TITLE: Universal graph homomorphisms QUESTION [7 upvotes]: By a graph I mean a pair $G = (V, E)$ where $V$ is a set and $E \subseteq \mathcal{P}_2(V) := \{\{a,b\}: a\neq b \in V\}$. A graph homomorphism between graphs $G, H$ is a map $f:V(G)\to V(H)$ such that $\{v, w\}\in E(G)$ implies $\{f(v), f(w)\} \in E(H)$. We say that a graph homomorphism $u:G\to H$ is universal if for every graph homomorphism $f:G\to H$ there is $v \in V$ such that $f(v) = u(v)$. Is the composition of two universal graph homomorphisms universal? REPLY [7 votes]: Ok, try again. Let $G_k=W_{2k+5}\cup W_{2k+1}$, where $W_k$ is the wheel with $k$ spokes. Then for $k\ge 3$, every homomorphism $G_k\to G_{k-1}$ fixes the hub of the smaller wheel as there is no homomorphism $W_{2k+1}\to W_{2k+3}$, and every homomorphism $W_{2k+1}\to W_{2k-1}$ fixes the hub. Now let $\phi_k:G_k\to G_{k-1}$ be injective homomorphisms (so the bigger wheel goes to the bigger wheel, the smaller to the smaller). Each of these is universal by the argument above. But if you you combine $\phi_{k+1}$ and $\phi_{k}$, the resulting homomorphism $G_{k+1}\to G_{k-1}$ is not universal: just use a homomorphism that switches the wheels.<|endoftext|> TITLE: The (un)reasonable (non-)ubiquity of the Grothendieck construction QUESTION [12 upvotes]: Is there a way to export the Grothendieck construction to different contexts than $Cat$? in theory, if you build $\int F$ out of $F\colon \mathcal C\to Cat$, or $F\colon \mathcal C\to Sets$, as a suitable (lax) pullback along the canonical forgetful $Cat_*\to Cat$, $Sets_*\to Sets$, you can do the same replacing $Sets_*$ with pointed categories, pointed spaces, pointed simplicial sets and their unpointed analogues. The central importance of the category of elements in both Category Theory and Geometry makes me wonder if there is a well-established, abstract way to perform these constructions, capturing the various examples. Is it possible to talk about a Grothendieck construction for suitable functors $F\colon \mathcal C\to $ "something", proveided this something is "nice"? More precisely, is there a "general Grothendieck construction in enriched category theory"? If not (I firmly believe the answer is no, given that weighted limits are seldom reducible to conical ones, and more importantly in non-cartesian situations "conical limits" make no sense), how can I enucleate a set of properties on $\cal V$ in such a way that there is a Grothendieck construction for $\cal V$-presheaves? Edit: apparently the answer to the previous questions is yes. But now the validity of this machinery (Definition 3.1 and the rest of section 3) makes me wonder if the following result remains true in enriched setting. The following properties for a functor $F\colon \mathcal C\to \bf Sets$ are equivalent: $F$ commutes with finite limits; The Yoneda extension $\text{Lan}_{y}F$ commutes with finite limits; $F$ is a filtered colimit of representable functors; The category of elements $\int F$ of $F$ is cofiltered. If yes, how's it proved? (Haven't thought about it yet, so the answer could be trivial.) REPLY [8 votes]: Given a $\mathbb V$-functor $F:X^\mathrm{op}\to \mathbb V$, I came up with the following answer some years back: Consider a $\mathbb V$-distributor $D:X^\mathrm{op}\otimes Y \to \mathbb V$. The Grothendieck construction $\mathrm{Gr}(D)$ is then the following $\mathbb V$-category. Objects are triples $\phi=(x,\phi,y)$ with $x\in X$, $y\in Y$ and $\phi:\mathbf 1\to D(x,y)$ a morphism in $\mathbb V$. Morphism objects are defined the following way. Given $\phi,\psi\in\mathrm{Gr}(D)$ we define $[\phi_1,\phi_2]$ to be the pullback $$[x_1,x_2]\times_{D(x_1,y_2)}[y_1,y_2].$$ As an alternative notation I found a notation remeniscent of set-comprehension usefull: $$\mathrm{Gr}(D)=:\{ x\in X \,\rangle\, D(x,y) \,\langle\, y\in Y \}$$ Showing that this is indeed a $\mathbb V$-category is lengthy but straightforward. The constrution exhibits the following universal property: First, we define a $\mathbb V\!-\!\mathrm{Cat}$-category of $\mathbb V$-distributors. Objects are triples $(X,D,Y)$ with $D:X^\mathrm{op}\otimes Y\to \mathbb V$. Morphism $\mathbb V$-categories are defined as $$\{ (L,R)\in [X_1,X_2]\otimes [Y_1,Y_2] \,\rangle\, [D_1(-,-),D_2(L-,R-)] \,\langle\, \mathbb{1}_{\mathbb V-\mathrm{Cat}} \}.$$ Denoting this category $\mathbb V-\mathrm{Dist}$ we find a $\mathbb V-\mathrm{Cat}$-equivalence (constituting a $\mathbb V-\mathrm{Cat}$-adjunction) $$\mathbb V-\mathrm{Cat}[Z,\mathrm{Gr}(D)]\simeq \mathbb V-\mathrm{Dist}(\mathrm{hom}(Z),D)$$ where $\mathrm {hom}$ is the functor assigning to a $\mathbb V$-category its $\mathrm {hom}$-distributor. In this sense one could call $\mathrm{Gr}(D)$ the covering category of $D$. This adjunction is a straightforward generalisation of the one given in Thomas Streichers notes "Distributors à la Jean Benabou". Another useful name/notation might be 'fibre product': $X\times_D Y$.<|endoftext|> TITLE: Moduli spaces of connections as representation spaces QUESTION [9 upvotes]: It is well known that the moduli space of flat connections over a closed manifold $M$ can be identified with the representation space $Hom(\pi_1(M), G) / G$. Furthermore, Atiyah and Bott (1983) showed that over a closed surface the moduli space of (central) Yang-Mills connections can be seen as the space of representations of a group $\Gamma_R$ (which is a central extension of the fundamental group by $\mathbb{R}$) to $G$. Are there other moduli spaces which can be identified with representation spaces? REPLY [5 votes]: Another interesting example, not mentioned yet, where character varieties come up is the moduli space of polygons. Moduli spaces of $(G,X)$-structures on compact manifolds are also related to character varieties via the Ehresmann--Thurston Theorem. See here for example. They are also related to spaces of phylogenetic trees and spaces of spin networks.<|endoftext|> TITLE: Nonperiodic points of piecewise-linear homeomorphisms QUESTION [11 upvotes]: Suppose $K$ is a compact polytope and $T$ is a piecewise-linear homeomorphism from $K$ to itself. Suppose also that $T$ is not of finite order (that is, for no $n \geq 1$ is it the case that $T^n(x)=x$ for all $x$ in $K$). Can we conclude that a nonperiodic point exists (that is, for some $x$ in $K$ it is the case that $T^n(x) \neq x$ for all $n \geq 1$)? Note that it need not be the case that for each $n \geq 1$ the set of $x$ in $K$ with $T^n(x) \neq x$ is dense. So my question is not just an application of the Baire category theorem (at least under the most simpleminded approach). Ideally I would like to draw stronger conclusions from the stated hypotheses (e.g. that the set of nonperiodic points has positive $d$-dimensional measure where $d$ is the dimension of $K$). Some of the hypotheses on $K$ and $T$ are likely to be red herrings, but I include them since they apply for the class of examples that motivated this question. I have posted a variant of this question in which the piecewise-linearity constraint is dropped; see Nonperiodic points of homeomorphisms of a ball. Montgomery's theorem solves that problem and therefore solves this one, though I'm still hoping for (and will award a bounty to) the best self- contained, clear, and complete proof for the piecewise-linear case. REPLY [4 votes]: The answer is yes. In other words, if the order of the orbit $x_n=T^n(x)$ of each point $x\in K$ is finite then $T^n$ is identity map for some $n$. Choose a simplex $\triangle$ of maximal dimension $m$. Set $$E_n=\{\,x\in\triangle\mid T^n(x)=x\,\}.$$ By Baire theorem, $E_n$ has nonempty interior for some $n$; fix such a value $n_1$. Note that $E_{n_1}$ is a polytope in $\triangle$. If $E_{n_1}\ne \triangle$, we can choose an $m$-simplex $\triangle'\subset \triangle$ such that (1) its base in $E_{n_1}$ (2) the remaining part is outside of $E_{n_1}$ and (3) $T^{n_1}$ is linear on $\triangle'$. Given $\ell_1$, we can choose a simplex $S_1$ in $\triangle'$ so that (1) $T^{n_1\cdot m}(x)\ne x$ if $x\in S_1$ and $m<\ell_1$ and (2) for some $m_1>\ell_1$, the set $E_{m_1\cdot n_1}\cap S_1$ has nonempty interior but $E_{m_1\cdot n_1}\not\supset S_1$. Indeed (1) follows if entire $S_1$ lies close to the base. Further by Bair theorem $S_1\cap E_{m_1\cdot n_1}$ has nonempty interior for some $m_1>\ell_1$. Note that $x\notin E_{m_1\cdot n_1}$ if $x$ lies terribly close to the base (depending on $m_1$) and we can arrange that there is such point in $S_1$. Set $n_2=m_1\cdot n_1$ and iterate the construction for an increasing sequence of numbers $\ell_i$. We get a nested sequence of compact set $S_1\supset S_2\supset\dots$ such that $T^n(x)\ne x$ for any $x\in S_n$. If $x\in\bigcap_nS_n$ then $T^n(x)\ne x$ for any $n$, a contradiction. We proved that the restriction $T^n|_\triangle$ is identity map for some $n$. Now remove the interior of $\triangle$ from $K$ and pass to the map $T^n$. Repeat this procedure recursively till no simplex is left in $K$. P.S. The same holds for any continuous map, see the answer in here.<|endoftext|> TITLE: Consistency of the collection axiom scheme compared to replacement QUESTION [6 upvotes]: Set theory ZFC- is ZFC without power set, but with replacement. It does not imply the collection axiom scheme, as discussed in http://jdh.hamkins.org/what-is-the-theory-zfc-without-power-set/ Does consistency of ZFC- imply consistency of that plus collection? I do not mind taking all of ZFC- as metatheory. REPLY [6 votes]: The answer is yes, because you can go to $L$. More specifically, if $M$ is a model of ZFC-, defined with replacement, then consider $L^M$, the class of sets that $M$ thinks arise in the constructible hierarchy. It is not difficult to see that $M$ thinks $L_\alpha$ exists for every ordinal $\alpha$ that it has, and ultimately one can see that $L^M$ is a model of ZFC-, since instances of replacement inside $L^M$ amount to instances of replacement in $M$, since $L^M$ is definable in $M$. But now, since $L^M$ has a definable well-ordering of the universe, it has a definable Skolem function and this means that $L^M$ models $\text{ZFC}^-$, defined with collection+separation. So the consistency of ZFC-, defined with replacement, implies the consistency of $\text{ZFC}^-$, defined with collection+separation. That said, the main point of our article is that ZFC-, defined with replacement, is the wrong version of this theory, since so many things go wrong in it.<|endoftext|> TITLE: Partitions of $\mathbb{R}^+$ into subset closed by sum and product QUESTION [12 upvotes]: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a "non trivial" solution of the Cauchy functional equation, i.e. $f$ is not of the form $f(x)=cx$ for any $c\in\mathbb{R}$ and satisfies the relation $$f(x+y)=f(x)+f(y)\quad\forall\,x,y\in\mathbb{R}$$ Define $$A=\{x\in\mathbb{R}:f(x)\geq 0\},\quad B=\{x\in\mathbb{R}:f(x)< 0\}$$ Now $\{A,B\}$ is a partition of $\mathbb{R}$ into two non empty, closed by sum subsets, and it's easy to see that it's different from the trivial ones $\left(\{\mathbb{R}_{\geq 0},\mathbb{R}^-\},\{\mathbb{R}_{\leq 0},\mathbb{R}^+\}\right)$. In particular $$\mathbb{R}^+=(A\cap\mathbb{R}^+)\cup(B\cap\mathbb{R}^+)$$ is a partition of $\mathbb{R}^+$ into two non empty, closed by sum subsets. My question: It's possible to partition $\mathbb{R}^+$ into two non empty set, both closed by sum and product? The analogous question for $\mathbb{R}$ has negative answer, and the argument is pretty easy: we can assume $0\in A$, and suppose that there exists $x\in B$. Now $-x\in A$ (otherwise $-x+x=0\in B$), and then $x^2=(-x)^2\in A$, which is a contradiction. I guess the answer to my question is negative (it seems related to the absence of nontrivial field automorphism of $\mathbb{R}$), but i don't find any efficient argument. Thank you for all suggestions. REPLY [8 votes]: Yes, it is possible to construct such a partition. See this note by Daniel M. Kane. Any proof would have to involve the axiom of choice, otherwise you can't partition the positive reals into subsets closed under addition, let alone both addition and multiplication. Added later: There is a recent paper, "Decomposing the real line into Borel sets closed under addition", where the authors show that the only partitions of $\mathbb R$ into countably many (Borel) sets closed under addition are of the form $\mathbb R_+\cup \{0\}\cup \mathbb R _-$, etc. And they mention that the structure of all partitions of $\mathbb R_+$ into two sets closed under addition and multiplication was determined in a paper by G. Kiss, G. Somlai, T. Terpai, which is still in preparation. (Here is G. Kiss's website.)<|endoftext|> TITLE: Are there nonisotrivial elliptic curves over $\mathbb{G}_m$? QUESTION [13 upvotes]: Is there an elliptic curve over $\mathbb{C}[t, t^{-1}]$ that has a nonconstant $j$-invariant? What is an equation for such a curve, if it exists? REPLY [6 votes]: It is an older topic, and very nice answers have been given. But I am new in MO and let me give another answer. The reason for nonexistence of such an elliptic curve is the same as for nonexistence of such an elliptic curve over $S:=$Spec$(\mathbb{C}[t])$. In order to have a universal elliptic curve one must remove the points from $S$ which correspond to elliptic curves with automorphism group larger than $\{ \pm1\}$, and these fibers have $j$-invariants 0 and 1728. This is explained in the book Arithmetic Moduli of Elliptic Curves by Katz and Mazur. So we must remove at least two points from $S$, i.e. inverting $t$ in $\mathbb{C}[t]$ is not enough. This argument works if we replace $\mathbb{C}$ by any algebraically closed field of characteristic $p \geq 5$. But for $p=2$ or 3 we have $0 = 1728$, so it is enough to remove only one point from $S$ in these cases and this explains the existence of examples given by Elkies above.<|endoftext|> TITLE: A polytope with a bound on the sum of any $k$ variables QUESTION [7 upvotes]: Let $2\le k\le n-1$ and define the polytope $$P_k(n) = \lbrace (x_1,\ldots,x_n) \in \mathbb{R}^n : -1\le x_{i_1}+\cdots +x_{i_k} \le 1 \text{ for all } 1\le i_1\lt\cdots\lt i_k\le n\rbrace.$$ There are $\binom nk$ constraints: the sum of any $k$ variables is in $[-1,1]$. Surely this is known, but I didn't manage to find it. Does this polytope have a name? What is known about it? Changing the lower bound from $-1$ to 0 is just a scale+translate so that would do instead. This problem arose in a graph reconstruction study but isn't needed for that any more. REPLY [4 votes]: After computing the vertices, I've reorganized this answer into four sections: Facts / Definitions, Conjectures / Observed Patterns, Polymake Computations, and the Table of Volumes Some Facts and Definitions: First, note that the defining inequalities of $P_k(n)$ and $P_{n-k}(n)$ can be related by an invertible linear transformation, so they are (combinatorially) isomorphic. In particular, $P_{n-1}(n)$ is isomorphic to $P_1(n)$, which is of course just an $n$-cube. Here's a hopefully not too headache-inducing animation of $P_2(3)$: Next, I define the polytopes $\mathfrak{A}^n$, from E.P. Baranovskii, "Partitioning of Euclidean space into L-polytopes" in the volume "Discrete Geometry and Topology", Proceedings of the Steklov Institute of Mathematics, vol. 196. The $n$-polytope $\mathfrak{A}^n$ is defined as the convex hull on the following $2n+2$ points: the vertices of a regular $n$-simplex $\Delta^n$ centered at the origin and the vertices of the inversion of $\Delta^n$ through the origin. Now recall that a point is a vertex of an $n$-polytope $P$ if and only if it is the unique solution of (at least) $n$ of the defining hyperplanes of $P$ and lies within $P$. $P_k(n)$ is symmetric under inversion through the origin as well as all permutations of the $n$ coordinates. From these two facts, we can prove that $P_k(n)$ always has the following four families of points as vertices: 1) The point $\left(\frac{1}{k},\dots,\frac{1}{k}\right)$; note that it lies at the intersection of the $\binom{n}{k}\geq n$ hyperplanes defined by $\sum_{i=1}^kx_{j_i}=1$ (with the same choice of sign, and over all choices $1\leq j_1\leq\cdots\leq j_k\leq n$). 2) Similarly, its inversion through the origin $\left(-\frac{1}{k},\dots,-\frac{1}{k}\right)$ is also a vertex. 3) The point $\left(\frac{2k-1}{k},-\frac{1}{k},\dots,-\frac{1}{k}\right)$ and the points whose coordinates are permutations of this one ($n$ points in total). This point lies at the intersection of $\binom{n-1}{k-1}+\binom{n-1}{k}=\binom{n}{k}$ hyperplanes defined by $x_1+\sum_{i=1}^{k-1}x_{j_i}=1$ and $\sum_{i=1}^kx_{l_i}=-1$ (over all choices $2\leq j_1\leq\cdots\leq j_{k-1}\leq n$ and $2\leq l_1\leq\cdots\leq l_{k}\leq n$). 4) The inversions of the previously defined points through the origin: $\left(-\frac{2k-1}{k},\frac{1}{k},\dots,\frac{1}{k}\right)$ and permutations. These will be called the four basic families of vertices. Finally, it will be useful to introduce the following notation: $\{x\}^n$ denotes the point $\left(x,\dots,x\right)$. $\{x\}^j\{y\}^{n-j}$ denotes the orbit of the point $\left(x,\dots,x,y,\dots,y\right)$ (where the first $j$ coordinates are equal to $x$ and the last $n-j$ coordinates are equal to $y$) under the permutation of all $n$ coordinates. The number of points in $\{x\}^j\{y\}^{n-j}$ is $\binom{n}{j}$, and the above results show that $\{1/k\}^n$, $\{1/k\}^{n-1}\{-\frac{2k-1}{k}\}^1$, $\{\frac{2k-1}{k}\}^1\{-1/k\}^{n-1}$, and $\{-1/k\}^n$ are always vertices of $P_k(n)$. Conjectures / Patterns observed in the computations below 1) Besides the four basic families of points, the vertices of $P_k(n)$ always take the form $\{x_j\}^j\{y_j\}^{n-j}$ with $x_j>0$ and $y_j<0$, and there is only one such pattern for each distinct value of $j$. Is there a nice formula for $x_j$ and $y_j$ (depending on $j$,$k$, and $n$)? The vertex data below exhibits various patterns: e.g. for $k=2$, $x_j=1/2$ and $y_j=-1/2$ except when $j=1,n-1$. Now let's define the spectrum of $P_k(n)$ to be the values of $j$ such that it has vertices of the above form. If the above conjecture about the form of the vertices is true, then the spectrum of the $n$-cube $P_{n-1}(n)$ must include all $0\leq j\leq n$ (to get the required $2^n$ vertices). Note that by the central symmetry of $P_k(n)$, if the spectrum includes $j$, it also includes $n-j$. The spectra of $P_k(n)$ and $P_{n-k}(n)$ seem to be identical in the computations below, but I haven't checked that the linear transformation between them transforms the $\{x_j\}^j\{y_j\}^{n-j}$ nicely. I have shown above that $0,1,n-1,n$ are always present in the spectrum. One can show that $2$ (and hence also $n-2$) is never present in the spectrum of $P_2(n)$ (I won't write the proof here, but in short it comes down to seeing that one cannot get $n$ linearly independent hyperplanes corresponding to a vertex of that form). It seems from the computations below that those are the only values excluded from the spectrum of $P_2(n)$, but I don't know how to prove this. Showing that other values are excluded or included in the spectra of various $P_k(n)$ seems to require some tricky and intricate arguments about the linearly independent sets of the $2\binom{n}{k}$ bounding hyperplanes. 2) The vertices of $P_m(2m)$, $P_{m}(2m+1)$ and $P_{m+1}(2m+1)$ consist of only the four basic families. In other words, the spectrum of $P_m(2m)$ is $2m,2m-1,1,0$, and the spectra of $P_m(2m+1)$ and $P_{m+1}(2m+1)$ are $2m+1,2m,1,0$. Furthermore, $P_{m}(2m+1)$ (and thus $P_{m+1}(2m+1)$) is isomorphic to $\mathfrak{A}^{2m+1}$, while the f-vector of $P_m(2m)$ agrees with that of $\mathfrak{A}^{2m}$ up to the $(m-1)$th component. 3) The number of $(n-1)$-faces of $P_k(n)$ is precisely $2\binom{n}{k}$; none of the defining inequalities are redundant. (I thought I had a proof of this before but I think it wasn't right). 4) The volumes of $P_2(n)$ are all equal to 4. Computations in polymake for $3\leq n\leq 9$ The polymake files I generated are here (34 MB ZIP). The files Pnk.poly correspond to $P_k(n)$ (note the ordering of the numbers is switched), and the files An.poly correspond to $\mathfrak{A}^n$. $n=3$ $P_2(3)$: Vertices: $\{1/2\}^3$, $\{1/2\}^2\{-3/2\}^1$, $\{3/2\}^1\{-1/2\}^2$, $\{-1/2\}^3$ Spectrum: 3,2,1,0 f-vector: [8 12 6] Isomorphic to $\mathfrak{A}^3$, which is also the 3-cube (as is clear in the above animation). $n=4$ $P_2(4)$: Vertices: $\{1/2\}^4$, $\{1/2\}^3\{-3/2\}^1$, $\{3/2\}^1\{-1/2\}^3$, $\{-1/2\}^4$ Spectrum: 4,3,1,0 f-vector: [10 28 30 12] Isomorphic to the bipyramid over the 3-cube (which I found from finding the f-vector on this page on permutation polytopes). Compare the f-vector of $\mathfrak{A}^4$: [10 40 60 30] $P_3(4)$: Vertices: $\{1/3\}^4$, $\{1/3\}^3\{-5/3\}^1$, $\{1\}^2\{-1\}^2$, $\{5/3\}^1\{-1/3\}^3$, $\{-1/3\}^4$ Spectrum: 4,3,2,1,0 f-vector: [16 32 24 8] Isomorphic to the 4-cube. $n=5$ $P_2(5)$ and $P_3(5)$ Vertices of $P_2(5)$: $\{1/2\}^5$, $\{1/2\}^4\{-3/2\}^1$, $\{3/2\}^1\{-1/2\}^4$, $\{-1/2\}^5$ Vertices of $P_3(5)$: $\{1/3\}^5$, $\{1/3\}^4\{-5/3\}^1$, $\{5/3\}^1\{-1/3\}^4$, $\{-1/3\}^5$ Spectrum: 5,4,1,0 f-vector: [12 60 120 90 20] Isomorphic to $\mathfrak{A}^5$. $P_4(5)$: Vertices: $\{1/4\}^5$, $\{1/4\}^4\{-7/4\}^1$, $\{3/4\}^3\{-5/4\}^2$, $\{5/4\}^2\{-3/4\}^3$, $\{7/4\}^1\{-1/4\}^4$, $\{-1/4\}^5$ Spectrum: 5,4,3,2,1,0 f-vector: [32 80 80 40 10] Isomorphic to the 5-cube. $n=6$ $P_2(6)$ and $P_4(6)$ Vertices of $P_2(6)$: $\{1/2\}^6$, $\{1/2\}^5\{-3/2\}^1$, $\{1/2\}^3\{-1/2\}^3$, $\{3/2\}^1\{-1/2\}^5$, $\{-1/2\}^6$ Vertices of $P_4(6)$: $\{1/4\}^6$, $\{1/4\}^5\{-7/4\}^1$, $\{1/2\}^3\{-1/2\}^3$, $\{7/4\}^1\{-1/4\}^5$, $\{-1/4\}^6$ Spectrum: 6,5,3,1,0 f-vector: [34 204 510 520 210 30] (c.f. f-vector of bipyramid over the 5-cube: [34 144 240 200 90 20]) $P_3(6)$ Vertices: $\{1/3\}^6$, $\{1/3\}^5\{-5/3\}^1$, $\{5/3\}^1\{-1/3\}^5$, $\{-1/3\}^6$ Spectrum: 6,5,1,0 f-vector: [14 84 240 330 200 40] (c.f. f-vector of $\mathfrak{A}^6$: [14 84 280 490 420 140]) $P_5(6)$ Vertices: $\{1/5\}^6$, $\{1/5\}^5\{-9/5\}^1$, $\{3/5\}^4\{-7/5\}^2$, $\{1\}^3\{-1\}^3$, $\{7/5\}^2\{-3/5\}^4$, $\{9/5\}^1\{-1/5\}^5$, $\{-1/5\}^6$ Spectrum: 6,5,4,3,2,1,0 f-vector: [64 192 240 160 60 12] Isomorphic to the 6-cube. $n=7$ $P_2(7)$ and $P_5(7)$ Vertices of $P_2(7)$: $\{1/2\}^7$, $\{1/2\}^6\{-3/2\}^1$, $\{1/2\}^4\{-1/2\}^3$, $\{1/2\}^3\{-1/2\}^4$, $\{3/2\}^1\{-1/2\}^6$, $\{-1/2\}^7$ Vertices of $P_5(7)$: $\{1/5\}^7$, $\{1/5\}^6\{-9/5\}^1$, $\{1/3\}^4\{-1/3\}^3$, $\{1/3\}^3\{-1/3\}^4$, $\{9/5\}^1\{-1/5\}^6$, $\{-1/5\}^7$ Spectrum: 7,6,4,3,1,0 f-vector: [86 742 2226 2800 1610 420 42] $P_3(7)$ and $P_4(7)$ Vertices of $P_3(7)$: $\{1/3\}^7$, $\{1/3\}^6\{-5/3\}^1$, $\{5/3\}^1\{-1/3\}^6$, $\{-1/3\}^7$ Vertices of $P_4(7)$: $\{1/4\}^7$, $\{1/4\}^6\{-7/4\}^1$, $\{7/4\}^1\{-1/4\}^6$, $\{-1/4\}^7$ Spectrum: 7,6,1,0 f-vector: [16 112 448 980 1120 560 70] Isomorphic to $\mathfrak{A}^7$. $P_6(7)$ Vertices: $\{1/6\}^7$, $\{1/6\}^6\{-11/6\}^1$, $\{1/2\}^5\{-3/2\}^2$, $\{5/6\}^4\{-7/6\}^3$, $\{7/6\}^3\{-5/6\}^4$, $\{3/2\}^2\{-1/2\}^5$, $\{11/6\}^1\{-1/6\}^6$, $\{-1/6\}^7$ Spectrum: 7,6,5,4,3,2,1,0 f-vector: [128 448 672 560 280 84 14] Isomorphic to the 7-cube. $n=8$ $P_2(8)$ and $P_6(8)$ Vertices of $P_2(8)$: $\{1/2\}^8$, $\{1/2\}^7\{-3/2\}^1$, $\{1/2\}^5\{-1/2\}^3$, $\{1/2\}^4\{-1/2\}^4$, $\{1/2\}^3\{-1/2\}^5$, $\{3/2\}^1\{-1/2\}^7$, $\{-1/2\}^8$ Vertices of $P_6(8)$: $\{1/6\}^8$, $\{1/6\}^7\{-11/6\}^1$, $\{1/3\}^5\{-2/3\}^3$, $\{1/2\}^4\{-1/2\}^4$, $\{1/3\}^3\{-2/3\}^5$, $\{11/6\}^1\{-1/6\}^7$, $\{-1/6\}^8$ Spectrum: 8,7,5,4,3,1,0 f-vector: [200 2160 8120 13048 10220 4032 756 56] $P_3(8)$ and $P_5(8)$ Vertices of $P_3(8)$: $\{1/3\}^8$, $\{1/3\}^7\{-5/3\}^1$, $\{1/3\}^4\{-1/3\}^4$, $\{5/3\}^1\{-1/3\}^7$, $\{-1/3\}^8$ Vertices of $P_5(8)$: $\{1/5\}^8$, $\{1/5\}^7\{-9/5\}^1$, $\{1/3\}^4\{-1/3\}^4$, $\{9/5\}^1\{-1/5\}^7$, $\{-1/5\}^8$ Spectrum: 8,7,4,1,0 f-vector: [88 704 2632 5684 7420 5040 1400 112] $P_4(8)$ Vertices: $\{1/4\}^8$, $\{1/4\}^7\{-7/4\}^1$, $\{7/4\}^1\{-1/4\}^7$, $\{-1/4\}^8$ Spectrum: 8,7,1,0 f-vector: [18 144 672 1876 3080 2800 1190 140] (c.f. $\mathfrak{A}^8$: [18 144 672 2016 3780 4200 2520 630]) $P_8(7)$ Vertices: $\{1/7\}^8$, $\{1/7\}^7\{-13/7\}^1$, $\{3/7\}^6\{-11/7\}^2$, $\{5/7\}^5\{-9/7\}^3$, $\{1\}^4\{-1\}^4$, $\{9/7\}^3\{-5/7\}^5$, $\{11/7\}^2\{-3/7\}^6$, $\{13/7\}^1\{-1/7\}^7$, $\{-1/7\}^8$ Spectrum: 8,7,6,5,4,3,2,1,0 f-vector: [256 1024 1792 1792 1120 448 112 16] Isomorphic to the 8-cube. $n=9$ $P_2(9)$ and $P_7(9)$ Vertices of $P_2(9)$: $\{1/2\}^9$, $\{1/2\}^8\{-3/2\}^1$, $\{1/2\}^6\{-1/2\}^3$, $\{1/2\}^5\{-1/2\}^4$, $\{1/2\}^4\{-1/2\}^5$, $\{1/2\}^3\{-1/2\}^6$, $\{3/2\}^1\{-1/2\}^8$, $\{-1/2\}^9$ Vertices of $P_7(9)$: $\{1/7\}^9$, $\{1/7\}^8\{-13/7\}^1$, $\{2/7\}^6\{-5/7\}^3$, $\{3/7\}^5\{-4/7\}^4$, $\{4/7\}^4\{-3/7\}^5$, $\{5/7\}^3\{-2/7\}^6$, $\{13/7\}^1\{-1/7\}^8$, $\{-1/7\}^9$ Spectrum: 9,8,6,5,4,3,1,0 f-vector: [440 5598 24912 49728 51660 29232 8736 1260 72] $P_3(9)$ and $P_6(9)$ Vertices of $P_3(9)$: $\{1/3\}^9$, $\{1/3\}^8\{-5/3\}^1$, $\{1/3\}^5\{-1/3\}^4$, $\{1/3\}^4\{-1/3\}^5$, $\{5/3\}^1\{-1/3\}^8$, $\{-1/3\}^9$ Vertices of $P_6(9)$: $\{1/6\}^9$, $\{1/6\}^8\{-11/6\}^1$, $\{5/18\}^5\{-7/18\}^4$, $\{7/18\}^4\{-5/18\}^5$, $\{11/6\}^1\{-1/6\}^8$, $\{-1/6\}^9$ Spectrum: 9,8,5,4,1,0 f-vector: [272 3078 15072 39228 58212 46872 18648 3192 168] $P_4(9)$ and $P_5(9)$ Vertices of $P_4(9)$: $\{1/4\}^9$, $\{1/4\}^8\{-7/4\}^1$, $\{7/4\}^8\{-1/4\}^1$, $\{-1/4\}^9$ Vertices of $P_5(9)$: $\{1/5\}^9$, $\{1/5\}^8\{-9/5\}^1$, $\{9/5\}^8\{-1/5\}^1$, $\{-1/5\}^9$ Spectrum: 9,8,1,0 f-vector: [20 180 960 3360 7560 10500 8400 3150 252] Isomorphic to $\mathfrak{A}^9$. $P_8(9)$ Vertices: $\{1/8\}^9$, $\{1/8\}^8\{-15/8\}^1$, $\{3/8\}^7\{-13/8\}^2$, $\{5/8\}^6\{-11/8\}^3$, $\{7/8\}^5\{-9/8\}^4$, $\{9/8\}^4\{-7/8\}^5$, $\{11/8\}^3\{-5/8\}^6$, $\{13/8\}^2\{-3/8\}^7$, $\{15/8\}^1\{-1/8\}^8$, $\{-1/8\}^9$ Spectrum: 9,8,7,6,5,4,3,2,1,0 f-vector: [512 2304 4608 5376 4032 2016 672 144 18] Isomorphic to the 9-cube. Table of volumes +-----+---+---------+-----+--------+--------+-------+----+ | n\k | 2 | 3 | 4 | 5 | 6 | 7 | 8 | +-----+---+---------+-----+--------+--------+-------+----+ | 3 | 4 | | | | | | | +-----+---+---------+-----+--------+--------+-------+----+ | 4 | 4 | 16/3 | | | | | | +-----+---+---------+-----+--------+--------+-------+----+ | 5 | 4 | 8/3 | 8 | | | | | +-----+---+---------+-----+--------+--------+-------+----+ | 6 | 4 | 16/9 | 2 | 64/5 | | | | +-----+---+---------+-----+--------+--------+-------+----+ | 7 | 4 | 32/27 | 8/9 | 8/5 | 64/3 | | | +-----+---+---------+-----+--------+--------+-------+----+ | 8 | 4 | 64/81 | 4/9 | 64/135 | 4/3 | 256/7 | | +-----+---+---------+-----+--------+--------+-------+----+ | 9 | 4 | 128/243 | 2/9 | 8/45 | 64/243 | 8/7 | 64 | +-----+---+---------+-----+--------+--------+-------+----+<|endoftext|> TITLE: Transitivity for Schutzenberger involutions on standard Young tableaux QUESTION [6 upvotes]: Let $\lambda$ be a partition of $ n$. Let $ SYT(\lambda) $ denote the set of standard Young tableaux of shape $ \lambda $. For $ i = 1, \dots, n $, let me define permutations $ S_i $ of the set $ SYT(\lambda) $. Let $ S_n(T) $ be the Schutzenberger involution of $ T $. For $ i < n $, let $ S_i(T) $ be the $i$th "partial" Schutzenberger involution. By this, I mean that we fix the part of $ T $ containing $ i+1, \dots, n $ and perform Schutzenberger involution of the part of $ T $ containing $ 1, \dots, i $. Berenstein and Kirillov studied these permutations, in their article "Groups generated by involutions, Gelfand-Tsetlin patterns and combinatorics of Young tableaux" (available at math.uoregon.edu/~arkadiy/bk1.pdf). They prove that these permutations give rise to the action of a certain group, called $ G_n $, on $ SYT(\lambda) $. Question Does the group $ G_n$ act transitively on $ SYT(\lambda)$? In other words, given two standard Young tableaux can I turn one into the other by applying a sequence of partial Schutzenberger involutions? REPLY [4 votes]: Yes. It has been a while since I did this calculation but I think this is the gist of it: Let $ s_{1q} $ be the permutation induced by the partial Schützenberger involution $ S_q(\cdot) $. Then define the permutations $$ s_{pq} := s_{1q} s_{1(q-p+1)} s_{1q}. $$ The $Q$-symbol of the RSK correspondence gives a bijection between words of length $n$ with shape $\lambda$ and $SYT(\lambda)$. One can check that on words the operators $s_{pq}$ when $p-q = 2$ induce the Knuth moves - they are in fact just the cactus operators, hence the choice of notation. Since the Knuth moves act transitively so does the group $G_n$. You can see this nicely using Speyer's cylindrical growth diagrams defined here.<|endoftext|> TITLE: How to prove that the following double sum is always an integer？ QUESTION [20 upvotes]: I have verified the following double sum is always an integer for $s$ up to $1000$ via Maple. But I can not prove it. Proofs, hints, or references are all welcome. Thanks! $$\sum_{m=s}^{2s}\sum_{k=0}^{s} {2s\choose s}{s\choose k}{m\choose k}{k\choose m-s} \frac{1}{(s+1)(2k-1)(2m-2k-1)}$$ What I have known is that: (1) Every term is not always an integer, but I can prove that ${2s\choose s}{s\choose k}{m\choose k}{k\choose m-s} \frac{1}{(2k-1)(2m-2k-1)}$ is always an integer. (2) $\sum_{k=0}^{s} {2s\choose s}{s\choose k}{m\choose k}{k\choose m-s} ={2s\choose s}^2{s\choose m-s}$. This combinatorial identity may be helpful to solve this problem. Note:- The problem has also been posted here. REPLY [17 votes]: Here is an attempt at an answer. We assume that the recurrence from my comment above holds [with a small correction] (a proof was obtained by Kevin using Zeilberger's algorithm, see the comment below): $$ (7s+8)(s+4)(s+3)^2 a_{s+3} - 4(56s^2+127s+57)(s+3)(s+2) a_{s+2} $$ $$ - 16(7s^4-6s^3-121s^2-210s-90) a_{s+1} + 128(7s+15)(2s+3)(2s+1)(s-1) a_s = 0 $$ Write $a_s = b_s/(s+1)$. According to Kevin's claim (1), $b_s \in \mathbb Z$. Now the sequence $(b_s)$ satisfies the recurrence $$ (7s+8)(s+3)^2(s+2)(s+1) b_{s+3} - 4(56s^2+127s+57)(s+2)^2(s+1) b_{s+2} $$ $$ - 16(7s^4-6s^3-121s^2-210s-90)(s+1) b_{s+1} + 128(7s+15)(2s+3)(2s+1)(s+2)(s-1) b_s = 0 .$$ We want to show that $s+1$ divides $b_s$. By the recurrence, we have that $$ (s+1) \mid 128(7s+15)(2s+3)(2s+1)(s+2)(s-1) b_s . $$ Since the gcd of $s+1$ with the factor in front of $b_s$ is a power of 2, we can conclude that the odd part of $s+1$ divides $b_s$. On the other hand, it is easy to see that each term in the sum for $a_s$ is 2-adically integral (it is a Catalan number times some binomial coefficients times a fraction with odd denominator), so there is no need to consider the 2-power part of $s+1$.<|endoftext|> TITLE: "Weight-monodromy" for open varieties QUESTION [7 upvotes]: Suppose that $X/\mathbb{Q}_p$ is a smooth, projective variety, and choose a prime $\ell\neq p$. Then the weight-monodromy conjecture says that the graded pieces $\mathrm{Gr}_k^M$ of the monodromy filtration on the $i$th cohomology groups $H^i_\mathrm{et}(X_{\overline{\mathbb{Q}}_p},\mathbb{Q}_\ell)$ are pure of weight $i+k$. I was wondering if there is a similar conjecture for open varieties, coming from looking at a good compactification $X\hookrightarrow \overline{X}$. If $X$ is a curve, then I can get my hands on what's happening explicitly, so suppose that $X$ is a curve with compactification $\overline{X}$ and $D$ is the set of missing points. Then the only interesting group is $H^1_\mathrm{et}(X_{\overline{\mathbb{Q}}_p},\mathbb{Q}_\ell)$ which by excision sits in an exact sequence $$ 0 \rightarrow H^1_\mathrm{et}(\overline{X}_{\overline{\mathbb{Q}}_p},\mathbb{Q}_\ell) \rightarrow H^1_\mathrm{et}(X_{\overline{\mathbb{Q}}_p},\mathbb{Q}_\ell) \rightarrow V \rightarrow 0 $$ where $V$ is a subspace of $H^0_\mathrm{et}(D_{\overline{\mathbb{Q}}_p},\mathbb{Q}_\ell)(-1)$, it is therefore pure of weight $2$. So (pretty tautologically) $V$ turns up in the 2nd graded piece $\mathrm{Gr}_2^W$ of the weight filtration, but with some basic linear algebra I think I've convinced myself that $V$ actually turns up in the 0th graded piece $\mathrm{Gr}_0^M$ of the monodromy filtration on $H^1_\mathrm{et}(X_{\overline{\mathbb{Q}}_p},\mathbb{Q}_\ell)$, and apart from this 'extra' $V$, the graded pieces of the monodromy filtration for $X$ are the same as those for $\overline{X}$. In other words, $\mathrm{Gr}^M_{-1}$ and $\mathrm{Gr}_1^M$ are pure of weights 0 and 2 respectively, but $\mathrm{Gr}^M_1$ is mixed with weights 1,2. So the naive weight monodromy conjecture fails, but you can still say something - the graded pieces $\mathrm{Gr}_k^M$ are mixed of weights $\geq i+k$. So I guess this is my question: is there a conjectural 'weight monodromy' for open varieties which says something like the above, e.g. the $k$th graded pieces is mixed with weights in $[i+k,2i+k]$? Maybe this would easily follow from the 'usual' weight monodromy by choosing a good compactification and then just using some linear algebra? REPLY [9 votes]: Tony Scholl gave a talk at a conference in Warwick in 2013 on exactly this topic (his talk was called "Remarks on monodromy and weights"). He explained how to formulate a precise version of weight-monodromy for arbitrary varieties over p-adic fields (not necessarily proper or smooth). The idea is that for any field $K$, and any finite-type $K$-scheme $X$, if we set $H^i(X) = H^i(X_{\overline{K}}, Q_\ell)$ for some $\ell \ne char(K)$, there is a canonical increasing filtration $W_n H^i(X)$ (the "geometric weight filtration") of $H^i(X)$ such that $gr_n^W H^i(X)$ is a subquotient of $H^n(Y)$ for some smooth proper $Y$. (This is due to Deligne and de Jong.) So it's reasonable to conjecture that if $K$ is a $p$-adic field and $\ell \ne p$, the pieces $gr_n^W H^i$ should be "quasi-pure of weight n" as representations of $G_K$, which means that all eigenvalues of Frobenius are Weil numbers of some weight, and for all $r$ the monodromy operator $N^r$ sends the weight $n + r$ generalized eigenspace isomorphically to the weight $n - r$ one. (Scholl just calls this "pure of weight $n$", but this conflicts with the more restrictive usage of "pure" in your question.) In Scholl's talk he announced a proof that the usual weight-monodromy conjecture (i.e. for $X$ proper and smooth over $K$) actually implies this more general conjecture. This implication is not obvious, because being quasi-pure is not preserved under passing to subquotients (and Tony didn't say how he gets around this, as far as I can recall). (It's not totally clear if this answers the question, because one would have to understand how the geometric weight filtration relates to the filtration on $H^i(X)$ given by the Grothendieck abstract-monodromy theorem -- they are definitely not the same, because the latter encodes information about the singularities of the special fibre.)<|endoftext|> TITLE: Can one make a category concrete by "enlarging the universe"? QUESTION [5 upvotes]: This is more or less a followup of this question. There, it was established that (it is well known that) the homotopy category of topological spaces is not concrete, in other words, there is no faithful functor $\bf{HoTop}$ $\to$ $\bf{Set}$ . In this blog post, Akhil Mathew explains that this lack of "concreteness" is due, more or less, to the objects of HoTop having a proper class of subobjects (or of quotients). But... what if one "does not believe" in proper classes? If I'm not mistaken, every "set vs proper class" phenomenon can be seen, in a suitable foundation that assumes the existence of Grothendieck universes, as a "small universe vs large universe" phenomenon. The non-concreteness of HoTop, translated in this language, would mean that there is no faithful functor $\bf{Ho(U}$-$\bf{Top)}$ $\to$ $\bf{U}$-$\bf{Set}$ where $\bf{U}$ is a Grothendieck universe and $\bf{U}$-$\bf{Set}$ (resp. $\bf{U}$-$\bf{Top}$) is the category of $\bf{U}$-small sets (resp. $\bf{U}$-small topological spaces). Now, what can happen if $\bf{U}\in\bf{V}$ for $\bf{V}$ a larger Grothendieck universe? My question is Are there faithful functors $\bf{Ho(U}$-$\bf{Top)}$ $\to$ $\bf{V}$-$\bf{Set}$ ? The same question, of course, can be asked for any other non-concrete category. REPLY [7 votes]: As already noted above, any category can be considered concrete after a base change to a suitably large universe. However, doing so would be completely missing the point of concreteness. The underlying topos of sets should really be considered not as a first-class object, but rather as a background reality within which you are doing mathematics. Some things are describable and studyable within this "reality", but some are hopelessly out of reach. Concreteness serves as a (vague) borderline between these cases. Basically, a concrete category is a one that can be reasonably studied once you fix the collection of objects (=sets) that interest you. It's not a strict rule, of course. For example, both of the categories $Set^{op}$ and $Rel$ (sets as objects, relations as morphisms) are not concrete, but are good enough to study. The reason is that they can be constructed from a concrete one in a simple way. Note that any algebraic category or a category of sheaves on a small site is concrete. In a sense that is a general example: some theory defined via a list of first-order axioms. It's not exhaustive of course, but it is what usually happens in practice. Going back to $Ho(Top)$, the theory essentially states that it is hopelessly complicated from the PoV of classical set theory. And it really is: it doesn't (usually) have limits and colimits, and they are badly behaved even when they exist, and doing algebra in this category, or studying its parametrized versions is a painful and fruitless experience. You can enlarge the universe, but this new universe on its own will be hopelessly complicated wrt your base set theory (at least as complex as the category you are embedding), so the net gain is zero, except for a formal concreteness property which no one really cares about. The moral is, you need some genuinely new ideas to do homotopy theory. You must either get a much better understanding of passage to the homotopy category (=invent model categories), or find some related, but entirely different and better behaved object (=higher categories). Both of these approaches work, with my love for the latter.<|endoftext|> TITLE: Lots of combinatorial interpretations of Catalan numbers QUESTION [13 upvotes]: During a lecture I gave on Catalan numbers, I pointed out that that it is possible to give a continuum number of combinatorial interpretations of these numbers. See the solution to (f$^5$) on page 54 of http://math.mit.edu/~rstan/ec/catadd.pdf. After the lecture someone from the audience (I don't know who) asked me if one can give more than a continuum number of combinatorial interpretations. For instance, for each subset of $\mathbb{R}$ can one give a different combinatorial interpretation? Can anyone shed some light on this question? REPLY [18 votes]: For a genuine answer, one needs a genuine definition of what counts as a combinatorial interpretation. One needs to exclude silly things like the following "interpretation $X$" for any set $X$ of real numbers: The number of pairs $(p,X)$ where the first component $p$ is a proper sequence of $2n$ parentheses and the second component is the fixed $X$. Although I haven't read all of the usual combinatorial interpretations of the Catalan numbers, I conjecture that they are all covered by the following set-up and that the set-up will look reasonable to combinatorialists. Fix a countably infinite collection of non-sets, which I'll call "atoms", and build an analog of the cumulative hierarchy of set theory over these atoms, but build only finite sets and don't iterate transfinitely. That is, let $V_0$ be the set of atoms, and let $V_{n+1}$, for each natural number $n$, be the union of $V_0$ with the collection of all finite subsets of $V_n$. Thus, $V_1$ consists of atoms and finite sets of atoms, $V_2$ consists of atoms and finite sets of (atoms and finite sets of atoms), etc. By a "combinatorial entity", I'll mean any element of the union of all the $V_n$'s. Note that standard codings from set theory allow you to represent, as combinatorial entities, any finite tuple of combinatorial entities, any function mapping a finite set of combinatorial entities to other combinatorial entities, etc. Von Neumann's coding also lets you represent each natural number as a combinatorial entity. Then by a "Catalan interpretation", I mean a function assigning to each natural number $n$ a combinatorial entity whose cardinality is the $n$-th Catalan number. If you buy this definition, then there are only continuum many Catalan interpretations, because there are only countably many combinatorial entities. If you don't buy this definition, tell me what you would buy.<|endoftext|> TITLE: Is there a simple direct proof of the Open Mapping Theorem from the Uniform Boundedness Theorem? QUESTION [15 upvotes]: The Open Mapping Theorem, the Bounded Inverse Theorem, and the Closed Graph Theorem are equivalent theorems in that any can be easily obtained from any other. The Closed Graph Theorem also easily implies the Uniform Boundedness Theorem. But is there a simple way to obtain any of the other three results from Uniform Boundedness, or is Uniform Boundedness really a "lower-level" result than the others? REPLY [6 votes]: I don't know whether you'll consider this "simple", but here is a proof. I distilled it from Eric Schechter's Handbook of Analysis and its Foundations, which has a proof of a more general statement at 27.35. The last part is from Folland's Real Analysis, Theorem 5.10. Suppose $X,Y$ are Banach spaces and $T : X \to Y$ is surjective. We wish to show $T$ is open. Let $B$ be the open unit ball of $X$; it suffices to show $T(B)$ contains a neighborhood of $0 \in Y$. The first step is to show that the closure $\overline{T(B)}$ contains a neighborhood of 0. The usual method is to use the Baire category theorem: if not, then $Y = \bigcup_{n=1}^\infty n \overline{T(B)}$ meaning that $Y$ is meager. We will use the uniform boundedness principle instead. For each $n$, construct a new norm $\|\cdot\|_n$ on $Y$ defined by $$\|y\|_n := \inf\{\|u\|_X+n\|v\|_Y : u \in X, v \in Y, v+Tu=y\}.\tag{*}$$ It is straightforward to verify this is a norm. Now let $Z$ be a countable direct sum of copies of $Y$, i.e., $Z$ is the vector space of all finitely supported functions $f : \mathbb{N} \to Y$, with the pointwise addition and scalar multiplication. Equip $Z$ with the norm $$\|f\|_Z := \sup_n \|f(n)\|_n.$$ Then for each $n$, define a linear operator $S_n : Y \to Z$ by $(S_n y)(n) = y$ and $(S_n y)(k) = 0$ for $k \ne n$. Note that $\|S_n y\|_Z = \|y\|_n$. Now by taking $u=0$, $v=y$ in (*), we obtain $\|y\|_n \le n \|y\|_Y$, so each $S_n$ is bounded. Moreover, by the surjectivity of $T$, for each $y \in Y$ there exists $x \in X$ with $Tx=y$. Taking $u=x$ and $v=0$ in (*) we see that $\| y\|_n \le \|x\|_X$ independent of $n$; hence $\{S_n\}$ is pointwise bounded. By the uniform boundedness theorem, there is a constant $C < \infty$ such that $\|S_n\|_{Y \to Z} \le C$ for all $n$. Fix $\delta < 1/C$. I claim that $\overline{T(B)}$ contains a ball of radius $\delta$ centered at 0. For suppose $\|y\|_{Y} < \delta$; then $$\|y\|_n = \|S_n y\|_Z \le \|S_n\|_{Y \to Z} \|y\|_Y \le C \delta < 1$$ for every $n$. Hence for each $n$ there exists $u_n \in X$, $v_n \in Y$ with $y = v_n + T u_n$ and $\|u_n\|_X + n\|v_n\|_Y < 1$. In particular, $\|v_n\|_Y < 1/n$, so we have $T u_n \to y$ where $u_n \in B$. Thus $y \in \overline{T(B)}$ and the proof of this step is complete. The rest of the proof proceeds as usual. We can show $T(B)$ contains a ball of radius $\delta/2$ centered at 0. Suppose $\|y\|_Y < \delta/2$, so that by the first step and scaling, $y \in \overline{T(B_{1/2})}$. Hence there is $x_1$ with $\|x_1\|_{X} < \frac{1}{2}$ and $\|y - Tx_1\|_Y < \delta/4$. Repeating this process inductively, we construct $x_n$ with $\|x_n\|_X < 2^{-n}$ and $\left\| y - \sum_{k=1}^n Tx_k\right\|_Y < \delta 2^{-(n+1)}$. Thus $\sum_{k=1}^\infty T x_k$ converges in $Y$ to $y$. Summing a geometric series, $\sum_{k=1}^\infty \|x_k\| < 1$, so by completeness of $X$, $\sum_{k=1}^\infty x_k$ converges in $X$ to some $x$ with $\|x\|_X < 1$. And by continuity of $T$, $Tx=y$. So we have shown $y \in T(B)$. Note that this proof works even if $Y$ is not Banach, so long as the uniform boundedness principle holds on $Y$. This happens iff $Y$ is barreled, and indeed the proof in Schecheter shows that both uniform boundedness and open mapping are equivalent to being barreled.<|endoftext|> TITLE: Non-closed geodesics on a convex polyhedron in $\mathbb{R}^3$ QUESTION [5 upvotes]: Let $P$ be the surface of a closed convex polyhedron in $\mathbb{R}^3$. Q. Does every non-closed geodesic $\gamma$ fill $P$ densely? Of course $\gamma$ cannot pass through a vertex of $P$, but it could come arbitrarily close. Here is an informal way to phrase the question. One can have non-closed geodesics on smooth convex surfaces in $\mathbb{R}^3$ that do not fill the surface. E.g., on ellipsoids, as in the MO question "Surfaces filled densely by a geodesic":             (Image from GeographicLib.) So: Is there a polyhedral analog? (One could ask the same question for nonconvex polyhedra.) REPLY [7 votes]: A simple triangulation of the previous example will work. Suppose we have a geodesic like the blue one. The key point is to look at the green angles (better the complementary ones, those respect to the horizontal plane). Each time the geodesic crosses a black edge, this angle decreases by a fixed (this is essential!) amount, namely the angle formed by two black vertical edges if extended. If the initial angle is small enough, then we may make sure that the geodesic will cross so many black edges as we want before it touches the top square, so in fact it is not touched at all. The blue geodesic goes down and the process goes on forever. One could answer: what happens if, when coming down, the angle is very large? For this not to happen, the polyhedron must be pointed enough. This is always feasible, as shown by this picture (which represents a periodic plane representation of the lateral faces): Of course non-periodicity is a generally satisfied.<|endoftext|> TITLE: Strongest large cardinal axiom compatible with $V = L$? QUESTION [16 upvotes]: What is the strongest known natural large cardinal axiom compatible with $V = L$ (strongest in the sense that it implies all known "small" large cardinal axioms, where a large cardinal axiom is said to be "small" if it doesn't imply $V \neq L$)? One candidate might be "there is an $\alpha$-Erdos cardinal for every countable ordinal $\alpha$." Using the ``instrumentalist dodge'' of Steel/Hamkins, it seems that we can "cheat" to obtain stronger large cardinal axioms within the confines of $\operatorname{ZF}+(V = L)$, such as "there is a transitive model of the theory ZFC+"$0^\sharp$ exists."" I will leave whether or not such axioms are "natural" for a matter of debate here. Disclaimer: I'm an algebraist and number theorist but am currently trying to grasp the big picture in set theory, large cardinals, inner model theory, and the like. Related MO posts: Is there a large-cardinal completeness theorem for $L$? If $0^{\sharp}$ exists, then every uncountable cardinal in $V$ is as large as it can be in $L$. Erdős cardinals and $0^\sharp$ REPLY [13 votes]: Maybe a comment: In the paper ``A large cardinal in the constructible universe'' Silver shows that if $\kappa\to (\alpha)^{<\aleph_0}$ for all countable $\alpha,$ then the same is true for $\kappa$ in $L$. On the other hand, by a result of Rowbottom, $\kappa \to (\omega_1)^{<\aleph_0} $ contradicts $V=L.$ Silver concludes with the following: It does not seem extravagant, then, to assert that, for all practical purposes, $\kappa \to (\omega)^{<\aleph_0}$ is the strongest strong axiom of infinity know to be consistent with $V=L,$ and therein lies its chief interest.<|endoftext|> TITLE: Is there a known primitive recursive upper bound on the nth "Zhang prime" QUESTION [14 upvotes]: (This question is pure curiosity. Feel free to close it if you feel it is not appropriate for mathoverflow.) In 2013 Zhang showed that there are infinitely many pairs of primes which are less that 70,000,000 apart. Since then, this result has been significantly improved. (See here for the state-of-the-art.) Let $z_n$ denote the $n$th "Zhang prime", that is the $n$th prime $p$ such that the next prime is not more than $p + 70,000,000$. My question is as follows. Does Zhang's proof or any of its improvements establish a primitive recursive upper bound on the $n$th Zhang prime, that is a primitive recursive function $f(n)$ such that $z_n \leq f(n)$? A few (trivial) comments for those who don't regularly think about primitive recursive functions: Zhang's result (regardless of the proof) gives a recursive (a.k.a. computable) upper bound, since one just needs to search for the $n$th Zhang prime until it is found. Zhang's result says this search will eventually terminate! By well-known facts of primitive recursive functions, if such a primitive recursive upper bound $f(n)$ exists, then the function $n \mapsto z_n$ is itself primitive recursive. The function $p(n)$, which returns the $n$th prime, is primitive recursive since Euclid's proof shows that there is a prime somewhere between $p(n) + 1$ and $p(n)! + 1$. REPLY [13 votes]: One does not have to dig too deep into the arguments to answer this. Maynard's formulation (see his preprint here) shows that if $x$ is sufficiently large, there is a Zhang prime (or even a prime $p$ so that $p+k$ is prime with $k \leq 600$) between $x$ and $2x$. It follows that there is a constant $C$ so that $z_{n} \leq C \cdot 2^{n}$, which is primitive recursive. Often in analytic number theory, it's difficult to show that something happens without showing that it happens "a lot".<|endoftext|> TITLE: Conjecture on maximum of symmetric combinatoric function QUESTION [14 upvotes]: A curious symmetric function crossed my way in some quantum mechanics calculations, and I'm interested its maximum value (for which I do have a conjecture). (The question was first asked at math.SE, where (even after 2.5months and 2 rounds of bounty) there is only one special-case-solution and one additional insight on concavity. So I'm hoping to get some more insights here). The problem There are $n$ different objects $A_1,...,A_n$, and there are sets containing $m$ different $A_i$s: $C_i=(A_{i_1}, A_{i_2}, ..., A_{i_m})$. There are $i_{max}=\binom{n}{m}$ different combinations $C_i$. Each combination $C_i$ has a probability $p_i$ (with $\sum_{i=1}^{i_{max}} p_i=1$). Defining the function For a given pair of objects $A_k$ and $A_l$: $f_1(k,l)$ contains all probabilities $p_i$ of the sets $C_i$, which contains both objects $A_k$ and $A_l$. $f_2(k,l)$ contains all probabilities $p_i$ of the sets $C_i$, which contains either object $A_k$ or $A_l$ (if it contains both elements, we add $p_i$ twice). $F(k,l)=\frac{f_1(k,l)}{f_2(k,l)}$ With that, we get the main-function $$D^{(n,m)}=\sum_{k=1}^{n-1} \sum_{l=k+1}^{n} F(k,l)$$ What is the maximum of $D^{(n,m)}$, given that the sum of all probabilities $p_i$ is 1? Alternative notation As pointed out by Wolfgang in a comment, the function $D^{(n,m)}$ can be written in a more intuitive way by the form: $$D^{(n,m)}=\sum_{ij$ is the same as the sum over $k TITLE: What is the number of equitriangulations of the n-cube? QUESTION [12 upvotes]: I wonder if this question has been considered before and if anything is known. My search attempts have failed so far. Let's consider the n-dimesnional cube, $[0,1]^n$, and let's call a simplex with vertices in $\{0,1\}^n$, nice if it has volume $\frac{1}{n!}$. How many triangulations does the unit cube have into nice simplices? REPLY [3 votes]: Here is a lower bound of $2^{\Omega(2^n)}$ for the number of unimodular triangulations. Let me start with the triangulation described by Włodzimierz Holsztyński, which is indeed quite classical and I will call the standard triangulation of the $n$-cube. Its $n$-simplices have as vertices the $n!$ monotone paths of vertices from $(0,\dots,0)$ to $(1,\dots,1)$. Observe that the standard triangulation $T$ has the following property: for each $2$-face of the cube, the two triangles in which $T$ divides that face have the same link (that is, they are joined to the same simplices). This implies we can perform a flip on that $2$-face: we can remove the two triangles in it together with all simplices containing them, and insert the other triangulation of the quadrilateral, with the same link as the old one had. (This is a higher dimensional analogue of changing the triangulation in a square pyramid, as described in David Eppstein's or TMA's answers) Now, if we find $k$ of these $2$-faces in which we can perform flips independently (that is, if no two of the flips affect a common simplex) then we get $2^k$ triangulations of the cube. One way to guarantee that the flips are independent is to use $2$-faces at the same level, by which I mean the following: every $2$-face is defined by $n-2$ equalities of the form $x_i=\epsilon_i$, where $\epsilon_i\in \{0,1\}$. I say a $2$-face has level $l$ if exactly $l$ of the $\epsilon$'s are equal to $1$. The number of $2$-faces at level $l$ is ${n \choose n-2}{n-2 \choose l}$ which, for $l\sim (n-2)/2$, grows as $2^n n^{3/2}$ (modulo a constant). Thus, we have at least $$ 2^{\Omega(2^n n^{3/2})} $$ unimodular triangulations of the $n$-cube. It is also easy to give a crude upper bound that is not that far from the lower bound. Observe that a unimodular triangulation is a set of $n!$ $n$-simplices taken out from a total of (at most) $2^n \choose n+1$ $n$-simplices. Thus, the number of unimodular triangulations (or, of all triangulations, for that matter) is at most $$ {{2^n \choose n+1} \choose n!} \le 2^{O(n^{n+2})} $$<|endoftext|> TITLE: Do all 0-dimensional Shimura Varieties show up (as CM points) in $\mathcal{A}_g$? QUESTION [8 upvotes]: Question: Let $S$ be a 0-dimensional Shimura variety. Does $S$ necessarily admit a morphism (in the category of Shimura varieties) to $\mathcal{A}_g$ for some $g\geq 1$? Here $\mathcal{A}_g$ is the moduli space of principally polarized abelian varieties. Motivation: Basically, I am trying to understand something for all Shimura varieties. This reduces to some statement about 0-dimensional Shimura varieties, and I wanna see if I can apply techniques from the theory of Abelian Varieties to do so. REPLY [3 votes]: You can always map $(T,h)$ to the trivial Shimura datum and then this into the Siegel one. I assume this isn't what you want however. May I therefore modify the question to ask whether a zero dimensional variety can be embedded in a Siegel variety. I.e. You want to know if all zero dimensional Shimura varieties are of Hodge type. I think this is certainly false and can give two quick reasons. 1) every Hodge type datum has weight defined over $\mathbb{Q}$ and has compact centre modulo the image of this weight. In general neither condition need hold for a zero dimensional Shimura datum. 2) consider the 'cyclotomic' Shimura datum $T=\mathbb{G}_m$ and $h(z) = z\bar{z}$. Any representation of T will induce a hodge structure of even weight but if it admits a Siegel embedding then this must yield a representation giving the weight 1 Hodge structure defining an abelian variety.<|endoftext|> TITLE: Existential statement without witness QUESTION [12 upvotes]: Are there existential theorems of ZFC, or PA say, with no witnesses? Ie does there exist a formula $\phi$ such that ZFC $\vdash\exists x \phi(x)$, but for all numerals $\underline{n}$, ZFC $\nvdash \phi(\underline{n})$? Can you give an example of such a formula? REPLY [10 votes]: $\def\epn{\mathrm{EP}_\mathbb N}\def\dp{\mathrm{DP}}$When a theory $T$ (with a distinguished definable set of “natural numbers” satisfying a suitable minimal arithmetic) has the property in the question, viz. $$\tag{$\epn$}T\vdash\exists x\in\mathbb N\,\phi(x)\text{ implies }T\vdash\phi(\underline n)\text{ for some $n\in\mathbb N$,}$$ $T$ is said to have the numerical existence property. As explained in Joel’s answer, classical theories can have the numerical existence property only if inconsistent or complete, which prevents any interesting examples by virtue of Gödel’s incompleteness theorem. However, the situation is quite different for theories over intuitionistic logic. Indeed, most natural constructive theories (e.g., Heyting arithmetic) do have $\epn$. Since $\phi\lor\psi$ is equivalent to $\exists n\,((n=0\land\phi)\lor(n=1\land\psi))$, $\epn$ implies the seemingly weaker disjunction property $$\tag{$\dp$}T\vdash\phi\lor\psi\implies T\vdash\phi\text{ or }T\vdash\psi.$$ By a result of Harvey Friedman, $\dp$ is in fact equivalent to $\epn$ if $T$ is recursively axiomatizable. We may stratify the definition by restricting the complexity of the formulas. A consistent recursively axiomatizable classical theory never has $\epn$ for $\Pi^0_1$ formulas, or $\dp$ for $\Pi^0_1$ sentences, or $\dp$ for $\phi\in\Sigma^0_1$ and $\psi\in\Pi^0_1$. This leaves $\Sigma^0_1$ as the only realistic possibility. Friedman’s argument shows that for recursively axiomatized $T$, the following are equivalent: $T$ has $\epn$ for $\Sigma^0_1$ formulas, $T$ has $\dp$ for $\Sigma^0_1$ sentences, $T$ is $\Sigma^0_1$-sound or inconsistent. Naturally occurring theories are $\Sigma^0_1$-sound (e.g., PA) or assumed to be $\Sigma^0_1$-sound (e.g. ZFC). See also Visser (§7) for an interesting connection of Friedman’s characterization to (non)interpretability of inconsistency.<|endoftext|> TITLE: Geometrically connected components of an algebraic group QUESTION [6 upvotes]: Suppose that $G$ is an algebraic group over a field $k$. Let $G^o$be the connected component of the identity. Since $G^0$ contains a $k$-rational point (the identity) therefore it is geometrically connected. I am wondering whether the remaining connected components of $G$ are also geometrically connected. REPLY [12 votes]: No, take for example $k=\mathbb{Q}$ and $G=\mu_p$ for some prime $p>2$. Then $G$ has only two connected components, whereas $G_\overline{\mathbb{Q}}$ has $p$.<|endoftext|> TITLE: Free subgroups in algebras of polynomial growth QUESTION [10 upvotes]: What is known about free non-abelian subgroups in finitely generated associative algebras of polynomial growth (e.g., over finite fields, to avoid finite-dimensional free subgroups)? For example, are there examples of ideals I in the group algebra A of the free group F such that A/I has polynomial growth but the natural map F\to A/I is injective? REPLY [10 votes]: Let $p$ be a prime number. Consider the algebra $A:=M_2((\mathbb Z/p\mathbb Z)[t])$ and the group $G:=(\mathbb Z/p \mathbb Z) \ast (\mathbb Z/p \mathbb Z) = \langle a,b \mid a^p,b^p \rangle$. Look at the homomorphism $\varphi \colon G \to A^{\times}$ which is defined by $$\varphi(a):=\left( \begin{matrix} 1& t \\ 0 & 1 \end{matrix} \right), \quad \mbox{and} \quad \varphi(b):=\left( \begin{matrix} 1& 0 \\ t & 1 \end{matrix} \right).$$ It is easy to see that $\varphi$ is injective. Indeed, if for example $w = a^{n_1}b^{m_1} \cdots a^{n_k}b^{m_k}$ with $n_i,m_i \in \{1,\dots,p-1\}$, then the polynomial in the upper left corner of the matrix $\varphi(w)$ has degree $2k$ and leading coefficient equal to $n_1 \cdots n_k m_1 \cdots m_k \neq 0.$ The other cases are similar. Note that $A$ is of polynomial growth and if $p \geq 3$, then $G$ contains a free subgroup.<|endoftext|> TITLE: Purely noncommutative algebra-Morita equivalence QUESTION [8 upvotes]: Morita equivalence of algebras certainly don't preserve commutativity: even if $A$ is commutative there are plenty of noncommutative algebras which are Morita equivalent with $A$---for example all algebras of the form $M_n(A)$ are good. What is the simplest example of an algebra which is not Morita equivalent to any commutative algebra? REPLY [19 votes]: An algebra is Morita equivalent to a commutative algebra iff it's Morita equivalent to its center, since the center is Morita invariant. So any representative of a nontrivial class in the Brauer group of the underlying field $k$ is a counterexample: for example, when $k = \mathbb{R}$ we can take the quaternions $\mathbb{H}$.<|endoftext|> TITLE: Fano variety of lines on the Segre and the Grassmannian QUESTION [8 upvotes]: Does every $\mathbb{P}^{19}\subset \mathbb{P}(\mathbb{C}^5\otimes\mathbb{C}^5)$ intersect the Segre variety of rank one matrices in at least a $\mathbb{P}^1$? A naive dimension count suggests this is possible. The intersection is a $3$-fold and I would be happy for any qualitative information about it. REPLY [5 votes]: This is a comment, but I cannot comment as my reputation is too low :/ I'm very sorry: why is this the top class? I.e. why we do not get additionally $pr^*_{Grass}[c_2(S^\vee)^2c_1(S^\vee)^2]$ (giving $30=5*6$) and $c_2(S^\vee)^3$ (giving $10$)? Did I miscompute that the top class is $(c_2(S^\vee)+c_1(T^\vee)c_1(S^\vee)+c_1(T^\vee)^2)^5$ or I have an error elswhere?<|endoftext|> TITLE: Why do sporadic simple groups have so few conjugacy classes? QUESTION [35 upvotes]: In finite group theory, there's a general intuition that the further away a group is from abelian, the fewer conjugacy classes it will have. So it is to be expected that non-abelian finite simple groups have a smallish number of conjugacy classes relative to size. Even with this intuition though, some of the numbers appearing in the list of sporadic simple groups are a bit surprising. For instance, the Monster group, with more than $10^{53}$ elements and $15$ prime divisors, has fewer than $200$ conjugacy classes. (An alternating group of comparable order has more than $30 000$ conjugacy classes.) $M_{22}$ has $443520$ elements and only $12$ conjugacy classes. What is going on here? Is there something about the special combinatorial structures that allow these groups to exist that also makes the centralisers exceptionally small? REPLY [18 votes]: This is also rather an expanded comment. -- Since for purely arithmetical reasons, $\ln(\ln(|G|))$ is a lower bound for the number $k(G)$ of conjugacy classes of a finite group $G$, maybe $$ f(G) := \ln(k(G))/\ln(\ln(\ln(|G|))) $$ is a better measure than $k(G)/|G|$ for how many or how few conjugacy classes a group $G$ has in comparison with its order. For example we have (examples ordered by group order): $f({\rm A}_5) \approx 4.68799$, $f({\rm PSL}(2,7)) \approx 3.64930$, $f({\rm A}_6) \approx 3.39930$, $f({\rm PSL}(2,8)) \approx 3.64187$, $f({\rm PSL}(2,11)) \approx 3.3204$, $f({\rm A}_7) \approx 3.04392$, $f({\rm PSL}(3,3)) \approx 3.23520$, $f({\rm M}_{11}) \approx 2.92936$, $f({\rm PSL}(2,31)) \approx 3.53994$, $f({\rm A}_8) \approx 3.17897$, $f({\rm PSL}(3,4)) \approx 2.77366$, $f({\rm Sz}(8)) \approx 2.83466$, $f({\rm M}_{12}) \approx 3.03727$, $f({\rm J}_1) \approx 2.96687$, $f({\rm A}_9) \approx 3.16283$, $f({\rm M}_{22}) \approx 2.63787$, $f({\rm J}_2) \approx 3.20085$, $f({\rm A}_{10}) \approx 3.23851$, $f({\rm M}_{23}) \approx 2.76986$, $f({\rm A}_{11}) \approx 3.31013$, $f({\rm Sz}(32)) \approx 3.39405$, $f({\rm HS}) \approx 3.01600$, $f({\rm J}_3) \approx 2.88256$, $f({\rm M}_{24}) \approx 3.00146$, $f({\rm PSL}(6,2)) \approx 3.55208$, $f({\rm O'N}) \approx 2.85566$, $f({\rm Fi}_{22}) \approx 3.36345$, $f({\rm HN}) \approx 3.18141$, $f({\rm B}) \approx 3.54764$, $f({\rm A}_{43}) \approx 6.61233$, $f({\rm M}) \approx 3.34883$, $f({\rm A}_{44}) \approx 6.69491$.<|endoftext|> TITLE: Producing finite objects by forcing! QUESTION [49 upvotes]: It is a trivial fact that forcing can not produce finite sets of ground model objects. However there are situations, where we can use forcing to prove the existence of finite objects with some properties. For example consider the following result of Shelah (I learned this example from the answer given by Prof. Komjath in Forcing as a tool to prove theorems): Theorem (Shelah) There exists a finite $K_4$-free graph which, when the edges colored by $2$ colors, always contains a monocolored triangle. Shelah's proof of the theorem is simply as follows: He constructs a forcing extension which adds a graph $X$ with the same property but with $\aleph_0$ colors. Then $X$ has the edge-coloring property for $2$ colors, as well. Then using compactness theorem, $X$ must contain a finite subgraph $Y$ with the same property. As forcing cannot create new finite graphs, $Y$ is already present in the ground model. By Godel, $ZFC$ proves that there is such a graph. Now my question is that if there are more examples of the above kind. To be more precise: Question 1. Are there any other examples for producing some finite object $Y$ (with some properties) in the following way: 1) We provide a suitable generic extension in which there is an infinite object $X$ with the required properties, 2) By compactness (or other devices), we can conclude that there must be a finite subset $Y$ of $X$ with the same properties, 3) So we can conclude that the object must exist in the ground model. Remark 1. I am mostly interested in examples where it is not known (or it is difficult) to produce such finite object directly Remark 2. It seems that some partition type theorems are of this type: It is possible to derive some kind of finite partition theorems (like finite Ramsey theorem) using infinite versions of them. So if we can prove the infinite version of these partition theorems, then we have produced examples of the required type (there are cases where we can prove infinite partition theorems using forcing). Question 2. Is Shelah's result the first non-trivial example of an answer to question 1? Are there othe non-trivial constructions of the above kind before him? Remark 3. Here by non-trivial I mean the above strategy is, in some sense, the only method for producing the required finite object. -- Question 3. Are there any results of the above type proved in other parts of mathematics other than logic? Of course Shelah's result is of this type, but the example given by Hamkins is in mathematical logic. REPLY [19 votes]: Here is another example. The theorem is: every finite partial order can be found as a suborder of the partial order of the Turing degrees. On the one hand, one can prove this by undertaking a priority construction, a detailed argument constructing the degrees directly. On the other hand, one can force to add countably many mutually generic Cohen reals $x_n$, and then noting that finite sums of them show that any finite power set algebra can be found in the degrees. But every finite partial order is a suborder of a finite power set order. Finally, the existence of any particular such pattern in the Turing degrees is a $\Sigma^1_1$ property and hence absolute to the ground model by Shoenfield absoluteness. So there was already such a pattern in the ground model. A similar argument finds every countable order as a suborder of the Turing degrees. In both cases, what the direct argument amounts to is the construction of sufficiently generic degrees. With forcing, we can just go all the way to a fully generic real, and this simplifies things.<|endoftext|> TITLE: Existence of internal toposes/inner models in a topos QUESTION [10 upvotes]: It has been known for some time that one can define a topos as a model of a (finitary) essentially algebraic theory (or in other words, can be defined internal to any category with finite limits). In particular, given a topos $E$ (for instance the topos of sets) one can define an internal topos in $E$ (e.g. a small topos). One can ask for stronger (and in fact is easier to define) in an internal universe: this is a category theoretic analogue of a Grothendieck universe. Such a thing gives rise to an internal topos $M$ in $E$ that is in addition a locally full subcategory. The nontechnical definition is that there is an $E$-object $e(x)$ of $M$-elements of any $M$-object $x$ (recall that elements are functions $1 \to x$, here taken in $M$), and (the $E$-object of) functions in $E$ between $e(x)$ and $e(y)$ correspond to the $E$-object of functions in $M$ between $x$ and $y$. There are toposes, given a metatheory of $ZFC$, that have no universes in this sense, namely the topos of sets in $V_{\omega+\omega}$, much like one cannot prove the existence of Grothendieck universes in vanilla ZFC. What I'm curious about is the existence of internal toposes that aren't locally full (equivalently, arise from universes). Shouldn't we get the free internal topos in a topos? If we have NNOs, can we get the free internal topos with NNO? Are there good descriptions of these? Given the topos of ZFC sets, the models of ZC give internal toposes, but I'm interesting in when we can say something about internal toposes without any material meta-theory. In essence: what can be said about "inner models" in topos theory? When we build inner models in ZFC (say), we leverage the extra structure that material sets have, for instance the cumulative hierarchy, which is not a priori available to the topos theorist. Postscript: Joyal defined (in work decades old by not yet published) a special sort of category called an arithmetic universe, which is much weaker than a topos, yet has enough structure to define the free internal arithmetic universe in it. I guess the free internal topos should be exist in a topos, but otherwise I can't be sure it's possible. I worry about erroneously "proving the existence of a model" out of nothing. REPLY [8 votes]: Your long question suggested at first that you were asking for something difficult (a locally full internal topos), but you ended up asking for the easy thing. Starting from your postscript, the purpose of an arithmetic universe is that it is is the least categorical structure in which one can construct the free or initial structure of a widely applicable kind, namely essentially algebraic. (See my answer here for a brief explanation of arithmetic universes.) Any elementary topos with natural numbers object is an arithmetic universe. The structure of an elementary topos (with natural numbers if you wish) is essentially algebraic, so There is an initial elementary topos with natural numbers in any arithmetic universe, in particular in any elementary topos with natural numbers. I would like to refer you to the books 2-Categories for the Working Categorist and Introduction to Arithmetic Universes, but regrettably nobody has written them yet. Talking about ZFC or cumulative hierarchies in this is really obfuscation. However, I indicated how to build the set-theoretic hierarchy in my paper Intuitionistic Sets and Ordinals, based on earlier ideas of Gerhard Osius. As François Dorais points out above, the initial model may be degenerate, just as it may be in any situation in logic or algebra, and so much more difficult questions of consistency or incompleteness arise. Indeed, André Joyal proved Gödel's incompleteness theorem categorically in the following form: In the initial arithmetic universe $\mathcal A$, the internal hom-object $A(1,0)$ of its internal initial arithmetic universe $A$ (where $0,1:{\mathbf 1}\rightrightarrows A$ are the internal initial and terminal objects of $A$) is not isomorphic to the initial object $\mathbf 0$ of $\mathcal A$. On the other hand, again as in traditional logic, if the outer structure is of a stronger kind than the inner one then there are methods of proving internal consistency. For example, the natural numbers provide a bare-hands model of set theory (without inifinity), in which $n\epsilon m$ if the $n$th binary digit of $m$ is $1$, and so a Boolean internal elementary topos. A particularly neat and powerful method is the gluing construction (known in computer science as logical relations); for my account of this see Section 7.7 of my book.<|endoftext|> TITLE: Intersection Cohomology and $L^2$ cohomology QUESTION [14 upvotes]: In the study of singular spaces, topological methods like intersection cohomology have played an important role. They have led to the development of technology like perverse sheaves and these find widespread application in, for instance, representation theory. There are, however, another approach to singular spaces that is analytical in nature (uses the Riemannian metrics defined on these spaces obtained with the singularity removed) and is called $L^2$ cohomology since it involves the study of differential forms that are $L^2$ complete and their associated chain complexes, cohomology etc. Now, there are standard conjectures of Cheeger-Goresky-Macpherson (back to 80s, I think), that relate the two theories. I would like to know the current status of these conjectures. In particular, are there some special cases for which it is already proven ? My interest is specifically in the case of the singular spaces that arise in representation theory. For a plethora of examples that satisfy this description, Lusztig's ICM (found here) is a good place (ex include Schubert varieties, Nilpotent Orbits etc.) REPLY [2 votes]: There are some other cases where the relation between intersection cohomology and L^2 cohomology are well understood. For example, see Hunsicker's "Hodge and signature theorems for a family of manifolds with fibre bundle boundary". Hunsicker mostly treats only the two-stratum case, but she has some nice results about what different intersection homology groups arise as the metric near the lower stratum varies.<|endoftext|> TITLE: The cohomology of an $S_{3}$ cover of an elliptic curve ramified in one point QUESTION [7 upvotes]: Let $E/\mathbb{C}$ be an elliptic curve. Let $C \to E$ be a Galois cover with group $G = S_{3}$ (symmetric group on $3$ elements), ramified in one point. (To clarify: there is a unique point in $E$ over which the cover is branched. Thanks Jason.) Then $C$ is a curve of genus $3$. I want to understand $H^{1}(C, \mathbb{Q})$. The $G$-invariants are $2$-dimensional, $H^{1}(C, \mathbb{Q})^{G} = H^{1}(E, \mathbb{Q})$. Using Chevalley–Weil one can compute the representation of $G$ on the complement $H^{1}(C, \mathbb{Q})^{\perp}$. It consists of twice the $2$-dimensional irrep $V$ of $G$. Consequently there is an action of $M_{2}(\mathbb{Q})$ on $H^{1}(C, \mathbb{Q})^{\perp}$, and in the category of $\mathbb{Q}$-Hodge structures, we obtain $H^{1}(C, \mathbb{Q})^{\perp} = H \otimes V$, where $H$ is a $2$-dimensional Hodge structure of type $(1,0) + (0,1)$, and $V$ is the $2$-dimensional irrep of $G$. Therefore, $H$ “is” an elliptic curve (up to isogeny). Q. Can we explicitly understand the isogeny class of $H$? Is it the same as $E$? One actually does not have to use the Hodge theory. The cover $C \to E$ gives rise to a map $\mathrm{Jac}(C) \to \mathrm{Jac}(E)$. The kernel $A$ is an abelian variety of dimension $2$, with an action of $M_{2}(\mathbb{Z})$. Up to isogeny, we get $A \cong \smash{E'}^{2}$. Note that $H^{1}(E', \mathbb{Q}) \cong H$ (the Hodge structure introduced above). So, equivalently, the question is: how does $E'$ relate to $E$? REPLY [2 votes]: I am going to take a shot at partially answering this question myself. (Parts of these arguments are due to Ben Moonen, my supervisor. All errors are mine.) Any feedback is welcome! The exact relation between $E'$ and $E$ is possibly difficult to describe more explicitly than the tautological: “given by some correspondence on $\mathcal{A}_{1} \times \mathcal{A}_{1}$”. What I can show: $E$ and $E'$ are not isogenous in general. $E'$ does depend on $E$ (i.e. is not constant in $\mathcal{A}_{1}$). Proof of claim 1 Note that the cover $C \to E$ factors as $C \to E_{1} \to E$, where $C \to E_{1}$ is the quotient of $C_{3} \subset S_{3}$, and $E_{1} \to E$ is induced $C_{2}$-cover, which is explicitly given by modding out some $2$-torsion point $P \in E_{1}[2]$. The study of the cover $C \to E$ may hence be replaced by the study of $C_{3}$-covers $C \to E_{1}$ ramified above the origin and a $2$-torsion points such that the corresponding monodromies are inverse to each other. If we compose $C \to E_{1}$ with the hyperelliptic quotient map $E_{1} \to \mathbb{P}^{1}$, we obtain a $C_{6}$-cover $C \to \mathbb{P}^{1}$ ramified above $4$ points. More precisely, above two of these points lies $1$ point, while above the other two there lie $3$ points. We thus have a family of cyclic covers of $\mathbb{P}^{1}$. Ben Moonen (my supervisor) has studied [M] the Jacobian of these covers, to see if they give rise to special subvarieties of $\mathcal{A}_{g}$. In his notation, this family corresponds to ramification data $(1,5,3,3)$, which is not in his list (p.10 (p.508 in print)). By the André–Oort conjecture (in low dimension a theorem) it follows that the Jacobian of $C$ is CM only finitely many times. Since $\mathrm{Jac}(C) \sim E \times \smash{E'}^{2}$, this implies that $E \not\sim E'$, proving the first claim. Proof of claim 2 I am not very well versed in the language of Shimura varieties, so the moduli space of elliptic curves with a marked point of order $2$ might have a standard notation. I'll just call it $\mathcal{M}$. Over this, we have a universal family $\mathcal{E}$, and over that a $C_{3}$-cover $\mathcal{C}$. We can compactify everything and look at the boundary. Obviously $\mathcal{M} \subset \bar{\mathcal{M}}_{1,2}$. One point on the boundary of $\mathcal{M}$ can be described as follows: mark $\mathbb{P}^{1}$ with the points $\{0,\infty\}$, attach a nodal curve to the $\mathbb{P}^{1}$ at $\{1\}$. The dual graph of this curve consists of two vertices joined by an edge; one vertex has a loop, and the other has two half-edges (corresponding to the marked points). This has a $C_{3}$-cover: take a $\mathbb{P}^{1}$ and attach three nodal curves to the $3$rd roots of unity. The map on $\mathbb{P}^{1}$ given by $z \mapsto z^{3}$ is ramified at $\{0,\infty\}$, and can be extended to the nodal curves. The dual graph contains $3$ loops (from the nodal curves), and hence the toric rank of the Jacobian is $3$. The procedure of assigning $E'$ to $E_{1}$ (described partly in the question, and partly in the proof of claim 1) gives a map $\mathcal{M} \to \mathcal{A}_{1}, E_{1} \mapsto E'$. The above example shows that if we pass to the closures $\bar{\mathcal{M}} \to \bar{\mathcal{A}}_{1}$, the map goes through the boundary of $\mathcal{A}_{1}$, and is in particular not constant. This proves the second claim. References [M]: B. Moonen — Special subvarieties arising from families of cyclic covers of the projective line. Documenta Math. 15 (2010), 793-819.<|endoftext|> TITLE: Is there a left-orderable profinite group? QUESTION [8 upvotes]: Is there a nontrivial profinite group $G$ with a binary transitive relation $<$ such that $x1$, then $x^n>1$ for all $n>0$ and $x^n<1$ for all $n<0$. But in the pro-finite topology, the sequence $x^{n!-1}$ converges to $x^{-1}$, so the set $\{x:x\geq 1\}$ is not closed. But this contradicts our assumption that the ordering is compatible with the topology.<|endoftext|> TITLE: Is forcing computable? QUESTION [13 upvotes]: By results similar to Tennenbaum's theorem we know that there exist no computable models of $ZF$. But suppose we are given, as a sort of oracle, access to some model of $ZF$ (e.g. we can make oracle answer queries of form $a\in b?$ for $\in$ being relation on $\Bbb N$ modelling $ZF$). My question then is: is it possible to, given this oracle representation of a model, compute a relation which would create model of $ZF$ with some property we desire, e.g. $AC$ and $\neg CH$? I hope I have explained my question well enough. Thanks in advance. REPLY [11 votes]: Dan's comment below answers the question that was actually asked. Let me explain one way to look at it. If $M$ is any model of ZF and $T$ is any consistent computably axiomatizable theory, then we may look at the version of $T$ inside $M$. There will be some longest initial segment of the axiomatization of $T$ that $M$ thinks is consistent, and this will include the actual $T$. By the completeness theorem applied in $M$, there is a complete consistent Henkin extension of this theory in $M$, and it is represented by a particular number in the presentation of $M$. Using the $\in^M$ relation as an oracle, we can now compute this theory outside $M$, and thereby construct a model of $T$ outside. So we can not only present the $\in$ relation of the model of $T$, but we can decide the entire satisfaction relation for the model. Let me now focus on the related question that I find to be implicitly suggested here, namely, given a model of set theory $M$, presented to us as an oracle, under what circumstances can we compute a structure for a forcing extension $M[G]$ of a particular kind? So let us suppose that $M=\langle\mathbb{N},\in^M\rangle$ has underlying set the natural numbers, in the style of computable model theory, and suppose that $P$ is a partial order in $M$ with which we want to force. We'd like to compute a presentation of $\langle M[G],\in^{M[G]}\rangle$ for some $M$-generic filter $G\subset P$. There are a number of interesting things to say. Theorem. If we are given the $\Delta_0$-elementary diagram of $M$ as an oracle, then we can compute a presentation for a forcing extension $M[G]$ via $P$, along with its $\Delta_0$-elementary diagram. Proof. The main idea is that the presentation of $M$ gives us a canonical enumeration of the dense sets of $P$ in $M$, and using that we can compute an $M$-generic filter $G$ an provide a presentation of $M[G]$. Specifically, let $\cal{D}$ be the object in $M$ that $M$ thinks is all the open dense subsets of $P$, and fix the objects coding the order $\leq_P$ and so on. We compute $G$ as follows. At any given stage, we will have committed ourselves to a certain finite number of compatible elements being in $G$. At stage $k$, we extend this set by searching for the first element of $P$ we can find that is below all of those elements and also in $D_k$, and then we put that element into $G$, and also all elements previously found in $P$ that are above it, and we put out of $G$ any elements of $P$ that we have found so far that are incompatible with that new element. All these questions are $\Delta_0$ in the data we have available, and so in this way, we'll compute an $M$-generic filter $G$. Next, we build a presentation of $M[G]$ using the $P$-names of $M$. We can computably decide whether a given object in $M$ is a $P$-name, because given $\tau$ we can find an object $A$ that $M$ thinks is a transitive set containing $\tau$ and $P$, and then it becomes a $\Delta_0$ property about $(A,\tau,P)$ whether $\tau$ is a $P$-name. Similarly, using the oracle we can computably decide the relations $p\Vdash\tau=\sigma$ and $p\Vdash\tau\in\sigma$, by searching for a large transitive set containing all that data, which thinks that it is true. We now build a presentation of $M[G]$ by enumerating all the $P$-names in $M$, and including the next name on the list just in case there is a condition in $G$ forcing that it is different from all the previous names on our list; otherwise, we find a condition in $G$ forcing that it is the same as one of our previous names. Similarly, we can decide any $\Delta_0$ statement for our presentation, since $p\Vdash\varphi(\tau)$ will be $\Delta_0$ in $M$ with respect to a large transitive set containing all the relevant data, and so we can go search for such a set and then consult our oracle. QED Theorem. From an oracle for the full elementary diagram of $M$, we can compute an oracle for the full elementary diagram of $M[G]$. Proof. This makes things even easier, since we don't have to worry about reducing things to $\Delta_0$. QED Observation. Using only $\in^M$ as an oracle, we can compute a set $G$ that is an $M$-generic filter. Further, for any large ordinal $\theta$ in $M$, we can compute a presentation of $V_\theta^M[G]$. Proof. The main point is that to construct $G$, we don't need a full oracle for the $\Delta_0$-elementary diagram of $M$. Rather, it would suffice to have an oracle for $\Delta_0$-truth in some large $V_\theta^M$, well above the rank of $P$. We can fix the number representing such a $V_\theta^M$, and another number representing the full satisfaction relation on $V_\theta^M$, since $M$ of course can compute a satisfaction relation for any of its sets. Now, using only $\in^M$ as an oracle, for any given $\Delta_0$ assertion $\varphi$ about $V_\theta^M$, we can search in $M$ for the thing that $M$ thinks is $\varphi$ and then look and see if it is in the corresponding thing that $M$ thinks is $\Delta_0$ truth in $V_\theta^M$, and thereby compute $\Delta_0$ truth relative to $V_\theta^M$. The point now is that this is all we needed in order to construct the filter $G$, since for that part of the construction, we needed only to know whether particular conditions were compatible, and so on. Similarly, using the $\Delta_0$ truth of $V_\theta^M$ (or perhaps we would want $\Delta_0$ truth for some much larger $V_\lambda^M$, we can compute a presentation of $V_\theta^M[G]$ as previously. QED Corollary. If $M$ is computably saturated, then using only an oracle for $\in^M$, we can computably provide a presentation of a forcing extension $M[G]$, where $G\subset p$ is $M$-generic for any desired $P$. Proof. If $M$ is computably saturated, then it follows, using a result of Harvey Friedman (see Ali Enayat's slides), that $M$ is isomorphic to some rank-initial segment $V_\theta^M$. Let $Q$ be the image of $P$ in that model. The previous theorem shows how to compute a presentation of a forcing extension $V_\theta^M[H]$, where $H\subset Q$ is $V_\theta^M$-generic. This will be isomorphic to a forcing extension $M[G]$, where $G\subset P$ is $M$-generic. QED In the corollary, we do not necessarily expect that the inclusion $M\subset M[G]$ is computable from the oracle, since I do not see that we can expect the isomorphism of $M$ with $V_\theta^M$ to be computable relative to $\in^M$. I am curious to know whether or not there could be a presentation of a model $M$ for which the oracle $\in^M$ does not compute any presentation of a particular kind of forcing extension $M[G]$. Question. Is there a presentation of a model $M=\langle\mathbb{N},\in^M\rangle\models\text{ZFC}$ such that no presentation of a forcing extension $M[G]$, for a particular forcing notion $P\in M$, is computable relative to oracle $\in^M$? I have a feeling one might be able to construct such a model $M$ by diagonalizing somehow against the possible computations of $M[G]$.<|endoftext|> TITLE: What is the universal property of quotienting a normaliser of the subgroup? QUESTION [10 upvotes]: Let $G$ be a group, $H$ a subgroup and $X$ a $G$-set. By taking orbits $X/H = X \times_H 1$ or fixed points $X^H = \mathrm{Hom}_H(1,X)$ we obtain a set on which $H$ acts trivially, and we've destroyed the $G$-action. What we have left instead is an action of $N_H/H$, where $N_H$ denotes the normaliser of $H$, the subgroup $\{ g \in G : gH = Hg \}$. What characterises this action abstractly? Is there a universal property? What other contexts does this construction exist in? Eg. rings $S \leq R$ and an $R$-module REPLY [10 votes]: The action is the action of the (natural) automorphism group of the relevant functor in each case. This is easiest to see for the case of $$X^H \cong \text{Hom}_H(1, X) \cong \text{Hom}_G(G/H, X)$$ since by the Yoneda lemma the automorphism group of this functor is the automorphism group of $G/H$ as a $G$-set, which is $N_H/H$. Similarly, $$X_H \cong X \times_H 1 \cong X \times_G G/H$$ also admits a natural action by the automorphism group of $G/H$ as a $G$-set, although I am less sure if there is a clean abstract nonsense proof that these are all the natural automorphisms. There is a universal property hiding here, which is that $G/H$ is the free $G$-set on an $H$-fixed point. Exactly the same words can be written down for endomorphism rings instead of automorphism groups in the context of rings and modules. In the context of linear representations of groups the endomorphism ring you end up writing down is a Hecke algebra. In more general contexts it's natural to look not only at the automorphism group or the endomorphism monoid but even the endomorphism Lawvere theory or endomorphism operad. Whereas the former give unary operations, the latter gives operations of higher arity. I give some examples here and here. Other keywords: Tannaka duality, the Barr-Beck theorem...<|endoftext|> TITLE: A proposition on cyclic group QUESTION [7 upvotes]: $G$ is a cyclic group iff $$ \forall H < G, \ \exists k, \ H = \{a^k : a \in G\}. $$ Is it right? REPLY [8 votes]: A bit of searching revealed the following reference for this statement in question F. Szasz, On cyclic groups, Fund. Math., 43(1956), 238-240 In the following more recent paper the authors proved a refinement. Let $k$ denote the number of subgroups of $G$ that are not of the form $\langle a^n,a\in G\rangle$. The main theorem says that $k=0$ iff $G$ is cyclic, $1\le k<\infty$ iff $G$ is finite non-cyclic, and $k=\infty$ iff $G$ is infinite non-cyclic. W. Zhoua, W. Shib, Z. Duan, A new criterion for finite non-cyclic groups Communications in Algebra, 34 (2006), 4453-4457 REPLY [6 votes]: $\def\ord{\mathop{\rm ord}}\def\dvds{\mathrel{|}}$Yes, this is true. The argument is similar to your `maximal order' argument; this shows again that it is helpful to provide your thoughts in the question. For $H1$, and $a=b^k$ for some $b\in G$. If $b\notin H$ then $\langle b\rangle > H$, which contradicts maximality. Thus $b=a^\ell$; this means that $\ord a$ is finite (denote $n=\ord a$) and $\gcd(n,k)=1$. For every $g\in G$, we have $g^{nk}=e$. Thus we see that the orders of elements are bounded, and we may assume that $\ord a$ is maximal (then $\langle a\rangle$ is still a maximal cyclic subgroup). If for every $g\in G$ we had $g^n=e$ (and $g^k\in H$) then we would have $g=g^{\gcd(n,k)}\in H$, which would imply $H=G$. Thus there exists some $g\in G$ with $g^n\neq e$ (and $\ord g\dvds nk$). Then $1<\ord g^n\dvds k$. Both cyclic subgroups $H$ and $\langle g^n\rangle$ are normal (due to the initial condition), and their orders are coprime, so they form a direct product, which contains an element of order $n\cdot \ord g^n>n$. A contradiction.<|endoftext|> TITLE: Least supersingular prime QUESTION [10 upvotes]: Given an elliptic curve over the rationals, what can one say about the size of the smallest supersingular prime? REPLY [6 votes]: edit I just saw that I misread the question. I leave my posting here since I hope that it is interesting for you anyway. /edit I don't know if you are still interested in the answer but I am currently working on exactly this problem. The result I have is explicit and does not depend on any unproven conjecture. It is one chapter of my PhD thesis (which is not finished yet) and details of the proof would probably go beyond the scope of this thread. My result ist this: Let $E$ be an elliptic curve over the rationals with $j$-invariant $j_E$ and conductor $N$. Let \begin{align*}B_{j_E} = \begin{cases} \left( \frac{\log j_E}{2 \pi}\right)^2 &\mbox{for } j_E>0,\\ \left(\frac{\log |j_E|}{\pi} + 1\right)^2 &\mbox{for } j_E<0\\ \end{cases},\end{align*} $M \in\mathbb{N}$, $q = 4 \text{rad} (6 N)$ and $n= \max (3, M, B_{j_E})$. For better readability we will write $A = \frac54q(n+2\log q)^2$. Then there exists a supersingular prime $p$ of $E$ such that $p \geq n$ and \begin{align*}\log p \leq 1.7 \cdot 10^{26} A^{13587505A} h^*(j_E).\end{align*} The "+1" in the definition of $B_{j_E}$ can probably be omitted but it would make one part of the proof more complicated. Since it only affects the result very very slightly I kept it there. The bound should be the same for number fields of odd degree. It may be possible to get a similar result on number fields with at least one real embedding (since Elkies proved the existance of inifinitely many supersingular primes also for elliptic curves over number fields with at least one real embedding) where the methods have to be adjusted slightly.<|endoftext|> TITLE: Conformal map of polygon with circle segments QUESTION [9 upvotes]: I am looking for a conformal map from a "polygon" to eg the upper half plane, which consists of circle segments instead of lines. So for example, it could be a quadrilateral ABCD, but where AB is a circle segment. The closest I can find is the Schwarz-Christoffel mapping. Anyone has any tips? REPLY [10 votes]: The mapping function is a solution of the Schwarz differential equation $$\frac{f'''}{f'}-\frac{3}{2}\left(\frac{f''}{f'}\right)^2=R(z),$$ where $R$ is a rational function with poles at the preimages of the vertices. The poles of $f$ are of second order, and the coefficients at the second order terms are determined by the angles. Schwarz equation can be reduced to a linear differential equation of the form $y''+Ry/2=0$. In the case of a triangle, this is a hypergeometric equation. In the case of a quadrilateral, this is a Heun equation. Literature: Courant, Geometrische Funktionentheorie, Caratheodory, Funktionentheorie, II, (There is an English translation), Golubev, Vorlesungen über Differentialgleichungen im Komplexen, (transled from Russian). The case of a triangle is completely understood (see any book on hypergeometric function) The case of a quadrilateral is already quite complicated, and there are many unsolved questions about quadrilaterals and Heun's equation. See for example arXiv:1409.1529 for some recent results about this, and literature there. EDIT. More concrete examples can be found here arXiv:1111.2296 and here arXiv:1110.2696 and in references to these papers.<|endoftext|> TITLE: Weil height of an Abelian Variety with everywhere (potentially) good reduction QUESTION [7 upvotes]: Background: Suppose that $E$ is an elliptic curve over $\mathbb{Q}$ with everywhere (potentially) good reduction. there are many ways to define the height of $E$, and I will be concerned with the height where one chooses an ample bundle on $X(1)$ and looks at the height of the point representing $E$. This is then well defined up to an $O(1)$ additive factor, and I am only concerned with asymptotics so this is fine. Now, the $j$ function gives a map $j:X(1)\cong \mathbb{P}^1$, and so we have the formula $$h(E) = \sum_{p\not\mid \infty} Max(|j(E)|_p,0) + Max(1,\log|j(E)|)$$ and since $E$ has potentially good reduction, $j(E)$ is an integer, so that (as long as its non-zero), $$h(E) = \log |j(E)|.$$ This is really nice, because it means we can compute the height from purely the infinite place. Question: Is there an analogue for heights of Abelian varieties with everywhere (potentially) good reduction? It can't quite hold as stated since $\mathcal{A}_g$ is not affine for any $g\geq 1$, but I am still wondering if one can use good reduction somehow... Thanks! REPLY [5 votes]: Here are two (almost three) comments that you might find useful. I don't know the answer to your question in general unfortunately. First, Autissier proved that for abelian varieties over $\overline{\mathbb Q}$ with good reduction, the Faltings height of $A$ equals the theta height of $A$ plus some contribution at infinity. See the main theorem on p. 1 of his paper Hauteur de Faltings et hauteur de Néron-Tate du diviseur thêta. So the difference of these two heights can be computed by a contribution at infinity (as you desire). Secondly, if you consider only Jacobians then you can say a little bit more. In fact, if $A = Jac(X)$ is the Jacobian of a curve $X$ over $\overline{\mathbb Q}$, then the arithmetic Noether formula (due to Faltings and Moret-Bailly, see for instance Section 1.6 in https://www.math.leidenuniv.nl/scripties/RdeJong.pdf) states that $$12 h_{Fal}(X) = \omega^2(X) + \Delta(X) + \delta_{Fal}(X) - 4g\log(2\pi).$$ If you assume the Jacobian to have good reduction over $\overline{\mathbb Q}$, then this doesn't imply $X$ has good reduction over $\overline{\mathbb Q}$. Nevertheless, it might be instructive to look at Jacobians of curves with good reduction over $\overline{\mathbb Q}$. So let's assume that $X$ has good reduction over $\overline{\mathbb Q}$ as well. Then, by definition of the discriminant $\Delta(X)$, the Noether formula reduces to $$12 h_{Fal}(X) = \omega^2(X) + \delta_{Fal}(X) - 4g\log(2\pi).$$ You now see that the Faltings height of $A$ is given by $\omega^2(X) +\delta_{Fal}(X) - 4g\log(2\pi)$, as $h_{Fal}(A) = h_{Fal}(X)$. The self-intersection of the dualizing sheaf is never zero (Bogomolov conjecture), so that the Faltings height will always involve $\omega^2$ (in some sense). You can go a bit further now and rewrite the above formula in terms of the height of the Weierstrass divisor and purely analytic contributions. To do so, use Prop. 2.5.2 in loc. cit. The end product is an expression for the Faltings height of $A$ in terms of the height of the Weierstrass divisor and the analytic invariants $R(X)$ and $\delta_{Fal}(X)$. Now your question reduces to: Can the height of the Weierstrass divisor with respect to the relative dualizing sheaf endowed with the Arakelov metric on the minimal regular semi-stable model of $X$ be computed in analytic terms solely? Thirdly, in the case of CM abelian varieties of abelian type (a case you most likely are interested in) there are also formulas relating the Faltings height to periods due to Colmez. I can also comment on this if you'd like. Note: I used some notation and terminology that I find convenient. Don't hesitate to let me know if elaboration needed.<|endoftext|> TITLE: Exotic "non-linear" (but "almost linear") automorphisms of symplectic vector space QUESTION [7 upvotes]: Let $V$ be a vector space over a field $k$ equipped with a symplectic form $\omega$. Let $f:V \rightarrow V$ be a bijective set map such that the following hold. For all $v \in V$ and $c \in k$, we have $f(c v) = c f(v)$. For all $v,w \in V$ such that $\omega(v,w)=0$, we have $f(v+w) = f(v)+f(w)$. Question: Must $f$ actually be linear? The answer is obviously false if $dim(V)=2$, so this question is only interesting for $dim(V) \geq 4$. If this question is too hard (or has a negative answer), I'd also be interesting in restricting myself to $f$ which also satisfy the following condition. For all $v,w \in V$ such that $\omega(v,w)=0$, we have $\omega(f(v),f(w))=0$. REPLY [5 votes]: Good question in symplectic linear algebra. $f$ must be linear (even if we do not assume it to be bijection) for $dim(V)\ge 4$. First remark is that the restriction of $f$ to any isotropic subspace is linear. V is symplectomorphic to $L^*\oplus L$ with the symplectic structure $\omega ((a,b),(x,y))=a(y)-x(b)$. We may assume (after subtraction of linear map $f|_{L^*\oplus 0}+f|_{0\oplus L}$) that $f$ equals to zero on $L^*\oplus 0$ and on $0\oplus L$. It remains to prove that $f$ is totally zero map. Note that $f$ is automatically zero on such $(a,b)$ that $a(b)=0$ (since (a,b)=(a,0)+(0,b)). Denote the set of all such vectors by $Ann$. Now I claim that any vector $(x,y)$ ($x(y) \ne 0$) can be decomposed as a sum of two skew-orthogonal vectors from $Ann$ and hence the value of $f$ is zero on $(x,y)$. There are plenty of opportunities to construct such a decomposition. Obviously, it is sufficient to prove the existence of such a decomposition for $x(y)=1$. We can choose a basis $e_1,..$ in $L$ and dual basis $f^1,..$ in $L^*$ such that $w=(x,y)=(f^1,e_1)$. Now take decomposition $(f_1,e_1)=1/2(f^1+f^2,e_1-e_2)+1/2(f^1-f^2,e_1+e_2)$. It is easy to check that vectors $1/2(f^1+f^2,e_1-e_2)$ and $1/2(f^1-f^2,e_1+e_2)$ belong to $Ann$ and skew-orthogonal. So we are done.<|endoftext|> TITLE: p-adic Stein spaces QUESTION [8 upvotes]: The higher cohomology of coherent sheaves vanish on Stein spaces (both complex and p-adic). In the case when the space ($X$) is a curve and we're working in the complex world, this shows that all holomorphic line bundles are trivial (The exponential exact sequence and the fact that $H^2(X,\mathbb{Z})= 0$). Is the same statement true in the p-adic case? That is, does the vanishing of $H^1(X, \mathcal{O})$ imply the vanishing of $H^1(X, \mathcal{O}^*)$? REPLY [11 votes]: The answer is no. In fact, a theorem of Lazard shows that the open unit disk has a non-trivial Picard group when the ground field is a non maximally complete ultrametric valued field. See Proposition 6 in Les zéros d'une fonction analytique d'une variable sur un corps valué complet. Publications Mathématiques de l'IHÉS, 14 (1962), p. 47-75.<|endoftext|> TITLE: Classifying set theories whose standard models sharing the same ordinals are equal QUESTION [14 upvotes]: Let's say that a (recursively axiomatizable) set theory $T$ extending ZF is "ordinal-categorical" if, whenever $M$ and $N$ are standard models of $T$ sharing the same ordinals, one has $M = N$. For example, if $T$ proves $V = L$, then $T$ is ordinal-categorical. I think the same is true if $T$ proves $V = L(0^\sharp)$. Is there anything else that can be said about such theories $T$? In particular, is there a way to find an axiom or axiom schema A such that all ordinal-categorical theories $T$ are precisely the (recursively axiomatizable) extensions of ZF+A? If this is not known, is finding such an A a goal of the inner model program? (Note: In the original post I used the word "canonical" instead of "ordinal-categorical.) REPLY [21 votes]: An old theorem of Harvey Friedman answers the question: Theorem. If $T$ is an r.e. extension of KP (Kripke-Platek set theory) and $T$ is $\alpha$-categorical for all countable ordinals $\alpha$, then $T$ proves $V = L$. In the above, "$T$ is $\alpha$-categorical" means that $T$ has at most one standard model of ordinal height $\alpha$. Friedman's result appears as Theorem 6.3 of his paper below: Countable models of set theories, In A. R. D. Mathias & H. Rogers (eds.), Cambridge Summer School in Mathematical Logic, Lecture Notes in Mathematics vol. 337, Springer-Verlag. pp. 539--573 (1973). Two Remarks: 1. The theory $ZF + V = L(0 ^{\#})$ need not be ordinal categorical since by Friedman's theorem, two standard models of set theory of the same height could both believe that $0 ^{\#}$ exists, and yet they might have distinct $0 ^{\#}$s. Recall that the complexity of $0 ^{\#}$ is $\Delta^1_3$. 2. It is open whether the conclusion of Friedman's theorem continues to hold if its hypothesis is weakened to "$T$ is $\delta$-categorical", where $\delta$ is the ordinal height of the Shepherdson-Cohen minimal model of set theory.<|endoftext|> TITLE: ULU Decomposition of a matrix QUESTION [16 upvotes]: Let $g \in GL_n(\mathbb{F}_q)$. Is it true that we can always write $g = u_1lu_2$, where $u_1$ and $u_2$ are upper-triangular and $l$ is lower-triangular? Note that I'm not requiring that the matrices be unipotent. This is equivalent to being able to write $g=u_1lu_2d$, where $u_1$ and $u_2$ are unipotent upper-triangular, $l$ is unipotent lower-triangular, and $d$ is a diagonal matrix (see Geoff's comment below). Remark: It suffices to show that this is true for $g$ a permutation matrix; the Bruhat decomposition will then guarantee that this will be true for arbitrary $g$. In particular, the statement is true for $n=2$ as $$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}. $$ REPLY [4 votes]: To atone for my misguided earlier attempt at an answer, I'd suggest a different way to get a general criterion of this sort (working first over an algebraically closed field, then adapting to finite fields via the BN-pair structure). While this is less elementary than the matrix theory over rings involved in the treatment of general linear groups, it explains the result conceptually in terms of the Bruhat decomposition. (Recall that this decomposition is a natural generalization to reductive groups of Gauss reduction in elementary linear algebra.) The ideas here all go back to the 1956-58 Chevalley seminar, but the three textbooks Linear Algebraic Groups by Borel, Springer, and myself may be easier to refer to. In a connected semisimple (or reductive) algebraic group $G$, the Borel subgroups are all conjugate, so we can fix a particular $B$. Working in $G$ or in the (projective) flag variety $G/B$, we get a disjoint decomposition indexed by elements of the Weyl group $W$, starting with $G= \bigcup_w BwB$. In the resulting Bruhat cell decomposition of $G/B$, the cell corresponding to $w \in W$ is isomorphic to an affine space of dimension $\ell(w)$. Then the longest element $w_\circ$ of $W$ (of length $= \dim G/B$, the number of positive roots) yields a dense open subset. The double coset $Bw_\circ B$ is similarly a dense open subset of $G$. Its left translate $\Omega:=w_\circ B w_\circ B$ is often called the big cell in $G$; it is in fact a principal open set in $G$. Relative to a fixed maximal torus lying in $B$, we have $B = TU$ with $U$ maximal unipotent. Similarly $B^- := w_\circ B w_\circ = TU^-$. (Note that $w_\circ$ is its own inverse.) Here $U, U^-$ would be the upper and lower triangular unipotent matrices in the general linear group. Now recall a trivial lemma (7.4 in my book): the product of two nonempty Zariski-open (hence dense) subsets in $G$ is all of $G$. Apply this to $Bw_\circ B$, along with the fact that $T$ can be moved past other factors to combine with $B$, to conclude: $$G = (B w_\circ B) (B w_\circ B) = B w_\circ B w_\circ B = U U^- B.$$<|endoftext|> TITLE: Is the space of real conics with a singular point an orientable manifold? QUESTION [5 upvotes]: Consider the space of non zero real homogeneous degree $2$ polynomials in three variables upto scaling. This space is $\mathbb{R} \mathbb{P}^5$. The zero set of such a polynomial gives a real curve in $\mathbb{R} \mathbb{P}^2$. Let me define the space $X$ to be the space of real curves $[f]$ and a marked point $p$, such that the curve has a singularity at $p$, i.e. $$ X := \{ ([f], p) \in \mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2: f(p) =0, ~~\nabla f|_p =0 \}. $$ $\textbf{Question $1$:}$ Is $X$ a smooth manifold of the expected dimension (four)? $\textbf{Question $2$:}$ If yes, is $X$ orientable or non orientable? $\textbf{Question $3$:} $ Is it obvious that $X \rightarrow \mathbb{RP}^2$ is a fiber bundle (i.e. locally trivial)? $\textbf{Comments on Question $3$:}$ Suppose we write a homogeneous polynomial explicitly as $$ f(X,Y,Z) := A_1 X^2 + A_2 Y^2 + A_3 Z^3 + A_4 X Y + A_5 X Z + A_6 YZ.$$ Suppose we want to trivialize the bundle near the point $[0,0,1]$. It is easy to see that if we ask for $f$ to have a singularity at $[0,0,1]$ then $A_3, ~A_5, ~A_6 =0$. Let $\mathcal{U}_{\epsilon_1, \epsilon_2}$ be an open set consisting of points of the form $[\epsilon_1, \epsilon_2,1]$. It is easy to see that the map $$ h: \pi^{-1} (\mathcal{U}_{\epsilon_1, \epsilon_2}) \rightarrow \mathcal{U}_{\epsilon_1, \epsilon_2} \times \mathbb{RP}^2 $$ given by $$ h([A_1, \ldots, A_6]; p):= (p, [A_1, A_2, A_4]$$ is not a trivialization. Of course, that doesn't prove anything; but I am just wondering if $\pi:X \rightarrow \mathbb{RP}^2$ is a fiber bundle? $\textbf{Comments:}$ I have two arguments giving me contradictory answers. First of all, note that (assuming $X$ is a manifold), the normal bundle to $X$ in $\mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2$ is given by $$ N_X:= \gamma_{5}^* \otimes \gamma_2^{* 2} \oplus \gamma_5^* \otimes T^*\mathbb{R P}^2 \otimes \gamma_2^{*2} \big{|}_X, $$ where $\gamma_n$ is the tautological line bundle over $\mathbb{RP}^n$. (I can justify this if someone is not convinced). Note that $$ TX \approx \frac{T (\mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2)}{N_X}. $$ It is now easy to see that the first stiefel whitney class of the tangent bundle of $X$ is zero, i.e. $w_1(TX) =0$. Hence $X$ is orientable. $\textbf{The above statement is incorrect:}$ One can check that $w_1(TX) \neq 0$, which is consistent with the fact that $X$ is non-orientable. On the other hand it seems to me that $~\pi: X \rightarrow \mathbb{RP}^2$ is a fibre bundle, with fibres $\mathbb{RP}^2$. One can show (using spectral sequences) that any $\mathbb{RP}^2$ bundle over $\mathbb{RP}^2$ is non orientable. $\textbf{Proof of why $X$ is a manifold:} $ Let us take $\mathbb{R}^6$ to be the space of real polynomials in two variables of degree at most $2$. Hence we can write such an element $\rho \in \mathbb{R}^6$ as $$ \rho(x,y) = \rho_{00} + \rho_{10} x + \rho_{01} y + \frac{\rho_{20}}{2} x^2 + \rho_{11} x y + \frac{\rho_{02}}{2} y^2. $$ Now consider the map $$ \psi : \mathbb{R}^6 \times \mathbb{R}^2 \rightarrow \mathbb{R}^3 $$ given by $$ \psi(\rho; x,y) := \Big( \rho_{00} + \rho_{10} x + \rho_{01} y + \frac{\rho_{20}}{2} x^2 + \rho_{11} x y + \frac{\rho_{02}}{2} y^2, \\ ~\rho_{10} + \rho_{20}x + \rho_{11} y, \\ ~\rho_{01} + \rho_{11}x + \rho_{02}y \Big). $$ It is easy to see that if $\psi(\rho,0,0) =0$, then the Jacobian matrix of $\psi(\rho,x,y)$ at $(x,y)=(0,0)$ has full rank. To see why, take the partial derivative of $\psi$ with respect to $\rho_{00}$, $\rho_{10}$ and $\rho_{01}$ and plug in $(x,y) = (0,0)$. That gives us a $3\times 3$ identity matrix. Hence $\psi$ is transverse to the zero set (it is easy to see taking $(x,y) = (0,0)$ was without any loss of generality). Hence, by Implicit Function Theorem, $\psi^{-1}(0,0,0)$ is a smooth manifold (even around double lines). Its now easy to see that the space $X$ defined will also be a manifold; by writing the evaluation map and the vertical derivative in a coordinate chart and trivialization, it reduces to the above calculation. REPLY [10 votes]: Yes, it is a smooth manifold. No, it is not orientable. For the first, just think geometrically, i.e., without bases: Fix a $3$-dimensional vector space $V$ and consider the homogeneous quadratic polynomials on it, which is a $6$-dimensional vector space isomorphic to $\mathsf{S}^2(V^\ast)$. Let $\mathbb{P}V$ be the projectivization of $V$, i.e., the space of $1$-dimensional subspaces of $V$. Given $L\in\mathbb{P}V$, the set of elements of $\mathsf{S}^2(V^\ast)$ that represent conics singular at $L$ is simply the $3$-dimensional space $\mathsf{S}^2(L^\perp)\subset \mathsf{S}^2(V^\ast)$, where $L^\perp\subset V^\ast$ is the space of linear functions on $V$ that vanish on $L$. In particular, the set $E\subset \mathbb{P}V\times \mathsf{S}^2(V^\ast)$ that consists of the pairs $(L,f)$ such that $f\in \mathsf{S}^2(L^\perp)$ is a smooth subbundle of rank $3$ over $\mathbb{P}V$ of the trivial bundle $\mathbb{P}V\times \mathsf{S}^2(V^\ast)$. Your space $X$ is simply $\mathbb{P}E$, the projectivization of this smooth, rank $3$ bundle. In particular, it is a smooth submanifold of $\mathbb{P}V\times \mathbb{P}\bigl(\mathsf{S}^2(V^\ast)\bigr)$ of dimension $4 = 2 + (3{-}1)$, and the projection $\mathbb{P}E\to\mathbb{P}V$ is a submersion that is a locally trivial fiber bundle. Second, $X=\mathbb{P}E$ cannot be orientable because no projectivization of a locally trivial, rank $3$ bundle $E$ over a smooth manifold $M$ is orientable. You don't need spectral sequences to see this, you just need to produce one closed loop around which the orientation bundle is nontrivial, but these are easy to find: Just take a loop in a single fiber $\mathbb{P}E_x$ that generates that fiber's fundamental group (which is isomorphic to $\mathbb{Z}_2$). This is clearly an orientation-reversing loop in $\mathbb{P}E$. Response to Question 3: The OP wondered why the bundle $\pi:E\to \mathbb{P}V$ is locally trivial. Here is one way to see this: Let $L_0\in \mathbb{P}V$ be a $1$-dimensional subspace and let $X,Y,Z\in V^\ast$ be a basis of the linear functions on $V$ such that $X$ and $Y$ vanish identically on $L_0$ (and hence are a basis of $(L_0)^\perp$) while $z$ does not identically vanish on $L_0$ (and hence only vanishes on $L_0$ at $0\in L_0$). Let $U\subset\mathbb{P}V$ be the open set consisting of those $L\in \mathbb{P}V$ such that $Z$ is not the zero linear functional when restricted to $L$. On $U$, there are two well-defined, smooth functions $x,y:U\to\mathbb{R}$ such that $x(L) = X(v)/Z(v)$ and $y(L) = Y(v)/Z(v)$ for some (and, hence, any) nonzero $v\in L$. Then, for each $L\in U$, the linear functions $X-x(L)\,Z$ and $Y - y(L)\,Z$ are a basis of $(L)^\perp\subset V^\ast$, and hence the quadratic forms $$ A(L) = \bigl(X-x(L)\, Z\bigr)^2,\quad B(L) = \bigl(X-x(L)\, Z\bigr)\bigl(Y-y(L)\, Z\bigr),\quad C(L) = \bigl(Y-y(L)\, Z\bigr)^2 $$ are a basis of $E_L = \mathsf{S}^2(L^\perp)$ for each $L\in U$. Thus, they define a smooth trivialization of $\pi:E\to\mathbb{P}V$ over $U$. Obviously, $\mathbb{P}V$ can be covered by such open sets $U$, and it is easy to see that the transitions on overlaps are smooth. Thus $E$ is a smooth bundle with local smooth trivializations.<|endoftext|> TITLE: Graduate program applications that require questionnaires and other non-letter material QUESTION [27 upvotes]: In the December 2014 AMS Notices, a letter to the editor (http://www.ams.org/notices/201411/rnoti-p1311.pdf) by Deconinck and Medlock addresses the problem of (math) graduate programs requiring letter writers to fill out questionnaires about applicants or do other tasks besides uploading a recommendation letter, unlike the quick process for job applications on Mathjobs where all a letter writer has to do is upload a letter (once). Deconinck and Medlock suggest that letter writers to graduate programs append a postscript to their recommendation letters urging faculty who read the letter to check how their own graduate program handles applications and get the process streamlined if it has any requirements like the despised questionnaires. Many faculty may not know what their own school makes letter writers do (esp. if they have not served on the graduate admissions committee). In that spirit, I propose that we collect here on MO a list of schools that make letter writers fill out questionnaires, as a kind of public service to encourage any faculty who see their own school listed here to go to their graduate school administration and insist that the math graduate program applicants -- if not all graduate school applicants -- avoid having any kind of survey as part of the application that letter writers have to fill out. If, in the future, a graduate program kills off the required questionnaire, the answer listing that school could be updated to make clear its passage from the dark side to the light side, so to speak. Of course the admission websites are designed for an entire graduate school, not just a math department, but math faculty could still tell their graduate school that they make no use of questionnaires, percentage rankings, etc. from the online form, so it is a waste of time and therefore they do not want to make letter writers for the math graduate program see any of that stuff. If you are writing graduate program recommendation letters this year and think this question is worthwhile, please keep a record of which graduate programs make you fill out questionnaires or do anything else besides upload a letter so that you can come back here and make a record of it if nobody else has yet. Post one school per answer, ideally with the school name first for ease of scanning, and describe what letter writers are asked to do besides upload a letter. (There are also some praise-worthy graduate programs where the letter writer has to do essentially nothing besides upload a letter, e.g., UIC and Wisconsin have no questionnaire, and Univ. Southern California only posts a questionnaire at the very end and makes it clear that filling it is not a requirement. Perhaps there should be a separate MO community wiki list for those schools.) REPLY [6 votes]: University of Minnesota Not bad -- I was Driven to Discover (SM) that I have to create a password just for using the system that one time. But was able to bypass the survey completely.<|endoftext|> TITLE: Any two bivariate algebraically dependent polynomials are always in the same ring generated by some bivariate polynomial? QUESTION [8 upvotes]: If $f(x,y)$ and $g(x,y)$ are two algebraically dependent polynomials over some field $k$, is it true that there exists a bivariate polynomial $p(x,y)$ such that both $f(x,y)$ and $g(x,y)$ are in the ring $k[p(x,y)]$? REPLY [10 votes]: Yes. Let $k[T]$ denote the $k$-subalgebra of $k(x,y)$ generated by a subset $T$, and let $k(T)$ denote its field of quotients. Luroth's Theorem says that every field $L$, $k \subset L \subset k(x,y)$, of transcendence degree one over $k$ is equal to $k(p)$ for some $p \in k(x,y)$. A sharpening of E. Noether (char $k$ = 0) and A. Schinzel (arbritrary $k$) says that if $L$ contains a nonconstant polynomial, then some $p \in k[x,y]$ suffices (Theorem 4, page 10 in "Selected Topics on Polynomials" , by A. Schinzel). The hypothesis that $f$ and $g$ are algebraically dependent implies that $k(f,g)$ has transcendence degree $\leq 1$ over $k$. Thus $f,g \in k(p)$ for some $p \in k[x,y]$. Unique factorization in $k[x,y]$ and $k[p]$ then implies that $f,g \in k[p]$.<|endoftext|> TITLE: Is every projective space curve a set-theoretic intersection of two surfaces? What is known about this question? QUESTION [18 upvotes]: I am sorry if this question has been already asked, couldn't find anything similar myself. I have recently recalled this long standing open problem of whether every irreducible curve in $\mathbb{P}^3(\mathbb{C})$ is a set-theoretic intersection of two surfaces in $\mathbb{P}^3(\mathbb{C})$. I know that there are results in characteristic $p > 0$. My question is what is the best result regarding this problem known today? Some references would be appreciated. REPLY [6 votes]: This question is wildly open still in characteristic zero. We know some examples of curves that are s.c.i.’s, for example Rao’s classification of self-linked ACM curves, Hartshorne and Polini’s examples of curves on ruled cubic surfaces. In Hartshorne and Polini’s recent paper on codepth, they reformulated the conditions for a curve to be an s.c.i. in terms of the new notion of codepth, and gave several necessary criteria. One interesting necessary condition for $C$ in $P^3$ to be an s.c.i. is that $C$ must be contained in a hypersurface $X$ such that $X-C$ is affine. Thus an example of a smooth curve $C$ in $P^3$ not contained in any such hypersurface would give a counter example to the s.c.i. conjecture. It is believed that the rational quartic is not likely an s.c.i, however no complete proof has been published yet. The best so far are results on the lower bounds of the degrees for the two equations should they exist.<|endoftext|> TITLE: Unstable homotopy groups of spheres beyond Toda's range QUESTION [30 upvotes]: In 1962 Toda published his book "Composition methods in homotopy groups of spheres", which contains computations of $\pi_{n+k}(S^n)$ for $k\le 19$ and $n\le 20$. The values of these groups are conveniently tabulated, and reproduced on the Wikipedia page Homotopy groups of spheres. The computations involve composition product, Toda brackets and the EHP sequence, as well as cohomology operations for the higher values of $k$. I am fairly certain that more values of $\pi_{n+k}(S^n)$ have been computed in the intervening years, perhaps with more modern methods such as the unstable Adams spectral sequence. Does anybody know of an up-to-date table of known unstable homotopy groups of spheres, beyond the range shown in Toda's tables? REPLY [19 votes]: By following Ryan's leads I've been able to find references computing $\pi_{n+k}(S^n)$ for $20\leq k \leq 30$ at the prime $2$ (and in some cases at odd primes as well). I thought I'd post these as an answer for ease of reference. M. Mimura, H. Toda, The (n+20)-th homotopy groups of n-spheres, J. Math. Kyoto Univ. 3 1963 37–58. [Contains $\pi_{n+20}(S^n)$ for all $n$ and at all primes] M. Mimura, On the generalized Hopf homomorphism and the higher composition, Parts I, II. J. Math. Kyoto Univ., 4, 171-190, 301-326 (1964/5). [Contains $\pi_{n+21}(S^n)$ and $\pi_{n+22}(S^n)$ for all $n$ and at all primes. For odd primes the author refers to work of Toda later published as a series of papers On iterated suspensions I--IV, J. Math. Kyoto Univ (1966/68).] M. Mimura, M. Mori, N. Oda, Determination of 2-components of the 23- and 24-stems in homotopy groups of spheres, Mem. Fac. Sci. Kyushu Univ. Ser. A 29 (1975), no. 1, 1–42. [Contains $\pi_{n+23}(S^n)$ and $\pi_{n+24}(S^n)$ for all $n$ at the prime $2$] N. Oda, On the 2-components of the unstable homotopy groups of spheres Parts I, II, Proc. Japan Acad. 53, Ser. A(1977), no. 6/7, 202-205/215-218. [Contains $\pi_{n+k}(S^n)$ for $25\leq k\leq 30$ for all $n$ at the prime $2$, and partial results for $31\leq k\leq 33$] The methods are as in Toda's book. I would be interested to learn of any later references, or errata to the above.<|endoftext|> TITLE: Algorithm to minimally connect line segments in Euclidean plane QUESTION [5 upvotes]: Suppose you have finitely many line segments in the Euclidean plane. How do you "connect them to form one chain of line segments of minimal length?" More formally and generally, what I'm looking for is an algorithm to solve the following modification of the TSP. Suppose you have a digraph $G = (V, A)$ with edge weights and an even number of vertices $V = \{ v_{2i}, v_{2i + 1} : i < n \}$. Find, if it exists, the closed walk $C$ of minimal weight such that for every $i < n$, either $v_{2i} \in C$ or $v_{2i+1} \in C$. (So in order to solve the above question, one would have two vertices for each line segment, one corresponding to each direction the line segment can be walked along.) I tried formulating the problem as a mixed integer linear program, but it seems making this program choose between two vertices for each $i < n$ inevitably results in something non-convex. It seems as if this should be a problem that has been studied before, but I cannot find any references. Edit: I think the following mixed integer linear program solves the problem, although maybe not as elegantly as one could. It's a modification of the standard way to turn a TSP into a MILP: Suppose $V = 2n = \{ 0, \ldots, 2n -1 \}$. Let $c_a$ be the weight of an arc $a \in A$. The intended meaning of the variables $x_a$ is that $x_a = 1$ iff the arc $a$ is in the tour. For a vertex $k \in V$, $\delta^-(k)$ and $\delta^+(k)$ denote the set of all incoming or outgoing arcs respectively. Minimize $$ \sum_{a \in A} c_a x_a $$ such that: \begin{eqnarray*} x_a \in \{0, 1\} &\qquad& \forall a \in A\\ \sum_{a \in \delta^+(2i)} x_a + \sum_{a \in \delta^+(2i + 1)} x_a = 1 &\qquad& \forall i < n\\ \sum_{a \in \delta^-(2i)} x_a + \sum_{a \in \delta^-(2i + 1)} x_a = 1 &\qquad& \forall i < n\\ \sum_{i \in K} \left( \sum_{a \in \delta^+(2i)} x_a + \sum_{a \in \delta^+(2i + 1)} x_a \right) \geq 1 &\qquad& \forall K \subset n\ \text{with}\ K \neq \emptyset, n \end{eqnarray*} Now the problem remains that the solution could go into a vertex 2i but exit vertex 2i+1. This is where I thought things would become non-convex. But adding artificial dimensions via helper variables $t_i$ with the intended meaning that $t_i = 0$ if the node $2i$ is picked, and $t_i = 1$ if node $2i + 1$ is picked, we can add as constraints \begin{eqnarray*} t_i \in \{ 0, 1 \} &\qquad& \forall i < n\\ \sum_{a \in \delta^+(2i)} x_a + \sum_{a \in \delta^-(2i)} x_a = 2 (1 - t_i) &\qquad& \forall i < n\\ \sum_{a \in \delta^+(2i + 1)} x_a + \sum_{a \in \delta^-(2i + 1)} x_a = 2 t_i &\qquad& \forall i < n \end{eqnarray*} I hope this makes sense and apologize if it doesn't, I basically have no background in any form of optimization. REPLY [2 votes]: Another way to solve the problem might be to reduce it to a traditional version of the TSP and then use a free, off-the-shelf solver like one of these. At the very least, it should be a relatively easy way of validating that your MILP construction is programmed correctly. Note that this problem can be described as a TSP problem where the solution must cross certain edges (i.e., the "line segments"). For each "forced" edge in the graph, replace the edge o-----o with a new node and two edges in series, like this: o--o--o Consider a traditional TSP solution on the new graph. In order to reach the new node, any valid TSP closed walk must cross the entire "o--o--o" structure. Therefore, a minimal closed walk on the new graph corresponds to a minimal closed walk on the original graph that crosses all the desired edges. There are no constant-time approximations for your general problem (because, with length zero links, the general TSP problem is a special case of your problem). However, there are constant-factor approximation algorithms for metric TSP, and even a PTAS for the TSP in the Euclidean plane (Arora 1998), so it's possible that there might be an approximation algorithm for your "connect the segments in the plane" special case. I spent a while unsuccessfully tinkering with Christofides' algorithm to try to find one for this problem; perhaps someone else might have more luck?<|endoftext|> TITLE: Connected components $0-1$ matrices QUESTION [12 upvotes]: Let $M$ be a $0-1$ matrix. Here a matrix has one component means we can traverse from a matrix entry $(i,j)$ which is $1$ to any other one by moving step of $(i\pm1,j),(i,j\pm1),(i\pm1,j\pm1)$ where each step you take you step on another $1$. Can every $0-1$ be converted to a matrix of one component by permutations of rows and columns? What classes of matrices cannot have one component? also posted: https://math.stackexchange.com/questions/1072461/connected-components-0-1-matrices (Say I proved it for $n$ components merged to one. Now say I have $n+1$ components. If I move the first $n$ components by induction argument, the last $(n+1)^{st}$ component may split up in exponentially many. Am I wrong about this?) Every $M$ can be given by $M=\sum_{i=1}^nM_i$ where $M_i$ are $0-1$ and rank $1$ and disjoint when placed on $M$. Row and column reduce each $M_i$ to $P_i$ with just one $1$ entry. Consider $P=\sum_{i=1}^nP_i$. There are various different ways to convert to $P_i$ for every given $M_i$. For each $M_i$ let $S_i$(area of rectangle) be the number of ways to reduce to point matrix $P_i$. Total ways is $S=\prod_{i=1}^nS_i$. I think the question can be thought as finding a permutation of $P$ given a point reduction out of $S$ ways such that when you expand each $P_i$ to its $M_i$, the resulting matrix should be one component $0-1$ matrix. I also think working with smallest possible $n$ should suffice. Is there an $M$ such that for all configuration of points $P_i$ from $S$ choices, any permutations on $P_i$ followed by expansions to $M_i$ would either keep $M$ disconnected or $M_i$ non-disjoint? I also think working with one candidate choice of $P_i$ out of $S$ many choices should suffice. How about using counting arguments? Will those help here? Given a connected matrix, we can count the number of permutations that permute the matrix to 'distinct' matrices. This can be done by looking at $M_i$s that admit the largest number permutation that change $M_i$ to something different from $M_i$. We probably can guess the number of ways different possibilities of $M_i$s that will give to 'distinct' $M$s that cannot be obtained from permutation of another. We know the total number of $0-1$ matrices is $2^{m^2}$. From this can we guess number of disconnected is bounded away from $0$ or bounded towards $0$? Christian's answer based on fedja's comment solves the problem. Infact the approach can be recursively used to get multiple component matrices for any constant number of components $k$. It would be nice to know how big $k$ can be for a given $n$. Is there any estimate for $k$ as a function of $n$ (posted as question in https://mathoverflow.net/questions/191248/maximum-connected-components-0-1-matrix)? REPLY [11 votes]: This is fedja's beautiful comment, posted as an answer for better visibility: Not all matrices can be brought to one component form by exchanging rows/columns. Consider large $n\times n$ matrices with all possible entries equal to $0,1$. By partitioning this into $3\times 3$ blocks, we see that the number of matrices with no isolated $1$'s is $\lesssim (2^9-1)^{n^2/9}=2^{cn^2}$, $c<1$, because one of the $2^9$ possible blocks is off-limits, the one with a lonely $1$ in the center. For each such matrix, there are at most $(n!)^2\lesssim 2^{dn\log n}$ row/column permuted matrices that can be obtained from it. So the number of matrices that can be brought to one component form is $\lesssim 2^{c'n^2}$ with $c'<1$, and this is $\ll 2^{n^2}$, the number of all matrices.<|endoftext|> TITLE: A More Advanced Version of Aluffi's Chapter 0 QUESTION [6 upvotes]: This is a crosspost of this question from MSE. Paulo Aluffi's Book, Algebra, Chapter 0 aims to teach basic algebra from a categorical viewpoint. The first chapters of the book, however, introduce groups and rings using only very basic categorical concepts (no general limits or adjoints). Is there a book which teaches algebra from a more advanced categorical point of view, in particular using general limits and adjoints? Is there a book which teaches algebra from the $n$POV? REPLY [6 votes]: The closest I know for what you're looking for is Bergman's An Invitation to General Algebra and Universal Constructions, which can be downloaded here. It is a simultaneous introduction to both universal algebra and universal constructions in category theory. (Universal algebra generalizes both groups and rings, and it has a natural category-theoretic translation.) Nothing about the nPOV, though.<|endoftext|> TITLE: Is a free group a product of f.g subgroups of infinite index? QUESTION [9 upvotes]: Let $F$ be a free group, and let $H,K \leq F$ be finitely generated subgroups of infinite index in $F$. Is it possible that for the set of products we have $HK = F$ ? REPLY [7 votes]: It's actually easier to prove the stronger statement that there are infinitely many double cosets $H\backslash F/K$. First, note that if $F'$ is a subgroup of finite index in $F$, with $H'$ and $K'$ its intersections with $H$ and $K$ respectively, then the natural map $H'\backslash F'/K'\to H\backslash F/K$ is finite-to-one. So we may pass to finite-index subgroups. EDIT: (I was a little glib in translating from topology to group theory before. Here's a corrected version of the final paragraph.) Therefore, by Marshall Hall's theorem, we may take $F$ to be the fundamental group of a graph $X$ and $H$ and $K$ to be carried by embedded subgraphs $Y$ and $Z$, say. But now it's easy. Indeed, let $a$ be a based loop not contained in $Y$ and $b$ a based loop not contained in $Z$. Then the double cosets $Ha^mb^nK$ are all distinct.<|endoftext|> TITLE: Modular forms and "too many symmetries" QUESTION [5 upvotes]: How do we interpret Barry Mazur's quote of Modular forms are functions on the complex plane that are inordinately symmetric. They satisfy so many internal symmetries that their mere existence seem like accidents. But they do exist. I figure the symmetries in a (elliptic) modular form are just what you get from the functional equation defining them. But $SL(2,Z)$ is the source of these symmetries, and while an infinite group, only has two generators. These correspond to the translation symmetry $f(z) = f(z+1)$ and "weaker" the inversion of a unit circle symmetry $f(-1/z) = z^k f(z)$ (weaker in that it's not an equality in images, there's that $z^k$ factor). I could create a function to a lattice exhibiting rotational symmetry as well as translational. Admittedly this would be boring, but it would have as many symmetries. Are there yet more symmetries to modular forms? Or is it that modular forms are highly non-trivial whilst maintaining this structure of symmetries? REPLY [15 votes]: My interpretation of Mazur's quote was in terms of the history of the discovery of modular forms. Of course trigonometric functions came first, and then a whole variety of other special functions in the eighteenth and nineteenth century, culminating in the study of elliptic functions. Since elliptic functions exhibit some kind of "universality" among special functions (most can be written in terms of evaluations of elliptic functions or limits thereof), when people first considered generalizations of their symmetries, it wasn't even clear if they would find anything. This is also illustrated by the section "Poincaré on discovery and his work on automorphic functions" in the wiki article on automorphic forms: One of Poincaré's first discoveries in mathematics, dating to the 1880s, was automorphic forms. He named them Fuchsian functions, after the mathematician Lazarus Fuchs, because Fuchs was known for being a good teacher and had researched on differential equations and the theory of functions. Poincaré actually developed the concept of these functions as part of his doctoral thesis. Under Poincaré's definition, an automorphic function is one which is analytic in its domain and is invariant under a discrete infinite group of linear fractional transformations. Automorphic functions then generalize both trigonometric and elliptic functions. Poincaré explains how he discovered Fuchsian functions: For fifteen days I strove to prove that there could not be any functions like those I have since called Fuchsian functions. I was then very ignorant; every day I seated myself at my work table, stayed an hour or two, tried a great number of combinations and reached no results. One evening, contrary to my custom, I drank black coffee and could not sleep. Ideas rose in crowds; I felt them collide until pairs interlocked, so to speak, making a stable combination. By the next morning I had established the existence of a class of Fuchsian functions, those which come from the hypergeometric series; I had only to write out the results, which took but a few hours.<|endoftext|> TITLE: Lurie's approach to the bar-cobar adjunction QUESTION [20 upvotes]: I've been trying to read Jacob Lurie's approach to the bar-cobar constructions (Higher algebra §5.2 in the 2014-09 version) but I don't recognize what I know about these constructions. I wonder if anybody has already digested this. Before I can ask my question, I need to recall some elements of Lurie's work. In 5.2.1 he considers the twisted arrow category $TwAr({\cal C})$ for ${\cal C}$ an $\infty$-category. Essentially the objects are maps $f:X\to Y\in {\cal C}$ and maps from $f$ to $f':X'\to Y'$ are pairs of maps $(X\to X',Y'\to Y)$ such that the obvious square commutes. $(f:X\to Y) \mapsto (X,Y)$ defines a functor $TwAr({\cal C})\to {\cal C}\times {\cal C}^{op}$. This functor is the right fibration classified by the mapping space functor : $Map:{\cal C}^{op}\times {\cal C}\to {\cal S}$ (where ${\cal S}$ is the $\infty$-category of Kan complexes). Then, Lurie develops the formalism of bimodules and adjunctions representing them. A pairing is a functor $M:{\cal C}\times {\cal D}\to {\cal S}$, we say that it is representable in the first variable if there exists a functor $f:{\cal C}\to {\cal D}$ such that $M(X,Y) = Map(f(X),Y)$. Same thing for the second variable with a functor $g:{\cal D}\to {\cal C}$. If $M$ is representable in both variables, the functor $f$ and $g$ are adjoint. In 5.2.2, ${\cal C}$ is assumed symmetric monoidal and Lurie considers monoids in $TwAr({\cal C})$. Such a monoid is a map $f:A\to C$ from a monoid $A$ to a comonoid $C$ satisfying some unusual condition (*): essentially, the map $\Delta_Cfm_A:A\otimes A\to A\to C\to C\otimes C$ must be equal to $f\otimes f$. Then, essentially because the functor $Map$ is always lax monoidal, we get a pairing $Mon(TwAr({\cal C}))\to Mon({\cal C})\times Mon({\cal C}^{op})$. The main result (Thm 5.2.2.17) is that this pairing is representable in both variables when ${\cal C}$ satisfies some mild assumptions. The corresponding adjunction is the bar-cobar adjunction for monoids and comonoids in ${\cal C}$. It is written $$ Map_{Mon({\cal C})}(A,Cobar(C)) = Map_{coMon({\cal C})}(Bar(A),C). $$ Remark : this result is perfectly symmetric in ${\cal C}$ and ${\cal C}^{op}$ replacing one by the another gives the exact same adjunction. Now, here is what I find strange: classically the bar-cobar adjunction is going the other way (Bar is right adjoint and Cobar left adjoint) so why the change here ? Variation on the same question : the classical the bar-cobar adjunction is related to "twisting cochains" which are Maurer-Cartan elements in the convolution dg-algebra $[C,A]$, hence some kind of maps from the comonoid $C$ to the monoid $A$. But in Higher Algebra the adjunction is related to maps from $A$ to $C$. Is there a way to relate twisting cochains $\alpha:C\to A$ to maps $f:A\to C$ satisfying the condition (*) ? Last remark : a version of the convolution algebra does appear in Lurie's approach, provided we consider comonoids in $TwAr({\cal C})$ instead of monoids. They correspond to maps $f:C\to A$ from a comonoid $C$ to a monoid $A$ which are idempotent for the convolution product in the (external) convolution algebra $Map(C,A)$ (this is the condition dual to (*)). I'm puzzled... I'll be glad if anybody has some insight into this. REPLY [8 votes]: I really like this question, I've been trying to sort out some of these ideas for a little while. I don't know the answer to your questions about conilpotence and twisting morphisms vs twisted arrows. I do have reason to believe that twisted arrows between A and C are the same as the twisted arrows from A to conil(C) but I don't know how to prove that. I think Gabriel's answer is worth expanding on. Lurie is describing an adjunction between infinity categories: Alg and Coalg. The bar and cobar construction you mention are between categories---let me denote them by ALG and COALG [and I mean conilpotent coalg]---and so must be equipped with weak equivalences in order to induce functors on the infinity categories. To model Alg, we equip ALG with quasi-isomorphisms. To model Coalg [or rather, conilCoalg], we must equip COALG with quasi-isomorphisms too. However, the classical bar and cobar construction are not homotopical between these relative (or model) categories. However, we have a second class of weak equivalences on COALG---you called them fancy---that makes this adjunction into a Quillen pair, and as you point out, this Quillen pair is an equivalence. Gabriel's point, though, is that (COALG, fancy) left localizes to (COALG, quasi-iso). Conjugating this localization by the ``bar-cobar as equivalence between (ALG, quasi-iso) and (COALG, fancy)" will show you that this left adjoint from (COALG, fancy) to (COALG, quasi-iso) models Lurie's infinity left adjoint from Alg to Coalg, and it is defined by something that looks like the classical bar construction.<|endoftext|> TITLE: Abelian $\ell$-adic representations in $\widehat{\mathrm{SL}(2,\mathbb{Z})}$ QUESTION [13 upvotes]: $\DeclareMathOperator\SL{SL}\DeclareMathOperator\Gal{Gal}\newcommand{\Z}{\mathbb{Z}}$In Grothendieck's Esquisse he claims that the action of $$\Gal(\mathbb Q)\to\text{Out}(\pi_1(M_{1,1})=\text{Out}(\widehat{\SL(2,\Z)})$$ obtained from the homotopy exact sequence of the étale fundamental group contains, after abelianization, all abelian $\ell$-adic representations defined by Jacobians and generalised Jacobians of algebraic curves defined over number fields. My questions (which might be hopelessly broad and misguided) are the following: How can one show this? Has any work been done extending this idea? A list of references is most welcome. The original text is the following paragraph: ...from the point of view of Galois-Teichmüller theory, $\SL(2,\Z)$ can be considered as the fundamental “building block” of the "Teichmüller tower". The element of the structure of $\SL(2,\Z)$ which fascinates me above all is of course the outer action of $\Gal(\mathbb Q)$ on its profinite compactification. By Bielyi’s theorem, taking the profinite compactifications of subgroups of finite index of $\SL(2,\Z)$, and the induced outer action (up to also passing to an open subgroup of $\Gal(\mathbb Q)$), we essentially find the fundamental groups of all algebraic curves (not necessarily compact) defined over number fields $K$, and the outer action of $\Gal(K)$ on them – at least it is true that every such fundamental group appears as a quotient of one of the first groups (*). Taking the “anabelian yoga” (which remains conjectural) into account, which says that an anabelian algebraic curve over a number field $K$ (finite extension of $\mathbb Q$) is known up to isomorphism when we know its mixed fundamental group (or what comes to the same thing, the outer action of $\Gal(K)$ on its profinite geometric fundamental group), we can thus say that all algebraic curves defined over number fields are “contained” in the profinite compactification $\widehat{\SL(2, \Z)}$, and in the knowledge of a certain subgroup $\Gal(\mathbb Q)$ of its group of outer automorphisms! Passing to the abelianisations of the preceding fundamental groups, we see in particular that all the abelian $\ell$-adic representations dear to Tate and his circle, defined by Jacobians and generalised Jacobians of algebraic curves defined over number fields, are contained in this single action of $\Gal(\mathbb Q)$ on the anabelian profinite group $\widehat{\SL(2, \Z)}$! REPLY [11 votes]: What Grothendieck is doing to get all the Galois representations is: 1) Take a finite index subgroup of $\pi_1(\mathcal M_{1,1})$ that is stable by the $\operatorname{Gal}(\overline{\mathbb Q}|\mathbb Q)$ action. 2) Take the maximal pro-$\ell$ quotient of its abelianization. 3) Look at the action of $\operatorname{Gal}(\overline{\mathbb Q}|\mathbb Q)$ on this $\mathbb Z_\ell$ module. The key points are, in step 1, that every such subgroup defines an algebraic curve over $\mathbb Q$, in step 2, that the maximal pro-$\ell$ quotient of the subgroup is the Tate module of its Jacobian, and in step 3, that the action of $\operatorname{Gal}(\overline{\mathbb Q}|\mathbb Q)$ by conjugation is the same action defined by Tate. This all follows from the basic categorical theory of the \'{e}tale fundamental group as worked out by Grothendieck. The only missing step is that every algebraic curve defined over $\mathbb Q$ arises this way. This is Belyi's theorem. In terms of your second question, here are some things that pop into my head. First, the fact that abelianizations of finite index subgroups of $\pi_1$ are homology of covers has been crucial in all of the progress on Grothendieck's anabelian conjecture for curves (that a hyperbolic curve is determined by its fundamental group) http://www.math.sci.osaka-u.ac.jp/~nakamura/zoo/rhino/NTM300.pdf. In addition, nilpotent quotients of fundamental groups have been connected to homology in works like Deligne's three points on $\mathbb P^1$ paper and the work coming out of that. However, I don't know any work that has been successful in using Grothendieck's observation to prove something deep about the Tate modules of Jacobians of curves. The information we get about them from other sources seems much easier to use.<|endoftext|> TITLE: Higher moments of information and Renyi entropy QUESTION [9 upvotes]: For a given discrete probability distribution, Shannon entropy can be though as an expectation value $\langle - \log p \rangle$ (see also: What is entropy, really?, What is the role of the logarithm in Shannon's entropy? - Stats.SE). Are higher moments of information $\langle (- \log p)^k \rangle$ ever used? (I can imagine some scenarios where we are not interested in the average information, but its fluctuations or 'pessimistic'/'optimistic' scenarios. However, I don't remember seeing it in applied anywhere.) Side note: they do carry the same, ehkm, information as Rényi entropy, as they are related by $$ 2^{z H_{1-z}(X)} = \sum_i p_i^{1-z} = \sum_i p_i 2^{-z \log p_i} = \sum_k \tfrac{z^k}{k!} \sum_i p_i (- \log p_i)^k. $$ EDIT: By "higher" I mean $k>1$. So, for example, is the variance of information used for something? REPLY [5 votes]: In H. Jürgensen, D. E. Matthews, "Entropy and Higher Moments of Information", Journal of Universal Computer Science vol 16, nr. 5 (2010) which is available here, the authors introduce the higher moments of information of memoryless sources (section 3) and general, resp. Markov sources in sections 5 resp. 6. They move from the Shannon entropy formalism to more general entropies in section 7. The main motivation behind the introduction of higher moments is cryptographic security, as stated at pag. 785 loc. cit in the "Final Observations" section.<|endoftext|> TITLE: Balanced binary code that "resists" local decoding? QUESTION [5 upvotes]: I am looking for a construction that can be stated as the following coding problem: a binary code with good distance ($d = \Omega(n)$ where codeword length is $n$) that "resists local decoding" in the sense that, for some $k$, reading $k$ bits is never sufficient to decode (even were the channel noiseless). In other words, for each subset of $k$ locations and each codeword $w$, there exists a codeword $w'$ that matches $w$ at those $k$ locations. To make the question harder, I'm actually most interested in "balanced" binary codes -- every word has Hamming weight $\frac{n}{2}$. But it seems independently interesting to get an answer even without the balanced condition. How large can $k$ be? Can it be $\Omega(n)$? (What is $k$ for the Hadamard code?) Notes: We can restate this as a combinatorics problem (up to a constant factor or two), treating each codeword as a subset of $\{1,\dots,n\}$ indicating the coordinates equal to $1$: Come up with a set of subsets of $\{1,\dots,n\}$, each of size $n/2$, such that pairwise intersection is $O(n)$ (this is the distance requirement) and for each subset $S$ and each $X \subseteq S$ with $|X| \leq k$, there exists an $S'$ such that $X \subseteq S'$. The problem of finding the largest $k$ seems too large for exhaustive search even over very small $n$. Examples: For $n=4$, take all of the weight-$2$ words: $\{1100,1010,1001,0110,0101,0011\}$. Here $d=2$ and $k=1$ (we must read at least two locations to have a hope of uniquely identifying a codeword). For $n=8$, I believe we can take the following weight-$4$ words: $\{11110000, 00001111, 11000011, 10100101, 10010110, 01101001, 01011010, 00111100\}$. Here $d=4$ and $k=2$. REPLY [3 votes]: For each $k$ take $n=2^k$ and the hyperplanes of the $k$-dimensional affine space over $\mathbb{F}_2$. There are $n$ points and $2^{k+1}-2$ hyperplanes; the codewords are the indicator functions of the latter. Then $k-1$ bits never suffice, but $k$ bits always do give you unique codeword.<|endoftext|> TITLE: When does Skolemization require the axiom of choice? QUESTION [7 upvotes]: Skolemization is often used for eliminating existential quantifiers, which is often useful for proving theorems, especially in automated resolution theorem proving. Skolemization in first order predicate calculus is often based on a second order identity: $\forall$x$\exists$y $\phi$(x,y) $\iff$ $\exists$f$\forall$x $\phi$(x, f(x)) I asked on the math.stackexchange site how to perform this operation in higher order logic, with a reasonably satisfactory answer, with one rather large caveat. Apparently, at higher levels of logic, exploiting this identity to use an operation like Skolemization to eliminate existential quantifiers requires the axiom of choice. Is this always the case? This seems like a fairly heavy assumption for using a proof calculus, so are there any restricted forms of higher order Skolemization or existential quantifier elimination that don't require the axiom of choice, or require weaker versions of choice like countable choice or dependent choice? REPLY [6 votes]: Upon rereading the question, I sense there may be a terminological or conceptual confusion in play. The terms “second-order logic” and “higher-order logic” are unfortunately used to denote two vastly different things: Proper second/higher-order logic. This is a semantically defined system; its models consist of a first-order structure to interpret the first-order sort, which is canonically expanded to higher sorts by taking power sets and/or sets of all functions with appropriate domains (depending on the exact language). In this case, validity of the $$\tag{$*$}\forall x\,\exists y\,\phi(x,y) \leftrightarrow \exists f\,\forall x \,\phi(x, f(x))$$ schema in this logic is equivalent to the axiom of choice, as noted in Asaf’s answer. This is not limited to the axiom of choice: a substantial number of other set-theoretic principles (e.g., (non)existence of various large cardinals) are equivalent to validity of particular formulas in second-order logic, so this logic is effectively set theory in disguise. For much the same reasons, it has no proof system (its complexity is much higher than recursively enumerable). Various first-order theories formulated in the multi-sorted language of second/higher-order logic. The fact that the question refers to “proof calculus” strongly suggests that this is the intended meaning here. In this case, the schema $(*)$ is just a convenient axiom that one might or might not choose to adopt in a particular system. The fact that it may be invalid in “standard models” as in 1. if the axiom of choice fails is neither here nor there: the logic is in any case not complete with respect to standard models, and there is no particular need for it to be sound either. Since these theories are generally recursively axiomatized, typical questions about their proof-theoretic properties (e.g., whether including the choice schema $(*)$ significantly strengthens a given system, or whether it is conservative for some class of formulas) are arithmetical statements of low quantifier complexity. As such, they are absolute, and in particular, they do not depend on whether the axiom of choice holds in the metatheory. However: to further terminological confusion, the schema $(*)$ is itself called the axiom of choice in the context of these logics. So, it is not actually clear to me what you are asking about: does the “require the axiom of choice” in the last paragraph of the question refer to the axiom of choice in metatheory, or to the schema $(*)$? Let me also give a possible answer to a possible reading of the question. Systematic application of skolemization makes any formula equivalent to an $\exists\forall$ formula. If you settle for $\exists\forall\exists$ instead, you can do away without any form of choice in either the axiom system or metatheory, using only full comprehension. Instead of introducing higher-order functions to simulate $\forall\exists$ quantifier alternations, introduce higher-order predicates for subformulas of the original formula: for example, $$\forall x_1\,\exists y_1\,\forall x_2\,\exists y_2\,\forall x_3\,\exists y_3\,\phi(x_1,y_1,x_2,y_2,x_3,y_3)$$ is equivalent to $$\begin{align} \exists P_1,P_2\,\Bigl(&\forall x_1,y_1,x_2,y_2,x_3\,(P_2(x_1,y_1,x_2,y_2)\to\exists y_3\,\phi(x_1,y_1,x_2,y_2,x_3,y_3))\\ {}\land{}&\forall x_1,y_1,x_2\,(P_1(x_1,y_1)\to\exists y_2\,P_2(x_1,y_1,x_2,y_2))\\ {}\land{}&\forall x_1\,\exists y_1\,P_1(x_1,y_1)\Bigr). \end{align}$$<|endoftext|> TITLE: "All retracts are closed" as separation axiom QUESTION [6 upvotes]: The starting point of this question is the fact that any retract of a $T_2$-space is closed. Let's say a topological space $(X,\tau)$ is $T_{\textrm{rc}}$ if all retracts of $X$ are closed. All $T_{\textrm{rc}}$-spaces are $T_1$, because singletons are always retracts, and a space is $T_1$ if and only if all singleton subsets are closed. So we have $T_2 \implies T_{\textrm{rc}}\implies T_1$. The second implication is not an equivalence: Consider $(\omega,\mathcal{P}_{\text{cf}}(\omega))$ where $\mathcal{P}_{\text{cf}}(\omega) = \{\emptyset\}\cup \{U\subseteq \omega: \omega\setminus U \text{ is finite}\}$. Let $S$ be the set of even numbers and $r:\omega \to S$ be defined by $2n\mapsto 2n$ and $2n+1 \mapsto 2n$ for all $n\in\omega$. Then $S$ is a retract of $(\omega,\mathcal{P}_{\text{cf}}(\omega))$ but it is not closed. So the implication $T_{\textrm{rc}}\implies T_1$ is not an equivalence. Question: Do we have $T_2 \Leftrightarrow T_{\text{rc}}$? REPLY [7 votes]: No. Let $X$ be a Hausdorff space which fails to be locally compact at precisely one point. Now take the Alexandroff compactification of $X$, adding exactly one new point, whose neighborhoods have compact complement in $X$. The new space is not Hausdorff, but has the property that it is compact, and each compact subspace is closed. In general the image of a compact space is compact, and hence each retract of the new space is a closed subspace of itself.<|endoftext|> TITLE: When is $\vartheta(x)>x$? [Skewes number analog] QUESTION [7 upvotes]: Let $\vartheta(x)=\sum_{p\le x}\log p$. What is known about the first time $\vartheta(x)>x?$ Bays & Hudson give good upper bounds (slightly improved by Chao & Plymen) on the first crossing $\pi(x)>\operatorname{li}(x)$, and Kotnik gives a lower bound, but I don't know what has been proved on the more fundamental (?) question of $\vartheta$. REPLY [11 votes]: Platt and Trudgian show in http://arxiv.org/abs/1407.1914 that $$ \theta(x)x$.<|endoftext|> TITLE: Which criteria for "uniformly splitting" polynomials? QUESTION [6 upvotes]: Let $P(x)$ be an irreducible monic polynomial of degree $\ge4$ with integer coefficients. We all know that over a finite field $\mathbb F_p$, $P$ will often split, and I am interested in polynomials with the property that, for any given $p$, all splitting factors in $\mathbb F_p[x]$ have the same degree. Call such polynomials "uniformly splitting". (This doesn't exclude the possibility that for certain $p$, they may remain irreducible.) E.g. it is not hard to see that $P(x)=x^4-x^2+1$ is uniformly splitting, because it will never split into two linear factors and one quadratic irreducible one. I suspect the situation is similar for $P(x)=x^8-x^6+x^4-x^2+1$. (Is it really?) On the other hand, e.g. $P(x)=x^8-x^6-x^4-x^2+1$ splits uniformly for all $p<43$, so far so good, but in $\mathbb F_{43}$, we have $$P(x)=(9+x) (19+x) (24+x) (34+x) (14+x^2) (40+x^2).$$ What are necessary or sufficient conditions for an irreducible polynomial to be uniformly splitting? It seems to me like such polynomials are very rare, at least for degree $>4$. I conjecture that cyclotomic polynomials $\Phi_n(x^k)$, as long as they are irreducible, are uniformly splitting. Given that the constant term must be $\pm1$, it seems reasonable to expect most if not all uniformly splitting polynomials to be symmetric (in the sense of having self-mirrored coefficients). Then all factors of a (uniform) split for a given $p$ look somewhat like "conjugates" of each other, in a yet-to-define broader way obviously involving the primitive elements of $\mathbb F_{p}$. Might Galois theory be of any help here? And: Are there uniformly splitting polynomials of degree $>4$ which are not of the form $\Phi_n(x^k)$? REPLY [11 votes]: Yes, there are uniformly splitting polynomials of all degrees which are not of the form $\Phi_{n}(x^{k})$. (For example, $f(x) = x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1$.) The Chebotarev density theorem implies that if $f(x)$ is irreducible and $p$ does not divide the discriminant of $f$, then the factorization of $f(x)$ mod $p$ corresponds to a cycle type of an element of the Galois group of $f(x)$ (acting on the roots of $f(x)$). Moreover, every element of the Galois group shows up "equally often" in this way. Modulo the small detail about primes dividing the discriminant of $f$, the question now is about transitive permutation groups $G$ with the property that every element consists of cycles of the same length. In such a permutation group, the only element that can have a fixed point is the identity, and this means the permutation group must be regular and indeed any regular permutation group arises from a group acting on itself by right multiplication. It is easy to see that in such a group, every element is a product of cycles of the same lengths. These are plentiful! In particular, given any field $K/\mathbb{Q}$ which is Galois, there is an element $\alpha \in K$ so that $K = \mathbb{Q}[\alpha]$ and the minimal polynomial of $\alpha$ is a uniformly splitting polynomial.<|endoftext|> TITLE: Diagonalization of the matrix $(1/(i+j+\rm{const}))_{i,j}$ QUESTION [6 upvotes]: Consider the following infinite matrix: $A_{i,j}=\frac1{i+j+\gamma}$, $0\leq i,j<\infty$, $\gamma>0$ is a constant. Is it known how to diagonalize $A$, or, say, calculate $(I+tA)^{-1}$ for parameter $t$? REPLY [10 votes]: See "On the Hilbert matrix II" by Rosenblum<|endoftext|> TITLE: Beginners Guide to Cartan for Beginners QUESTION [6 upvotes]: I am working through parts of Cartan For Beginners by Ivey and Landsberg. Thankfully some exercises have solutions, but, we would benefit from some additional guidance. Question: I am seeking further exposition of the material in Chapter 4 or 5 of the text. I have prolonged some tableaus, but, I have doubts and would like some additional examples. Can anyone point me to some such exposition on explicit PDEs and Tableaus ? Let me sketch the goal of the project as it might give insight into which resources we should seek. We have spent some time studying calculus over an associative real algebra. In short, when a function is $\mathcal{A}$-differentiable it must satisfy the $n^2-n$ generalized Cauchy Riemann equations, but, on the level of the algebra variables it's just calculus. On the other hand, we wonder, when can we take a given PDE or system of PDEs and gain insight into the solutions of the system by studying it in terms of the $\mathcal{A}$-calculus. For a basic example, obviously $\mathbb{C}$-variables gives insight into the solutions of Laplace's equation on $\mathbb{R}^2$. My naive hope was that the machine in Cartan for Beginners could be put to work in ciphering whether a given system of PDEs was compatible with an $\mathcal{A}$-calculus substitution. I hoped there was some natural way to use the Tableaus to judge compatibility. Maybe the hope is misguided. Added: $\mathcal{A} = \mathbb{R}^n$ paired with an associative multiplication over $\mathbb{R}$. To say $f: \mathcal{A} \rightarrow \mathcal{A}$ is $\mathcal{A}$-differentiable is to say $J_f$ is in the left-regular (matrix) representation. This is equivalent to insisting the algebra multiplication factors out of the differential; $df(v \star w) = df \star w$. For example, $\mathcal{A} = \mathbb{R} \oplus j \mathbb{R}$ with $j^2=1$ gives $$ J_f = \left[ \begin{array}{cc} a & b \\ b & a \end{array}\right]$$ I have a list of examples in: http://link.springer.com/chapter/10.1007/978-1-4614-9332-7_8 and to get a better idea of the PDE idea, see: is this a known method for solving PDEs which illustrates the idea. Thanks in advance for your help! REPLY [3 votes]: The following may help: Fueter tried to describe quaternionic analysis in the sense of $\mathcal A$-Analysis in the 20's or 30's, but only first order quaternionic polynomials were possible. The working generalization is hypercomplex Analysis or Clifford Analysis, where one generalizes the Laplacian from complex Analysis to more general algebras. I enclose references to two papers below. (I am not a specialist in this, I just listened to some talks during the years.) MR0822855 (87f:30102) Reviewed Bureš, J.(CS-CHRL); Souček, V. Generalized hypercomplex analysis and its integral formulas. Complex Variables Theory Appl. 5 (1985), no. 1, 53–70. MR3277680 Prelim Brackx, F.; Schepper, H. De; Eelbode, D.; Lávička, R.; Souček, V.; Fundaments of Quaternionic Clifford Analysis I: Quaternionic Structure. Adv. Appl. Clifford Algebr. 24 (2014), no. 4, 955–980.<|endoftext|> TITLE: invariant measures of the expanding maps on the circle QUESTION [5 upvotes]: I would be very happy to know about original references for the following results; For the expanding map $x \mapsto mx$ on the circle, (with $m$ some integer greater than 1) (1) There exist uncountably many ergodic invariant probability measures. (2) Atomic invariant measures are weak star dense in the set of all invariant probability measures. Thank you very much! REPLY [7 votes]: The proof of (2) is contained in the much more general theorem due to Sigmund. This is because the main result of Sigmund "Generic Properties Of Invariant Measures for Axiom A-Diffeomorphisms" Inventiones Math. 11 (1970), pp. 99-109 applies. One may argue that Sigmund considers only toral automorphisms, but his reasoning is based on the fact that the maps he considers have the periodic specification property. The proof that periodic specification sufficies for (2) to hold can be found in Ergodic theory on compact spaces (Volume 527 of Lecture notes in mathematics) by Manfred Denker, Christian Grillenberger, Karl Sigmund (Springer-Verlag, 1976). It is easy to see that maps $x\mapsto mx$ ($m>1$) have the periodic specification property (a theorem of Blokh stating that topological mixing implies specification for maps on graphs can also be invoked, and to a proof that $x\mapsto mx$ have topological mixing is an exercise). In any case, much more is known and Ian Morris above is right - the idea follows Parthasaraty's article. P.S. and (1) follows from another Sigmund paper ON THE CONNECTEDNESS OF ERGODIC SYSTEMS or can be deduced form the general result: if a nontrivial simplex of invariant measures (known to be a Choquet simplex see inverse problem for ergodic measures) has a dense set of ergodic measures (extreme points), then the set of extreme points must be arcwise connected and hence uncountable. This is a consequence of the Lindenstrauss, Olsen and Sternfeld result (The Poulsen simplex, Annales de l'institut Fourier 28.1 (1978): pp. 91-114).<|endoftext|> TITLE: Results about moduli of surfaces QUESTION [5 upvotes]: There are early successes of the moduli theory - the construction and compactification of the moduli spaces of curves $\overline{\mathcal{M}}_g$ . I want to study about the moduli of algebraic surfaces. What are known results about the moduli of surfaces? What kind of surface does have known moduli space? Do you give any reference? REPLY [11 votes]: As already said by Simon in his comment, this is a very vast topic. Let us stick for simplicity to the case of smooth surfaces $S$ of general type: in this case it is well known that $h^0(S, \, T_S)=0$, hence $\textrm{Aut}(S)$ is a finite group and the moduli functor $\textrm{Def}_S$ is prorepresentable. The existence of a quasiprojective coarse moduli space $\mathfrak{M}$ for such surfaces modulo birational equivalence was proven by Gieseker in [G77] using GIT theory and the birationality of $5$-canonical map proven by Bombieri in [B73]. It follows that for fixed values of $\chi(\mathcal{O}_S)$ and $K_S^2$ the space $\mathfrak{M}_{\chi, \, K^2}$ has a finite number of irreducible components. As in the case of $\mathcal{M}_g$, locally in a neighborhood of a point $[S]$ the moduli space $\mathfrak{M}_{\chi, \, K^2}$ is given by a quotient of the base $\textrm{Def}(S)$ of the Kuranishi family of $S$ by the finite group $\textrm{Aut}(S)$. However, in contrast with the case of curves, often patologies arise, for instence $\textrm{Def}(S)$ can be non reduced. This was first explained by Catanese ([C89]) in the case when $K_S$ is not ample, whereas Vakil later proved that such phenomena can be regarded as a particular case of a more general situation that he called the Murphy's law for moduli spaces ([V06]). In recent years, many people studied possible compactifications of $\mathfrak{M}_{\chi, \, K^2}$. For instance, Kollar and Shepherd-Barron proposed in [KSB88] to add surfaces with semi-log canonical singularities and ample $K_S$, that they called stable surfaces by analogy with stable curves. Using this construction, Alexeev and Pardini were for instance able to provide explicit compactifications of the moduli space of Burniat and Campedelli surfaces ([AP09]). Here is a (short) bibliography on the subject: [B73] E. Bombieri: Canonical models for surfaces of general type, Publications Mathématiques de l'IHES 42, Issue 1 (1973), 171-219. [G77] D. Gieseker: Global moduli for surfaces of general tipe, Invent. math 43 (1977), 233-282. [C84] F. Catanese: On the moduli spaces of surfaces of general type, J. Differential geometry 19 (1984), 483-515. [C89] F. Catanese: Everywhere non reduced moduli spaces, Invent. math. 98 (1989), 293-310. [V06] R. Vakil: Murphy's Law in algebraic geometry: Badly-behaved deformation spaces, Invent. Math. 164 (2006), 569-590. [KSB88] J. Kollar and N. I- Shepherd-Barron: Threefolds and deformations of surface singularities, Invent. Math. 91 (1988), 299-338. [AP09] V. Alexeev, R. Pardini: Explicit compactifications of moduli spaces of Campedelli and Burniat surfaces, arXiv:0901.4431.<|endoftext|> TITLE: Real algebraic solution QUESTION [5 upvotes]: Suppose a system of polynomial equations with rational coefficients has a real solution. Does necessarily there exists a real solution with algebraic coordinates? What about the simplest case of one polynomial equation in two variables? REPLY [4 votes]: The statement also follows from Tarski's theorem (1951) that the first-order theory of real closed fields is quantifier-eliminable. Namely, this theorem implies that if a first-order formula holds in some real closed field, then it holds in all real closed fields. One can apply this result to the field of real numbers and the field of real algebraic numbers (both of which are real closed fields). In particular, if a system of polynomial equations with rational coefficients has a real solution, then it also has a real algebraic solution. In fact this proof gives a bit more: the real algebraic solutions are dense among the real solutions (the topology is the usual one coming from the Euclidean metric). I know two more proofs, let me just give the key ideas. The first one is based on the fact (a consequence of the Tarski-Seidenberg principle) that the projection of a semialgebraic set in $\mathbb{R}^n$ to $\mathbb{R}^{n-1}$ (by forgetting the last coordinate) is semialgebraic. The second one is based on Theorem 3.1 in Chapter IX of Lang's algebra.<|endoftext|> TITLE: Joyal's letter to Grothendieck QUESTION [21 upvotes]: Mostly out of curiosity: Where do I find Joyal's letter to Grothendieck in which he defines a model structure on simplicial sheaves? The question was already asked in this MO post, but that particular part of the question has not been answered yet. REPLY [27 votes]: The letter may be found on Georges Maltsiniotis' webpage containing material related to Pursuing Stacks. (A direct link to the pdf.)<|endoftext|> TITLE: The letters of the word "ART" QUESTION [17 upvotes]: Edit: According to the Gelfand duality between topological spaces and commutative $C^{*}$algebras, I add some new tags. So the question is that what is the structure of $ Ext (A,A)$ where $A$ is $C_{0}(\mathbb{R})$. One can think to stabilization of this question, that is $A=C_{0}(\mathbb{R})\otimes \mathcal{K} $ where $\mathcal{K}$ is the algebra of compact operators. The following was the first version of my question: Are there only a finite number of connected topological spaces $X$ (up to homeomorphism) with the property that $X$ has an open subset $U$ such that $U$ and $X-U$ are homeomorphic to $\mathbb{R}$? I know three examples as I wrote in the title of this question. (We delete the end critical points from each letter.) Among capital alphabet, there are no other topological type with the above property. Is it true that any space $X$ with this property can be embedded in $\mathbb{R}^{2}$? REPLY [5 votes]: Yes there are infinitely many, by a version of Mike Jury's idea. In fact there are uncountably many embeddable in $\mathbb R^2$. Take the union of two real curves: the open one $y= x^{-1}\sin(1/x)$ the closed one $x= f(y)$ for $f$ a function that is $0$ on some closed set $S$ and negative elsewhere. The closure of the open curve is exactly $S$, so we can identify $S$ from the topology of the space. Clearly there are uncountably many closed subsets of $\mathbb R$ up to homeomorphism of $\mathbb R$ (e.g. encode a real number as a sequence of closed intervals, isolated points, and Cantor sets.) So there are uncountably many spaces.<|endoftext|> TITLE: Nonperiodic points of homeomorphisms of a ball QUESTION [15 upvotes]: Suppose $B$ is a $d$-dimensional ball (for some $d \geq 1$) and $T$ is a homeomorphism from $B$ to itself. Suppose also that $T$ is not of finite order (that is, for no $n \geq 1$ is it the case that $T^n(x)=x$ for all $x$ in $B$). Can we conclude that a nonperiodic point exists (that is, for some $x$ in $B$ it is the case that $T^n(x) \neq x$ for all $n \geq 1$)? Note that it need not be the case that for each $n \geq 1$ the set of $x$ in $B$ with $T^n(x) \neq x$ is dense. So my question is not just an application of the Baire category theorem (at least under the most simpleminded approach). Note that the answer becomes "No" if we replace the ball $B$ by a general compact set. For instance, consider the action on $\mathbf{R}/\mathbf{Z} \times \{0,1/2,1/3,1/4,1/5,…\}$ that sends ($x$, $y$) to ($x+y$ mod 1, $y$). This post is related to Nonperiodic points of piecewise-linear homeomorphisms . REPLY [13 votes]: Montgomery showed in 1938 that a "pointwise periodic" homeomorphism of a manifold without boundary is actually periodic (this is in Montgomery and Zippin, page 223, or thereabouts), so if your ball is a closed ball, the result you want follows from this by doubling, and if it's the open ball, you don't even need to double.<|endoftext|> TITLE: Complete resolutions of GCH QUESTION [14 upvotes]: Let's say that a "complete resolution of GCH" is a definable class function $F: \operatorname{Ord}\longrightarrow \operatorname{ Ord}$ such that $2^{\aleph_\alpha} = \aleph_{F(\alpha)}$ for all ordinals $\alpha$. It is known of course that $F(\alpha) = \alpha+1$ is a complete resolution of GCH (in the positive) that is relatively consistent with ZFC. I read that it's an unpublished theorem of Woodin that $F(\alpha) = \alpha+2$ is a complete resolution of GCH that is relatively consistent with ZFC plus some large cardinal hypothesis. My questions are: (1) What's the weakest known complete resolution of GCH in consistency strength other than $F(\alpha) = \alpha+1$ and what large cardinal axiom is required for it? (2) What are some other complete resolutions of GCH that are known to be consistent relative to specific large cardinal hypotheses, what are their respective large cardinal hypotheses, and how do these consistency strengths relate to one another? REPLY [8 votes]: Let me add more examples: If we consider the global behavior of the power function, then we have for example: (A) (Foreman-Woodin): $F$ can be such that $F(\alpha)>\alpha+\omega,$ all $\alpha$ (modulo a supercompact and infinitely many inaccessibles above it). Note that by a result of Patai, there is no $\beta>\omega$ such that $F(\alpha)=\alpha+\beta,$ all $\alpha$. Remark. In the above model, $F$ is not definable from the ground model, but we can go to intermediate submodel in which $F$ is definable. (B) (Cummings): $F$ can be such that $F(\alpha)=\alpha+1,$ all successor $\alpha,$ and $F(\alpha)=\alpha+2,$ all limit $\alpha$ (modulo a $\kappa+3$-strong cardinal $\kappa$. By work of Gitik-Mitchell, we need more than a $\kappa+2$-strong cardinal $\kappa$). (C) (Merimovich): Let $2\leq n < \omega.$ Then $F$ can be taken to be $F(\alpha)=\alpha+n,$ all $\alpha$ (modulo a $\kappa+n+1$-strong cardinal $\kappa$. By work of Gitik-Mitchell, we need more than a $\kappa+n$-strong cardinal $\kappa$). (D) (Firedman-G): We can have (B) or (C) just by adding a single real to a model satisfying $GCH$. More precisely, the final model can be of the form $V[R],$ where $V\models GCH$ and $R$ is a real. If we consider the local behavior of the power function, then we can say more: (E) (Gitik-Merimovich): Let $2\leq m <\omega,$ and let $\phi: \omega\to \omega$ be such that $\phi$ is increasing and $\phi(n)>n,$ for all $n$. Then we can have $F(n)=\phi(n)$ and $F(\omega)=\omega+m$ (modulo a $\kappa+m$-strong cardinal $\kappa$). (F) (Gitik): We can have $F$ defined on $\omega_1$ such that both sets $\{ \alpha<\omega_1: F(\alpha)=\alpha+2\}$ and $\{ \alpha<\omega_1: F(\alpha)=\alpha+3\}$ are stationary in $\omega_1$ (modulo suitable large cardinals. Some similar results are also proved by Gitik-Merimovich). If we avoid choice, then an Easton like theorem is valid for all cardinals: Let $\theta(\kappa)=sup\{\nu:$ there exists a surjection $f: p(\kappa)\to \nu \}.$ It is easily seen that $\theta(\kappa)>\kappa^+$ is a cardinal and it is increasing. The next theorem shows that these are the only restrictions that $ZF$ imposes on $θ(κ)$: (G) (Fernengel-Koepke, based on an earlier result of Gitik-Koepke) Let $M$ be a ground model of $ZFC + GCH +$Global Choice. In $M$, let $F$ be a function defined on the class of infinite cardinals such that i. $F(κ)$ is a cardinal > $κ^+$; ii. $κ < λ$ implies $F (κ)\leq F (λ)$. Then there is an extension $N$ of $M$ which satisfies $ZF$, preserves cardinals and cofinalities, and such that $θ (κ) = F (κ)$ holds for all cardinals in $N$.<|endoftext|> TITLE: Strong Morita equivalence and representation theory QUESTION [8 upvotes]: In the context of pure algebra we say that two algebras (in general: rings) $A,B$ are Morita equivalence when there are bimodules $_AP_B,_BQ_A$ such that $P \otimes_B Q \cong _AA_A$ and $Q \otimes_A P \cong _BB_B$. The remarkable theorem states that this condition is equivalent to the following fact: the categories of (for example) left modules over $A$ and over $B$ are naturally equivalent. There is a notion of so called strong Morita equivalence due to Rieffel: this notion is for the $C^*$-algebra context. I saw the following remark: two Morita equivalent $C^*$-algebras have the same representation theory. I wonder what exactly does it mean: I'm interested in $*$-representations. In particular, is it true that $*$-representations of the $C^*$-algebra somehow form the category and the notion of strong Morita equivalence is the natural equivalence of these categories? REPLY [11 votes]: Let $A$ and $B$ be two $C^{*}$ algebras. Then the category of $*$-representation of $A$ on Hilbert space is equivalent to that of $B$ if and only if their enveloping von Neumann algebra are morita equivalent (note that the notion of morita equivalence is slightly more restrictive for Von Neumann algebras because we expect the bi-modules to be self dual as module). In particular, two commutative algebras with the same enveloping von Neuman algebras are going to have the same representations categories but are not Morita equivalent (unless they are isomorphic of course) The good notion of module to have a similar result for $C^*$-algebra are the Hilbert Modules : two $C^*$-algebra are Morita equivalent if and only if they have equivalent categories of Hilbert modules (equivalent as C* categories). Oh, but the converse holds: two Morita equivalent $C^*$ algebras do have the same category of representations (for exemple, because the equivalence bi-module between them can be completed into a self-dual equivalence bi-module between their enveloping algebras) Some references: The first references are I think the two papers of Rieffel: Morita equivalence for C* algebras and W* algebras Morita equivalence for operator algebras In the first paper (and contrary to what the title suggest) he deals with morita equivalence for von Neumann algebras (what he calls morita equivalence of C* algebra in this paper corresponds to the weak notion and hence to morita equivalences of the enveloping W*-algebras) In the second paper he talks about strong morita equivalence of C* algebras. He didn't explicitly proves that it the same as the equivalence of the categories of Hilbert modules but He makes some remarks that essentially explain how it works. Bruce Blackadar's "Operator Algebras" give a nice and modern acount of the basic theory of Morita equivalence (section II.7) but didn't seems to prove that it is equivalent to the equivalence of the categories of C* module neither... Unfortunately, I haven't been able to find an actual complete and explicit proof in the literature of this last results. (I will edit again if I found one)<|endoftext|> TITLE: Is there a Riemann existence theorem for orbifolds? QUESTION [6 upvotes]: For smooth algebraic varieties $X$ over $\mathbb{C}$, the Riemann existence theorem establishes an equivalence of categories between the category of finite etale covers of $X$ and finite unramified covers of $X^\text{an}$. If $X$ is instead a Deligne Mumford stack over Spec $\mathbb{C}$ corresponding to a complex orbifold $X^\text{an}$, is there an analogous theorem providing an equivalence between finite etale morphisms from DM stacks to $X$ and finite orbifold coverings of $X^\text{an}$? (hence etale fundamental group of such a DM stack is just the profinite completion of the orbifold fundamental group?) In particular, I'm thinking of the situation where the stack is the moduli stack of elliptic curves over $\mathbb{C}$ and the orbifold is just $\mathcal{H}/SL_2(\mathbb{Z})$. Can anyone provide a reference to a precise statement? REPLY [4 votes]: The answer is yes. See Theorem 20.4 on p. 80 in Behrang Noohi's http://arxiv.org/pdf/math/0503247v1.pdf<|endoftext|> TITLE: Application of the Riemann hypothesis and the ABC conjecture to independence results QUESTION [8 upvotes]: In Old Home Page of Andreas Weiermann Andreas Weiermann has stated the following: Quite recently I submitted a preprint about an application of the Riemann hypothesis and the ABC conjecture to independence results for publication. Question 1. can someone give a short description of the stated work(s)? Question 2. Do such independence results say anything about the independence of Riemann hypothesis or ABC conjecture in $PA$ or some of its weaker sub-theories? The paper Unprovability, phase transitions and the Riemann zeta-function by Bovykin-Wiermann may be related. Also as Jaso Rute has suggested, the paper Phase transitions in logic and combinatorics might be also helpful. REPLY [14 votes]: Presumably this is my task: Question 1. can someone give a short description of the stated work(s)? The work centers around the "phase transition for Gödel incompleteness" program. The basis idea is to consider an assertion $A(f)$ depending on a function parameter $f$ so that $A(f)$ is true, $A(f)$ is PA-provable if $f$ is slow growing and $A(f)$ is PA-unprovable if $f$ is fast growing. We assume that $A(f)$ is monotone in $f$ with respect to unprovability. The goal is to classify the threshold region where the transition from provability to unprovability happens. The standard examples for $A(f)$ stem from Ramsey theory, wqo-theory and the theory of well-orders. Quite often analytic combinatorics (Tauberians, etc) has to be combined with proof theory to get decent results. The idea in using unproven hypotheses in this business is to sharpen bounds on the threshold region in some meaningful way. RH affects the numbers of primes in short intervals and ABC affects the number of square free numbers in short intervals. This allows in specific contexts to refine the threshold window. The drawback is that the resulting assertions $A(f)$ do not look very natural from the viewpoint of logic (and the referee was not too enthousiastic then). Question 2. Do such independence results say anything about the independence of Riemann hypothesis or ABC conjecture in PA or some of its weaker sub-theories? This research does not say much about the independence of the hypotheses in PA. Studying $A(f)$ could in principle be used to disprove hypotheses used for refining the threshold region. But this would be more sensible for hypotheses which are assumed to be false. The joint work with Andrey Bovykin is of a different nature. It offers a perspective to incorporate results on the value distribution of the $\zeta$-function to prove independence results. The basis idea is to model Ramseyan statements. Andrey has also some interesting ideas to get unprovability of some unproven number-theoretic hypotheses in PA. So he might be asked for further comments. Best, Andreas<|endoftext|> TITLE: What's so special about $1$-categories? QUESTION [5 upvotes]: I have been pretty thoroughly convinced for some time now that, when thinking about mathematics, one really should be thinking 'categorically', that is, one should always be thinking of the morphisms between objects instead of just the objects themselves. You might say that, as I have mathematically matured, my tendency has been to think more at the level of $1$-categories instead of at the level of $0$-categories. Phrasing it like that made me wonder: is this just the first step? Should I 'really' be thinking not about morphisms themselves, but about morphisms between morphisms? Indeed, of the couple of examples I can think of off the top of my head where such $2$-morphisms arise quite naturally, it does seem to be the 'right' thing to do to think about morphisms up to isomorphism (the best example I have in mind is the difference between isomorphism and equivalence of categories, the former not being quite as useful). Of course, once you've decided to think $2$-categorically, why not think $3$-categorically or $4$-categorically, or hell, why not $\infty$-categorically? I apologize for this being a subjective question, but: To what extent should I begin to train myself to think in terms of higher category theory, or, is there indeed something special occurring at the 'normal' level of $1$-categories that makes this level in particular the 'right' way to think about things? REPLY [2 votes]: Taking the homotopical point of view, a 1-category is a space where an imbedding of a more than one dimensional cell is determined by the boundary. On the one hand this makes it easier to calculate and make concrete statements, on the other it is often rather limiting and unnatural. A good example of this is the derived category, which can be seen as the homotopy category of a higher category.<|endoftext|> TITLE: How the modular theory of von Neumann algebras, deal with generating C*-algebras? QUESTION [7 upvotes]: Let $H$ be a separable infinite dimensional Hilbert space, $M \subset B(H)$ a von Neumann algebra and $A \subset M$ a separable ${\rm C}^*$-algebra such that $A''=M$. Suppose the existence of a bicyclic vector $\Omega$ for $M$ (i.e. cyclic and separating: $M\Omega$ and $M' \Omega$ dense in $H$). Let $\sigma_t^{\Omega} \in Aut(M)$ the modular action. Let $K(A,\Omega):= \{ t \in \mathbb{R} \ \vert \ \sigma_t^{\Omega}(A)\subset A \}$ a subgroup of $\mathbb{R}$. Question: Are they $A$ and $\Omega$ (as above) such that $K(A,\Omega) \neq \{0\}$? (If yes) such that $K(A,\Omega) = \mathbb{R}$? Example: Let $M \subset B(H)$ be a von Neumann algebra and $\Omega \in H$ bicyclic. Suppose there is a finite set $S \subset M$ such that $S'' = M$ and let $\mathcal{A}$ be the $\star$-algebra generated by $\sigma_{\mathbb{R}}^{\Omega}(S)$. What is $\mathcal{A}$? A separable ${\rm C}^*$-algebra or $M$ or something else? REPLY [8 votes]: All groups algebra are trivial examples because for them the modular time evolution is (can be chosen) trivial: the modular time evolution attached to a state is trivial if and only if the state is a trace and group algebra always have a trace. The algebras of the various Bost-Connes type system have this property (that $K=\mathbb{R}$) with a non trivial time evolution (see Hecke algebras, type III factors and phase transitions with spontaneous symmetry breaking in number theory for the original one, and Paugam et Ha for the most general version as well as references to other system of this kind). More generally, this situation (having $K=\mathbb{R}$) is equivalent to having a $C^*$-algebra with a one parameter group of automorphism and a KMS $1$ state. The KMS condition (for temperature $1$) is equivalent to the fact that in the GNS representation induced by this state the modular time evolution of the double commutant extend the initial group of automorhism. KMS state at other temperature corresponds to the same situation but with a linear change of parameter in the automorphism groups, so they are also examples. So this is a fairly frequent situation: basically any $C^*$-algebra that have interesting KMS states produces plenty of examples like this... and finding examples of this kind is roughly the same as finding algebra with an automorphism group which have KMS states. Edit : Due to Sebastien Palcoux precision, I'm now answering a different question: If $M$ is a von Neuman algebra with separable predual (equivalently acting on a separable Hilbert space) and $(\sigma_t)_{t\in \mathbb{R}}$ is some weakly continuous one parameter group of automorphism of $M$ then one can always find a weakly dense separabe sub $C^*$-algebra $A$ which is stable under the action of $\sigma_t$. In fact, any separable sub-algebra of $M$ is contained into a separable sub-algebra stable by $\sigma_t$. Indeed, start with $A_0$ some separable sub-$C^*$-algebra of $M$. and $a_1, \dots,a_n,\dots$ some countable dense subfamily. Pick also a countable dense subset $f_0, \dots,f_m,\dots$ of $L^1(\mathbb{R})$. Consider the set of element: $$k_{n,m} = \int_{\mathbb{R}} f_m(t) \sigma_t(a_n) dt$$ I claim that the $C^*$-algebra generated by the $k_{n,m}$ is a seprable sub-$C^*$-algebra of $M$ stable by $\sigma_t$ and containing $A_0$. I sketche the proof: 1) it is separable because the $*$-algebra generated by a countable set is countable, hence its closure (the $C^*$- algebra generated by the coutnable set) is separable. 2) it contains all the element of the form $\int_{\mathbb{R}} g(t) \sigma_t(a_n)$ for $g(t) \in L^1(\mathbb{R})$ simply by taking the limit of the $k_{n,m}$ for $f_n \rightarrow g$ 3) it is stable under $\sigma_t$ : $\sigma_x(k_{n,m})$ will be an integral of the form above (with $f_m$ shifted by $x$). 4) it contains $A_0$ : using the continuity of $t \mapsto \sigma_t(a)$ at $t=0$ and a function $g \in L^1(\mathbb{R})$ with support near $0$ and of integral $1$ one can approximate $a_n$ by some $\int g(t) \sigma_t(a_n) dt$. In particular if $A_0$ was weakly dense, $A$ is also weakly dense.<|endoftext|> TITLE: Prime labelling of graphs QUESTION [7 upvotes]: A prime labeling of a graph is an injective function $f: V(G) \to \{1, 2, ..., |V(G)|\}$ such that for every pair of adjacent vertices $u$ and $v$, $\text{gcd}(f(u), f(v)) = 1$ (labels of any two adjacent vertices are relatively prime). What can prime labeling be useful for? I was thinking of any applications like scheduling etc. but I'm not sure. REPLY [8 votes]: A somewhat fanciful "application." Suppose nodes represent museum guard stations, and arcs represent lines of sight between stations. The labels represent shift hours: hours at one station before a guard is replaced by a fresh guard. Then a prime labeling ensures that when there is a change of guard at one station, there is not simultaneously a change of guard at all the adjacent stations, until the lcm of the labels in the neighborhood is reached. So those guard(s) can monitor the change and ensure coverage. This schedule fails at $t=2 \cdot 3 \cdot 5 = 30$, when no guard covers $B$. Alternatively, the nodes can represent employees and the arcs employees with similar skills, so again when a shift change occurs, the skill set is covered during the switch.<|endoftext|> TITLE: Do geodesics in SL2R map to geodesics in the hyperbolic plane? QUESTION [6 upvotes]: I am looking for a reference/proof/disproof of the following statement. Equip the Lie group $SL_2(\mathbb{R})$ with the left-invariant Riemannian metric, whichis given on the Lie algebra by $\langle A,B\rangle_e :=tr(AB^*)$. Let $pr:SL_2(\mathbb{R})\rightarrow SL_2(\mathbb{R})/SO_2(\mathbb{R})=\mathbb{H}^2$ be the canonical projection. Is is true that $pr\circ \gamma$ is a (constant speed) geodesic in $\mathbb{H}^2$ for any geodesic $\gamma$ in $SL_2(\mathbb{R})$? The speed might even be $0$. Another similar example, where geodesics map to constant speed geodesics would just be the orthogonal projection $\mathbb{R^2}\rightarrow \mathbb{R}\times \{0\}$. REPLY [2 votes]: This is studied in the more general context of the unit tangent bundle in P. T. Nagy's 1977 paper. He shows that the geodesics have constant geodesic curvature in this setting. The most general version is discussed in a later paper @article {MR2027641, AUTHOR = {Berndt, J. and Boeckx, E. and Nagy, P. T. and Vanhecke, L.}, TITLE = {Geodesics on the unit tangent bundle}, JOURNAL = {Proc. Roy. Soc. Edinburgh Sect. A}, FJOURNAL = {Proceedings of the Royal Society of Edinburgh. Section A. Mathematics}, VOLUME = {133}, YEAR = {2003}, NUMBER = {6}, PAGES = {1209--1229}, ISSN = {0308-2105}, CODEN = {PEAMDU}, MRCLASS = {53C22 (53C25 53C35)}, MRNUMBER = {2027641 (2004k:53054)}, MRREVIEWER = {Andrea Spiro}, DOI = {10.1017/S0308210500002882}, URL = {http://dx.doi.org/10.1017/S0308210500002882}, }<|endoftext|> TITLE: Can a positive measure subset of a free group be nowhere dense? QUESTION [10 upvotes]: Let $F$ be a finitely generated free group and let $S \subseteq F$ be a subset for which there is some $\epsilon > 0$ such that for any epimorphism to a finite group $\phi \colon F \to G$ we have that $\frac{|\phi(S)|}{|G|} \geq \epsilon$ (that is, the closure of $S$ in the profinite completion of $F$ has positive Haar measure). Is it possible that the closure of $S$ with respect to the profinite topology on $F$ does not contain a coset of any finite index subgroup of $F$ ? This is the same as asking whether it is possible that $S$ is nowhere dense (the interior of the closure is empty) with respect to the profinite topology. REPLY [2 votes]: Yes, $S$ can be nowhere dense. The idea is to build a 'fat Cantor set' inside $F$. Let $H_0=F,H_1,H_2,\dots$ be finite-index subgroups of $F$ such that $H_{i+1}\leq H_i$ for each $i$ and such that $\bigcap_{i=0}^\infty H_i = \{1\}$. Start with $i_0=0$ and $C_0=\{H_0\}$, and then iteratively do the following. Having defined $C_n$ as a certain set of more than half the cosets of $H_{i_n}$, let $S_n\subset F$ consist of one member from each coset $gH_{i_n}\in C_n$, and then choose $i_{n+1}$ so large that $$\frac{|C_n|}{|F/H_{i_n}|} - \frac{|C_n|}{|F/H_{i_{n+1}}|} > 1/2,$$ and so that when we decompose each $gH_{i_n} \in C_n$ into a union of cosets of $H_{i_{n+1}}$ we can remove at least one of these cosets without uncovering any member of $S_0\cup \dots \cup S_n$. Let $C_{n+1}$ be the resulting set of cosets of $H_{i_{n+1}}$. In the end let $S=\bigcap_{n=0}^\infty \bigcup C_n$, and let $\bar{S}$ be the closure of $S$ in $\hat{F} = \lim F/H_i$. Then $S$ is closed in the profinite topology of $F$ and contains no coset of any $H_i$, so $\bar{S}$ is nowhere dense. On the other hand $S$ contains $\bigcup_{i=0}^\infty S_i$, so the image of $S$ in $F/H_i$ always has size at least $|F/H_i|/2$, so $\bar{S}$ has measure at least $1/2$.<|endoftext|> TITLE: Applications of $p$-adic Hodge theory QUESTION [16 upvotes]: I am trying to learn $p$-adic Hodge theory. I found some materials explaining main theorems (or aspects) of the theory. However, I could not find references which explaining how to use the theory. Especially, I would like to know geometric/arithmetic applications, if there are, of $p$-adic Hodge theory in number theory. Since I did not specify meaning of the words 'geometric/arithmetic', any reference would be appreciated. REPLY [7 votes]: For an example of an application of $p$-adic Hodge theory in a geometric setting, I thoroughly recommend reading the beautiful paper P. Berthelot, H. Esnault, K. Rulling, Rational points over finite fields for regular models of algebraic varieties of Hodge type $\geq 1$, Ann. of Math. (2) $\bf{176}$ 2012, no. 1, 413-508. The result is nice and concrete: they prove congruences for the number of rational points on such varieties. As you will see, the proof makes use of a nice range of big theorems from $p$-adic Hodge theory and is very clearly explained.<|endoftext|> TITLE: Is there a "good" reason that the universal central extension of $SL(2,\mathbb Z)$ is $Br_3$? QUESTION [10 upvotes]: Let $$ Br_3 := \langle \tau_1,\tau_2\ :\ \tau_1 \tau_2 \tau_1 = \tau_2 \tau_1 \tau_2 \rangle $$ be the braid group on three strands, and consider the surjection $$\phi : Br_3 \twoheadrightarrow SL_2(\mathbb Z), \qquad \tau_1 \mapsto \begin{pmatrix} 1&0\\ 1&1\end{pmatrix}, \quad \tau_2 \mapsto \begin{pmatrix} 1&-1\\ 0&1\end{pmatrix}. $$ According to Wikipedia, this is the universal central extension. Moreover, it arises as the fiber product of the inclusion $SL_2(\mathbb Z) \hookrightarrow SL_2(\mathbb R)$ and the universal cover $\widetilde{SL_2(\mathbb R)} \twoheadrightarrow SL_2(\mathbb R)$. Is there a satisfying reason why there should be a map $\phi$? Perhaps something Galois-theoretic, based on $Br_3$ being the fundamental group of the configuration space of $3$ points in the plane? Feel free to retag. If there are many satisfying reasons, I'll make it community wiki! ADDED: This is, if not the same question, at least really close to Details for the action of the braid group B_3 on modular forms , and Dylan Thurston's answer is pretty satisfying (even if he isn't satisfied). REPLY [12 votes]: This answer is essentially equivalent to Misha's and Dylan's, but phrased in terms of configuration spaces rather than mapping class groups. As you note, $Br_3$ is the fundamental group of the configuration space of 3 points in the (complex) plane, a 3-complex dimensional space. One may take a subspace whose center of mass is at the origin, splitting off $\mathbb{C}$. The conformally equivalent configurations are then related by the action of $\mathbb{C}^\times$. If we only quotient by $\mathbb{R}_+$ (say by normalizing the moment of inertia to be $1$), then we get a 3-dimensional space whose fundamental group is still $B_3$. However, if we quotient by $S^1$, then we get a 2-dimensional orbifold which is the conformal classes of the 4-punctured sphere with a single marked point at $\infty$ (or thrice-punctured $\mathbb{C}$). The 2-fold branched cover of a 4-punctured $\mathbb{CP}^1$ ramified over each puncture is a torus with a marked point, and covering translation given by the elliptic involution. So $SL_2(\mathbb{Z})$ acts on this space as the usual quotient of the affine action on $\mathbb{R}^2$ descending to $\mathbb{R}^2/\mathbb{Z}^2$. Note however that the orbifold fundamental group of the conformal classes of 3-punctured $\mathbb{C}$ is actually $PSL_2(\mathbb{Z})$, rather than $SL_2(\mathbb{Z})$, since we've quotiented by the center corresponding to the elliptic involution. However, the universal central extention of $PSL_2(\mathbb{Z})$ by $\mathbb{Z}$ certainly maps to the universal $\mathbb{Z}/2\mathbb{Z}$ central extension of $PSL_2(\mathbb{Z})$. But this point makes the geometric correspondence a bit less natural, since the hyperelliptic involution is the lift of the element of $\pi_1$ of the configuration space which rotates a triple of points by $2\pi$. Maybe another remark, that $PSL_2(\mathbb{Z}) \cong \mathbb{Z}/2\ast \mathbb{Z}/3$, so $H^2(PSL_2(\mathbb{Z}),\mathbb{Z})= \mathbb{Z}/6$. So I suppose there are two universal central extensions, but I guess these are differing by the outer automorphism of $\mathbb{Z}/3$ applied to $\mathbb{Z}/2\ast \mathbb{Z}/3$. If we think of $PSL_2(\mathbb{Z})$ again as the fundamental group of the modular orbifold, then its universal $\mathbb{Z}$-central extension is the fundamental group of the unit tangent bundle to the orbifold. I think the other central extension is also obtained by the action on the lower half-plane, and taking the unit tangent bundle to this orbifold. Here's another geometric description going in the other direction. Recall that a conformal class of torus is obtained from a point $z\in\mathbb{H}^2$ by taking the quotient $\mathbb{C}/\langle 1,z\rangle$. The unit tangent bundle then gives a unit tangent vector at $z$. Taking the quotient by the elliptic involution, this also gives a unit tangent vector at $\infty$ on the 4-punctured $\mathbb{CP}^1$. There is a unique conformal map to $\mathbb{C}$ which has this vector pointing in a fixed direction, and having center of mass at the origin and normalized moment of inertia. However, again this only gives a natural map to $PSL_2(\mathbb{Z})$, not $SL_2(\mathbb{Z})$.<|endoftext|> TITLE: Sierpinski's construction of a non-measurable set QUESTION [18 upvotes]: In the early 20th century there was a lot of fuss over the axiom of choice implying that there are Lebesgue non-measurable sets of reals. In his book about The Axiom of Choice, Gregory Moore points to the following paper: Sierpinski, W. "L’axiome de M. Zermelo et son rôle dans la théorie des ensembles et l’analyse." Bulletin de l’Académie des Sciences de Cracovie, Classe des Sciences Math., Sér. A (1918) (1918): 97-152. (And even more specifically, to pages 124-125.) Moore writes that Sierpinski proved that if $[\Bbb R]^\omega$ has cardinality $2^{\aleph_0}$, then there is a non-measurable set. Lebesgue argued that Sierpinski uses the axiom of choice, but Moore points out that while the assumption requires some amount of choice (as we full well know today), the implication does not. Being a historical book about the axiom of choice, Moore doesn't provide a sketch of the proof. However, I've been unable to locate the paper. Which brings me to my question. Question. Is there any accessible (preferably English) reference to what the argument of Sierpinski was? REPLY [17 votes]: Here's Sierpinski's argument: Let $h:[\mathbb{R}]^{\omega} \to \mathbb{R}$ be any injection. Define $f:\mathbb{R} \to \mathbb{R}$ by $f(x) = h(E_x)$ where $E_x$ is the set of all reals which are at a rational distance from $x$. Note that $x - y$ is rational iff $f(x) = f(y)$. Towards a contradiction, suppose $f$ is Lebesgue measurable. Then $g(x) = f(x) - f(-x)$ is also measurable and is nonzero precisely at irrationals. Let $N = \{x: g(x) > 0\}$. Note that for any rational $r$, $g(r - x) = -g(x)$ so $x \in N \iff r - x \notin N$. But this contradicts Lebesgue density theorem.<|endoftext|> TITLE: Anti-compactness QUESTION [6 upvotes]: Let $(X,\tau)$ be a topological space such that $\tau\ne\{\emptyset\ X\}.\ $ We call an open cover $\mathcal{U}$ of $(X,\tau)$ proper if $\ X\notin \mathcal{U}.\ $ Moreover we say that $(X,\tau)$ is anti-compact if it does not have a finite proper cover; anti-paracompact if for every proper cover $\mathcal{U}$ there is $x\in X$ such that every neighborhood intersects infinitely many members of $\mathcal{U}$; anti-metacompact if for every proper cover $\mathcal{U}$ there is $x\in X$ such that $x$ is a member of infinitely many members of $\mathcal{U}$. We have anti-metacompact $\Rightarrow$ anti-paracompact $\Rightarrow$ anti-compact. Do any of the converse implications hold? REPLY [3 votes]: Let me provide another proof of @EricWofsey's theorem prompted by Dominic's Question. Theorem (Eric Wofsay)   Let $\ X\ $ be an anti-compact space, and let $\mathcal V\ $ be a proper cover. Then every $\ x\in X\ $ belongs to infinitely many members of $\ \mathcal V$. Proof   Let $\ X\ $ be an anti-compact space, and let $\mathcal V\ $ be a proper cover, and let $\ x\in X\ $ be such that family $\ \mathcal K:=\{ G\in\mathcal V: x\in G\}\ $ is finite (a proof by contradiction). Next, let $\ \mathcal M:=\mathcal V\setminus\mathcal K,\ $ and also $\ K:=\bigcup\mathcal K\ $ and $\ M:=\bigcup\mathcal M.\ $ Then $\ K\ne X\ne M.\ $ We have a $2$-element open proper covering $\ \{K\ M\},\ $ which is a contradiction. End of Proof<|endoftext|> TITLE: Distance function to a submanifold QUESTION [7 upvotes]: Let $M$ be a compact Riemannian manifold and $\Sigma\subset M$ a closed submanifold. Given $x\in M$ we define the distance function to $\Sigma$ by $$d_\Sigma(x):=\inf\{d(x,y):y\in \Sigma\},$$ where $d$ is the metric on $M$. Of course, in a small tubular neighborhood of $\Sigma$ the function $d_\Sigma$ will be smooth. Rather, my questions have to do with global properties of $d_\Sigma$. Since $\Sigma$ is a closed subset of $M$, it is not hard to prove, using the triangle inequality, that $d_\Sigma$ is a Lipschitz-continuous function with respect to the metric $d$, with Lipschitz constant $1$. In fact, $d_\Sigma \in W^{1,\infty}(M)$ (see Section 5.8 in Evans' PDE book) and it is differentiable a.e. on $M$ by Rademacher's Theorem. My first question is the following: If $M=\mathbb{R}^n$ then $d_\Sigma$ is a solution to the Eikonal equation, i.e. $\|\nabla d_\Sigma\|=1$ a.e. Is this also true for a general manifold $M$? My second question is related to the behavior of $d_\Sigma$ when we vary the set $\Sigma$. Suppose $\Sigma_t$ are closed submanifolds of $M$ that vary continously in the Hausdorff distance $d_H$, with respect to $t$. Remember that $d_H$ is a metric in the set of compact subsets of $M$. In particular we have the triangle inequality $$d(x,\Sigma_t)\leq d(x,\Sigma_s) + d_H(\Sigma_s,\Sigma_t).$$ This implies that the functions $d(\cdot,\Sigma_t)$ form a continuous curve in $L^\infty(M)$. Is it also true that $d(\cdot,\Sigma_t)$ is a continuous curve in $W^{1,\infty}(M)$? i.e. does it's gradients vary continuously? If not, would it be continuous (perhaps under extra assumptions) in a less regular $L^p$-norm, e.g. $W^{1,2}(M)$? REPLY [5 votes]: I'm posting this sketch just for the sake of completeness. This question was already marked as answered by Anton. But I thought of this, more geometrical, argument after the discussion that answered the question in the first time. Notice that the fact that the $\Sigma$ are assumed to be submanifolds is of no relevance to this questions, the important feature is that they are compact subsets of $M$. Hence I will substitute $\Sigma$ and $\Sigma_n$ by $K$ and $K_n$ in what follows. One important observation is the following: Assertion If $d_K$ is differentiable at a point $x\in M$, then there exists a unique geodesic $\gamma$ starting at $x$ that minimize the distance to $K$. To see this, observe that we can always find at least one such geodesic by compacity. Along one of such geodesics the distance to $K$ decreases linearly with time (otherwise it would contradict the fact that is minimizing), then we can differentiate along it, in particular in $t=0$ to obtain $$\gamma'(0)\cdot \nabla d_K(x) = -1.$$ But $\|\gamma'(0)\|=1$ (because is geodesic) and $\|\nabla d_K(x)\|\leq 1$ (because of the Lipschitz bound). This implies $\gamma'(0)=- \nabla d_K(x)$, therefore $\gamma$ is unique and also $\|\nabla d_K(x)\|= 1$. This answers Question 1: The function is always a solution to the Eikonal equation. Next, suppose we have a sequence of compact sets $K_n$ converging to $K$ in the Hausdorff distance. As Anton pointed out in his answer, since the gradients of the functions $d_{K_n}$ are all bounded, we can obtain convergence in $L^p$, for high $p$, from a weaker form of convergence. In particular, by Lebesgue $L^p$-dominated convergence, it is enough to prove pointwise convergence almost everywhere. In almost every point, all the functions $d_{K_n}$ and $d_K$ are differentiable. By the Assertion above we have a unique minimizing geodesic $\gamma_n$ and $\gamma$ for each one, respectively. If $\gamma_n \nrightarrow \gamma$ we would find another geodesic minimizing the distance to $K$, contradicting the uniqueness. Then the geodesics converge, and therefore they velocities at $x$ too, i.e. the gradients of the functions $d_{K_n}$. This answers Question 2 for every $1\leq p<\infty$. The case $p=\infty$, does not hold, as Anton pointed out, even for distance functions to points in $\mathbb R$.<|endoftext|> TITLE: For any two noncrossing partitions $p, q$ of $n$, is the graph of geodesics from $p$ to $q$ in $NC(n)$ connected? QUESTION [8 upvotes]: Let $NC(n)$ denote the lattice of noncrossing partitions of $n$, and let $G$ denote the Hasse diagram of $NC(n)$ with respect to covering relations, viewed as an undirected graph. I'm interested in the connectivity of graphs of geodesics between vertices of $G$. If $p$ and $q$ are two noncrossing partitions of $n$, let $G_{p,q}$ be the graph whose vertices are minimal-length paths (i.e. geodesics) in $G$ from $p$ to $q$, and where an edge is drawn between two paths $\alpha$ and $\beta$ when $\alpha$ and $\beta$ differ in precisely one element. Is $G_{p,q}$ always connected, or is there an example of $(n,p,q)$ such that it's not? If, for example, $p = id$ and $q = (12\ldots n)$ (as noncrossing permutations), then geodesics from $p$ to $q$ are maximal chains from $p$ to $q$, and in this case $G_{p,q}$ is known to be connected (e.g. Bessis "The dual braid monoid" Prop. 1.6.1, http://arxiv.org/pdf/math/0101158.pdf). But how about if $p$ and $q$ are not comparable in the partial order? Is anything known about the connectivity of $G_{p,q}$? EDIT: A definition and basic discussion of $NC(n)$ can be found at http://en.wikipedia.org/wiki/Noncrossing_partition. EDIT 2: Using a computer program in Mathematica (with help from the package posets.m by Curtis Greene et al, http://www.haverford.edu/math/cgreene/posets.html), I've checked that $G_{p,q}$ is always connected for $n < 7$. REPLY [4 votes]: Third attempt, this time I try to prove that $G_{p,q}$ is connected. Before the proof we need a simple observation about NC partitions. Observation. Draw the elements of $\{1,\ldots,n\}$ on a circle. Any NC partition $p=(p_1,\ldots,p_k)$ of $\{1,\ldots,n\}$ either has a singleton part $|p_i|=1$ or a part that has at least two consecutive elements (such as $1$ and $2$). Proof. There are two very similar cases, we start with the simpler. It will be convenient to imagine that the path goes from $q$ to $p$. Case 1: $p$ has a singleton part $p_i=\{m\}$. Consider any path from $q$ to $p$. Suppose that it does not start with a partition move that separates $m$ from whichever part it belongs to in $q$. At some point, $p_i$ has to be created and thus $m$ separated from a part. But we could execute this separation already in the previous step and do the previous operation after this, as $p_i$ is a singleton and cannot cause any troubles (crossings). Thus, every such path is adjacent to another in $G_{p,q}$ that separates $m$ in an earlier step, and therefore, every path in $G_{p,q}$ is connected to another that starts with separating $m$. But then we can just delete $m$ from both $p$ and $q$ and use induction on $n$. Case 2: $p$ has a part $p_i$ that contains two consecutive elements. Without loss of generality, we can suppose $1,2\in p_i$. If $1$ and $2$ are in the same part of $q$, then we could consider them as one element (as in every shortest path they must stay in the same parts) and we are again done by induction. Otherwise, let $1\in q_1$ and $2\in q_2$. Now we consider the step when the parts containing $1$ and $2$ are unioned together. If this is not the first step, then we could do it one step earlier as this can create no crossings. But, like in the previous case, we can suppose that we start with taking the union of $q_1$ and $q_2$, thus we are done again (either by joining $1$ and $2$ into a single element, or by induction on the length of the shortest $p$-$q$ path).<|endoftext|> TITLE: What is a totient? QUESTION [11 upvotes]: In addition to the Euler totient function, there are a great many generalizations and related functions which go by the "totient", usually with some name: Jordan, Lehmer*, Schemmel, Nagell, Alder, Lucas, Stevens, Eugeni–Rizzi, Holden–Orrison–Vrable, Cohen, Menon, Garcia–Ligh, von Sterneck, etc. Is there some unifying reason that these are all called totient totient functions? I'm familiar with the (mainly historical) use of "totient" to mean "numbers less than or equal to and coprime to" but it's not obvious how that connects to the generalized functions. (Possibly there is no deeper connection than "related to Euler's function".) * Derrick Norman Lehmer, not to be confused with his son Derrick Henry Lehmer. REPLY [12 votes]: What is a totient? Here is one answer that might have the generality you seek: An arithmetical function is said to be a totient if it is the Dirichlet convolution between a completely multiplicative function and the inverse of a completely multiplicative function. For the origin of the name (Sylvester's nomenclature for Euler's counting function of relative primality), see this StackExchange Q&A, with a warning not to interpret the name too literally: as with most mathematical coinages, there is little that is meaningful in Sylvester's metaphor that corresponds to 'relative primality'<|endoftext|> TITLE: Are all well behaved "mean" functions on $\mathbb{R}^+$ equivalent? QUESTION [14 upvotes]: Given a set $S$, a function $M: S\times S \rightarrow S$ is a mean if it satisfies the properties: $M(a,a)=a\qquad$ (identity) $M(a,b)=M(b,a)\qquad$ (commutativity). and possibly $M(M(a,b),M(a,c))=M(a,M(b,c))\qquad$ (weak associativity) $M(M(a,b),M(c,d))=M(M(a,c),M(b,d))\qquad$ (strong associativity) $a\ne b \implies a\ne M(a,b)\ne b\qquad$ (sharpness). When $S$ is an abelian groupoid or an ordered set or a topological space, $M$ can have additional specific requirements, such as: $M(ac,bc)=cM(a,b)\qquad$ (homogeneousness - see UPDATE) $a < b \implies a\le M(a,b) \le b\qquad$ (order preservation - see UPDATE) continuity. This counteraxample is wrong. After fixing my gawk code I found that there (3) always implies (4) in a set of 5. Thanks to Eric Wofsey for noticing. In general (3) does not imply (4) as can be seen in this example for $S=\{a,b,c,d,e\}$: $$ \begin{array}{c|ccccc} M & a & b & c & d & e\\ \hline a & a & a & a & a & a\\ b & a & b & d & c & a\\ c & a & d & c & b & a\\ d & a & c & b & d & e\\ e & a & a & a & e & e\\ \end{array}$$ where $M(M(b,c),M(d,e)) \ne M(M(b,d),M(c,e))$. Here are some of the questions that come to mind. Q1. Is there a finite example where (3) and (5) hold, but not (4)? I know that $S$ will need to have at least 6 elements. Q2. Does $M$ in the above example naturally extend to a mean in $\mathbb{R}[a,b,c,d,e]$ where both (3) and (6) hold? Another example: if $A$ and $G$ are the arithmetic and geometric means on $\mathbb{R}^+$, it's easy to check that the mean function $M(x,y)=G(A(x,y),G(x,y))$ satisfies all the properties except (3) and (4). Q3. Assuming all of the above properties except (4) hold for $M$ on $\mathbb{R}^+$, does (4) follow? Q4. My starting point leading to this post: if all the above properties, including (4), hold for $M$ on $\mathbb{R}^+$, does it follow that $M$ is equivalent to the arithmetic mean, in the sense that $M(x,y)=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ for some continous strictly monotonic function $f: \mathbb{R}^+ \to \mathbb{R}$? I welcome suggestions for improvements to this post and references to relevant work. UPDATE. It turns out that neither (6) nor (7) are needed anywhere, at least for the questions explored in this thread. This is superseded by Eric Wofsey's answer UPDATE. If $M(x,y):=f^{-1}\big(\frac{f(x)+f(y)}{2}\big)$ then notice that we can translate and rescale $f$ and the equality still holds. Now we try to build $f$. To start, we are allowed to assume $f(1/2)=1/2$ and $f(2)=2$. The graph of $f$ can then be constructed in the following way, as per Eric Wofsey's comment below: for each 2 consecutive known points $(x_1, y_1=f(x_1))$ and $(x_2, y_2=f(x_2))$ build an intermediate point $\big(M_f(x_1,x_2), \frac{y_1+y_2}{2}\big)$. By density of the dyadics in $\mathbb{R}$ and properties (5), (7) and (8) of $M$, this procedure defines $f$ in the interval $I=[1/2,2]$. Associativity (hopefully in its weak form) should then be used to prove that the same procedure applied to 2 overlapping subintervals of $I$ yields identical results on the intersection. Finally, if that worked, conclude the proof by defining a second $f$ on $[1/4,4]$. This second $f$ can be translated and rescaled to satisfy $f(1/2)=1/2, f(2)=2$ and must then match the original $f$ in $[1/2,2]$. Repeating this step will extend $f$ to $\mathbb{R}^+$. REPLY [11 votes]: Define a mean algebra to be a set $S$ with an binary operation $M$ satisfying (1), (2), and (4). We can define $M(a,b,c,d)=M(M(a,b),M(c,d))$ and this will depend only on the multiset $\{a,b,c,d\}$. More generally, we can think of $M$ as an operation defined on multisets of size $2^n$ for any $n>0$ (and this is well-defined by an easy induction on $n$ using (2) and (4)). The free mean algebra on two generators can be identified with the dyadic rationals between $0$ and $1$ with the arithmetic mean (the generators being $0$ and $1$); freeness of this algebra is easy to see when you write $k/2^n$ as $M(0,0,\dots,0,1,1,\dots,1)$ (with $k$ $0$s and $2^n-k$ $1$s). Denote this algebra by $Q$. Write $I$ for the algebra $[0,1]$ with the arithmetic mean; this can be thought of as a sort of completion of $Q$. Suppose $A$ is a mean algebra with underlying set $[0,1]$ that further satisfies (5), (7), and (8). There is a unique mean-preserving map $f_0:Q\to A$ satisfying $f_0(0)=0$ and $f_0(1)=1$, and it follows easily from (5) and (7) that it is injective and order-preserving. The inverse of $f_0$ extends uniquely to an order-preserving surjection $g:A\to I$ (every element of $A$ defines a Dedekind cut in $Q$ via $f_0$). By (7) and (8) $g$ will also be mean-preserving, and so (5) implies $g$ is injective. We thus conclude that $A$ is isomorphic to $I$ as an ordered mean algebra, and furthermore this isomorphism is unique. Now suppose $A$ is a mean algebra with underlying set $\mathbb{R}$ satisfying (5), (7), and (8). Every compact subinterval of $A$ these has a unique order-preserving isomorphism to $I$, and by uniqueness we can glue these isomorphisms together to get an order-preserving isomorphism between $A$ and an open subinterval of $\mathbb{R}$ with the arithmetic mean. Note that there are actually 4 distinct isomorphism classes of such structures (with respect to both the order and the mean): $(-\infty,\infty)$, $(-\infty, 0)$, $(0,\infty)$, and $(0,1)$. In particular, this gives an affirmative answer to Q4 (without assuming (6)). Furthermore, the isomorphism in question is clearly unique up to composition with affine maps $\mathbb{R}\to\mathbb{R}$. As for your desire to only assume (3) and not (4), the only place where I used (4) is in asserting that $Q$ is free. Let $F$ be the free algebra on $\{0,1\}$ assuming (3) instead of (4); we wish to prove that the canonical map $F\to Q$ is an isomorphism. Every element of $F$ can be represented as a full binary tree of some height $n$ where each of the $2^n$ leaves is labelled by $0$ or $1$, and at each juncture of the tree we apply $M$. It suffices to show that if two such trees have the same number of leaves that are 1, then they represent equal elements of $F$. We prove this by induction on $n$; the cases $n\leq 1$ are trivial. Let $x\in F$ be represented by such a tree of height $n>1$ that has $i$ leaves that are $1$. WLOG $i\leq 2^{n-1}$ (otherwise we can just swap the roles of $0$ and $1$). Let $y$ be represented by a tree that looks the same as $x$'s, except that all the $0$s are on the left and all the $1$s are on the right. For instance, if $n=i=3$ and $$x=M(M(M(1,0),M(0,1)),M(M(0,1),M(0,0))),$$ then $$y=M(M(M(0,0),M(0,0)),M(M(0,1),M(1,1))).$$ It suffices to prove that $x=y$, since $y$ depends only on $n$ and $i$. Furthermore, by induction, it suffices to prove that $y=M(z,w)$ where $z$ and $w$ each have as many $1$s as the left and right halves of $x$, respectively (since $z$ and $w$ can then be transformed to look the same as the two halves of $x$). Let $j$ be the number of $1$s in the left half of $x$; WLOG $j\leq 2^{n-2}$ (if not, switch the two halves of $x$). By the induction hypothesis, we can write the right half of $y$ as $M(b,c)$, where $b$ has $j$ $1s$. Also, the left half of $y$ is $M(a,a)$, where $a$ is the tree of height $n-2$ consisting entirely of $0$s. We thus have $y=M(a,M(b,c))$, and so by (3) we can rewrite it as $y=M(M(a,b),M(a,c))$. Setting $z=M(a,b)$ and $w=M(a,c)$, the proof is complete. As a final remark, I'm not sure what happens when you additionally assume (6). Certainly most functions $f$ as in your Q4 will not give rise to a mean satisfying (6); the only ones I know of that do are $f(x)=x^p$ for $p\neq0$ and $f(x)=\log x$, up to composition with affine maps. Note that $f(x)=\log x$ (corresponding to the geometric mean) can be thought of as the $p=0$ case; indeed, the mean obtained from it is equal to the limit of the $x^p$ means as $p\to 0$, and these functions $f$ (including their compositions with affine maps) are exactly the solutions of the differential equation $(xf''/f')'=0$. Perhaps if you assume $f$ is sufficiently differentiable you could prove it must be of this form by differentiating the functional equation you get from associativity. EDIT: Here's a proof that if (6) holds, then $f(x)$ must be of the form $ax^p+b$ or $a\log x+b$ and hence $M$ is either $((x^p+y^p)/2)^{1/p}$ or $\sqrt{xy}$. Suppose that (6) holds and $f:\mathbb{R}_+\to U\subseteq\mathbb{R}$ is an isomorphism from $M$ to the arithmetic mean on some open interval $U$. For $c>0$, consider the map $x\mapsto f(cf^{-1}(x))$. This is a mean-preserving automorphism of $U$, so it must be of the form $x\mapsto A(c)x+B(c)$ for some constants $A(c)$ and $B(c)$ depending on $c$. That is, we have $f(cx)=A(c)f(x)+B(c)$. Write $r=A(e)$; then it is easy to show that if $c=e^q$ for $q\in\mathbb{Q}$, we have $A(c)=r^q$. By continuity, it follows that $A(c)=r^{\log c}=c^p$ for all $c$, where $p=\log r$. Suppose that $p=0$, so $f(cx)=f(x)+B(c)$. Then $B(cd)=B(c)+B(d)$ and it follows by continuity that $B(c)=a\log c$ for some $a$. Setting $b=f(1)$, we then get $f(c)=a\log c +b$, as desired. Now suppose that $p\neq 0$. Let $s=B(e)$; then by induction we have $$B(e^n)=s\frac{r^{n+1}-1}{r-1}$$ for $n\in \mathbb{N}$. But by the same argument using $e^{1/m}$ instead of $e$ for some integer $m>0$, we must have $$B(e^n)=B(e^{1/m})\frac{r^{n+1}-1}{r^{1/m}-1}.$$ It follows that the first formula is also valid when $n=1/m$, and it is now easy to show that it is valid for all rational $n$. By continuity, it holds for all real $n$, and we can rewrite it to get $$B(c)=s\frac{e^pc^p-1}{r-1}=tc^p+b$$ for some constants $t$ and $b$. Thus we have $f(cx)=c^pf(x)+tc^p+b$, and setting $x=1$ and $a=f(1)+t$ we get $f(c)=ac^p+b$, as desired.<|endoftext|> TITLE: Survey paper on isoperimetry QUESTION [6 upvotes]: I am searching for a comprehensive survey article (or more different articles) on the subject of isoperimetric problems from ancient Greece to modern mathematical physics. Could you point out some highlights? REPLY [5 votes]: There's been several articles in the comments that are "historical survey" articles. Its not totally clear if you're interested in "current research surveys," but if you are, here are several very nice ones: Osserman's article: http://www.ams.org/mathscinet-getitem?mr=500557 provides an excellent survey of older results. Ros's survey http://www.ugr.es/~aros/isoper.pdf contains a more modern perspective, including several very interesting open problems (for example, for dimension $n\leq8$, the isoperimetric problem in a slab is always solved by half-spheres and cylinders, while for $n\geq 10$, there are other shapes, the "unduloid" which are better for some volumes. It is unknown if this occurs for $n=9$!) Druet has some nice notes http://math.arizona.edu/~dido/presentations/Druet-Carthage.pdf on the isoperimetric problem for Cartan--Hadamard manifolds (a famous conjecture suggests that in a simply connected manifold of non-positive sectional curvature, the boundary area of an isoperimetric region enclosing a fixed volume should be greater than the corresponding question in Euclidean space---this is known in dimension 2,3,4, but no higher!) Eichmair and Metzger have collected an (amazingly short) list of all of the manifolds in which we know what the isoperimetric regions look like in Appendix H of their paper http://www.ams.org/mathscinet-getitem?mr=3127063<|endoftext|> TITLE: Generalization of Sylvester-Gallai theorem QUESTION [8 upvotes]: The Sylvester-Gallai theorem states that it is not possible to arrange a finite number of points so that a line through every two of them passes through a third unless they are all on a single line. Is there any modern research that generalizes this theorem or finds some unexpected relations between this theorem and other parts of mathematics? Could you point out some references (preferably survey papers)? REPLY [2 votes]: Here are links to some recent generalizations of the Gallai-Sylvester theorem. 1) B. Barak, Z. Dvir, A. Wigderson, A. Yehudayoff Fractional Sylvester-Gallai theorems, Proceedings of the National Academy of Sciences of the United States of America 2012. (Link to a journal proceeding.) 2) A. Ai, Z. Dvir, S. Saraf, A. Wigderson Sylvester-Gallai Type Theorems for Approximate Collinearity Forum of Mathematics, Sigma, vol. 2, 2014. See also this blog post on fractional Gallai-Sylvester.<|endoftext|> TITLE: Where is the Erdős–Rado theorem stated in Erdős and Rado's Bull AMS paper? QUESTION [12 upvotes]: This may be inappropriate for MO, but here goes: if I have understood the statement of the Erdős–Rado theorem correctly, then it contains as a special case the following result: if $\mu$ is an infinite cardinal then $(2^\mu)^+ \to (\mu^+)_\mu^2$ that is, every $\mu$-colouring of the $2$-elements subsets of a set $X$ of cardinality $> 2^\mu$ has a monochromatic subset of cardinality $>\mu$. Looking online, the usual citation given for the ER-theorem is P. Erdős, R. Rado, A partition calculus in set theory. Bull. Amer. Math. Soc. 62 (1956), 427–489 I am trying to track down where in this paper one can find the statement of the E-R theorem, or at least the special case described above. Unfortunately, partition calculus lies well outside my usual routes for mathematical excursions, and some of the notation E&R use seems slightly at odds with the notation I see in all the various online notes describing or proving the E-R theorem. I am hoping that the set theorists among the MO readership may be more familiar with the notation of the paper and hence be able to quickly locate the statement. REPLY [13 votes]: This is Theorem 39 in the paper (see Theorem 4.(i) for a user-friendly preview). But the fact that $(2^\kappa)^+\to(\kappa^+)^2_\kappa$ is older (1946) and due to Erdős, see here: Paul Erdős. Some set-theoretical properties of graphs, Univ. Nac. Tucumán. Revista A. 3 (1942), 363-367 MR5,151d. (Anyway, it is probably easier to read a more modern presentation, such as in Paul Erdős, Andras Hajnal, Attila Máté, and Richard Rado. Combinatorial set theory: partition relations for cardinals, volume 106 of Studies in Logic and the Foundations of Mathematics. North-Holland Publishing Co., Amsterdam, 1984.) There are still more modern presentations, using the language of elementary substructures, but if you are not familiar with it, this may not be too accessible. REPLY [9 votes]: It is also stated as Theorem 4.(i), I think, and again on pages 470 and 472 where the reference is given to earlier results. For a recent easy-to-read presentation of the proof, see Theorem 5.1.4 in David Marker's Introduction to Model Theory.<|endoftext|> TITLE: what is the intersection of all congruence subgroups of the profinite completion of SL(2,Z)? QUESTION [9 upvotes]: Let $\widehat{SL(2,\mathbb{Z})}$ be the profinite completion of $SL(2,\mathbb{Z})$. Let $\Gamma(N)$ denote the typical principal congruence subgroup of $SL(2,\mathbb{Z})$ (ie, all matrices congruent to the identity mod $N$). Let $\overline{\Gamma(N)}$ denote its closure in $\widehat{SL(2,\mathbb{Z})}$. Can we describe generators for $\bigcap_{N\ge 1}\overline{\Gamma(N)}$? (At first I thought this intersection is trivial, but since $SL(2,\mathbb{Z})$ has noncongruence subgroups, the congruence subgroups do not form a fundamental system of neighborhoods of the identity in $\widehat{SL(2,\mathbb{Z})}$, so now I'm rather uncertain...) What about $\bigcap_{N\ge 1}\overline{\Gamma_1(N)}$? ($\Gamma_1(N)$ is the subgroup consisting of matrices which mod $N$ are upper triangular unipotent). REPLY [17 votes]: It is a result of Melnikov that the congruence kernel $ker\{ \widehat{SL_2(\mathbb{Z})}\to SL_2(\hat{\mathbb{Z}})\} \cong \hat{F}_\omega$, the free profinite group on a countable number of generators.<|endoftext|> TITLE: Dual or pre-dual of BV QUESTION [8 upvotes]: Was there any relevant work to determine the dual (or more likely the predual) of the space of bounded variation functions $BV(\mathbb{R}^n)$ (I recall the definition : a function in $L^1(\mathbb{R}^n)$ such that the distributional gradient if a vector valued finite measure)? REPLY [4 votes]: Depending on the locally convex topology you take on $BV_c(\mathbb{R}^n)$ (BV functions with essentially compact support in $\mathbb{R}^n$), its dual is given by so called charges (or strong charges) as defined in "De Pauw, Pfeffer - Distributions for which div v = F has a continuous solution - Comm. Pure Appl. Math., 61(2), 2008. 230-260". An electronic version of this paper is here: http://webusers.imj-prg.fr/~thierry.de-pauw/preprints/div.v.equals.F.pdf A charge is a particular kind of distribution and its collection is given the structure of a Fréchet space. Moreover, a distribution in $\mathcal{D}'(\mathbb{R}^n)$ is a charge if and only if it can be realized as the distributional divergence of a continuous vector field, i.e. an element of $C(\mathbb{R}^n,\mathbb{R}^n)$. This result is also stated in Theorem 11.3.8 of the book "The Divergence Theorem and Sets of Finite Perimeter" by Pfeffer (some attention is needed since what is called a charge in this book is called a strong charge in the paper mentioned above). I hope this goes in the direction you are looking for...<|endoftext|> TITLE: Tate's thesis for Artin L-functions QUESTION [9 upvotes]: As far as I know, Tate's thesis has been successfully applied in two fronts: Hecke L-functions, by Tate and Iwasawa (and Teichmüller, Witt, Schmid) Automorphic L-functions, by Jacquet, Shalika, Shapiro etc. Of course the first approach works for everything below Hecke in the L-function food chain (Dedekind, Dirichlet, Riemann). My question is, given that the same ideas work for those extremes, abelian on the one side and... well, supposedly all of them on the other, why hasn't been any progress using harmonic analysis to study Artin L-functions? For example, Tate's thesis for 1-dimensional Artin representations is just... Tate's thesis again (this is, using $W_k$ instead of Galois). This detail is obvious enough, but I haven't seen the details ever worked out in any detailed (I don't think I've seen it mentioned at all). If the analysis goes through for general automorphic objects, I can't imagine the problem being in dealing with noncommutative locally compact groups. So, where is the problem exactly? REPLY [14 votes]: I think there is a misunderstanding on your side. Tate's thesis is not about special Artin $L$-functions, but about special automorphic $L$-functions. It is a reformulation of Hecke's ideas in a uniform adelic language. Of course the automorphic $L$-functions occurring in Tate's thesis (namely those associated to automorphic forms on ${\rm GL}_1$) agree with the $L$-functions of one-dimensional Galois representations, but this is not due to Tate but to class field theory established before his thesis. The short answer to your question "where is the problem exactly" is to show that Artin's $L$-functions are special automorphic $L$-functions. This is part of a bigger program formulated by Langlands.<|endoftext|> TITLE: homotopy fixed points and fixed points QUESTION [6 upvotes]: Let $X$ a smooth projective scheme over a field $k$. And let $THH(X)$ denotes the topological Hochschild homology of $X$. Recall that the spectra $THH(X)$ admits an action of the of circle $S^{1}$. Let $C_{p^n}$ the cycle subgroup of $S^{1}$ with $p^{n}$ elements. Question 1 when the map from homotopy fixed points to fixed points $$ THH^{C_{p}}(X)\rightarrow THH^{hC_{p}}(X)$$ is an equivalence (after $p$-compeltion)? Question 2 what is the interpretation in algebraic geometry of the the groups $$ \pi_{i}THH(X)^{hS^{1}}$$ up to $p$-completion? REPLY [5 votes]: The map (which actually goes from $THH^{C_p}$ to $THH^{hC_p}$) usually not an equivalence in the $p$-complete setting, at least if your input is genuinely a ring. You can detect the difference using a mapping cone; the mapping cone of this map is equivalent to the mapping cone of a map from $THH$ to the so-called Tate construction $THH^{tC_p}$ on it. For a ring $R$, the former is usually concentrated in nonnegative degrees while the latter rarely is (using homological grading). However, there are some useful cases where it is an equivalence in sufficiently high degrees. These include, for example, the cases of $\Bbb Z/p$ and $\Bbb Z_p$, and there is a theorem of Tsalidis that (under some relatively mild extra assumptions) if this holds, then it also holds for the maps $THH^{C_{p^k}} \to THH^{hC_{p^k}}$. (This result played a critical role in Hesselholt-Madsen's calculation of the K-theory of local fields.) If you were working rationally instead of in a $p$-complete setting, the homotopy groups of $THH^{hS^1}(R)$ correspond to "periodic" cyclic homology. In the $p$-complete setting instead, the homotopy groups of $THH^{hS^1}(R)$ are groups that are called $TF(R)$ in the literature. These are, at least in part, connected to an inverse limit of truncated portions of the de Rham-Witt complex under Frobenius maps. (My understanding of this is that, on the algebro-geometric side, there is some connection to Breuil-Kisin modules, but I understand nothing about these and only know this connection exists from watching a talk of Scholze's.)<|endoftext|> TITLE: Ruth-Aaron triples, etc QUESTION [5 upvotes]: A Ruth-Aaron pair is two numbers $(n,n+1)$ such that their sum of prime factors is equal, counting repeated prime factors. (The name refers to Hank Aaron's 715 homeruns surpassing Babe Ruth's 714!) So $$15 = 3 \cdot 5 \;;\; 3+5 = 8$$ $$16 = 2^4 \;;\; 2+2+2+2 = 2 \cdot 4 = 8$$ The notion can be generalized: $$417,162 = 2 \cdot 3 \cdot 251 \cdot 277 \;;\; 2+3+251+277 = 533$$ $$417,163 = 17 \cdot 53 \cdot 463 \;;\; 17+53+463 = 533$$ $$417,164 = 2^2 \cdot 11 \cdot 19 \cdot 499 \;;\; 2+2+11+19+499 = 533$$ My question is: Q. Are there $k$-tuples, for all $k \ge 2$, for which there exists at least one $n$ such that $(n, n+1, n+2, \ldots, n+k)$ has the same sum of prime factors (counting repeated prime factors as above)? Addendum. I was out of date, relying on a "claimed proof by Erdős" that there were an infinite number of R-A pairs, which turned out not to be sound, as Benjamin Dickman explained. So the problem is open even for pairs, let alone $k$-tuples. Apparently no Ruth-Aaron quadruple is known. David Stork wrote Mathematica search code and found there are no such quadruples up to $10^{10}$. REPLY [7 votes]: Observe that a positive answer to your question would imply that there are infinitely many Ruth-Aaron pairs; however, this infinitude already constitutes an open question. Consult: De Koninck, J. M. (2004). Computational Results and Queries in Number Theory. Annales Univ. Sci. Budapest., 23. pp. 149-161. Link (no paywall). Excerpt (p. 152): The paper does go on to provide a heuristic argument for a generalized problem in the same direction as your question, though it is in the context of $\beta$ rather than $S$. Another excerpt (pp. 153-154): (See the aforelinked citation for the argument.)<|endoftext|> TITLE: Are there non-trivial graphs that uniquely embed to hypercubes? QUESTION [10 upvotes]: The $n$-dimensional hypercube is the graph formed by $0$-$1$ sequences of length $n$ where two vertices are adjacent if they differ at only one place. The weight of a sequence is the number of $1$'s in it. For a (connected) subgraph of a hypercube we say that its level-vector is $(n_1,\ldots,n_k)$ if the number of its vertices with minimum weight is $n_1$, with minimum $+1$ weight is $n_2$ etc. In other words, $n_1,\ldots,n_k$ is the number of vertices on the non-empty (adjacent) levels of the hypercube. We say that a graph $G$ level-uniquely embeds to hypercubes if the level-vector of every induced subgraph of any hypercube that is isomorphic to $G$ is the same (and there is at least one hypercube that contains an induced subgraph that is isomorphic to $G$). Is there a graph that level-uniquely embeds to hypercubes but is not a hypercube itself? Update 2014.12.28 This has been answered in the negative by Flo, see his answer below. In fact, for my problem a slightly weaker condition would also suffice. Namely, I am also interested in graphs for which there is an $\alpha\ne -1$ such that for the level-vector $(n_1,\ldots,n_k)$ of every induced subgraph of any hypercube that is isomorphic to the graph it holds that $\alpha$ is the root of the level-polynomial $\sum n_ix^i$. Is there a graph and an $\alpha\ne -1$ such that $\alpha$ is the root of every level-polynomial? My motivation is that if there existed such a graph on a power of $2$ vertices, then I think it would give an example to the question posed here using this method. I've found some nice induced subhypercubes here, but they were too big for me to verify anything. The text below is an exposition of the question by Wlodzimierz, edited by me Let $\ \mathbb B:=\{0\ 1\}.\ $ Let $\ \mathbb B^n\ $ be the $n$-cube. The elements of the $n$-cube are called vertices. The weight $\ \sum x\ $ of a vertex $\ x:=(x_1\ldots x_n)\in\mathbb B^n\ $ is defined as the sum of the coordinates, $\ \sum x := \sum_{k=1}^n x_k.\ $ Given an arbitrary set $\ A\subseteq\mathbb B^n,\ $ define the weigram $\ w_A:\mathbb Z\rightarrow\mathbb Z,\ $ where $\ w_A(k)\ $ is the number of vertices $\ x\in A\ $ of weight $\ k,\ $ for every $\ k=0\ldots n;\ $ also $\ w_A(k):=0\ $ whenever $\ k<0\ $ or $\ k>n.\ $ Consider arbitrary $\ A\ \subseteq \mathbb B^n, \ B\ \subseteq \mathbb B^m. $ We say that $\ B\ $ is a weight-shift of $\ A\ $ $\ \Leftarrow:\Rightarrow\ \exists_{d\in\mathbb Z}\forall_{k\in\mathbb Z}\ w_B(k) = w_A(k+d)$. Now let's introduce the graph structure in the $n$-cube. Its edges are pairs $\ \{a\ b\}\subseteq\mathbb B^n\ $ such that $\ a\ b\ $ have exactly one coordinate different (i.e. $\ \sum |a-b| = 1$). Every $\ A\subseteq\mathbb B^n\ $ is considered to be a graph induced by the $n$-cube graph. Two subsets of the $n$-cube are considered to be isomorphic $\ \Leftarrow:\Rightarrow\ $ they are isomorphic as graphs. QUESTIONS (consider implicitly the versions for connected graphs too, or especially for the connected graphs): Is there set $A\ne \emptyset, \mathbb B^n$, such that every $\ B\subseteq \mathbb B^m\ $ which is isomorphic to $\ A\ $ is a weight-shift of $\ A?\ $ Can you describe all sets $\ A\ $ as above? Questions related to the roots of the polynomials $\ \sum_k w_A(k)\cdot t^k\ $ (see above the part one). REPLY [3 votes]: For your first question, I believe that only hypercubes can embed uniquely. First note that $G$ has to be $n_2$-regular, and $n_1=1$: take any embedding, and use coordinate flips so that your favorite vertex $v$ of $G$ is embedded in $(0,0,0,...,0)$. Then $n_1=1$, and all $n_2$ vertices of weight $1$ are neighbors of $v$. Now let $H$ be the smallest hypercube $G$ embeds uniquely into, say $|H|=2^n$. If $G$ is $n$-regular, then $G=H$. Otherwise, there is some embedding with $v=(0,...,0)$ and $(1,0,...0)\notin V(G)$. Switch the first coordinate on all vertices in $G$, giving a new embedding with minimum weight $1$. As $G$ embeds uniquely, every vertex of $G$ must have gained $1$ in weight, so all vertices had first coordinate $0$. But then $G$ embeds uniquely in the smaller hypercube, contradiction.<|endoftext|> TITLE: What does it mean when we say we have computed a number to a certain accuracy using a probabilistic algorithm? QUESTION [9 upvotes]: My intention is to ask a general question about probabilistic (Monte Carlo) algorithms. But to keep things simple, I will focus on a few specific examples. Let me start the discussion with deterministic algorithms. Suppose I use some iterative algorithm to compute say the $\sqrt{2}$ (for example, using Newton's method) and I claim that if I run $n$ iterations of the algorithm, then the answer I obtain has an error of at most $\epsilon_n$. The precise meaning of this statement is that $$ |x_n - \sqrt{2}| \leq \epsilon_n,$$ where $x_n$ is the answer I get in the $n^{th}$ iteration. Now suppose I have some iterative Monte Carlo algorithm. To take a simple example, suppose I want to compute the value of $\pi$ using the well known Monte Carlo algorithm: consider a square of side length $1$ and at each iteration, generate a pair of random numbers between $0$ and $1$. Check whether the point lies inside the circle quadrant or not. After $n$ iterations, calculate the fraction of points that are inside that circle quadrant. Call this number $\frac{\pi_n}{4}$. $\textbf{Question 1:}$ What is a meaningful question one can ask about $\pi_n$? Saying something like $$ |\pi - \pi_n| \leq \epsilon_n $$ doesn't make any sense, because there is a small chance that $\epsilon_n$ could we very big (even if $n$ is very large). $\textbf{Note:}$ To keep things simple let us assume that the random numbers we produce are "truly random" (whatever that means). $\textbf{Question 2}:$ I now have a more specific question. It seems that using a Monte Carlo algorithm (called the Pivot Method), one can numerically compute the connective constant of a lattice (for this discussion it doesn't matter too much what connective constant means except that its a real number, so I won't bother defining it). Now consider the following statement: The connective constant ($\mu$) of the square lattice upto two decimal places is $2.63$. What does this statement mean? If that $2.63$ was obtained by an ordinary algorithm, then it would have meant $$|\mu-2.63| \leq 0.005. $$ But this $2.63$ was obtained by a Monte Carlo Algorithm. So its not clear to me what is actually meant by saying a number has been computed to some accuracy. A discussion on connective constant (and their known numerical values) is available in the wikipedia page http://en.wikipedia.org/wiki/Connective_constant REPLY [2 votes]: I believe that under certain circumstances, you can do better than just getting a confidence interval. Suppose you decide to estimate $\int_0^1f(x)\,dx$ by $n^{-1}\sum_1^nf(x_i)$ for some random (or pseudorandom) sequence $x_1,x_2,\dots$. If your function $f$ is of bounded variation $V(f)$, then Koksma's Theorem says $$\left|\int f(x)\,dx-{1\over n}\sum_1^nf(x_i)\right|\le V(f)D(x_1,\dots,x_n)$$ where $D$ is the discrepancy of the initial segment of the sequence.<|endoftext|> TITLE: Continuity of central point operation QUESTION [5 upvotes]: Stanisław Mazur and Stanisław Ulam, in their joint paper, characterized the mid-point $\ \frac{a+b}2\ $ in a Banach space in pure metric terms (without algebra). This allowed them to show that any two isometric (not a priori isomorphic) Banach spaces are isometrically isomorphic. Thus in a sense a metric structure may imply an algebraic structure. Furthermore, what is here essential, the operation $\ s(a\ b) := \frac{a+b}2\ $ is continuous. After extracting the Mazur & Ulam construction as a definition of a central space for arbitrary metric spaces (see below), two challenges occur: which metric spaces are central? for which of the central spaces, is the central point operation $\ s\ $ continuous ? Let me focus here on the second question, and this will be THE question in this thread: QUESTION:   Is the central point operation $\ s\ $ continuous for arbitrary compact central space? Now let me provide the definition of the central point operation in an arbitrary metric space $\ (X\ d);\ $ let's do it more generally, not just for a pair $\ (a\ b)\ $ (or for $\ \{a\ b\})\ $ but for an arbitrary non-empty bounded $\ A\subseteq X$: First define $$\ S_1(A)\ \,:=\,\ \left\{x\in X: \forall_{a\in A}\ d(x\ a) \le\frac 12\cdot diam(A)\right\}$$ $$\forall_{n=1\ 2\ \ldots}\ S_{n+1}(A)\ :=\ S_n(A)\cap S_1(S_n(A))$$ Then there exists at the most one point $\ s(A)\in X\ $ such that $\ s(A)\ \in\ \bigcap_{n=1\ 2\ \ldots} S_n(A).\ $ Thus space $\ (X\ d)\ $ is called absolutely central if $\ s(A)\ $ is defined for every non-empty bounded $\ A\subseteq X;\ $ it's called strongly central if $\ s(A)\ $ is defined for every non-empty totally bounded $\ A\subseteq X;\ $ and it is simply called central if $\ s(A)\ $ is defined for every $1\!$- or $2$-element $\ A\subseteq X;\ $ above we are concerned about the last notion (just central). A partial positive answer, when $\ S_1(\{a\ b\})\ $ always consists of a single point, was obvious from the moment the strongly convex spaces were introduced by Karol Borsuk:     Operation $\ s\ $ is continuous for every compact strongly convex space. An application to fixed points:   Let $\ (X\ d)\ $ be an absolutely central space. Let $\ f:X\rightarrow X\ $ be an isometry. If $\ f(A)=A\ $ for a non-empty bounded set $\ A\subseteq X\ $ then $\ f\ $ has a fixed point. A more specific fixed point theorem is a corollary of the above theorem, together with the easier part (2) of the theorem below: THEOREM Every injective (i.e. hyperconvex) metric space is absolutely central. The central point operation $\ s\ $, defined for all non-empty bounded subsets, is continuous. Corollary   Let $\ f:X\rightarrow X\ $ be an isometry of an arbitrary injective metric space such that $\ f(A)=A\ $ for a certain non-empty bounded $\ A\subseteq X.\ $ Then there exists $\ p\in X\ $ such that $\ f(x)=x$. (I'd be willing to provide more material). REPLY [7 votes]: No, it's not: THEOREM (Example)   There exists a compact central metric space for which the central point operation is not continuous. PROOF (Construction)   Consider the spherical distance. For any two points that are not antipodals, $S_1$ is a singleton. For antipodal points, like the North and South poles, you get their equator. Our metric space will be a part of the sphere that contains the North and South poles, but no other antipodal points and contains the shortest path between any two of its points (except the North and South poles). It will be the set of points whose longitude is between, say, 0 and 20 degrees (including the North and South poles). Then for any two points other than the North and South poles $s=S_1$, while for them $s=S_2$ is the point of the equator with 10 degrees longitude. This shows that $s$ is not continuous as we can keep the South pole fixed and converge to the North pole along the 0 longitude line.<|endoftext|> TITLE: Questions about the $\mathbf{i}$-trails of Berenstein and Zelevinsky QUESTION [6 upvotes]: The $\mathbf{i}$-trails of Berenstein and Zelevinsky was introduced on page 5 (Definition 2.1) in this paper. It is defined as follows. Let $\gamma, \delta \in \mathfrak{h}^*$. Let ${\bf i}=(i_1, \ldots, i_l)$. Then $\pi=(\gamma = \gamma_0, \gamma_1, \ldots, \gamma_l=\delta)$ is called an ${\bf i}$-trail if for $k=1,\ldots, l$, $\gamma_{k-1}-\gamma_{k} = c_k\alpha_{i_k}$ for some $c_k \in \mathbb{Z}_{\geq 0}$ and $e_{i_1}^{c_1} \cdots e_{i_l}^{c_l}$ is a non-zero linear map from $V(\delta)$ to $V(\gamma)$. One main result in the paper is a polyhedral formula for tensor product multiplicities (Theorems 2.2 and 2.3 in the above paper). My question is: let $\mathfrak{g}$ be the simple complex Lie algebra of type $G_2$. Let $\lambda=k_1 \omega_1 + l_1 \omega_2$ and $\mu=k_2 \omega_1 + l_2 \omega_2$, where $\omega_1, \omega_2$ are fundamental weights. Are there some polyhedral formula which describe the decomposition of $V(\lambda) \otimes V(\mu)$ into irreducible $\mathfrak{g}$-modules using only $k_1, k_2, l_1, l_2$? The formula will be of the form $$ V(\lambda) \otimes V(\mu) = \oplus_{(r_1, r_2) \in A} V(r_1 \omega_1 + r_2 \omega_2), \quad (1) $$ where $$A = \left\{(r_1, r_2) \in \mathbb{Z}_{\geq 0}^2 \middle\vert \begin{aligned} r_1 &=g_1(k_1,k_2,l_1,l_2),\\ r_2 &=g_1(k_1,k_2,l_1,l_2),\\ 0\ &\le f_1(k_1,k_2,l_1,l_2),\\ &\quad\ldots,\\ 0\ &\le f_m(k_1,k_2,l_1,l_2) \end{aligned} \right\},$$ and where $f_1,\ldots, f_m,g_1,g_2$ are some (linear) polynomials. Maybe use the definition of ${\bf i}$-trail, we can translate Theorem 2.2 in the paper to the form $(1)$. But I don't know how to verify that $e_{i_1}^{c_1} e_{i_2}^{c_2}$ is a non-zero linear map from $V(s_i \omega_i^{\vee})$ to $V(w_0 \omega_i^{\vee})$ (in the case of type $G_2$, $w_0 = s_1s_2s_1s_2s_1s_2$). Are there some references which have some formula similar to $(1)$? Thank you very much. REPLY [7 votes]: There is a bit of a mistake in the question. If a formula like (1) held, then tensor product multiplicities for $G_2$ could only be at most 1. On the other hand, Theorem 2.2 of the paper does give a polyhedral formula of the form: $$ V(\lambda)\otimes V(\mu) = \oplus_{(t_1, t_2, t_3, t_4, t_5, t_6) \in A} V(\lambda + \mu - \sum_{k=1}^6 t_k \beta_k)$$ where $ \beta_1, \dots, \beta_6 $ are the positive roots of $G_2$ (ordered using a reduced word for $ w_0$) and $ A $ is a polyhedron. In Theorem 2.2, $ A$ is described using $\mathbb i$-trails. I agree that this description of $ A $ is a bit confusing. However, there is a simpler description given in Corollary 3.4 of the same paper. $$ A = \{ (t_1, \dots, t_6) : t_k \ge 0, t_1 \le \langle \lambda, \alpha_1^\vee \rangle, p_1 \le \langle \lambda, \alpha_2^\vee \rangle, t_6 \le \langle \mu, \alpha_2^\vee \rangle, p_6 \le \langle \mu, \alpha_1^\vee \rangle \} $$ where $p_1, p_6 $ are computed from $ t_1, \dots, t_6 $ using the tropicalization of the formula in Proposition 7.1.(4). Since $p_1, p_6 $ are tropical expressions in $ t_1, \dots, t_6 $, the above description of $ A $ consists only of inequalities. Unfortunately, the formulas for $ p_1, p_6 $ are pretty complicated, but this is an explicit description of the polyhedron.<|endoftext|> TITLE: Why does a tetracategory with one object, one 1-morphism and one 2-morphism give a symmetric monoidal category QUESTION [8 upvotes]: According to the periodic table of k-tuply monoidal n-categories, it should be the case that a tetracategory (= weak 4-category) with one object, one 1-morphism and one 2-morphism is effectively equivalent to a symmetric monoidal category. I understand geometrically why one should expect this, but I cannot see how it arises directly from an algebraic definition of tetracategory. I do know how we get a braided monoidal category: if $X$ and $Y$ are 3-morphisms, then the braiding $\gamma_{X,Y}$ is defined as the interchanger of $X$ and $Y$. What one then needs to show is that $\gamma_{X,Y} \circ \gamma_{Y,X} = \text{id}$. My understanding is that one would not expect an algebraic definition of tetracategory to carry this equation explicitly as an axiom; it would have to be derived. I am aware that, of course, there does not exist a stable easily-referenced definition of tetracategory. (Edit: John Baez links below to an explicit definition due to Todd Trimble, written up by Alex Hoffnung.) I would be happy if someone could give me an answer along the lines of ''this is the sort of way a tetracategory should be defined, and one would reasonably expect it to have such-and-such algebraic data associated to it, and hence $\gamma_{X,Y} \circ \gamma _{Y,X} = \text{id}$ would follow in such-and-such a way." REPLY [9 votes]: It will come from a compatibility between different ways of composing interchangers. (I'm going to use = to mean iso/homotopy in a HoTT-like way throughout, for ease of notation. I will also confuse proofs and homotopies throughout.) To get a better intuition, let's first think about the case of higher groupoids. As a warmup let's think about the braided case first. So there we have a homotopy 3-type which is 1-connected and we want to understand why there's a braiding on 2-loops. (Note that universally it's enough to just do this for the sphere $S^2$, so secretly we're trying to understand $\pi_3(S^2)$.) But, as you know, this is just the usual Eckmann-Hilton argument where we just braid the maps around each other. Translating this into compositions turns into the interchange law because the first half of the argument is: $$x \circ_1 y = (x \circ_2 1) \circ_1 (1 \circ_2 y) = (x \circ_1 1) \circ_2 (1 \circ_1 y) = x \circ_2 y.$$ Note carefully that there are actually two proofs of Eckmann-Hilton: one where you rotate clockwise and one where you rotate counterclockwise. These proofs are different and they correspond to the two generators of $\pi_3(S^2) = \mathbb{Z}$. It is the fact that these two proofs of Eckmann-Hilton are different that makes these categories braided and not symmetric. Alright, now let's move up one dimension higher. Now we're looking at 2-connected 4-groupoids. Now the algebraic topology question is the well-known fact that the generator of $\pi_3(S^2) = \mathbb{Z}$ has only order 2 when you stabilize it to put it in $\pi_4(S^3)$. Where does this come from? Well, it's clearly the same thing as asking why the two proofs of Eckmann-Hilton become the same when applied to 3-loops. Here the topological intuition is clear: take your clockwise proof and rotate it around the x-axis until it comes back into the plane becoming the counterclockwise proof. Let's think about this in more detail. You can think of Eckmann-Hilton as taking place on a big square made up of four little squares with two tiles labelled x and y and the blanks labelled by identities and then sliding them around. Now we're looking in 3-dimensions at cubes and we want to slide one proof into the other by moving into the third dimension. So we start with the proof: $$x \circ_1 y = (x \circ_2 1) \circ_1 (1 \circ_2 y) = (x \circ_1 1) \circ_2 (1 \circ_1 y) = x \circ_2 y = \ldots = y \circ_1 x.$$ Next we need to introduce a third dimension to give ourselves more room to work. So maybe: $$x \circ_1 y = (x \circ_2 1) \circ_1 (1 \circ_2 y) = (x \circ_1 1) \circ_2 (1 \circ_1 y) = ((x \circ_1 1) \circ_2 (1 \circ_1 y)) \circ_3 ((1 \circ_1 1) \circ_2 (1 \circ_1 1)) = \ldots$$ Then one gradually moves the clockwise proof around into the third dimension (so the interesting part is in the $\circ_2$ $\circ_3$ plane instead of the $\circ_1$ $\circ_2$ plane) and then back into the plane until it becomes the counterclockwise proof. I'm not going to write down the rest of this proof because it would take a while to get it entirely right and it would be hard to fit nicely on the screen anyway. But I'm confident that given a couple days I could write such a proof in Agda, and hopefully the idea is clear. Just like the key lemma in proving Eckmann-Hilton is the interchange law, in order to prove this result we're going to need a higher interchanger relating the three compositions and their pairwise interchangers. Once you have such a higher interchanger the above proof should work for any doubly monoidal 4-category without groupoid-ness appearing anywhere.<|endoftext|> TITLE: Packing disks on a cone, or: Garlands on a tree QUESTION [17 upvotes]: 'Twas the night before Christmas, and throughout the net Not a question was stirring, at least---not yet. The tree was all draped with garlands and lights, With presents beneath for morning delights. As I enjoyed this warm-hearted scene, A question arose as if in a dream: The tree is a cone, the garlands are O's. I wonder how many can fit of those? Has this been studied? Can you shed some light? Merry Christmas to all! And to all a good night!                 Q1. Have packings of congruent disks on (finite) cones been explored? Notation. Let $C$ be a (finite) right circular cone. A generator of $C$ is a segment connnecting the apex $a$ to a point on the rim of the base. Let $\alpha$ be the angle of the cone at $a$ when cut open along a generator; so $\alpha$ is the total surface angle incident to $a$. Let $L$ be the length of a generator. The surface area of $C$ is $A=A(L,\alpha)=\pi L^2 (\alpha/(2 \pi))$. Q2. For a fixed surface area $A$, which cone angle $\alpha$ permits the most unit-radius disks to be packed on $C$? In the example above, $L= 3 \sqrt{2} \approx 4.14$ and $\alpha = \frac{2}{3} \pi = 120^\circ$; so $A=6 \pi$, which is the area of $6$ unit-radius disks, although I only packed $4$ disks (and so achieved a density of $\delta=\frac{2}{3}$). As $\alpha \to 0$, no unit-radius disk can fit, and for $\alpha = 2 \pi$, we are packing disks in a circle of radius $L$, a heavily studied problem. In the absence of an answer to Q2, let me suggest: Q3. Can the behavior of the packing density along the $A(L,\alpha) = \mathrm{const}$ hyperbola-like curves be described qualitatively? E.g., is the density $\delta$ unimodal along those curves, perhaps for sufficiently large $A$? REPLY [3 votes]: Circle packings have been considered for $\alpha=2\pi$, which is the planar case; in that case the densest packing are the incircles of a hexagonal mesh. While the planar case is trivial, two non-planar cases with an exact answer can be derived from it: place the apex at a point, that is the common corner of three of the meshes' hexagons and cut out a sector of either $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$ and glue together the remains along the cuts. For arbitrary $\alpha$ one could take inspiration from nature's way of packing seeds in circular disks, the most prominent example being the arrangement of the sunflower's seeds. In the planar case, the $n$-th circle center would be placed at $\left(x(n),y(n)\right) := \left(c*n*\phi\cos(n*\phi),c*n*\phi*\sin(n*\phi)\right)$, where $\phi:=2\pi\left(1-\frac{\sqrt{5}-1}{2}\right), n\in\mathbb{N}, c\in\mathbb{R}^+$. In the case of a cone, the planar coordinate would have to be augmented by a z-coordinate, that lifts the planar point to the cone. If the density of the circle-centers shall be proportional to surface area, then the radius of the planar spiral must rather be proportional to the square root of the turning angle and not be an archimedean spiral.<|endoftext|> TITLE: What is the analogue of a Lefschetz Thimble for Morse-Bott critical components (sets of non-isolated critical points)? QUESTION [12 upvotes]: Small pre-face: I did an applied math PhD in the UK, but the problem I ended up studying has important ramifications in pure math, specifically to do with the Gauss-Manin connection in the presence of non-isolated critical loci. I worked on this from an applied math/asymptotics point of view and know absolutely nothing about homology or other advanced algebraic topology/geometry topics. I also build up to my question by explaining my work briefly so that replies can be in a mathematical language I understand; my eventual question is at the bottom of this post in the last paragraph. My research is in exponential asymptotic analysis, specifically on deriving hyperasymptotic expansions of multidimensional complex oscillatory integrals in the form \begin{equation} I(k)=\int_S {d\boldsymbol{z} \, g(\boldsymbol{z})e^{-kf(\boldsymbol{z})}} \end{equation} as $|k| \rightarrow \infty$, where $k \in \mathbb{C}$, $\boldsymbol{z}=(z_1,\dotsc,z_d) \in \mathbb{C}^d$, and $S \subset \mathbb{C}^d$ is a smooth unbounded $d$ real-dimensional surface of integration. The functions $f,g: \mathbb{C}^d \rightarrow \mathbb{C}$ are entire on $\mathbb{C}^d$ (this restriction can be relaxed but we keep it simple for now). It is well known that the asymptotic expansion of this type of integral can be expressed as functions of the elements of the critical set $C(f)$ of $f$. When $f$ has only non-degenerate isolated critical points (Morse singularities), rigorous results relying on the homology group of the allowable integration surfaces (chains) $S$ allows us to break up the integration surface into a linear combination of integration surfaces $S_n$ that are similar to $S$ but pass through only one critical point $\boldsymbol{z}_n$ of $f$ (due to Frédéric Pham; see for example Delabaere and Howls (2002) and references therein). The (holomorphic) Morse Lemma guarantees the existence of co-ordinates $\boldsymbol{s}=(s_1,\dotsc,s_d)$ local to $\boldsymbol{z}_n$ such that \begin{equation} f(\boldsymbol{s}) - f(\boldsymbol{z}_n) = s_1^2 + \dotsb + s_d^2, \end{equation} allowing us to realise the steepest descent conditions enforced on $S_n$ locally $\left(\left|f(\boldsymbol{s}) - f(\boldsymbol{z}_n)\right| < a\right)$ as the Lefschetz thimble \begin{equation} S_n^a = \{ \boldsymbol{s} \in \mathbb{C}^d \left| \, \Re(s_1)^2 + \dotsb + \Re(s_d)^2 \leq a, \ \Im(s_j) = 0 \ \forall j \right. \}. \end{equation} We can then locally parameterise $S_n$ as the union of the vanishing cycles \begin{equation} \gamma_n^a = \{ \boldsymbol{s} \in \mathbb{C}^d \left| \, \Re(s_1)^2 + \dotsb + \Re(s_d)^2 = a, \ \Im(s_j) = 0 \ \forall j \right. \}, \end{equation} and follow the flow of the vector field $\nabla(\Re(kf))$ to extend this Lefschetz thimble into the full integration surface $S_n$. This then allows the derivation of the relevant exponential (hyper)asymptotic expansion. If $f$ now has sets $\chi_j$ of non-isolated non-degenerate critical points (namely, non-degenerate critical components) of complex dimension and codimension $\mu_j$ and $q_j$ respectively, then there is currently no such rigorous homological justification that allows us to decompose $S$ into a linear combination of surfaces $S_n$ like before. I'm told this is in part due to a lack of study of the Gauss-Manin connection when non-isolated critical points are involved. I went ahead and assumed it was possible anyway (and all my numerical examples backed this up), then proceeded as before, this time using the (holomorphic) Morse-Bott lemma. This lemma guarantees the existence of coordinates $\boldsymbol{s} = (s_1,\dotsc,s_{q_n})$ such that local to every point within $\chi_n$, \begin{equation} f(\boldsymbol{s}) - f(\chi_n) = s_1^2 + \dotsb + s_{q_n}^2. \end{equation} This allows us to realise the steepest descent conditions enforced on $S_n$ local to $\chi_n$ - namely for $\left|f(\boldsymbol{s}) - f(\chi_n)\right| < a$ - as the higher dimensional analogue of a Lefschetz thimble \begin{equation} S_n^a = \{ \boldsymbol{s} \in \mathbb{C}^{q_n} \left| \, \Re(s_1)^2 + \dotsb + \Re(s_{q_n})^2 \leq a, \ \Im(s_j) = 0 \ \forall j \right. \}, \end{equation} that has as its boundary the union of the higher dimensional analogue of vanishing cycles \begin{equation} \gamma_n^a = \{ \boldsymbol{s} \in \mathbb{C}^{q_n} \left| \, \Re(s_1)^2 + \dotsb + \Re(s_{q_n})^2 = a, \ \Im(s_j) = 0 \ \forall j \right. \}. \end{equation} Again, following the flow of the appropriate vector field gives us a parameterisation of the whole surface $S_n$, leading to the desired asymptotic expansion. My question is this: in the non-isolated case, the Morse-Bott lemma allows us to parameterise $S_n$, but is clearly not a Lefschetz thimble; what is it called instead and what are the analogous vanishing cycles called? I erroneously used the term Lefschetz pencil, but I very quickly deduced that this means something else (I know it's related but I'm not sure what because I don't understand the homology/topology etc.). This picture shows - in 3D - a Lefschetz thimble for an isolated critical point and a set of non-isolated critical points forming a complex critical line. I may be completely overthinking this, but I have no experience with topology at this level and it seems like this type of object hasn't been defined before (or at least defined separately from the objects in the isolated case), so I wanted to find out for sure. REPLY [4 votes]: It seems that there currently exists no standard name for this type of object in this algebraic context. Given that a Lefschetz thimble is basically a "Lefschetz hyper-paraboloid" under the classification of quadratic surfaces (or quadrics), then I thought it made sense to call the surface \begin{equation} S_n^a = \{ \boldsymbol{s} \in \mathbb{C}^{q_n} \left| \, \Re(s_1)^2 + \dotsb + \Re(s_{q_n})^2 \leq a, \ \Im(s_j) = 0 \ \forall j \right. \} \end{equation} a Lefschetz hyper-parabolic cylinder (you could call it a Lefschetz seam but I don't particularly buy in to extending the sewing analogue).<|endoftext|> TITLE: Division algebras over extension fields / reducibility of $G$-modules QUESTION [9 upvotes]: Reformulation of the question (see below for the original question): Let $K$ be an algebraic number field and $D$ a finite-dimensional $K$-division algebra. Is there a description of the field extensions $L\supseteq K$ such that $L\otimes_K D$ is a division algebra? I have encountered this question in the context of representation theory of finite groups, see below. Original question: I have asked this question at math.stackexchange, but have not received an answer so far. Also, I'm not entirely sure that this is a suitable question for mathoverflow... See this link for the original question. Let $K$ be an algebraic number field, $G$ a finite group and $V$ an irreducible $KG$-module with character $\chi$. Since V is irreducible, $\chi$ is of the form $\chi=m(\vartheta_1+...+\vartheta_s)$, where the $\vartheta_i$ are the distinct conjugates of an absolutely irreducible character of G under the Galois group $\mathrm{Gal}(K[\vartheta]/K)$ of the character field $K[\vartheta]:=K[\vartheta(g) ~:~ g \in G]$ and m is the Schur index. We have $s=[K[\vartheta]:K]$. My question is: Is there a characterisation of field extensions $L\supseteq K$ such that $V_L := L\otimes_K V$ is a reducible $LG$-module? Since I asked this question at stackexchange I have found that for $m=1$ (and therefore $s\geq 2$, because otherwise we already have an absolutely irreducible representation) we have that $V_L$ is reducible if and only if $L$ contains an intermediate field $K\subseteq F \subseteq K[\vartheta]$ such that $\mathrm{Gal}(K[\vartheta]/F)\leq \mathrm{Gal}(K[\vartheta]/K)$ has more than one orbit on $\{\vartheta_1,...,\vartheta_s\}$. However, I am unsure about the case of $m\geq 2$, which I assume is more subtle. Is there a similarly succinct characterisation of field extensions $L|K$ such that $V_L$ is reducible? Edit: As Geoff Robinson points out, this problem may be restated as, "Which fields $L\supseteq K$ have the property that $\mathrm{End}_{LG}(V_L)$ is not a division algebra?" which is basically a question about division algebras over number fields. Considering the (small) example of the quaternion group $Q_8$ over the rationals and its irreducible four-dimensional representation $V$, a quick computation shows that $\mathrm{End}_{\mathbb{Q}Q_8}(V)$ becomes a matrix ring over $\mathbb{Q}(\sqrt{-1})$, $\mathbb{Q}(\sqrt{-2})$, $\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Q}(\sqrt{-5})$ but not over $\mathbb{Q}(\sqrt{-7})$. I have, at present, no idea how to characterize these fields, especially in terms of some group theoretic or character theoretic property of $Q_8$. Then again, this is probably not to be exprected, considering that $Q_8$ and $D_8$ have "equal" character tables, but the two-dimensional representations have distinct Schur indices... REPLY [4 votes]: Let $K$ be a number field and $D$ a finite dimensional central division algebra over $K$. I believe that one can characterize the field extensions $L$ of $K$ for which $D\otimes_K L$ is a division algebra. I am not sure how useful this particular characterization will be in the context of the representation theory of finite groups (which seems to be your primary interest), but perhaps if nothing else it will allow you to recast your problem in slightly different terms. Anyway, I believe that the characterization you are after is a more or less straightforward consequence of the Albert-Brauer-Hasse-Noether theorem and follows from the ideas in Chapters 31 and 32 of Reiner's book Maximal Orders. Given a prime $\mathfrak p$ of $K$ denote by $m_\frak{p}$ be the index of $D\otimes_K K_{\mathfrak p}$ (e.g., the degree of the division algebra part of $D\otimes_K K_{\mathfrak p}$). The following is Theorem 32.15 of Reiner. Theorem. Let $L$ be a finite extension of $K$ and suppose that $\dim_K D=d^2$. Then $D\otimes_K L\cong M_d(L)$ if and only if for every prime $\mathfrak p$ of $K$ and prime $\mathfrak P$ of $L$ lying above $\mathfrak p$, we have: \begin{equation} m_{\mathfrak p}\mid [L_\mathfrak P:K_\mathfrak p]. \end{equation} This theorem does not directly answer your question of course, but its proof contains all of the elements needed to do so. So it is worth saying something about how the theorem is proven. The idea is that the local Hasse invariant of $D$ (at a prime $\mathfrak p$ of $K$) is $\mathrm{inv}_{\mathfrak p}(D)=\frac{s_{\mathfrak p}}{m_{\mathfrak p}}$ where $s_{\mathfrak p}$ is an integer coprime to $m_{\mathfrak p}$ satisfying $1\leq s_{\mathfrak p}\leq m_{\mathfrak p}$. The Hasse invariant of $D\otimes_K L$ (at the prime $\mathfrak P$) on the other hand, is $[L_\mathfrak P:K_\mathfrak p]\cdot \frac{s_{\mathfrak p}}{m_{\mathfrak p}}$. The Albert-Brauer-Hasse-Noether theorem says that a central simple algebra is split globally if and only if it is split locally for all primes. Putting these facts together is how one proves the above theorem. It is not too hard to see how all of this must be modified to characterize not when $D$ is split by $L$ but instead when $D\otimes_K L$ is a division algebra. It goes without saying that you will need to ensure that $m_{\mathfrak p}$ does not always divide $[L_\mathfrak P:K_\mathfrak p]$. It is possible for this condition to hold and for $D\otimes_K L$ to still not be a division algebra though. It could be isomorphic to $M_{d'}(D')$ for some division algebra $D'$ over $L$ and integer $d'>1$, for instance. To ensure that $D\otimes_K L$ is a division algebra you will need to make use of the following result (which just cobbles together a bunch of results in Reiner and basically follows from the short exact sequence of Brauer groups that one gets in class field theory): Theorem. Let $S$ be a finite collection of primes of $L$ consisting of finite primes and real infinite places. Suppose that for each $\mathfrak P$ in $S$ we are given a reduced fraction $\frac{a_{\mathfrak P}}{b_{\mathfrak P}}$ such that: 1. $b_{\mathfrak P}>1$ and $a_{\mathfrak P}>0$; 2. $\frac{a_{\mathfrak P}}{b_{\mathfrak P}}=\frac{1}{2}$ whenever $\mathfrak P$ is real; and 3. $\sum_{\mathfrak P \in S} \frac{a_{\mathfrak P}}{b_{\mathfrak P}}\in \mathbb Z.$ Then there exists a unique division algebra $D'$ over $L$ having $S$ as its set of ramified primes, Hasse invariant $\frac{a_{\mathfrak P}}{b_{\mathfrak P}}$ at the prime $\mathfrak P$, and degree $d'$ for $d'=\mathrm{lcm}[b_{\mathfrak P}]$. So to make sure that $D\otimes_K L$ is a division algebra you just need to ensure that the least common multiple of the denominators of its local invariants $[L_\mathfrak P:K_\mathfrak p]\cdot \frac{s_{\mathfrak p}}{m_{\mathfrak p}}$ is equal to the degree of $D$. Finally, note that in certain cases one can recast this characterization in friendlier terms. Suppose for instance that $D$ is a quaternion algebra and that $L$ is a quadratic field extension of $K$. Then $m_{\mathfrak p}$ is equal to $1$ or $2$ and $[L_\mathfrak P:K_\mathfrak p]=1$ precisely when $\mathfrak p$ splits in $L/K$. So in this case the first theorem I stated just says that $D\otimes_K L$ is split if and only if no prime which ramifies in $D$ splits in $L/K$. You can come up with similar (though necessarily more complicated) characterizations in terms of the splitting behavior in $L/K$ of ramified primes in $D$ in the general case.<|endoftext|> TITLE: Properties of a finite random walk QUESTION [6 upvotes]: Consider the simplest random walk - $X_0 = 0$ and from there on (i.i.d), $X_i=X_{i-1}+1$ with probability $p$ or $X_{i-1}-1$ otherwise. Let $Y_N$ be the highest point $X$ have reached on the first $N$ steps and similarly, let $M_N$ be the furthest point, that is: $$Y_N=\max_{i\leq N} X_i\\ M_N=\max_{i\leq N}|X_i|$$ Is there a closed-form formula for the distribution of $Y_N$ and $M_N$? If not, what is $E(Y_N)$ and $E(M_N)$? If the answer in (1) is negative, how fast can we compute $\Pr(Y_N = y)$? REPLY [2 votes]: We have $P(Y_{n} = r) = {n \choose [\frac{n-r}{2}]}2^{-n}$. (for a proof see Theorem 2.4 from RANDOM WALK IN RANDOM AND NON-RANDOM ENVIRONMENTS of Pal Revesz, or Feller Vol I). A simple expression for $E(Y_{n})$ may also be found in the latter reference.<|endoftext|> TITLE: Who first introduced the functional definition of symmetry? QUESTION [10 upvotes]: Who first introduced the definition of symmetry using functions explicitly? (That is, for instance, a symmetry of a subset $X$ of the plane is a function $F$ from the plane to the plane that preserves distance and $F(X) = X$). Thanks! REPLY [8 votes]: From the introduction of Legendre's Revolution (1794): The Definition of Symmetry in Solid Geometry, Giora Hon and Bernard R. Goldstein, Archive for History of Exact Sciences, Vol. 59, No. 2 (January 2005), pp. 107-155. The modern scientific concept of symmetry is, in the first place, a mathematical idea. It refers to an intrinsic property of a mathematical object which makes that object invariant under certain classes of transformations, such as rotation, reflection, inversion, or other operations. These invariant properties are the subject of group theory - a mathematical theory which explores, systématises and formalises these permanent features that are preserved by the object despite the transformation that it undergoes. The pervasiveness of symmetry in modern mathematics and science is well known, and it will be taken for granted here. In antiquity the fundamental meaning of summetrian Greek was proportion which in mathematics also meant commensurability. Clearly, the meaning of symmetry (Latin: symmetrid) changed over time, and it will be the goal of this paper to describe the introduction of this term into solid geometry at the end of the 18th century with an entirely new meaning. We put aside for another occasion the aesthetic sense of summetria/symmetria in antiquity as well-proportioned (notably in Vitruvius's De architecture^ and its subsequent applications in architecture and art (and, to some extent, in scientific contexts). In fact, in the period from 1794 to 1815 three scientists claimed to use symmetry in a new way: Adrien-Marie Legendre (1752-1833) in solid geometry (1794), Sylvestre François Lacroix (1765-1843) in algebra (1797), and René- Just Haüy (1743-1822) in crystallography (1815). In a series of papers now in preparation we will discuss the introduction of symmetry in algebra and in crystallography, and then argue that the modern usages of symmetry in all scientific domains flow from these three seminal figures. ... In this paper we investigate Legendre's work on solid geometry where he introduced a new definition of symmetry that, we claim, has served as the basis for the modern scientific concept of symmetry. This text seems to be an extremely thorough account of the history of the modern concept of symmetry.<|endoftext|> TITLE: Any way to prove Prime Number Theorem using Hyperbolic Geometry? QUESTION [7 upvotes]: The prime number theorem says that the density of prime numbers is inverse as the number of digits of $n$: $$\displaystyle \frac{\{1 \leq k \leq n : \text{ prime } \}}{n} \approx \frac{1}{\log n}$$ Strategies to prove this result usually resort to estimating generating functions of various types. And it seems to be proven over and over again, e.g. in these notes on complex-analytic multiplicative number theory. my issue with the above proofs is they are vary analytic and have estimates that I am not comfortable with. In fact, instead of the prime number theorem let's try a much simpler estimate. Let $\Lambda(x)$ be the Van Mangoldt function. $$ \Lambda(n) = \begin{cases} \log p & \text{ if }n = p^k \\ 0 & \text{otherwise} \end{cases} = -\sum_{d|n} \mu(d) \log(d)$$ Logarithms and fractions naturally come up with discussions in the hyperbolic metric. Is it possible to lift classical proofs into hyperbolic geometry? In particular, I could try the result: $$ \sum_{n \leq x} \Lambda(x) = x + o(x)$$ Does estimates like these of the Van Mangoldt function have a geometric interpretation, e.g. in terms of geodesics? There seems to be already be a literature on the analogy between prime numbers and prime geodesics in $\mathbb{H}$. nLab seems to credit this to Sarnak and Selberg. $$\pi_\Gamma(x) = \# \{ \gamma \in SL(2,\mathbb{Z}): N(\gamma) = e^{\ell(\gamma)} \leq x \} = \int_0^x \frac{dt}{\log t} + \text{ error } $$ In this case, there is no analogue of the Van Mangoldt function. These results involve very difficult spectral arguments and I am worse off than I started. REPLY [4 votes]: It is not at all clear what you have in mind. True, there is an analogy between primes and closed geodesics on $\mathcal H/SL(2,\mathbb Z)$. But the analogy arises a posteriori when one knows they have similar distributions. The geodesics have no particular connection to prime numbers. Instead, lengths of geodesics are logarithms of fundamental units in real quadratic fields. Their multiplicities class numbers of orders. The generating function for the lengths of closed geodesics is the Selberg zeta function. Its zeros are parametrized by eigenvalues of the Laplacian; the corresponding eigenvectors are called Maass wave forms. This correspondence means that the analog of the Riemann Hypothesis for the Selberg zeta function is true. If one could prove the prime number theorem by an approach such as you suggest, one would expect to get optimal bounds on the error term, as good as the Riemann hypothesis would give. This seems unlikely to me.<|endoftext|> TITLE: What is a "Ramanujan Graph"? QUESTION [13 upvotes]: I noticed an apparent conflict in the definition in literature about what is a "Ramanujan graph, which I was wondering if someone could kindly clarify. (1) The Hoory-Linial-Wigderson review on expanders in its definition 5.11 calls a d-regular graph to be Ramanujan if the second highest (adjacency?) eigenvalue is bounded above by $2\sqrt{d-1}$. (2) The Batson-Spielman-Srivastava paper (arxiv:0808.0163v3) on page 2 seems to say that a d-regular graph is Ramanujan if the non-zero Laplacian eigenvalues lie between $[d-2\sqrt{d-1},d + 2\sqrt{d-1}]$. Aren't these two things different? Like the largest eigenvalue of a d-regular Laplacian can I guess be as large as $2d$ even for Ramanujan graphs (and will definitely be so for bipartite d-regular Ramanujan) but how come BSS is asking for a stronger upper bound than that? And why does Hoory-Linial-Wigderson not have any lower bound in their definition? Can the smallest eigenvalue of a d-regular adjacency matrix be arbitrarily low-right? (3) Also the BSS paper in the footnote on page 3 says that the largest (Laplacian?) eigenvalue is at least $d+2\sqrt{d-1}$ and the second-smallest (Laplacian?) eigenvalue is at most $d-2\sqrt{d-1}$. Are the above two bounds independent of what the Alon-Bopanna bound says as in the smallest possible value of the second-highest eigenvalue is $2\sqrt{d-1}$? REPLY [3 votes]: Like Sebi's answer, this should really be a comment but it's too long, and easier to read with proper formatting. The distinction between (1) and (2), once you've translated so that both talk about the same matrix, concerns whether we require one bound on the eigenvalues or two. This is a significant distinction which also caused some confusion for me (with this question), but it really just amounts to whether or not you care about "bipartiteness". For example, the rate of ordinary "forward" diffusion on a graph is governed by the smallest nontrivial eigenvalue of the Laplacian, $\lambda_2$ (the algebraic connectivity), but the rate of "reverse" diffusion is governed by the largest, $\lambda_n$ (the "bipartiteness"). Equivalently, a random walk on a bipartite graph (with $\lambda_n = 2d$) will alternate between the partite sets, and in this sense never truly converges to a uniform distribution on the vertices; what's more, for nonbipartite graphs, the closer $\lambda_n$ gets to $2d$, the more the random walk tends to alternate between two sets, corresponding to an eigenvector of $\lambda_n$. This is related to the expander mixing lemma that Sebi mentioned. The bound you mentioned in (3) is the Alon-Boppana bound, but it is only an asymptotic bound: if you prefer, it should really be $\lambda_2 \le d - 2\sqrt{d-1} + o(1)$. So a Ramanujan graph is one which is "better than any infinite family of graphs could be".<|endoftext|> TITLE: Collection of conjectures and open problems in graph theory QUESTION [21 upvotes]: Is there something similar to the Kourovka Notebook for graph theory (or anyway an organized, possibly commented, collection of conjectures and open problems)? REPLY [2 votes]: A list of about 60 algorithmic problems pertaining to trees are given in the paper by Stephen T. Hedetniemi. Some of the discussion in the paper can be extended to questions on graphs of bounded treewidth.<|endoftext|> TITLE: Mod sequences that seem to become constant; and the number 316 QUESTION [41 upvotes]: Define a "mod sequence" of nonnegative integers based on one start parameter $s$, its first term, as follows. $A(s)=(a_1,a_2,\ldots,a_n,\ldots)$ with $a_1 = s$ and $$ a_n = \left(\sum_{k=1}^{n-1} a_k \right) \bmod n\;.$$ For example, $A(13)=(13, 1, 2, 0, 1, 5, 1, 7, 3, 3, 3, \ldots)$. Here is how this is obtained. Let $S(s)$ be the sums, $s_n=\sum_1^{n} a_k$. Then $S(13)=(13, 14, 16, 16, 17, 22, 23, 30, 33, 36, 39, \ldots)$. In detail, $$a_2 = 13 \bmod 2 = 1$$ $$a_3 = 14 \bmod 3 = 2$$ $$a_4 = 16 \bmod 4 = 0$$ $$a_5 = 16 \bmod 5 = 1$$ $$a_6 = 17 \bmod 6 = 5$$ $$a_7 = 22 \bmod 7 = 1$$ $$a_8 = 23 \bmod 8 = 7$$ $$a_9 = 30 \bmod 9 = 3$$ $$a_{10} = 33 \bmod 10 = 3$$ $$a_{11} = 36 \bmod 11 = 3$$ $$a_{12} = 39 \bmod 12 = 3$$ And all remaining terms are $3$. Q1. For all $s \ge 1$, does $A(s)$ become a constant after some finite $n$, $a_n=c$? Here is a bit more data, showing for each $s$ (first row), the constant reached (second row): $$\left( \begin{array}{ccccccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ 97 & 97 & 1 & 1 & 2 & 2 & 2 & 2 & 316 & 316 & 2 & 2 & 3 \\ \end{array} \right)$$ The constant $316$ seems especially ubiquitous, always (apparently) reached at $a_{1241}=316$. For $s \le 50$, $316$ is reached for $$s=\{9,10,33,34,37,38,39,40,43,44,45,46,47,48,49,50\}\;.$$ For $s=9$, $s_{1241}=392472$ and $(392472 \bmod 1241) = 316$; then $s_{1242}=392788$ and $(392788 \bmod 1242) = 316$; etc. Q2. What is special (if anything) about $316$? Addendum. Here is an image that shows which $s$ map to which $a_n=c$, for all $s \le 100$. The upper-right cluster is $316$. The leftmost cluster is $13$, nearly as ubiquitous as $316$; so perhaps I misled to single out $316$.... REPLY [10 votes]: Before this becomes another forgotten open problem on MO, let me record here a comment. An equivalent way to state the sequence is as follows: Let $x(1)=2s-1$ and look at the recurrence equation: $$x(n+1)=x(n)+x(n)_{\operatorname{mod}n}$$ where I'm using $x_{\operatorname{mod}n}$ to mean the smallest nonnegative integer equal to $x$ mod n. You get essentially the same sequence if you start with $x(1)=2k$. Now the question in the OP is equivalent to: Question: Is this sequence eventually an arithmetic progression? The common difference of this progression is the convergent value of your sequence. This conjecture was mentioned in A117846. Moreover, the OEIS also includes a slightly more general conjecture in A074482: Suppose $x(1,k)=s$ and $$x(n+1,k)=x(n,k)+x(n,k)_{\operatorname{mod} n+k}$$ is this always eventually an arithmetic progression? What can we say about the common differences as a function of $s,k$? The mysterious 316 doesn't appear prominently anymore for different $k$, but the values do tend to concentrate around other seemingly random numbers. The oeis links have a lot of computed values in case anyone wants to hunt for patterns.<|endoftext|> TITLE: Calculus of variation QUESTION [5 upvotes]: This is probably simple but I'm stuck somewhere. I am trying to solve the calculus of variation problem that arise in an applied field: $$\min_{f \in C^1} \int^1_0 \int^1_0 (x-y)^2f(x,y)dxdy$$ $$\text{s.th: } f\geq 0 \text{, } \int^1_0 f(x,y)dy=1 \text{ and } \int^1_0 y \partial_x f(x,y) dy=0 \text{ }\forall\text{ } x.$$ I tried standard techniques - perturbations with test functions - but because of the bidimensionality I'm missing something. REPLY [2 votes]: For the moment, fix $$c=\int_0^1 yf(x,y)\,dy.$$ It is clear that we must have $0\le c\le 1$ to satisfy the constraints on $f$. We can write the integral to be minimized as $$\int_0^1\Bigl(x^2-2cx+\int_0^1 y^2f(x,y)\,dy\Bigr)\,dx.$$ Now we fix $x$ and minimize the inner integral. Since by Cauchy-Schwarz we have $$c^2\le \int_0^1 f(x,y)\,dy\int_0^1 y^2f(x,y)\,dy=\int_0^1 y^2f(x,y),$$with equality only if $f(x,y)=\delta(y-c)$, we find that the minimum is achieved when $f(x,y)=\delta(y-c)$. It remains to find $c$. With $f(x,y)=\delta(y-c)$, the outer integral becomes $\int_0^1 (x-c)^2\,dx$, which is minimized when $c=1/2$.<|endoftext|> TITLE: How close can one get to the missing finite projective planes? QUESTION [38 upvotes]: This question can be interpreted as an instance of the Zarankiewicz problem. Suppose we have an $n\times n$ matrix with entries in $\{0,1\}$ with no $\begin{pmatrix}1 & 1\\ 1& 1\end{pmatrix} $ minor. The problem asks for the maximum possible number of entries equal to $1$. When $n=q^2+q+1$, one may take the incidence matrix of (points vs. lines in) the finite projective plane of order $q$, giving the answer $(q^2+q+1)(q+1)$. Moreover one can prove that this answer is optimal, when a projective plane of the right order exists. Since there is no finite projective plane of order $6$ one may ask What is the maximum possible number of entries equal to $1$ in such a $43\times 43$ matrix? The upper bound of $(6^2+6+1)(6+1)=301$ can not be achieved, but is the answer close to it? The motivation here, is to understand if projective planes are "badly approximable" when they do not exist for a given order. One may speculate that the answer to the question above is given by cutting off from a projective plane of order $7$ a carefully chosen set of $13$ points and lines, but I'm not sure. REPLY [16 votes]: I tried this years ago (mid 1990's) and with much slower computers could never get over 290. Here's one example with 290. 0000000010000000010110000000000010011000000 0000001010010000000000000010000001000001001 1000000001000100000000001000010000000101000 0100000101000000000000010100000000010000001 0010000000000101100000000001000000010000000 1000100110000000101000000000000000100000000 1000011000000001000101010000000000000000000 1001000000000000000010000111000100000000000 0010100000000000000001000100000000001001000 0000000000100000000010010000100000100001000 1010000000001000010000000000101000000000001 0000000001100000100000100010001000001000000 1000000000010010000000100000000000010010000 0000000010100011000000001100000000000000010 0101010010000100000000100000100000000000000 0000000100010001000000000000100100001100000 0000000010000000000000010001001000000100100 0000100000000001000010100000010000000000101 0000000000000000100100000100110001000010000 0000000000100100001100000000000100000000001 0000001000000000000000000000011100110000010 0001000000000000100001000000000010000100011 0001000001000001010000000000000001100000000 0000000000000000001001001010100000010000100 0000010000000000000000001001000000101010001 0000100000000100010000010010000000000010010 0000000011001000000001000000000100000010000 0000101001000010000000000001100010000000000 0100000000000001001000000000001010000011000 0011001100100000000000000000000000000010100 0000000100000110000011000000001001000000000 0100001000001000100010001000000000000000000 0100000000110000010001000001010000000000000 0000000000011100000000000100000010100000100 0000000100001000000100100001000000000001010 0001000000001010001000010000010000001000000 0000110000101000000000000000000001010100000 0110000000000010000100000010000000100100000 0000010000000010110000000000000100000001100 0010000000000000000000111000000111000000000 1100000000000000000000000000000001001000110 0000001000000000011000100100000000000100000 0010010001010000001010000000000000000000010<|endoftext|> TITLE: On the elliptic curve $x(x+a^2)(x+b^2) = y^2$ QUESTION [5 upvotes]: Ajai Choudhry showed that special cases of the elliptic curve, $$x(x+a^2)(x+b^2)=y^2\tag1$$ can be used to prove that, $$u_1^7+u_2^7+\dots + u_9^7 = 0\tag2$$ has an infinite number of primitive integer solutions. Let $a,b$ be positive integers. Define non-torsion points to $(1)$ of forms, $$x_i=\Bigl(\frac{p\,\sqrt{a}}{q}\Bigr)^2,\quad\text{and}\quad x_k =\Bigl(\frac{r\,\sqrt{b}}{s}\Bigr)^2\tag3$$ for positive integer $p,q,r,s$. Questions: If $(1)$ has a solution of form $x_i$, does it also imply it has for $x_k$? If no, what are the conditions on $a,b$ such that $(1)$ has both $x_i,\, x_k$? (Especially for the case $-a+b = 1\;\text{or}\;2$.) For example, the curve with $a,b = 5,7$ has both, $$x_1 = \Bigl(\frac{7\cdot71\sqrt{5}}{361}\Bigr)^2,\quad x_2 = \Bigl(\frac{5\cdot47\sqrt{7}}{337}\Bigr)^2 $$ However, the curve with $a,b = 2^7\mp1 =127,129$ that inspired this post only has known, $$x_2 = \Bigl(\frac{r\sqrt{129}}{s}\Bigr)^2 $$ so I was wondering if it really does not have, or the solutions $p,q$ with $a=127$ to $(3)$ just enormous. REPLY [5 votes]: Yes, if $E : y^{2} = x(x+a^{2})(x+b^{2})$ has a point $P$ whose $x$-coordinate is $au^{2}$ for some nonzero rational number $w$, it also has a point $Q$ whose $x$-coordinate is $bv^{2}$ for some nonzero rational number $v$. In fact, we can take $Q = P + R$, where $R = (ab,a^{2} b + ab^{2})$ is a point of order $4$ on $E$. Silverman and Tate show (in Section III.5 of their book "Rational Points on Elliptic Curves") that the map $\alpha : E(\mathbb{Q}) \to \mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^{2}$ given by $\alpha(x,y) = x$, $\alpha(0,0) = a^{2} b^{2}$ and $\alpha(0:1:0) = 1$ is a homomorphism. Thus, $\alpha(P) = a$ makes it so that $\alpha(P + R) = \alpha(P) \alpha(R) = a(ab) = a^{2} b$, which is equivalent to $b$ in $\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^{2}$. In the case of $a = 127$, $b = 129$, we have a point $P$ with $x$-coordinate $129 \cdot (77684960/987263)^{2}$ and positive $y$-coordinate. The point $Q$ in this case has $x$-coordinate $127 \cdot (224312161/26352417)^{2}$.<|endoftext|> TITLE: A version of Wald identity QUESTION [6 upvotes]: Let $W$ be a standard one-dimensional Brownian motion. Let $T$ be a stopping time with $\mathbb{E}\sqrt{T}<+\infty$. Then $$\mathbb{E}W_T=0\quad \mathbb{E}W^2_T=\mathbb{E}T$$ I can prove these identies under the condition $\mathbb{E}T<+\infty$. But what about this one? REPLY [3 votes]: Well, I provide a proof using Burkholder-Davis-Gundy(BDG) inequality.(http://en.wikipedia.org/wiki/Quadratic_variation#Martingales) Let $M=\left\{M_t=W_{t\wedge T};0\le t<+\infty\right\}$, then $M$ is a martingale with $\langle M\rangle_t=t\wedge T$, so by BDG inequality, we have there exists a constant $K>0$ such that $$\mathbb{E}M_T^*\le K\mathbb{E}\sqrt{T}<+\infty$$ where $M_t^*=\sup_{0\le s\le t}|M_s|$. Since for all $t\ge0$ $$|M_t|\le\sup_{0\le s\le t}|M_s|=\sup_{0\le s\le t}|W_{s\wedge T}|\le\sup_{0\le s\le T}|W_{s\wedge T}|=\sup_{0\le s\le T}|M_s|=M_T^*$$ Hence $M$ is uniformly integrable, $\mathbb{E}W_T=\mathbb{E}M_T=0$. Hence $\left\{M_t;0\le t\le+\infty\right\}$ is a martingale with last element $M_T=W_T$, $\left\{M_t^2;0\le t\le+\infty\right\}$ is a submartingale. If $\mathbb{E}T<+\infty$, then we already know that $\mathbb{E}W_T^2=\mathbb{E}T$. If $\mathbb{E}W_T^2<+\infty$, then for all $t>0$, since $M_t = \mathbb{E}[W_T \mid \mathcal{F}_t]$, the conditional Jensen inequality yields $$\mathbb{E}M_t^2\le \mathbb{E}W_T^2<+\infty$$ hence $M$ is $L^2$ bounded, hence $L^2$ convergent. By the monotone convergence theorem, we have $$\mathbb{E}T=\mathbb{E}\lim_{t\to+\infty}(t\wedge T)=\lim_{t\to+\infty}\mathbb{E}(t\wedge T)=\lim_{t\to+\infty}\mathbb{E}W_{t\wedge T}^2=\lim_{t\to+\infty}\mathbb{E}M_{t}^2=\mathbb{E}M_T^2<+\infty,$$ so again $\mathbb{E}W_T^2=\mathbb{E}T$.<|endoftext|> TITLE: Unifying Geometry for Characteristic Classes QUESTION [31 upvotes]: When working with characteristic classes (more concretely Chern classes), one finds at least four essentially distinct approaches: Axiomatic Approach. See, for instance, Vector Bundles and K-Theory, page 78, th. 3.2. Characteristic Classes as Pullbacks of the Cohomology of Grassmannians. See, for instance, Vector Bundles and K-Theory, page 84, th. 3.9. Locus of Linear Dependency of General Sections (and translating into cohomology by intersection). See, for instance, 3264 & All That: Intersection Theory in Algebraic Geometry, page 59, prop-def. 1.36. Characteristic Polynomial of the Curvature Matrix (at least when working with smooth manifolds). See, for instance, Lecture Notes on Seiberg-Witten Invariants, page 19, sec. 1.5. The approaches 2, 3 and 4 are proven to be equivalent by comparing the obtained objects with the axiomatic definition. But I wonder How can be seen geometrically, in an intuitive way, that the approaches 2, 3 and 4, however different may seem, describe the same phenomena in the non-triviality of vector bundles? Any intuitive explanation is welcome. REPLY [5 votes]: (A lot of people here can write a better comment about connection b/w 2 and 4 — but per Jjm's request here is a rough sketch.) Algebraically the universal $G$-bundle is modelled by Weil (DG-)algebra $W(g)=\Lambda(g^\ast)\otimes S(g)$. For a principle $G$-bundle $E\to M$ a connection on $E$ gives a map $g^\ast\to\Omega^1(E)$ which together with curvature define a map $W(g)\to C_{de Rham}(E)$. This map agrees with filtrations so it induces a map of $E_2$-terms of corresponding spectral sequences, $H^q(g;S^p(g^\ast))\to H^{2p}(B;H^q(G))$. For $q=0$ we get a map from $S(g^*)^G=H(BG)$ to $H(M)$ — namely, an invariant polynomial $P$ goes to $P(\Omega)$. For $gl_n$ this gives the usual definition of Chern class via curvature.<|endoftext|> TITLE: Characterizing residually amenable groups QUESTION [5 upvotes]: Let $G$ be a finitely generated group. The amenability of $G$ is equivalent to the existence of a certain "weak measure" on $G$. Is there such a characterization for residually amenable groups as well? That is: Is the residual amenability of a discrete finitely generated group $G$ equivalent to the existence of a function $\mu \colon \mathcal{P}(G) \to \mathbb{R}$ satisfying certain properties? REPLY [5 votes]: It is easy to see that a finitely generated group is residually amenable if and only if there exists an bi-invariant ultra-metric on $G$ and a finitely additive $G$-invariant measure on open (with respect to the metric) subsets of $G$.<|endoftext|> TITLE: The generic fiber pullback for $p$-divisible groups in characteristic $p$ QUESTION [5 upvotes]: Let $R$ be a discrete valuation ring with the field of fractions $K$ and the residue characteristic $p$. If $K$ is of characteristic $0$, then a celebrated theorem of Tate says that the pullback functor $G\mapsto G_K$ is fully faithful on the category of $p$-divisible groups over $R$. I suppose that this full faithfulness fails if $K$ is instead of characteristic $p$; how can I see this failure? REPLY [10 votes]: The question of whether it is still fully faithful when the characteristic of $K$ is $p$ is mentioned in expose IX, SGA 7. The answer is yes, see "Homomorphisms of Barsotti-Tate groups and crystals in positive characteristic" by de Jong (Invent. math. 134, 301-333 (1998))<|endoftext|> TITLE: Cubic-exponential enumerative combinatorics QUESTION [10 upvotes]: There are many quantities in enumerative combinatorics that grow roughly exponentially, like the Fibonacci numbers, the Catalan numbers, and the factorials; indeed, most of the functions that arise in pre-19th century combinatorics -- even ones as large as the function $n^{n-2}$, which counts spanning trees of the complete graph on $n$ vertices among other things -- in a sense grow like an ordinary exponential, inasmuch as the logarithm can be asymptotically bounded above and below by linear functions of $n$. Over the past century enumerative combinatorialists have studied other quantities that grow faster than an ordinary exponential, associated with newer sorts of combinatorial objects, such as plane partitions, alternating sign matrices, and tilings of plane regions. In these situations one often encounters functions $f(n)$ that grow quadratic-exponentially, in the sense that $\log \log f(n)$ divided by $\log n$ converges to 2 rather than 1. My question is, are there any exact (non-trivial) combinatorial results concerning quantities that grow like a cubic-exponential function of $n$, with $(\log \log f(n)) / (\log n)$ converging to 3 (or maybe converging to something even larger, or diverging)? One can certainly find combinatorial problems that give rise to functions exhibiting this kind of growth, but in no cases that I am aware of (outside of trivial cases) can one actually exhibit closed formulas for these sequences, or even computationally useful recurrence relations. (If for instance a formula for $f(n)$ involves a sum of $2^n$ binomial coefficients, it's not going to enable you to compute $f(50)$, even if $f(50)$ has only a few million digits and hence is within the range of exact computer arithmetic.) Let me stress that I'm aware that a set with $n^3$ elements has exactly $2^{n^3}$ subsets, but this is the sort of case that I intended to rule out by my use of the phrase "trivial cases". REPLY [2 votes]: The number $m_n$ of matroids on $n$ elements is an interesting example, since a priori, it is unclear how fast $m_n$ grows. Knuth (1974) showed that $\log \log m_n$ is at least $n- \frac{3}{2} \log n -O(1)$. On the other hand, Piff (1973) showed that $\log \log m_n$ is at most $n-\log n + \log \log n + O(1)$. Recent work of Bansal, Pendavingh and van der Pol suggests that the true answer is closer to Knuth's lower bound.<|endoftext|> TITLE: Which known theorems of Lie algebras are still valid for Leibniz algebras? QUESTION [7 upvotes]: Leibniz algebras can be seen as a non-commutative generalization of Lie algebras. Thus, it is common to see a lot of papers which topic is about a generalization of a classic theorem of Lie algebras to Leibniz algebras. For instance, (1) gives a generalization of Engel's theorem to Leibniz algebras. Is there a survey of which known theorems of Lie algebras are still valid (and also not valid) for general Leibniz algebras? If not, could we make a community wiki to gather examples? I think it would be a nice idea to put these examples in the article of Wikipedia. (1): Ayupov, Sh A., and B. A. Omirov. "On Leibniz algebras." Algebra and operator theory. Springer, 1998. 1-12. REPLY [3 votes]: I don't know of such a survey, but a number of important properties are established in D.W.Barnes, Some theorems on Leibniz algebras, Comm. In Alg. 39, 2463-2472, (2011) (MSN), and D. W. Barnes, On Levi’s theorem for Leibniz algebras, Bull. Aust. Math. Soc. 86, 184-185, (2012) (MSN).<|endoftext|> TITLE: Measurable representations of semi simple Lie groups QUESTION [5 upvotes]: Let $G$ be a semi simple Lie group. I'm particularly interested in $SL(n,\mathbb{R})$. It is proved in I. E. Segal and J. von Neumann, A theorem on unitary representations of semisimple Lie groups, Annals of Mathematics 52 (1950), 509–517. that measurable unitary representations of $G$ are actually continuous. Is this also true for finite-dimensional non-unitary representations? REPLY [5 votes]: This is true and due to Béla von Szőkefalvi-Nagy, Über meßbare Darstellungen Liescher Gruppen (1936). Generalized to finite-dimensional representations of locally compact groups in A. Weil, L'intégration dans les groupes topologiques (1940, p. 66). Also exposed in Hewitt-Ross, Abstract harmonic analysis (1963, p. 346) or Fell-Doran, Representations of $*$-algebras, etc. (1988, p. 236).<|endoftext|> TITLE: Finding sparsest solution of a linear system QUESTION [5 upvotes]: I want to find the solution with most zero-components for the following problem: $Ax=b$ for $A\in \mathbb{R}^{k\times n}, b \in \mathbb{R}^{k},k TITLE: When is "independence of l" known? QUESTION [22 upvotes]: My question is for which varieties over local fields is "independence of l" known for etale cohomology. Say $X/{\mathbb Q}_p$ is a complete non-singular variety and $W_l$ is the (complex) Weil-Deligne representation associated to its etale cohomology group $H^i(\bar X,{\mathbb Q}_l)$. Is it known that $W_l$ is independent of $l\>\>(\ne p)$ a) when $X$ is an abelian variety (with bad reduction)? b) when $X$ is a variety with potentially good reduction? c) when $X$ is a variety with potentially semistable reduction? Any pointers to references or overviews would be very much appreciated! (Edit: To answer jmc and David Loeffler: by "independence of l" I mean that $W_l$ and $W_{l'}$ have isomorphic Frobenius-semisimplifications. I'd be happy with that!) REPLY [10 votes]: So maybe everything I'm about to say you already know, so apologies if I'm teaching my grandmother to suck eggs. This is discussed a bit at the end of a paper of Fontaine "Representations $\ell$-adiques potentiellement semistables" (section 2.4) where he describes a notion of independence that (kind of) doesn't depend on base changing to $\mathbb{C}$. It's a bit hard to find, but it's in the "Periodes $p$-adiques" Asterisque volume (for some reason, I can't get onto MathSciNet today, so I can't give you better links). UPDATE: The Imperial WiFi is now working properly, here's the MathSciNet link http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=fontaine&s5=representations%20l-adiques&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq From my understanding of it, he claims $\ell$-independence (in a reasonably strong form) for abelian varieties, curves (without conditions on the reduction), and anything with good reduction. As has been mentioned, the good reduction case is Deligne. (Also, as usual, the abelian variety case implies the case $i=1$ for any $X$, so you know $i=0,1,2d-1,2d$ when $X$ has dimension $d$.) I'm not exactly sure how the proof goes for abelian varieties in general, but with semistable reduction this basically follows from the explicit description of the weight/monodromy filtration in SGA 7, Expose IX, together with the fact mentioned by Fontaine that 'compatibility' of Weil-Deligne reps follows from equality of characters on the graded pieces of monodromy. I'm not sure about other cases you mention, like potentially good or potentially semistable reduction. If you knew the weight monodromy conjecture, then (I think) the weight spectral sequence would give you $\ell$-independence, again using the above check for 'compatibility' and the fact that you know $\ell$-independence for the $E_1$-page. But again, I'm not sure how to pass from semistable to potentially semistable. Finally, when $X$ has dimension $2$, since you know $i=0,1,3,4$, if you know independence results for the whole cohomology $H^*_\mathrm{et}(X,\mathbb{Q}_\ell)$ then you know it for $i=2$. There are results to this effect here: http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=176465&fileId=S1474748003000173 EDIT: On reflection, the claim that weight monodromy would imply $\ell$-independence for strictly semistable schemes is not true. What is true (and this is used in Saito's paper) is that for semistable schemes the existence of the weight spectral sequence implies that the alternating sum of traces of some element of the Weil group on $H^i(\overline{X},\mathbb{Q}_\ell)$ is independence of $\ell$. He then uses this and alterations to deduce independence of characteristic polynomials in dimension $\leq 2$.<|endoftext|> TITLE: Commutative spectral triples QUESTION [8 upvotes]: The corresponence between compact Hausdorff topological spaces and commutative unital $C^*$-algebras is rather well known: Gelfand Najmark theorem gives perfect correspondence between these categories. What is very easy: to define a structure of $C^*$-algebra (commutative and unital) on the algebra $C(X)$ where $X$ is compact Haudorff topological space. The real significance is the converse: namely, that each commutative unital $C^*$-algebra is of this form. This is the beginning of the whole story which culminates in the definition of spectral triple and Connes' reconstruction theorem. As I understood, the situation is the following: -if a triple $(A,H,D)$ (where $A$ is unital) is a commutative spectral triple it does not follow that there is a compact orientable manifold $M$ such that $A=C^{\infty}(M)$ -if we assume some additional technical conditions (about regularity, dimension and so on) then there is a compact orientable manifold $M$ such that $A=C^{\infty}(M)$-this is the celebrated and deep theorem of Connes. Furthermore, if I understood things correctly (see the discussion Noncommutative smooth manifolds) the main problem is to identyify the algebra $A$ as $C^{\infty}(M)$. Once you have that, it is less deep result (let us call it baby reconstruction theorem) that in this case there is also a hermitian vector bundle $E$ and essentially self adjoint elliptic operator $D_0$ such that $(A,H,D)=(C^{\infty}(M),L^2(M,E),D_0)$. After the formulation of reconstruction theorem, in the original paper of Connes there is also the following statement: each compact orientable smooth manifold arise in such way. I would like to focus on this part: what is known (for me at least), that once you have smooth, compact, spin$^c$ manifold it defines a spectral triple. So my first question is the following: Question 1 How it is always possible to obtain a commutative spectral triple from an arbitrary compact, smooth, orientable manifold (not necessary spin$^c$)? Furthermore: let us look for $C^*$-algebra situation: then there is a canonical way to obtain the $C^*$-algebra structure on $C(X)$. If we combine Connes' reconstruction theorem with baby reconstruction we obtain that if we have a commutative spectral triple $(A,H,D)$ such that $A=C^{\infty}(M)$ where now $M$ is spin$^c$ (compact, smooth) then in fact the hermitian bundle $E$ is in fact a spinor bundle (according to the cited discussion). But if the answer to the first question is affirmative then it is natural to ask: Question 2 What do we obtain if we start with an arbitrary smooth compact spin$^c$ manifold $M$: then form the spectral triple as in question 1 (as if $M$ would be not necessary spin$^c$) and after that apply reconstruction? More general question would be: Question 3 Let $M$ be a smooth compact orientable manifolds. Is it somehow possible to parametrize all pairs $(H,D)$ such that $(A,H,D)$ is a spectral triple? How does it depends from the manifold $M$ (from the fact of being spin$^c$ etc.) I would be grateful if anybody could clarify this issue for me. REPLY [10 votes]: From the perspective of the Gelfand–Naimark theorem, the heart of the reconstruction theorem is the following statement, Theorem 11.4 in Connes's paper: Let $\mathcal{A}$ be a commutative unital complex $\ast$-algebra. Then $\mathcal{A} \cong C^\infty(X)$ for some compact oriented smooth $p$-manifold $X$ if and only there exist a faithful $\ast$-representation $A \to B(H)$ on a complex Hilbert space $H$ and a self-adjoint unbounded operator $D$ on $H$ such that $(\mathcal{A},H,D)$ is a $p$-dimensional commutative spectral triple. Your Question 1 asks for the proof of the “only if” direction of this statement, which essentially goes as follows. Let $X$ be a compact orientable smooth manifold. Then you just pick your favourite Riemannian metric for $X$ and correspondingly form a commutative spectral triple as follows: if $X$ is even-dimensional, take $(C^\infty(X),L^2(X,\wedge T^\ast_{\mathbb{C}} X),d+d^\ast)$; if $X$ is odd-dimensional, take $(C^\infty(X),L^2(X,(\wedge T^\ast_{\mathbb{C}}X)^+),d+d^\ast)$, where $(\wedge T^\ast_{\mathbb{C}}X)^+$ is the $+1$ eigenbundle of the chirality operator on $\wedge T^\ast_{\mathbb{C}}X$. Of course, there's some checking to do, as is indicated in Connes's account. Now, if you combine the bare-bones reconstruction theorem with (the proof of) the “baby reconstruction theorem” in Gracia-Bondia–Varilly–Figueroa and a slight weakening of the orientability condition in the definition of commutative spectral triple, you can say even more (see Corollary 2.19 and its proof here, though the account is a bit out of date): Let $(\mathcal{A},H,D)$ be a $p$-dimensional commutative spectral triple. Then there exists a compact oriented Riemannian $p$-manifold $X$, a Hermitian vector bundle $E \to X$, and an essentially self-adjoint Dirac-type operator $D_E$ (i.e., a first-order differential operator such that $D_E^2 = -g^{ij}\partial_i \partial_j + \text{lower order terms}$) such that $$(\mathcal{A},H,D) \cong (C^\infty(X),L^2(X,E),D_E),$$ where $\cong$ denotes unitary equivalence of spectral triples. Conversely, if $X$ is a compact oriented Riemannian $p$-manifold, $E \to X$ is a Hermitian vector bundle, and $D_E$ is an essentially self-adjoint Dirac-type operator on $E$, then $(C^\infty(X),L^2(X,E),D_E)$ is a $p$-dimensional commutative spectral triple. This, in particular, answers your Question 3 as restricted to commutative spectral triples. If you drop commutativity, no such classification exists. Indeed, there come to mind a couple of extremely straightforward ways to get spectral triples with algebra $C^\infty(X)$ that aren't commutative spectral triples: Let $D$ be an essentially self-adjoint elliptic first-order differential operator on a Hermitian vector bundle $E \to X$ that isn't Dirac-type. Then $(C^\infty(X),L^2(X,E),D)$ is a perfectly good spectral triple that, in general, satisfies all the conditions for a commutative spectral triple except orientability and the additional “strong regularity” condition—the point is that orientability and strong regularity, together, would imply that $D$ was indeed Dirac-type. A somewhat silly example is $(C^\infty(X),L^2(X,E_1 \oplus E_2),D_1 \oplus D_2)$, where $D_1$ and $D_2$ are Dirac-type operators for distinct Riemannian metrics $g_1$ and $g_2$ on $X$. Let $(C^\infty(X),L^2(X,E),D)$ be an honest $p$-dimensional concrete commutative spectral triple. If $M$ is some bounded self-adjoint operator on $L^2(X,E)$ that isn't a bundle endomorphism, then $(C^\infty(X),L^2(X,E),D + M)$ is a spectral triple of metric dimension $p$ that cannot possibly be a commutative spectral triple; indeed, $D$ cannot even be a differential operator. In particular, if you take to be a smoothing pseudodifferential operator (e.g., the orthogonal projection onto the kernel of $D$, if it's nonzero), then your new spectral triple should satisfy all the conditions for a commutative spectral triple except order one and orientability. Finally, let me turn to your Question 2. Suppose you start with a concrete commutative spectral triple $(C^\infty(X),L^2(X,E),D)$ and feed it into the reconstruction theorem to get $(C^\infty(X^\prime),L^2(X^\prime,E^\prime),D^\prime)$. Then, in particular, you have an algebraic isomorphism $C^\infty(X) \cong C^\infty(X^\prime)$, which, by a theorem of Mrčun's, is necessarily given by composition by a diffeomorphism $\phi : X^\prime \to X$. Using the various axioms and the smooth Serre–Swan theorem, you can then check that the unitary $U : L^2(X,E) \cong L^2(X^\prime,E^\prime)$ is, in fact, given by a unitary isomorphism of Hermitian vector bundles $(E^\prime,X^\prime) \cong (E,X)$ covering $\phi : X^\prime \to X$, and hence, since $U^\ast D^2 U = (D^\prime)^2$, that $\phi$ was an isometry. Indeed, if you compare the Hermitian metrics on $E$ and $E^\prime$ with the inner products on the Hilbert spaces, you can even conclude that $\phi$ was orientation-preserving. Hence, if I'm not too mistaken, you can even refine the above refinement of the reconstruction theorem as follows: Let $(\mathcal{A},H,D)$ be a $p$-commutative spectral triple. Then there exists a compact oriented Riemannian $p$-manifold $X$, a Hermitian vector bundle $E \to X$, and an essentially self-adjoint Dirac-type operator $D_E$ (i.e., a first-order differential operator such that $D_E^2 = -g^{ij}\partial_i \partial_j + \text{lower order terms}$) such that $$(\mathcal{A},H,D) \cong (C^\infty(X),L^2(X,E),D_E),$$ where $\cong$ denotes unitary equivalence of spectral triples. Morever, the data $(X,E,D_E)$ is unique up to orientation-preserving isometry together with unitary bundle isomorphism covering the isometry that intertwines Dirac-type operators. Note, though that being or not being spin$^\mathbb{C}$ or spin is completely irrelevant to the machinery of the reconstruction theorem; the point is that you can form, for instance, the Hodge–de Rham spectral triple (i.e., the $d+d^\ast$ spectral triple) of a compact Riemannian spin$^\mathbb{C}$ manifold without choosing a spin$^\mathbb{C}$ structure. The only way this formalism can even see a spin$^\mathbb{C}$ structure is by choosing your Hermitian vector bundle to have been a spinor bundle—in light of Connes's Theorem 1.2, for an abstract $p$-dimensional commutative spectral triple $(\mathcal{A},H,D)$, this is equivalent to requiring that the representation of $\mathcal{A}^{\prime\prime}$ on $H$ have the correct spectral multiplicity $2^{\lfloor p/2 \rfloor}$, corresponding to the correct rank of a spinor bundle. In particular, if you start with a given spinor bundle, then the reconstruction theorem will necessarily spit out an isomorphic spinor bundle, so that if you do choose a spin$^\mathbb{C}$ structure by choosing a spinor bundle, then the reconstruction theorem can't change the spin$^\mathbb{C}$ structure. Again, the Hodge–de Rham spectral triple is completely agnostic about any spin$^\mathbb{C}$ structure, so that the reconstruction theorem, on its own, is completely agnostic about spin$^\mathbb{C}$ structures.<|endoftext|> TITLE: Completion of a local ring of a curve QUESTION [16 upvotes]: Let $X$ be a smooth projective irreducible curve defined over an algebraically closed field $\mathbb{K}$ (of arbitrary characteristic), and let $p\in X$ be a closed point. Denote by $\mathcal{O}_p(X)$ the local ring of rational functions which are regular at $p$. Then, is it true that the completion of $\mathcal{O}_p(X)$ (with respect to its maximal ideal) is isomorphic to the ring $\mathbb{K}[[x]]$ of formal power series in one variable ? I think that it should follow from Cohen theorem, but I cannot find a reference for this. Most of the results on this are in commutative ring theory books (for example Matsumura) but not really in the language of algebraic geometry. Can someone please give me a reference? I need this result in a research article. Thanks in advance. REPLY [3 votes]: One possible reference is Mumford, The red book of varieties and schemes, chap. III, § 6.<|endoftext|> TITLE: A question on rank-to-rank embeddings QUESTION [6 upvotes]: Consider a non-trivial elementary embedding $j:V_\lambda\to V_\lambda$ and, for each $A\subset V_\lambda$, set $j(A)=\bigcup_{\delta<\lambda}j(A\cap V_\delta)$. In Implications between strong large cardinals, Annals of Pure and Applied Logic 90 (1997) 79-90, Laver says that $j:(V_\lambda, \in,A)\to (V_\lambda, \in, j(A))$ is an elementary embedding. I think that this is equivalent to $j$ being $\Sigma_0^1$-elementary, i.e., it satisfies the definition of elementary embedding for formulas with second order variables but without second order quantifiers. Could someone provide a hint or a reference for this fact? I think I could prove it from a particular case, namely, that $\forall x\in V_\lambda\exists y\in V_\lambda\ (x,y)\in A$ implies $\forall x\in V_\lambda\exists y\in V_\lambda\ (x,y)\in j(A)$. Does this case follow easily from $j$ being elementary? I cannot see it. Edited: I have found a proof in chapter 2 of these notes (theorem 0.4), but I think that the proof has a gap, since it considers skolem functions $f_i$ and assumes tacitly that $j(f_i)$ are also total functions, but they could be just partial functions. REPLY [7 votes]: Here is the author of the notes you quoted. Thank you very much for pointing this out, there is indeed a gap in the proof: I truly took for granted that if $f$ is total then $j^+(f)$ is total (more specifically, that if $f$ is a Skolem function, then $j^+(f)$ is a Skolem function, but once one has totality, the rest is easy). Anyway, I think this is true. In fact, I think the following is true: Lemma If $f:V_\lambda\to V_\lambda$ is a partial function, then $j^+(dom(f))=dom(j^+(f))$. This is not immediate: by definition, $j^+(dom(f))=\bigcup_{n\in\omega}j(dom(f)\cap V_{\kappa_n})$; on the other hand, $dom(j^+(f))=\bigcup_{n\in\omega}j(dom(f\cap V_{\kappa_n}))$, and the single summands are not equal, for example if $f$ is the critical sequence then $j(dom(f\cap V_{\kappa_n}))=\{0,\dots,n-2\}$, while $j(dom(f)\cap V_{\kappa_n})$ is always $\omega$. The following is an attempt for proving this. I don't see any errors (but then again, I didn't in my notes). Sorry for the unpleasant wall of equations, that is just how I did it. Maybe there is another, more synthetic way to prove it. Proof of Lemma. The trick is to uniformize the division in pieces, i.e., to consider $dom(f\cap V_{\kappa_n})\cap V_{\kappa_m}$. Then $j^+(dom(f))=\bigcup_{m\in\omega}j(dom(f)\cap V_{\kappa_m})=\bigcup_{m\in\omega}j(\bigcup_{n\in\omega}dom(f\cap V_{\kappa_n})\cap V_{\kappa_m})$ because the dominion of the union of functions is the union of the dominions of the single functions. But now $\langle dom(f\cap V_{\kappa_n})\cap V_{\kappa_m}:n\in\omega\rangle$ is an $\omega$-sequence bounded in $V_{\kappa_m}$, therefore it is in $V_\lambda$, so we can carry the $j$ inside $\bigcup_{m\in\omega}j(\bigcup_{n\in\omega}dom(f\cap V_{\kappa_n})\cap V_{\kappa_m})=\bigcup_{m\in\omega}\bigcup_{n\in\omega}j(dom(f\cap V_{\kappa_n})\cap V_{\kappa_m})=$ $=\bigcup_{n\in\omega}\bigcup_{m\in\omega}j(dom(f\cap V_{\kappa_n})\cap V_{\kappa_m})$. $dom(f\cap V_{\kappa_n})$ is an element of $V_\lambda$, therefore there exists an $m$ such that it is in $V_{\kappa_m}$, so that is just a finite union. $\bigcup_{m\in\omega}\bigcup_{n\in\omega}j(dom(f\cap V_{\kappa_n})\cap V_{\kappa_m})=\bigcup_{n\in\omega}j(dom(f\cap V_{\kappa_n}))=j^+(dom(f))$.<|endoftext|> TITLE: Characterization of Frobenius complements QUESTION [13 upvotes]: I have learned that Frobenius complements are characterized (among finite groups) by having a fixed point free complex representation. That is, a finite group $G$ is a Frobenius complement if and only if there is a finite dimensional complex representation $\rho:G\to GL(V)$ so that the linear mapping $\rho(g)$ has no nonzero fixed points for any $g\in G\setminus\{1\}$. Is there a citable reference for this fact? I have found references that mention that this is well-known. For example: It is well-known that a finite group $H$ has the structure of a Frobenius complement if and only if it can act fixed-point-freely on some finite group $G$ (this means that non-trivial elements of $H$ fix only the identitty element of $G$). Equivalently, we may require that $H$ acts fixed-point-freely on some elementary abelian group $E$, or on a linear space $V$ over a field of characteristic zero. (Aner Shalev: A new characterization of frobenius complements) I'm aware of Passman's book Permutation groups, but it does not state this characterization. I would prefer a reference that states this characterization explicitly (and preferably includes a proof). My intended audience does not have a strong background in representation theory, so references of a more implicit kind ("this can be inferred from various results in this book" or "it is folklore/well-known/true that…") would be inconvenient. Edit: The references and descriptions given in the comments and the answer are very useful but do not contain an explicit reference for the representation theoretical characterization of Frobenius complements I'm looking for. The direction that Frobenius complements have fixed-point-free representations is given as Theorem 18.1v in Passman's book, but the other direction is still missing. REPLY [3 votes]: This is (part of) Theorem 6.13 in Serre's Finite Groups: An Introduction, which says the following. Say that an action of a group $H$ on another group $N$ is almost free if the action on $N \setminus \{ e \}$ is free. Then TFAE: $H$ is a Frobenius complement; equivalently, $H$ acts almost freely on another finite group $N$. $H$ acts almost freely on a finite-dimensional vector space over some field. $H$ acts almost freely on a finite-dimensional vector space over every field of characteristic not dividing $|H|$. $H$ acts almost freely and irreducibly on a finite-dimensional vector space over $\mathbb{C}$. $H$ acts freely and orthogonally on some sphere.<|endoftext|> TITLE: Is $n = p-q$ equivalent to Goldbach's Conjecture? QUESTION [10 upvotes]: One open conjecture is that every even integer greater than two is the difference of two primes. (Some superficial discussion here.) Goldbach's conjecture states that every even integer greater than two is the sum of two primes. The big question: are the two equivalent? That is to say, do these conjectures imply each other? I spent a bit of time pursuing this question, and I did not find a satisfactory answer. I now suspect that the two are actually not equivalent -- if they were, then I think it would suggest a symmetry on the prime numbers that I don't think they have. Anyway, I'd be glad to hear your input on the matter. REPLY [6 votes]: Close to nothing can be said rigorously about your question, but I believe the following heuristics: It's hard to imagine the exact solution of either conjecture not leading to substantial progress in the resolution of the other. This is because the current methods that handle structural results about primes have similar limitations for both problems. See Terence Tao's blog. There is a sense in which the two problems have different flavor, if you look at their "approximate" versions. We can show results about differences of primes falling in certain bounded intervals, while no analogous result is known about sums of primes. See Terence Tao's blog.<|endoftext|> TITLE: The relation between Hausdorff dimension of an $n$-manifold and $n$ QUESTION [7 upvotes]: It is known that for a topological space with different metrics, the Hausdorff dimensions may not be equal in general. For the case of manifolds, suppose $M$ is a $n$-manifold with a metric(distance), from https://math.stackexchange.com/questions/931628/hausdorf-dimension-of-a-manifold-of-dimension-n, we know that $dim_H M$ and $n$ may not be the same. But whether we can get a relation between $\dim_H M$ and $n$? I mean, whether we can prove that $\dim_H M\leq n$ or $\dim_H M\geq n$? Unfortunately, I have no idea how we can get it directly from the definition of Hausdorff dimension. REPLY [13 votes]: In a metrizable topological space, Hausdorff dimension is always larger or equal than the topological (covering) dimension. See Theorem 6.3.10 in Edgar's book "Measure, topology and fractal geometry". In particular, for an $n$-dimensional manifold $M$, if $\rho$ is any metric compatible with the Euclidean topology, then $(M,\rho)$ has Hausdorff dimension at least $n$. Because Hausdorff dimension is always at least the topological dimension, and both agree for "nice" spaces such as manifolds, Mandelbrot tentatively defined a fractal to be a metric space whose Hausdorff dimension is strictly larger than its similarity dimension, although this definition is not widely accepted today (the consensus is that you can't really define fractal). Finally, for any (at least separable) metric space $X$, the topological dimension equals the infimum (which is in fact a minimum) of the Hausdorff dimensions of $(X,\rho)$ where $\rho$ varies among the metrics compatible with the topology of $X$. This is a classical result due to Edward Marczewski.<|endoftext|> TITLE: Gaussian distributions as fixed points in Some distribution space QUESTION [10 upvotes]: I'm taking a course on topology and probabily. Today, the professor remarked something along the lines of: If you look at the space of probability distributions with $0$ mean and variance $1$, equipped with convolution, then the Gaussian distribution is characterized as the fixed point of each orbit." He also said this was a nice way to appreciate the importance of the gaussian distribution, and to gain insight for the central limit theorem. I asked for references on this point of view, but he said it's not standard and recalled only hearing about it in some seminar forty years ago. Where can I find a (preferably grad-level) reference for these ideas? Clarification: I am not asking about the fact 'the convolution of independent Gaussians is Gaussian'. REPLY [2 votes]: The introductory example in this video: https://www.youtube.com/watch?v=lTIchf0V9qo actually addresses this.<|endoftext|> TITLE: How do we show this matrix has full rank? QUESTION [12 upvotes]: I met with the following difficulty reading the paper Li, Rong Xiu "The properties of a matrix order column" (1988): Define the matrix $A=(a_{jk})_{n\times n}$, where $$a_{jk}=\begin{cases} j+k\cdot i&jk\\ 2(j+k\cdot i)& j=k \end{cases}$$ and $i^2=-1$. The author says it is easy to show that $rank(A)=n$. I have proved for $n\le 5$, but I couldn't prove for general $n$. Following is an attempt to solve this problem: let $$A=P+iQ$$ where $$P=\begin{bmatrix} 2&1&1&\cdots&1\\ 1&4&2&\cdots& 2\\ 1&2&6&\cdots& 3\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ 1&2&3&\cdots& 2n \end{bmatrix},Q=\begin{bmatrix} 2&2&3&\cdots& n\\ 2&4&3&\cdots &n\\ 3&3&6&\cdots& n\\ \cdots&\cdots&\cdots&\cdots&\cdots\\ n&n&n&\cdots& 2n\end{bmatrix}$$ and define $$J=\begin{bmatrix} 1&0&\cdots &0\\ -1&1&\cdots& 0\\ \cdots&\cdots&\cdots&\cdots\\ 0&\cdots&-1&1 \end{bmatrix}$$ then we have $$JPJ^T=J^TQJ=\begin{bmatrix} 2&-2&0&0&\cdots&0\\ -2&4&-3&\ddots&0&0\\ 0&-3&6&-4\ddots&0\\ \cdots&\ddots&\ddots&\ddots&\ddots&\cdots\\ 0&0&\cdots&-(n-2)&2(n-1)&-(n-1)\\ 0&0&0&\cdots&-(n-1)&2n \end{bmatrix}$$ and $$A^HA=(P-iQ)(P+iQ)=P^2+Q^2+i(PQ-QP)=\binom{P}{Q}^T\cdot\begin{bmatrix} I& iI\\ -iI & I \end{bmatrix} \binom{P}{Q}$$ REPLY [2 votes]: I use Christian Remling idea,In fact,I can find the matrix $$B_{ij}=\min{\{i,j\}}$$eigenvalue is $$\dfrac{1}{4\sin^2{\dfrac{j\pi}{2(n+1)}}},j=1,2,\cdots,n$$ proof: then we have $$B=\begin{bmatrix} 1&1&1&\ddots&1&1\\ 1&2&2&\ddots&\ddots&2\\ 1&2&3&3&\ddots&3\\ \vdots&\ddots&\ddots&\ddots&\ddots&\cdots\\ 1&\vdots&\ddots&\ddots&n-1&n-1\\ 1&2&\cdots&\cdots&n-1&n \end{bmatrix} $$ It is easy have $$C=B^{-1}=\begin{bmatrix} 2&-1\\ -1&2&-1\\ 0&\ddots&\ddots&\ddots\\ \vdots&\cdots&-1&2&-1\\ 0&\cdots&\cdots&-1&1 \end{bmatrix}$$ and consider $$b_{n}=|\lambda C-I|=\begin{vmatrix} \lambda-2&1&\cdots&\cdots&0\\ 1&\lambda-2&1&\cdots&0\\ \vdots&\ddots&\ddots&\ddots&\vdots\\ \cdots&\cdots&1&\lambda-2&1\\ 0&\cdots&\cdots&1&\lambda-2 \end{vmatrix} $$ so $$b_{n+1}=(\lambda-2)b_{n}-b_{n-1},b_{1}=\lambda-2,b_{2}=(\lambda-2)^2-1$$ let $\lambda-2=-2\cos{x}$, then $$b_{n+1}=-2\cos{x}\cdot b_{n}-b_{n-1},b_{1}=-2\cos{x},b_{2}=4\cos^2{x}-1$$ and induction have $$b_{n}=(-1)^n\cdot\dfrac{\sin{(n+1)x}}{\sin{x}}=0\Longrightarrow x=\dfrac{j\pi}{n+1},j=1,2,\cdots,n$$ so we $B^{-1}$ with eigenvalue is $$\lambda=2-2\cos{x}=4\sin^2{\dfrac{x}{2}}=4\sin^2{\dfrac{j\pi}{2(n+1)}}$$<|endoftext|> TITLE: Example of a saturated class of morphisms which is not _obviously_ saturated? QUESTION [10 upvotes]: By "saturated class of morphisms" in a category $\mathcal{C}$, I mean a subcategory $\mathcal{W} \subset \mathcal{C}$ such that the image of $\mathcal{W}$ in $\mathcal{C}[\mathcal{W}^{-1}]$ consists of exactly the isomorphisms. By the universal property of $\mathcal{C}[\mathcal{W}^{-1}]$, this is equivalent to saying that $\mathcal{W}$ is exactly the preimage of the class of isomorphisms under some functor. Saturated classes of morphisms are all over the place, but (in my very limited experience) I've never seen a saturated class which was not defined as the preimages of the isomorphisms under some explicit functor, so that the fact that the class is saturated follows immediately from the definition. (E.g. the weak equivalences in $\mathsf{Top_{*,\mathrm{conn}}}$ are the preimages of the isomorphisms under the functor $\prod_n \pi_n$; the quasi-isomorphisms of chain complexes are the preimages of the isomorphisms under the homology functor; I also include under this umbrella any class of morphisms defined as the saturation of some smaller class in light of the functor $\mathcal{C} \to \mathcal{C}[\mathcal{W}^{-1}]$.) I'd like an example which is naturally defined in a way that does not make the saturation obvious. I ask this because in the definition of a model category, we hypothesize that the weak equivalences satisfy 2-out-of-3 - a consequence of saturation - and then prove that the weak equivalences are saturated. Similarly, in a homotopical category, we hypothesize that the weak equivalences satisfy 2-out-of-6 - another consequence of saturation - and then prove in light of additional hypotheses that the weak equivalences are saturated. What does this extra work buy us? If saturation is obvious in all interesting examples, why not just hypothesize that the weak equivalences are saturated from the get-go? Is the motivation purely to keep the hypotheses elementary, or to avoid reliance on the existence of a possibly non-locally-small category like $\mathcal{C}[\mathcal{W}^{-1}]$? A couple of related questions would be: what is an example of a class of morphisms which satisfies 2-out-of-3 or 2-out-of-6 but is not saturated? REPLY [6 votes]: In the canonical model structure on $\omega\mathrm{Cat}$, the weak equivalences are not defined as preimages of isomorphisms under any functor, and are not even even closely related to any such preimage the way weak equivalences of unbased spaces are. And indeed, one of the hardest parts of the proof of the model structure is proving the 2-out-of-3 property.<|endoftext|> TITLE: Who first proved the fundamental theorem of projective geometry? QUESTION [11 upvotes]: The following theorem is often called the fundamental theorem of projective geometry: Let $k$ be a field and let $n \geq 3$. Let $X$ be the partially ordered set of nonzero proper subspaces of $k^n$. Then every poset automorphism of $X$ is induced by a semi-linear automorphism of $k^n$, i.e. a set map $f:k^n\rightarrow k^n$ for which there exists a field automorphism $\tau:k \rightarrow k$ such that $$f(c_1 \vec{v}_1+c_2\vec{v}_2) = \tau(c_1) f(\vec{v}_1) + \tau(c_2) f(\vec{v}_2)$$ for all $c_1,c_2 \in k$ and $\vec{v}_1,\vec{v}_2 \in k^n$. Question: who was the first person to prove this, and where does their proof appear? I know it has its origins in 19th century work of von Staudt, but I don't think that the above theorem appears in his work. On page 52 of Baer's book "Linear Algebra and Projective Geometry", he says that the first proof was due to Kamke, but he does not give a reference. REPLY [11 votes]: The version you state is definitely a 20th century development, only marginally related to Von Staudt's theorem. Here is a translation of the relevant section of Karzel & Kroll's Geschichte der Geometrie seit Hilbert, p. 51 (notation should be self-explanatory): Examples of "non-linear" collineations were given by C. Segre [Seg 1890] for projective geometries over the complex numbers and by Veblen and Bussey [VB 06] for projective geometries over finite fields. Thus arose at the beginning of this century the problem whether these are all affinities resp. collineations. The following realization theorems state that this is the case. (8.1) Let $(V,K)$ be a vector space with $\dim(V,K)\geqslant 2$ resp. $\dim(V,K)\geqslant 3$. a) For every affinity $a$ of the corresponding affine space $A(V,K)$ there is exactly one semilinear permutation $\sigma$ of $(V,K)$ and one $\mathbf a\in V$ such that $a = \mathbf a^+\circ\sigma$. b) For every collineation $\kappa$ of the corresponding projective space $\Pi(V,K)$ there is a semilinear permutation $\sigma$ of $(V,K)$ such that $\tilde\sigma = \kappa$; here $\sigma$ is unique up to a factor $\lambda\in K$, i.e. $\tilde\sigma = \tilde\sigma'$ for $\sigma' = \lambda_\ell\circ\sigma$, where $\lambda_\ell(\mathbf x) := \lambda\mathbf x$. Part a) was first proved by E. Kamke [Kam 27]. A proof for the case where $K$ is a commutative field is found in the textbook [SS 35], §14 of Schreier and Sperner. Further proofs over arbitrary fields are found in the textbooks [Bae 52], [Le 65], [KSW 73]. In [KSW 73] and [Le 65], part a) is proved first and part b) then deduced as a consequence, which makes the proof simpler and more transparent. Veblen [Ve 07] already gives a proof for projective geometries over finite fields. Theorem (8.1 b) is the generalization to [higher dimensional] space of Von Staudt's Theorem (9.1) on the group of projectivities. One might add that Darboux (1880, p. 59) already states and attributes the theorem: It is easy for example to recognize with v. Staudt (Geometrie der Lage § 121–122) that a projective or homographic correspondence in the plane or in space can be defined by the sole condition that aligned points in one figure correspond to aligned points in the other. He only sketches a proof, and Schur (1881, p. 254) comments: See v. Staudt, Geometrie der Lage, p. 60. Compare especially Möbius, der barycentrische Calcul, Chap. 6 and 7, where collineation in the plane and in space is first defined by the condition that straight lines correspond to straight lines. Further references, giving various versions of the theorem but apparently never tracing it beyond Baer, are Dieudonné (1955, p. 72), Artin (1957, p. 88), Bourbaki (1970, Exerc. II.9.16), Jacobson (1974, p. 470), Samuel (1986, p. 32), Berger (1987, 5.4.8), Bennett (1995, p. 203), Jeffers (2000, p. 810).<|endoftext|> TITLE: Isomorphism of Hilbert modules QUESTION [6 upvotes]: In Hilbert space theory if we have a linear bijection $U:H \to K$ which is an isometry, then it preserves inner products. Suppose now that you have two (right) Hilbert modules $X,Y$ (say, over the same $C^*$-algebra $B$) and a $B$-linear isomorphism $T:X \to Y$ which is an isometry (recall that the norm in Hilbert module is defined via $\|x\|^2_X:=\| \langle x,x\rangle\|_B$). Does it follows that $T$ preserves $B$-valued inner products? REPLY [11 votes]: Yes. The theorem was proven independently by Blecher, A new approach to C*-modules, Math. Ann. 307 (1997), 253-290 and Lance, Unitary operators on Hilbert C*-modules, Bull. London Math. Soc. 26 (1994), 363-366. (Google would have told you this.) [Correction: the result is due to Lance. However, the formula below which recovers the inner product is Blecher's.] In fact, the inner product can be recovered from the norm and $B$-module structure by the formula $\langle x,x\rangle = \sup\{r(x)^*r(x): r: X \to B$ is $B$-linear and $\|r\| \leq 1\}$, and then you get $\langle x,y\rangle$ by polarization.<|endoftext|> TITLE: Calculate Hausdorff measure with Frostman measures QUESTION [5 upvotes]: Fix a metrix space $(X,d)$ and consider the Hausdorff (outer) measures $\mathcal{H}^s$ on $X$. A Frostman measure on $X$ is a finite Borel measure $\mu$ such that there exists $C,t,r_0>0$ with $\mu(B_r(x)) \leq C r^t$ for all $x\in X$ and all $00\rbrace$$ Frostman himself proved that in fact this is essentially an equality: For any Borel set $A$ with $\mathcal{H}^t(A)>0$ there exists a Frostman measure of exponent $\geq t$ with $\mu(A)>0$. Now my question is: Can we also say something about the precise value of the Hausdorff measure in terms of Frostman measures? A naive conjecture might be that the above inequality is sharp in the sense that if $\infty>\mathcal{H}^t(A)>0$ then $$\sup\lbrace \mu(A) \mid \mu \text{ Borel measure with } \exists r_0 \forall 00$ there is a $r_0$ with $\mathcal{H}^t(A\cap E) \leq (1+\epsilon) diam(E)^t$ for all $r0:\sup\limits_{diam(E)<\delta} \mu(E)diam(E)^{-t}\leq 1\right\rbrace$$ The presence of the factor $2^t$ in the first inequality makes it implausible to me that my conjectural equation with the supremum over all Frostman measures really holds. Instead a similar density theorem for the spherical Hausdorff measure proves that the analogue of my conjecture holds for the spherical Hausdorff measure. (In particular: The idea with the hyperbolic manifolds will not work because for $t=n$ the usual and the spherical Hausdorff measure coincide) I'm very satisfied with this result although I wouldn't mind any additions regarding more general situations.<|endoftext|> TITLE: New research and re-discovering classic results in "basic" real analysis QUESTION [13 upvotes]: Sometimes, it happens that researchers publish a new proof of an old well-known result in "basic real analysis" (I'm referring to what some American people may call "honors calculus"). For instance, we can consider this article. I have two questions: (1) What are some examples recent novel proofs of old well-known results in "basic real analysis"? (2) Has it ever happened in recent times that such a proof had been particularly useful bringing about new insights into major problems? REPLY [7 votes]: Here is one I once saved: Peter D. Lax, Change of variables in multiple integrals (1999). It gave rise to two postscripts: Peter D. Lax, Change of variables in multiple integrals. II (2001), Nikolai V. Ivanov, A differential forms perspective on the Lax proof of the change of variables formula (2005).<|endoftext|> TITLE: Foliation with leaves which are and are not dense QUESTION [7 upvotes]: Do there exist a foliation on a closed surface (i.e. real dimension 2) which has a dense leaf and also a leaf which is not dense? REPLY [11 votes]: No, there doesn't exist such a foliation. The existence of any foliation would mean the Euler characteristic is zero, so the surface must be either a torus or a Klein bottle. Foliations for these surfaces are understood well enough to rule out having both dense and non-dense leaves. Any foliation will contain a "Reeb component" (for which no leaf is dense) or will be a suspension of a homeomorphism of $S^1$. If a circle homeomorphism has one dense orbit, all orbits are dense, so the situation you describe can't exist. If you allow surfaces with boundary, you need to consider a Mobius band or annulus, but it's still not possible to have a dense and non-dense leaf: the foliation will have pieces that are either Reeb components or suspensions of interval homeomorphisms, neither of which can have a dense leaf. For the definition of a Reeb component and complete details, see this page, especially the "Foliations of Surfaces" section: http://www.map.mpim-bonn.mpg.de/Foliations<|endoftext|> TITLE: What is the Schur multiplier of the affine linear group AGL(n,q)? QUESTION [7 upvotes]: What is the Schur multiplier of the $n$-dimensional affine linear group $\mathrm{AGL}(n,q)$ over the Galois field with $q$ elements? I am particularly interested in the simple case $n=1$. Computation using GAP shows that the Schur multiplier of $\mathrm{AGL}(1,q)$ is trivial for any prime power $q$ up to 19, except for $q=4$, in which case the multiplier has order 2. Is the Schur multiplier of $\mathrm{AGL}(1,q)$ always trivial except for $q=4$? Is the Schur multiplier of $\mathrm{AGL}(n,q)$ also trivial except for a few exceptions? REPLY [7 votes]: Let me concentrate on the ${\rm AGL}(1,q)$ case for now, where $q=p^k$ is a prime power. This group is a split extension $N:C$, where $N$ and $C$ are the additive and multiplicative groups of the field of order $q$, and the action of $C$ on $N$ is by multiplication in the field. So $N$ is elementary abelian and $C$ is cyclic. Since cyclic groups have trivial multiplier, it follows from standard results that the $r$-part of the multiplier is trivial for all primes $r$ except possibly $r=p$ when $k>1$. The Schur multiplier $M$ of $N$ is elementary abelian of order $p^{k(k-1)/2}$, and there is an induced action of $C$ on $M$. Again by standard theory, the $p$-part of the multiplier of $G$ is equal to the subgroup of $M$ that is fixed pointwise by $C$. (Since $|C|$ and $|M|$ are coprime, the action is completely reducible.) If we regard $M$ and $N$ as ${\mathbb F}_pC$-modules, then $M$ is equal to the exterior square of $N$. So we need to determine whether $M$ has any trivial constituents as ${\mathbb F}_pC$-module. If we extend to a splitting field for $N$ in characteristic $p$ (which we could take to be ${\mathbb F}_q$), then $N$ is isomorphic to the direct sum $\oplus_{i=0}^{p-1} N^{\sigma^i}$, where $N$ is now regarded as a module over ${\mathbb F}_q$, and $\sigma$ is the Frobenius field automorphism of ${\mathbb F}_q$ order $k$. The exterior square of $N$ is a section of the tensor product $N \otimes N$. So if we can show that $N \otimes N$ has no trivial constituents, then we will have shown that the Schur multiplier of the group is trivial. But, as a module over ${\mathbb F}_q$, $N \otimes N$ is the direct sum of $1$-dimensional submodules $N^{\sigma^i} \oplus N^{\sigma^j}$. The action of an element $g \in C$ on this submodule is by multiplication by $g^{\sigma^i+\sigma^j} = g^{p^i+p^j}$. So there will be a trivial constituent only if, for some $i$ and $j$ with $0 \le i,j < k$, we have $g^{p^i+p^j} = 1$ for all $g \in C$, which is equivalent to $p^k-1$ dividing $p^i+p^j$. It is easy to see that this happens only when $q=4$. (We are assuming that $k>1$.) Continued on 5 January: Turning now to the case $G={\rm AGL}(n,q)$ for $n \ge 2$, we have $G = N:H$ with $H = {\rm GL}(n,q)$ and $N$ the natural module for $H$. Let $\hat{G}$ be a Schur covering group of $G$, so $\hat{G}/M \cong G$, where $M$ is the multiplier of $G$. Then $M$ is generated by the inverse images in $\hat{G}$ of $[H,H]$, $[G,N]$ and $[N,N]$, so let's consider those three subgroups in turn. Firstly, (the inverse image of) $[H,H]$ is isomorphic to the Schur Multiplier of $H={\rm GL}(n,q)$. For $(n,q)=(2,2)$ and $(2,3)$, this is trivial. Otherwise $H' = {\rm SL}(n,q)$ is perfect, and $H/H'$ is cyclic, and it is not hard to see that the multiplier of $H$ is a quotient of that of $H'$. It is well-known that the multiplier of $H'$ is trivial except when $(n,q)$ = $(2,4)$, $(2,9)$, $(3,2)$, $(3,4)$ and $(4,2)$. It can be checked (for example by computer calculation) that the multiplier of $H$ has order $2$ when $(n,q)$ = $(2,4)$, $(3,2)$ or $(4,2)$, and is trivial otherwise. Secondly, the inverse image of $[H,N]$ is isomorphic to ${\rm Ext}(N,T)$, where $T$ is the trivial module for $G$ over ${\mathbb F}_q$, which is isomorphic to $H^1(G,{\rm Hom}(N,T)) = H^1(G,\hat{N})$, where $\hat{N}$ is the dual module of $N$. Again these are in the literature for ${\rm SL}(n,q)$, and in the few cases where this is nonzero, it can be calculated for $H = {\rm GL}(n,q)$. It turns out that the only nonzero case is $(n,q) = (3,2)$, where the dimension is $1$. Finally, the inverse image of $[N,N]$ is a quotient of the Schur Multiplier of $N$ and by a calculation similar to the case $n=1$, and using ${\rm AGL}(1,q^n) \le {\rm AGL}(n,q)$, we find that the only case when this is nontrivial is $(n,q)=(2,2)$ ($G \cong S_4$), when it has order $2$. Summing up, the cases when the Schur multiplier of ${\rm AGL}(n,q)$ is nontrivial are $(n,q) = (2,2)$, $(2,4)$, $(3,2)$ and $(4,2)$, and it has order $4$ for ${\rm AGL}(3,2)$, and $2$ in the other cases. I have checked this with a computer calculation.<|endoftext|> TITLE: Discovery and Study of Conic Sections in Ancient Greece QUESTION [14 upvotes]: Is there anything known about what drew the attention of ancient greek mathematicians to conic sections and, what were the models they used to study conic sections? What I would like to know, is whether ancient mathematician actually became aware of conic sections by looking at sections of circular cones or, whether they noticed them elsewhere, e.g. as the shadow of a coin or (highly speculative) even in the way that gardeners created interesting shapes with two sticks and a rope? I know that Apollonius of Perga wrote a thorough treatise about conic sections (cf e.g. http://en.wikipedia.org/wiki/Apollonius_of_Perga), but did he also discover them and how did he study them? REPLY [11 votes]: Euclid's Optics presents the visual cone with the apex at the eye as a geometric model for the appearance of things. In Optics various results are deduced about the appearance of flat surfaces below the eye and above the eye. See Figure 10 on page 359 in Burton’s translation, referenced below for a possible evidence that the visual cone might be intersected by a vertical plane in order to explain certain visual phenomena . Propositions 38-40 on pages 365-367 in Optics show how the circular base of a visual cone appears under certain circumstances. Although not expressed as "conic sections" the visual effects are described for acute and right and obtuse angle cones which correspond to whether the circle at the base of the visual cone is being viewed from a point on the hemisphere above the circle (right angle cone) , a point above the hemisphere (acute cone), or a point within the hemisphere (obtuse cone). The first four books of Apollonius' Conics are generally believed to have drawn heavily from an earlier lost work by Euclid, also called Conics. It is believed that like Euclid, Apollonious also studied astronomy and optics. Geometrical optics, and the model of the visual cone, was used to study relationships between the apparent size, position, or motion of an object and its actual size, position, or motion. Sources: http://philomatica.org/wp-content/uploads/2013/01/Optics-of-Euclid.pdf             (Image added by J.O'Rourke.)<|endoftext|> TITLE: Projective curves of constant curvature QUESTION [8 upvotes]: A nodal projective curve in $\mathbb{CP}^2$ inherits a Kähler metric from the Fubini-Study metric, and hence a Riemannian metric. In particular, with respect to this metric, a line has constant Gaussian curvature 1; according to Vitter (page 826), the Fermat conic $\{x^2+y^2+z^2=0\}$ has constant curvature 1/2, but no other Fermat curve has constant curvature. In fact, Vitter explicitly computes the Gaussian curvature of the zero set of a given homogeneous polynomial. Does there exist a flat plane cubic? More generally, Fix a degree $d>2$: does there exist a smooth, degree-$d$ plane curve with constant Gaussian curvature? Does there exist an irreducible degree-$d$ nodal curve with constant Gaussian curvature? If such a curve existed, and it had total Milnor number $\delta$, its curvature would be $(3-d)/2 + \delta/d$. REPLY [8 votes]: Actually, the rigidity result is local, and hence is even stronger: A very old result of Calabi implies that any connected (local) piece of a holomorphic curve in $\mathbb{CP}^n$ with constant Gaussian curvature in the induced metric from the Fubini-Study metric is a piece of a rational curve that is a rational normal curve in the smallest linear projective space that contains it. In fact, it must be an orbit of a representation of $\mathrm{SU}(2)$ into $\mathrm{SU}(n{+}1)$.<|endoftext|> TITLE: Divergence of a series similar to $\sum\frac{1}{p}$ QUESTION [15 upvotes]: Suppose we start with $k$ primes $p_1,p_2,\ldots ,p_k$ (not necessarily consecutive) and a residue class for each prime $r_1,r_2,\ldots ,r_k$. We denote the least integer not covered by the arithmetic progressions $r_i+m\cdot p_i$ as $r_{k+1}$ which is going to be the new residue class for a (random) prime $p_{k+1}$. We proceed in this way "covering" the natural numbers (without changing the $r_i$'s.) Question:Is it true that $\sum\limits_{n=1}^\infty\frac{1}{r_n}=+\infty$? Motivation: This would directly imply Dirichlet's theorem on primes in arithmetic progressions if we make use of this Lemma: Let $a_n$ be a sequence of natural numbers, strictly increasing with $\gcd(a_i,a_j)=1$. Then if $\sum\limits_{n=1}^\infty\frac{1}{a_n}=+\infty$ then the sequence contains infinitely many prime numbers. I tried to modify some proofs which show the divergence of $\sum\frac{1}{p}$ but without much success. Thank you very much in advance! EDIT: The primes are distinct and the residue classes are not reduced modulo $p$.Suppose we start with the primes $2,3,7$ and their residue classes are $(1)_2 , (2)_3 , (4)_7$ which means $r_1=1$, $r_2=2$ and $r_3=4$. The least number not covered by the progressions $1+2k , 2+3k , 4+7k$ is $6$.We define then $r_4=6$ and $6$ is going to be a new residue class for a new prime (random choice) let's say $p_4=17$.Then $r_5=10$ and we choose a new prime (Let's say $p_5=5$) and continue in this direction. The $r_i$'s could be much greater than the $p_i$'s REPLY [3 votes]: The question makes sense if the $r$'s are all positive. The answer is likely to be yes for the folllowing reason: if it were no, there would be an explicit sequence where the $r$'s grow faster than $O(1/f'(n))$, where $f(n)$ is smaller than any iterated $\log$ function, and thus smaller than $\log \log\log\log\log n$, call this $\log_5 n$ or $g(n)$. Current lower bounds of the conceivable maximal growth rate of the $r$'s are like $$\dfrac1{g'(n)(\log_3 n)^2},$$ see recent work of Ford, Konyagin, Green, Maynard, and Tao. Having such a sequence of $r$'s and $p$'s would allow one to radically improve on this bound, for one could leverage this to find really large gaps between primes. Of related interest may be Kanolds work in 1963-65 on such sequences of $r$'s. If the $r$'s grow slowly enough (something like $n^{2-c}$), one can get Linnik's theorem on the least prime in arithmetic progressions as well as a non-quantitative version of Dirichlet's Theorem through elementary means.<|endoftext|> TITLE: Is there a highly transitive action of a finitely generated torsion simple group? QUESTION [9 upvotes]: Is there a highly transitive action of a finitely generated torsion simple group $G$ on $\mathbb{Z}$ ? Highly transitive means $k$-transitive for each $k \in \mathbb{N}$, that is: for every two $k$-tuples $(x_1, \dots x_k) , (y_1, \dots, y_k) \in \mathbb{Z}^k$ of pairwise distinct elements, there is some $g \in G$ such that $g(x_i) = y_i$ for each $1 \leq i \leq k$. REPLY [4 votes]: Yes, such groups exist. Simple groups in my paper "Palindromic subshifts and simple periodic groups of intermediate growth" are like that.<|endoftext|> TITLE: Is residual finiteness a property of "many" finitely presented groups? QUESTION [8 upvotes]: Is there a reasonable random model for selecting a finitely presented group $G$ such that with positive probablity (or even with probability almost $1$) some of the following properties hold: $G$ is residually finite. $G$ is subgroup seperable (LERF). The first $l_2$ Betti number of $G$ is positive. The cost of $G$ is greater than 1. REPLY [4 votes]: A random group at density less than 1/6 is known to be the fundamental group of a compact, non-positively curved cube complex, by Ollivier--Wise. By Agol's theorem, all such hyperbolic groups are virtually special, and hence residually finite, QCERF, etc.<|endoftext|> TITLE: Minimum number of unit fractions to sum up a given positive rational QUESTION [6 upvotes]: For any positive $p,q\in\mathbb{N}$ there is a finite subset $S$ of $\{\frac{1}{n}:n\in\mathbb{N}, n\geq 1\}$ such that $\sum_{s\in S} s=\frac{p}{q}$, see this article by Paul Erdös and Sherman Stein (Sums of distinct unit fractions. Proceedings of the American Mathematical Society, 14(1), 126-131, 1963.) . Let $m(p,q)$ denote the minimal cardinality of such a subset $S$. Is there a polynomial-time algorithm to determine $m(p,q)$? REPLY [3 votes]: In fact, this is listed as an open problem in the Wikipedia page on Egyptian Fractions, presumably because they do use the output size as a parameter.<|endoftext|> TITLE: Density of multi-grade solutions to $x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k$ for $k = 5$ or $6$? QUESTION [5 upvotes]: Given the Diophantine equation, $$x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k\tag1$$ there is the rather curious observation that the smallest positive solutions for $k=5$ or $6$ is multi-grade. $$24^k+28^k+67^k=3^k+54^k+62^k,\quad k = 1,5$$ $$15^k + 10^k + 23^k = 3^k + 19^k + 22^k,\quad k = 2,6$$ Duncan Moore has exhaustively searched $(1)$ for all positive and primitive solutions below a bound $Z$. Table 1 is for $k=5$, while Table 2 is for $k=6$. We summarize the data below. I. Table 1: $$\begin{array}{|c|c|c||} \text{# of solns}&\color{blue}{A:=\text{(% of}\; k = 1,5)}&\text{diff}\\ 2^0\cdot168&63.7\text{%}& \\ 2^1\cdot168&65.8\text{%}&+2.7\\ 2^2\cdot168&65.6\text{%}&-0.3 \\ 2^3\cdot168&63.6\text{%}&-2.0\\ 2^4\cdot168&61.0\text{%}&-2.6\\ 2^5\cdot168&59.1\text{%}&-1.9\\ \end{array}$$ Note: To address one comment below, $A$ is the percentage of solns given in the first column that is valid for both $k=1,5$. For example, out of the first $2^5\cdot168 = 5376$ solns, then $59.1\text{%}$ are for $k=1,5$. Each row doubles the $\text{#}$. Since Moore's database has $5393$ solns, and $5393/2^5\approx168.53$, then I used that as the base value. II. Table 2: $$\begin{array}{|c|c|c|} \text{# of solns}&\color{blue}{B:=\text{(% of}\; k = 2,6)}&\text{diff}\\ 50&80\text{%}& \\ 100&85\text{%}&+5.0\\ 200&89\text{%}&+4.0\\ 400&91.7\text{%}&+2.7\\ \end{array}$$ Note: Thus, out of the first $400$ solns, then a whopping $91.7\text{%}$ of them are actually multi-grade for $k=2,6$. (I'm not sure if excluding non-primitive solutions below the bound $Z$ is relevant. Program-wise, it seems easier to just include them.) Questions: Why is the percentage of $A$ decreasing, while that of $B$ is apparently increasing? Or will $B$ eventually have a negative diff like $A$? (The data is too small to be conclusive.) If both are decreasing, will $A,B \to 0$? Or will it taper off to some constant? P.S. This answer to a related post might be informative. Incidentally, the smallest solutions to, $$x_1^k+x_2^k+x_3^k+x_4^k = y_1^k+y_2^k+y_3^k+x_4^k\tag2$$ are also multigrades as $k=1,5$, and $k=2,6$, though there are no exhaustive tables for these. REPLY [2 votes]: Not an answer, just a graphic illustrating the percentages $A$ and $B$ of multi-grade solutions vs. all solutions, sorted according to the biggest contained term. It is hard to believe they should drop to $0$ as $n$ grows, especially $B$, but sure enough, the available data after all only cover a tiny part of infinity...<|endoftext|> TITLE: Polynomial threading through a monotone corridor QUESTION [6 upvotes]: I have a need to find a polynomial of minimal degree that connects two points and stays within a given "corridor," by which I mean an $x$-monotone polygon. Here is an example:             The point $s=(0,3)$ needs to be connected to the point $t=(3,1)$ while remaining inside the $x$-monotone polygon $P$ illustrated. (Here "$x$-monotone" means that every vertical line meets the polygon in an interval or the empty set.) In the case illustrated, a cubic curve suffices, and no quadratic (or linear) curve suffices. Q. Given data as depicted (start & end points $s$ & $t$, an $n$-vertex $x$-monotone polygon $P$), is there an algorithm to find a polynomial of minimal degree that connects $s$ to $t$ and remains within $P$? Even an inefficient algorithm sketch would be appreciated, as I am not seeing any algorithm. REPLY [5 votes]: A much more efficient approach can be based on semidefinite programming feasibility testing. Namely, it is well-known that global nonnegativity of a degree $2d$ polynomial $g(x)$ is equivalent to existence of a decomposition of $g$ into a sum of squares of polynomials, which in turn is equivalent to existence of a postitive semidefinite $(d+1)\times (d+1)$ Hankel matrix $H$ so that $g(x)=X^\top HX$, where $X$ is the vector of monomials $X=(1,x,x^2,\dots,x^d)$. The condition $g(x)=X^\top H X$ is a positive semidefinite feasibility problem (each coefficient of $g$ gives a linear equation on entries of $H$, and $H$ must be p.s.d.). In our case we need certificates of nonnegativity (or non-positivity, which is the same up to sign thing) of polynomials of the form $g(x):=f(x)-ax-b$ on intervals $\alpha \leq x<\beta$. Change of variable $x\to \frac{\alpha+\beta bt^2}{1+t^2}$ allows one to specify the latter as global nonnegativity of the rational function $g\left(\frac{\alpha+\beta bt^2}{1+t^2}\right)$, which is equivalent, after clearing denominators, to global nonnegativity of certain polynomial $\tilde{g}(t^2)$, with coefficients linear functions of coefficients of $f$. Thus, for every segment $i$ of our region we get a polynomial $\tilde{g}_i(t^2)$, global nonnegativity (or nonpositivity) of which we encode as above, ending up with a semidefinite programming feasilility problem with block-diagonal matrix, each block Hankel, and free variables corresponding to coefficients of $f$.<|endoftext|> TITLE: Closure order on nilpotent orbits in exceptional Lie algebras QUESTION [9 upvotes]: Let $G$ be a simple algebraic group over the algebraically closed field $k$ of positive characteristic, and let ${\mathfrak g}={\rm Lie}(G)$. It is well known that there are finitely many nilpotent $G$-orbits in ${\mathfrak g}$ and that the classification of these orbits is the same as over the complex numbers, as long as the characteristic of $k$ is good, i.e. odd if $G$ is not of type $A$, greater than $3$ if $G$ is of exceptional type, and greater than $5$ if $G$ is of type $E_8$. There is quite a lot of history to the subject, but a uniform proof of the classification was given relatively recently by Premet. My question concerns the closure ordering on nilpotent classes: where (if anywhere) is it established that the closure ordering is also (in good characteristic) the same as over the complex numbers? This fact seems to be used in Varieties of nilpotent elements for simple Lie algebras I: good primes by the VIGRE group, but the reference is to Carter's book, and I can't find anywhere in the VIGRE paper where they mention a justification for why the order should be the same. (It may be in there, but I haven't found it.) For classical Lie algebras outside characteristic 2, I think it is reasonably straightforward to use the partition type classification to show that the closure order is also independent of the characteristic. So this is really a question about exceptional Lie algebras. REPLY [5 votes]: I'm assuming your question involves just good prime characteristic $p$. Much of the literature focuses on unipotent classes, but Springer's $G$-equivariant isomorphism (for good $p$) between the unipotent variety in $G$ and the nilpotent variety in $\mathfrak{g}$ shows that the classes and orbits are in bijection and also allows one to transfer the closure relationships. Thus the closure ordering graphs are the same for the group and the Lie algebra. (Here one has to be a bit careful about the isogeny type of $G$ in type $A_n$, however.) It turns out after some work that the closure orderings of unipotent classes are the same as those found much earlier by Gerstenhaber and others in characteristic 0. Much of this work was done by Spaltenstein (and those he cites including Mizuno, Shoji for exceptional types): see his Classes unipotentes et sous-groupes de Borel, Lect. Notes in Math. 946 (Springer, 1982), especially II.8 and the graphs for exceptional types in IV.2. [Carter's 1985 book follows this development, though for types $E_7, E_8$ on pages 442 and 444 the graphs lack several edges; this was probably a technical error made during the production of the book.] Though I've never checked all the details carefully, I've been assured that experts have done so and find Spaltenstein's results convincing. ADDED: Maybe it's useful to expand this answer and also respond to Ma's question. I should emphasize that the classification of nilpotent orbits or unipotent classes requires a lot of case-by-case work for each simple, simply connected algebraic group and its Lie algebra (obtained by Chevalley's process from a $\mathbb{Z}$-form of the corresponding simple Lie algebra over $\mathbb{C}$). Much of the literature here follows Chevalley's viewpoint, in which the unipotent classes are found to have representatives specified by certain parameters independent of good characteristic. (The bad primes, possibly $2,3,5$, may divide some of Chevalley's structure constants.) In particular, it's enough to work over finite fields of good characteristic, or over $\mathbb{Q}$. For exceptional Lie types, only $G_2$ can be done in a fairly direct way. For $F_4$ the unipotent classes in finite Chevalley groups were studied by Shoji. Then Mizuno made very detailed computations for types $E_6, E_7, E_8$. Spaltenstein drew the literature together in his work on the Springer correspondence, providing many refinements and a few corrections as well as an emphasis on the hidden duality of special classes which leads to a symmetry for these classes in the closure order diagram (generalizing the classical partial ordering of partitions in type $A_\ell$). The closure ordering itself is a relatively elementary byproduct of the classification when done in this spirit: roughly speaking, each parameter involved in a particular class is specialized to 0 and leads to other classes in the closure. But the classification itself is quite arduous to work out in detail as Shoji and Mizuno did. (It is probably best organized theoretically in the algebraic group context by the Bala-Carter method.)<|endoftext|> TITLE: Evidence that Graph Isomorphism problem is not $NP$-complete QUESTION [15 upvotes]: Graph isomorphism problem is one of the longest standing problems that resisted classification into $P$ or $NP$-complete problems. We have evidences that it can not be $NP$-complete. Firstly, Graph Isomorphism can not be $NP$-complete unless the polynomial hierarchy [1] collapses to the second level. Also, the counting[2] version of GI is polynomial-time Turing equivalent to its decision version which does not hold for any known $NP$-complete problem. The counting version of natural $NP$-complete problems seems to have much higher complexity since they are $\#P$-complete. Finally, the lowness result [3] of GI with respect to $PP$ ($PP^{GI}=PP$) is not known to hold for any $NP$-complete problem. The lowness result of GI has been improved to $SPP^{GI}=SPP$ after Arvind and Kurur proved that GI is in $SPP$ [4]. Edit: It is known that SAT is truth-table equivalent to USAT (set of Boolean formulas with exactly one satisfying assignment) while it is not known whether GI and UGI are equivalent under polynomial time truth-table reductions. What other (recent) results can provide further evidence that GI can not be $NP$-complete? [1]: Uwe Schöning, "Graph isomorphism is in the low hierarchy", Proceedings of the 4th Annual Symposium on Theoretical Aspects of Computer Science, 1987, 114–124 [2]: R. Mathon, "A note on the graph isomorphism counting problem", Information Processing Letters, 8 (1979) pp. 131–132 [3]: Köbler, Johannes; Schöning, Uwe; Torán, Jacobo (1992), "Graph isomorphism is low for PP", Computational Complexity 2 (4): 301–330 [4]: V. Arvind and P. Kurur. Graph isomorphism is in SPP, ECCC TR02-037, 2002. REPLY [4 votes]: A search done for a related question also turned up (as indicated in a comment) rjlipton.wordpress.com/2015/03/05/news-on-intermediate-problems: Now Eric and Bireswar complete the triad of relations to the other intermediate problems: Theorem 2 Graph Isomorphism is in ${\mathsf{RP}^{\mathrm{MCSP}}}$. Moreover, every promise problem in ${\mathsf{SZK}}$ belongs to ${\mathsf{BPP}^{\mathrm{MCSP}}}$ as defined for promise problems. Here $\mathsf{RP}$ stands for randomized polynomial time, MCSP is the Minimum Circuit Size Problem, and $\mathsf{SZK}$ stands for Statistical Zero Knowledge. So we learn that if MCSP is not NP-complete and RP!=NP, then graph isomorphism is not NP-complete either.<|endoftext|> TITLE: Correspondences as generalized morphism between $C^*$-algebras QUESTION [9 upvotes]: While reading about Morita equivalence in the category of $C^*$-algebras I met also the following notion: a correspondence between two $C^*$-algebras $A,B$ is a pair $(X,\varphi)$ where $X$ is a (right) $B$ Hilbert module and $\varphi:A \to \mathfrak{B}(X)$ is a homomorphism. Question 1 In all definitions which I met I found that the only requirment for $\varphi$ is that it is a homomorphism but it is somehow natural to require that it preserves the involution. Do we assume this? Identyfying isomorphic correspondences one can check that with the composition defined as tensor product we obtain the category: objects are $C^*$-algebras while morphism are isomorphism classes of correspondences. It is commonly stated that this definition is more flexible while it admits 'more' morphism: in fact given an ordinary morphism $f:A \to B$ one can define $X$ to be $B_B$ with the standard right Hilbert module structure and $\varphi(a)$ to be a multplication by $f(a)$. Question 2 Is it true that given any two $C^*$-algebras $A,B$ one can always find a correspondence between them? And finally I would like to know whether Question 3 isomorphism in this category is precisely the same as Morita equivalence. If $_AX_B$ is an equivalence $A,B$ bimodule then if we take a dual bimodule $_B\tilde{X}_A$ one can check the appropriate tensor products are isomorphic to $_AA_A$ and $_BB_B$-so we obtain isomorphisms in this category. I don't however see why the converse must be true. EDIT: it seems that I was too quick claiming that I see that two Morita equivalent algebras $A,B$ are isomorphic in the category described above. It is always true that $_AX_B \otimes _B\tilde{X}_A \cong _AA_A$ but I don't see why $_AA_A$ is and identity morphism in this category. To be more precise: for each $C^*$-algebra $A$ we put $X$ to be $A$ with the natural action of $A$ from the right and from the left by multiplication and the product $\langle a,a' \rangle_A=a^*a'$ (left action is the same as the homomorphism $\varphi:A \to \mathfrak{B}(X)$ which here is defined via $\varphi(a)(a')=aa'$). Suppose now that another correspondence $(Y,\psi)$ from $A$ to $B$ is given: so $Y$ is a right $B$-Hilbert module and left $A$-module thanks to $\psi$. If we compose our correspondences we obtain the module $A \otimes_{\psi} X$ which has the structure of left $A$-module by $a \cdot (a' \otimes x)=(aa') \otimes x$ and right $B$-module by $(a \otimes x) \cdot b=a \otimes (x \cdot b)$. The $B$-valued inner product on $A \otimes_{\psi} X$ is defined by $\langle a \otimes x,a' \otimes x' \rangle_B=\langle x,\psi(\langle a,a' \rangle_A)x' \rangle_B$. Put $\alpha:A \otimes_{\psi} X \to X, \ \alpha(a \otimes x)=\psi(a)x$. (in purely algebraic seeting, when one deals with genuine unital algebras and unitary modules in the sense that the unit of the algebra act as identity this is a natural isomorphism). One easily checks that it is correctly defined: moreover if we assume that $\psi$ is a *-homomorphism (namely that $\psi$ preserves involution) one can check that $\alpha$ preserves the products $\langle \cdot,-\rangle$. But if we don't assume that $A$ is unital and $\psi$ is unital, I don't see how to conclude that $\alpha$ is an isomorphism. Similar difficulties we get if we compose with $_AA_A$ from the right. MUCH LATER EDIT: This is rather funny coincidence: several days after I posed this question I found the following videos from the Trimester Programm on Noncommutative Geometry- here is the link for the first part https://www.youtube.com/watch?v=3ZGQxLy1ydk Now some my questions which I posed previously (and some which originated while discussing this topic) are no longer current. Let me briefly explain some details: in fact in the definition of a correspondence one assumes that the underlying Hilbert module is nondegenerate, namely that $\psi(A)X$ is dense in $X$-this assumption allows us to prove that $A \otimes_{\psi} X \cong X$ via the map $\alpha(a \otimes x)=\psi(a)x$. However, in order to prove the similar condition from the right one doesn't need any additional assumption: the natural isomorphism $X \otimes_{\varphi} A \cong X$ where $\varphi$ is left multiplication is of the form $\beta(x \otimes a)=x \cdot a$ and now $X \langle X,X \rangle$ is always dense in $X$ so $X \cdot A$ is also dense. This proves (together with my previous remarks) that two Morita equivalent $C^*$-algebras are isomorphic in this category. Now, compatibility with the composition of ordinary morhpsims goes as I explained provided that our morphisms are nondegenerate: $f:A \to B$ is called nondegenerate if $f(A)B$ is dense in $B$. This assumption is somehow natural: the category of all $C^*$-algebras (possibly nonunital) with nondegenerate morphisms as arrows is equivalent to thwe opposite category of locally compact Hausdorff spaces and proper maps. If we relax the assumption that our morphisms are nondegenerate one produces from $f:A \to B$ the correspondence in which $X=\overline{f(A)B}$ which is a right Hilbert $B$-module (with standard inner product) and now the nondegeneracy of the left action follows. Finally one should also note the following: the most natural notion for arrows at the topological level is the notion of continuos map (between locally compact spaces-without the assumption of being proper). One can therefore ask which notion at the $C^*$-algebraic level corresponds to the notion of continuity. It appears that the appropriate notion is the following: the morphism $\varphi:A \to B$ is a nondegenerate *-homomorphism from $A$ to the algebra of multipliers $M(B)$ which is nondegenerate in the sense that $\varphi(A)B$ is dense in $B$. So in the data defining a correspondence there is a $*$-homomorphism $\varphi:A \to \mathfrak{B}(X)$ which is nondegenerate in the sense that $\varphi(A)X$ is dense in $X$. This is equivalent to declaring that $\varphi(A)\mathfrak{K}(X)$ is dense in $\mathfrak{K}(X)$ (the algebra of ''compact'' operators on $X$) and if we recall that $M(\mathfrak{K}(X))-\mathfrak{B}(X)$ we see that the map $\varphi$ being the part of the data defining correspondence is nothing more than morphism $A \to \mathfrak{K}(X)$ is the above sense. I now suspect that given two arbitrary $C^*$-algebras one can always find a correspondence between them however I'm still interested in knowing some examples. Also, I still don't see how the isomorphism in this category implies Morita equivalence (I believe that it is true: I found some short note that this in fact is true, in Connes' Noncommutative Geometry). REPLY [10 votes]: (Apologies for resurrecting a year-old question. I thought it worth at least recording an answer to part 3, since I think this is something that comes up quite a lot.) Question 1: As you've already discovered, the appropriate definition (in this context) is that $\varphi$ should be a nondegenerate $*$-homomorphism from $A$ into the adjointable operators on $X$. Question 2: Yes. Take any correspondence $A\to \mathbb{C}$ (i.e. a nondegenerate $*$-representation of $A$ on a Hilbert space) and compose it with any correspondence $\mathbb{C}\to B$ (i.e. a right $B$-Hilbert module) to get a correspondence $A\to B$. The composition will be nonzero as long as both of the factors are. Question 3: Yes: a $C^*$-correspondence is invertible if and only if it is an imprimitivity bimodule in the sense of Rieffel. For a short proof see: Echterhoff, Kaliszewski, Quigg and Raeburn: A categorical approach to imprimitivity theorems for $C^*$-dynamical systems (Memoirs of the AMS 2006, arXiv:math/0205322, Lemma 2.4). Different proofs can be found in: Echterhoff, Kaliszewski, Quigg and Raeburn: Naturality and induced representations (Bulletin of the Australian Math. Soc. 2000, arXiv:math/0002039, Proposition 2.6) Blecher: A new approach to Hilbert $C^*$-modules (Math. Ann. 1997, Theorem 5.5) As you mention in the comments, for unital algebras one has the stronger result that $C^*$-algebra Morita equivalence coincides with the purely algebraic notion. For this, see: Beer: On Morita equivalence of nuclear $C^∗$-algebras (J. Pure Appl. Algebra 1982, Theorem 1.8).<|endoftext|> TITLE: Why is the CM closure of $\mathbb{Q}$ the "ultimate" coefficient field for motives? QUESTION [20 upvotes]: In a rough way, a category of motives over a field $k$ with coefficients in a field $K$ gives a universal cohomology theory with coefficients in $K$ for algebraic varieties defined over $k$. I had the impression that the choice of coefficients $K=\mathbb{Q}$ was the most standard one and at least the most natural one($\mathbb{Q}$ is the smallest field of characteristic zero). But in this video conference: https://www.youtube.com/watch?v=0M-jXPi_t1I around the time 8.00, Kontsevich says that the "ultimate coefficient field, that you can use in any situation" is $\mathbb{Q}^{CM}$, the union of CM fields, i.e. of imaginary quadratic extensions of totally real number fields. I don't understand why it is true so my question is: Why is $\mathbb{Q}^{CM}$ the "ultimate" coefficients fields for motives? More precisely: 1)I can understand that a bigger coefficient field than $\mathbb{Q}$ can be useful but I would like to know concrete examples of that. 2)I don't know why $\mathbb{Q}^{CM}$ should be enough to cover the eventual answers to 1). I know the usual appearance of CM fields in the theory of complex multiplication of abelian varieties but I don't see a direct relation with the question. Remark: I am deliberately unprecise on what I consider as "motives". If the answer depends on "details" of the notion (pure/mixed, equivalence relation on cycles...), I would like to know it for the various versions. REPLY [9 votes]: I recently learned a fact which seems to answer part of my question: the $CM$ extension $\mathbb{Q}^{CM}$ of $\mathbb{Q}$ is exactly the extension of $\mathbb{Q}$ generated by the Weil numbers (see for example Appendix D of this paper by Drinfeld: http://arxiv.org/abs/1007.4004 ). Recall that $\alpha \in \overline{\mathbb{Q}}$ is a Weil number if is a $q$-Weil number for some $q$ power of a prime, which means that $|\alpha^{\sigma}|^2 =\alpha^{\sigma} \overline{\alpha^{\sigma}}=q$ for every conjugate $\alpha^{\sigma}$ of $\alpha$ over $\mathbb{Q}$. The fact that the field generated by a Weil number $\alpha$ is $CM$ is elementary: $\beta = \alpha + \overline{\alpha}$ is totally real and $\alpha$ is a root of $T^2 - \beta T +q$, of discriminant $\beta^2-4q=(\alpha-\overline{\alpha})^2 \leq 0$. If $X$ is a smooth projective variety over a finite field $\mathbb{F}_q$ then the eigenvalues of the Frobenius acting on the $l$-adic cohomology of $X$ ($l$ prime different from the characteristic $p$ of $\mathbb{F}_q$) are Weil numbers independent of $l$ (Deligne) and equal to the eigenvalues of the Frobenius acting on the crystalline cohomology (Katz-Messing). From the Tate conjecture, all the motivic information about $X$ should be contained in these eigenvalues. This makes clear that $\mathbb{Q}^{CM}$ is a good coefficient for motives over finite fields. The same will be true for motives over global fields as such motive should be determined by its reductions to the various finite places.<|endoftext|> TITLE: A hyperbolic group with a small profinite completion QUESTION [5 upvotes]: Is there a finitely generated non-elementary word hyperbolic group the profinite completion of which is known (or conjectured) to be rather restricted, that is: abelian, pro-$p$, virtually prosolvable, etc. ? REPLY [12 votes]: It's a famous open question whether every word-hyperbolic group is residually finite. Kapovich--Wise showed that this is equivalent to asking whether every non-trivial word-hyperbolic group has non-trivial profinite completion. In the other direction, Agol--Groves--Manning showed that it's equivalent to asking whether every quasiconvex subgroup of every word-hyperbolic group is separable. In particular, since non-elementary word-hyperbolic groups always contain quasiconvex non-abelian free subgroups, if an example were known with virtually prosolvable profinite completion, the above questions would all be answered in the negative. The best conjectural candidates for non-elementary word-hyperbolic groups with restricted profinite completion (in some sense) are cocompact lattices in $Sp(n,1)$ (aka quaternionic hyperbolic lattices). It's unknown whether these have the congruence subgroup property, and if they do, then I believe it again follows that all the above questions are answered in the negative.<|endoftext|> TITLE: Itô's article "A measure-theoretic approach to Malliavin calculus" QUESTION [7 upvotes]: Apart from citations all over the internet, the following paper appears to be off-the-grid. K. Itô, A measure-theoretic approach to Malliavin calculus, in 'New Trends in Stochastic Analysis', Proc. Taniguchi Symposium, Sept. 1994, Charingworth, (eds. K. D. Elworthy, S. Kusuoka and l. Shigekawa), World Scientific, 1997 Does there exist any electronic copy of the article? Or does anyone have a physical copy that they are comfortable with scanning/sharing? REPLY [4 votes]: I have scanned my copy and sent it to the OP.<|endoftext|> TITLE: 2-category theory QUESTION [13 upvotes]: I know that we can do a lot of 2-category theory, seeing 2-categories as Cat-enriched categories. Yet, I know that there are some limitations of this approach. I also know that there are many articles which could help to understand 2-category theory... (I am only familiar with a few of the Lack's, Street's and Kelly's articles so far, but I know there are many more important articles). But always when I'm trying to deal seriously with 2-categories, I end finding serious difficulties. So I am sure I have to improve my 2-category theory knowledge in general. And just recently I have become aware of the Gray's Formal Category Theory. So my question is about basics of 2-category theory: which set of articles or books could be considered as a solid base to start thinking seriously about 2-category theory? I am just trying to avoid a common situation for me: being stuck in a well known (and basic) subject of 2-category theory, ignoring the existence of (classical) literature about this subject. Thank you in advance REPLY [4 votes]: It sounds like the new book from Johnson & Yau "2-Dimensional Categories" might be what you're looking for: https://arxiv.org/abs/2002.06055 While this is not a classical, or perhaps that well-known reference, it does seem to be a comprehensive and detailed take on the subject<|endoftext|> TITLE: Research and exposition: how does writing "basic" books affect your "serious" research work? QUESTION [15 upvotes]: I can see the benefit of writing a mathematical monograph: you revise and organize your own work and recollect the key ideas of your own research. But this applies only to books aimed at researchers and graudate students in a very specific area on which you've worked a lot during your career. I don't fully understand what kind of mathematical benefits writing a "basic" textbook on algebra or analysis could bring to young mathematicians. Could you share your insight and experiences on this matter? REPLY [21 votes]: Let's take three well know examples. Lawvere (set theory), Menger (calculus), MacLane (set theory,categories). All wrote textbooks. All of them indicated that they did it primarily to (a) make more accessible a particular way of approaching some basic material which they would like to see more of, (b) to determine, while writing, what is (in fact) the best current way to approach the basic material if you want to teach it to somebody, and (c) to attempt improving or revising basic notation used in mathematics more to their liking. I suggest that (c) followed by (a) followed by (b) is the ranking of how important these reasons usually are for motivation. One may also want to provide a book reintroducing all the basic material in the most general and systematic way known at the time of writing, as opposed to the usual or original way the concepts were discovered or taught. Since the state of mathematical knowledge evolves over time, authors end up periodically writing general basic monographs. This is a special case of motivation (b). I assume we are discussing books reintroducing all the usual basic material in perhaps a new way (or a way more concise or clear or abstract). There are also quasi-popular books (not the same thing as new general textbooks) meant to attract more people to mathematics (Courant, Penrose), which find a good way to introduce basic or intermediate material creatively to a wider (but still technical) audience; this another motivation (d), and usually goes hand in hand with motivations (a) and (c), but I assume you are not asking about quasi-popular books. In the end, it's the basic books by which an author is known to most individuals, so they improve one's reputation, in fact. This is to say that basic monograph writing is not merely done out of a sense of service to the community, although a desire to do a service is certainly a large part of all motivations (a)-(d) above. In a nutshell, it comes down to the author wanting more people interested in their field and doing work in the field more elegantly. UPDATE: A clarification of (c). Consider https://dx.doi.org/10.1017%2FS0305004100021162: not all notations are equally useful even if they are equally valid. They are different languages, and each makes it more or less difficult to write useless or uninteresting statements and more or less easy or automatic to write meaningful ones. This despite the fact that one could ultimately express what one means in any of these languages with sufficient effort (if they are all coherent ones). Dirac's thesis is that in good or better notations it's difficult to make certain serious errors---and significant ubiquitous statements are easy to process and construct without much effort. (And which errors these are varies together with the subject matter.) So (c) is closely tied with (a), (b), and (d) and the appropriate notations vary together with the subject matter that is the meaning. Using good notations one learns more of what one doesn't already know because one learns more easily. Consider how much reasoning is actually built into a given notation. (Mathematics itself is a constructed language with reasoning built into it.) Yes, we must agree with Popper, it's not useful to argue about words, names or symbols, if their meaning is recorded or communicated, known. But different notations and conventions are quintessentially different compression schemes. What is easy to communicate or to record by means of one is not so easy to communicate or to record by means of another. We recognize that language choice is meaningful only if we consider different languages as dynamical systems, each a more or less appropriate means of doing work that isn't already done, of communicating what isn't already communicated, of recording what isn't already recorded, not as entities that we find in work already done, in records or communications already made. Each communication or record is an output of a language. Which language we use isn't a meaningful choice if we have the output present. There where we've yet to make the output (or are in process of constructing it) the language we use as means to getting output (and so the language we don't use) is a meaningful choice. We get the same end with more or less effort by different means. So we get more or less work done or those we try to communicate a subject matter to learn more or less of what we desire to communicate. They have limited resources for learning or working and what means we use to communicate contributes to determining how easily they learn. We also have limited resources for doing work. Making different outputs is easier by means of different languages. It's easier to learn a subject matter or to do future work in a subject matter by means of one language and not another. So in this case the choice of language is meaningful. Once the work is done or the subject matter is learned the choice of language in which it is expressed isn't meaningful. Yes, which symbols are used to record or communicate a subject matter doesn't really matter once the content is recorded or communicated, constructed, ready. Only the content (meaning) of a statement ultimately determines its truth. Not its presentation. Which notation is used however does matter while constructing statements that communicate or record. Some symbol choices and symbol combination conventions make some subject matters easier to comprehend or work with, record or communicate. Different construction rules make different statements difficult to construct and also to record or to communicate. What is less easily done is less frequently done. So they appear less frequently. A set of notational rules are good or better relative a subject matter if and only if statements they make difficult to construct are typically not true or trivial ones considering this subject matter.<|endoftext|> TITLE: Given a set of 2D vertices, how to create a minimum-area polygon which contains all the given vertices? QUESTION [5 upvotes]: Not sure whether this question belongs here or math.stackexchange. You can assume that all the vertices are unique. The given vertices can be the vertices of the polygon, thus they do NOT have to be inside the polygon, but they must NOT be outside the polygon (and I think to get the minimum-area polygon, they must be the polygon vertices). The polygon can be concave. I am thinking of using convex-hull algorithm as the first step, and then from each edge of the convex-hull polygon, I "dig in" by removing an edge from the aforementioned polygon (let the edge connects vertex a and vertex b), and create 2 new edges (a-c and c-b) where c is a vertex which was previously located inside the polygon. And do it until there is no more edge remaining inside the polygon (i.e all vertices have become the polygon vertices). But I haven't got the "digging in" algo which is proven to minimize the polygon area. As a side question, is this an NP-complete problem? REPLY [13 votes]: I assume you intend the problem in which the polygon's vertices must be exactly the given set of points. If so, then, Yes, the problem is NP-hard: Fekete, Sándor P. "On simple polygonalizations with optimal area." Discrete & Computational Geometry 23.1 (2000): 73-110. (Journal link.)           Approximation algorithms have been explored: Taranilla, María Teresa, Edilma Olinda Gagliardi, and Gregorio Hernández Peñalver. "Approaching minimum area polygonization." (2011): 161-170. (Authors' link.) The key search term for this problem is polygonization.<|endoftext|> TITLE: Products of cyclotomic polynomials QUESTION [5 upvotes]: Is $\Phi_5(z) \Phi_6(z) = 1 + z^2 + z^3 + z^4 + z^6$ the only product of cyclotomic polynomials that has nonnegative coefficients and satisfies $p(\zeta)=0$, $p(\zeta^2)=2$, $p(\zeta^3)=3$, and $p(1)=5$ where $\zeta$ is a primitive 6th root of unity? This question arises from my continuing efforts to understand the cyclic sieving phenomenon (see Evaluating products of cyclotomic polynomials at roots of unity). Orbit-size data come nowhere close to determining the sieving polynomial $p(z)$, but I suspect that imposing extra constraints on the polynomial (specifically that $p(z)$ is a product of cyclotomic polynomials and that all coefficients are nonnegative) might do the trick (or at least reduce the set of candidates to a finite set). This special case seems like a good place to start (though I'd be interested in other cases as well). There are plenty of cyclotomic polynomials $q(z)$ that take the value 1 at $\zeta^2$ and $\zeta^3$ and 1, and any polynomial obtained by multiplying $\Phi_5(z) \Phi_6(z)$ by a product of such polynomials will evaluate to 0, 2, 3, and 5 at $\zeta$, $\zeta^2$, $\zeta^3$, and 1, respectively. But none of the combinations I've tried gives rise to a polynomial with nonnegative coefficients. REPLY [12 votes]: Try $x^{30}+x^{20}+x^{15}+x^{10}+1=\Phi_6\Phi_{30}\Phi_{25}$. For more general solutions see the extended comment below. One way to search for such examples is to solve for the set of polynomials with positive integer coefficients satisfying all the given conditions except being products of cyclotomic polynomials. Then, by studying which roots of unity could possibly be roots of such polynomials, you can start limiting your search. In this case, you get as your set of possible polynomials things of the form $p(x)=x^{n_1}+x^{n_2}+x^{n_3}+x^{n_4}+x^{n_5}$, with $n_1\equiv n_2\equiv 0\pmod{6}$, $n_3\equiv 2\pmod{6}$, $n_4\equiv 3\pmod{6}$ and $n_5\equiv 4\pmod{6}$ (where we might have $n_1=n_2$). Now from the fact that cyclotomic polynomials (besides $x-1$) have leading and constant coefficients which are $1$, we get that $n_1=0$ and $n_1\neq n_2$. From their palindromic nature, we see that $p(x)=1+x^{a}+x^{a+b}+x^{a+2b}+x^{2(a+b)}$. From our previous work we see that $a\equiv 2,4\pmod{6}$ and $a+b\equiv 3\pmod{6}$. Next, we see that the only sums of 5 (possibly non-distinct) roots of unity which sum to 0 are sums of 5th and 6th roots of unity (using the spacing given by $p(x)$, where one of the roots of unity is $1$, and the others are spaced accordingly). Let's deal with the 5th roots of unity first. Suppose when we plug in $\zeta_{m}$, the five terms of $p(\zeta_m)$ are the five distinct 5th roots of unity. This is only possible when $m=5m'$, $m'|n_i$ for each $i$, and $n_i/m$ runs through all of the classes modulo 5. Similarly if we plug in $\zeta_m$, and the five terms of $p(x)$ are 6th roots of unity, we achieve a similar divisibility criterion. So now let $\ell$ be the gcd of all the $n_i$ (or just $\gcd(a,b)$). If $\ell=1$, we get the solution $\Phi_5\Phi_6$ (and no others, since taking powers of $\Phi_5$ or $\Phi_6$ doesn't help). If $p(x)$ is a polynomial with only roots of unity as its roots, and all monomials are divisible by $\ell>1$, then $p(x^{1/\ell})$ has the same property. Further, since $\gcd(\ell,6)=1$, this new polynomial should still satisfy the other criteria. So you can reduce to $\ell=1$. So, a general solution should be $\Phi_5(x^{\ell})\Phi_6(x^{\ell})$ for any $\ell$ relatively prime to $6$.<|endoftext|> TITLE: Structure of the stabilizer of a vertex-neighborhood of a vertex-transitive graph QUESTION [5 upvotes]: Given a simple, undirected graph and a vertex $v$ of the graph, let $L_v$ denote the set of automorphisms of the graph that fixes the vertex $v$ and each of its neighbors. When the graph is vertex-transitive, the vertex-neighborhood stabilizer $L_v$ is independent of the choice of $v$. I noticed that for many vertex-transitive graphs, $L_v$ happens to be either trivial or is isomorphic to the direct product of copies of $C_2$. Thus $L_v$ happens to be isomorphic to $C_2^k$ for some $k \ge 0$. For example, for the modified bubble-sort graph on 24 vertices, $L_v$ is the Klein four-group $C_2 \times C_2$, and for many Cayley graphs generated by transposition sets, $L_v$ is trivial (so $k=0$ in this case) (cf. http://arxiv.org/abs/1205.5199). For the complete transposition graph, $L_v \cong C_2$ (cf. http://arxiv.org/abs/1404.7363). I also considered the Petersen graph, and again $L_v \cong C_2$. Is this a coincidence or is there some result that says that for some families of vertex-transitive graphs or for certain families of Cayley graphs, $L_v \cong C_2^k$ for some $k$? What are some counterexamples - for example, what are some (vertex-transitive) graphs for which $L_v \cong C_3$, say? REPLY [6 votes]: Consider the Kneser graph $K(v,k)$, with vertices the $k$-subsets of $V=\{1,\ldots,v\}$, where the $k$-subsets are adjacent if they are disjoint. Let $\alpha=\{1,\ldots,k\}$ and let $G$ be the subgroup of the symmetric group on $V$ that fixes each element of the complement of $\alpha$. So $|G|=k!$ and $G$ is a subgroup of the stablizer of $\alpha$ that fixes each neighbour of $\alpha$ in the Kneser graph. To get examples as you requested, take $k\ge 3$ (and $v>2k)$. Cayley graphs are not a good place to look, because the stabilizer of a vertex tends to consist of automorphisms of the underlying group, and any automorphism that fixes each element in a connection set is the identity.<|endoftext|> TITLE: PDF of the product of normal and Cauchy distributions QUESTION [5 upvotes]: I am having trouble in finding out the resulting PDF of the product of normal and Cauchy distributions. It turns out that we have a general formula for calculating the PDF of product of two random distribution but the integral is not converging . Also, I tried using Mellin Transform method but it is getting too complicated. If anybody has any idea about how to approach the problem, please share it with me. Thanks REPLY [3 votes]: Since the normal distribution is a $ t $ distribution with $\infty $ degrees of freedom, perhaps you could find the pdf of the product of two $ t $ distributed random variables and then take a limit.<|endoftext|> TITLE: Newton series and Fourier transform - is there an analogy? QUESTION [10 upvotes]: Fourier expansion for a function: $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}\int_{-\infty}^{+\infty}e^{i\omega t}f(t)dt \, d\omega$$ Newton series expansion of a function: $$f(x)=\sum_{m=0}^{\infty} \binom {x}m(-1)^{-m} \sum_{k=0}^\infty\binom mk(-1)^{k}f(k)$$ Is there an analogy? Is it possible to make with Newton's series the same things as with Fourier transform? Are the Fourier transform and inverse Fourier transform analoguous to difference delta and summation operators? REPLY [4 votes]: (This is a revamping of MSE-Q, which has examples of Newton series interpolation for particular functions and the relation to the Mellin transform.) With $ \; \; \displaystyle \bigtriangledown^{s}_n \; c_n=\sum_{n=0}^{\infty}(-1)^n \binom{s}{n} \; c_n \; \; $ and $\; \; \displaystyle m_j= \int_{0}^{\infty} h(x) \; x^j \; dx \; \; $, formally $$g(s)=\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{m_j}{j!}=\int_0^\infty h(x)\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{x^j}{j!} dx=\int_0^\infty h(x) \frac{x^{s-1}}{(s-1)!} \; dx,$$ so since $ \; \; \displaystyle m_j=j!\; g(j+1)=\int_0^\infty h(x) x^j \; dx \; \;$, $$ g(s+1)=\bigtriangledown^{s}_{n} \bigtriangledown^{n}_{j} g(j+1).$$ Identifying $g(x+1)$ with $f(x)$ in your formula, you can see that your Newton series is the Mellin transform in disguise. Performing the modified inverse Mellin transform $h(x)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{sin(\pi s)} g(s) \frac{x^{-s}}{(-s)!} ds$ . For the Fourier transform, the analogous formula is a sinc function interpolation of a bandlimited function $$\hat{g}(x) = \int_{-W/2}^{W/2} h(\omega) e^{i 2 \pi x \omega} d\omega.$$ Then $$\hat{g}(x) = \sum_{n=-\infty}^{\infty} \frac{sin[\pi W\; (n/W-x)]}{ \pi W\;(n/W-x)} \hat{g}(n/W)$$ with $$e^{i2 \pi x \omega}=\sum_{n=-\infty}^{\infty} \frac{sin[\pi W \; (n/W-x)]}{\pi W\; (n/W-x)} e^{i2 \pi n \omega /W} $$ for $-W/2<\omega0$. (This is the basis for the Whittaker-Shannon theorem.) An aside on a related Newton series and Ramanujan's Master Formula: Using umbral notation (MSE-Q) for the Taylor series, $f(t)=e^{a.t}$, the normalized Mellin transform formally gives $$g(s)= \displaystyle \int_{0}^{\infty} e^{-a.t} \frac{t^{s-1}}{(s-1)!}dt$$ $$ = \int_{0}^{\infty} e^{-t}\; e^{(1-a.)t} \frac{t^{s-1}}{(s-1)!}dt =\sum_{m=0}^{\infty} (-1)^m \binom{-s}{m}\sum_{k=0}^m(-1)^k \binom mk a_k \; .$$ Then $f(t)=e^{a.t}$ with $a_n=g(-n)$ since, for $s=-n$, the binomial transform here is an involution and the regularization of the normalized Mellin transform (MT) gives the Dirac delta function or its derivatives in the integrand. Both sides of the equation would converge simultaneously only over restricted values of $s$ as noted in the MO-Q link, but could be analytically continued. (Use $\binom{s+m-1}{m}=(-1)^m \binom{-s}{m}$ for the MT term-by-term of the Taylor series for $e^{(1-a.)t)}$.) This is the essence of Ramanujan's Master Formula/Theorem (cf. MO-Q). So, the Newton interpolation is more akin to the Fourier transform itself; in fact, you can regard the MT here as the interpolation of the coefficients of the Taylor series with the Newton series as one of its avatars. The inverse MT gives, for appropriate $ \sigma$ when convergent, $$\displaystyle \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} g(s) \frac{(-t)^{-s}}{(-s)!} ds = \sum_{n=0}^{\infty} g(-n) \frac{t^{n}}{n!} = f(t) \; .$$<|endoftext|> TITLE: Tilting the $d$-cube to vertically separate its vertices QUESTION [18 upvotes]: Let $C_d$ be a unit edge-length cube in $d$ dimensions. I would like to orient it ("tilt" it) so that the vertical (last) coordinates of its $2^d$ vertices are maximally separated, in the sense that the minimum vertical distance between any two vertices is maximized over all orientations. For $C_2$ (in standard orientation, edges parallel to Cartesian axes), tilting $\arctan \frac{1}{2} \approx 26.6^\circ$ separates the vertices by $\delta=1/\sqrt{5} \approx 0.447$:           For $C_3$ (in standard orientation), I believe that rotating the vector $(0,0,1)$ to lie on the vector $(\frac{1}{4},\frac{1}{2},1)$ results in a vertex separation of $\delta=1/\sqrt{21} \approx 0.218$:           My question is: Q. What is the generalization to $C_d$ for $d>3$? What is the largest vertex separation $\delta$ achievable? Can one always achieve a uniform vertex separation (the same $\delta$ between each vertically adjacent pair), as in $C_2$ and $C_3$? REPLY [6 votes]: Robert Israel's answer is best possible. As pointed out there, the problem is equivalent to finding a vector in $\mathbb{R}^d$ with minimal separation $1$ among sums of subsets of entries, and with the least possible length ($\ell_2$ norm). The vector Robert constructed, after rescaling to make the separation $1$, is $v_d=[2^0,2^1,\dots,2^{d-1}]$. Clearly this has the smallest possible $\ell_1$ norm. It turns out that it also has the smallest possible $\ell_2$ norm, but for $d>3$ not the smallest $\ell_\infty$ norm. Proof. It is a straightforward consequence of these 2 facts: Lemma. The following equality holds in $\mathbb{Z}[x_1,\dots x_d]$: $$\sum_{S,T\subseteq I_d}(x_S-x_T)^2=C_d \sum_{i=1}^d x_i^2$$ where $I_d=\{1,2,\dots d\}$, $\displaystyle x_S=\sum_{i\in S} x_i$ and the positive integer $C_d$ depends only on $d$. if $x=v_d$ then $\{x_S\}=\{0,1,2,\dots 2^d-1\}$ clearly has the tightest possible packing of values (among vectors with separation $1$, and uniquely up to permutations) and therefore must minimize the left hand side in the lemma. Proof of Lemma. There are 16 equally frequent cases for the occurrence of $x_ix_j$ in $(x_S-x_T)^2$: $$\begin{array}{@{}l|l|l|c|c@{}} \mbox{case} & \in S & \in T & x_S-x_T & x_ix_j \text{ coef in }\\ & & & & (x_S-x_T)^2\\ \hline \mbox{1} & \_ & \_ & \cdots & 0 \\ \mbox{2} & i & \_ & x_i\cdots & 0 \\ \mbox{3} & j & \_ & x_j\cdots & 0 \\ \mbox{4} & i,j & \_ & x_i+x_j\cdots & 2 \\ \mbox{5} & \_ & i & -x_i\cdots & 0 \\ \mbox{6} & i & i & \cdots & 0 \\ \mbox{7} & j & i & x_j-x_i\cdots & -2 \\ \mbox{8} & i,j & i & x_j\cdots & 0 \\ \mbox{9} & \_ & j & -x_j\cdots & 0 \\ \mbox{10} & i & j & x_i-x_j\cdots & -2 \\ \mbox{11} & j & j & \cdots & 0 \\ \mbox{12} & i,j & j & x_i\cdots & 0 \\ \mbox{13} & \_ & i,j & -x_i-x_j\cdots & 2 \\ \mbox{14} & i & i,j & -x_j\cdots & 0 \\ \mbox{15} & j & i,j & -x_i\cdots & 0 \\ \mbox{16} & i,j & i,j & \cdots & 0 \end{array}$$ the total coefficient of $x_ix_j$ over all the cases is 0. Therefore $\sum_{S,T\subseteq I_d}(x_S-x_T)^2$ only contains square terms and by symmetry must then be a multiple of $\sum_{i=1}^d x_i^2 \quad\blacksquare$ Regarding the fact that $v_d$ above is not minimal $\ell_\infty$-wise, counterexamples are provided via the Atkinson-Negro-Santoro sequence: $d=4: [3,5,6,7]$ $d=5: [6,9,11,12,13]$ $d=6: [11,17,20,22,23,24]$ etc.<|endoftext|> TITLE: Number of perfect matchings of the Dodecahedron QUESTION [6 upvotes]: This question seems just to be an elementary enumeration problem, but I believe something deeper might be involved: How many perfect matchings does a dodecahedron graph have? Here the dodecahedron graph has 20 vertexes, 12 faces and 30 edges. I know this graph is really a planar graph so one can draw its Kasteleyn orientation and then do the pfaffian calculation, but obviously this is not a clever way(one get a 20 x 20 matrix). This does not utilize the plenty symmetry of the graph. I wonder whether there is some other way that does not use the pfaffian aproach. For example a trick like graph transformation or of a divide-and-conquer style. I am interested in all kinds of solutions to this problem, thanks in advance. further explanations: here by "something deep" I mean: (but not limited to) 1. use the symmetry ($A_5$) to simplify the calculation of the pfaffian, or deduce some other way to enumerate the matching. 2. Tricks like the $Y-\Delta$ transformations that can be generalized to general cellular graphs; 3. By a dived-and -conquer way, I mean divide the matchings into subcases and then derive a recurrence relations. REPLY [2 votes]: Here's a proof I concocted, utilising our favourite graph: Consider the Petersen graph $P$, obtained from the dodecahedral graph $D$ by quotienting out by the antipodal map $\theta : D \rightarrow P$. Now suppose we have a matching $M \subset E(D)$, and label each edge $e \in P$ by the size of the preimage $M \cap \theta^{-1}(e)$. We have assigned each edge of $P$ a label $0$, $1$ or $2$, such that the sum of the labels around each vertex is $2$. Consequently, one can deduce that the $2$-edges are isolated and the $1$-edges form cycles. Since $P$ is non-Hamiltonian and girth-$5$, the only possibilities are: Two 5-cycles of 1-edges; One 8-cycle of 1-edges and one 2-edge; One 6-cycle of 1-edges and two 2-edges; Five 2-edges. In the first case, the preimages (under $\theta$) of the two 5-cycles must be centrally symmetric 10-cycles, which therefore intersect (contradiction). In the second case, removing an edge of $P$ gives a graph with a unique Hamiltonian 8-cycle $C$, which leads to two possible matchings in the preimage. Since $P$ has $15$ edges, this gives $30$ matchings overall. The third case is impossible, since after removing an edge of $P$, we cannot find a $6$-cycle. The fourth case reduces exactly to counting matchings of the Petersen graph. By complementation, this is equivalent to counting partitions of the edges of the Petersen graph into disjoint cycles (the only possibilities being pairs of disjoint $5$-cycles), which is just half of the number of $5$-cycles in the Petersen graph (namely $12$). Hence, we obtain another $6$ matchings in this manner. Consequently, there are $36$ matchings of the dodecahedron.<|endoftext|> TITLE: How to refer to plural of mathematical symbols - with or without an apostrophe QUESTION [13 upvotes]: Which one is correct, $x_i$s or $x_i$'s? Example sentence: The $x_i$s form a sequence. The $x_i$'s form a sequence. REPLY [7 votes]: Shall you use one of these, according to Oxford Dictionaries Online, you should only use an apostrophe "for the sake of clarity", therefore opting for the first option ($x_i$s). There are cases, like this in Statistics, where you can use the plural, for example, for a variable as the p value. You would then write ps to show the use of multiple p values. However, these are different variables, with possibly unrelated calculations (even though the formula is the same). In the case of a sequence, $x_i$ is already a generic term, representing each element or the sequence itself. You can then write: "the sequence $x_i$", or "the sequence $(x_i)$"<|endoftext|> TITLE: automorphisms of local rings vs local change of coordinates QUESTION [7 upvotes]: Let $R$ be a local (commutative, associative) ring over a field of zero characteristic. (My typical examples are $k[[x_1,..,x_p]]/I$, $k\{x_1,.,x_p\}/I$, $C^\infty(\Bbb{R}^p,0)$. If it helps one can assume $R$ to be Henselian.) I'd like to think of the ring automorphisms, $Aut(R)$, (those that act on the field as identity) as the local changes of coordinates, "$Aut(Spec(R))$". The two objects certainly coincide if $R$ is the localization of an affine ring. More generally, let $S=k[x_1,..,x_p]/I$, let $S\subseteq R\subseteq \hat{S}$, the completion with respect to $(x_1,..,x_p)$. Then the two objects coincide for $R$. But for $R=C^\infty(\Bbb{R}^p,0)$ there are endomorphisms which do not come from the local maps of coordinates. See Page 5. (In this particular example one has an endomorphism, not an automorphism. Still, it is not clear that here $Aut(R)=``Aut(Spec(R))"=Aut(\Bbb{R}^p,0)$, the later is the group of germs of local diffeomorphisms). Suppose $R$ is "geometric enough", so that one can speak of $Spec(R)$, its local coordinates, a local change of them. Does every local change of coordinates (that preserves the origin) extend to an automorphism of $R$? For which "geometric" rings rings $Aut(R)=``Aut(Spec(R))"$? (i.e. the group of all the automorphisms of $R$ vs the group of the local coordinate changes in $Spec(R)$.) what is the official notation for the "geometric" subgroup $``Aut(Spec(R))"$ of $Aut(R)$? (The notation $Aut(Spec(R))$ is somewhat heavy/lengthy.) Any paper/review on the state of the art in this direction? REPLY [2 votes]: About $C^\infty(\mathbb R^p,0)$, the ring of germs of smooth functions, I have the following remarks: It's ideal of flat functions is notoriously ill behaved. In the final topology on this ring for the mapping $C^\infty(\mathbb R^p)\to C^\infty(\mathbb R^p,0)$ it is in the closure of zero. There are results available classifying closed ideals, see the book [Tougeron: Ideaux des functions differentiable, Springer 1972]. The exotic automorphism that you describe involves a non-continuous part. The Whitney extension theorem gives an extension operator from Whitney jets to functions, and describes when this extension operator can be chosen continuous. If I remember correctly, this is the case for closed sets which are the closures of their open interiors. A point is not of this class. This is an indication that the exotic automorphisms all come from discontinuous constructions. If you do not insist on rings of germs but on the full rings $C^\infty(M)$ for smooth manifolds $M$, you have perfect duality between the category of manifolds and the these rings. See chapter 8 of this book. As spectrum you have to take the ideals of codimension 1. Ideals of finite codimension have interesting interpretations in terms of differential geometric constructions. See also the thorough treatment of $C^\infty$-rings in the first chapter of the book Moerdijk, Ieke; Reyes, Gonzalo E.: Models for smooth infinitesimal analysis. Springer-Verlag, New York, 1991. x+399 pp. and the characterization of rings of smooth functions of manifolds in this paper.<|endoftext|> TITLE: Mayer-Vietoris sequence for twisted R-homology QUESTION [5 upvotes]: In this paper Ando, Blumberg, Gepner, Hopkins and Rezk define the twisted $R$-Homology of a ring spectrum $R$ together with a map $f \colon X \to R$-$Line$ to be $$ R^f_n(X) = \pi_0(map_R(\Sigma^nR, Mf)) \cong \pi_n(Mf) $$ where $Mf$ is the Thom spectrum associated to the above map. The latter is defined as the colimit of $X \to R$-$Line \to R$-$Mod$ in an $\infty$-categorical sense. If this definition deserves to be called twisted $R$-homology, it should satisfy the corresponding version of the Mayer-Vietoris sequence (using $M(\left.f\right|_A)$, $M(\left.f\right|_B)$ for a decomposition $X = A \cup B$). Why is this true? REPLY [5 votes]: Here is a sketch. First, here is the argument for untwisted homology that I want to base the argument for twisted homology off of. Every categorical thing I say below is $\infty$-categorical by default, e.g. every colimit is an $\infty$-colimit and so forth. The untwisted $R$-homology spectrum of a space $X$ with coefficients in a ring spectrum $R$ is the colimit of the constant diagram $X \to \text{Mod}(R)$ with constant value $R$. Taking $R$-homology spectra defines a functor from spaces to $R$-module spectra which is itself cocontinuous (because colimits commute with colimits), and in particular which sends pushout squares to pushout squares. But $\text{Mod}(R)$, being stable, has the property that pushout squares are also pullback squares. A pullback square of $R$-module spectra can be converted into a fiber sequence of $R$-module spectra, and then we can apply the long exact sequence in homotopy. The argument works essentially without modification for twisted $R$-homology, except that the domain category is no longer spaces but, say, pairs of a space $X$ and a local system of $R$-module spectra on $X$ (there is no particular reason to restrict our attention to local systems of $R$-lines). Again taking $R$-homology is a cocontinuous functor and hence again sends pushout squares to pushout squares, which again are also pullback squares.<|endoftext|> TITLE: Is there a known Legendrian simple link? QUESTION [5 upvotes]: Several knots like unknot, $4_1$, $3_1$ are known to be Legendrian simple, i.e., Thurston-Bennequin number and rotation number determine Legendrian type completely. How about the same notion for link cases of more than two components? In this cases, of course, we may consider those numbers in a component-wise manner. Even for the most simplest link, i.e., Hopf link cases, I couldn't find a literature for Legendrian simplicity for that. Is there a reason not to consider Legendrian simple links? REPLY [7 votes]: The question for torus links has been studied by Jennifer Dalton in her PhD thesis, Legendrian torus links, in which she proves that, in fact, positive torus links are Legendrian simple, and all torus links are Legendrian simple if you allow reshuffling of the components. Similar questions have also been studied by Lenny Ng and his student, Wutichai Chongchitmate: they started the Legendrian link atlas. I would assume that they use Heegaard Floer invariants coming from grid diagrams to establish non-destabilisability of Legendrian links, but I can't find a reference to that in there...<|endoftext|> TITLE: Ring structure on the $K(1)$-local homotopy of $S^0$ at the prime 2 QUESTION [11 upvotes]: Let's write $S$ for the $K(1)$-local sphere at the prime 2. Then there is a cofibre sequence $$S \to KO \to KO$$ where I'm using $KO$ to denote the $K(1)$-localization of orthogonal K-theory, and the self map of $KO$ is given in terms of Adams operations: $\psi^3-1$. Drew Heard reminded me that on $\pi_1$, $\psi^3-1=0$ is the trivial endomorphism of $\mathbb{Z} / 2 = \pi_1 KO$; this yields the computation $$\pi_0(S) = \mathbb{Z}_2 \oplus \mathbb{Z} / 2,$$ where the 2-adic factor comes from $\pi_0 KO$, and the $\mathbb{Z} / 2$ gets carried around from $\pi_1 KO$ by the connecting map in the cofibre sequence. My question is: what is the ring structure on $\pi_0(S)$? It is certainly a $\mathbb{Z}_2$-algebra, so this describes most of the multiplication. If I write $x$ for an additive generator of $\mathbb{Z} / 2$ (the image of the connecting map), then the only remaining computation is a formula for $x^2$. Since $2x=0$, $2x^2=0$, so it must be the case that either $x^2=0$ or $x^2=x$, since these are the only 2-torsion elements of the ring. Is it known which of these two possibilities holds? REPLY [12 votes]: Here is another method which is in some sense more direct. Consider the diagram $$ \begin{array}{ccc} L_{K(1)}S & \xrightarrow{i} & KO \\ i \downarrow & & \downarrow (1,\psi^3) \\ KO & \xrightarrow{(1,1)} & KO\times KO \end{array} $$ This is a commutative diagram of maps of commutative ring spectra. Using the fact that $L_{K(1)}S$ is the fibre of $\psi^3-1$, one can check that it is homotopy cartesian. Whenever you have such a square $$ \begin{array}{ccc} A & \xrightarrow{i} & B \\ j \downarrow & & \downarrow k \\ C & \xrightarrow{l} & D, \end{array} $$ one can check that $\ker(i).\ker(j)=0$. In the relevant case we have $i=j$ and $x\in\ker(i)$ so $x^2=0$.<|endoftext|> TITLE: MathScinet reviewing: should I check the proofs? QUESTION [21 upvotes]: The guidelines do not clearly specify how much work one is assumed to do in writing a MathSciNet review. What is the consensus? Is one expected to read the whole paper and check the the proofs in detail? REPLY [18 votes]: I echo the answer that you are not expected to check the proofs as a MathSciNet reviewer. That said, I personally choose to do so, because I believe that this leads to better reviews. When I read a review I want it to tell me what to expect from the paper, and if there is a gaping hole I want to know that fact before I spend hours stuck on it. Of course, this is assuming that the paper is of somewhat moderate to small length, which has often been the case in the papers I've reviewed. I also feel like putting my name on a review gives it my stamp of approval, and for me that stamp means something more than just looking over a paper quickly. It might surprise you that I've found many errors in papers I've been asked to review. Some unfixable and central to the paper, some small and unfixable, lots of minor typos (which I don't mention in my reviews unless there are just gobs of them), and in one case an error that was quite subtle and it took a good week for me to find a fix. (I added the fix to my review, but only at the recommendation of the author of the paper.) Of course, you better be sure there is an error (perhaps by communicating with the author if appropriate) before mentioning it in a review. In all these cases, I hope that the people who read my reviews find them helpful.<|endoftext|> TITLE: Is there a manifold with fundamental group $\mathbb{Q}$? QUESTION [28 upvotes]: It is known that the fundamental group of a locally path connected, path connected compact metric space is finitely presented or uncountable. Furthermore the fundamental group of every manifold is countable so the fundamental group of every compact manifold must be finitely presented. Q1: Is there a non-compact manifold whose fundamental group is isomorphic to $\mathbb{Q}$, the group of rational numbers?   Q2: Or more general, for given countable group $G$, is there a manifold whose fundamental group is isomorphic to $G$? REPLY [31 votes]: This was answered in the comments. The answer to both questions is "Yes". As Fernando Muro mentioned, any countable group is the fundamental group of a manifold that can be embedded in $\mathbb{R}^5$. There is an open subset of $\mathbb{R}^3$ with fundamental group $\mathbb{Q}$, the complement of a particular solenoid. Let $C_1$ be the complement of an unknotted solid torus. $\pi_1(C_1) = \langle x_1 \rangle$. Consider an unknotted solid torus inside the first which wraps around it twice. Let the complement of this be $C_2$. There is an inclusion $\pi_1(C_1) \hookrightarrow \pi_1(C_2) = \langle x_2 \rangle$ so that $x_1 = x_2^2$, or in additive notation, $x_1 = 2x_2$. Let the $n$th solid torus wrap around the inside of the $n-1$st solid torus $m_n=n$ times. Let $C_n$ be the complement of this solid torus. There is an inclusion $\pi_1(C_{n-1}) \hookrightarrow \pi_1(C_n) = \langle x_n \rangle$ so that $x_{n-1} = n x_n$. The solenoid is the intersection of this infinite sequence of nested solid tori. The intersection of the solenoid with a meridianal disk of the first solid torus is a Cantor set. The solenoid can be viewed as a mapping cylinder of an automorphism of a Cantor set. The complement $C$ is the union of the complements. Any loop in the complement is contained in some $C_n$. The fundamental group is the direct limit, isomorphic to $\mathbb{Q}$ with $x_n = 1/n!$. If you choose $m_n=2$ instead of $m_n=n$, the solenoid complement has a fundamental group isomorphic to the dyadic rationals. You can select other subgroups of $\mathbb{Q}$ by varying the sequence $( m_n )$.<|endoftext|> TITLE: History of the Jaccard distance $d(A,B) = \mathbb P(\overline A\cup\overline B\mid A\cup B)$ QUESTION [11 upvotes]: I'm wondering where the relative probabilistic distance or Jaccard distance was first studied: $$d(A,B) =\mathbb P(\overline A\cup\overline B\mid A\cup B)$$ where $\overline A$ is the complement of $A$. A web search turned up this: @TechReport{Yianilos91, author = "Peter N. Yianilos", title = "Normalized Forms for Two Common Metrics", institution = "NEC Research Institute", year = {1991,2002} } which contains a detailed proof that $d $ obeys the triangle inequality, but surely that was discovered prior to 1991? On a seemingly related note, Cathy O'Neil mentions at 5:00 in Deciphering recommendation engines, http://youtu.be/lzavwJy1SgQ that $$\mathbb P(A\mid A\cup B)$$ is a mathematically interesting notion of closeness. REPLY [4 votes]: If I remember correctly, it was by Menger, in this series of papers. Menger, 1942(12), statistical metrics, PNAS 28(12):535-537. Menger, 1951(3), probabilistic theory of relations, PNAS 37(3):178-180. Menger, 1951(4), probabilistic geometry, PNAS 37(4):226-229. Menger et al, 1959(au), probabilistic metrics and numerical metrics with probability, czechoslovakian mathematical journal 9(3):459-466. If each pair of coordinates points is supposed to nowhere exist strictly a distance $x$ apart, for all possible $x$, and they only have a probability $z$ of being separated by such a distance (or by a lesser distance), then special cases exist, one per distinguishable $x$:---probabilities for one and only one $x$ are defined in each case and undefined for other distances. There the $z$ corresponding to each pair of points is itself FAPP a distance between any two such points. From each such metric, at most one further constraint is required for a metric of the sort in the question title to result. [The formal part of this answer, I agree it needs rewriting, temporarily gone. Also, I'm checking another possible source, slightly more recent, where the desired metric is explicitly stated unless my memory is wrong.]<|endoftext|> TITLE: On $e^{\pi\sqrt{4\cdot163}}$ and unusual connections QUESTION [13 upvotes]: We are familiar with the expansion of the j-function, $$j(\tau) = \tfrac{1}{q}+744+ 196884{q} + 21493760{q}^2 + \dots\tag1$$ and maybe with the approximation, $$e^{\pi\sqrt{652}} = (640320^3+744)^2-2\cdot196883.999999999918\dots$$ but can somebody give a short, non-specialist explanation (if it is even possible), why the relationship $a\approx b$ between, $$\begin{aligned} \log(196883)\; &\approx 12.19 = a\\ 4\pi\; &\approx 12.56 = b \end{aligned}$$ is suddenly mentioned, of all places, in quantum gravity? (Witten's paper is here.) $\color{brown}{Edit:}$ (To address possible comments) Witten defines a certain function $Z_k(q)$ in page 30, and for the first few $k$, $$\begin{aligned} J(q) = Z_1(q) &= q^{-1}+196884q+\dots\\ Z_2(q) &= q^{-2}+1+42987520q+\dots\\ Z_3(q) &= q^{-3}+q^{-1}+1+2593096794q+\dots\\ Z_4(q) &= q^{-4}+q^{-2}+q^{-1}+2+81026609428q+\dots \end{aligned}$$ On a hunch, I used Mathematica's Integer Relations and checked these coefficients with the coefficients $c_n>1$ of $J(q)$, $$c_n =196884, 21493760, 864299970, 20245856256,\dots$$ (OEIS A014708) and, sure enough, they were just simple linear combinations, $$\begin{aligned} 196884\; &=c_1\\ 42987520\; &= 2c_2\\ 2593096794\; &= c_1+3c_3\\ 81026609428\; &= c_1+2c_2+4c_4 \end{aligned}$$ Using the general formula at the bottom of p.34, $$\begin{aligned} \log(c_1)\;&\approx 12.19\\ 4\pi\sqrt{1}\; &\approx 12.56\\[2.5mm] \log(2c_2)\;&\approx 17.57\\ 4\pi\sqrt{2}\; &\approx 17.77\\[2.5mm] \log(c_1+3c_3)\;&\approx 21.67\\ 4\pi\sqrt{3}\; &\approx 21.76\\[2.5mm] \log(c_1+2c_2+4c_4)\;&\approx 25.12\quad\quad\quad\quad\\ 4\pi\sqrt{4}\; &\approx 25.13\\ \end{aligned}$$ and the paper states that "...agreement improves rapidly if one increases k..." for the Bekenstein-Hawking entropy. (Whatever that is.) REPLY [5 votes]: I'm not an expert on black holes, but I can give you a couple pointers. From work of Bekenstein and Hawking in the 1970s, we are pretty sure that macroscopic black holes in our 3+1 dimensional universe behave like thermodynamic objects. They have temperature, and entropy (reflecting some hidden microstates), and the entropy is proportional to the surface area of the event horizon. Since the derivation of their formula did not use quantum gravity, one expects that quantum corrections become relevant when one considers very small black holes. The black holes that appear in this question live in $AdS_3$ space, which is a 2+1 dimensional spacetime that has $SL_2(\mathbb{R})$-geometry (which is kind of negatively curved). While this universe is quite different from our own, one obtains an entropy versus surface area relationship for black holes by similar reasoning, and again one expects some quantum corrections to show up in the small entropy regime. Explicit black hole solutions to Einstein's equations were found by Bañados, Teitelboim, and Zanelli, and they were found to have event horizons with positive surface area (which is really circumference when we consider 2+1 dimensions). When quantum gravity is brought into the picture, the sizes of possible black holes become quantized. Following AdS/CFT, Witten conjectures that size corresponds to conformal weight of a primary field, and this is why you see the formula $4\pi\sqrt{k}$. The near-integer behavior of $e^{2\pi \sqrt{163}}$ does not seem particularly connected to any of this. It is basic class field theory, with some Hecke operators. See Chapter 3 (I think) of Silverman's Advanced Topics and this MathOverflow question.<|endoftext|> TITLE: Examples of component crossing between families of modular forms QUESTION [5 upvotes]: Is there a reference that contains explicit examples of component crossing of Hida families at height one primes? The paper of Emerton, Pollack, and Weston addresses component crossing obtained through level raising. I am interested in examples caused by other phenomena (e.g. a CM family meeting another CM family coming from different imaginary quadratics or a CM family meeting a non-CM family.) A question was previously asked that suggests such families exist: Example of a non-smooth irreducible component of the generic fibre of a Hida family? I am unable to find examples in the literature, but it's possible that I've looked in the wrong places. REPLY [5 votes]: You can write down explicit examples of such crossing with Eisenstein series. If one takes the $p$-adic family of Eisenstein series $E^{(p)}_k(\chi_1,\chi_2)$ and the family $E^{(p)}_k(\chi_2,\chi_1)$, then one sees explicitly (just look at $q$-expansions) that these families meet in weight 1 --- i.e. the order of the characters doesn't matter in weight 1. In the special case when $\chi_1$ is quadratic, $\chi_2$ is trivial and $\chi_1(p)=1$, there should even be some explicit CM family which also specializes in weight 1 to $E^{(p)}_1(\chi_1,1)$ coming from the the quadratic field cut out by $\chi_1$.<|endoftext|> TITLE: Inner and extendible automorphisms of C*-algebras QUESTION [12 upvotes]: If an automorphism $\alpha$ of a C*-algebra $A$ is inner then whenever $A$ is a subalgebra of another C*-algebra $B$, $\alpha$ obviously extends to $B$. Is the converse true: if an automorphism $\alpha$ of $A$ is such that whenever $A \subset B$ then $\alpha$ extends to an automorphism of $B$, is $\alpha$ necessarily inner? The analogous question for groups has a positive solution, see: Are the inner automorphisms the only ones that extend to every overgroup? REPLY [6 votes]: This is by no means a full answer, but Kishimoto has shown in Theorem 4.1 of his paper "Universally weakly inner one-parameter automorphism groups" that for an automorphism $\alpha$ of a separable $C^*$-algebra the following statements are equivalent: (1) $\alpha$ is extendible in every irreducible representation (2) $\alpha$ is universally weakly inner If I read the last property correctly, then this means that there is a $u \in A^{**}$ such that $\alpha(a) = uau^*$.<|endoftext|> TITLE: Sierpinski-like spaces QUESTION [6 upvotes]: Let $\mathbb{S}$ be the Sierpinski space, that is $\mathbb{S}$ has $\{0,1\}$ as a base set, and $\tau = \{\emptyset, \{0\}, \{0,1\}\}$ as a topology. The Sierpinski space $\mathbb{S}$ has the following property: $(\star)$ Given any topological spaces $X,Y$ and a function $f:X\to Y$, then $f$ is continuous if for every continuous map $z:Y\to \mathbb{S}$ the map $z\circ f: X\to\mathbb{S}$ is continuous. (Proof. If $f: X\to Y$ is not continuous, then there is $V\subseteq Y$ open such that $f^{-1}(V)$ is not open in $X$. Defining $z_V: Y\to \mathbb{S}$ as mapping $V$ to $0$ and $Y\setminus V$ to $1$ we immediately see that $z_V\circ f: X \to Z$ is not continuous.) If a topological space $S$ has property $(\star)$ then we call it Sierpinski-like. (Maybe there is some standard terminology, but I wasn't able to find it.) Soft question: How can Sierpinski-like spaces be characterized? Concrete question: Do Sierpinski-like spaces have to be $T_0$? Or do they necessarily contain a non-empty open set $U_0$ such that $U_0\subseteq U$ for all open non-empty $U$? REPLY [7 votes]: The following are equivalent: $S$ is Sierpiński-like. $\mathbb S$ embeds in $S$ as a subspace. $S$ is not symmetric (i.e., $R_0$). $2\to1$ follows from the facts that $\mathbb S$ is Sierpiński-like, and if $Y\subseteq Z$ is a subspace and $f\colon X\to Y$, then $f$ is continuous as a mapping $X\to Y$ iff it is continuous as a mapping $X\to Z$. $2\leftrightarrow3$ is a restatement of the definition of $R_0$. $1\to3$: If $S$ is $R_0$, then the points in the range of any continuous mapping $z\colon\mathbb S\to S$ are topologically indistinguishable, hence $z\circ f$ is continuous for any mapping $f\colon X\to\mathbb S$. Thus, taking any discontinuous mapping $X\to\mathbb S$ shows that $S$ is not Sierpiński-like.<|endoftext|> TITLE: A well-founded relation on lists QUESTION [11 upvotes]: Let $A$ be a set equipped with a well-founded relation $<$, let $LA$ be the set of finite lists of elements of $A$, and define a relation $\prec$ on $LA$ such that $\ell \prec m$ if $\ell$ is obtained from $m$ by replacing some element $x\in A$ occurring in $m$ with some finite list (perhaps empty) whose elements are all $ TITLE: Can homotopy pullbacks of spaces be checked on fibers? QUESTION [6 upvotes]: As should be clear, I would like to know if it is true that a given commmutative square of spaces (i.e. simplicial sets) is a homotopy pullback iff the induced map on each homotopy fiber is a weak equivalence. More precisely, consider the following diagram: $$\begin{array}{c}&&&&& A& \longrightarrow & B\\ &&&&&\downarrow && \downarrow \\ &&&&& C &\longrightarrow & D \\ &&&&\nearrow & &\nearrow\\ &&&1 & \longrightarrow & 1 \end{array}$$ Suppose that, after having taken homotopy pullbacks on both sides, the resulting front face is a homotopy pullback (i.e. the top horizontal arrow [between homotopy fibers] is a weak equivalence), and that this happens for any vertex of $C$. Is it true that the square involving $A,B,C,D$ is a homotopy pullback? This seems to appear quite often but I couldn't find a precise reference with a proof. Thanks in advance for your help. REPLY [2 votes]: If you are looking for a reference, Propostion 3.3.18 of Munson Volic's "Cubical homotopy theory" (available at http://palmer.wellesley.edu/~ivolic/pdf/Papers/CubicalHomotopyTheory.pdf) is doing what you want.<|endoftext|> TITLE: The periodic values in Bott periodicity QUESTION [10 upvotes]: After Bott periodicity is proved, one still has to compute the stable values. For the unitary group $U$, this is easy since you can get away with just $\pi_0$ and $\pi_1$. However, I'm having trouble doing this for the orthogonal group. It's easy to work out $\pi_i(O)$ for, say, $0 \leq i \leq 2$, but already for $i=3$ it takes some work. Can anyone either give me a reference or proof for the computation of $\pi_i(O)$ for $0 \leq i \leq 7$? REPLY [9 votes]: A nice way to do this calculation can be extracted from the papers [ABS] Atiyah, Bott, Shapiro: ''Clifford modules'' [AtKR] Atiyah: ''K-Theory and Reality''. Let $A_n$ be the group of $Cl^n$-modules, modulo $Cl^{n+1}$-modules, as introduced by Atiyah, Bott, Shapiro. The sum of these is a graded ring, and the main result of [ABS] is that the homomorphism $$abs:A_{\ast} \to KO^{-\ast}(\ast)$$ they construct is a ring isomorphism. That $A_{\ast} = Z [\eta, \lambda, \beta] /(2 \eta, \eta^3, \lambda \eta, \lambda^4- 2 \beta)$ is done using linear algebra in [ABS]. The original proof in [ABS] used the known structure of $KO^{-\ast}$. In [AtKR], a simple proof of real periodicity is given, and using ideas from [AtKR], one can give a proof that $abs$ is an iso without using that knowledge, and thereby compute the $KO$-groups. The ingredients one needs are $8$-periodicity, or more precisely, that multiplication by $abs (\beta )\in KO^{-8}$ is an isomorphism. The knowledge of the complex $K$-groups. That the complexification $KO^0 \to K^0$ is an isomorphism. That $KO^{-1} =Z/2$, generated by $abs(\eta)$. This is the easy fact that $O(n)$ has two components, in disguise. The long exact sequence (proven in [AtKR], § 3) $$\ldots KO^{1-q} \stackrel{\eta}{\to} KO^{-q} \to K^{-q} \to KO^{2-q} \ldots,$$ the map to complex $K$-theory is complexification, and we write $\eta:= abs (\eta)$ and use the same letter for the multiplication by $\eta$. First look at the piece $$ K^{-3} \to KO^{-1} \to KO^{-2} \to K^{-2} \to KO^0 \to KO^{-1} \to K^{-1} \to KO^1\to KO^0 \to K^0$$ As $KO^0 \to K^0$ is an iso and $K^{-1}=0$, you get that $KO^1 =0$. Since $KO^{-1}=Z/2$, the map $Z= K^{-2} \to KO^{0} = Z$ must be multiplication by $\pm 2$ and hence be injective. Therefore, multiplication by $\eta$ is surjective $KO^{-1} \to KO^{-2}$. Since $K^{-3}=0$, it is also injective and you get $KO^{-2} = Z/2$, the element $\eta^2 $ is nonzero. Continue with $$ 0= K^{-5} \to KO^{-3} \to KO^{-4} \to K^{-4} \to KO^{-2}\to KO^{-3} \to K^{-3}=0$$ Hence $\eta: KO^{-2} \to KO^{-3}$ is surjective, but as the only nonzero element of $KO^{-2}$ is $\eta^2$ and since $\eta^3 =0$ (this follows from the corresponding relation in the algebraic model $A_{\ast}$), $KO^{-2} \to KO^{-3}$ is also null. Thus $KO^{-3}=0$. It follows that $KO^{-4}=Z$, and that $KO^{-4} \to K^{-4}$ is multiplication by $\pm 2$. The last two portions of the long exact sequence are $$ 0=K^{-7} \to KO^{-5} \to KO^{-6} \to K^{-6} \to KO^{-4} \to KO^{-5} \to K^{-5}=0 $$ and $$ KO^{-8} \stackrel{\cong}{\to} K^{-8} \to KO^{-6} \to KO^{-7} \to K^{-7}=0. $$ We have seen that $KO^{-7}=KO^1 =0$, thus $KO^{-6}=0$ by the second sequence. This shows that $KO^{-5}=0$ (use first sequence) and completes the argument.<|endoftext|> TITLE: Elliptic operator on non compact manifolds with ends of the type $\Omega\times (r,\infty)\times\mathbb{R}$ QUESTION [9 upvotes]: A smooth manifold $M$ is a manifold with a cylindrical end if there exists a compact subset $K\subset M$ such that $M\backslash K$ is diffeomorphic to $\Omega\times (r,\infty)$ where $\Omega$ is a compact manifold. This is a special class of non-compact manifolds where elliptic theory of differential operators is well-understood, Sobolev embeddings, Hodge theory, Fredholm theory are already know, standard references are: i) R.B. Lockhart and R.C. McOwen. Elliptic differential operators on noncompact manifolds. Ann.Scuola Norm. Sup. Pisa Cl. Sci., 12 (1985), 409–447. ii)R.B. Lockhart. Fredholm, Hodge and Liouville theorems on noncompact manifolds.Trans. Amer. Math. Soc., 301:1–35, 1987. iii) V.G. Maz’ya and B.A. Plamenevski ̆ Estimates in Lp and Holder classes and Miranda-Agmon maximum principle for solutions of elliptic boundary value problems in domains with singular points on the boundary. Math. Nachr., 81:25–82, 1978. I am looking for references of elliptic operators on non compact manifolds such that outside a set $K$ (possibly non compact) the space $M\backslash K$ is diffeomorphic to $\Omega\times (r,\infty)\times \mathbb{R}$. This case is more general than the cylindrical case but close to it and it seems natural to consider such spaces, however I have not been able to find any references. Any reference or comment you could provide about elliptic operators on this class of manifolds would be very useful. Thanks!! REPLY [5 votes]: After looking carefully for this type of manifolds I found that these manifolds are called manifolds with edges. Consider the manifold $M=\Omega\times[0,\infty)\times\mathbb{R}^{N}$ the boundary of this manifold is a fibration over $\mathbb{R}^{N}$ i.e. $\partial M=\Omega \times \left\{ 0 \right\}\times\mathbb{R}^{N}$. This is the total space of a fibration over $\mathbb{R}^{N}$ with fiber $\Omega$. Also each $y\in\mathbb{R}^{N}$ has a conical fiber $\Omega\times[0,\infty)$ so at infinity the manifold $M$ looks like a fibration of conical submanifold over $\mathbb{R}^{N}$. The stretched manifold associated to $M$ is the blow up of the tip of the cone i.e $\tilde M=\Omega\times(0,\infty)\times\mathbb{R}^{N}$ and at infinity it is a fibration of cylindrical submanifolds. It seems there are several approaches to this kind of manifolds and differential operators on them. On one hand we have the approach of Richard Melrose: i) Differential analysis on manifolds with corners. Richard Melrose. (unfinished book). This book explain the setting of manifold with corners and edges. Unfortunately it is not finished and it does have the analysis of edge PDO yet. However some theorems are spread in many paper of R. Mazzeo. On the other hand we have the approach of B-W Schulze. ii) B.-W. Schulze. Pseudo-Differential Operators on Manifolds with Singularities. North-Holland, Amsterdam, 1991. This looks to me a more complete and developed theory of PDO (actually $\psi$DO) on manifolds with edge singularities. Moreover B.-W. Schulze has many other books dealing with BVP and lots of applications of his theory: iii)Ju.V. Egorov and B.-W. Schulze. Pseudo-Differential Operators, Singularities, Applications. Birkhäuser Verlag, Basel, 1997. iv) B.-W. Schulze. Boundary Value Problems and Singular Pseudo-Differential Operators. J. Wiley, Chichester, 1998. v)G. Harutyunyan and B.-W. Schulze. Elliptic Mixed, Transmission, and Singular Crack Problems. EMS Tracts in Mathematics Vol.4, European Math. Soc Just to mention some of them<|endoftext|> TITLE: Forcing as a replacement of induction and diagonal arguments QUESTION [33 upvotes]: Let me give some examples motivating the question. The use of forcing instead of induction: For this consider Cantor's theorem: Theorem 1. Any two countable dense linear orders $I, J$ without end points are order isomorphic. Proof. Let $\mathbb{P}$ be the partial order consisting of partial finite order preserving maps from $I$ to $J$. For $i\in I, j\in J$ the sets $D_i=\{ p: i\in dom(p)\}$ and $R_j=\{p: j\in range(p) \}$ are dense, and since we have only countably many dense sets, we can get a filter $G$ which meets all of these dense sets. Then $\bigcup G: I \to J$ is the required isomorphism. The use of forcing instead of diagonal arguments: This time let's consider another result of Cantor: Theorem 2. For any regular cardinal $\kappa, 2^\kappa > \kappa.$ Proof. Let $F$ be a family of functions from $\kappa\to 2$ of size $\kappa.$ Let $Add(\kappa, 1)$ be the Cohen forcing for adding a new subset of $\kappa,$ and for $f\in F,$ let $D_f=\{p: p\neq f \}.$ Each $D_f$ is easily seen to be dense in $Add(\kappa, 1)$, and since the forcing is $\kappa-$closed, there is an $F-$generic filter $G$ over $Add(\kappa, 1).$ Then $\bigcup G: \kappa \to 2$ is different from all $f\in G.$ Theorem 3. If $\lambda=cf(\kappa) < \kappa,$ then $\kappa^\lambda > \kappa.$ Proof. Let $F$ be family of size $\kappa$ of functions from $\lambda\to \kappa.$ Let $\mathbb{P}=\{p:p$ is a partial function of size $<\lambda$ from $\lambda\to \kappa \}.$ It is $\lambda-$closed. Let $(\kappa_\alpha: \alpha<\lambda)$ be increasing cofinal in $\kappa,$ let $F_0 \subseteq F_1 \subseteq ... (\alpha<\lambda)$ with $|F_\alpha|=\kappa_\alpha$ and $F=\bigcup_\alpha F_\alpha.$ Now consider dense sets $D_\alpha=\{p: \forall f\in F_\alpha, p\neq f \}.$ If a filter $G$ meets all $D_\alpha$'s, then $\bigcup G$ is different from all elements of $F$. Below are a few more examples of the results that can be proved in a similar way: (A) If $\mathcal{A}$ and $\mathcal{B}$ are countable families of subsets of $\mathbb{N}$ such that $A \cap B$ is finite for each $A\in \mathcal{A}$ and $B\in \mathcal{B}$, then there is a $C$ such that $A \subseteq^* C$ for every $A\in \mathcal{A}$ and $B\cap C$ is finite for every $B\in \mathcal{B}.$ (B) There exists a continuous, nowhere dieffrentiable function on $[0,1].$ (C) Let's call a set $A$ of natural numbers (not including 0) is small, if $\Sigma_{n\in A}1/n < \infty.$ Given a countable family $\mathcal{F}$ of small sets, there exists a small set $J$ such that $I\subseteq^* J,$ for all $I\in \mathcal{F}.$ (ِِD) All consequences of $MA(\aleph_0),$ in particular the Baire category theorem,.... Question. Are there any more non-trivial results which are proved using forcing in the following way: we produce a forcing notion $\mathbb{P}$ and a family $\mathcal{D}$ of dense subsets of it. Then we argue that there must be a $\mathcal{D}-$generic filter $G$ over $\mathbb{P},$ and use $G$ to conclude our required result. REPLY [6 votes]: There is a forcing proof of Sierpinski's theorem that $\mathbb{Q}$ with its usual topology is the unique countable, metrizable space without isolated points. The argument is analogous to the proof of Cantor's theorem given in the OP. In both cases, one uses a forcing poset to mimic a back-and-forth construction. Theorem (Sierpinski). If $X$ and $Y$ are countable, metrizable spaces without isolated points, then $X$ is homeomorphic to $Y$. Proof. Fix metrics for $X$ and $Y$. Since these spaces are countable, most of their closed balls are also open. Hence we may fix a base of clopen sets $\langle O_i: i \in \omega \rangle$ for $X$ and similarly $\langle N_j: j \in \omega \rangle$ for $Y$. The idea is to define a poset whose conditions approximate some bijection $h: X \rightarrow Y$, and arranges that $h[O_i]$ will be open in $Y$ and $h^{-1}[N_j]$ will be open in $X$ for every $i,j$. Let $\mathbb{P}$ be the poset with conditions of the form $(f, m, P, Q)$, where $P$ is a finite partition of $X$ into clopen sets, $Q$ is a finite partition of $Y$ into clopen sets with $|Q| = |P|$, $m$ is a bijection of $P$ with $Q$, $f$ is a finite partial injection from $X$ into $Y$ that respects the matching $m$ (i.e. if $x \in$ dom($f$) and $p$ is the unique piece of the partition $P$ containing $x$, then $f(x)$ is in $m(p)$). A condition $(f', m', P', Q')$ extends $(f, m, P, Q)$ if $P'$ refines $P$, $Q'$ refines $Q$, $m'$ refines $m$ (in the obvious sense), and $f'$ extends $f$. I claim that for every $x \in X$, $y \in Y$, and $i, j \in \omega$, the set of conditions $D_{x, y, i, j} = \{(f, m, P, Q)$: $x \in$ dom($f$), $y \in$ ran($f$), $O_i$ is a union of pieces in $P$, $N_j$ is a union of pieces in $Q\}$, is dense in $\mathbb{P}$. To see this, fix a condition $(f, m, P, Q)$. Given $x \not \in$ dom($f$), there is a unique piece $p_x$ in $P$ containing $x$. Since $Y$ is without isolated points, $m(p_x)$ is infinite. Pick $z \in m(p_x)$ not already in ran($f$). Similarly, given $y \not \in$ ran($f$), we may find $w \in m^{-1}(q_y)$ not already in the dom($f$), where $q_y$ is the unique piece of $Q$ containing $y$. Then $f' = f \cup \{(x, z), (w, z)\}$ extends $f$ and still respects $m$. Now, any fixed $O_i$ splits every partition piece $p$ into two clopen sets, $p_1 = O_i \cap p$ and $p_2 = (X \setminus O_i) \cap p$. Let $P'$ be the partition consisting of all sets of this form (ignoring those intersections which are empty). Then $P'$ refines $P$ and $O_i$ is a union of elements in $P'$. We must find a matching refinement of $Q$. Given $q \in Q$ we have $q = m(p)$ for some $p \in P$. We split $q$ into two pieces $q_1$ and $q_2$ as follows. If $p_1 = p$ and $p_2 = \emptyset$, let $q_1 = q$ and $q_2 = \emptyset$. And vice versa. Now suppose both $p_1$ and $p_2$ are nonempty. If dom($f'$) $\cap \, p_1$ is empty, then let $q_1$ be a clopen ball strictly contained in $q$ that does not intersect ran($f'$). Let $q_2 = q \setminus q_1$. If instead dom($f'$) $\cap \, p_1$ is nonempty, for every $x \in$ dom($f'$) $\cap \, p_1$ pick a clopen ball around $f(x)$ that is strictly contained in $q$ and contains no other points in ran($f'$). Let $q_1$ be the union of these balls, and $q_2 = q \setminus q_1$. Let $Q'$ be the collection of the $q_i$ (again ignoring any empty sets), and $m'$ be the function that sends $p_i$ to $q_i$ for every $p$, $i$. Then $Q'$ refines $Q$ and $m'$ (which refines $m$) is a matching of $P'$ with $Q'$ that respects $f'$. Similarly, given any $N_j$ we may refine $P'$ to $P''$, $Q'$ to $Q''$ and $m'$ to $m''$ so that $N_j$ is a union of elements in $Q''$ and $m''$ still respects $f'$. Then $(f', m'', P'', Q'')$ is in $D_{x, y, i, j}$ and extends $(f, m, P, Q)$, as desired. Since there are only countably many $D_{x, y, i, j}$, we can find a filter $G$ meeting all of them. Taking the union over the $f$'s in $G$ yields a bijection $h: X \rightarrow Y$. This $h$ is actually a homeomorphism. For if $O_i$ is a fixed element of our base for $X$, there is a condition $(f, m, P, Q)$ in $G$ where $O_i$ is a union of pieces in $P$. These pieces are matched by $m$ to pieces in $Q$, and one may check that the union of these pieces (which is open, in fact clopen, in $Y$) is exactly $h[O_i]$. Similarly $h^{-1}[N_j]$ is open in $X$ for every $j$, so that $h$ is a homeomorphism, as claimed. Cantor's theorem and Sierpinski's theorem have the following generalizations that are useful in certain contexts, and can be similarly proved by forcing arguments by modifying the respective posets in the evident way: Theorem (Skolem): Fix some $k$, $1 \leq k \leq \omega$. Let $X, Y$ be countable dense, linear orders without endpoints. Fix a partition $X = \bigcup_{i < k} X_i$ such that each $X_i$ is dense in $X$, and similarly $Y=\bigcup_{i < k}Y_i$. There is an isomorphism $f: X \rightarrow Y$ such that $f[X_i]=Y_i$ for every $i < k$. Theorem: Fix some $k$, $1 \leq k \leq \omega$. Let $X, Y$ be countable, metrizable spaces without isolated points. Fix a partition $X = \bigcup_{i < k} X_i$ such that each $X_i$ is dense in $X$, and similarly $Y=\bigcup_{i TITLE: Reference for puzzle on dividing piles and scoring products QUESTION [8 upvotes]: There is a pile of $n$ items. Every time you divide a pile into two piles, you get a score being the product of the number of items in the two piles. Show that the sum of your scores at the end is always $\binom{n}{2}$. My question: What are some (preferably well-known) books/articles that discuss or at least mention this puzzle? Is there a name to the puzzle? I guess it's quite famous, and as I want to mention it in an article, I would like to cite it properly instead of just saying that it is "a famous puzzle". Searching Google for "pile product score" doesn't yield useful results. REPLY [2 votes]: A name that has been published for this puzzle is pile splitting. Bill Marion wrote about it as an example of strong induction in a discrete mathematics pedagogy collection I edited. He presented three other versions, given below, and referenced a James Tanton article with many more. Also, he mentioned that the puzzle is included in Rosen's textbook. Bill Marion, Pile Splitting Problem: Introducing Strong Induction, in Resources for Teaching Discrete Mathematics: Classroom Projects, History Modules, and Articles, MAA Notes #74, ed. Hopkins, Mathematical Association of America, 2009, 7-10. James Tanton, A Dozen Questions About: Pile Splitting, Math Horizons 12 (2004) 28-31. Here are the other versions from Bill's article. Write $n = r+s$ for the split (and require $r, s \ne 0$). See also Allen Knutson's comment. Take the sum of all scores $rs(r+s)$. Take the product of all scores $\frac{1}{r} + \frac{1}{s}$. Take the product of all scores $\binom{r+s}{r}$.<|endoftext|> TITLE: What is known about global well ordering of classes in Gödel-Bernays? QUESTION [9 upvotes]: I would like to have something like a linear order on classes, such that every instantiated predicate of classes has a minimal instance in that order. For my purposes, it is fine to assume V=L for sets. REPLY [10 votes]: In the context of models of second-order arithmetic, Mostowski showed that starting with the theory $Z_2$ (the second-order arithmetic analogue of Kelley-Morse) together with the scheme of dependent choices, one can use forcing to add a well-ordering relation on classes so that the full second-order comprehension scheme continues to hold in the language with the new relation. The argument appeared in his paper "Models of second order arithmetic with definable Skolem functions" [Fund. Math. 75 (1972), 223–234] and was later corrected in "Erratum to the paper `Models of second order arithmetic with definable Skolem functions'" [Fund. Math. 84 (1974), 173]. Initially, Mostowski used only the choice scheme to prove the result. Mostowski's argument generalizes to the theory Kelley-Morse together with the scheme of dependent choices. The scheme of dependent choices states that if a second-order definable relation has no terminal nodes, then it has an $\omega$-chain. The forcing is supposed to be akin to the forcing argument that a global well-ordering class can be added to a model of ZFC, but in this case the condition are classes, and so we have a hyperclass forcing (in Sy Friedman's terminology). I only know about this construction from conversations with Ali Enayat, who will probably give many more details when he is around.<|endoftext|> TITLE: Non-abelian Grothendieck group QUESTION [21 upvotes]: By general nonsense the forgetful functor from groups to monoids has a left adjoint. It maps a monoid $(X,\cdot,1)$ to the free group on $\{\underline{x} : x \in X\}$ modulo the relations $\underline{1}=1$, $\underline{x \cdot y}=\underline{x} \cdot \underline{y}$. Notice that the elements of this group have the form $\underline{x_1} \cdot \underline{x_2}^{-1} \cdot \underline{x_3} \cdot \underline{x_4}^{-1} \cdot \dotsc$. Question 1: Does this group have a name? It is analogous to the Grothendieck group, which is the left adjoint of the forgetful functor from abelian groups to commutative monoids, so perhaps we may call it the non-abelian Grothendieck group? Question 2: Is there any criterion when an element of this group vanishes? REPLY [9 votes]: I'd just like to point out that in the many-object case this has been called the fundamental groupoid of a category by Paré, which would suggest that it could be called the fundamental group of a monoid, a possibility strengthened by the fact that, as Benjamin Steinberg points out in the comments, it's the fundamental group of the classifying space of the monoid. Another possibility might be classifying group, but that's probably ill-advised since it's not clear (to me) what it classifies.<|endoftext|> TITLE: Schrodinger equation with magnetic vector potential QUESTION [6 upvotes]: In many papers dealing with the Schrodinger equation with magnetic potential $$u_t=i(\nabla+iA(t,x))^2u$$ the authors say that this equation can be studied with Kato's methods for abstract evolution equations. Is there someone who can suggest some reference in which this approach is used? REPLY [2 votes]: Maybe the following references will be helpful: http://link.springer.com/article/10.1007%2FBF01682741 (Remarks on schrödinger operators with vector potentials, by Tosio Kato), https://projecteuclid.org/euclid.dmj/1077313102 (Schrödinger operators with magnetic fields. I. general interactions, by J. Avron, I. Herbst, and B. Simon). Many relevant references can be found through these two more recent articles: http://arxiv.org/abs/math-ph/0510055 (Recent developments in quantum mechanics with magnetic fields, by Laszlo Erdos) and http://arxiv.org/abs/1410.8210 (Magnetic Schrödinger operators and Manes critical value, by Peter Herbrich).<|endoftext|> TITLE: What invariants distinguish these three-folds? QUESTION [9 upvotes]: Let $a,b$ be two positive integers and everything is over $\mathbb C$. Let's denote by $S_{a,b}$ the hypersurface $$\{x^ay+z^bt=1\}\subset \mathbb A^4.$$ Can we distinguish the $S_{a,b}$'s up to isomorphism? What invariants would be useful in this case? REPLY [6 votes]: Consider the morphism $S_{m,n}\to \mathbb{A}^2_{*}=\mathbb{A}^2\setminus \{(0,0)\}$ given by the projection to $(x,z)$. Then, it is a locally trivial $\mathbb{A}^1$-bundle. This latter is trivial for the Euclidean topology so $S_{m,n}$ is diffeomorphic to $\mathbb{A}^2_{*}\times \mathbb{A}^1$. However, these can give non-isomorphic varieties. In fact $S_{a,b}$ is isomorphic to $S_{a',b'}$ if $a+b=a'+b'$, but apart from this, the question is open (except maybe for $S_{1,1}\simeq\mathrm{SL}(2,\mathbb{C})$ which is probably not isomorphic to the other ones). For more details on this, look at the paper "On exotic affine $3$-spheres" of A. Dubouloz and D. Finston http://www.ams.org/journals/jag/2014-23-03/S1056-3911-2014-00612-3/home.html<|endoftext|> TITLE: counterexample regarding quotient algebras in forcing QUESTION [11 upvotes]: Suppose $A$ and $B$ are complete subalgebras of a complete boolean algebra $C$. Let $G \subseteq A$ be generic. In the extension $V[G]$, we can define the quotient algebras $B/G$ and $C/G$ in the usual way. Is $B/G$ a regular subalgebra of $C/G$? The answer is yes in certain cases, such as if $A \subseteq B$ or $B \subseteq A$, or if $C$ is the completion of product of three partial orders, and $A$ and $B$ correspond to two of them. I know a counterexample under MM, and I thought I had one in ZFC but I've lost it. Can you construct a counterexample? REPLY [3 votes]: Here's an example using random and Cohen forcing, denoted respectively by $\mathcal R$ and $\mathcal C$. Consider the two-step iteration $\mathcal C * \dot{\mathcal R}$. If $(c,r)$ are generic reals for this iteration, then $r$ is random over the ground model $V$. Therefore the random forcing as constructed in $V$ completely embeds into $\mathcal C * \dot{\mathcal R}$. The map takes the following form. If $x$ is a code for a positive measure Borel set, and $A_x$ denotes the set coded, then we map $e : A_x \mapsto (1,\dot{A_{\check x}})$. An important point is that while the generic Cohen real changes the interpretation of $A_x$, $c$ does not exclude any such $\dot{A_{\check x}}^c$ from being in $\dot{\mathcal R}^c$, since the statement that $x$ codes a set of positive measure is absolute. So to show this is a counterexample, we just need to show that the ground-model measure algebra $\mathcal R_0$ (given in terms of Borel codes), is not a regular subalgebra of the random forcing in $V[c]$. Suppose it were, then: (1) The real $s$ given by $e^{-1}$ applied to the generic filter determined by $(c,r)$ would be random over $V[c]$. This is because a regular embedding of a partial order $\mathbb P$ into a complete boolean algebra $B$ extends uniquely to a complete embedding of $\mathcal B(\mathbb P)$ into $B$, and any complete subalgebra of a measure algebra is a measure algebra. (2) Forcing with $\mathcal R$ over $V[c]$ is equivalent to forcing with $\mathcal R_0 * \dot{\mathbb Q}$, where $\mathbb Q$ is some further (possibly trivial) forcing. Thus $(c,s)$ is (Cohen $\times$ Random)-generic over $V$, and so $c$ is Cohen-generic over $V[s]$. By a well-known argument, $c$ constructs a Borel code $y$ for a measure zero set that covers all ground model reals. Thus in $V[s][c]$, $s \in A_y$, and therefore there is a code $\neg y \in V[c]$ for a measure one set such that $s \notin A_{\neg y}$, meaning $s$ is not random over $V[c]$. I also have an example involing Suslin trees and collapsing $\omega_1$, but it seems more interesting if we can stick to random and Cohen. EDIT: Here's the other example. The idea is very similar to my MM example. First note that it suffices to show the situation can be forced by some $\mathbb P$, because then if $\dot A, \dot B, \dot C$ are the algebras in $V^{\mathbb P}$, we can use $\mathbb P * \dot A, \mathbb P * \dot B, \mathbb P * \dot C$ in the ground model. So assume there is a Suslin tree $T$ (which can always be forced), and let $G \subseteq Col(\omega,\omega_1)$ be generic. Since $\omega_1^V$ and $T$ are now countable, we can add a countable top level $l$ to $T$ such that every node in $T$ is below something in $l$. Further, since $Col(\omega,\omega_1) \cong Col(\omega,\omega_1) \times Add(\omega,1)$, and a Cohen real adds a Suslin tree, there is a Suslin tree $S$ in $V[G]$. Now above each node at level $l$, put a copy of $S$, and call this $T'$, which is also a Suslin tree. The set of predecessors of any node at level $l$ determines a $V$-generic branch through $T$. Therefore, by general forcing lemmas, the map $e : T \to Col(\omega,\omega_1) * \dot S$ given by $p \mapsto || p \in \dot H ||$ is a complete embedding, and in this case we can see that $e(p) = (1,p)$ for all $p \in T$. So $\mathcal B(T)$ appears as a complete subalgebra of $\mathcal B(Col(\omega,\omega_1) * \dot S)$. If $G$ is the collapse generic as before, then $T / G = T$. But $T$ is not a regular suborder of $S$ in $V[G]$ since $T$ is countable and thus adds a real over $V[G]$, while $S$ doesn't add reals.<|endoftext|> TITLE: How large must $A$ be if $\{1, \ldots, N\} \subseteq A-A$? QUESTION [15 upvotes]: Given a positive integer $N$, what is the size of the smallest set of integers $A$ such that, for any integer $1 \leq k \leq N$, we can find two integers $x, y \in A$ such that $x - y = k$? (An alternative way to write this condition is to ask that $\{1, \ldots, N\} \subseteq A-A$.) For example, for $N=9$, we could take $A=\{-3, -2, -1, 0, 3, 6\}$, which achieves $|A|=6$. It easy to see that $|A| \geq \sqrt{2N}$, as at most $\binom {|A|} 2 \leq |A|^2/2$ differences can be formed from the elements of $A$. I can also construct suitable sets $A$ with $|A| = 2 \sqrt N$. Is the lower bound $\sqrt{2N}$ asymptotically correct? If not, what is the correct lower bound? REPLY [10 votes]: These are not Golomb rulers, but difference bases. The contrast is that Golomb rulers are a packing problem (differences must be distinct, but can be as sparse as you like), and difference bases are a covering problem (differences must cover an interval, but can overlap as much as you like). Difference bases have been studied at least from 1948, and it seems most of the research is in asymptotic bounds. In fact the list of known concrete values is conspicuously short: OEIS A239308 lists values up to $a(37)=10$, indicating a basis of $10$ elements whose differences cover $[1,37]$. For $N=9$ one can take $A=\{0, 1, 4, 7, 9\}$, achieving $|A|=5$. The current question asks about lower bounds on $|A|$ with respect to $\sqrt{N}$. Answer: The proposed bound $\sqrt{2N}$ is not tight. I think the first nontrivial lower bound was $(1.5570\ldots)\sqrt{N}$ by Rédei and Rényi in 1948 (cited by Erdős and Gál; I haven't seen the original). Leech (1956) improved it to $(1.5602\ldots)\sqrt{N}$. Recently Bernshteyn and Tait (2019) showed that Leech's bound could be improved further (at least "by $\epsilon$") but they did not explicitly show how much. Further note: A similar MO question is here. In that question, it is required that the differences cover $N$ consecutive integers (not necessarily $[1,N]$). But note that if $[1,N]$ is covered by $A-A$, then in fact $2N+1$ consecutive integers $[-N,N]$ are covered, so the problems are very closely linked. (See also Fedor Petrov's comment there.) Bibliography Erdős, Pál; Gál, S. A., On the representation of $1,2,\ldots,N$ by differences, Proc. Akad. Wet. Amsterdam 51, 1155-1158 (1948). ZBL0032.01302. Leech, John, On the representation of $1, 2,\ldots, n$ by differences, J. Lond. Math. Soc. 31, 160-169 (1956). ZBL0072.03401. Bernshteyn, Anton; Tait, Michael, Improved lower bound for difference bases, J. Number Theory 205, 50-58 (2019). ZBL07101901.<|endoftext|> TITLE: Does k(X) have a k-basis for every set X, without AC? QUESTION [12 upvotes]: This question is inspired by Pace Nielsen's recent question Does a left basis imply a right basis, without AC?. For any field $k$, the field $k(x)$ of rational functions in one variable has an explicit $k$-basis given by partial fractions: the set $$B(x,k)=\left\{x^i:i\geq 0\right\}\cup \left\{\frac{x^i}{m(x)^j}:\mbox{$m(x)$ monic irreducible, } 0 TITLE: Plane partitions not containing (1,1,1) QUESTION [14 upvotes]: A plane partition is a subset of $\mathbb Z_{\geqslant0}^3$ s.t. if it contains $(i+1,j,k)$ or $(i,j+1,k)$ or $(i,j,k+1)$ it also contains $(i,j,k)$. What is the generating function $R(q)$ of (volumes of) plane partitions not containing the cell $(1,1,1)$? Since a plane partition containing $(1,1,1)$ contains a $2\times2\times2$ cube, mod $q^8$ this g.f. coincides with MacMahon’s $$ M(q)=\frac1{\prod(1-q^i)^i}=1+q+3q^2+6q^3+13q^4+24q^5+48q^6+86q^7+160q^8+282q^9+500q^{10}+\ldots $$ and computing a couple more terms $$ R(q)=1+q+3q^2+6q^3+13q^4+24q^5+48q^6+86q^7+159q^8+279q^9+488q^{10}+\ldots $$ OEIS doesn’t know this sequence, but in the form $$ R(q)=1+\frac{q-3q^3+6q^6-10q^{10}+\ldots}{[(1-q)(1-q^2)(1-q^3)\ldots]^3}. $$ a pattern is evident. So actually the question is How to prove that $R(q)=1+\frac1{(q)_\infty^3}\sum\limits_i(-1)^i\frac{i(i-1)}2q^{\tfrac{i(i-1)}2}$? AFAIK the statement can be found somewhere in the literature on representation theory of quantum toroidal something. But surely there is a well-known combinatorial (in the broad sense, not necessarily bijective) proof? Context 1. G.f. for plane partitions not containing $(1,1,0)$ is $1/(q)_\infty^2\cdot\sum(-1)^iq^{\frac{i(i+1)}2}$. See the subsection about «V-shaped partitions» in Stanley's Enumerative Combinatorics for a (simple) combinatorial proof. Perhaps there is something like that proof in «(1,1,1)-case». Context 2. G.f. for plane partitions not containing $(1,1,1)$ but intersecting axes at $(i,0,0)$, $(0,j,0)$ and $(0,0,k)$ resp. is $\genfrac[]{0pt}{}{i+j}i\genfrac[]{0pt}{}{j+k}j\genfrac[]{0pt}{}{k+i}kq^{i+j+k+1}$. So maybe it’s possible to compute $R(q)$ as $1+\sum_n q^{n+1}\sum_{i+j+k=n}\genfrac[]{0pt}{}{i+j}i\genfrac[]{0pt}{}{j+k}j\genfrac[]{0pt}{}{k+j}k$. And for the internal sum I know the answer at least at $q=1$: https://math.stackexchange.com/q/177209/ REPLY [8 votes]: It is indeed possible to get the result from your "Context 2". Correcting typos, we have $$R(q)=1+\sum_{i,j,k=0}^\infty\frac{(q)_{i+j}(q)_{i+k}(q)_{j+k}}{(q)_i^2(q)_j^2(q)_k^2}\,q^{i+j+k+1} $$ (where $(q)_k=\prod_{j=1}^k(1-q^j)$). By the $q$-binomial theorem, $$\frac{(q)_{i+j}}{(q)_i(q)_j}=\sum_{x=0}^i\frac{1}{(q)_x(q)_{i-x}}(-1)^xq^{\binom x2+(j+1)x}. $$ Using this, and the expansions obtained by cyclically permuting $(i,j,k)$, we get \begin{align*}R(q)-1&=\sum_{i,j,k=0}^\infty\sum_{x,y,z=0}^{i,j,k}\frac{(-1)^{x+y+z}q^{\binom x2+\binom y2+\binom z3+(j+1)x+(k+1)y+(i+1)z+i+j+k+1}}{(q)_x(q)_{i-x}(q)_y(q)_{j-y}(q)_z(q)_{k-z}}\\ &=\sum_{x,y,z=0}^\infty \frac{(-1)^{x+y+z}q^{\binom x2+\binom y2+\binom z2+(y+2)x+(z+2)y+(x+2)z+1}}{(q)_x(q)_y(q)_z}\\ &\quad\times\sum_{i=0}^\infty\frac{q^{(1+z)i}}{(q)_i}\sum_{j=0}^\infty\frac{q^{(1+x)j}}{(q)_j}\sum_{k=0}^\infty\frac{q^{(1+y)k}}{(q)_k}\\ &=\frac 1{(q)_\infty^3}\sum_{x,y,z=0}^\infty {(-1)^{x+y+z}q^{\binom x2+\binom y2+\binom z2+(y+2)x+(z+2)y+(x+2)z+1}} ,\end{align*} where we first replaced $(i,j,k)$ by $(i+x,j+y,k+z)$ and then computed the inner sums using again a special case of the $q$-binomial theorem. We now observe that $$\binom x2+\binom y2+\binom z2+(y+2)x+(z+2)y+(x+2)z+1=\binom{x+y+z+2}2. $$ Since there are $\binom n2$ solutions $(x,y,z)$ to $x+y+z+2=n$, we get indeed $$R(q)=1+\frac 1{(q)_\infty^3}\sum_{n=0}^\infty(-1)^n\binom n2q^{\binom n2}. $$<|endoftext|> TITLE: Minimum number of edges to remove to have low degree QUESTION [5 upvotes]: I have the following problem, where $k$ is a fixed integer. Input: Graph $G$. Output: Minimum number of edges to remove from $G$ to obtain a graph such that every node has degree at most $k$. Do you know the complexity of this problem? REPLY [10 votes]: If you also insist that the bounded-degree subgraph is connected, then your problem is NP-Hard, since it includes the Longest Path problem when $k=2$. On the other hand, without the connectivity constraint, your problem can be solved in polynomial-time using standard techniques from matching theory. See this paper of Amini, Peleg, Pérennes, Sau and Saurabh, where they give the book Matching Theory by Lovász and Plummer as one reference to the polynomial result.<|endoftext|> TITLE: Is there a "simplification" functor in algebraic topology? QUESTION [38 upvotes]: Recall that a space (=CW complex) is called simple if it is connected, the fundamental group is abelian, and the fundamental group acts trivially on all higher homotopy groups. Call Simp(X) a simplification of X if it is universal for maps from X to a simple space. Does Simp(X) exist for any connected space X? This would give a higher analogue of abelianization of groups. (A natural guess would be the loop suspension of X, but I'm pretty sure that doesn't work as the following example shows. Let X be $\mathbb{R}\mathrm{P}^2$, and note that the $2$-type of $X$ can be described completely by saying that $\pi_1(X) = \mathbb{Z}/2$, $\pi_1(X) = \mathbb{Z}$, the action is the nontrivial one sending $n \mapsto -n$, and the gluing is by the nontrivial element of $H^3(\mathbb{Z}/2,\mathbb{Z})$. There is a simple space $Y$ which is a $2$-type where $\pi_1(X) = \mathbb{Z}/2$, $\pi_2(X) = \mathbb{Z}/2$, the action is trivial, and the gluing is by the nontrivial element of $H^3(\mathbb{Z}/2,\mathbb{Z}/2)$. There's a natural map from X to Y since the nontrivial action on $\mathbb{Z}$ becomes trivial when you kill $2$. On the other hand, $\pi_2(\mathbb{R}\mathrm{P}^2) = \mathbb{Z}$ maps to $\pi_2(\Omega \Sigma \mathbb{R}\mathrm{P}^2) = \mathbb{Z}/4$ sending $1 \mapsto 2$, and so the surjective map $\mathbb{Z} = \pi_2(X) \rightarrow \pi_2(Y) = \mathbb{Z}/2$ cannot factor through $\pi_2(\Omega \Sigma \mathbb{R}\mathrm{P}^2)$.) REPLY [3 votes]: Sorry to resurrect an old question, but I just wanted to expand here on an observation that Qiaochu made above (in the form of a question, but I think he was just playing Jeopardy!). Let $X$ be a connected space, and suppose that it has a simplification $X \to X^s$. Since Eilenberg-MacLane spaces are simple, the map $X \to X^s$ is a (co)homology equivalence in all degrees. Since every local coefficient system on $X^s$ is trivial, this implies that $X \to X^s$ is an acyclic map. It also must kill the perfect radical of $\pi_1(X)$. But the plus construction $X \to X^+$ is the unique acyclic map killing the perfect radical of $\pi_1(X)$. Therefore $X^s = X^+$. If $X^+$ happens to be simple, then $X \to X^+$ is a simplification, simply because any map $X \to Y$ with $Y$ simple must kill the perfect radical (which in this case must coincide with the commutator) of $\pi_1(X)$, and $X \to X^+$ is the universal map which does this. So among the spaces $X$ that do have a simplification are: $X$ such that $\pi_1(X)$ is perfect $X$ such that $X^+$ is an $H$-space (e.g. all the cases used in algebraic $K$-theory) We can also conclude that if $X$ has a simplification $X \to X^s$, then it factors through the plus construction $X \to X^+ \to X^s$, and $X^s$ is also the simplification of $X^+$. So in asking which spaces $X$ admit a simplification, we can reduce to the case where $\pi_1(X)$ is solvable (at least under suitable finiteness assumptions). Warning: From here on, this discussion grows increasingly aimless. Because Eilenberg-MacLane spaces are simple, a simplification is a (co)homology isomorphism. Assume $X^s$ exists, and write $G = \pi_1(X), G^s = \pi_1(X^s)$. If $G \to G^s$ is not surjective, then it lifts to some cover $\overline{X^s}$ of $X^s$. On homology, $X^s$ is a retract of $\overline{X^s}$, but both spaces have abelian fundamental group so it's also a retract on $\pi_1$. But the map on $\pi_1$ is injective, so it is an isomorphism. Hence $G \to G^s$ is surjective after all, and is precisely the quotient by the commutator subgroup, i.e. we have $G^s = G^{ab}$. So the cofiber of $X \to X^s$ is acyclic and is also simply-connected, so is contractible. Since the 1-truncation functor preserves cofiber sequences, the map $BG \to BG^{ab}$ likewise has trivial cofiber. So $BG \to BG^{ab}$ is a homology isomorphism. I'm not sure where to go from there, so instead, let's pick off a special case. Assume that $G$ is abelian, so $G = G^{ab}= G^s$. Then by comparing fiber sequences $\tilde X \to X \to BG$ and $\tilde X^s \to X^s \to BG$ (where tilde's denote universal covers), we see that $H_\ast(\tilde X^s) = H_\ast(\tilde X)_G$ and in addition, the map $H_\ast(\tilde X) \to H_\ast(\tilde X)_G$ is an equivalence on $G$-homology, for any constant coefficients. So for example, if $G$ is a finitely-generated torsion-free abelian group and $H_\ast(X)$ is finitely-generated in each degree over $G$, then I think we can conclude that $X$ was already at least weakly simple (i.e. has trivial $\pi_1$-action on $H_\ast$).<|endoftext|> TITLE: quasicrystal and penrose tiling, mathematical introduction QUESTION [9 upvotes]: Starting to research on quasicrystal from material science, I want to know more about how to understand quasicrystal from a purely mathematical (especially tiling) perspective (probably start from Penrose tiling). I am more familiar with Wang Tile from my previous experience, do you have any suggestion on where shall I start reading on this topic? Thank you. REPLY [6 votes]: If you really aim for substantial mathematical facts I also recommend "Aperiodic Order" by Baake and Grimm. (My account is so new that I cannot "comment" or "Vote up" or something.) The first 6 or 7 chapters are easy to understand for anyone with some basic knowledge on calculus and algebra. The next chapters are tougher. Already in the first 6-7 chapters you learn a lot not only on tilings but on all the relevant mathematics.<|endoftext|> TITLE: The function algebra $C^{\infty}(M\#N)$ of the connected sum of two spaces QUESTION [6 upvotes]: Operations such as taking union or Cartesian products of spaces have direct analogues in term of algebra of functions on them (direct sum and tensor product, respectively), my question is: Is there a way to determine the structure of $C^{\infty}(M\#N)$ in terms of the function spaces $C^{\infty}(M)$ and $C^{\infty}(N)$? OR How is the operation of taking the connected sum of two closed manifold reflected in the algebra of observables? Are there any similar methods used in algebraic geometry or homotopy theory? REPLY [7 votes]: In full generality, the connected sum of $M$ and $N$ depends on more data than just $M$ and $N$; you also need to choose a disk in $M$ and a disk in $N$ to cut out. For example, this choice clearly matters if either $M$ or $N$ has more than one connected component. So to me it's better to think about the connected sum as a special case of the composition of two cobordisms, one from the empty manifold to $S^{n-1}$, and one from $S^{n-1}$ to the empty manifold. The composition of cobordisms is in particular a kind of pushout, and smooth functions on a pushout is a pullback. And of course pushouts are used all the time in algebraic geometry and homotopy theory.<|endoftext|> TITLE: Is a normal matrix satisfying $A^TA=...$ circulant? QUESTION [6 upvotes]: Let $A=\{a_{ij}\}$ be a normal matrix such that $a_{ij}\geq 0$ with equality iff $i=j$. Suppose that $$ A^TA=\begin{pmatrix} a & b & \cdots & b\\ b & a & \ddots & \vdots\\ \vdots & \ddots & a & b\\ b & \cdots & b & a\\ \end{pmatrix},\ where\ b>0. $$ Does it follow that $A$ is a circulant matrix? Note: There is a partial classification of non-negative normal matrices posted here, which seems like it can be used to attack this problem. There is a geometric interpretation as well: both the set of rows and the set of columns of $A$ form equidistant sets of vectors on a sphere, and basic geometry appears to severely restrict the possibilities. Reposted from math.SE REPLY [5 votes]: NO. Let $A$ satisfy the assumptions. If $P$ is a permutation matrix, then $B:=PA$ satisfies the assumptions too: on the one hand, we have $B^TB=A^TP^TPA=A^TA$. On the other hand (remark that the matrix in the question is permutation-invariant) $$BB^T=PAA^TP^T=PA^TAP^T=A^TA=B^TB.$$ If the claim is true, we find that $B=PA$ is circulant for every permutation matrix $P$. We deduce that $a_{ij}$ not only depends upon $j-i$ (modulo $n$), but also depends only upon $j-\sigma(i)$, for every $\sigma\in{\frak S}_n$. We conclude that $a_{ij}$ is constant. Therefore every matrix $A$ satisfying the assumptions, such that $a_{ij}$ is not constant, provides a counter-example of the form $PA$ for some (many) permutation matrices $P$.<|endoftext|> TITLE: Does every embedded 2-sphere in $\mathbb{R}^n$ bound an embedded ball? QUESTION [9 upvotes]: Fix $n \geq 3$ and let $S \subset \mathbb{R}^n$ be a smoothly embedded $2$-sphere. Must there exist a smoothly embedded $3$-ball $B \subset \mathbb{R}^n$ such that $\partial B = S$? This is true for $n=3$ by the Schoenflies theorem. Also, it is true for $n \geq 7$; indeed, since $\mathbb{R}^n$ is contractible we can find a continuous map $B \rightarrow \mathbb{R}^7$ taking $\partial B$ homeomorphically onto $S$, and by Whitney's theorem we can jiggle this continuous map a little to make it an embedding. This might also work for $n=6$, but the homotopy to make $B$ embedded might be complicated and mess up $S$ (I don't know the details of Whitney's strong embedding theorem), so I'm not sure. What about $4 \leq n \leq 6$? REPLY [11 votes]: Haefliger proved that $S^k$ in $\mathbb{R}^n$ is unknotted if $2n > 3(k+1)$. Therefore, any 2-sphere embedded in $\mathbb{R}^n$ with $n\ge 5$ is unknotted. On the other hand, as Mark Grant points out in his comment, there are knotted 2-spheres in $\mathbb{R}^4$ (see this paper by Andrews and Curtis, for example).<|endoftext|> TITLE: Preservation of properness QUESTION [15 upvotes]: Suppose $\mathbb P$ is a countably closed forcing, and $\mathbb Q$ is a c.c.c. forcing that adds reals. Is $\mathbb P$ still proper in $V^{\mathbb Q}$? REPLY [12 votes]: Here is the short version of a negative answer. Let $P$ be the collapse of $\omega_2$ to $\omega_1$ with countable conditions. Fix a tree $(N_\eta:\eta\in 2^{<\omega})$ of quite different models, increasing along each branch. Specifically, make sure that along each branch in $2^\omega$ the union of the models will get a different intersection with $\omega_2$. Now go to any extension $V^Q$ where there is a new real. The new real will give a new branch, yielding a new model; no condition in $P$ can be generic for this model, since a generic condition has to know the model. (I could not find a counterexample for a week. But since I am in Jerusalem at the moment, I eventually gave up and asked Shelah. It took him only a minute or so to come up with this answer. He also said that the reason he is using this specific $P$ is just to understand generic conditions better; many others would work as well.) Here is the more explicit version. (Perhaps too explicit? Perhaps too complicated? It is quite possible that I have overlooked a simpler solution.) Let $P$ be the set of countable partial 1-1 functions from $\omega_1$ to $\omega_2$. Note that if a condition $q$ is $N$-generic it must satisfy the following properties: $dom(q) \supseteq \omega_1\cap N $. (Otherwise $q\cup \{(\alpha,\beta)\}$ for some $\alpha\in \omega_1\cap N$, $\beta\in \omega_2\setminus N$ will force that $N[G]$ has the new ordinal $\beta$.) $q[\omega_1\cap N]\subseteq \omega_2\cap N$. Obviously. $q[\omega_1\cap N] \supseteq \omega_2 \cap N$. (If $\beta\in \omega_2\cap N$ is outside the range of $q[\omega_1\cap N]$, then $q$ would force that $g^{-1}(\beta)\in \omega_1\cap N$ is an ordinal outside $N$. Here I write $g$ for the generic function, i.e. the union of all conditions in the filter.) Summarizing: $q\restriction(\omega_1\cap N)$ is a bijection between $\omega_1\cap N$ and $\omega_2\cap N$. (If I have not overlooked anything, this is also sufficient for being generic.) So a generic condition must really know a lot about $N$. We will use this to get a counterexample. Let $(N_\eta:\eta\in 2^{<\omega})$ be a family of countable elementary submodels with the following properties: For alle $\eta\subseteq \nu$ we have $N_\eta \subseteq N_\nu$. For all $\eta,\nu\in 2^\omega$ with $\eta\not=\nu$ we have $N_\eta \cap \omega_2 \not=N_\nu\cap \omega_2$. ($N_\eta$ is defined naturally as $\bigcup_n N_{\eta\restriction n}$.) Moreover: For every $\eta\in 2^{<\omega}$ there is an ordinal $\alpha_\eta\in \omega_2$ witnessing that the two branches that split at $\eta$ will yield models whose intersections with $\omega_2$ are different: $\alpha_\eta\in N_{\eta^\frown 0}$, but not in $N_{\eta^\frown1}$, and also $\alpha_\eta\notin N_\nu$ for any $\nu$ extending $\eta^\frown1$. There is a $\delta$ such that $\omega_1\cap N_\eta=\delta$ for all $\eta\in 2^\omega$. (We will in fact have $\omega_1\cap N_\eta=\delta$ also for all $\eta\in 2^{<\omega}$.) Now let $Q$ be a (ccc, or just proper) forcing adding a new real $\nu\in 2^\omega$. In the extension $V^Q$ the family $(N_{\nu\restriction n}[G]:n\in \omega)$ is an increasing sequence of elementary submodels; its union $N_\nu[G]$ is again an elementary submodel, and $N_\nu[G]\cap \omega_1=\delta$. Assume that $q$ is generic for this model. Then $S:=q[\delta]=\omega_2 \cap N_\nu[G]$ is a set in the ground model. But now $\nu$ can be computed from $S$: $\eta(0)=1 $ iff $\alpha_{\langle\rangle}\in S$, $\eta(1)=1 $ iff $\alpha_{\langle\eta(0)\rangle}\in S$, etc. Contradiction. It remains to find a tree as described above. (I took some of the following arguments from chapter XV of "Proper and Improper Forcing". I suspect there might be a much easier construction.) Start with a Namba tree $(N_s:s\in \omega_2^{<\omega})$ of countable models which are increasing along each branch and satisfy $s\in N_s$ for all $s\in \omega_2^{<\omega}$. Thin out the tree to make all $N_s\cap\omega_1$ equal to some $\delta$ for all $s\in \omega_2^\omega$, still keeping $\aleph_2$ many successors of each node. (Proper and improper, XV 2.12, based on XI 3.5, going back to a partition theorem in Rudin+Shelah, APAL 1987, RuSh:117. Use determinacy of a low level Borel game.) Replace each model $N_s$ ($s\in \omega_2^{<\omega}$) by the Skolem closure of $N_s\cup\delta$. Now all $\omega_1\cap N_s=\delta$ for all $s\in \omega_2^{\le \omega}$. Now find a binary tree $(s_\eta:\eta\in 2^{<\omega})$ with corresponding models $M_\eta:=N_{s_\eta}$ as follows. Start with $s_{\langle\rangle}:=\langle\rangle$, $M_{\langle\rangle}=N_{\langle\rangle}$. Given $M_\eta=N_{s_\eta}$, first pick an ordinal $\alpha_\eta$ which appears in one of the successors of $M_\eta$ in the $N$-tree (i.e., in some $N_{s_\eta^\frown i}$, but not in coboundedly many of them). Let $M_{\eta^\frown 0}$ be a successor of $M_{\eta}$ containing $\alpha_\eta$. Let $M_{\eta^\frown 1}$ be a successor of $M_{\eta}$ not containing $\alpha_\eta$, but containing some larger ordinal $\beta_\eta<\omega_2$. Note that no model $M\supseteq M_{\eta^\frown1}$ with $M\cap \omega_1=\delta$ can contain $\alpha_\eta$ as an element, since already $M_{\eta^\frown1}$ sees a bijection between $\omega_1$ and $\beta_\eta$; if a new element appeared in the range of this bijection, a new element would have to appear also in the domain. This finishes the inductive step, and thus completes the construction of the tree of models.<|endoftext|> TITLE: Exact Definition of Dirac Operator QUESTION [19 upvotes]: Many definitions of the Dirac operator in the tradition of the Physics literature are hard to grasp for a mathematician. I would like to ask for a precise, general, definition of the Dirac operator in the setting of pure mathematics: Which underlying structure (metrics, bundles, etc.) a manifold have in order to be able to define a Dirac? What are the domain a range of the Dirac operator (perhaps the set of sections of a certain bundle?) Is there an intuitive explanation of what the Dirac operator does? (say, along the lines of "the Laplacian represents diffusion, as in the heat equation $\partial_tu=\Delta u$). I will be grateful for any suggestions. REPLY [11 votes]: I would think that these these notes by Akhil Mathew provide the "exact definition" you are asking for: In response to the follow-up question "which is the first Clifford module used in the physics context": Two different representations (modules) of the Clifford algebra were studied in early work on the Dirac equation. Paul Dirac himself used a quaternionic representation, while Ettore Majorana used a real representation. The physics implications were entirely different: Dirac had a complex field equation, and concluded that the field and its complex conjugate described two different particles. One was the electron, the complex conjugate was unknown at the time. Dirac hypothesised that it described a positively charged "antiparticle", with the same mass as the electron. This "positron" was discovered shortly afterwards, a triumph of mathematical physics. Majorana, in contrast, had a real wave equation and hypothesised that it would describe charge-neutral particles that were their own antiparticle. We still do not know whether such particles exist in nature (the neutrino may or may not be of this type). Historical note: the real representation of the Dirac equation is called the Majorana equation, but this was actually a rediscovery: Eddington had published it a decade before Majorana, so "Eddington-Majorana equation" would be a more appropriate name. REPLY [11 votes]: Let $(M,g)$ be an orientable pseudo-riemannian manifold. Each tangent space $T_xM$ is a pseudo-euclidean space and hence has an associated Clifford algebra $CL(T_xM)$, which is the fibre at $x\in M$ of the Clifford bundle $Cl(TM)$. If the manifold is spin (a topological condition which says that the oriented orthonormal frame bundle lifts to a spin bundle) then it admits a vector bundle $\Sigma$ whose fibre at $x$ is an irreducible Clifford module of $Cl(T_xM)$. This means that we have a bundle map $$Cl(TM) \times \Sigma \to \Sigma$$ given fibrewise by the Clifford action of $Cl(T_xM)$ on $\Sigma_x$. (Please allow me the notational abuse of confusing bundles with their sections, to save me some writing.) Because we have a metric, we have musical isomorphisms $$\flat: TM \to T^*M \qquad\text{and}\qquad \sharp: T^*M \to TM~,$$ the latter of which can be composed with the above Clifford action to arrive at $$Cl(T^*M) \times \Sigma \xrightarrow{\operatorname{cl}} \Sigma~.$$ The Spin group $Spin(T_xM)$ is contained in $Cl(T_xM)$ and hence acts on $\Sigma_x$ by restriction. This representation need not be irreducible; although if $M$ is odd-dimensional it is. This allows us to view $\Sigma$ as a bundle of spinors, associated to a (not necessarily irreducible) spinor representation of the spin group. In fact, it is an associated vector bundle of the spin bundle. The Levi-Civita connection on $TM$ defines an Ehresmann connection on the bundle of oriented orthonormal frames. That connection lifts to a connection on the spin bundle and hence induces a Koszul connection on $\Sigma$, called the spin connection. We thus get a bundle map $$ \Sigma \xrightarrow{\nabla} T^*M \otimes \Sigma $$ The Dirac operator is now simply the composition $$ \Sigma \xrightarrow{\nabla} T^*M \otimes \Sigma \xrightarrow{\operatorname{cl}} \Sigma~.$$ Thus the domain and range are the sections of $\Sigma$, the so-called spinor fields. If $M$ is even-dimensional (and depending on signature perhaps after complexifying) then one can refine this. Under spin, $\Sigma$ splits into eigenbundles of the Clifford action by the volume form: $$ \Sigma = \Sigma_+ \oplus \Sigma_-$$ Then $\nabla : \Sigma_\pm \to \Sigma_\pm$, but $$ \operatorname{cl}: T^*M \otimes \Sigma_\pm \to \Sigma_\mp$$ whence now the Dirac operator, defined in the same way, breaks up into two maps $\Sigma_\pm \to \Sigma_\mp$. Added I'm adding more details to answer Igor's question. If we are in signature $(s,t)$, with $n=s+t$, being the dimension, then the answer depends on $s-t \mod 8$: $s-t = 0 \mod 8$: $\Sigma$ is real and $\Sigma = \Sigma_+ \oplus \Sigma_-$, where $\Sigma_\pm$ are real. $s-t = 1 \mod 8$: $\Sigma$ is complex but has a spin-invariant real structure: it does not split. $s-t = 2 \mod 8$: $\Sigma$ is quaternionic and it splits if we complexify: $\Sigma_{\mathbb{C}} = \Sigma_+ \otimes \Sigma_-$, with $\Sigma_\pm$ complex conjugates. $s-t = 3 \mod 8$: $\Sigma$ is quaternionic and remains so under spin: it does not split. $s-t = 4 \mod 8$: $\Sigma$ is quaternionic and it splits $\Sigma = \Sigma_+ \oplus \Sigma_-$, where $\Sigma_\pm$ are also quaternionic. $s-t = 5 \mod 8$: $\Sigma$ is complex but has a spin-invariant quaternionic structure: it does not split. $s-t = 6 \mod 8$: $\Sigma$ is real and if we complexify then it splits $\Sigma_{\mathbb{C}} = \Sigma_+ \otimes \Sigma_-$, with $\Sigma_\pm$ complex conjugates. $s-t = 7 \mod 8$: $\Sigma$ is real, does not split and remains real.<|endoftext|> TITLE: What is the longest recorded gap between "proof" of a "theorem" and discovery that the result is false QUESTION [11 upvotes]: I hope this question is not a duplicate. I am motivated by wondering when widely accepted results may be considered have a secure place in the mathematical literature. The question is intended to refer to temporal gap. I am particularly interested in cases where a result generally recognised to be of interest has later turned out to be definitely false (eg by explicit demonstration of a counterexample). However, cases where a widely accepted "proof" has later been shown to be incorrect, yet the result has later been correctly proved are also of some interest (such as the 19th century example of a "proof" of the four colour theorem whose incorrectness went undetected for 11 years, although the theorem is now known to be true). The nature of this question may change over time as formal proof checking becomes more advanced. REPLY [6 votes]: There have been several historic 'proofs' of the parallel postulate in terms of Euclid's other four postulates, and it would be unsurprising if one of those holds the record: http://en.wikipedia.org/wiki/Parallel_postulate#History In particular, Ptolemy 'proved' it about 300 years before Proclus discovered a flaw and (amusingly) replaced it with his own (equally invalid, naturally!) 'proof' of the parallel postulate. And I have no idea how long Proclus's proof subsequently survived before being debunked. REPLY [2 votes]: There is the proof put forward by Koenig in 1904 regarding a proof of the falsity of the continuum hypothesis was at first taken as possibly correct and depending on the thesis of Felix Bernstein, especially in the circle surrounding Hilbert. Cantor himself could not judge whether this proof was correct, and it wasn't until Ernst Zermelo found the next the day that Koenig's results were not valid even based off of Bernstein's thesis. I am not sure if this gap between Koenig's purported proof and Zermelo's correction satisfies your request, but here is a link to my reference: http://www-history.mcs.st-andrews.ac.uk/Biographies/Konig_Julius.html<|endoftext|> TITLE: Extensions of an abelian variety by a torus vs. extensions of their $\ell$-adic Tate modules QUESTION [6 upvotes]: Let $K$ be a number field, let $A$ be an abelian variety over $K$, and let $H$ be a torus over $K$. For a prime $l$, we have the natural map $$\mathrm{Ext}^1(A, H) \otimes_{\mathbb{Z}} \mathbb{Z}_l \rightarrow \mathrm{Ext}^1_{\mathrm{Gal}(\overline{K}/K)}(T_l A, T_l H),$$ where the first $\mathrm{Ext}$ is in the category of commutative algebraic groups over $K$ and the second one is in the category of continuous representations of $\mathrm{Gal}(\overline{K}/K)$ on finite free $\mathbb{Z}_l$-modules. Is the displayed map injective and, if so, why? REPLY [6 votes]: It is affirmative (up to a compatibility check at the end of the argument given here). If $K'/K$ is a finite Galois splitting field of $H$ and $T' = H_{K'}$ is the associated split $K'$-torus then we have an exact sequence of $K$-tori $$0 \rightarrow H \rightarrow {\rm{R}}_{K'/K}(T') \rightarrow T \rightarrow 0.$$ But ${\rm{Hom}}(A,T)=0$ since $T$ is affine and ${\rm{Hom}}_K(T_{\ell}(A),T_{\ell}(T))=0$ by consideration of Frobenius eigenvalues. Thus, the natural maps $${\rm{Ext}}^1_K(A,H) \rightarrow {\rm{Ext}}^1_K(A,{\rm{R}}_{K'/K}(T')),\,\,\, {\rm{Ext}}^1_K(T_{\ell}(A),T_{\ell}(H)) \rightarrow {\rm{Ext}}^1_K(T_{\ell}(A),T_{\ell}({\rm{R}}_{K'/K}(T'))$$ are injective. It therefore suffices to treat ${\rm{R}}_{K'/K}(T')$ in place of $H$. Using pushout along ${\rm{R}}_{K'/K}(T')_{K'} \rightarrow T'$ and pullback along $A \rightarrow {\rm{R}}_{K'/K}(A_{K'})$, the Ext's are compatibly identified with ${\rm{Ext}}^1_{K'}(A_{K'},T')$ and ${\rm{Ext}}^1_{K'}(T_{\ell}(A_{K'}),T_{\ell}(T'))$, so this reduces us (upon renaming $K'$ as $K$) to treating the case when $H$ is a split torus, and then even $H = {\rm{GL}}_1$ by direct sum compatibility in $H$. So far, so good. But in the special case when $H={\rm{GL}}_1$ we naturally have $${\rm{Ext}}^1_K(A,{\rm{GL}}_1) \simeq A^{\vee}(K)$$ for the dual abelian variety $A^{\vee}$ (Barsotti's formula, valid over any field), and likewise the Galois Ext is ${\rm{H}}^1(K,T_{\ell}(A^{\vee}))$. The $\ell^n$-power Kummer sequence of $A^{\vee}$ defines an injective map $$\delta:A^{\vee}(K)/(\ell^n) \rightarrow {\rm{H}}^1(K,A^{\vee}[\ell^n]).$$ Letting $S$ be a non-empty finite set of places containing the archimedean places and the bad places of $A$ over $K$ and the $\ell$-adic places, $\delta$ is seen to factor through ${\rm{H}}^1(K_S/K,A^{\vee}[\ell^n])$ where $K_S/K$ is the maximal extension unramified outside $S$. Since ${\rm{Gal}}(K_S/K)$ has good cohomological finiteness properties, it is legitimate to identified the injective map between inverse limits over $n$ with a map $$A^{\vee}(K) \otimes_{\mathbf{Z}} \mathbf{Z}_{\ell} \rightarrow {\rm{H}}^1(K_S/K,T_{\ell}(A^{\vee})).$$ The correctness of the left side uses that $A^{\vee}(K)$ is finitely generated (so its $\ell$-adic completion is given by tensoring against $\mathbf{Z}_{\ell}$). Composing with the injective inflation map then defines an injection $${\rm{Ext}}^1_K(A,{\rm{GL}}_1) \otimes_{\mathbf{Z}} \mathbf{Z}_{\ell} \rightarrow {\rm{Ext}}^1_K(T_{\ell}(A), T_{\ell}({\rm{GL}}_1)).$$ The final step is to check that this injection is actually the original map of interest (in the special case $H = {\rm{GL}}_1$); this is left to the reader to sort out (maybe it is correct up to a harmless sign or something like that).<|endoftext|> TITLE: Which surfaces have only a finite number of connecting geodesics? QUESTION [7 upvotes]: Q1. For a smooth, closed (compact) surface $S$ embedded in $\mathbb{R}^3$, under which conditions is it true that, for every pair of points $a,b \in S$, there are an infinite number of geometrically distinct geodesics connecting $a$ to $b$? By "geometrically distinct" (my terminology) I mean that there is a point on one geodesic not on the other. So, if $S$ is the geometric unit-radius sphere $\mathbb{S}^2 \subset \mathbb{R}^3$, there are only two geodesics connecting $a$ to $b$, the two subarcs of the great circle through $\{a,b\}$. I cannot think of another $S$ that has only a finite number of distinct geodesics connecting each pair of points. For example, any two points on a torus are connected by winding geodesics [left, below], as well as other less obvious geodesics [right below].             Images from: Irons, Mark L. "The curvature and geodesics of the torus." 2008. (PDF Download.) An infinite cylinder would be a counterexample, but it is not closed & compact. A narrow version of Q1 is: Q2. Could it be that only $\mathbb{S}^2$ has a finite number of connecting geodesics? That every other $S$ has an $\infty$# between every point pair? REPLY [13 votes]: Any smooth compact surface smoothly embedded in $\mathbb{R}^3$ that is not the $2$-sphere must have an infinite fundamental group and hence must have infinitely many distinct (in your sense) geodesics joining any two distinct points. This result follows from Morse theory: If $S$ is the surface and $a$ and $b$ are points on it, then each fixed-endpoint homotopy class of curves $c:[0,1]\to S$ with $c(0)=a$ and $c(1)=b$ contains at least one constant speed (minimal) geodesic. When $S$ is not the $2$-sphere, the set of such fixed-endpoint homotopy classes is infinite because it is isomorphic (though not canonically) to the fundamental group of the surface, which is infinite. Now, it is easy to see that these minimal geodesics in each fixed-endpoint homotopy class cannot fall into a finite number of geodesics that are distinct in your sense. Note that this does not depend on the surface being embedded isometrically into $\mathbb{R}^3$, just on having the fundamental group be sufficiently 'large', thus, the only potential compact counterexamples are $S^2$ and $\mathbb{RP}^2$, and then probably only for very special metrics on these. Added remark: It is, perhaps, an interesting question as to which Riemannian metrics $g$ on the $2$-sphere have the property that, for the generic pair of points $a$ and $b$, there are only a finite number of distinct geodesics joining $a$ to $b$. (Note that even the standard unit sphere in $3$-space does not have the property that each pair of distinct points lie on only a finite number of distinct geodesics, since antipodal points lie on a whole circle of geodesics in this case. This is why I'm considering the weaker condition of asking finiteness only for generic pairs.) One general class in which this is true is that of Zoll metrics, i.e., when all of the geodesics are closed. In that case, an application of Sard's Theorem shows that the generic pair of points on the surface lie on only a finite number of geodesics, in the sense that the set of pairs that lie on an infinite number of geodesics is a set of measure zero in the $4$-manifold that consists of (ordered) pairs of points on the surface.<|endoftext|> TITLE: Axiom of choice and vector spaces over a given field QUESTION [8 upvotes]: It is my impression that the following question is open: Does the existence of a basis for every vector space over the field K = the reals having a basis imply the axiom of choice? I saw an answer from several years ago that indicated it was open. There was also a somewhat vague comment about the status of the question for the 2-element field. Does anyone know what is known for various fields K? REPLY [10 votes]: Allow me to steal my answer to a question from math.SE on the same topic. With minor changes In their paper, Howard and Tachtsis discuss these sort of questions. The paper was published rather recently, and I suspect that there hasn't been any significant progress since then. Paul Howard and Eleftherios Tachtsis, On vector spaces over specific fields without choice, MLQ Math. Log. Q. 59 (2013), no. 3, 128--146. In general, it seems that the axiom of regularity might play an important part in such proof. Which means that there might be more than just the structure of the vector spaces involved (although this might be mitigated by going at it as Blass did, by showing that you can prove $\sf MC$ rather than $\sf AC$). If we slightly extend the result to "Every spanning set includes a basis", which in particular means that every vector space has a basis, then Keremedis showed in his paper cited below, that over $\Bbb Q$ if every generating set includes a basis, then the axiom of choice holds. Kyriakos Keremedis, Extending independent sets to bases and the axiom of choice, Math. Logic Quart. 44 (1998), no. 1, 92--98. I don't know the proof, but it might be possible to extend it to $\Bbb R$ or $\Bbb C$. All in all, it seems that the questions you particularly address are still very much open, and perhaps new techniques are needed before we can find answers.<|endoftext|> TITLE: How to roll a $p$ QUESTION [14 upvotes]: Let $p$ be a positive integer (which is not a power of $2$), and suppose we want to generate a number uniformly randomly in the set $\{ 0, 1, \dots , p-1 \}$ (to emulate a dice roll). We are given access to an unbiased coin (where successive coin flips are iid Bernoulli random variables with probability $\frac{1}{2}$ of heads and $\frac{1}{2}$ of tails). The challenge is to minimise the expected number of coin flips. Strategy 1: Naive method Choose the smallest $n$ such that $2^n \geq p$, and flip $n$ coins to obtain a number uniformly distributed over $\{ 0, 1, \dots, 2^n - 1 \}$. If the number is less than $p$, we are done; otherwise forget everything and start again. Calculating the expected number of coin flips for this strategy is easy: $$\mathbb{E}[N] = \dfrac{n 2^n}{p}$$ For the case $p = 3$, this gives an expectation of $\frac{8}{3}$. For $p = 6$, corresponding to an ordinary die, the expectation is $\frac{24}{6} = 4$. Strategy 2: Divide and conquer The naive method is clearly suboptimal for $p = 6$, since it gives an expectation of $4$, whereas we can reduce it to $\frac{11}{3}$ by applying the naive strategy for $p = 3$ followed by another coin flip (to determine the parity of the roll). This suggests an alternative approach: Factorise $p$ into a product of prime factors, and apply some strategy to each factor in isolation. Unfortunately, this isn't optimal either. For $p = 125$, this would advocate performing three instances of the strategy for $p = 5$, each of which must require at least three flips in the best-case scenario (since after flipping two coins, the probability of each outcome is greater than $\frac{1}{5}$, so we require a further flip). Hence we would need at least $9$ flips for strategy 2, whereas the naive strategy requires merely $7.168$ flips on average. Strategy 3: Greedy method The idea behind this is that the naive method 'wastes' lots of information if we get a number in $\{ p, \dots, 2^n - 1\}$. We can attempt to reuse this information to a certain extent. We keep track of a 'state', which will be an ordered pair $(a, b)$, initially set to $(1, 0)$, where $b$ is guaranteed to be uniformly distributed in the set $\{ 0, 1, \dots, a-1\}$ at all times. Now repeatedly apply whichever condition holds: If $a < p$, then flip a coin. If heads, move to state $(2a, 2b + 1)$. If tails, move to state $(2a, 2b)$. If $a \geq p$ and $b < p$, then output the value $b$ and terminate. If $a \geq p$ and $b \geq p$, then move to state $(a - p, b - p)$. I suspect that the greedy method is optimal, although I haven't proved it. It's more difficult to calculate expected values since we're essentially dealing with a finite Markov chain with a separate state for each value of $a$. Punchline The three strategies mutually coincide if and only if $p$ is a Mersenne prime. REPLY [6 votes]: Here is an optimal method that I think is equivalent to your greedy method. We start with a constant random variable, say $0$. Inductively, after $n$ steps we have a random element uniformly distributed on a set of $k$ elements where $k \equiv 2^n \mod p$ and $k \lt p$. At each step, we flip the coin and produce a random element in a set of size $2k \equiv 2^{n+1} \mod p$. If $2k \ge p$ then we take $p$ elements from the set, label them $0$ through $p-1$, and stop with those, continuing with the remaining $2k$ or $2k-p$ elements. This is optimal because we minimize the probability of not stopping by the $n$th step, and the expected number of flips is the sum of the probabilities that we have to make the $n$th flip for $n = 1, 2, 3,...$. If the probability of stopping were any higher, then by the pigeonhole principle some outcome would be weighted too highly. It generalizes to the optimal way to roll a $p$-sided die with a $q$-sided die. I put this in my August 2013 backgammon column, "Rolling Coins." That link requires a subscription but I can send a copy to anyone who is interested. I don't know who came up with this method first but I'd guess it was no later than Turing's work on communication.<|endoftext|> TITLE: Two groups that are the automorphism groups of each other QUESTION [33 upvotes]: Let $H,K$ be two non-isomorphic groups such that $H\cong Aut(K)$ and $K\cong Aut(H)$. Is there any example of such groups ? Note: I had asked the question there. REPLY [6 votes]: Computer search in sage/gap didn't found any solutions for orders up to $120$. It didn't assume the orders are equal.<|endoftext|> TITLE: Characterization of the exterior derivative QUESTION [15 upvotes]: This is a cross-post of someone else's question. I am cross-posting this question from MSE since it hasn't received any answers. In the paper Natural Operations on Differential Forms, the author R. Palais shows that the exterior derivative $d$ is characterized as the unique "natural" linear map from $\Phi^p$ to $\Phi^{p+1}$ (Palais' $\Phi^p$ is what is perhaps more commonly written as $\Omega^p$, and "commutes with all diffeomorphisms", I believe, means $f^*(d\omega) = d(f^*\omega)$): the exterior derivative on $p$-forms is determined to within a scalar factor by the condition that it be a linear mapping into $p+1$ forms which commutes with all diffeomorphisms. I've tried to read the proof in the paper, but I'm struggling to follow the details and missing a sense of the big picture of the proof. I'm interested in Palais' claim because this characterization is the most compelling one I have seen $-$ it seems far more appropriate as an axiomatic definition of $d$ than the definitions found in many textbooks, which often define $d$ based on properties such as $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^k \alpha \wedge d\beta$ (where $\alpha$ is a $k$-form), $d^2=0$, or $d(\sum \omega_{i_1...i_k}dx_{i_1} \wedge \cdots \wedge dx_{i_k}) = \sum d\omega_{i_1...i_k} \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k}$. While these are indeed quite basic properties of $d$, they are more appropriate as theorems than as a priori assumptions. (Of course, what is more "natural" is a matter of opinion, so please don't belabor the issue.) Since it's such an innocent-looking and natural characterization, I would like to see a clear, motivated, reasonably elementary proof of it. Why ought it to be true that there is only one natural linear map from $\Omega^p$ to $\Omega^{p+1}$, up to a constant multiple? What are the key steps of a proof? REPLY [12 votes]: Here is a sketch of one possibility how to prove this: The first key step is to see that the operator you are looking at has to be a differential operator. One usual way to ensure this is to require that it commutes with local diffeomorphisms (which include open embeddings). This implies that the operator is local and hence a differential operator by the Peetre theorem. Having this at hand, the operator has to be induced by a vector bundle map $J^k\Lambda^pT^*M\to\Lambda^{p+1}T^*M$, where $J^k$ is the $k$th jet prolongation, and this bundle map has to be compatible with the natural action of all local diffeomorphisms. Such bundle maps are then determined by linear maps between the standard fibers of these bundles, which are equviariant for the actions of the so-called jet groups. (Formally, this is proved via higher order frame bundles.) In this case, the relevant group will be $G^k_n$, where $n$ is the dimension of $M$. This is an extension of $GL(n,\mathbb R)$, formally defined as the group of $k$-jets at $0\in\mathbb R^n$ of local diffeomorphisms of $\mathbb R^n$ fixing $0$. Now on the target space $\Lambda^{p+1}\mathbb R^{n*}$, the jet group acts just via the usual action of $GL(n,\mathbb R)$ and this is an irreducible representation. On the other hand, restricted to the subgroup $GL(n,\mathbb R)$ the representation inducing $J^k\Lambda^pT^*M$ is the direct sum of the representations $S^{\ell}R^{n*}\otimes\Lambda^p\mathbb R^{n*}$ for $\ell=0,\dots,k$. Now there is only one summand which contains a copy of $\Lambda^{p+1}\mathbb R^{n*}$, namely the one for $\ell=1$. Hence your operator must be of first order, and its principal symbol must be the one of the exterior derivative (up to a constant multiple). Hence you have a multiple of the exteiror derivative up to adding an operator of order zero, i.e. a bundle map $\Lambda^pT^*M\to\Lambda^{p+1}T^*M$ which commutes with the action of all diffeomorphisms. Representation theory of $GL(n,\mathbb R)$ immediately implies that such a bundle map does not exist. There are lots of examples like this one discussed in the book Natural operations in differential geometry by Kolar, Michor and Slovak (MR 94a:58004)<|endoftext|> TITLE: A vector space associated with a vector field on a symplectic manifold QUESTION [6 upvotes]: $\DeclareMathOperator\Div{Div}$Edit: The correct formulation of the vector space $S(X)$ which is defined in this question is the following:$$S(X)=\{Y\in \chi^{\infty}(M)\mid X.\omega(X,Y)=(1/n)\Div(X)\omega(X,Y)\}.$$ This mistake (typos)had been occurred in remark 6, page 7 of Taghavi - On periodic solutions of Liénard equations. Let $(M,\omega)$ be a $2n$ dimensional symplectic manifold and $X$ a smooth vector field on $M$. Consider the following subvector space of $\chi^{\infty}(M)$: $$S(X)=\{Y\in \chi^{\infty}(M)\mid X.\omega(X,Y)=n\Div(X)\omega(X,Y)\}.$$ Here $\Div$ is the divergence corresponding to the volume form $\omega^{n}$ This vector space contains the Lie algebra $C(X)=\{Y\in \chi^{\infty}(M)\mid [X,Y]=0\}$. It also contains the Lie algebra $M(X)=\{fX\mid f\in C^{\infty}(M)\}$. Note that, according to the above definition of $S(X)$, the inclusion $C(X)\subset S(X)$ sensitively depends on the scalar $n$. If we replace $n$ by another scalar, this inclusion is no longer true. (Nevertheless the inclusion $M(X)\subset S(X)$ is not sensitive to this scalar, that is, it is valid for every other scalar.) Questions: What other interesting Lie algebras are contained in $S(X)$? Is $S(X)$ a Lie subalgebra of $\chi^{\infty}(M)$? If the answer is yes, what are some interesting ideals of $S(X)$? If the answer is no, is the Lie algebra generated by $S(X)$ equal to the Lie algebra generated by $C(X)$ and $M(X)$? Motivated by the usual dynamical question "Is the triviality of centralizer a generic situation?", we ask: Is it true to say that for a generic vector field $X$ we have $S(X)=M(X)$? Note: At the international workshop on dynamical system in ICTP, Italy, 2001, I heard from a specialist of dynamical system that "the centralizer problem has various aspects both in discrete and continuous dynamics, but I think that the symplectic version of this problem is interesting and unknown". So this my post is a try for a possible symplectization of "centralizer problem". REPLY [2 votes]: For $n>1$, and the standard symplectic structure $\omega=\sum dx_i\wedge dy_i$ of $\mathbb{R}^{2n}=\{(x_1,x_2,\ldots,x_n,y_1,y_2,\ldots,y_n)\}$ and for the vector field $X=\partial/\partial_{x_1}$ it is easy to observe that the following vector space is not a Lie algebra, since $Div(\partial/\partial_{x_1})=0$ $$S_{\lambda}(X)=\left\{Y\in \chi^{\infty}(\mathbb{R}^{2n})\mid X.\omega(X,Y)=\lambda Div(X)\omega(X,Y)\right \}$$ But for $n=1$ and $\lambda=1$ it is always a Lie algebra. In fact we have the following obvious fact: Obvious Fact: Let $(M,\omega)$ be a $2$- dimensional symplectic manifold(i.e: $\omega$ is a volume form on $M$) and $X$ is a vector field on $M$. Then the vector space $$S(X)=\left\{Y\in \chi^{\infty}(M)\mid X.\omega(X,Y)= Div(X)\omega(X,Y)\right \}$$ is a Lie algebra. Moreover it contains the centralizer $C(X) $ Proof: We apply the well known formula $$d\alpha(X,Y)=X.\alpha(Y)-Y.\alpha(X)-\alpha([X,Y])$$ to $\alpha=i_X(\omega)$. So we conclude that the $S(X)$ in the Obvious Fact is equal to $\{Y\in \chi^{\infty}(M)\mid \omega(X,[X,Y])=0\}$. The later is obviously a Lie algebra containing the centralizer $C(X).$ Remark: For a symplectic manifold $N$ of arbitrary dimension $2n$ it can be shown that the centralizer $C(X)$ of a vector field $X$ is contained in the following vector space: $$\left\{Y\in \chi^{\infty}(N)\mid X.\omega(X,Y)=(1/n) Div(X)\omega(X,Y)\right \}$$ So in the question of this post one should replace $n$ by $1/n$. Proof of Remark: Assume that $[X,Y]=0$. We prove that $X.\omega(X,Y)=(1/n)Div X\omega(X,Y)$. But we need only to prove this formula at all points $p\in N$ with $\omega(X(p),Y(p))\neq 0$. For any such a point $p$, there exist locally a $2$ dimensional symplectic manifold $M$ containing $p$ such that $X,Y$ are tangent to $M$. Now we apply the Obvious Fact above to $M$. We have $X.\omega(X,Y)=Div_{\omega}X.\omega(X,Y) $, where $Div_{\omega} X$ is the divergence of $X$ as a vector field on $M$ with the volume form $\omega$. On the other hand $Div X=(1/n)Div_{\omega} X$ where $Div X$ is the divergence of $X$ as a vector field on the whole manifold $N$ with volume form $\omega^n$.This completes the proof of "Remark".<|endoftext|> TITLE: Gabriel's theorem over a commutative ring QUESTION [10 upvotes]: Is Gabriel's theorem on the indecomposables of representations of quivers of finite type true over a commutative ring, i.e. not necessarily a field? REPLY [12 votes]: Certainly it doesn't generalize in a really obvious way. One way to think about this is the following: Gabriel's theorem uses in really deep way that the category of quiver representations over a field is hereditary (in particular, $\mathrm{Ext}^2(M,N)=0$ for any representations). Over an arbitrary commutative ring, this is false, and so lots of things go out the window. For example, my preferred proof of Gabriel's theorem is from Crawley-Boevey's notes (http://www1.maths.leeds.ac.uk/~pmtwc/quivlecs.pdf); perhaps the most important lemma there is that a Dynkin quiver can't have any indecomposable quiver representations that have automorphisms other than scalars, or any self-extensions. Of course, this is totally false over a non-semisimple ring.<|endoftext|> TITLE: Compatibility of two definitions of Koszul dual QUESTION [7 upvotes]: Let $k$ be a field and $A$ a nonnegatively graded ring over $k$. Assume $A_0 = k.$ We have a bigrading on $\operatorname{Ext}(k,k)$ (one corresponding to homological degree, one corresponding to the grading on $A$). We assume $A$ is Koszul, i.e., that $\operatorname{Ext}^{ij}(k,k)=0$ whenever $i\neq j$. Following Beilinson-Ginzburg-Soergel, "Koszul duality patterns in representation theory", we can define the algebra $E(A)=\oplus\operatorname{Ext}^{ii}(k,k).$ They also define the quadratic dual, $A^{!}$. Their Theorem 2.10.1 then states that $E(A)=A^{!\operatorname{opp}}$. On the other hand, the start of chapter $2$ of "Quadratic Algebras", Polishchuk&Positselski states instead that $E(A)=A^{!}$. Why is there this disparity? This is probably something stupid, but I can't figure it out. (On a related note: I'm also confused as to how BGS is using the functor $\operatorname{RHom}(k,-)$ to go from $D(A-\operatorname{mof})$ to $D(A^{!}-\operatorname{mof});$ I would have assumed that applying that functor would give you a module over $\operatorname{RHom}(k,k)$, which is formal and thus should just correspond to $E(A)$, not $A^{!}$. Could someone clarify this as well?) REPLY [8 votes]: Let $A$ be a quadratic graded algebra over a field $k$ with finite-dimensional components $A_n$ and $A_0=k$. Then the construction of the quadratic dual algebra $A^!$ involves setting $A^!_1$ to be the dual $k$-vector space to $A_1$ and the subspace of quadratic relations in $A^!$ to be the orthogonal complement to the subspace of quadratic relations $R\subset A_1\otimes A_1$. The orthogonal complement is taken with respect to a natural nondegenerate pairing between the vector spaces $A_1\otimes A_1$ and $A_1^*\otimes A_1^*$. Accordingly, the definition of the algebra $A^!$ depends on the choice of this pairing, or in other words, on the choice of a natural isomorphism $(A_1\otimes A_1)^*\simeq A_1^*\otimes A_1^*$. There are, basically, two options: one can set $\langle f\otimes g, u\otimes v\rangle=\langle f,u\rangle\langle g,v\rangle$ or $\langle f\otimes g, u\otimes v\rangle=\langle f,v\rangle\langle g,u\rangle$. Switching between these two choices replaces the algebra $A^!$ with the opposite algebra. In our book, the convention is to use the first of these two pairings, which leads to $E(A)=A^!$. This looks like the natural choice for as long as you work with quadratic algebras over a field. However, when you replace a field with a noncommutative ring (e.g., a noncommutative semisimple algebra) in the role of $A_0$, the choice-of-pairing issue becomes the one of constructing a natural isomorphism $$Hom_{A_0}(A_1\otimes_{A_0}A_1,A_0)\simeq Hom_{A_0}(A_1,A_0)\otimes_{A_0} Hom_{A_0}(A_1,A_0),$$ where $Hom_{A_0}$ denotes the homomorphisms of (say) left $A_0$-modules. Then it turns out that the only such isomorphism is a particular case of the natural isomorphism $$Hom_{A_0}(U\otimes_{A_0}V,A_0)\simeq Hom_{A_0}(V,A_0)\otimes_{A_0} Hom_{A_0}(U,A_0)$$ for $A_0$-$A_0$-bimodules $U$ and $V$ that are finitely generated and projective as left $A_0$-modules. Notice that $U$ and $V$ switch places from the left-hand side to the right-hand side of the latter formula, which basically means that the only one of the above two commutative pairings that admits a noncommutative generalization is the second one. A simple way to distinguish between the two opposite versions of the quadratic dual algebra in the case of a noncommutative ring $A_0$ is to look on the degree-zero components. Notice that $Ext^0_A(A_0,A_0) = Hom_{A_0}(A_0,A_0)=A_0^{opp}$ is the opposite ring to $A_0$ (if $Ext_A$ denotes the $Ext$ of left $A$-modules), while the algebra $A^!$ would be probably defined in such a way that $A^!_0=A_0$. Then the algebras $E(A)$ and $A^!$ cannot possibly be isomorphic, but only anti-isomorphic. One of the consequences of these complications is that the functors $A\longmapsto A^!$ and $A\longmapsto A^{opp}$ do not commute with each other when the ring $A_0$ is noncommutative. Furthermore, if the functor $A\longmapsto A^!$ is defined in such a way that $A^!_0=A_0$ rather than $A_0^{opp}$, then this functor is not self-inverse when $A_0$ is noncommutative. Instead, one has $(A^{!opp})^{!opp}=A$. One can also say that the functor $A\longmapsto A^!$ comes in two versions, the left and the right one, $A\longmapsto A^!$ and $A\longmapsto {}^!\!A$, with $(A^!)_0=A_0=({}^!\!A)_0$ and ${}^!\!A=((A^{opp})^!)^{opp}$, which are inverse to each other, ${}^!(A^!)=A=({}^!\!A)^!$.<|endoftext|> TITLE: ODE properties true in finite dimension but not in Banach spaces of infinite dimension QUESTION [14 upvotes]: Some properties of Ordinary Differential Equations - ODE are true in finite dimension spaces but not in Banach spaces of infinite dimension. The first one I know is the Peano existence theorem. I give a counterexample here for infinite dimension. The second one states that is the maximal solution of a differential equation is defined in an interval smaller than the one of definition of the map of the unique value problem, then the solution is "exploding". I give a counterexample here for infinite dimension. Both are from the mathematician Jean Dieudonné. Do you know other ODE properties valid in finite dimension spaces but not in Banach spaces of infinite dimension? REPLY [6 votes]: A nice survey is the paper of Lobanov and Smolyanov: Sergey Grigorievich Lobanov and Oleg Georgievich Smolyanov, Ordinary differential equations in locally convex spaces. Russian Mathematical Surveys 49.3 (1994): 97–175. The paper lists several counterexamples in the infinite dimensional setting for classical properties and theorems of ODEs in finite-dimensional spaces, such as Peano's theorem, Kneser's theorem, continuous dependence on initial data, continuation of solutions and Picard's theorem.<|endoftext|> TITLE: An old conjecture of M.Newman QUESTION [5 upvotes]: M.Newman raised several questions in his 1957 paper on modular forms. Definition: $H_n$ is the subclass of all zero-free weakly modular forms of weight 0 on $\Gamma_0(n)$, where $n$ is a composite number. Additionally, any $h\in H_n$ is holomorphic at cusps other than $i\infty$ and meromorphic at $i\infty$. One of the most interesting conjecture is: Conjecture: Every weakly modular form of weight 0 on $\Gamma_0(n)$ holomorphic at cusps other than $i\infty$ and meromorphic at $i\infty$ is a linear combination of functions in $H_n$. Question: Does this conjecture have a positive answer? REPLY [7 votes]: First, I think your definition of $H_{n}$ does not agree with Newman's definition. Newman says the following: "Let $H_{n} \subset G_{n}$ be the set of functions of $G_{n}$ with non-negative valence at all parabolic points of $Q_{n}$ other than $\tau = i\infty$." Here $G_{n}$ is the set of modular functions that are expressible in terms of the Dedekind eta-function (which I will refer to as eta-quotients). So Newman is asking if the set of level $n$ eta-quotients of weight $0$ span the space of modular functions holomorphic everywhere except at $i\infty$. If $n$ is squarefree, any weight zero modular function that is non-vanishing on the upper half plane is a constant multiple of an eta quotient (see Winfried Kohnen's paper). However, this is not true when $n$ is not squarefree. The reason is that any weight zero eta-quotient has rational Fourier coefficients, and hence corresponds to an element of $\mathbb{Q}(X_{0}(n))$. However, when $n$ is not squarefree, the cusps of $X_{0}(n)$ need not be rational. Given any two cusps $p_{1}$ and $p_{2}$, the divisor class $p_{1} - p_{2}$ is torsion in $J_{0}(n)$ (by a Theorem of Drinfeld and Manin) and as a consequence, there is a modular function all of whose zeroes are at $p_{1}$ and all of whose poles are at $p_{2}$. In general, this modular function will not have rational Fourier coefficients, and this modular function would be included in $H_{n}$ (via your definition), but not via Newman's definition. I've written a paper with John Webb (on arXiv here) where we study some related questions to Newman's conjecture and do some more computations. We show that if $n$ is composite, the span of $H_{n}$ has finite codimension in the space of all modular functions holomorphic everywhere except infinity. Also, Newman's conjecture is true for all composite $n \leq 300$ with the possible exceptions of $n = 121$ and $n = 209$. However, it seems likely that $n = 121$ may be a genuine exception to his conjecture. The form $f(z) = \frac{\eta(121z)^{22}}{\eta(11z)^{2}}$ has weight $10$ and has a zero of order $110$ at $i \infty$ and is nonzero everywhere else. As a consequence, if $g(z)$ is a modular function holomorphic everywhere except at infinity, then $f(z)^{r} g(z)$ is a holomorphic modular form of weight $10r$ provided $110r$ is $\geq$ the order of pole of $g(z)$ at infinity. For $2 \leq r \leq 8$, the subspace of $M_{10r}(\Gamma_{0}(121))$ generated by eta quotients has codimension $90$ - this suggests that $H_{121}$ may have codimension $90$ in the space of all modular functions with poles only at $i\infty$.<|endoftext|> TITLE: Condition for a certain subset being a subgroup QUESTION [9 upvotes]: For any finite group $G$ and $n$ a divisor of $|G|$, consider the following subset of elements of "co-order" dividing $n$: $$G(n) = \{ g \in G \mid g^{|G|/n} = 1 \}$$ By a classical theorem of Frobenius, $\frac{|G|}{n} \mid |G(n)|$. A conjecture of Frobenius (which was proved via the classification of finite simple groups) states that if there's equality in the above divisibility, $G(n)$ is a normal subgroup. $G(n)$ contains the identity and is closed to taking inverse and conjugation. I'm interested in the following question: For a prime $p$, when is $G(p)$ a subgroup of $G$? By Lagrange's Theorem, if $G(p)$ is a subgroup of $G$, then $|G(p)| \in \{ |G|, \frac{|G|}{p} \}$. When $G$ is abelian, $G(p)$ is a evidently a subgroup. When $G$ is a non-cyclic $p$-group, $G(p)=G$. In fact, as long as the $p$-Sylow subgroup of $G$ is not cyclic, $G(p)=G$. The case $p=2$ is very elegant: composing the natural maps $G \to S_G \to \{\pm 1\}$, we obtain $G(2)$ as the kernel of this composition, hence it's a normal subgroup. The case $p=3$ is false in general (as seen from the example $G=S_3$), and so I guess an additional condition should be imposed - perhaps $p$ should be the smallest prime dividing $|G|$. An additional question is: When $G(p)$ is a subgroup, does is have an alternative characterization? For example, for $p=2$, if the 2-Sylow subgroup of $G$ is cyclic, $G(2)=\langle g^2 \mid g\in G\rangle$. REPLY [5 votes]: You have noted (accurately) that $G(p) =G$ unless $G$ has a cyclic Sylow $p$-subgroup. However, it is also clear that when $G$ has a cyclic Sylow $p$-subgroup and $G(p) \neq G,$ the group $G$ has a normal $p$-complement. For, otherwise, we have by Burnside's transfer theorem, there is a $p$-regular element $x \in N_{G}(P) \backslash C_{G}(P)$. Now we have $P = [P,x] \times C_{P}(x).$ Since $P$ is cyclic and $P \neq C_{P}(x),$ we must have $P = [P,x]$ and $C_{P}(x) = 1.$ But then $P \leq G^{\prime},$ a contradiction, since $[G:G(p)] =p$ and $G(p)$ is clearly a normal subgroup when it is a subgroup. On the other hand, if $G$ has a normal $p$-complement and cyclic non-trivial Sylow $p$-subgroup, then clearly $G$ has a normal subgroup of index $p$, and that that subgroup is indeed $G(p)$. The conclusion (when $|G|$ has order divisible by $p,$ is that $G(p)$ is a subgroup of $G$ if and only if one of the following cases occur: $G(p) = G$, or, $G$ has a normal $p$-complement and a cyclic Sylow $p$-subgroup. Notice that when $p =2$, every finite group with a cyclic Sylow $2$-subgroup is known to have a normal $2$-complement, so the condition given for $p$ odd to guarantee that $[G:G(p)] = p$ is somewhat analogous to what you observed when $p =2$. It is indeed sufficient that $G$ should have a cyclic Sylow $p$-subgroup when $p$ is the smallest prime divisor of $|G|$ to ensure that $[G:G(p)] = p$. It is not, however, necessary.<|endoftext|> TITLE: Colimit density and monads QUESTION [5 upvotes]: Let $C$ be a cocomplete category, and suppose that it has an object that is colimit dense. Is $C$ automatically monadic over $Set$? And if not, is there an explicit counterexample? REPLY [4 votes]: I just want to add an argument that $\mathsf{Cat}$ is also a counterexample (partly because I suspect that Todd chose to use $\mathsf{Pos}$ instead on account of it not being obvious that $\mathsf{Cat}$ has a dense generating object :). The same examples that show $\mathsf{Pos}$ is not regular should show that $\mathsf{Cat}$ is not regular. So the question is how to find a dense generating object for $\mathsf{Cat}$. In fact $\sum_n \Delta^n$, the coproduct of the dense generating set of finite linear orders / simplices is a dense generator in $\mathsf{Cat}$ / $\mathsf{sSet}$. This is not immediate, as Zhen Lin points out above -- the coproduct of the objects of a dense generating set need not even be a generating object in general, as evidenced by sheaves over any nontrivial space (with the representables as the dense generating family). Nonetheless, $\sum_n\Delta^n$ is a dense generator in $\mathsf{Cat}$ or in $\mathsf{sSet}$ because every representable is a retract of $\sum_n \Delta^n$, and a subcategory is dense iff its closure under retracts is dense (since if $i: C \to \tilde C$ is a full subcategory such that every object of $\tilde C$ is a retract of an object of $\tilde C$, then $\mathsf{Hom}(i,1): [\tilde C^\mathrm{op}, \mathsf{Set}] \to [C^\mathrm{op}, \mathsf{Set}]$ is an equivalence, i.e. $i$ is a Morita equivalence). Actually, in $\mathsf{Cat}$, it's clear that the three-element set $\{[0],[1],[2]\} = \{1,2,3\}$ consisting of the simplices of dimension $\leq 2$ / ordinals $\leq 3$ is dense, but these objects are all retracts of the single simplex $[2]$ / ordinal 3. So this object gives an even simpler dense generator for $\mathsf{Cat}$. The (nerve of) the ordinal $\omega$ would also do for a dense generator in either category. As Todd observed, by the theorem discussed by Vitale in the notes he linked to, A category is monadic over $\mathsf{Set}$ if and only if it is exact and contains a regular projective generating object. (here's the link again), it follows (once we observe that the representable simplicial sets are projective, so their coproduct is too, and that a dense generator is a regular generator) that $\mathrm{Hom}(\sum_n \Delta^n, 1): \mathsf{sSet} \to \mathsf{Set}$ is monadic, surprising as that seems! And categories and simplicial sets are just certain types of $M$-set where $M$ is the endomorphism monoid of $\sum_n \Delta^n$ (via full, reflective inclusions)! I haven't thought about how to characterize the image of these inclusion functors. In any 2-valued Grothendieck topos (i.e. a topos category such that every non-initial object has a point), the coproduct of a dense generating set similarly contains every member of the generating set as a retract, and so serves as a dense generating object. If this topos is a presheaf category where the base category has a terminal object, then the representables are a projective dense generating family, so their coproduct is a projective dense generating object, so 2-valued presheaf toposes where 1 is projective are monadic over $\mathsf{Set}$.<|endoftext|> TITLE: Analytic diffeomorphisms of the circle from complex domains QUESTION [6 upvotes]: Let $\gamma \subset \mathbb C$ be a simple closed analytic curve and let $\Delta$ be the closure of the disk it bounds. The Riemann mapping theorem gives two biholomorphisms: $$\phi : (D^2,S^1) \to (\Delta,\gamma)$$ and $$\psi : (\mathbb{CP}^1 - \text{Int}\,\Delta,\gamma) \to (D^2,S^1)\,,$$ where $\text{Int}$ means interior and $D^2 \subset \mathbb C$ is the closed unit disk. We have the induced maps $$\alpha:= \phi|_{S^1}\,,\quad \beta:=\psi|_\gamma\,.$$ Let $$\Gamma(\gamma):=\overline \beta \circ \alpha : S^1 \to S^1$$ where the bar denotes complex conjugation. This $\Gamma(\gamma)$ is an orientation-preserving analytic diffeomorphism of $S^1$. Moreover, $\alpha$ and $\beta$ are well-defined up to precomposition and postcomposition with elements of the biholomorphism group $G$ of $(D^2,S^1)$, restricted to $S^1$, respectively, which implies that $\Gamma(\gamma)$ is also well-defined up to pre- and postcomposition with elements of $G$, since conjugation can be moved inside an automorphism of $D^2$ by changing the automorphism. Therefore the double coset of $\Gamma(\gamma)$ in the analytic diffeomorphism group of $S^1$ by $G$ is well-defined. For instance if $\gamma = S^1$, this is the double coset of the identity. Probably this can also be computed for ellipses in terms of elliptic functions. Question: What can be said about this double coset in general? Can the curve $\gamma$ be reconstructed from it? What are its dynamical properties? I have to admit that I'm asking this out of sheer curiosity, although this does tie with a line of research I tried not so long ago. REPLY [9 votes]: This is the so-called conformal welding problem. One can ask the same question for any Jordan curve $\gamma$ (non necessarily analytic). With this domain of definition, your map $\Gamma$ is well-known to be neither injective nor surjective. There are even orientation-preserving homeomorphisms of the circle analytic everywhere except at one point that are not the welding homeomorphism of any Jordan curve. However, the image of your map $\Gamma$ contains all quasisymmetric orientation-preserving homeomorphisms of the circle, this is sometimes referred to as the fundamental theorem of conformal welding and it was first proved by Pfluger in 1960. A simpler proof was later given by Lehto and Virtanen. In general, it is difficult to explicitely reconstruct the curve $\gamma$ from the welding homeomorphism. In some cases though, the associated curve $\gamma$ has a special form. For instance, if the homeomorphism is the $n$-th root of a Blaschke product of degree $n$, then the corresponding curve $\gamma$ is a proper polynomial lemniscate of the same degree. Conversely, the welding homeomorphism of a proper polynomial lemniscate of degree $n$ is a $n$-th root of a Blaschke product. This was proved by Ebenfelt, Khavinson and Shapiro in the paper "Two-dimensional shapes and lemniscates", arXiv:1003.4567. See also arXiv:1406.3545 for a simpler proof and a generalization to rational lemniscates. Furthermore, there are numerical methods to compute the curve from its welding homeomorphism. Probably the most efficient one is Marshall's Zipper algorithm, see his paper "Conformal Welding for Finitely Connected Regions". There is also Sharon and Mumford's paper "2d-shape analysis using conformal mapping". If you're interested in conformal welding, a good survey is the one by Hamilton MR1966191 (2005e:30012) Hamilton, D. H.(1-MD) Conformal welding. Handbook of complex analysis: geometric function theory, Vol. 1, 137–146, North-Holland, Amsterdam, 2002. 30C35 EDIT Perhaps I should add more details to what I mean exactly by the fact that the map $\Gamma$ is in general not injective. It is easy to see that if $T$ is a Möbius transformation, then $\gamma$ and $T(\gamma)$ have the same welding homeomorphism. The map $\Gamma$ is not injective even modulo Möbius transformations. The easiest way to see this is to consider a curve $\gamma$ of positive area and use the measurable Riemann mapping theorem to obtain an infinite-dimensional family of homeomorphisms of the sphere conformal outside $\gamma$. If $f$ is any such map, then it is easy to see that $\gamma$ and $f(\gamma)$ give rise to same welding homeomorphism. However, a dimension argument shows that the image of $\gamma$ under such a map $f$ cannot be always Möbius-equivalent to $\gamma$. A sufficient condition for the uniqueness of the curve $\gamma$ from its welding homeomorphism is if $\gamma$ is conformally removable, i.e. if every homeomorphism of the sphere conformal outside $\gamma$ is a Möbius transformations.<|endoftext|> TITLE: Did Brouwer evade uncountability? QUESTION [6 upvotes]: I have the distinct memory of having often heard and read that intuitionism was inter alia geared to avoid Cantor's uncountable sets, and it may be that this was Brouwer's plan. But are there accounts which demonstrate that early intuitionism (i.e. before the advent of modern intuitionistic set theories, which either do not have the power set or do accept Cantor's conclusion) had some intuitionistically reasonable way of evading Cantor's uncountable sets? REPLY [9 votes]: You are probably referring to Brouwer's considerations of the Creative Subject, which can be formulated mathematically as Kripke's schema. It implies that all subsets of $\mathbb{N}$ are countable, for example. I am having trouble finding good references, maybe these two will get you started: Göran Sundholm: "Constructive recursive functions, Church's thesis and Kripke's schema" "The Use of Kripke's Schema as a Reduction Principle" D. Van Dalen The Journal of Symbolic Logic Vol. 42, No. 2 (Jun., 1977), pp. 238-240 I know little about the history of Brouwer's mathematics, but I would be very much surprised to hear that he set out to demolish Cantor's set theory. I thought his criticism was pointed at Hilbert's purely existential proofs, not at Cantor. I also never heard that uncountability was considered a problem, it was rather methods of proof. From a purely mathematical point of view (i.e., ignoring history) it makes no sense to "avoid uncountability" because the usual diagonalization proofs of uncountability of Baire space $\mathbb{N}^{\mathbb{N}}$, Cantor space $\{0,1\}^{\mathbb{N}}$ and powerset $\mathcal{P}(\mathbb{N})$ are intuitionistically valid, and Brouwer would have of course known that. It would be hard for Brouwer to avoid these spaces, especially the Baire and the Cantor spaces, as these correspond to the totalities of all paths through a spread and through a fan, respectively. At best some limited form of "everything is countable" is tenable, for instance "every subset of $\mathbb{N}$ is countable" – which is of course valid classically but not in pure intuitionistic logic.<|endoftext|> TITLE: Is the restriction of a representation semisimple? QUESTION [5 upvotes]: Let $F$ be local field of characteristic zero and $\pi$ be a irreducible admissible representation of $GL_n(F)$. Let us consider its restriction to $GL_{n-1}(F)$. Then I want to know whether $\pi|_{GL_{n-1}(F)}$ is completely reducible, namely, $\pi|_{GL_{n-1}(F)}=\oplus_{i\in A}m_i \cdot \tau_i$ where $\tau_i$ is an irreducible representation of $GL_{n-1}(F)$ and $m_i$'s are non-negative integers. Is it true? If so, from which theorem and how does it follow? REPLY [7 votes]: The anwser to your question is "no in general" since you already have a counter-example in the case $n=2$. Take for $\pi$ an irreducible supercuspidal representation of ${\rm GL}(2,F)$. Then its restriction to ${\rm GL}(1,F)\simeq F^\times$ is known by the Kirillov model. This is the space $S(F^\times )$ of locally constant functions with compact support on $F^\times$, where the action is given by $\pi (t)f(x) = f(tx)$, $t\in F^\times$. If the restriction of $\pi$ were completely reducible then $S(F^\times )$ would have an invariant subspace spanned by a non-zero function $f_0\in S(F^\times )$. So for some abelian character $\chi_0$ of $F^\times$, one would have: $$ f_0 (x) =\chi_0 (x) f_0 (1)\ . $$ But such a function $f_0$ cannnot have compact support; a contradiction.<|endoftext|> TITLE: What is the smallest 4-chromatic graph of girth 5? QUESTION [13 upvotes]: It is known that the smallest 4-chromatic graph of girth 4 is the Grötzsch graph (11 vertices). What happens for girth 5? The Brinkmann graph (21 vertices) has chromatic number 4, girth 5 and is 4-regular. Moreover it is the smallest graph (in terms of the order) with these three properties. Dropping the 4-regularity constraint, is there a graph of smaller order than the Brinkmann graph that is 4-chromatic and of girth 5? I haven't found anything in the literature but I may have missed it. REPLY [9 votes]: My computer tells me that there 195291625 graphs on 20 vertices with minimum degree at least 3 and maximum degree at most 6 and girth at least 5. Sadly none of them have chromatic number 4, I just get 48 bipartite ones and the remainder being 3-chromatic. Independent verification would be useful. Added Here's a 21-vertex graph that is not the Brinkmann graph, but is 4-chromatic and has girth equal to 5, also in graph6 format for direct input to Sage. It has the same number of edges as a 4-regular graph but has two vertices of degree 3 and two of degree 5. I bet that it is obtained from the Brinkmann graph by some small edge-swap move. g = Graph("T???C@?K@OA_A_b?AWAQ_?kPCGc`OFCG?da?") Graph 1, order 21. 0 : 7 13 18 19; 1 : 8 15 17 19; 2 : 9 10 19 20; 3 : 9 11 14 18; 4 : 10 12 13 15; 5 : 11 13 17 20; 6 : 12 14 16 19; 7 : 0 14 15 20; 8 : 1 16 18 20; 9 : 2 3 15 16; 10 : 2 4 17 18; 11 : 3 5 19; 12 : 4 6 20; 13 : 0 4 5 16; 14 : 3 6 7 17; 15 : 1 4 7 9; 16 : 6 8 9 13; 17 : 1 5 10 14; 18 : 0 3 8 10; 19 : 0 1 2 6 11; 20 : 2 5 7 8 12; More added There are smaller 4-chromatic girth-5 graphs than Brinkmann's graph, because some of the examples that are appearing have fewer edges even though they have the same number of vertices. The search is about 10% complete and there are 7 graphs found so far. The prettiest one has just 40 edges and an automorphism group of order 5 consisting of a single fixed point and 4 5-cycles (which is vaguely reminiscent of Mycielski's construction). Final Addition After a little over 31 hours on the same multi-core machine, the computation of the full list of graphs with minimum degree at least 3, maximum degree at most 6 and girth at least 5 terminated. Of the 5006797077 graphs constructed, just 18 had chromatic number 4 and therefore have an equal claim to being the smallest 4-chromatic graphs of girth 5 (at least in terms of counting vertices). Among these graphs, 1 has 40 edges, 3 have 41, 12 have 42 and 2 have 43 edges. One is the 4-regular Brinkmann graph. The automorphism groups have orders 14 (Brinkmann graph), 5 (the element of order 5 has one fixed point and 4 x 5-cycles), 2 (8 times) and 1 (8 times). Some of the graphs are obtained from the others by adding/deleting an edge, but mostly not. For those interested, the graphs (in graph6 format) are as follows: T???C@?K@OA_A_b?AWAQ_?kPCGc`OFCG?da? T???C@?GC_B?@_p?@W@cADS@`CCDg@HP@GY? T???C@?GC_B?@_p??T@cAOp?Po@Y@AOsA_e? T???C@?GC_B?@_p??T@cAAE_SoAEgD@K?l?_ T???C@?GC_B?@_p??T@cAAE_So@Q@AEGA_e? T???C@?GC_B?@_p??T@cAAE_So@Q@AEgA_e? T???C@?GC_B?@_b?@WAIGCWaac?HpBOS@QP? T????A?OD?B?P_[?Ac@W?@F?I`@BKAQPAROO T????A?O@?Q?F?S_HC@S?HGQGWFAC?b__FK_ T???C@?g?o?os?PCKP?d_AoOpCCPG@WC?Dg? T???C@?G?oA_A__`IGB?QH?jK?C`c@RG?FW? T???C@?GC_@_E_QOOS?M?CX?s@_PobCWBCQ? T????A?O@_@_g_Y?BC?H_CD_R@QDWCQo@DA_ T???C@?G?_P?R?IOOa?i?SDAqAACR?T__ck? T????A?O@?Q?R?[?BO@P?AF?E`?bSAEGadO_ T????A?O@?Q?R?[?BO@P?AF?E`GbSAEGadO_ T????A?WA_D?@_`CGSCc?@J?JH?pcEAK?t?_ T???C@?GC_H?@_T??[BCOAIac_CSa@aK?C[_<|endoftext|> TITLE: Which properties of a variety are detected by its derived category of coherent sheaves? QUESTION [29 upvotes]: Context: I'm giving an informal seminar/reading group collection of talks on derived categories, following on from earlier talks giving the abstract definition. I am starting to talk about $\mathcal{D}^{b}(\mathrm{Rep}\ kQ)$, i.e. derived categories for representations of quivers, and am going on later to talk about $\mathcal{D}^{b}(\mathrm{Coh}\ X)$, i.e. coherent sheaves. Question: I want to motivate studying $\mathcal{D}^{b}(\mathrm{Coh}\ X)$ by saying, analogously to the representation-theoretic setting, "derived equivalence is good because it's not too strong but it still detects key properties such as..." - and here, I need some help. What properties of a variety are detected by the derived category of coherent sheaves? I'm aware of the Bondal-Orlov reconstruction theorem, which is obviously as good as one could hope for but doesn't always hold. But one expects some invariants to be picked up by the derived category, especially cohomological ones, but are there others too? Of course I'm deliberately being vague about what I mean by "variety" here, as I expect the answer to vary depending on what one asks for. Disclaimers: IANAAG (as they say), as you'll have guessed from the question. I'm aware of this question on MSE and this one on MO and a few other similar ones but I feel that none tackle my particular query - if I missed one that does, mea culpa. I also know of some survey-type articles on the topic by various authors, where one might have expected to find this question addressed, but I haven't seen it. I'd be very open to a good reference in lieu of an answer. I also don't see a community wiki button but surely this should be, if the powers that be could make it so, please. REPLY [6 votes]: One can see some initial answers in Huybrechts' book "Fourier-Mukai Transforms in Algebraic Geometry" ([Huybrechts], Prop 4.1) If two smooth projective varieties are derived equivalent, then they have the same dimension. ([Huybrechts], Prop 3.10) If $X$ is Noetherian, then $X$ is connected if and only if $D^b(Coh X)$ is indecomposable. I heard about the first part of the following argument as a folklore, someone should correct me if I am wrong: (EDIT: The following is WRONG. However, there is some interesting discussion in the comments to this answer.) The compact objects in $D^b(Coh X)$ are precisely those quasi-isomorphic to bounded complex of locally free sheaves, so $X$ is regular if and only if every object in $D^b(Coh X)$ is compact.<|endoftext|> TITLE: How many cospectral graphs available for a given number of nodes? QUESTION [6 upvotes]: Two graphs are said to be cospectral if they have same eigenvalues wrt adjacency matrix, Normalised or Signless laplacian matrix. How many graphs has cospectral mates for a given number of nodes? We know answer to this question when number of nodes is less than $12$. I did not see any research paper till now where author has shown any algorithmic approach to compute those statistics. Either it is too simple to say or they do not wish to disclose it. For a given number of nodes I like to compute number of graphs with at least one cospectral mate. Is there any algorithmic way to do so? If available please give me some references. Thank you for your help. REPLY [9 votes]: This appears an open problem according to a paper. In connection with the graph isomorphism problem, it is of interest what fraction of all graphs is uniquely determined by its spectrum. Haemers onjectures that the fraction of graphs on n vertices with a cospectral mate tends to zero as n tends to infinity. Numerical data for n ≤ 9 was given in [2], and for n = 10, 11 in [3]. Here we do n = 12, and also take the opportunity to correct a few earlier values. OEIS A082104 Number of distinct characteristic polynomials among all simple undirected graphs on n nodes. has some more references.<|endoftext|> TITLE: Are there spaces in which there are no fibered knots? QUESTION [5 upvotes]: I am looking for orientable closed 3-manifolds in which there are no fibered knots. Although I know little about this, I think for links the answer to the question above is "no", and the result is usually formulated as the existence of open book decompositions. Can this result be upgraded to knots? If not, I would be interested in looking at examples, in which it is proved, that they contain no fibered knots. Preferably, I would like these spaces to be homology spheres, even better if Seifert fibered. REPLY [8 votes]: The answer for knots is still "no", because if you have an open book decomposition with disconnected binding then you can stabilize it (see section 2 of Etnyre's lecture notes) by attaching a handle to the page with feet on different binding components. This reduces the number of binding components by one, and you can repeat until the binding is connected, in which case it is a fibered knot.<|endoftext|> TITLE: Relationship between Hochschild cohomology and Drinfeld centers QUESTION [13 upvotes]: Let $HH_*(A,N)$ (or $HH^*(A,N)$) be the Hochschild homology (or cohomology) of an associative algebra $A$ with coefficients in an $A$-bimodule $N$. I was reading nlab's entry on Hochschild cohomology and I saw this term: Hochschild homology object of any bimodule over an monoid in a monoidal ($\infty ,1$)-category , which, when $N = A$, is also called the ($\infty ,1$)- or derived center of $A$. In this paper (http://arxiv.org/pdf/0805.0157v5.pdf) they say that for an associative algebra object $A$ in a closed symmetric monoidal $\infty$-category $\mathcal{S}$, the derived center or Hochschild cohomology $Z(A) = HH^*(A) \in \mathcal{S}$ is the endomorphism object $End_{A \otimes A^{op}} (A)$ of $A$ as an $A$-bimodule. Then I found this on page 45: 6.2. Deligne-Kontsevich conjectures for derived centers. The notion of Drinfeld center for monoidal stable categories is a categorical analogue of Hochschild cohomology of associative (or $A_{\infty}$ algebras)... Honestly I had never heard of Hochschild cohomology objects (or derived centers?) or Hochschild (co)homlogy used in this type of context but mostly I wanted to know: What is the relationship between Hochschild cohomology and Drinfeld centers? REPLY [12 votes]: The key formula is, as you note, $End_{A\text{-mod-}A} (A)$. We can then interpret this same formula in lots of settings. In the category of sets, $A$ is an ordinary algebra, and $End_{A\text{-mod-}A} (A)$ is the ordinary center of A. (The only things that commute with the left action are right multiplication by elements, and those only commute with right multiplication if the element is central.) In the derived setting you get the derived version of the center, namely Hochschild cohomology. In the $(2,1)$-category of categories, an associative algebra is exactly a monoidal category, and $End_{A\text{-mod-}A} (A)$ is the Drinfel'd center. To see this you need to be a bit careful unpacking what a functor of bimodule categories means. In particular instead of a condition saying $f(am) = af(m)$ there's a natural isomorphism $f(a \otimes m) \rightarrow a f(m)$ satisfying a coherence condition (see Ostrik for more details). Again left $A$-module endofunctors of $A$ correspond exactly to tensoring on the right with some object of $A$ (with the associator for $A$ giving the natural transformation). In order for tensoring on the right with an object to commute with the right action, we need to pick isomorphisms $\eta_b: a \otimes b \rightarrow b \otimes a$. In other words, $End_{A\text{-mod-}A} (A)$ consists of objects together with half-braidings and so is the Drinfeld center.<|endoftext|> TITLE: This inequality why can't solve it by now (Only four variables inequality)? QUESTION [24 upvotes]: I asked a question at Math.SE last year and later offered a bounty for it, only johannesvalks give Part of the answer; A few months ago, I asked the author(Pham kim Hung) in Facebook, he said that now there is no proof by hand.and use of software verification is correct,and I try it sometimes,and not succeed.Later asked a lot of people (such on AOPS 1,AOPS 2) have no proof interesting inequality: Let $a,b,c,d>0$, show that $$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}$$ In fact,we have $$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \underbrace{\sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}\ge \sqrt{\dfrac{a^2+b^2+c^2+d^2}{4}}}_{\text{Generalized mean}}$$ Now we only prove this not stronger inequality: $$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt{\dfrac{a^2+b^2+c^2+d^2}{4}}$$ Proof:By Holder inequality we have $$\left(\sum_{cyc}\dfrac{a^2}{b}\right)^2(a^2b^2+b^2c^2+c^2d^2+d^2a^2)\ge (a^2+b^2+c^2+d^2)^3$$ and Note $$a^2b^2+b^2c^2+c^2d^2+d^2a^2=(a^2+c^2)(b^2+d^2)\le\dfrac{(a^2+b^2+c^2+d^2)^2}{4}$$ Proof 2:(I hope following methods(creat is Mine) will usefull to solve my OP inequality,So I post it): \begin{align*}&\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^2-4(a^2+b^2+c^2+d^2)\\ &=\sum_{cyc}\dfrac{3a^4b^2d+5a^4c^3+24a^3cd^3+3a^2b^3c^2+10ab^3d^3+15bcd^5-60a^2bcd^3}{15a^2bcd}\\ &\ge 0 \end{align*} NoW I use computer $$\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^4-64(a^2+b^2+c^2+d^2)=\dfrac{a^{12}c^4d^4+4a^{10}b^3c^3d^4+4a^{10}bc^6d^3+4a^9bc^4d^6 +6a^8b^6c^2d^4+12a^8b^4c^5d^3+64a^8b^4c^4d^4+6a^8b^2c^8d^2+12a^7b^4c^3d^6+12a^7b^2c^6d^5+4a^6b^9cd^4+\cdots+4ab^4c^6d^9+b^4c^4d^{12}}{a^4b^4c^4d^4}$$ REPLY [8 votes]: $\newcommand{\res}{\operatorname{res}}$ Here is a complete proof of the inequality in question. By symmetry, without loss of generality (wlog) $d$ is the maximum of $\{a,b,c,d\}$. By homogeneity, wlog $d=1$. So, the inequality reduces to \begin{equation*} r(a,b,c) :=\frac{\frac{a^2}{b}+\frac{b^2}{c}+c^2+\frac{1}{a}}{\sqrt[4]{a^4+b^4+c^4+1}}\ge2\sqrt2 \tag{1} \end{equation*} for $a,b,c$ in $(0,1]$. Consider the following polynomials in $a,b,c$: \begin{align*} D_ar(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial a}\, a^2 b c \left(a^4+b^4+c^4+1\right)^{5/4} \\ D_br(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial b}\, a b^2 c \left(a^4+b^4+c^4+1\right)^{5/4} \\ D_cr(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial c}\, a b c^2 \left(a^4+b^4+c^4+1\right)^{5/4}, \\ p_1(a,b)&:=a^{16} b^4+2 a^{16} b^2+4 a^{13} b^3+3 a^{12} b^8+4 a^{12} b^6 \\ & +4 a^{12} b^4+4 a^{12} b^2+4 a^{10} b^7+2 a^{10} b^4+6 a^{10} b^2 \\ &+8 a^9 b^7+4 a^9 b^5+8 a^9 b^3+3 a^8 b^{12}+2 a^8 b^{10} \\ &+6 a^8 b^8+4 a^8 b^6+3 a^8 b^4+2 a^8 b^2+4 a^7 b^8 \\ &+4 a^7 b^3+4 a^6 b^{11}+4 a^6 b^8+4 a^6 b^7+6 a^6 b^6 \\ &+4 a^6 b^4+6 a^6 b^2+4 a^5 b^{11}+8 a^5 b^7+4 a^5 b^3 \\ &+a^4 b^{16}+4 a^4 b^{12}+3 a^4 b^8+2 a^4 b^4+4 a^3 b^{12} \\ &+4 a^3 b^8+4 a^3 b^7+4 a^3 b^3+2 a^2 b^{12}+4 a^2 b^8 \\ &+2 a^2 b^4+4 a^{13} b+4 a^9 b+a^{16}+a^{12}+b^8+b^4, \end{align*} which latter is manifestly $>0$. Then \begin{equation*} p_{11}(a,b):=\frac1{a b^3 p_1(a,b)}\,\res_c(D_ar(a,b,c),D_br(a,b,c)) \end{equation*} is a polynomial in $a,b$ of respective degrees $35,25$, where $\res_c(D_ar(a,b,c),D_br(a,b,c))$ is the resultant with respect to $c$ of the polynomials $D_ar(a,b,c)$ and $D_br(a,b,c)$ in $a,b,c$. Similarly, \begin{equation*} p_{21}(a,b):=\frac1{a^4 b^3 p_1(a,b)}\,\res_c(D_br(a,b,c),D_cr(a,b,c)) \end{equation*} is a polynomial in $a,b$ of respective degrees $38,39$. The key observation is that all the roots $(a,b)\in(0,1)\times(0,1)$ of the polynomials $p_{11}$ and $p_{21}$ satisfy the inequalities $b TITLE: Generalized Thom spectra QUESTION [9 upvotes]: I am currently trying to wrap my head around all kinds of different definitions of the notion of (generalized) Thom spectrum. My setup is as follows: Suppose I have a commutative (symmetric) ring spectrum $R_{\bullet}$ and a group $G$, which acts on each of the spaces $R_n$ in a way that is compatible with all structure maps and the $\Sigma_n$-action. Moreover, let $P \to X$ be a principal $G$-bundle classified by the map $f \colon X \to BG$. I can associate a Thom spectrum to this in different ways: I can form the bundle of spectra $\mathcal{R}_n = P \times_{G} R_n \to X$, which has a zero section $\sigma_n \colon X \to \mathcal{R}_n$. The first Thom spectrum is then $Mf^{(1)}_n = \mathcal{R}_n / \sigma_n(X) = r_!\mathcal{R}_n$, where $r \colon X \to \{pt\}$ and $r_!$ is the shriek map as defined by May and Sigurdsson. I can use the two-sided bar construction $B(P, G, R)$ and define $Mf^{(2)}_n = r_!B(P,G,R_n)$ as in Definition 23.5.1 of May and Sigurdsson. I can form the smash product $Mf^{(3)} = \Sigma^{\infty}P_+ \wedge_{\Sigma^{\infty} G_+} R$. I can view $f$ as a map $f \colon BG \to BGL_1(R)$ and use the $\infty$-categorical construction in the work of Ando, Blumberg, Gepner, Hopkins and Rezk to get $Mf^{(4)}$. What are the precise conditions on $R,G,P$ and $X$ that are needed such that these definitions agree? Here is what I got so far: I think I understand how $Mf^{(1)}$ and $Mf^{(2)}$ are related (even though I am not sure, if I need any cofibrancy conditions) and it is proven in the paper mentioned in the fourth point that $Mf^{(3)}$ and $Mf^{(4)}$ are equivalent. Does the bar construction in this case always give a representation of the smash product? REPLY [4 votes]: The Thom spectrum $Mf^{(1)}$ obtained from the group action of $G$ on $R$ agrees with the (non-derived!) smash product of $\Sigma^{\infty}_+ P$ and $R$ over $\Sigma^{\infty}_+G$. The only difference between this and $Mf^{(3)}$ is that the latter uses the derived smash product. Therefore the two agree if either $R$ or $\Sigma^{\infty}_+P$ is cofibrant as $\Sigma^{\infty}_+G$-module spectrum. This happens for example, if $X$ is a finite CW-complex. Then $P$ is (homeomorphic to) a finite (free) $G$-CW-complex. Then $\Sigma_+^{\infty}P$ is cofibrant, since we can lift $G$-equivariant maps to the base space of $G$-equivariant trivial fibrations.<|endoftext|> TITLE: Buildings associated to generalized $BN$ pairs QUESTION [6 upvotes]: I'll begin by asking a general question, and then specializing to the situation I really care about. Let $G$ be a group and let $(B, N)$ be a $BN$-pair in $G$ (see, for instance, page 39 of Tits' "Buildings of Spherical Type and Finite BN-pairs). Then $(B,N)$ has associated to it a chamber complex $X$ with a number of nice properties; most specifically, $X$ is connected, each apartment $\Sigma$ is a Coxeter complex, and $G$ acts transitively on the set $\{(\Sigma,\, C)\}$ of pairs where $\Sigma$ is an apartment and $C \in \Sigma$ is a chamber. When $G$ is a semisimple group over a $p$-adic field $K$, there is a $BN$-pair where $B$ is an Iwahori subgroup. More generally, when $G$ is reductive, there is no such $BN$-pair; there is, however, an generalized $BN$-pair. My general question is: Is there a connected building $X$ associated to a generalized $BN$-pair; that is, a connected chamber complex $X$ with an action of $G$ that is transitive on the set $\{(\Sigma,\, C)\}$ as above, and where each apartment $\Sigma$ is a Coxeter complex? More specifically, if $G$ is a reductive group over a $p$-adic field, is there a canonical way to construct a chamber complex $X$ with the above properties, such that each Iwahori subgroup fixes a unique chamber? For example, if $G = GL_2(\mathbb{Q}_p)$, then there is a generalized $BN$-pair where $B$ is the standard Iwahori subgroup, the Weyl group $W$ is generated by $w_1 = \pmatrix{0 & 1 \\ 1 & 0}$ and $w_2 = \pmatrix{0 & p \\ p^{-1} & 0}$, and $\Omega$ is generated by $s = \pmatrix{0 & p \\ 1 & 0}$ (here I am using the notation given at the beginning of Iwahori's Generalized Tits Systems on $p$-adic Semisimple Groups). One way I can think to build a building is as follows: $X$ is one-dimensional, where the set of vertices is $G/I_1 \cup G/I_2$, where $I_j$ is the parahoric generated by $Bw_jB$. The set of edges corresponds to $G/B$, and there is the obvious $G$-action. This, however, is disconnected since it is a countable disjoint union of trees. REPLY [4 votes]: You may either work with an extended building as L. Spice suggests in his answer, or consider the following based on the fact that a generalized $BN$-pair contains a genuine $BN$-pair. With you notation, the group $N$ writes as a semidirect product $\Omega\ltimes N_0$ for some invariant subgroup $N_0$ of $N$. Then $G_0 = BN_0 B$ is a group with a true $BN$-pair $(B,N_0 )$, and we have $G = \Omega \ltimes G_0$. We may form the building $X$ for $G_0$ with respect to the $BN$-pair $(B,N_0 )$. Recall that $X$ is a simplicial complex whose vertices are in $G_0$-equivariant bijection with the parahoric subgroups $P$ of $G_0$. The action of $G$ by conjugation permutes the parahoric subgroups of $G_0$, whence induces a simplicial action of $G$ on $X$.<|endoftext|> TITLE: Smooth 4-manifolds with $E_8$ intersection form QUESTION [16 upvotes]: Does there exist a closed orientable smooth 4-manifold $M$ whose intersection form is the $E_8$-form? Here by the intersection form I mean the $\mathbb{Z}$-valued bilinear form on $H^2(M;\mathbb{Z})/\text{torsion}$ induced by the cup product map $$H^2(M;\mathbb{Z}) \times H^2(M;\mathbb{Z}) \longrightarrow H^4(M;\mathbb{Z}) \cong \mathbb{Z}.$$ By Rochlin's theorem, this is not possible if $M$ is spin. It is almost the case that if the intersection form on a 4-manifold $M$ is even, then $M$ is spin. However, this is not necessarily the case if $H_1(M;\mathbb{Z}/2)$ is nonzero. So we're looking for a manifold with an interesting fundamental group. It's not clear to me whether or not such a thing exists. I'd also be interested in realizing any other even form whose signature is nonzero modulo $16$. REPLY [5 votes]: This post is a summary of the comments above. No, such a manifold doesn't exist. Donaldson proved in 1982 that if the intersection form of a simply connected, closed, orientable, smooth 4-manifold is definite, then it is diagonal. Since then, the theorem has been extended to arbitrary fundamental groups: see (for example) Ozsváth and Szabó's 2003 paper Absolutely graded Floer homologies and intersection forms for four-manifolds with boundary in Adv. Math. In particular, $E_8$ can't be the intersection form of a closed, orientable, smooth 4-manifold.<|endoftext|> TITLE: What is the idea behind interpolation spaces? QUESTION [7 upvotes]: I am working through a text on Numerics for SPDEs and there the concept an interpolation (Hilbert-)space associated to an operator is used. To be specific: Definition. Let $H$ be an $\mathbb{R}$-Hilbert space and $A: D(A) \rightarrow H$ a diagonal linear operator on $H$ with point spectrum $\sigma_P(A) \subseteq (0,\infty)$ and $\inf \sigma_P(A)>0$. Then the family $(H_r)_{r\in \mathbb{R}}$ of $\mathbb{R}$-Hilbert spaces with the properties $\forall$ $r\geq s$ : $H_r \subset H_s = \overline{H_r}^{H_s}$ $\forall$ $r\in [0,\infty)$ : $(H_r, \langle \cdot,\cdot\rangle_{H_r}) = (D(A^r), \langle A^r(\cdot), A^r(\cdot)\rangle_H)$ $\forall$ $r\in (-\infty,0]$,$v\in H$ : $\|v\|_{H_r} = \|A^r v\|_{H}$ is called a family of interpolation spaces associated to $A$. Now, I am searching for a concrete explanation of how this concept is useful in the analysis of SPDEs. By looking through the literature I found some hints, but no concise answer: It seems the interpolation spaces play a role in dealing with the nonlinearity F of the SPDE $dX_t = [AX_t+F(X_t)] dt+ B(X_t)dW_t$. They are somewhat of an analogue to Sobolev spaces and reveil a certain kind of regularity. I would be very greatful if somebody has a good explanation/motivation of this idea. Thanks in advance! REPLY [10 votes]: In the definition, "diagonal" does not make much sense, you probably mean "self-adjoint", although this can be relaxed to "m-sectorial" (powers of $A$ still make sense then). Also, it is irrelevant for the definition whether or not $A$ has some essential spectrum. Anyway, the main reason why these interpolation spaces are useful for the analysis of SPDEs (and deterministic PDEs for that matter) is that they give us good short-time control over the behaviour of the semigroup $S(t) = e^{At}$ for small times. More precisely, one has $$ \|S(t)u\|_\alpha \le C t^{-(\alpha-\beta)}\|u\|_\beta\;, $$ for any $u \in H_\beta$ and any $\alpha \ge \beta$, uniformly over $t \in (0,1]$, say. When you want to show that a given equation admits local solutions, this allows you to play a game whereby you rewrite the equation in its mild formulation and then trade some of the loss of regularity by the nonlinearity $F$ (and / or $B$) against some negative power of $t$ via the action of the semigroup. You can find more details in my lecture notes on SPDEs available here.<|endoftext|> TITLE: The elliptic curve for $x_1^9+x_2^9+\dots+x_6^9 = y_1^9+y_2^9+\dots+y_6^9$ QUESTION [6 upvotes]: I. Theorem: "If there are $a,b,c,d,e,f$ such that, $$a+b+c = d+e+f\tag1$$ $$a^2+b^2+c^2 = d^2+e^2+f^2\tag2$$ $$3u^3-3uv+w=-def\tag3$$ where $u=a+b+c,\; v = ab+ac+bc,\;w = abc$, then, $$(a + u)^k + (b + u)^k + (c + u)^k + (d - u)^k + (e - u)^k + (f - u)^k = \\ (a - u)^k + (b - u)^k + (c - u)^k + (d + u)^k + (e + u)^k + (f + u)^k\tag4$$ for $k=1,2,3,9$." A rational point implies that, $$\small14(a^6+b^6+c^6-d^6-e^6-f^6)^2-9(a^4+b^4+c^4-d^4-e^4-f^4)(a^8+b^8+c^8-d^8-e^8-f^8) = t^2$$ The first solution to $\sum\limits^6 x_i^9 = \sum\limits^6 y_i^9$ was found by Lander in 1967. In a 2010 paper, Bremner and Delorme realized that it had the form of $(4)$, was good for $k = 1,2,3,9$, a rational point on a homogeneous cubic, and thus one can find an infinite more. (The first soln had $a,b,c,d,e,f = 9, 14, -19, 17, -18, 5$.) II. An alternative method is to directly solve $(1),(2)$ with simple identities such as, $$a,b,c = 3 + 3 m + n - 3 m n + x,\; -6 m - 2 n + x,\; -3 + 3 m + n + 3 m n + x$$ $$d,e,f = 3 - 3 m + n + 3 m n + x,\quad 6 m - 2 n + x,\,\; -3 - 3 m + n - 3 m n + x$$ which incidentally also obeys, $$-a+nb+c = -d+ne+f$$ Then substitute it into $(3)$, and end up only with a quadratic in $m$ whose discriminant $D$ must be made a square. After some algebra, let $c_1 = \tfrac{1}{8}(n^2+3),\; c_2 = \tfrac{1}{2}(n^3-9n)$, then one is to find rational $x$ such that, $$Poly_1:= 7c_1x+c_2$$ $$Poly_2:= -7x^3-21c_1x+c_2$$ and, $$D:=Poly_1 Poly_2 = \text{square}\tag5$$ a situation similar to the linked post for sixth powers. If a rational point $x$ can be found for some constant $n$, it is a simple matter to transform it into an elliptic curve. (The first soln used $n = 1/2$.) Question: Is it possible to find a non-trivial polynomial solution to $(5)$ as a non-zero square $y$? P.S. To clarify re comments, a polynomial solution would be $x,y$ as rational functions of $n$. Or $x,y,n$ as rational functions of a variable $v$. REPLY [9 votes]: For the equation, \begin{equation*} y^2=(-7x^3-21c_1x+c_2)(7c_1x+c_2) \end{equation*} where $c_1=(n^2+3)/8$ and $c_2=(n^3-9n)/2$, and assuming $n$ is rational, we have a rational point $(0,c_2)$, so the quartic is birationally equivalent to an elliptic curve. Using an ancient Ms-Dos version of Derive, it is easy to use Mordell's method to find the curve \begin{equation*} v^2=u^3+49((n^2+3)u+4(3n^6-5n^4+145n^2+49))^2 \end{equation*} which has $2$ points of order $3$ at \begin{equation*} ( \, \, 0 \, , \, \pm \, 28(3n^6-5n^4+145n^2+49)\, \, ) \end{equation*} and numerical experimentation suggests that $\mathbb{Z}/3\mathbb{Z}$ is the torsion subgroup. The reverse transformation is \begin{equation*} x=\frac{8n(n^2-9)u}{(252n^6-420n^4+7n^2(u+1740)+3(7u+v+1372))} \end{equation*} Further numerical experimentation suggests that, for $1 \le n \le 19$, the curve has rank one for $n=6,7,8,9,12,15,16,17$, rank two for $n=18$ and rank zero otherwise - the curve is singular for $n=3$. The presence of rank $0$ curves means (I think!!) that there cannot be a polynomial solution in $n$, valid for all $n$. It might be possible to find a parametric subset of n-values by investigating the curves with strictly positive rank. The problem is that the curves do not have a point of order $2$, so finding generators can take time. Allan MacLeod<|endoftext|> TITLE: Common sizes of intersections QUESTION [5 upvotes]: I'm trying to come up with the largest family of sets that obeys the following properties: Consider $X = \{1,\dots,n\}$ and take $\mathcal{F} \subset 2^X$ such that for any three subsets $A,B,C \in \mathcal{F}$ we have that (at least) two of the numbers $|A \cap B|, |B \cap C|, |A \cap C|$ are the same size. There is a trivial example here, $\{1\}, \{1,2\}, \{1,2,3\},\dots$ which gives a family $\mathcal{F}$ of size $n$, but I don't see how to do better (I've tried using modular arithmetic, which doesn't seem to help all that much.) In this problem it doesn't particularly matter (to me at least) what $n$ is, eg, for the modular arithmetic way I was trying I set $n = L^2$ for some integer. So, my question is as follows, is there a better choice for $\mathcal{F}$, eg, one that is larger than the trivial? REPLY [8 votes]: Consider $\ \binom A2\ $ for an arbitrary finite set $\ A.\ $ Then you get a proper family with $\ \binom n2\ $ members for $\ n:=|A|.\ $ You get an improvement for every $\ n>3$. EDIT:   Let me coin the name "Three Is a Crowd" (or TIsC for short) for the families introduced in the Question by James Kilbane. Now thanks to the comments let me record certain improvements: The Masked Avenger has proposed to add the whole set $\ A,\ $ and the empty set too--let's be greedy. Thus the enlarged Three Is a Crowd family looks like this: $$ \binom A2\cup\{A\,\ \emptyset\} $$ for a new record $\ \binom n2+2\ $ for $\ n=|A|$. Tony Huynh has successfully added the singletons (and the empty set). This gives the following increased Three Is a Crowd family: $$\binom{A}{2} \cup \binom A 1 \cup \binom A 0$$ for a new record $\ \binom {n+1}2$. The semi-dual example is just as good: $$\binom A{n-2}\cup\binom A{n-1}\cup\{\emptyset\}$$ for $\ n:=|A|.\ $ We get $\ \binom {n+1}2\ $ different members again. It'd be nice to find the maximum.<|endoftext|> TITLE: How did Ramanujan discover this identity? QUESTION [35 upvotes]: Let $$\small F_n=(a+b+c)^n+(b+c+d)^n-(c+d+a)^n-(d+a+b)^n+(a-d)^n-(b-c)^n$$ and $ad=bc$, then $$64 F_6 F_{10}=45 F_8^2$$ This fascinating identity is due to Ramanujan and can be found in "Ramanujan for Lowbrows", by B.C. Berndt and S. Bhargava. Would anyone have any idea how Ramanujan discovered this identity? The proofs of the identity offered so far in the papers "A Note on an Identity of Ramanujan", by T. S. Nanjundiah, "Two or Three Identities of Ramanujan", by M.D. Hirschhorn and "A remarkable identity found in Ramanujan's third notebook", by B.C. Berndt and S. Bhargava make the identity less mysterious, but how Ramanujan found the identity in the first place still remains a mystery. As Berndt and Bhargava remarked, it is also not clear whether this is an accidental, isolated result (along with the 3-7-5 counterpart discovered by Hirschhorn), or if there is some deeper theorem lurking behind it. REPLY [2 votes]: How Ramanujan found the identity is, of course, a mystery, but with the intelligent use of a computer algebra system, many such identities can be found. I wrote an article about this. Two of the simpler identities that were not listed above include $$ F_{-2}F_3^2 = -3 F_{-1}^2 F_6 $$ and $$ 245F_3 F_{11} + 330F_7^2 = 539F_5 F_9 $$<|endoftext|> TITLE: Location of maximum of Brownian motion with rough drift QUESTION [7 upvotes]: I am interested in the distribution of the $\text{argmax}_{t \in [0,1]} \{B(t) + f(t)\}$, where $B$ is a Brownian motion (or Brownian bridge) and $f:[0,1] \to \mathbb{R}$ is a continuous function. There are good results when the function $f$ is constant, linear or parabolic. However, I am particular interested in the case when $f$ is a sample path of another Brownian motion. Consider the following problem: Let $B(t)$ and $\tilde B,(t)$, $t \in [0,1]$ be two independent standard Brownian motions (or Brownian bridges). Define $I := \text{argmax}_{t \in [0,1]} \{ B(t) + \tilde B(t) \}$. Consider the following two questions about the conditional distribution of $I$ given $\tilde B$: (1) Is this conditional distribution absolutely continuous with respect to the Lebesgue measure on $[0,1]$, for almost every sample path of $\tilde B$? In another words, is there a regular conditional density function $g( \cdot \mid \tilde B)$ ? (2) If the conditional density function $g( \cdot \mid \tilde B)$ exists, do we have $E[ \int_0^1 g^2(t\mid \tilde B) dt ] = \infty$, where the expectation $E[\cdot]$ is with respect to the law of $\tilde B$ ? REPLY [2 votes]: They are almost surely singular to each other. First of all, here is a basic observation from measure theory that is useful in the proof. Consider two random measures, $M_1$ and $M_2$ on the same space - say, $[0,1]$, depending on some randomness $\omega \in \Omega$. With these measures one associates measures $\mathsf{Q}_1$, $\mathsf{Q}_2$ on $\Omega \times [0,1]$, defined as $$\mathsf{Q}_i(d\omega, dt) := \mathsf{P}(d\omega) M_i(\omega, dt)$$ In other words, $M_i$ can be obtained from $\mathsf{Q}_i$ by disintegration w.r.t. the $t \in [0,1]$ coordinate. On the other hand, one can do the disintegration w.r.t. the other coordinate, i.e. $\omega$, and define the corresponding functions $t \mapsto \mathsf{Q}_i(t)$ with values in measures on $\Omega$, as follows: $$\mathsf{Q}_i(d\omega, dt) = \mathsf{Q}_i(t, d\omega) \mu(dt),$$ where $\mu$ is a measure on $[0,1]$, such that $\mathsf{E} M_i \ll \mu$. The basic observation that I'm talking about is this: The following are equivalent: $M_1 \ll M_2$ almost surely $\mathsf{Q}_1 \ll \mathsf{Q}_2$ (as measures on $\Omega \times [0,1]$) $\mathsf{Q}_1(t) \ll \mathsf{Q}_2(t)$ (as measures on $\Omega$) for $\mu$-almost all $t$ Now we turn back to our problem. Consider the random measure $M := \delta_I$, and the measure $M_1 := \mathsf{E} \left[ M \middle| \tilde B \right]$. $M_1$ is exactly the conditional distribution of the maximum of $B + \tilde B$ given a realization of $\tilde B$. I'm proving that $M_1$ is almost surely singular, that is, singular to the deterministic measure $M_2 := \mathrm{Lebesgue}$. By the above observation, this is equivalent to saying that the distribution of $(B, \tilde B)$ according to the measure $\mathsf{Q}_1(t, d\omega)$ is singular to $\mathsf{Q}_2(t) = \mathsf{P}$, for Lebesgue-almost all $t$. Now we describe the distribution of $(B, \tilde B)$ according to $\mathsf{Q}_1(t)$. It turns out to be easier to describe the distribution according to $\mathsf{Q}(t)$ first --- that is, the one that corresponds to $M = \delta_I$. This description (and a little bit more) is given by the Williams' path decomposition theorem (e.g. Theorem 4.9 in Revuz-Yor). It follows from there that the distribution of the process $\frac{1}{\sqrt 2}(B(t) + \tilde B(t) - B(t + \cdot) - \tilde B(t + \cdot))$ according to $\mathsf{Q}(t)$ is absolutely continuous w.r.t. that of the Bessel(3) process started from $0$. $\frac{1}{\sqrt 2} (B - \tilde B)$ is still the Brownian motion according to $\mathsf{Q}_1(t)$, since it was $\mathsf{P}$-independent of $\frac{1}{\sqrt 2} (B + \tilde B)$ and $M_1$ is measurable w.r.t. $\frac{1}{\sqrt 2}(B + \tilde B)$. Now, another purely measure-theoretic observation: the fact that $M_1 = \mathsf{E} \left[ M \middle| \tilde B \right]$ corresponds, in terms of the $\mathsf{Q}$ measures, to the following: the distribution of $\tilde B$ according to $\mathsf{Q}_1(t)$ is the same as that according to $\mathsf{Q}(t)$. But from the paragraph above it follows that $\tilde B(t + \cdot) - \tilde B(\cdot)$, according to $\mathsf{Q}_1(t)$, can be decomposed as $\frac{1}{\sqrt 2}$ times the sum of a Brownian motion and an independent Bessel(3) process (up to absolute continuity). It remains to prove that if $B$ is a Brownian motion and $U$ is an independent Bessel(3) process starting from $0$ then the distribution of $\frac{1}{\sqrt 2}(B + U)$ is singular to that of the Brownian motion. There are many ways to do that; the simplest one that I know is the following. The Bessel(3) process is essentially "the Brownian motion conditioned to stay positive", in particular, its distribution is the limit of Brownian motions conditioned on some convex sets. This implies that it's strictly log-concave (w.r.t. the covariance of the Brownian motion) --- i.e. "more log-concave than the Gaussian distribution of the Brownian motion". It follows from the inequality of Harge (Theorem 1.1 in Harge "A convex/log-concave correlation inequality for Gaussian measure and an application to abstract Wiener spaces") that for any linear functional $\psi$ on $C[0,1]$ we have $\mathsf{E} (\psi(U - \mathsf{E} U))^2 \le \mathsf{E} \psi(B)^2$; by approximation, this extends to measurable linear functionals, i.e. stochastic integrals of deterministic $L^2$ functions. In other words, for any deterministic function $f \in L^2[0,1]$ the stochastic integral $\intop f d (U - \mathsf{E} U)$ "converges" (i.e. is well-defined as an appropriate limit...). For the Brownian motion, however, $\intop f d B$ converges without the need to recenter $B$. Since $\mathsf{E} U(t) \sim t^{1/2}$ is not in the $W^{1,2}$ Sobolev space, this means that there are some functionos $f \in L^2$, such that $\intop f d B$ is well-defined bub $\intop f d U$ is not --- thus, $\intop f d ((B + U) / \sqrt 2)$ is also not defined. This proves that the law of $(B + U) / \sqrt 2$ is singular to that of $B$.<|endoftext|> TITLE: Intersections of $B$ and $B^-$ orbits in the flag variety $G/B$ QUESTION [5 upvotes]: Let $G = SL_n(\mathbb{C})$, $B$ be a Borel subgroup, and $B^-$ be the opposite Borel. Both the $B$ and $B^-$ orbits on the flag variety $G/B$ are indexed by the Weyl group $W$. Let $S_{w_1}$ and $S^-_{w_2}$ denote the $B$ and $B^-$ orbit corresponding to $w_1, w_2 \in W$ respectively. So how much is known about the intersections of $B$ and $B^-$ orbits $S_{w_1} \cap S^-_{w_2}$ in the flag variety $G/B$? Are these intersections affine? Are they equi-dimensional? What are their dimensions? REPLY [6 votes]: Everything good happens: they are smooth, irreducible, affine, of the expected dimension $\ell(w_1)-\ell(w_2)$; the standard reference is Kleiman 1973. Even their closures, "Richardson varieties", are nice (normal, C-M, rational singularities), which one can blame on similar results for Schubert varieties: nearby any T-fixed point a Richardson variety is locally isomorphic to a product of two Schubert varieties (up to factoring out a vector space), due to Knutson-Woo-Yong in http://arxiv.org/abs/1209.4146 .<|endoftext|> TITLE: Reference for Mod 2 cohomology of $BZ_{2r}$ in terms of Stiefel-Whitney Classes QUESTION [5 upvotes]: I was hoping for an explicit reference to the description of the mod 2 cohomology of a cyclic group $C_{2r}=\langle t \rangle$ of even order in terms of Stiefel-Whitney classes, i.e., that $H^*(BZ_{2r};\mathbb{F}_2)=\mathbb{F}_2[x,y]/(x^2-ry)$, where $x=w_1(\chi)$ is the 1st Stiefel-Whitney class of the representation $\chi:Z_{2r} \rightarrow \mathbb{R}$ sending $t$ to -1, and $y=w_2(\rho)$ is the 2nd Stiefel-Whitney class (1st Chern class reduced mod 2) of the standard representation $\rho: \mathbb{Z}_{2r}\rightarrow \mathbb{C}$. Thanks. REPLY [2 votes]: It is true that statements like this are completely classical and yet hard to find stated just like that in the literature! If noone has a reference, I would suggest to go as follows : -- the cohomology groups are given in the book by Cartan and Eilenberg (oldest reference I can think of!) -- the cohomology ring is obtained by looking at the exact sequence $G \to S^1 \to S^1$ where $G$ is your cyclic group of order $n$ and the map $S^1 \to S^1$ is multiplication by $n$; there is a fibration $G \to S^1 \to S^1 \to BG \to BS^1 \to BS^1$, and the "Gysin exact sequence" of the $S^1 \to BG \to BS^1$ part gives you the multiplicative structure. You have to use that $BS^1 = \mathbb{P}^1(\mathbb{C})$. In a nutshell: it is standard algebraic topology. --this next step is probably more relevant to your question. It is always the case that $H^1(G, \mathbb{F}_2) = Hom(G, \mathbb{F}_2)$ (reference: any book on cohomology, such as Cartan-Eilenberg). This identifies the elements in degree $1$ as Stiefel-Whitney classes by definition (of course $\mathbb{F_2} = O_1(\mathbb{R})$...) Then, the exponential exact sequence $0\to \mathbb{Z} \to \mathbb{C} \to \mathbb{C}^\times \to 1$ gives you $H^2(G, \mathbb{Z}) = Hom(G, S^1) ~~~(= Hom(G, GL_1(\mathbb{C})))$ when $G$ is finite. From there you can identify the elements in degree $2$. OK this is way too long. I agree that a good reference would be better.<|endoftext|> TITLE: Is there a standard name for this poset QUESTION [5 upvotes]: I've run into the following poset and I would expect it has a standard name. Let $n\geq k\geq 0$. Then $P_{n,k}$ consists of all $k$-element subsets of $\{1,\ldots,n\}$ ordered by $X\leq Y$ if $X=\{x_1< \ldots TITLE: Convention about "long" roots for simple Lie algebras of types ADE? QUESTION [6 upvotes]: The classification of simple Lie algebras (over $\mathbb{C}$ or other sufficiently large field of characteristic 0) correlates these Lie algebras with the irreducible reduced root systems (in Bourbaki's language). Here there are at most two possible lengths of roots, relative to a euclidean metric arising from the nondegenerate Killing form restricted to a fixed but arbitrary Cartan subalgebra. In the "simply-laced" types ADE, all roots have equal length, whereas in other types BCFG there are both "long" and "short" roots. At least since the early work of Dynkin, it has been conventional to say that all roots in types ADE are "long" (though it would be possible to say all are "short", or else just avoid the distinction here). For example, Dynkin's method for drawing his diagrams uses open circles for long simple roots, filled-in circles for short simple roots. Is there a most natural rationale for the convention that all roots are long in types ADE? One situation in which the convention seems handy involves the classification of the finitely many nilpotent orbits in the Lie algebra (which also goes back to Dynkin). Here one finds a unique minimal nonzero orbit, characterized as the set of root vectors corresponding to long roots (for any choice of Cartan subalgebra). But I'm uncertain about the main motivation behind the convention (and its history). ADDED: I was thinking at first just about the algebraic theory of simple Lie algebras, which can be studied over any splitting field of characteristic 0 in terms of root systems and Weyl groups. Over $\mathbb{C}$ these arise classically as complexifications of various real Lie algebras belonging to Lie groups. In this direction the answer (resp. comment) by Andre (resp. Henrik) are both useful. The paper by Jeff Adams is a unification of previous case-by-case study, working in a traditional Lie group setting, whereas the KNR paper also brings in some more modern ideas as well as tools from algebraic geometry. I can accept such an answer but should wait a little longer in case someone else can suggest an answer in the purely algebraic setting I've sketched. (By the way, in his papers from about 1946 to 1950, Dynkin's convention for labelling vertices seems to have evolved. Maybe this was influenced by his classification of nilpotent orbits, including the description of the minimal nonzero orbit?) REPLY [8 votes]: Let us define a coroot $\alpha^\vee \in \mathfrak g$ to be `short' if the corresponding group homomorphism $SU(2)\to G$ generates $\pi_3(G)$ (the homomorphism has Dynkin index $1$). With the above definition, in the ADE case, all coroots are short. Dually, it is then reasonable to call all roots `long' in the ADE case.<|endoftext|> TITLE: Why does this sequence converges to $\pi$? QUESTION [35 upvotes]: One of my daughters was having a small programming exercise. Let's consider following algorithm: Take a list of length $n$: $\ (1\,\ 2\,\ \ldots\,\ n)$. Remove every $2$nd number. From the resulting list, remove every $3$rd number. From the resulting list, remove every $4$th number. ... Follow on until the list remains unchanged and let $u_n$ be the number of remaining elements. Example with $n=11$ $(\ 1\,\ 2\,\ 3\,\ 4\,\ 5\,\ 6\,\ 7\,\ 8\,\ 9\,\ 10\,\ 11\ )\quad \Rightarrow\quad (\ 1\ *\ 3\ *\ 5\ *\ 7\ *\ 9 \ *\ 11\ )$ $(\ 1\,\ 3\,\ 5\,\ 7\,\ 9\,\ 11\ )\quad \Rightarrow\quad (\ 1\,\ 3\ *\ 7\,\ 9\ *\ )$ $(\ 1\, 3\,\ 7\,\ 9\ )\quad \Rightarrow\quad (\ 1\,\ 3\,\ 7\ *\ )$ $(\ 1\,\ 3\,\ 7\ )\ $ -- will not be modified anymore, and therefore $u_n=3$. QUESTION:   why do we have $\lim\limits_{n \to +\infty} \frac{n}{u_n^2}=\frac{\pi}{4}$ ? Thanks! REPLY [4 votes]: This is an extended comment: Interestingly enough, displaying the differences of consecutive terms of A000960 shows an amazing degree of fluctuation.<|endoftext|> TITLE: Products of Cohen forcings QUESTION [15 upvotes]: Let $\lambda$ be an infinite cardinal. Does the full support product of $\lambda$ copies of $Add(\omega, 1)$ collapse $2^\lambda$ to $\aleph_0$? For $\lambda = \omega$, it is known to be true (it is an exercise in Kunen's book), but what about uncountable $\lambda$? REPLY [6 votes]: The answer is yes. Lemma: There is a function $F: 2^{\lambda} \rightarrow 2^{\lambda}$ such that for every $S \in [\lambda]^{\lambda}$ and function $g: \lambda \setminus S \rightarrow 2,$ $F$ restricted to extensions of $g$ is surjective. Fix a well-ordering of all ordered pairs of the form $(g, h),$ $g$ as above and $h \in 2^{\lambda},$ of length $2^{\lambda}.$ We construct $F$ in $2^{\lambda}$ stages, each stage determining $F$ at one input. Let $(g,h)$ be the $\alpha$th pair in the ordering. At stage $\alpha,$ we have determined $F$ at less than $2^{\lambda}$ inputs, so there is some extension $g'$ of $g$ for which $F(g')$ has not been determined. We declare $F(g')=h.$ This completes the construction. Let's consider the case where $\text{cf}(\lambda)>\omega.$ Notice that for every $h \in 2^{\lambda}$ and $p \in \mathbb{P}=\Pi_{\alpha<\lambda} Add(\omega,1),$ there are $q \le p$ and $n<\omega$ such that $n \in \text{dom}(q_{\alpha})$ for every $\alpha,$ and $F(\alpha \mapsto q_{\alpha}(n))=h.$ Here $n$ is any number such that $|\{\alpha<\lambda: n \not \in \text{dom}(p_{\alpha})\}|=\lambda.$ Thus, in $V[G],$ there is a surjection from $\omega$ to $(2^{\lambda})^V$ definable from $F$ and $G.$ Now suppose $\text{cf}(\lambda)=\omega$ and $\lambda>\omega.$ Here we use a generalization of the matrix argument in Hamkins' answer. Let $\kappa_0=0,$ and fix a cofinal $\omega$-sequence $\langle \kappa_n: 0\max(\text{dom}(p_{\kappa_k}))$ be such that $|\{\alpha \in [\kappa_k, \kappa_{k+1}): n_k \not \in \text{dom}(p_{\alpha})\}|=\kappa_{k+1},$ and let $n=n_0.$ It is easy to construct $q$ as desired.<|endoftext|> TITLE: Define Turing machine with algebraic concepts/structures QUESTION [10 upvotes]: Usually, during lectures Turing Machines are firstly introduced from an informal point of view (for example, in this way) and then their definition is formalized (for example, in this way). Is it possible to give another 'equivalent' definition that relies more precisely on algebraic concepts (i.e. algebraic structures: semigroups, monoids, etc; just like, for instance, regular languages are recognized by finite monoids and context-free languages are recognized by a product of a free group and a finite monoid)? REPLY [7 votes]: Yes, there is now Pavlovic's characterization of Turing computability in terms of the monoidal computer, based on monoidal categories. http://arxiv.org/abs/1208.5205<|endoftext|> TITLE: Embedding Euclidean buildings into products of trees QUESTION [8 upvotes]: A Euclidean building has a natural metric space structure. (A definition of Euclidean building can be found on Wikipedia, or, more expansively, in Section 4 of Kleiner-Leeb.) Question: Is it true that every (finite rank) Euclidean building has a biLipschitz embedding into a finite product of metric $\mathbb{R}$-trees? It follows from the results of Lang-Schlichenmaier that a Euclidean building $X$ of rank $n$ admits an embedding $f$ into the product $T$ of $n+1$ trees, with the property that for some $C>1$ and $p>0$, $$ C^{-1}d_X(x,y)^p \leq d_T(f(x),f(y)) \leq Cd_X(x,y)^p.$$ So what I am asking is if there is any obstruction to taking $p=1$ (maybe increasing the number of trees in the product if necessary). Apologies if the question is foolish, I am not an expert on buildings. REPLY [6 votes]: The problem of existence/nonexistence of quasi-isometric embeddings $X\to Y$ between symmetric spaces and locally compact Euclidean buildings of rank $\ge 2$ is wide-open in the case when $rank(Y)> rank(X)$ (assuming, of course, that $dim(X) TITLE: Are two forms of the Dual Schroeder-Bernstein property equivalent? QUESTION [6 upvotes]: We know the Shroeder-Bernstein (SB) theorem can be proved in ZF, while the Dual Schroeder-Bernstein (DSB) can be proved in ZF+AC but not in ZF. Define as ISB the property that whenever there are both an injection and a surjection $X \to Y$, then there must be a bijection too. Trivially ZF+DSB $\implies$ ZF+ISB (because an injection $X\to Y$ allows to define a surjection $Y\to X$, without using AC). Is the converse true? I'm assuming here that ZF $\nRightarrow$ ISB, but I don't know a proof of that either (UPDATE: this is the case, as pointed out by Asaf Karagila in a comment). REPLY [5 votes]: What you call $\sf ISB$, is better known as $\sf WPP$ (Weak Partition Principle) and can be formulated as "If there is a surjection from $X$ onto $Y$, then $X$ cannot have a strictly smaller cardinality than $Y$" (alternatively, $|X|\leq^*|Y|\leq|X|\rightarrow |X|=|Y|$ or $|X|\leq^*|X|\rightarrow |X|\nless|Y|$). In their paper Banaschewski and Moore write that this is still open, and Higasikawa makes no mention of a possible answer in his paper from 1995. I am unaware of any new papers since then that have dealt with these topics, meaning that this problem is probably still open. Bibliography. Bernhard Banaschewski, Gregory H. Moore, The dual Cantor-Bernstein theorem and the partition principle, Notre Dame J. Formal Logic 31 (3), (1990), 375–381. Masasi Higasikawa, Partition principles and infinite sums of cardinal numbers. Notre Dame J. Formal Logic 36 (1995), no. 3, 425–434. Additional reading: A question about the Axiom of Choice<|endoftext|> TITLE: The Metrizability of Symmetric Products of Metric Spaces QUESTION [9 upvotes]: The (infinite) symmetric product of a based topological space $(X,e)$, denoted by $\mathrm{SP}(X,e)$, can be viewed as the topological space of ''multisets'' in $X$ containing the base point $e$ infinitely many times (please see http://en.wikipedia.org/wiki/Infinite_symmetric_product for the precise definition). The following is my question: Provided that $X$ is metrizable, can we say that $\mathrm{SP}(X,e)$ is metrizable in general? Due to my poor knowledge I have absolutely no idea as to how the symmetric product of simple spaces look like. However, I suppose I can explain how I come up with this question at least. Let $(X,d)$ be a metric space with a base point $e$. An arbitrary element $S$ in $\mathrm{SP}(X,e)$ admits a representation $S = [s_1,s_2,\ldots]$, where $(s_i)_{i \in \mathbb{N}}$ is a sequence in $X$ with the property that all but finitely many terms of the sequence are the base point $e$. We can now define a metric on $\mathrm{SP}(X,e)$ by $$ \mathrm{dist}([s_1,s_2,\ldots],[t_1,t_2,\ldots]) := \inf_{\pi} \sum_i d(s_i,t_{\pi(i)}), $$ where the inf is taken over all permutations $\pi$. In my masters thesis (on functional analysis) I had to show that the fundamental group of $\mathrm{SP}(X,e)$, equipped with the above metric topology, is the first homology group of $X$ (provided that $X$ is both path connected and locally simply connected). I have met one algebraic-topologist who made a somewhat interesting remark that this result is analogous to the Dold-Thom theorem (please see http://en.wikipedia.org/wiki/Dold%E2%80%93Thom_theorem). This is how I started finding a relationship between the topology induced by this metric and the standard topology on $\mathrm{SP}(X,e)$. The Dold-Thom theorem seems pretty famous, and so I thought the metrizability of the symmetric product of a metric space might have been well studied. Sadly, I know nothing about topology, so this is how I ended up using this website. Cheers. REPLY [7 votes]: The infinite symmetric product of a pointed metric space $(X,e)$ is metrizable iff the basepoint $e$ is isolated. First, if $e$ is isolated and $X=Y\coprod \{e\}$, then $SP(X,e)=\coprod SP^n(Y)$ and each finite symmetric product $SP^n(Y)$ is metrizable (by the metric you define). In fact, in this case it is not hard to see that your metric induces the topology on $SP(X,e)$. Conversely, suppose $e$ is not isolated; then I claim $SP(X,e)$ is not first countable and hence not metrizable. Indeed, suppose $SP(X,e)$ is first countable. Since $e$ is not isolated, $[e]=[e,e,\dots]\in SP(X,e)$ is in the closure of $SP^n(X,e)\setminus SP^{n-1}(X,e)$ for each $n>0$. By first countability, we can find elements $x_n\in SP^n(X,e)\setminus SP^{n-1}(X,e)$ such that the sequence $(x_n)$ converges to $[e]$ in $SP(X,e)$. But the set $\{x_n\}$ has finite and hence closed intersection with each $SP^n(X,e)$, so the entire set $\{x_n\}$ is closed in $SP(X,e)$. This contradicts the assumption that $(x_n)$ converges to $[e]$. More generally, this argument shows that if a $T_1$ space $A$ is a colimit of subspaces $A_0\subset A_1\subset \dots$ such that the interiors of the $A_n$ do not cover $A$, then $A$ cannot be first countable. As a final note, let $SP_d(X,e)$ be the symmetric product with the topology induced by your metric. Then the identity $i:SP(X,e)\to SP_d(X,e)$ is continuous, and in fact is a homeomorphism when restricted to each $SP^n(X,e)$. Since every compact subset of $SP(X,e)$ is contained in some $SP^n(X,e)$ (by essentially the same argument as the previous paragraph), the map $i$ is a weak equivalence (i.e., induces an isomorphism on homotopy groups). In particular, when $(X,e)$ is a pointed metric space that is homotopy equivalent to a connected CW-complex (relative to the basepoint), the Dold-Thom theorem implies that $\pi_n(SP_d(X,e))=\pi_n(SP(X,e))=H_n(X,e)$.<|endoftext|> TITLE: Embedding $G$ in a $Z(G)$ extension of $\operatorname{Aut}G$ QUESTION [5 upvotes]: This question follows up a question I asked on math.SE. This is a refinement and a reference request. For what groups $G$ does there exist a $Z(G)$-extension of $\operatorname{Aut}G$ (call it $\tilde G$) that contains $G$ normally, in such a way that $\tilde G\rightarrow \operatorname{Aut}G$ describes $\tilde G$'s conjugation action on $G$? Do you know if this question has been studied? If so, can you point me to references? Comments It is trivial that the class $\mathscr{C}$ of groups admitting such an embedding includes centerless groups (take $\tilde G = \operatorname{Aut} G$). Also trivially, $G\in\mathscr{C}$ if $\operatorname{Out}G = 1$. Take $\tilde G = G$. Also easily, $G\in\mathscr{C}$ if $G$ is abelian. Take $\tilde G$ to be $G$'s holomorph. It seems to me that also $G\in\mathscr{C}$ if $G$ has any faithful irreducible representation of a unique dimension and $\operatorname{Aut}G$ has trivial Schur multiplier, for the following reason. Let $\zeta:G\rightarrow GL(V)$ be the representation in question. $\operatorname{Aut}G$ acts on $G$'s representations but must fix $\zeta$ since it is the unique irreducible representation of its dimension; so any automorphism of $G$ is induced by conjugation by an element of $GL(V)$, which is determined up to a scalar factor; this gives us a faithful projective representation of $\operatorname{Aut}G$ on $V$, which lifts to an ordinary representation $\xi$ because the Schur multiplier of $\operatorname{Aut}G$ is trivial. Then it seems to me that the subgroup of $GL(V)$ generated by the images of $\zeta$ and $\xi$ can be taken to be $\tilde G$. Actually in this construction (when $G$ has a faithful irreducible representation $\zeta$ of a unique dimension), the assumption of trivial Schur multiplier for $\operatorname{Aut} G$ is overly restrictive. All we need is that the particular projective representation of $\operatorname{Aut} G$ arising from considering its action on the image of $\zeta$ (as above) represents the trivial class in the Schur multiplier. I am not sure how to tell when this happens. Even this latter condition seems too much to ask for this construction to work, because it is actually asking for $\tilde G \rightarrow\operatorname{Aut}G$ to split. It seems to me that the construction will work as long as the class of $H^2(\operatorname{Aut}G,\mathbb{C}^\times)$ corresponding to the projective representation of $\operatorname{Aut}G$ lies in the subgroup $H^2(\operatorname{Aut}G, \zeta(Z(G)))$. Again, I am not sure how to tell when this happens. REPLY [7 votes]: As Derek Holt points out, the existence of such an extension is equivalent to the universal cohomology class in $H^3(\text{Out}(G);Z(G))$ being zero. The Eilenberg-MacLane paper is the original reference, a modern reference is K. Brown: Cohomology of Groups. Some time ago I wrote some Magma programs to check whether a cohomology class in H^3 vanishes or not. To sum up, while the desired extension exists for many small groups, it does not exist for all finite groups. The smallest counterexamples are the dihedral group $D_{16}$ and the quaternion group $Q_{16}$. There is also an old theorem of de Siebenthal which says that if $G$ is a compact connected Lie group, then the epimorphism $\text{Aut}(G)\rightarrow \text{Out(G)}$ splits, so in this case the desired extension always exists.<|endoftext|> TITLE: Reference for homotopy (co)limits of (co)chain complexes via totalization of double complexes QUESTION [7 upvotes]: It seems to be a well-known fact that homotopy (co)limits of (co)simplicial diagrams of nonnegatively graded (co)chain complexes in (Grothendieck) abelian categories can be computed by using the Dold-Kan correspondence to pass to double complexes and then applying the totalization functor for double complexes, possibly applying the truncation functor afterward. For example, this is claimed (without proof) in Dugger's notes on homotopy colimits, see Section 16.8 in http://math.uoregon.edu/~ddugger/hocolim.pdf. In the above, “homotopy (co)limit” is used in the abstract ∞-categorical sense, i.e., the homotopy terminal (respectively initial) object in the ∞-category of (co)cones. It can be presented as the appropriately derived functor of the ordinary (co)limit functor in the setting of (stable) model categories or as the quasicategorical (co)limit in the setting of stable quasicategories. Is there a written proof of this result in the literature? What about the case of unbounded chain complexes? REPLY [3 votes]: Rodríguez González, Beatriz(E-CSIC-IM) Simplicial descent categories. (English summary) J. Pure Appl. Algebra 216 (2012), no. 4, 775–788<|endoftext|> TITLE: Contexts and notations for composing asymmetric simplices QUESTION [14 upvotes]: Imagine the elements of a group-like structure as puzzle pieces with essential two sides, an IN-side and an OUT-side. You can compose two such pieces in two obvious ways: Now consider triangular puzzle pieces with at least one IN- and one OUT-side. These are 2-simplices with a non-trivial partition of their sides. As long as two sides of the same kind are not distinguished (i.e. the simplices are symmetric), there are again two ways to compose two such pieces: But when two sides of the same kind are distinguished: a single operator + doesn't suffice anymore. One has to specify which of the (eventually) two OUT-sides of the first piece is to be plugged into which of the (eventually) two IN-sides of the second piece: I wonder: (1) In which specific (algebraic or simplicial resp. topological) contexts do such asymmetric pieces appear? (2) How then is the problem of notation solved, especially: how are "words" (conglomerates) of such pieces symbolically written down (which is trivial for group-like structures and symmetric simplices by the use of + or $\circ$ or even no symbol at all). Note that the composition is supposed to be in a natural way associative. A related question concerns the possibility that cycles are allowed. For group-like structures, cycles are not allowed (and in the rigid picture of puzzle pieces cycles are not even possible), for simplex-based structures cycles are supposed to be allowed: (3) How is the problem of notation solved for possibly circular conglomerates? REPLY [6 votes]: Here is an attempt. Not a full answer, rather a suggestion of an approach. Instead of triangles let us look at nodes with three strings coming out of them. So, for the triangle of type "A" let us write and for the triangle of type "B" let us write The bottom line is that the strings that go upwards from the node correspond to OUT-sides, and strings that go downwards from the node correspond to IN-sides. Let us regard these pictures as string diagrams http://ncatlab.org/nlab/show/string+diagram. Then we can compose them as we compose string diagrams. Some of the possible compositions are: We allow strings to intersect each other (as is the case in the leftmost diagram). Such compositions should correspond to the "triangle compositions". So as the leftmost diagram above is "A1 + 0B", the second diagram is "A1 + 1B" and the third diagram is "B + A" as in OP. Triangle tilings which are obtained inductively by adjoining one tile by a single edge at a time are representable by string diagrams. On the other hand, a tiling can be constructed from a string diagram that has no cycles. These can be shown by induction on number of tiles/nodes. I believe that there is a one-to-one correspondence between tilings and string diagrams of these types. For more complicated tilings things get more difficult. Anyway, the "cycle" in the OP would be After this category theory and string diagrams tell us how to solve the problem of notation. We can imagine that we are inside a symmetric monoidal category where we have an object $X$ and two morphisms $f : X\otimes X \rightarrow X$ and $g : X \rightarrow X\otimes X$ corresponding to the triangles "A" and "B" respectively (more concretely say that we are in the symmetric monoidal category free on such a data). Then the notation comes from the string calculus for a symmetric monoidal category. Thus for example, "A1 + 0B" would be the composite $$X\otimes X \xrightarrow{1_X\otimes g} X\otimes X\otimes X \xrightarrow{s\otimes 1_X} X\otimes X\otimes X \xrightarrow{1_X\otimes f} X\otimes X,$$ where $s$ stands for the symmetry of the monoidal category.<|endoftext|> TITLE: Why do we need localization by Leftschetz motive? QUESTION [10 upvotes]: Definition of the Grothendieck group and Leftschetz motive. The Grothendieck group of varieties is a free abelian group generated by classes of algebraic varieties with the following relation: $$ [X]=[X\smallsetminus Y]+[Y] $$ for $Y\subset X$ a closed subvariety. Let $\mathbb L$ denote the class of an affine line (the Leftschetz motive). For the base field $\mathbb C$ the Grothendieck group forms a ring ($[X\times Y]=[X]\times[Y]$). Cancelation conjecture. This conjecture says the Lefschetz motive $\mathbb L$ is not a zero divisor in the Grothendieck ring, i.e., $\mathbb L^{-1}$ exists. A counterexample. Paper [L. Borisov: Class of the affine line is a zero divisor in the Grothendieck ring, arXiv:1412.6194] provides two varieties $X$ and $Y$ (they are Calabi--Yau varieties with the same derived category of coherent sheaves, but they are not birationally equivalent) with the following relation in the Grothendieck ring: $$ ([X]-[Y])(\mathbb L^2-1)(\mathbb L-1)\mathbb L^7=0, $$ i.e., $\mathbb L$ is a zero divisor (of course, paper shows that all other factors are not zero). There are papers that use the Cancelation conjecture, say [S. Galkin, E. Shinder: The Fano variety of lines and rationality problem for a cubic hypersurface, arXiv: 1405.5154]. My question is (1) Is the Cancelation conjecture is important and why? or, more precise, (2) Which theorem and theories use the Cancelation conjecture? And which do not? and also (3) How does the results of Lev Borisov affect on these theories? Do we need the existence of $\mathbb L^{-1}$ or it is just formal requirement? Maybe, my questions show my misunderstanding the difference between Motives and the Grothendieck group. I'll be glad if you correct me and help me understand the questions in both cases. REPLY [8 votes]: This is more of an opinion than a complete answer but it is too long to leave as a comment. Let $$H : K_0(\mathcal{M}_{\mathbb{C}}) \to K_0(MHM_{\mathbb{C}})$$ be the Hodge realization from the Grothendieck ring of Chow motives to the Grothendieck ring of mixed Hodge modules, then $H$ kills all $(\mathbb{L}-1)$-torsion. However, a conjecture concerning a weight filtration on $\mathcal{M}_{\mathbb{C}}$ implies that $K_0(\mathcal{M}_{\mathbb{C}})$ does not have $(\mathbb{L}-1)$-torsion, and hence, the closure $\bar K$ of $K_0(Var_{\mathbb{C}})[\mathbb{L}^{-1}]$ in the completion of $K_0(Var_{\mathbb{C}})[\mathbb{L}^{-1}]$ along the dimensional filtration implies that the natural morphism $K_0(Var_{\mathbb{C}})[\mathbb{L}^{-1}]\to K_0(\mathcal{M}_{\mathbb{C}})$ factors through $\bar K$. Independently of whether this conjecture is true or not, the composition $K_0(Var_{\mathbb{C}})[\mathbb{L}^{-1}] \to K_0(MHM_{\mathbb{C}})$ does factor through $\bar K$. Therefore, generically speaking, types of problems which concern stabilization of a sequence of elements in $K_0(Var_{\mathbb{C}})[\mathbb{L}^{-1}]$ giving rise to invariants in $K_0(MHM_{\mathbb{C}})$ and beyond will be unaffected. So, results concerning counting points or hodge numbers for example via ideas from motivic integration would remain valid. So, if $\mathbb{L}$ is a zerodivisor, then the natural morphism $K_0(Var_{\mathbb{C}})[\mathbb{L}^{-1}]\to \bar K$ is not injective, yet this does not imply that the conjecture on a weight filtration of chow motives is wrong. In the above, I am mostly following some remarks in the following two papers: J. Denef and F. Loeser, Germs of arcs on singular algebraic varieties and motivic integration, Inventiones Mathematicae 135 (1999), 201-232. J. Denef and F. Loeser, Motivic Igusa zeta functions, Journal of Algebraic Geometry 7 (1998), 505-537.<|endoftext|> TITLE: Mathematical value of constructing sphere eversions QUESTION [8 upvotes]: I am extremely impressed by the work that has been done constructing sphere eversions, and other similar explicit geometrical proofs. In particular, surely nobody can fail to be impressed by the famous video 'Outside In' by Bill Thurston and collaborators, and book (Amazon, free pdf) by Scott Carter. However, we already know when such eversions can be constructed, as the embeddings of $k$-spheres up to regular homotopy in $\mathbb{R}^n$ have been characterized by Smale's theorem. So here is my question: What is the mathematical value, aside from the aesthetic quality of the proofs, of finding explicit regular homotopies? Let me be clear: these proofs are extraordinarily beautiful, and exhibiting beauty is (in my opinion) absolutely valid as a primary goal of mathematics. But, just for the purposes of this question, I would like to put this aside. To put my question in a different way: suppose I exhibited in an ingenious way some new explicit regular homotopy $M \sim M'$ of immersed manifolds, such that $M$ and $M'$ were already known in a nonconstructive sense to be regular homotopic. Other than aesthetic appreciation, why would a homotopy theorist be interested in studying my result? Are there, for example, important open conjectures about the minimal length of regular homotopies? REPLY [4 votes]: Inspired by Thurston's "on proof an progress in mathematics", one can say that the goal of mathematics is to improve our understanding of mathematical objects. The value of any construction which is simple enough to be made explicit, or visualized, of a phenomenon that has been abstractly known to exist, is then obvious: it improves our understanding of the phenomenon. This applies to sphere eversions as well as $C^1$ embeddings of flat tori in $\mathbb{R}^3$, for example. It also applies to new, simpler proofs of known theorems.<|endoftext|> TITLE: The 'class version' of almost disjoint sets: can it fail? QUESTION [5 upvotes]: I have a question about 'class versions' of almost disjoint sets. To even state what I'm after, I need to go beyond what one can state in theories like NBG or MK. I'm wondering about the status of the following: There is a class C of proper classes of sets with the following two properties (1) C is almost disjoint in the sense that every two different members of C intersect in a set, and yet (2) The size of C is larger than the size of the class V of sets in the sense that there is no injection of C into V. I don't have a preferred theory in mind in which to discuss these, I'm interested in hearing about what the possible approaches could be. It is clear that (1) + (2) is consistent from an inaccessible, but my question is whether these are provable in any natural system. Even more, I'd be interested in whether there are models where there is a class of classes C where (1) + (2) fails. The reason that I'm interested in these is that they are connected to results on functors on the category of classes that I'm thinking about. Once the matter of almost disjoint classes is cleared up, I might be back with a follow up on the question that motivates this. REPLY [7 votes]: Here is a way to formalize the almost-disjointness phenomenon in GBC. You don't need an inaccessible cardinal. Theorem. In any model of GBC, there is definable transformation of classes $X\mapsto X^*$, such that if $X\neq Y$, then $X^*$ and $Y^*$ are almost disjoint, and there is no definable map from sets $a\mapsto X_a$ such that every $X^*$ is some $X_a$. (What I mean by a transformation of classes is that there is a formula $\phi(X,Y)$ with only first-order quantifiers and class variables $X$ and $Y$ such that GBC proves that every class $X$ has one and only one class $Y=X^*$ for which $\phi(X,X^*)$ holds.) Proof. Let $X^*=\left\{X\cap V_\alpha \mid\alpha\in\text{Ord}\right\}$. If $X\neq Y$, then $X\cap V_\alpha\neq Y\cap V_\alpha$ for all sufficiently large ordinals $\alpha$, and so $X^*\cap Y^*$ is a set. Thus, they are almost disjoint. But I claim that there is no definable map $a\mapsto X_a$ that is surjective onto the classes $X^*$. For any map $a\mapsto X_a$, let $A=\{\ a\ \mid a\notin X_a\ \}$, and it is easy to see that $A\neq X_a$ for any $a$. Since $X=\bigcup X^*$ for any $X$, it follows that $A^*\neq X_a^*$ for any set $a$. QED In particular, there can be no "injection" of the classes $X^*$ into the sets, since by inverting that we would get such a forbidden surjection $a\mapsto X_a^*$. REPLY [7 votes]: There is an easy way to construct a bijection between the collection of all classes and a certain collection of almost-disjoint classes: send each class $X$ to the class $\{X \cap V_\alpha : \alpha \in \mathrm{Ord}\}.$ Theories with collections of classes should support this construction with relative ease since this argument makes no essential use of collections of classes. So the question becomes whether the collection of classes is larger than $V$. This is true, provided some relatively basic combinatorics. By the usual diagonal argument, no map from $V$ into the collection of classes can be onto: given such a map $F$, consider the class $R = \{x \in V : x \notin F(x)\}$. This requires comprehension with higher-order parameters, so this could potentially fail in a very weak theory with collections of classes. Since the second fact talks about surjections rather than injections, we also need the fact that if there is an injection from the collection of all classes into $V$ then there is a surjection from $V$ onto the collection of all classes. This is trivial, given such an injection, map each set to its preimage if there is one, or to $\varnothing$ (say) if there is none. Without further knowledge about your theory of collections of classes, there is no way to know whether these three pieces work. However, I suspect whatever you have in mind satisfies all three pieces.<|endoftext|> TITLE: A categorical method to, say, determine the cardinality of a group QUESTION [10 upvotes]: I am trying to figure out how much one can figure out about an object using category theory. Ideally, any property that is well defined up to isomorphism should be computable using only category theory. Let us say that we are trying to figure out how many elements are in a group? For a set, we could "simply" count the morphisms to it from the terminal object. Obviously, this wouldn't work. Is there a "categorical" method to find the cardinality of a group? If one wants a rigorous definition of what I mean by "categorically", here is your compass and straightedge: You have an abstract symbol for the object in question. For any two abstract symbols for objects $G$ and $H$, you can get abstract symbols for $Hom(G,H)$ For any abstract symbol for an object $G$, you can get the abstract symbol for $id_G$. For any two abstract symbols for morphisms $f$ and $g$ such that $f \circ g$ is defined, you can get the abstract symbol of $f \circ g$ In terms of the objects and morphisms for abstract symbols you already have, you can get the abstract symbol of any object that uniquely (up to isomorphism) satisfies a given universal property (if needed, I can clarify this.) Using a finite number of steps, I am trying to find the cardinality of the group. Note for example, you can't take the forgetful functor from $Grp$ to $Set$ (since you can't compute a functor on an abstract symbol.) I am thinking one method would be to count how many automorphisms there are, but am I not sure how to get the cardinality of the group from this. If these set of "rules" are too restrictive, it would be interesting to see how they could be lightened to make it possible. REPLY [6 votes]: A bunch of bits and pieces from a bunch of people: Let us say we are trying to find the cardinality of the Group $G$. First, we select the group, $\bullet$ (unique up to isomorphism) such that for any other group, there is a unique arrow in and out of $\bullet$. Now, we select a group $\mathbb{Z}$, unique up to isomorphism, such that it only has two idempotent homomorhpisms, and it admits at least two morphisms to any other group besides $\bullet$ (thanks Todd Trimble). Now we find how many morphisms there are from $\mathbb{Z}$ to $G$ (special thanks to Steven Gubkin). This is $|G|$. Proof The group $\bullet$ is the trivial group. Now, the group $\mathbb{Z}$ of integers satisfies the properties above via the proof here. Now, for any other group $H$ which satisfies the properties, we know that there is a nontrivial morphism $f : H \rightarrow \mathbb{Z}$. Now the image of $f$ will be a group of integers, and so must be multiples of a given integer $n$ (which won't be zero since $f$ is nontrivial.) We can take a map from this to all the integers, so that we can turn $f$ into a surjection $\bar f$. Now, we make a morphism from $\mathbb{Z}$ to $H$, $i$, such that $i(1) = x$ for some $\bar f(x) = 1$. Since $\bar f(i(n)) =\bar f(x+x+x+\dotsb)=\bar f(x)+ \bar f(x)+ \bar f(x)+\dotsb=1+1+1+\dotsb=n$, $\bar f \circ i = id_\mathbb{Z}$. This means that $i \circ \bar f$ is idempotent, and since $i(\bar f(x)) = x$, it is not the zero morphism. Since $H$ has only two idempotents (the zero morphism and $\operatorname{id}_H$), and $i \circ \bar f \neq 0$, $i \circ \bar f = id_H$. (Thanks Slade.) Therefore they are inverses. Therefore we can select $\mathbb{Z}$ up to isomorphism. For each element of $G$, $x$ we make a morphism from $\mathbb{Z}$ to $G$, $h$, such that $h(n) = n * x$. Also, for any morphism $h: \mathbb{Z} \rightarrow G $, $h(n) = n * h(1)$, where $h(1)$ is an element of $G$. Therefore, the morphisms between the integers and $G$ are in one to one correspondence with the elements of $G$. Therefore $|\operatorname{Hom}(\mathbb{Z}, G)| = |G|$. $\square$<|endoftext|> TITLE: Some questions about the ring Z((x)) QUESTION [13 upvotes]: $\newcommand{\ZZ}{\mathbb{Z}}$ $\newcommand{\dim}{\text{dim }}$ Let me begin by apologizing for the length of this question, but I thought this might be interesting to some of you. This ring isn't exactly contrived and yet it seems sort of mysterious. This is a follow up to an earlier question What is the etale fundamental group of Spec Z((x))? where it was proved that $\ZZ((x)) := \mathbb{Z}[[x]][x^{-1}]$ has no nontrivial etale extensions, so its etale fundamental group is trivial. From basic formulas we know that the units are precisely the laurent series whose leading coefficient (ie, ``first nonzero coefficient'') is a unit in $\mathbb{Z}$ (ie, is $\pm1$). Since $\mathbb{Z}$ is a PID, $\mathbb{Z}((x))$ is a unique factorization domain. Question 1: Is there a nice way to describe the fraction field of $\mathbb{Z}((x))$? Note that for any $f(x) = x^{-n}(a_0 + a_1x + a_2x^2+\cdots)\in\mathbb{Z}((x))$, we have $\frac{1}{f(x)}\in\mathbb{Z}[a_0^{-1}]((x))$, so Frac $\mathbb{Z}((x))\ne \mathbb{Q}((x))$ (for example, $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\notin$ Frac $\mathbb{Z}((x))$). On the other hand, a series like $$\sum_{n\ge 0}\frac{x^n}{r^{2^n}}\in\mathbb{Z}[r^{-1}]((x))$$ shouldn't be the inverse of anything in $\mathbb{Z}((x))$, since the denominators grow too fast, and the inverse of $f(x)$ described above has denominators growing roughly like $\frac{1}{r},\frac{1}{r^2},\frac{1}{r^3},\cdots$. Question 2: What is the dimension of $\mathbb{Z}((x))$? Clearly $\dim\mathbb{Z} = 1$, and I'm pretty sure $\dim\mathbb{Z}[[x]] = 2$ with all prime ideals having the form $(0),(p)$, or $(p,x)$. Thus I'm similarly pretty sure that $\dim\mathbb{Z}((x)) = 1$ again, but I don't have a proof (partially due to the fact that the localizations of $\mathbb{Z}((x))$ seem hard to describe). If one is to believe that $\dim\mathbb{Z}((x)) = 1$, then since it's a UFD, it must also be a PID. A natural question is: Question 3: Are the rings $\mathcal{O}_K((x))$ Dedekind domains? ($\mathcal{O}_K$ is the ring of integers of some number field $K$). Anyway, back to $\ZZ((x))$. This ring seems to have at least two types of primes ideals. Clearly rational primes $p\in\mathbb{Z}$ give maximal ideals with residue fields $\mathbb{F}_p((x))$. On the other hand, any power series of the form $f(x) = p + a_1x + a_2x^2 + \cdots$ for some prime $p$ is also irreducible, hence prime (any factorization must include a factor of the form $\pm 1 + b_1x + b_2x^2+\cdots$, which is a unit), and is not equivalent to a rational prime $p$ as along as $p\nmid f(x)$ . If one believes that $\dim\mathbb{Z}((x)) = 1$, then these also give maximal ideals. One can prove that any power series $a_0 + a_1x + a_2x^2 + \cdots$ with $a_0$ divisible by at least 2 primes is composite. On the other hand, if $a_0 = p^2$, then for it to be composite, in which case it must factor as $(p + b_1x + \cdots)(p + c_1x + \cdots)$, it is necessary but not sufficient that $p\mid a_1$. Question 4: What are the residue fields of the form $\ZZ((x))/f(x)$, where $f(x) = p^n + a_1x + a_2x^2 + \cdots$? (assuming $f$ is irreducible and $p\nmid f$) An easy case is $f(x) = p - x$, in which case it seems pretty clear that the quotient field is $\mathbb{Q}_p$. Another example is $p^2 - x$, in which case it seems like the quotient field is also $\mathbb{Q}_p$. For now, lets restrict to the case $n = 1$, so $f(x) = p + a_1x + a_2x^2 + \cdots$. If $p\nmid f(x)$, then the residue field must be characteristic 0, so it must be an extension of $\mathbb{Q}$. One may ask, what does $x$ map to? Since $x$ is a unit in $\ZZ((x))$, it can't map to 0. Furthermore, it must map to something such that arbitrary power series in $x$ makes sense, so the image should be some kind of field which is complete w.r.t. a nonarchimedean valuation, in which $x$ gets sent to something of positive valuation, and such that $a_1x + a_2x^2 + \cdots = -p$. The only extensions of $\mathbb{Q}$ that I can think of which fit this description are $\mathbb{Q}((t))$ and $\mathbb{Q}_l$. The first can't be a residue field, since $a_1x + a_2x^2 + \cdots \ne -p$, so lets suppose the residue field is some $\mathbb{Q}_l$. Since $v(a_1x + a_2x^2 + \cdots) = v(a_1x) = v(-p)$, and $v(a_1x)\ge v(x) > 0$, $v(-p) > 0$, so $l = p$. Just by considering $f(x) = p - x^n$, we get all totally ramified extensions of $\mathbb{Q}_p$ as residue fields. My intuition says that the residue fields of $\ZZ((x))$ are either $\mathbb{F}_p((x))$ or finite extensions of $\mathbb{Q}_p$. However, it's unclear to me if you can get unramified extensions as well as ramified extensions, and it's unclear why there can't be other weird complete fields that could show up. REPLY [11 votes]: Question 1: The fraction field is the same as that of $\mathbb{Z}[[x]]$. It can be gotten by inverting all irreducibles. The irreducibles of the UFD $\mathbb{Z}[[x]]$ are described in Theorem 1.4 of http://arxiv.org/abs/1107.4860 Question 2: You missed a lot of prime ideals of $\mathbb{Z}[[x]]$. (It has uncountably many.) In particular, its height one primes are principal (since it is a UFD) and generated by an arbitrary irreducible power series (as described in Question 1), and none of them are maximal. The maximal ideals are all of height two and of the form $(p,x)$, and they are killed by inverting $x$, so the Krull dimension of $\mathbb{Z}((x))$ is indeed $1$ (hence $\mathbb{Z}((x))$ is a PID). Question 3: I believe user74230 answered that correctly. Question 4: This is answered by Theorem 1.2 of http://arxiv.org/pdf/1107.4860v4.pdf In particular, the integral domains $\mathbb{Z}[[x]]/(f)$ for irreducible $f$ (besides $x$ and primes $p$) are precisely $\mathbb{Z}_p[\alpha]$ for $\alpha$ in the unique maximal ideal of the integral closure of $\mathbb{Z}_p$ in $\mathbb{C}_p$ (or equivalently for $\alpha$ in $\overline{\mathbb{Q}_p}$ with positive $p$-adic valuation), where $p$ is any prime. Since $\alpha$ is the image of $x$ in these isomorphisms and localization commutes with quotient rings, the answer to your question is all fields of the form $\mathbb{Q}_p(\alpha) = \mathbb{Z}_p[\alpha][\alpha^{-1}]$, where $\alpha$ is as described earlier.<|endoftext|> TITLE: exceptional cases in Kazhdan-Lusztig QUESTION [6 upvotes]: The Kazhdan-Lusztig story doesn't apply to the four exceptional cases $(E_6)_1$, $(E_7)_1$, $(E_8)_1$, $(E_8)_2$ (see this earlier question of mine). What's special about those cases? REPLY [2 votes]: If I remember correctly, Kazhdan and Lusztig first pick an appropriate irreducible rigid object/module ${\bf V}^{\kappa}_{\lambda}$ in the tensor category $\tilde{\mathcal{O}}_{\kappa}$ of $\tilde{\frak{g}}$-modules (here $\kappa-h^{\vee}$ is the level). Then they use this object to prove rigidity of the whole tensor category. But ${\bf V}^{\kappa}_{\lambda}$ is by definition an induced representation (Weyl module), where you induce from the finite-dimensional $\frak{g}$-module $L(\lambda)$. As you know, induced modules are reducible in general, so you have to choose $\ell$ negative enough in order to be irreducible (there is a simple condition for that). For example, for $E_6$ you choose $L(\lambda)$ to be a smallest non-trivial irreducible representation (there are two of dimension $27$). Of course, you don't pick the trivial trivial representation; ${\bf V}_0^{\kappa}$ is the unit object in the category. The irreducibility condition implies $\kappa \leq -14$ [There is something wrong here; see Remark 1]. Finkelberg reverses $\kappa$ to $-\kappa$, so you get the level to be at least $14-12=2$. Remark 1: What I said above is probably incorrect. In KL (Part I, Proposition 2.12) it was proved that ${\bf V}^\kappa_{\lambda}$ is irreducible if $\langle \lambda+ 2\rho, \lambda \rangle < -2\kappa$, where $\kappa \in \mathbb{Q}_{<0}$. For $E_6$ there are two $27$-dimensional modules $L(\omega_1)$ and $L(\omega_5)$ of dimension $27$. For $\omega_1$ (same for $\omega_5$) I'm getting $\langle \omega_1+2 \rho,\omega_1 \rangle=17.33..$ so $\kappa \leq -9$ already guarantees irreducibility. I guess one also has to consider the rigidity condition (Part IV). Remark 2: There seems to be a mismatch between the values of $\kappa$ for which $\tilde{\mathcal{O}}_{-\kappa}$ is rigid in Finkelberg's paper and in KL. In 3.15 (Corollary 2) of KL, Part IV, they list $\kappa \geq 14,20,{\bf 26}$ for $E_{6,7,8}$ resp. while Finkelberg (in Section 2.6) claims $\kappa \geq 14,20,{\bf 33}$, referring to the same result of KL. It's either a typo or Finkelberg for some reason chooses more conservative lower bound $3+h^{\vee}$ which applies to all types. If $\kappa \geq 26$ for $E_8$ is correct, level 1 and 2 are non-issues.<|endoftext|> TITLE: Conditions for positivity of Fourier transform QUESTION [10 upvotes]: Assume you are given a non-negative, continuous, radial function $f\in L^q(\mathbb{R}^3)$ (for any $q\geq 1$). Are there any conditions which would guarantee that the Fourier transform of $f$, that is $\hat{f}(p)$, is also non-negative? REPLY [9 votes]: the three-dimensional Fourier transform $F(\vec{p})$, of the radial function $f(r)$ has Fourier transform $$F(\vec{p})=\int_0^\infty dr\int_0^\pi d\theta \int_0^{2\pi}d\phi\;e^{ ipr\cos\theta}f(r) r^2\sin\theta$$ $$\qquad=\frac{4\pi}{p}\int_0^\infty rf(r)\sin(pr)\,dr,\;\;{\rm with}\;\;p=|\vec{p}|.$$ so you're asking when the Fourier-sine-transform $S(p)$ of $rf(r)$ will be (pointwise) positive for $p>0$. A sufficient condition is that $rf(r)$ is a decreasing function of $r>0$, see On positivity of Fourier transforms. For a related question, see this MO post. For the connection to Bochner's theorem: the OP's question amounts to finding a function that is both positive and positive-definite, since positive functions have positive-definite Fourier transforms.<|endoftext|> TITLE: Subadditivity of the square root for matrices QUESTION [6 upvotes]: For positive numbers $a$ and $b$ we have the inequality $\sqrt{a+b} \leqslant \sqrt{a} + \sqrt{b}$. Is it true that the same holds if we take $a$ and $b$ to be positive semidefinite matrices? If not, there is a weaker statement that I am interested in: Is it true that the inequality $\sigma_n( \sqrt{A} - \sqrt{B}) \leqslant \sigma_n(|A-B|^{\frac{1}{2}})$, where by $\sigma_n$ I denote the $n$-th singular value (they are listed in decreasing order), holds? If even this fails, maybe the following is true: $\sigma_{2n}(\sqrt{A} - \sqrt{B}) \leqslant \sigma_{n} (|A-B|^{\frac{1}{2}})$? I suspect that analogous inequalities should hold for any $0 0$. Here is an explicit example: \begin{equation*} A = \begin{pmatrix} 19 & 17 & 9\\ 17 & 17 & 11\\ 9 & 11 & 11\end{pmatrix},\quad B = \begin{pmatrix}19 & 11 & 21\\ 11 & 9 & 15\\ 21 & 15 & 27\end{pmatrix},\quad A-B = \begin{pmatrix}0 & 6 & -12\\ 6 & 8 & -4\\ -12 & -4 & -16\end{pmatrix}. \end{equation*} Now, $\sigma_n(\sqrt{A}-\sqrt{B}) = 0.1853...$, while $\sigma_n(|A-B|^{1/2})=0$. However, a weaker claim that holds is described in this MO post, namely a weak majorization relation: \begin{equation*} \|f(A) - f(B)\| \le \|f(|A-B|)\|, \end{equation*} for any symmetric (i.e., unitarily invariant) norm $\|\cdot\|$ and where $f(t) = t^r$, for $0< r < 1$, and more generally, $f$ is a nonnegative concave function.<|endoftext|> TITLE: Why is this group called "The Holomorph of a group" QUESTION [7 upvotes]: Many years ago I found in google the notation "Holomorph of group". It is the semi direct product of $G$ with $Aut(G)$. Why is the term "Holomorph" used here, while it is usually used for complex analytic functions? More information on this object is very appreciated. REPLY [8 votes]: I'm not a history expert, but according to Miller, Blichfeldt and Dickson: "Theory and applications of finite groups" (1916), footnote p. 46: "The concept of holomorph was used by many early writers, but the term was introduced by W. Burnside in the first edition of his Theory of Groups, 1897, p. 228."<|endoftext|> TITLE: on the local structure of schemes QUESTION [9 upvotes]: Let $X$ be an integral finite type scheme over $\mathbb{C}$. Let $x\in X$, such that there exists a neighborhood $U$ of $x$, such that the sheaf of differentials $\Omega^{1}_{U}$ decomposes into: $\Omega_{U}=E\oplus T$, where $E$ is a locally free sheaf of rank $d$ and $T$ just coherent. Can we prove that locally for étale topology around $x$, $U$ is isomorphic to $\mathbb{A}^{d}\times Y$, for a certain scheme $Y$? REPLY [6 votes]: Edit, 12/2016. The original post was long ago, but I had a reason to revisit this some months back when I was asked a similar question. I am including below the positive answer to this question that I found in a special case. There is a little setup required. Flatness over $\mathbb{A}^r$. First there is a flatness lemma. Let $k$ be a commutative, unital ring (for instance, $k=\mathbb{Z}$). Let $A$ be a commutative, unital $k$-algebra that is Noetherian. Let $a\in A$ be an element. By the Noetherian hypothesis, the following chain of ideals stabilizes, $$\{0\}=\text{Ann(1)}\subset \text{Ann}(a^1)\subset \text{Ann}(a^2) \subset \dots \subset \text{Ann}(a^n) \subset \text{Ann}(a^{n+1}) \subset \dots $$ There exists a least nonnegative integer $n$ such that $\text{Ann}(a^n) = \text{Ann}(a^{n+1})$. Of course $n$ equals $0$ if and only if $a$ is a nonzerodivisor in $A$, and in that case $n$ is a zerodivisor in $A$ (since $n$ equals $0$). The following lemma proves that Moret-Bailly's examples above are definitely a positive characteristic phenomenon. Lemma. If the integer $n$ is a nonzerodivisor in $A$, then for every $k$-derivation $\delta:A\to A$, $\delta(a)$ is a zerodivisor. Proof. This is basically the same proof as in the existence of primary decompositions for Noetherian modules. Let $b$ be an element in $\text{Ann}(a^n)$ that is not in $\text{Ann}(a^{n-1})$. Since $a^nb$ equals $0$ in $A$, $-na^{n-1}b\delta(a)$ equals $a^n\delta(b)$. Since $a^nb$ equals $0$, $$a^{n+1}\delta(b) = -n(a^nb)\delta(a) = 0.$$ Since $\delta(b)$ is in $\text{Ann}(a^{n+1}),$ already $\delta(b)$ is in $\text{Ann}(a^n)$. Thus, $a^n\delta(b)$ equals $0$, i.e., $(-na^{n-1}b)\delta(a)$ equals $0$. By hypothesis, $a^{n-1}b$ is nonzero. Since also $n$ is a nonzerodivisor in $A$, $-na^{n-1}b$ is nonzero. Thus, $\delta(a)$ is a zerodivisor in $A$. QED Let $r>0$ be an integer (the letter "d" is needed for differentials). Denote by $S_r$ the polynomial $k$-algebra $k[s_1,\dots,s_r]$. Let $(a_1,\dots,a_r)\in A^r$ be a collection of elements, and let $f:S_r\to A$ be the unique $k$-algebra homomorphism with $f(s_i)=a_i$ for every $i=1,\dots,r$. There is an associated $A$-module homomorphism $f^*:\Omega_{S_r/k}\otimes_{S_r} A \to \Omega_{A/k}$. Proposition. If $A$ contains $\mathbb{Q}$, and if there exists an $A$-linear retraction $\rho:\Omega_{A/k}\to \Omega_{S_r/k}\otimes_{S_r} A$ of $f^*$, then $(a_1,\dots,a_r)$ are a regular sequence. In that case, assuming also that $k$ is Noetherian and that $A$ is $k$-flat, then $f$ is a flat ring homomorphism. Proof. This is proved by induction on $r$. First of all, composing the retraction $\rho$ with projection onto the summand $Ads_r$, there exists a $k$-derivation $\delta:A\to A$ with $\delta(a_r)$ equal to $1$. Thus, by the previous lemma, $a_r$ is a nonzerodivisor. When $r$ equals $1$, this proves that the sequence is a regular sequence. Assuming further that $k$ is Noetherian and $A$ is $k$-flat, by applying the Local Flatness Theorem, Theorem 22.2, p. 174 of Matsumura's "Commutative ring theory", also $k[t_1]\to A$ is flat. Thus, by way of induction, assume that $r>1$, and assume that the result is proved for $r-1$. Form the quotient ring $\overline{A}=A/\langle a_r \rangle$ and the image sequence $(\overline{a}_1,\dots,\overline{a}_{r-1})\in \overline{A}^{r-1}$. Form the projection onto the first $r-1$ summands, $$\Omega_{S_r/k}\otimes_{S_r} A \to \Omega_{S_{r-1}/k}\otimes_{S_{r-1}} A \to \Omega_{S_{r-1}/k}\otimes_{S_{r-1}}\overline{A}.$$ Using the fundamental exact sequence, $$ \overline{A}\xrightarrow{d(a_r)} \Omega_{A/k}\otimes_A \overline{A} \to \Omega_{\overline{A}/k} \to 0, $$ the projection above factors through a retraction, $$\overline{\rho}:\Omega_{\overline{A}/k} \to \Omega_{S_{r-1}/k}\otimes_{S_{r-1}}\overline{A}.$$ Thus, by the induction hypothesis, $(\overline{a}_1,\dots,\overline{a}_{r-1})$ is a regular sequence in $\overline{A}$. Thus, by definition, also $(a_1,\dots,a_r)$ is a regular sequence in $A^r$. Assuming further that $k$ is Noetherian and $A$ is flat, then $A/\langle a_r\rangle$ is flat over $S_r/\langle s_r \rangle = S_{r-1}$. Thus, by the Local Flatness Criterion once more, $A$ is flat over $S_r$. So the propostion is proved by induction on $r$. QED For any rank $r$, free summand of $\Omega_{A/k}$, since $\Omega_{A/k}$ is generated by symbols $da$ for $a\in A$, at least up to passing to a finite Zariski open cover of $\text{Spec}(A)$, we can assume that there exist elements $a_1,\dots,a_r$ such that the retractions of $d(a_1),\dots,d(a_r)$ form a basis for the summand. Let $f$ be the associated ring homomorphism. By the proposition, $f$ is flat. That is enough to conclude the local product structure, at least when the fibers of $f$ have isolated singularities (as in my proposed counterexample). Hypothesis. Assume that $k$ is a field containing $\mathbb{Q}$, assume that $f$ is locally finitely presented, and assume that the closed fibers of $f$ have only isolated singularities. Here is the e-mail I sent to a colleague about this situation. Dear ***, As soon as you left, I wrote out the computation for 4 concurrent lines, and I saw that you are (of course) correct: the extension class is zero if and only if the cross-ratio of the four lines is constant. Then I had to teach my class. I just finished, and I remembered another reference. That reference quickly leads to a proof of the formal local splitting whenever you have vanishing of the Ext class, at least in characteristic 0, at least for germs of isolated singularities. The reference is in the following book of Michael Artin. Lectures on Deformations of Singularities Volume 54 of mathematics: Tata Institute of Fundamental Research lectures Michael Artin, C. S. Seshadri, Allen Tannenbaum Tata Institute of Fundamental Research, 1976 http://www.math.tifr.res.in/~publ/ln/tifr54.pdf In Theorem 8.1, Artin establishes that there is a versal deformation space $\text{Spf}(R)$ for a germ of an isolated singularity $(X,0)$ with $X$ locally finitely presented over a field. Earlier, on pp. 26, 27, in Theorem 6.2 and Proposition 6.1, Artin proves that for a flat, finitely presented morphism, $$f : X \to \text{Spec}(A),$$ with a section, $$s: \text{Spec}(A) \to X,$$ that is a family of isolated singularities of $f$, for the induced morphism to the versal deformation space, $$g : \text{Spec}(A)^{\wedge} \to \text{Spf}(R), $$ the induced derivative is given by the map from the global Ext^1 you were describing to a natural target space T^1 introduced by Schlessinger. In particular, if your Ext class is zero, the derivative is zero. Of course a nonzero morphism can have a zero derivative at one point. However, by generic smoothness / Sard's Theorem, in characteristic $0$ a $k$-morphism from a smooth $k$-scheme to an arbitrary (locally finite type) $k$-scheme is constant if the derivative map is everywhere zero. Perhaps you can find a counterexample in positive characteristic $p$ by considering the family of concurrent lines, $\text{Zero}(xy(y-x)(y-t^p x))$ in $\mathbb{A}^2 \times \mathbb{A}^1$ with coordinates $((x,y),t)$, where $\mathbb{A}^1$ is the base of the deformation. For non-isolated singularities, I am not sure what is the best way to proceed. Often there is no versal deformation space (at least not as the formal completion of a finite type scheme at a point), but perhaps the proofs of Theorem 6.2 and Proposition 6.1 still apply. You may be able to find something in the book on deformations of singularities by Greuel, Lossen, and Shustin. Jason Starr, April 2016. Original answer. I am adding below my original (wrong) answer, as well as the edit I made right after posting the answer. I believe this is false. There are equisingular families that are not étale locally, nor even formally locally, trivial. Let $(z,w)$ be coordinates in $\mathbb{A}^2$. Let $t$ be a coordinate on some dense Zariski open subset $V$ of $\mathbb{A}^1$. Consider the zero scheme $X$ of $zw(w-z)(w-tz)$ in $\mathbb{A}^2\times V$. I believe that $\Omega_X$ will have a direct sum decomposition as above for $V$ a suitable open subset. However, I very much doubt there is any étale decomposition of $X$. If there were, then the fiber of the projection to $Y$ would correspond to the zero locus of $(z,w)$. However, the formal completion of $X$ along this zero scheme is not a product, not even after making a further étale base change of $V$. Edit. I just did the computation for my polynomial above. There is not a direct sum decomposition of $\Omega_X$. I was wrongly assuming that the existence of a direct sum decomposition would be preserved under small deformations. Thus, deforming from a trivial family to an equisingular family would give a counterexample. However, because the Ext group is nonzero, existence of a direct sum decomposition is not preserved under small deformations.<|endoftext|> TITLE: James' theorem—going from the separable case to the general case QUESTION [7 upvotes]: Consider the following famous theorem by Robert C. James (1964): Let $X$ be a Banach space over $\mathbb R$ and $C$ a non-empty, bounded, weakly closed subset. Then, $C$ is weakly compact if and only if every continuous linear real-valued functional on $X$ attains its supremum on $C$. The “only if” part is trivial, while known proofs of the “if” part are notoriously involved. Some texts prove it only for the case in which $X$ is assumed to be separable (see, for example, Holmes, 1975, pp. 157–161). Some sources do not make this extra assumption, yet they tend to be still too involved for my taste and patience (see, for example, Pryce, 1966). Using Holmes (1975), who treats only the separable case, I can prove the following statement: If $C\subseteq X$ is non-empty, bounded, weakly closed, but not weakly compact, then there exists a separable norm-closed subspace $Y\subseteq X$ and a continuous real-valued functional $f:Y\to\mathbb R$ on it such that $C\cap Y$ is not empty and $f$ doesn't attain its supremum on $C\cap Y$. My question is: Do you think the second statement can be used to prove the existence of a continuous real-valued functional $F:X\to\mathbb R$ that doesn't attain its supremum on $C$? Do you think trying extending $f$ onto the whole space by using the Hahn–Banach theorem, or Zorn's lemma, could lead to anything useful? References James, R. C. (1964): “Weakly Compact Sets,” Transactions of the American Mathematical Society 113, 129–140. Holmes, R. B. (1975): Geometric Functional Analysis and Its Applications, New York: Springer-Verlag. Pryce, J. D. (1966): “Weak Compactness in Locally Convex Spaces,” Proceedings of the American Mathematical Society 17, 148–155. REPLY [6 votes]: I got it. I just read Pryce (1966), which is more accessible than I had thought. The proofs by Pryce (1966) and by Holmes (1975) use almost identical ideas, it's just that the latter refrains from dealing with the non-separable case for some reason. Both proofs start out with assuming that $C$ is non-empty, weakly closed, bounded, but not weakly compact. In this case, there exist sequences $\{x_n\}_{n\in\mathbb N}\subseteq C$ and $\{f_m\}_{m\in\mathbb N}\subseteq X'$ with the amusingly pathological property that $\lim_{n\to\infty}\lim_{m\to\infty}f_m(x_n)$ and $\lim_{m\to\infty}\lim_{n\to\infty}f_m(x_n)$ both exist in $\mathbb R$ but they are not equal. Holmes (1975) actually constructs these sequences in a transparent way, but Pryce (1966) does not, just gives a quite obscure reference. Pryce (1966) claims also that the family $\{f_m\}_{m\in\mathbb N}$ is equicontinuous, while the construction of Holmes (1975) states that this sequence is in the closed unit ball of $X'$. What I was intimidated about is that I hadn't recognized immediately the fact that any non-empty family of functionals in the closed unit ball of $X'$ is actually equicontinuous. However, once I recognized this, I could delve into the proof by Pryce (1966) with more confidence and enthusiasm. By the way, the trick Pryce (1966) uses is to consider the separable (with respect to a certain seminorm-topology) subspace of $X'$ given by the linear span of $\{f_m\}_{m\in\mathbb N}$, while Holmes (1975) uses the assumption that $X$ is separable to ensure that the closed unit ball in $X'$ be metrizable in the weak* topology. The point in both cases is to ensure the existence of a subsequence of $\{f_m\}_{m\in\mathbb N}$ with certain properties. I don't really understand why Holmes (1975) makes the separability assumption—certainly restrictive for a textbook that is supposed to make the proofs in published papers more accessible. The trick by Pryce (1966), ingenious as it is, is actually quite elementary and does not require the separability assumption. At any rate, I suggest anyone interested in an accessible proof of James' theorem read first the section in Holmes (1975) that constructs the pair of sequences mentioned above, and then immediately switch to Pryce (1966) and put those sequences into action. The end result will be the construction of a linear functional in $X'$ that fails to attain its supremum on $C$!<|endoftext|> TITLE: Continuous functions with convex level sets QUESTION [10 upvotes]: Assume that $f:\mathbb{R}^{2}\to \mathbb{R}$ is a continuous function such that each level set $f^{-1}(c)$ is a convex set. To what extent such functions are studied? In particular: Is there a partial or total ordering on $\mathbb{R}^{2}$ such that all functions with this property, must be monotone? 2.Is it true to say that $f$ is differentiable, almost every where? These two questions are motivated by the fact that this property for continuous functions $f:\mathbb{R} \to \mathbb{R}$ is equivalent to monoton property and every monoton function from $\mathbb{R}\to \mathbb{R}$ is almost every where differentiable. Note: As usual, a convex set in the plane is a subset $C$ such that $\forall a,b\in C$ and $\forall t\in [0,1]$ we have $ta+(1-t)b\in C$ REPLY [10 votes]: Let's call the functions defined by Ali Taghavi to be sliced functions: a continuous function $\ f:\mathbb R^2\rightarrow\mathbb R\ $ is called sliced $\ \Leftarrow:\Rightarrow\ \ \forall_{c\in\mathbb R}\ f^{-1}(c)\ $ is convex. NOTATION: $$\ [x;y]\ :=\ \{(1\!-\!t)\cdot x\ +\ t\cdot y\ :\ 0\le t\le 1\}\ $$ for arbitrary $\ x\ y\in \mathbb R^n\ $ and $\ n=0\ 1\ 2\ \ldots.\ $ Thus $\ [x;y]=[y;x]\ $ in every dimension including $\ n=1$. THEOREM 0   Let $\ f:\mathbb R^2\rightarrow \mathbb R\ $ be sliced. Then $\ f^{-1}(C)\ $ is convex for every convex $\ C\subseteq\mathbb R$. PROOF   Let $\ C\subseteq \mathbb R\ $ be convex. Let $\ a\ b\in C.\ $ We want to show that $$\ [a;b]\ \subseteq\ C $$ If $\ f(a)=f(b)\ $ then the above holds due to the convex part of the definition of a sliced function. Now, assume $\ f(a)\ne f(b).\ $ Then $\ f^{-1}([f(a);f(b)])\ \cap\ [a;b]\ $ is a closed subset of $\ [a;b].\ $ Next, consider arbitrary $\ x\ y \in [a;b]\cap f^{-1}([f(a);f(b)],\ $ and $\ x\ne y.\ $ If $\ f(x)=f(y)\ $ then again $$ [x;y]\ \subseteq f^{-1}(x)\ \subseteq\ f^{-1}([f(a);f(b)]) $$ And if $\ f(x)\ne f(y)\ $ then, due to continuity of $\ f\ $ (and of the nature of $\ \mathbb R$) we have $$ f([x;y])\ \supseteq\ [f(x);f(y)] $$ Thus there exists $\ w\in[x;y]\ $ such that $$ f(w)\ =\ \frac {f(x)+f(y)}2 $$ We see that $\ w\in(x;y)\ $ belongs to the interior of the interval, and $\ w\in f^{-1}[f(a);f(b)].\ $ This shows that $[a;b]\cap f^{-1}([f(a);f(b)])\ $ is dense in $\ [a;b],\ $ hence $\ [a;b]\subset f^{-1}(C).\ $ END of PROOF After this exercise we get: THEOREM 1   Let $\ f:\mathbb R^2\rightarrow \mathbb R\ $ be sliced. Then there exists $\ (a\ b)\in\mathbb R^2\setminus\{(0\ 0)\}\ $ such that: $$\forall_{(x\ y)\,\ (x'\ y')\,\in\,\mathbb R^2 }\ \ \left(\ a\cdot x+b\cdot y = a\cdot x'+b\cdot y'\ \ \Rightarrow\ \ f(x\ y)=f(x'\ y')\ \right)$$ PROOF   The case of a constant function is trivial. Otherwise there exists $\ h\in\mathbb R\ $ such that both sets $\ f^{-1}((-\infty\;h))\ $ and $\ f^{-1}((h;\infty))\ $ are non-empty. Then these two sets are disjoint open half-planes, i.e. they are non-empty, open and convex, and the complement of each of them is non-empty, closed and convex. Thus there exist $\ s\ t\in\mathbb R\ $ and $\ (a\ b)\in\mathbb R^2\setminus\{(0\ 0\}\ $ such that $\ s TITLE: Surjectivity of curl QUESTION [14 upvotes]: Let: $\mathbb R^3\ni x\mapsto v(x)\in\mathbb R^3$ be a vector field with null divergence belonging to the Schwartz class such that $$ \int_{\mathbb R^3} v(x) dx=0. $$ Is it true that there exists a vector field $w$ in the Schwartz class such that $$\text{curl } w=v\quad?$$ In other words, this is a regularity question for a Poincaré lemma: let $u$ be a closed two-form on $\mathbb R^3$. Then, there exists a one-form $a$ such that $u=da$. If $u$ is smooth, $a$ can be chosen smooth; the above question can be reformulated: if $u$ belongs to the Schwartz class, is it possible to choose $a$ in the Schwartz class? REPLY [3 votes]: I think that the answer is Yes. 1st step. Because Fourier transform is an automorphism of the Schwartz class, the problem is equivalent to show that every vector field $v(x)\in{\mathcal S}({\mathbb R}^3)^3$ verifying $x\cdot v(x)\equiv0$ can be ``divided'', that is can be written $$v(x)=x\wedge w(x), \qquad w\in{\mathcal S}({\mathbb R}^3)^3.$$ 2nd step. The jets at the origin. Let $$v(x)\sim \sum_kv^k(x)$$ be the formal power series of $v$ at the origin, with $v_k$ a homogeneous polynomial vector field of degree $k$. Each $v_k$ satifies $x\cdot v_k(x)=0$. It is an algebraic fact that there exists a polynomial vector field $w_k$, homogeneous of degree $k-1$, such that $v_k=x\wedge w_k$. Choose a $\phi\in{\mathcal D}({\mathbb R}^3)$ be such that $\phi\equiv1$ in $B(0;1)$, and form $$w^0(x)=\sum_k\phi(kx)w_k(x).$$ This is a locally finite series, therefore convergent to a ${\cal C}^\infty$-function away from the origin, compactly supported. It is actually ${\cal C}^\infty$ everywhere, because $w^0$ differs from a ${\cal C}^k$ field by an $O(|x|^{k+1})$. Therefore $w^0$ is in the Schwarz class. Then the jet of $v^0:=x\wedge w^0$ equals that of $v$ at the origin. 3rd step. Away from the origin. Define $v^1:=v-v^0$, which is of Schwartz class and is flat at the origin. Define $$w^1=\frac{x}{|x|^2}\wedge v^1$$ is of Schwarz class and satisfies $v^1=x\wedge w^1$. Finally, $w=w^0+w^1$ is the solution of the problem.<|endoftext|> TITLE: How many knots are there with hyperbolic volume less than a given constant QUESTION [9 upvotes]: Are there any known upper bounds on: $$\#\left\{\text{hyperbolic knots }K\subseteq S^3\middle|\operatorname{Vol}(S^3\setminus K)0$, there can be only finitely many hyperbolic knot complements (or more hyperbolic generally orientable hyperbolic manifolds) that have volume at most $v_8 -\epsilon$. Finding this number would involve classifying manifolds in this set which do not come from filling the Whitehead link. Second, one could ask: for a specific volume, how many hyperbolic manifolds have this volume? Hodgson & Masai investigated this question and found examples of closed manifolds that are distinguished by their volume and link complements that share a volume with at least $f(v)$ other manifolds, where $f(v)$ is a function that grows exponentially in $v$. More relevant to your question, Millichamp followed up on this work by considering (amongst a number of other things) lower bounds on how many hyperbolic knot complements have specific given volumes.<|endoftext|> TITLE: periods of Mixed Hodge Structures QUESTION [7 upvotes]: Two Questions: First. As I know the notion of periods comes when one has two vector spaces over a subfield $k$ of $\mathbb{C}$ (usually given by two cohomology theories) and an isomorphism between their extensions to $\mathbb{C}$. But I'm confused when this notion is used for a mixed Hodge structure. What are two vector spaces over $\mathbb{Q}$ and the isomorphism between their tensor with $\mathbb{C}$? Or we need additional data for attaching periods to a MHS? What is the geometric meaning of periods for cohomology groups of a variety over $\mathbb{C}$? Second. What is the relation between periods of MHS's and Ext groups: $\mathrm{Ext}^i(\mathbb{Q}(0),\mathbb{Q}(n))$? Thanks for your answers and references suggestions. Edit. I found a reference: Goncharov, Volumes of hyperbolic manifolds and mixed Tate motives for the definition of periods for mixed Hodge-Tate structures. It seems that the definition only works for Hodge-Tate structures and cannot be generalized for all MHS's. But the question of "geometric meaning" also remains in this case. REPLY [4 votes]: The classical notion of period, is the (set of) integral(s) of an algebraic differential form along cycles on a variety. As you said yourself, this can be expressed as an entry of the change of basis matrix from algebraic de Rham to singular (or Betti) cohomology. This still makes sense for a singular or open variety. E.g. if $X$ is smooth variety, then we can find a smooth proper compactification $\bar X$ such that $X= \bar X-D$, where $D$ is a divisor with normal crossings. So we have, by Deligne, a comparison map $$H^*(X,\mathbb{Z})\otimes \mathbb{C}\cong H^*(\bar X,\Omega_{\bar X}(\log D))$$ So you get a period matrix as above (after choosing bases). For the second, I am not entirely sure what "period" means either. However, there should certainly be a map from $Ext^*$ in any reasonable category of mixed motives, to $Ext^*$ in the category of polarizable MHS. But note that $Ext^i=0$ for $i>1$ in PMHS (Beilinson), so there is no information in this case.<|endoftext|> TITLE: Ring of differential operators of a quotient ring QUESTION [10 upvotes]: All rings are assumed to have unity. Let $k$ be a field. Recall the definition of Grothendieck's ring of ($k$-linear) differential operators $D(R;k)$ of a commutative $k$-algebra $R$: Definition. Set $D_0(R;k)$ to be $R$ viewed as a subring of $\operatorname{End}_k(R)$ via multiplication, then for all $i>0$, define $$ D_{i+1}(R;k) = \left\{P\in\operatorname{End}_k(R)\,\middle|\, [P,r]\in D_i(R;k)\text{ for all }r\in D_0(R;k)\right\}.$$ Then we set $$D(R;k) = \bigcup_{i=0}^\infty D_i(R;k).$$ Question Context Remark 17.7 of Twenty-Four Hours of Local Cohomology by Iyenger, et al. states the following without proof or citation: Let $I$ be an ideal of a commutative $k$-algebra $R$, and set $S=R/I$. One can identify $D(S;k)$ with the subring of $D(R;k)$ consisting of operators that stabilize $I$, modulo the ideal generated by $I$. I've interpreted this as follows: Let $\operatorname{Stab}(I) = \{P\in D(R;k) \mid P(I) \subseteq I\}$, and let $\Phi\colon \operatorname{Stab}(I) \to \operatorname{End}_k(S)$ be the natural map. Then (i) $\operatorname{im}\Phi = D(S;k)$, and (ii) $\ker\Phi$ is the (two-sided) ideal generated by $I$. Question Can anyone point me to and/or outline a proof of this fact? Specifically, how does one prove that $\operatorname{im}\Phi \supseteq D(S;k)$? (The other inclusion is immediate). The proof of (ii) is not difficult: one direction is immediate. For the other direction, equip $\operatorname{End}_k(R)$ with the left $R\otimes R$-mod structure $(x\otimes y)\varphi = x\varphi + \varphi y$. Let $N=I\otimes R + R\otimes I$ be the kernel of the natural map $R\otimes R\to S\otimes S$. Let $P\in \operatorname{ker}\Phi$, and look at the $R\otimes R$-module $M$ it generates. Then $$ 0 = (S\otimes S)\Phi(P) = M/NM,$$ so $M=NM=NP=IP + PI$, which is contained in the ideal of $\operatorname{Stab}(I)$ generated by $I$. In particular, $P$ is contained in this ideal. REPLY [2 votes]: I found a reference, with proof: it's Theorem 15.5.13 in Noncommutative Noetherian Rings by J.C. McConnell and J.C. Robson; here it is made explicit that $R$ is a polynomial ring. They mention that the result was first proved in Differential operators on an affine curve by S.P. Smith and J.T. Stafford, but that "it was known earlier".<|endoftext|> TITLE: Convex subcomplexes of CAT(0) cubical complexes QUESTION [6 upvotes]: Is the following statement true? If so, can anyone provide a reference? Let $X$ be a CAT(0) cubical complex, and let $Y$ be a connected subcomplex of $X$. Then the following are equivalent: $Y$ is convex in $X$. For every cube $C$ in $X$, the intersection $C\cap Y$ is a face of $C$. Here the empty set is considered a face of every cube, and each cube is a face of itself. Clearly (1) implies (2), and in all the examples I can think of (2) also implies (1). REPLY [4 votes]: In addition to Anton Petrunin's answer, I would like to mention that a more combinatorial argument is possible. Indeed, in a CAT(0) cube complex $X$, a full subcomplex $Y$ (i.e. a subcomplex which contains a cube if it already contains its vertices) is convex with respect to the CAT(0) metric if and only if it is convex with respect to the combinatorial metric; see Theorem 2.13 in Haglund's article Finite index subgroups of graph products. In other words, it suffices to show that the one-skeleton of $Y$ is convex in $X$ endowed with the graph metric. Now, the proposition is a straightforward consequence of the following two observations: Proposition 1: Let $X$ be a CAT(0) cube complex and $\alpha,\beta$ two combinatorial paths with the same endpoints. Assume $\beta$ is a geodesic. There exists a sequence of combinatorial paths $\gamma_1=\alpha, \gamma_2,\ldots, \gamma_{n-1},\gamma_n=\beta$ such that, for every $1 \leq i \leq n-1$, $\gamma_{i+1}$ is obtained from $\gamma_i$ by removing a backtrack or flipping a square. Here, by flipping a square, I mean replacing two consecutive edges of a square by the other two consecutive edges. Proposition 2: Let $X$ be a CAT(0) cube complex and $\alpha,\beta$ two combinatorial geodesics with the same endpoints. There exists a sequence of combinatorial paths $\gamma_1=\alpha, \gamma_2,\ldots, \gamma_{n-1},\gamma_n=\beta$ such that, for every $1 \leq i \leq n-1$, $\gamma_{i+1}$ is obtained from $\gamma_i$ by flipping a square. Proposition 2 is Theorem 4.6 in Sageev's thesis Ends of group pairs and non-positively curved cube complexes, and Proposition 1 can be proved similarly. We deduce that: Let $X$ be a CAT(0) cube complex and $Y \subset X$ a full subcomplex. Then $Y$ is convex if and only if it is connected and every square containing two consecutive sides in $Y$ lies entirely in $Y$. Fix two vertices $a,b \in Y$. Because $Y$ is connected, there exists a combinatorial path in $Y$ between $a,b$. As a consequence of Proposition 1, $\alpha$ can be turned into a combinatorial geodesic $\gamma$ in $Y$. As a consequence of Proposition 2, every geodesic between $a$ and $b$ can be obtained from $\gamma$ by flipping squares, and so must lie in $Y$. We conclude that $Y$ is (combinatorially) convex.<|endoftext|> TITLE: Is the functor of points of a scheme cofinally small? QUESTION [10 upvotes]: Background: In functorial algebraic geometry one would like to consider the category of all functors $\mathsf{CRing} \to \mathsf{Set}$ and define/characterize the category of schemes as a full subcategory. However, there are serious set-theoretic difficulties: this is not a category, it is too large. One possible solution is suggested in Demazure-Gabriel's book. They use three nested universes. I wonder if we may stay in one universe. In general, one constructs the cocompletion $\widehat{\mathcal{C}}$ of a category $\mathcal{C}$ (not assumed to be small) as the category of functors $F : \mathcal{C}^{\mathrm{op}} \to \mathsf{Set}$ which are cofinally small, i.e. the category of elements $\int F$ is cofinally small. Then $\widehat{\mathcal{C}}$ is a category. We may apply this to $\mathcal{C}^{\mathrm{op}}=\mathsf{CRing}$ and develop functorial algebraic geometry inside $\widehat{\mathcal{C}}$. But the question is if the usual category of (geometric) schemes embeds into it. This leads to the following questions: Question. Let $X$ be a scheme. Is there a set of commutative rings $\{A_i\}$ with the property that for every commutative ring $A$, every morphism $\mathrm{Spec}(A) \to X$ factors through some $\mathrm{Spec}(A_i)$? The answer is yes if $X$ is quasi-compact quasi-separated. In fact, in that case $X(-) : \mathsf{CRing} \to \mathsf{Set}$ commutes with filtered colimits so that we may take a skeleton of all finitely generated commutative rings (which is a set). If $X$ is arbitrary, I guess that $X(-)$ commutes with $\lambda$-directed colimits for some cardinal number $\lambda$, which would answer the question. But I'm not sure about that. Question. In case the answer to the first question is "Yes": Can we choose that set in such a way that for every $\mathrm{Spec}(A) \to X$ the category of morphisms $\mathrm{Spec}(A) \to \mathrm{Spec}(A_i)$ over $X$ is connected? REPLY [6 votes]: Maybe I'm missing something, but it seems to me like the first question is just a simple definition chase. A morphism $\mathrm{Spec}(A)\to X$ is determined by finitely many affine open subsets $\mathrm{Spec}(B_n) \subseteq X$ and some ring maps $B_n\to A_{f_n}$ to localizations of $A$ satisfying certain compatibility properties. There are not so many elements of $A$ that are needed to witness all of this. Specifically, we need to witness the $f_n$, numerators of everything in the images of the maps $B_n\to A_{f_n}$, some elements annihilated by powers of the $f_n$ that witness that the maps $B_n\to A_{f_n}$ are actually homomorphisms and are appropriately compatible, and elements of $A$ that witness that some $f_n$ are in the ideal generated by others (and that $1$ is in the ideal generated by all of them). In total, it is clear that some subring $A_0\subseteq A$ of cardinality at most $\aleph_0+\sum |B_n|$ contains all the needed witnesses, and our morphism will factor through $\mathrm{Spec}(A_0)$. For the second question, consider two factorizations $\mathrm{Spec}(A)\to\mathrm{Spec}(A_i)\to X$ ($i=0,1$) where $|A_i|<\kappa_0$ for some cardinal $\kappa_0$ depending only on $X$. Let $Y=\mathrm{Spec}(A_0)\times_X \mathrm{Spec}(A_1)$; this may not be affine, but it will still satisfy some cardinality bound depending on $X$. Applying the previous paragraph with $Y$ in place of $X$, we can factor the map $\mathrm{Spec}(A)\to Y$ through $\mathrm{Spec}(A_2)$, and we can bound $A_2$ by some cardinal $\kappa_1$ depending only on $X$. Now repeat this argument allowing $A_0$ and $A_1$ to have cardinality up to $\kappa_1$, and get some new bound $\kappa_2$ on the cardinality of $A_2$. Repeating this inductively, we get a sequence of cardinals $\kappa_n$; let $\kappa=\sup \kappa_n$. Then whenever $|A_0|<\kappa$ and $|A_1|<\kappa$, we can find $A_2$ with $|A_2|<\kappa$ that connects them. (Actually, I'm pretty sure that if you choose $\kappa_0$ in the obvious way, $\kappa_1$ will just be the same as $\kappa_0$, so none of this iteration is actually necessary.)<|endoftext|> TITLE: Soft question: mathematics about truchet tiles QUESTION [11 upvotes]: It seems that this is the first question on Truchet tiles on MO. Shown above is a picture of a random tile, which you can see the resulting configuration is much like many membranes of cells. I wonder if there may be some interesting and deep math behind random Truchet tiles, but I do not know much about this topic. As I can guess, there might be questions like these: Given a random tile in finite grid, what is the expectation length of a closed curve? Questions like percolation on this graph, for example, give an infinite tiling of the plane, what is the probability that there is an infinite connected area? (2-color the basic pattern with blue and red, then the red area is like cells.) Connections with the $O(1)$ loop model. (I'm sorry, I know very little about this, but I do think it is in the picture.) Of course this list is very incomplete, and I hope someone can fill it and show me the connections with other branches of mathematics. Thanks! REPLY [8 votes]: To answer your question 3: Indeed, in statistical mechanics such a model is known as a "loop model". In general an "O(n) loop model" (this is not the place to rant on why this is a misnomer especially in the present context) means instead of drawing truchet tile/loop configurations equiprobably, you assign them a probability which is proportional to n^(number of closed loops), where n is a fixed parameter (not necessarily an integer). These loop models have become very popular in the 80s, and are a subject of active research up to now. The particular model corresponding to Truchet tiles is sometimes known as Temperley--Lieb loop model, sometimes also "CPL" (Completely Packed Loops). It has the nice property of being exactly solvable, in the sense of quantum integrability, which means various quantities can be computed exactly as the size of the sample goes to infinity. If you further specialize the loop weight to be 1, then this model has even more remarkable properties, allowing exact finite-size results, and interesting connections to combinatorics. Concerning questions 1 & 2, the model in question is critical, which means such questions have a definite answer in the limit of infinite size, where they're governed by critical exponents. (which I don't know off the top of my head). In some cases, such scaling behaviors have been proven rigorously using SLE; I don't think it is so for CPL, so from a mathematical standpoint I suppose they're only "conjectures". Ref: the best I can think of is Bernard Nienhuis' lectures "Loop models" in "Les Houches 2008: Exact methods in low-dimensional statistical physics and quantum computing", Eds Jacobsen, Ouvry, Pasquier, Serban, Cugliandolo. (you can also check out my lectures in the same volume, as well as my HDR on my webpage)<|endoftext|> TITLE: Examples of Polyhedra with Large Shadows QUESTION [5 upvotes]: Let $P \subseteq \mathbb{R}^n$ be a polyhedron described by $\mathcal{O}(n^{c_1})$ inequalities, where $c_1$ is a constant. Moreover, let $M\colon P \to \mathbb{R}^2$ be a linear mapping. I'm looking for $P$ and $M$ such that the polyhedron $M(P)$ has $\mathcal{O}({c_2}^n)$ vertices, where $c_2$ is a constant. Are there any other known examples besides 1, 2 (Theorem 4.4)? REPLY [5 votes]: If I understood your question right, you are looking for a polygon that has a small so-called extended formulation. This problem is studied for example here: http://arxiv.org/abs/1107.0371.<|endoftext|> TITLE: higher reciprocity theorems from ratios of Gauss sums QUESTION [8 upvotes]: One recent proof of quadratic reciprocity involves computing various rations of the Gauss sum. In Quadratic reciprocity and the sign of the Gauss sum via the finite Weil representation Gurevich, Hadani and Howe discuss the Gauss sums, which are finite field analogues of the Gamma function $\Gamma(n+1)$: $$ G_p = \sum_{x \in \mathbb{F}_p} e^{\frac{2\pi i x^2}{p}} = \left\{ \begin{array}{rc} \sqrt{p} & \text{if }p \equiv 1 \mod 4 \\ i\sqrt{p} & \text{if }p \equiv 3 \mod 4\end{array}\right.$$ with the relationship $G_p^2 = (\tfrac{-1}{p})\cdot p$. Using properties of the Weil representation of $SL(2,\mathbb{Z})$, and the fact that the Finite Fourier Transform corresponds to $$ S = \left( \begin{array}{cr} 0 & -1 \\ 1 & 0 \end{array} \right)$$ they are able to prove quadratic reciprocity and correctly get the sign of the Legendre symbol. In fact, they compute the ratio: $$ (\tfrac{p}{q}) (\tfrac{q}{p}) = \frac{G_p \cdot G_q}{G_{pq}}$$ The expression on the right looks like a Mobius function for the divisors of $pq$. So I started trying other combinations to see if I got meaningful symbols. Using three variables: $$ \frac{G_p G_q G_r G_{pqr}}{G_{pq}G_{qr}G_{pr}} \equiv 1$$ Computer experiments showed this ratio always to be $1$. In order to fine more interesting behavior I found: $$ \frac{G_p G_q G_r }{G_{pqr}} \equiv \pm 1$$ And there is no reason to stop there since Mobius functions can be defined for any lattice using incidence algebra. In particular, the poset of numbers and their divisors. Is there a name for these generalized symbols, or are these trivial extensions of the original Legendre symbol? Finally, what is the rule determining the sign in the second example? REPLY [7 votes]: If you look at bit further in the preprint you will see that the authors also state the generalisation \begin{equation*} \frac{G_{n_1} G_{n_2}}{G_{n_1 n_2}} = \left(\frac{n_1}{n_2}\right) \left(\frac{n_2}{n_1}\right) \end{equation*} for any coprime odd numbers $n_1$, $n_2$. From this it easy to see that indeed \begin{equation*} \frac{G_p G_q G_r G_{pqr}}{G_{pq} G_{qr} G_{pr}}=1, \end{equation*} and that \begin{equation*} \frac{G_p G_q G_r}{G_{pqr}}= \begin{cases} +1 & \mbox{if at most one of $p$, $q$ and $r$ is $\equiv 3\pmod{4}$,}\\ -1 & \mbox{otherwise.} \end{cases} \end{equation*} In general the formula also shows that such "balanced" quotients as above always can be converted to a symmetric product of Legendre symbols which in turn only depends on the congruence classes mod $4$ for the primes involved.<|endoftext|> TITLE: Is every $S^3$ block bundle over $S^4$ a fiber bundle? QUESTION [6 upvotes]: I am interested in the difference between block bundle and fiber bundle. Let $K$ be a simplicial complex and $p: E\to |K|$ be a continuous map. A block diffeomorphism of $\Delta^p\times M$ is a diffeomorphism $\Delta^p\times M\to \Delta^p\times M$ which for each face $\sigma \subset \Delta^p$ restricts to a diffeomorphism of $\sigma\times M$. A block chart for $E$ over a simplex $\sigma\in K$ is a homeomorphism $h_{\sigma}:p^{-1}(\sigma)\to \sigma\times M$ which for every face $\tau$ restricts to a homeomorphism $p^{-1}(\tau)\to \tau\times M$. A block atlas is a set $\mathcal{A}$ of block charts, at least one over each simplex of K, such that if $h_{\sigma_i}:p^{-1}(\sigma_i)\to \sigma_i\times M$ for $i=0,1$ are two elements of $\mathcal{A}$ then the composition $h_{\sigma_1}\circ h_{\sigma_0}^{-1}$ from $(\sigma_0\cap\sigma_1)\times M$ to itself is a block diffeomorphism. A block bundle structure is a maximal block atlas. The resulting structure is a block bundle. This notion is very close to fiber bundle. I am wondering if there exists a block bundle s.t. both fiber and base are manifolds but it does not admit fiber bundle structure.Is every $S^3$ block bundle over $S^4$ a fiber bundle? (This may be reduced to a lifting problem,since the fiber bundle has classifying space $BO(4)$ and the concordance class of such block bundle has classifying space $B\widetilde{Cat}(S^3)$.some knowledge about the homotopy group of $B\widetilde{Cat}(S^3)$ and $\widetilde{Cat}(S^3)/Cat(S^3)$ would surely be helpful here.$Cat=Diff,Top,PL$) REPLY [4 votes]: In my paper ``Generalised Miller--Morita--Mumford classes for block bundles and topological bundles" with Johannes Ebert, we construct a block $\mathbf{HP}^2$-bundle $\pi: E^{20} \to S^{12}$ which cannot admit a fibre bundle structure (even up to concordance). After constructing $\pi$, the property that guarantees that it cannot be a fibre bundle is that a certain Miller--Morita--Mumford class does not vanish, but it must vanish for trivial reasons on any fibre bundle.<|endoftext|> TITLE: Explicit Isomorphism between $Cl(8)$ and $\mathbb{R}(16)$ QUESTION [5 upvotes]: I am looking for a explicit isomorphism between $Cl(8)$ (Clifford algebra over $\mathbb{R}^8$ with standard Euclidean metric) and $\mathbb{R}(16)$ (algebra of $16\times 16$ matrices over $\mathbb{R}$). More concretely, it would be very useful to know: The image of the basis elements $e_k$ The image of the volume element $v=e_1e_2e_3e_4e_5e_6e_7e_8$ The equations of the $\pm$-eigespaces of $v$, $\Delta_{\pm}$ REPLY [2 votes]: In order to establish isomorphism of Clifford $C_8$ with $M_{16}\mathbb R$ it is enough to define 8 letters generators which square to $-1$ and anticommute. Define $e_k$=$$\begin{pmatrix} L_k & \\ & R_k \end{pmatrix} $$ where $L_k$ and $R_k$ are left and right multiplication by imaginary base octonions, $k$=1..7. Last letter $e_8$=$$\begin{pmatrix} & C \\ -C & \end{pmatrix} $$. In this presentation we can see that first 7 letters generate $C_7$=$M_8\mathbb R$+ $M_8\mathbb R$. Alternatively you can use $\bigl( \begin{smallmatrix} L & \\ & -L \end{smallmatrix} \bigr)$ and $\bigl( \begin{smallmatrix} & -I \\ I & \end{smallmatrix} \bigr)$.<|endoftext|> TITLE: Two-sided ideals of $B(H)$ are hereditary QUESTION [6 upvotes]: I seem to recall that (not necessarily closed) two-sided ideals of $B(H)$ are hereditary. Is that true? If it is, can anyone post a proof/reference? REPLY [6 votes]: Maybe it is in order to add that the same holds for arbitrary von Neumann algebras. Indeed, let $I \lhd M$ be a two-sided ideal in a von Neumann algebra $M$. We want to show that $0\leqslant x \leqslant y \in I$ implies that $x \in I$. It follows that $\sqrt{x} = a \sqrt{y}$, where $a \in M$ is a contraction that vanishes on $(Ran(\sqrt{y}))^{\perp}$ (this is the generalised polar decomposition); the fact that $a$ can be taken to be an element of $M$ follows from the bicommutant theorem. Now, using self-adjointness of $\sqrt{x}$, we can write $x = a \sqrt{y} (\sqrt{y} a^{*}) = aya^{*} \in I$.<|endoftext|> TITLE: What is a good introduction to cluster algebras from surfaces? QUESTION [11 upvotes]: What is a good reference for cluster algebras from surfaces, with a view to their connection to Teichmuller theory? In my view, that means it should start off with unpunctured surfaces (and in fact, it would be fine with me if it never went further). So far as I understand, this means that the results involved might well predate the invention of cluster algebras, but I still think that it would be nice to have an exposition of them from a cluster algebras perspective. I am hoping someone else agrees (and has consequently been inspired to write something along these lines). My ideal answer (while I'm dreaming) would not assume familiarity with cluster algebras, and as little knowledge of Teichmuller theory as possible. REPLY [3 votes]: Belatedly, per comments: Lauren Williams gave a survey course with recorded lectures, referencing her paper "Cluster algebras: an introduction." More recently: Introduction to Cluster Algebras: Chapter 6 (2020) Some background/motivational material: "The Positive Grassmannian (from a mathematician’s perspective)" slides by Williams "Combinatorics of KP solitons from the real grassmannian" by Kodama and Williams KP solitons, total positivity, and cluster algebras" presentation slides by Williams "A mathematician's unanticipated journey through the physical world," a popularizing article by Hartnett article in Quantamagazine on Williams and her work.<|endoftext|> TITLE: Is being reduced a generic property of schemes? QUESTION [7 upvotes]: (Naive formulation:) Let $X$ be an (irreducible) affine variety (over an algebraically closed field $k$) and $I$ be an ideal of the coordinate ring $R$ of $X$. Assume $Y = V(I)$ is equidimensional. The question is, can we detect if $Y$ is reduced as a subscheme on generic points of $Y$? More precisely, assume there is a (proper) closed subvariety $Z$ of $Y$ such that for all $y \in Y \setminus Z$, the localization of $I$ at the maximal ideal of $y$ is a radical ideal. Then is it true that $I$ is a radical ideal? The answer to the naive question is false, even in the simplest case that $I$ is principal, e.g. take $X = Spec~ k[x,xy,y^2]$ and $I$ to be the ideal generated by $x$. Then $I$ is radical on $X\setminus\{O\}$, where $O$ is the "origin". A perhaps more natural set of counter-example can be constructed as follows: Take a subvariety $Y'$ of $X$ which is not a complete intersection, take a minimal collection of generators $g_1, \ldots, g_k$ of the ideal of $Y'$, and set $I$ to be the ideal generated by $g_1, \ldots, g_{k-1}, g_k^2$. So the `real question', which is perhaps too broad, is: Are there 'natural' conditions on $X$ and $I$ under which the reducedness of $V(I)$ can be detected at generic (closed) points on $Y$? A 'natural' situation that I have in mind (and that disallows both counter-examples above) is the following: Is the answer to Question 2 affirmative when $X$ is normal and $I$ is generated by a regular sequence, i.e. $Y$ is a set-theoretic complete intersection? (The answer is affirmative in the simplest case, i.e. when $I$ is principal.) REPLY [8 votes]: I don't think normal is quite enough, but something similar should work. So, normal is equivalent to $S_2$ and $R_1$ reduced is equivalent to $S_1$ and $R_0$ If $Y$ is a regular hypersurface in something normal and it is not entirely singular which follows from you generically reduced assumption, then $Y$ is reduced. (This is just restating what you are saying in #3). A simple sufficient condition is to assume that $X$ is at least $S_{t+1}$ where $t=\mathrm{codim}_X Y$. That way $Y$ will be $S_1$ and your generically reduced assumption implies that it is $R_0$. Here is an example that the question in #3 only holds if the codimension of $Y$ is $1$: Let $X_0 = \mathrm{Spec}\ k[x,y,z,t,w]/(xz,xt,yz,yt)$. This is a threefold in $\mathbb A^5$. Near any point with any of $x,y,z,t$ non-zero, this is locally isomorphic to $\mathbb A^3$, hence smooth. When $x=y=z=t=0$, that's a line in $\mathbb A^5$, so $X_0$ is smooth away from that line, so it is $R_1$. We will see below that it is also $S_2$. In case you would like your $X$ be integral, then do this: This $X_0$ is really just two copies of $\mathbb A^3$ intersecting in a line. So, take two skew lines in $\mathbb A^3$ and glue them together using the local structure of $X_0$ near its singular line. That way you get an irreducible $X$ which is locally like $X_0$. Next let $Z = Z(w)\subset X_0$. This is the union of two planes meeting in a single point, the famous example of something not normal, not Cohen-Macaulay, etc.. It is easily seen to be $S_1$: For instance $x+z$ is a regular element, but modding out by that we get $Y=\mathrm{Spec} k[x,y,z,t]/(x^2,xy,xt,yt)$ which is two lines intersecting in a fat point, which is $R_0$, but not reduced at the fat point and hence non-reduced and cannot be $S_1$. Now in order to work on the irreducible $X$, take the $Z$ and $Y$ to be what you get while gluing the two lines of $\mathbb A^3$ together. Both $Z$ and $Y$ are linear away from the origin, so we can do the same gluing on them as on $X$. So, naming the new "glued" objects $X$, $Y$, and $Z$, we have that $Y$ is $S_0$ and $R_0$, in particular non-reduced, $Z$ is $S_1$ and $R_1$, in particular reduced, but not normal $X$ is $S_2$ and $R_1$, in particular normal. So, $Y$ is the complete intersection of a regular sequence, $w, x+z$ in the normal $X$, it is generically reduced, but not everywhere reduced. I think you can adapt this example to show that if $X$ is not $S_{t+1}$, then there is a codimension $t$ complete intersection which is generically reduced but not everywhere reduced. Otherwise there is still taking general hypersurface sections, those retain even normality.<|endoftext|> TITLE: Dual cell structures on manifolds QUESTION [10 upvotes]: Suppose that $M$ is a compact manifold without boundary (smooth if you like), and suppose further that $M$ is equipped with a regular CW-complex structure. Denote the face poset of this CW-complex by $P$. Is it true that there is a dual cell structure, also a regular CW complex, and whose face poset is the opposite (dual) poset $P^{\mbox{op}}$? So far I can't find a reference for this in the literature. One often starts with a triangulation of $M$, and I am looking for a more general statement. UPDATE: I accepted the answer to the original question, which points out that this is not true in the generality stated. But I would still like to understand under what conditions something like this holds. Suppose instead that instead of a regular CW-complex structure on a compact manifold, we have a polyhedral (convex cell) complex which is a PL-sphere. Under these conditions, is there a dual polyhedral structure with opposite poset as above? REPLY [4 votes]: In "Combinatorial cell complexes and Poincare duality", T. Basak defines a combinatorial cell complex (c.c.c.), which seem to meet your "polyhedral (convex cell) complex" description. For a c.c.c. $S$, the author defines the opposite c.c.c. $S^\circ$ as the reversal of the partial order. This construction of $S^\circ$ appears to yield the "dual polyhedral structure with opposite poset" you desire.<|endoftext|> TITLE: Limits of determinacy on reals QUESTION [6 upvotes]: For $X\subseteq\mathbb{R}^\omega$, say that $X$ is determined if the associated game on $\mathbb{R}$ of length $\omega$ (players I and II alternate playing reals, player I wins iff the sequence built is in $X$) is determined. My question is: what is known about the consistency of determinacy principles for games played with reals? For example, is it consistent with large cardinals that every $X\subseteq\mathbb{R}^\omega$ is determined? (I suspect not, but I'm having trouble coming up with a counterexample.) EDIT: This actually splits into two questions; I'm interested in each: How much determinacy on $\mathbb{R}$ is consistent (relative to large cardinals) with ZF? Andreas' answer completely settles this question. However, I'm also interested in: How much determinacy on $\mathbb{R}$ is consistent (relative to large cardinals) with ZFC? REPLY [3 votes]: I know this is an old question, but here's a little information about the $\textsf{ZFC}$ case to anyone finding this question. In my MO question, Juan kindly pointed out that Woodin has a result stating that it's consistent relative to a sharp for a Woodin limit of Woodins that every $\textsf{OD}(\mathbb R)$ game of length $\omega_1$ on the reals is determined. This result can be found in Neeman's book on long games, exercise 7F.15. Beyond that, we can find a non-determined definable game of length $\omega_1+\omega$ on the reals (equivalently the integers). After $\omega_1$ many rounds, player I has played a sequence $X$ of reals. If $X$ contains a perfect subset then we can define a well-ordering of the reals via $X$ and thus a non-determined set of reals $A\subseteq \omega^\omega$. Player I is then supposed to spend his last $\omega$ moves to land in $A$, making the game non-determined. If $X$ did not contain a perfect subset then the last $\omega$ rounds is spent playing the perfect set game on $X$, which is non-determined as $X$ doesn't have the perfect set property. It seems like this game is $\Delta^2_2$, so that's at least a lower definability bound for inconsistency. When we get to length $\mathfrak c+\omega$ then we only have to consider the first case above, so that the game is (at most) $\Pi^1_2$ (for all strategies there's a move which makes the strategy non-winning). I'm not sure what happens below these lower bounds, except that Borel determinacy at least ensures that all Borel games are determined, no matter the length (just play elements of ${^\alpha}\omega$ for $\omega$ many rounds instead of playing elements of $\omega$ for $\alpha\omega$ many rounds). Here's an attempt to illustrate the situation, where red indicates inconsistencies and blue indicates consistency relative to large cardinals. I'm not sure if more is known.<|endoftext|> TITLE: Number of critical points of smooth functions on $S^1$ QUESTION [8 upvotes]: Let $u$ be a smooth function on the unit circle $S^1$ such that $\int_{S^1}ux_j=0$, for $j=1,2$. Is the number of critical points of $u$ strictly bigger than 2? REPLY [13 votes]: Yes, in fact the number of critical points of $u$ is at least four. Expand $u:S^1\to\mathbb R$ as a Fourier series: $$u(\theta)=a_0+\sum_{i=1}^\infty a_j\cos(j\theta)+b_j\sin(j\theta)$$ Then your condition may be stated as $a_1=b_1=0$ (I assume by $x_1$ and $x_2$, you mean to be identifying $S^1$ with $\{x_1^2+x_2^2=1\}$, so $x_1=\cos\theta$ and $x_2=\sin\theta$ in the notation above). Now let $u(t,\theta)$ be the solution to the heat equation $u_t=u_{\theta\theta}$ with initial value $u(0,\theta)=u(\theta)$. By the maximum principle, we have: $$\#\operatorname{crit}(u(\cdot))\geq\#\operatorname{crit}(u(t,\cdot))$$ for all $t\geq 0$. On the other hand, we have the explicit solution: $$u(t,\theta)=a_0+\sum_{i=1}^\infty a_je^{-j^2t}\cos(j\theta)+b_je^{-j^2t}\sin(j\theta)$$ Let $j_0=\min\{j\geq 1:(a_j,b_j)\ne(0,0)\}$. If $j_0=\infty$ (i.e. only $a_0$ is nonzero), then $u$ is constant and there is nothing to prove, so therefore we have $1 TITLE: Equivalence of Gaussian measures QUESTION [5 upvotes]: Let $H$ be a separable Hilbert space and $N(0, C)$ and $N(0, D)$ be Gaussian measures on it. Further, for each $v \in H$, define $R_v = \frac{\left\langle v,Cv \right\rangle}{\left\langle v,Dv \right\rangle}$. Basically, $R_v$ is the ratio of Covariance of one dimensional Gaussian measures induced by $N(0, C)$ and $N(0, D)$ along $v$ respectively. Are the following statements equivalent- 1) $N(0, C)$ and $N(0, D )$ are equivalent to each other. 2) $ 0 < \inf_{v \in H} R_v \leq \sup_{v \in H} R_v < \infty.$ REPLY [7 votes]: They are not equivalent. For an explicit counterexample, let $\{e_1, e_2, \dots\}$ be an orthonormal basis for $H$, and let $C$ be the diagonal operator $C e_n = \frac{1}{n^2} e_n$. Let $D =2C$. Then $R_v = 1/2$ for every $v$ so (2) is satisfied. Define random variables $X_n$ on $H$ by $X_n(v) = {n} \langle v, e_n \rangle$. Then under $N(0,C)$, the $X_n$ are iid $N(0,1)$, and under $N(0,D)$ they are iid $N(0,2)$. So by the strong law of large numbers, we have $\frac{1}{n} (X_1(v)^2 + \dots + X_n(v)^2) \to 1$ for $N(0,C)$-a.e. $v$, but likewise we have $\frac{1}{n} (X_1(v)^2 + \dots + X_n(v)^2) \to 2$ for $N(0,D)$-a.e. $v$. So $N(0,C)$ and $N(0,D)$ are mutually singular and (1) fails. In H. H. Kuo's book Gaussian Measures in Banach Spaces, section II.3, you can find the following theorem: Theorem. $N(0,C)$ is equivalent to $N(0,D)$ iff there exists a bounded operator $T$ which is positive definite and invertible and such that $I-T$ is Hilbert–Schmidt, with $D = \sqrt{C} T \sqrt{C}$. Otherwise, $N(0,C)$ and $N(0,D)$ are mutually singular. (Kuo credits this result to Feldman–Hajek, but the reference given is to Varadhan's Stochastic Processes.) This shows that we do have (1) implies (2), since if $D = \sqrt{C} T \sqrt{C}$ we can write $R_v = \frac{\|\sqrt{C} v\|^2}{\|\sqrt{T} \sqrt{C} v\|^2}$, and from that you can conclude $\|T\|^{-1} \le R_v \le \|T^{-1}\|$.<|endoftext|> TITLE: Mean value of Maass forms QUESTION [7 upvotes]: Let $X = SL_2(\mathbb{Z}) \backslash \mathbb{H}$ be the modular surface. Consider a basis of $L^2$-normalized Hecke-Maass cusps forms $\phi_j$ on $X$ with $-\Delta$-eigenvalue $\lambda_j$. Hejhal-Rackner (1991, on the Topography of Maass forms) seems to claim that for any fixed set $A \subset X$ with finite measure, $$\int_A \phi_j (z) d\mu z \to 0$$ as $\lambda_j \to \infty$. I am probably overlooking something, but is there a simple proof of this? One of the even/odd Maass forms case should be trivial by sign considerations, but unless I am missing something the other case does not seem as apparent. REPLY [9 votes]: This is an analog of the Riemann-Lebesgue lemma. Let $f$ be a smooth function on $X$. Then by self-adjointness of the Laplacian $$ \lambda_j \langle \phi_j, f \rangle = \langle \Delta \phi_j ,f \rangle = \langle \phi_j, \Delta f\rangle , $$ and by Cauchy-Schwarz this is bounded in size by $\Vert \phi_j \Vert \Vert \Delta f\Vert$. Therefore $\langle \phi_j, f\rangle$ goes to zero as $\lambda_j$ gets large. Now approximate the characteristic function $\chi_A$ of the set $A$ by suitable smooth functions $f$, and bound $\langle \phi_j, f-\chi_A \rangle$ by $\Vert \phi_j \Vert \Vert f-\chi_A \Vert$.<|endoftext|> TITLE: What would be an infinity-groupoid analogue of the duality between sets and complete atomic boolean algebras? QUESTION [16 upvotes]: Consider the object classifier of the $\infty$-topos of $\infty$-groupoids. For the role it plays in homotopy type theory as the type of types, let’s denote it as $Type = \coprod_{[F]} B Aut(F)$, the coproduct of the automorphism ∞-groups of all (small) homotopy types $[F]$. Now attempting to imitate the duality between the category of sets and the category of complete atomic Boolean algebras, we might consider the map taking an $\infty$-groupoid $A$ to $[A, Type]$. Similarly to how the internal Boolean algebra structure on $\mathbf{2}$ induces a Boolean algebra structure on $[X, \mathbf{2}]$, for a set $X$, the internal $\infty$-topos structure on $Type$ induces an $\infty$-topos structure on $[A, Type] = \infty-Grpd/A$, the slice $\infty$-topos. Question: Just as $[X, \mathbf{2}]$ is a special kind of Boolean algebra, being complete and atomic, how can I characterise those $\infty$-toposes of the form $[A, Type]$? The answer will probably involve Mike Shulman's suggestions to me: having a set of tiny generators and being a Boolean presheaf $\infty$-topos. Further questions: with such a characterisation in hand, should we expect a duality between such $\infty$-toposes and the $\infty$-topos of $\infty$-groupoids? What would be the equivalent of Stone Space-Boolean algebra duality? REPLY [19 votes]: Let $\mathcal{S}$ denote the $\infty$-category of spaces. For any $\infty$-topos $\mathcal{X}$, there is an essentially unique geometric morphism $\pi^{\ast}: \mathcal{S} \rightarrow \mathcal{X}$. The $\infty$-topos $\mathcal{X}$ has the form $\mathcal{S}_{/A}$ if and only if the geometric morphism $\pi^{\ast}$ is etale. This is true if and only if the following three assertions hold: $(1)$ The functor $\pi^{\ast}$ admits a left adjoint $\pi_{!}$ (equivalently: $\pi^{\ast}$ preserves small limits). $(2)$ The functor $\pi_{!}$ is conservative (that is, if $\alpha: X \rightarrow Y$ is a morphism in $\mathcal{X}$ for which $\pi_{!}(\alpha)$ is an equivalence, then $\alpha$ is an equivalence). $(3)$ There is a projection formula $$\pi_{!}( \pi^{\ast} X \times_{ \pi^{\ast} Y } Z ) \simeq X \times_{Y} \pi_{!} Z.$$ The construction $A \mapsto \mathcal{S}_{/A}$ gives a fully faithful embedding from the $\infty$-category of spaces to the $\infty$-category of $\infty$-topoi. However, I wouldn't be inclined to see this as analogous to Stone duality: as you point out, Boolean algebras of the form $[X, \mathbf{2} ]$ are rather special. It's more analogous to the observation that any set can be regarded as a topological space by equipping it with the discrete topology. To get a closer analogue of Stone duality, note that the $\infty$-category of $\infty$-topoi admits small limits. Consequently, the functor $A \mapsto \mathcal{S}_{/A}$ extends formally to a functor $F$ from Pro-spaces to $\infty$-topoi which commutes with filtered inverse limits. This functor $F$ is not fully faithful, but it is fully faithful when restricted to profinite spaces (that is, Pro-spaces which can be represented by filtered diagrams of spaces which have only finitely many homotopy groups, each of which is required to be finite). You therefore get an embedding $\theta:$ { Profinite spaces } $\hookrightarrow$ { $\infty$-topoi } which is a better analogue of Stone duality. In fact, it generalizes Stone duality: the RHS contains the ordinary category of sober topological spaces as a full subcategory, the LHS contains the ordinary category profinite sets as a full subcategory, and on the latter subcategory $\theta$ restricts to the usual fully faithful embedding { Profinite sets } $\simeq$ { Stone Spaces } $\hookrightarrow$ {sober topological spaces}.<|endoftext|> TITLE: Thom conjecture in CP3 QUESTION [18 upvotes]: Thom conjecture, that was originally asked in $\mathrm{CP}^2$, and is now proven for symplectic 4 manifolds, states that complex curves (symplectic surfaces) are genus minimizing in their homology class. (The $\mathrm{CP}^2$ case was proven by Kronheimer and Mrowka '94, the symplectic case by Ozsvath and Szabo '00.) Does anybody know if there is any analogous work being done in $\mathrm{CP}^3$? More concretely, is there any reason to think that complex hypersurfaces in $\mathrm{CP}^3$ are topologically the simplest oriented 4 manifolds in their homology classes. The simplest perhaps in a sense that the sum of betti numbers is minimal. REPLY [18 votes]: This question was addressed in higher dimensions by Mike Freedman, Surgery on codimension 2 submanifolds. Mem. Amer. Math. Soc. 12 (1977), no. 191. (I think this was his PhD thesis.) Earlier work of Thomas and Wood (On manifolds representing homology classes in codimension 2. Invent. Math. 25 (1974), 63–89) had given some bounds for the homological complexity of codimension-2 submanifolds, generalizing those given by Hsiang and Szczarba for surfaces in 4-manifolds. My recollection is that Freedman shows that the Thomas-Wood bounds are close to optimal. The technique is a very clever variation of surgery theory, rather different from the typical uses of surgery theory in constructing embeddings in higher codimensions. I think the upshot of Freedman's work is that the analogue of the Thom conjecture in higher dimensions does not hold. Update 9/15/21: After this MO exchange, we started discussing how to reduce $b_2$ for the degree $d$ hypersurface $V_d$ in $CP^3$. In recent preprint (On the Thom conjecture in $CP^3$), Marko (the OP), Sašo Strle, and I showed that when $d \geq 5$, there is a smooth $4$-manifold in the homology class of $V_d$ with strictly smaller $b_2$. So the analogue of the Thom conjecture doesn't hold in $CP^3$ either. (As noted in my comment below, it does actually hold for $d \leq 4$.) It's likely that our construction doesn't produce the smallest possible $b_2$ so there is still some room for improvement.<|endoftext|> TITLE: What is modular representation theory for groups good for? QUESTION [27 upvotes]: I am an absolute beginner in modular representation theory of finite groups. I know some things in representation theory in characteristic zero. My questions are regarding the main goals of this part of representation theory. For example, does the modular representation theory give information back regarding the structure of the group algebra or of the group itself? I would like to see some specific examples where this theory can be applied. I know that it studies the blocks of the group algebra but I don't know too many things about this. What are the other applications of the theory? I have seen there are some interesting things related with modular forms also discussed here on the forum. Besides that, what are the other applications of this theory? REPLY [16 votes]: As Geoff explains, modular representations of finite groups do play an important role in some of the standard group-theoretic developments (although they play less of a role in the classification of simple groups than some people had expected initially). In fact, I think Richard Brauer began to develop these ideas, including block theory, in the 1940s because he saw connections with structural questions about finite groups. By now of course the subject is also pursued just because of its intrinsic beauty, but it has applications. As Dylan Wilson points out in his comment, some aspects of the modular theory played a key role for Quillen in his work on algebraic K-theory. Brauer's results in both ordinary and modular representation theory are essential to Quillen's proof of the Adams conjecture here. These ideas such as "Brauer lifting" are emphasized in Serre's lecture notes (later translated as Springer GTM 42). I'd add that for finite groups of Lie type defined over fields of prime characteristic $p$, the modular theory relative to $p$ has been helpful at times in the study of ordinary (characteristic 0) representations of those groups. Moreover, it has implications for the rational representations in characteristic $p$ of ambient algebraic groups, due to the close connections between irreducible representations of the finite groups and algebraic groups (Curtis, Steinberg, ...) In a different direction, the study of Galois representations (and $p$-adic representations in general) has often required information about the modular representations of the finite groups of Lie type which occur naturally as quotients when reduction mod $p$ is applied to rings of $p$-adic integers. All of this is complicated to explain, but is part of the broader study of certain matrix groups over local fields and their representations. Which in turn has implications for the unifying program laid out by Langlands.<|endoftext|> TITLE: The Minkowski sum of two curves QUESTION [7 upvotes]: Let $\gamma$ be a continuous curve in the complex plane without self-intersections and let $\lambda$ be a complex non-real number less than 1 in modulus. Put $\gamma'=\lambda\gamma$. Question. Is it true that the sumset $\gamma+\gamma'$ has a non-empty interior? REPLY [3 votes]: Well, I cannot call it a complete answer, but it seems an idea that can work. First: take any loop in the (s,t) coordinates, and consider its image on the complex plane under $\gamma(s)+\gamma'(t)$. Any point having nonzero index w.r.t. this loop is then contained in $\gamma+\gamma'$. Second: take the loop to be the boundary of a rectangle, $[s_1,s_2]\times [t_1,t_2]$, and assume that along the second coordinate it is very thin, $t_1$ is very close to $t_2$. Then, the boundary consists of two translated images $\gamma([s_1,s_2])+\gamma'(t_1)$ and $\gamma([s_1,s_2])+\gamma'(t_2)$, and two very small curves $\gamma(s_j)+\gamma'([t_1,t_2])$. Third: I would say that the two translated images should more or less coincide, up to what happens at the endpoints. As if they do not, there is a part of one "somewhere in the middle", where it is far away from the other one, and then, crossing the curve, we change the index -- so on one of two sides there will be an open set of points with nonzero index. Last: making $t_2$ go towards $t_1$, in the limit of the almost-translation-invariance above we get that $\gamma([s_1,s_2])$ is a straight line. And this allows to conclude easily, as $\lambda$ is not real. Well, as I've said, it's a sketch, but it has a good chance of working. Upd.: I'm now sure that this argument works. For "somewhere in the middle", I mean the following. Lemma. Let $\delta$ be the diameter of $\gamma'([t_1,t_2])$, and assume that there is no point with nonzero index w.r.t. the loop described in the "Second." step. Then, the sets $\gamma([s_1,s_2])+\gamma'(t_1)$ and $\gamma([s_1,s_2])+\gamma'(t_2)$ coincide outside $\delta$-neighborhoods of $\gamma'(t_{j})$. Sketch of the proof. Assume the contrary and let $z$ be a point of a curve $\gamma([s_1,s_2])+\gamma'(t_1)$ that lies outside the above neighborhoods and that does not belong to the $\gamma([s_1,s_2])+\gamma'(t_2)$. Due to the continuity, there is $\epsilon$-neighborhood of $z$ that the latter curve does not intersect. Now, in this neighborhood one can find two points "on different sides" with respect to the first curve (yes, this phrase should also be more formally stated with an appropriate reference to the Jordan curve theorem). Such two points have thus different index w.r.t. all the closed loop (image under $\gamma(s)+\gamma'(t)$ of the boundary of the rectangle $[s_1,s_2]\times [t_1,t_2]$). Hence, at least one of them in an internal point of the sum $\gamma+\gamma'$. $\square$ Passing to the limit as $t_2\to t_1$ says that the curve $\gamma([s_1,s_2])$ should admit (if there is no internal point) arbitrarily small translational symmetries outside arbitrarily small neighborhoods of its endpoints. Hence, it is a straight segment.<|endoftext|> TITLE: A weak analytic version of the valuative criterion of properness QUESTION [5 upvotes]: EDIT: Let $f\colon X\to Y$ be a morphism of complex analytic spaces (not necessarily smooth or reduced). Assume that (a) $f$ is injective on points; (b) $f$ is local imbedding near each point $x\in X$; (c) for any holomorphic germ $\gamma\colon (\mathbb{C},0)\to Y$ with image contained in $f(X)$ there exists a holomorphic germ $\tilde\gamma\colon (\mathbb{C},0)\to X$ such that $f\circ \bar \gamma=\gamma$. (d) $f(X)$ is a closed subset of $Y$. Question. Is it true that $f$ is proper? Remark. (1) In the algebraic situation the answer is positive by the standard valuative criterion of properness. (2) A priori it is not clear to me that locally on $Y$ the image $f(X)$ is an analytic subset. If there is any analytic version of the valuative criterion of properness, I would be curious to know anyway. A reference would be helpful. EDIT: Conditions (a),(c), (d) imply the following property which looks closer to the standard valuative criterion of properness. For any holomorphic germs $\gamma\colon (\mathbb{C},0)\to Y$ and $\beta\colon (\mathbb{C},0)\backslash\{0\}\to X$ such that $f\circ \beta=\gamma|_{(\mathbb{C},0)\backslash\{0\}}$, there exists a holomorphic germ $\bar\gamma\colon (\mathbb{C},0)\to X$ such that $$\bar\gamma|_{(\mathbb{C},0)\backslash\{0\}}=\beta,\, f\circ \bar\gamma=\gamma.$$ REPLY [4 votes]: Here is a counterexample: $Y=\mathbb C^2$, $X=(\mathbb C\setminus\{0\})\cup\mathbb C$, and $f$ is given by $z\mapsto (z,e^{1/z})$ on $\mathbb C\setminus\{0\}$ and is given by $w\mapsto(0,w)$ on the second copy of $\mathbb C$. This map is certainly not proper. It clearly satisfies conditions (a),(b),(d). Let's check condition (c). Let $\gamma:(\mathbb C,0)\to Y=\mathbb C^2$ be a germ whose image is contained in $f(X)$. If the germ sends $0$ to a point whose first coordinate is nonzero, then it clearly lifts, so it suffices to check the case when $\gamma(0)=(0,w)$ for some $w\in\mathbb C$. Consider the first coordinate of $\gamma$: if this first coordinate is not identically zero, then it covers an open neighborhood of zero by the open mapping theorem. On the other hand, the function $e^{1/z}$ takes arbitrarily large values near zero since it has an essential singularity there, and this contradicts the fact that the image of $\gamma$ is contained in the image of $f$. Thus the first coordinate of $\gamma$ is identically zero, and thus it trivially lifts to $X$.<|endoftext|> TITLE: When was the "arrow notation" for functions first introduced? QUESTION [15 upvotes]: When was the "arrow notation" $f: X \to Y$ for functions first introduced? Who introduced it and with which motivation? I ask this question in order to understand whether it was, in part, this notation to suggest that there could be "higher morphisms" (in analogy with oriented paths and homotopies between them, and homotopies between homotopies and so on), or if it went the other way around (with category theorists first realizing that many constructions involving paths and homotopies thereof in Homotopy Theory could be generalized to other more abstract settings, and then setting up a notation that suggested the analogy "1-morphisms $\sim$ paths"). REPLY [4 votes]: Regarding element-to-element arrows, Emili Bifet writes on The Math Forum that Riemann used the notation $a \to b$ as follows: Ist $a$ ein Verzweigunspunkt der Loesung einer linearen Differentialgleichung zweiter Ordnung und geht, waehrend $x$ sich im positiven Sinn um $a$ bewegt, $z_1$ ueber in $z_3$ und $z_2$ in $z_4$, was kurz durch $z_1 \to z_3$ und $z_2 \to z_4$ angedeutet werden soll... II. Die Integrale einer linearen Differentialgleichung zweiter Ordnung in einem Verzweigungspunkt. (Aus einer Vorlesung Wintersemester 1856/57.) Or in translation by Michael Bächtold: If $a$ is a branching point of the solution to a linear differential equation of second order and, as $x$ moves clockwise around $a$, $z_1$ goes over to $z_3$ while $z_2$ goes over to $z_4$, what shall be denoted with $z_1 \to z_3$ and $z_2 \to z_4$ for short...<|endoftext|> TITLE: Irreducible/prime/indivisible elements QUESTION [10 upvotes]: in what follows all the rings are commutative, nontrivial, with unit. Recall the following definitions: 1) $\pi\in A$ is prime if $(\pi)$ is a nonzero prime ideal 2) $\pi\in A$ is irreducible if $\pi$ is nonzero, non invertible, and for all $a,b\in A$, $\pi=ab$ implies that $a$ or $b$ is a unit. 3) $\pi\in A$ is indivisible if $\pi$ is nonzero, non invertible, and for all $a,b\in A$, $\pi=ab$ implies that $\pi\mid a$ or $\pi\mid b$. The following facts are known: if $A$ is a domain, prime$\Rightarrow$ irreducible. This is not true anymore if $A$ has zero divisor (e.g. $A=\mathbb{Z}/6\mathbb{Z}$: $A$ has prime elements, but no irreducible elements) irreducible $\Rightarrow$ indivisible there exist indivisible elements which are not irreducible: $A=\mathbb{Z}/6\mathbb{Z}$, $\pi=3$ . However, this one is prime. if $A$ is a domain, irreducible $\iff$ indivisible if $A$ is noetherian, $A$ has indivisible elements if $A$ is a noetherian domain which is not a field, $A$ has irreducible elements. After this lengthy introduction, let me ask the following questions: Q1: can we find an example of an indivisible element which is neither prime or irreducible ? If possible, I would like $A$ to be noetherian or, even better, finite. Q2: can we find an example of a noetherian ring which is not a field, which has no prime elements AND no irreducible elements ? (so necessarily, $A$ has zero divisors) Q3: can we find an example of a noetherian domain which is not a field which has no prime elements ? Q4: if the answer to Q3 is NO, can we find an example of a domain which has irreducible elements, but which has no prime elements ? Thanks for your time. Greg REPLY [5 votes]: It looks like these questions have been taken care of, but I just thought I would mention a few references and talk about other similar concepts in between the definitions you provide here! I find this kind of thing very interesting, so I am always excited to see others interested in it as well. If you are interested in even more of these kinds of irreducible properties, I would suggest taking a look at the paper by D.D. Anderson and S. Valdez-Leon http://projecteuclid.org/euclid.rmjm/1181072068 Many of these concepts of irreducible/indivisible diverge with zero-divisors present. It may be a bit confusing because what you have called indivisible is what is called irreducible in this paper. Your definition of irreducible is very strongly irreducible. In fact there are two other notions that are in between these two concepts. $\pi$ is m-irreducible if it is maximal among principal ideals. $\pi$ is strongly irreducible if $\pi=bc$ implies $\pi \approx b$ or $\pi \approx c$ where $a\approx b$ means there is a unit $\lambda$ such that $a=\lambda b$. Then for non-zero elements you have very strongly irreducible $\Rightarrow$ m-irreducible $\Rightarrow$ strongly irreducible $\Rightarrow$ irreducible. Lastly, prime $\Rightarrow$ irreducible (your indivisible) but none of the others. If you look at example 2.3 in the article I linked, they provide the example $F[X,Y,Z]/(X-XYZ)$ where $F$ is a field. Then $x,y,z$ are the image of $X,Y,Z$, respectively. Then $x$ is prime, and hence irreducible (indivisible in your terms) but not even strongly irreducible and hence neither m-irreducible nor very strongly irreducible. Other examples showing these are distinct in rings with zero-divisors are provided in this article following Theorem 2.12. Regarding the last few questions I would point you towards (ON INTEGRAL DOMAINS WITH NO ATOMS) which are cleverly called "anti-matter domains" https://www.ndsu.edu/pubweb/~coykenda/paper13.pdf<|endoftext|> TITLE: Computing Dolbeault cohomology of some simple domains QUESTION [10 upvotes]: I have seen computations of the Dolbeault cohomology groups on compact Kahler manifolds using Hodge theory. I have never seen the computation of Dolbeault cohomology for simple domains in $\mathbb{C}^n$, aside from showing that they are trivial (for domains of holomorphy). For example, I would like to see a computation of the dimension of $H^{(0,1)}\left(B(2)-B(1)\right)$ (which is not a domain of holomorphy by Hartog's extension phenomenon), where $B(r)$ is the ball of radius $r$ in $\mathbb{C}^2$. I can produce some $\bar{\partial}$ closed but not exact forms by hand, but I am not seeing a good way to write down all of them. REPLY [8 votes]: $\def\CC{\mathbb{C}}\def\cO{\mathcal{O}}$Here is a computation of the Dobault cohomology of $X:=B(\infty) \setminus B(0) = \CC^2 \setminus \{ (0,0) \}$. I think that balls of finite radius should be behave basically the same way, but the details will be messier and it sounds like you just want to see an example. Set $$U_1 = (\CC \setminus \{ 0 \}) \times \CC,\ U_2 = \CC \times (\CC \setminus \{ 0 \}),\ U_{12} = (\CC \setminus \{ 0 \})^2.$$ So $\CC^2 \setminus \{ (0,0) \} = U_1 \cup U_2$ and $U_{12} = U_1 \cap U_2$. Each of $U_1$, $U_2$ and $U_{12}$ is a product of open sets in $\CC$, so they are Stein and thus any coherent sheaf on them has vanishing cohomology. We deduce by Leray's theorem that the Cech complex on $U_1$, $U_2$ computes Dolbeault cohomology of $\CC^2 \setminus \{ (0,0) \}$. Let $\cO$ be the sheaf of holomorphic functions. So $$\cO(U_1) = \left\{ \sum\nolimits_{j \geq 0} a_{ij} z_1^i z_2^j : \ \mbox{the sum is convergent} \right\}$$ $$\cO(U_2) = \left\{ \sum\nolimits_{i \geq 0} a_{ij} z_1^i z_2^j : \ \mbox{the sum is convergent} \right\}$$ $$\cO(U_{12}) = \left\{ \sum a_{ij} z_1^i z_2^j : \ \mbox{the sum is convergent} \right\}.$$ We deduce that $$H^0(X, \cO) = \left\{ \sum\nolimits_{i,j \geq 0} a_{ij} z_1^i z_2^j : \ \mbox{the sum is convergent} \right\}$$ $$H^1(X, \cO) = \left\{ \sum\nolimits_{i,j < 0} a_{ij} z_1^i z_2^j : \ \mbox{the sum is convergent} \right\}.$$ The sheaf $\Omega^1$ is a free $\cO$-module of rank two with basis $d z_1$, $d z_2$; the sheaf $\Omega^2$ is free of rank one with basis $d z_1 \wedge d z_2$. Thus, $$H^0(X, \Omega^1) = \left\{ \sum\nolimits_{i,j \geq 0} a_{ij} z_1^i z_2^j d z_1 + \sum\nolimits_{i,j \geq 0} b_{ij} z_1^i z_2^j d z_2 \right\}$$ $$H^0(X, \Omega^2) = \left\{ \sum\nolimits_{i,j \geq 0} c_{ij} z_1^i z_2^j d z_1 \wedge d z_2 \right\}$$ $$H^1(X, \Omega^1) = \left\{ \sum\nolimits_{i,j < 0} a_{ij} z_1^i z_2^j d z_1 + \sum\nolimits_{i,j < 0} b_{ij} z_1^i z_2^j d z_2 \right\}$$ $$H^1(X, \Omega^2) = \left\{ \sum\nolimits_{i,j < 0} c_{ij} z_1^i z_2^j d z_1 \wedge d z_2 \right\}$$ where I have stopped writing down that the sums have to converge. Let's see how this works with the Dolbeault-deRham spectral sequence. The $(p,q)$ term is $H^q(\Omega^p)$. The maps on the first page go $(p,q) \to (p+1,q)$ and are induced by $d$. So we need to compute the cohomology of $$H^0(X, \cO) \to H^0(X, \Omega^1) \to H^0(X, \Omega^2)$$ $$H^1(X, \cO) \to H^1(X, \Omega^1) \to H^1(X, \Omega^2).$$ Put a $\mathbb{Z}^2$-grading on these vector space where $z_1$ and $dz_1$ are in degree $(1,0)$ and $z_2$ and $dz_2$ are in degree $(0,1)$. Ignoring the convergence conditions for a moment, these complexes are direct sums of finite dimensional complexes in each degree $(i,j)$, and we compute that they are exact except when $(i,j) = (0,0)$. The complexes in degree $(0,0)$ is $\CC (1) \to 0 \to 0$ and $0 \to 0 \to \CC (z_1^{-1} z_2^{-1} dz_1 \wedge dz_2)$. I am going to omit the argument for length, but the convergence conditions don't change anything: The next page of the spectral sequence is $\CC(1)$ in position $(0,0)$ and $\CC (z_1^{-1} z_2^{-1} d z_1 \wedge d z_2)$ in $(2,1)$. There can be no arrows from $(0,0)$ to $(2,1)$ on future pages, so the complex collapses, and we compute that $H^0(X) \cong \CC$ and $H^3(X) \cong \CC$. This is of course correct, since $X$ retracts onto the $3$-sphere of radius $1$.<|endoftext|> TITLE: Invariant Laurent polynomials under cyclic group action QUESTION [5 upvotes]: Start with the cyclic group $G:=\mathbb{Z}/p$ of prime order $p$ and and an integer lattice $P:=\mathbb{Z}^p$. Let $G$ act on $P$ by cyclic permutation of coordinates. There is an induced action on the group ring $\mathbb{Z}[P]$ of the lattice $P$ which is isomorphic to the ring $\mathbb{Z}[\mathbf{x^{\pm 1}}]$ of Laurent polynomials in $p$ variables $\mathbf{x}=(x_1,\dots,x_p)$ over $\mathbb{Z}$. How do I find a (hopefully) finite/minimal set of invariants which generates the ring of $G$-invariants $\mathbb{Z}[\mathbf{x^{\pm 1}}]^G$? Since $\mathbb{Z}[\mathbf{x^{\pm 1}}]^G=\mathbb{Z}[\mathbf{x}]^G[\sigma_p^{-1}]$ where $\sigma_p=x_1\dots x_p$, the question about Invariant Polynomials under a Group Action (hidden GIT) is of some help but not much is said there about what happens over $\mathbb{Z}$. From the Lie theory point of view this is related to the weight lattice of the root system $A_{p-1}$ where $G$ is a subgroup of the Weyl group $S_p$ and the fundamental domain I'm considering is some sort of "larger Weyl chamber". REPLY [2 votes]: I think there is little hope that your question has a good answer. Even over $\mathbb{Q}$, the invariants (polynomial invariants, not rational invariants, which are easier) of the cyclic group are messy. I don't see a reason why localizing by $\sigma_p$ should make things easier. Now doing the computation over $\mathbb{Z}$ adds to the difficulty since computing over $\mathbb{Z}$ means that all anomalies of the modular situation will show up. To give evidence to my first claim, here are the numbers $r$ of minimal generators of the invariant rings $\mathbb{Q}[x_1,\ldots,x_n]^{C_n}$ for some $n$, computed by MAGMA: $n = 3$: $r = 4$; $n = 4$: $r = 7$; $n = 5$: $r = 15$, $n = 7$: $r = 48$; $n = 9$: $r = 119$; $n = 11$: $r = 348$.<|endoftext|> TITLE: Analogues of Primitive Recursive Functions QUESTION [18 upvotes]: Let $\mathbf{A}$ be an admissible set (possibly with urelements). I am wondering if there is some good notion of "primitive recursive arithmetic" relative to $\mathbf{A}$. More precisely, I would like to single out a class of $\Pi_{2}$-sentences (with parameters) about $\mathbf{A}$ which reduces to the class of $\Pi_{2}$-theorems of PRA when specialized to the case where $\mathbf{A}$ is the set of hereditarily finite sets. In particular, I would like to single out a special class of "provably total" $\Sigma_1$-definable functions on $\mathbf{A}$, which reduces to the class of primitive recursive functions when $\mathbf{A} =$ hereditarily finite sets. I would be grateful for any pointers to relevant literature. If it helps, I am primarily interested in the case where $\mathbf{A}$ is the smallest admissible set containing some mathematical structure $M$ (that is, $\mathbf{A} = HYP_M$, in the notation of Barwise's book). REPLY [2 votes]: (This is more of a comment than an answer, but it's a bit too long to be split into comments so I'll post it as an answer.) I don't know about functions defined on an arbitrary admissible set, but at least for admissible levels of the constructible hierarchy, you might be interested in what are called "$(\infty,0)$-recursive functions" in chapter VIII ("Recursion on Ordinals") of Peter G. Hinman's book Recursion-Theoretic Hierarchies (1978, available here), and also on this related question I asked a while ago while trying to make sense (without much success) of the various definitions. Hinman writes (op.cit., p.378) that: The $(\infty,0)$-recursive functions will play somewhat the role here of the primitive recursive functions of ordinary recursion theory. Perhaps even more relevant to your question would be the $(\infty,\lambda)$-recursive functions in Hinman's terminology, or even more the primitive $(\infty,\lambda)$-recursive functions in the terminology of the question I linked to, where $\lambda$ is the height of the admissible set considered (at least for a $L_\lambda$). But as I noted, the precise relation between these concepts escapes me. Also somewhat relevant to your question might be Stephen G. Simpson's paper titled "Short Course on Admissible Recursion Theory" on p. 355–390 of Fenstad, Gandy & Sacks (eds.), Generalized Recursion Theory II (1978), proceedings of a symposium held in Oslo in 1977. It contains the clearest (if terse) explanation I found so far of how primitive recursive ordinal functions are defined and how they relate to more general recursion on ordinals.<|endoftext|> TITLE: Intuitive Aproach to Dolbeault Cohomology QUESTION [12 upvotes]: (Duplicated from math.stackexchange) I would like to understand an intuitive approach to the definitions of Dolbeault Cohomology (using $\partial$ and $\bar{\partial}$) similar to the one given here. It is not enough to know that it works, but to know what it means. All suggestions are welcome. REPLY [17 votes]: Here is some nonsense that I find useful: On a complex manifold, $$\frac{\mbox{locally constant functions}}{\mbox{smooth functions}} \approx \frac{\mbox{locally constant functions}}{\mbox{holomorphic functions}} \cdot \frac{\mbox{holomorphic functions}}{\mbox{smooth functions}}$$ The left hand side is where good geometric intuition lies, and de Rham cohomology measures how complicated it is. For example, $H^0$ measures how many locally constant functions there are. On a contractible space, where $H^{\ast}$ vanishes, smooth maps are homotopic to locally constant maps. Dolbeault cohomology measures how complicated the second term on the RHS is. This isn't as geometric, since it is "the square root of geometry". However, it is useful to think about when it is large or small. On a Stein space, Dolbeault vanishes. This means that smooth functions and holomorphic functions are close to being the same, and all the interesting geometry is in the first fraction. Indeed, on a Stein space, there are lots of holomorphic functions, and every cohomology class has a holomorphic representative. On a compact Kahler manifold, on the other hand, Dolbeault cohomology is large. This should mean that there are many fewer holomorphic functions than smooth functions, and that only a small part of the geometry can be seen in holomorphic terms. Indeed, in this case, all holomorphic functions are locally constant, and only a small number of cohomology classes have holomorphic representatives. To actually say something precise, there are three exact sequences of sheaves that come up everywhere in algebraic geometry. Write $\underline{\mathbb{C}}$ for the locally constant $\mathbb{C}$-valued functions, $\mathcal{H}^p$ for the holomorphic $(p,0)$ forms and $\Omega^{(p,q)}$ for the $C^{\infty}$ $(p,q)$-forms. Set $\Omega^n = \bigoplus \Omega^{p, n-p}$, the smooth $n$-forms. Then we have exact sequences: $$0 \to \underline{\mathbb{C}} \to \Omega^0 \overset{d}{\longrightarrow} \Omega^1 \overset{d}{\longrightarrow} \Omega^2 \overset{d}{\longrightarrow} \cdots$$ $$0 \to \underline{\mathbb{C}} \to \mathcal{H}^0 \overset{\partial}{\longrightarrow} \mathcal{H}^1 \overset{\partial}{\longrightarrow} \mathcal{H}^2 \overset{\partial}{\longrightarrow} \cdots$$ $$0 \to \mathcal{H}^p \to \Omega^{(p,0)} \overset{\bar{\partial}}{\longrightarrow} \Omega^{(p,1)} \overset{\bar{\partial}}{\longrightarrow} \Omega^{(p,2)} \overset{\bar{\partial}}{\longrightarrow} \cdots$$ The LHS of the nonsense equation refers to things related to the first sequence. The two fractions on the RHS refer respectively to the things related to the second and third sequences.<|endoftext|> TITLE: Étale cohomology versus classical cohomology QUESTION [10 upvotes]: Let $X$ be an algebraic variety over $\mathbb{C}$. If $X$ is smooth, the étale cohomology $H^p_{\textrm{ét}}(X,\mathbb{Z}/n)$ is isomorphic to the singular cohomology $H^p(X(\mathbb{C}),\mathbb{Z}/n)$. What is the situation if $X$ is not smooth? Are there counter-examples? REPLY [6 votes]: In Katz's review of $\ell$-adic cohomology in the first "Motives" volume, this isomorphism is stated without any smoothness assumptions, with a reference to SGA4, XVI 4.1 (which I don't have easily available).<|endoftext|> TITLE: Krein Milman theorem without the axiom of choice QUESTION [13 upvotes]: The Krein-Milman theorem asserts that in a locally convex topological vector space, a nonvoid compact convex subset is the closed convex envelope of its extreme points. But I would like to know when it is possible to avoid using Zorn's lemma, in more simple settings (outside of $\mathbb{R}^n$), because for example, in ergodic theory, one can prove quite easily with this theorem that if $X$ is a compact metric space, and $f:X\rightarrow X$ is continuous, and $M_f(X)$ is the set of $f$-invariant probability measures, then the $f$-invariant, ergodic probability measures are the extreme points of $M_f(X)$. The point is that one can prove the existence of $f$-invariant probability measures whithout the axiom of choice. REPLY [6 votes]: The solution is in the comment of Asaf Karagila, which points to the paper "A geometric form of the axiom of choice" by Bell and Fremlin. In fact, this paper asserts that the assertion : "For every normed vector space $X$ over the reals, there exists at least one extreme point in the unit ball of the continuous dual of $X$" is equivalent to the axiom of choice. So it doesn't matter how nice the space $X$, we can't prove the Krein-Miman for $M(X)=C(X)'$ in ZF. As a corollary, the ultrafilter lemma and the Krein Milman theorem imply (so is equivalent to) the axiom of choice, so the Krein Milman theorem cannot be provable only with the ultrafilter lemma, as the ultrafilter lemma is not equivalent to the axiom of choice. In particular, it shows that we can't prove the Krein Milman theorem without at least a weak form of axiom of choice, and as the ultrafilter is not enough, maybe we really need the axiom of choice in full generality, but the paper doesn't answer this question.<|endoftext|> TITLE: Is there a solvable point on any variety over the field of complex rational functions? QUESTION [10 upvotes]: Let $K = \mathbb{C}(T)$ be the field of complex rational functions in one variable, and let $V$ be a variety defined over $K$. Must $V$ have a solvable point? The variety $V$ is assumed geometrically irreducible. A solvable point is a point in $V(L)$ where $L/K$ is a finite Galois extension with $\mathrm{Gal}(L/K)$ solvable. REPLY [10 votes]: No, $V$ need not have a solvable point. The proofs I know construct such $V$ via deformation theory. The basic idea is in my paper. MR2579389 (2011g:14095) Reviewed Starr, Jason Michael(1-SUNYS) A pencil of Enriques surfaces of index one with no section. (English summary) Algebra Number Theory 3 (2009), no. 6, 637–652. 14J28 (14D06 14G05) Let $d$ and $n$ be positive integers with $d>n$. Let $H= \mathbb{P}\text{Sym}^d(V^\vee)$ be the projective space parameterizing degree $d$ hypersurfaces in $\mathbb{P}V = \mathbb{P}^n$. Let $$\mathcal{X}\subset H\times \mathbb{P}V$$ be the universal hypersurface. Let $S\subset H$ be the subvariety parameterizing hypersurfaces that are unions of $d$ hyperplanes; this is the same as the image of the natural morphism from the Segre variety $$(\mathbb{P}(V^\vee))^d \to \mathbb{P}\text{Sym}^d(V^\vee)$$ induced by the multiplication map. Let $$\phi: \mathbb{P}^1 \to S,$$ be a sufficiently general morphism such that $\phi^*\mathcal{O}_H(1)$ has degree at least $d$. It is straightforward to compute that the pullback $$\mathcal{X}_\phi := \mathbb{P}^1\times_{\phi,H} \mathcal{X}$$ has no solvable multisection over $\mathbb{P}^1$ as soon as $d \geq 5$. Simply consider the monodromy among the components of the strata in the stratification of the geometric generic fiber as a normal crossings variety. Of course the geometric generic fiber is not irreducible, as required. However, since the base field is the uncountable field $\mathbb{C}$, for a sufficiently general deformation of the image curve $\phi(\mathbb{P}^1)$ inside $H$, the same result holds true. Using the valuative criterion of properness, and analysis of the finite flat covers of curves, solvable multisections specialize to solvable multisections. Thus, if a sufficiently general deformation always has solvable multisection, since there are countably many parameter spaces for solvable multisections and since the parameter space $M$ for deformations of $\phi(\mathbb{P}^1)$ has uncountably many closed points, it follows that the original family $\mathcal{X}_\phi$ also has a solvable multisection. In fact, a similar argument works over fields that have large transcendence degree over their prime subfield. However, that excludes fields such as finite fields. That is why this argument does not apply to global function fields. However, as you know, Ambrus Pal does prove the result over function fields of surfaces over finite fields, the next best result possible.<|endoftext|> TITLE: Canonical immersion of the double torus QUESTION [20 upvotes]: It is easy to check that the immersion $\mathbb{T}^2=\mathbb{S}^1\times \mathbb{S}^1\longrightarrow\mathbb{R}^4$, $(\alpha,\beta)\longmapsto(\cos\alpha,\sin\alpha,\cos\beta,\sin\beta)$ induces the flat metric in the torus, and then, from the mathematical point of view, this immersion is highly preferred to the standard 'doughnut' one into $\mathbb{R}^3$. I have always wondered whether there exists something similar for higher genus surfaces, for instance a genus 2 surface – double torus. I imagine that, since the hyperbolic plane is the space used as universal cover for the double torus with constant curvature $-1$, one should search some smooth functions from the hyperbolic plane (say, the Poincaré disc in $\mathbb{C}$) to a higher Euclidean or hyperbolic space which are periodic respect to the action of some Fuchsian group. Having these functions, the point is that the immersion induces the $-1$ constant curvature. Of course, there must be some such embedding, according to the theorems of isometric embeddability. But the aim of this question is to ask whether there exist some nice and natural functions which do the job as in the case of the torus. Any suggestion is welcome. REPLY [6 votes]: This is not really an answer, but rather a longish comment and a suggestion about how one might focus the question a bit better. First, when one asks for a 'canonical' isometric embedding into some Euclidean space of a genus $2$ surface endowed with a metric of curvature $K=-1$, one might want to consider in what sense even the Clifford torus is 'canonical'. In fact, there are many flat metrics on the 1-holed torus: If $\Lambda\subset\mathbb{C}$ is a lattice (i.e., a discrete subgroup of rank $2$, say generated by $1$ and $\tau\in\mathbb{C}$ with positive imaginary part), then the standard metric $\mathrm{d}z\circ\mathrm{d}\bar z$ induces a flat metric $g_\Lambda$ on $T = \mathbb{C}/\Lambda$, and these flat metrics are not globally isometric as $\tau$ varies. Nevertheless, as Montiel and Ross showed in this paper, there is a $g_\Lambda$-isometric immersion $f_\Lambda:T\to\mathbb{E}^6$ by first eigenfunctions (that actually maps into a round $5$-sphere) that, moreover, is equivariant with respect to the identity component of the isometry group $G_\Lambda$ of $g_\Lambda$ in the sense that there is an embedding $G_\Lambda\to\mathrm{SO}(6)$ so that $f_\Lambda(g\cdot p) = g\cdot f_\Lambda(p)$ for all $p\in T$. In particular, this isometric embedding is as 'homogeneous' as possible. When $\tau=i$ (so that $\Lambda$ is the square lattice), this reduces to the Clifford torus example that the OP listed. Of course, these examples are simply quotients of global equivariant isometric immersions of $\mathbb{E}^2$ into $\mathbb{E}^n$ for $n\ge 2$. Meanwhile, it is well-known that the isometry group of the Poincaré disk is isomorphic to $\mathrm{PSL}(2,\mathbb{R})$ and that this group does not have any nontrivial homomorphism to the isometry group of Euclidean space in any (finite) dimension. Thus, there is no hope to construct an equivariant isometric immersion of the Poincaré disk into any (finite dimensional) Euclidean space, much less to find one that 'closes up' under some Fuchsian subgroup to give a 'canonical' isometric embedding of the $2$-holed torus into Euclidean space. Thus, one must look elsewhere for a notion of 'canonical isometric embedding'. One possibility one might try to generalize the above flat case would be to ask for an isometric embedding by the eigenfunctions of the Laplace operator associated to a particular eigenvalue. A related (non-existence result) is the following: If $(M^2,ds^2)$ is a surface of constant negative Gauss curvature then there is no nontrivial map $f:M\to S^n\subset\mathbb{E}^{n+1}$ that satisfies $\Delta f = -\lambda f$ for any $\lambda\not=0$. This follows by the same techniques used to prove Theorem 2.3 in this paper of mine. Thus, some other kind of extra system of equations would be needed to determine what you mean by 'canonical'.<|endoftext|> TITLE: Whether the manifold part of an Alexandrov space is connected? QUESTION [6 upvotes]: The title is my question. Alexandrov space here means finite dimensional Alexandrov space with curvature bounded below ,denoted by CBB. Let $\gamma$ be a simple curve in a $n$ dimensional CBB $M$ with two end points in the manifold part of $M$. Analysis. $\gamma$ can be covered by a finite collection $\mathcal U$ of cone neighborhoods, where the cone neighborhood here means a subspace of $M$ homeomorphic to $\mathbb{R}^m\times cone$. Since every cone nbhd in $\mathcal{U}$ must contain two manifold points. If we are able to find a new path contained in the manifold part of every cone nbhd in $\mathcal{U}$, then we done. So my question is reduced to the the restriction of the cone nbhd. However, since I don't know the the structure of the $cone$ part in the cone nbhd very well. I don't know how to get such a new path. Maybe the density of the manifold points in $M$ would help, but I don't know how to use this condition. REPLY [8 votes]: Yes. Assume $A$ is an $m$-dimensional Alexandrov space and $\Omega\subset A$ be the maximal open subset which is a topological $m$-manifold and $A'\subset A$ be the subset of all points with tangent space isometric to Euclidean space. From Perelmans paper "Beginning of Morse...", we get that $A'\subset \Omega$. The set $A'$ is a dense convex subset in $A$, this follows from Burago--Gromov--Perelman paper and from my paper on parallel translation. (Convexity means that any geodesic with the ends in $A'$ lies completely in $A'$.) In particular $A'$ is connected; hence the result follows.<|endoftext|> TITLE: An old paper of S.Chowla on unit equations QUESTION [10 upvotes]: It is referenced that in Chowla, S., Proof of a conjecture of Julia Robinson, Norske Vid. Selsk. Forh. (Trondheim) 34, 100–101 (1961), it is shown that the equation $\epsilon_1 + \epsilon_2 = 1$ has only finitely many solutions, when $\epsilon_1, \epsilon_2$ are units of a given number field. Unfortunately, the article seems to be unavailable electronically. I hence wonder if the argument given in the article is available from another source or if it could be reproduced here, if not too lengthy. Thank you. REPLY [4 votes]: For what it is worth, I reproduce below the relevant paragraphs in Chowla's aforementioned paper. The number of solutions of the equation $$\epsilon - \epsilon^{\prime} =1 \qquad \qquad \qquad(1)$$ where $\epsilon$ and $\epsilon^{\prime}$ are units of a fixed algebraic number field $R(\theta)$ is finite. Proof. Let $n$ be the degree of the field $R(\theta)$. With the usual notation we write $n=r_{1}+2r_{2}$, $r=r_{1}+r_{2}-1$ (see e.g. the Chelsea reprint of Landau's Algebraische Zahlen) and (in accordance with the Dirichlet-Minkowski theorem) $$\epsilon = \rho\eta_{1}^{a_{1}} \cdots \eta_{r}^{a_{r}}, \quad \epsilon^{\prime} = \rho^{\prime} \eta_{1}^{a_{1}^{\prime}} \cdots \eta_{r}^{a_{r}^{\prime}}$$ where $\eta_{1}, \ldots, \eta_{r}$ is a system of fundamental units of $R(\theta)$ and $\rho, \rho^{\prime}$ are certain roots of unity belonging to $R(\theta)$. Write for $m\in\{1,\ldots, r\}$ $$a_{m} = q_{m}(2n+1)+t_{m}, \quad a_{m}^{\prime} = q_{m}^{\prime}(2n+1)+t_{m}^{\prime}$$ where the $q$'s and $t$'s are rational integers and $0\leq t_{m}, t_{m}^{\prime} \leq 2n$. Then $(1)$ becomes $$\rho \, \eta_{t_{1}}\cdots \eta_{r}^{t_{r}} \alpha^{2n+1}-\rho^{\prime} \, \eta_{1}^{t_{1}^{\prime}} \cdots \eta_{r}^{t_{r}^{\prime}} \beta^{2n+1}=1$$ where $\alpha, \beta$ are integers (in fact, units) of $R(\theta)$. By a well-known extension of the Thue-Siegel-Roth theorem (see, for example, vol. 2 of LeVeque's "Topics in number theory", pp. 150-154) the equation $$\lambda \alpha^{2n+1} - \mu\beta^{2n+1} = 1$$ where $\lambda, \mu$ are fixed integers of $R(\theta)$ and $\alpha, \beta$ are the unknowns (also integers of $R(\theta)$) has only a finite numbers of solutions. Hence our assertion regarding $(1)$. [Note. $\rho$ and $\rho^{\prime}$ can assume only a finite number of values.]<|endoftext|> TITLE: Are "Unions" of small exotic $\mathbb{R}^4$'s small? QUESTION [5 upvotes]: Suppose $M$ is a smooth 4-manifold, and $U,V \subset M$ are exotic $\mathbb{R}^4$'s, i.e. homeomorphic to standard $\mathbb{R} ^4$. Further more suppose $U$ an $V$ intersect nicely sucht that $U \cup V \subset M$ is again homeomorphic to $\mathbb{R}^4$. An exotic $\mathbb{R}^4$ is called small if it embeds into $S^4$. Now my question: If $U$ and $V$ are both small, is $U \cup V $ then also small? I thought about that for quite a long time, somehow I believe it should be true, but do not see any way to prove it. REPLY [3 votes]: The answer is yes. You can embed U and V disjointly in $S^4$, and then do the connected sum at infinity, or `end-connected sum', using an arc going from one to the other. This construction is, I guess, what you intend by $U \cup V$, and is discussed for example in Gompf's paper, Three exotic R4's and other anomalies, J. Differential Geom. 18 (1983), no. 2, 317–328.<|endoftext|> TITLE: Is there a compendium of the consistency strength between the most important formal theories? QUESTION [16 upvotes]: Preliminar Notions: A formal system is a tuple $(\Sigma,G,A,R)$ where $\Sigma$ is an alphabet (set of symbols), $G$ is a formal grammar on $\Sigma$ that generates a formal language $L$ (set of well formed formulas -wffs-). $A$ is a formal theory (subset of $L$) called the set of axioms and $R$ is a set of inference rules that determine the mathematical theory (formal theory) derived from the axioms. A mathematical theory $\tau$ is consistent if it does not have any contradictions. Given two mathematical theories, $s$ and $\tau$, it is said that $s$ is consistent relative to $\tau$ if the consistency of $\tau$ implies the consistency of $s$ ($Con(\tau)\to Con(s)$). We also say that $s$ has greater consistency strength than $\tau$ if $\tau$ is consistent relative to $s$ but $s$ is not known to be consistent relative to $\tau$. Motivation: In logic it is common practice to analyze the consistency strength of a theory relative to another, and the most popular comparison is made against $ZFC$. There are a lot of theorems in the literature showing $Con(\tau)\to Con(s)$ for $\tau$ and $s$ formal theories and also it is a typical question to ask about the consistency strength of a mathematical theory whenever its formal system is presented/introduced. Based on this, I believe a diagram or a compendium of the consistency strength between the most important formal theories would be useful to have (in order to locate a new formal system quickly in the "hierarchy of consistency strength") but I have not been able to find any on the internet. The Question: Do you know of a diagram or a compendium of the consistency strength of the most important formal theories studied in mathematical logic? I'm searching for theories that contain enough arithmetic to prove Gödel's Incompleteness Theorems and by important I mean that the formal system that generates the theory has been studied by several mathematical logicians (let's say 25). I'm also searching for diversity, not just relations between large cardinal hypothesis but also systems like Martin-Löf Type Theory or the systems used to study Homotopy Type Theory or Reverse Mathematics. Extra: If there is none, we can contribute by adding consistency relations as answers and referencing the paper/book where the result comes from. An Example: $$Con(ZFC)\leftrightarrow Con(ZF) \to Con(PA) \to Con(PRA)$$ On the answer to the question: The answer that fits the most to the requirements of the question is a combination of Joel David Hamkins' answer and the comment on it by Ali Enayat about Harvey Friedman's paper. It should be easy for someone to have a good general picture of the consistency strength between formal theories combining these two. If you wish to add more theories from other fields such as HoTT or Category Theory you are welcome to do so as a comment in that answer. REPLY [2 votes]: This is an old question, but for interested readers, I might add the charts on this webpage, which I find quite good: https://dansnielsen.wordpress.com/2018/06/28/a-travel-guide-to-the-large-cardinals/<|endoftext|> TITLE: Decidability of diophantine equation in a theory QUESTION [9 upvotes]: Given a theory $T \subseteq \operatorname{Th}(\mathbb{N})$, define the decision problem $D_T$ as follows: Given a polynomial $p$ with integer coefficients and variables $\bar{x}$, decide whether $$ T \vdash \forall \bar{x}\, p(\bar{x}) \neq 0, $$ i.e., whether $T$ shows that $p$ does not have a root. By the MDRP theorem, we know that $D_T$ is undecidable for $T = \operatorname{Th}(\mathbb{N})$. Using the same theorem, we can also show that for any computably enumerable $T$ there is a $p$ such that $p$ has no root in $\mathbb{N}$, but $T$ does not show this, so $D_T$ is not equivalent to solving diophantic equations. My question is for which smaller $T$ the problem $D_T$ is still undecidable, especially $T$ = Peano arithmetic? $T$ = Robinson arithmetic? Some computably enumerable $T$? Any computably enumerable $T$ containing, say, P.A. or R.A.? (this is my intuition, but I haven't been able to show it) Edit: Had "... has a root" before, which is kinda trivial (thanks to @NoahS for pointing this out) REPLY [13 votes]: If $T$ contains $I\Delta_0+\mathit{EXP}$ (or a strong theory like Peano arithmetic for that matter), then by the formalized MRDP theorem [1], $D_T$ is essentially the $\Pi^0_1$ fragment of $T$, which is of course undecidable if $T$ is consistent. Curiously, this holds even for some weaker theories that do not prove the MRDP theorem: in particular, Kaye [2] showed that $D_T$ is undecidable for any consistent extension of $IE_1$ (induction for bounded existential formulas). On the other hand, it is a long-standing open problem whether $D_T$ is decidable for the theory of quantifier-free induction (IOpen); by results of Wilkie [3], this is equivalent to the same problem for the theory of $\mathbb Z$-rings. Concerning Robinson’s arithmetic $Q$: the problem is not quite well defined here, as $Q$ does not prove the usual semiring axioms like distributivity, hence different terms representing the same polynomial may give different answer. Worse yet, the question only makes sense for integer polynomials, whereas the theory only has nonnegative integers, and its models cannot be extended with negatives in a coherent way. So, for definiteness, let me formulate it as $$\let\ob\overline\let\ub\underline\def\N{\mathbb N}\let\LOR\bigvee D_T=\{(p,q):p,r\text{ are $L$-terms, }T\vdash\forall\ob x\,p(\ob x)\ne q(\ob x)\},$$ where $L=(0,S,+,\cdot)$ is the usual language of arithmetic. EDIT: I realized that the “exercise” in Claim 2 below was incorrect, so let me formulate the result for the slightly stronger theory $$Q^+=Q+\forall x\,0\cdot x=0$$ where it works. While $D_Q$ is also decidable, this is considerably more difficult to prove; see [4]. Proposition: $D_{Q^+}$ is decidable (in fact, coNP-complete). Proof: The main point is that $Q^+$ has weird models where one can satisfy next to any equation $p(\ob x)=q(\ob x)$, and the exceptions are easy to deal with. Let $\ub n=S^n(0)$ denote the numeral for $n\in\N$. Consider the model $\N^\infty$ of $Q^+$ with domain $\mathbb N\cup\{\infty\}$, where $x+\infty=\infty+x=x\cdot\infty=\infty\cdot x=\infty$, except that $\infty\cdot0=0\cdot\infty=0$. By induction on the complexity of the term $p$, we show Claim 1: If the value of $p(\ob\infty)$ in $\N^\infty$ is $n\in\N$, then $Q^+$ proves $\forall\ob x\,p(\ob x)=\ub n$. Given terms $p,q$, we can compute the value of $p(\ob\infty)$ and $q(\ob\infty)$ in $\N^\infty$. If both equal $\infty$, we have a witness that $(p,q)\notin D_{Q^+}$ and we are done. By Claim 1, the remaining cases are when one of the terms is provably equal to a standard constant, hence we can assume without loss of generality that $q$ is $\ub n$ for some $n\in\N$. As a little exercise with the axioms of $Q$, one can show by induction on $n$: Claim 2: $Q$ proves $$\begin{align*} x+y=\ub n&\to\LOR_{k,l\colon k+l=n}(x=\ub k\land y=\ub l),\\ x\cdot y=\ub n&\to x=0\lor\LOR_{k,l\colon kl=n}(x=\ub k\land y=\ub l)\qquad\text{for }n\ne0,\\ x\cdot y=0&\to x=0\lor y=0 \end{align*}$$ for every $n\in\N$. Let me write $x\le\ub n$ as a shorthand for the formula $x=0\lor x=\ub1\lor\dots\lor x=\ub n$. (In fact, this agrees with the usual definable ordering in $Q$, but I will not need this.) In $Q$, define $$x^{\le n}=\begin{cases}x&\text{if }x\le\ub n,\\\ub n&\text{otherwise.}\end{cases}$$ Claim 3: $Q^+$ proves $p(\ob x)\le\ub n\to p(\ob x)=p(\ob x^{\le n})$. Again, we prove this by induction on the complexity of $p$. For example, assume that $p=p_0\cdot p_1$. Reason in $Q^+$, and let $p(\ob x)=\ub m$ for some $m=0,\dots,n$. If $m\ne 0$, then $p_0(\ob x)=\ub k$ and $p_1(\ob x)=\ub l$ for some $k,l\le n$ such that $kl=m$ by Claim 2 (the case $p_0(\ob x)=0$ is impossible by the extra axiom). By the induction hypothesis, we have also $p_0(\ob x^{\le n})=\ub k$ and $p_1(\ob x^{\le n})=\ub l$, hence $p(\ob x^{\le n})=\ub m$. If $m=0$, then Claim 2 gives $p_0(\ob x)=0$ or $p_1(\ob x)=0$, hence $p_0(\ob x^{\le n})=0$ or $p_1(\ob x^{\le n})=0$ by the induction hypothesis, thus $p(\ob x^{\le n})=0$ as well. Now, resuming the proof of decidability of $D_{Q^+}$: if there is a model of $Q^+$ in which $p(\ob x)=\ub n$ is satisfiable, then by Claim 3, it is also satisfiable by a tuple of elements of $\{0,\dots,n\}$. The value of $p(\vec x)$ on standard tuples is evaluated the same in all models, hence we can as well do it in the standard model. Thus, $$(p(\ob x),\ub n)\in D_{Q^+}\iff\forall\ob x\in\{0,\dots,n\}\,p(\ob x)\ne n,$$ which is a decidable condition. References: [1] Haim Gaifman and Constantinos Dimitracopoulos, Fragments of Peano’s arithmetic and the MRDP theorem, in: Logic and Algorithmic, Monographie No. 30 de L'Enseignement Mathématique, Université de Genève, 1982, pp. 187–206. [2] Richard Kaye, Hilbert's tenth problem for weak theories of arithmetic, Annals of Pure and Applied Logic 61 (1993), no. 1–2, pp. 63–73. [3] Alex Wilkie, Some results and problems on weak systems of arithmetic, in: Angus Macintyre (ed.), Logic Colloquium ’77, North-Holland, 1978, pp. 285–296. [4] Emil Jeřábek, Division by zero, Archive for Mathematical Logic 55 (2016), no. 7, pp. 997–1013.<|endoftext|> TITLE: Does the Legendre-Hadamard condition imply a generalized Gårding inequality? QUESTION [14 upvotes]: For simplicity, we restrict to constant coefficients. Let $A^{ij}_{ab} \in \mathbb{R}$, $1 \le i, j \le n$ and $1 \le a, b\le m$, satisfy the Legendre-Hadamard condition: $$ A^{ij}_{ab}\xi_i\xi_jv^av^b \ge \lambda |\xi|^2|v|^2 $$ for some $\lambda > 0$ and any $\xi \in \mathbb{R}^n$ and $v \in \mathbb{R}^m$. Let $B$ be the unit ball in $\mathbb{R}^n$. It is straightforward to use the Fourier transform to prove that there exists $c > 0$ such that given any $u \in C^\infty_0(B,\mathbb{R}^m)$, $$ \int_{B} A^{ij}_{ab}\partial_iu^a\partial_ju^b \ge c\int_B |\partial u|^2. $$ If the (stronger) Legendre condition $$ A^{ij}_{ab}p_i^ap_j^b \ge \lambda\sum_{i,a}(p_i^a)^2 $$ for any $p_i^a \in \mathbb{R}$ holds, it is easy to use an extension operator to extend the inequality to any function $u \in C^\infty(\overline{B},\mathbb{R}^m)$. Question: Does the Legendre-Hadamard condition imply a Garding inequality of the form $$ \int_{B} A^{ij}_{ab}\partial_iu^a\partial_ju^b \ge \int_B c|\partial u|^2 - c'|u|^2, $$ for any $u \in C^\infty(\overline{B},\mathbb{R}^m)$? If not, what is a counterexample? REPLY [4 votes]: Gui-Qiang Chen suggested me to look at your question. I assume that your notation $\partial u$ means the gradient $\nabla u$. In the following I use $\bar\partial u$ to denote the standard first order operator for complex functions. There is another simple counter-example in the $2\times 2$ case. Consider the convex functional $I(u)=\int_B|\bar\partial u|^2dx$ for $u=u_1+iu_2$ and write the quadratic form in terms of $u_1$ and $u_2$. Since $|\bar\partial u|^2=|P_{E_{\bar\partial}}\nabla u|^2$, where $P_{E_{\bar\partial}}$ is the orthogonal projection to the subspace of anti-conformal matrices and $\nabla u$ is the gradient of $(u_1,u_2)$, we have, for every rank-one matrix $\xi\otimes\eta$ that $|P_{E_{\bar\partial}} \xi\otimes\eta|^2=|\xi|^2|\eta|^2/2$. It is easy to see that the corresponding coefficient tensor $A$ satisfies the strong L-H condition. However, if one takes any holomorphic function $u$, then $I(u)=0$. This leads to easy counter-examples for every fixed pair of constants $c>0$ and $c^\prime>0$, e.g. $u_n=e^{nz}$. This type of constructions by using subspaces of conformal and anti-conformal matrices in the space of $2\times 2$ real matrices are commonly used in the vectorial calculus of variations related to quasiconvexity and material microstructure. If $A$ is not a constant tensor, say, $A\in L^\infty$, even under the homogeneous Dirichlet boundary condition, Garding's inequality fails in general. I constructed such an example some years ago: `A counterexample in the theory of coerciveness for elliptic systems. J. Partial Differential Equations 2 (1989), no. 3, 79–82 (MR1026095)'. I also have some recent results on the so-called universal coercivity problem: 'On coercivity and regularity for linear elliptic systems. Calc. Var. PDEs 40 (2011), no. 1-2, 65–97 (MR2745197)'.<|endoftext|> TITLE: Examples of Kan extensions, adjunctions, and (co)monads in analysis, Lie theory, and differential geometry? QUESTION [26 upvotes]: In introductory texts on category theory, it seems like the majority of examples come from algebraic topology, algebra, and logic. What are some good examples of Kan extensions, adjunctions, and (co)monads in analysis, Lie theory, and differential geometry? Since limits and colimits can be characterized as Kan extensions or adjunctions, we have the obvious standard constructions: (co)products, (co)equalizers, etc., but these are common to a lot of categories we work with. Are there any examples more specific to analysis and differential geometry? (Crosspost from Math.SE) REPLY [3 votes]: The Schwartz kernel theorem, e.g, that all continuous linear maps from the space $\cal S(\mathbb R^m)$ of Schwartz functions on $\mathbb R^m$ to tempered distributions $\cal S'(\mathbb R^n)$ on $\mathbb R^n$ are given by "kernels" $K(,)$ in $\cal S'(\mathbb R^{m+n})$, is ${\rm Hom}(\cal S\otimes \cal S,\mathbb C)\approx {\rm Hom}(\cal S, \cal S')$, assuming/when the tensor product exists in the category of locally convex topological vector spaces. (Emphatically, the usual "projective" and/or "injective" tensor products are not tensor products in the full categorical sense.) This adjunction is an instance of the Cartan-Eilenberg ${\rm Hom}(A\otimes B,C)\approx {\rm Hom}(A,{\rm Hom}(B,C))$. Existence of the genuine tensor product uses the nuclearity of $\cal S$, that is, the demonstrable fact that $\cal S$ is a countable projective limit of Hilbert spaces with transition maps that are Hilbert-Schmidt. (And "trace-class" might best be characterized as being the composition of two Hilbert-Schmidt operators.)<|endoftext|> TITLE: Do hom-sets really live in the category Set? QUESTION [7 upvotes]: This isn't really a research-level question (sorry!), but I asked on math.se (link), and though the question was upvoted a few times, I didn't get any answers. So since there may well be more category theorists hanging out here, let me try again! In familiar introductory books on category theory, one of the very first examples of a category given is Set. And what category is that? Typically no explanation is given at this stage. But of course which category we are dealing with depends on our set theory. For an NF-iste, the category of NFsets has very different properties from the usual category Set (for a start NFsets is not cartesian closed). But fair enough, in an intro book you aren't going to mention that in Ch. 1! No: the authors are, surely, intending to point to stuff that their beginning readers can be assumed to know about, and are saying, "Hey you in fact already know about some categories, for example ..." The charitable reading, then, is that authors are relying on their readers to think of Set as comprising the sets they already know and love from their standard intro set-theory course. Which are pure sets of the cumulative hierarchy -- pure in that there are no urlements, no memberless entities in the universe of sets other than the empty sets. [If set theories with urelements are mentioned in passing in an intro course, it is usually only to be dismissed and forgotten about.] OK, then: in the absence of special explicit signals to the contrary, it seems (doesn't it?) that we might reasonably take the category Set mentioned in the very early pages to be a category of pure sets of the usual hierarchy. What else? But then what are we to make, a bit later in the book, of e.g. the usual presentation of the Yoneda embedding as $\mathcal{Y}\colon \mathscr{C} \to [\mathscr{C}^{op}, \mathbf{Set}]$. Putting it this way assumes that hom-collections $\mathscr{C}(A, B)$ for $A, B \in \mathscr{C}$ actually live in $\mathbf{Set}$. And since such a hom-collection is a set of $\mathscr{C}$-arrows, that assumes that the $\mathscr{C}$-arrows must live in the world of pure sets too. [We may want the relevant hom-collections to be set-sized in the Yoneda embedding case -- but being no bigger than set-sized is one thing, living in the universe of pure sets is something else!] But do we really want to assume that arrows [in the small-enough categories] are always pure sets? Isn't category theory supposed to be a story about how different bits of the mathematical universe hang together which doesn't presuppose some over-arching, all-in, set-theoretic reductionism, and so in particular doesn't presuppose from day one that all morphisms are pure sets?? Now, the foundational sections you often meet early in category theory books worry away about questions of size (sets vs classes etc.). But the present worry is orthogonal to all that, and is in a way more basic. If we want to make no assumption that the denizens of different bits of the mathematical universe are all cut from the same cloth, we won't want to slip into assuming that sets of these denizens are all pure sets. So in particular, do we really want to assume that a collection of arrows (hom-set) must live in $\mathbf{Set}$ -- where that's the category mentioned back almost on p.1 of the book -- (as opposed, perhaps, to being fully faithfully mappable into that world? I guess there must be good discussions of this sort of thing in the literature somewhere, and I'm no doubt showing my ignorance by asking where! But, please, any pointers would be most gratefully received. REPLY [4 votes]: I think the question is vague and probably does not have a unique answer. I would says that this kind of concern is actually not so different from the "size issue " generally presented by the set of objects, it's just that because this problem only arise under very weak set theoretical foundation this question is not mentioned in books. In the same way that you don't care whether the "set of object" is actually a set or not, you don't really care whether the "sets of morphisms" are sets or not. what you need is "a notion of object" and a "a notion of morphism" that you can compose, but if you want to devise categories as algebraic structures then it's convenient to say that you have "a set of objects" and "a set of morphisms" but those "sets" of objects and morphisms does not really have to be elements of what you call the category of 'sets'. Although you would have to be careful at some point: certain theorem of category theory require to take limit and co-limit indexed by hom sets and might become false if you are not assuming that "hom-set" are actually sets. (for example the special adjoint theorem) Now, any category theory book I've ever opened was working in ZFC and hence had no reason to care about this kind of questions. In fact, if we just want to drop the axiom of choice then a good thing to do would be to replace the notion of "functor" by the notion of "anafunctor" (which is a generalization of functor where for each object $X$, '$F(X)$' is well defined only up to canonical isomorphism) for which most theorems of category theory, like the fact that a fully faithfull and essentially surjective functor is an equivalence of category, remains true without the axiom of choice. And I don't know of any book of category theory which follows this point of view.<|endoftext|> TITLE: Induced subgraphs of small strongly regular graphs QUESTION [7 upvotes]: Consider a strongly regular graph $G$ with parameters $(76,30,8,14).$ Hoffman's bound tells us that $\overline{G}$ has an independent set of size at most $4$ and its not hard to see there are indeed $4$ independent vertices in $\overline{G}.$ So if $x \in V(G)$ then the graph $H$ induced by $N(v)$ is a $8$-regular, $K_4$-free graph on $30$ vertices and known bounds tell us that $\alpha(H) \ge 5.$ Hence $G$ must must contain $K_{1,5}$ as an induced subgraph. I do not see any way to extend this subgraph without introducing cases. What I am wondering is Question 1. Can someone construct larger graphs that must be present as induced subgraphs of $G$ and Question 2. Can someone find large induced subgraphs for some of the missing SRG's on less than 100 vertices? The motivating factor for this problem is that getting an induced subgraph of order 19 not having $2$ as an eigenvalue is enough to reconstruct $G.$ REPLY [2 votes]: On strategies of reconstructing bigger subgraphs: suppose for a vertex $v$ we have we know the set $M$ of (possible) subgraphs $G(v,w)$ induced on $N(v,w):=N(v)\cap N(w)$, for a vertex $w$ at distance 2 from $v$, and the set $\Lambda$ of (possible) subgraphs $G(v,u)$ induced on $N(v,u)$, for $u\in N(v)$. Take $H\in M$ and $F\in\Lambda$ so that they match on $N(u,v)$ and on $N(u,w)$, for $w\in N(u)-N(v)$ (i.e. we identify $H$ with $G(v,w)$ and $F$ with $G(v,u)$). Now, let us pick $w\neq w_1\in N(u)-N(v)$, and go though all the possible $H_1\in M$ so that in addition $H_1$ can be identified with $G(v,w_1)$; for each success we can further pick $w_2\in N(u)-N(v)-\{w,w'\}$ and go through all the possible $H_2\in M$ so that $H_2$ can be identified with $G(v,w_2)$, etc. One possible variation here is to switch the roles of $u$ and $v$ and reconstruct subgraphs in $N(u)$ in the same vein. If such a search tree does not get too big, one can think of what to do next.<|endoftext|> TITLE: How are the Walker-Wang TQFT and the Crane-Yetter TQFT related? QUESTION [16 upvotes]: Mathematical physicists in solid state physics and topological insulators talk a lot about Walker-Wang models, which are a family of Hamiltonians defined on a 3d lattice. Unfortunately, the original paper is lacking a lot of mathematical details which were promised to appear in a later article, but (to my knowledge) never did. The model looks a lot like the 20 years older Crane-Yetter model, in that it needs a ribbon fusion category (also called "premodular, in that it doesn't need to be modular) as labelling data. An article treating (amongst many other things) the ground state degeneracy of Walker-Wang models suggests that for modular categories, the ground state is in fact nondegenerate for all spatial topologies, that is, the topological state space is 1-dimensional. This is the same behaviour as in Crane-Yetter for modular categories, where $CY(N) = n^{\sigma(N)} \implies CY(S^1 \times M) = 1$ also suggests 1-dimensional state spaces (the general case is, I think, unknown). The article briefly mentions the Crane-Yetter model for modular categories, but I fail to find a definitive statement like "Walker-Wang and Crane-Yetter are the same TQFT" or "they're different" in the article or elsewhere. Are they the same (as Turaev-Viro and Levin-Wen seem to be related as well) and Walker-Wang is just the hamiltonian formulation? REPLY [17 votes]: Yes, the Walker-Wang model is related to the Crane-Yetter-Kauffman TQFT in the same way the Levin-Wen model is related to the Turaev-Viro TQFT. See, for example, the table on page 14 of the notes from the talk "Premodular TQFTs" found on this page In general, given an $n$-category with the right sort of duality, there is a standard procedure for constructing A fully extended $n{+}1$-dimensional TQFT A state sum computing the path integral for an $n{+}1$-manifold equipped with a cell decomposition A commuting projection hamiltonian whose ground state is isomorphic to the Hilbert space for an $n$-manifold equipped with a cell decomposition. Here are three main examples of this procedure. Input: a fairly general pivotal 2-category A (generalized) Turaev-Viro type TQFT (full extended) Generalized Turaev-Viro state sum Levin-Wen model (again generalized) Input: premodular category (i.e. a 3-category with trivial 0- and 1-morphisms) Premodular (also known as Crane-Yetter-Kauffman) TQFT Crane-Yetter state sum Walker-Wang model Input: $\pi_{\le n} BG$, for $G$ and finite group $n{+}1$-dimensional Dijkgraaf-Witten TQFT DW state sum Kitaev finite group model I've suppressed a few details above.<|endoftext|> TITLE: Transcendence of products of certain real algebraic numbers QUESTION [11 upvotes]: Let \begin{equation} z := \prod_p p^{1/p^2}, \end{equation} where the product is over all prime numbers $p$, and we always take the positive real root. Is $z$ transcendental or algebraic, or (as I suspect) is the answer not at all clear (meaning it's "probably" transcendental)? More generally, let $(p_i)$ be an increasing sequence of prime numbers, and $(e_i)$ a sequence of integers such that the infinite product $\prod_{i=1}^\infty (p_i)^{1/e_i}$ converges. Are there some numbers of this form where we can say anything, other than those where the convergence is so fast that we can use classical Liouville-type arguments? Edit: Just to clarify: I'm not sure if there's any natural reason to look at such numbers, except that I had a vague idea (which is too long to get into here) that I might be able to prove something about them. I guess that was the case in the beginning of transcendence theory (a subject I don't usually think about) -- the first numbers to be proved transcendental were not numbers anyone cared about for any other reason. Another comment about these numbers: they "look" transcendental, because the partial products live in larger and larger extensions of the rationals. However, a power of each partial product is an integer, so in some sense we are not getting too far away from the rationals -- or away from $\mathbb{Q}^\times \otimes \mathbb{Q}$. REPLY [3 votes]: I'm not sure this really constitutes an answer, but I think that it may not be fruitful to study the type of generalized question you discussed at the start of your post; you should be able to prove rather easily that for any real number $r>1$ you can find a sequence of primes $p_i$ with exponents $e_i=p_i$ such that the infinite product $\prod_{i=1}^\infty (p_i)^{1/e_i}$ converges to exactly $r$. Just pick the primes in a greedy fashion. Or maybe you use every prime, and choose the exponents in a greedy fashion. Or mix and match, and end up with many ways to represent your real in this fashion (if it matters, or you are diligent enough, you can show that there are uncountably many such representations for each $r>1$). Given that there is so much choice and so little effect, it's my opinion that you're going to have to rely on the incredibly specific structure of your product to go anywhere fast, and perhaps this isn't so unreasonable of an opinion given the nature of almost all results we have about transcendence (namely that they exploit the very special definitions, structures, relations, etc., that their subject numbers enjoy, like $e$ and $\pi$). But hey, I'm no expert. Edit: New account, can't make comments, so this space is for Wojowu's request (and perhaps others in the future). You're right in questioning me on that point; this is not a proof that I have done myself, but it seems fairly straight forward given that the product $\prod_{i=1}^\infty (p_i)^{1/p_i}$ diverges if you take it over all primes. You can't fail to make it past $r$ since you can always just add more primes in sequence (there is always a "partial tail" that is as large as you want), but by selectively (greedily) deleting certain primes we can lower the partial products below $r$. So you should converge exactly to $r$. If you go about formalizing that and run into problems, let me know.<|endoftext|> TITLE: Tannakian fundamental group of two explicit tensor categories QUESTION [6 upvotes]: Let $K/k$ is a field extension and $G$ an affine group scheme over $K$. What are the Tannakian fundamental groups of these two $k$-tensor categories (with trivial fiber functors over $k$): 1. The category of pairs of finite vector spaces over $k$ with an isomorphism of their extensions to $K$, 2. The category of finite vector spaces $V$ over $k$ with a representation $\rho: G\to \mathrm{GL}(V\otimes_k K)$. Thanks! REPLY [7 votes]: One possibly useful way of describing these groups is by their universal properties. The first group has the property that homomorphisms from it to any other pro-algebraic group $H$ are in bijection with $H(K) / H(k)$, and the second group has the property that homomomorphisms from it to $H$ are in bijection with homorphisms from $G$ to $H(K)$. So the second one is the reverse of the universal property of the Weil restriction. We can check these properties easily using Tanaka-Krien duality. Homomorphisms from $G$ to $H$ up to conjugacy are the same as homomorphisms from the category of $H$-reps to the category of $G$-reps, and actual homomorphisms are those plus an isomorphism of fiber functors. Any functor from the category of $H$-reps to the category of pairs of $k$-vector spaces inside the same $K$-space must send an $H$-rep to a pair of vector spaces where the second differs from the first by an element of $H(K)$. This is well-defined up to automorphisms of the second space, which are $H(k)$. A similar argument works for the second one. Assume $K$ is a Galois extension of $k$ of degree $d$. We can use this unversal property to compute the groups. Let's see what happens when we pull the groups back to $K$. Pullback is adjoint to Weil restriction. So the universal functor now sends $H$ to $\operatorname{Res}_{K/k} H (K) / \operatorname{Res}_{K/k} H (k) = H(K)^d / H(K) = H(K)^{d-1}$. So this functor is equivalent to the free group on $d$ generators. So the group is the free pro-algebraic group on $d-1$ generators, or the pro-algebraic envelope of the free group on $d-1$ generators. How do we descend this from $K$ to $k$? We need a natural $\operatorname{Gal}(K/k)$ action. From our construction, it is more natural to identify our free group with the subgroup of the free group on $d$ generators that alternates element-inverse-element-inverse, which is free on the $d-1$ generators $x_2x_1^{-1}, x_3x_1^{-1}, \dots, x_d x_1^{-1}$. The $d$ generators correspond to the automorphisms of $K$, so $\operatorname{Gal}(K/k)$ acts on them naturally by left multiplication. The second one is the same thing except with the pro-algebraic envelope of the free product of all the Galois conjugate copies of $G$.<|endoftext|> TITLE: Global Affine Flag Variety and Affine Flag Variety QUESTION [8 upvotes]: There is a construction of a global affine flag variety over $\mathbb{A}^1$ (or another curve) $Fl_{\mathbb{A}_1}$ such that each fiber above $\epsilon \neq 0$ is isomorphic to a direct product of the affine Grassmannian $Gr$ with the ordinary flag variety $G/B$, $Gr \times G/B$. The fiber above $\epsilon = 0$ is the affine flag variety $Fl$. This first appeared in Gaitsgory's paper 'Construction of central elements in the affine Hecke algebra via nearby cycles' http://arxiv.org/abs/math/9912074 If we have some projective varieties in $Gr \times G/B$, how do we find out more about their images in the affine flag variety $Fl$, as $\epsilon \rightarrow 0$? More precisely in page 5, section 1.2.3 of the paper above, there is this example where for $G = GL_2$, a family of $\mathbb{P}^1 \subset Gr$ degenerates to two copies of $\mathbb{P}^1$ glued at a point, in the affine flag variety $Fl$ for $GL_2$. How do we verify this? Could we calculate things like this by some concrete methods? REPLY [3 votes]: Let $G$ be a split reductive group over a field $k$. Let $T\subset B$ a maximal split torus contained in a Borel. Denote by $X_*(T)$ the group of cocharacters of $T$. To every $\mu\in X_*(T)$ there is associated an element $t^\mu\in G(k((t)))$. Let $Gr_\mu$ be the reduced closure of the $G(k[[t]])$-orbit of $t^\mu\cdot e_0$ in $Gr$, where $e_0$ denotes the base point of $Gr$. In this case the image of the closed subvariety $Gr_\mu\times\{e\}\subset Gr\times G/B$ in the affine flag variety $Fl$ as $\epsilon\to 0$ was first determined by X. Zhu in his paper "On the coherence conjecture of Pappas and Rapoport", Theorem 3.8: http://arxiv.org/pdf/1012.5979v3.pdf Let me sketch his result. Let $I$ be the preimage of the Borel $B$ under the reduction map $G(k[[t]])\to G$, $t\mapsto 0$. For every $\mu\in X_*(T)$, let $Fl_\mu$ be the reduced closure of the $I$-orbit of $t^\mu\cdot e_0$ in $Fl$. Then Zhu shows that the image of $Gr_\mu\times\{e\}$ in $Fl$ is the union of the $Fl_\lambda$, where $\lambda$ runs over all translates of $\mu$ under the finite Weyl group acting on $X_*(T)$. In the case $G=Gl_n$, $B$ the upper triangular Borel and $T$ the diagonal torus, we have $X_*(T)=\mathbb{Z}^n$ and the finite Weyl group is just the symmetric group on $n$ elements acting on $\mathbb{Z}^n$ by permuting the coordinates. Note that Zhu's family $Fl_{\mathbb{A}^1}$ slightly differs from Gaitsgory's family: there is not extra $G/B$-factor. In fact his results work for more general reductive groups and use the theory of Bruhat-Tits group schemes.<|endoftext|> TITLE: Is every degree 1 self-map a homotopy equivalence? QUESTION [23 upvotes]: In a rather obscure article, I found (without proof) the following statement: If $M$ is a closed orientable manifold, every degree $1$ map $f: M \rightarrow M$ is a homotopy equivalence. Is this really true? Using Poincare duality, it is easy to see that $f$ is a homology equivalence. But has $f$ to induce an isomorphism on $\pi_1$? Another (maybe related) result is Hopf's theorem: The degree classifies maps $M \rightarrow S^n$ up to homotopy equivalence. (I am sorry if this question is too basic. Feel free to delete it in this case.) REPLY [25 votes]: I believe that this is an open question in general, and the assertion is an old conjecture of Hopf. Some special cases were considered by Jean-Claude Hausmann, Geometric Hopfian and non-Hopfian situations. Geometry and topology (Athens, Ga., 1985), 157–166, Lecture Notes in Pure and Appl. Math., 105, Dekker, New York, 1987. I don't think there's been much progress since then, at least not that I could find via Mathscinet or Google Scholar. It is easy to see that $f$ induces a surjection on $\pi_1$; if not, then $f$ factors through a non-trivial covering space of $M$, contradicting the degree-$1$ assumption. So if $\pi_1$ is Hopfian (any surjection from G to G is an isomorphism) then you get that $f_*$ is an isomorphism on $\pi_1$. There are, however, some non-Hopfian groups. Even when you know that $f_*$ is an isomorphism, you need more to show $f$ to be a homotopy equivalence; you'd need $f$ to induces homology isomorphisms with coefficients in $Z[\pi_1]$.<|endoftext|> TITLE: Infinite graphs isomorphic to their line graph QUESTION [7 upvotes]: The only finite connected graphs $G$ that are isomorphic to their line graph $L(G)$ are the cycle graphs $C_n$ (see this link for example). There are connected countable graphs that are isomorphic to their line graph: $G=(\omega,E)$ where $E=\{\{k,k+1\}: k\in\omega\}$; $G=(\mathbb{Z},E)$ where $E=\{\{k,k+1\}: k\in\mathbb{Z}\}$. Note that in the second graph, all vertices have degree 2, which makes it into a kind of "infinite cycle". (An interesting side question would be whether these are the only connected countable graphs (up to isomorphism) that are isomorphic to their line graph.) Question. Is there a connected graph $G=(V,E)$ with $V$ uncountable such that $G\cong L(G)$? REPLY [6 votes]: Note first that $L$ is naturally an endofunctor on the category of graphs and injective graph-homomorphisms that commutes with filtered colimits. Let $G$ be any graph such that there is an embedding $i:G\to L(G)$. This gives rise to an embedding $L(i):L(G)\to L(L(G))$, an embedding $L(L(i)):L(L(G))\to L(L(L(G)))$, and so on. Let $L^\omega(G)$ be the colimit of $G\to L(G)\to L(L(G))\to \dots$. Since $L$ commutes with filtered colimits, there are canonical isomorphisms $$L(L^\omega(G))\cong L(\varinjlim(G\to L(G)\to\dots))\cong \varinjlim( L(G)\to L(L(G))\to\dots)\cong L^\omega(G).$$ Furthermore, if $G$ is connected, so is $L^\omega(G)$, and if $G$ is infinite, $L^\omega(G)$ has the same cardinality as $G$. Thus to get a connected graph of a given infinite cardinality isomorphic to its line graph, it suffices to give a connected graph $G$ of that cardinality that embeds in $L(G)$. But this is easy; for instance, complete graphs work. In fact, any connected graph whose cardinality $\kappa$ has uncountable cofinality works, since such a graph must contain a vertex of degree $\kappa$, and thus $L(G)$ contains a clique of size $\kappa$.<|endoftext|> TITLE: Open access journals in number theory QUESTION [12 upvotes]: I'm a phd student in number theory, and I'm required (by the funding council) to publish any article I write in open access journals. The problem is that all journals I can find are either out of my league (but open access), or at the right level (but not open access). So I was wondering: Is there a list of good open access journals specialised in or biased towards number theory? If not, what are good open access journals (with a rough ranking in terms of quality and/or prestige) for number theorists? REPLY [6 votes]: The New York Journal of Mathematics (http://nyjm.albany.edu/) is a peer reviewed and free online general math journal that has published lots of papers in number theory. It has been publishing since the mid-1990s. It is not an ArXiv overlay journal. Published articles appear on the journal website as free downloadable PDF files. However, it is an electronic only journal, so if your funding requires a hard-copy print journal, it would not be allowed.<|endoftext|> TITLE: What are the products $\prod_{A\subset{\mathbb F}_p\colon |A|=n} \sum_{a\in A} \zeta^a$ equal to? QUESTION [18 upvotes]: This is a somewhat more explicit version of a question I have recently asked. Let $p$ be an odd prime, and write $\zeta:=\exp(2\pi i/p)$ (any other primitive $p$th root of unity will do as well). For integer $n\in[0,p]$, the products $$ {\mathcal P}_p(n) := \prod_{\substack{A\subseteq{\mathbb F}_p \\ |A|=n}}\ \sum_{a\in A} \zeta^a $$ are rational integers; can they be found "explicitly"? It is immediately seen that ${\mathcal P}_p(0)=0$ and ${\mathcal P}_p(1)=1$, and I can show that ${\mathcal P}_p(2)=\left(\frac2p\right)$ (it is not difficult to see that ${\mathcal P}_p(2)=\pm 1$, but to determine the sign is trickier). Also, we have ${\mathcal P}_p(p-n)=(-1)^{\binom{p}{n}}{\mathcal P}_p(n)$, so that ${\mathcal P}_p(p)=0$, ${\mathcal P}_p(p-1)=-1$, and ${\mathcal P}_p(p-2)=\pm 1$. $\quad$ What are the values of ${\mathcal P}_p(n)$ for $n\in[3,p-3]$? Some numerical data (thanks to Talmon Silver for the programming): $\quad {\mathcal P}_5(3)=-1$ $\quad {\mathcal P}_7(3)=-2^7$ $\quad {\mathcal P}_{11}(3)=23^{11}$ $\quad {\mathcal P}_{13}(3)=159^{13}$ $\quad {\mathcal P}_{17}(3)=-24617^{17}$ $\quad {\mathcal P}_{19}(3)=-611009^{19}$ $\quad {\mathcal P}_{23}(3)=1265401351^{23}$ $\quad$ If finding the individual values ${\mathcal P}_p(n)$ is difficult, can we at least find explicitly the product $$ {\mathcal P}_p(1){\mathcal P}_p(2)\dotsb{\mathcal P}_p(p-2){\mathcal P}_p(p-1) = \prod_{\varnothing\ne A\subsetneq{\mathbb F}_p} \sum_{a\in A} \zeta^a \ ?$$ Denoting this product by ${\mathcal P}_p$, $\quad {\mathcal P}_3=-1$ $\quad {\mathcal P}_5=-1$ $\quad {\mathcal P}_7=-2^{14}$ $\quad {\mathcal P}_{11}=-(3\cdot 23^4 \cdot 67\cdot 89)^{22}$ $\quad {\mathcal P}_{13}=-(3^{12}\cdot 5\cdot 53^6 \cdot 79^4\cdot 131^2 \cdot 157^2 \cdot 313\cdot 547\cdot 599\cdot 911)^{26}$ The problem can be restated in a purely combinatorial way, as hinted to in Ofir's comment below. Write $N:=\binom pn$, let $A_1,\dotsc,A_N$ be all the $n$-element subsets of ${\mathbb F}_p$, and for $z\in{\mathbb F}_p$ denote by $r_n(z)$ the number of representations $z=a_1+\dotsb+ a_N$ with $a_1\in A_1,\dotsc, a_N\in A_N$. We have then ${\mathcal P}_p(n)=\sum_{z\in{\mathbb F}_p}r_n(z)\zeta^z$, and from the fact that ${\mathcal P}_p(n)$ is an integer, it follows that $r_n(z)$ are actually equal to each other for all $z\in{\mathbb F}_p\setminus\{0\}$; as a result, we have, say, ${\mathcal P}_p(n)=r_n(0)-r_n(1)$. On the other hand, $$ r_n(0)+(p-1)r_n(1) = \sum_{z\in{\mathbb F}_p} r_n(z) = |A_1|\dotsb|A_N| = n^{\binom pn}. $$ This yields $$ {\mathcal P}_p(n) = \frac1{p-1} \left( p\,r_n(0)-n^{\binom pn}\right). $$ Thus, the problem boils down to finding $r_n(0)$, the number of all zero-sum $N$-tuples $(a_1,\dotsc,a_N)$ with the components $a_i$ representing each of the $n$-element subsets of ${\mathbb F}_p$. REPLY [2 votes]: Only a partial answer: I can prove that ${\cal P}_p(3)=\epsilon_p v_p^p$, with $\varepsilon_p^2=1$, for every integer $p\ge 3$ (not necessarily prime). The proof is too technical to present here. The values of $v_p$ are given by $$ v_p^6=\frac{9W_p}{(2^p-(-1)^p)^2(1-(-2)^p)}, $$ where $$ W_p=\prod_\zeta\{(1+\zeta)^p-(-1)^p\}; $$ except for the sign these $W_p$ are the numbers of sequence A096964 in OEIS (Wendt's determinant). Also $W_p=\det(M+(1-(-1)^p) I)$ where $M$ is the circulant matrix with first line $\binom{p}{k}$ for $0\le k\le p-1$. The sequence $\varepsilon_p$ appear to have a simple periodic structure, but this I have not proved.<|endoftext|> TITLE: integral p-adic Hodge theory and de Rham representations QUESTION [6 upvotes]: $p$-adic Hodge theory gives us some comparison theorems between several cohomology theories. It also provides a hierarchy in the category of $p$-adic representations of the absolute Galois group of a finite extension $K$ of the field of $p$-adic numbers. Because those representations always admit a lattice stable under the action of the Galois group, it is natural to ask if one can produce an integral $p$-adic Hodge theory. Satysfying answers for such a theory are given by the work of several people, including Fontaine-Laffaille, Breuil, Kisin, Liu among many others. They provided a theory for crystalline or semistable representations. Now the question, hoping it is not complete non sense : is there an integral theory for de Rham representations ? And another question : is there a "nice" integral avatar (with which we can deal with torsion representations) of the field of $p$-adic periods $B_{dR}$ introduced by Fontaine ? REPLY [3 votes]: All de Rham representations are potentially semistable, so if you have a satisfactory theory for semistable representations, it should allow you to deal with the de Rham ones.<|endoftext|> TITLE: When does the continuous Galois(=etale) cohomology of fields coincide with the naive one? Often true by the Bloch-Kato conjecture? QUESTION [8 upvotes]: For a field $F$ I am interested in its $l$-adic (Galois=\'etale) cohomology; here $l$ is a prime distinct from the characteristic of $F$ (for simplicity one may assume that the latter is $0$). For $i,j\in \mathbb{Z}$ one can define the corresponding "naive" cohomology group as $\varprojlim_n H^i(F, \mathbb{Z}/ l^n\mathbb{Z}(j))$. This is not a very "good" definition; the "correct" cohomology groups of $F$ are the continuous ones (defined by Jannsen) that take into account the (first) derived projective limit (functor) for the system $H^i(F, \mathbb{Z}/ l^n\mathbb{Z}(j))$. My question is: when does the latter $\varprojlim{}^1$-group necessarily vanishes or (at least) torsion? It seems that the Bloch-Kato conjecture yields: the transition maps are surjective (and so, $\varprojlim{}^1$ vanishes) if $j=i$. Is this correct? Are there any general results of this sort available for $j\neq i$ (and if $K$ does not contain all $\mu_{l^n}$)? For example, what is known for $j=1$, $i=2$ (this case is closely related to the Brauer group)? REPLY [4 votes]: There are two questions here: when does the $\varprojlim\nolimits^1$ of a sequence of abelian groups $M_1 \longleftarrow M_2\longleftarrow M_3\dotsb$ vanish, and what can be said about the Galois cohomology with cyclotomic coefficients. Concerning the first question: this is called the Mittag-Leffler condition, and I recall it being formulated as follows. The $\varprojlim\nolimits^1$ vanishes if, and perhaps also only if, for every $i>0$ the decreasing sequence of subgroups $\operatorname{im}(M_i\leftarrow M_j)$ eventually stabilizes in $M_i$ as $j$ grows to infinity. Concerning the second question: certainly, it follows from the Milnor-Bloch-Kato conjecture that $\varprojlim_n{}^1 H^i(F,\mathbb Z/l^n\mathbb Z(j))=0$ when $i=j$, because this is a sequence of surjective maps of abelian groups. For $i\ne j$, this $\varprojlim\nolimits^1$ vanishing should not be true in general.<|endoftext|> TITLE: nonstandard models and mathematical theorems QUESTION [13 upvotes]: Is there a first order statement about the natural numbers (not nonstandard analysis) such that the truth of the statement is easier to see in a nonstandard model? In other words, do nonstandard models offer "more" mathematical insight in some way about the standard model? Compare this with the use of the compactness theorem to prove certain facts about infinite objects which follows in a straightforward manner from that about finite objects; whereas a direct "mathematical" argument tends to be more tedious (eg. 4 colorability of an infinite planar graph). REPLY [17 votes]: I view nonstandard models (particularly those produced by ultraproducts) as a convenient way to take limits of discrete structures to obtain continuous structures that capture all of the asymptotic first-order properties of the discrete structure. This of course can also be done by the compactness theorem, but ultraproducts also offer good saturation properties, which tends to make the continuous structure "complete" or "locally compact" in various senses. This allows one to usefully deploy various tools from continuous mathematics (e.g. measure theory, ergodic theory, topological group theory) in a manner which would be inconvenient to do back in the discrete setting, or even in a continuous model that was constructed purely through a compactness argument. I give some examples of this at https://terrytao.wordpress.com/2013/12/07/ultraproducts-as-a-bridge-between-discrete-and-continuous-analysis/ . Some more sophisticated examples include the proof of the inverse conjecture for the Gowers norms, or the inverse theorem for approximate groups; at present the only known proofs of these results proceed via nonstandard methods (though I am sure that if one really wanted to, one could unpack the nonstandard arguments and obtain a fully standard, but very messy, proof of these results). Model theorists have become quite good at exploiting additional properties of (suitably constructed) nonstandard models, such as the existence of a very large group of automorphisms, which (when combined with sufficiently strong saturation properties) then leads to the ability to describe "generic" elements, "indiscernible" sequences of elements, representatives of various "types", and similar useful gadgets. I'm not so familiar with this aspect of nonstandard methods though.<|endoftext|> TITLE: Multiplicity of $Ext^{d-t}(M,\omega_R)$, ($d=\dim R, t=\dim M$) QUESTION [9 upvotes]: Let $R=\bigoplus_{i \geq 0} R_i$ be a Cohen-Macaulay graded ring ($R_0$ is a field and $R$ is generated by $R_1$) of dimension $d$ with canonical module $\omega_R$, and $M$ a graded Cohen-Macaulay $R$-module of dimension $t$. Assume that we know the Hilbert series, Hilbert polynomial and all Betti numbers of $M$. What can be said about the multiplicity of $\textrm{Ext}^{d-t}(M,\omega_R)$? Can we compute it from the given data? REPLY [4 votes]: It is equal to the multiplicity of $M$. In fact, you do not need graded or even Cohen-Macaulayness of $M$. Let $N= \textrm{Ext}^{d-t}(M,\omega_R)$. Let $S(M) := \{P \in \textrm{Supp}(M), \dim R/P = t\}$. Then we have the so-called associativity formula: $e(M) = \sum_{P \in S(M)} \textrm{length}_{R_P}(M_P)e(R/P)$ Now we just need to observe that $S(M) =S(N)$ and for each $P\in S(M)$, $\textrm{length}(M_P)= \textrm{length}(N_P)$. Both are consequence of Grothendieck Local Duality applied to the ring $R_P$.<|endoftext|> TITLE: Elliptic curves and connected components QUESTION [14 upvotes]: Are there elliptic curves of positive rank with two real connected components in which all the rational points lie only on one component? Concrete examples are really appreciated. REPLY [19 votes]: Yes. It is not hard to find an example: Take $$E \colon y^2 = x^3 - 12 x - 1\,.$$ Then $E(\mathbb Q) \cong \mathbb Z$ and $P = (5, 8)$ is a generator (according to Magma). Since $P$ is on the component of the identity, all rational points are on that component.<|endoftext|> TITLE: Can a graph be reconstructed from its cycle lengths? QUESTION [11 upvotes]: All graphs discussed are finite and simple. The cycle sequence of a graph $G$, denoted $C(G)$, is the nondecreasing sequence of the lengths of all of the cycles in $G$, where cycles are distinguished by the vertices they contain, not by the edges they contain. For example, $C(K_{3,2})=4,4,4$ and $C(K_4)=3,3,3,3,4$. Two graphs are isoparic if they have the same number of vertices and the same number of edges. Main question: If $G$ and $H$ are 2-connected nonisoparic graphs, can $C(G)=C(H)$? The 2-connected condition is so we can't just make a bunch of edge-disjoint cycles that share a vertex. The nonisoparic condition is so we can ignore situations like the following: These graphs are not isomorphic but are isoparic. Both graphs have the cycle sequence $3,3,4,5,5,6$ and can be viewed as just a square surrounded by two triangles. Perhaps there's a better way to ignore this trick besides the nonisoparic condition. I'm interested more generally in finding out exactly what the cycle sequence can tell us. When is a cycle sequence realizable by a 2-connected graph? Is such a realization ever unique? I've looked at a couple dozen graphs on fewer than seven vertices and the only duplicate cycle sequences have been for the graphs shown above. Thank you. REPLY [10 votes]: Second Answer I'm adding this as another separate answer, rather than editing the first "answer" because otherwise anyone coming late to this discussion will end up doubly confused. So let's try again, and say that the answer to your question is still "Yes". If you type the following into Sage g1 = Graph("G?rFf_") g2 = Graph("H??EDz{") and then show them as before, we get then I think that they each have exactly 11 4-blobs and 4 6-blobs (using "blob" rather than overloading the word cycle) but one has 8 vertices and 12 edges and the other has 9 vertices and 13 edges. Here's a list of the blobs for the first graph (preceded by the size) 4 5 4 1 0 4 6 4 1 0 4 6 5 1 0 4 7 4 1 0 4 7 5 1 0 4 7 6 1 0 4 7 6 2 0 4 7 6 2 1 4 7 6 3 0 4 7 6 3 1 4 7 6 3 2 6 7 6 4 2 1 0 6 7 6 4 3 1 0 6 7 6 5 2 1 0 6 7 6 5 3 1 0 and here's the ones for the second graph 4 8 6 1 0 4 8 7 2 0 4 8 7 3 0 4 8 7 3 2 4 8 7 4 0 4 8 7 4 2 4 8 7 4 3 4 8 7 5 0 4 8 7 5 2 4 8 7 5 3 4 8 7 5 4 6 8 7 6 2 1 0 6 8 7 6 3 1 0 6 8 7 6 4 1 0 6 8 7 6 5 1 0<|endoftext|> TITLE: A question on the twistor space of a manifold QUESTION [9 upvotes]: Let $M$ be either (a) self-dual conformal 4-manifold, or (b) hypercomplex $4n$-manifold. In either case one can construct the twistor space $Z$ (in the case (b) $Z=\mathbb{C}\mathbb{P}^1\times M$ as a smooth manifold) which admits a natural structure of a complex analytic manifold. Let $D(Z)$ denote the Douady space of $Z$, though we will be interested only in the space of rational curves in $Z$. $D(Z)$ is a complex analytic space. Let $p\colon Z\to M$ be the natural smooth map (in case (b) $p$ is the obvious projection). The fibers of $p$ are complex curves isomorphic to $\mathbb{C}\mathbb{P}^1$. Consider the map $q\colon M\to D(Z)$ defined by $q(x)=p^{-1}(x)$. It is well known in the literature (and uses a Kodaira theorem) that the image $q(M)$ is contained in the smooth part $U$ of $D(Z)$. I need a reference to the following fact which seems to be well known to experts: The map $q\colon M\to U$ is an infinitely differentiable map of smooth manifolds. The earliest mentioning of this fact in literature I was able to find is in the paper Atiyah, M. F.; Hitchin, N. J.; Singer, I. M. Self-duality in four-dimensional Riemannian geometry. Proc. Roy. Soc. London Ser. A 362 (1978). This paper treats only the case (a) (while I need (b)) and states the result without proof in a somewhat different language (see p. 438). REPLY [2 votes]: I think that this follows from basic properties of the moduli space $D(Z)$, since one knows that, for each $x\in M$, the normal bundle $\nu_x$ in $Z$ of each fiber $q(x) = p^{-1}(x)\simeq \mathbb{CP}^1$ is isomorphic to $\mathcal{O}(1)^{2n}$. Specifically, since one then has $H^1\bigl(q(x),\mathcal{O}(1)^{2n}\bigr) = (0)$ for each $x\in M$, the deformation space is unobstructed (by Kuranishi) so this gives that $$ T_{q(x)}D(Z) = \Gamma\bigl(q(x),\nu_x\bigr) \simeq H^0\bigl(q(x),\mathcal{O}(1)^{2n}\bigr) \simeq \mathbb{C}^{4n}. $$ Now consider the map $Q:Z\to D(Z)$ defined by $Q(z) = q(p(z))$. I think you'll agree that $Q$ is smooth, as this holds for any complex manifold $Z$ that has a locally trivial fibration by compact complex submanifolds each of which is unobstructed. (Probably you can find a proof of this general fact in Morrow and Kodaira's Complex Manifolds, which I don't have with me.) To prove that $q$ (which is injective) is smooth, it will be enough to show that $Q$ is a submersion onto its image in $D(Z)$. For this, you need to compute the differential $Q':TZ\to TD(Z)$ and show that $Q'(z):T_zZ\to T_{Q(z)}D(Z)$ has (real) rank $4n$ everywhere. However, this differential is easy to compute: Choose a (smooth, not holomorphic) splitting $TZ = K \oplus P$, where $K\subset TZ$ is complex line bundle that is the kernel of $p'$, i.e., the tangent vectors to the fibers of $p$. Now, for each $v\in T_zZ$, let $X_v$ be the vector field along the fiber $Q(z) = q(p(z))$ such that $p'(X_v) = p'(z)(v)$ and $X_v(y)$ lies in $P_y$ for all $y\in Q(z)$. Then $X_v$, when regarded as a section of $\nu_{p(z)}$ in the natural way, is holomorphic. Thus, one has the formula $$ Q'(z)(v) = [X_v]\in \Gamma\bigl(q(x),\nu_x\bigr)= T_{Q(z)}D(Z). $$ Moreover, it is clear that $Q'(z)(v) = 0$ if and only if $v$ lies in $K_z$. Thus, $Q$ is a submersion onto its image and the rest follows by elementary differential topology, since $p:Z\to M$ is also a submersion with the same fibers as $Q$.<|endoftext|> TITLE: Statistical distance between discrete and continuous distributions QUESTION [5 upvotes]: Are there any statistical distance functions that are capable of comparing a continuous and a discrete distribution? From reading this list http://en.wikipedia.org/wiki/Statistical_distance the only such distances seems to be the Wasserstein metric, a.k.a. the earth mover's distance, and Kolmogorov-Smirnov in the one-dimensional case, but none of the others (e.g. f-divergence) seems to work. Have I missed some? I find it surprising that no such f-divergences exist. REPLY [4 votes]: TL;DR Look for metrics which metricize weak convergence. After the comments above, I think I can answer your question to a reasonable degree. I am not a practicing statistian, so I can't comment too authoritatively on "usefulness". However, I am a computability theorist who specializes in measure theory, so I have a good sense of what is useful in theory (but maybe not in practice). One way to think of a metric is to think of the topology it corresponds to. In my experience the weak topology (or the weak-$*$ topology in functional analysis terminology) is by far the correct topology to talk about measures/distributions. The other topologies are too strong. (I had originally misunderstood your question as wanting a strong topology---which separates more stuff. Instead, you wanted a weak topology---which puts things close together if they are close in a practical sense.) First of all, the weak topology is good because the finite discrete measures are dense in this topology. Indeed, as @usul mentioned, you want the distributions of finitely many independent samples (the empirical distribution?) to converge to the true distribution. This happens almost-surely in the weak topology. (There is probably a result that says that no stronger topology works, but I don't know of a reference off hand.) Now, the Wasserstein Metric isn't exactly a metric for the weak topology. It is a metric for weak convergence and convergence in the $p$th moment. In particular it only works if both the distributions have finite p-th moments. For example, if $\mu$ does not have $p$th moments, then it will have infinite distance to any finite discrete measure. Now, if you are only considering measures on a bounded metric space (and you can always bound a metric $d$ by just using $\rho(x,y) = d(x,y) \wedge 1$ or $\rho(x,y) = \frac{d(x,y)}{d(x,y) + 1}$) then the Wasserstein metric is a metric for weak convergence. Now, (contrary what I said in the comments) the total variation distance separates things too much. Consider the uniform measure on $[0,1]$ and the uniform measure on $\{0,\ldots, k2^{-i}, \ldots, 1\}$. This has distance $1$ in the total variation metric, making it completely impractical for approximating a continuous measure with a discrete one. Moreover, it is even worse, the total variation metric is not separable, that is there is no countable dense set of measures $\mu_i$ in this metic. Therefore, there are measures that are not approximable in any practical way in this metric. (By practical, I mean you have some countable list of approximations to choose from.) Now, what metrics topology weak convergence? The Levy-Prokhorov metric (any complete separable metric space). The Levy metric ($\mathbb{R}$). The Wasserstein metric (a bounded metric space). If your distribution is on the space of infinite sequence of symbols $A^\mathbb{N}$ from an countable alphabet $A$, then one can use the metric $\sum_{w \in A^*} 2^{-|w|} |\mu([w]) - \nu([w])|$, where $A^*$ is the set of finite words of $A$ and $|w|$ is the length of $w$, and $[w]$ is the cylinder set of all inifinite sequences starting with $w$. (The space $\{0,1\}^\mathbb{N}$ is often a useful approximation to $[0,1]$ where the words correspond to dyadic intervals.) (You can ignore this...) Like I said, I work in computability theory and I find the weak convergence is the right topology to work with. Why? Well, separable topologies have a computability theoretic nature. In this case, given sufficiently good approximations to a measure in the weak topology, one can compute information about the measure. In particular, the weak topology gives exactly the information needed to compute $f \mapsto \int f\, d\mu$ for a bounded continuous function $f$. In the case of $A^\mathbb{N}$, it is exactly the information needed to compute $\mu{[w]}$ for each word $w \in A^*$. If I were to represent a distribution on a computer (with no worries about running times or memory usage) then I would represent it by approximating it by finite discrete measures which are close enough in some metric for weak convergence. (In the same way that reals are represented by rational approximations.)<|endoftext|> TITLE: Vanishing of certain periodic series: A question related to $L(1 , \chi) \neq 0$. QUESTION [12 upvotes]: Fix $q$ to be a positive integer. Let $$f : \mathbb{N} \to \{-1 ,0, 1\}$$ be a $q$-periodic arithmetic function such that $$\sum_{n = 1}^q f(n) = 0.$$ If $f$ is not identically zero, is it true that $$\sum_{n=1}^{\infty} \frac{f(n)}{n} \neq 0?$$ I ask this question in attempt to understand how general of a statement it is that $L(1 , \chi) \neq 0$ where $\chi$ is a non-principal real character. REPLY [16 votes]: On the other hand, a variant of the question has a positive answer. This question was raised by Chowla in 1964 in the case that $q = p$ is prime and $f(p) = 0$ (but with $f$ taking arbitrary rational values). This was settled affirmatively in a paper by A. Baker, B. Birch, and E. Wirsing ("On a problem of Chowla," Journal of Number Theory, 1973), using Baker's theory of logarithmic linear forms. For a general modulus $q$ they prove: If $f : \mathbb{N} \to \mathbb{Q}$ has a period $q$, satisfies $f(r) = 0$ for $(r,q) > 1$, and is not identically zero, then $\sum_{n} f(n)/n \neq 0$. The condition $f(r) = 0$ for $(r,q) > 1$ is only natural as it is met by a primitive mod $q$ character.<|endoftext|> TITLE: On the boundary of the twindragon QUESTION [13 upvotes]: Let $\mathcal T$ be the famous twindragon, i.e., $$ \mathcal T=\left\{\sum_{n=0}^\infty a_n\left(\frac{1+i}2\right)^n : a_n\in\{0,1\}\right\}. $$ Then, as is well known, $\mathcal T$ has a non-empty interior, whereas $\partial\mathcal T$ is indeed a fractal whose Hausdorff dimension is known as well - see, e.g., this survey (it's on the Heighway dragon, but the twindragon is just two of those placed back to back). Now let $X\subset\{0,1\}^{\mathbb N}$ be defined as follows: $$ \partial \mathcal T=\left\{\sum_{n=0}^\infty b_n\left(\frac{1+i}2\right)^n : b_n\in X\right\}. $$ Clearly, $X$ is a subshift. QUESTION. Is there a closed description of $X$? In particular, is $X$ a sofic subshift (or even a subshift of finite type)? The closed formula for its dimension - $\log\lambda/\log\sqrt2$ with $\lambda$ being a root of $2x^3-x+1$ - suggests so. The proof from the link uses the Hutchinson formula for some self-similar IFS whose attractor is precisely $\partial\mathcal T$, which is nice, but I'd like it to be in the form $h(X)/\log\sqrt2$, where $h(X)$ is the topological entropy of the subshift $X$. REPLY [6 votes]: The link that you refer to does not describe the boundary as the attractor of a simple IFS, rather it describes a collection of portions of the boundary as the invariant list of a digraph IFS - or directed graph iterated function system. I am not an expert on sofic systems, but I believe that an analysis of that digraph IFS yields the type of description that you want. Also, I can't agree with your dimension computation as your polynomial factors $$2 x^3-x+1=(x+1) \left(2 x^2-2 x+1\right)$$ revealing only one real root, which is negative. The analysis below computes the dimension using something that looks like an entropy computation. Digraph self-similarity Here's an image of the twin dragon surrounded by 6 copies of itself. The boundary consists of 6 parts, each of which is an intersection between the original twin dragon and one of the translated copies. That collection of 6 sets is exactly the invariant list of the following digraph IFS: Thus, for example, the second piece consists of one copy of itself together with two copies of the first. To each edge corresponds one of two possible functions, namely $$f_0(z)=\frac{1+i}{2}z \: \text{ or } \: f_1(z)=\frac{1+i}{2}(z + 1).$$ To compute the dimension of the boundary, we count the number $N_n$ of walks of length $n$ through the graph. The graph is strongly connected so it's adjacency matrix is irreducible. The Perron-Frobenius theorem guarantees that there is a positive eigenvalue $\lambda$ that is strictly larger than the absolute values of all the other eigenvalues. Furthermore, $N_n$ grows like $\lambda^n$. The size of a neighborhood generated by a path of length $n$ is $2^{-n/2}$. The dimension of the boundary is then given by $$\lim_{n\rightarrow\infty}\frac{\log(N_n)}{\log(2^{n/2})} = 2\frac{\log(\lambda)}{\log(2)}.$$ The characteristic polynomial of the adjacency matrix is $$\lambda ^6-2 \lambda ^5+\lambda ^4-4=(\lambda +1) \left(\lambda ^2-2 \lambda +2\right) \left(\lambda ^3-\lambda ^2-2\right).$$ The largest root of the characteristic polynomial coincides with the largest root of $\lambda^3-\lambda ^2-2$ and is approximately $\lambda=1.69$. The dimension of the boundary is $$2\frac{\log(\lambda)}{\log(2)} \approx 1.52363.$$ As a sanity check, we should be able to generate a portion of the boundary using your formulation. It takes just a few lines of Mathematica code to do so and it seems worth a look. Here's the whole set: n = 15; TPic = ListPlot[{Re[#], Im[#]} & /@ (Tuples[{0, 1}, n].Table[((1 + I)/2)^k, {k, 0, n - 1}]), PlotStyle -> RGBColor[1/5, 1/3, 4/5]] Note that Tuples[{0, 1}, n] generated a list of all tuples of zeros and ones of length $n$. We then took a dot product of that with a list of powers of $(1+i)/2$. Let's now try to do the same thing but, rather than using all tuples, we'll use only those tuples generated by a walk through the directed graph above starting at position 6. The code is a bit more involved than I should post here but that's the basic idea behind how I generated the the following pic: Looks promising. The description of the boundary of a self-similar tile as a collection of digraph self-similar sets was formulated the paper of Strichartz and Wang below. I wrote an exposition with Mathematica implementation here. Strichartz, R. and Wang, Y. Geometry of self-affine tiles I. Indiana University Mathematics Journal. 1999 . 7:1-2 3.<|endoftext|> TITLE: Identity for Power Series and Binomial Coefficients QUESTION [7 upvotes]: This question concerns a combinatorial identity obeyed by power series coefficients. Throughout we let $[x^{M}]\{\phi(x)\}$ denote the coefficient of $x^{M}$ in a power series $\phi(x)$. Let $k$ be a positive integer, and consider the function $F(k,x)$ defined as the following power series in $x$: \begin{equation} F(k,x)=\sum_{s=1}^{\infty} \frac{(-1)^{s-1}}{s^{2}}\binom{s \ k}{s+1}(s+1)\ x^{s}. \end{equation} I am interested in the series coefficients of the function $\exp(N F (k,x))$ for positive integer $N.$ Through comparison of various formulas that arose in a research project, I have been lead to the following identity for the case $N=M+1$: \begin{equation} [x^{M}]\{e^{(M+1)F(k,x)}\}= \frac{k(M+1)}{k+(k-1)M}\binom{(k-1)^{2}M+k(k-1)}{M}~. \end{equation} Although I am convinced that this identity is true, I have no idea how to demonstrate it, nor do I have any idea why this power series coefficient has such a simple expression. Thus, my main question is how can this identity be motivated and proven ? More generally, can we determine the coefficient $[x^{M}]\{e^{N F(k,x)}\}?$ I am also interested in a generalization which depends on an additional positive integer $j$. Specifically, set \begin{equation} F(k,j,x)=\sum_{s=1}^{\infty} \frac{(-1)^{s-1}}{s^{2}}\binom{s \ k}{s\ j+1}(s\ j+1)\ x^{s}~. \end{equation} The previous function is recovered for the special case $j=1.$ Can the coefficients $[x^{M}]\{e^{NF(k,j,x)}\}$ be similarly determined? REPLY [6 votes]: Let $\mathscr{B}_t(z)$ be a generalized binomial series $$\mathscr{B}_t(z)=\sum\limits_{k=0}^{\infty}{tk+1\choose k} \dfrac{1}{tk+1}z^k.$$ The answer follows from these two formulae \begin{gather} \tag{1}\mathscr{B}_t^r(z)=\sum\limits_{k=0}^{\infty}{tk+r\choose k} \dfrac{r}{tk+r}z^k,\\ \tag{2}\log\mathscr{B}_t(z)=\sum\limits_{k=1}^{\infty}{t k\choose k}\frac{z^k}{tk} \end{gather} because $$F(k,x)=k(1-k)\log\mathscr{B}_k(-x)=k(k-1)\log\mathscr{B}_{1-k}(x).$$ You can find (1) (and some other formulae) in (see Ch.5.4) Graham, R. L.; Knuth, D. E. & Patashnik, O. Concrete mathematics. Addison-Wesley Publishing Company, 1994. Formula (2) follows from calculations performed in Bizley, M. Derivation of a new formula for the number of minimal lattice paths from $(0,0)$ to $(k m, k n)$ having just $t$ contacts with the line $m y = n x$ and having no points above this line; and a proof of Grossman's formula for the number of paths which may touch but do not rise above this line. J. Inst. Actuaries 80, 55-62 (1954). The proof of (1) (see Concrete Mathematics) is also based on combinatorics of paths. Probably it can give some tips for $j>1$. See also Donald Knuth's 20th Annual Christmas Tree Lecture: (3/2)-ary Trees for additional connections and for the history of (2). I applied these formulae in the theory of formal groups. Can you give us some background information about your question?<|endoftext|> TITLE: Choosing a metric in which homeomorphism is Holder continuous QUESTION [15 upvotes]: Let $X$ be a compact metrizable space, and let $f:X \to X$ be a homeomorphism. Is it always possible to choose a compatible metric on $X$ in which $f$ is Holder continuous? I've tried some simple tricks like taking any metric $d$ and then defining a new one $d'$ by \begin{equation*} d'(x_1,x_2) = d(f^{-1}(x_1),f^{-1}(x_2)) + d(x_1,x_2) \end{equation*} or \begin{equation*} d'(x_1,x_2) = d(x_1,x_2) + d(f(x_1),f(x_2)) \end{equation*} and I can see no reason for them to be Holder continuous. We can also try to take any metric $d$ and set \begin{equation*} d'(x_1,x_2) = \sup_{n \in \mathbb{Z}} d(f^n(x_1),f^n(x_2)). \end{equation*} but this $d'$ being compatible with the topology is equivalent to the family of iterates $(f^n)_{n \in \mathbb{Z}}$ being equicontinuous. REPLY [7 votes]: This is not a complete answer, but topological entropy is not enough to rule out Hölder continuity. The following Hölder continuous automorphism of the Cantor space has infinite topological entropy. Infinite topological entropy is only enough to rule out Lipschitz continuity. Thus the example gives a Hölder continuous function $\varphi: X \to X$ that is not Lipschitz continuous with respect to any equivalent metric on $X$. Consider $X=\{0,1\}^\mathbb{Z}$ together with the usual metric $d(x,y)=2^{-\min\{|k|\: :\: x_k \neq y_k\}}$. Define $\varphi:X \to X$ by $\varphi(x)_i=\begin{cases} x_{2i} & i \geq 0\\ x_{-i/2} & i<0, 2\mid i\\ x_{(i-1)/2} & i<0, 2\not\mid i\end{cases}.$ (If one only needs an endomorphism one can take $\psi: 2^{\mathbb{N}_0} \to 2^{\mathbb{N}_0}, \psi(x)_i=x_{2i}$). This is a homeomorphism. It is Hölder continuous wrt. $d$, since $d(x,y)=2^{-(2n+1)}$ or $d(x,y)=2^{-(2n+2)}$ implies $(x_{-2n},\dots,x_{2n})=(y_{-2n},\dots,y_{2n})$, hence $(\varphi(x)_{-n},\dots,\varphi(x)_{n})=(\varphi(y)_{-n},\dots,\varphi(y)_{n})$ and $d(\varphi(x),\varphi(y))\leq 2^{-(n+1)}$. Therefore $d(\varphi(x),\varphi(y)) \leq d(x,y)^{\frac{1}{2}}$. To see that $\varphi$ has infinite topological entropy consider the cardinality $S_{n,\ell}$ of $2^{-2n}$ seperated sets with respect to $d_\ell(x,y)=\max_{k=0,\dots,\ell-1}d(\varphi^k(x),\varphi^k(y))$. If $x_i \neq y_i$ for some $i \in I:=\{0,\dots,2n-1\} \cup 2\cdot\{0,\dots,2n-1\} \cup \dots \cup 2^{\ell-1} \cdot\{0,\dots,2n-1\}$, then $d_{\ell}(x,y)>2^{-2n}$. Hence $S_{n,\ell}\geq |I| = 2^{2n+(\ell-1)n}\geq 2^{\ell n}$. Therefore $h_{\text{top}}(\varphi)\geq\lim_{n \to \infty} \limsup_{\ell \to \infty} \frac{1}{\ell} \log 2^{\ell n}=\infty$.<|endoftext|> TITLE: Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module? QUESTION [6 upvotes]: Does anyone know if the right Banach $\mathcal{B}(H)$-module $\mathcal{K}(H)$ is injective? This module is not dual, so standard arguments via flatness do not work. Injectivity is understood in the following sense. Definition #1. A bounded morphism of $A$-modules is called admissible if it admits a left inverse bounded linear operator. Definition #2. A Banach module $J$ over Banach algebra $A$ is called injective if for any admissible bounded morphism of $A$-modules $\xi:X\to Y$ and any bounded morphism of $A$-modules $\varphi:X\to J$ there exists a bounded morphism of $A$-modules $\psi:Y\to J$ such that $\psi\xi=\varphi$. Note, The standard tool for this type of problems is the following fact. Fact #1. A Banach module $J$ over a unital algebra $A$ is injective iff the map $$ \rho_J:J\to\mathcal{B}(A,J):x\mapsto (a\mapsto x\cdot a) $$ admits a left inverse map $\tau$ which is a morphism of $A$-modules. Here are my ideas that eventually failed. Idea #1. Construct $\phi$ and $\xi$ such that $\psi$ does not exist. The problem is that $\mathcal{K}(H)$ is not complemented in $\mathcal{B}(H)$ for infinite dimensional $H$. The only "interesting" admissible morphism I'm aware of is $\xi=\rho_{\mathcal{K}(H)}$. By interesting I mean admissible morphism without obvious left inverse bounded module map. Idea #2. Solve commutative analogue of the problem and modify solution for the non-commutative case. Strangely, the $\ell_\infty$-module $c_0$ is injective (see my proof below), but one cannot adopt this proof for $\mathcal{K}(H)$ since $\mathcal{B}(H)$ does not possess the Dunford-Pettis property. Idea #3. Note that $\mathcal{B}(H)$-module $\mathcal{B}(H)$ is injective, and modify proof for the case of $\mathcal{K}(H)$. Indeed, $\rho_{B(H)}$ has a left inverse module map $$ \tau:\mathcal{B}(\mathcal{B}(H), \mathcal{B}(H))\to \mathcal{B}(H): T\mapsto (\xi\mapsto T(\xi\bigcirc e)(e)) $$ where $e$ is a unit vector in $H$ and $\eta\bigcirc\zeta:H\to H:\xi\mapsto\langle\xi,\zeta\rangle\eta$ is a generic rank one operator. Unfortunately, for infinite dimensional $H$ this expression for $\tau$ does not define a left inverse for $\rho_{\mathcal{K}(H)}$ since $\tau(T)=1_H\notin\mathcal{K}(H)$ for $T:\mathcal{B}(H)\to\mathcal{K}(H):a\mapsto a\circ (e\bigcirc e)$. Idea #4. Look for more sophisticated right inverses. One can check that the linear map $$ \tau:\mathcal{B}(\mathcal{B}(H), \mathcal{K}(H))\to \mathcal{K}(H): T\mapsto \left(\xi\mapsto \sum_{i=1}^{n}\langle T(\xi\bigcirc e_i)(e_i), e_i\rangle e_i\right) $$ is a left inverse for $\rho_{K(H)}$ when $H$ is finite dimensional with orthonormal basis $(e_i)_{i=1}^n$. Unfortunately $\Vert \tau\Vert=n^{1/2}$, so this operator is not even well defined for infinite dimensional $H$. Idea #5. Look for general description of left inverses of $\rho_{K(H)}$ when $H$ is finite dimensional and hope for new insights. Let $(e_i)_{i=1}^n$ be an orthnormal basis of $H$. Denote $$ E_{k,l,p,q}:\mathcal{B}(H)\to\mathcal{K}(H):a\mapsto \langle a(e_l),e_k\rangle(e_p\bigcirc e_q) $$ then $(E_{k,l,p,q})_{k,l,p,q=1}^n$ form a basis of $\mathcal{B}(\mathcal{B}(H),\mathcal{K}(H))$. Also denote $$ \tau_{k,l,p,q,i,j}=\langle\tau(E_{k,l,p,q})(e_j),e_i\rangle $$ For example, given this notation the map $\tau$ from idea #4 is well defined by $$ \tau_{k,l,p,q,i,j}=\delta_k^j\delta_l^i\delta_p^i\delta_q^i $$ Then one can show that $\tau$ is a left inverse module map for $\rho_{K(H)}$ iff $$ \tau_{k,l,p,q,i,j}=\delta_k^j\mu_{i,l,p,q} $$ for some array of numbers $(\mu_{i,l,p,q})_{i,l,p,q=1}^n$ satisfying $$ \sum_{l=1}^n \mu_{i,l,p,l}=\delta_p^i $$ Interesting, but not very useful. REPLY [2 votes]: This is a partial answer that solves the commutative analogue of the original question. Consider $\ell_\infty$-morphisms $$ \rho:c_0\to\mathcal{B}(\ell_\infty, c_0):x\mapsto(a\mapsto x\cdot a), $$ $$ \tau:\mathcal{B}(\ell_\infty,c_0)\to c_0:T\mapsto (T(\delta_1)(1), T(\delta_2)(2),\ldots) $$ It is really easy to check $\tau\rho=1_{c_0}$, but the only thing left to prove is that the range of $\tau$ is actually the $c_0$. Since $\ell_\infty$ is a Grothendieck space, then any $T\in\mathcal{B}(\ell_\infty, c_0)$ is weakly compact. Since $\ell_\infty$ has the Dunford-Pettis property, then any weakly compact operator is completely continuous. Therefore any $T\in\mathcal{B}(\ell_\infty, c_0)$ maps weakly null sequences to the norm null sequences. It is remains to note that $(\delta_i)_{i=1}^\infty$ is weakly null to get that $\lim_i\Vert T(\delta_i)\Vert=0$. Therefore $\operatorname{Im}(\tau)\subset c_0$.<|endoftext|> TITLE: 1-wasserstein distance v.s. total variation distance QUESTION [5 upvotes]: Suppose that $\mu_1$ and $\mu_2$ are two distributions defined on $\mathbb{R}^n$ and $\gamma$ is a symmetric distribution (around $0$) on $\mathbb{R}^n$ with compact support. Let $\gamma_x$ denote the resulting distribution by translating the centre of $\gamma$ from $0$ to $x$, $d_{TV}(\cdot,\cdot)$ denote the total variation distance and $d_W(\cdot,\cdot)$ 1-Wasserstein distance. Question: Does it hold that $$ d_{TV}(\mu_1\ast\gamma,\mu_2\ast\gamma) \leq \left(\sup_{x\neq y} \frac{d_{TV}(\gamma_x,\gamma_y)}{\|x-y\|} \right)\cdot d_W(\mu_1,\mu_2)? $$ REPLY [3 votes]: Yes. I presume that your "1-Wasserstein" distance is what is otherwise called the transportation metric. Let $M$ be any measure with marginals $\mu_1$ and $\mu_2$, so that $\mu_1=\int \delta_x\,dM(x,y)$ and $\mu_2=\int\delta_y\,dM(x,y)$, whence $\mu_1-\mu_2=\int (\delta_x-\delta_y)\,dM(x,y)$, which, after taking convolution with $\gamma$ and passing to the total variation norm, yields $$ \| \mu_1*\gamma-\mu_2*\gamma \| \le \int \| \gamma_x-\gamma_y \|\,dM(x,y) \le K \int \|x-y\|\,dM(x,y) \;, $$ where $K$ is the $\sup$ from your question (none of the additional conditions on $\gamma$ are required). Taking for $M$ a measure which realizes the transportation distance between $\mu_1$ and $\mu_2$, one gets the claim.<|endoftext|> TITLE: References for general Hasse-Weil zeta function QUESTION [11 upvotes]: Most research on the Hasse-Weil zeta function focuses on some particular type of algebraic variety, and general surveys usually deal mostly with the better understood elliptic curve case. I am looking for references about the Hasse-Weil zeta for arbitrary variety and number field, particularly analytic continuation and functional equation (this is, not focused on special values or zeroes). Also, I have Serre's "Facteurs locaux", so probably anything published before 1970 would be redundant. Edit. For future reference, Ivan Fesenko's research, particularly "Analysis on arithmetic schemes" parts I, II and III, fit perfectly in this context. Also, some older papers like Alexei Parshin, for example, "Chern classes, adeles and L-functions". REPLY [2 votes]: Kahn has quite a panoramic view of L-functions in arithmetic geometry. He has written a small book in french, which has been translated to english. Bruno Kahn makes a clear distinction between zeta-functions which count points in fibers of arithmetic varieties, and L-functions which are associated to varieties "indirectly" via their motives and depend only on the generic fiber of arithmetic varieties, on varieties over fields.<|endoftext|> TITLE: Recent trends in effective analysis QUESTION [6 upvotes]: The references listed at http://en.wikipedia.org/wiki/Computable_analysis have all been published 30-15 years ago. Are the approaches which these references expose still up-to-date and relevant to the current paths of research in computable analysis? (In case they are not, where can I find a good introduction to the current trends?) REPLY [5 votes]: (At François's request, my comment in now an answer.) Yes, it is still an active research area. It however is spread out throughout a number of camps (traditions): The Weihrauch camp, the reverse math camp, the computability theory camp, the randomness camp, the proof theory camp, and a few different constructive math camps. (Also, see many of the quantitative results in classical analysis.) Most researchers span two or more camps, and I don't mean to imply there is a feud or anything. However, there isn't necessarily an organized central list of open problems or a central agenda. You may want to check out cca-net.de as a starting point. Edit: You may also want to check out this survey by Brattka and Avigad. It does a good job of explaining many of the different traditions.<|endoftext|> TITLE: Divisibility of the degree of an extension by the degree of its residual field QUESTION [6 upvotes]: Let $A$ be an integrally closed domain whose quotient field is $K$, $L$ be a finite Galois extension of $K$, and $B$ be the integral closure of $A$ in $L$. Let $M_A$ be a maximal ideal of $A$, and $M_B$ be a maximal ideal of $B$ that lies above $M_A$ (that is, $M_B\cap A = M_A$). Denote by $F_A$ the field $A/M_A$, and by $F_B$ the field $B/M_B$; so $F_B/F_A$ is an algebraic field extension, and it is normal. 1) Must $F_B/F_A$ be finite ? (this is the case if, say $A$ is Noetherian, and I can show that this is also true if $F_B/F_A$ is separable) 2) The most important question for me : Assuming $F_B/F_A$ finite, must the residual degree $[F_B : F_A]$ divide $[L : K]$ (this is actually the case if $A$ is a Dedekind ring, or $A$ and $B$ are discrete valuation rings). REPLY [3 votes]: As pointed out above, the answer of Count Dracula is also valid, mutatis mutandis, if one desires counter-examples where the field extensions are of characteristic $p>2$. It turns out that it can also be adapted to provide an example where the characteristic of the base field extension is $0$, and is $2$ for the residual field extension. Here is how: Let $R={\mathbb Z}[t_i, x_i, i\in \mathbb N]$, and $L$ the fraction field of $R$. Define the order 2 automorphism $\sigma$ of $R$ by $\sigma(z\in \mathbb Z)=z$, $\sigma(t_i) = t_i$ and $\sigma(x_i) = t_i-x_i$. Let $S$ be the fixed sub-ring under the action, and $K$ its fraction field. Then $\sigma$ extends to a $K$-automorphism of $L$ in an obvious manner, hence $[L:K]=2$ by Galois theory. $R$ is integrally closed (since factorial), and is integral over $S$ because every $f\in R$ satisfies an integral dependance relation over $S$ of the form $(X-f)(X-\sigma f)=0$. It follows easily that $S$ must be integrally closed too. We consider the ideals ${\frak P} =(2,t_i)\subset R$, and ${\frak p} = S\cap {\frak P}\subset S$. It is not difficult to see that $\frak P$, and hence $\frak p$, is prime. Now, as in the proof of Count Dracula, we see that $f\in R$ can be written $$f_0 + \sum_i t_i f_i + \hbox{terms multiples of $t_j t_k$},$$ where $f_0, f_i$ depends only on the $x_j$ (eventually, consider the polynomial $f_0$ lacunary in some variables, and add some null $f_i$ in order for the variables in all the $f_0,f_i$ to be the same). If $f\in S$, then $$ \sigma f(\star) = f_0(-\star) + \sum_i t_i(f_i(-\star) + {\partial f_0/\partial x_i}(-\star)) + \hbox{terms multiples of $t_j t_k$}.$$ Therefore $$\sum_i t_i \partial f_0/\partial x_i(-\star) = f_0(\star)-f_0(-\star) + \sum_i t_i(f_i(\star) - f_i(-\star)).$$ But $-1 = 1 \pmod2$ hence $ \partial f_0/\partial x_i = 0 \pmod2$. It follows, as in the proof of Count Dracula, that $A/{\frak p}={\mathbb F_2}(x_i^2)$, and that $B/{\frak P}={\mathbb F_2}(x_i)$ is infinite over $A/\frak p$. Considering finitely many variables, it can be shown similarly that $[L:K]=2$ is not divisible by $[B/{\frak P}:A/\frak p]$. NOTE: to obtain a more general counter-example, it suffices to consider the order $p$ automorphism $$\sigma(t_i)=t_i,\quad \sigma(z\in {\mathbb Z}[\zeta])=z,\quad \sigma(x_i)= t_i+\zeta x_i,$$ where $\zeta$ is a p-root of unity. The adaptations are straightforward, except, perhaps, the fact that the ideal $(2,t_i)$ must be replaced by the ideal ${\frak P}= (1-\zeta, t_i)$ and not $(p,t_i)$: this is because $(1-\zeta)$ is the only maximal ideal above $(p)$ in ${\mathbb Z}[\zeta]$.<|endoftext|> TITLE: Trigonometric polynomials on non-compact and non-abelian groups QUESTION [7 upvotes]: I asked this initially in math.stackexchange, but it disappeared almost immediately, so I hope it will be proper to aks this here. Hewitt and Ross define trigonometric polynomial on a locally compact group $G$ as a (finite) linear combination of matrix elements of continuous unitary irreducible representations $\pi_i:G\to B(H_i)$ (not necessarily finite dimensional) of $G$: $$ f(t)=\sum_{i=1}^n \lambda_i\cdot\langle\pi_i(t)x_i,y_i\rangle,\qquad t\in G, \quad \lambda_i\in{\mathbb C},\quad x_i,y_i\in H_i. $$ If $G$ is abelian or compact then the space ${\tt Trig}(G)$ of trigonometric polynomials on $G$ is an algebra with respect to the pointwise multiplication. This is strange, I can't find mentionings of the same proposition in the non-abelian and non-compact case. Is it possible that in general case (for arbitrary locally compact group $G$) the space ${\tt Trig}(G)$ is not an algebra? I would be grateful, if somebody could advice reading on this theme. REPLY [3 votes]: It isn't an algebra: the tensor product of irreducibles does not always decompose as a finite sum of irreducibles, although in some of those cases it may decompose as a direct integral of irreducibles. For instance, take $G=H_3({\bf R})$, the group of upper-triangular matrices with real entries and with $1$ on the diagonal. This has very nice representation theory from the point of view of harmonic analysis (it is not only Type I, but it is what Kaplansky called CCR, that is $\pi(L^1(G))\subseteq {\mathcal K}(H_\pi)$ for every continuous irreducible unitary representation $\pi: G \to {\mathcal U}(H_\pi)$). There is a continuous quotient homomorphism $q:G \to G/Z(G) \cong {\bf R}^2$. Now for $t\neq 0$ there is an infinite irreducible unitary representation $\pi_t: G \to {\mathcal U}(L^2({\bf R}))$ (in some realizations called the Schrödinger representation). It is true that if $s$, $t$ and $s+t$ are all non-zero, and $f$ is a coefficient function of $\pi_s$ and $g$ is a coefficient function of $\pi_t$, then $fg$ is the norm-limit inside $B(G)$ of sums of coefficient functions of $\pi_{s+t}$. I don't know if $fg$ is actually a coefficient function itself. However, the representation $\pi_t\otimes \pi_{-t}$ is equivalent to the representation $\lambda_{\bf R{^2}} \circ q: G \to {\mathcal U}(L^2({\bf R}^2))$ which is not (quasi-)equivalent to a direct sum of irreducibles. Consequently there should exist $f$ a coefficient function of $\pi_t$ and $g$ a coefficient function of $\pi_{-t}$ such that $fg$ does not belong to the closure of ${\rm Trig}(G)$ inside $B(G)$. Similar example should exist for e.g. ${\rm SL}(2,{\bf R})$ but I don't remember the ``fusion rules'' for its irreducible representations off the top of my head. On the positive side: if $G$ is a Moore group then the completion of ${\rm Trig}(G)$ inside $B(G)$ is indeed a subalgebra of $B(G)$. This may be folklore: I learned of it from the paper http://arxiv.org/abs/1208.1519 which has more information on "negative examples".<|endoftext|> TITLE: Geodesics and Riemannian submersions QUESTION [9 upvotes]: Let $X,Y$ be Riemannian manifolds, and $f\colon X\to Y$ be a Riemannian submersion. Let $\gamma$ be a geodesic on $X$ starting at a point $x\in X$ and which is orthogonal to the fiber $f^{-1}(f(x))$. Questions. (1) Is it true that $\gamma$ is orthogonal to any fiber of $f$ which it intersects? (2) If this is not true in general, is it true under the extra assumption that $f$ has totally geodesic fibers? Remark. I think this is true in the special case when $X=G$ is a compact Lie group with a bi-invariant metric, and $Y=G/H$ is its homogeneous space with the only metric such that the canonical map $f\colon G\to G/H$ is a Riemannian submersion. (In this case $f$ does have totally geodesic fibers.) REPLY [7 votes]: Take horizontal lift $\gamma$ of the minimal geodesic $\bar\gamma$ in Y, connecting two points $p$ and $q$ close to each other. It is (minimal) geodesic, since any other curve connecting them is not shorter than its horizontal projection, which in turn not shorter than minimal $\bar\gamma$. Since in each direction we can issue only one geodesic - any geodesic normal to the fiber is a horizontal lift. Which in turn implies that such geodesics stay normal to fibers. So, the answer is "yes". REPLY [4 votes]: Yes. For a printed proof see 26.12 of here.<|endoftext|> TITLE: Probability of a giant component existing in a $G(n,p)$ random graph with $p=\omega(\frac 1n)$ QUESTION [5 upvotes]: Consider a random $G(n,p)$ graph where $p=\omega(\frac 1n)$, and let $x$ denote the probability that the graph has a connected component of size linear in $n$. It is well known that $x$ tends to $1$ as $n\to \infty$, even if $p\geq \frac {1+\epsilon}{n}$. However, I am looking for a more exact statement: Something of the form $$x\geq 1-O(f(n,p))$$ where $f$ is some small function of $n$ and $p$. I'm sure that a reference exists somewhere, but I have yet to find it. Thanks! REPLY [4 votes]: In "On tree census and the giant component in sparse random graphs, B. Pittel proves the following (see Gap Theorem and Lemma 4): for $c>1,$ any $a>0$ and any $\omega \to \infty$ however slowly, \begin{equation*} P(|\# \text{ of vertices in largest component}-\theta n |> \omega \sqrt{ n \ln n})\leq n^{-a}, \end{equation*} where $\theta$ is the unique positive root of $1-\theta = e^{-c \theta}$. So definitely you can choose $f(n,p)=n^{-a}$ for whichever $a>0$ you want. I imagine that one can mimic the proof to get a better bound when $np \to \infty$.<|endoftext|> TITLE: Open problems in Berkovich geometry QUESTION [19 upvotes]: I would like to know if there is a state of the art recent reference on non-archimedean analytic spaces mentioning/listing open problems, conjectures, unresolved questions in the theory (*). I have looked for such a reference a bit, but didn't find anything. In case there is really nothing, I would be glad if experts in the field present here could enlighten me. (*) By "theory" I mean "mainstream" theory over a non-archimedean complete valued field (trivial valuations are accepted), including cohomological questions concerning the spaces themselves, but also theory over Banach rings, as mentioned in Berkovich's AMS monogograph. I must precise that even if dynamics and potential theory do perfectly fit for an answer, they are not my first target of interest, I am indeed looking for open problems, conjectures, unresolved or partially resolved questions that are more linked to Berkovich's or Temkin's works, that is, to works developing the theory or its links with schemes and formal schemes, cohomology, vanishing cycles. REPLY [13 votes]: This is a very broad question and it is difficult to know where to start. Remember that Berkovich's theory is a theory of analytic geometry, hence it makes sense to look for the counterpart of anything you have in complex analytic geometry: does there exist a good notion of Kähler manifold, for instance? I will try to be somehow more specific though and concentrate on the work of Berkovich as you suggest. In the recent years, there were several attempts to understand the topology of Berkovich spaces better. In [Berkovich, Smooth $p$-adic analytic spaces are locally contractible, Inventiones, 1999], Berkovich proves that every smooth space is locally contractible. In [Hrushovski-Loeser, Non-archimedean tame topology and stably dominated types], they prove that the result holds for quasi-projective spaces. There are also some results by Thuillier, but, as far as I know, there are no written notes yet, and I am afraid that I cannot remember the exact level of generality he deals with. Anyway, I think that the question is open in full generality: are Berkovich spaces locally contractible? In another direction, Berkovich has written a book called [Integration of One-forms on P-adic Analytic Spaces]. He basically constructs sheafs of primitives of one-forms and their iterates on smooth spaces. He proves that the resulting de Rham complex is exact in degree 0 and 1 but in higher degree the question is open. Edit: I have just realized that, at the end of the introduction of the book, Berkovich gives a list of six open questions, the one I mentioned above being the first one. You may want to have a look at those. Last, I would like to add a few words about Berkovich's paper [A non-Archimedean interpretation of the weight zero subspaces of limit mixed Hodge structures]. The title says clearly what it is about. The weight zero subspace of some limit of mixed Hodge structures is given an interpretion using the Betti numbers of some Berkovich analytic space. It would certainly be very interesting to say similar things for higher weights. Edit: I have just realized that I forgot to say something about general Banach rings. Over an arbitrary Banach ring, almost nothing is done and it is even doubtful that one can actually do something in such a gereral setting. There are some results over $\mathbb{Z}$ (or rings of integers of number fields): properties of local rings, mainly, but this is definitely only the very beginning of the theory. In particular, at the topological level (local arcwise connectedness, local contractibility, etc.) or at the cohomological level (finiteness of cohomology in the proper case, GAGA, etc.), there is nothing (yet). And here, I am only speaking of the usual transcendental topology and of the coherent cohomology. As far as I know, étale morphisms have not even been defined...<|endoftext|> TITLE: Some calculus in the orthogonal group $O(n)$ QUESTION [6 upvotes]: How can one compute each of the following matrices, explicitly: $$\int_{O(n)} e^{g}dg$$ or $$\int_{O(n)} g^{n}dg \;\;\;\;n\in \mathbb{N} \;\;n>1$$ What is the explicite entries of the resulting matrices, for $n=2$? Moreover, for $n,m\in \mathbb{Z}$ define the linear operator $T_{n,m}$ on $M_{n}(\mathbb{R})$ as follow: $$T_{n,m}(A)=\int_{O(n)}g^{n}Ag^{m}$$ Under what suficcient and necessary condition, $T_{n,m}$ is conjugate to $T_{n',m'}$? Is there an explicit formulation for $T_{n,m}$? The integration is based on the Haar measear defined on orthogonal group $O(n)$. REPLY [13 votes]: let me work out the comments a bit further, starting from the identity (equation 8.2 from Diaconis and Evans, correcting an earlier paper by Diaconis and Shahshahani) $$\int_{{\rm O}(n)}{\rm tr}\,g^p\,dg=\begin{cases} 0&{\rm if}\;\;p\;\;\text{is an odd integer}\\ 1&{\rm if}\;\;p\;\;\text{is an even integer} \end{cases} $$ so the second integral of the OP evaluates to $$\int_{{\rm O}(n)}g^p\,dg=\begin{cases} 0&{\rm if}\;\;p\;\;\text{is an odd integer}\\ n^{-1}\mathbb{1}&{\rm if}\;\;p\;\;\text{is an even integer} \end{cases} $$ Taylor expansion of the exponent in the first integral of the OP and term-by-term integration gives $$ \int_{{\rm O}(n)}e^g\,dg=n^{-1}\mathbb{1}\sum_{p=0}^\infty \frac{1}{(2p)!}=\frac{\cosh 1}{n}\mathbb{1} $$ now for the third integral of the OP, we need the fourth-order tensor $$\int_{{\rm O}(n)}(g^p)_{ij}(g^q)_{kl}\,dg=a_{pq}(n)\delta_{ij}\delta_{kl}+b_{pq}(n)\delta_{ik}\delta_{jl}+c_{pq}(n)\delta_{il}\delta_{jk}$$ so that the required integral takes the form $$\int_{{\rm O}(n)}g^p Ag^q\,dg=a_{pq}(n)A+b_{pq}(n)A^{\rm t}+c_{pq}(n)\mathbb{1}\,{\rm tr}\,A $$ by taking traces I find, using again equation 8.2 from Diaconis and Evans, that $$na_{pq}(n)+nb_{pq}(n)+n^2 c_{pq}(n)=\int_{{\rm O}(n)}{\rm tr}\,g^{p+q}\,dg=\begin{cases} 0&{\rm if}\;\;p+q\;\;\text{is an odd integer}\\ 1&{\rm if}\;\;p+q\;\;\text{is an even integer} \end{cases} $$ $$n^2a_{pq}(n)+nb_{pq}(n)+nc_{pq}(n)=\int_{{\rm O}(n)}({\rm tr}\,g^p)({\rm tr}\,g^q)\,dg=\begin{cases} {\rm min}\,(p,2n)&{\rm if}\;\;p=q\;\;{\rm odd}\\ {\rm min}\,(p,2n)+1&{\rm if}\;\;p=q\;\;{\rm even}\\ 1&{\rm if}\;\;p\neq q\;\;\text{both even}\\ 0&{\rm otherwise} \end{cases} $$ $$na_{pq}(n)+n^2 b_{pq}(n)+nc_{pq}(n)=\int_{{\rm O}(n)}{\rm tr}\,g^{|p-q|}\,dg=\begin{cases} 0&{\rm if}\;\;p+q\;\;\text{is an odd integer}\\ 1&{\rm if}\;\;p+q\;\;\text{is an even integer} \end{cases} $$ Three equations with three unknowns solves the third integral.<|endoftext|> TITLE: A positivity problem involving the number of ways of expressing $n$ as a product of $k$ factors QUESTION [7 upvotes]: Let $d_k(n)$ denote the number of ways of expressing $n$ as a product of $k$ factors, and let $$D_k(x)=\sum_{n\leq x}d_k(n)$$ be the summatory function. During a study of Mertens' function I was lead to consider the zero distribution of the sequence of polynomials $$P(z,x)=\sum_{j=0}^{m}\left(\sum_{k=j}^{m}{m+1\choose k-j}(-1)^{m-k}D_{m+1-k}(x)\right)z^j$$ of degree $m=m(x)=\max \{\Omega(n):n\leq x\}\sim \log_2 x$. I am interested in the question of whether the zeros of the sequence of polynomials $P(z,x)$ have strictly negative real parts-the truth of which depends, among other things, on the non-negativity of the coefficients $$c(j,x)=\sum_{k=j}^{m}{m+1\choose k-j}(-1)^{m-k}D_{m+1-k}(x).$$ I have checked this up to $x=50$ and for each such $x$ the $c(j,x)$ are positive and unimodal. For instance, for $1\leq x \leq 20$, the coefficients are: [1], [1, 2], [1, 3], [1, 4, 4], [1, 5, 5], [1, 4, 6], [1, 5, 7], [1, 6, 12, 8], [1, 6, 13, 9], [1, 5, 13, 10], [1, 6, 15, 11], [1, 6, 13, 12], [1, 7, 15, 13], [1, 6, 15, 14], [1, 5, 15, 15], [1, 6, 20, 30, 16], [1, 7, 23, 33, 17], [1, 7, 21, 32, 18], [1, 8, 24, 35, 19], [1, 8, 22, 34, 20]. Is there a good reason why these numbers should be positive? I know this would follow if it could be shown that the sequence $${m(x)+1\choose k}D_{m(x)-j+1-k}(x)$$ is unimodal for $k\in\{0,m(x)-j\}$, but I don't see how this could be proved. REPLY [3 votes]: The answer is no, on both counts. Certainly I was hasty to post this question based on the observation that the zeros have negative real parts and the coefficients are positive in the observed ranges. Here's a simple explanation of why: First note that $D_k(x)$ may be written as $$D_k(x)=\sum_{n\leq x}\prod_{p|n}{\alpha_p+k-1\choose \alpha_p}$$ where $\alpha_p$ is the exponent of $p$ in the canonical factorisation of $n$. $D_k(x)$ is thus a polynomial in $k$ of degree $m(x)$, so $$\sum_{k=0}^{m+1}{m+1\choose k}(-1)^kD_{m+1-k-j}(x)=0.$$ Therefore $$c(j,x)=-\sum_{k=m-j+1}^{m+1}{m+1 \choose k} (-1)^{m-k-j}D_{m+1-k-j}(x).$$ For example then, $$c(1,x)=m(x)+1-D_{-1}(x)$$ where $D_{-1}(x)=M(x)$ is Mertens' function. It follows from known results that $M(x)$ is $\Omega_{\pm}(x^{1/2})$, so $c(1,x)$ will eventually be negative of square root order. As for the zeros, by virtue of the fact that the coefficients are real and $c(0,x)=1$, we have $$c(1,x)=-\sum_{\rho:P(\rho,x)=0}\frac{1}{\rho}=-\sum_{\rho:P(\rho,x)=0}\frac{\Re{\rho}}{|\rho|^2},$$ so the same argument applies. On the other hand, from the linear relation above, it is easy to see that $c(m(x),x)=x$, so it is certainly not so obvious that $c(m(x)-1,x)$, $c(m(x)-2,x)$, etc, will also change sign eventually. These questions are tied up with the Riemann hypothesis.<|endoftext|> TITLE: Certain signed sum over $S_n$ QUESTION [9 upvotes]: The following question appeared in my research: Let $G_1,G_2,G_3$ all be subgroups of $S_n$, and consider the sum $$ \sum_{g_i \in G_i, g_1g_2g_3 = id} \epsilon(g_1) $$ that is, we only consider triplets whose product is the identity permutation, and we sum all the signs of $g_1$. The question is: is this sum always non-negative? EDIT: What if we restrict $G_1$ to be the entire $S_n$? What if we restrict $G_2$ and $G_3$ to be groups of type $G_T=\{\sigma \in S_n : c(\sigma) \leq T \}$ where $c$ indicates the cycle type, and $T$ is a fixed set partition, and the inequality indicates refinement. Update Even with the updated version of subgroups, there is a counter-example. The three set partitions $$S=((1, 2, 3, 4), (5, 6, 7, 8)) \quad T=((1, 5),(2, 8),(3, 6),(4, 7)) \quad U=((1, 6),(2, 4, 5),(3, 7, 8))$$ give rise to a sum with value $-4$. REPLY [19 votes]: If $G_1 = S_n$ as in your edit, then every pair $(g_2,g_3) \in G_2 \times G_3$ occurs exactly once in the sum, and for each such summand we have $\epsilon(g_1) = \epsilon(g_2g_3)$. So the sum simplifies to $$\sum_{g_2 \in G_2}\sum_{g_3 \in G_3} \epsilon(g_2g_3) = \left( \sum_{g_2 \in G_2} \epsilon(g_2)\right)\left(\sum_{g_3 \in G_3} \epsilon(g_3)\right)$$ which is nonnegative by the following observation: if $G \subset S_n$ is a subgroup, then the sum $\sum_{g \in G} \epsilon(g)$ is either $0$ or $\vert G \vert$, depending on whether the map $\epsilon \colon G \to \{\pm 1\}$ is trivial or not.<|endoftext|> TITLE: Do canonical stacks exist over Spec(Z)? QUESTION [13 upvotes]: Suppose a scheme $X$ has tame quotient singularities. Does there exist a smooth DM stack $\mathcal X$ with coarse space $X$ so that the coarse space morphism $\mathcal X\to X$ is an isomorphism away from codimension 2? If $X$ is finite type over a field, the answer is yes, by the following argument. Theorem 4.6 of Fantechi-Mann-Nironi's Smooth toric DM stacks shows that such stacks have a universal property, so it suffices to show that they exist étale-locally (the universal property ensures the locally constructed canonical stacks will glue). Étale locally, we may assume $X=U/G$, where $U$ is smooth and $G$ is a finite group acting tamely on $U$—that's the definition of "tame quotient singularities". Let $H\subseteq G$ be the (normal) subgroup acting by pseudoreflections (through a given closed point $x\in U$; shrinking $U$ if necessary). Then by the Chevalley-Shephard-Todd theorem, $T_x/H$ is smooth, where $T_x$ is the tangent space at $x\in U$. If $X$ is defined over the residue field $k(x)$, then we can construct an $H$-equivariant morphism $U\to T_x$ sending $x$ to the origin, which is étale at $x$. Since $T_x/H$ is smooth, so is $U/H$. (If $X$ isn't defined over $k(x)$, base change to $k(x)$ and get smoothness of $U/H$ by descent.) The action of $G/H$ on $U/H$ is free away from codimension 2, so $\mathcal X = [(U/H)/(G/H)]$ does the trick. As far as I can tell, the only place we had to use that $X$ is defined over a field was in showing smoothness of $U/H$. Without it, we can't construct the étale morphism $U/H\to T_x/H$ we need to deduce smoothness of $U/H$ from smoothness of $T_x/H$. Can this "over a field" condition be relaxed to get the existence of canonical stacks in an absolute setting (i.e. over $\mathrm{Spec}(\mathbb Z)$)? Context: my more general goal is to understand if canonical stacks exist over an arbitrary base $S$. That is, suppose $X$ is a scheme (probably locally of finite type) over a base $S$ which has an étale cover by a disjoint union of schemes of the form $U/G$, where $U$ is smooth over $S$, and $G$ is an abstract group acting on $U$ (over $S$), with order relatively prime to all residue fields of $U$. Then does there exist a stack $\mathcal X$ which is smooth over $S$ and has coarse space $X$, such that the coarse space morphism is an isomorphism away from codimension 2 (on both $X$ and $\mathcal X$)? REPLY [7 votes]: In the smooth case, I think that the answer is positive over an arbitrary regular excellent base. The argument was in my PhD thesis; it was done over a field, but I think it adapts to this case. Cover your $X$ in the étale topology with schemes of the form $U/G$, where $G$ is prime to all the residue characteristics of $U$, and $U$ is smooth over $S$. Choose a geometric point $p$ of $U$; we can restrict to the stabilizer of $G$, and assume that $G$ leaves $p$ fixed. Call $H$ the subgroup generated by the elements of $G$ that are pseudoreflections, or the identity, when restricted to the fiber along $p$; this is normal in $G$. The scheme $U/H$ is flat over $S$, because of the tameness hypothesis. Furthermore taking quotients by $H$ commutes with base change on $S$, again because of tameness; hence the geometric fiber of $U/H \to S$ along the image of $p$ is the quotient of the geometric fiber, which is smooth, because of Cartan’s and Chevalley’s theorems. In this way we can assume that $G$ stabilizes $p$, and the restriction of $G$ to the fiber through $p$ contains no pseudoreflexions. Now look at the locus in which $U \to X$ is étale. Notice that if the restriction of $U \to X$ to the fiber is étale at $p$, then $U \to X$ is étale at $p$, by the local criterion of flatness. The locus on which $U \to X$ is étale is open in X; its complement must have codimension larger than $1$, because otherwise it would intersect the fiber in codimension $1$. Now take two of these charts $U \to X$ and $V \to X$; the normalization of the part of the fibered product $U \times_X V$ that dominates $X$ is étale over $U$ and $V$, by purity of the branch locus. These data give a Q-variety, in the sense of Mumford; from them you get an étale groupoid that defines the stack that you are looking for. When S is not regular, I am really not sure, I suspect it might be false.<|endoftext|> TITLE: Can a symplectic manifold be recovered from its Lagrangians? QUESTION [11 upvotes]: Something I have wondered idly about from time to time is: If $(M,\omega), (M',\omega')$ are symplectic manifolds, and you "know what the Lagrangians $L \subset M$ resp. $L' \subset M'$ are," can you determine whether $M$ and $M'$ are symplectomorphic? The sort of thing I have in mind is this corresponding statement for coherent sheaves: Theorem (Gabriel). Suppose $X$ and $Y$ are smooth projective varieties. If there exists an equivalence $\mathrm{\bf Coh}(X) \simeq \mathrm{\bf Coh}(Y)$, then $X$ and $Y$ are isomorphic. (That's proven in Huybrecht's book on FM transforms using Orlov's theorem and skyscraper sheaves.) One issue is, what's the right way to formalize "know all $L \subset M$"? (Maybe you mean some version of the Fukaya category up to $A_\infty$-equivalence?) A bigger issue is, how the heck do you prove it?? For certain $M$, there's an analogue of skyscraper sheaves coming from mirror symmetry, but I don't know how to ape the argument beyond that point; for other kinds of $M$, I know how to get some bits of information from the Fukaya category. But it would be awfully nice to have an argument for fairly general $M$, e.g. all compact $M$. REPLY [2 votes]: I'd like to add two more examples. The first one is in the negative direction. Let $\pi:E^{2n}\rightarrow\mathbb{C}$ be an exact Lefschetz fibration, by the work of Seidel one can associate its Fukaya category $\mathscr{F}(\pi)$, which can be realized as a directed $A_\infty$ algebra or a diagonal bimodule. Denote by $\mathscr{F}(\pi)^\vee$ its dual bimodule. Seidel constructed the following bimodule map $\beta:\mathscr{F}(\pi)^\vee[-n]\rightarrow\mathscr{F}(\pi)$ which describes the natural transformation of degree $n$ from the Serre functor to the identity. It's an observation due to Maydanskiy that there are examples of exact Lefschetz fibrations with quasi-isomorphic $\mathscr{F}(\pi)$, but with different bimodule maps $\beta$. This shows that one should include $\beta$ as an additional piece of information to distinguish exact symplectic manifolds which admit Lefschetz fibrations. The second example is in the positive direction. Weiwei Wu proved the following: The special isogenous tori are symplectomorphic if and only if they have equivalent derived Fukaya categories. A special isogenous torus, by definition, is obtained by taking products of tori which are finite group quotients of a standard product torus. The proof is a combination of the result of Abouzaid-Smith on homological mirror symmetry for standard $T^{2n}$ and finite group actions on Fukaya categories which generalize the case of a $\mathbb{Z}/2$-action considered by Seidel in the definition of $\mathscr{F}(\pi)$. This result is also expected to be true for symplectic tori equipped with linear symplectic forms. Addendum Of course you may also think of monotone Fukaya categories, and that suggests you to take into consideration the scaling of the symplectic form. The reconstruction is most likely to be possible in the case when there is a well-defined mirror functor, which in turn means that the SYZ picture of mirror symmetry should somehow be compatible with homological mirror symmetry. Up to my knowledge, this is true only for $T^{2n}$, $(\mathbb{C}^\ast)^n$, Kodaira-Thurston manifold, or things like that.<|endoftext|> TITLE: Legendre transform and Lipschitz approximation QUESTION [6 upvotes]: Let $(X,d)$ be a compact metric space and $f:X \to \mathbb{R}$ a real valued continuous function. Let us agree that a modulus of continuity means concave, nondecreasing, uniformly continuous function $\omega$ such that $\omega(0)=0$ and $|f(x)-f(y)| \leq \omega(d(x,y))$ for $x,y \in X$. Let $\delta(s)$ be a distance of $f$ from the set of all functions satisfying Lipschitz condition with a constant $s$. According to wikipedia, it can be shown, using Legendre transform, that if $\omega$ is a minimal concave modulus of continuity for $f$ then the following formulas holds: $$2\delta(s)=\sup_{t \geq 0}\{\omega(t)-st\},$$ and $$\omega(t)=\inf_{s \geq 0}\{2\delta(s)+st\}.$$ I would be grateful if anybody could give me reference, where I can find the proof of that fact. PS. I need this result in order to justify some construction in the world of $C^*$-algebras-otherwise this construction would be somehow artificial. REPLY [4 votes]: These are somehow folk results. Here's a proof. Let $(X,d)$ be a metric space, $f$ a function on $X$ which admits a concave modulus of continuity, and $g$ any $s$-Lipschitz function on $X$ with $\|f-g\|_\infty<\infty$. Then, for all $x$ and $y$ in $X$ $$|f(x)-f(y)|\le|f(x)-g(x)|+|g(x)-g(y)|+|g(y)-f(y)|\le$$ $$\le2\|f-g\|_\infty+sd(x,y)\,. \qquad(1) $$ The minimum concave modulus of continuity of $f$ is the inferior envelope of all increasing affine functions whose graphs are above the set of points $$\{(d(x,y),|f(x)-f(y)|)\in [0,\infty]\times[0,\infty] \;:\; x\in X,y\in X\}$$ that is, for any $t\ge0$ $$\omega(t)=\inf\{at+b\; :\;a\ge0,\,b\ge0,\forall x\;\forall y\; |f(x)-f(y)|\le a|x-y|+b \}\, .$$ So by (1) also $\omega(t)\le 2\|f-g\|_\infty +st$, and since this is true for all $s$ and $g$ $s$-Lipschitz, $$\omega(t)\le\inf_{s\ge0} \{ 2\delta(s) +st\}\,.\qquad(2)$$ Keeping $f,g,s$ as above, it is straightforward to check that the inferior envelope $$f^*_s(x):=\inf_{y\in X}\{f(y)+sd(x,y)\}$$ is the largest $s$-Lipschitz function below $f$ (note that the infimum is finite because it is larger than $g-\|f-g\|_\infty$, which is itself an $s$-Lipschitz function below $f$). This implies that the minimal distance $\delta(s)$ is realized by the $s$-Lipschitz function $f^*_s+\delta(s)$ and that $$2\delta(s)=\|f-f^*_s\|_\infty= \sup_{x,y\in X}\{f(x)-f(y)-sd(x,y)\}\,.$$ Since $f(x)-f(y)\le\omega(d(x,y))$, we also have $$2\delta(s)\le\sup_{t\ge0}\{\omega(t)-st\}\,.\qquad(3)$$ One may analogously prove that (2) and (3) are indeed equalities, by similar arguments, but note that, introducing the convex function $\alpha(t):=-\omega(-t)$ for $t\le0$ and $+\infty$ for $t>0$, and also putting $\delta(t)=+\infty$ for $t<0$, the stated inequalities (2) and (3) reads, in terms of Legendre transforms $$\alpha\ge (2\delta)^*,\qquad 2\delta\le \alpha^*.$$ Since the Legendre transform reverses the inequalities and is involutory we conclude that these inequalities are indeed equalities, that is, $\alpha$ and $2\delta$ are a pair of convex conjugate functions.<|endoftext|> TITLE: Does a semistable curve descend to a regular base? QUESTION [9 upvotes]: Let $f\colon X \rightarrow S$ be a semistable curve of genus $g \ge 0$. Being a semistable curve means that $f$ is a morphism of schemes such that $f$ is proper, flat, and of finite presentation; The geometric fibers of $f$ are reduced, connected, one-dimensional, of arithmetic genus $g$, and smooth away from finitely many points at which the singularities are required to be ordinary double points. Does $f$, Zariski locally on $S$, come as a base change of a semistable curve $f' \colon X' \rightarrow S'$ with $S'$ regular? A lemma of this sort seems to be used in the proof of 9.4/1 of "Neron models," i.e., in the proof of the representability of $\mathrm{Pic}^0_{X/S}$ by a scheme. The authors cite the paper of Deligne and Mumford for this reduction to a regular base. I can't see how the citation justifies the claim because Deligne and Mumford seem to deal with a narrower class of curves, namely, they impose an additional condition: If a geometric fiber of $f$ has an irreducible component isomorphic to $\mathbb{P}^1$, then that component meets the other components in $\ge 3$ points. EDIT. The real question is: for a semistable curve $f$ as above, is $\mathrm{Pic}^0_{X/S}$ a scheme, as claimed in 9.4/1 of "Neron models"? The proof given there seems incomplete (see the comments below for some discussion). Any ideas are welcome. REPLY [2 votes]: The answer to the original problem (not the one in the edit) is: No. Let $S$ be a nodal curve over the complex numbers with normalization $T$. Say $t, t' \in T$ are the inverse images of the node. Let $E$ be an elliptic curve. Take $E \times T$ and glue the fibres $E \times t$ and $E \times t'$ by translating with a nontorsion point to get $X \to S$. This is the standard example of a smooth genus $1$ fibration $X \to S$ which is not locally projective. Such a thing can never come from $X' \to S'$ with $S'$ regular. If it did then you would be able to Zariski shrink $S'$ and assume that $X' \to S'$ is smooth and proper. Then you would take a divisor on the generic fibre and take the closure $D'$ in $X'$. A small argument then shows $D'$ is relatively ample hence the pullback to $X$ is relatively ample over $S$. Contradiction.<|endoftext|> TITLE: Can all unit-distance graphs have their vertices at algebraic integers? QUESTION [18 upvotes]: A graph $G$ is described as a unit-distance graph if there exists a function $f:G \rightarrow \mathbb{C}$ such that for every edge $(u,v) \in E(G)$, we have $|f(u) - f(v)| = 1$. Obviously, we can necessarily find an embedding into the algebraic numbers $\bar{\mathbb{Q}}$. In other words, if $G$ is a unit-distance graph, we can find $g:G \rightarrow \bar{\mathbb{Q}}$ such that for all $(u,v) \in E(G)$, we have $|g(u) - g(v)| = 1$. But what about if we further restrict ourselves to the algebraic integers $\mathcal O(\bar{\mathbb{Q}})$? Can all unit-distance graphs be embedded in such a way? REPLY [13 votes]: $\let\eps\varepsilon$No. I will present a graph whose realization necessarily contains a pair of vertices at distance $1/2$. THis cannot happen if the vertices are algebraic integers. Firstly, we note that we may force a graph to contain a given piece of triangular lattice. It will be clear after we understand how to enforce the two opposite vertices of a $60^\circ-120^\circ$ unit-sided rhombus to be distinct. This is made by augmenting this rhombus to a Mosers' spindle: the vertices of its realization cannot coincide. Thus we may enforce two vertices $A$ and $B$ to have any distance realized in the triangular lattice. Now consider five points $A$, $B$, $C$, $D$, $E$ such that $AB=AC=AD=2$, $BC=DC=BE=DE=1$. If $B\neq D$ and $C\neq E$ then we have $CE=1/2$ as required. Thus it remains to enforce these relations. To enforce $B\neq D$ it suffices to introduce a point $X$ with $BX=2$ and $DX=1$. To enforce $C\neq E$ it suffices to introduce $Y$ with $CY=2$ and $EY=\sqrt 3$. REMARK (expanded). Maehara showed that any algebraic number can be realized as a distance between two vertices of a rigid unit-distance framework. This inspired the answer; I just needed to make the graph "more rigid". A similar reinforcing may be applied to any rigid framework to make it "absolutely rigid". To perform this, for every two points $A$, $B$ at distance $d$ in the fixed rigid realization, one needs to ensure that $d-\epsd-\eps$, find in the triangular lattice two distances $\ell_1$ and $\ell_2$ with $d-\eps <\ell_1-\ell_2 TITLE: Can any object in a presentable category be written as a colimit of generators? QUESTION [9 upvotes]: Let $\mathcal{C}$ be a presentable category, and let $S$ be a set of objects such that $S$ generates $\mathcal{C}$ under colimits, i.e., such that the smallest cocomplete subcategory of $\mathcal{C}$ containing $S$ is all of $\mathcal{C}$. Under what conditions is it true that for every object $x \in \mathcal{C}$, there exists a functor $f: \mathcal{I} \to \mathcal{C}$ whose image is contained in $S$ and whose colimit is isomorphic to $x$? This is true for example if $\mathcal{C}$ is a presheaf category and $S$ the canonical set of generators (the representables). I suspect the answer is no in general, but I don't know how to construct an example. I am also interested in the analogous question for presentable $\infty$-categories. Edit: After posting this question, I became aware of Mike Shulman's answer here which addresses the same question with the counterexample being the category of compact Hausdorff spaces. But is there a natural presentable example that one might expect to come across (some type of structured sets, for instance)? I'd like to get some intuition for this phenomenon. REPLY [6 votes]: Here is a short note by Mike Shulman with some relevant material. He remarks that the arrow category $\mathbb{2}$ inside $Cat$ (which of course is presentable) is a colimit generator in your sense (see his definition 3.6), but it is not a colimit-dense generator (see his definition 3.5) in the sense that any object of $C$ is a colimit of a functor into the full subcategory of $Cat$ containing $\mathbb{2}$. So this answers at least an implicit question of the post. I think I'll give him a poke, because he's undoubtedly better placed to handle this query.<|endoftext|> TITLE: Generic filters of inverse limits QUESTION [7 upvotes]: Maybe this doubt is silly, but I do not understand the final step of the proof of Lemma 5.2 in Hamkins' paper Fragile measurability, Journal of Symbolic Logic 59 (1994) 262-282. There, $\mathbb P_\lambda$ is the $\lambda$-th step of an iterated forcing construction which is the inverse limit of the previous steps, and $G$ is a $V$-generic filter. More specifically, $\mathbb P_\lambda$ is the canonical Easton support iteration which forces the GCH below $\lambda$, and $\lambda$ is a strong limit cardinal of countable cofinality. We have a condition $r\in\mathbb P_\lambda$ (in the proof it is called $j(p)$) such that for all $\beta<\lambda$, $r(\beta)\in G(\beta)$. My doubt is why this implies that $r\in G$. I can prove that $r|_\beta\in G_\beta=i_{\beta\lambda}^{-1}(G)$ for all $\beta<\lambda$, and maybe the conclusion is trivial, but I do not see it. In fact, I know very few about generic filters on inverse limits and their relation with previous steps. REPLY [7 votes]: We can justify this with a general fact about forcing. Fact: Let $G$ be $\mathbb P$-generic, where $\mathbb P$ is separative. If $X$ is a subset of $\mathbb P$ in the ground model, $X \subseteq G$, and $m = \inf X$, then $m \in G$. Proof: Consider the set $\{ p \in \mathbb P : p \leq m$ or $(\exists x \in X)p \perp x \}$. Using separativity we show this is a dense set. Since $X \subseteq G$, this implies $m \in G$. Now I claim that in this situation $r = \inf_{\beta<\lambda} r \restriction \beta$. Certainly $r \leq r \restriction \beta$ for all $\beta$. If $q$ also has this property, then $q \leq r$ by the definition of the ordering at limit stages of iterations. ($q \leq r$ just means $q \restriction \beta \leq r \restriction \beta$ for all $\beta < \lambda$.)<|endoftext|> TITLE: Parametric solutions of Pell's equation QUESTION [28 upvotes]: Given a positive integer $n$ which is not a perfect square, it is well-known that Pell's equation $a^2 - nb^2 = 1$ is always solvable in non-zero integers $a$ and $b$. Question: Let $n$ be a positive integer which is not a perfect square. Is there always a polynomial $D \in \mathbb{Z}[x]$ of degree $2$, an integer $k$ and nonzero polynomials $P, Q \in \mathbb{Z}[x]$ such that $D(k) = n$ and $P^2 - DQ^2 = 1$, where $a = P(k)$, $b = Q(k)$ is the fundamental solution of the equation $a^2 - nb^2 = 1$? If yes, is there an upper bound on the degree of the polynomials $P$ and $Q$ -- and if so, is it even true that the degree of $P$ is always $\leq 6$? Example: Consider $n := 13$. Putting $D_1 := 4x^2+4x+5$ and $D_2 := 25x^2-14x+2$, we have $D_1(1) = D_2(1) = 13$. Now the fundamental solutions of the equations $P_1^2 - D_1Q_1^2 = 1$ and $P_2^2 - D_2Q_2^2 = 1$ are given by $P_1 := 32x^6+96x^5+168x^4+176x^3+120x^2+48x+9$, $Q_1 := 16x^5+40x^4+56x^3+44x^2+20x+4$ and $P_2 := 1250x^2-700x+99$, $Q_2 := 250x-70$, respectively. Therefore $n = 13$ belongs to at least $2$ different series whose solutions have ${\rm deg}(P) = 6$ and ${\rm deg}(P) = 2$, respectively. Examples for all non-square $n \leq 150$ can be found here. Added on Feb 3, 2015: All what remains to be done in order to turn Leonardo's answers into a complete answer to the question is to find out which values the index of the group of units of $\mathbb{Z}[\sqrt{n}]$ in the group of units of the ring of integers of the quadratic field $\mathbb{Q}(\sqrt{n})$ can take. This part is presumably not even really MO level, but it's just not my field -- maybe someone knows the answer? Added on Feb 14, 2015: As nobody has completed the answer so far, it seems this may be less easy than I thought on a first glance. Added on Feb 17, 2015: Leonardo Zapponi has given now a complete answer to the question in this note. REPLY [4 votes]: Here are some results concerning the degree of the polynomial $P$. We only treat the cases where $n$ is a positive, square-free integer congruent to $3$ modulo $4$, showing that the degree of $P$ is less than or equal to $2$. We start by the weaker assumption that, given the positive integer $n$ and a solution $(a,b)$ of Pell's equation $a^2-nb^2=1$, there exist polynomials $D,P,Q\in\Bbb Q[X]$ and a rational number $k$ such that $P^2-DQ^2=1$, with $D(k)=n,P(k)=a$ and $Q(k)=b$. Let $d$ be the degree of $P$. As mentioned in my other post, and following the conventions and results in David Speyer's answer, there exist constants $\alpha\in\Bbb C$ and $\beta,\gamma\in\Bbb Q$, with $\alpha^2,\gamma^2\in\Bbb Q$ such that $$\begin{cases} P=\pm T_d\left(\alpha(X+\beta)\right),\\ Q=\gamma U_{d-1}\left(\alpha(X+\beta)\right),\\ D=\gamma^{-2}\left(\alpha^2(X+\beta)^2-1\right), \end{cases}$$ where $T_d$ (resp. $U_d$) denotes the degree $d$ Chebyshev polynomial of the first kind (resp. of the second kind). From the hypothesis on $P$, we can assume $\beta=0$. We have the identity $$T_d+U_{d-1}\sqrt{X^2-1}=\left(X+\sqrt{X^2-1}\right)^d,$$ which leads to $$P+Q\sqrt{D}=\left(\alpha X+\sqrt{\alpha^2X^2-1}\right)^d.$$ If $d$ is odd then the explicit expression of $T_d$ shows that $\alpha$ (and therefore $\gamma$) is rational. Evaluating at $k$, we then obtain the identity $$a+b\sqrt{n}=\left(u+v\sqrt{n}\right)^d,$$ with $u,v\in\Bbb Q$. It then follows that $u+v\sqrt{n}$ is a unit in the ring of integers of $\Bbb Q(\sqrt{n})$ and, since we are assuming $n\equiv3\pmod4$, the elements $u$ and $v$ are integers. In particular, for $d>1$, the couple $(a,b)$ cannot be a fundamental solution of Pell's equation. From now on, we assume $d=2m$ even. For $\alpha\in\Bbb Q$, we proceed as above and deduce that $(a,b)$ cannot be a fundamental solution for $d>1$. Suppose then that $\alpha=\sqrt{w}$, we $w\in\Bbb Q$ not a square, so that $\gamma=t\alpha$, with $t\in\Bbb Q$. We have the relation $$T_d+U_{d-1}\sqrt{X^2-1}=\left(X^2-1+2X\sqrt{X^2-1}\right)^m$$ and thus $$P+Q\sqrt{D}=\left(\alpha^2X^2-1+2\alpha\sqrt{\alpha^2X^2-1}\right)^m.$$ Evaluating at $k$, we then easily obtain the identity $$a+b\sqrt{n}=\left(u+v\sqrt{n}\right)^m,$$ with $u,v\in\Bbb Q$. Once again, for $m>1$, it then follows that the couple $(a,b)$ cannot be a fundamental solution, leading to the result. A final remark: for general $n$, if $(a,b)$ is a fundamental solution then $$a+b\sqrt{n}=\left(u+v\sqrt{n}\right)^r,$$ where $u+v\sqrt{n}$ is a fundamental unit of $\Bbb Q(\sqrt{n})$. For example, in Stefan Kohl's example for $n=13$, we have $$649+180\sqrt{13}=\left(\frac{11}2+\frac32\sqrt{13}\right)^3.$$ I'm definitely not an expert on this subject, but the above discussion combined with an explicit bound for the integer $r$ would lead to a bound for the degree of $P$ and therefore give a complete answer to the second question.<|endoftext|> TITLE: Adding a real with infinite conditions QUESTION [13 upvotes]: Consider the forcing $\Bbb P$ whose conditions are partial functions $p\colon\omega\to2$ with $\operatorname{dom}(p)$ a co-infinite subset of $\omega$, ordered by reverse inclusion. Does $\Bbb P$ collapse the continuum? REPLY [14 votes]: While Prikry-Silver forcing satisfies Axiom A and hence does not collapse $\omega_{1}$, it is independent of $MA + \neg CH$ whether Prikry-Silver forcing preserves the continuum. Under $MA + \neg CH$, this question is equivalent to whether the additivity of the Silver ideal is the continuum. So, for example, PFA implies that Prikry-Silver forcing does indeed preserve the continuum. For the corresponding question for Sacks forcing, see: H. Judah, A. W. Miller and S. Shelah, Sacks forcing, Laver forcing, and Martin's axiom. Arch. Math. Logic 31 (1992), 145–161.<|endoftext|> TITLE: Irreducibility of a polynomial QUESTION [7 upvotes]: For $n\ge 1$, let $g(x_1,x_2,\ldots,x_n)$ be an irreducible homogeneous polynomial in $n$ variables over a field $k$ and $f(x)$ an irreducible polynomial of $k[x]$. Is $f(g(x_1,x_2,\ldots,x_n))$ necessarily irreducible? For instance this holds when $n=1$ (since then $g(x_1)=\lambda x_1$), or when $f$ has degree 1 (by a simple argument). REPLY [5 votes]: I believe that the answer is yes. Put $c:=g(x_1,\ldots, x_n)$, which is irreducible in the UFD $R:=k[x_1,\ldots, x_n]$. Assume $f$ is irreducible, but also assume by way of contradiction that $f(c)=\alpha\beta$ in $R$, with $\alpha,\beta\notin k$. Write $\alpha=\sum \alpha_i$ and $\beta=\sum \beta_j$, where the $\alpha_i$ and $\beta_j$ are homogeneous (of the appropriate degrees). [Note: We implicitly only consider terms with nonzero support in the remainder.] We can write $\alpha_i=c^{e_i}\alpha_i'$ with $e_i$ maximal. Similarly write $\beta_j=c^{e_j'}\beta_j'$. Fix $m_1$ maximal with $\deg(\alpha_{m_1}')$ maximal among $\{\deg(\alpha_i')\}$. Similarly, fix $m_2$ maximal with $\deg(\beta_{m_2}')$ maximized. Case 1: $\deg(\alpha_{m_1}')=0$ and $\deg(\beta_{m_2}')=0$. In this case $\alpha$ and $\beta$ are polynomials in $c$, which contradicts the irreducibility of $f$. Case 2: Without loss of generality, $\deg(\alpha_{m_1}')>0$. Consider the degree $m_1+m_2$ coefficient of $f(c)=\alpha\beta$. On the one hand, since $f(c)$ is a polynomial in $c$, it is either zero or a $k$-multiple of a power of $c$, say $c^{e}$. On the other hand, this term on the right-hand side is $$c^{e_{m_1}+e_{m_2}'}\alpha_{m_1}'\beta_{m_2}' + \sum_{(i,j)\neq (m_1,m_2)\ :\ i+j=m_1+m_2}\alpha_i\beta_j.$$ Each term in the big sum is divisible by strictly more powers of $c$ than $e_{m_1}+e_{m_2}'$, by maximality of degrees. Also $e>e_{m_1}+e_{m_2}'$, since $\deg(\alpha_{m_1}')>0$. This gives us the necessary contradiction.<|endoftext|> TITLE: Can we extend a multiplicative linear functional of a closed left ideal on whole of the algebra? QUESTION [7 upvotes]: Let B be a closed left ideal of a Banach algebra A. Also, B has a right approximate identity (in B). If g is a nonzero multiplicative linear functional on B, can we always extend g to a multiplicative linear functional on A ? REPLY [4 votes]: Let $H$ be a complex Hilbert space with $\dim(H)\geq 2$. Denote by $B(H)$ the Banach algebra of all bounded linear operators on $H$. Since there is no closed ideal of codimension $1$ in $B(H)$ we see that (1) there is no nonzero multiplicative linear functional on $B(H)$. Choose and fix a vector $e\in H$, $\| e\|=1$. Let $K=\{ x\in H;\quad \langle x,e\rangle=0\}$. Then $K^\perp ={\mathbb C}\, e$ and $H={\mathbb C}\, e \oplus K$. Let $$ {\mathcal I}=\{ T\in B(H);\quad K\subseteq \ker T\}. $$ It is easy to see that this is a closed left ideal in $B(H)$. If $P\in B(H)$ is the orthogonal projection onto ${\mathbb C}\, e$ along $K$, then, of course, $P\in {\mathcal I}$. Since any $x\in H$ has a unique decomposition $x=\alpha e+y$ with $\alpha \in {\mathbb C}$ and $y\in K$ we have $$ TPx=\alpha Te=Tx $$ for any $T\in {\mathcal I}$. Hence $P$ is a right identity in ${\mathcal I}$. Define $\varphi: {\mathcal I} \to {\mathbb C}$ by $$ \varphi(T)=\langle Te,e\rangle \qquad (T\in {\mathcal I}). $$ It is obvious that $\varphi $ is a linear functional and because of $\varphi(P)=\langle Pe,e\rangle=1$ it is nonzero. Let $S, T\in {\mathcal I}$ be arbitrary. Let $Te=\lambda_T e+y_T$ with $\lambda_T\in {\mathbb C}$ and $y_T\in K$. Then $\varphi(T)=\langle Te,e\rangle=\lambda_T$. Hence $$ \varphi(ST)=\langle STe,e\rangle=\langle S(\varphi(T)e+y_T),e\rangle=\varphi(S)\varphi(T). $$ Thus, $\varphi$ is a nonzero multiplicative linear functional on ${\mathcal I}$. However it cannot be extended to a multiplicative linear functional on $B(H)$ because of (1).<|endoftext|> TITLE: line bundle descends? QUESTION [5 upvotes]: Let the permutation group $S_4$ act on $\mathbb C^4$ by permuting the coordinates. Consider the categorical quotient $\mathbb P(\mathbb C^4)/S_4$. It is a projective variety by a theorem of Mumford. Does the line bundle $\mathcal O(1)^{\otimes 12}$ on $\mathbb P(\mathbb C^4)$ descend to the quotient $\mathbb P(\mathbb C^4)/S_4$ ? REPLY [11 votes]: Yes, and the good news are that there isn't anything to compute. By a result of Kempf a line bundle $L$ on $\mathbb{P}^3$ descends to the quotient if and only if the stabilizer $S_x$ of each point $x\in \mathbb{P}^3$ acts trivially on the fiber $L_x$. Now $S_4$ acts on $\mathcal{O}(1)$, and $S_x$ act on $\mathcal{O}(1)_x$ through a character $\chi _x:S_x\rightarrow \mathbb{C}^*$, and on $\mathcal{O}(1)^{\otimes k}$ through the character $\chi _x^{k}$. The abelian subquotients of $S_4$ have order $2,3$ or $4$, hence taking $k=12$ kills all the characters, and therefore $\mathcal{O}(1)^{\otimes 12}$ descends.<|endoftext|> TITLE: Set with small internal radius, small perimeter and prescribed area QUESTION [8 upvotes]: Given a regular set $E\subset \mathbb R^2$ define $$ R(E) = \sup\{r\colon \exists x,\ B(x,r)\subseteq E\} $$ to be the radius of the largest circle contained in $E$ and let $|\partial E|$ be the length of the perimeter of $E$. Among all planar sets $E$ with fixed area, what is the one which minimizes the product $|\partial E| \cdot R(E)$? Do you have any keyword to look for? edit: Actually I'm looking for an estimate of the kind: $$ R(E) \ge c \frac{|E|}{|\partial E|} $$ for $E$ simply connected. And even a non optimal constant $c$ would be useful to me... I didn't realized that simply connectedness played a role. Sorry for that. REPLY [4 votes]: For the simply connected version of your question: yes, such an estimate exists. Namely: take a radius $R=R(E)$ (closed) "coloring" disk, and let us move its center along the boundary $\partial E$. Note that while the center of the disk makes a path of length $l=|\partial E|$, the disk "colors" the area that is at most $2\pi Rl + \pi R^2$. Indeed, you count the initial area $\pi R^2$ plus estimate the increase in area per $\Delta l$ displacement of the disk as its boundary length $2\pi R$ times the displacement $\Delta l$. (In fact, this upper bound can be improved to $2 R l$, but it requires a bit more words to be pronounced. Namely, you say that there are two disjoint discs, so you can count only increase terms -- area of difference between displaced and not displaced discs, -- and you estimate such an increase term more carefully as $2R\Delta l$. See also: analogous formulae in the theory of Minkowski sums/convex polyhedra, like the one for the area or the volume of $\varepsilon$-neighborhood.) On the other hand, by definition of $R$ any point of $E$ is at distance at most $R$ from the boundary $\partial E$. Thus, all the set $E$ will be colored. Hence, $$ |E|\le 2\pi Rl + \pi R^2 $$ and as $2\pi R\le l$, the $|E|\le (2\pi + \frac{1}{2}) R l$. So you can take $c=(2\pi + \frac{1}{2})^{-1}$. In fact, the improved estimate mentioned earlier tells us that you can even take $c=1/2$: $$ |E|\le 2Rl= 2 R(E) \cdot |\partial E|. $$<|endoftext|> TITLE: Is the space $S'(\mathbb{N})$ of slowly increasing sequences the projective limit of Hilbert sequence spaces? QUESTION [8 upvotes]: Let $S(\mathbb{N})$ be the space of rapidly decreasing sequences and $S'(\mathbb{N})$ its topological dual, the space of sequences bounded by a polynomial. For $m\in \mathbb{Z}$, we also define $\ell_2^m (\mathbb{N})$ as Hilbert spaces of sequences such that $(u_n (n+1)^m)_{n\in \mathbb{N}} \in \ell_2 (\mathbb{N})$. It is known than $S(\mathbb{N})$ is the projective limit of the spaces $\ell_2^m (\mathbb{N})$. As such, $S(\mathbb{N})$ is a Frechet space with a nuclear topology. Its dual $S'(\mathbb{N})$ is hence the inductive limit of the spaces $(\ell_2^m)' (\mathbb{N}) = \ell_2^{-m} (\mathbb{N})$. So: Fact 1: $S'(\mathbb{N})$ has a natural complete nuclear topology defined as a countable inductive limit of Hilbert spaces. It is also known that any complete nuclear space is isomorphic with the projective limit of a suitable family of Hilbert spaces. See for instance Corollary 3, Section 7.2 of Topological Vector Spaces. Fact 2: $S'(\mathbb{N})$ has an abstract complete nuclear topology as a projective limit of Hilbert spaces. Question: Is it possible to describe the topology of $S'(\mathbb{N})$ as a countable projective limit of Hilbert spaces $H_m$, meaning that $S'(\mathbb{N}) =\bigcap_{m\in \mathbb{N}} H_m$, such that the $H_m$ are described as sequence spaces (bigger than $S'(\mathbb{N})$ of course)? REPLY [5 votes]: I can now answer my own question thanks to the following discussion: Which Fréchet spaces have a dual that is a Fréchet space? As explained in the comments, $\mathcal{S}'(\mathbb{N})$ can be defined as the projective limit of a family of semi-norms $(p_u)$ indexed by $u \in \mathcal{S}(\mathbb{N})$. It is however impossible to extract a countable family from the $(p_u)$ due to the much more general fact that the strong dual of a Frechet space $E$ is metrizable if and only if $E$ is normable. Here, $E = \mathcal{S}(\mathbb{N})$ is not normable, so $E' = \mathcal{S}'(\mathbb{N})$ is not metrizable, hence cannot be defined as a projetive limit of a countable family of sequence spaces.<|endoftext|> TITLE: Can ZFC prove it cannot derive an inconsistency in $n$ steps? QUESTION [19 upvotes]: Let $Con(\mathtt{ZFC}, n)$ denote the statement "$\mathtt{ZFC}$ cannot prove the contradiction within $n$ steps (or better within $n$ symbols) within a given proof system (say a natural deduction to avoid trivialities)". Suppose that ($\mathtt{ZFC}$ is consistent and) $\mathtt{ZFC} \models Con(\mathtt{ZFC}, n)$, then since the sentence could be represented as a first order statement using codings for proofs, by Godel's completeness theorem, it can be proved as well. Or more directly (without the assumption of the consistency of $\mathtt{ZFC}$) one could produce an algorithm that would enumerate all the possible statements provable within $n$ steps from $\mathtt{ZFC}$ and then it would check if such a statement is a contradiction or not and output the corresponding proof in $\mathtt{ZFC}$ of $Con(\mathtt{ZFC}, n)$. But any such proof for all statements derivable in $n$ steps would be very large, at least of size $O(n)$. Therefore, is there a large natural number $n > 2^{100}$ and a proof in no more than $n$ steps (symbols) in $\mathtt{ZFC}$ of the sentence $Con(\mathtt{ZFC}, 2^n)$? REPLY [42 votes]: It’s not very clear to me what you mean by “steps”. One might interpret it as the number of lines in a Hilbert/natural deduction proof, but then there are infinitely many proofs with a fixed number of steps, so this is inconsistent with the argument outlined in the question. So, let me use a measure that has the property that there are only finitely many proofs of length $n$, namely the total number of symbols in the proof when written down as a string. The choice of proof system does not matter that much, but natural deduction will do. Then, for any consistent theory $T$, and $n\in\mathbb N$, $\mathrm{Con}(T,n)$ is a true bounded sentence, and as such it is provable already in Robinson arithmetic. However, the length of its shortest proof is a nontrivial problem. By a result of Pudlák, for any mildly reasonable theory $T$, the shortest proof of $\mathrm{Con}(T,n)$ in $T$ has length between $n^\epsilon$ and $n^c$ for some constants $0<\epsilon<1 TITLE: Besse p134 Riemann tensor in dimension 4 QUESTION [5 upvotes]: Does someone have a reference for the proof of 4.72 page 134 of Einstein Manifolds? It is said that $$\check{R}-\vert R\vert^2g/4=S/3 (Ric-S/4) +2\mathring{W}(Ric -S/4) $$ because we are in dimension 4, where$\check{R}_{ab}=R_{ajkl}R^{jkl}_b$ and $\mathring{W}$ is Weyl acting on symmetric tensor. Is there is a simple idea to get it, except to develop everything? In fact I am trying to get it from a moving frame point of view, and more generally, I am looking for any reference making computation about curvature functional, Einstein metric and Bach tensor from the moving frame point of view. Thanks in advance. REPLY [9 votes]: I don't have a reference, but there is a conceptual way to see that you only have to check a few constants to get this formula. Notice that it's a statement about a quadratic mapping from the space of curvature tensors in 4D to the space of traceless quadratic forms in 4D (because both sides are clearly traceless quadratic forms). Also, this mapping has to be invariant under change of frame, i.e., under the action of $\mathrm{O}(4)$, and, in particular, $\mathrm{SO}(4)$, so you could expect representation theory to help. Now, the space of curvature tensors in 4D, when considered as an $\mathrm{SO}(4)$ representation space, is actually a representation of $\mathrm{SO}(4)/\{\pm I_4\} = \mathrm{SO}(3)\times\mathrm{SO}(3)$, which is a product of two simple groups of rank $1$, so the irreducible representations are of the form $V^{p,q} = V^{p,0}\otimes V^{0,q}$, where $V^{p,0}$ is the irreducible representation of $\mathrm{SO}(3)\times\{1\}$ of dimension $2p{+}1$ and $V^{0,q}$ is the irreducible representation of $\{1\}\times\mathrm{SO}(3)$ of dimension $2q{+}1$. With this notation, a curvature tensor $R$ in dimension $4$ is uniquely expressed as a sum $$ R = R^{0,0} + R^{1,1} + R^{2,0} + R^{0,2} $$ where $R^{0,0}$, with values in $V^{0,0}$, is the scalar curvature piece; $R^{1,1}$, with values in $V^{1,1}$, is the traceless Ricci tensor; $R^{2,0}$, with values in $V^{2,0}$, is the self-dual Weyl curvature; and $R^{0,2}$, with values in $V^{0,2}$, is the anti-self dual Weyl curvature. Now, an $\mathrm{SO}(4)$-equivariant quadratic expression in $R$ taking values in $V^{1,1}$, which is what the left hand side of your equation is, must come from an $\mathrm{SO}(4)$-equivariant map $$ \mathsf{S}^2\bigl(V^{0,0} \oplus V^{1,1} \oplus V^{2,0} \oplus V^{0,2}\bigr)\longrightarrow V^{1,1} $$ But, from the Clebsch-Gordan formula for tensor products of representations of $\mathrm{SO}(3)$, we see that the left hand representation has four $V^{1,1}$-components, which come from the terms $$ \begin{aligned} V^{1,1} &\simeq V^{0,0}\otimes V^{1,1}\\ V^{1,1} &\subset \mathsf{S}^2\bigl( V^{1,1}\bigr) \simeq V^{1,1} \oplus V^{2,2}\oplus V^{0,2}\oplus V^{2,0}\oplus V^{0,0}\\ V^{1,1} &\subset V^{1,1}\otimes V^{2,0} \simeq V^{3,1}\oplus V^{2,1}\oplus V^{1,1}\\ V^{1,1} &\subset V^{1,1}\otimes V^{0,2} \simeq V^{1,3}\oplus V^{1,2}\oplus V^{1,1} \end{aligned} $$ The first line gives you a bilinear pairing between the scalar curvature and the traceless Ricci, which, up to a constant factor, is the first term on the right hand side of your equation. The second line gives a quadratic map from the traceless Ricci tensors into traceless symmetric tensors (but that potential term turns out not to show up in the final formula). The third and fourth lines give you pairings between the self-dual and anti-self-dual parts of the Weyl curvature and the traceless Ricci, which, up to a constant factor, is the second term on the right hand side of your equation. (Because the mapping actually has to be equivariant under $\mathrm{O}(4)$, which exchanges the self-dual and anti-self dual parts of the Weyl curvature, the two constants for those two terms have to enter symmetrically, which is why there is only one term there.) To determine the correct constant multiples, it suffices to simply check the formula on curvature tensors that are made up of very simple cases: Take a curvature tensor $R$ that has no scalar or Weyl curvature, just a diagonalized traceless Ricci tensor, plug it into the left hand side, and you'll get zero, so that means that the potential quadratic in traceless Ricci term (from the second line) must be zero. Next, take a curvature tensor $R$ that has no Weyl curvature part and plug it in to determine the constant multiple for the first term in the right hand side of the above formula . Finally, take a tensor $R$ with no scalar curvature term and as simple as possible Ricci to check the coefficient for the second term. While this method does not avoid all calculation, it does reduce the calculation to a few simple checks of constants. Also, this method illustrates the usefulness of representation theory in these calculations, which are generally extremely useful in getting explicit formulas such as the one you are trying to understand.<|endoftext|> TITLE: Why is "The Higman Rope Trick" thus named? QUESTION [14 upvotes]: I'm studiyng Higman's Embedding Theorem, and a fundamental part of the proof is the following lemma: If R is a benign normal subgroup of finitely generated group F, then F/R can be embedded in a finitely presented group. A proof of the lemma in context could be found on page 16 of this paper. I understand the lemma on technical levels, and to some degree also on an intuitive level. But I have no idea why Lyndon & Schupp chose the name "The Higman Rope Trick" for the lemma. I'm hoping someone could enlighten me because my feeling is that I'm missing something important about the meaning of the lemma...otherwise I'd get the meaning of the name, not so? Thank you very much! Shlomi. P.S: This is my first time posting a question. I read the different guides and I'm pretty sure I stuck to the rules. If not though then I apologize and will correct upon notice. REPLY [12 votes]: Well, Sam Nead and Flounderer got it right but in different ways... :-) I took up Sam's idea and asked Schupp directly. He said Flounderer was right, and I'll quote: "The answer on mathoverflow is exactly correct - the reference is to the Indian Rope Trick. I always think of the lemma in the following way: As one is concentrating on the details of the proof, Higman the magician makes the infinitely many relations disappear. The Higman Embedding Theorem is certainly one of the great theorems." So all credit to Schupp and Flounderer, but I decided to post this as an answer for the sake of closure. Thanx all, Shlomi.<|endoftext|> TITLE: Categorical proof subgroups of free groups are free? QUESTION [20 upvotes]: This is a crossport of this question from MSE. Is there a categorical proof that subgroups of free groups are free? How about the result that subgroups of free abelian groups are free abelian? What is it about $\mathsf{Grp}$ and $\mathsf{Ab}$ that makes subobjects of free objects free? Can one characterize such categories? In particular, are there any other ones apart from $\mathsf{Grp}$ and $\mathsf{Ab}$? I have received some comments on MSE, one claiming such a proof exists, one doubting it, and another claiming it cannot be proved in $ZF$. REPLY [12 votes]: Let me elaborate on my comment. I think freeness is a red herring. The content of the standard topological proof is that a covering space of a $1$-dimensional CW complex is again a $1$-dimensional CW complex. Of course this naturally generalizes to the case where $1$ is replaced by any positive integer $k$, which suggests the following definition. Say that a group $G$ has homotopy dimension at most $k$ if $BG$ can be presented by a $k$-dimensional CW complex (e.g. take $BG$ to be a compact hyperbolic $k$-manifold). Then the same argument about covering spaces shows that every subgroup of a group with homotopy dimension at most $k$ again has homotopy dimension at most $k$. Homotopy dimension at most $1$ is equivalent to freeness in the presence of choice, but using it instead of freeness in the argument removes the need to choose a spanning tree and that should address concerns about uses of choice. As the long list of examples in Todd's answer also suggests, we should be looking at some property other than freeness in general. I think some kind of cohomological dimension is a better candidate; as stated in Matthias Wendt's comment below, the nontrivial free groups are precisely the groups of cohomological dimension $1$. And $1$-dimensionality also shows up in Ben Webster's answer.<|endoftext|> TITLE: Papers about decentralized search and cluster QUESTION [11 upvotes]: I just start an independent study about small world network and clusters and I try to find papers about decentralized search and clusters. Can anyone give me some references? Thanks! EDIT (David White): This question is from my student, and I encouraged her to ask it here. We are just starting research on decentralized search in social networks, and she had the idea to form a graph where the nodes are sets of individuals and you put an edge between two sets if there is any overlap. We'd like to write down some decentralized search algorithms (e.g. greedy, random) and analyze them. I think this is a great starting project, but I don't know enough about the literature. Has someone else already done this? Is there a name for graphs of the form above? Has anyone studied a version where you weight each edge by the size of the intersection of the two sets? Joonas is correct that googling "decentralized search and cluster" yields many articles. I can't seem to find any which do what we're thinking of doing, but that doesn't mean they don't exist. I'd hate to steer my student in a bad direction in her first research project. Any help you can provide would be much appreciated. REPLY [5 votes]: Those graphs are called intersection graphs. Perhaps you can find literature relevant to your problem using that name. For instance I found the following paper where they do something similar to what you want, if I understood correctly: M. Bloznelis, Degree and clustering coefficient in sparse random intersection graphs, The Annals of Applied Probability, Vol.23, No.3, 2013.<|endoftext|> TITLE: An integral with respect to the Haar measure on a unitary group QUESTION [6 upvotes]: Let $A,D\in \mathbb{C}^{n \times n}$ be diagonal matrices. I need to calculate $$\int_{U(n)}\det{(A-HDH^\dagger)}\,\mathrm{d}H$$ where $dH$ is the unit invariant Haar measure on the group of unitary matrices and $H^\dagger$ is the conjugate transpose of $H$. (If $A=I$ this is very easy to solve, but I want the answer for $A\neq I$ in terms of $A$ and $D$.) REPLY [6 votes]: Let $A={\rm diag}[a_1,\dots,a_n]$ and $B={\rm diag}[b_1,\dots,b_n]$. Let $\Delta(a)$ be the Vandermonde product in the $a_j$, and similarly $\Delta(b)$ be the Vandermonde product in the $b_k$. Suppose ${\rm min} \, a_j \ge {\rm max} \, b_k$. Let ${\rm d}U$ denote the normalised Haar measure on the unitary group $U(n)$. Then by Eq. (3.21) in Gross and Richards, "Total positivity, spherical series, and hypergeometric functions of matrix argument", J. Approx. Th. vol. 59 (1989): 224-246, $$ \int_U \det(A - U B U^\dagger)^p {\rm d}U = c_{n,p} {\det [ (a_j - b_k)^{p+n-1}]_{j,k=1}^n \over \Delta(a) \Delta(b)}, $$ where $c_{n,p} = \prod_{j=0}^{n-1} \binom{p+n-1}{j}^{-1}$. Setting $p=1$, subject to the condition ${\rm min} \, a_j \ge {\rm max} \, b_k$, this is the sought evaluation. See Theorem 2.3 in Kieburg, Kuijlaars and Stivigny, "Singular value statistics of matrix products with truncated unitary matrices", IMRN vol. 2016 (2016) 3392-3424 for a generalisation.<|endoftext|> TITLE: Maximal compact subgroup of $\mathrm{SL}(2,\mathbb{H})$ QUESTION [9 upvotes]: $\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SL{SL}$The exceptional isomorphism $\Spin(5,1)\simeq \SL(2,\mathbb{H})$ is well-known, and I can find references that say the maximal compact of $\Spin(5,1)$ is $\Spin(5) \simeq \Sp(2)$. So I know the answer to the question, but not the how or why. In particular, is there a proof that $\Sp(2)$ is maximal compact in $\SL(2,\mathbb{H})$ not via the exceptional isomorphisms? Perhaps some sort of analogue of Gram-Schmidt or other explicit factorisation? REPLY [3 votes]: Here is a different argument from Robert's. Observation: given a compact Lie group $G$ and a proper closed subgroup $H$ the inclusion $H\to G$ is not a homotopy equivalence, else the homogeneous space $G/H$ would be a closed contractible manifold. Next, it's a general fact that the inclusion of a maximal compact subgroup $K$ into a semisimple Lie group $G$ is a homotopy equivalence and the quotient space is contractible. Putting these two things together immediately implies that if $G$ is a semisimple Lie group and $H\subset G$ is a compact subgroup such that $G/H$ is contractible then $H$ is maximal compact in $G$. It remains to observe that $SL(2,\mathbb H)/Sp(2)$ is contractible. From Robert's answer the homogeneous (in fact, symmetric) space $SL(2,\mathbb H)/Sp(2)$ can be identified with the 5-dimensional hyperbolic space which is of course contractible. Alternatively this can also be seen as follows. Look at the standard transitive action of $Sp(2)$ on the unit sphere $S^7$ in $\mathbb H^2$. The stabilizer of $(1,0)$ is easily seen to be equal to $Sp(1)=S^3$. Similarly look at the action of $SL(2,\mathbb H)$ on $\mathbb H^2\setminus (0,0)$ ( which deformation retracts onto $S^7$). the stabilizer $K$ of $(1,0)$ consists of matrices of the form $\begin{pmatrix} 1&a\\0&b\end{pmatrix}$ where $|b|=1$ and $a\in \mathbb H$ is arbitrary. So $K=S^3\times \mathbb R^4$ topologically. Comparing long exact homotopy sequences of the homogeneous space bundles $Sp(1)\to Sp(2)\to Sp(2)/Sp(1)=S^7$ and $K\to SL(2,\mathbb H)\to SL(2,\mathbb H)/K=\mathbb H^2\setminus (0,0)$ and using the natural map between them it now immediately follows by the 5-lemma that the inclusion $Sp(2)\subset SL(2,\mathbb H)$ is a homotopy equivalence. Hence by above $Sp(2)$ is maximal compact in $SL(2,\mathbb H)$. By induction this generalizes to show that $Sp(n)$ is maximal compact in $SL(n,\mathbb H)$.<|endoftext|> TITLE: What is the original reference for disorientations on tangle diagrams? QUESTION [6 upvotes]: There are several invariants whose "natural" domain is a category of disoriented tangles, that is tangles which are piecewise-oriented, but which contain points called `disorientations' at which the orientation is reversed. For example: Khovanov homology HERE and HERE. Kauffman's extended bracked polynomial HERE. I was unable, however, to track down the origin of the idea- where were disorientations first used, and what is the correct citation to use to reference it outside the context of one of these invariants? Further, I would like to ask whether there are other known contexts in which disoriented tangles of some flavour appear, other than the contexts mentioned above. REPLY [3 votes]: The term "disoriented tangle" first appeared in arxiv:0701339v2 (your first link). That is also the earliest reference I know for bordisms of disoriented tangles. But the idea of disoriented tangles is much older. It emerges naturally when one considers string diagrams for $Rep(U_q sl_2)$. See, for example, Figures 3.22 and 4.8 of Kirby and Melvin's paper here. I strongly suspect there are earlier examples in papers of Reshetikhin and/or Kirillov and/or Turaev.<|endoftext|> TITLE: singularize the least inaccessible? QUESTION [9 upvotes]: Is it consistent that there is some partial order $\mathbb P$ and some inaccessible cardinal $\kappa$, which is the least inaccessible, such that $\mathbb P$ forces $\kappa$ to be singular while preserving all cardinals? REPLY [10 votes]: In the paper "On Lowenheim-Skolem-Tarski numbers for extension of first order logic", by Magidor and Vaananen, in Theorem 21 they state that it is consistent, relative to the existence of a supercompact cardinal, that the Lowenheim-Skolem-Tarski number of $L(I)$ is the first inaccessible cardinal, were $L(I)$ is the extension of the first order logic with a quantifier of "equicardinality". An important ingredient in the proof is the following observation (which I state in a slightly different way from the original paper): Let $\kappa$ be a measurable cardinal. Let $\mathbb{NM}$ be the forcing that adds a club $D$ through the singular cardinals below $\kappa$ using bounded approximations. I will assume that the non limit points of the club are inaccessible cardinals. Let $Col$ be the forcing that collapses all cardinal between the any successor of a point in $D$ and the next point of the $D$, with Easton support. So $\mathbb{NM}\ast Col$ will force $\kappa$ to be the first inaccessible. On the other hand, in $V$ let $\mathbb{P}$ be a Prikry type forcing that adds a cofinal $\omega$-sequence at $\kappa$, $\{\eta_0, \eta_1, \dots\}$ and pick a sequence of conditions in $\mathbb{NM}$, $\{p_0, p_1,\dots\}$ such that $p_i \subseteq [\gamma_i, \gamma_{i+1})$. So $\mathbb{P}$ singularizes $\kappa$ and adds a club $\tilde{D} = \bigcup p_n$ through the singular cardinals below $\kappa$. Let $\tilde{Col}$ be the forcing that collapses any cardinal between point in $\tilde{D}$ as above, again with Easton support. Note that this time, as $\kappa$ is singular when we define $\tilde{Col}$, the support is unbounded in $\kappa$. Lemma: $\mathbb{NM}\ast Col$ forces $\kappa$ to be the first inaccessible cardinal. $\mathbb{P}\ast \tilde{Col}$ forces $\text{cf } \kappa = \omega$, and doesn't collapse cardinals above $\kappa$. Theorem: $\mathbb{P}\ast \tilde{Col}$ adds a generic filter for $\mathbb{NM}\ast Col$, and the quotient forcing, $\mathbb{R}$, changes the cofinaly of $\kappa$ to $\omega$ and doesn't add bounded subsets of $\kappa$. A variant of this theoerm is proved in the paper of Magidor an Vaananen, during the proof of Theorem 21. The main difference is that they assume there that $\kappa$ is $\kappa^+$-strongly compact in order to obtain a cleaner definition on the generic for $\mathbb{NM}$ that is obtained from the Prikry forcing. Magidor showed me that the same can be achieved using only a measurable, and this is the version that I sketched above.<|endoftext|> TITLE: Examples of Symplectic Questions Solved by ``Mirror Symmetry Translation'' to Complex Questions QUESTION [10 upvotes]: According to the proponents of homological mirror symmetry, when a complex and symplectic manifold are mirror symmetric, we can take difficult questions about the symplectic space and transfer them across the symmetry into more tractable complex geometry questions. Can anyone provide an example of an actual problem that was solved in this way. More specifically, one of the better understood examples of mirror symmetry is between the flag manifolds and certain Landau--Ginzburg models, see the papers of Rietsch. Can anyone provide an example of a difficult question about Landau-Ginzburg spaces that was answered in this way using mirror symmetry. Finally, can anyone point me to the papers where the equivalence between the flag manifolds and Landau--Ginzburg models was originally established? REPLY [12 votes]: A perfect example is Abouzaid-Smith's classification of genus 2 Lagrangian surfaces in $(T^4,\omega_{std})$. In this paper (http://arxiv.org/pdf/0903.3065v2.pdf), they proved that any Lagrangian genus 2 surface is Floer cohomologically indistinguishable from the Lagrangian surgery of two linear Lagrangian tori meeting transversely. The method is roughly to prove homological mirror symmetry for $T^4$ first, then establish the corresponding result on the B-side, finally translate it in the A-side. Recently, Abouzaid developed Fukaya's family Floer theory to produce a faithful mirror functor. The motivation for developing this is to study the symplectic topology of the Thurston manifold, which is the first example of a non-Kahler symplectic manifold. The difficulty is that the usual method of resolving the diagonal is not available here to prove that a finite set of Lagrangians split-generate the Fukaya category. However, since this space admits a Lagrangian fibration, family Floer theory can be used here to produce a mirror functor. Showing this functor is an equivalence will yield homological mirror symmetry in this case. The mirror is a rigid analytic Kodaira surface, therefore one can speculate from looking at the mirror that there are not so many non-trivial Lagrangians in the Fukaya category. This is very different from the case of $T^4$. Currently, there is no well-established mirror transformation in the general case, which makes it difficult to explicitly transform something to the mirror side, therefore mirror symmetry, in many cases, just helps in the philosophical level. For example, a Landau-Ginzburg model is in many cases mirror to a Fano. For many Fano manifolds, a full excpetional collection of the derived category of coherent sheaves can be found by the work of Beilinson and others. The analogous theory on the A-side is Seidel's work on Lefschetz fibrations. In the converse direction, one has Dehn twist along a Lagrangian sphere on the A-side, then it's natural to expect similar algebraic twist on the B-side, this is established by Seidel and Thomas. These works are not aimed to solve explicit problems, but they really comes from the philosophy provided by mirror symmetry, and they are of course helpful in solving classical problems in symplectic topology. For example, Arnold's nearby Lagrangian conjecture, see Fukaya-Seidel-Smith's very beautiful paper: http://arxiv.org/pdf/math/0701783v2.pdf.<|endoftext|> TITLE: Orders in Central Simple Algebras. Applications QUESTION [9 upvotes]: It is known that orders in quaternion algebras (over a number field) are used for constructing geometric objects like hyperbolic orbifolds and Shimura curves. Moreover, if one knows embedding properties (selectivity) of the orders one can construct isospectral but non isometric manifolds. I wonder what happen in higher dimensions? I mean, are there known connections between orders in central simple algebras of dimension bigger than 4 and higher dimensional manifolds? is there a known connection with the analytic theory (for instance autmorphic forms)? I appreciate any reference. Thank you! REPLY [6 votes]: I believe that the answer to your question about whether or not connections are known in higher dimensions is yes. Before I get into the details however, it might be useful to recall what happens for quaternion algebras. Let $k$ be a totally real field and $B$ be a quaternion division algebra over $k$ which is split at a unique real place of $k$. This split place, call it $\mathfrak p$, furnishes us with an isomorphism $B_{\mathfrak p} \cong M_2(\mathbf{R})$. Let $\mathcal O$ be a maximal order of $B$ and $\mathcal O^1$ be the multiplicative subgroup consisting of elements with reduced norm $1$. Let $\Gamma_\mathcal O$ denote the image in $PSL_2(\mathbf{R})$ of $\mathcal O^1$. Recalling that $PSL_2(\mathbf{R})$ is the group of orientation-preserving isometries of the hyperbolic plane, the work of Borel and Harish-Chandra imply that the quotient $\mathbf{H}^2/\Gamma_\mathcal O$ is a compact hyperbolic $2$-orbifold of finite area. Note that everything in the last paragraph works equally well for other number fields. If $k$ were a number field with a unique complex place and $B$ was ramified at all real places of $k$, then we would end up getting lattices in $PSL_2(\mathbf{C})$, hence hyperbolic $3$-orbifolds. More generally one gets lattices in products of $PSL_2(\mathbf{R})$ and $PSL_2(\mathbf{C})$. These manifolds (or orbifolds) will only be hyperbolic in the two special cases I mentioned. Everything that I have written thus far generalizes to arbitrary dimensions in the obvious way to yield lattices in groups like $PSL_d(\mathbf{R})$, $PSL_d(\mathbf{C})$ and their products. The manifolds (or orbifolds) one gets will therefore be quotients of the symmetric spaces of these groups. You also asked about the spectral theory of these higher dimensional manifolds and the embedding theory of orders in the relevant division algebra. In this context the selectivity theory you would want to use is due to Luis Arenas-Carmona (http://arxiv.org/abs/1403.5826). Geometrically however, things are a bit trickier. I think that the first person to seriously do anything with the spectral theory of orbifolds arising from these sort of constructions was Marie-France Vignéras (Variétés riemanniennes isospectrales et non isométriques, Ann. of Math. (2) 112 (1980), no. 1, 21–32. ) In this paper Vignéras constructed arbitrarily large families of isospectral but not isometric manifolds with universal cover the symmetric space of products of $PSL_2(\mathbf{R})$ and $PSL_2(\mathbf{C})$. Note that this stuff is also written up in Chapter 12 of Maclachlan and Reid's book "The arithmetic of hyperbolic 3-manifolds". The point is that what is actually going on behind the scenes is that a trace formula is being used. In this case, it follows from the Selberg Trace Formula that two hyperbolic surfaces are isospectral (with respect to the Laplace operator) if and only if they have the same geodesic length spectra (e.g. sets of lengths of closed geodesics). It is the latter that is intimately connected with orders in quaternion algebras, for geodesic lengths on the orbifold obtained from $\Gamma_\mathcal O $ correspond to conjugacy classes of elements in $\Gamma_\mathcal O$ with a given minimal polynomial, which in turn correspond to conjugacy classes of embeddings of certain quadratic $\mathcal O_k$-orders into $\mathcal O$. Thus for $\Gamma_\mathcal O$ and $\Gamma_{\mathcal O'}$ to yield isospectral orbifold quotients, one would need to know that $\mathcal O$ and $\mathcal O'$ admit embeddings of precisely the same set of quadratic orders. And this of course is what the notion of selectivity in quaternion algebras was designed to address. If you wanted to use this sort of method to produce isospectral orbifolds from higher dimensional division algebras and other variants thereof (for instance, central simple algebras equipped with certain types of involutions), you would need a trace formula to tell you that it sufficed to produce orbifolds with the same geodesic length spectra. Unfortunately these trace formulas are only known in a small number of cases and there are definitely cases in which the Laplace spectrum of a (locally-symmetric) manifold determines the lengths which occur in the geodesic length spectrum but not their multiplicities. So you would probably want to study geodesics on your manifolds. This is still doable in this more general setting, but more difficult. Also keep in mind the paper "Division algebras and non-commensurable isospectral manifolds" (http://arxiv.org/abs/math/0501064) by Lubotzky, Samuels and Vishne. In this paper they construct isospectral manifolds from higher dimensional division algebras. In fact they manage to do so without studying orders and embeddings at all. The idea is based upon the fact that a division algebra and its opposite algebra will have the exactly the same maximal subfields but will not be isomorphic when the dimension is greater than $4$. The authors consider lattices in a division algebra and its opposite algebra, and then use the global Jacquet-Langlands correspondence to deduce that their manifolds have equal Laplace spectra. Note that when you use the Vigneras style construction of isospectral manifolds you always get manifolds that are commensurable, hence "less interesting" than the Lubotzky, Samuels and Vishne examples. Of course there are still plenty of very interesting things that one can say about the geometry of the manifolds obtained from orders in higher dimensional division algebras. No references come to mind immediately, though for some inspiration you might try looking at: Chapter 10 of Maclachlan and Reid The papers of Prasad and Rapinchuk. You might start with: http://arxiv.org/abs/0809.2401 and the references therein. Note that Prasad and Rapinchuk are experts on algebraic groups and so work in a lot more generality than you would want to. If you are not an expert in this style of thinking, it might be a good exercise to try to work out exactly what happens when the groups being considered are obtained from division algebras defined over a number field.<|endoftext|> TITLE: Dual of Banach-valued $L^p$ QUESTION [6 upvotes]: Let $X$ be an infinite-dimensional Banach space and let $p\in(1,+\infty)$. We may define $L^p(\mathbb R;X)$. Is it always true that the topological dual of $L^p(\mathbb R;X)$ is $L^{p'}(\mathbb R;X^*)$? Maybe some reflexivity or separability is needed for the Radon-Nikodym argument to work. Is there a simple realization of the dual of $L^1(\mathbb R;X)$? REPLY [9 votes]: Diestel-Uhl, Vector measures, Section IV.1., Theorem 1: Let $(\Omega,\mu)$ be a $\sigma$-finite measure space, $1\leq p<\infty$ and $\frac{1}{p}+\frac{1}{p'}=1$. The dual of $L^p(\mu;X)$ is $L^{p'}(\mu,X^\ast)$ if and only if $X^\ast$ has the Radon-Nikodym property. This is especially the case if $X$ is reflexive.<|endoftext|> TITLE: Surveys of the items of Erdős' "toolbox" QUESTION [10 upvotes]: Could you point out some survey papers and monographs that highlight the kernel of tricks, techniques, and tools that Paul Erdős employed the most in his research work (in particular in graph theory, combinatorics, and number theory)? REPLY [6 votes]: Thank you @Fry, @so-calledfriendDon, and @DanPetersen (see comments) for these interesting references: Graham, Ronald L., Nešetřil, Jaroslav, Butler, Steve (eds.), The Mathematics of Paul Erdős I and II, 2nd edition, Springer, 2013. Lovász, László, Ruzsa, Imre, Sós, Vera T. (eds.), Erdös Centennial, Springer, 2013. Alon, Noga, Spencer, Joel H., The probabilistic method, 2nd edition, Wiley-Interscience, 2000.<|endoftext|> TITLE: Antoine's Necklace and positive Hausdorff/Lebesgue measure QUESTION [5 upvotes]: I have the following question: The usual construction of the Antoine's Necklace produces a Cantor of $1$-dimensional Hausdorff measure in $\mathbb R^3$. I would like to know whether one could adapt the construction to produce a larger Cantor set, namely a antoine's necklace (or similar constructions) with positive $3$-dimensional Hausdorff measure or Lebesgue measure. This looks very plausible for me since we have freedom to determine the sizes of the linked chain torus at each step and topologically the construction can be proceeded as in the usual Cantor cube constructions. If this is already known, precise references are greatly appreciated. Comments and suggestions are also warmly welcome. Thanks. REPLY [2 votes]: It can be done for the very same reason you mention: given a torus, one can find inside it four linked tori of arbitrarily large relative measure. So one can proceed as in the standard construction of a Cantor set of positive Lebesgue measure. More explicitly, fix a sequence $r_n\in (0,1)$ such that $\prod_{n=1}^\infty r_n=1/2$ (say). Start with a torus $T_0\subset\mathbb{R}^3$ of unit volume. Replace $T_0$ by the union $T_1$ of four linked tori $T_{1,1},\ldots, T_{1,4}$ whose union has measure $r_1$. Then replace each $T_{1,i}$ by four linked tori $T_{1,i,j}$ contained in $T_{1,i}$ whose union has measure $r_1 r_2$; let $T_2$ be the union of these 16 tori. If you continue inductively, the intersection $\cap_n T_n$ is an Antoine necklace of measure $1/2$. Edit: I'm not so sure this actually works. One needs to be a little careful to ensure that the diameters of the tori that make up $T_n$ go to zero as $n\to\infty$ (which is needed to ensure the resulting set is totally disconnected).<|endoftext|> TITLE: What is the purpose of the flat/fppf/fpqc topologies? QUESTION [48 upvotes]: There have been other similar questions before (e.g. What is your picture of the flat topology?), but none of them seem to have been answered fully. As someone who originally started in topology/complex geometry, the étale topology makes some sense to me. It's sort of like the complex topology, in that there are enough "open sets" that things like the inverse function theorem work. But I don't really understand what these other more "exotic" topologies represent. From the (naive) background of topology, it sounds like we're just "adding more open sets" to our topology (which I know isn't right, since this isn't a topology in the standard sense). So what do they represent? What do they do for us? REPLY [21 votes]: The question is what kind of structure you are interested in. For example, the étale topology sees more than you might like as a topologist/complex geometer. If you look at Spec(k) where k is not algebraically closed, the étale topology will see the Galois group. As a topologist/complex geometer you might also like the Nisnevich topology. It has enough covers to make a smooth variety look locally like affine space, but it doesn't see the "arithmetic stuff" that only depends on the ground field and not on the variety itself. But if you are interested in the properties of the ground field, too, you will choose the étale topology. As for the flat/fppf/fpqc topologies, I think you already got some very convincing answers. I can't add anything to them. Maybe this is rather an overlong comment than an answer.<|endoftext|> TITLE: Do free profinite groups satisfy Howson's theorem? QUESTION [6 upvotes]: Let $F$ be a free profinite group, and let $A,B \leq F$ be finitely generated closed subgroups. Must $A \cap B$ be finitely generated? REPLY [7 votes]: The following example shows that the answer is no. Let $p$ and $q$ be two different primes. First I want to construct two generated profinite group $G=A\ltimes H$ isomorphic to semi direct product of a infinitely generated free pro-$p$ group $H$ and a cyclic free pro-$q$ group $A=\langle a \rangle$. It can be done, for example, in the following way. Start with a two generated free profinite group $F$ and a normal subgroup $N$ such that $F/N\cong A$. Let $H=N/N_p$ be the maximal pro-p quotient of $N$. Clearly $H$ is infinitely generated. Then $G=F/N_p\cong A\ltimes H$ satisfies the requiered conditions. Now, let $T=$ be a two generated free pro-$q$ group. Define its action on $H$ in such way that $g_1$ and $g_2$ act as it does $a$. Form a semidirect product $T\ltimes H$ and put $G_i= $ ($i=1,2$). It is clear that $G_i\cong G$ are two generated. $T\ltimes H$ is a 3-generated profinite group. Since its Sylow subgroups are free, $T\ltimes H$ has cohomological dimension 1, and so it is a subgroup of a free profinite group. However $G_1\cap G_2=H$ is not finitely generated.<|endoftext|> TITLE: How to make the Capelli's identity less mysterious? QUESTION [29 upvotes]: The formulation of the Capelli's identity is very elementary; it has important applications in invariant theory and representation theory, see http://en.wikipedia.org/wiki/Capelli%27s_identity To remind it, it requires some notation. Let $x_{ij}$, $1\leq i,j\leq n$, be commuting variables. Define differential operators on the space of functions on $n\times n$ matrices: $$E_{ij}=\sum_{a=1}^n x_{ia}\frac{\partial}{\partial x_{aj}},\, 1\leq i,j\leq n.$$ The Capelli identity states that $$\det\left[\begin{array}{cccc} E_{11}+n-1&\dots&E_{1,n-1}&E_{1n}\\ \vdots&\vdots&\dots&\vdots\\ E_{n-1,1}&\dots&E_{n-1,n-1}+1&E_{n-1,n}\\ E_{n,1}&\dots&E_{n,n-1}&E_{n,n}+0 \end{array}\right]=\\\det\left[\begin{array}{ccc} x_{11}&\dots&x_{1n}\\ \dots&\dots&\dots\\ x_{n1}&\dots&x_{nn} \end{array} \right]\cdot \det\left[\begin{array}{ccc} \frac{\partial}{\partial x_{11}}&\dots&\frac{\partial}{\partial x_{1n}}\\ \dots&\dots&\dots&\\ \frac{\partial}{\partial x_{n1}}&\dots&\frac{\partial}{\partial x_{nn}} \end{array}\right]. $$ Note that in the right hand side of the equality the two matrices have commuting entries, while in the left hand side the entries do not commute. Hence one has to be careful to define the determinant. The convention for the determinant of such matrices is $$\det(a_{ij})=\sum_{\sigma\in S_n}sgn(\sigma) a_{1\sigma(1)}a_{2\sigma(2)}\dots a_{n\sigma(n)},$$ where the order of terms is important. This order of terms in the determinant is the most mysterious point for me. Is there any reason for it? What happens if one chooses some other ordering of terms: will the Capelli identity be modified somehow or it will not work at all? There are more recent generalizations of the Capelli identity, see e.g. the above link to Wikipedia. However unfortunately they do not clarify to me the original Capelli idenity, but rather use it as a basic inspiration for further extensions (may be I am missing something). UPDATE. So far the most conceptual approach to the Capelli identity I was able to find in the literature is due to I. Gelfand and V. Retakh, Funktsional. Anal. i Prilozhen. 25 (1991), no. 2, 13--25 (Section 3.4). This is their first paper on their general theory of non-commutative determinants. They use it to rewrite the Capelli identity in their language. I have not studied their proof in detail, but apparently they do not use this strange and seemingly arbitrary definition of det as a sum over all permutation of terms with prescribed ordering. I still have to understood how difficult and natural it is to pass from their language to the classical one. REPLY [8 votes]: I am somewhat late for a year, sorry, but since the wiki-article is mainly written by me and I somehow worked on the subject, it is difficult to resist writing an answer. Currently there is some understanding about non-commutative determinants which makes identity less mysterious: Point 1. There are determinants (and whole linear algebra) for very special classes of matrices with noncommutative elements. Although for generic matrices with non-commutative elements there is NO distingueshed determinant, however for some special classes of matrices with noncommuting entries which are in certain sense not very far from commutativity, there are natural definitions of the determinant and moreover many theorems of linear algebra can be naturally extended to such matrices. Probably q-determinant for quantum groups is the most well-known example. Point 2. Capelli matrix "E" is of that special type. The matrix "E" actually of this type, i.e. it can be related to "Manin matrices", although this relation is not unique. For Manin matrices everything is perfect: determinant, linear algebra ... everything ... - clear concept with no mysteries. Point 3. In particular your questions: the determinant is the most mysterious point for me. Is there any reason for it? For Manin matrices there are clear and simple reasons, for Capelli matrix "E" we can obtain the determinant above from determinant for Manin. (If you find relation Capelli <-> Manin "natural", you might not - look below). What happens if one chooses some other ordering of terms: will the Capelli identity be modified somehow or it will not work at all? For Manin matrices defining determinant you use arbitrary ordering. It implies that there is control on what happens when you change ordering in the determinant for "E". Let me sketch how to relate Capelli matrix $E$ with Manin matrices. We must introduce auxiliary formal variable "u" And consider: $$ E(u) = E + u Id$$ So we can restore $E$ by taking $u=0$. That is not a Manin matrix, but it is almost Manin matrix. The idea that is that sometimes matrix $T_0$ may not be Manin, but it is possible to find corrections $T_i$, such that $T(u) = T_0+ u T_1 + u^2 T_2 + ...$, would be (almost) Manin matrix. The Capelli case is the most simple where correction is the most simple $T_1 = uId$. Finally I need to explain what I mean saying "almost Manin matrix", short way of saying it is that if one introduces new element $K = exp(-d/du)$, which commute with $u$ as a shift operator $K f(u) = f(u+1) K$, then the matrix $K E(u)$ would be Manin matrix. Longer, but more explicit way is look on the defining relations for Manin matrices, for example column-commutativity: $ac = ca$ are substituted by $a(u+1) c(u) = c(u+1) a(u) $ i.e. shift in $u$ to $u+1$ is introduced.<|endoftext|> TITLE: An example of two cofibrant dg categories whose tensor product is not cofibrant QUESTION [9 upvotes]: I have been reading the paper by Toën "The homotopy theory of dg categories and derived Morita theory" where in chapter 4 it is stated that the tensor product of two cofibrant dg categories $C$ and $D$ (defined by $obj(C\otimes D)=objC\times obj D$ and with space of morphisms obtained by tensoring the two chain complexes in $C$ and $D$) is not in general cofibrant, and this is a problem in both the "standard" and the Morita model structures on the category $\mathbf{dgCat}_k$ (here $k$ is a commutative ring with 1 and with arbitrary characteristic). It is also known (proposition 2.3 (2) in the same article) that a cofibrant replacement functor can be chosen in such a way that it is the identity on objects, therefore the problem seems to lie exclusively in what happens to morphisms. The same proposition contains the result that a cofibrant dg category has morphisms spaces which are cofibrant complexes wrt the projective model structure on $\mathbf{Ch}(k)$. A cofibrant chain complex $X_\bullet$ has $X_n$ projective $\forall n$ and the converse is true only if $X_\bullet$ is bounded below. Nevertheless, the tensor product (and the direct sum) of two projective modules is again projective so I think I am missing some essential point: given $X_\bullet$ and $Y_\bullet$ cofibrant chain complexes, $(X\otimes Y)_\bullet$ will be made of projective modules, so I think I need to find an unbounded cofibrant complex which tensored with some other cofibrant complex loses cofibrancy. This result was cited in several other places, but I was unable to find even a discussion longer than the statement. I would be glad if someone could give me such an example, or point out some reference where I can find it. Thank you in advance REPLY [7 votes]: You can consider DGAs, which are DG-categories with only one object. The polynomial ring $k[x]$ concentrated in degree $0$ (with trivial differential) is cofibrant, since it is free as a (graded) $k$-algebra. Take another copy, $k[y]$. The tensor product $k[x]\otimes k[y]=k[x,y]$ is not cofibrant as a DGA. Indeed, you can construct a cofibrant resolution $C\stackrel{\sim}\twoheadrightarrow k[x,y]$ such that $C_0=k\langle x,y\rangle$ is a free non-commutative $k$-algebra on the same two generators, and if $k[x,y]$ were cofibrant, it would be a retract of $C_0=k\langle x,y\rangle$ by the lifting axiom, but that's impossible because $k[x,y]$ should then also be free as a non-commutative $k$-algebra. REPLY [2 votes]: Let $\Delta^1_k$ be the $k$-linear dg-category with two objects $0$ and $1$, mapping complexes $$ Map(0,0) = [k], $$ $$ Map(0,1) = [k], $$ $$ Map(1,0) = [0], $$ $$ Map(1,1) = [k] $$ where $[k]$ means the complex with $k$ in degree zero, and a composition law defined in the obvious way. This is a cofibrant dg-category, and one can show that the tensor product $$ \Delta^1_k \otimes \Delta^1_k $$ is not cofibrant. For an argument, see Exercise 3.2.2 in [Bertrand Toën, Lectures on DG-categories, in "Topics in algebraic and topological K-theory", pp. 243-302 (2011)], available here (labelled as Exercise 14 in the latter pdf).<|endoftext|> TITLE: Can a Brownian motion be fast at its extrema? QUESTION [5 upvotes]: After pondering this MO question > Location of maximum of Brownian motion with rough drift <, I wonder whether a Brownian motion can be fast (i.e. beats the law of the iterated logarithm) at its extrema? Is it necessarily fast? REPLY [4 votes]: Heuristically, a point $(t,B_t)$ being a extremum is antithetical to fast oscillation, since there is no oscillation on one side of (above/below) $B_t$. However, This is only a heuristic, and One may wonder whether there are so many fast times and so many maxima as to force an overlap. At least we can rule out (2) as follows: Theorem. If $W$ and $B$ are two independent Brownian motions then the set of maxima of $W$ is disjoint from the set of fast times of $B$. Proof: Let $M$ be the set of local maxima of $W$ on a given closed interval. Then $M$ is a union of finite discrete, hence closed, sets $M_n$, where $M_n$ consists of those $t$ such that $B_s\le B_t$ for all $s$ with $|s-t|\le 1/n$. Since $M_n$ is countable, the packing dimension $d_P(M_n)=0$. So by Theorem 10.22 of http://research.microsoft.com/en-us/um/people/peres/brbook.pdf we have almost surely that $$\sup_{t\in M_n} \limsup_{h\downarrow 0} \frac{|B_{t+h}-B_t|}{\sqrt{2h\log(1/h)}}=\sqrt{d_P(M_n)}=0.$$ Given $a>0$ we call a time $t\in [0,1]$ an $a$-fast time if $$\limsup_{h\downarrow 0} \frac{|B_{t+h}-B_t|}{\sqrt{2h\log(1/h)}}\ge a.$$ And a time is fast if it is $a$-fast for some $a>0$. Thus, a.s. no fast time of $B$ is a local maximum of $W$.<|endoftext|> TITLE: Expectation of trace of nth power of unitary matrices QUESTION [8 upvotes]: I am trying to find the answer of $$\int dU \ |Tr(U^m)|^2$$ where $m\in\mathbb{N}$ and $U$'s are unitary matrices in $\textit{U}(n)$ and $dU$ is a normalized Haar measure. In the case $m=1$, the answer seems to be $1$. I don't know where to start. Does anyone have an idea? Is there a clean answer for $m>1$ ? REPLY [3 votes]: This is an elaboration on my comment. Let's start in a more general setting since the comment was not about the groups $U(n)$: Let $G$ be a compact group and $H$ a normalized Haar measure on it. Let $p_m:G\to G$ be defined by $p_m(x)=x^m$, $m\in\mathbb Z$. Now for any continuous function $f:G\to\mathbb R$ we have $$ \int_Gf(p_m(x))dH(x) = \int_Gf(x)d(p_m)_*H(x). $$ Since $p_m$ is a homomorphism1, the pushforward $(p_m)_*H$ is a normalized Haar measure on $p_m(G)$. If $p_m$ is surjective, it follows from the uniqueness of Haar measures that $p_{m*}H=H$ and thus $$ \int_Gf(p_m(x))dH(x) = \int_Gf(x)dH(x). $$ This idea made me think that the integral would be independent of $m$. 1 The map is not a homomorphism. The argument works (I think) for homomorphisms, but typically $p_m$ is not a homomorphism in a nonabelian group. If $p_m$ is not a homomorphism, the pushforward doesn't have to be a Haar measure and the proof falls apart. Maybe I'll let this answer stay here as a warning example...<|endoftext|> TITLE: de Rham cohomology of smooth affine varieties QUESTION [9 upvotes]: Let $U$ be a smooth variety over $\mathbb{C}$. We know that there exists a smooth compactification $X$ such that $X-U$ is a normal crossings divisor $D$ and that the de Rham cohomology of $U$ can be computed using the complex of sheaves with logarithmic differentials on $X$: $$ H_{dR}^\ast(U)=\mathbb{H}^\ast(X, \Omega^\bullet_X(\log D)) $$ In general, one needs hypercohomology here but I am wondering if, when $U$ is affine, one simply has $$ H^\ast_{dR}(U)=H^\ast(\Gamma(\Omega^\bullet_X(\log D)) $$ I know that this is true if one uses instead the de Rham complex $\Omega^\bullet_U$ to compute cohomology and I know that $j_\ast \Omega_U^\bullet$ and $\Omega_X^\bullet(\log D)$ are quasi-isomorphic (here $j: U \hookrightarrow X$). Is this enough to conclude? REPLY [4 votes]: The de Rham cohomology of an affine variety $U$ is simply $H^\bullet(\Gamma(U,\Omega_U^\bullet))$ where $\Omega_U^\bullet$ is the sheaf of algebraic forms. See Grothendieck (1966), « On the de Rham cohomology of algebraic varieties », Inst. Hautes Études Sci. Publ. Math., (29), p. 96. In general, you cannot get rid of hypercohomology AND use logarithmic forms only.<|endoftext|> TITLE: A possibly surprising appearance of Lucas numbers QUESTION [17 upvotes]: Let $S$ be the set of polynomials defined as follows: $0$ is in $S$, and if $p$ is in $S$, then $p + 1$ is in $S$ and $x \cdot p$ is in $S$, so that $S$ "grows" in generations: $g(0)=\{0\}$, $g(1)=\{1\}$, $g(2)=\{2,x\}$, $g(3)=\{3,2x,x+1,x^2\}$, and so on, with $|g(n)|=2^{n-1}$. Let $S^*$ be the set obtained from $S$ by substituting $r=\sqrt{2}$ for $x$ and keeping only the first appearance of each duplicate. Successive generations $G(n)$ now begin with $\{0\}$, $\{1\}$, $\{2,r\}$, $\{3,2r,r+1\}$, with $|G(n)|$ starting with $1,1,2,3,4,7,11,18,29,...$; i.e., Lucas numbers beginning at the 4th term, and checked for 30 generations. Can someone prove that $|G(n)|=L(n-1)$ for $n \geq 4$? REPLY [15 votes]: For a number of the form $a + b\sqrt{2}$ with nonnegative integers $a$ and $b$, define its length to be the minimal number of steps needed to obtain it from zero, where the allowed steps are $x \mapsto x+1$ and $x \mapsto \sqrt{2}x$. Then the problem asks to count the numbers of given length. Now it should be easy to show the following by induction: If $a$ is odd, then the last step must be $x \mapsto x+1$ (this is clear). If $a$ is even, then the last step can be taken to be $x \mapsto \sqrt{2}x$, unless $a + b\sqrt{2} = 2$. EDIT: Here is a proof. Let $\ell(x)$ be the length of $x$. Lemma. Write $x = a_0 + a_1 \sqrt{2} + a_2 \sqrt{2}^2 + \ldots + a_k \sqrt{2}^k$ with $a_j \in \{0,1\}$ and $a_k = 1$. If $x > \sqrt{2}$, then $\ell(x) = k + \#\{j : a_j = 1\} - (1 - a_{k-1})$. Proof. Write $\ell'(x) = k + \#\{j : a_j = 1\} - (1 - a_{k-1})$. The proof is by induction on $x$ (the set of all possible $x$ has the order type of the natural numbers). We have $\ell(1 + \sqrt{2}) = 3 = \ell'(1 + \sqrt{2})$. So assume $x \ge 2$. If $a_0 = 1$, then $\ell(x) = \ell(x-1) + 1 = \ell'(x-1) + 1 = \ell'(x)$ (this needs that $0 \neq k-1$, so that changing $a_0$ from 1 to 0 does not affect $a_{k-1}$). If $a_0 = 0$, then $\ell(x) \le \ell(x/\sqrt{2}) + 1 = \ell'(x/\sqrt{2}) + 1 = \ell'(x)$, so we need to show that $\ell'(x-1) \ge \ell(x/\sqrt{2}) = \ell'(x) - 1$. We can assume $x > 2$, since the claim of the lemma holds for $x = 2$. The rational part (denoted $a$ above) of $x$ must be at least 2 (it must be positive, otherwise $x-1$ is not of the required form, and even since $a_0 = 0$). Then in $x-1$, either $k$ stays the same and we have the same or a larger number of 1's among the $a_j$, which shows $\ell'(x-1) \ge \ell'(x) - 1$. (Note that $a_{k-1}$ could change from 1 to 0.) Or else $k$ gets smaller; then $a = 2^m$ and $k = 2m$ in $x$, and $a = 2^m-1$, $k \le 2m-1$ in $x-1$. So we reduce $k$ by at least 1, but increase the number of 1's by $m-1$. Note also that $a_{k-1}$ can only change from 1 to 0 if $k$ goes down by 2. So $\ell'(x-1) \ge \ell'(x) + m - 2 \ge \ell'(x) - 1$ as before. $\Box$ Assuming this, we can arrange our numbers $a + b\sqrt{2}$ into a rooted tree with zero at the root and where an odd number (meaning odd $a$) has only one (even) child, whereas an even number has two children, one odd and one even. There are two exceptions: the odd node $1$ has the two children $2$ and $\sqrt{2}$, and the even node $\sqrt{2}$ has only one child $1 + \sqrt{2}$. So the $n$th level of the tree, for $n \ge 3$, consists of the $n$th level of the usual Fibonacci tree (think of the rabbits) (with root $1$), together with the ($n-2$)nd level of the Fibonacci tree (with root $1 + \sqrt{2}$). This gives $F_n + F_{n-2} = L_{n-1}$ for the number of nodes at distance $n$ from the root (the root has level $1$). REMARK. Replacing $\sqrt{2}$ by $2$ gives a somewhat simpler problem that leads to the Fibonacci tree (starting at $1$) and gives Fibonacci numbers. The golden ratio $(1+\sqrt{5})/2$ gives the sequence $(1,2,3,5,8, 12, 18, 25, 35, 51,\ldots)$ (starting at generation 1), apparently satisfying $a_n = a_{n-1}+a_{n-3}$ for $n \ge 12$. [Previously, I claimed that one gets Fibonacci numbers, but this is wrong.] If we take any number $r > 1$ instead of $\sqrt{2}$, then I would expect that for any $x \in R = {\mathbb Z}_{\ge 0}[r]$ that is divisible by $r$ and sufficiently large, the shortest path to zero is via division by $r$. For any other $x$, one is forced to go via $x-1$. So in the relevant tree, if we are at a sufficiently high level, nodes $x$ such that $x+1$ is divisible by $r$ in the semiring $R$ have one child $rx$ and the other nodes have two children $x+1$ and $rx$. I would expect this to lead to a linear recurrence with constant coefficients, but since we are dealing with a semiring and not with a ring (where we could consider the quotient by the ideal generated by $r$), this is not clear to me.<|endoftext|> TITLE: What is the difference between p-adic Lie groups and linear algebraic groups over p-adic fields? QUESTION [14 upvotes]: I thought they were the same, just different names. Let me make question more precise: Let $G$ be any linear algebraic group over a p-adic field $\mathbb{Q}_p$, is $G$ a p-adic Lie group w.r.t. the analytic topology from $\mathbb{Q}_p$ in the sense of Peter Schneider? If this is the case, Does the Lie algebra from the algebraic group coincide with the Lie algebra from the Lie group? As far as I can see this is true for real number case. But I'm not familiar with p-adic Lie group theory. p-Adic Lie Groups: Peter Schneider: http://books.google.de/books?id=bjWU3GF93YQC&printsec=frontcover&dq=p-adic%20lie%20groups&hl=de&sa=X&ei=Ml83UcOILpS-9gSLnICYDA&ved=0CDQQ6AEwAA#v=onepage&q=p-adic%20lie%20groups&f=false REPLY [14 votes]: The development of both Lie groups and linear algebraic groups is rather complicated, starting with the definitions over various fields. For example, working over $\mathbb{Q}_p$ or a finite extension is one classical setting, and many of the ideas (though not all) carry over well to "local fields" in prime characteristic. While ACL and Venkatarama have correctly pointed to Serre's lectures for a conventional treatment of Lie groups over local fields (including $\mathbb{R}$ and "ultrametric" fields), it may help to look more broadly at the development of the ideas over time. Chevalley set out to write a six volume series of books on Lie groups (and "linear algebraic groups"), but abandoned that after three books in order to develop an improved theory of linear algebraic groups using a recent version of algebraic geometry. But he had already realized that compact (real) Lie groups carry a natural algebraic group structure (which yields the same Lie algebra), even though most other Lie groups do not. His student Robert Hooke at Princeton wrote a thesis in 1942, soon published in the Annals here (available through JSTOR). In this work the standard dictionary between Lie groups and Lie algebras is adapted to $p$-adic groups. In the next decades Bourbaki (no doubt actively influenced by Serre) began to issue chapters of their treatise Groupes et algebres de Lie, later published in English translation by Springer. Chapter II (Hermann, 1972) lays the foundations for a unified theory of Lie groups and their Lie algebras over $\mathbb{R}, \mathbb{C}$, and complete local fields; sometimes the field is required to be of characteristic 0, so one has to be aware of this. It's very useful to consult their historical notes on Chapters I-III. Already Serre had lectured at Harvard in 1964 on Lie algebras and Lie groups, borrowing some of the not yet published Bourbaki approach. These lectures were published by W.A. Benjamin (1965) and later reissued as Springer Lecture Notes 1500 here. Though the theory of $p$-adic Lie groups is now well grounded and to some extent unified with traditional Lie group theory, it remains at some distance from linear algebraic groups. However, as in the real and complex Lie group cases, there are much closer connections when the groups are connected and reductive: this has led to a rich literature, including the work of Iwahori-Matsumoto and Bruhat-Tits on structure theory over local fields. By the time Conrad-Gabber-Prasad wrote their recent book Pseudo-reductive Groups, the group scheme approach allowed further refinements in the study of reductive groups, especially in prime characteristic. (But these developments move farther away from the traditional study of manifolds and Lie groups, including the close relationship between Lie groups and their Lie algebras.)<|endoftext|> TITLE: Co-Hausdorffification QUESTION [8 upvotes]: Given a topological space $(X,\tau)$ we can define the "$T_2$-ification" of $X$ by setting $T_2(X,\tau) = X/\simeq$ where $x\simeq y$ in $X$ if and only if for every open neighborhood of $x$ has nonempty intersection with every open neighborhood of $y$. Then $T_2(X,\tau)$ has the following universal property: For every $T_2$-space $Z$ and continuous function $f: X\to Z$, there is $\bar{f}: T_2(X,\tau) \to Z$ such that $f = \bar{f} \circ pr$ where $pr: X\to X/\simeq$ is the canonical projection. A co-$T_2$-ification would be a space with a similar property as above, but all arrows reversed. (I hope this description is clear enough.) Does every space have a co-$T_2$-ification? (There are similar constructions for $T_i$-ifications at least for $i\in \{0,1\}$, so the same question could be asked for those.) REPLY [13 votes]: While johndoe gave the right answer to the question I believe you meant to ask, the question that you did ask is slightly different and also has a negative answer. Specifically, johndoe's answer addresses the version of your question in which the factorization $\bar{f}$ is required to be unique. However, even if you do not require the factorization to be unique, then a co-Hausdorffification still does not exist. (One might call such a thing a weak co-Hausdorffification, by analogy with the term weak limit.) Specifically, let $X=\{0,1\}$ be the Sierpinski 2-point space (with $1$ closed), and suppose $f:T\to X$ is a map from a Hausdorff space to $X$. Then I claim that there exists a Hausdorff space $Z$ and a map $g:Z\to X$ that does not factor through $f$. To show this, let $\kappa$ be any ordinal of cofinality greater than $|T|$, let $Z=\kappa+1$, and let $g:Z\to X$ send $\kappa$ to $1$ and everything else to $0$. Suppose $h:Z\to T$ is such that $fh=g$. Since $\operatorname{cf}(\kappa)>|T|$, there is an unbounded set $S\subseteq\kappa$ on which $h$ is constant. By continuity of $h$, we must then have $h(\alpha)=h(\kappa)$ for all $\alpha\in S$. But then $g(\alpha)=f(h(\alpha))=f(h(\kappa))=g(\kappa)$ for all $\alpha\in S$, a contradiction. (In the language of the adjoint functor theorem, this is saying that a co-Hausdorffification fails to exist not only because Hausdorff spaces are not closed under coequalizers, but also because the solution set condition fails. Nevertheless, it turns out that the (large) colimit of all Hausdorff spaces mapping to $X$ does exist in the category of all spaces: it is just $X$ itself. This is not hard to prove; it essentially amounts to showing that any topology is determined by which maps from Hausdorff spaces are continuous.) Update: As johndoe noted, this argument also rules out a weak co-$T_1$-ification and a weak co-$T_3$-ification, or more generally a weak co-$P$-ification if $P$ is a property which implies $T_1$ and is satisfied by any ordinal under the order topology ($T_1$ is the lower bound because my argument required $T$ to be $T_1$ in order to conclude that $h(\alpha)=h(\kappa)$ for $\alpha\in S$). Here is a slightly different argument that shows there is no weak co-$T_0$-ification: Let $X=\{0,1\}$ with the indiscrete topology, and suppose $f:T\to X$ is a map from a $T_0$ space to $X$. Then I claim there is a $T_0$ space $Z$ and a map $g:Z\to X$ that does not factor through $f$. As before, let $\kappa$ be an ordinal of cofinality greater than $|T|$; let $Z$ be $\kappa$ equipped with the Alexandrov topology. Let $g:Z\to X$ be the characteristic function of any unbounded and co-unbounded subset of $\kappa$. Suppose $h:Z\to T$ is such that $fh=g$; then as before, $h$ must be constant on an unbounded set. But $h$ must preserve the specialization order, and so since $T$ is $T_0$ this implies that $h$ is eventually constant. Since $g$ is not eventually constant, this is a contradiction.<|endoftext|> TITLE: Homogeneous polynomial vector fields tangent to the unit sphere QUESTION [5 upvotes]: This question has something to do with that one. Let $n\ge1$ and $d\ge1$ be two given integers. Consider the polynomial vector fields $v=(v_1,\ldots,v_n)$ whose components $v_j$ are homogeneous of degree $d$ in the indeterminates $X_1,\ldots,X_n$. What is the dimension $D(n;d)$of the subspace defined by the equation $$X_1v_1(X)+\cdots+X_nv_n(X)=0\quad ?$$ I computed this dimension for small dimensions: $D(1;d)=0$, $D(2;d)=d$ and $D(3;d)=d(d+2)$. I suspect that the problem has been solved a while ago and the formula is simple in terms of binomials. A solution might come by considering a the sequence of morphisms $$\cdots\rightarrow\Lambda_{n-2}({\mathbb R}^n)\otimes {\rm Hom}_n^{d-1}\rightarrow\Lambda_{n-1}({\mathbb R}^n)\otimes {\rm Hom}_n^{d}\rightarrow\Lambda_{n}({\mathbb R}^n)\otimes {\rm Hom}_n^{d+1},$$ where ${\rm Hom}_n^d$ denotes the space of homogeneous polynomials of degree $d$, and each arrow is of the form $V\mapsto X\wedge V$. REPLY [4 votes]: Isn't the answer just $$ D(n;d) = n{{n+d-1}\choose{d}}- {{n+d}\choose{d+1}}= d{{n+d-1}\choose{d+1}}\quad ? $$<|endoftext|> TITLE: Examples of Brody hyperbolic affine varieties which are not Kobayashi hyperbolic QUESTION [13 upvotes]: Let $X$ be a complex space. We say that $X$ is Brody hyperbolic if there is no non-constant holomorphic map $f\colon\mathbb C\to X$. We say that $X$ is Kobayashi hyperbolic if the Kobayashi pseudo-distance $d_X$ on $X$ is a genuine distance. Since holomorphic maps are distance-decreasing withe respect to the Kobayashi pseudo-distance, and since $d_\mathbb C\equiv 0$, we immediately see that any Kobayshi hyperbolic complex space is also Brody hyperbolic: if $f\colon\mathbb C\to X$ is any holomorphic map then $$ d_X(f(z),f(w))\le d_\mathbb C(z,w)=0, $$ so that $f$ must be constant, $d_X$ being a true distance. Now, by Bordy's reparametrization lemma, if $X$ is moreover compact, then we also have that Brody hyperbolicity implies Kobayashi hyperbolicity. But in the non-compact case it is well known that Kobayashi hyperbolicity is actually stronger. A standard example is given by the domain $X$ in $\mathbb C^2$ described as follows: $$ X=\{(z,w)\in\mathbb C^2\mid|z|<1,|zw|<1\}\setminus\{(0,w)\mid|w|\ge 1\}. $$ This domain $X$ can be quite straightforwardly seen to be not Kobayashi hyperbolic (the origin has zero distance from points of the form $(0,b)$) but nevertheless there is no non-constant holomorphic map $f\colon\mathbb C\to X$. Another exemple of algebraic nature (a reference of which can be found in S. Lang's book "Introduction to complex hyperbolic spaces" as pointed out to me by W. Cherry) is the following. Take four lines in general position in $\mathbb P^2$ and let $\ell_1, \ell_2,\ell_3$ the diagonal lines of this configuration. Now, choose three distinct points $P_1,P_2,P_3$, such that $P_i\in\ell_i$ but $P_i$ is not on the starting four lines. Then, $$ W:=\mathbb P^2\setminus\{\text{the four lines}\}\setminus\{P_1,P_2,P_3\} $$ is Brody hyperbolic but not Kobayashi hyperbolic. Of course, $W$ is quasi-projective, but not affine. Here is my question (which is indeed already asked by S. Lang right after the above example): Question 1. Is there an example of a complex affine variety $X$ which is Brody hyperbolic but not Kobayashi hyperbolic? Or do you at least know a Stein example? If such an example exists, then any compactification $\overline X$ of $X$ is not Kobayashi hyperbolic and so must contain non-constant holomorphic images of the complex plane. But then, all these images must cut the boundary $\overline X\setminus X$ of $\overline X$ in at least two points... Possibly, a natural place to look at, might be the complement of an ample divisor in an abelian variety: by a result of G.-E. Dethloff and S. S.-Y. Lu this complement is always Brody hyperbolic. Moreover, it is of course affine. So, Question 1 would be answered in the affirmative if one were able to answer positively to the following. Question 2. Is there an example of a pair $(A,D)$, where $A$ is an abelian variety and $D\subset A$ an ample divisor, such that $A\setminus D$ is not Kobayashi hyperbolic? Thanks in advance! REPLY [6 votes]: Here is an answer to the second part of the first question. It has been communicated to me by Leandro Arosio. The answer is: yes, there does exist a Stein manifold which is Brody hyperbolic but not Kobayashi hyperbolic (it is constructed as a pseudoconvex domain in $\mathbb C^2$). The example can be found in this paper by T. J. Barth.<|endoftext|> TITLE: Sets of points containing permutations - a Ramsey-type question QUESTION [16 upvotes]: The following question arised as a side-question in a geometric problem. It has a "feel" similar to problems in Ramsey-theory, but I have not found any mention of it (also I'm not very familiar with the field). Was this problem considered before? Does it have an easy answer? Consider the set of grid points $[n] \times [n]$, and color each point either black or white, giving rise to the sets $B,W$ (such that $B \cup W = [n] \times [n]$, and $B \cap W = \emptyset$). Are the following true? (stronger): Either $B$ or $W$ contains every permutation of $[n/2]$. (weaker, implied by stronger): Every permutation of $[n/2]$ is contained in either $B$ or $W$. if true, holds also for $k$ colors and $n/k$? if not true, what is largest $m$ for which it holds? A set of points $X$ containing a permutation $\sigma$ of $[n]$ means that: there are points $(x_1, y_1), \dots, (x_n,y_n) \in X$, such that $y_1 TITLE: Statistics of strongly connected components in random directed graphs QUESTION [7 upvotes]: I'm interested in the statistics of strongly connected components in random directed graphs. However, I'm unable to find any results on this, partly because I don't know the terminology to search for. By "random directed graph" I mean a digraph with $N$ nodes, where every possible edge has a uniform probability $p$ of being included in the graph, independently of the other edges. For undirected graphs this is sometimes known (possibly erroneously) as the Erdös-Rényi model, but for digraphs I don't know the correct term. Equivalently, given a random (in general non-symmetric) matrix where every element has an independent probability of being non-zero, one can do a change of basis using a permutation matrix to put it in block upper triangular form; I'm interested in the statistics of the number and size of the resulting blocks on the diagonal. I'm most interested in knowing the expected number of strongly connected components of size $>1$, for a given $N$ and $p$, but any results concerning the statistics of the strongly connected components (e.g. the size of the largest one) would be enormously helpful. I'm interested primarily in the limit of large $N$, for non-extreme values of $p$. REPLY [2 votes]: First, this model is typically denoted $D(n,p)$. I'm not aware of a formal name, but personally I like to call it the Bernoulli random digraph. For $p=c/n$, note that the number of vertices with out-degree zero is going to be on the order of $n$. So in this range, the number of strong components will also be on the order of $n$. The phase transition for this model where a large strongly connected component emerges is due to Richard Karp and Tomasz Łuczak (Łuczak considers the analogous model $D(n,m)$ where $m$ directed edges are uniformly at random chosen to be present). Karp shows that for $p=c/n$ and $c<1$, and any $\omega \to \infty$ however slowly, w.h.p. each strong component of $D(n,p)$ has at most $\omega$ vertices. For $c>1$, there is a unique strong component with roughly $\Theta^2 n$ vertices where $\Theta$ is the unique positive root of $1-\Theta = e^{-c \Theta}$; in this case, w.h.p. all other strong components contain at most $\omega$ vertices. It is not a coincidence that the number of vertices in the giant component of $G(n,p=c/n)$ is roughly $\Theta n$ vertices. Boris Pittel and I recently proved that for $D(n, p=c/n), c>1$, the number of vertices in the largest strong component is asymptotically normal. In fact, we show that the pair of the numbers of vertices and arcs in the strong giant is jointly asymptotically 2-dim normal.<|endoftext|> TITLE: How to calculate the sum of remainders of N? QUESTION [9 upvotes]: I'm trying to sum the remainders when dividing N by numbers from $1$ up to $N$ $$\sum_{i = 1}^{N} N \bmod i$$ It's easy to write a program to evaluate the sum if N is small in $O(N)$ but what if N is large ~ 1e10 so I was wondering if there is a formula or an algorithm to get the sum in a better way than $O(N) $. I searched online and found a paper which drives some interesting properties of the sum yet doesn't state an efficient way to compute it although it contains a recursive formula yet that formula isn't practical since it reduced the problem from $N$ to $N-1$ and the relationship contains $\sigma (n)$ which can be evaluated at best in $O$(#prime factors of $N$) resulting in a much worse time than $O(N)$ on the other hand the paper contains the formula $$ \sum_{i = 1}^{N} N \bmod i = N^{2} - \sum_{i = 1}^{N} \sigma (i) $$ so if I can evaluate the sum of the sum-of-divisors function I would get the answer, yet I searched online and found only this discussion which had a formula for the sum of the sum-of-divisors function yet it's not exact and contains the sum of fractions which introduce rounding errors so I was wondering if there is a way to evaluate any of the two sums exactly and in a better time complexity than $O(N)$. REPLY [6 votes]: I found an article from Richard Sladkey, "A Successive Approximation Algorithm for computing the Divisor Summary Function", June 18, 2012 arXiv:1206.3369v1 [math.NT] 15 June 2012 where he shows that the solution can be computed in $\mathcal{O}(N^{1/3})$ steps. The paper introduces restricted divisor sums for part of this calculation.<|endoftext|> TITLE: Formal group law over $\mathbb{F}_p$ QUESTION [15 upvotes]: Let $p$ be a prime. For each $n > 0$ there is a unique 1-dimensional commutative formal group law $F$ over $\mathbf{Z}$, $F(X, Y) = X + Y + \dots \in \mathbf{Z}[[X, Y]]$, whose logarithm function is given by $$l(x) = \sum_{k \ge 0} \frac{x^{p^{nk}}}{p^k}.$$ Let $\bar{F} \in \mathbf{F}_p[[X, Y]]$ be the formal group over $\mathbf{F}_p$ given by reduction of $F$ modulo $p$. (Cf. Prop. 9.25 in http://neil-strickland.staff.shef.ac.uk/courses/formalgroups/fg.pdf.) Is $\bar{F}$ an element of $\mathbf{F}_p[X] [[Y]]$? Thank you for your answers. REPLY [19 votes]: Now multinomial-free, I believe that Ghassan Sarkis and I have a proof of the following Theorem. Let $h\ge2$, and let $L(x)=x + x^{p^h}/p + x^{p^{2h}}/p^2+\cdots$ be the logarithm of the formal group $F(x,y)\in\Bbb Z_p[[x,y]]$. Then $F(x,y)\in\Bbb Z_p\{\{x\}\}[[y]]$, where $\Bbb Z_p\{\{x\}\}$ is the ring of convergent power series: those whose coefficients go to zero. A word about this ring: it’s the completion of the polynomials with respect to the “Gauss norm”, i.e. the uniform norm on the closed unit disk; or, if you like, the $p$-adic completion of the ring of polynomials. Since you get $\Bbb F_p[x]$ when you tensor the ring of convergent series with $\Bbb F_p$, Neil Strickland’s guess turns out to be correct, in a very strong way. Now for an outline of the proof, which depends entirely on $L'(x)$ being a convergent series, but the proof I found depends also on the particular form of the logarithm. (Perhaps I should say that the cognoscenti may look at all this and say, C’mon, it’s all clear ’cause the invariant differential is a convergent series, and it all drops out automatically from general facts. But I’m no cognoscente in anything, so I have to go through at least some of the motions. I add that Ghassan wonders whether the present result may be in Hazewinkel already, though in some indecipherable formulation.) Treat $F(x,y)$ as an element of $\Bbb Z_p[[x]][[y]]$, so write it as $$ F(x,y)=x +\sum_{m\ge 1}f_m(x)y^m\,. $$ The aim is to show that each $f_m$ is in $\Bbb Z_p\{\{x\}\}$, not just in $\Bbb Z_p[[x]]$. The argument is by induction, starting with $f_1$, which we already know to be $1/L'(x)$, so convergent. We write out the fundamental property of the logarithm: $$ L\bigl(F(x,y)\bigr)=L(x)+L(y)\,, $$ and arrange the pieces differently: $$ 0=\sum_{N\ge0}\Bigl[F(x,y)^{p^{Nh}} - y^{p^{Nh}}\Bigr]\Big/p^N-L(x)\,. $$ In the above, we want to look at the total coefficient-function of $y^s$, knowing inductively that all $f_m(x)$ for $m TITLE: A congruence involving binomial coefficients QUESTION [19 upvotes]: The following open problem was shown to me by Maxim Kontsevich. I state it in a different but equivalent form. Let $a(n)$ be the sequence at http://oeis.org/A131868, that is, $$ a(n) =\frac{1}{2n^2}\sum_{d|n}(-1)^{n+d}\mu(n/d){2d\choose d}. $$ Is it true that $6a(n)/n$ is always an integer? The sequence begins (starting at $n=1$) $(1,1,1,2,5,13,35,100,300,925,2915,9386,\dots)$. If $n$ is prime then the result is true. It is equivalent to the well-known fact that ${2p\choose p}\equiv 2\,(\mathrm{mod}\,p^3)$ for a prime $p>3$. See for instance Enumerative Combinatorics, vol. 1, second ed., Exercise 1.14(d). REPLY [2 votes]: I'd like to draw your attention to a recent paper On p-adic approximation of sums of binomial coefficients where we briefly discuss (in Section 4) the divisibility of this kind. Namely, we claim that for any integers $a\geq b>0$ and $m>0$, $$m^3\mid M(a,b)\cdot\sum_{d\mid m}\mu\left(\frac md\right)\binom{ad}{bd}$$ and $$m^3\mid M'(a,b)\cdot\sum_{d\mid m}(-1)^{m+d}\mu\left(\frac md\right)\binom{ad}{bd},$$ where $M(a,b) = \frac{12}{\gcd(12,ab(a-b))}$ and $M'(a,b) = \frac{3}{\gcd(3,ab(a-b))}\cdot 2^\delta$, where $$\delta = \begin{cases} \min\{1,\nu_2(b)\}, & \text{if}\ \nu_2(a-b)=\nu_2(b),\\ 2, & \text{otherwise}. \end{cases} $$ In particular, for the case $(a,b)=(2,1)$ (questioned here), we have $M(2,1)=6$ and $M'(2,1)=3$. We don't give a proof in the paper (as this is only loosely related to the main subject), but we can share it if anyone is interested. In fact, the paper is mainly devoted to generalization of the congruence ${2p\choose p}\equiv 2\pmod{p^3}$ ($p>3$) to higher powers of $p$.<|endoftext|> TITLE: Reference request: Topology on the space of smooth compact submanifolds QUESTION [11 upvotes]: In Allen Hatchers short exposition of the Madsen-Weiss Theorem he defines the topology on the space $\mathcal{C}^n$ of smooth oriented properly embedded $d$-dimensional submanifolds of $\mathbb{R}^n$ on page 6 with the help of the "standard $C^\infty$ topology on the space of d-dimensional smooth compact submanifolds of B that are properly embedded (meaning that the intersection of the submanifold with $\partial B$ is the boundary of the submanifold, and this is a transverse intersection)", where $B$ is a closed ball centered in the origin. I am searching for a reference or an explicit definition of the "standard topology" Hatcher is referring to. REPLY [11 votes]: It's in Hirsch's "Differential Topology" textbook. Specifically, given a compact manifold $M$, the weak $C^k$-topology on the set of $C^k$-smooth embeddings $Emb(M,\mathbb R^n)$ is the one that requires uniform closeness of not only the maps themselves, but all derivatives up to order $k$. For $C^\infty$ mapping spaces you demand uniform agreement up to some arbitrary order $k$ -- it is not a normable space anymore so you think of the topology as being induced by a countable sequence of (semi)norms, one for each $k$. That's the space of embeddings. So the space of submanifolds of $\mathbb R^n$ diffeomorphic to $M$ is the space $$Emb(M,\mathbb R^n) / Diff(M) $$ where $Diff(M)$ is given the same "weak" topology as in Hirsch. $Diff(M)$ is the space of $C^k$-diffeomorphisms of $M$. Notice two embeddings $M \to \mathbb R^n$ have the same image if and only if they differ by a diffeomorphism of $M$. Then the space of all $m$-dimensional manifolds in $\mathbb R^n$ is the disjoint union: $$\sqcup_M Emb(M, \mathbb R^n) / Diff(M)$$ where you take the disjoint union over all diffeomorphism types of $m$-dimensional manifolds $M$. It's a jazzed-up version of the Whitney embedding theorem that states $Emb(M,\mathbb R^n)$ is highly-connected for $n$ much larger than $m$.<|endoftext|> TITLE: What is a Frobenioid? QUESTION [141 upvotes]: Since there will be a long digression in a moment, let me start by reassuring you that my intention really is to ask the question in the title. Recently, there has been a flurry of new discussion surrounding Shinichi Mochizuki's interuniversal Teichmueller theory (IUTT). I've been personally quite ashamed about the state of affairs. Regardless of the eventual correctness of the paper in all detail, this is an earnest attempt by an esteemed colleague to present a serious vision of mathematics. Not to have given it proper attention for so long reflects poor manners on my part, to say the least. After an initial attempt to organise a workshop, I've essentially let things slide, assuming things would get sorted out somehow. Clearly, it hasn't happened until now. Some dedicated young people have been reading the papers (for example, I've had a number of illuminating exchanges with Chung Pang Mok recently), but my impression is that they also need some support. Some might ask why I, for example, don't just read the papers carefully myself. Well, as you know, for most people chugging along amid the cares of everyday life, it's quite painful to read long difficult papers without the aid of human discussion and interaction. Mochizuki, by the way, is one of those rare people gifted with the incredible powers of concentration and stamina necessary to sit and do mathematics for long periods in solitude. This was true ever since he was a young student. Perhaps this is a key reason he can't understand why the rest of us are so reluctant. Anyway, I'm moving towards suggesting another workshop in the near future. However, to increase the chance that good mathematicians will participate, it seems sensible to disseminate some more background material on the relevant mathematical structures. My plan is to do this in a series of questions on MO. How many there will be, I don't have a clear sense of at the moment. Of course, I have myself just a vague general idea of the notions in the IUTT papers, and even what I knew two years ago, I've forgotten. So my plan is to outline something, quite possibly incorrect and certainly incomplete, and then invite others to improve on my exposition. It's my hope that the people who have read the papers much more carefully will contribute and that some new readers will be motivated. For some time, I've been deeply impressed by the energy and integrity of the community surrounding this site. It seems quite appropriate as a venue for unraveling some of the mystery surrounding IUTT. To read the IUTT papers, a new concept we need to understand well is that of a Frobenioid. To explain why, let me remind you that the main object of study in IUTT is the log-theta lattice associated to an elliptic curve over a number field. The name refers to a collection of oriented paths (Panoramic Overview, Figure 3.1) going between points arranged like a two-dimensional lattice, which represent copies of a $\Theta^{\pm ell}NF$-Hodge theatre. It might be worth emphasising that once you understand this lattice, most of the conceptual background is in place. The extent to which completely new structures are supposed to be endlessly multiplied in these papers has, in my opinion, been greatly exaggerated. What kind of a thing is a $\Theta^{\pm ell}NF$-Hodge theatre? It is a category, itself glued out of two other categories, a $\Theta^{\pm ell}$-Hodge theatre and a $\Theta NF$-Hodge theatre, but for now, we will deemphasise this particular decomposition. (Once again, in case you're worried, there are no other kinds of Hodge theatres.) But in view of the apparently complicated structure whose details might require some guiding principle to grasp, it is reasonable to ask about the main goal, that is, what exactly the point might be of a $\Theta^{\pm ell}NF$-Hodge theatre. Well, as is stated in a variety of ways by Mochizuki himself, it is supposed to be a categorical model of the spectrum of an algebraic number field $F$, together with some extra structure of a crystalline nature. In some sense, it's the same kind of combinatorial encoding of $F$ as the category of finite étale $F$-algebras, or an `abstract combinatorialization of scheme-theoretic arithmetic geometry.' If you are not used to this point of view, you should try to visualise a category as something like a one-dimensional abstract simplicial complex where even quite complicated objects are reduced to points and their structure encoded in a network of arrows. In the present context, this can be a bit confusing because the Hodge theatres are represented as points connected by paths in the log-theta lattice, but when observed closely, they also resolve into a network. The 'crystalline nature' of the extra structure is used in a very vague sense. It's just an elliptic curve over the number field, which determines a set of so-called `theta data'. On the one hand, this extra structure is there because we're interested in Szpiro's conjecture (=ABC conjecture). However, a far more essential goal of the theory is to deform the number field in a canonical fashion, in a manner analogous to lifting a curve over a perfect field of positive characteristic to characteristic zero. There as well, one normally doesn't have a canonical lift. However, the $p$-adic Teichmueller theory was developed exactly to deal with this difficulty: If you equip the curve with an ordinary nilpotent indigenous bundle, which exists for a generic curve, then there is a canonical lift. Because these kinds of indigenous bundles are finite in number, one might say that the curve itself has been canonically lifted up to finite indeterminacy. (Even naively, this is much better than the uncountably infinite indeterminacy associated with choosing any lift.) There is by now a long tradition of crystalline philosophy, whereby an object that doesn't itself admit a canonical deformation acquires one through the addition of some natural extra structure. The simplest example of this phenomenon is the structure of a crystal on the universal extension of an abelian variety. The elliptic curve and the attendant theta data are supposed to be exactly this kind of extra structure that allows us to deform the field in a canonical fashion. That is to say, from a certain point of view, it's really the field that's of interest, and the elliptic curve is just auxiliary structure one has to incorporate in order to arrive at a crystal-like situation. In fact, the log-theta lattice is itself the deformation. At the risk of boring you with repetition, I will restate that a log-theta lattice is made out of copies of a single $\Theta^{\pm ell}NF$-Hodge theatre, which, in turn, is a categorical model of a number field together with a crystalline structure. If you would like to see why a deformation might be made out of the object that it tries to deform, the simplest example to consider is the way in which $\mathbb{Z}_p$ is made out of copies of $\mathbb{F}_p$, especially when considered as the ring of Witt vectors. Perhaps a more elementary remark is that many reasonable deformations in geometry are fibre bundles over the deformation space, so that all fibres are somewhat the same (with varying degrees of rigidity depending on the situation) as the central fibre that we started out with. With the Hodge theatres, Mochizuki's analogy is that they all have the same real-analytic structure, while the holomorphic structure varies. Before getting to the main point, let me say a word about `glueing'. Now, there may be other notions involved, but one basic one is that of grafting, If $A$ and $B$ are categories equipped with functors $\phi$ and $\psi$ to $C$, then we define the (directed) graft $$A\vdash_C B$$ by taking the union of $Ob(A)$ and $Ob(B)$, then simply defining new morphisms from an object $a$ of $A$ to an object $b$ of $B$ as morphisms $\phi(a)\rightarrow \psi(b)$. This construction is quite simple, and I am very far from understanding the different ways of gluing the categories that occur in IUTT, so more on this later. For the moment, I wish to concentrate just a little bit on the internal structure of a $\Theta^{\pm ell}NF$-Hodge theatre, leaving aside for now the nature of the paths connecting the copies. A $\Theta^{\pm ell}NF$-Hodge theatre is also glued in various way out of smaller categories, and this brings us to back to our title. The basic building blocks of everything in sight are categories called Frobenioids. Among the prerequisites for studying IUTT, this notion is the really new one. My feeling is that getting a concrete grip on it will already take us a good way towards understanding the whole picture. The other papers on absolute anabelian geometry and so forth are also hard, but still belong to more or less familiar sorts of anabelian geometry, since many people will have heard of the Neukirch-Uchida theorem or Grothendieck's conjectures. (However, the main focus of the absolute anabelian geometry papers is to prove such reconstruction theorems algorithmically. We will return to this as well in a later question.) The first question in this series is then What is a Frobenioid? In the usual spirit of Mathoverflow, I will set up some rudimentary language, give a few examples to show that I'm a serious participant, and wait for contributions from people with more expertise. I start with a quote from Mochizuki, which we can hope to understand better as more contributions come in: a Frobenioid may be thought of as a sort of a category-theoretic abstraction of the theory of divisors and line bundles on models of finite separable extensions of a given function field or number field. The simplest Frobenioid is constructed out of a commutative monoid $M$ (satisfying some natural condition of being divisorial). One can then form a semi-direct product with $\mathbb{N}_{\geq 1}$, the multiplicative monoid of positive integers: $$\mathbb{F}_M:=M\ltimes \mathbb{N}_{\geq 1},$$ which is a non-commutative monoid with composition defined by $$(a,n)(b,m)=(a+nb, nm).$$ For example, $M$ might be a monoid of effective Cartier divisors on a normal variety $B$ (or line bundles equipped with sections). The $\mathbb{N}_{\geq 1}$-action encodes the tensor power operation on line bundles, which is simply absorbed into the structure of $\mathbb{F}_M$. In positive characteristic, the Frobenius map induces the $p$-th tensor power map on line bundles, and this monoid structure enables the construction of a substitute of sorts. Note also that we can regard a monoid as a category with a single object. The next case starts with a family of monoids. By this, we mean a contravariant functor $$\phi: \mathcal{D}\rightarrow Mon$$ from some category $\mathcal{D}$ to the category of commutative (divisorial) monoids. This $\mathcal{D}$ notation seems to occur rather often and consistently throughout many papers as an indexing category (called a base category in the papers) for some family of monoids. The key example to keep in mind is where $\mathcal{D}$ is a category of finite separable geometrically integral normal covers $X\rightarrow B$ of a normal variety $B$, that is, the usual kind of category of Galois type that occurs in anabelian geometry. In that case, we can take $\phi(X)$ to be a suitable family of effective $\mathbb{Q}$-Cartier divisors on $X$. For example, they might be divisors that lie over a specific monoid of divisors on $B$, say generated by a specific set of prime divisors. (I am going to ignore here the subtleties surrounding pull-backs and types of singularities.) The earlier semi-direct product construction can now be applied to the functor $\phi$. The objects of $\mathbb{F}_{\phi}$ are just the objects of $\mathcal{D}$ (as an extension of considering a monoid a category with a single object) but a morphism $$ X\rightarrow Y$$ is a triple $$(f, S , n),$$ where $f:X \rightarrow Y $ is a morphism in $\mathcal{D}$, $S\in \phi(X)$, and $n\in \mathbb{N}_{\geq 1}$. If $Y\rightarrow Z$ is given by the triple $(g, T, m)$, then the composition is defined by $$(g, T, m)(f, S, n)= (g\circ f, f^*(S)+mT, mn).$$ Up to here is pretty elementary. But some confusion may arise from the fact that the general Frobenioid involves yet another category $\mathcal{C}$ equipped with a functor $ \mathcal{C}\rightarrow \mathbb{F}_{\phi}$. In fact, $\mathcal{C}$ is in many ways more fundamental than $\mathbb{F}_{\phi}$ and should be thought of as a fiber bundle over the base $\mathbb{F}_{\phi}$, as $\mathbb{F}_{\phi}$ is fibered over $\mathcal{D}$. That is, we have a composition of fibrations $$\mathcal{C}\rightarrow \mathbb{F}_{\phi}\rightarrow \mathcal{D}.$$ The nature of $\mathcal{C}$ is clarified by the main example, whereby $\mathcal{C}$ is associated to multiplicative groups of rational functions on varieties. The abstract framework for this example, is that of a model Frobenioid, which one constructs out of yet another functor $\psi: \mathcal{D}\rightarrow Ab$ to abelian groups together with a map of functors $$Div:\psi \rightarrow \phi^{gp},$$ where $\phi^{gp}$ denotes the group completion of $\phi$ in an obvious sense. (As the notation suggests, the example to keep in mind is the homomorphism from rational functions to divisors.) Out of this data, we form the category $\mathcal{C}$ whose objects are pairs $$(X, \alpha),$$ with $X$ an object of $\mathcal{D}$ and $\alpha\in \phi(X)^{gp}$. A map from $(X, \alpha)$ to $(Y, \beta)$ is then a quadruple $$(f, S, n, u)$$ where $f: X\rightarrow Y$ is a morphism in $\mathcal{D}$, $I\in \phi(X)$, $n \in \mathbb{N}_{\geq 1}$, $u\in \psi(X)$, and $$n\alpha+S=f^*(\beta)+Div(u).$$ We have merely added the components $\alpha$ to the objects and the components $u$ to the morphisms. There is thus an obvious projection functor to $\mathbb{F}_{\phi}$. The number $n$, by the way, is referred to as the Frobenius degree of the morphism, and seems to be eventually very important. For the geometric $\phi$ above, we can take $\psi(X)\subset K(X)^*$, a subgroup of the multiplicative group of rational functions on $X$ whose supports are controlled by the divisors in $\phi(X)$. If we view the monoids themselves as being morphisms, this formalism would suggest that $u$ is kind of a 2-morphism, and that $\mathcal{C}$ might better be described as a 2-category in some way. Anyway, I hope you'll agree at this point that the constructions really are not overly exotic. It's worth working out the morphisms for which all components but one are trivial. We will denote the domain of a morphism by $(X, \alpha)$ and the codomain by $(Y,\beta)$. --A morphism of the form $(f, 0, 1,1)$, where $f:X\rightarrow Y$ is a morphism of $\mathcal{D}$. For it to map $(X,\alpha)$ to $(Y,\beta)$, we must have $\alpha=f^*\beta$. We might, for example, choose a divisor $\gamma$ on the base $B$,and let $\alpha$ and $\beta$ be pullbacks of $\gamma$. This determines a faithful embedding of $\mathcal{D}$ into $\mathcal{C}$. --Now consider a morphism of the form $(Id_X, S, 1, 1)$. Of course we must have $Y=X$ and $\alpha+S=\beta$. Thus, this is the `tensor product map' from $(X,\alpha)$ to $(X, \alpha+S)$. --A morphism of the form $(Id_X, 0, n, 1)$ from $(X,\alpha)$ to $(X,\beta)$ imposes $n\alpha=\beta$. So we have formally adjoined a map from a line bundle to its $n$-th tensor power. --$(Id_X, 0, 1, u)$ goes from $(X,\alpha)$ to $(X,\beta)$ such that $\alpha=\beta+Div(u)$. This is a map of line bundles in the usual sense. Obviously, the intention is that as these morphisms intertwine, something interesting will happen. Hopefully, we will soon see more precise clarifications of my approximate account. But the astute reader will already have noticed something not quite right. Earlier on, it was stated that a Frobenioid is a category, whereas we have indicated that it is a category $\mathcal{C}$ equipped with a functor $\mathcal{C}\rightarrow \mathbb{F}_{\phi}$ for some family of monoids $\phi.$ That some conceptual ambiguity is all right is essentially the main theorem of the first Frobenioid paper: In all natural cases, the category $\mathbb{F}_{\phi}$ and the functor to it are canonically determined by the category $\mathcal{C}$. Thus, it is safe to refer to $\mathcal{C}$ itself as the Frobenioid. Certainly, in the geometric example above, it's clear that $\mathcal{C}$ contains the information for $\mathbb{F}_{\phi}$. But the point of the theorem is that with certain rigid assumptions, this encoding can be detected purely category-theoretically. There are a number of other theorems of intrinsic interest about categoricity. For example, the base-category $\mathcal{D}$ can be recovered from $\mathcal{C}$. (What conditions are exactly necessary for this, I'm not able to say at the moment.) If $\mathcal{C}_1$ and $\mathcal{C}_2$ are Frobenioids and $E: \mathcal{C}_1\simeq \mathcal{C}_2$, then $E$ must preserve the Frobenius degrees of morphisms. I think this is all I wish to say for now. Allow me to stress again that I still don't know what a Frobenioid really is and eagerly await corrections and elaborations. There are clearly numerous subtleties and points of emphasis that I am missing. In particular, if someone could give a good account of the main theorem alluded to above, I would be very grateful. I suspect that there are consequences of a rather concrete nature that we can appreciate within the realm of usual arithmetic geometry. This, of course, is the kind of thing that will convince a greater number of people to invest time in understanding the various papers. However, I hope even these superficial paragraphs will provide some indication that the kind of mathematical language developed by Mochizuki is interesting and natural. Indeed, to my untrained mind, the geometric Frobenioids appear very much to be in the spirit of $p$-adic Hodge theory, being subtle composites of structures of étale and De Rham type. Since the earlier Hodge-Arakelov papers had started with the intention of developing a global $p$-adic Hodge theory, maybe this association is not too far from correct. REPLY [41 votes]: About two years ago, I spent the better part of a month trying to get to the bottom of this question: What is a Frobenioid? After all, the IUTer's are fond of insisting that it's incumbent upon other mathematicians to invest time in understanding the foundations of IUT simply because Mochizuki's claimed theorems are, if true, so remarkable. I'm not a number theorist, so I had little hope of understanding the main body of IUT, but the foundational material on Frobenioids is pure category theory, so why not give that part a go? I came away feeling that my time spent on this project was wasted, for several reasons: It was around the same time that Scholze and Stix met with Mochizuki. Apparently they were able to agree from the outset of their discussions that the whole apparatus of Frobenioids was unneccessary for understanding IUT. In this light, the insistence of the IUT camp that one can only understand the theory by plowing through the foundational papers from the bottom up (dubious as it may ever have seemed) is downright insulting. Essentially, I was promised mathematical insight for reading Mochizuki's papers and then told that I was a sucker for taking this promise at face value. The definitions found in Mochizuki's Geometry of Frobenioids I, II are byzantine. My opinion by the end of the project (echoing suggestions that other category theorists had already made publicly) was that the whole apparatus, with a few minor tweaks, could be much more cleanly expressed, with much less need for ad hoc terminology, by using basic properties of the language of Grothendieck fibrations. Once re-expressed this way, the definitions appear clearly to consist of a somewhat-reasonable core, encrusted with numerous ad hoc extra conditions. I find it bewildering that Mochizuki did not care to use this clarifying language when writing these papers -- are not Grothendieck fibrations standard fare for anybody who is familiar with the material of SGA? It might be argued that Mochizuki's definitions as stated are "more correct" than the tweaked versions I ended up formulating for theoretical reasons to do with how they are used in IUT. But my distinct impression is that this is not the case. To the extent that Frobenioids are used at all (and as discussed above, their use at all is apparently inessential), it is only certain well-behaved Frobenioids which are ever invoked, and these fall squarely within the parameters of the tweaked definitions I ended up using. The avoidance of fibrational language is not only obuscatory, but downright harmful. In the course of this project, I found a counterexample to Proposition 4.4.ii in The Geometry of Frobenioids I, while also finding an amended statement of the theorem which is in fact true. I wrote to Mochizuki about it, and he kindly took my objection seriously. However, his response was to add a note to one of his many online manuscripts detailing errors in his work which corrected the statement, without engaging in a serious way with the material. In fact, I assert that the claimed theorem, in amended form, takes more work to prove than appears in the paper or the errata. Anyway, here's what a Frobenioid is, in slightly tweaked form: Definition: A commutative monoid $\Phi$ is said to be divisorial if $\Phi$ is cancellative (i.e. $x+y = x+z \Rightarrow y = z$), group-free (i.e. $M$ contains no nonzero invertible elements), and saturated (i.e. $M$ is closed under division in its group completion $\Phi^{gp}$). If $\Phi$ is a commutative monoid, let $\Phi \rtimes \mathbb N_{\geq 1}$ denote the semidirect product of $\Phi$ with the multiplicative monoid of positive integers, with respect to its canonical action. If $\mathcal D$ is a category, then a divisorial monoid in $\mathcal D$ is defined to be a functor $\Phi: \mathcal D^{op} \to \mathsf{CMon}$, where $\mathsf{CMon}$ is the category of commutative monoids) taking values in divisorial monoids. [1] If $\Phi: \mathcal{D}^{op} \to \mathsf{CMon}$ is a divisorial monoid in $\mathcal D$, then the elementary Frobenioid associated to $\Phi$ is the Grothendieck fibration $\mathbb F_\Phi \to \mathcal D$ associated to the functor $\Phi \rtimes \mathbb N_{\geq 1}: \mathcal D^{op}\to \mathsf{Mon}$ (where $\mathsf{Mon}$ is the category of monoids, regarded as 1-object categories).[2] If $\Phi: \mathcal D^{op} \to \mathsf{CMon}$ is a divisorial monoid in $\mathcal D$, with group completion $\Phi^{gp}$, and if $0 \to \Phi^{birat} \to \Phi^{gp} \to Pic \to 0$ is a short exact sequence of prehseaves of groups, then there is an action of the presheaf of monoids $\Phi \rtimes \mathbb N_{\geq 1}$ on $Pic$. The presheaf of transport categories for this action has a Grothendieck construction denoted $\mathbb U_\Phi \to \mathcal D$, with canonical map to $\mathbb F_\Phi \to \mathcal D$. This data is called the skeletal Frobenioid of unit-trivial type associated to $(\mathcal D, \Phi, \Phi^{birat})$. Note that the map $\mathbb U_\Phi \to \mathbb F_\Phi$ is a Grothendieck opfibration fibered over $\mathcal D$. A skeletal pseudo Frobenioid of isotropic type consists of a series of functors $\mathcal C \to \mathbb U_\Phi \to \mathbb F_\Phi \to \mathcal D$ such that $\mathbb U_\Phi \to \mathbb F_\Phi \to \mathcal D$ is a skeletal Frobeniod of unit-trivial type, and additionally $\mathcal C \to \mathcal D$ is a Grothendieck fibration, and $\mathcal C \to \mathbb U_\Phi$ is a Grothndieck opfibration fibered over $\mathcal D$, fibered in groups, such that reindexing along a monoid element $Z \in \Phi(S)$ is always a group isomorphism. A model Frobenioid is one arising via a similar construction to the construction of $\mathbb U_\Phi$ using only a right-exact sequence $B \to \Phi^{gp} \to Pic \to 0$ of presheaves of abelian groups, where we set $\Phi^{birat}$ to be the image of $B$ in $\Phi^{gp}$. That is, we take $\mathcal C = ``\mathbb U_B"$; the map $\mathcal C \to \mathbb U_\Phi$ comes from the map $B \to \Phi^{birat}$. So essentially, a Frobenioid is just a way to package the data of a presheaf of exact sequences of abelian groups, plus a bit of positivity data and paying particular attention to the action of $\mathbb N_{\geq 1}$ on everything, into a single category $\mathcal C$. From a categorical perspective, this is certainly something that one is free to do, but there's not really anything forcing one to package things this way. And apparently from the number-theoretic perspective, there's nothing particularly militating for this way of packaging the data. So there doesn't really seem to be anything more to say about them than the definition. [1] Actually Mochizuki imposes additional conditions on $\mathcal D$ and $\Phi$, but the additional conditions don't seem to play much role in the theory. [2] Actually Mochizuki requires only that $\Phi$ be a so-called pre-divisorial monoid here, but only the divisorial case plays much role in the theory.<|endoftext|> TITLE: A dual version of a theorem of Øystein Ore in group theory QUESTION [12 upvotes]: This post is a dual version for the Generalization of a theorem of Øystein Ore in which it's proved: Theorem: Let $[H, G]$ be a distributive interval of finite groups. Then $\exists g \in G$ such that $\langle H,g \rangle = G$. Definition: Let $W$ be a representation of $G$, $K$ a subgroup of $G$, and $X$ a subspace of $W$. Let the fixed-point subspace $W^{K}:=\{w \in W \ \vert \ kw=w \ , \forall k \in K \}$. Let the pointwise stabilizer subgroup $G_{(X)}:=\{ g \in G \ \vert \ gx=x \ , \forall x \in X \}$. Definition: $[H,G]$ is called linearly primitive if $\exists V$ irred. complex repr. of $G$ with $G_{(V^H)} = H$. Remark: $[1,G]$ is linearly primitive iff $G$ is linearly primitive. Question: Let $[H, G]$ be a distributive interval of finite groups. Is $[H,G]$ linearly primitive? Remark: The case $H = 1$ is true because $[1,G]=\mathcal{L}(G)$ is distributive iff $G$ is cyclic, but an abelian group is linearly primitive iff it is cyclic. It follows that it's also true if $H \triangleleft G$. It's true by GAP for $|G:H|<32$ or $\vert G \vert \le 1000$ or $G$ perfect with $\vert G \vert < 10080$ (except $7680$). REPLY [2 votes]: Yes. This was proved in the planar algebra framework, see arXiv:1704.00745, Corollary 6.10. For a self-contained group-theoretic proof, see arXiv:1708.02565.<|endoftext|> TITLE: Geometric dominating set: NP-complete? QUESTION [6 upvotes]: Let $G=(V,E)$ be a geometric graph, a graph embedded in the plane whose edge lengths are the Euclidean distance between its endpoint vertices. Say that a set of vertices $D \subseteq V$ is a geometric dominating set if for every $v \in V$, the shortest path in the graph from $v$ to some vertex in $D$ is at most length $1$. For example, a pentagonal wheel with spokes of length $1$ can be dominated with one vertex, but if the shortest edge length is greater than $1$, then $D$ must equal $V$: The traditional dominating set problem has been known to be NP-complete since at least Garey & Johnson's 1979 book. My question is: Q. Is finding a smallest geometric dominating set for a given geometric graph also NP-complete? Conceivably it is not intractable, if the geometric structure can be exploited. Perhaps plane graphs $G$ would especially allow that exploitation. If anyone knows of references, I would appreciate pointers. Thanks! REPLY [7 votes]: I came up with an explicit construction that showed that your problem was NP-hard, even on the real line, but then I realized there's an even simpler argument: Minimum dominating set is known to be NP-hard even for bipartite graphs, and those embed nicely. For example, you could send one part of the partition to $(0,0)$ and the other to $(1,0)$. (If you insist that distinct vertices be represented by distinct points, then you may choose different points like $(\varepsilon,0)$ and $(1-\varepsilon,0)$ instead.)<|endoftext|> TITLE: Are all smooth functions composites of 0-, 1-, and 2-ary functions? QUESTION [14 upvotes]: I will formalize my question in terms of algebraic theories. Background: Recall that an algebraic theory (in the sense of Lawvere) is a category $\mathcal{C}$ which is closed under taking finite products, and whose set of objects can be identified with the set $\mathrm{Ob}(\mathcal{C})\cong\{T^0,T^1,\ldots\}$, where $T^i=T^1\times T^1\times\cdots\times T^1$ is the $i$-fold product of $T^1$. We denote $T^1$ by $T$. The interesting aspect of an algebraic theory $\mathcal{C}$ are the morphisms $\mathcal{C}(T^i,T^1)$ and their compositions. For any algebraic theory $\mathcal{C}$, a $\mathcal{C}$-model is defined to be a functor $\mathcal{C}\to\mathbf {Set}$ that preserves all finite products. Denote the category of $\mathcal{C}$-models (and natural transformations between them) by $\mathbf{Mod}(\mathcal{C})$. Let $n\in\mathbb{N}$ be a natural number, and let $\mathcal{C}$ be an algebraic theory. Its $n$-truncation, denoted $\mathcal{C}_{\leq n}\subseteq\mathcal{C}$, is the smallest algebraic sub-theory that has the same $k$-ary functions $$\mathcal{C}_n(T^k,T^1):=\mathcal{C}(T^k,T^1),$$ for all $k\leq n$. full subcategory spanned by the objects $\{T^0,\ldots,T^n\}$. We say that $\mathcal{C}$ is $n$-truncated if it is equivalent (as a category) to its $n$-truncation. Setup: Let $R$ denote the algebraic theory whose morphisms $T^n\to T$ are the smooth functions ${\mathbb R}^n\to{\mathbb R}$. This of course defines the set of morphisms $T^n\to T^m$ for any $m$. Endow $R$ with the usual formula for composing smooth functions. To me it would be quite surprising if $R$ were 2-truncated. But I've never heard of a 3-ary function $\mathbb{R}^3\to\mathbb{R}$ that wasn't constructed from a combination of 0-, 1-, and 2-ary functions. Question: Can you prove that $R$ is not 2-truncated? More interestingly, can you name, i.e., construct a 3-ary function that isn't also constructible by a combination of 0-, 1-, and 2-ary functions? (Note: thanks to Todd Trimble for suggestions on how to clean up this question.) REPLY [28 votes]: For any $n$, there is an $n$-ary smooth function that is not a composition of smooth functions of lower arity; according to this answer to a very similar question, this is due to Vitushkin (at least for $n=3$). Here is a simple but rather inexplicit proof (adapted from this paper of Akashi and Kodama). Suppose that a smooth n-ary function $f$ can be expressed as a composition of smooth functions $g_i$ of lower arity (WLOG, all of arity $n-1$). Then for each $k$, the $k$-jet of $f$ at any point is determined by the $k$-jets of the $g_i$ at corresponding points. But dimension of the space of $n$-ary $k$-jets grows like $k^n$, while the dimension of the space of $(n-1)$-ary $k$-jets grows like $k^{n-1}$. It follows that for any particular way to compose lower arity functions to get an $n$-ary function, most $k$-jets of $n$-ary functions cannot be so obtained for $k$ sufficiently large. Since there are only countably many such ways to compose, we can use bump functions to construct a single $f$ that has jets at different points that fail for all of them. Actually, by a diagonal argument we can find a single $n$-ary $\infty$-jet (i.e., power series) that is not a composition of jets of lower arity, and every $\infty$-jet can be realized by a smooth function.<|endoftext|> TITLE: Homomorphisms from projective modules QUESTION [6 upvotes]: Let $B$ be a $A$-algebra which is free of finite rank as $A$-module. Let $X$ be a finitely generated projective left $B$ module. (So $X$ is also a f.g. projective $A$ module.) Are these homomorphism groups naturally isomorphic as right $B$-modules? $$\mathrm{Hom}_A(X,A)\stackrel{?}{\cong} \mathrm{Hom}_B(X,B).$$ By dimension count (e.g. in special case of fields) one can see easily that these two are really isomorphic. After a lot of work I found a long proof of natural equivalence which only works for the case of division rings with an assumption on characteristic, which doesn't give an explicit isomorphism. So my question is: Are these two $B$-modules always naturally isomorphic? (At least for the case of a division ring $B$ over a field $A$) I also wonder if there is a simple explicit natural morphism in general (without above restrictions on $B$ and $X$ ) from one side to the other which gives a natural isomorphism in good cases. Edit. I simplified my proof and found an explicit natural morphism from the right side to the left. Let $f \in \mathrm{Hom}_B(X,B)$, I define the corresponding element $t_f\in \mathrm{Hom}_A(X,A)$ as follows: For every $x\in X$ define a $A$-linear morphism $T_f(x):X\to X$ by $(T_f(x))(y) := f(y)x$. Then $t_f(x):= \mathrm{tr}_A(T_f(x)).$ I can show that $t:\mathrm{Hom}_B(X,B)\to \mathrm{Hom}_A(X,A)$ is an isomorphism for separable finite algebras over fields. But the question remains unsolved in the general case. REPLY [2 votes]: To make abx's comment more explicit, consider the following example: Let $A=k$ be a field. Let $B=k[s,t]/(s^2,t^2,st)$, which is not Gorenstein. We see that $B$ is an $A$-algebra which is free of rank $3$ as an $A$-module. Set $_BX=\!_BB$. We compute that ${\rm Hom}_B(X,B)\cong B_B$. Now, consider the hom set ${\rm Hom}_A(X,A)$. This hom set has the structure of a $B$-module via the following action: Given $\varphi\in {\rm Hom}_A(X,A)$, $b\in B$, and $x\in X$, we have $(\varphi\cdot b):x\mapsto \varphi(bx)$. Now, suppose by way of contradiction, that there is an isomorphism of $B$-modules $\theta:B_B\cong {\rm Hom}_A(X,A)$. Let $\varphi:=\theta(1)$. Thus ${\rm Hom}_A(X,A)=\varphi\cdot B$. Write $\varphi(1)=a_1,\varphi(s)=a_2,\varphi(t)=a_3$ with $a_1,a_2,a_3\in k$. Given any $b_1+b_2s+b_3t\in B$ (with $b_1,b_2,b_2\in k$) then $$(\varphi\cdot b)(s)=\varphi(bs)=\varphi(b_1s)=a_2b_1$$ and $$(\varphi\cdot b)(t)=\varphi(bt)=\varphi(b_1t)=a_3b_1.$$ Thus, if $\psi\in {\rm Hom}_A(X,A)=\varphi\cdot B$, we must have $\begin{pmatrix}\psi(s)\\ \psi(t) \end{pmatrix}\in \begin{pmatrix}a_2\\ a_3 \end{pmatrix}k$, which contradicts the fact that elements in ${\rm Hom}_A(X,A)$ can be defined arbitrarily on the set $\{1,s,t\}$.<|endoftext|> TITLE: Proper continuous image of metrizable space QUESTION [5 upvotes]: Motivated by the following post, "Gelfand duality" and the fact that "a Hausdorff continuous image of a compact metric space is metrizable", we ask: What is a counter example of two locally compact Hausdorff spaces $X$ and $Y$ and a surjective proper continuous map $f:X\to Y$ such that $X$ is metrizable but $Y$ is not? REPLY [9 votes]: If $X$ is metrisable and $f$ is a perfect surjection onto $Y$, where perfect means continuous, closed and pre-images of singletons are compact, then $Y$ is metrisable. This is proved e.g. in Engelking's General Topology (Thm. 4.4.15), and due independently to K. Morita and S. Hanai ("Closed mappings and metric space", Proc. Japan Acad. 32 (1956), 10-14) and A.H. Stone ("Metrizability of decomposition spaces", Proc. Amer. Math. Soc 7 (1956), 690-700). The proof in Engelking is based on the Bing-Nagata-Smirnov metrisation theorem. A proper (in the sense of inverse images of compact sets being compact) continuous map with a $k$-space as codomain is closed, if the codomain is Hausdorff (for a simple proof see this paper) so the above theorem applies, as all locally compact spaces are $k$-spaces (or compactly generated); for a proof for locally compact see this question. This means that no counterexample exists. As an afterthought, I looked a bit more at the question whether a proper continuous image of a metric space is metrisable, without conditions. We saw above that with $Y$ being a Hausdorff $k$-space (which are necessary conditions to be metrisable anyway) the answer is yes, as we then have a perfect map. But without a condition on $Y$ we do have an example: let $X$ be the reals in the discrete topology, and $Y$ the set of reals with the topology where all subsets of $\mathbb{R} \setminus \{0\}$ are open, and a set $O$ containing $0$ is open iff $O$ is co-countable (so it's essentially the one-point Lindelöfication of the reals in the discrete topology). One checks that $Y$ is hereditarily normal (but not perfectly normal) and is not a $k$-space because the only compact subsets of $Y$ are the finite ones (and so all subsets are compactly closed, but some are not closed). The last fact also implies that $f(x) = x$ is continuous and proper, but its image is not metrisable. And $X$ is locally compact metrisable.<|endoftext|> TITLE: Random partitions with prescribed pairwise membership probabilities QUESTION [5 upvotes]: Let $(p_{ij}) \in [0,1]^{n \times n}$ be a given symmetric matrix, with $1$ on the diagonal. Suppose $\pi$ is a partition of $[n]=\{1,\dots,n\}$ and let us write $i \stackrel{\pi}{\sim} j$ if $i$ and $j$ belong to the same component of $\pi$. Is there a random partition $\pi$, such that $\mathbb{P}( i \stackrel{\pi}{\sim} j ) = p_{ij}$ for all $i \neq j$? In other words, given a matrix $(p_{ij})$, is there a distribution on partitions of $[n]$ with the above property, and if so is there a way to sample from such distribution? Also, assuming existence, is it unique? EDIT: As has been pointed out, not all such matrices produce a distribution. But what are the conditions on $(p_{ij})$ that allows for such distribution? REPLY [2 votes]: The $p \in \mathbb R^{n(n-1)/2}$ corresponding to distributions on partitions form a convex polytope, the extreme points of which correspond to individual partitions. I don't know if there's a simple way to characterize the faces of this polytope in general. For $n=3$ the polytope is the convex hull of $[0,0,0], [1,0,0],[0,1,0],[0,0,1], [1,1,1]$, and this is the intersection of the half-spaces $p_{12} \ge 0$, $p_{13} \ge 0$, $p_{23} \ge 0$, $p_{12} + p_{13} - p_{23} \le 1$, $p_{12} - p_{13} + p_{23} \le 1$, and $-p_{12} + p_{13} + p_{23} \le 1$. For $n=4$, if my programming is correct, there are $22$ half-spaces: in addition to the six $p_{ij} \ge 0$ and the twelve $p_{ij} + p_{ik} - p_{jk} \le 1$, there are four of the form $\sum_{\{j,k\}: i \in \{j,k\}} p_{jk} - \sum_{\{j,k\}: i \notin\{j,k\}} p_{jk} \le 1$, $i = 1 \ldots 4$.<|endoftext|> TITLE: On Grothendieck's idea on his Standard Conjecture B QUESTION [31 upvotes]: Let me recall the Standard Conjecture B (see [1,2] below): The $\Lambda$-operation of Hodge theory is algebraic. It more or less says that the partial inverse to “cupping with the class of a hyperplane” comes from an algebraic cycle. This is true, for example, for abelian varieties. Question On the fifth page of his article on the Standard Conjectures [2] (page 197 of the journal) in the second paragraph from the top Grothendieck writes: I have an idea of a possible approach to Conjecture $B$, which relies in turn on certain unsolved geometric questions, and which should be settled in any case. Q1: Did Grothendieck write/speak anywhere else about this idea? Did he work it out? Q2: What are the “unsolved geometric questions” that he mentions? What is the status of these questions? References https://en.wikipedia.org/wiki/Standard_conjectures_on_algebraic_cycles http://www.math.jussieu.fr/~leila/grothendieckcircle/StandardConjs.pdf REPLY [10 votes]: My guess is that he was thinking about crystalline cohomology. It fits rather nicely in Grothendieck research at that time. He had obviously in mind the success of Dwork's p-adic approach to the Weil conjectures, and the limitations of the other cohomologies avaible at the time (see sections 1.5 to 1.8 of "Crystals and the de Ram cohomology of schemes"). The standard conjectures were worked out in 1965 (according to Grothendieck's 1968 paper), so he had to have them in mind while working on his p-adic cohomology, that he presented to Bourbaki on december 1966. He then gave it to his student Pierre Berthelot to develop. The intro of the Bourbaki notes reads: The content of the notes are by no means intended to be a complete theory. Rather, they outline the start of a program of work which has still not been carried out (*). (*) For a more detailed exposition and progress in this direction, we refer to the work of P. Berthelot, to be developped presumably in SGA 8. Berthelot's complete exposition was not presented as SGA 8, but as in independient work in 1974. So even if the cohomology was ready long before that, Grothendieck had to regard it as unsolved in 1968, when he wrote about the standard conjectures. It is also reasonable to imagine that he had hopes at those early stages for crystalline cohomology to be an important tool in the yoga of motives. Again, this is just a guess. I'm not sure that he would refer to this as "unsolved geometric questions" (perhaps in the sense of its aplication to Hodge/Betti coefficients?). Maybe someone who knows more about all of this can add some details. Some relevant references: "On the de Rham cohomology of algebraic varieties" (Grothendieck, written in 1963) "Crystals and the de Rham cohomology of schemes" pp. 254-306 (Grothendieck, written in 1966) "Letter to Tate" (Grothendieck, written in 1966)<|endoftext|> TITLE: Rings satisfying the polynomial equation $x^4=x^2$ QUESTION [6 upvotes]: There are many results concerning the commutativity of rings satisfying a polynomial equation. I want to know if there is any result/reference about (finite) rings that satisfy the polynomial equation $x^4=x^2$, which seems to be much more complicated than equations like $x^n=x$, etc. REPLY [14 votes]: Alfred Foster introduced a notion of Boolean-like ring in a 1946 paper http://www.ams.org/tran/1946-059-01/S0002-9947-1946-0015045-5/S0002-9947-1946-0015045-5.pdf He calls elements of a ring that satisfy $x^4=x^2$ weakly idempotent. Boolean-like ring is a commutative ring of characteristic two with identity in which $(1— a)a(1— b)b=0$ holds for all elements $a,b$ of the ring. The following properties of Boolean-like rings are well known: Each element is weakly idempotent; The nilpotent elements form an ideal; The idempotent elements form a subring; Each element can be uniquely written as the sum of an idempotent element and a nilpotent element. Omitting the commutativity and the existence of identity in Boolean-like rings, Iwao Yakabe defines generalized Boolean-like ring as a ring of characteristic two and in which $(a—a^2)(b—b^2)=0$ holds for all elements $a,b$ of the ring: http://catalog.lib.kyushu-u.ac.jp/handle/2324/1449033/13_2_p079.pdf Each element of a generalized Boolean-like ring is also weakly idempotent. The notion of (m,n)-Boolean ring ($m>n\ge 1$) was introduced by Maurer and Szigeti as a ring in which every element satisfies $x^m=x^n$ in https://eudml.org/doc/229780 It is claimed in the paper that the structure of (m,n)-Boolean rings heavily depends on the parity of the difference $m-n$: if this difference is odd, some reduction theorems are proved in the paper. In the case of even $m-n$ difference, no such reduction theorems are expected. The ring of $2\times 2$ upper-triangular matrices over a Boolean ring is is an example of a (4,2)-Boolean ring which is not commutative.<|endoftext|> TITLE: Compact operators on Lebesgue spaces QUESTION [7 upvotes]: Let $K:{\rm L}^p({\bf R}^d)\to {\rm L}^p({\bf R}^d)$ be a bounded linear operator for every $p\in(1,\infty)$. Assume that for some $r\in(2, \infty)$ it holds that $K$ is compact on ${\rm L}^q({\bf R}^d)$ for every $q\in[2,r)$. My question is: does it hold that $K$ is compact on ${\rm L}^{r}({\bf R}^d)$? I was thinking about the counterexample, but unsuccessfully. Remark: I would like to apply such result (or not to apply it) on the K that is a commutator of Fourier integral operator and simple multiplication operator. Thank You in advance. REPLY [5 votes]: More is true: if K is compact on $L^{p_1}$ for some $1\leq p_1<\infty$ and bounded on $L^{p_2}$ for some other $1 \leq p_2 \leq \infty$, then it is compact for all $p$ in the interval $[p_1,p_2)$ or $(p_2,p_1]$. This is a results of Krasnoselʹskiĭ. See http://www.ams.org/mathscinet-getitem?mr=119086 Interestingly it is an open question whether this holds in complete generality for complex interpolation, see http://arxiv.org/abs/1410.4527<|endoftext|> TITLE: Asymptotic dimension of $C'(1/6)$ small cancellation groups QUESTION [11 upvotes]: Do there exist finitely presented $C'(1/6)$ small cancellation groups with arbitrarily high asymptotic dimension? To offer a little more motivation, Roe proves that all hyperbolic groups have finite asymptotic dimension, a result that was particularly interesting at the time as it ensured that such groups satisfy the Novikov conjecture due to the work of Yu. Small cancellation groups are a rich and interesting subclass of hyperbolic groups - surface groups (genus $\geq 2$) and random groups in the few relator or low density models are almost surely examples. Calculations and computations are often easier for these groups - for instance they have cohomological dimension at most 2. Unfortunately I cannot find an argument which says either that such groups have uniformly bounded asymptotic dimension, or proving that this is unbounded. REPLY [6 votes]: The asymptotic dimension is bounded by 2. I don't know the original proof of this, but I found some references on this. Torsion-free $C'(\frac16)$ small-cancellation groups have cohomological dimension 2 (see Theorem 6.5 (5) due to Bestvina and Mess and the following discussion of $C'(\frac16)$ groups, and any such group is virtually torsion-free), so their boundaries have dimension 1. Thus, they have asymptotic dimension bounded by 2.<|endoftext|> TITLE: Is there a description of the moduli space of elliptic surfaces? QUESTION [7 upvotes]: In this question elliptic surface means a smooth projective complex surface $X$, such that there is an elliptic fibration $\pi \colon X \to C$. (I.e., there is a curve $C$ and a proper map $\pi$, such that almost all fibres are elliptic curves.) I am aware of 1 and 2. These describe the moduli space of rational elliptic surfaces (unless I stupidly overlooked som parts; I read quite a bit of them, but not every letter). Moreover they assume that the fibration $\pi$ has a section. I would like to know if there is more known about the other cases, in particular those where $C$ is not rational. First some general questions, asking for literature/references: Q1: Is there literature on the moduli space of minimal elliptic surfaces? Q1.i:  In general? (With or without assuming that $\pi$ has a section.) Q1.ii: In special cases, say when $p_g = q = 1$? I am particularly interested in whether the Hodge structure of such elliptic surfaces ($p_g = q = 1$) vary when one varies the surface. I want to do this by exhibiting (for every connected component) two elliptic surfaces with different Picard number. However, I have no clue about how the moduli space looks. Q2.i:  How many components are there in this case ($p_g = q = 1$)? Q2.ii: What are their dimensions? Let me finally remark that Remke Kloosterman shows in 2 that there exists extremal elliptic surfaces with these invariants (i.e., maximal Picard number). References 1 Gert Heckman and Eduard Looijenga, The moduli space of rational elliptic surfaces. www.math.ru.nl/~heckman/Heck_14.pdf 2 Remke N. Kloosterman, Arithmetic and Moduli of Elliptic Surfaces. www.math.hu-berlin.de/~klooster/proefschrift-kloosterman.pdf REPLY [4 votes]: For the case with section and $q=0$ see http://www.math.colostate.edu/~miranda/preprints/weierstrassfibrations.pdf A similar construction should work in the case (with section, $q$ fixed and $p_g$ sufficiently large) you should get a moduli space together with a morphism to $M_g$. In the case (with section; $q=p_g=1$) then $p_g$ is ``sufficiently large" and you have that the Weierstrass equation of the elliptic surface depends on the base curve $C$, a choice of a line bundle $L$ of degree 1 on $C$ and two sections in $H^0(L^4)$ and $H^0(L^6)$. You easily can get the dimension of the corresponding moduli space from this. Also this moduli space is obviously connected. Moreover, I believe that the period map is locally an isomorphism in this case, hence different Picard numbers occur.<|endoftext|> TITLE: Citation: earliest incidence of the Borel localization theorem QUESTION [9 upvotes]: The Borel localization theorem in (Borel) equivariant cohomology states that if $T$ is a torus and $M$ a smooth $T$-manifold, with fixed point set $M^T$, then upon localizing the coefficient ring $H^*_T := H^*(BT)$ (that is, tensoring with its field of fractions), the restriction map $$H^*_T(M) \to H^*_T(M^T)$$ becomes an isomorphism. The earliest reference to this result I know of is in Hsiang Wu-Yi's classic book on transformation groups, where he refers to the result as a "localization theorem of Borel–Atiyah–Segal type." It seems from Hsiang's presentation that Borel only proved this result for $T = S^1$; in any event, I've never been able to find the more general result in the Seminar on Transformation Groups. But I'd like to attribute it to someone. Who originated this result? Failing that, what is the earliest citation you know? REPLY [9 votes]: Here is the reference trail, according to this source: Borel made the key observation [1] that the cohomology of the fixed point set was closely related to a torsion-free quotient. In the 1960’s, this was formalized as the “localization theorem” of Borel-Atiyah-Segal-Quillen [2,3]. A. Borel, Seminar on transformation groups, Annals of Math. Studies 46, Princeton (1960) [Sect. XII Theorem 3.4] M.F. Atiyah and G. Segal, Equivariant cohomology and localization, lecture notes, 1965, Warwick. D. Quillen, The Spectrum of an Equivariant Cohomology Ring I, Annals of Mathematics 94, 549–572 (1971). [Theorem 4.4]. A more extensive overview of the literature leading up to, and following after the localization theorem can be found here (section 1.7). In addition to the unpublished 1965 lecture notes of Atiyah and Segal, there is a 1968 publication [4], crediting the theorem to Segal [5] (who writes that "The theory was invented by Professor Atiyah, and most of the results are due to him."). M.F. Atiyah and G.B Segal: Index of elliptic operators II, Ann. Math. 87, 531–545 (1968). G.B. Segal, Equivariant K-theory, Publ. Math. Inst. Hautes Etudes (Paris) 34, 129-151 (1968).<|endoftext|> TITLE: Is this variant of the balanced bracket language context free? QUESTION [7 upvotes]: Consider the language generated by the following context free grammar: $$ S \to SS \quad S \to () \quad S \to (S) \quad S \to [] \quad S \to [S] $$ There is a one-to-one correspondence between this language and rooted planar trees where each edge is either dashed or solid ([] corresponds to a dashed edge and () corresponds to a solid edge). Call a tree $T$ good if when you remove all the solid edges, the remaining dashed forest is a connected tree where all the vertices have valence $ \leq 2 $ (i.e a path). Let $L$ be the sublanguage consisting of all good trees. Question: is $L$ context free? It is easy to check that $L$ satisfies the pumping lemma for context free languages. My instincts tell me that it shouldn't be context free because you can only add square brackets (which correspond to dashed edges) in certain contexts, but I don't know if this intuition can be turned into a proof. REPLY [4 votes]: As Bjørn says, it is context-free, but I don't think his solution is quite right. Here's a set of rules I think does work, where $S$ is start and $e$ is the empty word: $S\to TST$ $T\to(T)$ $T\to TT$ $T\to e$ $S\to(S)$ $S\to U$ $U\to TUT$ $U\to [U]$ $U\to e$ Here, $S$ is the start, and represents any word in $L$. $T$ represents a tree with only solid lines. $U$ represents a tree in $L$ whose dashed path begins at the root (the dashed path may be empty). The part with the $T$'s and the $U$'s is essentially identical to Bjørn's solution; the additional trickery with the $S$'s is to allow the dashed path to begin below the root.<|endoftext|> TITLE: What is known about the reverse mathematics of algebraic number fields? QUESTION [12 upvotes]: I know work on the reverse mathematics of countable algebraic field extensions including Galois theory, notably including Dorais, Hirst, and Shafer http://arxiv.org/pdf/1209.4944v2.pdf. But algebraic number fields are a very restrictive subset of countable fields and will not support all the codings used in the cited paper. I suspect they will not support any codings of such flexible expressive power, and that results on algebraic number fields actually have less proof theoretic strength, but I do not know that is true. What is known about this? For example, Dorais, Hirst, and Shafer reverse the theorem that an automorphism of an algebraic extension of countable fields extends to an automorphism of any algebraic closure of the extension. They reverse it over $\mathsf{RCA}_0$ to $\mathsf{WKL}_0$. Has anyone reversed the number field case of this? If I am not mistaken, their Thm 8 shows $\mathsf{RCA}_0$ itself proves every automorphism of a number field $L$ extends to an automorphism of any larger Galois number field $K/L$. There has been some thought about Galois theory of number fields in the weaker $\mathsf{RCA}^*_0$. See Reverse mathematics below RCA. Is there recent progress on that? Is the matter sensitive to the distinction between coding algebraic number fields as extensions of $\mathbb{Q}$ which have finite bases, versus coding them as extensions with given bases? Since many people up voted a request to clarify what reverse math is, I'll add a bit to the tag description: It is a branch of proof theory which calibrates the strength of classical mathematical theorems in terms of the axioms, typically of set existence, needed to prove them. First you show some theorem is provable in some fragment of second order arithmetic (e.g. every countable field has an algebraic closure). So the theorem is no stronger than that fragment. Then you "reverse the theorem" by showing the axioms of that fragment are actually provable from the theorem, assuming some standard base axioms. So, relative to the base axioms, the theorem is exactly as strong as the axioms of the fragment. Usually the way to reverse a theorem by coding every situation that the axiom addresses in terms of what the theorem says can be done, so the theorem itself implies whatever the axiom did. It originated in its modern form in the 1970s by H. Friedman and S. G. Simpson (see R.A. Shore, "Reverse Mathematics: The Playground of Logic", 2010). REPLY [6 votes]: In short, algebraic number theory ``flies below the radar'' of current Reverse Mathematics. First, after working on this a while, I think it is fair to say most familiar theorems on algebraic number fields that do not refer to the algebraic closure, or real closure, of the rationals are provable in EFA, exponential function arithmetic. (This is roughly because polynomials are themselves finite sequences of coefficients, and EFA gives an adequate theory of finite sequences.) Second, the answer by Bjørn Kjos-Hanssen to What is the reverse mathematical strength of the fundamental theorem of algebra? notes that Tanaka and Yamazaki have proved the fundamental theorem of algebra as well as quantifier elimination for the theory of real closed fields in $\mathrm{RCA}_0$. So the reverse mathematics of these topics lies below the base theory of most of Reverse Mathematics today. On one hand, reverse math results about this will await the creation of a reverse math over EFA. On the other hand, provability (without reversals) of various theorems that deal with the algebraic or real closures of $\mathbb{Q}$ can be explored now by catch as catch can methods.<|endoftext|> TITLE: Gauss-Bonnet invariant Ω: explicit intrinsic expression for Π in Ω=dΠ? QUESTION [5 upvotes]: Let the Gauss-Bonnet form be $\Omega\propto\text{Pf}(\Omega^i{}_j)$ with $\Omega^i{}_j$ the curvature 2-form of an even-dimensional manifold with dim=$n$. The Gauss-Bonnet form is exact, as shown in the explicit construction in Chern 1944 (also available here). We may write $\Omega = d\Pi$ with $\Pi$ a rank $n-1$ form. Chern gave a construction for $\Pi$ in terms of an arbitrary vector field. Compare this to the Chern-Simons form, as developed in Chern and Simons 1974. The Chern-Simons form is also exact, and Chern and Simons 1974 give an explicit construction without any arbitrary vector field—just in terms of the connection 1-form and the curvature 2-form. Question: is there a modern version of Chern 1944 which, similar to Chern and Simons 1974, gives an explicit construction for $\Pi$ without reference to an arbitrary choice of vector field, but rather is only in terms of the connection 1-form and curvature 2-form? REPLY [11 votes]: I see that the OP may not be entirely convinced by my comments, so let me try this, which may help. It's understandable that reading the older literature can be confusing; the classical language is often quite different from ours. Here is a slightly different interpretation of Chern's construction of the transgressed form $\Pi$ that may make it clearer what is going on. Let $n=2p>0$ be an even integer and let $(R^n,g)$ be an oriented Riemannian $n$-manifold. Let $\pi:B\to R$ be the principal right $\mathrm{SO}(n)$-bundle consisting of the oriented $g$-orthonormal frames on $R$, i.e., each $e\in B$ with $\pi(e)=x\in R$ is of the form $e = (e_1,\ldots e_n)$ where $(e_1,\ldots,e_n)$ form an oriented, $g$-orthonormal basis of $T_xR$. As usual, one has the tautological $1$-forms $\omega_i$ defined by $\omega_i(v) = e_i\cdot \pi'(v)$ for $v\in T_eB$. There exist unique connection $1$-forms $\omega_{ij}=-\omega_{ji}$ on $B$ satisfying the first structure equations, $$\mathrm{d}\omega_i = - \omega_{ij}\wedge\omega_j.$$ (NB: My $\omega_{ij}$ are the negatives of Chern's $\omega_{ij}$. I'm sorry about that, but I can't switch back to Chern's conventions without getting confused. On the bright side, the first structure equations actually play no role in the sequel, which is why the Gauss-Bonnet formula for the Euler class works for any $\mathrm{SO}(n)$-connection on any oriented orthogonal bundle over $R$, not just the tangent bundle.) The curvature $2$-forms are defined by the second structure equations $$\Omega_{ij} = \mathrm{d}\omega_{ij} + \omega_{ik}\wedge\omega_{kj} = \tfrac12 R_{ijkl}\,\omega^k\wedge\omega^l, $$ and they satisfy the Bianchi identities $\mathrm{d}\Omega_{ij} = \Omega_{ik}\wedge\omega_{kj}-\omega_{ik}\wedge\Omega_{kj}$. The Gauss-Bonnet $n$-form on $B$ is $$ \tilde\Omega = \frac1{2^{n}\pi^{n/2}(n/2)!}\ \epsilon_{i_1i_2\cdots i_n} \Omega_{i_1i_2}\wedge\Omega_{i_3i_4}\wedge\cdots\wedge \Omega_{i_{n-1}i_n}\ , $$ where the sum is over all permutations in $S_n$. The Bianchi identities imply that $\mathrm{d}\tilde\Omega=0$, and, since $\tilde\Omega$ is a multiple of $\omega_1\wedge\cdots\wedge\omega_n$, it follows that there exists a unique $n$-form $\bar\Omega$ on $M$ such that $\pi^*\bar\Omega = \tilde\Omega$. All this is standard, except that, for clarity, I am distinguishing $\tilde\Omega$ and $\bar\Omega$. (Chern uses the same letter $\Omega$ for both.) Similarly, below, I will try to distinguish forms that Chern identifies when they differ via pullback under a submersion with connected fibers. What Chern does next, though, is what seems to be causing the confusion about 'arbitrary vector fields'. He lets $u = (u_i):S^{n-1}\to \mathbb{R}^n$ denote the inclusion of the $(n{-}1)$-sphere into $\mathbb{R}^n$, and he defines a submersion $\bar\pi: B\times S^{n-1}\to M^{2n-1}$, where $M^{2n-1}\subset TR$ is the unit sphere bundle, by $\bar\pi(e,u) = u_ie_i$. He then constructs an $(n{-}1)$-form $\tilde\Pi$ on $B\times S^{n-1}$ that has the property that $\mathrm{d}\tilde\Pi = \tilde\Omega$ and has the property that there exists an $(n{-}1)$-form $\Pi$ on $M$ (not $R$) satisfying $\bar\pi^*\Pi = \tilde\Pi$. It is important to recognize that Chern's $u_i$ are not the components of a vector field on anything, arbitrary or otherwise. N.B.: Actually, Chern says that he is going to work locally on an open subset (say) $O\subset R$ and choose a section of $B$ over $U$, i.e., an 'oriented orthonormal frame field' on $O$, he identifies the part of $M$ that lies over $O$ with $O\times S^{n-1}$ and constructs $\Pi$ on $O\times S^{n-1}$, and then he says that the resulting $\Pi$ doesn't depend on the local choice of frame field. However, he never actually uses the local section in any of his calculations, so they are perfectly valid on $B\times S^{n-1}$. Now, there's a way to avoid introducing the $u_i$ at all, and you may like this better. Consider the submersion $\pi_1:B\to M$ defined by $\pi_1(e) = e_1$. The fibers of this map are (connected) $\mathrm{SO}(n{-}1)$-orbits that are the leaves of the system $$ \omega_1=\omega_2=\cdots=\omega_n=\omega_{12}=\omega_{13}=\cdots=\omega_{1n}=0. $$ Then one constructs an $(n{-}1)$-form $\tilde\Pi$ directly on $B$ that is a polynomial with constant coefficients in the forms $$ \{\omega_{12},\omega_{13},\cdots,\omega_{1n}\}\cup \bigl\{\ \Omega_{ij}\ \bigl|\ 1 TITLE: Vanishing theorem for big divisors QUESTION [7 upvotes]: Let $X$ be a projective, smooth variety over $\mathbb{C}$, and let $D$ be a irreducible, big Cartier divisor(notice, I do not assume nefness). Then is it true that $${\rm{H}}^1(X, K_X + D) = 0\quad?$$ On the one hand, I doubt it because the result seems to be written nowhere; on the other hand, I feel the conditions here are much nicer than many vanishing theorem, and I was wondering if some modification would reduce it to one of the vanishing theorem. Besides, any references/results/comments on vanishing theorem for big divisors are great welcome! REPLY [6 votes]: Let $X\to \mathbb P ^1$ be a family of smooth quadric surfaces degenerating to $X_0=\mathbb F _2=\mathbb P (\mathcal O _{\mathbb P ^1}\oplus \mathcal O _{\mathbb P ^1} (2))$ where $0\in \mathbb P ^1$. We denote by $E$ the $-2$ curve in $X_0$, $F$ the fiber of $X_0\to \mathbb P ^1$ and by $f:X\to Z$ the corresponding small contraction (following the notation of arXiv:0901.0389 remark 4.4 and remark 4.9). For $b>>0$ consider $L$ a line bundle whose restriction to the general fiber is $\mathcal O _{\mathbb P ^1} (b+1)\boxtimes \mathcal O _{\mathbb P ^1} (b-1)$ and whose restriction to the special fiber is $\mathcal O _{X_0}(b(E+2F)+E)$ and $D\in |L|$ a general member. We claim that $R^1f_*\omega _X(D)\ne 0$. To see this consider the short exact sequence $0\to \omega _X(D-X_0)\to \omega _X(D)\to \omega _{X}(D)|_{X_0}\to 0$. Since $R^if_*\omega _X(D-X_0)\cong R^if_*\omega _X(D)\otimes \mathcal O_{Z}(-Z_0)$ (by the projection formula), then it follows that $R^1f_* \omega _X(D)\to R^1f_*\omega _{X}(D)|_{X_0}\cong R^1f_* \omega _{X_0}(D|_{X_0})$ is surjective. Consider now the short exact sequence $0\to \omega _{X_0}\to \omega _{X_0}(D|_{X_0})\to \omega _{D|_{X_0}}\to 0$. Since $R^1f_*\omega _{X_0}=0$ (by Grauert-Riemanschneider), it follows that $R^1f_* \omega _{X_0}(D|_{X_0})\to R^1f_*\omega _{D|_{X_0}}$ is surjective. Since $|L|_{X_0}|=|b(E+2F)|+E$ where $E+2F$ is the pull back of an ample divisor $H$ on $X_0$, we have $R^1f_*\omega _{D|_{X_0}}\cong R^1f_*\omega _E\otimes \mathcal O _{Z_0}(bH)$. Since $H^1(\omega _E)\cong H^0(\mathcal O _E)^\vee\ne 0$, we have $R^1f_*\omega _{D|_{X_0}}\ne 0$ and the claim follows. It should be easy to see that $D$ is big, it is not nef because $D\cdot E =E^2=-2$, moreover we can arrange that $H^1(f_*\omega _X(D))=0$ (if necessary replace $D$ by $D$ plus the pull-back of a sufficiently ample divisor on $Z$ and apply Serre vanishing). Then, by the Leray spectral sequence $H^1(\omega _X (D))\cong H^0(R^1f_*\omega _X (D))\ne 0$. Maybe there are easier examples, but this is the first one that came to mind. I do not think there can be surface examples, because if $D$ is irreducible and not nef, then $D^2<0$, but as $D$ is big, $D\sim _{\mathbb Q}D'\neq D$ subtracting common components of $D$ we obtain $\lambda D\sim _{\mathbb Q}D''$ with $0<\lambda$ and the support of $D''$ does not contain $D$. But then $D^2=\frac 1 {\lambda ^2} D\cdot D''\geq 0$ a contradiction.<|endoftext|> TITLE: Comparing a Chevalley basis with the canonical basis of the adjoint module? QUESTION [12 upvotes]: First some background: Given a simple Lie algebra $\mathfrak{g}$ over an algebraically closed field of characteristic 0 such as $\mathbb{C}$, fix a Cartan decomposition $\mathfrak{g} = \mathfrak{h} \oplus \sum_\alpha \mathfrak{g}_\alpha$. Here $\mathfrak{h}$ is a Cartan subalgebra and the sum runs over the roots $\alpha$ of $\mathfrak{g}$ relative to $\mathfrak{h}$. One can further fix a set of simple roots, leading to positive and negative roots. Then a basis of $\mathfrak{h}$ can be chosen to consist of coroots of simple roots. All of this is determined up to conjugacy under the adjoint group and leads to various compatible choices for a basis of the adjoint module $\mathfrak{g}$, including a basis vector for each 1-dimensional root space $\mathfrak{g}_\alpha$. It was realized fairly early that a careful choice would lead to structure constants in $\mathbb{Q}$. Chevalley's 1955 paper Sur certains groupes simples went a step further, showing in a uniform way how to produce a basis over $\mathbb{Z}$ whose structure constants are uniquely determined up to sign by the root data. (Some questions on MO deal with such a Chevalley basis, for instance here.) Notation of course differs a lot in the literature. In a 1966 paper Tits here created an algorithm for consistent sign choices. Here he realized that the conventional commutation product $[X_\alpha X_{-\alpha}] = H_\alpha$ works better if written with reversed factors. His paper was contemporaneous with SGA 3. (More recently Demazure revisited Tits' paper here along with Remark 6.7 in his Exp. XXIII of SGA 3.) The adjoint module for $\mathfrak{g}$ is a simple finite dimensional module. Each such module is characterized up to isomorphism by its highest weight (once a system of simple roots is fixed), which in this case is the unique highest root. Since a typical simple module has much more complicated weight space structure than the adjoint module, it was long thought that no "canonical" choice of basis (even up to signs) was possible. But in 1990 Lusztig here showed the existence of a canonical basis by an indirect procedure which specializes from a quantized enveloping algebra. At first this is done just for simply-laced types, but later in general. The idea is to show the existence of a canonical basis in the part of the quantized enveloping algebra corresponding to positive (or equally well negative) roots, then apply this to a lowest (or highest) weight vector in the module. As Lusztig indicates in his Example 3.4, direct computation of his canonical basis is possible in a few low rank cases but will get extremely difficult in general. Soon afterward Kashiwara discovered another method based on the use of crystal bases (shown by Lusztig to be equivalent to his own method); this is exposed by Jantzen in his 1996 AMS book Lectures on Quantum Groups. I've been unable to find in the literature an explicit discussion of what happens in the special case of the adjoint module, though the proof of Jantzen's Lemma 9.6b) is suggestive. It does seem to be understood by experts that the canonical basis produces a Chevalley basis up to signs. My basic question is: Is there a published comparison of Lusztig's canonical basis of the adjoint module (once the choices above have been made) with a Chevalley basis? REPLY [4 votes]: I think there is an answer to your question, and it is contained in Lusztig's work; see references and further details in my recent preprint at http://arxiv.org/abs/1602.04583 Best regards, Meinolf<|endoftext|> TITLE: Is the Cayley graph of Thompson's group isolated in the space of vertex-transitive graphs? QUESTION [12 upvotes]: Consider Thompson's group (the one commonly referred to as $T$), which is finitely presentable. Consider the Cayley graph, but then forget the coloring and direction on edges. So now we just have an undirected graph. Call this $G$. Let $B \subset G$ be a ball around some vertex in $G$ of large radius. Since $G$ is vertex-transitive, the choice of vertex is irrelevant. The radius must be large enough that $B$ includes the cycles corresponding to the relations in the presentation. If $G'$ is a (connected) vertex-transitive graph with a ball isomorphic to $B$, is $G'$ necessarily isomorphic to $G$? If we make the additional assumption that $G'$ is itself the Cayley graph of some group $H$, then I believe the answer is yes. My reasoning is that since $B$ is large enough to witness both relations, $H$ must be a quotient of $T$, but $T$ is simple. I'm a little uncertain about the claim that $H$ must be a quotient, though. REPLY [6 votes]: I'll suggest an approach. I'll not focus on $T$ especially: let $H$ be a finitely presentable group, pick a generating set $S$; how can we prove that the unlabeled Cayley graph (denoted $(H,S)$) is isolated among vertex-transitive graphs? 1) a necessary condition for this isolation statement is that $(H,S)$ of $H$ is isolated among labeled Cayley graphs. This condition is simply called, for a group, "isolated", because it does not depend on the generating subset. It is equivalent to be finitely presented and finitely discriminable, where the latter means: having a finite subset $\Omega\subset H\smallsetminus\{1\}$ such that every nontrivial subgroup $N$ of $H$ satisfies $N\cap\Omega\neq\emptyset$. Simple groups, finite groups, are finitely discriminable, but there are many others; for instance Thompson's group $F$. 2) Now here is a condition (X) for $(H,S)$ which implies, if $H$ is isolated, that $(H,S)$ is isolated among vertex-transitive graphs: (X): there exists $R_0\ge 1$ such that every automorphism of the $R_0$-ball in $(H,S)$ fixing 1 is the identity. Indeed, assume that $(H,S)$ satisfies (X). Let $Y$ be a (non-empty) vertex-transitive graph whose $R_0$-balls are isomorphic to the balls of $(H,S)$. Let $G$ be the automorphism group of $Y$. We claim that $G$ acts freely on $Y$. Indeed, assume otherwise, hence there exists a vertex $v$ and $g\in G$ fixing $v$ but not fixing some neighbor $w$ of $v$. Since $g$ is an automorphism of the $R_0$-ball around $v$ and by (X), we deduce that $g$ fixes $w$, a contradiction. Now let us be more precise on the labeling: $(H,S)$ has oriented edges labeled by generators (we can assume wlog $1\notin S$). Now let $v$ be a vertex in $Y$ and $e$ an oriented edge in $Y$ (here I just mean an ordered pair of vertices in $Y$ linked by an edge, since $Y$ is not an oriented graph!), contained in the $R_0$-ball around $v$. Then there exists a unique isomorphism $f_v$ from $B_Y(v,R_0)$ to $B_1((G,S),R_0)$ mapping $v$ to 1, let $s(v,e)\in S$ be the label of the image of $e$ by this isomorphism. Let us show that $s(v,e)$ does not depend on $v$: it is enough to show that $s(v,e)=s(v',e)$ if $v,v'$ are neighboring vertices such that $e$ is in the $R_0$-ball around both. $f_v$ maps $v'$ to some element $t\in S$. Hence, writing $L_g(h)=gh$ in $H$, we see that $L_t^{-1}f_v$ maps $v'$ to 1, and by uniqueness we get $f_{v'}=L_t^{-1}f_v$. $L_t$ preserves the labeling, we deduce that $s(v,e)=s(v',e)$. Hence with this labeling, $Y$ has the same $R_0$-balls as $(H,S)$. On the other hand, the definition of the labeling was canonical and therefore it is invariant by isometries of $Y$. Since, after fixing some vertex 1 in $Y$, $Y$ is the Cayley graph of $G$ with respect to its 1-sphere (call it $S'$), this shows that this is the Cayley labeling of $S'$, modulo the canonical bijection $S'\to S$ mapping $s'$ to the label of $(1,s')$. If now $H$ is isolated and $R_0$ has been chosen large enough, this implies that $S'\to S$ extends to an isomorphism $G\to H$. So you should check (X). You first need to check that the isometry group of the Cayley graph is reduced to $H$, I don't know if it holds for $T$, and maybe this requires a good choice of generating set to avoid obvious symmetries.<|endoftext|> TITLE: Classification of $SU(2)$ principal fibre bundles over four-dimensional manifolds QUESTION [11 upvotes]: I would like to find a pedagogical reference where the classification, up to isomorphism, of principal $SU(2)$ bundles over a four-dimensional compact, oriented manifold is explained. In particular I am interested in the torus $T^4$ case. Is there a similar classification when the base manifold is non-compact, in particular in the case when it is $\mathbb{R}^4$ or $\mathbb{R}^4-\left\{ 0\right\}$? A pedagogical answer explaining this problem would be also very welcome. Thanks. REPLY [8 votes]: Let $X$ be an $(n-1)$-connected CW complex with $\pi_n(X) = G$. By attaching cells of dimensions at least $n+2$, we obtain a CW complex $Y$ the same $(n+1)$-skeleton as $X$, but with $\pi_i(Y) = 0$ for $i > n$. For $i \leq n$, we have, by cellular approximation, $$\pi_i(Y) = \pi_i(Y^{(n+1)}) = \pi_i(X^{(n+1)}) = \pi_i(X),$$ so $Y$ is a $K(G, n)$ with $X$ as a subcomplex. If $M$ is a CW complex of dimension at most $n$, then $$[M, X] = [M, X^{(n+1)}] = [M, Y^{(n+1)}] = [M, Y] = [M, K(G, n)] = H^n(M, G).$$ Now consider the case $X = BSU(2)$. As $\pi_i(BSU(2)) = \pi_{i-1}(SU(2)) = \pi_{i-1}(S^3)$, $BSU(2)$ is $3$-connected and $\pi_4(BSU(2)) = \mathbb{Z}$, so for a CW complex $M$ of dimension at most four $$\operatorname{Prin}_{SU(2)}(M) = [M, BSU(2)] = [M, K(\mathbb{Z}, 4)] = H^4(M, \mathbb{Z}).$$ The isomorphism $[M , BSU(2)] \to [M, K(\mathbb{Z}, 4)]$ is given by $[f] \mapsto [\iota \circ f]$ where $\iota : BSU(2) \to K(\mathbb{Z}, 4)$ is the inclusion map. As $\mathbb{Z} \cong \pi_4(K(\mathbb{Z}, 4)) \cong H_4(K(\mathbb{Z}, 4); \mathbb{Z})$, the identity map $\operatorname{id} : \mathbb{Z} \to \mathbb{Z}$ gives rise to an element of $\operatorname{Hom}(H_4(K(\mathbb{Z}, 4); \mathbb{Z}), \mathbb{Z})$ and hence an element $\alpha$ of $H^4(K(\mathbb{Z}, 4); \mathbb{Z})$. The isomorphism $[M, K(\mathbb{Z}, 4)] \cong H^4(M; \mathbb{Z})$ is given by $[g] \mapsto g^*\alpha$. The composition of these two isomorphisms is an isomorphism $[M, BSU(2)] \to H^4(M; \mathbb{Z})$ given by $[f] \mapsto (\iota\circ f)^*\alpha = f^*(\iota^*\alpha)$. The map $\iota^* : H^4(K(\mathbb{Z}, 4); \mathbb{Z}) \to H^4(BSU(2); \mathbb{Z})$ is an isomorphism because $BSU(2)^{(5)} = K(\mathbb{Z}, 4)^{(5)}$. As $\alpha$ is a generator of $H^4(K(\mathbb{Z}, 4); \mathbb{Z})$, $\iota^*\alpha$ is a generator of $H^4(BSU(2); \mathbb{Z}) \cong \mathbb{Z}c_2$, so $\iota^*\alpha = \pm c_2$. Therefore the isomorphism $\operatorname{Prin}_{SU(2)}(M) \to H^4(M, \mathbb{Z})$ constructed above is either $P \mapsto c_2(P)$ or $P \mapsto -c_2(P)$. Either way, we see that for a CW complex $M$ of dimension at most four, principal $SU(2)$-bundles over $M$ are completely determined by their second Chern class; moreover, every element of $H^4(M; \mathbb{Z})$ arises as the second Chern class of some $SU(2)$-principal bundle on $M$.<|endoftext|> TITLE: Simple Hurwitz Groups of order less than 10^7 QUESTION [6 upvotes]: I'm trying to calculate a table of all simple hurwitz groups of order less than 10^7. None of the tables I found went further than 10^6, so I decided to use the tables of all simple groups up to 10^7 (which is easy to find), and then remove the one which are not hurwitz. Using some of the papers I have read on hurwitz groups I have managed to reduce the list down to only the projective special linear groups of degree 2 (that are hurwitz), the Janko groups J1 and J2 (which I know are hurwitz) and four others, which I am not sure about. The steinberg groups 2A(3, 9), 2A(2, 49), 2A(2, 64), and the chevalley group C(3, 2). Which of these four groups are hurwitz (I suspect none of them)? Added later: I now know all hurwitz groups with orders less than 10^10. These are the PSL(2,q) groups (which satisfy the conditions for it to be hurwitz), J1, J2, 3D(4, 2), He, G(2, 5), and 2G(2, 27). There is only one group that is unknown up to 10^12, and that is C(4, 2) (which is isomorphic to B(4, 2)). REPLY [6 votes]: In the paper M.C. Tamburini and M. Vsemirnov, Irreducible $(2,3,7)$-subgroups of ${\rm PGL}_n(F)$, $n \le 7$, J. Algebra 300 (2006), 339–362 the Hurwitz groups with absolutely irreducible projective representations of degrees up to $7$ over any field are determined (although the results in the smaller dimensions (up to $5$ I think) were not new. Here is a list of the simple groups that arise - I hope I have copied this correctly! Added later: I am sorry, I misunderstod the paper. This is not a complete list - this is a list of so-called rigid triples - and the authors say that they will complete the classification in a leter paper, which has now appeared in J. Algebra 321 (2009), no. 8, 2119–2138. In the second paper, they have not completely identified the groups that arise in dimension $7$, but they remark that some of the groups $G_2(q)$ arise - these had been found earlier by Malle. $n=2$: ${\rm PSL}(2,p^m)$, where $m=1$ if $p \equiv 0,\pm 1 \mod 7$, and $m=3$ otherwise. $n=3$ and $n=4$: nothing new. $n=5$: ${\rm PSL}(5,p^m)$ and ${\rm PSU}(5,p^m)$ with $p \ne 5$ for certain values of $m$. See the survey paper by Conder for details. $n=6$: ${\rm PSL}(6,p^m)$, with $p \ne 3$ and $m$ odd, and ${\rm PSU}(6,p^m)$, with $p \ne 3$ and $m$ even where, in both cases, $m$ is the order of $p$ mod $9$. $n=7$: ${\rm PSL}(7,p^m)$, with $p \ne 7$ and $m$ odd, and ${\rm PSU}(7,p^m)$, with $p \ne 7$ and $m$ even where, in both cases, $m$ is the order of $p$ mod $49$. Added later: I checked with a computer calculation that none of the three groups of order less than $10^9$ that you are uncertain about are Hurwitz. These are $C_2(7) = {\rm PSp}(4,7)$, $D_4(2) = {\rm P \Omega}^+(8,2)$ and $^2D_4(4) = {\rm P \Omega}^-(8,2)$. The first of these has a projective representation of degree $4$ and is covered by known results: none of the $4$-dimensional symplectic groups are Hurwitz. As far as I know, the $8$-dimensional orthogonal groups are not covered by published results, but it is very likely that somebody has dome these calculations already! The computer checks I used are more or less brute force, and they work easily for groups of order up to $10^9$, but will start to become impractical with group orders much higher than that. I checked also that ${\rm He}$ is not an image of $(2,3,7;10)$.<|endoftext|> TITLE: Quotient of metric spaces QUESTION [14 upvotes]: Let $(X,d)$ be a compact metric space and $\sim$ an equivalence relation on $X$ such that the quotient space $X/\sim$ is Hausdorff. It is well known that in this case the quotient is metrizable. My question is, can we choose a compatible metric on $X/\sim$ so that the quotient map does not increase distances? As in this question Is there a conceptual reason why topological spaces have quotient structures while metric spaces don't? we can define a pseudometric $d'$ on the quotient $X/\sim$ by letting \begin{equation*} d'([x],[y]) = \inf\{ d(p_1,q_1) + \cdots+ d(p_n,q_n):p_1 = x,q_i \sim p_{i+1},q_n=y\} \end{equation*} where $[x]$ is the equivalence class of $X$. With this $d'$ the quotient map clearly does not increase distances. So do we know when $d'$ is a metric and when it is compatible with the quotient topology? Wikipedia says that $X$ being compact implies that $d'$ is compatible with the quotient topology; I would appreciate a reference or a quick proof. REPLY [2 votes]: Also in the topological transformation group the existence of one $G$-invariant metric by a $G$-metric space $X$ is open. You can thing your problem in this case like a particular case in orbit space $X/_G$.<|endoftext|> TITLE: Mountain Pass theorem for minimization problems with constraints QUESTION [7 upvotes]: Let $I[u]$ be a functional on a (possibly infinite dimensional) Hilbert space. Then, under some conditions, the Mountain Pass theorem guarantees the existence of a saddle point (see http://en.wikipedia.org/wiki/Mountain_pass_theorem for the precise statement of the theorem). I wonder if there is a generalization of this theorem for minimization problems with constraints. REPLY [4 votes]: You can generalize the mountain pass theorem if the constraints consist of a Banach manifold. This is because you can construct a pseudo gradient flow on a Banach manifold, which is required in the proof of the theorem. Check Chapter 27 of 'Nonlinear Functional Analysis' by K. Deimling. For example, in the paper 'Standing waves of some coupled nonlinear Schroedinger equations' by A. Ambrosetti and E. Colorado (2007), the authors used the mountain pass theorem on a Nehari manifold.<|endoftext|> TITLE: Fourier transform of compactly supported distribution is smooth QUESTION [8 upvotes]: My advisor made the comment that if $u\in \mathcal{E}'$ is a compactly supported distribution, then $\hat{u}(\xi)\in C^{\infty}(\mathbb{R}^n)$ is actually a smooth function (not merely a distribution or generalized function). I'm trying to prove this, but I'm realizing I'm not even sure how we should properly define the F.T. of a distribution in $\mathcal{E}'$. When $u\in \mathcal{S}'$, then the standard definition $$\left<\hat{u},\phi \right> := \left$$ makes perfect sense, since $\phi\in\mathcal{S}\Rightarrow \hat{\phi}\in\mathcal{S}$, and so pairing $\hat{\phi}$ with an element of $\mathcal{S}'$ is well defined. The problem I'm having with applying this same definition to $u\in\mathcal{E}'$ is that while $\phi\in C^{\infty}$, we don't necessarily have $\hat{\phi}\in C^{\infty}$; and so the statement $\left$ might not make sense, as $u$ should only be acting on test functions from $C^{\infty}$. I've been toying around with the idea of a smooth cutoff function $\rho \equiv 1$ one on $\text{supp}(u)$, and taking $$\left<\hat{u},\phi \right> := \left,$$ to ensure that $\widehat{\rho\phi}$ is at least well-defined. But I'm not sure if that is even close to the right idea, though. So some clarification on this matter would be much appreciated. That aside, considering that $e^{-ix\cdot\xi}\in C^{\infty}(\mathbb{R}^n_x)$, I'm at least able to show that $\left$ is a function of $\xi$ (and not a generalized function). This of course heuristically matches up with the idea that $$\left = \int u(x)e^{-ix\cdot\xi}\;dx = \hat{u}(\xi),$$ if $u(x)$ had in fact been a function. Since $\exists\{u_k(x)\}\subset C^{\infty}_0(\mathbb{R}^{n})$ that converge weakly to $u\in\mathcal{E}'$, then we have \begin{align*} \mathbb{C} \ni \left &= \lim_{k\to\infty}\left\\ &= \lim_{k\to\infty}\int u_k(x)e^{-ix\cdot\xi}\;dx\\ &= \lim_{k\to\infty}\hat{u_k}(\xi)\\ &= U(\xi), \end{align*} some function of $\xi$. But when it comes to showing that $U = \hat{u}$ as distributions, we would need to prove that $$\left = \int U(\xi)\psi(\xi) \;d\xi= \left,$$ for all $\psi$ in some appropriate space of test functions, which leads to the same issues with $\hat{\psi}$ I was having before. I also haven't been able to figure out how I would prove that $U(\xi)$ is smooth. So any help with that would also be welcomed. REPLY [7 votes]: The Fourier transform is the same as the FT $\mathcal{S}'\to\mathcal{S}'$ since $\mathcal{E}'$ is canonically contained in $\mathcal{S}'$ (which boils down to $\mathcal{S}$ being dense in $\mathcal{E}$). It is therefore perfectly well-defined to talk about Fourier transforms of compactly supported distributions. One way of making sense the statement $\mathcal{F}(u) = \xi\mapsto\langle{u,e^{-i\xi x}}\rangle$ would therefore be to find an approximation of $e^{-i\xi x}$ with Schwartz functions. Another is the way with cutoff functions you described. Note that the support of a distribution is defined exactly so that "$f_1 = f_2$ in an open set containing $supp(u)$ $\implies \langle u,f_1 \rangle = \langle u,f_2\rangle$" holds. This proves independence of the cutoff-function. Actually proving it is another question. There is an ugly (but standard) way with approximations and an elegant way. The elegant way is this proof: $$\langle \mathcal{F}u,\phi\rangle = \langle u,\mathcal{F}\phi\rangle = \langle u(x),\int \phi(\xi)e^{-i\xi x}d\xi\rangle = \int \langle u(x),\phi(\xi)e^{-i\xi x}\rangle d\xi = \int \phi(\xi) \langle u(x),e^{-i\xi x}\rangle d\xi$$ To make this work one has to show that the first integral exists as a $\mathcal{S}$-valued Pettis-integral and that this integral equals the Fourier transform. It is probably easiest to do this if $\phi$ has compact support because there are general existence theorem for the Pettis integral of continuous functions with compact support. Now the smoothness is a thing that can either be done by brute force or with elegance (i.e. some more functional analysis). My favorite way: Show that $\xi \mapsto (x\mapsto e^{-i\xi x})$ is sufficiently smooth as a map $\mathbb{R}^n \to \mathcal{E}(\mathbb{R}^n)$. Because $\langle u,-\rangle: \mathcal{E}(\mathbb{R}^n) \to \mathbb{C}$ is analytic (because it is linear and continuous) you then find that $\mathcal{F}(u)$ is at least as smooth as you have proved before. Now what kind of smoothness you can easily prove certainly depends on your previous knowledge. Say you prove it continuous (should be possible directly with the definition) then you can use the formulas for derivatives of fourier transform to prove continuity of all derivatives. If you can prove $C^k$ (also possible with careful application of the definition and some Taylor-ing) then you get directly that $\mathcal{F}(u)$ is $C^k$. But in fact more is true: $\mathcal{F}(u)$ is real analytic! It is in fact the restriction of the complex analytic map $\mathbb{C}^n \to \mathbb{C}, \zeta\mapsto \langle u,e^{-i \zeta x}\rangle$ to $\mathbb{R}^n$. So one should really show that $\zeta \mapsto (x\mapsto e^{-i \zeta x})$ is complex analytic as a map $\mathbb{C}^n\to\mathcal{E}(\mathbb{R}^n)$. This can be done by showing that the usual power series converges w.r.t. the $\mathcal{E}$-topology. And the cherry on top is the Paley-Wiener theorem that gives you the converse: Every holomorphic function $\mathbb{C}^n \to \mathbb{C}$ with a certain asymptotic behaviour (which is also satisfied by $\mathcal{F}(u)$) is the Fourier transform of some distribution with compact support. This can even be generalized to a similar characterizations about distributions with support contained in some fixed convex subset $K\subseteq\mathbb{R}^n$. REPLY [6 votes]: If $u \in \mathcal E'$ is compactly supported, then your smooth cutoff extends it to a tempered distribution. The fact that the Fourier transform of a compactly supported distribution is analytic (not just $C^\infty$) is part of Schwartz's version of the Paley-Wiener theorem.<|endoftext|> TITLE: Schemes over topological rings QUESTION [12 upvotes]: I have recently been interested in studying an extension of 'usual' algebraic geometry to take into account the topology of $R$ in the definition of the affine scheme $\mathrm{Spec}\, (R)$ when the ring comes equipped with a given topology. For example, I am curious to investigate schemes over $C ^\infty (\mathbb{R}^d)$, where it is obvious from the beginning that it would be 'wrong' to ignore the canonical Fréchet space topology (for example, the 'correct' definition of $\mathrm{Spec}\, (R)$ in this case should probably be the collection of all closed prime ideals). (Part of the motivation for this is, in the spirit of Grothendieck's famous quote, to replace a bad category of only good objects (the smooth category) with a good category containing some bad objects.) I can't imagine that people haven't investigated things like this before. On the other hand, I don't know what this subject would be called, and so I don't know where to begin looking to read up on the subject. Could someone help point me in the right direction? REPLY [5 votes]: Although I agree with your statement that it would be wrong to forget the topology of $R$, I would say that there is little evidence to support the idea that the set of closed prime ideals is the `right' definition for Spec(R). This is certainly not the case for analytic geometry over a non-Archimedean field, which has many interesting points not related to prime ideals at all. In general what you often do is try to define a topos and then identify which objects in the topos are sufficiently `geometric' to deserve your attention. If you really must, you can also try to identify a good family of points. Some nice examples are Spivak's derived manifolds and $C^\infty$-schemes (which are dual to the smooth algebras that Qiaochu mentioned). Both of these are essentially some purely formal extension of the category of manifolds, and do not really involve any understanding of the topological algebra of $C^\infty$ functions. In a different direction, Alain Connes's flavour of non-commutative geometry really depends on functional analysis of Dirac operators on nuclear Fréchet spaces - but as far as I know, there is no underlying topological space (or topos) in this case. Perhaps the the theory of schemes you are looking for will eventually involve a combination of these ideas.<|endoftext|> TITLE: Torsors under the group scheme over projective line QUESTION [5 upvotes]: Consider the group scheme $\mathcal T$ over $\mathbf P^1$ given locally (variable $t$) by the equation $x^2 - f(t)y^2 = 1$ where $f(t)$ is a polynomial of degree $r$ with distinct roots (assume that $r$ is even and the field is algebraically closed). What is the group of torsors for $\mathcal T$, i.e. $\mathrm H^1(\mathbf P^1, \mathcal T)$? Using the machinery of spectral sequences I was able to deduce that my group is an extension of the Jacobian $J(C)$, where $C$ is the double cover that splits $\mathcal T$: $1 \to (\mathbb Z/2\mathbb Z)^{r-1} \to \mathrm H^1(\mathbf P^1, \mathcal T) \to J(C) \to 1$ Unfortunately, I cannot confirm whether my answer is correct. It is difficult to believe that such example hasn't been computed anywhere in the literature, but I couldn't find any reference. Would be happy if someone could give a hint or a reference. REPLY [2 votes]: Taking cohomology of the short exact sequence $\mathcal T \to f_* \mathbb G_m \to \mathbb G_m$ gives a long exact sequence: $H^0(C, \mathbb G_m) \to H^0 (\mathbb P^1, \mathbb G_m) \to H^1(\mathbb P^1, \mathcal T) \to H^1(C, \mathbb G_m) \to H^1(\mathbb P^1, \mathbb G_m) $ We can identify most of the terms: $k^\times \to k^\times \to H^1(\mathbb P^1, \mathcal T) \to \mathcal J \times \mathbb Z \to \mathbb Z $ Assuming $k$ is algebraically closed, the arrow $k^\times \to k^\times$, which is squaring, is an isomorphism. The map $\mathcal J \times \mathbb Z \to \mathbb Z$ is the norm map, which sends $\mathcal J$ to $0$ and doubles $\mathbb Z$, so the kernel is $\mathcal J$. Hence I think your group is just $\mathcal J$. $\mathcal J$ is an extension of itself by $(\mathbb Z/2)^{r-2}$, but I don't know where the $-1$ comes from. It may depend on how you extend your group scheme to $\infty$, which you didn't really specify.<|endoftext|> TITLE: Pseudo-Prikry sequences vs Prikry sequences QUESTION [7 upvotes]: Definition: Let $V\subseteq W$ be two transitive models of $ZFC$. A pseudo-Prikry sequence, $s$, at a cardinal $\kappa$ for $(V, W)$ is an $\omega$-sequence, cofinal at $\kappa$ such that for every club $D\subseteq \kappa$ from $V$, $s\setminus D$ is finite. In the paper "On squares, outside guessing of clubs and $I_{ TITLE: exp(S) exp(T) = exp(S+T) for commuting operators QUESTION [11 upvotes]: The standard way to prove the exponential law for two bounded commuting operators $S, T$ $$ \exp(S)\exp(T) = \exp(S+T) $$ is to pass by the binomial formula and the power series of $\exp(.)$. I wonder whether one can prove it alternatively by a (Dunford-Riesz) functional calculus argument i.e. a resolvent-based proof, either directly, via $$ \exp(S)\exp(T) = \tfrac{1}{(2\pi i)^2 } \int_{|z|=r} \int_{|\lambda|=R} e^{\lambda+z} R(\lambda, S) R(z, T)\,d\lambda dz $$ for some $r> \|T\|, R > \max(r, \|S\|)$ or maybe by an operator matrix approach (for example the functional calculus of $A= \left( \begin{array}{rr} S & I \\ 0 & T \end{array}\right)$). REPLY [4 votes]: Also: for bounded operators, $(I+T/n)^n$ converges to $e^T$ uniformly on bounded sets (however, uniform convergence on compact sets will suffice). So if $T$ commutes with $S$, $$e^Te^S=\lim_{n\to+\infty}\Big(I+\frac{T}{n}\Big)^n\Big(I+\frac{S}{n}\Big)^n=\lim_{n\to+\infty}\bigg(I+\frac{T+S+\frac{TS}{n}}{n}\bigg)^n=e^{S+T}\, .$$<|endoftext|> TITLE: How important is Weil's decomposition theorem today? QUESTION [17 upvotes]: Andre Weil's Apprenticeship of a Mathematician (p. 46) tells how he as a student realized that all of Fermat's uses of descent are unified in one principle: "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." Serre agrees with Weil's estimate of the result and says that Weil put an end to the time when "a different little miracle" seemed to happen in each use of descent (André Weil. 6 May 1906-6 August 1998. Biographical Memoirs of Fellows of the Royal Society, 45 (1999), 520--529). One special case of this is omnipresent in undergraduate number theory: when $x$ and $y$ are relatively prime integers then the GCD of $x+y$ and $x-y$ is either 1 or 2. The theorem is widely praised but it is not widely given. Books that mention it often go on to say they will not use the full strength. Lang's Fundamentals of Diophantine Geometry (p. 263) gives a "reformulation of the existence of Weil functions in the words of Weil's decomposition theorem" and I believe the two formulas he gives after that actually make up a form of Weil's theorem but am not sure of that. Anyone who does not already know these formulas would need considerable explanation of notation to read them, so I will not type them here. The passage can be found by searching "Weil's decomposition theorem" in Google books . Is Lang's Theorem 3.7 a modern form of Weil's theorem? Are there other modern expositions of the theorem today? Weil's own dissertation is not too hard to read but he was inventing a lot of ideas and the terminology is not what we use today. And anyway I wonder how important the theorem is today. Has some other insight tended to replace it? Vesselin Dimitrov's answer led me to see that the suggestive statement Weil gives in his autobiography is precisely proven in many books today in the following form. Define the height function $H$ for rational numbers so $H(m/n)$ is $\mathrm{Max}(|m|,|n|)$ for relatively prime $m,n$. Then for relatively prime polynomials $f(T),g(T)$ and $d$ the maximum of their degrees there is some real $C>0$ with $$H(x)^d \leq C\cdot H(g(x)/f(x))$$ whenever $f(x)\neq 0$. This seems to suffice for Fermat's uses of descent. It is never called Weil's decomposition theorem to my knowledge. Weil's dissertation proves the theorem for algebraic points on curves rather than for rational numbers. That is called his decomposition theorem and it has various modern forms which are only given in rather advanced texts as Vesselin Dimitrov describes. REPLY [11 votes]: Yes, this is the modern statement of Weil's theorem of decomposition. It is a basic component of the theory of heights. For a more recent exposition see 2.7.15 in Bombieri and Gubler's Heights in Diophantine Geometry. If you look for a specific application of the theorem and its point of view, you should be aware of Bombieri's paper [On Weil's "Theoreme de decomposition," Amer. J. Math., 1983]. There, Bombieri employs the theory of heights (Weil's theorem of decomposition and a theorem of Neron, cf. 9.3.10 in Bombieri-Gubler), to extend work of Sprindzhuk on the Hilbert irreducibility theorem and deduce a generalization of an old theorem of Runge stating the finiteness of solutions $(x,y) \in \mathbb{Z} \times \mathbb{Q}$ to $G(x,y) = 0$ for an irreducible $G \in \mathbb{Z}[x,y]$ whose leading homogeneous part is not proportional to a power of an irreducible polynomial. Bombieri's main result in that paper is roughly that if $f : C \to \mathbb{P}^1$ is a morphism from a curve over a number field $K$, then for $P \in C(\bar{K})$, the contribution from every given pole of $f$ to the height of $f(P)$ can be approximately read from the factorization of the function $f \in K(C)$. This exactly captures the spirit of the theorem of decomposition. To get at the precise statement, let $\iota : C \hookrightarrow \mathbb{P}_K^N$ a projective embedding of $C$ and $d_v(\cdot,\cdot)$ the $v$-adic chordal distance on $\mathbb{P}^N(\mathbb{C}_v)$. Then, for all $P \in C(K) \setminus f^{-1}(\infty)$, and for each pole $Q$ of $f$, it holds $$ \sum_{\substack{v \\ d_v(\iota(P),\iota(Q)) < 1}} \log^+{|f(P)|_v} = \frac{\mathrm{ord}_Q (1/f)}{\deg{f}} \sum_v \log^+{|f(P)|_v} + O \Big( \sqrt{\sum_v \log^+{|f(P)|_v}} \Big), $$ with the sum being over the places $v$ of $K$ at which $\iota(P)$ belongs to the open unit disk centered at $\iota(Q)$ (which for almost all $v$ means that $P$ is closer to $Q$ than to any other pole), and with an implied constant depending only on $f$ and the embedding $\iota$. With some care in the notation, the statement moreover extends to all $P \in C(\bar{K})$. Here is the intuitive meaning of this. Note that the local height of $f(P)$ at $v$ is large if and only if $P$ is $v$-adically close to some pole of $f$. Then as $v$ varies, each pole is approached with frequency proportional to the order of the pole. A full treatment of this result of Bombieri's and its consequent generalization of Runge's theorem is presented in the chapter 9 on Neron-Tate heights in Bombieri and Gubler's book. There, it is used to give an essentially algebro-geometric proof of the Hilbert irreducibility theorem.<|endoftext|> TITLE: Making spheres shellable QUESTION [8 upvotes]: This is equivalent to my earlier question A question about something like "shelling" in a PL manifold, but maybe more comprehensible and to the point. Given a triangulation of the PL sphere $S^n$, is there always a subdivision (a.k.a. refinement, a.k.a. finer triangulation) that makes it shellable? Put this way, I'm guessing that the answer is well-known. EDIT: I quickly got two different answers, each of which seems to give just what I need. I'm more or less arbitrarily accepting Allan's. REPLY [7 votes]: According to the reviewer of Bruggesser, H.; Mani, P., Shellable decompositions of cells and spheres. Math. Scand. 29 (1971), 197–205 (1972), MR0328944, "The authors provide a rather ingenious proof of the following proposition: For every triangulation of an n-cell and every triangulation of an n-sphere there exists a subdivision of the triangulation that is sellable."<|endoftext|> TITLE: Lower bound for spectral radius on $\operatorname{GL}(n,\mathbb{Z})$ QUESTION [8 upvotes]: Consider the group of matrices $G =\operatorname{GL}(n,\mathbb{Z})$ with integer entries and determinant $\pm 1$. For each matrix $D \in G$, the product of the eigenvalues of $D$ is equal to $\det D =\pm 1$, and so the spectral radius $\rho(D)$, which is the size of the largest eigenvalue, is at least one. Moreover, if $\rho(D) =1$, then all the eigenvalue have size 1 and can be proven to be roots of unity. My question is this: is there a lower bound $B>1$ such that if $\rho(D) \neq 1$ then $\rho(D) \geq B$? If so, does this $B$ depend on $n$? REPLY [6 votes]: Lehmer's conjecture (which is what you'll want to Google) says that the spectral radius $R$ of a non-cyclotomic polynomial of degree $n$ satisfies $$ R > 1 + \frac{c}{n} $$ for an absolute constant $c$. There is even a conjecture for the value of $c$ corresponding to a certain 10th degree polynomial. Dobrowolski proved that $$ R > 1 + \frac{c'}{n}\left(\frac{\log\log n}{\log n}\right)^3, $$ and as Doug Lind noted, it is known one can take $c'=1/2$, but no one has improved the form of Dobrowolski's result in more then 30 years. For non-reciprocal polys, Chris Smyth proved Lehmer's conjecture with best possible constant in that case. REPLY [5 votes]: Using Qiaochu's terminology, there is an explicit lower bound for the spectral radius $R$ of a non-cyclotomic polynomial of degree $n$, namely $$ R>1+\frac{1}{2n}\Bigl(\frac{\log\log n}{\log n}\Bigr)^3 $$ (see A. Dubickas, On a conjecture of A. Schinzel and H. Zassenhaus, Acta Arith. 63 (1993), 15-20 for a slightly sharper result).<|endoftext|> TITLE: Matching polynomials and Ramanujan graphs QUESTION [5 upvotes]: Is it purely coincidental that the same number $2\sqrt{d-1}$ appears in these two following apparently disparate concepts? A $d-$regular graph is said to be called Ramanujan if its adjacency eigenvalues except the highest and the lowest are inside the interval, $[-2\sqrt{d-1}, 2\sqrt{d-1}]$. (one considers only one copy of the highest and the lowest if either has multiplicities) The largest root of the matching polynomial of a graph with largest degree is $d$ over its vertices, is $2\sqrt{d-1}$ A side question : anyone knows of a pedagogic rewriting/exposition of the second result apart from its original proof in this paper, http://projecteuclid.org/euclid.cmp/1103857921 ? A related fact that has been shown recently is that if one assigns elements of $\mathbb{Z}_k$ to the set of oriented edges of a graph ( such that the group element assigned to he edge $(u,v)$ is inverse of the group element assigned to $(v,u)$ ) then over all such signings $s$, one has, $\mathbb{E}_s [ det ( xI - A_{s,i} ) ] = \mu (x)$ where $A_{s,i}$ is the $i-$fold Hadamard product of the signed adjacency matrix $A_s$ and $\mu$ is the matching polynomial of the graph. REPLY [8 votes]: One approach that goes some way to explaining this is through the path-tree of a graph. This is defined as follows. Choose a vertex $u$ in the graph $G$, The vertices of the path-tree $T(G,u)$ are the paths in $X$ that start at $u$; two paths are adjacent if one is a maximal proper subpath of the other. If we use $\phi$ to denote the characteristic polynomial and $\mu$ for the matching polynomial and abbreviate $T(G,u)$ to $T$, then we have \[ \frac{\mu(G\setminus u,x)}{\mu(G,x)} = \frac{\mu(T\setminus u,x)}{\mu(T,x)}. \] Here I am using $u$ to denote the one-vertex path in $T(G,u)$ (as well as the vertex in $G$). This identity is useful because the matching and characteristic polynomials of a tree are equal, and so we can study the right side using linear algebra. One consequence is that the largest zero of the matching polynomial of $G$ is equal to the largest zero of the characteristic polynomial of $T$. This yields the bound stated on the largest eigenvalue of the matching polynomial, because the largest eigenvalue of a tree with maximum valency $d$ is $2\sqrt{d-1}$. The source for this is my paper "Matchings and walks in graphs", J. Graph Theory 5 (1981) 285--297. There is also an exposition in Chapter 6 of my book "Algebraic Combinatorics". (Sorry about the self promotion, but I am not aware of other treatments.) REPLY [4 votes]: The moments of the adjacency matrix eigenvalues count closed walks in the graph, while the moments of the matching polynomial roots count tree-like closed walks. When the graph has few short cycles, as in a Ramanujan graph, these sets of walks are both rather similar to closed walks on an infinite $d$-regular tree. This is far from the whole story; see Chris Godsil's book for much more.<|endoftext|> TITLE: Variational formulation of second order equations of the divergence form QUESTION [5 upvotes]: Consider the second order operator $Lu=\partial_i(a_{ij}\partial_j)u+b_i\partial_iu+cu$. Can we find a functional $I[u]$ such that $Lu$ is the variation of $I[u]$ with respect to $u$? I have successfully dealt with the first and third term, but found difficulties with the second term. Is it even possible to do so? REPLY [7 votes]: Actually, Math604 is closer to the right answer. To see the correct condition, you need to pay attention to the placement of your indices. Your operator should be written in the form $$ Lu = \partial_i(a^{ij}\partial_ju) + b^k\partial_ku + c u = 0, $$ where one sums over repeated indices in opposition (i.e., `one up, one down'). Also, you didn't say this, but, usually, when one writes the operator this way, one assumes that the coefficients $a^{ij}$, $b^k$ and $c$ are functions of the independent variables (say, $x^i$), and that the matrix $A = (a^{ij})$ is both symmetric and invertible. If you didn't intend for us to make these assumptions, you should say so, and we can work out the more general case. Under these conditions, there are unique functions $f_j$ satisfying the equations $$ a^{ij}f_j = b^i. $$ Then the condition you want is simply that the $1$-form $\phi = f_i\,\mathrm{d} x^i$ should be exact, i.e., that there should exist a function $f$ on your $x$-domain such that $\partial_i f = f_i$ for all $i$. Then, the Lagrangian you want is $$ I[u] = \frac12\int_D e^{f(x)}\bigl(-a^{ij}(x)\,(\partial_iu)(\partial_ju) + c(x)\, u^2\bigr)\,\mathrm{d} x $$ If the $1$-form $\phi$ is not exact, the equation $Lu=0$ is not the Euler-Lagrange equation of any first-order functional.<|endoftext|> TITLE: Finite Nontrivial Unramified Towers of Number Fields QUESTION [6 upvotes]: Let $F$ be a number field and $L=F^{un}$ its maximal unramified extension. By Class Field Theory, $$Gal(L/F)^{ab}\cong Cl(F).$$ It's well-known that we can have $[L:F]=1$ (e.g. $F=\mathbb{Q}$), and $[L:F]=\infty$ (by Golod-Shafarevich we can have infinite towers of Hilbert Class Fields; alternatively see Maire's "On Infinite Unramified Extensions."). What about the intermediate case? That is, can we have $1<[L:F]<\infty$? And is there an easy way to determine $[L:F]$, or at least, which of these 3 categories it falls into? REPLY [7 votes]: It is certainly possible that $1 < [L:F] < \infty$, i.e. that the extension $F^{\mathrm{un}}/F$ be finite and non-trivial. The simplest example of this is $F = \mathbb{Q}(\sqrt{-5})$. Its Hilbert class field is $F(\sqrt{-1}) = \mathbb{Q}(\sqrt{-1},\sqrt{-5})$, and it can be shown that this field has a smaller root discriminant than any number field of degree at least eight, and hence that it has no unramified extensions. Therefore $F^{\mathrm{un}} = F(\sqrt{-1})$ in this case, and $[L:F] = 2$. Continuing this, note that if $B(2d)$ is a lower bound on the root discriminants of totally imaginary number fields of degrees $\geq 2d$, and if $F$ is an imaginary quadratic field with root discriminant $< B(2d)$, then $[F^{\mathrm{un}}:F]$ is finite and $< d$. Using the available lower bounds on the $B(2d)$ (Odlyzko; Diaz), Yamamura [Maximal unramified extensions of imaginary quadratic numer fields of small conductors, Proc. Japan Acad. 1997] has verified in this way that $[F^{\mathrm{un}}:F] < \infty$ for all imaginary quadratic fields of conductors $\leq 420$. In fact, for these fields the Hilbert class field tower terminates at most at the third step $K_3$, and $K^{\mathrm{un}} = K_3$. Thus, if you exclude the nine fields of class number one, all imaginary quadratic fields of conductor $\leq 420$ have $1 < [F^{\mathrm{un}}:F] < \infty$. Regarding your second question, in general it is not easy to determine $[F^{\mathrm{un}}:F]$; in particular, it is a difficult open problem to decide whether there are infinitely many $F$ with no unramified extensions. On the other hand, Golod-Shafarevich yields criteria for the Hilbert class field tower to be infinite and hence for $[F^{\mathrm{un}}:F] = \infty$; this is so for instance for all quadratic fields whose conductor has at least eight prime factors.<|endoftext|> TITLE: Maps which induce the same homomorphism on homotopy and homology groups are homotopic QUESTION [35 upvotes]: I am interested in the following question. Are maps which induce the same homomorphism on homotopy and homology groups homotopic? I am sure the answer is no, however I cannot imagine how to construct counterexamples. REPLY [5 votes]: Phantom maps provide a large class of examples of maps which are not homotopic to the constant map, but which induce the zero map on both homology and homotopy groups. There are uncountably many distinct homotopy classes of phantom maps $\mathbb{C}P^\infty\to S^3$, for example. Edit: In many cases one can determine whether nontrivial phantom maps can be found between two spaces by investigating rational homotopy invariants. Some great references include Phantom maps and Rational Equivalences by Roitberg and McGibbon, or McGibbon's survey in the Handbook of Algebraic topology.<|endoftext|> TITLE: Condition(s) for the full autormophism group $\operatorname{Aut}(C(G, S))$ of the Cayley graph of $G$ to be isomorphic to $G$ QUESTION [5 upvotes]: If $\Gamma = C(G, S)$ is the (undirected) Cayley graph of a finite group $G$ with generating set $S$, then $G \le \operatorname{Aut}(\Gamma)$, the "full" automorphism group of $\Gamma$. When is it true that $G \cong \operatorname{Aut}(\Gamma)$? In other words, when is a group $G$ (isomorphic to) the automorphism group of its own Cayley graph (via a specified generating set)? Here, by "full" automorphism group, I mean the group of permutations of the vertex set that preserve adjacency (with no regard to the coloring that is usually associated with Cayley graphs). REPLY [2 votes]: A group $G$ is said to admit a graphical regular representation (GRR) if there is a graph $\Gamma$ such that $Aut(\Gamma) \cong G$ and $G$ acts regularly on the vertices of $\Gamma$. Such a graph is called a GRR of the group $G$. A graph $\Gamma$ is said to be a GRR if $Aut(\Gamma)$ acts regularly on the vertices of $X$. Sabidussi showed that a connected graph $\Gamma$ is isomorphic to a Cayley graph iff $Aut(\Gamma)$ contains a regular group. A Cayley graph $C(G,S)$ is said to be a GRR if its automorphism group is $G$ (which is the smallest possible full automorphism group for a Cayley graph $C(G,S)$). The question of determining, for a given $G,S$, whether $C(G,S)$ is a GRR, is a difficult and generally open problem. The problem of which groups admit GRRs was resolved in the 1970s by the work of several authors and finally by (Godsil, 1978). It was proved that the only two infinite families of finite groups which do not admit GRRs are the abelian groups of exponent at least 3 and the generalized dicyclic groups. Each Cayley graph $C(G,S)$ on these two groups has a nontrivial automorphism so that $Aut(C(G,S))$ is strictly larger than $G$. Hence, these two families of groups do not admit GRRs. For all other groups, it is possible to construct at least one GRR for the group. Recall that if $G$ is any group besides the two exceptions mentioned above, there is an $S$ such that $C(G,S)$ is a GRR. It was conjectured in (Babai and Godsil, "On the automorphism group of almost all Cayley graphs", Eur. J. Comb., 1982, Conjecture 2.1) that almost all Cayley graphs of such non-exceptional groups are GRRs. Both the right regular representation $R(G)$ and the set $Aut(G,S)$ of automorphisms of $G$ that fixes $S$ setwise are automorphisms of the Cayley graph $C(G,S)$. $Aut(G,S)$ also fixes the identity vertex $e$. Observe that $Aut(C(G,S))$ equals $R(G) Aut(C(G,S))_e$, which contains $R(G)Aut(G,S)$. Thus, an equivalent condition for the Cayley graph $C(G,S)$ to be a GRR is that $Aut(C(G,S))_e =1$, and a necessary condition for $C(G,S)$ to be a GRR is that $Aut(G,S)=1$. Since $Aut(G,S)$ might be easier to determine than $Aut(C(G,S))_e$, an open problem posed by (Godsil, "The automorphism groups of some cubic Cayley graphs", European Journal of Combinatorics, Eur. J. Comb., 25–32, 1983) is to determine conditions under which $C(G,S)$ is a GRR iff $Aut(G,S)=1$. Partial solutions to this problem include (Godsil, 1981, Combinatorica), (Li and Sim, Eur. J. Comb, 2000) and (Godsil, Eur. J. Comb, 1983). If $G$ is abelian of exponent larger than 2 or generalized dicyclic, then $C(G,S)$ admits $G$ and a particular nontrivial map $i$ as automorphisms, and hence, $C(G,S)$ admits $G \rtimes \langle i \rangle$ (a group strictly larger than $G$) as a subgroup of automorphisms. Hence, $G$ does not admit a GRR. An open question was to determine whether almost all Cayley graphs of $G$ have the smallest possible automorphism group $G \rtimes \langle i \rangle$. This question was answered in the affirmatively recently by (Dobson, Spiga, and Verret, arXiv, 2014) for abelian groups and by (Morris, Spiga and Verret, 2015) for generalized dicyclic groups. A Cayley graph $C(G,S)$ is said to be normal if $Aut(C(G,S))$ is exactly equal to $R(G) Aut(G,S)$. An open problem in the literature is to determine which Cayley graphs $C(G,S)$ are normal; see (Xu, "Automorphism groups and isomorphisms of Cayley digraphs", Discrete Math., 309–319, 1998). A conjecture is that almost all Cayley graphs are normal. More specifically, the conjecture is that if $G \ne Q_8 \times C_2^m$ is a group of order $n$, then the probability that a random Cayley graph of $G$ is normal approaches 1 as $n \rightarrow \infty$. A Cayley graph which is a GRR is also a normal Cayley graph, and so the conjecture that almost all Cayley graphs are normal is weaker than the conjecture that almost all Cayley graphs are GRRs. Non-normal Cayley graphs are rare, and to determine, for a given $G,S$, whether $C(G,S)$ is normal, is an open problem.<|endoftext|> TITLE: Numerical approximation to the Wasserstein metric? QUESTION [9 upvotes]: Are there numerical methods for approximating/calculating the Wasserstein metric in particular cases? Suppose that $f$ and $g$ are two density functions with the same support. How can I calculate the Wasserstein metric for these two models? REPLY [5 votes]: Yes, there are. First note that the Wasserstein metric is, after discretization, the solution of a linear program (LP) that can be fed into any LP solver. Moreover, there are specialized algorithms, try googling for "Earth-Movers-Distance" (mostly Wasserstein-1). Then is the Benamou-Brennier framework which puts optimal transport for Wasserstein-2 into the framework of fluid-mechanics, see here. Also you may want to look at the numerical method here for the Wasserstein-1 distance. In the end, it really depends, what your goals are, and how you discretize your density functions.<|endoftext|> TITLE: What is an étale theta function? QUESTION [75 upvotes]: Let me start out by urging you to take seriously that whatever I write about the papers surrounding IUTT really are questions. If you would like to use it as a guide to the mathematics in any way, they should be regarded as 'introductions to introductions'. I've wondered about the wisdom of writing in such a loose and ambiguous style based upon a few days worth of cursory skimming. However, I believe the circumstance justifies the effort somewhat. In any case, caveat lector. This might be especially true when it comes to the étale theta function, which touches on many technical aspects of arithmetic geometry that require careful checking to feel confident about. The main reason I am posting this question hastily is to keep myself on the job, since worry and inertia tend to build up as I let things sit. Much of anabelian geometry deals with recovering arithmetic geometry from group theory. The situation is somewhat reminiscent of Klein's programme, with the important difference that the geometry there was of a far more homogeneous nature. However, that the analogy is not so far fetched is captured by the notion of rigidity that was employed in Grothendieck's letter to Faltings, and which has been ubiquitous also in Mochizuki's papers. Grothendieck's usage took its inspiration from rigidity theorems of Mostow-Margulis type, whereby an isomorphism between fundamental groups of hyperbolic manifolds of dimension at least 3 is necessarily induced by an isometry. The original anabelian conjectures, proved by Nakamura, Tamagawa, and Mochizuki in the 1990's, stated that hyperbolic curves over number fields are similarly rigid, so that isomorphisms between their arithmetic fundamental groups are necessarily induced by maps of schemes. In slightly fanciful terms, the arithmetic structure endows the curve with a rigidity similar to a hyperbolic metric in higher dimensions. This makes a modicum of sense since (some ring of integers in) the base field is adding dimensions, creating a bundle of which the geometric curve is a fibre. Even when the geometry can't be entirely recovered from group theory, one might still ask about certain geometric invariants or characteristics. For example, an isomorphism of groups $$\pi_1(X_1)\simeq \pi_1(X_2)$$ induces isomorphisms on group cohomology $$H^n(\pi_1(X_1), \hat{\mathbb{Z}})\simeq H^n(\pi_1(X_2), \hat{\mathbb{Z}}).$$ On $K(\pi,1)$-spaces, one might consider the subgroups $$Ch(X_i)\subset \oplus_{n}H^n(\pi_1(X_i), \hat{\mathbb{Z}})$$ arising as Chern classes of vector bundles and ask if these classes are preserved under the isomorphism. (Some discussion of Tate twists is necessary for this to make sense.) This specific question was an important one in the resolution of the anabelian conjectures. Here is another example of a more category-theoretic nature. Suppose $G_1$ and $G_2$ are absolute Galois groups of local fields $F_1$ and $F_2$ that are finite extensions of $\mathbb{Q}_p$. An interesting phenomenon is that $G_1$ and $G_2$ can be isomorphic even when the fields are not. Nevertheless, other refined questions are possible. For example, denote by $Rep_i$ the category of continuous finite-dimensional $\mathbb{Q}_p$-representations of $G_i$. An isomorphism between the $G_i$ will of course induce an equivalence of categories $$F: Rep_1\simeq Rep_2.$$ We might then consider various subcategories of interest, for example, the sub-categories $$HT_i\subset Rep_i$$ of Hodge-Tate representations. The definition of this subcategory depends explicitly on the field, so that it's not at all obvious that the functor will preserve it. In fact, one of Mochizuki's easier theorems says that $F$ will carry $HT_1$ to $HT_2$ if and only if the fields are isomorphic. Mochizuki's recent papers have generally been concerned with refining the dictionary geometry $\leftrightarrow$ group theory into something like geometry $\leftrightarrow$ category theory Sometimes, this is straightforward, such as replacing the fact that a number field $F$ can be recovered from its absolute Galois group $\mbox{Gal}(\bar{F}/F)$ by the statement that $F$ can be recovered from the Galois category (opposite to that) of finite separable extensions of $F$. However, the category can be made more elaborate, or 'stretched out', so that more refined operations can be performed on it. Roughly speaking, this idea informs the theory of Frobenioids. The theory of the étale theta function, as expounded in the paper The etale theta function on its Frobenioid-theoretic manifestations, carries both group-theoretical and category-theoretical aspects. We will mostly ask about the former, putting off even a superifical description of the category theory until it is needed for IUTT. The group theory is that of the tempered etale fundamental group $\Pi^{tp}_X$ of a once punctured curve $X$ of geometric genus 1 over a finite extension $K$ of $\mathbb{Q}_p$ with split stable reduction. (That is, an elliptic curve minus the origin.) The tempered fundamental group is quite similar to the profinite one, except in the use of either rigid analytic or formal geometry to allow infinite covers. In particular, it admits a surjection $$\Pi^{tp}_X\rightarrow \mathbb{Z}$$ to the discrete group of integers corresponding to a formal cover $$Y\rightarrow X,$$ which is essentially the Tate uniformization. I should warn you that it's rather important to move efficiently between various formal schemes and their (algebraized) generic fibers, but I will mostly ignore the distinction to avoid clutter in this exposition. Also important is that Mochizuki is using log curves rather than punctured ones. But these as well I will pretend are the same, since most people will not be familiar with log geometry. The analytic theta function is a function on $Y$, to which one can associate a cohomology class $$\ddot{\eta}^{\Theta}\in H^1(\Pi_{\ddot{Y}}^{tp}, \hat{\mathbb{Z}}(1)),$$ on a double cover $\ddot{Y}$ of (some base-change of) $Y$, the étale theta function. It's not unreasonable to say that the main goal of the paper is to (1) characterize this class group-theoretically; and (2) give it a Frobenioid theoretic interpretation. A precise characterization doesn't quite work, and much of the paper is devoted to formalizing the indeterminacy via the notion of the mono-theta environment, a sophisticated invariant of the theta function that does have a group-theoretic construction. As a small exercise, you might try to characterize the cyclotomic character that appears in the coefficient group-theoretically. One way is to use the exact sequence $$1\rightarrow \Delta^{tp}_X\rightarrow \Pi^{tp}_X\rightarrow G_K\rightarrow 1,$$ which can be constructed group-theoretically from $\Pi^{tp}_X$ by a theorem from Mochizuki's paper 'Absolute anabelian geometry of hyperbolic curves.' (We will discuss such matters more carefully later.) One then has the theta quotient $$(\Delta^{tp})^{\Theta}_X:=\Delta^{tp}_X/\overline{ [ \Delta^{tp}_X , [\Delta^{tp}_X ,\Delta^{tp}_X ] ] },$$ which fits into an exact sequence $$1\rightarrow \Delta_{\Theta}\rightarrow (\Delta^{tp}_X)^{\Theta}\rightarrow (\Delta^{tp}_X)^{ell}\rightarrow 1,$$ where the superscript `$ell$' denotes the abelianization. The group $(\Delta^{tp}_X)^{ell}$ is essentially the fundamental group of the elliptic curve compactification of $X$, except for the 'tempered' nature. The group $ \Delta_{\Theta}$, on the other hand, equipped with the conjugation action of $\Pi^{tp}_X$, is canonically isomorphic to $\hat{\mathbb{Z}}(1)$ (the fundamental group of the punctured tangent space to the elliptic curve at the missing point). Note that $(\Delta^{tp}_X)^{\Theta}$ is a group of Heisenberg type. We outline briefly the construction of the class $\ddot{\eta}^{\Theta}$. There are compatible towers of covers $$\begin{array}{ccc} Z_N & \rightarrow & Z_M \\ \downarrow & & \downarrow \\ Y_N & \rightarrow & Y_M\end{array}$$ for $M|N$, where $Y_1=Z_1=Y$. All these covers come from the theta quotient $(\Delta^{tp})^{\Theta}_Y$ of $\Delta^{tp}_Y$ which fits into an exact sequence $$1\rightarrow \Delta_{\Theta}\rightarrow (\Delta^{tp}_Y)^{\Theta}\rightarrow (\Delta^{tp}_Y)^{ell}\rightarrow 1$$ with $(\Delta^{tp}_Y)^{\Theta}$ now profinite abelian of rank 2. It's important not to be too intimidated by the notation for all the covers in the paper. The towers just mentioned are the main ones. The other covers are $$\ddot{Z}_N\rightarrow \ddot{Y}_N\rightarrow Y_N,$$ where $$\ddot{Y}_N=Y_{2N}\otimes \ddot{J}_N,$$ and the field extension $\ddot{J}_N$ is obtained by adjoining suitable $2N$-th roots. Base changes of this sort will also be ignored for the purpose of this question. The cover $\ddot{Z}_N\rightarrow \ddot{Y}_N$ is obtained as a compositum of $\ddot{Y}_N\rightarrow Y_N$ and $Z_N\rightarrow Y_N$. The importance of these variants to $Y_N$ and $Z_N$ has to do with eliminating sign ambiguities of the sort that come up quite frequently even in the classical geometric theory of theta functions (as in Mumford's papers on algebraic theta functions). In the paper, the entire discussion and all objects are generalized after a pull-back to $$\underline{\underline{X}}\rightarrow \underline{X} \rightarrow X,$$ a sequence of geometrically $l$-cyclic covers that appear to be important in the eventual applications, especially the question of labeling the components of the special fiber on the regular minimal model of a semi-stable elliptic curves. All the objects under consideration acquire a double underline, like $\underline{\underline{\ddot{Y}}}$. We will stick to the case of $l=1$ for now. (I'm uncertain, however, if some of the results require $l$ to be strictly larger. There are also portions where we need to assume that $X$ is not arithmetic.) There is a sequence of line bundles $L_N$ on $Y_N$ corresponding to the divisor of special fibers (on the formal schemes) together with natural isomorphisms $$L_M^{\otimes M/N}\simeq L_N|Y_M$$ for $N|M$. Over the tower $\ddot{Z}_N$, there are two compatible sequences $s_N$ and $\tau_N$ of sections, one associated to the special fiber and the other to the cusps. These two trivializations give rise to actions of $\Pi^{tp}_{\ddot{Y}}$ on the line bundles, and the difference between them determines a class $$\ddot{\eta}^{\theta}\in H^1(\Pi^{tp}_{\ddot{Y}}, \Delta_{\Theta}),$$ which is the étale theta function. The 'functional' nature can be thought of in terms of evaluation on points $y\in \ddot{Y}(L)$ over field extensions $L$. We get a class $$\ddot{\eta}^{\theta}|_y\in H^1(G_L, \hat{\mathbb{Z}}(1))\simeq \widehat{L^*},$$ and a comparison with the usual analytic theory can be used to show that the value actually lies in $L^*\subset \widehat{L^*}$. Various choices were made in the construction of the class, and there are two natural group actions on the cohomology accounting for the indeterminacy: One is the conjugation action of $\Pi^{tp}_X/\Pi^{tp}_Y\simeq \underline{\mathbb{Z}}$ (here, I've followed Mochizuki and underlined the $\mathbb{Z}$, to indicate that it's not canonically isomorphic to $\mathbb{Z}$, a point that is eventually important), and the other coming from the translation by the constant integral classes $$\mathcal{O}_{\ddot{K}}^*\hookrightarrow H^1(G_{\ddot{K}}, \hat{\mathbb{Z}}(1))\hookrightarrow H^1(\Pi^{tp}_{\ddot{Y}}, \Delta_{\Theta}).$$ (Here, I believe $\ddot{K}$ is the field obtained by adjoining a square root of $q_X$, the Tate period of $X$, but I'm not entirely sure.) If we denote the orbit of $\ddot{\eta}^{\Theta}$ under these two actions by $$\mathcal{O}_{\ddot{K}}^*[\ddot{\eta}^{\theta}]^{\underline{\mathbb{Z}}}\subset H^1(\Pi^{tp}_{\ddot{Y}}, \Delta_{\Theta}),$$ this set of classes turns out to be a group-theoretic invariant of $\Pi^{tp}_X$. That is to say, if $X_{\alpha}$ and $X_{\beta}$ are two curves defined over fields $K_{\alpha}$ and $K_{\beta}$ with corresponding covers $\ddot{Y}_{\alpha}$ and $\ddot{Y}_{\beta}$, then any isomorphism of topological groups $$\Pi^{tp}_{X_{\alpha}}\simeq\Pi^{tp}_{X_{\beta}}$$ will induce isomorphisms $$\Pi^{tp}_{\ddot{Y}_{\alpha}}\simeq\Pi^{tp}_{\ddot{Y}_{\beta}}$$ and $$(\Delta_{\Theta})_{\alpha}\simeq (\Delta_{\Theta})_{\beta}$$ in such a way that the isomorphism $$H^1(\Pi^{tp}_{\ddot{Y}_{\alpha}}, (\Delta_{\Theta})_{\alpha})\simeq H^1(\Pi^{tp}_{\ddot{Y}_{\beta}}, (\Delta_{\Theta})_{\beta})$$ takes $\mathcal{O}_{\ddot{K}}^*[\ddot{\eta}^{\theta}]^{\underline{\mathbb{Z}}}$ on one side to the other. By using the fundamental group of the orbifold quotient $C=X/\pm 1$, it seems to be possible to designate a class of $\theta$ classes of 'standard type', which eliminate the $\mathcal{O}_{\ddot{K}}^*$ ambiguity up to a sign. It is with this class that one constructs the theta environment for each $\Pi_{X}$. The basic 'ambient group' is the cyclotomic envelope $$\Pi^{tp}_{Y}[\mu_N]:= \mu_N\rtimes \Pi^{tp}_{Y}$$ of $\Pi^{tp}_{Y}$, where $\Pi^{tp}_Y$ acts on $\mu_N$ through the projection $\Pi^{tp}_Y\rightarrow G_K$. There is an action of $H^1(\Pi^{tp}_Y, \mu_N)$ on $\Pi^{tp}_Y[\mu_N]$ whereby a cocycle $c$ sends $(\zeta, g)$ to $(\zeta c(g), g)$. This is independent of the cocycle up to inner automorphism, giving us a homomorphism $$H^1(\Pi^{tp}_Y, \mu_N)\rightarrow \mbox{Out} (\Pi^{tp}_Y\mu_N]).$$ Hence, $K^*$ also acts on $\Pi^{tp}_Y[\mu_N]$ via the map $$K^*\rightarrow K^*/(K^*)^N\simeq H^1(G_K, \mu_N)\rightarrow H^1(\Pi^{tp}_Y, \mu_N).$$ There is also a canonical action of the group $\mbox{Gal}(Y/X)$. Denote by $$\mathcal{D}_{Y}\subset \mbox{Out} (\Pi^{tp}_{Y}[\mu_N])$$ the subgroup generated by the image of $K^*$ and $\mbox{Gal}(Y/X)$. The theta class $\ddot{\eta}^{\Theta}$ gives rise to a homomorphism $$s^{\Theta}:\Pi^{tp}_{\ddot{Y}}\rightarrow \Pi^{tp}_{Y}[\mu_N]$$ whose image is independent of the various choices up the $\mu_N$ conjugacy. (This is one of the many points I'm not so clear about.) Denote by $s^{\Theta}_{\Pi}$ the conjugacy class of subgroups of $\Pi^{tp}_{Y}[\mu_N]$ obtained as the image. The triple, $$(\Pi^{tp}_{Y}[\mu_N], \mathcal{D}_{Y}, s^{\Theta}_{\Pi})$$ is then the $\mod N$ mono-theta environment of $\Pi^{tp}_{X}$. Just to be clear, we note that it consists of: (1) A group $\Pi$; (2) A subgroup $\mathcal{D}\subset \mbox{Out}(\Pi)$; and (3) A set of subgroups $s^{\Pi}$ of $\Pi$. For the triple given above, there is a compatibility over $N$, and the main group-theoretical theorem is that this system is determined group theoretically by $\Pi^{tp}_{X}$ (or perhaps $\Pi^{tp}_C$). The Frobenioid-theoretic construction of the mono-theta environment is more or less built into the objects already discussed, although the details are quite elaborate. The 'tempered' Frobenioid is made out of mildly technical functors that assign divisors and meromorphic functions to covers of the formal scheme corresponding to $X$ in the manner of the geometric Frobenioids desrcibed in the previous question. The trivializations discussed above can be viewed as morphisms inside the Frobenioid between $\ddot{Z}_N$ with the 0 divisor and the line bundle $L_N|\ddot{Y}_N$. From this, using the automorphism groups of $L_N|{\ddot{Z}_N}$, there is a complicated but natural way to rebuild the mono-theta environment mod $N$. The main categorical theorem makes a precise statement to the effect that this construction agrees with the group-theoretic one, and depends only on the category theory of the tempered Frobenioid. As far as I know, the category theoretical formulation is to be used in the process of somehow 'globalising' the local theory of theta functions, a goal that came up already in the Hodge-Arakelov papers. It should be obvious that this question is based on an understanding even more superficial than the previous one. Nevertheless, perhaps it will be useful at least in eliciting answers that provide more background and motivation. Actually, my hope is that even the little bit I explained here will help people read the elaborate explanatory paragraphs in Mochizuki's original papers, which are often considered difficult. One more word of technical advice for now (probably well-known to everyone but me): I find it useful to go through the papers with both a paper copy and an electronic version in hand. This way, you can combine the comfort of the hardcopy with the ease of searching for definitions. REPLY [25 votes]: Update (27 December, 2015): A number of corrections have been made in the two linked PDF documents below, Update (15 December, 2015): I note that Brian Conrad has written a thoughtful summary available here: http://mathbabe.org/2015/12/15/notes-on-the-oxford-iut-workshop-by-brian-conrad/ This makes my original note a bit superfluous. Nevertheless, since the goal there was to state the main point in a very short page, I've reenabled the link below. The key statement has been made somewhat more vague. I will make it more precise when more information becomes available. Update (12 December, 2015): I've written a brief summary of the Oxford workshop on IUTT rather rapidly, so as to save people the trouble of circulating rumours. This seemed to be a reasonable place to put it. All errors in it are my own: http://people.maths.ox.ac.uk/kimm/papers/iutt=clay.pdf After much hesitation, I've decided to post an answer to this question. Rather, the answer is not to this question, but a flawed response to this as well as two previous entries on MO: Philosophy behind Mochizuki's work on the ABC conjecture What is a Frobenioid? I've put it here because this was the most recent post. On the other hand, I am giving something of a physicist's answer to this question as well by directing attention to how the etale theta is used, rather than what it is. It might be reasonable to put on record a strong disclaimer. I have by no means even begun to understand the papers (with the corollary that I can hardly vouch for its correctness). In fact, my understanding is not much deeper than when I wrote the first superficial answer three years ago. Perhaps the state of knowledge can be best summarised by an honest appraisal of the amount of time invested. Over the last three years, I estimate a total of 4 or 5 days looking through the papers with a modicum of concentration, and perhaps a total of 10 hours conversing with Mohamed Saidi and Shinichi Mochizuki. Essentially all of this time, instead of studying systematically, I was doing what most experienced (and lazy) mathematicians do in such a situation: reading a bit, asking a bit, and then sporadically trying to figure out myself what might be going on. Nonetheless, the Oxford workshop on Interuniversal Teichmueller Theory is fast approaching. So last week, I felt obligated to begin some mathematical preparations myself. For this, I went to Exeter to talk to Saidi on Thursday and Friday, and then had a Skype conversation with Mochizuki yesterday. At this point, I decided it might not be a bad idea to summarise my hypotheses in a short exposition. This is what I recommend to my Ph.D. students: As soon as possible, try to describe a mathematical theory or result in one's own words, faulty as they may be. This is the spirit in which my 'answer' should be read. In fact, I didn't check either with Saidi or Mochizuki if what I wrote is even halfway reasonable, for fear that careful re-examination would send me into an infinite regress and/or paralysis. There are parts that are already sloppy or that I know to be a bit wrong, but it seemed rather painful to try to get it right just now. Needless to say, any serious misunderstanding is my own fault. It's entirely possible that I'll produce revisions both before and after the workshop. However, they will be posted in addition to what I've written now, which will remain available so other people might benefit from taking note of my errors. With that long prologue out of the way, here it is: http://people.maths.ox.ac.uk/kimm/papers/pre-iutt.pdf 17/11/2015: A correction about reconstruction of theta values has been added.<|endoftext|> TITLE: Computing intersection of subrings QUESTION [13 upvotes]: Let $R$ be a finitely generated commutative ring over a field, for concreteness. If $S,T \leq R$ are two finitely generated subrings, is their intersection also finitely generated? (Certainly this isn't true for infinite intersections.) If so, How can one compute such an intersection using a computer algebra package? It's easy to reduce to the case that $R = {\mathrm k}[s_1,\ldots,s_n,t_1,\ldots,t_m]/I$, the subring $S$ is generated by $s_1,\ldots,s_n$, and $T$ is generated by $t_1,\ldots,t_m$. REPLY [4 votes]: A couple of years ago, David Speyer showed me the following counterexample. Let $E$ be an elliptic curve defined over a field, with points $P$, $Q$ that are linearly independent under the group law. Let $p:X\to E$ be the total space of the rank two vector bundle $\mathcal{O}([P])\oplus\mathcal{O}(-[Q])$ on $E$. Then $X$ is finite type, but it can be checked that the ring $\Gamma(\mathcal{O}_X)$ of global regular functions on $X$ is not finitely generated. If we pick an affine open cover $\{U,V\}$ of $E$, then $\{p^{-1}(U),p^{-1}(V)\}$ is an affine open cover of $X$. The rings of regular functions on $p^{-1}(U)$ and $p^{-1}(V)$ are finitely generated, but their intersection inside the function field of $X$ (or inside the finitely generated subalgebra they generate) is $\Gamma(\mathcal{O}_X)$, which is not finitely generated.<|endoftext|> TITLE: Ordinary mathematics in Chang's model QUESTION [13 upvotes]: This question is prompted by a paper by Andre Kornell that just appeared on the arXiv. A large portion of the paper is devoted to showing that a surprising amount of ordinary mathematics can be developed in Chang's model. Although the axiom of choice does not hold in this model, weaker choice principles are available which apparently suffice for most mainstream uses. The development is motivated by the author's view that a general theory of "quantum sets" works better in Chang's model. I suppose the fact that every set of reals is Lebesgue measurable could already be cited as a way in which the model is better from the point of view of mainstream math. Has there been any previous work in this direction, specifically about ordinary mathematics in Chang's model? (Comments about what is known to follow from known properties of the model, e.g., the axiom of determinacy plus dependent choice, are also welcome.) REPLY [13 votes]: I don't have enough reputation to comment, so I'm posting my reply to François as an answer. It is possible that these results can be established from $ZF+AC_{ae}+DC+LM+BP+PSP$, but I don't make this claim. I work in a transitive model of this theory that is closed under countable sequences, or equivalently under countable unions, which implies that certain functional-analytic properties, such as the completeness of a metric space, are absolute. We can often verify a familiar fact by appealing to absoluteness, and this is typically much faster than checking its proof down to first principles. For most individual results, this is a matter of convenience, but for research in an established research area, this is effectively a necessity.<|endoftext|> TITLE: What do you do if you believe a problem is undecidable? QUESTION [22 upvotes]: While the title of this question is subjective, I hope to make what I'm looking for quite concrete. The first, and main question is this: If you believe that a problem you are working on is formally undecidable, but you are not a logician, what are some ways to begin learning the tools necessary for you to try to prove undecidability? Where should one begin? Now, for an explicit example. In ring theory one of the big open problems is called Koethe's conjecture. There are many equivalent ways to state the conjecture such as "The sum of two nil left ideals is still nil." Let me give another equivalent description. Let $R=\mathbb{Z}\langle a,b\rangle$ be the free ring on two non-commuting generators. Suppose that $I$ is a left ideal of $R$, containing some power of $b$, and containing some power of every element of $Ra$. Then the conjecture asserts $I$ contains some power of $a+b$. Note that $R$ is countable, and in fact we can easily enumerate the elements of $Ra$ as $f_1,f_2,\ldots$. Thus, given any sequence of positive integers $\vec{n}=\{n_0,n_1,n_2,\ldots\}$, the conjecture asserts that the left ideal containing $b^{n_0},f_1^{n_1},f_2^{n_2},\ldots$ also contains a power of $a+b$. For some sequences $\vec{n}$ (such as eventually constant sequences), it is not difficult to compute that the conjecture is true. But in general the computations get extremely difficult. Lam says in his "First Course in Noncommutative Rings" p. 171, that there has been a (long-held) suspicion that the Conjecture is false. If so, then it is false for a sufficiently fast growing sequence $\vec{n}$. However, one of the troubles here is that nobody knows how to control the behavior of $I$ for fast growing sequences $\vec{n}$, and there are also ways to modify $I$ only slightly and provably not produce a counter-example. My current hunch is that the conjecture is false, but to see this we need tools to assert that (for fast growing vectors $\vec{n}$) the ideal $I$ stays complicated. I'm not confident that the regular tools from ZFC are sufficient to the task; but have no idea where to go from here. One last comment: There is always the danger of fearing that a problem one is working on is formally undecidable, just because you can't immediately solve it. Because of this I haven't given up hope that there is such a solution. However, there is also the chance that some stronger set theory may provide tools to look at the problem differently, and really that's what I'm after. REPLY [4 votes]: Joel David Hamkins has given a good answer in the case of (2), where a certain notion when expressed formally in the language of a theory $T$ (and also the negation of this notion) may not be provable from $T$ using the appropriate proof system. I would like to touch on the other case (1), where it may be possible to show not only that the conjecture is false, but that it can be false in a noncomputable way. This would be a good resolution from the standpoint of the original poster, because it would shed light on the nature of algorithmic questions in general algebra. (Disclaimer: In spite of my proximity to the subject, I am no expert on decidability. I recommend Algorithmic Problems in Varieties by Sapir and Kharlampovich, and also section 6 and earlier of Ross Willard's "An Overview of Modern Universal Algebra", both available on the Web. I had the privilege of being one of the first to see McKenzie present his work on Tarski's Finite Basis Problem: that work influences this post.) Consider the possibility that Koethe's conjecture is false. As Pace Nielsen points out in his question, one way that it could be false is that there exists a fast growing sequence $n_i$ and a (hopefully recursive) enumeration $f_1,f_2, \ldots$ of $Ra$ such that $a+b$ and all of its powers avoids all the left-ideals generated by a finite subsequence of $b^{n_0}, {f_1}^{n_1}, {f_2}^{n_2}, \ldots$ . (Unless I missed something, this should be equivalent to his formulation.) It would be nice to be able to characterize those ideals which fail the conjecture, or given a particular ring, to characterize those sequences and enumerations which lead to failure. As mentioned in the comments, to show that no recursive characterization is possible, one method would be to reduce an undecidable problem to this question. In other words, construct a recursive map $\phi$ so that for every instance $I$ of a problem known to be undecidable, $\phi(I)$ is a sequence (say) of exponents $n_j$ such that a certain enumeration of $Ra$ raised to exponents $n_j$ produces an ideal which contains a power of $a+b$ iff $I$ satisfies the property in the problem known to be undecidable. The Boone-Novikov result is well known (a finitely presented group with undecidable word problem), and the article of Sapir and Kharlampovich contains many more examples, and if the Koethe problem "looks" enough like one of the undecidable problems, one might then construct a recursive $\phi$ as suggested above. Alternatively, one could have a mastery of the situation that one could try to build a sequence of ideals with such a property that "encoded" Turing machine computations. In the case of Tarski's Finite Basis problem (is there a computer program that can take as input the function tables of a finite algebra A of finite type, and output "yes" if the equational theory of V(A) is finitely-based, and "no" otherwise), there was a series of related problems that McKenzie was working on that allowed him to express a Turing program as an algebra with certain operations corresponding to the state transition table of the program, such that if the halting state were reached, that would correspond to a certain algebraic property in an infinite power of the base algebra, and not correspond otherwise. The properties involved had to do with residual smallness, and McKenzie told me that part of his thought process was to arrange all of the properties he needed in small groups, and make small changes in the algebraic construction to get the group of properties he needed (my wording based on a poor remembering of an event over twenty years in the past). This is an attempt then to answer the question in a different way: if you suspect the answer is not just no, but no in a non-recursive way, try two approaches: reduction from something in Kharlampovich-Sapir (or other sources), or understand the properties of the ideals that satisfy Koethe's conjecture as well as those that don't, and see if those properties are recursively inseparable (can be related somehow to computations halting or not). Gerhard "That's How I Would Start" Paseman, 2015.03.24<|endoftext|> TITLE: Algebraic characterization of commutative rings of Krull dimension 1,2, or 3 QUESTION [6 upvotes]: A commutative ring $R$ (with $1$) is $0$-dimensional if and only if $R/\sqrt 0$ is von Neumann regular. Besides this result, there is a wealth of information about the algebraic structure of zero-dimensional rings. I could not find any information about the algebraic structure of rings of higher Krull dimension. If the problem for general commutative rings is too hard, how about some special cases, such as when $R$ is Noetherian and/or an integral domain? REPLY [3 votes]: A remark about the one-dimensional case. By the characterization of Krull dimension given by T. Coquand and H. Lombardi, dim($R$)$\,\leq 1$ iff $\forall_{x,y\in R}\exists_{a,b\in R,m,n\in\mathbb{N}}\,x^{m}(y^{n}(1+by)+ax)=0$. Here we may assume that neither $x$ nor $y$ is nilpotent or a unit. If $R$ is reduced, we can require $m=1$. If $R$ is a domain, the condition simplifies to $\exists_{b\in R,n\in\mathbb{N}}\,y^{n}(1+by)\in Rx$ for all non-zero non-units $x$ and $y$ in $R$. If, in addition, $R$ is local with maximal ideal $\mathfrak m$, one gets dim($R$)$\,\leq 1\iff\forall_{0\neq x\in \mathfrak m}\,\mathfrak m=\sqrt {Rx}$. But this is of course quite trivial.<|endoftext|> TITLE: Abelianization of Hilbert modular group QUESTION [6 upvotes]: Let $d>0$ be a square free positive integer and let $\mathcal{O}_d$ be the ring of integers in $\mathbb{Q}[\sqrt{d}]$. What is the abelianization of the Hilbert modular group $\text{SL}_2(\mathcal{O}_d)$? If this is too hard, is at least the rank of the abelianization known? I'd also be interested in knowing this for finite-index subgroups of $\text{SL}_2(\mathcal{O}_d)$. These groups are lattices in $\text{SL}_2(\mathbb{R}) \times \text{SL}_2(\mathbb{R})$. I believe that this implies that they don't have property (T), so there isn't a cheap way of seeing that the rank of the abelianization is $0$. But they do have some higher-rank behavior; for instance, Serre proved that they do have the congruence subgroup property. REPLY [6 votes]: The question you ask (at least for ${\rm SL}_{2}(\mathcal{O}_{d})$) is the subject of a paper by Hatice Boylan and Nils-Peter Skoruppa (available on arXiv here). Basically, all the abelian quotients of ${\rm SL}_{2}(\mathcal{O}_{d})$ come from prime ideals above $2$ and $3$ in $\mathcal{O}_{d}$. In cases when there are none (like $d = 5$), the group ${\rm SL}_{2}(\mathcal{O}_{d})$ is perfect.<|endoftext|> TITLE: assumptions on local rademacher complexities QUESTION [5 upvotes]: A lot of the work on Local Rademacher complexities of Koltchinskii, and Bartlett for fast rates of convergence is based on Bousquet's version of Talagrand's inequality [1] (Theorem 2.11). However the assumption used by Bousquet is that the space of functions used is countable. Nevertheless, all the results by Koltchinskii and Bartlett ignore this assumption and claim that it holds for arbitrary function spaces. My question is does the general inequality follows from the countable inequality? I have been trying to show this but it is not immediate without some continuity assumptions on the empirical process. The proof of Bousquet also uses the fact that the set can be approximated with finite sets so I do believe it is necessary to have a countable space of functions. [1] Concentration inequalities and Empirical Processes Theory Applied to the Analysis of Learning Algorithms. Olivier Bousquet, 2002. REPLY [2 votes]: I do not think the general inequality holds without further qualifications. Separability of the function class is one assumption which enables application of Bousquet's version of Talagrand's inequality, since separability implies that the function admits a countable dense subset. I recommend looking at "General nonexact oracle inequalities for classes with a subexponential envelope" by Lecué and Mendelson, which can be found here: http://projecteuclid.org/download/pdfview_1/euclid.aos/1338515139 Specifically, look at at sentence after equation (2.1) on page 843, which mentions that the version of Theorem 2.1 stated there (which is a version of Talagrand's inequality) can be extended from countable classes to those satisfying a separability condition. They go on to mention that one such condition is condition (M) in the paper "Risk bounds for statistical learning" by Massart and Nédélec. Also, note that Koltchinskii mentions "measurable" when he mentions function classes, which might already impose some restrictions making things go through. Also, Bartlett, Bousquet, and Mendelson discuss measurability of the supremum as a condition in the context of how to treat the sup (search for "measurability" in the journal version their paper on local Rademacher complexities). You might also get clarification by looking in Talagrand's book "Upper and Lower Bounds for Stochastic Processes" (which entirely subsumes his book "The Generic Chaining"). On page 13, Talagrand comments "A side issue (in particular when $T$ is uncountable) is that what is meant by the quantity $\mathsf{E} \sup_{t \in T} X_t$ is not obvious." He then offers one patch-up, which is to define it as follows: $\mathsf{E} \sup_{t \in T} = \sup \left\{ \mathsf{E} \sup_{t \in F} X_t ; F \subset T , F \text{ finite} \right\}$. This is a very good question, and it is hard to get answers, because as explained by Talagrand, issues like this were covered all the time in the pre-1950's era, to the point where people who are well-versed in it no longer have the patience to continue mentioning these issues in papers in any detail.<|endoftext|> TITLE: Which polygons have *simple* periodic billiard paths? QUESTION [10 upvotes]: I know (or, rather, believe) that it remains unknown whether every polygon has a periodic billiard path. But Howard Masur proved in the 1980's that every rational polygon (vertex angles rational multiples of $\pi$) has (many) periodic billiard paths.             (Image from: W. Patrick Hooper, "Some irrational polygons have many periodic billiard paths." PDF download link.) My question concerns simple paths: non-self-intersecting, i.e., embedded paths: Q. Is it known that certain classes of rational polygons have simple periodic billiard paths? If so, which? That certain classes (of rational polygons) have no such simple periodic paths? If so, which? For example, in an acute triangle, its pedal triangle is a simple periodic path, and the only such. To be narrowly specific: Might every regular polygon have a simple periodic path? REPLY [7 votes]: Consider the simple polygonal billiard path itself. It is a polygon and it is convex, because all interior angles are less than $\pi$. Now start with an arbitrary convex polygon. It is a billiard path on some polygonal table. Indeed, the lines from which the reflections happened are uniquely determined by our billiard path. Continue these lines until they intersect and form a table. Of course this table is not unique, because you can add arbitrarily many sides which lie outside of the polygonal path and do not intersect our polygonal path. Neither the resulting table, nor the path itself has to have any symmetry. EDIT. If the table is a triangle with acute angles, then a simple billiard path exists, is unique, and it is the triangle whose vertices are the bases of the three heights of the table-triangle. This is due to H. A. Schwarz. (See Courant-Robbins, What's mathematics, Ch. VII, sect. 4). On the other hand, a triangle whose one angle is $\geq\pi/2$ violates the inequality noticed in Will Sawin's remark: the angles of the triangular table must satisfy $\alpha+\beta-\gamma\in(0,\pi)$ for a simple billiard part to exist. Thus we have a complete description of triangular tables with a simple billiard path, and more generally, of those tables which have a triangular billiard path. REPLY [2 votes]: An example that may be difficult for Aaron's line-of-symmetry argument:<|endoftext|> TITLE: Is there an analog of compactified moduli spaces(/stacks) for smooth manifolds? QUESTION [8 upvotes]: This question has been inspired by an answer to the question Reference request: Topology on the space of smooth compact submanifolds; I've asked it in a comment to that answer but then decided to make it a separate question, as it is also related to one of my previous questions, "Naïve"cobordism?. The question I refer to is about the space of $m$-dimensional submanifolds of $\mathbb R^n$, and the above answer describes it as the disjoint sum of spaces of the form $\mathrm{Emb}(M,\mathbb R^n) /\mathrm{Diff}(M)$. In particular, each diffeomorphism type of $m$-manifolds produces a separate connected component of this space. My question is whether any kind of compactification of this space is known which would tie together all these components. Two things that come to mind in connection to this are (a) the Deligne-Mumford compactification of the moduli spaces of curves via stable curves; (b) Vassiliev invariants which are constructed via adding to the space of smooth 1-submanifolds of $\mathbb R^3$ new points corresponding to certain non-embedding immersions. Is there actually a relationship between these two? Are there any higher dimensional analogs of these constructions known? My personal motivation: a map from a (say, connected) space $X$ to the above disjoint sum should give a fibration over $X$ with fibre an $m$-manifold (plus some additional structure describing this fibre as a submanifold of $\mathbb R^n$); whereas a map to that purported compactification would instead give a family over $X$ which "mostly" consists of such manifolds but allows for "jumping" between different diffeomorphism types through (more or less) controllable "mild" singularities. This would give a picture which I was asking about in my above question about "naïve" cobordism. REPLY [5 votes]: I have a piece of juvenilia on this topic, considering the case of zero-dimensional submanifolds. See Section 9 of O. Randal-Williams, Embedded cobordism categories and spaces of submanifolds, IMRN 3 (2011) 572-608. It concerns how close the relationship between the cobordism category having oriented 0-manifolds inside $M$ as objects, and oriented 1-dimensional cobordisms in $M \times [0,t]$ as morphisms, and the fundamental (topological) groupoid of McDuff's space of annihilating positive and negative particles in $M$, is. The main theorem in this direction is that while the categories are in no sense equivalent as topological categories (the circle as a cobordism $\emptyset \leadsto \emptyset$ is contractible as a loop in McDuff's space), they do nontheless have homotopy equivalent classifying spaces.<|endoftext|> TITLE: Anything between vector bundles and sphere bundles? QUESTION [18 upvotes]: There are two extremities: on the "easy end" one has vector bundles which are classified by maps to the (more or less) well understood spaces like Grassmanians; on the "hard end" there are spherical fibrations/sphere bundles classified by maps to the classifying space of the monoid of self-homotopy equivalences of spheres. There are ways to move "just slightly down" from the hard end, by considering self-homeomorphisms, self-PL-isomorphisms, self-diffeomorphisms, ... On the other hand I have not seen any ways to move "just slightly up" from vector bundles. What I mean here is this: one may switch from a vector bundle to an associated sphere bundle; the corresponding sphere bundle is one whose gluing maps are linear in one sense or another ($\mathbb R$-linear, or $\mathbb C$-linear, $\mathbb H$-linear, ...). Thus there is a huge gap where on one end we have something linear and on the other something "as nonlinear as it gets". My question is whether it is possible to squeeze in between some groups consisting of controllably nonlinear maps. Say, maps of degree $n$, with composites truncated back to degree $n$ or something similar. Many years ago I was asking around about this, and Leonid Makar-Limanov suggested to look at maps of "finite codegree" - that is, something describable by series with vanishing low degree coefficients - since these readily form a group under composition; but this would still mean "moving down from the hard end" rather than "moving up from the easy end"... ...After some hesitation decided to add more speculative possibility that occurred when commenting to the answer below. Maybe reducing the structure group of a bundle from some linear group (say, $SU(n)$) to its increasingly highly connective covers could be interpreted as switching from linear gluing maps to ones which are "linear up to homotopy" in some sense, together with (coherent) choices of "linearizing homotopies". This brings up 2-vector bundles in the sense of Kapranov-Voevodsky or Baas-Dundas-Rognes, I wonder if this has been looked at from this angle? Actually I am confused here by another thing, and maybe this needs separate question, but why not ask it right away? What I don't understand is this - reducing structure group to more and more highly connective covers sort of "moves towards a contractible "structure group"", while the ultimate "structure group" should be the aforementioned monoid of self-homotopy equivalences of the sphere rather than contractible. I seem to mix up different things here but how exactly? REPLY [2 votes]: Just as $ku$ is quite close to $H\mathbb Z$ in the sequence $S \to ku \to H\mathbb Z$ of commutative ring spectra, and $KU$ is quite close to $H\mathbb Q$, the classifying space $BGL(ku)$ for $2$-vector bundles is quite close to $BGL(\mathbb Z)$ in the sequence $BGL(S) \to BGL(ku) \to BGL(\mathbb Z)$. Here $GL = GL_\infty$. I do not know how to get closer to $H\mathbb R$ and $BGL(\mathbb R) \simeq BO$, or $H\mathbb C$ and $BGL(\mathbb C) \simeq BU$, but you could apply Quillen's plus construction and look at $A(*) = K(S) \to K(ku) \to K(\mathbb Z)$. Even closer to $H\mathbb Z$ is the second Postnikov section $P^2ku$ of $ku$. It is the nerve of a symmetric bimonoidal category $R$ with objects the integers and $R(m, n) = \mathbb C^\times$ for $m=n$, $\emptyset$ for $m \ne n$. I think Thomas Kragh wrote this out in his paper in Math. Scand. (2013). So $P^2 ku$-bundles are quite close to $\mathbb Z$-bundles, as reflected in the map $BGL(P^2 ku) \to BGL(\mathbb Z)$ of classifying spaces. If you prefer to compare with (stable) spherical fibrations, classified by $BF = BGL_1(S)$, the classifying spaces $BGL_1(ku)$ and $BGL_1(P^2 ku)$ are close to $BGL_1(\mathbb Z)$. Note that $BSL_1(P^2 ku) = K(\mathbb Z, 3)$ is the classifying space for gerbes with band $U(1)$. So an early "non-linearity" to consider is that of gerbes.<|endoftext|> TITLE: Equation between the two branches of the lambert w function QUESTION [5 upvotes]: My question: Is there an equation connecting the two branches $W_0(y)$ and $W_{-1}(y)$ of the Lambert W function for $y \in (-\tfrac 1e,0)$? For example the two square roots $r_1(y)$ and $r_2(y)$ of the equation $x^2=y$ fulfill the equation $r_1(y)=-r_2(y)$. So if one has computed one root, he already knows the second one by taking the negative of the computed root. It is also possible to calculate $W_0(y)$ by knowing $W_{-1}(y)$ and vice versa? Note: I asked the question two years ago on math.stackexchange.com. Unfortunately I didn't get an answer or comment there. That's why I decided to reask the question here and I hope that's okay. I read that questions shall not be migrated when they are older than 60 days. That's why I did not ask for migration on MSE... REPLY [2 votes]: The two branches are related in a trascendental way. Their difference can be used to solve other trascendental equations. If $y(x)=\frac{x}{1-e^x}$ then : $x=W_0(ye^y)-W_{-1}(ye^y)=y-W_{-1}(ye^y)$ for $-1< x <0 $ $x=W_{-1}(ye^y)-W_{0}(ye^y)=y-W_{0}(ye^y)$ for $x < -1 $ See http://www.apmaths.uwo.ca/~djeffrey/Offprints/SYNASC2014.pdf<|endoftext|> TITLE: Which group algebras in analysis are "true group algebras"? QUESTION [8 upvotes]: Let $G$ be a group, $A$ a unital associative algebra over ${\mathbb C}$, and let us call a representation of $G$ in $A$ an arbitrary map $\pi:G\to A$ such that $$ \pi(1)=1,\qquad \pi(a\cdot b)=\pi(a)\cdot\pi(b),\qquad a,b\in G. $$ Consider the group ring, or, what is the same, the group algebra ${\mathbb C}[G]$ of $G$ over ${\mathbb C}$, and let $\delta:G\to{\mathbb C}[G]$ be the corresponding embedding (which is, of course, a representation of $G$). It is obvious that every representation $\pi:G\to A$ can be (uniquely) extended to a homomorphism of algebras $\varphi:{\mathbb C}[G]\to A$: $$ \pi=\varphi\circ\delta, $$ (and vise versa, every such $\varphi$ generates $\pi$). Moreover, this characterizes the group algebra ${\mathbb C}[G]$: If $\delta:G\to B$ is a representation with the same property, then $B\cong {\mathbb C}[G]$. This is what is called group algebra in Algebra. In Analysis the situation becomes completely different. The algebras playing the role of "classical" group algebras of topological groups, like $L^1(G)$, or $C^*(G)$, or $W^*(G)$ seem to do not have characterizations like that. Am I right? Are there any constructions of "group algebras" in Analysis (for some classes of topological groups $G$) that can be caracterized by this (or similar) universality property (so that they indeed have a right to be called "group algebras")? The only examples that come to me are group algebras from the stereotype theory: ${\mathcal C}^\star(G)$, ${\mathcal E}^\star(G)$, ${\mathcal O}^\star(G)$, ${\mathcal R}^\star(G)$ (in these constructions the homomorphisms $\varphi$ must be continuous, and the representations $\pi$ must be continuous, smooth, holomorphic, regular, respectively -- see Theorem 10.12 here). Is it possible that I miss something? Yemon Choi inspired me some doubts in our discussion here. REPLY [5 votes]: Just to avoid prolonged discussion in comments I'll put a partial answer for the discrete case here — at the time of writing, I am less sure about the precise picture for non-discrete groups, although some parts carry over since one can use the Bochner integral to convert suitable continuous representations $\pi:G\to A$ to continuous algebra homomorphisms $L^1(G)\to A$. So, let $G$ be a group (considered as having the discrete topology). Then every bounded representation of $G$ in a unital Banach algebra $A$ extends uniquely to a unital algebra homomorphism $\ell^1(G)\to A$. As Tobias Fritz has observed in comments: every $*$-representation of $G$ in a unital ${\rm C}^*$-algebra $A$ extends uniquely to a unital $*$-homomorphism ${\rm C}^*(G)\to A$, and since the original representation must map $G$ into a subgroup of ${\mathcal U}(A)$, one can indeed view the universal property as showing ${\rm C}^*$ is a left adjoint to a suitable forgetful functor (by the "initial object" description of left adjoints. (For locally compact groups that are non-discrete, I think Ernest's $W^*(G)=C^*(G)^{**}$ has an analogous description but where one uses von Neumann algebras and normal $*$-homomorphisms as the target category; however I have not checked the details.) It is less obvious how $\ell^1(G)$ might be viewed as a left adjoint since general Banach algebras, since the natural analogues of the "unitary group" do not have such good properties as in the ${\rm C}^*$-world. However, I claim that the $\ell^1$-monoid algebra construction (which when applied to a monoid that happens to be a group) does have a reasonable interpretation as a left adjoint). In detail: let us define a representation of a monoid $S$ in a (Banach) algebra $A$ as being a function $\pi:S \to A$ satisfying $\pi(e_S)=1_A$ and $\pi(st)=\pi(s)\pi(t)$ for all $s,t\in S$. Let ${\rm ball}(A)$ denote the closed unit ball of a Banach algebra $A$, so that if $A$ is unital then ${\rm ball}(A)$ is a monoid. Then ${\rm ball}$ is a functor from the category of unital Banach algebras and unital homomorphisms of norm $\leq 1$ to the category of monoids and monoid homomorphisms; and the left adjoint to this functor is the $\ell^1$-monoid algebra.<|endoftext|> TITLE: A more natural proof of Dold-Kan? QUESTION [36 upvotes]: The Dold-Kan correspondence gives an equivalence of categories between $SAb$, the category of simplicial abelian groups, and $Ch_{\geq 0}$, the category of non-negatively graded chain complexes of abelian groups. Recall that the equivalence in the direction $C: SAb \rightarrow Ch_{\geq 0}$ is most naturally given $C(A)_{n}=A_n/(s_{0}A_{n-1}+\cdots s_{n-1}A_{n-1})$ (simplices modulo degenerate simplices) and taking the differential to be $d=\sum (-1)^{i}d_{i}$. The equivalence in the other direction $\Gamma: Ch_{\geq 0} \rightarrow SAb$ is most naturally given by $\Gamma(C)_{n}=Ch_{\geq 0}(C(\mathbf{Z}\Delta[n]),C)$ (the group of chain maps) where $\Gamma(C)$ gets a simplicial structure by pulling back the comsimplicial structure on the collection of complexes $C(\mathbf{Z}\Delta[n])$. The reason that I say this is the most natural is because fits the standard pattern of constructing an adjunction between a presheaf category and another category. Namely, there is a functor $\Delta \rightarrow Ch_{\geq 0}$ sending $[n]$ to the complex of non-degenerate simplices in $\Delta[n]$ with differential given as the alternating sum of face maps. Then $C: SAb \rightarrow Ch_{\geq 0}$ is the left Kan extension of $\Delta \rightarrow Ch_{\geq 0}$ and $\Gamma$ is the obvious right adjoint to that (it can't be anything else). (Once upon a time I learned this kind of construction from Maclane-Moerdijk, but it can also be found on the nLab :http://ncatlab.org/nlab/show/nerve+and+realization.) One would then like to show that $C$ and $\Gamma$ are inverse equivalences by showing that the (obvious) unit $Id \rightarrow \Gamma C$ and counit $C\Gamma \rightarrow Id$ are isomorphisms. By constrast, in every textbook treatment that I have seen, one constructs instead of a quotient complex $C$ an isomorphic subcomplex $N$, complementary to degenerate simplices, but you can do this in at least two different ways. Moreover, one constructs an isomorphism $\oplus_{[n] \rightarrow [k]} NA_{k} \rightarrow A_{n}$, where the sum is over surjections $[n] \rightarrow [k]$ in $\Delta$, and defines $\Gamma$ similarly, as $\Gamma(C)=\oplus_{[n] \rightarrow [k]} C_{k}$. Ultimately, it seems that one then shows $\Gamma N(A)\simeq A$ for $A \in SAb$ and $N\Gamma(C) \simeq C$ for $C \in Ch_{\geq 0}$, but the isomorphism $\Gamma N(A)\simeq A$ is somehow in the wrong direction (before you know it's an isomorphism), since a priori you would only expect $N,\Gamma$ to form an adjunction and the isomorphism should be given by the unit $A \simeq \Gamma N(A)$. So my question is whether there is a 'more natural' treatment somewhere in the literature, phrased in terms of the quotient complex $C$, rather than the subcomplex $N$, and in terms of $\Gamma$ given by the obvious right adjoint, not as a sum over surjections, which at first seems to be pulled out of a hat. Of course one can unwind the usual proof and see that everything matches up. For example, one should be able to identify $Ch_{\geq 0}(C(\mathbf{Z}\Delta[n],C)$ with $\oplus_{[n]\rightarrow [k]} C_{k}$, but one can see at least two ways to do this. If there is no such treatment, can one explain how to construct one or why one shouldn't bother? REPLY [25 votes]: The proof of the Dold-Kan theorem basically amounts to the following. Let $\mathbb{Z}\Delta$ denote the pre-additive category generated by the simplicial indexing category, so that $s\mathrm{Ab}=\mathrm{Fun}^{\mathrm{add}}(\mathbb{Z}\Delta^\mathrm{op}, \mathrm{Ab})$, the category of additive functors. Let $\mathcal{C}$ be the "Karoubi envelope" of $\mathbb{Z}\Delta$, the category obtained by freely adjoining splittings of idempotents. You can identify the Karoubi envelope with a full subcategory $s\mathrm{Ab}$, namely the closure of the image of the Yoneda embedding under splitting idempotents. Then it is trivial that $\mathrm{Fun}^{\mathrm{add}}(\mathcal{C}^\mathrm{op},\mathrm{Ab}) \approx \mathrm{Fun}^{\mathrm{add}}(\mathbb{Z}\Delta^\mathrm{op}, \mathrm{Ab})=s\mathrm{Ab}$. The Dold-Kan theorem amounts to observing that every object in $\mathcal{C}$ is a direct sum of objects $G(n)$, and that the full subcategory of $G(n)$s is the "indexing category" for chain complexes. You also have that $G(n)\approx \mathbb{Z}\Delta[n]/\mathbb{Z}\Lambda^0[n]$. The usual formulas for normalized chains arise by explicitly identifying $G(n)$ with either a (split) subobject or a (split) quotient object of $\mathbb{Z}\Delta[n]$. In fact, there is only one way (up to sign) to identify $G(n)$ as a split subobject of $\mathbb{Z}\Delta[n]$, and this translates into the "quotient by degeneracies" construction of normalized chains. There are several ways to identify $G(n)$ as a split quotient: the usual presentation of normalized chains as subobjects is induced by the projection $\Delta[n]\to\Lambda^0[n]$. Of course, you could also be perverse: $G(n)$ is a summand of $\mathbb{Z}\Delta[k]$ for any $k\geq n$, so you could define normalized chains using your favorite idenification of $G(n)$ as a summand of $\mathbb{Z}\Delta[n+42]$, if you want! Added. Let me try to answer Karol's question, by giving a proof which doesn't use much computation. Set $F(n):=\mathbb{Z}\Delta[n]$ and $G(n):=\mathbb{Z}\Delta[n]/\mathbb{Z}\Lambda^0[n]$. The key observation is to show that the projection $F(n)\to G(n)$ admits a section. I don't have a clever way to do that ... you just have to write down the formula. (Which is: send the "generator" $\langle0,1,\dots,n\rangle$ of $G(n)$ to $\sum \pm\langle 0,a_1,\dots,a_n\rangle$, where the sum is over sequences with $a_k\in \{k-1,k\}$, and the sign is the parity of $\lvert\{k\,|\,a_k\neq k\}\rvert$.) The nice thing is that the section turns out to be unique (this is hardly obvious to begin with). Now filter $\Delta[n]$ by subcomplexes $\Delta^k[n]=$ the union of all $k$-simplices which contain the vertex $0$. It's easy to see that $$ \mathbb{Z}\Delta^k[n] /\mathbb{Z}\Delta^{k-1}[n] \approx G(k)^{\oplus \binom{n}{k}}. $$ But we just showed that the $G(k)$ are projective, so the filtration lifts to a direct sum decomposition $F(n)\approx \bigoplus_k G(k)^{\oplus\binom{n}{k}}$. Now you want to compute $\hom(G(m), G(n))$. Here is a trick. It's clear that $\hom(F(m),F(n))=\mathbb{Z}^{\binom{n+m+1}{m+1}}$. Since the $G(k)$ are summands of the $F(n)$, we have $\hom(G(i),G(j))\approx \mathbb{Z}^{a_{i,j}}$ for some $a_{i,j}\geq0$. We want to show $a_{i,i}=a_{i,i+1}=1$ and all other $a_{i,j}=0$. It's easy to exhibit that $a_{i,i}\geq1$ and $a_{i,i+1}\geq1$. Plugging the direct sum decomposition into $\hom(F(m),F(n))$ gives $$ \binom{n+m+1}{m+1} = \sum_{i,j} \binom{m}{i}\binom{n}{j} a_{i,j}. $$ But you can use standard combinatorial identities to show this is already an equality when we set $a_{i,i}=a_{i,i+1}=1$ and other $a_{i,j}=0$. So that must be the answer.<|endoftext|> TITLE: Grothendieck on polyhedra over finite fields QUESTION [20 upvotes]: In Grothendieck's Sketch of a Programme he spends a few pages discussing polyhedra over arbitrary rings and concludes with some intriguing remarks on specializing polyhedra over their "most singular characteristics". I am having trouble understanding what he means or finding other references. Here is a link to the version of Sketch I was reading. Sketch of a Programme More specifically, Grothendieck begins (p. 252): In 1977 and 1978, in parallel with two C4 courses on the geometry of the cube and that of the icosahedron, I began to become interested in regular polyhedra, which then appeared to me as particularly concrete “geometric realizations” of combinatorial maps, the vertices, edges and faces being realised as points, lines and plans respectively in a suitable 3-dimensional affine space, and respecting incidence relations. This notion of a geometric realisation of a combinatorial map keeps its meaning over an arbitrary base field, and even over an arbitrary base ring. It also keeps its meaning … My understanding is that he is viewing the polyhedron as a configuration of affine subspaces—what does he mean by "combinatorial maps"? This seems to be the key to understanding the last line, on the concept making sense over an arbitrary ring. The most curious remark comes at the end of this section where he writes (p. 255): … on the already known cases. Thus, examining the Pythagorean polyhedra one after the other, I saw that the same small miracle was repeated each time, which I called the combinatorial paradigm [underlined in original] of the polyhedra under consideration. Roughly speaking, it can be described by saying that when we consider the specialisation of the polyhedra in the or one of the most singular characteristic(s) (namely characteristics 2 and 5 for the icosahedron, characteristic 2 for the octahedron), we read off from the geometric regular polyhedron over the finite field ($\mathbb F_4$ and $\mathbb F_5$ for the icosahedron, $\mathbb F_2$ for the octahedron) a particularly elegant (and unexpected) description of the combinatorics of the polyhedron. It seems to me that I perceived there a principle of great generality, which I believed I found again for example in a later reflection on the combinatorics of the system of 27 lines on a cubic surface, and its relations with the root system $E_7$. Whether it happens that such a principle really exists, and even that we succeed in uncovering it from its cloak of fog, or that it recedes as we pursue it and ends up vanishing like a Fata Morgana, I find in it for my part a force of motivation, a rare fascination, perhaps similar to that of dreams. No doubt that following such an unformulated call, the unformulated seeking form, from an elusive glimpse which seems to take pleasure in simultaneously hiding and revealing itself — can only lead far, although no one could predict where… How is he determining the "most singular characteristics" of these polyhedra? If I can understand the combinatorial maps comment above, it may make more sense how he is considering the specializations of icosahedra over finite fields. What is the elegant description of their combinatorics Grothendieck refers to? If anyone can explain Grothendieck's comments or point to other references, I would be appreciative. REPLY [3 votes]: Not an exact answer to your question but maybe it can be helpful to know: La théorie combinatoire de l'icosaèdre by V Diekert This was a "Rapport pour un DES d'université de l'année 1977/78" under Grothendieck.<|endoftext|> TITLE: An example of an object in $D^b_{\text{coh}}(\mathbb{P}^2)$ which is not formal QUESTION [12 upvotes]: We know that for a curve $X$, any object $\mathcal{E}^{\bullet}$ in the derived category $D^b_{\text{coh}}(X)$ is formal, i.e. $\mathcal{E}^{\bullet}$ is quasi-isomporphic to the direct sum of its cohomology sheaves. The reason is that the cohomological dimension of $X$ is $1$. We can see Corollary 3.15 of Daniel Huybrechts' book "Fourier–Mukai transforms in algebraic geometry" for details. Now could we find an "easy" example of object in $D^b_{\text{coh}}(\mathbb{P}^2)$ which is not formal? In particular, could we find a complex of sheaves on $\mathbb{P}^2$ of length $2$ with coherent cohomology which is not quasi-isomorphic to the direct sum of its cohomology sheaves? REPLY [13 votes]: Yes; whenever you have two objects in an abelian category such that $Ext^2(M,N)$ is not equal to 0, we have a nonformal object given by coning with this morphism. More down-to-earthly, the element of $Ext^2(M,N)$ is given by some complex $N \to K \to L\to M$; the non-formal complex is just $\cdots \to 0\to K \to L \to 0\to \cdots$. EDIT: Thanks for the example below. I was too lazy to provide one, but I also think it risks camouflaging the actual point here, since there's nothing special about coherent sheaves on $\mathbb{P}^2$, this happens in any abelian category with global dimension $>1$.<|endoftext|> TITLE: Mazur secret Bourbaki report "Analyse p-adique" QUESTION [19 upvotes]: Does anyone happen to know if a scan of Mazur's report exists, and, if so, where to find it? It appears in the references for Katz's "Higher congruences" and "Eisenstein measure" papers. REPLY [70 votes]: I have it. Mazur gave me a xerox copy off his shelf when I asked him (in grad school) if a copy exists. It's 56 pages and the first sentence is: L'objet de ce rapport est de construire la série L p-adique de Kubota-Leopold et d'établir quelques propriétés fondamentales. It was in my office and I was going to try scanning it this evening, but literally as soon as I stepped into the elevator to go to the scanning machine a fire alarm went off in the building. So I had to leave. On my way out someone said he saw smoke in the hallway of the physics wing of the building. But even if the whole place burns down, fear not! I took Mazur's paper with me. You'll get a scan in the next few days. Update 1: A scan has now been made, and my department's building did not burn down. Update 2: I sent the scan to Mazur, who has posted it on his homepage at the "Older Material" link, so everyone can get a copy for themselves. A direct link to the scan on his homepage is here.<|endoftext|> TITLE: Is the center of the automorphism group of a von Neumann algebra M trivial whenever M is a factor? QUESTION [9 upvotes]: Question: Is the center of the automorphism group of a von Neumann algebra $\mathscr{M}$ trivial (=$\{\mathrm{id}\}$) whenever $\mathscr{M}$ is a factor (=$\mathscr{M}$ has center $\{\lambda I; \lambda \in \mathbb{C}\}$)? It is true in the finite dimensional case. A finite dimensional factor is $M_n(\mathbb{C})$ for some $n \in \mathbb{N}$. We proceed in two steps. First, every automorphism of $M_n$ is of the form $a \mapsto u^*a u$ for some unitary $u$. We will show that if an automorphism is in the center of $\mathop{\mathrm{Aut}}$, then $u$ is in the center of its unitary group. Finally, we show that the center of the unitary group is "trivial". There is a surjective group homomorpism $\varphi\colon U(n) \to \mathop{\mathrm{Aut}}(M_n)$ with $\varphi(u)(a)=u^*au$. The kernel of $\varphi$ are the unitaries in the center of $M_n$. Thus $\ker \varphi = \{\lambda I;\ |\lambda|=1\}$. Now, given any $u$ such that $\varphi(u)$ is in the center of $\mathop{\mathrm{Aut}}(M_n)$. Then for any other $v$ we have $vuv^{-1}u^{-1} \in \ker\varphi$. Thus $vuv^{-1}u^{-1}=\lambda I$ for some $\lambda \in \mathbb{C}$ with $|\lambda|=1$. In particular, when $u=v$ we find $I=\lambda I$. Apparently $vuv^{-1}u^{-1}=I$. Thus $u$ is in the center of $U(n)$. The embedding $e\colon U(n) \to GL(n)$ is an irreducible group representation. A unitary $u$ in the center of $U(n)$, is an intertwining map from this representation to itself. Thus, by Schur's lemma, it is $\lambda I$ for some $\lambda \in \mathbb{C}$. Clearly $|\lambda| = 1$. Thus the center of $U(n)$ is the kernel of $\varphi$. REPLY [6 votes]: Suppose that $\phi $ is in the centre of $\text {Aut}(M) $. Fix a unitary $u\in\mathcal M $. Then $$\tag {1}\phi (uxu^*)=u\phi(x)u^*$$ for all $x $. In particular, when $x=u $, we have $\phi ( u)=u\phi (u)u^*,$ or $\phi (u)u=u\phi (u)$. As $u $ is unitary, this also implies that $u^*\phi (u) =\phi (u)u^*$. Replacing $x $ with $xu $ in $(1) $, we get $\phi (ux)=u\phi (xu)u^*$, so $$u^*\phi (u)\phi (x)=\phi (x)u^*\phi (u). $$ As $\phi $ is onto, we deduce that $u^*\phi (u) $ is in the centre of $\mathcal M $. Since $\mathcal M $ is a factor, $\phi (u)=\lambda_u u $ for some $\lambda_u \in\mathbb C $. It is easy to see that the map $u\longmapsto\lambda_u $ is a group homomorphism. (Thanks to Jesse Peterson for the following argument) From $\phi(u)=\lambda_u u$, we get that $\sigma(u)=\lambda_u\,\sigma(u)$, since $\phi$ preserves the spectrum. So, if the spectrum of $u$ has no rotational symmetry, then $\lambda_u=1$. The set of unitaries with no rotational symmetry is dense in the set of unitaries of $\mathcal M$ (via the Spectral Theorem, since we can use unitaries with finite spectrum and tweak the eigenvalues slightly so that there is no rotational symmetry). By continuity, it turns out that $\phi$ is the identity on all the unitaries, and so $\phi$ is the identity.<|endoftext|> TITLE: Non-reflexive Banach space s.t. X,X*,X**,... are separable QUESTION [5 upvotes]: Is there an infinite-dimensional Banach space $X$, which is not reflexive, such that all the spaces $X,X^{\ast},X^{\ast\ast}, X^{\ast\ast\ast},\dots$ are separable? REPLY [7 votes]: I believe the James space is an example. It is isomorphic to its double dual (but not by the canonical embedding).<|endoftext|> TITLE: Can two fibered knots have the same exterior? QUESTION [7 upvotes]: Suppose I have two distinct fibered knots in a homology sphere. Is it possible for them to have (orientation-preservingly) homeomorphic exteriors? See Oriented knot complement conjecture for fibered knots for another version of this question. REPLY [10 votes]: Here we are assuming "distinct" means "non-isotopic". Here is an example, found using SnapPy: http://www.math.uic.edu/t3m/SnapPy/ Consider the SnapPea manifold M = Manifold('m390'). This has first homology group Z. We Dehn fill M along the slope (2,-3), via the command M.dehn_fill((2,-3)). Note that the filled manifold is still called M. We check M.homology() to see M is a homology sphere. By examining M.length_spectrum(2) we find M has a pair of geodesics of length 1.169 + 1.937*I. Let's call these the "twins" K and K'. The twins are tied for second and third shortest curves in the manifold. We now look at the list M.dual_curves(). These are the curves that SnapPy can drill for us. Luckily for us, the zeroth dual curve is one of the twins, K. So we drill K via N = M.drill(0). Just to be on the safe side, we take P = N.filled_triangulation(). Now, P.identify() tells us that P is the SnapPea manifold v2869. If we check Nathan Dunfield's list http://www.math.uiuc.edu/~nmd/snappea/tables/mflds_which_fiber we find that v2869 is fibered. So, K is a fibered knot in the homology sphere M. It is left to show that the complement of K' in M is homeomorphic to that of K. We examine P.length_spectrum(). The second shortest curve has length 0.870 + 2.070*I. This will be K'; its length has changed because we drilled K. Again we check the dual curves to find that K' can be drilled. [There is some junk in the collection of dual curves. I don't know why that happens?] So form Q = P.drill(2). Now, if we fill either cusp of Q using the meridional slope (1,0), we get manifolds isomorphic to P. Here is the list of commands: Q.dehn_fill((1,0),0) # fill K P.is_isometric_to(Q) # check Q.dehn_fill((0,0),0) # drill K Q.dehn_fill((1,0),1) # fill K' P.is_isometric_to(Q) # check Since the isometry checker returned true in both cases, we are done. Just as a final remark - I found m390 by looking in the cusped census for homology solid tori, with symmetries, with fundamental group of rank at least three (according to SnapPy). The first requirement is because we want to fill to get a homology sphere. The second and the third help the twins appear.<|endoftext|> TITLE: Automorphisms of generic complete intersections QUESTION [9 upvotes]: This question concerns a seemingly folk lore result, which states that automorphism groups of generic complete intersections are trivial, under certain assumptions. To state the question, let $r \geq 1$ and let $d_1,\ldots,d_r, n \in \mathbb{N}$ be such that $n \geq 3$. $d_i \geq 2$ for each $i$. $(d_1,\ldots,d_r;n) \neq (2;n)$ nor $(2,2;n)$. Let $X$ be a generic smooth complete intersection of dimension $n$ in projective space $\mathbb{P}^{n+r}$ over $\mathbb{C}$ with equations of degrees $d_1,\ldots,d_r$. Is $\mathrm{Aut}(X)$ trivial? Remarks: The conditions (1), (2), (3) imply that $\mathrm{Aut}(X)$ is finite and preserves the hyperplane class, for each such smooth complete intersection $X$ (not necessarily generic). The result is not true without condition (3), as here the generic such complete intersection has a non-trivial automorphism group (hopefully I did not miss any other ''bad'' cases). The answer is well-known to be yes when $r=1$, i.e. for generic hypersurfaces. I need a version of this result in a paper I am currently writing. We are able to use this to show that the action of $\mathrm{Aut}(X)$ on $H^n(X,\mathbb{C})$ is faithful for any such smooth complete intersection (not necessarily generic). I would be most interested if anyone has any ideas on alternative approaches to this problem as well. We are able to handle many special cases, such as when the degrees $d_i$ are all different or $X$ is of general type. A critical case is for example when all the degrees are the same and $X$ is Fano of high codimension. REPLY [3 votes]: I think the result you are looking for is worked out in Theorem 1.3 of Chen-Pan-Zhang "Automorphism and Cohomology II: Complete intersections" (+ Remark 1.4), see: https://arxiv.org/pdf/1511.07906.pdf<|endoftext|> TITLE: Jordan decomposition of the tensor product of two matrices QUESTION [6 upvotes]: I asked this question on Math.SE here, but did not get a lot of attention. I am interested in the problem of determining the Jordan decomposition of the tensor product of two unipotent matrices over a field of positive characteristic $p$. It is enough to consider the case of single Jordan block. So let $u$ and $v$ be unipotent Jordan blocks, of sizes $n \times n$ and $m \times m$ respectively. In characteristic $0$, if $n \geq m$ then $u \otimes v$ has Jordan blocks of sizes $n + m - 1, n + m - 3, \cdots, n - m + 1$. This formula no longer works in positive characteristic. But apparently it is valid if $p \geq m+n$. What is known about the Jordan decomposition of the tensor product in general? Is this open? I would like some useful description of the decomposition in terms of $n, m$ and $p$. REPLY [7 votes]: The situation (over a field of characteristic $p >0$) is well understood, and spelled out in : Srinivasan, Bhama The modular representation ring of a cyclic p-group. Proc. London Math. Soc. (3) 14 1964 677–688. The key points (for $p >2$) are that $J_{2}(1) \otimes J_{m}(1) = J_{m+1}(1) \oplus J_{m-1}(1)$ as long as $m TITLE: Does pointwise convergence of holomorphic functions on the boundary imply pointwise convergence in the interior? QUESTION [5 upvotes]: Let $\Omega$ be a simply connected open set in the complex plane and $\gamma$ be a simple path inside $\Omega$. Suppose $f_n$ is a sequence of holomorphic functions converging pointwise to 0 on $\gamma$. Does it imply that $f_n$ converges pointwise on the region enclosed by $\gamma$? REPLY [10 votes]: For a counterexample, let $\gamma$ be the unit circle. Let $$A_n = \{z \in \gamma:\; \text{Im}(z) \in [-1,0] \cup [1/n, 1]\}$$ By Runge's theorem there is a polynomial $f_n$ such that $|f_n| < 1/n$ on $A_n$ but $f_n(0) = (-1)^n$. We then have $f_n \to 0$ pointwise on $\gamma$ but $f_n(0)$ does not converge.<|endoftext|> TITLE: Can any finite lattice be realized as an intermediate subgroups lattice? QUESTION [10 upvotes]: Let $G$ be a finite group and $H$ a subgroup. Let $\mathcal{L}(H \subset G )$ be the lattice of all the intermediate subgroups between $H$ and $G$. Question: Can any finite lattice be realized as an intermediate subgroups lattice? Remark: It's true for all the finite distributive lattices (see theorem 2.1 here). REPLY [7 votes]: The more recent paper I've found about this problem is by Michael Aschbacher in 2013: Overgroup lattices in finite groups of Lie type containing a parabolic His introduction is a short survey of the last advanced, he recall Palfy–Pudlak theorem and question, and focus on the following John Shareshian's conjecture. His paper is a first step for a proof of this conjecture: Let $B_n$ be the subgroups lattice of the cyclic group $C_m$ with $m=p_1p_2 \dots p_n$ square free and $p_i$ prime number. Let $\mathcal{L}$ be a finite lattice and $\mathcal{L}'$ the poset $\mathcal{L}-\{l,g \}$ with $l$ and $g$ the least and greatest elements of $\mathcal{L}$. Shareshian's Conjecture: If $\mathcal{L}'$ is a disconnected graph with connected components $(B'_{n_i})_{i=1, \dots , k}$ and $n_i \ge 3$, then $\mathcal{L}$ is not an intermediate subgroups lattice. The smallest lattice $\mathcal{L}$ coming from this conjecture is the following: with $k=2$, $n_1=n_2=3$<|endoftext|> TITLE: Why does inconstructibility of $\sqrt[3]{2}$ imply impossibility of cube doubling? QUESTION [11 upvotes]: In this question "constructing" and "doubling" is meant in the compass-and-straightedge sense. On my desk I have five Basic Algebra texts treating constructability in the plane $\mathbb{C}$ or $\mathbb{R}^2$ as an application of basic field theory. After appropriate definitions of the possible construction steps, four of these, namely Hornfeck, Jacobson, Lorenzen, and Meyberg, prove that $\sqrt[3]{2}$ is inconstructible starting from $\{0,1\}$ or $\{(0,0),(1,0)\}$, respectively, but then conclude without further justification that the duplication of the cube is impossible. For a while I believed this last step to be obvious. But now, having to teach this for the first time the day after tomorrow, I have doubts: Being given a cube, say in $\mathbb{R}^3$, should mean being given its eight corners, and then I could use these to do constructions in space, using lines through two points and circles around one point and through two points. Or, to put it differently, I could take any three noncollinear points given or already constructed and do plane constructions in the plane spanned by these. Restricting the constructions initially to one particular coordinate plane containing one face of the cube appears unjustified to me. My specific questions are: How does one treat this problem honestly and elegantly, with a minimum of coordinate computations? Is the problem I see perhaps the reason why the fifth of my books, by M. Artin, does not mention cube doubling? REPLY [14 votes]: Disclaimer: The following perhaps isn't an answer to your question as stated, so my apologies if this answer is useless to you. However, you're asking for how to treat this problem "honestly", and I think that adding the right kind of historical perspective falls under the heading of honesty. Anyway, I think it is important to observe here that the ancient Greeks themselves did not limit their solutions to plane constructions. As can be read in Sir Thomas L. Heath's A History of Greek Mathematics, Vol. 1, pp. 246-9, Archytas proposed a solution to the problem where he intersected three surfaces of revolution in Euclidean $3$-space (a cone, a cylinder, and a torus) to obtain a point whose coordinates generate the field extension $\mathbb{Q}(\sqrt[3]{2})$. It is somewhat misleading, I think, to keep referring to these problems as "the three famous unsolved problems of Greek mathematics", because the Greeks in fact solved them many times over, Archytas' solution being only one out of a multitude. Moreover, they even recognized that the solution could not be achieved by plane methods, in a way: Pappus has it that the Greeks classified construction problems as "plane", "solid", and "linear", according to the methods with which the problem could be solved. Of course, they never tried to make this very precise, let alone tried to prove it, but then they weren't trying to do the impossible either.<|endoftext|> TITLE: Why is a matrix pencil called a pencil? QUESTION [18 upvotes]: I'm trying to understand the historical context behind the word pencil in matrix pencils, or pencil of curves so on. I am aware that even Gantmacher 1959 has this terminology however I don't know where it originates from. I am also curious what he uses in the original Russian version in place for that word (though I don't know any Russian, I can handle a literal translation ala Körper etc.). EDIT Since there are answers given towards the meaning of the word "pencil" which is really good to know, I would appreciate if the context is also taken into account. It is from the definition of the pencil forms that some sort of bundling or parameterization is involved. However the definition itself of the word pencil does not introduce the context. Compare it with the word affine which comes from the similar meaning (Latin affinis) "adjacent,connected" but this is not preferred for some reason although the structure of matrix pencils resembles $a\lambda - b$ more an affine transformation in my opinion. Obviously, it might be a nonlinear function of $\lambda$ but that context looks like long forgotten (until recently the computational tools for quadratic and nonlinear eigenvalue problems started to emerge). Thus, I would speculate that some circles deliberately avoided either pencil or the affine word at some point. That's what I would like to understand. REPLY [14 votes]: The Oxford English Dictionary has an example from 1665 of "pencil" in the sense of "A group of rays or a beam of radiation converging to or diverging from a point." And one from 1840 in the geometric sense of "A set of lines meeting in a point"<|endoftext|> TITLE: What is the explicit ideal (wrt the Lazard ring) generated by the associativity of formal group laws? QUESTION [6 upvotes]: Quick Preliminaries: A commutative formal group law is a formal power series $F(x,y)=\sum_{ij}c_{ij}x^iy^j$ that satisfies: Commutativity: $F(x,y) = F(y,x)$ Identity: $F(x,0)=x=F(0,x)$ Associativity: $F(F(x,y),z) = F(x,F(y,z))$ The restrictions imposed by the formal group law axioms on the coefficient ring of the formal power series generate an ideal $I$. The Lazard ring is the result of modding out our coefficient ring $c_{ij} \in Z[c_{ij}]$ by this ideal. $L=\mathbb{Z}[c_{ij}]/I$. Explicitly, the relations imposed on our coefficients are: $F(x,y)=F(y,x)$ => $c_{ij}=c_{ji}$ $F(x,0)=x=F(0,x)$ $F(x,0)=\sum_{ij}c_{ij}x^i0^j=x$ => $c_{10}=1$ $F(0,x)=\sum_{ij}c_{ij}0^ix^j=x$ => $c_{01}=1$ => $c_{10}=c_{01}=1$ $F(F(x,y),z) = F(x,F(y,z))$ => ???? Here's my question: What do the relations imposed by the associativity condition explicitly look like? Thusfar, in every source I've looked at (including Lazard's original papers), I've been unable to find this! It's driving me up the wall! I derived the following. However, I'm unsure if it is correct, and would be deeply appreciative if someone could point out a simpler way express it. My sad attempt: Looking for the coefficient of $x^{\alpha}y^{\beta}z^{\gamma}$ in $F(x,F(y,z))$: We start by expanding the coefficients of $F(x, -)$ (with indices $i$ and $j$), then we must have $i = \alpha$ to get $x^\alpha$ (this is why $i$ does not appear). Next we expand the coefficients of $F(y,z)$ (with indices $k$ and $\ell$) -- this whole thing gets raised to the power $j$. I think we need $j$ copies of it that multiply out correctly (which corresponds to the exponents adding up correctly): $b_{\alpha\beta\gamma} = \sum_{j=0}^{\infty}\sum_{(k_{1}+...+k_{j}=\beta)}$ $\sum_{(\ell_{1}+...+\ell_{j}=\gamma)}c_{\alpha j}c_{k_{1}\ell_{1}}...c_{k_{j}\ell_{j}}$ Repeating the procedure with the appropriate indices, we get the coefficients associated to $F(F(x,y), z)$: $b_{\alpha\beta\gamma} = \sum_{i=0}^{\infty}\sum_{(k_{1}+...+k_{i}=\alpha)}$ $\sum_{(\ell_{1}+...+\ell_{i}=\beta)}c_{i\gamma}c_{k_{1}\ell_{1}}...c_{k_{i}\ell_{i}}$ A related question: I get that we must make sure that each of our 3 constraints is homogeneous (this ensures that you still have a grading when you mod out by the ideal generated by them). However, I don't understand why we impose the following grading: $$\text{deg}c_{ij} = i + j - 1$$ My only hint is looking at the coefficients of $F(x,F(y,z)$ (i.e. $c_{\alpha j}c_{k_{1}\ell_{1}}...c_{k_{j}\ell_{j}}$). If we add the degrees of each term in the product, we get: $(\alpha + j - 1) + (k_1 + l_1 - 1) + ... + (k_j + l_j - 1)$ $\begin{align*} &= \alpha + (k_1 + ... + k_j) + (l_1 + ... + l_j) + j - j - 1 &= \alpha + \beta + \gamma - 1 \end{align*}$ I'm guessing the two expressions for $b_{ijk}$ that we have are homogeneous of the same degree, so that setting them equal to each other doesn’t ruin the homogeneity of the ideal. REPLY [12 votes]: Your computation of the relations looks correct to me. The grading is that way because it comes from the symmetry where you multiply the variable by a constant. What I mean is that you take a formal group law: $$z = F(x,y)$$ and you multiply every variable by $\lambda$: $$\lambda z = F(\lambda x, \lambda y) $$ $$ z = \frac{ F( \lambda x, \lambda y)}{\lambda}$$ and you get another formal group law. So $c_{i,j}$ is sent to $\lambda^{i+j-1}c_{i,j}$. This $\mathbb G_m$-action gives you a grading.<|endoftext|> TITLE: History of $\frac d{dt}\tan^{-1}(t)=\frac 1{1+t^2}$ QUESTION [86 upvotes]: Let $\theta = \tan^{-1}(t)$. Nowadays it is taught: 1º that $$ \frac{d\theta}{dt} = \frac 1{dt\,/\,d\theta} = \frac 1{1+t^2}, \tag1 $$ 2º that, via the fundamental theorem of calculus, this is equivalent to $$ \theta = \int_0^t\frac{du}{1+u^2}, \tag2 $$ 3º that, expanding the integrand in a geometric series and integrating term by term, this becomes the Nilakantha Madhava-Gregory-Leibniz formula $$ \theta = t - \frac{t^3}3+\frac{t^5}5-\frac{t^7}7+\dots. \tag3 $$ Question: Who first proved $(1)$ in print as we do, by deriving an inverse function? I can’t find it in Nilakantha: According to Ranjan Roy (1990, p.300), Nilakantha first published $(3)$ without proof in his Tantrasangraha (1501); a later commentary known as Yuktibhasa contains a proof by rectification of an arc of circle, which is beautiful but certainly not quite the same as $(1)$. I can’t find it in Gregory: According to Dehn & Hellinger (1943, p.149), Gregory communicated $(3)$ in a 1671 letter to Collins, and never published a proof; speculation exists that he found it by deriving $\tan^{-1}$ enough times to figure out its Taylor series at $0$, but in any event, he left no trace of how he may have computed these derivatives. I can’t find it in Leibniz: According to González-Velasco (2011, p.347), Leibniz communicated $(3)$ in 1674 letters to Oldenburg and Huygens, and later published the case $t=1$ in Acta Eruditorum (1682, pp.41-46); his unpublished proof is available (many times over) in the 700+ pages of his Collected Works, Vol. VII,6. There, or in the nice exposition given in Hairer & Wanner (1996, 2nd printing, pp.49-50), one sees that he was squaring the circle in an elaborate way which has nothing to do with $(1)$. Of course Leibniz must have become aware of $(1)$ and $(2)$ at some point, as (later!) inventor of the notation that makes them almost automatic. Unfortunately, I can’t find any written evidence of that. Maybe someone else will have better luck! (The closest I can find is a 1707 letter of Wolff to Leibniz, where the new notation is used to write in effect that $d\theta = dt:(1+t^2)$, and then deduce $(2)$ and $(3)$. The two correspondents may well have had in mind the modern proof $(1)$ of this differential relation, but neither says so.) I can’t find it in Jacob Bernoulli: With Leibniz notation spreading, one might think that a disciple would write $(1)$ at the first opportunity. But that’s not what Jacob B. does to rectify a unit circle in Positionum de Seriebus Infinitis Pars Tertia (Basel, 1696, Prop. XLV): instead, he parametrizes one with $(x,y)=$ $\bigl(x,\sqrt{2x-x^2}\bigr)$ and then expresses the resulting arc length differential — also seen in Leibniz (1686) — as $$ d\theta =\sqrt{\smash{dx^2+dy^2}\vphantom{a^2}} =\frac{dx}{\sqrt{2x-x^2}} =\frac{2d\mathsf t}{1+\mathsf t^2} \tag{$*$} $$ $(=\mathrm{LH}$ on his Fig. 3$)$ by introducing a “diophantine” (a.k.a. Weierstraß) substitution $\smash{\mathsf t=\frac xy}$ $=\smash{\tan\frac\theta2}$ $(=\mathrm{BI}$ on the figure, as he notes; so his $\mathsf t=\tan\mathrm{BAI}$ is not our $t=\tan\mathrm{BAH})$. While this still proves $(2)$ and $(3)$ for the halved angle and its tangent, the argument definitely isn’t $(1)$. $\hspace{8.5em}$ I can’t find it in Johann Bernoulli: When faced with the task of integrating $\smash{\frac{dt}{1+t^2}}$ in his paper on rational integrals (1702), Johann B. proposes two substitutions: The first (in Probl. I, Corol.) comes from the partial fraction decomposition $\frac1{1+t^2}=\frac{1/2}{1+it} + \frac{1/2}{1-it}$, and consists in putting $u = \frac{1+it}{1-it}$ so that $ \frac{dt}{1+t^2}=\frac{du}{2iu}=d\left[\frac1{2i}\log\frac{1+it}{1-it}\right] $. The second (in Probl. II) consists in putting $u=\frac1{1+t^2}$ so that $\frac{dt}{1+t^2} = \frac{-du}{\sqrt{4(u-u^2)}}$, which he knows (perhaps by recognizing half $(*)$ with $x=2(1-u)$?) is a “circular sector or arc differential”. Neither of these is the substitution $\theta = \tan^{-1}(t)$, which via $(1)$ would have led directly to $\frac{dt}{1+t^2} = d\theta$. And in later papers (1712, 1719) Bernoulli is content to describe this relation as “well-known”. I can’t find it in de Moivre: Schneider (1968, footnotes 248 & 250) seems to claim that a 1708 letter of de Moivre to Bernoulli has the proof $(1)$ and also the “Euler” formula $\theta=\smash{\frac1i}\log(\cos\theta+i\sin\theta)$. But this is not borne out by the letter’s text in (1931, pp.241-257): there, as also in his paper (1703, p.1124) and book (1730, p.44), de Moivre simply states $\smash{d\theta=\frac{dt}{1+t^2}}$ without proving it anew. I can’t find it in Euler: Euler was of course well aware of $(2)$, which appears for instance in his fifth paper (1729, pp.93, 95) and in his later E60, 65, 66, 125, 129, 130, 162, 217, 391, 475, 482, etc. But when it comes to proving $(2)$ or $(3)$, then again he seems to eschew $(1)$: In his precalculus book (1748, §§139-140) he chooses to first establish Bernoulli’s above formula $$ \theta = \frac1{2i}\log\frac{1+it}{1-it} \tag4 $$ (this he does by multiplying numerator and denominator by $\cos\theta$ so they become $e^{\pm i\theta}$), and then to deduce $(3)$ by plugging $it$ in the series for $\smash{\log\frac{1+x}{1-x}}$. None of this requires $(1)$, $(2)$, or any calculus. In his differential calculus book (1755, §§194-197) he first differentiates a similar logarithmic formula for $\theta = \sin^{-1}(s)$, namely $\theta = -i\log(\sqrt{1-s^2} + is)$, to obtain $$ d\theta = \frac{ds}{\sqrt{1-s^2}}. \tag5 $$ Plugging $s = t/\sqrt{1+t^2}$ into $(5)$ then gives him $(2)$. He might as well have differentiated $(4)$ directly! Either way, $(1)$ is not used, although to be fair, Euler at least gives (§195) an alternative proof of $(5)$ using the argument $(1)$, but applied to $\smash{\sin^{-1}}$ instead of $\smash{\tan^{-1}}$. So where can you find it? It’s in Lacroix (1797, pp.113-114) and its progeny. Still I have trouble believing it took over 100 years for $(1)$ to become the standard proof — hence my question. REPLY [51 votes]: I now believe that my question (and suggestion that proof $(1)$ should have become standard before Lacroix) relied on the misconception that tangent was easier to differentiate than arctangent. In fact the calculus of inverse trigonometric functions took off earlier, as has been explained by G. Eneström (1905), C. Boyer (1947), or V. J. Katz (1987): it was quite common [at first] to deal with what we call the arcsine function rather than the sine. C. Wilson (2001, 2007) concurs and stresses that our trig functions with periodic graphs weren’t much seen or differentiated until Euler “found” them to solve 2nd order linear differential equations (1741); so much so that he still wrote in (1749, p.15): as this way of operating is not yet commonplace, it will be apropos to warn that the differentials of the formulas $\sin.$ : $\cos.$ : $\mathrm{tang}.$ : $\cot.$ are $d\,\cos.$ : $-d\,\sin.$ : $\smash[b]{\frac{d}{\cos.\,^2}}$ & $\smash[b]{-\frac{d}{\sin.\,^2}}$ — and e.g. in (1796, p.163) L’Huilier still computed $\tan'$ from $\arctan'$ rather than vice versa. In this vein, $d = \frac{dt}{1+t^2}$ was not proved like $(1)$, but by a differential triangle argument similar to $(*)$ but simpler and attributed to Cotes (Aestimatio errorum, 1722): in modern notation, parametrize the unit circle with $(x,y)=\frac{(1,\,t)}{\sqrt{1+t^2}}$ and obtain $$ d\theta =\sqrt{\smash{dx^2+dy^2}\vphantom{a^2}} =\frac{dt}{1+t^2} \tag6 $$ ($=\mathrm{CE}$ in Cotes’ figure, which became standard in many books even before his own — the list could almost be described as “everyone but Euler”): 1708 Charles-René Reyneau §590 fig. 41   1718 John Craig pp.52–54   1722 Roger Cotes (posthumous) Lemma II   1730 Edmund Stone p.63 fig. 13   1736 James Hodgson p.230   1736 John Muller §247 fig. 153  1737 Thomas Simpson §143   1742 Colin MacLaurin §195 fig. 52   1743 William Emerson pp.171–172 fig. 76   1748 Maria Agnesi p.639 fig. 4   1749 Charles Walmesley (credits Cotes) pp.3,53 fig. 10   1749 William Emerson p.29 fig. 6   1750 Thomas Simpson §142   1754 Louis-Antoine de Bougainville p.24 fig. 9   1761 Abraham Kästner  §299 fig. 18   1765 Jean Le Rond D’Alembert p.640 fig. 25   1767 Étienne Bézout p.146 fig. 46   1768 Thomas Le Seur & François Jacquier p.63 fig. 7   1774 Jean Saury pp.25,63 fig. 3   1779 Samuel Horsley pp.298–299   1786 Simon L’Huilier pp.103–104 fig. 20   1795 Simon L’Huilier §76 fig. 17  As to our usual proof $(1)$, it appears before Lacroix in 18-year-old Legendre’s Theses mathematicæ (1770), then in a book by their common teacher J.-F. Marie and several others: 1770 Adrien-Marie Legendre pp.10,16   1772 Joseph-François Marie §904   1777 Jacques-Antoine Joseph Cousin p.81   1781 Claude Bertrand p.140   1781 Louis Lefèvre-Gineau p.31   1795 Simon L’Huilier §76   1797 Sylvestre-François Lacroix pp.113–114   1801 Joseph-Louis Lagrange p.81   1810 Sylvestre-François Lacroix pp.lii,203–204<|endoftext|> TITLE: Is the Ford disk packing optimal? QUESTION [14 upvotes]: Given two unit-diameter disks tangent to a given line and to each other, determining a region bounded by two circular arcs and a line segment, is the Ford disk packing of that region the unique packing that covers as much total area as possible, among all ways of packing the region with disks tangent to the line? As usual, disks in a packing must have disjoint interiors. For background on Ford disk packings, see http://en.wikipedia.org/wiki/Ford_circle . Note that I am not asking about Apollonian packings; in my packings, all disks must be tangent to the bounding line segment. I would also be interested in links to existing literature on other packing problems of a similar nature, where the, um, tiles in the packing (is there a more apt generic word than "tiles"?) are required to be tangent to one of the sides of the region being packed. REPLY [2 votes]: There is an affirmative answer to a related question in which we view these disks as horodisks in the upper half-plane model of the hyperbolic plane. In this setting, it is natural to extend the Ford disk packing by applying (integer) horizontal translation and to add one more horodisk to the packing, namely, the shifted upper half plane ${\rm Im} \ z \geq 1$. We superimpose a dissection of the plane into ideal triangles whose edges are geodesics joining the points at infinity of mutually tangent horodisks. Each triangle in this hyperbolic Delaunay triangulation has some finite (hyperbolic) area $A$, and its overlap with the three disks that intersect it has area $3A/\pi$. Laszlo Fejes-Toth, in his 1953 article "Kreisausfullungen der hyperbolischen Ebene" (Acta Mathematica Hungarica 4(1-2), 103-110), showed that $3/\pi$ was the largest fraction of an ideal triangle that can be covered by disjoint horodisks. If one applies the ideas of Bowen and Radin (see for instance their article "Densest packing of equal spheres in hyperbolic space", https://www.ma.utexas.edu/users/radin/papers/hyper.pdf), one can make rigorous sense of the assertion that these horodisks, taken together, occupy exactly $3/\pi$ of the hyperbolic plane, and that no horodisk packing can beat this bound. I do not understand Toth's article to the extent of being able to assert that Ford packings are the only packings that achieve Toth's bound, but I expect that this is true. Thanks to Henry Cohn for pointing out this variant of my question.<|endoftext|> TITLE: Counterexample for associativity of smash product QUESTION [15 upvotes]: In Section 1.7 of Parametrized Homotopy Theory by May and Sigurdsson it is stated that the smash product of pointed topological spaces is not associative (which is just another hint that $\mathrm{Top}$ is the "wrong category"). Specifically, they claim that $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q})$ is not isomorphic to $(\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ in $\mathrm{Top}_*$. But actually they just prove that the canonical bijection $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q}) \to (\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ is not an isomorphism. Equivalently, there is no isomorphism in the slice category $(\mathbb{N} \times \mathbb{Q} \times \mathbb{Q}) / \mathrm{Top}_*$. Therefore, my question is as follows: How to prove that there is no isomorphism between $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q})$ and $(\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ in $\mathrm{Top}_*$? I had asked the same question on math.SE. REPLY [9 votes]: Building on Fernando's answer, here is a proof that they are not homeomorphic. By Fernando's answer, it suffices to show that if a sequence $(x_k,y_k,n_k)$ converges to the basepoint in $\mathbb{Q}\wedge(\mathbb{Q}\wedge\mathbb{N})$, it also converges to the basepoint in $(\mathbb{Q}\wedge\mathbb{Q})\wedge\mathbb{N}$. Let $(x_k,y_k,n_k)$ be such a sequence. We may assume that the points $(x_k,y_k,n_k)$ are all distinct from the basepoint. I claim $\{n_k\}\subseteq\mathbb{N}$ must be a finite set; the result then follows easily. Suppose for a contradiction that $\{n_k\}$ is infinite. Passing to a subsequence, we may assume that for each $n\in\mathbb{N}$, there are only finitely many $k$ such that $n_k=n$. For definiteness of notation, let us say that $0$ is our chosen basepoint in both $\mathbb{Q}$ and $\mathbb{N}$. For each $n>0$, let $\epsilon_n>0$ be such that $\epsilon_n<|x_k|$ and $\epsilon_n<|y_k|$ for all $k$ such that $n_k=n$. Now let $$U=\{(x,y,n):|x|<\epsilon_n\text{ or }|y|<\epsilon_n\}\subset \mathbb{Q}\wedge(\mathbb{Q}\wedge\mathbb{N}).$$ Then $U$ contains the basepoint and is disjoint from our sequence. Furthermore, the pullback of $U$ to $\mathbb{Q}\times(\mathbb{Q}\wedge\mathbb{N})$ is open: it is the union of the open sets $$V=\mathbb{Q}\times\{(y,n):|y|<\epsilon_n\}$$ and $$W_n=\{x:|x|<\epsilon_n\}\times \{(y,n):y\neq0\},$$ the latter being a separate open set for each fixed $n>0$. Thus $U$ is open, contradicting the assumption that our sequence converges to the basepoint.<|endoftext|> TITLE: Definition and sigularity of Ramified covers QUESTION [5 upvotes]: Let $X$ be a normal variety over $\mathbb{C}$. In their book Birational geometry of algebraic varieties, Kollár and Mori define [Definition 2.50 and 2.51] a ramified m-th cyclic cover associate to a line bundle $L$ ramified along $D \subseteq |mL|$ to be the relative spec $$Spec_X(\oplus_{i=0}^{m-1}L^{-i}).$$ Or more generally, for a rank $1$ torison free sheaf $L$, the m-th cyclic ramified cover is $$Spec_X(\oplus_{i=0}^{m-1}L^{[-i]}),$$ where $L^{[i]}$ is the double dual of $L^{\otimes{i}}$. In the book Singularities of the Minimal Model Program, Kollár and Kovács give a definition of ramified cover [see Definition 2.39], which is roughly as follows: A finite morphism of normal schemes $\pi: \tilde{X} \to X$ is called a ramified cover of degree m if there is a dense open subset $U \subseteq X$ that contains every codimension $1$ point of Sing$X$ such that the restriction $\pi_U: \tilde{U}\to U$ is etale and has degree m. My question: Is the (general) ramified m-th cyclic cover in the sense of Kollár and Mori a special case of the ramified cover in the sense of Kollár and Kovács? In my case, $X$ has canonical singularities, and $D$ can contain the codimension 1 point of Sing$X$. So, when choose $U$, it is inevitable to intersect $D$, hence I worry if the resulting morphism etale? My interest in the problem is because I want to know the singularity of the ramified cyclic cover $\tilde{X}$ (in the sense of Kollár and Mori). Again, $X$ has canonical singularities, I want to know if $\tilde{X}$ has the same singularities. By the book of Kollár and Kovács (See Page 65-65), it claims that the discrepancy does not get worse by taking a finite ramified cover (in their definition). I looked at the proof, and feel it could go through without any change for the (general ) cyclic ramified cover case. Did I miss something? REPLY [5 votes]: Ok, so based on the discussion in the comments, maybe I should put this into an answer. I think the confusion comes from the phrase every codimension 1 point of $\text{Sing }X$. What the authors Kollár and Kovács mean here is to consider every point of $\text{Sing }X$ that is also a codimension $1$ point of $X$. I agree that this could be interpreted in other ways, but this is what the authors mean (I'm sure Sándor will agree if he sees this). Given this, and that $X$ is normal over $\mathbb{C}$ (or any field of characteristic zero), it is easy to see that every cyclic cover is a ramified cover in the sense of Kollár-Kovács (as I think you already see). As pointed out in the comments, if $X$ is non-normal, or if we are working in characteristic $p > 0$, life becomes more complicated. Your particular situation You had $X$ with canonical singularities and $\widetilde{X}$ a ramified cyclic cover. Then it is easy to see that $(X, -\text{(Ramification Divisor)})$ also has canonical singularities (notice we have a non-effective divisor here). For a proof simply see Kollár-Mori 5.20(3).<|endoftext|> TITLE: For a quasicategory $C$, why is $\mathrm{Fun}(\Lambda^2_0,C) \to \mathrm{Fun}(\Delta^{\{2\}},C) \cong C$ a cocartesian fibration? QUESTION [6 upvotes]: More generally, I expect that the following is true: Let $D$ be a diagram quasicategory, let $d \in D$ be a vertex, and use this to define $D' = D \amalg_{\Delta^{\{0\}}} \Delta^1$. Then $\mathrm{Fun}(D',C) \to \mathrm{Fun}(\Delta^{\{1\}},C) \cong C$ is a cocartesian fibration. (This follows from Corollary 2.4.7.12 of Higher Topos Theory when $D$ is a point.) Note that $D'$ will not generally be a quasicategory, but it will generate a quasicategory $\tilde{D'}$ (i.e. $D' \stackrel{\approx}{\rightarrowtail} \tilde{D'}$ in the Joyal model structure) such that the restriction map $\mathrm{Fun}(\tilde{D'},C) \to \mathrm{Fun}(D',C)$ is a categorical equivalence (probably even a categorical fibration). Either of these computes the Joyal-homotopy pushout of $D \xleftarrow{d} \Delta^{\{0\}} \to \Delta^1$, which should be thought of as "freely adding a morphism to $D$ with source $d$". This fact is sufficiently obvious in 1-categories that people generally assert it without proof, but the intuition is the same: given a map $x \xrightarrow{f} y$ in $C$ and a functor $F : D' \to C$ restricting to some map $\Delta^1 \to C$ selecting an edge $a \xrightarrow{g} x$, a cocartesian lift should be given by the functor $f_*F$ which agrees with $F$ on $D$, takes $\Delta^{\{1\}}$ to $y$, and takes the edge $\Delta^1$ to a composite $a \xrightarrow{fg} y$. I have an incomplete attempt at a proof of the statement in this form. Really I would prefer an "invariant" argument, i.e. one that runs in the quasicategory of quasicategories, but of course anything would be better than nothing. The key observation is that this is true in case that $D$ is a point, i.e. it's true of the cocartesian fibration $\mathrm{Fun}(\Delta^1,C) \to \mathrm{Fun}(\Delta^{\{1\}},C) \cong C$: more precisely, given an edge $x \xrightarrow{f} y$ of $C$ and a point $a \xrightarrow{g} x$ of the fiber over $x$, there is a cocartesian edge $\Delta^1 \to \mathrm{Fun}(\Delta^1,C)$ described by a commutative square $$\require{AMScd} \begin{CD} a @= a \\ @VV{g}V @VV{f g}V \\ x @>>{f}> y \end{CD}$$ (where the un-drawn lower-left $\Delta^2$ is a composition and the un-drawn upper-right $\Delta^2$ is degenerate, and the outer and inner copies of $\Delta^1$ correspond to the horizontal and vertical directions, resp.). Let me write $\tilde{f}$ for this edge of $\mathrm{Fun}(\Delta^1,C)$. Now, note that $$ \mathrm{Fun}(D',C) \cong \mathrm{Fun}(D \amalg_{\Delta^0} \Delta^1,C) \cong \mathrm{Fun}(D,C) \times_C \mathrm{Fun}(\Delta^1,C). $$ The edge $id_{F|D}$ of the left factor agrees in $C$ with the edge $\tilde{f}$ of the right factor (both restrict to the edge $id_a$), so they define an edge of the fiber product, which I'll call $\tilde{F}$. The claim is that this edge is a cocartesian edge over $f$ with source $F$. By definition, this is the assertion that the map $$ \mathrm{Fun}(D',C)_{\tilde{F}/} \longrightarrow \mathrm{Fun}(D',C)_{F/} \times_{C_{x/}} C_{f/}$$ is a trivial Kan fibration. Unwinding the definitions, the source is $$ \mathrm{Fun}(D',C)_{\tilde{F}/} \cong \mathrm{Fun}(D \amalg_{\Delta^0} \Delta^1 , C)_{\tilde{F}/} \cong \mathrm{Fun}(D,C)_{id_{F|D}/} \times_{C_{id_a/}} \mathrm{Fun}(\Delta^1,C)_{\tilde{f}/} , $$ while the target is $$ \mathrm{Fun}(D',C)_{F/} \times_{C_{x/}} C_{f/} \cong \left( \mathrm{Fun}(D \amalg_{\Delta^0} \Delta^1 , C)_{F/} \right) \times_{C_{x/}} C_{f/}$$ $$ \cong \left( \mathrm{Fun}(D,C)_{(F|D)/} \times_{C_{a/}} \mathrm{Fun}(\Delta^1,C)_{g/} \right) \times_{C_{x/}} C_{f/} $$ $$ \cong \mathrm{Fun}(D,C)_{(F|D)/} \times_{C_{a/}} \left( \mathrm{Fun}(\Delta^1,C)_{g/} \times_{C_{x/}} C_{f/} \right) $$ (where this last reassociation comes from the fact that on both lines, the map from the inner fiber product to the base of the outer fiber product factors through the projection to the middle term $\mathrm{Fun}(\Delta^1,C)_{g/}$). Under these decompositions, the map which must be shown to be a trivial Kan fibration is given by the induced map from the pullback of the top row to the pullback of the bottom row in the diagram $$\begin{CD} \mathrm{Fun}(D, C)_{id_{F|D} /} @>>> C_{id_{a} /} @<<< \mathrm{Fun}(\Delta^1, C)_{\tilde{f} /} \\ @VVV @VVV @VVV \\ \mathrm{Fun}(D, C)_{(F|D) /} @>>> C_{a /} @<<< \mathrm{Fun}(\Delta^1, C) \times_{C_{x /}} C_{f /} \end{CD}$$ (If we consider these as fitting into a cube where the left square here gives the front square of the cube and the right square here gives the right square of the cube, then I'm interested in showing that if we take the pullbacks of the cospans in the top and bottom faces, then the induced back-left vertical arrow is a trivial Kan fibration.) Now, by definition of $\tilde{f}$ being cocartesian, the right vertical arrow is a trivial Kan fibration. On the other hand, I'm pretty sure the left and middle vertical arrows are trivial categorical fibrations, and that both left-hand horizontal arrows are categorical fibrations. But I don't quite see how to leverage this all into a proof of the statement I'm after. It would follow from pullback-pasting if the front square were a pullback, but I don't think that's quite true. REPLY [5 votes]: This follows from (the coCartesian version of) HTT Corollary 2.4.7.12, applied to the "evaluation at $d$" functor $\mathrm{Fun}(\mathcal{D}, \mathcal{C}) \rightarrow \mathcal{C}$, since we can identify $\mathrm{Fun}(\mathcal{D}', \mathcal{C})$ with $\mathrm{Fun}(\mathcal{D}, \mathcal{C}) \times_{\mathcal{C}} \mathrm{Fun}(\Delta^1, \mathcal{C})$, where the fibre product is via the evaluation at $0$, and the projection to $\mathcal{C}$ is via evaluation at $1$. (In fact, this is the free coCartesian fibration on the evaluation at $d$ functor.)<|endoftext|> TITLE: Are $L^\infty(\Bbb R)$ and $L^2(\Bbb R)$ homeomorphic? QUESTION [13 upvotes]: It's easy to see that, for $1\le p,q< \infty$ the spaces $L^p(\Bbb R)$ and $L^q(\Bbb R)$ of $p$-th and $q$-th power integrable functions on the real line are homeomorphic as topological spaces. In fact, the map $f(x)\mapsto sgn(f(x))|f(x)|^{q/p}$ provides an explicit homeomorphism $L^p(\Bbb R) \to L^q(\Bbb R)$. However this argument cannot be applied to the case where $p$ or $q$ equals infinity. Thus, I'm asking whether there is a homeomorphism from $L^\infty(\Bbb R)$ to any (and hence every) $L^p(\Bbb R)$ with $p<\infty$. REPLY [18 votes]: Paul is right. $L^2(\mathbb{R})$ is separable. (The rational simple functions ought to be one example of something ctbl. and dense.) However, $L^{\infty}(\mathbb{R})$ isn't separable. By Jones' Lemma, if $L^{\infty}(\mathbb{R})$ were separable then any closed discrete (i.e., nonclustering) set must be of size less than continuum. But the characteristic functions $\chi_{[0,x]}$, $x\in[0,\infty)$, are pairwise of distance 1 from eachother.<|endoftext|> TITLE: Is the Thom diagonal co-$E_\infty$? QUESTION [5 upvotes]: Given a map of spaces $f:X\to BGL_1(R)$ for $R$ an $E_\infty$-ring spectrum (of course this can be done more generally) one can produce a Thom spectrum $Mf$ by a number of methods. Let's denote such a datum by $(X,f)$ and let $(X,\ast)$ denote the datum $X\to \ast\to BGL_1(R)$, whose associated Thom spectrum is $R\wedge\Sigma^\infty_+X$, which we will denote by $R[X]$. There is a morphism in $Top_{BGL_1(R)}$, which we might denote $(\Delta,-\times\ast):(X,f)\to (X\times X,f\times\ast)$ which Thomifies to the morphism of spectra $Mf\to Mf\wedge R[X]$ (note that the target of this map is an object of $Top_{BGL_1(R)}$ because we can multiply the two maps together using the multiplicative structure on $BGL_1(R)$). It is not hard to show that the map $(X,\ast)\to (X,\ast\times\ast)$ induces a co-$E_\infty$ comultiplication on $R[X]$. However, it seems much less clear to me that the Thom diagonal in general exhibits $Mf$ as an $E_\infty$-comodule over $R[X]$. This would follow from showing that the above diagonal map exhibits $(X,f)$ as a co-$E_\infty$-comodule in $Top_{BGL_1(R)}$. Intuitively, I believe this to be true because all of the necessary coherences exist on the diagonal map, and on the morphism $f$ we're simply crossing with the trivial map. Note that the above is not the trivial coaction, which would be given by $X\simeq X\times\ast\overset{id_X\times\ast}\to X\times X$. For me, being a co-$E_\infty$ comodule simply means being an $E_\infty$-module over the $E_\infty$-algebra $R[X]$ in the opposite of the category of interest. Note also that in the category $Top_{BGL_1(R)}$ the symmetric monoidal structure is given by the infinite loop space structure of $BGL_1(R)$. I have been playing with this for some time but have not been able to prove it with the tools I have. REPLY [4 votes]: The answer to this is yes, it is co-$E_\infty$. Let $\iota\colon BGL_1(R)\to Mod_R$ be the inclusion. Since colimit is left adjoint to the strong monoidal diagonal functor, it's oplax monoidal. Note that the constant functor $\kappa_R\colon X\to Mod_R$ is the monoidal unit for the pointwise monoidal structure in $Mod_R^X$ (hence a coalgebra over which every other functor is a comodule). So there is an equivalence $\iota\circ f\xrightarrow{\sim} (\iota\circ f)\otimes_{pw} \kappa_R$ in $Mod_R^X$, where $\otimes_{pw}$ is the pointwise tensor product. Taking the colimit of this, and using that colimit is oplax monoidal gives us a coaction of $colim(\kappa_R)\simeq R[X]$ on $colim(\iota\circ f)\simeq Mf$. This is all as monoidal (e.g. $\mathbb{E}_k$-monoidal) as its various moving parts allow it to be. The only thing one needs to check is that the resulting morphism $Mf\to Mf\otimes_R R[X]\simeq Mf\otimes X$ is the "Thom diagonal" in the literature. This follows from Theorem 4.15 in my preprint.<|endoftext|> TITLE: Defining functions pointwise vs. almost everywhere (w.r.t. uncountably many mutually singular measures) QUESTION [9 upvotes]: My question is motivated by a general measure-theoretic problem that one frequently encounters in probability: the need to work with uncountably many mutually singular measures at once, and with functions that are initially only defined almost everywhere, not pointwise. Whenever such a situation arises, it seems like a common practice to resolve the problem in ad hoc ways, imposing requirements like continuity or quasi-continuity to pin down the "good enough" pointwise versions of the relevant functions. Broadly put, my question is whether there is a chance for a purely measure-theoretic (i.e. non-topological) way to construct a pointwise version of a function from its almost-everywhere "shadows" w.r.t. different measures. Let $(X, \mathcal B(X))$ be a standard Borel space, and let $M$ be the Banach lattice of (signed, finite) Borel measures on it. We equip $M$ with the standard $\sigma$-algebra $\mathcal B(M)$ generated by the evaluation maps $\mu \mapsto \mu[A], A \in \mathcal B (X)$. Let $I \subset M$ be a norm-closed ideal (equivalently: a subspace that, together with every measure $\mu \in I$ contains all measures that are $\mu$-absolutely continuous). Assume that, furthermore, $I$ is a $\mathcal{B}(M)$-measurable set (note that $\mathcal{B}(M)$ is much smaller than the Borel $\sigma$-algebra of the norm topology, so this doesn't follow from being norm-closed). $I$ is closed under taking barycenters. That is, if $\mu : [0,1] \to I$ is a measurable map and $\intop_0^1 \Vert \mu_t \Vert dt < \infty$ then $\intop_0^1 \mu_t dt \in I$. Now assume that with every measure $\mu \in I$ we associate a $\mu$-equivalence class of a function: $f_\mu \in L^0(|\mu|)$. Call the family $(f_\mu, \mu \in I)$ consistent if: If $\nu \ll \mu$ then $f_\nu = f_\mu$ $\nu$-almost everywhere. For any measurably parametrized family $\mu : [0,1] \to I$ as above, the equivalence classes of functions $f_{\mu_t}$ are related to $f_\nu, \nu := \intop_0^1 \mu_t dt$ in the natural way: there is a Borel function $\tilde f : X \to \mathbb{R}$, such that $\tilde f = f_{\nu}$ $\nu$-almost everywhere and $\tilde f = f_{\mu_t}$ $\mu_t$-almost everywhere for Lebesgue-almost all $t$. The latter condition allows to recover $f_{\mu_t}$ for Lebesgue-almost all $t$ from $f_{\intop |\mu_t| dt}$ and vice versa. Note that nothing really depends on the choice of the version $\tilde f$. The point is merely its existence, which imposes a strong compatibility requirement on $f_{\mu_t}$, even when $\mu_t$ are singular to each other. Question: Obviously, if there is a universally measurable function $F : X \to \mathbb{R}$ that is a version of $f_\mu$ for all $\mu \in I$ simultaneously then the family $(f_\mu)$ is consistent. Is the converse true? If not, should I replace $I \in \mathcal{B}(M)$ by some better regularity condition in order for the answer to be positive? Actually, I don't know the answer even for the ideal of all atomless measures. REPLY [2 votes]: While I can't answer the bulk of your question, I can point you toward a "purely measure-theoretic" construction, as you requested in the first paragraph. A paper of S. Cohen (http://arxiv.org/abs/1110.2592) identifies a condition (which he calls the "Hahn property", Definition 10) on the family of measures which permits such a construction (Theorem 4, which requires an unfortunate amount of notational back-tracking). In the financial math community, this is known as the "aggregation problem".<|endoftext|> TITLE: What is known about the boundary between Richardson's theorem and the Tarski-Seidenberg theorem? QUESTION [6 upvotes]: Tarski proved that equalities and inequalities in can be decided over $\mathbb{R}[x].$ Richardson proved that adding composition with the sine and exponential functions caused the problem to become undecidable. What are the best results sharpening this gap? Laczkovich (improving Wang) was able to show undecidability for the ring generated by the integers, $x$, $\sin x^n,$ and $\sin(x\sin x^n)$ which is the best result I know of in this direction. I don't know of any results on the decidable side. In particular, is it known whether inequalities over the ring $\mathbb{Z}[x,\sin x]$ are decidable? REPLY [5 votes]: This seems unlikely to have a simple or known decision procedure. Consider the question: For $x>1$, is $x^4$ always greater than $1 + x^4 \sin(x)$? This is a question of Diophantine approximation. It comes close for $x$=2 or 8 or 33 or 573204. The last one is half a numerator in a continued fraction approximant of $\pi$, and the evaluations of the two terms start with the same 13 digits. So the concrete question might be nice for people to ponder, and I doubt we'll get an algorithm to solve the question in the original post. A proof of undecidability seems more likely, and difficult.<|endoftext|> TITLE: Proving a certain $ C^{*} $-algebraic inequality QUESTION [8 upvotes]: Let $ A $ be a non-unital $ C^{*} $-algebra. Is there an ‘elementary’ way to prove, for all $ (a,\lambda) \in A \times \mathbb{C} $, the inequality $$ |\lambda| \leq \sup_{b \in A, ~ \| b \| \leq 1} \| a b + \lambda \cdot b \|_{A}? $$ I have a proof of this, but it is simply overkill. Proof Firstly, define a linear functional $ \phi $ on the unitization $ A^{\sim} $ of $ A $ by $$ \forall (a,\lambda) \in A \times \mathbb{C}: \quad \phi(a,\lambda) \stackrel{\text{df}}{=} \lambda. $$ As all positive elements of $ A^{\sim} $ must have a non-negative scalar component, it follows that $ \phi $ is a positive linear functional, in which case, by a well-known result about positive linear functionals on $ C^{*} $-algebras, $ \phi $ is automatically continuous and $ \| \phi \| = \phi(0,1) = 1 $. Hence, for all $ (a,\lambda) \in A \times \mathbb{C} $, we have $$ |\lambda| = |\phi(a,\lambda)| \leq \| (a,\lambda) \|_{A^{\sim}} \stackrel{\text{df}}{=} \sup_{b \in A, ~ \| b \| \leq 1} \| a b + \lambda \cdot b \|_{A}. \quad \blacksquare $$ I believe that one can avoid theorems about positive linear functionals on $ C^{*} $-algebras and invent a proof that is mostly Banach $ * $-algebraic in nature, with the finishing blow provided by the $ C^{*} $-identity. However, I do not see the light. I hope that my request is not too vague. Thanks! REPLY [4 votes]: The following argument seems easier, but there might be a still more fundamental one. Notice that $ \phi: A^{\sim} \to \mathbb{C} $ above is also a $ C^{*} $-algebraic homomorphism. As $ C^{*} $-algebraic homomorphisms are automatically contractive (which is a consequence of a not-too-difficult spectrum argument), we have $$ \forall (a,\lambda) \in A \times \mathbb{C}: \quad |\lambda| = |\phi(a,\lambda)| \leq \| (a,\lambda) \|_{A^{\sim}} = \sup_{b \in A, ~ \| b \|_{A} \leq 1} \| a b + \lambda \cdot b \|_{A}. $$ No hard facts about positive linear functionals on $ C^{*} $-algebras were used. Additional Information: The inequality is valid if we only assume that $ A $ is a Banach algebra that satisfies the following conditions: $ A $ has no two-sided identity. $ A $ has a right approximate identity (r.a.i.) norm-bounded by $ 1 $, which we denote by the net $ (e_{i})_{i \in I} $. Let $ A $ be a Banach algebra satisfying Conditions (1) and (2). Let $ \mathbb{L}(A) $ denote the Banach algebra of all bounded homomorphisms from $ A $ to itself, where multiplication is defined by composition and the norm is simply the operator norm. The only leap (not too great, I hope!) of imagination required is to notice that for $ (a,\lambda) \in A \times \mathbb{C} $, $$ \sup_{b \in A, ~ \| b \|_{A} \leq 1} \| a b + \lambda \cdot b \|_{A} $$ represents the operator norm of $ L_{a} + \lambda \cdot \text{id}_{A} \in \mathbb{L}(A) $, where $ L_{a} $ denotes left-multiplication by $ a $. Define $ A^{\sim} \stackrel{\text{df}}{=} \{ L_{a} + \lambda \cdot \text{Id}_{A} \mid (a,\lambda) \in A \times \mathbb{C} \} $, $ L_{A} \stackrel{\text{df}}{=} \{ L_{a} \mid a \in A \} $. Then clearly $ A^{\sim} $ is a sub-algebra of $ \mathbb{L}(A) $. Claim 1: $ L_{A} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $. Proof of Claim 1 Let $ a \in A $. We already know that $ \| L_{a} \| \leq \| a \|_{A} $. However, $ \| e_{i} \|_{A} \leq 1 $ for all $ i \in I $ and $$ \lim_{i \in I} \| {L_{a}}(e_{i}) \|_{A} = \lim_{i \in I} \| a e_{i} \|_{A} = \| a \|_{A}, $$ so we actually have $ \| L_{a} \| = \| a \|_{A} $. The map $ x \mapsto L_{x} $ is therefore a bijective isometry from $ A $ to $ L_{A} $, so $ L_{A} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $. $ \quad \blacksquare $ Define a (not a priori continuous) linear functional $ \phi $ on $ A^{\sim} $ by $$ \phi(L_{a} + \lambda \cdot \text{Id}_{A}) \stackrel{\text{df}}{=} \lambda. $$ To prove that $ \phi $ is well-defined, we must show that given $ (a,\lambda) \in A \times \mathbb{C} $, if $ L_{a} + \lambda \cdot \text{Id}_{A} = 0_{\mathbb{L}(A)} $, then it must follow that $ \lambda = 0 $. Assume the contrary. If we define $ e \stackrel{\text{df}}{=} - \dfrac{1}{\lambda} \cdot a $, then $ L_{e} = \text{Id}_{A} $, which means that $ e $ is a left identity of $ A $. Now, for all $ x \in A $, we have \begin{align} x e - x & = \lim_{i \in I} ~ (x e - x) e_{i} \qquad (\text{As $ (e_{i})_{i \in I} $ is an r.a.i..}) \\ & = \lim_{i \in I} ~ (x e e_{i} - x e_{i}) \\ & = \lim_{i \in I} ~ (x e_{i} - x e_{i}) \qquad (\text{As $ e $ is a left identity.}) \\ & = \lim_{i \in I} ~ 0_{A} \\ & = 0_{A}. \end{align} Hence, $ e $ is a right identity of $ A $ as well. This contradicts our assumption that $ A $ has no two-sided identity, so we indeed have $ \lambda = 0 $. Claim 2: $ A^{\sim} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $. Proof of Claim 2 As $ L_{A} $ is a complete linear subspace of $ A^{\sim} $, it is automatically closed. The quotient vector space $ A^{\sim} / L_{A} $ can thus be given a norm, which then has to be complete because $ \ker(\phi) = L_{A} $ and so $$ A^{\sim} / L_{A} = A^{\sim} / \ker(\phi) \cong \mathbb{C}. $$ Exploiting the result that a normed vector space is complete if its quotient by a complete linear subspace is complete, we conclude that $ A^{\sim} $ is complete w.r.t. the operator norm on $ \mathbb{L}(A) $. $ \quad \blacksquare $ We now see that $ A^{\sim} $ is a unital Banach algebra w.r.t. the operator norm on $ \mathbb{L}(A) $. As $ \phi $ is a multiplicative linear functional on $ A^{\sim} $, it follows that $ \phi $ must be bounded with norm $ \leq 1 $ (this is an easy fact whose proof can be found in Rudin’s Real and Complex Analysis). The inequality is therefore established. Notes: If Condition (1) is violated, i.e., $ A $ has a two-sided identity $ e $, then the inequality is false because then $$ 1 \nleq \sup_{b \in A, ~ \| b \|_{A} \leq 1} \| e b - 1 \cdot b \|_{A} = 0. $$ I do not know what happens if Condition (2) is violated, however. If $ A $ is a $ C^{*} $-algebra, then we only need Condition (1) because Condition (2) automatically holds. This can be misleading, however, because it is not at all obvious that non-unital $ C^{*} $-algebras should have an r.a.i. norm-bounded by $ 1 $. Fortunately, it turns out that in proving Claims 1 and 2, the $ C^{*} $-identity is all that is needed as it takes over the pivotal role played by Condition (2).<|endoftext|> TITLE: Centralizers in the universal central extensions of the alternating groups? QUESTION [5 upvotes]: For $n \ge 8$ the Schur multiplier $H_2(BA_n, \mathbb{Z})$ (where $A_n$ denotes the alternating group) stabilizes to $\mathbb{Z}_2$, and hence there is a universal central extension $\widetilde{A}_n$ of $A_n$ by $\mathbb{Z}_2$. Question 1: Do any interesting groups (e.g. central extensions of sporadic simple groups) occur as centralizers (of elements) in $\widetilde{A}_n$? Question 2: What are their Schur multipliers? Motivation (feel free to ignore): Write $X = B \widetilde{A}_n$. In trying to find a relatively concrete answer to this question, I was led to a map $$H_3(X, \mathbb{Z}) \to \pi_3(\mathbb{S}) \cong \mathbb{Z}_{24}$$ which I believe is an isomorphism for sufficiently large $n$. In any case, this map, regarded as an element of $H^3(X, \mathbb{Z}_{24})$, transgresses to a class in $$H^2(LX, \mathbb{Z}_{24})$$ where for a group $G$, $LBG$ is the (classifying space of the) adjoint quotient $G/G$, the groupoid whose objects are elements of $g$ and whose morphisms come from conjugation. In particular, this transgressed class restricts to a distinguished family of classes in $$H^2(C(g), \mathbb{Z}_{24})$$ where $C(g)$ denotes the centralizer and $g \in \widetilde{A}_n$. This gives a distinguished family of central extensions of the groups $C(g)$ by $\mathbb{Z}_{24}$, and it seems interesting to ask what these central extensions are. In particular (feel extremely free to ignore) the Schur multipliers of the sporadic simple groups are all subgroups of $\mathbb{Z}_{24}$, and it would be nice if this construction in some sense explained that.... REPLY [2 votes]: The answer to Q1 is no, all these centralisers are boring, and look much the same as these in $A_n$ itself. Indeed, think what happens to them under the homomorphism squashing the central $\mathbb{Z}_2$.<|endoftext|> TITLE: Homologically distinct infinite loop structures on a space QUESTION [7 upvotes]: Let $X$ be a connected pointed topological space equipped with two different actions of $E_\infty$-operad. Each action provides a collection of deloopings $X_i$, where $X_0 = X$ and $\Omega X_i$ is homotopy equivalent to $X_{i-1}$, so there are two $\Omega$-spectra having $X$ as a zeroth space. Can these spectra have different cohomology or different cohomology operations? I would like an explicit example or just a better reason than "why not". I don't have any applications for such spectra. This question is motivated by thinking about D. Rector's paper Loop Structures on the Homotopy Type of $S^3$, which produces uncountably many different deloopings of three-dimensional sphere. They can be distinguished using cohomology operations. REPLY [12 votes]: There are easier ways to distinguish connective spectra with the same underlying space, if that's all you want to do. Like spaces, spectra have a theory of Postnikov towers, and in the same way that the Postnikov tower of a (simply connected, say) space $X$ measures the extent to which it differs from the product $$\prod_n B^n \pi_n(X)$$ of Eilenberg-MacLane spaces, the Postnikov tower of a connective spectrum $E$ measures the extent to which it differs from the product $$\prod_n B^n \pi_n(E)$$ of Eilenberg-MacLane spectra. Suppose $E$ is a spectrum with exactly two nontrivial homotopy groups $\pi_n(E), \pi_m(E), n < m$. Then $E$ naturally fits into a fiber sequence of spectra $$B^m \pi_m(E) \to E \to B^n \pi_n(E)$$ and such fiber sequences are classified by a $k$-invariant given by a homotopy class of maps $B^n \pi_n(E) \to B^{m+1} \pi_m(E)$ of spectra, or equivalently by a stable cohomology operation $$H^{\bullet + n}(-, \pi_n(E)) \to H^{\bullet+m+1}(-, \pi_m(E)).$$ If $n \ge 1$, then taking underlying spaces produces from the above fiber sequence of spectra a fiber sequence of connected spaces $$B^m \pi_m(E) \to E \to B^n \pi_n(E)$$ which is a principal $B^m \pi_m(E)$-bundle, and in particular which is classified by a $k$-invariant given by a homotopy class of maps $B^n \pi_n(E) \to B^{m+1} \pi_m(E)$ of spaces, or equivalently by an unstable cohomology operation $$H^n(-, \pi_n(E)) \to H^{m+1}(-, \pi_m(E)).$$ In particular, we can find nontrivial infinite loop space structures on the space $B^n \pi_n(E) \times B^m \pi_m(E)$ by finding stable cohomology operations whose corresponding unstable cohomology operation in some degree is trivial. Example. The simplest nontrivial case where the underlying space is connected occurs when we pick the $k$-invariant to be the second Steenrod square $$\text{Sq}^2 : B \mathbb{Z}_2 \to B^3 \mathbb{Z}_2$$ (taking $\text{Sq}^1$ just gives $\mathbb{Z}_4$ as a nontrivial extension of $\mathbb{Z}_2$ by $\mathbb{Z}_2$). This gives the unique nontrivial infinite loop space structure on the space $B \mathbb{Z}_2 \times B^2 \mathbb{Z}_2$, which arises naturally as the $2$-truncation of the connected component of the identity of the sphere spectrum; in particular, we know that the underlying space is $B \mathbb{Z}_2 \times B^2 \mathbb{Z}_2$ because $\text{Sq}^2$ as an unstable cohomology operation $$H^1(-, \mathbb{Z}_2) \to H^3(-, \mathbb{Z}_2)$$ vanishes. Equivalently, $\text{Sq}^2$ is nontrivial as a map of spectra but trivial as a map of spaces. The product spectrum structure on $B \mathbb{Z}_2 \times B^2 \mathbb{Z}_2$ gives the cohomology theory which is $$H^1(X, \mathbb{Z}_2) \times H^2(X, \mathbb{Z}_2)$$ in degree $0$, with the product abelian group structure. By comparison, the nontrivial spectrum structure gives the cohomology theory which in degree $0$ has the same underlying set, but with the following modified abelian group structure, where $(w_1, w_2) \in H^1(X, \mathbb{Z}_2) \times H^2(X, \mathbb{Z}_2)$: $$(w_1, w_2) + (w_1', w_2') = (w_1 + w_1', w_2 + w_2' + w_1 \cup w_1').$$ This group is in general a nontrivial extension of $H^1(X, \mathbb{Z}_2)$ by $H^2(X, \mathbb{Z}_2)$; it arises as a kind of super Brauer group of $X$. The multiplication can be thought of as the cup product on the invertible elements $1 + w_1 + w_2 \in H^{\bullet}(X, \mathbb{Z}_2) / H^{\bullet + 3}(X, \mathbb{Z}_2)$ of a truncated version of the cohomology ring over $\mathbb{Z}_2$.<|endoftext|> TITLE: Torsion in profinite groups QUESTION [6 upvotes]: Is there a finitely generated profinite group $G$ with a closed subgroup of infinite index $K \leq G$ such that for every $g \in G$ there exists some $n \in \mathbb{N}$ for which $g^n \in K$ ? Can $G$ be pro-$p$, for some prime number $p$? REPLY [2 votes]: Too long for a comment. I note first that I made an attempt to reduce the problem to the case where $K$ is normal, but it turned out to be false; I'm thankful to Ian Agol for his discussion. The case where $K$ is normal follows at once from a theorem of Zelmanov stating that every periodic torsion group is locally finite. This second attempt is not a complete answer; however, it reduces the problem considerably: Let $G$ be as above, then, virtiually, $L_p(G)$ satisfies a PI, for every prime $p$; with $L_p(G)$ denotes the Lie algebra associated to the dimension subgroups (over the field of $p$ elements). Indeed, let $K_n$ denote the set of elements of $G$ satisfying $x^n\in K$. As $K$ is closed, each $K_n$ is closed. By assumption, $\cup_{n\geq1} K_n=G$, so by the standard Baire category theorem, there exists $n$ such that $K_n$ contains an open subset, so there is an open normal subgroup $N$ and $t\in G$ such that $tN \subseteq K_n$. Let $H=\langle t \rangle N$, then $H$ is open. Consider the subgroup generated by $X$, the set of the elements $(tx)^n$, with $x\in N$. Then $X$ is a normal subset of $M$, and $X \subseteq K$ by the above paragraph. It follows that the closed subgroup $L$ generated by $X$ lies $K$. Let us work now in the finitely generated profinite group $M/L$ ($K$ may be identified with $K/L$. We have $M/L$ satisfies the coset identity $X^n=1$ with respect to $N/L$ (see Wilson and Zelmanov's http://www.sciencedirect.com/science/article/pii/0022404992901386), by the main result in the previous paper, the Lie algebra $L_p(M/L)$ satisfies a polynomial identity. This proves the claim. Remark. If $G$ is a pro-$p$ group, then we can find a finite generating set of $M/L$ in which every element satisfies the identity $X^n=1$ (take $t$ together with $tx_1,..,tx_s$, where $x_1,..,x_s$ generate $N$; or actually thier images in $M/L$). I wished to deduce from this (using the remark by Professor Yiftach Barnea in his answer Elements of infinite order in a profinite group) that $M/L$ is finite, from which it follows that $K$ has a finite index in $G$. Unfortunately, it seems that this remark is incorrect.<|endoftext|> TITLE: Can any finite distributive weighted lattice be realized by inclusion of groups? QUESTION [5 upvotes]: By theorem 2.1 here, any finite distributive lattice $\mathcal{L}$ can be realized as an intermediate subgroups lattice. A weighted lattice $(\mathcal{L},\tau)$ is a lattice $\mathcal{L}$ with a weight $\tau: \mathcal{L} \to \mathbb{N}$ satisfying; - $b \le b' \Rightarrow \frac{\tau(b')}{\tau(b)}\in \mathbb{N}$ - $\tau(l) = 1$ for $l$ the least element of $\mathcal{L}$. Let $(H \subset G)$ be an inclusion of finite groups, then it realizes $(\mathcal{L}(H \subset G),\tau)$, with $\tau(K) = [K:H]$ and $\mathcal{L}(H \subset G)$ the lattice of intermediate subgroups $H \subseteq K \subseteq G $. Question: Can any finite distributive weighted lattice be realized by inclusion of groups? Remark: It's obviously true for the lattice with two elements and any weight thanks to the inclusion $(S_{n-1} \subset S_{n})$. I don't know if it's true for the lattice with three elements, weighted $(mn,n,1)$ or even just $(n^{2},n,1)$, but I have checked by GAP that it's true for $mn < 32$. REPLY [4 votes]: Let me hastily summarise my comments above: a subgroup inclusion chain of length 3 corresponds precisely to an imprimitive permutation group on a set of size $mn$ with a unique system of imprimitivity (and we require that the blocks in this system have size $m$). The corresponding lattice will then be $(mn,n,1)$. Such a group is given by $Sym(m) \wr Sym(n)$, although for particular $m$ and $n$ there are no doubt many others. This answers the specific question given by the OP. As for generalizations: one naturally wonders how to deal with $(\ell mn, mn, n, 1)$. I think for this you can use $(Sym(\ell) \wr Sym(m)) \wr Sym(n)$, with similar iterations working for longer chains. Indeed one can probably generalize this construction even more so that given any intermediate subgroup lattice, you can stick a chain "on the top of it" by taking wreath products. This at least reduces the general question somewhat.<|endoftext|> TITLE: A result attributed to Whitney QUESTION [10 upvotes]: One of the basic results of real analysis says that any closed subset of a smooth ($C^\infty$) manifold $M$ is the set of zeros of some map $\lambda\in C^\infty(M;[0,1])$. This result (or some equivalent variations of it) is usually attributed to Hassler Whitney. Question: exactly where (title of Whitney's paper and page) is this stated? REPLY [2 votes]: According to J.-C. Tougeron, page 73 (Théorème 2.2) of Idéaux de fonctions différentiables, Ergebnisse, Band 71, Whitney's paper is Analytic extensions of differentiable functions defined in closed sets. Trans. Amer. Math. Soc. 36 (1934), pp 63-89. Actually, Whitney's Theorem is stronger in two ways. On the one hand, every $C^m$-function $F:K\rightarrow{\mathbb R}$ ($K$ the closed set) can be matched at order $m$ by a $C^m$-function $W(F):M\rightarrow{\mathbb R}$ which is $C^\infty$ away from $K$. On the other hand, the operator $W$ can be chosen linear and continuous (for the $C^m$-norms).<|endoftext|> TITLE: Relation between Different Definitions of Induced Representation QUESTION [5 upvotes]: I've seen two different ways to define induced representation. One is as in the book Introduction to representation theory: If $G$ is a group, $H$ is a subgroup of it, and $V$ is a representation of $H$, then the induced representation $Ind^G_H V$ is the representation of $G$ with $$Ind^G_H V=\{f:G\longrightarrow V|f(hx)=\rho_V(h)f(x)\forall x\in G, h\in H\}$$ and the action $g(f)(x)=f(xg)$, $\forall g\in G$. If we choose a representative $x_{\sigma}$ from each right $H$-coset $\sigma$ of $G$, then any $f\in Ind^G_H V$ is uniquely determined by $\{f(x_\sigma)\}_{\sigma}$. I've also seen another way to define induced representation. Let $G$ be a group, $H$ a subgroup of it, and $V$ a representation of $H$. The underlying vector space of $Ind^G_H V$ is the direct sum $$\bigoplus_{\tau\in G/H}\tau V$$ with $\tau$ going over all the left cosets in $G/H$. It multiplication operation is defined by choosing a set $\{g_\tau\}_{\tau\in G/H}$ of coset representatives and setting $$g(\tau v)=\beta (hv)$$ where $\tau v$ is an element in $\tau V$, $\beta$ is the unique left coset containing $gg_{\tau}$, and $gg_{\tau}=g_{\beta}h$ for some $h\in H$. It's easy to verify this definition of $Ind|^G_HV$ does not depend on the choice of the representatives $\{g_\tau\}_{\tau\in G/H}$. Induced representation is the left adjoint to the restricted representation. So any definition of it should be unique up to unique isomorphism. But I cannot see why the two definitions above are equivalent. Can anybody show me where the problem is? Thanks. REPLY [4 votes]: These two versions of the induced representation are not the same in general. You get isomorphic objects only if you add finiteness conditions. Indeed your second definition corresponds to the subspace of functions supported on a finite number of $H$-cosets. The two definition agree if e.g. $H$ is of finite index in $G$, or if you add topological conditions. For instance if $G$ is totally disconnected and $H$ is open, then you have the notion of compactly induced representation ${\rm c-Ind}_H^G \, V$ which agrees with your second definition. This compactly induced representation is the subspace of ${\rm Ind}_H^G \, V$ formed of those function which have compact support modulo $H$. Note that in that case ${\rm Ind}_H^G$ is right-adjoint to restriction and ${\rm c-Ind}_H^G $ is left-adjoint ...<|endoftext|> TITLE: Generating primes with floor of a polynomial $[p(n)]$ QUESTION [6 upvotes]: Is there a polynomial $p(x)$ with real coefitients and degree at least one that $[p(n)]$ for everey natural number like $n$ be a prime? If yes, what is such a polynomial $p(x)$ and if no, how to prove.(I have asked this problem in math.stackexchange.com in this question but has'nt given any solution to the problem yet.) REPLY [5 votes]: With Vesselin's idea in the comments proof is ready as below: If $p(x)-p(0) \in \mathbb{Q}[x]$ then the problem isn't so hard. If $p(x)-p(0) \not \in \mathbb{Q}[x]$ then there is an irrational coefficient for a term of degree bigger than or equal one. There is a problem in ergodic theory that says that the sequence $p(n) \text{ mod 1}$ is equidistributed in $[0,1)$. It can prove similarly that the sequence $p(n) \text{ mod 2}$ is equidistributed in $[0,2)$, and with this observation we find that there are a lot of $n \in \mathbb{N}$ that for them $[p(n)]$ is even and thus isn't prime. Also with the equidistributivity of $p(n) \text{ mod m}$ in $[0,m)$ for all $m \in \mathbb{N}$ we find that natural density of $$\{n\ |\ m\ |[p(n)]\}$$ is $\frac{1}{m}$.<|endoftext|> TITLE: Origins and Industrial Applications of stochastic processes (eg. Brownian motion) on Riemannian manifolds QUESTION [5 upvotes]: I am studying BM on Riemannian manifolds and I am curious how this theory started. In the references below (esp. in Hsu's exposition), you will find many applications of that theory such as a probabilistic proof of the Atiyah-Singer index theorem. I am also curious about the industrial applications (if any yet) given that Riemannian geometry is used in Machine learning and statistics. Thanks anyhow In case you are interested [1]https://www.math.kyoto-u.ac.jp/probability/sympo/PSS03abstract.pdf [2]http://www.math.northwestern.edu/~ehsu/Brownian%20Motion%20and%20Riemannian%20Geometry.pdf (proof of Atiyah Singer index thm) REPLY [4 votes]: The earliest "industrial" application I know is in the context of microwave engineering: the eigenvalues of the transmission matrix through a waveguide with random scatterers perform a Brownian motion in hyperbolic space as a function of the length of the waveguide. Waveguides with Random Inhomogeneities and Brownian Motion In the Lobachevsky Plane (1959). REPLY [3 votes]: A somewhat recent engineering application is in study of dynamics and control on manifolds such as SO(3), which is the manifold on which the (position) state of a rotating rigid body (such as a satellite) evolves. For inference/filtering problems (i.e. trying to estimate the position on SO(3) given some noisy measurements), it is required to perform uncertainty propagation on these manifolds. Here's a reference: http://arxiv.org/abs/0803.1515<|endoftext|> TITLE: Is there a simple combinatorial characterization for when a direct limit of ultrapowers of $V$ is well-founded? QUESTION [7 upvotes]: I want to know if there are fairly simple combinatorial necessary conditions for when a direct limit of ultrapowers of $V$ is well-founded similar to $\sigma$-completeness. By combinatorial, I mean that these conditions are conditions on the ultrafilters instead of the elementary embeddings they produce. For simplicity, we shall formulate the question with regards to the following ultrapower construction. Assume that $X$ is a set and $(P_{n})_{n}$ is a sequence of partitions of $X$ such that $P_{n+1}$ refines $P_{n}$ for all $n$ and whenever $x,y\in X,x\neq y$ there is some $n$ where $x,y$ belong to different blocks of the partition $P_{n}$. Furthermore, assume that whenever $R_{n}\in P_{n}$ and $R_{n+1}\subseteq R_{n}$ for all $n$ then $\bigcap_{n}R_{n}$ is non-empty. Let $B=\bigcup_{n\in\omega}\{\bigcup\mathcal{R}|\mathcal{R}\subseteq P_{n}\}$. Then $B$ is a Boolean algebra. Let $U$ be an ultrafilter on $B$. We shall use the ultrafilter $U$ to construct an ultrapower. Suppose $\mathcal{A}$ is a structure. Then let $\mathcal{A}^{(P_{n})_{n}}$ be the collection of all functions $f:X\rightarrow\mathcal{A}$ where there is some natural number $n$ where $P_{n}\preceq\{f^{-1}[\{a\}]|a\in\mathcal{A}\}$ (the ordering $\preceq$ is refinement of partitions). Now let $\simeq_{U}$ be the equivalence relation on $\mathcal{A}^{(P_{n})_{n}}$ where $f\simeq_{U}g$ if and only if $\{x\in X|f(x)=g(x)\}\in U$. One can define the fundamental operations, constants, and relations on $\mathcal{A}^{(P_{n})_{n}}/\simeq_{U}$ as one does with the classical ultrapower construction. Let the structure $\mathcal{A}^{(P_{n})_{n}}/\simeq_{U}$ be denoted by $\mathcal{A}^{(P_{n})_{n}}/U$. Is there a fairly simple combinatorial characterization of when the ultrapower $V^{(P_{n})_{n}}/U$ well-founded? For instance, in order for $V^{(P_{n})_{n}}/U$ to be well-founded, the ultrafilters $\{\mathcal{R}\subseteq P_{n}|\bigcup\mathcal{R}\in U\}$ must be $\sigma$-complete. On the other hand, if whenever $R_{n}\in\mathcal{U}$ for all $n$, we have $\bigcap_{n}R_{n}\neq\emptyset$, then the ultrapower $V^{(P_{n})_{n}}/U$ is well-founded. Even though I just listed necessary conditions and sufficient conditions for when an ultrapower is well-founded, I do not know of any conditions that are both necessary and sufficient. Can someone give a reference or a proof of combinatorial necessary and sufficient conditions for when such an ultrapower is well-founded? If $V^{(P_{n})_{n}}/U$ is well-founded, then do we necessarily have $\bigcap_{n}R_{n}\neq\emptyset$ whenever $R_{n}\in U$ for all $n$? $\textrm{This last question was answered affirmatively by Joel David Hamkins}$ $\textbf{Remark}$ Although I phrased this question in terms of a direct limit of a countable sequence of ultrapowers, this characterization also holds for any kind of direct limit of ultrapowers since any possible infinite descending sequence in an ultrapower of a well-founded set must be in a countable direct limit of ultrapowers. Furthermore, the condition that $\bigcap_{n}R_{n}\neq\emptyset$ whenever $R_{n}\in P_{n},R_{n+1}\subseteq R_{n}$ for all $ n$ does not restrict the types of ultrapowers that one can form since one can always extend the set $X$ to a larger set $\hat{X}$ so that $\bigcap_{n}R_{n}\neq\emptyset$ whenever $R_{n}\in P_{n},R_{n+1}\subseteq R_{n}$. For instance, give $X$ the uniformity where the uniform covers are generated by the uniform partitions $P_{n}$. Then the appropriate set $\hat{X}$ is the completion of the uniform space $X$. Equivalently, one can let $\hat{X}=\varprojlim_{n}P_{n}.$ By replacing $X$ with $\hat{X}$ one obtains an isomorphic ultrapower but where $\bigcap_{n}R_{n}\neq\emptyset$ whenever $R_{n}\in P_{n},R_{n+1}\subseteq R_{n}$ for all $n$. REPLY [7 votes]: The answer is yes. Theorem. The direct limit ultrapower you describe is well-founded if and only if $\bigcap_n A_n\neq \emptyset$ whenever $A_n\in U$ for all $n$. Proof. You've already noted the converse direction, since any instance of ill-foundedness amounts to $[f_{n+1}]\in_U [f_n]$, which gives measure one sets $A_n=\{x\mid f_{n+1}(x)\in f_n(x)\}\in U$, with empty intersection. For the forward implication, suppose that we have $A_n\in U$ with $\bigcap_n A_n=\emptyset$. Each $A_n$ lives on some level of the sequence of partitions, but let us assume for convenience (by padding if necessary) that $A_n$ respects the partition $P_n$. We may also assume that $A_{n+1}\subset A_n$. There is a tree structure $T$ in this situation, whose nodes are the elements of any of the partitions $P_n$ that are in $A_n$, ordered by inclusion. That is, if $b\in P_n$ and $b\subset A_n$, then we place $b\in T$, and we say that $b$ is a parent to any of the corresponding $c\in P_{n+1}$ with $c\subset b$ and $c\subset A_{n+1}$, if there are any such $c$. The key insight to make is that this tree is well-founded. Any infinite descending sequence of nodes in the tree amounts to a descending sequence $R_n\in P_n$ of partition elements, and you had stated that $\bigcap_n R_n\neq\emptyset$ in any such case. But any element of that intersection would be in $\bigcap_n A_n$, since $R_n\subset A_n$. So there can be no such infinite descending paths through the tree, and so the tree is well-founded. Because of this, we may define an ordinal ranking function on the tree. Namely, the rank of any $b\in T$ is the supremum of the $\text{rank}(c)+1$ for all $c\in T$ below $b$. This is well-defined by recursion on the tree. Note that the minimal nodes in $T$ get rank $0$, and the rank of any child node is strictly smaller than the rank of its parent. Consider the ranking function of the nodes of $T$ that happen to be in $P_n$. Let $f_n(x)=\text{rank}(b)$ whenever $x\in b\in T\cap P_n$ be the corresponding rank function. This is an ordinal-valued function with support in $P_n$. But since the rank of a child node is strictly less than the rank of a parent, we see that $f_{n+1}(x) TITLE: Etale spaces using Kan extensions QUESTION [16 upvotes]: Mac Lane - Moerdijk's "Sheaves" gives this cryptic hint in page 91 that the equivalence between etale spaces and sheaves on a space $X$ can be cooked up using formal methods. More precisely, we are in the following nerve-realization situation: $$\begin{matrix} \mathcal{O}(X)\xrightarrow{A}&\mathbf{Top}/X\\ \downarrow^y&\\ \mathbf{Set}^{\mathcal{O}(X)^{op}} \end{matrix}$$ where the left Kan extension $\text{Lan}_yA$ has a right adjoint $N_A\colon \mathbf{Top}/X\to \mathbf{Set}^{\mathcal O(X)^\text{op}}$, which is defined precisely taking the (pre)sheaf of sections of $(p\colon E\to X)\in \mathbf{Top}/X$: $$N_A(p)\colon U\mapsto \mathbf{Top}/X\left(AU, p \right) = \mathbf{Top}/X\left(\left[\begin{smallmatrix} U \\ \downarrow \\ X\end{smallmatrix}\right], \left[\begin{smallmatrix} E \\ \downarrow \\ X\end{smallmatrix}\right] \right) = \{s\colon U\to E\mid ps\colon U\subseteq X\}$$ I am trying to work out the details of this construction, in particular I would like to "Prove formally" by the Kan formula for $\text{Lan}_yA(F)$ that it is precisely the etale space of the sheaf $F$; "Prove formally" that this adjunction restricts to an equivalence $\mathbf{Sh}(X)\cong \mathbf{Et}(X)$ (this can be done appealing Lemma 4 right before page 91) A nice consequence of adjoint nonsense would be that the reflection obtained in this way is also exact ($A$ commutes with finite limits, which exist in $\mathcal O(X)$). I'm stuck in trying to make $\text{Lan}_yA(F)$ an explicit object; one can appeal the Kan formula to obtain $$ \text{Lan}_yA(F)\cong\int^{U\colon \mathcal O(X)} FU\otimes AU $$ where $\otimes$ denotes the canonical $\bf Set$-tensoring of ${\bf Top}/X$ which acts like $S\otimes \left[\begin{smallmatrix} E \\ \downarrow \\ X\end{smallmatrix}\right] = \left[\begin{smallmatrix} \coprod_SE \\ \downarrow \\ X\end{smallmatrix}\right]$. The shape of colimits in $\mathbf{Top}/X$ gives that this space consists of $\left[\begin{smallmatrix} \big(\coprod_U FU\times U\big)/\simeq \\ \downarrow \\ X\end{smallmatrix}\right]$ where I am modding out by a suitable equivalence relation. It would be nice to deduce that $\big(\coprod_U FU\times U\big)/\simeq = \coprod_{x\in X}F_x$, with the topology... Well, I'm beginning to suspect this is a too-painful alternative to the old explicit method. This is why I'm asking you if this can really be done. REPLY [9 votes]: I actually did exactly this when I taught a course on topos theory in Bonn. See my lecture notes: Lecture 3 and Lecture 4 here: http://www.math.ubc.ca/~carchedi/topos.html I give a very detailed proof of 1. and 2. in these notes, using precisely the left Kan extension you describe as the definition, and then explicitly computing it, and arriving at precisely the classical definition of etale-space construction. The course was for masters students, so everything is spelled out, but it could be slightly condensed. Now that I have some extra time, let me give you some more details: You asked how to deduce formally that the etale space of a sheaf $F$, as a set, is the disjoint union of its stalks. For this, make a couple observations: 1.) For each $x \in X,$ the functor $Set^{\mathcal{O}\left(X\right)^{op}}\to Set$ sending $F$ to $F_x$ is colimit preserving (one way of seeing this is that if $i_x:\mathcal{O}\left(X\right)_x \hookrightarrow \mathcal{O}\left(X\right)$ is the inclusion of the poset of open neighborhoods of $x$ into all open subsets of $X$, $F_x$ is simply the colimit of $F \circ i_x.$ 2.) The composite $$Set^{\mathcal{O}\left(X\right)^{op}} \stackrel{A}{\longrightarrow} \mathbf{Top}/X \to Set/X \stackrel{\sim}{\longrightarrow} Set^{X} \stackrel{ev_x}{\longrightarrow} Set$$ is colimit preserving. (Also notice that the composite $Set/X \stackrel{\sim}{\longrightarrow} Set^{X} \stackrel{ev_x}{\longrightarrow} Set$ sends a map $Y \to X$ to its fiber $Y_x$ over $x.$) A simple check shows that both functors agree on representables, that is, the fiber of $U \hookrightarrow X$ over $x$ is exactly the stalk $y(U)_x$ of the representable sheaf. It hence follows that the underlying set of the etale space of $F$ is the disjoint union of its stalks. It also follows that the topology is the final topology with respect to all maps of the form $U \to \coprod F_x$ sending $y \mapsto germ_y \lambda$ for $\lambda \in F\left(U\right)$, and one can show pretty easily that this topology maps the map down to $X$ a local homeomorphism. One can also check that the unit of the adjunction induced by $A$ is an iso precisely on sheaves, and the co-unit is an iso precisely on local homeomorphisms into $X$. You can see http://www.math.ubc.ca/~carchedi/Lecture4.pdf for the glorious details.<|endoftext|> TITLE: Waring problem for binomial coefficients (generalization of Gauss' Eureka Theorem) QUESTION [6 upvotes]: Is there a number k such that every natural number can be written as $\sum_{i=1}^k \binom{a_i}{3}$ for some natural numbers $a_i$'s? REPLY [2 votes]: In more general case this problem was studied by Nechaev, see On the question of representing natural numbers by a~sum of terms of the form $x(x+1)\ldots (x+n-1)/n!$ On the representation of natural numbers as a sum of terms of the form $x(x+1)\ldots (x+n-1)/n!$<|endoftext|> TITLE: Weisinger's thesis QUESTION [8 upvotes]: I am currently reading Atkin and Li's paper on Twists of newforms and Atkin-Lehner pseudo eigenvalues and one of the references there is to Weisinger's thesis: Weisinger J., Some results on classical Eisenstein series and modular forms over function fields. Harvard thesis. (1977) Does anyone have a scan or digital copy of it? Or an alternative reference that treats Atkin-Lehner operators $W_Q$ acting on Eisenstein series? (I know how $W_N$ acts on level $N$ Eisenstein series) REPLY [11 votes]: A scanned copy of the thesis is available here (Wayback Machine).<|endoftext|> TITLE: stationary tower forcing QUESTION [5 upvotes]: It is known that if $\delta$ is a Woodin cardinal and $\kappa < \delta$, then the stationary tower forcing $\mathbb Q^\kappa_{<\delta}$ preserves cardinals up to $\kappa$ and forces $\delta = \kappa^+$. Thus if there is a Woodin cardinal $\delta$ then there is a forcing preserving cardinals up to $\aleph_\omega$ and making $\delta = \aleph_{\omega+1}$. But it is also known that $\mathbb Q^\kappa_{<\delta}$ is not $\delta$-c.c. Question: Is there some large cardinal assumption that implies the existence of a cardinal $\kappa > \aleph_{\omega+1}$ and a $\kappa$-c.c. forcing $\mathbb P$ which preserves $\aleph_n$ for finite $n$ and makes $\kappa = \aleph_{\omega+1}$? I'm no expert on these things, but naively I would suggest two possible approaches: (a) Find a large cardinal $\delta$ that implies the existence of a $\delta$-saturated tower of ideals with similar effects as the stationary tower. (b) Find an inaccessible cardinal $\delta$ with a precipitous tower of ideals of height $\delta$ that preserves the $\aleph_n$'s but actually collapses $\delta$, so that $\delta^+$ is the witness. Update: (b) is ruled out by Mohammad's result here. Note: It is consistent relative to large cardinals that there is some $\kappa$-c.c. forcing collapsing a regular $\kappa$ to be $\aleph_{\omega+1}$ while preserving cardinals below $\aleph_\omega$. Namely an $\aleph_{\omega+2}$-saturated ideal on $\aleph_{\omega+1}$, which can be forced from a huge cardinal. But I want to see if it is outright implied by large cardinals, because then it is much easier to combine with other things. New Idea: Foreman-Magidor-Shelah show in "Martin's Maximum Part I" that if $\mu$ is regular and $\kappa > \mu$ is supercompact, then $\mathrm{Col}(\mu,<\kappa)$ forces that $NS_\mu$ is precipitous. I believe this was improved by Goldring to a Woodin cardinal. So perhaps for large $\kappa$, $\mathrm{Col}(\aleph_{\omega+1},<\kappa) * \dot{\mathcal{P}(\aleph_{\omega+1}) / NS}$ does the trick. If we force below $cof(\omega_n)$ for $n > 0$, then we are sure to collapse $\kappa = \aleph_{\omega+2}$ (by a theorem of Shelah), and the whole forcing is $\kappa$-dense, so $\kappa^+$ should be the witness. But the problem is, what happens below $\aleph_\omega$? Despite being precipitous, could forcing with $NS_{\aleph_{\omega+1}}$ actually make $\kappa$ countable? The proof of precipitousness given in the paper is a bit abstract so I have no idea how the generic ultrapower compares to the generic extension. REPLY [3 votes]: This may not be an answer but it is longer than be a comment: Starting from a supercompact cardinal $\kappa$ an inaccessible $\lambda$ above it, we can construct a model $M$ of ZFC in which $\kappa=\aleph_\omega$ and $\lambda=\aleph_{\omega+1}.$ The model in an intermediate submodel of a supercompact Prikry forcing with suitable collapses. I think using an analysis similar to Foreman-Woodin's paper "GCH can fail everywhere", we can show that this intermdiate submodel $M$ is a $\lambda-c.c.$ extension of the ground model. Now find another intermediate submodel $N \subset M$ of the ground model which is essentially the Prikry extension of the ground which makes $\kappa$ into $\aleph_\omega,$ preserves cardinals above $\kappa$ and it makes the cardinal structure below $\aleph_\omega$ the same as in $M$. Now consider $M$ as a generic extension of $N$, which makes $\lambda$ into $\aleph_{\omega+1}$ and is $\lambda-c.c.$ extension of $N$.<|endoftext|> TITLE: Fractal dimension of scaling limits of discrete structures QUESTION [7 upvotes]: Let $S$ be the set of positive integers whose base-three expansion contains only the digits 0 and 2. The discrete set $S$ in a sense has (negative) fractal dimension $(\log 1/2)/(\log 3)$, since if you dilate the set by a factor of 3 you get a set that's half as big (two translates of the dilated set tile the original set). Meanwhile, the Cantor set (which can be constructed as the scaling limit of initial segments of $S$) has fractal dimension $(\log 2)/(\log 3)$, since if you dilate the set by a factor of 3 you get a set that's twice as big (two translates of the Cantor set tile the dilated Cantor set). Lastly note that $(\log 1/2)/(\log 3) = - (\log 2)/(\log 3)$. I'd ike to know where in the literature this sort of connection is spelled out in appropriate generality, with clear definitions of the different notions of fractal dimension that are involved, and theorems asserting numerical relationships between those quantities in a more general setting (applying for instance to the locations of the odd entries in Pascal's triangle and Sierpinski's gasket). REPLY [3 votes]: Martin Barlow And S. James Taylor published a number of papers on this topic, for example: Fractional dimension of sets in discrete spaces, Journal of Physics A: Mathematical and General, Volume 22 (1989) Defining fractal subsets of ${\mathbb Z}^d$, Proc. London Math. Soc. Volume 64 1992 The second is available online. Both authors did plenty of important work studying the fractal properties of stochastic processes. Taylor proved the seminal result that brownian motion in $\mathbb R^d$ almost surely has dimension 2 when $d>1$. As such, they are primarily interested in definitions of fractal dimension that yield expected results when studying random walks on $\mathbb Z^d$. The basic idea is to define the lattice cube $$V_n = \{x\in{\mathbb Z}^d: -n/2 TITLE: Integrals of pullbacks and the Inverse function theorem(s?) QUESTION [17 upvotes]: The usual story goes like this: Smooth picture (?): For a smooth bijection $\phi: M \to N$ between $n$-manifolds the following is true: $\phi^{-1}$ is a local diffeomorphism a.e. Given an open set $U \subset N$ and a form $\omega \in \Omega^k(U)$ we have the equality: $\int_U \omega= \int_{\phi^{-1}(U)} \phi^*\omega$. Satisfied with this simple and very intuitive picture I slowly came to believe that this is the most general change of variables theorem there could be. That is, until I met the following theorem in a measure theoretic context. Theorem 1: Let $U\subset\mathbb{R}^n$ be a measurable subset and $\phi :U\to \mathbb{R}^n$ be injective and differentiable. $\implies$ $\int_{\phi(U)} f = \int_U f\circ\phi |\det D\phi|$ for all real valued functions $f$. Then upon searching the internet I came by a weaker version of the inverse function theorem for everywhere differentiable functions: Theorem 2: Let $U \subset \mathbb{R}^n$ and $f:U \to \mathbb{R}^n$ be differentiable s.t. $Df_{x_0}$ has full rank for all $x_0 \in U$ $\implies$ $f$ is a local differentiable homeomorphism. Which leads to the following generalization: Differentiable picture (conjecture): For a differentiable bijection $\phi: M \to N$ between $n$-manifolds the following is true: $\phi^{-1}$ is a local differentiable homeomorphism a.e. Given an open set $U \subset N$ and a form $\omega \in \Omega^k(U)$ we have the equality: $\int_U \omega= \int_{\phi^{-1}(U)} \phi^*\omega$. In search of a unifying picture i listed the properties a function $\phi$ must have so that the pullback wont change the value of the integral. I. $\phi$ must be locally absolutely continuous. (otherwise it could send a null set to a positive measure set). This also establishes the almost everywhere differentiability of $\phi$. II. $D \phi$ must have full rank almost everywhere. If we can preform change of variables with $\phi$. What more properties must it have? Following the connection with lipschitz functions i arrived at the following unifying conjecture: The great conjecture: For a locally lipschitz bijection $\phi: M \to N$ between $n$-manifolds the following is true: $\phi^{-1} $ is locally bi-lipschitz (a.e.?) Given an open set $U \subset N$ and a form $\omega \in \Omega^k(U)$ we have the equality: $\int_U \omega= \int_{\phi^{-1}(U)} \phi^*\omega$. Is this true? REPLY [5 votes]: If you consider continuous injections (resp. homeomorphisms onto their range) instead of locally Lipschitz bijections (resp. locally bi-Lipschitz), then the modified conjecture is true because of Brouwer's theorem on invariance of domain, with the proviso that in (2) you should consider $n$-forms instead of $k$-forms because the domain $U$ of integration is assumed open and hence is $n$-dimensional. Formula (2) in this case is simply the change of variables formula for integrals with respect to a measure $\mu$ and its pushforward $\phi_*\mu$, once you write $n$-forms in terms of the volume measure associated to a Riemannian metric on $M$ (recall that a continuous map is always Borel measurable). Here I'm assuming $M$ and $N$ oriented for simplicity. Otherwise (assuming back your original hypotheses), you should consider locally $k$-rectifiable subsets $U$ of the target manifold $N$ in (2) instead of $U$ open ("locally" equals "countably" if your manifolds are second countable). All that remains is to show that the inverse of $\phi$ is locally Lipschitz. This, however, needs an additional hypothesis as follows. By Rademacher's theorem, any locally Lipschitz map $\phi:M\rightarrow N$ is differentiable almost everywhere; more precisely, the (Clarke sub)differential $D\phi$ of $\phi$ is a locally bounded set-valued map which is single-valued almost everywhere. That being said, the additional hypothesis you need in your conjecture to settle part (1) is that $D\phi$ takes only non-singular values, for in this case you can invoke Clarke's inverse function theorem for Lipschitz maps (Theorem 1 of F.H. Clarke, "On The Inverse Function Theorem". Pac. J. Math. 64 (1976) 97-102). Part (2) follows immediately by using the $k$-dimensional Hausdorff measure with respect to some Riemannian metric on $M$. Some of the comments addressed the possibility of doing away with the hypothesis of $D\phi$ being non-singular by invoking Sard's theorem. However, this does not work with so little regularity assumed from $\phi$ - a necessary hypothesis for Sard's theorem to hold true, as pointed in Sard's original paper ("The Measure of Critical Values of Differentiable Maps", Bull. Amer. Math. Soc. 48 (1942) 883-890) is that $\phi$ should be at least $\mathscr{C}^1$. $\phi$ being locally Lipschitz is not good enough - as shown by J. Borwein and X. Wang ("Lipschitz functions with maximal subdifferentials are generic", Proc. Amer. Math. Soc. 128 (2000) 3221–3229), the set of 1-Lipschitz functions $f$ such that all points in its domain are critical (in the sense that $Df$ contains zero everywhere) is generic with respect to the uniform topology. In particular, generically one cannot expect the inverse of a locally Lipschitz bijection, albeit certainly continuous, to be locally Lipschitz (even if only up to a subset of measure zero).<|endoftext|> TITLE: Are polynomials dense in holomorphic $L^p(\mathrm{Gauss})$ for $p < 1$? QUESTION [5 upvotes]: Let $\mu$ be standard Gaussian measure on $\mathbb{C}^n$, i.e. $d\mu = \frac{1}{(2 \pi)^n} e^{-|z|^2/2}\,dz$, and fix $0 < p < 1$ (note carefully). Suppose $g$ is holomorphic on $\mathbb{C}^n$ and satisfies $\int |g|^p\,d\mu < \infty$. Do there exist holomorphic polynomials $g_n$ such that $\int |g-g_n|^p\,d\mu \to 0$? Stated another way, let $\mathcal{H}$ be the space of holomorphic functions on $\mathbb{C}^n$ and $\mathcal{P}$ be the holomorphic polynomials. Let $\mathcal{H}L^p(\mu) = \{ g \in \mathcal{H} : \int |g|^p\,d\mu < \infty\}$, equipped with the usual metric $d(f,g) = \int |f-g|^p\,d\mu$ (which of course is not a norm). Clearly $\mathcal{P} \subset \mathcal{H} L^p(\mu)$. Is $\mathcal{P}$ dense in $\mathcal{H} L^p(\mu)$? Even a proof or counterexample for $n=1$ would shed some light. This is true for $1 \le p < \infty$ and seems to be well known, but the proof I've seen does not go through for $p < 1$, since it needs $t^p$ to be a convex function. It goes like this. First you show that as $\theta \to 0$ we have $g(e^{i\theta} \cdot) \to g$ in $L^p(\mu)$; this part still works for $p < 1$. Now let $F_n$ be the Fejér kernel and set $g_n(z) = \int_{-\pi}^\pi F_n(\theta) g(e^{i \theta} z)\,d\theta$, which can be shown to be a polynomial of degree at most $n-1$. If $p \ge 1$ we can use Jensen's inequality on the probability measure $F(\theta) \,d\theta$ to show $$\int_{\mathbb{C}^n} |g(z)-g_n(z)|^p\,\mu(dz) \le \int_{\mathbb{C}^n} \int_{-\pi}^\pi F_n(\theta) \left|g(z)-g(e^{i\theta} z)\right|^p\,d\theta \,\mu(dz).$$ After interchanging the integrals, properties of the Fejér kernel imply this converges to $\lim_{\theta \to 0} \int_{\mathbb{C}^n} \left|g(z)-g(e^{i\theta} z)\right|^p\,d\mu = 0$ as previously argued. But for $p < 1$, Jensen's inequality goes the wrong way. One idea is to try to find a dominating function for $|g-g_n|$ and use dominated convergence. For instance, $g_n$ is controlled by $\sup_{\theta} |g(e^{i \theta} z)|$, and maybe we could bound that in terms of $g$ or some related $L^p$ function. But I don't see how to do that. REPLY [3 votes]: Yes, they are dense. The key result comes from a paper of R. Wallstén [1], in which Theorem 3.1 implies that the set $\mathcal{E}$ of functions of the form $$f(z) = \sum_{j=1}^m a_j e^{\langle z, w_j \rangle}, \qquad a_j \in \mathbb{C}, \, w_j \in \mathbb{C}^n$$ is dense in $\mathcal{H} L^p(\mu)$ for any $0 < p < 1$. Clearly $\mathcal{E} \subset \mathcal{H} L^1(\mu)$, and we already know that $\mathcal{P}$ is dense in $\mathcal{H} L^1(\mu)$. So given $g \in \mathcal{H} L^p(\mu)$ and $\epsilon > 0$, we may choose $f \in \mathcal{E}$ and $h \in \mathcal{P}$ with $$\int |g-f|^p\,d\mu < \epsilon, \qquad \int |f-h|\,d\mu < \epsilon.$$ But by Jensen's inequality, $\int |f-h|^p \,d\mu \le \left(\int |f-h|\,d\mu\right)^p$, so by the triangle inequality in $L^p$ we have $$\int |g-h|^p\,d\mu \le \int |g-f|^p\,d\mu + \int |f-h|^p \,d\mu \le \epsilon + \epsilon^p.$$ [1] Wallstén, Robert. The $S^p$-criterion for Hankel forms on the Fock space, $0 TITLE: Is the space of Radon measures a Prohorov space? QUESTION [5 upvotes]: Consider the spaces $C_c(\mathbb{R})$ of compactly supported continuous functions equipped with the inductive limit topology and the Banach space $C_0(\mathbb{R}) = \overline{C_c(\mathbb{R})}^{\, _{||.||_\infty}}$ of continuous functions vanishing at infinity equipped with the sup norm. The dual $M = C_c'$ is the space of Radon measures on $\mathbb{R}$ and $M_f = C_0' \subseteq M$ the subspace of finite Radon measures. Equip $M$ and $M_f$ with their weak-* topology. Now consider the space of probability measures on $M$ resp. $M_f$ equipped with the weak topology. I want to know whether $M$ and $M_f$ are (sequential) Prohorov spaces, i.e. if every compact set (or sequence) of probability measures on $M$ resp. $M_f$ is tight. Bogachev, Measure Theory II, Remark 8.10.15 mentions that $\mathcal{D}'(\mathbb{R})$ (the space of distributions) is Prohorov. So, since $M \subseteq \mathcal{D}'$ is closed, it follows that $M$ is also Prohorov. On the other hand, Proposition 8.10.19 says that the dual of any infinite-dimensional Banach space equipped with its weak-* topology is never Prohorov (by applying the Baire category theorem). So, it follows that $M_f$ is not Prohorov, which is somehow confusing. I don't have any intuition for that fact. I always thought that Banach spaces are not such beasts. Is it somehow related to the distinction between Banach spaces and nuclear spaces? Correction: $M$ is dense in $\mathcal{D'}$! (Thanks to @weather for the correction). [I have misleadingly thought of the continuous injection $M \to \mathcal{D}'$ as an isomorphism.] So there is no confusion anymore. REPLY [4 votes]: The theorem of Prohorov states that polish spaces, i.e. complete separable metric spaces, satisfy your condition. Another result states that the space of probability measures on a polish space when provided with the weak topology generated by the bounded, continuous functions. Hence your first question has a positive answer if you are prepared to use a stronger weak topology. There seems to be some confusion in your other question. The weak topologies in the two spaces involved---distributions and measures---are distinct and, far from being closed, $M$ is dense in your space of distributions. This leaves the question of whether it has the Prohorov property open. Given the result on dual spaces you mention, I suspect that the answer will be negative---the (presumably crucial) difference to the case of distributions is that the latter space is nuclear.<|endoftext|> TITLE: Distance function from a topological submanifold QUESTION [5 upvotes]: Let $(M,g)$ be a Riemannian manifold, and let $N\subset M$ be an embedded sphere that is everywhere smooth except for a single point at which the embedding will only be $C^0$. How much regularity can I obtain for the square of the distance function $$ F(x) := \inf\{d^2(x,y)| y\in N\}? $$ I would be surprised if this extra information had some effect on the answer, but you never know: The embedding is in high codimension; $M$ is dimension $2n+1$ and $N$ is dimension $n-1$. Thank you very much for every comment, help or reference. Best wishes Klaus REPLY [3 votes]: welcome to MO! It seems to me that you cannot expect anything more than the obvious, which is local Lipschitz regularity. First, observe that even with a smooth embedding, there are problems at the some points (where the level hypersurface of the square distance fonction has a "double point"). In the $C^1$ case, you can avoid this in a neighborhood of the submanifold, but I think that if you lack regularity even at one point, then it can happen that these problematic points accumulate to the submanifold. Take the following examples for a curve in $\mathbb{R}^2$ (or, taking a product, $\mathbb{R}^n$ for any $n$): $$\gamma(t)=\sqrt{|t|}\sin(1/t)$$ It defines a topological embedding of a line (which of course can be made into an embedding of a circle, or a higher-dimensional sphere easily) which only fails to be smooth at one point. It needs a bit of computation, but I think that there is no open set of values for which the square distance function is differentiable.<|endoftext|> TITLE: Banach-Zarecki theorem - who was Zarecki? QUESTION [11 upvotes]: I'm writing a paper for real analysis seminar, a paper about Banach-Zarecki theorem and I need some information about the authors. Stefan Banach - there is no problem to find information about him. What about Zarecki? I only found that he was russian mathematician, and in "Theory of functions of a real variable" Natanson writes "M. A. Zarecki". This is all I know - I cannot find any information about this mathematician or his life. REPLY [13 votes]: Here's an excerpt from Integration and Modern Analysis by Benedetto and Czaja (pp. 202-203): Moisej A. Zaretsky (1903–1930) is not a household name in mathematics, which is why we feature him now because of his beautiful result, independent of Banach, of the Banach–Zaretsky theorem (1925) (Theorem 4.6.2). Zaretsky was a Russian, and he died in the Caucasian resort of Batumi. The English translation of the Russian Isidor P. Natanson’s text uses the spelling “Zarecki” As a matter of fact, Zarecki is a Polish transliteration of his name -- most likely it was used by Banach and his group.<|endoftext|> TITLE: Accuracy of the formulas for angles between almost colinear vectors QUESTION [5 upvotes]: Assume $x$ and $y$ are two vectors in $\mathbb{R}^3$ and we want to compute the acute angle $\alpha\in(0,\pi/2]$ between these two (noncolinear) vectors. There are (at least) two possibilities: In the naive approach, we compute the absolute value of the dot product of the normalized vectors $x$ and $y$ $$\frac{x^Ty}{\|x\|\|y\|}$$ and take the inverse cosine of the result. The less naive approach is based on the fact that $$\|x\times y\|=\|x\|\|y\|\sin\alpha\quad\text{and}\quad|x^Ty|=\|x\|\|y\|\cos\alpha$$ so $\alpha$ is equal to an angle in a right triangle with legs of the length $\|x\times y\|$ and $|x^Ty|$ (for convenience, one can use a variant of the inverse tangent implemented in the atan2 function which is available in most programming languages; the function takes the side lengths of the legs as two arguments). Now assume that $x$ is given and $y=x+\Delta x$ where $\|\Delta x\|\leq\tau\|x\|$, $\tau\ll 1$, that is, the vectors are almost colinear (for simplicity also of almost same norms). Assume that $\tilde\alpha_1$ and $\tilde\alpha_2$ are, respectively, the angles computed by the naive and the less naive approaches. Recently, I've run several tests which suggest that $$\tag{1} \frac{|\alpha-\tilde\alpha_1|}{\alpha}\leq \epsilon\mathcal{O}(\tau^{-2}) \quad\text{and}\quad \frac{|\alpha-\tilde\alpha_2|}{\alpha}\leq \epsilon\mathcal{O}(\tau^{-1}), $$ where $\epsilon$ is the machine precision. I understand that both algorithms suffer from a certain inaccuracy when $y\approx x$; in particular, both computing the dot and cross products. I suppose the inverse trigonometric functions are not an issue as these are usually implemented to give very accurate results. I'm not asking anybody for performing any kind of analysis. I was just wondering whether there is a known reference where the accuracy of the two approaches is considered, if possible revealing why (1) (probably) holds. Thanks a lot in advance. REPLY [3 votes]: For some background on these sort of issues, this might be interesting: R.W. Sinnott, "Virtues of the Haversine", Sky and Telescope, vol. 68, no. 2, 1984, p. 159<|endoftext|> TITLE: Kan extensions in concrete 2-categories QUESTION [7 upvotes]: Kan extensions make sense in any 2-category. I am interested in Kan extensions in "concrete" 2-categories consisting of actual categories with some sort of structure (e.g., finite products, finite limits, pretoposes, toposes, etc.), with 1- and 2-cells structure-preserving functors and any natural transformations. First of all, when do such extensions exist? The usual Kan extensions exist whenever the target category is co-complete, but these functors do not typically preserve the relevant structure (aside: what are some relatively simple examples of this phenomenon). Is there some way to "fix up" this problem and provide a "structure-preserving Kan" given co-completeness? If not, are there other conditions which might allow us to do that? Failing either of these, are there at least conditions under which the ordinary Kan extensions will preserve structure X? This leads to a second question. If (some) extensions do exist in a concrete 2-category, how are they related to the usual extensions in $\textbf{Cat}$? Intuitively, I would expect the forgetful functor to preserve right Kan extensions, as these are akin to limits, and the free completion to preserve left Kan extensions. Thanks! REPLY [4 votes]: One situation in which a Kan extension can be "fixed up" is if the category of structure-preserving maps between two structured categories is a reflective or coreflective (full) subcategory of the category of all maps. For instance, if $C$ is small and $D$ is well-behaved (e.g. locally presentable) and both have finite limits, then $\mathrm{Lex}(C,D)$ is a reflective subcategory of $[C,D]$. Thus, applying the reflector to a left Kan extension in $\mathrm{Cat}$ will "fix it up" into a left Kan extension in the 2-category $\mathrm{Lex}$. I spent a few minutes thinking about whether anything general can be said about for what sorts of "structure" this is the case. A natural question to ask is whether it is always true when our structured categories are the algebras for a lax-idempotent or colax-idempotent 2-monad. However, I don't have an answer.<|endoftext|> TITLE: Orlicz Norm and A result on expectation QUESTION [6 upvotes]: I am reading paper which is mainly about Dobrushin's contraction coefficient and its generalization. In page 27, the following is defined: Consider arbitrary, non-negative, convex function $\psi:\mathbb{R}\mapsto \mathbb{R}$ with $\psi(0)<1$. Then for an integrable random variable $X$ $$||X||_{\psi}:=\inf \{c>0:~\mathbb{E}[\psi(\frac{X}{c})]\leq 1\}.$$ Then letting $\psi^*$ be the convex conjugate of $\psi$, we can write the following for arbitrary random variables $X$ and $Y$ by Young's inequality: $$XY\leq \psi(X)+\psi^*(Y)$$ and hence, $$\mathbb{E}[XY]\leq 2||X||_{\psi}||Y||_{\psi^*}.$$ First of all, in the typical definition of Orlicz norm, it is assumed that $\psi(0)=0$ and also $\psi$ is non-decreasing. Why here those properties are removed? Secondly, the last inequality is not clear to me. Can anybody give me a hint on how to show that? Thanks REPLY [2 votes]: It seems that the author applies the result with $\psi_1(x):=e^x /2$ and $\psi_2(x):=e^{x^2}/2$ in order to exploit the identity $\lVert X^2\rVert_{\psi_1}=\lVert X\rVert_{\psi_2}^2$. It turns out that with these definition, $\lVert X\rVert_{\psi_j}$ is equal to the norm associated to the true Orlicz functions $\phi_1(x):=e^x-1$ and $\phi_2(x)=e^{x^2}-1$. As suggested by Christian Remling, we have for fixed $c,c'\gt 0$ the inequality $$\frac Xc\frac Y{c'} \leqslant \psi\left(\frac Xc\right)+\psi^*\left(\frac Y{ c'} \right),$$ hence if $\psi\left(\frac Xc\right)\leqslant 1$ and $\psi\left(\frac Y{c'} \right)\leqslant 1$, the inequality $$\mathbb E[XY] \leqslant 2cc'$$ takes place. We conclude by taking the infimum over those $c$ and $c'$.<|endoftext|> TITLE: A combinatorial identity generalizing identity (3.111) from Gould's book QUESTION [8 upvotes]: I am trying to prove the following identity, which I am sure is correct: $$ \sum_{m=0}^{K}(-1)^m{n \choose m}{(n-m)r-1 \choose n-1}=1, $$ where $K:=\left[n\frac{r-1}{r}\right]$ for some integer $r$, and $[x]$ denotes the largest integer contained in $x$. In the special case $r=2$ this is identity (3.111) from H. Gould's book "Combinatorial identities" (1972). For $r=2$ it was first stated by B. C. Wong in Amer. Math. Monthly in May 1930 as an open question. I am wondering if someone has seen a proof of (or can prove) this identity for any $r$. The upper summation limit$\left[n\frac{r-1}{r}\right]$ also occurs in Gould's book in identity (3.113), so it is not entirely unheard of. REPLY [10 votes]: As Brendan McKay commented, this looks like inclusion-exclusion. I just happened to write up such an inclusion-exclusion. Suppose we want to count the number of ways to roll a total of $t$ on $n$ $r$-sided dice (with sides $1, 2, ..., r$). By inclusion-exclusion (or, via generating functions, by using the binomial theorem twice to expand the powers in $[x^t](x^n (x^r-1)^n (x-1)^{-n}$), this is $$\sum_{m=0}^{[(t-n)/r]} (-1)^m {n \choose m}{t-1 - m r \choose n-1}.$$ Choose $t=nr$. There is just one way to roll a total of $nr$, by getting the maximum $r$ on each of the $n$ dice. So, $$\sum_{m=0}^{[(n(r-1)/r]} (-1)^m {n \choose m}{(n-m)r-1 \choose n-1}=1.$$<|endoftext|> TITLE: The groupoid of algebraic expressions and proofs QUESTION [13 upvotes]: Fix a set of variables $V$, and suppose we're given a presentation of a monosorted algebraic theory, with variable symbols taken from $V$. For the sake of example, suppose the presentation consists of a sort symbol $U$, a function symbol $+ : U \times U \rightarrow U,$ as well as an "explicitly named identity," expressing the commutativity of addition. $$\mathsf{CoA}(x,y) : x+y = y+x$$ We think of our identities as being morphisms, whose domain and codomain are well-formed algebraic expressions. Then there is a corresponding groupoid, whose objects are the well-formed algebraic expressions, and whose arrows are proofs of equality between two expressions. Question. What is this construction called, how can we rigorously define it, and where can I read about it? Let me be a little more explicit about what the groupoid looks like in the aforementioned example. Objects are well-formed algebraic expressions such as $x+y$ and $a+(b+c)$. The variable symbols have to be taken from $V$, of course; I'm assuming that $\{x,y,a,b,c\} \subseteq V$. Morphisms are proofs that one expression equals another. So for example, we have a morphism $$\mathsf{CoA}(x,y) : x+y = y+x$$ expressing that $x+y$ equals $y+x$; and another morphism $$a+\mathsf{CoA}(x,y) : a+(x+y) = a+(y+x)$$ expressing that $a+(x+y)$ equals $a+(y+x)$; and another morphism $$\mathsf{CoA}(x+y,a) : (x+y)+a = a+(x+y).$$ Furthermore, we can compose these to get a proof $$(\mathsf{CoA}(x+y,a)) \circ (a+\mathsf{CoA}(x,y)) : (x+y)+a = a+(y+x)$$ of the fact that $(x+y)+a=a+(y+x)$. Note that we can also take inverses; this allows us to get a proof that $\sigma = \tau$ from a proof that $\tau = \sigma$. REPLY [6 votes]: The construction you describe seems more like the the category of reductions generated by the abstract rewrite system given by an algebraic theory. I suggest you take a look to section 8.2("Rewrite systems revisited") of Term Rewriting System where these concepts are defined. Here a short summary of basic the idea: you can consider a rewrite system as a graph whose vertex are terms of the signature and arrows are step of reduction. From this data enclosing by some operations you get what in the reference is called an abstract rewrite system with compositions, whose objects are terms for the signature and whose morphisms are proof of reductions. Quotienting this structure for axioms of category you get the category of reductions. Hope this helps.<|endoftext|> TITLE: Matrix-convexity of inverse of the cofactor matrix QUESTION [6 upvotes]: Consider the matrix-valued function $f(A) = \frac{A}{\det(A)}$ on the set of $3\times 3$ positive-definite matrices. Is this function matrix-convex ? (i.e., is $tf(A) + (1-t)f(B) - f(tA+(1-t)B)$ positive semi-definite $\forall \ t \in [0,1]$?) REPLY [2 votes]: Well, this is not an answer. But I cannot resist to mention the following equivalent property. Let $A\mapsto \hat A$ denote the cofactor map, and $B\mapsto \check B$ its inverse. For positive definite symmetric matrices, $\hat A=(\det A)A^{-1}$ and $\check B=(\det B)^{\frac1{n-1}}B^{-1}$. Then the harmonic mean is less than or equal to the cofactor mean over the cone of positive definite matrices: $$\langle A^{-1}\rangle^{-1}\le\widehat{\langle\hat A\rangle}.$$<|endoftext|> TITLE: Primes that are sums of two squares with constraints on the squares QUESTION [21 upvotes]: It is well known that there are infinitely many primes of the form $a^2+b^2$ (namely all primes congruent to $1$ modulo $4$). On the other hand, Euler raised the problem as to whether there are infinitely many primes of the form $a^2+1$, which is still open (a positive answer is a special case of Bunyakovsky's conjecture). I am wondering whether there are known results in between these cases. To be precise, I am interested in the following problem: Are there infinitely many primes $p$ of the form $a^2+b^2$ such that $b$ is bounded by a (slowly growing) function of $p$, for example $b = o(\sqrt{p})$? REPLY [26 votes]: Hecke showed that there are infinitely many primes of the form $p=a^2+b^2$ with $a = o(\sqrt{p})$. Ankeny improved this to $o(\log p)$, conditional on the Extended Riemann Hypothesis. Harman and Lewis obtain $o(p^{0.119})$ unconditionally. (Disclaimer: I copied the Hecke reference from Ankeny's bibliography without checking it.) The latter two papers cite other related work. I'd like to point out that I am not an expert here -- this was the result of googling "Gaussian prime" angle and then chasing references through MathSciNet and ResearchGate.<|endoftext|> TITLE: two-dimensional sections of polyhedral cones QUESTION [6 upvotes]: Given a polyhedral cone, its intersection with any two-dimensional plane is either a polygon or a region enclosed by a polygonal curve. Is it a characterization of polyhedral cones? Does there exists a convex cone, which is not polyhedral, but such that all its two-dimensional sections are regions enclosed by polygonal curves? REPLY [2 votes]: Victor Klee, "Some characterizations of convex polyhedra." Acta Mathematica 1959, Volume 102, Issue 1-2, pp 79-107. I haven't read the whole paper, but Klee proved that if all $j$-dimensional sections of an $n$-dimensional convex region are polyhedral for $2\le j \le n-1$, then the region is polyhedral. His definition for polyhedral includes polyhedral cones and other unbounded polyhedra. For $j=2$ and a closed convex cone, Klee cites Mirkil for a slightly earlier proof.<|endoftext|> TITLE: History of the orientation of Cartesian coordinates in drawing QUESTION [5 upvotes]: Is there any actual historical example in which a Cartesian plane with all four quadrants has been used, but with all axes marked with positive numbers? [Please see Sawyer's paper below for a "made-up" example] I am writing the result of a research the point of which was to teach children negative numbers in an algebraic context. There were many interesting similarities between their conceptions and difficulties and the so-called historical ones. For the one in the question (suggested by one of the children) I have claimed that "It has an interesting historical counterpart". Has it? Or, I have to delete my claim? Added: In this nice paper of Warwick Sawyer, The Importance of the Unbelievable, he writes"without negative numbers the equation of a line would depend upon which quarter it was in." The question is whether this possibility has ever been practiced in history or Sawyer just made it up for educational purposes? PS. Following Todd's comments, I changed the wording of the question. I hope it is now clearer. REPLY [3 votes]: A very detailed historical account of the debates about clarifying the concept of negative numbers can be found in the book http://link.springer.com/book/10.1007%2F0-387-28273-4 (Conflicts between Generalization, Rigor, and Intuition, by Gert Schubring). It is stated in the book (p. 82) that "Reyneau (1736) was probably the first to explicitly introduce the four quadrants in the coordinate system of the plane, a novelty that appears self-evident to us today" and that (p. 289) in De Prony's lectures (1795) Cours d'Analyse appliquée à la mécanique "for the first time in France the four quadrants of a coordinate system are explicitly assigned to the respective positive and negative values of the x– and y–axes in the plane".<|endoftext|> TITLE: When do two lattices have the same stabilizer in the diagonal torus? QUESTION [6 upvotes]: This is moved from MSE, where I asked and didn't receive an answer (see https://math.stackexchange.com/questions/1145151/lattices-in-mathbbq-pn-with-the-same-stabilizer) Let $T$ be the diagonal torus in $G = GL_n(\mathbb{Q}_p)$, and consider the action of $G$ on the set of (full-rank) lattices $\Lambda\subset \mathbb{Q}_p^n$. Let $\Lambda$ and $\Lambda'$ be full-rank sublattices ($\mathbb{Z}_p$-submodules of rank $n$) such $Stab_T(\Lambda) = Stab_T(\Lambda')$. Is it true that $\Lambda' = t\cdot \Lambda$ for some $t\in T$? This is true when $n = 2$ and $p > 2$; to see this, if $\{e_1,\, e_2\}$ is the standard basis and $\Lambda$ any lattice, we can always take a basis of the form $$\{p^re_1,\, \alpha e_1 + p^s e_2\}$$ where $r,\, s\in \mathbb{Z}$ and $\alpha$ is chosen modulo $p^s$; at this point you can check directly that $\Lambda$ and $\Lambda'$ have the same stabilizer if and only if $v_p(\alpha) - r = v_p(\alpha') - r'$, and any two such lattices are in the same $T$-orbit. If this is known, I'd love a reference. EDITED: When $p = 2$, this is false, even when $n = 2$, in view of JWitte's example below. I'm still interested in the case where $p > 2$. REPLY [2 votes]: First a counter example: Look at the case $\mathbb{Q}_2$ and $n=2$: Take $\Lambda = \langle e_1,e_2\rangle$ and $\Lambda'= \langle pe_1, e_1+e_2\rangle$. Let $t_{\alpha,\beta}$ be the diagonal matrix with $\alpha$ and $\beta$ on the diagonal: $t_{\alpha,\beta}e_1 = \alpha e_1$ and $t_{\alpha,\beta}e_2 = \beta e_2$. The stabilizer of $\Lambda$ in $T$ is $\{ t_{\alpha,\beta} : v(\alpha)=v(\beta)=0\}$. Now we show that this is also the stabilizer of $\Lambda'$: Assume that $v(\alpha)=v(\beta)=0$, then $\alpha\beta^{-1} = 1 + \gamma p$ with $v(\gamma)\geq 0$ (p=2!). Now $t_{\alpha,\beta}\Lambda' = \langle p\alpha e_1, \alpha e_1+\beta e_2\rangle = \langle pe_1, \alpha\beta^{-1} e_1+e_2\rangle = \langle pe_1, \gamma p e_1+e_1+e_2\rangle=\langle pe_1,e_1+e_2\rangle=\Lambda'$. Let $\pi_i$ be the projection to $e_i$ for $i\in\{1,2\}$. Then $\pi_i(\Lambda') = \mathbb{Z}_p e_i$. Now $\pi_1(t_{\alpha,\beta} \Lambda') = \alpha\mathbb{Z}_p$ and $\pi_2(t_{\alpha,\beta}\Lambda') = \beta\mathbb{Z}_p$. Therefore if $t_{\alpha,\beta}$ fixes $\Lambda'$, then $v(\alpha)=v(\beta)=0$. Moreover if $t_{\alpha,\beta}\Lambda' =\Lambda$, then $\pi_i(t_{\alpha,\beta}\Lambda') = \pi_i(\Lambda) = \mathbb{Z}_p e_i$. Thus $v(\alpha)=v(\beta)=0$ by the preceding calculations. But then $t_{\alpha,\beta}$ is in the stabilizer of $\Lambda$. Thus $\Lambda$ and $\Lambda'$ are not in the same $T$ orbit. A general comment: If you know something about buildings, you could see that this example is based on 3.6.1 of Reductive groups over local fields, J. Tits. The stabilizer of a point in the apartment of $T$ is the set of diagonal matrices with units. And 3.6.1 states that if the residual characteristic is 2 this stabilizer fixes also points in the building that are not in the apartment of $T$. Because $T$ acts on its own apartment the points outside the apartment of $T$ are not in the same orbit as a point in the apartment. So the answer to your question is negative for all $n$ in the case $p=2$. However for odd $p$ I don't know the answer.<|endoftext|> TITLE: $\chi(\omega_X)>0$ implies that $X$ is of general type QUESTION [12 upvotes]: If $X$ is a smooth (complex) projective variety of maximal Albanese dimension such that $\chi(\omega_X)>0$, how does one show that $X$ is of general type? I've seen this used but I can't find a proof. It is a result of Ein-Lazarsfeld that if $X$ is a smooth projective variety of maximal albanese dimension such that $\chi(\omega_X)=0$, then the albanese image is fibered by tori, so in partixular if $X$ is birational to its image under the albanese map, then it can't be of general type. REPLY [14 votes]: If it is not of general type, then the Iitaka fibration $X\to Z$ is fibered by tori say $F$ (the fibers of the Iitaka fibration have Kodaira dimension 0 and maximal Albanese dimension, and so by a Theorem of Kawamata, they are abelian varieties). For general $P\in Pic^0(A)$, we have that $P|_F\ne 0$ and so $h^0(K_F+P)=h^0(P|_F)=0$ where $F$ is the general fiber. But then $h^0(K_X+P)=0$ and by the generic vanishing theorems of Green and Lazarsfeld $h^i(K_X+P)=0$ for $i>0$ so that $\chi (K_X)=\chi(K_X+P)=0$.<|endoftext|> TITLE: Does homotopy invariance of homology follow from the structure of the simplex category $\Delta$? QUESTION [6 upvotes]: Explicitly: Let $\Delta$ denote the simplex category, and $\mathscr{C}$ any small category, and fix a functor $F:\Delta \rightarrow \mathscr{C}$ such that $F\Delta^0$ is terminal. Also, assume $\mathscr{C}$ has products. Using $F$, we can define a homology theory in $\mathscr{C}$ letting $C^n(X) = \mathbb{Z}(Hom(F\Delta^n,X))$, the free abelian group generated by morphisms from $F\Delta^n$ to $X$. The boundary map is constructed from the face maps in $\Delta$, and one can prove that $\partial^2 = 0$- this is exactly what is done to find singular homology of topological spaces. Now, suppose $f,g:X \rightarrow Y$ are maps in $\mathscr{C}$. We say they are homotopy equivalent if there is a map $h:F\Delta^1 \times X \rightarrow Y$ such that $h_0 = f$ and $h_1 = g$. Here $h_0 = X \rightarrow F\Delta^0 \times X \stackrel{F\delta^0 \times 1_X}{\longrightarrow} F\Delta^1 \times X \stackrel{h}{\rightarrow} Y$, where $\delta^0:\Delta^0 \rightarrow \Delta^1$ is a face map. $h_1$ is defined similarly. My question is: If $f$ and $g$ are homotopic, does it follow that the induced maps of chain complexes $C^{\bullet}(X) \rightarrow C^{\bullet}(Y)$ are homotopic as well? REPLY [6 votes]: Chris's comment suggests that very little about the target category $C$ is being used in the standard argument, but I still think there's something interesting to check, namely what exactly is being used. The question concerns first of all a "singular chains" functor $$F_{\bullet}: X \mapsto \left( \Delta^n \mapsto \mathbb{Z}[\text{Hom}(F \Delta^n, X)] \right)$$ from $C$ to the category $\widehat{\Delta}$ of presheaves of abelian groups on $\Delta$ (equivalently, of simplicial abelian groups). After taking singular chains, the rest of the argument will take place almost entirely in $\widehat{\Delta}$. The question concerns second of all a notion of homotopy in $C$ coming from taking $F \Delta^1$ to be the interval object. The only relevant data in such a homotopy is the induced map $$F_{\bullet}(F \Delta^1) \times F_{\bullet}(X) \to F_{\bullet}(Y)$$ in $\widehat{\Delta}$. This is the only place where we use that $C$ has finite products and that $F_{\bullet}$, by construction, preserves them. This is also the only place where we use that $F_{\bullet}(X)$ and $F_{\bullet}(Y)$ have anything to do with $C$. Next, note that for every $n$ we have natural maps $$\Delta^n \to F_{\bullet}(F \Delta^n)$$ (where $\Delta^n$ denotes the corresponding representable presheaf of abelian groups on $\mathbb{Z}[\Delta]$) which arise as follows. By definition, a map $\Delta^n \to F_{\bullet}(F \Delta^n)$ is an element of $\mathbb{Z}[\text{Hom}(F\Delta^n, F\Delta^n)]$. But there is a distinguished such element, namely $\text{id}_{F \Delta^n}$. These maps in fact organize themselves into a natural transformation from the Yoneda embedding $\Delta \to \widehat{\Delta}$ to $F_{\bullet}(F(-))$, which is (a restriction of) the unit map of an adjunction. This implies in particular that they respect the face maps $\Delta^0 \to \Delta^1$. Now it's clear that pulling back along these natural maps gives a natural transformation from homotopies involving $F_{\bullet}(F \Delta^1)$ to simplicial homotopies $$\Delta^1 \times F_{\bullet}(X) \to F_{\bullet}(Y).$$ Checking that this natural transformation respects sources and targets as defined in the OP is the only place where we use that $F \Delta^0$ is the terminal object. The remaining question is why simplicial homotopies of simplicial abelian groups induce chain homotopies on the corresponding chain complexes. This is where the work in Hatcher's proof is, and it has absolutely nothing to do with $C$ or $F$. It exhibits part of the monoidal Dold-Kan correspondence, namely the part having to do with the Eilenberg-Zilber map.<|endoftext|> TITLE: Center of a simply-connected simple compact Lie group and McKay correspondence QUESTION [20 upvotes]: Let $G$ be a simply-connected simple compact Lie group. Its center $Z(G)$ is a finite abelian group, say $Z(G) = \mathbb Z/k\mathbb Z$ for $G=SU(k)$. I find the following interpretation of $Z(G)$ in terms of McKay correspondence: Let $\Gamma$ be a finite subgroup of $SU(2)$. Let $\{ \rho_i\}_{i\in I}$ be the set of (isomorphism classes of) irreducible representations of $\Gamma$. Let $Q$ be the $2$-dimensional representation of $\Gamma$ given by the inclusion $\Gamma\subset SU(2)$. Consider the tensor product decomposition $\rho_i\otimes Q = \bigoplus \rho_j^{\oplus a_{ij}}$. McKay correspondence says $2\delta_{ij} - a_{ij}$ is an affine Cartan matrix of type ADE. Let $J$ be the subset of $I$ consisting of $1$-dimensional irreducible representations. It consists of $i$ such that the coefficient of $\alpha_i$ in the imaginary root $\delta$ is $1$. (Sometimes they are called special vertices.) We put $J$ an abelian group structure by the tensor product as representations of $\Gamma$. Then $J$ is isomorphic to $Z(G)$. This observation can be checked by case-by-case analysis. It is easy for type $A_{k-1}$. The group $\Gamma$ is $\mathbb Z/k\mathbb Z$, $J$ is the whole $I$, and is isomorphic to $\Gamma$ itself as a group. For type $D_n$, $J$ consists of $4$ extremal vertices. One needs to know the dual representation of $\rho_j$ for $j\in J$ in order to determine the group $J$. It is given in Gozalez-Sprinberg Verdier paper, and the answer is: $J$ is either $\mathbb Z/2\mathbb Z\oplus \mathbb Z/2\mathbb Z$ or $\mathbb Z/4\mathbb Z$ according to $n$ is even or odd. This coincides with the center of $Spin(2n)$. Exceptional cases can be determined in the same way. My questions are Did anybody find this observation before ? Is there a conceptual explanation of this observation ? In particular, is there a proof without case-by-case check ? REPLY [8 votes]: Dave Morrison give me the following answer. The center $Z(G)$ is known to be isomorphic to $P/Q$, where $P$ is the weight lattice and $Q$ is the root lattice. On the other hand, let $X$ be the minimal resolution of $\mathbb C^2/\Gamma$. This is the place where McKay correspondence is realized by geometry. The configuration of the exceptional set is the finite Dynkin diagram, thus $Q$ is $H_2(X,\mathbb Z)$ with the intersection matrix. The weight lattice $P$ is isomorphic to $H^2(X,\mathbb Z)$ and the inclusion $Q\to P$ is realized by the intersection pairing, in other words, $Q\cong H^2_c(X,\mathbb Z)$ and $Q\to P$ is the natural homomorphism $H^2_c$ to $H^2$. Thus the quotient $P/Q$ is the cohomology of the `boundary' $H^2(S^3/\Gamma,\mathbb Z)$. McKay correspondence a la Gonzalez-Spinberg Verdier gives a homomorphism from the representation ring $R(\Gamma)$ to $P$ by the first Chern classes of tautological bundles. The first Chern class is computed by the connecting homomorphism $H^1(S^3/\Gamma, U(1))\to H^2(S^3/\Gamma,\mathbb Z)$, which is an isomorphism in our case. Thus $H^2(S^3/\Gamma,\mathbb Z)$ is given by 1-dimensional characters of $\Gamma$.<|endoftext|> TITLE: Factorization of polynomials in two variables QUESTION [11 upvotes]: I have read, from the question Irreducibility of polynomials in two variables, that all polynomials $f(x)-g(y)$, where $f, g$ are indecomposable polynomials, and there are no $a, b$ such that $g(ax+b)=f(x)$, are irreducible, unless the degrees of $f$ and $g$ are $$7, 11, 13, 15, 21,\ \ \text{or} \ \ 31$$ Is there an example of the exceptional case in degree $7$, with the factorization? REPLY [10 votes]: An example for $n=7$ is given in J. W. S. Cassels, Factorization of polynomials in several variables, Proc. Fifteenth Scandinavian Congress (Oslo, 1968), vol. 118, Lecture Notes in Mathematics, Springer, Berlin, pp. 1-17. This reference is from https://oeis.org/A112090 and is quite technical (using topology of Riemann surfaces). here I type the example from the paper in a sort of computer-readable format l=(1+sqrt(-7))/2 m=(1-sqrt(-7))/2 # t a nonzero(?) parameter f(x)-g(y)=(x^3+l*x^3*y-m*x*y^2-y^3-(3*l+2)*t*x+(3*m+2)*t*y+t)* (x^4-l*x^3*y-x^2*y^2-m*x*y^3+y^4+2*(m-l)*t*x^2- 7*t*x*y+2*(l-m)*t*y^2+(3-l)*t*x-(3-m)*t*y-7*t^2) Sage script (with input f and (-)g taken from the paper z=QQ['z'].0 K.=NumberField(z^2-z+2) m=l.conjugate() R.=K[] # t a nonzero(?) parameter f=x^7-7*l*t*x^5+(4-l)*t*x^4+(14*l-35)*t^2*x^3-(8*l+10)*t^2*x^2+(3-l+7*(3*l+2)*t)*t^2*x g=-y^7+7*m*t*y^5+(4-m)*t*y^4-(14*m-35)*t^2*y^3-(8*m+10)*t^2*y^2-(3-m+7*(3*m+2)*t)*t^2*y-7*t^3 (f+g).factor() outputs (x^3 + (l)*x^2*y + (l - 1)*x*y^2 - y^3 + (-3*l - 2)*x*t + (-3*l + 5)*y*t + t) * (x^4 + (-l)*x^3*y - x^2*y^2 + (l - 1)*x*y^3 + y^4 + (-4*l+2)*x^2*t- 7*x*y*t + (4*l-2)*y^2*t + (-l + 3)*x*t + (-l-2)*y*t - 7*t^2) which is the factorisation that should be like the one above. But it is not - there is a typo above (and in the paper: in the 1st factor the monomial l*x^3*y should be l*x^2*y. After this change everything checks out.<|endoftext|> TITLE: Structure of symplectic group over finite fields QUESTION [6 upvotes]: We are working over the finite field $\mathbb{F}_{q}$ of odd prime characteristic $p$ and of cardinality $q$ some power of $p$. We recall the symplectic group $Sp(4,\mathbb{F}_{q})$ as the group of transformations over $\mathbb{F}_{q}^{4}$ preserving a non degenerate alternate bilinear form, and we denote by $PSp(4,\mathbb{F}_{q})$ the quotient by its center. I would like to know all the possible maximal subgroups of this projective group or the initial one. A good reference may be of precious help. REPLY [12 votes]: You can find tables of the maximal subgroups of all (almost) simple classical groups in dimensions up to $12$ in the book: The Maximal Subgroups of the Low-Dimensional Finite Classical Groups, John N. Bray,Derek F. Holt, Colva M. Roney-Dougal, Cambridge University Press, 2013. The maximal subgroups of ${\rm Sp}_4(q)$ for odd $q$ were first classified in: H. H. Mitchell. The subgroups of the quaternary abelian linear group. Trans. Amer. Math. Soc. 15 (1914), 379–396. Here is a complete list. The notation is similar to that used in the ATLAS. There is one conjugacy class of each type except where otherwise stated. $q^{1+2}\!:\!((q-1) \times {\rm Sp}_2(q))$ (reducible) $q^3\!:\!{\rm GL}_2(q)$ (reducible) ${\rm Sp}_2(q)^2\!:\!2$ (imprimitive) ${\rm GL}_2(q).2$ (imprimitive) ${\rm Sp}_2(q^2)\!:\!2$ (semilinear) ${\rm GU}_2(q).2$ (semilinear) ${\rm Sp}_4(q_0).(2,r)$, $(2,r)$ classes, $q=q_0^r$, $r$ prime (subfield) $2^{1+4}.S_5$, $2$ classes, $q$ prime, $q \equiv \pm 1 \bmod 8$ $2^{1+4}.A_5$ $q$ prime, $q \equiv \pm 3 \bmod 8$ $2.A_6$, $q$ prime, $q \equiv \pm 5 \bmod 12$, $q \ne 7$ $2.S_6$, $2$ classes, $q$ prime, $q \equiv \pm 1 \bmod 12$ $2.A_7$, $q=7$ ${\rm SL}_2(q)$, $p \ge 5$, $q>7$.<|endoftext|> TITLE: Localization principle in supersymmetry QUESTION [5 upvotes]: In $\S$ 9.3 of the book "Mirror symmetry" (Vafa, Zaslow eds.) the authors formulate the following general localization principle for computation of integrals with respect to both even and odd variables: Assume there is some supersymmetry transformation of variables. Then the integral becomes localized on the field configurations for which the fermionic variables are invariant under the supersymmetry. I would like to understand the precise meaning of this sentence. Though several non-trivial and beautiful applications are discussed in the book, I do not understand how they are deduced from this principle due to the fact that the principle itself seems to be too vague. The explanation on pp.158-159 how it works is too concise for me, and it is not clear how that example can be generalized to other situations. Question: how to formulate a precise statement behind this principle (the situation of finite dimensional integrals would be enough for me for the moment)? Can I read a proof of it somewhere? REPLY [2 votes]: I think what you are looking for is Theorem 1 in "Supersymmetry and Localization" by Schwarz and Zaboronsky.<|endoftext|> TITLE: Uniform convergence of convex functions QUESTION [9 upvotes]: It is a well-known result that if a sequence of convex function $f_n(\cdot)$ converges on a dense set $C'$ of an open set $C$, then the limit function $f$ exists on $C$, and the converge is uniform over any compacta within $C$. I am concerned with the uniform convergence around the boundary. In $1$-dimension, the question is a classical analysis: https://math.stackexchange.com/questions/126142/uniform-convergence-of-sequence-of-convex-functions However I don't find any results in higher dimension, i.e. if a sequence of converx function $f_n(\cdot)$ converges to another continuous convex function $f$ pointwise on a compact convex set $D$, can we obtain uniform convergence over the whole region $D$? (Or, under what conditions on $D$ does the uniform convergence over the whole region is valid?) @Pietro Majer: I cannot comment due to my current low reputation...What I am looking at is uniform convergence over the whole compact convex $D$ instead of any compacta within the interior. If we are interested in the latter, then Rockafellar has already established the theory. The interesting part is trying to understand boundary convergence property given that the function $f$ is continuous on $D$. Ideally, I don't want a uniform Lipshitz estimate of the sequence since it is often not dorable in practice.... REPLY [5 votes]: If you have a bound on the uniform norm, say $\|f\|_{\infty, C}\le M$, the sequence has a uniform Lipschitz estimate $ 2M/r$ on the set $C_r \subset C$ of all points with distance at least $r$ from $\partial C$, so by Ascoli-Arzelà a subsequence does converge uniformly on compact sets of the open sets $C$. $$*$$ In general, a uniformly bounded sequence $f_n$ of continuous convex functions on a compact convex set $D\subset\mathbb{R}^2$, converging point-wise to a continuous function, need not converge uniformly on $D$. Consider, on the closed unit disk $D$ $$f_n(x,y):=2n^2\Big(x+\frac{y}{n}-1\Big)_+$$ The convex function $f_n$ vanishes on the whole $D$ but a small circular segment $S_n:=\{f_n>0\}$, cut off by the straight line $x+\frac{y}{n}=1$. Note that $\cap_n S_n=\emptyset$. So for any $u\in D$, we have $f_n(u)=0$ eventually. But this convergence to zero is not uniform, because e.g. on the medium point $(x_n,y_n)$ of the arc that bounds $S_n$, $$x_n:=\frac{n}{\sqrt{n^2+1}},\quad y_n:=\frac{1}{\sqrt{n^2+1}}$$ we have $$f_n(x_n,y_n):=2n^2\bigg(\sqrt{1+\frac{1}{n^2}}-1\bigg)=1+o(1)\, .$$<|endoftext|> TITLE: How to proceed with a type-theoretic proof that $\Sigma \mathbb{S}^1 \simeq \mathbb{S}^2$? QUESTION [10 upvotes]: The circle in homotopy type theory $\mathbb{S}^1$ is a higher inductive type freely generated by the following constructors: $\mathsf{b} : \mathbb{S}^1$ and $\mathsf{loop} : \mathsf{b} = \mathsf{b}$. The sphere $\mathbb{S}^2$ is freely generated by the following constructors $\mathsf{b'} : \mathbb{S}^2$ and $\mathsf{surf} : \mathsf{refl_{b'}} = \mathsf{refl_{b'}}$. We could also define the sphere as the suspension of the circle i.e. $\Sigma \mathbb{S}^1$. I want to show that these two definitions of the sphere, define the same shape up to homotopy i.e. $\Sigma \mathbb{S}^1 = \mathbb{S}^2$ . I plan to do this by univalence, specifically by defining two quasi-inverse functions $f : \Sigma \mathbb{S}^1 \rightarrow \mathbb{S}^2$ and $g : \mathbb{S}^2 \rightarrow \Sigma \mathbb{S}^1$. First I have to construct the functions using the recursor on their respective types. I will first define $f$ using recursion on $\Sigma \mathbb{S}^1$, in order to do this I need to map $\mathsf{N} : \Sigma \mathbb{S}^1$ to some point in $\mathbb{S}^2$ and I have to do the same with $\mathsf{S}$. Finally I have to define a function $m : \mathbb{S}^1 \rightarrow (f(\mathsf{N}) = _{\mathbb{S}^2} f(\mathsf{S}))$. Let's start with the simple definitions first: \begin{equation} f(\mathsf{N}) := \mathsf{b'} \end{equation} \begin{equation} f(\mathsf{S}) := \mathsf{b'} \end{equation} We then have to define $m : \mathbb{S}^1 \rightarrow (\mathsf{b'} =_{\mathbb{S}^2} \mathsf{b'})$, we define it by circle recursion such that: \begin{equation} m(\mathsf{b}) := \mathsf{refl_{b'}} \end{equation} \begin{equation} \mathsf{ap}_m(\mathsf{loop}) := \mathsf{surf} \end{equation} Since we require $\mathsf{ap}_m(\mathsf{loop}) : \mathsf{refl_{b'}} = \mathsf{refl_{b'}}$ and $\mathsf{surf}$ has exactly this type. We have now defined all the data require to have a function $f : \Sigma \mathbb{S}^1 \rightarrow \mathbb{S}^2$. Where I get stuck is in defining $g$. What I've tried is the following definition by sphere recursion: \begin{equation} g(\mathsf{b'}) := \mathsf{N} \end{equation} but then I need a path of the type $\mathsf{refl_{N}} = \mathsf{refl_{N}}$ and I have no idea how to define a non-trivial path of this type. I can get a non-trivial two-dimensional path like this $\mathsf{ap}_{\mathsf{merid}}(\mathsf{loop}) : \mathsf{merid}(\mathsf{b}) =_{\mathsf{N} =_{\Sigma \mathbb{S}^1} \mathsf{S}} \mathsf{merid(b)}$ but I have really no idea, how to turn this path into a path of the type $\mathsf{refl_{N}} = \mathsf{refl_{N}}$. I guess I could use path induction but that seems it would be inelegant and tedious. So I have two questions: 1) Is the $f$ I've defined a good way to prove that $\Sigma \mathbb{S}^1 \simeq \mathbb{S}^2$ via the approach I've outlined above? 2) If so how I define $f$'s quasi-inverse $g$ in the most elegant way? REPLY [4 votes]: I'm not sure if it is the most elegant way, but it is certainly the most direct. So, we need to define the image of $\mathrm {surf}$ as an element of $refl_{\mathrm N} = refl_{\mathrm N}$. We proceed as follows. Let $p \equiv m(\mathrm b) : \mathrm N = \mathrm S$. Consider the inverse path $p^{-1}$. We have a proof (via path induction) that $\psi: p^{-1} \circ p = refl_{\mathrm N}$. We define $g(\mathrm {surf}) \equiv \psi \circ \mathrm{transport}_{p^{-1}} (\mathrm{surf}) \circ \psi^{-1}$. Here $\mathrm{transport}_{p^{-1}}$ maps $p=p$ to $p^{-1} \circ p = p^{-1} \circ p$. Thus by sphere recursion $g$ is defined. It is a bit more tricky to construct the homotopies. As usual, we define them by higher induction. They are trivial to define on points and paths. I denote the corresponding homotopy by $h : g\circ f = \mathrm{id}_{\Sigma S^1}$. We define $h(\mathrm N) \equiv refl_{\mathrm N}$, $h(\mathrm S) = p^{-1}$. We also choose $h(p) \equiv \psi$ (note that the types are correct). Now we need to construct a 2-path lying over $p^{-1}$ and $\psi$ between $m(\mathrm {loop})$ and $g \circ f( m(\mathrm {loop}) )$. Note that by definition $g \circ f( m(\mathrm {loop}) )$ is constructed from $m(\mathrm {loop})$ by transport along $p^{-1}$ and $\psi$ (for loops transport by paths and conjugation by paths is the same) and an element is always connected to its transport by any path by a path lying over it (by definition 6.2.2, essentially). Thus $h$ is constructed. To construct $t: f\circ g = \mathrm{id}_{S^2}$ we define $t(\mathrm b^{\prime}) \equiv refl_{b^\prime}$ and $t(refl_{b^\prime}) \equiv refl_{refl_{b^\prime}}$. To construct a 2-path between $\mathrm{surf}$ and $f\circ g(\mathrm{surf})$, note that $\mathrm{surf} \equiv f(m(\mathrm{loop}))$ and that we have a correctly defined 2-path between $g(\mathrm{surf})$ and $m(\mathrm{loop})$ by construction as above. We define the required 2-path by application of $f$. It is easy to check that its type is indeed correct. Q.E.D.<|endoftext|> TITLE: In a closed monoidal abelian category, are the compact projectives a monoidal subcategory? QUESTION [9 upvotes]: Question: In a closed monoidal abelian category such that the unit object is compact projective, must the tensor product of compact projective objects be compact projective? Recall that an object $X\in \mathcal C$ for an abelian category $\mathcal C$ is compact projective if $\hom(X,-) : \mathcal C \to \mathrm{AbGp}$ preserves colimits. If $\mathcal C$ is monoidal with unit $\mathbb 1$, then $\hom(\mathbb 1,-)$ is a "global sections" functor, so I am assuming that it is exact. Note that if you do not assume that $\mathbb 1$ is compact projective, then there are plenty of examples of closed monoidal categories in which compact projectives do not tensor to compact projectives. For example, let $H$ be any infinite-dimensional Hopf algebra (over a field for convenience), and $\mathcal C = \mathrm{Mod}_H$ its category of (right, say) modules with the usual tensor structure coming from Hopf comultiplication. Then the rank-one free module $H$ itself is compact projective, but Hopfness provides an isomorphism $H \otimes H \cong H^{\oplus \dim H}$, and the latter is not compact projective if $\dim H = \infty$. On the other hand (at least in the case of infinite group algebras and universal enveloping algebras) the trivial module is not compact projective. For comparison, in the category of comodules for a Hopf algebra, if the trivial is compact projective, then the compact projectives are precisely the dualizables. It feels like there should be some trivial argument using the closedness of the monoidal functor. For example, call an object $X \in\mathcal C$ "compact projective over $\mathcal C$" if the inner hom functor $\underline{\hom}(X,-) : \mathcal C \to \mathcal C$ is cocontinuous. (For example, $\underline{\hom}(\mathbb 1,-) = \mathrm{id}$, so $\mathbb 1$ is always compact projective over $\mathcal C$.) Then a hom-tensor adjunction verifies that the compact-projectives-over-$\mathcal C$ are closed under $\otimes$, and that if $X$ is compact projective over $\mathcal C$ and $Y$ is compact projective in the usual sense (i.e. over $\mathrm{AbGp}$), then $X \otimes Y$ is compact projective in the usual sense. In particular, if $\mathbb 1$ is compact projective, then compact projectivity over $\mathcal C$ implies usual compact projectivity, but the converse can fail. Perhaps there is some counterexample... REPLY [7 votes]: In general, the answer to this question is no. Counterexamples can be found by using Day's notion of a promonoidal category, see http://ncatlab.org/nlab/show/promonoidal+category. To give a promonoidal structure on a small (additive) category is equivalent to giving biclosed monoidal structure on the category of (additive) presheaves. The compact projective objects in the latter are well understood, they are precisely the retracts of finite direct sums of representables. For an explicit example, take your domain category to consist of two objects and no non-trivial morphisms. Its presheaf category is simply the category $\mathrm{Ab} \times \mathrm{Ab}$ of pairs of abelian groups $(A,B)$. The representables are $(\mathbb{Z},0)$ and $(0,\mathbb{Z})$, so the compact projective objects in this category are precisely the pairs $(P,Q)$ where both $P$ and $Q$ are projective. For any fixed abelian group $M$ there is a promonoidal structure whose resulting tensor product is given by $(A,B) \otimes (A^{\prime},B^{\prime})=(A\otimes A^{\prime},A\otimes B^{\prime} + B \otimes A^{\prime} + B \otimes B^{\prime} \otimes M)$. This has unit $(\mathbb{Z},0)$ and the associativity isomorphism is built from associator and symmetry of the standard symmetric monoidal structure on $\mathrm{Ab}$ (you can even get it to be strictly associative if you take a skeleton of $\mathrm{Ab}$). Now if you take $M$ to be any abelian group that is not finitely generated projective, then $(0,\mathbb{Z}) \otimes (0,\mathbb{Z}) \cong (0,M)$ is not compact projective.<|endoftext|> TITLE: Hard Lefschetz Theorem for the Flag Manifolds QUESTION [8 upvotes]: In the case of a generalized flag manifold $G/P$, we have an explicit description of their cohomology groups due to Borel.(See herehere for a description.) I would like to know what the hard Lefschetz theorem looks like in this presentation. REPLY [13 votes]: To add to David's answer a bit: The Borel description of $H(G/P)$ is the $W_P$-invariant subring of the $W$ coinvariants, $S^{W_P}_W$, where $W$ is the associated Weyl group and $W_P\subseteq W$ is the parabolic subgroup associated to $P$. As David indicates above, we can choose a Schubert basis for $S^{W_P}_W$, and compute the matrices for the Lefschetz maps in that basis. The Pieri rule allows us to interpret these Lefschetz matrices as so-called ''weighted path matrices'' with respect to the Bruhat order restricted to the set of maximal coset representatives of $W/W_P$. A well known ''lemma'' of Gessel and Viennot is then helpful in computing these determinants (the original paper is available here). These ideas are used in this paper (now published in Journal of Algebra) to give a purely algebraic proof of the hard Lefschetz theorem for $S_W$ where $W$ is any finite Coxeter group, as well as for $S^{W_P}_W$ for certain maximal parabolics $W_P\subseteq W$. One could probably also use these ideas to prove that $S^{W_P}_W$ is hard Lefschetz for any parabolic subgroup, although the Bruhat order for $W/W_P$ is more complicated.<|endoftext|> TITLE: Set of integers having finite intersection with the image of any polynomial of degree $\geq 2$ QUESTION [8 upvotes]: Is there a set $A$ of positive integers such that $\sum_{n \in A} \frac{1}{n} = \infty$, and there is no polynomial $f \in \mathbb{Z}[x]$ of degree at least $2$ which takes infinitely many values in $A$? Added on Feb 16, 2015: Seva answered this question completely. He proved even a great deal more -- namely, that there is a partition of $\mathbb{N}$ into a set $A$ of asymptotic density $1$ of 'non-values of non-linear polynomials', which has finite intersection with the image of any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$, and a set $B$ of asymptotic density $0$ of 'values of non-linear polynomials', which contains all but finitely many positive values taken by any polynomial $f \in \mathbb{Z}[x]$ of degree $\geq 2$. REPLY [14 votes]: There are countably many polynomials with integer coefficients of degree at least $2$; write them all in a sequence as $P_1,P_2,P_3,...\ $ For every $k\ge 1$, the set $A_k$ of all positive integers which are not the values of any of $P_1,...,P_k$ has a divergent sum of reciprocals: $\sum_{a\in A_k} 1/a=\infty$. Consequently, we can find pairwise disjoint, finite subsets $A_k'\subset A_k$ such that $\sum_{a\in A_k'} 1/a>1$. Now let $A:=\cup_{k\ge 1} A_k'$. By the construction, the series $\sum_{a\in A}1/a$ diverges, and for each $k\ge 1$, the image of $P_k$ is disjoint with $A_k'\cup A_{k+1}'\cup\ldots$; hence, has a finite intersection with $A$. Indeed, a slight variation of this argument shows that there is a subset $A\subset{\mathbb N}$ of asymptotic density $1$ such that every non-linear polynomial with integer coefficients represents a finite number of elements of $A$ only. To see this, for every $k\ge 1$ fix an integer $N_k$, and let $B_k$ be the set of all positive integers larger than $N_k$ that are representable by $P_k$. We can choose $N_k$ large enough to have $|B_k\cap[1,x]|<2^{-k} x^{2/3}$ for every positive integer $x$, and then the set $B:=\cup_{k\ge 1} B_k$ will have zero asymptotic density. Consequently, the set $A:={\mathbb N}\setminus B$ will have asymptotic density $1$, and we will also have $A\cap{\rm Im} P_k\subseteq[1,N_k]$ for each $k\ge 1$.<|endoftext|> TITLE: What is obstructing two stably-isomorphic vector bundles from being isomorphic? QUESTION [19 upvotes]: The specific situation is the following: Let $n>0$ be a natural number, let $X$ be a finite CW complex of dimension $n$, and let $\xi_0,\xi_1$ be oriented real vector bundles of rank $n$ over $X$ such that $\epsilon\oplus\xi_0 \cong \epsilon\oplus \xi_1$ (where $\epsilon$ is the trivial rank 1 bundle). Here the vector bundles $\xi_i$ are considered as maps $X\to BSO(n)$ and the operation "$\epsilon\oplus\cdot$" is interpreted as compostion with the natural map $BSO(n)\to BSO(n+1)$. The stable isomorphism asserts that there is a homotopy $H:X\times I\to BSO(n+1)$ between the stabilized bundles, and in oder to determine if $\xi_0\cong \xi_1$ one is lead to the commutative diagram \begin{array}{ccc} X\times\{0,1\}&\to &BSO(n)\\ \downarrow & & \downarrow \\ X\times [0,1]&\to &BSO(n+1) \end{array} where the upper map restricted to $X\times\{i\}$ is $\xi_i$. The obstructions to lifting $H$ in this diagram find themselves in the groups $H^k(X\times[0,1],X\times\{0,1\};\pi_{k-1}(S^n))$, and so (using the suspension iso) the only potentially non-zero obstruction, say $o(\xi_0,\xi_1)$, lies in $H^n(X;\mathbb{Z})$. So the question itself is "Can we identify this obstruction with something familiar?" (Maybe restricting to the case of $X$ a manifold or Poincare complex is easier) In the case where $n$ is even, a guess would be $o(\xi_0,\xi_1)=e(\xi_1)-e(\xi_0)$, the difference of the Euler classes. This is certainly a necessary condition, and at least in the case of $X=S^2$ it is also sufficient. For $n$ odd this guess is definitely not correct: the euler class of any oriented odd-rank real vector bundle is 2-torsion, so must vanish for any rank $n$ bundle over $S^n$; but the tangent bundle of $S^n$ is stably-trivial and not actually trivial for $n\neq 1,3,7$. In this case it's not as clear what a description of the obstruction should be. (Remark: if $rank(\xi_i)>dim(X)$, then $\xi_i\cong\xi'_i\oplus\epsilon^k$ where $\xi'_i$ has rank equal to $dim(X)$ (this can be shown with an obstruction argument using the connectivity of Stiefel manifolds). Then by another obstruction argument one can show that in fact $\xi'_0\oplus\epsilon \cong \xi'_1\oplus\epsilon$, and so the case $rank(\xi_i)>dim(X)$ is handled by the case $rank(\xi_i)=dim(X)$.) REPLY [4 votes]: I know, I am late to the party (as always) but I could provide some more information. In this article I show that two $5$-dimensional vector bundles with $w_2=w_4=0$ over a spin $5$-manifold which are stably isomorphic, are isomorphic if and only if their generalized Kervaire semi-charactersitics agree. Moreover it is possible to show that if $w_4\neq 0$ then the vector bundle is uniquely determined by its stable class. I think, it should be possible to generalize this to higher dimensions.<|endoftext|> TITLE: Embedding the group von Neumann algebra into an injective von Neumann algebra on the same Hilbert space QUESTION [9 upvotes]: Let $\Gamma$ be a discrete group, $\newcommand{\VN}{\rm VN}$ and let $\VN(\Gamma)$ denote its von Neumann algebra, regarded as a subalgebra of ${\sf B}(\ell^2(\Gamma))$. It is well known that $\VN(\Gamma)$ is injective as a von Neumann algebra if and only if $\Gamma$ is amenable. I'm looking for von Neumann subalgebras of ${\sf B}(\ell^2(\Gamma))$ which are injective and contain $\VN(\Gamma)$. Of course, ${\sf B}(\ell^2(\Gamma))$ is one itself, but does the family of such subalgebras have a least element with respect to inclusion? If not, is there something smaller than ${\sf B}(\ell^2(\Gamma))$? Something that sees the group structure more closely, and which is reasonably canonical? I'd be happy with an answer for $\Gamma={\rm SL}(n, {\bf Z})$, or even for $\Gamma$ the free group on two generators. The corresponding question for ${\rm C}^*$-algebras seems to have a nice answer, at least for discrete exact groups, in recent work of Kalantar and Kennedy http://arxiv.org/abs/1405.4359 However, it isn't clear to me if their crossed product construction embeds into ${\sf B}(\ell^2(\Gamma))$. REPLY [13 votes]: Since a von Neumann algebra is injective if and only if its commutant is injective, this is the same as finding injective von Neumann subalgebras of $VN(\Gamma)' \cong VN(\Gamma)$. Maximal injective subalgebras of $VN(\Gamma)$ have been studied quite a bit. In particular, they always exist and for $\Gamma$ a free group, the von Neumann subalgebra generated by one of the group generators gives an explicit example [Popa: Maximal injective subalgebras in factors associated with free groups. Adv. Math., 50 (1983), 27-48.]. For $\Gamma = SL(n, \mathbb Z)$, it's a recent result that the subgroup of upper triangular matrices generates a maximal injective von Neumann subalgebra [Boutonnet, Carderi: Maximal amenable von Neumann subalgebras arising from maximal amenable subgroups, arXiv:1411.4093]. Related to boundaries, I should mention that if $\mu \in {\rm Prob}(\Gamma)$ then the corresponding Furstenberg boundary $(B, \eta)$ is amenable in Zimmer's sense and hence gives rise to an injective von Neumann algebra $L^\infty(B, \eta) \rtimes \Gamma$ [Zimmer, Hypernite factors and amenable ergodic actions, Invent. Math. 41 (1977), no. 1, 23-31.] This crossed product typically does not embed as an intermediate von Neumann algebra between $VN(\Gamma)$ and $\mathcal B(\ell^2\Gamma)$. ($L^\infty(B, \eta) \rtimes \Gamma$ is typically type $III$ while intermediate von Neumann algebras are always type $I$ or type $II$). However, $L^\infty(B, \eta) \rtimes \Gamma$ does always embed between $VN(\Gamma)$ and $\mathcal B(\ell^2\Gamma)$ as an operator system. You can find details on this in the appendix to [Izumi: E0-semigroups: around and beyond Arveson's work. J. Operator Theory 68 (2012), no. 2, 335–363.]<|endoftext|> TITLE: Optimal shape for stabbing balls in $\mathbb{R}^3$ QUESTION [10 upvotes]: I have radius $r < \frac{1}{2}$ congruent balls with centers randomly distributed uniformly within a region, say, within a unit-radius sphere $S$. I shoot a ray/path through $S$, hoping to intersect many balls. The cost of my probe is the length of the ray within $S$; the payoff is the number of balls intersected. My question is: Q0. What is the optimal shape of the ray, as a function of $r$?           For example, above shows $100$ balls of radius $r=\frac{1}{10}$. On the left I shoot a (green) straight-line ray of length $2$, and pierce $3$ (red) balls, for a hit/length ratio of $1.5$. On the right I shoot a spiral ray of length $3.57$ through the same distribution of balls (but displayed rotated), and pierce $4$ balls, for a hit/length of $1.1$. Intuition might suggest that the shape of the ray probe is irrelevant. But this cannot be correct, for a jittery path with deviations much smaller than $r$ is inefficient in that such oscillations increase the path length without increasing the likelihood of intersecting more balls. It may be possible to answer the following 'No' without entirely resolving Q0: Q1. Is the optimal shape in fact a straight-line probe, independent of $r$? An answer 'Yes' to Q1 answers Q0 as well. REPLY [3 votes]: From HenrikRüping's comment, we want to consider the volume of an $r$-neighborhood of the topological arc $\gamma$. Hotelling proved that the volume of an embedded tubular neighborhood of an arc only depends on the length. The $D^2 \times \gamma$ can fail to be embedded due to local or nonlocal self-intersections, both of which decrease the volume. As long as it does not come back near itself or have curvature greater than $1/r$, the $r$-neighborhood of the arc has volume $\pi r^2 |\gamma| + \frac{4}{3} \pi r^3$. There is a generalization due to Weyl for neighborhoods of higher dimensional submanifolds. The surprise is that the volume depends on intrinsic properties of the submanifold and not on the embedding otherwise. H. Hotelling. "Tubes and spheres in n-spaces, and a class of statistical problems." Amer. J. Math., vol. 61 (1939), pp. 440-460. H. Weyl. "On the volume of tubes." Amer. J. Math., vol. 61 (1939), pp 461-472.<|endoftext|> TITLE: Homology equivalence and isomorphism on $\pi_1$ not enough for homotopy equivalence? QUESTION [16 upvotes]: It is a standard consequence of Hurewicz's theorem that a homology eqivalence between simply connected spaces is a weak equivalence (and hence a homotopy equivalence, if the spaces are CW-complexes). What is more, it is even enough to assume that the map is a homology equivalence with local coefficients and an iso on $\pi_1$, see e.g. this question. (On the other hand, a map which is only a homology equivalence does not need to be an isomorphism on $\pi_1$ and hence not a weak equivalence.) What is a concrete example of a map that is an isomorphism on homology with $\mathbb Z$ coefficients, and also an isomorphism on $\pi_1$, but not a weak equivalence? REPLY [22 votes]: Let $X$ be the CW complex obtained from $S^1 \vee S^n$, $n>1$, by attaching an $(n+1)$-cell via a map $S^n\to S^1\vee S^n$ representing the element $2t-1$ in $\pi_n(S^1\vee S^n) \cong {\mathbb Z}[t,t^{-1}]$, so $\pi_n(X)\cong{\mathbb Z}[t,t^{-1}]/(2t-1)\cong {\mathbb Z}[1/2]\subset{\mathbb Q}$. The inclusion map $S^1 \to X$ then induces an isomorphism on homology and on $\pi_i$ for $i TITLE: Does a surjective measurable map induce a surjective pushforward operator? QUESTION [9 upvotes]: I hope it is OK to post a question that is basically the same as the months old currently unanswered question at math stackexchange Suppose X, Y are Polish spaces (without loss of generality, we may assume that X, Y are both the unit interval for the purpose of this question). Let $f: X \to Y$ be a measurable surjective function. Is it true that the pushforward operator $f_*: P(X) \to P(Y)$, where $P(X)$ denotes the set of all (Borel) probability measures on $X$, is also surjective? In other words, does an arbitrary probability measure $\nu \in P(Y)$ lift to some preimage $\mu \in P(X)$ such that $\nu$ is the pushforward measure of $\mu$ under $f$? Remark 1: If we additionally assume that $f$ is continuous and that $X$ is compact, then the answer is positive for this special case, as is shown in: https://math.stackexchange.com/questions/593821/prove-the-existence-of-a-measure-mu Remark 2: As tomasz pointed out in the original thread, we may assume that $f$ is continuous by replacing the topology of X by adding more open sets so that the resulting topology is still Polish but has the same Borel sets. I don't think the resulting topology is guaranteed to be locally compact, let alone, compact or being the usual topology of the unit interval. Remark on the selected answer: The proof in the referenced book proceeds by using the existence of a universally measurable selector (w.r.t. $f$) in the obvious way. REPLY [4 votes]: This holds even for $X$ an analytic space and $Y$ separable metric space. One reference for this result is this book by Doberkat, in Proposition 1.101. Actually, he proves it for subprobabilities, but I think this should go through.<|endoftext|> TITLE: Descent properties of spaces QUESTION [5 upvotes]: I am trying to make sense of what is written in Rezk's draft http://www.math.uiuc.edu/~rezk/i-hate-the-pi-star-kan-condition.pdf In particular, I am referring to Proposition 2.3, which is there stated without proof. According to different models for the homotopy colimit functor on $\text{sSet}^J$, we may or may not find maps $V_j \to \text{hocolim}_J V$ for any $j \in J$ and for any simplicial presheaf $V \in \text{sSet}^J$. FIRST PROBLEM:In any case, it seems to me that it is impossible to get a strict cocone over $V$ with vertex $\text{hocolim}_J V$, as in the case of ordinary colimits. Hence, given a map $E \to \text{hocolim}_J V$ it makes no sense to speak of the functor $U(\cdot):J \to \text{sSet}$ defined by $U(\cdot):=V(\cdot) \times_{\text{hocolim}_J V} E$. Is it possible that the author meant to use as a model the colimit of a cofibrant replacement $QV \to V$ in $\text{sSet}^J$, and so he is thinking of $QV_j \to \text{colim}_J QV\simeq \text{hocolim}_JV$ (thus committing an abuse of language)? In such case the definition of $U$ makes sense, and I think I can get a proof by using point (1) of Thm.1.4 in http://www.math.uiuc.edu/~rezk/rezk-sharp-maps.pdf . SECOND PROBLEM: in the proof of point (1) of Thm.1.4 from the last link (page 14), I can't see why $\tilde{X}_i$ should be the pullback of $\text{colim}X'$ along the composite map $\tilde{Y}_i→ Y_i → \text{colim}Y$, and why it should imply that the right hand square is a pullback. Thanks in advance for any help, which will of course be highly appreciated. REPLY [3 votes]: For the first problem, as I wrote in the comments, the author is implicitly using the language of $(\infty,1)$-categories, where it does make sense to speak of such a functor. To understand why, you can consult any of the many introductions to $(\infty,1)$-category theory. The rest of this answer deals with the second problem. $\newcommand{colim}{\mathop{\mathrm{colim}}} \newcommand{hocolim}{\mathop{\mathrm{hocolim}}}$ A fundamental property of any topos is universality of colimits, which means that colimits commute with base change: $$ (\colim_i X_i) \times_B A \simeq \colim_i (X_i \times_B A) $$ for any morphism $A \to B$. What Rezk calls the "distributive law" (Proposition 3.7) in his paper "Fibrations and homotopy colimits..." is just the special case of universality of colimits where $B = \colim_i X_i$. In any "homotopy topos" or $\infty$-topos, there is the analogous property of commutativity of homotopy colimits with homotopy base change, i.e. $$ (\mathrm{hocolim}_i X_i) \mathop{\times}^h_B A \simeq \mathrm{hocolim}_i (X_i \mathop{\times}^h_B A) $$ for any morphism $A \to B$. In particular this holds in the archetypal example of a homotopy topos: the homotopy theory of simplicial sheaves. As a special case one has a "homotopy distributive law". Then Rezk's Theorem 1.4(1) in "Fibrations..." is a direct application of this homotopy distributive law. Indeed, in the notation of loc. cit., one sees $$\begin{align} \hocolim_i X_i &\simeq \hocolim_i (\colim X \mathop{\times}^h_{\colim Y} Y_i)\\ &\simeq \hocolim_i (\colim X \mathop{\times}^h_{\hocolim Y} Y_i) \\ &\simeq \mathrm{colim} X \end{align}$$ when $\colim Y \simeq \hocolim Y$. For a reference for this in the language of model categories, see paragraph 6.5 in these notes of Rezk (this is the property P1). In fact, the proof of the homotopy distributive law is the same as the proof of the ordinary distributive law as described by Rezk, using as a starting point the claim for spaces instead of for sets. Since your question was rather about the proof in Rezk's older paper, let me try to say something about that. Let me emphasize that I have not really read the paper, so I may say something wrong in what follows. As far as I understood, the idea of the argument there is to use the distributive law for ordinary toposes, plus the explicit description of homotopy colimits of simplicial sheaves due to Bousfield-Kan. For clarity, the proof can be divided into two steps: first, one demonstrates the claim assuming that $\colim X \to \colim Y$ is a sharp map. Then in general one reduces to the above case by factoring this map as a weak equivalence followed by a sharp map. The sharp version is: Lemma: Let $Y : I \to s\mathscr{E}$ be a diagram and let $p : W \to \colim Y$ be a sharp map in $s\mathscr{E}$. Let $X : I \to s\mathscr{E}$ be the diagram defined by $X_i = Y_i \times_{\colim Y} W$ for each $i$. If $\hocolim Y \to \colim Y$ is a weak equivalence, then $\hocolim X \to \colim X$ is a weak equivalence. Proof: First of all, the distributive law says there is an isomorphism $$ \colim_i X_i \approx \colim_i (Y_i \times_{\colim Y} W) \approx W. $$ Using the assumptions that $p$ is sharp and $Y$ is a homotopy colimit diagram, one gets the chain of weak equivalences $$\begin{align} \colim X &\simeq \colim X \mathop{\times}^h_{\colim Y} \hocolim Y \\ &\simeq \colim X \mathop{\times}_{\colim Y} \hocolim Y \\ &\simeq \colim X \mathop{\times}_{\colim Y} \colim \tilde{Y} \\ &\simeq \colim_i (\colim X \mathop{\times}_{\colim Y} \tilde{Y}_i) \\ &\simeq \colim_i \tilde{X}_i \\ &\simeq \hocolim X \end{align}$$ using at the end $\colim X \mathop{\times}_{\colim Y} \tilde{Y}_i \approx \tilde{X}_i$ (this is where one needs to open up the Bousfield-Kan constructions $\tilde{X}$ and $\tilde{Y}$, I believe, making use of the isomorphism $X_i \approx Y_i \times_{\colim Y} \colim X$ from above).<|endoftext|> TITLE: Symmetric L-groups of integral group ring of finite cyclic groups QUESTION [7 upvotes]: Where can i find the results about $L^{\ast}(\mathbb{Z}\pi)$ for $\pi$ a finite cyclic group? REPLY [8 votes]: The symmetric $L$-groups $L^*(Z[\pi])$ for finite cyclic groups $\pi$ have never been computed, perhaps for lack of applications: do you have any? In principle, it is possible to extend the known calculations (mainly due to Wall himself) of the quadratic $L$-groups $L_*(Z[\pi])$ to the symmetric $L$-groups. The failure of 4-periodicity in the symmetric $L$-groups was studied by Gunnar Carlsson in his 1979 paper Desuspension in the symmetric $L$-groups. In his 1985 paper Surgery and the generalized Kervaire invariant Michael Weiss interpreted the relative $L$-groups $\widehat{L}^*(Z[\pi])$ (which are 8-torsion) in the long exact sequence $$\dots \to L_*(Z\pi) \to L^*(Z[\pi]) \to \widehat{L}^*(Z[\pi]) \to L_{*-1}(Z[\pi]) \to \dots$$ as the twisted $Q$-groups $Q_*(B,\beta)$ of the universal chain bundle $(B,\beta)$ of $Z[\pi]$. The twisted $Q$-groups are homological in nature and so much more computable then either the quadratic or the symmetric $L$-groups, which by definition are generalized Witt groups. See Chapters 2 and 9 of my 1992 book Algebraic $L$-theory and topological manifolds for the general theory. Example 9.17 computes $L^0(Z[Z_2])$, which is a start!<|endoftext|> TITLE: A specific Model of ZFC QUESTION [6 upvotes]: In his paper "Some Second Order Set Theory", Joel Hamkins asked whether there is a model of set theory $V$ that is elementary equivalent to $V[G]$, Whenever $G$ is $V$-generic for the collapse of a cardinal $\delta$ to $\omega$? It seems that the existence of such a model will have large cardinal strength. Assume such model exists, then ; 1.There is no switches(a statement $\phi$ in language of set theory such that both $\phi$ and $\neg \phi$ can be forced by set forcing-here by collapse forcing)for collapse forcing over this model. 2. $\diamondsuit\phi \longleftrightarrow \phi \longleftrightarrow\square \phi$(By collapse forcing). 3. $V \models$ MP(Hamkins' maximality principle) Assume such a model exists, my questions are : Question 1. What is valid principle for collapse forcing? Is it modal logic of provability? Question 2. What are other straightforward consequences? Question 3. In general, is this model strange? Is it ugly or no, it's nice? will we have nice consequences or not(not necessarily related to modal logic of forcing)? Question 4. Can this question be asked about other forcing notions, I mean are they interesting? REPLY [12 votes]: In a joint work with Mitchell, which is under preparation, we succeeded to give a full answer to Hamkins-Löwe question. Edit: The paper is ready. See On a Question of Hamkins and Löwe on the modal logic of collapse forcing. The proof uses Radin forcing with interleaved collapses. Also it is shown (by Mitchell) that some large cardinals are needed to obtain a consistent answer.<|endoftext|> TITLE: Maps with Hopf invariant zero are suspensions QUESTION [11 upvotes]: Let $h:\pi_{2n-1}(S^n) \rightarrow \mathbb{Z}$ be the Hopf invariant. I believe that in the same paper that proves his suspension theorem, Freudenthal proved that if $x \in \pi_{2n-1}(S^n)$ satisfies $h(x)=0$, then $x$ is in the image of the suspension map $\pi_{2n-2}(S^{n-1}) \rightarrow \pi_{2n-1}(S^n)$. Observe that the usual Freudenthal suspension theorem says that the map $\pi_{2n-1}(S^n) \rightarrow \pi_{2n}(S^{n+1})$ is a surjection. Can anyone either describe a proof of this or point me in the direction of a modern source for it? The only one I know of is Pontryagin's book "Smooth manifolds and their applications in homotopy theory", where this is Theorem 16, but I find his proof very hard to follow. I guess I would particularly appreciate a proof in the spirit of Pontryagin's proof, though more algebraic proofs are welcome too. REPLY [8 votes]: The integral statement is most generally this: For a based CW space $X$ one has a suspension map $E: X\to \Omega \Sigma X$ and a Hopf invariant $H: \Omega \Sigma X \to \Omega \Sigma (X\wedge X)$ (construction outlined below). The composite $H\circ E$ is canonically null and the sequence $$ X \overset E \to \Omega \Sigma X \overset H \to \Omega \Sigma (X \wedge X) $$ is a homotopy fiber sequence in a range, roughly $3r$, where $r$ is the connectivity of $X$. Your result will follow from this easily (take $X = S^n$). The map $H$ is often constructed as follows: let $JX$ be the reduced free monoid on the points of $X$. Using say the Moore loops model for $\Omega \Sigma X$, one can construct a monoid homomorphism $JX \to \Omega \Sigma X$ that extends the map $E$ using the universal property of the free monoid. A homology calculation shows that this map his a weak equivalence (James did this calculation when $X$ is a sphere). Finally, the proof the above sequence is a fibration in the range roughly thrice the connectivity of $X$ can be deduced from the following four facts: 1) $J_2 X \to JX$ is $(3r+2)$-connected, where $J_2X = X \cup (X \times X)$ is filtration two. 2) The quotient $J_2X/X$ is $X\wedge X$, so we have a cofiber sequence $$ X \to J_2 X \overset q\to X \wedge X $$ 3) By the Blakers-Massey Theorem, the above sequence is a homotopy fibration in the range roughly $3r$. 4) The diagram $$ \require{AMScd} \begin{CD} J_2X@>q>>X\wedge X\\ @VVV @VVV\\ JX@>>H>\Omega\Sigma(X\wedge X) \end{CD} $$ is commutative, where the left vertical maps is the inclusion and the right one is the suspension map for $X\wedge X$.<|endoftext|> TITLE: English translation of Witt's paper on the Mathieu groups? QUESTION [5 upvotes]: Does anyone know of an English (or French) translation of Witt's paper Die 5-fach transitiven Gruppen von Mathieu ? (It's the one in which the Witt design is introduced. Well, I guess.) Here's an auxilliary question. Assuming there is no translation available and I end up writing one (after all it's only 9 pages long and there's "google translate" to help...), where would be a nice place to put it on the Internet? I thought maybe on the arxiv, with the title in the meta-data being "A translation of Witt's paper Die 5-fach...", while the title in the actual PDF file would be a "Mathieu's 5-transitive groups" or something. Thanks, Pierre REPLY [2 votes]: In light of this book http://www.amazon.co.uk/Collected-Papers-Abhandlungen-Ernst-Witt/dp/3642150950 (Collected Papers - Gesammelte Abhandlungen by Ernst Witt) it seems very few (if any) of Witt's papers were translated in English, because only comments are given in English and not the papers themselves. If you translate the article, arXiv will be an excellent choice to put it on the Internet, in my opinion. "At our days arXiv is considered as a synonym of wisdom, scientific freedom, enlightenment and progress!" - from here: http://arxiv.org/abs/1203.6789<|endoftext|> TITLE: A possibly surprising appearance of Fibonacci numbers QUESTION [23 upvotes]: Let $H(n) = 1/1 + 1/2 + \dotsb + 1/n,$ and for $i \leq j,$ let $a_1$ be the least $k$ such that $$H(k) > 2H(j) - H(i),$$ let $a_2$ be the least k such that $$H(k) > 2H(a_1) - H(j),$$ and for $n \geq 3,$ let $a_n$ be the least $k$ such that $$H(k) > 2H(a_{n-1}) - H(a_{n-2}).$$ Prove (or disprove) that if $i = 5$ and $j = 8$, then $(a_n)$ is the sequence $(5,8,13,21,\dotsc)$ of Fibonacci numbers, and determine all $(i,j)$ for which $(a_n)$ is linearly recurrent. REPLY [29 votes]: The statement is true. Write $F_n$ for the $n$-th Fibonacci (my indexing starts at $(F_0, F_1, F_2, F_3, \dots) = (0,1,1,2,\dots)$). We are being asked to show that $$\frac{1}{F_{j+1}} > \sum_{m=F_{j}+1}^{F_{j+1}} \frac{1}{m} - \sum_{m=F_{j-1}+1}^{F_{j}} \frac{1}{m} > 0\ \mbox{for}\ j \geq 6.$$ Computer computations easily check this for $6 \leq j \leq 20$, so we only need to check large $j$. We know that $$H(n) = \log n + \gamma + \frac{1}{2n} + O(1/n^2)$$ where the constant in the $O( \ )$ can be made explicit -- something like $1/12$. So $$\sum_{m=B+1}^C \frac{1}{m} - \sum_{m=A+1}^B \frac{1}{m} = \log \frac{AC}{B^2} + \frac{1}{2} \left(\frac{1}{A}-\frac{2}{B}+\frac{1}{C} \right) + O(1/A^2)$$ where the constant in the $O( \ )$ is something like $1/3$. Now, $F_{j-1} F_{j+1} = F_j^2 \pm 1$. So $$\log \frac{F_{j-1} F_{j+1}}{F_j^2} = \log \left( 1 \pm \frac{1}{F_j^2} \right) = O(1/F_j^2).$$ So, up to terms with error $O(1/F_j^2)$ and a fairly small constant in the $O( \ )$, we are being asked to show that $$\frac{1}{F_{j+1}} > \frac{1}{2} \left( \frac{1}{F_{j+1}} - \frac{2}{F_j} + \frac{1}{F_{j-1}} \right) > 0.$$ Set $\tau = \frac{1+\sqrt{5}}{2}$, and recall that $F_j = \frac{\tau^j}{\sqrt{5}} + O(\tau^{-j})$. Then, up to errors of $O(\tau^{-3j}) = O(1/F_j^3)$, this turns into the inequalities $$\frac{1}{\tau^2} > \frac{1}{2} \left( 1 - \frac{2}{\tau} + \frac{1}{\tau^2} \right) > 0.$$ The latter is obviously true; the former can be checked by calculation. It looks like the same analysis should apply sufficiently far out any linear recursion with solution of the form $G_j = c_1 \theta_1^j + \sum_{r=2}^s c_r \theta_r^j$ where $1 < \theta_1 < 1+\sqrt{2}$ and $|\theta_r|<1$ for $r>1$. (So $\theta_1$ should be a Pisot number.) The inequality $|\theta_r|<1$ makes $\log \frac{G_{j+1} G_{j-1}}{G_j^2} \approx \frac{\max_{r \geq 2} |\theta_r|^j \theta_1^j}{\theta_1^{2j}}$ be much less than $1/G_j \approx 1/\theta_1^j$. The inequality $\theta_1 < 1+\sqrt{2}$ makes $1/\theta^2 > (1/2)(1-2/\theta+1/\theta^2)$. REPLY [8 votes]: I would say that an asymptotic expansion formula for H(n) would suffice. That is, one has $$ H(n)=\log n + \gamma + \frac{1}{2n}+... ., $$ and the error terms are quite controllable. Assuming all i,j,k of the same order of magnitude, you can rewrite the condition $H(k)>2H(j)-H(i)$ as $H(k)-H(j)>H(j)-H(i)$ and thus as $$ \log \frac{k}{j} + (\frac{1}{2k}-\frac{1}{2j})+ \dots > \log \frac{j}{i} + (\frac{1}{2j}-\frac{1}{2i})+ \dots $$ If there would be only the first terms, and no integer restriction, the equality in the above formula would correspond to the geometric sequence: $k/j=j/i$. Now, roughly speaking, you get the Fibonacci sequence, because it is so close to the geometric one (error at $F_n$ is of magnitude $\sim 1/F_n$). Well, you will need some fine analysis to check that for $i=F_{n-1}$ and $j=F_n$ the inequality holds at $k=F_{n+1}$ and does not at $k=F_{n+1}-1$, but the difference between the left hand sides are "quite large" (that is, $1/k$), and you control the error terms as precise as you'd like to (see formula (13) here -- http://mathworld.wolfram.com/HarmonicNumber.html ). The rest should be pure (and not so difficult) technique: just check the $\sim 1/k$ terms for $k=F_{n+1}$ and for $k=F_{n+1}-1$, check that they are of opposite signs, and control the error terms.<|endoftext|> TITLE: What are iterated cobar constructions? QUESTION [5 upvotes]: In Beck's paper "On H-spaces and Infinite Loop Spaces", he states that every algebra over the monad $\Omega^k$$\Sigma^k$ is a $k$-fold loop space. He proves the trivial case k = 0 when this is the identity monad, and says the remaining cases follow from "iterated cobar constructions". I'm hoping someone can elaborate on what exactly is meant by this statement. REPLY [2 votes]: There is a map of mondas $D_k\longrightarrow \Omega^kS^k$ where $D_k$ is the little disks operad (see May's original lecture notes, for example), so any algebra over the target monad is one over the little disks, and hence a $k$-fold loop space.<|endoftext|> TITLE: Which power series have bounded integral coefficients and have an inverse given by a series having bounded integral coefficients QUESTION [31 upvotes]: Let $A=1+\sum_{n=1}^\infty \alpha_nx^n\in\mathbb Z[[x]]$ and $B=\frac{1}{A}=1+\sum_{n=1}^\infty\beta_n x^n$ two mutually inverse power series having bounded integral coefficients (ie. $\vert \alpha_n\vert,\vert \beta_n\vert TITLE: fake $S^{2k}\times S^{2k}$ QUESTION [23 upvotes]: Let $X$ be a fixed closed manifold,$S(X)$ the structure set and $Aut(X)$ the group of self homotopy equivalence of $X$. surgery theory tells us that $\mathcal{M}(X):=S(X)/Aut(X)$ is in bijection with the set of $h$-cobordism classes of manifolds homotopy equivalent to $X$. If $X$ is simply connected and dim$X\geq 5$,then by the $h$-cobordism theorem,$S^{Top}(X)/Aut(X)$ is in bijection with the homeomorphism classes of manifolds homotopy equivalent to $X$. We call manifold $M$ a fake $X$ if $M$ is homotopy equivalent but not homeomorphic to $X$. $S^{Top}(S^{2k+1}\times S^{2k+1})=0$,hence there is no fake $S^{2k+1}\times S^{2k+1}$. For $S^{4k}\times S^{4k}$,we have $S^{Top}(S^{4k}\times S^{4k})\cong\mathbb{Z}\oplus\mathbb{Z}$ and $Aut(S^{4k}\times S^{4k})$ is finite group. This means $\mathcal{M}(S^{4k}\times S^{4k})$ is infinite. How much do we know about fake $S^{4k}\times S^{4k}$? what is the general procedure of constructing fake $S^{4k}\times S^{4k}$? For $S^{4k+2}\times S^{4k+2}$, $S^{Top}(S^{4k+2}\times S^{4k+2})\cong\mathbb{Z}_2\oplus\mathbb{Z}_2$ and $Aut(S^{4k+2}\times S^{4k+2})$ is still finite.since i do not know the action of $Aut(S^{4k+2}\times S^{4k+2})$ on $S^{Top}(S^{4k+2}\times S^{4k+2})$,i have no idea if $\mathcal{M}(S^{4k+2}\times S^{4k+2})$ is trivial or not,so Is there a fake $S^{4k+2}\times S^{4k+2}$? REPLY [7 votes]: You may want to have a look of the following paper by Kreck and Lueck: Topological Rigidity for Non-aspherical Manifolds where they showed that a necessary condition for $S^d\times S^d$ ($d>2$) being Borel (which means $Aut(S^d \times S^d)$ acts on $S^{Top}(S^d \times S^d)$ transitively,or equivalently,there is no "fake" $S^d\times S^d$ in your sense) is that $d$ is odd or $2d+2=2^l$ for some $l$. Now for $S^{4k+2}\times S^{4k+2}$ ($k>0$),we know the necessary condition above is not satisfied,hence,there is some fake $S^{4k+2}\times S^{4k+2}$. For $S^2\times S^2$,it is indeed Borel,i.e.Every manifold which is homotopy equivalent to $S^2\times S^2$ is actually homeomorphic to it.<|endoftext|> TITLE: Concise mathematical definition of the fusion product on the Verlinde ring? QUESTION [7 upvotes]: The Verlinde ring of a (let us say) simply connected simple compact Lie group has as underlying additive group the Grothendieck group of representations of the central extension $\widehat{LG}$ of the loop of $G$ with the 'positive energy' condition. I'm trying to find a concise mathematical definition of the product on this ring, the so-called fusion product. In Freed-Hopkins-Teleman's Loop groups and twisted K-theory III they define an $R(G)$-module structure on this additive group using induction of representations (hence a functorial description). All references I'm reading just mention 'fusion rules', and cite Verlinde, whose 1988 paper is in the journal Nuclear Physics B. Surely we have a more recent discussion along the lines of the FHT construction mentioned above, and not in terms of linear combinations of coefficients of some irreps considered as generators...? REPLY [6 votes]: For any triple $(M_0,M_1,M_\infty)$ of positive energy representations of $\widehat{LG}$ at a fixed level, restriction to small punctured discs produces a canonical (up to some scaling which doesn't matter here) action of the Lie algebra $\mathfrak{g}_X = \Gamma(X, \mathfrak{g} \otimes \mathscr{O}_X)$, where $X$ is the thrice punctured line $\mathbb{P}^1 \setminus \{0,1,\infty\}$. The multiplicity of $M_\infty^\vee$ in $M_0 \boxtimes M_1$ (i.e., the fusion rule) is given by the dimension of the space of coinvariants. Since you are only asking about dimension, you get the same answer from the dual space $\mathrm{Hom}_{\mathfrak{g}_X}(M_0 \otimes M_1 \otimes M_\infty, \mathbb{C})$ of conformal blocks. Looijenga has a brief treatment of the theory.<|endoftext|> TITLE: Symplectic K-theory QUESTION [14 upvotes]: For a ring $R$ consider symplectic K-theory defined as follows: let $\operatorname{Sp}(R) = \lim_n \operatorname{Sp}_{2n}(R)$, let $\operatorname{ESp}(R)$ be the subgroup generated by elementary matrices, let $\operatorname{BSp}(R)$ be a classifying space, let $\operatorname{BSp}(R)^+$ be the result of applying Quillen's $+$-construction (with respect to $\operatorname{ESp}(R)$) and define $\operatorname{KSp}_*(R) := \pi_*(\operatorname{BSp}(R)^+)$. [Is this the right definition?] Is the following statement true (and is there a reference)? Let $\tilde{\operatorname{Sp}}_{2n}(R)$ be the Steinberg group of $\operatorname{Sp}_{2n}(R)$. There is an exact sequence $$ 1 \to \operatorname{KSp}_2(R) \to \tilde{\operatorname{Sp}}_{2n}(R) \to \operatorname{Sp}_{2n}(R) \to 1 $$ for $n$ large enough. [Is the kernel always $\pi_2(\operatorname{BSp}_{2n}(R)^+)$?] What is the right analogue of the ``fundamental theorem of K-theory'', i.e. how can $\operatorname{KSp}_*(R[t])$ and $\operatorname{KSp}_*(R[t,t^{-1}])$ be expressed in terms of $\operatorname{KSp}_*(R)$? So essentially I am wondering what changes if one replaces $\operatorname{GL}(R)$ by $\operatorname{Sp}(R)$ in the definition of algebraic K-theory. I am mostly interested in the case where $R$ is a field. In that case $\operatorname{KSp}_1(R)$, $\operatorname{KSp}_1(R[t])$, and $\operatorname{KSp}_1(R[t,t^{-1}])$ all vanish at least if the characteristic is not $2$. REPLY [9 votes]: Matthias has already pointed out that the map $\widetilde{\mathrm{ESp}}_{2n}(R)\rightarrow Sp_{2n}(R)$ need not be surjective. In fact, the same phenomenon arises already for $\mathrm{SL}_n$ (the image of $\widetilde{\mathrm{E}}(n, R) = \mathrm{St}(n, R)$ is exactly the elementary subgroup $\mathrm{E}(n, R)$ which is typically strictly smaller than $\mathrm{SL}(n, R)$). I would also like to add that from Lavrenov's result it follows that an unstable version of Gersten's formula holds (i.e. $\mathrm{H}_3(\widetilde{\mathrm{ESp}}_{2n}(R), \mathbb{Z}) = \pi_3(B\mathrm{Sp}_{2n}(R)^+)$ for $n\geq 4$). The proof is the same as in the stable case (it is only relevant that $\widetilde{\mathrm{ESp}_{2n}}$ is a universal central extension of $\mathrm{ESp}_{2n}$). There is a theorem by Grunewald, Mennicke and Vaserstein which states the following: If $R$ is a "locally principal" ring (i.e. localization $R_\mathfrak{M}$ at every maximal ideal $\mathfrak{M}\triangleleft R$ is a principal ideal domain) then $\mathrm{K_1Sp}_{2n}(R[x_1,\ldots x_n]) = \mathrm{K_1Sp_{2n}}(R)$, $n\geq 2$. If $R$ is a p.i.d. then $\mathrm{K_1Sp}_{2n}(R[x, x^{-1}])=\mathrm{K_1Sp}_{2n}(R)$, $n\geq 2$. Both statements apply to a rather narrow class of rings (either assumption implies that Krull dimension of $R$ is $\leq 1$). There exist also theorems asserting triviality of $\mathrm{K_1Sp}_{2n}(R[t_1,\ldots, t_n])$ for certain classes of rings (e.g in this paper of Kopeiko such theorem is proved for rings of geometric type).<|endoftext|> TITLE: Applications of cohomology to probability and statistics QUESTION [10 upvotes]: Are there interesting/useful applications of cohomology (and homological algebra in general) to probability and statistics, or information theory? By "interesting/useful", I mean "not merely descriptive", that is, they can actually say something new and not just formalize well known concepts. For example, I have recently found this paper, which addresses dually flat manifolds (and so, indirectly, information geometry). Any other examples I have missed? Thanks! (Feel free to edit tags appropriately.) REPLY [10 votes]: There is a very nice interpretation of entropy as a cohomology class by Baudot and Bennequin which you can read about HERE. In general, I strongly believe that there is an underlying topological content to parts of information theory- as there is information geometry, there will be information topology.<|endoftext|> TITLE: $R^{\dim X-\dim Y}f_{\ast}\omega_X \simeq \omega_Y$ in positive characteristic? QUESTION [7 upvotes]: In Proposition 7.6 of his paper "Higher Direct Images of Dualizing Sheaves", Kollár shows that if $X,Y$ are smooth complex projective varieties and $f:X\rightarrow Y$ is a proper surjective morphism with connected fibers, then there is an isomorphism $R^df_{\ast}\omega_X \simeq \omega_Y$, where $d=\dim X-\dim Y$. I was wondering whether this is true in positive characteristic. By Grothendieck duality one has $$Rf_{\ast}\mathcal{O}_X \simeq Rf_{\ast}R\mathcal{H}om(\omega_X^{\bullet},\omega_X^{\bullet}) \simeq R\mathcal{H}om(Rf_{\ast}\omega_X^{\bullet}, \omega_Y^{\bullet})[-d]$$ so taking 0-th cohomology we get $$f_{\ast}\mathcal{O}_X \simeq \mathcal{E}xt^{-d}(Rf_{\ast}\omega_X^{\bullet}, \omega_Y^{\bullet})$$ From the spectral sequence $$\mathcal{E}xt^p(\mathcal{H}^{-q}(\mathcal{F}^{\bullet}), \mathcal{G}^{\bullet}) \Longrightarrow \mathcal{E}xt^{p+q}(\mathcal{F}^{\bullet},\mathcal{G}^{\bullet})$$ it follows that the RHS is isomorphic to $\mathcal{H}om(R^df_{\ast}\omega_X,\omega_Y)$ and the LHS is just $\mathcal{O}_Y$ by the connected fibers assumption so we have $$\mathcal{O}_Y \simeq \mathcal{H}om(R^df_{\ast}\omega_X,\omega_Y)$$ Finally since $R^df_{\ast}\omega_X$ is locally free we conclude that $R^df_{\ast}\omega_X \simeq \omega_Y$. REPLY [6 votes]: So a paper of Blickle, myself and Tucker handles some questions really closely related to this, see the arXiv version here Ok, throw away the connected fibers hypothesis, I'm going to assume my base field is $F$-finite (although we can also work with integral $F$-finite separated schemes instead of varieties). Suppose that $f : X \to Y$ is a proper surjective morphism of varieties in characteristic $p > 0$. Then we have a map $R^d \pi_* \omega_X \to \omega_Y$. This map is never zero (see Prop 2.13 in the aforementioned paper). The nice thing is that the image of this morphism in $\omega_Y$ always contains the parameter test submodule $\tau(\omega_Y) \subseteq \omega_Y$ (see Prop 2.21 and Definition 2.33). For smooth varieties, $\tau(\omega_Y) = \omega_Y$ (and even for varieties with mild singularities). It's worth remarking that this is in some sense optimal: For non-smooth varieties, we also show that there always exists some $X \to Y$ such that $\tau(\omega_Y)$ is the image of $R^d \pi_* \omega_X \to \omega_Y$. Anyway, the upshot of the above discussion is that if $Y$ is smooth, then $R^d \pi_* \omega_X \to \omega_Y$ is surjective. Obviously without the connected fibers hypothesis, it can't be injective. EDIT: Ok, so then we need the injectivity under the hypothesis of connected fibers (or maybe also $X$ smooth). In an earlier version of this ansewr I was incorrectly speculating that maybe we could use something like the argument in the question to help here. As Sándor points out, that won't work...<|endoftext|> TITLE: How many hamiltonian cycles can be removed from a complete directed graph before it becomes disconnected? QUESTION [5 upvotes]: The question started from a problem brought home by a friend's 5th grader: "How many ways can you seat 5 people around a round table so that the people sitting to the left of any person is different in each seating arrangement?" Here is a snap solution: For seating arrangements of $N$ people consider a directed graph with $N$ vertices and two opposite arcs connecting every pair of distinct vertices. Every hamiltonian cycle in that graph corresponds to a seating arrangement, where a directed edge corresponds to ''sitting to the left''. Call two hamiltonian cycles "intersecting" if they share a directed edge. The problem boils down to finding the maximal set of pairwise non-intersecting hamiltonian cycles. Since every vertex has $N-1$ outgoing edges the upper bound is $N-1$. And here comes the error: Since hamiltonian cycles don't break a complete graph into disconnected components, we conjectured that the same is true for a set of non-intersecting hamiltonian cycles and therefore the above upper bound is exact. However, a small experiment with $N=4$ demonstrates that the above conjecture is not true: removing any 2 hamiltonian cycles from a complete directed graph with 4 nodes breaks it into 2 disconnected components with 2 nodes each. Therefore the answer for $N=4$ is $2$, not $3$, and the above arguments leads only to an upper bound rather than an exact answer. Thus the question: how does one compute the maximum number of non-intersecting hamiltonian cycles in a complete directed graph that can be removed before the graph becomes disconnected? REPLY [10 votes]: I will rephrase your question slightly. Let $K_{n}^{*}$ be the directed graph with $n$ vertices and two oppositely directed edges for each pair of vertices. Your question is then the following. What is the maximum number of edge-disjoint directed Hamiltonian cycles of $K_{n}^{*}$? For $n=2k+1$ odd, it is an old theorem of Walecki that $K_n$ can be decomposed into $k$ Hamiltonian cycles, and hence $K_n^*$ can be decomposed into $2k$ directed Hamiltonian cycles. For $n=2k$ even, you are right to note that for $n=4$ we cannot achieve the upper bound of $n-1.$ One can also check that we cannot achieve the upper bound for $n=6$. However, Tilson proved that for even $n \geq 8$, $K_n^*$ can de decomposed into $n-1$ directed Hamiltonian cycles. This completely answers your question. Namely, $n=4$ and $n=6$ are the only exceptions.<|endoftext|> TITLE: Line bundles over Kähler–Hodge manifolds QUESTION [6 upvotes]: A Kähler–Hodge manifold $M$ can be defined as a Kähler manifold whose Kähler form $\omega$ is integral, namely $\omega\in H^{2}(M,\mathbb{Z})$. It is known then that there always exists a Hermitian line bundle $L\to M$ whose Chern class satisfies $c_{1}(L) = [\omega]$. I have the following questions: Does anyone know a reference where the statements above are proven? How can this line bundle $L$, given a particular integral Kähler form $\omega$, be explicitly constructed? What happens if $\omega$ is not integral? Can one still construct some kind of $\mathbb{R}$-bundle whose curvature is $\omega$? Thanks. REPLY [6 votes]: At least when $M$ is compact, which I assume here, the standard reference for this subject is probably the book Algebraic Geometry, Griffiths & Harris but the following books are excellent references as well : Differential analysis on complex manifolds, Wells Complex Analytic and Differential Geometry, Demailly Hodge Theory and Complex Algebraic Geometry, Voisin A proof could run as follows: Hodge's theorem implies that there exists a complex line bundle $L$ on $M$ with first Chern class $c_1(L)=\omega$; Endow $L$ with an hermitian metric $(L,\|\cdot\|_0)$; its curvature form $c_1(L,\|\cdot\|_0)$ is a $(1,1)$-form cohomologous to $\omega$. By the $dd^c$-lemma, there exists a function $\phi$ such that $\omega= c_1(L,\|\cdot\|_0)+dd^c\phi$; then $\|\cdot\|=e^{-\phi}\|\cdot\|_0$ is a hermitian metric on $L$ with curvature form equal to $\omega$. For an explicit construction of $L$: the standard proof of Hodge's theorem is cohomological, and uses a long exact sequence in cohomology. Tracing down the arguments gives an actual construction if one is given an open covering of $M$ by contractible open sets. The arguments really require that $\omega$ be integral. A generic complex torus of dimension $>1$ carries no complex line bundle, but has many Kähler forms.<|endoftext|> TITLE: Codimension of the set of topologically singular points of an Alexandrov space. QUESTION [5 upvotes]: I am reading Burago, Burago and Ivanov's book A course in metric geometry. In chapter 10 the mention that Alexandrov spaces of curvature bounded below have a stratification into topological manifolds. On further reading, I found that on Perelman's article Elements of Morse theory on Alexandrov spaces that the stratification comes from his result that says that an Alexandrov space is a Multiple Conical Singularities Space (MCS). The stratification is as follows: Let $X$ be an Alexandrov space. The $l$-dimensional stratum is the subset of points $x\in X$ whose conical neighborhood admit a topological splitting $\mathbb{R}^m \times \mathrm{Cone}(\Sigma_{n-m-1})$, where $\Sigma_{n-m-1}$ is a compact MCS-space of dimension $n-m-1$ for all $m\leq l$. In Burago's book the mention that it is possible to prove that there is no codimension $2$ stratum for Alexandrov spaces. This implies that the set of topologically singular points (i.e. points whose space of directions is not homeomorphic to the sphere) which are not boundary points, has codimension at least $3$. Could you help me prove this? I have tried to give a proof by induction on the dimension of $X$ but I am finding some difficulties. The base of the induction would be clear since $1$ and $2$ dimensional Alexandrov spaces are topological manifolds. Then I assume that the affirmation is true for every Alexandrov space of dimension $\leq n$. Now if $X$ has dimension $n+1$, the singular points are not in the top-dimensional stratum since the said stratum is the set of topologically regular points. Therefore it has at least codimension 1. Further if we exclude boundary points, since the closure of the $n-1$ stratum is the boundary, we obtain that the set of topologically singular points which are not on the boundary is at least $2$. Here is were I find trouble. I can't seem to be able to use my induction hypothesis. My intuition tells me that essentially I have to use that the space of directions is an $n$-dimensional Alexandrov space. Could you give me a hint? REPLY [4 votes]: Your intuition is correct. Since the Alexandrov space is locally homeomorphic to the cone on the space of directions, your induction hypothesis implies that the statement is true locally, and hence globally.<|endoftext|> TITLE: Change of time variable in Wiener process QUESTION [7 upvotes]: I'm following a solution of an SDE from here http://www.math.ethz.ch/~delbaen/ftp/preprints/CEV.pdf Start with the SDE $$ dX_t = \delta dt + 2\sqrt{X_t} dW_t $$ consider a deterministic time change $$ \tau = \frac{\sigma^2}{2\nu(2-\delta)}\left(1-\exp\left(-\frac{2\nu t}{2-\delta}\right)\right) $$ the process $Y_t$ is defined as $$ Y_t = exp(\nu t) X_{\tau}^{1-\delta/2} $$ then, using Ito's lemma, we get $$ dY_t = \nu Y_t dt + \sigma Y_t^{\frac{1-\delta}{2-\delta}} dW_t $$ I'd like to understand how the Ito's lemma is applied here. The problem that I have is that when you write it, the $Y$ is derivated with respect to the $X$, but in the definition of $Y_t$ is used $\tau$, not $t$, in $X$, so there's a mixture between different times. How is this handled? REPLY [8 votes]: The time change described in the question may be handled as follows. Recall that if $W(t)$ is a standard Brownian motion then $$ W(\tau(b))-W(\tau(a)) $$ has the same distribution as $$ \int_a^b \sqrt{\tau'(t)} dW(t) $$ where $a \le b$. Therefore the time-changed process $\tilde X_t = X_{\tau(t)}$ satisfies: $$ d \tilde X_{t} = \delta \tau'(t) dt + 2 \sqrt{ \tilde X_{t} } \sqrt{\tau'(t)} dW(t) \;. $$ Now apply Ito's chain rule to $Y_t = \exp(\nu t) \tilde X_t^{1 - \delta/2}$ to obtain $$ d Y_t = \nu Y_t dt + \exp(\nu t) (1-\frac{\delta}{2}) \tilde X_{t}^{-\delta/2} d \tilde X_{t} +\underbrace{\exp(\nu t) \left( 2 ( -\frac{\delta}{2} ) (1-\frac{\delta}{2}) \tau'(t) \tilde X_{t}^{-\delta/2} \right) dt}_{\text{Ito correction to standard chain rule}} \;. $$ Eliminate $\tilde X(t)$ to obtain the desired SDE for $Y_t$.<|endoftext|> TITLE: Different styles of writing/reading articles QUESTION [36 upvotes]: Recently, I discovered a rather unexpected thing. We are writing an article in collaboration and we permanently have some discussions about how to write, in which order, how to organize material etc. Today we have understood that we are reading articles in different maners. I start from the abstract, then I'm reading the introduction where I expect all results stated clearly and the motivation is explained. If I don't understand the introduction, I don't read this acticle. Then, I am reading the text in the article which is kind of "water". Probably, at the end, I start to carefully check the details in theorems and proofs. My friend usually proceeds in an opposite way. He skips the introduction, reads only definitions, propositions and theorems, and some stuff around which he could understand. Then, if he is really interested, he starts to read the usual text. These two approaches result in writing: I care about the introduction, beginings and ends of each chapter, making proofs as short as possible and explaning motivation only in the introduction. I suppose that the reader reads from the beginning till the end. He only cares about all the important thing being stated in propositions and theorems, no matter in which part (in which order) of the paper. He also does not care a lot about the logical structure, but more about motivation explained and repeated. So, what are possible ways to structure an article? Do you normally suppose that the reader reads from the beginning till the end or just skimming? Does it correlate with your writing style? It is a big vague and personal, so I expect also rather personal opinions and strategies. REPLY [5 votes]: I do not think you are asking the right question. I think the question you should be asking is "How should I structure my article, given the possible ways my audience will read it?". The answer is to make it clear, simple and easy to understand. Naturally, this is easier said than done. I think it impossible to understand how each person reads a paper - how are you going to put a measure on that? Take someone like me - I have no fixed structure when reading papers. How I read papers depends on the type of paper, the journal that it is in, the author and most importantly what I am trying to achieve from reading it. Sometimes, I dive straight into the paper and get a high level understanding of what is going on. Then I attempt to replicate the results, only to get confused with something - so I look back and start learning the definitions. Then I start thinking what the point of the paper is - so I look at the abstract. And so on. Other times I make a concise effort to read the abstract, understand the structure and the author's intentions. I can think of three texts that would be impossible (in my opinion) to read if this approach were not taken: Brownian Motion and Stochastic Calculus by S. Shreve. Stochastic Equations in Infinite Dimensions by G. da Prato. Measure Theory Volume I and Volume II by V. Bogachev. It is worth reading those books just to see how the authors structure their content - they know the reader is in for a long road (of torture) and take their time to ease them in - they provide intuition, give examples and make you excited about the subject.<|endoftext|> TITLE: homotopy tensor product of functors and bar construction QUESTION [6 upvotes]: I'm trying to take familiarity with homotopy theory and I have the following questions. Let $\mathcal{C}$ be a small category, and let $F\: : \:\mathcal{C}\to \mathcal{M}$ that take values in a (simplicial) model category. For any simplicial presheaves $K\: : \: \mathcal{C}^{op}\to sSet$, I denote with $$ K\otimes_{\mathcal{C}} F $$ the functor tensor product. Let * be the constant simplicial presheaf, it is easy to note that $$ *\otimes_{\mathcal{C}} F=colim F $$ Now choose a cofibrant replacement P of * in the projective model structure of simplicial presheaves. If we assume that $F$ is pointwise cofibrant, then the homotopy colimit is given by $$ P\otimes_{\mathcal{C}} F $$ A good candidate for the cofibrant replacement for $*$ is the nerve functor $N(-/ \mathcal{C})$. Another strategy to compute the homotopy colimit is given by the bar construction $B(K, \mathcal{C}, F)$ (see for example www.math.harvard.edu/~eriehl/hocolimits.pdf or the book categorical homotopy theory). In particular it is possible to show that there is a natural isomorphism $$ B(*, \mathcal{C}, F)\to N(-/ \mathcal{C})\otimes_{\mathcal{C}} F $$ Thus in some sense the bar construction contains the data of a cofibrant replacement for the simplicial presheaf $*$. Here are my questions: 1) Is it possible to generalize the above facts for a general simplicial presheaf $K\: : \: \mathcal{C}^{op}\to sSet$, i.e there exists a cofibrant replacament QK of K and a natural isomorphism $$ B(K, \mathcal{C}, F)\to QK\otimes_{\mathcal{C}} F? $$ 2) If not, it is possible to find at least a weak equivalence? Another formulation of this question is as follows: since $K\otimes_{\mathcal{C}} F$ may be intepretated as a weighted colimit, does the bar construction compute the homotopy weighted colimits? 3)(i don'think make sense) If $\mathcal{M}$ is not a simplicial category, it is possible to find a relation between the homotopy weighted colimit and the Bar construction? REPLY [5 votes]: Yes. (Of course for these constructions to be homotopically well-behaved you need $F$ to be levelwise cofibrant.) In fact, assuming that such $QK$ exists we can express it by an explicit formula which then proves that it indeed exists. All we need to know is that tensoring over $\newcommand{\C}{\mathcal{C}}\C$ with a functor represented by $x \in \mathcal{C}$ is just evaluation at $x$. Hence $QK_x = QK \otimes_\C \C(x,-) = B(K, \C, \C(x,-))$ and that's how we construct $QK$. Now the formula for $B(K, \C, F)$ holds for general $F$. $$QK \otimes_\C F = B(K, \C, \C(-,-)) \otimes_\C F \\ = |\coprod_{y_0, \ldots, y_m} \C(-, y_0) \times \C(y_0, y_1) \times \ldots \times \C(y_{m-1}, y_m) \times K_{y_m}| \otimes_\C F \\ = |\coprod_{y_0, \ldots, y_m} F \otimes_\C \C(-, y_0) \times \C(y_0, y_1) \times \ldots \times \C(y_{m-1}, y_m) \times K_{y_m}| \\ = |\coprod_{y_0, \ldots, y_m} F y_0 \times \C(y_0, y_1) \times \ldots \times \C(y_{m-1}, y_m) \times K_{y_m}| = B(K, \C, F)$$ Moreover, $QK$ is projectively cofibrant. Indeed, we can explicitly write it as a cell complex in the projective model structure on diagrams $\C^{\mathrm{op}} \to \mathsf{sSet}$. For each $m$ there is a functor $G_m K \colon \C^{\mathrm{op}} \to \mathsf{Set}$ where $G_m K_x$ is the set $$\{ (y, z) \mid y \colon [m] \to x \downarrow \C, z \in K_{y_m}, (y,z) \text{ is a non-deg. simplex of } N(x \downarrow \C) \times K_{y_m} \}$$ and a pushout square $$ \begin{array}{ccccccccc} G_m K \times \partial\Delta[m] & \xrightarrow{} & \mathrm{Sk}^{m-1} QK & \newline \downarrow & & \downarrow & & \newline G_m K \times \Delta[m] & \xrightarrow[]{} & \mathrm{Sk}^m QK .& \end{array} $$ It remains to see that the map $QK \to K$ is a weak equivalence, but this follows from the fact that $\C(x,-)$ is projectively cofibrant since that map coincides with $K \otimes_\C B(\C(-,-),\C,\C(x,-)) \to K \otimes_\C \C(x,-)$ by a computation similar to the one above.<|endoftext|> TITLE: The difference between an étale finite group scheme and a finite group QUESTION [16 upvotes]: I am trying to understand the statement that a Deligne-Mumford stack is locally a quotient $[U/G]$, where $G$ is a finite group. I don't understand why you can make $G$ a finite group, instead of a general group scheme. For example, if $G$ is a nonconstant étale group scheme over $S$, and $\mathcal{X}=[S/G]$ a quotient stack over $S$, how to make $\mathcal{X}$ a group quotient? Intuitively, I feel the difference between a finite group and a finite étale group scheme is like the difference between a constant sheaf and a local system. But I am not sure whether it is true. Hope someone can help me to understand these group scheme things. It would be helpful if you can provide some reference. REPLY [30 votes]: The way I think about finite etale group schemes over a connected scheme S, is that it's a group object in the category of schemes finite etale over S. This latter category is well-known to be equivalent to the category of finite sets X together with a action of $\pi_1(S,s)$ (pick some geometric point $s\in S$). This is described in many sources, though the canonical reference would probably be SGA 1, Expose V. Thus, a finite etale group scheme is just a finite (abstract) group $G$, together with an action of $\pi_1(S,s)$ acting on $G$ via group automorphisms. Expose XI of SGA 1 also talks a little about group schemes. A finite group is just the group $G$ without the action of $\pi_1(S,s)$. Of course, you can always have $\pi_1(S,s)$ act on any group trivially, giving $G$ the structure of a constant group scheme over $S$ (ie, just $|G|$ disjoint copies of $S$, corresponding to a trivial degree $|G|$ cover of $S$), so in that sense any abstract group can be thought of as a constant group scheme. As an example, you can consider the difference between $\mathbb{Z}/N\mathbb{Z}$ (integers mod $N$) and $\mu_N$ ($N$th roots of unity). They can both be thought of as group schemes over $\mathbb{Q}$, but whereas the former doesn't have a natural action by $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, the latter does. Indeed, usually people think of $\mu_N$ over $\mathbb{Q}$ as $\text{Spec }\mathbb{Q}[x]/(x^N-1)$, and $x^N-1$ will only factor into degree 1 polynomials over $\mathbb{Q}$ if $N \le 2$. When $N\ge 3$, the irreducible factors of degree $> 1$ correspond to connected components of the cover $\mu_N\rightarrow \text{Spec }\mathbb{Q}$ which themselves are nontrivial covers of $\mathbb{Q}$, also showing that $\mu_N$ is nonconstant. The action of $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ can be seen as acting on the geometric fiber of $\mu_N$ over $\overline{\mathbb{Q}}$, where you can see explicitly that the action is just the typical action sending primitive $N$th roots of unity to other primitive $N$th roots of unity. Contrast this with the situation over $\mathbb{C}$, where for any $N$, $x^N-1$ will factor into linear polynomials (corresponding to the fact that $\pi_1(\text{Spec }\mathbb{C})$ is trivial), making $\mu_N/\mathbb{C}$ the trivial degree $N$ cover, hence constant. A DM stack can be thought of as locally being $[U/G]$ ($G$ a finite group) because etale locally, any finite etale group scheme is constant (and quotienting out by a constant group scheme is the same as quotienting by a group - think about what the quotient means on functors of points). To see that every finite etale group scheme is locally constant, note that an action of $\pi_1(S)$ on $G$ corresponding to some group scheme $\mathcal{G}$ is the same as a homomorphism $\pi_1(S)\rightarrow S_{|G|}$. The kernel of this homomorphism is a finite index subgroup of $\pi_1(S)$, corresponding to a scheme $T$ finite etale over $S$. The pullback of this homomorphism gives an exact sequence $$\pi_1(T)\rightarrow\pi_1(S)\rightarrow S_{|G|}$$ since $T$ was defined by the property that $\pi_1(T)$ is the kernel, so the composition is 0. Thus the homomorphism $\pi_1(T)\rightarrow S_{|G|}$ thought of as the pullback of the homomorphism $\pi_1(S)\rightarrow S_{|G|}$ to $T$ corresponds to the pullback of $\mathcal{G}$ to $T$, which is constant because the associated $\pi_1(T)$-action is trivial.<|endoftext|> TITLE: Asymptotic behavior of $X_n$ in a Dirichlet vector $(X_1, ..., X_n)$ QUESTION [5 upvotes]: Let $(\alpha_k)$ be a sequence of positive numbers and let $(Y_k)$ be a sequence of independent random variables $Y_k \sim \text{Gamma}(\alpha_k,1)$. Set $X_n=\dfrac{Y_n}{\sum_{i=1}^nY_i}$. (edit) The title is not appropriate: $(X_1, ..., X_n)$ is not a Dirichlet vector, because $X_k=\dfrac{Y_k}{\sum_{i=1}^kY_i} \neq \dfrac{Y_k}{\sum_{i=1}^nY_i}$. Sorry for the confusion. First question: Is it possible that these two conditions simultaneously occur: $\sum X_n < +\infty$ almost surely? $\sum \mathbb{E}[X_n] = \infty$? (edit) Answer to question 1: No. One has $\sum_{i=1}^nY_i =\dfrac{1}{\prod_{i=2}^n(1-X_i)}$, hence $\sum X_n < +\infty$ $\iff$ $\sum Y_n < \infty$ $\iff$ $\sum \alpha_n < \infty$ (because $\sum Y_n \sim \text{Gamma}(\sum \alpha_n,1))$. Now, $E[X_n]=\dfrac{\alpha_n}{\sum_{i=1}^n\alpha_i}$, and by a similar reasoning, $\sum E[X_n] < \infty$ $\iff$ $\sum \alpha_n < \infty$. Second question: Is it possible that these three conditions simultaneously occur: $\sum X_n = +\infty$ almost surely? $\sum X_n^2 < \infty$ almost surely? $\sum(\mathbb{E}[X_n])^2 = \infty$? REPLY [2 votes]: The answer to the second question is no too. Set $D_k^n=\dfrac{Y_k}{\sum_{i=1}^nY_i}$. The neutrality property of Dirichlet vectors says that $D_n^n=:X_n$ is independent of $(D_1^{n-1}, \ldots, D_{n-1}^{n-1})$. But $X_k=\dfrac{D_k^{n-1}}{D_1^{n-1}+\ldots+D_k^{n-1}}$ for every $k=1, \ldots, n-1$, therefore $(X_n)$ is a sequence of bounded independent random variables. Then, by a variant of Borel-Cantelli's lemma, $\sum X_n^2 < \infty$ $\iff$ $\sum E(X_n^2)<\infty$. But $E(X_n^2)\geq E(X_n)^2$, hence the second and third bullets are opposite.<|endoftext|> TITLE: Super-cobordisms QUESTION [20 upvotes]: One can construct the $d$-dimensional bordism category by declaring the objects to be the $(d-1)$-dimensional compact manifolds without boundary and the morphisms the $d$-dimensional bordisms between them. Call it $\mathcal{Cob}_d$. It is well known that the connected components of the geometric realization of this category are in one-to-one correspondence with $\pi_d(MO)$, where $MO$ is the Thom spectrum for the orthogonal group. This is the classical Thom-Pontryagin theorem. One can think of constructing a similar category with supermanifolds. Namely, $\mathcal{Cob}_{(d|k)}$ is the category whose objects are $(d-1|k)$-dimensional supermanifolds and the morphisms the $(d|k)$-dimensional bordisms. Does anyone know of a Thom-Pontryagin type result for this category? Is there a spectrum $MO_{|k}$, whose homotopy groups recover the connected components of the geometric realization of $\mathcal{Cob}_{(d|k)}$? REPLY [16 votes]: There are a number of technical issues with making what you describe precise, for example: what precisely is a supermanifold with boundary? how can you glue/compose bordisms? etc. I am going to ignore these technicalities because I don't think it makes much of a difference to your question. Many of these technical issues have been addressed by other people, for example I would suggest looking at the work of Stolz-Teichner and the reference therein to see what sort of things people have tried to do. In any case in the smooth category we have: Batchelor's theorem: every $(d|k)$-dimensional supermanifold is (non-canonically!) of the form $\pi E$ for $E$ a rank $k$ vector bundle over a $d$-manifold. Moreover the isomorphism class of the vector bundle is uniquely determined by the super manifold. Here $\pi E$ is the super manifold whose ring of functions is the global sections of the exterior algebra bundle $\wedge^* E^*$. So the morphisms from $\pi E$ to $\pi E'$ come from all algebra maps and are more than just the vector bundle maps (which correspond to homogeneous algebra maps). If $(X, \mathcal{O}_X)$ is a super manifold, the vector bundle $E$ can be obtained by considering $\mathcal N / \mathcal N^2$ where $\mathcal N$ is the subsheaf of $\mathcal{O}_X$ of nilpotent elements. So there are functors in both directions but there is no natural isomorphism from the identity functor on supermanifolds to $\pi(\mathcal N / \mathcal N^2)$. This is what the "non-canonical" means. But we still get a bijection between isomorphism classes of supermanifolds and isomorphism classes of manifolds with vector bundles. Thus, unless you do something more fancy like work in families, when you pass to bordism classes you will just get the bordism group of $d$-manifolds equipped with rank $k$-vector bundles. This has two names: $$ \pi_d( MO \wedge BO(k)) = MO_d(BO(k))$$ either as the d-th homotopy group of the smash of the spectrum $MO$ and space $BO(k)$ (which is also another Thom spectrum) or equivalently as the d-th $MO$-homology group of $BO(k)$. So that identifies the bordism group. I think it is a very interesting question whether the whole Pontryagin-Thom construction can be carried out inside the world of supermanifolds, but that is a different question.<|endoftext|> TITLE: Interesting integral QUESTION [24 upvotes]: Numerical evidence shows the validity of the following identity $$\int\limits_0^z\frac{xdx}{\sin{x}\sqrt{\sin^2{z}-\sin^2{x}}}=\frac{\pi}{4\sin{z}}\ln{\frac{1+\sin{z}}{1-\sin{z}}},\tag{1}$$ if $0< z< \pi/2$. How can it be proved? An indirect proof can be found in the paper http://link.springer.com/article/10.1134%2FS1547477113010044 (Potential of multiphoton exchange in the scattering of light charged particles of a heavy target, by Yu.M. Bystritskiy, E.A. Kuraev and M.G. Shatnev). Formula (1) is equivalent to (12) from the paper where it is called "the marvelous identity". REPLY [16 votes]: This integral is due to Lobachevskii. He gave it in more general form as follows which can be found in his work "Application of imaginary geometry to certain integrals" (1836). Also see equation 3.842.7 in Gradsteyn and Ryzhik. The integral in OP's post follows from this formula in the limit $~\alpha\to i\infty$.<|endoftext|> TITLE: What is known about primes of the form $x^2-2y^2$? QUESTION [16 upvotes]: David Cox's book Primes of The Form: $x^2+ny^2$ does a great job proving and motivating a lot of results for $n>0$. I was unable to find anything for negative numbers, let alone the case I am interested in, $n=-2$. What is the reason for this? Maybe I am missing something. Any references to results involving primes of this form will be greatly appreciated. REPLY [4 votes]: Ummm. Just so you know, the same type of conclusion holds for $$ x^2 - p y^2 $$ for prime $p,$ $$ 5 \leq p \leq 197, \; \; \; \; p \equiv 1 \pmod 4. $$ For that matter, one may switch to the forms $$ x^2 + xy - \left( \frac{p-1}{4} \right) y^2 $$ For these forms, a number $n$ is represented if and only if $-n$ is represented. There is a solution to $x^2 - p y^2 = -1,$ a result in Mordell's book. Since every odd prime $q$ that satisfies $(p|q) = 1$ is represented by some form of the discriminant, and there is only one class of this discriminant, then all odd primes $q $ with $(q|p) = 1$ are integrally represented. Representation of the prime $2$ is a different matter, as we need $(2|p) = 1,$ so this works only when we further demand $p \equiv 1 \pmod 8.$<|endoftext|> TITLE: Can Enriques Surfaces have non-trivial TWISTED Fourier-Mukai partners? QUESTION [12 upvotes]: It is a well-known fact that for an Enriques surface $Y$, if $D^b(Y)\cong D^b(X)$ for some smooth projective variety $X$, then $X\cong Y$. In other words, Enriques surfaces have no non-trivial Fourier-Mukai Partners. I was wondering if the answer to this question is known if we consider twisted derived categories instead. In particular, what is known about $(X,\alpha)$ such that $D^b(Y)\cong D^b(X,\alpha)$, where $Y$ is an Enriques surface as above and $\alpha$ is a Brauer class on the smooth projective variety $X$. REPLY [8 votes]: There is a natural pair $(X,\alpha)$ that you can construct. Only in this pair $X$ is not a smooth projective variety but is a smooth orbifold surface. If you choose a genus one pencil $Y \to \mathbb{P}^{1}$ on the Enriques surface, then the relative Picard stack $\mathcal{X} = \mathcal{P}ic^{0}(Y/\mathbb{P}^{1})$ of the pencil is a Fourier-Mukai partner. The stack $\mathcal{X}$ is a $\mathbb{G}_{m}$ gerbe over an orbifold surface $X$. The moduli space of $X$ is the relative Jacobian $J$ of the chosen genus one fibration. $J$ is a rational elliptic surface, and $X$ is $J$ equipped with a $\mathbb{Z}/2$ orbifold structure along two smooth fibers of the anticanonical map $J \to \mathbb{P}^{1}$, i.e. along the two fibers dual to the double fibers of $Y \to \mathbb{P}^{1}$. The gerbe $\mathcal{X} \to X$ gives you a non-trivial $2$-torsion element $\alpha \in Br(X)$ and the derived category of weight one sheaves on $\mathcal{X}$ is equivalent to the $\alpha$-twisted derived category of $X$. So you get an equivalence $D^{b}(Y) \cong D^{b}(X,\alpha)$.<|endoftext|> TITLE: Geodesics on manifolds with boundary QUESTION [7 upvotes]: Let $(M,g)$ be a Riemannian manifold with non-empty boundary. Is there any notion of injectivity radius on $(M,g)$ in points away from the boundary? By this I mean points lying in $M- \partial M$. What about geodesics or the exponential map in points away from the boundary? I am actually intrested in the description of the injectivity radius if there is one. Do you know a good reference? Hope for answers. Greetings, Phillip REPLY [10 votes]: You might want to have a look at the following paper and its references: MR1226829 (94c:53053) Alexander, Stephanie B.(1-IL); Berg, I. David(1-IL); Bishop, Richard L.(1-IL) Cut loci, minimizers, and wavefronts in Riemannian manifolds with boundary. Michigan Math. J. 40 (1993), no. 2, 229–237. 53C22 (53C20) This has some basic information about geodesics on manifolds with boundary that you might find good for starting out in finding out what is known on these questions.<|endoftext|> TITLE: Axiom of choice and the equality between second-order constructible universe and HOD QUESTION [6 upvotes]: I try to prove $L_{SO}=\mathrm{HOD}$, where $L_{SO}$ is second-order constructible universe which has similar definition with $L$ but it uses second-order definability rather than the first-order definability, and I found the answer in MO. Also, I found the referred article in the answer which is written by Myhill and Scott. The proof of $L_{SO}=\mathrm{HOD}$ in the answer mentioned previoisly and the article uses axiom of choice. (Exactly, it uses trichotomy for cardinals and they use it to prove $\mathrm{HOD}\subset L_{SO}$). My question is: using the axiom of choice is essential to prove $\mathrm{HOD}\subset L_{SO}$? Thanks for any information or clarification. REPLY [8 votes]: The equality $L_{SO}=HOD$ can not be proved just in $ZF$. This is proved in the paper ``The consistency of the theory $ZF+L^1\neq HOD$'' by Szczepaniak. Here $L^1$ refers to what you named $L_{SO}$. The idea of the proof is as follows: $(1)$ If two models of $ZF$ have the same sets of ordinals, then they have the same classes $L^1,$ $(2)$ There are models $N_1 \subset N_2$ with the same sets of ordinals, such that there is a real $a\in N_1$ such that $a\notin HOD^{N_1}$ but $a\in HOD^{N_2}.$ Now the result follows from $(1)$ and $(2)$. The paper can be find here<|endoftext|> TITLE: A question on Collatz's conjecture:proportion of "low flying" orbits QUESTION [22 upvotes]: Let $C$ : ${\mathbb N}\longrightarrow {\mathbb N}$ be Collatz's map defined by $C(n) = 3n+1$ if $n$ is odd, and $C(n)=n/2$ if $n$ is even. Then according to Collatz's conjecture, we should have $C^k (n)=1$, for all $n>0$ and $k$ large enough. Assume that the conjecture holds and for $n>0$ define the top of the orbit of $n$ by $$ t(n) = \text{max} \{ C^k (n), \ k\geq 0\} $$ Define $$ T(N) = \text{ Card}\{ n\ , \ t(n) \leq N\} /N $$ Had the assymptotic behaviour of T(N) been studied ? REPLY [7 votes]: (Without any intended answer:) Two further pictures. First, a bit more detailed curve up to $N=100 000$. I supressed the datapoints $T(N)$ at $N$, where the frequencies are below $6$ here to have the curve a bit more sharpened/showing the peaks in the underlying frequencies-list: And it seems to me, that the actual frequency of occurence of N as maximum of a trajectory gives some more intuition what's going on here. While the above cumulative curve tends to "iron out" any variability when going to the right, the systematic spikes of the frequencies-table here indicate, that such spikes do not wear out in the long run but (likely) even get higher counts (again I left out the datapoints of smaller frequencies in the picture):<|endoftext|> TITLE: When are toral orbits in buildings the difference of fixed-sets? QUESTION [6 upvotes]: Let $L$ be a $p$-adic field, let $G$ be a reductive group over $L$ (I'm even okay assuming semisimplicity for now). Let $T$ be a maximal torus of $G$. Let $B$ be the building for $G(L)$. (Edit 1: "split" has been removed from "maximal torus"; the building of $U_3(L/F)$, where $L/F$ is a ramified quadratic extension of $p$-adic fields, provides a counterexample to the question if we require $T$ to be split). I'm interested in the relationship between, on the one hand: $T(L)$-orbits in $B$ and, on the other hand: fixed sets $B^{T_0}$ where $T_0 \subseteq T(L)$ is an open-compact subgroup. Clearly, since $T_0$ is commutative, any set $B^{T_0}$ is closed under multiplication by $T$ and is therefore a union of $T(L)$-orbits. It is, of course, too much to hope that every $T(L)$-orbit is of the form $B^{T_0}$ for some fixed compact-open subgroup $T_0$ of $T(L)$, for two reasons: the first is that all elements of a $T(L)$-orbit are of the same type, which is not true of sets of the form $B^{T_0}$. The second is that for any open-compact subgroup $T_0$, $B^{T_0}$ contains the apartment of $T(L)$ (and possibly contains more). My question is as follows: Let $X$ be a $T(L)$-orbit in $B$, and let $Typ(x) = t$ for every $x\in X$. When are there compact-open subgroups $T_0 \subseteq T_0'$ such that $X = (B^{T_0} - B^{T_0'})\cap Typ^{-1}(t)$? and, more, specifically: For a reductive (semisimple) group over a global field $F$, are there only finitely many primes at which the above fails? Let me give a little bit of clarity by way of example. Let's take $G = SL_2$, let $T$ be the diagonal torus, and let $A_0$ be the standard apartment. Then one can check by direct computation that two vertices $v,\, v'$ are in the same $T$-orbit if they are of the same type and if $d(v, \,A_0) = d(v', \,A_0)$ (i.e. they are the same distance from the standard apartment). On the other hand if define the open-compact subgroups $$T_{r} := \bigg\{\begin{pmatrix} t & 0 \\ 0 & t^{-1}\end{pmatrix}: t \in 1 + \varpi^r \mathcal{O}_L\bigg\}.$$ Then if the residue characteristic of $L$ is at least $3$, the fixed set $B^{T_r}$ consists of all vertices that are at most distance $r$ from the standard apartment $A_0$. As such, in the case where the residue characteristic is $> 2$, we can answer in the affirmative; if an orbit $X$ consists of vertices that are distance $r$ from $A_0$, then we can write $X = B^{T_r} - B^{T_{r-1}}$. On the other hand if the residue characteristic of $L$ is $2$, and $T_0$ is the maximal compact subgroup of $T_0$, then by the notation above $T_0 = T_1$ so $T_0$ fixes the standard apartment and all vertices of distance $1$. Therefore, we cannot write the standard apartment as a difference of fixed sets. This question is extremely open-ended, and I'm really interested in anything that can be said in this area. Even some illumination of examples where $G$ is split-rank $1$ (so $B$ is a tree) beyond $SL_2$ would be of great interest to me. REPLY [4 votes]: (Editted: a "weaker" example about $GL_3$ at the end) If I didn't make a mistake in my computation, then the second question doesn't hold for $G=Sp_4$, as it doesn't hold for any $\mathbb{Q}_p$. Allow me to use $F$ as my p-adic field and $k$ its residue field. Let $V/_F$ be spanned by $e_2,e_1,f_1,f_2$ with the symplectic form $(e_i,f_j)=\delta_{ij}$, $(e_i,e_j)=(f_i,f_j)=0$, so that we identify $G(F)=Sp(V)$. The hyperspecial vertices on the building correspond to self-dual lattices of $V$. Let $T$ be the diagonal torus which acts on $V$ by $(c_1,c_2):e_i\mapsto c_ie_i, f_i\mapsto c_i^{-1}f_i$. Let $\pi\in\mathcal{O}_F$ be a fixed uniformizer. Consider the lattices $$\Lambda_{a,b}=\mathcal{O}_F\langle e_2+\pi^{-1} e_1+\pi^{-2}af_1+\pi^{-3}bf_2, e_1+\pi^{-1}f_1+\pi^{-2}(a-1)f_2,f_1-\pi^{-1}f_2,f_2\rangle,$$ where we let $a,b$ runs over a fixed set of representatives of $k$ in $\mathcal{O}_F$. One checks that all $\Lambda_{a,b}$ are in different $T(F)$-orbits by essentially showing that they are in different $T(\mathcal{O}_F)$-orbits. Next one compute the stabilizer of $\Lambda_{a,b}$. What's necessary then is to check the following: Claim. (1) $\text{Stab}_{T(\mathcal{O}_F)}(\Lambda_{a,b})\supset T_3=T(1+\pi^3\mathcal{O}_F)$. (2) If $ab+b-a+1\not\in\pi\mathcal{O}_F$, then $\text{Stab}_{T(\mathcal{O}_F)}(\Lambda_{a,b})\subset T_2=T(1+\pi^2\mathcal{O}_F)$. However, there are only $q+3$ ($q=\#k$) subgroups between $T_2$ and $T_3$. In other words, we have $q^2-q+1$ such lattices, but only $q+3$ possible choices for stabilizers for them; an open compact subgroup of $T(F)$ have to correspond to about $O(q)$ orbits. Note. I did some brute force (using Iwasawa decomp.) and it seems that such an assertion holds for $GL_3$ in most or all cases of $GL_3$. However it also seems to me that if one fix a (semisimple) split rank, then it will be difficult for similar assertion to holds for larger $\dim(G)$; in other words, I'd say such an assertion probably only holds for $GL_n$ (and not for any other classical groups or non-split groups), if it does. -- Editted: Here is an example about $GL_3$ where things fail. Let $T$ again be the diagonal torus. Let $a>0$ be any integer. Consider the two lattices in $F^3$ $$\Lambda_a=\mathcal{O}_F\langle e_1+\pi^{-a}e_2+\pi^{-a}e_3,e_2,e_3\rangle$$ $$\Lambda_a'=\mathcal{O}_F\langle e_1+\pi^{-a}e_3,e_2+\pi^{-a}e_3,e_3\rangle$$ These two lattices correspond to two hyperspecial vertices which are not in the same $T(\mathcal{O}_F)$-orbit and thus $T(F)$-orbit. But they have the same stabilizer $Z(F)\cdot T(1+\pi^a\mathcal{O}_F)$. This is however a weaker example because in this case a stabilizer correspond to $2$ orbits. It's probably the case that for $GL_3$ every possible stabilizer corresponds to at most $2$ orbits, but I have no good intuition about why it should be true. (Note that the number of orbits with distant $r$ from the apartment no longer have a uniform bound independent of $p$.)<|endoftext|> TITLE: Survey of Engineering Problems for Mathematicians QUESTION [6 upvotes]: I am looking for survey-books on open math (esp. probability) problems from engineering fields but phrased in mathematical language. There are hundreds of specialized math-engineering books out there, but I haven't found an introductory survey of math-engineering problems. Given the diversity of problems in engineering, feel free to post survey books of specific engineering fields eg. "survey of math problems in civil engineering". The main fields are: chemical engineering, civil engineering, electrical engineering, mechanical engineering and systems engineering. So a survey book from each, will be nice. One excellent related post is Computer Science for Mathematicians . Thank you PS Even though engineering is diverse, there are certain mathematical problems that keep repeating such as removing noise, estimating integrals etc. This is the kind of list I had in mind. Math people are usually far removed from engineering problems so at least having a sense of the themes will be highly beneficial to both math and engineering. If there is no such survey, then at least list of big open math problems from various engineering fields is good enough. Mathoverflow is a great place to have this list since many people visit it and can add their suggestions. Who knows maybe someone will get inspired to compile a list of the themes in mathematical-engineering problems. REPLY [4 votes]: There is a (perhaps somewhat dated) nice little book devoted to open problems in communications (coding and information theory), system and control theory, and computer science where open problems are motivated. Some of these problems were actually solved by the participants at the 3 workshops in the 1980s, on which this book was based. Open Problems in Communication and Computation, edited by Thomas M. Cover, B. Gopinath. http://www.amazon.com/Problems-Communication-Computation-Thomas-Cover/dp/1461291623 If your institutions subscribes to Springerlink, you may be able to access this title in softcopy.<|endoftext|> TITLE: Is $ACA_0$ + `True Arithmetic exists' interpretable in $ACA$? QUESTION [7 upvotes]: Maybe someone here can help me with a question concerning second-order arithmetic. Consider the system $ACA_T := ACA_0 + \exists X \forall x (x \in X \leftrightarrow T(x))$, where $T(x)$ is a $\Pi_1^1$ truth definition for first-order arithmetic (as given e.g. in Takeuti, Proof Theory, p. 183-188). So $ACA_T$ is $ACA_0$ plus a statement that says that the set of (Gödel numbers of) of true arithmetic exists. The question I have is: Is $ACA_T$ interpretable in $ACA$? I know that one can show that the arithmetical consequences of $ACA_T$ are precisely the consequences of $ACA$. But I would like to know whether $ACA_T$ can be directly interpreted in $ACA$. REPLY [8 votes]: $\def\aca{\mathit{ACA}}\aca_T$ is finitely axiomatized, hence whenever it is interpretable in some theory, it is also interpretable in its finite subtheory. On the other hand, $\aca$ has full induction, hence it is (uniformly essentially) reflexive, that is, it proves the consistency of all its finite subtheories. It follows that if $\aca$ interpreted $\aca_T$, it would actually prove the consistency of $\aca_T$. (The converse also holds.) Thus, assuming that your claim that $\aca$ and $\aca_T$ have the same arithmetical consequences is correct, the answer is negative, as $\aca_T$ does not prove its own consistency by Gödel’s theorem.<|endoftext|> TITLE: Primitive recursive arithmetic via universal algebra QUESTION [7 upvotes]: From the Wikipedia article on Primitive recursive arithmetic: "Primitive recursive arithmetic, or PRA, is a quantifier-free formalization of the natural numbers. It was first proposed by Skolem[1] as a formalization of his finitist conception of the foundations of arithmetic, and it is widely agreed that all reasoning of PRA is finitist. " Later: "It is possible to formalise PRA in such a way that it has no logical connectives at all - a sentence of PRA is just an equation between two terms." This suggests it may be possible to present primitive recursive arithmetic as a purely algebraic system, in the style of universal algebra. Is this possible? References? REPLY [10 votes]: According to unpublished notes by Gavin Wraith ("Notes on arithmetic universes and Gödel incompleteness theorems" (1985)), PRA can be described as an equational theory or as a Lawvere theory, and is abstractly characterized as initial among all Lawvere theories whose generating object is a parametrized natural numbers object. (For some details on Wraith's notes, see Alan Morrison's Master's Thesis; see particularly chapter 5.) This statement may take a bit of unpacking. Lawvere theories are a way of doing universal algebra categorically. Formally, a Lawvere theory is a category $T$ with finite products together with a product-preserving functor $\Phi: \text{Fin}^{op} \to T$ that is a bijection on objects. Here $\text{Fin}$ is the category whose objects are finite cardinals $k = \{0, \ldots, k-1\}$ and whose morphisms are functions between them. The cardinal $1$ generates (the objects of) $\text{Fin}$ by taking finite coproducts of copies of $1$; similarly it generates $\text{Fin}^{op}$ under finite products. Thus the special object $x = \Phi(1)$ generates the objects of $T$ by taking finite products: every object of $T$ is of the form $x^n$. General morphisms $x^n \to x^m$ are $m$-tuples of morphisms $x^n \to x$; we think of the morphisms $x^n \to x$ in $T$ as parametrizing the definable $n$-ary operations of an equational theory. [This categorical description of equational theories is closely related to the concept of clone. If you accept that every equational theory gives rise to a finitary monad on $Set$, then the corresponding Lawvere theory is the category opposite to the finitary Kleisli category consisting of finitely generated free objects. But this is a somewhat hurried discussion which I'll cut short here.] A morphism $F: S \to T$ is a product-preserving functor which is compatible with the given product-preserving functors $\text{Fin}^{op} \to S$ and $\text{Fin}^{op} \to T$. A parametrized natural numbers object in a category with finite products $\mathbf{C}$ is an object $N$ that comes equipped with maps $z: 1 \to N$ (here $1$ denotes the terminal object, and read '$z$' as 'zero') and $s: N \to N$ (successor), such that given any objects $A, X$ of $\mathbf{C}$ and maps $f: A \to X$, $g: X \to X$, there exists a unique map $h: N \times A \to X$ such that the following diagram commutes: $$\begin{array}{ccc} A & \stackrel{\langle z \circ !, 1_A\rangle}{\to} & N \times A & \stackrel{s \times 1_A}{\leftarrow} & N \times A \\ & f \searrow & \downarrow h & & \downarrow h \\ & & X & \underset{g}{\leftarrow} & X \end{array}$$ (here $!$ denotes the unique map $A \to 1$). This axiom is what you need to internalize primitive recursion in a category with finite products. So now the Lawvere theories we are interested in are those for which the generator $x = \Phi(1)$ is a parametrized natural numbers object. A concrete example of such is the full subcategory of $Set$ whose objects are finite powers $\mathbb{N}^n$ of the set of natural numbers. For that matter, for any category with finite products and a natural numbers object $N$ (for example, a Grothendieck topos), you can cook up a Lawvere theory by considering the full subcategory consisting of finite powers of $N$. Or, the subcategory needn't be full: just retain enough arrows to retain finite product structure and primitive recursive structure guaranteed by the axiom of parametrized NNO's. Finally, we are interested in the initial such Lawvere theory. As is the case with any initial algebraic object, the explicit construction is syntactic: we start with $N$ and $z: 1 \to N$ and $s: N \to N$ and use the axiom of parametrized NNO's together with finite cartesian product structure (products of copies of $N$, projection maps, diagonal maps) and categorical composition to generate formally all the arrows. (As a simple exercise, show how to construct formal addition and formal multiplication on $N$.) One thing to check is that morphisms $1 \to N$ (the definable constants of the equational theory, considered up to provable equality) correspond bijectively to standard natural numbers. If $T$ is the initial such theory (which according to Wraith is PRA), with generator denoted $N$, and if $f: T \to Set$ is the unique Lawvere theory morphism sending $N$ to $\mathbb{N}$, then the functorial map $\hom_T(N^n, N) \to \hom(f(N^n), f(N)) \cong \hom(\mathbb{N}^n, \mathbb{N})$ is a surjection onto the total $n$-ary primitive recursive functions on the standard natural numbers. Is is an injection? No. For example, consider the primitive recursive "function" $G: N \to N$ defined by $G(n) = 1$ if $n$ codes the proof of a contradiction in ZFC, and $G(n) = 0$ otherwise. (This is primitive recursive because no unbounded searches are required to verify the validity of a proof.) In the standard model of arithmetic living in the ZFC model $Set$, $G$ would be sent to the constant $0$ function mapping $\mathbb{N} \to \mathbb{N}$. But we could equally consider a model $\mathcal{M}$ of $ZFC + \neg Con(ZFC)$ and its natural number object $\mathbb{N}_\mathcal{M}$, in which $G$ would not be sent to the constant $0$ function. Thus $G$ and the constant $0$ function must be distinct in the category $T$. (Thanks to Zhen Lin Low for supplying this argument, here.)<|endoftext|> TITLE: Do all algebraic number fields arise from Eisenstein polynomials? QUESTION [6 upvotes]: This question came up while going through the application of Eisenstein criterion: The $p$-th cyclotomic polynomial after changing the variable $x$ to $(x+1)$ satisfies Eisenstein criterion. That is the minimal polynomial of $\zeta_p-1$ is an Eisenstein polynomial. Now let us take a general algebraic number field Q$[\alpha]=K$. Can one find another primitive element $\beta$ for $K$ such that its minimal polynomial is Eisenstein? As all quadratic fields arise from $\sqrt d$ which has equation $x^2-d$, it is true. Since the evidence so far is from cyclotomic and quadratic, is this true for all abelian extensions? REPLY [13 votes]: As already indicated by the comment of Mostafa, the criterion is that $K$ is totally ramified at some prime $p$. Mostafa's comment shows that this is necessary. To see that it is also sufficient, take any integral element $\alpha$ of $K$ whose $p$-adic valuation is $1/n$ (if the valuation is normalizied to be 1 at $p$), where $n = [K : \mathbb Q]$; then the minimal polynomial of $\alpha$ will be a $p$-Eisenstein polynomial (of degree $n$).<|endoftext|> TITLE: Is there a formula for the total Chern Class of the tangent space of a projectivized vector bundle? QUESTION [5 upvotes]: Let $V\rightarrow M$ be a complex vector bundle (of rank $k$) over a complex manifold $M$ (you can assume $M$ is compact if that helps, but it may not be relevant to my question). Let $\pi:\mathbb{P}V \rightarrow M$ be the projectivization of $V$. $\textbf{Question}:$ Is there a formula for $c(T\mathbb{P}V)$, the total Chern class of the Tangent space of $\mathbb{P}V$? My naive guess would be that it should be $\pi^*(c(TM))(1+c_1(\gamma^*))^{k+1}$, where $\gamma \rightarrow \mathbb{P}V $ is the tautological line bundle over $\mathbb{P}V$. I think my guess is correct if $M$ was just a point, or more generally if $V$ was a trivial bundle. But I do not know if this is correct in general. The specific case for which I need an answer is when $M:= \mathbb{P}^1 \times \mathbb{P}^1$ and $V:= \mathcal{O}(d_1) \oplus \mathcal{O}(d_2)$. $\textbf{Added Later}:$ It has been pointed out my guess is wrong in general. The correct answer is $$\pi^*(c(TM))c(\pi^*V \otimes \gamma^*).$$ REPLY [8 votes]: No, your formula is not correct. You have to take into account the Chern classes of $V$. The relative tangent bundle $T_{\mathbb{P}V/M}$ is given by the so-called Euler exact sequence $$0\rightarrow \mathscr{O}_{\mathbb{P}V}\rightarrow \pi ^*V\otimes \gamma^* \rightarrow T_{\mathbb{P}V/M}\rightarrow 0\ ,$$ while $$0\rightarrow T_{\mathbb{P}V/M}\rightarrow T_{\mathbb{P}V}\rightarrow \pi ^*T_M\rightarrow 0\ .$$Putting things together we find $c(T_{\mathbb{P}V})=c(\pi ^*V\otimes \gamma^* )\,\pi ^*c(T_M)$. Then use the standard formula for $c(\pi ^*V\otimes \gamma^* )$. REPLY [3 votes]: For any smooth fiber bundle $$ F\hookrightarrow P \stackrel{\pi}{\to} M $$ we have a short exact sequence of vector bundles over $P$ $$ 0\to VTP\to TP \to \pi^* TM\to 0, $$ where $VTP$ denotes the vertical tangent bundle defined as the kernel of the differential of $\pi$. If the bundle is holomorphic then the above is a short exact sequence of complex vector bundles and we deduce $$ c(TP)= c(VTP)\cdot \pi^* c(TM). $$ The classical Euler exact sequence argument shows that when $P=\mathbb{P}(V)$ that $\newcommand{\bC}{\mathbb{C}}$ $$ \gamma^*\otimes \pi^*V \cong \underline{\bC}\oplus VTP, $$ where $\underline{\bC}$ denotes the trivial line bundle. Hence $$ c(TP)= c(\gamma^*\otimes \pi^*V)\cdot \pi^* c(TP). $$ In Section I.3 of Fulton-Lang Riemann-Roch algebra you can find an explicit formula for $c_k(L\otimes E)$, $L$ line bundle and $E$ vector bundle of rank $m$. More precisely $$ c_k(L\otimes E)=\sum_{j=1}^k \binom{m-j}{k-j} c_j(E)c_1(L)^{k-j}. $$ Note. The original answer had an error that I have now corrected. (Hat tip to abx).<|endoftext|> TITLE: regular tiling of a surface of genus 2 by heptagons QUESTION [11 upvotes]: Let $S$ be a Riemann Surface of genus 2. Is there a picture in the literature for a regular tiling of $S$ by 12 heptagons (where three heptagons meet at each vertex)? Also, apart from the obvious restriction given by the Euler characteristic $2-2g=f-nf/2+nf/v$ (where $g$ is the genus, $f$ is the number of faces, $v$ is the number of faces meeting at each vertex) are there any obstructions for the existence of a tiling of a surface of genus $g$ by $n$-gons (where the same number $v$ of $n$-gons meet at each vertex). I know that such a tiling exists (for a surface of genus 2 by heptagons), but I am unable to make a drawing. The construction goes like this (thanks to Mladen Bestvina): Start with a sphere. View this as a octahedron so that you have 8 triangles. Pick a hexagon (start from the top go down, left, down come back up from the opposite side) We fatten up these 6 sides to look like circles with one ray coming from the middle. You should think of these as the quotient of 2 triangles attached along one edge <|> via rotation by 180 degrees (so you get two edges meeting in a point A and 2 edges plus a ray meeting in the other point B). We call the center of the circle C so that the ray goes from C to B. You have to view the point $B_i$ (of the $i$-th side, $i=1,2,3,4,5,6$) as being attached to $A_{i+1}$ for $i=1,2,3,4,5,6$ (of course mod 6) so each of these vertices has valence $5+2=7$. Now you take the double cover with fixed points $C_1,...,C_6$ and you get a surface of genus 2 with 28 triangles (28=2x8+2x6 where 8 is the number of triangles and 6 are the circles with a ray that unwrap in to two triangles). The vertices still have valence 7 (because the cover is étale here). Take the dual tiling and you are done! This is just a curiosity. I tried asking the first part of this on stackexchange with no luck. I am also aware of the beautiful pictures of the famous tiling of a surface of genus 3 by 24 heptagons but you can not use that to obtain the tiling of a genus two surface in an obvious way. REPLY [4 votes]: Addressing the second question positively, this is the Main Theorem of this paper: Edmonds, Allan L.; Ewing, John H.; Kulkarni, Ravi S., Regular tessellations of surfaces and (p,q,2)-triangle groups, Ann. Math. (2) 116, 113-132 (1982). ZBL0497.57001. MR0662119 In fact, they show this also works for non-orientable surfaces, except for the non-existence of a degree 3 triangulation of the projective plane.<|endoftext|> TITLE: Deformations of a blowup QUESTION [11 upvotes]: Let $S$ be a smooth projective surface over $\mathbb{C}$. (I guess this can be more general—higher dimension, other ground fields, non-projective, maybe even singular?—and I'dd like to hear that.) Let $s \in S$ be a point. Let $\beta \colon X \to S$ be the blowup of $s \in S$. Suppose that $H^{i}(S, T_{S})$ is known for $i \in \{0,1,2\}$, as well as $\mathrm{Def}(S)$. If I am not mistaken, the exceptional divisor $E$, which is a $(-1)$-curve, is rigid, in the sense that every deformation of $X$ also has a $(-1)$-curve. Therefore $$\mathrm{Def}(X) \cong \mathrm{Def}(X,E) \cong \mathrm{Def}(S,s) \cong \mathrm{Def}(S) \times T_{S,s},$$ where the last isomorphism is not canonical. (Rather $T_{S,s}$ is the kernel of the forgetful map $\mathrm{Def}(S,s) \to \mathrm{Def}(S)$.) This question is about the cohomological side of the picture, i.e. $H^{i}(X,T_{X})$ for $i \in \{0,1,2\}$. My intuition says that $H^{1}(X,T_{X})$ should also increase with dimension two, whereas the obstruction space $H^{2}(X,T_{X})$ should stay the same. I've tried to fiddle around with the spectral sequence $$ H^{p}(S, R^{q}\beta_{*}T_{X}) \Longrightarrow H^{p+q}(X, T_{X}) $$ but I could not really come to the desired conclusions. For $H^{0}(X, T_{X})$ we get the term $H^{0}(S, \beta_{*}T_{X})$. For $H^{1}(X, T_{X})$ we get the terms $H^{1}(S, \beta_{*}T_{X})$ and $H^{0}(S, R^{1}\beta_{*}T_{X})$. Now $R_{1}\beta_{*}T_{X}$ is a skyscraper sheaf supported on $s$, and if I'm not mistaken, and vague geometric intuition makes me think that it is the tangent space $T_{S,s}$. Finally, $R^{2}\beta_{*}T_{X} = 0$, so for $H^{2}(X, T_{X})$ we get the terms $H^{2}(S, \beta_{*}T_{X})$ and $H^{1}(S, R^{1}\beta_{*}T_{X})$. But maybe this isn't the right way to approach the question… So the main question is: What are the $H^{i}(X,T_{X})$ for $i \in \{0,1,2\}$? I've not been able to find this via google, though I guess this is pretty basic knowledge in deformation theory. But I'm pretty new to this field, so please bear with me. REPLY [3 votes]: Let $S$ be a surface and $Z=\{p_1,...,p_n\}\subset S$ be a reduced subscheme of dimension zero. Let $\epsilon:\widetilde{S}\rightarrow S$ be the blow-up of $S$ at $Z$. Consider the exact sequence $$0\mapsto \epsilon^{*}\Omega_{S}\rightarrow \Omega_{\widetilde{S}}\rightarrow i_{*}\Omega_{E/Z}\mapsto 0$$ where $i:E\hookrightarrow\widetilde{S}$ is the exceptional divisor. Note that $\mathcal{H}om(i_{*}\Omega_{E/Z},\mathcal{O}_{\widetilde{S}}) = 0$ and by Grothendieck duality $\mathcal{E}xt^{1}(i_{*}\Omega_{E/Z},\mathcal{O}_{\widetilde{S}})\cong i_{*}T_{E/Z}(E)$. So dualizing the above exact sequence we get $$0\mapsto T_{\widetilde{S}}\rightarrow \epsilon^{*}T_{S}\rightarrow i_{*}T_{E/Z}(E)\mapsto 0.$$ Since $R^{1}\epsilon_{*}T_{\widetilde{S}} = 0$ we have $$0\mapsto\epsilon_{*}T_{\widetilde{S}}\rightarrow T_{S}\rightarrow T_{S|Z}\mapsto 0.$$ Now, $R^{i}\epsilon_{*}T_{\widetilde{S}} = 0$ for any $i > 0$. So $H^{i}(\widetilde{S},T_{\widetilde{S}})\cong H^{i}(S,\epsilon_{*}T_{\widetilde{S}})$ for any $i\geq 0$ and we get the following exact sequence in cohomology $$ \begin{array}{l} 0\mapsto H^{0}(\widetilde{S},T_{\widetilde{S}})\rightarrow H^{0}(S,T_S)\rightarrow K^{2n}\rightarrow H^{1}(\widetilde{S},T_{\widetilde{S}})\rightarrow H^{1}(S,T_S)\rightarrow 0 \rightarrow \\ \rightarrow H^{2}(\widetilde{S},T_{\widetilde{S}})\rightarrow H^{2}(S,T_S)\rightarrow 0\\ \end{array} $$ Since the map between the tangent spaces $H^{1}(\widetilde{S},T_{\widetilde{S}})\rightarrow H^{1}(S,T_S)$ is surjective and the map between the obstruction spaces $H^{2}(S,\epsilon_{*}T_{\widetilde{S}})\rightarrow H^{2}(S,T_S)$ is injective the map $Def_{\widetilde{S}}\rightarrow Def_S$ is smooth of relative dimension $2n-\dim H^{0}(S,T_S) + \dim H^{0}(\widetilde{S},T_{\widetilde{S}})$. This means that the obstructions to deforming $\widetilde{S}$ are exactly the obstructions to deforming $S$. The vector space $K^{2n}$ parametrizes the deformations of $Z$ inside $S$ and the spaces $H^{0}(\widetilde{S},T_{\widetilde{S}})$, $H^{0}(S,T_S)$ parametrize the infinitesimal automorphisms of $\widetilde{S}$ and $S$ respectively. If the map $$H^{0}(S,T_S)\rightarrow K^{2n}$$ is surjective then $H^{1}(\widetilde{S},T_{\widetilde{S}})\cong H^{1}(S,T_S)$ and the deformations of $\widetilde{S}$ are induced by deformations of $S$. Otherwise the deformations of $Z$ inside $S$ induce non-trivial deformations of $\widetilde{S}$. For instance, take $S = \mathbb{P}^{2}$. Then $H^{0}(S,T_S)\cong T_{Id}Aut(\mathbb{P}^{2})$ has dimension $8$. If $n\leq 4$ the map $$T_{Id}Aut(\mathbb{P}^{2})\rightarrow K^{2n}$$ is surjective and $H^{1}(\widetilde{S},T_{\widetilde{S}})\cong H^{1}(\mathbb{P}^{2},T_{\mathbb{P}^{2}})$. Indeed if $n\leq 4$ there is an automorphism mapping $Z$ to any other set of $n$ points in general position and moving $Z$ inside $\mathbb{P}^{2}$ just induces trivial deformations of $\widetilde{S}$. Furthermore, since $\mathbb{P}^{2}$ itself is rigid we have $H^{1}(\mathbb{P}^{2},T_{\mathbb{P}^{2}})=0$. More generally, let $X$ be a smooth variety and $Z\subseteq X$ be a smooth subvariety. Then $$T^{1}Def_{(X,Z)} = H^{1}(X,T_{X}(-log (Z))) = H^{1}(Bl_{Z}X,T_{Bl_{Z}X}) = T^{1}Def_{Bl_{Z}X}.$$<|endoftext|> TITLE: Tensor product of dendroidal sets: counter-examples QUESTION [12 upvotes]: For any smal category $A$, I shall write $\widehat A$ for the category $[A^{\text op}, \mathbf{Set}]$ of presheaves on $A$, and $y_A\colon A \to \widehat A$ for the Yoneda embedding relative to $A$. I denote by $\Delta$ the category of simplices, by $\Omega$ the category of symmetric dendroids (or symmetric finite rooted trees) and by $\mathbf{Oper}$ the category of symmetric operads. The last two categories were introduced by I. Moerdijk and I. Weiss in WeissPhD and MW07; in the same articles they introduced the notion of Boardman-Vogt tensor product of operads $\otimes_{\text BV}\colon\mathbf{Oper}\times\mathbf{Oper}\to\mathbf{Oper}$, which endows the category of small operads with a symmetric closed monoidal structure. As $\mathbf{Oper}$ is a Bénabou cosmos with the Boardman-Vogt tensor product, we can define a pair of adjoint functors $\tau_d\dashv N_d$ with the nerve-realization paradigm (using the fact the $\mathbf{Oper}$ is cocomplete) and a tensor product of dendroidal sets (i.e. presheaves on $\Omega$) $\otimes \colon\widehat \Omega\times \widehat\Omega\to \widehat\Omega$ defined as the left Kan extension of $N_d(-\otimes -)\colon \Omega\times \Omega\to \widehat\Omega$ along $y_\Omega$. In §3.5 of the article HHM15 it is stated that the category of dendroidal sets $\widehat \Omega$ is weakly enriched over the cartesian closed category of simplical sets $\widehat \Delta$. Thanks to the nerve-realization paradigm $\tau_d\dashv N_d$ (for the dendroidal-operadic case), it is enough to prove that there exists a natural isomorphism $$ \varphi \colon y_\Delta\!\cdot\otimes(y_\Delta\!\cdot\otimes\, y_\Omega\cdot) \to N_d\bigl(\,\cdot\otimes_{\text BV}(\,\cdot\otimes_{\text BV}\cdot\,)\bigr):\Delta\times (\Delta\times \Omega) \to \widehat\Omega $$ and hence the weak enrichement follows by astract non-sense and the associativity of the Boardman-Vogt tensor product. More generally, fixed an object $R$ of $\Omega$, the functor $\Omega^R\otimes\cdot\,\colon \widehat\Omega \to \widehat\Omega$ can be express as the left Kan extension of $N_d\circ R\otimes_{\text BV}\cdot$ along $y_\Omega$. Thus $$ \bigl( \Delta^m\otimes (\Delta^n \otimes \Omega^T)\bigr)_S \cong \int^R \widehat \Omega(\Omega^R, \Delta^n \otimes \Omega^T)\times N_d([m]\otimes_{\text BV} R)_S\\ \phantom{AALA\bigl( \Omega^R\otimes (\Omega^S \otimes \Omega^T)\bigr)_Q}\cong \int^R \mathbf{Oper}(R, [n]\otimes_{\text BV} T) \times \mathbf{Oper}(S, [m]\otimes_{\text BV} R)\\ $$ for any $[m], [n]$ in $\Delta$, $T$ in $\text{Ob}\Omega$, and where we have denoted by $\Omega^T$ the presheaf $y_\Omega T$ on $\Omega$. Amazingly enough, $$ \int^{R\in \Omega} \mathbf{Oper}(R, [n]\otimes_{\text BV} T) \times \mathbf{Oper}(S, [m]\otimes_{\text BV} R) \cong \int^{\mathcal P \in \mathbf{Oper}} \mathbf{Oper}(\mathcal P, [n]\otimes_{\text BV} T) \times \mathbf{Oper}(S, [m]\otimes_{\text BV} \mathcal P) $$ where the last term is exaclty the coend form (Yoneda expansion) of the functor $N_d([m]\otimes_{\text BV}\cdot\,)_S\colon \mathbf{Oper} \to \mathbf{Set}$ evaluated in $[n]\otimes_{\text BV} T$. Using the adjunction $\tau_d\dashv N_d$ and the fact that $\tau_d$ distributes over the tensor product of dendroidal sets, it is possible to get then a natural transformation $$ \alpha\colon (y_\Delta\cdot\times y_\Delta\cdot\,)\otimes y_\Omega\cdot \to y_\Delta\cdot\otimes (y_\Delta\cdot\otimes y_\Omega\cdot\,) : \Delta \times \Delta \times \Omega \to \widehat\Omega\ . $$ Question 1. What is an example of a pair of operads $\mathcal{P, Q}$ in $\text{Ob}\mathbf{Oper}$ such that $N_d(\mathcal P\otimes_{\text BV}\mathcal Q)$ is not isomorphic to $N_d(\mathcal P) \otimes N_d(\mathcal Q)$, i.e. for which the nerve does not distribute over the Boardman-Vogt tensor product? It is stated in §3.5 of HHM15 that the category $\widehat\Omega$ of dendroidal sets is not a monoidal category with the tensor product, as the associativity fails. Indeed, they prove a weak associativity for representable presheaves, as I worked out above, and then it is stated that extending by colimits to all triple of presheaves on $\Omega$ that particular natural isomorphism between triple of representable presheaves, then the comparison morphism is not an isomorphism in general. Question 2. What is an example of a triple of presheaves $X, Y, Z$ on $\Omega$ such that the comparison morphism built in HHM15 is not an isomorphism? An example for which $X, Y$ are actually presheaves on $\Delta$ would be appreciated even more. Question 3. Why cannot exist any other associator $\alpha$ (natural isomorphism for the associativity) making $(\widehat \Omega, \otimes, \alpha, \lambda, \rho, \underline{hom}_\Omega)$ a closed symmetric monoidal category? For any presheaves $X, Y$ on $\Omega$ and any object $T$ of $\Omega$, we define $\underline{hom}_\Omega(X, Y)_T = \widehat\Omega(\Omega^T\otimes X, Y)$. REPLY [5 votes]: Associativity of the tensor product in fact does not hold even for representable presheaves, which addresses your Questions 2 and 3. Proposition 3.6.9 and its proof (in HHM, as you cite) give a typical counterexample, worked out explicitly there. As to Question 1: consider the associative operad $\mathbf{A}$. Tensoring that with itself gives the commutative operad (essentially by the Eckman-Hilton argument), but the tensor product $N_d(\mathbf{A}) \otimes N_d(\mathbf{A})$ is actually a model for the homotopy-coherent nerve of the $\mathbf{E}_2$-operad.<|endoftext|> TITLE: Number of Dyck paths with prescribed number of edges QUESTION [5 upvotes]: I am trying to find a formula for the trace of certain matrix. To do that I was forced to determine the number of Dyck paths with prescribed number of edges. By a Dyck path I mean a lattice path from $(0,0)$ to $(2n,0)$ consisting of $n$ steps of type $(1,1)$ and $n$ steps of type $(1,-1)$ never going below the $x$-axis (y=0). It is well known the number of Dyck paths equals the $n$th Catalan number. If a multiindex $m=(m_{1},\dots,m_{\ell})\in\mathbb{N}^{\ell}$ is given I consider the set of Dyck paths which encounter edges $(j,j+1)$ and $(j+1,j)$ exactly $2m_{j}$ times. Let me denote the set of such paths by $\Omega(m)$. I can prove (and it is not very difficult) that $$ |\Omega(m)|=\prod_{j=1}^{\ell-1}\binom{m_j+m_{j+1}-1}{m_{j+1}}. $$ For sure, this has to be known. I would like to know some references on works dealing with these paths (depending on a multiindex). Since the problem I am working on is not in its nature combinatorial, I would like to only use the existing terminology and results and cite other papers. Perhaps someone knows the original work with the above formula. Thanks a lot. REPLY [8 votes]: This formula is Proposition 3B in P. Flajolet, Combinatorial Aspects of Continued Fractions, Discrete Math 32 (1980), 125–161.<|endoftext|> TITLE: What is $\sum_{i=0}^{n}\binom{n}{i}^3$? QUESTION [9 upvotes]: We know that $$\sum_{i=0}^{n}\binom{n}{i}=2^n$$ and that $$\sum_{i=0}^{n}\binom{n}{i}^2= \binom{2n}{n}$$ what about $$\sum_{i=0}^{n}\binom{n}{i}^3$$ ? REPLY [4 votes]: These are called the Franel Numbers As Michael noted, see oeis.org/A000172 for a summary of known facts and references...<|endoftext|> TITLE: Alternate proof of Morley's theorem? QUESTION [7 upvotes]: I'm trying to understand the result given in the first box at slide 45 of this talk. Specifically: 1) What is the source cited? I have not been able to find any article by Keisler, Chudnovsky and/or Shelah corresponding to the situation. 2) Is this an alternate proof of the classical Morley's theorem (using $\mathcal{L}_{\omega_1\omega}$ as a tool, I guess like how you can use cut-elimination in $\omega$-logic to prove consistency of ordinary first-order PA) or an approach to proving the $\mathcal{L}_{\omega_1\omega}$ version of Morley's theorem? Thanks. REPLY [5 votes]: Regarding (2), some evidence that Baldwin refers to some sort of $L_{\omega_1 \omega}$ version of Morley's theorem, rather than just an alternate proof making use of $L_{\omega_1 \omega}$ machinery, comes from a 1970 survey by Keisler himself. He mentions that "various forms" of Morley's theorem were extended to $L_{\omega_1 \omega}$ by "Choodnovsky [sic], Keisler, and Shelah, 1969" (p.149) though no citation is included in the references. And a look through the Shelah archive seems to turn up no relevant joint work with either of the other two. I don't have a copy on hand, but one promising source for clarification (beyond inquiring with Baldwin about the content of his slides) is Keisler's 1971 book Model Theory for Infinitary Logic, which likely covers the result(s) in question such as they are; and though perhaps only a coincidence, that does match the year Baldwin's slides assign to the matter. ETA: Baldwin's Categoricity book confirms both the nature of the result and his direct source: "Keisler [Kei71] generalized Morley’s categoricity theorem to sentences in $L_{\omega_1 \omega}$, assuming that the categoricity model was $\aleph_1$-homogeneous" (p.22). Though Baldwin points to Keisler's book as the basis for transferring Morely's theorem to infintary logic, he also attributes most of the machinery to Shelah (p.xi). Having now gotten ahold of Keisler's book, the main generalization of Morley's theorem there (see Section 23) is as Baldwin describes: Theorem. Let $T$ be a set of sentences from a countable fragment $L$ of $L_{\omega_1 \omega}$, and let $\kappa,\lambda > \omega$. Assume: $T$ is $\kappa$-categorical. For every countable model $M$ of $T$ there are models $N$ of $T$ of arbitrarily large size such that $M \prec_{L} N$. Every model $M$ of $T$ of size $\kappa$ is $(\omega_1,L)$-homogeneous. Then $T$ is $\lambda$-categorical, and every model of $T$ of cardinality $\lambda$ is $L$-homogeneous. When $L$ is first-order logic, (2) is just upward Lowenheim-Skolem and (1) implies (3), so Morley's theorem really is a special case. Keisler explicitly notes that the special cases where either $\kappa=\omega_1$ or $\lambda=\omega_1 \alpha$ for $\alpha\ge 1$ follow from results due independently to Chudnovsky, Shelah and himself, so that seems to clarify everything.<|endoftext|> TITLE: Are there only countably many compact topological manifolds? QUESTION [38 upvotes]: Up to homeomorphism, there are 2 one-dimensional topological manifolds and countably many 2- and 3-dimensional compact manifolds, respectively, since each manifold in these dimensions can be triangularized and hence be described by a finite amount of combinatorial data. In higher dimensions this argument doesn't work anymore. So it still true for $n\ge 4$ that the set of homeomorphism classes of compact, connected topological $n$-manifolds (without boundary) is countable? (I'd be also interested in the same question for diffeomorphism classes of compact smooth manifolds.) REPLY [41 votes]: It was shown in J. Cheeger and J. M. Kister, Counting topological manifolds. Topology 9, 1970 149–151. that there are only countably many compact manifolds up to homeomorphism (even allowing boundaries). Here is a link to the article.<|endoftext|> TITLE: Is a distributive lattice planar iff it admits no B3 sublattice? QUESTION [11 upvotes]: A finite lattice is planar if it admits a Hasse diagram which is a planar graph (we want the Hasse diagram to be represented in the plane so that the $y$-coordinate of each element respects the order). Remark: See the paper Planar lattices and planar graphs (1976) by C.R Platt : It is shown that a finite lattice is planar if and only if the (undirected) graph obtained from its (Hasse) diagram by adding an edge between its least and greatest elements is a planar graph. The $B_3$ lattice (below) is not planar. Question: Is a finite distributive lattice planar iff it admits no sublattice isomorphic to $B_3$? REPLY [9 votes]: It's already been answered positively, but here's another argument that shows something a little stronger: every finite distributive lattice either contains a B3 (and is not planar) or it can be drawn as a planar grid graph. By Birkhoff's representation theorem, every finite distributive lattice is isomorphic to the lattice of lower sets of a finite partially ordered set. If this partial order has width three or more, (that is, if it has an antichain of three elements), then they generate a B3 and the lattice is not planar. And if the partial order has width two, then by Dilworth's theorem it can be decomposed into two chains. This decomposition can be used to embed the lattice as a grid graph, by setting the two coordinates of each lower set to be the numbers of elements it has in each of the two chains (and then rotating the whole thing by 45 degrees to make it an upward drawing).<|endoftext|> TITLE: Translates of null sets QUESTION [32 upvotes]: Does there exist a null set of reals $N$ such that every null set is covered by countably many translations of $N$? REPLY [8 votes]: Here is an answer for the Cantor space $C$, the set of functions from $\omega$ to $2$. The plan is to show that for each null set $X \subseteq C$ there is a measure $0$ set $C_{a} \subseteq C$ such that $C_{a}$ is homeomorphic with $C$, and for any translate $X'$ of $X$, $X' \cap C_{a}$ is a null set in the measure induced by this homeomorphism. I haven't thought about whether this example can be converted into one for the real reals. Basic open sets in the Cantor space are represented by functions $\sigma \colon n \to 2$, for some $n \in \omega$, where $[\sigma]$ denotes the set of $f \in C$ such that $\sigma \subseteq f$. For each such $\sigma$, the measure of $[\sigma]$, $\mu([\sigma])$, is $2^{-n}$. Given a set $a \subseteq \omega$, let $C_{a}$ be the set of $f \in C$ such that $f(n) = 0$ for each $n \in a$. If $a$ is infinite, $\mu(C_{a}) = 0$. Each $C_{a}$ is naturally homeomorphic to $C$, via deleting the coordinates in $a$. This homeomorphism induces the measure $\mu_{a}$ on $C_{a}$, where, for $\sigma$ as above, $\mu_{a}([\sigma] \cap C_{a})$ is $2^{|n \cap a|}\mu([\sigma])$ (which is $2^{-|n \setminus a|}$) if $\sigma(m) = 0$ for all $m \in a \cap n$, and $0$ otherwise. Given a null set $X$, we may fix for each rational $r \in (0,1)$ a sequence $\langle \sigma^{r}_{i} : i \in \omega \rangle$ such that (1) each $\sigma^{r}_{i}$ is a function from some $s^{r}_{i} \in \omega$ to $2$ (2) $X \subseteq \bigcup_{i \in \omega} [\sigma^{r}_{i}]$ and (3) $\sum\{ 2^{-s^{r}_{i}} : i \in \omega\} < r$. For any $a \subseteq \omega$, and any translate $X'$ of $X$, $\mu_{a}(X' \cap C_{a})$ is at most $$\sum\{ 2^{-|s^{r}_{i} \setminus a|} : i \in \omega \}.$$ It suffices then to find an infinite $a \subseteq \omega$ and a sequence $\langle r_{k} : k \in \omega \rangle$ of rationals from $(0,1)$ such that the sequence of values $$\sum\{ 2^{-|s^{r_{k}}_{i} \setminus a|} : i \in \omega \}$$ goes to $0$. Given a sequence $\bar{r} = \langle r_{k} : k \in \omega \rangle$ of rationals in $(0,1)$, let $a_{\bar{r}} \subseteq \omega$ (enumerated in increasing order as $\langle a_{j} : j \in \omega \rangle$) be such that for each $j\in \omega$, and each $k \leq j$, $$\sum\{ 2^{-s^{r_{k}}_{i}} : i \in \omega \setminus a_{j}\} < r_{k}/2^{2j}.$$ Then for each $k \in \omega$, $\sum\{ 2^{-|s^{r_{k}}_{i} \setminus a|} : i \in \omega \}$ is at most $$(2^{k}r_{k}) + \sum\{ (2^{j+1}r_{k})/2^{2j} : k \leq j < \omega\}$$ which is at most $2^{k+2}r_{k}$ (if I've done the math correctly; the first term comes from considering the terms for $i < a_{k}$, the rest for $i \geq a_{k}$). Then if we choose the $r_{k}$'s so that $2^{k+2}r_{k}$ goes to $0$, $C_{a_{\bar{r}}}$ is the desired null set.<|endoftext|> TITLE: Square-free grows as $6n/\pi^2$: $k$-th free? QUESTION [5 upvotes]: The asymptotic number of square-free numbers $\le n$ is $Q(n) = 6n/\pi^2 + O(\sqrt{n})$. Because $\zeta(2)=\pi^2/6$, $Q(n) \approx n/\zeta(2)$. OEIS A004709 says that cube-free numbers have asymptotic density of $1/\zeta(3)$, "the reciprocal of Apery's constant" ("the probability that three randomly chosen integers are relatively prime": link here). Q. Are there analogous results or conjectures for $k^{\textrm{th}}$-power-free numbers?, $k>3$? Does the density continue to $1/\zeta(k)$? Is that conjectured or proven or disproven? I ask this in (obvious) number-theoretic naiveté. Answered by Gjergji Zaimi and Douglas Zare and Noam Elkies: the density indeed grows as $1/\zeta(k)$, and this has been established. REPLY [9 votes]: Yes, it works in much the same way for any $k$. Here's an elementary proof. Let $Q_k(n)$ be the number of $k$-th power free integers $\leq n$. Then $$ Q_k(n) = \sum_{d^k \leq n} \mu(d) \lfloor n/d^k \rfloor = \sum_{d^k \leq n} \mu(d) \, (n/d^k + \theta_d) $$ for some $\theta_d \in [0,1)$. Hence $$ \Bigl| \, Q_k(n) - \sum_{d^k \leq n} \frac{\mu(d)}{d^k} n \, \Bigr| < n^{1/k}. $$ But $\sum_{d^k \leq n} \mu(d)/d^k$ is a partial sum of a series that converges to $1/\zeta(k)$, with error bounded by $\sum_{d^k > n} 1/d^k \ll n^{1/k}/n$. Therefore $Q_k(n) = n/\zeta(k) + O(n^{1/k})$, QED.<|endoftext|> TITLE: Asymptotic expansion of $\zeta(s \mid a,b)= \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s}(n+b)}$ QUESTION [7 upvotes]: I'm interested in an asymptotic expansion of the following Riemann zeta-type function $$ \begin{align} \displaystyle \zeta(s \mid a,b) := \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re a >-1, \, \Re b >-1, \, s>0, \tag1 \end{align} $$ as $s \to 0^+$. The case $a=b$ in $(1)$ leads to the Riemann Hurwitz zeta function with the Laurent expansion near $0$: $$ \begin{align} \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}} = \frac{1}{s}+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a+1)s^{k}, \quad s>0, \tag2 \end{align} $$ where $\displaystyle \gamma_{k}(a+1)$ are the generalized Stieltjes constants with $\displaystyle \gamma_{0}(a+1)=-\Gamma'(a+1)/\Gamma(a+1)$. What is an asymptotic expansion, as $s \to 0^+$, of $\displaystyle \zeta(s \mid a,b)$ when $a\neq b$? REPLY [4 votes]: We may follow Euler's lead. Euler was the first to define a constant of the form (1734) $$ \begin{align} \gamma & = \lim_{N\to\infty}\left(1+\frac12+\frac13+\cdots+\frac1N-\log N\right)=0.577215\ldots. \tag1 \end{align} $$ Later Stieltjes found (1885) that the Laurent series expansion around $1$ of the Riemann zeta function, $$ \zeta(1+s) = \frac{1}{s} + \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\gamma_k s^k, \quad s \neq 0,\tag2 $$ is such that the scaled coefficients of the regular part of the expansion, now called the Stieltjes constants, are given by $$ \begin{align} \gamma_k& = \lim_{N\to \infty}\left(\sum_{n=1}^N \frac{\log^k n}{n}-\frac{\log^{k+1} \!N}{k+1}\right). \end{align} \tag3 $$ In the same vein, J.B. Wilton (1927) and B. Berndt (1972) established that the Laurent series expansion in the neighbourhood of $1$ of the Hurwitz zeta function $$ \begin{align} \zeta(1+s,a) = \frac1s+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a)\:s^{k}, \quad \Re a>0, \,s\neq 0, \tag4 \end{align} $$ is such that the scaled coefficients of the regular part of the expansion, called the generalized Stieltjes constants, are given by $$ \begin{align} \gamma_k(a)& = \lim_{N\to \infty}\left(\sum_{n=0}^N \frac{\log^k (n+a)}{n+a}-\frac{\log^{k+1} (N+a)}{k+1}\right), \quad \Re a>0. \end{align} \tag5 $$ Do we have a form resembling the original definition of Euler's constant for our coefficients? Theorem. Let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$. Consider the Riemann zeta type function initially defined as $$ \begin{align} \zeta(s\mid a,b) := \sum_{n=1}^{+\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re s>0. \tag6 \end{align} $$ Then the meromorphic extension of $\displaystyle \zeta(\cdot\mid a,b)$ admits the following Laurent series expansion around $0$, $$ \zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{+\infty} \frac{(-1)^{k}}{k!}\gamma_k(a,b) s^k, \quad s \neq 0,\tag7 $$ and $$ \begin{align} \gamma_k(a,b)& = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log^k (n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right). \end{align} \tag8 $$ To see this, let $a,b$ be complex numbers such that $\Re a >-1, \, \Re b >-1$. We first assume $\Re s>0$. Observing that, for each $n \geq 1$, $$ \left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right| \leq \sum_{k=0}^{\infty}\left|\frac{\log^k(n+a)}{n+b}\right|\frac{|s|^k }{k!}<\infty $$ and that $$ \sum_{n=1}^{\infty}\left|\sum_{k=0}^{\infty}\frac{\log^k(n+a)}{n+b}\frac{(-1)^{k}}{k!}s^k\right|=\sum_{n=1}^{\infty}\left|\frac1{(n+a)^s(n+b)}\right| = \sum_{n=1}^{\infty}\frac1{|n+a|^{\Re s}|n+b|}<\infty,$$ we obtain $$ \begin{align} &\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) s^k \\\\ &= \lim_{N\to+\infty}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!}\left(\sum_{n=1}^N\frac{\log^k(n+a)}{n+b}-\frac{\log^{k+1} \!N}{k+1}\right) s^k \\\\ &=\lim_{N\to+\infty}\sum_{k=0}^{\infty}\left(\sum_{n=1}^N\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\ &=\lim_{N\to+\infty}\left(\sum_{n=1}^N\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^k(n+a)}{n+b}s^k -\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!}\frac{\log^{k+1} \!N}{k+1}s^k\right) \\\\ &=\lim_{N\to+\infty}\left(\sum_{n=1}^N\frac1{(n+a)^s(n+b)} +\frac1{N^s}-\frac1s\right) \\\\ &=\zeta(s \mid a,b)-\frac1{s}. \end{align} $$ Then we extend the preceding identity by meromorphic continuation to all $s \neq 0$.<|endoftext|> TITLE: Distinguishing combinatorial maps by their linearizations QUESTION [10 upvotes]: Every (not-necessarily invertible) map $f$ from $[n]:=\{1,2,,,,.n\}$ to itself determines a linear map $L_f$ from ${\bf R}^n$ to itself that sends the basis vector $e_k$ to $e_{f(k)}$ for $1 \leq k \leq n$. If $f$ and $g$ are two self-maps of $[n]$ for which the associated linear maps $L_f$ and $L_g$ are conjugate in the sense that there exists $A$ in $GL(n)$ with $L_f \circ A = A \circ L_g$, must $f$ and $g$ be conjugate in the sense that there exists a permutation $\pi$ in $S(n)$ with $f \circ \pi = \pi \circ g$? Note that if we restrict to the case where $f$ and $g$ are permutations, the answer is Yes. In the unrestricted case, I'm fairly sure the answer is No; Jordan canonical form (IIRC) classifies linear maps up to conjugation by $GL(n)$, and the decomposition into Jordan blocks is too combinatorially simple to capture the kind of tree structure self-maps of sets can possess. On the other hand, I don't see how to turn this intuition into a counterexample. In the event that the answer is No (as I expect), then I ask a follow-up question: Is there some more refined way to linearize $f$ so that the resulting map, considered up to linear-map conjugacy, captures all the combinatorics of $f$? REPLY [4 votes]: As Benjamin Steinberg comments, the problem reduces to the following: represent $f$ by the digraph with vertices $[n]$ and edges $i\to f(i)$. Delete all the cycles. What remains is an acyclic digraph $D$ that defines a nilpotent matrix $N$ whose rows and columns are indexed by the vertices of $D$. Namely, $N_{uv}=1$ if there is an edge $u\to v$, and $N_{uv}=0$ otherwise. We need to find the Jordan form of $N$. Now more generally let $E$ be any finite acyclic digraph and for each edge $u\to v$ let $x_{uv}$ be an indeterminate. Define $N_{uv}=x_{uv}$ if $u\to v$, and $N_{uv}=0$ otherwise. The Jordan form of this "generic nilpotent matrix" was determined by M. Saks and E. R. Gansner (independently). For Gansner's paper see http://www2.research.att.com/~erg/pdf/JADM-Gansner81.pdf. Namely, the sum of the sizes of the largest $j$ Jordan blocks is equal to the largest number of vertices in a union of $j$ paths of $E$. If we replace $x_{uv}$ by 1, then the problem is no longer combinatorial. For instance, let $A$ be any $n\times n$ $(0,1)$-matrix, and place it in the upper-right corner of a $(2n)\times (2n)$ matrix $B$ whose other entries are 0. Then $A$ and $B$ have the same rank so the Jordan form of the nilpotent matrix $B$ depends on the rank of an arbitrary $(0,1)$-matrix $A$, for which there is not a purely combinatorial description. However, going back to $D$, because $N$ has at most one 1 in each row it is easy to see that the Gansner-Saks theorem will continue to hold for the matrix $N$. That is, the sum of the sizes of the largest $j$ Jordan blocks is equal to the largest number of vertices in a union of $j$ paths of $D$.<|endoftext|> TITLE: Wavefront sets of irreducible representations with non-integral infinitesimal characters QUESTION [6 upvotes]: Let $G$ be a complex reductive algebraic group (connected, simply connected etc), viewed as a real group. We study the representations of $G$, and we follow the notations in the paper of Barbasch and Vogan: "Unipotent representations of complex semisimple groups", Annals of Math. Vol.121, 41-110, 1985. Consider the irreducible representations $\overline{X}(\lambda, \mu)$ of $G$ (following Zhelobenko's notation or the notation in the above paper of Barbasch-Vogan, which means the irreducible constituent of principal series $X(\lambda, \mu)$ containing the extremal weight $\lambda-\mu$) and their wavefront sets, we know their wavefront sets are closures of nilpotent orbits of the Lie algebra. We have the following results (Definition 1.10 in the above paper): If $\lambda, \mu$ are integral, and the wavefront set of $\overline{X}(\lambda, \mu)$ is the closure $\overline{O}$ of the nilpotent orbit $O$, then the orbit $O$ is special, i.e. under the Springer correspondence, it correspond to a special representation of the Weyl group $W$ (in the sense of Lusztig). My question is: Is there any example of irreducible representation, with the closure of a non-special orbit as its wavefront set? For integral infinitesimal characters and special orbits, I have a bunch of examples. But I have never seen any non-special orbits as wavefront sets. In Theorem 3.20 of the above paper of Barbasch-Vogan, the authors show an effective way to calculate wavefront sets of any irreducible representations with integral infinitesimal characters. So for non-integral infinitesimal characters, is there any generalization of the theorem 3.20, that we can use to calculate their wavefront sets? Thanks! REPLY [3 votes]: 1) An example is the metapletic representation in the complex symplectic group. They have half-integral infinitesimal characters, and the wavefront set is the minimal orbit which is non-special. 2) Check out the book by Monty McGovern (here). In Section 5. He mentioned how to find out the wavefront set/associated variety (which are the same by a deep theorem of Schmid and Vilonen) for $U(g)/J(\lambda)$, where $\lambda$ can be any infinitesimal character.<|endoftext|> TITLE: Free Loop-Space Recognition Principle QUESTION [16 upvotes]: It is well-known that one can detect based loopspaces using the machinery of operads. Namely, given a group-like space $X$ with an action of $\mathbb{E}_n$-operad, then it is homotopy equivalent as an $\mathbb{E}_n$-space to $\Omega^n Y$ for some space $Y$. Is anyone aware of a similar recognition principle for free loopspaces? Namely, is there some operad-like structure that if a space $X$ admits an action of this structure, then it is an $n$-fold free loopspace $\mathcal{L}^n Y$ for a space $Y$, which up to a weak equivalence is recovered from this action? REPLY [6 votes]: Edit, 2/25/15: Well, after discussing it with Nerses in the comments, it seems I was wrong; taking homotopy fixed points does work! (If you're already guaranteed that the space is a free loop space, that is.) The argument is at the end. Most of my original answer is below. First, there is a general machine you can try to run in situations like this to see what structure is available: in any category with finite coproducts, the functor $\text{Hom}(c, -)$ naturally acquires the structure of a model of the Lawvere theory whose $n$-ary operations are given by maps $c \to nc$, where $$nc = \bigsqcup_{i=1}^n c$$ denotes the coproduct of $n$ copies of $c$. This is because $\text{Hom}(nc, -) \cong \text{Hom}(c, -)^n$, and by the Yoneda lemma, natural transformations $\text{Hom}(c, -)^n \to \text{Hom}(c, -)$ are naturally in bijection with homomorphisms $c \to nc$. If you run this machine in the homotopy category of pointed spaces with $S^1$ the circle, you get that $\text{Hom}(S^1, -)$ naturally acquires the structure of a model of a Lawvere theory which turns out to be the Lawvere theory of groups, as expected. Similarly, for $S^n, n \ge 2$ you get the Lawvere theory of abelian groups. (Details here; other examples here and here.) Unfortunately, any version of this machine in unpointed spaces suffers from the same problem: in unpointed spaces, the coproduct is the disjoint union, so e.g. morphisms from a circle $S^1$ into a disjoint union $n S^1$ of $n$ circles factor through the inclusion of one of the connected components. This means that the resulting Lawvere theory has no interesting $n$-ary operations, and that everything is determined by the unary operations, or by morphisms $S^1 \to S^1$. And this is just not a lot of data. Consider the case that $Y = BG$ is the classifying space of a discrete group $G$. Then $LY = LBG$ is the classifying space of the groupoid $G/G$, the adjoint quotient of $G$, which has objects the elements of the group and morphisms given by conjugation. This is equivalent to the disjoint union of the classifying spaces $B C_G(g)$ of the centralizers of a representative of each conjugacy class of $G$. A map $S^1 \to S^1$ of degree $k$ sends the conjugacy class of $g \in G$ to the conjugacy class of $g^k$, but otherwise the different conjugacy classes don't interact, so I don't see any hope for gluing them back together via a procedure like taking homotopy fixed points. Edit, 2/25/15: Now for the argument. I was being silly. It's very straightforward: Theorem: Let $S$ be a topological group and let $Y$ be a space. Then the space of $S$-homotopy fixed points for the natural action of $S$ on the mapping space $[S, Y]$ is weakly equivalent to $Y$. Sketch. By the universal property, $S$-homotopy fixed points for the natural action of $S$ on $[S, Y]$ is weakly equivalent to the mapping space $[S/S, Y]$, where $S/S$ now denotes the homotopy quotient of the natural action of $S$ on itself by translation (not conjugation). But $S/S$ is a point. $\Box$ But because I was so confused about the case $S = S^1, Y = BG$ I want to go through it in detail. First let's describe explicitly what the action of $S^1$ on $$LBG = B(G/G) = \bigsqcup_{[g]} B C_G(g)$$ is, where the union runs over all conjugacy classes $[g]$ of $G$. One of my confusions is that I forgot that taking $g$ to be the identity gave a copy of $BG$ sitting inside $LBG$ to aim for; what happens when we take homotopy fixed points is that all of the other connected components just disappear, rather than becoming spliced back together. Giving an action of $S^1$ on each component amounts to writing down a map $S^1 \to \text{Aut}(B C_G(g))$ for each $[g]$, hence (after taking $\pi_1$) writing down a map $\mathbb{Z} \to Z(C_G(g))$. Of course there is only one natural candidate for such a map, which is the map sending a generator $1 \in \mathbb{Z}$ to $g \in Z(C_G(g))$. The action of $S^1$ on $B C_G(g)$ obtained in this way is encoded in a fiber sequence $$B C_G(g) \to E \to B S^1$$ where $E$ is the homotopy quotient of $B C_G(g)$ by this action, and the homotopy fixed points of the action are homotopy sections of the map $E \to BS^1$. The long exact sequence in homotopy applied to this fiber sequence is $$1 \to \pi_2(E) \to \pi_2(BS^1) \cong \mathbb{Z} \xrightarrow{g} C_G(g) \to \pi_1(E) \to 1$$ and we conclude that $\pi_2(E) \cong |g| \mathbb{Z}$ if $g$ has finite order $|g|$ in $C_G(g)$ and that $\pi_2(E) \cong 0$ if $g$ has infinite order. It follows that the map $\pi_2(E) \to \pi_2(BS^1)$ has no homotopy sections unless $g$ is the identity, so in fact only the conjugacy class of the identity contributes to homotopy fixed points. In this case the action of $S^1$ is trivial, $E = BG \times BS^1$, and hence the space of homotopy sections is the space of maps $BS^1 \to BG$. But since $\pi_1(BS^1)$ vanishes this is just $BG$ itself.<|endoftext|> TITLE: Torsion elements in the mapping class group QUESTION [5 upvotes]: Let $S$ be an orientable surface of genus $g$ with $b>0$ boundary components, and let $\mathrm{Mod}(S)$ be its mapping class group, that is, the group of isotopy classes of its homeomorphisms modulo isotopy. Homeomorphisms are allowed to reverse orientation or permute boundary components. What can one say about its torsion elements? Is it true that every mapping class of order 2 is orientation reversing? Now assume that $S$ also has marked points on the boundary, and homeomorphisms are also allowed to permute them. Is it true that every mapping class of order 2 is orientation reversing? It is well-known that if a mapping class preserves orientation and fixes each point of $\partial S$, then it cannot have finite order. It is not clear to me what happens when we allow orientation reversing or permutations. REPLY [5 votes]: Collecting some of the answers above, and editing a bit: What can one say about its torsion elements? $\newcommand{\Mod}{\mathrm{Mod}}$As with lattices in Lie groups, and in word-hyperbolic groups, the mapping class group of a surface has finitely many torsion subgroups, up to conjugacy. This means, in principle, if you fix $g$ and $b$ then you can list all finite subgroups of $\Mod(S_{g,b})$. This follows (for example) from "Nielsen realization" stating that for any finite subgroup $G$ in $\Mod(S_{g,b})$ there is a hyperbolic metric on $S$ where $G$ acts via isometries. Despite this one should note that for any finite group $H$ there is a $g > 0$ so that $H$ is a subgroup of $\Mod(S_g)$. Also, note that if you require that the surface has boundary, and require all mapping classes and isotopies to fix the boundary pointwise, then the finite order elements go away. For example, the braid group has no torision. Is it true that every mapping class of order 2 is orientation reversing? $\newcommand{\ZZ}{\mathbb{Z}}$No. In the closed case the most famous examples are the "hyperelliptic involutions". As a somewhat different source of examples, take any surface $S$ and take any double cover $T$. Then the deck group of $T$ over $S$ is a copy of $\ZZ/2\ZZ$ inside of $\Mod(T)$. Now assume that $S$ also has marked points on the boundary, and homeomorphisms are also allowed to permute them. Is it true that every mapping class of order 2 is orientation reversing? Again no, as above.<|endoftext|> TITLE: Boardman-Vogt tensor product QUESTION [6 upvotes]: Let $\mathbf{sSet}$ be the model category of simplicial sets and $\mathbf{Op}$ the model category of symmetric operads. Equipped with Boardman-Vogt tensor product $ \otimes_{BV}$, the category $\mathbf{Op}$ is symmetric monoidal. Questions: 1) is $(\mathbf{Op}, \otimes_{BV})$ a symmetric monoidal model category ? 2) let $P$ be a $E_{n}$-operad and $Q$ be a $E_{m}$-operad, is $P\otimes_{BV}Q$ a $E_{n+m}$-operad ? Edit 1) $P$ and $Q$ are cofibrant symmetric operads. 2) a weak equivalence (fibration) of symmetic operads $f: A\rightarrow B$ is a level-wise weak equivalence (fibration) of simplicial sets $f_{n}: A(n)\rightarrow B(n)$. 3) the Boardman-Vogt tensor product $A\otimes_{BV} B$ is a (tricky) quotient of $A\sqcup B$ (the coproduct in $\mathbf{Op}$). 4) a $A\otimes_{BV}B$-algebra is a $A$-algebra in the category of $B$-algebras, or similarly a $B$-algebra in the category of $A$-algebras. 5) it is natural to ask if the category of $E_{n}$-algebras in the category of $E_{m}$-algebras is equivalent (in homotopical sense) to $E_{n+m}$-algebras. The question can be formulated as follows. It is true that $$E_{n}\simeq E_{1}^{\otimes_{BV}^{n}}$$ REPLY [10 votes]: The answer to 2) is yes for cofibrant operads, see http://arxiv.org/abs/1102.1311 by Fiedorowicz and Vogt. The answer to 1) is no; the Boardman-Vogt tensor product does not interact well with cofibrations. EDIT: As an answer to Chris' comment, here is a counterexample to 2) in the setting of colored operads, although you can adapt it to the one-color case as well. In fact, the example just concerns simplicial categories (i.e. simplicial operads with only unary operations), where the tensor product coincides with the Cartesian product of simplicial categories. Therefore it also shows why the Bergner model structure on simplicial categories is not Cartesian. Write $I$ for the category with objects 0 and 1, with one non-trivial morphism from 0 to 1, and write $\partial I$ for the disjoint union of 0 and 1 (with just their identity morphisms). Clearly $\partial I \rightarrow I$ is a cofibration. However, the pushout-product $\partial I \times I \cup I \times \partial I \rightarrow I \times I$ is not a cofibration; it's not even a monomorphism. On the left-hand side, the (discrete) simplicial set of morphisms from $(0,0)$ to $(1,1)$ consists of two points, on the right-hand side there is only one. If you tweak this example to apply to the one-object case (and identifying categories with one object with monoids), you run into the morphism from the free monoid on two generators to the free commutative monoid on two generators, which again is not a monomorphism. By the way, this is not the only issue: it is also possible to cook up counterexamples to the pushout-product axiom for the Boardman-Vogt tensor product by playing around with nullary operations, using an Eckmann-Hilton style argument.<|endoftext|> TITLE: Reflection of light from function graph QUESTION [20 upvotes]: Let a positive convex decreasing differentiable function $f(x)$ be defined on $\mathbb{R}$ and $\lim_{x \to +\infty}f(x)=0.$ Let the point light source be placed at $ P(x_0,y_0)$ with $ y_0>0,\,y_0 R$ is not lightened? The model example $f(x):=e^{-x},\,P(0,0.5)$ suggests the answer is yes. The question is migrated from SE. REPLY [3 votes]: In the paper The existence of unbounded oscillating trajectories in a problem of billiards (1962) Leontovich proved that under bell-like curve (it must be zero at $\pm\infty$) each trajectory oscillates, i.e. it crosses y-axis infinitely often. Also he proved that among all trajectories do exist finite and infinite ones.<|endoftext|> TITLE: Inequality of the norm of the convolution in $L^p(\mathbb{R}^n)$ with symmetric decreasing rearrangement? QUESTION [7 upvotes]: Is it true that $$ ||f*g||_p \le ||\,|f|^* * |g|^*||_p\quad ? $$ where $|f|^*$ and $|g|^*$ are the symmetric decreasing rearrangements of the functions $|f|$ and $|g|$. Under what conditions on $f$ and $g$ this is true? REPLY [7 votes]: Yes, this follows from the Riesz rearrangement inequality: $$ \int_{\mathbb R^n} h(x)(f*g)(x)\, dx \le \int_{\mathbb R^n} h^*(x) (f^* *g^*)(x)\, dx $$ (We can assume that all functions are $\ge 0$.) Since $\|h^*\|_q=\|h\|_q$, this shows that $$ \|f*g\|_p = \sup_{\|h\|_q=1} \int |h(f*g)| \le \|f^* *g^*\|_p , $$ as desired (with $1/p+1/q=1$).<|endoftext|> TITLE: New series for $1/\pi$ based on Ramanujan's ideas QUESTION [7 upvotes]: In his classic paper "Modular Equations and Approximations to $\pi$ (1914)", Ramanujan gives a standard technique to obtain a general family of series for $1/\pi$ based on series for $(2K/\pi)^{2}$ in terms of $k$ (see the details here). Most of the series which Ramanujan provides are based on the following expressions for $(2K/\pi)^{2}$: \begin{align}\left(\frac{2K}{\pi}\right)^{2}&= 1 + \left(\frac{1}{2}\right)^{3}(2kk')^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}(2kk')^{4} + \cdots\tag{1}\\ \left(\frac{2K}{\pi}\right)^{2}&= (1 + k^{2})^{-1}\,_{3}F_{2}\left(\frac{1}{4},\frac{3}{4},\frac{1}{2}; 1, 1; \left(\frac{g^{12} + g^{-12}}{2}\right)^{-2}\right)\tag{2}\\ \left(\frac{2K}{\pi}\right)^{2}&= (1 - 2k^{2})^{-1}\,_{3}F_{2}\left(\frac{1}{4},\frac{3}{4},\frac{1}{2}; 1, 1; -\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2}\right)\tag{3}\\ \left(\frac{2K}{\pi}\right)^{2}&= \{1 - (kk')^{2}\}^{-1/2}\,_{3}F_{2}\left(\frac{1}{6},\frac{5}{6}, \frac{1}{2};1;1; \frac{27G^{24}}{(4G^{24} - 1)^{3}}\right)\tag{4}\end{align} (In the above $G = (2kk')^{-1/12}, g = (2k/k'^{2})^{-1/12}$. I have left the series related to functions ${}_{3}F_{2}(1/3, 2/3, 1/2; 1; 1; a(k))$ based on elliptic functions to base 3). Next Chudnovsky brothers (around 1989) use the following series $$\left(\frac{2K}{\pi}\right)^{2} = \{1 - 4G^{-24}\}^{-1/2}\,_{3}F_{2}\left(\frac{1}{6},\frac{5}{6}, \frac{1}{2};1;1; \frac{-27G^{48}}{(G^{24} - 4)^{3}}\right)\tag{5}$$ to give their famous series based on $G_{163}$. Another series can be obtained from $(5)$ above by changing the nome $q$ into $(-q)$: $$\left(\frac{2K}{\pi}\right)^{2} = \{k'^{4} + 16k^{2}\}^{-1/2}\,_{3}F_{2}\left(\frac{1}{6},\frac{5}{6}, \frac{1}{2};1;1; \frac{27g^{48}}{(g^{24} + 4)^{3}}\right)\tag{6}$$ The general family of series for $1/\pi$ based on equation $(6)$ is as follows: $$\frac{1}{\pi} = \sum_{m = 0}^{\infty}\frac{(1/6)_{m}(5/6)_{m}(1/2)_{m}}{(m!)^{3}}(A + mB)X_{n}^{m}\tag{7}$$ where \begin{align} X_{n} &= \frac{27g_{n}^{48}}{(g_{n}^{24} + 4)^{3}}\notag\\ A &= \frac{1}{2k\sqrt{g_{n}^{24} + 4}}\left(\frac{\sqrt{n}}{3}(1 - 2k^{2}) - \frac{R_{n}(k, k')}{6}\right) - \sqrt{n}\cdot\frac{g_{n}^{12}(k^{2} + 7)}{4(g_{n}^{24} + 4)^{3/2}}\notag\\ B &= \sqrt{n}(g_{n}^{12} + k)\cdot\frac{g_{n}^{24} - 8}{(g_{n}^{24} + 4)^{3/2}}\notag\\ n &> 4\notag\\ k &= k(q) = k(e^{-\pi\sqrt{n}})\notag \end{align} I am currently trying to work out a series based on the above equation $(7)$ (using $n = 58$) and bit bogged down into calculations with various radicals. I don't know if the equation $(7)$ above has been used anywhere to generate series for $1/\pi$ (let me know of any references if this has already been in literature). I want to know whether this approach would lead a genuine new family of series for $1/\pi$. REPLY [5 votes]: See page 5 in http://arxiv.org/abs/1112.3259 Are the series presented there somewhat related to your series (7)?<|endoftext|> TITLE: Does the stable category of a nice exact category embed in (the underlying category of) a derivator? QUESTION [6 upvotes]: In Derivators, Pointed Derivators, and Stable Derivators, Moritz Groth gives as an example of a non-invertible morphism with trivial cone an inclusion $f:X\to I$. Here $X$ is an object of injective dimension $1$ in an exact category $\mathcal{E}$ with enough injectives, $I$ is an injective, and the stable category $\underline{\mathcal{E}}$ has a "suspended structure" given in the same way as the triangulation of the stable category of a Frobenius category but without the invertibility of suspension. This is fine as far as it goes, but it's not obvious to me that this is an example of a cone in a pointed derivator, what he's just defined, and Groth doesn't give any indication of how to see it as such. Question: how can I realize this example as a cone in a derivator? Having looked around, I don't see any evidence that $\underline{\mathcal{E}}$ is known to be the underlying category of a derivator, even with extra assumptions a la a paper of Stovicek to make $\mathcal{E}$ more like a Grothendieck category. Is it, in fact, such an underlying category, given some extra assumptions? Or can we, perhaps, give it an exact embedding in a homotopy category, and extend to a derivator in that way? REPLY [2 votes]: This is not exactly answering your question but: a Frobenius category $\mathcal{E}$ (with suitable assumptions) can be given the structure of a model category whose homotopy category is the stable category $\underline{\mathcal{E}}$. See James Gillespie's http://arxiv.org/abs/1009.3574. Now take the derivator associated with this model category structure (again, I am not so sure about the required assumptions). Its underlying category will be the stable category $\underline{\mathcal{E}}$.<|endoftext|> TITLE: A variant to the Hadwiger-Nelson problem QUESTION [5 upvotes]: Consider the following graph $G=(V,E)$ where $V=\mathbb{R}^2$ and $E = \{\{x,y\}: x,y \in \mathbb{R}^2 \text{ and } |x-y|\in \mathbb{Q}\}$. What is $\chi(G)$? (This is a variant of the Hadwiger-Nelson problem.) REPLY [10 votes]: By considering all the rational numbers on the $x$-axis we can see that we need at least countably many colors. This is also sufficient, that is the chromatic number of the rational-distances graph is countable. This is due to Erdos and Hajnal in the case of $\mathbb R^2$. They show that the rational-distances graph in the plane does not contain a copy of the complete bipartite graph $K(2,\omega_1)$, and that any such graph must have countable chromatic number. P. Erdos and A. Hajnal, "On chromatic number of graphs and set systems", Acta Math. Hungar. 17(1966), 61-99. The result was generalized to rational distances graphs in $\mathbb R^n$ by Peter Komjath. However, the previous method doesn't generalize since now the graph contains even a copy of $K(\omega, 2^\omega)$. Instead, Komjath uses a clever transfinite induction argument. P. Komjath, "A decomposition theorem for $\mathbb R^n$" Proc. Amer. Math. Society (1994): 921-927.<|endoftext|> TITLE: Does independence of the sequence $f(A_i, B)$ imply the sequence is independent of $B$? QUESTION [5 upvotes]: Suppose $B, \{A_i: i \in \omega\}$ are i.i.d. random variables with uniform distributions on $[0,1]$. If $f$ is a map such that $\{f(A_i, B): i \in \omega\}$ are independent, must $\{f(A_i, B): i \in \omega\}$ also be independent of $B$? REPLY [5 votes]: It is true. Let $\chi_Z(x,y)$ be the indicator function of the set $\lbrace (x,y) \,|\, f(x,y)\in Z\rbrace$ for some set $Z$. The condition that the events $f(A_1,B)\in Z$ and $f(A_2,B)\in Z$ are independent is that $$ \int_{[0,1]^3} \chi_Z(x,y)\chi_Z(x'y)\,dxdx'dy = \int_{[0,1]^2} \chi_Z(x,y) \,dxdy \ \ \int_{[0,1]^2} \chi_Z(x',y') \,dx'dy'.$$ Define $g(y) = \int_0^1 \chi_Z(x,y)\,dx$. Then the above condition can be rearranged into $$ \int_{[0,1]^2} g(y) (g(y)-g(y'))\, dy dy' = 0.$$ That's just a change of variables from $$ \int_{[0,1]^2} g(y') (g(y)-g(y'))\, dy dy' = 0,$$ so we can subtract the two to get $$ \int_{[0,1]^2} (g(y)-g(y'))^2 \,dydy' = 0$$ which by positivity means $g(y)$ is constant a.e.. Harking back to the definition of $\chi_Z$ and noting that the set $Z$ was arbitrary, we see that $f(A_1,B)$ is independent of $B$. I have only considered pairwise independence, but once you find that $g(y)$ is independent of $y$ it makes the whole lot of them independent. At least, it seems so at 3am... ADDED after waking up: At the moment I can't see how uniform distributions are needed for this argument. Just consider any vertical measure $\mu$ and horizontal measure $\nu$ and replace $dx$ by $d\mu$, $dy$ by $d\nu$, etc, in the above argument. Still works, doesn't it? MORE ADDED: Nate's question leads me to note the following, which surely must be well-known in the theory of exchangeable random variables (but I didn't know it). Theorem. Let $X,Y$ be random variables such that $(X,Y)$ and $(Y,X)$ have the same distribution. Then $X$ and $Y$ are independent iff for all measurable sets $S$, $P(X\in S \wedge Y\in S)=P(X\in S)^2$. Proof. Let $S,T$ be measurable sets. First assume $S$ and $T$ are disjoint. Since by the exchangeability assumption $P(X\in T\wedge Y\in S)=P(X\in S\wedge Y\in T)$, we have \begin{align} 2 P(X\in S\wedge Y\in T)&=P(X,Y\in S\cup T) - P(X,Y\in S) - P(X,Y\in T) \\ &= P(X\in S\cup T)^2 - P(X\in S)^2 - P(X\in T)^2 \\ &= (P(X\in S) + P(X\in T))^2 - P(X\in S)^2 - P(X\in T)^2 \\ &= 2 P(X\in S) P(X\in T) \\ &= 2 P(X\in S) P(Y\in T). \end{align} If $S$ and $T$ are not disjoint, break $P(X\in S\wedge Y\in T)$ into four disjoint cases like $P(X\in S\wedge Y\in T\setminus S)$ and apply the above to each. It works.<|endoftext|> TITLE: Why would the roots of the generating functions of the number of k-almost primes less than x have negative real parts? QUESTION [18 upvotes]: Specifically, I find it appealing to count only squarefree numbers having $k$ prime factors, so I define $$\pi_k(x)=\#\{n\leq x: \omega(n)=k;\mu(n)\neq0 \}$$ and consider the generating functions \begin{eqnarray}f(z,x)&=&\sum_{k=0}^{m(x)}\pi_k(x) z^k\\ &=&\sum_{n\leq x}|\mu(n)|z^{\omega(n)}. \end{eqnarray} On one hand, these generating functions are polynomials in $z$ of degree $$m(x)=\max \{\omega(n): n\leq x;\mu(n)\neq 0\}\sim\log x/\log\log x.$$ On the other, they are the inverse Mellin transform of $$F(z,s)=\prod_p1+zp^{-s}=H(z,s)\zeta^z(s)$$ where $H(z,s)$ is an analytic function of $s$ for fixed $z$ which is bounded above and away from zero in any half plane $\sigma\geq\sigma_0>1/2$. I have checked that the roots of $f(z,x)$, as polynomials in $z$, do indeed have negative real parts for all $x\leq 10^6$. Why would the roots have negative real parts? What would it say about $\zeta(s)$ or the numbers of $k$-almost primes less than $x$ if the roots were to have negative real parts? On the Riemann hypothesis one would expect to find the roots---in some sense---closer to the non positive integers as $x\rightarrow\infty$ because these are the only values of $z$ for which $\zeta^z(s)$ is analytic throughout a neighbourhood of $s=1$ and, therefore, the only points at which $f(z,x)\in O(x^a)$ for some $a<1$. However, I am interested in the less-restrictive conjecture that they have negative real parts. I am aware of the notion of 'stability' of linear translation invariant systems and in dynamical systems, and it's equivalence with the positivity of the principle minors of the associated Hurwitz matrices, Routh tables, Sylvester's criterion, etc. I find that these equivalences serve more as a test than to provide reasoning but, if you can say something in this regard, I would be pleased to hear about that too. It appears this may be related to the fact that $\zeta^z(s)$ tends to infinity or zero as $s\rightarrow 0^+$ depending on whether $\Re z$ is positive or negative. However, for small $\Re z$ and $s=1$ the convergence is very slight, and making the distinction appears to be a tricky problem. One can generalize this question and ask about the locations of the roots of $$f(z,x;s)=\sum_{n\leq x} \frac{|\mu(n)|z^{\omega(n)}}{n^s}.$$ Computationally, it appears that the same conjecture holds whenever $s$ is real and, moreover, there is a good reason for this to be true when $s>1$ is real because $$\lim_{x\rightarrow\infty}f(z,x;s)=F(z,s)$$ and this limit certainly does have it's zeros in the left half plane. One only needs then an approximation theorem on the distribution of the zeros of a uniformly convergent sequence whose limit has a prescribed distribution of zeros. One possible approach to the original problem might then be via partial summation $$f(z,x;s)=x^{-s}f(z,x)+s\int_{1}^{x}\frac{f(z,y)dy}{y^{s+1}}$$ but I have not had any success with this yet. REPLY [8 votes]: In any bounded region, for large $x$, the polynomial can only take on zeros very near the negative real axis (and indeed near the non-positive integers). This follows from the work of Selberg (Note on a paper by L.G. Sathe, see Theorem 2 there) which shows that, for bounded $z$ and large $x$, $$ f(z,x) = x C(z) \frac{(\log x)^{z-1}}{\Gamma(z)} + O(x (\log x)^{z-2}), $$ with $$ C(z) = \prod_{p} \Big(1+\frac{z}{p}\Big) \Big(1-\frac 1p\Big)^z. $$ If $z$ is bounded, but not too close to a non-positive integer, then the main term above dominates the error term, and shows that $f(z,x)$ cannot be zero. Further note that $1/\Gamma(z)$ becomes zero at the non-positive integers, and the absolutely convergent Euler product $C(z)$ can only be zero when $z$ equals the negative of a prime number. Therefore, if $-k$ is a non-positive integer with $k$ not being prime, then we see from the asymptotics that $f(z,x)$ changes sign when passing from $z=-k+\epsilon$ to $z=-k-\epsilon$ producing a real zero in this interval. If $-k$ is a non-positive integer with $k$ a prime number, then take a small circle centered at $-k$ with radius $\epsilon$. Computing the change in argument around this circle for the main term in our asymptotic, we find that there are two zeros in this small circle. Note that these zeros may not be real, but just very close to the negative real axis. Of course this asymptotic does not explain why all the zeros of the polynomial should have negative real part. I don't quite see why that holds; perhaps somewhat relevant to this would be the unimodality of the integers below $x$ with $k$ prime factors (see the work of Balazard establishing a conjecture of Erdos).<|endoftext|> TITLE: Structure of the automorphism group of a Riemann surface QUESTION [5 upvotes]: I was wondering if anything is known about the possible structure of $\mathrm{Aut}(S)$ for a Riemann surface $S$. More precisely, are there known obstructions for a finite group $G$ to be such an automorphism group? REPLY [12 votes]: There are no obstructions. In fact, every finite group is isomorphic to the full automorphism group $\textrm{Aut}(S)$ of some compact Riemann surface $S$ (of genus at least $2$). Moreover, $S$ may be chosen so that the quotient Riemann surface $S/\textrm{Aut}(S)$ has any preassigned genus. For a reference, see Theorem 4 in the paper by L. Greenberg Maximal Fuchsian groups, Bull. Amer. Math. Soc. Volume 69, Number 4 (1963), 569-573.<|endoftext|> TITLE: Quest for a human proof of a $q-$binomial identity QUESTION [9 upvotes]: Let $$f(n,k) = \sum\limits_{j = - k}^k {{{( - 1)}^{k - j}}} \binom{n-j}{k-j}\binom{n+j}{k+j}.$$ Then $f(n,k)=\binom{n}{k}$ because it satisfies $f(n,k)=f(n-1,k)+f(n-1,k-1)$ and the obvious boundary values. Let $ {n\brack {k}}$ be a $q-$binomial coefficient. I want to know if there is a similar proof for the identity $$f(n,k,q) = \sum\limits_{j =-k}^k {(-1)^{k-j}} q^{\binom{j}{2}+\binom{k+1}{2}-{(n+1)k}} {{n-j}\brack{k-j}}{{n+j}\brack{k+j}}={n\brack{k}}.$$ There is an easy computer proof using the $q-$Zeilberger algorithm and perhaps it also follows from a simple $q-$ hypergeometric summation. But I am interested in a direct proof. REPLY [8 votes]: At first, we use a formula $\binom{u}{m}=(-1)^m\binom{m-u-1}{m}$. Thus $$\binom{n\pm j}{k\pm j}=(-1)^{k-j}\binom{k-n-1}{k\pm j}.$$ Denote $k-n-1=x$, and $k-j$ by $s\in \{0,\dots,2k\}$, we have to prove that $$ \sum_{s=0}^{2k} (-1)^s \binom{x}{s}\binom{x}{2k-s}=(-1)^k \binom{x}{k}. $$ LHS counts the coefficient of $t^{2k}$ in $(1-t)^x\cdot (1+t)^x$, as is seen immediately immediately from expanding both multiples $(1\pm t)^x=\sum (\pm 1)^s\binom{x}{s} t^s$. Now we have to $q$-ify this argument, and this seems to be dooable. Indeed, instead of $(1\pm t)^x$ we consider the $q$-power $(1\pm t)(1\pm qt)\dots (1\pm q^{x-1}t)$. The coefficients are known by $q$-binomial theorem, and their values allow to define this $q$-power for non-natural $x$ (each specific coefficient is a polynomial in $q^x$). The product of two such $q$-powers is $q^2$-power of $t^2$, which again expands by $(q^2)$-binomial theorem.<|endoftext|> TITLE: Deformation with fixed ramification QUESTION [5 upvotes]: Suppose that $f : X \to Y$ is a finite, surjective morphism of normal varieties. I want to know about the space of first-order deformations of $X$ over $Y$ with fixed ramification, i.e. the deformations of $f$ with fixed target $Y$, such that the (set-theoretic) image of the ramification divisor $f(R_t)$ is constant. I think that when $X$ and $Y$ are smooth, there are no such deformations: the normal sheaf $N_f$ is supported on the ramification divisor $R \subset X$ (Sernesi, "Deformations...", pg 171), and a deformation of $f$ forces the ramification to move too. But I am not sure if I can dispense with the smoothness assumption (or indeed whether my question is well-posed without it?) REPLY [4 votes]: The moduli problem you are interested in is zero-dimensional in characteristic zero. (In this answer we will work over $\mathbb{C}$.) Indeed, for all $d\geq 1$, and all normal varieties $X$ and $Y$, the set of isomorphism classes of finite degree $d$ surjective morphisms $X\to Y$ ramified over a fixed closed subset $B\subset Y$ is finite. This follows from the fact that $\pi_1(Y^{an})$ is finitely generated. (The fact that $d$ is fixed is because you are deforming a fixed finite map $X\to Y$. Thus, $d$ equals $\deg(X\to Y)$.) The fact that $\pi_1(S^{an})$ is finitely generated holds for any variety $S$ over $\mathbb{C}$; see SGA7.I Théorème 2.3.1 Expose II. In case you are interested: there is a (small) difference in studying finite etale covers of a variety $U = Y\setminus D$ and studying finite surjective maps $X\to Y$ ramified only over $D$. Indeed, there is a fully faithful functor from the category of finite etale covers of $U$ to the category of finite surjective morphism $X\to Y$ ramified only over $D$. (To $V\to U$ one associates the normalization of $Y$ in the "function field" of $V$.) This functor is not essentially surjective (because you can sometimes extend a given $V\to U$ to a finite surjective map $X\to Y$ with $Y$ normal but $X$ non-normal. Think of a rational function on a nodel curve.) However, if you stick to normal varieties, the category you are interested in is indeed equivalent to the category of finite etale covers of $Y\setminus B$.<|endoftext|> TITLE: liftings of principal bundles QUESTION [7 upvotes]: I would like to know what structure has the category of liftings of a principal bundle. Let me be more precise. Fix $k$ an algebraically closed field and $X$ a smooth projective variety over it (for eg. a curve) as well as a short exact sequence of smooth connected groups $1\to K\to G\stackrel{\pi}{\to} H\to 1$ (if one prefers one can work aver $\mathbb{C}$ and take the groups to be connected complex Lie groups). Since we're dealing with principal bundles pick your favourite topology, say $\tau$ (fpqc, etale, classical analytic - I hope that the answer will not depend in an essential way on this). Now let us also fix $F_H$ an $H$-bundle on $X$. I'm interested in understading the category $\mathcal{K}_{F_H}$ of liftings of $F_H$ to $G$ (it is a stackover $X$, right?). This is defined as follows: for an $X$-scheme $f:T\to X$ we have: objects: pairs $(E,\alpha)$ where $E$ is a $G$-bundle on $T$ and $\alpha:\pi_*(E)\to f^*(F_H)$ an isomorphism of $H$-bundles. morphisms: from a pair $(E,\alpha)$ to $(E',\alpha')$ are isomorphisms of $G$-bundles $u:E\to E'$ that are compatible with $\alpha$ and $\alpha'$, i.e. $\alpha' \pi_*(u) = \alpha$. In case $K=\ker(\pi)$ is abelian $H$ acts naturally by conjugation on $K$ and one can form the following group scheme over $X$: $K_{F_H} = F_H\stackrel{H}{\times}K$. One can see rather easily (for e.g. using local charts) that $\mathcal{K}_{F_H}$ is a gerbe over $X$ which is, locally on $X$, isomorphic to $BK_{F_H}\times X$. The obstruction of $F_H$ to have a lift to a $G$-bundle is an element $\xi\in H^2_{\tau}(X,K_{F_H})$. Moreover, if $\xi=0$ the gerbe is trivial and hence the stack $\mathcal{K}_{F_H}$ is equivalent to the stack of $K_{F_H}$-bundles on $X$. I hope I haven't messed things up yet. My question is: what if $K$ is not abelian? then there's no natural action of $H$ on $K$ (even if the sequence is split the conjugation action depends on the splitting). Where lives the obstruction of lifting an $H$-bundle? And, what interests me more actually, what is the structure of the category of liftings of $F_H$ to $G$ once one knows it has a lift? I'm pretty sure the answer should be in Giraud's book somewhere but I find it quite hard to read so I'm having troubles detecting the right place to look. Any references or comments would be helpful. REPLY [6 votes]: The broad outlines of how this business works don't depend on the fact that you're working with varieties so let me work with spaces instead, by which I mean homotopy types. Let $f : X \to BG$ be a principal bundle and let $BH \to BG$ be a map along which you'd like to lift. Then the space of lifts of $f$ to $BH$ (up to homotopy) is precisely the space of (homotopy) sections of the pullback $X \times_{BG} BH$, or equivalently the space of sections of the $G/H$-bundle associated to $f$. (Here by $G/H$ I mean the homotopy fiber of $BH \to BG$, which may or may not be the actual quotient $G/H$, whatever that means in your setting.) The nicest case is when $BH \to BG$ is itself a homotopy fiber, or equivalently when it fits into a fiber sequence $$\Omega Y \to BH \to BG \to Y$$ for some space $Y$ (which must necessarily deloop $G/H$; in this case the associated $G/H$-bundle is a "principal $\Omega Y$-bundle"). Then the space of lifts of $f$ to $BH$ is, by the universal property of the homotopy fiber, precisely the space of nullhomotopies of the composite map $X \to BG \to Y$. In the very nicest cases $Y$ is an Eilenberg-MacLane space. In particular, if $G, H$ fit into a short exact sequence $$K \to H \to G$$ then $BK, BH, BG$ fit into a fiber sequence $$BK \to BH \to BG.$$ If $K$ is not only abelian but central in $H$, then this fiber sequence deloops to a fiber sequence $$BK \to BH \to BG \to B^2 K$$ and so we can take $Y = B^2 K$, in which case the existence of a nullhomotopy of the composite $X \to BG \to Y$ is equivalent to the corresponding cohomology class in $H^2(X, K)$ vanishing. If $K$ is abelian but not central then this cohomology group needs to be taken with local coefficients. But sometimes $Y$ is more complicated than this. For example, if $G = \text{Spin}$ and $H = \text{String}$ then $Y = B^4 \mathbb{Z}$ and the existence of a lift is controlled by a cohomology class $\frac{p_1}{2} \in H^4(X, \mathbb{Z})$. In general, though, $G/H$ just isn't a loop space, and so can't be delooped, which means that $Y$ doesn't exist and the obstruction theory gets harder. The obstructions are a sequence of classes in cohomology $H^{k+1}(X, \pi_k(G/H))$ with local coefficients, each of which is only well-defined provided that the previous one vanishes. A typical case here is $G = O(2n), H = U(n)$.<|endoftext|> TITLE: Chains of forking extension in stable theories QUESTION [6 upvotes]: Let $T$ be an stable theory. Further we work in the monster model of $T^{eq}$. We say that a chain of types of the form $$tp(a_1/A_1)\subset tp(a_2/A_2) ... \subset tp(a_n/A_n)$$ is a forking chain if for every $1< i \le n$ the type $tp(a_i/A_i)$ forks over $A_{i-1}$. What can we say about the length of forking chains? For example in strongly minimal theories such a chain (in the home sort) has an maximum length of 1. Moreover, there exists no chain of length $|T|^+$ if and only if $T$ is simple. This is since forking has local character (every type does not fork over a set of size $<|T|$) if and only if $T$ is simple. For theory of Morley rank $N$ such a chain is bounded by $N$. But what about the lower bounds? Can we find a chain of length $n$ in a theory with Morley rank $\ge n$? Does any type $p$ of Morley rank $n$ start an forking chain of length $n$? Do theories without Morley rank have chains of length $|T|$? REPLY [3 votes]: The answer is no to all three questions. First notice that if there exists a maximal finite forking chain, then its length is the SU-rank (or U-rank in the stable context) of $tp(a_1/A_1)$. Since there are theories of Morley-rank $>1$ and U-rank $=1$, this gives negative answer to 1 and 2. Then note that in a supersimple theory there is no infinite forking chain. As $\bigcup_i tp(a_i/A_i)$ would not fork over some finite $B$ and this would be contained in $A_N$ for $N$ sufficiently big. Hence strictly superstable theories are a counter-example to question 3.<|endoftext|> TITLE: Number of solutions to equations in finite groups QUESTION [17 upvotes]: Suppose $G$ is a finite group and that $E$ is an equation of the form $x_1 x_2 ... x_n = e$, where each $x_i$ is in the set of symbols $\{x, y, x^{-1}, y^{-1}\}$. Is it always true that the number of ordered pairs $(g, h)$ of elements of $G$ satisfying $E$ is a multiple of $|G|$? I know that it is true in many cases and I can't find any counterexamples. Does anybody know a simple counterexample, or a suitable reference for this question? REPLY [18 votes]: Yes. This is a special case of results in http://arxiv.org/pdf/1205.2824.pdf<|endoftext|> TITLE: Combinatorial designs textbook recommendation QUESTION [6 upvotes]: Good evening, I am currently taking a class which has combinatorial designs as the first topic, we are using Peter Cameron's book Designs, Graphs, Codes and their Links which I am finding extremely interesting. I am really interested in a rigorous study of combinatorial designs without their links now.Of course I plan to continue taking the class, but I would really like to read a textbook just on designs (I am especially interested in realizable designs (like the social golfer and Kirkman girl problem)). Thank you very much in advance. Regards. REPLY [7 votes]: You could try "Stinson - Combinatorial design theory" too. Also, just in case it might interest you: I am half-learning design theory myself and trying to implement the constructive proof of existence that I find in the software Sage. The goal is to be able to actually build all the combinatorial designs which exist in the litterature. I'd be delighted to not be alone in the attempt. You can see what is already implemented on This page Also, you will find pointers from our implemented constructions to the paper/books that provide the instructions, so that this also partly answers your question. Finally, you may want to find a copy of the "Handbook of Combinatorial Designs" somewhere. It is not a textbook at all, but there is just no way to know what exists and what does not without this ;-) Good luck ! Nathann REPLY [7 votes]: There is a list of references in http://www.maths.qmul.ac.uk/~pjc/design/resources.html#books and http://en.wikipedia.org/wiki/Block_design#References From the latter, I know books by Beth et al (a long and detailed) and by Hughes and Piper (shorter and more readable). All of it is outdated in several important aspects, e.g. the huge progress on existence questions made by P.Keevash in http://arxiv.org/abs/1401.3665 is understandably not there.<|endoftext|> TITLE: A group whose automorphism group is cyclic QUESTION [10 upvotes]: Is there an Abelian group $A$ which is not locally cyclic whose automorphism group is cyclic ? This question was first posted here. REPLY [17 votes]: There's a construction of a rank two (and therefore not locally cyclic) abelian group with endomorphism ring $\mathbb{Z}$, and therefore automorphism group cyclic of order 2, in "On the cancellation of modules in direct sums over Dedekind domains" by L. Fuchs and F. Loonstra, Indagationes Mathematicae, Volume 74, (1971), 163-169 (link)<|endoftext|> TITLE: What is the Hausdorff dimension of this fractal? QUESTION [12 upvotes]: Let $\sum_{i=h}^\infty d_i/b^i $ be the base $b$ representation of $x \geq 0,$ where $b>1$ and the $d_i$ are uniquely determined by the greedy algorithm. For fixed $c>1,$ let $f(x)= \sum_{i=h}^\infty d_i/c^i .$ Since $cf(x)=f(bx),$ the graph of $f$ is self-similar; e.g., its shape is the same on $[0,b^k]$ for all integers $k$. What is its Hausdorff dimension? The graph shares some characteristics with some fractals in Wikipedia's List of fractals by Hausdorff dimension. For $(b,c)=(3,2)$, it looks like so: REPLY [9 votes]: Strictly speaking, the graph is not self-similar. It is (nearly) self-affine. More precisely, if $b>1$ is an integer and $c>1$ is real, then the graph $G$ of $f$ over the interval $[0,b]$ agrees with the invariant set $K$ of an iterated function system containing $b$ affine functions up to a countable set. Specifically, $$K=\bigcup_{i=1}^b T_i(K),$$ where $$T_i(\pmb{x}) = \left( \begin{array}{cc} 1/b & 0 \\ 0 & 1/c \end{array}\right) + \left( \begin{array}{c} i-1 \\ i-1 \end{array}\right),$$ for $i=1,\ldots,b$. Furthermore, the graph $G=K \setminus C$, where $C$ is countable; thus, conclusions about the dimension of $K$ can be extended to the dimension of $G$ for any $\sigma$-stable notion of dimension, such as Hausdorff dimension or modified box-counting dimension. Here's an illustration for $(b,c)=(3,2)$. The three smaller rectangles indicate the action of the IFS on the larger rectangle. The invariant set $K$ of that IFS is, by definition, compact. The graph $G$ of the function $f$ is not closed, however, as $f$ is not continuous; I think it's a mistake to connect the dots as in your linked image. The points of discontinuity are exactly those that have multiple base $b$ representations. The set $K$ contains points with both possible $y$ values but, again those countably many points won't affect the dimension. Falconer has a formula to compute an estimate to the dimension of a self-affine set that often yields the exact result. It's quite a bit more complicated than the corresponding result for strictly self-similar sets and there are special cases where it gives only an upper bound. For a set in the plane whose dimension is known to be at least 1, we use the so-called singular value function: $$\varphi ^s(f) = \alpha \beta ^{s-1},$$ where $\alpha \geq \beta$ are the singular values of the linear part of $f$. Let $J_k$ denote the set of all sequences of integers chosen from $\{1,\ldots ,m\}$. Thus if $\left(i_1,\ldots ,i_k\right)$ is such a sequence then $f_{i_1}\circ \text{$\cdots $f}_{i_k}(E)$ is a small copy of $E$ and the set of all such sets covers $E$ with small sets. Falconer proved that there is a unique number $s$ so that $$\lim_{k\to \infty } \left(\underset{\left(i_1,\ldots ,i_k\right)\in J_k}{\sum }\phi ^s\left(f_{i_1}\circ \cdots \circ f_{i_k}\right)\right){}^{1/k}=1$$ and, furthermore, that this number $s$ is an upper bound for the box dimension of the set $E$. In the case we have here, the expression inside the limit on the left simplifies considerably, since all the functions have the same, diagonalizable linear part. In fact, it's just $$\left(b^k/\left(c^kb^{k(s-1)}\right)^{1/k}=b\left/\left(c b^{s-1}\right)\right.\right..$$ Setting this equal to one and solving for $s$ we get $s=2-\log(c)/\log(b)$, in agreement with Martin's answer. Recall that the singular values satisfied $\alpha \geq \beta$. That is, they are ordered. Thus, an implicit assumption here was that $b\geq c$. Otherwise, the formula returns a result smaller than 1, which is clearly incorrect. In the case where $b TITLE: Massey products and $A_{\infty}$ structures QUESTION [5 upvotes]: I know the general theorem of Kadeishvili which says that, for a DGA $C$, when $H^{i}(C)$, $i\geq 0$, is free, $H(C)$ can be made into an $A_{\infty}$ algebra. If my understanding is correct, the proof essentially uses the freeness of $H^{i}(C)$ to produce a section $$s:H(C)\to C$$ of the projection $p:C\to H(C)$. Then a choice of chain homotopy between the identity and $sp$ can be used to define the maps $$m_{i}:\otimes^{i}H(C)\to H(C).$$ It is not hard to see that these give you an element of the higher Massey products (when defined). Here is my question: When $H^{i}(C)$ is not free, is there any good way to define an $A_{\infty}$ structure on $H(C)$? It seems that the fact that one has an actual chain homotopy is crucial to the construction and in general $p$ may only be a quasi-iso. I hope this question is not too elementary (I am still a lowly graduate student). It just seems odd to me that the higher Massey products can be defined in general, even for torsion $H(C)$ (as long as the lower ones vanish) but one may not be able to identify these with a full $A_{\infty}$ structure. REPLY [7 votes]: Let $C = \Bbb Z[x,y] \otimes \Lambda[u,v]$, with $x$ and $y$ in (homological) degree 2 and $u$ and $v$ in degree 3, with $dx = dy = 0$ and $du = 2x$, $dv = 2y$. Then $H_5(C)$ is $\Bbb Z/2$, generated by the Massey product $x v - u y = \langle x,2,y\rangle$ (with no indeterminacy in this case). This structure does not come from the homology equipped with an $A_\infty$ structure, because then we would have $$\langle x,2,y \rangle = m_3(x \otimes 2 \otimes y) = 2 \cdot m_3(x \otimes 1 \otimes y) = 0.$$ The usual generalization clarifies that the original could be viewed as, instead of a theorem about homology, a theorem about chain equivalence: if $C \to D$ is a chain homotopy equivalence and $C$ is a DGA, then $D$ gets an $A_\infty$ structure.<|endoftext|> TITLE: Most dense subset of numbers that avoids arbitrarily long arithmetic progressions QUESTION [11 upvotes]: The famous Green-Tao theorem says that there exist arbitrarily long sequences of primes in arithmetic progression. I am wondering: How dense can a subset $S \subset \mathbb{N}$ be and still avoid arbitrarily long sequences of elements of $S$ in arithmetic progression? To make this more precise (following a comment by Robert Israel), Q. What is the cardinality of the largest subset $S_n$ of $[1,n]=\{1,2,3,\ldots,n\}$ that avoids $k$-term arithmetic progressions of elements in $S_n$, as a function of $n$ and $k$? As $n \to \infty$, can the density be significantly more dense than the primes density, $n / \log_e n$? I suspect this is a well-studied question, in which case quotes and/or pointers would suffice. Thanks! REPLY [16 votes]: You are essentially asking for quantitative estimates on Szemerédi's theorem, which states that the largest subset of $[1,n]$ without a k-term arithmetic progression has size $o(n)$. To be precise, let us define $r_k(n)$ to be the largest subset of [1,n] with no k-term arithmetic progression. Then a construction due to Behrend (essentially projecting a high-dimensional sphere onto the integers) shows that $$ r_3(n) = \Omega\left(n e^{-c \sqrt{\log n}}\right), $$ while a result of Bloom (moderately improving on a result of Sanders), shows that $$ r_3(n) = O\left(n \frac{(\log \log n)^4}{\log n}\right). $$ For general $k$, the best known upper bound is due to Gowers and says that $$ r_k(n) = O\left(\frac{n}{(\log \log n)^{c_k}}\right) $$ for an appropriate $c_k$. Behrend's construction clearly provides a lower bound in this case as well, but may be improved a little by projecting a collection of concentric spheres. There is some evidence (see, for example, http://arxiv.org/pdf/1408.2568.pdf) to believe that the lower bound is closer to the truth.<|endoftext|> TITLE: Bound on the sum of arguments QUESTION [6 upvotes]: Problem: Show that for all real $s,t,u$ and all complex $z$ with $|z|<1$ one has $$(*)\qquad \arg\frac{1-zf(s-u)}{1-zf(s+u)} +\arg\frac{1-zf(t+u)}{1-zf(t-u)}<\pi, $$ where $f$ is the characteristic function of a probability distribution $\mu$, so that $f(t)=\int_{-\infty}^\infty e^{itx}\mu(dx)$ for all real $t$, and, as usual, $\arg w\in(-\pi,\pi]$ for any nonzero complex number $w$. Comments: Letting $(t_1,\dots,t_4):=(s,t,u,-u)$, note that the matrix $M:=(f(t_k-t_j))_{k,j=1}^4$ is Hermitian and nonnegative definite or, equivalently, its principal minors (or just the leading principal minors) are nonnegative. This condition on the minors can be rewritten as a system of polynomial inequalities in the real variables $a_{k,j}:=\Re f(t_k-t_j)$ and $b_{k,j}:=\Im f(t_k-t_j)$, with $k,j=1,\dots,4$ and $k>j$. Note that $f(s-u),f(s+u),f(t+u),f(t-u)$ are elements of the matrix $M$. Note also that inequality (*) can be expressed as a logical formula whose terms are polynomial inequalities in the real and imaginary parts of the complex numbers $z,f(s-u),f(s+u),f(t+u),f(t-u)$. Numerical experiments suggest that the mentioned condition on the minors together with the condition $|z|<1$ are enough for inequality (*), on the sum of the arguments, to hold. In principle, this can of course be verified purely algorithmically, but it appears to take just too much computing resources for the calculation to complete. Any other idea? I think tools provided in arXiv:0907.2960 [math.GN] may be useful here. Fedor Petrov: your first inequality, $$\operatorname{arg}\frac{1−zf(s−u)}{1−zf(s+u)}<π/2−\operatorname{arg}(1+z\bar{z}f(2u))$$ (as well as the second one, of course) is incorrect, in general. E.g., take $s=0$, $u=\pi/2-h$, $h\downarrow0$, $00 $$ in your last display is incorrect as well, and even the inequality you apparently wanted to get from that identity is also in general incorrect. E.g., for $A=-i$, $B=1+i$, and $C=i$, one has $2\Re (1-A)(1-\bar{B})(1+C)=-4<-1=\det M$. What I found somewhat fascinating here, though, is that the above identity would be true if one could change just the sign of one monomial in the polynomial expansion (in the real and and imaginary parts of $A,B,C$) of the difference between the left-hand side and the right-hand side of this incorrect identity. Of course, one can rewrite the nonstrict version of the inequality (*) that I proposed as $\Psi_{f,z}(u)+\Psi_{f,z}(-u)\le\pi$, where $$\Psi_{f,z}(u):=\sup_{s\in\mathbb{R}}\arg\frac{1-zf(s-u)}{1-zf(s+u)}, $$ and then try to find a manageable upper bound $\Theta_{f,z}(u)$ on $\Psi_{f,z}(u)$ such that $\Theta_{f,z}(u)+\Theta_{f,z}(-u)\le\pi$. However, I would be greatly surprised if such a simple bound as your $\pi/2- \arg(1+z\bar{z}f(2u))$ would do. I still think the best bet so far is to try to use tools provided in arXiv:0907.2960 [math.GN]. As for your question about what real algebraic geometry has to do with this problem: real algebraic geometry is to a large extent about solving systems of polynomial inequalities over $\mathbb{R}$ (including various Positivstellensatzen), which is clearly relevant here. REPLY [5 votes]: Let's prove that $$ \arg \frac{1-zf(s-u)}{1-zf(s+u)}< \pi/2- \arg(1-z\bar{z}f(2u)). $$ Then summing this up with an analogous inequality $$ \arg \frac{1-zf(t+u)}{1-zf(t-u)}< \pi/2- \arg(1-z\bar{z}f(-2u)) $$ we get what we need. Denote $zf(s-u)=A$, $zf(s+u)=\bar{B}$, $z\bar{z}f(2u)=C$. Then for functions $\varphi_1(x)=e^{isx}$, $\varphi_2(x)=\bar{z}e^{iux}$, $\varphi_3(x)=\bar{z}e^{-iux}$ in $(L^2,\mu)$ we have $A=\langle\varphi_1,\varphi_2\rangle$, $\bar{B}=\langle\varphi_1,\varphi_3\rangle$, $C=\langle\varphi_2,\varphi_3\rangle$. Thus the following matrix is non-negative definite as the Gram matrix of our functions $$ \pmatrix{\|\varphi_1\|^2&A&\bar{B}\\\bar{A}&\|\varphi_2\|^2&C\\B&\bar{C}&\|\varphi_3\|^2}. $$ We may increase diagonal elements upto 1. Of course, matrix remains non-negative definite and, moreover, becomes positive definite (since $|z|<1$, we strictly increase two diagonal elements of our matrix, and its determinant becomes strictly positive): $$ M=\pmatrix{1&A&\bar{B}\\\bar{A}&1&C\\B&\bar{C}&1},\,\det M>0. $$ What we have to prove is that $\arg (1-A)(1-B)=\arg \frac{1-A}{1-\bar{B}}<\pi/2-\arg(1-C)$. Assume the contrary, i.e. $\arg(1-A)+\arg(1-B)\geq \pi/2-\arg(1-C)$. Since $\arg(1-A),\arg(1-B),\arg(1-C)\in (-\pi/2,\pi/2)$ we get $\theta:=\arg (1-A)+\arg (1-B)+\arg (1-C)\in [\pi/2,3\pi/2]$, i.e. $(1-A)(1-B)(1-C)=e^{i\theta}r$, $r>0$, $\Re (1-A)(1-B)(1-C)=r\cos \theta\leq 0$. After this point there should be more elegant way to finish the proof, but let me provide at least some way. First of all, $$ \det M=1-|A|^2-|B|^2-|C|^2+2\Re(ABC)=(1-|A|^2)(1-|B|^2)-|C-\bar{A}\cdot \bar{B}|^2> 0. $$ Thus $C=\bar{A}\cdot \bar{B}+w$, $|w|< R:=\sqrt{(1-|A|^2)(1-|B|^2)}$. Under these conditions the best lower estimate for $\Re (1-A)(1-B)(1-C)$ is $$ \Re (1-A)(1-B)(1-C)>\Re (1-A)(1-B)(1-\bar{A}\cdot \bar{B})-R\cdot |1-A|\cdot |1-B|. $$ Now we have \begin{align*} X:=\Re (1-A)(1-B)(1-\bar{A}\cdot \bar{B})=\Re (1-A)(1-B)\left((1-\bar{A})+\bar{A}\cdot (1-\bar{B})\right)=\\ =|1-A|^2(1-\Re B)+|1-B|^2(\Re A-|A|^2) \end{align*} Analogously $X=|1-A|^2(\Re B-|B|^2)+|1-B|^2(1-\Re A)$. Taking half sum of two expressions for $X$ and applying AM-GM for two summands we get $$ X=\frac{|1-A|^2(1-|B|^2)+|1-B|^2(1-|A|^2)}2\geq \sqrt{(1-|B|^2)(1-|A|^2)}\cdot |1-A|\cdot |1-B|, $$ hence $\Re (1-A)(1-B)(1-C)>0$, a contradiction.<|endoftext|> TITLE: A proper smooth surface is projective QUESTION [10 upvotes]: My question is a reference request for the following fact: if $k$ is a field and $X$ a proper smooth surface over $k$, then $X \rightarrow \mathrm{Spec}\, k$ is projective. Where is this well-known fact proved (in the stated generality)? REPLY [7 votes]: Quoting from a very nice paper by Stefan Schroeer we have: "The criterion of Zariski [3, Cor. 4, p. 328] tells us that a normal surface $Z$ is projective if and only if the set of points $z \in Z$ whose local ring $\mathcal{O}_{Z,z}$ is not $\mathbf{Q}$-factorial allows an affine open neighborhood." The reference [3] is the following: S. Kleiman: Toward a numerical theory of ampleness. Annals of Math. 84 (1966), 293–344. The paper of Stefan is here.<|endoftext|> TITLE: Maximal volume of a simplex inscribed in a spherical cap QUESTION [6 upvotes]: Let $B_n$ be the $n$-dimensional unit ball, and $B_n(\varepsilon)$ be the spherical cap with height $\varepsilon$ I am interested in the quantity $$\Gamma:=\sup_{\Delta:\textrm{ inscribed simplex in }B_n(\varepsilon)}\mathrm{vol}(\Delta)$$ I think this should be a classical result and have very nice upper and lower bounds on the order $\varepsilon^{(n+1)/2}$. The thing I am interested in knowing some good bounds for the constants in terms of $n$, but unfortunately I wasn't able to settle down with a good reference for this. Any help would be greatly appreciated. REPLY [2 votes]: I don't know any reference for this, and I don't know if this should be a "classical result", but let me give a lower bound, which might even be tight. Let's denote the base of the cap by $BS$. It is a sphere of dimension $n-2$ with radius $$r=\sqrt{1-(1-\varepsilon)^2}=\sqrt{2\varepsilon-\varepsilon^2}.$$ Let $\Delta_{n-1}^R$ denote a regular simplex of dimension $n-1$ inscribed in a sphere of dimension $n-2$ with radius $R$. If I am not mistaken, the volume can be calculated as follows: $$\text{vol}_{n-1}(\Delta_{n-1}^1)=\frac{n^\frac{n}{2}}{(n-1)^{\frac{n-1}{2}}(n-1)!}$$ and $$\text{vol}_{n-1}(\Delta_{n-1}^R)=R^{n-1}\text{vol}_{n-1}(\Delta_{n-1}^1).$$ Now if we define an $n$-simplex $P_n(\varepsilon)$ as the convex hull of the apex of the cap together with the vertices of a regular $(n-1)$-simplex $\Delta_{n-1}^r$ inside the sphere $\partial BS$, we can calculate its $n$-dimensional volume as follows: $$\begin{align}\text{vol}_n(P_n(\varepsilon))&=\frac{\varepsilon}{n}\text{vol}_{n-1}(\Delta_{n-1}^r)\\ &=\frac{\varepsilon}{n} r^{n-1}\text{vol}_{n-1}(\Delta_{n-1}^1)\\ &=\frac{\varepsilon}{n} (\sqrt{2\varepsilon-\varepsilon^2})^{n-1}\text{vol}_{n-1}(\Delta_{n-1}^1)\\ &=\frac{\varepsilon}{n} (\sqrt{2\varepsilon-\varepsilon^2})^{n-1}\frac{n^\frac{n}{2}}{(n-1)^{\frac{n-1}{2}}(n-1)!} \end{align} $$ Here is an illustration for $n=3$: This agrees with the order $\varepsilon^\frac{n+1}{2}$ for $\varepsilon\rightarrow 0$ that you expected and you can easily get complete asymptotics to all orders. Clearly this is a lower bound: $$\Gamma\geq\text{vol}_n(P_n(\varepsilon)).$$ I find it plausible that this bound is tight. For this you would need to show two things are true for small $\varepsilon$: a largest simplex in the cap has all but one vertex in the base of the cap (see comment by Joseph O'Rourke.) for a largest simplex those vertices in the base of the cap form a regular simplex.<|endoftext|> TITLE: Can you write $\mathbb R^2$ as a disjoint union of two totally disconnected sets? QUESTION [24 upvotes]: Can you write $\mathbb R^2$ as a disjoint union of two totally disconnected sets? REPLY [13 votes]: Here's a proof that if $X$ is any simply connected Hausdorff space such that $X\setminus \{p\}$ is path connected for all $p\in X$, then the complement of any totally disconnected subset is connected. In particular, if $X$ has more than one point then it cannot be the disjoint union of two totally disconnected subsets. Taking $X=\mathbb{R}^2$ answers the question. [This is inspired by Włodzimierz Holsztyński's first answer, by trying to prove the result in a similar way, but using as few advanced properties of $\mathbb{R}^2$ as possible. I expect that the connectedness properties can be replaced by a statement about the Čech cohomology, such as $\check{H}^1(X)=0$ and $X\setminus\{p\}$ is connected for each $p$]. I will use proof by contradiction. So, suppose that $A\subseteq X$ is totally disconnected such that $B\equiv X\setminus A$ is not connected. Then, there are open $U,V\subseteq X$ such that $B\cap U$, $B\cap V$ are disjoint and nonempty and such that $B\subseteq U\cup V$. So, $W= U\cap V$ is disjoint from $B$ and, hence, $W\subseteq A$. If $W$ was nonempty, then choosing points $p\in W$ and $q\in X\setminus\{p\}$, there is a continuous $\gamma\colon[0,1]\to X$ with $\gamma(0)=p$, $\gamma(1)=q$. Letting $t$ be maximal such that $[0,t)\subseteq\gamma^{-1}(W)$ then $t > 0$ and the Hausdorff hypothesis implies that $\gamma$ is not constant on $[0,t)$. So, $\gamma([0,t))$ is a subset of $A$ containing more than one point, and is connected, contradicting the fact that $A$ is totally disconnected. Hence, $W=\emptyset$. So, we have constructed a disjoint nonempty pair $U,V$ of open subsets of $X$ such that $C=X\setminus (U\cup V)$ is contained in $A$ and, therefore, is totally disconnected. That, is $C$ is a totally disconnected closed set which disconnects $X$. I'll prove that this is impossible using a bit of simple intersection theory. Let $S$ be a closed subset of $C$ such that $C\setminus S$ is closed. If $\gamma\colon[0,1]\to X$ is a path with $\gamma(0),\gamma(1)\in X\setminus S$, then we can define the intersection number of $\gamma$ with $S$ as follows. Choose $0=t_0\le t_1\le\cdots\le t_n=1$ such that each $\gamma(t_k)\in X\setminus S$ and $\gamma([t_{k-1},t_k])$ is contained in one of the open sets $X\setminus S$ or $X\setminus (C\setminus S)$. On the interval $[t_{k-1},t_k]$ we can assign an intersection number of $0$ if $\gamma([t_{k-1},t_k])\subseteq X\setminus S$, otherwise we assign the number $F(\gamma(t_k))-F(\gamma(t_{k-1}))$, where $F=1$ on $V$ and $F=0$ on $U$. Sum these up to get the intersection number of $\gamma$ with $S$. It can be seen that adding additional points to the $t_k$ does not change the intersection number, so it is independent of the choice of the $t_k$. It can also be seen that the intersection number will be unchanged under small changes in the path, so homotopic paths have the same intersection numbers. Now choose an arc $\gamma\colon[0,1]\to X$ with $\gamma(0)\in U$ and $\gamma(1)\in V$, and let $t$ be the supremum of $\gamma^{-1}(U)$. By the hypothesis, there is a path $\tilde \gamma\colon[0,1]\to X\setminus\{\gamma(t)\}$ with $\tilde\gamma(0)=\gamma(0)$ and $\tilde\gamma(1)=\gamma(1)$. As $\gamma^{-1}(C)$ is totally disconnected, there are points $t_0 < t < t_1$ arbitrarily close to $t$ such that $\gamma(t_0)\in U$, $\gamma(t_1)\in V$ and, choosing them close enough, $\gamma([t_0,t_1])$ will be disjoint from the image of $\tilde\gamma$. Then, as $T_1=\gamma([t_0,t_1])\cap C$ and $T_2=(\gamma([0,t_0])\cup\gamma([t_1,1])\cup\tilde\gamma([0,1]))\cap C$ are disjoint compact subsets of $C$, we can use total disconnectedness to find a closed $S\subset C$ containing $T_1$ with $C\setminus S$ closed and containing $T_2$. Then, the intersection number of $\gamma$ with $S$ is 1 and the intersection number of $\tilde\gamma$ with $S$ is $0$, so the paths are not homotopic, contradicting simply connectedness of $X$.<|endoftext|> TITLE: Is factorial definable using a $\Delta_0$ formula? QUESTION [16 upvotes]: The factorial function is primitive recursive, and therefore definable by a $\Sigma_1$ formula. Is it also definable by a $\Delta_0$ formula (i.e. bounded quantifiers)? If not, why? REPLY [8 votes]: I didn't look at the above references, and I took this as an exercise for myself. Here is a way to express "$x! = y$" as a $\Delta_0$ formula with two free variables $x$ and $y$. The idea is to check that, for each prime $p \le x$, $p$ divides $y$ the right number of times. The number of times the prime $p$ divides $x!$ is $f(x,p) = \sum_{i\ge 1} \lfloor x/p^i \rfloor$. For example, the number of times $7$ divides $1000!$ is $142+20+2=164$. Note that each element $a_{i} = \lfloor x/p^i \rfloor$ in this sum is obtained from the previous one by $a_{i} = \lfloor a_{i-1}/p \rfloor$. Now, let us recall a method by Nelson of encoding sets using $\Delta_0$ formulas. Consider for example the set $S = \{6,13\}$. Since $6=110_{2}$ and $13=1101_2$ in binary, we encode $S$ by the number $2110211012_4$ (or $2110121102_4$). The $2$'s serve as separators between the elements of the set. Nelson shows how to express "$x \in S$" as a $\Delta_0$ formula. Also, recall the pairing formula $(a,b) = (a+b)(a+b+1)/2+a$. Hence, we can express "$c=(a,b)$" by the formula $2c=(a+b)(a+b+1)+2a$. Given $x$ and a prime $p$, let $S(x,p)$ be the set of pairs $S(x,p) = \{(u_1,v_1), \ldots, (u_j,v_j)\}$ where $u_i = a_{i}$ as defined above, $v_i$ is the partial sum $v_i = \sum_{k\le i} u_i$, and $j$ is the least element for which $u_j TITLE: Analytic perturbation of eigenfunctions QUESTION [5 upvotes]: Consider a domain $\Omega_0 \subset \mathbb{R}^n$, and deformations of $\Omega_0$, called $\Omega_t$, obtained by a one-to-one mapping $x \mapsto x + t\varphi (x)$, where $\varphi$ is smooth. It is known that the Dirichlet and Neumann eigenvalues of the Laplacian on $\Omega_t$ vary real analytically with respect to $t$ near $t = 0$. My question is: do the eigenfunctions vary real analytically as well? Any reference would be appreciated. REPLY [2 votes]: See this recent paper for answers to this and similar questions. You have to translate the parameter dependence of the domain to the parameter dependence of the operator first.<|endoftext|> TITLE: Irrationality of Dedekind zeta values QUESTION [7 upvotes]: For Riemann's zeta function, one knows that: $\zeta(2n)$ is irrational (because a rational multiple of $\pi^{2n}$ is) $\zeta(3)$ is irrational (proved by Apéry) and a few other results like "there are infinitely many irrational values at odd integers" (Ball-Rivoal). Is there something known for the values of the Dedekind's zeta function $\zeta_K(s)$ of a number field? For instance, do we know the irrationality of $\zeta_K(2)$ in some cases? REPLY [7 votes]: As with all motivic L-functions, irrationality of special values breaks into two rather different problems: Critical values. In the case of Dedekind zeta functions, those are the even positive integers, and the strategy is to show that $\zeta_K(-2n-1)$ is rational, so that from the functional equation follows that $\zeta_K(2n)$ is irrational. Of course, this only works for totally real number fields, since that's the only case where the functional equation doesn't vanish at the odd integers. This is precisely the Siegel-Klingen theorem, $\zeta_K(2n)$ is a rational multiple of $\pi^{2n[K:\mathbb{Q}]}$, and therefore irrational. The actual value of that algebraic number is in general not known, see the Lichtenbaum conjectures. Non-critical values. Odd positive integers. This is the (even more) difficult part, because you have to actually prove that something non-trivial is irrational. As you mention, some results are known for Riemann's zeta, but as far as I know, there are no known results for number fields other than $\mathbb{Q}$. (On a side note, a generalized "infinitely many irrational numbers" type result for Dirichlet L-functions was published a few years ago by Masaki Nishimoto (paper here)).<|endoftext|> TITLE: Is there a quotient or exact sequence of symmetric, premodular (ribbon fusion) and modular categories? QUESTION [5 upvotes]: In Walker and Wang's article about (3+1)-TQFTs from premodular categories, they say on page 14 that you can take a quotient of a premodular category $\mathcal{C}$ by its symmetric fusion subcategory $\mathcal{S_C}$ consisting of transparent objects. That quotient is promised to be modular. I'm willing to believe this and I'd like to use it, but I can't find a mathematical reference for it, nor have I managed to work it out in all detail. Is this a known fact? Is it treated somewhere? My attempt of making sense of it, is the following: Take a premodular category $\mathcal{C}$. That's just a ribbon category which is fusion (finitely semisimple etc.). Take all the transparent (trivially braiding) objects. They form a subcategory $\mathcal{S_C}$. Form an "exact sequence" $0 \to \mathcal{S_C} \to \mathcal{C} \to \mathcal{Q_C} \to 0$. I have no idea what exactly an exact sequence of ribbon fusion categories is or whether we even have to consider this. I'm vaguely expecting something like this: $\mathcal{Q_C}$ has the same objects as $\mathcal{C}$, but somehow the morphisms to and from the objects in $\mathcal{S_C}$ are identified with 0. What I don't understand about it is: How does it work exactly? How do I prevent the monoidal unit to get killed? It's in $\mathcal{S_C}$ after all. Are the simple objects in the quotient also simple in $\mathcal{C}$? Do they braid the same, i.e. is the quotient obviously modular? REPLY [6 votes]: Short answer: Look for papers on "deequivariantization". (I think the original references are by Müger and Brugières, but I am not sure whether they used the term "deequivariantization".) Longer answer: The procedure mentioned in the paper is actually a sort of predual to the deequivariantization procedure found in the literature, in the sense that $Rep(C/S) \cong Deeq(Rep(C))$. So it is very closely related to deequivariantization, but not exactly the same thing. You attempt was on the right track. One starts with $C$, then adds isomorphisms between objects of $S$ and the trivial object. This has to be done carefully, of course. One uses the fact that $S \cong Rep(G)$ for some finite group $G$, and $Rep(G)$ has a fiber functor. A fancier way of thinking of it goes as follows. Since $S$ is symmetric monoidal we can think of it as an $n$-category for any $n$, and in particular we can think of it as a 4-category. The 3-category $C$ then becomes a module 3-category for the 4-category $S$. We can construct another module 3-category $F$ for $S$ by using the isomorphism $S\cong Rep(G)$ and the fiber functor for $Rep(G)$. Now we can now define $$ C/S = C \otimes_S F $$ $F$ is actually a $S$-$G_4$ bimodule 3-category, where $G_4$ denotes the finite group $G$ thought of as a 4-category. It follows that $C/S$ has a $G$ action. (A $G$ action on a 3-category is the same thing as a $G_4$-module structure on that 3-category.) I have a preprint which fills in the details of the above arguments, but it is not yet ready for the arXiv. If you go to this page and look at the slides from talks at Vienna and Princeton (Feb 2014), you can find some of the details.<|endoftext|> TITLE: Geometric interpretation of Cusps for general groups? QUESTION [9 upvotes]: Let $\mathrm{G}$ be a reductive group over a number field $F$, but for simplicity we can think about $\mathrm{G}=\mathrm{GL_n}$ for $n>2$ and $F =\mathbb{Q}$. Then for an automorphic form, $\varphi : \mathrm{G}(F)\backslash\mathrm{G}(\mathbb{A}) \to \mathbb{C}$ is said to be cuspidal if for each parabolic subgroup $\mathrm{P}=\mathrm{MU}$, $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\int_{\mathrm{U}(F)\backslash\mathrm{U}(\mathbb{A})} \varphi(ug)du = 0$, where $\mathrm{U}$ is the unipotent radical of $\mathrm{P}$. In the case that $\mathrm{G}=\mathrm{GL_2}$, then we can unravel the adelic language to the more classical setting, where this becomes $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\int_0^1f(x+iy)dx = 0$, (ie: the constant term vanishes) where $f$ is a Maass form which for simplicity I will assume has level 1. In this setting, we view $f$ as a function on the upper half plane, $\qquad\qquad\qquad\qquad\qquad\qquad\qquad H \cong Z\backslash\mathrm{GL_2}(\mathbb{R})/\mathrm{SO}(2)$, and $H$ comes equipped with a "canonical" hyperbolic metric $ds^2 = \frac{dxdy}{y^2}$. Passing to the quotient, $\mathrm{SL_2}(\mathbb{Z})\backslash H$, we get a punctured sphere with one geometric cusp at infinity, and the requirement that $f$ be a cusp form reduces to $f$ "vanishing at the cusp". For more general $\mathrm{G}$ (eg: $\mathrm{G}=\mathrm{GL_3}$), I am curious about analogous geometric understanding of unipotent radicals of various parabolic subgroups corresponding to the "cusp at infinity". I don't know if this is better to think about this on the adelic quotient, $\mathrm{G}(F)\backslash\mathrm{G}(\mathbb{A})$, or perhaps on a locally symmetric space like $ Z\backslash\mathrm{G}/\mathrm{K}$. Specifically, Is there a geometric way of understanding how the unipotent radicals correspond to "cusps" beyond simply as failure of a fundamental domain to be compact, perhaps with respect to some nice metric, for groups beyond $\mathrm{GL_2}$? I am interested in answers in either the classical or adelic language. REPLY [2 votes]: It's actually a lot easier to write a short comment than to say something more precise. In any case, I can only say something in the classical language of locally symmetric spaces. The shortest answer is that the geometric understanding of the cusps is revealed in the various compactifications of locally symmetric spaces that can be considered. The book of Borel and Ji (the link was already given in user40276's answer) gives a good overview of the compactifications that have been considered and their relations. For the question, there seem to be at least two compactifications which are relevant. On the one hand, the metric setting you described generalizes to the geodesic compactification. Every locally symmetric space $\Gamma\backslash G/K$ inherits a metric, and you can compactify it by adding geodesics going off to infinity (but probably this works better if you consider torsion-free arithmetic groups $\Gamma$). The structure at infinity that you add in the compactification is the Tits building for the algebraic group $G(\mathbb{Q})$ modulo the action of $\Gamma$. In the case of $SL_2\mathbb{Z}$ you end up adding one point, because $\mathbb{Z}$ having class number one implies that the action of $GL_2\mathbb{Z}$ is transitive on the parabolic subgroups of $GL_2\mathbb{Q}$. However, the unipotent radicals are not really visible in this compactification. To see some relation to unipotent subgroups, you can consider the Borel-Serre compactification. One of the main features of the Borel-Serre compactification is that the inclusion of the locally symmetric space into its compactification is a homotopy equivalence. The compactification adds part of the Langlands decomposition of the $\mathbb{R}$-points of parabolic subgroups modulo the action of the arithmetic subgroup $\Gamma$. In the case of $SL_2\mathbb{Z}$, the locally symmetric space is a punctured sphere, and the Borel-Serre compactification adds an $S^1$ at infinity. This $S^1$ is given as the $\mathbb{R}$-points of the unipotent radical of the (unique up to $SL_2\mathbb{Z}$-conjugacy) Borel subgroup of $SL_2\mathbb{Q}$ modulo the action of the $\mathbb{Z}$-points. Some of this generalizes to the higher rank cases, but it is better to look up the exact definitions in the book of Borel and Ji.<|endoftext|> TITLE: Is Scholl construction of modular motives related to Deligne's construction of $\ell$-adic representations? QUESTION [8 upvotes]: First of all, I need to declare my extreme ignorance on the topic of modular forms, so, please, does not assume that I know Deligne's construction in details. In Motives for modular forms, Scholl constructs motives associated to modular forms . Deligne's also constructed $\ell$-adic representations attached to modular forms of weight $k \geq 2$. For $k=2$, Shimura constructed an abelian variety $A_f$ and then obtained a $\ell$-adic representation from it's Tate module. It's known that Scholl construction coincides with Shimura construction, i.e., the motive attached to a modular form of weight $k = 2$, namely $M_f$ coincides with $M(A_f)$. My question is if Scholl construction is compatible with Deligne's construction in the sense that these motives are the underlying geometric objects of Deligne's constructions. More explicitly, I would like to know if from these motives $M_f$ I can create an $\ell$-adic representation with values in some object of cohomological nature arising from $M_f$ (like motivic cohomology) such that this representation is the one constructed by Deligne. REPLY [7 votes]: More explicitly, I would like to know if from these motives $M_{f}$ I can create an $\ell$-adic representation with values in some object of cohomological nature arising from $M_{f}$ (like motivic cohomology) such that this representation is the one constructed by Deligne. The answer is: of course, and in some sense almost by definition. When we say a Galois representation $\rho_{f}$ is attached to an eigencuspform $f$, this means that the $L$-function of $\rho_f$ and of $f$ are equal and, likewise, when we say that a motive $M(f)$ is attached to $f$, this means that the motive and the modular form have the same $L$-function (where the $L$-function of a motive $M$ is the $L$-function of the $p$-adic étale representation attached to the $p$-adic étale realization of $M$ at a prime $p$ of good reduction of $M$). Hence, $\rho_f$ and $M(f)_{\operatorname{et},p}$ have the same $L$-function by definition and so (as $\rho_{f}$ is semi-simple) the $\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)$-representations $\rho_{f}$ and $M(f)$ are isomorphic. But let me be more precise, because we actually know much more in this case. Let $k≥2$ be an integer and let $M=(KS_k,\Pi)$ be the (Chow) motive constructed by Scholl, where $KS_{k}$ is the Kuga-Sato variety (that is to say the canonical desingularization of the $k-2$-fold fiber product of the universal generalized elliptic curve $\pi:\bar{E}\longrightarrow X(N)$ over the modular curve $X(N)$) and where $\Pi$ is the projector arising from the action of the wreath product of $((\mathbb Z/2\mathbb Z)^2\rtimes\{\pm 1\})$ with $\mathfrak S_{k-2}$ on $\bar{E}^{k-2}$ (and hence on $KS_k$). Let $f\in S_{k}(\Gamma(N))$ be a normalized eigencuspform. Let $M(f)$ be the (Grothendieck) motive cut out of $M$ by the $f$-component of the action of the Hecke algebra. Finally, denote by $H^{1}_{\operatorname{et}}(X(N)\times_{\mathbb Q}\bar{\mathbb Q},\operatorname{Sym}^{k-2}R^1\pi_*\mathbb Q_{p})$ the Galois representation constructed by Deligne and by $V(f)$ its quotient corresponding to $f$. Then there is a canonical Galois equivariant isomorphism \begin{equation} M_{\operatorname{et},p}:=H^{k-1}(KS_k(\mathbb C),\mathbb Q)(\Pi)\otimes_{\mathbb Q}\mathbb Q_{p}\simeq H^{1}_{\operatorname{et}}(X(N)\times_{\mathbb Q}\bar{\mathbb Q},\operatorname{Sym}^{k-2}R^1\pi_*\mathbb Q_{p}) \end{equation} inducing a Galois equivariant isomorphism $M(f)\simeq V(f)$. So there is a geometric description of Deligne's Galois representation for $k\geq 2$ just as you suspected.<|endoftext|> TITLE: Implicit Function Theorem on Singular Varieties QUESTION [8 upvotes]: Let $X$ and $Y$ be two complex reduced affine algebraic or analytic varieties, possibly singular. Take a regular proper function $$f\colon X \to Y $$ and assume that it is bijective at the level of $\mathbb{C}$-points. Moreover, assume that for every point $x$ of $X$ the differential $$df(x)\colon T_xX \to T_{f(y)}Y $$ is an isomorphism, where the tangent spaces are Zariski tangent spaces. Can we conclude that $X$ and Y are isomorphic? Do we need any assumption on the singularities? In particular, should $Y$ be normal? In other words, I am asking under which conditions on the singularities the implicit function theorem holds. I remember some notes by Kollar, where this issue was related with Canonical singularities, but I could not find them anymore. Any reference or example is welcome. thanks REPLY [3 votes]: In Joe Harris' book, Algebraic Geometry, this is theorem 14.9.<|endoftext|> TITLE: Ultraweak topology on B(X): Is the map X\otimes X* -> B(X)* isometric? QUESTION [16 upvotes]: Let $X$ be a Banach space. Consider the map $$ \alpha\colon X\hat{\otimes} X^* \to B(X)^*, $$ defined one simple tensors as $$ \alpha(\xi\otimes\eta)(a) = \eta(a(\xi)).\quad (\xi\in X, \eta\in X^*, a\in B(X)) $$ Put differently, we consider the pairing between $X\hat{\otimes} X^*$ (the projective tensor product of $X$ and its dual space $X^*$) and $B(X)$ (the algebra of bounded linear operators on $X$) induced by $$ \langle \xi\otimes\eta, a\rangle = \langle a(\xi), \eta \rangle. $$ Question: Is $\alpha$ isometric? This question arose when I tried to make sense of the ultraweak topology on $B(X)$ for a (nonreflexive) Banach space $X$. The question asks if $B(X)$ always has a distinguished part of its dual, which could be thought of as a generalization of a predual. However, even if $\alpha$ is isometric, I don't think it necessarily implies that $X\hat{\otimes} X^*$ is a predual of $B(X)$. One reformulation of the question is as follows: Consider the map $$ \beta\colon B(X) \to B(X^*) $$ which sends an operator $a\in B(X)$ to its transpose $a^*\in B(X^*)$. This map is isometric and we can therefore think of $B(X)$ as a subspace of $B(X^*)$. Then $\alpha$ is isometric if and only if $B(X)$ is $1$-norming in $B(X^*)$ - meaning that the unit ball of $B(X)$ is weak*-dense in the unit ball of $B(X^*)$, for the weak*-topology on $B(X^*)$ induced by the isometric isomorphism $B(X^*)\cong (X\hat{\otimes} X^*)^*$. Some more remarks: Originally I had written: "We know that $B(X)$ is always weak*-dense in $B(X^*)$. (In general, this is weaker then being $1$-norming.) Translating back to $\alpha$, this means that $\alpha$ is always injective." However, as pointed out by Bill Johnson below, not even that is the case. So the answer to my question is a big "NO". If $X$ is reflexive, then $\beta$ is surjective, and so $\alpha$ is isometric. (This can also be seen more directly.) If $X$ has the metric approximation property, then already the unit ball of $F(X)$, the finite rank operators on $X$, is weak*-dense in the unit ball of $B(X^*)$. Thus, also in this case, $\alpha$ is isometric. REPLY [6 votes]: I think the OP’s point (1) is not correct; i.e., $B(Z)$ need not be weak$^*$ dense in $B(Z, Z^{**}) (= B(Z^*)$. You don’t need a Pisier space to show this. Instead, take a separable reflexive space $X$ that fails the approximation property (AP) (reflexivity is not so important, but it makes the notation and reasoning slightly simpler). As in my first answer, let $(E_n)$ be an increasing sequence of subspaces of $X$ with $E_n$ having dimension $n$ so that $\cup_{n} E_n$ is dense in $X$, let $Y$ be the $\ell_1$ sum of $(E_n)$, and let $Q: Y\to X$ be the natural quotient map, defined by $Q(e_n)_n : = \sum _n e_n$. As was hinted at in my first answer, $Y^* = W^\perp + Q^*X$ is a direct sum decomposition of $Y^*$ with the projection onto $Q^*X$ having norm one and where $W$ is the kernel of $Q$; i.e., $W $ is the subspace of all $(e_n)$ in $Y$ so that in $X$ $\sum_n e_n = 0$. Since $X$ fails the AP, by Grothendieck’s duality theory there is a tensor $T$ in $X^* \hat{\otimes} X$, $T = \sum_n x_n^* \otimes x_n$ with $x_n^* \in X^*$, $x_n \in X$, $\sum_n \|x_n^*\| \cdot \|x_n\| < \infty$, and for all finite rank operators $S$ on $X$ we have $\langle S, T\rangle = \sum_n \langle x_n^*, Sx_n \rangle =0$ but $\langle I_X, T\rangle = \sum_n \langle x_n^*, x_n \rangle =1$. Consider now the tensor $T_1$ in $Y^* \hat{\otimes} X$ defined by $T_1 := \sum_n Q^* x_n^* \otimes x_n$. If $J$ denotes the natural isometric embedding from $X$ into $Y^{**} $ induced by $Q$, we see that $\langle T_1, J \rangle = \sum_n \langle Q^* x_n^*, Jx_n \rangle = \sum_n \langle x_n^*, x_n \rangle =1$. Now consider any $S$ in $B(X,Y)$. The operator $S$ is compact because $X$ is reflexive and $Y$ has the Schur property. We claim that $\langle T_1, S\rangle $, which is $\sum_n \langle Q^* x_n^*, Sx_n \rangle$, is zero. If not, since $\| P_kS-S\| \to 0$ as $k \to \infty$ by compactness of $S$ (here $P_k$ is the natural norm one projection from $Y= (\sum_{n=1}^\infty E_n)_1$ onto $(\sum_{n=1}^k E_n)_1$) we can choose $k$ s.t. $\langle T_1, P_k S\rangle \not= 0$. But $$ \langle T_1, P_kS\rangle = \sum_n \langle Q^* x_n^*, P_k Sx_n \rangle = \sum_n \langle x_n^*, QP_k Sx_n =0 $$ because $QP_k S$ has finite rank. To see that (1) is false, consider the space $Z:= Y\oplus X$.<|endoftext|> TITLE: How do small central extensions drop the dimension of a faithful representation? QUESTION [9 upvotes]: Apologies in advance that this is a very soft question. I might be talking complete nonsense. But I hope I am talking about something that has even been studied... I am interested in the phenomenon in representation theory that a group whose faithful representations (over $\mathbb{C}$, say) have minimum dimension $n$ often has a central extension with faithful representations in dimensions smaller than $n$. For example, $S_4$ has no faithful representation in dimension less than 3, but it has a double cover $\tilde S_4$ with a faithful 2-dimensional representation. This phenomenon is striking to me because it takes "more information" to write down the elements of the bigger group, since an element of $\tilde A_5$ specifies an element of $A_5$, but not vice versa. So the dimension of the representation does not correspond to the amount of information it contains. One might speculate that the "extra information" in the smaller-dimensional representation of the bigger group is housed in the (smallest) field over which the representation is defined; in my example, the faithful three-dimensional representations of $S_4$ are defined over $\mathbb{Q}$, but the faithful two-dimensional representations of $\tilde S_4$ I believe require $\mathbb{Q}(i)$. But I'm unsatisfied by this explanation. First of all, although I do not have an example on hand, I am told (by Derek Holt) that one observes the same phenomenon with permutation representations, with degree replacing dimension, where there is no question of the field of definition. Secondly, it begs the question (to me anyway) of "how" the central extension "causes" the information to be moved from the size of the space to the size of the field. My question is this: Has this phenomenon attracted any research attention? Can you point me to any exploration of it in the literature? Thanks in advance. REPLY [6 votes]: I think the basic reason that this happens is quite simple. Matrix groups, such as ${\rm GL}(n,{\mathbb C})$ have a centre, consisting of the scalar matrices and so it natural for groups with nontrivial centres to have representations in which their centre maps to the scalars. More to the point, the perfect group ${\rm SL}(n,{\mathbb C})$ has a cyclic centre of order $n$. So finite groups $G$ with cyclic subgroups $Z < Z(G) \cap [G,G]$ of order dividing $n$ are good candidates for mapping into ${\rm SL}(n,{\mathbb C})$. So, for example, ${\rm SL}(2,5)$ with centre of order $2$ maps into ${\rm SL}(2,{\mathbb C})$, whereas the smallest dimension of a faithrful representation of $A_5={\rm PSL}(2,5)$ is $3$. Or $3.A_6 < {\rm SL}(3,{\mathbb C})$ and $2.A_6 = {\rm SL}(2,9) < {\rm SL}(4,{\mathbb C})$, but the smallest degree nontrivial representation of $A_6$ is $5$.<|endoftext|> TITLE: When does a map in the stable homotopy group gets killed when smashed with cone of itself? QUESTION [7 upvotes]: Consider an element $e \in \pi_n^s(S^0)$, in the stable homotopy groups of sphere. Let $C$ denote the spectrum which is cone of $e$, i.e. $C$ fits in the cofiber sequence $$ S^n \to S^0 \to C.$$ $\textbf{Question:}$ I want to know under what condition $e$ viewed as a self map of $C^{\wedge k}$ (the $k$-fold smash product of C) is null-homotopic, in other words how do we detect if $e \wedge 1_{C^{\wedge k}}$ is homotopically trivial. Let me expand on the case when $k=1$. We have a cofiber sequence $$S^n \wedge C \to S^0 \wedge C \to C \wedge C $$ where the left most map is $e \wedge 1_C$. If $e \wedge 1_C$ is trivial then the $C$ splits of $C \wedge C$. I believe that if $e \cup_1 e \neq 0$ then $e \wedge 1_C$ is non-trivial. The reason being (I think) $e \cup_1 e$ is the non-trivial attaching map for the top-cell of $C \wedge C/\Sigma_2$ to the $0$-cell. Let me know if I am stating something incorrect. I think the converse should be true as well. $\textbf{Question:}$ If $e \cup_1 e \neq 0$ then can we say that $e \wedge 1_{C^{\wedge k}}$ is non-trivial for $k>1$ as well? Further let me give an example, let $e = 2: S^0 \to S^0$. We know that $2$ is a non-trivial as a self-map of $M_2$ (the cone of $2$), the reason being $2 \cup_1 2 = \eta$. $\textbf{Question:}$ Is $ 2 \wedge 1_{M_2^{\wedge k}}$ non-trivial for all $k$? I believe the answer should be yes. Does anyone know of a proof? $\textbf{Question:}$ Is there an example of an $e$ such that $e \wedge 1_{C^{\wedge k}}$ is non-trivial for $k 1? REPLY [4 votes]: The answer to your third question is "yes". A little more generally, if $Sq^{n+1}$ acts nontrivially in the mod $2$ cohomology of $C$, then $Sq^{(k+1)(n+1)}$ acts nontrivially in the mod $2$ cohomology of $C^{\wedge k+1}$, by the Cartan formula. If $e$ smashed with the identity on $C^{\wedge k}$ were null-homotopic, then its mapping cone, which is $C^{\wedge k+1}$, would split in a way that is not compatible with this nontrivial cohomology operation. A similar argument works for $e = \alpha_1$, using the Steenrod operations in mod $p$ cohomology. I think this kind of argument goes back to Michael Barratt. There is a Cartan formula for secondary cohomology operations, due to Leif Kristensen, which might give similar results for some classes $e$ of Adams filtration $2$, but I have not looked at the details.<|endoftext|> TITLE: When is the tensor product of two graphs planar? QUESTION [7 upvotes]: Given two graphs $G=(V_1,E_1)$ and $H=(V_2,E_2)$, the tensor product of $G$ and $H$ is the graph $G \times H = (V,E)$, where $V=V_1 \times V_2$ is the Cartesian product of the $V_i$ and $ (u,v) \ E \ (u',v') \Leftrightarrow u E_1 u' \wedge v E_2 v'$. Is anyone aware of a characterization of which $G,H$ give rise to a planar tensor product $G \times H$? REPLY [6 votes]: Yuichiro Fujiwara's comments seem to answer the question in full so I am making it an answer. I quote below from Kronecker products and local joins of graphs by M. Farzan and D. A. Waller, Can. J. Math. 29 (1977), 255–269. By a 1-contraction of $G$ we mean the removal from $G$ of each vertex of degree 1 (and its incident edge). 5.3 THEOREM. Let $G_1$ and $G_2$ be connected graphs with more than four vertices. Then $G_1\times G_2$ is planar if and only if either (i) one of the graphs is a path and the other one is 1-contractible to a path or a circuit, or (ii) one of them is a circuit and the other is 1-contractible to a path. This leaves open only the case in which at least one of the graphs has fewer than five vertices. Proposition 5.4, which I won't bother reproducing here, covers some of these cases. Beaudou et al. addresses the case $G=K_2$. A complete solution appears to be still an open problem.<|endoftext|> TITLE: Who coined "mob" and "clan" and why these words? QUESTION [6 upvotes]: A mob is a word used for a topological semigroup which is a Hausdorff space. A clan is a compact connected mob with a two-sided identity element. Who used these words with these meanings first and when? Why were these words chosen? I'm guessing it's a play on "group", but is it really? REPLY [8 votes]: The mob/clan terminology goes back to Alexander Wallace: A note on mobs (1952). The structure of topological semigroups (1953).<|endoftext|> TITLE: Piecewise linear (PL) structures on $\mathbf R^4$ QUESTION [12 upvotes]: One can read in Wikipedia that the 4-dimensional affine space $\mathbf R^4$ has uncountably many piecewise linear structures (in contrast with other dimensions, where it has exactly one). A reference is given to Milnor's paper, Differential Topology Forty-six Years Later where this fact is indeed printed on the bottom of the first column of page 2. However, I could not follow the arguments that lead to this conclusion, nor find them in the litterature (eg, Freedman-Quinn's book). Results of the 60s imply (Theorem 2 of Milnor's quoted paper) that any PL structure on $\mathbf R^4$ gives rise to a well-defined, compatible, smooth structure. Then Freedman classified topological closed 4-manifolds, and Donaldson proved that many of them cannot be smoothed. But how does this help constructing exotic PL structures on $\mathbf R^4$? REPLY [13 votes]: There are three facts: existence of uncountably many non-diffeomorphic exotic $\mathbf R^4$'s. any smooth manifold has a PL structure. Any PL manifold of dimension $<7$ has a smooth structure which is unique up to diffeomorphism. For the latter two facts see 1.5 and 1.8 in this survey where explicit references are given.<|endoftext|> TITLE: Product of binary Boolean operators QUESTION [6 upvotes]: I asked this question a day ago on math.stackoverflow but figured it could have an interest here. I'm interested in the set $\mathcal{P}_N$ of boolean functions of boolean variables $p_1, p_2, \ldots, p_N$ that can be written as products of operators of 2 variables only: $$ \phi(p_1, \ldots, p_N) = \bigwedge\limits_{i TITLE: How does one compute the first Chern class of a Line bundle defined as the Kernel of a linear map? QUESTION [5 upvotes]: Let $M$ and $N$ be compact complex manifolds of the same dimension ($m$) and $\mu: M \rightarrow N$ a holomorphic map. Let $D \subset M$ be the subset of points of $M$, where $d\mu|_p$ fails to be injective. Assume that $D$ is a smooth complex submanifold of $M$ of the expected dimension $m-1$ (more precisely, $d\mu|_p$ is not injective means a certain determinant is zero, assume that determinant vanishes transversally). Furthermore assume that on all points of $D$, the Kernel of $d\mu|_p$ is $\textit{exactly}$ one dimensional. $\textbf{Question:}$ Define the line bundle over $D$, given by $L:= Ker(d\mu) \rightarrow D$. How does one compute $c_1(L)$? The specific example where I need to compute $c_1(L)$ is as follows: $M:= \mathbb{P}^1 \times \mathbb{P}^1$, $N:= \mathbb{P}^2$ and $\mu:M \rightarrow N$ is a map of type $(d,k)$, i.e. $\mu^*\mathcal{O}(1) = \mathcal{O}(d,k)$. $\textbf{Added Later:}$ My main interest is in the specific example I asked. Its being pointed out that in general there may not be any explicit/reasonable formula for $c_1(L)$. REPLY [3 votes]: On $M$ there is an exact sequence $$ 0 \to \mu^*\Omega_N \to \Omega_M \to i_*L^\vee \to 0, $$ where $i:D \to M$ is the embedding. This allows to understand the class of $D$ since $D = c_1(i_*L) = c_1(\Omega_M) - \mu^*c_1(N)$. In your case it is equal to $$ (-2,-2) - (-3d,-3k) = (3d-2,3k-2). $$ Thus $D$ is a curve of bidegree $(3d-2,3k-2)$. In particular, its genus is equal to $g = 9(d-1)(k-1)$. To understand the degree of $L$ you can use Riemann--Roch: $$ \deg(L^\vee) + 1 - g = \chi(i_*L^\vee) = \chi(\Omega_M) - \chi(\mu^*\Omega_N). $$ In your case $\chi(\Omega_M) = -2$, while the pullback of the Euler sequence $$ 0 \to \mu^*\Omega_N \to O(-d,-k)^{\oplus 3} \to O \to 0 $$ allows to compute $\chi(\Omega_N) = 3(d-1)(k-1) - 1$. In the end $$ \deg(L^\vee) = -2 + 1 - 3(d-1)(k-1) + 9(d-1)(k-1) - 1 = 6(d-1)(k-1)-2. $$<|endoftext|> TITLE: Foliations of Lorentzian manifolds by Spacelike Hypersurfaces QUESTION [10 upvotes]: Suppose that $M$ is a Lorentzian manifold (not necessarily satisfying Einstein's equations). What conditions do we need in order to guarantee that $M$ admits a foliation by codimension-$1$ spacelike submanifolds? Geroch showed that global hyperbolicity is equivalent to admitting a foliation by Cauchy hypersurfaces, and Bernal and Sánchez showed this foliation can be taken to be smooth (arXiv:gr-qc/0306108). But a Cauchy hypersurface can contain a null geodesic segment (unless I'm quite confused about that). My question then is under what conditions we can take this foliation to consist of spacelike Cauchy hypersurfaces? REPLY [8 votes]: In the case of a globally hyperbolic spacetime, what you want is a smooth Cauchy temporal function (the gradient is everywhere timelike, not just causal, and each level set is a Cauchy surface that is necessarily spacelike). That global hyperbolicity is also sufficient the the existence of a smooth temporal function was also shown by Bernal and Sanchez, in the followup [arXiv:gr-qc/0512095] to their original paper that you cite. I'm not sure if you are also interested in foliations of non globally hyperbolic spacetimes. I'm not completely sure what the right conditions would be then. See comment about stable casuality below.<|endoftext|> TITLE: Comparing cardinalities of the spectrum of two masas in $B(H)$ QUESTION [6 upvotes]: If I imagine that (the self-adjoint part of) a C*-algebra $A$ represents the algebra of observables of some quantum system, then certain perspectives on algebraic quantum theory would ask me to imagine that each (maximal) commutative C*-subalgebra $C \subseteq A$ provides a (maximal) "classical snapshot" of this quantum system. Gelfand duality yields that $C \cong C(X)$ for the compact Hausdorff space $X = \mathrm{Spec}(C)$, so I would picture $X$ as a classical state space that (maximally) "approximates" the would-be quantum state space corresponding to $A$. I would like to know how different these spaces $X$ can be as the maximal commutative $*$-subalgebra $C \subseteq A$ varies. Specifically, can it happen that these have different cardinalities? I'm interested in the particular case $A = B(H)$ for a separable Hilbert space $H$ and two of its well-known masas: the continuous one $C \cong L^\infty[0,1] \subseteq A$ and discrete one $D \cong \ell^\infty(\mathbb{N}) \subseteq A$. Thus I ask: Q: Is there a bijection between the Gelfand spectra $\mathrm{Spec}(C)$ and $\mathrm{Spec}(D)$ for the continuous and discrete masas $C,D \subseteq B(H)$? It's possible to describe these spectra in more explicit terms using Boolean algebra. Note that each of these masas is an (A)W*-algebra. By a combination of Gelfand and Stone dualities (see section 2 of this paper for a bit more detail), the spectrum of a commutative AW*-algebra $K$ is the Stone space of the complete Boolean algebra $\mathrm{Proj}(K)$ of projections in $K$, whose points are the ultrafilters of $\mathrm{Proj}(K)$. The continuous masa $C \cong L^\infty[0,1]$ has $\mathrm{Proj}(C)$ isomorphic to the Boolean algebra of measurable subsets of $[0,1]$ modulo the null sets. I have just learned through the magic of Wikipedia that this is called the random algebra; I will denote it by $B$. The discrete masa $D \cong \ell^\infty(\mathbb{N})$ has $\mathrm{Proj}(D)$ isomorphic to the power set Boolean algebra $2^\mathbb{N}$. (Note that an ultrafilter on the Boolean algebra $2^\mathbb{N}$ is alternatively referred to as an ultrafilter on the set $\mathbb{N}$.) Thus my question is equivalent to: Q': Is there a bijection between the sets of ultrafilters on the random algebra $B$ and the power set algebra $2^\mathbb{N}$? I am aware that $\mathrm{Spec}(D)$ has spectrum homeomorphic to the Stone-Cech compactification $\beta\mathbb{N}$ of the discrete space $\mathbb{N}$, and that this space has various properties that depend on set-theoretic assumptions. Now that I know what the random algebra is called, I see that it bears a relationship to forcing. Thus I can imagine that the answer to my question could be independent of ZFC. Nevertheless, as I am not asking exactly what the cardinality of this spectrum is, but whether it is in bijection with some other (possibly complicated) spectrum, I have an ounce of hope that this can indeed be decided in ZFC. (By the way, the classification of the possible masas of $B(H)$ implies that, if the answer to my question is affirmative, then the spectra of all masas of $B(H)$ are in bijection with one another.) REPLY [5 votes]: Yes. The spectra of $\ell_\infty$ and $L_\infty$ have the same cardinality, namely $2^{\mathfrak{c}}$. Indeed, every infinite, compact $F$-space space (in particular, an extremely disconnected compact space such as the spectrum of $L_\infty$) contains a copy of $\beta \mathbb{N}$ (which happens to be the spectrum of $\ell_\infty$). This is 14N(5) in L. Gillman and M. Jerison, Rings of continuous functions, van Nostrand Reinhold, New York, 1960. and it is quite elementary. The Hausdorff–Pospíšil theorem implies that $|\beta\mathbb{N}|=2^{\mathfrak{c}}$. Thus, $2^{\mathfrak{c}}\leqslant |{\rm spec}\, L_\infty|$. I claim that the cardinality of ${\rm spec}\, L_\infty$ cannot be bigger than $2^{\mathfrak{c}}$. Indeed, $L_\infty^*$ is a bidual of a separable Banach space, hence by Goldstine's theorem, it is separable in the weak*-topology. Every separable space has cardinality at most $2^{\mathfrak{c}}$. Thus $|{\rm spec}\, L_\infty|\leqslant |L_\infty^*|\leqslant 2^{\mathfrak{c}}.$<|endoftext|> TITLE: Residual finiteness: why do we care? QUESTION [26 upvotes]: Residually finite groups have been studied for a long time. However, I am struggling to work out why we care, or perhaps, why they continue to be of interest. Let me explain. Magnus, in his 1968 survey article, motivates residually finite groups by saying that residual finiteness allow us to extract information about the group in an algebraic manner. I understand and agree with this, and that was a fine motivation during the golden age of group theory. However, what about in today's world? How can we apply this property of groups to other settings? So, I have two concrete questions. Why do we care whether hyperbolic groups are residually finite or not - we have soluble word problem, soluble isomorphism problem, Hofian, and so on. These properties arguably imply that Magnus' motivation does not hold. I should say that "because we don't know and it is an interesting question" is not really the answer I am looking for...(EDIT: I am aware that this implies that fundamental groups of hyperbolic $3$-manifolds are LERF, but, in a certain sense, this is still a group-theoretic property.) What are examples of theorems which say "this group is residually finite and therefore that amazing theorem in number theory holds!", or "this class of groups are residually finite so that class of rings have this wonderful property". That is, how does residual finiteness fit in to the big picture? REPLY [5 votes]: I addressed your first question in the comments a while ago (we would like to know whether hyperbolic groups have a finite-index torsion-free subgroup, realizing the vcd, which is equivalent to the question of residual finiteness). Another reason that I'm interested in this question is the relation to QCERF, as mentioned by Henry Wilton. I proved that hyperbolic groups which are cubulated are residually finite, and the proof is by an intricate induction on the dimension of the cube complex (in concert with the QCERF property, or rather the stronger local retract property). If hyperbolic groups in general are residually finite, this would obviously subsume my result. Moreover, the proof of such a result would have to be completely different, since there are e.g. hyperbolic groups with property (T), so cannot be cubulated or have the local retract property. So it seems likely that if true, the proof of such a result would entirely subsume my proof in the cubulated case (i.e., it seems unlikely that the proof would depend or build on my proof as a special case). Thus I have spent a bit of time trying to show the existence of non residually-finite hyperbolic groups, to no avail. I think this is an important question: what does residual finiteness tell us that is not internal to the theory of groups? I don't have an explicit answer to this question. However, certain polynomial diophantine equations can be solved, by first considering solutions on a finite-sheeted covering space, e.g. the primitive solutions to $x^2+y^3=z^7$ were found this way. So in some sense residual-finiteness of the fundamental group comes into play when solving the equation by descent. Other applications of residual-finiteness (not exactly of the sort that you request, but not mentioned in the other comments or answers): the Kaplansky conjecture holds for group rings of residually finite groups (so "that class of rings have this wonderful property" - ha). iterated monodromy groups (coming from branched covers associated to a rational map) encode information about the dynamics of rational maps, and were used to solve Hubbard's "twisted rabbit problem". These groups are inherently residually finite. Golod and Shafarevich showed that class towers can be infinite by showing that the pro-p completion (quotient?) of a certain Galois group can be infinite. This is related to residual finiteness, although not precisely in the form of your question. Moreover, the question of whether there are infinitely many number fields of class number one would be answered if we knew the existence of infinitely many number fields whose absolute Galois group had a finite maximal solvable quotient.<|endoftext|> TITLE: When is $X \rightarrow \text{Spec}(C(X))$ a homeomorphism? QUESTION [12 upvotes]: Let $X$ be compact Hausdorff topological space. Consider the ring $C(X)$ of continuous functions $X \rightarrow \mathbb C$ (we do not consider the C* algebra structure, just consider $C(X)$ as a ring) and its (purely algebraic) spectrum $\text{Spec}(C(X))$. There is a injective continuous map $X \rightarrow \text{Spec}(C(X))$ sending $x$ to the maximal ideal $\mathfrak m_x = \{f \in C(X) : f(x) = 0\}$. Under which conditions on $X$ is this map surjective? Under which conditions is it a homeomorphism to its image? REPLY [15 votes]: The map $i:X\to\operatorname{Spec}(C(X))$ is a homeomorphism onto its image iff $X$ is completely regular; this is essentially the definition of complete regularity. However, it is very rarely surjective. Indeed, suppose $X$ is completely regular and $i:X\to\operatorname{Spec}(C(X))$ is surjective and hence a homeomorphism. Then for any $f\in C(X)$, the subset $D(f)\subset\operatorname{Spec}(C(X))$ of prime ideals that do not contain $f$ is compact, being naturally homeomorphic to $\operatorname{Spec}(C(X)_f)$. But this means that every cozero subset of $X$ is compact and thus clopen. It follows easily that $X$ must be finite. More generally, a similar argument shows that if the spectrum of a ring is Hausdorff, it must be totally disconnected. Note, however, that for $X$ compact Hausdorff, the image of $i$ is exactly the maximal ideals of $C(X)$: the closure of any proper ideal is proper since every function sufficiently close to $1$ is invertible, so every maximal ideal is closed. Furthermore, any prime ideal is contained in a unique maximal ideal: for any two points $x,y\in X$, we can find functions $f,g\in C(X)$ such that $fg=0$ but $f(x)\neq0$ and $g(y)\neq 0$, and so no prime can be contained in both $\mathfrak{m}_x$ and $\mathfrak{m}_y$. So in some sense, $X$ really isn't that far from being the same as $\operatorname{Spec}(C(X))$. Edit: Here's an example of what these non-maximal prime ideals might look like. Let $X$ be the 1-point compactification of a countable discrete space and identify $C(X)$ with the ring of convergent sequences. Fix a nonprincipal ultrafilter $U$ on $\mathbb{N}$ and let $P$ be the set of sequences $(x_n)$ which converge to $0$ and for which $(n^kx_n)$ converges to $0$ with respect to $U$ for all $k$. Then I claim $P$ is a prime ideal (which is strictly contained in the maximal ideal corresponding to the point at infinity). It is easy to see it is an ideal; suppose $(x_n)\not\in P$ and $(y_n)\not\in P$. Then for $k$ sufficiently large, $(n^kx_n)$ and $(n^ky_n)$ both go to infinity with respect to $U$. It follows that for large $k$, $(n^{2k}x_ny_n)$ also goes to infinity with respect to $U$, and thus $(x_ny_n)\not\in P$.<|endoftext|> TITLE: Homological criterion for $A(B \cap C) = AB \cap AC$? QUESTION [7 upvotes]: Is there a homological criterion for the condition $A(B \cap C) = AB \cap AC$ for ideals in a ring $R$? I mean a statement such as "the given equation holds if and only if (some $\operatorname{Tor}$, $\operatorname{Ext}$, local cohomology, etc) group vanishes/does not vanish". Note that this is a local question, since we are asking when the inclusion $A(B \cap C) \to AB \cap AC$ is surjective. So I'm happy to assume $R$ local. This is coming from the similar-looking fact that $AB = A \cap B$ if and only if $\operatorname{Tor}^1(R/A,R/B) = 0$. (I asked this on StackExchange and did not receive a response.) REPLY [5 votes]: Ah, here it is: we have $A(B\cap C) = AB \cap AC$ if and only if the natural map $$\operatorname{Tor}_1^R(R/A,B) \oplus \operatorname{Tor}_1^R(R/A,C) \to \operatorname{Tor}_1^R(R/A,B+C)$$ is surjective. This is not very pretty, but it's still entirely homological (for example, it can be computed in the completion.) For a more 'geometric' formulation, we may use the natural isomorphism $$\operatorname{Tor}_2^R(M,R/I) \to \operatorname{Tor}_1^R(M,I)$$ to express this in terms of $R/A$, $R/B$, $R/C$, and $R/(B+C)$, so this does indeed relate (somehow) to how $A$ intersects with $B$ and $C$. Proof: Take $0 \to B \cap C \to B \oplus C \to B + C \to 0$ and tensor with $R/A$. Note that $$\frac{AB \cap AC}{A(B \cap C)} = \ker\big(\frac{B \cap C}{A(B \cap C)} \to \frac{B}{AB} \oplus \frac{C}{AC} \big)$$ and so is the cokernel of the map above. (In particular, we are asking for the first boundary map in the long exact sequence for $\operatorname{Tor}$ to be zero.)<|endoftext|> TITLE: Flooding a cycle digraph via chip-firing: $n^{k-1} + n^{k-2} + \cdots + 1$ bound (a Norway 1998-99 problem generalized) QUESTION [7 upvotes]: Let $k > 1$ and $n$ be positive integers. Let $\mathbb{N} = \left\{0,1,2,\ldots\right\}$. Let $D$ be a digraph which has exactly $k$ vertices $v_0$, $v_1$, ..., $v_{k-1}$ and exactly $k$ arcs $v_0 \to v_1$, $v_1 \to v_2$, ..., $v_{k-2} \to v_{k-1}$, $v_{k-1} \to v_0$. (That is, $D$ is a directed cycle on $k$ vertices.) Let $V$ be the set of vertices of $D$. A chip configuration on $D$ will mean a map $f : V \to \mathbb{N}$. Visually, we represent a chip configuration by imagining that $f\left(v\right)$ "chips" (or coins or whatever tokens) are placed at each vertex $v \in V$. If $v$ is a vertex of $D$, then kicking $v$ will mean an operation which takes a chip configuration $f$ on $D$ satisfying $f\left(v\right) \geq n$, and returns another chip configuration $f'$ on $D$ which is defined as follows: $f'\left(v\right) = f\left(v\right) - n$; $f'\left(w\right) = f\left(w\right) + 1$ for every vertex $w$ such that $v \to w$ is an edge of $D$ (note that there exists only one such vertex $w$ for our graph $D$); $f'\left(w\right) = f\left(w\right)$ for all remaining vertices $w$. (In visual language, $f'$ is obtained from $f$ by removing $n$ chips from the vertex $v$ and adding one chip at the vertex that follows $v$ on the cycle.) Notice that kicking $v$ is similar to the "firing $v$" operation from chip-firing (aka sandpile) theory; indeed, it can be seen as a particular case of the latter if we add a new vertex to $D$, proclaim it the sink, and add $n-1$ arcs from each $v \in V$ to this sink. A kicking operation will mean an operation which is kicking $v$ for some $v \in V$. If $J$ is a subset of $V$, then a chip configuration $f$ is called $J$-flooding if every $v \in J$ satisfies $f\left(v\right) > 0$. If $f$ is a chip configuration on $D$, then $\left|f\right|$ will mean the sum $\sum\limits_{v \in V}f\left(v\right)$. Conjecture 1. If $f$ is a chip configuration on $D$ satisfying $\left|f\right| \geq n^{k-1} + n^{k-2} + \cdots + 1$, then one can transform $f$ into a $V$-flooding chip configuration by a sequence of kicking operations. Conjecture 2. If $J$ is a subset of $V$, and if $f$ is a chip configuration on $D$ satisfying $\left|f\right| \geq n^{k-1} + n^{k-2} + \cdots + n^{k-\left|J\right|}$, then one can transform $f$ into a $J$-flooding chip configuration by a sequence of kicking operations. I have proven Conjecture 1 for all $k \leq 4$ and Conjecture 2 for all $k \leq 3$ by casework; at least the bound of Conjecture 1 is sharp. I am wondering if the conjectures are true in general. The motivation was a number theory problem from Norway's Abelkonkurransen 1998-99 (final problem 2b) which I tried to generalize back in 2006. The actual number theory problem, even in my generalization, is rather easy, as divisibility of integers boils down to comparing the exponents of prime powers in them; but it appears to me that there is much more substance to the combinatorial generalization that is Conjecture 1 above. Conjecture 2 was my attempt to make Conjecture 1 more amenable to induction by generalizing it even further; but so far it has just resulted in two conjectures rather than one. The next question will probably surprise noone: Question 3. If $D$, instead of being a cycle, is an arbitrary strongly connected digraph, what can we say about the smallest $N \in \mathbb{N}$ such that every chip configuration $f$ on $D$ satisfying $\left|f\right| \geq N$ can be transformed into a $V$-flooding chip configuration by a sequence of kicking operations? REPLY [4 votes]: It now looks to me that conjecture 2 is only easier. We induct on $|J|$. Base $|J|=1$, say, $J=\{k\}$. If $f(k)>0$ there is nothing to prove, so assume that $f(k)=0$. We have $n^{k-1}$ coins in vertices $1,\dots,k-1$ (there may be more coins, but we use only $n^{k-1}$) and perform operations until $f(k)$ becomes positive or we can not proceed. In the second case note that $I=f(1)+nf(2)+\dots+n^{k-2}f(k-1)$ remains invariant under our operations, and, since $f(i)\leq n-1$ (we can not proceed), we have $I\leq (n-1)(1+n+\dots+n^{k-2})=n^{k-1}-1$, though initially $I$ was not less then $n^{k-1}$. A contradiction. Assume that for lesser values of $|J|$ the statement holds, and prove it for $J$. At first, we may suppose that $f(j)=0$ for all $j\in J$, else reserve a coin in $j$, replace $J$ to $J\setminus j$ (do not touch reserved coin, so $|f|$ decreases by 1) and induct on $|J|$. The set $V$ of vertices is a union of disjoint segments $\Delta=\{t-p,t-p+1,\dots,t\}$, where $\Delta\cap J=\{t\}$. Of course, $p\leq k-|J|$ for any such a segment. If $f(t-p)+f(t-p+1)+\dots+f(t-1)\geq n^p$, we may make $f(t)$ positive by our operations by above argument, the cost is at most $n^p\leq n^{k-|J|}$ (at most $n^p-1$ coins are lost and 1 coin in $t$ must be preserved). Do it, remove $t$ from $J$ (do not forget to preserve a coin in $t$) and induct on $|J|$. If not, then summing by all segments we see that $|f|$ is too small.<|endoftext|> TITLE: Maximizing Frobenius Norm of Commutator (an opposite Procrustes problem) QUESTION [5 upvotes]: I was wondering if anybody has any suggestions on the following problem: Let $S$ be an $n\times n$ positive definite symmetric matrix. I wish to find an $n\times n$ orthogonal matrix $R$ which MAXIMIZES the Frobenius norm of the commutator, i.e. $$ R = \arg\max_{R' \in O(N)}||[R',S]||_F^2 = \arg\max_{R' \in O(N)}||R'S - SR'||_F^2, $$ where $O(N)$ is the set of all orthogonal matrices. Clearly $||[R,S]||_F^2 \leq 2||S||_F^2$, provides an upper bound. I've found some papers which are relevant to this problem, in particular Commutators with maximal Frobenius norm, which identify matrices which satisfy this upper bound, but the resulting matrices will not be orthogonal. I'm not confident the results there are so applicable. Any ideas, help or suggestions would be greatly appreciated. REPLY [7 votes]: Unless I'm mistaken, the following argument provides a solution. Since the Frobenius norm is orthogonally invariant we can assume without loss of generality that $S$ is diagonal. I'll write $Q$ instead of $R'$ to avoid confusion with matrix transposition. \begin{equation*} \|QS-SQ\|_F^2 = \|QS\|^2 + \|SQ\|^2 - 2\text{tr}(SQ^TSQ). \end{equation*} So the task is to minimize the trace term (as the first two terms are independent of $Q$). But we know that \begin{equation*} \text{tr}(Q^TSQS) \ge \langle \lambda^{\uparrow}(Q^TSQ),\lambda^{\downarrow}(S)\rangle, \end{equation*} where $\lambda^\downarrow$ denotes the sorted vector of eigenvalues. From this it follows that $Q$ is a suitable permutation matrix that reorders the largest diagonal entries of $S$ to match up with the smallest ones. Note: We did not require $S$ to be positive definite (the original $S$ in your question can be any symmetric matrix, not just a positive definite one).<|endoftext|> TITLE: How close to an integer can a polynomial root be? QUESTION [7 upvotes]: Suppose I have a polynomial $p(x) = a_n x^n + ... + a_0$ where $a_n, \dots, a_0$ are integers. I would like to show that any root of this polynomial is either an integer or is far from an integer. That is, if $r$ denotes the fractional part of a root $x$, I want to show that either $r = 0$ or $r > w$ for some real valued $w$. If the polynomial $p$ has degree one this is easy --- I know either $w = 0$ or $w \geq \frac{1}{|a_1|}$. Is there a similar bound that can be shown for arbitrary polynomials, where the minimum non-zero value of $w$ can be bounded in terms of the sizes of $a_0, \dots, a_n$? REPLY [10 votes]: This is a standard question in diophantine approximation. See for example Chapter 3 of Waldschmidt's book Diophantine Approximation of Linear Algebraic Groups. Here is the simplest bound that Waldschmidt gives. Assuming $a_0\ne 0$, let $H := \max_{0\le i\le n} |a_i|$. We will show that if $\alpha$ is any nonzero root of your polynomial, then $|\alpha| > 1/(H+1)$. If $|\alpha| \ge 1$ then the inequality trivially holds. Otherwise, if $|\alpha|<1$, then $${1\over |\alpha|} \le \left|{a_0\over \alpha}\right| = \left| a_1 + a_2\alpha + \cdots + a_n \alpha^{n-1} \right| \le H(1+|\alpha| + \cdots+|\alpha|^{n-1}) < {H\over 1-|\alpha|},$$ which rearranges to $|\alpha| > 1/(H+1)$. Edit: See David Lampert's comment below for how to use the above bound to answer the question that David Harris asked.<|endoftext|> TITLE: When does a perverse sheaf occur in the decomposition theorem? QUESTION [14 upvotes]: Suppose I am in the setting of the decomposition theorem, i.e., we have the decomposition of the direct image $f_*\mathbb Q_\ell$, where $f:X\to Y$ is proper. Then the direct image decomposes into a direct sum of shifted $IC$ extensions of semisimple local systems. I would like to know a kind of converse to this question, that is, given a (semi)simple perverse sheaf, say some intermediate extension of a local system on $S\subset Y$, how can I determine whether it occurs in the decomposition above? REPLY [16 votes]: In general it is a difficult problem. For example, the core of Ngô's proof of the fundamental lemma is his support theorem which implies that in the context of the Hitchin fibration, all simple constituents of the direct image have full support. If the morphism is semismall, then one has a direct sum of IC's without shift and the situation is simpler. Only relevant strata (for which the semismall inequality is an equality) can have local systems whose IC extension can contribute to the direct image. This was discussed in the other question. The top cohomology of a stratum has an action of the fundamental group of the stratum, by permutation of the top dimensional irreducible components (hence the action factors through the finite group of permutation of the components, which implies easily the semisimplicity in this case). For a relevant stratum, this representation corresponds to the local system you have to take the IC of. So the irreducible constituents of the representation tell you which simple IC's appear. For example, for a relevant stratum, the trivial local system always appears. Example: the Springer resolution. The simple IC's that appear are the image of the Springer correspondence. I don't know a general argument to compute the image without actually computing the correspondence. In the general case, I assume you know the stalks the IC's which don't have full support (assuming you are doing an induction and you don't already know the big IC's: otherwise, the explanations below might or might not be more relevant). Then you want to spot the factors which occur with a negative shift (so they're shifted to the right). Take the biggest stratum for which the perverse conditions are not satisfied on the right, take the highest nonvanishing cohomology sheaf of the restriction of the direct image to that stratum (it is a local system). Then you know the IC of this (maybe reducible) local sytem, with the appropriate shift, is a direct summand. You may split it off the direct image, and by duality also the IC of the dual local system with the opposite shift. Go on like this until you are left with a perverse sheaf, and do as in the preceding paragraph (the IC's of the local systems which are on the maximally allowed cohomological degrees). You can try this in small examples, but if you are doing something general it does not seem very nice. Are you working in a particular situation? Then try to use the particularities of the situation. For example, see the proof that the KL basis of the Hecke algebra corresponds to IC sheaves on the flag variety, in Springer's Bourbaki seminar. On the other hand, if you already know all the stalks of all IC's (including the ones which are fully supported), and if you can keep track of degrees in the Grothendieck group (maybe exploiting some purity properties, like in the KL setting), then all you need to do is to compute the stalks of the direct image (the cohomology of the fibers), and find the decomposition in the (enriched by degree) Grothendieck group. If you can't keep track of the degrees, you will only know the alternating sum of the contributions, some cancellations might occur. It's hard to explain all this without a blackboard, but I hope it helps! Don't hesitate to ask for more explanations. I (and others) might be able to give you more specific answers, depending on the particular situation.<|endoftext|> TITLE: On an example of an eventually oscillating function QUESTION [50 upvotes]: For $x\in(0,1)$, put $$f(x):=\sum_{n=0}^{\infty}(-1)^{n}x^{2^{n}}.$$ This function possesses interesting properties. It grows monotonically from $0$ up to certain point. Then it starts to oscillate around the value $1/2$ on a left neighborhood of $1$, see Figure 1. So there is at least a numerical evidence that $f(x)>1/2$ for some $x\in(0,1)$, for instance $f(0.995)\dot{=}0.500882$. Nevertheless, I did not find an analytic way to prove this observation. Thus, I have two questions concerning the function $f$. First, I would like to ask if there is an analytic proof (simple at best $\sim$ understandable for undergraduate students) that there exists $x\in(0,1)$ such that $f(x)>1/2$. Second, since the limit of $f(x)$ as $x\to1-$ does not exists, I would like to know at least something about the value $$\limsup_{x\to1-}f(x).$$ Many thanks. REPLY [6 votes]: A reference to this answer is Hardy's 'Divergent Series'. Although we do not obtain a magnitude of oscillation, the divergence of $\lim_{x\rightarrow 1-} f(x)$ follows from the following method. Here, we do not need numerical values, nor the Tauberian theorems, just a little bit of Complex Analysis. After solving this problem, I realized that the same method applies here. Let $f(x)=\sum_{n=0}^{\infty} (-1)^n x^{2^n}$ as given. Let $g(x)=\sum_{n=0}^{\infty} \frac{(\log x)^n}{(2^n+1)n!}$. Then $$ f(x)+f(x^2)=x, \ \ g(x)+g(x^2)=x. $$ Let $\Phi(x)=f(x)-g(x)$. Then we have $$ \Phi(x)=-\Phi(x^2)=\Phi(x^4). $$ If we prove that $\Phi$ is not a constant, then it follows that $\lim_{x\rightarrow 1-} f(x)$ does not exist. Taking the principal branch of logarithm, we see that $\Phi(z)=f(z)-g(z)$ is analytic on $D=\{z: |z|<1, \ \ z\notin (-1,0]\}$. Let $z=re^{2\pi i /3}$, and let $r\rightarrow 1-$. Then we have $$ \Im(f(z))\rightarrow \infty, \textrm{ hence }|f(z)|\rightarrow\infty, $$ $$ g(z) \textrm{ is bounded.} $$ Thus, $\Phi(z)$ is not a constant on $D$. Hence, $\Phi(x), \ \ 0 TITLE: Symmetry type of non-cohomological automorphic forms QUESTION [5 upvotes]: By Katz-Sarnak philosophy a family of $L$-functions would have a symmetry type which would reflect the statistics of $L$-functions, such as low lying zeros and moments. Shin-Templier's paper on Sato-Tate theorem calculated the symmetry type of many families of automorphic L-functions of varying weight or level, with the assumption that the family of automorphic representations involved are cohomological at the $\infty$-place. What is the expected answer for the symmetry type of Maass forms and their symmetric powers? (level and weight aspect) I am aware of the paper by Alpoge-Miller that Maass forms on $GL(2)/\mathbb{Q}$ of a fixed level and increasing Laplacian eigenvalue should be an orthogonal family. Is there anything known about their symmetric powers and in particular, do we expect them to behave like modular forms? (where it is known that say in fixed level and changing weight, (fixed) even symmetric power of modular forms of weight $k$ form a symplectic family, whereas odd symmetric power of modular forms would form an orthogonal family; cf Duenez-Miller, arXiv:math/0607688) REPLY [2 votes]: You should read the following preprint of Sarnak, Shin, and Templier: http://arxiv.org/abs/1401.5507 In particular, they study "nice" families $\mathfrak{F}$ of automorphic representations. They assume RH and that there exists $A < \infty$, $\delta < 1$, such that uniformly in $n \geq 1$, \[\sum_{\pi \in \mathfrak{F}(x)} \lambda_{\pi}(n) = t_{\mathfrak{F}}(n) |\mathfrak{F}(x)| + O(n^A |\mathfrak{F}(x)|^{\delta}),\] where $\mathfrak{F}(x)$ denotes the set of $\pi \in \mathfrak{F}$ of analytic conductor at most $x$. Then they define invariants \[i_1(\mathfrak{F}) = \lim_{x \to \infty} \frac{1}{x} \sum_{p \leq x} |t_{\mathfrak{F}}(p)|^2 \log p,\] \[i_2(\mathfrak{F}) = \lim_{x \to \infty} \frac{1}{x} \sum_{p \leq x} t_{\mathfrak{F}}(p)^2 \log p,\] \[i_3(\mathfrak{F}) = \lim_{x \to \infty} \frac{1}{x} \sum_{p \leq x} t_{\mathfrak{F}}(p^2) \log p.\] Then $i_1(\mathfrak{F}) \geq 1$ with equality iff almost every $\pi$ is cuspidal. We have $0 \leq i_2(\mathfrak{F}) \leq 1$, with $i_2(\mathfrak{F}) = 1$ iff almost every $\pi$ is selfdual and $i_2(\mathfrak{F}) = 0$ iff almost every $\pi$ is not selfdual (i.e. unitary); in this case, $i_3(\mathfrak{F}) = 0$. We have $-1 \leq i_3(\mathfrak{F}) \leq 1$, with $i_2(\mathfrak{F}) = 1$ iff almost every $\pi$ is orthogonal and $i_3(\mathfrak{F}) = -1$ iff almost every $\pi$ is symplectic. Here we say that a single automorphic representation $\pi$ is unitary if it is not selfdual, while if it is self-dual, then the completed Rankin-Selberg $L$-function $\Lambda(s,\pi \times \widetilde{\pi})$ factorises as $\Lambda(s,\mathrm{sym}^2 \pi) \Lambda(s,\wedge^2 \pi)$; if the former has a pole at $s = 1$, then $\pi$ is said to be orthogonal, while $\pi$ is symplectic if the latter has a pole at $s = 1$. So the way to determine the symmetry type of a family of automorphic representations is to use the trace formula to work out what $t_{\mathfrak{F}}(n)$ ought to look like, then use this to calculate $i_1(\mathfrak{F}), i_2(\mathfrak{F}), i_3(\mathfrak{F})$, with this telling you the symmetry type of the family.<|endoftext|> TITLE: Translates of meager sets QUESTION [8 upvotes]: Does there exist a meager set of reals M such that every meager set can be covered by countably many translates of M? This is the category analogue of the following. REPLY [6 votes]: No, there is no such set. The situation for meager sets is dual to that described by Pietro Majer in a comment on Translates of null sets, "I was vaguely thinking to Hausdorff measures w.r.to gauge functions. One needs to know that, given $N$, there is $\phi=o(t)$ (for $t\rightarrow 0$) such that $H^\phi(N)=0$. So there is still room for a $\psi$, $\phi(t)<\psi(t)0$. Now, let $M$ be any meager set of reals (potentially very large) and let $A$ be the complement of $M$ (so $A$ is very small). Nevertheless, by (2) $A$ is not that small: we can let $h$ be such that $H^h(A)>0$. Now let $g$ be another dimension function which is sufficiently far from $h$. By (1) let $B$ be a comeager set with $H^g(B)=0$. Then the complement $N$ of $B$ is too large to be covered by countably many translates of $M$. (*) I apparently proved this for the Cantor space using Kolmogorov complexity. Source: Lecture notes for MATH 788 , 2006. At some point my notes say "details missing here, given in class".<|endoftext|> TITLE: Classification of ergodic measures for circle expanding maps QUESTION [7 upvotes]: Let us consider the classical self-covering of the circle $S^1=\mathbb{R}/\mathbb{Z}$ given by $$\times_d(x) = dx \mod 1$$ where the degree $d$ is any integer greater than $1$. There are a wealth of ergodic invariant measures; I know at least of uniform measures on periodic orbits, the Lebesgue measure, Gibbs measures for Hölder potentials, Bernoulli and Markov measures. My question is: To which extend do we know a kind of classification of all ergodic measures of $\times_d$? For example, is there a precise classification of ergodic measure of positive entropy? I am pretty sure no complete, very explicit classification is known, because it would probably answer Furstenberg's conjecture (Lebesgue measure is the only non-atomic measure which is invariant under both $\times_2$ and $\times_3$). Any classification, however rough, or examples of invariant measures not cited above would be welcome. REPLY [3 votes]: You're right. The set of measures for these maps is a zoo! There is the obvious map (base $d$ expansion) $\pi$ from $\{0,1\ldots,d-1\}^{\mathbb N}$ to $[0,1)$ which is a bijection off a countable set. $\pi$ is then a factor map from the full one-sided $d$-shift to $\times_d$. There are only two ergodic invariant measures that give positive measure to the set where $\pi$ fails to be 1-1, namely the $\delta$-measures at $\overline 0$ and $\overline{d-1}$. This means that you can study invariant measures for $\times_d$ by studying invariant measures for the full shift on $d$ symbols, and there is a 1-1 correspondence between these sets of measures except for the additional fixed point that I mentioned. There is also a natural bijection between one-sided and two-sided invariant measures on the full $d$-shift. Maybe a nice way to dramatize the fact that there are lots of measures is to quote the Krieger generator theorem: for any invertible ergodic process at all (say on a Lebesgue space) with entropy less than $\log d$, there exists a generator: a finite partition such that the set of points that have a `twin' (another point whose entire forward and backward orbit follows the same sequence of partition elements) has measure 0. There is then a measure-theoretic isomorphism between your arbitrarily chosen ergodic transformation with entropy less than $\log d$ and $\times_d$ equipped with an appropriate ergodic invariant measure. This property of $\times_d$, that it contains copies of all processes with entropy less than its own topological entropy is called universality. Some other universal processes are known.<|endoftext|> TITLE: 3D models of the unfoldings of the hypercube? QUESTION [24 upvotes]: There are (apparently) 261 distinct unfoldings of the 4D hypercube, a.k.a., the tesseract, into 3D.1 These unfoldings (or "nets") are analogous to the 11 unfoldings of the 3D cube into the plane.2 Usually only one hypercube unfolding is illustrated,                     (Image from this link.) the one made famous in Salvador Dali's painting Corpus Hypercubus. My question is: Q. Has anyone made models/images of the 261 unfoldings as solid objects in $\mathbb{R}^3$? (If not, I might do so myself.) 1Peter Terney, "Unfolding the Tesseract." Journal of Recreational Mathematics, Vol. 17(1), 1984-85. 2 Update. See also the followup question, "Which unfoldings of the hypercube tile 3-space: How to check for isometric space-fillers?." REPLY [13 votes]: I used sage to make a 3d animation of all 261 unfoldings. Here is a screenshot of the first few: The file cube-unfoldings.txt contains all the unfoldings, each line contains a list of 8 points. Edit: By popular, I add the (poorly commented) code: unfolding the hypercube.ipynb: a jupyter notebook with sage code to generate the pairings for the unfoldings together and find the embeddings. To view the code easily, checkout the file at github The animation on the website is made with threejs, and all the code is contained in the unfoldings.html, which you can also view on github.<|endoftext|> TITLE: Non-Forcing and Independence QUESTION [11 upvotes]: I asked this question about two weeks ago on MSE and haven't gotten an answer, so I thought I would post the question here. Do there exists sentences which are independent of ZFC, cannot be shown to be independent through some method of forcing, and do not increase the consistency strength of ZFC (e.g. so Large Cardinal Axioms are out)? If there does exist such a sentence, I would love to know a concrete example. One with a combinatorial flavor would be ideal. Edit: Feel free to increase the consistency strength of ZFC (say ZFC+ $\Delta$) in the "meta-"sense (i.e. work where you want to work). Does there exist a "non-forcible" independent sentence that does not increase the consistency strength of ZFC? REPLY [8 votes]: If we consider $ZF$ instead of $ZFC$, then we can say more. There are examples of such results which are obtained by Krivine, using his method of realizability. For example in the paper Realizability algebras II: New models of ZF+DC the following is stated: Using the proof-program (Curry-Howard) correspondence, we give a new method to obtain models of $ZF$ and relative consistency results in set theory. We show the relative consistency of $ZF + DC$ + there exists a sequence of subsets of $\mathbb{R}$ the cardinals of which are strictly decreasing + other similar properties of $\mathbb{R}$. As it is stated in the introduction of the paper: These results seem not to have been previously obtained by forcing. see also 50 years after forcing, the Curry-Howard correspondence gives new models of ZF<|endoftext|> TITLE: Is the ring of invariants Noetherian? QUESTION [5 upvotes]: Let $R$ be a complete regular local ring whose residue field is perfect. Suppose that a finite group $G$ acts on $R$ by ring automorphisms in such a way that the induced action on the residue field is trivial. Is the ring of invariants $R^G$ necessarily Noetherian? REPLY [8 votes]: Yes, and regularity isn't needed (assuming noetherian). By the Eakin-Nagata Theorem (3.7, Matsumura CRT), it is enough that $R$ is $R^G$-finite. For the Cohen ring $W$ of the perfect residue field, the unique local map $W\rightarrow R$ lifting the identity on residue fields is $G$-invariant. Pick a surjection $W[\![x_1, \dots, x_n]\!]\rightarrow R$. Let $t_{i1},\dots$ in $R^G$ be the indexed sequence of $\#G$ elementary symmetric "functions" in the $G$-orbit of $x_i$. Then the unique local $W$-algebra map $W[\![T_{ij}]\!] \rightarrow R$ carrying $T_{ij}$ to $t_{ij}$ is module-finite by completeness of $R$ since each $x_i$ is nilpotent in $R/(T_{ij})R$, yet this factors through $R^G$, so $R$ is also $R^G$-finite. QED<|endoftext|> TITLE: For which quiver varieties is Kirwan surjectivity known? QUESTION [6 upvotes]: The cohomology of Nakajima quiver varieties is a quite interesting object. It's equipped with some natural classes given by the Chern classes of the tautological bundles associated to the spaces in the quiver representation. We say that a quiver variety satisfies Kirwan surjectivity if these classes generate the cohomology ring. I know that in type A, Nakajima quiver varieties are homotopic to Spaltenstein varieties, and I can see surjectivity using a presentation of that cohomology. If all the entries in the dimension vector are 1, then the quiver variety is a hypertoric variety, and we know Kirwan surjectivity for those. There are also hyperpolygon spaces. Are there any other large classes of examples where this is proven? I'm particularly interested in what's in the literature from the finite case. It seems that this case is "obvious" based on work of Kodera and Naoi, but no one seems to have actually written it down (at least not anyone citing their paper). REPLY [4 votes]: As an update, Kevin McGerty and Thomas Nevins prove Kirwan surjectivity for Nakajima quiver varieties in this recent paper.<|endoftext|> TITLE: Tree property and singular strong limit cardinals QUESTION [5 upvotes]: I heard that the following theorem is proved recently by Foreman-Magidor, which answers a famous old open question: Theorem. It is consistent, relative to the existence of large cardinals, that there is a singular strong limit cardinal $\kappa,$ such that the tree property hold at both $\kappa^+$ and $\kappa^{++}.$ Here are some questions related to the above theorem: 1) Does anyone know the main idea of the proof? 2) Is the $\kappa$ in the theorem large, or it can be also small, like $\aleph_\omega?$ 3) What kind of large cardinal(s) are used in the proof? 4) Are there any notes or slides presenting the proof (as far as I know the paper itself is not written yet, as I asked Magidor for the paper, and did not receive anything). REPLY [2 votes]: The following theorem from Sinapova's paper ''THE TREE PROPERTY AT THE FIRST AND DOUBLE SUCCESSORS OF A SINGULAR'' answers your main question. Theorem. Suppose that $\langle\kappa_{n}:n<\omega\rangle$ is an increasing sequence of super-compact cardinals and that $\lambda$ is a weakly compact cardinal with $\lambda>\sup\{\kappa_n:n<\omega\}$. Then there is a generic extension in which $\kappa_0$ is strong limit singular with $cf(\kappa_0)=\omega$ and $\lambda=\kappa_0^{++}$, and the tree property holds at both $\kappa_0^{+}$ and $\kappa_0^{++}$.<|endoftext|> TITLE: Adding sets not containing arithmetic progressions of length three by forcing QUESTION [6 upvotes]: Consider the following forcing notion: conditions in $\mathbb{P}$ are pairs $(s, N),$ where: 1) $s\in 2^{<\omega}$, 2) $N\in \mathbb{N}$, 3) (by identifying $s$ with a subset of $lh(s)$) $s$ contains no arithmetic progressions of length $3$, and $\Sigma_{n\in s}1/n \geq N$. The ordering is defined in the natural way: $(t, M)\leq (s,N)$ iff $t$ extends $s$ and $M\geq N.$ Now let $G$ be $\mathbb{P}$-generic over $V$, and let $R=\bigcup\{s: \exists N, (s,N)\in G \}.$ We can imagine $R$ as a subset of $\mathbb{N}$. The following is clear: Claim 1. $R$ contains no arithmetic progressions of length three. Given any finite $M$, consider the set $D_M=\{(s, N)\in \mathbb{P}: N\geq M \}.$ Then Claim 2. If for each $N$, the set $D_N$ is dense in $\mathbb{P},$ then $R$ is a large set of natural numbers, i.e., $\Sigma_{n\in R}1/n=\infty$ (and hence $R$ witnesses a counterexample to the famous Erdos-Turan conjecture). Question. Suppose $(s, N)\in \mathbb{P}.$ is there $t\in 2^{<\omega}$ extending $s$ such that $(t, N+1)\in \mathbb{P}?$ Of course a positive answer to the above question implies that all $D_M$'s are dense. Remark. If such an $R$ exists in an extension, then it already exists in the ground model. REPLY [10 votes]: A "yes" answer to your question is equivalent to the statement "there exists a large set of natural numbers that admits no arithmetic progression of length three." I'm submitting the proof of this equivalence as an answer since I don't expect to see an actual answer unless it shows up in Annals too :) So, to the proof. You've already noted the forward direction. For the backward direction, suppose $A$ is a large set of natural numbers with no arithmetic progression of length three, $k\in\mathbb{N}$, and $s\subseteq k$ has no arithmetic progressions of length three. We'll show that there's a large set $B$ such that $B\cap k = s$, and $B$ has no arithmetic progressions of length three. This is clearly sufficient. The "naiive" choice for $B$ is the set $s\cup (A\setminus k)$, which is large and has no length-3 AP's which are entirely below $n$ or entirely above $k$; but of course there may be an AP of length 3 which crosses $k$. There are only finitely-many APs of length 3 with two points in $s$, so we may remove the corresponding points from $A$ (if they exist) and still have a large set. So we'll assume that we've already done this, i.e. $A\cap k = \emptyset$, and there are no APs of length 3 with two points in $s$. If the AP has two points in $A$, then it's more complicated. Suppose $i\in s$. Let $B_i$ be the set you get from $A$ by removing the possible third points, i.e. $$ B_i = A\setminus\{2n - i \;|\; n\in A\}$$ We'll show that $B_i$ is still large. Let $N$ be a large natural number. Note that $$ \sum_{n\in B_i\cap N} \frac{1}{n} \ge \sum_{n\in A\cap N} \frac{1}{n} - \frac{1}{2n-i}$$ If $n$ is large enough, say $n > 3i$, then $\frac{1}{2n - i} < \frac{2}{3n}$, so if $A\cap 3k = \emptyset$ then the above sum is at least $$ \sum_{n\in A\cap N} \frac{1}{3n} = \frac{1}{3} \sum_{n\in A\cap N} \frac{1}{n} $$ Hence, as the sums on the right go to $\infty$ as $N\to\infty$, it follows that $B_i$ is large. Applying this process multiple times, once for each member of $s$, we eventually get a large set $B$ such that $s\cup B$ has no AP's of length 3.<|endoftext|> TITLE: Group laws in class field theory QUESTION [6 upvotes]: In the case of a quadratic imaginary number field one can construct its maximal abelian extension using torsion points of an elliptic curve with complex multiplication by this field. In the case of a local field, Lubin-Tate theory provides an explicit construction of its maximal abelian extension using torsion points of a formal group law on the maximal ideal. Are there any examples of similar arguments, or even general facts? REPLY [5 votes]: As mentioned in the comments, this is precisely Hilbert's twelfth problem, for the simple reason that any solution to that problem can be turned into a "group law argument" (or any solution to it is a "group law argument" in disguise in the first place). For example, you can think of the Kronecker-Weber theorem as saying that abelian extensions of $\mathbb{Q}$ are generated by torsion points of its multiplicative group. The most general result of this type is Shimura's complete solution for CM-fields (see Complex multiplication of Abelian varieties and its applications to number theory). You can complete the global field picture as mentioned in the comments (Carlitz-Hayes-Drinfeld) and the local one with Lubin-Tate. PS. I corrected the question: "In the case of quadratic number field...". Only the imaginary case is know. For real quadratic fields this is wide open (the best way to attack it seems to be via the Stark conjectures).<|endoftext|> TITLE: Regularized sums of Mobius sequence QUESTION [10 upvotes]: Do $\lim_{s \rightarrow \infty} \sum_{n \geq 1} \mu(n) e^{-n/s}$ and $\lim_{s \rightarrow \infty} \sum_{n \geq 1} \mu(n) e^{-n^2/s^2}$ both equal $-2$? Experimentally this seems plausible (up through $s=10^6$). On a related theme, does the Dirichlet series $\sum_{n \geq 1} \mu(n) n^{-s}$ converge to $1/\zeta(s)$ for all real $s$ between 0 and 1? If so, then sending $s \rightarrow 0^+$ would assign the divergent sum $\sum_{n \geq 1} \mu(n)$ the same regularized value as the first two regularization procedures, since $\zeta(0) = -1/2$. REPLY [17 votes]: No! This is very badly false -- the Riemann zeta function has non-trivial zeros. For example, suppose that $M(x) = \sum_{n=1}^{\infty} \mu(n) e^{-nx}$ tends to $-2$ as $x\to 0$ (I've rewritten your hypothesis with $x=1/s$). In particular, you're assuming that $M(x)$ is always bounded. But in that case note that $$ \int_0^{\infty} M(x) x^{s}\frac{dx}{x} = \sum_{n=1}^{\infty}\mu(n) \int_0^{\infty} e^{-nx} x^{s} \frac{dx}{x} = \frac{\Gamma(s)}{\zeta(s)}, $$ where the integral a priori converges for Re$(s)>1$ (because as $x\to \infty$ clearly $M(x)\ll e^{-x}$ decreases exponentially, and trivially as $x\to 0^+$ we can use $|M(x)| \ll x^{-1}$). The assumption that $M(x)$ is bounded as $x\to 0^+$ now implies that the integral actually makes sense in Re$(s)>0$, or in other words that $\zeta(s)$ has no zeros! See also my answer to Is it possible to show that $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges? . Alternatively, one can write down an explicit formula for $M(x)$ in terms of zeros of $\zeta(s)$. Namely for some $c>1$ $$ M(x) = \frac{1}{2\pi i} \int_{c- i\infty}^{c+i\infty} \frac{1}{\zeta(s)} x^{-s} \Gamma(s) ds, $$ and moving the line of integration to the left, we find $$ M(x) = \sum_{\rho} \frac{\Gamma(\rho)}{\zeta^{\prime}(\rho)} x^{-\rho}+ \frac{1}{\zeta(0)} + O(x), $$ where the sum is over non-trivial zeros of zeta (assumed to be simple for convenience). The second term above arises from the pole of the Gamma function at $s=0$, and note that it equals $-2$. The error term can be made explicit in terms of the poles at $-1$, $-2$, etc. This shows why $M(x)$ will have to be of size at least $x^{-1/2}$ occasionally, and further explains why the numerical evidence is misleading: the first zero of $\zeta$ has large ordinate (about $14.1\ldots$), and $\Gamma(1/2+14.1\ldots i)$ is very small in size.<|endoftext|> TITLE: contractible configuration spaces QUESTION [7 upvotes]: Let $F(M,k)=\{(x_1,\cdots,x_k)\mid x_1\cdots,x_k\in M,x_i\neq x_j, \text{ for } i\neq j \}$. It is known that $F(\mathbb{R}^\infty,k)$ is contractible for each $k$. My question: is $F(S^\infty,k)$ contractible for each $k$? For what kind of space $X$ can we have $F(X,k)$ is contractible for each $k$? REPLY [15 votes]: The space $S^\infty$ is actually homeomorphic to $\mathbb{R}^\infty$. To see this, put \begin{align*} B_n &= \{x\in\mathbb{R}^\infty\::\: \|x\|\leq n, x_k = 0\text{ for } k \geq n\} \\ C_n &= \{x\in S^\infty\::\: x_n \leq 1/2,\; , x_k = 0\text{ for } k > n\}. \end{align*} Then $\mathbb{R}^\infty$ is the colimit of the spaces $B_n$, and $S^\infty$ is the colimit of the spaces $C_n$. Moreover, $B_n$ and $C_n$ are both homeomorphic to the $n$-ball, and they are contained in the interiors of $B_{n+1}$ and $C_{n+1}$ respectively. One can cook up compatible homeomorphisms $f_n\:B_n\to C_n$, and pass to the colimit to obtain a homeomorphism $\mathbb{R}^\infty\to S^\infty$. One can prove along similar lines that various other standard contractible spaces are also homeomorphic to $\mathbb{R}^\infty$, for example $\mathbb{R}^\infty\setminus A$ (for any finite subset $A$), or $F(\mathbb{R}^\infty\setminus A,n)$, or the Stiefel manifold $V_n(\mathbb{R}^\infty)$. However, the linear isometry space $L(\mathbb{R}^\infty,\mathbb{R}^\infty)$ is homeomorphic to $\prod_{i=0}^\infty\mathbb{R}^\infty$, which is different.<|endoftext|> TITLE: On the coherence theorem for bicategories QUESTION [12 upvotes]: The coherence theorem for bicategories, as usually stated, reads Any bicategory $B$ is biequivalent to a (strict) 2-category. It is possible to give an explicit construction of the strictification as the full image of its Yoneda embedding $y:B\rightarrow [B,\text{Cat}]$, see for instance this reference. This seems like a natural construction, so I would expect an equivalence of tricategories $$ \text{Bicat} \leftrightarrows 2\text{-Cat}$$ where the rightgoing functor is the full image of the yoneda embedding, and the leftgoing functor is the inclusion. However, I cannot find such a statement in the literature. If it is true, a reference would be appreciated. REPLY [12 votes]: Probably you had some trouble finding this because the search term $2$-$\text{Cat}$ is not accurate enough; you want not the cartesian monoidal product on $2$-$\text{Cat}$, but rather what is called the Gray monoidal product; the tricategory you want then is denoted $\text{Gray}$, the tricategory of strict 2-categories, strict 2-functors, pseudonatural transformations, and modifications between them, but most usefully considered as equipped with the Gray tensor product. Anyway, see this paper by Steve Lack: Bicat is not triequivalent to Gray. This actually corrects a slight misstatement in the monograph by Gordon-Power-Street. See the end of Lack's paper for the punchline. I'm not sure what the state of the art is, but perhaps you could email Lack (or perhaps Street).<|endoftext|> TITLE: Decomposing $(\mathbb C^n)^{\otimes m}$ as a representation of $S_n\times S_m$ QUESTION [13 upvotes]: $V=\mathbb C^n$ is a $\mathbb CS_n$-module, where $S_n$ is the symmetric group of degree $n$, via the representation sending a permutation to the corresponding permutation matrix. The tensor power $V^{\otimes m}$ is therefore also a $\mathbb CS_n$-module via the action $\sigma(v_1\otimes\cdots\otimes v_m) = \sigma v_1\otimes\cdots \otimes \sigma v_m$ on elementary tensors. But $V^{\otimes m}$ is also a $\mathbb CS_m$-module where $S_m$ acts by permuting the tensor factors. These two actions commute and hence $V^{\otimes m}$ is a representation of $S_n\times S_m$ in a natural way. I would like a pointer to the literature on the decomposition of $V^{\otimes m}$ into irreducible representations of $S_n\times S_m$. REPLY [8 votes]: Let $\nu$ be a partition of $n$ and $\lambda$ a partition of $m$. Mark Wildon points out, the multiplicity of $S^\nu \otimes S^\lambda$ in $V^{\otimes m}$ is equal to the multiplicity of an irreducible $S_n$ module $S^\nu$ in the irreducible $Gl_n$ module indexed by the partition $\lambda$. The latter restatement of this multiplicity is sometimes referred to as "the restriction problem" since it is determining the decomposition of the restriction of an irreducible $Gl_n$ module to $S_n$. People say (me included) that there does not yet exists a satisfactory combinatorial method or positive integral formula for computing this multiplicity (and this is what it means when it is said that the problem is "considered open"), but there is more to say about it because formulae for computing it do exist. This multiplicity can be expressed in terms of inner plethysm of symmetric functions. If we denote $s_\lambda\{s_\nu\}$ as the Frobenius image of the character of the symmetric group module $\Delta^\lambda(S^\nu)$ (the Schur functor indexed by the partition $\lambda$ applied to an irreducible $S_n$ module indexed by the partition $\nu$), then $$\left< s_\lambda\{s_{(n-1,1)} + s_{(n)}\}, s_\nu \right> = \sum_{\gamma \vdash n} \chi^\nu(\gamma)\frac{s_\lambda[\Xi_\gamma]}{z_\gamma}$$ because the Frobenius characteristic of $V$ is $s_{(n-1,1)} + s_{(n)}$. The right hand side of this expression comes from character theory of the symmetric group $S_n \subseteq Gl_n$, where I have used here the notation $s_\lambda[\Xi_\gamma]$ to denote the evaluation of a Schur function at the eigenvalues of a permutation matrix with cycle structure $\gamma$. There is an additional computational formula due to Littlewood [1] (for more modern treatment see [2]) that can be used to compute the multiplicity of an irreducible $S_n$ module in an irreducible $Gl_n$ module in terms of operation of outer plethysm of symmetric functions. The formula is (Theorem XI in [1]$^{(*)}$ and Theorem 4.1 in [2]): \begin{equation} [ \Delta^\lambda(V), S^\nu ] = \left< s_\lambda, s_\nu[1+s_1+s_2+s_3+\cdots] \right> \end{equation} where the square bracket $f[g]$ represents the operation of outer plethysm (in notation from Macdonald's book this is denoted $f \circ g$). This result is proved and reproved in the literature: [1] D. E. Littlewood, Products and Plethysms of Characters with Orthogonal, Symplectic and Symmetric Groups, Canad. J. Math., 10, 1958, 17–32. [2] T. Scharf, J. Y. Thibon, A Hopf-algebra approach to inner plethysm. Adv. in Math. 104 (1994), pp. 30–58. (*) Technically Littlewood's result (in modern notation) says $\left< s_\lambda\{ s_{(n-1,1)} \}, s_\nu \right> = \left< s_\lambda, s_\nu[ s_1 + s_2 + \cdots ] \right>$, but as Scharf and Thibon point out, this is equivalent to $\left< s_\lambda\{ s_{(n-1,1)} + s_{(n)} \}, s_\nu \right> = \left< s_\lambda, s_\nu[ 1 + s_1 + s_2 + \cdots ] \right>$. If you wish to compute any of these multiplicities in Sage for a fixed $\lambda$ and $\nu$ (below $\lambda = (3,2)$ and $\nu = (2,1)$), here are three functions which implement the formulae that I have stated above. sage: s = SymmetricFunctions(QQ).s() sage: def eq1LHS(la, nu): ....: n = sum(nu) ....: return s(la).inner_plethysm(s[n-1,1]+s[n]).scalar(s(nu)) sage: def eq1RHS(la, nu): ....: n = sum(nu) ....: return s(la).character_to_frobenius_image(n).scalar(s(nu)) sage: def eq2RHS(la, nu): ....: m = sum(la) ....: return s(nu).plethysm(1+sum(s[r] for r in range(1,m+1))).scalar(s(la)) sage: [eq1LHS([3,2],[2,1]), eq1RHS([3,2],[2,1]), eq2RHS([3,2],[2,1])] [5, 5, 5]<|endoftext|> TITLE: Approximation of convex body by polytopes QUESTION [7 upvotes]: In a recent survey paper, Approximation of convex sets by polytopes, Bronstein claims that under Hausdorff distance $\rho_H$, for every convex body $U$, $$\rho_H(U,\mathcal{P}_n)\leq c(U) n^{-2/(d-1)},$$ where $\mathcal{P}_n$ is the set of all polytopes with at most $n$ vertices. It is claimed in the survey paper (pp.729, section 4.1) that the same estimate is also valid if we replace $\mathcal{P}_n$ with other sets, including (1) $\mathcal{I}_n(U)$, the set of all inscribed polytopes in $U$ with at most $n$ vertices; (2) $\mathcal{I}_{(n)}(U)$, the set of all inscribed polytopes in $U$ with at most $n$ facets; (3)$\mathcal{O}_n(U)$, the set of all polytopes circumscribing $U$ with at most $n$ vertices, and (4) $\mathcal{O}_{(n)}(U)$, the set of all polytopes circumscribing $U$ with at most $n$ facets. My questions are: (1) Is there any result concerning the symmetric difference metric with the same estimate? In particular I am interested in the case $\mathcal{O}_{(n)}(U)$. An upper bound would be fine. Preferably with explicit constants such as $c_1\cdot c_2(d)c_3(U)$ with $c_2,c_3$ explicitly defined. No smoothness of the boundary is assumed and I am currently not interested too much about the asymptotic expansion the error estimate (usually involving the Gaussian curvature somewhere) (2) Similarly, do we have results that give an estimate of how well can polyhedral functions can approximate (good) convex function on a fixed domain $\Omega$? REPLY [2 votes]: If by the "symmetric difference metric" you mean the Nikodym Metric, there are plenty of references in Section 4.2 in Bronstein's 2008 survey. In particular, very precise upper bounds for $\mathcal{P}_n = \mathcal{I}_n$ and lower bounds for $\mathcal{P}_n= \mathcal{O}_{(n)}$. Regarding upper bounds for $\mathcal{P}_n = \mathcal{O}_{(n)}$, you could use [1, Section 5]. In that paper, Reisner, Schütt and Werner show that for any convex body $C$ in $\mathbb{R}^n$ such that $0 \in C$, there is a polytope $R \subset C$ of at most $N$ vertex such that $R$ approximates $C$ well (under the symmetric difference metric) and the polar $R^*$ approximates $C^*$ well (i.e. $error = O(N^{-2/(d-1)}$). [1] S. Reisner, C. Schu ̈tt, and E. Werner. Dropping a vertex or a facet from a convex polytope. In Forum Math., volume 13, pages 359–378, 2001.<|endoftext|> TITLE: A balls and urns model for a hashing problem QUESTION [7 upvotes]: Fix $N \in \mathbb{N}$. Suppose we throw $N$ numbered balls into $N$ numbered urns, so that for each $b \in \{1,\ldots,N\}$, ball $b$ lands in urn $j$ with equal probability $1/N$. Choose a number $c \in \{1,\ldots, N\}$ uniformly at random. Then choose further $b_1, \ldots, b_r \in \{1,\ldots, N\}$, so that $b_i$ is chosen uniformly at random from $\{1,\ldots,N\} \backslash \{b_1,\ldots,b_{i-1}\}$, stopping as soon as ball $b_r$ and ball $c$ are in the same urn. What is the expected value of $r$? I can get some fairly crude upper and lower bounds. I would like an asymptotically correct answer. One possible approach to the problem is to approximate the number of balls in urn $j$ by a Poisson random variable with mean $1$. So I would also be interested in the answer to the following question. Let $B_1,\ldots, B_N$ be independent Poisson random variables with mean $1$. What is the expected value of $r$ if we start with $B_j$ balls in urn $j$, for each $j$? Motivation. Suppose $\{1,\ldots, N\}$ are permitted passwords, and that passwords are hashed using an idealized hash function $h : \{1,\ldots, N\}\rightarrow \{1,\ldots, N\}$, constructed so that each $h(b)$ is chosen uniformly at random from $\{1,\ldots, N\}$. Then $r$ is the expected number of hashes we must compute to obtain a password $b \in \{1,\ldots,N\}$ with the same hash as a randomly chosen $c \in \{1,\ldots, N\}$. Very possibly the answer to my question is out there in the cryptography literature, but if so, I'm finding it hard to find among all the papers dealing with the birthday paradox or other types of hash collision. REPLY [3 votes]: General solution: assume there are $n$ passwords, $k$ hashes and $x_i$ passwords hashing to $i$. The expected time (drawing without replacement) for the drawing of the first password with hash $i$ is ${n+1 \over x_i +1}$. the probability that a randomly chosen $c$ hashes to $i$ is ${x_i \over n}$. Thus $$\mathbb{E}(r)={n+1 \over n}\sum_{i=1}^k {x_i \over 1+x_i}$$ If the hash function is a random mapping from $[n]$ to $[k]$ each $X_i$ is $Bin(n,{1 \over k})$, and $$\mathbb{E}(R)={k \over n}\big(n+1-k+k(1-{1 \over k})^{n+1}\big)$$<|endoftext|> TITLE: Frequency of a representation of SO(3) QUESTION [5 upvotes]: When generalizing the basic tenets of Fourier Theory to the symmetric group $S_n$, we can define a notion of the frequency of a basis function (i.e. an irreducible representation of $S_n$). In particular, the authors of [1] consider the Young tableaux of the permutation and define a semi-ordering of such permutations based on their relative 'dominance'. To define this dominance ordering, the authors decompose a permutation on a set of $n$ elements into its constituent cycles. This decomposition is expressed as a partition of the cardinality $n$ of the set. Definition (Dominance Ordering). Let $\lambda,\mu$ be partitions of $n$. Then $\lambda$ dominates $\mu$ if, for each $i$, $$\sum_{k=1}^i\lambda_k\geq \sum_{k=1}^i \mu_k.$$ Thus for a set of size $n$, the third-order partition $\lambda=(n-3,1,2)$ would dominate the fourth-order partition $\mu=(n-4,1,1,1)$. We think of the representations corresponding to $\lambda$ as being of 'higher frequency' than those corresponding to $\mu$. To clarify, it associates each irreducible representation of $S_n$ to a dominance ranking. This (partial) ordering establishes a foundation upon which we can band-limit a function, etc. It seems natural that we could construct a basis of irreducible representations of $SO(3)$ in a similar manner. **Has an analogous definition of the 'frequency' of Fourier basis representations been established for $SO(3) $? Thank you! REPLY [3 votes]: Though your interests are partly separate from the purely mathematical framework here, the basic theory is well developed. Notation and terminology vary, of course: e.g., your "semi-ordering" is usually called a "partial ordering". In the case of the symmetric group $S_n$, a convenient modern treatment is given by Gordon James in The Representation Theory of Symmetric Groups, Lect. Notes in Math. 682, Springer, 1978. Here the dominance ordering of partitions already comes into play in the early sections, where he constructs explicit models over the integers of the irreducible representations as "Specht modules" starting with a family of natural induced representations ("permutation modules"). Note however that in his later sections James is mostly interested in how all of this machinery over the integers behaves under reduction modulo a prime. Here the dominance ordering plays an even more crucial role in organizing the submodule structures. In any case, the dominance ordering for $S_n$ is natural in many situations. Also, it generalizes to all finite (real) reflection groups as the Chevalley-Bruhat ordering, which makes sense for any Coxeter group. This kind of partial ordering played a key role, for example, in Chevalley's older treatment of Schubert varieties for special linear groups, where $S_n$ occurs as the Weyl group. Turning to the compact real Lie groups $SO(3)$, there is again a purely mathematical development of the finite dimensional representations in an algebraic framework. Among many textbook treatments, see for instance section II.5 of Representations of Compact Lie Groups by Brocker and tom Dieck, Grad. Texts in Math. 98, Springer, 1985. Here the representations are naturally labelled by highest weights, identifiable with ordinary non-negative even integers: the representation $\rho_n$ has dimension $n+1$ for $n=0,2,4, \dots$. The twofold cover $SU(2)$ then has representations parametrized by all non-negative integers. Here the parametrization by integers actually gives a total ordering of highest weights, though for higher rank compact Lie groups there is only a partial ordering. Such an ordering isn't immediately essential in the $SO(3)$ setting, but as in the work of James and others for $S_n$ it plays a much larger role when the representations are constructed algebraically over the integers and then reduced modulo a prime. (They tend to lose irreducibility, which creates a hierarchy of interesting submodules.) Aside from the language involved, your notion of "frequency" does correlate with the standard notion of "highest weight" in the representation theory of compact Lie groups. Here as in the classical theory of finite group representations, Fourier analysis is often used as a tool in manipulating the resulting characters. But probably the most direct connection between the partial orderings of partitions and of highest weights comes in the relationship of Weyl groups to compact Lie groups of Lie type $A$ and arbitrary rank (not just $SO(3)$).<|endoftext|> TITLE: Vanishing of Dolbeault cohomologies and Steinness QUESTION [17 upvotes]: That Stein manifolds have all $(p,q), p \geq 0, q \geq 1$ vanishing Dolbeault cohomology groups is more or less standard. I am a little bit confused about the reverse implication: whether the vanishing of all Dolbeault cohomologies ($p \geq 0, q \geq 1$) implies Steinness? I could not find a reference, yet I noticed the following: In the recent (and quite authoritative) book by Forstnerič ''Stein Manifolds and Holomorphic Mappings'' Theorem 2.4.6 says: On any Stein manifold $X$ the Dolbeault cohomology groups vanish: $H^{p,q}_{\bar\partial}(X) = 0$ for all $p \geq 0, q \geq 1$. So the reverse is not even mentioned. In the older book by Grauert and Remmert ''Theory of Stein spaces'' on page 81 one reads: The assumptions of Theorem 5 are always satisfied by Stein manifolds. It is not known if there exist non-Stein manifolds of this type (i.e. $H^q(X,\Omega^p)=0, p \geq 0, q \geq 1$) So I assume this was open at least as of 2004 (?) On the other hand a recent paper by Chakrabarti and Shaw entitled ''The $L^2$-cohomology of a bounded smooth Stein domain is not necessarily Hausdorff'' (on arXiv, the author's website says that it is to appear in Math. Ann.) begins with: For each bidegree $(p,q)$, with $p\geq0,q >0$, the Dolbeault Cohomology group $H^{(p,q)}(\Omega)$ of a Stein manifold $\Omega$ in degree $(p,q)$ vanishes, and indeed this property characterizes Stein manifolds among complex manifolds. The provide references for that fact are the books of Hörmander and Gunning-Rossi (where I couldn't find it). REPLY [7 votes]: Sorry for the misleading remark in the paper. Indeed we were thinking about domains in Stein manifolds, and were careless in making the side remark, which of course has nothing to do with the main topic of the paper. Thanks for pointing out!<|endoftext|> TITLE: Etale local fibrations in the Grothendieck ring of varieties QUESTION [7 upvotes]: Let $k$ be a field and $K_0(Var_k)$ the Grothendieck ring of varieties over $k$. This is the ring generated by isomorphism classes of varieties over $k$ with multiplication given by $$ [X \times_k Y] = [X][Y] $$ and the relation that $[X] = [X \setminus Z] + [Z]$ for any closed subvariety $Z \subset X$. It is well known that if $\pi :X \to Y$ is a Zariski local fibration with fiber $F$, then $[X] = [F][Y]$. I assume this result is false if $\pi$ is only etale locally a fibration. Is there a counterexample? REPLY [10 votes]: To clarify what's happening, let us introduce the etale Grothendieck ring varieties $K^{et}(Var/k)$ by imposing the scissor congruence relation AND the relation $[X] = [F][Y]$ for every finite etale covering $X \to Y$ with (finite) fiber $F$. Let us prove that at least when $k$ has characteristic zero, we have $K^{et}(Var/k)$ isomorphic to $\mathbb{Z}$ via the Euler characteristic with compact supports. The proof goes in three steps. First we show by induction on dimension and using Noether Normalization that every element of $K^{et}(Var/k)$ is represented by a polynomial $P(\mathbf{L})$ in the Lefschetz class. The key observation is that any $n$-dimensional variety $X$ has an open subset $U$ which is a finite etale covering of an open subset of a projective space $U' \subset \mathbf P^n$. Then we show that $\mathbf{L} = 1$; this follows using the $2:1$ covering $\mathbf G_m \to \mathbf G_m$, $z \mapsto z^2$ as in Jim Bryan's comment. Thus every element in $K^{et}(Var/k)$ is represented by an integer. Finally, since the Euler characteristic respects etale coverings, it gives a well-defined homomorphism $K^{et}(Var/k) \to \mathbb Z$, and we just showed that this is an isomorphism!<|endoftext|> TITLE: Does the ring of invariants inherit normality? QUESTION [8 upvotes]: Let $A$ be a normal ring (in the sense that its localizations at prime ideals are normal domains), and suppose that a finite group $G$ acts on $A$ by ring automorphisms. Form the subring $A^G \subset A$ of $G$-invariant elements. Is the ring $A^G$ also normal? What if $A$ is Noetherian (so that it is a product of Noetherian normal domains, which, however, need not be preserved by the $G$-action)? I know that the answer is 'yes' if $A$ is an integral domain, but I am curious about the general case. REPLY [5 votes]: This is the answer of user74230 using localization instead of henselization. Namely, if $\mathfrak p$ is a prime of $A^G$, then we can replace $A^G$ by $(A^G)_\mathfrak p$ and $A$ by $A_\mathfrak p$ (same explanation as in the answer of user74230). Thus we may assume $A^G$ is local. Then $A$ has finitely many maximal ideals, each lying over the maximal ideal of $A^G$ (same reason as in the answer of user74230). Since $A$ is normal, each of the local rings has a unique mimimal prime. Hence $A$ has finitely many minimal primes. A normal ring with finitely many minimal primes is a finite product of normal domains. The case of a product of normal domains is easy.<|endoftext|> TITLE: Do the real numbers "know" that they are countable in a larger model? QUESTION [15 upvotes]: (This was first posted to math.stackexchange but had no answers there after several days): Let ${\mathbb R}$ be the set of real numbers in whatever is your favorite model of $ZFC$. Then (by Levy collapse) there is some larger model in which ${\mathbb R}$ is countable. I wonder whether there are any mathematically interesting facts about ${\mathbb R}$ that are, on the one hand, entirely internal to the original model, but on the other hand, best proven (or perhaps best discovered) by reference to the countability in the larger model. In other words, I am looking for arguments of the following form: "Choose an enumeration of ${\mathbb R}$ in some larger model $M'$. Then ..... and therefore $X$'', where $X$ is a statement about the real numbers that would both make sense and be interesting to a person who had never heard of model theory. Is there a reason to believe that no such arguments are likely to exist? REPLY [12 votes]: In the theory of Borel equivalence relations, one can define something called a pinned equivalence relation. Roughly speaking, a Borel equivalence relation $E$ on a Polish space $X$ is unpinned if there is a forcing which adds $E$-classes consisting only of new elements of $X$. (For a more precise definition, see http://users.math.cas.cz/~zapletal/f2note.pdf. Kanovei's book also has a good discussion, though I don't have my copy on me currently to give a more specific reference than that.) So a Borel equivalence relation is pinned if any new elements of $X$ added by forcing are $E$-equivalent to old elements of $X$ (again, roughly speaking). Many Borel equivalence relations are pinned. For example, every countable Borel equivalence relation is pinned. Also notably, the relation $E_{l^\infty}$ on $\mathbb{R}^\mathbb{N}$, which makes two sequences of reals equivalent if their $l^\infty$ distance is finite, is pinned. It's also not too hard to show that being pinned is closed downward under Borel reducibility, i.e. if $E$ Borel reduces to $F$ and $F$ is pinned, then $E$ is pinned. On the other hand, let $E_{ctble}$ be the equivalence relation on $\mathbb{R}^\mathbb{N}$ given by $(x_n) E_{ctble} (y_n)$ iff $(x_n)$ and $(y_n)$ enumerate the same set of reals. One can see that $E_{ctble}$ is unpinned by doing a collapse forcing. If you collapse the continuum to be countable, then you get elements of $\mathbb{R}^\mathbb{N}$ in the new model which enumerate all of the old reals. Clearly these sequences are not $E_{ctble}$-equivalent to any of the old elements of $\mathbb{R}^\mathbb{N}$! This means that $E_{ctble}$ does not Borel reduce to $E_{l^\infty}$. I would say that this is a fact about (equivalence relations on) $\mathbb{R}^\mathbb{N}$ which is completely internal to the model, but as far as I know, right now the only way we know to prove it is through this forcing argument. REPLY [4 votes]: Probably the interesting examples involve large cardinals. Instead of just collapsing $\mathbb R$ to be countable, force with $Col(\omega,<\kappa)$, where $\kappa$ is an inaccessible cardinal. This is the direct limit of repeatedly making the reals countable. So for each ordinal $\alpha \leq \kappa$, there is a new version of the reals $\mathbb R_\alpha$, and for $\alpha < \kappa$ these all become countable in the end. This feature is essential to the landmark theorem of Solovay that in the final model, every definable set of reals--definable meaning is in $L(\mathbb R)$--is Lebesgue measurable, equal to an open set modulo a meager set, and if uncountable contains a perfect subset. The proof is presented in Chapter 26 of Set Theory by Thomas Jech. It is a bit involved, but a key step that relates to your question is the following. Suppose $\mathbb R$ is the version of the reals in the final model, and $r \in \mathbb R$. Then in the final model, the intersection of all measure one Borel sets coded in $L(r)$ is measure one in $L(\mathbb R)$. This is because in the final model there are only countably many of these. Now with larger cardinals this enables us to actually say things about "our own universe." If there is a Woodin cardinal $\delta$, then there is a forcing called the stationary tower that turns $\delta$ into $\aleph_1$ and has some special features. Say we start in the model $V$ and $G$ is the generic filter for the stationary tower. Then: (1) There is a submodel $M$ of $V[G]$ that is closed under countable subsequences (thus contains all the new reals), and an elementary embedding $j : V \to M$. (2) There is another forcing extension $V[H]$ with the same reals as $V[G]$, where $H$ is generic for $Col(\omega,<\delta)$. Thus Solovay's theorem may be applied to say that every set of reals in $L(\mathbb R)^{V[G]}$ has all the regularity properties. By the elementarity of $j$, this holds for $L(\mathbb R)^V$ as well. There are set theorists on MO who are far more expert than me on these topics, so probably even more fascinating results can be given.<|endoftext|> TITLE: Definition of internal field objects QUESTION [11 upvotes]: Let $\mathcal{C}$ be a category with finite limits and a strictly initial object $\mathbf{0}$. The final object is denoted by $\mathbf{1}$. I propose the following definition of a field object internal to $\mathcal{C}$: Definition. A field object internal to $\mathcal{C}$ is a commutative ring object $R=(X,+,0,*,1)$ internal to $\mathcal{C}$ with the following property: Let $X^*$ be the equalizer of the maps $X \times X \rightrightarrows X$, defined by $(x,y) \mapsto x \cdot y$ resp. $(x,y) \mapsto 1$ on points. We may write $X^* = \{(x,y) \in X \times X : x \cdot y = 1\}$. Let $p : X^* \to X$ be the projection $(x,y) \mapsto x$. Then, we require that a morphism $T \to X$ factors through $p$ if and only if $T \to X$ and $\mathbf{1} \xrightarrow{0} X$ are disjoint, i.e. $$\begin{array}{cc} \mathbf{0} & \to & \mathbf{1} \\ \downarrow && ~~\downarrow {\scriptsize 0} \\ T & \to & X \end{array}$$ is a pullback square. Notice that this implies, in particular, that $1$ and $0$ are disjoint. Examples. Field objects internal to $\mathsf{Set}$ coincide with fields in the usual sense, right? Also, field objects internal to $\mathsf{Top}$ should coincide with topological fields in the usual sense. This is a bit subtle, because notice that, for any commutative ring object, we have a well-defined inversion map $X^* \to X^*$, $(x,y) \mapsto (y,x)$. But in the definition of a topological field, one usually assumes continuity of the inversion map on $X \setminus \{0\}$, which is not automatic. The solution is that for topological rings whose underlying ring is a field, the map $X^* \to X \setminus \{0\}$, $(x,y) \mapsto x$ has no reason to be a homeomorphism, unless the inversion map on $X \setminus \{0\}$ is continuous. This is somewhat built into the definition of a field object, namely applied to the inclusion map $X \setminus \{0\} \to X$. Questions. So this definition of a field object seems reasonable to me. But I have never seen it before. In fact, on mathoverflow there was some discussion that such a definition is either impossible or useless. Nevertheless, do you also think that my definition is correct? Is it natural? What other definitions can you think of? Is there some literature about this? Edit. In a comment Laurent Moret-Bailly points out that $\mathbb{A}^1_S$ carries a field structure internal to the category of $S$-schemes in the above sense. This is because a global section $s$ of an $S$-scheme $T$ is invertible if and only if the fiber product of $s : T \to \mathbb{A}^1_S$ with the zero section $0 : S \to \mathbb{A}^1_S$ is empty. REPLY [8 votes]: It's probably worth pointing out that there are various definitions that are equivalent in $\text{Set}$ but not in general. Let me call your definition invertible-elements. Here is another: Free-modules: A field $k$ is a commutative ring, not isomorphic to the terminal commutative ring, such that every $k$-module is free (that is, we require that the forgetful functor from $k$-modules down to the underlying category has a left adjoint, and free objects are objects in the essential image of this left adjoint). This definition doesn't agree with yours in topoi. For example, in the topos of $G$-sets, any field $k$ with trivial $G$-action is an invertible-elements field, but almost never a free-modules field: $k$-modules in $G$-sets are externally $k[G]$-modules, and in general not every $k[G]$-module is the free $k[G]$-module on a $G$-set. This suggests that free-modules is a bad definition because it doesn't do the right thing intuitionistically (probably a standard observation to the topos theory cognoscenti), but in any case it means that what definition you pick depends on what you want to do with it. REPLY [3 votes]: Martin's claim seems to be that he can define a field in a category with products and an initial object, but I am skeptical of this. I am uneasy with the language of the definition, in particular with the occurrences of $X\setminus\lbrace 0\rbrace$. This seems to be a mixture of categorical diagrams and a naive mathematical language that possibly assumes excluded middle. I think it would be a good exercise first to try to define an integral domain in this kind of language. Yves Diers did a lot of work in the 1980s on disjunctive theories in his categorical study of commutative algebra. A useful survey paper that will give you a good introdction to this is A syntactic approach to Diers’ localizable categories by Peter Johnstone in M. P. Fourman, C. J. Mulvey, and D. S. Scott, editors, Applications of Sheaves, volume 753 of Lecture Notes in Mathematics pp 466–478. Springer Verlag, 1979. Following the ideas in these papers, one can define a field in an extensive category with products, that is, one can express axioms such as $$ 0=1 \ \vdash\ \bot, \qquad x y = 0 \ \vdash\ x=0 \lor y=0, \qquad \ \vdash\ x=0 \lor \exists y. x y = 1. $$ Any category that one would be likely to regard as a "category of spaces" (in a very general sense, including locales and affine varieties) is extensive. So, if Martin wants to define fields more generally than this, I would first ask why? Does he have in mind some candidate for a "field" in a particular concrete category with products and an initial object but not coproducts? The foregoing comments assume that we working with fields where equality makes sense, such as algebraic number fields. In the case of $\mathbb R$, by contrast, inequality is the more natural relation. I could say a lot more about this from the point of view of different approaches to constructive analysis, but I haven't studied or thought about is in the context of categorical logic.<|endoftext|> TITLE: Publication in proceedings QUESTION [9 upvotes]: Why and how publishing a paper in proceedings? What are the difference with a "classical" journal? What's the list of the main proceedings in which one can publish? Do proceedings papers (never, sometimes, often or always) appear on mathscinet? REPLY [22 votes]: Proceedings of conferences are often published as special issues of "classical" journals. But even those that are not are usually included in MathSciNet if they include a statement (often a footnote on the first page of each paper) to the effect that the papers are in final form and will not be published elsewhere. Some but not all conference proceedings are refereed less thoroughly than reputable journal articles. As a result, mathematicians are sometimes suspicious about results published in conference proceedings, and administrators sometimes assign less value to such publications. Some conference proceedings have responded to this problem by explicitly saying (usually in the preface of the proceedings volume) that the papers have been refereed to the standards of such-and-such journal. Nevertheless, I would advise young (= not yet tenured) mathematicians to publish most if not all of their work in regular (and reputable, of course) journals. Once you have tenure, so that administrators' opinions are less critical for your life, it becomes reasonable to contribute more to conference proceedings. REPLY [8 votes]: I agree completely with Andreas' answer. One further consideration is publicity. It is easy for papers published in conference proceedings to become lost to general knowledge, or known only to very specialized groups. By publishing in a regular and reputable journal, the chances others will read your paper goes up. Further, it is not only administrators who hold the opinion that many conference proceeding volumes are of lower quality (at least in mathematics, but not, say, in computer science).<|endoftext|> TITLE: Completed and uncompleted operations for Morava $E$-theory QUESTION [13 upvotes]: Let $E = E_n$ be the $n$-th Morava $E$-theory with coefficient ring $$ E_* = \mathbb{W}(\mathbb{F}_{p^n})[\![u_1,\ldots,u_{n-1}]\!][u^{\pm 1}]. $$ It is usual to consider the completed co-operations $$ E^\vee_* E := \pi_*L_{K(n)}(E \wedge E) $$ rather than the 'ordinary' co-operations $E_*E$. The latter is shown by Hovey (although this was a folklore result) to be isomorphic to $\operatorname{Hom}^c(\mathbb{G}_n,E_*)$ the ring of continuous functions from the $n$-th Morava stabilizer group to $E_*$ under the $\mathfrak{m} = (p,u_1,\ldots,u_{n-1})$-adic topology. Let $L_0$ be the $L$-completion functor (for example see Appendix A of Hovey-Strickland). It is known that $E^\vee_*E = L_0(E_*E) = (E_*E)^\wedge_{\mathfrak{m}}$. There is a map $\phi:E_*E \to E^\vee_*E$ which can be identified with the usual $\frak{m}$-adic completion map. Is the map $\phi:E_*E \to E^\vee_* E$ injective? Standard commutative algebra tells us that the kernel of this map is isomorphic to $\bigcap_{n=0}^\infty \mathfrak{m}^n E_*E$, although it seems difficult to approach this problem in this way. Here is a closely related question. Let $E(n)$ be the Johnson-Wilson cohomology theory with coefficient ring $$ E(n)_* = \mathbb{Z}_{(p)}[v_1,\ldots,v_{n-1},v_n^{\pm 1}]. $$ Now there is an isomorphism $L_0(E(n)_*E(n)) = E(n)_*^\vee E(n) \simeq \operatorname{Hom}^c(\mathbb{G}_n,\widehat{E(n)}_*)$, where $$\widehat{E(n)}_* \simeq \mathbb{Z}_{(p)}[v_1,\ldots,v_{n-1},v_n^{\pm 1}]^\wedge_I$$ for $I = (p,v_1,\ldots,v_{n-1})$. In this case by work of Johnson it is known that there is an injection of $E(n)_*E(n)$ into $\operatorname{Hom}^c(\mathbb{G}_n \times\mathbb{G}_n,A)$ where $A$ the ring of integers in an unramified degree $n$ extension of the p-adic numbers. It is known that the latter is $L$-complete, and since $E(n)_*E(n) \to L_0(E(n)_*E(n))$ is initial amongst maps with $L$-complete target, in this case it follows that $E(n)_*E(n) \to E(n)^\vee_*E(n)$ is injective. REPLY [12 votes]: No, this map is not injective. To see this, put $W=\mathbb{W}(\mathbb{F}_{p^n})$, which is a free module of finite rank over $\mathbb{Z}_p$. It is standard that $\mathbb{Z}_p\otimes\mathbb{Z}_p$ contains a rational vector space of uncountable dimension, so the same is true of $W\otimes W$. (Here tensor products are implicitly taken over $\mathbb{Z}$ or $\mathbb{Z}_{(p)}$; you get the same anwswer either way.) This rational vector space can only map trivially to the group $E^\vee_*E$. Thus, it will suffice to check that $E_*E$ contains a copy of $W\otimes W$. The left and right unit maps provide a map $W\otimes W\to E_0E$. In the opposite direction, we can define a map $\phi:E_0\to W$ sending $u_i$ to $0$ for $0 TITLE: Are isometric homorphisms of C* algebras *-homorphisms QUESTION [6 upvotes]: Here is my precise question: Let $A$ and $B$ be two $C^*$ algebras. Let $f: A \rightarrow B$ a homomorphism of $C^*$ algebras which is isometric (on its image). Is $f$ necessary a $*$-homomorphism ? It sounds like a basic question, but I haven't found any counterexample nor any basic reference mentioning this kind of result, so I hope it is non trivial and suitable for MO. Another equivalent question is the following: Let $A$ be a $C^*$-algebra and $x$ an element of $A$ such that: 1) the spectrum of $x$ is included in $\mathbb{R}$ 2) for any polynomial $P$ (with coefcients in $\mathbb{C}$) the norm of $P(x)$ is the supremum of $|P(t)|$ for $t \in \text{Spec}(x)$. Is $x$ necessarily self-adjoint ? Indeed if the answer to this second question is yes, then any isometric algebra homorphism send self adjoint element to self adjoint element hence is a $*$-homorphism, and conversely if the answer to the first question is yes then for such an element $x$ one can construct an isometric morphism from $\mathcal{C}(\text{Spec}(X))$ to $A$ which is hence a $*$-homomorphism and hence $x$ is self adjoint as the image of a self adjoint element. Moreover, If I'm not mistaken the answer to these two questions is yes at least for finite dimensional algebras. I don't really have a precise motivation for this question except curiosity, but it might be interesting to have such a "$*$-free" characterization of morphisms of $C^*$-algebras if one want to develop a satisfying analogue to $C^*$ algebras for other valued field than $\mathbb{R}$ and $\mathbb{C}$. REPLY [5 votes]: This fact is proved in greater generality in propositions A.5.8 and A.5.9 in Operator algebras and their modules. An operator space approach. D. P. Blecher, C. Le Merdy<|endoftext|> TITLE: A reference for $\mathbb{A}^1_R$ being a coarse moduli space of the stack of elliptic curves QUESTION [13 upvotes]: Let $R$ be a ring and let $\mathcal{M}_{1, R}$ be the algebraic $R$-stack of elliptic curves (over $R$-schemes as bases). One knows that the coarse moduli space of $\mathcal{M}_{1, R}$ is supposed to be $\mathcal{M}_{1, R} \rightarrow \mathbb{A}^1_R$, with the map being given by the $j$-invariant. Is there a reference for this fact for an arbitrary $R$? I know that Deligne and Rapoport treat the case $R = \mathbb{Z}$ in Theorem VI.1.1 of their article, but they note that even this case is slightly painful to prove directly. The general case does not seem to follow formally because the formation of a coarse moduli space does not commute with non-flat base change. For completeness, I recall the definition of the coarse moduli space $M$ of an algebraic $R$-stack $\mathcal{M}$ as is given in Definition I.8.1 of Deligne-Rapoport: $M$ is an $R$-algebraic space equipped with a morphism $\mathcal{M} \rightarrow M$ that induces a bijection on (isomorphism classes of objects of) $\overline{k}$-points for every algebraically closed $R$-field $\overline{k}$ and that is initial among morphisms from $\mathcal{M}$ to $R$-algebraic spaces. REPLY [3 votes]: Here are some modern references: [1, Theorem 13.1.15] and [2, Lemma 2.1] [1] Olsson, "Algebraic Spaces and Stacks", Colloquium Publications 62, AMS (2016) [2] Fulton, Olsson, "The Picard group of $\mathscr{M}_{1,1}$", Algebra & Number Theory, vol. 4, no. 1 (2010) link<|endoftext|> TITLE: Function and its Gradient with Prescribed Norms QUESTION [5 upvotes]: I'm not sure if the following question is too elementary for Mathoverflow. I'm sorry if it is the case. Question: Let $n\in\mathbb{N}$ and let $1\leqslant p<\infty$. Let $\alpha,\beta>0$. What is the necessary and sufficient condition $\alpha,\beta$ for which there exists a $u\in C^\infty_{0}(\mathbb{R}^n)$ such that $$ \|u\|_p=\alpha\text{ and }\|Du\|_p=\beta? $$ What happens if we replace $\mathbb{R}^n$ with an open (not necessarily bounded) set $U\subseteq\mathbb{R}^n$? Probably, it has something to do with the Poincare Inequality or the first eigenvalue of the domain. But I'm unable to make it precise. Thank you. Stefan REPLY [6 votes]: Regarding $\mathbb{R}^n$, that's a simple matter of scaling. Assume w.l.o.g. that $\alpha = 1$ (the quantity that does matter in your problem is the ratio $\frac{\beta}{\alpha}$). Let $u$ be your favorite smooth cut-off function and assume w.l.o.g. that its $L^p$ norm is equal to $1$. Denote by $N$ the quantity $\|Du\|_p$. Define $u_{\lambda}(x) := \lambda^{-\frac{n}{p}} u(\frac{x}{\lambda})$. Then, notice that $u_{\lambda}$ has still an $L^p$ norm equal to $1$, while that of its derivative is $\lambda^{-1}N$. Choosing $\lambda$ such that $\lambda^{-1}N = \frac{\beta}{\alpha}$, you are done. Regarding the case of a general open subset $U \subset \mathbb{R}^n$, if you choose $\|u\|_p = 1$ (say), then the possible choices for $\frac{\beta}{\alpha}$ should lie in something like $\left[\sqrt{\mu_1(p)}, + \infty \right[$, where $\mu_1(p)$ is the first eigenvalue of the Dirichlet p-Laplacian on $U$. I'm no expert on this subject, but if things go like in the $p=2$ case, then having $U$ bounded in one direction gives a nontrivial range, contrary to what happens with $\mathbb{R}^n$.<|endoftext|> TITLE: Density of smooth functions in Sobolev space, respecting nonnegative traces QUESTION [5 upvotes]: I am looking for a reference for the following result (if it is true, which I would expect): Let $\Omega\subset\mathbb{R}^n$ be a bounded Lipschitz domain. Let $\Gamma_0\subset\partial\Omega$ be sufficiently regular. Let $V_1:=\left\{\phi\in C^\infty\left(\overline{\Omega}\right):\phi\geq 0 ~\text{on}~ \Gamma_0\right\}$. Let $V_2$ be the set of functions $u\in W^{1,p}(\Omega)$ with non-negative trace on $\Gamma_0$. Then $V_1$ is dense in $V_2$. If we consider the case of "$\phi = 0$" instead of "$\phi \geq 0$", of course an analogous result for $\Gamma_0=\partial\Omega$ is well known, and there is a result that it stays true if $\Gamma_0\subset\partial\Omega$ is relatively open and has relative Lipschitz boundary. It would already be very helpful to have a reference for the result with $\phi \geq 0$ and $\Gamma_0=\partial\Omega$. REPLY [4 votes]: More of a comment than an answer, but... I don't know a reference, but perhaps you could prove it as follows. For $u \in W^{1,p}(\Omega)$ with nonnegative trace, write $u = u^+ - u^-$ which are both in $W^{1,p}(\Omega)$. Since $u^+$ is nonnegative everywhere, you should be able to approximate it by nonnegative smooth functions $\phi_n$. And since $u^-$ has zero trace, by the result you quote, you can approximate it by smooth functions $\psi_n$ vanishing on the boundary (maybe even compactly supported). Then $\phi_n - \psi_n$ are nonnegative on the boundary and converge to $u$.<|endoftext|> TITLE: Is every connected reductive group over a local field already defined over a global field? QUESTION [8 upvotes]: Let $K$ be a local field, e.g. $\mathbb{Q}_p$ or $\mathbb{F}_p((t))$. Let $G$ be a connected reductive group over $K$. Is it true that $G$ is already defined over a global field? More precisely, does there exist a global field $F$, a place $v$ in $F$ with $F_v\simeq K$ and a connected reductive group $\tilde{G}$ over $F$ such that as groups $$\tilde{G}\otimes_F F_v\simeq G?$$ I am particularly interested in the case where $K$ is of equal characteristic. Thanks in advance! REPLY [14 votes]: Pick a global field $E$ and finite place $w$ with $E_w=K$. The fraction field $k$ over $E$ of the henselization of the "algebraic" local ring at $w$ is the direct limit of finite separable sub extensions $F/E$ for which the place $v$ on $F$ from the valuation on $k$ satisfies $F_v=K$. Thus, it suffices to "algebraize" $G$ over $k$. We therefore forget about number field and focus on a henselian valued field $k$ with completion denoted $K$ and aim to descend a connected reductive $K$-group $G$ to a connected reductive $k$-group. The Galois groups of $k$ and $K$ are naturally isomorphic, so if $R$ denotes the root datum of $G_{K_s}$ then ${\rm{H}}^1(k, {\rm{Aut}}(R)) \rightarrow {\rm{H}}^1(K, {\rm{Aut}}(R))$ is bijective. This says that every quasi-split connected reductive $K$-group descends to a quasi-split connected reductive $k$-group which is moreover unique up to isomorphism (since these H$^1$'s classify quasi-split forms with a given geometric root datum). Every connected reductive $K$-group $G$ has a (unique) quasi-split inner form $G_0$, so $G$ is obtained from $G_0$ via twisting against a class in ${\rm{H}}^1(K, G_0^{\rm{ad}})$ for the adjoint semisimple $G_0^{\rm{ad}} := G_0/Z_{G_0}$. Thus, it suffices to show that ${\rm{H}}^1(k, H) \rightarrow {\rm{H}}^1(K,H_K)$ is surjective for every smooth connected affine $k$-group $H$ (such as $H$ being the quasi-split $k$-descent of $G_0^{\rm{ad}}$). Even better: Theorem: For any smooth affine $k$-group $H$, the natural map $${\rm{H}}^1(k,H) \rightarrow {\rm{H}}^1(K,H)$$ is bijective. Proof: By Galois-twisting, injectivity reduces to triviality of the kernel. In other words, if $E$ is an $H$-torsor over $k$ which has a $K$-point then it has a $k$-point. More generally, if $X$ is a smooth $k$-scheme then $X(k)$ is dense in $X(K)$ for the valuation topoloy. This is Zariski-local on $X$, so we can assume there is an etale map $f:X \rightarrow \mathbf{A}^n_k$. The open image $V=f(X)$ is dense open in $\mathbf{A}^n_k$, so $V(k)$ is dense in $V(K)$ due to density of $k$ in $K$. By the Zariski-local structure theorem for etale morphisms and the $K$-analytic inverse function theorem, for each $x \in X(K)$ and $v=f(x)\in V(K)$, every $v'$ sufficiently near $v$ admits $x'\in f^{-1}(v')$ near $x$ in $X(K)$. By openness of $X(K) \rightarrow V(K)$, for any $x \in X(K)$ and open $\Omega \subset X(K)$ around $x$ we can find an open $U \subset V(K)$ around $v=f(x)$ such that every $u \in U$ is the image of a $K$-point in $\Omega$. Consider such $u \in V(k) \cap U$ (as exists by density of $V(k)$ in $V(K)$). The fiber scheme $f^{-1}(u)$ is finite etale over $k$, so the equivalence of Galois theories of $k$ and $K$ shows that every $K$-point in $f^{-1}(u)$ comes from a unique $k$-point of $f^{-1}(u)$. Hence, we can find a $k$-point in $\Omega$. This completes the proof of injectivity. For surjectivity, choose a closed $k$-subgroup inclusion $j:H \hookrightarrow {\rm{GL}}_n:=G$ and let $X=G/H$ (a smooth $k$-scheme). Thus, there is a natural surjection $$G(k)\backslash X(k) \rightarrow {\rm{H}}^1(k,H)$$ and likewise for $K$ (since ${\rm{H}}^1(k, {\rm{GL}}_n)=1$ and likewise for $K$). It therefore suffices to show that the natural map $$X(k) \rightarrow G(K) \backslash X(K)$$ is surjective. Since $X(k)$ is dense in $X(K)$ by the above, it suffices to show that all $G(K)$-orbits in $X(K)$ are open. But each orbit map $G_K \rightarrow X_K$ through a $K$-point is a smooth map since $H$ is smooth, so the induced map on $K$-points is open (hence has open image) by the $K$-analytic implicit function theorem (using the Zariski-local structure of smooth morphisms). QED<|endoftext|> TITLE: Limits at infinity of fellow-travelling sequences in Teichmuller space, QUESTION [9 upvotes]: I have a question concerning limits of sequences of points in Teichmuller space, and how this notion is preserved under fellow-travelling. Suppose that we have closed surface of genus $g\geq 2$, and a sequence $\{X_n\}_{n\in \mathbb{N}}$ of points in its Teichmuller space $T(S)$. Suppose furthermore that, for $n\rightarrow \infty$ this sequence converges to a point $\lambda\in \mathbb{P}ML(S)$ in Thurston's compactification of Teichmuller space. For example, we can take a diverging sequence of points on a Teichmuller geodesic associated to a pseudo-Anosov automorphism of $S$. Now suppose that we have another sequence $\{Y_n\}_{n\in \mathbb{N}}$ of points in $T(S)$ which fellow travels the first, meaning that $$d(X_n,Y_n)\leq r$$ for all $n$ and some $r>0$. Here, $d$ denotes the Teichmuller distance. Here are the questions: Does the sequence $\{Y_n\}$ converge to some point in $\mathbb{P}ML(S)$? If (1) is true, is the limit point of $\{Y_n\}$ the same as the limit point of $\{X_n\}$? It seems to me that the answer to both questions should be positive, but I am having trouble in proving the above statements, as convergence to a point in the boundary is expressed in terms of ratios between hyperbolic lengths and transverse measures of curves, while Teichmuller distance is related to the stretch factor of Teichmuller maps. These notions are (at least for me) not easily related. Any help is kindly appreciated! REPLY [7 votes]: There are some limited situations where your question has a positive answer, which can be stated in the setting of "convex cocompact subgroups of the mapping class group" $\text{MCG}(S)$, as developed in a paper of myself and Farb with title as in the quotes. Namely, if $\Gamma < \text{MCG}(S)$ is convex cocompact, and if $Z \in T(S)$ is any point, and if there exists $s>0$ such that the sequence $X_n$ stays within distance $s$ of the $\Gamma$-orbit of $Z$, then both of your questions 1 and 2 are answered affirmatively.<|endoftext|> TITLE: Replacing functors by topologically or simplicially enriched functors QUESTION [11 upvotes]: I believe it is true that if a functor $F:Top\to Top$ preserves weak homotopy equivalences then it is related by a natural weak equivalence to some other such functor that is in some sense continuous, perhaps continuous on morphisms in the sense that when a set map $K\to Top(X,Y)$ corresponds to a continuous map $K\times X\to Y$ then the resulting set map $K\to Map(F(X),F(Y))$ corresponds to a continuous map $K\times F(X)\to F(Y)$ when $K$ is a cell. Question 1: Is this true? If so, what is a reference? There are also simplicial analogues of this: instead of seeking to replace $F$ by a continuous functor, one could seek to replace it by a simplicial functor, i.e. a map of simplicially enriched categories from $Top$ to $Top$ (so that in addition to assigning an object $F(X)$ to each object $X$ it also assigns a map of ``simplicial hom-sets'' to each pair of objects). I believe this can be done by realizing the simplicial object $$n\mapsto F(X^{\Delta^n}).$$ Perhaps another way is to use the cosimplicial object $$n\mapsto F(X\times \Delta^n).$$ Here $F$ might be a functor from simplicial sets to simplicial sets rather than from spaces to spaces. Question 2: Again, references please. Question 3: Can a simplicially enriched replacement for $F$ be used to make a continuous replacement? I get a bit muddled here by the thought that while continuity is a property simplicial enrichment is a structure. ADDED LATER: To clarify: I'm quite sure that I see how to replace a functor (that preserves weak equivalences) by a weakly equivalent one that is simplicially enriched, in both the topological and the simplicial setting. I think that this is "well known", and I would be grateful for a good reference. I would be even more interested in a method of replacing such a functor by a weakly equivalent one that is topologically enriched (in the topological setting, of course). I think that somebody told me years ago that this can be done. And to address a possible point of confusion: Although realizations of "simplicial hom-sets" can be used as "hom-spaces", that's not quite the same as putting a topology on the ordinary hom-set. Answers to any of this would help both me and a student of mine with things that we are writing. REPLY [2 votes]: Regarding Question 2, it seems to me that Proposition 6.4 in the paper Simplicial structures on model categories and functors by Rezk, Schwede and Shipley pretty much gives you what you want (in a somewhat more general setting of simplicial model categories), using the same idea that you outlined. For Question 3, can't we use derived Kan extensions to pass from simplicial enrichment to topological one? To avoid set-theoretic issues you may want take Top to be a skeletally small category of CW complexes of some bounded cardinality. Having fixed a category of spaces Top, let Top$^s$ be the category with the same objects as Top, equipped with the topological enrichment that is the geometric realization of the standard simplicial enrichment. I.e., $$\mbox{Top}^s(X, Y)=|S_*\mbox{map}(X, Y)|$$ where $S_*$ is the singular set, and $\mbox{map}(X, Y)$ is the usual mapping space. There is a canonical functor $\mbox{Top}^s\to$ Top, that is a weak equivalence in the sense that it is a bijection between sets of objects, and a weak equivalence on each mapping space. Suppose $F\colon \mbox{Top}\to \mbox{Top}$ is a simplicially enriched functor. Then $F$ can be regarded as a continuous functor from Top$^s$ to itself. From here we easily get a continuous functor $F\colon$ Top$^s\to $Top (which I continue denoting by $F$). Finally to get $F$ to be a continuous functor from Top to Top, replace $F$ with the homotopy left Kan extension of the functor $F\colon \mbox{Top}^s\to \mbox{Top}$ along the canonical functor $\mbox{Top}^s\to \mbox{Top}$. Explicitly, you can construct the Kan extension as the realization of the bar construction $$ \tilde F(Y)\leftarrow\coprod_{X_0\in \mbox{Top}} F(X_0)\times \mbox{map}(X_0, Y)\Leftarrow \coprod_{X_0, X_1\in \mbox{Top}} F(X_0)\times |S_*\mbox{map}(X_0, X_1)| \times \mbox{map}(X_1, Y)\cdots $$ This new functor $\tilde F$ is enriched over Top. That $\tilde F$ is equivalent to $F$ follows from the fact that the canonical functor $\mbox{Top}^s\to \mbox{Top}$ is a weak equivalence.<|endoftext|> TITLE: Establishing Duality in Tannakian Categories QUESTION [6 upvotes]: I sometimes need to prove a category is Tannakian. Part of the definition of a Tannakian category is that it is rigid. However, I find the definition of rigid categories somewhat difficult. I don't know how to show that these morphisms are identities. Is there a way of looking at this that makes it more clear when these identities hold? In my cases it is obvious that there exists a functor $D$ and isomorphisms: $$Hom(X,Y) = Hom(D(Y),D(X))$$ $$Hom(X,Y)=Hom(1, Y \otimes D(X) ) $$ $$ X = D(D(X)) $$ $$ D(X \otimes Y) = D(X) \otimes D(Y)$$ So in particular I get maps $1 \to X \otimes D(X)$ and $X \otimes D(X) \to 1$. Is there a simple way of expressing duality in terms of a property of this functor? REPLY [8 votes]: As Mostafa points in the comments, it suffices to have a canonical isomorphism $$ \hom(X\otimes Y,Z) \overset?= \hom(X,Z\otimes D(Y)). $$ But if I am not mistaken, you have $$ \begin{aligned} \hom(X\otimes Y,Z) & = \hom(1,D(X\otimes Y)\otimes Z) \\ & = \hom(1,D(X)\otimes D(Y)\otimes Z) \\ & = \hom(X, D(Y) \otimes Z)\end{aligned}$$ using what you know (and symmetric monoidality, which I assume you have, as otherwise I would have expected $D(X\otimes Y) = D(Y) \otimes D(X)$). Note that I only used your second and fourth properties. I'm pretty sure that the second property alone is not good enough: you need some way to deal with tensor products in the first slot of $\hom$.<|endoftext|> TITLE: Which unfoldings of the hypercube tile 3-space: How to check for isometric space-fillers? QUESTION [26 upvotes]: Recently Mark McClure constructed and displayed the 261 unfoldings of the hypercube (tesseract) in response to the question, "3D models of the unfoldings of the hypercube?": The first 9 unfoldings in Mark McClure's display Each of the 11 unfoldings of the cube form monohedral tilings of the plane, as so well illustrated in the "Etudes" video to which Igor Pak pointed:           A polyhedron that is the prototile of a monohedral tiling is called an isometric space-filler: $\mathbb{R}^3$ can be tiled by congruent copies of that one shape (rotated and translated but not reflected). Now that we have the unfoldings of the hypercube, analogy with the cube raises the question: Q1. Which (if any) of the 261 unfoldings of the hypercube are isometric space-fillers? Asking this question raises another: Q2. How can one determine if a given shape, in this case a polycube / 3D polyomino, is an isometric space-filler? Update (7Dec2015). Aside from the two hypercube unfoldings that Steven Stadnicki showed tile space (below), with a student I found two more that tile $\mathbb{R}^3$, including the Dali hypercube cross unfolding, confirming Steven's intuition ("I don't know if the 'Dali unfolding' tiles space, though I'd be surprised if it didn't.") We posted an arXiv note on the topic. REPLY [39 votes]: Answer to Q1: All of the 261.  I looked at this question because of a video of Matt Parker and wrote an algorithm to find solutions. See here for an example of how a solution would look like. I dumped all solutions on github. For some cases I list multiple solutions. The files start with a number followed by an underscore and the number is between 0 and 260 corresponding to one of the unfoldings in this list. 129 <--- Number of unfolding (this time 0-indexed) [[-2, -2, 0], [1, 1, 1]] <---- the box, we put it in Boolean Program (maximization, 1152 variables, 428 constraints) <--- some information to ignore 32.0 <--- How many of the voxels in the cube are covered (here it happens to be all of them) [(-2, -2, 0), (-1, -2, 0), (-1, -2, 1), (-1, -1, 0), (-1, 0, 0), (-1, 1, 0), (0, -2, 1), (1, -2, 1)] [(-2, -2, 1), (-2, -1, 0), (-2, -1, 1), (-2, 0, 0), (-2, 1, 0), (-1, -1, 1), (0, -1, 1), (1, -1, 1)] [(-2, 0, 1), (-1, 0, 1), (0, 0, 1), (1, -2, 0), (1, -1, 0), (1, 0, 0), (1, 0, 1), (1, 1, 1)] [(-2, 1, 1), (-1, 1, 1), (0, -2, 0), (0, -1, 0), (0, 0, 0), (0, 1, 0), (0, 1, 1), (1, 1, 0)] (the last rows are then the coordinates of the copies of the unfolding, whatever sticks out of the box fits in another one..) Here is an example of one of the two one that fit in a 4x4x2 box ( in an exploded view, you can shift them together and they neatly fit into a 4x4x2 box, which can then be stacked to obtain a tiling.: Answer to Q2: One way is to use integer programming. The basic idea is to take a box with volume divisible by 8 as a fundamental domain and to cover it as much as possible with non-overlapping copies of an unfolding and make sure that the stuff that sticks out of the bottom actually fits into gaps at the top and so on. Here's a two-dimensional example: The tile: A 5x6 box filled with copies of that tile: A tiling produced from this: This can be formulated a as an integer program and it turns out that those 261 unfoldings all have feasible solution for some smallish box (for most 4x4x4 is enough). Setting up the integer program is just a few lines of sage: from sage.combinat.tiling import Polyomino import itertools def get_mod_points(point, diff): for coeffs in itertools.product([-1, 0, 1], repeat=diff.length()): yield vector([diff[i]*coeffs[i] + point[i] for i in range(diff.length())]) def milp_mod_box(tile, inner_box, outer_box, solver=None): # tile, inner_box and outer_box are `Polyomino`. begin, end = inner_box.bounding_box() diff = vector(end) - vector(begin) diff += vector([1]*diff.length()) tiles = [Q for Q in tile.isometric_copies(outer_box)] p = MixedIntegerLinearProgram(solver=solver, maximization=True) x = p.new_variable(binary=True) for i in inner_box: cons = p.sum(sum([int(j in T) for j in get_mod_points(i, diff)])*x[T] for T in tiles) == 1 p.add_constraint(cons) for i in outer_box: p.add_constraint(p.sum(x[T] for T in tiles if i in T) <= 1) p.set_objective(p.sum(x[T] for i in inner_box for T in tiles if i in T)) return p, x Here we optimize for having as much voxels filled in the box as possible, but only so that the solutions look nicer, it would give valid tilings even without setting that objective. A similar integer programming approach could be used to prove that a certain polyomino is not a space filler: We can check for larger and larger boxes that if it is not possible to cover them completely with non-overlapping copies of the original polyomino, but this wasn't necessary. Update: I added 3d-plots of all solutions to 3d-renderings of all 261 unfoldings page. When you click on the number of each one you get a 3d-rending of copies of that unfolding mostly in a box (drawn in grey), which then can be used to tile the plane. Note this is a different order than the one above (and it starts indexing with $1$. #213 is the Dalí-unfolding, needs 8 pieces in the prototile. #3 is the only one that only needs 3 pieces (in a 4x3x2 box) (what a coincidence..) #72 and #159 are the ones that perfectly fit in a 4x4x2 box. #139 needs a long 8x2x2 box.<|endoftext|> TITLE: Product of "prime" topological spaces QUESTION [5 upvotes]: We call a topological space $(X,\tau)$ product-decomposable if there is an index set $I$ and subsets $X_i\subseteq X$ for $i\in I$ such that $|X_i| > 1$ and $X \cong \prod_{i\in I} X_i$ where each $X_i$ is endowed with the subspace topology. If a space $(X,\tau)$ is not product-decomposable and $|X|>1$, we call it prime. (Is there established terminology for this?) If $X_1, X_2$ are prime spaces, is it possible that there are prime topological spaces $Y_1, Y_2$ such that $X_i \not\cong Y_j$ for $i,j\in\{1,2\}$ but $(X_1\times X_2) \cong (Y_1\times Y_2)$? REPLY [2 votes]: This is just a cheap extension of PVAL's comment, but I think it works: Taking for granted that $S^2$ and $TS^2$ are prime, which I believe is true, you can write $$TS^2 \times (\mathbb R \amalg S^2) = S^2 \times (\mathbb R^3 \amalg TS^2),$$ and now all four spaces appearing are prime and pairwise not homeomorphic.<|endoftext|> TITLE: Errata for Getzler-Kapranov "Cyclic operads and Cyclic homology" QUESTION [7 upvotes]: Do you know if anyone made an errata for the Getzler-Kapranov paper "Cyclic operads and Cyclic homology"? I was trying to read it, and I have found (at least I think I did) quite a few typos not all of which are obvious. I have looked at the published version of this paper in the "Geometry, topology, & physics" but seems to be exactly the same as the preprint. This paper has a lot of citations, so I thought it might be reasonable to ask. Thank you very much! REPLY [9 votes]: I am using this version of the paper, and it seems like the published version looks exactly like that. Here is what I have found. First, a warning: Getzler and Kapranov use the term "quadratic operad" for what Loday-Vallette call "binary quadratic". In example 4.5 (a), there is one more relation missing. Namely, we also need to include $B(x_1x_0,x_2)=B(x_0,x_2x_1)$ corresponding to the transposition $\sigma\in k[S_2]=Ass(2)$. In the beginning of section 4.7, on page 17, the map called $B_n$ should come from the action of $\mathcal{P}$ on $A$. In proposition 4.9, $A$ is the free $\mathcal{P}$-algebra $F(\mathcal{P},V)$ on $V$. In the second formula in the proof of proposition 4.9, one should swap $p$ and $q$. In the beginning of section 5.8, $\mathcal{Z}$ not necessarily a cyclic cooperad, can also be anti-cyclic, because the construction applies to the bar construction $\mathcal{BP}$, which is anti-cyclic. In lemma 5.11(2) instead of the first $CA(\Phi,A)$ should be $CB$. In lemma 5.12 in the RHS of the formula, $A$ should be replaced by $V$. In section 6.6, there is a formula $CC_n(\Phi,A)\simeq CHarr_{n+1}(A,A)$, which in the diagram of section 6.9 becomes $CC_n(\Phi,A)\simeq CHarr_{n-1}(A,A)$. I think the first one is correct, but I am not sure here. In the definition of $\lambda(\mathcal{Z},C)$ in section 6.11 on p.28, I think the map should be modified to be $C\otimes C\to \mathcal{Z}(n)\otimes C^{\otimes (n+1)}\to \mathcal{Z}(n)\otimes C^{\otimes (n+1)}$, where the last map is $1-\sigma$, for every $\sigma\in S_{n+1}$. Otherwise, the definition of Getzler-Kapranov for cooperads does not give a result dual to the usual definition for operads. If you have found some more, please, add to this list! There are still plenty of places where I don't know what is going on.<|endoftext|> TITLE: Global existence for infinite dimensional ODE QUESTION [5 upvotes]: Let us consider the ODE $\hskip3pt \dot x=F(t,x)\hskip3pt $ in an infinite-dimensional Banach space $E$, where the flux $F$ is defined and continous from the whole $\mathbb R\times E$ into $E$. (1) Question 1. Assuming $$ \Vert{F(t,x)}\Vert\le \alpha(t)\Vert{x}\Vert,\quad \text{with $\alpha\in L^1_{loc}(\mathbb R)$}, \tag{$\ast$} $$ is probably not sufficient for global existence: it should be a variation on J. Dieudonné's counterexamples for infinite dimensional ODEs. (2) Question 2. However I do believe that solutions of linear equations (i.e. $F(t,x)=A(t) x$, $A(t)$ bounded endomorphism of $E$, depending continuously on $t$) do exist globally. Why? Note that it is of course obvious for a constant $A$, since we have in this case the explicit solution $ e^{tA}x(0). $ REPLY [2 votes]: The answer to (2) is yes, and it's very standard: the corresponding $F$ satisfies the Cauchy-Lipschitz-Picard-Lindelöf hypotheses: being continuous $A$ is locally bounded, so $F$ is locally uniformly Lipschitz in the variable $x$. In fact, for a continuous $A:[a,b]\to L(E)$ you can directly solve the Cauchy problem for the ODE with parameter: $$\begin{cases}G(s,s)=I \\ \partial_1 G(t,s)=A(t)G(t,s) \end{cases}$$ solving the equivalent integral equation $$ G(t,s)- \int_s^t A(\tau)G(\tau,s)d\tau=I \ , $$ which consists in inverting a quasinilpotent perturbation of the identity on the Banach space $C^0([a,b]\times [a,b],L(E))$. This is easily done in terms of the Neumann series: $$ G(t,s)=\sum_{k=0}^\infty W_k(t,s) , $$ $$\begin{cases}W_0(t,s)=I \\ W_{k+1}(t,s)=\int_s^t A(\tau)W_k(\tau ,s)d\tau \end{cases}\ .$$ The variation of constant formula holds too, and produces the unique solution of the non-homogeneous problem $$\begin{cases}u(s)=u_0 \\ \dot u(t)=A(t)u(t)+b(t) \end{cases}$$ as $$u(t)=G(t,s)u_0+\int_s^tG(t,\tau)b(\tau)d\tau \ .$$<|endoftext|> TITLE: Category of Gödel Codings? [Reference Request] QUESTION [8 upvotes]: Consider computation with the integers $\mathbb{Q}$. The traditional theory of recursive functions on $\mathbb{N}$ applies to $\mathbb{Q}$ by the identification of $\frac{a}{b} \in \mathbb{Q}$ with $2^{a}3^{b} \in \mathbb{N}$. Similarly, if I have any similar unnamed structure I can proceed likewise, as follows: First, note that in this process of "Gödel numbering" a given object may have more than one representative Gödel number in $\mathbb{N}$. However the set of valid Gödel numbers is primitive recursive and there is a primitive recursive equivalence relation that determines whether two numbers represent the same object. This leads to the following category $\textsf{PRA}$: The objects of $\textsf{PRA}$ objects are pairs $(A, \sim _{A})$ where $A \subseteq \mathbb{N}$ is a primitive recursive set and $\sim _{A} \subseteq A \times A$ is a primitive recursive equivalence relation. The morphisms of $\textsf{PRA}$ are are functions $f: (A, \sim _A) \rightarrow (B, \sim _B)$ induced by primitive recursive functions $F: \mathbb{N} \rightarrow \mathbb{N}$ such that $x \sim _{A} y \Rightarrow F(x) \sim _{B} F(y)$. This is all rather rudimentary and seems like the correct definition (to me) of a Gödel numbering and leads naturally to the category just described. However I have never seen this category described or studied. Is this category simply Uninteresting, or perhaps Unimportant? Any references would be appreciated. [See comment below.] REPLY [7 votes]: Andrej and Nate have given good introductions to a number of "sophisticated" ideas in computability theory, but I would like to point out something elementary that ought perhaps to be said beforehand. We have been doing practical computatation for at least seven decades now, so it mystifies me why theoretical compubaility still insists on basing all of its encodings on $\mathbb N$. We can all do actual computation in $\mathbb Q$, but it becomes absurdly infeasible if $p/q$ has to be encoded as $2^p 3^q$. This is unnecessary -- why do it? The axioms for $\mathbb N$ use zero, successor and recursion. If we simply replace unary successor in this with a binary operation (pairing), written $[-|-]$, then we obtain a type $\mathbb T$ of finite binary trees that is sufficient to encode a large variety of computational objects in a way that is computationally feasible. From the definition it clearly satisfies $$ {\mathbb T} \equiv {\mathbb 1} + {\mathbb T} \times {\mathbb T}, $$ whilst also $$ {\mathbf{List}} ({\mathbb T}) \equiv {\mathbb T} $$ by, for example, $[a,b,c,d] \equiv [a|[b|[c|[d|0]]]]$. This form of encoding has been used in functional and logic programming languages at least since Lisp. Given that it differs so little from the natural numbers I don't understand why it is not adopted in coomputability theory. Of course $\mathbb T$ is isomorphic to $\mathbb N$, but the functions are ridiculously infeasible.<|endoftext|> TITLE: Variety acquiring rational point over any quadratic extension QUESTION [25 upvotes]: Does there exist a variety $X$ over $\mathbb{Q}$ (or a number field) such that it has no rational points over $\mathbb{Q}$ but acquires points over any quadratic extension $\mathbb{Q}(\sqrt{d})$? If there is an example, how often can this happen? Can we generalize to an extension of fixed degree? (This might be a stupid question and I'm guessing answer is no, but I couldn't show it) REPLY [8 votes]: According to this answer by Laurent Moret-Bailly, the set of rational non-squares is diophantine over the rationals (a result of Bjorn Poonen): there is a polynomial $P(a,x_1,\dots,x_n)$ which for $a\in\mathbb{Q}$ has a rational solution iff $a$ is a rational non-square. For the variety, take the hypersurface with equation $P^2+(x_{n+1}^2-a)^2=0$ in $n+2$ coordinates $(a,x_1,\dots,x_{n+1})$.<|endoftext|> TITLE: Averages over integer points of the sphere QUESTION [10 upvotes]: A paper of William Duke proves that integer points on the sphere are equidistributed: $$ V_n = \{ (x,y,z) \in \mathbb{Z}^2 : x^2 + y^2 + z^2 = n \}. $$ Up to reflections across the $x$, $y$ and $z$ axes, it seems intuitive these integer points should not accumulate in any octant. The proof involves estimating averages over the points on the sphere, where a slow decay is obtained: $$ \frac{1}{r_3(n)} \sum_{\xi \in \frac{1}{\sqrt{n}}V_n} u(\xi) \ll_{u,\epsilon} n^{-1/28+\epsilon},$$ where $u \in L^2(SO(3))$ is a spherical harmonic. This is enough to show the rescaled integer points become equidistributed. If I understood the Siegel bound $r_3(n) \gg_\epsilon n^{1/2-\epsilon}$ and the difficult estimates of Ivaniec for holomorphic cusp forms, I would understand Duke's proof. Is there really no more direct way of understanding this? It looks like we have taken a Riemann sum. We could try to measure the discrepancy $$\left| \frac{1}{r_3(n)} \sum_{\xi \in \frac{1}{\sqrt{n}}V_n} f(\xi) - \int_{S^2} f(x) \, dS \right| \leq K_f D(V_n)$$ This is like the Erdős-Koksma-Hlawka inequality or something. REPLY [9 votes]: First of all note that $n$ has to satisfy some trivial local conditions, otherwise $r_3(n)$ will be zero or too small. Let us assume that $n\equiv 1\pmod{4}$ for simplicity, then the theorem holds. Let me also remark that the best known exponent of $n$ in your second display is $-\frac{1}{12}+\epsilon$, see the work of Conrey-Iwaniec (Annals, 2000) and Matt Young (arXiv:1405.5457). The Grand Lindelöf Hypothesis (a consequence of the Grand Riemann Hypothesis) would yield the exponent $-\frac{1}{4}+\epsilon$. As far as I know, Duke's proof (Inventiones, 1988) is still the only proof, and it was a real breakthrough at the time, based on the work of Iwaniec (Inventiones, 1987). More precisely, the Weyl-type sum in your second display is directly related to a central value of a twisted modular $L$-function, and now there are several methods and results about the subconvexity of these $L$-values (see the work of Duke, Friedlander, Iwaniec, Bykovskii, Conrey, Harcos, Blomer, Michel, Sarnak, Cogdell, Piatetski-Shapiro, Venkatesh, Wu, Maga, Hoffstein, Hulse, Petrow, Young). Linnik could prove the theorem under the condition that $n$ is a quadratic residue for any fixed odd prime, which of course implies (by throwing in more and more odd primes) that 100% of the $n$'s with $n\equiv 1\pmod{4}$ satisfy the equidistribution property. The work of Einsiedler, Lindenstrauss, Michel, Venkatesh builds on Linnik's ideas, and they develop them further, but I don't think they got rid of Linnik's congruence condition in the original Linnik problem. Correct me if I am wrong.<|endoftext|> TITLE: Application of Fraïssé construction in set theory QUESTION [15 upvotes]: As you know Fraïssé limit construction and its generalization, Hrushovski's construction, have many applications in model theory to build models with interesting property. Now I would like to know the application of Fraïssé construction in set theory. Question: What are the major applications of Fraïssé construction in set theory? Any reference will be appreciated. REPLY [7 votes]: This is a comment on something Nate Ackerman posted two days ago in the context of Shelah's categoricity conjecture for Abstract Elementary Classes. As I according to math overflow I don't have the necessary credentials to comment directly, I encloed this. At present even the consistency of that conjecture is open. The best ZFC result is for AECs that satisfy extra conditions assuming that the class is categorical in a successor (Grossberg & VanDieren). Boney managed to derive the above "extra conditions" from a class-many strongly compact cardinals. In all approximations in the last 25 years to the categoricity conjecture for AECs one had to assume categoricity in \lambda^+ removal for the successor assumptionis considered one of teh major open problems. Last week Sebastien Vasey posted on the arXiv a rather long paper, his Theorem 1.7(2) establishes the consistency of the eventual categoricity conjecture (removing the successor). In his proof he is using several major recent results (total of about 500-600 pages) that I am sure are correct. However he is also using a Theorem that Shelah announced several years ago, the draft of that paper of Shelah is more than 150 pages long and he is working on it several years now.<|endoftext|> TITLE: Abstract connectedness QUESTION [15 upvotes]: Is there an abstract structure that characterizes connectedness, analogously to how topological spaces characterize continuity? Here's one way to make this question more precise: if $(X,T_X)$ is a topological space with underlying set $X$ and topology $T_X$, then consider the pair $(X,C_X)$ where $C_X$ is the set of connected subsets of $X$. A continuous map $f:(X,T_X) \to (Y,T_Y)$ has the property that the direct image of a connected set is connected. Thus, we have a functor from Top to the category whose objects are pairs $(X,C_X)$ of a set equipped with a set of subsets, and whose maps $f:(X,C_X)\to (Y,C_Y)$ are functions $f:X\to Y$ such that $f(U) \in C_Y$ for all $U\in C_X$. Is there a naturally defined full subcategory of the latter category that is "close" to the image of Top? There are undoubtedly other ways that one could make this question more precise; if you have a better suggestion feel free to make it. REPLY [2 votes]: Let me answer as to how well the connected subspaces of a space determine the topology. I claim that for a large class of spaces, the collection of all connected subsets of the space determines the topology on $X$. However, the functions between nice connected spaces that map connected sets to connected sets are generally not continuous. The following lemma states that the closure of a connected subspace of a regular space $X$ is determined by the collection of all connected subspaces of $X$. In particular, the closed connected subspaces of a regular space $X$ are determined by the collection of all the connected subspaces of $X$. $\mathbf{Lemma}$ Suppose that $X$ is a regular space and $A\subseteq X$ is a connected subset. Then $\{a\}\cup A$ is connected if and only if $a\in\overline{A}$. Furthermore, if $A\subseteq B\subseteq X$, then $B\subseteq\overline{A}$ if and only if whenever $A\subseteq C\subseteq B$ then $\overline{C}$ is connected. The proof of the above lemma is straightforward. We shall say that a point $x$ is a connected space $X$ is a nbd-cut-point if there is some open neighborhood $U$ of $x$ such that whenever $V$ is an open set with $x\in V\subseteq U$, then $X\setminus V$ is not connected. Then it is easy to show that a connected space $X$ has no nbd-cut-points if and only if every closed subspace of $X$ is the intersection of connected closed subspaces of $X$. I should also mention that I made up the notion of a nbd-cut-point in order to answer this question, but the notion of a nbd-cut-point is closely related to the notion of a cut-point since every cut point is a nbd-cut-point. $\mathbf{Proposition}$ Suppose that $X,Y$ are connected regular spaces with no nbd-cut points. If there is a bijection $f:X\rightarrow Y$ so that if $C\subseteq X$ then $C$ is connected if and only if $f[C]$ is connected, then $X$ and $Y$ are homeomorphic. However, I must mention that even for connected regular spaces with no nbd-cut-points, there are non-continuous maps where the image of a connected space is connected. For example, define an equivalence relation on $\mathbb{R}$ where $x\simeq y$ iff $x-y\in\mathbb{Q}$. Then there is a surjective function $f:\mathbb{R}/\simeq\rightarrow\mathbb{R}^{2}$ since $|\mathbb{R}/\simeq|=|\mathbb{R}^{2}|$. Define a mapping $g:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ by letting $g(x,y)=f([x])$ and define $h:\mathbb{R}\rightarrow\mathbb{R}^{2}$ by letting $h(x)=f([x])$. Then $h[U]=\mathbb{R}^{2}$ for each nonempty open set $U$. I claim that $g$ maps connected subsets to connected subsets. Let $\pi_{1},\pi_{2}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the projection maps where $\pi_{1}(x,y)=x,\pi_{2}(x,y)=y$. Suppose that $C\subseteq\mathbb{R}^{2}$ is connected. Then $\pi_{1}[C]$ is also connected, so $\pi_{1}[C]$ is either a singleton or an interval. If $\pi_{1}[C]$ is a singleton, then $g[C]$ is a singleton in $\mathbb{R}^{2}$. However, if $\pi_{1}[C]$ is an interval, then $g[C]=\mathbb{R}^{2}$. In either case, the function $g$ maps connected sets to connected sets, but the function $g$ is very far from being continuous. It therefore seems as if the notion of continuity cannot be described in terms of connectedness unless one describes the open or closed sets in terms of connectedness and directly translates the notion of continuity into a notion of a connected mapping.<|endoftext|> TITLE: Maryam Mirzakhani's works QUESTION [56 upvotes]: Maryam Mirzakhani has made several contributions to the theory of moduli spaces of Riemann surfaces. Mirzakhani was awarded the Fields Medal in 2014 for "her outstanding contributions to the dynamics and geometry of Riemann surfaces and their moduli spaces." She died July 15, 2017. I'm not expert in these areas of mathematics, but I am eager to know her main ideas and the importance of her results. Can one draw a general picture of her works? (Any expository reference will be appreciated). REPLY [4 votes]: Following up on Tom Church's answer, Alex Wright has recently written an overview for Mirzakhani's work on surfaces. A tour through Mirzakhani's work on moduli spaces of Riemann surfaces<|endoftext|> TITLE: Explict form of $E_\infty$-morphisms between differential graded commutative algebras QUESTION [9 upvotes]: This is a partial duplicate to this MO question, I apologize for that. I'm asking since the answers there still do not allow me to work out an answer to my question, which is a bit more specific. Informally speaking, an $E_\infty$-algebra is an algebra which is associative and commutative up to coherent homotopies, a notion that can be properly formalized via the language of operads. However, in characteristic zero, one can can obtain an equivalent category to $E_\infty$ algebras by taking commutative differential graded algebras and inverting the quasi-isomorphisms. This is the same mechanism by which one can define the category of $L_\infty$ algebras in characteristic zero by starting with differential graded Lie algebras and inverting quasi-isomorphisms. However, in the $L_\infty$-case it is well known that $L_\infty$-morphisms between differential graded Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$ can be also given an "explicit" definition as sequences of multilinear maps $\varphi_k\colon \wedge^k \mathfrak{g}\to \mathfrak{h}[1-k]$ which satisfy a set of quadratic equations. This explicit definition obscures the theory behind it but has nice "practical" consequences. For instance one immediately sees that an $L_\infty$-morphism between two Lie algebras (seen as differential graded Lie algebras concentrated in degree zero) is precisely the same thing as a Lie algebra morphism between them, or that an $L_\infty$-morphism from a Lie algebra $\mathfrak{g}$ to a differential graded Lie algebra $\mathfrak{h}$ concentrated in nonegative degree is the same thing as a morphism of differential graded Lie algebras from $\mathfrak{g}$ to $\mathfrak{h}$. Or even, if $\mathfrak{g}$ is a Lie algebra and $\mathfrak{h}=(\mathfrak{h}^{-1}\to \mathfrak{h}^0)$ is a differential graded Lie algebra concentrated in degrees -1 and 0, one has a complete, simple and explicit control on what an $L_\infty$ morphism from $\mathfrak{g}$ to $\mathfrak{h}$ would be. I'm wondering whether something similar can be said for $E_\infty$-morphisms between differential garded commutative algebras. For instance: can it be given an explicit description of the $E_\infty$-morphisms between a commutative algebra $A$ (seen as a differential graded commutative algebra concentrated in degree zero) and the differential graded commutative algebra $\mathbb{K}[1]\oplus \mathbb{K}$, where $\mathbb{K}$ is the base field? (the differential on $\mathbb{K}[1]\oplus \mathbb{K}$ is the zero differential and the multiplication $(\mathbb{K}[1]\oplus \mathbb{K})^0\otimes (\mathbb{K}[1]\oplus \mathbb{K})^{-1}\to (\mathbb{K}[1]\oplus \mathbb{K})^{-1}$ is the multiplication in $\mathbb{K}$). I guess such an $E_\infty$-moprhism should reduce to the datum of two $\mathbb{K}$-linear maps $\varphi_0\colon A\to \mathbb{K}$ and $\varphi_1\colon A\otimes A\to \mathbb{K}$ satisfying suitable equations which shuld precisely encode the fact that $\varphi_0\colon A\to \mathbb{K}$ is a morphism of commutative algebras up to a homotopy given by $\varphi_1$. But I'm not able to find a rigorous statement along these lines anywhere in the literature. REPLY [4 votes]: For future readers' convenience, I'm summarizing the comments to the original question in the form of an answer. Many thanks to Adeel, Sean Tilson, Akhil Mathew and Fernando Muro for their inputs. I'm obviously the only one to blame for eventual inaccuracies I'll be including in this answer. So, the answer is as follows: Let $A$ and $B$ two differential graded commutative $\mathbb{K}$-algebras, where $\mathbb{K}$ is a characteristic zero field. An $E_\infty$-morphism $\varphi\colon A\to B$ is precisely a morphism of $C_\infty$-algebras between $A$ and $B$. Equivalently, it is a morphism of $A_\infty$-algebras form $A$ to $B$ such that all of its Taylor coefficients $\varphi_k\colon A^{\otimes k}\to B[1-k]$ vanish on the shuffle products.<|endoftext|> TITLE: Topological Grothendieck Construction QUESTION [10 upvotes]: Let $C$ be a small category and $F\colon C^{op}\rightarrow Set$ a functor. The Grothendieck construction is the category $F\wr C$ with objects being pairs $(c,x)$ where $c$ is a object of $C$ and $x\in F(c)$. An arrow from $(c,x)\rightarrow (c',x')$ is an arrow $f\colon c\rightarrow c'$ with $F(f)(x')=x$. This comes with a natural functor $F\wr C\rightarrow C$ forgetting the second coordinate. In other words it is a category $F\wr C$ together with a functor $F\wr C\rightarrow C$, such that the diagram of nerves $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} N_1(F\wr C) & \ra{d_0} & N_o(F\wr C) \\ \da{} & & \da{} \\ N_1(C) & \ra{d_0} & N_o(C) \end{array} $$ is cartesian. Now topologize the whole situation, i.e. let $C$ be a topological category (a category internal to $Top$) and $F\colon C^{op}\rightarrow Top$ a functor. The grothendieck construction of this situation should give me a topological category $F\wr C$ with a continuous functor $F\wr C\rightarrow C$, such that the diagram above is cartesian. However, I ran into difficulties making this precise. Taking the non-topological definition for the underlying sets $ob(F\wr C)$ and $mor(F\wr C)$, one can topologize the set of morphisms as subspace of $mor(C)$, but what is the right topology on $ob(F\wr C)$? In some situations, the way to go is obvious : If $F$ takes values in subspaces of a fixed topological space $X$, one could topologize $ob(F \wr C)$ as subspace of $ob(C)\times X$. If $C$ has a discrete set of objects, $ob(F\wr C)$ can be topologized as a subspace of $\coprod\limits_{c\in ob(C)}\{c\}\times F(c)$. REPLY [4 votes]: Let me expand a bit on what Oscar Randal-Williams is saying and mix it with what Geoffroy Horel is saying: You can't naively speak about an "internal functor" $F:C^{op} \to Top.$ But lets do a small exercise: Suppose that $C$ is an ordinary small category. Define a right $C$-set to be a set $X$ together with a map $$\mu:X \to C_0$$ (its moment map) together with an "action map" $$\rho:C_1 \times_{C_0} X \to X$$ where an object of $C_1 \times_{C_0} X$ consists of an arrow $f:D \to C$ and an object $x \in X$ such that $\mu\left(x\right)=C.$ The action map must satisfy the obvious axioms for a right action. There is an obvious notion of morphism of such $C$-sets (equivariant maps) and this category is caonically equivalent to $Set^{C^{op}}$. How does this work? Given a functor $F:C^{op} \to Set,$ we can let $X=\coprod_\limits{C \in C_0} F(C) \to C_0$ and if $f:D \to C$ and $x \in F(C)$ we let $\rho\left(f,x\right)=F\left(f\right)\left(x\right).$ Here's how to go the other way around: If $X$ is a $C$-set, we can form the "action category" $X \rtimes C$. Its objects are $X$ and the arrows are the fibered product $C_1 \times_{C_0} X$ where a pair $\left(f,x\right)$ is an arrow from $F\left(f\right)\left(x\right)$ to $x$. There is a canonical functor $X \rtimes C \to C$ which is a discrete fibration, and since discrete fibrations (Grothendieck fibrations which encode a functor into sets rather than categories) are an equivalent category to presheaves of sets, we recover a presheaf of sets. More importantly however, notice that the action category $X \rtimes C$ applied to the $C$-set coming from a functor $F:C^{op} \to Set$ as above is exactly the Grothendieck construction $F \wr C$ of $F$. Now, when $C$ is a small internal categeory to $Top$, then we can't define a presheaf as a functor $C^{op} \to Top$, but we CAN define it as a continuous right $C$-space, i.e. as a topological space $X$ together with a continuous moment map $\mu:X \to C_0,$ with a continuous right $C$-action... and if do, we can define the "grothendieck construction" of such an internal presheaf as the associated "action category" $X \rtimes C$ which carries a natural topology, which in particular, makes the canonical functor $X \rtimes C \to C$ continuous (and in fact is an internal split fibration in the $2$-category of categories internal to $Top$)<|endoftext|> TITLE: Infinite sequence avoiding a countable set of words QUESTION [5 upvotes]: As an application in group theory, I would need an infinite sequence over a finite alphabet, that avoids a sequence of words $w_i$, where the length of $w_i$ is such that $l(w_i) > 10^8 l(w_{i-1})$. I have found several results about avoiding patterns, but here I would really just need to avoid the words themselves. Do there exist results along these lines? REPLY [8 votes]: I am not sure that it is exactly what you need, but the following is true: For an alhabet with $q\geq 4$ letters and a sequence of forbidden words with lengths $n_1 TITLE: Relations between functors in a recollement QUESTION [6 upvotes]: Consider a recollement situation like the following by the very definition of the various functors it follows that $i^* j_*=0$, and $j^! i_* = 0 = j^* i_!$. Also, $j^! i_! = 0 = j^* i_*$ by inspection. Are these "kernel properties" true in a general recollement situation? More precisely, let $$ \mathbf{D}^0 \underset{\underset{i_R}\leftarrow}{\overset{\overset{i_L}\leftarrow}\to} \mathbf{D} \underset{\underset{q_R}\leftarrow}{\overset{\overset{q_L}\leftarrow}\to} \mathbf{D}^1 $$ be a recollement where $i_L\dashv i\dashv i_R$ and $q_L\dashv q\dashv q_R$. From the axioms of recollement it follows that $qi=0$ implies $i_L q_L = 0 =i_R q_R$. Is it true that also $i_R q_L = 0 = i_L q_R$? REPLY [5 votes]: I'll write my comment again here, so it appears as an answer. If you have a recollement as in the question with the property that $i_Rq_L=0=i_Lq_R$, then the category $D$ splits as an orthogonal sum of $D^0$ and $D^1$; i.e. every object $d$ of $D$ can be written as a direct sum $d_0 \oplus d_1$ with $d_i \in D^i$, and there are no nonzero morphisms between $D^0$ and $D^1$ either direction. You can see this by looking at one of the distinguished triangles associated to the recollement and observing that the connecting morphism must be zero, and thus the triangle is split. In particular, it is not true that $j^\ast i_\ast = 0$ in the derived category of sheaves on a space (where $i: U \hookrightarrow X$ is an open embedding and $j: F \hookrightarrow X$ the closed complement). In fact, the (hyper)cohomology of the complex $j^\ast i_\ast \mathbb Z_U$ computes the cohomology of the link of $F$ inside $X$ - an important invariant of the stratification. This is never zero, unless $U$ and $F$ are disjoint components.<|endoftext|> TITLE: Lemma 2.1.1.4 in Lurie's HTT QUESTION [13 upvotes]: I have encountered a problem in understanding Lurie's proof of the following fact: "Given a left fibration between simplicial sets $q:X \to S$, there exists a functor $$ho(S) \to Ho(sSet)$$ which is defined as follows: a point $s \in S$ is sent to the fiber $X_s$ of $q$ over $s$, and an arrow $f:s \to s'$ in $S'$ is sent to the homotopy class of a lift in restricted to $\{1\} \times X_s$, which will be denoted by $f_!:X_s \to X_{s'}$". The author wants to characterize the map $$[f_!]\circ (\cdot): Ho(sSet)(\cdot, X_s) \to Ho(sSet)(\cdot, X_{s'})$$ so as to obtain the desired result (except for the independence from the homotopy class of $f$, which I guess is left to the reader). The problem is I don't see why it holds that, given a $K\in sSet$ and arrows $\eta:K \to X_s, \ \eta':K \to X_{s'}$, $$\eta'\simeq f_! \circ \eta \iff \exists p:K \times \Delta[1] \to X$$ such that $p_{|K \times \{0\}}\simeq \eta, \ p_{|K \times \{1\}}\simeq \eta'$ and $$q \circ p= K \times \Delta[1] \to \Delta[1] \to S$$ where the last arrow is $f$. Thanks in advance for any help, which will of course be highly appreciated. REPLY [2 votes]: Consider the diagram where the top left horizontal arrow is given by $\eta$. Observe that homotopy classes of lifts of the map $X_s \times \Delta^1 \rightarrow S$ are the same as homotopy classes of lifts of the map $K \times \Delta^1 \rightarrow S$. This follows from the same argument showing that the homotopy class of $f_!$ only depends on $f$, together with the fact that the left hand square is a pullback. (I'll include the argument below for completeness). Given that, we're done. A lift of $K \times \Delta^1 \rightarrow S$ is precisely the map $p$ as stated, and the above observation says the lift is homotopic to the composition $K \times \Delta^1 \rightarrow X_s \times \Delta^1 \rightarrow X$, which is the same as saying that $[\eta'] = [f_!] \circ [\eta]$. Okay, now for the promised argument which says that homotopy classes of lifts in a diagram like only depend on the map $f$. But such a lift is equivalent to a section of $X \times_S (Y \times \Delta^1)$ over $Y \times \Delta^1$ which is specified at $Y \times \{0\}$. This pullback may be obtained by first pulling back $X$ to $\Delta^1$, where the fibration trivializes up to homotopy. Thus $X \times_S(Y \times \Delta^1)$ is trivial up to homotopy so that sections are determined up to homotopy by maps $Y \times \Delta^1 \rightarrow X_s$. Since we've already specified the map on $Y \times \{0\}$, there is only one extension up to homotopy as $Y \times \{0\} \rightarrow Y \times \Delta^1$ is an equivalence.<|endoftext|> TITLE: Structure of sign changes under the heat flow QUESTION [5 upvotes]: Let $f$ be a smooth function on $R^2$, and define $N_f$ to be the set of points $p$ such that the nodal set of $f$ ($\{x\in R^2: f(x)=0\}$) divided every neighborhood of $p$ into four regions. Indeed, the nodal set of $f$ looks like $"\times"$ around $p$ and $f$ changes sign four times on every small enough circle centered at $p$. Let $u(t,x)\in C^{\infty}(R\times R^2)$ be a solution of the heat equation $u_t=\Delta u$ with $u(0,x)=u_0(x)$, and assume that $N_{u_0}=\emptyset$. I wonder if the heat flow can stably generate such points in time, i.e. can $ \cup_{t>0} N_{u(t,x)}$ contain an unbounded continuous curve $C$ in $R^3$ such that $C \cap N_{u(t,x)} \neq \emptyset$ for all $t>t_1$ for some $00$, which is the usual context for the heat equation, not for all $t \in {\bf R}$ as implied by "$u(t,x) \in C^\infty({\bf R} \times {\bf R}^2)$"): (source: harvard.edu) Here $u(t,x)$ is antisymmetric about the vertical axis $x_2=0$; thus the zero-set $V_{u(t,\cdot)} = \{x: u(t,x) = 0\}$ is always symmetric about that axis and contains it. The curved contours of that Sage plot, in black, red, orange, green, blue, purple, and gray, show the other component(s) of $V_{u,(t,\cdot)}$ for $t = t_1/8$, $t_1/4$, $t_1/2$, $t_1$, $2t_1$, $4t_1$, $8t_1$. The zero set $V_{u,(t,\cdot)}$ contains a $\ast$-shaped triple point for $t=t_1$, and two $+\,$-shaped double points for all $t > t_1$, at height proportional to $\pm \sqrt{t-t_1}$. To obtain this function, start from the usual heat kernel $g(t,x) = (4\pi t)^{-1} \exp (-|x|^2/4t)$, and set $u(t,x) = \Delta_x(g(t,x-e_1)-g(t,x+e_1))$ where $e_1$ is the unit vector $(1,0)$. (To check that $u$ is a solution of the heat equation $u_t = \Delta_x u$, note that $g$ satisfies the heat equation and that the differential operators $\partial / \partial t$ and $\Delta_x$ commute with $\Delta_x$ and with translations by $\pm e_1$.) The nodal set of each term $\Delta_x(g(t,x\mp e_1))$ of $u$ is the circle of radius $2t^{1/2}$ about $\pm e_1$. For small $t$, the nodal set $V_{u(t,\cdot)}$ of the difference consists of the vertical axis and very close approximations to those circles. As $t$ increases, the circles grow and distort, eventually meeting to form a figure-eight at $t=t_1$ and a single closed curve for all $t > t_1$. The two double points for $t>t_1$ can be located as the zeros on the vertical axis $x_2=0$ of the partial derivative of $u(t,x)$ with respect to $x_1$.<|endoftext|> TITLE: Sum of consecutive cubes QUESTION [10 upvotes]: I'm investigating when the sum of $n$ consecutive cubes equals a cube, i.e., for which $n$ does $$\sum_{i=0}^{n-1} (k+i)^3 = k^3 + (k+1)^3 + \cdots + (k+n-1)^3 = Y^3 $$ have nontrivial solutions $(k,Y)$ for $k, Y \in \mathbb{N} $. I have found (using programs) that if this equation has non-trivial solutions, n is not squarefree (for $n > 3$). Now I'm trying to prove that $n > 3$ cannot be squarefree. Here is a link to my proof and what I've done so far but I've reached a wall. I have three equations that I believe contradict each other (I am almost certain they contradict each other). I just can't see how they contradict each other and I might need a new set of eyes to look at it. The three equations are given in the link but, if you like, I've put them below. I'm trying to show the following: For natural numbers $ x, y, k, d \in \mathbb{N} $ and $ d > 1,$ $ d^2y = 2k + dx - 1 $ $ xy(d^4y^2+d^2x^2-1) = cube $ $ x {\space} | {\space} k(k-1) $ cannot all be true. Please let me know if you have any questions or suggestions for me! Thanks in advance! REPLY [10 votes]: Found some bigger numbers: {n,k} as you call them: {4913 , 11368} {6591 , 305} {6859 , 18171} {8000 , 22534} {10648 , 33558} {12167 , 40381} {13923 , 3010} {14161 , 1624} {25201 , 46690} {33124 , 18551} {63001 , 11170} {48841 , 967190} {277729 , 711785} Most 'n' are squares, but a strange ones are 6591, 13923 and 25201 pipo<|endoftext|> TITLE: Model-theoretic accounts of feasibility in bounded arithmetic and related systems QUESTION [9 upvotes]: Various weak theories of arithmetic have been partially motivated by a concern with numbers (or functions/proofs) that are feasible. This concern is sometimes connected to an interest in strictly finitistic approaches to arithmetic. Examples of work where this concern is apparent are Parikh's work on bounded arithmetic, Sazonov's systems of feasible numbers, or Nelson's work on Predicative Arithmetic. While the precise account of feasibility varies across these systems, the general idea is that the theory should prove that 0 is feasible if $n$ is feasible then so is its successor $S(n)$ the feasible numbers are in some sense 'bounded' There are various ways of making this last statement precise: e.g. we can see it as a statement of the form $\exists xy \neg\exists z (z= \exp(x,y))$ stating that some fast-growing function is not total (or, as in the case of $I\Delta_{0}$, it may suffice to know that exponentiation -- and, by Parikh's Theorem, any function with superpolynomial growth -- is not provably total, so that the above is at least consistent with $I\Delta_{0}$). A different approach is to give some explicit upper bound on feasibility, and require the theory to prove $\forall x (\log_{2}\log_{2}x<10)$, as in Sazonov's $\texttt{FEAS}$ system. My question: is there any model-theoretic, or more broadly 'semantic', account of feasible numbers? Preferably, an account that would be (1) helpful in providing a clear mathematical picture of the structure of feasible numbers and/or (2) acceptable by the strict finitist's own lights? A word on the two desiderata: in the above systems, characterisations of feasibility are rather implicit, as well as very sensitive to the underlying language and proof systems. Moreover, models of those theories (when they exist) seem to fail both (1) and (2). For instance, models of $I\Delta_{0}$ where $\exp$ is not total (say, obtained via cuts of nonstandard models of PA) are not, presumably, objects to be taken seriously by the strict finitist as `concrete' objects, be it only due to their size. In addition, they hardly seem to be good models of 'intuitively feasible' numbers: their domains are basically given by (possibly nonstandard) integers bounded above by a power of some infinite nonstandard integer. The link to feasibility, or counting, or smallness, is very unclear, and it does not help building a mental picture consistent with the strict finitist's motivations. Sazonov's theory is downright inconsistent in the classical sense (i.e. if we allow unbounded proof length), so it admits no (classical) models. So: is there a serious mathematical account of 'feasibility' of this kind? Some additional remarks: Many logicians, like Gaifman, suggest a connection with vagueness, as the feasible numbers can be seen as forming a vague set. But do we really need to resort to vagueness to provide semantics for feasible numbers? One possibility is to attempt an account in modal terms, where we imagine a Kripke frame where states are finite sets of integers representing 'the numbers we've counted to so far', and accessibility relations represent something like reaching further numbers via applying 'feasible' functions to the current (finite) domain. Of course, the Kripke frame would have infinite domain, but one could at least argue that it models feasibility in a way that gets things right 'locally', in providing an intuitive mental picture of the process of constructing numbers. But it is difficult to see how any construction of this sort could account in any way for the role played by particular notation systems or induction axioms (bounded induction). I understand that most strict finitists are not concerned with giving a semantic account of arithmetic; some (like Nelson) are explicit formalists, and regard `semantics' as an unnecessary, or perhaps even misleading, distraction. At the very least, the idea seems to be that feasibility depends on the notational system used. This makes good sense from a constructivist perspective; feasible numbers are not a finished collection that our formal theory describes; instead, the theory describes the rules that we can employ to 'construct' numbers. Nonetheless, there may be some intrinsic interest in the question of whether an elegant 'semantic' mathematical account of feasible numbers exists, or can be provided at all. REPLY [3 votes]: Did you look at Manucci and Cherubin's preprint about the model theory of ultrafinitism? It wasn't that well developed, but I thought the envisioned approach was interesting. http://arxiv.org/abs/cs/0611100<|endoftext|> TITLE: Symmetric inequality on generalized means QUESTION [5 upvotes]: Do there exist two functions $f$ and $g$ continuous and strictly increasing $[0,1] \to \mathbf{R}$ such that $$ f^{-1}\left(\frac{1}{3} f(x) + \frac{2}{3} f(y)\right)g^{-1}\left(\frac{2}{3} g(x) + \frac{1}{3} g(y)\right)$$ for all $0\le x < y \le 1$? This question, related to generalized means (characterized by Kolmogorov-Nagumo and de Finetti), arises from the fact that such system has a [infinitely many] solution in the case that the sign of the inequalities are the same, e.g. $(f(t),g(t))=(t^\alpha,t^\beta)$ for some $\alpha \neq \beta$. What happens if the signs are reversed? REPLY [6 votes]: No. Denote $f(x)=a$, $f(y)=b$, then $g(f^{-1}(t))=h(t)$. Applying $g$ to both parts of first inequality we get $h(\frac13 a+\frac23 b)<\frac13 h(a)+\frac23h(b)$. It means that $h$ is convex (proof below), analogously the second inequality means that $h$ is concave. Lemma. If $h$ is continuous on $[0,1]$ and $h(tx+(1-t)y) TITLE: Properties of coefficients of ring spectra QUESTION [6 upvotes]: This is an awkwardly backwards question, but bear with me here: Suppose I have a graded ring $R$ with unit, which has an invertible element $u$ in degree $2$. The multiplicative formal group law $f(x,y) = x + y + u\,x\,y$ yields a homomorphism $MU_* \to R$. Suppose that I know the following things The functor $X \mapsto MU^*(X) \otimes_{MU_*} R$ is a multiplicative generalized cohomology theory (in particular, it is exact). The above functor is represented by an $E_{\infty}$-ring spectrum. What properties can I deduce from this about the ring $R$? For example: Is it true that $R$ is torsion-free (in each degree)? I vaguely seem to remember that $K$-theory with mod p-coefficients does not have an $E_{\infty}$-ring structure, so this seems at least plausible. Does $R_0$ have to be a subring of $\mathbb{Q}$ if I throw into the mix that $R_0$ is countable? REPLY [8 votes]: You have given yourself an invertible element $u\in\pi_2(R)$ and a coordinate $x\in R^2(\mathbb{C}P^\infty)$ with $\psi(x)=x\otimes 1 + 1\otimes x + ux\otimes x$. This means that the class $m=1+ux\in R^0(\mathbb{C}P^\infty)$ satisfies $\psi(m)=m\otimes m$. In other words, $m$ can be regarded as a map of ring spectra from the ring spectrum $T=\Sigma^\infty_+\mathbb{C}P^\infty$ to $R$. Snaith defined an element $v\in \pi_2(T)$ and proved that $T[v^{-1}]$ is equivalent to $K$. You are also implicitly assuming that $x$ restricts to the standard generator of $\widetilde{R}^2(\mathbb{C}P^1)\simeq R^0(\text{point})$. I think that this means that $m_*(v)=u$, which is invertible, so $m$ induces a ring map $K=T[v^{-1}]\to R$. Under plausible additional assumptions, this map $K \to R$ will in fact be $E_\infty$. Thus, $R$ will always be a $K$-algebra, and probably an $E_\infty$ $K$-algebra. The non-$E_\infty$ version could also be extracted from the standard Landweber exactness technology, at least up to phantoms. If we have an $E_\infty$ map $K\to R$ and $R$ is $p$-complete then we can check that $R^0(B\Sigma_p)/\text{tr}(1)$ is isomorphic to $R^0$, and we can use this to define a power operation $\psi^p$ on $R^0(X)$ for all spaces $X$, which is a ring map satisfying $\psi^p(t)=t^p\pmod{p}$. The existence of $\psi^p$ excludes many possibilities for $R^0$ (or the $p$-completion of $R^0$, if $R^0$ is not already $p$-complete). For example, the ring $A=\mathbb{Z}[e^{2\pi i/p}]$ does not admit a map $\psi^p$ of the required type.<|endoftext|> TITLE: Probability of a graph procedure QUESTION [12 upvotes]: We are going to build $K_n$ one edge at a time. Begin with the empty graph on $n$ vertices. Take a random permutation of the edges of $K_n$ and, one at a time, place the edges onto the graph (so, after the $k$th edge in the list is placed, the so-far created graph will have exactly $k$ edges). Let $H_k$ be the graph induced by the first $k$ edges. Let $p(n)$ be probability that $H_k$ is connected for all $k \in \{1,2,\dots,{n \choose 2}\}$. As a corollary to a completely different problem I'm working on, and through a very round-about way, we were able to show that the answer for $p(n)$ is beautiful. Indeed, it is simply $$\frac{2^{n-2}}{C_{n-1}},$$ where $C_{m}$ is the $m$th Catalan number. I'd like to find a good combinatorial reason why this is the answer. (This is the easiest case of a more general problem I'm working on. I do have one semi-combinatorial argument but unfortunately it doesn't generalize well.) Any thoughts would be appreciated! Thanks!! REPLY [11 votes]: Here's one. You can think of the graph construction process as gradually building a set $S$ of vertices that have been touched so far, beginning with a random two vertices. Let $S_k$ be the set of the first $k$ vertices in this process. Now $p(n)$ is exactly the probability that, after the first two vertices, each additional edge we select either has both endpoints in the current $S_k$, or else crosses the cut between $S_k$ and the rest of the graph. (In this case, the other endpoint becomes the next vertex to add to our set.) Thus $p(n)$ is the probability that, at each set size $k$, conditioned on picking an edge that does not have both endpoints in $S_k$, the edge selected crosses the cut. Well, the number of edges crossing the cut is exactly $k(n-k)$, and the number of edges entirely outside of $S_k$ is exactly ${n-k \choose 2}$. Therefore the probability of picking an edge crossing the cut conditioned on one of these occurring is exactly \begin{align} \frac{k(n-k)}{k(n-k) + {n-k \choose 2}} &= \frac{2k}{2k + n-k-1} \\ &= \frac{2k}{n + k - 1} \end{align} As base cases, immediately $p(2) = 1$. For $n \geq 3$, we get \begin{align} p(n) &= \prod_{k=2}^{n-1} \frac{2k}{n-1 + k} \\ &= 2^{n-2} \prod_{k=2}^{n-1} \frac{k}{n-1 + k} \\ &= 2^{n-2} \frac{1}{C_{n-1}} . \end{align}<|endoftext|> TITLE: When are (Abelian) Cayley graphs also expanders? QUESTION [7 upvotes]: I want to ask the question in two parts, (1) Is there some fundamental distinguishing property between Abelian and non-Abelian Cayley graphs? (say some specific proof technique which distinguishes them?) (2) Are there any set of (constant degree) (Abelian) Cayley graphs which are expanders? Do they have any distinguishing property? For reference one can see chapter 5, starting on page 30 of these notes, http://www.eecs.berkeley.edu/~luca/books/expanders.pdf REPLY [9 votes]: To add to Anthony's comment, one can make an explicit connection between the large number of walks between vertices and the spectra of Abelian Cayley graphs. It turns out that constant-degree Abelian Cayley graph are not only bad expanders,but they tend to be disconnected (they have a positive proportion of their eigenvalues close to their valency). See: http://www.math.udel.edu/%7Ecioaba/inpress_version.pdf for a short proof. Alon and Roichman showed earlier that Abelian Cayley graphs have large diameter (power of the order of the graph); the diameter of an expander should be logarithmic in the order of the graph. A proof (using character sums) that the Abelian Cayley graphs have large 2nd eigenvalue is given by Friedman, Murty and Tillich: http://www.mast.queensu.ca/~murty/f-m-t.pdf<|endoftext|> TITLE: History of Tarski's problems on free groups QUESTION [11 upvotes]: As is known, Tarski posed his questions about first-order theories of non-abelian free groups around 1945. However, the questions were not published in his papers or books. What is the original published source of reliable information about posing Tarski's problems on free groups? When exactly had Tarski posed these questions at first time? Did it happen at some seminar or conference? REPLY [5 votes]: Why did I guess that Tarski's problems about first-order theories of non-abelian free groups were not published in his papers or books, and he posed the problems around 1945? Here are the reasons of that: $\bullet$ In the papers of Kharlampovich-Myasnikov [J. Algebra 302 (2006), no. 2, 451-552] and Sela [Geom. Funct. Anal. 16 (2006), no. 3, 707-730] with solutions of Tarski's problems there is no references to his publications; $\bullet$ In the Kharlampovich-Myasnikov paper it is written that the Tarski's conjectures were formulated around 1945; $\bullet$ Lyndon in his paper [Problems in combinatorial group theory, in: Combinatorial group theory and topology, Ann. Math. Stud., Princeton Univ. Press, 1987] formulated the problem on elementary equivalence of all non-abelian free groups and called it a folklore problem of Alfred Tarski. I checked Tarski's papers and books and now I know that his problems on the elementary theory of free groups had been formulated in some of his publications; so these problems were not `folklore'. Also, it seems the problems had been posed not around 1945 but later. The first Tarski's publication where the decision problem for the theory of non-abelian free groups was mentioned is the Tarski's book [Undecidable theories, North-Holland, 1953]. On p. 77 he informed that he stated his result of undecidability of the elementary theory of groups at a conference in Princeton in December, 1946. On p. 85, he wrote: 'For many extensions of the elementary theory of groups, e.g. for elementary theories of finite groups and non-Abelian free groups, the decision problem remains open'. Note that it was not a conjecture on decidability of the elementary theory of non-abelian free groups; he just said that it was not known whether the theory is decidable or not. Clearly, the question had been raised between 1945 and 1953, and I don't know when exactly. In that Tarski's book the problem of elementary equivalence of all non-abelian free groups was not mentioned. First time it was published in Vaught's abstract [Bull. AMS, 61 (1955), N 2, 173-174], where he formulated his theorem implying that standard embeddings of free groups of infinite rank are elementary and wrote: 'These investigations arose from a still unresolved conjecture of Tarski's that any two free groups with at least two generators are arithmetically equivalent'. That theorem was a result from Chapter 3 of Vaught's PhD thesis (Berkeley, 1954), the supervisor of which was Tarski. Later the Vaught's theorem and the Tarski's conjecture were published in Tarski-Vaught's paper [Compositio Math. 13 (1957), 81-102]; see pages 82 and 98. Also, in the introduction of the paper it is written that the decision problem for the elementary theory of free non-abelian groups is a closely related open problem.<|endoftext|> TITLE: Blow-ups and blow-downs QUESTION [5 upvotes]: Can $\mathbb P \times \mathbb P \times \mathbb P$ be obtained from $\mathbb P^3$ by a finite succession of blow-ups and blow-downs along non-singular centers? REPLY [4 votes]: One can obtain $\overline{M}_{0,n}$ (the moduli space of $n$-pointed rational curves) in at leat two ways: Blowing-up $n-1$ points $\{p_1,...,p_{n-1}\}$ in $\mathbb{P}^{n-3}$ and then all the strict transforms of linear spaces generated by subsets of $\{p_1,...,p_{n-1}\}$ in order of increasing dimension. Blowing-up, in order of increasing dimension, suitable smooth subvarieties of the diagonals of $(\mathbb{P}^1)^{n-3}$. The first construction is due to Kapranov, you can find it for instance in Section 6.2 "B. Hassett, Moduli spaces of weighted pointed stable curves, Advances in Mathematics 173 (2003) 316–352". The second one is due tue Keel, see Section 6.3 of the same paper. Putting togheter 1 and 2 you see that $(\mathbb{P}^1)^{n-3}$ can always be obtained from $\mathbb{P}^{n-3}$ by a sequence of blow-ups and blow-downs along smooth centers. In your case, $n =6$, Kapranov constrction is as follows: begin with $p_1,...,p_5\in\mathbb{P}^{3}$ in general position; blow-up the points $p_{1},...,p_{5}\in\mathbb{P}^{3}$; blow-up the strict transforms of the lines $\left\langle p_{i},p_{j}\right\rangle$, $i,j=1,...,5$. In this way you get a morphism $\phi:\overline{M}_{0,6}\rightarrow\mathbb{P}^3$. While Keel construction is as follows: Let $X_1$ be the blow-up of $\mathbb{P}^1_1\times\mathbb{P}^1_2\times\mathbb{P}^1_3$ in $p_1 = ([0:1],[0:1],[0:1])$, $p_2 = ([1:0],[1:0],[1:0])$, and $p_3 = ([1:1],[1:1],[1:1])$. Consider the projections $\pi_i:\mathbb{P}^1_1\times\mathbb{P}^1_2\times\mathbb{P}^1_3\rightarrow\mathbb{P}^1_i$, and define $F_0 = \bigcup_{i=1}^3\pi_i^{-1}([0:1])$, $F_1 = \bigcup_{i=1}^3\pi_i^{-1}([1:0])$, $F_{\infty} = \bigcup_{i=1}^3\pi_i^{-1}([1:1])$. Let $\Delta_2$ be the union of the $2$-dimensional diagonals of $\mathbb{P}^1_1\times\mathbb{P}^1_2\times\mathbb{P}^1_3$. Then we have $X_2$ the blow-up of $X_1$ along the strict transform of $\Delta_2\cap (F_0\cup F_1\cup F_{\infty})$. Finally, the blow-up $X_3$ of $X_2$ along the strict transform of the $1$-dimension diagonal $\Delta_1$ of $\mathbb{P}^1_1\times\mathbb{P}^1_2\times\mathbb{P}^1_3$. This is $\overline{M}_{0,6}$. Now, let $\psi:\overline{M}_{0,6}\rightarrow (\mathbb{P}^1)^3$. You can take your sequence of blow-ups and blow-downs as $f = \phi^{-1}\circ \psi$.<|endoftext|> TITLE: Expected value of the minimum with limited independence QUESTION [14 upvotes]: Imagine you sample $n$ number with replacement uniformly from the integers $1,\dots, n$. Let $X$ be the minimum of these samples. I am interested in $\mathbb{E}(X)$ but with a twist. All I know is that the samples are uniform and pairwise independent. Assuming $n$ is large, what bounds can one get for $\mathbb{E}(X)$? If we generalize this to $k$-wise independence, for $k \geq 2$, what can we say? [Also asked at https://math.stackexchange.com/questions/1179943/expected-value-of-the-minimum-with-limited-independence previously. ] REPLY [9 votes]: Let $u_k$ be the number of variables with value exactly $k$. If you pick a distribution of the $u_k$ such that $E[u_k]=1$, $E[u_k^2] = 2-1/n$, $E[u_ku_l] = 1-1/n$ for $k \neq l$ and $\sum_{k=1}^n u_k$ is always $1$, then by choosing a random $u_1, \dots ,u_n$ according to the distribution and then choosing a random ordering, you get a random set of variables with the desired independence property. On the other hand if you do the same thing for $n+1$ pairwise independent random variables you get $E[u_k]=1+1/n$, $E[u_k^2] = 2+ 2/n$, $E[u_k u_l] = 1+1/n$. Here is a way to construct a distribution on $u_k$ that satisfies these conditions. To get a distribution for $n$ pairwise independent random variables, we just delete the last variable. The formula is: With probability $(1+1/n) /k(k+1)$, we have $u_k=k+1$, $u_l =0$ for $l k$. To compute the three moments, we ignore a factor of $1+1/n$: Mean: $$ \frac{(k+1)}{k(k+1)} + \sum_{l < k} \frac{1}{l(l+1) } = \frac{1}{k} + 1- \frac{1}{k} = 1 $$ Mean of square: $$ \frac{(k+1)^2}{k(k+1)} + \sum_{l < k} \frac{1}{l(l+1) } = \frac{k+1}{k} + 1- \frac{1}{k+1} = 2 $$ Mean of product is the same as mean, since if $l0$. Then put back in the factor of $(1+1/n)$ to get the right answer. The expected value of the minimum of the $n+1$ variables at least $\sum_{k=1}^n (1+1/n) /(k+1) \approx \log n$. Removing a variable won't make the minimum any smaller.<|endoftext|> TITLE: Upper bound for the number of integral points in a convex set QUESTION [7 upvotes]: Let $K \subset \mathbb{R}^3$ be a bounded convex set such that the points with integer coordinates in $K$ are not all coplanar. Is it true that $|K \cap \mathbb{Z}^3| \leq 6{\rm Vol}(K) + 3$? REPLY [3 votes]: It suffices to prove the bound for (non-planar) polytopes $K$ with integer vertices. Let $v$ be a vertex of $K$. By triangulating faces of $K$ which does not contain $v$ and considering the corresponding simplexes with vertex $v$ based at these triangles one can obtain a decomposition of $K$ to simplexes with integer vertices.For such simplexes one can easily prove that the number of lattice points inside and on the faces except the vertices is at most six times the volume of the simplex minus 1: Proof. By a well known result the number of lattice points in the parallelepiped generated by independent integer vectors $v_1,v_2,v_3$: $$\{\lambda_1 v_1+\lambda_2 v_2 + \lambda_3 v_3 : 0\leq \lambda_1,\lambda_2,\lambda_3<1 \}$$ is equal to the volume of parallelepiped which is six times the volume of the simplex with vertices $0,v_1,v_2,v_3$ and the origin with other lattice points of the simplex except vertices are in this parallelepiped. So the number of all integer points in $K$ is at most $6 \mathrm{Vol(K)}$-$F$(the number of triangles = simplexes) + 1 (for $v$) + $V$ (the number of other vertices of $K$). But one can consider the triangulation as a connected triangulation in plane. By easy induction (adding triangles one by one) $E$ (the number of edges) is at most $2F+1$. So by Euler's formula: (we don't consider the unbounded face) $$V+F = E+1 \leq 2F+2 \Rightarrow V \leq F +2.$$ Therefore the total number of integer points of $K$ is $\leq 6\mathrm{Vol}(K)+3$.<|endoftext|> TITLE: When is a linear combination of permutation matrices unitary? QUESTION [6 upvotes]: Question: Let $P_\pi$ denote the matrix representation of permutation $\pi$. Consider a linear combination of all $n \times n$ permutation matrices $$U := \sum_{\pi \in S_n} c_\pi P_\pi$$ where $c_\pi$ are arbitrary complex coefficients. When is the matrix $U$ unitary? It would be great to have a simple parametrization of all tuples of coefficients $(c_\pi : \pi \in S_n)$ for when this happens. Example: In the $n = 2$ case there are only 2 permutations, so the matrix $U$ looks like this: $$U = \begin{pmatrix} c_0 & c_1 \\ c_1 & c_0\end{pmatrix}$$ where the constants $c_i$ must obey $$|c_0|^2 + |c_1|^2 = 1 \quad \text{and} \quad c_0 c_1^* + c_1 c_0^* = 0$$ for $U$ to be unitary. We can parametrize the solution of these equations as follows: $$c_0 = e^{i\varphi} \cos t \quad c_1 = e^{i\varphi} i \sin t$$ for any $\varphi \in [0,2 \pi)$ and $t \in [0,\pi/2]$. It would be nice to have something similar for general $n$. For example, I can write down the equations for $n = 3$ but I don't know any nice way to parametrize the solutions. Note: The only reference on this topic I could find is Orthogonal matrices as linear combinations of permulation matrices. REPLY [7 votes]: I think this can be done(in principle) in general. The $n \times n$ permutation matrices span a $\mathbb{C}$-vector space of dimension $1 + (n-1)^{2}$ since the natural permutation representation of $S_{n}$ is the sum of the trivial representation and an irreducible representation of degree $n-1.$ It is necessary for $\sum_{\pi} c_{\pi} P_{\pi}$ to be unitary that $\sum_{\pi} c_{\pi} \in S^{1}$ (just consider the effect on the vector $v$ with every component $\frac{1}{\sqrt{n}}).$ The non-trivial irreducible character comes from the action of $S_{n}$ on $v^{\perp}$, and for the moment it isn't clear me how to give a concise description to characterize which linear combinations of permutation matrices act as a unitary transformation on $v^{\perp}.$ Note that this is not really an issue when $n =2.$ Later edit: Thanks to Sean Eberhard's comment, it becomes clear that the unitary matrices which are linear combinations of permutation matrices are precisely those unitary matrices which have the vector $v$ above as an eigenvector- any unitary matrix which has $v$ as an eigenvector necessarily leaves $v^{\perp}$ invariant, so any linear combination of permutation matrices both has $v$ has an eigenvector and leaves $v^{\perp}$ invariant. By a dimension count, the space of matrices which leave both span($v$)and $v^{\perp}$ invariant, which has dimension $1 + (n-1)^{2}$, is precisely the span of the permutation matrices. In conclusion, the unitary matrices which are linear combinations of permutation matrices are precisely the unitary matrices which have $v$ as an eigenvector.<|endoftext|> TITLE: Continuous functions and 2-bushy trees QUESTION [12 upvotes]: The following problem was asked by Joe Miller in the fall of 2010 at a bar in Madison. A subtree $T \subseteq 4^{< \omega}$ is $2$-bushy if for some node $\sigma \in T$, every node above $\sigma$ has two immediate successors. Is the following true: For every continuous function $f: 4^{\omega} \to 2^{\omega}$, there exists a $2$-bushy tree $T \subseteq 4^{< \omega}$ such that $f \upharpoonright [T]$ is either one-one or constant? Here $[T]$ is the set of branches through $T$. I am tagging it set-theory + recursion-theory. REPLY [2 votes]: Yes. Let me use the reduction to labelings suggested by Rohit. Let's call a two-bushy tree where $\sigma$ is the root a $2$-branchy tree. For any continuous function $L$ from $4^{<\omega}$ to $\{0,1\}$, there is a either a three-branchy tree $T$ where $L$ restricted to $[T]$ is the constant function $0$ or a two-branchy tree $T$ where $L$ restricted to $[T]$ is the constant function $1$. This follows by backwards induction on the nodes, starting from nodes where $L$ is constant. If a node has all four children the base of one of these two types of trees, then either three have the first type of tree or two have the second, so it has one of the two types of threes. As a corollary there is either a $2$-branchy tree $T$ such that $L$ restricted to $T$ is $0$, or a $2$-branchy tree $T$ such that $L$ restricted to $T$ is $1$. Without loss of generality, assume it is $0$. Then let $k$ be the maximum number such that there is a $2$-branchy tree $T$ where $f$ restricted to $[T]$ lands on functions that begin with $k$ zeros. By assumption $k\geq 1$. If $k=\infty$ we win - there is a tree where $f$ is constant. So assume $k$ is finite. Suppose there exists a $2$-branchy tree $T'$ such that $f$ restricted to $[T']$ lands on functions that do not begin with $k$ zeros. Then we win. Indeed the first $k$ values of $f(x_0x_1\dots)$ are determined by $x_0x_1\dots x_{N-1}$ for some $N$. Then choose a two-branchy tree of sequences $x_0x_1\dots$ such that if $x_i$ is the leftmost of the two options at $x_0\dots x_{i-1}$ then $x_{Ni}x_{N{i+1}} \dots x_{Ni+N-1}$ is in $T$, and if $x_i$ is the rightmost of the two options then $x_{Ni}x_{N{i+1}} \dots x_{Ni+N-1}$ is in $T'$ (and we might have to switch left and right for $x_0$). Then $f$ restricted to that tree is injective, as we can reconstruct $x_i$ by looking at the $Ni$ through $Ni+k-1$ digits of $f(x_0x_i\dots)$. If there does not exist such a $2$-branchy tree, then there exists a $3$-branchy tree $T^*$ where $f$ restricted to $T^*$ always begins with $k$ zeroes. Then consider the tree of sequences $x$ that are in $T^*$ and that, after removing the first digit, remain in $T^*$. This is $2$-branchy tree, and $f$ applied to everything in it has $k+1$ zeroes. This contradicts our assumption on $k$. So we win.<|endoftext|> TITLE: Analysis of first-order methods for constrained convex optimization with approximate oracles QUESTION [5 upvotes]: In many first-order optimization methods an oracle is needed whose action enforces the constraint/regularizations. For example, in projected gradient descent, conditional gradient method, and proximal methods, these oracles are the projection oracle, the (constrained) linear optimization oracle, and the prox. operator, respectively. I could find some papers that analyze the convergence of the first-order methods considering inexact oracles with additive error. However, I could not find results on convergence analysis of these methods when we have inexact oracles with multiplicative error. Let's take the projected gradient descent for example. Suppose that we are given an oracle $P_{\mathcal{C},\gamma}(x)$ that returns an approximate projection of any point $x$ onto the compact convex set $\mathcal{C}$ in the sense that $$ \begin{align*} \|P_{\mathcal{C},\gamma}(x)-x\|&\leq(1+\gamma)\min_{y\in\mathcal{C}}\|y-x\|, \end{align*} $$ for a given $\gamma\geq 0$. I would appreciate any pointer to references that analyze (approximate) convergence under this kind of oracle model. REPLY [2 votes]: Building on Nesterov's work, in his Ph.D thesis, Peter Richtarik considers first-order methods with relative error of approximation guarantees. I haven't looked in too closely, but I am sure that a large part of this analysis can be extended to the case of inexact oracles too.<|endoftext|> TITLE: Is there a version of Miller forcing "guided by" an ultrafilter? QUESTION [8 upvotes]: It’s well known that Mathias forcing factors as a two-step iteration $P(\omega)/\mathrm{fin}\ast \mathbb{M}_{\dot U}$, where $\mathbb{M}_{\dot U}$ is Mathias forcing guided by the generic ultrafilter added by the first poset. That is, Mathias forcing is the two-step iteration of a countably closed poset and a $\sigma$-centered poset. In a similar way, Laver forcing completely embeds into a closed $\ast$ centered poset. It’s natural to wonder whether their cousin Miller forcing (aka rational perfect set forcing) has the same property. While Laver and Mathias forcings have natural ccc versions guided by an ultrafilter, it’s not clear (to me) how to guide Miller forcing by an ultrafilter. Here is my question: Does Miller forcing completely embed into the two-step iteration of a $\sigma$-closed poset and a $\sigma$-centered poset? This question allows for slightly more than the vague question of the title; I suppose the first closed poset might be something other than $P(\omega)/\mathrm{fin}$. One reason to ask this is that a "Yes" would imply that finite products of Miller, Laver, and Mathias forcing are all proper. [Why? Every poset completely embeddable into a closed $\ast$ centered poset is proper, and the class of such posets is closed under finite products.] Note that Spinas has already established the properness of finite powers of Miller & Laver forcing. Of course, I’d be grateful for whatever references—for negative or positive results—you’re willing to provide. REPLY [5 votes]: The answer is no. Like Sacks forcing, Miller forcing adds a minimal degree, meaning that if $G$ is the Miller generic and $x\in M[G]$, then either $x\in M$ or $G\in M[x]$. Hence we cannot have Miller forcing $\mathbb{Q}$ equivalent to a two-step iteration $\mathbb{P}_0*\mathbb{P}_1$ since the generic for $\mathbb{P}_0$ contradicts the minimality of the degree corresponding to the generic for $\mathbb{P}_0*\mathbb{P}_1$. The minimality of the generic added by $\mathbb{Q}$ is a consequence of Theorem 2 in "Combinatorics on ideals and forcing with trees" by Marcia Groszek. I think this is also covered in Miller's original paper, Rational perfect set forcing, though I don't have access right now.<|endoftext|> TITLE: Erdős-Szekeres game QUESTION [9 upvotes]: Given $n$. Two players in turn mark points on the plane. No three may be collinear, no $n$ may form a convex $n$-gon. The player who does not have legal move loses. Who has a winning strategy? REPLY [10 votes]: The problem was raised and discussed in Parikshit Kolipaka and Sathish Govindarajan, Two player game variant of the Erdős-Szekeres problem, Discrete Math. Theor. Comput. Sci. 15 (2013), no. 3, 73–100, MR3141828. It says the second player wins for $n=5$.<|endoftext|> TITLE: Factorization when a factor is partially known QUESTION [15 upvotes]: Let's say that I have a very large number of the order ($10^{250+}$) which is composite. I have been given one of its factor partially to a significant amount of digits (say 75+). Then, how can I figure out both its factors completely? That is, $$X = a b$$ where $X$ is known and $a$ is known to 75+ digits. REPLY [14 votes]: You can use Coppersmith's algorithm [1] (or Howgrave-Graham's [2] simplification) to find the factor, which will be efficient if the number of remaining bits is not too large. The PARI/GP documentation http://pari.math.u-bordeaux.fr/dochtml/html-stable/Arithmetic_functions.html#zncoppersmith has an explicit example. Coron, Faugère, Renault, & Zeitoun [3] give an improved version with impressive speed improvements, though I don't know if their code has been released or re-implemented. [1] D. Coppersmith. Finding a small root of a univariate modular equation. In U. Maurer, ed., Advances in Cryptology - EUROCRYPT '96, Springer, 1996, pp. 155-165. [2] Nicholas Howgrave-Graham, Finding small roots of univariate modular equations revisited. In Cryptography and Coding (Lecture Notes in Computer Science volume 1355), 1997, pp. 131-142. [3] Jean-Sébastien Coron and Jean-Charles Faugère and Guénaël Renault and Rina Zeitoun, A variant of Coppersmith's algorithm with improved complexity and efficient exhaustive search, Cryptology ePrint 2013/483. REPLY [10 votes]: I assume you mean you know the leading 75 digits of a roughly 125-digit factor. Then you can reconstruct the factorization using lattice basis reduction. For an explicit algorithm, see the paper Small solutions to polynomial equations, and low exponent RSA vulnerabilities by Coppersmith. All it requires is that the number of digits you know of a factor is a little more than a quarter of the number of digits in the product, although it gets slightly less efficient if you don't know how large the factors are. Coppersmith's algorithm is not just fast in theory, but also in practice. In your case, 75 is enough greater than 62.5 that it should be easy.<|endoftext|> TITLE: Decision problem on triviality of intersection of two subgroups QUESTION [7 upvotes]: What is known about the following decision problem? Given two finite sets in a finitely generated group G, decide whether the subgroups generated by them have trivial intersection. Is this problem decidable for a free non-abelian group G? REPLY [3 votes]: Allow me to elaborate on Benjamin's answer: Proposition: There is a fixed finitely presented group $H$, and fixed finite set of words $S$ on the generators of $H$ for which $\langle S\rangle \leq H$ has solvable subgroup membership problem, such that the problem of determining if the subgroup generated by one word $\langle w \rangle$ intersects $\langle S \rangle$ non-trivially in $H$ is algorithmically undecidable. Proof: There exists a finitely presented group $G=\langle X|R \rangle$ for which the word problem is solvable, but for which the torsion problem, of determining if a word has infinite order, is unsolvable. This is a consequence of Theorem A in D. Collins, The word, power, and order problem in finitely presented groups, in Word Problems, eds. W. W. Boone, F. B. Cannonito and R. C. Lyndon, 401–420 (1973). (On top of this theorem, one needs the additional observation that having solvable word problem means one can effectively compute the order of a torsion element). So now take the group $H$ to be $F_{X} \times F_{X}$, and take the set $S$ to be the finite set $\{(x_{1}, x_{1}), \ldots, (x_{n}, x_{n}), (r_{1}, 1) \ldots, (r_{m}, 1)\}$, and thus $\langle S \rangle =\{(u,v) \in F_{X}\times F_{X}\ |\ u=v \ \ in \ \ G \}$. Since $G$ has solvable word problem, we have that $\langle S \rangle \leq H$ has solvable membership problem. Moreover, as Benjamin observed, for an arbitrary word $z \in G$, we have $\langle (1,z) \rangle$ intersects $\langle S \rangle$ trivially if and only if $z$ has infinite order in $G$ (the latter being undecidable). I suppose the above should really be a comment to Benjamin's answer, but at the time I couldn't add comments as I had <50 reputational points. -Maurice<|endoftext|> TITLE: Yet another Erdős–Szekeres game QUESTION [8 upvotes]: Given $n$. Two players in turn write different real numbers $x_1,x_2,x_3,\dots$ The player after whose turn there is a monotone subsequence of length $n$ loses. I guess that the question 'who wins' may be hopeless (nice if not so), but possibly there exists a better estimate of the number of turns than Erdős–Szekeres bound $(n-1)^2+1$ (which works not only for optimal, but for all strategies)? REPLY [8 votes]: As noted in the comments (but with not quite the right reference) the game is a first player win for $n \geq 4$. The question here is about the misere form, so this is a combination of Proposition 7, Theorem 10 and a bit of Proposition 9 in the paper Monotonic Sequence Games by Albert (who he?), Aldred, Atkinson, Handley, Holton, McCaughan and Sagan. This also appeared in Games of no chance 3. We did not address the question of the length of the game when one player is playing using a winning strategy, and the other to lose as slowly as possible (obviously, if the players cooperate to lengthen the game then it reaches the Erdos-Szekeres bound). The proof uses a "strategy stealing" argument, but one which is not entirely clear cut. In common with most such arguments it does not actually yield a strategy for the first player to win, only a proof that such a strategy exists.<|endoftext|> TITLE: Relation between dualization complex, cotangent complex and Deligne-Du Bois complex? QUESTION [8 upvotes]: Given a smooth variety $X$, one can define the cotangent sheaf $\Omega_X$, the canonical sheaf $\omega_X$ and the deRham complex $\Omega_X^\bullet$. These three object has obvious relations. For general variety $X$ (say over $\mathbb{C}$), one generalize cotangent sheaf to cotangent complex $L_X$ of Illusie, canonical sheaf to dualization comlex and deRham complex to complex of Deligne-Du Bois (I don't know if this is the standard terminology, which is defined to be the derived pushforward of the deRham complex of a cubical hyperresolution. [c.f. below Theorem 7.22, Mixed Hodge structures by Peters and Steenbrink]). My question is for general $X$, Is there relation between dualization complex, cotangent complex and Deligne-Du Bois complex? REPLY [4 votes]: Here is a counterexample that shows the relationship doesn't hold in general. For the variety $X = {\rm Spec}~\mathbb C [x,y]/xy$, the cotangent complex is just the usual module of differentials $$L_X = \frac{\mathcal O_X dx \oplus \mathcal O_X dy}{ y ~dx + x ~dy},$$ but since $X$ is a complete intersection (and therefore Gorenstein), the dualizing complex is a line bundle $\omega_X = \mathcal O_X[1]$<|endoftext|> TITLE: In how many steps a random walk visits all the elements of a finite group, with a probability 1/2? QUESTION [15 upvotes]: This question is a variation of the return to the origin problem. Let $G$ be the finite group $\mathbb{Z}/n \times \mathbb{Z}/n$ and let the random transformation $T: G \to G$ such that $T(a,b) = (a \pm 1, b)$ or $(a , b \pm 1)$, with a uniform probability (i.e. a uniform random walk on $G$). Let $P_n(r)$ be the probability that $\{T^{s}(0,0) \ \vert \ 1 \le s \le r \} = G$. So of course $P_n(r) = 0$ iff $r < n^{2}$ and $P_n(r)$ is increasing with $r$. Let $R_n$ be the number such that $P_n(R_{n}-1)<1/2$ and $P_n(R_{n}) \ge 1/2$. Note that $R_n$ is hard to compute by brute-force: $R_1 = 1$, $R_2 = 6$, $R_3 = 13$(?). Question: What is the value of $R_n$? A formula would be great, but I'm ok with an evaluation. Remark: This question generalizes to any finite group together with a presentation, and we get an invariant of the group as the maximum for all the presentations, of such number of steps. For $G = \mathbb{Z}/n$ with the presentation $\langle a \vert a^{n} \rangle$ we get $R_n=1,2,3,6,10,14,19 \dots$ a general formula? What do we get for the groups $D_n$, $S_n$, $A_n$, $B_n \dots$ (see the common examples)? REPLY [25 votes]: The quantity $R_n$ is asymptotic to ${4\over \pi}(n\log n)^2$, see "Cover times for Brownian motion and random walks in two dimensions" by Dembo, Peres, Rosen and Zeitouni. This was previously conjectured by Aldous.<|endoftext|> TITLE: Combinatorial\Probabilistic Proof of Stirling's Approximation QUESTION [10 upvotes]: Stirling's approximation is the following well-known asymptotic result: $$n! \approx \left(\frac{n}{e}\right)^n \sqrt{2 \pi n}$$ This result has several analytical proofs, for example via Laplace's method, the trapezoidal rule (and Euler–Maclaurin formula for an asymptotic expansion) or Hayman's method (essentially Cauchy's formula). There are also other (usually ad-hoc) proofs. The result is applied often in combinatorics and probability, especially in the study of random walks. I want a result which is the other way around - a combinatorial\probabilistic proof for Stirling's approximation. I'm not sure if this is possible, but to convince you that it might be I'll give some partial results. First, write $n! = \left(\frac{n}{e}\right)^n f(n)$. We'll obtain bounds related to $f(n)$: Consider $\{S_n\}_{n\ge 0}$, the random walk on $\mathbb{Z}^{d}$, with $S_0=\vec{0}$. It is recurrent iff $\sum_{n\ge 1} P(S_{2n} = \vec{0})$ diverges. This series is $\sum_{n\ge 1} (2d)^{-2n} \sum_{a_1+\cdots+a_d=n} \binom{2n}{a_1,a_1,a_2,a_2,\cdots,a_d,a_d}$. For $d=1$, this becomes $\sum_{n\ge 1} \frac{\binom{2n}{n}}{4^n} = \sum_{n\ge 1} \frac{f(2n)}{f^2(n)}$. There's is Stirling-free proof for the recurrence of the 1-dimensional random walk, using a criterion due to Nash-Williams. Similarly, for $d=2$, there's a rotation trick that shows $\sum_{n\ge 1} P(S_{2n} = \vec{0})=\sum_{n\ge 1} \left( \frac{\binom{2n}{n}}{4^n} \right)^2= \sum_{n\ge 1} \left( \frac{f(2n)}{f^2(n)} \right)^2$. Again, the Nash-Williams criterion proves the recurrence of the 2-dimensional random walk without Stirling, and hence the divergence of said series. Back to the 1-dimensional case. Using the reflection principle, the probability that the random walk returns to the origin during the first $2n$ steps is $1-\frac{\binom{2n}{n}}{4^n} = 1-\frac{f(2n)}{f^2(n)}$, and this tends to 1 as the walk is recurrent. Still with the 1-dimensional case. One can calculate $E|\frac{S_n}{\sqrt{n}}|$ explicitly - it is $2^{1-n} \sqrt{n}\binom{n-1}{\lfloor \frac{n-1}{2} \rfloor}$. On the other hand, the CLT together with the notion of uniform integrability justifies the following limit: $\lim_{n\to\infty} E|\frac{S_n}{\sqrt{n}}| = E|N(0,1)| = \frac{2}{\sqrt{2\pi}}$. When $n$ is odd this implies $\sqrt{n} \frac{f(n-1)}{f^2(\frac{n-1}{2})} \to \frac{2}{\sqrt{2\pi}}$. So if we assume $f(n) \approx C \cdot n^{\alpha}$, the first results give bounds on $\alpha$ ($\alpha \le 1, \alpha \le 0.5, \alpha >0$) and the last one gives the true value of $\alpha$ and $C$. The fact that the 3-dimensional random walk is transient implies $\alpha \ge \frac{1}{3}$, but I don't know how to prove it without tools that prove Stirling's approximation. Questions: Can we remove\relax the assumption $f(n) \approx C \cdot n^{\alpha}$ somehow? Can we use less analysis than we used in the 3rd result? Are there other combinatorial approaches to the problem that give partial results (or even the full one)? Can the fact that the 3-dimensional random walk is transient be proved Stirling-free (and preferably combinatorially)? REPLY [3 votes]: Meanwhile, I'll answer the other question about random walks: Yes, it is quite easy to prove random walks in $\geq 3$ dimensions are transient without Stirling's approximation. Indeed, this is the standard proof I know, and was used in the second hit when I googled for it here. See Section 2 here. In summary, let $p_n$ be the probability that the random walk returns to $0$ at time $n$. One shows that the walk is transient if and only if $\sum p_n < \infty$. Then one writes $$p_n = \frac{1}{(2 \pi)^d} \int_{-\pi}^{\pi} \cdots \int_{-\pi}^{\pi} \left( \frac{1}{d} \sum \cos \theta_j \right)^n d \theta_1 \cdots d \theta_n$$ and approximates the integral by $$\frac{1}{(2 \pi)^d} \int_{- \infty}^{\infty} \cdots \int_{- \infty}^{\infty} \exp\left(- \frac{n}{2} \sum \theta_i^2 \right) d \theta_1 \cdots d \theta_n$$ to show that $p_n \sim \frac{c}{n^{d/2}}$, so $\sum p_n$ converges for $d>3$ and diverges for $d=1$, $d=2$.<|endoftext|> TITLE: A set of integers whose factorial can be written as a product of two factorials QUESTION [17 upvotes]: I am trying to collect informations concerning the set $$\mathcal{A}=\left\{n\in\mathbb{N} \mid (\exists k,l\in\{2,3,\dots,n-2\})(n!=k!l!)\right\}.$$ It seems not much is known about the set $\mathcal{A}$. Below, I mention all what I have. Any other result relating properties of $\mathcal{A}$ will be appreciated. First, it is not known (to my best knowledge), if $\mathcal{A}$ is finite or not. Note that the range for factors $k$ and $l$ excludes values $1,n$, and also $n-1$. Clearly, I want to avoid trivial identities of the form $n!=1!n!$. Less trivially, nevertheless it not an interesting case if one factor equlas $n-1$ since for any integer $n$ which is a factorial of another integer, say $n=m!$, we have $n!=m!(n-1)!$. For example, $6!=3!5!$, but $6\notin\mathcal{A}$. The least number and the only one I know belonging to $\mathcal{A}$ is 10 since $$10!=6!7!.$$ I can prove (and it is not hard at all) that if $n\in\mathcal{A}$ and $n!=k!l!$, then $k+l>n+1$. Mathematicians studied a more general set allowing decomposition of $n!$ into a product of more then only two factorials (hence $\mathcal{A}$ is a subset). In $\S$B23 (page 80) of R. K. Guy, Unsolved Problems in Number Theory, 2nd ed., New York, Springer-Verlag, 1994, one finds the following: With the aid of computer, J.Shallit and M. Easter showed that, between numbers $1,2,\dots,18160$, only the number 10 belongs to $\mathcal{A}$. This computation comes from early 90's and, as I indicated above, it treats even more general problem. So I thing that at least this result could be extended significantly with today's computers. I do not know the algorithm used by J.Shallit and M. Easter but certainly the main obstacle is that one has to check a lot of equalities between huge numbers. This could be simplified by taking a logarithm of both sides in $n!=k!l!$, for instance. Next, it has been proved by Paul Erdős that if $$\lim_{n\to\infty}\frac{P(n(n+1))}{\ln n}=\infty,$$ where $P(n)$ is the largest prime factor of $n$, then $\mathcal{A}$ has to be finite. Finally, by personal communications with some colleges (mathematicians, though not number theorists) I encountered also a curios opinion that $\mathcal{A}$ contains only number 10. REPLY [3 votes]: UPDATE The algorithm have been improved, and c up to 100,000,000,000 have been tested. with a billion per hour on a laptop. The reason why this can be done so quickly looping in c, is You can stop checking b's if b=p-1: a TITLE: Are Sharps in Countable Models Really Sharps? QUESTION [5 upvotes]: Suppose $M$ is a countable transitive model of $\mathsf{ZFC}$ (maybe more). Suppose $x, y \in {}^\omega\omega \cap M$ and $M \models y = x^\sharp$. Is it true (possible with additional assumptions), that $V \models y = x^\sharp$? From Kanomori's book, $y = x^\sharp$ is $\Pi_2^1$. However, I can not apply Shoenfield absoluteness since $M$ is countable and does not have the countable ordinals of $V$. An even stronger general question may be: are there conditions on $M$ or $V$ such that $\Pi_2^1$ absoluteness can hold between the universe and the countable model $M$. Thanks for any insight on these questions. REPLY [9 votes]: The answer is no, not necessarily. Countable transitive models can have fake sharps. To see this, observe simply that the assertion, $$\hbox{There is a countable transitive model of $\text{ZFC}+x^\sharp$ exists}$$ is itself a $\Sigma^1_2(x)$ assertion, and thus it is absolute to $L[x]$, where the real $x^\sharp$ does not exist. So there will be a countable transitive model $M\in L[x]$ that thinks $x^\sharp$ exists, but it must have a fake $x^\sharp$. In particular, if there is a transitive model of ZFC+$0^\sharp$ exists at all, then there is such a model inside $L$. And any such model inside $L$ cannot have the real $0^\sharp$.<|endoftext|> TITLE: Time for Langton's ant to cover a "square" torus QUESTION [23 upvotes]: Langton's ant is a cellular automaton running as follows: Squares on a plane are colored variously either black or white. We arbitrarily identify one square as the "ant". The ant can travel in any of the four cardinal directions at each step it takes. The ant moves according to the rules below: At a white square, turn 90° right, flip the color of the square, move forward one unit At a black square, turn 90° left, flip the color of the square, move forward one unit We consider Langton's ant on a torus $n$ by $n$ gridded such that all the squares are white. Preliminary question: Is it true that Langton's ant will visit every square, for all $n$? Remark: I've checked it's true for $n \le 1000$. In fact Langton's ant could enter into a local cycle without having visiting every square (see here), so the fact that such phenomena can't appear must be proved. If so, let $s_n$ be the number of steps Langton's ant needs for visiting all the squares. Question: what's the asymptotic of $s_n$? $\small{ \begin{array}{c|c} n &2&3& 4& 5& 6& 7& 8& 9& 10& 50 & 100 & 500 \newline \hline s_n &3&12&41&62&166&113&318&281&692&57672 & 225905 & 12740527 \end{array} } $ Remark: Following the table above, this asymptotic seems to be $\frac{4}{\pi}(nln(n))^2$, as for the random walk. $\small{ \begin{array}{c|c} n &1000&2000 & 5000 &6000& 10000& 11000 & 12000 & 13000 & 14000 \newline \hline \frac{4(nln(n))^2}{s_n} &2.919&2.196&2.177&1.770&1.506&2.067&1.734&1.502&1.911 \end{array} } $ Remark: This new data suggests that there is no asymptotic, because for $n$ large $\frac{4(nln(n))^2}{s_n}$ seems bounded in $[1,4]$ but not convergent. For $n=50$ and the ant starting "up" at position $(25,25)$, the grid looks as follows at step $s_n$: Now by encoding square's color by how many times it was visited (no effect on the rules) we get: And for $n=500$ at step $s_n = 12740527$: These pictures was computed online at http://www.turnerbohlen.com/langtonsant/ And for $n=5000$ at step $s_n = 3331448985$: For comparison, this last picture was generated from a uniform random walk for $n=5000$ (covered after $2410514205$ steps): These two last pictures were computed by Sage, with this code. REPLY [4 votes]: I have been playing with Langtons ant for a while now, Have a look at this video which shows 10^30 iterations. https://www.youtube.com/watch?v=LXDBJ4zKWvI And the wave equations are here with octave script and a link to octave online where you can run the math. https://sites.google.com/site/extendedlangtonsant/ Hope someone finds this useful. Graham<|endoftext|> TITLE: Borel cross section QUESTION [6 upvotes]: It is known from metric space topology that a closed equivalence relation on a Polish space has either countably many or $\mathfrak{c}$ many equivalence classes. A short elementary proof is given in Srivastava "A course on Borel sets", Theorem 2.6.7. This proof does not proceed through a Borel or analytic cross section for the equivalence relation. Is it possible that a closed equivalence relation can have no analytic cross section? (A "cross section", or "transversal", is a set with exactly one point in each equivalence class. An equivalence relation on a space $X$ is "closed" if it is a closed subset of the product space $X\times X$.) Several results asserting the existence of a Borel or analytic cross section assume additional conditions, such as requiring the equivalence classes to be closed, for example. The stated theorem is a particular instance of a much deeper theorem of J. Silver: every coanalytic equivalence relation has either countably many or $\mathfrak{c}$ many equivalence classes. For a proof, see Jech "Set Theory", 3rd ed. 2002/2006, Chapter 32. REPLY [3 votes]: Shashi Srivastava suggests taking the equivalence relation on $\omega^\omega$ induced by a continuous $f:\omega^\omega\to \omega^\omega$ with non-Borel range. This is a closed equivalence relation with no Borel cross-section.<|endoftext|> TITLE: Permutation polynomials mod $p$ of the form $(x+1)^n-x^n$ QUESTION [5 upvotes]: Among some permutation polynomials I've been studying, $f(x)=(x+1)^n-x^n$ is one of the polynomials that I cannot grasp. Question : For a given odd prime $p$, how can we find every positive integer $n$ such that $f(x)=(x+1)^n-x^n$ is a permutation polynomial mod $p$ ? The answer seems $n=(p-1)m+2\ \ (m=0,1,2,\cdots)$, but I'm facing difficulty in proving that. This question has been asked previously on math.SE without receiving any answers. The followings are what I've got. $f(0)\equiv 1.$ $f(p-1)\equiv -(-1)^n\Rightarrow \text{$n$ has to be even}\Rightarrow f(p-1)\equiv p-1$. For $n=(p-1)m+r$, $f(x)\equiv (x+1)^r-x^r$ because $a^{p-1}\equiv 1$ for $a$ which is coprime to $p$. $f\left(\frac{p-1}{2}\right)\equiv 0$. $f\left(\frac{p-1}{2}+a\right)+f\left(\frac{p-1}{2}-a\right)\equiv 0$ for any $a$. I would like to know any relevant references as well. REPLY [10 votes]: The expected answer is correct, it is a theorem by Norman Johnson, see here. Note that $(X+1)^n-X^n$ is a permutation polynomial if and only if $(X+a)^n-X^n$ is a permutation polynomial for each fixed nonzero $a$. This latter condition is a special case of planarity: A function $f:K\to K$ on a field $K$ is called planar, if $x\mapsto f(x+a)-f(x)$ is bijective for every nonzero $a$. So you ask for planar monomials on $\mathbb F_p$ for odd primes $p$. Actually, it was later proved that any planar function on $\mathbb F_p$ is given by a quadratic polynomial, see this MO page for more information and references.<|endoftext|> TITLE: Are there integral representations of the Mertens constant? QUESTION [14 upvotes]: It is well-known that the Euler constant $$\gamma=\lim\limits_{n\to \infty}\biggl( \sum\limits_{k\le n}\dfrac{1}{k}-\ln{n}\biggr ) $$ has a bunch of integral representations, e.g. $$\gamma=-\int\limits_0^ \infty e^{-t}\ln t\; dt=-\int\limits_0^ 1\ln(\ln t)dt ,$$ all of them containing $\log$ and/or $\exp$ (or some special functions). I am wondering if there are non trivial integral representations of similar styles for the Mertens constant $B_1$, defined by $$B_1=\lim\limits_{n\to \infty}\biggl(\sum\limits_{p\le n}\dfrac{1}{p}-\ln\ln{n}\biggr ) .$$ Otherwise stated: Is $B_1$ a period in a broader sense (meaning: allowing also $\log$ and/or $\exp$ under the integral), which is sometimes called an "exponential period"? REPLY [3 votes]: Given that there's been no other answer to this question, I'll leave some indications from my earlier (failed) attempt to answer it. I think the answer to the question as stated is yes. If you include absolutely convergent integral of arbitrary products of algebraic, exponential and logarithmic functions, then the Meissel-Mertens constant is a period in your sense, and it shouldn't be very hard to prove. In fact, Zagier and Kontsevich mention something very close at the end of their seminal paper on the topic: M. Kontsevich & D. Zagier, Periods (2001) "There have been some recent indications that one can extend the exponential motivic Galois group still further, adding as new period the Euler constant $\gamma$, which is, incidentally, the constant term of $\zeta(s)$ at $s=1$. Then all classical constants are periods in an appropiate sense." To see this, take the classic expression: $$B_1=\gamma+\sum_p\biggl[\ln\biggl(1-\frac{1}{p}\biggl)+\frac{1}{p}\biggl]$$ As you mention, $\gamma$ has an integral representation. If you can prove that the sum in the previous expression is also a period, you are obviously done. Now, I suspect that is a relatively easy exercise of Chen integration. This is the standard method used to prove that the Riemann zeta function (and more generally the multiple zeta function) at the integers is a period. I'm not familiar with the technique, but perhaps someone else not necessarily familiar with periods can complete the argument.<|endoftext|> TITLE: Area of hyperbolic triangle in terms of Lengths of its sides QUESTION [5 upvotes]: Let a, b and c denote the cosh of the lengths of the sides of an hyperbolic triangle and A, B, and C its angles. Its area is well knwon to be S = pi - A - B - C . What is S in terms of a, b, c ? In J. Smorodinskij, Fortschritte der Physik, 18 (1965) 157 -- 173 I find (without reference or proof) cos(S/2) = (1 + a + b + c) / (4 (a' b' c')^2) where a' is the cosh of half the lenght of the side a. My cumbersum calculations yield cos(S/2) = (1 + a + b + c) / (4 (a' b' c')) Where do I find a simple proof of this simple formula? REPLY [2 votes]: Your version of the formula is correct. The proof can be found, for example, on pp. 102-103 in http://arxiv.org/abs/1102.0462 (The Hyperbolic Theory of Special Relativity, by J.F. Barrett).<|endoftext|> TITLE: The action of the center on the extended Dynkin diagram QUESTION [6 upvotes]: Let $R$ be an irreducible root system with a basis $\Pi$. We obtain the Dynkin diagram $D$ and the extended Dynkin diagram ${\widetilde{D}}$ of $R$ with respect to $\Pi$. Let $Q^\vee\subset P^\vee$ denote the coroot lattice and the coweight lattice, respectively. It is known that the finite abelian group $P^\vee/Q^\vee$ (which is isomorphic to the center of the corresponding simply connected compact Lie group) acts on ${\widetilde{D}}$. I am looking for references where the action of $P^\vee/Q^\vee$ on ${\widetilde{D}}$ is described in detail (preferably with examples, say for a root system of type $D_n$). REPLY [4 votes]: The question is perhaps best answered not in the context of Lie theory but in the related setting of affine Weyl groups, where an irreducible root system in Bourbaki's sense leads to an extended Dynkin diagram and corresponding Coxeter group. Here the isomorphic finite abelian groups $P/Q$ and $P^\vee/Q^\vee$ may be interpreted in Lie theory as fundamental groups of related adjoint compact Lie groups or as centers of their simply connected covers. But this interpretation may not shed much light on the question asked. Probably the earliest detailed reference is the first part of the 1965 IHES paper by Iwahori and Matsumoto, which is usually (though not at the moment) available online through numdam.org: Iwahori-Matsumoto. In particular, they work out the complicated details for each Lie type, though this requires a large amount of notation. Here as elsewhere in the literature (Bourbaki for instance), notation varies a lot but is essential for understanding the situation. Another (harder-to-locate) source, inspired by Iwahori-Matsumoto but using somewhat different notation due to the suggested applications in modular representation theory, is the first part of the conference write-up by D.N. Verma. This was published only in 1975 but reflects some of the talks given at the 1971 Budapest summer school on Lie groups: Verma. One advantage of Verma's exposition is that he gives a thorough account of how the finite abelian group acts on the extended Dynkin diagram, though without examples. In both of these accounts the focus is on an affine Weyl group (a Coxeter group) along with a usually larger extended affine Weyl group. The resulting finite quotient group $\Omega$ is isomorphic to the above fundamental group but has the merit of being an explicit group of affine transformations which preserves the fundamental alcove (or simplex). The closure of this alcove is a fundamental domain for the affine Weyl group, and the vertices are in bijection with the set of simple reflections along with 0, or equivalently with the $\ell+1$ generators of the group (where $\ell$ is the rank of the underlying root system). While $\Omega$ preserves this alcove, it permutes the $\ell+1$ vertices in a way described in both papers cited. This realizes the action of $\Omega$ on the vertices of the extended Dynkin diagram. Here the elements of $\Omega$ are in natural bijection with the minuscule fundamental weights (along with 0): such weights correspond to simple roots having coefficient 1 in the expression of the highest root and come up frequently in representation theory. ADDED: Concerning type $D_n$, the summary in Iwahori-Matsumoto (see page 19) naturally varies for $n$ even or odd (where the structure of $\Omega$ differs), but the permutation of vertices in each case is fairly natural even though hard to guess in advance.<|endoftext|> TITLE: Area of metric spheres in Riemannian manifolds QUESTION [5 upvotes]: I am trying to estimate the integral $\int \mathbb{e} ^{-d(x_0,x)^2} \mathbb{d}x$ on a Riemann manifold $(M,g)$, for some arbitrary fixed $x_0 \in M$ and $d$ the usual distance. The only thing that I can think of is to use some coarea theorem, leading to $\int _0 ^\infty \mathbb{e} ^{-r^2} A(x_0, r) \mathbb{d}r$, where $A(x_0, r)$ is the area of the metric sphere of center $x_0$ and radius $r$. The issue now is to have some estimate of $A$. Surprisingly, even though there are plenty of comparison theorems for the volume of metric balls, I haven't been able to find anything usable on the area of spheres. Therefore, the questions: how would you approach this integral (if not by the coarea formula)? is it finite? (I assume so) is $A$ a polynomial in $r$ (of degree $n-1$)? If so, I would expect it not to have a free term, but what about its leading coefficient? can $A$ be computed explicitly in space forms (other than Euclidian, of course)? are there comparison theorems for $A$? Thank you. REPLY [9 votes]: The integral might be infinite. Indeed, the function $A(r)$ can grow as fast as you want, but in this case the Ricci curvature in some of the radial direction have to go to $-\infty$ as $r\to\infty$. (Needless to say $A$ can grow faster than any polynomial.) If Ricci curvature is bounded below then $A(r)$ has at most exponential growth --- this follows from the Bishop--Gromov inequality. In this case you integral has to be finite.<|endoftext|> TITLE: Existence of an invariant measure on an infinite dimensional space via Lyapunov functional QUESTION [5 upvotes]: Set-up. Assume that we have a complete separable metric space $\mathcal{X}$ that is not locally compact. Let $V: \mathcal{x} \to [0; +\infty]$ be a functional such that $K_r :=\{x \in \mathcal {X} : V (x) \leq r \}$ is compact in $\mathcal{X}$ for any $r >0$. Let $(X_t)$ be a Markov process in $\mathcal{X}$ such that $P\{ V(X _t) < \infty \text{ for all } t \geq 0 \} = 1$ and $$ V(X _t) - \int ^t _0 g(X _s) ds \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \quad \quad \quad \quad \quad \quad \quad \quad (1) $$ is a martingale, where $$ g(x) \leq c - V(x), \quad \text{ for all } x \in \{z \in \mathcal{X}: V(z) < \infty \} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) $$ for some $c >0$. Furthermore, assume that $(X_t)$ is a Feller process, that is, if $f$ is a bounded continuous function on $\mathcal{X}$, then so is $P_t(\cdot, f)$, where $\{P_t, t \geq 0 \}$ is the family of transition kernels corresponding to the process $(X_t)$. Under these general conditions, is it possible to deduce the existence of an invariant measure for $(X _t)$? An attempt to prove the existence of an invariant measure. My idea would be to try to prove that the family of distributions of $\mu _t (\Gamma):=\frac 1T \int _0 ^T P \{X _s \in \Gamma \}ds$ is tight and obtain an invariant measure as a limit point. Imitating the proof of Lemma 9.7 of Chapter 4 of the book by Ethier and Kurtz, we can show that for any $t \geq 1$, $\mu _t (K _r) \to 1$ as $r \to \infty$. Namely, we have $$ 0 \leq E V (X _t) = E V (X _0) + E \int ^t _0 g (X _s ) ds \\ = E V (X _0) + E \int ^t _0 g (X _s ) I\{ X_s \in K _r \} ds + E \int ^t _0 g (X _s ) I\{ X_s \notin K _r \} ds \\ \leq E V (X _0) + c t \mu _t (K _r) + (c-r)t[1-\mu _t (K _r)], $$ hence $$ \mu _t (K _r) \geq 1 - \frac{c + \frac 1t E V (X _0)}{r}. $$ We see that the family $\{ \mu _n , n \in \mathbb{N} \}$ is tight and thus relatively compact by Prokhorov's theorem, therefore there exists a limit point $\mu$. The measure $\mu$ should be invariant for $(X_t)$ since $(X_t)$ is a Feller process: let $t>0$, $\mu _{T_n} \to \mu$, $T_n \to \infty$, and let $\nu _0 = Law (X _0)$, then for every bounded continuous $f$ $$ \int P _t (x,f) \mu (dx) = \lim _n \frac{1}{T_n}\int _0 ^{T_n} ds \int P _t (x,f) P _s (y,dx) \nu _0 (dy) \\ = \lim _n \frac{1}{T_n}\int _0 ^{T_n} ds \int P _{s+t} (y,f) \nu _0 (dy) = \lim _n \frac{1}{T_n}\int _t ^{T_n+t} ds \int P _{s} (y,f) \nu _0 (dy) \\ = \lim _n \Big[ \frac{1}{T_n}\int _0 ^{T_n} + \frac{1}{T_n}\int _{T_n} ^{T_n+t} - \frac{1}{T_n}\int _{0} ^{t} \Big] = \lim _n \langle \mu _{T_n}, f \rangle =\langle \mu , f \rangle $$ Question. Does the proof above work? I am asking because in the mentioned book by Ethier and Kurtz the state space is assumed to be locally compact in corresponding statements, and although it seems for me that in the proof above it is not necessary, I am not sure that I have not missed something. Also, I would appreciate any reference where a similar question is addressed. Motivation. I would like to prove the existence of an invariant distribution for some interacting particle systems represented by an $\mathbb{N}^{\mathbb{Z^d}}$-valued process, so that $\mathcal{X} = \mathbb{N}^{\mathbb{Z^d}}$, with $V(x) = \sum _{q \in \mathbb{Z} ^d} x(q)s(q) $, where $s$ is some summable positive function. REPLY [3 votes]: As far as references for these things, have a look at this webpage for lecture notes by Martin Hairer. In particular, the lectures on "Ergodic properties of Markov processes" are I think exactly what you need (see discussion around Theorem 4.21 therein). The important property for $\mathcal{X}$ is to be a Polish space. Local compactness is not necessary.<|endoftext|> TITLE: Square root of normal distribution QUESTION [10 upvotes]: Let $X$ and $Y$ be independent random variates with the same probability distribution, $P(x)$. Assuming that the product $Z=XY$ is a random variate with normal distribution, say $$f_Z(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} x^2}$$ what is the form of $P(x)$? Does it exist in closed form? EDIT: I just realized that a product distribution is necessarily divergent at zero, which means that the normal distribution cannot be a product distribution (at least when $P(x)$ is interpreted as a normal function; maybe there is a generalized function that solves the problem?). Do you agree? REPLY [9 votes]: First: No helvio, the density of the product of two iid random variables (r.v.'s) is not necessarily divergent at 0. E.g., take the product of two iid r.v.'s each having a Gamma distribution with shape parameter 2. Second: Let $U:=\ln|Z|$ and $W:=\ln|X|$. Then the characteristic function (c.f.) $f$ of $U$ is given by the formula $f(t)=2^{i t/2} \Gamma((1+i t)/2)/\sqrt{\pi}$ for real $t$. So, the c.f. of $W$ (say $g$) would be a square root of $f$, and then the density (say $q$) of $W$ would be given by the formula $q(x)=\frac1{2\pi}\,\int_{-\infty }^{\infty } g(t) e^{-i t x} \, dt$ for real $x$. I have tried to evaluate the latter integral numerically, and that seems to suggest that $q(3)<0$; however, I have not done a rigorous estimate of the approximation error. On the other hand, numerical experiments using Bochner's theorem on the positive definite functions suggest that $g$ is a c.f.! If so, then the corresponding r.v. $W$ exists, and then it suffices to let $X$ have the distribution of $\varepsilon e^W$, where $\varepsilon$ is a Rademacher r.v. independent of $W$. Addendum: The calculation of $q(3)$ mentioned previously was likely mistaken, as I forgot to make sure that $g(t)$ be taken to be the value of the square root of $z:=f(t)$ on the Riemann surface for $\sqrt z$, so as for $g$ to be a continuous function -- as it must be. Another approach to the problem: Note that $\mu_j:=E|Z|^j=2^{j/2} \Gamma \left(\frac{j+1}{2}\right)/\sqrt{\pi }$ and then one would have $\nu_j:=E|X|^j=\sqrt{\mu_j}$ for $j>0$. To prove that such a desired r.v. $X$ exists, it is enough to show that for all nonnegative integers $n$ the determinants, say $d_n$ and $e_n$, of the matrices $(\nu_{i+j})_{i,j=0}^n$ and $(\nu_{i+j+1})_{i,j=0}^n$ are nonnegative (see e.g. Ch. V in Kreǐn, M. G. and Nudel'man, A. A., The Markov moment problem and extremal problems). These determinants are indeed positive for $n\le9$. One can hopefully find a recursion for $d_n$ and $e_n$ to show that they are indeed nonnegative for all $n$. Note here that $\mu_j$ is an integer or an integral multiple of $\sqrt{2/\pi}$.<|endoftext|> TITLE: Arranging numbers from $1$ to $n$ such that the sum of every two adjacent numbers is a perfect power QUESTION [38 upvotes]: I've known that one can arrange all the numbers from $1$ to $\color{red}{15}$ in a row such that the sum of every two adjacent numbers is a perfect square. $$8,1,15,10,6,3,13,12,4,5,11,14,2,7,9$$ Also, about two weeks ago, a colleague taught me that one can arrange all the numbers from $1$ to $\color{red}{305}$ in a row such that the sum of every two adjacent numbers is a perfect cube. $$256,87,129, 214, 298, 45, 171, 172, 44, 299, 213, 130, 86, 257, 255,$$ $$88, 128, 215, 297, 46, 170, 173, 43, 300, 212, 131, 85, 258, 254, 89, 127, 216, 296,$$ $$ 47, 169, 174, 42, 301, 211, 132, 84, 259, 253, 90, 126, 217, 295, 48, 168, 175, 41, 302, $$ $$210, 133, 83, 260, 252, 91, 125, 218, 294, 49, 167, 176, 40, 303, 209, 134, 82, 261, 251,$$ $$ 92, 33, 183, 160, 56, 287, 225, 118, 98, 245, 267, 76, 140, 203, 13, 14, 202, 141, 75, 268,$$ $$ 244, 99, 26, 190, 153, 63, 280, 232, 111, 105, 238, 274, 69, 147, 196, 20, 7, 1, 124, 219,$$ $$ 293, 50, 166, 177, 39, 304, 208, 135, 81, 262, 250, 93, 32, 184, 159, 57, 286, 226, 117, 8,$$ $$ 19, 197, 146, 70, 273, 239, 104, 112, 231, 281, 62, 154, 189, 27, 37, 179, 164, 52, 291, 221,$$ $$ 122, 3, 5, 22, 194, 149, 67, 276, 236, 107, 109, 234, 278, 65, 151, 192, 24, 101, 242, 270,$$ $$ 73, 143, 200, 16, 11, 205, 138, 78, 265, 247, 96, 120, 223, 289, 54, 162, 181, 35, 29, 187,$$ $$156, 60, 283, 229, 114, 102, 241, 271, 72, 144, 199, 17, 108, 235, 277, 66, 150, 193, 23,$$ $$ 4, 121, 222, 290, 53, 163, 180, 36, 28, 188, 155, 61, 282, 230, 113, 103, 240, 272, 71, 145,$$ $$ 198, 18, 9, 116, 227, 285, 58, 158, 185, 31, 94, 249, 263, 80, 136, 207, 305, 38, 178, 165,$$ $$ 51, 292, 220, 123, 2, 6, 21, 195, 148, 68, 275, 237, 106, 110, 233, 279, 64, 152, 191, 25,$$ $$100, 243, 269, 74, 142, 201, 15, 12, 204, 139, 77, 266, 246, 97, 119, 224, 288, 55, 161,$$ $$ 182, 34, 30, 186, 157, 59, 284, 228, 115, 10, 206, 137, 79, 264, 248, 95$$ Here, I have a few questions. Question 1 : For each $N\ge 2\in\mathbb N$, does there exist at least one positive integer $n\ge 2$ satisfying the following condition ? Condition : One can arrange all the numbers from $1$ to $n$ in a row such that the sum of every two adjacent numbers is of the form $m^N$ for some $m\in\mathbb N$. Question 2 : Can we find at least one concrete $n$ with an arragement for a given $N$? Question 3 : How about cyclic arrangements where the sum of the first and last numbers is also a perfect power? I would like to know any relevant references as well. Remark : Question 1 has been asked previously on math.SE without receiving any answers. Additional information : On math.SE, a user Micah commented, "For fixed $n$ and $N$, this is equivalent to asking whether some graph on $n$ vertices with $O(n^{1+1/N})$ edges has a Hamiltonian path. This is substantially above the threshold for a random graph to have a Hamiltonian path (which happens when the expected number of edges is $O(n\log n)$ or so), so the answer is probably "yes" unless there's some interesting structure in this specific graph that interferes with your chances." Also, a user MJD showed a square-cyclic arrangement for $n=32$ : $$\small1,8,28,21,4,32,17,19,30,6,3,13,12,24,25,11,5,31,18,7,29,20,16,9,27,22,14,2,23,26,10,15$$ REPLY [5 votes]: If we do not fix the exponent, the sum of every two adjacent numbers can be any perfect power, then we obtain a larger class of solutions. I modified the Sage code provided by Moritz Firsching to handle this case, an applet can be found here: https://cloud.sagemath.com/projects/43540988-5a9c-473c-b2f0-d5adf4168301/files/2015-08-03-161507.sagews for given $n$ it looks for an appropriate arrangement. E.g. if $n=17,$ then one has the following cyclic arrangement $$ 1,3,5,4,12,13,14,11,16,9,7,2,6,10,15,17,8. $$<|endoftext|> TITLE: Scott-Solovay unpublished paper on ``Boolean valued models of set theory'' QUESTION [16 upvotes]: I have read some papers from 1970$^{th}$, and in some of them, the paper of Scott and Solovay on ``Boolean valued models of set theory'' is given as a main reference, with many references to the results from it. Unfortunately the paper never published. Question 1. Does anyone know the historical reasons for not publishing the paper? Of course I know there are some papers and books covering the topic, in particular: 1) Boolean-valued set theory and forcing by Richard Mansfield, John Dawson. 2) Set Theory: Boolean-Valued Models and Independence Proofs by John Bell. The second reference gives some historical points about the creation of Boolean valued method. Though the above references are very good for learning the method, I am mainly interested in the original paper. Question 2. Is there any typed or scanned version of the Scott-Solovay paper available? how can I find a version of the paper? Of course, maybe the simplest answer is: send an email to one of the authors, and ask them about the paper. But I would rather first try the Mathoverflow. -- As it is stated in the answer below (and I was aware of it), the paper by Scott ``A proof of the independence of the continuum hypothesis'' presents some aspects of the theory. But it does not give answer to my questions. REPLY [5 votes]: It is worth reading the following paragraphs from the G. Moore's paper " The origins of forcing". The paper may lead you to other useful resources. In his model Solovay had used Borel sets of positive measure as forcing conditions. Solovay pointed out to Scott, at Stanford in September 1965, that Cohen's definition of forcing could be regarded as a way of assigning Boolean values to formulas [Scott 1967, 108; 1978, xiv]. Moreover, Solovay noticed that if he associated with each formula the pair of sets consisting first of those conditions forcing the formula and, second, of those forcing the negation of the formula, then he obtained a complete Boolean algebra. Solovay did not think of the Boolean algebra topologically, and when he showed it to Scott at Stanford, Scott pointed out that it consisted of regular open sets [Solovay 1982]. Scott reversed the process: Since it was possible to start with forcing and obtain a Boolean algebra, then it was possible to start with Boolean-valued logic and obtain forcing from it. Meanwhile, Solovay had independently come to the same conclusion [Scott 1967, 109; 1978, xv]. At that point there occurred a dispute between Scott and Solovay over Boolean-valued models. Solovay was afraid that Scott, whom he then still regarded as a "big name", was going to steal his result from him. Then their differences were resolved, and they agreed to prepare a joint paper, which was partially written but never completed [Solovay 1981]. Instead, Scott's lecture notes on Boolean-valued models, given in 1967 at the U.C.LA. set theory conference, circulated widely among the cognoscenti.<|endoftext|> TITLE: When is the endofunctor category of a monoidal category braided? When is it ribbon? Fusion? Modular? QUESTION [5 upvotes]: Given a category $\mathcal{C}$, we can define the category of endofunctors $\operatorname{Cat}(\mathcal{C})$, with objects functors $F: \mathcal{C} \to \mathcal{C}$ and morphisms natural transformations. Since $\mathrm{Cat}$ is a 2-category, $\operatorname{Cat}(\mathcal{C})$ is naturally endowed with a strict monoidal product, which is functor composition. I'm interested in additional structure on such categories. Notice that if we set $\mathcal{C} = \mathrm{Vect}_{\text{fin.dim.}}$ and restrict to linear functors, we find that an endofunctor is given by a choice of vector space, so the endofunctors have a natural symmetric monoidal structure. I think one can convince oneself that the endofunctor category is rigid if all functors have adjoints. What can we say about braided structures? Do they occur on endofunctor categories of monoidal categories? It's not even clear why endofunctors should commute up to isomorphism if we consider more complicated examples than $\mathrm{Vect}$. For (linear) endofunctors of a fusion category, we could study the image of each simple object $X_i$, decompose it into simples and study the matrix with the entries $\dim \mathcal{C}(FX_i, X_j)$. When do all such matrices commute? It's unclear to me, but possibly one can understand it if one restricts the functors further (to make them pivotal, for instance). One can go on and ask when the endofunctors form a ribbon or a fusion category. When is it modular? When is it symmetric? Is there anything known about that? A similar question has been asked by Ben Sprott: symmetric monoidal dagger endofunctor categories Edit: I specialised to monoidal categories, which makes more sense from the perspective of the periodic system of higher categories. Remark: Here is more background to my question. I'm interested in actions of braided monoidal categories on monoidal categories. I don't know whether this has been worked out explicitly (and I'd be happy for a reference) but it's not too hard to write down the axioms, you just have to think of the braided category as a special kind of tricategory. Now an action $\mathcal{B} \times \mathcal{M} \to \mathcal{M}$ is about the same as a functor $\mathcal{B} \to \mathrm{MonCat}(\mathcal{M})$, the latter denoting some category of endofunctors. (Of course we have to take care of what the precise axioms are and what kind of endofunctors to allow, but let's assume we can make that precise.) Then there has to be some compatibility structure (not just data) for the braiding $\mathcal{B}$ and the monoidal structure on $\mathcal{M}$. I thought that this compatibility structure must be related to a braiding on $\mathrm{MonCat}(\mathcal{M})$, and the functor $\mathcal{B} \to \mathrm{MonCat}(\mathcal{M})$ would have to be a braided functor. REPLY [6 votes]: This is in some sense an unnatural question to ask. Endofunctors only form a monoidal category in general, and if you want a braiding, that's not just an extra property: it's extra structure. Where would it come from? In what sense would it be unique or natural? You've been misled by the example of vector spaces. Here is some context to put it in: pick a commutative ring $k$, and consider the ("Morita") 2-category whose objects are $k$-algebras $A$, to be thought of as standing in for their cocomplete $k$-linear categories $\text{Mod}(A)$ of right $A$-modules; 1-morphisms are $(A, B)$-bimodules $M$ over $k$, to be thought of as standing in for cocontinuous $k$-linear functors $\text{Mod}(A) \to \text{Mod}(B)$ (by the Eilenberg-Watts theorem), and 2-morphisms are morphisms of bimodules. This 2-category is naturally symmetric monoidal under the tensor product of $k$-algebras, and the symmetric monoidal category $\text{Mod}(k)$ of $k$-modules itself naturally appears as endomorphisms of the object $k$. What's special about $k$ as an object in this 2-category is that it is the monoidal unit for the tensor product. In general, endomorphisms of the unit object in an $E_n$ monoidal (higher) category naturally form an $E_{n+1}$ monoidal (slightly lower) category. This is a version of the Deligne conjecture. You can think of this construction as a "(based) loop space" construction. Edit: I'm interested in actions of braided monoidal categories on monoidal categories. I don't know whether this has been worked out explicitly (and I'd be happy for a reference) but it's not too hard to write down the axioms, you just have to think of the braided category as a special kind of tricategory. There is a notion of action of a braided monoidal category $B$ on a monoidal category $M$, but the objects of $B$ are not acting as endofunctors, monoidal or otherwise, of $M$. (Think about the decategorified question: what does it mean for a commutative ring to act on a ring? Certainly not by ring endomorphisms, which only form a monoid in general, not even a noncommutative ring.) Instead, the objects of $B$ act as endomorphisms of the identity functor $\text{id}_M : M \to M$. Explicitly, the collection of all such endomorphisms is the Drinfeld center $Z(M)$ of $M$. This is the object associated to $M$ that naturally has a braided monoidal structure (a special case of the assertion I made above: $\text{id}_M$ is the unit object in the monoidal 2-category of monoidal endofunctors $M \to M$), and an action of $B$ on $M$ is a braided monoidal functor $B \to Z(M)$. (So, the answer to the decategorified question: an action of a commutative ring $k$ on a ring $R$ is a morphism $k \to Z(R)$.)<|endoftext|> TITLE: How do solutions of a PDE depend on parameters? QUESTION [6 upvotes]: Let $\Omega\subset\mathbb R^n$ be a bounded smooth domain and $\sigma_1,\sigma_2:\Omega\to(c^{-1},c)$ measurable (for some constant $1 TITLE: The Maximal $\ell_2$ norm of a signed sum of vectors QUESTION [5 upvotes]: Suppose we have $n$ vectors in $\mathbb{R}^n.$ Consider the signed sum of these vectors: $$U(s_1,\ldots,s_n)=s_1 v_1+s_2 v_2 + \ldots + s_n v_n$$ where $s_j$'s can only take values of $+1$ or $-1.$ I am interested in the maximal $\ell_2$ norm of the vector $U$ over all possible values of $(s_1,\ldots,s_n).$ This maximal $\ell_2$ norm of $U$ is certainly a function of $v_1,\ldots,v_n.$ For example, when $n=2,$ an easy argument in geometry shows that this maximal $\ell_2$ norm is proportional to the largest singular value of the matrix $V$ with $v_1,\ldots,v_n$ as columns. However, this is not true for $n=3$. I was wondering whether there is any existing result on this maximal $\ell_2$ norm as a function of the matrix $V,$ or is there an algorithm that solves this problem in linear time. In particular, I was wondering what this function is for $n=3.$ Is it some function of the singular values? Thanks! REPLY [6 votes]: This is, in essence, the most general form of the zero-one quadratic programming problem, and is known to be NP-complete. (see, for example, Computational Aspects of a Branch and Bound Algorithm for Quadratic Zero-One Programming by Pardalos and Rogers in Computing, 1990). Of course, this has not stopped mankind from developing reasonably efficient algorithms in practice.<|endoftext|> TITLE: Complete L-function and FE of Rankin-Selberg on GL(2)? QUESTION [7 upvotes]: Let $f$ be a Maass cusp form of $\Gamma_0(N)$ on the upper half plane with character $\chi$ mod $N$ and eigenvalue $1/4+\mu^2$. What is the complete $L$-function of the Rankin-Selberg product $L(s,f\times \tilde f)$? I am having trouble writing down the local part at $p|N$. I can't find good references either. REPLY [7 votes]: See here for the answer (in particular, it's in section 1 and 2). Basically, the answer isn't pretty - it depends on the local components of the automorphic representation $\pi$ of $\mathrm{GL}_2(\mathbb{A}_{\mathbb{Q}})$ associated to the Maass form $f$. So if $p \nmid N$, then the local component $\pi_p$ of $\pi$ is a principal series representation $\omega_{p,1} \boxplus \omega_{p,2}$, where $\omega_{p,1}, \omega_{p,2}$ are unramified characters of $\mathbb{Q}_p^{\times}$. We have that \[L_p(s,\pi_p) = \frac{1}{(1 - \omega_{p,1}(p) p^{-s})(1 - \omega_{p,2}(p) p^{-s})}\] where $\omega_{p,1}(p) + \omega_{p,2}(p) = \lambda_f(p)$, the Hecke eigenvalue of $f$ at $p$, and $\omega_{p,1}(p) \omega_{p,2}(p) = \chi(p)$. Then the local Rankin-Selberg $L$-function is \[L_p(s,\pi_p \times \widetilde{\pi_p}) = \frac{1}{(1 - |\omega_{p,1}(p)|^2 p^{-s})(1 - \omega_{p,1}(p) \overline{\omega_{p,2}(p)} p^{-s})(1 - \overline{\omega_{p,1}(p)} \omega_{p,2}(p) p^{-s})(1 - |\omega_{p,2}(p)|^2 p^{-s})}.\] If $p \mid N$ and $\pi_p$ is a principal series representation $\omega_{p,1} \boxplus \omega_{p,2}$, where at least one of $\omega_{p,1}, \omega_{p,2}$ is ramified, then the same result holds as for the case $p \nmid N$ with the minor change that if $|\omega_{p,1}|^2$ is a ramified character of $\mathbb{Q}_p^{\times}$, we replace $|\omega_{p,1}(p)|^2$ with $0$, and similarly if $\omega_{p,1} \overline{\omega_{p,2}}$, $\overline{\omega_{p,1}} \omega_{p,2}$, or $|\omega_{p,2}|^2$ is ramified (see Jeremy's comment below). Note that if both $\omega_{p,1}$ and $\omega_{p,2}$ are ramified then $p^2 \mid N$. This is part of Proposition 1.4 of the linked paper. If $p \mid N$ and the local component $\pi_p$ of the automorphic representation is supercuspidal (which can only happen if $p^2 \mid N$), then $\lambda_f(p) = 0$ and \[L_p(s,\pi_p) = 1,\] and either \[L_p(s,\pi_p \times \widetilde{\pi_p}) = \frac{1}{1 - p^{-s}}\] or \[L_p(s,\pi_p \times \widetilde{\pi_p}) = \frac{1}{1 - p^{-2s}}.\] The former occurs if $\pi_p \ncong \pi_p \otimes \eta_p$, where $\eta_p$ is the unramified quadratic character of $\mathbb{Q}_p^{\times}$, while the latter occurs if $\pi_p \cong \pi_p \otimes \eta_p$. This is Corollary 1.3 of the linked paper. If $p \mid N$ and the local component $\pi_p$ is a Steinberg representation $\omega_p \mathrm{St}_p$, where $\omega_p$ is a unitary character of $\mathbb{Q}_p^{\times}$, then $\widetilde{\pi_p} = \omega_p^{-1} \mathrm{St}_p$. We have that \[L_p(s,\pi_p) = L_p\left(s + \frac{1}{2}, \omega_p\right)\] and \[L_p(s,\pi_p \times \widetilde{\pi_p}) = \zeta_p(s + 1) \zeta_p(s) = \frac{1}{(1 - p^{-s-1})(1 - p^{-s})}.\] This is part of Proposition 1.4. Note that if $\omega_p$ is unramified, we must have that $p \parallel N$, $\lambda_f(p) = \omega_p(p) p^{-1/2}$, and so \[L_p(s,\pi_p) = \frac{1}{1 - \omega_p(p) p^{-s - 1/2}},\] If $\omega_p$ is ramified, then $p^2 \mid N$, $\lambda_f(p) = 0$, and \[L_p(s,\pi_p) = 1.\] In particular, if $N$ is squarefree and $\chi$ is the principal character, then for $p \mid N$, $\pi_p$ must be $\omega_p \mathrm{St}_p$ with $\omega_p$ the trivial character, so $\lambda_f(p) = p^{-1/2}$. Finally, the archimedean $L$-function is \[L_{\infty}(s,\pi_{\infty}) = \pi^{-s - \kappa} \Gamma\left(\frac{s + i\mu + \kappa}{2}\right) \Gamma\left(\frac{s - i\mu + \kappa}{2}\right),\] where $\kappa = 0$ if $f$ is even and $1$ if $f$ is odd, and \[L_{\infty}(s,\pi_{\infty} \times \widetilde{\pi_{\infty}}) = \pi^{-2s} \Gamma\left(\frac{s + 2i\mu}{2}\right) \Gamma\left(\frac{s}{2}\right)^2 \Gamma\left(\frac{s - 2i\mu}{2}\right).\] The completed $L$-function $\Lambda(s,f \times \widetilde{f})$ is then defined to be \[\Lambda(s,f \times \widetilde{f}) = L_{\infty}(s,\pi_{\infty} \times \widetilde{\pi_{\infty}}) \prod_p L_p(s,\pi_p \times \widetilde{\pi_p}).\] Theorem 2.2 of the linked paper states that this extends meromorphically to the entire complex plane with only simple poles at $s = 0$ and $s = 1$.<|endoftext|> TITLE: Choice of fibrations is like a choice of a basis of a module QUESTION [5 upvotes]: In some notes on derived stacks, in describing categories of fibrant objects, the author drops this parenthetical: (Grothendieck said in his famous letter to Quillen that the choice of $\mathscr F$ is like the choice of a basis of a module.) Question 1: Where can I find the quote that this refers to? What is the wording of this quote? Question 2: Can someone explain this analogy? My understanding of model category related stuff isn't good enough for me to understand this. REPLY [5 votes]: You can also find this quote in the first paragrah of section 1 of the letter of Grothendieck to Thomason (a pdf can be found at http://webusers.imj-prg.fr/~georges.maltsiniotis/groth/ps/lettreder.pdf). "Les constructions homotopiques essentielles sont indépendantes de toutes structures supplémentaires, tel un ensemble C de cofibrations ou un ensemble F de fibrations ou les deux à la fois. De telles structures supplémentaires sont utiles, dans la mesure où elles permettent d'expliciter les constructions essentielles, et d'en établir l'existence. Mais elles ne sont pas plus essentielles pour le sens intrinsèque des opérations (qu'elles auraient tendance plutôt à obscurcir, jusqu'à présent) que le choix d'une base plus ou moins arbitraire d'un module, en algèbre linéaire."<|endoftext|> TITLE: Classification of $SU(2)$-bundles versus the classification of $SO(3)$-bundles QUESTION [6 upvotes]: As explained in: Classification of $SU(2)$ principal fibre bundles over four-dimensional manifolds principal $SU(2)$ bundles $P_{SU(2)}$ over a four-dimensional manifold $M$ are classified by their second Chern-class $c_{2}(P_{SU(2)})\in H^{4}(M,\mathbb{Z})$. Usually, when solving the Yang-Mills equations $\nabla F =0$, $\nabla\ast F =0$ in $P_{SU(2)}$, where $F$ is the curvature of the connection, one goes to a local patch in the base manifold $U$ and using a local section $s$ one works with the pullback $s^{\ast}A$ of the connection to $U$, and then one solves the equations locally in $U$. The pullback of the connection $s^{\ast}A$ is a one-form in $M$ with values in $\mathfrak{su}(2)$. But of course we have $\mathfrak{su}(2)\simeq so(3)$, so locally one could not distinguish if one is dealing with a $SU(2)$-bundle over $M$ or a $SO(3)$-bundle $P_{SO(3)}$ over $M$, at least at the level of locally solving the Yang-Mill equations for the connection. My question is then, how different is the topological classification of $SO(3)$ bundles over $M$ from the classification of $SU(2)$ bundles? In addition, it looks that one can use a local solution to the Yang-Mill equations of motions for a $SU(2)$-bundle in a $SO(3)$-bundle an vice-versa. Thanks. REPLY [9 votes]: You are correct that locally, there is no difference between SU(2) and SO(3) bundles, but there are important global differences. In particular, SO(3) bundles may have non-trivial $w_2$, which obstructs their lifting to an SU(2) bundle. The Dold-Whitney theorem (Classification of oriented sphere bundles over a 4-complex. Ann. of Math. (2) 69 1959 667–677) says that the topological classification of SO(3) bundles is given exactly by $w_2$ and the first Pontrjagin class $p_1$. The only constraint on the two classes is that $p_1$ is the (mod 4 valued) Pontrjagin square of $w_2$; compare The Stiefel-Whitney and Pontryagin classes of SO(3)-bundles. There are also some subtle differences in the topology of the gauge group. In mathematical gauge theory, as applied to 4-manifold topology, this is an important distinction. In many circumstances, one can prove that the moduli space of anti-self-dual SO(3) connections is compact, leading to somewhat simpler proofs of some of Donaldson's fundamental theorems. This was observed by Fintushel and Stern (SO(3)-connections and the topology of 4-manifolds, J. Differential Geom. 20 (1984), 523-539).<|endoftext|> TITLE: Functor generalisation QUESTION [5 upvotes]: In an article I am writing, I am led to the following generalization of the notion of functor. Let $C$ and $D$ and be two categories. A generalized functor $F : C \to D$ is given by: a function $f : C_0 \to D_0$ for all $c \in C_0$, a set $F_c$ for all $c_1,c_2 \in C_0$, a function $F_{c_1,c_2}:C(c_1,c_2) \times F_{c_2} \to F_{c_1} \times D(f(c_1),f(c_2))$. All this must be compatible with the product and units, that is: for all $c \in C$ and all $x \in F_c$, $F_{c,c} (1_c, x) = (x,1_{f(c)})$ for all $c_1,c_2,c_3 \in C$, the following composites are equal: $$C(c_1,c_2) \times C(c_2,c_3) \times F_{c_3} \to C(c_1,c_3) \times F_{c_3} \to F_{c_1} \times D(f(c_1),f(c_3)) $$ and $$C(c_1,c_2) \times C(c_2,c_3) \times F_{c_3} \to C(c_1,c_2) \times F_{c_2} \times D(f(c_2),f(c_3)) \to F_{c_1}\times D(f(c_1),f(c_2))\times D(f(c_2),f(c_3)) \to F_{c_1}\times D(f(c_1),f(c_3))$$ In particular, a functor is a generalized functor, where every $F_c$ is the singleton set. Are you aware of a similar structure arising somewhere? One way to see this structure is the following. One can see objects $a,b$ of a category as points, and then see $C(a,b)$ as an arrow form $a$ to $b$. If you say that the composition of two arrows is given by the Cartesian product, then you can see the composition in $C$ as a $2$-arrow from $C(a,b) \times C(b,c) \to C(a,c)$. Now how do you represent a functor in this setting? You can't represent the function $C(a,b) \to D(f(a),f(b))$ by a $2$-arrow, because the 'arrows' $C(a,b)$ and $D(f(a),f(b))$ are not parallel (the first one goes from $a$ to $b$, while the second goes from $f(a)$ to $f(b)$). One way to deal with that is to add two arrows: one called $F_{a}$ from $a$ to $f(a)$ and the other called $F_{b}$ from $b$ to $f(b)$. Now we have a square that we can fill with a $2$-arrow $F_{a,b}$. If we keep in mind the correspondence 'arrows are sets' and '$2$-arrows are functions' from the previous paragraph, we obtain the definition of a generalized functor. Two more things to note: One can define generalized natural transformations between generalised functors. The usual naturality arrows $\eta_a : f(a) \to g(a)$ are then replaced by functions $G_a \to F_a \times D(f(a),g(a))$: This can be extended to $n$-functors by replacing the sets $F_a$ by $(n-1)$-categories and the functions $F_{a,b}$ by $(n-1)$-functors. REPLY [2 votes]: Here is one way to look at it: if V is a monoidal category and $\mathbf{B} V$ is the corresponding one-object bicategory, then a V-category in the usual sense is the same thing as a lax functor $\mathrm{c}(C_0) \to \mathbf{B} V$, where $C_0$ is a set and $\mathrm{c}(\cdot)$ is the fully faithful functor that takes a set to the codiscrete category on it (i.e. the one with $C_0$ as object set and exactly one morphism between each pair of objects). If C and D are V-categories then a V-functor from C to D is given by a function $f \colon C_0 \to D_0$ together with an identity-component oplax natural transformation (an icon) $\hat C \to \hat D \circ \mathrm{c}(f)$, where $\hat C, \hat D$ are the lax functors corresponding to C and D. What you have seems to be a general oplax transformation, without the condition that its components be identities. For that reason it's probably a not-unreasonable definition to write down, although I for one have never seen an example in the wild. At the end of your question you are getting somewhere close to the notion of a category enriched in a bicategory, see here.<|endoftext|> TITLE: Are the p-adics a direct summand of the direct product of the groups $\mathbb{Z}/p^n\mathbb{Z}$? QUESTION [23 upvotes]: The p-adic integers $\mathbb{Z}_p$ can be thought of as a subgroup of the direct product group $P = \prod_{n \geq 1} \mathbb{Z}/p^n\mathbb{Z}$. Are they a direct summand of this group? That is, is the inclusion $\mathbb{Z}_p \hookrightarrow P$ split? REPLY [11 votes]: Let me give a bit of an algebro-geometric perspective on Will's excellent answer, which I note gives not just a splitting of abelian groups but actually a ring homomorphism as well. Let's write $R=\prod \mathbb{Z}/p^n$ and try to understand $\operatorname{Spec} R$. Suppose we have a field $k$ and a homomorphism $f:R\to k$. For any subset $A\subseteq \mathbb{N}$, let $1_A\in R$ be the element that is $1$ on coordinates in $A$ and $0$ on other coordinates. This element is an idempotent, so $f(1_A)$ must be $0$ or $1$. If we write $U$ for the set of all $A$ such that $f(1_A)=1$, then it is easy to see that $U$ is an ultrafilter. This defines a continuous map $\operatorname{Spec} R\to \beta\mathbb{N}$, where $\beta\mathbb{N}$ is the space of all ultrafilters on $\mathbb{N}$. The fiber over any principal ultrafilter is easily seen to be a single point, corresponding to the projection from $R$ to a single coordinate followed by the quotient map to $\mathbb{F}_p$. Over a nonprincipal ultrafilter $U$, the fiber can be identified with $\operatorname{Spec} R_U$, where $R_U$ is the quotient of $R$ by the elements $1-1_A$ for $A\in U$; alternatively, $R_U$ is the ultraproduct $\prod_U \mathbb{Z}/p^n$. For any $n$, the canonical map $\mathbb{Z}/p^n\to R_U/p^n$ is an isomorphism (because this property can be expressed in the first-order theory of rings and is true on all sufficiently large coordinates of the product), and so there is a unique map $R_U\to \mathbb{Z}_p$. This map is exactly the splitting which Will describes in his answer, and this shows that Will's splittings are the only possible splittings which are ring homomorphisms. In particular, it is surjective, so we obtain a canonical copy of $\operatorname{Spec} \mathbb{Z}_p$ as a closed subset of the fiber over each nonprincipal ultrafilter $U$. Note, however, that $\operatorname{Spec} R_U$ is a lot larger than $\operatorname{Spec}\mathbb{Z}_p$. In particular, the kernel of the map $R_U\to \mathbb{Z}_p$ includes all elements such that the divisibility of their coordinates by $p$ goes to infinity with respect to $U$. Any such element for which the divisibility goes to infinity slower than $n^{1/k}$ for all $k$ will be non-nilpotent, and in fact it is not hard to construct uncountably many prime ideals in $R_U$ corresponding to different rates of convergence to infinity (compare with my example at the end of this answer). However, we can at least say that every prime ideal of $R_U$ either comes from $\mathbb{Z}_p$ or is contained in the kernel of $R_U\to\mathbb{Z}_p$: any element not in the kernel is of the form $p^nu$ where $u$ is a unit, and $p$ generates the unique maximal ideal of $R_U$.<|endoftext|> TITLE: Help with the integral $\int_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx$ QUESTION [5 upvotes]: We have the integral : $$\int_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx$$ Where s is a complex parameter, and n is a positive integer. The integral converges by virtue of the exponential factor. I tried to deform the path of integration such that we avoid the branch cut(s) of the logarithm. But here is where i got stuck, the internal complex log makes it confusing to do so ! A different version of this problem was posted here with no answers. REPLY [8 votes]: $$I_n(s)=\int_{0}^{\infty}\log\left(1+\frac{s^{2}}{4\pi^{2}} \log^{2}(1+ix)\right ) e^{-2\pi nx}dx$$ a closed-form evaluation of this integral does not look promising, but small and large-$|s|$ asymptotics is doable: small $|s|$ (with $\gamma$ Euler's constant and $_3F_3$ the generalized hypergeometric function): $$I_n(s)=\frac{s^2}{48\pi^3 n}\left(24 i \pi n \; _3F_3(1,1,1;2,2,2;2 i n \pi )-\tfrac{3}{2} [\pi +2 i \log (2 \pi n)] [2 i \log (2 \pi n)+4 i \gamma +\pi ]+6 \gamma ^2+\pi ^2\right)+{\rm order}(|s|^4)$$ large $|s|$: $$I_n(s)=\frac{1}{\pi n}\log(s)+{\rm order}(|s|^0)$$ and for large $n$ the integral decays as $1/n^3$, $$\lim_{n\rightarrow\infty} n^3 I_n(s)=-\frac{s^2}{16\pi^5}$$<|endoftext|> TITLE: Why does the Gluck twist on a spun knot give the standard $S^4$? QUESTION [5 upvotes]: Given a knotted arc $A \subset D^3$ (whose endpoints are, say, at $(\pm 1,0,0)$), the spun knot on this arc is $$\partial\left((D^3, A) \times D^2\right), = (\partial(D^3,A) \times D^2) \cup ((D^3,A) \times \partial D^2),$$ a smoothly embedded 2-sphere in $S^4$. The Gluck twist on a smoothly embedded 2-sphere $S^2 \hookrightarrow S^4$ deletes a tubular neighborhood $\nu(S^2)$, giving a closed manifold with boundary $S^2 \times S^1$. Reglue $\nu(S^2)$ via the diffeomorphism $S^2 \times S^1 \to S^2 \times S^1, (x,t) \mapsto (\alpha(t)x, t)$, where $\alpha$ is the nontrivial element of $\pi_1(SO(3))$ (ie, $\alpha(t)$ is rotation by the angle $t$). This gives a homotopy 4-sphere. Gluck claims, in "Embeddings of two-spheres in the four-sphere", that on a spun knot, this construction results in $S^4$. I don't understand his argument, which is (essentially) as follows. "If the two-sphere $S^2 \times 0$ on the boundary of $S^4 \setminus \nu(S^2)$ is itself the boundary of a three-sphere with handles, "nicely" situated in $S^4 \setminus \nu(S^2)$, then $M^4$ is homeomorphic to $S^4$." He then claims spun knots as a particular case. (It should be possible to replace homeomorphic here with diffeomorphic, if we can see this geometrically - i.e., without invoking Freedman's theorem. References to this particular case agree that the resulting smooth manifold is the standard $S^4$.) What he calls $S^2 \times 0$, I assume, just means $S^2 \times \{pt\}$, the latter some point in $S^1$. I do not understand what he means by a "three-sphere with handles" - and this term does not appear elsewhere - nor do I see why he should conclude that $M^4$ is homeomorphic to $S^4$. (I know there are more general results involving twist-spun knots and 0-concordant knots, but I'd like to understand this argument.) REPLY [8 votes]: This was shown by Gluck, in the cited paper; see section 22. The basic point is explained in section 17. (What we now call) the Gluck twist will produce an equivalent knot if the circle action on the 2-sphere extends over some 3-manifold that the knot bounds; this is Theorem 17.1 but I recommend that you try to prove it yourself before looking. For the spin of a knot K in $S^3$, you can cook up such a 3-manifold by `spinning' any Seifert surface that K bounds. I didn't see the phrase you mention (three-sphere with handles) but I presume that it means a punctured connected sum of copies of $S^1\times S^2$. Note that you can readily find the extension of the circle action over such a manifold. Indeed, $S^1\times S^2$ has a circle action with two circle fixed points. Forming equivariant connected sums gives you such an action on $\#^n (S^1\times S^2)$, and you can just remove an invariant neighborhood of any point on a circle of fixed points. Finally, a spun knot always bounds a punctured $\#^n (S^1\times S^2)$.<|endoftext|> TITLE: Brandt's definition of groupoids (1926) QUESTION [17 upvotes]: The definition of a category is usually attributed to Mac Lane and Eilenberg (1945). What seems to be less known is that the german mathematician Heinrich Brandt has developed the notion of a groupoid already in 1926 (motivated by questions on quadratic forms). The paper "Über eine Verallgemeinerung des Gruppenbegriffes" introduces, names and studies (connected) groupoids explicitly. The object-free definition is used, so that groupoids look very much like groups except that the product is only defined partially. Well, first of all I have to say that it is quite amazing that the definition of a category is basically already there in Brandt's paper. Just erase the inverse elements from Axiom III. The paper doesn't really mention the arrow picture stressed by Mac Lane and Eilenberg, although the notions "einander rechts" (sharing the same codomain) and "einander links" (sharing the same domain) are introduced for groupoid elements aka morphisms. See here for a list of further early publications on groupoids. Nowadays, groupoids are usually seen as special categories. (Curiously, in homotopy type theory, categories are seen as special $\infty$-groupoids.) But the definition of a groupoid has appeared 20 years before the definition of a category. This leads to many (related) questions: Why did it take 20 years? Does the study of groupoids have influenced the development of the notion of a category? Did Mac Lane and Eilenberg know the work on groupoids? Is Brandt's work rather unknown? (Why?) Or have you heard of it before? REPLY [2 votes]: I would like to address a mathematical, and not historical, point. Theer is a vast difference between the theories of infinity-groupoids (which may, for example, be realized as Kan complexes) and of infinity-groupoids (which may, for example, be realized as Rezk's complete Segal spaces). The former takes just a few pages (at most a chapter) to set up, the latter a whole book. I doubt that the prior investigation of groupoids has much relevance to the development of category theory, either historically or from the point of view of what the important questions are.<|endoftext|> TITLE: Can a finite von Neumann algebra be strongly morita equivalent to a properly infinite von neumann algebra? QUESTION [7 upvotes]: Can a finite (by finite I mean when the projection $1$ is finite) von Neumann algebra be strongly morita equivalent to a properly infinite von Neumann algebra? (Strong morita equivalence is the same as Morita equivalence as a ring for $C^*$-algebras according to the theorem on page 253 of http://www.sciencedirect.com/science/article/pii/0022404982901098, see remark 1.5 for definition of strong Morita equivalence) REPLY [7 votes]: Assume $A$ and $B$ are two unital $C^*$-algebras which are strongly Morita equivalent. This means that there exists a Hilbert $A$-module $H$ such that $B$ is the algebra of compact operators on $H$. But as $B$ is unital it implies that the unit of $B$ has to act as the identity of $H$ (because $B$ contains all "rank one operators") hence the identity of $H$ is a compact operator. One now has the following lemma: Lemma : For any $C^*$-algebra $A$, If the identity of $H$, a Hilbert $A$-module, is compact then $H$ is a direct (orthogonal) summand of $A^n$ for some integer $n$ This implies that $B$ is a corner of $M_n(A)$. Hence if $A$ is a finite von Neuman algebra, $B$ also is: $M_n(A)$ is finite* and any projection of a finite von Neuman algebra is finite hence $B \simeq p.M_n(A).p$ is also finite. Proof of the lemma: The identity of $H$ is compact, hence it can be approximated by a "finite rank operator" and if this approximation is close enough it will be invertible. Hence there exists an operator $k:A^n \rightarrow H$ such that $kk^*$ is invertible. Let $v = (kk^*)^{-1/2}$, then $p=vk$ is an operator $A^n \rightarrow H$ such that $pp^* = Id$ and hence $p^* p $ is a symetric projection and $H \simeq_p (p^*p).A^n$ is a direct summand. *: Note that the fact that $M_n(A)$ is finite is well known but non trivial. It fails for general $C^*$-algebras, and corresponds to the fact (proved in Murray and Von neuman original paper) that in von Neuman algebras the sum of two orthogonal finite projection is again a finite projection.<|endoftext|> TITLE: Characterising subsets of the reals as ordered spaces QUESTION [6 upvotes]: There are concise and elegant characterisations of the real line as a topological space and as an ordered space in the literature. I am interested in the harder case of characterising subsets of the reals in this manner. There are satisfactory answers to the topological version (e.g., de Groot, Mary Ellen Rudin) which are, as to be expected, more complicated and inticrate in proof than for the whole space. I recall reading a solution for the corresponding result in the category of ordered spaces but the standard search methods have failed to locate it. Can anybody on this site assist me with a reference? REPLY [4 votes]: Just for the record, the result seems to be due to Isodore Fleischer ("Numerical representation of utility", Jour. Soc. Ind. Appl. Math, 9 (1961) 48-50). As the title indicates, it is useful in determining when a suitable ordering on a state space is induced by a numerical function (price, utility function, temperature, entropy) and unifies many such results in fields such as economics, thermodynamics, philosophy ... This (surprisingly late) reference is the earliest one that I can trace.<|endoftext|> TITLE: How do you *state* the Classification of finite simple groups? QUESTION [47 upvotes]: From the point of view of formal math, what would constitute an appropriate statement of the classification of finite simple groups? As I understand it, the classification enumerates 18 infinite families and 26 sporadic groups and asserts that a finite group is simple iff it is in one of these families. Now the 18 infinite families are all fairly clearly defined as cyclic groups, permutation groups, matrix groups over finite fields, etc. so I don't think there is much difficulty in defining these precisely. Much more problematic are the sporadic groups, which are "known" and hence apparently need no definition. To give an example, since the monster group is some finite object we could just write down its Cayley table and define that to be the monster group. There are two big problems with this: (1) this table is huge and redundant, and (2) it's not easy to work with this table to prove properties of it. The main problem is that we don't think about the monster group in terms of its Cayley table, nor even as the group generated by a certain pair of $196882^2$ matrices. Instead we view it as a specific group which satisfies some properties and is uniquely defined by those properties; presumably it is in this context that a given sporadic group will show up in the course of the classification proof. My problem is that I have no idea what those characterizing properties are. Indeed under some definitions it would rather weaken or trivialize the statement of classification, for example if I defined the sporadic groups as the simple groups that are not in the 18 families. What definition of these objects is actually used in the proof? (Side question: 16 of the 18 families are usually collected under one label, the "groups of Lie type". Is this class definable in some uniform way, or are the definitions individualized and the name is just due to some commonalities we recognize between these families?) REPLY [35 votes]: There are really two separate questions that you seem to be conflating here. The first is how to state the CFSG in a way that could be mechanically formalized. The second is how to state the CFSG that adequately reflects how human mathematicians think about it. For the former question, one straightforward possibility for the sporadic groups, since we know their orders, is simply to state something like, "There exists a unique simple group, not in one of the aforementioned families, of each of the following orders: 7920, 95040," etc. This is the barest possible statement that could count as a classification theorem, and for a computer, it provides (in principle) enough information to reconstruct the groups in question. For the second question, though, there's no sharp boundary demarcating where the classification theorem ends and the detailed study of the properties of the sporadic groups begins. There's also no canonical way of describing a particular group of interest in a way that satisfies a human that he or she now "knows what the group is." But there's nothing unique about group theory here. Any sufficiently large and complicated mathematical object is going to suffer from this problem. There will be some bare-minimum way of referring to it that in principle picks it out from the amorphous universe of all mathematical objects but that fails to answer basic questions about it. There will be a continuum of theorems that answer other basic questions, shading off into questions that we can't answer. It is a matter of opinion how many questions we have to be able to answer before we can claim to have "adequately described" the object.<|endoftext|> TITLE: Covering of schemes and flatness QUESTION [5 upvotes]: Let $f:X \to Y$ be a finite surjective morphism of quasi-projective schemes over $\mathbb{C}$, $X$ is reduced and $Y$ is integral. Suppose that there exists an integer $n$ such that for every closed point $y \in Y$, the fiber $f^{-1}(y)$ is reduced and consists of $n$ distinct closed points. Is it true that $f$ is flat? REPLY [2 votes]: Apply Hartshorne, Exercise II.5.8 to $f_*\mathcal{O}_X$.<|endoftext|> TITLE: applications of Berkovich spaces QUESTION [12 upvotes]: What are applications of the theory of Berkovich analytic spaces? The analytification $X \mapsto X^{\mathrm{an}}$ REPLY [7 votes]: Let me add a few more applications to what has already been mentioned. Relation with Bruhat-Tits buildings (Berkovich, then Rémy/Thuillier/Werner). If $G$ is a reductive group over a non-archimedean valued field, then the Bruhat-Tits building $\mathscr{B}(G)$ of $G$ embeds into the analytification $G^{an}$ of $G$. If you choose a parabolic subgroup $P$, you have a map to $(G/P)^{an}$, which is the analytification of a proper variety, hence a compact space. This can help you compactify the building, describe the strata of the compactification, etc. Complex dynamics. Favre and Jonsson used Berkovich spaces (actually some instance of it that they call "valuative tree") in order to study the dynamics of a polynomial endomorphism of $\mathbb{C}^2$ near a superattracting fixed point or at infinity. Resolution of singularities (Thuillier). Let $X$ be an algebraic variety over a perfect field $k$ with an isolated singular point $x$. Let $f \colon Y \to X$ be a resolution of it such that $f^{-1}(x)$ is a normal crossing divisor $E$. Then the homotopy type of the incidence complex of $E$ is independent of the choice of the resolution. (Remark here that $k$ is any perfect field and that the Berkovich spaces that come up in the proof are over the field $k$ endowed with the trivial valuation.) $p$-adic differential equations. André used Berkovich spaces in order to prove a conjecture of Dwork on the logarithmic growth of the solutions of $p$-adic differential equations (he actually needs Berkovich spaces only when the base field is not locally compact). In a different direction, Baldassarri (then Pulita and myself, and also Kedlaya) took up the study of $p$-adic differential equations on Berkovich curves. For instance, in my work with Pulita, we manage to give conditions for the finiteness of the de Rham cohomology of a curve with coefficients in some module with a connection. (Here, my point is that even if you can easily state the results in any theory you like, rigid geometry for instance, you will have a hard time proving them without Berkovich spaces.)<|endoftext|> TITLE: Hamkins infinite time Turing machines: dovetailing ordinal time QUESTION [5 upvotes]: It is claimed in the Hamkins and Lewis founding article "Infinite time Turing machines" (proof of the gap existence theorem 3.4) that for $\omega$ steps of a computation of a machine performing a dovetailing on input $0$, $\omega$ steps are also performed in each of the simulated machines. How does this simulation work precisely? I would really appreciate to see the details of such a trick! Is it possible to generalise the technique to any clockable ordinal $\alpha > \omega$? If so, I wonder what kind of processes we can add to the dovetailing (e.g., checking properties, taking some additional time) so that "same time" still holds. What could be the limitations of such processes? Thanks for explanations in advance! REPLY [6 votes]: I'm glad to hear that you're reading my paper. This particular method of dovetailing infinitely many computations into one, however, is relatively standard in computability theoretic constructions, and doesn't have anything essentially to do with infinitary computability. Let me explain it a little. The idea is that you want to simulate infinitely many programs $p_0, p_1, p_2,\ldots$ on some specified input. First, by using a pairing function, you may think of the one-dimensional linear tape as encoding a matrix of infinitely many linear tapes, one for each program, with some extra space to organize the whole simulation. In this way, you effectively reserve plenty of space specifically for each program. Next, you begin the simulation, by starting to perform a few steps of $p_0$, then a few more steps of $p_0$ plus a few steps of $p_1$, then a few more steps of $p_0$, of $p_1$ and then of $p_2$, then a few steps of $p_0, p_1, p_2$ and $p_3$, and so on. Proceeding in this way, after infinitely many ($\omega$ many) steps of computation, you've performed $\omega$ many simulated steps in each of the computations $p_i$. Although during the simulation, the earlier programs seem always a little bit ahead, everybody catches up in the limit. The same idea works for any limit ordinal, because the universal computation that simulates all computations (on fixed input $0$, say) catches up at every limit ordinal. The infinitary context does require, however, that one must take care with a few issues in order that one may iterate such simulations past $\omega$. For example, one should perform the simulated computations in such a way that the limit configuration of the simulated tape is simulated properly by the limit process of the simulation limit. That is, at a limit stage, the tape of the simulator is updated by the limsup rule, and it should be arranged that this is also correctly updating the simulated computations at this limit as well. Also, there should be an appropriate flag set at limits so that the machine can recognize that the scratch tapes may have accumulating garbage at limit stages and especially at limits-of-limits. But these issues are easy enough to handle.<|endoftext|> TITLE: $\mathfrak{p}=\mathfrak{b}=\mathfrak{a}=\aleph_1$ and $\mathfrak{d}=\mathfrak{c}=\kappa$ QUESTION [5 upvotes]: Asumme tha in $M$, $CH$ holds and $\kappa>\aleph_0$ and $\kappa^{\aleph_0}=\kappa$. Let $K$ be $Fn(\kappa,2)$-generic over $M$. Question: Then we can say in $M[K]$ that: $(i)$ $\mathfrak{p}=\mathfrak{b}=\mathfrak{a}=\aleph_1$ and $\mathfrak{d}=\mathfrak{c}=\kappa$ ? Where $\mathfrak{c}=2^{\aleph_0}$ the size of the continuum. $\mathfrak{d}$, is the least size of a $\mathfrak{d}$ominating family. $\mathfrak{b}$, is the least size of an un$\mathfrak{b}$ounded family. $\mathfrak{p}$, is the least size of family $\mathcal{E}\subseteq [\omega]^\omega$ such that $\mathcal{E}$ has the SFIP and there does not exist any $\mathfrak{p}$seuod-intersection of $\mathcal{E}$. $\mathfrak{a}$, is the least size of an infinite m$\mathfrak{a}$d family. REPLY [8 votes]: Yes, these values are correct for the Cohen model. (Self-promotion: See the table in Section 11 of my chapter in the Handbook of Set Theory. The pre-publication version is on my web site at http://www.math.lsa.umich.edu/~ablass/hbk.pdf .)<|endoftext|> TITLE: How to calculate one Cauchy type determinant QUESTION [7 upvotes]: As we know, a Cauchy determinant of size n admits the following explicit formula: $$\det \left(\frac{1}{x _i+y _j}\right) _{1\le i,j \le n}=\frac{\prod _{1\le i < j\le n} (x _j-x _i)(y _j-y _i)}{\prod _{i,j=1}^n (x _i+y _j)}.$$ Is there something known about the following generalized Cauchy determinant? $$\det \left(\frac{A_i+B_j}{x _i+y _j}\right) _{1\le i,j \le n}.$$ More specially, how does it go for $$\det \left(\frac{A_i+A_j}{x _i+x _j}\right) _{1\le i,j \le n}.$$ A simple case is for $x_i=i$. I wonder if there are some references about them. Thank you. REPLY [2 votes]: A determinant of this type is discussed in these two publications on the two-dimensional square lattice Ising model on the rectangle: Hucht, Alfred, The square lattice Ising model on the rectangle. I: Finite systems, J. Phys. A, Math. Theor. 50, No. 6, Article ID 065201, 23 p. (2017). ZBL1357.81154. https://arxiv.org/abs/1609.01963 Hucht, Alfred, The square lattice Ising model on the rectangle. II: Finite-size scaling limit, J. Phys. A, Math. Theor. 50, No. 26, Article ID 265205, 23 p. (2017). ZBL1369.82005. https://arxiv.org/abs/1701.08722<|endoftext|> TITLE: Parodies of abstruse mathematical writing QUESTION [85 upvotes]: Perhaps under the influence of a recent question on perverse sheaves, in conjunction with the impending $\pi$-day (3/14/15 at 9:26:53), I recalled a long-ago parody of abstruse mathematical language that I can no longer remember in detail nor find by searching. I am not seeking merely "examples of colorful language," as in that earlier MO question, but rather parodies almost in the Alan Sokal Fashionable Nonsense sense (although I don't think he parodied abstract mathematics directly). I am partly motivated by the possible educational advantage of self-mockery (or self-awareness), tangentially related to an MESE question, "Wonder as Motivation." But I ask here to tap into the likely greater density of mathematicians working in abstract fields ripe for parody. Q. Can you provide examples of (or pointers to) intentionally comic parodies of abstruse mathematical language, written by knowledgeable mathematicians so that they could (in another universe) make mathematical sense. REPLY [4 votes]: The Qualifying Exam by Richard Roth (Mathematics Magazine 38(3):166–167 (May, 1965)) mentioned in Zhen Lin's comment seems worth posting as an answer. Excerpt: ALPHA: The candidate will please define what is meant by a continuous denominator. CANDIDATE: Consider the set of all doubly evocative singly homologous functions on the unit sphere. Introducing a continuous group structure in the usual way we may define the Skolem uniformity of automorphic cycles to be the theta relation on all sets of measure zero and the zeta function on left ideals whose valuation is Gaussian, uniformly on compacta. Then given any cardinal predicate, the continuous denominator is the corresponding normal quaternion for which the problem vanishes almost everywhere. BETA: Could the candidate please give an example of a non-Skolem uniformity? CANDIDATE: I believe the inversion of the reals under countable intersections is non-Skolem… at least almost everywhere. BETA: That’s correct. Now could you… OMICRON: (Interrupting) I wish to contradict. It isn’t a non-Skolem uniformity since the third axiom concerning the density of the seventh roots of unity is not in fact satisfied. BETA: Ah, yes, but you see, in my paper on toxic algebras… 1957… Journal of Refined Mathematics and Statistical Dynamics of the University of Lompoc… I showed that the third axiom need not be satisfied if the basis is countably finite and the metric is Noetherian, hence…<|endoftext|> TITLE: Motivational ideas for the Gelfand-Graev character of a finite group of Lie type QUESTION [6 upvotes]: I've been studying the Gelfand-Graev character's general construction for a finite group of Lie type. I wish to discuss its particularization in a seminar for the general linear group over a finite field, but I'm lacking some motivational ideas for introducing the concept in a simple fashion. In I.M Gelfand and M.I Graev's 1962 motivational paper "Construction of irreducible representations of simple algebraic groups over a finite field", Gelfand refers back to the general construction of irreducible representations of Dickson-Chevalley groups, which I would prefer to avoid since discussing it would easily go beyond the scope of the seminar. Also, I can't seem to relate the ideas discussed in Gelfand's paper with the construction presented in R. W Carter's extensive book Finite Groups of Lie Type: Conjugacy Classes and Complex Characters in chapter 8.1. I think that an answer to the following question will help me (I will be using the same notation as in Carter's book that I mentioned before): Let $G$ be a reductive group with connected center, $F$ a Frobenius morphism of $G$, $B$ an $F$-stable Borel subgroup, and $U$ its unipotent radical. How are the ideas from Gelfand's paper related to the fact that one is interested in inducing a linear nondegenerate character from $U^F$ to $G^F$? (I've asked about the nature of these nondegenerate characters in another post for the case where $U^F$ is the group of unitriangular matrices over a finite field). Thank you in advance. REPLY [9 votes]: Motivation or intuition is usually difficult to recover from older published work, but I think it helps a lot here to put the work of I.M. Gelfand and his collaborators in perspective. Finite groups of Lie type were never their main focus, and in any case by the 1970s the ideas of Deligne and Lusztig (developed much further by Lusztig and others) largely took over that subject. Gelfand looked much more broadly at representations of Lie groups and at linear algebraic groups over many kinds of local fields, along with finite fields. For Gelfand's own work it's useful to look at the English translations of various papers in volume 2 of the Collected Papers (Springer, 1988), especially those gathered in Part IV: Models of representations; representations of groups over various fields. This includes the short 1962 papers with Graev which introduced Gelfand-Graev characters. (Note that the Russian papers are now freely available online, but usually it's harder to get access to the translation journals -- and some of the translations are less reliable than others.) It's worth quoting the opening lines of a 1974 note by Bernstein-Gelfand-Gelfand A new model for representations of finite semisimple algebraic groups (in a sometimes non-idiomatic English translation): "An important problem in representation theory is to construct the so-called model, i.e. representations of the given group $G$ that contain almost every irreducible representation of $G$ exactly once." This theme was pursued for various classes of Lie groups (compact and noncompact) and finite groups of Lie type, where the technical methods naturally vary a lot. But roughly speaking the emphasis is placed on working with induced representations from a maximal unipotent subgroup. In the finite case, this presumably led to experimentation with inducing the simplest types of characters (those of degree 1) from a Sylow $p$-subgroup such as the upper triangular unipotent subgroup in a finite general linear group. Since the subgroup is relatively small in this case, it's of course not to be expected that such an induced character is irreducible. Indeed, if one induces the trivial character of a Borel subgroup such as the full triangular subgroup of a general linear group, the decomposition of the induced character is complicated to work out. But if the ambient algebraic group has a connected center, the more subtle idea of Gelfand-Graev is remarkably successful: inducing a "regular" (meaning most non-trivial) character of the unipotent group yields an induced character whose constituents all have multiplicity 1 and exhaust the "regular" characters of $G$. Connectedness of the center is essential for getting a unique character of $G$ regardless of which regular character of the unipotent group is used in the construction. (Gelfand-Graev could only prove the multiplicity 1 property in a special case, but Steinberg then provided a general proof given in Carter's book. It's worth looking at Steinberg's discussion toward the end of his last section 14; his 1967-68 Yale lectures are still available online here.) This notion of "regular" character was later shown by Lusztig to fit well with the Deligne-Lusztig theory, in terms of a more precise analogue of the notion of "regular" element in an algebraic group. See also Chapter 14 in the concise textbook by Digne and Michel Representations of Finite Groups of Lie Type (Cambridge, 1991).<|endoftext|> TITLE: Sets of natural numbers with finite intersections and divergent sums of reciprocals QUESTION [5 upvotes]: Does there exist an uncountable collection $\Lambda$ of infinite subsets of the set of natural numbers such that (i) any two distinct subsets in the collection have a finite intersection and (ii) the sum of the reciprocals is divergent for each $A \in \Lambda$? REPLY [2 votes]: First we partition $\mathbb N$ into a disjoint union of intervals each consisting of finitely many consecutive integers such that the sum of reciprocals in each interval exceeds 1. The collection of these intervals is a countably infinite set, which we denote by $\tilde{\Sigma}$. Next we consider the set $\Sigma'$ of square-free integers. (An integer $ n$ is square-free if $p^2$does not divide $n$ for any prime $p$. ) For every infinite subset $A = \{p_1 TITLE: When is $1+a+a^2+\dotsb+a^{{\rm ord}_n(a)-1}$ divisible by $n$? QUESTION [6 upvotes]: I posted this question on math.SE 10 days ago and had no answer, comment or vote. If an answer is not available, I could really use a reference point as well. For the sake of completeness, I am restating the essence of my question below, omitting details that can be found on the link above. I am interested in the following problem: For $n\in\mathbb{N}$, define the function $S_n:\mathbb{Z_n^*}\to\mathbb{Z_n}$ by $$ S_n(\bar a) := \bar 1 + \bar a + \bar a^2 + ...+ \bar a^{\left(ord_n(a)\right)-1}, $$ where $\mathbb{Z_n^*}$ is the set of all invertible elements of $\mathbb{Z_n}$, and $ord_n(a)$ the order of $a$ modulo $n$. Can a characterization for the sets $$ A_n:=\{ \bar a \in \mathbb{Z_n^*} : S_n(\bar a) = \bar 0 \} $$ be found? Now, considering the $\amalg {\mathbb{Z_n} }$ as consisting of the elements $(\bar a, n)$ where $n \in \mathbb{N}$ and $\bar a \in \mathbb{Z_n}$, we define the function $S : \mathbb{N} \to \amalg {\mathbb{Z_n} } $ such as \begin{align} S(n) = \left( \sum_{\bar a \in \mathbb{Z_n^*}} {S_n(\bar a)}, n \right)\end{align} thinking of $S(n)$ as an element of $\mathbb{Z_n}$ when there is no danger of confusion. This bring us to the second question. Can we find (or know more about) the set \begin{align} A:= \{ n \in \mathbb{N} : S(n) \in \mathbb{Z_n^*} \}\end{align} Thank you in advance. REPLY [4 votes]: Here is a partial answer to your first question: A necessary condition for $S_n(\bar a)=0$ to hold is that $\gcd(a-1,n)\mid{\rm ord}_n(a)$; moreover, if $n$ is square-free, then this condition is also sufficient. To see this, let $k:={\rm ord}_n(a)$ and $d:=\gcd(a-1,n)$, and write $a-1=db$ and $n=dm$. Then $S_n(\bar a)=0$ can be equivalently re-written as $$ 1+a+\dotsc+a^{k-1}\equiv 0\pmod{dm}; \tag{$\ast$} $$ this yields $1+a+\dotsc+a^{k-1}\equiv 0\pmod d$, and necessity follows now by observing that $a\equiv 1\pmod d$, resulting in $1+a+\dotsc+a^{k-1}\equiv k\pmod d$. This argument also shows that, conversely, $1+a+\dotsc+a^{k-1}\equiv 0\pmod{d}$ holds whenever $d\mid k$. Since $\gcd(d,m)=1$ for $n$ square-free, to prove sufficiency it suffices to show that for such $n$, we have $1+a+\dotsc+a^{k-1}\equiv 0\pmod{m}$. This is equivalent to $m\mid\frac{a^k-1}{db}$, or $dbm\mid a^k-1$, which further splits into the two conditions $db\mid a^k-1$ and $m\mid a^k-1$ in view of $\gcd(db,m)=1$ (following from $\gcd(d,m)=\gcd(b,m)=1$). Recalling that $db=a-1$ and and that $m\mid n$ and $k={\rm ord}_n(a)$, we see that, in fact, both $db\mid a^k-1$ and $m\mid a^k-1$ hold in a trivial way.<|endoftext|> TITLE: Does $E^{x,t}(f(X_T))$ solve a PDE if $f$ is not continuous? QUESTION [5 upvotes]: Many books [see below for references] explore the connections between partial differential equations and expectation values. Assume $X$ is a diffusion with generator $A$, then they conclude, that under certain conditions, the function $u(t,x):=E^{t,x}(f(X_T))$ is the solution to the following Cauchy problem: $$\begin{gather} \frac{\partial u}{\partial t}+Au=0, \text{ on } [t,T)\times\mathbb R \\ u(T,x) = f(x), \text{ on } \mathbb R\end{gather} \tag{1}$$ More genereal versions of this are usually called the Feynman-Kac theorem. Questions Do you have a reference where $f$ is not required to be continuous? I have never (expect for the case where $X$ is a Brownian Motion) found any theorem that did not require $f$ to be continuous. Why? Apart from rigorous treatments (as in the mentioned books), most people seem not to care about $f$'s continuity. (e.g. the Wikipedia entry on the Kolmogorov backward equations, and many more). Are they wrong? What about piecewise continous $f$? Is the problem to trivial and to specialized for the cited books? My understanding If for the process $X$ has a transition density $p(t,x,s,y)$ that is in $C^{1,2}$ for fixed $s,y$ (which seems to be the case under similar conditions as for the Feynman-Kac theorem), then $$ u(t,x) = \int f(y) p(t,x,T,y) dy $$ and $u$ is also in $C^{1,2}$ by Leibniz's integral rule for bounded $f$ and thus should also solve (1). Is this correct? References Oksendal, Stochastic Differential Equations Karatzas and Shreve, Brownian Motion and Stochastic Calculus Friedman, Stochastic differential equations and applications REPLY [4 votes]: For a probabilistic and analytic treatment to your questions check out: Chapter 1 of Second Order PDE's in Finite and Infinite Dimension. S. Cerrai. Springer, 2001 Chapter 2 of Analytical Methods for Markov Semigroups. L. Lorenzi, M. Bertoldi, CRC Press, 2006. respectively. I believe both references treat the Kolmogorov equation where time flows forward, not backwards as in (1), but that is a minor difference. For example, Cerrai shows that if the coefficients of the generator are of class $C^k$, then for any $t>0$ and for any $f \in B_b(\mathbb{R}^n)$, the function $P_t f(x) = \mathbb{E}_x (f(X_t))$ is $k$-times differentiable and its derivatives up to order $k$ are bounded in the supremum norm; for precise assumptions see Hypotheses 1.1-1.3. Let me emphasize the proof of this regularizing property of the semigroup $P_t$ relies on the fact that the diffusion term is not degenerate, see Hypothesis 1.3. In addition, Theorem 1.6.2 states that the function $P_t f(x)$ solves (1), however, it requires that $f \in C_b(\mathbb{R}^n)$.<|endoftext|> TITLE: Consequences of the Inverse Galois Problem QUESTION [5 upvotes]: Are there any papers written about the consequences of the Inverse Galois Problem in case it is proved to be true or false? We know a lot of things that would be true if the Riemann Hypothesis holds. What results would the Inverse Galois problem imply? REPLY [5 votes]: I once saw an application of a solved case of the inverse Galois problem. It is well known, that the Dedekind $\zeta$-function of a number field does not determine the number field up to isomorphy. In the talk it was shown that the $\zeta$-function together with a certain number of twists by characters do determine the number field. Let $K$ be the number field in question, $L$ be its normal closure. To define the right twist an abelian extension $M$ of $K$ was considered, which is as independent from $L$ as possible, that is, the Galois group of the normal closure of $M$ is a wreath product of the Galois group of $L$ and a cyclic group. The existence of such an $M$ is a special case of the inverse Galois problem, which had been solved before. Sorry, but I have no name or further detail.<|endoftext|> TITLE: Independence of the countable axiom of choice QUESTION [10 upvotes]: How does one proove that the Countable axiom of choice is not provable in ZF?Is there any brief proof?Does the Independence of the countable axiom of choice implies the independence of the axiom of choice? REPLY [13 votes]: I'm going to play fast and loose with details here, but the outline is correct. To answer your new questions: no, there is no short proof. And this only shows one direction of the indpenedence of AC, that ZF doesn't prove AC; in the other direction, we need to show that ZF doesn't disprove AC. This is proved by showing that any model of ZF contains a really nice "inner model," called "$L$," in which ZFC (and much more) is true. OK, so the details of the proof are quite complicated, but let me give an outline: First, let's make it clear what we're going to prove. You've got your ZF axioms over here, and you want to show that they don't prove Countable Choice (CC) - assuming, of course, that ZF is consistent (otherwise it proves everything, including CC). To do this, we're going to show that if ZF is consistent, then ZF+$\neg$CC is consistent. Even thought this is a question about proofs, though, we're going to argue semantically: we'll show how, if you give me a model of ZF, I can give you a model of ZF+$\neg$CC. By Completeness + Soundness, this means that if ZF is consistent then ZF+$\neg$CC is consistent. So this is going to be a relative construction: we'll start with some $V$ in which the ZF axioms (in fact, the ZFC axioms) are true, and we'll build a $W$ in which the ZF axioms are still true, and CC is false. In order to do this, though, we need some method to build models of ZF which gives us some measure of control over their properties. And this is really hard, since it's not even clear how to build models of ZF at all! The trick is forcing. Given a partial order $\mathbb{P}$, we can think of the elements of $\mathbb{P}$ - conditions - as approximations to some bigger object. For example, if $\mathbb{P}$ is the set of finite binary strings, then an element of $\mathbb{P}$ is an approximation of a real number - an infinite binary string. Think of a condition as providing incomplete information about something we're trying to build; specifically, the thing the condition is approximating is a maximal filter through $\mathbb{P}$. This isn't by itself a new idea - this is how the Baire category theorem goes, and Cantor's diagonal proof. OK, so what do posets have to do with models of ZF? To answer that question, let's talk about something completely different - the idea of an conditional name for a set. Remember that a model of ZF is built in layers - we start with $\emptyset$ as our 0th layer, and take powersets and limits to get successive layers, continuing all the way up the ordinals. (This can actually be formalized and proved as a theorem of ZF - the crucial axiom is Foundation.) This means we can think of a set as a well-founded tree, or as an instruction for how to build a set which is appropriately non-circular, e.g. the instructions "This set has $\omega$-many elements;" "The first element has one element;" "That element has no elements;" "The second element has one element;" "That element has one element;" "That element has no elements;" . . . describe the set $\{\{\}, \{\{\}\}, \{\{\{\}\}\}, . . .\}.$ Now, we could imagine a similar set of instructions which were conditional - e.g., "This set is the emptyset if my name is Steve, and otherwise is the set containing the emptyset." These things feel like sets in a lot of ways, but they aren't quite - in the same way that a formula isn't a sentence without evaluating the free variables, or a polynomial isn't a set of points without describing the field its over, these names aren't really referring to individual sets. Still, they're not far off: we can evaluate names, once we supply sufficient information - for example, once I tell you my name isn't Steve you know the name above refers to the set $\{\{\}\}$. So here's an idea for how to, given a model $V$ of ZF, build a new "thing" $W$: take all the names in $V$ which refer to only a certain kind of information - say, "names of people" - provide some "complete information" - say, everyone's name - and then evaluate all those names to produce actual sets; and say $W$ is the collection of all the sets produced in that way. Oh hey, I used the word "complete" - that sounds like what I said about posets! In fact, for our purposes, a "name" will be "an ambiguous description of a set, which depends on questions of the form "Is $x$ in $G$?" where $G$ is only known to be a maximal filter of some fixed poset $\mathbb{P}$." Formally, the class of $\mathbb{P}$-names is defined inductively, as: $$\mbox{A $\mathbb{P}$-name is a set of ordered pairs $(p, \nu)$, where $p\in\mathbb{P}$ and $\nu$ is a $\mathbb{P}$-name;}$$ $$\mbox{the *evaluation* $\mu[G]$ of a name $\mu$ on a max. filter $G\subseteq\mathbb{P}$ is $\mu[G]=\{\nu[G]: \exists p\in G((p, \nu)\in \mu)\}$.}$$ That is, thinking of $\mu$ as a set of instructions for how to build a set, we're looking at instructions of the form "If $p\in G$, then put (the thing defined by the following sub-instructions) into our set; otherwise, ignore (the following sub-instructions)." Now here's the neat bit: it turns out that if $V$ is a countable (not a problem, use Lowenheim-Skolem) model of ZFC, $\mathbb{P}\in V$ is a poset, and $G$ is a maximal filter through $\mathbb{P}$ which meets every dense subset of $\mathbb{P}$ living in $V$ - called a generic filter; note that this means $G\not\in V$, probably - then $V[G]$, the set of all $\mathbb{P}$-names in $V$ evaluated at $G$, is a model of ZFC! (We call such a $V[G]$ a generic extension of $V$.) So this is great . . . . . . except we don't want choice to hold. And this is a problem - if we assume merely "ZF is consistent," the only model we can easily describe is $L$, which satisfies ZFC. So we need some way to start with a model of ZFC, and turn it into a model where choice fails. To do this, we go back to the description of $V[G]$, and make it smaller - rather than throw in all the names, well restrict attention to good names for some value of "good." It turns out that the right notion of "good" here has to do with families of groups of automorphisms of $\mathbb{P}$ - roughly, a name is good if it is fixed by one of the groups of automorphisms. Different families of automorphisms give rise to different notions of good - and in particular, if our family includes the trivial group, then every name is good. It turns out that we preserve the axioms of ZF by doing this, so this is a viable construction to try. OKAY, THAT WAS LONG. So let me very briefly describe how the actual construction goes: Start with $V\models ZFC$ countable. Let $\mathbb{P}$ be the poset $((2^{<\omega})^\omega)^\omega$ - basically, an element of $\mathbb{P}$ is an approximation to a countable family of countable families of reals - or, we're building countably many "blocks" of countably many rows, and each row represents a real. For $F$ a finite set of rows, let $S_F$ be the set of automorphisms of $\mathbb{P}$ which only act by swapping rows, don't swap rows between blocks, and don't move the rows in $F$. Then the resulting "good names" give the model we're looking for - our generic filter $G$ is (basically) a countable collection of sets of reals, but no choice function has a good name. REPLY [4 votes]: First of all, the easy answer. We can prove that the axiom of choice implies the axiom of countable choice, quite easily. So by showing that the axiom of choice is consistent with the axioms of $\sf ZF$, and the negation of the axiom of countable choice is consistent as well with the rest of the axioms, we essentially prove the independence of both at once. Of course, the axiom of choice is strictly stronger and we can use other weakened versions of the axiom of choice to prove the independence of the axiom of choice from the axiom of countable choice as well. The harder answer is just going to be a broad strokes argument of the technical part. Godel proved that if the axioms of $\sf ZF$ are consistent, then you can define a subclass of the universe of $\sf ZF$ which satisfy $\sf ZFC$ and much much more. This class is called $L$ in modern times, and it has many many wonderful properties. This means that if there is a model of $\sf ZF$, then there is a model of $\sf ZFC$, which alternatively means that if $\sf ZF$ is consistent then $\sf ZFC$ is consistent. On the other hand, Fraenkel proved that if you allowed urelements, or atoms (which are non-sets objects) to exist, then the axiom of choice is not provable. His methods were later corrected by Mostowski and improved by Specker (which is why these models are often called FM(S) models). When Paul Cohen first described forcing, he showed how to transfer the core ideas from the existence of atoms to forcing extensions. He constructed two basic models which mimicked Fraenkel's original constructions. In both the axiom of countable choice fails pretty bad. The basic idea is that you add to the model, using forcing, a sequence of sets which are "generic" and from the original model's point of view, they are "almost" the same. Then using one of two clever (but often equivalent) constructions, you can keep those new sets, but forget the enumeration of the sequence, in a way that leaves them un-well orderable, and in fact without a countably infinite subset. You can find the proofs for all those, and much much more in Jech "The Axiom of Choice", as well in most modern set theory 'big books' (Jech, Kunen, Halbeisen).<|endoftext|> TITLE: Inaccessible cardinal and $\Sigma_1$ reflection QUESTION [5 upvotes]: A theorem of A. Levy says that, if $\kappa$ is an inaccessible cardinal, then $V_\kappa\prec_{\Sigma_1}V$ namely $V_\kappa$ is an elementary submodel when considering only $\Sigma_1$ formulas. Where can I find a proof of this theorem ? Is this property true also for some other (non inaccessible) cardinals ? REPLY [7 votes]: Here's one way to prove it. Let $H_\kappa = \{x: |tc(\{x\})|<\kappa\}$. Then we have: Theorem 1 If $\kappa$ is an uncountable cardinal, then $H_\kappa\prec_1 V$. Proof. Let $\phi$ be $\Delta_0$ with free variables among $y,x$. Since $\Sigma_1$ formulas are upward absolute for transitive models, if $H_\kappa\vDash \exists y\phi$, then $\exists y\phi$. So suppose $\exists y\phi$ for $x\in H_\kappa$. Now let $M\prec_1 V$ with $tc(\{x\})\subseteq M$ and $|M|<\kappa$ (such an $M$ exists because $\kappa$ is uncountable and $|tc(\{x\})|<\kappa$ by definition of $H_\kappa$). By the Mostowski collapse lemma, there is an isomorphism $j:M \to M'$ for some transitive $M'$. Since $tc(\{x\})\subseteq M$, $j(x) = x$ and thus $M'\vDash \exists y\phi$. Finally, we note that because $M'$ is transitive and has cardinality less than $\kappa$, $M'\in H_\kappa$ and so $H_\kappa\vDash \exists y\phi$ (again by upward absoluteness). $\Box$ Our result then follows from the fact that $H_\kappa = V_\kappa$, for $\kappa$ inaccessible. It is furthermore optimal for inaccessibles since ``$\kappa$ is inaccessible" is $\Pi_1$. But stronger results hold for other large cardinals. For example: Theorem 2 If $\kappa$ is supercompact (or even strong), then $V_\kappa\prec_2 V$. Proof. See Kanamori The Higher Infinite (2003) p. 299 and p. 359. $\Box$ Theorem 3 If $\kappa$ is extendible, then $V_\kappa\prec_3 V$. Proof. See Kanamori The Higher Infinite (2003) p. 318. $\Box$ This result is also optimal for extendibles since ``$\kappa$ is an extendible" is $\Pi_3$.<|endoftext|> TITLE: Why is the first chern class of a line bundle $c_1(L) = 1-L$ in complex K-theory? QUESTION [9 upvotes]: I'm trying to understand why on earth the first chern class of a line bundle in K-theory $c_1(L) = 1-L$. I understand that the first Chern class of the trivial bundle is zero, and that $H-1$ generates the reduced K-theory of $CP^1$ (where H is the canonical line bundle over $CP^1$), but there must be more reasoning behind this! Additionally, I have seen $c_1(L)$ as both $L-1$ and $1-L$, and I'm curious if this has to do with choosing the universal line bundle to be $\mathcal{O}(1)$ or $\mathcal{O}(-1)$, respectively. REPLY [14 votes]: This comes from the choice of the $K$-theory Thom class for complex vector bundles. Firstly, recall that $K$-theory $K^0(X)$ can be described as the group of bounded chain complexes of vector bundles on $X$, modulo the relation of forcing short exact sequences of such chain complexes to split. This is related to the usual description of $K$-theory by sending a chain complex of vector bundles to its Euler characteristic. However, in this model $K^0(X,A)$ is described by the bounded chain complexes of vector bundles on $X$ which are exact over $A$. If $\pi : E \to B$ is an $n$-dimensional complex vector bundle, then we can pullback the bundle $\pi^*E \to E$, and get chain complex $$0 \to \wedge^0 \pi^* E \to \wedge^{1} \pi^* E \to \wedge^{2} \pi^* E \to \cdots \to \wedge^n \pi^* E \to 0$$ where over a point $v \in E$ the maps are each given by wedge with $v$. By basic linear algebra, for $v \neq 0$ this is exact. The class $\lambda_E \in K^0(E, E-0)$ is then that represented by the linear dual of this complex. This is a possible choice of Thom class. In particular, if $L \to B$ is a complex line bundle then $\lambda_L$ is the chain complex $$0 \to \pi^*L^* \to \underline{\mathbb{C}} \to 0,$$ where $\underline{\mathbb{C}}$ lies in degree zero. The Euler class of a bundle is by definition the pullback of the Thom class along the zero-section: pulling this back along the zero-section $s : B \to E$ we obtain the complex $$0 \to L^* \overset{0}\to \underline{\mathbb{C}} \to 0$$ which, on taking Euler characteristics, gives $1-L^* \in K^0(B)$. On the other hand, the Euler class and first Chern class of a line bundle are the same thing (by definition), which gives the familiar $$c_1^K(L) = 1-L^* \in K^*(B).$$ It is an important point that the convention for the Thom class is only a convention. You can choose perfectly reasonable other conventions, by dualising E and replacing wedge with contraction, or not dualising the first complex above, leading to any of $L-1$, $1-L$, $L^*-1$, and $1-L^*$ as the first Chern class. Of these $1-L^*$ above and $L-1$ are slightly preferable as then the perverse equation $c_1(c_1^K(L)) = c_1(L)$ holds. I forget which way round they go, but one of these is standard in algebraic topology and the other is standard in algebraic geometry. This makes combining formulas from papers in these two subjects a uniquely frustrating experience.<|endoftext|> TITLE: fixpoint algebras of a permutation action QUESTION [5 upvotes]: Let $D$ be an infinite UHF algebra, e.g. the infinite tensor product of the matrix algebra $M_k(\mathbb{C})$. The permutation group $\Sigma_n$ acts on the $n$-fold tensor product $D^{\otimes n}$ in a canonical way. What is known about the fixpoint algebra of this action? Is it simple? Is it again a UHF algebra? REPLY [2 votes]: Here is my take on answering this question. It more-or-less provides the same as David's answer, but by somewhat different methods. 1) Let us first address the easier question: As David mentioned in his answer, the fixed point algebra is simple and has a unique trace. A permutation action of this sort is always pointwise strongly outer. In this case, the extension of this dynamical system to the weak closure with respect to the unique trace yields the analogous permutation action on the hyperfinite II${}_{1}$-factor, which is outer. By some well-known results due to Kishimoto, the crossed product is then simple and has a unique trace. Since the fixed point algebra sits inside the crossed product as a corner, we get the same statement for the fixed point algebra by Brown's theorem. 2) Now the harder question: For $n\geq 2$, the fixed point algebra is never UHF. In fact, this holds even if you consider the fixed point algebra of any faithful action of a non-trivial finite group $G$ by tensorial permutations. Here is why: Let $D$ be a UHF-algebra of infinite type and let $\sigma: G\curvearrowright D^{\otimes n}\cong D$ be a faithful action of a non-trivial finite group via tensorial permutations. Claim: The $K_0$-group of the fixed point algebra $D^\sigma$ has more than one direct summand. In fact, we can compute the $K$-theory. As mentioned above, the fixed point algebra is a corner in the simple crossed product, and so it suffices to look at the $K_0$-group of the crossed product. By the main result of [1], the crossed product is $D$-absorbing. In particular, $$ D\rtimes_\sigma G \cong (D\rtimes_\sigma G)\otimes D \cong (D\otimes D)\rtimes_{\sigma\otimes\operatorname{id}_D} G. $$ Now by Proposition 4.5 of [2], the action $\sigma\otimes\operatorname{id}_D: G\curvearrowright D\otimes D$ is homotopic to the trivial action. But this immediately yields a homotopy between $D\rtimes_\sigma G$ and $D\rtimes_{\operatorname{id_D}} G \cong D\otimes C^*(G)$. Since $G$ is a non-trivial finite group, its group algebra $C^*(G)$ is a direct sum of at least 2 matrix algebras (it has finite dimension at least 2 and admits a character). Combining all of this, we get that $D^\sigma$ is $KK$-equivalent to $D\otimes C^*(G)$, which is the direct sum of at least two infinite-dimensional UHF-algebras, and thus its K-theory has at least two direct summands. 3) As David already mentioned, the fixed point algebra should be a simple AF-algebra. Instead of getting into detailed calculations, one could also appeal to classification results in order to see this. Because of the above observations, the fixed point algebra $D^\sigma$ is a separable, nuclear, simple, unital, quasidiagonal and $D$-stable C*-algebra with a unique tracial state. By Theorem 6.1 of [3], it is thus TAF in the sense of Lin. It also satisfies the UCT by the above $KK$-equivalence. By Lin's classification theory of TAF-algebras, the only thing left to show is that some simple AF-algebra has the same ordered $K$-theory as $D\otimes C^*(G)$ - and this should be well-known, if I am not mistaken. [1] I. Hirshberg, W. Winter: Permutations on strongly self-absorbing C*-algebras, Internat. J. Math., 19(9):1137–1145, 2008. (http://arxiv.org/abs/0708.0213) [2] I. Hirshberg, N. C. Phillips: Rokhlin dimension: obstructions and permanence properties, http://arxiv.org/abs/1410.6581 [3] H. Matui, Y. Sato: Decomposition rank of UHF-absorbing C*-algebras, Duke Math. J. 163, no. 14 (2014), 2687-2708. (http://arxiv.org/abs/1410.6581) EDIT: My original argument, as outlined above, has a crucial flaw. Namely, it was recently brought to my attention that two homotopic group actions on the same C*-algebra do not necessarily give rise to two homotopy-equivalent C*-dynamical systems . (The terminology is very confusing here) Apparently, there are examples of homotopic actions of a finite group such that the crossed products are not even $KK$-equivalent.<|endoftext|> TITLE: Surjective (strong) reducibility of Borel equivalence relations QUESTION [6 upvotes]: Suppose $E$ and $F$ are Borel equivalence relations on Polish spaces $X$, $Y$, resp. Say that $E$ is surjectively Borel reducible to $F$ iff there is a Borel surjection $f:X \to Y$ such that $xEy$ iff $f(x) F f(y)$. This is (at least on the surface) stronger than normal Borel reducibility, weaker than Borel isomorphism. What do we know about this reducibility? If $E$ is effectively Borel with only countably many classes, is $E$ effectively surjectively reducible to the trivial equivalence relation on $\omega$? edit: Probably this is the version I am interested in: Say $E$ is strongly Borel reducible to $F$ iff $E$ reduces to $F$ via a Borel $f$ such that the range of $f$ is Borel. When $Y=\omega$ and $E$ has infinitely many classes, "$E$ is effectively stongly Borel reducible to the relation $F$ on $Y$" exactly means "$E$ is effectively surjectively Borel reducible to $F$". edit again: Actually, the arXiv version of Fokina-Friedman-T¨ornquist "The Effective Theory of Borel Equivalence Relations" probably has a (minor) mistake. On page 9, the equivalence relation $E^*$ is $\Pi^1_1$, not Hyp. The $\Sigma^1_1$ line $$(\exists x_0,x_1) ( x_0 \in H(d_0) \wedge x_1 \in H(d_1) \wedge x_0 E x_1)$$ fails when $H(d_0) = \emptyset$. The proof of Proposition 2 can be easily repaired, but the issue of empty set is worth attention when we seriously require the range being $\Delta^1_1$. REPLY [2 votes]: The answer is no. The question is basically asking for a $\Delta^1_1$ transversal for a $\Delta^1_1$ Borel equivalence relation with only countably many classes. Let $A$ be a closed subset of $\mathbb{R}\times \omega$ whose projection to the second coordinate is not $\Delta^1_1$. Put $(x,m)E(y,n)$ iff either $(x,m)\notin A \wedge (y,n)\notin A$ or $(x,m)\in A \wedge (y,n)\in A\wedge m=n$. Then $E$ does not have a $\Delta^1_1$ transversal.<|endoftext|> TITLE: Formula for the distance in noncommutative geometry QUESTION [8 upvotes]: Probably the most famous formula in noncommutative geometry is the following formula allowing one to compute distance of two points using the operator theoretic data: $$(1) \ \ d(p,q)=\sup\{|f(p)-f(q)| f \in \mathcal{A}, \|[D\!\!\!/,f]\| \leq 1\}$$ where everything takes place on spin (or spin^c) manifold and $D\!\!\!/$ is the Dirac operator (but at this moment I would not specify what is $\mathcal{A}$). It is defined by the formula $D\!\!\!/ =ic \circ \nabla^S$ where $c$ is the Clifford action and $\nabla^S$ is the spin connection. This definition allows us to prove that, for a function $f$ one has $$(2) \ \ [D\!\!\!/,f]=ic(df)$$ and further it follows from this identity that $\|[D\!\!\!/,f] \|=\|grad(f) \|$. One we have that, we can show that $|f(p)-f(q)| \leq \| grad(f)\| \ell(\gamma)$ where $\gamma$ is piecewise smooth curve joining $p$ and $q$. Taking into account only $f$ such that $\|grad(f)\| \leq 1$ one then obtains $\sup\{|f(p)-f(q)| f \in \mathcal{A}, \|[D\!\!\!/,f]\| \leq 1\} \leq d(p,q)$. In order to get equality one considers the function $d(p,\cdot)$---but the problem is that this function is only continuos but is not smooth. So the first problem is: Question 1 Is the formula $(1)$ true when one takes supremum only over smooth functions? And the second thing which I would like to know is the following: as I said to get formula $(1)$ it is enough to know formula $(2)$ so Question 2 Suppose that we have Riemannian manifold $M$ (not necessarily spin$^c$). Is it always possible to find operator $D$ such that formula (1) holds? REPLY [5 votes]: Let $\|\cdot\|_{d_g} : L^\infty(M) \to [0,+\infty]$ be the Lipschitz seminorm induced by the geodesic metric $d_g$ on $M$, defined by $$ \forall f \in L^\infty(M), \quad \|f\|_{d_g} := \sup_{x \neq y} \frac{\lvert f(x)-f(y) \rvert}{d_g(x,y)}, $$ so that the Lipschitz algebra of the metric space $(M,d_g)$ is $$ \operatorname{Lip}(M,d_g) := \{f \in L^\infty(M) \mid \|f\|_{d_g} < +\infty\} = \{f \in C(M) \mid \|f\|_{d_g} < +\infty\}, $$ which is a Banach algebra for the norm $\|\cdot\|_{\operatorname{Lip}(M,d_g)}$ defined by $$ \forall f \in \operatorname{Lip}(M,d_g), \quad \|f\|_{\operatorname{Lip}(M,d_g)} := \|f\|_\infty + \|f\|_{d_g}. $$ Proving the famous formula, then, boils down to proving the following two claims, where your Question 1 pertains to the first claim, whilst your Question 2 pertains to the second: Claim 1: For all $x,y \in M$, $$d_g(x,y) = \sup\{\lvert f(x) - f(y) \rvert \mid f \in \operatorname{Lip}(M,d_g), \; \|f\|_{d_g} \leq 1\}.$$ Claim 2: Suppose that $M$ is spin$^\mathbb{C}$ and that $D$ is a spin$^\mathbb{C}$ Dirac operator on a spinor bundle $S \to M$. Then, for any $f \in L^\infty(M)$, $[D,f] \in B(L^2(M,S))$ if and only if $f > \in \operatorname{Lip}(M,d_g)$, in which case $$\|[D,f]\| = \|f\|_{d_g}. $$ Let me first turn to your Question 1. In principle, the algebra you're really taking a supremum over in Claim 1 is $$ \{f \in L^\infty(M) \mid df \in L^\infty(M,T^\ast M)\} = \{f \in C(M) \mid df \in L^\infty(M,T^\ast M)\} = \operatorname{Lip}(M,d_g), $$ where, for given $x$, $y \in M$, the supremum, which involves the Lipschitz seminorm $\|\cdot\|_{d_g}$, is attained by $d_g(x,\cdot) \in \operatorname{Lip}(M,d_g)$. Since $C^\infty(M)$ is not dense in $\operatorname{Lip}(M,d_g)$ endowed with the Banach algebra norm induced by $\|\cdot\|_{d_g}$—indeed, the closure, from what I understand, will be the proper closed subalgebra $C^1(M)$ of $C^1$ functions on $M$—I suspect the answer is no, unless there's a result out there that lets you approximate Lipschitz functions by smooth functions in the Lipschitz seminorm alone (whilst necessarily doing violence to the uniform norms). Edit: As Nik Weaver points out in his comment, the answer is indeed yes. Fix $x$, $y \in M$. Let $r = d_g(x,\cdot)$, which is smooth on the complement of $\{x\} \cup \text{cut locus of $x$}$ and has Lipschitz constant $\|r\|_{d_g} = 1$. Then, by mollification on geodesic neighbourhoods of $x$ and of the cut locus of $x$ (cf. the approximation results of Greene and Wu, 1972, 1976, 1979), one can construct for any $\epsilon > 0$ a smooth function $f_\epsilon$ such that $\|r-f_\epsilon\|_\infty < \epsilon/2$ and $\|f_\epsilon\|_{d_g} \leq 1$, so that $$ \lvert f_\epsilon(x) - f_\epsilon(y) \rvert = \lvert (r(x)-r(y)) + (r(x)-f_\epsilon(x)) + (r(y) - f_\epsilon(y)) \rvert \geq d_g(x,y) - \epsilon, $$ and hence $$ d_g(x,y) - \epsilon \leq \sup\{\lvert f(x) - f(y) \rvert \mid f \in C^\infty(M), \; \|f\|_{d_g} \leq 1 \} \leq d_g(x,y). $$ Note that in context, in Connes's book and in the 1995 paper Noncommutative geometry and reality, $\mathcal{A}$ is either $L^\infty(M)$ or $$ \{f \in L^\infty(M) \mid [D,f] \in B(L^2(M,S))\} = \operatorname{Lip}(M,d_g), $$ where $D$ is the spin Dirac operator on the spinor bundle $S \to M$ on the spin manifold $M$. Let me now turn to your Question 2. The essential point is that Claim 2 holds for any essentially self-adjoint Dirac-type operator on $M$, i.e., a first-order differential operator $D$ on a Hermitian vector bundle $E \to M$ such that $$ D^2 = -g^{ij}\partial_i \partial_j + \text{lower order terms}; $$ spin$^\mathbb{C}$ and spin Dirac operators are certainly Dirac-type operators, but so is the Hodge–de Rham operator $d+d^\ast$ on $\wedge^\ast T^\ast M$, which only requires an orientation and a Riemannian metric on $M$. The reason is that an essentially self-adjoint Dirac-type operator $D$ always defines a self-adjoint Clifford action $c : T^\ast M \to \operatorname{End}(E)$ on $E$ by $$ \forall f \in C^\infty(M), \quad c(df) := i[D,f], $$ so that, in general, if $f \in L^\infty(M)$, then $[D,f] \in B(H)$ if and only if $f \in \operatorname{Lip}(M,d_g)$, with $$ \|[D,f]\|_{B(H)} = \|c(df)\|_{B(H)} = \|c(df)^2\|^{1/2}_{B(H)} = \|g^{-1}(df,df)\|^{1/2}_\infty = \|f\|_{d_g}. $$<|endoftext|> TITLE: Upperbounding the number of regions induced by a set of unit disks QUESTION [5 upvotes]: Given a set $D$ of $n$ same radius disks, embedded in the plane, their arrangement induces a number $k$ of connected regions in $\mathbb{R}^2 \setminus \cup_{d \in D}$ . I am interested in an upper bound on $k$ as a function of $n$. Does anybody know (a reference for) a good upperbound on $k$? Since the Union Complexity, i.e., the number of arcs on the boundary of $D$ is at most $6n-12$ (if $n \geq 3$) and each connected region is bounded by at least 3 disks, it follows that $k \leq 2n - 4$, but I feel that this bound should be much closer to $n$ than to $2n$. REPLY [3 votes]: If you take a triangular packing of discs and slightly increase the radius of each disc then enclose the packing in a large regular square and remove all discs outside the square. Then inside the square the ratio of discs to regions outside the discs will be two to one since there is a hexagonal tiling with a three coloring such that one color is assigned to the discs and two colors are assigned to regions not in the discs and each coloring has the same number of hexagons see here and look at the one uniform three coloring. There may be a disparity near the sides the square but that will be linear and the number of discs inside a square will be quadratic so any bound other than $2n$ will be exceeded by increasing the size of the square. So the upper bound will not be $n$or any constant less than $2$ and greater than $n$ plus another constant. I don't know how close to $2n-4$ you can get though or if there is an improvement to the triangular lattice.<|endoftext|> TITLE: If $\kappa$ is weakly inaccessible and $A\subset\kappa$, can $L[A]$ violate $\kappa^{\lt\kappa}=\kappa$? QUESTION [10 upvotes]: In some current work, my co-authors and I had wanted in a certain argument to appeal to $\kappa^{\lt\kappa}=\kappa$ in $L[A]$, in a situation where $A\subset\kappa$ and $\kappa$ was weakly inaccessible in $V$, but $\kappa$ was below the continuum in $V$ (and so $\kappa^{\lt\kappa}\neq\kappa$ in $V$). But we have lost confidence in this statement. Is it consistent that there is a counterexample? Question. Is it (relatively) consistent that $\kappa$ is weakly inaccessible, but there is $A\subset\kappa$ with $L[A]\models\kappa^{\lt\kappa}>\kappa$? The model $L[A]$ here is the relative constructible universe, defined as in the constructible universe, but with a predicate for the set $A$. We had at first thought a condensation argument might show $\kappa^{\lt\kappa}=\kappa$ in every $L[A]$, but upon looking at it closely, I can't quite make it work. Condensation works fine above $\kappa$, but I don't see why, for example, every real number in $L[A]$ must be added by a stage before $\kappa$. It would be interesting to me even to resolve the question of how big $2^\omega$ must be or can be in comparison with $\kappa$ in $L[A]$, where $A\subset\kappa$ and $\kappa$ is a regular limit cardinal in $V$. (Meanwhile, we've saved our application by finding another route for our argument that does not rely on this issue.) REPLY [12 votes]: By Lemma 2.2. of my paper Shelah's strong covering property and CH in V[r], we can show that: Claim. If $A\subseteq \kappa,$ and if $Y\in L[A]$ is a bounded subset of $\kappa,$ then there exists a proper initial segment $A'$ of $A$ such that $Y\in L[A'].$ From this, $\kappa^{<\kappa}=\kappa$ in $L[A].$<|endoftext|> TITLE: power laws emerging from the sandpile model QUESTION [5 upvotes]: Is there a rigorous proof that the abelian sandpile model generates a power law distribution of avalanche lengths? REPLY [2 votes]: A recent paper http://arxiv.org/abs/1602.06475 claims a proof of lower estimates for the sizes of toppling clusters.<|endoftext|> TITLE: Should the Grothendieck ring of varieties be K_0 of numerical motives? QUESTION [6 upvotes]: Assuming the Standard Conjectures, should the Grothendieck ring of varieties be the $K_0$ of the abelian category of numerical motives? REPLY [8 votes]: As explained by Dan Petersen, the answer is no. There are additional issues though. For example, multiplication by the class $\mathbf L$ of the affine line $\mathbf A^1_k$ corresponds to Tate twist on the motivic level, hence is “motivically regular”. However, it has been proved recently by Lev Borisov that $\mathbf L$ is a zero-divisor in the Grothendieck group of varieties.<|endoftext|> TITLE: Mixed Hodge structure and cup product QUESTION [6 upvotes]: I'm looking for a reference for the answer to the following questions. Let $X$ be an algebraic variety over C. When is the cup product a morphism of Mixed Hodge structures? Does $X$ have to be smooth? REPLY [9 votes]: This is true with no hypothesis on X: see Corollaire 8.2.11 in Deligne Théorie de Hodge III, Pub. Math. IHES 44 (1974), p. 5-77.<|endoftext|> TITLE: Singular models of K3 surfaces QUESTION [5 upvotes]: Let us work over a ground field of characteristic zero. As is well-known, a K3 surface is a smooth projective geometrically integral surface $X$ whose canonical class $\omega_X$ is trivial and for which $\operatorname{H}^1(X,\mathscr{O}_X)$ vanishes. A bit of folklore (proven e.g. in Beauville's Complex Algebraic Surfaces) is that if $X$ is a smooth complete intersection of a quadric and a cubic in $\mathbb{P}^4$, or a smooth complete intersection of three quadrics in $\mathbb{P}^5$, then $X$ is K3. The question Suppose I have a (non-smooth) complete intersection $X_{2,3}$ of a quadric and a cubic in $\mathbb{P}^4$ all of whose singularities are rational double points, and a (non-smooth) complete intersection $X_{2,2,2}$ of three quadrics in $\mathbb{P}^5$, again with at most rational double points. Do these two surfaces have K3 surfaces as their minimal regular models? I suspect the answer is yes; for example, since the singularities are assumed to be rational double points, they should admit a crepant resolution. However, I can't really locate this fact in the literature. As for the vanishing of the appropriate $\operatorname{H}^1$, I really haven't a clue. REPLY [4 votes]: For what it's worth, I wrote up a proof (pretty detailed) following the hints in Francesco Polizzi's answer. It's in an unpublished preprint found here (p. 38 onwards). I am not a geometer, so the exposition may be a little clunky, but at least I think all the crucial steps are there.<|endoftext|> TITLE: Associativity of Kontsevich's star product up to second order QUESTION [15 upvotes]: In Deformation quantization of Poisson manifolds, Kontsevich gives the quantization formula $$f \star g = \sum_{n=0}^\infty \hbar^n \sum_{\Gamma \in G_n} w_\Gamma B_{\Gamma,\alpha}(f,g).$$ He gives an explicit formula up to $O(\hbar^3)$: $$f \star g = fg + \hbar \sum_{i,j} \alpha^{ij} \partial_i(f)\partial_j(g) + \frac{\hbar^2}{2} \sum_{i,j,k,l} \alpha^{ij}\alpha^{kl} \partial_i\partial_k(f)\partial_j\partial_l(g) + \frac{\hbar^2}{3}\left(\sum_{i,j,k,l} \alpha^{ij}\partial_j(\alpha^{kl})(\partial_i\partial_k(f)\partial_l(g) - \partial_k(f)\partial_i\partial_l(g))\right) + O(\hbar^3).$$ This is an equality up to gauge equivalence. Other than that, the equality $w_\Gamma B_{\Gamma,\alpha} = w_{\Gamma'} B_{\Gamma',\alpha}$ for graphs that differ only in the labeling (by virtue of skew symmetry in the weight and the Poisson bivector) has been used to collect like terms, and one term vanishes (also explained in the preceding link). Anyhow, We are assured that this star product is associative up to second order, but I don't see it: We can write the above star product up to $O(\hbar^3)$ in terms of graphs without losing any information: Here it is understood that the order of the ground vertices is fixed (first is $f$, second is $g$), and all arrows correspond to independent indices; the labels $L$ and $R$ just indicate whether the contraction is with the first or second upper index of $\alpha$. Computing $f \star (g \star h)$ and $(f \star g) \star h$ is an exercise in the Leibniz rule, which can also be done graphically: Of course, many terms do match identically (struck out with green). Some terms match because $1/2 + 1/2 = 1$ (red underline 1 and 2). Some other terms match because swapping $L$ and $R$ once gives a sign change (black underline). But what happens with the boxed terms? (I've verified using a computer program that the expressions above are correct.) The desired equality is the following: In terms of bidifferential operators, this is: $$\sum_{i,j,k,l} \alpha^{ij}\partial_i(\alpha^{kl})\partial_k(f)\partial_l(g)\partial_j(h) = \sum_{i,j,k,l} \alpha^{ij}\partial_j(\alpha^{kl})\partial_i(f)\partial_k(g)\partial_l(h).$$ We can relabel the indices on the right hand side (such that the last 3 match) and use skew symmetry: $$\sum_{i,j,k,l} \alpha^{ij}\partial_i(\alpha^{kl})\partial_k(f)\partial_l(g)\partial_j(h) = \sum_{i,j,k,l} \alpha^{ik}\partial_i(\alpha^{jl})\partial_k(f)\partial_l(g)\partial_j(h).$$ The only difference is the position of $j$ and $k$ in the first two factors. If I'm not mistaken, the Jacobi identity for the Poisson bivector is $$0 = [\alpha,\alpha]^{ijk} = \alpha^{li}\partial_l(\alpha^{jk}) + \alpha^{lj}\partial_l(\alpha^{ki}) + \alpha^{lk}\partial_l(\alpha^{ij}).$$ Using this on the factor in the right hand side gives $$\alpha^{ik}\partial_i(\alpha^{jl}) = -\alpha^{ij}\partial_i(\alpha^{lk}) - \alpha^{il}\partial_i(\alpha^{kj}) = \alpha^{ij}\partial_i(\alpha^{kl}) + \alpha^{il}\partial_i(\alpha^{jk}),$$ but this forces $\alpha^{il}\partial_i(\alpha^{jk}) = 0$, which is nonsense. What am I doing wrong? REPLY [6 votes]: The answer is that the black underlined terms do not cancel. Instead, they contribute an extra term which gives precisely the Jacobi identity (times $2/3$). (The reason I missed this is that I previously made a sign error, in which case they did cancel.) The desired equality is as follows (new terms drawn in black): In terms of bidifferential operators this is $$\sum_{i,j,k,l} \alpha^{ij}\partial_i(\alpha^{kl}) \partial_k(f)\partial_l(g)\partial_j(h) = \sum_{i,j,k,l} \alpha^{ij}\partial_j(\alpha^{kl})\partial_i(f)\partial_k(g)\partial_l(h) + \sum_{i,j,k,l} \alpha^{ij} \partial_j(\alpha^{kl})\partial_k(f)\partial_j(g)\partial_l(h).$$ To see that it is true, relabel the indices on the right hand side (as before): $$\sum_{i,j,k,l} \alpha^{ij}\partial_i(\alpha^{kl}) \partial_k(f)\partial_l(g)\partial_j(h) = \sum_{i,j,k,l} \alpha^{ik}\partial_i(\alpha^{jl})\partial_k(f)\partial_l(g)\partial_j(h) + \sum_{i,j,k,l} \alpha^{il} \partial_i(\alpha^{kj})\partial_k(f)\partial_l(g)\partial_j(h).$$ Indeed, it follows from the Jacobi identity with the appropriate indices: $$\alpha^{ij}\partial_i(\alpha^{kl}) = -\alpha^{ik}\partial_i(\alpha^{lj}) - \alpha^{il}\partial_i(\alpha^{jk}) = \alpha^{ik}\partial_i(\alpha^{jl}) + \alpha^{il}\partial_i(\alpha^{kj}).$$ QED. Thanks for watching.<|endoftext|> TITLE: Expected centered entropy of the binomial distribution QUESTION [9 upvotes]: In short, the function I am interested in is the following: $$I_n(p) = \sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} \left[h(p) - h\left(\tfrac{k}{n}\right)\right],$$ where $h(x) \triangleq -x \log x - (1-x) \log (1-x)$ is the standard binary entropy function. I am able to prove that for large $n$ this function attains a local maximum around $p \approx \frac{1.338}{n}$ at which $I_n(p) \approx \frac{0.84}{n}$ in case binary logarithms are used. By assuming that $p = \frac{\alpha}{n}$ for $\alpha = O(1)$ and using a Poisson approximation of the binomial distribution, the value $\alpha \approx 1.338$ is a (numerically found) maximizing value of $$\max_{\alpha > 0} \left\{\sum_{z=1}^{\infty} \frac{\alpha^z}{z!} e^{-\alpha} (z \log z) - \alpha \log \alpha\right\}.$$ Note that the latter expression does not contain $n$ anymore; I am interested in large-$n$ asymptotics of $I_n(p)$ which leads to the above, $n$-independent expression. What I would like to show is that this point $p = \frac{\alpha}{n}$ is actually not just a local, but a global maximum, i.e., there are no values $p = \omega(\frac{1}{n})$ with even higher asymptotic values of $I_n(p)$ for large $n$. So my question can be formulated as: Can we prove that for large $n$, $\max\limits_{p \in [0,1]} I_n(p) \to I_n(\frac{\alpha}{n})$ with $\alpha \approx 1.338184$? Numerical inspection of $I_n(p)$ for say $n = 100\,000$ shows that $I_n(p) \propto \frac{1}{n}$ is nice and smooth, has two global maxima at $\frac{\alpha}{n}$ and $1 - \frac{\alpha}{n}$ with value $\frac{0.84}{n}$ and a local minimum at $p = \frac{1}{2}$ with value $\frac{0.72}{n}$ (and two trivial minima at $0$ and $1$ with value $0$). The figure below sketches what $I_n(p)$ (scaled with a factor $n$) looks like for small $n$. As $n$ increases, the function might be best described as a flat line at height $0.72$ with two peaks close to $0$ and $1$ of height $0.84$. I've tried various approaches using e.g. Taylor expansions of the entropy, but I am not able to prove that "order terms" are actually small for arbitrary $p$. Also, a bound given here using Jensen's inequality is not sharp enough; it's roughly a factor $2$ too loose to prove the above statement. Any help would be appreciated! Attempt 1 For large $n$, the dominating terms in the summation in $I_n$ come from values $k \approx np$. Expanding $h(\frac{k}{n})$ around $k = np$ leads to $$h\left(\frac{k}{n}\right) = h(p) + \left(\frac{k}{n} - p\right) h'(p) + \frac{1}{2}\left(\frac{k}{n} - p\right)^2 h''(p) + \frac{1}{6}\left(\frac{k}{n} - p\right)^3 h^{(3)}(p) + \dots.$$ Substituting this back into the definition of $I_n(p)$, this leads to $$I = \sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} \left[-\left(\tfrac{k}{n} - p\right) h'(p) - \tfrac{1}{2}\left(\tfrac{k}{n} - p\right)^2 h''(p) - \tfrac{1}{6}\left(\tfrac{k}{n} - p\right)^3 h^{(3)}(p) - \dots\right].$$ Now the term $\frac{k}{n}$ leads to a factor $\frac{np}{n} = p$ when pulled out of the summation, so the first term disappears. As the second term is simply the variance (scaled by a factor $1/n^2$), this term results in $\frac{1}{2n} p(1-p)h''(p)$. As $h'(p) = \log_2(\frac{1-p}{p})$ and $h''(p) = -1/[p(1-p)\ln 2]$, the second term contributes $1/(2n \ln 2) \approx \frac{0.72}{n}$ as expected. So we have: $$I_n(p) = \frac{1}{2n \ln 2} + \sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} \left[- \tfrac{1}{6}\left(\tfrac{k}{n} - p\right)^3 h^{(3)}(p) - \dots\right].$$ At this point, what remains is to show that all remaining order terms are small, if $np(1-p) = \omega(1)$. Attempt 2 Using the approach described here, consider the sum $$S = \sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} \frac{k}{n} \log_2 \frac{k}{n}.$$ First, we can cancel the $\frac{k}{n}$ with factors in the binomial coefficient, and pull out a factor $p$ to obtain $$S = p \sum_{k=1}^n \binom{n-1}{k-1} p^{k-1} (1-p)^{n-k} \log_2 \frac{k}{n} = p \sum_{k=0}^{n-1} \binom{n-1}{k} p^{k} (1-p)^{n-k-1} \log_2 \frac{k+1}{n}.$$ Applying Jensen's inequality, we obtain $$S \leq p \log_2\left(\sum_{k=0}^{n-1} \binom{n-1}{k} p^{k} (1-p)^{n-k-1} \frac{k+1}{n}\right) = p \log_2 \frac{(n-1)p + 1}{n}.$$ Pulling out the term $\log_2 p$, we obtain $$S \leq p \log_2 p + p \log_2\left(1 + \frac{1 - p}{np}\right).$$ Applying the same procedure to the other term in $I_n(p)$, this leads to $$I_n(p) \leq h(p) + \sum_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} \left[\frac{k}{n} \log_2 \frac{k}{n} + \frac{n-k}{n} \log_2 \frac{n-k}{n}\right] \\ = h(p) + \left[p \log_2 p + p \log_2\left(1 + \frac{1 - p}{np}\right)\right] + \left[(1-p) \log_2 (1-p) + (1-p) \log_2\left(1 + \frac{p}{n(1-p)}\right)\right] \\ = p \log_2\left(1 + \frac{1 - p}{np}\right) + (1-p) \log_2\left(1 + \frac{p}{n(1-p)}\right)$$ Using $\ln(1 + x) \leq x$ for all $x$, this leads to $$I_n(p) \leq \frac{1 - p + p}{n \ln 2}.$$ Unfortunately this is a factor $2$ off, as $I_n(p) \sim \frac{1}{2n \ln 2}$ for almost all $p$. REPLY [4 votes]: By symmetry I can assume $p\le\frac12$. I will use natural logs. Put $f(x)=h(p)-h(x)$ and $b_k = \binom{n}{k} p^k(1-p)^{n-k}$. Define $k_0=\lceil pn/2\rceil$ and $k_1=n-k_0$. The plan is: find polynomials $f_0(x),f_1(x)$ such that $f_0(x)\le f(x)\le f_1(x)$ for $k_0/n\le x\le k_1/n$. Then make bounds on the various parts of $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_0(k/n)) + \sum_{k=0}^n b_k f_0(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_0(k/n)) \le \sum_{k=0}^n b_k f(k/n) \le \sum_{k=0}^{k_0-1} b_k(f(k/n)-f_1(k/n)) + \sum_{k=0}^n b_k f_1(k/n) + \sum_{k=k_1+1}^n b_k (f(k/n)-f_1(k/n)) .$$ Since $f^{(iv)}(x) = 2/x^3+2/(1-x)^3$, we can use Taylor's theorem with remainder to get $$f_j(x) = (\ln p -\ln(1-p))(x-p) + \frac{(x-p)^2}{2p(1-p)} - \frac{(1-2p)(x-p)^3}{6p^2(1-p)^2} + \frac{j(8-8p+2p^2+p^3)(x-p)^4}{3p^3(2-p)^2}$$ for $j=0,1$. Maple now tells us $$\sum_{k=0}^n b_k f_j(k/n)=\frac{1}{2n}-\frac{(1-2p)^2}{6p(1-p)n^2} + \frac{Bj}{n^2},$$ where $$B = \frac{(1-p)^2(8-8p+2p^2+p^3)}{p(2-p)^2} + \frac{(1-p)(8-8p+2p^2+p^3)(1-6p+6p^2)}{3p^2(2-p)^2n}.$$ This much is $1/(2n)+O(1/(pn^2))$. In the range $0\le k\le k_0-1$, $b_k$ is dominated by a geometric series with ratio $\frac12$ and $f(x)-f_j(x)=O(p\ln p)$. Using the Stirling approximation for $b_k$, we find that $$\sum_{k=0}^{k_0-1} b_k(f(k/n)-f_j(k/n))=o(1/n)$$ for $p\ge (2+\epsilon)\ln\ln n/n$. In the range $k_1+1 \le k\le n$, $b_k\le 2^{-3n/4}$ (using the assumption $p\le \frac12$) and $f(x)-f_j(x)=O(p^{-3})$, so this part is negligible compared to the previous part. In summary, $$\sum_{k=0}^n b_k f(k/n) = \frac{1+o(1)}{2n}$$ for $(2+\epsilon)\ln\ln n/n\le p\le 1-(2+\epsilon)\ln\ln n/n$, any $\epsilon\gt 0$. For the case $p=a/n$ with $a=O(\ln\ln n)$, use $\binom{n}{k}=\frac{n^k}{k!}\left(1-\binom{k}{2}/n -O(k^3/n^2)\right)$ and simple bounds on the tail to find $$\sum_{k=0}^n b_k f(k/n) = \sum_{k=0}^{(\ln n)^2} \frac{a^k}{k!}\left(1-\frac{\binom{k}{2}}{n}-\frac{ak}{n}\right)f(k/n) + O((\ln n)^{O(1)}/n^2).$$<|endoftext|> TITLE: Rounding errors in images of Julia sets QUESTION [12 upvotes]: One typically computes Julia sets by iterating a complex function, such as a polynomial or rational function. How do rounding errors affect the results? I'm looking for references on this issue, especially but not exclusively for the escape time method. The only ones I have found so far are: "The [inverse iteration] method is very insensitive to round-off errors, since $f$ tends to be expanding on its Julia set, so that $f^{-1}$ tends to be contracting.'' [Milnor, Dynamics in one complex variable, page 49] "Theorem 1 makes it plausible that in almost all cases rounding errors do not affect the computer graphics of Julia sets.'' [Steinmetz, Rational iteration, page 175] REPLY [5 votes]: When calculating Julia Sets and performing iterative operations with real numbers, round-off errors manifest themselves at the boundary of the Julia set. (They may or may not manifest themselves elsewhere, but my efforts were principally at the boundary.) The boundary is referred to as the Julia set by Kenneth Falconer in Fractals: A Short Introduction (NY: Oxford University Press, 2013). Falconer demonstrates z= z*z+c where c=(0,0) as a nice starting place. It is a nice starting place and the most elementary means of generating a Julia set. Points inside the Julia set converge to (0,0); points outside diverge to infinity; points on the Julia set stay at the unit circle. However, when performing computer calculations, most the points that should remain on the unit circle do not behave well due to round-off errors. They either run off to infinity (with the postman) or converge to (0,0). The problem is exactly with round-off error. To convince yourself, open a spreadsheet, and set up a couple of columns for unit circle points, and do the tedious iterative calculations: x2-y2, 2.0*x*y Computationally, nothing behaves well on the unit circle. You can switch to polar coordinates when the starting point c=(0,0). This approach is nice because it shows that points on the unit circle converge piecewise to the unit circle. Now plot the points for everything I've mentioned above - not the iterations, but the points themselves. It will give you a clear visual perception of the gap between mathematics as it is and what really happens with your computer calculations. Lynn Wienck<|endoftext|> TITLE: Is it true that for each bounded continuous function we can find a set of analytic functions to uniformly converge it? QUESTION [8 upvotes]: Is it true that for each bounded continuous function $f:\mathbb R \to \mathbb R$, we can find a set of analytic functions $g_i:\mathbb R \to \mathbb R, i=1,2,...$ such that $g_i$ uniformly converges to $f$ ? REPLY [8 votes]: In the paper MR0098847 (20 #5299) Grauert, Hans: On Levi's problem and the imbedding of real-analytic manifolds. Ann. of Math. (2) 68 1958 460–472. it is proved (Proposition 8) that real analytic functions are dense in continuous functions for the Whitney $C^0$-topology, for any paracompact real analytic manifold. The sup-norm gives a coarser topology, so this also follows.<|endoftext|> TITLE: A kind of uniqueness for the double of a manifold QUESTION [8 upvotes]: Given two smooth, connected manifolds, M, N, with the same boundary. If their doubles, D(M) and D(N) are diffeomorphic, does it follow that M and N are diffeomorphic ? The condition on the boundary is to exclude Mazur manifolds.. REPLY [5 votes]: Here's an example in dimension 4: let $M$ and $N$ be the 4-dimensional 2-handlebodies associated to $-1$-surgery along the trefoil knot and $+1$-surgery along the figure-eight knot respectively. They have the same boundary, as the two surgeries are diffeomorphic. On the other hand, Corollary 5.1.6 in Gompf and Stipsicz's 4-manifolds and Kirby calculus tells us that $D(M)$ and $D(N)$ are both diffeomorphic to $\mathbb{CP}^2\#\overline{\mathbb{CP}}^2$. However, $M$ and $N$ can't even be homeomorphic, since the intersection form on $M$ is negative-definite and the intersection form on $N$ is positive-definite.<|endoftext|> TITLE: What are known examples of a 3-manifold $Y$ embedded into $Y'\times I$ where $Y'$ is another 3-manifold? QUESTION [7 upvotes]: The question I have is the following: Let $Y,Y'$ be two integer homology 3-spheres. Can we embed $Y'$ into $Y\times I$ such that $Y'$ separates the two boundary components apart? Do we know any nontrivial examples of this type? (For example for $Y'=S^{3}$ or Brieskorn spheres?) Also, we can consider the similar problem for knot concordance: are there two knots $K,K'$ with concordance $K\rightarrow K'$ and $K'\rightarrow K$ such that if we compose them, we get trivial concordance $K\rightarrow K$? Anyone can tell me examples of this kind? (especially for $K'$ a torus knot). REPLY [8 votes]: To expand on Ian's answer to the first question: I proved many years ago (Seifert surfaces of knots in $S^4$, Pac. J. Math. 145 (1990), 97–116) that for any 3-manifold Y, there is a hyperbolic 3-manifold $Y'$ embedded in $Y \times I$ separating the boundary components, so that both pieces are homology cobordisms. This gives many non-trivial examples related to the first question. Budney and Burton have an excellent survey of the state of the art on embedding 3-manifolds in $S^4$, which explains most of the known obstructions, and exhibits lots of interesting examples. Note that it's much harder to specify $Y'$ and find $Y$; I would bet that (maybe assuming the Schoenflies conjecture) a `generic' $Y'$ can't be embedded in this fashion. The construction is related to the second question, which is in part answered by the existence of doubly-slice knots, ie the special case when $K$ is the unknot. Examples can be found in Fox's Quick Trip through Knot Theory, and the subject was expanded quite a bit by De Witt Sumners (Invertible knot cobordisms. Comment. Math. Helv. 46 1971 240–256). There's a fair amount of literature on the subject, mostly giving obstructions to a given knot $K'$ being a slice of $\mathcal{O} \times I$ (i.e. $K'$ being doubly slice). I would make the same bet as above (without any assumptions) in the knot-theory setting: for a `generic' $K'$, there is no $K$ with $K'$ splitting $K \times I$.<|endoftext|> TITLE: Two H-space structures on S^3 and [X,S^3] different as groups for each: Explicit Example? QUESTION [10 upvotes]: There are twelve continuous maps $S^3\times S^3\to S^3$ up to homotopy that make the three-sphere $S^3$ into an H-space. This follows from a result of James [1], which says that if there exists one such multiplication on $S^n$ then the homotopy classes of multiplications that make $S^n$ into an H-space are in bijection with $\pi_{2n}(S^n)$. For $n=3$, we have $\pi_6(S^3) \cong \mathbb{Z}/12$. Not all of these other multiplications are homotopy associative, but eight of them are, and hence necessarily also have homotopy inverses. Question. Is there an example in the literature of a space $X$, and two homotopy associative multiplications $m,m':S^3\times S^3\to S^3$, such that the groups $[X,(S^3,m)]$ and $[X,(S^3,m')]$ have been calculated explicitly and are not isomorphic? Necessarily, $X$ cannot be a suspension. I am also interested in the same question but with $S^3$ replaced with any other H-space. [1] James, I. M. "Multiplication on spheres (II)." Transactions of the American Mathematical Society (1957): 545-558. REPLY [5 votes]: This is semiexplicit. For any H-space $G$ with multiplication $\mu$, the projection maps $p_1, p_2: G\times G\to G$ have the property that $$ [p_1] \cdot [p_2] = [\mu] \in [G\times G, G]. $$ So if you have two different multiplications $\mu_1$ and $\mu_2$ on $S^3$ with induced multiplications $*_1$ and $*_2$ on $[-,S^3]$, you'll have $$ [p_1]*_1 [p_2] = [\mu_1] \neq [\mu_2] = [p_1]*_2 [p_2] $$ in $[S^3\times S^3, S^3]$.<|endoftext|> TITLE: How to compute class number of a torus QUESTION [7 upvotes]: Let $T$ be an algebraic torus over a number field $K$. Following notations in Ono's The Arithmetic of Tori, http://www.jstor.org/discover/10.2307/1970307?sid=21105671135711&uid=3739888&uid=2&uid=3739256&uid=4 we define $T_A$ the adele points, $T_K$ the $K$-rational point of $T$ and $$T_{A,S_{\infty}}=\prod_{v \in S_{\infty}}T(K_v)\times \prod_{v \notin S_{\infty}}T_v^c $$ The class number of $T$ is defined as $$ h_T := [{T_A}:{T_{A,S_{\infty}}T_K}] $$ My question is how do we compute $h_T$ in pratice. If $T=\mathbb{G}_m$, $h_T$ is just the class number of $K$ and we compute it using Minkowski's bounds. But I don't know how to compute $h_T$ in general. For an example, if $T$ is the torus $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1))$, what is $h_T$ ? Thank you very much. REPLY [2 votes]: You can use relationships between different tori. For instance, your torus lives in a long exact sequence $T\to \operatorname {Res} _{\mathbb Q(i) / \mathbb Q } \mathbb G_m \to \mathbb G_m $ This gives you a cohomology long exact sequence: $ H^0( \mathbb Z, \operatorname {Res} _{\mathbb Q(i) / \mathbb Q } \mathbb G_m) \to H^0( \mathbb Z, \mathbb G_m) \to H^1(\mathbb Z, T) \to H^1 ( \mathbb Z, \operatorname {Res} _{\mathbb Q(i) / \mathbb Q }\mathbb G_m ) \to H^1(\mathbb Z, \mathbb G_m)$ Equivalently: $ H^0( \mathbb Z[i], \mathbb G_m) \to H^0( \mathbb Z, \mathbb G_m) \to H^1(\mathbb Z, T) \to H^1 ( \mathbb Z[i], \mathbb G_m ) \to H^1(\mathbb Z, \mathbb G_m)$ Evaluating the $H^0$s as unit groups and $H^1$s as class groups: $ \mu_4 \to \mu_2 \to H^1(\mathbb Z, T) \to 0 \to 1$ The map $\mu_4 \to \mu_2$ is the norm map, which is trivial, hence $H^1(\mathbb Z, T) = \mu_2$, and the class number is $2$. You can probably do this in general but you need a spectral sequence.<|endoftext|> TITLE: A question about Mirzakhani et. al.'s algebraicity theorem QUESTION [9 upvotes]: While the geodesic flow on a complete hyperbolic surface is ergodic, the closure of an individual orbit (a geodesic line) can take a complicated fractal-like shape. Nonetheless, there is an affirmative result in this direction. One higher-dimensional generalization of a geodesic line is a totally geodesic immersed submanifold, and for these, Nimish Shah has proved (Closures of totally geodesic immersions in manifolds of constant negative curvature) that similarly to what happens on a flat torus, the closure of a complete immersed totally geodesic submanifold of dimension at least $2$ in any compact hyperbolic manifold $M$ is always a totally geodesic immersed submanifold of $M$. (I think this result has subsequently been extended to cover the weaker assumption that $M$ is complete, not necessarily compact.) In moduli space $\mathcal{M}_g$ with the Teichmuller metric, where again the closure of a geodesic can take a wild fractal shape, Mirzakhani and her coworkers have proved that the closure of a complex geodesic is always an algebraic subvariety. This setting, while technically rather more involved, has been frequently compared to the homogeneous one of the first paragraph. In light of this, here is my question, which I will split into two variants as I am unsure which of them (if either) makes more sense: For a totally geodesic immersed submanifold $N \to \mathcal{M}_g$ of dimension at least $2$, is it reasonable to expect that the closure is still an immersed submanifold? It is easy to imagine a definition of a higher-dimensional complex totally geodesic immersed submanifold in $\mathcal{M}_g$. Extending Mirzakhani et. al.'s theorem, can it be shown that the closure of such an immersed complex submanifold is an algebraic subvariety? REPLY [7 votes]: All of these results are about the dynamics of group actions. There is no group action on the hyperbolic manifold itself, rather there is a group action on the frame bundle, which is a homogeneous space G/Gamma. Here G is a Lie group, and Gamma is a lattice. For hyperbolic n-manifolds, G is the isometry group of the n dimensional hyperbolic plane. In the homogenous setting, Ratner's Theorem classifies orbit closures for homogeneous group actions, when the group is generated by unipotents. SL(2,R) is generated by unipotents, but the one parameter subgroup giving rise to geodesic flow is not. The orbits of SL(2,R) in G/Gamma project to totally geodesic planes in the hyperbolic manifold. For hyperbolic manifolds, since the dynamics takes place on G/Gamma, there are many groups which act. Indeed, any Lie subgroup of G acts naturally on G/Gamma. For moduli space, the situation is partially analogous, but also substantively different. The Hodge bundle over M_g, i.e. the set of pairs (X,omega) where X is a Riemann surface and omega is a holomorphic one form on X, admits a SL(2,R) action, even though M_g itself doesn't admit any such action. So the Hodge bundle is the analogue of G/Gamma, and indeed all of the work of Eskin-Mirzakhani-Mohammadi takes place on the Hodge bundle, using the dynamics of the SL(2,R) action. The difference is that since SL(2,R) is a small group, there are very few subgroups that can act. For the diagonal subgroup, orbit closures can be complicated fractal objects. Orbit closures of unipotent flow, unlike in the homogenous case, are not currently understood. Orbit closures for all of SL(2,R), and for the upper triangular subgroup of SL(2,R), were studied by Eskin-Mirzakhani-Mohammadi. The connection to your question is the SL(2,R) orbits map to complex geodesics. Since the group GL(2,R) is so small, there are no other actions we can consider. Other totally geodesic submanifolds of M_g are not, as far as I know, tied to a group action with well understood dynamics. Let me end with two side notes. 1) Stergios Antonakoudis has shown that the situation for complex disks is quite special; other bounded symmetric domains don't admit isometric maps to M_g: http://www.math.harvard.edu/~stergios/papers/TeichBSD.pdf. 2) I wrote a short piece for the Bulletin of the AMS introducing the work of Eskin-Mirzakhani-Mohammadi from a more elementary perspective: http://web.stanford.edu/~amwright/BilliardsToModuli.pdf.<|endoftext|> TITLE: Obstruction and rational points on curves QUESTION [6 upvotes]: Is etale-Brauer the only obstruction to the existence of rational points on projective plane curves over number fields? REPLY [10 votes]: As a supplement to Daniel's answer, I'd like to mention that for (smooth projective geometrically irreducible) curves over number fields, "Brauer-Manin is the only obstruction to the existence of rational points" is a question formulated in Skorobogatov's book (page 133) and a conjecture of mine. See my paper, in particular sections 8 and 9. The statement is true when the Tate-Shafarevich group and the Mordell-Weil group of the Jacobian are both finite, see Thm. 8.6 in the paper. Poonen in this paper (Experiment. Math. 2006) has a conjecture based on heuristic considerations that would imply the general statement.<|endoftext|> TITLE: Geometric intersection with incompressible surfaces QUESTION [13 upvotes]: Let $M$ be a oriented compact $3$-manifold, closed or with boundary. For any incompressible surface $F$, define a function $i_F$ on the set of homotopy classes of closed curves in $M$ by $$i_F (\alpha) = \alpha \cap F $$ the geometric intersection number of $\alpha$ with $F$. Is it true that two incompressible, $\partial$-incompressible surfaces $F$ and $F'$ are isotopic if $i_F =i_{F'}$? REPLY [11 votes]: Yes, this is true (with an appropriate definition of $i_F(\alpha)$). An incompressible surface gives rise to an action of $\pi_1(M)$ on a tree (see for example Chapter 1 of Shalen's notes). For a homotopy class of loops $\alpha$, define $i_F(\alpha)=\|\alpha\|$ to be the translation length of $\alpha$ acting on this tree (well-defined up to conjugacy). In fact, $\alpha$ can be homotoped so that the geometric intersection number is $\|\alpha\|$. Then this defines a length function on conjugacy classes in $\pi_1(M)$, satisfying the Culler-Morgan axioms (1.11): Culler and Morgan show that their axioms uniquely characterize a minimal action of a group on a tree, if the action is semi-simple. It's not hard to see that if the surface is not a fiber or semi-fiber, that the action is semi-simple, essentially from the cocompactness of the action. In the fiber or semi-fiber case, the intersection number determines a homomorphism to $\mathbb{Z}$ or $D_\infty$, whose kernel is finitely generated and determines the surface group. In the semifibered case, I don't think $I_F(\alpha)$ distinguishes the two non-orientable surfaces. However, your assumption of incompressible surface usually means the surface is orientable, and thus this case won't appear. Thus, suppose two surfaces induce the same length function. Then they induce the same action on a tree, and thus the edge stabilizer will be the same, implying that the surfaces are homotopic. But homotopic incompressible surfaces are isotopic, by a result of Waldhausen (Corollary 5.5; Waldhausen assumed that the manifold is irreducible, but I believe this was only to avoid the pitfall of a potential counterexample to the Poincaré conjecture as a connect summand).<|endoftext|> TITLE: Reference or explanation: Cup products, deformations of complex structure and Mirror Symmetry QUESTION [10 upvotes]: In section 0.3. of their paper "Frobenius Manifolds and Formality of Lie Algebras of Polyvector Fields," Barannikov and Kontsevich discuss the fact that Kontsevich's formality morphism (from his paper on the deformation quantization of Poisson manifolds) respects the cup products on the cohomologies of the tangent complexes. They write: "The Formality theorem (see [K2]) identifies the germ of the moduli space of $A_\infty$ deformations of the derived category of coherent sheaves on $M$ [a Calabi-Yau] with the moduli space $\mathcal{M}_{\mathbf{t}}$ [associated with the algebra of holomorphic polyvector fields]. The tangent bundle of this moduli space after the shift by $[2]$ has natural structure of the graded commutative associative algebra. The multiplication arises from the Yoneda product on Ext-groups. The identification of moduli spaces provided by the Formality theorem respects the algebra structure on the tangent bundles of the moduli spaces. This implies, in particular, that the usual predictions of numerical Mirror Symmetry can be deduced from the homological Mirror Symmetry conjecture proposed in [K1]. [My emphasis.] We hope to elaborate on this elsewhere." I am thouroughly familiar with Kontsevich's Formality theorem and while I'm not really up to speed regarding the technical details of how to deform the derived category of coherent sheaves as an $A_\infty$ category, I trust that such deformations are classified by the relevant Hochschild cohomology. It is the second to last sentence (boldfaced) that I do not understand. My question: Did Barannikov and Kontsevich elaborate on it elsewhere? Did someone else? Edit: The reason I ask is that I can prove that the two tangent complexes in question are at general base points not quasi-isomorphic as $A_\infty$ algebras (though their cohomologies are isomorphic as associative algebras, forgetting higher multiplications), and I'm trying to figure out if this has any interesting implications for any Mirror Symmetry calculations. REPLY [3 votes]: Yes, this story is heavily expanded upon :-) As far as I understand it, the genus-zero Gromov-Witten invariants of the A-side and the Hodge theory of the B-side can be arranged into a gadget called 'Variations of semi-infinite Hodge structures" introduced by Barannikov. Mirror symmetry predicts that if $X$ and $Y$ are mirrors, then there should be an isomorphism between their respective VSHS's. Kontsevich proposed the Homological Mirror Symmetry conjecture which sees mirror symmetry as an equivalence between two non-commutative spaces - the (derived) Fukaya category of the A-side, and the derived category of coherent sheaves on the B-side. The question is how to get from this very sophisticated statement to the classical one. This expectation was made precise by Barannikov and Katzarkov-Kontsevich-Pantev: the idea is that the cyclic homology of a 'nice' $A_\infty$-category (proper, smooth, Hodge-to-deRham degeneration conjecture holds) carries a VSHS. That's the bridge you need to make the connection, i.e., taking cyclic homology of the Fukaya category should recover the A-side VSHS and taking cyclic homology of the bounded derived category should recover the B-side VSHS. Proving this is exactly the content of Gantara-Perutz-Sheridan work ... see: Mirror symmetry: from categories to curve counts and Formulae in noncommutative Hodge theory.<|endoftext|> TITLE: Are all rational exactly solvable differential equations known? QUESTION [5 upvotes]: Are there known necessary and sufficient conditions that specify in terms of an algorithm in a real arithmetic model (where real operations, elementary functions, and comparisons are elementary steps) when a first order differential equation $x'(t)=F(x(t),t)$ with a real-valued rational function $F(x,t)$ in real scalars $x$ and $t$ is exactly solvable by elementary functions and finitely many integrations? Useful (positive and negative) partial results? In particular, are there strong sufficient conditions for rational functions with low numerator and denominator degree? REPLY [5 votes]: I don't think a general answer to your question is known. I personnaly doubt that it could be positive. Partial decidable answer: if the variational linear differential system obtained along a given (algebraic) trajectory is not solvable by quadrature (condition expressed as the virtually solvability of its Picard-Vessiot group) then the equation is not solvable by quadrature. This is a theorem by Casale (an improvement on the works by Morales and Ramis) available here: «Morales-Ramis Theorems via Malgrange pseudogroup, Ann. Institut Fourier 59 n 7 (2009)» If I remember correctly such a condition can be algorithmtically checked (although you need to address the "equality-to-$0$" problem). Partial undecidability answer: if your real arithmetic cannot decide the equality to $0$, then the Poincaré question is undecidable. The question asks for the existence of a rational first-integral of the differential equation. This is easily checked by considering the linear system $$t\dot x=\lambda x$$ with constant $\lambda$: such a rational first-integral exists only if $\lambda$ is rational (which is not decidable under the assumption). There does exist «elementary closed-form» first-integral, though, e.g. $H(t,x)=xt^{-\lambda}$, which are totally acceptable for answering your question (as you point out in the comments). I just thought that mentionning this fact could shed some light on "concrete" diffculties if you try to deal with actual equations.<|endoftext|> TITLE: Decidability of the Hilbert lattice and quantum logic QUESTION [9 upvotes]: What is known about the decidability of (first-order formulas in) the structure $(\mathcal{L}(H),\leq)$, where $\mathcal{L}(H)$ is the collection of all closed linear subspaces of a (separable) Hilbert space $H$, and $X\leq Y$ means $X\subseteq Y$? (Clearly meets and joins always exist and are first-order definable, so you can throw those in too.) I can find some reference to the fact that this problem is open for infinite dimensional $H$, but known to be answered in the affirmative for finite dimensional $H$ (see: http://arxiv.org/pdf/math/0412144.pdf). I'm unfamiliar with the quantum logic literature, so references dealing with this question would be appreciated! REPLY [7 votes]: A year after you posted the question, Fritz showed the common theory of all such lattices is undecidable: https://arxiv.org/abs/1607.05870 In reponse to @MattF's query I'll post an example of how infinite dimension differs from finite. Namely, the lattice of closed subspaces is modular only in the finite-dimensional case. This is due to Birkhoff and von Neumann 1936 although they gave fewer details. Let $e_1,e_2,\dots$ be a countable orthonormal basis for a separable subspace of your Hilbert space. Let $A$, $B$ be respectively the closed subspace spanned by the vectors $$a_n=e_{2n}+10^{-n}e_1+10^{-2n}e_{2n+1},\qquad b_n=e_{2n},$$ and let $C$ be spanned by $A\cup\{e_1\}$. Then $B$ is incomparable with both $A$ and $C$ (under inclusion), and $A$ is a proper subspace of $C$. Let's prove that $A$ is a proper subspace of $C$. If not, $e_1$ should be expressible as a combination of vectors from $A$, i.e., $$e_1=\sum k_na_n=\sum k_n e_{2n}+\sum k_n10^{-n}e_1+\sum k_n10^{-2n}e_{2n+1}=0$$ since first and third terms must be 0, so that all the coefficients $k_n=0$. This gives the contradiction $e_1=0$. The modular law requires that $$A\subset C\implies A\vee (B\cap C)=(A\vee B)\cap C$$ but in this case we have $$A=A\vee (B\cap C)\quad\text{and}\quad (A\vee B)\cap C=C,$$ so we have a counterexample. On the other hand, in finite-dimensional Hilbert spaces the modular law does hold.<|endoftext|> TITLE: Which kind of subsets of primes one needs to generate a positive ratio of the natural numbers? QUESTION [7 upvotes]: Not knowing elementary number theory well, I ask this one, which is not very clear to answer, rather I am looking for some results around this question or known theorems. The problem is the following: The set of prime numbers $\mathbb{P}=2,3,5,7,11...$ generates $\mathbb{N}$ by multiplication. Now I am interested in subsets $S$ of $\mathbb{P}$, which generate a positive fraction of all numbers, that means: There is $\epsilon>0$ which holds the following equation for all $N \in\mathbb{N}$ big enough: $$\frac{\|span(S)_{\leq N}\|}{N}\geq \epsilon N$$ where $span(S)_{\leq N}$ is the subset of $\mathbb{N}_{\leq N}$, which consists of the numbers, whose prime factors are all elements of $S$. Clearly $S$ has to be infinite to have this property, but what can one say about $S$ more specificely? Is there any known criterions or examples for $S$ being too small to generate a positive ratio of the natural numbers? REPLY [12 votes]: Let $T=\mathbb P\setminus S$. If $\sum_{p\in T}1/p=\infty$, then $\prod_{p\in T}(1-1/p)=0$ and for any $\epsilon>0$, there exist $p_1,\ldots,p_n\in T$ such that $\prod_{i=1}^n(1-1/p_i)<\epsilon$. Now modulo $P=p_1\cdots p_n$, the fraction of integers that have no factor of the form $p_i$ with $i\le n$ is $\prod(1-1/p_i)<\epsilon$. Hence in the entire set of natural numbers, those with no factor in $T$ has density less than $\epsilon$. Conversely, if $\sum_{p\in T} 1/p<\infty$, then let $\alpha=\prod_{p\in T}(1-1/p)$. Then $0<\alpha<1$. Let $M$ be chosen such that $\sum_{p\in T; p>M} 1/p<\alpha/2$. Then up to $N$, the number of integers that has a factor in $T$ that exceeds $M$ is at most $\sum_{p\in T; p>M} N/p<\alpha N/2$. For large $N$, the number of $n\le N$ having no factor in $T$ below $M$ is close to $\alpha N$, and hence the density of $n$'s having no factor in $T$ is at least $\alpha/2$.<|endoftext|> TITLE: Is a locally free sheaf projective in the category of $\mathcal{O}_X$-modules when $X$ is an affine scheme? QUESTION [15 upvotes]: Let $X$ be an affine scheme and $\mathcal{E}$ a finitely generated locally free sheaf on $X$. It is obvious that $\mathcal{E}$ is a projective object in the category Qcoh$(X)$ since we can pass to rings and modules. My question is: is $\mathcal{E}$ still projective when we consider it in the larger category $\mathcal{O}_X$-mod which consists of all $\mathcal{O}_X$-modules? If it is true, how to prove it in a "cohomological way"? I'm not sure whether this question is suitable for mathoverflow. Please feel free to move it if necessary. REPLY [19 votes]: This answer is inspired by the discussion at this question. Let $X$ be an integral affine scheme admitting an open cover $X=U\cup V$ with $U$, $V$ and $X$ all distinct. I claim that $\mathscr O_X$ is not projective in $\operatorname{Mod}(\mathscr O_X)$. Let $i\colon U\hookrightarrow X$ and $j\colon V\hookrightarrow X$ be the inclusion morphisms. There is an obvious surjection $p\colon i_! \mathscr O_U\oplus j_! \mathscr O_V\twoheadrightarrow \mathscr O_X$ (check surjectivity on stalks). If $\mathscr O_X$ were projective, $p$ would split, so we would have $\Gamma(\mathscr O_X)\hookrightarrow \Gamma(i_!\mathscr O_U)\oplus \Gamma(j_! \mathscr O_V)=0$, a contradiction.<|endoftext|> TITLE: Is the ring $\mathbb{Z}_p [[x]]\otimes_{\mathbb{Z}_p} \overline{\mathbb{Q}}_p$ Noetherian? QUESTION [5 upvotes]: Is the ring $\mathbb{Z}_p [[x]]\otimes_{\mathbb{Z}_p} \overline{\mathbb{Q}}_p$ Noetherian? REPLY [11 votes]: Yes. Consider any nonzero element of the ring. Using the norm, we can see that it divides some nonzero element of $\mathbb Z_p [[x]] \otimes_{\mathbb Z_p} \mathbb Q_p$. By multiplication by $p$, it divides some nonzero element of $\mathbb Z_p[[x]]$. Applying Weierstrass preparation, we may factor it as a unit times a power of $p$ times a monic polynomial. If $f$ is a monic polynomial in $x$, then $\mathbb Z_p [[x]] \otimes_{\mathbb Z_p} \mathbb Q_p / f$ is finite over $\mathbb Q_p$, so tensoring it with $\overline{\mathbb Q}_p$ it will still be finite, hence Noetherian. Since the quotient by any nonzero element is Noetherian, the ring is clearly Noetherian.<|endoftext|> TITLE: Maximal abelian subgroup of general linear groups QUESTION [6 upvotes]: Thanks for any help or comments. Is it possible to recognize all maximal abelian subgroups of general linear group on finite field $F$ of order $q$, $GL_n(F)$. By maximal abelian I mean if $A$ is maximal abelian and $B$ is abelian such that $A\subseteq B$, then $B=A$. REPLY [3 votes]: Questions of this type have been raised about various finite groups of Lie type at MathOverflow previously, for example here. As Nick Gill's comment indicates, the work of E. Vvodin is worth consulting, along with an earlier paper by M. Barry, etc. Naturally the general (or special) linear group over a finite field is somewhat easier to study directly, using a mixture of techniques from linear algebra and finite group theory. But there is some advantage in looking at all finite groups of Lie type from the perspective of algebraic groups. Two basic questions tend to arise: (1) determine (up to conjugacy) all maximal abelian subgroups, (2) find the largest order of any such subgroup. From either the linear algebra or the algebraic group viewpoint, a natural tool here is the Jordan decomposition of elements. It turns out that semisimple elements (those of order not divisible by $p$) are the easiest to study systematically, largely because their centralizers are again reductive -- and even connected when the algebraic group is simply connected (true here for $\mathrm{SL}_n$). In particular, an abelian subgroup $A$ of the finite group $G$ consisting of semisimple elements always lies in a maximal torus of the algebraic group defined over $\mathbb{F}_q$. This is one of the results developed for all finite Chevalley groups by Springer and Steinberg in their extensive notes on conjugacy classes: Part E in Seminar on Algebraic Groups and Related Finite Groups (Springer Lecture Notes in Math. 131, 1970), II, 5.8-5.12. (But there are nuances for some primes in types other than the special linear groups.) In particular, the groups of rational points (or fixed points under Steinberg's endomorphism $\sigma$) in the various maximal tori are easily seen to be maximal abelian subgroups and have orders specified in terms of data from the Weyl group, here $S_n$: II, 1.7. These orders are approximately $q^n$ for the finite general linear groups ($n$ being the overall rank). Inductive methods like those suggested by Geoff Robinson are often helpful when only semisimple elements are discussed. The complication is that the maximum order of an abelian subgroup is approximately $q^{n^2/4}$, typically much larger than a finite torus. Since the centralizers in the algebraic group of nontrivial unipotent elements (= elements having $p$-power orders) are usually far from being reductive, it is tricky to work out the orders of all maximal abelian subgroups of $G$ which involve such elements.<|endoftext|> TITLE: When is the pullback in Chow groups defined? QUESTION [6 upvotes]: This is the first time I ask a question on Mathoverflow, so I apologize in advance if it is not suitable/a duplicate/otherwise inappropriated. I am thinking about Voevodsky's category of motives and I realized that in his presheaves with transfers formalism pullbacks for Chow groups are defined for arbitrary maps of smooth schemes. Precisely if $f:X\to Y$ is a map of smooth schemes and $\alpha\in CH_*(Y)$ he defines $$f^*\alpha = (pr_1)_*(\Gamma_f \cdot (pr_2)^*\alpha)$$ where $pr_1,pr_2$ are the projection maps from $X\times Y$ and $\Gamma_f$ is the closed subscheme of $X\times Y$ determined by the graph of $f$ (note that $(pr_1)_*$ is well defined more or less because the restriction of $pr_1$ to $\Gamma_f$ is an isomorphism). After researching a bit I found a paper by Bloch ("Algebraic cycles and Higher K-theory") he seems to define the pullback for all maps with smooth target. However in classical treatments of intersection theory I've only seen $f^*$ defined for flat maps. Q: In what generality is the pullback of cycle classes defined? REPLY [10 votes]: You have to distinguish between pullbacks of cycles, pullbacks on Chow groups, and pullbacks of relative cycles. You cannot always pull back cycles. If $f: Y\to X$ is a morphism of (arbitrary) schemes and $Z\subset X$ is an elementary cycle, $f^*(Z)$ is defined provided that $Z$ is ``in good position with respect to $f$'', which simply means that $f^{-1}(Z)$ has the same codimension in $Y$ as $Z$ had in $X$; this is always true if $f$ is flat. The formula for $f^*(Z)$ is the one you wrote, where the intersection uses Serre's multiplicities. Pullbacks on Chow groups are defined at least for arbitrary maps between smooth schemes over fields (see below for various generalizations). This is because of Chow's moving lemma, which says that you can always replace a cycle by a rationally equivalent one which is in good position with respect to a given map. If $X$ is of finite type over $S$ regular, an equidimensional relative cycle on $X/S$ is a cycle on $X$ which is equidimensional and dominant over (a connected component of) $S$. For instance a finite correspondence from $X$ to $Y$ over $S$ is such a cycle on $X\times_SY/X$, which is moreover finite over $X$. Equidimensional relative cycles can be pulled back along any map $T\to S$. There is a notion of relative cycle which need not be equidimensional and these notions can be extended to singular schemes, but it gets technical, see Suslin-Voevodsky http://www.math.uiuc.edu/K-theory/0035/susvoe2.pdf or Cisinski-Déglise http://arxiv.org/pdf/0912.2110v3. With hard work, the functoriality on Chow groups can be extended to Bloch's cycle complexes $z^r(X,*)$, and hence to higher Chow groups. This was done by Bloch for smooth schemes over a field, by Levine for smooth schemes over a Dedekind domain (https://www.uni-due.de/~bm0032/publ/ChowMovLemFinal.pdf), and recently by Spitzweck for smooth schemes over variable Dedekind domains (http://arxiv.org/pdf/1207.4078.pdf). A huge advantage of Voevodsky's definition of the motivic complexes is that the functoriality is apparent, since these complexes are defined in terms of equidimensional cycles. At the end of the day Voevodsky's complex is equivalent to Bloch's, but that requires a moving lemma (which is known for fields but not for more general Dedekind domains). The basic idea for the comparison is that, up to controlled rational equivalence, a codimension $r$ cycle on $X$ is as good as a codimension $r$ cycles on $X\times\mathbb{A}^r$ (by homotopy invariance), and any such can be moved to be of relative dimension zero over $X$.<|endoftext|> TITLE: Stokes theorem with corners QUESTION [8 upvotes]: I've found the following version of Stokes' theorem in G. Stolzenberg's lecture notes 19: Notation: for $1 \le n \le m$ $\Lambda(m, n) = \{ \lambda: \{1,...,n\} \rightarrow \{1,...,m\} \ | \ \lambda(1) < ... \lambda(n) \}$ $p_{\lambda}: \mathbb{R}^m \ni (x_1, ..., x_m) \rightarrow (x_{\lambda(1)}, ..., x_{\lambda(n)}) \in \mathbb{R}^n$ $\mathcal{L}^n$ is $n$-dimensional Lebesgue measure and $\mathcal{H}^n$ is Hausdorff measure Now, the theorem itself: Assume that $(1)$ $N$ is an $n$-dimensional, oriented class $C^1$ submanifold of $\mathbb{R}^m$, $2 \le n \le m$ $(2) \ M \subset N$ is open, $int _N (\overline{M}^N) = M$ and $\partial _N M$ is either empty or is an $n-1$-dimensional class $C^1$ submanifold in $\mathbb{R}^{m}$ $(3) \ \overline{M}$ is compact, $\mathcal{H}^n(M) < \infty$, $\mathcal{H}^{n-1}(\partial_NM) < \infty$ $(4)$ Let's define $\delta: = \overline{M} \setminus N$ and assume that $(*) \ \ \forall \lambda \in \Lambda(m, n-1) : \ \mathcal{L}^{n-1}(p_{\lambda}(\delta)) = 0$ $(5) \ \omega \in \Omega_{n-1}^{(1)} (\mathbb{R}^m)$ $(6)$ We induce the orientation on $M$ and on the boundary: $\partial _NM$ from $N$ Then we have that: $d \omega $ absolutely integrable on $M$, $\omega$ on $\partial_NM$ if it is not empty and $\int_M d \omega = \int_{\partial_NM} \omega$ While proving this theorem, we assume, wlog, that $p_{\lambda} : \mathbb{R}^m \ni (x_1, ..., x_m) \rightarrow (x_1, ..., x_{n-1}) \in \mathbb{R}^{n-1}$ My question is: Apparently, we can replace $(*)$ with $\mathcal{H}^{n-1}(\delta)=0$ which implies $(*)$. I was wondering whether you could explain to me why this is true. Also, could you recommend a book or a paper in which I can find something more about Stokes' theorem with corners (singular points)? I've already read Stokes theorem for manifolds with corners? and consequently: Brian Conrad's notes on differential geometry: math.stanford.edu/~conrad/diffgeomPage/handouts.html (but the problem is that I need a source which has been published) a chapter dedicated to Stokes' theorem in Sauvigny's "Partial Differential Equations" (here, we simlarly consider the set of singular boundary points which has capacity zero, although at the moment I'm not able to decide which assumption is more general ) I've also read this article but it's not connected to my main theorem (presented above) Could you recommend some books, papers in which I can find something about Stokes theorem which will "agree" with the theorem I wrote down here? I would be extremely grateful for all your insight. REPLY [2 votes]: This book http://www.math.wustl.edu/~sk/books/root.pdf (Geometric Integration Theory, by S.G. Krantz and H.R. Parks) is a self-contained introduction to geometric measure theory. See also Hassler Whitney's classic "Geometric Integration Theory". http://www.ams.org/journals/bull/1993-29-02/S0273-0979-1993-00429-4/ (Stokes' theorem for nonsmooth chains, by J. Harrison) provides a generalization of the Stokes' theorem.<|endoftext|> TITLE: Reference for higher order Campanato Lemmas, e.g. `Sufficiently fast L^2 decay on balls to affine functions implies C^{1,\alpha}' QUESTION [5 upvotes]: Whence can I reference the following fact (I have seen it quoted as `standard' in respectable places, so I hope it is so)?: Let $f : B_2(0) \to \mathbb{R}$, say $f \in L^2(B_2(0))$ . Suppose that there exists $\alpha \in (0,1)$ and $c > 0$ such that the following is true: For every $y \in B_1(0)$, there exists an affine function $l_y : \mathbb{R}^n \to \mathbb{R}$ such that $\rho^{-n-2}\int_{B_{\rho}(y)}|f(x) - l_y(x)|^2dx \leq c\rho^{2\alpha}\int_{B_2(0)}|f(x)|^2dx$ for all $\rho \in (0,1/4)$. Then $f \in C^{1,\alpha}(B_1(0))$ with $\|f\|_{C^{1,\alpha}(B_1(0))} \leq c\|f\|_{L^2(B_2(0))}$. REPLY [5 votes]: First you can rescale so $\|f\|_{L^2(B_2)} = 1$ and you see that you are basically talking about the boundedness of the first order Campanato norm of your function $f$. Your conclusion then is a classical theorem concerning the comparison of Campanato and Holder spaces. See e.g. http://link.springer.com/chapter/10.1007%2F978-3-0348-0537-7_15 (Theorem 4.4) and references therein.<|endoftext|> TITLE: Symmetric product of a vector bundle QUESTION [6 upvotes]: Let $E$ be a vector bundle of rank $r$ and degree $d$ over a smooth curve $X$. Is there any canonical exact sequence for $Sym^k(E)$? in particular what is the degree of $Sym^k(E)$? Suppose $E$ is stable, is $Sym^k(E)$ is stable ? Is there any interesting information about $Sym^k(E)$? Thanks REPLY [5 votes]: There is no "canonical exact sequence", whatever that means. The determinant of $Sym^k(E)$ is $(\det E)^m$, with $m=\binom{r+k-1}{r}$; this follows from the analogous equality of $GL(V)$-modules $\det(Sym^k(V))=(\det V)^m$ for a vector space of dimension $r$ (which one gets easily by looking at the action of the scalars). Therefore the degree of $Sym^k(E)$ is $\ d\binom{r+k-1}{r}$. Finally, if $E$ is stable and the characteristic is 0, $Sym^k(E)$ is semi-stable: this follows from the Narasimhan-Seshadri description of (semi-) stable bundles in terms of unitary representations. It is easy to find examples where $Sym^k(E)$ is not stable; there are also examples in characteristic $p$ where it is not semi-stable.<|endoftext|> TITLE: Images of $\{0,1\}^\kappa$ QUESTION [7 upvotes]: Is there a compact topological space $(X,\tau)$ such that for no cardinal $\kappa$ there is a surjective continuous map $e:\{0,1\}^\kappa \to X$? (We assume that $\{0,1\}$ is endowed with the discrete topology, and $\{0,1\}^\kappa$ has the product topology.) REPLY [3 votes]: I'd say that the simplest counter-example is the $1$-point compactification of a non-countable discrete space (of course any compactification of any non-countable discrete space would do).<|endoftext|> TITLE: Minimal expected absolute value of linear combinations of Gaussian random variables QUESTION [11 upvotes]: I am interested in the following question. Consider $n$ independent standard normal random variables $g_i$. Cosider a linear combination $w_1g_1+\cdots+w_ng_n$. Can one give a "decent" upper bound for \begin{equation} \mathbb{E}\min_{w_i \in \left\{-1,1\right\}}|w_1g_1+\cdots+w_ng_n| \text{?} \end{equation} Basically, I am asking about the minimum expected absolute value of a family of correlated gaussian random variables. If a good bound can be obtained, what about the same question for more general linear combinations, such as $w_1a_1g_1+\cdots+w_na_ng_n$ in term of $n$ and some norm of $a_i$, say $l_{2}$? REPLY [3 votes]: Experimentally, a constant bound of $2/3$ should do, while the bounds above grow with n. We can get a reasonable bound by using the Thue-Morse sequence to select $w_i$'s. So we start with a weight of $+1$ for the largest $|g|$, and then weight smaller $|g|$'s with the inverse of the signs so far. \begin{equation} \mathbb{E}\min_{w_i \in \left\{-1,1\right\}}|w_1g_1+\cdots+w_ng_n| \ \le\ \mathbb{E}\left|\sum_{i=1}^n s_{n-i}\ |g|_{(i)}\right| \end{equation} where $|g|_{(i)}$ is the $i^{th}$ element after sorting the $|g|$'s, and $s_i$ is the $i^{th}$ element of A106400. E.g. if the $w$'s are 1.31, -0.25, 2.59, 0.68, -0.77, then this bound is |2.59 - 1.31 - 0.77 + 0.68 - 0.25|. This gives an expectation of $(4-2\sqrt{2})/\sqrt{\pi}$ for $n=2$, using reasoning like Bjorn Kjos-Hanssen's. Here is some Mathematica code for experimenting with 100 sets of $n$ random variables: I got expectations for this bound around 0.18 with $n$ of 100,000 or 1,000,000. $\\$ [Update: We can use the same notation to prove that the expectation in the question is less than $E[\max|g_i|]$. Let $v_n = 1$, let $v_{j-1} = -\text{sign}( \sum_{i=j}^n v_i |g|_{(i)} )$, and then indeed $\left|\sum v_i |g|_{(i)} \right| < \max |g(i)|$.]<|endoftext|> TITLE: Is the Poincaré metric continuous with respect to the domain? QUESTION [6 upvotes]: Suppose $K \subset \mathbb{C}$ is a Cantor set and let $u:\mathbb{C} \setminus K \to \mathbb{R}$ be the maximal smooth function such that the conformal metric $e^{2u}(\mathrm{d}x^2 + \mathrm{d}y^2)$ has constant curvature $-1$ on $\mathbb{C} \setminus K$. Suppose we put the Hausdorff distance on the set of compact subsets of $\mathbb{C}$. I'm interested in how $u$ varies with respect to the Cantor set $K$. Question: Does $u$ vary continuously in the smooth topology on compact subsets of $\mathbb{C} \setminus K$ with respect to $K$? Some more background: The existence of a unique maximal hyperbolic metric (sometimes called the Poincaré metric of $\mathbb{C} \setminus K$) is proved for example in Ahlfors' book. I've looked around and found several references estimating the metric in terms of distance to the Cantor set (for example this one). I haven't found any result on how the metric varies with respect to the domain save the easy observation of domain monotonicity (i.e. $u$ grows if $K$ does). REPLY [8 votes]: In "D.A. Hejhal, Universal covering maps for variable regions, Mathematische Zeitschrift 137 (1974), 7--20." it is shown that the universal covering map of a hyperbolic domain depends locally uniformly continuously on the domain. The convergence of the derivatives follows from Cauchy's integral formula. Since the hyperbolic metric on a domain is the push-forward of the Poincare metric by the universal covering map, the hyperbolic metric also depends continuously on the domain. This is all with respect to the Caratheodory topology on pointed domains, which is a weaker topology than the Hausdorff topology on the complements, so it implies what you want.<|endoftext|> TITLE: When do two non-degenerate quadratic forms give rise to isomorphic Lie algebras? QUESTION [7 upvotes]: Let $V$ be a vector space over some number field $k$. (I'm fine with $\mathbb{Q}$.) Let $\phi \colon V \to k$ be a non-degenerate quadratic form. Associated with $\phi$ is the orthogonal group $\mathrm{O}(V,\phi) \subset \mathrm{GL}(V)$ of linear automorphisms preserving $\phi$. The connected component of the identity is $\mathrm{SO}(V,\phi)$. Suppose $\psi \colon V \to k$ is another non-degenerate quadratic form. If $\phi \sim \psi$ (equivalent as quadratic forms), then definitely $\mathrm{SO}(V,\phi) \cong \mathrm{SO}(V,\psi)$. The same is true if $\phi \sim \lambda \cdot \psi$, for some scalar $\lambda \in k^{*}$. Question: Is there a general statement about when $\phi$ and $\psi$ have special orthogonal groups in the same isogeny class? Remarks: I am asking about isogeny classes, not isomorphism classes of groups. I don't know if this makes the question harder or easier. (With isogeny, I mean a homomorphism between algebraic groups of the same dimension (trivial in this case) such that the kernel is finite.) [Edit] I changed the question, so that it is no longer about isogenous groups, but about isogeny classes. In particular, I would like to know what the conditions on $\phi$ and $\psi$ are, so that there exists a group $G$, with isogenies $G \to \mathrm{SO}(V,\phi)$ and $G \to \mathrm{SO}(V,\psi)$. [/Edit] I would prefer a statement similar to the classification of quadratic forms. So in terms of data similar to local Hasse invariants or such. (But maybe this is optimistic, because, for example, my above remark shows that the discriminant can be changed to anything, by twisting with a scalar $\lambda$.) [Edit2] As YCor points out in the comments below, the current version of the question is equivalent to When do $\phi$ and $\psi$ induce isomorphic Lie algebras $\mathfrak{so}(\phi)$ and $\mathfrak{so}(\psi)$ over $k$? [/Edit2] Somehow the literature (which most likely exists) cannot be found easily via Google and the likes. REPLY [6 votes]: Here's a proof assuming $n\ge 3,n\neq 8$ that the Lie algebras are isomorphic only when the quadratic forms are equivalent up to rescaling (I assume $K$ has characteristic zero and fix an algebraically closed extension $C$). Let $f:\mathfrak{so}(\phi)\to \mathfrak{so}(\psi)$ be a $K$-defined isomorphism. We can assume that both $\mathfrak{so}(\phi)$ and $\mathfrak{so}(\psi)$ are $K$-defined subalgebras of $\mathfrak{sl}_n$ preserving a nondegenerate quadratic form on the $n$-dimensional space. Then both are $C$-conjugate to $\mathfrak{so}(n)$. Since I assume $n\neq 8$ odd, over $C$, all automorphisms of $\mathfrak{so}(n)$ can be realized by some element of $\mathrm{GL}_n(C)$ (actually, of $\mathrm{O}_n(C)$). It follows that $f$ can be realized by a conjugation, namely there exists $A\in\mathrm{GL}_n(C)$ satisfying: $f(g)A=Ag$ for all $g\in \mathfrak{so}(\phi)$. The set of $A$ satisfying this condition is a $K$-defined linear subspace on which the determinant map does not vanish; hence it contains a $K$-point with nonzero determinant. That is, $A$ can be found in $GL_n(K)$. Hence $\mathfrak{so}(\psi)$ preserves the $K$-defined quadratic form $x\mapsto \phi'(x):=\phi(A^{-1}x)$. Since the set of $K$-defined invariant forms is 1-dimensional (because the standard representation of $\mathfrak{so}(\psi)$ is absolutely irreducible), it follows that $\phi'$ and $\psi$ are collinear. I don't know what's going on for $n=8$. For $n=2$, while there's only one 1-dimensional Lie algebra over $K$ so it's not enough to classify. Nevertheless it's still true that two quadratic forms are $K$-isomorphic up to rescaling [equivalently, have same determinant in $K^*/(K^*)^2$] iff they have isogenous SO(-). The point is that for 1-dimensional $K$-tori, $K$-isogenous is the same as $K$-isomorphic, and we can run the same proof as the above Lie-algebra-theoretic one, where we need to use the fact that every automorphism of $\mathrm{SO}_2(C)$ can be realized by some element of $\mathrm{GL}_2$. Actually this latter proof works for all $n\ge 2$ to show that if $\mathrm{SO}(\phi)$ and $\mathrm{SO}(\psi)$ are $K$-isomorphic then $\phi$ and $\psi$ are $K$-equivalent up to rescaling.<|endoftext|> TITLE: Smallest $k$ so that $k$-wise independence guarantees a constant expected minimum QUESTION [13 upvotes]: Imagine you sample $n$ numbers with replacement uniformly from the integers $1,\dots, n$ (we can assume $n$ is large). Let $X$ be the minimum of these samples. I am interested in $\mathbb{E}(X)$ but with a twist. All I know is that the samples are uniform and $k$-wise independent for some $k$. What is the smallest $k$ so that there is a constant upper bound for $\mathbb{E}(X)$? We know from the very nice answer of Will Sawin at Expected value of the minimum with limited independence that for pairwise independence, that is for $k=2$, $ \mathbb{E}(X)$ can be as large as approximately $\log {n}$. Obviously if $k=n$ then there is a constant upper bound on the expected minimum. What can we say for $k$ between $2$ and $n$? REPLY [9 votes]: I can do $k\geq 4$. This is done using a method similar to the upper bound from last time. Let $N_m$ be the number of samples that are at most $m$. Then we wish to upper bound the probability that $N_m=0$. We can do this using the fourth moment method, because the first four moments are the same as for a totally independent and uniform distribution. $E(N_m-m)^4=3m^2(n-m)^2(n-1)/n^3 + (m^3+(n-m)^3) m(n-m)/n^4= O(m^2)$ So the probability that $N_m-m= -m$ is $O(1/m^2)$. Summing over all $m$ and adding $1$ to find the expectation of the minimum gives something $O(1)$. (Close to $1+ \pi^2/2$, I think.) I'm not sure about $k=3$, but I'll think about it<|endoftext|> TITLE: Morgan Shalen compactification of $\mathbb C^2$ QUESTION [10 upvotes]: I'm reading the Otal's survey on the compactification of Morgan Shalen. (available here) He claims in an example (page 8) that the compactification of $\mathbb C^2$ is $S^4$, which sounds completely natural, but I'm not able to understand why, and if I do the calculations I obtain a different object. Here the definition: Let $X$ be an affine algebraic set and take a finite (or countable) generator set $F$ of the ring of the regular functions on $X$. In the example I'm interested in $X=\mathbb C^2$ and the coordinates $z,w$ are my generating set. Let $[0,\infty)^F$ and denote by $\mathbb P^F$ its projectivized, with $\pi$ the natural projection. In my example, $F$ has two elements so $\mathbb P^F$ is a closed segment. Define $\theta_0:X\to [0,\infty)^F$ by $\theta_0(x)=(\log(|f(x)|+2))_{f\in F}$ and $\theta=\pi\circ\theta_0$. Let $\hat X$ be the one-point compactification of $X$. The MS compactification of $X$, w.r.t. $F$ is the closure of the graphic of $\theta(X)$ in $\hat X\times\mathbb P^F$ In the example the function $\theta$ is $$\theta(z,w)=\dfrac{\log(|z|+2)}{\log(|w|+2)}$$ By studying the level sets of such function it seems to me that the compactification of $\mathbb C^2$ is a singular object, while in the survey is claimed that it is $S^4$. In particular the closure of a level set in $\hat X\times \mathbb P^F$ is the one-point compactification of the level set itself. Level sets are $|z|+2=(|w|+2)^c$ so at infinity they are of the form $T^2\times[a,\infty)$ whose one-point compactification is the cone over $T^2$, which is singular. As level sets are disjoint this would show that the compactification of $\mathbb C^2$ is not a manifold. My question is: Is there any place where i can find the proof that the MS compactification of $\mathbb C^2$ is $S^4$? Or, can anyone give some hint? REPLY [5 votes]: Not true. Let $S^1$ be the unit circle in $\mathbb C$. Let $\Delta$ be the closure, in the projective plane, of the quadrant $Q=[log\ 2,\infty)\times [log\ 2,\infty)\subset\mathbb R^2$. It is topologically a $2$-simplex. Map the product $S^1\times S^1\times\Delta$ to the one-point compactification of $\mathbb C\times \mathbb C$ by giving a proper map from the dense open subset $S^1\times S^1\times Q$ to $\mathbb C\times \mathbb C$, namely $$ (a,b,(s,t))\mapsto ((e^s-2)a,(e^t-2)b). $$ Map $S^1\times S^1\times\Delta$ also to the projective segment $\mathbb P^F$ by $$ (a,b,(s,t))\mapsto t/s. $$ The combined map $S^1\times S^1\times\Delta\to \mathbb C\times \mathbb C\times \mathbb P^F$ displays the space you are asking about as a quotient of $S^1\times S^1\times\Delta$. Now look at which points have been identified. For a point $p$ in the interior of the ``infinity'' side of $\Delta$, $S^1\times S^1\times p$ goes to one point, and a neighborhood of that point in the quotient space looks like the product of an open interval and a cone on $S^1\times S^1$. Thus a neighborhood looks like a cone on the suspension of $S^1\times S^1$, and is not a manifold.<|endoftext|> TITLE: Homotopy Type Theory: What is it? QUESTION [39 upvotes]: My question is, broadly, what is the main project of Homotopy Type Theory (HoTT). I asked a professor who is likely to be correct and he say the following: There are three directions: Topologists are seeing type theory as a concise and convenient ways to reason about topology, where equalities are interpreted as paths (and higher-dimensional variants thereof). Type theorists are seeing topology as a way to get new insights on type theory and variants thereof. People are pushing univalence and constructive type theory as a new foundations for mathematics. The only one of those I have any grip on is 2. My understanding, gleaned from some talks I didn't understand goes like this: Some people would like to write a proof checking/generating client that is sufficiently expressive so that you can actually do mathematics in it. The way that they get this high level of expressiveness is through a very rich type theory. It was noticed that you could intepret these types topologically: An object is a point, a proof of equality is a path, a proof that two proofs of equality are "the same" is a homotopy. Now, people use attractive features of the model, to guide further development of the proof environment, which is approximately point 2. Somehow though, there is a hope that you can use this system to compute homotopy groups of spheres? Why is this credible? Also, this is being pushed as a new foundation? Why? What advantages does it have over set theory? REPLY [27 votes]: That description of the three directions is not too bad, although they are not of course completely separate, nor does everything called HoTT fall under one of them. Also, I'm not sure whether this applies to you, but I always point out that it is potentially confusing to say "topology" when what is meant is "homotopy theory", i.e. "$\infty$-groupoid theory" — all "spaces" appear in HoTT only up to homotopy. Some people would like to write a proof checking/generating client that is sufficiently expressive so that you can actually do mathematics in it. I'm not sure what use a proof checking program would be if you couldn't do mathematics in it... (-: But yes, I think your description of point 2 is roughly accurate. Point 1 is just about using the model in the other direction: if you prove something in the type theory, then it's true as a statement about spaces. In particular, if you calculate something like a homotopy group of a sphere in the type theory, then it's also a true statement about the homotopy groups of spheres in classical algebraic topology. And in fact we've done this in a few simple cases, like $\pi_n(S^n)$ and $\pi_3(S^2)$, so it better ought to be credible. These proofs are different from the classical ones, elegant in a certain sense, and relatively easily formalizable in a proof assistant. (Notice that I'm talking about different proofs of known results here; it's not inconceivable that the new methods of HoTT might lead to proofs of new results even in classical homotopy theory, but I expect that day is far in the future.) As for point 3, type theory in general has a lot of advantages over set theory as a foundation for mathematics. Among other things, it requires less arbitrary "encoding" to represent math, it more closely matches mathematical practice, and it can be more easily programmed in a computer, producing small "proof term" certificates that can be easily verified. UF/HoTT potentially fixes some of the disadvantages of traditional type theory, such as a lack of well-behaved quotients and the problem of transporting structures across equivalences, so it's even better. And in addition to traditional mathematics, it also includes new mathematics such as the "synthetic homotopy theory" calculations of homotopy groups mentioned above. Finally, a point that I always emphasize: HoTT has many models other than the "standard" one in spaces. Roughly we can use any "$(\infty,1)$-topos" (although there are coherence questions still work in progress). This means that when we prove something in UF/HoTT (such as a homotopy group calculation, or any part of traditional mathematics that we might formalize using it as a foundation) the result is more general than in classical homotopy theory, since it is true in any of the models.<|endoftext|> TITLE: Cricket and the Hardy-Littlewod maximal function QUESTION [9 upvotes]: I'v read somewhere that one motivation for Hardy to define his maximal function is the game of cricket. But I can't see how they are related. Could anyone provide some more information on their connections? REPLY [36 votes]: One hesitates to explain a joke, but this is quite a nice joke, and I cannot resist answering a cricket question (especially now). Suppose a batsman scores 20, 100, 30, 40, 70 and 0 in his last six innings (0 being the most recent). Being upset at scoring 0 in his last innings, he might say to himself -- at least I am averaging 35 in my last two innings; or going further, that I'm averaging 36.67 in my last three; or averaging 35 in my last four; or 48 in my last five; or 43.33. Probably he would most prefer the fifth statement which gives the largest average, or the most satisfaction. Suppose now the batsman does this over an entire season of cricket, after each innings computing his satisfaction until then. Call the total satisfaction the sum of the satisfactions after each innings in the season. Then for a given stock of scores in a season, the Hardy-Littlewood maximal theorem gives that the batsman's total satisfaction is a maximum if his scores are in descending order throughout the series -- or in other words, his satisfaction is a maximum precisely when his batting has been in decline throughout the season! There is a nice discussion of this in Bela Bollobas's problem book "The art of mathematics: coffee time in Memphis" -- see problem 85 on Satisfied Cricketers: the Hardy-Littlewood maximal theorem there. The English cricket team has clearly taken this Theorem a little too seriously!<|endoftext|> TITLE: Understand rough path iterated integral and how to compute it numerically? QUESTION [9 upvotes]: The "signature" of rough path theory is defined by iterated integral as $s(k)=\int_{0 \le u_1 \le \cdots \le u_k \le t} \mathrm{d}X_{u_1} \otimes \cdots \otimes \mathrm{d}X_{u_k}$ in witch $X(t)$ is a $d$ dimensional rough path. I'm new to tensor calculus, and still in struggle to figure out what the above equation actually means, and further more, to implement it numerically? Let me make the problem more specific: 1) $X(t)$ is a $d$ dimensional rough path, which will be represented by a matrix with $d$ rows and $N$ columns, each column is actually $X(t_k)$; 2) $s(k)$ is a sequence of size $d^k-1$, of which each element is actually an integral of the following form: $s^{(j_1,j_2,...j_k)}=\int_{0 \le u_1 \le \cdots \le u_k \le t} \mathrm{d}x_{j_1}(u_1)\mathrm{d}x_{j_2}(u_2)\cdots\mathrm{d}x_{j_k}(u_k)$, $j_k\in(1,...,d)$ What confuses me is what $s^{(j_1,j_2,...j_k)}$ actually means? And, since $X(t)$ is a $d*N$ matrix, how come integrate along one dimension affect the integration along another dimension? REPLY [8 votes]: $s(k)$ is made of $d^k$ numbers. They are labelled by the k-tuple $(j_1,j_2,\dots j_k)$ - and there are $d^k$ possibilities. For example, if $d$ is 3 and $k$ is 2, there are 9 values: $s^{(1,1)}$, $s^{(3,2)}$ etc. For a concrete example, the value of $s^{(2,3)}$ is the integral $\int_{01$, and is specified as a $d \times N$ matrix $M$. $M$ is enough to calculate the signature - we don't need to know the exact speed the path is traversed, we don't need $t$ or $X$. Then let the signature of the straight path from the $i$th point to the $(i+1)$th point be $a_i$. For any $k$, and any $(j_1,\dots,j_k)$, we know that the value of $(a_i)^{(j_1,\dots,j_k)}$ is $\frac1{k!}\prod_{h=1}^k(M_{j_h,i+1}-M_{j_h,i})$ by explicitly doing the integrals. Let the signature of the whole of the path from the first point up to the $(i+1)$th point be $b_i$. We can calculate the value of $b_i$ "up to level $K$" (i.e. for all tuples $(j_1,\dots,j_k)$ with $k\le K$) cumulatively in $i$, from the fact that $b_1=a_1$ and Chen's identity, which says that, for each $k$, and any $(j_1,\dots,j_k)$, $(b_{i+1})^{(j_1,\dots,j_k)}=\sum_{h=0}^k(b_i)^{(j_1,\dots\,j_h)}(a_{i+1})^{(j_{h+1},\dots,j_k)}$ . We then get the signature of the whole path, $b_{N-1}$, up to level $K$. Edit 2018 to add a specifically requested explicit example. (Note that we are still using some of the notation of the original question, which may not be the friendliest to read. Newcomers to the signature may prefer this or the documents linked from this.) Consider the two-dimensional path from $(1,5)$ straight to $(2,9)$ straight to $(3,4)$. We have $N=3$ and $d=2$. Let's calculate its signature up to level $K=2$. We have the following $a_1^{()}=1$, $a_1^{(1)}=1$, $a_1^{(2)}=4$, $a_1^{(11)}=\frac12$, $a_1^{(12)}=a_1^{(21)}=2$, $a_1^{(22)}=8$. Also we calculate $a_2^{()}=1$, $a_2^{(1)}=1$, $a_2^{(2)}=-5$, $a_2^{(11)}=\frac12$, $a_2^{(12)}=a_2^{(21)}=-\frac52$, $a_2^{(22)}=\frac{25}{2}$. We have $b_1=a_1$. With Chen's identity we calculate $b_2^{()}=b_1^{()}a_2^{()}=1$, $b_2^{(1)}=b_1^{()}a_2^{(1)}+b_1^{(1)}a_2^{()}=2$, $b_2^{(2)}=b_1^{()}a_2^{(2)}+b_1^{(2)}a_2^{()}=-1$, $b_2^{(11)}=b_1^{()}a_2^{(11)}+b_1^{(1)}a_2^{(1)}+b_1^{(11)}a_2^{()}=2$, $b_2^{(12)}=b_1^{()}a_2^{(12)}+b_1^{(1)}a_2^{(2)}+b_1^{(12)}a_2^{()}=-\frac{11}{2}$, $b_2^{(21)}=b_1^{()}a_2^{(21)}+b_1^{(2)}a_2^{(1)}+b_1^{(21)}a_2^{()}=\frac72$, $b_2^{(22)}=b_1^{()}a_2^{(22)}+b_1^{(2)}a_2^{(2)}+b_1^{(22)}a_2^{()}=\frac{1}{2}$. The signature we are aiming for is $b_2$.<|endoftext|> TITLE: Algorithm for computing the Arf invariant of a knot QUESTION [6 upvotes]: According to "The knot book", by Colin Adams, two knots are pass equivalent if they are related by a finite sequence of pass-moves. Moreover every knot is pass-equivalent to either the unknot or the trefoil knot and these two knots are not pass-equivalent. We define Arf invariant of a knot to be 0 if the know is pass equivalent to unknot and to be 1 if it is pass equivalent to the trefoil knot. Now here is my question: I am not interested to know how a knot is pass-equivalent to unknot or the trefoil knot. I just want to know if there is an easy algorithm for computing the Arf invariant of a knot based on its projection (for example, like the algorithm that we use to compute the linking number of links)? Can we move along a knot and compute its Arf invariant by taking its crossings somehow into consideration (maybe by counting its positive and negative crossings)? REPLY [3 votes]: Arf invariant is modulo two reduction of the Vassiliev invariant $v_2$ (the coefficient of $z^2$ in the Conway polynomial). Thus a well-known Gauss diagram formula for $v_2$ can be used for its computation. Namely, fix a knot diagram and choose a base point distinct from the crossings. Traverse the diagram starting from the base point. Each crossing is visited twice; write down the sequence in which we visit these crossings, denoting a passage on an overpass by $O_i$ and on an underpass by $U_i$. The result is the Gauss code of the knot diagram. For example, the standard diagram of a trefoil will be encoded by $U_1 O_2 U_3 O_1 U_2 O_3$. Then the Arf invariant is the number of "linked" pairs of crossings $...U_i...O_k...O_i...U_k...$ in the Gauss code.<|endoftext|> TITLE: homology of a mapping spectrum QUESTION [5 upvotes]: If $X$ and $Y$ are two spectra, I denote by $F(X,Y)$ their mapping spectrum. This is uniquely determined by the existence of a natural isomorphism $[X\wedge Y, Z]\cong [X,F(Y,Z)]$. I denote by $H_*$ the rational homology functor. If $X$ is a finite spectrum, I have an isomorphism $H_*(F(X,Y))\cong Hom(H_*(X),H_*(Y))$. Indeed, each side is a generalized cohomology theory in the $X$ variable that takes the value $H_*(Y)$ on the sphere. I would like to extend this result to a general $X$. What I can do is write a general spectrum $X$ as a filtered homotopy colimit of spectra $X_i$ that are finite. Then $F(X,Y)$ is the inverse limit of the spectra $F(X_i,Y)$. I get a Milnor exact sequence: $$0\to \mathrm{lim}_i^1Hom_{n-1}(H_{*}(X_i),H_{*}(Y))\to H_nF(X,Y)\to \mathrm{lim}_i Hom_n(H_*(X_i),H_*(Y))\to 0$$ where $Hom_n$ denotes the set of homomorphisms of degree $n$. I want to look at an example of the previous exact sequence with $X=\bigvee_{\mathbb{N}}S^0$ and $Y=S^0$. I can write $X$ as the homotopy colimit of $X_i=\bigvee_{1\leq k\leq i}S^0$. Then, $F(X,Y)$ is equivalent to $(S^0)^{\mathbb{N}}$, thus $H_0(F(X,Y))\cong \mathbb{Q}\otimes\mathbb{Z}^{\mathbb{N}}$. On the other hand, we have $$\mathrm{lim}_i\;Hom_0(H_*(X_i),H_*(Y))\cong \mathrm{lim}_i\;\mathbb{Q}^i\cong\mathbb{Q}^{\mathbb{N}}$$ The obvious map $\mathbb{Q}\otimes\mathbb{Z}^{\mathbb{N}}\to\mathbb{Q}^{\mathbb{N}}$ is not even surjective ! My question has two parts: (1) Explain where the mistake is in the previous calculation. My guess is that I misunderstood the Milnor exact sequence. (2) Is there a method for computing $H_*(F(X,Y))$ knowing $H_*(X)$ and $H_*(Y)$ ? REPLY [5 votes]: Since you are doing everything rationally and stably, homotopy and homology are the same. So $H_* F(X,Y) = [X,Y]_* \otimes \mathbb{Q}$ If $Y$ is itself a rational spectrum, then this is the same as $\text{Hom}_{\mathbb{Q}}(H_* X, H_* Y)$ but of course in general it will not be. Said another way, $\text{Hom}_{\mathbb{Q}}(H_* X, H_* Y) = H_*F(X,LY)$ where $LY$ is the rational localization of $Y$.<|endoftext|> TITLE: Centralizers of reflections in special subgroups of Coxeter groups QUESTION [6 upvotes]: Let us consider a (not necessarily finite) Coxeter group $W$ generated by a finite set of involutions $S=\{s_1,...,s_n\}$ subject (as usual) to the relations $(s_is_j)^{m_{i,j}}$ with $m_{i,j}=m_{j,i}$ and $m_{i,j}=1$ if and only if $i=j$ (if necessary you may also assume that $m_{i,j}<\infty$ for all $i,j$ or even that $W$ is an affine reflection group). Let $P\leq W$ be a subgroup generated by all but one of the $s_i$, say wlog $P=\langle s_1,...,s_{n-1}\rangle$. I am interested in the centralizer of $s_n$ in $P$. In particular I would like to know if $C_P(s_n)=C_W(s_n) \cap P=\langle s_i~|~ 1\leq i\leq n-1, m_{i,n}=2\rangle=:Z$ always holds. Obviously this is true if $n=2$ and I believe (though I have not written it down rigorously) I can prove it for reflection groups of type $A_n$ by using the standard isomorphism to $S_n$. On the other hand the centralizer of $s_n$ in $W$ is not necessarily a standard parabolic subgroup (look at the dihedral group of order $8$ for example). There are some results on centralizers of reflections in Coxeter groups and on normalizers/centralizers of parabolic subgroups (which is the same in this special case) to be found in the literature but most deal with the centralizer in $W$. In principle it should be possible to obtain the centralizer in $P$ from these results by simply taking the intersection but the results I found so far are not explicit/ simple enough for this to be a feasible solution. Here are some thougts so far: I can show that elements of $C_P(s_n)$ of length $1$ or $2$ already lie in $Z$ (the case of length $1$ being trivial) and that elements of $C_P(s_n)$ of length $3$ where all three occurring simple reflections are pairwise distinct already belong to $Z$. On the other hand look at $s_1s_2s_1 \in P$ which centralizes $s_n$ if and only if $s_2$ centralizes $s_1s_ns_1$. I don't see any reason why this should not be the case so I tried constructing a counterexample consisting of $s_1,s_2$ and $s_3$ such that $s_1,s_2$ do not commute and $s_1,s_3$ do not commute but $s_2$ and $s_1s_3s_1$ do. Any ideas on how to do that? Edit: I should note that I already posted this question to math.StackExchange (https://math.stackexchange.com/questions/1193740) but did not get any helpful feedback. Edit 2: Regarding the question whether $s_1s_2s_1$ can centralize $s_3$ (all reflections pairwise distinct; $s_1$ neither centralizing $s_2$ nor $s_3$) in the case of $W$ being an affine reflection group I did a case by case check on the possible Dynkin-diagrams. The cases to consider are $A_3,B_3,\tilde{A_2},\tilde{B_2}$ and $\tilde{G_2}$ and using the standard representation on the root space I found that $s_1s_2s_1$ never centralizes $s_3$. Edit 3: Here is a proof in the case that $m_{i,n} \neq 3$ for all $1 \leq i \leq n-1$: Assume $C_P(s_n) \neq Z$ and take an element $w \in C_P(s_n) - Z$ of minimal length. Then each reduced expression for $w$ neither starts nor ends with one of the simple reflections in $Z$. Let $w=s_{i_1}...s_{i_r}$ be such a reduced expression. Since $s_{i_1}...s_{i_r}s_ns_{i_r}...s_{i_1}=s_n$ there is a sequence of braid- and nil-moves that reduces $s_{i_1}...s_{i_r}s_ns_{i_r}...s_{i_1}$ to $s_n$. Since $m_{i_r,n} >3$ we cannot start with a braid- or nil-move involving $s_n$ and since we chose a reduced expression for $w$ there are certainly no nil-moves possible at all. Hence all we can start with is a braid-move in the reduced expression for $w$ (or $w^{-1}$). But after finitely many of such braid-moves the expression we get for $w$ still ends with a simple reflection $s_{i_k}$ which does not commute with $s_n$ (since this would yield an element of shorter length in $C_P(s_n) - Z$). Furthermore $m_{i_k,n} \neq 3$ so we still are unable to perform a braid-move involving $s_n$ and we still have a reduced expression for $w$ so there are no possible nil-moves. In conclusion: After finitely many steps we will never have performed any nil-moves and hence we cannot reduce $ws_nw^{-1}$ to $s_n$ which is a contradiction and hence such a $w$ does not exist. I hope one can use an analogous argument in the case $m_{i_r,n}=3$. REPLY [2 votes]: I believe I finally found a proof. Surprisingly it does only use standard facts on Coxeter-groups (exchange condition, solving the word problem via braid-moves,...). Let me first make the notation a bit easier: Claim: Let $P\leq W$ be a special subgroup of $W$ generated by some subset $S'\subsetneq S$ of $S$ and $s \in S-S'$. Then the centralizer of $s$ in $P$ is generated by those involutions in $S'$ which commute with $s$, $C_P(s)=\langle s' \in S'~|~ s's=ss' \rangle$. Proof: Let $w \in C_P(s)$ and $w=s_1...s_r$ be a reduced expression. By induction it is enough to prove that $s_rs=ss_r$ since the elements of length $1$ in the centralizer are precisely the simple involutions commuting with $s$. We have $\ell(ws)=\ell(w)+1$ and since $\ell(wsw^{-1})=\ell(s)=1$ we conclude that $\ell(wss_r)<\ell(ws)$, so $\ell(wss_r)=\ell(w)$. By the exchange condition there is a reduced expression for $ws$ ending in $s_r$ and since $s_1...s_rs$ is already a reduced expression for $ws$ there exists a finite series of braid-moves connecting these two expressions. The expression $s_1...s_rs$ contains $s$ only once and no simple involution that does not commute with $s$ shows up to the right of $s$. Consider now any braid-move in this situation. If $s$ is not involved in the move the two conditions obviously still hold afterwards. If $s$ is involved the other simple involution involved must commute with $s$ since any braid-move involving $s$ and a non-commuting $s'$ requires either at least two occurrences of $s$ (to the left and to the right of $s'$) or an occurrence of $s'$ to the right of $s$ neither of which happens. Hence any braid-move fixes our two conditions and after finitely many braid moves there is still no simple involution to the right of $s$ which does not commute with $s$. On the other hand there is, as noted above, a finite series of braid-moves after which the expression ends in $s_r$ so $s_r$ has to commute with $s$ as asserted.<|endoftext|> TITLE: Is the upper boundary of a Schubert variety Cartier? QUESTION [5 upvotes]: On $G/B$, the divisor $\bigcup_\alpha X_{r_\alpha}$ is Cartier (where $X_w := \overline{B_- w B}/B$, and $\alpha$ varies over simple roots), not least because $G/B$ is smooth. Is the same true for the divisor $\bigcup_{w' \gtrdot w} X_{w'} \subset X_w$, where $\gtrdot$ is the covering relation in strong Bruhat order? Definitely some combination $\sum_{w'\gtrdot w} c_{w'} [X_{w'}]$, all $c_{w'} > 0$, is Cartier; restrict a $G$-equivariant ample line bundle $\mathcal L_\lambda$ from $G/B$ and consider the unique $T$-weight section $\sigma$ that doesn't vanish at $wB/B \in X_w$. This vanishes along the set $\bigcup_{w'\gtrdot w} X_{w'}$ and for $w' = w r_\beta$, its order of vanishing $c_{w'}$ will be $\langle \lambda, \check\beta \rangle$. If the first question fails, how could one find the minimal Cartier combinations $\sum_{w'\gtrdot w} c_{w'} [X_{w'}]$? REPLY [4 votes]: The answer to the first question is No. Up to a twist by the restriction of $\mathcal{L}_{-\rho}$, this divisor is the canonical divisor, so the question is equivalent to whether the Schubert variety is Gorenstein. In my paper with Alex Yong on the Gorenstein Schubert varieties, there probably is enough combinatorics to tell you how to work out the answer to the second question in type A. (AFAIK, no explicit answer is known.) For other types, you need to use the appropriate Chevalley formula instead of the Monk formula to tell you what the Cartier divisors are, and the combinatorics are not as nice.<|endoftext|> TITLE: Lipschitz-free spaces of $\mathbb R^n$ QUESTION [10 upvotes]: We define $$ \text{Lip}_0(\mathbb R^n)=\{f:\mathbb R^n\rightarrow \mathbb R, \text{such that $f(0)=0$ and } \sup_{x\not=y}\frac{\vert f(x)-f(y)\vert}{\vert x-y\vert}<+\infty. \} $$ It is well-known that $\text{Lip}_0(\mathbb R^n)$ is a Banach space, which is the dual space of the so-called $\mathcal F(\mathbb R^n)$, a.k.a. the Lipschitz-free space of $\mathbb R^n$. Claim: $\text{Lip}_0(\mathbb R^n)$ is the dual space of $X/N$ where $X$ is the space of $L^{1}({\mathbb R}^{n})$ vector fields and $N$ is the subspace of vector fields with null divergence. In other words, with $$ X=(L^{1}({\mathbb R}^{n}))^{n},\quad N=\{(f_{j})_{1\le j\le n}\in X, \ \sum_{1\le j\le n}\frac{\partial f_{j}}{\partial x_{j}}=0\}, $$ we have $ \text{Lip}_0(\mathbb R^n)=(X/N)^{*}. $ Note that in the easy case $n=1$, we find the familiar $\mathcal F(\mathbb R)=L^1(\mathbb R)$. The derivatives above are taken in the distribution sense. Questions. (1) Is the statement of this claim well-known? (2) Could it be useful to describe more explicitly the properties of $\mathcal F(\mathbb R^n)$ when $n\ge 2$? REPLY [6 votes]: Yes, this is essentially contained in the paper M. Cúth, O. F. K. Kalenda, P. Kaplický: Isometric representation of Lipschitz-free spaces over convex domains in finite-dimensional spaces, Mathematika, 63 (2) (2017), 538–552. where it is proved that for any non-void, open subset $\Omega$ of $\mathbb R^n$, the space ${\rm Lip}_0(\Omega)$ is isometric to the dual of the quotient of $L_1(\Omega, \mathbb R^n)$ by the vector fields with zero divergence in the distributional sense.<|endoftext|> TITLE: generating set for symmetric group $S_n$ QUESTION [12 upvotes]: Say that $a_1, \cdots, a_{n-1}$ is an independent generating set for $S_n$. Let $b$ be any element in $S_n$. Is it true that $b$ can replace one of the generators, i.e. that there exists an index $i$, such that we have that $a_1,\cdots, \hat{a_i},\cdots, a_{n-1}, b$ generate $S_n$? If $a_1, \cdots, a_{n-1}$ is the standard (n-1)-tuple that generates $S_n$, $(12),(13),...,(1n)$, then it's true and it can easily be shown. Does it hold in general? REPLY [10 votes]: The answer is yes. There is a paper by Cameron and Cara which describes all maximal generating sets of length $n-1$ of $S_n$. They are not very hard to describe and basically are variants of the standard $n-1$ length generating sets. http://www.maths.qmul.ac.uk/~pjc/preprints/igsgsn.pdf Cameron and Cara say build a tree with the vertices being $\{1,\ldots,n\}$. Add transpositions corresponding to edges into your generating set. These are (about) half the maximal length generating sets. The other set is found by taking one of those generating sets, picking a transposition and multiplying against all the others. In both cases it is pretty clear that given a $b$ we can find an edge to replace. Basically $b$ will connect some vertices of the tree, and we can remove an edge to have the tree be connected, and since everything is basically a transposition this is all you really need. I can supply more details if this was unclear, but I think just given the Cameron paper things become clear.<|endoftext|> TITLE: Is $\mathcal M _{g,n}$ anabelian? QUESTION [7 upvotes]: Are the moduli spaces $\mathcal M _{g,n}$ expected to be anabelian? Is there anything known in that direction? Thank you very much for your help in any case! REPLY [10 votes]: Grothendieck expected the moduli spaces $\mathcal{M}_{g,n}$ over $\mathbb{Q}$ to be the basic examples of anabelian varieties (besides hyperbolic curves, which was proved by Mochizuki, even over number fields, or more generally over sub-$p$-adic fields). The first non-trivial case is $(g,n)=(0,4)$, where we have $\mathcal{M}_{0,4}=\mathbb{P}^1_{\mathbb{Q}}\setminus \{0,1,\infty\}$. The only other moduli space of dimension $1$ is $\mathcal{M}_{1,1}$, the moduli space of elliptic curves. The geometry of these spaces has been described explicitly by cutting them into cells, enumerated by "fatgraphs" (equivalent to dessins d'enfants), and they are conjectured to be anabelian (I don't know further results, but I am not in this field. Certainly the question is very ambitious).<|endoftext|> TITLE: How many subsets of $[0,1)$ are there modulo null sets? QUESTION [19 upvotes]: For subsets $A$ and $B$ of $[0,1)$, say $A\sim B$ iff $\lambda(A\Delta B)=0$ where $\lambda$ is Lebesgue measure. Question: How many equivalence classes of subsets of $[0,1)$ are there given AC? I would guess the answer is $2^c$ given AC, but I haven't got a proof. What got me thinking about this was trying to find a way to say that there are more nonmeasurable sets than measurable ones. There are, of course, $2^c$ of each, but modulo null sets there are only $c$ measurable ones (at least given AC), so if there were more than $c$ subsets modulo null sets, we could say that modulo null sets there are more nonmeasurable sets than measurable ones. REPLY [8 votes]: Edit. I had posted this answer to complement Eric's original answer, which showed that the number of classes was at least ${\frak c}^+$, since at that time we didn't quite yet know whether there were $2^{\frak c}$ classes. Afterwards, however, Eric improved his answer to get $2^{\frak c}$ directly. Following the comments, though, I have left this answer up. Let me complement Eric's answer by showing that it is relatively consistent to have strictly more than ${\frak c}^+$ many equivalence classes. Indeed, it is relatively consistent with ZFC to have $2^{\frak c}$ many equivalence class, in a case where this is larger than ${\frak c}^+$. Specifically, I claim that if the continuum hypothesis holds and there is a thick Kurepa tree (an $\omega_1$ tree with $2^{\omega_1}$ many branches), then there are $2^{\omega_1}=2^{\frak c}$ many equivalence classes. Indeed, I shall construct an almost-disjoint family of $2^{\omega_1}$ many Vitali sets. To see this, let $T$ be a thick Kurepa tree, and let $\langle A_\alpha\mid\alpha<\omega_1\rangle$ enumerate the equivalence classes of reals under translation-by-a-rational. Label the $\alpha^{th}$ level of $T$ with the countably many elements of $A_\alpha$. For any path $s$ through $T$, the set $A_s$ of labels appearing on the nodes of $s$ will be a Vitali set, and therefore non-measurable. Further, any two distinct paths $s\neq t$ will have $A_s\cap A_t$ being countable, and so $A_s\not\sim A_t$. Since $T$ is a thick Kurepa tree, we therefore have $2^{\omega_1}$ many branches and thus this many equivalence classes modulo your relation. The collection $\{\ A_s\mid s\in[T]\ \}$ is an almost-disjoint family of $2^{\omega_1}$ many Vitali sets. Finally, let me explain that it is relatively consistent from an inaccessible cardinal that there is a thick Kurepa tree, yet CH holds and $2^{\omega_1}$ is very large. One way to do this is as follows. Start with $\kappa$ inaccessible in $V$ and $2^\kappa$ very large (by forcing if necessary). Let $V[G]$ be the forcing extension by the Levy collapse, so that $\kappa=\omega_1^{V[G]}$. Consider the tree $T=(2^{<\kappa})^V$ in the model $V[G]$. Since every ordinal less than $\kappa$ was made countable, this has become an $\omega_1$-tree. Yet, since $2^\kappa$ was very large and cardinals $\kappa$ and above were preserved, we have $(2^\kappa)^V$ many branches through this tree. So it is thick.<|endoftext|> TITLE: Bound on the ratio of top 2 eigenvalues QUESTION [9 upvotes]: Let $P$ be a $n \times n$ stochastic matrix such that $P_{ij}=\tau$ if $i \neq j$ and $P_{ii} = 1 - (n-1)\tau$ where $0<\tau < \frac{1}{n}$. It is clear that the largest eigenvalue of $P$ is 1, and the second largest eigenvalue is $(1-n\tau)$, hence $$\frac{\lambda_{2}}{\lambda_{1}} = 1-n\tau \leq 1 - 2\tau.$$ Let $D$ be a $n \times n$ diagonal matrix such that $D_{ii} \geq 1$ for all $i$. Consider the matrix $PD$ and let $\lambda_{1}',\lambda_{2}'$ be the top two eigenvalues. Prove that $$\frac{\lambda_{2}'}{\lambda_{1}'} \leq 1-2\tau.$$ I have verified that it's true for $n=2,3$ by brute force calculations. Also using Horn's inequalities I can find a bound which is much worse. Thanks REPLY [3 votes]: I claim that $\lambda_2'/\lambda_1'\leqslant 1-\frac{2\tau}{1-(n-2)\tau}$ which is bit stronger than you ask for. This is sharp as the example $D_{11}=D_{22}=1\gg \max(D_{33},\dots,D_{nn})$ shows. Of course it is not important that $D_{ii}\geqslant 1$, only that $D_{ii}>0$ (since everything is homogeneous in $D$). We may suppose that all $D_{ii}$'s are distinct, since $\lambda_1'$ and $\lambda_2'$ are continuous with respect to $D$. Denote $D_{ii}=d_i$, $\beta=1-n\tau$, $\rho=\beta/\tau$. Then we should prove that $\lambda_2'/\lambda_1'\leqslant \rho/(\rho+2)$. Rewriting the eigenvector equation in the form $(\tau J+\beta I)Dx=\lambda x$, for the coordinates $x_i$ of the eigenvector $x$ with eigenvalue $\lambda$ we get $\beta d_ix_i+\tau(\sum_j d_j x_j)=\lambda x_i$. Denote $\mu=\sum d_j x_j$. If $\mu=0$, we get $x_i(\lambda-\beta d_i)=0$ for all $i$, so $\lambda$ must be equal to certain $\beta d_j$ and $x_i=0$ for $i\ne j$. But this contradicts to $\mu=0$. So $\mu\ne 0$, we have $x_i=\tau \mu (\lambda-\beta d_i)^{-1}$ and $$\sum d_i(\lambda-\beta d_i)^{-1}=(\tau \mu)^{-1}\sum d_i x_i=\tau^{-1}.$$ This equation (for $\lambda$) has $n$ roots in total. If $d_1\rho/(\rho+2)$, then also $\theta_n/(1+\theta_n)=\beta d_n>\rho/(\rho+2)$, $\theta_n>\rho/2$. Denoting $m=\lambda_2'$ we get $$ \rho=\sum \frac{\beta d_i}{m-\beta d_i}=\sum \frac{\theta_i}{m-(1-m)\theta_i}=- \frac{\theta_n}{m-(1-m)\theta_n}+\sum_{i=1}^{n-1} \frac{\theta_i}{m-(1-m)\theta_i}\leqslant \\ - \frac{\theta_n}{m-(1-m)\theta_n}+\frac{\sum_{i=1}^{n-1} \theta_i}{m-(1-m)\sum_{i=1}^{n-1} \theta_i}= \frac{\theta_n}{m-(1-m)\theta_n}+\frac{\rho -\theta_n}{m-(1-m)(\rho-\theta_n)}=:f(m) $$ (the last denominator is positive since $m/(1-m)>\rho/2>\rho-\theta_n$.) The function $f(m)$ is decreasing between $\alpha:=(\rho-\theta_n)/(1+\rho-\theta_n)$ and $\beta:=\theta_n/(1+\theta_n)$. We have $\alpha<\rho/(\rho+2)<\beta$ and $m\in (\rho/(\rho+2),\beta)$. Therefore $\rho\leqslant f(m)\leqslant f(\rho/(\rho+2))=-(\rho+2)$, a contradiction.<|endoftext|> TITLE: Pseudomanifolds and Poincaré duality QUESTION [13 upvotes]: 1) A $n$-dimensional homology manifold is a topological space $X$ such that for any $x\in X$, the homology groups $$H_p(X,X-x,\mathbb{Z})$$ are trivial unless $p=n$ where $$H_n(X,X-x,\mathbb{Z})\cong \mathbb{Z}$$ 2) A $n$-dimensional pseudomanifold is a topological space together with a triangulation such that each simplex is the face of a $n$-simplex, every $(n – 1)$-simplex is a face of exactly two $n$-simplices for $n > 1$, any two -dimensional simplices can be joined by a "chain" of n-dimensional simplices in which each pair of neighbouring simplices have a common $(n-1)$-dimensional face. I am searching for examples of pseudomanifolds that satisfy Poincaré duality but that are not homeomorphic to homology manifolds. I would like to know if there exist examples of complex projective varieties that satisfy these properties: satisfy poincaré duality but are not homology manifolds. Edit (Ben Wieland's example): Let $M^3$ be a manifold and let us consider $[0,1]\subset M$, glue the interval to itself by an involution $f$ that reverses the endpoints. Then we get a pseudomanifold $M'$ whose homotopy type is the same as $M$ then it satisfies Poincaré duality but it is not a homology manifold, you can look at the local homology of the equivalence class $[0]\in M'$ of $0\in [0,1]\subset M$. This is due to the fact that the link of $[0]$ in $M'$ is not connected ($M'$ is not normal). The normalization of $M'$ is $M$ itself, thus I would like to know an example where the pseudomanifold is normal. REPLY [2 votes]: 1. Introduction 2. Easy example 3. Normality, Duality 4. Normal example Introduction The pseudomanifold and homology manifold conditions are both local conditions, while Poincaré duality is a global condition. It is possible for a pseudomanifold to fail the homology manifold conditions in several places, but so that the local deviations globally cancel, so it retains duality. Simplicial complexes are "locally cone-like": every point $x$ has a neighborhood that is the cone on another space $L$, the link, with $x$ the cone point. Then $$H_*(X,X-\{x\})=H_*(CL,L\times I)=\tilde H_*(SL)=\tilde H_{*-1}(L)$$ So the homology manifold condition is that the links have the homology of spheres. Easy example The interval admits an involution reversing the endpoints. The quotient by this involution is again contractible. Think of this quotient operation as gluing one half of the interval to the other half. Embed the interval in an $n$-manifold. Form a quotient of the manifold by gluing the one half of the interval to the other half. If $n\geqslant 3$, this leaves the cells in degree $n$ and $n-1$ unchanged, so does not affect the pseudomanifold condition. Since we have replaced one contractible subspace by another, the homotopy type of the total space is unchanged and satisfies duality (with the pseudomanifold fundamental class, etc). But the the links are no longer spheres at any point along the interval. At the image of the ends of the interval, the link is two spheres glued in one point. At general points along the interval, it is two spheres glued in two points. At the image of the midpoint, the fixed point of the involution, the link is a single sphere with two of its points glued to each other. Normality, Duality That example involved changing a manifold in low dimensions, which doesn't violate the high dimensional pseudomanifold condition. To rule out such examples, one has the concept of a normal pseudomanifold (cf normal variety). The normalization of a pseudomanifold is produced by taking a disjoint union of $n$-simplices parameterized by those of the original space and gluing them along their $n-1$-faces, according to how the original was glued. Thus the normalization of the above example is the unmodified manifold. A normal pseudomanifold is one isomorphic to its normalization. It may be useful to use the Verdier dualizing sheaf. The cohomology with coefficients in the dualizing sheaf matches the homology: $H^*(X; D)\cong H_*(X)$. An oriented pseudomanifold yields a map of sheaves $\mathbb Z\to D$. We seek Poincaré duality, that is, $H^*(X; \mathbb Z)\cong H_*(X)$. So we seek spaces where the map of sheaves induces an isomorphism on cohomology $H^*(X; \mathbb Z)\cong H^*(X; D)$, even though it is not an isomorphism of sheaves. That is, we seek the cone to be a nontrivial sheaf with no cohomology in any dimension. There are such sheaves, such as a local system supported on a circle with appropriate monodromy. That motivates the construction. It also gives immediate proofs of the claims, but it should not be hard to check them without using sheaves. Normal example Take an $n$-manifold manifold $M$ and an automorphism $\phi$ so that the action of on the homology in intermediate degrees is sufficiently mixing so that if we consider it an action of the group $\mathbb Z$ and take group homology with those coefficients, the homology is trivial. For example, $M=T^2$ and $\phi=\left(\begin{matrix}2&1\\1&1\\ \end{matrix}\right)$. Then form the mapping torus ($T_\phi=M\times I/(x,0)\sim(\phi(x),1)$). This is an $M$-bundle over the circle, homology computed by a spectral sequence $H_*(S^1;H_*(M))\Rightarrow H_*(T_\phi)$. By assumption on $\phi$, the spectral sequence degenerates and $H_*(T_\phi)=H_*(S^1\times S^n)$. Then cone off each copy of $M$, forming a circle of cone points. To put it another way: form the mapping cylinder of the map to the circle $T_\phi\to S^1$. Yet another way: the mapping torus of the self-map of the cone $\tilde\phi\colon CM\to CM$. This space is a normal pseudomanifold (with boundary) because those properties are preserved by cones and products. Its set of singular points is a circle, and the dualizing sheaf twists about it with the prescribed monodromy, so it contributes nothing to the sheaf cohomology. If you prefer a closed example, double the space along the boundary (or equivalently take the double mapping cylinder of the map to the circle; or the mapping torus of the automorphism of the suspension). This gives a pseudomanifold that is not a homology manifold, but which satisfies Poincaré duality. Just as the mapping torus had the homology of a manifold $S^1\times S^n$, this space has the homology of $S^1\times S^{n+1}$. Actually, that is only correct if we restrict to trivial coefficients. If we define Poincaré duality to be for all local systems, then this space fails, for some unwind the twist around the singular circles. However, we can eliminate them by killing the fundamental group by surgery. That is, cut out a neighborhood of a circle, $S^1\times D^{n+1}$, leaving a boundary $S^1\times S^n$ and fill it in with $D^2\times S^n$. Now local systems are trivial, so it satisfies full Poincaré duality.<|endoftext|> TITLE: Did Bourbaki write a text on algebraic geometry? QUESTION [65 upvotes]: Certainly Bourbaki never wrote an introduction to algebraic geometry: we would have heard about it, right? REPLY [110 votes]: Wrong! Here is Bourbaki document on algebraic geometry, taken from the now available Master's Archives: click on Autres rédactions, then on Chap.I Théorie globale élémentaire (91 p.) This preliminary draft was apparently written (according to a penciled annotation on the first page) for Bourbaki in 1954 by Samuel, a distinguished algebraic geometer and number theorist. Alas, it is hard to conceive a worse timing for a book on algebraic geometry: one year later Serre would publish his paradigm shifting FAC, shortly followed by Grothendieck's theory of schemes, a vast development of Serre's article (as acknowledged in the Preface to the EGA), which would forever change our vision of algebraic geometry. Samuel's point of view is that of Weil: at the forefront is a "universal domain", a field extension $k\subset K$ with $K$ algebraically closed and of infinite transcendency degree over $k$. Geometry would happen in $\mathbb A^n(K)$ or $\mathbb P^n(K)$, whereas algebra and number theory would take place inside $k[T_1,\dots,T_n]$ or $k[T_0,\dots,T_n]$. A variety in Weil's vision could have a multitude of generic points, essentially points such that a polynomial vanishing on them must be zero. It is quite moving to see the author struggling with, for example, the product of varieties: he notices the difficulty due to the tensor product $E\otimes_k F$ of two field extensions $k\subset E,F$ having non-zero nilpotents but doesn't envision incorporating these in his foundational text. Grothendieck would soon show the world how considering nilpotents in the very foundations of scheme theory would enrich and beautify algebraic geometry. I encourage every algebraic geometer to browse this nostalgic and unacknowledged witness of a bygone era of our beloved science. Edit (May 27th, 2016) Browsing the fascinating Grothendieck-Serre Correspondence ( a review of which is here) I found this excerpt from the very first letter of the Correspondence (page 3, dated January 28th, 1955), written by Grothendieck then in Lawrence, Kansas, USA : "You said that Bourbaki wanted to send me a draft by Samuel on algebraic geometry (and commutative algebra?). I would be happy to get it..." This confirms that the document mentioned above was indeed authored by Samuel.<|endoftext|> TITLE: Well-posedness of Fokker-Planck equation QUESTION [9 upvotes]: Consider the following equation on $[0,T]\times\mathbb{R}^n$ \begin{eqnarray} &\partial_t\rho=\mathrm{div}(\rho\nabla V)+\Delta\rho\\ &\rho|_{t=0}=\rho^0, \end{eqnarray} where $V\in C^2(\mathbb{R}^n;\mathbb{R})$. Typical $V$ are of the form $V(x)=x^2$. Further the initial data $\rho^0\in L^1(\mathbb{R}^n;\mathbb{R})$. Question: How would one prove the existence of solutions of this equation by using standard techniques? My feeling is that this should work since it is a nice enough parabolic equation, however with somewhat nasty initial data. Such a question has been answered in Jordan-Kinderlehrer-Otto-98 where they use the gradient flow structure of the equation. They assume additionally that the initial data has bounded moments which could be done here as well. It should also be noted that the above PDE is the Fokker-Planck equation associated to the stochastic differential equation $$\begin{equation} dX_t=-\nabla V(X_t)dt+\sqrt{2} dW_t, \end{equation}$$ where $W_t$ denotes an $n$-dimensional Wiener-process. And since the drift $-\nabla V$ is sufficiently regular, standard results suggest that this equation has a solution. However, I do not yet understand what this result implies for the corresponding Fokker-Planck equation. REPLY [6 votes]: The argument in JKO (short for Jordan-Kinderlehrer-Otto-98) is based on a Lyapunov function for the Fokker-Planck equation. As such, it requires that this Lyapunov function evaluated at $\rho^0$ be finite. (The main result of JKO assumes this condition.) This is a strong constraint on $\rho^0$: it seems to require, e.g., that the support of $\rho^0$ be $\mathbb{R}^n$. For your convenience here is the Lyapunov function from JKO: $$ F(\rho) = \int_{\mathbb{R}^n} \rho \; ( V - \underbrace{(- \log \rho )}_{\text{free energy of $\rho$}} ) dx $$ To answer your question and as far as I can tell: it does not seem possible to prove using standard techniques existence of solutions to the Fokker-Planck equation with what you call "nasty initial data." Recall, in the standard (probabilistic) approach, the Fokker-Planck equation is "well-posed" if: $V$ is of class $C^2$; the transition probability of $X(t)$ admits a probability density function that is twice differentiable for $t>0$; and, the probability law of $X(0)$ has a continuous probability density function on $\mathbb{R}^n$.<|endoftext|> TITLE: notion of $\mathrm{Gal}(\overline{\mathbb{Q}}_p / \mathbb{Q}_p)$ representation with complex multiplication QUESTION [5 upvotes]: In usual Hodge theory, there is the notion of Hodge structure $H$ with complex multiplication, that can be defined in several ways, i.e. asking that there exists a CM number field $E$ such that $\dim H=[E: \mathbb{Q}]$ and an embedding of $E$ into the endomorphisms of Hodge structures of $H$. Is there a similar notion in $p$-adic Hodge theory, that is, a definition of $\mathrm{Gal}(\overline{\mathbb{Q}}_p/ \mathbb{Q}_p)$-representation with complex multiplication? The easiest example to the notion should apply is the following: $E$ is a CM elliptic curve over $\mathbb{Q}$ and one considers $H^1_{dR}(E / \mathbb{Q}_p)$. REPLY [3 votes]: I think you want to consider virtually abelian representations - i.e., representations that factor through some virtually abelian group. CM abelian varieties are exactly those whose Galois representations are virtually abelian as a representation of the full Galois group of $\mathbb Q$. So they will still be abelian as representations of the Galois group of $\mathbb Q_p$. But many non-CM abelian varieties will be CM in this sense as well. This is to be expected as p-adic Hodge theory carries much less information than usual Hodge theory.<|endoftext|> TITLE: Is there a way to define a Lie derivative of a connection? QUESTION [5 upvotes]: I've been reading a little bit about the definition of symmetries on General Relativity, and they are related with the concept of Killing vector, i.e., vectors along which the Lie derivative of the metric vanishes $\mathcal{L}_X g =0$. However, afaik the most symmetric geometrical object is the Ricci tensor (see the post), and the a vector $X$ satisfying $\mathcal{L}_X \text{Ric} = 0$ is known as a collineation of the Ricci tensor. I'd like to know whether is possible to define a sort of Lie derivative for a (general) connection, or a way to somehow define the symmetries of a connection. REPLY [2 votes]: I was expecting the Lie derivative of a connection to be another connection but Vladimir's answer about the difference of two connections being a tensor makes a lot of sense. I would say a coordinate free algebraic formula for the Lie derivative of a connection is: $\left(\mathcal{L}_{X}\nabla \right)_{Y}Z:=\mathcal{L}_{X}\left( \nabla_{Y}Z\right)-\nabla_{\mathcal{L}_{X}Y}Z-\nabla_{Y}\mathcal{L}_{X}Z$. This is just the formula for the Lie derivative of a tensor and in this case it does give a $C^{\infty}(M)$-linear object in the variables $Y$ and $Z$. Since this is a tensor field probably a better notation would be $\left(\mathcal{L}_{X}\nabla\right)\left( Y,Z \right)$.<|endoftext|> TITLE: Continuous maps which send intervals of $\mathbb{R}$ to convex subsets of $\mathbb{R}^2$ QUESTION [67 upvotes]: Let $f : \mathbb{R} \longrightarrow \mathbb{R}^2$ be a continuous map which sends any interval $I \subseteq \mathbb{R}$ to a convex subset $f(I)$ of $\mathbb{R}^2$. Is it true that there must be a line in $\mathbb{R}^2$ which contains the image $f(\mathbb{R})$ of $f$? Yes, this question seems rather elementary, but I have already spent (or lost?) too much time on this devilish problem, and I have communicated this question to sufficiently many people to know that it is far from trivial... REPLY [3 votes]: I asked [a very similar question] (convexity of images of space-filling curves) here once. Suppose $f:[0,1]\to[0,1]^2$ is continuous and for each $t\in[0,1]$, the area of $\lbrace f(s) : 0\le s\le t \rbrace$ is $t$. For what sets of values of $t\in[0,1]$ can $\lbrace f(s) : 0\le s\le t \rbrace$ be convex? All $t$? Only countably many $t$? If so, which countable sets? Topologically discrete ones? Dense ones? Perhaps Pietro Majer's answer to that question will shed some light on this one as well.<|endoftext|> TITLE: For a partition of $\mathbb{R}$ into countably infinite sets, must there be an almost-disjoint family of $2^{\frak c}$ many selectors? QUESTION [8 upvotes]: My question arises from a construction I gave in my recent answer to a question of Alexander Pruss concerning large families of independent non-measurable sets of reals. In that argument, using the continuum hypothesis and the existence of a thick Kurepa tree $T$, I produced a family of $2^{\frak c}$ many Vitali sets $\{\ A_s\mid s\in[T]\ \}$, which was almost disjoint in the sense that $A_s\cap A_t$ was countable whenever $s\neq t$. The only aspect of the Vitali relation that was used in the construction was that the Vitali equivalence classes (equivalence under rational translation) are countably infinite. Thus, the construction proves: Theorem. If the CH holds and there is a thick Kurepa tree, then for every partition of $\mathbb{R}$ into countably infinite sets, there is an almost-disjoint family of selectors of size $2^{\frak c}$. By almost-disjoint here, I mean that any two distinct elements of the family have countable intersection; by selector, I mean that each set in the family has exactly one element from each equivalence class; and by a partition into countably infinite sets, I mean that we have an equivalence relation on $\mathbb{R}$ with every equivalence class countably infinite. To prove the theorem, simply label the nodes on the $\alpha^{th}$ level of $T$ with distinct members of the $\alpha^{th}$ equivalence class in the partition. Being thick, the tree has $2^{\frak c}$ many branches, each of which provides a selector, and any two such selectors can agree only up to some countable height in the tree, where those branches separate. My question is whether I really needed those set-theoretic assumptions in order to make the conclusion. Question. How much can one weaken the hypotheses of the theorem and still prove the conclusion? For example, can we drop the thick Kurepa tree assumption? Can we omit CH? Can we prove it in ZFC? Can one show the consistency with ZFC of a counterexample? REPLY [7 votes]: If CC (Chang's Conjecture) holds, then there are no $\aleph_2$ pairwise almost different $\omega_1\to\omega$ functions. For this, assume that $\{f_\alpha:\alpha<\omega_2\}$ are as described. By CC there is an elementary submodel $N$ such that $|N|=\aleph_1$ and $\delta=N\cap\omega_1<\omega_1$. (Technically, we have to consider the function $F(\alpha,\xi)=f_\alpha(\xi)$ for $\alpha<\omega_2$, $\xi<\omega_1$ to make the model of countable length. Then, apply CC to $M=(\omega_2;F,\omega_1,\in,\dots)$.) If $\alpha\neq \beta$ are in $N$, then they differ from some $\gamma<\omega_1$ on, by elementarity, there is such a $\gamma<\delta$, i.e., any two functions in $\{f_\alpha:\alpha\in N\}$ differ from some $\gamma<\delta$. Specifically, $\{f_\alpha(\delta):\alpha\in N\}$ are distinct, a contradiction.<|endoftext|> TITLE: $\text{ZFGC}^{\text{−f}}+\text{BAFA}+\exists\kappa(κ \text{ is Reinhardt})$ and its implication QUESTION [6 upvotes]: A. S. Daghighi, M. Golshani, J. D. Hamkins, and E. Jeřábek proved in "The foundation axiom and elementary self-embeddings of the universe" that, working in ZFGC$^{\text{−f}}$+BAFA, there are nontrivial automorphisms and elementary embeddings of the universe $V$ into itself. Accordingly, Kunen inconsistency is circumscribed for this class of ill-founded theories. Does it follow that $\text{ZFGC}^{\text{−f}}+\text{BAFA}+\exists\kappa(κ \text{ is Reinhardt})$ is a non inconsistent extension of $\text{ZFGC}^{\text{−f}}+\text{BAFA}$? If so, is it known which of the large cardinal properties would $\text{ZFGC}^{\text{−f}}+\text{BAFA}+\exists\kappa(κ \text{ is Reinhardt})$ imply? REPLY [5 votes]: I'm glad to hear you're reading our paper, which can be found here: The foundation axiom and elementary self-embeddings of the universe. Click through to the arxiv for a pdf — and I note that the title you mention is from an earlier draft of this article, so you may want to look at the updated version of the article. In theorem 1 of the article, we prove in $\text{ZFC}^{-f}$ that any $\Sigma_1$-elementary embedding $j:V\to V$ must fix every well-founded set, and in particular, every ordinal. This is the residue of the Kunen inconsistency in this foundation-ness context. It follows, however, that although as you mention BAFA proves the existence of nontrivial elementary embeddings $j:V\to V$, we do not get nontriviality on the ordinals, and so there are no Reinhardt cardinals here to be found. The embeddings provided by BAFA have no critical points. Further, since the consistency strength of BAFA is no greater than that of ZFC, it follows also that consistency-wise, one cannot provably get any large cardinals from such an embbedding in $\text{ZFC}^{-f}$.<|endoftext|> TITLE: Blow-Up for Semi-Linear Wave Equations QUESTION [5 upvotes]: I am reading C. D. Sogge's book "Lectures on Non-Linear Wave Equations". As an exercise, I attempted to fill out the details of the proof of Theorem 5.1 (Local Existence of Solutions for Semilinear Wave Equations), but I got stuck in the last part of the proof regarding the blow-up... Allow me to first state the theorem and then my question. Consider the equation $$ \left\{\begin{array}{ll}\square\, u(t,x) = F(u(t,x)),\; t>0\\ u(0,x) = f(x),\; \partial_tu(0,x)=g(x) \end{array}\right.\qquad\qquad(a)$$ Theorem 5.1. Assume that $F\in C^k$, $F(0)=0$, and that $f\in C_0^{k+1}(\mathbb{R}^3)$, $g\in C_0^k(\mathbb{R}^3)$, with $k = 1,2,\ldots$. then there is a $T > 0$ so that $(a)$ has a unique solution $u\in C^k([0,T]\times\mathbb{R}^3)$. If the supremum, $T_\ast$, of such times $T$ is finite then $\sup_x \lvert u(t,x)\rvert\to \infty$ as $t \to T_\ast$. Question. I am having difficulties understanding the blow up part of the theorem. What is the best way of proving this blow-up phenomenon (the second half of Theorem 5.1)? To this end, let us make the following assumption (this is proved in the book): Assumption. Given a $C^k$ solution $u$ of ($a$) in $[0,T)\times\mathbb{R}^3$. If $\sup_{\{(t,x)\colon 0\leq t T_\ast$ such that $u$ extends to $[0,T_\ast]\cup [T_\ast,T)$. My idea was then to use the first half of the theorem (local existence) on the following problem: $$ \left\{\begin{array}{ll}\square\, \tilde{u}(t,x) = F(\tilde{u}(t,x)),\; t>T_\ast\\ \tilde{u}(T_\ast,x) = u(T_\ast,x),\; \partial_t\tilde{u}(T_\ast,x)=u(T_\ast,x) \end{array}\right.\qquad\qquad(b)$$ This is not a viable strategy, as can be seen easily: The local existence of a solution $u$ in $C^k$ solving $(a)$ requires that $f$ be $C^{k+1}$-smooth, but the local existence theorem guarentees no more than $u(T_\ast,x)$ being of class $C^k$, that is, the initial data in $(b)$ does not have the sufficient regularity needed. This is migrated from: https://math.stackexchange.com/questions/1198583/blow-up-for-semi-linear-wave-equation REPLY [2 votes]: The regularity mentionned in the theorem is not accurate, for two reasons. The first is that the spaces ${\cal C}^k$ don't behave well with PDEs. Often, it is better to work with ${\cal C}^{k,\alpha}$ with $\alpha\in(0,1)$. The second and deeper reason is that the solutions have what is called a co-normal regularity. Physicists say that the solution is polarized. Let me explain this with the one-dimensional case ($x\in\mathbb R$). Then the equation reads $$(\partial_t+\partial_x)(\partial_t-\partial_x)u=F(u).$$ from this, you deduce that the quantity $w:=(\partial_t-\partial_x)u$ has a better regularity in the direction of $\partial_t+\partial_x$ than in other directions. In other words, $(\partial_t-\partial_x)w$ is more regular than $\partial_tw$ and $\partial_x w$ separately. In three space dimensions, this can be quantified using pseudo-differential operators. But again, appropriate regularity results are directional, or polarized. The important theorems for propagation of regularity are due to Egorov and Taylor.<|endoftext|> TITLE: Homeomorphism historically: When did it reach its modern formulation? QUESTION [21 upvotes]: Q. When did the notion of homeomorphism reach its modern formulation as a bicontinuous bijection, i.e., a continuous bijection between topological spaces whose inverse is also continuous? Was this present in Riemann's work (1826-1866)? Or in the work of Möbius (1790-1868); or Jordan (1838-1922)? Was it clearly formulated by 1900? Or only later, perhaps by Heegaard (1871-1948), or Dehn (1878-1952), or Whitney (1907-1989)? I am trying to understand the history behind the classification theorem for compact surfaces, and it seems developing a precise notion of "homeomorphism" was a key step. REPLY [42 votes]: Well, all I did was a search on "homeomorphism history", but... I tried to extract some points that are made in conjunction to your question (Riemann, Möbius, Jordan), though feel free to edit it down if it is too long (and apologies to those who think this should be remapped to a History of Math Q/A). The evolution of the concept of homeomorphism, by Gregory H. Moore Historia Mathematica, Volume 34, Issue 3, August 2007, Pages 333–343 http://www.sciencedirect.com/science/article/pii/S0315086006000863 tl,dr: By 1930, the modern concept had been reached. From 1910-30 the definition (or common understanding of "topological invariance") moved definitively to bicontinuous (Fréchet, Hausdorff, and encyclopedists Zoretti and Rosenthal aiding this), whereas previously merely "continuous" was more popular (starting from von Dyck in 1888). Prior to that, the less-restrictive "deformations" (homotopies) were more prevalent in the thinking of Möbius, Jordan, and Klein (1860/70s), and while Poincaré (1895) coined the term homeomorphism, his meaning was that of diffeomorphism. Incidentally, Riemann had no notion of a topological mapping in his work. Specifically regarding the question (unless otherwise indicated, all the below is from Moore's article): Section 2. The evolution of the concept of “homeomorphism” was essentially complete by 1935 when Pavel Aleksandrov (Paul Alexandroff) at the University of Moscow and Heinz Hopf at the Eidgenossische Technische Hochschule in Zurich published their justly famous book Topologie, aiming to unify the two major branches of topology, the algebraic and the set-theoretic. They took as their fundamental undefined concept “topological space,” based on the closure axioms of Kazimierz Kuratowski [1922]. And they defined a homeomorphism between topological spaces in the way that is now standard: “A one–one continuous mapping $f$ of a space $X$ into a space Y is called a topological mapping or a homeomorphism (between $X$ and $f (X) = Y ⊆ Y$ ) if the inverse of $f$ is a continuous mapping of $Y$ to $X$. Two spaces... are called homeomorphic if they can each be mapped topologically onto each other” [Aleksandrov and Hopf, 1935, 52]. Concerning the origins of topology, Aleksandrov and Hopf wrote: “We must regard Poincaré and Cantor as the immediate founders of topology” [1935, 5]. So the reader might think that he should read the works of Poincaré and Cantor if he wished to find the origin of the concept of homeomorphism. However, the reader would then find that Cantor’s published works contain nothing at all about homeomorphisms, and very little about continuous functions, ... [Quoting Johnson [1979, 127] about Riemann: What we find conspicuously lacking in Riemann’s work is the notion of a topological mapping. For modern mathematicians topology is inseparable from homeomorphisms. Riemann never contemplated these in his programme of analysis situs [i.e. topology].] As for Poincaré, there is a particularly interesting discovery to be made. Although Poincaré coined the word “homeomorphism” in [1895], he meant by it something quite different from and more restricted than what is meant nowadays. ... his “homeomorphism” had strong requirements of differentiability and smoothness that have nothing directly to do with topology ... for Poincaré a “homeomorphism” is not a homeomorphism in the modern sense, but rather a diffeomorphism. ... [Also noted is his 1892 article which spoke of deformations (homotopy), while an 1900 article of his might have used the word "homeomorphism" in the modern sense.] Section 3. Since Poincaré’s 1895 article is the origin of the word “homeomorphism” but not of the concept of homeomorphism, how did that concept originate? At the time that Poincaré wrote, there was a second and broader concept of homeomorphism in use. This second concept was clearly stated by Walther von Dyck in an 1888 article on analysis situs: “Absolute properties [in analysis situs] can also be characterized as those, for whose agreement on two manifolds, it is necessary and sufficient that there exists a one–one continuous function [umkehrbar eindeutiger stetiger Beziehung] between all elements of the two manifolds” [1888, 457]. [Footnote 4 notes that the German phrase could conceivably be interpreted either as "(umkehrbar eindeutiger) und stetiger" or "umkehrbar (eindeutiger und stetiger)", loosely "reversible mapping and continuous" and the question is whether "umkehrbar" (reversible) modifies "stetiger" (continuous); Moore notes that later mathematicians preferred the first construal.] For the rest of the present note, a one–one continuous function from one subset of Euclidean space onto another will be called a “$D$-homeomorphism,” after von Dyck, since a $D$-homeomorphism is a broader concept (both for Euclidean spaces and more generally) than a homeomorphism. But more than three decades were to pass before $D$-homeomorphisms were clearly distinguished from homeomorphisms. ... It is a plausible conjecture that many of those who used $D$-homeomorphisms (something which continued to occur at least until 1924) implicitly assumed that every $D$-homeomorphism is a homeomorphism, i.e., that the inverse function is continuous. Such a result was necessary, in particular, if the inverse of a $D$-homeomorphism was to be a $D$-homeomorphism. And such a result would have been true if they had only considered, in $n$-dimensional Euclidean space, sets which are both closed and bounded. But prior to Camille Jordan’s work [1893], discussed in Section 4 below, this condition was never made explicit. As we shall see when discussing Hurwitz [1898], there was a good deal of murkiness around $D$-homeomorphisms. ... In his article of 1863, Möbius wrote: Two geometric figures are said to be elementarily related to each other if each infinitesimal element (in all dimensions) of the one figure corresponds to a similar element of the other in such a way that if any two bounding elements of the one figure are contiguous, then the same is true of the corresponding elements in the other; or, what expresses the same thing, if a point of the one figure corresponds to a point of the other, then any two infinitely close points of the one figure correspond to infinitely close points of the other. In this regard, a curve can only be elementarily related to a curve, a surface only to a surface, and a solid body only to a solid body. [1863, 435] Nothing in Möbius’s article settles the question of whether he was thinking of his “elementary relationships” as homeomorphisms or as diffeomorphisms or as something different from both, such as deformations. All of his diagrams dealt with smooth two-dimensional manifolds. His use of infinitesimals suggests that he regarded his relations as differentiable. (In any case, at the time he wrote, examples of continuous nowhere differentiable functions were known only to a few mathematicians.) The theorems stated in Möbius’s article would all remain true whether he was thinking of his elementary relationships as homeomorphisms, diffeomorphisms, or deformations. A typical such theorem was that a curve could be elementarily related only to a curve, a surface to a surface, and a solid body to a solid body. Another theorem was that two surfaces in a plane are elementarily related if and only if they have the same finite number of boundary curves [1863, 439]. ... [Later in the section, it is noted Möbius introduced the "rubber sheet" geometry (deformations) in this article/manuscript, similarly Jordan had deformations in his 1866 paper, and 10 years later Klein appeared to think in terms of deformations (which are homotopies, so less restrictive than homeomorphisms). Some comments are made about Schonflies (1906-8), Zoretti (1912), and Rosenthal (1924) the latter who finally distinguished between $D$-homeomorphism and homeomorphism: "For F. Klein [1872] and A. Hurwitz [1898] analysis situs [topology] is the study of those properties of figures (or point-sets) which are preserved by all one–one continuous mappings. It is altogether preferable to require here invariance with respect to one–one continuous mappings whose inverses are also continuous. However, this distinction is only relevant for those figures which are not closed and bounded." Moreover, the article states "[t]here is evidence that even Felix Hausdorff, a decade before he introduced his version of topological spaces, shared the erroneous view of Hurwitz [1898] and Schoenflies that topology is about the invariants of $D$-homeomorphisms."] As late as 1924 some eminent algebraic topologists continued to use the older and inadequate definition of homeomorphism, i.e., of $D$-homeomorphism. Among those who did so was Solomon Lefschetz. ... However, six years later Lefschetz adopted the modern definition of homeomorphism [1930, 3], and he retained this definition in later works [1942, 7]. In their 1934 textbook of algebraic topology, Herbert Seifert and William Threlfall (Dresden) wrote that “topology has to do with those properties of geometric figures that are unaltered by topological mappings, i.e., one–one functions such that they and their inverses are continuous” [1934, 1]. Thus Seifert and Threlfall accepted the modern definition of homeomorphism. From about 1930, the modern definition has been dominant both in general topology and in algebraic topology. Section 4. (Homeomorphisms in abstract spaces) Already by 1906, spaces more abstract than $n$-dimensional Euclidean space had been proposed. In his doctoral dissertation, Maurice Fréchet had been the first to introduce metric spaces (though under another name) as well as his still more general L-spaces, which were based on an abstract notion of the limit of an infinite sequence of points [1906]. In this context, Fréchet carried over from analysis the concept of a continuous function defined in terms of sequences. ... Although in 1906 Fréchet did not discuss homeomorphisms between two of his L-spaces, he did so four years later in an article in Mathematische Annalen on the concept of topological dimension. Given two sets $E_1$ and $E_2$, each part of an L-space, he wrote “that they are the image of each other, or that they are homeomorphic, if there exists between them a one–one correspondence which is bicontinuous [une correspondance biunivoque qui est bicontinue]” [1910, 146]. And he was quite explicit that such a function was bicontinuous if and only if it and its inverse were continuous. This was the first time that the concept of homeomorphism was formulated in a more general context than $n$-dimensional Euclidean space. [Footnote 6: Fréchet adopted the term “homeomorphism” from his teacher Hadamard, rather than directly from Poincaré.] When Hausdorff formulated the concept of a Hausdorff topological space in his 1914 book, he generalized the concept of a continuous function to such spaces. His definition was explicitly motivated by the epsilon–delta definition of continuity of a real function due to Weierstrass. Since Hausdorff’s axioms for a topological space were in terms of neighborhoods, his new generalized definition of continuous function was also formulated in terms of neighborhoods: ... He then isolated the essential relationship between homeomorphisms and one–one continuous functions by giving the condition under which a one–one continuous function is actually a homeomorphism, although he did so without ever using the term homeomorphism: “If $B$ is the one–one continuous image of a set $A$ that is compact-in-itself [every infinite subset $A$ of $B$ has a limit point that belongs to $B$] then $A$ is also the continuous image of $B$." ... It is surprising, under these circumstances, that nowhere in his 1914 book does Hausdorff define the concept of homeomorphism, and nowhere in it does he use the word “homeomorphism” or an equivalent term. Only in the second edition of the book in 1927 did he define the concept of homeomorphism. There he gave the definition in a way that suggests that he may have gotten it from Fréchet’s article of 1910. Hausdorff was thinking of a set $A$ and a set $B$, together with a function $φ : A → B$ and its inverse function $ψ : B → A$ (he explicitly allowed both $φ$ and $ψ$ to be many-valued). Then he wrote: If, however, both functions $y = φ(x)$ and $x = ψ(y)$ are single-valued and continuous, then each of them will be called reversibly continuous or continuous from both sides or doubly continuous (fonction bicontinue). $B$ is also called a homeomorphic image of $A$.., and the one–one mapping between both sets is called a homeomorphism. ... [1927, 195–196] The context in which Hausdorff gave this definition was that of metric spaces, since topological spaces were only discussed, very briefly, later in the book [1927, 226–232]. The French phrase “fonction bicontinue” that Hausdorff included in his original German text shows that he had in mind some French author as the source of the underlying idea that a function and its inverse are both continuous. Apparently this use of “bicontinue” originated with Fréchet [1910]. ... Intriguingly, by 1921 the Polish school of topologists had ... surpassed French and German authors in understanding the difference between homeomorphisms and $D$-homeomorphisms, even in Euclidean spaces. [Kuratowski's 1921 solution of a 1920 problem posed by Sierpinski is noted: if $P$ is a 1-1 continuous image of $Q$ and vice-versa, must they be homeomorphic?] ... Both Sierpinski and Kuratowski wrote important topology textbooks, Sierpinski in Polish [1928], which was then translated into English [1934], and Kuratowski in French [1933]. In both of these books, homeomorphisms were given the modern meaning. Thus by the early 1930s, homeomorphisms in the modern sense were the standard within English, French, German, and Polish textbooks of topology.<|endoftext|> TITLE: Upper bound for a Selberg-type integral over a rectangular region QUESTION [6 upvotes]: (Cross-posted from math-SE). I am trying to estimate the values of the following integral for large $n$, $$\frac{1}{n!}\intop_{\Omega}\prod_{1\leq i\lambda,\,x_2<\lambda,\,x_3<\lambda\ldots,\,x_n<\lambda\},$$ $$\Omega_2^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2>\lambda,\,x_3<\lambda\ldots,\,x_n<\lambda\},$$ $$\ldots$$ $$\Omega_n^{(n)}=\{(x_1,\ldots,x_n):x_1>\lambda,\,x_2>\lambda,\,x_3>\lambda\ldots,\,x_n>\lambda\},$$ and $\lambda$ is a large positive fixed number. Based on the context where this question comes from, my guess is that the integral over $\Omega_k$ decays like $b^{-k}$ (with some $b>1$) when $\lambda$ is large enough. Moreover, this $b$ can supposedly be made as large as one wants by increasing $\lambda$. The original question: This question was motivated by Estimating a Selberg-type integral (or a Fredholm determinant) REPLY [7 votes]: Mathematica can do these integrals explicitly, for small $n$. One gets $$ \frac1{n!}\int_{\Omega_1^{(n)}}\prod_{1\leq i0$, $$ \begin{aligned} \int_{\Omega_1^{(n)}}\prod_{1\leq i\lambda$ introduces another factor $c\,e^{-\lambda^2}\lambda^N$ in the asymptotics as $\lambda\to\infty$, for some constant $c>0$ and integer $N$ (with varying $c$, $N$). This probably can be worked out, too, but it will get messy rather quickly if one wants to know these $c$, $N$.<|endoftext|> TITLE: Infinite Hausdorff space that is not homeomorphic to any proper quotient QUESTION [5 upvotes]: Let $S$ be a set and $\vartheta$ be an equivalence relation on $S$. We say that $\vartheta$ is proper if there are $x\neq y\in S$ with $(x,y)\in\vartheta$. Is there an infinite Hausdorff space $(X,\tau)$ such that for every proper equivalence relation $\vartheta$ on $X$ we have $X\not\cong X/\vartheta$? REPLY [3 votes]: Let $X$ be a strongly rigid infinite Hausdorff space. Let $\vartheta$ be an equivalence relation on $X$. Let $q \colon X \to X/\vartheta$ be the quotient map. If there is a homeomorphism $f \colon X/\vartheta \to X$, then $(f \circ q)$ is a continuous map from $X$ to itself; hence constant or the identity. Since $X$ is infinite, we deduce that $\vartheta$ is trivial.<|endoftext|> TITLE: Disjoint curves in an algebraic surface QUESTION [6 upvotes]: Let $X$ be an algebraic surface (over the complex) with $p_g=q=0$. Is it possible to have disjoint curves $C_1,\ldots, C_b$, of positive genus, spanning $H_2(X,{\mathbb Q})$, $b=b_2(X)$? (When $X$ is the blow-up of the projective plane at some points, the exceptional divisors and a curve coming from the projective plane not passing through the points, give an answer if we do not assume the condition on the genus.) REPLY [3 votes]: Thank you. The following gives an answer in the case that all the curves are assume to have genus $g=1$, and $b\geq 2$. One of the curves, say $C=C_1$ has $C^2>0$. As $p_g=q=0$ we have $\chi(O_X)=1-q+p_g=1$. By Riemann-Roch, we have $\chi(C)=1+ \frac12 (C^2-K\cdot C)$. As $C^2+K\cdot C=2g(C)-2=0$ (by adjunction), we have $\chi(C)=1+C^2$. Also as $h^0(K)=p_g=0$ and $C$ is effective, we have $h^2(C)=h^0(K-C)=0$. From the exact sequence $0\to O_X\to O_X(C) \to O_C(C)\to 0$, taking the long exact sequence in cohomology, we have $H^1(X,O(C))=H^1(C, O_C(C))$. If $C^2>0$ then the last group vanishes, hence $h^1(C)=0$. So $\chi(C)=h^0(C)=1+C^2>1$, and there is a pencil of curves of genus $1$. If necessary, blow up at points of $C$ to arrange $C^2=1$. The bundle $O_C(C)$ is of the form $O_C(p_0)$, for some point $p_0\in C$. The pencil defined by the linear series $|C|$ is a pencil of genus $1$ curves passing through $p_0$. Blowing up at this point we get a surface $X'$ and a map $\pi:X'\to P^1$. Let $\sigma$ be the exceptional divisor, which is a rational curve of self-intersection $-1$. This gives the structure of an elliptic surface, possibly non-minimal, and containing all our genus $1$ curves. Blow down any $(-1)$-curve in a fiber of $\pi$ to get a minimal elliptic surface $X''\to P^1$. This has $q=0, p_g=0$ and it has a section (the image of $\sigma$). By the Enriques-Kodaira classification, this has to be a rational surface. Any section of a minimal elliptic surface has to have self-intersection $-1$ (let $s$ be a section; the canonical bundle is $K=-f$; then $s^2+K\cdot s=-2$, so $s^2=-1$). So $\sigma$ cannot intersect any rational curve being blown-down when going from $X'$ to $X''$, since otherwise its self-intersection increases. All the genus one surfaces $C_i\subset X$ give genus one surfaces $C_i'\subset X'$ which map down to $C_i''\subset X''$. These must be fibers (any curve not intersecting $C$ is contained in a fiber, so it is a rational curve or a fiber). This means that $C_i''$ intersect $\sigma$, and hence $C_i'$ intersects $\sigma$. When going to $X$, $C_i$ intersects $C$. Contradiction!<|endoftext|> TITLE: Is there a natural bijection from $\mathbb{N}$ to $\mathbb{Q}$? QUESTION [22 upvotes]: In a conversation where it came up that the Pythagoreans probably found an enumeration of the rational numbers I erroneously remarked that Georg Cantor found a natural bijection from $\mathbb{N}$ to $\mathbb{Q}$ with his pairing function. Is there a natural bijection bethween these sets? Naturalness is of course not a precise criterion. But we may distinguish between degrees of naturalness and say that a bijection $f$ between $\mathbb{N}$ and $\mathbb{Q}$ is more natural than another bijection $g$ between these sets if for the identity statements $f(n)=\alpha(n)$ and $g(n)=\beta(n)$ the formula $\alpha(n)$ is lower in the arithmetical hierarchy than formula $\beta(n)$. Also, $f$ is more natural than $g$ if the formula $\alpha(n)$ is shorter than the formula $\beta(n)$. REPLY [8 votes]: There is an interesting bijection between the factorial numbering system and the interval of rationals $[0,1)$. Factorials replace powers in the factorial system. $321_! = 3 \cdot 3! + 2 \cdot 2! + 1 = 23_{base10}$. The numerals in each position are limited by the factorial. Only 0 and 1 can be in the first position; 0, 1, or 2 in the second position; 0-3 in the third, etc. Similarly, we can write fractions using the inverses $\frac{1}{2!}, \frac{1}{3!}, \frac{1}{4!}$. $0.123_! = \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} = \frac{23}{24}$. Every rational number has a unique finite representation in the factorial numbering system. We can count in factorial, 1, 10, 11, 20, 21, ..., and take the "inverse" to get a bijection with the rationals in $[0,1) : .1 = \frac{1}{2}, .01 = \frac{1}{6}, .11= \frac{2}{3}, .02 = \frac{1}{3}, .12 = \frac{5}{6}, .001 = \frac{1}{24}$ etc. The factorial numbering system is just one of many product based numbering systems that have a unique finite representation for every rational number.<|endoftext|> TITLE: Dimensions of a vector space akin to modular symbols QUESTION [5 upvotes]: The group $\operatorname{SL}_2(\mathbb Z)$ acts on polynomials in two variables $\mathbb C[x,y]$ via $A\cdot f(x,y)\mapsto f(A^{-1}.(x,y))$ where $(x,y)$ is regarded as a column vector. There are two standard matrices $S=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ and $T=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ which generate $\operatorname{PSL}_2(\mathbb Z)$. Let $\epsilon = \begin{bmatrix}-1&0\\0&1\end{bmatrix}\in \operatorname{GL}_2(\mathbb Z)$. In the course of studying the Johnson homomorphism, I defined a space $\Omega_2(V)$ whose specialization to the case $V=\mathbb C$ in degree $2n$ is defined as follows. Let $R=\mathbb C[x,y]_{2n}$ and let $\langle a_1,\ldots, a_m\rangle=a_1R+\cdots+a_mR$. $$\Omega_2(\mathbb C)_{2n}=\dfrac{R}{\langle 1+ST+(ST)^2,1-\epsilon S\rangle+\mathbb C\{x^{2n},y^{2n}\}}.$$ Computer calculations give the dimensions as $0,1,1,2,3,3,4,5,5,6,\ldots$ starting at $n=1$, and I am looking for a proof of the claim that this pattern continues. If you add the additional relation $\langle 1-\epsilon\rangle$ to this presentation, then it is not hard to see, using for example modular symbols, that the quotient is isomorphic to $H^1(\operatorname{GL}_2(\mathbb Z);R)$, which is known to be isomorphic to the space of cusp forms for $\operatorname{SL}_2(\mathbb Z)$ of weight $2n+2$. So we have an exact sequence: \begin{multline*}0\to\dfrac{\langle 1-\epsilon\rangle}{\langle 1+(ST)+(ST)^2,1-\epsilon S\rangle\cap \langle 1-\epsilon\rangle+\mathbb C\{x^{2n},y^{2n}\}}\\ \to \Omega_2(\mathbb C)_{2n}\to H^1(\operatorname{GL}_2(\mathbb Z);R)\to 0\end{multline*} All the dimensions work out correctly if $$\langle 1+(ST)+(ST)^2,1-\epsilon S\rangle\cap\langle 1-\epsilon\rangle= \langle 1-\epsilon S\rangle\cap\langle 1-\epsilon\rangle=\mathbb C\{x^{2m}y^{2n-2m}-x^{2n-2m}y^{2m}\,:\, 0\leq m\leq n\},$$ but I have not been able to prove this! Remarks: It might help to use the fact that $\langle 1+(ST)+(ST)^2\rangle=\ker(1-(ST))$. I asked an equivalent version of this question at math.stackexhange, but have gotten no useful responses. The dimensions of $\Omega_2(\mathbb C)$ resemble those of the space of cusp forms for the congruence subgroup $\Gamma_0(3)$, but the polynomial degrees are off by one, and so it seems unlikely that they are related. REPLY [3 votes]: Martin Kassabov sent me an elegant solution to this problem. Start with the observation that $\epsilon S$ and $ST$ generate a copy of the symmetric group $S_3$. Now decompose $R$ into a direct sum of irreducible $S_3$-modules. Modding out by $1+ST+(ST)^2$ and $1-\epsilon S$ kill both $1$ dimensional representations and reduce the dimension of the $2$ dimensional representation to $1$. So $\Omega_{2n}(\mathbb C)$, if you ignore the $x^{2n},y^{2n}$, will have the same dimension as the multiplicity of the two dimensional representations in $R$. It's also not hard to show that further modding out by these two monomials will reduce the dimension by $1$. Now to calculate the $S_3$ decomposition, calculate the character. It's not too hard to see that $\chi_R=\begin{bmatrix} 2n+1&1&1\end{bmatrix}, \begin{bmatrix} 2n+1&1&-1\end{bmatrix}$ or $\begin{bmatrix} 2n+1&1&0\end{bmatrix}$ depending on the congruence class of $n$ modulo $3$. From here it follows that the multiplicity of the 2D representation in $R$ is $\frac{2n+1}{3}, \frac{2n+2}{3}$ or $\frac{2n}{3}$ again depending on the congruence class modulo 3.<|endoftext|> TITLE: Obtain any 3-manifold from repeating surgeries on knots in $S^3$ QUESTION [14 upvotes]: In Witten's “QFT and Jones Polynomials” paper, page 383, it states that: "It is a not too deep result that every 3-manifold can be obtained from or reduced to $S^3$ (or any other desired 3-manifold) by repeated surgeries on knots. What are the methods to show this? In the simplest intuitive level? I understand this may be a relevant post, but I hope there are better illuminations. Thanks. p.s. I am a QM/QFT theorist trying to understand the topology better. REPLY [5 votes]: This is really an extended comment. Firstly, as pointed out by Neil Hoffman, the result was proved by Wallace two years before Lickorish, and Lickorish was well aware of this (Wallace's paper is cited in Lickorish's). I really can't understand why Lickorish's name is attached to this result at all. Lickorish's reasonable claim was that his argument was more elementary than Wallace's. Indeed, this is true, since Lickorish's entire paper is eight pages (Wallace's is 28), BUT Lickorish (as suggested in my comment re Sam Nead's answer) does NOT prove that a finite number of Dehn twists suffice (that result is proved in a 1964 paper - two years after, and if you look at the math review, the reviewer is not so sure that it is really proved). Of course, had Lickorish been aware of Dehn's work, his 1962 paper would be down to around two pages. So, another (somewhat complicated) example of Arnold's principle at work. REPLY [4 votes]: Math Reviews says of the paper "A type of homology-preserving surgery on 3-manifolds", J. London Math. Soc. 17 (1978), 183–185 by Norris Weaver: "Let $M$ be a 3-manifold and let $T \subset {\rm int}\,M$ be a tame submanifold homeomorphic to $S^1\times D^2$. Let $g: \partial T \to \partial T$ be an orientation-preserving homeomorphism, and let $n$ be an integer. Glue $T$ or $M - {\rm int}\, T$ by $g^n$ to obtain a manifold $M_1$. Call this construction an $n$th power surgery. The following result is obtained. Theorem: If $n\equiv 0 \, ({\rm mod}\,6)$ and $0\leq n\leq 36$, then $H^1(M_1:\mathbb{Z}_n) \cong H^1(M:\mathbb{Z}_n)$. If $n \not\equiv 0\,({\rm mod}\,6)$ then every closed, orientable 3-manifold can be obtained from $S^3$ by successively performing a finite number of $n$th power surgeries."<|endoftext|> TITLE: Illumination of a convex body QUESTION [5 upvotes]: If $\mathbf{K}$ is a compact, convex set with nonempty interior in $d$-dimensional Euclidean space $\mathbb{E}^d$, and $\mathbf{p}$ is an exterior point outside of $\mathbf{K}$, we say that $\mathbf{p}$ 'illuminates' point $\mathbf{q} \in \mathbf{K}$ if the line from $\mathbf{p}$ to $\mathbf{q}$ intersects the interior of $\mathbf{K}$. Suppose we let $\mathbf{K}_1$ be the set illuminated by $\mathbf{p}$ and $\mathbf{K}_2=\delta \mathbf{K} \setminus \mathbf{K}_1$ (where $\delta \mathbf{K}$ is the boundary of $\mathbf{K}$). My question is: are these two sets ($\mathbf{K}_1$ and $\mathbf{K_2}$) linearly separated by some hyperplane? REPLY [6 votes]: If I have understood your question correctly, the answer is No already in $\mathbb{R}^3$, for the "shadow" boundary is, in general, quite irregular:     (Figure 2.14 from Discrete and Computational Geometry, p.54. (a) Convex hull $Q$. (b) Hull of $p \cup Q$, with $p$ exterior to $Q$.) REPLY [2 votes]: The answer depends on the point $p$ and on the body. Here is a simple counterexample in $R^3$. Take two regular hexagons in parallel planes, so that the line $L$ connecting their centers is perpendicular to these planes. One hexagon is obtained from another by rotation by 30 degrees about the center and parallel translation along the line $L$. The body is the convex hull of the union of the hexagons. The illumination source is somewhere on $L$. I suppose this is simpler that Joseph's example.<|endoftext|> TITLE: Enumeration of $0-1$ matrices with determinant $1$ QUESTION [13 upvotes]: Has the number $f(n)$ of $n \times n$, $0{-}1$ matrices whose determinant is $+1$ been enumerated? E.g., for $n{=}2$, there are $f(2)=3$ such matrices: $$ \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) \;,\; \left( \begin{array}{cc} 1 & 0 \\ 1 & 1 \\ \end{array} \right) \;,\; \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right) \;. $$ For $n{=}3$, I count $f(3)=84$ such matrices, from $$ \left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right) \;, $$ to $$ \left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{array} \right) \;. $$ These matrices are a subset of $SL(n,\mathbb{R})$. Update. Oh, I see $f(n)$ is OEIS A086264: $$ 1, 3, 84, 10020, 4851360, 9240051240. $$ No substantive information is provided in OEIS besides those six computed values. Addendum. Unrevealing, but just as a curiosity, here is an overlay of the $84$ equal-volume parallelepipeds that result by applying the $n{=}3$ matrices to the $3 \times 2 \times 1$ box with lowerleft corner at the origin: REPLY [2 votes]: Will Orrick might have a good guess for this one. As far as I know the answer has only been determined for n up to 8. The number of matrices with odd determinant is known: it is $$\prod_{i=0}^{n-1}(2^n - 2^i)$$, which is about $0.3 * 2^{n^2}$. Noam Elkies has the best guess, but since the number of matrices achieving larger determinants drops off rapidly, I would guess more like $2^{n^2 -cn}$ for a small positive value of $c$. From the arxiv paper of Zivkovic in a comment above, one has for matrices with absolute determinant value 1 : n=6, 18480102480; n=7,135491563468800; n=8, 3766962568171582080 . It is foolish to conjecture a value for $c$ based on this small amount of data, so I will only guess that $c \in [1/2 , 1]$. One can use certain methods to guess at a better lower bound. I like the adjunction method of adding a row and column to a matrix of ADV 1. This gives a lower bound for $a_{n+1}$ of roughly $n^2*2^n*a_n$, with $a_n$ being the number of matrices of ADV=1 and order $n$. Gerhard "It's Nice To Be Referenced" Paseman, 2015.03.23<|endoftext|> TITLE: What are the applications of operator algebras to other areas? QUESTION [30 upvotes]: Question: What are the applications of operator algebras to other areas? More precisely, I would like to know the results in mathematical areas outside of operator algebras which were proved by using operator algebras' techniques, or which are corollaries of operator algebras' theorems. I ask this question for seeing how operator algebras are connected to the other mathematical areas and for better understanding what concrete role it is currently playing in the mathematical world. REPLY [27 votes]: I'm a little puzzled by the tone of the original question. My personal view is that operator algebras are intrinsically interesting, and if there are good applications to other fields, so much the better ... I think this is a pretty common attitude, probably among people in most areas of pure math. Anyway, I am not the most qualified to describe some of these applications, but here are a few of the main ones. (1) Connes' index theorem for foliated manifolds. For instance see here and here. Connes 1982 Fields medal was awarded in part for his work on foliations. (2) Jones' work connecting von Neumann algebras and geometric topology, which gave rise to a new knot invariant. See here for a nice overview. Jones was awarded the Fields medal in 1990 in part for this work. (3) Mathematical physics. Many connections, some more established and some more conjectural. The KMS theory is surely one of the standouts. (4) An early motivation was the theory of group representations, e.g., J. Dixmier, Anneaux d'operateurs et representations des groups, Seminaire Bourbaki, Vol. 1 (1995), 331-336. (5) As mkreisel mentions, there are applications to the Novikov conjecture. See G. G. Kasparov, Equivariant KK-theory and the Novikov conjecture, Invent. Math. 91 (1977), 147-201. (6) The Kadison-Singer problem originally arose as a problem in operator algebras. It now has connections to many other areas (harmonic analysis, Banach space theory, signal analysis, ...). I'm sure I am forgetting some important ones.<|endoftext|> TITLE: When are the adjacency matrices of non-isomorphic graphs similar? QUESTION [8 upvotes]: From Wikipedia. In linear algebra, two n-by-n matrices A and B are called similar if $$ B = P^{-1} A P$$ for some invertible n-by-n matrix $P$. If $P$ is a permutation matrix, $A$ and $B$ are permutation similar. Two graphs $G,H$ are isomorphic iff their adjacency matrices $A_G,A_H$ are permutation similar. If $A_G$ and $A_H$ are not similar then $G$ and $H$ are not isomorphic. It is possible $A_G$ and $A_H$ to be similar and $G$ and $H$ are not isomorphic. Experimentally this doesn't happen often. Recognizing matrix similarity is polynomial, $O(n^3)$. $A$ and $B$ are similar if and only if they have the same rational canonical form (AKA Frobenius form). Is there characterization of the graphs for which $A_G \sim A_H \implies G \cong H$? Consider the following algorithm for graph isomorphism. Recursively delete vertex from $G$ and from $H$ and check the matrices for similarity. If we don't hit bad case we will find isomorphism in polynomial time. The exponential case is if we hit bad cases very often. Is there construction of non-isomorphic graphs for which the above algorithm is exponential? Added The sequence of non-similar matrices of graphs of order $n$ starts 1 , 2 , 4 , 11 , 33 , 151 , 988 , 11453 , 247357 This coincides with OEIS A082104 Number of distinct characteristic polynomials among all simple undirected graphs on n nodes If true, this would imply that symmetric matrices with $0,1$ entries are similar iff their characteristic polynomials are equal (which is likely known). REPLY [3 votes]: There is an old result by L.Babai and D.Grigoriev saying that graph isomorphism for graphs with bounded dimension of eigenspaces can be decided in polynomial time. So yes, a variation of this approach - matching eigenspaces of $A_H$ and $A_G$ - is well-known. For the second question, IMHO, the Cai-Furer-Immerman graphs look like an example.<|endoftext|> TITLE: Why Jacobson, but not the left (right) maximals individually? QUESTION [5 upvotes]: I firstly asked the following question on MathStackExchange a couple of months ago. I did not receive any answers, but a short comment. So, I decided to post it here, hoping to receive answers from experts. It ended in a nice argument, again about the Jacobson Radical, proposed an proved in the following, but the main question remained untouched. Working with Path Algebras, it does not need sophisticated tools to prove for a finite, connected, acyclic quiver $Q$, the Jacobson Radical of $KQ$ is the arrow ideal. But, I have never seen any description of the left (right) maximal ideals of the path algebra for a given quiver, even under the assumptions above (finite, connected and acyclic). Expect for some simple examples, which repeatedly appear in the literature and talks, I am inclined that textbooks and notes intentionally skip this classification, may be due to complexity. It is puzzling to me why this question is not even addressed! In the aforementioned setting, intersection of a certain class of ideals (left maximals) of $KQ$ is the arrow ideal. What about an explicit description of each element of this class, in the sense of the description we have for simples, indecomposable projectives and injectives? i.e., could one classify all the maximal (right) ideals of such a path algebra, in the above or a bit more general setting? Any reference which might address this question would be highly appreciated. REPLY [4 votes]: Dag has already answered the case where the quiver is finite and acyclic, and given a conjecture in the case that cycles are allowed. I will prove his conjecture. Suppose we have an element $x$ of the Jacobson radical. We want to show that it is generated by arrows not lying on any cycle. We can therefore throw away term in $x$ which includes such an arrow, leaving some $x'$. Suppose that $x'$ is non-zero. We want to show that $x'$ is, in fact, not in the Jacobson radical. Choose some term in $x'$, which we can identify as a path $p$ in $Q$. Choose it so that $p$ is maximal length among paths appearing in $x'$. Extend $p$ to a cycle $c$. (This is possible because every arrow of $p$ lies on a cycle, so if $p$ is $a_1\dots a_r$, for each $a_i$ there exists a path $f_i$ which completes $a_i$ to a cycle, and then $a_1\dots a_rf_r \dots f_1$ is a cycle.) Note that this cycle may pass more than once through some vertices. Denote this cycle by $b_1\dots b_m$. Define a representation of dimension $m$, where we define the vector space at vertex $v$ to be the sum of one-dimensional vector space $V_i$ for $i$ such that $b_i$ starts at vertex $v$. Then, we define $b_i$ to be the identity map from $V_i$ to $V_{i+1}$ and the zero map on all other $V_k$. (We let $V_{m+1}$ stand for $V_1$.) This representation has no subobjects (any non-zero element of the representation generates the whole thing), so it is simple. $p_i$ acts non-trivially on it, so $x'$ acts non-trivially on it, because no shorter path than $p_i$ could act like $p_i$ to cancel it out, and $x'$ contains no longer paths. Since this representation is simple, it is isomorphic to the algebra modulo a maximal ideal, and since $x'$ acts non-trivially on the representation, $x'$ is not in the maximal ideal. This contradicts our assumption that $x'$ was in the Jacobson radical.<|endoftext|> TITLE: Root criterion for polynomial over number fields QUESTION [8 upvotes]: It's well known that if $\alpha $ is a rational root to an integer coefficient polynomial, then its denominator divides the leading coefficient and its numerator divides the constant term. I'm asking if there is an analog if the coefficients of the polynomial live in a number field and I'm looking for solutions in that number field, which doesn't necessarily have unique factorization. I'm asking if there is a criterion that reduces the problem to a finite amount of checking just like the integral case. REPLY [8 votes]: Let $f(x) = a_nx^n+\cdots+a_0$ be a polynomial with coefficients in the ring of integers $\mathcal{O}_K$ of a number field $K$. Then for every nonzero root $\alpha$ of $f$ in $K$ one has $a_n \alpha, \frac{a_0}{\alpha} \in \mathcal{O}_K$. This can be seen by the simple observation that $a_n \alpha$ and $\frac{a_0}{\alpha}$ are roots of the monic polynomials $$x^n+a_{n-1}x^{n-1}+a_{n}a_{n-2}x^{n-2}+\cdots+a_{n}^{n-1}a_0,\\ x^n + a_1 x^{n-1}+ a_0 a_2 x^{n-2}+\cdots+a_0^{n-1}a_n, $$ and $\mathcal{O}_K$ is integrally closed. These conditions impose bounds on the valuations of $\alpha$ but in the cases that $K$ has infinitely many units, these conditions doesn't restrict to only finite number of candidates. REPLY [4 votes]: Write $\alpha=a/b$. The ideal $(a)/(a,b)$ divides the last term and $(b)/(a,b)$ divides the first. To prove this, plug in $a/b$ and clear denominators. Each term is in the ideal generated by the other terms. Divide both sides by $(a,b)^n$ and apply unique factorization of prime ideals. There are finitely many possible ideals dividing the first and last coefficients. Given a pair in the same ideal class, that determines $a/b$ up to a unit. However there may be infinitely many units. To deal with thus, you can give an upper bound on the norm of $\alpha$ at each place, reducing it to a finite check.<|endoftext|> TITLE: Are there any natural differential operators besides $d$? QUESTION [22 upvotes]: Let $\lambda = (\lambda_1, \ldots, \lambda_r)$ and $\mu = (\mu_1, \ldots, \mu_r)$ be partitions such that $\mu_j = \lambda_j +1$ for one index $j$ and $\mu_i = \lambda_i$ for all other $i$. Then there is a natural transformation $\alpha_{\mu/\lambda}: \mathbb{S}_{\lambda}(V) \otimes V \to \mathbb{S}_{\mu}(V)$, where $\mathbb{S}_{\kappa}$ denotes the $\kappa$-Schur functor; $\alpha_{\mu/\lambda}$ is unique up to scaling. For a smooth manifold $X$, let's define a $\mu/\lambda$-differential operator to be a map $\delta$ from sections of $\mathbb{S}_{\lambda} T^{\ast} X$ to sections of $\mathbb{S}_{\mu} T^{\ast} X$ such that, for any smooth function $f$ and section $v$, we have the Leibniz rule $$\delta(f v) = \alpha_{\mu/\lambda}(df \otimes v) + f \delta(v).$$ Let's define a natural $\mu/\lambda$-differential operator to be a choice $\delta_X$ of a $\mu/\lambda$-differential operator on each manifold $X$ such that, if $\phi: X \to Y$ is a smooth map, then $\phi^{\ast} \circ \delta_Y = \delta_X \circ \phi^{\ast}$. Are the only natural differential operators scalar multiples of the exterior derivative $d$ with $\mu= 1^{k+1}$ and $\lambda = 1^k$? Motivation: Just curiosity. I've been trying to make $d$ sound natural this term, and one thing that I've said a lot is that it is the only thing that commutes with pullback, so I'm curious if this formalization of that this is true. REPLY [18 votes]: I think your question, the way it is stated, makes one want to classify unary and binary (depending how far you generalise the question as written) invariant differential operators on tensor fields. This has been done for unary operators by an awful lot of people, and the statement indeed is that $d$ is the only operator of that sort. More interestingly, there exists a full classification of binary invariant operators, this was done by Grozman around 1980, and is documented in http://arxiv.org/abs/math/0509562 .<|endoftext|> TITLE: Generalizing a result of Kreisel on $\omega$-consistency QUESTION [6 upvotes]: In (reference)The following result is attributed to Kreisel: Lemma1(Kreisel) If $T$ is an $\omega$-consistent theory in the language of arithmetic and $\pi$ is a true $\Pi_1$ sentence, then $T+\pi$ is also $\omega$-consistent. My question is: Question: If $T$ is an $\omega$-consistent theory in the language of arithmetic, is $T+Th_{\Pi_1}(\mathbb{N})$ also $\omega$-consistent? ($Th_{\Pi_1}(\mathbb{N})$ is the set of all true $\Pi_1$ sentences). The best i could do is the following result: lemma2) If $T\supset I\Sigma_1$ is an $\omega$-consistent theory in the language of arithmetic and $A$ an r.e. subset of $Th_{\Pi_1}(\mathbb{N})$, then $T+A$ is also $\omega$-consistent. proof: if $T_1=T+A$ was $\omega$-inconsistent, then there was a sentence $\exists x \alpha(x)$ such that $T+A\vdash \exists x \alpha(x)$ and also $T+A\vdash \neg \alpha(\overline{n})$ (for all $n$). Let $A=\{\pi_{i}\}_{i \in \mathbb{N}}$. By the compactness theorem, there are indexes $j_{1},\ldots,j_{m}$ such that: $T+\{ \pi_{j_{1}},\ldots,\pi_{j_{m}}\}\vdash \exists x \alpha(x)$. It follows from conservation theorem(reference, theorem 5.2.1) that $T+con(T_{1})\vdash \pi_{i}$ (for every $i\in \mathbb{N}$). Then $T+con(T_{1})\vdash \exists x \alpha(x)$, and by similar reason: $T+con(T_{1})\vdash \neg \alpha(\overline{n})$ (for every $n\in \mathbb{N}$). Then $T_1=T+con(T_{1})$ is an $\omega$-inconsistent theory and it contradicts the lemma1. REPLY [9 votes]: The property does not hold in general. First, I’ll recall some basic properties of $\omega$-consistency. Let $T\vdash_1\phi$ denote the relation that $\phi$ is derivable from $T$ using rules of first-order logic, and unnested instances of the $\omega$-rule. If $T\vdash_1\phi$, then $\phi$ is derivable from $T$ using a single instance of the $\omega$-rule. In particular, $T$ is $\omega$-inconsistent if and only if $T\vdash_1\bot$. If $T$ is r.e., then $T\vdash_1\phi$ is a $\Sigma_3$ property of $\phi$. Thus, the $\omega$-consistency of $T$ is a $\Pi_3$ statement. On the other hand, $Q\vdash_1\phi$ for every true $\Sigma_3$ sentence $\phi$: write $\phi=\exists x\,\forall y\,\psi(x,y)$, where $\psi\in\Sigma_1$, and fix $n\in\omega$ such that $\mathbb N\models\forall y\,\psi(\bar n,y)$. Then $Q\vdash\psi(\bar n,\bar m)$ for every $m$ by $\Sigma_1$-completeness of $Q$, hence $Q\vdash_1\forall y\,\psi(\bar n,y)$ using the $\omega$-rule. Similarly, $Q+\mathrm{Th}_{\Pi_1}(\mathbb N)\vdash_1\phi$ for every true $\Sigma_4$ sentence $\phi$: this follows by the same argument as above, with $\psi\in\Sigma_2$. If $T$ is an r.e. extension of $I\Delta_0+\mathit{EXP}$ (this can be negotiated down with a bit of care), then $T\vdash_1\phi$ satisfies the Bernays–Löb derivability conditions: that is, if we write $\Box_{T,1}\phi$ for the natural arithmetization of $T\vdash_1\phi$, we have $T\vdash_1\phi\implies T\vdash_1\Box_{T,1}\phi$: this is a consequence of 2 and 3. $T\vdash\Box_{T,1}(\phi\to\psi)\to(\Box_{T,1}\phi\to\Box_{T,1}\psi)$: we can concatenate two proofs. $T\vdash\Box_{T,1}\phi\to\Box_{T,1}\Box_{T,1}\phi$; more generally, if $\psi\in\Sigma_3$, then $T\vdash\psi\to\Box_{T,1}\psi$. This follows by formalizing the argument in 3, using the ordinary formalized $\Sigma_1$-completeness of $Q$. See [1] for more information about the provability logic of $\vdash_1$ and related provability predicates. Now, let $T_0$ be an $\omega$-consistent r.e. extension of $I\Delta_0+\mathit{EXP}$ (such as $I\Sigma_1$ or $\mathit{PA}$), and put $$T=T_0+\Box_{T_0,1}\bot$$ (that is, $T_0$ + its own formalized $\omega$-inconsistency). The standard proof of the second Gödel incompleteness theorem using the derivability conditions shows that $T$ is $\omega$-consistent. On the other hand, $\neg\Box_{T_0,1}\bot$ is a true $\Pi_3$ sentence, hence $$T_0+\mathrm{Th}_{\Pi_1}(\mathbb N)\vdash_1\neg\Box_{T_0,1}\bot$$ by 4, thus $T+\mathrm{Th}_{\Pi_1}(\mathbb N)$ is $\omega$-inconsistent. Kreisel’s lemma can be strengthened in a different direction, namely it holds for a larger class of formulas than $\Pi_1$: Proposition 1: If $T\supseteq Q$ is $\omega$-consistent, and $\phi$ is a true $\Sigma_3$ sentence, then $T+\phi$ is $\omega$-consistent. Proof: Otherwise $T+\phi\vdash_1\bot$, hence $T\vdash_1\neg\phi$. On the other hand, $Q\vdash_1\phi$ by property 3 above, hence $T\vdash_1\bot$, i.e., $T$ is $\omega$-inconsistent. Corollary 2: If $T\supseteq Q$ is $\omega$-consistent, and $A$ is an r.e. set of true $\Pi_2$ sentences, then $T+A$ is $\omega$-consistent. Proof: We may assume $T\supseteq I\Delta_0+\mathit{EXP}$ by Proposition 2, as $I\Delta_0+\mathit{EXP}$ is a finitely axiomatizable true $\Pi_2$ theory. It suffices to find a true $\Pi_2$ (or $\Sigma_3$) sentence $\phi$ such that $T+\phi\vdash A$. So, let $\alpha(x)$ be a $\Sigma_1$ definition of $A$ in $\mathbb N$, and $\mathrm{Tr}_{\Pi_2}(x)$ a universal $\Pi_2$ formula. Then we can take $\phi=\forall x\,(\alpha(x)\to\mathrm{Tr}_{\Pi_2}(x))$. Notice that both statements are optimal with respect to arithmetic complexity: The Proposition may fail for true $\Pi_3$ sentences $\phi$: for example, take the $\omega$-consistent theory $T=T_0+\Box_{T_0,1}\bot$ considered above, and $\phi=\neg\Box_{T_0,1}\bot$, which is a true $\Pi_3$ sentence. Then $T+\phi$ is inconsistent. The Corollary may fail for r.e. sets $A$ of true $\Sigma_2$ sentences: continuing the previous example, write $\phi=\forall x\,\psi(x)$ with $\psi\in\Sigma_2$, and put $A=\{\psi(\bar n):n\in\omega\}$. Then $T+A\vdash_1\phi$, while $\neg\phi\in T$, hence $T+A$ is $\omega$-inconsistent. For yet another extension of Kreisel’s lemma, I recall that Smoryński proved that for any r.e. theory $T\supseteq Q$, $$T\vdash_1\phi\iff\mathrm{Tr}_{\Sigma_3}(\mathbb N)+\mathrm{RFN}_T\vdash\phi,$$ where $\mathrm{RFN}_T$ is the uniform reflection principle for $T$: the schema $$\forall x\:\bigl(\Box_T\psi(\dot x)\to\psi(x)\bigr)$$ for all formulas $\psi(x)$, where $\Box_T$ denotes the usual provability predicate for $T$. It is also known that if $T\supseteq Q$ is finite, then $Q+\mathrm{RFN}_T\equiv T+\mathrm{PA}$. Thus: Proposition 3: If $T\supseteq Q$ is $\omega$-consistent, then $T+I\Sigma_n$ is $\omega$-consistent for every $n\in\omega$. Moreover, if $S\subseteq T$ is r.e., and $\phi(x)$ is any formula, then $T+\forall x\,\bigl(\Box_S\phi(\dot x)\to\phi(x)\bigr)$ is $\omega$-consistent. References: [1] George Boolos, The Logic of Provability, Cambridge University Press 1993. [2] Craig Smoryński, Self-reference and modal logic, Springer, 1985.<|endoftext|> TITLE: Lagrangian submanifold of a Calabi-Yau manifold QUESTION [5 upvotes]: In the paper 'Special Lagrangians, stable bundles and mean curvature flow' by R. P. Thomas and S.-T. Yau, page 2. They said A Lagrangian submanifold $L$ of the Calabi-Yau manifold $(X,\Omega)$, we get an induced volume form $vol$ on $L$, and by a short calculation $\Omega_L=e^{i\theta}dvol_L$. And By Lagrangian we will always mean graded Lagrangian (thus the Maslov class of the Lagrangian, which is the class of $d\theta$ in $H^1(L; 2\pi Z)$, is assumed to vanish, and we have chosen a lift of $\theta$). Can anyone tell we how a short calculation to getting $\Omega_L=e^{i\theta}dvol_L$. How to lift and getting the zero Maslov class. REPLY [3 votes]: These facts are a standard part of the lore of Lagrangian submanifolds of Kähler manifolds, but they are often not that explicitly explained in research papers. Probably your best source will be Calibrated Geometries, by Harvey and Lawson (Acta Mathematica 148 (1982), 47–157), which is where the original argument was given. Here is the basic line of argument though: The first part is really a linear algebra fact: Let $V = \mathbb{C}^n$, endowed with the standard symplectic form $$ \omega = \tfrac{\sqrt{-1}}{2}\left(\mathrm{d}z_1\wedge\mathrm{d}\overline{z_1} + \cdots + \mathrm{d}z_n\wedge\mathrm{d}\overline{z_n}\right), $$ which is also the Kähler form for the Kähler metric $$ g = \mathrm{d}z_1\circ\mathrm{d}\overline{z_1} + \cdots + \mathrm{d}z_n\circ\mathrm{d}\overline{z_n}\,. $$ Let $\mathrm{Sp}(\omega) \simeq \mathrm{Sp}(n,\mathbb{R})$ be the group of linear transformation of $V$ that preserve $\omega$. Then, by the usual symplectic linear algebra, $\mathrm{Sp}(\omega)$ acts transitively on the manifold $\mathrm{Lag}^+(\omega)\subset \mathrm{Gr}^+_n(V)$ of oriented Lagrangian $n$-planes in $V\simeq\mathbb{R}^{2n}$ and hence its maximal compact subgroup does as well. Now, the subgroup $\mathrm{U}(\omega,g)\subset \mathrm{Sp}(\omega)$ consisting of the linear transformations that preserve both $\omega$ and $g$ is a maximal compact subgroup of $\mathrm{Sp}(\omega)$ and $\mathrm{U}(\omega,g)$ is isomorphic to $\mathrm{U}(n)\subset \mathrm{O}(2n)$. Thus, $\mathrm{U}(n)$ acts transitively on $\mathrm{Lag}^+(\omega)$. Now, while $\mathrm{U}(n)$ preserves the volume form on an oriented $n$-planes (since it preserves the metric), it does not preserve the complex valued $n$-form $$ \Omega = \mathrm{d}z_1\wedge\cdots\wedge\mathrm{d}z_n $$ on the nose, but only up to a 'phase factor', i.e., $$ g^*\Omega = \det(g)\,\Omega $$ for $g\in\mathrm{U}(n)$, and, for such $g$, we have $\bigl|\det(g)\bigr| = 1$, i.e., $\det(g)\in S^1$. Since $L = \mathbb{R}^n\subset\mathbb{C}^n$ is a Lagrangian $n$-plane and since, on it, we have that $$ \Omega_{\mathbb{R}^n} = \mathrm{d}x_1\wedge\cdots\wedge\mathrm{d}x_n = dvol_{\mathbb{R}^n} $$ when $\mathbb{R}^n$ is given its standard orientation, it follows that when an oriented Lagrangian plane $L$ is written in the form $L = g(\mathbb{R}^n)$ for some $g\in \mathrm{U}(n)$, we must have $$ \Omega_{L} = \det(g)\,dvol_{L}\,. $$ In particular, there is a (smooth) map $\lambda:\mathrm{Lag}^+(\omega)\to S^1$ for which $$ \Omega_{L} = \lambda(L)\,dvol_{L}\,. $$ For the second fact, if you now assume that your manifold $X$ has an $\mathrm{SU}(n)$ structure defined by a nondegenerate $2$-form $\omega$ and a compatible complex-valued $n$-form $\Omega$, then, on the bundle $\mathrm{Lag}^+(X,\omega)$ of oriented Lagrangian $n$-planes, there is a well-defined function $\lambda:\mathrm{Lag}^+(X,\omega)\to S^1$ such that $$ \Omega_{L} = \lambda(L)\,dvol_{L}\, $$ for all $L\in \mathrm{Lag}^+(X,\omega)$. In particular, for any oriented Lagrangian submanifold $P\subset X$, we have a well-defined map $\lambda_P:P\to S^1$ that satisfies $\lambda_P(p) = \lambda(T_pP)$ for all $p\in P$. If $\mu\in H^1(S^1,\mathbb{R})$ is the fundamental generator (i.e., the dual to the fundamental class), then the Maslov class of $P$ is the element $$ \mu_P = \lambda_P^*\bigl(\mu\bigr)\in H^1(P,\mathbb{R}). $$ It vanishes if and only if there is a smooth function $\theta:P\to \mathbb{R}$ such that $$ \lambda_P = \mathrm{e}^{i\theta}. $$<|endoftext|> TITLE: rational cohomology of finite real grassmannian QUESTION [7 upvotes]: Let $p_j$ to be the $j$-th Pontryagin class of the universal $n$-plane bundle $E_n(\mathbb{R}^\infty)\to G_n(\mathbb{R}^\infty)$. Then according to Theorem 1.6, The Cohomology of BSO n and BO n with Integer Coefficients, Proceedings of the American Mathematical Society 1982 Vol 85-2, Edgar H.Brown JR., $H^*(G_n(\mathbb{R}^\infty);\mathbb{Q})=\mathbb{Q}[p_1,p_2,\cdots,p_{[n/2]}]$ . Is there any reference giving $H^*(G_n(\mathbb{R}^{n+k});\mathbb{Q})=$ ? REPLY [9 votes]: I could not find the explicit formulas in the Algebraic models book (they seem to only do infinite Grassmannians and Stiefel varieties) or Mimura-Toda (they do the complex and symplectic case but not the orthogonal one). Generally, I found it annoyingly difficult to dig up explicit formulas and simple explanations in the literature. So, as a service for those who feel the same way, I provide a short discussion of the rational cohomology of real Grassmannians here and apologize for bouncing an old question on which the dust had already settled. The main point (for understanding why cohomology of Grassmannians is the way it is) is to note that the homogeneous space description of the Grassmannians as ${\rm O}(n)/{\rm O}(k)\times{\rm O}(n-k)$ implies that there is a fiber bundle $$ {\rm Gr}_k(\mathbb{R}^n)\to {\rm BO}(k)\times{\rm BO}(n-k)\to{\rm BO}(n), $$ which can be interpreted as saying that the Grassmannian is the universal space with a pair of real vector bundles $(\mathcal{E},\mathcal{F})$ of ranks $k$ and $n-k$, respectively, such that $\mathcal{E}\oplus\mathcal{F}$ is the trivial rank $n$ bundle. The classifying map ${\rm Gr}_k(\mathbb{R}^n)\to {\rm BO}(k)\times{\rm BO}(n-k)$ induces a homomorphism $$ {\rm H}^\bullet({\rm BO}(k)\times{\rm BO}(n-k),\mathbb{Q})\to {\rm H}^\bullet({\rm Gr}_k(\mathbb{R}^n),\mathbb{Q}) $$ and so we get two sets of characteristic classes: the Pontryagin classes $p_i$ for $\mathcal{E}$ and the Pontryagin classes $\overline{p}_i$ for $\mathcal{F}$. Moreover, since $\mathcal{E}\oplus\mathcal{F}$ is trivial, there is a natural relation $p\cdot\overline{p}=1$ for the product of total Pontryagin classes (this is essentially the Whitney sum formula). The main result is then that for the even-dimensional Grassmannians this is already the description: $$ {\rm H}^\bullet({\rm Gr}_{2k}(\mathbb{R}^{2n}),\mathbb{Q})\cong {\rm H}^\bullet({\rm Gr}_{2k}(\mathbb{R}^{2n+1}),\mathbb{Q})\cong {\rm H}^\bullet({\rm Gr}_{2k+1}(\mathbb{R}^{2n+1}),\mathbb{Q})\cong \mathbb{Q}[p_1,\dots,p_k,\overline{p}_1,\dots,\overline{p}_{n-k}]/(p\cdot\overline{p}=1). $$ (Note that the Euler classes don't appear here: we are talking about the non-oriented Grassmannians and therefore the Euler classes naturally live in cohomology with twisted coefficients; they appear when passing to the oriented Grassmannians.) For the odd-dimensional Grassmannians, there is an additional class $r$ in degree $2n+1$: $$ {\rm H}^\bullet({\rm Gr}_{2k+1}(\mathbb{R}^{2n+2}),\mathbb{Q})\cong \mathbb{Q}[p_1,\dots,p_k,\overline{p}_1,\dots,\overline{p}_{n-k},r]/(p\cdot \overline{p}=1,r^2=0). $$ The class $r$ is detected after pullback along ${\rm O}(2n+2)\to{\rm Gr}_{2k+1}(\mathbb{R}^{2n+2})$, it's not in the image of ${\rm H}^\bullet({\rm BO}(k)\times{\rm BO}(n-k),\mathbb{Q})$. As far as I understand, the results for the even-dimensional Grassmannians were first proved by Borel, the results for the odd-dimensional ones by Takeuchi. A. Borel. Sur la cohomologie des espaces fibre principaux et des espaces homogenes de groupes de Lie compacts. Ann of Math (2) 57 (1953), 115-207. M. Takeuchi. On Pontryagin classes of compact symmetric spaces. J. Fac. Sci. Univ. Tokyo Sect I 9 (1962), 313-328. The results were also proved by He (arXiv paper here), Carlson (arXiv paper here), Sadykov (paper in PJM here) and probably I am missing some other references...<|endoftext|> TITLE: When are (weak) homotopy equivalence testable on open covers? QUESTION [12 upvotes]: I asked this question on math.stackexchange, but did not get an answer. Let $f\colon X\rightarrow X'$ be a continuous map between two spaces $X,X'$, which might be arbitrary wild, especially I don't want to work in any convenient category of topological spaces. Let $X=U\cup V$ and $X'=U'\cup V'$ be open covers such that $f(U)\subseteq U'$ and $f(V)\subseteq V'$ holds. Consider the following claim. If the three restrictions $f\colon U\rightarrow V$, $f\colon V\rightarrow V'$ and $f\colon U\cap U'\rightarrow V\cap V'$ are weak homotopy equivalences, then so is $f\colon X\rightarrow X'$. Is this claim true in general? If not, are there mild assumptions on $X$, $X'$ or $X$ and $X'$, such that the claim holds, e.g. does the claim hold if $X$ and $X'$ are Hausdorff spaces? What about the corresponding claim with homotopy equivalences instead of weak equivalences? REPLY [2 votes]: The claim about weak equivalences follows as soon as one proves that the cocartesian squares generated by U←U∩V→V and U'←U'∩V'→V' are also homotopy cocartesian. To this end one can use Lurie's Seifert-van Kampen theorem (Theorem A.3.1 in Higher Algebra) to establish that these squares are always homotopy cocartesian: in Lurie's notation, take C={1←0→2} and the functor χ sends C to {U←U∩V→V}. The fact that {U,V} is an open cover of X establishes the required property (*). (Of course, this particular fact had been established long before Lurie's book came out and can be found in many older, less accessible references.)<|endoftext|> TITLE: Maximum occupancy balls in bins with limited independence QUESTION [12 upvotes]: Throw $n$ balls into $n$ bins and let $X_n$ be the maximum occupancy. That is the maximum number of balls found in any bin. If you throw the balls uniformly and independently it is known that $\mathbb{E}(X_n) = \Theta(\log{n}/\log{\log{n}})$. If the process is merely pairwise independent (but still uniform) then it is known that $\mathbb{E}(X_n)$ can grow as quickly as $\Theta(\sqrt{n})$. If the random process is $k$-wise independent, then for each $k$ there will be some tight asymptotic upper bound for $\mathbb{E}(X_n)$. How do the asymptotics of a tight upper bound for $\mathbb{E}(X_n)$ depend on $k$? This question may be hard to answer precisely but approximate answers and even conjectures would be very helpful too. Sometimes $k=4$ is a transition point as you can then use more powerful moment methods but I am not sure how to do that usefully here. What are the asymptotics for a tight upper bound for $\mathbb{E}(X_n)$ when $k=4$? REPLY [6 votes]: Consider the probability that $k$ particular balls go into the same bin. By $k$-wise independence, this is $n^{-k}$ for each bin, or $n^{-k+1}$ when we sum over all bins. On the other hand, if we choose a random $k$-tuple of indices, the probability that these are all sent to the same bin is at least the probability that all $k$ are sent to the bin with the highest load $X_n$, so if we condition on $X_n$, the probability is at least ${X_n \choose k} / {n \choose k}$. Unfortunately, $x \choose k$ is not always convex as a function of $x$, so we have to modify it slightly to use convexity. For $k \in \mathbb{N}, x\in \mathbb{R}$, define ${x \choose k}^+ = {\begin{cases} {x \choose k}, x \gt k-1 \newline 0, x \le k-1\end{cases}} = \frac{1}{k!}\prod_{i=0}^{k-1} \max (0,x-i).$ Since each factor $\max (0,x-i)$ is convex and nonnegative, the scaled product ${x \choose k}^+$ is convex. By Jensen's inequality, the probability $k$ balls go to the same bin is at least ${E[X_n] \choose k}^+ / {n \choose k}$. So, $k$-wise independence implies $$\begin{eqnarray}{E[X_n] \choose k}^+ &\le & {n \choose k}n^{-k+1} \newline &\le & \frac{n}{k!} \newline \prod_{i=0}^{k-1} \max(0,X_n-i) &\le & n \newline \max(0,E[X_n]-k+1)^k &\le & n \newline E[X_n] &\le & \sqrt[k]{n} + k-1. \end{eqnarray}$$<|endoftext|> TITLE: What is the value of the infinite product: $(1+ \frac{1}{1^1}) (1+ \frac{1}{2^2}) (1+ \frac{1}{3^3}) \cdots $? QUESTION [5 upvotes]: What is the value of the following infinite product? $$\left(1+ \frac{1}{1^1}\right) \left(1+ \frac{1}{2^2}\right) \left(1+ \frac{1}{3^3}\right) \cdots $$ Is the value known? REPLY [3 votes]: I'm not sure what the criterium for a full answer is here, so here a technique for $(1+c_k)$ kind of products, turning the infinite product into an infinite sum: Via telescoping, for friendly $a_n$ and any $m$, we have $$\lim_{n\to\infty}a_n=a_m+\sum_{n=m}^\infty\left(\dfrac{a_{n+1}}{a_n}-1\right)\,{a_n}.$$ So define $$a_n:=\prod_{k=1}^{n-1}\left(1+c_k\right)\hspace{.5cm}\implies\hspace{.5cm}\dfrac{a_{n+1}}{a_n}-1=c_n,$$ and then $$\prod_{n=1}^\infty\left(1+c_n\right) = \lim_{n\to\infty}a_n = \prod_{k=1}^{m-1}\left(1+c_k\right)+\sum_{n=m}^\infty c_n\prod_{k=1}^{n-1}\left(1+c_k\right).$$ For $c_n=\dfrac{1}{n^n}$, that's $$\frac{1^1+1}{1^1}\,\frac{2^2+1}{2^2}\frac{3^3+1}{3^3}+\sum_{n=4}^\infty\frac{1}{n^n}\prod_{k=1}^{n-1}\left(1+\frac{1}{k^k}\right)=2.603\dots$$ The first term is the lower bound $\frac{70}{27}=2.592\dots$ that's been pointed out in the comment and the remaining sum $\frac{1}{4^4}\dots+\frac{1}{5^5}\dots$ collects some $\mathcal{O}(10^{-2})$. Truncation of the product after $k=1$ reveals the infinite product is almost two times Sophomore's dream: $$\prod_{n=1}^\infty\left(1+n^{-n}\right)\approx 2\sum_{n=1}^\infty \frac{1}{n^n}=2.582\dots$$<|endoftext|> TITLE: On a drawing in Dixmier's Enveloping Algebras QUESTION [20 upvotes]: This image comes from Dixmier's book, 'Enveloping Algebras' ('Algèbres enveloppantes'). Dixmier writes that The curves shown on p. XIV have their origin in the study of U(sl(3)). They are due to Professor W. Borho, who kindly authorized me to reproduce them. What do these curves represent ? REPLY [15 votes]: I want to elaborate on the answer by Jim Humphreys. I encountered the same question in my mind a few months back when I was going through the book by Dixmier. At that point, I read the above answer, but the concise explanation by Borho didn't make a lot of sense to me. A few weeks back, Prof. Victor Ginzburg suggested the following approach which seemed quite appropriate from the perspective of the above explanation: The weight lattice for $\mathfrak{sl}(3,\mathbb{C})$ is given by the following picture: On a heuristic level, one sees that that the shaded region looks similar to the curves that we are trying to study, and that motivates considering the Weyl group action. Now, in the notation of the picture, if we let $x,y,z$ denote the dual functionals of the vectors $\epsilon_1,\epsilon_2,\epsilon_3$, respectively, we have that $x,y$ and $z$ span the Cartan subalgebra $\mathfrak{h}$, and also, $x+y+z=0$. As the Weyl group $S_3$ acts by permuting the $\epsilon_i$'s, by the Harish Chandra isomorphism, we have that : $$Z(\mathfrak{sl}( 3,\mathbb{C})) \cong S(h)^W = \Big(\frac{\mathbb{C}[x,y,z]}{(x+y+z)}\Big)^{S_3} = \Big(\frac{\mathbb{C}[x+y+z,xy+yz+zx,xyz]}{(x+y+z)}\Big)\cong\frac{\mathbb{C}[A,B]}{(x+y+z)},$$ where $A=xy+yz+zx$ and $B=xyz$. One sees that $A$ and $B$ are algebraically independent, and so, the above becomes a polynomial ring in $2$ variables implying that the maximal spectrum of $Z(\mathfrak{sl}(3,\mathbb{C}))$ is isomorphic to $\mathbb{C}^2$ as an algebraic variety. It is into this algebraic variety that we want to project our weight lattice. Now, if we call the origin of the weight lattice $O$, the $3$ lines passing through the origin together have the equation $(x-y)(y-z)(z-x)$. More generally, for each natural number $n$, we have 6 lines in the lattice that are at a distance $n$ from $O$ which can be given by the joint equation: $$P_n:=(x-y-n)(x-y+n)(y-z-n)(y-z+n)(z-x-n)(z-x+n)=[(x-z)^2-n^2][(y-z)^2-n^2][(z-x)^2-n^2],$$ which can be seen to be invariant under the action of $S_3$. (The fact is obvious geometrically too!) Hence, if we project $P_n$ onto $Z(\mathfrak{sl}(3,\mathbb{C}))$, under the above isomorphism, we can write (after a brief computation) $P_n=27B^2-4A^3+9n^2B^2-6n^4B+n^6$. Now, a point to be noted here is that the underlying field is $\mathbb{C}$, and so, the above equation can not, technically, be plotted on the Cartesian plane. But, as all the coefficients involved are real, we can plot the above cubic equations in the $AB$-plane to get: Close enough!<|endoftext|> TITLE: Can you decide whether the commutator subgroup of a f.p. group is f.g? QUESTION [9 upvotes]: Is the following algorithmic problem known to be decidable/undecidable? Input: a finite group presentation $P$. Decide: is the commutator subgroup of the group presented by $P$ finitely generated? REPLY [16 votes]: It's undecidable. Lemma: a group $G$ is nontrivial if and only if the free product $H=G\ast\mathbf{Z}$ has an infinitely generated derived subgroup. Proof: assume $G$ finitely generated and nontrivial. The kernel $N$ of the canonical epimorphism $H\to\mathbf{Z}$ is isomorphic to $G^{\ast\mathbf{Z}}$ and hence is infinitely generated. Since $H/[H,H]$ is finitely generated abelian, so is $N/[H,H]$; if $[H,H]$ were finitely generated, so would be $N$, a contradiction. Hence $[H,H]$ is infinitely generated. In case $G$ is infinitely generated and if by contradiction $[H,H]$ is finitely generated, then its generators belong to $G_1\ast\mathbf{Z}$ for some proper subgroup $G_1$ of $G$, which is obviously a contradiction.$\Box$ The result then from the fact that if we have a Turing machine $X$ whose input is a finite presentation $P$ and whose answer is yes or no according to whether the group presented by $P$ has a finitely generated derived subgroup, then if we input in $X$ the presentation obtained from $P$ by adding a generator, the output is yes or no according to whether the group presented by $P$ is trivial. Thus the resulting machine solves the triviality problem, and it is known that there is no such machine.<|endoftext|> TITLE: Isometries of some simple Cayley graphs QUESTION [12 upvotes]: Consider a Cayley graph of a group $G$ with respect to a symmetric finite generating set $S$. There are some obvious candidates to isometries of this graph - for example, translation by elements of $G$, and group automorphisms which preserve $S$. In some simple cases, these are everything one has (up to composition) - for example, take $G$ be a the free abelian group $G=\mathbb{Z}^{d}$ with $S$ being $\{\pm e_i\}$, with $\{e_i\}$ being the standard basis. This is true for other generating sets as well. However, it is far from correct, say, for the free group with its standard generators. I am more inclined toward "$\mathbb{Z}^d$-esque" situations in my research, so I would like to try and know more on the first case example. For the questions themselves: (a) Consider $G=\mathbb{Z}^d$ with some generating set. Can all isometries of the Cayley graph be seen as a composition of translations and group Automorphisms (preserving S)? (b) Is this claim true for "nice" nilpotent groups, e.g. the Heisenberg groups? (c) Must it be false for hyperbolic groups, considering all possible generating sets? REPLY [10 votes]: Question (b) (and, therefore, also Question (a)) was answered affirmatively (for all torsion-free nilpotent groups) in 1998 by R.G.Möller and N.Seifter [Europe J. Comb. 19, 597-602] (see Theorem 4.1(1)). The answer to Question (a) was rediscovered in 2007 by A.A.Ryabchenko [Siberian Math. J. 48 (2007) 919–922] (see the proof of Theorem 2). This short proof is reproduced in Theorem 5.3 of a preprint of J.Morris (arXiv:1502.06114). A more elementary proof of the answer to Question (b) (for all torsion-free nilpotent groups) is in a preprint by J.Morris, G.Verret, and me (arXiv:1603.01883). I have no information about Question (c).<|endoftext|> TITLE: Evaluation maps for moduli of stable maps QUESTION [5 upvotes]: Let $\overline{M}_{0,n}(\mathbb{P}^N,d)$ be the moduli space of stable maps of degree $d$ from curves of genus zero with $n$-marked points to $\mathbb{P}^N$. Consider the product of the evaluation maps: $$ev:=ev_1\times ...\times ev_n:\overline{M}_{0,n}(\mathbb{P}^N,d)\rightarrow (\mathbb{P}^N)^{n},\quad [C,f,(x_1,...,x_n)]\mapsto (f(x_1),...,f(x_n)).$$ Is is true that $ev$ is proper and $ev_{*}\mathcal{O}_{\overline{M}_{0,n}(\mathbb{P}^N,d)} = \mathcal{O}_{(\mathbb{P}^N)^{n}}$? REPLY [4 votes]: For $N>1$ and large $n$ the map $ev$ will not be surjective for dimension reasons: $$\text{dim} \overline{M}_{0,n}(\mathbb{P}^N,d) = Nd + N + d + n-3$$ whereas $$\text{dim} (\mathbb{P}^N)^{n} = N n.$$ Thus the pushforward of the structure sheaf cannot be the structure sheaf on the target. On the other hand for $n=1$ the map $ev=ev_1$ is surjective and flat (see for example the book "An invitation to quantum cohomology" by Kock and Vainsecher, Lemma 2.5.1). So all one would need to show in this case is that the fibres of $ev$ are geometrically connected (by Exercise 28.1.H. in Vakil's Foundations of Algebraic Geometry). Idea of a probably much too complicated proof: Given two closed points $\mu,\nu$ in the fibre of a point $p \in \mathbb{P}^N$ they can be connected by a rational curve $g: C \to \overline{M}_{0,n}(\mathbb{P}^N,d)$ because this variety is rational. Via $ev_1$ this induces a curve $f: C \to \mathbb{P}^N$. Choose fixed $N+1$ points $q_i \in \mathbb{P}^N$ such that for all but finitely many points $c\in C$ the points $f(c),q_1, \ldots, q_{N+1}$ are in general linear position. Let $\tilde C \subset C$ be the set of points satisfying this, then $\tilde C$ is still connected. There exists a unique $A(c) \in \text{PGL}(N)$ satisfying $A(f(c))=p, A(q_i)=q_i$. By postcomposing the family induced by $\tilde C \to \overline{M}_{0,n}(\mathbb{P}^N,d)$ with the map $A(c)$ we obtain a family connecting $\mu,\nu$ contained in the fibre of $ev$ over $p$. Edit: I think that for $n\leq N+2$ and $ev$ surjective, the above arguments can be expanded to conclude the desired property of $ev$ when restricting to the preimage of the set $U \subset (\mathbb{P}^N)^n$ of tuples of points $(p_1, \ldots, p_n)$ in general linear position. The action of $\text{PGL}(N)$ on $\mathbb{P}^N$ induces a transitive action on $U$ and also an action on $\overline{M}_{0,n}(\mathbb{P}^N,d)$ (by postcomposition of the stable map) such that $ev$ is equivariant. Then by generic flatness it should hold that $ev$ is flat over $U$ (see the arguments in the book of Kock and Vainsecher). It is also still proper as the restriction (on the target) of a proper map. For the connectedness of its fibres one copies the argument above, where now one considers the maps $p_1=ev_1 \circ g, \ldots, p_n=ev_n \circ g : C \to \mathbb{P}^N$ and chooses general $q_1, \ldots, q_{N+2-n}$ fixed. For a general point in $C$ the points $p_1, \ldots, p_n, q_1, \ldots, q_{N+2-n}$ will be in general linear position and the argument goes as before.<|endoftext|> TITLE: Uninteresting questions with interesting answers QUESTION [33 upvotes]: What are best examples of questions in mathematics that are not interesting until one knows the answers, whose answers themselves are what is interesting? The thing that prompts me to post this is just one example. I've seen others, but they escape me at the moment. Here it is: A torus is embedded in just the usual way in $\mathbb R^3$. It has parallels of latitude and meridians of longitude. A curve that meets every parallel of latitude at the same angle, or, equivalently, meets every meridian of longitude at the same angle, is a loxodrome. Suppose that angle is so chosen, given the shape of the particular torus, that the loxodrome goes through all $360^\circ$ of longitude in just the length it takes to go through all $360^\circ$ of latitude, returning there to its starting point. (There must be some conventional terminology for describing these windings, but I don't know it.) The question is: What are the curvature and torsion at the various points along this curve? Doubtless some will consider this question interesting, but to me, and, I suspect, to many, the answer, because it is so unexpected, is where this starts to get interesting. The answer is that the curvature is constant --- the same at all points on the curve --- and the torsion is everywhere $0$. (And it's really easy to deduce from that the precise value of the curvature.) I believe this was discovered in the 1890s and is stronger than the celebrated theorem of Villarceau, published in 1848. Villarceau's theorem says that a plane bitangent to a torus intersects the torus in two circles. This proposition does not assume as a hypothesis, but rather has as a (trivial corollary of its) conclusion, that the curve lies in a plane. REPLY [4 votes]: Watson's integral. Seemingly uninteresting question: calculate $$W_S=\frac{1}{\pi^3}\int\limits_0^\pi\int\limits_0^\pi\int\limits_0^\pi \frac{dx\,dy\,dz}{3-\cos{x}-\cos{y}-\cos{z}},$$ produces truly amazing answer: $$W_S=\frac{\sqrt{6}}{96\pi^3}\Gamma\left(\frac{1}{24}\right)\Gamma\left(\frac{5}{24}\right)\Gamma\left(\frac{7}{24}\right)\Gamma\left(\frac{11}{24}\right)= \frac{\sqrt{3}-1}{96\pi^3}\left[\Gamma\left(\frac{1}{24}\right) \Gamma\left(\frac{11}{24}\right)\right]^2.$$ See http://link.springer.com/article/10.1007%2Fs10955-011-0273-0 (70+ Years of the Watson Integrals, by I. J. Zucker).<|endoftext|> TITLE: reflexive banach space QUESTION [6 upvotes]: I want to ask this non-expert question: What does it mean geometrically for a Banach space to be reflexive? Well, we could say a Banach space is reflexive iff unit ball is weakly compact. Or some other theorems may be. But this doesn't give me a geometric intuition so far. REPLY [9 votes]: This is not exactly an answer, and not exactly giving a condition equivalent to reflexivity, but I want to give a geometric example warning for the development of the geometric intuition. Let me start from a well-known characterization that a Banach space $X$ is super-reflexive if and only if $X$ can be equivalently renormed with a uniformly convex norm. If you are not familiar with these definitions, please check Wikipedia. Intuitively uniform convexity says that the ball is "uniformly" round. One has that if $x, y$ belong to the unit sphere of $X$ and are $\varepsilon>0$ apart (i.e. $\|x-y\|>\varepsilon$), then the midpoint has to be inside the unit ball, and not on the sphere, and it has to be uniformly "deep", i.e. $$\|\frac{x+y}2\|\le 1- \delta_X(\varepsilon),$$ where $\delta_X(\varepsilon)>0$ and depends only on $\varepsilon$. Note that the condition is that $X$ can be renormed to satisfy this condition, not that every norm satisfies it. In fact, it is possible to equivalently renorm the Hilbert space $\ell_2$ to have a positive face of the unit sphere of $\ell_1$ (that is a very "flat" set) inside the positive face of renormed $\ell_2$, which is "the most reflexive space". In fact this is possible in every infinite dimensional Banach space. To see this, let $(x_i, x_i^*)$ in $X\times X^*$ be a biorthogonal system with $\|x_i\| = 1$ and $\|x_i^*\|\leq 2$. (Such a biorthogonal system exists by applying a theorem of Ovsepian and Pelczynski, see for example the book J. Diestel, Sequences and Series in Banach Spaces,page 56, to a separable subspace of $X$ and then extending to functionals on all of $X$ via the Hahn-Banach theorem.) Then let $$|||x||| = \max\{ |x_1^*(x)|, \frac12 \|x\|, \displaystyle\sup_{i\ne j; \, i, j\geq 2} (\,|x_i^*(x)| + |x_j^*(x)|\, ) \}.$$ This defines an equivalent norm on $X$ with $||| x_1 + x_n|||=1$ and for all $\alpha_n\geq 0$ we have $$|||\, \sum_{n=1}^\infty \alpha_n (x_1 + x_n)\,||| = \sum_{n=1}^\infty \alpha_n.$$ I first learned about this fact from Vitali Milman. I think it goes back to Ptak. The above norm was defined by A. Pelczynski and M. Wojciechowski.<|endoftext|> TITLE: Connectedness in the language of path-connectedness QUESTION [33 upvotes]: Is there a topological space $(C,\tau_C)$ and two points $c_0\neq c_1\in C$ such that the following holds? A space $(X,\tau)$ is connected if and only if for all $x,y\in X$ there is a continuous map $f:C\to X$ such that $f(c_0) = x$ and $f(c_1) = y$. Is there also a Hausdorff space satisfying the above? REPLY [4 votes]: I hope that my example below is as elegant as the continuous long line provided by Goldstern above, while my example is less expected. Also, while long line is simpler in itself, the proof is simpler in my case. Finally, perhaps logicians will find some advantages (I'll do a little of it--I am not confident to do it well). Let $\ A\ $ be an arbitrary set. The following ordered triple $\ (\mathbf S_A\ \mathbf 0\ \mathbf 1)\ $, where $\ \mathbf S_A:=(S_A\ T_A)\ $ is a topological space--call it a skeleton, and $\ \mathbf {0\ 1}\in S_A,\, $ is to be defined below, while first (ahead of time) let's formulate THEOREM   For every connected subset $\ X\subseteq S_A,\ $ such that $\ \mathbf {0\ 1}\in X,\ $ the inequality of cardinalities $\ |X|\ge|A|\ $ holds. This instantly gives a simple negative answer to the question of this thread posed by Dominic. DEFINITION $\ S_A\ :=\ \{(x_a)_{a\in A}\in[0;1]^A\ :\ \forall_{a\ b\in A} [x_a\ x_b\in(0;1)\ \Rightarrow\ a=b]\ \}$ $\ \mathbf 0\ :=\ (0)_{a\in A}\ $ and $\ \mathbf 1\ :=\ (1)_{a\in A}$ $\ T_A\ $ is the topology in $\ S_A\ $ induced by the Tikhonov toplogy in cube $\ [0;1]^A$ PROOF (of the theorem)   The connected component of $\ \mathbf 0\ $ in $\ S_A\ $ is dense in $\ S_A,\ $ which means that its closure, i.e. space $\ S_A\ $ itself, is connected too. Next, let: $$H^a\ :=\ \{x\in[0;1]^A\ :\ x_a=\frac 12$$ Let $\ X\subseteq S_A\ $ be a connected subset such that $\ \mathbf {0\ 1}\in X.\ $ Then $\ p_a(X)=[0;1],\ $ hence $\ H\,^a\cap X\ne \emptyset.\ $ Thus $$ |X|\ \ge\ \left|\left\{H^a\ :\ a\in A\ \right\}\right|\ =\ |A|$$ Indeed, sets $\ H^a\ $ are disjoint (hence different).   END of Proof             G E N E R A L I Z A T I O N We may replace the topological interval $\ [0;1],\ $ and its three points $\ 0\ \frac 12\ 1,\ $ by an arbitrary connected space $\ S\ $, and its three points $\ a\ h\ b,\ $ such that $\ h\ $ separates $\ a\ b\ $ (meaning that there are open sets $\ G\ $ and $\ H:=(S\setminus\{h\})\setminus G\ $ of $\ S\ $ such that $\ a\in G\ $ and $\ b\in H$. Etc. The theorem still holds. Logical considerations I am not using ordinal numbers. My construction is free of any complications, especially when $\ S\ $ of the generalization is a proper 3-point space $\ \{a\ h\ b\}.\ $ Thus I am worried only about axioms like the axiom of choice or continuum hypothesis, and similar, about their relation to the cartesian product, and to the ordinary $\ [0;1]\ $ of my main example. EXTRA. Compactness. Another connectedness proof. Space $\ \mathbf S_A\ $ is compact, it is a closed subset of the Tikhonov cube $\ [0;1]^A:\ $ indeed, let $\ x\in [0;1]^A\setminus S_A.\ $ Then there exist two different indices $\ a\ b\ \in\ A\ $ such that $\ (x_a\ x_b)\in (0;1)^{\{a\ b\}}.\ $ Thus the inverse image of this open square under the canonical projection $\ p_{a\ b} : [0;1]^A\rightarrow(0;1)^{\{a\ b\}}\ $ is disjoint from $\ S_A\ $ (one could say that $\ [0;1]^A\setminus S_A\ $ is open because it is a union of the inverse images of the open squares). Thus indeed $\ S_A\ $ is compact. Now $\ \mathbf S_A\ $ is connected because it is an inverse limit of spaces $\ \mathbf S_B\ $ for all finite $\ B\subseteq A,\ $ under the canonical projections. (One could also use som other similar arguments). This inverse limit nature of $\ \mathbf S_A\ $ shows its covering 1-dimensionality: $$\dim \mathbf(S_A)\ =\ 1$$<|endoftext|> TITLE: Strange problem about triplets of differential forms QUESTION [6 upvotes]: Suppose we have the following map: $$(\Omega^1(\mathbb{R}^n))^3\longrightarrow(\Omega^2(\mathbb{R}^n))^3$$ $$(\alpha,\beta,\gamma)\longmapsto(\mathrm{d}\alpha+\beta\wedge\gamma,\mathrm{d}\beta+\gamma\wedge\alpha,\mathrm{d}\gamma+\alpha\wedge\beta)$$ Is it injective / surjective? Which is its kernel / cokernel? Does it depend on $n$? Any suggestion is welcome. REPLY [6 votes]: Your map is not onto for $n>2$, even locally. If $(A,B,C)$ is a triple of $2$-forms on $\mathbb{R}^n$ that can be written in the form $$ (A,B,C) = \bigl(\mathrm{d}\alpha + \beta\wedge\gamma, \mathrm{d}\beta + \gamma\wedge\alpha, \mathrm{d}\gamma + \alpha\wedge\beta\bigr), $$ then, taking the exterior derivative of these equations, we find $$ (\mathrm{d}A,\mathrm{d}B,\mathrm{d}C) = \bigl(B\wedge\gamma-\beta\wedge C,\ C\wedge\alpha-\gamma\wedge A,\ A\wedge\beta-\alpha\wedge B\bigr). $$ In particular, if $A$, $B$, and $C$ vanish at a point $x\in\mathbb{R}^n$ at which $(\mathrm{d}A,\mathrm{d}B,\mathrm{d}C)$ does not vanish, then the $1$-forms $\alpha$, $\beta$, and $\gamma$ cannot exist on a neighborhood of $x$. (Such examples are trivial to construct.) Moreover, when $n>4$, this map does not contain the generic triple $(A,B,C)$ in its image. The cases $n=3$ and $n=4$ are special, and, for suitable genericity hypotheses, one can prove surjectivity in special cases and under the right conditions, but it is, indeed, somewhat delicate. By the way, this is not a 'strange problem'. It is known as the problem of prescribed curvature for $\mathrm{SU}(2)$ connections. For more information, you might look at my answer to this question.<|endoftext|> TITLE: Interesting triple integral QUESTION [8 upvotes]: Some time ago I stumbled on an alleged identity $$\int\limits_0^\infty \frac{dx}{x} \int\limits_0^x \frac{dy}{y} \int\limits_0^y \frac{dz}{z} [\sin{x}+\sin{(x-y)}-\sin{(x-z)}-\sin{(x-y+z)}]= -\frac{\pi^3}{12}.$$ How this identity can be proved? The context under which this integral has emerged is described in Multiple Integral (American Mathematical Monthly problem 11621 and its generalization) . REPLY [10 votes]: We shall denote the integral with the letter $I$, i.e. : $$I:=\int_0^\infty \frac{\mathrm{d}x}{x} \int_0^x \frac{\mathrm{d}y}{y} \int_0^y \frac{\mathrm{d}z}{z} [\sin{x}+\sin{(x-y)}-\sin{(x-z)}-\sin{(x-y+z)}]$$ The key point is to insert a damping exponential $e^{-\alpha z}$ (where $\alpha \geq 0$) into the integrand, this makes the integral directly dependent on the $\alpha$ factor, then apply the Leibnitz formula for differentatiaon under the integral sign, finally integrate again recalling $I(\infty)=0$ and $I(0)=I$. Albeit one can view on this procedure, respectively it can be done, using replacement $$\frac{1}{z}=\int_0^\infty \! e^{-\alpha z} \; {\mathrm{d}\alpha}$$ Then, the change of the order of integration would imply this reformulation to appear : $$I=\int_0^\infty \! {\mathrm{d}\alpha} \int_0^\infty \frac{\mathrm{d}x}{x} \int_0^x \frac{\mathrm{d}y}{y} \int_0^y {\mathrm{d}z} \, [\sin{x}+\sin{(x-y)}-\sin{(x-z)}-\sin{(x-y+z)}] \, e^{-\alpha z}$$ The inner integral is now directrly evaluable, holding the line : $$I=\int_0^\infty \! {\mathrm{d}\alpha} \! \int_0^\infty \! \frac{\mathrm{d}x}{x} \int_0^x \frac{\mathrm{d}y}{y} \! \frac{1-e^{-\alpha y}}{\alpha(1+\alpha^2)} (\sin{x}+\sin{(x-y)}) +\frac{1+e^{-\alpha y}}{1+\alpha^2}(\cos{x}-\cos{(x-y)}) $$ As you can fine in this article (http://arxiv.org/pdf/1201.1975v1.pdf), integral similar to one contained in $I$ is computed there, namely : $$\zeta{(2)} = \int_0^\infty \! \frac{\mathrm{d}x}{x} \int_0^x \frac{\mathrm{d}y}{y} (\cos{(x-y)}-\cos{x}) $$ Althought, from representation of $I$ we are able to derive this result (denoted $\tilde{I}$ in the article) regardless of using the final one as a fact or from a different source. The actual approach is however very similar as the one described in the article : Just consider a parametric integral $\tilde{I}(\alpha)$ defined as: $$\tilde{I}(\alpha) := \int_0^\infty \! \frac{\mathrm{d}x}{x} \int_0^x \frac{1+e^{-\alpha y}}{y} (\cos{(x-y)}-\cos{x}) \, \mathrm{d}y $$ Evidently : $\tilde{I}(0)=2\tilde{I} \wedge \tilde{I}(\infty)=\tilde{I}$, using Fundamental Theorem of Calculus we can write $$\tilde{I}=-\left( \tilde{I}-2\tilde{I} \right)=-\left( \tilde{I}(\infty)-\tilde{I}(0) \right)=-\int_0^\infty \tilde{I'}(\alpha) \, \mathrm{d}\alpha$$ Then differentiation under the integral sign and some tricky integration will lead us to the value $\zeta{(2)}$ as well. Nevertheless, we will now continue in evaluation of $I$, the effort done on the discussion of the $\zeta{(2)}$ integral is not meaningless at all, because now we will separate this result from our integral representation of $I$, because of the relation : $$1+e^{-\alpha y}=2-(1-e^{-\alpha y})$$ we now split the $I$-integral into two pieces : $$I=J-2\int_0^\infty \mathrm{d}\alpha \frac{\zeta{(2)}}{1+\alpha^2} =J-\frac{\pi^3}{6}$$ where $$J:=\int_0^\infty \! {\mathrm{d}\alpha} \! \int_0^\infty \! \frac{\mathrm{d}x}{x} \int_0^x \frac{\mathrm{d}y}{y} \! \frac{1-e^{-\alpha y}}{\alpha(1+\alpha^2)} (\sin{x}+\sin{(x-y)})-\frac{1-e^{-\alpha y}}{1+\alpha^2}(\cos{x}-\cos{(x-y)})$$ We are doing this for preparation to make an unified replacement (another), namely : $$\frac{1-e^{-\alpha y}}{y}=\int_0^\alpha \! e^{-\beta y} \; {\mathrm{d}\beta}$$ After change of the order of integration in $\beta$ : $$J=\int_0^\infty \! \frac{\mathrm{d}\alpha}{\alpha(1+\alpha^2)} \! \int_0^\alpha \! {\mathrm{d}\beta} \! \int_0^\infty \! \frac{\mathrm{d}x}{x} {\int_0^x} \mathrm{d}y \,[\sin{x}+\sin{(x-y)}-\alpha (\cos{x}-\cos{(x-y)})] e^{-\beta y}$$ Again, by direct integration of inner integral : $$J\!=\!\int_0^\infty \!\!\!\! \frac{\mathrm{d}\alpha}{\alpha(1+\alpha^2)} \!\! \int_0^\alpha \!\! {\mathrm{d}\beta} \! \int_0^\infty \!\! \mathrm{d}x \frac{\sin{x}\!-\!\alpha\cos{x}}{\beta}\frac{1\!-\!e^{-\beta y}}{x}\!+\!\frac{\alpha \beta-\!1}{1\!+\!\beta^2}\frac{\cos{x}\!-\!e^{-\beta y}}{x}\!+\!\frac{\alpha\!+\!\beta}{1\!+\!\beta^2}\frac{\sin{x}}{x}$$ Resulting integrals are somewhat elementary (i.e. Dirichlet integral etc.), however we can use that dumb replacement rule again using $$\frac{1}{x}=\int_0^\infty \! e^{-\gamma x} \; {\mathrm{d}\gamma}$$ The other replacement however could also be done, i.e. $$\frac{1-e^{-\beta x}}{x}=\int_0^\beta \! e^{-\gamma x} \; {\mathrm{d}\gamma}$$ Define $$ \begin{split} S(\beta) & :=\int_0^\infty \mathrm{d}x \frac{1-e^{-\beta y}}{x}\sin{x} \\ C(\beta) & :=\int_0^\infty \mathrm{d}x \frac{1-e^{-\beta y}}{x}\cos{x} \\ L(\beta) & :=\int_0^\infty \mathrm{d}x \frac{\cos{x}-e^{-\beta y}}{x} \\ D & :=\int_0^\infty \mathrm{d}x \frac{\sin{x}}{x} \end{split}$$ By definition then $$ J = \int_0^\infty \frac{\mathrm{d}\alpha}{\alpha(1+\alpha^2)} \int_0^\alpha {\mathrm{d}\beta} \; \frac{1}{\beta}S(\beta)-\frac{\alpha}{\beta} C(\beta)+\frac{\alpha \beta-1}{1+\beta^2} L(\beta)+\!\frac{\alpha+\beta}{1+\beta^2} D $$ Now on the evaluation - for the first two integrals $S$ and $C$ the quickest way to compute them is via second type of replacement, The third an fourth one are done (due to the convergence issues) by the "dumb" replacement. For each integral we have: $$S(\beta)=\int_0^\beta {\mathrm{d}\gamma} \int_0^\infty \mathrm{d}x \sin{x} e^{-\gamma x} = \int_0^\beta {\mathrm{d}\gamma} \frac{1}{1+\gamma^2}=\arctan{\beta} \\ \\ C(\beta)=\int_0^\beta {\mathrm{d}\gamma} \int_0^\infty \mathrm{d}x \cos{x} e^{-\gamma x} = \int_0^\beta {\mathrm{d}\gamma} \frac{\gamma}{1+\gamma^2}=\frac{1}{2} \ln{(1+\beta^2)} \\ L(\beta)=\int_0^\infty {\mathrm{d}\gamma} \int_0^\infty \mathrm{d}x (\cos{x}\!-\!e^{-\beta x}) e^{-\gamma x} = \int_0^\infty {\mathrm{d}\gamma} \frac{\gamma}{1\!+\!\gamma^2}\!-\!\frac{1}{\beta\!+\!\gamma} = \ln{\frac{\sqrt{1 \!+\!\gamma^2}}{\beta \! + \! \gamma}} {\Big|}_0^\infty = \ln{\beta} \\ D=\int_0^\infty {\mathrm{d}\gamma} \int_0^\infty \mathrm{d}x \sin{x} e^{-\gamma x} = \int_0^\infty \frac{\mathrm{d}\gamma}{1+\gamma^2} = \frac{\pi}{2} $$ Substituting these results to the $J$ integral : $$ J = \int_0^\infty \frac{\mathrm{d}\alpha}{\alpha(1+\alpha^2)} \int_0^\alpha {\mathrm{d}\beta} \; \frac{\arctan{\beta}}{\beta}-\frac{\alpha}{2\beta} \ln{(1+\beta^2)}+\frac{\alpha \beta-1}{1+\beta^2} \ln{\beta}+\!\frac{\alpha+\beta}{1+\beta^2} \frac{\pi}{2} $$ Next step will turn out to be really important. Just consider how far we are, we ended up with a double integral with its domain of integration - infinite triangle, i.e. half of the first quadrant. Let us perform a variable change in such manner it transforms to rectangle. There are many candidates (e.g. polar one), however we chose this particular : $$\alpha=t \\ \beta=tk$$ With Jacobian of transformation equal to $J=t$, domain itself now consists of a semiinfinite strip $$(t,k)\in\{(0,\infty)\times(0,1)\}$$ After this transformation : $$ J \!=\! \int_0^1 \!\mathrm{d}k \int_0^\infty \!\mathrm{d}t \; \frac{\arctan{kt}}{kt(1+t^2)}-\frac{\ln{(1\!+\!k^2t^2)}}{2k(1\!+\!t^2)}\!-\!\frac{1-kt^2}{(1\!+\!t^2)(1\!+\!k^2t^2)} \ln{(kt)}\!+\!\!\frac{t(1+k)}{(1\!+\!t^2)(1\!+\!k^2t^2)} \frac{\pi}{2} $$ Defining ($0 TITLE: Which way for reading the proofs? QUESTION [25 upvotes]: I am a master student in mathematics. For me a large part of doing mathematics is thinking about, reading and verifying the proof of theorems that I find them in my field of study. I can do this action in 3 ways: When I see a theorem I get a paper and think to prove it: this action takes time a lot and maybe I couldn't prove it after thinking for a lot of time. Finding the proof of the theorem in a book or in the internet and begin reading, going step by step with proof,understanding and verifying all steps: this action may takes time a lot and maybe it is not necessary that I read all steps and it's better that I jump from not important steps (but how I can find that a step is not important?). Finding the proof of the theorem and just read it like reading a newspaper for finding the sketch of the proof: this action is good because of its speed but maybe there be some important details in the proof that I couldn't see them in this type of reading. My questions: What is the way that famous mathematicians like Fields medalists take for reading the proofs usually? Which way is the the best for which proofs? (For example classifying proofs and saying that the first way is good for the first class and...) REPLY [19 votes]: Here is a quote of Poincaré (one of the most accomplished mathematicians of all time) regarding the reading of mathematics: I am used, when I read a memoir, to glance over first quickly so as to have a general impression, then come back to the points which seem to me obscure. I find it more convenient to do proofs over than to examine thoroughly those of the author. My proofs are generally far poorer, but they have for me the advantage that they are mine. (Letter from Poincaré to Mittag-Leffier, 5 February 1889--IML; cited as per the article "The Poincaré-Mittag-Leffler Relationship" by Philippe Nabonnand, Mathematical Intelligencer, 1999.)<|endoftext|> TITLE: Automatically generate BibTeX item from arxiv QUESTION [8 upvotes]: I'm looking for a tool which generates a BibTeX item for a given arxiv id. I only found http://www.crcg.de/arXivToBibTeX/ using Google but this tool always tells me that the arxiv ids I enter don't exist which is of course not true. Edit: For example http://www.crcg.de/arXivToBibTeX/?q=1503.06747&format=bibtex gives the error message "No paper with the ID “1503.06747” could be found on the arXiv." even though it exists: http://arxiv.org/abs/1503.06747 Edit (by author of the discussed arXiv2BibTeX tool): Sorry for the inconvenience! The feedback website didn’t send us the e-mails with notes about the problem, so it took a while to discover your feedback. The tool is working again now at https://arxiv2bibtex.org/ with a proper github issue tracker at https://github.com/ssp/arXivToBibTeX/issues . REPLY [2 votes]: If you use a Mac, you may find the following arXiv to BibDesk applescript based solution useful. In summary, suppose you are at an arXiv page for a paper. Then, an execution of the above script does the following: Downloads the PDF from arXiv into a folder managed by BibDesk (to a running instance) Adds a bibliographic entry to BibDesk Moreover, since this is BibDesk you can generate a "cite key" of your choice by pressing CMD-K, and also rename the arXiv PDF as per paper title / author names if you want.<|endoftext|> TITLE: Abelian varieties with good reduction everywhere over function fields QUESTION [8 upvotes]: There is a famous theorem due to J.-M. Fontaine, Il n'y a pas de variété abélienne sur Z (and independently by V.A. Abrashkin) that there are no abelian varieties over Z. I was wondering whether there is a function field analog of this result. More precisely: Let $k$ be a finite field. Is it true that there are no non-isotrivial abelian varieties over $\mathbb{P}^1_{k}$ with good reduction everywhere? The case of elliptic curves is elementary. There are related results for families of curves but my question really focuses on the case of higher dimensional abelian varieties over $k(T)$. REPLY [12 votes]: There are non-isotrivial families of supersingular abelian varieties of dimension $g$ over $\mathbb P^1_{\overline {\mathbb F_p}}$ if $g\geq 2$; see Goren, E. Z.(3-MGL); Oort, F. Stratifications of Hilbert modular varieties. (English summary) J. Algebraic Geom. 9 (2000), no. 1, 111–154. There are many other papers of Goren and Oort with explicit constructions. One could now ask whether there exist non-isotrivial families of non-supersingular abelian varieties over the projective line. I can make one remark about this. A result of Moret-Bailly shows that the locus of ordinary abelian varieties is quasi-affine. In particular, any non-isotrivial family of abelian varieties over $\mathbb P^1$ contains a non-ordinary fibre. You can find this result in Moret-Bailly, Asterisque 129, Pinceaux de varieties abeliennes, p. 237 , Theoreme XI.5.2<|endoftext|> TITLE: Sufficient conditions for establishing a total order on a family of probability distributions? QUESTION [5 upvotes]: Let $\mathcal{X}$ be some set of independent random variables. Define the ordering on $\mathcal{X}$ by $X_i \prec X_j$ if and only if $\mathcal{P}\left\{X_i \le X_j\right\} \ge \frac{1}{2}$. Are there known weak conditions on $\mathcal{X}$ such that this ordering is a total ordering? A partial ordering? (I am guessing there are multiple answers to this question.) The counterexample of non-transitive dice illustrates that not all $\mathcal{X}$ have partial orderings, much less total orderings. However, if, for example, $\mathcal{X}$ consisted of Gaussian random variables, a total ordering is established by comparing the means. ${}\qquad{}$ REPLY [2 votes]: The random variable $X$ stochastically dominates the random variable $Y$, written $X\succeq Y$, if $\Pr(X\ge y)\ge\Pr(Y\ge y)$ for all $y$. This relation is transitive. Let's write $X\gtrsim Y$ if $\Pr(X\ge Y)\ge 1/2$. This relation is not transitive as the example of intransitive dice shows. However it is "total" in the sense that $X\lesssim Y$ or $Y\lesssim X$ holds. Theorem. $X\succeq Y$ implies $X\gtrsim Y$. Proof: In the notation of continuous random variables with pdfs $f_X$, $f_Y$, \begin{align} \Pr(X\ge Y)&=\int_{-\infty}^\infty \int_y^\infty f_X(x)f_Y(y)\,dx\, dy =\int_{-\infty}^\infty \Pr(X\ge y) \,f_Y(y)\, dy\\ &\ge\int_{-\infty}^\infty \Pr(Y\ge y) \,f_Y(y)\, dy=\int_{-\infty}^\infty \int_y^\infty f_Y(x)f_Y(y)\,dx\, dy= \frac12. \end{align} By the Theorem, the intransitive dice phenomenon cannot occur when stochastic domination totally orders the elements of $\mathcal X$.<|endoftext|> TITLE: Represent matrix immanants using Schur functions QUESTION [7 upvotes]: For each irreducible character $\chi^\lambda$ of the symmetric group $S_n$, the immanant of an $n\times n$ square matrix $A$ is defined as \begin{equation*} d_\lambda(A) := \sum_{\sigma \in S_n} \chi^\lambda(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}. \end{equation*} Observe that for $\lambda=(1^n)$, $\chi^\lambda(\sigma)=(-1)^{\text{sgn}(\sigma)}$, so that $d_\lambda(A)=\det(A)$, while for $\lambda=(n)$, $d_\lambda=\text{per}(A)$. I've gone through numerous papers in multilinear algebra (including those of Minc, Marcus, Pate, Merris, Watkins), as well as a superficial skimming of a few in representation theory. However, I have been unable to find (except for the implicit form (9) in this paper) an explicit representation of $d_\lambda$ using Schur functions. Question. Does anybody know of an explicit formula that represents $d_\lambda$ using Schur functions $s_\lambda$ (or as a sum of Schur functions). I'd also be happy to know about the converse direction. REPLY [6 votes]: I don’t know if this is proved anywhere. It looks like one can obtain a formula for the Schur functions in terms of immanants of replicated matrices. Let me introduce the following notation. Define a type $T$ as an ordered set $(i_1,\dots, i_n)$ such that $\sum_k i_k=n$. This defines a partition $\nu(T)$ obtained by rearranging the $i_k$ in decreasing order. For a type $T$, define $A_T$ as the matrix $A$ with the first row and column repeated $i_1$ times, second row and column repeated $i_2$ times etc. This is still an $n\times n$ matrix. Then we can obtain \begin{equation} s_\lambda(\mu_1\dots,\mu_n)=\sum_T d_\lambda(A_T)\,, \end{equation} where $\mu_i$ are the eigenvalues of $A$, $m_\lambda$ is the dimension of the symmetric group irrep $\lambda$ and the sum is over all types $T$ such that the partitions $\nu(T)$ have a non-zero Kostka number $K_{\nu,\lambda}\neq 0$. To obtain this, we can use Schur-Weyl duality. First, write the immanant as \begin{equation} d_\lambda(A) = \sum_\sigma \chi_\lambda(\sigma) \langle 1,\dots,n| A^{\otimes n} |\sigma(1),\dots,\sigma(n)\rangle \end{equation} This is equal to \begin{equation} d_\lambda(A) = \frac{n!}{m_\lambda}\langle 1,\dots,n| A^{\otimes n}\Pi_\lambda | 1,\dots,n\rangle\,, \end{equation} where $\Pi_\lambda$ is the projector on to the isotypic space of the symmetric group irrep $\lambda$. This can now be written as \begin{equation} d_\lambda(A) = \frac{n!}{m_\lambda} \text{Tr}(A^{\otimes n}\Pi_\lambda |1,\dots,n\rangle\langle 1,\dots,n|) \,. \end{equation} Finally, this can be written as \begin{equation} d_\lambda(A) = \frac{1}{m_\lambda}\text{Tr}(A^{\otimes n}\Pi_\lambda \Pi_{1^n}) \,, \end{equation} where $\Pi_{1^n}$ is the projector onto the induced representation from the trivial representation of the Young subgroup corresponding to the partition $1^n$. This induced representation is sometimes called the permutation module of the partition $1^n$. The multiplicity of the symmetric group irrep $\lambda$ in this module is the Kostka number $K_{1^n,\lambda}$. In the last step, we have also used the fact that \begin{equation} \frac{1}{n!}\sum_\sigma \sigma |1,\dots,n\rangle\langle 1,\dots,n |\sigma^{-1} = \Pi_{1^n}\,. \end{equation} Let us also block diagonalize $A^{\otimes n}$. We then obtain \begin{equation} d_\lambda(A) = \frac{1}{m_\lambda}\text{Tr}((I_\lambda\otimes A_\lambda)\Pi_\lambda \Pi_{1^n}) \,, \end{equation} where $I_\lambda$ is the identity on the symmetric group irrep space and $A_\lambda$ is the part in the $GL_n$ irrep space. Now it is clear what to do. If we add all the immanants for different types, we get \begin{equation} \sum_T d_\lambda(A_T) = \sum_T\frac{1}{m_\lambda}\text{Tr}((I_\lambda\otimes A_\lambda)\Pi_\lambda \Pi_{\nu(T)})= s_\lambda(\mu_1,\dots,\mu_n)\,. \end{equation} For the determinant, the sum over the other partitions do not matter since their projections onto the alternating irrep are zero. So we obtain that it is the Schur function in that case.<|endoftext|> TITLE: Is there a nonabelian finite simple group with Grothendieck ring of multiplicity one? QUESTION [5 upvotes]: Let $G$ be a finite group. It admits finitely many irreducible complex representations $H_1, \dots, H_r$ which generate, for $\oplus$ and $\otimes$, the Grothendieck ring $\mathcal{G}(G)$ of $G$ (also called its fusion ring). The tensor product of irreducible representations decomposes into direct sum as follows: $$H_i \otimes H_j = \bigoplus_k M_{ij}^k \otimes H_k$$ with $M_{ij}^k$ the multiplicity space. The Grothendieck ring is of multiplicity $m$ if $max_{i, j, k}(\dim(M_{ij}^k)) = m$. For $G = A_5$ we get the following dimensions: and for $G = A_1(7)$, we get: We observe that $\mathcal{G}(A_5)$ is of multiplicity two, and $\mathcal{G}(A_1(7))$ of multiplicity three. I've checked (with GAP) that for $\vert G \vert < 10^4$ (nonabelian simple), then $\mathcal{G}(G)$ is not multiplicity one. Question: Is there a nonabelian finite simple group with Grothendieck ring of multiplicity one? If no by using CFSG, can we expect a direct proof? Remark: Suppose $\dim(H_1) \le \dim(H_2) \le \dots \le \dim(H_r)$, then there is the necessary condition: If the multiplicity is one then $\dim(H_r)^2 \le \sum_i \dim(H_i)$. Perhaps there is no nonabelian finite simple group verifying this weaker condition, which is easier to check (I've used it for $\vert G \vert < 10^4$). REPLY [5 votes]: There may be some clever trick to do this by elementary arguments, but I don't see it at the moment. With CFSG, the following argument shows that it is likely to be rare to have a Grothendieck ring with multiplicity $1$, if it happens at all. Let $b$ the the largest character degree of the non-Abelian finite simple group $G$, and suppose that $G$ has $k$ conjugacy classes. It is conjectured that $|G| |G|$, so we must have $k > |G|^{0.295}$. This last inequality can be achieved ( eg in ${\rm SL}(2,2^{n})$), but it seems likely to be relatively rare ( though precise checking may be painful). Later edit: In fact, it is interesting to note that when $G = {\rm SL}(2,2^{n}) (n \geq 2)$, if we let $\chi$ denote the Steinberg character (which has degree $2^{n}$, whereas the largest irreducible character degree is $2^{n}+1$), we always .have $\chi \chi = \sum_{ \mu \in {\rm Irr}(G)} \mu$. This is because the projective cover of the (characteristic $2$) trivial module has dimension $2^{2n}-2^{n}$, and occurs as a summand of ${\rm St} \otimes {\rm St}.$<|endoftext|> TITLE: Homotopy of orthogonal groups in the unstable range QUESTION [7 upvotes]: We fix an integer $n$ and consider the stabilization map $O(n)\to O$. Using rational methods one can easily check that the map $\pi_{4i-1}(O(n))\to \pi_{4i-1}(O)\cong\mathbb{Z}$ vanishes for sufficiently large $i$. Is a similar fact known for the map $$\pi_{8i}(O(n))\to \pi_{8i}(O)\cong\mathbb{Z}/2\mathbb{Z}\ ?$$ REPLY [7 votes]: This is false. In fact, for all sufficiently large n the maps $\pi_{8i}(O(n)) \to \pi_{8i}(O) \cong \Bbb Z/2$ are surjective for all $i$. One possible tool for proving this has to do with $v_1$-periodic homotopy theory. For sufficiently large $d$, the mod-2 Moore space $W = \Sigma^{d-1} \Bbb{RP}^2$, which is the cofiber of a degree-2 map $S^d \to S^d$, has a certain type of "periodic" self-map $\Sigma^8 W \to W$. Any $2$-torsion element $\alpha \in \pi_{d+k} X$ extends from a map $S^{d+k} \to X$ to a map $\Sigma^k W \to X$, and we can precompose to get maps $\Sigma^{8+k} W \to X$ which determine a new $2$-torsion element $\alpha' \in \pi_{d+k+8} X$. It so happens that, in the case of the orthogonal group $O$, this construction $\alpha \mapsto \alpha'$ gives an isomorphism $\pi_{8i} O \to \pi_{8i+8} O$, despite the possible indeterminacy in these definitions. Therefore, if one of these classes lifts to $\pi_{8i} O(n)$ and $i$ is bigger than $d$, the next one lifts to $\pi_{8i+8} O(n)$.<|endoftext|> TITLE: The maximal eigenvalue of a symmetric Toeplitz matrix QUESTION [9 upvotes]: Let $0\le x\le 1$ be a real number. Denote by $A_n(x)=(a_{ij})$ the $n$ by $n$ matrix such that $a_{ij}=x^{|i-j|}$ and let $\lambda_n(x)$ be the maximal eigenvalue of $A_n(x)$. Is there any formula for $\lambda_n(x)$? I am particularly interested in the following question. Let $y_n$ be such that $\lambda_n(y_n)=\frac{n}4$. What is the value of $$\lim_{n\to \infty} (y_n)^n$$ if the limit exists? REPLY [6 votes]: Using the results in pages 59-63 of Rosenblum and Rovnyak (p. 62 in particular, and noting that your Toeplitz matrices are symmetric, hence normal, so the operator norm is what you want), it follows that $\lim_n \lambda_n(x) = \frac{1+x}{1-x}$. The non-asymptotic behavior is probably hard, but numerics clearly indicate (and it is probably not hard to show) that $\lambda_n(1) = n$. It might be possible to combine these observations, or more detailed ones along similar lines, to get good estimates for $\lambda_n(x)$.<|endoftext|> TITLE: How can we interpret the eigenvalues and eigenvectors of Euclidean Distance Matrices? QUESTION [18 upvotes]: I asked this question in Math Stack Exchange earlier here: https://math.stackexchange.com/questions/1199380/what-is-the-intuition-behind-how-can-we-interpret-the-eigenvalues-and-eigenvec and since I felt this was a better forum for the question, I have posted it here. Given a set of points $x_1,x_2,...,x_m$ in the euclidean space $R^n$, we can form a $m$ x $m$ Euclidean Distance Matrix $D$ where $D_{ij}=||x_i−x_j||_2$. We know a little bit about these matrices like : It is symmetric. Its Trace=0. It has (at most) n+2 non-zero eigenvalues; It has exactly n+2 non-zero eigenvalues whenever m>n; Source : https://math.stackexchange.com/questions/1198895/relationship-between-eigenvalues-of-two-related-euclidean-distance-matrices What is the intuition behind the eigenvalues and eigenvectors of such matrices ? In the case of a covariance matrix formed from data points, we can say that the eigenvectors are the directions in the the spread of data is maximum and these are called as principal components. In the case of adjacency matrices of graphs also, there seems to be an interpretation for the eigenvectors as given here : http://daylateanddollarshort.com/math/pdfs/spectral.pdf Is there a similar interpretation for these Euclidean Distance Matrices (EDM's)? Thank You. Even partial answers, ideas and references are welcome. REPLY [11 votes]: If $D_{ij}=|v_i-v_j|^2$ encodes the squared distances between a set of vectors $\{v_1,v_2,\ldots v_n\}$ in $d$-dimensional Euclidean space, with $V=\sum_{i=1}^{n}v_i=0$, then [*] $G=-\tfrac{1}{2}JD J$ with centering matrix $J_{kl}=\delta_{kl}-1/n$ gives the inner-product matrix $G_{ij}=v_i\cdot v_j$, called the Gram matrix. The positive eigenvalues of $G$ are those of the covariance matrix, which have the usual interpretation of principal component analysis. [*] check $$-2G_{ij}=(JDJ)_{ij}=D_{ij}+n^{-2}\sum_{k,l=1}^n D_{kl}-n^{-1}\sum_{k=1}^n (D_{ik}+D_{kj})$$ substitute $D_{ij}=v_i^T v_i+v_j^T v_j-2v_i^T v_j$, $$-2G_{ij}=v_i^T v_i+v_j^T v_j-2v_i^T v_j+n^{-2}\left(2n\sum_{k}v_k^T v_k-2V^T V\right)$$ $$\qquad\qquad -n^{-1}\left(nv_i^T v_i+nv_j^T v_j+2\sum_k v_k^T v_k-2V^T v_j-2v_i^TV\right)$$ $$\qquad =-2v_i^T v_j-2n^{-2}V^T V+2n^{-1}(V^T v_j+v_i^T V)$$ under the assumption $V=\sum_{i=1}^{n}v_i=0$, this gives the desired result $G_{ij}=v_i^T v_j$.<|endoftext|> TITLE: Differential forms along the fiber QUESTION [6 upvotes]: Let $E \to M$ be a smooth fiber bundle. Instead of differential forms defined on the whole tangent bundle $TE$ one could also consider forms on the vertical tangent bundle $VE$, i.e. forms defined on every fiber $E_m$ varying smoothly with $m \in M$. Is this notation of 'differential form along the fiber' discussed somewhere in the literature? Clearly the basic notations of inner product and wedge product still make sense (with some modification). However, differentiation is now possible in two directions (differentiation along the fiber and with respect to the base) and thus should lead to two exterior differentials. Remark: One approach would be to choose a connection on the bundle and then extend the fiber differential form by $0$ to the horizontal space and thereby get a bona-fide form on $E$. Besides the dependence on the connection this construction has another disadvantage. A fiber differential form, which is closed in fiber direction, does not need to have a closed extension. REPLY [6 votes]: The differential forms on the total space that vanish when restricted to the fibers give you a differential ideal of $\Omega^*(E)$, the powers of this ideal give a filtration which give rise to a spectral sequence. Some people call this the differential Leray-Serre spectral sequence, but I'm not shure if this is the standard name. Unfortunately I also don't know any written references, I learned this from Luca Vitagliano and Alexandre Vinogradov.<|endoftext|> TITLE: Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation? QUESTION [7 upvotes]: Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra, and let $\phi_1:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{g}$ and $\phi_{2}:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{g}$ be complex Lie algebra morphisms. By composing $\phi_1$ and $\phi_2$ with the adjoint representation of $\mathfrak{g}$, we obtain two representations of $\mathfrak{sl}_2(\mathbb{C})$ on $\mathfrak{g}$. My question is then the following. Question: If these two $\mathfrak{sl}_2(\mathbb{C})$-representations are isomorphic, does it follow that $\phi_1$ and $\phi_2$ are related by an element of the Lie algebra automorphism group $\text{Aut}(\mathfrak{g})$? To provide some context, suppose that $\mathfrak{g}=\mathfrak{so}_{4n}(\mathbb{C})$. If $\lambda$ is a partition of $4n$ having only even parts with each part appearing an even number of times, then $\lambda$ corresponds to exactly two distinct nilpotent orbits, $\mathcal{O}_1$ and $\mathcal{O}_2$, in $\mathfrak{so}_{4n}(\mathbb{C})$. Now, let $\phi_1:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{so}_{4n}(\mathbb{C})$ and $\phi_2:\mathfrak{sl}_2(\mathbb{C})\rightarrow\mathfrak{so}_{4n}(\mathbb{C})$ be Lie algebra maps satisfying $\phi_1(e)\in\mathcal{O}_1$ and $\phi_2(e)\in\mathcal{O}_2$. As discussed above, each map gives a representation of $\mathfrak{sl}_2(\mathbb{C})$ on $\mathfrak{so}_{4n}(\mathbb{C})$. These two representations are actually isomorphic. While $\phi_1$ and $\phi_2$ cannot be related by an inner automorphism of $\mathfrak{so}_{4n}(\mathbb{C})$ (as $\mathcal{O}_1\neq\mathcal{O}_2$), they are nevertheless related by a Lie algebra automorphism. REPLY [9 votes]: No, here is a counterexample. Let $\mathfrak{g}_1 = \mathfrak{sl}_2(\mathbb{C}) \oplus \mathfrak{sl}_2(\mathbb{C}) \oplus \mathfrak{sl}_2(\mathbb{C})$, and let $\phi_1 \colon \mathfrak{sl}_2(\mathbb{C}) \to \mathfrak{g}_1$ be the diagonal embedding, so $\phi(x) = (x,x,x)$. Then the representation obtained by composing $\phi_1$ with the adjoint representation is the direct sum of three $3$-dimensional irreducible modules. Let $\mathfrak{g}_2 = \mathfrak{so}_5(\mathbb{C})$, and let $\phi_2$ be the embedding onto $\langle \mathfrak{u}_\alpha, \mathfrak{u}_{-\alpha} \rangle$, where $\alpha$ is a short root. Then the representation obtained by composing $\phi_2$ with the adjoint representation is the direct sum of a trivial representation and precisely three $3$-dimensional irreducible modules. Now, if we let $\mathfrak{g} = \mathfrak{g}_1 \oplus \mathfrak{g}_2$, then we can think of $\phi_1$ and $\phi_2$ as homomorphisms into $\mathfrak{g}$ and, from the above, we see that their compositions with the adjoint representation are isomorphic. However, the image of $\phi_2$ is contained in the unique ideal of $\mathfrak{g}$ that is isomorphic to $\mathfrak{so}_5(\mathbb{C})$, but the image of $\phi_1$ is not contained in this ideal. So no element of $\mathrm{Aut}(\mathfrak{g})$ can take $\phi_1$ to $\phi_2$. REPLY [3 votes]: REVISED VERSION: On further reflection, I think the answer to your question is always "yes" (if $\mathfrak{g}$ is simple, to avoid complications of the type Dave indicates). At first I was confused by the example discussed in the question, but I think the main issue is how the automorphism group interacts with Dynkin diagrams of nilpotent orbits. In the classical Dynkin-Kostant treatment of nilpotent orbits in a simple Lie algebra $\mathfrak{g}$, the basic strategy is to embed a (nonzero) nilpotent element in a copy of $\mathfrak{sl}_2(\mathbb{C})$. Such an embedding isn't unique, but there is a resulting bijection between conjugacy classes of nilpotents (under the adjoint group of $\mathfrak{g}$) and conjugacy classes of such subalgebras. In turn, the adjoint action of the subalgebra leads to a decorated Dynkin diagram (with vertices labelled $0, 1, 2$) which determines uniquely the given nilpotent orbit. (Some references to textbooks by Carter and Collingwood-McGovern which include Dynkin diagrams are given in my old notes here.) In your type $D$ example, there are typically pairs of orbits interchanged by an outer automorphism of $\mathfrak{g}$ that comes from a graph automorphism. These orbits have many common properties, and corresponding copies of $\mathfrak{sl}_2(\mathbb{C})$ do act equivalently on $\mathfrak{g}$ even though they lead to distinct Dynkin diagrams. The key fact is that these Dynkin diagrams just involve a permutation of labels induced by the outer automorphism. Only in such limited cases can two subalgebras of type $\mathfrak{sl}_2(\mathbb{C})$ act equivalently in the adjoint representation of $\mathfrak{g}$: this is clear from the determination of Dynkin diagrams for each $\mathfrak{g}$. [Concerning terminology, it's fairly conventional to call two representations equivalent but the associated modules isomorphic. The choice of either representation or module language is usually optional, at least in finite dimensional cases.]<|endoftext|> TITLE: Freiling's Axiom of Symmetry Concretized QUESTION [8 upvotes]: Freiling's Axiom of Symmetry says that for any function $f:[0,1]\to \mathcal{P}([0,1])$ such that for every $x\in [0,1]$ we have $|f(x)|=\aleph_0$, then there exist $y,z\in [0,1]$ such that $z\notin f(y)$ and $y\notin f(z)$. This is equivalent to the negation of the continuum hypothesis (see this wiki page). My question is whether one can produce an uncountable set $A$, using the axioms in ZFC, so that if we plug $A$ in place of $[0,1]$, it satisfies Freiling's Axiom. REPLY [4 votes]: For any function $f:X\to P(X)$, a set $F\subseteq X$ is called free (or $f$-free) if $y\notin f(z)$ and $z\notin f(y)$ for all distinct $y,z$ in $F$. Hajnal's "free set theorem" says the following: If $f:\lambda \to [\lambda]^{<\mu}$ for some $\mu<\lambda$, then we cannot only find a 2-element free set but even a free set of size $\lambda$. (Chapter 26 in the wonderful book by Komjáth and Totik is called "Set mappings". Hajnal's theorem is problem 8. For regular $\lambda$ this was proved earlier by Sierpiński.) If you plug in $\lambda\ge\aleph_2$, $\mu=\aleph_1$, then you get: If all values of $f$ are at most countable, then there is free set of size $\lambda$.<|endoftext|> TITLE: Symplectic form/Kahler metric on a toric manifold QUESTION [7 upvotes]: I have a standard question about symplectic forms on toric manifolds: Let $P$ be an $n$-dimensional Delzant polytope and let $X_P$ be the corresponding symplectic toric manifold with symplectic form $\omega_P$ (usual Delzant construction via symplectic reduction). In the algebraic geometry side, we can also construct $X_P$ by embedding the torus $(\mathbb{C}^*)^n$ into a projective space $\mathbb{C}\mathbb{P}^{N_k-1}$ using monomials corresponding to the lattice points in $kP$ for a sufficiently large integer $k$ (here $N_k$ is the number of lattice points in $kP$). Then $X_P$ is the closure of the image of the torus in the projective space $\mathbb{C}\mathbb{P}^{N_k-1}$. My question is, how one can realize the (canonical) symplectic form $\omega_P$ on $X_P$ as the pull-back of a Fubini-Study Kahler form to $X_P$, for an embedding of $X_P$ in a projective space corresponding to the lattice points in some $k\Delta$? REPLY [5 votes]: Take $P$ be a polyhedral or polytope as follows. Let $G$ be a Torus with Lie algebra $\mathfrak g$ and integral lattice $\mathbb Z_G$. Take primitive vectors $u_1,u_2,...,u_N\in \mathbb Z_G$ spans $\mathfrak g$ and Let $$P=\{\eta\in\mathfrak g^*\mid<\eta,u_i>-\lambda_i\geq 0 , 1\leq j\leq N\}$$ . Let $P$ be compact then $M_P=(\mathbb CP^n,k\omega_{FS})//_vK$ the Kahler $G$-space with moment map $\phi_P:M_P\to \mathfrak g^*$ where $k$ is a bound of standard simplex. i.e $$\Delta_R=\{l\in (\mathbb R^N)^*\mid \geq 0 \; \; \text{and}\;\; \sum\leq k\}$$ where $\{e_1,e_2,...,e_n\}$ is the standard basis of $\mathbb R^n$. Note that this symplex is image of $\mathbb CP^n$ under the moment map for the standard action of $\mathbb T^n$ with Kahler form of $\mathbb CP^n$ being the appropriate multiple of the standard Fubini-Study form<|endoftext|> TITLE: Etale cohomology approach on $\tau(n)$ QUESTION [6 upvotes]: Ramanujan's $\tau$ conjecture states that $$\tau(n)=O_\epsilon(n^{\frac{11}2+\epsilon}),$$ which is a consequence of Deligne's proof of Weil conjectures. Answers in https://math.stackexchange.com/questions/1205419/status-of-taun-before-deligne/1205516 tell that best exponent before Deligne reached $\frac{29}5$. What was it that cohomological approach gave that broke the traditional approach barrier to reach $5.5$ in exponent? Posted originally in here https://math.stackexchange.com/questions/1206558/etale-cohomology-approach-on-taun where comments suggested to post in MO. REPLY [12 votes]: One of the goals of the development etale cohomology was to generate a cohomology theory that could successfully count points on varieties over finite fields, with one of the main goals of proving the conjecture that Andre Weil laid out in 1949 (see his paper here). Indeed, the formulation of the conjectures already suggests Weil was thinking along these lines. The most striking of these conjectures is the Riemann hypothesis for the zeta function of a variety over a finite field. This conjecture was proven by Deligne in 1974 (available here). The simplest case had been known for a quite while, namely that if $E/\mathbb{F}_{p}$ is an elliptic curve, then $$ p + 1 - 2 \sqrt{p} < |E(\mathbb{F}_{p})| < p + 1 + 2\sqrt{p}.$$ This result is as sharp as it gets. For every integer $k$ satisfying $p+1 - 2 \sqrt{p} < k < p+1 + 2 \sqrt{p}$, there is an elliptic curve $E/\mathbb{F}_{p}$, with exactly $k$ points. In general, the bounds on the number of points on a variety (and character sums in general) that one obtains using the Riemann hypothesis are very powerful. (For example, in 1981 Hooley proved an asymptotic formula for the number of representations of an integer as a sum of two squares and three non-negative cubes. Hooley's work required estimates sharp bounds on certain character sums, and Milne proved these estimates by relating them to etale cohomology and using Deligne's proof of the Weil conjectures. Hooley and Milne's papers appear back-to-back in the same issue of Crelle.) What does this have to do with the Fourier coefficients of modular forms? Deligne constructed algebraic varieties whose etale cohomology groups afford the Galois representations (conjectured to exist by Serre and Swinnerton-Dyer) attached to modular forms. Applying the Riemann hypothesis to these gives the "Deligne bound" that $|\tau(n)| \leq d(n) n^{11/2}$, where $d(n)$ is the number of divisors of $n$. (This application is explicitly mentioned in Deligne's 1974 paper - see Theoreme 8.2.) Most interestingly, Rankin's 1939 bound $\tau(n) = O(n^{29/5})$ played a role in Deligne proving the Riemann hypothesis for varieties over finite fields. In particular, Langlands observed that knowledge of the poles of symmetric power $L$-functions attached to $\Delta$ would be sufficient to conclude that $|\tau(p)| \leq 2p^{11/2}$, and Deligne observed that there was a nice translation of this idea into Grothendieck's cohomology theory. By making this work, Deligne overcame the fact that (at the time), knowledge was only had of the poles of $L({\rm Sym}^{2} \Delta, s)$. Finally, I will note that Kapil Paranjape and Dinakar Ramakrishnan have constructed an explicit Calabi-Yau 11-fold that gives rise to the Galois representation attached to $\Delta$. (See their paper here.)<|endoftext|> TITLE: UFD and fundamental group QUESTION [6 upvotes]: Let $C$ be the curve $x^2+y^2-1$, defined over $\mathbb R$. It is easy to see that $\mathbb R[C]$ is not a UFD, as witnessed by the identity $(1-x)(1+x)=y^2$. On the other hand, the real locus $C(\mathbb R)$ is a circle, which is not smply connected. I'm then wondering if there is some more general connection between the fundamental group of $C(\mathbb R)$ and the class group of $\mathbb R[C]$. I vaguely heard that the class group of a number field can be thought of as its fundamental group (though I don't understand the details). Can this analogy be carried over to function fields? REPLY [13 votes]: It's the absolute Galois group that can be thought of as a fundamental group, since it is the étale fundamental group of $\text{Spec } K$. The ideal class group is instead a Picard group of line bundles, which for topological spaces and real line bundles is the cohomology group $$H^1(X, \mathbb{G}_m(\mathbb{R})) \cong H^1(X, \mathbb{Z}_2)$$ and which for varieties is the étale cohomology group $H^1(X, \mathbb{G}_m)$ (here we need $X = \text{Spec } \mathcal{O}_K$). In fact the Serre-Swan theorem implies that if $X$ is a smooth manifold then the algebraic Picard group of $C^{\infty}(X)$, defined in terms of invertible modules, is the smooth Picard group $H^1(X, \mathbb{Z}_2)$ of real line bundles on $X$. That's why you aren't surprised to get a group of order $2$ when $X$ is the circle, as KConrad says in the comments. (But don't expect the relationship between the topology of the real points of a curve over $\mathbb{R}$ and the Picard group of its ring of functions to be this nice in general. For example, sometimes there may not be any real points. You should instead be looking, at the very least, at the complex points together with the action of complex conjugation on them.) Here is an attempt I made to sort all this out in more detail.<|endoftext|> TITLE: Lovasz's Path removal conjecture QUESTION [9 upvotes]: The Lovász Path Removal Conjecture states: For any positive integer $k$, there exists a minimum positive integer $f(k)$ such that, for any two vertices $x$, $y$ in any $f(k)$-vertex-connected graph $G$, there is an $x$-$y$ path $P$ in $G$ such that $G\backslash P$ is $k$-vertex-connected. It is commonly stated that $ f(1)=3 $ due to a theorem in Tutte's paper "How to draw a graph" (Proc. London Math. Soc. 13.3 (1963), 743–768). However, I can't find any theorem in Tutte's paper that directly states that a 3-vertex-connected graph contains a non-separating path between any two vertices. Can someone point out to me the relevant theorem in Tutte's paper and show how it implies the above fact? REPLY [3 votes]: Tutte in that paper shows that if the connectivity of a graph is at least 3, then for any two distinct vertices in that graph, there exists a non-separating path between them. Any cycle plus a single chord shows $f(1) \geq 3$ and Tutte's Wheel Theorem shows that $f(1) \leq3$. Thus, $f(1)=3$.<|endoftext|> TITLE: When does the Borel construction have the homotopy type of a CW-complex? QUESTION [6 upvotes]: Suppose that $G$ is a Lie group acting smoothly on a manifold $M,$ does the Borel $M \times_G EG$ construction have the homotopy type of a CW-complex? If not, under what conditions would this be true? (I mostly care about the case of discrete stabilizers). More generally, if $G$ is any topological group acting on a CW-complex $X$, under what conditions does $X \times_G EG$ have the homotopy type of a CW-complex? REPLY [8 votes]: Yes, this is true. It suffices for $X$ to have the homotopy type of a CW-complex (this is true of smooth manifolds; see e.g. here or here). I'm going to assume that you're using a definition of $EG$ that includes something like: $EG$ is a $G$-CW-complex, so that it is built by iteratively taking pushouts of diagrams of the form $$ D^n \times G \leftarrow S^{n-1} \times G \rightarrow Y. $$ As a result, the space $EG \times_G M$ is formed by an iterated sequence of pushouts $$ D^n \times M \leftarrow S^{n-1} \times M \rightarrow Z. $$ (This comes with the standard warnings about probably having to use compactly generated spaces so that products, quotients, and the direct limit topology interact well.) Each of these pushouts is the mapping cone of the map $S^{n-1} \times M \to Z$. This mapping cone would not necessarily be a CW-complex even if $M$ and $Y$ were (the map would have to be cellular for that), but if $M$ and $Z$ both have the homotopy type of CW-complexes, the cone does have the homotopy type of a CW-complex (it's homotopy equivalent to a cellular map, and that equivalence carries across to an equivalence on mapping cones). By inducting on the cell structure of $EG$, we can assume that $Z$ has the homotopy type of a CW-complex and find that this next pushout map $Z \to Z'$ is homotopy equivalent to a cell inclusion of CW-complexes. Taking colimits we get the desired result.<|endoftext|> TITLE: Cyclic structure on a balanced (or ribbon) monoidal category QUESTION [9 upvotes]: As it is well known, a balanced (and in particular ribbon) monoidal category is an algebra over the framed little 2-discs operad. The latter is homotopy equivalent to the operad of moduli space of genus 0 surfaces with boundaries, hence has a natural cyclic structure. What additional structure correspond to a cyclic algebra in categories over this operad ? Usually, in vector spaces, a cyclic algebra over a cyclic operad has a non-degenerate pairing compatible with the operadic operations. One marvelous property of categories is that the already have a canonical pairing given by hom spaces, and under mild conditions it is non-degenerate. So I guess that a cyclic structure in that case should be a certain compatibility between the twist/ribbon element and the home spaces. REPLY [5 votes]: In https://arxiv.org/abs/2010.10229 we characterize cyclic framed little 2-disks algebras in any symmetric monoidal bicategory. In the case that this symmetric monoidal bicategory is given by finite categories, left exact functors and their transformations, it turns out that they amount precisely to balanced Grothendieck-Verdier categories in the sense of Boyarchenko-Drinfeld, i.e. braided monoidal categories with a weaker form a rigidity that also have a compatible balancing. Note: Our article tries to shed some light on the relation of cyclicity of certain operads in low-dimensional topology to duality properties of representation categories appearing in quantum algebra, and your question was extremely helpful for that. Thank you!<|endoftext|> TITLE: How singular can the Stein factorization of a proper map between smooth varieties be? QUESTION [7 upvotes]: A little bit of motivation (the question starts below the line): I am studying a proper, generically finite map of varieties $X \to Y$, with $X$ and $Y$ smooth. Since the map is proper, we can use the Stein factorization $X \to \hat{X} \to Y$. Since the composition is generically finite, $X \to \hat{X}$ is birational, and therefore a sequence of blowups. I am currently interested in the other map: $\hat{X} \to Y$. I would like to apply Casnati–Ekedahl's techniques from “Covers of algebraic varieties I” (Journal of alg. geom., 1996). For this, I need $\hat{X} \to Y$ to be Gorenstein. (Since $Y$ is Gorenstein (since it is smooth), this is equivalent with $\hat{X}$ being Gorenstein.) When is this true? Specifically, in my case $X \to Y$ is the albanese morphism of a smooth projective surface: so $Y$ is an abelian surface, and I am in the situation that the albanese morphism is surjective. Let $f \colon X \to Y$ be a proper map between two varieties $X$ and $Y$ over a field $k$. Assume $X$ and $Y$ are smooth (and proper, if you want). Let $\pi \colon X \to \hat{X}$ and $\hat{f} \colon \hat{X} \to Y$ be the Stein factorization ($f = \hat{f} \circ \pi$). Of course, in general $\hat{X}$ is not smooth. However: Q1: Does $\hat{X}$ have some other nice properties? I am thinking in the direction of, e.g., Gorenstein or Cohen–Macaulay. If not, does it help if we assume a bit more on $f$? Or, alternatively: Q2: Under what conditions is $\hat{X}$ Gorenstein? REPLY [2 votes]: There are three good answers to this question, and together they have more or less answered what I wanted to know. I find it hard to choose one of them as best, but nevertheless I think this question should have an accepted answer to move it from the unanswered list. Hence a CW-answer summarizing the (in my eyes) most important points made. Laurent Moret-Bailly points out that $\hat{X}$ must be normal. Sasha then says that besides that, it can get as bad as you want. Take a normal subvariety $\hat{X} \subset \mathbb{A}^{N}$. By Noether's lemma we get a finite map $\hat{X} \to \mathbb{A}^{n} = Y$. A resolution of singularities $X \to \hat{X}$ has connected fibres. The composition $X \to Y$ is generically finite. Karl Schwede remarks that the pair $(\hat{X}, -\mathrm{Ram})$ is log-Gorenstein (where $\mathrm{Ram}$ is the ramification divisor). He also states the slogan “if $\hat{X}$ has really bad singularities at some points, then $\mathrm{Ram}$ also has really bad singularities at those points too. Another way to say this is if the ramification divisor has mild singularities, then $\hat{X}$ does too.”<|endoftext|> TITLE: How to pack 3D boxes into a bigger box? QUESTION [9 upvotes]: Given a box of given size $L\times M\times N$ and a list of smaller boxes of given sizes $(l_i,m_i,n_i)$, decide whether the smaller boxes altogether fit into the big box (and produce such a packing if possible). The problem is NP-complete...so I am looking for a good heuristic algorithm...the algorithm should allow for (the obvious possible) rotations of the boxes. What are currently good/best heuristic algorithms and codes? Links to papers or webpages are also welcome. REPLY [5 votes]: It would probably take some work to turn this into an algorithm that can deal with rotations of the boxes, but you might be able to modify the three weight algorithm (a variation of ADMM) by Derbinsky, Bento, Elser, and Yedidia, which is a fairly simple algorithm that has recently beaten various records for circle and sphere packing in boxes.<|endoftext|> TITLE: Are there any algebraic geometry theorems that were proved using combinatorics? QUESTION [19 upvotes]: I'm collaborating with some algebraic geometers in a paper, and when writing the introduction I mentioned the interaction of combinatorics and algebraic geometry, and gave some examples like the combinatorial Nullstellensatz, the affirmative answer to the conjecture of Read and Rota-Heron-Welsh and the graph-theoretic analogue of the Riemann-Roch theorem, but then, it seems that all these interactions are one way. I would like to know of an important algebraic geometry theorem proved using combinatorics, or at least used combinatorics in a critical part. REPLY [2 votes]: Disclaimer: I am quite ignorant of algebraic geometry. I am also most opinionated about subjects that I do not know. Caveat lector! Ehrhart theory is a fundamental tool in toric geometry. In my opinion Ehrhart polynomials and Ehrhart generating functions belong to the land of combinatorics. When Eugene Ehrhart invented these beautiful things, he was doing combinatorics.<|endoftext|> TITLE: A classic cardinal characteristic of the continuum in disguise? QUESTION [12 upvotes]: We believe the answer to the following question, that is relevant to a joint research project with Piotr Szewczak, should be known. We would appreciate any help or pointer. Needed definitions may be found in, e.g., Blass's chapter in the Handbook of Set Theory. For $f\in\omega^\omega$, let $K_f:=\{g\in \omega^\omega : g\le^* f\}$. Let $\mathfrak{tmp}$ be the minimal cardinality of a set $Y\subseteq \omega^\omega$ such that the set $\bigcup_{f\in Y}K_f$ is not meager. Since every set $K_f$ is meager, we have that $\mathfrak{b}\le \mathfrak{tmp}\le \mathrm{non}(meager)$. Is $\mathfrak{tmp}$ (provably) equal to either of these cardinals? REPLY [7 votes]: For every meager $M \subseteq \omega^{\omega}$ there exists $f_M \in \omega^{\omega}$ such that for every increasing $f \in \omega^{\omega}$, $K_f \subseteq M$ implies $f \leq^{\star} f_M$. For a proof, see Theorem 2.2.2 in Bartozynski, Judah book. It follows that your invariant is the bounding number.<|endoftext|> TITLE: A property of the Frechet filter and every ultrafilter QUESTION [6 upvotes]: (Joint question with Piotr Szewczak.) Definitions and notation. By filter we mean a filter on $\omega$ containing the cofinite sets at least. For a filter $\mathcal{F}$, let $\mathcal{F}^+:=\{A\subseteq\omega : A^c\notin \mathcal{F}\}$. For an infinite set $A\subseteq\omega$ and a natural number $n$, define $\mathrm{next}(A,n)=\min\{k\in A:n TITLE: Anything known about the Grundy Ordinal of Sylver's Coinage QUESTION [6 upvotes]: Sylver's coinage is an example of an unbounded finite (if slightly modified) combinatorial impartial game. Quoth wikipedia: The two players take turns naming positive integers that are not the sum of nonnegative multiples of previously named integers. After 1 is named, all positive integers can be expressed in this way: 1 = 1, 2 = 1 + 1, 3 = 1 + 1 + 1, etc., ending the game. The player who named 1 loses.* This can be made to have the normal play convention if we make $1$ an illegal move. In Conway's ONAG, it is shown that Grundy's number can be generalized to ordinal numbers for unbounded impartial games. Is anything known about Grundy's ordinal for Sylver's Coinage and its various positions? *Wikipedia contributors. "Sylver coinage." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 21 Oct. 2014. Web. 27 Mar. 2015. REPLY [3 votes]: One thing we do know about the Grundy number of a position is that it is less than or equal to the ordinal representing the maximum length of the game starting from that position; to be precise, define $\ell(p)$ to be 0 if the position $p$ is an ending position, and $$ \ell(p) = \sup_{q \in p}\{\ell(q)+1 \} $$ otherwise. (by $q \in p$ I mean that from position $p$ the acting player can move to position $q$) Then the Grundy number $G(p) \le \ell(p)$. For Sylver Coinage, $\ell(p)$ is rather easy to determine. $\ell$ of the starting position is $\omega^2$; for any other position, the players will have selected some nonzero finite set of positive integers. Let $d$ be the GCD of all the numbers in the set, and let $m$ be the number of (not necessarily distinct) primes dividing $d$. Let $n$ be the number of possible next moves that are divisible by $d$. Then $\ell(p) = \omega m + n$. The Grundy number must be less than or equal to this ordinal.<|endoftext|> TITLE: Is the derived category of perfect complexes idempotent complete? QUESTION [7 upvotes]: Let $\mathcal{C}$ be a category. We call a morphism $\alpha: X\rightarrow X$ an idempotent if $\alpha^2=\alpha$ in $\mathcal{C}$. We call $\mathcal{C}$ is $\textit{idempotent complete}$ if any idempotent $\alpha: X\rightarrow X$ has a splitting in $\mathcal{C}$, i.e. there exists an object $Y$ together with morphisms $i: Y\rightarrow X$ and $p: X\rightarrow Y$ such that $i\circ p=\alpha$ and $p\circ i=id_Y$. A famous result by in Bokstedt&Neeman in "Homotopy limits in triangulated categories" is that if $\mathcal{C}$ is a triangulated category with (possibly infinite) direct sums, then it is idempotent complete. See Proposition 3.2 of that paper. It is clear from their result that the derived category of complexes sheaves of $\mathcal{O}_X$-modules $D(X)$ is idempotent complete. However, the derived category of perfect complexes, $D_{\text{perf}}(X)$ does not allow infinite direct sum. $\textbf{My question}$ is: is $D_{\text{perf}}(X)$ still idempotent complete? By the way, Bokstedt and Neeman has proved a very similar result, which claims that the derived category of finite complexes of finitely generated projective modules of a ring is idempotent complete. This illustrate that the infinite direct sums condition is not always necessary. REPLY [13 votes]: The derived category of perfect complexes is idempotent complete, because it is the sub category of compact objects in the derived category of quasi coherent sheaves (which is idempotent complete by the result you mention) and compact objects are stable under retracts.<|endoftext|> TITLE: Combinatorics of palindromic decompositions QUESTION [15 upvotes]: This is sort of a companion to my question Number of trivializations of a trivial word in the free group (which in turn is motivated by my earlier question here). It turns out that that question may be reduced to this one (well to a certain extent at any rate). A $\text{palindromic decomposition}$ (paldec for short) of a word $w$ (in letters, say, $a_1,...,a_n$) is any equality $$ w=p_1\cdots p_k $$ in the free semigroup on $a_1$, ..., $a_n$ such that each of the $p_i$ is a $\text{palindrome}$, i. e. coincides with itself read backwards. Obviously each word has at least one such decomposition since each single letter word is a palindrome according to this definition. For many words this is the only one, but for quite a few there are several others. For example, the word referee has seven paldecs: refer·ee refer·e·e r·efe·r·ee r·efe·r·e·e r·e·f·ere·e r·e·f·e·r·ee r·e·f·e·r·e·e Thus for each $n$ and $N$ the set of $n^N$ words of length $N$ in $n$ letters decomposes into classes, with the class $P_n^{(N)}(m)$ containing all such words having exactly $m$ paldecs. These may be further subdivided according to various structures, but I cannot really judge which of these structures are more significant. For example, paldecs of a given word form a poset since some paldecs are subdecompositions of some others - say, the decomposition into single word letters is obviously the smallest element of this poset. Have the numbers $\#P_n^{(N)}(m)$ or any of those corresponding to the above further subdivisions been considered in the literature? There are all kinds of papers on palindromes, too many for me to sort them out. Maybe somebody knows? Any generating functions, or statistics, or anything at all? Later - collected some statistics: here are some of the distributions of words according to the numbers of paldecs: $$\frac{\#P_n^{(2)}(m)}{n^2}\\ \begin{array}{r|llllllllll} m\backslash n &1&2&3&4&5&6&7&8&9&10\\ \hline 1 & 0 & 0.5 & 0.666667 & 0.75 & 0.8 & 0.833333 & 0.857143 & 0.875 & 0.888889 & 0.9 \\ 2 & 1. & 0.5 & 0.333333 & 0.25 & 0.2 & 0.166667 & 0.142857 & 0.125 & 0.111111 & 0.1 \end{array} $$ $$\frac{\#P_n^{(3)}(m)}{n^3}\\ \begin{array}{r|lllllllll} m\backslash n &1&2&3&4&5&6&7&8&9\\ \hline 1& 0 & 0 & 0.222222 & 0.375 & 0.48 & 0.555556 & 0.612245 & 0.65625 & 0.691358 \\ 2& 0 & 0.75 & 0.666667 & 0.5625 & 0.48 & 0.416667 & 0.367347 & 0.328125 & 0.296296 \\ 3& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 4& 1. & 0.25 & 0.111111 & 0.0625 & 0.04 & 0.0277778 & 0.0204082 & 0.015625 & 0.0123457 \\ \end{array} $$ $$\frac{\#P_n^{(4)}(m)}{n^4}\\ \begin{array}{r|llllllll} m\backslash n &1&2&3&4&5&6&7&8\\ \hline 1& 0 & 0 & 0.0740741 & 0.1875 & 0.288 & 0.37037 & 0.437318 & 0.492188 \\ 2& 0 & 0 & 0.37037 & 0.46875 & 0.48 & 0.462963 & 0.437318 & 0.410156\\ 3& 0 & 0.5 & 0.296296 & 0.1875 & 0.128 & 0.0925926 & 0.0699708 & 0.0546875 \\ 4& 0 & 0.375 & 0.222222 & 0.140625 & 0.096 & 0.0694444 & 0.0524781 & 0.0410156\\ 5& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 6& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 7& 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 8& 1. & 0.125 & 0.037037 & 0.015625 & 0.008 & 0.00462963 & 0.00291545 & 0.00195313 \end{array} $$ $$\frac{\#P_n^{(5)}(m)}{n^5}\\ \begin{array}{r|lllllll} m\backslash n &1&2&3&4&5&6&7\\ \hline 1& 0 & 0 & 0.0246914 & 0.09375 & 0.1728 & 0.246914 & 0.31237\\ 2& 0 & 0 & 0.148148 & 0.304688 & 0.384 & 0.416667 & 0.424823\\ 3& 0 & 0 & 0.246914 & 0.234375 & 0.192 & 0.154321 & 0.124948\\ 4& 0 & 0.125 & 0.246914 & 0.210938 & 0.1664 & 0.131173 & 0.104956\\ 5& 0 & 0.375 & 0.148148 & 0.0703125 & 0.0384 & 0.0231481 & 0.0149938\\ 6& 0 & 0.1875 & 0.0740741 & 0.0351563 & 0.0192 & 0.0115741 & 0.00749688\\ 7& 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 8& 0 & 0.25 & 0.0987654 & 0.046875 & 0.0256 & 0.0154321 & 0.00999584\\ 9& 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 10& 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 11& 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 12& 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 13& 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 14& 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 15& 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 16& 1. & 0.0625 & 0.0123457 & 0.00390625 & 0.0016 & 0.000771605 & 0.000416493 \end{array} $$ $$\frac{\#P_n^{(6)}(m)}{n^6}\\ \begin{array}{r|llllll} m\backslash n &1&2&3&4&5&6\\ \hline 1& 0. & 0. & 0.00823045 & 0.046875 & 0.10368 & 0.164609 \\ 2& 0. & 0. & 0.0576132 & 0.1875 & 0.288 & 0.349794 \\ 3& 0. & 0. & 0.106996 & 0.169922 & 0.1728 & 0.156893 \\ 4& 0. & 0. & 0.18107 & 0.228516 & 0.21504 & 0.187757 \\ 5& 0. & 0. & 0.17284 & 0.123047 & 0.08064 & 0.0540123 \\ 6& 0. & 0.09375 & 0.139918 & 0.0908203 & 0.0576 & 0.0379372 \\ 7& 0. & 0.15625 & 0.0411523 & 0.0146484 & 0.0064 & 0.00321502 \\ 8& 0. & 0.1875 & 0.148148 & 0.0878906 & 0.05376 & 0.0347222 \\ 9& 0. & 0.09375 & 0.0246914 & 0.00878906 & 0.00384 & 0.00192901 \\ 10& 0. & 0.15625 & 0.0411523 & 0.0146484 & 0.0064 & 0.00321502 \\ 11& 0. & 0.0625 & 0.0164609 & 0.00585938 & 0.00256 & 0.00128601 \\ 12& 0. & 0.0625 & 0.0164609 & 0.00585938 & 0.00256 & 0.00128601 \\ 13& 0. & 0. & 0. & 0. & 0. & 0. \\ 14& 0. & 0. & 0. & 0. & 0. & 0. \\ 15& 0. & 0. & 0. & 0. & 0. & 0. \\ 16& 0. & 0.15625 & 0.0411523 & 0.0146484 & 0.0064 & 0.00321502 \\ 17& 0. & 0. & 0. & 0. & 0. & 0. \\ 18& 0. & 0. & 0. & 0. & 0. & 0. \\ 19& 0. & 0. & 0. & 0. & 0. & 0. \\ 20& 0. & 0. & 0. & 0. & 0. & 0. \\ 21& 0. & 0. & 0. & 0. & 0. & 0. \\ 22& 0. & 0. & 0. & 0. & 0. & 0. \\ 23& 0. & 0. & 0. & 0. & 0. & 0. \\ 24& 0. & 0. & 0. & 0. & 0. & 0. \\ 25& 0. & 0. & 0. & 0. & 0. & 0. \\ 26& 0. & 0. & 0. & 0. & 0. & 0. \\ 27& 0. & 0. & 0. & 0. & 0. & 0. \\ 28& 0. & 0. & 0. & 0. & 0. & 0. \\ 29& 0. & 0. & 0. & 0. & 0. & 0. \\ 30& 0. & 0. & 0. & 0. & 0. & 0. \\ 31& 0. & 0. & 0. & 0. & 0. & 0. \\ 32& 1. & 0.03125 & 0.00411523 & 0.000976563 & 0.00032 & 0.000128601 \end{array} $$ REPLY [2 votes]: To get started you can use oeis.org to investigate this. For instance, from plugging in the numerators corresponding to some of your data it seems that $$\#P_n^{(2)}(1)=n(n-1)$$ ("the oblong numbers") and $$\#P_n^{(6)}(1)=n(n-1)(n-2)^4$$ ("the number of $n$-colorings of the Triangle Graph of order 3").<|endoftext|> TITLE: Why not develop a Hamiltonian-based Morse theory? QUESTION [9 upvotes]: I have begun to learn the basics of Morse theory and Floer homology. I understand that Floer homology is the natural theory for symplectic manifolds, but from my preliminary knowledge of Morse theory am curious why the seemingly more natural analogue of this can not be applied to the study of symplectic manifolds (or indeed maybe it can but is of no use?). By this I mean a sort of finite-dimensional "Hamiltonian Morse theory" where the gradient-like vector field is replaced with a Hamiltonian-gradient-like vector field $X_H:i_{X_H} ω=dH$, and it is the critical points of H that are relevant rather than those of the action functional (as in Floer homology). Any comments would be welcomed. REPLY [11 votes]: To any continuous dynamical system $\Phi_t$ on a reasonable space $X$ (say a compact metric space) there is an associated Morse like theory, namely the Conley index theory, C. Conley: Isolated Invariant Sets and the Morse Index, CBMS Regional Conf. Series, vol. 38, Amer. Math. Soc., 1978. Conley and Zehnder, in Sect. 3 in their paper Morse type index theory for flows ..., Comm. Pure Appl. Math. (37) (1984), have shown how to extend the Morse inequalities to this more general case. This extension is based on a certain increasing filtration of the space $X$ compatible in a certain way with the flow $\Phi_t$. Such a filtration produces a spectral sequence in homology and the Morse inequalities relate the Betti numbers of the $E_1$ page of this spectral sequence to the Betti numbers of $X$. $\newcommand{\bZ}{\mathbb{Z}}$ In the very special case when $X$ is a compact smooth manifold, and the flow in question is the (negative) gradient flow of a self-indexing Morse function satisfying the Smale transversality conditions several nice things happen. The above Conley-Zehnder filtration can be identified with the filtration by the sublevel sets $\bigl\lbrace f\leq k+\frac{1}{2}\bigr\rbrace$, $k\in\bZ_{\geq 0}$. (I recall that the critical values of $f$ are nonnegative integers.) The $E_1$ page of the canonical spectral sequence of this filtration is the Floer complex; see sec. 2.5 of these lectures. I have not seen applications of the Conley index applied directly to Hamiltonian flows as suggested in your question, though there might exist. At a first sight this seems difficult but, as Arnold liked to say, you never know how hard a problem is until you try to solve it.<|endoftext|> TITLE: Are all quotients of a weakly contractible space via a free group action classifying spaces of the group? QUESTION [10 upvotes]: I asked this question on math.stackexchange a week ago, but did not get an answer. First of all, I don't want to restrict to any kind of "nice spaces" since I am interested in the most general situation. Especially I do not work in any "convenient" category of spaces. Wikipedia defines a classifying space of a topological group $G$ as the quotient space of a weakly contractible space $EG$ with respect to a free action of $G$ on $EG$, which confuses me. For me, a classifying space $BG$ should be a space for which there exists a principal $G$-bundle $$G\rightarrow EG\rightarrow BG,$$ such that for sufficiently nice spaces $X$ the homotopy classes from $X$ to $BG$ are in bijection (via pullback of the bundle $EG\rightarrow BG$) to the set of isomorphism classes of principles $G$-bundles on $X$. I think it was Milnor (Can you give me a reference?) that showed that universality of the bundle $EG\rightarrow BG$ is equivalent to the (weak?) contractibility of $EG$. From this viewpoint the definition of wikipedia suggests the quotient map of any weakly contractible space to the quotient via a free group action of $G$ being a principle $G$-bundle. This sounds strange to me as I would except, that I will need more topological restrictions on my spaces for this to work. So here is the question: Which are the exact most general topological restrictions to each space, such that the situations I described work out? To be more specific: When is the quotient map of a space with respect to a free action of a topological group $G$ a principal $G$-bundle? Does the definition of wikipedia gives me always the right thing? What is the canonical reference for the claim that a principal $G$-bundle is universal iff it's total space is (weakly?) contractible? Which topological restrictions on $X$ do I need such that $[X,BG]$ classifies what it should classify? REPLY [7 votes]: I'll try to say as much as I can about each of the four questions above. If $X$ is a space with an action of a topological group $G$, then the quotient map $X\to X/G$ is a principal bundle if and only if a) the action admits local slices and b) the map $X\times_{X/G} X\to G$, sending $(x, y)$ to the unique element of $G$ satisfying $xg = y$, is continuous. Husemoller's book on fiber bundles discusses this. The Wikipedia article is a bit sloppy, as shown by the example of $X = G^{i}$, the set $G$ with the indiscrete topology. I think Husemoller's book proves that if $E$ is weakly contractible and $E\to E/G$ is a principal $G$-bundle, then $E/G$ is universal in the sense that for every CW complex $X$, there is a bijection between $[X, E/G]$ and Prin$_G (X)$, the set of isomorphism classes of principal $G$-bundles over $X$. There is a discussion of this in my course notes: http://www.math.iupui.edu/~dramras/601.html You're also interested in the converse statement, that if $P\to B$ is a principal $G$-bundle that is universal in an appropriate sense, then $P$ has to be weakly contractible. I don't recall seeing this sort of statement in the literature. Here's a simple version: if there exists a CW complex $B$ with a principal $G$-bundle $P\to B$ such that $P$ is contractible, then a Yoneda-type argument says that if $P'\to B'$ is a principal $G$-bundle over a CW complex which is universal in the sense that for all CW complexes $X$, $[X, B']$ is in bijection with Prin$_G (X)$, then there's a homotopy equivalence $f: B'\to B$ such that $f^*(P) = P'$. Then the LES in homotopy shows $\pi_* (P') = \pi_* (P) = 0$. I don't know whether Milnor wrote anything about this. I don't see it in his 1956 Annals papers "Construction Universal Bundles, I and II." One answer is "take $X$ to be a CW complex." But you probably want more than this. Segal discusses something more general in Section 4 of his paper Classifying Spaces and Spectral Sequences. He puts some conditions on the group, so as to obtain a classifying space $BG$ such that for all paracompact spaces $X$ he gets the desired bijection between $[X, BG]$ and the set of isomorphism classes of principal $G$-bundles over $X$. (Note that on a paracompact space all covers are numerable in the sense Segal discusses, which is just to say they admit subordinate partitions of unity.) There may be related ideas in old papers of Dold.<|endoftext|> TITLE: Simply connected Lie groups homeomorphic to R^n are solvable QUESTION [5 upvotes]: I have found the following claim in many proofs "Simply connected Lie groups homeomorphic to $\mathbb{R}^n$ are solvable". But the universal covering of $SL(2,\mathbb{R})$ satisfies the hypothesis of this claim and it is far from being solvable. Could someone give me the right topological hypothesis on a Lie group which lead to its solvability and where I can find a proof of this fact. REPLY [3 votes]: I think you just need the extra hypothesis that the Lie group is a matrix group. That is, Proposition: Let $G$ be a Lie group with a faithful finite-dimensional complex representation $V$. If $G$ is homeomorphic to $\mathbb{R}^k$ (which implies that it's simply connected), then it is solvable. In fact we can use the (apparently?) weaker hypothesis that $G$ has trivial homology. Proof. It suffices to show that the semisimple part $\mathfrak{g}_{ss}$ of the Levi decomposition of $\mathfrak{g}$ vanishes. First, recall that every connected Lie group $G$ deformation retracts onto its maximal compact subgroup, which, being in particular a compact oriented manifold, has nontrivial top homology. Hence if $G$ has trivial homology then its maximal compact vanishes, and so it can have no compact subgroups. Now consider the Cartan decomposition $\mathfrak{g}_{ss} = \mathfrak{k} \oplus \mathfrak{p}$. By hypothesis, $\mathfrak{g}_{ss} \otimes \mathbb{C}$ acts on $V$, and hence so does its compact real form $\mathfrak{g}_c = \mathfrak{k} \oplus i \mathfrak{p}$, which integrates to a compact Lie group $G_c$ also acting on $V$. In particular, $\mathfrak{k}$ integrates to a compact subgroup of $G_c$ and hence of $G$. But $G$ has no nontrivial compact subgroups, and so $\mathfrak{k}$ vanishes. It follows that the Killing form of $\mathfrak{g}_{ss}$ is positive definite, and so $\mathfrak{g}_{ss}$ must also vanish. $\Box$<|endoftext|> TITLE: Applications of set theory in physics QUESTION [20 upvotes]: In the introduction of the paper "Links between physics and set theory", the following quote of Eris Chric is stated: "Set theory perhaps is too important to be left just to mathematicians." I know several papers on connections between set theory and physics, but I don't know if these connections are important in physics. So my question is: Question. Are there any applications of set theory in physics, which are of interest to physicists, and have important applications in physics? REPLY [4 votes]: Are there any applications of set theory in physics, which are of interest to physicists, and have important applications in physics? I'm a physicist. Any answer to this question is going to depend completely on your definitions of "application" and "set theory." If you consider only some trivial corner of naive set theory, and count cases where its use is purely a matter of convenience or ease of notation, then certainly there are many applications. For example, we often solve quadratic equations in physics, and it's convenient to talk about the set of real solutions. But every physics experiment that has ever been done was performed with finite physical and computational resources, which means that all of our experience of physics can be described within finitism. There are even rigorous arguments (Krauss 1999) that, given the cosmological facts we observe, any hypothetical future physical process will be able to harness only finite energy and finite computation. (This is nontrivial; before the discovery of dark energy, the opposite conclusion was reached by Dyson.) Because of these physical limitations, it is not possible, even in principle, for us to measure an irrational number or to demonstrate affirmatively that spacetime has the structure of a manifold. I only have access to the abstract of the Augenstein paper, but it seems to argue that "physical reality," specifically quantum mechanics, provides direct realizations of set-theoretical constructs, including "individual axioms" of ZFC. This sounds bogus to me on both physical and mathematical grounds. Physically, Krauss's result tells us that we can never harness more than a finite amount of energy, and via the de Broglie relation, this puts a a cutoff on the wavelengths that we will be able to probe. The region within our cosmological event horizon will always be finite as well. The combination of these two cutoffs means that any quantum-mechanical experiment, even in principle, can be described by a finite-dimensional Hilbert space. Mathematically, the vast majority of mathematics is carried out without any consideration of any underlying foundational issues, such as the use of ZFC as opposed to some other framework. The sphere within which physicists operate is even more restricted than that of normal mathematics. We can also consider the role of computation, through which a physical machine (such as a computer, a brain, or a slide rule) can prove things about mathematics. I can use an analog computer to compute the square root of 2, e.g., by constructing two pendulums with lengths in a 2:1 ratio and measuring the ratios of their periods. If my analog computer says that the second decimal place of the decimal expansion of $\sqrt{2}$ is a 1, and your computation says that it's a 3, then my physics experiment has successfully demonstrated something to you about your mathematical theory. If the axioms of ZFC were to be directly realized in physical experiments, as Augenstein seems to propose, then I ought to be able to do the same kind of thing with ZFC. Does anyone really expect that a physicist will do an experiment that will prove the axiom of choice?<|endoftext|> TITLE: Texts about Dwork's work QUESTION [6 upvotes]: I want to ask about references to papers, that probably exist, which explain the articles of Bernard Dwork starting from "The rationality of the zeta function of an algebraic variety" to "On the Boyarsky principle". REPLY [3 votes]: You could start with the book by Dwork, Gerotto and Sullivan, "An introduction to G-functions", published by Princeton University Press in the collection Annals of math studies. It contains a full account of the rationality of the zeta function, and the beginnings of the study of p-adic differential equations.<|endoftext|> TITLE: Scotts Theorem for one ended Fuchsian groups QUESTION [5 upvotes]: in Peter Scott's work "Subgroups of Surface groups are almost geometric" is proven that for a closed surface $S$ and any fin. gen. subgroup $U$ of $G:=\pi_1(S,x)$ there exists a finitely sheeted covering map $p: (\tilde{S},\tilde{x})\to (S,x)$ such that U is realized by a subsurface in $\tilde{S}$. This is Theorem 3.3 in his work. Is there a generalization of this theorem for the case where $G$ is a one-ended Fuchsian group? In this case $S$ is a two-orbifold and $U\leq G=\pi_1(S,x)$. And the question is: Does there exist a finite sheeted cover $\tilde{S}$ such that the subgroup $U$ corresponds to a sub-two-orbifold of $\tilde{S}$. And if not, does anyone knows a counterexample? Thanks. REPLY [4 votes]: Scott explicitly claims that Fuchsian groups are LERF on the first page of his paper. Lemma 1.4 shows that LERF is equivalent to the "geometric condition". He assumes the cover is regular, but I think the proof goes through for a regular orbifold cover. Thus Fuchsian groups also have the "geometric condition". Finally, orbifolds (with finitely generated orbifold fundamental group) have compact cores, completing the proof. NB: There is an erratum for this paper, published in 1985.<|endoftext|> TITLE: $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m) \simeq \mathcal S(\mathbb R^{n+m})$? QUESTION [6 upvotes]: If $S(\mathbb R^n)$ is the Scwartz space of smooth rapidly decaying functions equipped with the topology generated by the family of semi-norms $$\mathcal N_p (\varphi)= \sum_{|\alpha|, |\beta| \leq p} \sup_{x\in \mathbb R^n} |x^\alpha \partial^\beta \varphi (x) |\, ,$$ Is it true that $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m)\simeq \mathcal S(\mathbb R^{n+m})$, where $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m)$ is the completion of $\mathcal S(\mathbb R^n) \otimes \mathcal S(\mathbb R^m)$ for the topology $\pi$ defined by the family of semi-norms $$\mathcal N_{p,q} (\varphi\otimes \psi)=\mathcal N_p(\varphi)\mathcal N_q (\psi)\, .$$ REPLY [8 votes]: Yes, if you understand the tensor product topology in the right way. Since the spaces are nuclear, inductive and projective tensor products coincide. The result is theorem 51.6 of Treves: Topological vector spaces, distributions, and kernels. Added later: Attention: The description of seminorms on the tensor-product given in the question is not sufficient to specify a locally convex topology on the tensor product. There are many satisfying this description; all between the the projective one and the the inductive one and even more. See the source I have given, or many other books.<|endoftext|> TITLE: Is $\sum_{k=1}^{n} \sin(k^2)$ bounded by a constant $M$? QUESTION [40 upvotes]: I know $\sum_{k=1}^{n} \sin(k)$ is bounded by a constant. How about $\sum_{k=1}^{n} \sin(k^2)$? REPLY [28 votes]: Terry Tao has already given an excellent answer to this, but I want to point out that much more is known about the partial sums of $\sin(\pi k^2 x)$ with $x$ being an irrational (such as $1/\pi$ in the question). This and related problems were studied extensively in the classical paper of Hardy and Littlewood Some Problems of Diophantine approximation. II from 1914. Hardy and Littlewood exploit the connection with the transformation formulae for $\theta$-functions, and show a number of $\Omega$ and $O$ results for such partial sums. In particular, Theorem 2.30 of their paper proves that for any irrational $x$ the series $$ \sum n^{-\alpha} \cos(n^2 \pi x), \ \text{and} \ \sum n^{-\alpha} \sin(n^2 \pi x) $$ are not convergent when $0< \alpha \le 1/2$ (and moreover are not summable by any Cesaro means). By partial summation, this implies that there are arbitrarily large values $N$ with $$ \Big| \sum_{k\le N} \sin (k^2 \pi x) \Big| \ge \frac{\sqrt{N}}{(\log N)^2}, $$ say (otherwise the series in the Hardy-Littlewood result would converge for $\alpha=1/2$), and in fact one can probably get $\gg \sqrt{N}$ from their paper (they do this explicitly for partial sums of $e^{i\pi n^2 x}$). Quadratic Weyl sums continue to be of interest -- see this very recent paper of Cellarosi and Marklof.<|endoftext|> TITLE: Enriques surfaces over $\mathbb Z$ QUESTION [29 upvotes]: Does there exist a smooth proper morphism $E \to \operatorname{Spec} \mathbb Z$ whose fibers are Enriques surfaces? By a theorem of, independently, Fontaine and Abrashkin, combined with the Enriques-Kodaira classification of complex surfaces, the only possible surface of Kodaira dimension $0$ or $1$ that can appear this way is an Enriques surface. In particular, no K3 surfaces are smooth over $\mathbb Z$. Hence the K3 double cover of $E$ is not smooth. The cover is etale away from $\mathbb F_2$, hence must be singular over $\mathbb F_2$. This means $E$ most be a classical Enriques surface (a $\mu_2$-torsor rather than a $\mathbb Z/2$-torsor or $\alpha_2$-torsor). It is possible to get some other information about $E$: The Picard group of this surface has rank $10$. The Galois action on the Picard group must be unramified at each prime, hence trivial, so the full lattice of cycles is defined over $\mathbb Q$. Thus by the Lefschetz trace formula, $E$ has exactly $25$ $\mathbb F_2$-points. Some other questions that might be helpful to solve this one are: How many K3 surfaces are there with good reduction away from $2$ (and Picard rank at least $10$, and a fixed-point-free involution, etc.)? Given an Enriques surface over $\mathbb Q_2$, what are obstructions to good reduction over $\mathbb Z_2$, other than ramification of the cohomology? Which classical Enriques surfaces over $\mathbb F_2$ are liftable to $\mathbb Z_2$? Can something be said about the singularities and Galois representations of their $K3$ double covers? Can we compute the discriminants of explicit families of Enriques surfaces and try to solve the Diophantine equation $\Delta=1$? One example of an Enriques surface over $\mathbb Z[1/2]$ whose cohomology is unramified at $2$ can be constructed as the quotient of a Kummer surface with good reduction away from $2$. Let $E_1$ and $E_2$ be two elliptic curves that are either $y^2=x^3-x$ or $y^2=x^3-4x$. Let $e_1$ and $e_2$ be $2$-torsion points on $E_1$ and $E_2$ respectively. Then we can construct (Example 3.1) a fixed-point free involution on the Kummer surface of $E_1 \times E_2$, giving an Enriques surface. Because $E_1$ and $E_2$ have good reduction away from $2$, this surface has good reduction also. $H^2$ of the Kummer surface comes from $H^2(E_1 \times E_2)$ plus the exceptional classes of the 16 blown-up $2$-torsion points. Because these points are defined over $\mathbb Q$, the cohomology classes are unramified. $H^2(E_1)$ and $H^2(E_2)$ are unramified as well, so the only ramified part of the cohomology of the Kummer surface is $H^1(E_1) \times H^1(E_2)$. Because the involution acts as reflection on $E_1$ and translation on $E_2$, it acts as $-1$ on $H^1(E_1) \times H^1(E_2)$, so that does not descend to the Enriques surface, hence its cohomology is unramified. This surface also has $\mathbb Q$-points, thus $\mathbb Q_2$ points. Thus I cannot see any obstruction to good reduction at $2$. However, the construction certainly does not produce a smooth model of the surface over $2$. Does this surface have good reduction at $2$? REPLY [12 votes]: A preprint by Stefan Schröer came out today with the answer to this question: arXiv:2004.07025. No such Enriques surface exists. In fact, there is no classical Enriques surface over $\mathbb F_2$ with 25 $\mathbb F_2$-points (and with the extension of $\mathbb Z^{10}$ by $\mathbb Z/2$ split in the Picard group, which you can also deduce).<|endoftext|> TITLE: Reference for proof that consistency of $\omega_1$-Erdos cardinal implies Con(Chang's Conjecture) QUESTION [8 upvotes]: What is a good source for Silver's proof (or a more modern version) that Con($\exists \omega_1$-Erdos cardinal) implies Con(Chang's Conjecture)? Silver's original proof seems to have never been published and I didn't find a proof in the set theory books I looked at (i.e. Jech's "Set Theory: the 3rd Millennium Edition" and Kanamori's "Higher Infinite") REPLY [6 votes]: A sketch of Silver's original argument appears in section 19 here: http://math.bu.edu/people/aki/e.pdf<|endoftext|> TITLE: Must we know $MU^*(X)$ in order to compute $Ell^*(X)$? QUESTION [9 upvotes]: Let $Ell^*(X)$ be the elliptic cohomology theory (associated to a given elliptic curve $E$) of a nice space $X$. Recall the Landweber-Ravenel-Stong construction: $MU^*(X) \otimes_{MU^*} R \simeq Ell^*(X)$, where $R \simeq Ell^*$. How does the Landweber-Ravenel-Stong construction of $Ell^*(-)$ tell us how to actually compute $Ell^*(X)$ with the Atiyah-Hirzebruch spectral sequence? Let $L$ be the Lazard ring, and $R$ is flat over $L$. It seems like the Landweber-Ravenel-Stong construction tells us that: Given a map $L \to R$, $$MU^* \simeq L \longrightarrow R \simeq Ell^*$$ This gives us a map between their Atiyah-Hirzebruch SS? $$MU^*-AHSS(X) \longrightarrow Ell^*-AHSS(X)$$ The formal group law associated to our elliptic curve $\hat{E} \simeq \text{Spf } Ell^*(\mathbb{CP}^\infty)$. It was told that $Ell^*(X)$ is computationally simpler than computing $MU^*(X)$. But we seem to need to know $MU^*(X)$ before we can compute $Ell^*(X)$, so I must be missing something! Do we need to know the AHSS of $MU^*(X)$ before we can compute the AHSS of $Ell^*(X)$? Edit wrt computing the AHSS of $Ell^*(X)$: It was explained to me that we don't need $MU^*(X)$, we need $Ell^*(pt)$ and $H^*(X)$ (plus differentials) to compute $Ell^*(X)$. We know $MU^*(pt)$ and thus our choice of elliptic curve (or better, it's ground ring) determines $Ell^*(pt)$. REPLY [15 votes]: The AHSS is unlikely to be a good method for computing either $MU^*(X)$ or $Ell^*(X)$ except in cases where the AHSS collapses for easy reasons (the $E^2$ term is torsion free, or concentrated in even total degree). Even in those cases, it is usually desirable and possible to use other methods (usually Chern classes) to produce some generators and relations for $MU^*(X)$, and then use the collapsing AHSS to show that no more generators or relations are required. In these cases one can compute both $MU^*(X)$ and $Ell^*(X)$. The basic case where $Ell^*(X)$ is easier than $MU^*(X)$ is the case $X=BG$, where $G$ is a finite group. Even here, the group $Ell^*(X)$ itself is not so easy. It is better to consider the $I_2$-adic completion of $Ell$, which we could call $E$. Then $E$ is essentially a version of Morava $E$-theory, and $E^*(BG)$ is a finitely generated module over the Noetherian ring $E^*(\text{point})$, which is often a free module. Everything can be computed quite explicitly in the case where $G$ is abelian. Even if $G$ is nonabelian one can use HKR character theory, Chern classes of representations, transfers and power operations. The AHSS is almost never useful in this context.<|endoftext|> TITLE: Weak convergence in $W^{1,p}_0$ QUESTION [5 upvotes]: Note from the answerer : this question stems from this article. I ask this question in https://math.stackexchange.com/questions/1206617 I have a bounded sequence $(u_n)$ from $W^{1,p}_0(\Omega)$ so it weakly converge to $u\in W^{1,p}_0(\Omega)$ and strongly converge to $u$ in $L^p(\Omega).$ We define a function $f:\Omega\times \mathbb{R}\rightarrow \mathbb{R}$ a bounded Caratheodory function such that $\lim_{s\rightarrow+\infty} f(x,s)=f^{+\infty}(x)$ My question is why $$\lim_{n\rightarrow +\infty} \int_{\Omega}f(x,u_n)(u_n-u) dx=0$$ and $$\lim_{n\rightarrow +\infty}\int_{\Omega} |u_n|^{p-2} u_n(u_n-u) dx=0$$ For the first integral, I'm trying to apply Lebesgue dominated convergence, but I have no idea. For the second integral, when $p=2$ I have no problems, because in this case we have not $|u_n|^{p-2}$ it is equal to 1 and then I just have to do $u_n(u_n-u)=(u_n-u+u)(u_n-u)$ and I do the Cauchy–Schwarz inequality, but when $p$ is not equal to 2 I have no idea. Thank you REPLY [5 votes]: Let's call $I_1$ and $I_2$ your two integrals respectively. You know that $u_n \to u$ strongly in $L^p$. Because $\Omega$ is bounded, $u_n$ also converges to $u$ strongly in $L^1$. As you assume that $f$ is bounded in its two arguments, $$|I_1| \leq \|f\|_{L^{\infty}(\Omega \times \mathbb{R})} \|u_n-u\|_{L^1(\Omega)} \to 0.$$ Regarding the second integral, by Hölder inequality, you have $$|I_2| \leq \| |u_n|^{p-2} u_n \|_{L^q(\Omega)} \|u_n-u\|_{L^p(\Omega)} $$ where $\frac 1p + \frac 1q = 1$, i.e. $q = \frac{p}{p-1}$. Thus $$|I_2| \leq \|u_n\|_{L^p(\Omega)}^{p-1} \|u_n-u\|_{L^p(\Omega)} \to 0 $$ thanks to the strong convergence of $u_n$ in $L^p$ and its boundedness in $L^p$.<|endoftext|> TITLE: Is an open map with open relative diagonal necessarily a local homeomorphism? QUESTION [9 upvotes]: Let $f : X \to Y$ be an open (and continuous) map of locales. Suppose the relative diagonal $\Delta_f : X \to X \times_Y X$ is an open embedding of locales. Does it follow that $f : X \to Y$ is a local homeomorphism? The answer is yes if I replace "locale" with "topological space". Indeed, by the definition of the product topology, there must exist an open covering of the image of $\Delta_f$ by "rectangles", i.e. open subspaces of the form $U \times_Y V \subseteq X \times_Y X$ for open subspaces $U \subseteq X$, $V \subseteq X$; but if $U \times_Y V \subseteq \operatorname{im} \Delta_f$, then the restriction $f : U \cap V \to Y$ must be injective, hence is an open embedding (because $f : X \to Y$ is an open map). But $\{ U \cap V : U \times_Y V \subseteq \operatorname{im} \Delta_f \}$ is an open covering of $X$, so we are done. In fact, everything in the above argument goes through for locales, except for the very last step where I assert that we have an open covering of $X$. There, I have used the fact that a collection of open subspaces is an open covering if and only if every point is contained in some member of the collection. So the argument would work if $X$ is a spatial locale. But can we avoid the use of points? For reference, here are some standard definitions: The image of a locale morphism $f : X \to Y$ is the locale $\operatorname{Im} f$ corresponding to the frame $$\Omega (\operatorname{Im} f) = \{ v \in \Omega (Y) : f_* (f^* (v)) = v \}$$ where $f^* : \Omega (Y) \to \Omega (X)$ is the frame homomorphism corresponding to $f : X \to Y$ and $f_* : \Omega (X) \to \Omega (Y)$ is the right adjoint; the "inclusion" $\operatorname{Im} f \hookrightarrow Y$ corresponds to the frame homomorphism $v \mapsto f_* (f^* (v))$. An open sublocale of a locale $Y$ is a locale $Y_v$ that corresponds to a frame of the form $$\Omega (Y_v) = \{ v' \in \Omega (Y) : v' \le v \}$$ for some $v \in Y$; the "inclusion" $Y_v \hookrightarrow Y$ corresponds to the frame homomorphism $v' \mapsto v' \land v$. An open embedding of locales is a morphism $f : X \to Y$ that is isomorphic to the inclusion of some open sublocale of $Y$. An open map of locales is a morphism $f : X \to Y$ such that the image of every open sublocale of $X$ is an open sublocale of $Y$. A local homeomorphism of locales is a morphism $f : X \to Y$ for which there is a set $\mathfrak{U} \subseteq \Omega (X)$ such that: $\sup \mathfrak{U} = \bigvee_{u \in \mathfrak{U}} u = \top$. For each $u \in \mathfrak{U}$, the composite $X_u \hookrightarrow X \to Y$ is an open embedding. REPLY [2 votes]: The answer is yes. It appears for example as lemma C3.1.15 in Johnstone's sketches of an elephant. Roughly, it can be proved by working in the internal logic of the target (hence assuming that the target is a point) and considering open subspaces $U$ of $X$ such that $U \times U \subset \Delta$, on can then show that such $U \times U$ cover the diagonal, hence the $U$ cover $X$ and that if they are positive they correspond to isolated points of $X$. The proof in Johnstone's book is an external formulation of this argument and hence looks a bit more technical. Edit : I didn't read your post well enough and didn't see you already proposed a proof. The trick you are missing is I think the following: If the $U \times V$ form a covering of $\Delta$ then the $\Delta^{-1}(U \times V)$ form a covering of $X$, and $\Delta^{-1}(U \times V) = U \cap V$.<|endoftext|> TITLE: How can dimension depend on the point? QUESTION [7 upvotes]: Let $M$ be a metric space. For any subset $A\subset M$ let $\dim(A)$ denote its Hausdorff dimension. For $x\in M$, define the dimension of $M$ at $x$ by $\dim(x)=\lim_{r\to0}\dim(B(x,r))$; this limit exists because dimension depends monotonously on the set. What can this dimension function look like? Are there any results about dimension at a point? I haven't heard or found any. For example, what is known about the regularity of the dimension function? Is $\dim:M\to[0,\infty]$ always upper semicontinuous? Lower semicontinuity can fail, as can be seen by attaching a stick to a ball. Or is it at least Borel measurable? I gave one answer below to give an idea of what I might be interested in. If you need to assume something more (like $M$ being a compact subset of a Euclidean space) or can say something about some other kind of dimension, feel free to do so. REPLY [8 votes]: Lars Olsen [1], [2] (2005, 2005) has proved some results about this notion. Let $E \subseteq {\mathbb R}^{n}$ and $x \in {\mathbb R}^{n},$ where $n$ is a fixed positive integer. Let $\dim_{H}(E,x)$ and $\dim_{P}(E,x)$ denote the local Hausdorff and local packing dimensions of $E$ at $x,$ defined as in your question. Discussion of Results in Olsen [1] In [1] Olsen proved that, given any continuous function $f:{\mathbb R}^n \rightarrow [0,n],$ there exists $E \subseteq {\mathbb R}^{n}$ such that for all $x \in {\mathbb R}^{n}$ we have $f(x) = \dim_{H}(E,x) = \dim_{P}(E,x).$ Olsen observed (p. 214) that it is easy to see that some local dimension functions can be discontinuous, giving the example $f:{\mathbb R} \rightarrow [0,1]$ defined by $f(x) = \frac{\ln 2}{\ln 3}$ for $x \in C$ and $f(x) = 0$ if $x \notin C,$ where $C$ is the Cantor middle thirds set. In fact, the characteristic function of a compact interval gives a simpler example, but I suppose Olsen gave the example he did because then the discontinuities occur at every point of the Cantor set. Olsen also observed (p. 214) that it is easy to see that some very simple discontinuous functions $f:{\mathbb R} \rightarrow [0,1]$ cannot be a local dimension function, giving the example $f(x) = 0$ if $x \neq 0$ and $f(0) = 1.$ Olsen posed the problem (p. 214) of characterizing those functions $f:{\mathbb R}^n \rightarrow [0,n]$ that can be the local dimension function (Hausdorff and/or packing) of some subset of ${\mathbb R}^{n}.$ The result Olsen actually proved was a bit sharper than I stated above. Let $M(f,x,r)$ denote the supremum of $|f(x_{1}) – f(x_{2})|$ as $x_1$ and $x_2$ vary over the open ball $B(x,r)$ of radius $r$ centered at $x.$ Olsen's Theorem 1.1 in [1] states that, given any function $f:{\mathbb R}^n \rightarrow [0,n]$ (continuous or not), there exists $E \subseteq {\mathbb R}^{n}$ such that for all $x \in {\mathbb R}^{n}$ and for all $r > 0$ we have $$| f(x) \; - \; \dim_{H}\left(E \cap B(x,r) \right)| \;\; \leq \;\; M(f,x,r)$$ $$| f(x) \; - \; \dim_{P}\left(E \cap B(x,r) \right)| \;\; \leq \;\; M(f,x,r)$$ Olsen observed that if $f$ is continuous, then we obtain the result I gave earlier. Discussion of Results in Olsen [2] In [2] Olsen provided an answer to the problem he posed in the earlier paper by characterizing those functions $f:{\mathbb R}^n \rightarrow [0,n]$ that can be the local dimension function (in both the Hausdorff and the packing sense) of some subset of ${\mathbb R}^{n}.$ For Hausdorff dimension, the characterization Olsen gave is that $f:{\mathbb R}^n \rightarrow [0,n]$ satisfies: (1) For each $x \in {\mathbb R}^{n},$ we have $\limsup_{x' \rightarrow x} f(x') = f(x).$ (2) For each $y$ with $0 \leq y < \sup f$ and for each $x \in \{f > y\} \; = \; \{x' \in {\mathbb R}^{n}: \; f(x') > y \},$ we have $\dim_{H}\left( \{f > y\}, \;x\right) \; > \; y.$ Here, "$\sup f$" denotes the supremum of $f$ over ${\mathbb R}^{n}.$ The characterization for packing dimension is identical except that $\dim_{P}$ replaces $\dim_{H}$ in (2). Before proving this result, Olsen gave an example (2nd example on p. 233) to show that there exist functions $f:{\mathbb R} \rightarrow [0,1]$ satisfying (1) but not satisfying (2). The example Olsen gave is the function $f$ equal to $s$ (a constant) at each point of the Cantor middle thirds set $C$ and equal to $0$ elsewhere, where $\frac{\ln 2}{\ln 3} < s \leq 1.$ To see that (2) does not hold, note that if $y$ is chosen so that $\frac{\ln 2}{\ln 3} \leq y < s,$ then $\{f > y\} = C.$ Incidentally, Olsen uses Hausdorff dimension throughout in this example, but since the packing and Hausdorff dimensions of the Cantor middle thirds set are both equal to $\frac{\ln 2}{\ln 3},$ the same example shows that (1) can hold and (2) can fail for packing dimension as well. As was the case with Olsen's earlier paper, Olsen actually proved a bit more than I've stated thus far. Rather than limiting himself to the Hausdorff or packing dimensions, he proved the characterization for the local dimension function of any "regular dimension index", which is an assignment $\dim$ of a non-negative real number to each subset of ${\mathbb R}^{n}$ such that $\dim$ is monotone with respect to set inclusion, $\dim$ is countably stable, $\dim$ assigns the value of $0$ to every finite subset, and $\dim$ is regular in the sense that given any Borel set $E$ and any real number $t$ with $0 \leq t < \dim E,$ then there exists a compact set $K$ such that $K \subseteq E$ and $\dim K = t.$ Also, Olsen proved that the set for which the function is to be a local dimension function of can always be chosen to be an $F_{\sigma}$ set. In a Remark on p. 235 (that Olsen attributes to a referee -- see Acknowledgements at the end of the paper), he shows that $F_{\sigma}$ cannot be strengthened to "closed". Olsen's punctured upper limit and punctured upper semicontinuous notions Olsen uses a non-deleted notion of $\limsup$ in which the value of the function is taken into account, and he uses the phrase punctured upper limit for the usual $\limsup$ notion in which the value of the function is not taken into account. For a simple example, if $f(x) = 0$ for $x \neq 0$ and $f(0) = 1,$ then the non-deleted $\limsup$ of $f$ at $x=0$ is $1$ and the delted $\limsup$ of $f$ at $x=0$ is $0.$ For a less simple example, if $T$ is the Thomae function, then at each nonzero rational $x$ the non-deleted $\limsup$ of $T$ is equal to $T(x) \neq 0$ and the deleted $\limsup$ of $T$ is equal to $0.$ In the case of upper semicontinuous, Olsen uses the standard definition in which at each point the value of the function is greater than or equal to the deleted $\limsup$ at that point. However, for his characterization of the local dimension function, Olsen requires a refinement of this, which he calls punctured upper semicontinuous. This is the property in which at each point the value of the function is equal to the deleted $\limsup$ at that point. In my summary above I have avoided this terminology and stated (1) directly in terms of the $\limsup$ operation (deleted version being understood, as that is the standard notion). Incidentally, one sometimes sees the phrase upper boundary function used for functions that satisfy (1). Finally, if anyone is interested in continuity issues, I believe that any $F_{\sigma}$ first category (in the Baire sense) subset of ${\mathbb R}^{n}$ can be the discontinuity set for a dimension function, but I have not looked at this carefully. I do know that any such set can be the discontinuity set for a function satisfying (1) above. For some possible properties of sets that are $F_{\sigma}$ and first category, see my answer to How discontinuous can a derivative be?. The less precise result that any such set can be the discontinuity set for an upper semicontinuous function is proved in Proof Sketch at Oscillation of a Function, but I believe the proof there needs to be modified to actually show the result for (1). However, the result for (1) is a consequence of the more precise results proved by Zbigniew Grande in Quelques remarques sur la semi-continuité supérieure [Fundamenta Mathematicae 126 (1985), pp. 1-13] and by Tomasz Natkaniec in On semicontinuity points [Real Analysis Exchange 9 (1983-844), pp. 215-232]. The issue I have not investigated is whether any such set can be the discontinuity set for a function satisfying both (1) and (2) above. [1] Lars Olsen, Applications of divergence points to local dimension functions of subsets of ${\mathbb R}^{d}$, Proceedings of the Edinburgh Mathematical Society (2) 48 #1 (February 2005), 213-218. MR 2005m:28023; Zbl 1061.28004 [2] Lars Olsen, Characterization of local dimension functions of subsets of ${\mathbb R}^{d}$, Colloquium Mathematicum 103 #2 (2005), pp. 231-239. MR 2006j:28020; Zbl 1105.28007<|endoftext|> TITLE: Coherent sheaves on $\mathbb C^2$ and commuting matrices QUESTION [6 upvotes]: Let $V$ be an $n$-dimensional complex vector space. The stack $Coh^n(\mathbb C^2)$ of coherent sheaves on $\mathbb C^2$ supported on $n$ points (not necessarily distinct) is equivalent to the stack quotient $C_n/GL_n$, where $C_n\subset End(V)^2$ is the variety of couples of commuting matrices. In $Coh^n(\mathbb C^2)$ there lives the substack $Coh^n(\mathbb C^2)_0$ of sheaves supported at $0\in\mathbb C^2$. Question 1. What locus does $Coh^n(\mathbb C^2)_0\subset Coh^n(\mathbb C^2)$ correspond to in $C_n/GL_n$? When I pick a coherent sheaf $F\in Coh^n(\mathbb C^2)$, I can thus let it correspond to a couple of commuting matrices $(A,B)$, up to $GL_n$. Now, let $s\in\textrm{Supp}(F)$ be a point of multiplicity $i$, say. If I restrict $F$ to $s$ I get a new sheaf $$ F|_s\in Coh^i(\mathbb C^2)_s\cong Coh^i(\mathbb C^2)_0, $$ which will correspond to a point $(A',B')\in C_i/GL_i$. Question 2. How is the point $(A,B)\in C_n/GL_n$ related to $F|_s$? In other words, how are $(A,B)$ and $(A',B')$ related? Thank you for any help! REPLY [8 votes]: For any commuting pair of matrices there is a basis in which both are upper triangular. The eigenvalues give you $n$ points of $\mathbb{C}^2$ and this recovers the support of the corresponding sheaf. This leads to the following answers: Question 1. Both $A$ and $B$ should be nilpotent. Question 2. $(A',B')$ is a subquotient of $(A,B)$.<|endoftext|> TITLE: Relative Picard functor for the Zariski topology QUESTION [18 upvotes]: I'm trying to understand better the relative Picard functor, as defined, for example, in Kleiman's article. Let $X \to S$ be a smooth projective morphism of schemes whose geometric fibres are integral. Let $$\mathrm{Pic}_{X/S}(T) = \mathrm{Pic}(X_T)/\mathrm{Pic}(T)$$ denote the relative Picard functor of $X$, with sheafications $$\mathrm{Pic}_{X/S,Zar} \quad \mbox{and} \quad \mathrm{Pic}_{X/S,et}$$ for the Zariski and etale topology, respectively. Kleiman shows that $\mathrm{Pic}_{X/S}(S) \neq \mathrm{Pic}_{X/S,et}(S)$ in general, e.g. when $X$ a conic over a field without a rational point. My question concerns what happens for $\mathrm{Pic}_{X/S,Zar}$. What is an explicit example for which $$\mathrm{Pic}_{X/S}(S) \neq \mathrm{Pic}_{X/S,Zar}(S)$$ ? There are various conditions on $X$ if one wants such an inequality hold. For example, $S$ cannot be the spectrum of a local ring, and $X/S$ cannot admit a section for the Zariski topology. REPLY [16 votes]: Let $Y$ be Cayley's nodal cubic surface over the complex numbers, given in $\mathbb{P}^3$ by $X_0X_1X_2 + X_0X_1X_3 + X_0X_2X_3 + X_1X_2X_3 = 0.$ This surface has four simple double points. It also contains six straight lines, each of which passes through two of the double points and is not locally principal there. Let $f \colon X \to Y$ be the minimal desingularisation. Then $X$ is a smooth rational surface, isomorphic to the blow-up of $\mathbb{P}^2$ in six points in a rather special configuration. The six exceptional curves of $X \to \mathbb{P}^2$ are the strict transforms of the six lines on $Y$. I claim that $\DeclareMathOperator{\Pic}{Pic}\Pic_{X/Y}$ is not a sheaf in the Zariski topology. To prove this, let $P$ be one of the singular points of $Y$, let $U$ be a neighbourhood of $P$ not containing any of the other three singular points, and let $V$ be $Y \setminus P$. Let $U'$ and $V'$ be the inverse images of $U$ and $V$ respectively in $X$. Let $L$ be one of the lines passing through $P$, and let $L'$ be its strict transform in $X$. The restriction of $L'$ to $U' \cap V'$ lies in $f^{-1}(\Pic(U \cap V))$, since it's the inverse image of the line $L$ which is locally principal on $U \cap V$. Thus the classes $[L'] \in \Pic_{X/Y}(U)$ and $0 \in \Pic_{X/Y}(V)$ agree on $U \cap V$. However, these two classes do not glue to give a section of $\Pic_{X/Y}$ on $U \cup V = Y$, which we can see by looking at intersection numbers with the four $(-2)$-curves $E_1 = f^{-1}(P),E_2,E_3,E_4$. If the classes did glue, then that section would be represented by a class in $\Pic X$ having intersection number $1$ with $E_1$ and intersection number $0$ with $E_2,E_3,E_4$. But $\Pic X$ is generated by the classes of the strict transforms of the six lines on $Y$ together with the pull-back of the hyperplane class on $Y$. Because each line on $Y$ passes through precisely two singular points, the sum $\sum_{i=1}^4 (D \cdot E_i)$ is even for every $D \in \Pic X$. So what's actually going on here? Have a look at the Leray spectral sequence for the sheaf $\newcommand{\O}{\mathcal{O}}\O_X^\times$ in the Zariski topology. Using $f_* \O_X = \O_Y$, we get an exact sequence $0 \to \Pic Y \to \Pic X \to \mathrm{R}^1 f_* \O_X^\times\to \mathrm{H}^2(Y,\O_Y^\times) \to \mathrm{H}^2(X, \O_X^\times).$ So the obstruction to gluing local sections of $\Pic_{X/Y}$ is given by $\mathrm{H}^2(Y,\O_Y^\times)$, the "Zariski Brauer group" of $Y$. This group vanishes if $Y$ is regular: in that case, the Weil-divisor exact sequence $0 \to \O_Y^\times \to R_Y^\times \to \bigoplus_{Z \in Y^{(1)}} \mathbb{Z}_Z \to 0$ gives a flabby resolution of $\O_Y^\times$, showing that it has no cohomology in degrees $>1$. (In particular, this shows $\mathrm{H}^2(X,\O_X^\times)=0$ in the sequence above.) So, when $f_* \O_X = \O_Y$ holds and $Y$ is regular, your $\Pic_{X/Y}$ will be a Zariski sheaf. If $Y$ is allowed to be singular, there are counterexamples. For rational singularities, this is all controlled by intersection numbers with the exceptional divisors. See arXiv:1202.4299.<|endoftext|> TITLE: A decision problem for clones QUESTION [10 upvotes]: E. Post proved that there are only countably many clones on a two-element set (classes of operations closed under superposition and containing all projections). All these clones are finitely generated. Also, each such clone is computable in the sense that we can computably decide does an n-ary function (given its value table) belong to the clone or not. Moreover, the procedure is uniform under arbitrary set of generators. The situation is different for clones on a finite set of size greater than two: there are continuum many clones on a three-element set (J. I. Janov, A. A. Muchnik) and, therefore, not all of them are finitely generated. Of course, every finitely generated clone is computably enumerable. Are there any known examples of finitely generated clones which are not computable? REPLY [7 votes]: Every finitely generated clone on a finite set is computable. Indeed, fix $k$. If we want to determine which $k$-ary functions belong to the clone $\mathcal C$, we can start generating functions by composition. There are only finitely many functions $f\in F_d$ with a given depth $d$ of the composition tree. Eventually we find a $d$ such that no new $k$-ary functions were added in going from depth $d$ to $d+1$, i.e., $F_d=F_{d+1}$. Then we know that we have found all the $k$-ary functions in $\mathcal C$, by extensionality (i.e., the principle that if we anywhere substitute expressions representing the same function then we don't change the function represented).<|endoftext|> TITLE: Classifying spaces of topological groups whose underlying spaces are homotopy equivalent QUESTION [6 upvotes]: Let $G$, $H$ be topological groups and $f:G\rightarrow H$ a continuous group homomorphism which happens to be a homotopy equivalence of the underlying topological spaces. Let us assume that $G$, $H$ are well-pointed compactly generated Hausdorff as topological spaces, where well-pointedness means that the inclusions of basepoints are closed cofibrations. (By well-pointedness, there is no difference between a homotopy equivalence and a based homotopy equivalence.) Let $B$ be the classifying space functor. My question is: Is $Bf: BG \rightarrow BH$ a homotopy equivalence? (Again, there is no difference between based and non-based homotopy equivalence since $BG$ is well-pointed if $G$ is.) I understand that $Bf$ is a weak homotopy equivalence even without the assumptions made above on the topologies of $G$ and $H$, by this post. I would like it to be a homotopy equivalence with those extra assumptions. Can we show $BG$ and $BH$ have the homotopy type of CW complexes or something? REPLY [6 votes]: As John Klein remarked, the answer to this question will depend on the classifying space functor $B$ one uses. Let me present one case for which the question can be answered positive which is basically the case Dan Ramras mentioned. We define $BG$ for a topological group $G$ to be the fat realization of the simplicial space obtained by applying the topological nerve construction to the topological category one obtains by regarding $G$ as a category in the usual way and including its topology. (See Segal's 'Classifying Spaces and Spectral Sequences' §3) A continuous group homomorphism $G\rightarrow H$ which is a homotopy equivalence will induce a morphism of the corresponding simplicial spaces which is a degreewise homotopy equivalence. (This is easy to check, right from the definitions.) Since we have chosen the fat realization the induced map $BG\rightarrow BH$ will be a homotopy equivalence by Proposition A.1 in Appendix A in Segal's 'Catgeories and Cohomology Theories'. Up here, no point-set topological restrictions are needed, it even all works if the spaces are not compactly generated. In the last paper cited, you'll also find information about the question, in which cases the fat realization is homotopy equivalent to the usual one. In the case of well pointed compactly generated groups, their simplicial nerves will be "good" in the sense of Segal. A survey on the results of that manner is given here.<|endoftext|> TITLE: Interplay between Loop Quantum Gravity and Mathematics QUESTION [25 upvotes]: It is known that there are many interesting connections between String Theory and modern Mathematics, with a rich feedback going on in both directions: there have been advances in mathematics thanks to research in String Theory and vice-versa. For the interested reader, some of the interactions between String Theory and Mathematics have been reviewed in http://rsta.royalsocietypublishing.org/content/roypta/368/1914/913.full.pdf by Atiyah, Dijkgraaf and Hitchin. However, String Theory is not the only quantum theory of gravity on the market: it has a much less popular competitor called Loop Quantum Gravity. My question is, are there any interesting interactions between mathematics and LQG? Has there been any advances in mathematics thanks to LQG research, as it has happened with ST? Thanks. REPLY [3 votes]: One topic which started in the general area of LQG but now has taken a life of its own is the theory random tensors and group field theories. For its connections to mathematics you can have a look at the talks and courses from the program I co-organized last summer in Vienna: http://www.mat.univie.ac.at/~kratt/esi3/<|endoftext|> TITLE: Average probability that a random cosine polynomial with bernoulli coefficients is small QUESTION [5 upvotes]: Let $P_{n}(t)=\sum_{k=0}^{n}\varepsilon_{k}\cos(kt)$ where $\varepsilon_{i}$ are independent random variables taking values in $\left\{-1,1\right\}$ with equal probability. Is is true that for any $\delta\rightarrow 0$ we have \begin{equation} \int_{-\pi}^{\pi}\mathbb{P}(|P_{n}(t)|<\delta)\,dt \approx \frac{\delta}{\sqrt{n}}? \end{equation} For Gaussian random variables this is trivial, but for discrete ones it becomes hard when $\delta \ll n^{-1/2}$. Can the averaging effect smooth things out so that it is still true in the discrete case? REPLY [6 votes]: In general this is false, if $\delta$ is small enough. Indeed, let $Q_n:=\sum_{k=0}^{n}\varepsilon_{k}$ and $R_{n}(t):=P_{n}(t)-Q_n$. Then $\mathsf{E}R_{n}(t)=0$ and $\mathsf{Var}R_{n}(t)\le\frac1{20}(n+1)^5t^4$. Suppose now that $n$ is odd. Then $$\mathsf{P}(|P_{n}(t)|<\delta)\ge\mathsf{P}(Q_n=0)-\mathsf{P}(|R_{n}(t)|\ge\delta) \ge\sqrt{\frac2\pi}\,\frac1{\sqrt{n+2}}-\frac{(n+1)^5t^4}{20\delta^2} \ge p_*$$ if $|t|\le t_*:=\Big(\frac{20}{\sqrt{2\pi}}\,\frac{\delta^2}{(n+2)^{11/2}}\Big)^{1/4}$, where $p_*:=\frac1{\sqrt{2\pi}}\,\frac1{\sqrt{n+2}}$. So, if (say) $20\delta^2<1$, then $$\int_{-\pi}^{\pi}\mathsf{P}(|P_{n}(t)|<\delta)\,dt \ge2t_*p_* >>\frac{\delta}{\sqrt{n}} $$ if $\delta<< n^{-11/4}$.<|endoftext|> TITLE: Representability of morphism of stacks QUESTION [7 upvotes]: A morphism of Artin stacks $f:X\to Y$ over $\mathbb Q$ is representable by algebraic spaces if and only if its geometric fibres are algebraic spaces. I would like to know if one can use this to prove the following statement. Let $f:X\to Y$ be a morphism of finite type separated DM stacks over $\mathbb Q$. Suppose that, for any geometric point $x$ of $X$ with $y= f(x)$, the induced morphism on stabilizers $Stab(x)\to Stab(y)$ is injective. Then $f:X\to Y$ is representable by algebraic spaces. REPLY [4 votes]: This is http://stacks.math.columbia.edu/tag/04Y5 . I quote : " lemma Let $S$ be a scheme contained in $Sch_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of algebraic stacks over $S$. The following are equivalent: 1 for $U \in Ob((Sch/S)_{fppf})$ the functor $f : \mathcal{X}_U \to\mathcal{Y}_U$ is faithful, 2 the functor $f$ is faithful, and 3 $f$ is representable by algebraic spaces. " See also http://stacks.math.columbia.edu/tag/04YY for a fancy reformulation.<|endoftext|> TITLE: Identities for power series like $\sum_n z^{n^3}$ QUESTION [26 upvotes]: Probably, one of the first power series that every mathematician encounter is the geometric series $$\sum_{n=0}^\infty z^n = \frac1{1-z}, \quad z \in \mathbb{C},\; |z| < 1 .$$ Also, a particular case of the Jacobi triple product gives a beautiful identity for the power series with $z$ raised to the square numbers $$\sum_{n=-\infty}^\infty z^{n^2} = \prod_{m=1}^\infty \left(1-z^{2m}\right)\left(1+z^{2m-1}\right)^2, \quad z \in \mathbb{C},\; |z| < 1 .$$ However, I have never seen an equally beautiful identity involving $\sum_n z^{n^3}$. Clearly, such series appear, for example, in papers regarding the positive integers which are sum of $k$ cubes, since the fact that $(\sum_{n=0}^\infty z^{n^3})^k = \sum_{n=0}^\infty r_k(n) z^n$, where $r_k(n)$ is the number of ways to write $n$ as a sum of $k$ cubes, can be exploited. But this is a general generating-functions property that has nothing special to do with the cubes. So my question is: Are there known other nice identities involving series similar to $\sum_n z^{n^k}$, for some integer $k \geq 3$? REPLY [2 votes]: One nice relation can be written as follow: $$\sum\limits_{n = - \infty }^ \infty z^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n z^{n^3}= \sum\limits_{n = - \infty }^ \infty (-1)^n z^{2n^3} \phi(z^{6n}) $$ $\quad z \in \mathbb{C},\; |z|=1 $ Where $$\phi(q)=\sum\limits_{m = - \infty }^ \infty (-1)^m q^{m^2}$$ More generally if we write $$\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}= \sum\limits_{n = - \infty }^ \infty \sum\limits_{m = - \infty }^ \infty (-1)^{n+m} z^{2n} q^{2(n^2+m^2)} h^{2n(n^2+3m^2)} \tag{1} $$ The proof of relation (1): $$F(z)=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$ $$F(-z)=\sum\limits_{n = - \infty }^ \infty (-z)^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n (-z)^n q^{n^2} h^{n^3}$$ $$F(-z)=\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n (-1)^n z^n q^{n^2} h^{n^3}$$ $$F(-z)=\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3}$$ Because of $F(-z)=F(z)$ , we can write that: $$F(z)=\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} A_n(q,h)$$ We need to find $A_n(q,h)$ $$\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} A_n(q,h)=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$ Let's use the transformation $z=ZQ^{2}h^{3}$ $q=Qh^{3}$ $$\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{2n} Q^{2n^2+4n} h^{2n^3+6n^2+6n} A_n(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^n Q^{n^2+2n} h^{n^3+3n^2+3n} .\sum\limits_{n = - \infty }^ \infty (-1)^nZ^n Q^{n^2+2n} h^{n^3+3n^2+3n}$$ If we multiply both side by $Z^2Q^2h^2$ $$\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{2n+2} Q^{2n^2+4n+2} h^{2n^3+6n^2+6n+2} A_n(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^{n+1} Q^{n^2+2n+1} h^{n^3+3n^2+3n+1} .\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{n+1} Q^{n^2+2n+1} h^{n^3+3n^2+3n+1}$$ $$\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{2(n+1)} Q^{2(n+1)^2} h^{2(n+1)^3} A_n(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^{n+1} Q^{(n+1)^2} h^{(n+1)^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nZ^{n+1} Q^{(n+1)^2} h^{(n+1)^3}$$ $$\sum\limits_{n = - \infty }^ \infty (-1)^{n-1}Z^{2n} Q^{2n^2} h^{2n^3} A_{n-1}(Qh^3,h)=\sum\limits_{n = - \infty }^ \infty Z^{n} Q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^{n-1}Z^{n} Q^{n^2} h^{n^3}$$ $$\sum\limits_{n = - \infty }^ \infty (-1)^{n}z^{2n} q^{2n^2} h^{2n^3} A_{n-1}(qh^3,h)=\sum\limits_{n = - \infty }^ \infty z^{n} q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^{n}z^{n} q^{n^2} h^{n^3}$$ $$A_n(q,h)=A_{n-1}(qh^3,h)$$ $$A_{-1}(q,h)=A_{0}(qh^{-3},h)$$ $$A_1(q,h)=A_{0}(qh^3,h)$$ $$A_2(q,h)=A_{1}(qh^3,h)=A_{0}(qh^6,h)$$ $$A_{-2}(q,h)=A_{-1}(qh^{-3},h)=A_{0}(qh^{-6},h)$$ $$A_3(q,h)=A_{2}(qh^3,h)=A_{1}(qh^6,h)=A_{0}(qh^9,h)$$ $$A_n(q,h)=A_{0}(qh^{3n},h)$$ We need to find $A_{0}(q,h)$ to complete the proof: We need to focus on $z^0$ terms $$F(z)=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$ If we multiply terms by terms for $z^0$ terms $$A_{0}(q,h)=\sum\limits_{m = - \infty }^ \infty (-1)^m q^{2m^2}$$ $$A_{n}(q,h)=A_{0}(qh^{3n},h)=\sum\limits_{m = - \infty }^ \infty (-1)^m q^{2m^2}h^{6nm^2}$$ Thus We can write that $$\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} \sum\limits_{m = - \infty }^ \infty (-1)^m q^{2m^2}h^{6nm^2}=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$ $$\sum\limits_{n = - \infty }^ \infty \sum\limits_{m = - \infty }^ \infty (-1)^{n+m}z^{2n} q^{2n^2+2m^2} h^{2n^3+6nm^2}=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$ $$\sum\limits_{n = - \infty }^ \infty (-1)^nz^{2n} q^{2n^2} h^{2n^3} \phi(q^2h^{6n}) =\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^nz^n q^{n^2} h^{n^3}$$ If we put that $z=1$ and $q=1$ $$\sum\limits_{n = - \infty }^ \infty (-1)^n h^{2n^3} \phi(h^{6n}) =\sum\limits_{n = - \infty }^ \infty h^{n^3} .\sum\limits_{n = - \infty }^ \infty (-1)^n h^{n^3} $$ Finally, the same method can be used for $\sum\limits_{n = - \infty }^ \infty z^{n^k}$ to get similiar identies where $k>3$<|endoftext|> TITLE: Example of a ring $R$ such that $\dim(R[[X]])<\dim(R[X])$ QUESTION [14 upvotes]: Dimension refers to the Krull dimension of a commutative ring. In the paper "Prime ideals in power series rings" J. Arnold gives an example of such a ring: Let $k$ be a field and $K=k(t)$ a simple transcendental extension of $k$. Suppose that $V=K+M$ is a discrete valuation ring with maximal ideal $M$. Let $D=k+M$. Then $\dim(D)=1$ and $\dim(D[X])=3$. Here is a proof: $M$ is the the only nonzero prime ideal of $D$. Thus $dim(D)=dim(D_M)=1$. For a $1$-dimensional integral domain $R$, $R[X]$ is $2$-dimensional iff every localization of the integral closure of $R$ is a valuation ring. Since $D$ is integrally closed and $D_M$ is not a valuation ring, $dim(D[X])=3$. Arnold claims that $\dim(D[[X]])=2$. If fail to see why this is true. Are there any other examples of such rings? REPLY [2 votes]: The ring $D$ given in the question indeed has $\dim(D[[X]])$=2. This is proved in detail in the paper "Power series rings over Pruefer domains" by J. Arnold.<|endoftext|> TITLE: Tauberian theorem with better error term QUESTION [11 upvotes]: This is a fairly vague question. Suppose we have a sequence of positive numbers $(c_n)_n$ and we want to find an asymptotic formula for $S(x) = \sum_{n \leq X} c_n$. In favorable circumstances, standard Tauberian theorems, e.g. the one in Appendix A of Chambert-Loir and Tschinkel's paper "Fonctions zeta des hauteurs des espaces fibres", available at http://www.math.nyu.edu/~tschinke/papers/yuri/01zeta/zeta.pdf, give an asymptotic formula of the form $X^a P(\log X) + O(X^{a - \delta})$, with $P$ a polynomial of degree $b-1$, with $b$ the order of pole at $a$ ($a>0$ is the right most pole). The leading term of the polynomial $P$ is explicit. Question. Can one explicitly compute the polynomial $P$? In my particular problem, I have a zeta function $Z(s)= \sum_n c_n.n^{-s}$ which I understand fairly well. One can show that $Z(s) = f(s) + h(s)$ with $f$ written in terms of Eisenstein series with a pole at $a=1$ of order $b=2$, and $h(s)$ harmless. In the Chambert-Loir–Tschinkel theorem $\delta = 1/2$ works, and I have an asymptotic formula of the form $(3/\pi) X \log X + B X + O(X^{1/2})$. Now I'm trying to find the value $B$. REPLY [6 votes]: If your zeta function $Z(s)$ has analytic continuation to the left of your pole $a$ (with some reasonable bounds etc.), then you can just use an inverse Mellin transform, no? The main term is the residue of $X^s Z(s)$ at $s=a$, so all you need is the Taylor series expansion of $X^s$ at $s=a$ (which gives the powers of $\log X$) and the Laurent series expansion of $Z(s)$ at $s=a$. Multiply the two expansions and pick the coefficient of $(s-a)^{-1}$.<|endoftext|> TITLE: Curvature of a principal bundle and the exterior covariant derivative QUESTION [6 upvotes]: I am sorry if this is too elementary; I had posted it on math.stack but no one answered. Let $P\to M$ a principal fibre bundle with fibre $G$, and let $A\in \Omega^{1}(P)\otimes\mathfrak{g}$ be a connection on $P$, where $\mathfrak{g}$ is the Lie algebra of $G$. Associated to every connection there is a curvature $F\in\Omega^{2}(P)\otimes\mathfrak{g}$ defined as $F = DA$ where $D\colon \Omega^{r}(P)\otimes\mathfrak{g}\to \Omega^{r+1}(P)\otimes\mathfrak{g}$ defined as $D\Omega(X_{1},\dots, X_{r+1}) = d\Omega(X^{H}_{1},\dots, X^{H}_{r+1})$. The superscript $H$ denotes the projection to the horizontal distribution given by the connection and $d\colon \colon \Omega^{r}(P)\to \Omega^{r+1}(P)$. Now, $F$ induces (using local sections), a two form $\mathcal{F}\in\Omega^{2}(M;\mathrm{ad} P)$ taking values in the adjoint bundle of $P$, which is a vector bundle of rank $dim\, \mathfrak{g}$. The two-form $\mathcal{F}$ satisfies $\mathcal{D}\mathcal{F} = d\mathcal{F} + [\mathcal{A},\mathcal{F}] = 0$ where $\mathcal{A}$ is the local form of the connection one-form $A$. The two-form $\mathcal{F}$ is a particular case of a form taking values in a vector bundle with a connection. For this kind of forms there is a natural derivative $d_{\nabla}$, the exterior covariant derivative. In the case of $\mathcal{F}$ the vector bundle is $\mathrm{ad}\, P$ and the connection $\nabla$ is the induced on the adjoint bundle by $A$. My question is, what is the relation between $\mathcal{D}$ and $d_{\nabla}$? Are they the same? Thanks. REPLY [4 votes]: First, there is a canonical isomorphism $\Phi$ between $\Omega^k(M; Ad P)$ and $\Omega^k_{Ad, h}(P; \mathfrak{g})$, where the subscripts signify that the form is horizontal and of type $Ad$ (i.e. equivariant). This isomorphism is canonical and does not need a local trivialization. To illustrate the idea, consider the case $k=0$. A section $\varphi$ of the adjoint bundle defines a function $\Phi(\varphi)$ on $P$ with values in $\mathfrak{g}$ by $\varphi(m) = [p, \Phi(\varphi) (p)]$ where $\pi(p) = m$. In your notation, $\Phi^{-1}(F) = \mathcal{F}$. Next, the exterior differential in the adjoint bundle (indeed in every associated bundle) is defined by making the following diagram commutative: $$\begin{matrix} \Omega^k_{Ad, h}(P; \mathfrak{g}) & \overset{D}{\to} & \Omega^{k+1}_{Ad, h}(P; \mathfrak{g}) \\ \downarrow & & \downarrow \\ \Omega^k(M; Ad P) & \overset{d_{\nabla}}{\to} & \Omega^{k+1}(M; Ad P) \end{matrix}$$ (the vertical arrows being the isomorphism $\Phi$) Thus $D F = 0$ coincides with $d_{\nabla}\mathcal{F} = 0$ under the isomorphism $\Phi$. Finally, there is also the viewpoint via local trivializations. Locally, we can identify the connection form with a Lie algebra valued 1-form $\tilde{A}$ on $M$ and the curvature is a 2-form $\tilde{F}$ on $M$ with values in the Lie algebra. It is an easy calculation to see that $D$ locally takes the form $\widetilde{D F} = d \tilde{F} + [\tilde{A}, \tilde{F}]$, i.e. the local representant of $DF$ is given by the right-hand side. Note that this expression only makes sense locally, since the differential of $\mathcal{F}$ as a vector-valued 2-form is not well-defined (without a connection). Nonetheless, $d_\nabla$ is always defined and can be applied to $\mathcal{F}$ without any problems.<|endoftext|> TITLE: Proof that no differentiable space-filling curve exists QUESTION [26 upvotes]: Could someone provide a reference or a sketch of a proof that no differentiable space-filling curve exists? Or piecewise differentiable? Must every continuous space-filling curve be nowhere differentiable? REPLY [2 votes]: Using the Sard's theorem. Assume that your curves exists. Then Each point in the domain is a critical point and so each point in the image is a critical value. But this is a contradiction with the sard's theorem since the set of critical values has measure zero.<|endoftext|> TITLE: What's an example of 2 elliptic curves with the same ground ring s.t. their associated cohomology theories detect different things? QUESTION [14 upvotes]: My understanding is that a complex-oriented spectrum is a ring spectrum $E$ with a map $MU \to E$. Analogously, a ring with a formal group law is a ring $R$ with a map $L \to R$ (where $L$ is the Lazard ring). This analogy can be made explicit (assuming that $L \to R$ is Landweber-exact for all primes): Let $E$ be an elliptic spectrum. We know the ring spectrum $MU$, and thus our choice of elliptic curve determines $E$. In other words, the formal group law (associated to the completion of our elliptic curve about the origin) gives us a map $L \to R$, which determines $E^*(-) \simeq MU^*(-) \otimes_{L} R$. Despite this relatively clear path from elliptic curve to cohomology theory, the following is frustratingly unclear to me! How does changing the elliptic curve actually affect what its associated cohomology theory detects? More explicitly: What's an example of 2 elliptic curves over the same ring s.t. their associated cohomology theories detect different things? REPLY [9 votes]: Let me try to be very concrete. (This is more or less just an elucidation of part of Charles's answer.) Consider the elliptic spectrum $TMF_0(2)$ with $\pi_*TMF_0(2) = \mathbb{Z}[\frac13][b_2,b_4,\Delta^{-1}]$. This is the modern name for the elliptic cohomology theory defined by Landweber, Ravenel and Stong; today one can define it as the $E_\infty$-ring spectrum $\mathcal{O}^{top}(\mathcal{M}_0(2))$ where $\mathcal{M}_0(2)$ is the moduli stack of elliptic curves with a level-$2$-structure. For simplicity, let us localize everything in sight at the prime $3$. We want to complete at one point corresponding to a height $1$ elliptic curve over $\mathbb{F}_3$ and at one point corresponding to a height $2$ elliptic curve over $\mathbb{F}_3$. An elliptic curve over $\mathbb{F}_3$ has height $2$ if and only if it has $j$-invariant $0$ in $\mathbb{F}_3$. (See either http://en.wikipedia.org/wiki/Supersingular_elliptic_curve or Hartshorne, Algebraic Geometry, 4.23.1.) The $j$-invariant of an elliptic curve $y^2 = x^3+ b_2x^2+b_4x$ is $(b_2^2-24b_4)/\Delta \equiv b^2/\Delta \mod 3$ where $\Delta = -8b_4^3$. Thus, the elliptic curves $$E_1: y^2 = x^3+x^2+x $$ and $$E_2: y^2 = x^3 + x$$ are of height $1$ and $2$ respectively. These elliptic curves correspond to the closed points $P_1$ and $P_2$ in $\mathcal{M}_0(2)$ defined by the homogeneous ideals $I_1 = (3, b_2-1, b_4-1)$ and $I_2 = (3, b_2, b_4-1)$ respectively. The completions $R_1 = TMF_0(2)^\widehat{}_{I_1}$ and $R_2TMF_0(2)^\widehat{}_{I_2}$ have both homotopy groups isomorphic to $\mathbb{Z}_3[[x]][u, u^{-1}]$ for a formal coordinate $x$ at the points $P_1$ and $P_2$ respectively. But $R_1$ and $R_2$ are very different as $R_1$ has height $1$ and $R_2$ has height $2$. In particular, $v_1$ is invertible on $R_1$, but it is not on $R_2$ (because in the first case $v_1$ is non-zero modulo the maximal ideal, in the second case it is zero). At the prime $3$, there is a $v_1$-self map on the Moore spectrum $S^0 /3$ (that is build as the cofiber of the map $S^0\xrightarrow{\cdot 3}S^0$. Denote the cofiber of this map $S^0/3 \xrightarrow{v_1} S^0/3$ by $X$. Then $(R_1)_*(X) = 0$, but $(R_2)_*(X) \not\cong 0$.<|endoftext|> TITLE: Notions of infinity in $\mathsf{ZF}$ without choice QUESTION [7 upvotes]: Consider the following statements about a given set $X$ in in $\mathsf{ZF}$: (1) There is $x_0\in X$ such that there is a surjective map $\varphi: X\setminus\{x_0\}\to X$. (2) There is an injective map $\iota:\mathbb{N}\to X$. It is easy to see that (2) implies (1) in $\mathsf{ZF}$, but are they equivalent? REPLY [14 votes]: No, they are not equivalent. It is a nice theorem that if there exists an infinite Dedekind-finite set (which is a set which satisfies the negation of (2)), then there is one which satisfies the first condition. If $D$ is a Dedekind-finite set, then $S(D)$ which is the set of all injective finite sequences from $D$ is also Dedekind-finite (because the sets are injective, every collection of them is uniformly enumerated, so if there was a countable infinite set of these sequences, their union would be a union of uniformly enumerated sets, which would be a countably infinite subset of $D$). Now simply consider the projection from $S(D)\setminus\{\varnothing\}$ onto $S(D)$ where you remove the last coordinate of the sequence. You might be interested in the following paper: Truss, J. Classes of Dedekind finite cardinals. Fund. Math. 84 (1974), no. 3, 187–208. PDF In which the author takes seven definition of finiteness (proposed by Azriel Levy) and investigates the relations between them. Your first condition is one of the properties considered there.<|endoftext|> TITLE: If $R$ is generated by idempotents, then $\text{Ann}(R)=0$? QUESTION [5 upvotes]: Let $R$ be a ring (not necessarily commutative or unital) that is generated by idempotents. I'd like to know if $\text{Ann}(R)=0$ must hold. Here I use $\text{Ann}(R)$ to denote the set of all elements $r\in R$ such that $rR=Rr=0$. All I knew is that it holds when $R$ is commutative. REPLY [9 votes]: No, $\mathrm{Ann}(R)$ does not necessarily hold when $R$ is generated as a ring by idempotents. Let $K$ be a field, or more generally any commutative (associative) ring with 1. Let $R$ be the (associative, non-unital) $K$-algebra of matrices $m(e,a,b,c)=\begin{pmatrix}0 & a & c\\0 & e & b\\ 0 & 0 & 0\end{pmatrix}$ with $a,b,c,e\in A$. Then $m(0,0,0,1)$ belongs to $\mathrm{Ann}(R)$. On the other hand, all elements of the form $m(1,a,b,ab)$ are idempotents, and the idempotents $m(1,0,0,0)$, $m(1,1,0,0)$, $m(1,0,1,0)$, $m(1,1,1,1)$ form a basis of $R$ as a $K$-module; in particular they generate $R$ as a ring (and even as an additive group) when $K=\mathbf{Z}$ or $\mathbf{Z}/n\mathbf{Z}$.<|endoftext|> TITLE: Constructing normal crossing varieties QUESTION [5 upvotes]: Let $X_i$ be a smooth projective variety with a smooth divisor $D_i$ for $i=1,2$. Suppose that $D_1$ is isomorphic to $D_2$. Then does it make sense to construct a normal crossing variety $X=X_1 \cup_D X_2 $ by 'pasting' $D_1$ and $D_2$, where '$\cup_D$' means pasting along $D_1$ and $D_2$? If the answer is yes. Suppose that there is ample divisor $H_i$ on $X_i$ such that $H_1|_{D_1}$ is lineary equivalent to $H_2|_{D_2}$. Then is the normal crossing variety $X$ necessarily projective? REPLY [6 votes]: Q1: Yes. For affine schemes, the gluing construction you need is just the fiber product of rings. Because gluing two affines in this way produces an affine, you can glue two schemes and get a scheme by taking an affine cover of each. Q2: Yes, and the proof is more or-less completely explicit. Choose a high enough $n$ such taht $\mathcal O(n H_i)$ is a projextive embedding of $X_i$ and $\mathcal I_{D_i} \otimes \mathcal O(n H_i)$ is globally generated with no higher cohomology, for both $i$. Then in the embedding $X_i \to \mathbb P^{N_i}$, $D_i$ is contained in some linear subspace cut out by the functions in $H^0(X_i, \mathcal I_{D_i} \otimes \mathcal O(n H_i))$. Because they generate, in fact $D_i$ is the intersect of $X_i$ with that linear subspace. Because the map $H^0(X_i, \mathcal O(nH_i)) \to H^0(X_i, \mathcal O_{D_i} (nH_i))$ is surjective, the emedding of $D_i$ into that linear subspace is just the embedding coming from the very ample line bundle $nH_i$. This embedding is the same whether $i=1$ or $2$. So $X_1$ and $X_2$ are expressed as two projective varieties with an isomorphism between their restrictions to two linear subspaces. We can easily embed the two projective spaces in a higher-dimensional projective space such that their intersection is exactly that linear subspace. Then the union of $X_1$ and $X_2$ in that higher space will clearly be $X_1 \cup_D X_2$, and hence it is a projective variety.<|endoftext|> TITLE: Monoidal structure on simplicial sheaves QUESTION [6 upvotes]: Let $\mathcal{C}$ be a site and let $sPh(\mathcal{C})_{proj}$ be the category of simplicial presheaves equipped with the projective model structure. This category is a closed monoidal model category (with the ordinary tensor product of functor $-\times -$, and an internal hom $sPh(\mathcal{C})[-,-]$) and a simplicial model category. Now let $sPh(\mathcal{C})_{proj, Cech}$ be the model structure obtained from the the global model structure on simplicial presheaves on $\mathcal{C}$ by left Bousfield localizations at Cech covers. Here my questions. 1) Is it again a simplicial model category? tensored and cotensored? 2) Is it again a closed monoidal model category? 3) In particular from nlab I know that for any cofibrant object $X$, the functor $X\times -$ is again a Quillen left in $sPh(\mathcal{C})_{proj, Cech}$ with right adjoint $sPh(\mathcal{C})[X,-]$. See http://ncatlab.org/nlab/show/model+structure+on+simplicial+presheaves#MonoidalStructure Now let $\{D_{i}\}\to D$ be a cover of $D\in\mathcal{C}$ and let $C\{D_{i}\}$ its Cech nerve. Let $A$ be a fibrant object in $sPh(\mathcal{C})_{proj, Cech}$, then does the map between cofibrant objects $$C\{D_{i}\}\to y(D), $$ where $y$ is the Yoneda embedding, induces a weak equivalence $$ sPh(\mathcal{C})[y(D),A]\to sPh(\mathcal{C})[C\{D_{i}\},A] $$ in $sPh(\mathcal{C})_{proj, Cech}$? I ask that because $C\{D_{i}\}\to y(D)$ is not in general a weak equivalence in $sPh(\mathcal{C})_{proj}$. REPLY [6 votes]: Yes. This is well-known and can be deduced, for example, from the general statements in the last section of this paper of Barwick. Take $V$ to be the symmetric monoidal model category of simplicial sets. Note that it is tractable because every object is cofibrant. By Corollary 3.33, the projective model structure on simplicial presheaves in $C$ is symmetric monoidal. Then by Theorem 3.36 the Cech-local projective model structure is simplicial, and by Theorem 3.38 it is symmetric monoidal.<|endoftext|> TITLE: (Non)existence of mirrors with more than two foci QUESTION [14 upvotes]: Do there exist any mirrors $M$ in $d$-dimensional Euclidean space $\mathbb{R}^d$ for which there exist three different points $x_1$, $x_2$, $x_3 \in \mathbb{R}^d$ such that if any ray of light passes through one of the points $x_i$ it automatically passes through the other two after a number of reflections on $M$? The question needs some formalisation for what is meant by $M \subset \mathbb{R}^d$ and “rays of light”, although it is easy to guess what they should be: $M$ is a $(d-1)$-dimensional differentiable submanifold of $\mathbb{R}^d$. A ray of light is a continuous function $g:\mathbb{R} \rightarrow \mathbb{R}^d$ for which $g^{-1}(g(\mathbb{R}) \cap M)$ is a discrete set. Moreover, $g$ is locally smooth, linear and unit speed at any point $t\in \mathbb{R}$ for which $g(t) \notin M$: $\left\|g'(t)\right\|=1$ and g''(t)=0. At any point $t^* \in \mathbb{R}$ for which $g(t^*)\in M$ we require $g$ to satisfy the reflection law: denoting $a=\lim_{t \to t^* -} g'(t)$ and $b = \lim_{t \to t^* +} g'(t)$ \begin{equation} b + a \perp n(g(t)) \end{equation} and \begin{equation} b - a \propto n(g(t)) \end{equation} (n is a local unit-normal vector field on $M$). REPLY [12 votes]: Here is a way that works for a set of full measure in any dimension, for any number of points, not just $3$. We can construct it for dimension $2$, then rotate it about the $x$-axis. Let the points be $x_i=(i,0)$ for $i=1,...,n$. Attach semi-ellipses $\gamma_i$ whose major axes and end points are along the $x$-axis whose foci are $x_i,x_{i+1}$ so that for $i$ odd we take the intersection of the ellipse with the upper half plane, and for $i$ even we take the intersection with the lower half plane. Extend this by adding semicircles $\gamma_0$ and $\gamma_n$ about the end points. If we choose the semiellipses so that the total distance to $x_i$ and $x_{i+1}$ is $2$, and use semicircles with radius $1/2$, this is even a simple closed curve, although there are points where it is not smooth. Light passing through $x_1$ going up bounces off $\gamma_1$ and is reflected to $x_2$ going down, so it reflects off $\gamma_2$ to $x_3$ going up, etc. After it passes through $x_n$ it bounces off $\gamma_n$ and reverses. Light passing through $x_1$ going down bounces off $\gamma_0$ and reverses through $x_1$, going up. The behavior of the horizontal ray through the nonsmooth points and all $x_i$ is not really defined, but you could argue that it passes through all of the points, too. Mathematica code for drawing this picture: ell[i_, x_] := (-1)^(i + 1) Sqrt[3/4 (1 - (x - (i + 1/2))^2)] circ[i_, x_] := Sqrt[1/4 - (x - i)^2] Plot[{-circ[1, x], ell[1, x], ell[2, x], ell[3, x], ell[4, x], circ[5, x]}, {x, 0, 6}, PlotRange -> {-1.5, 1.5}, AspectRatio -> 1/2]<|endoftext|> TITLE: Relating different topologies on $C^{\infty}_c(M)$ QUESTION [8 upvotes]: This is somehow connected to this question. I can think of at least four topologies to put on $C_c(M)$: Topologize $C^{\infty}_c(M)\subseteq C^{\infty}(M)$ as a subspace with the weak Whitney $C^\infty$ topology on $C^{\infty}(M)$. Topologize $C^{\infty}_c(M)\subseteq C^{\infty}(M)$ as a subspace with the strong Whitney $C^\infty$ topology on $C^{\infty}(M)$. Topologize $C^{\infty}_c(M)=colim_{K\subseteq M} C^{\infty}_K(M)$ as direct limit over compact subsets $K\subseteq M$ where $C^{\infty}_K(M)$ are the functions, which have support in $K$ with the Whitney topology. (The strong and the weak one should coincide in that case.) Take $h\in C^{\infty}_c(M)$ a smooth function $\epsilon\colon M\rightarrow (0,\infty)$, vector fields $X_1,...,X_k$ and declare the subsets of the shape $\{g\in C^{\infty}_c(M)\colon\text{ }|X_1..X_k(h-g)(x)|<\epsilon(x)\forall x\in M\}$ as a subbase varying over $h,k\in\mathbb{N}_0, X_1,...,X_k$ and $\epsilon$. How are these topologies related? My vague guesses are: Topology 2.) is finer than topology 1.), but those are in general not equal. Topology 2.) is finer than topology 3.), but in general not the same. I have no idea how to relate topology 4.) to the others. REPLY [2 votes]: A lot of detailed information is also contained in sections 3 and 4 of Peter W. Michor: Manifolds of differentiable mappings. Shiva Mathematics Series 3, Shiva Publ., Orpington, (1980), iv+158. (pdf)<|endoftext|> TITLE: Finding joint probability from double marginals QUESTION [5 upvotes]: Consider three probability distributions in the form $p_1(y,z),p_2(x,z),p_3(x,y)$. When does a global joint probability $p(x,y,z)$ (possibly not unique) exist? The first compatibility condition to check is of course that the first order marginals check out: $p_2(x)=p_3(x)$, and so on. Is this the only condition, is it necessary and sufficient? Where can I find it? Thanks! PS. I would also be curious about what happens in any order, not just 2 and three...if it's possible! Note: cross-posting from MSE. REPLY [4 votes]: The problem is one of linear programming. Indeed, suppose that $X$, $Y$, and $Z$ are finite sets. Then the problem is whether there exist nonnegative real numbers $p(x,y,z)$ such that $\sum_{x\in X}p(x,y,z)=p_1(y,z)$ for all $(y,z)\in Y\times Z$, $\sum_{y\in Y}p(x,y,z)=p_2(x,z)$ for all $(x,z)\in X\times Z$, and $\sum_{z\in Z}p(x,y,z)=p_3(x,y)$ for all $(x,y)\in X\times Y$. More generally, this may be a problem of infinite-dimensional linear programming. The fundamental paper "The Existence of Probability Measures with Given Marginals" by Strassen (1965) in The Annals of Mathematical Statistics deals with the existence of a probability measure $\mu$ on the product space $X\times Y$ given the marginals $\mu\pi_X^{-1}$ and $\mu\pi_Y^{-1}$ (where $\pi_X$ and $\pi_Y$ are the projections from $X\times Y$ to $X$ and $Y$, respectively) plus further affine restrictions on $\mu$. At least in principle, the case of the product of more than two spaces should be reducible to Strassen's setting. For instance, suppose that, given three spaces $X$, $Y$, and $Z$ with probability measures $\mu_1$, $\mu_2$, and $\mu_3$ over $Y\times Z$, $X\times Z$, and $X\times Y$, respectively, one has to say whether there is a measure $\mu$ over $X\times Y\times Z$ such that $\mu\pi_{Y\times Z}^{-1}=\mu_1$, $\mu\pi_{X\times Z}^{-1}=\mu_2$, and $\mu\pi_{X\times Y}^{-1}=\mu_3$, where $\pi_{Y\times Z}$ is the projection from $X\times Y\times Z$ to $Y\times Z$, etc. This problem can be restated as follows. Let $U:=Y\times Z$. Then one has to say whether there is a measure $\mu$ over $X\times U$ such that $\mu\pi_X^{-1}=\mu_2\pi_{X\times Z\to X}^{-1}$ and $\mu\pi_{X\times U\to U}^{-1}=\mu_1$, with the additional affine restrictions specifying the $\mu$-distributions of the maps $X\times U\ni(x,u)\mapsto(x,\pi_{U\to Y}u)$ and $X\times U\ni(x,u)\mapsto(x,\pi_{U\to Z}u)$, where $\pi_{A\to B}$ is the projection from $A$ to $B$.<|endoftext|> TITLE: Consecutive numbers with mutually distinct exponents in their canonical prime factorization QUESTION [10 upvotes]: Is it possible to find 23 consecutive positive integers each of which has mutually distinct exponents in its canonical prime factorization? Such numbers are sequence A130091 in OEIS. 24 such numbers are impossible because of $36n-6$ and $36n+6$. REPLY [9 votes]: We prove that there are only finitely many such intervals. Suppose $[36n + 7, 36n + 29]$ is one such interval. Now, $36n + 10$ and $36n + 15$ cannot both be divisible by $5$ (since one must occur with multiplicity $1$), hence $n \neq 0 \mod{5}$. We can say the same about $36n + 21$ and $36n + 26$ (so $n \neq 4 \mod{5}$). And about $36n + 14$ and $36n + 24$ (so $n \neq 1 \mod{5}$). And indeed about $36n + 12$ and $36n + 22$ (so $n \neq 3 \mod{5}$). This implies that $n \equiv 2 \mod{5}$, so $36n + 18$ is divisible by $5$ (and therefore by $25$). So our set must actually be of the form: $$[900m + 439, 900m + 461]$$ Irrelevant aside: We can do a similar analysis modulo $7$, enabling us to deduce that precisely one of $900m + 445$, $900m + 450$ and $900m + 455$ is divisible by $49$. This doesn't seem to help as much, though. Indeed, no generalisation of this argument will work, because only the primes $p \leq 23$ are pertinent (we can just shift the interval with the help of the Chinese Remainder Theorem to avoid large primes completely), and it's possible to find a $23$-element interval in which the multiplicities of each of the primes $2, 3, 5, 7, 11, 13, 17, 19, 23$ are distinct for each element (left as an exercise to the reader). Let us return to the task of proving that there are only finitely many such intervals. For each interval, either: $900m + 440$ and $900m + 456$ are both congruent to $4 \mod{8}$. $900m + 444$ and $900m + 460$ are both congruent to $4 \mod{8}$. In which case we take those two numbers and divide them by $4$ to obtain odd coprime integers $b, b+4$. One of them is divisible by $3$ (but not $3^2$) and the other is divisible by $5$ (but not $5^2$). Any other primes dividing them must do so with multiplicity $\geq 3$, because $2$ divided each of the original numbers with multiplicity $2$. Whence $(4, b, b+4)$ has a radical less than $k b^{2/3}$ for some universal constant $k$ which I can't be bothered to calculate. Since the $abc$ conjecture is true, it follows there are only finitely many intervals with your property.<|endoftext|> TITLE: Combinatorial identity and Fuss-Catalan numbers QUESTION [5 upvotes]: I would like to show that $$ \lim_{N\to\infty}\frac{1}{N^{np+1}}\frac1{p!}\sum_{j=0}^{p-1}(-1)^j\binom{p-1}{j} \left(\frac{\Gamma(N+p-j)}{\Gamma(N-j)}\right)^{n+1} =\frac1{np+1}\binom{(n+1)p}{p}, $$ for $p,n=1,2,\ldots$. Background The reason we expect this equality to hold is the following: The left hand side appear in [arXiv:1307.7560] as the moments of the squared singular values of a product of $n$ Gaussian non-Hermitian $N\times N$ matrices, while the right hand side (Fuss-Catalan numbers) is the asymptotic prediction obtained using techniques from free probability, see e.g. [arXiv:0710.5931]. Thus the results should agree at leading order in $N$. REPLY [4 votes]: Note that $$ f(N):=\left(\frac{\Gamma(N+p)}{\Gamma(N)}\right)^{n+1} $$ is a unitary polynomial in $N$ of degree $pn+p$: $f(N)=N^{pn+p}+\dots$. Its $(p-1)$-st finite difference $$ \Delta^{p-1}f(N)=\sum_{j=0}^{p-1} (-1)^j\binom{p-1}{j} f(N-j) $$ is a polynomial in $N$ of degree $(pn+p)-(p-1)=pn+1$ and leading coefficient $(pn+p)(pn+p-1)\dots(pn+p-(p-2))$. It yields your limit relation. REPLY [3 votes]: From taking derivatives of $(x-1)^{p-1}$ and setting $x = 1$, we have $$\sum_{i=0}^{p-1} j^k (-1)^j \binom{p-1}{j} = 0$$ for $k < p - 1$. The same method gives $$(-1)^{p-1}(p- 1)! = \sum_{i=0}^{p-1} j^{p-1} (-1)^j \binom{p-1}{j}.$$ Now consider $$\left(\frac{\Gamma(N+ p - j)}{\Gamma(N - j)}\right)^{n+1} = (N + p - j - 1)^{n+1}\cdot \dots \cdot(N - j)^{n+1}.$$ This can be considered a polynomial in two variables: $j$ and $N$. Consider a term $Kj^aN^b$ in the decomposition into monomials. If $a < p - 1$, then the term will cancel via the above identity. If $b < np + 1$, then the term will contribute a negligible amount asymptotically. Then we can focus on $a = p - 1$, $b = np + 1$. But the $j^{p-1}N^{np+1}$ term of this polynomial will equal the $j^{p-1}N^{np+1}$ term of the homogenous $(N - j)^{(n+1)p}$. The coefficient of the term is $(-1)^{p-1}\binom{(n+1)p}{np+1}$. Combining with the above identity, we find that the left hand side is $$\frac{(p-1)!}{p!} \binom{(n+1)p}{np+1} + O(N^{-1}) = \frac{1}{np+1} \binom{(n+1)p}{p} + O(N^{-1})$$ which gives you what you need.<|endoftext|> TITLE: Singular/Smooth locus of Schubert variety of the affine grassmannian QUESTION [5 upvotes]: Let $G$ be a connected, simply connected, semisimple, complex linear algebraic group with maximal torus $T$ and affine Grassmannian $\mathcal Gr$. It is well known that $\mathcal Gr$ admits a Bruhat decomposition $$ \mathcal Gr = \bigsqcup_{\lambda\in X_*(T)} \mathcal B \lambda $$ where $\mathcal B$ is the Iwahori subgroup, and the union is taken over coweights of $T$. Let $X_\lambda = \overline{\mathcal B\lambda}$ be the corresponding Schubert variety. What is the singular/smooth locus of $X_w$? It is known that under the Bruhat order, $$ X_\lambda = \overline{\mathcal B\lambda} = \bigsqcup_{\mu\leq \lambda} B\mu,$$ and I have some sneaking suspicion that the smooth locus should just be $\mathcal B\lambda$, but I cannot think of how to conclude that this is the case. REPLY [7 votes]: The smooth locus of a spherical orbit (i.e. $G(\mathcal O)$-orbit) on the affine Grassmannian is just the spherical orbit itself (but this orbit consists of several Iwahori-orbits). See Malkin-Ostrik-Vybornov, The minimal degeneration singularities in the affine Grassmannians, Duke Math. J. 126 (2005), no. 2, 233–249 for a more precise result: they describe the singularity of each orbit closure along each spherical orbit open in the boundary. They quote Evens-Mirković Characteristic cycles for the loop Grassmannian and nilpotent orbits, Duke Math. J. 97 (1999), no. 1, 109–126 for the result that the smooth locus of a spherical orbit is just the orbit itself. I have an alternative modular representation-theoretic fun proof using the geometric Satake correspondence (see a preprint of mine from 2008). So the answer to your question was clearly no as stated on general grounds: for a general Kac-Moody Schubert variety, if w is maximal modulo some finite parabolic subgroup $W_I$, then the smooth locus of $\overline{BwB}/B$ contains $P_IwB$, which is the union of the Schubert cells corresponding to the elements of the coset $W_I w$, so more than one if $W_I$ is nontrivial (for $W$ finite, the case of the full flag variety is $W_I = W$). If one considers $G/P$ instead of $G/B$, I guess one has to consider double cosets... But the result above tells you that for $\lambda$ dominant, the smooth locus does not get larger than that (considering the finite Weyl group as a maximal parabolic subgroup of the affine Weyl group). I don't know the answer for general $\lambda$, but there is an extensive literature about smooth (resp. rationally smooth, even p-smooth) loci of Schubert varieties. A similar result holds for nilpotent cones: the smooth locus of a nilpotent orbit is just the orbit itself (this is a consequence of results by Namikawa and Kaledin; again, one can describe all minimal degeneration singularities: this was done by Kraft-Procesi in classical types, and by Fu-J.-Levy-Sommers in exceptional types).<|endoftext|> TITLE: Rate of convergence of Bayesian posterior QUESTION [7 upvotes]: Suppose a data generating process (DGP) is parameterized by some unknown parameter $\theta_0$, say $P_{\theta_0}$, and we want to estimate the value of $\theta_0$ using Bayesian method. Let $\pi(\theta)$ be the prior over the possible values of $\theta$, $\Theta$. I understand that for almost all priors, if the data observed are iid, then the Bayesian posterior will eventually concentrate on $\theta_0$, as the number of observations ($n$) tends to infinity. (Correct me if I'm wrong.) My question: Are there any results that predict differential rates of convergence of the posterior distribution based on different priors? For example, suppose $\Theta=\{\theta_0,\theta_1\}$ and the DGP is $P_{\theta_0}$. Consider two priors on $\Theta$: $$\pi_1(\theta_0)=0.2,\qquad \pi_1(\theta_1)=0.8$$ and $$\pi_2(\theta_0)=\pi_2(\theta_1)=0.5.$$ Are there any results that says the posterior based on $\pi_2$ converges faster than that based on $\pi_1$? (or the other way around?) Intuitively I would expect the posterior based on $\pi_2$ to have faster convergence, as it is "closer" to $\theta_0$. But in my experience, intuition is hardly reliable when it comes to probability theory. Note: I've asked this quesiton on Math.SE but received no answer. I thought I would try my luck here. REPLY [4 votes]: One can measure the rate of convergence of the posterior distribution with density $p_n$ to the Dirac probability distribution at $\theta_0$ by how large the ratios $p_n(\theta_0)/p_n(\theta)$ are for $\theta\ne\theta_0$, where $\theta_0$ is the "true" value of the parameter. One has $$ \frac{p_n(\theta_0)}{p_n(\theta)}= \frac{\pi(\theta_0)}{\pi(\theta)}\,\frac{L_n(\theta_0)}{L_n(\theta)}, $$ where $\pi$ is the prior density and $L_n(\theta):=\prod_1^n f_\theta(X_i)$ is the likelihood value based on the first $n$ of the iid observations $X_1,X_2,\dots$ with common density $f_{\theta_0}$. It is clear from the above display that, for the same "data" $X_1,X_2,\dots$, the greater is the prior ratio $\pi(\theta_0)/\pi(\theta)$ the greater is the posterior "contrast" ratio $p_n(\theta_0)/p_n(\theta)$. In particular, in the extreme case when (say) $\pi(\theta)$ is $1$ if $\theta=\theta_0$ and is $0$ otherwise, the prior ratio is $\infty$ for any $\theta\ne\theta_0$, and hence so is the "contrast" ratio $p_n(\theta_0)/p_n(\theta)$. One may want to measure the posterior contrast by one number, say by averaging the posterior contrast over the data space and over the parameter space, to get something like $$\int\mu(d\theta)\,\mathsf{E}\ln\frac{p_n(\theta_0)}{p_n(\theta)}= \int\mu(d\theta)\,\ln\frac{\pi(\theta_0)}{\pi(\theta)} +n\int\mu(d\theta)\,\mathsf{E}\ln\frac{f_{\theta_0}(X_1)}{f_{\theta}(X_1)},$$ where $\mu$ is a measure on the parameter space. So, the greater is the average prior contrast $\int\mu(d\theta)\,\ln\frac{\pi(\theta_0)}{\pi(\theta)}$ the greater is the average posterior contrast $\int\mu(d\theta)\,\mathsf{E}\ln\frac{p_n(\theta_0)}{p_n(\theta)}$. In particular, your prior $\pi_2$ should result, by this logic, in a greater posterior contrast than $\pi_1$. Addendum: The average prior contrast and hence the average posterior contrast may behave somewhat surprisingly, though. E.g., suppose that the parameter space is $\Theta=\mathbb{R}$, $\theta_0=0$, $\pi$ is the normal density with mean $\theta_1$ and variance $\tau^2$, $f_\theta$ is the normal density with mean $\theta$ and variance $\sigma^2$, and $\mu$ is the normal distribution with mean $0$ and variance $\gamma^2$. Then for the average prior and posterior contrasts one has $$\int\mu(d\theta)\,\ln\frac{\pi(\theta_0)}{\pi(\theta)}=\frac{\gamma^2}{2\tau^2}\quad\text{and}\quad \int\mu(d\theta)\,\mathsf{E}\ln\frac{p_n(\theta_0)}{p_n(\theta)}= \frac{\gamma^2}2\Big(\frac1{\tau^2}+\frac n{\sigma^2}\Big),$$ respectively. As could be expected, both these average contrasts are the greater the smaller is the variance $\tau^2$ of the prior distribution. However, neither of these contrasts depends on the prior mean $\theta_1$ ! This will be so for any distribution $\mu$ symmetric about $0$ (or, more generally, about $\theta_0$) -- because, for the normal prior $\pi$ as in this addendum, the symmetrized prior contrast $$\ln\frac{\pi(0)}{\pi(\theta)}+\ln\frac{\pi(0)}{\pi(-\theta)} =\frac{\theta^2-2\theta \theta _1}{2 \tau ^2} +\frac{\theta^2+2\theta \theta _1}{2 \tau ^2} =\frac{\theta^2}{\tau^2}$$ does not depend on $\theta_1$.<|endoftext|> TITLE: When is the category of small (pre)sheaves a(n elementary) topos? QUESTION [10 upvotes]: When $C$ is essentially small, the presheaf category $[C^\mathrm{op},\mathsf{Set}]$ is the free cocompletion of $C$. The presheaf category $[C^\mathrm{op},\mathsf{Set}]$ is also a topos. When $C$ is truly large, the presheaf category $[C^\mathrm{op},\mathsf{Set}$] is not even locally small. But the full subcategory $\mathcal{P}C \subset [C^\mathrm{op},\mathsf{Set}]$, consisting of the presheaves which are small colimits of representables, is still the free cocompletion of $C$. When $C$ is small, $\mathcal{P}C$ is the whole category $[C^\mathrm{op},\mathsf{Set}]$, so it is a topos. But when $\mathcal{C}$ is large, is $\mathcal{P}C$ a topos? Day and Lack, following Rosicky, study when $\mathcal{P}C$ is complete, and when it is cartesian closed. Certainly cartesian closedness is necessary to be a topos; I believe that a cocomplete topos is also complete, so that completeness is also necessary. But I suspect that $\mathcal{P}C$ might only have a subobject classifier if $\mathcal{C}$ is essentially small. If it makes a difference to restrict to sheaves in some Grothendieck topology, that would be interesting to know. I suppose my motivating example is when $C$ is the category of schemes (or just affine schemes). Algebraic geometers are very interested in sheaves on this category in various topologies; I suspect most such interesting sheaves are small, but I don't know if that restriction gives one a topos to work with. EDIT As David Carchedi points out in the comments, I should emphasize that I'm interested in when $\mathcal{P}C$ is an elementary topos (a finitely complete, cartesian closed category with a subobject classifier). As Eric Wofsey argues in the comments, if $C$ is essentially large, then $\mathcal{P}C$ is never a Grothendieck topos (it doesn't have a small generating set). EDIT The weakest of observations: the Yoneda embedding $C \to \mathcal{P}C$ is still continuous, so preserves monos, and is fully faithful; so it induces injections on subobject lattices. So if $C$ is not well-powered to begin with, then $\mathcal{P}C$ is also not well-powered, so can't be a topos (given that it's locally small: look at the size of $\mathrm{Hom}(a,\Omega)$). EDIT It looks like the canonical way to give a general answer to this problem would be to generalize the work of Menni which characterizes when the exact completion of a finitely complete category is a topos. Taking the category of small sheaves is put in the same framework as exact completions in Mike Shulman's paper which Adeel linked to in the comments. So the way forward ought to be clear. I'm not sure how easy criteria along Menni's lines would be to check in particular cases like schemes, though. REPLY [5 votes]: As you seem to suspect, $\mathcal{P}(C)$ is rarely a topos if $C$ is large. As a simple example, if $C$ is large discrete (identified with a large set $X$) and $\mathbf{Set}$ is the category of small sets, then $\mathcal{P}(C)$ is equivalent to $\mathbf{Set}/X$, which does not have a terminal object. You are correct that a cocomplete topos $E$ is complete. For that we would just need small products. If $\{X_i\}_{i \in I}$ is a small family of objects of $E$, then under cocompleteness there is an obvious fibered object $p: \sum_i X_i \to \Delta I$ where the codomain $\Delta I$ is the coproduct of $|I|$ copies of the terminal object $1$. The object of sections of $p$, namely $\Pi_! p$ where $!: \Delta I \to 1$ is the unique map and $\Pi_!: E/\Delta I \to E/1 \simeq E$ is right adjoint to the pullback $I^*: E \to E/\Delta I$, is the product $\prod_i X_i$ in $E$. So completeness is certainly a necessary condition for $\mathcal{P}(C)$ to be a topos. Much of the study of small cocompletions is bound up with the study of exact completions. Just to cite one result (Lemma 3 in this paper of Rosický): if $C$ is finitely complete, then $\mathcal{P}(C)$ is equivalent to $(\Sigma C)_{ex}$ where $\Sigma C$ is the coproduct completion of $C$ and the subscript denotes exact completion. One expert on properties of exact completions, besides Mike Shulman, is Matías Menni, and so you should probably consult some of his papers for properties of exact completions, starting with his thesis. See also here.<|endoftext|> TITLE: "Epicycles" (Ptolemy style) in math theory? QUESTION [17 upvotes]: By analogy: The epicycles of Ptolemy explained the known facts in the sun system and in this sense were not "wrong". But they distracted from a better insight. From another viewpoint, everything fell neatly into place. Do you have examples from pure math? Again, you have a theory, it's OK in the sense that it gives the answers you want, but it's hopelessly contrieved, and suddenly someone sees the stuff from another viewpoint and everything becomes ridiculously simple? It's quite probable a comparable question with other wording has already been asked, a link then would suffice. :-) (QUICK EDIT as long as it's hot: While I already upped an answer with nice examples, I also would be interested which "current" theories you see as epicyclic and in need of a Kopernikus/Kepler/Newton. Remember, soft question, a lot of esthetics is involved in the judgment.) REPLY [2 votes]: Hamilton's quaternions made physical calculations a good deal easier, but vector calculus and vector algebra have proven to be far superior tools in mathematical physics.<|endoftext|> TITLE: Applications of small Kakeya sets over finite fields QUESTION [6 upvotes]: It was proved by Dvir that a Kakeya set in $\mathbb{F}_q^n$ has size at least $q^n/n!$, a bound which was later improved to $q^n/2^n$. For $n = 2$ and $q$ odd the exact bound is $q(q+1)/2 + (q-1)/2$ as given by Blokhuis and Mazzocca in [1], where they also classified the sets attaining this lower bound. For $q$ even the bound is $q(q+1)/2$ and all Kakeya sets attaining this bound are known. Some more examples of "small" Kakeya sets are given in [2] and [3]. My question is, what are some possible applications of constructing these small Kakeya sets in $\mathbb{F}_q^2$? Moreover, what is the state of the art for $n > 2$? What are the best known bounds and the examples that achieve those bounds? Would it be worthwhile to construct explicit examples attaining the bounds there? (which is of course subjective) [1] A. Blokhuis and F. Mazzocca. The finite field kakeya problem. In Building Bridges, pages 205–218, 2008. http://link.springer.com/chapter/10.1007%2F978-3-540-85221-6_6 [2] A. Blokhuis, M. De Boeck, F. Mazzocca, L. Storme. The Kakeya problem: a gap in the spectrum and classification of the smallest examples. Des. Codes Cryptogr., 72 (1) (2014), 21–31. [3] J. M. Dover and K. E. Mellinger. Small Kakeya sets in non-prime order planes. European J. of Combin. Volume 47 (2015), pages 95–102. REPLY [4 votes]: I became interested in Kakeya sets because they have the interesting property that a Kakeya set in a projective plane cannot be a subset of a blocking set, and with the exception of the full plane with two lines removed, Kakeya sets are the only sets with this property. As far as state of the art in higher dimensions, not much seems known. The article http://onlinelibrary.wiley.com/doi/10.1002/jcd.21507/full, by Maarten De Boeck makes some strides in 3-space, and is an interesting paper. UPDATE: Shouldn't post so late in the day. I should have said that the affine part of a Kakeya set has the property of not being contained in a blocking set, and that they are almost the only sets not containing a line with that property.<|endoftext|> TITLE: Mathematicians wearing hats on arbitrary total orders QUESTION [42 upvotes]: I've been pondering the following generalisation of a famous problem (the special case where $T = \mathbb{N})$: Question: We have some totally-ordered set $T$ of mathematicians, each wearing a hat which is either red or blue. The mathematician $x$ can see the colour of $y$'s hat if and only if $x < y$. The mathematicians each simultaneously guess the colour of his or her own hat. Under what conditions on $T$ can the mathematicians pre-arrange a strategy such that, independently of the colouring, only finitely many guess incorrectly? I've made the following progress, which suggests the problem is quite tractable: Clearly $T$ must be well-ordered, since otherwise we would have an infinite decreasing sequence $a_1 > a_2 > a_3 > \cdots$ of mathematicians. Then we choose all other hat colours arbitrarily (for concreteness, all blue) and sequentially set the hat colour of $a_i$ to make the mathematician guess incorrectly. So we're only interested in the case where $T$ is an ordinal. Clearly if $T' > T$, then a winning strategy for $T'$ implies a winning strategy for $T$. So the problem reduces to: What is the minimal ordinal $\alpha$ such that the mathematicians do not have a winning strategy? It is trivial that $\alpha \geq \omega$, since for finite ordinals, it is impossible for the mathematicians to lose. It's not too difficult (indeed, it is a famous exercise) to show in ZFC that the mathematicians have a winning strategy for $\omega$. This implies $\alpha \geq \omega^2$ since the sum of two 'winning ordinals' is necessarily another winning ordinal (so $\alpha$ is a power of $\omega$). So far I have no upper bound on $\alpha$ (indeed, it could be the case that the mathematicians have a winning strategy for any ordinal, in which case $\alpha$ is undefined), and I have no lower bound beyond $\omega^2$. Thus an easier (and much more concrete, since you only have two possible answers instead of a proper class) question is: Do the mathematicians have a winning strategy for $\omega^2$? REPLY [48 votes]: It's a great problem! Theorem. The mathematicians have a winning strategy in the game for every ordinal $\alpha$. Proof. Let's prove the theorem by transfinite induction. Suppose that the mathematicians have winning strategies in the games of any particular length $\beta$ less than $\alpha$, and let us fix agreed-upon strategies for those games. Consider now the case of $\alpha$ many mathematicians. I shall describe a winning strategy. In the space of all possible assignments of hat colors to the $\alpha$ many mathematicians, define that two such assignments are equivalent $s\sim t$, if they agree from some ordinal onward; that is, they agree on a tail segment $[\beta,\alpha)$ of $\alpha$. Let the mathematicians agree on a choice of representative for each $\sim$-equivalence class. Now, suppose that the hats are given out. Each mathematician (except the last one, in the case that $\alpha$ is a successor ordinal) is able to observe a tail segment of the actual assignment given, and thus knows the $\sim$-class of the actual assignment. Let the mathematician at any particular position $\gamma$ compare the observed assignment beyond $\gamma$ with the assignment of the pre-chosen representative from that class. If they agree perfectly beyond $\gamma$, then let mathematician $\gamma$ announce the color of the hat that he or she would wear according to the chosen representative assignment. Otherwise, mathematician $\gamma$ observes some errors at positions beyond $\gamma$. Let $\beta$ be the supremum of the positions of these errors, so that $\gamma<\beta$ but also $\beta<\alpha$ since the observed assignment definitely agrees with the representative from some point on. In this case, let mathematician $\gamma$ ignore the part of the assignment beyond $\beta$, and instead use the fixed strategy for the game of assignments of length $\beta$, using only the information about the hats up to $\beta$. Notice that if $\beta$ is the supremum of the places where the actual assignment differs from the representative, then everyone beyond $\beta$ will guess correctly, and everyone before $\beta$ will compute $\beta$ correctly and therefore use the agreed-upon strategy for the length $\beta$ game. So by the induction hypothesis, only finitely many will be wrong. QED Let me also describe another strategy, in the style of Alan Taylor, from what I recall from a talk he gave at our seminar in New York several years ago. See also Christopher S. Hardin and Alan D. Taylor, A Peculiar Connection Between the Axiom of Choice and Predicting the Future, which was mentioned in the comments. We prove directly that there is a strategy for $\alpha$ many mathematicians. First of all, let the mathematicians agree upon a fixed well-ordering $\triangleleft$ of the space of all hat $\alpha$-assignments. Now, suppose the hats are given out. Let each mathematician observe the portion of the hat assignment given to the mathematicians ahead, and let each mathematician compute the $\triangleleft$-least total assignment that agrees with the portion of the actual assignment that they observe. Each mathematician should predict that their own hat color is the same as it in in the $\triangleleft$-least assignment that they compute. We now argue that only finitely many mathematicians are wrong. The main point is that if $\gamma<\beta$, then the $\triangleleft$-least assignment that mathematician $\gamma$ computes is at least as high in the $\triangleleft$ order as the $\triangleleft$-least assignment that mathematician $\beta$ computes, since mathematician $\gamma$ observes all the information that $\beta$ observes and more about the actual assignment of hats. That is, as you move up higher in the mathematicians, the computed $\triangleleft$-least approximation can only move down in the $\triangleleft$ order if it changes. Every time there is an incorrect guess as we move up in the mathematicians, we drop strictly lower in the $\triangleleft$-well-ordering. And since $\triangleleft$ is a well-order, this can happen only finitely many times. Thus, only finitely many mathematicians guess incorrectly. The argument is extremely general, and leads to the conclusion that in any partial order, where the mathematicians are looking upward, then the collection of incorrect guesses forms a converse well-founded subset. And one can even generalize further than this, as Hardin and Taylor do.<|endoftext|> TITLE: Can the first ordinal in which $V\neq HOD$ be $\aleph_\omega$? QUESTION [8 upvotes]: Assume that $V\neq HOD$ and let $\kappa = \min \{\alpha\in On \mid \mathcal{P}(\alpha) \not\subseteq HOD\}$. Clearly, $\kappa$ is a cardinal. Question: Is it consistent that $\kappa = \aleph_\omega$? Note that it is consistent that $\kappa$ is a regular cardinal: start with $V=L$ and force with $Add(\kappa,1)$. Since this forcing is weakly homogeneous, its generic filter is not in $HOD$. Since we don't add any bounded subsets to $\kappa$, for every $\alpha < \kappa$, $\mathcal{P}(\alpha) \subseteq L \subseteq HOD$. Similarly, it is consistent that $\kappa$ is singular with countable cofinality. Let $\kappa$ be a measurable cardinal and let $V = L[\mu]$ ($\mu$ is a normal measure for $\kappa$), the canonical inner model for one measurable cardinal. Let $C$ be a Prikry sequence. Then $HOD^{V[C]}\cap \kappa^{<\kappa} = L[\mu]\cap \kappa^{<\kappa}\subseteq HOD$, but since the Prikry forcing is weakly homogeneous, $C\notin HOD^{V[C]}$. REPLY [8 votes]: Assume $GCH$ and let $\kappa$ be $(\kappa+2)-$strong. Force with extender based Prikry forcing $P$ with interleaved collapses to make $\kappa=\aleph_\omega$ and $2^{\aleph_\omega}=\aleph_{\omega+2}.$ Call the resulting extension $V[H].$ Also let $V[G]$ be an intermediate submodel, which just adds the Prikry sequence related to the normal measure and adds collapses, so that the following holds: (1) $V[G] \subset V[H]$ have the same cardinals and bounded subsets of $\kappa=\aleph_\omega,$ (2) $V[G] \models GCH.$ Note that there are many new $\omega$-sequences through $\kappa=\aleph_\omega$ in $V[H]\setminus V[G].$ A much stronger version of the following lemma will appear in my paper "$HOD, V$ and the $GCH$" (where extender based Prikry forcing is replaced by extender based Radin forcing): Homogeneity lemma. Assume $p,q\in P$ are such that $\pi(p)$ is compatible with $\pi(q),$ where $\pi$ is the projection map. Then there are $p' \leq p, q' \leq q$ and an isomorphism $\Phi: P/p' \simeq P/q'.$ It follows from the above lemma that $HOD^{V[H]} \subseteq V[G].$ Now force over $V[H]$ to code any bounded subset of $\aleph_\omega$ into $HOD$ using a homogeneous forcing, call the extension $V[H][K].$ Now $V[G] \subset V[H][K],$ and any new $\omega-$sequence cofinal in $\kappa$ which is in $V[H]\setminus V[G]$ witnesses $\min\{\alpha: P(\alpha) \nsubseteq HOD\}=\aleph_\omega$. Remark. In fact it seems we need much weaker assumption. All we need is to be able to add a new cofinal $\omega$-sequence in $\kappa$ which is not in $V[G],$ and it seems to me that for this just a measurable is sufficient.<|endoftext|> TITLE: Integration currents vs Poincaré dual QUESTION [5 upvotes]: Let $M$ be a complex manifold of dimension $n$ and $S \subset M$ a closed complex submanifold of complex codimension $r$. Let $[S] \in H_{2r}(S)$ be the fundamental class of $S$. We have the integration current $T_{S}$ associated to $S$ defined by $$ \langle T_{S}, \omega \rangle = \int_{S} \omega $$ for any $2(n-r)$-form $\omega$ defined on $M$. The current $T_{S}$ on $M$ has order $2r$. We can look at this current as a $2r$-cohomology class, i.e. $T_{S} \in H^{2r} (M)$. On the other hand, associate to $S$ the Poincaré dual $\eta_{S} \in H^{2r}(M)$. Question: What is the relation between $T_{S}$ and $\eta_{S}$? REPLY [8 votes]: Here is briefly the story. More details can be found in DeRham's monograph Variétés differentiables. Let $M$ be a smooth, compact, oriented, $m$-dimensional manifold. Denote by $\Omega^k(M)$ the space of smooth degree $k$-forms on $M$ and by $\Omega_k(M)$ its dual space, namely the space of $k$-dimensional currents. Let $\newcommand{\bR}{\mathbb{R}}$ $$ \langle-,-\rangle :\Omega^k(M)\times \Omega_k(M)\to\bR,\;\; (\eta,C)\mapsto\langle \eta,C\rangle, $$ denote the natural pairing between topological vector space and its dual. We have a natural map $$\cap [M]: \Omega^k(M)\to \Omega_{m-k},\;\;\Omega^k(M)\ni \alpha \mapsto \alpha\cap [M] \in \Omega_{m-k}(M), $$ determined by $\newcommand{\lan}{\langle}$ $\newcommand{\ran}{\rangle}$ $$\lan \eta, \alpha\cap [M]\ran :=\int_M \alpha\wedge \eta ,\;\;\forall \eta\in\Omega^{m-k}(M). $$ If we denote by $\newcommand{\pa}{\partial}$ $\pa:\Omega_k(M)\to\Omega_{k-1}(M)$ the boundary operator on $\Omega_\bullet(M)$ defined by $$ \lan\alpha ,\pa C\ran:= \lan d\alpha, C\ran,\;\;\forall \alpha\in\Omega^{k-1}(M),\;\;C\in \Omega_k(M), $$ then we obtain a cochan complex $$(\Omega_{m-\bullet}(M), \pa):\;\; \cdots \stackrel{\pa}{\to} \Omega_{m-k}(M)\stackrel{\pa}{\to}\Omega_{m-(k+1)}(M)\stackrel{\pa}\cdots . $$ We then have a morphism of cochain complexes $$PD_M :(\Omega^\bullet(M), d) \to (\Omega_{m-\bullet}(M),\pa),\;\; \alpha \mapsto \alpha\cap [M]. $$ Georges DeRham proved that this morphism induces isomorphism in cohomology. This is one incarnation of Poincare duality. The cohomology of $(\Omega_{m-\bullet}(M),\pa)$ with the homology of $M$ with real coefficients. In your question you have identified $T_S$ with a $2r$-cohomology class using using $PD_M^{-1}$. This identification implicitly uses Poincare duality.<|endoftext|> TITLE: Nilpotence of the stable Hopf map via framed cobordism QUESTION [30 upvotes]: The Pontryagin-Thom construction shows that the stable homotopy groups of spheres are the same as the groups of stably framed manifolds up to cobordism. Specifically the Hopf map corresponds to the circle with its Lie group framing. It is well known that this element $\eta$ in stable homotopy (like all elements of positive degree) is nilpotent in the stable homotopy ring, more specifically $\eta^4$ is trivial. On the cobordism side, multiplication in the ring is just the product of manifolds. So the corresponding statement about manifolds is that the 4-dimensional torus with its Lie group framing is nullcobordant. Is there a geometric way to understand this? What makes the 4d torus so different from the 2d and 3d ones? REPLY [24 votes]: Answer Summary Let $\eta$ be the framed 1-manifold which is the Lie group framing on the circle and let $\nu$ be the Lie group framing on $S^3 = Spin(3)$. I am probably going to conflate these framed manifolds with their classes in frame cobordism. I hope you forgive me. There are lots of geometric ways to see that $2 \eta = 0$ in stably framed bordism so I will take that as given. One thing which makes dimension four different from dimensions two and three is the existence of the K3 surface. What I will show below is that there is a way to cut the K3 surface in half to get a 4-manifold, which I will call $X$, with boundary the 3-torus $T^3$. We will see that if we remove 12 points from $X$ then it admits an (unstable) framing, which moreover restricts to the Lie group framing $\nu$ at each of the 12 points and restricts to the Lie group framing $\eta^3$ on the 3-torus. This is a geometric incarnation of the relation $\eta^3 = 12 \nu$, which, as you observed, immediately shows that $\eta^4 = 0$ as well since $2 \eta = 0$. This is closely related to the fact that $24 \nu = 0$ and I will start there. 24 Torsion in the 3rd stable stem, via geometry The 3rd stable stem ( = the 3rd stable homotopy group of spheres = group of stably framed 3-manifolds up to cobordism) is $\mathbb{Z}/24$ and it is a very natural question to ask for a geometric description for the source of this 24. In fact this has been asked and answered before on MO. What I am about to describe is really just a synthesis of the answers given by Tilman and Mrowka for that previous question, which gives a geometric reason why $24 \nu = 0$. The key geometric ingredient is the existence of the K3 surface. This complex surface is famous for its many interesting and useful properties. The two salient facts about this 4-manifold are the following: The cohomology groups of the K3 surface are: $\mathbb{Z}, 0, \mathbb{Z}^{22}, 0, \mathbb{Z}$. In particular it is simply connected and the Euler characteristic is 24. The K3 surface is hyperkahler. This means that there is not just one complex structure on the tangent bundle of the K3 but that there are three different ones $I, J, K$ and they generate a quaternion algebra bundle. This second property is critical because it means that if we are given a non-zero vector $v$ in the tangent space of the K3, then we actually get a frame $(v, Iv, Jv, Kv)$. On a K3 surface (and submanifolds thereof) we can turn non-vanishing vector fields into 3-framings. The Euler class obstructs the existence of a non-vanishing section of the tangent bundle, however since the Euler characteristic is 24, if we puncture the K3 surface 24 times (introducing 24 3-sphere boundary components) then we can find a non-vanishing vector field. Applying the hyperkahler structure to this gives us a 3-framing of the punctured K3 surface. This is now a framed bordism which shows that the sum of those 24 framed 3-spheres is zero in the framed bordism group. In fact I think we could find a non-vanishing vector field on the K3 after puncturing just one time. However by puncturing 24 times we can arrange for each of the punctures to be a simple source for the vector field. This is important because then when we apply the hyperkahler structure near the source we see that it does indeed give the Lie group framing on each of those 24 3-spheres. We conclude that $24 \nu = 0$. Half the K3 surface and $\eta^3 = 12 \nu$, via geometry There is an important famous way to construct the K3 surface called the Kummer construction. I will just discuss the topological aspects first. In this construction you first consider the 4-torus $T^4$, which we will think of as a Lie group. Next we take the quotient by the involution $\sigma: a \mapsto -a$. This results in a quotient space/orbifold $T^4/ \sigma$ which has 16 singularities. To get the k3 surface we "blow-up" those 16 points. The result is the K3 surface. To get half the K3 surface we will modify this construction. Consider one of the circle factors of the 4-torus. The $Z/2$ action comes from a product of an action on each of these factors. It is simply reflecting the circle through the x-axis. It has two fixed points in the circle at (1,0) and (-1,0). The product of these fixed points in each circle give us the 16 fixed points in $T^4$. We will take one circle factor and consider cutting it in half into the semicircles with either positive or negative x-coordinate. The action is compatible with this splitting (it is an equivariant decomposition) and each semicircle now has only one critical point. This decomposition gives us a decomposition of of the 4-torus $$ T^4 = (I \times T^3) \cup_{\partial I \times T^3} (I \times T^3) $$ The boundary of each half consists of two disjoint $T^3$s, and the $\mathbb{Z}/2$-action exchanges them. The quotient $$(I \times T^3)/\sigma $$ is an orbifold with boundary $T^3$ and with eight singular points on the interior. If we blow-up those eight points, then we get a manifold $X$ which is half the K3 surface in the sense that $$X \cup_{T^3} X = K3.$$ Since the Euler characteristic of $T^3$ is zero, this means that $\chi(X) = \frac{1}{2} \chi(K3) = 12$. Now just as the Euler characteristic obstructs the existence of a non-vanishing vector field on a closed manifold, on a manifold with boundary it obstructs the existence of a non-vanishing vector field which further restricts to the inward pointing normal vector field near the boundary. Thus if we take the 4-manifold $X$ and puncture it 12 times, then we may choose a non-vanishing vector field which is pointing inward at each of the punctures (i.e. they are sources) and which is inward pointing along the boundary 3-torus. Then we apply the hyperkahler structure to extend this to a 3-framing. I already mentioned that near each of the 3-spheres we get the Lie group framing. This is believable because we just need to understand what is going on near an isolated point and so some standard flat model of $R^4$ is a good enough approximation, at least to identify the homotopy class of the induced framing. However I claim that the induced framing on the 3-torus agrees with $\eta^3$. If so then this bordism witnesses the relation $\eta^3 = 12 \nu$, and we are done. Why is this framed 3-torus $\eta^3$? There might be an easy way to see this, but I at least found one way to do this, by looking more closely a the geometric structure on the K3 surface. Much of this comes from D. Joyce's book Compact Manifolds with Special Holonomy. One of the claimed properties of the K3 surface is that it has a hyperkahler structure, but how can we construct such a structure explicitly? Here specifically I mean a Kahler metric which has special holonomy given by $SU(2)$. We know that the moduli space of such metrics is 20 dimensional, how to we construct points in there? One method is to go back to the Kummer construction. On the 4-torus we have a flat Kahler metric, viewing it as the quotient of $\mathbb{C}^2$ by a lattice. This is invariant under $\sigma$ and so descends to give a flat metric on $T^4/\sigma$, which we can think about as being a singular degenerate metric on the K3 surface. We can think of this as being a point on the boundary of some compactification of the space of hyperkahler metrics and the idea is to deform this metric to get the desired one. Away from the singular points the flat metric is a good approximation, but near the singular points it is bad, so you replace it with the "Eguchi-Hanson" metric. Then you have to interpolate. The details of this seem difficult and Joyce attributes it to Page, LeBrun-Singer, and Topiwala. So in the end we see that the metric (and hence the hyperkahler structure) looks like the flat metric when you are far away from the singular points. The 3-torus is, by construction, just about as far away from these singular points as possible. We can also choose the punctures to be in the "flat region" as well. Once the torus and the punctures have a nearby hyperkahler structure which is approximately flat, it is easy to see that our construction of a framing produces $\nu$ near each of the punctures and produces $\eta^3$ on the 3-torus. Addendum Now that I write this I see that there is probably a much easier topological argument. All you need to know is that you can glue the flat topological almost-hyperkaher structure (not necessarily integrable) to some standard hyperkaher structure near the blow-ups. This was probably a necessary first step in the more difficult geometric results I mentioned.<|endoftext|> TITLE: Conservativity of multiplicative linear logic over intuitionistic multiplicative linear logic QUESTION [6 upvotes]: It is well known that multiplicative linear logic (MLL) is conservative over intuitionistic multiplicative linear logic (IMLL). In other words, if an IMLL formula is provable in MLL then it is already provable in IMLL. Who first proved this, and how? It doesn’t seem to be in Girard’s original Linear Logic paper, yet I've never seen a reference given when this fact is referred to. REPLY [6 votes]: This is (a piece of) Proposition 3.8 in Harold Schellinx, Some Syntactical Observations on Linear Logic, J. Logic Compututa., Vol. 1 No. 4, pp. 537-559, 1991. (pdf at oxford journals) His proof-theoretic argument is that the only way to use a sequent with multiple conclusions when giving a cut-free MLL proof of an intuitionistic sequent $\Gamma \Rightarrow C$ is as the left premise (of the form $\Gamma_1 \Rightarrow A,C$) of the ${\multimap}L$ rule. But then the right premise (of the form $\Gamma_2,B \Rightarrow \cdot$) must have an empty right hand side, and by induction it is easy to see that such sequents are unprovable in the absence of 0 or $\bot$. Hence any cut-free MLL proof of an intuitionistic sequent $\Gamma \Rightarrow C$ is limited to sequents with exactly one conclusion, and can be replayed as an IMLL proof.<|endoftext|> TITLE: Is there any good survey on the hook length formula and related topics? QUESTION [8 upvotes]: I am recently doing some research related to the hook length formula. The hook formula counts the number of Young tableaux of certain type. I find there are plenty of research already been done and I can observe a lot extension and generalization of hook length formula types. But is there a good survey on those research and open problems? What I am interested are those extensions purely combinatorial, but it is still good if anybody can provide some survey on the representation theory side. Please correct me if there are easy-to-find survey or the problem is too general for asking. REPLY [5 votes]: Some references you might find interesting: Proctor classify certain posets (d-complete posets), that admit hook formulas. There are hook formulas for forests, as well as some other types. (Victor Reiner, P-partitions revisited, Triangle Lectures in Combinatorics slides, 2011.) Also, I believe there is some recent results on hook formulas for skew shapes. (Morales, Pak, Panova, Hook formulas for skew shapes I, 2015.)<|endoftext|> TITLE: Universal coefficient theorem for group homology and cohomology QUESTION [14 upvotes]: I've been looking for any kind of universal coefficient theorem for group homology and cohomology, including dual universal coefficient theorems. However, the only things I can find are ones where the group action on the coefficients is trivial. As such, my question is: Let $G$ be a group, $M$ be a $G$-module. Is there any universal coefficient theorem for $H^*(G, M)$ or $H_*(G, M)$? My specific interest, as in Cohomology of lattice with coefficients in field of rational functions, is where $G = \mathbb{Z}^n$, but I'd love if there were a more general answer. REPLY [10 votes]: You're only finding UCT in the literature for trivial group actions, because there is no general UCT for nontrivial group actions: The general Kunneth formula does not hold for arbitrary groups and actions. But it does hold a good amount of times, and I elaborated on this here: Kuenneth-formula for group cohomology with nontrivial action on the coefficient. Now the UCT, which relates $H_*(G,M)$ to $H_*(G,\mathbb{Z})$, only follows from the Kunneth formula for trivial group actions. The Kunneth theorem considers the tensor product $C_*\otimes D_*$ of two chain complexes, and the special case for UCT is $D_*=M$ for some $R$-module. To guarantee that the images of the boundary maps are $R$-projective, some extra assumptions are needed (like $R$ is a PID). For group homology, we work with $F_*\otimes_{\mathbb{Z}G}M$ where $F_*$ is a free resolution of $\mathbb{Z}$ as a $\mathbb{Z}G$-module. We cannot take $R=\mathbb{Z}G$ (otherwise the assumption about the boundaries would imply that all homology groups are trivial), so we take $R=\mathbb{Z}$. But then $M$ must be trivial as a $\mathbb{Z}G$-module in order to express $F_*\otimes_{\mathbb{Z}G}M$ as $C_*\otimes_\mathbb{Z}M$. In this case, $F_*\otimes_{\mathbb{Z}G}M=(F_*\otimes_{\mathbb{Z}G}\mathbb{Z})\otimes_\mathbb{Z}M$ and we can apply the UCT. Morally, in lieu of Qiaochu's remark you must ask: Given any data involving the $G$-action, how would the operators $\oplus,\otimes,\text{Tor},\text{Ext}$ encode such information? And as shown above, you can't use that information to pass from $M$ to $\mathbb{Z}$. For example, let $\mathbb{Z}_2$ act on $M=\mathbb{Z}_2\oplus\mathbb{Z}_2$ by swapping the generators of the summands. Where would this maneuver exist on the coefficient $\mathbb{Z}$ or on any homological object? We can't simply "forget" the action, because $H^1(\mathbb{Z}_2,M_\text{nontriv})=0$ while $H^1(\mathbb{Z}_2,M_\text{triv})=M$.<|endoftext|> TITLE: NCG with all noncommutativity in a nilpotent ideal QUESTION [10 upvotes]: While in general non-commutative geometry behaves rather differently from commutative geometry when it comes to local-to-global properties (descent), there are versions of "mild" noncommutative geometry that behave very much like commutative geometry in this respect. The archetypical example here is supergeometry. One may argue that the reason that the theory of supergeometry proceeds in close analogy with ordinary differential geometry is simply because a supercommutative algebra is just a commutative algebra, but internal to a nontrivially braided symmetric monoidal category. On the other hand when it comes to local properties and the fact that Grothendieck topologies work well in supergeometry, this is to do more specifically with the fact that the non-commutativity is all in the nilpotent ideals of supercommutative superalgebras, and hence irrelevant to coverings and descent. This leads one to wonder whether there is something to be gained in developing a geometry based on formal duals to those noncommutative algebras for which "all the noncommutativity is in the nilpotent ideals", e.g. such that when quotienting out the maximal two-sided nilpotent ideal they become commutative. Supergeometry would be a special case of this, but it would be more general. Has anything like this been investigated somewhat systematically anywhere? Is there any names attached to this that one could search for to find more? REPLY [4 votes]: Lieven le Bruyn kindly points out M. Kapranov, Noncommutative geometry based on commutator expansions, J. reine und angew. Math. 505 (1998), 73-118, math.AG/9802041 which develops pretty much exactly the idea that I was asking about. With that in hand, Google tells me to my surprise that my own wiki had a hidden entry on this all along nLab -- Kapranov's noncommutative geometry which Zoran Škoda once started, thankfully. This has a few more links. Good stuff.<|endoftext|> TITLE: Consistency strength of $\aleph_2$-Souslin hypothesis QUESTION [15 upvotes]: Question 1. What is known about the consistency strength of $\aleph_2$-Souslin hypothesis? Question 2. What is known about the consistency strength of having both $\aleph_2$-Souslin hypotheis and $\aleph_3$-Souslin hypothesis? Remark 1. By $\kappa$-Souslin hopothesis, I mean there are no $\kappa$-Souslin trees. Remark 2. By Laver-Shelah, the existence of a weakly compact cardinal implies the consistency of $\aleph_2$-Souslin hypothesis. On the other hand by results of Shelah-Stanly, if we assume some instances of $GCH$+ $\aleph_2$-Souslin hypothesis (having $CH$ is sufficient), then some large cardinals (at least Mahlo) are required. In the above question I do not take care of preserving instances of $GCH$. REPLY [11 votes]: Answer to 1, without CH: Mitchell and Silver, 1973: Weakly compact is an upper bound. Answer to 1, with CH: Laver and Shelah, 1981: Weakly compact is an upper bound. Shelah and Stanley, 1982: Inaccessible is a lower bound. Answer to 1, with GCH: Gregory, 1976: Mahlo cardinal is a lower bound. Rinot, 2016: Weakly compact is a lower bound. Answer to 2, with GCH: Rinot, 2016 (building on recent work of Schindler and Steel): AD holds in $L(\mathbb R)$ is a lower bound.<|endoftext|> TITLE: Embedding of classical into intuitionistic linear logic QUESTION [5 upvotes]: Following on from this recent question, there is another construction that is well-known, but I don’t know a good primary source for: the Kolmogorov-style double-negation embedding of classical into intuitionistic linear logic (where the translation of a sequent is intuitionistically provable iff the original was classically provable). I would be very happy to learn of suitable references for this, too. I’d rather not specify a specific theorem, because I’d be interested to learn about any early published result of this sort. REPLY [6 votes]: There are actually many negative translations of classical linear logic into intuitionistic linear logic, just as there are many negative translations of classical logic into intuitionistic/minimal logic (by Kolmogorov, Gentzen, Gödel, Kuroda, etc.). A good way of understanding this is in terms of polarities, which were introduced into linear logic by Girard in the early 1990s, following Andreoli's work on focusing proof search. Polarized linear logic can be seen as a refinement of classical linear logic, in the sense that there is a forgetful translation $$|{-}| : \text{polarized } LL \to CLL$$ which erases polarities. On the other hand, there is also a deterministic translation $$(-)^\dagger : \text{polarized } LL \to ILL$$ that interprets polarized formulas into a negative fragment of intuitionistic linear logic. This fragment of ILL has been called tensorial logic by Melliès, and is essentially equivalent to polarized linear logic but in an asymmetric presentation. Then, different negative translations of CLL into ILL can be understood as the composition of an ad hoc polarization $$(-)^* : CLL \to \text{polarized } LL$$ (which must be a section of $|{-}|$, i.e., the equivalence $A \Leftrightarrow |A^*|$ holds for all CLL formulas $A$) followed by the deterministic translation $(-)^\dagger$. Unfortunately, I don't know of a place in the literature where this is all explained very clearly. The early work on focusing and polarities includes Jean-Marc Andreoli, Logic Programming with Focusing Proofs, JLC 2(3), 1992. Jean-Yves Girard, "On the unity of logic", APAL 59(3), 1993. A more recent paper discussing polarities from the point of view of negative translation (and linear continuations) is Paul-André Melliès and Nicolas Tabareau, Resource modalities in tensor logic, APAL 161(5), 2009.<|endoftext|> TITLE: Decay of solutions to Schrodinger equation with local minimum in potential QUESTION [6 upvotes]: Consider the one-dimensional Schrodinger operator on the real line $\mathbb{R}$ given by $$ L = - \partial_x^2 + V $$ where $V$ is a potential with the following properties: $V$ is non-negative, and infinitely differentiable $|V| = \frac{1}{|x|^2}$ for $|x| \gg 1$ (so in particular it is in $L^p$ for any $p\geq 1$). Question: Are there general theorems concerning the (pointwise or local energy) decay of the Schrodinger ($i\partial_t \Phi = L\Phi$) equations for such operators? I am particularly interested in the case where $V$ has more than one local maximum. Motivation: First let us consider the case where $V$ has exactly one local (and hence global) maximum. In the classical particle picture we see that the global maximum corresponds to an unstable fixed point of the dynamics, and for most energies a particle either has large energy so it flies over the hump, or has small energy so it comes in, turns around, and bounces off. Only at a very specific energy can you achieve the balancing act of sending in a particle that comes eventually to rest on the top of the hump. This classical picture is essentially sufficient for analyzing the quantum picture. If we assume that the local maximum is non-degenerate, then we can prove relatively simply an integrated local energy decay estimate for solutions. One version of which states that, assuming the local maximum is at the origin, for every positive $\delta$ there exists some $b$ such that $$ \int_0^\infty \int_{\mathbb{R}} \frac{1}{(1 + (bx)^2)^{\frac12 + \delta}} \Phi_x^2 ~\mathrm{d}x ~\mathrm{d}t \tag{*}$$ is bounded by some universal constant times some quantity depending only on the initial data. In the case where $V$ has more than one local maximum, the classical picture is drastically different. We see that between two consecutive local maxima of the potential, we expect classically there to be stable trapped particles, since classical particles cannot jump over the hump. So the classical picture would contradict local energy decay, since a purely classical picture would admit spatially localised solutions over a non-trivial range of energies. On the other hand, when dealing with quantum phenomenon there should be tunneling where particles can escape through finite barriers. So I expect the intuition to be that in the quantum case some (possibly weaker) energy decay is still available. What is known about this problem? Is my intuition okay? Related question: At the present I just don't have a starting point to search from. So if someone can give me a few names and/or papers to start looking, or some keywords to search for, that would be appreciated. Edit: As Christian Remling pointed out, there is the standard RAGE theorem which in particular implies that energy will escape from any compact set. What I seek is something a bit stronger. The RAGE theorem (as far as I know) does not give explicit rates, and I am hoping for some sort of result giving either the localised energy has an explicit rate of decay $\leq t^{-\alpha}$ for some $\alpha > 0$ or that the decay can be made explicit in the integral sense (something like equation (*) above but with possibly the $L^1$ integration in $t$ replaced by some higher $L^p$ for $p < \infty$). REPLY [2 votes]: Your assumptions on $V$ imply that $L=-d^2/dx^2 + V(x)$ on $L^2(\mathbb R)$ has purely absolutely continuous spectrum. This implies decay estimates of the type $\| Ke^{-itL}\psi\|_2 \to 0$ as $|t|\to\infty$ for relatively compact $K$; the case of interest here is $K=$ multiplication by $\chi_{(-R,R)}(x)$. (Some people call this the RAGE Theorem; it's really the Riemann-Lebesgue lemma in disguise.) This argument is rather general; I only used that $V\ge 0$ and $V$ has sufficiently rapid decay at $\pm\infty$. There is also much work on more specific estimates in more specialized situations. I'm not very familiar with this, but this paper might provide an entry point to the literature.<|endoftext|> TITLE: What's the name of this geometric mathematical modeling problem? QUESTION [8 upvotes]: There is a right angle corner with width 1 in both directions. One wants to find the largest area shape which can pass through this corner. I know that this is a famous problem, but what is it called? REPLY [9 votes]: A supplement to Ian's answer: Here is the largest-area sofa known, due to Gerver: Gerver, Joseph L. (1992). "On Moving a Sofa Around a Corner". Geometriae Dedicata 42 (3): 267–283. (Springer link.) Added (triggered by @GeraldEdgar's remark). The computational complexity of algorithms grows exponentially in the dimension, about $n^5$ for polyhedral objects with $n$ vertices moving in $\mathbb{R}^3$. Here is an algorithm moving an $n{=}4500$-triangle piano through a challenging apartment requiring several tricky maneuvers:             Kuffner, James J., and Steven M. LaValle. "RRT-connect: An efficient approach to single-query path planning." Robotics and Automation, 2000. Proceedings. ICRA'00. IEEE International Conference on. Vol. 2. IEEE, 2000. (IEEE link.) Not surprisingly, the problem is also called The Piano Mover's Problem. REPLY [8 votes]: The Moving sofa problem, I believe.<|endoftext|> TITLE: Character table of $\mathrm{SL}_2(\mathbb{Z}/p^n\mathbb{Z})$ QUESTION [5 upvotes]: Is there any reference where I can find the character table of $\mathrm{SL}_2(\mathbb{Z}/p^n\mathbb{Z})$? A simple search in google gave me this paper of Philip C. Kutzko on "The characters of the binary modular congruence group". But this is just an announcement and I couldn't find the complete paper. I also know that Kloosterman has a paper which discuss representation theory of $\mathrm{SL}_2(\mathbb{Z}/p^n\mathbb{Z})$ but as far as I see he hasn't computed the characters. I am looking for a reference where I can find the actual character table. I would be thankful if you please let me any reference about this. REPLY [3 votes]: Unlike the well-known case of these groups over a prime field, it's probably asking too much to exhibit full character tables over all such finite rings. (Note too that Kutzko limits his discussion to odd primes.) There have been related discussions over the years of representations of the finite groups coming from $\mathrm{GL}_2$ and $\mathrm{SL}_2$, typically in the context of $p$-adic rings and representations of the resulting infinite groups. Maybe it would help to look at older papers by Alexandre Nobs and his collaborators, such as here. There is also some more recent work in this direction by Alexander Stasinski here. I'm unaware of any published "tables", though I'm not a specialist in this field.<|endoftext|> TITLE: Between Tietze's and Dugundji's Extension Theorems QUESTION [12 upvotes]: The celebrated Tietze Extension Theorem asserts that any continuous real-valued function defined on a closed subset of a normal space, can be extended to a continuous function on the whole space. Seen as a property of the target space $\mathbb{R}$, this leads to the important notion of absolute neighborhood retract, or AR(normal), in Dugundji's notation; Tietze Extension Theorem can thus be rephrased saying that $\mathbb{R}$ is an AR(normal) space. If in Tietze theorem we restrict the class of domains from normal to metric spaces, by the Dugundji Extension theorem, at least all locally convex topological vector spaces are suitable codomains: any continuous LCTVS-valued function on a closed subset of a metric space can be extended to a continuous function on the whole space. Of course, this situation in principle allows a wide variety of intermediate situations. The first natural questions, that I would be glad to learn an answer of, are: Q1. Does Dugundji's theorem hold true for normal spaces, namely, can any continuous LCTVS-valued function on a closed subset of a normal topological space be extended to a continuous function on the whole space? I guess the answer is no, but I can't imagine a counterexample. In case of a negative (or not known) answer: Q2. Are Banach spaces absolute retract for Hausdorff compact spaces, namely, can any continuous Banach-valued function on a closed subset of a Hausdorff compact space be extended to a continuous function on the whole space? edit After Bill Johnson's answer to question 2, and the other useful comments, I would like to focus on the following question, that should have some good reference in the (wide) literature. Q3. Let $X$ be a Hausdorff compact topological space, $Y\subset X$ a closed set, $E$ a Banach space. Does there always exist a bounded linear extension operator $C(Y,E)\to C(X,E)$? REPLY [7 votes]: The following is a rather well-known theorem of Haydon that might be useful for your purpose. Theorem: The following are equivalent for a compact Hausdorff space $Y$: ($Y$ is a Dugundji space): For every compact $X$ and embedding $e:Y \to X $ there is a bounded linear extension operator $T:C(Y) \to C(X)$ such that $T(1)=1$ and $T(f)\geq0$ whenever $f\geq0$. ($Y$ is $\mathrm{AE}(0)$): For every compact zero-dimensional $Z$, every closed $A \subseteq Z$ and every map $f:A \to Y$ there is an extension $f:Z \to Y$. $Y$ is the inverse limit of a Haydon system. See for example Todorcevic´s book "Topics in Topology" for a proof of this theorem (and the definition of Haydon system). For example any product of compact metrizable spaces is Dugundji and compact topological groups are Dugundji.<|endoftext|> TITLE: Historical (personal) examples of teaching-based research QUESTION [33 upvotes]: The phrase "teaching-based research" brings to mind research about teaching, though important, it is not what I mean. Unfortunately, I couldn't come up with a better phrase, thus please bear with me while I explain the intended meaning. I have taught multi-variable calculus several times. As usual of such repetition, I had the feeling that I know the concepts involved and how they are connected to each other and so on. But, when last week I was preparing for one of my sessions - in which I decided to use a bathymetric map (depth contours) rather than a topographic map (height contours) - a problem occurred to me for the first time. Imagining myself swimming to the shore while looking at the bathymetric map, it seemed "obvious" that if I wanted to take the shortest path the to shore (from where I was), moving in the opposite direction of the gradient would not be my choice! Prompted by my observation, I came to this quite "recent" paper "When Does Water Find the Shortest Path Downhill? The Geometry of Steepest Descent Curves" addressing the very same problem that whether gradient curves are geodesics. Now here is the question: Do you know any personal (or historical) examples of such "teaching-based research"? And, here is why I think the question is suitable for MO: Many mathematician friends of mine, for obvious reasons, prefer spending their time on research rather than teaching. Having a collection of such examples could be encouraging in particular for early career mathematicians. There is a recent movement to encourage "teaching inquiry" the point of which is to "teach students to ask and explore mathematical questions". For that aim, it seems that lecturers should be ready to be faced with some problems never posed before in the subjects that are too familiar to them, and better, be ready to pose such genuine questions in such contexts. Finally, it goes without saying that, it is a habit of mind to pose such questions in everyday research practice. It seems that what makes it difficult in teaching is rooted in an all-knowing feeling. If we know how to bypass such feeling, we could understand how students might develop such a habit beyond procedural fluency and conceptual understanding. REPLY [3 votes]: Students' standard topological reasoning went often along this line: a set $S$ is not closed hence it's open (and vice versa). Thus I defined the topological student spaces, where each subset of a space is open or closed (or both). Actually, there was already an exercise in the standard Kelley's topology textbook where such spaces were called (a bit not precisely) the door spaces (a door is either opened or closed; however the door space name ignored the fact that a door cannot be both). The said exercise was restricted to the trivial case of Hausdorff spaces only. Later, when A. Mishchenko visited Warsaw, he and I wrote a paper about the student spaces without using this name for technical reasons though--we have treated more general spaces hence the student spaces name didn't fit.<|endoftext|> TITLE: Should one post a paper on the arXiv if it is not intended to be published? QUESTION [37 upvotes]: A brief description: I have written a paper which contains a new result which I believe is somewhat important but not vital to the field. It is a generalization of an existing proof to get significant new information, in a framework that did not exist at the time of the original paper (not by me). I do not believe the result can be recovered from the original result, only from the proof, which is lengthy. The paper as it stands is self-contained but a significant portion is a reproduction of the original paper (recovering additional information from various lemmas). Some lemmas are new but many are old. A copy of it which does not reproduce original work is only about 5 pages long but is very difficult to read. It is my adviser's opinion (and I agree) that neither form of the paper is suitable for publication, for different reasons. We both believe the result on its own is significant enough, though. So, the question: would posting the long form on the arXiv as is be appropriate? To head off the obvious question: yes, the author of the original paper is prominently pointed out in the abstract and the introduction, where it is pointed out that we follow the original proof closely in most cases. REPLY [2 votes]: Given that arxiv is a pre-print publication site and many of the hosted papers are not published elsewhere, yes i would say you should post it there. Note that arxiv (and various similar sites like vixra) is good for other functions as well. For example we know the journals are sometimes heavily refereed and not everything published might be the best (for example if by an otherwise important author) and/or not everything important is published (for example by a relatively unknown author or sth very contrived) Apart from that, personaly i think that sites like arxiv enable research and technology to go further by allowing partial or otherwise un-published results (see above) be published even in this format i make heavy use of arxiv and of course i am aware that sometimes invalid results get publishd there. Furthermore this would give you something like a copyright license (e.g "i published it on arxiv on that date") or to be more precise since i am not fan of copyright. It would give you the chronological advantage if you like<|endoftext|> TITLE: Rate-Distortion theory: What is the distribution of distortion on an optimal Gaussian encoder? QUESTION [5 upvotes]: If we wish to encode a gaussian source, $X\sim\mathcal{N}(0,\sigma^2)$ at rate $R$, then decode it to create an estimate $\hat{X}$, rate-distortion theory tells us that the lowest mean-squared-error we can achieve through the encoder-decoder is $\sigma^2 2^{-2R}.$ That is, using an optimal encoder and decoder, we can achieve $E(X-\hat{X})^2=\sigma^2 2^{-2R}.$ (possibly asymptotically as we let code block lengths increase) What might the distribution of $\hat{X}|_{X=x}$ be, for such encoders and decoders? That is, how is our estimate distributed around the right answer? For illustration, we know obviously that no choice of enc/dec could yield a deterministic $\hat{X}|_{X=x}=x+\sqrt{\sigma^22^{-2R}},$ even though this distribution would give us the correct mean-squared-error. (Otherwise we could design a lossless decoder!) I believe that it should be possible to design some random coding scheme where, as the code block lengths get large, eventually each $i$th message in our block, $\hat{X_i}|_{X_i=x_i}$, becomes close in distribution to $\mathcal{N}(x_i,\sigma^22^{-2R}).$ Equivalently, I am trying to find a sequence of encoders and decoders where: $$D(p^n||q^n)\rightarrow 0, \text{ as }n \uparrow \infty$$ where $x^n$ denotes a block of $n$ messages, $\hat{X}^n$ is the decoded estimate of all these $n$ messages, $p^n$ is the distribution of ${\hat{X}^n|_ {X^n=x^n}}$ and $q^n$ is the distribution of $\mathcal{N}(x^n,\sigma^22^{-2R}I_{n\times n})$. Finding such an encoder/decoder pair would essentially allow us to properly say that quantization of a gaussian source can be modeled as gaussian noise. REPLY [5 votes]: Yes, you can design a countable-alphabet quantizer for Gaussian RVs where quantization noise approaches Gaussian noise in relative entropy, as codeword length increases. 'On Lattice Quantization Noise' by Zamir and Feder in Transactions on Information Theory Vol. 42 No. 4, July 1996: http://www.eng.tau.ac.il/~zamir/papers/lqn.pdf<|endoftext|> TITLE: Do we know any bound on $\operatorname{lcm}(2^1-1, 2^2-1,\dots,2^n-1)$? QUESTION [9 upvotes]: $\DeclareMathOperator\lcm{lcm}$We know that $\operatorname{lcm}(1,\dotsc,n)$ is approximately $e^n$ and we also know that $\gcd(2^a-1, 2^b-1)=2^{\gcd(a,b)}-1$. I wonder if there exists an upper bound/lower bound/approximation for $\operatorname{lcm}(2^1-1, 2^2-1,\dotsc,2^n-1)$. REPLY [5 votes]: The Szymiczek paper that I mentioned in comments some time ago is now available on the web. From Theorems 7 and 8 in the paper, we get $$\log{\rm lcm}(2^1-1,\dots,2^{[x]}-1)={3\log2\over\pi^2}x^2+O(x\log x)$$<|endoftext|> TITLE: The intersection of two $l_1$ balls QUESTION [10 upvotes]: Let $B_1$ and $B_2$ be two balls in $\mathbb{R}^n$ with respect to the $l_1$ norm that have different radii and different centers. Is there an upper bound for the number of vertices that $B_1\cap B_2$ has? REPLY [9 votes]: There is an exponential upper bound of $9^n$, since every vertex of $B_1 \cap B_2$ is the intersection of a $k$-face of $B_1$ and a $(n-k)$-face of $B_2$ for some $k$, and the $\ell_1$-ball has $3^n$ faces (all dimensions counted together). You cannot do better for large $n$ except for the value of the constant $9$. Indeed, let $B_1$ be $\ell_1$ ball of center $(\frac 1n,\cdots,\frac 1n)$ and radius $1$, and $B_2=-B_1$. Then $B_1 \cap B_2$ is the $(n-1)$-dimensional polytope defined as $$ \{ (x_1,\dots,x_n) \, : \, \sum x_i=0, \ |x_i| \leq 1/n \} .$$ If (for simplicity) $n$ is even, it has $\binom{n}{n/2} \geq (2-o(1))^n$ extreme points, namely those vectors with $|x_i|=1/n$ and evenly distributed signs. Remarkably, with rotated balls centered at the origin, you get the same phenomenon. Indeed the intersection $K$ of the standard $\ell_1$ ball with a randomly rotated copy of itself has (with high probability) the property that $\frac{1}{\sqrt{n}} B \subset K \subset \frac{C}{\sqrt{n}} B$ for some constant $C$ (here $B$ is the Euclidean unit ball). This is known as the "global form of Kashin's theorem", see Theorem 5.5.4. in [1]. That sandwiching forces $K$ to have at least $\exp(cn)$ vertices for some other constant $c$ (essentially because a spherical cap of fixed angle less than $\pi/2$ covers an exponentially small proportion of the sphere as $n \to \infty$). [1] S. Artstein-Avidan, A. Giannopoulos, V. Milman, Asymptotic Geometric Analysis, Part I<|endoftext|> TITLE: divisible by all standard prime numbers QUESTION [8 upvotes]: This question is about prime numbers in nonstandard models of Peano Arithmetic. Every such model looks like N+AxZ, where A is a dense linear order without end points. There are many nonstandard numbers that are divisible by all standard prime numbers (there is exactly one standard number with this property). Question: does every copy of Z contain a number with this property? REPLY [9 votes]: The answer is no. Proposition: Let $f$ be a recursive function. If $M\models\mathrm{PA}$, then every interval $[a,b]$ in $M$ of nonstandard length contains an $x$ such that $$x\equiv f(p)\pmod p$$ for all standard primes $p$. Proof: Let $\phi(u,v)$ be a $\Sigma_1$ formula representing $f$. The formula $$\psi(w)=\exists x\in[a,b]\,\forall u TITLE: Natural probability on integers QUESTION [14 upvotes]: This is a follow-up to this classical question asked recently here: we know (e.g. using the second Borel-Cantelli Lemma) that no probability measure on $\mathbb{Z}$ has the property that $n\mathbb{Z}$ has mass $\frac1n$. The argument I know of makes use of an independence property induced by the formulation of the problem: $p_1\dots p_k \mathbb{Z} = p_1\mathbb{Z} \cap\dots\cap p_k\mathbb{Z}$ has measure $1/(p_1\dots p_k)= (1/p_1)\dots(1/p_k)$. This question was looking to a weakened notion of "uniform" measure on $\mathbb{Z}$: one would even like better that every class $\mod n$ has mass $\frac1n$, but this is trivially impossible. Here comes my question: Does there exist a probability measure on $\mathbb{Z}$, such that for all $n\in\mathbb{N}$ there is at least one class $\mod n$ which has mass $\frac1n$? this weakening should destroy the independence above, and thus prevents applying Borel-Cantelli. REPLY [5 votes]: Here are examples showing that unlike in the previous problem, here it does not suffice to simply use the fact that the harmonic series / the sum of the reciprocals of the primes diverges. In fact for all $s > 1$ there exists a probability measure on $\mathbb{Z}$ such that for all $n \in \mathbb{N}$ there is at least one class $\bmod n$ which has mass $$\frac{1}{n} + O \left( \frac{1}{n^{1 + \frac{1}{s}}} \right)$$ where the constant depends on $s$ (and the same will be true for other big-Os appearing in this argument). Note that the exponent $1 + \frac{1}{s}$ can get arbitrarily close to $2$. With some more annoying examples I can do very slightly better than this but I still can't reach an exponent of $2$. So let me propose the following companion problem, a negative answer to which would also answer your problem in the negative: Does there exist a probability measure on $\mathbb{Z}$ such that for all $n \in \mathbb{N}$ there is at least one class $\bmod n$ which has mass $\frac{1}{n} + O \left( \frac{1}{n^2} \right)$? Now for the examples. What we'll actually work with is a probability distribution on $\mathbb{N}$, but this can straightforwardly be converted to a probability distribution on $\mathbb{Z}$ by setting the mass of a nonpositive integer to be $0$. The distribution in question is just the zeta distribution which also appears in the other question; that is, if $X$ denotes a random positive integer then we have $$\mathbb{P}(X = k) = \frac{1}{\zeta(s) k^s}.$$ The zeta distribution has the property that the numbers $\mathbb{P}(X \equiv a \bmod n)$ are monotonically decreasing, where $a \in \{ 0, 1, \dots, n - 1 \}$, and since they sum to $1$ it follows that they begin at a value $\ge \frac{1}{n}$ and end at a value $\le \frac{1}{n}$. We'll try to find a value of $a$ for which $\mathbb{P}(X \equiv a \bmod n)$ is close to $\frac{1}{n}$ as follows. First, we'll bound the differences between the sums $$\zeta_{a, n}(s) = \sum_{k \equiv a \bmod n} \frac{1}{k^s} = \mathbb{P}(X \equiv a \bmod n) \zeta(s).$$ We have $$\zeta_{a, n}(s) - \zeta_{a+1, n}(s) = \sum_{k \equiv a \bmod n} \left( \frac{1}{k^s} - \frac{1}{(k+1)^s} \right).$$ Each term in the sum is bounded by $$\begin{eqnarray*} \frac{1}{k^s} - \frac{1}{(k+1)^s} &=& \frac{1}{k^s} \left( 1 - \left( \frac{k}{k+1} \right)^s \right) \\ &=& \frac{1}{k^s} \left( 1 - e^{-s \log \frac{k+1}{k}} \right) \\ &\le& \frac{1}{k^s} \left( s \log \left( 1 + \frac{1}{k} \right) \right) \\ &\le& \frac{s}{k^{s+1}} \end{eqnarray*}$$ where we use the inequality $e^x \ge 1 + x$ twice, first in the form $1 - e^x \le -x$ and second in the form $\log (1 + x) \le x$. It follows that $$\zeta_{a, n}(s) - \zeta_{a+1, n}(s) \le \sum_{k \equiv a \bmod n} \frac{s}{k^{s+1}} \le \frac{s}{a^{s+1}} + \sum_{n \mid k} \frac{s}{k^{s+1}} \le \frac{s}{a^{s+1}} + \frac{s}{n^{s+1}} \zeta(s + 1).$$ This bound is monotonically decreasing as a function of $a$, and now our goal is to find out how large we can take $a$ and still guarantee that $\mathbb{P}(a, n) \ge \frac{1}{n}$. Using the very crude lower bound $$\zeta_{a, n}(s) \ge \frac{1}{a^s}$$ we see that it suffices to require $$\frac{1}{a^s} \ge \frac{1}{n} \zeta(s) \Leftrightarrow a \le \left( \frac{n}{\zeta(s)} \right)^{ \frac{1}{s} }.$$ For $a \ge \left( \frac{n}{\zeta(s)} \right)^{ \frac{1}{s} }$ it now follows that $$\zeta_{a, n}(s) - \zeta_{a+1, n}(s) \le \frac{s \zeta(s)^{ \frac{s + 1}{s}}}{n^{ \frac{s+1}{s} } } + \frac{s}{n^{s+1}} \zeta(s+1) = O \left( \frac{1}{n^{1 + \frac{1}{s}}} \right)$$ and hence that $$\mathbb{P}(X \equiv a \bmod n) - \mathbb{P}(X \equiv a + 1 \bmod n) = O \left( \frac{1}{n^{1 + \frac{1}{s}}} \right).$$ Hence for $a \ge \left( \frac{n}{\zeta(s)} \right)^{ \frac{1}{s} }$ the sequence $\mathbb{P}(X \equiv a \bmod n)$ begins at a value $\ge \frac{1}{n}$, ends at a value $\le \frac{1}{n}$, while changing by at most the RHS above. It follows that there is some $a$ in this range for which $$\left| \mathbb{P}(X \equiv a \bmod n) - \frac{1}{n} \right| = O \left( \frac{1}{n^{1 + \frac{1}{s}}} \right)$$ as desired.<|endoftext|> TITLE: A variant of Nelson-Hadwiger Problem on the chromatic number of the plane QUESTION [9 upvotes]: The famous Nelson-Hadwiger problem asks about the chromatic number of the graph $G$, with the vertex set $V(G)={\mathbb R}^2$ where $a_1=(x_1,y_1), a_2=(x_2,y_2) \in V(G) \ $ form an edge iff $a_1-a_2$ has Euclidean length $1$, or equivalently, $$(x_1-x_2)^2+(y_1-y_2)^2=1.$$ It is relatively easy to see that the chromatic number is one of the number $4,5,6,7$, and there is a vast literature on this and some closely related problems. The question I have come across is the following: consider the graph $G'$ with the same vertex set $V$, but assume that $a_1=(x_1,y_1), a_2=(x_2,y_2) \in V \ $ form an edge iff $$ (x_1-x_2)^2-(y_1-y_2)^2=1.$$ What is the chromatic number of this graph? Since the set $x^2-y^2=1$ misses a neighborhood of $0$, it is clear that $ \chi(G') \le \aleph_0$. But I cannot even show that $ \chi( G')$ is finite. I would be thankful if anyone can help by an idea of pointing me to any existing literature on this question. REPLY [5 votes]: It's apparently an open problem as to whether it's finite: http://www.cs.utep.edu/vladik/2008/tr08-42.pdf<|endoftext|> TITLE: products of conjugates in free groups QUESTION [23 upvotes]: While trying to carry out some technical arguments in free groups, I have encountered the following problem, to which I don't know the answer. Let $F$ be a free group and let $g,a_1,\ldots,a_n \in F$. Suppose that $g$ is not equal to a product of conjugates of $a_1,a_2,\ldots,a_n$, in that order. That is, there do not exist $x_1,x_2,\ldots,x_n \in F$ with $g = a_1^{x_1}a_2^{x_2}\cdots a_n^{x_n}$. Is there necessarily a finite quotient $F/N$ of $F$ in which the same is true of the images of $g,a_1,\ldots,a_n$. That is, there do not $x_1,x_2,\ldots,x_n \in F$ such that $g^{-1}a_1^{x_1}a_2^{x_2}\cdots a_n^{x_n} \in N$? REPLY [27 votes]: The answer to your question is no; there need not be a finite quotient like that. This also answers the question of Lev Glebsky and Luis Manuel Rivera Martinez mentioned in a comment. I learned this argument from Jakub Gismatullin (who presented it in a similar form at a workshop at the Erwin-Schrödinger-Institute in April 2013). The key is the following deep theorem, proved by Nikolay Nikolov and Dan Segal in [Nikolay Nikolov, Dan Segal, Generators and commutators in finite groups; abstract quotients of compact groups, Invent math (2012) 190:513–602]. Theorem (Nikolov-Segal): There exists a constant $n$ such that the following holds. For every $2$-generator finite group $G$ with generators $g_1,g_2$, every element of the subgroup $[G,G]$ is a product of $n$ factors of the form $[g,g_i^{\pm 1}]$ with $g \in G$, $i =1,2$. Let now $F$ be the free group on two generators $g_1,g_2$. Observe that $[g,g_i^{\pm 1}] = (gg_i^{\pm}g^{-1})g_i^{\mp}$, so that any commutator with a generator is a product of two conjugates of the generators. Choose $N=2n \cdot 4^n-1$ and find a sequence $a_0,\dots,a_N$, where $a_j \in \{g_1,g_1^{-1},g_2,g_2^{-1}\}$ and $a_{2j+1} = a_{2j}^{-1}$, and such that any possible choice of a sequence of length $n$ appears among sequences $(a_{2j}, a_{2j+2}, \dots, a_{2(j+n-1)})$. (Existence is easy, just concatenate all possible sequence on the even indices and choose appropriate elements for the odd indices.) Use the well-known fact that the commutator width of $F$ (free group on two generators) is infinite and choose some element $g \in [F,F]$, whose commutator length is strictly larger than $n \cdot 4^n$. It is clear that $g$ is not of the form $a_0^{x_0}a_1^{x_1} \cdots a_N^{x_N}$, since $a_{2j}^{x_{2j}} a_{2j+1}^{x_{2j+1}}$ is a commutator by construction. However, in any finite quotient $G=F/N$, we have $gN \in [G,G]$ and thus (by the theorem above) we can find $x_0,x_1,\dots,x_N$ such that $a_0^{x_0}a_1^{x_1}a_2^{x_2} \cdots a_N^{x_N} \in gN$. Indeed, we write $gN$ as a product of $n$ commutators with generators, locate a suitable segment of the sequence of the $a_i$'s, choose appropriate $x_i$'s there and set all other $x_i$'s equal to the neutral element. As a remark, this does not provide an example of a group which is not weakly sofic and it does not disprove Conjecture 2.1 in the paper by Glebsky-Rivera Martinez. It only proves that products of conjugacy classes need not be closed in the pro-finite topology.<|endoftext|> TITLE: Element with unique representation in A+B QUESTION [5 upvotes]: Let $A, B \subseteq \mathbb{Z}$ be finite subsets of the integers. Then there exists an element in $A+B$ with a unique representation as a sum of an element in $A$ and an element in $B$, namely $\max(A+B)$ (or $\min(A+B)$). If $A, B \subseteq \mathbb{Z}_m$ with $|A+B| < |A|+|B|-1$ or $|A|+|B| \geq m+2$ then $r_{A+B}(x) \geq 2$ for all $x \in A+B$. So, suppose $A, B \subseteq \mathbb{Z}_m$, $|A+B| \geq |A| + |B| -1$ and $|A|+|B| < m+2$, under which conditions can we expect the existance of an element having unique representation in $A+B$? Clearly, if $A$ and $B$ are union of cosets of the same subgroup $H$ of $\mathbb{Z}_m$, such an element cannot exist, so I am interested in sets $A$ and $B$ with low density in every coset, say $|A \cap x+H| < |H|/2$ for all $x \in \mathbb{Z}_m$ and subgroup $H$ of $\mathbb{Z}_m$. REPLY [6 votes]: First, a technical remark: you do not need to assume $|A|+|B|