Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2014-04-24T04:02:10Zhttp://mathoverflow.net/feedshttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://mathoverflow.net/q/1641760Does mapping each of a countable collection of disjoint closed contractible sets to a point give a homotopy equivalent space?Brian Rushtonhttp://mathoverflow.net/users/279332014-04-24T01:15:16Z2014-04-24T01:15:16Z
<p>This question assumes a space with the homotopy extension property, such as a CW complex.</p>
<blockquote>
<p>Does mapping each of a countable collection of disjoint contractible closed sets to a point give a homotopy equivalent space?</p>
</blockquote>
<p>This is true for a finite collection of closed contractible sets (I believe; I'd be happy to be shown wrong).</p>
<p>In case the question is worded vaguely, each contractible set is mapped to a point.</p>
http://mathoverflow.net/q/1641724"topological" Ochanine genus?Urs Schreiberhttp://mathoverflow.net/users/3812014-04-24T00:07:41Z2014-04-24T00:07:41Z
<p>The Witten genus has famously been lifted to the <a href="http://ncatlab.org/nlab/show/string%20orientation%20of%20tmf" rel="nofollow">string orientation of tmf</a> ("topological Witten genus"). For the Ochanine genus I am aware of a lift to a "<a href="http://ncatlab.org/nlab/show/spin%20orientation%20of%20Tate%20K-theory" rel="nofollow">spin orientation of Tate K-theory</a>", namely to a map of ring spectra $M \mathrm{Spin}\to KO[ [q] ]$: lemma 5.4, 5.8 in </p>
<ul>
<li>Matthias Kreck, Stefan Stolz, <em>$HP^2$-bundles and elliptic homology</em>, Acta Math, 171 (1993) 231-261</li>
</ul>
<p>At least naively one would hope to see a further refinement of the Ochanine genus to an orientation of something like $tmf_0(2)$. Is there some reason why this should not work, or has it just not been considered?</p>
http://mathoverflow.net/q/1641710If $f(x) = Ax$, show that for all $t \in \mathbb{R}$, the extreme $x_n = x_n(t)$ of polygon converges to $e^{At} x_0$Crooshttp://mathoverflow.net/users/497752014-04-23T23:54:50Z2014-04-23T23:54:50Z
<p>Let $f$ is a vector field in $\mathbb{R}^n$, $x_0 \in \mathbb{R}^n$ and $x_{k+1} = x_k + f(x_k)\Delta t$, $k= 0,1,...,n-1$, where $\Delta t = \frac{t}{n}$. A polygon whose points are the $x_i$ vérices, called Euler's polygon . If $f(x) = Ax$, show that for all $t \in \mathbb{R}$, the extreme $x_n = x_n(t)$ of polygon converges to $e^{At} x_0$.</p>
<p><strong>Comments:</strong></p>
<p>The hint of the exercise is to show that $x_n(t) = (I + \frac{At}{n})^n x_0$</p>
<p>Have I proved that $$\lim_{n \rightarrow \infty} (I + \frac{At}{n})^n = e^{At}$$
But I can not show that $x_n$ is the term that way. I did the following:</p>
<p>$x_1 = x_0 + f(x_0) \Delta t$,</p>
<p>$x_2 = x_1 + f(x_1) \Delta t = x_0 + (f(x_0) + f(x_1)) \Delta t$, ...</p>
<p>$x_n = x_{n-1} + f(x_{n-1}) \Delta t = x_0 + (\sum_{i=0}^{n-1} A x_i)) \frac{t}{n}$</p>
http://mathoverflow.net/q/1641702A Special Pair of Formulasuser49933http://mathoverflow.net/users/499332014-04-23T23:36:15Z2014-04-24T01:12:03Z
<p>Consider the first order language $\mathcal{L}=\{\in,\subseteq\}$ and $\{\in\}$-theory $\text{ZFC}$. Is there a formula $\psi (x,y) \in \{\subseteq\}-Form$ with the following property?</p>
<p>$ZFC\cup\{\forall x\forall y~(x\subseteq y\leftrightarrow \forall z (z\in x\rightarrow z\in y))\}\vdash$$ ~\forall x\forall y~(x\in y\leftrightarrow \psi (x,y))$</p>
http://mathoverflow.net/q/1641690A noncommutative analogy of the tube lemmaAli Taghavihttp://mathoverflow.net/users/366882014-04-23T23:18:25Z2014-04-24T03:33:02Z
<p>Assume that $A$ and $B$ are two unital commutative Banach algebras. Assume that $\phi \in \mathcal{M} (A)$ is an element of the maximal Ideal space. Define $\alpha: A\hat{\otimes} B \to \mathbb{C}\otimes B\simeq B$ with $\alpha=\phi \otimes Id_{B}$. $\;$Put $I=\ker \alpha$. Assume that $J$ is an ideal in $A\hat{\otimes} B$ with $J+ I=A\hat{\otimes} B$. </p>
<blockquote>
<p>Is there an ideal $K$ in $A$ with $K\otimes B \subset J$ and $(K\hat{\otimes} B )+ I= A\hat{\otimes} B$?</p>
</blockquote>
<p>This question is motivated by the "tube lemma" in general topology. So the answer to this question is "yes" for commutative $C^{*}$ algebras $A$ and $B$</p>
<p>We can consider the same question for (noncommutative) $C^{*}$ algebras $A$ and $B$, an irreducible representation $\phi: A \to B(H)$ and $\alpha: A\otimes B \to B(H) \otimes B$ and $I=\ker \alpha$.</p>
<p><strong>I explain that why I think that this is a noncommutative version of the tube lemma:</strong></p>
<p>In general, assume that $X$ is a compact hausdorf space and $F$ and $K$ are two closed sets in $X$, then they are two disjoint set iff $I_{F}+I_{K}=C(X)$ where $I_{F}$ is the ideal in $C(X)$ which consists all $g\in C(X)$ with $g(F)=0$ . Now assume that $F=\{x_{0}\}\times Y$ is a slice in $X\times Y$. Then $I_{F}$ has an algebraic description as the above $I=\ker \alpha$, in my question. If $U$ is an open set containing this slice, then $U^{c}$ and the slice are two disjoint closed set. So the inclusion $\{x_{0}\}\times Y \subset U$ implies that $J+I=C(X)$ where $J=I_{U^{c}}$.</p>
http://mathoverflow.net/q/164168-3Riemann Hypothesis and Kahr, Moore and WangMichael McGradyhttp://mathoverflow.net/users/499312014-04-23T23:05:19Z2014-04-23T23:05:19Z
<ol>
<li>Is there an expression of the Riemann Hypothesis in the First-Order Logic?</li>
<li>Is there a conversion of this expression to the Kahr, Moore, Wang AEA reduction class for satisfiability?</li>
</ol>
http://mathoverflow.net/q/1641670Riemannian metric on complexification of Lie groupXiaoyang Chenhttp://mathoverflow.net/users/497402014-04-23T22:14:40Z2014-04-23T22:14:40Z
<p>Let $G$ be a compact linear group and $G^c$ be its complexification. Then there is a diffeomorphism $f: G^c \to G \times Lie(G) $ given by $$ x e^{iA} \to (x,A).$$
Let $h$ be the pull back metric of the product metric on $G \times Lie(G)$.
Then $h$ has nonnegative sectional curvature. However, $h$ may not be left invariant under $G^c$ since $f$ in general is not a group homomorphism.</p>
<p>My question is: Is there a way to construct a left invariant metric on $G^c$ with nonnegative sectional curvature?</p>
http://mathoverflow.net/q/1641632Efficiently factorize a KKT system with block diagonal upper cornermangledorfhttp://mathoverflow.net/users/230642014-04-23T21:23:51Z2014-04-23T21:48:48Z
<p>I have a system resulting from a quadratic energy minimization with linear equality constraints enforced with Lagrange multipliers which has the form:</p>
<p>\begin{equation}
A =
\left[\begin{array}{c|c}
\tilde{Q}&B^T \\
\hline
B&*\\
\end{array}\right] =
\left[\begin{array}{ccccc|c}
Q & *& *& *&* \\
*& Q &*&*&* \\
*&*& \ddots&*&*& B^T\\
*&*&*& Q& * \\
*&*&*& *&Q \\
\hline
& &B& & &*\\
\end{array}\right] ,
\end{equation}
where $*$ means all zeros.</p>
<p>Let's say that $Q$ is a symmetric positive-definite $n$ by $n$ matrix and $B$ is $n*k$ by $m$ and full column rank. Both are <strong>sparse</strong> (small number of non-zeros per row).</p>
<p>I can factorize $A$ using sparse LDL decomposition but this doesn't take advantage of the repeated nature of the upper left corner.</p>
<p>Knowing that a factorization of $\tilde{Q}$ will just be a repeated block diagonal made up of factorizations of $Q$,
I tried using the <a href="http://en.wikipedia.org/wiki/Schur_complement" rel="nofollow">Schur complement method</a>, but then it seems I need to factorize $B A^{-1} B^T$ which could be <strong>dense</strong>.</p>
<p>Is there a way to build a LU-style factorization for $A$ (for efficient solving) which takes advantage of both the sparsity of $Q$ and $B$ and the repeated block diagonal in $\tilde{Q}$?</p>
http://mathoverflow.net/q/164161-5Infinite Limits at Infinity [on hold]user49929http://mathoverflow.net/users/499292014-04-23T21:08:02Z2014-04-23T21:08:02Z
<p>
You have to use the defintion for any N>0 , there exist an M>0 such that if x>M then f(x)>N
I am having trouble proving lim x-> infinity (3x^3-4x + 1)/(2x^2 + 3)=infinity</p>
<p></p>
http://mathoverflow.net/q/1641603Unexpected interaction between limits and colimitsDavid Spivakhttp://mathoverflow.net/users/28112014-04-23T20:49:14Z2014-04-23T21:34:42Z
<p>Let $D$ be a limit-complete category. My vague question is: given two diagrams in $D$, what comparisons between them induce a morphism of their limits? I'm especially interested in the case that the comparison has something to do with colimits.</p>
<p>Here is a well-known case in which the comparison has nothing to do with colimits. Let $I,J$ be small categories, and let $X:I\to D$ and $Y:J\to D$ be functors. Given a functor $f:J\to I$ and a natural transformation $\alpha:X\circ f\to Y$, there is an induced morphism between their limits $lim(f,\alpha):lim(X)\to lim(Y)$ </p>
<p>But today I came across another case in which one gets such a map between limits. Let $D={\bf Set}$ be the category of sets. Suppose $I$ is the W-shaped category $M\to N\leftarrow P\to Q\leftarrow R$ in $D$. Suppose $J$ is the cospan $M\to S\leftarrow R$, and let $g:I\to J$ be the functor sending $M\mapsto M, R\mapsto R$ and $N,P,Q\mapsto S$.
\begin{align}
I:\hspace{.5in}M\to \fbox{$N\leftarrow P\to Q$}\leftarrow R\\\\
g\downarrow\hspace{.8in}\\\\
J:\hspace{.5in}M\xrightarrow{\hspace{.5in}} S\xleftarrow{\hspace{.5in}} R
\end{align}
Given a functor $X:I\to D$, let $Y=Lan_gX:J\to D$ be the left Kan extension of $X$ along $g$. To my (perhaps very naive) surprise, I get a morphism of limits in this case too, $lim(X)\to lim(Y)$. It is only because we have an isomorphism
$$Y(u)=colim(X(w)\leftarrow X(x)\to X(y))$$
that we get such a morphism $lim(X)\to lim(Y)$. To be explicit, this map sends a tuple $(m,p,r)\in lim(X)$ to the pair $(m,r)\in lim(Y)$.</p>
<p><strong>Question 1 (Sets)</strong>: Is it true for any small categories $I, J$, and $X:I\to{\bf Set}$ that a functor $g:I\to J$ induces a function $lim(X)\to lim(Lan_gX)$?</p>
<p><strong>Question 2 (General)</strong>: Is it true for general $D$ that for any categories $I, J$, and $X:I\to D$ that a functor $g:I\to J$ induces a morphism $lim(X)\to lim(Lan_gX)$?</p>
<p><strong>Question 3 (Vague)</strong>: Is this part of any more general result about the interaction between limits and colimits?</p>
<p>Any references would be appreciated.</p>
http://mathoverflow.net/q/1641580Two-dimensional Perron formulaBodigrimhttp://mathoverflow.net/users/499282014-04-23T19:49:12Z2014-04-24T01:00:44Z
<p>There is a well-known Perron formula, which connects a mean value of certain arithmetic function with its Dirichlet series:
$$ \sum_{n\le x} f(n) = {1\over 2\pi i} \int_{c-i\infty}^{c+i\infty} F(s) x^s s^{-1} ds, $$
where $F(s)=\sum_{n=1}^\infty f(n) n^{-s}$, $c>\sigma_f$, $F(s)$ is absolutely convergent for $\Re s > \sigma_f$.</p>
<p>Is there any two-dimensional analog? Something of form
$$ \sum_{m,n\le x} f(m,n) = {1\over 2\pi i} \int_{c-i\infty}^{c+i\infty} \int_{c-i\infty}^{c+i\infty} F(z,w) x^{z+w} (zw)^{-1} dz dw $$
for $F(z,w) = \sum_{m,n=1}^\infty f(m,n) m^{-z} n^{-w}$.</p>
http://mathoverflow.net/q/1641570Decomposing connections on extensions of the frame bundleDavid Hornshawhttp://mathoverflow.net/users/370722014-04-23T19:28:14Z2014-04-23T21:53:12Z
<p>I have posted <a href="http://math.stackexchange.com/questions/761585/how-to-decompose-connections-on-the-complexified-orthonormal-frame-bundle">this question</a> on math.stackexchange, without success. I'll make it brief:</p>
<p>Let $E\rightarrow M$ be an orientable vector bundle of rank n equipped with some Riemannian metric, $P:=F_{SO(n)}(E)$ its orthonormal frame bundle. Set $P^{c}:=F_{SO(n)}(E)\times_{SO(n)} SO(n,\mathbb{C})$.</p>
<p>I am trying to understand why the following statement is true:</p>
<p><em>If we have a principal connection $\omega^{c}$ on $P^{c}$, we can decompose it uniquely into $(\omega,\phi)$, with $\omega$ a principal connection on $P$ and $\phi\in \Omega^{1}_{M}(iad P)$.</em></p>
<p>What I understand is how to decompose the connection so that $\phi\in\Omega^1_M(iadP^c)$. It should be possible in general, as long as we have a setting where $W\subset Q$ is a reduction of principal bundles s.t. the quotient of their respective lie groups form a reductive space.</p>
<p>I do not understand how we can reduce $\phi$ even further. Is it possible that $w^c$ needs to be flat?</p>
<p>Thank you very much.</p>
http://mathoverflow.net/q/16414816Is there a computable ordinal encoding the proof strength of ZF? Is it knowable?Scott Aaronsonhttp://mathoverflow.net/users/25752014-04-23T18:10:36Z2014-04-23T21:21:59Z
<p>In comments on Quora (see, for example, <a href="http://www.quora.com/Physics/What-is-physics/answer/Ron-Maimon/comment/1983479">here</a>, <a href="http://www.quora.com/How-can-a-theorem-or-conjecture-or-hypothesis-be-proved-to-be-unprovable/answer/Ron-Maimon">here</a>, <a href="http://www.quora.com/Are-there-any-mathematical-statements-which-have-been-proven-to-be-unprovable/answer/Ron-Maimon">here</a>), Ron Maimon has repeatedly expressed the strong opinion that Hilbert's program was <i>not</i> killed by Gödel's results in the way typically claimed. More precisely, he argues that the consistency of axioms sufficient for all meaningful mathematics should be provable from "intuitively self-evident" statements about various computable ordinals being well-defined. Here, of course, he points to Gentzen's 1936 consistency proof, which proves Con(PA) from primitive recursive arithmetic plus induction up to the ordinal ε<sub>0</sub>; as well as more recent results from the field of <a href="http://en.wikipedia.org/wiki/Ordinal_analysis">ordinal analysis</a>, which show that the consistency of various weaker set theories than ZF (for example, Aczel's constructive ZF and Kripke-Platek set theory) can also be reduced to the well-definedness of various computable ordinals. Maimon then goes on to say that Con(ZF) should similarly be reducible to the well-definedness of some "combinatorially-specified," computable ordinal, although the details haven't been worked out yet. (And indeed, the Wikipedia page on ordinal analysis says that it hasn't been done "as of 2008.") This sounds amazing, especially since I'd never heard anything about this problem before! So, here are my questions:</p>
<ul>
<li><p>Is there a general theorem showing that Con(ZF) must be reducible to the well-definedness of <i>some</i> computable ordinal, i.e. something below the Church-Kleene ordinal (even if we don't know how to specify such an ordinal "explicitly")?</p></li>
<li><p>Is finding an "explicit description" of a computable ordinal whose well-definedness implies Con(ZF) a recognized open problem in set theory? Do people work on this problem? Or is there some reason why it's generally believed to be impossible, or possible but uninteresting? Or does it just come down to vagueness in what would count as an "explicit description"?</p></li>
</ul>
<p><b>Addendum:</b> There's a connection here to a <a href="http://mathoverflow.net/questions/67214/pi1-sentence-independent-of-zf-zfconzf-zfconzfconzfconzf-etc">previous MO question of mine</a>, about the existence of Π<sub>1</sub>-statements independent of ZF with lots of iterated consistency axioms. In particular, using an observation from Turing's 1938 PhD thesis, I now see that it's indeed possible define a "computable ordinal" whose well-definedness is equivalent to Con(ZF), but only because of a "cheap encoding trick." Namely, one can define the ordinal ω via a Turing machine which lists the positive integers one by one, but which simultaneously searches for a proof of 0=1 in ZF, and which halts and outputs nonsense if it ever finds one. What I suppose I'm asking for, then, is a computable ordinal α such that Con(ZF) can be reduced to the statement that there's <i>some</i> Turing machine that correctly defines α.</p>
http://mathoverflow.net/q/1641412Coercivity for functional and complete orthonormal systemGiorgiohttp://mathoverflow.net/users/499232014-04-23T16:51:26Z2014-04-23T19:49:22Z
<p>Consider with $\rho \in W^{1,2}([0,\pi])$ the following functional
$$J(\rho)=\frac{1}{2}\int_{0}^{\pi}{\rho^2\,dx}$$
I know that in the $L^{2}([0,\pi])$ the coercivity condition is satisfied, but i'm asking if that condition is also true $W^{1,2}([0,\pi])$, because i don't find find an example for wich this property is not verified.</p>
<p>If i consider a othonormal complete system $\{\phi_{i}\}_{i\in N}$ for our Sobolev space, and also fixed the parameter $i \in N$, for each sequence $u_n \in span\{\phi_i\}$ the condition of coercivity it satisfied, can i conclude that the condition it's in general true?</p>
<p>Similar question (coercivity) for the following
$$
I(\rho)=\int_{0}^{\pi}{\sqrt{\dot\rho^2+\rho^2}\,dx}
$$
with $\rho \in W^{1,2}([0,\pi])$.</p>
http://mathoverflow.net/q/1641385A nilpotent quotient of free groupsSlava Krushkalhttp://mathoverflow.net/users/250112014-04-23T16:17:08Z2014-04-23T22:03:14Z
<p>Let $F$ denote the free group on $n$ generators $g_1,\ldots, g_n$. Consider its quotient $Q$ by the universal relation $[x,[x,y]]$ (a "Serre relation" familiar from Lie theory). This group is nilpotent of class $\leq n$. Denote by $Q_k$ the k-th term of its lower central series. It appears that all commutators of length $\geq 3$ are torsion elements in $Q$, while $Q_2/Q_3$ is the abelian group freely generated by $\{ [g_i, g_j], i<j\}$. Is the precise structure of $Q$ known?</p>
http://mathoverflow.net/q/1641372Results about the existence of solutions in groupsJosé Siqueirahttp://mathoverflow.net/users/466382014-04-23T16:08:13Z2014-04-24T00:09:58Z
<p>Let $G$ be a group. Consider an arbitrary equation given by $w(\vec{g})=e$, where $w: G^n \to G$ takes an $n$-tuple $(g_1,...,g_n)$ to some expression involving products of the $g_i$, their inverses and other elements of $G$.</p>
<p>Are there any good methods to determine whether such equation has a solution $\vec{g} \in G^n$ or not? For which classes of groups do they work? I'd appreciate references on related results. </p>
http://mathoverflow.net/q/1641241A question for the inverse orbit in the construction of conformal measurecomplex dynamicshttp://mathoverflow.net/users/499162014-04-23T14:41:15Z2014-04-23T19:55:23Z
<p>Recently, I read a theorem of existence of conformal measure for the rational map.</p>
<p>I did not understand two places in the proof. The author claims that
there exists an open set $V\subset \hat{C}\setminus J(R)$, such that each inverse branch $R_{j}^{-n}$ of $R^{n}$ is a single valued function. And also the inverse orbits
$$R_{j_1}^{-1}(V), R_{j_2}^{-1}R_{j_1}^{-1}(V),\dots R_{j_k}^{-1}R_{j_(k-1)}^{-1}(V)\dots R_{j_1}^{-1}(V)$$ are disjoint and the orbits will uniformly converge to $J(R)$, if $V$ is a subset in Siegel disk and Herman ring. there is at most one exceptional inverse orbit.</p>
<p>I was confused with this argument for a very long time, I did know how to give a complete proof.
any reference and comments will be appreaciated.</p>
<p>EDIT(Thanks for professor Eremenko's advice.): this arguement is form theorem 3 (page 740) in Sullivan's paper. <a href="http://download.springer.com/static/pdf/268/chp%253A10.1007%252FBFb0061443.pdf?auth66=1398441124_3dd347091ebaae9a952145910da7220b&ext=.pdf" rel="nofollow">http://download.springer.com/static/pdf/268/chp%253A10.1007%252FBFb0061443.pdf?auth66=1398441124_3dd347091ebaae9a952145910da7220b&ext=.pdf</a> </p>
http://mathoverflow.net/q/1640995Are there infinitely many commensurable classes of finite-covolume hyperbolic Coxeter groups?Hao CHENhttp://mathoverflow.net/users/205952014-04-23T08:50:42Z2014-04-24T00:53:38Z
<p><a href="http://arxiv.org/abs/0903.0138">Allcock(2006)</a> proved that </p>
<blockquote>
<p>there are infinitely many finite-covolume (resp. cocompact) Coxeter groups acting on hyperbolic space $H^n$ for every $n\le 19$ (resp. $n\le 6$).</p>
</blockquote>
<p>His main technique of construction is a "doubling trick". If a wall of the Coxeter polyhedron meets all the neighbor walls at even submultiples of $\pi$, reflection in this wall creates a larger polyhedron. Such walls are called doubling walls. If there are disjoint doubling walls, reflections in them generate infinitely many hyperbolic Coxeter polyhedra.</p>
<p>I noticed that the doubling trick constructs hyperbolic Coxeter subgroups <strong>of finite index</strong>. In terms of Coxeter complex (with polyhedral cells), it correspond to a subcomplex with the same vertices. I then wonder, what if we quotient the set of finite-covolume hyperbolic Coxeter groups by commensurable classes? Therefore the questions:</p>
<blockquote>
<p><strong>Up to commensurability</strong>, are there infinitely many finite-covolume (resp. cocompact) Coxeter groups acting on lower dimensional hyperbolic spaces?</p>
</blockquote>
<p>It forces another proof of Allcock's theorem without using the doubling trick. For Coxeter groups whose fundamental domain is a hyperbolic simplex, <a href="http://www.sciencedirect.com/science/article/pii/S0024379501004773">Johnson et al.(2002)</a> found all the commensurable classes (in the wide sense).</p>
http://mathoverflow.net/q/1640882Blow-ups in Motivic Homotopy TheoryJesse Wolfsonhttp://mathoverflow.net/users/95812014-04-23T05:46:00Z2014-04-23T22:47:11Z
<p>Let $X$ be a smooth scheme over a field $k$, and let $Z\subset X$ be a closed sub-scheme of codimension $2$. Let $Bl_Z(X)$ denote the blow-up of $X$ at $Z$, and let $\pi\colon Bl_Z(X)\to X$ denote the projection. Suppose I have a section $\sigma\colon Z\to Bl_Z(X)$ of $\pi$ over $Z$ (i.e. $\pi\sigma=1_Z$). </p>
<p>Question: (when) is the map $Bl_Z(X)\setminus\sigma(Z)\to X$ a weak equivalence?</p>
<p>The references to blow-up theorems which I have found (e.g. Voevodsky's Seattle lectures) suggest that it becomes an equivalence after suspension, but I'd like to avoid suspending if possible. </p>
<p>I'm also happy to restrict the choice of $Z$ and $X$. The case I'm most interested has $X$ being an iterated blow-up of affine space at (proper transforms) of linear sub-spaces, and $Z$ being a (proper transform of a) linear sub-space.</p>
http://mathoverflow.net/q/1640652Can we prove an open affine subscheme of a noetherian scheme is noetherian without Axiom of Choice?Makoto Katohttp://mathoverflow.net/users/376462014-04-22T22:35:02Z2014-04-23T21:11:05Z
<p>I'm interested in proving basic results of algebraic geometry without Axiom of Choice.
As for why I think this is interesting, please see Pete L. Clark's answer to <a href="http://mathoverflow.net/questions/22927/why-worry-about-the-axiom-of-choice">this question</a>.
To state my problem, I need some basic definitions.</p>
<p><strong>Definition 1</strong>
A ring $A$ is called noetherian if every nonempty set of ideals of $A$ has a maximal element.</p>
<p><strong>Definition 2</strong>
A scheme $X$ is called noetherian if it is a finite union of affine open subschemes Spec $A_i$, where each $A_i$ is noetherian. </p>
<p><strong>My Question</strong>
Let $X$ be a noetherian scheme.
Let $U =$ Spec $A$ be a <em>nonempty</em> affine open subscheme of $X$.
Can we prove that $A$ is noetherian without Axiom of Choice?</p>
<p><strong>Remark</strong>
I came up with this problem when I tried to solve <a href="http://mathoverflow.net/questions/163618/can-we-construct-cohomolgy-theory-on-noetherian-separated-schemes-without-axiom">this problem</a>.</p>
<p>If the answer is negative, what conditions are needed to make it affirmative?
$X$ should be separated, of finite type over a noetherian ring, etc?</p>
<p>The usual proof uses the following fact.</p>
<p>Let $X =$ Spec $A$ be an affine scheme.
Suppose $X$ is a finite union of open affine subschemes Spec $A_{f_i}$, where each $A_{f_i}$ is noetherian. Then $A$ is noetherian.</p>
<p>To prove this, the usual proof uses the fact that the set $\{f_i\}$ generates $A$, which can be easily proved using Axiom of Choice.</p>
<p><strong>Remark 2</strong>
The following observations might help.</p>
<p>Let $X$ be a noetherian scheme.
Let $U =$ Spec $A$ be an affine open subscheme of $X$.</p>
<p>It can be proved without Axiom of Choice that the underlying topological space of $X$ is noetherian.
See my answer to <a href="http://math.stackexchange.com/questions/712691/can-we-prove-that-a-quasi-compact-locally-noetherian-space-is-noetherian-without">this question</a>.</p>
<p>It is easy to prove withhout Axiom of Choice that every subspace of a noetherian topological space is noetherian.
Hence the underlying topological space of $U =$ Spec $A$ is noetherian.</p>
<p>See also my answer to <a href="http://math.stackexchange.com/questions/681413/noetherianess-of-a-locally-noetherian-affine-scheme-without-axiom-of-choice">this question</a>.</p>
http://mathoverflow.net/q/1640574Can we sometimes define the parity of a set?domotorphttp://mathoverflow.net/users/9552014-04-22T20:35:23Z2014-04-23T20:46:10Z
<p>Suppose that ${n\choose k}, {n-1\choose k-1}, \ldots, {n-k+1\choose 1}$ are all even. (This happens for example if $k=2^\alpha-1$ and $n=2k$.) In this case, can we select ${n\choose k}/2$ sets of size $k$ from an $n$ element set such that any $i<k$ elements are contained in exactly ${n-i\choose k-i}/2$ of the selected sets?</p>
<p>This would be somekind of parity for the sets. If true, I am also interested in the natural generalization for $p\ne 2$ to define some $\!\!\mod p$ of a set.</p>
<p>The only nontrivial example that I know is for $k=3$ and $n=6$ which I have just learned from Douglas Zare who might have heard it from Robin Chapman, see <a href="http://mathoverflow.net/questions/163689/what-is-the-best-lower-bound-for-3-sunflowers#comment417968_163689">his comment here</a>.</p>
http://mathoverflow.net/q/16404813What can be said about the Fourier transforms of characteristic functions?Joni Teräväinenhttp://mathoverflow.net/users/230082014-04-22T18:52:50Z2014-04-24T01:07:29Z
<p>What can be said about the Fourier transform of the characteristic function $1_A$, where $A\subset \mathbb{R}^n$ is of finite Lebesgue measure? In particular, </p>
<blockquote>
<p>What properties are common to Fourier transforms of all characteristic functions?</p>
</blockquote>
<p>Here are a few trivial properties; what other properties are known?</p>
<ul>
<li>$\widehat{1_A}$ is a bounded continuous function converging to zero.</li>
<li>$\widehat{1_A}$ is in $L^2$.</li>
</ul>
<blockquote>
<p>What interesting functional analytic properties does the set
$\{\widehat{1_A}: A \in \mathbb{R}^n\}$ of Fourier transforms of characteristic functions have?</p>
</blockquote>
<p>At least it is closed in $L^2$ (just apply Plancherel's formula and use the fact that an $L^2$-limit of characteristic functions is a characteristic function). Is it closed in other norms? Is it dense in some interesting spaces (if we are allowed to multiply the functions by a constants)? For which $p$ is the Fourier transform a bounded operator from our set to $L^p$?</p>
<blockquote>
<p>How does the regularity (in a vague sense) of $A$ affect on the decay of $\widehat{1_A}$? Are these Fourier transforms always entire?</p>
</blockquote>
<p>Here are a few easy remarks:</p>
<ul>
<li>If $A$ is a finite union of intervals, then $\hat{1}_A(\xi)$ is a trigonomteric polynomial divided by $\xi$, so it is in every $L^p,p>1$ but not absolutely integrable.</li>
<li>If $A$ is bounded, the Fourier transform is an entire function in $\mathbb{C}^n$.</li>
</ul>
http://mathoverflow.net/q/1639723Numerical Evaluation of Some Triple Integral involving Negative PowersRay Baihttp://mathoverflow.net/users/379872014-04-21T21:35:35Z2014-04-23T19:50:47Z
<p>Let $\beta_i\in (-1/2,0)$, $i=1,2,3,4$. I'm interested in obtaining numerical value of the following integrals:
$$
\int_{0<u_1<u_2<u_3<1} (1-u_1)^{\beta_1}(1-u_2)^{\beta_2} (u_3-u_1)^{\beta_3}(u_3-u_2)^{\beta_4} d\mathbf{u}
$$
and
$$
\int_{0<u_1<u_2<u_3<1} (1-u_2)^{\beta_1}(1-u_3)^{\beta_2} (u_2-u_1)^{\beta_3}(u_3-u_1)^{\beta_4}d\mathbf{u}.
$$</p>
<p>I'm able to use MATLAB function "integral3" to compute it, but the time cost is too much for me. The singularities in the integrand seem to slow down the computation substantially. </p>
<p>Although my question in general would be "can anyone help me to compute them efficiently?", one specific question is:</p>
<p>Are the integrals above related to some known special function (e.g, gamma, beta, hypergeometric...), which I could make use of?</p>
http://mathoverflow.net/q/1639340Fixed point problem with a monotone vector as a fixed point? [on hold]TomHhttp://mathoverflow.net/users/498312014-04-21T11:26:32Z2014-04-23T21:08:05Z
<p>Suppose $F : [0,1]^n \to [0,1]^n$ is continuously differentiable and $0 < \frac{\partial F_1}{\partial x_i} \leq \dots \leq \frac{\partial F_n}{\partial x_i} < \beta < 1$ for all $i =1,\dots,n$. Conjecture: there exists unique $x^* = F(x^*)$, and moreover, $x_1^* \leq \dots \leq x_n^*$.</p>
<p>Proof of the first part is quite straightforward: one can easily verify that $F$ is a contraction mapping and then apply the contraction mapping theorem. I would need some help with the second claim.</p>
http://mathoverflow.net/q/1636725Characterizing and counting boolean functions with all influences 1/2karpasihttp://mathoverflow.net/users/356602014-04-17T15:50:50Z2014-04-23T22:01:25Z
<p>Is there a characterization of boolean functions $f:\{-1,1\}^n \longrightarrow \{-1,1\}$,
so that $\mathbf{Inf_i}[f]=\frac{1} {2}$, for all $1\leq i\leq n$? Is it known how many such functions there are? </p>
http://mathoverflow.net/q/1634825Can eta invariant be written in terms of topological data?Zitao Wanghttp://mathoverflow.net/users/456002014-04-15T17:50:03Z2014-04-23T21:33:57Z
<p>The eta invariant was introduced by Atiyah, Patodi, and Singer. It roughly measures the asymmetry of the spectrum of a self-adjoint elliptic operator with respect to the origin. In <a href="http://link.springer.com/article/10.1007%2FBF01394348#page-1" rel="nofollow">http://link.springer.com/article/10.1007%2FBF01394348#page-1</a> Stolz showed that the eta invariant $\eta(M,g,\phi)$ of the twisted Dirac operator on a smooth closed 4 manifolds with Riemannian metric $g$ and $pin^{+}$ structure $\phi$ is a $pin^{+}$ bordism invariant. </p>
<p>As we know, for an oriented 4 manifold M, the Hirzebruch Signature Theorem implies that $\text{index}(D)=\text{sign}(M)$, where the LHS is the index of the signature operator of M (the analytic signature), and the RHS is the topological signature (the signature of a quadratic form on $H^{2k}(M)$ defined by the cup product), and moreover, we have $\text{sign}(M) = \int_M L(p_1,\dots,p_n)$, where $L$ is the Hirzebruch $L$-Polynomial, and $p_i$ the Pontryagin numbers of $M$.</p>
<p>In my understanding, the twisted Dirac operator is the unoriented generalizaton of the signature operator, and the eta invariant is the unoriented generalization of the signature of a manifold. So I was wondering if one can have a similar topological description of the eta invariants $\eta(M,g,\phi)$ of a $pin^{+}$ manifold in terms of topological data like the Stiefel Whitney numbers or Pontryagin numbers of M.</p>
http://mathoverflow.net/q/16333318Is the diameter of a centrally symmetric convex body realized by a pair of antipodal points?Alfredo Hubardhttp://mathoverflow.net/users/169592014-04-14T13:34:33Z2014-04-23T20:00:32Z
<p>Let $S \subset \mathbb{R}^n$ be the boundary of a centrally symmetric convex body and provide $S$ with the geodesic metric given by its embedding in Euclidean space (i.e., the distance between two points is the infinimum of the Euclidean lengths of all rectifiable curves on $S$ that join them). </p>
<blockquote>
<p><strong>Question.</strong> Is the diameter of $S$ realized by a pair of antipodal points?</p>
</blockquote>
<p>I am curious about this question when the metric is the induced metric from euclidean space, but I'm mostly inetrested in the case of a polytope provided the graph metric on the 1-skeleton. In the discrete case I do not care too much about constants.</p>
http://mathoverflow.net/q/1631044Symmetric matrices with $\rho(A)\gg\|A\|_\infty$Sevahttp://mathoverflow.net/users/99242014-04-11T10:25:31Z2014-04-23T19:16:24Z
<p>For a symmetric real matrix $A$, denote by $\rho(A)$ the spectral radius of $A$, and by $\sigma(A)$ the largest absolute row sum of $A$; that is, $\sigma(A)=\max_i \sum_j |a_{ij}|$, where $a_{ij}$ are the elements of $A$. We thus have $\rho(A)\le\sigma(A)$.</p>
<h3>A general question:</h3>
<blockquote>
<p>What can we say about $A$ given that $\sigma(A)\le K\rho(A)$ with a real $K>0$?</p>
</blockquote>
<h3>A pinpointed version (this is what I eventually need):</h3>
<blockquote>
<p>Given that $\sigma(A)\le K\rho(A)$, does there necessarily exist a unit-length vector $x=(x_1,\ldots,x_n)$ (where $n$ is the order of $A$) such that $\|Ax\|\ge 0.1\rho(A)$ and $(|x_1|+\dotsb+|x_n|)\max_i|x_i|\le 10K^{10}$?</p>
</blockquote>
<p>(It is not difficult to see that a unit-length vector $x=(x_1,\ldots,x_n)$ with $\|Ax\|=\rho(A)$ and $(|x_1|+\dotsb+|x_n|)\max_i|x_i|\le 10K^{10}$ may <em>fail</em> to exist.)</p>
<h3>An ideologically motivated restatement:</h3>
<p>For a (not necessarily square) matrix $A\ne 0$, define the <em>height</em> of $A$ by $h(A):=\|A\|_1\|A\|_\infty/\|A\|^2$, where $\|A\|_1,\|A\|_\infty$, and $\|A\|$ are induced operator norms (see <a href="http://en.wikipedia.org/wiki/Matrix_norm" rel="nofollow">Wikipedia</a> for the definitions). Notice, that $1\le h(A)\le\sqrt{mn}$, where $m$ and $n$ are the dimensions of $A$.</p>
<p>If $m=n$ and $A$ is symmetric, then $\|A\|_1=\|A\|_\infty=\sigma(A)$ and $\|A\|=\rho(A)$. If $A$ is actually a vector, then $$ h(A)=\frac{(|x_1|+\dotsb+|x_n|)\max_i|x_i|}{x_1^2+\dotsb+x_n^2},$$ where $x_1,\ldots x_n$ are the coordinates of $A$. My question can be equivalently restated as follows:</p>
<blockquote>
<p>If $A$ is a symmetric real matrix with $h(A)\le K$, does there necessarily exist a vector $x$ with $\|Ax\|\ge 0.1\|A\|\|x\|$ and $h(x)\le 10K^{10}$?</p>
</blockquote>
<p>The following special cases may be worth mentioning.</p>
<ul>
<li>if $A$ is diagonal, then the coordinate vector corresponding to the absolutely largest eigenvalue has height $1$; thus, denoting this vector by $x$, we have $\|Ax\|=\|A\|\|x\|$ and $h(x)=1$.</li>
<li>if $K=1$, then $\rho(A)=\sigma(A)$. I have a complete characterization of matrices with this property, and it turns out that every principal eigenvector of such a matrix has height $1$. Hence, the assertion is true in this case, too.</li>
<li>If $A$ is rank-$1$ then, in view of $A^t=A$, we have $A=\pm xx^t$ for some vector $x$. It is easy to see that in this case $\sigma(A)=h(x)\|x\|^2$, $\rho(A)=\|x\|^2$, and $x$ is a principal eigenvector of $A$. Thus, $\|Ax\|=\|A\|\|x\|$, and $h(x)=\sigma(A)/\rho(A)\le K$. </li>
</ul>
<hr>
<h2>Update 23.04</h2>
<p>I doubt this will result in a major breakthrough, but I noticed recently that my original requirement $|\langle Ax,x\rangle|\ge 0.1\rho(A)$ can be relaxed to $\|Ax\|\ge 0.1\rho(A)$, and I update the problem accordingly, for a good record-keeping</p>
http://mathoverflow.net/q/1553653Generalisation of the Grothendieck construction for presheaves as a lax pullbackTom Hirschowitzhttp://mathoverflow.net/users/382582014-01-22T10:09:53Z2014-04-23T19:05:25Z
<p>It is well-known that for any presheaf $F \colon \mathcal{C}^{\mathrm{op}} \to \mathrm{Set}$, the category of elements (obtained by the so-called Grothendieck construction) of $F$ is a comma category $y/\ulcorner F \urcorner$ in the category $\mathrm{CAT}$ of </p>
<p>$$\mathcal{C} \xrightarrow{y} \widehat{\mathcal{C}} \xleftarrow{\ulcorner F \urcorner} 1.$$</p>
<p><b>Question:</b> does this generalise to presheaves $F \colon \mathcal{C}^{\mathrm{op}} \to \mathrm{Gpd}$ of groupoids (or even categories)?
Mere comma categories yield discrete fibrations, hence won't give the expected answer. So the question is whether some other construction with a similar flavour could do.</p>
<p>Probably, if the construction does generalise, then it will also work for pseudo-functors to $\mathrm{Gpd}$ or $\mathrm{Cat}$, but I'm really interested in strict functors.</p>
<p>Note: I'm half-aware of another universal property of the Grothendieck construction as an oplax colimit. Is it related? </p>
http://mathoverflow.net/q/108276If associated-graded of a filtered bialgebra is Hopf, does it follow that the original bialgebra was Hopf?Theo Johnson-Freydhttp://mathoverflow.net/users/782010-01-05T17:52:58Z2014-04-23T20:59:41Z
<p>Warning: older texts use the word "Hopf algebra" for what's now commonly called "bialgebra", whereas now "Hopf" is an extra condition. So as to avoid any confusion, I'll give my definitions before concluding with my question.</p>
<h2>Definitions</h2>
<p>Let $C$ be a category with symmetric monoidal structure $\otimes$ and unit $1$ (and either strictify, or decorate all the following equations with associators and unitators and so on). An (associative, unital) <em>algebra</em> in $(C,\otimes)$ is an object $V$ along with maps $e: 1\to V$ and $m: V\otimes V \to V$ satisfying associativity and unit axioms: $m\circ(m\otimes \text{id}) = m\circ (\text{id}\otimes m)$ and $m\circ (\text{id}\otimes e) = \text{id} = m\circ (e\otimes \text{id})$. A (coassociative, counital) <em>coalgebra</em> is an object $V$ along with maps $\epsilon: V\to 1$ and $\Delta: V \to V\otimes V$ satisfying coassociativity and counit axioms. A <em>bialgebra</em> is any of the following equivalent things:</p>
<ul>
<li>A coalgebra in the category of algebras and algebra-homomorphisms ($1$ has its canonical algebra structure coming from the $\otimes$ axioms that $1\otimes 1 = 1$; in the tensor product of algebras, elements in the different multiplicands commute)</li>
<li>An algebra in the category of coalgebras and coalgebra-homomorphisms</li>
<li>An object $V$ with maps $e,m,\epsilon,\Delta$ satisfying the axioms above and a compatibility axiom:
$$ \Delta \circ m = (m\otimes m) \circ (\text{id} \otimes \text{flip} \otimes \text{id}) \circ (\Delta \otimes \Delta) $$</li>
</ul>
<p>A bialgebra can have the property of being <em>Hopf</em> (it is a property, not extra data): a bialgebra $V$ is <em>Hopf</em> if there exists an <em>antipode</em> map $s: V\to V$ satisfying
$$ m \circ (s\otimes \text{id}) \circ \Delta = e\circ \epsilon = m \circ (\text{id} \otimes s) \circ \Delta $$
Naturally, it's better to see these definitions than read them; check e.g. <a href="http://en.wikipedia.org/wiki/Hopf_algebra">the Wikipedia article</a>. If an antipode exists for a bialgebra, it is unique (justifying considering Hopfness a property rather than a structure) and it is an antihomomorphism for both the algebra and coalgebra structures.</p>
<p>Let VECT be the category of vector spaces (over your favorite field), with $\otimes$ the usual tensor product and $1$ the ground field. A ($\mathbb N$-)<em>filtered vector space</em> is a sequence $V = \{V_0 \hookrightarrow V_1 \hookrightarrow V_2 \hookrightarrow \dots\}$ in VECT. A morphism of filtered vector spaces $V \to W$ is a sequence of morphisms $V_n \to W_n$ so that every square commutes: $\{V_n \hookrightarrow V_{n+1} \to W_{n+1}\} = \{V_n \to W_n \hookrightarrow W_{n+1}\}$. Equivalently, a <em>filtered vector space</em> is a space $V \in $VECT along with an increasing sequence of subspaces $V_0 \subseteq V_1 \subseteq \dots \subseteq V$ such that $V = \bigcup V_n$, and a linear map of filtered vector spaces $V \to W$ is <em>filtered</em> if the image of $V_n$ lies in $W_n$ for each $n$.</p>
<p>Because $\otimes$ is exact in VECT (because every monomorphism splits), to a pair $V,W$ of filtered vector spaces we can define an $\mathbb N^2$-filtered space with $(p,q)$-part $V_p\otimes W_q$, and then we can define the $\mathbb N$-filtered space $V\otimes W$ by setting $(V\otimes W)_n$ to be the colimit of the diagram given by all $V_p\otimes W_q$ with $p+q \leq n$. Equivalently, we can take the tensor product in VECT of the unions $V = \bigcup V_n$ and $W = \bigcup W_n$, and then filter it by declaring that the $n$th part is the union of the $(p\otimes q)$th parts for $p+q = n$.</p>
<p>A ($\mathbb N$-)<em>graded</em> vector space is a sequence $\{V_0,V_1,V_2,\dots\}$ in VECT, or equivalently a space $V$ along with a direct sum decomposition $V = \bigoplus V_n$. A morphism of graded vector spaces preserves the grading.</p>
<p>Let $V$ be a filtered vector space. Its <em>associated graded</em> space $\text{gr}V$ is given by $(\text{gr}V)_n = V_n / V_{n-1}$, where $V_{-1} = 0$, of course. Then $\text{gr}$ is a symmetric monoidal functor, and so takes filtered bialgebras to graded bialgebras.</p>
<h2>Question</h2>
<p>Let $V$ be a filtered bialgebra, i.e. a bialgebra in the category of filtered vector spaces. Then $\text{gr}V$ is a graded bialgebra. Suppose that $\text{gr}V$ is Hopf. Does it follow that $V$ is Hopf? I.e. suppose that $\text{gr}V$ has an antipode map. Must $V$ have an antipode map?</p>
<p>(Or perhaps it requires additional hypotheses, e.g. that we be in characteristic 0, or that $V$ is <em>locally finite</em> in the sense that each $V_n$ is finite-dimensional?)</p>