that the extension goes well on a small neighborhood of the 2-sphere bounding Delta_0 and Delta_1.<|endoftext|> TITLE: Request for an exact formula related to a partition in number theory QUESTION [5 upvotes]: The Frobenius equation is the Diophantine equation $$ a_1 x_1+\dots+a_n x_n=b,$$ where the $a_j$ are positive integers, $b$ is an integer, and a solution $$(x_1, \dots, x_n)$$ must consist of non-negative integers, i.e. $$ x_j \in \mathbb{N} $$ as Natural numbers. For negative $b$, there are no solutions. My question: Is there any known formula that counts the number of solutions, by giving $a_1, \dots, a_n$, $b$, and $n$? Let us call this function as $F (a_1, \dots, a_n; b, n)$, what is known for this: $$F (a_1, \dots, a_n; b, n)=?$$ For all the $a_j=1$, we can simplify the above Frobenius equation to: $$ x_1+\dots+x_n=b, \tag{1}$$ where $b \in \mathbb{Z}^+$ is a positive integer. Here is another simpler question: Is there a general formula for Eq.(1) counting all the possible solutions $$(x_1, \dots, x_n)$$ for given the positive integer $n \in \mathbb{Z}^+$ and $b \in \mathbb{Z}^+$? This should be related to the Partition, but I am not sure the exact forms are known? Say, can we find the total number of soultions as a function $f(n,b)$, and what is $$ f(n,b)=? $$ It seems the answer is known: $$ f(n,b)= \binom{b+n-1}{n-1}. $$ p.s. Sorry if this question is too simple for number theorists. But please provide me answer and Refs if you already know the answer. Many thanks! REPLY [4 votes]: You can take generating function $$f(z):=\frac{1}{1-z^{a_1}}\frac{1}{1-z^{a_2}}\cdots \frac{1}{1-z^{a_n}}$$ as in Max Alekseyev's answer and calculate $F (a_1, \dots, a_n; b, n)$ as $$ \frac{1}{2 \pi i} \int_{|s|=\rho} f(s) \frac{d s}{s^{b+1}} \quad (0<\rho<1). $$ It gives the answer $$ F (a_1, \dots, a_n; b)=\frac{b^{n-1}}{(n-1) ! a_{1} \ldots a_{n}}+\sum_{k=0}^{n-2} c_{k} b^{k}. $$ It is a classical applications of contour integration taken from the book "Residues and their applications" by A.O. Gelfond (1966, pp. 98-99, Russian). If $(a_j,a_k)=1$ ($j\ne k$) then all poles (excepting $s=1$) are simple and formula can be simplified: $$ F (a_1, \dots, a_n; b)=\frac{(-1)^{n-1}}{(n-1) !} \frac{d^{n-1}}{d s^{n-1}}\left[s^{-b-1} \prod_{k=1}^{n} \frac{1-s}{1-s^{a_{k}}}\right]_{s=1}+R $$ where $|R| TITLE: Bounding and domination numbers for relation $\leq$ modulo $\omega$-nullsets QUESTION [7 upvotes]: We say that $A\subseteq \omega$ is a nullset if $$\lim\sup_{n\to \infty} \frac{|A\cap n|}{n+1} = 0.$$ Let $\omega^\omega$ denote the set of functions $f:\omega\to\omega$. We define a pre-ordering relation $\leq^0$ on $\omega^\omega$ by saying that $f\leq^0 g$ if $f(x) \leq g(x)$ for all $x\in\omega\setminus N$ where $N\subseteq \omega$ is a nullset. Similarly to the bounding number and the dominating number respectively, we define ${\frak b}^0 = \min\{|B|: B\subseteq \omega^\omega \land \forall f\in \omega^\omega\; \exists b\in B(b\not\leq^0 f)\}$, and ${\frak d}^0 = \min\{|D|: D\subseteq \omega^\omega \land \forall f\in \omega^\omega\; \exists d\in D(f\leq^0 d)\}$. Do we have ${\frak b}^0={\frak b}$? And what about ${\frak d}^0={\frak d}$? REPLY [4 votes]: The answer seems to be positive according to this paper: Barnabás Farkas, Lajos Soukup: The zero density ideal, cardinal invariants and related forcing problems.1 Theorem 2.3. If $\mathcal I$ is a rare ideal on $\mathbb N$ then $\mathfrak b = \mathfrak b_{\mathcal I}$ and $\mathfrak d = \mathfrak d_{\mathcal I}$. Just before this theorem the authors mention that the ideal $\mathcal Z_0$ of the sets with zero density is a rare ideal. A similar result is shown for analytic P-ideals in Corollary 5.5 of More on cardinal invariants of analytic P-ideals by the same two authors (arXiv, eudml). Again, this class of ideals includes $\mathcal Z_0$. 1I wasn't able to find whether the paper was published somewhere, but a preprint is available here (Wayback Machine). The same paper was also mentioned in this answer: Are these two quotients of $\omega^\omega$ isomorphic?<|endoftext|> TITLE: When is the model structure on functors correct, i.e. when does localization commute with taking functor categories? QUESTION [7 upvotes]: Let $C$ be a small category and $M$ a model category. Then there are various "global" model structures (projective, injective, Reedy) on the category $Fun(C,M)$ of functors from $C$ to $M$, all with the same (levelwise) weak equivalences. The whole point of having such a model structure is that it should present the $\infty$-category $Fun(C,\tilde M)$, where $\tilde M$ is the $\infty$-category presented by $M$. But I'm not sure when this is actually the case. Of course, by "the $\infty$-category presented by $M$", I mean $\tilde M = M[W^{-1}]$ is $M$ localized at the weak equivalences in the $\infty$-categorical sense, and similarly "the $\infty$-category presented by $Fun(C,M)$" is the $\infty$-categorical localization $Fun(C,M)[Fun(C,W)^{-1}]$. Questions: If $C$ is a small category and $M$ is a model category, then under what conditions do the standard model structures on $Fun(C,M)$ present the $\infty$-category $Fun(C,\tilde M)$, where $\tilde M = M[W^{-1}]$ is the $\infty$-category presented by $M$? More generally, if $C$ and $M$ are relative categories, then under what conditions does the mapping relative category $\widetilde{Fun(C,M)} = Fun(\tilde C, \tilde M)$ where $\tilde{(-)}$ denotes taking the associated quasicategory? In a more model-independent direction, when does localization of $\infty$-categories commute with taking functor categories? That is, when does $Fun(C,M[W^{-1}]) = Fun(C,M)[Fun(C,W)^{-1}]$ where $C,M$ are $\infty$-categories and $W \subseteq M$ is a subcategory? REPLY [10 votes]: If your model structures are assumed to have small limits or colimits, the answer to the question of the title is: always. For any model category $M$ and any small category $C$, inverting levelwise weak equivalences in $Fun(C,M)$ is equivalent to considering the $\infty$-category of functors from $C$ to $M[W^{-1}]$. This is a special case of Theorem 7.9.8 (and Remark 7.9.7) in my book on higher categories. It is even possible to take for $M$ a model $\infty$-category in the sense of Mazel-Gee. In fact, Theorem 7.5.8 gives sufficient conditions on $M$ which are much more general: essentially, the mere existence of a class of well behaved fibrations is good enough (this includes Brown's categories of fibrant objects, but, more generally, a version where we do not assume all objects to be fibrant; in particular, all the variations on semi-model structures are OK) if we assume further properties: we mainly need this extra structure to exist on functor categories $Fun(C,M)$ as well (which is automatic in practice, as explained in Example 7.9.6 and Remark 7.9.7 of loc. cit.). If we restrict ourselves to those $C$ whose nerve is a finite simplicial set (e.g. finite partially ordered sets), this kind of properties is true in a much greater level of generality; see Theorem 7.6.17. Observe that, if $M$ is good enough in the sense that $\widetilde{Fun(C,M)}\cong Fun(C,\tilde M)$ for any small category $C$, then, it is automatic that, for any subcategory $S\subset C$, the localization of the full subcategory of $Fun(C,M)$ whose objects are those functors sending the maps of $S$ to weak equivalences will autmatically be a model of the $\infty$-category of functors from $C[S^{-1}]$ to $\tilde M$. All of this is obviously model independent.<|endoftext|> TITLE: Infinite simple group of exponent a power of 2 QUESTION [6 upvotes]: Let G be a simple group of exponent 2^n for n>1. Is G necessarily finite? If not, what is an example of an infinite simple group of exponent a power of 2? REPLY [6 votes]: For every large enough $n$, there exist infinite simple groups of exponent dividing $n$. (I call a group aperiodic if it has no nontrivial finite quotient.) Indeed, by the solution to the restricted Burnside problem the Burnside group $B(2,n)$ is known to be virtually aperiodic; let $B(2,n)^\circ$ be its (unique) minimal finite index subgroup, it is thus finitely generated. For $n$ large enough, $B(2,n)$ is known to be infinite. Hence $B(2,n)^\circ$ is infinite too, hence has a simple quotient, also aperiodic and hence infinite. $\Box$ The above proof produces groups of exponents dividing $n$, probably with some more efforts we can get exponent exactly $n$ but I think it's unimportant. I don't know if one can obtain quasi-finite groups in this context, when the exponent is $2^m$ (quasi-finite means infinite with all proper subgroups finite); these are close variants of Tarski monsters. Also probably one can produce infinitely generated countable examples, and uncountable too, but this requires other/additional arguments. Nevertheless: $\forall n$, there exists no infinite locally finite simple group of exponent $n$. (Nor even in which every element has order dividing some power of $n$.) For $n=2^k$ (or more generally $p^aq^b$ with $p,q$ prime, these would be locally solvable and indeed Malcev proved that there is no infinite, locally solvable simple group. In general, this is essentially due to Hartley (1995) (Springerlink behind paywall). Namely he proved that for every finite subset $F$ in a simple locally finite group $G$, there exists a finite subgroup $H$ containing $F$, and a normal subgroup $N$ of $H$ such that $H/N$ is simple and $F$ projects injectively into $H/N$. Since (by classification) finite simple groups of given exponent have bounded order, this implies the result. (Hartley also quotes Meierfrankelfeld for the same result.) The statement is explicit in a 2005 paper by Cutolo-Smith-Wiegold (ScienceDirect), Lemma 4.<|endoftext|> TITLE: Book on manifolds from a sheaf-theoretic/locally ringed space PoV QUESTION [11 upvotes]: I'm looking for an introductory (or rather, non-advanced) book on manifolds as locally ringed spaces, i.e., from the algebraic geometric viewpoint. Most introductory texts only introduce manifolds from the differentiable viewpoint; I wonder if a text introducing differential manifolds from an algebraic viewpoint exists. REPLY [4 votes]: In Introduction to differential geometry (see the review) by R.Sikorski the author introduces the concept of (what is now called) Sikorski space. Sikorski spaces are "affine, reduced differential spaces" and hence they can be approached algebraically by looking at their coordinate rings. Differentiable manifolds are important examples Sikorski spaces. Unfortunately the book was not translated to english. Luckily there are some publications (in english and perhaps in french) by Sikorski in which he explains this very natural concept. One of them is Differential modules. A book $C^{\infty}$ Differentiable Spaces by Navarro González and Sancho de Salas develop theory of differentiable spaces by first constructing real spectra for smooth algebras and then glue them in order to obtain general spaces. This is analogical to the development of algebraic geometry (scheme theory) by Grothendieck and his school. The book might be a bit advanced.<|endoftext|> TITLE: Is the imaginary part of $t\mapsto\zeta(1/2+it)$ close to the derivative of its real part? QUESTION [5 upvotes]: Plotting $t\mapsto\zeta(1/2+it)$ on Wolfram alpha, it seems that the maxima of its real part are close to the zeros of its imaginary part, while the maxima of the latter seem close to the inflection points of the former. Can this be made precise? For example, is there a canonical notion of distance between those two functions that attains only small values? REPLY [3 votes]: (An extended comment.) The derivative of the real part does not really seem to be close to the imaginary part, as seen in the following picture generated by Mathematica: Code: Plot[{Re[I Zeta'[1/2 + I t]], Im[Zeta[1/2 + I t]]}, {t, 0, 80}] The corresponding zeroes of the two functions indeed seem to be reasonably close to each other. This is no surprise, however: $\zeta(\tfrac{1}{2} + i t)$ essentially circles around (mostly in the right half-plane). In each "circle" the distance between a maximum or minimum of the real part and the corresponding zero of an imaginary part is roughly as large as the distance froth the "center" to the real axis. And most "circles" seem to be centered near the real axis.<|endoftext|> TITLE: Why is the Fourier transform so ubiquitous? QUESTION [63 upvotes]: Many operations and equivalences in mathematics arise as some sort of Fourier transform. By Fourier transform I mean the following: Let $X$ and $Y$ be two objects of some category with products, and consider the correspondence $X \leftarrow X \times Y \to Y$. If we have some object (think sheaf, function, space etc.) $\mathcal{P}$ over $X \times Y$ and another, say $\mathcal{F}$ over $X$, assuming the existence of suitable pushpull and tensoring operations, we may obtain another object over $Y$ by pulling $\mathcal{F}$ back to the product, tensoring with $\mathcal{P}$, then pushing forward to $Y$. The standard example is the Fourier transform of functions on some locally compact abelian group $G$ (e.g. $\mathbb{R}$). In this case, $Y$ is the Pontryagin dual of $G$, $\mathcal{P}$ is the exponential function on the product, and pushing and pulling are given by integration and precomposition, respectively. We also have the Fourier-Mukai functors for coherent sheaves in algebraic geometry which provide the equivalence of coherent sheaves on dual abelian varieties. In fact, almost all interesting functors between coherent sheaves on nice enough varieties are examples of Fourier-Mukai transforms. A variation of this example also provides the geometric Langlands correspondence $$D(Bun_T(C)) \simeq QCoh(LocSys_1(C))$$ for a torus $T$ and a curve $C$. In fact, the geometric Langlands correspondence for general reductive groups seems to also arise from such a transformation. By the $SYZ$ conjecture, two mirror Calabi-Yau manifolds $X$ and $Y$ are dual Lagrangian torus fibrations. As such, the conjectured equivalence $$D(Coh(X)) \simeq Fuk(Y)$$ is morally obtained by applying a Fourier-Mukai transform that turns coherent sheaves on $X$ into Lagrangians in $Y$. To make things more mysterious, a lot of these examples are a result of the existence of a perfect pairing. For example, the Poincaré line bundle that provides the equivalence for coherent sheaves on dual abelian varieties $A$ and $A^*$ arises from the perfect pairing $$A \times A^* \to B\mathbb{G}_m.$$ Similarly, the geometric Langlands correspondence for tori, as well as the GLC for the Hitchin system, arise in some sense from the self-duality of the Picard stack of the underlying curve. These examples seem to show that non degenerate quadratic forms seem to be fundamental in some very deep sense (e.g. maybe even Poincaré duality could be considered a Fourier transform). I don’t have a precise question, but I’d like to know why we should expect Fourier transforms to be so fundamental. These transforms are also found in physics as well as many other “real-world” situations I’m even less qualified to talk about than my examples above. Nevertheless I have the sense that something deep is going on here and I’d like some explanation, even if philosophical, as to why this pattern seems to appear everywhere. REPLY [7 votes]: A signature property of the fourier transform is that it converts convolution into multiplication. (This is crucial for real world applications such as image processing especially as in the discrete case the fourier transform can be computed very quickly using the FFT). Another signficant property of the fourier transform is that it measures how randomly a finite set is distributed via the size of its fourier coefficients. This is applied in Roth's proof of Szemeredi's as one can show that if all the fourier coefficients of a set are small then the set has approximately the right number of arithmetic progressions of length three. Similarly, counting solutions to algebraic equations over finite fields uses the fact that the values of polynomials including simple powers are randomly distributed and this can be measured using the fourier transform and exponential sum estimates.<|endoftext|> TITLE: Is the symplectic quotient $\mu^{-1}(0)/G$ unique up to something? QUESTION [6 upvotes]: Given a Hamiltonian action of a compact Lie group $G$ on a symplectic manifold $(M,\omega)$, we may choose a moment map $\mu \colon M\to \mathfrak{g}^* $ and obtain the symplectic reduction $M/\!\!/G = \mu^{-1}(0)/G$. This construction clearly depends on the choice of moment map. However, I wonder if it is still unique up to some sort of (very?) weak equivalence in the symplectic category? REPLY [5 votes]: Given a Hamiltonian group action, moment maps may only differ by constant addition. So you seem to be comparing the reduced spaces at different levels. Let me state the two extreme cases. When $G$ is a torus, any constant addition to a moment map is also a moment map. In a paper "Birational equivalence in the symplectic category (1989)" by Guillemin and Sternberg, authors showed that reduced spaces at regular levels are related by blowing up and down. I do not know the recent progress though. It might be helpful to read papers citing their paper. The other extreme case is when $G$ is semisimple. In this case, the moment map, which is $G$-equivariant, is unique by the semisimplicity. Then the reduced space is unique and there is nothing to do.<|endoftext|> TITLE: How often a random walk with irrational increments is close to 0? QUESTION [8 upvotes]: Let $\omega$ be an irrational number, and $X$ a random variable taking values $1,-1,\omega,-\omega$ each with probability $1/4$. Let then $X_i$ be iid variables with the same law as $X$ and $S_n=\sum_{i=1}^n X_i,n\in \mathbb N$ be the corresponding random walk. Is it possible to have a precise asymptotics for $P(|S_n|<\epsilon)$ for $\epsilon>0$? Ultimately I would like to know the behaviour of $$\sum_{n=1}^\infty n^{-3/2} P(|S_n|<\epsilon)$$ as $\epsilon\to 0$. I feel like the diophantine properties of $\omega$ are relevant for this asymptotics. How would you proceed to get such an estimate? Ideally I would like to consider $X$ with any discrete law, with eventually infinitely many atoms. EDIT: to be clear, I think there are ad-hoc methods to solve this kind of problems, as Mateusz shows below. I want to make sure not to miss any kind of general theory that solves this kind of problems in the theory of random walks. REPLY [4 votes]: Just an extended comment. Let $X_n$ be the simple random walk in $\mathbb{Z}^2$. Then $$\mu(\{x\}) = \sum_{n = 1}^\infty n^{-3/2} P(X_n = x)$$ is comparable with $$\sum_{n = 1}^\infty n^{-3/2} \times n^{-1} \exp(-|x|^2 / (2 n)) \approx (1 + |x|)^{-3}.$$ So your question boils down to estimating $$\sum_{k \in \mathbb{Z}} \frac{1}{(1 + |k|)^3} \, \mathbb{1}_{(-\epsilon, \epsilon)}(k \omega - \lfloor k \omega\rfloor) . $$ This indeed seems closely related to how well one can approximate $\omega$ with rationals.<|endoftext|> TITLE: When is a large graph with a given degree sequence likely to be connected? QUESTION [10 upvotes]: Are there any results on whether a large random graph with a given degree distribution is likely to be connected? In Erdős-Rényi graphs with $n$ vertices and $m$ edges, we have two sudden transitions (for large $n$): A giant component appears above the threshold $m/n = 1/2$. The graph becomes connected above the threshold $m/n = (\ln n)/2$. There is a result analogous to (1) above by Molloy and Reed for random graphs with a given degree distribution. If $d$ denotes the vertex degree and $\langle \cdot \rangle$ denotes the average, then the quantity of interest is $Q = \langle d^2 \rangle - 2\langle d \rangle$. A giant component suddenly appears above the threshold $Q > 0$. Question: Is there a result analogous to (2) for random graphs with a fixed degree sequence, in the large graph limit? Is there a quantity that can be computed from the degree distribution, and when it crosses a threshold, the graph suddenly becomes connected (in the $n\rightarrow\infty$ limit)? Let us assume that there are no isolated vertices ($d\ne 0$). Clarification update: Let me try to give a more precisely specified version of the problem. Suppose we have $n$ vertices. Of these, precisely $n_d = f_d n$ have degree $d$: thus we have a degree sequence $$( \overbrace{0,\dots,0}^{\text{$n_0$ times}},\; \overbrace{1,\dots,1}^{\text{$n_1$ times}},\; \overbrace{2,\dots,2}^{\text{$n_2$ times}}, \dots). $$ Choose one simple (labelled) graph with this degree sequence uniformly at random. What conditions do we need to have on the $f_d$ (the degree distribution), or on $n_d$, so that in the $n \rightarrow \infty$ limit the graph is connected with probability 1? Clearly, if the $f_0 \ne 0$, then there are isolated vertices and the graph is not connected. Therefore, one condition is that $f_0 = 0$. REPLY [2 votes]: Second edition This is a partial answer to the question per the "Clarification Update", but first I'll generalize a little. Suppose that for each $n$ we have a degree sequence $n_0,n_1,n_2,\ldots$, where $n_d=n_d(n)$ means the number of vertices of degree $d$. Also let the number of edges be $m=m(n)$ and the maximum degree be $\varDelta=\varDelta(n)$. Now we take a random simple graph $G=G(n)$ with this degree sequence, each such graph being equally likely. We seek to know if $G$ is connected. Take $n_0=0$ from now on. This type of random graph has been extensively studied. I'll just make some simple observations using Theorem 2.1 of this paper. By Theorem 2.1 the expected number of isolated edges is $$\binom{n_1}{2}\frac{1+O(\varDelta/m)}{2m}$$ if $\varDelta=o(m)$. Assuming the latter condition, the expected number of isolated edges goes to $0,\infty$ according as $n_1^2/m$ goes to $0,\infty$, respectively. This doesn't imply instantly that $n_1\approx \sqrt{m}$ is the threshold for having an isolated edge, but it is true (use the second moment method). So now assume $n_1=o(m)$. I thought a combination of degrees 1 and 2 might be an issue, but the most likely isolated component, a path of two edges, is unlikely if $n_1=o(\sqrt m)$. (So, if these components are likely, so are isolated edges.) Now consider isolated cycles. The expected number of isolated cycles of length $k$ is $$\frac{(n_2)_k(1+O(k\varDelta/m))}{2k\,(m)_k},$$ where $(x)_k=x(x-1)\cdots(x-k+1)$, provided $k\varDelta=o(m)$. Since $n_2\le m$, this never goes to infinity for fixed $k$, but the sum over an increasing number of $k$ values does go to infinity if $n_2=(1-o(1))m$. In the other direction, if $n_2=o(m)$ then the expectation goes to 0 for each $k$ and moreover the terms appear to be decreasing exponentially as $k$ increases. Here there is a gap in the proof because $k\varDelta=o(m)$ might not be true for very large $k$ unless also $\varDelta=O(1)$. This gap can be filled but I won't go into it. Modulo some things I haven't quite proved, the probability of connectedness goes to 1 if $n_2=o(m)$ and to 0 if $n_2=(1+o(1))m$. In the intermediate ranger, for example if $n_2=cm$ for $0\lt c\lt 1$, I believe that the distribution of the number of isolated cycles will be Poisson with constant mean. Beyond this, I'm reluctant to reinvent the wheel because someone must have done this before except possibly in the case that some degrees are very low and others very high. There are no component types that are likely to occur under conditions when isolated edges or cycles are unlikely to occur. The fact that random regular graphs of degree at least 3 are almost always connected was proved by Wormald in the 1970s. I hypothesize that $n_0=0$, $n_1=o(\sqrt m)$ and $n_2\le cm$ for some $c\le 1$ are necessary and sufficient conditions for almost sure connectivity. The question also asks us to consider the case that there constants $f_0,f_1,\ldots$ independent of $n$ such that $n_d(n)=f_d\,n$ for all $n,d$. Translating what is above, the condition for connectivity is $f_0=f_1=f_2=0$. Clearly forcing $f_d$ to be independent of $n$ loses a lot of detail.<|endoftext|> TITLE: The geometry of the action of the semidirect product QUESTION [9 upvotes]: I'm going by the maxim Groups, like men, are known by their actions This naturally leads one to ask "given groups $G, H$ which act on sets $S, T$ and the semidirect product $G \rtimes H$, how does one visualize the action of $G \rtimes H$? What does it act on? Some combination of $S$ and $T$? ($S \times T$ perhaps?) I know some elementary examples, likr $D_n \simeq \mathbb Z_n \rtimes \mathbb Z_2$. However, given an unknown situation, I am sure I cannot identify whether it is a semidirect product that is governing the symmetry. The best responses on similar questions like intuition about semidirect product tend to refer to this as some kind of "direct product with a twist". This is shoving too much under the rug: the twist is precisely the point that's hard to visualize. Plus, not all "twists" are allowed --- only certain very constrained types of actions turn out to be semidirect product. I can justify the statement by noting that: the space group of a crystal splits as a semidirect product iff the space group is symmorphic --- this is quite a strong rigidity condition on the set of all space groups. This question on the natural action of the semidirect product identifies one choice of natural space for the semidirect product to act on, by introducing an unmotivated (to me) equivalence relation, which "works out" magically. What's actually going on? The closest answer that I have found to my liking was this one about discrete gauge theories on physics.se, where the answer mentions: If the physical space is the space of orbits of $X$ under an action $H$. Ie, the physical space is $P \equiv X / H$. Then, if this space $P$ is acted upon by $G$. to extend this action of $G \rtimes H$ onto $X$ we need a connection. This seems to imply that the existence of a semidirect product relates to the ability to consider the space modulo some action, and then some action per fiber. I feel that this also somehow relates to the short exact sequence story(though I don't know exact sequences well): Let $1 \rightarrow K \xrightarrow{f}G \xrightarrow{g}Q \rightarrow 1$ be a short exact sequence. Suppose there exists a homomorphism $s: Q \rightarrow G$ such that $g \circ s = 1_Q$. Then $G = im(f) \rtimes im(s)$. (Link to theorem) However, this is still to vague for my taste. Is there some way to make this more rigorous / geometric? Visual examples would be greatly appreciated. (NOTE: this is cross posted from math.se after getting upvotes but no answers) REPLY [2 votes]: Suppose $K\circlearrowright X$ is an action of $K$ on some set. We have the structure homomorphism $\varphi:K\rightarrow \text{Sym}(X).$ Let $H$ act on $K$ by automorphisms, and suppose these automorphisms can be realized as inner automorphisms within $\text{Sym}(X)$. That is, the action is given by some $\theta:H\rightarrow\text{Sym}(X)$ such that conjugations by $\theta(H)$ leave $K$ invariant. Equivalently, the $H$-action takes stabilizer subgroups of $K\circlearrowright X$ to stabilizers (it may permute them nontrivially). Then we can construct an action of $K\rtimes H$ on $X$ by $(k,h).x = \varphi(k)\cdot\theta(h).x$, where the product $\varphi(k)\theta(h)$ is just taken in $\text{Sym}(X)$. Here, the multiplication rule for $K\rtimes H$ is $$(k_1, h_1)\cdot (k_2,h_2) = (k_1 \cdot (h_1.k_2), h_1 h_2). $$ Any action arises in this way, since $H$ acts on $K$ by conjugation in the semidirect product $K\rtimes H$ and therefore it also acts by conjugation in the image under the structure morphism $K\rtimes H \rightarrow \text{Sym}(X)$ of an action of the semidirect product on $X$. This explains when an action of $K$ can be extended to an action of $K \rtimes H$. It is not clear how to visualize the above. So let's pass to a nice special case. Given another action of $H$ on a set $Y$, we can extend $H\circlearrowright Y$ to an action $K\rtimes H \circlearrowright Y$ (let the latter act via the quotient $K\rtimes H\twoheadrightarrow H$). Then we can produce an action of $K\rtimes H \circlearrowright Y \times X$. This action descends to the quotient $H\circlearrowright Y$ so that $K$ fixes each fiber $\{y\}\times X$, and $H$ acts by permuting fibers and "twisting" $X$. If the $H$-action is faithful, the action of any element $(k,h)$ can be nicely separated into an $h$-part and a $k$-part, the $h$-part being uniquely identified by the action on $Y$. Thus given a permutation $\sigma$ of $Y\times X$ we can write that it is of the form $(k,h)$ for a known $h$, and then compute the permutation $(y,x)\mapsto \sigma\left((e,h^{-1}).(y,x)\right)$, which acts the same as $(k,h)\cdot(e,h^{-1}) = (k,e)$. One gets some examples which appear different, but are isomorphic to these, by choosing different identifications between fibers than $\text{id}_X : \{y_1\}\times X \rightarrow \{y_2\}\times X$. These identifications may be analogous to the connection described in the question.<|endoftext|> TITLE: Is the separability of the space needed in the proof of the Prohorov's theorem? QUESTION [6 upvotes]: The Section 5 of the book: Billingsley, P., Convergence of Probability Measures, 1999, studies Prohorov's theorem. A short reminder is given below. Let $\Pi$ be a family of probability measures on $(S,\mathcal{F})$. We call $\Pi$ relatively compact if every sequence of elements of $\Pi$ contains a weakly convergent subsequence. The family $\Pi$ is tight if for every $\epsilon$ there is a compact set $K$ such that $P(K)>1-\epsilon$ for every $P$ in $\Pi$. The direct half of the Prohorov's theorem is given in the Theorem 5.1: If $\Pi$ is tight, then it is relatively compact. The converse half of Prohorov's theorem is given in the Theorem 5.2: Supose that $S$ is separable and complete. If $\Pi$ is relatively compact, then it is tight. My question: In the proof of the Theorem 5.2 (i.e. relatively compact $\Rightarrow$ tight), we use separability and completness of the space $S$. On the other hand, in the proof of the Theorem 5.1 (i.e. tight $\Rightarrow$ relatively compact), I know that we do not need completness of $S$, but I do not know if we do need separability. I didn't find the place where separability is used in the proof of the Theorem 5.1. So my question is do I need or not the separability of the space $S$ in the direct part of the Prohorov's theorem? Remarks: I know the proofs of the same theorem that use separability (e.g. Note). Prohorov's theorem in most books is given as one theorem on Polish spaces, so they assume separability in both halfs. It goes like this usually: Let $S$ be a Polish space and $\Pi$ a collection of probability measures on $S$. Than $\Pi$ is tight if and only if it is relatively compact. The reason I am asking is that I would like to use the direct half of Prohorov's theorem on the problem I am currently working. Space $S$ in my case is complete but not separable. Help with this would be great and needed. Thanks in advance. REPLY [4 votes]: Separability is not necessary. In fact, tightness of a family of Borel probability measures implies relative compactness in the vague/weak-* topology on any completely regular space. For instance, this can be found in volume 4 of Fremlin's Measure Theory. Specifically Proposition 437U (b) shows that tight families are compact in the narrow topology, and 437K (c) shows that for completely regular spaces, the narrow topology agrees with the weak-* topology. My original answer below answers the wrong question - the question is about whether tight implies relatively compact, rather than the other way. Let $\kappa$ be a real-valued measurable cardinal, and $\mu : \mathcal{P}(\kappa) \rightarrow [0,1]$ a probability measure vanishing on singletons. Consider $\kappa$ to be a discrete metric space. Then the 1-element family $\{\mu\}$ is compact, because it is a singleton, but it is not tight because all compact subsets of $\kappa$ are finite sets, so have measure zero. You say that completeness is not necessary, but (unless you are making an extra assumption) it is, for essentially the same reason - there are separable metric spaces with Borel measures on them that are not tight. REPLY [3 votes]: You are correct that separability is not needed. However, there is also not really any loss of generality in assuming it. For suppose that $\Pi$ is tight. Then for every $n$ there exists a compact set $K_n$ such that $\mu(K_n) > 1-\frac{1}{n}$ for all $\mu \in \Pi$. So if we set $S_0 = \bigcup_{n=1}^\infty K_n$, then $S_0$ is separable and $\mu(S_0) = 1$ for all $\mu \in \Pi$. We can now view $\Pi$ as a set of probability measures on $S_0$, and it is still tight (since the $K_n$ are also compact in $S_0$). The separable case of the theorem then implies that $\Pi$ is weakly relatively compact in $\mathcal{P}(S_0)$, i.e. every sequence in $\Pi$ has a subsequence converging weakly in $\mathcal{P}(S_0)$, and you can easily check that such a subsequence also converges weakly in $\mathcal{P}(S)$. So $\Pi$ is weakly relatively compact in $\mathcal{P}(S)$, as desired. In other words, once you have a tight family, then all those measures live on a separable subset of $S$ anyway, so the rest of the space is irrelevant and might as well not be there.<|endoftext|> TITLE: Does Higman's embedding theorem hold inside group varieties? QUESTION [7 upvotes]: Suppose $\mathfrak{U}$ is a variety of groups. Let's define $F_n(\mathfrak{U})$ as relatively free groups in $\mathfrak{U}$. Suppose $G \in \mathfrak{U}$ is a finitely generated group. We call $G$ finitely presented in $\mathfrak{U}$ iff $\exists n \in \mathbb{N}$ and finite $A \subset F_n(\mathfrak{U})$ such that $G \cong \frac{F_n(\mathfrak{U})}{\langle \langle A \rangle \rangle}$. We call $G$ recursively presented in $\mathfrak{U}$ iff $\exists n \in \mathbb{N}$ and recursively enumerable $A \subset F_n(\mathfrak{U})$ such that $G \cong \frac{F_n(\mathfrak{U})}{\langle \langle A \rangle \rangle}$. My question is: Is it true, that a finitely generated group is recursively presented in $\mathfrak{U}$ iff it is isomorphic to a finitely generated subgroup of a group finitely presented in $\mathfrak{U}$? This fact is true for the varieties of abelian groups due to linear algebra, proved for the variety of all groups by Higman, and for the Burnside varieties by Olshanski. However, I do not know, whether it is true in general. This question on MSE REPLY [10 votes]: Kharlampovich proved that there exist two finitely based varieties of solvable groups $ {\mathfrak A} \subset {\mathfrak B}$ such that the word problem is not solvable in the groups f.p. in the smaller variety but solvable in the groups that are f.p. in the bigger variety (the result can be found in our joint survey "Algorithmic problems in varieties", see also Kharlampovich, O. G. The word problem for solvable Lie algebras and groups. Mat. Sb. 180 (1989), no. 8, 1033–1066, 1150; translation in Math. USSR-Sb. 67 (1990), no. 2, 489–525). Now if you take a group $G$ that is finitely presented in the smaller variety and has undecidable word problem then $G$ is finitely generated, recursively presented in the class of all groups since the varieties are finitely based, but cannot be embedded into any group $H$ which is finitely presented in the bigger variety. So variety ${\mathfrak B}$ does not have the property in the OP. Varieties with that property are sometimes called Higman varieties. In my comment above I mentioned some known Higman varieties in addition to those in the OP. It is not known if the variety of solvable groups of class $\le 3$ is Higman. In fact all known Higman varieties have been mentioned in the OP and in my comment.<|endoftext|> TITLE: When is the homotopy category of an accessible $\infty$-category accessible? QUESTION [9 upvotes]: Let $\mathcal C$ be an accessible $\infty$-category, and let $ho(\mathcal C)$ be its homotopy category. I can think of two "trivial" reasons for $ho(\mathcal C)$ to be accessible: $ho(\mathcal C) = \mathcal C$; $ho(\mathcal C)$ is small with split idempotents. Otherwise, I am aware of very few examples where $ho(\mathcal C)$ is accessible. Indeed, given that $ho(Spaces)$ is very far from accessible, I think I should expect that it is quite rare for $ho(\mathcal C)$ to be accessible. However, I know of one interesting class of examples where $ho(\mathcal C)$ is "nontrivially" accessible. Let $k$ be a field. Write $D(k)$ for "derived $\infty$-category of $k$", i.e. the category of chain complexes over $k$ localized ($\infty$-categorically) at the quasi-isomorphisms. This is a presentable $\infty$-category. Then $ho(D(k))$ is the usual derived category of $k$, which is equivalent to the usual 1-category of graded $k$-vector spaces (though of course the triangulated structure is different), and so is obviously accessible (locally presentable, in fact). This one class of examples has me second-guessing my expectation that taking homotopy categories rarely preserves accessibility. Questions: What are some other examples of accessible $\infty$-categories $\mathcal C$ such that $ho(\mathcal C)$ is also accessible (which do not satisfy (i) or (ii) above)? Given such an example, is the functor $\mathcal C \to ho(\mathcal C)$ accessible (i.e. does it preserve $\kappa$-filtered colimits for some $\kappa$?) Can we (partially) characterize this condition, giving necessary and / or sufficient conditions for $ho(\mathcal C)$ and $\mathcal C \to ho(\mathcal C)$ to be accessible? Is there anything to be said in particular about the case where $\mathcal C$ is stable and presentable, or even more particularly the case where $\mathcal C$ is the derived $\infty$-category of a ring? More broadly, at the moment it seems very mysterious to me that $ho(D(k))$ comes out to be accessible. I'd appreciate any perspective which makes this fact look less mysterious. REPLY [2 votes]: Here's a further generalization: Claim: Let $\mathcal T$ be a triangulated category and with a $t$-structure such that $\mathcal T^\heartsuit$ is has coproducts, which are exact. Suppose there exists $M \in \mathcal T^\heartsuit$ and a monic endomorphism $M \overset r \rightarrowtail M$ in $\mathcal T^\heartsuit$ which is not an isomorphism, and such that $Hom(M,M/r)$ is an $r$-torsion $End(M)$-module. Then $\mathcal T$ is not concrete, and in particular not accessible. Proof: For most of the proof, we work in $\mathcal T^\heartsuit$. Similar to before, for every ordinal $\alpha$, we define $W_\alpha$ to be the set of finite subsets of $\alpha+1$, and set $M_\alpha$ to be the cokernel of the map $\oplus^{W_\alpha} M \xrightarrow{1-s} \oplus^{W_\alpha} M$, where $s$ carries the $[\alpha_0<\dots< \alpha_n]$ -copy of $M$ to the $[\alpha_1<\dots<\alpha_n]$ copy of $M$ via the map $r$. A straightforward transfinite induction shows that the $[\alpha_0<\dots<\alpha_n]$'th structure map $M \to M_\alpha$ is in $r^{\alpha_0} Hom(M,M_\alpha)$, and in particular the $[\alpha]$th structure map is in $r^\alpha Hom(M,M_\alpha)$. It is straightforward to see that the natural map $M_\alpha \to M_{\alpha+1}$ actually splits, and its cokernel $Q$ sits in an exact sequence $M/r \to Q \to M_\alpha \to 0$, where the map to $M_\alpha$ is obtained by deleting the last letter of each word (which is always $\alpha$ here), and the copy of $M/r$ corresponds to the generator $[\alpha]$ (moreover, the $\alpha$th structure map $M \to M_\alpha$ is nonzero and factors through $M/r$). We claim now that $r^{\alpha+1}Hom(M,M_\alpha) = 0$. This is proved by induction on $\alpha$ using the following Lemma: Let $M$ be an object of an abelian category and $r: M \to M$ a map. If $0 \to A \to B \to C$ is exact, and if $Hom(M,C)$ is $r$-torsion, then any map $\phi \in r^{\alpha+1} Hom(M,B)$ factors (necessarily uniquely) through $A$, and the factored map lies in $r^\alpha Hom(M,A)$. Dually, if $A \to B \to C \to 0$ is exact and $Hom(M,A)$ is $r$-torsion, then if $r^\alpha Hom(M,C) = 0$ it follows that $r^\alpha Hom(M,B) = 0$, for $\alpha \geq 1$. We apply the lemma to the exact sequences $0 \to M_\alpha \to M_{\alpha+1} \to Q \to 0$ and $M/r \to Q \to M_\alpha \to 0$ at the successor steps of a transfinite induction to conclude that indeed $r^{\alpha+1} Hom(M,M_\alpha) = 0$. Now we conclude as before: the $\alpha$th structure map $M \to M_\alpha$ doesn't factor through the $\beta$th structure map $M \to M_\beta$ for $\beta > \alpha$ because then it would lie in $r^\beta Hom(M,M_\alpha) = 0$, and it is in fact nonzero. Since $M\to M_\alpha$ are weak cokernels in $\mathcal T$, they form a proper class of pairwise-inequivalent weak cokernels out of $M$, so that $\mathcal T$ is not concrete.<|endoftext|> TITLE: Is it possible to express the functional square root of the sine as an infinite product? QUESTION [5 upvotes]: Cross-post from MSE. It is known that the sine can be expressed as an infinite product: $$\sin(x) = x \prod_{n=1}^{\infty} \Big{(} 1 - \frac{x^{2}}{n^{2}{\pi}^{2}} \Big{)} .$$ We can define that functional square root of a function $g(\cdot)$ to be the function $f(\cdot)$ that satisfies $f(f(x)) = g(x)$. The square root of the sine function with respect to function composition has been discussed previously on MO on a number of occasions. For instance, here the formal power series is considered. I wonder whether the functional square root of the sine also has an infinite product representation. If not, has any research been done on this question? REPLY [4 votes]: Functional square root of sine is not unique, and it cannot be defined in the whole complex plane (according to a theorem of I. N. Baker). You can define it on $(0,\pi)$ with $f(0)=0$, for example. The resulting function is analytic in the component of Fatou set of sine adjacent to zero on the right (see, for example the first figure here). But this function $f$ has no zeros in its domain, therefore a representation as infinite product does not make much sense. I mean that $\log f$ can be defined and any breaking of this log into summands gives you an infinite product representation. Of course this answer is related to a true product, it does not exclude that there is some "formal" product representing this square root, whatever a "formal product" may mean.<|endoftext|> TITLE: "Sub-logarithmic" zero-free regions from Deuring-Heilbronn/Linnik's repulsion theorem QUESTION [6 upvotes]: For each $n\in\mathbb{N}$, let: $\chi_n\pmod{q_n}$ a real non-principal Dirichlet character ($q_1 < q_2 < \cdots$), $\beta_n$ the largest real zero of $L(s,\chi_n)$, $\delta_n := (1-\beta_n)\log(q_n)$. Let $\chi\pmod{q}$ be a Dirichlet character, and consider $s = \sigma + it$ with $|t| < 1$. In p. 206 of Heath-Brown's "Prime twins and Siegel zeros", it is mentioned that the Deuring-Heilbronn phenomenon implies that there is some absolute constant $C>0$ such that, for each $n\in \mathbb{N}$, the region $$ \bigg\{ s ~\bigg|~ \sigma \geq 1 - \frac{C\log(\delta_n^{-1})}{\log(q)},\ |t| < 1 \bigg\} $$ has no zeros (besides $\beta_1,\ldots,\beta_n$) of $L(s,\chi)$. (At least, that is how I interpreted the assertion "$r_0 \gg L^{-1}\log \eta$" at the mentioned page). Assuming this statement, it follows that: (Sub-logarithmic zero-free regions (ZFR)) If there exists a sequence of Siegel zeros $\beta_n$ with $\delta_n \to 0$, then all the other zeros $\sigma + i\gamma$ of $L(s,\chi)$ with $|\gamma| < 1$ for Dirichlet characters $\chi\pmod{q}$ satisfy: $$ \sigma < 1 - \frac{1}{o(\log(q))}. $$ It appears to me that this (or slight variations of this) statement is often used in the literature [the only example I have in mind at the moment is Remark 1 in p. 515 (p. 6 in the link) of Granville & Stark's $ABC$ implies no "Siegel zeros" for $L$-functions of characters of negative discriminant, where it is mentioned that $\delta_n \to 0$ implies $\frac{L'}{L}(1,\chi_n) = (1-\beta_n)^{-1} + o(\log(q_n))$]. However, I am having trouble following the deduction of these "sub-logarithmic ZFRs" from the Deuring-Heilbronn phenomenon alone. Using the Deuring-Heilbronn (Linnik's repulsion theorem) as in Théorème 16, Sec. 6 of Bombieri's "Le grande crible", we get that there are absolute constants $c_1,c_2 >0$ such that, fixing $n\in\mathbb{N}$, it holds: $$ \sigma < 1 - c_1\log\left(c_2\frac{\delta_n^{-1}}{\log(q_n q)/\log(q_n)}\right) \cdot \frac{1}{\log(q_n q)} $$ (I am just taking $T = q_n q$ in Théorème 16). Assuming we have an infinite sequence $q_n \to +\infty$ with $\delta_n \to 0$, for a given $q\in\mathbb{N}$ we may take $q_k \leq q < q_{k+1}$, so that $\log(q_{k+1} q)/\log(q_{k+1})< 2$, and hence: $$ \sigma < 1 - c_1\frac{\log(\frac{c_2}{2}\delta_{k+1}^{-1})}{\log(q_{k+1} q)}. $$ It appears, then, that to derive the "sub-logarithmic" ZFRs, it is necessary to have $\log(q_{k+1}) \ll \log(q_k)$ as $k\to \infty$, i.e.: the gaps between the conductors of consecutive exceptional characters need to be polynomially bounded, even if we assume $\delta_n \to 0$. I do not think my conclusion is correct (e.g., I believe I misinterpreted some aspect of Heath-Brown's paper), but I have not been able to get rid of this condition on the growth of the $q_n$. In short, my question is the following: Q. Is it really possible to derive "sub-logarithmic" ZFRs from Linnik's repulsion theorem (i.e., without additional growth conditions on the $q_n$)? REPLY [4 votes]: Let $\chi$ be a non-principal real Dirichlet character modulo $q$. Let $$\beta_0=1-\frac{1}{\eta\log q}$$ be a real zero of $L(s,\chi)$ satisfying $\eta\geq 100$ for convenience (Heath-Brown's condition is $\eta\geq 3$). Let $\rho=\beta+i\gamma$ be any zero of $L(s,\chi)$ such that $\rho\neq\beta_0$ and $|\gamma|\leq 1$. We strengthen Heath-Brown's claim $r_0\gg L^{-1}\log\eta$ to (note that $L=\log q)$ $$1-\beta\gg\frac{\log\eta}{\log q}.$$ We shall deduce this from Theorem 2 of Jutila's 1977 paper "On Linnik's constant", which is also Heath-Brown's reference. Let us write $\delta$ for the left hand side. If $\delta>1/60$, then the statement is trivial by $\eta\ll q$. So we shall assume that $\delta\leq 1/60$. Then, Jutila's theorem yields readily (using $D\leq 2q$) that $$1-\beta_0\geq\frac{1}{10}\cdot\frac{q^{-3\delta}}{\log q}.$$ In other words, $\eta\leq 10 q^{3\delta}$. By the assumption $\eta\geq 100$, this implies that $\eta\leq q^{6\delta}$, or equivalently that $$\delta\geq\frac{1}{6}\cdot\frac{\log\eta}{\log q}.$$ We have verified Heath-Brown's claim.<|endoftext|> TITLE: Hecke algebra relation versus $\operatorname{SL}_2$ trace relation QUESTION [7 upvotes]: The quadratic relation in the (type $A$) Hecke algebra is $(T-t)(T+t^{-1}) =0$, which can be rewritten as $$ T-T^{-1} = t-t^{-1}$$ Suppose $A \in \operatorname{SL}_2(\mathbb{Q})$ with eigenvalues $a,a^{-1}$. Then it isn't hard to show the following identity $$A + A^{-1} = (a+a^{-1}) \mathrm{Id}$$ If $i^2=-1$ then we can "convert" between these two formulas by saying $T=iA$ and $t = ia$. Is there a conceptual reason for this? REPLY [3 votes]: Interesting observation! Unfortunately, this seems to be a coincidence. The original, algebraically motivated definition of the Hecke algebra gives the quadratic relation as $$(T-q)(T+1)=0.$$ For $SL_2$, the Hecke algebra has a faithful two dimensional representation. The eigenvalues of $T$ are $q$ and $-1$, and the quadratic relation simply becomes the Cayley--Hamilton theorem. But any 2x2 matrix $A$ also satisfies its characteristic equation $$(A-\lambda_1)(A-\lambda_2)=0,$$ where $\lambda_1,\lambda_2$ are the eigenvalues of $A$. If $A$ is nonsingular, then the rescaling $A\mapsto \sqrt{\lambda_1\lambda_2}A$ transforms this into $$(A-a)(A-a^{-1})=0,$$ where $a=\sqrt{\frac{\lambda_1}{\lambda_2}}$. Thus your observation reduces to the fact that $i$ is a "nice" value of $\sqrt{\lambda_1\lambda_2}$ for the Hecke algebra. But in fact the niceness here is rather circular. As mentioned above, the more motivated definition of the Hecke algebra is not symmetric. To reach the modern presentation, we symmetrize the definition by replacing $T\mapsto q^{1/2} T$ to find (with $t=q^{1/2}$) $$(T-t)(T+t^{-1}) =0. $$ The goal of this rescaling is to make this relation "nice", i.e. transforming nicely under $t\mapsto t^{-1}$. We could have alternatively rescaled $T\mapsto iq^{1/2}$ to get $$(T-t)(T-t^{-1})=0. $$ These are the only two rescalings if we want the relation to be nice. Thus a priori the only possible values of $\sqrt{\lambda_1\lambda_2}$ for a nice Hecke algebra presentation are $1$ or $i$, explaining your observation. (We pick the $i$ presentation over the $1$ presentation to preserve the behavior of the Hecke algebra as $t\rightarrow 1$.) Thus the fundamental reason for the similarity between the Hecke algebra presentation and the trace relation in $SL_2$ is that the Hecke algebra is two dimensional. But the dimension of the Hecke algebra is the size of the Weyl group $S_2$. The fact that $|S_2|=2=\mathrm{rank}(SL_2)$ is a coincidence.<|endoftext|> TITLE: Parametrization of real-valued SU(N) QUESTION [5 upvotes]: I want to construct a $SU(N)$ matrix $V$, with the following property: All the elements of the first row are given, i.e. $V_{1,j}=a_i$ (with $\sum_i a_i^2=1$) All matrix elements are real, i.e. $V_{i,j} \in \mathbb{R}$ How can I find a matrix $V$ that satifies the criteria? Specifically, how can I find the matrix elements as a function of $a_i$, i.e. $V_{i,j}(a_i)$? Special case: SU(2) $$ V= \left[ {\begin{array}{cc} a_1 & a_2 \\ V_{2,1} & V_{2,2} \\ \end{array} } \right] $$ We easily find $V_{2,1}=-a_2$ and $V_{2,2}=a_1$. Special case: SU(3) $$ V= \left[ {\begin{array}{cc} a_1 & a_2 & a_3 \\ V_{2,1} & V_{2,2} & V_{2,3} \\ V_{3,1} & V_{3,2} & V_{3,3} \\ \end{array} } \right] $$ Here already I cannot find any feasible way to represent $V_{i,j}$ as a function of $a_1, a_2, a_3$. I have tried to use the generators of SU(3), the Gell-Mann matrices $\lambda_i$. In particular, $\lambda_2$, $\lambda_5$, $\lambda_7$ are the generators for real-valued SU(3) matrices. However, the resulting equation system involves multiple trigonometric functions for which I cannot solve $V_{i,j}(a_1, a_2, a_3)$. The matrix $V$ is not unique, I just want any solution. REPLY [3 votes]: In addition to the comments I made above about continuous solutions, I thought I'd point out a solution that works for all $n$ with only one point of discontinuity, namely $$ (a_1\ a_2\ \ldots\ a_n) = (1\ 0\ \ldots\ 0). $$ Away from this point, one can start with the following formulae: $$ V_{i,1}= V_{1,i} = a_i\qquad\text{and}\qquad V_{i,j} = \delta_{ij} - \frac{a_ia_j}{(1-a_1)}\quad\text{when}\ 1 TITLE: Why is modular forms applicable to packing density bounds from linear programming at $n\in\{8,24\}$? QUESTION [10 upvotes]: Sphere packing problem in $\mathbb R^n$ asks for the densest arrangement of non-overlapping spheres within $\mathbb R^n$. It is now know that the problem is solved at $n=8$ and $n=24$ using modular forms. I understand some sphere packing and issues going with it but my understanding is most upperbounds come from linear programming and the bounds that are currently proven optimal (including at $n=8$ and $n=24$) already come from linear programming bounds. How do modular forms become part of the story that provide the lower bound (do they arise naturally from some packing related structure)? Is there a bigger story that this is just a chapter of that may apply to other upper bounds generated from linear programming? What makes modular forms click for this class of linear programming bounds (perhaps Sphere packing and quantum gravity is of utility?)? REPLY [12 votes]: This is a tough question, and I don’t think there’s a definitive answer yet. For some mathematical details, see the following survey articles: https://arxiv.org/abs/1611.01685 https://arxiv.org/abs/1603.05202 Instead, I’ll focus on the big picture here. Why modular forms? I can see a couple of potential answers: (1) Why not modular forms? Before Viazovska’s proof, numerical experiments indicated that there were remarkable special functions in 8 and 24 dimensions that would prove the optimality of $E_8$ and the Leech lattice. However, nobody had any idea how to construct them explicitly, or prove existence at all. Modular forms are by far the most important class of special functions related to lattices (in higher dimensions, since arguably trig functions and the exponential function are the most important special functions related to lattices), so lots of people expected that the magic functions for sphere packing should be connected somehow to modular forms. The proof had to wait for Viazovska to come up with a beautiful integral transform, but the fact that it used modular forms wasn’t such a great surprise. I.e., her contribution wasn’t the idea that modular forms should play a role, but rather figuring out how to use them, which was quite subtle and ingenious. You’re right that nobody has any idea how to use modular forms to optimize the linear programming bound in other dimensions. However, it’s possible that they will continue to play a role. For example, see the example Felipe Gonçalves and I found at the end of Section 2.1 of our paper https://arxiv.org/abs/1712.04438 (which is not a sphere packing bound, but closely related). It really looks like a small perturbation of a function based on modular forms (see https://arxiv.org/abs/1903.05737), and I wouldn’t be surprised if the optimal function has a nice series expansion based in some way on modular forms. From this perspective, the remarkable thing about 8 and 24 dimensions wouldn’t be the appearance of modular forms, but rather the fact that the series collapses to a single term, with a matching sphere packing. However, this is all speculative. (2) The other perspective is that we have very little understanding of why 8 and 24 dimensions are special in the first place. For example, why shouldn’t sphere packing in 137 dimensions also admit an exact solution via linear programming bounds? It sure doesn’t look like it does, but perhaps we just don’t know the right sphere packing to use, and some currently unknown packing might match the upper bound. That would be very surprising, since our experience is that exceptional phenomena occur in clusters. We would expect to see some sort of remarkable symmetry group, probably a finite simple group, and there aren’t any candidates acting on 137 dimensions. However, this expectation is just based on our limited experience, and mathematics can confound our expectations. So far, nobody has found even a convincing heuristic argument for why there shouldn’t be an exact solution in 137 dimensions, and that’s a major gap in our understanding. The most we can say is that it would have to differ in some important ways from 8 and 24 dimensions, which is far from an explanation of why it can’t happen. I guess I’d summarize it like this. If you accept that lattices in 8 and 24 dimensions play a special role, then modular forms feel naturally connected. However, we’re missing a deeper explanation of the role of these special dimensions. Let me add one more specific mathematical comment. The magic functions in 8 and 24 dimensions fit into a general picture of building radial functions that vanish on all but finitely many vector lengths in a lattice, and whose Fourier transforms vanish on all but finitely many vector lengths in the dual lattice. If you can do this in full generality, then Poisson summation lets you solve for the number of lattice vectors of each length. These are the coefficients of a modular form, namely the theta series of $E_8$ or the Leech lattice, so the conclusion is that this family of functions somehow “knows about” the theta series. In other words, you can’t expect to construct the whole family without running into modular forms in some way. This leaves a couple of possibilities: maybe the magic functions for sphere packing are simpler than most functions in this family, and could be constructed without modular forms, or maybe these functions are deeper than modular forms (and require some mysterious special functions not yet known to mathematicians). What we know now is that modular forms suffice, and in a sense are necessary because the magic functions are unique.<|endoftext|> TITLE: How flexible is the infinite-dimensional torus? QUESTION [14 upvotes]: Let $\mathbb T=\mathbb R/\mathbb Z$ be the circle group and $\mathbb T^\omega$ be the infinite-dimensional torus, considered as an abelian compact topological group. Problem 1. Is it true that for any finite set $F\subset\mathbb T^\omega$ and any neighborhood $U\subseteq \mathbb T^\omega$ of zero there exists an automorphism $\alpha$ of $\mathbb T^\omega$ such that $\alpha(F)\subset U$? This problem can be reformulated in the language of the special linear groups $SL(n,\mathbb Z)$. Problem 2. Is it true that for any $n\in\mathbb N$, neighborhood of zero $U$ in $\mathbb R^n$ and vectors $x_1,\dots,x_n$ in $\mathbb R^\omega$ there exists $m>n$ and a matrix $A\in SL(m,\mathbb Z)$ such that $\mathrm{pr}_n\circ A\circ \mathrm{pr}_m(x_i)\in U$ for all $i\in\{1,\dots,n\}$? Here $\mathrm{pr_k}:\mathbb R^\omega\to\mathbb R^k$, $\mathrm{pr}_k:x\mapsto x{\restriction}k$, is the projection onto the first $k$ coordinates. Remark 1. For any field $\mathbb F$ and vectors $x_1,\dots,x_n\in\mathbb F^{2n}$ there exists a linear transformation $A\in SL(2n,\mathbb F)$ of $\mathbb F^{2n}$ such that $A(\{x_1,\dots,x_n\})\subset\{0\}^n\times\mathbb F^n$. Remark 2. For any vector $(x,y)\in\mathbb R^2$ and any $\varepsilon>0$ there exists a matrix $A\in SL(2,\mathbb Z)$ such that $(x,y)\cdot A\in (-\varepsilon,\varepsilon)\times\mathbb R$. Such matrix $A$ can be constructed by finding relatively prime integer numbers $p,q$ such that $|xp+yq|<\varepsilon$ and then finding integer numbers $a,b$ such that $pb-qa=1$ (using the extended Euclidean algorithm). Then the matrix $A=\left(\begin{array}&p&a\\q&b\end{array}\right)\in SL(2,\mathbb Z)$ has the required property. Taking into account Remarks 1 and 2, I would expect that the following stronger form of Problem 2 has an affirmative answer. Problem 3. Is it true that for any $n\in\mathbb N$ and vectors $x_1,\dots,x_n\in\mathbb R^{2n}$ there exists a linear transformation $A\in SL(2n,\mathbb Z)$ such that $A(\{x_1,\dots,x_n\})\subset(-\varepsilon,\varepsilon)^n\times\mathbb R^n$? REPLY [4 votes]: This is a draft proof of an affirmative answer to Problem 3. Proposition. For any $n\in\mathbb N$, $\varepsilon>0$ and vectors $x_1,\dots,x_n\in\mathbb R^{2n}$ there exists a linear transformation $A\in SL(2n,\mathbb Z)$ such that $A(\{x_1,\dots,x_n\})\subset[-\varepsilon,\varepsilon]^n\times\mathbb R^n$. Proof. Let $m=2n$, $x_i=(x_{i1},\dots,x_{im})$ for each $i$ and $X=\|x_{ij}\|$. We shall call a column of a matrix small, provided all its entries have absolute value at most than $\varepsilon$, and big, otherwise. Let $k$ be the maximal number of small columns in a matrix $XB$, where $B\in SL(m,\mathbb Z)$ and $C$ be an arbitrary matrix in $SL(m,\mathbb Z)$ such that $XC$ has $k$ small columns. It suffices to show that if $k(2lM)^n$ and define a map $f$ from the subset $Q^l$ of points of the set $[-K, K]^l$ with all integer coordinates to $\mathbb R^n$ by putting $f(d)=d_1y_1+\dots + d_ly_l$ for each $d=(d_1,\dots,d_l)\in Q^l$. Since all $|d_i|\le K$ and all entries of columns $y_i$ have absolute value at most $M\varepsilon$, each coordinate of a vector $f(d)$ is at most $lKM\varepsilon$. Therefore the image $f(Q^l)$ can be covered by $(2lKM)^n$ axis-parallel cubes with side $\varepsilon$. Since $|Q^l|=(2K+1)^l>(2K)^{n+1}>(2lKM)^n$, there exist two distinct elements $d’$ and $d’$ of $Q^l$ such that each coordinate of a vector $y=f(d’)-f(d’’)$ has an absolute value at most $\varepsilon$. Put $d=d’-d’’$. Dividing entries of $d$ by their greatest common divisor, if needed, we can assume that the greatest common divisor of entries of $d$ is $1$. It is well-known (see, for instance, this MSE thread]) that there exists a matrix $D’\in SL(l,\mathbb Z)$ whose first column is $d$. Put $D=\begin{pmatrix} D’ & 0\\ 0 & I\end{pmatrix}$, where $I$ is the $(m-l)\times (m-l)$ identity matrix. Then the first column of the matrix $XCD$ is small, whereas its last $k$ columns are the same as in the matrix $XC$. $\square$<|endoftext|> TITLE: Non-polynomial splines, a non-linear problem QUESTION [5 upvotes]: I'm looking for references on how to construct spline-like functions from a basis that does not include piecewise polynomials. To be specific, given a class of functions such as "decaying exponentials" or "sines and cosines" (which are parameterized by a single parameter, e.g. the decay rate or the frequency), is there an efficient and numerically stable method to construct a function that is piecewise a linear combination of $N$ such functions (whose parameters are to be determined), interpolates given data $(x_k,f_k)$ [which are assumed to be such that they can be interpolated using the given function class, i.e. e.g. monotonically decreasing for decaying exponentials] and has continuous derivatives at $x_k$ up to order $2N-2$ (in order to fix $N$ linear coefficients and $N$ non-linear parameters)? I can of course write down the explicit equations needed to satisfy these conditions, but directly solving those using a non-linear equation system solver does not look too promising as an approach. What literature I could find so far on splines with non-polynomial components referred to spaces spanned by polynomials and some given non-polynomial. Here I'm looking for the case where there are no polynomials and the non-polynomial functions are parameterized by a parameter whose values are to be determined by the interpolation and smoothness conditions. REPLY [2 votes]: The interpolant $$A\left(e^{\frac{a}{A}\xi}-1\right)+B\cdot\left(\cosh\left(\frac{a}{A}\xi\right)-1\right)\ =\ (A+B)\mathbf{e^{\frac{a}{A}\xi}}+B\cdot \mathbf{e^{-\frac{a}{A}\xi}}-(A+B)\\ a=\frac{d}{dx}S(x_i),\\ \xi=x_{i+1}-x_i,\\ \eta=y_{i+1}-y_i,\\ B=\frac{\eta-A\left(e^{\frac{a}{A}\xi}-1\right)}{\cosh\left(\frac{a}{A}\xi\right)-1}$$ The construction of the spline $S(x)$ happens from left to right and requires knowldege of the slope at $x_0$ to be able to calculate the Interpolant that connects $(x_0,y_0)$ with $(x_1,y_1)$; knowing the interpolant we can determine the slope at $x_1$ and we are in the same situation as before so that eventually $S(x)$ is determined. The underlying ideas that led to identifying the interpolant for $C^1$ continuous interpolation can most likely be generalized to higher degrees of smoothness but I'm still on my way. Addendum: parameter $A$ provides limited control over the shape of the interpolant: if we want the "purest" expontial functions, the $A$ should be chosen to minimize $B$ if we strive for preserving shape we can chose $A$ to control the abscissa of the (unique) local extremum in cases of $y_i\lt y_{i+1}\land y_{i+2}\lt y_{i+1}$ align it with the vertex of the parabola through $\left(\,(x_i,y_i),\,(x_{i+1},y_{i+1}),\,(x_{i+2},y_{i+2})\,\right)$ align it with the abscissa of the intersection of the line with slope $a$ that contains $(x_i,y_i)$ and the line through $(x_{i+1},y_{i+1})$ and $(x_{i+2},y_{i+2})$<|endoftext|> TITLE: What's an illustrative example of a tame algebra? QUESTION [7 upvotes]: A finite-dimensional associative $\mathbf{k}$-algebra $\mathbf{k}Q/I$ is of tame representation type if for each dimension vector $d\geq 0$, with the exception of maybe finitely many dimension vectors $d$ representations*, the indecomposable representations of $\mathbf{k}Q/I$ with that dimension vector can be described up to isomorphism as finitely many one-parameter families, the parameter coming from $\mathbf{k}$. What is an illustrative example of a tame algebra? Specifically, what's an example of a quiver $Q$ and admissible ideal $I$ such that (1) for some dimension vectors $d$ the indecomposables can't be described as finitely many one-parameter families*, and (2) for some dimension vectors there is more than just one one-parameter family. I ask because my go-to example of a tame algebra now is the path algebra of the Jordan quiver, the quiver having one vertex and one loop, over an algebraically closed field. But this example doesn't utilize all the wiggle-room that the definition of a tame algebra allows. So I'm hoping there is a better quintessential example to keep in mind. * Note that, as it was originally written, this was not the correct definition of an algebra having tame representation type, and actually condition (1) is not possible. See the comments below for the correct definition. REPLY [6 votes]: I think the following is an example of a tame algebra where there is more than one component to a moduli space of fixed dimension. I don't know any examples where there are dimension vectors with moduli of dimension $>1$. Take a quiver with two vertices $1$ and $2$, two arrows $x_1$ and $x_2$ from $1$ to $2$ and two arrows $y_1$ and $y_2$ from $2$ to $1$. Impose the relations $x_i y_j = 0$ and $y_j x_i = 0$ for $1 \leq i,j \leq 2$. I believe every indecomposable representation either satisfies $x_1=x_2=0$ or $y_1=y_2=0$. Thus, every indecomposable representation is a representation of either the Kronecker quiver $1 \rightrightarrows 2$ or else $1 \leftleftarrows 2$, both of which are tame, so this is tame. For each dimension vector of the form $(n,n)$, we get two families of indecomposable representations, coming from choosing whether to make $x_1=x_2=0$ or else $y_1=y_2=0$. REPLY [5 votes]: For the Kronecker quiver (two vertices, two arrows in the same direction) and dimension vector (1,1), over an algebraically closed ground field, the indecomposables are naturally parameterized by points in $\mathbb P^1(k)$. (The representation with the two maps given by $a$ and $b$ is sent to $[a:b]$. For other tame quivers with no relations over an algebraically closed ground field, the situation is slightly worse: the natural indexing set for the representations whose dimension vector is the null root is $\mathbb P^1(k)$ with some points (up to three of them) counted more than once (but finitely many times). This happens in the example Bugs gave: there are three inhomogeneous tubes, each of width two, each containing two representations of dimension vector the null root, whereas the other points of $\mathbb P^1(k)$ each correspond to one representation. (With the all-inward orientation, the reason for the indexing by $\mathbb P^1(k)$ is that the moduli space of 4 points on $\mathbb P^1$—equivalent to representations with dimension vector $(1,1,1,1,2)$, i.e., the null root—is again $\mathbb P^1$.) I am not quite sure what you mean by the extra wiggle room of type (1). Are these supposed to be dimension vectors that have only finitely many indecomposables? I would usually think that in that case, they can also be described by one-parameter families: just make the families constant.<|endoftext|> TITLE: Fermat's Last Theorem for integer matrices QUESTION [41 upvotes]: Some years ago I was asked by a friend if Fermat's Last Theorem was true for matrices. It is pretty easy to convince oneself that it is not the case, and in fact the following statement occurs naturally as a conjecture: For all integers $n,k\geq 2$ there exist three square matrices $A$, $B$ and $C$ of size $k\times k$ and integer entries, such that $\det(ABC)\neq 0$ and: $$A^n + B^n = C^n$$ Of course, the case $k=1$ is just Fermat's Last Theorem, but in that case the conclusion is the opposite for $n>2$. I think that I read somewhere that it is known that the above assertion is true (I do not remember exactly where, and haven't seen anything on Google, but this old question on MSE, on which there is an old reference, that I think does not answer this). Two observations that are pretty straightforward to verify are the following: the case $2\times 2$ and $3\times 3$ solve the general case $k\times k$ by putting suitable small matrices on the diagonal. Also, as it is stated in a comment on that question, if the exponent $n$ is odd then the case $2\times 2$ can be solved by this example: $$\begin{bmatrix} 1 & n^\frac{n-1}{2}\\ 0 & 1\end{bmatrix}^n + \begin{bmatrix} -1 & 0\\n^\frac{n-1}{2} & -1\end{bmatrix}^n = \begin{bmatrix} 0 & n\\1& 0\end{bmatrix}^n$$ Does anybody know of such examples in the $2\times 2$ case for even $n$, and the general $3\times 3$ case? More clearly: are there easy and explicit examples for each $n$ and $k$ for the above conjecture? REPLY [18 votes]: This problem is addressed in "On Fermat's problem in matrix rings and groups," by Z. Patay and A. Szakács, Publ. Math. Debrecen 61/3-4 (2002), 487–494, which summarizes previous work on the topic and gives some new results. It seems that the problem is not completely solved. When $k=2$, Khazonov showed that there are solutions in $SL_2(\mathbb Z)$ if and only if $n$ is not a multiple of 3 or 4, but I couldn't immediately find any statement anywhere about the case $4\mid n$ and $2\times 2$ integer matrices with nonzero determinant. Khazanov also proved that $GL_3(\mathbb Z)$ solutions do not exist if $n$ is a multiple of either 21 or 96, and $SL_3(\mathbb Z)$ solutions do not exist if $n$ is a multiple of 48. Patay and Szakács give explicit solutions for $SL_3(\mathbb Z)$ when $n=\pm 1\pmod 3$ as well as for $n=3$. Here's a solution for $n=3$: $$\pmatrix{0& 0&1\\ 0 &-1& 1\\ 1 & 1 & 0}^3 + \pmatrix{0&1&0\\ 0&1&-1\\ -1&-1&0}^3 = \pmatrix{0&1&1\\0&0&1\\1&0&0}^3.$$<|endoftext|> TITLE: Curious identity between the two kinds of Chebyshev polynomials QUESTION [14 upvotes]: I have found, by accident, an identity that relates a sum of Chebyshev polynomials of the first kind to a Chebyshev polynomial of the second kind. It goes as follows: Given an integer partition of $n$, let $g_a$ be the number of times $a$ appears in said partition.${}^1$ Then the following identity holds for all $n \in\mathbb{N}$: $$ U_n(x) = \sum_{\substack{n_i>0\\ \sum_i n_i = n}} \frac{1}{\prod_{a\in \{n_i\}} g_a!} \prod_{i=1} \frac{2}{n_i} T_{n_i}(x)\,. $$ The sum is over all integer partitions of $n$, the product is on all $n_i$'s in the partition, with repetitions. I have a very roundabout way to prove this identity (I'll skip the details). The left hand side is obtained by contracting two symmetric traceless tensors of $SO(4)$. That is, letting $|x|=|y|=1$ and $x,y\in \mathbb{R}^4$ then $$ (x^{i_1}\cdots x^{i_n} - \mathrm{traces}) (y_{i_1}\cdots y_{i_n} - \mathrm{traces}) \propto U_n(x\cdot y)\,. $$ The right hand side instead comes from the same contraction but in spinor notation. Namely we let $$ \mathrm{x} = \left(\begin{matrix}x_3-x_4 & x_1 - i x_2 \\ x_1 + i x_2 & -x_3-x_4\end{matrix}\right)\,,\quad \bar{\mathrm{x}} = \epsilon\, \mathrm{x}\, \epsilon^T\,, $$ and $\mathrm{y}$ in a similar way ($\epsilon$ is the Levi Civita tensor). Then we introduce two dimensional spinors $\eta,\tilde{\eta}$ let $\partial_{\eta^\alpha}\eta^\beta = \delta_\alpha^\beta$ (similar for $\tilde{\eta}$) and finally $$ (\partial_\eta \mathrm{x} \partial_{\tilde{\eta}})^n (\eta \mathrm{y}\tilde{\eta})^n \sim \sum_{\substack{n_i>0\\ \sum_i n_i = n}} \# \prod_{i} \mathrm{tr}\,((\mathrm{x}\bar{\mathrm{y}})^{n_i})\,. $$ The sum over partitions comes from a combinatoric argument. Then it's a simple exercise to show that $\mathrm{tr}\,((\mathrm{x}\bar{\mathrm{y}})^n) \propto T_n(x\cdot y)$. My questions are Is this identity known already? If not, could you come up with some more direct argument to prove it? $\;{}^1$ For example, $(1,1,1,2,2,3)$ is an integer partition of $n=10$ with $g_1 =3,\, g_2=2,\,g_3=1$. REPLY [5 votes]: I think I can sketch a shorter proof. Let $z_j = x_j+x_j^{-1}$, and let $p_m$ and $h_m$ denote the power-sum and complete homogeneous symmetric polynomial. Then (see e.g p.3 in this preprint) $$ 2 T_m(z_j/2) = p_m(x_j,x_j^{-1}) \text{ and } U_m(z_j/2) = h_m(x_j,x_j^{-1}) $$ Now, we can use the Newton identities, to express $h_m$ in terms of the power-sum symmetric functions. This gives a relation between the $U_m$ and the $T_m$. Looking at your formula, it is very similar to the Newton identity.<|endoftext|> TITLE: Resolution graphs in the sense of Némethi QUESTION [7 upvotes]: The following definitions are from lecture notes of Némethi. A surface singularity $(X,0)$ is defined by $$(X,0) = (\{ f_1 = \ldots = f_m=0 \}) \subset \mathbb (\mathbb{C}^n,0),$$ where $f_i : (\mathbb{C}^n ,0) \to (\mathbb{C},0)$ are germs of analytic functions with $$r(p) = \mathrm{rank} \left [ \frac{\partial f_i}{\partial z_i} (p) \right ]_{i=1, \ldots, m; j=1, \ldots, N} = N-2$$ for any generic or smooth point $p$ of $X$. If $r(0) = N-2$, then $(X,0)$ is analytically isomorphic to $\mathbb (\mathbb{C}^2,0)$. The singularity $(X,0)$ is called normal, if any bounded holomorphic function $f: X - \{ 0\} \to \mathbb C$ can be extended to a holomorphic function defined on $X$. Then there is an algorithm in Appendix 1 for finding the resolution graphs of singularities in $\mathbb{C}^3$ with equation of the form $g(x,y) + z^n$ (i.e. suspensions of curve singularities in $(\mathbb C^2,0)$). I couldn't understand the steps of that algorithm. My question is that Is there anyone to describe this algorithm for example Brieskorn sphere $\Sigma(2,3,4)$? Is it possible to find any Magma or Sage code for this purpose? REPLY [3 votes]: About question 2: I think that the software Singular has this feature; it's well-documented, and if you look for resolution graph you should find the reference. About question 1: well, I must admit that that algorithm is not pleasant, and that it took me a while to work out the case of $\Sigma(2,3,4)$. Anyway, here we go. I can't draw graphs here (but if anyone can tell me that we can embed \xygraphs, next time I can make an effort), so things will have to be less pictorial than I'm comfortable with. I'll use greek letters for vertices to avoid clashing with Némethi's letters. So, we fix $g(x,y) = x^2+y^3$ and $n=4$. First off, the resolution graph of $g$ has three vertices $\alpha, \beta, \gamma$, and an arrowhead $\delta$. The self-intersections are $e_\alpha = -3$, $e_\beta=-2$, $e_\gamma = -1$. ($\gamma$ is the central vertex, corresponding to the last blow-up, $\alpha$ was the first blow-up, and $\beta$ the second.) From the formula (*) in the appendix, relating multiplicities with self-intersections, and using that $m_\delta = 1$, we obtain that $(m_\alpha, m_\beta, m_\gamma) = (2,3,6)$. We the compute $(d_\alpha, d_\beta, d_\gamma) = (2,1,3)$. Step 1(a): from the computations above, $\alpha$ is covered by two vertices of multiplicity 1, while $\beta$ and $\gamma$ are covered by one vertex of multiplicity 3 each. Call them $\tilde\alpha_1, \tilde\alpha_2, \tilde\beta, \tilde\gamma$. The genus formula says that they're all rational ($\tilde g = 0$). Step 1(b): the edge $(\alpha,\gamma)$ lifts to strings connecting $\tilde\alpha_i$ to $\tilde\gamma$, each of type $G(1,3,2)$; which means just a single vertex $\varepsilon_i$ with $(e_{\varepsilon_i}, m_{\varepsilon_i}) = (-2,2)$ (and two arrowheads ending at $\tilde\alpha_i$ and $\tilde\gamma$). The edge $(\beta,\gamma)$ lifts to a string of type $G(6,3,4)$, which is a single vertex $\zeta$ with $(e_{\zeta}, m_{\zeta}) = (-2,3)$ (and two arrowheads ending at $\tilde\beta$ and $\tilde\gamma$. Step 1(c): the arrowhead $\delta$ lifts to a string of type $G(6,1,4)$, which is again a single vertex $\eta$ with $(e_{\eta}, m_{\eta}) = (-2,2)$ (and two arrowheads, one ending at $\tilde\gamma$ and the other one free). At this point, the graph is star-shaped with four legs: its center is at $\tilde\gamma$ (with weight unspecified); two legs with weights $(-2,?)$ (ending at $\tilde\alpha_1, \tilde\alpha_2$), one with weights $(-2,?)$ (ending at $\tilde\beta$), one with weight $-2$ and an arrowhead. Step 2: we now compute the missing self-intersections using (*). We obtain $e_{\tilde\alpha_i} = -2$, $e_{\tilde\beta} = e_{\tilde\gamma} = -1$. Step 3: we drop the arrowhead, and blow down $\tilde\gamma$ and then $\zeta$ (which, after blowing down $\tilde\gamma$, can be blown down). The graph we obtain is $E_6$, as expected.<|endoftext|> TITLE: Gaussian measure on function spaces QUESTION [6 upvotes]: I'm reading this classic work and I'd like to get deeper inside some of its techniques. In particular, the authors state: "We construct a Gaussian measure $d\mu_{0}(\phi)$ on a measure space of continuous functions $\phi(x), x\in \Lambda \subset \mathbb{R}^{3}$ with covariance $u$: \begin{eqnarray} \int d\mu_{0}(\phi)e^{i\int f\phi} = e^{-\frac{1}{2}\int f u f} \tag{1}\label{1} \end{eqnarray} It is then straightforward to show that: \begin{eqnarray} e^{-\beta U} = \int d\mu_{0}(\phi) e^{i\sqrt{\beta}\sum_{\alpha}e_{i(\alpha)}\phi(x_{\alpha})}" \tag{2}\label{2} \end{eqnarray} First of all, how to construct such a Gaussian measure $d\mu_{0}$ on a space of continuous functions? Is it defined by condition (\ref{1}) or does (\ref{1}) follow as a consequence? Besides, how can we prove existence? Does anyone know any reference on this construction? Second, equation (\ref{2}) seems to follow by taking $f = \sum e_{i(\alpha)}\delta(x_{\alpha})$. But how can we take such an $f$ is $f$ must be a continuous function rather than a distribution? REPLY [4 votes]: You should have a look at the book by Gelfand and Vilenkin Generalized functions. Vol. 4: Applications of harmonic analysis where they describe how to construct Gaussian measures on (duals) of nuclear spaces. Thus, given an open set in $\newcommand{\bR}{\mathbb{R}}$ $D\subset \bR^n$ one begins by constructing a measure on the space $C^{-\infty}(D)$ of generalized functions on $D$. If the covariance kernel is sufficiently regular then this measure is concentrated one on a much smaller subspace. Also, if you read French, I recommend this 1967 paper by Xavier Fernique. It is not the most comprehensive but I found it very helpful. Finaly, there is V. Bogachev's book Gaussian Measures.<|endoftext|> TITLE: Distribution of signs of automorphic forms QUESTION [6 upvotes]: Let's say we have an automorphic form $f$ on $GL(2)$ that is self-dual. In particular, the associated L-function $L(s,f)$ satisfies a functional equation with sign $\varepsilon_F = \pm 1$. Is it known that the proportion of such automorphic forms with given sign (say $-1$) is exactly $1/2$? I know many results about distributions of signs for coefficients and eigenvalues of automorphic forms, however when I think of this question I wonder whether it is well-known or difficult? REPLY [8 votes]: For simplicity, let's consider the case of holomorphic modular forms over $\mathbb Q$ of squarefree level and trivial nebentypus. Then one knows from Iwaniec, Henryk; Luo, Wenzhi; Sarnak, Peter. Low lying zeros of families of $L$-functions. Publications Mathématiques de l'IHÉS, Tome 91 (2000) pp. 55-131. an asymptotic formula for the dimensions of the subspaces of cusp forms with root number $+1$ and root number $-1$. In particular, the proportion of forms with root number $+1$ tends to $\frac 12$ as you take some combination of the weight and the level to infinity. As Peter mentions in the comments, I also consider this in my paper Refined dimensions of cusp forms, and equidistribution and bias of signs. Journal of Number Theory, Vol. 188 (2018), 1-17. Specifically, I get an exact formula for the dimensions of subspaces with prescribed root number (or prescribed Atkin-Lehner signs), and observe a "strict bias" phenomenon for root number +1: while the proportion of newforms with root number +1 is asymptotically $\frac 12$, in any given space, there are always at least as many newforms with root number +1 as with -1, and it is strictly greater except in a few special situations. One should similarly be able to prove that the proportion is $\frac 12$ in more general families of automorphic forms, say using a trace formula and simply bounding error terms, though I don't know the most general situation in which this has already been done in the literature. Essentially one needs to know that the trace of the Fricke involution is not large compared to the dimension.<|endoftext|> TITLE: Foliation of $\mathbb R^n$ by connected compact manifolds QUESTION [7 upvotes]: Does there exist a smooth nontrivial fiber bundle $p: F \hookrightarrow \mathbb R^n \to B$ such that $F$ and $B$ are connected manifolds with $F$ compact? "Nontrivial" here means the fiber $F$ is not a point. REPLY [14 votes]: On the other hand, if you only mean "foliation" as in your title, and not "fibration", then there is Vogt's foliation of R^3 by circles! (But it is not C^1, only differentiable). Vogt, Elmar, "A foliation of R3 and other punctured 3-manifolds by circles", Publications Mathématiques de l'IHÉS, Tome 69 (1989), p. 215-232 http://www.numdam.org/item/PMIHES_1989__69__215_0/<|endoftext|> TITLE: Area of a elliptic surface confined by a sphere QUESTION [5 upvotes]: Let $S$ be a surface enclosed inside the unit sphere in $R^3$. If every point of S is elliptic, then must $\operatorname{Area}(S)≤\operatorname{Area}(S^2)$? REPLY [4 votes]: I do not think so. Just imagine that you peel a large orange whose surface area is much bigger than that of a unit sphere. Then you "spiral" the peel to make it arbitrarily small and place it inside a unit sphere. Imagine a surface that looks more or less as this one: It has positive curvature so very point is elliptic. Here you remove a small cylinder around the vertical axis so there is no problems with the curvature. Since you can have as many "turns" as you want, its surface area can be arbitrarily large while the surface occupies a small region in space.<|endoftext|> TITLE: Positively curved metric with uniformly positive scalar curvature QUESTION [5 upvotes]: Can we find a complete noncompact Riemannian manifold $(M^n,g)$ with bounded geometry satisfying the following conditions? the curvature operator $Rm>0$; the scalar curvature $R \ge 1$. Notice that any such manifold must be diffeomorphic to $\mathbb R^n$. REPLY [4 votes]: Yes, this is possible. Note that a strictly convex hypersuface in $\mathbb R^{n+1}$ has positive $Rm$. To get an example, consider the following graph $H\subset \mathbb R^{n+1}$ over the open unit $n$-disk in $\mathbb R^n$: $$H:=\left(x_{n+1}=\frac{1}{1-\sum_{i=1}^n{x_i^2}}\right).$$ Clearly, $H$ is convex. Moreover, the scalar curvature of $H$ tends to the scalar curvature of the unit $n-1$-sphere as $x_{n+1}$ tends to infinity. In particular the scalar curvature is greater than a certain positive $c>0$ on $H$. So if one scales down this hypersurface by a constant (i.e takes the hypersurface $\varepsilon\cdot H\subset \mathbb R^{n+1}$), we get $R(\varepsilon\cdot H)>1$. The example works if $n\ge 3$.<|endoftext|> TITLE: Primes mod 4 and integer polynomials QUESTION [6 upvotes]: I have asked these questions as comments here (these are related to the question there). The questions are: Let $S$ be one of the following sets of primes: All primes of the form $4k+1$ ; All primes of the form $4k+3$; All primes of the form $4k+1$ except $5, 13$; Is there a monic integer polynomial which is reducible mod prime $p$ iff $p\in S$. REPLY [5 votes]: Here is a way to argue without showing directly that the polynomial must have degree $2$. It was explained to me by Borys Kadets (all further mistakes are, of course, my contribution). Lemma. If a set of primes $S$ of density $\frac{1}{2}$ admits such polynomial then some subset $S'\subset S$ with $\#(S\setminus S')<\infty$ admits a monic quadratic polynomial that is reducible precisely at $S'$. Proof. Suppose that $f$ is a polynomial of degree $n$ satisfying the condition for the set $S$. Let $G$ be the Galois group of its splitting field coming with an embedding $G\subset S_n$. By Chebotarev density, exactly $\frac{1}{2}\# G$ elements of this group must be cycles of length $n$. Since the centralizer of a length $n$ cycle $\sigma\in S_n$ is the subgroup generated by $\sigma$, the number of conjugacy classes of length $n$ cycles in $G$ is $\frac{n}{2}$. In particular, $n$ is even and $G\cap A_n$ has index $2$ in $G$ with cycles of length $n$ forming the non-trivial coset. The subgroup $G\cap A_n\subset G$ corresponds to a degree $2$ extension $K/\mathbb{Q}$. If a prime $p$ is ramified in the splitting field of $f$ then $f$ is reducible modulo $p$. For any unramified prime $p$ the polynomial $f$ is reducible modulo $p$ iff the Frobenius element of a prime above $p$ in the splitting field is not a length $n$ cycle, the latter condition being equivalent to the fact that $p$ is split in $K$. Thus, the set of primes split (including ramified) in $K$ is equal to $S$ with the possible exception of a finite set of ramified primes. The minimal polynomial of a generator of $\mathcal{O}_K$ satisfies the conclusion of the lemma. $\square$ Starting with any of the three sets $S$ the lemma gives a quadratic polynomial $x^2+ax+b$ with $a,b\in \mathbb{Z}$ that is reducible precisely at the primes from a set $S'$. Since we want it to be irreducible mod $2$, both $a$ and $b$ have to be odd. This polynomial is irreducible modulo $p>2$ if and only if $D:=a^2-4b$ is not a square mod $p$. Set 2: The number $(-D)$ is supposed to be a non-residue modulo all but finitely many primes, but that's impossible. This can be shown by a counting argument: if there was a finite set of primes $p_1,\dots, p_k$ such that $D+n^2$ is a product of powers of $p_i$'s then there would be $O((\log N)^k)$ numbers of the form $D+n^2$ in the interval $[1,\dots, N]$. Sets 1 and 3: Here we want $(-D)$ to be a square modulo all but finitely many primes. That forces it to be a square in $\mathbb{Z}$. However, setting $-D=c^2$ gives $a^2+c^2=4b$. That is impossible for odd $a$.<|endoftext|> TITLE: Pulling back a functor, it becomes monadic QUESTION [8 upvotes]: $\require{AMScd}$I am in the following situation: the diagram $$ \begin{CD} \cal M @>r>> [{\cal B},Set] \\ @VuVV @VVf^*V \\ \cal D @>>N_g> [{\cal A}^\text{op},Set] \end{CD} $$ is a (strict) pullback in $\bf Cat$; moreover, $f : \cal A^\text{op}\to B$ is bijective on objects (and $f^*$ is the "inverse image" functor given by precomposition). In turn, $g$ is the "nerve" induced by a functor $g : \cal A \to D$. $N_g$ is fully faithful because $g$ happens to be dense, and so $r$ is f.f. as well. If needed, $\cal D$ is complete and cocomplete. This gives $\cal M$ a very explicit description: it is a reflective subcategory of $[{\cal B},Set]$ made by the functors $F : {\cal B}\to Set$ such that $F(fA)={\cal D}(gA,D)$ for some $D\in\cal D$. Is all this sufficient to imply that $u$ is monadic? If not, what additional assumptions are needed? REPLY [10 votes]: If $A$ and $B$ are small and $D$ is locally presentable then $u$ is a monadic right adjoint. More generally, what is non-trivial is the construction of a left adjoint of $u$. As soon as $u$ has a left adjoint, $u$ is monadic. The proof is relatively simple: the forgetful functor $f^*$ is monadic, so it satisfies all the conditions of Beck monadicity criterion, and all the conditions of the Beck criterion except the existence of an adjoint are automatically satisfied for its pullback ($f^*$ is an isofibration, so the square is both a strict and pseudo-pullback). Now when all the category involved are locally presentable as both $f^*$ and $N_g$ are accessible right adjoint functors, the square is also a pullback in the category of locally presentable categories and accessible right adjoint functor (which is known to have all limits, and these limits are preserved by the forgetful functor to set by an old results, which appears in Bird's phd thesis, though I think it was known before) For more details, you can also have a look to John Bourke and Richard Garner Monads and theories this is how they construct their functor from "pre-Theory to Monads" (which is left adjoint to the Kleisli category functor)<|endoftext|> TITLE: The "Chaos Game" as a particular series of i.i.d. random variables QUESTION [6 upvotes]: Fix a parameter $\alpha\in(0,1)$ and take an i.i.d. sequence $X_0,X_1,\ldots$ of $\mathbb{R}^n$ valued random variables. Construct the limiting random variable $X_\infty = (1-\alpha)\sum_{k=0}^\infty \alpha^k X_k.$ Is any general result known about this kind of limit? What if the $X_i$ follow a well known distribution like uniform/Rademacher? I was motivated by this sum after running into: https://en.wikipedia.org/wiki/Chaos_game. For example if the $X_i$ are uniformly distributed on the 3 vertices of a triangle and $\alpha = 1/2$ the limiting distribution is supported on the associated Sierpinski Triangle. The fact that finite support distributions can give fractal shapes from this construction leads me to believe this is a non-trivial question. My apologies if this ends up being an exercise in some well known textbook on probability theory. If it is, I'd appreciate a reference for that textbook. Edit: I was able to locate http://u.math.biu.ac.il/~solomyb/RESEARCH/Bernotes.pdf REPLY [10 votes]: $\newcommand\al{\alpha}$Let us drop the factor $1-\al$, by considering $$Y:=X_\infty/(1-\al)=\sum_{k=0}^\infty\al^k X_k.$$ By Kolmogorov's three-series theorem, this series will converge almost surely (a.s.) unless at least one of the tails of the distribution of $X_0$ is too heavy. Assume that the series indeed converges a.s. Then, obviously, $$Y\overset D=X+\al Y,$$ where $\overset D=$ denotes the equality in distribution and $X$ is an independent copy of the $X_k$'s. So, we have the functional equation for $F_Y$: $$F_Y(y)=\int_{-\infty}^\infty F_Y((y-x)/\al)\,dF_X(x)$$ for real $y$, where $F_Z$ denotes the cdf of $Z$. Equivalently, we have the functional equation for $f_Y$: $$f_Y(t)=f_Y(\al t)\,f_X(t)$$ for real $t$, where $f_Z$ denotes the characteristic function of $Z$. Of course, we can also write $$f_Y(t)=\prod_{k=0}^\infty f_X(\alpha^k t)$$ for real $t$. In the particular case when $X$ is Rademacher, the distribution of $Y$ is the well-studied Bernoulli convolution. In the particular case when $X$ is $U(0,1)$ and $\alpha=1/2$, $F_Y$ is the well-studied Fabius function.<|endoftext|> TITLE: Pseudo-intersections, splitting families, and ultrafilters QUESTION [7 upvotes]: Suppose $U$ is a non-principal ultrafilter on $\omega$, and let us define $\tau(U)$ to be the minimum cardinality of a family $\mathcal{X}\subseteq U$ such that $\mathcal{X}$ does not have an infinite pseudo-intersection, that is, there is no infinite $A$ such that $A\setminus B$ is finite for all $B\in \mathcal{X}$. Claim: $\tau(U)\leq\mathfrak{s}$ for any $U$. The point is that $U$ will contain a splitting family of cardinality $\mathfrak{s}$, and this splitting family has no infinite pseudo-intersection. (To see the first statement, note that if $\mathcal{X}$ is a splitting family and we replace some of the elements of $\mathcal{X}$ by their complements, then the resulting collection is still a splitting family. Since an ultrafilter will contain one of $X$ and $\omega\setminus X$ for each $X\in\mathcal{X}$, we may as well assume $\mathcal{X}\subseteq U$.) Question: Is there (in ZFC) an ultrafilter $U$ on $\omega$ for which $\tau(U)=\mathfrak{s}$? I conjecture that the answer is "no", and that this negative answer will be witnessed in the original Blass-Shelah model of NCF from the paper below. Blass, Andreas; Shelah, Saharon, There may be simple $P_{\aleph _ 1}$- and $P_{\aleph _ 2}$-points and the Rudin-Keisler ordering may be downward directed, Ann. Pure Appl. Logic 33, 213-243 (1987). ZBL0634.03047. REPLY [7 votes]: The answer is no -- it is consistent that every $U \in \omega^*$ has $\tau(U) < \mathfrak{s}$. I had an idea for proving this earlier today, using the Mathias model. I couldn't quite make things work, and I ended up talking about the problem with Alan Dow for a good part of the afternoon. (1) We still think the Mathias model could work, but it seems tricky. (2) There's a fix: if you interleave Laver forcings with Mathias forcings in a certain way, then the resulting iteration does work. (I'll sketch this below.) (3) I learned from Alan that your question has been studied already. The consistency of "every $U \in \omega^*$ has $\tau(U) < \mathfrak{s}$" is already known (via a different argument than the Mathias-Laver iteration sketched below), and the characteristic $\tau(U)$ has been studied quite a bit. A rich source of information on $\tau(U)$ is the following paper by Brendle and Shelah: Jörg Brendle and Saharon Shelah, ``Ultrafilters on $\omega$ -- their ideals and their characteristics,'' Transactions of the AMS 351 (1999), pp. 2643-2674. (available here) What you call $\tau(U)$ is in this paper called $\pi \mathfrak{p}(U)$. They prove, among other things, that $\bullet$ If $U$ is not a $P$-point, then $\tau(U) \leq \mathfrak{b}.$ $\bullet$ If $U$ is not a $P$-point, then the cofinality of $\tau(U) \leq \mathfrak{b}$ is uncountable. (This provides a partial answer to Santi's question in the comments.) $\bullet$ A characterization of $\tau(U)$ is given in terms of an ideal defined from Ramsey-null sets. $\bullet$ It is consistent that $\tau(U) < \mathfrak{s}$ for all $U \in \omega^*$. The first two results are in Section 2, the next in Section 3, and the last in Section 7. The last result answers your question, of course, but I should also mention another relevant paper: Alan Dow and Saharon Shelah, ``Pseudo P-points and splitting number,'' Archive for Mathematical Logic 58 (2019), pp. 1005-10027. (available here) In the Brendle-Shelah paper, they prove $\sup_{U \in \omega^*}\tau(U) < \mathfrak{s}$ is consistent, but the gap between $\sup_{U \in \omega^*}\tau(U)$ and $\mathfrak{s}$ is only one. In the Dow-Shelah paper, they use a more complicated matrix iteration to make the gap between $\sup_{U \in \omega^*}\tau(U)$ and $\mathfrak{s}$ arbitrarily large. Finally, let me sketch the idea I mentioned above. The idea is to do a countable support iteration that uses Laver forcing at limit steps of cofinality $\omega_1$, and uses Mathias forcing everywhere else. The iteration is of length $\omega_2$ and CH holds in the ground model. Let $V[G]$ denote the result of such an iteration, and let $U \in \omega^*$ in $V[G]$. By reflection, there is some intermediate model $V[G_\alpha]$ with $\alpha < \omega_2$ where $U \cap V[G_\alpha]$ is an ultrafilter in $V[G_\alpha]$, and where $\alpha$ has cofinality $\omega_1$. At this stage, we force with Laver forcing, and this adds a length-$\omega_1$ tower to $U \cap V[G_\alpha]$. All the subsequent Laver forcings and Mathias forcings preserve the fact that this tower has no pseudo-intersection, and so this tower is a witness to the fact that $\tau(U) = \aleph_1$ in the final extension. (See Theorem 7.11 from this paper of Alan's for more detail on the last two sentences.) Finally, the Mathias forcings enable us to get $\mathfrak{s} = \mathfrak{c}$. This is well-known to be true in the Mathias model (although not in the Laver model), and it's true in this model for essentially the same reasons.<|endoftext|> TITLE: Calabi-Yau threefold with an automorphism of infinite order QUESTION [13 upvotes]: I am looking for a (hopefully simple) example of a Calabi-Yau threefold (projective, simply connected, with trivial canonical bundle) admitting an automorphism of infinite order. REPLY [6 votes]: Another nice example is by Oguiso and Truong, "Explicit Examples of rational and Calabi-Yau threefolds with primitive automorphisms of positive entropy". Briefly, take $E$ to be an elliptic curve with an order $3$ automorphism $\tau$ and form the abelian variety $A = E \times E \times E$. The quotient of $A$ by the diagonal action of $\tau$ is mildly singular, but it has only isolated singularities and it's easy to check/standard that there's a crepant resolution $X$ that's a CY3. Then it's easy to get automorphisms of infinite order: $SL(3,\mathbb Z)$ acts on $X$ in an obvious way. In fact, [OT] show that some of these automorphisms are "primitive", meaning they don't preserve any nontrivial fibration (unlike in the elliptically fibered examples). This is done using the theory of dynamical degrees. (Note that this obviously gives honest automorphisms, not just birational maps; there's nothing subtle to check on singular fibers, unlike what you usually run into looking at elliptically fibered examples, as Bort alludes in the comments above.)<|endoftext|> TITLE: The abc-conjecture over the positive rationals and Levy-Schoenberg kernels? QUESTION [6 upvotes]: I am continuing the "abc-adventure" and have a specific question, which needs some explanation: In this paper by Gangolli, the term "Levy-Schoenberg" kernel is defined (Definition 2.3). Consider the group of $G = (\mathbb{Q}_{>0},\times)$ of positive rationals. Then, in this paper by Boudreaux & Beslin, the $\gcd$ is extended to $G$, which I will call for short $\gcd^*$ Having a multiplicative function $f:\mathbb{N} \rightarrow \mathbb{N}$, one might extend it to $G$ via: $$f^*\left(\frac{a}{b}\right) = \frac{f\left(\dfrac{a}{\gcd(a,b)}\right)}{f\left(\dfrac{b}{\gcd(a,b)}\right)}$$ Using: $$\gcd^*(a,b)=1 \iff a,b \in \mathbb{N} \text{ and } \gcd(a,b)=1$$ then $f^*$ is multiplicative on $G$. I will look at $f=\operatorname{rad}$, hence $\operatorname{rad}^*$ is the extension to $G$. Using these "extensions" one might formulate the abc-conjecture over $G$. It is not difficult to show, that it is equivalent to the abc-conjecture of the natural numbers. My question is, if $k(a,b) = \frac{\gcd^*(a,b)}{a+b}$ is positive definite $\ge 0$. Let $d(a,b) = \sqrt{1-2k(a,b)}$ and $$f(a,b) = \frac{1}{2}\big(d(a,1)^2+d(b,1)^2-d(a,b)^2\big)$$ If $k(a,b)$ is positive definite over $G$, then $d$ is an Euclidean metric, and by the characterization of Schoenberg, $f$ is posivite definite. Furthermore: $$f(a,b) = f(b,a)$$ $$f(a,1) = 0 \quad \forall a \in G$$ $$r(a,b) := f(a,a)+f(b,b)-2f(a,b) = d(a,b)^2$$ is invariant under the action of $G$: $$r(qa,qb) = r(a,b) \quad \forall q,a,b \in G$$ This makes $f$ by the definition (2.3) of the paper at the begining of the question to a "Levy-Schoenberg" kernel. Of course replacing $k(a,b)$ with $k(a,b) := \frac{1}{\operatorname{rad}^*\left(\frac{ab(a+b)}{\gcd^*(a,b)^3}\right)}$ and using the abstract invariance property: $$ k(qa,qb) = k(a,b) \quad \forall q,a,b \in G $$ we can construct more of these Levy-Schoenberg kernels, if the "rad"-function above is a positive definite kernel... which seems difficult to prove. Why the question, if $k(a,b)$ is positive definite: If we go back to the natural numbers, and define: $X_a := $ set of divisors of $a$. Then $\mu(X) = \sum_{x \in X} \phi(x)$ for every finite subset $X \subset \mathbb{N}$, and hence $\mu(X_a) = a$ and $X_a \cap X_b = X_{\gcd(a,b)}$, where $\phi$ is the Euler totient function. Using this, one can prove that over the natural numbers, with the help of this paper by Nader, Bretto, Mourad and Abbas, that: $$ \frac{\gcd(a,b)}{a+b} = \frac{\mu(X_a \cap X_b)}{\mu(X_a)+\mu(X_b)}$$ is positive definite. My idea was to do the same in the case $G$: Let for $a \in G$ be defined $X_a := \{ d | \gcd^*(a,d) = d \}$ be the set of divisors of $a$, which does not need to be finite. Then $X_a \cap X_b = X_{\gcd^*(a,b)}$. Hence it remains (?) to find a measure $\mu$ on $G$ such that for all $X_a$ we have: $$\mu(X_a) = a$$ Then we would have that $k(a,b)$ is positive definite! Of course, one does not need to follow this route, to prove the positive-definiteness of $k$, this is just an idea. Thanks for your help! Related: The abc-conjecture as an inequality for inner-products? REPLY [5 votes]: I think I found an answer to the question above: Let $k(a,b)$ be a (positive definite $\ge 0$, symmetric) kernel on $\mathbb{N}\times \mathbb{N}$ such that if $k^*(a,b)$ is a function on $G \times G$ then we have: $$k^*(a,b) = k(a',b')$$ where $a'=\frac{a}{\gcd^*(a,b)}, b'=\frac{b}{\gcd^*(a,b)}$, then $k^*$ is a kernel on $G\times G$. Proof: Since $k$ is positive definite on $\mathbb{N}\times \mathbb{N}$, it follows that for $a_i',b_i' \in \mathbb{N}$ (which are pairwise coprime), the matrix: $$k(a_i',b_i')=k^*(a_i,b_i)$$ is positive definte ($i=1,\cdots,n$ for some $a_i,b_i \in G$, $n$ a natural number). Hence $k^*$ is positive definite. Since $k^*(a,b) = \frac{\gcd(a,b)}{a+b}$ satisfies the assumption $k^*(a,b) = k(a',b')$, it follows that $k^*$ is a positive definite kernel. Thanks for your patience, with my never-ending questions! ;)<|endoftext|> TITLE: Does this equation have more than one integer solution? QUESTION [7 upvotes]: Consider the following diophantine equation $$n = (3^x - 2^x)/(2^y - 3^x),$$ where $x$ and $y$ are positive integers and $2^y > 3^x$. Does $n$ have any other integer solutions besides the case when $x=1$ and $y=2$, which give $n=1$? REPLY [15 votes]: This follows quickly from the observations of user44191. Check each $1 \leq x \leq 66$ and note that, for $x \geq 67$, we have $y < 1.6x$. Applying lower bounds for linear forms in two complex logarithms (as in, say, Theorem 5.2 of de Weger's thesis), we have that $$ 2^y - 3^x \geq 3^{0.9x}, $$ since $3^x > 10^{15}$. From the fact that $2^y-3^x \mid 2^{y-x}-1$, it follows that $$ 2^{0.6x}-1 \geq 3^{0.9x}, $$ a contradiction.<|endoftext|> TITLE: HNN-extension as a 2-colimit QUESTION [9 upvotes]: In the spirit of this question, it would be interesting to give a characterization of HNN extensions as a 2-colimit. If $G$ is a group and $\alpha:H \xrightarrow{\cong} K$ is an isomorphism between two subgroups of $G$, then I think that the HNN extension $G_{\alpha}$ has the following universal property: if we let $i_1:H \hookrightarrow G,i_2:K \hookrightarrow G$ be the canonical inclusions, then the set of group homomorphisms $G_{\alpha} \to T$ are in natural bijection to pairs $(f,t)$ where $f:G \to T$ is a group homomorphism and $t:f \circ i_2 \circ \alpha \Rightarrow f \circ i_1$ is a 2-morphism. Here we're considering $\mathbf{Grp}$ as a full subcategory of $\mathbf{Cat}$ which carries a standard 2-category structure. I'm just wondering if this universal property can be phrased as some kind of 2-colimit. REPLY [10 votes]: Assuming your universal property is true, it exactly says that the HNN extension is the coinserter of $(i_2 \circ \alpha,i_1) : H \rightrightarrows G$.<|endoftext|> TITLE: Conceptual reason that monadic functors create limits? QUESTION [16 upvotes]: Let $U: Alg_T \to C$ be the forgetful functor from the category of algebras of $T: C \to C$ ($T$ could be a monad; I'm happy to think about the simpler case where $T$ is just an endofunctor or pointed endofunctor). Then a very simple diagram chase shows that $U$ creates limits. In other words, if I have a diagram in $Alg_T$ whose image in $C$ has a limit, there is an obvious way to lift the limit cone to $Alg_T$, and a straightforward diagram chase verifies that indeed, this is a limit in $Alg_T$. This is all very easy, but I'm unhappy with the state of affairs for a few reasons: I'm very lazy and I hate doing diagram chases. This proof does not obviously generalize to other contexts. For one thing, I have to rehash variations on the same diagram chase for each of the cases where $T$ is a monad, an endofunctor, a pointed endofunctor, etc. For another, even if I stick with just monads, say, I have to rehash the same diagram chase if I want to generalize to other contexts such as enriched or internal category theory. Of course, doing the same work over and over is supposed to mean there's a bigger picture I'm missing. I find it remarkable that in order for $U$ to create limits, one need not assume any kind of limit-preservation hypotheses about $T$. There's something to be explained, and the proof via diagram chase doesn't accomplish that. The second point may have real weight -- I haven't checked very diligently, but it seems that it might not actually be known whether this this theorem remains true in full generality in the enriched context, for example! (Although a moment's reflection makes me think I could run the same diagram chase with no fuss if it weren't for point (1) above.) So here's my Question: What is the conceptual reason for which monadic functors (and other forgetful functors from "categories of algebras") create limits? REPLY [6 votes]: From an abstract point of view, the reason is that the monad $T$ always preserves any limits that exist colaxly and colax preservation is what is required. (This answer is closely related to Peter's answer, but describes some published results on the topic.) The following is Proposition 4.11 of Limits for Lax morphisms by Steve Lack. If $T$ is a $2$-monad on a $2$-category $C$ then the forgetful $2$-functor $U:T-Alg_c \to C$ from strict algebras and colax morphisms to the base creates lax limits. Note the switch between lax limits and colax morphisms. The sense of creation is that the projections from the limit should be strict maps. The Eilenberg-Moore object of a monad or the category of algebras for a pointed endofunctor are both examples of lax limits. One can take $T$ the $2$-monad for categories with a class $D$ of limits and the result then applies to your setting. Another instance would be to take as $T$ the $2$-monad for monoidal categories. Then the result becomes that if you have an opmonoidal monad on a category, it lifts to a monoidal structure on the category of algebras. REPLY [3 votes]: This is just to flesh out the approach using inserters and equifiers discussed in the answers, in a way that doesn't quite go down to the level of diagram chasing. I fear, however, that some things implicit in this argument woule require a diagram chase to carefully check. Also, there's something I still find mysterious: what is it about inserters and equifiers which makes it so there is a particular leg of the limit cone and particular elements of the diagram such that certain hypotheses on these elements in the diagram ensure the forgetful functor down the special leg creates limits? To put a finer point on it: can we give a better description of the class of restriction functors among limits which (under certain partial limit preservation conditions) create limits? A better description than "whatever can be built up from inserters and equifiers"? Lemma: Let $F,G: C\rightrightarrows D$ be functors, and let $Ins(F,G)$ be the inserter. Then the forgetful functor $Ins(F,G) \to C$ creates any limits that $G$ preserves. For the proof, note that in in general, if $(c',\phi'), (c,\phi) \in Ins(F,G)$, then $$ Hom((c',\phi'), (c,\phi)) = Hom(c',c) \times_{Hom(Fc',Gc)^2} Hom(Fc',Gc)$$ where the pullback is over the diagonal map. Proof: Consider a diagram $(c_i, F(c_i) \xrightarrow {\phi_i} G(c_i))_{i \in I}$ in $Ins(F,G)$ such that $G(\varprojlim_i c_i) = \varprojlim_i G(c_i)$. Then $(F(\varprojlim_i c_i) \to F(c_i) \xrightarrow{\phi_i} G(c_i))_{i \in I}$ is a cone, and so induces a map $\phi: F(\varprojlim_i c_i) \to \varprojlim_i G(c_i) = G(\varprojlim_i c_i)$. We claim that $(\varprojlim_i c_i , \phi)$ is a limit of our diagram. Indeed, $$Hom((c',\phi'), (c,\phi)) = Hom(c',c) \times_{Hom(Fc',Gc)^2} Hom(Fc',Gc) \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad = \varprojlim_i Hom(c',c_i) \times_{\varprojlim_i Hom(Fc',Gc_i)^2} \varprojlim_i Hom(Fc',Gc_i) \\ \qquad \qquad \qquad \qquad \qquad \qquad = \varprojlim_i (Hom(c',c_i) \times_{Hom(Fc',Gc_i)^2} Hom(Fc',Gc_i)) \\ \qquad \quad = \varprojlim_i Hom((c',\phi'),(c_i,\phi_i))$$ where we have used that limits commute with limits. Lemma: Let $\phi,\psi: F \rightrightarrows G : C \rightrightarrows D$ be a diagram of categories, and let $Eq(\phi,\psi)$ be its equifier. Then the full subcategory inclusion $Eq(\phi,\psi) \to C$ is closed under any limits preserved by $G$. Proof: This boils down to the limit of equal morphisms being equal.<|endoftext|> TITLE: Approximating ring maps of finite Tor-dimension QUESTION [6 upvotes]: Let $R$ be a commutative ring, and let $S$ be a finitely presented $R$-algebra of finite Tor-dimension over $R$. Can $R \to S$ be realized as the base change, along some ring map $R_0 \to R$, of a finitely presented ring map $R_0 \to S_0$ of finite Tor-dimension with $R_0$ Noetherian? Can we furthermore arrange that $R_0$ is actually of finite type over $\mathbf{Z}$? REPLY [2 votes]: First of all, I strongly suspect that the answer to the question is: no. Secondly, I want to give an example to show that the answer by Riza Hawkeye is incorrect as it stands (see also the comment by Denis Nardin for why the result is quoted incorrectly from Lurie). Namely, consider the system of rings $$ A_0 \to A_1 \to A_2 \to A_3 \to \ldots $$ where $A_i = \mathbf{Z}[x, z_i]/(xz_i)$ and where the maps send $x$ to $x$ and $z_i$ to zero. Then we see that the colimit of the system is $A = \mathbf{Z}[x]$. Now set $B_0 = A_0/xA_0$. Clearly, we see that $B_i = B_0 \otimes_{A_0} A_i$ is equal to $A_i/xA_i$ which has infinite tor dimension over $A_i$ for all $i$. On the other hand, we have $B = A/xA$ and this has tor dimension $1$ as a module over $A = \mathbf{Z}[x]$.<|endoftext|> TITLE: Intuition about how Voronoi formulas change lengths of sums QUESTION [6 upvotes]: In reading the literature one encounters countless examples of Voronoi formulas, i.e., formulas that take a sum over Fourier coefficients, twisted by some character, and controlled by some suitable test function, and spits out a different sum over the same Fourier coefficients, twisted by some different characters, and this time controlled by some integral transform of the test function. The reason one wants to do this in practice is that the second sum is somehow better, of course, which in my (admittedly limited) experience tends to boil down to the length of the second sum having changed significantly to the better. I'll give an example (from Xiaoqing Li's Bounds for GL(3)×GL(2) L-functions and GL(3) L-functions, because it is what I happen to have in front of me). In this case we have the GL(3) Voronoi formula $$ \sum_{n > 0} A(m, n) e\Bigl( \frac{n \bar d}{c} \Bigr) \psi(n) \sim \sum_{n_1 \mid c m} \sum_{n_2 > 0} \frac{A(n_2, n_1)}{n_1 n_2} S(m d, n_2; m c n_1^{-1}) \Psi \Bigl( \frac{n_2 n_1^2}{c^3 m} \Bigr), $$ where $\psi$ is some smooth, compactly supported test function, $\Psi$ as suggested above is some integral transform of it, $A(m, n)$ are Fourier coefficients of (in this case) an SL(3) Maass form, $(d, c) = 1$, and $d \bar d \equiv 1 \pmod{c}$. (I've omitted lots of details here, but the details, I think, aren't relevant to my question.) Doing so essentially transforms the $n$-sum into the $n_2$-sum, where, as is evident in the formula, the $n_2$-sum has a very different argument in its test function. What happens in practice now is that, once we get to a point where applying the Voronoi formula is appropriate, we transform the sum and study the integral transform, chiefly by means of stationary phase analysis in order to find what length of the new $n_2$-sum is. In the particular example at hand, this, after identifying the stationary phase and playing along, this takes us from an $n$-sum on $N \leq m^2 n \leq 2 N$, i.e., $n \sim \frac{N}{m^2}$, to an $n_2$-sum on $$ \frac{2}{3} \frac{N^{1/2}}{n_1^2} \leq n_2 \leq 2 \frac{N^{1/2}}{n_1^2}, $$ i.e., $n_2 \sim \frac{N^{1/2}}{n_1^2}$, which then means that the arguments in the test function are now of size $\frac{N^{1/2}}{c^3 m}$. I can go through the motions of performing this stationary phase analysis and so on, but my question is this: is there any intuition to be had about how and in what way these Voronoi formulas alter the lengths of sums? REPLY [4 votes]: First of all, the description of $\psi$ after the first display is confusing (assuming OP meant $\psi$ is supported around $N$, otherwise conclusion form the first display does not make sense). I went to the relevant part (end of p.318) of Li's paper and found that $\psi$ is not just any test function, it has a weight of $n^{-3/4}$, and, more importantly, it has oscillation. A necessary display of the LHS would be (taking $m=d=1$) $$\sum_{n}A(1,n)e\left(\frac{n}{c}+2\sqrt{n}-\frac{1}{\sqrt{n}}\right)\psi(n/N),$$ where $\psi$ is a test function supported on $[1,2]$. To have an intuition about what the length of the dual sum would be one can follow the general heuristic formula (HF) below. $$\text{length of the dual sum} = \frac{\text{total conductor}}{\text{length of the original sum}}.$$ Here total conductor is the conductor of the oscillating object taken all the twisting into account. For example, the total conductor of $L(1/2+it,\pi'\otimes\pi)$ where $\pi$ and $\pi'$ are fixed $\mathrm{GL}(n)$ and $\mathrm{GL}(m)$ automorphic representations, is $t^{nm}$. Conductor of $e_q(x):=e(x/q)$ is $q$. In this case the denominator in the entry of $e()$ is of size $c\sqrt{N}$. But there is twist by a $\mathrm{GL}(3)$ Hecke eigenvalue. So the total conductor is $c^3N^{3/2}$. Applying the above formula one obtains the length of the dual sum. One can check that the above HF also occurs in the formula of the approximate functional equation of the central $L$-value. There are two sums in the formula corresponding to the representation and its contragredient. One can check that the product of the length of the sums equals to the conductor of the $L$-function. (The main reason of this HF to work is the automorphy under some suitable Weyl element of the underlying automorphic form, which is, indeed, the key ingredient to prove the approximate functional equation and Voronoi formula.)<|endoftext|> TITLE: Moments of a positive random variable QUESTION [7 upvotes]: Suppose one is handed a list of $K$ numbers, with a claim that these numbers are the first $K$ moments of a positive random variable $X$ (meaning there is 0 probability that $X<0$). What is the strongest possible test that one could run on this list to test this claim? (We do not know any additional information about $X$.) The most obvious thing to check first is that all the moments are positive. A better test would involve checking that Jensen’s inequalities are satisfied. What is the most powerful test? In general, there is a convex "allowed region” in the $K$-dimensional space of possible moments of $X$. Is there a good way to characterize this space? REPLY [6 votes]: This is known as the truncated Stieltjes moment problem, and there is a necessary and sufficient condition taking the form of a semidefinite program. See Section 5 of the classic paper by Curto and Fialkow.<|endoftext|> TITLE: Examples of plane algebraic curves QUESTION [12 upvotes]: There are many interesting sequences of polynomials which contain polynomials of arbitrarily high degree, for example classical orthogonal polynomials. Most of them arise as characteristic polynomials of some sequences of operators, or as polynomial solutions of some differential equations. What are some natural specific sequences of plane (affine or projective) algebraic curves which contain curves of arbitrarily high degree and genus? One such example is Fermat's curves $x^n+y^n=1$. Lissajous (a.k.a. Chebyshev) curves are of arbitrary degree but they have zero genus. Sequences of hyperelliptic curves occur in the theory of integrable systems. What else? I looked to the Catalog of Plane curves by D. Lawrence (Dover, 2014) and to the book of Brieskorn and Knörrer, Plane algebraic curves, and found only Lissajous curves, epitrochoids and hypotrochoids (all of genus zero) as examples of arbitrarily high degree. I understand that many examples can be constructed. But I am asking on some naturally occuring sequences, whatever it can mean. Of some historical significance or appearing in applications. EDIT. Thanks to all who answered or commented. I am not marking this question as "answered" for a while, hoping for more examples. Of course, TITLE: Lifting a complete intersection in $\mathbb{P}^n_{\mathbb{F}_p}$ to $\mathbb{Z}_p$ QUESTION [7 upvotes]: Suppose that you are given a (not necessarily smooth) projective variety $X \subseteq \mathbb{P}^n_{\mathbb{F}_p}$ of codimension $d$ that is a complete intersection. In other words, it can be defined by exactly $d$ homogeneous polynomials and no set of polynomials of cardinality less than $d$ has $X$ as its zero set. Let $f_1$, ..., $f_d$ be a set of polynomials defining $X$. If I lift these polynomials arbitrarily to $\mathbb{Z}_p$ (call the lifts $F_1$, ..., $F_d$), I will get a projective algebraic set $X' \subseteq \mathbb{P}^n_{\mathbb{Z}_p}$. My question is this: will this $X'$ be flat over $\mathbb{Z}_p$? Equivalently, is the module $\mathbb{Z}_p[x_0,...x_n]/(F_1, ..., F_d)$ flat over $\mathbb{Z}_p$? I know that $\mathbb{Z}_p$ is a DVR, so flatness is equivalent to being torsion-free but I just cannot see how to prove that it is either. If $X$ is smooth, then $X'$ is definitely flat: $X'$ will also be smooth by the Jacobian criterion and hence flat. REPLY [3 votes]: This is an expanded version of my comment: The main claim is that the question can be answered from basic facts in (local) commutative algebra. In particular, the same statement holds for a complete intersection in affine space, which I explain below. We work over any dvr $R$ with residue field $k$. Let $X$ be of codimension $d$ in $\mathbb{A}^n_k$ defined by equations $f_1,f_2,\dots,f_d$. Let $F_1,F_2,\dots,F_d$ be elements of $R[x_1,x_2,\dots,x_n]$ such that $F_i$ is a lift of $f_i$ and let $X'$ be the subscheme of $\mathbb{A}^n_R$ defined by the ideal $(F_1,F_2,\dots,F_d)$. Then $X'$ is a local complete intersection scheme since the uniformizer of $R$ is a nonzero divisor in the polynomial ring and $X$ is a local complete intersection. In particular, $X'$ is Cohen-Macaulay, so it has no embedded points. Now suppose the coordinate ring $A$ of $X'$ has non-zero $R$-torsion elements. The set of all such elements is a non-zero ideal in $A$ and the support of this ideal (as an $A$-module) is contained in $X'$ (viewed as a subset of $X$). This implies that $A$ has an embedded prime, so we get a contradiction.<|endoftext|> TITLE: Greatest common divisor in $\mathbb{F}_p[T]$ with powers of linear polynomials QUESTION [11 upvotes]: Let $n>1$ and $p$ be an odd prime with $p-1 \mid n-1$ such that $p^k - 1 \mid n-1$ does not hold for any $k>1$. Notice that, since $p-1 \mid n-1$, we have $T^p - T \mid T^n-T$ in $\mathbb{F}_p[T]$ and hence also $T^p - T = (T+u)^p - (T+u) \mid (T+u)^n - (T+u)$ for all $u \in \mathbb{F}_p$. Question. Is $T^p - T$ actually the gcd of $\{(T+u)^n - (T+u) : u \in \mathbb{F}_p\}$ in $\mathbb{F}_p[T]$? I have verified this with computer algebra software for $n \leq 7000$ (code link). For many $n$ actually $u=0,1$ are sufficient. I tried to find a proof, but my first idea didn't work. The only thing I know so far is that the gcd is invariant under $T \mapsto T+1$ and therefore contained in $\mathbb{F}_p[T^p-T]$. I expect that there are two proofs (if the statement is true at all), namely one using finite fields $\mathbb{F}_{p^m}$, and one using a direct calculation with polynomials. I am more interested in a direct calculation here. The background is a new proof of Jacobson's theorem I am working on. Notice that the statement is false for $p=2$ (but still true for many $n$ in this case) and that it is clearly false without the $p^k-1$-requirement. REPLY [11 votes]: This is false. Let $p$ be an odd prime, let $\ell$ be another prime, and let $m$ be a small prime divisor of $p^{\ell}-1$, that doesn't divide $p-1$. Let $n= 1 + \frac{ p^{\ell}-1}{m}$. Then $n-1$ is a multiple of $p-1$, is not a multiple of $p^{\ell}-1$, and is not a multiple of $p^{k}-1$ for any other $k$ because $p^{\ell}-1$ is not a multiple of $p^{k}-1$ for any $1 < k < \ell$. Then $x \in \mathbb F_{p^\ell}$ is a root of $T^{n } - T$ if and only if $x$ is an $m$'th power in $\mathbb F_{p^\ell}$. So roots of the gcd of $(T+u)^n - (T+u)$ for all $u$ in $\mathbb F_p$ are exactly those $x \in \mathbb F_{p^\ell}$ such that $x = y_0^m, x+1 = y_1^m, \dots, x+p-1 = y_1^{m}$ for some $y_0, \dots, y_{p-1}$ in $\mathbb F_{p^\ell}$. Thus, to find a counterexample, it suffices to check that the number of $\mathbb F_{p^\ell}$ points of the curve $C_m$ with variables $x,y_0,\dots, y_{p-1}$ and equations $x +i = y_i^m$ is greater than the number $p m^{p-1}$ of solutions with $x \in \mathbb F_p$. Then the $x$ coordinates of the extra points will be roots of the gcd but not roots of $T^p- T$. By Riemann-Hurwitz, the genus $g$ of $C$ satisfies $$2-2g = 2 m^p - (p+1) m^{p-1} (m-1)$$ since the degree over $\mathbb P^1$ (under the map $x$) is $m^p$, there are $p+1$ branch points $0,1\dots, \infty$, and each branch point has ramification of order $m$ on each point lying above it. There are $m^{p-1}$ missing points at $\infty$. So by Weil's theorem, the number of $\mathbb F_{p^\ell}$-points of $C$ is at least $$p^\ell - p^{\ell/2} ((p+1) m^{p-1} (m-1) + 2 - 2 m^p) +1 - m^{p-1}$$ with the first term the main term, the second term coming from the Frobenius eigenvalues, and the last term coming from the missing points. Thus, as long as $$ p^\ell > p^{\ell/2} ((p+1) m^{p-1} (m-1) + 2 - 2 m^p) + m^{p-1} + p m^{p-1},$$ there is a counterexample. Plugging into Wolfram Alpha, this inequality fails for $p=3, \ell=11, m=23$ but succeeds for $p=3,\ell=23, m=47$ and $p=3, \ell=29, m=59$. I found these by looking at the sequence of multiplicative orders of $3$ modulo primes and looking for prime values, which became $\ell$, with the modulus prime becoming $m$. It seems there are many examples with $\ell$ not much smaller than $m$, which as long as both are much larger than $p$, means this inequality is easily satisfied, since anything of the form $p^\ell$ beats anything of the form $m^p$. In the comments, François Brunault found an explicit example: The polynomial $$T^{23}-T^{22}-T^{21}-T^{20}-T^{19}+T^{18}-T^{16}+T^{13}+T^{12}+T^{11}-T^{10}+T^8+T^6+T^4-T^2-T-1$$ divides $$\gcd( T^{n} - T, (T-1)^n - (T-1) , (T+1)^n - (T+1))$$ in $\mathbb F_3[T]$ when $n= 1 + \frac{3^{23}-1}{47}$ and provided the following Pari/GP code to check it: P = Mod(x^23-x^22-x^21-x^20-x^19+x^18-x^16+x^13+x^12+x^11-x^10+x^8+x^6+x^4-x^2-x-1, 3); n = 1 + (3^23-1)/47; t = Mod(x, P); print(t^n == t & (t+1)^n == t+1 & (t-1)^n == t-1);<|endoftext|> TITLE: Is this a new strange attractor? QUESTION [7 upvotes]: I recently made some experiments in programming strange attractors, and I found this (very simple) equations, which create a nice strange attractor: xn=x+dt*(z-y) yn=y+dt*(x/2-1) zn=z+dt*(-xy/2-z) You can see it in action on my Youtube channel: https://youtu.be/Bm_M6mUGjtg My question: Is this a variation of the Lorenz- or Rössler attractor - or did I stumble upon something new? EDIT: Meanwhile I programmed a little 3D View for this attractor: You can see/move the view on this applet (Java needed): https://cerumen.de.cool/attractor/index.html (Here you also find the source code for Processing) And here the Javascript-Version: https://cerumen.de.cool/attractor/js/index.html (with processing.js... bit slow) Perhaps my question was also asked too amateurishly. I was simply surprised by the simplicity of the system of equations I have found. Therefore I would like to know if this strange attractor is a descendant of one of the well known ones (Lorenz / Rössler). Edit 2: I have now brought the system of equations into a more general form: xn=x+dt*(z-y) yn=y+dt*(ax-b) zn=z+dt*(-axy-z) with a in range [0 to 1], b in range [0.5 to 1] This makes it more interesting. Here some sample images for different values for a and b: Edit 3: Here a video with the generalized equations and constantly changing parameters a and b: https://youtu.be/gxusM8pmNwU I think you can see here quite well how the system goes from order through bifurcation into chaos... REPLY [3 votes]: In another forum a user drew my attention to the publication of J. C. Sprott, Some simple chaotic flows, Phys. Rev. E 50, R647-R650 (1994), doi:10.1103/PhysRevE.50.R647, author pdf. This shows that there are many very simple chaotic systems of equations. I guess this answers my question... You can see some of the equations here, and here are the corresponding graphs. But thank you all for your interest.<|endoftext|> TITLE: Significance of the length of the Perron eigenvector QUESTION [5 upvotes]: Let $A$ be a positive square matrix. Perron-Frobenius theory says that there exist $\lambda,v$ with $Av=\lambda v$ and $\lambda$ equals the spectral radius of $A$, $\lambda$ is simple, and $v$ is positive. Now consider also the left Perron eigenvector $u^T A=\lambda u^T$. Another result of Perron-Frobenius theory is that $$\lim_{m\to \infty} \frac{A^m}{\lambda^m} = \frac{v u^T}{u^T v}.$$ Suppose $\|v\|=1$. The above result says that the "correct" normalization for u is $u^T v=1$ rather than the more usual $u^T u=\|u\|^2=1$. This motivates the question: what is the significance of the ratio $$\frac{u^T v}{u^T u} ?$$ Are there matrices $A$ for which this ratio is arbitrarily large? Arbitrarily small? Does this ratio determine any properties of $A$? Note that if $A$ is symmetric, then $u=v$ and this ratio is always equal to $1$, but that's not the case in general for arbitrary $A$. Could it be the case that this ratio is measuring how far $A$ is from being symmetric? Note too that this normalization is necessary so that the limit $\frac{v u^T}{u^T v}$ is a projection matrix (i.e. that its only non-zero eigenvalue is one). In this context, I understand why the normalization is necessary, but I'm interested in the amount of normalization necessary with respect to the length of $u$. Any pointers appreciated. Thanks! EDIT In the comments, it is argued that the real quantity of interest in this setup is $$\frac{\left( u^T v \right)^2}{\left(u^T u \right) \left( v^T v \right)}.$$ This quantity is also of interest to me, and an acceptable replacement for my original question. REPLY [3 votes]: That quantity $s = \frac{|u^Tv|}{\|u\|\|v\|}$ is the inverse of the eigenvalue condition number. The smaller it is, the more sensitive to perturbation the Perron value is. More precisely, any perturbed matrix $A+E$ with $\|E\| \leq \varepsilon$ has a Perron value $\tilde{\lambda}$ that satisfies $|\tilde{\lambda}-\lambda| \leq \frac{\varepsilon}{s} + \mathcal{O}(\varepsilon^2)$. See e.g. Section 7.2.2 of Golub and Van Loan's Matrix Computations 4th ed. In addition, note that if $A$ is normal then $s=1$ (its maximum possible value) and the Perron value is perfectly conditioned; while if $\lambda$ is a defective eigenvalue (e.g. $A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$) then $s=0$. So rather than a "distance from symmetric" I'd say that $1-s$ is a "distance from normal" or $s$ is a "distance from defective".<|endoftext|> TITLE: Calculate the $\downarrow$, $\downarrow\uparrow$ and $\uparrow\downarrow$ cofinalities of the poset of nontrivial finitary partitions of $\omega$ QUESTION [5 upvotes]: Let $(P,\le)$ be a poset. For a point $x\in P$ let $${\downarrow}x=\{p\in P:p\le x\}\quad\text{and}\quad{\uparrow}x=\{p\in P:x\le p\}$$be the lower and upper sets of the point $x$, and for a subset $S\subset P$, let $${\downarrow}S=\bigcup_{s\in S}{\downarrow}s\quad\text{and}\quad{\uparrow}S=\bigcup_{s\in S}{\uparrow}s$$be the lower and upper sets of the set $S$ in $P$. Now consider the following cardinal characteristics of $P$: $\bullet$ the $\downarrow$-cofinality ${\downarrow}(P)=\min\{|C|:C\subseteq P\;\wedge \;{\downarrow}C=P\}$; $\bullet$ the $\uparrow$-cofinality ${\uparrow}(P)=\min\{|C|:C\subseteq P\;\wedge\;{\uparrow}C=P\}$; $\bullet$ the $\uparrow\downarrow$-cofinality ${\uparrow}{\downarrow}(P)=\min\{|C|:C\subseteq P\;\wedge \;{\uparrow\downarrow}C=P\}$; $\bullet$ the $\downarrow\uparrow$-cofinality ${\downarrow}{\uparrow}(P)=\min\{|C|:C\subseteq P\;\wedge\;{\downarrow\uparrow}C=P\}$. Proceeding in this fashion, we could define the $\downarrow\uparrow\downarrow$-cofinality ${\downarrow\uparrow\downarrow}(P)$ and $\uparrow\downarrow\uparrow$-cofinality ${\uparrow\downarrow\uparrow}(P)$ and so on. It is clear that $$\max\{{\uparrow\downarrow}(\mathfrak P),{\downarrow\uparrow}(\mathfrak P)\}\le \min\{{\downarrow}(\mathfrak P),{\uparrow}(\mathfrak P)\}.$$ I would like to know the values of the $\downarrow$, $\downarrow\uparrow$ and $\uparrow\downarrow$ cofinalities of the poset $\mathfrak P$ of nontrivial finitary partitions of $\omega$. By a partition I understand a cover $\mathcal P$ of $\omega=\{0,1,2,\dots\}$ by pairwise disjoint sets. A partition $\mathcal P$ is defined to be $\bullet$ finitary if $\sup_{P\in\mathcal P}|P|$ is finite (i.e., the cardinalities of the cells of the partition are upper bounded by some finite cardinal); $\bullet$ nontrivial if the subfamily $\{P\in\mathcal P:|P|=1\}$ is finite (i.e., $\mathcal P$ contains infinitely many cells of cardinality $\ge 2$). The family $\mathfrak P$ of all nontrivial finitary partitions of $\omega$ is endowed with the refinement partial order $\le$ defined by $\mathcal P_1\le\mathcal P_2$ if each cell of the partition $\mathcal P_1$ is contained in some cell of the partition $\mathcal P_2$. It can be shown that $${\uparrow\downarrow\uparrow}(\mathfrak P)=1={\downarrow\uparrow\downarrow}(\mathfrak P),$$ so only four cofinalities (with at most two arrows) can be infinite. Using almost disjoint families of cardinality continuum, it can be shown that ${\uparrow}(\mathfrak P)=\mathfrak c$. Problem 1. Calculate the $\downarrow$-cofinality ${\downarrow}(\mathfrak P)$ of the poset $\mathfrak P$. In particular, is ${\downarrow}(\mathfrak P)=\mathfrak c$? Or ${\downarrow}(\mathfrak P)=\mathfrak d$? Remark 1. It can be shown that ${\downarrow}(\mathfrak P)\ge\mathfrak d$. Problem 2. Evaluate the cardinal characteristics ${\downarrow\uparrow}(\mathfrak P)$ and ${\uparrow\downarrow}(\mathfrak P)$ of the poset $\mathfrak P$. REPLY [2 votes]: At the moment we have the following information on the cofinalities of the poset $\mathfrak P$ (see Theorem 7.1 in this preprint). Theorem. 1) ${\downarrow}\!{\uparrow}\!{\downarrow}(\mathfrak P)={\uparrow}\!{\downarrow}\!{\uparrow}(\mathfrak P)=1$. 2) ${\downarrow}(\mathfrak P)={\uparrow}(\mathfrak P)=\mathfrak c$. 3) ${\downarrow}\!{\uparrow}(\mathfrak P)\ge \mathrm{cov}(\mathcal M)$. 4) $\mathsf \Sigma\le{\uparrow}\!{\downarrow}(\mathfrak P)\le\mathrm{non}(\mathcal M)$. Here $\mathrm{non}(\mathcal M)$ is the smallest cardinality of a nonmeager set in the real line, and $\mathsf \Sigma$ is the smallest cardinality of a subset $H$ in the permutation group $S_\omega$ of $\omega$ such that for any infinite sets $A,B\subseteq \omega$ there exists a permutation $h\in H$ such that $h(A)\cap B$ is infinite. By Theorem 3.2 in this preprint, $$\max\{\mathfrak b,\mathfrak s,\mathrm{cov}(\mathcal N)\}\le\mathsf\Sigma\le\mathrm{non}(\mathcal M).$$ The cardinal $\mathsf\Sigma$ is equal to the cardinal $\mathfrak j_{2:2}$, discussed in this MO-post. However I do not know the answer to the following Problem. Is ${\downarrow\!\uparrow}(\mathfrak P)\le\mathrm{non}(\mathcal N)$? Here $\mathrm{non}(\mathcal N)$ is the smallest cardinality of a subset of the real line, which is not Lebesgue null.<|endoftext|> TITLE: Linear combination of sine and cosine QUESTION [26 upvotes]: I was explaining to my students the other day why $\cos(2x)$ is not a linear combination of $\sin(x)$ and $\cos(x)$ over $\mathbb{R}$. Besides the canonical method of using special values of sine and cosine, I noticed something interesting. In the following, all vector spaces are over $\mathbb{R}$. Consider the linear space $C^\infty_b(\mathbb{R})$ of real-valued bounded smooth functions on $\mathbb{R}$, and take any $c > 0$. We say a function $f \in C^\infty_b(\mathbb{R})$ has property $P(c)$, if for all $k \in \mathbb{N}$ (including $0$), we have $$\sup f^{(k+1)} = c \sup f^{(k)} = -\inf f^{(k+1)} = -c \inf f^{(k)}.$$ Here, the supremum and infimum are of course taken over $\mathbb{R}$, and $f^{(k)}$ is the $k$-th derivative of $f$, with the convention that $f^{(0)}=f$. Define $$S(c) = \{f \in C^\infty_b(\mathbb{R}) \,\vert\, f \text{ has property } P(c)\}.$$ Since for fixed $a,b$ and all $x$, we have $a \sin(x) + b \cos(x) = \sqrt{a^2 + b^2} \sin(x + \theta) $ for some fixed $\theta$, it is clear that all linear combinations of $\sin(cx)$ and $\cos(cx)$ belong to $S(c)$. In particular, linear combinations of $\sin(x)$ and $\cos(x)$ are all in $S(1)$, while $\cos(2x)$ is not. Question: is it true that $S(c) = \operatorname{Vect}\bigl(\sin(cx), \cos(cx)\bigr)$? If $f \in S(1)$ and $f$ is periodic, then using Fourier series, I can prove that $f$ is indeed $2 \pi$-periodic and $f \in \operatorname{Vect}\bigl(\sin(x), \cos(x)\bigr)$ with some work. Although I haven't checked this yet, I also believe that the periodic case for $S(c)$ where $c>0$ is arbitrary could be established by a more elaborate Fourier series argument (of course, I could be wrong). So the real interest lies in treating the non-periodic case, i.e., answering the following Special case: does $f \in S(c)$ imply $f$ is periodic? At first, I suspect the answer to the above special case is negative. But after some experiment, I am not so sure. Note that the radius of convergence of the Taylor series (say around $0$) for all $f \in S(c)$ is infinite, so the Taylor series of $f$ converges to $f$ itself. In particular, all functions in $S(c)$ are automatically analytic, so one does not have much freedom when trying to construct a (counter-)example. If the answer turns out to be negative, then can one at least assert that $S(c)$ is a linear subspace? What if we only consider periodic functions for some fixed period in case $S(c)$ is not a linear subspace? Of course, these probably depend on the explicit form of the answer which is not yet known to me, and all of these are just some (perhaps stupid and naive) speculation on an old exercise of a first-year undergraduate. But it seems interesting, and any thought is appreciated. Edit: I was a bit careless in formulating the question since the questions for all different $c$ are equivalent merely by rescaling, so one can simply assume $c = 1$, in which case we still have much work to do. Edit 2: Proof of the periodic case can be found here in case anyone is interested. REPLY [7 votes]: As noted by the OP we can replace $f$ by $af(bx)$ for suitable $a,b\in\mathbb{R}$ so that wlog we can take $c=1$ and ensure that $\sup f=-\inf f=1$. Firstly we note that $f(z)$ is infinitely differentiable on $\mathbb{R}$ so we can form the taylor series at 0, $f(z)=\sum_{i=0}^{\infty}\frac{f^{(i)}(0)}{i!}z^i$. Since $|f^{(i)}(0)|\leq 1$ for all $i\geq 0$ we have $\sum_{i=0}^{\infty}|\frac{f^{(i)}(0)}{i!}z^i|=\sum_{i=0}^{\infty}\frac{|f^{(i)}(0)|}{i!}|z|^i\leq \sum_{i=0}^{\infty}\frac{|z|^i}{i!}$ which converges for all $z$ to $e^{|z|}$. Hence $F(z)=\sum_{i=0}^{\infty}\frac{a_i}{i!}z^i$ is an absolutely convergent series defining an entire function on $\mathbb{C}$ agreeing with $f$ on $\mathbb{R}$ s.t. $\sup f^{(k)}=-\inf f^{(k)}=1$ for all $k\in \mathbb{Z}_{\geq0}$ and $|F(z)|\leq e^{|z|}$ for all $z\in \mathbb{C}$. We now determine the form of $F$ given the preceding conditions. First note that Bernstein proved the following (see Rahman and Tariq$^1$) as an extension of his related inequality for polynomials: Theorem Let $g$ be an entire function of exponential type $\tau>0$ such that $|g(x)|\leq M$ on the real axis. Then $$\sup_{-\infty TITLE: Is $\mathfrak j_{2:1}=\mathfrak{j}_{2:2}$ in ZFC? QUESTION [11 upvotes]: A function $f:\omega\to\omega$ is called $\bullet$ 2-to-1 if $|f^{-1}(y)|\le 2$ for any $y\in\omega$; $\bullet$ almost injective if the set $\{y\in \omega:|f^{-1}(y)|>1\}$ is finite. Let us introduce two critical cardinals, related to $2$-to-$1$ functions: $\mathfrak{j}_{2:1}$ is the largest cardinal $\kappa\le\mathfrak c$ such that for any family $F\subset \omega^\omega$ of $2$-to-$1$ functions with $|F|<\kappa$ there exists an infinite subset $J\subset\omega$ such that for any $f\in F$, the restriction $f{\restriction}J$ is almost injective; $\mathfrak{j}_{2:2}$ is the largest cardinal $\kappa\le\mathfrak c$ such that for any family $F\subset \omega^\omega$ of $2$-to-$1$ functions with $|F|<\kappa$ there are two infinite sets $I,J\subset\omega$ such that for any $f\in F$ the intersection $f(I)\cap f(J)$ is finite. It can be shown that $\max\{\mathfrak s,\mathfrak b\}\le\mathfrak j_{2:1}\le\mathfrak j_{2:2}\le\mathrm{non}(\mathcal M)$. I would like to have more information on the cardinals $\mathfrak j_{2:1}$ and $\mathfrak j_{2:2}$. Problem 0. Is $\mathfrak j_{2:1}=\mathfrak j_{2:2}$ in ZFC? Problem 1. Is $\mathfrak j_{2:2}=\mathrm{non}(\mathcal M)$ in ZFC? Problem 2. What is the value of the cardinals $\mathfrak j_{2:1}$ and $\mathfrak j_{2:2}$ in the Random Model? (In this model $\mathfrak b=\mathfrak s=\omega_1<\mathfrak c=\mathrm{non}(\mathcal M)$, see $\S$11.4 in this survey paper of Blass). Remark. It can be shown that the cardinal $\mathfrak j_{2:1}$ (resp. $\mathfrak j_{2:2}$) is equal to the smallest weight of a finitary coarse structure on $\omega$ that contains no infinite discrete subspaces (resp. contains no infinite asymptotically separated sets). In this respect $\mathfrak j_{2:1}$ can be considered as an asymptotic counterpart of the cardinal $\mathfrak z$, defined as the smallest weight of an infinite compact Hausdorff space that contain no nontrivial convergent sequences. The cardinal $\mathfrak z$ was introduced by Damian Sobota and deeply studied by Will Brian and Alan Dow. The similarity between $\mathfrak j_{2:1}$ and $\mathfrak z$ suggests another Problem 3. Is $\mathfrak j_{2:1}=\mathfrak z$ in ZFC? REPLY [2 votes]: I can answer problems 2 and 3, although I still don't know the answer to problems 0 and 1. The main point is that $\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model. I'll sketch a proof of this below. (It's a bit long, but I've tried to make it readable.) The proof actually shows a little more: it gives you $\mathrm{cov}(\mathcal{N}) \leq \mathfrak{j}_{2:1}$. This result also answers problem 3, because we know that $\mathfrak{z} = \aleph_1$ in the random model. (This was first proved by Alan Dow and David Fremlin here. It is also a corollary to Theorem 4.2 in this paper by me and Alan.) Therefore $\mathfrak{z} < \mathfrak{j}_{2:2},\mathfrak{j}_{2:1}$ is consistent. On the other hand, Koppelberg proved that $\mathfrak{z} \leq \mathrm{cov}(\mathcal{M})$. (Actually, she proved the dual statement in the category of Boolean algebras here. Stefan Geschke wrote a purely topological proof here.) Because you have proved that $\mathfrak{j}_{2:2},\mathfrak{j}_{2:1} \leq \mathrm{non}(\mathcal{M})$, and because $\mathrm{non}(\mathcal{M}) < \mathrm{cov}(\mathcal{M})$ in the Cohen model, it follows that $\mathfrak{j}_{2:2},\mathfrak{j}_{2:1} < \mathfrak{z}$ is consistent. Thus there is no inequality between $\mathfrak{z}$ and either of $\mathfrak{j}_{2:2}$ of $\mathfrak{j}_{2:1}$ that is provable in $\mathsf{ZFC}$. (I know I gave a different argument for this in the comments. I don't like that argument as much because it relies on Alan's unpublished -- and mostly unwritten -- argument that $\mathfrak{z} = \aleph_1$ in the Laver model. I'm sure he's right. But I like that the argument here relies on the fact that $\mathfrak{z} = \aleph_1$ in the random model, and you can go read one or two proofs of this if you like.) Now let's sketch the proof that $\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model. For the sake of clarity, I'm going to avoid forcing jargon and give a probabilistic argument that (I hope) will give you the right idea. To show that $\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model, let's first recall how random real forcing works. Roughly, we imagine ourselves to live in a universe $V$ of sets, containing real numbers, subsets of $\mathbb N$, lots of $2$-to$1$ functions, and whatever else. But we know that our universe is about to get bigger -- this is the forcing -- by the introduction of a "truly random" real number $r$. The new, bigger universe is called $V[r]$. The first observation I'd like to make is that all continuous measures on uncountable Polish spaces are essentially isomorphic. This means that it doesn't matter whether we view $r$ as a random element of $\mathbb R$, or of $[0,1]$, or of $2^\omega$ with the standard product measure, or whatever. For this problem, we want to view $r$ as an infinite sequence of random selections from larger and larger finite sets $I_n$, where $I_n$ has size $n!$. We select, at random, only a single element from each set. (This can be formalized by saying that we'd like $r$ to be a random element of the Polish space $\prod_{n = 0}^\infty \{1,2,\dots,n!\}$, equipped with the usual product measure. But let's keep it informal.) So our universe is about to get bigger by introducing a truly random sequence of selections from some sets $I_0, I_1, I_2, \dots$ with $|I_n| = n!$. Within $V$, we can try to anticipate objects that will be constructible from $r$ in $V[r]$. For example, we can anticipate that once we get $r$, we can build a set $J \subseteq \mathbb N$ according to the following recipe: first identify $I_n$ with the interval $[1+1+2+\dots+(n-1)!,1+1+2+\dots+(n-1)!+n!) \subseteq \mathbb N$, and then let the $n^{\mathrm{th}}$ element of $J$ be whatever $r$ randomly selects from this interval. Now I claim that this set $J$ described above has the following property: if $f$ is any $2$-to-$1$ function in the ground model $V$, then the restriction of $f$ to $J$ is almost-injective. To prove this, it suffices to argue that it's true with probability $1$, given that $r$ makes its selections randomly. This suffices because this is precisely what we mean when we say that $r$ is a "truly random" addition to $V$: if there is a randomness test defined in $V$ (such as one defined from any $f \in V$), then $r$ is random with respect to that test. So let's argue probabilistically. Fix a $2$-to-$1$ function $f \in V$. If $f(a) = f(b)$, we may view this as a "guess" that $f$ is making about our set $J$: the guess is that $a$ and $b$ are both in $J$. In other words, $f$ gets to guess at pairs from $J$ infinitely many times, and it is our job to prove that, with probability $1$, only finitely many of these guesses are correct. So what is the probability that $f$ correctly guesses a pair of elements from $J$? If $f$ identifies a member of some $I_m$ with a member of some $I_n$, where $m \neq n$, then there is a probability of exactly $\frac{1}{m!n!}$ that $f$ will have correctly guessed a pair from $J$. When $f$ makes other kinds of guesses (not identifying some member of some $I_m$ with a member of some $I_n$, where $m \neq n$), then the probability is $0$ that $f$ will have correctly guessed a pair from $J$. If $m < n$, then $f$ gets at most $|I_m| = m!$ chances to guess a pair from $J$ with one member in $I_m$ and the other in $I_n$. By the previous paragraph, the probability of one of these guesses being correct is $\leq\! m!\frac{1}{m!n!} = \frac{1}{n!}$. Summing over all $n > m$, it follows that the probability of $f$ correctly guessing any pair of elements from $J$ with one member in $I_m$ and the other in $I_n$ for some $n > m$ is $$\leq\! \sum_{n = m+1}^\infty \frac{1}{n!} < \frac{1}{(m+1)!} \sum_{k = 0}^\infty \frac{1}{(m+1)^k} = \frac{1}{(m+1)!}\frac{m+1}{m} < \frac{1}{m!m}.$$ Now fix $k > 0$. Summing over all $m > k$, we see that the probability of $f$ correctly guessing a pair of elements from $J$ with one member in $I_m$ and the other in $I_n$ for some $n > m > k$ is $$\leq\! \sum_{m = k+1}^\infty \frac{1}{m!m} < \sum_{m = k+1}^\infty \frac{1}{m!} < \frac{1}{k!k}.$$ Therefore the probability of $f$ correctly guessing a pair of elements from $J \setminus (I_0 \cup \dots \cup I_k)$ is at most $\frac{1}{k!k}$. For any fixed $\varepsilon > 0$, we can choose $K$ large enough that $\sum_{k = K}^\infty \frac{1}{k!k} < \varepsilon$. This means that for $K$ large enough, the probability of $f$ correctly guessing more than ${K+1} \choose 2$ pairs of elements of $J$ is less than $\varepsilon$. Therefore the probability of $f$ correctly guessing infinitely many pairs of elements of $J$ is less than $\varepsilon$. As $\varepsilon$ was arbitrary, the probability of $f$ correctly guessing infinitely many pairs of elements of $J$ is $0$. This shows that our set $J$ in $V[r]$ "should" (probabilistically) have the property that $f \restriction J$ is almost injective for every $f \in V$. But as we said earlier, this means $J$ really does have this property. Why does this mean $\mathfrak{j}_{2:1} = \mathfrak{c}$ in the random model? The random model is $V[G]$, where $G$ is a "random" element of the measure algebra $2^{\aleph_2}$. If $\mathcal F$ is any set of $\aleph_1$ $2$-to-$1$ functions in $V[G]$, then a standard "nice names" argument shows that there is some weight-$\aleph_1$ subalgebra $X$ of $2^{\aleph_2}$ such that $\mathcal F$ is already in the intermediate model $V[X \cap G]$. Because $|X| = \aleph_1$, there will be random reals added in moving from the intermediate model $V[X \cap G]$ to the final model $V[G]$ -- random over $V[X \cap G]$, not just over $V$. We've just showed that the addition of these random reals adds some $J$ that "works" for every $f \in \mathcal F$. Why does this mean $\mathrm{cov}(\mathcal{N}) \leq \mathfrak{j}_{2:1}$? There are a few ways to see this. The easiest is probably just to go through the above argument and convince yourself that what we've really proved is that every $2$-to-$1$ function $f$ is "solved" by a measure-$1$ set of $J$'s in the Polish space $\prod_{n = 0}^\infty \{1,2,\dots,n!\}$. Equivalently, the set of $J$'s that fail to work for a given $2$-to-$1$ function $f$ is a null set $N_f$. Therefore, if $\mathcal F$ is any size $<\! \mathrm{cov}(\mathcal{N})$ family of $2$-to-$1$ functions, $\bigcup_{f \in \mathcal F}N_f$ does not cover our Polish space, and so there is some $J$ that works for every $f \in \mathcal F$.<|endoftext|> TITLE: Near permutation $n\mapsto n+1$ not conjugate to its inverse on the Stone-Čech remainder? QUESTION [12 upvotes]: Let $\beta\omega$ be the Stone-Čech compactification of the discrete infinite countable space $\omega$, and $\beta^*\omega=\beta\omega\smallsetminus \omega$ is the Stone-Čech remainder. The map $j:n\mapsto n+1$ extends to an self-injection of $\beta\omega$, which itself restricts to a self-homeomorphism $\phi$ of $\beta^*\omega$. In ZFC+CH, is it true that $\phi$ and $\phi^{-1}$ are not conjugate in $\mathrm{Homeo}(\beta^*\omega)$? Indeed in Shelah's model ("forcing axiom"), in which CH fails, there exists a homomorphism $\mathrm{Homeo}(\beta^*\omega)\to\mathbf{Z}$ mapping $\phi$ to $1$. So the non-conjugacy of $\phi$ with $\phi^{-1}$ is consistent. But under CH, the group $\mathrm{Homeo}(\beta^*\omega)$ is simple (Rubin) so the non-conjugacy couldn't be attested by a homomorphism to $\mathbf{Z}$ as above. Note: Boolean algebraic translation through Stone duality: consider the endomorphism of the Boolean algebra $2^\omega$ of subsets of $\omega$ given by $A\mapsto \{a\in\omega:a+1\in A\}$. It induces an automorphism $\Phi$ of the quotient Boolean algebra $2^\omega/\mathrm{fin}$, where $\mathrm{fin}$ is the ideal of finite subsets. Is (under ZFC+CH) $\Phi$ non-conjugate to its inverse in $\mathrm{Aut}_{\mathrm{Ring}}(2^\omega)$? Indeed Stone duality yields (in ZFC) an isomorphism $\mathrm{Homeo}(\beta^*\omega)\to\mathrm{Aut}_{\mathrm{Ring}}(2^\omega)$ mapping $\phi$ to $\Phi$. Further comments: A side question is whether it is consistent with ZFC that $\phi$ and $\phi^{-1}$ are conjugate, I don't know either (but I'm primarily interested in the CH case). Also in ZFC it is easy to check that $\phi$ is not conjugate to $\phi^n$ for any $n\ge 2$. REPLY [12 votes]: This is a great question -- and it's wide open. Here's what I know about it: $\bullet$ As you mentioned, it is consistent that $\phi$ and $\phi^{-1}$ are not conjugate. This observation was first made by van Douwen, soon after the publication of Shelah's result that you mention in your question. You mentioned forcing axioms, so let me point out that the non-conjugacy of $\phi$ and $\phi^{-1}$ follows from $\mathsf{MA}+\mathsf{OCA}$, which is a weak form of $\mathsf{PFA}$. This is due to Boban Velickovic. $\bullet$ If it is consistent with $\mathsf{ZFC}$ that $\phi$ and $\phi^{-1}$ are conjugate, then it is consistent with $\mathsf{ZFC}+\mathsf{CH}$. (Proof sketch: If $\phi$ and $\phi^{-1}$ are conjugate in some model, then force with countable conditions to collapse the continuum to $\aleph_1$ and make $\mathsf{CH}$ true. Because this forcing is countably closed, it won't change much about the Boolean algebra $\mathcal P(\omega)/\mathrm{fin}$, and will preserve the fact that $\phi$ and $\phi^{-1}$ are conjugate.) $\bullet$ Even better, the existence of certain large cardinals implies that if it is possible to force "$\phi$ and $\phi^{-1}$ are conjugate" then this statement is already true in every forcing extension satisfying $\mathsf{CH}$. This follows from a theorem of Woodin concerning what are called $\Sigma^2_1$ statements about the real line (explained further here). The assertion "$\phi$ and $\phi^{-1}$ are conjugate" is an example of such a statement. (Very roughly, this theorem seems to suggest that if this statement is consistent, then it should follow from $\mathsf{CH}$. At any rate, trying to prove it from $\mathsf{CH}$ seems like a reasonable strategy.) $\bullet$ In fact, Paul Larson has pointed out to me that the statement "$\phi$ and $\phi^{-1}$ are conjugate" is a now very rare example of a $\Sigma^2_1$ statement about the real line whose status we do not know under $\mathsf{ZFC}+\mathsf{CH}$ (plus large cardinal axioms). $\bullet$ I proved a partial result a few years ago, showing that $\mathsf{CH}$ implies $\phi$ and $\phi^{-1}$ are semi-conjugate: $\qquad$Theorem: Assuming $\mathsf{CH}$, there is a continuous surjection $Q: \omega^* \rightarrow \omega^*$ such that $$Q \circ \phi = \phi^{-1} \circ Q.$$ The paper is "Abstract $\omega$-limit sets," Journal of Symbolic Logic 83 (2018), pp. 477-495, available here. In the same paper, I show that the forcing axiom $\mathsf{MA}+\mathsf{OCA}$ implies $\phi$ and $\phi^{-1}$ are not semi-conjugate. (Or rather, I show that this is a corollary to a deep structure theorem of Ilijas Farah.) $\bullet$ Finally, in a more recent paper (to appear in Topology and its Applications, currently available here), I show that there is no Borel set separating the conjugacy class of $\phi$ and the conjugacy class of $\phi^{-1}$ (in the space of self-homeomorphisms of $\omega^*$ endowed with the compact-open topology). Roughly, this shows that if $\phi$ and $\phi^{-1}$ fail to be conjugate, it's not "for any real reason" -- or at least not for any nicely definable topological reason.<|endoftext|> TITLE: Are $f\sqrt{1+g^2}$ and $fg\sqrt{1+g^2}$ smooth if $f,fg,fg^2$ are smooth? QUESTION [6 upvotes]: Suppose that $f$ and $g$ are functions from $\mathbb R$ to $\mathbb R$ such that the functions $f,fg,fg^2$ are smooth, that is, are in $C^\infty(\mathbb R)$. Does it then necessarily follow that the functions $f\sqrt{1+g^2}$ and $fg\sqrt{1+g^2}$ are smooth? Of course, the problem here is that the function $g$ does not have to be smooth, or even continuous, at zeroes of the function $f$. One may also note that the continuity of the functions $f\sqrt{1+g^2}$ and $fg\sqrt{1+g^2}$ (at the zeroes of $f$ and hence everywhere) follows easily from the inequalities $|f\sqrt{1+g^2}|\le|f|+|fg|$ and $|fg\sqrt{1+g^2}|\le|fg|+|fg^2|$. REPLY [14 votes]: No. Set $$ f(x) = \exp(-2/|x|^2) \operatorname{sign} x, \qquad g(x) = \exp(1/|x|^2) \sqrt{|x|} \operatorname{sign} x $$ for $x \ne 0$, and, of course, $f(0) = g(0) = 0$. Then clearly $$ \begin{aligned} f(x) & = \exp(-2/|x|^2) \operatorname{sign} x , \\ f(x) g(x) & = \exp(-1/|x|^2) \sqrt{|x|} , \\ f(x) (g(x))^2 & = x \end{aligned} $$ are infinitely smooth, but $$ f(x) g(x) \sqrt{1 + (g(x))^2} = \sqrt{|x| \exp(-2/|x|^2) + |x|^2} = |x| (1 + o(1)) $$ is not even differentiable at $0$.<|endoftext|> TITLE: Homotopy Gerstenhaber algebras: description via operads vs derivations QUESTION [7 upvotes]: There are at least a couple of definitions in the literature for an $E_2$-algebra, also known as a homotopy Gerstenhaber algebra, also known as $G_{\infty}$-algebra. Suppose $V$ is a graded vector space. Let $S(V), \text{Lie}(v)$ denote the graded symmetric algebra and the graded free Lie algebra of $V$ respectively and $S_+(V)$ be the symmetric algebra in strictly positive degree. $\textbf{Definition (Tamarkin)}$ A $G_{\infty}$-algebra structure on $V$ is a degree $+1$ map $$ \delta: S_+(\text{Lie}(V^*[1])[1]) \rightarrow S_+(\text{Lie}(V^*[1])[1]) $$ which behaves as a derivation of both $\cdot$ and $[\,,]$ such that $\delta^2 = 0$. Unpacking this leads to saying that we have a collection of maps $$ m_{k_1, \dots, k_n}: V^{\otimes k_1} \otimes \dots \otimes V^{\otimes k_n} \rightarrow V$$ of degree $3-(k_1 + \dots k_n + n)$ obeying appropriate symmetry and associativity relations. The second definition pertains to algebras over the little disk operad. Let $D_2(k)$ denote the configuration space of $k$ little disks inside a big disk and let $\text{Chains}_{\bullet}(D_2(k))$ be the singular chain complex. Letting $\mathcal{P}(k) = \text{Chains}_{\bullet}(D_2(k))$, one can define an operadic structure on this collection of vector spaces. This leads one to the $\textbf{Definition (Getzler-Jones?)}$ An $E_2$-algebra is an algebra over the operad $\text{Chains}(D_2)$. How can one show that these two definitions are equivalent? Short of a full proof of equivalence, it would be nice to understand a description of the cycle in $D_2(k_1 + \dots +k_n)$ which corresponds to the map $m_{k_1, \dots, k_n}$. For example: if one were working in $H_{\bullet}(D_2)$, the homology operad, one associates to the point class in configuration space of two disks the operation $\cdot = m_2$, and to the cycle involving one little disk going around the other the bracket $[\,,] = m_{1,1}$ in the Gerstenhaber algebra. Is there an explicit description of the cycle corresponding to $m_{k_1, \dots, k_n}$? REPLY [4 votes]: Here are the operads that are involved in that game: the operad $D_2$ of little disks, which is a topological operad. its chain operad $C_{-*}(D_2,\mathbb{k})$, which is an operad in cochain complexes (of $\mathbb{k}$-modules). its homology operad $H_{-*}(D_2,\mathbb{k})$, which is known to be isomorphic to the Gerstenhaber operad $G^{\mathbb{k}}$, which is itself a binary quadratic operad satisfying the Koszul property. the minimal resolution of $G^{\mathbb{k}}$ is the operad $G^{\mathbb{k}}_\infty$ governing ($\mathbb{k}$-linear) $G_\infty$-algebras. Being a resolution, $G^{\mathbb{k}}_\infty$ is obviously quasi-isomorphic to $G^{\mathbb{k}}$. As Phil Tosteson mentions in his answer, the difficult part relies on proving that $C_{-*}(D_2,\mathbb{k})$ is formal. It's only proven over a field $\mathbb{k}$ of characteristic zero, and it's actually not formal for $\mathbb{k}=\mathbb{F}_p$ (see e.g. the introduction of https://arxiv.org/pdf/1903.09191.pdf). To my knowledge, there are essentially two different proofs: one by Tamarkin: https://sites.math.northwestern.edu/~tamarkin/Papers1/Formality.pdf. another one by Kontsevich, which generalizes to higher dimension: https://arxiv.org/abs/math/9904055 (see also https://arxiv.org/abs/0808.0457 for more details). Note that the formality quasi-isomorphisms from these two proofs happen to coincide, after the correct "choice of associator" has been made. See https://arxiv.org/abs/0905.1789. Finally, I don't think it is meaningful to ask which cycle corresponds to the map $m_{k_1,\dots,k_n}$. The reason is that $m_{k_1,\dots,k_n}$ is not closed. You may ask if they are represented by nice chains... I don't know the answer to that question (and I suspect that it is close to be as hard as proving the formality itself), but for the $m_{1,\dots,1}$ the answer is known: first observe that $D_2$ is weakly homotopy equivalent to the operad of compactified configuration spaces of points in the plane. $m_{1,\dots,1}$ (with $n$ "$1$"s) can be represented by the top-dimensional/fundamental cell of the compactified configuration space $\overline{C}_n\simeq D_2(n)$ of $n$-points in the plane.<|endoftext|> TITLE: Existence of a strange measure QUESTION [36 upvotes]: The answer to this question must be known, but I do not know where to find it. It is related to the Ulam measures I believe. Question. Is there a finitely additive measure defined on all subsets of positive integers $\mathbb{N}$, with values into $\{0,1\}$, (only two values) that is $$ \mu:2^{\mathbb{N}}\to \{0,1\} $$ such that $\mu(\{n\})=0$ for every $n\in\mathbb{N}$ and $\mu(\mathbb{N})=1$ ? I believe such a result is needed in a proof of the co-area inequality that is stated in Coarea inequality, Eilenberg inequality. I am working with my student on some generalizations of that result so this is a research related question. REPLY [5 votes]: This can be proved without introducing ultrafilters by name, by doing "finitary measure theory" and using Zorn's lemma. An algebra $A$ on a set $X$ is just a $\sigma$-algebra without the $\sigma$, i.e. $\newcommand{\powerset}{\mathcal{P}}A \subseteq \powerset(X)$ and is closed under finite unions and complements (and therefore all other Boolean operations). Let $\newcommand{\N}{\mathbb{N}}F \subseteq \powerset(\N)$ be the set of finite sets and their complements (so-called cofinite sets). This is an algebra. Furthermore, we can define a 2-valued finitely-additive measure $\mu : F \rightarrow \{0,1\}$ to be $0$ on the finite sets and $1$ on the cofinite sets. The existence of the required 2-valued finitely-additive measure on $\powerset(\N)$ then follows from: Proposition For any algebra $A \subseteq \powerset(X)$ and finitely-additive 2-valued measure $\mu : A \rightarrow \{0,1\}$, there exists a finitely-additive measure $\overline{\mu} : \powerset(X) \rightarrow \{0,1\}$ extending $\mu$. Proof: Most of the difficulty is in believing that it's true. We use Zorn's lemma. The poset consists of pairs $(B,\nu)$ where $B \supseteq A$ is an algebra of sets, and $\nu : B \rightarrow \{0,1\}$ is a finitely-additive measure extending $\mu$. The order relation $(B_1,\nu_1) \leq (B_2,\nu_2)$ is defined to hold when $B_1 \subseteq B_2$ and $\nu_2$ extends $\nu_1$. Every chain in this poset has an upper bound - we just take the union of algebras (this is the step that fails for $\sigma$-algebras) and define the measure on the union in the obvious way. Let $(B,\nu)$ be a maximal element in the poset. Suppose for a contradiction that $B \neq \powerset(X)$, so there is some $U \in \powerset(X) \setminus B$. We contradict the maximality of $B$ by extending $\nu$ to a larger algebra $B'$ including $U$. Define $B' = \{ (U \cap S_1) \cup (\lnot U \cap S_2) \mid S_1, S_2 \in B \}$. It is clear that $B \subseteq B'$ and $U \in B'$, and with a little Boolean reasoning we can prove that for all $S_1,S_2,T_1,T_2 \in B$: $$ ((U \cap S_1) \cup (\lnot U \cap S_2)) \cup ((U \cap T_1) \cup (\lnot U \cap T_2))\\ = (U \cap (S_1 \cup T_1)) \cup (\lnot U \cap (S_2 \cup T_2)) $$ and $$ \lnot ((U \cap S_1) \cup (\lnot U \cap S_2)) = (U \cap \lnot S_1) \cup (\lnot U \cap \lnot S_2) $$ This proves that $B'$ is an algebra. Now, define $d \in \{0,1\}$ to be the "outer measure" of $U$, i.e. $d = 0$ if there exists $S \in B$ such that $U \subseteq S$ and $\nu(S) = 0$, otherwise $d = 1$. Without loss of generality we can take $d = 1$, because we can exchange the roles of $U$ and $\lnot U$. We define $\nu'((U \cap S_1) \cup (\lnot U \cap S_2)) = \nu(S_1)$. This is well-defined because if $(U \cap S_1) \cup (\lnot U \cap S_2) = (U \cap T_1) \cup (\lnot U \cap S_2)$, then $U \cap S_1 = U \cap S_2$, so $U \subseteq \lnot (S_1 \triangle S_2)$, so as $d = 1$, $\nu(\lnot (S_1 \triangle S_2)) = 1$, and therefore $\nu(S_1) = \nu(S_2)$. The identities we used to prove that $B'$ is an algebra can then be used to prove that $\nu'$ is finitely additive, and it follows directly from the definition that it extends $\nu$. So we successfully contradicted the maximality of $B$. $\square$ Of course, I actually think ultrafilters are a good thing to know about, both in topology and logic. There is also no metamathematical benefit in doing it this way - over ZF the existence of a non-principal ultrafilter on $\N$ and a finitely-additive 2-valued measure on $\powerset(\N)$ are equivalent. The above proof is based on something I came up with while reproving Stone duality in the case where the points of the Stone space are defined to be Boolean homomorphisms into $2$, rather than ultrafilters. The proposition above is a special case of the fact that complete Boolean algebras (such as $2$) are injective objects in the category of Boolean algebras.<|endoftext|> TITLE: Number of regions formed by $n$ points in general position QUESTION [9 upvotes]: Given $n$ points in $\mathbb{R}^d$ in general position, where $n\geq d+1$. For every $d$ points, form the hyperplane defined by these $d$ points. These hyperplanes cut $\mathbb{R}^d$ into several regions. My questions are: (1) is there a formula in terms of $d$ and $n$ that describes the number of regions? (2) the same question for the number of bounded regions? I tried many key words on google but found nothing helpful. Any reference or ideas will be appreciated. Thanks REPLY [2 votes]: This non-answer completes Joseph O'Rourke's nice non-answer, for the case of $n$ hyperplanes in $\mathbb{R}^d$ in general position. But it also suggests that the OP situation may also well have unique answers. Define: $U_{d,n}=$ number of unbounded regions cut by $n$ hyperplanes in $\mathbb{R}^d$ $B_{d,n}=$ number of bounded regions $T_{d,n}=$ total number of regions $=U_{d,n}+B_{d,n}$ $S_{d,n}=$ number of regions cut on the sphere $S^d$ by $n$ great $S^{d-1}$-circles Then $U_{d,n}, B_{d,n}$, $T_{d,n}$ and $S_{d,n}$ are unique with these formulas: $U_{d,0}=1$, $\quad B_{d,0}=0$, $\quad T_{d,0}=1$, $\quad S_{d,0}=1$ $U_{1,n}=2$, $\quad B_{1,n}=n-1$, $\quad T_{1,n}=n+1$, $\quad S_{1,n}=2n$ and for $n>0$ $U_{d+1,n}=S_{d,n}$ $S_{d,n}=U_{d,n}+2B_{d,n}$ $T_{d,n+1}=T_{d,n}+\sum_{i=0}^{i=d-1}{n\choose i}$ Proof. In $\mathbb{R}^{d+1}$ take a huge and growing $S^d$ sphere, so that all the bounded regions zoom down to a point at the center of the sphere, the hyperplanes become great circles on the sphere and the unbounded regions corresponds to regions cut by the circles on the sphere. Therefore if the numbers are unique (as will be proved at the end) 1. follows. Centrally project $\mathbb{R}^d$ onto a half $S^d$ (tangent to it). Complete the semisphere to a sphere by central symmetry. Then the hyperplanes become great circles, the $B_{d,n}$ bounded regions in $\mathbb{R}^d$ become $2B_{d,n}$ regions in $S^d$ and the $U_{d,n}$ unbounded ones become $U_{d,n}$ regions stretching across the suture line (equator) of the sphere. Again by unicity 2. follows. Start with $\mathbb{R}^d$ and $n$ hyperplanes in generic position inside it. Now add a new hyperplane in generic position the following way: first chose a point inside one region: no matter how that point is eventually stretched to a hyperplane, to it will split the region in two, for a gain of 1, or $n \choose 0$. Now stretch that point to a line: since it is a generic line it will meet each of the $n$ hyperplanes once and at each meeting the line will cross into one one new region and split it - with a gain of $n \choose 1$ new regions. Next stretch the line to a generic 2-plane, which will meet once each of the $(d-2)$-dimensional intersections of two hyperplanes; at each meeting the growing plane will arrive from having already crossed 3 of the 4 regions, to cross into the fourth and cut it; this a gain of another $n \choose 2$ regions. In general as a generic $m-1$-plane grows to a generic $m$-plane it will meet all the $n \choose m$ $(n-m)$-dimensional intersections of $m$ hyperplanes, and each time it will go from cutting $2^m-1$ regions before crossing the intersection to cutting all $2^m$ after crossing, for a total gain of $n \choose m$ regions. This continues up to $m=d-1$, proving 3. Proof of Unicity. By induction: $U_{d,0}$, $\quad B_{d,0}$, $\quad T_{d,0}$, $\quad S_{d,0}$ are unique; $T_{d,n}$ unique $\implies$ $T_{d,n+1}$ unique (by the proof of 3.); $U_{d,n}$ and $B_{d,n}$ unique $\implies$ $S_{d,n}$ unique (by the proof of 2. and the fact that the construction can be reversed in a non-unique way to show that $S_{d,n}=U_{d,n}+2B_{d,n}$ for some values of $U_{d,n}$ and $B_{d,n}$); $S_{d,n}$ unique $\implies$ $U_{d+1,n}$ unique (by the proof of 1.); $U_{d+1,n}$ and $T_{d+1,n}$ unique $\implies$ $B_{d+1,n}$ unique (as $B=T-U$).<|endoftext|> TITLE: The meaning and purpose of "canonical'' QUESTION [23 upvotes]: This question is jointly formulated with Neil Barton. We want to know about the significance of canonicity in mathematics broadly. That is, both what it means in some detail, and why it is important. In several mathematical fields, the term 'canonical' pops up with respect to objects, maps, structures, and presentations. It's not clear if there's something univocal meant by this term across mathematics, or whether people just mean different things in different contexts by the term. Some examples: In category theory, if we have a universal property, the relevant unique map is canonical. It seems here that the point is that the map is uniquely determined by some data within the category. Furthermore, this kind of scheme can be used to pick out objects with certain properties that are canonical in the sense that they are unique up to isomorphism. In set theory, L is a canonical model. Here, it is unique and definable. Furthermore, its construction depends only on the ordinals-- any two models of ZF with the same ordinals construct the same version of L. In set theory, other models are termed 'canonical' but it's not clear how this can be so, given that they are non-unique in certain ways. For example, there is no analogue of the above fact for L with respect to models of ZFC with unboundedly many measurable cardinals. No matter how we extend the theory ZFC + "There is a proper class of measurables," there will not be a unique model of this theory up to the specification of the ordinals plus a set-sized parameter. See here. Presentations of objects can be canonical: The most simple being that of fractions, whose presentation is canonical just in case the numerator and denominator have no common factors (e.g. the canonical presentation of 4/8 is 1/2). But this applies to other areas too; see here. Sometimes canonicity seems to be relative. Given a finite-dimensional vector space, there is a canonical way of defining an isomorphism between V and its dual V* from a choice of a basis for V. This determines a basis for V*, and thus the initial choice of basis for V yields a canonical isomorphism from V to V**. But two steps can be more canonical than one: The resulting isomorphism between V and V** does not vary with the choice of basis, and indeed can be defined without reference to any basis. See here. Other examples can be found here. Our soft questions: (a) Does the term 'canonical' appear in your field? If so what is the sense of the term? Is it relative or absolute? (b) What role does canonicity play in your field? For instance, does it help to solve problems, help set research goals, or simply make results more interesting? REPLY [2 votes]: Some of the examples given in the question are a careless misuse of the word. Who writes "canonical basis" for $K^n$ when they mean "standard basis", and who writes of a "canonical presentation of a fraction" when they mean a "fraction in its lowest terms", which isn't even canonical unless everything is positive. In my field, arithmetic geometry, "canonical" has a well-understood meaning even if it is difficult to write down a precise definition. In his 1980 book, Milne was comfortable assuming that his readers would know what it meant (in his later writings, he has switched to using $\simeq$ for "canonically isomorphic"). Roughly, it means that the object can be constructed without making any arbitrary choices. There is a huge difference between saying two objects are isomorphic and saying they are canonically isomorphic. Barr botched this in his translation of Grothendieck's Tohoku paper by replacing "=" (meaning canonically isomorphic) with isomorphic. I agree that the use of "canonical" is problematic in the Langlands program. There are major conjectures saying that some set (of representations) is bijective to some other set. After Serre pointed out that this only means that the two sets have the same cardinality, the word "canonical" was added. It is part of the problem to figure out what that means.<|endoftext|> TITLE: Brown representability in slice category QUESTION [7 upvotes]: Brown's representability theorem gives us a very nice set of conditions to check that a (contravariant) functor $Hot^{op}\rightarrow Set$ is representable. Choose an object $X$ in $Hot$. Then it is seems natural to ask whether or not an analogue of the Brown representability theorem is true for the slice category, i.e. if there exists a nice set of conditions to check whether or not a contravariant functor $$F:(Hot/_X)^{op} \rightarrow Set$$ is representable. Is there such an analogue? I've searched online, but couldn't find an comments on the subject. My hope would be that this functor is representable if $F$ respects coproducts and sends homotopy pushouts to weak pullbacks. One way I would hope to recover such a criterion is to consider the functor $$Hot\rightarrow Hot/_X, Y\mapsto (Y\rightarrow \{*\})$$ which by composition gives us a functor $Hot^{op}\rightarrow Set$, for which we know we can apply Brown's Theorem. I don't know however if this is enough to test representability of the functor $F$. REPLY [6 votes]: Edgar Brown wrote two papers on this topic: one in the Annals in 1962 that focused on the category of topological spaces, and a second one "Abstract homotopy theory" Trans. Amer. Math. Soc. 119 (1965). His second paper is very axiomatic, and I believe your situation is easily checked to satisfy his properties. (Check this!) This paper obviously predated Quillen's work on model categories (it likely partially inspired Quillen), so of course he doesn't use that language. I confess that I have always been a bit disturbed by the number of papers on Brown Representability written by authors who show no indication that they have ever looked at the original papers. Yes, people have written about more general versions (and less general, when they assume a triangulated category!) but I encourage folks to use their library resources to look at Brown's own work.<|endoftext|> TITLE: Positively curved manifold with collapsing unit balls QUESTION [7 upvotes]: Can we find a complete connected noncompact Riemannian manifold $(M^n,g)$ such that the curvature operator $Rm>0$ and $$ \inf_{p \in M} \text{Vol}_gB(p,1)=0? $$ REPLY [6 votes]: The answer is negative if $\dim M=2$ and positive otherwise, as shown in the paper: Croke, C. B., & Karcher, H. (1988). VOLUMES OF SMALL BALLS ON OPEN MANIFOLDS: LOWER BOUNDS AND EXAMPLES. AMERICAN MATHEMATICAL SOCIETY (Vol. 309). https://www.ams.org/journals/tran/1988-309-02/S0002-9947-1988-0961611-7/S0002-9947-1988-0961611-7.pdf The negative result is an immediate consequence of Theorem A, while the positive one follows from Example 1. Their construction for the latter is as follows: Consider the $d$-dimensional hypersurface given by the paraboloid $\mathcal P = \{x\in \mathbb R^{d+1}\mid x_{d+1}=x_1^2+\ldots+x_d^2\}$. At any point $A$ of $\mathcal P$ one can consider its tangential cone $\mathcal C$, i.e., the Euclidean cone of vertex $V$ with axis along the line $AV$ and that intersects $\mathcal P$ tangentially. One then can consider $\tilde {\mathcal P}$ to be $\mathcal P$ with $\mathcal C$ "attached" (see the Figure at p.765). This is a piecewise $C^\infty$ hypersurface, with a conical singularity at $V$ and that is $C^1$ at the intersection $\mathcal P\cap \mathcal C$. Outside of these regions, the curvature of $\tilde {\mathcal P}$ is $K>0$, since the curvature of $\mathcal P$ is positive and the one of the cone is zero. Moreover, it is always possible to smooth a conical singularity preserving the $K>0$ bound, and the same is true for the $\mathcal P\cap \mathcal C$ part. (You can maybe look at this thesis, e.g., Lemma 3.2.4.) So we have constructed a smooth hypersurface with $K>0$, and where we can estimate the volume of balls contained in the smoothed conical region by the volumes of the corresponding regions in the non-smoothed cones. Moreover, this operation can be done with a sequence of tangent cones $\mathcal C_n$, as soon as they are sufficiently distant one from the other. The estimation of the volumes of the balls on the tangent cones is the object of the claim at the end of p.765. In particular, there the authors show that if the vertex of the cone is $V = (k+1,\ldots, 2k+1)$ for $k\ge 5$ (I'm fixing $r=1$), this volume is bounded by the volume of a "spherical spindle" : $$ \operatorname{vol}(B(V,1))\le C_d\sin^{d-2}\theta , $$ where $\theta$ is an angle bounded by $\tan^2\theta\le 6/k$. (Here I chose $\varepsilon =1\le k^2/(2k+2)$.) This shows that it is possible to attach to $\mathcal P$ a sequence of cones $\mathcal C_n$ with vertices $V_n\to +\infty$, and smooth them out so that $$ \lim_{n\to +\infty}\operatorname{vol}(B(V_n,1))\le (C_d+1) \lim_{n\to +\infty} \sin^{d-2}\left(\theta_n\right) = 0. $$ This proves that the obtained hypersurface (which has positive curvature) satisfies your requirement.<|endoftext|> TITLE: What is the etale homotopy type of the Witt group of braided fusion categories? QUESTION [6 upvotes]: The Witt group $\mathcal{W}$ of braided fusion categories (see also the sequel paper) can be defined over any field; I am happy to restrict to characteristic $0$ if it matters. Is $\mathbb k \mapsto \mathcal W(\mathbb k)$ an (affine?) algebraic group scheme? Assuming $\mathcal W$ is sufficiently scheme-like for the following question to make sense, what I really want to know is: What is the etale homotopy type of $\mathcal{W}$? REPLY [2 votes]: The answers to your questions are essentially in the first paper you cite. The second paper has more information on finer structure, like torsion, but the basic properties are all we need. In the first paragraph of the introduction, the authors mention that any embedding of algebraically closed fields of characteristic zero yields an isomorphism on rational points, so it is geometrically a zero dimensional object. In particular, it is perhaps more profitable to think of it as a big Galois module. By Corollary 5.23, $\mathcal{W} \otimes \mathbb{Q}$ is a vector space of countably infinite dimension, so (assuming one manages to define $\mathcal{W}(R)$ for commutative $\mathbb{Q}$-algebras $R$) it is in fact an ind-affine group ind-scheme, rather than an affine group scheme.<|endoftext|> TITLE: Volume of solution sets for polynomials in $\mathbb{C}[x]$ QUESTION [7 upvotes]: Denote $\pmb{a}=(a_1,\dots,a_d)\in\mathbb{R}^d$ and consider the set $$\mathcal{E}_d=\{\pmb{a}\in\mathbb{R}^d: \text{each root $\xi$ of $x^d+a_dx^{d-1}+\cdots+a_2x+a_1=0$ lies in $\vert\xi\vert<1$}\}.$$ In the reference shown below, Fam proved that the $d$-dimensional Lebesgue measure satisfies $$\lambda_d(\mathcal{E}_d)=2^d\prod_{k=1}^{\lfloor\frac{d}2\rfloor}\left(1+\frac1{2k}\right)^{2k-d}.$$ I'd like to propose a complex version here. Denote $\pmb{c}=(c_1,\dots,c_d)\in\mathbb{C}^d$ and consider the set $$\mathcal{S}_d=\{\pmb{c}\in\mathbb{C}^d: \text{each root $\xi$ of $x^d+c_dx^{d-1}+\cdots+c_2x+c_1=0$ lies in $\vert\xi\vert<1$}\}.$$ Now, let's ask: QUESTION. What is the $2d$-dimensional Lebesgue measure $$\lambda_{2d}(\mathcal{S}_d)?$$ For contrast, $\lambda_1(\mathcal{E}_1)=2$ while $\lambda_2(\mathcal{S}_1)=\pi$. Reference. A. T. Fam,The volume of the coefficient space stability domain of monic polynomials, Proc. IEEE Int. Symp.Circuits and Systems, 2 (1989), pp. 1780–1783. REPLY [5 votes]: $\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}$The answer is $\tfrac{\pi^n}{n!}$. It is certainly a surprise to have the answer come out so simple! Let $\phi : \CC^n \to \CC^n$ be the map which takes $(z_1, z_2, \ldots, z_n)$ to the elementary symmetric functions $(e_1, e_2, \ldots, e_n)$ where $e_k = \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq n} z_{i_1} z_{i_2} \cdots z_{i_n}$. Let $D$ be the unit disc in $\CC$. So you want to compute the volume of $\phi(D^n)$, also known as $\int_{\phi(D^n)} \mathrm{Vol}$. The map $D^n \to \phi(D^n)$ is $n!$ to $1$ and, since $\phi$ is complex analytic, $\phi$ is orientation preserving. So $$\int_{\phi(D^n)} \mathrm{Vol} = \frac{1}{n!} \int_{D^n} \phi^{\ast}(\mathrm{Vol}) = \frac{1}{n!} \int_{D^n} \det J^{\RR}_{\phi}$$ where $J^{\RR}_{\phi}$ is the Jacobian of $\phi$ considered as a smooth map $\RR^{2n} \to \RR^{2n}$. I will write $J^{\CC}_{\phi}$ when I instead want the $n \times n$ matrix of complex numbers coming from thinking of $\phi$ as a complex analytic map $\CC^n \to \CC^n$. The relation between these two notions is this: $\det J^{\RR}_{\phi} = |\det J^{\CC}_{\phi}|^2$. (This is just linear algebra -- if $L/K$ is a degree $d$ field extension, $f: L^n \to L^n$ is a linear map and $g: K^{dn} \to K^{dn}$ is the linear map gotten by identifying $L$ with $K^d$, then $\det g = N_{L/K}(\det f)$.) We have the well known identity $$\det J^{\CC}_{\phi}(z_1, \ldots, z_n) = \prod_{i TITLE: Relative logarithmic cotangent bundle QUESTION [5 upvotes]: Let $\mathcal X \rightarrow S$ be a flat family of projective varieties over a discrete valuation ring $S$ such that the generic fibre $\mathcal X_{\eta}$ (say) is smooth projective variety and the special fibre $\mathcal X_0$ (say) is a normal crossing divisor in $\mathcal X$. Question: Does there exist a vector bundle $\mathcal E$ over $\mathcal X$ such that $\mathcal E|_{_{\mathcal X_{\eta}}}\cong \Omega^1_{_{\mathcal X_{\eta}}}$ and $\mathcal E|_{_{\mathcal X_0}}\cong \Omega^1_{_{\mathcal X_0}}(\mathrm{Log} D)$, where $D$ is the singular locus of $\mathcal X_0$? It will be very helpful if anyone can explain this construction or provide a reference. Thank you.. REPLY [6 votes]: First of all, it's unclear what you mean by $\Omega^1_{X_0}(\log D)$ since $X_0$ is singular. Second, if you make up such a definition then most probably such a vector bundle will not exist. Note that the Euler characteristics $\chi(X_\eta, \Omega^1_\eta)$ and $\chi(X_0, \Omega^1_{X_0}(\log D)$ have to agree. I suggest checking this for an elliptic curve degenerating to a node. Finally, what people usually consider in your situation is the vector bundle $\Omega^1_{X/S}(\log X_0)$, so differentials on $X$ with log poles along the components of $X_0$, divided by the pull-back $t^{-1}dt$ where $t$ is a uniformizer of $S$. This is indeed locally free and restricts to $\Omega^1_{X_\eta}$.<|endoftext|> TITLE: What is known about the non-existence of strongly regular graphs srg(n,k,0,2)? QUESTION [6 upvotes]: Only few strongly regular graphs with parameters $\lambda=0$ (triangle-free) and $\mu=2$ (any two non-adjacent vertices have exactly two common neighbors) are known, see the wikipedia page: the 4-cycle, the Clebsch graph and the Sims-Gewirtz graph. I am looking for any information about the potential existence of more such graphs. For which values of $n$ and $k$ are they known not to exist? REPLY [5 votes]: Example 1 in A.Neumaier paper says in partcular that the vertex degree in this case must be $k=t^2+1$, for $t$ not divisible by 4. As well, the number of vertices is $v=1+k+\binom{k}{2}$. The examples you list correspond to $t=2,3$. The next possible parameter set corresponds to $t=5$, so you have $v=352$, $k=26$. A.Brouwer's database lists this tuple of parameters as feasible, but no examples known. Similarly for $t=6,7$ you have feasible sets of parameters $v=704,1276$, resp. $k=37,50$, but no examples known. To see that $k=t^2+1$, note that the 2nd eigenvalue of the adjacency matrix is $$ r:=\frac{1}{2}\left[(\lambda-\mu)+\sqrt{(\lambda-\mu)^2 + 4(k-\mu)}\right]=-1+\sqrt{k-1},\quad \text{i.e. $t^2:=(r+1)^2=k-1.$} $$ Similarly, the 3rd eigenvalue is $s:=-1-\sqrt{k-1}$, and one can compute their multiplicites, see e.g. Brouwer-van Lint, p.87 to rule out the case $t$ divisible by 4. Namely, the multiplicity of $r$ is given by $$ -\frac{k(s+1)(k-s)}{(k+rs)(r-s)}=\frac{k\sqrt{k-1}(k+1+\sqrt{k-1})}{4\sqrt{k-1}}=\frac{(t^2+1)(t^2+2+t)}{4}, $$ which cannot be an integer if $4|t$.<|endoftext|> TITLE: Conceptual proof of classification of surfaces? QUESTION [28 upvotes]: Every compact surface is diffeomorphic to $S^2$, $\underbrace{T^2\#\ldots \#T^2}_n$, or $\underbrace{RP^2\#\ldots \#RP^2}_n$ for some $n\ge 1$. Is there a conceptual proof of this classification theorem? Here, the term "conceptual" a little bit up for interpretation... One may take it to mean: that doesn't rely on unintuitive facts. For example, a proof that doesn't need $T^2\#RP^2\cong RP^2\#RP^2\#RP^2$ as an ingredient, but which has the property that this diffeomorphism comes out as a consequence would definitely count as conceptual. REPLY [12 votes]: This is more an extended comment than an answer to the question. The first thing to note is that there are different strenghts of the classification theorem for surfaces. Of course, there are the differentiable, triangulated and topological setting. But even if we choose such a setting, there are two statements one has to prove (at least in one approach): Every closed surface is isomorphic to a sphere with handles or cross-caps attached. The isomorphism type only depends on the number of handles and cross-caps attached. Both the Zip-proof and the Zeeman proof refered to above in the comments only prove the first part and not the second part. The second part is essentially equivalent to the well-definedness of the connected sum. Especially in the topological setting, this is a subtle point, requiring even in dimension 2 a kind of Schönflies theorem (and being a very difficult theorem in dimension 4). Out of frustration about this state of affairs I wrote up a variant of Zeeman's proof (based on a treatment by Thomassen), but including the well-definedness of attaching handles/cross-caps. (It took me a long time to thus actually understand the argument of Zeeman.) This is not really an answer to the original question, both because this proof is in the triangulated setting and does not avoid the isomorphism of the original question. But I really want to stress the point that well-definedness of attaching handles or connected sum is something one has to prove and not just hide by abuse of notation.<|endoftext|> TITLE: Geometry of algebraic curve determined by point counts over all number fields? QUESTION [24 upvotes]: Let $C$ be a smooth (geometrically irreducible) projective curve of genus $g>1$ over a number field $K$. The Mordell conjecture (first proved by Faltings) says that for any finite field extension $L/K$ (say, inside a fixed algebraic closure $\overline{K}$) there is a function $\{$(isomorphism classes of) finite extensions $L/K$ inside $\overline{K}\} \rightarrow\ \{$non-negative integers$\}$ defined by $L \mapsto \#C(L).$ Does this function determine (the isomorphism class of) $C/\overline{K}$, or otherwise any geometric properties of $C$ such as the genus? REPLY [6 votes]: Heuristically, we can compute the genus. This follows essentially the srategy of damiano and David Lampert in the comments. First note that it is possible to compute from this function the number $\#'C(L)$ of points over $L$ with field of definition exactly $L$. (Here the field of definition is the smallest field containing $K$ that the point is defined over.) This follows from inclusion exclusion. Now observe that if we let $$ r_d = \lim \sup_{X\to \infty} \frac{\log \sum_{L/K, |\Delta_{L/K}|0$, I think it is reasonable to expect that $d$ is the gonality of $C$ and that $r_d = \frac{1}{ g+ d-1}$, so that the genus of $C$ is $\frac{1}{r_d} +1-d$. Why do I think this? Consider first a degree $d$ map $\pi: C \to \mathbb P^1$. Associated to a point of $x\in \mathbb P^1(K)$, of which there are $\approx Y^2$ of height $ TITLE: What types are to mathematical proofs as types à la Martin-Löf are to constructive proofs, and what's wrong with them? QUESTION [7 upvotes]: The question is motivated by this surprising sentence from Freek Wiedijk's The QED Manifesto Revisited. I agree that the QED-like systems that exist today are not good enough to start developing a library as is described in the QED manifesto. Surprising to me, that is, who had not heard of the QED project before. (Edited after A. Bauer's answer) OK, time I mopped up. I was 1st introduced to the Curry-Howard correspondence back in school days long, loooooong ago; I understood it as s/th along the lines of "Any computer program is a constructive proof that inhabitation of its input type entails inhabitation of its output type"; my very next question was, "OK then, if instead of any program we restrict ourselves to programs that model mathematical proofs in classical logic, what would be the proper type system(s) for them?". I guessed it had to be existential types, guarded subtypes (meaning types restricted by any formula; so that you would write e. g. "let x: {r : Real | !(r : Rational)}" if all you mean is to express that x is irrational), or such like stuff; then I let the matter rest and turned to more pressing ones, to wit, getting a degree. If I had pressed it further at the time &, instead of the title question, asked "What types are to classical logic as simple types are to constructive logic?", maybe I'd have been answered "There are no such types: there is no need to extend or otherwise modify simple types for classical logic, since it is encompassed in constructive logic. Go get some understanding of type theory, and learn how to do with them. " If I had asked again at the time the QED manifesto was issued, a plausible answer might have turned to "They're called refinement types. What's wrong with them is, they're a pointless gadget: the problem they solved was not a foundational one, and introducing them did not present type theory with any new challenge. So all the papers they deserve have been written already. " Returning to it now, it seems to me the natural way to attack the QED challenge would be: 1. to set forth a serious answer to my old question, 2. design the corresponding language, 3. build the proper language tools, including, in the end, a full-fledged proof assistant (probably posing as a type checker), 4. look what happens. In a programming language with refinement types, the C-H correspondence now works as: subprograms returning (refinements of) void model conjectures, their bodies model proof outlines, assertions in the bodies model steps in the proof, Skolem functions arise naturally if handlers for failed assertions are allowed. Then, verified theorems in predicate logic are those subprograms in which the type checker has been able to reduce all assertions to tautologies. What hopefully would happen if a suitable extension to such a language had become popular is, useful tools could start developing, such as doclets to turn fragments of source code to academic papers; people might get used to writing in it just to convert formulas from .pdf to .ps, or to save themselves the hassle of adjusting to the style guides of every other journal they submit to; later, we might dream of replacing arXiv with a global GitHub repository of their works. This could take place long before any proof assistant worth mentioning existed for the language. Given Wiedijk's paper though, the real story must have been quite different. It reads like people trying to endow dependent types with refinement soon hit some snag and, instead of developing proof assistants (Edit: read formal languages and tools) for classical logic, they went happily to rewrite classical math for constructive logic. (Edit) With predictable results: mathematicians have been accepting a certain style of proof outline for some time now and non-mathematicians are used to it, so if it is too arcane work to make it understandable by a automated checker, they will rather dispose of the proof checker than of the proof. Turn away for a minute from what dependent types mean, and look how they would be used in a language purporting to be strongly typed. The user is student X with a major in control systems, a hobby in Java programming, currently at grips with an intro to topological groups. So far, he has managed to translate "pick any dyadic integer " as "let x: Integer(2)" for proof-checking freeware he downloaded a minute ago, then "consider first the case of a rational integer" to "let x1:Integer(2)*Rational"; now attacking "pick some irrational number x and consider the fractional parts of the p.x's, p ranging over Z: obviously, blah blah blah". Not understanding a word of type theory, he first looks for a way to say "let x: Real - Rational", finds none, looks harder, still finds none, then let go & turns back to more pressing matters, to wit, control systems. Of course, if student X had access to such a language as I sketched, he would have given up long before he got any proof automated. No harm done, though: what he wants is to understand TG's, not that a computer program understand it in his stead, nor to understand type theory. All he needs is to keep his fingers busy while reading, and I advocate giving him a productive way to do just that. My read of Wiedijk's paper is, if you bar student X from modelling things the way he understands them, he will turn to what he considers more pressing matters, regardless of how pressing your need for computer-checkable proofs, and how deep your understanding of what the modelling language guarantees. It explains how we still lack a free-access, widely-known, low-quality, multi-audience, computer-interpreted codebase of mathematical theories the way we have a free-access, widely-known, low-quality, multi-audience, human-readable textbase of common lore. Edit Just to clarify: the codebase of mathematical theories in question would still be as far away from being computer-checked, as is WP from auto-generating ontologies for the subject matter of its articles; light-yrs away. The (hopeful) progress lies in the rest of the road being mostly for metamathematicians & software engineers to tread, with the part of "ordinary" mathematicians perceived as non-problematic. Hence my question: what type systems are the natural ones to express assertions in classical logic? Or maybe I should have titled this post "What types are to dependent types, as refinement types à la Freeman-Pfenning are to simple types?". Anyway, what have they that makes their theory so unsavory? REPLY [12 votes]: A good starting point to learn about type theories for classical logic is the $\lambda\mu$-calculus introduced in 1992 by Parigot in λμ-Calculus: An algorithmic interpretation of classical natural deduction, which extend the $\lambda$-calculus to give a computational interpretation of classical natural deduction. For the next step, I would recommend reading the research summary on Hugo Herbelin's home page. Chasing some of the references listed there, you will learn how incorporating control operators into type theory yields a computational understanding of excluded middle. For a modern take on the role of excluded middle in Martin-Löf type theory I recommend reading at least Section 3.4 of the HoTT book. The important bit to take away from there is the fact that excluded middle can coexist with the rest of homotopy type theory quite naturally. You speak of proving correctness of programs, possibly using classical logic. For that you can use a system based on refinement types, such as F*. Also note that proving correctness of programs is an activity unlike formalization of traditional mathematics, so do not expect a single tool to work equally well for both of them. In my opinion the presence or lack of excluded middle in a proof assistant has very little to do with it being useful for formalization. Of course, if you want to formalize classical mathematics then you need excluded middle (but much less frequently than most mathematicians expect). Most modern proof assistants let you postulate excluded middle quite directly and use it to your heart's content, so that cannot be the show stopper. One of the most popular proof assistants is Isabelle/HOL, which has excluded middle built in, but that does not make it significantly more successful in comparison with other formalization tools. It is a bit difficult to understand what precisely you are getting at in your long discussion, but I would venture to say that you've misidentified negation and excluded middle as the culprit. I am not even sure whether you think that Martin-Löf type theory has no notion of negation – of course it does! In fact, the problems faced by your student X has nothing whatsoever to do with type theory. Student X would face essentially the same obstacles if they used any other formal system, such as first-order logic and set theory, or higher-order classical logic. Your student X is quite naive to think that formalization of mathematics is just a simple exercise in translation from English prose. It may be true that usability is decided by the users, but the users' abilities play a role as well.<|endoftext|> TITLE: Precise probability that $m$ random vectors in $n$ dimensional space are nearly-orthogonal QUESTION [5 upvotes]: Consider $m$ vectors $v_1,\dots,v_m$ in $\mathbb R^n$, drawn uniformly and independetly from unit sphere. It is pretty straightforward from Chebyshev inequality that $$ \mathrm P (\forall i\ne j \ |v_i \cdot v_j|\leq \varepsilon) \to 1\ \text{as} \ n \to \infty. $$ But what about quantitative version of this limit, i.e. if we define $$ f(m, \varepsilon, \delta) = \min\{n : \mathrm P(\forall i\ne j\ |v_i \cdot v_j| \leq \varepsilon)\geq 1 - \delta\} $$ what can we say about asymptotic behavior of $f$? REPLY [4 votes]: $\newcommand{\ep}{\varepsilon} \newcommand{\Ga}{\Gamma} \newcommand{\de}{\delta}$ For $\ep\in(0,1)$, let \begin{equation*} P_{m,n}:=P\Big(\bigcap_{1\le i\ep\}\Big). \end{equation*} By Bonferroni inequalities, \begin{equation*} Mp\ge Q_{m,n}\ge Mp-R/2, \end{equation*} where \begin{equation*} p:=P(|v_1\cdot v_2|>\ep), \end{equation*} \begin{equation*} M:=m(m-1)/2, \end{equation*} \begin{equation*} R:=\sum_{1\le i\ep,|v_k\cdot v_l|>\ep). \end{equation*} If $\{i,j\}\cap\{k,l\}=\emptyset$, then $P(|v_i\cdot v_j|>\ep,|v_k\cdot v_l|>\ep)=P(|v_i\cdot v_j|>\ep)\,P(|v_k\cdot v_l|>\ep)=p^2$. If $\{i,j\}\cap\{k,l\}\ne\emptyset$ but $\{i,j\}\ne\{k,l\}$, then, using the iid condition on the $u_i$'s and the spherical symmetry, for (say) the unit vector $e_1$ of the standard orthonormal basis of $\mathbb R^n$, we have \begin{multline} P(|v_i\cdot v_j|>\ep,|v_k\cdot v_l|>\ep) =P(|v_1\cdot v_2|>\ep,|v_1\cdot v_3|>\ep) \\ =P(|e_1\cdot v_2|>\ep,|e_1\cdot v_3|>\ep) =P(|e_1\cdot v_2|>\ep)\,P(|e_1\cdot v_3|>\ep)=p^2. \end{multline} So, $P(|v_i\cdot v_j|>\ep,|v_k\cdot v_l|>\ep)=p^2$ for any $i,j,k,l$ such that $1\le i\ep)=K_nI_n, \end{equation*} where, with $n\to\infty$, \begin{equation*} K_n:=\frac{\Ga(n/2)}{\Ga(1/2)\Ga((n-1)/2)(n-1)^{1/2}}\to1/\sqrt\pi, \end{equation*} \begin{equation*} I_n:=(n-1)^{1/2}\int_{\sqrt c}^\infty(1+t^2)^{-n/2}\,dt=e^{-nc/(2+o(1))}, \end{equation*} \begin{equation*} c:=\frac{\ep^2}{1-\ep^2}. \end{equation*} Collecting the pieces, we see that \begin{equation} Q_{m,n}=Me^{-nc/(2+o(1))}. \end{equation} Setting now $\de=Q_{m,n}\to0$, we find the asymptotics of the needed $n$: \begin{equation} n\sim2\frac{1-\ep^2}{\ep^2}\,\ln\frac{m(m-1)}{2\de}. \end{equation}<|endoftext|> TITLE: Free product decompositions of the fundamental group of Hawaiian Earrings QUESTION [11 upvotes]: This is a spin-off of my question here, separated from the older question following Jeremy's suggestion. Definition. Call a group $G$ essentially freely indecomposable if in every free product decomposition $G\cong G_1\star G_2$, one of the free factors $G_1, G_2$ is free of finite rank. Question. Let ${\mathbb H}$ denote the Hawaiian Earrings. Is the fundamental group $\pi_1({\mathbb H})$ essentially freely indecomposable? The only relevant result I could find in the literature is the theorem (due to Higman) which (according to "The combinatorial structure of the Hawaiian earring group" by Cannon and Conner) implies that that every freely indecomposable free factor of $G=\pi_1({\mathbb H})$ is either trivial or infinite cyclic. Maybe Higman's methods prove more, but his paper ("Unrestricted Free Products, and Varieties of Topological Groups", Journal of LMS, 1952) is behind the paywall. REPLY [9 votes]: This answer is courtesy of Sam Corson who kindly pointed out the following. Theorem: The Hawaiian earring group $\pi_1(\mathbb{H})$ is essentially freely indecomposable, i.e. if $\pi_1(\mathbb{H})\cong G_1\ast G_2$, then one of $G_1$ or $G_2$ must be a finitely generated free group. The key is to apply a special case of Theorem 1.3 in K. Eda, Atomic property of the fundamental groups of the Hawaiian earring and wild locally path-connected spaces, J Math. Soc. Japan 63 (2011), 769-787. Let $C_n$ be the $n$-th circle of $\mathbb{H}$. Take $\mathbb{H}_{\geq n}=\bigcup_{k\geq n}C_k$ to be the smaller copies of the Hawaiian earring and $\mathbb{H}_{\leq n}=\bigcup_{k=1}^{n}C_k$ to be the union of the first $n$-circles. One of the defining properties of the Hawaiian earring group is that there is a canonical isomorphism $\pi_1(\mathbb{H})\cong \pi_1(\mathbb{H}_{\geq n+1})\ast \pi_1(\mathbb{H}_{\leq n})$ for all $n\in\mathbb{N}$. Suppose that $\phi:\pi_1(\mathbb{H})\to G_1\ast G_2$ is an isomorphism. In the case of the Hawaiian earring, Eda's theorem cited above implies that for any homomorphism $\phi:\pi_1(\mathbb{H})\to G_1\ast G_2$, there exists an $n\in\mathbb{N}$, $i\in\{1,2\}$, and $w\in G_1\ast G_2$ such that $\phi(\pi_1(\mathbb{H}_{\geq n+1}))\leq w G_i w^{-1}$. Suppose, without loss of generality, that $i=1$. Now if $\gamma(g)=w^{-1}gw$ is conjugation in $G_1\ast G_2$, then $\gamma\circ \phi:\pi_1(\mathbb{H})\to G_1\ast G_2$ is an isomorphism mapping $\pi_1(\mathbb{H}_{\geq n+1})$ into $G_1$. Let $\psi:G_1\ast G_2\to G_2$ be the projection, which is surjective. Then $\psi\circ\gamma\circ\phi:\pi_1(\mathbb{H})\to G_2$ is a surjection and since $\pi_1(\mathbb{H})\cong \pi_1(\mathbb{H}_{\geq n+1})\ast \pi_1(\mathbb{H}_{\leq n})$ and $\psi\circ\gamma\circ\phi(\pi_1(\mathbb{H}_{\geq n+1}))=1$, we must have that $\psi\circ\gamma\circ\phi$ maps $\pi_1(\mathbb{H}_{\leq n})$ onto $G_2$. Therefore, $G_2$ is finitely generated. Moreover, since $G_2$ is isomorphic to a subgroup of the locally free group $\pi_1(\mathbb{H})$, it follows that $G_2$ is free.<|endoftext|> TITLE: Maximal dimension of abelian subalgebra of exceptional simple Lie algebra in positive characteristic QUESTION [6 upvotes]: For complex semisimple Lie algebras, the maximal dimension of an abelian subalgebra was determined by Mal'cev in 1945. For $E_7$, for example, it is $27$, and is the radical of the $E_6$ parabolic. What about in characteristic $p$ for $p>0$? I suspect the answer is nearly, but not quite, the same. Maybe you have that it is bounded by $29$ or something. Is there any literature on this? All of the papers and references I have found so far are in characteristic $0$. I don't need the exact bound, just something close will do. (I have a Lie algebra $L$ with an abelian subalgebra of dimension, say, $43$, and I want to know that $L$ cannot be embedded in $E_7$, and in particular is not $E_7$. But I'm in characteristic $7$, for example, or worse, $2$ or $3$.) REPLY [4 votes]: Let $\ell\ge0$ be the characteristic of the (algebraically closed) ground field. Let $G$ be semisimple with Lie algebra $\mathfrak g$. First, the maximal dimension of an abelian subalgebra of $\mathfrak g$ can increase for small $\ell$. Let, e.g., $\mathfrak g=sl(2)$ and $\ell=2$. Then the Borel subalgebra is abelian, hence the maximal dimension is $2$ instead of $1$. The same holds for $\mathfrak g=pgl(2)$ and $\ell=2$ where $\langle e,f\rangle$ is abelian. Secondly, this is a phenomenon of very small characteristics. The following statement is not the most general one but covers most cases: Assume that $G$ is of adjoint type and $\ell>3$. Then the dimension of a maximal abelian subalgebra is the same as in characteristic $0$. Proof: Key is the following deformation argument: Being an abelian subalgebra is a closed condition in the Grassmannian of $\mathfrak g$. In other words, there is a closed subscheme $A_d\subseteq Gr_d(\mathfrak g)$ classifying abelian subalgebras of dimension $d$. The group $G$ acts on $A_d$ by conjugation. Assume that $A_d\ne\emptyset$. Since $A_d$ is projective, any Borel subgroup $B$ of $G$ has a fixed point. This shows, that if $\mathfrak g$ contains an abelian subalgebra $\mathfrak a$ of dimension $d$ then it will also contain one which is normalized by $B$. Assume from now on that this is the case. Then, in particular, $\mathfrak a$ will be normalized by a maximal torus $T\subseteq B$. Let $\mathfrak g=\mathfrak t\oplus\bigoplus_\alpha\mathfrak g_\alpha$ be the root space decomposition. Then any $T$-stable subspace, so also $\mathfrak a$, has the form $$ \mathfrak a=\mathfrak a_0\oplus\bigoplus_{\alpha\in S}\mathfrak g_\alpha $$ where $S$ is a set of roots. Since $\mathfrak a$ is ablian, its subset of semisimple elements $\mathfrak a_0$ inside $\mathfrak b$ is also normalized by $B$. Hence $\mathfrak a_0$ consists of fixed points of $B$ and therefore of $G$. So $\mathfrak a_0$ sits in the schematic center of $\mathfrak g$. Because $G$ is of adjoint type we infer $\mathfrak a_0=0$. Thus everything depends on $S$. I argue that the condition on $S$ making $\mathfrak a=\bigoplus_{\alpha\in S}\mathfrak g_\alpha$ abelian is independent of $\ell$, proving our assertion. For this let $e_\alpha\in\mathfrak g_\alpha$ be a Chevalley generator. Then $[e_\alpha,e_\beta]$ must be zero for all $\alpha,\beta\in S$. If $\alpha+\beta=0$ then $h_\alpha=[e_\alpha,e_\beta]\ne0$ since $\ell\ne2$. So this case must not occur. If $\gamma=\alpha+\beta$ is a root then a famous formula of Chevalley asserts $$ [e_\alpha,e_\beta]=\pm N_{\alpha\beta}e_\gamma\quad\text{with }N_{\alpha\beta}\in\{1,2,3\}. $$ Because $\ell>3$, by assumption, we get $[e_\alpha,e_\beta]\ne0$. So this case must not occur either. In the remaining cases we have $[e_\alpha,e_\beta]=0$. The condition on $S$ is therefore: $\mathfrak a$ is an abelian subalgebra if and only if $\alpha+\beta$ is not zero and not a root for all $\alpha,\beta\in S$. This condition is clearly characteristic free.<|endoftext|> TITLE: Characterisation of essentially algebraic theories as monads QUESTION [7 upvotes]: The following correspondence between algebraic theories and monads on $\mathbf{Set}$ is well-known (see, for example, Algebraic Theories: A Categorical Introduction to General Algebra). The category of (finitary) $S$-sorted algebraic theories is equivalent to the category of (finitary) monads on $\mathbf{Set}/S$. The category of models for a fixed algebraic theory $T$ is equivalent to the category of algebras for the corresponding monad. I would like to know if there is a known analogous result in the setting of essentially algebraic (i.e. finite limit) theories. That is, some statement like the following. The category of (finitary) $S$-sorted essentially algebraic theories is equivalent to the category of [some class of] monads on [some category]. The category of models for a fixed algebraic theory $T$ is equivalent to the category of algebras for the corresponding monad. There are many characterisations of categories of models of essentially algebraic theories (i.e. locally presentable categories), but I have not been able to find one in terms of a category of algebras for some monad. Given the many generalisation of Linton's result, for example in Notions of Lawvere theory and Monads with arities and their associated theories, it would seem like this result should be a straightforward application of an existing theorem, but, as far as I can tell, the case of essentially algebraic theories is never explicitly treated. (I originally read Theorem 6.7 of Notions of Lawvere Theory as stating that one-sorted essentially algebraic theories also correspond to finitary monads on $\mathbf{Set}$, but this seems unlikely to be correct.) If the $S$-sorted case is unknown, I am also interested in the correspondence specifically in the one-sorted setting. REPLY [3 votes]: I'm going to give a partial answer to my question, which addresses a misconception I had and illustrates why many of the existing generalisations of theory–monad correspondence are not sufficient to provide a monadic correspondence with essentially-algebraic theories. As far as I know, this indicates that a correspondence is open. Let $\mathbb C$ be a small category, and denote by $\widehat{\mathbb C}$, $\mathbf{Rex}(\mathbb C)$, $\mathbf{Ind}(\mathbb C)$ and $\mathbf{Lex}(\mathbb C)$ the free cocompletion, finite cocompletion, filtered cocompletion and finite completion of $\mathbb C$, respectively. We have $\mathbf{Ind}(\mathbf{Rex}(\mathbb C)) \simeq \widehat{\mathbb C}$. We'll instantiate the results of Bourke & Garner's Monads and theories for $K : \mathbf{Rex}(\mathbb C) \hookrightarrow \widehat{\mathbb C}$. This is an example of the so-called "presheaf context". The results follow directly, so I won't spell everything out. $K$ preserves finite colimits and so a $\mathbf{Rex}(\mathbb C)$-theory is a finite colimit-preserving identity-on-objects functor $J : \mathbf{Rex}(\mathbb C) \to \mathcal T$. A model is a finite-limit preserving functor $F : \mathcal T^{\mathrm{op}} \to \mathbf{Set}$. A $\mathbf{Rex}(\mathbb C)$-nervous monad is a monad on $\mathbf{Rex}(\mathbb C)$ whose underlying endofunctor preserves filtered colimits. The categories of $\mathbf{Rex}(\mathbb C)$-nervous monads and $\mathbf{Rex}(\mathbb C)$-theories are equivalent by Theorem 17 of ibid. (and similarly their categories of algebras and categories of models by Theorem 34 of ibid.) We can now take opposites appropriately to generalise the classical algebraic theory–finitary monad correspondence: the categories of finite limit-preserving identity-on-objects functors from $\mathbf{Lex}(\mathbb C^{\mathrm{op}})$ (which I'll dub "category-sorted lex theories") are equivalent to finitary monads on $\widehat{\mathbb C}$. When $\mathbb C = S$ is discrete, category-sorted lex theories are equivalent to sorted algebraic theories (this can be seen as a consequence of sifted colimits and filtered colimits coinciding on indexed sets, or of the coincidence of finite completion and finite product completion on sets). When $\mathbb C$ is a non-discrete category, the two constructions differ. This essentially gives a classification of finitary and sifted colimit-preserving monads on presheaf categories (at least on small categories), which was suggested by Simon Henry. I was previously assuming that category-sorted lex theories (at least for discrete categories) was the right notion of essentially algebraic theory. However, I was mistaken: category-sorted lex theories are less expressive than sorted algebraic theories. In particular, the requirement for theories to be identity-on-objects is too restrictive, as the codomains may no longer be finitely complete. An analogous definition of sorted essentially algebraic theory to sorted algebraic theory will look different (as far as I'm aware, no such characterisation has been given in the literature) and, as such, the existing theory–monad correspondences are insufficient to describe a correspondence for essentially algebraic theories. I'll continue to pursue this question according to this line of thinking, but unless anyone can see that I've missed something obvious, I'm satisfied that such a correspondence is at least not known (or easily derivable from results in the literature).<|endoftext|> TITLE: An "analytic continuation" of power series coefficients QUESTION [43 upvotes]: Cauchy residue theorem tells us that for a function $$f(z) = \sum_{k \in \mathbb{Z}} a(k) z^k,$$ the coefficient $a(k)$ can be extracted by an integral formula $$a(k) = \frac{1}{2\pi i}\oint f(z) z^{-k-1},$$ with a contour around zero. Now, there is nothing to prevent us from thinking of $a\colon \mathbb{Z} \to \mathbb{C}$ as a function over the complex domain, defined by the above integral. In this way, we have naturally "continued" the function $a(k)$ over the integers to one over the complex numbers. Is there any meaning to this beyond just something amusing? For example, consider $f(z)=\exp(z)$. In this case, $a(k)=1/k!$ which one would think would be continued to $1/\Gamma(1+k)$. But at least if you attempt to plot what happens above, this is not the case. We get a function with Bessel-like oscillations (blue) which nevertheless coincides with $1/\Gamma(1+k)$ (orange) on integral points! One can experiment with other choices of $f(z)$. When $f(z)=(z+1)^n$, it seems like the continuation actually coincides with what one would naturally think (the Beta function). Which is actually quite strange. For example for $n=5$ and $f(z) = 1 + 5 z + 10 z^2 + 10 z^3 + 5 z^4 + z^5$, how does this computation "learn" that the sequence $1,5,10,10,5,1$ actually corresponds to the binomial coefficients and thus should be continued to the Beta function? If we perturb one of the coefficients, say change $5 z^4$ to $2 z^4$, the result would look like a "perturbed" Beta function (blue below vs the Beta function in orange). Any explanation (or references in the literature) for what is going on here? REPLY [6 votes]: Edit (1/21/21): (Start) For your first example, a classic method of interpolation is related to fractional calculus. $$k!\; a(k) = k! \; \oint_{|z|=r} \frac{e^z}{z^{k+1}} \; dz = e^{-1}k! \; \oint_{{|z|=r}} \frac{e^{z+1}}{z^{k+1}} \; dz $$ $$= e^{-1}k! \; \oint_{|z-1|=1} \frac{e^{z}}{(z-1)^{k+1}} \; dz =e^{-1} D^k_{z=1} e^z.$$ Interpolating using a standard fractional integroderivative, of which there are several reps, $$\lambda! \; a(\lambda) = \; e^{-1} D_{z=1}^{\lambda} \; e^z = e^{-1} \; \sum_{n \ge 0} \frac{z^{n-\lambda}}{(n-\lambda)!}\; |_{z=1}$$ $$ = e^{-z} \; \sum_{n \ge 0} \frac{z^{n-\lambda}}{(n-\lambda)!}\; |_{z=1} = e^{-z} z^{-\lambda}\; E_{1,-\lambda}(z) \; |_{z=1},$$ where $E_{\alpha,\beta}(z)$ is the Mittag-Leffler function (general definition in Wikipedia, MathWorld; some history and applications), encountered very early on by anyone exploring fractional calculus. This method of interpolation gives the entire function $$ a(\lambda) = e^{-1} \; E_{1,-\lambda}(1) \frac{1}{\lambda!} = e^{-1} \; \sum_{n \ge 0} \frac{1}{(n-\lambda)!} \; \frac{1}{\lambda!}, $$ which matches the OP's first graph and extends it to any real or complex argument, giving, of course, $a(k) = 1/k!$ for $k=0,1,2, ...$. Similarly, the fractional calculus can be applied to usefully interpolate coefficients generated by $ D_z^n \; f(z)$ to those of $D_z^{\lambda} f(z)$, with care taken with defining branch cuts and contours of integration for the complex function for the point of evaluation. The fundamental example is Euler's integral rep of the beta function as depicted below. Another is the interpolation of the associated Laguerre polynomials to the confluent hypergeometric functions and, therefore, interpolations of the associated coefficients. (See links below.) Other productive methods of interpolation--all can be related to fractional calculus--are sketched below. (End) Edit 1/23/21: (Start) The interpolation you desire is obtained by applying Ramanujan's favorite Master Formula, a.k.a. Mellin transform interpolation, as discussed below and illustrated in four other MO-Qs--Q1, Q2, Q3, and Q4. This, naturally, amounts to replacing $k$ by $s$ only outside your Cauchy integral rather than within for the reason Terry Tao notes, which is equivalent to noting the Cauchy contour integration is not equal to the Mellin transform integration. (End) The Mellin transform pair allows for interpolation of the coefficients of generating functions, often directly connected to sinc and/or Newton interpolation. Basically, the following is a sketch of the analytics of Ramanujan's Master Formula/Theorem, which he so profoundly and elegantly wielded. Here $(a.)^n := a_n$ in the Taylor series $e^{a.x}$ is interpolated as $a_{-s}$ via the Mellin transform of the Taylor series $f(x) = e^{-a.x}$. First consider the normalized Mellin transform and its inverse $$F(s) = MT[f(x)] = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx$$ $$f(x) = MT^{-1}[F(s)] = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} F(s) \frac{x^{-s}}{(-s)!} \; ds .$$ Then the RMT holds for a class of functions such that the simple poles of the inverse sine in the inverse Mellin transform give $$f(x) = e^{-a.x} = \sum_{n \geq 0} \frac{(-a.x)^n}{n!} = \sum_{n=0} a_n \frac{(-x)^n}{n!} = \sum_{n=0} F(-n) \frac{(-x)^n}{n!} \; ,$$ that is, such that we may close the complex contour to the left (e.g., in the sense of the limit of a semicircle with its radius expanding to infinity) for $0 < \sigma < 1$ and $0 < x < 1$ when $F(s)$ has no singularities/poles within the contour. This rep allows an extension of the RMT (and the Mellin transform) to cases in which poles are present in $F(s)$ and to other ranges of $x$. Also note (see, e.g., Gelfand and Shilov's "Generalized Functions") the relation $$D_x^{m+n+1} \; H(x) \frac{x^m}{m!} = H(x) \frac{x^{-n-1}}{(-n-1)!} = \delta^{(n)}(x),$$ reflected in the two (of several) reps of the fractional differintegro op equivalent under analytic continuation $$\frac{x^{\alpha-\beta}}{(\alpha-\beta)!} = \frac{d^{\beta}}{dx^\beta}\frac{x^{\alpha}}{\alpha!}=\int_{0}^{x}\frac{z^{\alpha}}{\alpha!}\frac{(x-z)^{-\beta-1}}{(-\beta-1)!} dz = \frac{1}{2\pi i} \oint_{|z-x|=|x|}\frac{z^{\alpha}}{\alpha!}\frac{\beta!}{(z-x)^{\beta+1}}dz ,$$ with $H(x)$ the Heaviside step function and $\delta(x)$, the Diraac delta. (This is equivalent through binomial expansion to a sinc function/cardinal series interpolation of the binomial coefficients $\binom{s}{\alpha} = \sum_{n \geq 0} \binom{s}{n} \frac{\sin(\pi(\alpha-n))}{\pi(\alpha-n)}=\sum_{n \geq 0} \binom{s}{n} \binom{0}{\alpha-n} $, an instance also of the Chu-Vandermonde identity. The contour integral for this beta function integral is easily transformed into a bandlimited Fourier transform. See, e.g., the relation to Newton interpolation in the MSE_Q "Why is the Euler gamma function the best extension of the factorial function to the reals?" For more detail and connections, see my post "Fractional calculus and interpolation of generalized binomial coefficients." The integral reps of the confluent (see this MO-Q) and non-degenerate hypergeometric functions can be expressed effectually as this or easily related differintegral ops acting on simple functions. Years ago I found the Danish masters Niels Nielsen and Niels Norlund to be most informative on contour integrals in interpolation. There is also a connection to Riemann surfaces via Pochhammer's contour for the beta function integral.) So, under the conditions above, $$F(-n) = \int_{0}^{\infty} f(x) \; \frac{x^{-n-1}}{(-n-1)!} \; dx = \int_{0}^{\infty} e^{-a. x} \; \delta^{(n)}(x) \; dx = a_n,$$ and this suggests the analytic continuation and relation to umbral calculus $$F(s) = \int_{0}^{\infty} f(x) \; \frac{x^{s-1}}{(s-1)!} \; dx = \int_{0}^{\infty} e^{-a.x} \; \frac{x^{s-1}}{(s-1)!} \; dx = (a.)^{-s} = a_{-s}.$$ The iconic guiding example is the Euler gamma function integral rep with $(a.)^n = a_n = c^n$ $$ (a.)^{-s} = a_{-s} = c^{-s} = F(s) = MT[f(x)= e^{-c\; x}] = \int_{0}^{\infty} e^{-c \; x} \; \frac{x^{s-1}}{(s-1)!} \; dx = \frac{1}{c^{s}},$$ giving the interpolation of the coefficients of the Taylor series of $e^{cx}$, i.e., $a_n = c^n$, as $a_{-s}=c^{-s}=F(s)$. Another useful example, which vividly illustrates the relation to the Appell Sheffer sequences of umbral calculus (of which the $x^n$ with e.g.f. $e^{x}$ is the basic example), is the integral rep for (what I call) the Bernoulli function, simply related to the Hurwitz zeta function and generalizing the Bernoulli polynomials, $$ B_{-s}(z) = (B.(z))^{-s} = \int_{0}^{\infty} e^{-B.(z)t} \; \frac{t^{s-1}}{(s-1)!} \; dt $$ $$ = \int_{0}^{\infty} \frac{-t}{e^{-t}-1} \; e^{-zt} \frac{t^{s-1}}{(s-1)!} \; dt = s \; \zeta(s+1,z)$$ where the e.g.f. for the Bernoulli polynomials with $(b.)^n = b_n$ the Bernoulli numbers is $$e^{B.(x)t} = e^{(b.+x)t} = e^{b.t} e^{xt} = \frac{t}{e^t-1} \; e^{xt}.$$ Note that $$B_n(z) = -n \; \zeta(1-n,z),$$ $$B_n(1) = -n \; \zeta(1-n,1) =-n \; \zeta(1-n) (Riemann) = (-1)^n B_n(0) = (-1)^n b_n.$$ Through this characterization, it is not too difficult to show that the Bernoulli function inherits all the elegant properties of a regular Appell sequence, such as $D_z \; B_{s}(z) = s \; B_{s-1}(z)$. Hankel contour deformations, Hadamard finite part regularization of the Mellin transform guided by the inverse Mellin transform, the Mellin-Barnes contour integral, and other methods of analytic continuation can be used to extend the range of $s$ and other parameters, as have been done for the integral reps of the Euler gamma and beta functions and Riemann and other zeta functions and their generalizations. In addition, interchanging summation and integration often gives rise to useful asymptotic expansions of functions a la Borel, Heaviside, Hardy, and Poincare. Riemann knew all this stuff. Ramanujan intuited it. Hardy formalized it. (I stumbled across it on a journey starting from the ladder ops of QM and a brief comment by my old math prof Stallybrass about the sequence $D^{m+n} H(x) \frac{x^m}{m!}$ in his integral transforms class an eon ago.) For application to defining fractional powers of operators, see my answer and comments therein to the MO-Q "What does the inverse Mellin transform really mean?" and several of my blog posts, such as "The Creation / Raising Operators for Appell Sequences."<|endoftext|> TITLE: Smallest $S\subset \mathbb C$ on which no degree $k$ polynomial always vanishes? QUESTION [21 upvotes]: Say $p$ is a polynomial of degree $k$ in $\mathbb C[x]$. Then $p$ can have at most $k$ distinct roots. A somewhat obtuse way to state that is to say that among any set of $k+1$ distinct complex numbers, there must exist a value $a$ for which $p(a)\neq 0$. The question here has to do with generalizing the above fact to multivariate polynomials. Given some $p \in \mathbb C[x_1,\ldots,x_n]$, it turns out that whenever we have enough different values to choose from, we can always pick from these some $a_1,\ldots,a_n$ so as to give $p(a_1,\ldots,a_n)\neq 0$. Moreover, how many is "enough" is a function just of $n$ and the degree of $p$. I want to emphasize that, in the above and in what follows, the $a_1,\ldots,a_n$ so chosen are distinct. To make things more precise and specific, we have the following. Theorem.  Let $p \in \mathbb C[x_1,\ldots,x_n]$ be a polynomial of total degree $k$. Then any set of $n+k$ distinct complex numbers contains a subset of $n$ that can be assigned to the variables $x_1,\ldots,x_n$ so as to give a nonzero value for $p$. (Again, we are talking about choosing from the set $n$ distinct values that we assign to $x_1,\ldots,x_n$.) The proof of the above is a not-too-difficult exercise. But the question is: Question.  In the above theorem, can the $n+k$ be replaced with some other function of $n$ and $k$ so as to give a better upper bound? In fact, let's define $M(n,k)$ to be the smallest positive integer such that for every $p\in \mathbb C[x_1,\ldots,x_n]$ of total degree $k$, every set $S$ of at least $M(n,k)$ distinct complex numbers contains some subset $\{a_1,\ldots,a_n\}$ (of cardinality $n$, so that these are distinct values) such that $p(a_1,\ldots,a_n)\neq 0$. The theorem above states that $M(n,k) \le n+k$, but I don't even see how to meet that bound for $n=k=2$. REPLY [11 votes]: This is probably just another way to present Fedor Petrov's solution: Expand $$\frac{(1-t \alpha_1) (1-t \alpha_2) \cdots (1-t \alpha_{n+k-1})}{(1-t \beta_1)(1 - t \beta_2) \cdots (1-t \beta_n)}$$ as a formal power series in $t$. The coefficient of $t^k$ is a degree $k$ polynomial in the $\beta$'s, which vanishes whenever $\{ \beta_1, \beta_2, \ldots, \beta_n \}$ is a $n$-element subset of the $\alpha$'s.<|endoftext|> TITLE: Internal vs. external definability of inner models QUESTION [6 upvotes]: Suppose $\kappa$ is an inaccessible cardinal. Is the following situation consistent? There is $p \in V_\kappa$ and a formula $\phi(x)$ such that there is exactly one $M \subseteq V_\kappa$ such that $M$ is a transitive set of size $\kappa$ and $M \models \phi(p)$. The $M$ above is not a definable class in $V_\kappa$, meaning there is no $q \in V_\kappa$ and formula $\psi(x,y)$ such that $M = \{ x \in V_\kappa : V_\kappa \models \psi(x,q) \}$. REPLY [5 votes]: I will try to partially answer your question. I claim that if $\kappa$ is weakly compact then this situation is inconsistent: I will show that such an $M$ is necessarily definable in $V_\kappa$. Define the finite set of formulae $\Lambda:=\{\phi(x)\} \cup \text{tc}(\phi(x))$, where tc denotes the transitive closure (with respect to the 'proper sub-formula' relation). Use reflection in $M$ to find a transitive $q \prec_{\Lambda} M$ such that $p \in q$ and $q \in M$. For every $x \in V_\kappa$ we shall inductively construct a ${<}\kappa$-branching tree $T_x$. It will consist of sequences (always with a last element) of transitive, $\Lambda$-elementary submodels containing $x$ and satisfying $\phi(p)$: Set $\langle q \rangle$ to be the root of $T_x$. Assume that $\bar{y} \in T_x$ and $\bar{y}=\langle y_0, ... , y_\alpha\rangle$. Define the set of successors of $\bar{y}$ in $T_x$ as follows: $$\text{succ}_{T_x}(\bar{y}):=\{\bar{y}^{\frown} z \,\,\colon \,z \in V_{f(\bar{y})} \land y_\alpha \prec_{\Lambda} z \land x, y_\alpha \in z \land z \, \text{is transitive} \, \}$$ where $f(\bar{y}):=\max(\vert y_\alpha \vert^+ , \vert x\vert^+)$. In the limit case let $(\bar{y}_\alpha)_{\alpha < \gamma}$ ($\bar{y}_\alpha$ has length $\alpha +1$) be an increasing chain and set $\bar{y}_\gamma:=\langle y_0,...,y_\alpha,... \rangle ^\frown y_\gamma$, where $y_\gamma:= \bigcup_{\alpha < \gamma} y_\alpha$. I claim that $x \in M \Longleftrightarrow T_x \,\, \text{has height} \,\, \kappa$. Assume that $x \in M$. Using reflection in $M$, Löwenheim-Skolem and Mostowski collapse (and $q \in M$) one can easily show that $T_x$ must have height $\kappa$. On the other hand, assume that $T_x$ has height $\kappa$. Any branch through $T_x$ defines an increasing chain of transitive, $\Lambda$-elementary submodels. The direct limit of this chain is a transitive model $M' \subseteq V_\kappa$ of size $\kappa$, satisfying $\phi(p)$ and containing $x$. By your assumption $M'=M$ follows and so $x \in M$.<|endoftext|> TITLE: Nonbraided rigid monoidal category where left and right duals coincide QUESTION [5 upvotes]: In a braided rigid monoidal category $(\mathcal{M},\otimes)$ left and right duals coincide. What is an example of a rigid monoidal category where left and right duals coincide but there exist no braiding for the category? REPLY [6 votes]: The simplest example is G-graded vector spaces where G is a non-abelian group.<|endoftext|> TITLE: An upper bound for the G.C.D. of $\binom{a}{3}$ and $\binom{b}{3}$ QUESTION [6 upvotes]: I can't seem to find anything in the literature on how to estimate the g.c.d. of $\binom{a}{k}$ and $\binom{b}{k}$. In particular, I would like to know why $\gcd(\binom{a}{3}, \binom{b}{3})\leq b \binom{a-b}{3}$ for $a-b\geq 4$. I'm confident it's true (computer search). REPLY [2 votes]: I will hopefully prove this statement below modulo a finite check (I believe this will also work to show that $\gcd(\binom{b}{3}, \binom{a}{3}) \le \varepsilon b(a-b)^3$ for any $\varepsilon > 0$ except for a finite set of counterexamples). Denote for simplicity $a - b = c$ and $\gcd(\binom{b}{3}, \binom{a}{3}) = d$. First we will show that $d \ll c^5$. Indeed, each divisor of $d$ comes from one of the pairs $(a-k, b - l)$, where $k, l = 0,1, 2$. Each such pair gives us as a contribution a divisor of $a - b + l - k$ and there are five such numbers from $a - b - 2$ to $a - b + 2$. Note also that (up to a finite number of divisors like extra $6$ or so) different pairs with the same $l - k$ will contribute to the different prime divisors of $a - b + l - k $ because it will otherwise divide $\gcd(a - k, a - k_1) | (k - k_1)$. So we get $\ll (a - b - 2)\ldots (a - b + 2) \ll c^5$. Therefore, if $c \le \delta \sqrt{a}$, then we are done. Indeed, in that case $d \ll c^5 \le c^3\delta^2 a$ while $b\binom{c}{3} \asymp ac^3$. Thus, $c \ge \delta \sqrt{a}$ for some fixed positive $\delta$. Now assume that $c \ge C a^{2/3}$ for big enough $C$. Then $b\binom{c}{3} \gg C^3ba^2 \gg C^3b^3$ which is bigger than $\binom{b}{3}$ for big enough $C$ and we are done (here we used the obvious fact that $d\le \binom{b}{3}$). So $c \le Ca^{2/3}$ for some fixed positive $C$. In particular, $c = o(a)$ so $b \sim a$. Now we will use the ingenious observation of GH from MO that $$(2a - b - 1)\binom{b}{3} - (2b - a - 1)\binom{a}{3} = (a + b - 2)\binom{c + 1}{3}.$$ Note that if $\gcd(2a-b-1, 2b - a - 1) \ge C$ then we are done. Indeed, in that case we can divide by this $\gcd$ and get that $d \ll \frac{(a+b-2)c^3}{C}$. Since now $a\sim b$ we get that this is $\ll \frac{bc^3}{C}$ which is what we need for big enough $C$ (being a bit more careful we can show that it is enough to consider cases with $\gcd(2a-b-1, 2b-a-1) \le 2$). We have $$\gcd(2a-b-1, 2b - a - 1) = \gcd(2a-b-1, 3(a-b)) \ge \gcd(2a-b-1, a-b)=\\ =\gcd(a-1, a - b) =\gcd(a-1, b-1).$$ Thus, $a-1$ and $b-1$ are almost coprime. Similar to the above computation we can get that (up to the division by some uniformly bounded number) $d = (a+b-2)\binom{c+1}{3}$. We have here four factors: $(c-1), c, c+1, a+b-2$. $c-1$ can come from gcd of $a-1,b$ or gcd of $a-2, b-1$. $c$ can come from gcd of $a, b$; $a-2, b-2$ or $a-1, b-1$, but the last case can be excluded since $\gcd(a-1, b-1) \le C$. $c + 1$ can come from gcd of $a, b-1$ or gcd of $a-1, b-2$. Finally, $a+b-2$ can come from gcd of $a, b-2$; $a-2, b$ or $a-1, b-1$, but the last case we can again exclude. Assume that $c-1$ splits between gcd of $a-1, b$ and gcd of $a-2, b-1$ as $r_1$ and $s_1$. Similarly for the next ones we would have $r_2, s_2$, $r_3, s_3$ and $r_4, s_4$. Observe that different numbers among $r_1, \ldots , s_4$ corresponding to the same number among $a, a-1, a-2, b, b-1, b-2$ will be almost coprime in a sense that their $gcd$ is at most some constant (for $r_1, \ldots , s_3$ it is obvious because $c-k, c-l$ are almost coprime and with $r_4, s_4$ e.g. when we look at $r_1$ and $s_4$ corresponding to $b$ we have $\gcd(r_1, s_4) \mid \gcd(a-b-1, b, a+b-2) = 1$. Other cases are similar). Note that $r_1s_1 \sim r_2s_2 \sim r_3s_3 \sim c$ and $r_4s_4 \sim a$. Let's look at the numbers $b, b-2, a, a-2$: For $b$ we have by almost coprimeness $r_1r_2s_4 \ll a$, For $b-2$ we have $s_2s_3r_4 \ll a$, For $a$ we have $r_2r_3r_4 \ll a$, For $a-2$ we have $s_1s_2s_4 \ll a$. Multiplying these inequalities we get $c^4a^2 \ll a^4$, that is $c \ll \sqrt{a}$. So, after all this reasoning, we got that $\sqrt{a} \ll c \ll \sqrt{a}$ so $c\sim \sqrt{a}$! I guess one can reach the same conclusion from the assumption $d \ll \varepsilon b\binom{c}{3}$. Now, we have two numbers which are (almost) divisible by our gcd: $\binom{c+2}{5}$ and $(a+b-2)\binom{c+1}{3}$. Note that both of them are proportional to the required $b\binom{c}{3}$. Thus, they are almost the same in a sense that $n\binom{c+2}{5} = m(a+b-2)\binom{c+1}{3}$ for some integers $n, m \le C$. Dividing by $\binom{c+1}{3}$ we get $r(c+2)(c-2) = (a+b-2)$ for some $r\in \mathbb{Q}$, $r > 0$ with bounded denominator and numerator. So $a$ and $b$ are some second degree polynomials of $c$: $$a = \frac{r(c+2)(c-2) + c + 2}{2}, b = \frac{r(c+2)(c-2)-c+2}{2}.$$ Therefore $\binom{a}{3}$ and $\binom{b}{3}$ are polynomials of degree $6$ in $c$. It remains to show that their $\gcd$ (as polynomials in $\mathbb{Q}[x]$) has degree at most $4$ – in that case $\gcd(\binom{a}{3}, \binom{b}{3}) \ll c^4$ while $b\binom{c}{3} \sim c^5$. Since $\binom{a}{3}$ is a product of three quadratic polynomials, if $\gcd$ has degree at least $5$ then all factors should be used, in particular $a-1 = \frac{r(c+2)(c-2) + c}{2}$. If it is not coprime to some $b-k$ then their $\gcd$ divides their difference which is $\frac{2c + 2k - 2}{2}$. Thus, $c = 1 - k$ is a root of $a-1$ for $k = 0$, $k = 1$ or $k = 2$. Case $c = 0$ gives us $r = 0$, $c = -1$ gives us $r = -\frac{1}{3}$ and $c = 1$ gives us $r = \frac{1}{3}$. In all three cases the degree of gcd is at most $4$ and the proof is complete.<|endoftext|> TITLE: Inner automorphisms of group algebras vs. inner automorphisms of the group QUESTION [6 upvotes]: In a recent question on MSE I asked about conditions under which the canonical morphism $Out(G) \to Out(k[G])$ is injective. Is it true that this morphism is indeed injective if $G$ is finite and $k=\mathbb{Z}$ ? What I already know: It's not injective if $k$ is a field in non-modular characteristic. The kernel is $Out_c(G)$, the group of conjugacy class preserving automorphisms. That is Captain Lama's answer to my question. It is true if $k=\mathbb{Z}$ and $G$ is nilpotent. That's my own answer to the question. In the mean time I have also found out that the morphism is injective if $G$ is rational, i.e. if $g$ is cojugated to $g^k$ for every $g\in G$ and every $k\in\mathbb{Z}$ with $gcd(k,ord(g))=1$, e.g. $G$ a symmetric group. Proof: If $\alpha\in Aut(G)$ is in the kernel of the morphism, say $\alpha(g)=ugu^{-1}$ for some $u\in\mathbb{Z}G^\times$, then $$u^\ast u g=u^\ast \alpha(g) u = u^\ast \alpha(g^{-1})^{-1} u = (u^\ast \alpha(g^{-1}) u)^\ast = (u^\ast u g^{-1})^\ast = g u^\ast u$$ where $^\ast$ is the antiautomorphism of the group algebra with $g\mapsto g^{-1}$. Hence $u^\ast u \in\mathbb{Z}G$. We want to prove that $u^\ast u = 1$. Consider all the complex characters $\chi\in Irr(G)$ and their central characters $\omega_\chi: Z(\mathbb{C}G) \to \mathbb{C}$. We can pick a matrix representations $\rho_\chi: G\to GL_n(\mathbb{C})$ affording $\chi$ with $\rho(G)\subseteq U_n(\mathbb{C})$ so that $u^\ast u$ is mapped to some self-adjoint and positive definite matrix so that $\omega_\chi(u^\ast u)\in\mathbb{R}_{>0}$. Furthermore $u^\ast u\in Z(\mathbb{Z}G)$ is integral over $\mathbb{Z}$ so that $\omega_\chi(u^\ast u)$ is an algebraic integer. Moreover it must be in $\omega_\chi(Z(\mathbb{Q}G)) =\mathbb{Q}(\chi)$. Now if $G$ is rational, then all characters have rational values so that $\omega_\chi(u^\ast u)$ is a real, positive, rational, invertible integer. In other words $\omega_\chi(u^\ast u) = 1$, i.e $\rho_\chi(u^\ast u) = 1_{n\times n}$. Since $\chi$ was arbitrary, $u^\ast u=1$. The only units of $\mathbb{Z}G$ with $u^\ast u=1$ are elements of the form $\pm g$ so that $\alpha\in Inn(G)$ as we wanted. QED. Note that we can get the same conclusion under weaker conditions. For example if $\mathbb{Q}(\chi)=\mathbb{Q}(i)$, then the only real, positive, integral unit is also 1. REPLY [4 votes]: The question for finite $G$ and $k = \mathbb{Z}$ is the normalizer problem, see [1, Section 1]. By a result of Jan Krempa, the kernel of the cannonical morphism is in that case always an elementary abelian $2$-group. As far as I know, there is basically only one example known where the kernel is non-trivial [1, Theorem A]. This example was constructed by Martin Hertweck and used to provide a counterexample to the isomorphism problem for integral group rings. [1] Martin Hertweck; A counterexample to the isomorphism problem for integral group rings. Ann. of Math. (2) 154 (2001), no. 1, 115–138, MR1847590.<|endoftext|> TITLE: Tight apartness relations in toposes QUESTION [8 upvotes]: A tight apartness relation on a set is a binary relation $\#$ such that the following conditions hold: $x = y$ if and only if $\neg (x \# y)$. If $x \# y$, then $y \# x$. If $x \# z$, then either $x \# y$ or $y \# z$ for every $y$. I want to understand this notion better. Classically, it is completely trivial. Thus, it makes sense to look at it in various toposes. I tried to use the Kripke-Joyal semantics to get the external interpretation of an arbitrary object of a topos with a tight apartness relation, but it seems that it does not give anything particularly interesting in general. Thus, I've got the following question: Question: What are examples of objects in toposes with a tight apartness relation which externally correspond to some interesting or useful notion? Since constructively, a set can have more than one tight apartness relation on it, I'd like to see examples of an object with two different tight apartness relations which both have interesting interpretations. Edit: There are several "generic" examples of objects with tight apartness relations, i.e., objects that can be defined in every topos (e.g., Dedekind reals). I'm particularly interested in "non-generic" examples, i.e., objects that can be constructed only in a specific topos. REPLY [9 votes]: I'm not precisely sure what you're looking for. Here is an example for the external interpretation of an apartness relation: Recall that the object of Dedekind reals $\mathbb{R}$ in a sheaf topos $\mathrm{Sh}(X)$ is the sheaf $\mathcal{C}$ of continuous (Dedekind-)real-valued functions on $X$. The apartness relation on $\mathbb{R}$, defined by $x \mathrel{\#} y \Longleftrightarrow x - y \text{ is invertible}$, is then the following subsheaf $\mathcal{E}$ of $\mathcal{C} \times \mathcal{C}$: $$ \mathcal{E}(U) = \{ (f,g) \,|\, \text{for all $x \in U$: $f(x) - g(x) \in \mathbb{R}$ is invertible} \} $$<|endoftext|> TITLE: Does every $SL_2\mathbb{C}$ representation of a closed oriented surface extend over a compact oriented three-manifold? QUESTION [14 upvotes]: Let $F$ be a compact oriented surface and $\rho:\pi_1(F)\rightarrow SL_2\mathbb{C}$ be a representation. Does there exist a compact oriented three-manifold $M$ with $\partial M=F$ and a homomorphism $\tilde{\rho}:\pi_1(M)\rightarrow SL_2\mathbb{C}$ so that the restriction of $\tilde{\rho}$ to $\pi_1(F)$ is equal to $\rho$? If not is there an obstruction that allows you to identify the representations that do extend? For instance let $BSL_2\mathbb{C}^\delta$ denote the classifying space of $SL_2\mathbb{C}$ as a discrete group. Corresponding to $\rho:\pi_1(F)\rightarrow SL_2\mathbb{C}$ is a continuous map $f:F\rightarrow BSL_2\mathbb{C}^\delta$. If $\rho$ extends over a three-manifold $M$ then the homology class represented by $f_*[F]$ is zero. Is there a computable way to detect this? REPLY [4 votes]: This is an extended comment on the algorithmic question. As Moshe points out, the lack of realization as a boundary follows from the Baire category theorem. On the other hand, how does one recognize when an element is not in this infinite countable union of subspaces? Let's consider the genus 1 case $F=T^2$. Then a rep $\rho:\pi_1(F)\to SL_2(\mathbb{C})$ is determined by two eigenvalues of generators $(\mu, \lambda)$ (assuming the representation is not unipotent). In turn if $F=\partial M$, the boundary of a 3-manifold, then there is an associated $A$-polynomial $A(x,y) \in \mathbb{Z}[x,y,x^{-1},y^{-1}]$ such that $A(\mu,\lambda)=0$. I suspect that $[\rho]=0\in H_2(SL_2(\mathbb{C})^\delta)$ iff $[\rho]$ extends to a representation of a 3-manifold, but I haven't checked this. In any case, one sees that $\mu, \lambda$ are algebraically related in the bounding case. However conversely, if $\mu,\lambda$ satisfy an algebraic relation, it's not clear to me that this implies that $[\rho]=0$, since in general A-polynomials satisfy some non-trivial conditions. I suspect there could be a formulation in terms of algebraic K-theory, but I don't know enough about this. One may also ask if $H_2(SL_2(\mathbb{C})^\delta)$ is generated by representations of $T^2$? I suspect this might be true. It's not hard to see that a representation of a closed surface of genus $>2$ is cobordant to a sum of representations of genus 2 surfaces (since the commutator map in $SL_2(\mathbb{C})$ is onto). Then I think that genus 2 reps. may be cobordant to a pair of genus 1 reps. (at least the numerology works out, but I haven't checked it). Then one could ask for when a sum of genus 1 reps. is homologically trivial? In turn, this should be realized by a zero of an A-variety. But I'm not sure how one recognizes such points.<|endoftext|> TITLE: Geometry of Level sets of elliptic polynomials in two real variables QUESTION [5 upvotes]: Updated: A polynomial $P(x,y)\in \mathbb{R}[x,y]$ is called an elliptic polynomial if its last homogeneous part does not vanish on $\mathbb{R}^2\setminus\{0\}$.The two answers to this post provide a proof for the following theorem: Theorem: If $p$ is an elliptic polynomial whose last homogeneous part is positive definitive, then for $c$ sufficiently large , $p^{-1}(c)$ is a simple closed curve. Moreover if the centroid of interior of $p^{-1}(c)$ is denoted by $e_c$ then $e_c$ is convergent as $c$ goes to $+\infty$. The limit $\lim_{c\to \infty} e_c$ can be written in terms of coefficients of $p$. If we drop the ellipticity condition then this convergence result is not necessarily true. The previous version of the post: Is there a polynomial function $P:\mathbb{R}^2 \to \mathbb{R}$ with the following property? For sufficiently large $c>0$, $P^{-1}(c)$ is a simple closed curve $\gamma_c$, homeomorphic to $S^1$, but as $c$ goes to $+\infty$. the centroid $e_c$ of the interior of $\gamma_c$ does not converge to any point of $\mathbb{R}^2$. Motivation: The answer is negative if we consider this question for polynomials $p:\mathbb{R} \to \mathbb{R}$ whose eventual level sets are $2$-pointed set, i.e. $S^0$.(Namely a polynomial of even degree). The motivation comes from line -3, item III, page 4 of Taghavi - On periodic solutions of Liénard equations, which can be generalized to every even degree polynomial with one variable. REPLY [7 votes]: Concerning homogeneous polynomials: Let $P(x,y)=\sum_{j=0}^n a_j x^j y^{n-j}$ be such a polynomial, of degree $n$ such that $C:=P^{-1}(\{c\})$ is a simple closed curve for all large enough $c>0$. If $n$ is odd, then every line through the origin will have at most one point of intersection with $C$. So, then $C$ cannot be a simple closed curve for any real $c$ -- because every line through any point interior to a simple closed curve must intersect the curve in at least two points. It remains to consider the case when $n$ is even. Then $C$ is symmetric about the origin, and hence so is the interior of $C$. Then the centroid of the interior is the origin, and it does not depend on the level $c$. Consider now the case of an elliptic polynomial \begin{equation*} P(z)=P(x,y)=\sum_{j=0}^n a_j x^j y^{n-j}+\sum_{j=0}^{n-1}b_j x^j y^{n-1-j}+K|z|^{n-2} \end{equation*} of (necessarily even) degree $n$, where $z:=(x,y)$ and $K=O(1)$ (as $|z|\to\infty$). The ellipticity here is understood as the following condition: \begin{equation*} \min_{|z|=1}\sum_{j=0}^n a_j x^j y^{n-j}>0. \end{equation*} For any $d_*\in(0,1)$ and any real $D>0$, let $\mathcal P_{n,d_*,D}$ denote the set of all polynomials $p(x)=\sum_{j=0}^n d_j x^j$ such that $d_n\ge d_*$ and $\sum_{j=0}^n|d_j|\le D$. Then it is not hard to see that there is a real $c_*(n,d_*,D)>0$, depending only on $n,d_*,D$, such that for any polynomial $p(x)=\sum_{j=0}^n d_j x^j$ in $\mathcal P_{n,d_*,D}$ and for all real $c\ge c_*(n,d_*,D)$ the equation $p(x)=c$ has exactly two roots $x_\pm:=x_\pm(c)$ such that $x_-<00. \end{equation*} So, by (1.6), wlog (1.5) holds indeed. Let us now turn back to the elliptic polynomial $P(x,y)$. For each real $t$ consider the polynomial \begin{equation*} p_t(r):=P(r\cos t,r\sin t). \end{equation*} By the ellipticity of the polynomial $P(x,y)$, there exist $d_*\in(0,1)$ and a real $D>0$ such that $p_t\in\mathcal P_{n,d_*,D}$ for all real $t$. Take now any real $c\ge c_*(n,d_*,D)$. Then, by the paragraph right above, for each real $t$ the equation $p_t(r)=c$ has exactly two roots $r_\pm(t):=r_\pm(c;t)$ such that $r_-(t)<00&\text{ for }t\in[0,\pi],\\ |r_-(t-\pi)|>0&\text{ for }t\in[\pi,2\pi]. \end{cases} \end{equation*} So, the level curve $C$ is closed and simple, and its interior is \begin{equation*} I(c):=\{r\,(\cos t,\sin t)\colon0\le r0$ between the rays $t$ and $t+dt$ is at distance $\frac23\,r$ from the origin, (ii) the area of this sector is $\frac12\,r^2\,dt$, and (iii) the centroid of the union of the two sectors is the weighted average of the centroids of the two sectors, with weights adding to $1$ and proportional to the areas of the sectors, and thus proportional to the squared radii of the sectors. Simplifying (2), we get \begin{equation*} d(t)\sim-\frac{2b(t)}{na(t)}. \end{equation*} Averaging now over all the pairs of opposite infinitesimal sectors, we see that the centroid converges to \begin{align*} &-\int_0^\pi dt\,\frac{2b(t)}{na(t)}(\cos t,\sin t)\frac12\,\Big(\frac c{a(t)}\Big)^{2/n} \Big/\int_0^\pi dt\,\frac12\,\Big(\frac c{a(t)}\Big)^{2/n} \\ &=-\int_0^\pi dt\,\frac{2b(t)}{na(t)}(\cos t,\sin t)\Big(\frac1{a(t)}\Big)^{2/n} \Big/\int_0^\pi dt\,\Big(\frac1{a(t)}\Big)^{2/n}. \tag{3} \end{align*} I have checked this result numerically for $P(x,y)=x^4 + y^4 + 3 (x - y)^4 + y^3 + x y^2 + 10 x^2$, getting the centroid $\approx(-0.182846, -0.245149)$ for $c=10^4$ and $\approx(-0.189242,-0.25)$ for the limit (as $c\to\infty$) given by (3). From the above reasoning, one can see that the distance of the centroid from its limit is $O(1/c^{1/n})$; so, the agreement in this numerical example should be considered good, better than expected. One may also note that in general the level sets $P^{−1}([0,c])$ will not be convex, even if $P$ is a positive elliptic homogeneous polynomial. E.g., take $P(x,y)=(x−y)^2(x+y)^2+h(x^4+y^4)$ for a small enough $h>0$. Here is the picture of this level set for $c=1$ and $h=1/10$: Clearly, the shape of this level set does not depend on $c>0$. This non-convexity idea can be generalized, with $$P(x,y)=P_{k,h}(x,y) :=\prod_{j=0}^{2k-1}\Big(x\cos\frac{\pi j}k-y\sin\frac{\pi j}k\Big)^2+h(x^{4k}+y^{4k})$$ for natural $k$ and real $h>0$. Here is the picture of the curve $P_{k,h}^{-1}(\{1\})$ for $k=5$ and $h=(3/10)^{4k}$:<|endoftext|> TITLE: Embedding of a group into a simple group in which every element is a commutator QUESTION [8 upvotes]: It is known that any group $G$ can be embedded into a simple group $S$, see, e.g., the discussion at Can any group be embedded in a simple group? My question is whether one can get an embedding such that the ambient group $S$ is of commutator width $1$ (i.e., each element of $S$ can be represented as a single commutator). REPLY [12 votes]: Here is a construction. Every finite group with $n$ elements embeds into $A_{2n}$ (and even $A_{n+2}$) which is simple if $n>2$ and of commutator width 1 (as any other finite simple group by the Ore conjecture proved by Martin W. Liebeck, E. A. O'Brien, Aner Shalev, Pham Huu Tiep, although for $A_n$ it was probably known to Frobenius and certainly to Ore in 1951). Suppose that the group $G$ is infinite. Using HNN extensions embed your group into a group $S$ where all pairs of elements of the same order are conjugate (this is the standard application of HNN extensions: use HNN extensions with cyclic associated subgroups to make $G TITLE: Is $\mathbb{C}^n$ rigid? QUESTION [8 upvotes]: Let $\pi:X\to S$ be a smooth family of complex manifolds (in the sense of deformation theory) such that $\pi^{-1}(0)\cong\mathbb{C}^n$ and $S\subset \mathbb{C}$ is a neighborhood of $0$. Is $\pi$ trivial? That is, is $X\cong \mathbb{C}^n\times S$ possibly after shrinking $S$? I know that a smooth family of compact complex manifolds with $H^1(M,\mathcal{T}_M)=0$ is trivial (where $M=\pi^{-1}(0)$) but I'm not sure whether this extends to this non-compact situation. REPLY [17 votes]: Example. $\pi: \{(z,w)\in {\mathbb C}^2: |zw|<1\}\to {\mathbb C}$, $\pi(z,w)=z$. Edit. Similarly, to get a nontrivial deformation of ${\mathbb C}^n$, consider $$ X=\{(z_0, z_1,...,z_n)\in {\mathbb C}^{n+1}: |z_0 z_1|<1\} $$ and let $\pi$ be the projection of $X$ to ${\mathbb C}$ which the 1-st coordinate line in ${\mathbb C}^{n+1}$. Then $\pi^{-1}(t)$ (for $t\ne 0$) will be biholomorphic to $B \times {\mathbb C}^{n-1}$, where $B\subset {\mathbb C}$ is an open disk. Hence, these fibers are not biholomorphic to $\pi^{-1}(0)={\mathbb C}^{n}$. A better question, I think, is if every Stein manifold of positive dimension admits a nontrivial deformation. I suspect that this is known.<|endoftext|> TITLE: Every complex number has a square root via LLPO without weak countable choice QUESTION [6 upvotes]: Is it possible to prove that every complex number has a square root using analytic LLPO, but avoiding Weak Countable Choice or Excluded Middle? Unique Choice is allowed. (Analytic LLPO is the statement that given any pair of real numbers $x$ and $y$, either $x \leq y$ or $x \geq y$. This statement is non-constructive, but still weaker than other statements like LPO or Excluded Middle.) It is true in Johnstone's Topological Topos. This is because the Fundamental Theorem of Algebra is true for Cauchy Real numbers, and Cauchy reals are isomorphic to Dedekind reals in the Topological Topos. But this reasoning doesn't seem to work unless Cauchy is iso to Dedekind. REPLY [7 votes]: Yes it is but there is no extensional square root function unless we also have LPO. Note that the squaring function is a bijection from $Q_{+} = \{x + iy \mid x \geq 0, y \geq 0\}$ onto $H_{+} = \{x + iy \mid y \geq 0\}$. Similarly it is a bijection from $Q_{-} = \{x + iy \mid x \geq 0, y \leq 0\}$ onto $H_{-} = \{x + iy \mid y \leq 0\}$. Given LLPO, $H_{+} \cup H_{-} = \mathbb{C}$ and that is enough to prove the existence of square roots. The confusing part is that for a negative real number $x$, we obtain the square root $+ i\sqrt{-x}$ or $-i\sqrt{-x}$ depending on whether LLPO gives $x \in H_{+}$ or $x \in H_{-}$. So this does not give an extensional square root function. Interestingly, this argument gives the stronger conclusion that every complex number has a square root in $$Q_{+} \cup Q_{-} =\{ x + iy \mid x \geq 0\}. $$ This stronger statement turns out to be equivalent to LLPO. Indeed, given a small enough real number $x$, if the square root of $-1+ix$ in $Q_{+} \cup Q_{-}$ is close to $i$ then $x \geq 0$ and similarly if that square root is close to $-i$ then $x \leq 0$.<|endoftext|> TITLE: Is there a strictly increasing differentiable function maps positively measurable set to zero measure set? QUESTION [10 upvotes]: Let $g(t)$ be a strictly increasing differentiable function. Can it map positively measurable set to zero measurable set? It's obviously that $\{g'>0\}$ is dense. If I can prove that the Lebesgue measure $m(\{g'=0\}) = 0$, then for every set with positive measure, there will be a positively measurable subset with $g'>\epsilon$ on it, and consequently maps the set to nonzero measure set(It's a theorem and I forget it's name). The question derives from my textbook, which says if $g(t)$ is a strictly increasing differentiable function and Riemann integrable, and $f(x)$ is Riemann integrable, then $$\int f(x) = \int f(g(t))g'(t)$$ All functions defined on suitable closed set. It seems that $f(g(t))$ may even be not integrable and that is totally a typo. But with my intuition of measure theory, this might be true since $g$ is differentiable. REPLY [17 votes]: There are strictly increasing $C^1$ functions that map sets of positive measure to sets of measure zero. Here is a construction: Let $C\subset [0,1]$ be a Cantor set of positive measure. For a construction, see https://en.wikipedia.org/wiki/Smith-Volterra-Cantor_set. Let $g(x)=\operatorname{dist}(x,C)$. The function $g$ is clearly continuous and equal zero on $C$. In fact $g$ is a $1$-Lipschitz function. Let $$ f(x)=\int_0^x g(t)\, dt. $$ The function $f$ is $C^1$ and it is strictly increasing. Indeed, if $y>x$, then $$ f(y)-f(x)=\int_x^y g(t)\, dx>0 $$ because the interval $[x,y]$ is not contained in the Cantor set $C$ and therefore it contains an interval where $g$ is positive. On the other hand $f'=g=0$ on $C$ which has positive measure and $f(C)$ has measure zero since $m(f(C))=\int_C f'(t)\, dt=\int_C g(t)\, dt=0$. As was pointed out by Mateusz Kwaśnicki in his comment, this construction gives the following result: Theorem. Let $f$ be as above. Then there is a Riemann integrable function $h$ such that $h\circ f$ is not Riemann integrable. Proof. The set $f(C)$ is homeomorphic to the Cantor set ($f$ is strictly increasing so it is a homeomorphism) and has measure zero as explained above. Let $$ h(x)=\begin{cases} 1 & \text{if $x\in f(C)$}\\ 0 & \text{if $x\not\in f(C)$.} \end{cases} $$ The function $h$ is Riemann integrable with the integral equal zero since it is bounded and continuous outside the set $f(C)$ of measure zero (because $\mathbb{R}\setminus f(C)$ is open and $h=0$ there). However, $$ (h\circ f)(x)=\begin{cases} 1 & \text{if $x\in C$}\\ 0 & \text{if $x\not\in C$.} \end{cases} $$ is not Riemann integrable since it is discontinuous on a set $C$ of positive measure. $\Box$<|endoftext|> TITLE: When are projective modules closed under highly-filtered colimits? QUESTION [5 upvotes]: Let $R$ be a ring. Let $Mod(R)$ be the category of left $R$-modules, and let $Proj(R) \subseteq Mod(R)$ be the full subcategory of projective $R$-modules. Let's say that $R$-projectives are closed under highly-filtered colimits if there exists a cardinal $\kappa$ such that $Proj(R)$ is closed under $\kappa$-filtered colimits in $Mod(R)$. For example, When $R$ is a field (or division ring), we have $Proj(R) = Mod(R)$, and so vacuously we have that $R$-projectives are closed under highly-filtered colimits. When $R = \mathbb Z$, the question (perhaps surprisingly) depends on set theory Under the anti-large cardinal hypothesis $V=L$, Eklof and Mekler showed that there exist aribtrarily large non-free abelian groups all of whose smaller subgroups are free and it follows that $\mathbb Z$-projectives are not closed under highly-filtered colimits. Whereas if there exists a strongly compact cardinal, then it follows from the powerful image theorem of Makkai and Pare that free abelian groups are an accessible, accessibly embedded subcategory of all abelian groups, and in particular $\mathbb Z$-projectives are closed under highly-filtered colimits. The positive result of Makkai and Pare generalizes immediately to show that if $R$ is a PID (in the noncommutative case I'm not sure of the terminology, but the hypothesis is that every submodule of a projective $R$-module is free), then assuming the existence of a sufficiently-large strongly compact cardinal (I think the cardinal just has to be bigger than the cardinality of $R$) we have that $R$-projectives are closed under highly-filtered colimits. In fact, Rosicky and Brooke-Taylor's "$\lambda$-pure" version of the powerful image theorem shows that if there is a sufficiently large strongly compact cardinal, then projective $R$-modules are closed under highly-filtered colimits provided that $R$ is of global projective dimension $\leq 1$ (every submodule of a projective module is projective rather than free), so this result holds even for some $R$ which are not domains. Question: What are some other rings $R$ for which we can say (perhaps under set-theoretical hypotheses) whether $R$-projectives are closed under highly-filtered colimits? Are there any rings $R$ other than division rings for which the question "are $R$-projectives closed under highly-filtered colimits?" is decidable in ZFC? Is the precise large-cardinal strength of "$\mathbb Z$-projectives are closed under highly-filtered colimits" -- or (equivalently, I think) "any sufficiently-large abelian group all of whose smaller subgroups are free, is free" -- known? REPLY [8 votes]: Let $R$ be a ring and $\kappa$ be a strongly compact cardinal such that $|R|<\kappa$. Then the class of all projective $R$-modules is closed under $\kappa$-filtered colimits. This is Theorem 3.3 in the recent preprint of J. Šaroch and J. Trlifaj "Test sets for factorization properties of modules", https://arxiv.org/abs/1912.03749 For any left perfect ring $R$ (which includes both all left Artinian and all right Artinian rings), all flat left $R$-modules are projective, so the class of projective left $R$-modules is closed under $\aleph_0$-filtered colimits. For example, this applies to all the rings $\mathbb Z/n\mathbb Z$, $n\ge2$ (many of which are not division rings).<|endoftext|> TITLE: Unexpected behavior involving √2 and parity QUESTION [24 upvotes]: This post makes a focus on a very specific part of that long post. Consider the following map: $$f: n \mapsto \left\{ \begin{array}{ll} \left \lfloor{n/\sqrt{2}} \right \rfloor & \text{ if } n \text{ even,} \\ \left \lfloor{n\sqrt{2}} \right \rfloor & \text{ if } n \text{ odd.} \end{array} \right.$$ Let $f^{\circ (r+1)}:=f \circ f^{\circ r}$, consider the orbit of $n=73$ for iterations of $f$, i.e. the sequence $f^{\circ r}(73)$: $$73, 103, 145, 205, 289, 408, 288, 203, 287, 405, 572, 404, 285, 403, 569, 804, 568, 401, \dots$$ It seems that this sequence diverges to infinity exponentially, and in particular, never reaches a cycle. Let illustrate that with the following picture of $(f^{\circ r}(73))^{1/r}$, with $2000$, together with the miniman such $r$ (in red): In fact all these numbers (as first terms) reach the following cycle of length $33$: $$(15,21,29,41,57,80,56,39,55,77,108,76,53,74,52,36,25,35,49,69,97,137,193,272,192,135,190,134,94,66,46,32,22)$$ except the following ones: $$7, 8, 9, 10, 12, 13, 14, 18, 19, 20, 26, 27, 28, 38, 40, 54,$$ which reach $(5,7,9,12,8)$, and that ones $1, 2, 3, 4, 6$ which reach $(1)$, and $f(0)=0$. If the pattern continues like above up to infinity, they must have infinity many such $n$. Bonus question: Are there infinitely many $n$ reaching a cycle? Do they all reach the above cycle of length $33$ (except the few ones mentioned above)? What is the formula of these numbers $n$? Below is their counting function (it looks logarithmic): REPLY [4 votes]: Here is a heuristic answer inspired by this comment of Lucia. First, let assume that the probabilty for an integer $n$ to be odd is $\frac{1}{2}$, and that the probabilty for $f(n)$ to be odd when $n$ is even (resp. odd) is also $\frac{1}{2}$. We will observe that (surprisingly) it is no more $\frac{1}{2}$ for $f^{\circ r}(n)$ when $r \ge 2$ (in some sense, the probability does not commute with the composition of $f$ with itself). if $n$ and $m=f(n)$ are even: note that $\frac{n}{\sqrt{2}} = m+\theta$ (with $0 < \theta < 1$) so that $m=\frac{n}{\sqrt{2}}- \theta$, then $$f^{\circ 2}(n) = f(m) = \left \lfloor{\frac{m}{\sqrt{2}}} \right \rfloor = \left \lfloor{\frac{\frac{n}{\sqrt{2}}- \theta}{\sqrt{2}}} \right \rfloor = \left \lfloor \frac{n}{2} - \frac{\theta}{\sqrt{2}}\right \rfloor$$ but $\frac{n}{2}$ is even with probability $\frac{1}{2}$, so in this case, $f^{\circ 2}(n)$ is odd with probability $\frac{1}{2}$. if $n$ is even and $m=f(n)$ is odd: $$f^{\circ 2}(n) = f(m) = \left \lfloor\sqrt{2}m \right \rfloor = \left \lfloor \sqrt{2}(\frac{n}{\sqrt{2}} - \theta) \right \rfloor = \left \lfloor n - \sqrt{2} \theta) \right \rfloor$$ but $n$ is even and the probability for $0<\sqrt{2} \theta<1$ is $\frac{\sqrt{2}}{2}$ (because $\theta$ is assumed statistically equidistributed on the open interval $(0,1)$), so $f^{\circ 2}(n)$ is odd with probability $\frac{\sqrt{2}}{2}$. if $n$ is odd and $m=f(n)$ is even: $$f^{\circ 2}(n) = f(m) = \left \lfloor{\frac{m}{\sqrt{2}}} \right \rfloor = \left \lfloor{\frac{\sqrt{2}n-\theta}{\sqrt{2}}} \right \rfloor = \left \lfloor n - \frac{\theta}{\sqrt{2}} \right \rfloor $$ but $n$ is odd and $0 < \frac{\theta}{\sqrt{2}}<1$, so $f^{\circ 2}(n)$ is even. if $n$ is odd and $m=f(n)$ is odd: $$f^{\circ 2}(n) = f(m) = \left \lfloor \sqrt{2} m \right \rfloor = \left \lfloor \sqrt{2} (\sqrt{2}n-\theta) \right \rfloor = \left \lfloor 2n - \sqrt{2} \theta \right \rfloor $$ but $2n$ is even and the probability for $0<\sqrt{2} \theta<1$ is $\frac{\sqrt{2}}{2}$, so $f^{\circ 2}(n)$ is odd with probability $\frac{\sqrt{2}}{2}$. By combining these four cases together, we deduce that the probability for $f^{\circ 2}(n)$ to be odd is $$\frac{1}{2} \times \frac{1}{2} \times (\frac{1}{2} + \frac{\sqrt{2}}{2} + 0 + \frac{\sqrt{2}}{2}) = \frac{2\sqrt{2}+1}{8}$$ By continuing in the same way, we get that the probability for $f^{\circ 3}(n)$ to be odd is: $$ \frac{1}{4} (\frac{1}{2}\frac{1}{2} + \frac{1}{2}\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2} + 1\frac{1}{2} + \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}+7}{16}$$ For $2 \le r \le 24$, we computed the probability $p_r$ for $f^{\circ r}(n)$ to be odd (see Appendix). It seems (experimentally) that $p_r$ converges to a number $\simeq 0.532288725 \simeq \frac{8+3\sqrt{2}}{23}$ by Inverse Symbolic Calculator. This leads to the following question/conjecture: $$\lim_{r \to \infty}p_r = \frac{8+3\sqrt{2}}{23} \ \ ?$$ If so, consider the number $\alpha$ mentioned in the main post, then $$\alpha = 1-\frac{8+3\sqrt{2}}{23} = \frac{15-3\sqrt{2}}{23} \simeq 0.467711,$$ which matches with the computation in the main post. And next, we would have: $$ \delta = \frac{\sqrt{2}}{2^{\alpha}}= 2^{\frac{1}{2}-\alpha} = 2^{\frac{6\sqrt{2}-7}{46}} \simeq 1.022633$$ Appendix Computation sage: for i in range(3,26): ....: print(sq2(i)) ....: [1/4*sqrt(2) + 1/8, 0.478553390593274] [1/16*sqrt(2) + 7/16, 0.525888347648318] [3/32*sqrt(2) + 13/32, 0.538832521472478] [15/64*sqrt(2) + 13/64, 0.534581303681194] [5/128*sqrt(2) + 61/128, 0.531805217280199] [39/256*sqrt(2) + 81/256, 0.531852847392776] [93/512*sqrt(2) + 141/512, 0.532269260352925] [51/1024*sqrt(2) + 473/1024, 0.532348527032254] [377/2048*sqrt(2) + 557/2048, 0.532303961432938] [551/4096*sqrt(2) + 1401/4096, 0.532283123258685] [653/8192*sqrt(2) + 3437/8192, 0.532285334012406] [3083/16384*sqrt(2) + 4361/16384, 0.532288843554459] [3409/32768*sqrt(2) + 12621/32768, 0.532289246647030] [7407/65536*sqrt(2) + 24409/65536, 0.532288816169701] [22805/131072*sqrt(2) + 37517/131072, 0.532288667983386] [24307/262144*sqrt(2) + 105161/262144, 0.532288700334941] [72761/524288*sqrt(2) + 176173/524288, 0.532288728736551] [159959/1048576*sqrt(2) + 331929/1048576, 0.532288729880941] [202621/2097152*sqrt(2) + 829741/2097152, 0.532288725958633] [639131/4194304*sqrt(2) + 1328713/4194304, 0.532288724978704] [1114081/8388608*sqrt(2) + 2889613/8388608, 0.532288725350163] [1825983/16777216*sqrt(2) + 6347993/16777216, 0.532288725570602] [5183461/33554432*sqrt(2) + 10530125/33554432, 0.532288725561857] Code def sq2(n): c=0 for i in range(2^n): l=list(Integer(i).digits(base=2,padto=n)) if l[-1]==1: cc=1/4 for j in range(n-2): ll=[l[j],l[j+1],l[j+2]] if ll==[0,0,0]: cc*=1/2 if ll==[0,0,1]: cc*=1/2 if ll==[0,1,0]: cc*=(1-sqrt(2)/2) if ll==[0,1,1]: cc*=sqrt(2)/2 if ll==[1,0,0]: cc*=1 if ll==[1,0,1]: cc=0 break if ll==[1,1,0]: cc*=(1-sqrt(2)/2) if ll==[1,1,1]: cc*=sqrt(2)/2 c+=cc return [c.expand(),c.n()]<|endoftext|> TITLE: Evaluation of the quality of research articles submitted in mathematical journals: how do they do that? QUESTION [25 upvotes]: I would like to know as curiosity how the editorial board or editors* of a mathematical journal evaluate the quality, let's say in colloquial words the importance, of papers or articles. Question. I would like to know how is evaluate the quality of an article submitted in a journal. Are there criteria to evaluate it? Many thanks. I think that it must be a difficult task to evaluate the quality of a mathematical paper due how is abstract (the high level of abstraction of research in mathematics) the work of professional mathematicians. Are there criteria to evaluate it, or is it just the experience, knowledges and good work of the people working as publishers? I'm curious about it but I think that this is an interesting and potentially useful post for other. As soon as I can I should accept an answer. If there is suitable references in the literature about how they do this work, feel free to refer the literature, answering this question as a reference request, and I try to search and read it from the literature. *I don't know what is the role of each person working in the edition of a mathematical journal. REPLY [35 votes]: Assume we are talking about a good journal with a large editorial board representing a wide scope of mathematical interests. I will describe both the role of the editors and the role of the referees. This is my personal viewpoint and others might have different opinion/experience. The role of the editors. Good journals can accept only about 20% of submitted papers. This is not easy to reject 80% of papers and it often results in rejecting really good papers. Everyone understands that. The procedure of evaluating the papers by the editors is more or less as follows: The editors have many years of research experience and (hopefully) developed a good mathematical taste. If the paper is close to the research interest of the editor then he or she can relatively easily identify the papers that are not particularly interesting. The reason for not being interesting can be based (for example) on the following criterion: The result is not well motivated. It is very technical and follows more or less standard arguments. The authors simply take a known result, and prove a new result by slightly modifying the given assumptions. Usually it means that they make the statement more complicated and in a sense more general. Often, they neither have interesting examples supporting such generalizations nor indication of possible applications. Unfortunately, most of the papers fell into this category. If the editor is sure that this is the case, then he or she rejects the paper without sending it to a referee. Then the authors usually get a rejection notice similar to this one: We regret that we cannot consider it, in part because at present we have a large backlog of excellent articles awaiting publication. We are thus forced to return articles that might otherwise be considered. If the editor is not sure about the quality of the paper, then they ask an expert (or several experts) for a quick opinion: I wonder if you could make a quick, informal assessment of it. Are the results strong enough to warrant sending the article to a referee? Because of our backlog, we like to send to referees only articles that appear to be of very good to outstanding quality. In that case the expert evaluating the paper is not asked to check all the proofs but to make a judgement based on the criterion explained above. This is an easy task for an expert. If the expert writes a negative opinion, then the authors receive a rejection notice often phrased the way as the rejection notice listed above. For top journals all experts have to write a positive opinion before the paper is sent to referees. If the experts' opinion is positive, then the paper is sent to a referee or to many referees. The most extreme case that I know of was a panel of 12 referees who took several years to evaluate the paper (this was when Thomas Hales proved the famous Kepler conjecture). For top quality journals all referees must write positive reports before the paper is accepted (once I received 6 reports, 4 positive and 2 not so positive and the paper was rejected). Let me also add that the editorial boards are structured basically in two different ways. (1) The authors are asked to choose an editor from the list of editors and submit the paper directly to them. Then the editor who receives the paper handles the submission process according to the rules explained above. (2) The authors submit the paper to the main editor or just to the journal and then the main editor either rejects the paper by themselves or he/she sends it to one of the editors from the editorial board and that editor applies the rules listed above. Of course some of the journals might have a slightly different approach than the one explained here. There is no a canonical solution and what I wrote is a somewhat a simplified version of the process that is applied in reality. The role of referees. A paper passed through an initial screening and it was sent to a referee. This is the most unpleasant part of the process. A referee spends a lot of time to read the paper, they are not paid for this job and since their work is anonymous, they do not get any recognition for what they do. What is the referee required to do? First of all, the referee has to assess originality of the results and whether the results are interesting enough. This part is the same as the one in the initial screening when the paper is sent to an expert for a quick opinion. Secondly, the referee is required to read the paper and check details. Let's be clear about that. Unless the paper is directly related to the research of the referee and he or she really wants to understand the details, there is no way the referee can check all details. Since I cannot speak for other people, let me say what I do in this situation. My answer will only be a simplified version of the real process of the refereeing a paper, just a main idea of what I do. I go through the whole paper (or most of the paper) to have a good idea of what it is all about, to see a big picture not only of the meaning of the theorems, but also a big picture of the techniques used in the proof. Then I check carefully details of many/some arguments while for other arguments I briefly skim over. If the argument seems reasonable and believable to me I do not bother checking it very carefully. If all details that I check are correct and if all other arguments seem reasonable I am content. In this case, if I like the statement of the main result, I accept the paper. If however, some arguments seem fishy to me, then I check them carefully. This is a point where often I ask the authors for further clarifications. If I really cannot pass through the paper, because I think it has mistakes or if it is written in an unreadable way, I often reject the paper. The biggest problem is when I am convinced that the result proved in a paper is of an outstanding quality, but the paper is very difficult and for that reason not very easy to read. Then, hmm... Then, it is not easy and I often struggle with making a right decision. REPLY [6 votes]: (i) I think almost every journal has a declared "Aims and scope" policy, and how well the submitted paper fits that policy is perhaps the important criterion in the editors' judgment of how good the paper would be for the readers of the journal. (ii) Obviously, for any reputable mathematical journal, the mathematics in the paper must be correct and nontrivial. Nowadays, in the era of citation indexes, the main criterion, after the criteria (i) and (ii) above, seems to be how "topical" the paper is, that is, how high is the level of current interest the paper may attract.<|endoftext|> TITLE: Generalized "Homology Whitehead" -- How much does stabilization remember? QUESTION [8 upvotes]: Classically, the (non-local-coefficients) homology Whitehead theorem says that if $X \xrightarrow f Y$ is a map of simple spaces, and if the induced map $H_\ast(X;\mathbb Z) \to H_\ast(Y;\mathbb Z)$ is an isomorphism, then $f$ is a weak homotopy equivalence. Conceptually for me, the essence of this theorem is that we have a stable invariant ($H_\ast(-;\mathbb Z)$), and we identify a (reasonably large) class of spaces (the simple spaces) such that our invariant detects equivalences when restricted to this class. I'm wondering how generally a statement of this form holds in a fairly general $\infty$-category $\mathcal C$ in place of $Spaces$. For my purposes, I'm not particularly concerned with which stable invariant we use, so we might as well restrict attention to the universal case. Moreover, there are two general forms of "stabilization" which come to mind -- the category of spectrum objects $Sp(\mathcal C) = \varprojlim (\cdots \xrightarrow \Omega \mathcal C_\ast \xrightarrow \Omega \mathcal C_\ast)$, and the Spanier-Whitehead category $SW(\mathcal C) = \varinjlim (\mathcal C_\ast \xrightarrow \Sigma \mathcal C_\ast \xrightarrow \Sigma \cdots)$ (I use $\mathcal C_\ast$ to denote the $\infty$-category of pointed objects in $\mathcal C$). But we're eventually passing to some subcategory anyway, so we can reduce the $SW$ notion to the $Sp$ notion if we start out by replacing $\mathcal C$ with $Ind(\mathcal C)$ via the equation $Ind(SW(\mathcal C)) = Sp(Ind(\mathcal C))$. Thus we are led to the following formulation: Question: Let $\mathcal C$ be a presentable $\infty$-category. Can we identify a (reasonably large) full subcategory $\mathcal D \subseteq \mathcal C$ such that the composite functor $\mathcal D \to \mathcal C \xrightarrow {\Sigma^\infty_+} Sp(\mathcal C)$ is conservative? In particular, is this the case for $\mathcal D$ being one of the following? The 1-fold suspension objects? The 1-fold loop objects? The 1-connected objects? Here, a 1-fold suspension object is simply an object of the form $X = \Sigma Y$ for some $Y \in \mathcal C$; a 1-fold loop object is an object of the form $X = \Omega Y$ where $Y \in \mathcal C_\ast$ is a pointed object of $\mathcal C$. A 1-truncated morphism $W \to Z$ is a morphism such that for every $C \in \mathcal C$, the map $\mathcal C(C,W) \to \mathcal C(C,Z)$ has 1-truncated fibers, a morphism is 1-connected if it is left orthgonal to the 1-truncated morphisms, and an object $X$ is 1-connected if the map $X \to 1$ is 1-connected, where $1$ is the terminal object. As a sanity check, I think each of my candidates for $\mathcal D$ are trivial when $\mathcal C$ has discrete hom-spaces, which is a good thing because in this case $Sp(\mathcal C)$ is also trivial. REPLY [5 votes]: Putting together my and Dan's comments deserves to be called an answer. Namely: If $\mathcal{C}$ is an $(\infty,1)$-topos, then the statement is true when $\mathcal{D}$ is the class of hypercomplete, pointed, nilpotent objects. Hypercompleteness is the usual $(\infty,1)$-categorical notion, pointed is obvious, while "nilpotent" here means that the internal group object $\pi_1(X)$ is nilpotent and acts nilpotently on each internal abelian group object $\pi_n(X)$, in an appropriate internal sense. For a proof, see section 3 of Nilpotent Types and Fracture Squares in Homotopy Type Theory by Luis Scoccola, which proves in homotopy type theory that any cohomology isomorphism between pointed nilpotent types is $\infty$-connected — hence an equivalence if the types are hypercomplete. Then we get the result for $(\infty,1)$-toposes by interpreting homotopy type theory internally therein, as shown here for universes and here for higher inductive types. (Those papers don't yet quite complete the interpretation by showing that the universe is closed under HITs, but I doubt that this proof depends crucially on that.) Note that the assumption used here is weaker than yours, namely that an isomorphism is induced only on internal cohomology group objects with coefficients in all internal abelian group objects. It seems likely that with your stronger assumption hypercompleteness could be removed from the result, but probably a different method would be required.<|endoftext|> TITLE: Birkhoff's completeness theorem put into practice QUESTION [8 upvotes]: Birkhoff's completeness theorem (see here, Theorem 14.19) states that an equation which is true in all models of an algebraic theory can be proven in equational logic. Question. Does the proof of Birkhoff's completeness theorem actually produce for each specific equation a proof in equational logic? If yes, can you please demonstrate this with an instructive example? Actually, I suspect that the answer is "No", but I am not entirely sure. Let us look at the following well-known statement: If $R$ is a ring in which every element $r \in R$ satisfies $r^2=r$ (i.e. $R$ is boolean), then $R$ is commutative. There is the following overkill proof: $R$ is reduced, hence a subdirect product of domains $R_i$. Since the maps $R \to R_i$ are surjective, each domain $R_i$ satisfies the same equation and then must be isomorphic to $\mathbb{F}_2$. In particular, $R_i$ is commutative. Since $R \to \prod_{i \in I} R_i$ is injective, it follows that $R$ is commutative. So by Birkhoff's theorem, there must be some equational proof of this. My question is not how an equational proof looks like - this is just a basic algebra exercise. I would like to see how (if possible) it can be extracted from the overkill proof. I think the proof of Birkhoff's theorem in this special case works as follows: Consider the free ring on two generators $\mathbb{Z}\langle X,Y\rangle$ and take the quotient with respect to the relations $r^2=r$ for all elements $r$. This is the free boolean ring $R$ on two generators. By the overkill proof, $XY=YX$ holds in $R$. This means that $XY=YX$ can be derived from the relations $r^2=r$ in $\mathbb{Z}\langle X,Y\rangle$. But we don't get a derivation, right? How to produce, for example, the following equation? $$\begin{align*} XY - YX &= \bigl((X+Y)^2 - (X+Y)\bigr) - \bigl(X^2-X\bigr) - \bigl(Y^2-Y\bigr) \\ &\phantom{=}+ \bigl((YX)^2 - (YX)\bigr) - \bigl((-YX)^2 - (-YX)\bigr), \end{align*}$$ As far as I can tell, we are not even guaranteed a priori to have a proof which works without the axiom of choice, since this is used in the structure theorem for reduced rings in the overkill proof? (At first this might be confusing since the axiom of choice is surely not allowed in an equational proof, but actually the axiom of choice is used to show the existence of an equational proof.) I am interested in more complicated applications of Birkhoff's theorem. This here is just an example to get started. You can also choose other examples if they are more instructive. REPLY [10 votes]: Let me try to restate the question. I consider an identity to be a pair, written $(s,t)$ or $s\approx t$. I also consider a set of identities to be a set of pairs. Birkhoff's Theorem compares three things, namely (1) $\Sigma\models s\approx t$, (2) The pair $(s,t)$ belongs to the fully invariant congruence $\Theta^{T(X)}(\Sigma)$ on the term algebra $T(X)$, and (3) $\Sigma\vdash s\approx t$. I think the question is asking whether it is possible to extract from the proof of Birkhoff's Theorem and from a specific instance of $\Sigma\models s\approx t$ a proof witnessing that $\Sigma \vdash s\approx t$. The proof of Birkhoff's theorem does explain why (1) $\Leftrightarrow$ (2). Also, if we are told HOW the pair $(s,t)$ is generated as a member of $\Theta^{T(X)}(\Sigma)$, that can be translated into a $\Sigma$-proof of $s\approx t$. What is missing is: given the knowledge that $(s,t)\in\Theta^{T(X)}(\Sigma)$, do we know how the pair $(s,t)$ got into $\Theta^{T(X)}(\Sigma)$? The proof of Birkhoff's Theorem does not require this missing ingredient, it only requires that if $(s,t)$ is in $\Theta^{T(X)}(\Sigma)$, then it was generated in some way. So the answer to the question in the form it was asked (can we extract a $\Sigma$-proof of $s\approx t$ from the proof of Birkhoff's Theorem?) has to be No. You could try asking a slightly different question: Suppose, given finite $\Sigma$, that we are told that $\Sigma \models s\approx t$. By B's Theorem we know there must exist a $\Sigma$-proof of $s\approx t$. Question: can we estimate an upper bound on the complexity of such a proof? An affirmative answer would imply that any finitely axiomatizable variety has a decidable equational theory. But we know this to be false, since in the 1940's Tarski found a finitely axiomatizable variety of relation algebras with an undecidable equational theory. Many other examples are known now.<|endoftext|> TITLE: Lipschitz-continuity of convex polytopes under the Hausdorff metric QUESTION [6 upvotes]: Recently, I proved the following Lipschitz-continuity like result for convex polytopes: Let $A\in\mathbb R^{m\times n}$ and $b,b'\in\mathbb R^m$ be given such that $\{x\,:\,Ax\leq 0\}=\{0\}$ (which is equivalent to boundedness of all induced polytopes) and that $\{x\in\mathbb R^n\,:\,Ax\leq b\},\{x\in\mathbb R^n\,:\,Ax\leq b'\}$ are non-empty. Then $$ \delta\big(\{x\in\mathbb R^n\,:\,Ax\leq b\},\{x\in\mathbb R^n\,:\,Ax\leq b'\}\big)\leq \Big( \max_{A_0\in\operatorname{GL}(n,\mathbb R),A_0\subset A}\|A_0^{-1}\| \Big)\|b-b'\|_1 $$ where the operator norm $\|\cdot\|$ as well as the Hausdorff metric $\delta$ are taken with respect to $(\mathbb R^n,\|\cdot\|_1)$. Also $A_0\subset A$ is short for "every row of $A_0$ is also a row of $A$" so the above maximum is taken over all invertible submatrices of $A$. What this intuitively means is that if two polytopes have parallel faces (i.e. they are both described by the same $A$ matrix), but the location of these faces differs ($b\neq b'$), then the distance between the polytopes is upper bounded by the distance between the vectors $b,b'$ times a "geometrical" constant coming from $A$. Hence the function $b\mapsto\{x\in\mathbb R^n:Ax\leq b\}$ (with suitable domain such that the co-domain equals all non-empty subsets of $\mathbb R^n$) is Lipschitz continuous with a constant determined by $A$. This came up as a lemma to something only vaguely related which is why I don't have a problem with posting it publicly. Actually if this was a known result then my manuscript could be shortened by 3 pages. Thus my quesiton is: Is the above result known and, if so, where can in be found in the (convex polytope-)literature? I would be surprised if nobody has thought about this until now. While I haven't seen this result in the books of Grünbaum and Schrijver or the few papers on convex polytopes I am aware of, this is not the field I usually work in; hence this might very well be known but beyond my mathematical horizon. Thanks in advance for any answer or comment! REPLY [2 votes]: This is a classic question in the literature on linear programming, since it is related to the stability of the feasible set (and hence the solutions) under perturbation of the parameters. The classic work in this field is: A. J. Hoffman, Approximate solutions of systems of linear inequalities, J. Res. Nat. Bur. Standards, 1952. A more recent work, which essentially states your result as a special case (see Theorem 2.4 and the Lipschitz constant on p.19), is: W. Li, The sharp Lifshitz constants for feasible and optimal solutions of a perturbed linear program, Linear Algebra and Its Applications, 1993.<|endoftext|> TITLE: Is this subset of matrices contractible inside the space of non-conformal matrices? QUESTION [5 upvotes]: Set $\mathcal{F}:=\{ A \in \text{SL}_2(\mathbb{R}) \, | \, Ae_1 \in \operatorname{span}(e_1) \, \, \text{ and } \, \, A \, \text{ is not conformal} \,\}$, and $\mathcal{NC}:=\{ A \in M_2(\mathbb{R}) \, | \det A \ge 0 \, \,\text{ and } \, A \text{ is not conformal} \,\}$. By a non-conformal matrix, I mean a matrix whose singular values are distinct. (i.e. I allow non-zero singular matrices in $\mathcal{NC}$). Is each connected component of $\mathcal{F}$ contractible in $\mathcal{NC}$? $\mathcal{F}$ has two connected components, both homeomorphic to an open half-plane with one point removed. Indeed, $Ae_1 \in \operatorname{span}(e_1)$ and $A \in \text{SL}_2(\mathbb{R})$ imply that $$ A=\begin{pmatrix} \lambda & y \\\ 0 & \lambda^{-1} \end{pmatrix} \, \, \, \text{for some }\, \lambda \neq 0.$$ $A$ is conformal if and only if $\lambda=\pm 1$ and $y=0$. So, $A$ is not conformal if f $\lambda \neq 1,-1,0$ or $\lambda=\pm 1$ and $y \neq 0$. Thus, one connected component of $\mathcal{F}$ is homeomorphic to $$\{ 0<\lambda \neq 1\} \times \mathbb{R} \cup \{1\} \times \mathbb{R}\setminus\{0\}.$$ (The second component corresponds to $\lambda <0$.) Here is what I know about the topology of $\mathcal{NC}$: Let $\mathcal D=\{ (\sigma_1,\sigma_2) \, | 0 \le \sigma_1 < \sigma_2\}$. Then the map \begin{align*} \mu: SO_2\times \mathcal D\times SO_2\to \mathcal{NC}\\ (U,\Sigma,V)\mapsto U\Sigma V^T \end{align*} is a $2$-fold smooth covering map*. (i.e. $\mu(U,\Sigma,V)=\mu(-U,\Sigma,-V)$, and this is the only ambiguity in $U,V$ for a pre-image of a given point in $\mathcal{NC}$. Since $SO_2 \cong \mathbb{S}^1$, and since after identifying antipodal points in $\mathbb{S}^1 \times \mathbb{S}^1$, we get the $2$-torus $\mathbb{T}^2$ again, it follows that $\mathcal{NC} \cong \mathbb{T}^2 \times D$. *I am not entirely sure regarding the behaviour at the boundary points where $\sigma_1=0$, but I don't think this creates a serious problem. REPLY [2 votes]: Edited. In the first version of the answer I was assuming that the space in which the contraction was taking place was not $\cal NC$ but the complement to non-conformal matrices in $SL(2,\mathbb R)$. I'll suggest a fix for this now. Note, that we have a natural continuous map $u: {\cal NC}\to S^1=\mathbb RP^1$. Namely, to each matrix $A$ from ${\cal NC}$ we can associate the following one-dimensional subspace $u(A)\in \mathbb R^2$. Take the matrix $AA^{*}$ and take the eigenspace corresponding to the maximal eigenvalue of $AA^*$ (there will be two distinct eigenvalues since $A$ is not conformal). So, if we find a closed path $\gamma$ in $\cal F$, such that its image $u(\gamma)\subset S^1$ is not contractible, we are done. How to find such a path is explained in the previous answer to this question, which the path $\gamma(t)$ constructed in the previous answer below. ( I believe that what I suggest works for several reasons but I don't have time to work out all the details now. By the way, it is also funny that $\pi_1(\cal NC)$ seem to be equal $\mathbb Z^2$, moreover it deformation retracts to $T^2$, I believe.) Previous answer. It is not contractible. Let us associate to each matrix $A\in SL_2(\mathbb R)$ the following vector $v(A)$. Take an orthogonal matrix $O\in SO_2(\mathbb R)$ such that $OA(e_1)$ is proportional to $e_1$ with a positive coefficient. Then set $v(A)=OA(e_2)$. We get a map to the upper half plane: $$V:SL(2,\mathbb R)\to \{y>0\}$$ Note that the image of confromal matrices is the point $(0,1)$, and the image of any component $\cal F$ is the complement to $(0,1)$. And so each component can be identified with this puncutred half-plane. Hence it is enough to construct a path in $\cal F$ whose image under $V$ is not contractible in $\{y>0\}\setminus \{(0,1)\}$. This is easy, just take a non-contractible path $\gamma(t)\subset \{y>0\}\setminus \{(0,1)\}$ (that has a non-zero winding number around $(0,1)$), and consider the unique path of matrices $A_t\subset \cal F$ such that $A_t(e_2)=\gamma(t)$.<|endoftext|> TITLE: Is the inclusion of its 2-skeleton into the walking idempotent homotopy cofinal? QUESTION [6 upvotes]: Let $Idem = Idem^{(\infty)}$ be the walking idempotent [1], and let $Idem^{(n)}$ be its n-skeleton. Note that $Idem$ has one nondegenerate simplex in each dimension. Let $\iota_n^m: Idem^{(n)} \to Idem^{(m)}$ be the inclusion. Lurie has shown [2] the following: If $X$ is a quasicategory, and if $Idem^{(3)} \xrightarrow f X$ is a map, then there exists a map $Idem \xrightarrow g X$ such that $g\iota_1^\infty = f\iota_1^3$. That is, although a homotopy-coherent idempotent involves infinitely many pieces of coherence data corresponding to the infinitely many nondegenerate simplices of $Idem$, nevertheless all of this data is guaranteed to exist once the first 3 have been found. However, the resulting coherence data might be different from the original data in dimensions 2 and 3 [3]. The proof is a bit involved, and I have not studied it in detail, but what I do understand seems to suggest an affirmative answer to the following Question: Is the inclusion $Idem^{(n)} \to Idem$ homotopy cofinal for certain $n$ (i.e. is it cofinal in the $\infty$-categorical sense -- depending on how one defines this, it may be necessary to Joyal-fibrantly replace $Idem^{(n)}$ before the question makes sense)? As noted in the comments, this is definitely not true for $n$ odd, nor is it true for $n=0$. Somehow it seems unlikely for $n=2$; hence the title question, which asks this for $n=4$. I don't believe the inclusion $Idem^{(n)} \to Idem$ is left or right anodyne, so any proof will encounter some complications. One idea would be to use the usual map $N \to Idem^{(n)}$, where $N \subseteq \mathbb N$ is the graph on which the poset $\mathbb N$ of natural numbers is free -- the 0-cells are natural numbers, and there is a 1-cell from $n$ to $n+1$ for each $n$. For the inclusion $N \to \mathbb N$ is a categorical equivalence, and it's easy to show that $\mathbb N \to Idem$ is homotopy cofinal using Quillen's Theorem A (since the relevant slice categories are just ordinary 1-categories). By composition, $N \to Idem$ is homotopy cofinal, but this factors through $Idem^{(n)}$. By a cancellation property of cofinality, in order to show that $Idem^{(n)} \to Idem$ is homotopy cofinal, it will suffice to show that $N \to Idem^{(n)}$ is homotopy cofinal, which sounds straightforward, since $N$ and $Idem^{(n)}$ are finite-dimensional. But I'm not sure the relevant slice objects are still finite-dimensional... [1] That is, $Idem$ is the category with one object and one non-identity morphism $i$, satisfying the equation $i^2 = i$. In this question I identify 1-categories with their nerves, which are quasicategories. [2] HTT 4.4.5.20 in the current version. This does not appear in the published version of HTT. It appears in older versions of HA as 7.3.5.14, but was moved to HTT when Lurie rewrote the section on idempotents in HTT. [3] For example, consider the inclusion $Idem^{(3)} \to \widetilde{Idem^{(3)}}$ given by a Kan fibrant replacement such as $Ex^\infty$. This map is nontrivial on homology, so it cannot actually be extended to a map $Idem \to \widetilde{Idem^{(3)}}$ since this would amount to a trivialization. Thus the extension produced by Lurie's theorem must be changing the image of the 3-cell of $Idem^{(3)}$, at least. In particular, it's not the case that all quasicategories have the right lifting property with respect to $\iota_3^\infty$. REPLY [2 votes]: I don't know what's wrong with the following computation, but the answer is clearly no: if there were a cofinal functor from a finite simplicial set to $Idem$, then any $\infty$-category with finite colimits would have split idempotents, which is not the case (witness finite spaces). Somewhat surprisingly, this seems to work for even $n>0$, even for $n=2$! That is, Claim: Let $n \in \mathbb N$. Then the inclusion $Idem^{(n)} \to Idem$ is homotopy cofinal (equivalently, since everything is self-dual: co-cofinal) if and only if $n$ is positive and even. Proof: We have seen that this can't happen when $n=0$ or $n$ is odd. Otherwise, we verify the hypotheses of the Joyal-Lurie version of Quillen's Theorem A, i.e. we check that the simplicial set $X^{(n)} = Idem^{(n)} \times_{Idem} Dec(Idem)$ is weakly contractible. Here $Dec(Idem)$ is the decalage contruction $Dec(C)_n = C_{n+1}$ where we forget the 0th degeneracy. So in terms of simplices, we have $X^{(n)}_m = Idem^{(n)}_m \times_{Idem_m} Idem_{m+1}$. For $m \geq n+1$, an $m$-simplex in $X^{(n)}_m$ is a string of morphisms in $Idem$ (each either $i$ or $1$) of length $m+1$, such that among the last $m$ morphims in the string, all but at most $n$ are $1$. Any such simplex is a degeneracy of a simplex obtained by deleting one of the copies of $1$. That is, $X^{(n)}$ is $n$-skeletal. Now, the $n$-skeleton of $X^{(n)}$ agrees with that of $X^{(\infty)}$, which is weakly contractible. Therefore, since $n \geq 2$, we have $\pi_1(X^{(n)}) = \tilde H_{\leq n-1}(X^{(n)}) = 0$. So it will suffice to show that $H_n(X^{(n)}) = 0$. There are two nondegenerate simplices of degree $n$: the string $1,i,\dots,i$ (a $1$ followed by $n$ $i$'s) and the string $i,i,\dots,i$ (a string of $n+1$ $i$'s). The boundaries of these (remember that $n$ is even and we are omiting the $\partial_0$ term of the boundary map) are $1,i,\dots,i$ and $i,i,\dots,i$ (where now we have one fewer term in each string) respectively, which are linearly independent. Thus there are no nondegenerate cycles and $H_n(X^{(n)}) = 0$ as desired.<|endoftext|> TITLE: Thickness and hierarchical hyperbolicity QUESTION [7 upvotes]: Thick metric spaces were introduced by Behrstock, Drutu and Mosher, see here. Hierarchically hyperbolic spaces were introduced by Behrstock, Hagen and Sisto, see here. I've heard that it is open whether hierarchically hyperbolic groups are thick. Of course, one has to restrict to hierarchically hyperbolic groups that are not non-elementary relatively hyperbolic (NERH for short), for it was proved in Behrstock, Drutu, Mosher that thick groups cannot be NERH. See also there, where Levcovitz proved that thick groups have trivial Floyd boundary, which cannot happen for NERH groups, since the Floyd boundary covers the Bowditch boundary. Question : Do we know if this is true among all known examples of hierarchically hyperbolic groups (that are not NERH) ? Note for example that we know that mapping class groups and Artin groups that are not NERH are indeed thick. Special emphasis on CAT(0) cube complexes. As far as I know, we still don't know if all cubical groups are hierarchically hyperbolic, although to my knowledge, it's likely to be true. On the other hand, do we know if all (not NERH) cubical groups are thick ? REPLY [4 votes]: This is not an answer, just some sketchy thoughts that are too long for the comment box. I and some other HHS enthusiasts are very interested in this question being answered; we've tried a fair bit and have set it aside, so I don't think they'll mind me trying to recall what some of the strategies and issues are. It's indeed open for cocompact special groups, as far as I know. (Part of the motivation for the question comes from Coxeter groups, which are known to be hyperbolic relative to thick subgroups [Theorem VII here], and many of which are virtually compact special.) There's a very similar question: for the classes of groups in question (HHG, cocompactly cubulated, virtually compact special), is it true that any group $G$ in the class is hyperbolic relative to a collection of subgroups, each of which has at most polynomial divergence (we're allowing $G$ to be hyperbolic relative to itself)? In particular, does super-polynomial divergence imply exponential divergence for such $G$? (A nontrivially relatively hyperbolic group has at least exponential divergence [Theorem 6.13 here].) It's also interesting to ask whether $G$ is hyperbolic relative to NRH subgroups (whether because the peripheral subgroups are thick, or NRH for some other reason). Here is a naive approach that has been tried a couple of times (once for cubical groups, once for HHG) unsuccessfully. I think it's worth sketching because it's probably the first thing one might try, so it might save someone some work to see what the problems with it are. (Maybe they are surmountable.) The idea is to build candidate thick/NRH subspaces inductively by "brute force", show that this construction terminates, and then show that the resulting subspaces give a valid peripheral structure. More precisely: let $G$ act geometrically on a proper CAT(0) cube complex $X$, or a proper hierarchically hyperbolic space $X$. In either case, either $X$ is already hyperbolic, or there is a nontrivial product region $P$. (In the cubical case, $P$ is a convex subcomplex decomposing as the product of two unbounded CAT(0) cube complexes. In the HHS case, $P$ is a hierarchically quasiconvex subspace admitting a coarse-median-preserving quasi-isometry to the product of two unbounded hierarchically hyperbolic spaces. The HHS fact follows from Corollary 2.16 here.) Let $\mathcal P_0$ be the ($G$--invariant) set of such product regions. Call $P,P'\in\mathcal P_0$ "elementary equivalent" if they have unbounded coarse intersection. Taking the transitive closure gives an equivalence relation on $\mathcal P_0$. For each equivalence class, one can take the union $U$ of the subspaces in the class. These $U$ are your candidate thick-of-order-at-most-$1$ pieces. Now iterate this process: take the set of such $U$, impose the same equivalence relation, and make candidate thick-of-order-at-most-$2$ pieces, etc. (If I remember right, the equivalence relation used at the $n^{th}$ stage, $n\geq 1$, is that $P,P'\in\mathcal P_{n-1}$ are elementary equivalent if their hierarchically quasiconvex hulls have unbounded coarse intersection. These are described nicely by Russell-Spriano-Tran in Section 5 of this paper; in the cubical case, you use cubical convex hulls.) Suppose the process terminates, i.e. at some point you have a $G$--invariant collection $\mathcal P_n$ of subspaces, no two of which have hierarchically quasiconvex hulls with unbounded coarse intersection. At this point, you should: (1) Verify that each $P$ is "quasiconvex" in the sense appropriate to the category you're working in. In the cubical case, you should check that $P$ is at finite Hausdorff distance from its cubical convex hull; in the HHS case, you should check that $P$ is \emph{hierarchically quasiconvex} --- Proposition 5.11 in this paper by Russell-Spriano-Tran is probably the best way to go about this. (2) Hope that the stabiliser of each $P\in\mathcal P_n$ acts coboundedly. ($G$ acts on $X$ coboundedly, $\mathcal P_n$ is $G$--invariant; you want to use properness of $X$ and boundedness of the coarse intersections between the elements of $\mathcal P_n$ to show that only finitely many elements of $\mathcal P_n$ intersect any given ball.) Already there is something tricky here. (3) Verify that adding combinatorial horoballs over the elements of $\mathcal P_n$ in $X$ gives you something hyperbolic. I think one approach was to verify hyperbolicity by finding a hierarchically hyperbolic structure with no interesting "product regions". (4) Hope that, more or less by construction, each $P\in\mathcal P_n$ is thick of order at most $n$. At that point, you'd have what you want. Difficulties include: (a) It's not remotely clear that this process terminates. (But I think it's worth nailing down whether or not it does). (b) If it doesn't terminate, one can still hope to be able to pass to some sort of limit, and find a collection $\mathcal P_\infty$ of subspaces, all with bounded coarse intersection, to get candidate peripherals in a relatively hyperbolic structure. If this works, one doesn't get that the peripherals are thick, but the hope is that they are at least not relatively hyperbolic. (c) If I remember correctly, one big difficulty is that, at each (finite) stage of the induction, one might have to pass to larger and larger neighbourhoods to get something convex/hierarchically quasiconvex, and this creates problems if the process doesn't terminate. More vaguely, at each step, there are various constants to control, and they blow up if the process doesn't terminate. There was a quite different approach (for CAT(0) groups) discussed by many people at the AIM in 2016. There was some subsequent work on it and some useful ideas, but I'll have to check with those involved to see if it's okay to mention the idea/subsequent developments here --- it's possible that one or more of them are still actively working on it, although I'm not. So, the short answer, is that it's a very interesting question for which the naive approach is a bit of a mess. My feeling is that if there's a counterexample, then there's a counterexample where $G$ is $\pi_1$ of a CAT(0) square complex.<|endoftext|> TITLE: Maximal subgroups of odd index in $\mathrm{PSL}(3,q)$ QUESTION [5 upvotes]: Let $G = \mathrm{PSL}(3,q)$ for $q$ odd. I am trying to understand a question that involves understanding the subgroups that contain a Sylow $2$-subgroup, and in particular, are subgroups of odd index in $G$. I need to find a complete description of the maximal subgroups of odd index in the group $G = \mathrm{PSL}(3,q)$ REPLY [8 votes]: The subgroups of ${\rm PSL}_3(q)$ for odd $q$ were first enumerated by H.H. Mitchell in 1911. (The case $q$ even was done by R.W. Hartley in 1925/6.) Table 8.3 of the book "The Maximal Subgroups of Low-Dimensional Finite Classical Groups" by Bray, Holt and Roney-Dougal provides a convenient list. Using that it is not hard to answer your question. For $q$ odd, the maximal subgroups of ${\rm SL}_3(q)$ of odd index are as follows: (i) Two classes of maximal parabolic subgroups with structure ${\rm E}_q^2:{\rm GL}_2(q)$ and index $q^2+q+1$, where ${\rm E}_q$ denotes an elementary abelian group of order $q$ (the additive group of the field). These two classes are interchanged by the graph (inverse-transpose) automorphism of ${\rm SL}_3(q)$. (ii) When $q \equiv 1 \bmod 4$, we have one class of imprimitive subgroups with structure $(q-1)^2:S_3$. (iii) When $q = q_0^r$ for some odd prime $r$, we have $\gcd(\frac{q-1}{q_0-1},3)$ classes of subgroups with the structure ${\rm SL}_3(q_0).\gcd(\frac{q-1}{q_0-1},3)$. When there are three such classes, they are all conjugate under a diagonal outer automorphism of ${\rm SL}_3(q)$.<|endoftext|> TITLE: What is the winning strategy in this pebble game? QUESTION [19 upvotes]: Consider the following two-player pebble game. We have finitely many stones on a finite linear track of squares. We take turns, and the allowed moves are: move any one stone one square to the left, if that square is empty, or remove any one stone, or remove any two adjacent stones. Whoever takes the last stone wins. Question. What is the winning strategy? And which are the winning positions? The game will clearly end always in finitely many moves, and so by the fundamental theorem of finite games, one of the players will have a winning strategy. So of course, I know that there is a computable winning strategy by computing with the game tree, and we have a computable algorithm to answer any instance of the question. What I am hoping for is that there will be a simple-to-describe winning strategy. This is what I know so far: Theorem. It is a winning move to give your opponent a position with an even number of stones, such that the stones in each successive pair stand at even distance apart. By even-distance, I mean that there are an odd number of empty squares between, so adjacent stones count as distance one, hence odd. Also, I am only concerned with the even distance requirement within each successive pairs, not between the pairs. For example, it is winning to give your opponent a position with stones at ..O...OO.O....O.....O........... We have distance 4 in the left-most pair, distance 2 in the next pair, distance 6 in the third pair, ignoring the distances in front and between the pairs. Proof. I claim that if you give your opponent a position like that, then he or she cannot give you back a position like that, and furthermore, you can give a position like that back again. If your opponent removes a stone, then you can remove the other one in that pair. If your opponent moves the lead stone on a pair, then you can move the trailing stone. If your opponent moves the trailing stone on a pair, then either you can move it again, unless that pair is now adjacent, in which case you can remove both. And if your opponent removes two adjacent stones, then they must have been from different pairs (since adjacent is not even distance), which would cause the new spacing to be the former odd number plus another odd number plus 2, so an even number of empty squares between, and so you can move the trailing end stone up one square to make an odd number of empty squares between and hence an even distance between the new endpoints. Thus, you can maintain this even-distance property, and your opponent cannot attain it; since the winning move is moving to the empty position, which has all even distances, you will win. $\Box$ What I wonder is whether there is a similarly easy to describe strategy that solves the general game. REPLY [3 votes]: We can go further and find the nim-value of any position of this game. Theorem. The nim-value of a position in this game is the Nim-sum of the contributions of its pebbles, where Pebbles in odd-numbered squares contribute $1$. Pebbles in even-numbered squares contribute $2$. (Here we count the leftmost square as the $1^{st}$ square.) Corollary. As in Gro-Tsen's answer, the winning positions (with nim-value $0$) are the ones with both an even number of pebbles in odd-numbered squares and an even number of pebbles in even-numbered squares. Corollary. All positions have nim-value $0, 1, 2,$ or $3$. The proof relies on the following idea, which is encoded in the lemma below. Key Insight. A position in this pebble game is equivalent to the sum of an individual game for each pebble with its own private track. Lemma. Given a position $P$ of pebbles $p_1, p_2, ... p_n$, consider the games $P_i$ which each consist of only the single pebble $p_i$ in the same square. Then $P$ is equivalent to $\sum_i P_i$. Proof. To check this, we need only show that the game $P + \sum_i P_i$ is a zero game, by demonstrating a winning strategy for the second player. So consider the possible moves for the first player in the game $P + \sum_i P_i$: If the first player removes two adjacent pebbles $p_i$ and $p_{i+1}$ in $P$, the second player can respond by moving the pebble in $P_{i+1}$ to the left, leaving behind two copies of $P_i$, which nim-cancel. If the first player removes any one pebble, the second player responds by removing the corresponding pebble. If the first player moves a pebble to the left in $P$, the second player responds by moving the pebble to the left in $P_i$. If the first player moves a pebble to the left in a $P_i$, the second player responds by moving the corresponding pebble to the left in $P$. If that move is blocked by a pebble directly to the left, that's fine -- the second player instead removes those two pebbles, noting that there are now two copies of $P_{i-1}$ that nim-cancel out. Proof of Theorem. Because of the Lemma, we need only consider the case of a single pebble in position $i$ (where position $1$ is the leftmost square). By induction on $i$, There is always the option to remove the pebble, an option with nim-value $0$. If $i$ is odd, there may also be the option to move the pebble to the left, which is an even-numbered square with nim-value $2$. Thus the nim-value in the odd case is $mex(0,2) = 1$. If $i$ is even, there is always also the option to move the pebble to the left, which is an odd-numbered square with nim-value $1$. Thus the nim-value in the even case is $mex(0,1) = 2$. How to Evaluate Positions. Example 1. The nim-value of the position in the question . . O . . . O O . O . . . O O . . . . . O is found quickly by seeing that a pebble in an odd-numbered position contributes nim-value 1, and a pebble in an even-numbered position contributes nim-value 2. So using this mask: . . O . . . O O . O . . . O O . . . . . O 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 yields the Nim-sum 1 +1+2 +2 +2+1 +1 which equals 2, so removing any of the even-numbered pebbles (the "2s") is a win. Example 2. You are playing the sum of the following two positions: O . O . . . O O . O + O . O O Evaluate each one with the mask: O . O . . . O O . O + O . O O 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 +1 +1+2 +2 + 1 +1+2 This has a Nim-sum of 3, so moving any pebble to the left or removing any two adjacent pebbles (in either component!) is a win.<|endoftext|> TITLE: 2D closed convex shape which minimizes average distance between points QUESTION [6 upvotes]: For a 2D closed convex shape, with metric $d$ and fixed area $A$, we can calculate the average distance between random (interior) points. For different shapes, we will get different values for this average. We can ask, for which shape is this average a minimum (in terms of $A$). For the Euclidean metric, the answer is the circle. If we pick a different metric, what shape do we get? REPLY [3 votes]: For an arbitrary metric and measure, this is going to be a very difficult question to answer. The shape of the minimizer will depend on the choice of metric and measure, if it exists at all. Joseph O'Rourke discussed the $L^1$ distance already, so I'm going to focus on distance functions which are either symmetric, or else induced by a smooth Riemannian metric. These sorts of questions are known as extremal problems [1] and are closely related to isoperimetric inequalities, which are an active area of research. As noted in the comments, we need to specify which measure we use to compute averages. It appears that there are at least two natural choices of measure. the Lebesgue measure on $\mathbb{R}^2$ the measure induced by a Riemannian metric As I mentioned earlier, the particular shape of the minimizer will depend on the choice of distance function and measure. However, from a variational standpoint, there is a fairly general principle which is very helpful. A convex region $S$ is a critical point for the average distance between interior points iff the integral $\int_S d(p,x) \, d \mu(x)$ is constant for all $p \in \partial S$. Here, $\mu$ is the measure used to determine averages. In other words, a region $S$ is a possible minimizer for the average distance between points if the average distance between a point $p$ on the boundary of $S$ and interior points in $S$ does not depend on $p$. It's possible to prove this rigorously using variational techniques or using Crofton's differential equation (for a good exposition, see [2]). However, the basic intuition is that if this property does not hold, we can slightly change the shape of $S$ so as to reduce the total average distance. From this, there are a few immediate consequences. If the distance function is translation and rotationally symmetric and the measure we use is the Lebesgue measure on $\mathbb{R}^2$, the disk is a critical point. Oftentimes, this will be the unique critical point (and thus also the minimum), but any proof of this seems likely to depend on the particular details of the distance function. If we consider a surface of constant curvature with its natural distance function and induced measure, the critical points for the average distance between points are geodesic balls. There is no guarantee that geodesic balls will be convex in the choice of coordinates you use, but they will be the minimizers nonetheless. For more general (non-symmetric) Riemannian metrics (with their induced distance function and measure), this question is much more difficult, and the variational principles are less useful. However, with a lower bound on the sectional curvature, it is possible to get a lower bound on the areas of geodesic balls. Using this, it should be possible to establish uniform lower bound on the average distance between points in a region of area $A$. However, these sorts of estimates don't tell you what the minimizer looks like (or even if it exists at all). It is possible to construct distance functions where no minimizer exists, by incorporating patches where the curvature is more and more negative. [1] Bauer, Christina; Schneider, Rolf, Extremal problems for geometric probabilities involving convex bodies, Adv. Appl. Probab. 27, No. 1, 20-34 (1995). ZBL0827.52004. [2] Eisenberg, Bennett; Sullivan, Rosemary, Crofton’s differential equation, Am. Math. Mon. 107, No. 2, 129-139 (2000). ZBL0986.60011.<|endoftext|> TITLE: Proving a specific case of Robin's Inequality QUESTION [5 upvotes]: Edit: It turns out that this is equivalent to the RH which gives the idea that this might a a little difficult to show. As such we could consider an even simpler case in which the number $n$ is squarefree (all values $k_j$ are equal to $1$. In previous papers it has been shown that squarefree numbers satisfy Robin's Inequality, but is this still the case for $2^kn$? If we make this loose condition we find our simpler form of $$ \dfrac{2^{k+1}-1}{2^{k}}\prod_{j=1}^m\dfrac{p_j+1}{p_j} < e^\gamma\log\log\left(2^k\prod_{j=1}^m p_j\right) $$ with a minima of $f(k)$ at $$ k_{min} = -\dfrac{W_{-1}\left(-e^\gamma\prod_{j=1}^m \frac{1}{p_j+1}\right)+\log\left(\prod_{j=1}^m p_j\right)}{\log2} $$ if we plug this in to our inequality we need to show that $$ e^\gamma\log\left(-W\left(-e^\gamma\prod_{j=1}^m\frac{1}{p_j+1}\right)\right) > \dfrac{e^{-W\left(-e^\gamma\prod_{j=1}^m\frac{1}{p_j+1}\right)-\log\left(\prod_{j=1}^m p_j\right)+1}-1}{e^{-W\left(-e^\gamma\prod_{j=1}^m\frac{1}{p_j+1}\right)-\log\left(\prod_{j=1}^m p_j\right)}}\prod_{j=1}^m\dfrac{p_j+1}{p_j} $$ which I will admit is disgustingly messy, but looks (at least naively to me) potentially tractable since prime product and inverse prime product series are well studied. One reformulation of the Riemann Hypothesis is Robin's Inequality which states that for $n>5040$ the following holds iff the RH holds: $$ \sigma(n) < e^\gamma n\log\log(n) $$ where $\sigma$ is the sum of divisors function and $\gamma$ is the Euler Mascheroni Constant. Now for my specific case I want to show that given given some number $n=p_1^{k_1}p_2^{k_2}\ldots p_m^{m} > 5040$ where $p_j \neq 2$ is a prime number, if Robin's Inequality holds for $n$, then it must also hold for $2^k\cdot n$. Performing some algebra on the inequality we can see that this is the same as showing that if the following inequality holds $$ \prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)} < e^\gamma\log\log\left(\prod_{j=1}^m p_j^{k_j}\right) $$ then this inequality must hold as well $$ \dfrac{2^{k+1}-1}{2^{k}}\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)} < e^\gamma\log\log\left(2^k\prod_{j=1}^m p_j^{k_j}\right) $$ I'll admit that I am not well versed in Analytic Number Theory, so this might be obvious and I have no idea, but so far I have only been able to show three fairly trivial things According to numerical computations this seems to hold true. For large values of $n$ it appears that the R.H.S. is strictly larger that $2$ times the L.H.S. in the assumed inequality. Since the left side is bounded with respect to $k$ and the right side is not, there must exist some $N$ for which if $k\geq N$ then the inequality holds. Therefore there are only finite cases for which this inequality may not hold. Taking the difference of the left and right sides as this $$ f(k) = e^\gamma\log\log\left(2^k\prod_{j=1}^m p_j^{k_j}\right) - \dfrac{2^{k+1}-1}{2^{k}}\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)} $$ has a derivative where $f'(0) < 0$ and $f'(N) > 0$, as such there exists a local minima of $f(k)$ which we can find to be the following value $$ k_{min} = -\dfrac{W_{-1}\left(-e^\gamma\prod_{j=1}^m \frac{p_j-1}{p_j^{k_j+1}-1}\right)+\log\left(\prod_{j=1}^m p_j^{k_j}\right)}{\log2} $$ where $W_{-1}$ is the second, more negative, solution of the Lambert W function when the argument is between $0$ and $-\frac{1}{e}$. The derivative also appears to be strictly positive past $k_{min}$ as $$ f'(k) = \dfrac{e^\gamma \log(2)}{k\log 2 + \log\left(\prod_{j=1}^m p_j^{k_j}\right)} - \dfrac{\log(2)}{2^k}\prod_{j=1}^m\dfrac{p_j^{k_j+1}-1}{p_j^{k_j}(p_j-1)} $$ as this will behave like $\frac{1}{k} - \frac{1}{2^k}$ where the negative part decreases significantly faster than the positive part. If anyone could offer some potential insight that would be much appreciated! REPLY [6 votes]: By Theorem 1.2 in this paper, Robin's Inequality is true for every odd integer $n>10$. If we knew what the OP wants to prove, then we would also know Robin's Inequality for every integer $n$ whose odd part exceeds $5040$. In particular, we would know Robin's Inequality for every colossally abundant number exceeding $5040$, because each colossally abundant number divides the second next one (cf. Proposition 4 in this paper). So, by Proposition 1 in Section 3 of Robin's paper, we would know Robin's Inequality for every integer exceeding $5040$, which is equivalent to the Riemann Hypothesis. In short, it is hopeless to prove what the OP wants to prove, because it implies the Riemann Hypothesis.<|endoftext|> TITLE: Trivial homology with local system QUESTION [5 upvotes]: Let $X$ be the classifying space of the Higman group $G$. It is well known that $G$ is an acyclic group $$H_{\ast}(X;\mathbb{Z})=H_{\ast}(pt;\mathbb{Z}).$$ Now, suppose that $\mathcal{M}$ is a local system on the space $X$ such that $$H_{i}(X;\mathcal{M})=0, \textrm{ for all $0\leq i$}.$$ Does such local system $\mathcal{M}$ on $X$ exist (other then $\mathcal{M}= 0$) REPLY [14 votes]: For $X = BG$ local systems on $X$ can be identified with $G$-modules, and homology with the derived tensor product $-\otimes^L_{\mathbb ZG}\mathbb Z$, i.e. $H_i(X;M) \cong \operatorname{Tor}^i_{\mathbb Z G}(M,\mathbb Z)$. One way to see this is to take the definition $H_i(X;M):= H_i(\mathcal S_*(\widetilde X)\otimes_{\mathbb Z\pi_1(X)} M)$, where $\pi_1(X)$ acts on (singular) chains on the universal cover $\widetilde X$ via deck transformations, and replace $\mathcal S_*(\widetilde X)$ with the cellular complex $C_*(\widetilde X)$ of the CW structure of the realization of the nerve of the groupoid $G//G$; then $C_*(\widetilde X)\otimes_{\mathbb Z G} M = \dots\to \mathbb Z[G^2]\otimes M\to \mathbb Z[G]\otimes M\to M$ is the bar complex computing group homology. Let $G$ be an arbitrary group, and let $M = \operatorname{ker}(\mathbb Z G\to \mathbb Z)$ be the reduced group algebra, so that we have a short exact sequence $$ 0\to M\to \mathbb ZG\to\mathbb Z\to 0 $$ of $\mathbb ZG$-modules. This gives rise to a long exact sequence of Tor-groups, in particular a boundary operator $\partial:\operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z,\mathbb Z)\to \operatorname{Tor}^i_{\mathbb Z G}(M,\mathbb Z)$. Its kernel is the image of the map $0 = \operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z G,\mathbb Z)\to \operatorname{Tor}^{i+1}_{\mathbb Z G}(\mathbb Z,\mathbb Z)$, so it is always injective; its cokernel is the kernel of the map $\operatorname{Tor}^{i}_{\mathbb Z G}(\mathbb Z G,\mathbb Z)\to \operatorname{Tor}^{i}_{\mathbb Z G}(\mathbb Z,\mathbb Z)$, which is an isomorphism for $i = 0$ and has the zero group as its codomain for $i > 0$, so it is always injective, so $\partial$ is always surjective and thus always an isomorphism. In your example of the Higman group, we know that $H_i(X;\mathbb Z)$ is $\mathbb Z$ concentrated in degree $0$, so that $\partial$ is an isomorphism with the zero group, so that $H_i(X;M) = 0$ for all $i \ge 0$.<|endoftext|> TITLE: Problems in advanced calculus QUESTION [18 upvotes]: I have been teaching Advanced Calculus at the University of Pittsburgh for many years. The course is intended both for advanced undergraduate students and the first year graduate students who have to pass the Preliminary Exam. This is a difficult course as you can see from the problems that we have on our exam: http://www.mathematics.pitt.edu/graduate/graduate-handbook/sample-preliminary-exams What bothers me quite a lot is the lack of a good collection of problems for functions of several variables. There are plenty of excellent collections for functions of one variable and for metric spaces, but there is almost nothing regarding good problems for functions of several variables. The only exception that I know is: P. N. de Souza, J.-N. Silva, Berkeley problems in mathematics. Third edition. Problem Books in Mathematics. Springer-Verlag, New York, 2004. This is an amazing collection of problems covering many areas of mathematics and what is important the problems have complete solutions. Question. Do you know a good collection of problems for functions of several variables? By this I mean a collection of problems that require deep understanding of the problem rather than a standard application of formulas and theorems. I believe that most of the problems in our Preliminary Exam in Analysis fell into this category. Ideally, I would prefer to have a collection with solutions or hints, as it would be very helpful for students (and for me as well). There are many non published collections of problems available online. I am also interested in links to such collections. While this might seem as a question that is not research level, I think otherwise. We teach Advanced Calculus to students and if we want them to be ready to do research in Analysis, we need to teach them with such problems. Edit. I actually knew all the references listed in the answers. Clearly, the answers show that there is no good source of "ready to use" problems in Advanced Calculus of several variables. REPLY [2 votes]: By this assumptions, I think you can find many interesting problems in the real analysis books that have chapters about several variable calculus. For example, I suggest the book: Problems and Solutions in Real Analysis, 2nd Edition, by Masayoshi Hata. There are a lot of good problems with detailed solutions about several variable calculus in this book. Also, you can see the section of problem and solution of the Mathematical Monthly journals and select the relevant problems.<|endoftext|> TITLE: Motivation for Karoubi envelope/ idempotent completion QUESTION [5 upvotes]: This is the second part of my venture to become more comfortable with the concept of idempotent elements and idempotent splittings from category theoretical viewpoint. In the first part we considered the interpretation of idempotent elements & splitting from viewpoint of commutative algebra. The most fruitful analogy (at least for me) was that if we consider the category $\text{$R$-ModFree}$ of free $R$-modules, taking its completion means making it closed under taking direct summands. As the direct summands of free modules are exactly the projective modules completing means "to add some objects" which occur naturally as building blocks. Now in case of commutative algebra projective objects allows to deal with projective resolutions and provide a framework for direct calculations of derived functors of right exact functors. I read that there are a lot of constructions spreaded in a relatively wide areas of mathematics where one starts with a certain category $C$, construct from this one another say $F(C)$, and then pass to its idempotent completion $\widehat{F(C)}$. Probably the most prominent example is the construction of pure motives where we start with category $(\operatorname{Sm}/k)$ of smooth varieties over a field $k$, then pass to category of correspondences $\operatorname{Cor}_k$, build its idempotent completion $\widehat{(\operatorname{Cor}_k)} $ and go ahead with the construction to build the category of Motives $\operatorname{Mot}_k$ and then, by trying to mimic the procedure of building the derived category, we arrive at the category of pure motives (of course that's just a very coarse overview). The point of my interest is the necessity of taking idempotent completion in the intermediate step. Of course, that's just an example, but similar strategies occur for example in $K$-theory when one study vector bundles or in constructions dealing with triangulated categories. My Question: Can there be extracted a common motivation in these examples making the step that takes idempotent completion necessary or does it in every construction almost everywhere strongly depend on "what one wants"? The only one "general mantra" that I found up to now having the $\text{$R$-Mod}$ example in mind was the necessity of projective objects in order to study right exact functors. Question: Is this the only motivation or are there some other common deep reasons for the importance of taking idempotent completions? REPLY [3 votes]: My understanding of the use of Karoubian completion for motives is that one would really like to have an abelian category of pure motives (modulo homological equivalence, say). However, we don't know how to adjoin all kernels and cokernels, and the Karoubian completion is the best we can do. There is a hope for an abelian category of pure motives that has all the nice properties we want. There are many flavours of motives around (Chow, André, Nori, Voevodsky, ...), and each of them satisfies some but not all of the desired properties. You use whichever one is most convenient for your problem. (As Mikhail Bondarko pointed out: Chow motives modulo numerical equivalence¹ are semisimple abelian, and this is basically the only way we know how to prove Chow motives form an abelian category. However, this result of Jannsen was only proven in 1992, so I don't think it was the original motivation.) ¹The problem with Chow motives modulo numerical equivalence is that it does not have a cohomological realisation, unless we prove standard conjecture D.<|endoftext|> TITLE: Homology of perfect complexes QUESTION [6 upvotes]: I apologize in advance if this question is basic. If $P_{\bullet}$ is a perfect complex over say a ring $R$ such that $H_{i}(P_{\bullet})=0 $ if $i\neq n$ $H_{i}(P_{\bullet})=E$ if $i=n$ is $E$ a finitely generated $R$-module ? What can we say about the homology of a generic perfect complex in general? REPLY [2 votes]: It is even finitely presented See Lemma 14.1.27 of the book Derived Categories (also available at the arXiv at https://arxiv.org/abs/1610.09640).<|endoftext|> TITLE: Sum of squares and partitions QUESTION [7 upvotes]: This is an off-shot from my previous post on MO. Given an integer partition $\lambda=(\lambda_1,\dots,\lambda_{\ell(\lambda)})$ of $n$, denote $\ell(\lambda)$ to be the length of $\lambda$. Let $r_2(n)$ denote the number of ways of expressing $n$ as a sum of two squares of integers, look it up on OEIS. Here is perhaps a "new formulation" of $r_2(n)$: QUESTION. Is this true? If it is, please either provide a reference or a proof. $$r_2(n) =\sum_{\lambda\vdash n}(-1)^{n-\lambda_1}{\prod_{j=1}^{\ell(\lambda)}} \,4\,(\lambda_j-\lambda_{j+1})$$ where $\lambda_{\ell(\lambda)+1}=0$ and the product excludes $\lambda_j=\lambda_{j+1}$. Example. Take $n=4$. The solutions to $4=x^2+y^2$ are $(\pm2,0), (0,\pm2)$ and hence $r_2(4)=4$. On the other hand, $\lambda=(4,0), (3,1,0), (2,2,0), (2,1,1,0), (1,1,1,1,0)\vdash 4$ so that $$(-1)^{4-4}4\cdot4+(-1)^{4-3}4\cdot2\cdot4\cdot1+(-1)^{4-2}4\cdot2 +(-1)^{4-2}4\cdot1\cdot4\cdot1+(-1)^{4-1}4\cdot1=4.$$ REPLY [15 votes]: Start by checking that the following formal product can be expanded as a sum over partitions $$\prod_{i\geq 1}\left(1+\sum_{r\geq 1}a_r(x_1x_2\cdots x_i)^r\right)=\sum_{\lambda}\left(\prod_{j\geq 1}a_{\lambda_j-\lambda_{j+1}}\right)\left(\prod_{j\geq 1}x_j^{\lambda_j}\right)$$ with the convention that $a_0=1$. The proof of this is really just a weighted version of the usual generating function for partitions. If we set $x_1=t$ and $x_i=-t$ for $i\geq 2$, and all $a_r=4r$ for $r\geq 1$ then the right side becomes $$\sum_{n\geq 0} t^n\sum_{\lambda\vdash n}(-1)^{n-\lambda_1}{\prod_{j=1}^{\ell(\lambda)}} \,4\,(\lambda_j-\lambda_{j+1})$$ and the identity says that this is equal to the product $$\prod_{i\geq 1} \left(1+\frac{4(-1)^{i-1}t^i}{(1-(-1)^{i-1}t^i)^2}\right)=\frac{(1+t)^2}{(1-t)^2}\cdot \frac{(1-t^2)^2}{(1+t^2)^2}\cdot \frac{(1+t^3)^2}{(1-t^3)^2}\cdots$$ $$=\prod_{k\geq 1}\frac{(1-t^{2k})^{10}}{(1-t^k)^4(1-t^{4k})^4}.$$ This last final expression is a well known product formula for $r_2(n)$: $$\sum_nr_2(n)t^n=\frac{\eta(t^2)^{10}}{\eta(t)^4\,\eta(t^4)^4}, \qquad \text{where $\eta(t)=t^{\frac1{24}}\prod_{k=1}^{\infty}(1-t^k)$ is the Dedekind eta function}.$$ So your identity follows.<|endoftext|> TITLE: Is there such a thing as a weighted Kan extension? QUESTION [6 upvotes]: The title pretty much sums it up. More in detail. Let $C$, $D$ and $E$ be categories, let $F:C\to D$ and $G:C\to E$ be functors, and let $P:C^{op}\to \mathrm{Set}$ be a presheaf. The colimit of $F$ in $D$ satisfies $$ D(\mathrm{colim} \,F, d) \cong [C,D](F, d) $$ for each object $d$ of $D$, where in the right-hand side $d$ denotes the constant functor, and $[C,D]$ the functor category. This can be seen as a special case of a Kan extension, which satisfies $$ [E,D](\mathrm{Lan}_G F, K) \cong [C,D](F-,K\circ G-) $$ for each functor $K:E\to D$. Namely, by setting $E$ the terminal category we get exactly a colimit. Just as well, a colimit is a special case of a weighted colimit, which satisfies $$ D(\mathrm{colim}_W \,C, d) \cong [C^{op}, \mathrm{Set}](W-, D(F-, d)) $$ for each object $d$ of $D$. We get an ordinary colimit by setting $W$ to be the constant presheaf at the singleton. Now, is there a common generalization? Note that In the Kan extension, the "dependent variable" of $F$ is paired to $K\circ G$, while in the weighted colimit, it is paired to $W$. So it's unclear how to fit both dependencies together. One can express Kan extensions as particular weighted colimits - this is not what I'm asking. (I could ask the same question for the enriched case.) Any reference would also be welcome. REPLY [6 votes]: Yes. Given $F:C\to D$ and a profunctor $H:E$ ⇸ $C$, i.e. a functor $H : C^{\rm op}\times E\to \rm Set$ (or to the enriching category $V$), the $H$-weighted colimit of $F$ is the functor $L : E \to D$ such that each value $L(e)$ is the $W(-,e)$-weighted colimit of $F$ (in a coherent way). Of course, if $E$ is the unit category this reduces to an ordinary weighted colimit. On the other hand, if $G:C\to E$ and $H(c,e) = E(G(c),e)$ is the corresponding representable profunctor, this reduces to a (pointwise) Kan extension. There are real advantages of viewing weighted colimits and Kan extensions in this profunctory light. In particular, this is the natural definition of "weighted (co)limit" that makes sense in the abstract generality of a proarrow equipment or a Yoneda structure. In this paper I found it very useful to obtain a good notion of (co)limit in a new kind of category. It also has good formal properties for relating limits and colimits; see for instance Prop. 8.5 of ibid.<|endoftext|> TITLE: Understanding fundamental group of Poincare homology sphere QUESTION [7 upvotes]: I'm currently reading Knots, Links, Braids, and 3-Manifolds by V. V. Prasolov and A. B. Sossinsky. I have trouble understanding the following picture. The dashed line denotes a trefoil whose tubular neighborhood is to be cut out, and the thickened line denotes a longitude with frame number 1 around the dashed trefoil. Therefore, after pasting back the solid torus, the meridian would be pasted onto the thickened line therefore bounding a disk in the new 3-manifold. Hence we would have a nontrivial relation in the fundamental group. What I don't understand is the specific order of elements depicted in the circle on the right. Any help is greatly appreciated. Thank you very much. REPLY [8 votes]: It turns out that the order is actually not that important: choose a vertex on the circle to start from and a direction to travel in, and then read letters. If the orientation of the edge disagrees with your direction of travel, invert the letter (so save yourself some trouble by traveling counter-clockwise). If you and I happen to make identical choices except for a choice of starting vertex, then the words we write down will differ by conjugation (in the free group on the set of generators). If additionally we disagreed about the direction of travel, then one of us will have to invert our word. However, the relations we each come up with will be true in both presentations, because they only differ by conjugation and inversion in the free group.<|endoftext|> TITLE: Sum of degree differences for simple graphs QUESTION [8 upvotes]: For a simple graph $G$ on $n$ vertices, let us define $$\mathcal{I}_{n}(G)=\sum_{i,j=1}^{n}|\deg\ x_{i}-\deg\ x_{j}|^{3}.$$ I know that there are many different topological indices defined and studied for graphs. Have You ever seen such that was defined similar as above? Can You provide any references? I am highly interested in finding $\sup \mathcal{I}_{n}$ over all graphs with $n$ vertices (or at least some tight upper bound). What I have tried myself, was noticing that $\mathcal{I}_{n}$ must be maximized by a threshold graph - these graphs produce degree sequences that are extreme points of he convex hull of all degree sequences. But this didn't lead me too far. I will be glad for any insight. REPLY [5 votes]: I will guess that the optimum occurs for $k$ isolated vertices and a complete graph on the other $n-k$ where $k=\lfloor\frac{n+1}5\rfloor.$ The same count occurs for $k$ vertices of degree $n-1$ and no other edges so the other $n-k$ have degree $k.$ Past that I have these observations: A graph $G$ and the complement $\bar G$ give the same value to the sum. If the maximum degree in an optimal $G$ is $\Delta$ then any degree $\Delta$ vertex is connected to any other. This is because connecting two such increases some of the $|\deg(x_i)-\deg(x_j)|$ but decreases none. Similarly two vertices with the minimum degree are non-adjacent. For the type of graph I defined above, the count is $k(n-k)(n-k-1)^3.$ The maximum over the reals occurs at $$k=\frac{3\,n-\sqrt {4\,{n}^{2}-n+1}-1}5\approx \frac{n}{5}-\frac3{20}.$$ As commented, the exponent of $3$ is relevant. Take the conjectured optimal case of a $K_{4t}$ and $t$ isolated vertices. Deleting one edge reduces $2t$ degree difference from $4t-1$ to $4t-2$ and increases $2(4t-2)$ other differences from $0$ to $1.$ If one is summing the square or cubes of the differences that is worse. But with exponent $1$ that is an improvement. NOTE Based on limited calculations, The same things seem maximal if we replace the exponent of 3 by 2<|endoftext|> TITLE: Inductive definition of Bernstein polynomials QUESTION [8 upvotes]: For $n\in \mathbb{N}$ let $B_n$ be the linear operator taking a function $f$ on the unit interval $I=[0,1]$ to its $n$-th Bernstein polynomial $B_nf$, $$ B_nf(x):=\sum_{k=0}^n\binom{n}{k} f\Big(\frac{k}{n}\Big)x^k(1-x)^{n-k}\label{1}\tag{1}$$ The polynomial $B_nf(x)$ has a natural probabilistic interpretation, namely, it is the expected value of $f(\xi)$, where $\xi=\frac{1}{n}\sum_{j=1}^n \omega_j$ is the average of $n$ independent random variables with identical Bernoulli distribution of parameter $x$, that is, $\mathbb{P}(\omega_j=1)=x$. In fact, this is the starting point in the beautiful Bernstein's proof of the Weierstrass' density theorem via the WLLN. However, this question is about an alternative definition of the sequence $(B_n)_{n\ge0}$. Let $D:C^1(I)\to C^0(I)$ be the derivative operator, and for all $n\ge1$, let $D_n:C^0(I)\to C^0(I)$ be the approximate discrete derivative given by the incremental ratio $$D_nf(x):=\frac{f\big( \frac{n-1}{n} x+\frac{1}{n}\big)-f\big( \frac{n-1}{n} x\big)}{\frac{1}{n}}, $$ (which is well-defined for $f\in C^0(I)$ and $x\in I$). It is easy to check that definition \eqref{1} implies $$DB_n=B_{n-1}D_n\label{2}\tag{2}$$ together with: $$B_0f(x)=B_nf(0)=f(0)\label{3}\tag{3}$$ Conversely these two imply formula \eqref{1}, as it follows immediately by induction, at least, if we already have it (quite a common situation of formulas proven by induction). Thus, since \eqref{2} and \eqref{3} characterize $(B_n)_n$, we may even take them as an inductive definition of $(B_n)_n$. Note that replacing $D_n$ with $D$ in \eqref{2} gives the analogous inductive definition for the Taylor polynomials in $0$. (Incidentally, formula \eqref{2} is relevant in the approximation theory, in that it implies that for $f\in C^k(I)$ one has $B_nf\to f$ in $C^k$: this by induction from the case $k=0$, since $D_n$ converges strongly to $D$. Also, it says that if some derivative $f^{(k)}$ is non-negative on $I$, so is $(B_nf)^{(k)}$.) Question: How can we deduce naturally formula \eqref{1} (i.e., assuming we don't know it, and we do not have a crystal ball to guess it) from \eqref{2} and \eqref{3}? REPLY [4 votes]: A comment on Josif Pinelis' formula $(b)$ for $\Delta_{n-k+1} \dots\Delta_{n-1}\Delta_{n}$, which is a main point of the computation. Let $\{\tau_{a}\}_{a\in\mathbb{R}}$ and $\{\delta_{b}\}_{a\in\mathbb{R}_+}$ denote respectively the linear group of translations on functions (that we may think defined on the whole real line w.l.o.g.), $f(\cdot)\mapsto f(\cdot+a)$, and the linear group of dilations, $f(\cdot)\mapsto f(\cdot b)$. So $$\tau_{a+b}=\tau_a\tau_b,$$ $$\delta_{ab}=\delta_a\delta_b,$$ $$\tau_{ab}=\delta_b^{-1}\tau_a\delta_b$$ Since $\Delta_n:=\delta_{\frac{n-1}{n}}\big(\tau_{\frac{1}{n}}-\mathbb{1}\big)$, moving all dilations on the left by the above relations imply nicely $$\Delta_{n-k+1} \dots\Delta_{n-1}\Delta_{n}=\delta_{\frac{n-k}{n}}\big(\tau_{\frac{1}{n}}-\mathbb{1}\big)^{k},$$ whence $$\frac{1}{k!} D^kB_n=\frac{1}{k!}B_{n-k} D _{n-k+1} \dots D _{n-1} D _{n}=\Big({n\atop k}\Big)B_{n-k}\delta_{\frac{n-k}{n}}\big(\tau_{\frac{1}{n}}-\mathbb{1}\big)^{k},$$ which we can expand to formula $(b)$. edit. In fact we may skip the last expansion too, keeping all Josif's formulas on the level of operators. Since the $D_k$'s lower the degree of polynomials, $(2)$ and $(3)$ imply that $B_n$ takes values on polynomials of degree less than or equal to $n$, as said. So, for any $x$, denoting $e_x$ the evaluation form, $$ e_xB_n=e_0\bigg[\sum_{k=0}^n \frac{x^k}{k!}D^kB_n\bigg]=e_0\bigg[\sum_{k=0}^n x^k \Big({n\atop k}\Big)B_{n-k}\delta_{\frac{n-k}{n}}\big(\tau_{\frac{1}{n}}-\mathbb{1}\big)^{k}\bigg]=$$ $$=e_0\bigg[\sum_{k=0}^n \Big({n\atop k}\Big)x^k\big(\tau_{\frac{1}{n}}-\mathbb{1}\big)^{k}\bigg]=e_0\bigg( \mathbb{1} + x \big(\tau_{\frac{1}{n}}-\mathbb{1}\big) \bigg)^n =e_0\bigg( x \tau_{\frac{1}{n}} + (1-x)\mathbb{1} \bigg)^n$$ $$=e_0\bigg(\sum_{k=0}^n \Big({n\atop k}\Big)x^k(1-x)^{n-k}\tau_{\frac{k}{n}} \bigg) $$ which indeed takes $f$ to the original $(B_nf)(x)$ given by $(1)$.<|endoftext|> TITLE: Example of closed 4 manifold with $\mathbb{S}^1$ action with 1 fixed point and free away from it QUESTION [5 upvotes]: I am looking for a smooth closed 4-manifold $M$ with a distinguished point $x\in M$, endowed with an $\mathbb{S}^1$ action such that the stabilizer of $p\in M\setminus\{x\}$ is trivial and $x$ is fixed. A naive attempt: If we consider the action given by $\mathbb{S}^1$ on $\mathbb{S}^3\subset \mathbb{C}^2$ that gives the Hopf fibration we can extend this action to the 4-ball (it is an unitary action), and it will have the origin as single fixed point. Unfortunately this manifold is not closed. REPLY [15 votes]: Such a closed $4$-manifold does not exist, and this follows from: Church, P., & Lamotke, K. (1974). Almost free actions on manifolds. Bulletin of the Australian Mathematical Society, 10(2), 177-196 Let me present the argument anyway. The answer breaks down into a local and global part. The local question is to understand what happens near the single fixed point of the $S^1$-action. The global question is about whether we can find, for each local model near the fixed point, a $4$-manifold which closes off the boundary of the model. Claim 1: The only local model near the fixed point is the "naive attempt" from your question statement, i.e. the standard $S^1$-action on $\mathbb{C}^2$. Proof of Claim 1: Up to coordinate change, we may assume the $S^1$-action acts by orthogonal transformations on $\mathbb{R}^4$. In particular, it preserves and is free on $S^3$, and the only free $S^1$-action on $S^3$ is the Hopf (or anti-Hopf if we invert one coordinate) fibration. So your naive attempt really is the only local model. Claim 2: There is no closed $4$-manifold with the property you ask for. Proof of Claim 2: Suppose there were such a closed $4$-manifold. Removing a ball around $p$ corresponding to the local model, $\widetilde{M} := M \setminus B_p$. Then we obtain a fibration $$S^1 \rightarrow \widetilde{M} \rightarrow X,$$ such that on the boundary the fibration is just the Hopf fibration $\partial\widetilde{M} = S^3$ over $\partial{X} = S^2$. The fibration over $X$ comes with its classifying map $X \rightarrow BS^1$, and the composition $$S^2 = \partial X \hookrightarrow X \rightarrow BS^1$$ classifies the Hopf fibration. But the image of $S^2$ under the classifying map for the Hopf fibration $S^2 \rightarrow BS^1$ (which in more down-to-earth language is just $\mathbb{C}\mathbb{P}^1 \hookrightarrow \mathbb{C}{P}^{\infty}$) represents a nontrivial homology class, while $X \rightarrow BS^1$ is a nullhomology of this class. So we arrive at a contradiction. If you prefer characteristic classes, it suffices to consider the first Chern class in this argument. In addition, we see that if we have a $4$-manifold with an "almost free" $S^1$-action, then the number of fixed points is even, since each fixed point either adds or subtracts $1$ from the first Chern class, depending upon orientations. Conversely, this argument can be boosted to prove you can find a closed $4$-manifold with any even number of fixed points. EDIT (responding to comment) You can ask this question for other group actions, and again, there’s the local and global parts to consider. Let me do this for the case of $\mathbb{Z}_p$ (the cyclic group, not the $p$-adics for anybody who might be confused) which was asked in a comment. We claim that in this case, again, there is no $4$-manifold $M$ with a $\mathbb{Z}_p$ action which is free except for a single fixed point. (Hopefully I haven't made a mistake, which is entirely possible, so feel free to be skeptical. I'm sure an algebraic topologist on this site has a better argument for the last part.) Again, we start with the local models, which arise from the representation theory of $\mathbb{Z}_p$. For any finite cyclic group, the irreducible real representations are either 1-dimensional (act by the +1 or -1, the latter if p is even) or 2-dimensional (a rotation of order $p$). Since you want every point to be free except the fixed point, you can't use the 1-dimensional representations, since they have order 1 and 2 respectively, so you have a direct sum of two rotations by some 2$\pi k_1/p$ and $2\pi k_2/p$ where $k_1$ and $k_2$ are coprime to $p$. Up to $\mathrm{Aut}(\mathbb{Z}_p)$, we may assume $k_1 = 1$, and we will simply write $k_2 = k$. At the boundary $S^3$ of this local model, we obtain the fibration $$\mathbb{Z}_p \rightarrow S^3 \rightarrow L(p;k)$$ where $L(p;k)$ is a Lens space. Now for the global part. Assume there exists such a $4$-manifold $M$. Then we have the classifying map for the bundle $S^3 = \partial \widetilde{M} \rightarrow \partial X = L(p;k)$ factors through the classifying map for $X$ (which is now itself a $4$-manifold): $$\phi \colon L(p;k) = \partial X \hookrightarrow X \rightarrow B\mathbb{Z}_p = K(\mathbb{Z}_p,1).$$ The cohomology ring $H^*(K(\mathbb{Z}_p,1);\mathbb{Z}_p)$ can be completely understood in terms of Steenrod squares and the Bockstein homomorphism, and in particular, we find that for $u \in H^1(K(\mathbb{Z}_p,1);\mathbb{Z}_p)$ the fundamental class, the class $$v:= u \smile \beta(u) \in H^3(K(\mathbb{Z}_p;1)\mathbb{Z}_p)$$ is a nontrivial element (in fact a generator). Then one can check (e.g. from the cohomology ring structure of $L(p;k)$) that $\phi^*v \neq 0$ as well. Dually, the pushforward of the mod $p$ fundamental class represents a nontrivial element $$\phi_*[L(p;k)] \neq 0 \in H_3(K(\mathbb{Z}_p,1);\mathbb{Z}_p),$$ and so $\phi$ cannot factor through a $4$-manifold $X$. So no such $4$-manifold $M$ exists.<|endoftext|> TITLE: Cofinality for coends? QUESTION [13 upvotes]: Recall that a functor $I \xrightarrow u J$ is cofinal if it has the property that for any functor $J \xrightarrow F C$, we have that $\varinjlim F \cong \varinjlim Fu$ via the canonical map, either side of the equation existing if the other does. This notion is very useful: it admits various reformulations which can be checked directly; there are numerous practical examples, and once a functor is known to be cofinal, the property which I've just treated as a definition becomes a great tool for computing colimits. When passing from colimits to coends, it would be nice to have similarly powerful tools available. Coends can be reduced to colimits -- e.g. we have $\int^{j \in J} F(j,j) = \varinjlim(Tw(J) \to J^{op} \times J \xrightarrow F C)$, where $Tw(J)$ is the twisted arrow category. But from examples I've tried in the past, my impression is that it's relatively uncommon for a functor $I \xrightarrow u J$ to induce a cofinal functor $Tw(I) \xrightarrow{Tw(u)} Tw(J)$. Question 1: What are some examples of functors $I \xrightarrow u J$ which induce cofinal functors $Tw(I) \xrightarrow {Tw(u)} Tw(J)$? Are they really quite rare? Even if I'm right in thinking that such functors are rare, we shouldn't be deterred. After all, a coend is not the colimit of an arbitrary functor out of $Tw(J)$ -- but rather one which factors through $J^{op} \times J$. So there may still be functors $I \xrightarrow u J$ out there which always induce equivalences of coends, even if $Tw(I) \xrightarrow{Tw(u)} Tw(J)$ is not cofinal. Question 2: What are some examples of functors $I \xrightarrow u J$ such that for any $J^{op} \times J \xrightarrow F C$, we have $\int^{j \in J} F(j,j) \cong \int^{i \in I} F(ui,ui)$ (via the canonical map), either side existing if the other does? Finally, a more systematic question about these functors: Question 3: Is it possible to give a direct combinatorial characterization of functors $u$ of the form described in Question 1 or Question 2 (analogous to the usual characterization of cofinal functors via connectedndess of slice categories)? I'm also interested in versions of these questions other settings like enriched category theory or $\infty$-category theory. Obviously, everything should have a dual story about limits and ends, too. REPLY [8 votes]: Offline, Alex Campbell independently suggested a similar approach to the one Roald mentions in the comments, and worked it out. Here are the results -- we work with ends rather than coends for simplicity: We observe that if $I^{op} \times I \xrightarrow F C$ is a functor, then the end $\int_{i \in I} F(i,i)$ is precisely the limit $\varprojlim_{Hom_I} F$ of $F$ weighted by the Hom-functor $Hom_I: I^{op} \times I \to Set$. Thus, we can apply the weighted version of initiality (the "limit" version of cofinality -- the more modern thing seems to be to say "final" for what I called "cofinal" above), which says in general that Initiality for Weighted Limits: Let $I \xrightarrow u J$ be a functor, let $\phi: I \to Set$ and $\psi: J \to Set$ be functors (which we regard as "weights" for weighted limits), and let $\eta: \phi u \Rightarrow \psi$ be a natural transformation. Then the following are equivalent: For any $C$ and any functor $J \xrightarrow F C$, we have $\varprojlim_\psi F \cong \varprojlim_\phi F u$ via the canonical map induced by $\eta$, either side existing if the other does. $\eta$ exhibits $\psi$ as the Left Kan extension $\psi = Lan_u \phi$ of $\phi$ along $u$. In particular, we can apply this in the case where $\phi = Hom_I$, $\psi = Hom_J$, and $\eta$ is given by the action of the functor $u$. The left Kan extension can be computed explicitly via a coend formula, and the result is the following: Proposition (Initiality for Ends): Let $I \xrightarrow u J$ be a functor. The following are equivalent: For every functor $F: J^{\mathrm{op}} \times J \to C$, we have $\int_{j \in J} F(j,j) \cong \int_{i \in I} F(ui,ui)$ via the canonical map, either side existing if the other does. For every $j,j' \in J$, the canonical map $\int^{i \in I}Hom_J(j,ui) \times Hom_J(ui,j') \to Hom_J(j,j')$ is an isomorphism. There are various ways to reformulate (2). For instance, The composite of profunctors $Hom_J(1,u) \circ_I Hom_J(u,1)$ is canonically isomorphic to $Hom_J$. For every $j\xrightarrow \beta j' \in J$, the "category of $u$-factorizations" of $\beta$ -- whose objects consist of triples $i \in I, j \xrightarrow \alpha ui \xrightarrow {\alpha'} j'$ composing to $\beta$ (morphisms are the obvious thing) -- is connected. [ABSV] For any $C$, the functor $Fun(u,C): Fun(J,C) \to Fun(I,C)$, given by precomposition with $u$, is fully faithful. [ABSV again] The functor $u$ is absolutely dense, i.e. for any $j \in J$ we have $j = \varinjlim (u / j \to J)$ and the colimit is absolute. In the ABSV paper linked to above, such functors are called "lax epimorphisms" in light of (5) above (the idea being that a "pseudo-epimorphism" is a functor $u$ such that $F(u,C)$ is always a pseudo-monomorphism, which has something to do with the core of the categories involved, but here we take into account non-invertible 2-cells of $Cat$). In light of (5) above, one might also say "co-fully-faithful" or something like that. Any localization is an example of such a functor. So is any composite or transfinite composite of localizations. The transfinite composites of localizations form the left half of a factorization system on $Cat$ whose right half is the conservative functors, and it's not hard to see that if $u$ is co-fully-faithful, in the factorization $u = wv$ with $v$ being a transfinite composite of localizations and $w$ being conservative, both $v$ and $w$ are co-fully-faithful. Thus when we look beyond localizations, it seems the appropriate thing to ask is "which conservative functors are co-fully-faithful?". For example, I think the functor from a category to its idempotent completion is co-fully-faithful (while also being fully faithful and in particular conservative). I think that's about all there is to say about co-fully-faithful functors which are also fully-faithful -- any such functor induces an equivalence of idempotent completions (one way to see this is to use the absolute density condition above with the Yoneda embedding). But of course, there may be quite a lot of daylight between co-fully-faithful functors which are conservative and those which are fully faithful. The co-fully-faithful functors also seem related to the "liberal" functors (functors $u$ such that $Fun(u,C)$ is always conservative) of CJSV: co-fully-faithful implies liberal but not conversely. Of course, the enriched and $\infty$-categorical counterparts of all of this should be clear at a conceptual level, at any rate. Note also that everything is self-dual: a functor $u$ is co-fully-faithful iff $u^{op}$ is, so the questions about ends and coends are actually equivalent (and not just dual).<|endoftext|> TITLE: Necessary conditions for the existence of solution of Sylvester equation AX=XB QUESTION [13 upvotes]: Let's consider square matrices $A_{n \times n}$, $B_{n \times n}$ and $X_{n \times n}$ with elements from $\mathbb{R}$. Could you tell me please, what would be the necessary conditions for the existence of solution (may be not unique) of Sylvester equation: $$ AX=XB. $$ As I know, sufficient condition looks like (but probably it is a necessary and sufficient condition) $$ \sigma_p(A) \cap \sigma_p(B) \neq \varnothing, $$ here $\sigma_p(A)$ and $\sigma_p(B)$ are the spectra of matrices $A$ and $B$. REPLY [32 votes]: This equation always has a solution: $X = O$. I'll assume throughout this answer that you're interested in a non-zero solution. The equation $AX = XB$ is equivalent to $(A \otimes I - I \otimes B^T)\mathbf{x} = \mathbf{0}$, where $\otimes$ denotes the Kronecker product and $\mathbf{x}$ is the vectorization of $X$. Your question is thus equivalent to asking when the matrix $A \otimes I - I \otimes B^T$ is not invertible (i.e., when $0$ is not an eigenvalue of $A \otimes I - I \otimes B^T$). Since the eigenvalues of $A \otimes I - I \otimes B^T$ are exactly the sums of the eigenvalues of $A$ and $-B$, the condition that you wrote ($\sigma_p(A) \cap \sigma_p(B) \neq \varnothing$) is in fact both necessary and sufficient.<|endoftext|> TITLE: Connectedness, loops and formal moduli problems QUESTION [10 upvotes]: Let $k$ be an algebraically closed field of characteristic zero. Formalizing a classical folk concept, Pridham and (in a different way,) Lurie defined a formal moduli problem (over $k$) to be a functor from local Artin CDGA's to homotopy types satisfying a certain sheaf condition. If the commutativity condition is weakened to an $E_n$ condition, any formal moduli problem is (uniquely) representable by an $E_n$ algebra; in the commutative case, a formal moduli problem is not necessarily representable by a CDGA, but rather by a (derived) Lie algebra. There's an intuitive picture I like for this in a special case, and I want to understand how it fits into the general picture. Namely, say that $G$ is an affine algebraic Lie group. Let $BG$ be its classifying stack. Then the deformation problem of maps to $BG$ (relative to a choice of point $*\to BG$) is classified by the Lie group $\mathfrak{g}$. Of course this picture involves a group rather than a "formal stack", which is my intuition for (the opposite category to) formal moduli problems, and I am curious how hard it is for a formal moduli problem to be representable by a group object. From the point of view of spaces, going from groups to spaces "is not very hard": the category of groups is equivalent (via the delooping functor) to the category of connected pointed spaces. On the other hand, for $E_n$ algebras, the category of cogroup objects is equivalent to the category of (suitably defined) $(n,1)$-commutative Hopf algebras, which is more closely related (via Koszul duality) to the category of $E_{n+1}$ algebras than to $E_n$ algebras; in particular, the delooping functor from group objects to $E_n$ algebras is far from fully faithful. Now by formal nonsense, there is a loop-deloop pair of adjoint functors $$B:GFMP\leftrightarrows FMP:\Omega,$$ where $FMP$ is the category of ($E_\infty$) formal moduli problems and $GFMP$ is the category of cogroup objects in $FMP$. It seems natural to ask the following questions, to which I don't know the answer: Is $B$ fully faithful, and if so what is its image? Is there an analogue on the level of (co)group objects in Lie algebras to the Koszul duality functor from $(n,1)$ Hopf algebras to $E_{n+1}$ algebras? A natural further question, which I suspect is harder, is to ask whether there is an "algebraic point of view" via Lie-like objects for $E_\infty$ objects in $FMP$. REPLY [10 votes]: The presentation of the formal moduli problems story in Gaitsgory-Rosenblyum A Study in Derived Algebraic Geometry, Vol 2 may be what you are looking for. We review it here (in the case over $\mathrm{Spec}\, k$ for a field $k$ of characteristic zero, that the question concerns): 1. Looping/delooping equivalence in formal DAG Just like the familiar adjoint equivalence in the homotopy theory of spaces $$\mathrm B: \mathrm{Grp}_{\mathbb E_1}(\mathcal S)\simeq \mathcal S_*^{\ge 1}:\Omega,$$ there is an analogous adjoint equivalence in formal DAG $$\mathrm B:\mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k) \simeq \mathrm{FMP}_k:\Omega,$$ where the loop space functor is in both cases given by $\Omega X = \mathrm{pt}\times_X \mathrm{pt}$, as per usual in homotopical/$\infty$-categorical things. Note that formal moduli problems are already inherently pointed, by the assumption that $X(k)$ is contractible. 2. Formal groups and Lie algebras Now, $\mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k) = \mathrm{FGrp}_k$ is (an incarnation) of the $\infty$-category of (derived) formal groups over $k$. Thanks to the characteristic zero assumption, there is a further equivalence $$ \mathrm{Lie}:\mathrm{FGrp}_k \simeq \mathrm{LieAlg}_k : \exp $$ with derived Lie algebras (as modelled for instance by dg Lie algebras). Just like expected, the derived Lie algebra $\mathfrak g$ corresponding to the formal group $G$ is $\mathfrak g = T_{G, e}$, the tangent fiber at the unit. 3. Formal moduli problems and Lie algebras The celebrated Lurie-Pridham identification between formal moduli problems and derived Lie algebras is precisely the composite of these two equivalences of $\infty$-categories $$ \mathrm{FMP}_k\simeq \mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k)=\mathrm{FGrp}_k\simeq \mathrm{LieAlg}_k. $$ That is It sends a formal moduli problem $X$ to $$ \mathrm{Lie}(\Omega X) = T_{\Omega X, e} = T_{X, x_0}[-1], $$ where $x_0$ is the base-point of $X$, unique up to a contractible space of choices $X(\kappa)$. This shifted tangent fiber carries a canonical Lie algebra structure coming from (i.e. as the Lie algebra of) the $\mathbb E_1$-group structure of $\Omega X$. The inverse equivalence $\Psi: \mathrm{LieAlg}_k \simeq \mathrm{FMP}_k$ is then given by $\Psi(\mathfrak g)= \mathrm B\exp(\mathfrak g)$. 4. Lurie's formula for $\Psi$ You may justifiably complain that this description of the inverse functor $\Psi$ does not look the same as the one in Lurie's writing. Let's see how to get it in that form. Let's assume that the formal moduli problem $\Psi(\mathfrak g)$ is formally affine (true under some finiteness assumptions on $\mathfrak g$), in the sense that $$ \mathrm B\exp(\mathfrak g)= (\mathrm{Spf}\,k)/\exp(\mathfrak g)\simeq \mathrm{Spf} \,k^{\mathfrak g}. $$ Here of course the formal spectrum is the usual functor $\mathrm{Spf}:(\mathrm{CAlg}^\mathrm{aug}_k)^\mathrm{op}\to \mathrm{FMP}_k$. The derived Lie algebra invariants may be computed via the Chevalley-Eilenberg complex, thus $k^\mathfrak g\simeq {C}^*(\mathfrak g)$. Then for any Artinian $k$-algebra $A$ we have $$ (\Psi(\mathfrak g))(A) \simeq \mathrm{Map}_{\mathrm{CAlg}^\mathrm{aug}_k}({C}^*(\mathfrak g), A). \qquad \quad(1) $$ This is why Lurie tells us to consider the "Koszul duality functor" $\mathfrak D: (\mathrm{CAlg}_k^\mathrm{aug})^\mathrm{op}\to \mathrm{LieAlg}_k,$ right-adjoint to the Chevalley-Eilenberg cochains functor. Indeed, (through a little use of the finiteness of $\mathfrak g$) we get $$(\Psi(\mathfrak g))(A) \simeq \mathrm{Map}_{\mathrm{LieAlg}_k}( \mathfrak D(A), \mathfrak g),\qquad \qquad(2) $$ which is how Lurie tells us to define $\Psi$. Note that this is going the other way than the $C^*\dashv\mathfrak D$ adjunction. Indeed, without finiteness assumptions on $\mathfrak g$, the formal moduli problem $\Psi(\mathfrak g)$ will not necessarily be formally affine, and the formula (1) will not necessarily work. On the other hand, as Lurie teaches us, formula (2) will always work. 5. Loose ends If I understand the original question correctly, this is presenting the formal moduli problem story precisely like the intuitive picture mentioned. Indeed: any formal moduli problem may be written as $X\simeq \mathrm B G$ for a derived formal group $G$, and the formal moduli problem (of mapping into) $\mathrm BG$ is classified by the Lie algebra $\mathfrak g$. That said, the question makes analogy with the $\mathbb E_n$-algebra version of this story too. I am a little confused about the points raised - in particular, it seems like the following is asserted: a formal moduli problem on $\mathbb E_n$-algebras is represented by an $\mathbb E_n$-algebra. That is, so far as I understand, incorrect. Instead, any FMP on $\mathbb E_n$-algebras in classified by an $\mathbb E_n$-algebra, in a way somewhat analogous to the way that the functor $\mathfrak D$ presents formal moduli problems in the commutative case with Lie algebras. In particular, there seems to be no need to think about cogroup objects, as the equivalence of $\infty$-categories $\mathrm{FMP}_k^{\mathbb E_n}\simeq \mathrm{Alg}^\mathrm{aug}_{\mathbb E_n}$ is covariant. $\qquad$ Edit: A question was raised in the comments whether group objects in $\mathrm{FMP}_k^{\mathbb E_n}\simeq \mathrm{Alg}_{\mathbb E_n}^\mathrm{aug}$ correspond to $\mathbb E_{n+1}$-algebras. Unless I am misunderstanding the question, the answer is negative. A monoid structure (of which a group structure is a particular example of) on an $\mathbb E_n$-algebra $A$ is given by a map $A\times A\to A$, plus coherence data. On the other hand, an additional $\mathbb E_1$-algebra structure on $A$ (which is equivalent to making it into an $\mathbb E_{n+1}$-algebra by Dunn Additivity) is given by a map $A\otimes_k A\to A$, plus coherence data. So a group object in $\mathbb E_n$-algebras, and an $\mathbb E_{n+1}$-algebra are different structures. This might feel a little weird because we're used to tensor products corresponding to products of schemes. Alas, unlike the contravariant $\mathrm{Spec}$, the equivalence $\Psi:\mathrm{Alg}^{\mathrm{aug}}_{\mathbb E_n}\simeq \mathrm{FMP}_k^{\mathbb E_n}$  is covariant. (PS: another difficulty: the tensor product only becomes the coproduct on the level of $\mathbb E_\infty$-algebras, not $\mathbb E_n$-algebras, so even a contravariant equivalence wouldn't necessarily do the trick). Rather, the situation is the same as in the $\mathbb E_\infty$-case above: there is an equivalence $$\mathrm B:\mathrm{Grp}_{\mathbb E_1}(\mathrm{FMP}_k^{\mathbb E_n})\simeq \mathrm{FMP}_k^{\mathbb E_n}:\Omega.$$ This is a special case of the following general phenomenon: for any operad $\mathcal O$, the loops functor $\Omega: \mathrm{Alg}_{\mathcal O}(\mathrm{Mod}_k)\to \mathrm{Grp}_{\mathbb E_1}(\mathrm{Alg}_{\mathcal O}(\mathrm{Mod}_k))$ is an equivalence of $\infty$-categories (A Sudy in Derived Algebraic Geometry, Vol 2, Chapter 6, Proposition 1.6.4). The above claims are special cases for $\mathcal O$ the Lie operad and the $\mathbb E_n$-operad respectively.<|endoftext|> TITLE: Interpreting proper elementarily equivalent end extensions? QUESTION [6 upvotes]: Is there a tuple of parameter-free formulas $\Phi$ and a nonstandard $M\models PA$ such that $\Phi^M\models PA$, the induced $M$-definable initial segment embedding $j_\Phi^M:M\rightarrow\Phi^M$ is non-surjective, and $M\equiv \Phi^M$? (Here by "$\Phi^M\models PA$" I mean "$\Phi$ defines an interpretation of a structure in the language of arithmetic in $M$, and that structure satisfies $PA$." Per Joel's comment below, we may freely require additionally that $j^M_\Phi$ be elementary.) If we allow parameters the answer is yes, but in both the arguments there parameters are absolutely essential. I vaguely recall$^1$ a not-too-hard negative proof via Kripke's notion of fulfillability (see Putnam or Quinsey) but I can't reconstruct it at the moment. $^1$I also recalled seeing this about the version with parameters, which turned out to be bogus; I think the parameter-free version is what I was actually thinking of, but now I'm much less sure I'm not just making stuff up. REPLY [4 votes]: This answer is an attempt at explaining my critical posted comments on Hamkins' proposed answer; it also expands my posted comments to the MO question. I will explain: (a) the gap in Hamkins' answer, (b) how it can be fixed (at the cost of considerably strengthening the hypotheses of the question), and (c) relevant history and literature. (a) [the gap] Suppose $M$ is a model of $PA$ and $\Phi$ consists of pair of arithmetical ternary formulae $\phi_{\rm{add}}(x,y,z)$, $\phi_{\rm{mult}}(x,y,z)$ for how to re-interpret (the graphs of) addition and multiplication such that (1) through (3) below hold: (1) The model $\Phi^M$ satisfies $PA$, i.e., ($M$, $\phi^{M}_{\rm{add}}(x,y,z)$, $\phi^{M}_{\rm{mult}}(x,y,z)) \models PA$. (2) $M$ and $\Phi^M$ are elementarily equivalent. (3) The $M$-induced initial segment embedding $j^M_{\Phi}:M \rightarrow \Phi^M$ is not surjective. Then as observed by Hamkins in his answer, by replacing $M$ by its elementary submodel $M_0$ of $M$ consisting of definable elements of $M$, the above conditions remain true. So far, so good. The gap in the proposed proof occurs at the end of its first paragraph, where it is asserted that $j^{M_0}_{\Phi}:M_0 \rightarrow \Phi^M_0$ is an elementary embedding. The reasoning given is: "Since $M_0$ is pointwise definable, however, elementary equivalence will imply full elementarity, since all parameters are definable". However, for $j^{M_0}_{\Phi}$ to be an elementary embedding of $M_0$ into $\Phi^{M_0}$ we need to know that if $a \in M_0$ is definable in $M$ (and also in $M_0$) as the unique $x$ satisfying $\psi(x)$, then $j(a)$ is the unique $x$ satisfying $\psi(x)$ in $\Phi^M_0$, which is not warranted by the mere assumption that $j$ is an initial embedding. (b) [the fix]. The gap explained above can be readily seen to be fillable if WE ASSUME, FURTHERMORE, THAT $M$ DOES NOT CONTAIN ANY NONSTANDARD DEFINABLE ELEMENTS (equivalently: $M$ is elementarily equivalent to the standard model of arithmetic $\Bbb{N}$, which is often paraphrased as "$M$ is a model of true arithmetic"). The reasoning: initial embeddings are automatically $\Delta_0$-elementary, so this extra assumption guarantees that $ a\in M_0$ is definable in $M$ via "the least $x$ such that $\delta(x)$ holds" for some $\Delta_0$-formula $\delta(x)$, which in turn guarantees that $j(a)$ is the unique $x$ satisfying $\delta(x)$ in $\Phi^M_0$. (c) [history and literature] A few years before the appearance of Tennenbaum's famous characterization of the standard model of arithmetic as the only model of $PA$ (up to isomorphism) whose addition and multiplication are recursive, Feferman published an abstract in 1958 (in the Notices of AMS) to announce his result that there is no nonstandard model of true arithmetic (i.e., a model elementarily equivalent to the standard model of arithmetic $\Bbb{N}$) whose addition and multiplication are arithmetically definable, i.e., he proved: Theorem (Feferman, 1958) The standard model of arithmetic $\Bbb{N}$ cannot interpret a nonstandard model of true arithmetic. An early exposition of the above theorem of Feferman can be found in this 1960 paper of Scott. The theorem is also exposited as Theorem 25.4a in this edition of the textbook Computability and Logic (by Boolos, Burgess, and Jeffrey), but mysteriously with no reference to Feferman. Many years later, Feferman's theorem was rediscovered and fine-tuned in by K. Ikeda and A. Tsuboi, in their paper Nonstandard models that are definable in models of Peano Arithmetic, Math. Logic Quarterly, 2007. Theorem 1.1 of this paper says the following (in the language used here). Theorem. (Ikeda and Tsuboi, 2007) Suppose $M$ is an elementary extension of $\Bbb{N}$ (i.e., $M$ is a model of true arithmetic), and suppose that $M$ interprets a model $\Phi^M$ such that $M$ and $\Phi^M$ are elementarily equivalent. Then the $M$-induced initial segment embedding $j^M_{\Phi}:M \rightarrow \Phi^M$ is an isomorphism. The proof of Ikeda and Tsuboi for Theorem 1.1 is essentially the one obtained by gluing Hamkins' answer (to the MO question) to the fix (b) above. Ikeda and Tsuboi also pose the MO question and answer discussed here as an open question in their paper in item 2 of Remark 3.6. To my knowledge the problem remains open. Finally, let me emphasize that the above discussion all pertained to parameter-free interpretations, since it is well-known that if $M$ is a recursively saturated model of $PA$, then there is some $\Phi(x)$ and some parameter $m \in M$, such that $M$ and $\Phi(m)^{M}$ are elementarily equivalent, and the $M$-induced initial segment embedding $j^M_{\Phi(m)}:M \rightarrow \Phi(m)^{M}$ is a nonsurjective embedding. Here is the proof outline: by recursive saturation, Th($M$) is in the standard system of $M$, and coded by some $m \in M$ such that, in the eyes of $M$, $m$ codes a consistent theory. So by the arithmetized completeness theorem, $M$ can build a model of $m$ whose elementary diagram is definable in $M$; the induced initial embedding is not surjective by Tarski's undefinability of truth.<|endoftext|> TITLE: Multiplying all the elements in a group QUESTION [27 upvotes]: Let $G = \{ g_i | i = 1, ...,n \}$ be a finite group and denote by $G!$ the multiset consisting of all the products of all different elements of $G$ in any order, that is $$ G! = [ \prod_i g_{\sigma(i)} | \sigma \in S_n] $$. I'm interested in knowing how $G!$ behaves as a set, and also how often does every element appear (i.e., how it behaves as a multiset). In the case of an abelian $G$, $G!$ is either 1 or (iff exists) the single element of order 2 in $G$. We can use the abelian case to get a (seemingly tight) upper bound on $G!$ as a set, by considering modding by $[G,G]$: projecting every such product to the quotient, which is abelian, we get $a^{\#[G,G]} \in G/[G,G]$ where $a$ is the single element of order 2 in $G/[G,G]$ if it exists, and the identity otherwise. Thus, if either $a$ is the identity or $\#[G,G]$ is even, $G!$ is entirely contained in $[G,G]$, and otherwise contained in the coset corresponding to $a$. The reason this bound seems tight is that it's generally easy to get elements in the commutator group (up to the element of order 2): if $a,b,a^{-1},b^{-1}$ are distinct, $[a,b] \in G!$ by putting every other element of $G$ right next to it's inverse except for $a^{\pm1}, b^{\pm1}$ and likewise for products of commutators and so on. Is it always the case that $G!$ is a coset of the commutator group? How often does each element appear? It might also be useful to look at the action of $Aut(G)$ on $G!$, but I'm not totally sure what can that tell us. REPLY [41 votes]: Yes, your $G!$ (as a set) is always either $[G,G]$ (if the order of $G$ is odd, or its Sylow $2$-subgroup is non-cyclic) or $z[G,G]$ if $G$ has cyclic Sylow $2$-subgroup, where $z$ is the involution in the Sylow $2$-subgroup. This was apparently a question/conjecture of Golomb (see p. 973) and independently of Fuchs (in a 1964 seminar), proved by Dénes and Hermann. (Addressing Mark Sapir's comment, I could not find a published reference for Fuchs, but did manage to track down this paper by Dénes and Keedwell which contains a discussion of some of the history of the question.)<|endoftext|> TITLE: Aleksandrov's proof of the second order differentiability of convex functions QUESTION [12 upvotes]: Aleksandrov [A], proved a remarkable property of convex functions. Theorem. If $f:\mathbb{R}^n\to\mathbb{R}$ is convex, then for almost every $x\in\mathbb{R}^n$ there is $Df(x)\in\mathbb{R}^n$ and a symmetric $(n\times n)$ matrix $D^2f(x)$ such that $$ \lim_{y\to x} \frac{|f(y)-f(x)-Df(x)(y-x)-\frac{1}{2}(y-x)^TD^2f(x)(y-x)|}{|y-x|^2}=0. $$ I know two proofs of this result. One based on the theory of maximal monotone functions and one based on the fact that the second order distributional derivatives of a convex function are Radon measure. Both proofs are mentioned in Second order differentiability of convex functions. Since these proofs use relatively modern techniques not available during Aleksandrov's time, his argument must have been very different. Question 1. Can you briefly explain what was the idea of the original proof due to Aleksandrov? My guess would be that his proof was based on methods of differnetial geometry. What else could he use in those days? Question 2. In there any textbook where I can find the original proof due to Aleksandrov? [A] A. D. Alexandroff, Almost everywhere existence of the second differential of a convex function and some properties of convex surfaces connected with it. (Russian) Leningrad State Univ. Annals [Uchenye Zapiski] Math. Ser. 6, (1939), 3–35. REPLY [10 votes]: The paper On the second differentiability of convex surfaces by Bianchi, Colesanti, and Pucci (Geometriae Dedicata volume 60, pages 39–48 (1996)) concerns the proof of the Busemann-Feller-Alexandroff Theorem on the second order differentiability of convex functions. Its introduction gives brief synopses on several different methods of proof, including the original argument of Busemann-Feller and later Alexandroff, the two methods you mentioned in the other question (the monotone operator method of Rockafeller (using a result of Mignot); and the measure/distribution method of Reshetnyak), as well as a different one by Bangert (using almost purely differential geometric methods). The paper gives also a new proof of the theorem, which is claimed to be in the same spirit of the original arguments of Busemann-Feller and Alexandroff. The authors considered the second order difference quotient of the convex function based at a point $x$, which they show has a limit a.e. as a convex function. This new convex function is related to the argument of Busemann-Feller in that the indicatrices constructed by Busemann-Feller are the 1-level-sets of this limited convex function.<|endoftext|> TITLE: Which free strict $\omega$-categories are also free as weak $(\infty,\infty)$-categories? QUESTION [7 upvotes]: There are a number of formalisms available for presenting free strict $\omega$-categories -- Street's parity complexes, Steiner's directed complexes, computads, polygraphs,... Typically one has a certain category $\mathcal C$ of combinatorial data, a straightforward notion of "map" from $C \in \mathcal C$ to a strict $\omega$-category $X$, and a free functor $F: \mathcal C \to \omega Cat$ with an explicit combinatorial description of $F$. The free functor will have the universal property that strict $\omega$-functors $F(C) \to X$ are in bijection with "maps" $C \to X$ whenever $X$ is a strict $\omega$-category. Depending on one's choice of model, there may still be a clear notion of "map" from $C \in \mathcal C$ to a weak $(\infty,\infty)$-category $Y$. I'm interested to know conditions on $C \in \mathcal C$ guaranteeing that "maps" $C \to Y$ are in bijection with morphisms $i(F(C)) \to Y$, where $i: \omega Cat \to (\infty,\infty)Cat$ is the inclusion from strict $\omega$-categories to weak $(\infty,\infty)$-categories. Of course, this may be model-dependent. I find myself needing a result of this form in a particular setting, but I'd be interested seeing results of this kind for any choice of $\mathcal C$ and any model of weak $(\infty,\infty)$-categories -- I'd be particularly happy if the model of weak $(\infty,\infty)$-categories is nonalgebraic in nature. Question: What is an example of a category $\mathcal C$ of "presentations of certain strict $\omega$-categories", a free functor $F: \mathcal C \to \omega Cat$ from $\mathcal C$ to strict $\omega$-categories with an explicit combinatorial description, a 1-category $(\infty,\infty)Cat$ which "models" the $\infty$-category of weak $(\infty,\infty)$-categories (e.g. via a model structure or whatever) with inclusion functor $i: \omega Cat \to (\infty,\infty)Cat$, a straightforward notion of "map" from objects $C \in \mathcal C$ to objects of $(\infty,\infty) Cat$, and a (not completely vacuous) condition $\Phi$ on the objects of $\mathcal C$ such that Objects $C \in \mathcal C$ satisfying $\Phi$ have the property that $Hom(iF(C),Y)$ is naturally isomorphic to the set of maps from $C$ to $Y$, for all (suitably fibrant, perhaps) $Y \in (\infty,\infty)Cat$? I'm happy to see quite restrictive conditions $\Phi$; in fact in my case I don't need to understand much more than Gray tensor powers of the arrow category $\bullet \to \bullet$. I suspect that something along the lines of "$F(C)$ is gaunt" or "loop-free" or something may often do the trick, but I'd be happy with something more or less restrictive. I suppose I'd also be happy to see examples with "$n$" replacing "$\omega$" and "$(\infty,n)$" replacing "$(\infty,\infty)$". And let me stress that I'm not looking for some kind of fancy $\infty$-categorical freeness -- when I say $Hom(i(F(C)),Y)$ above, I mean $Hom$ in whatever 1-category is being used to model $(\infty,\infty)Cat$. Although if there are results showing something fancier, that would be interesting to hear about too. REPLY [2 votes]: Important progress on this question has been made by Yuki Maehara, who shows in Orientals as free $\omega$-categories that the complicial nerve of the $n$th oriental is a complicial anodyne extension of the $n$-simplex (providing a fibrant replacement thereof in Verity's model structure on complicial sets). In other words, at least in the complicial set model, the orientals -- which a priori have a strict universal property -- have the expected weak universal property too.<|endoftext|> TITLE: How far does this restricted definition on $\mathcal{O}$ goes? QUESTION [5 upvotes]: $\mathcal{O}$ notation describes an onto function $f:\mathcal{O} \rightarrow \omega_{CK}$. In calculating all values $n \in \mathbb{N}$ such that $f(n)=\alpha$, when $\alpha$ is a limit, all indexes $e$ of ordinary programs are considered such that $\phi_e(i)=n_i$ (with $i \in \mathbb{N}$). The values $n_i$ must satisfy the condition that $f(n_i)=\alpha_i$ with the $\alpha_i$'s forming a (fundamental) sequence for $\alpha$. I am just interested in the variation where we don't consider indexes of normal programs but instead consider indexes of primitive recursive functions/programs (given a suitable indexing for them). (Q1) How far would such a variation go? That is, what is the smallest ordinal it can't represent. (Q2) Consider the notation system (described by a 1-1 function $g:\beta \rightarrow \mathbb{N}$) assigning a unique number to each ordinal in which on limit values $\alpha<\beta$ we seek the p.r. function with smallest index $e$ satisfying $\phi_e(i)=g(\alpha_i)$ where the $\alpha_i$'s must form a (fundamental) sequence for $\alpha$. At what value $\beta$ would such a system stop? I don't have proficiency with the underlying theory so if the question seems posed in a strange manner then that might be the reason. REPLY [5 votes]: For Q2, the answer is $\omega^2$, for both recursive and primitive recursive notations. It's not hard to see that every ordinal below $\omega^2$ can be reached. To show that $\omega^2$ cannot be reached, the argument is the same for both primitive recursive and full recursive. For each $a$, we construct a $b$ ensuring that $a$ is not a notation of this sort for $\omega^2$. We will construct $b$ knowing its own index. For recursive notations, this is an application of the recursion theorem. For p.r. notations, even though we don't have the recursion theorem, this is still possible so long as we're willing to specify an a priori p.r. bound on the runtime. Since we intend to kill time by enumerating successors, an exponential bound will be fine. Our first goal is to arrange that $b$ codes $\gamma+\omega$, where $\gamma$ is greatest such that some $c < b$ with $c <_\mathcal{O} a$ codes $\gamma$. So we unwrap the notation $a$ while enumerating successors. As soon as we see some $c < b$ with $c <_{\mathcal O} a$ and $|c|$ larger than any of the others yet seen, a c.e. event, we include $c$ in the sequence we're enumerating. As soon as we see $b <_\mathcal{O} a$, we do something to ensure that $b$ is not a valid notation, e.g. break the monotonicity of our sequence. If $a$ is a notation for $\omega^2$ with $b \not <_\mathcal{O} a$, then $\gamma+\omega < |a|$, and $b$ is the least notation for $\gamma+\omega$ (for any smaller notation would be included in the definition of $\gamma$, contradicting our choice of $\gamma$). So if $a$ is a notation of the desired sort for $\omega^2$, then $b <_\mathcal{O} a$. But then $b$ is not a notation, by construction, so $a$ is not a notation.<|endoftext|> TITLE: Finite dimensional algebras over $\mathbb{Q}$ QUESTION [6 upvotes]: It is known that a finite dimensional basic algebra over an algebraically closed field is isomorphic to the path algebra of a finite quiver modulo an admissible ideal. Question 1: Is the same true for algebras over $\mathbb{Q}$? If not, are there suitable further assumptions that would guarantee this? Question 2: Are there some general useful tools to compute the dimension of the algebra, given the quiver and the ideal? (In concrete, easy cases it might be doable by hand, but I am looking for methods that are more broadly applicable) REPLY [7 votes]: Question 1: No, take any finite field extension of $\mathbb{Q}$. It is basic but has a simple module that is not 1-dimensional and thus it is not of the form $KQ/I$ for $I$ admissible (since all simple modules are 1-dimensional for algebras of the form $KQ/I$). For any field $K$, a basic algebra is isomorphic to a quiver algebra with admissible ideal if and only if the algebra $A$ is split (also called elementary sometimes), meaning that $A/J$ is isomorphic to matrix algebras over $K$. You can find this result in most representation theory books (of finite dimensional algebras) such as in the book of Auslander,Reiten and Smalo or the book by Kirichenko and Drozd. Question 2: The best (computer) tool is the GAP-package qpa: https://folk.ntnu.no/oyvinso/QPA/ . Given a quiver and admissible ideal, you can calculate many information for the algebra, including the dimension.<|endoftext|> TITLE: Understanding the adjunction $\mathfrak{C}:\mathbf{Set}^{\Delta^{op}}\rightleftharpoons \mathbf{Cat}_\Delta:\mathcal{N}$ QUESTION [6 upvotes]: Let $\mathbf{Cat}_\Delta$ denote the category of simplicially enriched categories (ie simplicial objects in $\mathbf{Cat}$ with all face and degeneracy maps bijective on objects). There is a canonical functor $\mathfrak{C}:\Delta\rightarrow\mathbf{Cat_\Delta}$ that takes $[n]$ to the simplicial category $\mathfrak{C}[n]$ defined as follows: For $i, j\in[n]$, let $P_{ij}$ be the poset with objects $J\subseteq[n]$ containing $i$ as a least element and $j$ as a greatest element. (In particular $P_{ij}$ is empty if $i>j$.) Then $\mathfrak{C}[n]$ has objects the set $[n]$, and mapping spaces $\text{Map}_{\mathfrak{C}[n]}(i, j)$ given by the nerve of $P_{ij}$. The morphisms        $\text{Map}_{\mathfrak{C}[n]}(i_0, i_1)\times \text{Map}_{\mathfrak{C}[n]}(i_1, i_2)\times \cdots \times \text{Map}_{\mathfrak{C}[n]}(i_{k-1}, i_k)\rightarrow \text{Map}_{\mathfrak{C}[n]}(i_0, i_k)$ are induced by the poset maps $P_{i_0 i_1}\times\cdots P_{i_{k-1} i_k}\rightarrow P_{i_0 i_k}$ given by $(J_0, \cdots, J_k)\mapsto J_0\cup\cdots\cup J_k$. Finally, morphisms in $\Delta$ are mapped to morphisms in $\mathbf{Cat}_\Delta$ in the obvious way. We can now define a functor (the "homotopy coherent nerve") $\mathcal{N}:\mathbf{Cat_\Delta}\rightarrow \mathbf{Set}^{\Delta^{op}}$ as follows: for $\mathcal{C}$ a simplicially enriched category, let the $n$-simplices of $\mathcal{N}(\mathcal{C})$ be the set $\text{Hom}_{\mathbf{Cat_\Delta}}(\mathfrak{C}[n], \mathcal{C})$. Face and degeneracy maps are given by post-composition with the co-face and co-degeneracy morphisms on the $\mathfrak{C}[n]$ in $\mathbf{Cat_\Delta}$, and $\mathcal{N}(f)$ is given by pre-composition with $f$ for any morphism $f$ in $\mathbf{Cat_\Delta}$. This functor is not too difficult to wrap one's head around; essentially the $n$-simplices of $\mathcal{N}(\mathcal{C})$ can be thought of as "homotopy-coherent diagrams" in $\mathcal{C}$. For instance, $0$-simplices are simply objects of $\mathcal{C}$, and $1$-simplices are vertices of the mapping spaces of $\mathcal{C}$ (ie "morphisms" of $\mathcal{C}$). $2$-simplices are specified by a trio $x, y, z$ of objects of $\mathcal{C}$, a pair of vertices (ie "morphisms") $f\in \text{Map}_\mathcal{C}(x, y)_0$ and $g\in \text{Map}_\mathcal{C}(y, z)_0$, and a 1-simplex $\sigma\in \text{Map}_\mathcal{C}(x, z)_1$ with one of its faces the composition $g\circ f$. $\sigma$ can be thought of as giving a "homotopy" from $g\circ f$ to some other morphism in $\text{Map}_\mathcal{C}(x, z)$. This idea generalizes naturally for higher values of $n$. The issue for me now is in defining a left adjoint to $\mathcal{N}$. This is done is a purely formal way by extending the functor $\mathfrak{C}:\Delta\rightarrow\mathbf{Cat}_\Delta$. Indeed, every simplicial set $S$ is a colimit of a small diagram of $\Delta^i$s in $\mathbf{Set}^{\Delta^{op}}$, and the category $\mathbf{Cat}_\Delta$ admits small colimits, so define $\mathfrak{C}:\mathbf{Set}^{\Delta^{op}}\rightarrow\mathbf{Cat}_\Delta$ by letting $\mathfrak{C}(S)$ be the colimit of the corresponding diagram of $\mathfrak{C}[i]$s. It is immediate by construction that $\mathfrak{C}$ and $\mathcal{N}$ will be adjoint. While I understand this on a formal level, I am having an enormous amount of difficult visualizing what the functor $\mathfrak{C}$ actually "looks like". In particular I am not able to follow computations involving $\mathfrak{C}$ at all. Does anyone know of a way of better understanding this functor on an intuitive level? I've tried to give a more explicit construction of a general $\mathfrak{C}(S)$ but have thus far failed; clearly it should have as its objects the vertices of $S$, but I have come up short in trying to describe its mapping spaces and composition laws. I know that morally $\mathfrak{C}$ should be the "free simplicial category generated by $S$", but it is really unclear to me how this should look. Any insight or references would be greatly appreciated, thank you so much. (Also was not sure whether to post this here or on math.stackexchange; if it's better suited for the latter site feel free to close and I will repost there.) Edit: in light of the comments below perhaps the right question to ask is this; what are some simplicial sets $S$ for which it will be particularly illuminating to compute $\mathfrak{C}(S)$ explicitly? (Rather than trying to do a general computation.) REPLY [2 votes]: To a fairly crude approximation: $\newcommand{\C}{\mathfrak{C}}$think of the functor $\C(-)$ as like geometric realisation, but realising the basic simplices as the categories $\C[n]$ instead of as the topological simplices. Lots of functors out of $\hat{\Delta}$ are defined analogously by left Kan extension of some functor on $\Delta$, and I find thinking of such functors as “roughly like geometric realisation” is always a good first approximation. Of course, this depends on first understanding the categories $\C[n]$ — but you should be already described how to do that, since you describe the simplices appearing in $\mathcal{N}(-)$ as “not too difficult to wrap one’s head around […] homotopy coherent diagrams”, and the categories $\C[n]$ are just the categories that represent such diagrams, i.e. $\C[n]$ consists of essentially just a “homotopy coherent $n$-diagram” and nothing more. The other thing that can be helpful for concrete computations is noticing that the functor $\C(-)$ preserves cofibrations — in particular, it maps the the boundary inclusions of simplicial sets to injective-on-objects functors, and colimits of such functors are fairly well-behaved in $\mathrm{Cat}$. So given a small combinatorial simplical sets, write it out explicitly as a “cell complex”, i.e. a composite of pushouts of the boundary inclusions; then the realisation functor will take this to a corresponding “cell complex” in $\mathrm{Cat}$, which should typically give a clear combinatorial description of the realisation.<|endoftext|> TITLE: Reason to apply the Koszul sign rule everywhere in graded contexts QUESTION [5 upvotes]: The Koszul sign rule is a sign rule that arises from graded-commutative algebras. For instance, let $\bigwedge(x_1,\dots, x_n)$ be the free graded-commutative algebra generated by $n$ elements of respective degrees $\lvert x_i\rvert$. Then, the sign $\varepsilon(\sigma)$ of a permutation $\sigma$ on $(x_1,\dotsc, x_n)$ is given by $$x_1\wedge\dotsb\wedge x_n=\varepsilon(\sigma)x_{\sigma(1)}\wedge\dotsb\wedge x_{\sigma(n)},$$ which comes from the fact that in a graded-commutative algebra one has by definition $a\wedge b = (-1)^{\lvert a\rvert\lvert b\rvert}b\wedge a$. There is also an antisymmetric Koszul sign rule which arises from graded-anticommutative algebras and it's just the previous sign times the sign of the permutation. Both signs are used for instance in Lada and Markl - Symmetric brace algebras. However, I've been seeing the Koszul sign rule used in any graded context and even for operations that are not products in some algebra. For example, from Roitzheim and Whitehouse - Uniqueness of $A_\infty$-structures and Hochschild cohomology, given graded maps of graded algebras $f,g:A\to B$, if we want to evaluate $f\otimes g$ in an element $x\otimes y$, apparently we need to apply the sign rule to get $$(f\otimes g)(x\otimes y)=(-1)^{\lvert x\rvert\lvert g\rvert}f(x)\otimes g(y),$$ but I see no mathematical reason to do that, it just seems to be a convention. A more complex example of application of the Koszul sign rule is in the definition of brace algebra (also in the Lada and Markl paper). I could give many more examples. In some of them I can understand the reason. For instance, the differential of a tensor product of complexes $C$ and $D$ cannot simply be $d_C\otimes 1_D+ 1_C\otimes d_D$ (it can be defined this way if we use the sign rule when we apply it to elements, but in any case it needs the sign). But maps in general need not be differentials. In other cases, the signs appear in nature and one use this sign rule to justify them, as in $A_{\infty}$-algebras, but this feels too artificial for me and doesn't really explain why we should use that sign rule. So, in the end, every time there is a sequence $(x_1,\dotsc, x_n)$ of graded objects of any kind and not necessarily all of them of the same kind (elements, maps, operations, …), and related in any way (they can be multiplied, or applied, or whatever), we use the Koszul sign rule to permute the sequence. To me all of this seems more philosophical than mathematical, and as I said it feels to be just a convention. But, is there some general mathematical reason to use the sign rule in such an extensive way? And if it's just a convention, why should we use it? From my experience, it gets very messy when it comes to applying the sign rule to larger formulas, and in the end everything is just a $+$ or $-$ sign, so I see no advantage. REPLY [16 votes]: A precise statement of the conventions (which Jesse is referring to) is that the authors are using the symmetric monoidal structure on graded vector spaces, where the braiding map,, $\tau: V \otimes W \to W \otimes V$, is given by $$v \otimes w \mapsto (-1)^{{\rm deg} ~w ~{\rm deg}~ v} w \otimes v.$$ Roughly, what it means to use this symmetric monoidal structure is that you have to make all of your definitions diagramatically, using only $\tau$ to exchange symbols. For instance suppose we have two algebras $A, B$ and an $A$ module $M$ and a $B$ module $N$. Then if $A,B,M,N$ were ordinary vector spaces, we are used to the fact that $M \otimes N$ is an $A \otimes B$ module. In the graded context, under the Koszul conventions, we define the action $$A \otimes B \otimes M \otimes N \to A \otimes M \otimes B \otimes N \to M \otimes N,$$ where in the first step we have used $1 \otimes \tau \otimes 1.$ Something quite similar is happening in the example your mention. So far, this is more of a unified answer to how rather than why people use this convention. For the question of why, the main reason the Koszul convention is useful in homological algebra has to do with the origin of homological algebra---topology. Consider $\mathbb R^{p +q}$ with its standard orientation. Then the switching map $$\tau: \mathbb R^p \times \mathbb R^q \to \mathbb R^{q} \times \mathbb R^p$$ multiplies this orientation by $(-1)^{p q}$. This fundamental fact manifests itself in several ways. One is that homology functor $H_*(-, k)$ from topological spaces to graded vector spaces is symmetric monoidal, but only with respect to the Koszul sign rule. This means that if one has an algebraic structure on a topological space $X$, then $H_*(X)$ naturally carries the same algebraic structure, but with respect to the Koszul sign rule. For instance, $X$ is always a co-commutative coalgebra, so $H_*(X)$ becomes a graded co-commutative coalgebra (with sign conventions according to the Koszul rule). Something similar happens with the $A_\infty$ operad. Namely, the $A_\infty$ operad is the $dg$ operad obtained by taking the cellular homology of a (cellular) operad in topological spaces. The orientations of the cells of this operad explain the signs which arise. There is also the monoidal Dold Kan correspondence, which you can read about on the nLab. At the end of the day, it is just a convention (and not always the right one) but the relation with topology explains why people like to use it systematically.<|endoftext|> TITLE: Shrinking and stretching of vector bundles QUESTION [5 upvotes]: Let $M$ be a manifold, $p:E\to M$ a rank $d$ vector bundle. Suppose that $U \subset E$ is an open subset such that $U \cap p^{-1}(x)$ is nonempty and convex for all $x \in M$. Is it true that $U \to M$ is a fiber bundle with fiber $\mathbb R^d$? And that $U \cong E$ as fiber bundles? We may assume with no loss of generality that $U$ contains the zero section. This seems like a statement that could be a lemma in any number of textbooks (if true), e.g. in connection with the tubular neighborhood theorem, but I haven't seen it anywhere. Lang proves in his differential geometry book that any vector bundle over a manifold is what he calls compressible, meaning that any open neighborhood of the zero section of $E$ can be shrunk to a smaller open neighborhood which is diffeomorphic to $E$ as a bundle over $M$. REPLY [3 votes]: Since there are no references so far, let me give a sketch proof along the lines of my comment. I'll assume that $M$ is compact. Let's show first that there is a smooth section of $E$ lying in $U$. Indeed, for any point $x\in M$ there is a neighbourhood $U_x$ with a section $s_x$. Take a finite cover $U_i$ of $M$ by such neighbourhoods and take the corresponding partition $1=\sum f_i$ of unity. Then by convexity $\sum s_i f_i$ is a smooth section lying in $U$. Clearly we can assume that $s$ is the zero section (by taking an appropriate fiberwise diffeo), we will assume this from now on. Now we will construct an exhaustion of $U$ by an increasing sequence of fiber-wise compact convex subsets $0\subset {\cal B_1}\subset ... \subset {\cal B_i}\subset ...$ so that $U=\cup_i {\cal B_i}$. Let me show first how to construct one such subset ${\cal B_1}\subset U$. For every point $p\in M$ let us choose some covex compact subset $B_p$ with smooth boundary in the fiber $U_p$. Then, since $U$ is open, there is an open neighbourhood $V_p$ of $p$ in $M$ such that over this neighbourhood there is a smoothly varying family of $B_x$ ($x\in V_p$), such that $B_x\subset U_x$. Take a finite cover of $M$ by such $V_i's$, let $\phi_i$ be the partition of unity. Then the sum $${\cal B_1}=\sum_i \phi_i B_i (x)$$ is the desired subset $B\subset U$. Here by sum I mean the Minkowski sum. It is clear that the interior of $\cal B_1$ is diffeomorphic to the bundle of vectors of length less than $1$ in $E$ (for some fiber-wise Euclidean metric). So the only need to construct a family of $\cal B_i$ that will exhaust $U$. This can be done as in 1).<|endoftext|> TITLE: Concerning $k \subset L \subset k(x,y)$ QUESTION [9 upvotes]: The following is a known result in algebraic geometry: Let $k$ be an algebraically closed field of characteristic zero (for example, $k=\mathbb{C}$). Let $L$ be a field such that $k \subset L \subset k(x,y)$ and $L$ is of transcendence degree two over $k$. Then there exist $h_1,h_2 \in k(x,y)$ such that $L=k(h_1,h_2)$. Is it possible to find $g_1,g_2 \in k[x,y]$ such that $L=k(g_1,g_2)$? The motivation is the following result: If $k \subset L \subset k(x,y)$ is of transcendence degree one over $k$, then $L=k(h)$, where $h \in k[x,y]$; see this answer. Perhaps the arguments in that answer are also applicable here? I have asked the above question in MSE, with no comments. See also this question. Thank you very much! REPLY [5 votes]: No. $M\mathrel{:=}k(x,y)$ has a $k$-automorphism $\sigma:x\mapsto 1/x,\,y\mapsto 1/y$, of order 2. Let $G\mathrel{:=}\langle\sigma\rangle$, and put $L\mathrel{:=}M^{G}$, the fixed field. The elements $x+1/x$ and $y+1/y$ of $L$ are algebraically independent over $k$, hence $L$ has transcendency degree 2. However, no non-constant polynomial $g\in k[x,y]$ can be in $L$ (that is, be invariant under $\sigma$). By the same token, the automorphism $\tau:x\mapsto 1/y,\,y\mapsto 1/x$, fixes the field $k(x/y)$, which has transcendency degree 1 over $k$, but it cannot fix any non-constant polynomial. So $k(x/y)\neq k(h)$ for every $h\in k[x,y]$. (Here, $M$ is again quadratic over the fixed field of $\tau$, so that the latter is of tr. deg. 2 over $k$ again. A transcendental element independent of $x/y$ is $x+1/y$, for instance).<|endoftext|> TITLE: About the number of primes which are the sum of 3 consecutive primes (OEIS A034962) QUESTION [13 upvotes]: I made some numerical simulations about the number of primes which are the sum of 3 consecutive primes (OEIS A034962), that is for instance: $$5+7+11=23$$ $$7+11+13=31$$ $$11+13+17=41$$ $$17+19+23=59$$ $$19+23+29=71$$ $$23+29+31=83$$ $$29+31+37=97$$ $$...$$ The number of such triplets, till a certain integer n, seems to be well approximated by the following function: $$\frac{e\cdot\pi(n)}{\log(n)}$$ If the previous was true, the density of such primes in the whole set of prime numbers would be comparable to that of primes inside the set of naturals. I ask if this estimate is correct and some references about this matter. Thanks. REPLY [24 votes]: The question asks how many primes $p_n \le x$ are there such that $p_n + p_{n+1}+p_{n+2}$ is also prime. This is beyond our reach to answer, but one can use Hardy-Littlewood type heuristics to attack this. Since $p_n + p_{n+1}+ p_{n+2}$ is roughly of size $x$, it has about $1/\log x$ chance of being prime, and so one should expect the answer to be on the scale $\pi(x)/\log x \approx x/(\log x)^2$. But the asymptotic will be a little different, since $p_n + p_{n+1} +p_{n+2}$ is not quite random -- for example it will always be odd (omitting $2+3+5$). Let's flesh out the usual heuristic. To be prime, a number must be coprime to each prime number $\ell$. A random integer has a chance $(1-1/\ell)$ of being coprime to $\ell$. What is the chance that $p_n+ p_{n+1} + p_{n+2}$ is coprime to $\ell$? Each of the primes $p_n$, $p_{n+1}$, $p_{n+2}$ itself lies in some reduced residue class $\mod \ell$. Assuming these possibilities are equally distributed, one should get the chance $$ \frac{1}{(\ell-1)^3} \sum_{\substack{a, b, c=1 \\ (a+b+c,\ell)=1}}^{\ell-1} 1. $$ If $a$ and $b$ are given with $a\not\equiv -b \mod \ell$, then $c$ has $\ell-2$ possible residue classes, and if $a\equiv -b \mod \ell$ then $c$ has $\ell-1$ possible residue classes. So the answer is $$ \frac{1}{(\ell-1)^3} \Big( (\ell-1)(\ell-2)(\ell-2) + (\ell-1)(\ell-1)\Big)= \frac{\ell^2-3\ell+3}{(\ell-1)^2}. $$ So we adjust the random probability at $\ell$ by the factor $$ \Big(1-\frac {1}{\ell}\Big)^{-1} \frac{(\ell^2-3\ell+3)}{(\ell-1)^2}= 1 +\frac{1}{(\ell-1)^3}. $$ For example, when $\ell=2$, this adjustment factor is $2$ reflecting the fact that $p_n+p_{n+1}+p_{n+2}$ is guaranteed to be odd. Then the analog of the Hardy--Littlewood conjecture will be $$ \sim \prod_{\ell} \Big(1+\frac{1}{(\ell-1)^3} \Big) \frac{x}{(\log x)^2}. $$ The constant in the product is approximately $2.3$; definitely not $e$. There are interesting biases concerning why this conjecture would be an underestimate. Work of Lemke Oliver and Soundararajan makes precise conjectures on the distribution of consecutive primes in arithmetic progressions. For example, if we look at $\ell=3$, the sum $p_n + p_{n+1} +p_{n+2}$ can be a multiple of $3$ only if all three primes are congruent to each other $\mod 3$. This configuration is not preferred, and there are significant biases against it for small numbers! Similarly for other $\ell$ also, there are biases that indicate that the actual probability would be a tiny bit larger (with the effect wearing off slowly for large $x$). In other words, one might be able to identify lower order terms on the scale of $x(\log \log x)/(\log x)^3$, which might explain why the numerics suggest a larger value for the constant.<|endoftext|> TITLE: Relevant mathematics to the recent coronavirus outbreak QUESTION [16 upvotes]: I would like to ask about (old* and new) reliable mathematical literature relevant to various mathematical aspects of the recent coronavirus outbreak: In particular, standard statistical/mathematical models that are used to predict the spread, mathematical studies of effectiveness of various strategies, etc. *(Added) By old I also mean well-established models. REPLY [5 votes]: The following paper is extremely important because it has informed the decisions of the UK government that realised (announced) on Monday 16/03/2020 that it can not afford "Herd immunity". The paper only shows the outcomes of the model and speaks about its parameters. It would of course be extremely interesting to know what exactly is the mathematics behind it. Mathematicians should try to read it. https://www.imperial.ac.uk/media/imperial-college/medicine/sph/ide/gida-fellowships/Imperial-College-COVID19-NPI-modelling-16-03-2020.pdf?fbclid=IwAR2Ca5Ki23DWn-EGWeB3yaNE4f9GmnUcEWU_S60lsDC230AKUg4v_w82qeE<|endoftext|> TITLE: Which compact metrizable spaces have continuous choice functions for non-empty closed sets? QUESTION [9 upvotes]: Let $X$ be a compact metrizable space and let $\mathcal{K}_{ne}(X)$ be the collection of non-empty closed subsets of $X$ with the Vietoris topology (i.e. the topology induced by the Hausdorff metric for any compatible metric on $X$). Question: When does there exist a continuous function $f: \mathcal{K}_{ne}(X) \rightarrow X$ such that for every $G \in \mathcal{K}_{ne}(X)$, $f(G) \in G$? This feels like it should have been studied before, but I am unable to find a reference. Some easy observations: If $X$ has a continuous choice function for non-empty closed sets and $Y$ is a closed subspace of $X$, then $Y$ has a continuous choice function for non-empty closed sets. $\inf : \mathcal{K}_{ne}([0,1])\rightarrow [0,1]$ is a continuous choice function for non-empty closed subsets of $[0,1]$. So we also have this for any closed subspace of $[0,1]$, such as Cantor space and any countable compact metrizable space. The circle and the tripod (three copies of $[0,1]$ glued together at $0$) both do not have continuous choice functions for non-empty closed sets (in both spaces given a set with two points there is a continuous path that makes the points switch places while keeping them separate). So no spaces in which these embed do either. Any finite disjoint union of spaces with continuous choice functions for non-empty closed sets also has a continuous choice function for non-empty closed sets (having elements in a clopen subset is a clopen condition in $\mathcal{K}_{ne}(X)$, so we can patch together the choice functions by cases). A reasonable conjecture is that any such space embeds into $[0,1]$, but I could also see something tricky like the pseudo-arc having a continuous choice function for non-empty closed sets. REPLY [6 votes]: It's an old (1981) theorem by Jan van Mill and Evert Wattel (see this paper) that a compact space has a continuous selection iff it is orderable. (So has a linear order whose order topology is the topology on $X$). $F \to \min(F)$ and $F \to \max(F)$ are then the two only continuous selection functions IIRC. Even a continuous selecting function for $[X]^2$, the subspace of doubletons, is enough to get orderabilility.<|endoftext|> TITLE: Solution to differential equation $f^2(x) f''(x) = -x$ on [0,1] QUESTION [14 upvotes]: I'd like to solve a differential equation $$ f^2(x) f''(x)=-x $$ where $f(x)$ is defined on $[0,1]$ and has a boundary condition $f(0)=f(1)=0$. I somehow found out that the solution is fairly close to $f(x) = x^{1/3} \phi^{2/3}(\Phi^{-1}(1-x))$ where $\phi$ and $\Phi$ are pdf and cdf of a standard normal distribution, but it fails to solve the differential equation exactly. Thank for all comments! Based on the solution structure of Emden–Fowler Equation, I was able to identify the values of constants that satisfy the boundary conditions. The followings are the details: Define \begin{equation} Z_R(\tau) \triangleq \sqrt{3} J_{1/3}(\tau) - Y_{1/3}(\tau) , \quad Z_L(\tau) \triangleq - \frac{2}{\pi} K_{1/3}(\tau) \end{equation} where $J, Y, K$ are Bessel functions. Further define \begin{equation} \bar{\tau} \triangleq \inf\{ \tau > 0; Z_R(\tau) = 0 \} \approx 2.3834 , \quad a \triangleq \frac{1}{ \bar{\tau}^{4/3} Z_R'(\bar{\tau})^2 } \approx 0.2910 , \quad b \triangleq a \left( \frac{9}{2} \right)^{1/3} \approx 0.1763. \end{equation} Then, the solution curve $\{ (x, f(x)) \}_{x \in [0,1]}$ is characterized by \begin{equation} \left\{ \left( x_R(\tau), y_R(\tau) \right) \right\}_{\tau \in [0, \bar{\tau}]} \bigcup \left\{ \left( x_L(\tau), y_L(\tau) \right) \right\}_{\tau \in [0, \infty]} \end{equation} where \begin{equation} x_R(\tau) \triangleq a \tau^{-2/3}\left[ \left( \tau Z_R'(\tau) + \frac{1}{3} Z_R(\tau) \right)^2 + \tau^2 Z_R(\tau)^2 \right] , \quad y_R(\tau) \triangleq b \tau^{2/3} Z_R(\tau)^2. \end{equation} \begin{equation} x_L(\tau) \triangleq a \tau^{-2/3}\left[ \left( \tau Z_L'(\tau) + \frac{1}{3} Z_L(\tau) \right)^2 - \tau^2 Z_L(\tau)^2 \right] , \quad y_L(\tau) \triangleq b \tau^{2/3} Z_L(\tau)^2. \end{equation} In addition to this analytic solution, I also obtained a numerical solution by repeatedly computing $$ f_{k+1}(x) \gets \left[ \left( f_k(x-2h) + f_k(x+2h) \right) + 4 \left(f_k(x-h)+f_k(x+h)\right) + \frac{8 x h^2}{f_k^2(x)} \right] \big/ 10 $$ on the grid $x \in \{2h,3h,\ldots,1-3h,1-2h\}$ for small $h$ with an initialization $f_0(x) \triangleq 0.5(1-(1-2x)^2)$. The following figure shows these solutions: REPLY [24 votes]: Surprisingly, this case of the Emden-Fowler equation is explicitly solvable: see formula (2.3.27) in A. Polyanin and V. Zaitsev, Handbook of exact solutions of ordinary differential equations, Chapman & Hill, 2003. I copy the formula, without verifying it. Let $$Z=C_1J_{1/3}(\tau)+C_2Y_{1/3}(\tau),$$ or $$Z=C_1I_{1/3}(\tau)+C_2K_{1/3}(\tau),$$ where $J,Y$ are Bessel and $I$, $K$ are modified Bessel functions. Then $$x=a\tau^{-2/3}[(\tau Z^\prime+(1/3)Z)^2\pm\tau^2Z^2],\quad y=b\tau^{2/3}Z^2$$ satisfy $d^2y/dx^2=Axy^{-2}$ with $A=-(9/2)(b/a)^3.$ For the $+$ sign in $\pm$ take the first formula for $Z$, and for the $-$ the second one. Remark. Emden-Fowler equation appears for the first time in the famous book by R. Emden, Gaskugeln (1907) and since then frequently arises in the study of stars and black holes.<|endoftext|> TITLE: Does there exist a non-zero signed finite borel measure which is zero on all balls? QUESTION [12 upvotes]: Let $(X,d)$ be a compact separable metric space. Let $\mu$ be a Borel, regular, finite, signed measure on $X$ such that for all $x\in X$, for all $r>0$, $\mu(B(x,r))=0$, where $B$ denotes the (either open or closed) ball w.r.t $d$. Is $\mu$ zero? If $\mu$ is positive one can show that $\mu=0$ using the Borel-Lebesgue theorem, but what if $\mu$ is signed? REPLY [11 votes]: Stealing an answer from "user940" on Math.SE, the answer is yes, such measures can exist. In the paper Davies, Roy O., Measures not approximable or not specifiable by means of balls, Mathematika, Lond. 18, 157-160 (1971). ZBL0229.28005. the author constructs a compact metric space $X$ and two distinct Borel probability measures $\mu_1, \mu_2$ that agree on every closed ball. (They must therefore also agree on open balls, because an open ball is a countable increasing union of closed balls.) Taking $\mu = \mu_1 - \mu_2$ provides your desired signed measure.<|endoftext|> TITLE: Algebraic operation corresponding to "taking residues at roots of unity" QUESTION [7 upvotes]: I'm not sure if this is more appropriate for MO or MSE, this is a question that came up in actual research, but it's of a somewhat elementary nature. Let $R$ be the ring of rational functions with poles at roots of unity, is there a sufficiently nice way to algebraically describe the operation on $R$ of "taking the sum of residues at roots of unity of $f(x)dx/x$? This operation takes $\frac{x^a}{x^m-1}$ to $1$ if $a$ is a multiple of $m$ and 0 otherwise, and this determines it on the entirety of $R$. By "sufficiently nice", I mean I'd like to be able to describe it purely algebraically, and hopefully without making reference to complex numbers. What I envisioned was something like "taking the algebraic residue at $x^m-1$", where $m$ is the lcm of the orders of the roots of unity in the denominator of $f$. However this operation is not well defined, as to express an arbitrary polynomial as a series in $x^m-1$ I need to to take $m$th roots. For example, consider the case of $x/x^m-1$, after renaming $x^m-1$ to $\alpha$, the result becomes $Res_{\alpha}\frac{(\alpha+1)^{1/m}}{\alpha}\frac{d\alpha}{\alpha+1}$. I still get the correct answer if I take the appropriate residues at each branch and sum them (resulting in 0), but this still requires complex analysis. I could alternatively just declare such functions have 0 residue, but I don't have any justification for doing that. EDIT: Because of the particulars of my situation, I'm specifically looking for an interpretation somehow in terms of the functions $x^\alpha-1$. REPLY [6 votes]: The sum of the (finite) residues of a rational function $F(x)$ is equal to $$ -\mathrm{Res}(F(x),x=\infty) = \mathrm{Res}\left( \frac{1}{x^2}F(1/x), x=0\right). $$ This is just the coefficient of $x$ in the Laurent expansion with finitely many negative exponents of $F(1/x)$ at $x=0$.