Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2015-05-26T16:10:30Zhttp://mathoverflow.net/feedshttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://mathoverflow.net/q/2076540Four kinds of generalized hypergeometric formulas for $\pi$Tito Piezas IIIhttp://mathoverflow.net/users/129052015-05-26T15:43:41Z2015-05-26T15:43:41Z
<p>Given,</p>
<p>$$\begin{array}{|c|c|c|c|}
\hline
n&a_n&b_n&c_n\\
\hline
1 & 6541681608 & 163096908 & -640320^3\\
\hline
2 & 85840 & 4492 & -14112^2\\
\hline
3 & 28302 & 1654 & -300^3\\
\hline
4 & 48 & 8 & -2^9\\
\hline
\end{array}$$</p>
<p>using those $a_n,\, b_n,\, c_n$, how do we rigorously show that the corresponding four <em>Ramanujan-type</em> pi formulas,</p>
<p>$$\begin{aligned}
\frac{1}{\pi}&=\sum_{n=0}^\infty (-1)^n \frac{(6n)!}{(3n)!n!^3}\frac{a_1n+b_1}{(-c_1)^{n+1/2}}\\
\frac{1}{\pi}&=\sum_{n=0}^\infty (-1)^n \frac{(4n)!}{n!^4}\frac{a_2n+b_2}{(-c_2)^{n+1/2}}\\
\frac{1}{\pi}&=\sum_{n=0}^\infty (-1)^n \frac{(2n)!(3n)!}{n!^5}\frac{a_3n+b_3}{(-c_3)^{n+1/2}}\\
\frac{1}{\pi}&=\sum_{n=0}^\infty (-1)^n \frac{(2n)!^3}{n!^6}\frac{a_4n+b_4}{(-c_4)^{n+1/2}}
\end{aligned}$$</p>
<p>can be transformed to,</p>
<p>$$\pi = \frac{\sqrt{-c_1}}{b_1\Big(\,_3F_2\big(\tfrac{1}{6},\tfrac{5}{6},\tfrac{1}{2};1,1;\frac{1728}{c_1}\big)+120\frac{a_1}{b_1\,c_1} \,_3F_2\big(\tfrac{7}{6},\tfrac{11}{6},\tfrac{3}{2};2,2;\frac{1728}{c_1}\big) \Big)}$$</p>
<p>$$\pi = \frac{\sqrt{-c_2}}{b_2\Big(\,_3F_2\big(\tfrac{1}{4},\tfrac{3}{4},\tfrac{1}{2};1,1;\frac{256}{c_2}\big)+24\frac{a_2}{b_2\,c_2} \,_3F_2\big(\tfrac{5}{4},\tfrac{7}{4},\tfrac{3}{2};2,2;\frac{256}{c_2}\big) \Big)}$$</p>
<p>$$\pi = \frac{\sqrt{-c_3}}{b_3\Big(\,_3F_2\big(\tfrac{1}{3},\tfrac{2}{3},\tfrac{1}{2};1,1;\frac{108}{c_3}\big)+12\frac{a_3}{b_3\,c_3} \,_3F_2\big(\tfrac{4}{3},\tfrac{5}{3},\tfrac{3}{2};2,2;\frac{108}{c_3}\big) \Big)}$$</p>
<p>$$\pi = \frac{\sqrt{-c_4}}{b_4\Big(\,_3F_2\big(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};1,1;\frac{64}{c_4}\big)+8\frac{a_4}{b_4\,c_4} \,_3F_2\big(\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};2,2;\frac{64}{c_4}\big) \Big)}$$</p>
<p>The first one I found in Mathworld's <em><a href="http://mathworld.wolfram.com/PiFormulas.html" rel="nofollow">Pi Formulas</a></em>, (eq.85) but was not in the form above. Hence did not transparently show its connection to the <em><a href="https://en.wikipedia.org/wiki/Chudnovsky_algorithm" rel="nofollow">Chudnovsky algorithm</a></em>. </p>
<p>The other generalized hypergeometric formulas I only found empirically, so it would be nice to know if someone knows of a slick way to transform it from one kind to the other.</p>
<p><strong>P.S.</strong> The four $c_n$ are integer values of certain eta quotients of the <em>Dedekind eta function</em>, hence the "modular forms" tag.</p>
http://mathoverflow.net/q/2076530Characterize polytopes resulting from cutting a convex polytope by a hyperplanemathttp://mathoverflow.net/users/514692015-05-26T15:30:03Z2015-05-26T15:49:08Z
<p>We have a convex polytope $P$ for which we know its set of vertices. Using this set we characterize the H-representation of $P$ as: $\mathbf{A}\mathbf{x} < \mathbf{b}$.<br>
If a hyperplane defined by the equation $\mathbf{c}.\mathbf{x}-d=0$ cuts the polytope in two. How we can characterize (i.e. find the H-representation or V-representation) the two resulting polytopes?</p>
http://mathoverflow.net/q/207652-4How to find the matrix of adjoint transformation R* according to the usual scalar productmonday25http://mathoverflow.net/users/741572015-05-26T15:27:53Z2015-05-26T15:27:53Z
<p>transformation R3 -> R3 is a" circle" around the line x / z = -y = z / 2 </p>
http://mathoverflow.net/q/2076510Intersection of an irreducible curve with an exceptional divisoruser74178http://mathoverflow.net/users/02015-05-26T15:01:54Z2015-05-26T15:01:54Z
<p>Suppose $A$ is an irreducible curve on the blowup of $\mathbb{P}^2$. Then if $A$ is not equal to $E$ (where $E$ is the exceptional divisor) it has to intersect it positively. However, $A.E = c_1 ([E]). A$ which is the integral of a non positive form (at least if one chooses the metric on $E$ chosen by Griffiths and Harris in their proof of the Kodaira embedding theorem) over $A$ which is less than or equal to zero. How does one resolve this paradox ?
Apologies for the simple question but it is driving me crazy.</p>
http://mathoverflow.net/q/2076500Why the composition of planar tangles is well-defined?Sébastien Palcouxhttp://mathoverflow.net/users/345382015-05-26T14:30:38Z2015-05-26T14:30:38Z
<p>In the planar algebra theory (see <a href="http://arxiv.org/abs/math/9909027" rel="nofollow">here</a>), a planar tangle is an isotopy class; then to define the composition of two tangles, we need to choose a representative in each classes. </p>
<p><em>Question</em>: why this composition does not depend on the choice of representatives? </p>
http://mathoverflow.net/q/2076490What is the probability that a Gaussian random variable is the largest of three Gaussian random variables?Alastair Smithhttp://mathoverflow.net/users/741752015-05-26T14:25:21Z2015-05-26T14:25:21Z
<p>Consider three normally distributed RVs: $YA \sim N(a,\sigma ^{2}),$ $%
YB\sim N(b,\sigma ^{2})$ and $YC\sim N(c,\sigma ^{2})$.</p>
<p>What is the probability that $YA$ is the maximum?: $\Pr (YA>YB$ and $YA>YC)=IA=\int_{-\infty }^{\infty }\Phi (\frac{Y-b}{%
\sigma })\Phi (\frac{Y-c}{\sigma })\phi (\frac{Y-a}{\sigma })dY$</p>
<p>My first thought was to differentiate $IA$ with respect to $b$ and $c$, complete the square and exploit that $\int_{-\infty }^{\infty }\frac{\sqrt{3}}{\sigma}\phi (\frac{Y-\frac{a+b+c}{2}}{%
\sigma /\sqrt{3} })dY=1$ to remove the integral over $Y$.
Therefore, $\frac{d^2IA}{dbdc}\propto \exp \left(-\frac{a^2-2 a y+b^2-2 b y+c^2-2 cy+3 y^2}{2 \sigma ^2}\right)$.
Unfortunately, I cannot figure how to integrate with respect to $b$ and $c$ to get back to $IA$.
Any suggestions would be much appreciated. </p>
http://mathoverflow.net/q/2076421Is Eilenberg-Maclane $\wedge$ Moore space the spectrum of the cohomology theory $H^*(\ ,G)$?Paula Cartagena Ataráhttp://mathoverflow.net/users/741682015-05-26T12:28:12Z2015-05-26T15:10:18Z
<p>In the web page <a href="http://www.encyclopediaofmath.org/index.php/Moore_space" rel="nofollow">http://www.encyclopediaofmath.org/index.php/Moore_space</a> it can be found the following statement:</p>
<blockquote>
<p>If $K(\mathbb Z,n)$ is the Eilenberg–MacLane space of the group of integers $\mathbb Z$ and $M_k(G)$ is the Moore space with $\tilde{H}_k(M_k(G))=G$, then</p>
<p>$$lim_{N\rightarrow\infty}[\Sigma^{N+k}X,K(\mathbb Z, N+n)\bigwedge M_k(G)]\cong
H^n(X,G)$$,</p>
<p>that is, $\{K(\mathbb Z,n)\wedge M_k(G)\}$ is the spectrum of the cohomology theory $H^*(\ ,G)$.</p>
</blockquote>
<p>The reference of the webpage is the articule of Moore ´´On homotopy groups of spaces with single non-vanishing homotopy group´´ but I don´t find anything like that on the article. </p>
<p>It would be very helpful if you could give a reference where I can see the explanation or you can explain me that.</p>
http://mathoverflow.net/q/2076413Polish by compact is Polish?biringerhttp://mathoverflow.net/users/741692015-05-26T12:16:05Z2015-05-26T13:26:13Z
<p>Let $X,Y$ be separable and metrizable, with $Y$ Polish, and suppose there is a topological quotient map $f:X\to Y$ with compact fibers. Is $X$ Polish?</p>
<p>I have a specific space in mind, so if the answer is no and some extra criterion comes to mind ($Y$ is not locally compact, and $f$ is not a fiber bundle), I would appreciate it. </p>
http://mathoverflow.net/q/2076401Intersection of complements of connected components (2)Dominic van der Zypenhttp://mathoverflow.net/users/86282015-05-26T12:07:23Z2015-05-26T14:24:46Z
<p>Let $(X,d)$ be a non-compact, complete metric space and $K\subseteq X$ compact. Pick $x^* \in X\setminus K$.</p>
<p>Let $E$ be the connected component of $X\setminus K$ that contains $x^*$. Let ${\cal C}$ be the collection of connected components of $K$. For each $C\in {\cal C}$ let $E_C$ be the connected component of $X\setminus K$ that contains $x^*$.</p>
<p>Do we have $E=\bigcap\{E_C: C\in{\cal C}\}$?</p>
http://mathoverflow.net/q/2076384Isomorphism problem on the class of induced subgraphs of a hypercubeJernejhttp://mathoverflow.net/users/17372015-05-26T11:46:29Z2015-05-26T12:49:22Z
<p>A problem that I am currently studying translates to the problem of deciding whether two induced subgraphs of the hypercube $Q_k$ are isomorphic.</p>
<p>Now it feels to me that this class of graphs is "too rich" for there to be an efficient isomorphism algorithm yet I don't see any easy argument that this problem is GI-complete. The subclassess of bipartite graphs that are proven to be GI-complete do not seem to match this class.</p>
<p>Hence I am wondering</p>
<blockquote>
<p>Given two induced subgraphs of order $k$ that are induced subgraphs of $Q_n$ is there a polynomial algorithm (in $k$) to decide whether they are isomorphic?</p>
</blockquote>
<p>Or, is there perhaps an "obvious" subclass of bipartite graphs that is $GI$-complete?</p>
<p>For this specific problem we can assume that the embedding of the graphs into $Q_n$ is given though the second question is interesting as well.</p>
http://mathoverflow.net/q/2076354application of factorization theoremInquisitivehttp://mathoverflow.net/users/330182015-05-26T11:23:24Z2015-05-26T11:42:55Z
<p>Young's inequlity tells us that $L^{1}(\mathbb R)\ast L^{p}(\mathbb R) \subset L^{p}(\mathbb R)$ with norm inequality
$$\|f\ast g\|_{L^{p}} \leq \|f\|_{L^1}\|g\|_{L^p};$$</p>
<p>and of course this inequality has lot of importance in Analysis and PDEs. (every body know this, and perhaps all the time we are using these kind of inequality)</p>
<p>On the other hand, <a href="http://en.wikipedia.org/wiki/Cohen%E2%80%93Hewitt_factorization_theorem" rel="nofollow">Cohen- Hewitt factorization theorem</a> states that, </p>
<p>$L^{p}\subset L^{1}\ast L^{p}, (1\leq p <\infty.)$</p>
<blockquote>
<p><strong>My Question are</strong>: (i) Can you illustrate some application of this factorization theorem in Analysis and PDEs ? (or some other math branch)
(ii) What is an importance of this factorization theorem?</p>
</blockquote>
<p>Edit: Some references [paper or books (which contains some application)] will be o.k. for me.</p>
http://mathoverflow.net/q/2076340Existence of Solution steady navier stokes with do nothing outflow conditionMax Behrhttp://mathoverflow.net/users/741652015-05-26T11:19:08Z2015-05-26T11:53:58Z
<p>We consider the stationary navier stokes equation with mixed boundary conditions
$$
\begin{align}
-\nu\Delta u +(u\cdot\nabla) u + \nabla p &= 0\ \textrm{in}\ \Omega \\
div\ u&=0\ \textrm{in}\ \Omega\\
u &=g\ \textrm{on} \ \Gamma_d \\
pn-\nu\nabla u\cdot n&=0\ \textrm{on}\ \Gamma_n \\
\partial \Omega &= \Gamma_d \stackrel{\cdot}{\cup} \Gamma_n
\end{align}
$$
i.e. No external forces $f$. Inhomogenous Dirichlet boundary Data and
a "do nothing" condition on another part of the boundary.
Are for that kind of boundary conditions any existence results for solutions?</p>
http://mathoverflow.net/q/2076334finite approximation equation on free groupuser182085http://mathoverflow.net/users/604832015-05-26T11:16:13Z2015-05-26T11:46:46Z
<p>An equation in a free group $F$ is an identity of the form $W(x_1,..,x_n,a_1,...a_n)$ where $x_1,...,x_n$ are variables, $a_1,...,a_n$ free generators of $F$ and $W$ a word in the free group on $x_1,..,x_n,a_1,..a_n$.
G.Sabbagh asked the following question: does an equation in $F$ have a solution if and only if it has a solution in any finite quotient?
The answer is "no", and $[x^pa,y^{-1}z^pby]=t^q$ where $p$ and $q$ are two distinct primes is a counterexample.
My question is: Are there more counterexamples? </p>
http://mathoverflow.net/q/2076300Does the equality of product of integers modulo prime p holds in a given interval?Shivarajhttp://mathoverflow.net/users/706522015-05-26T10:19:13Z2015-05-26T14:30:11Z
<p>For any given prime $p$, does there exist $a_1,a_2,\dots,a_k,$ (not necessarily distinct) $b_1,b_2,\dots,b_m$ (not necessarily distinct) and $y_1$, $y_2$ such that </p>
<p>$$\left(\frac{p-x_1}{y_1}\right)\cdot a_1\cdot a_2\cdots a_k\equiv\left(\frac{p-x_2}{y_2}+1\right)\cdot b_1\cdot b_2\cdots b_m \pmod p.$$</p>
<p>Where $a_1,a_2,...,a_k$, $b_1,b_2,...,b_m$, $y_1, y_2$ are the elements of $S=\{1,2,...,\lceil c_{\log\log\log p}\,p^{\frac{1}{\log\log\log p}}\rceil\}$ (p is a prime and $c_{\log\log\log p}$ is some constant), $x_1<y_1$ and $x_2<y_2$ such that $\frac{p-x_1}{y_1}$ and $\frac{p-x_2}{y_2}$ are integers.</p>
http://mathoverflow.net/q/207629-1Finite groups whose non-trivial elements have no fixed points [migrated]user50982http://mathoverflow.net/users/509822015-05-26T10:11:23Z2015-05-26T10:11:23Z
<p>I am interested in finite groups $G$ acting on a finite set $X$ with the following property:</p>
<p>(*) fix(g)=$\emptyset$ for all $g\in G\setminus\{1\}$,</p>
<p>where</p>
<p>fix(g):=$\{x\in X|gx=x\}$</p>
<p>denotes the set of fixed points, i. e. (*) means: <strong>no non-trivial group element has a fixed point on $X$</strong>. (Equivalently: All non-trivial elements are derangements on $X$.)</p>
<p>I'd like to see (interesting) examples / classes of examples of such group actions. How rare are they?</p>
http://mathoverflow.net/q/2076281Limiting absorption principleFelice Iandolihttp://mathoverflow.net/users/457292015-05-26T10:09:29Z2015-05-26T10:49:48Z
<p>I would like to know if there is a book (or a paper) which can give me an introduction to LAP. I tried to read some papers by myself, but I don't feel comfortable. I think that I need the basic ideas behind this tool.
I apologize for being too vague.
Thank you for any suggestions.</p>
http://mathoverflow.net/q/2076260Sheaves whose restriction maps are monomorphisms?Trivialitieshttp://mathoverflow.net/users/741602015-05-26T10:00:07Z2015-05-26T10:00:07Z
<p>When the restriction maps of a sheaf of $\mathcal O_X$-modules are epimorphisms,
the sheaf is flasque and we have a whole theory of that. Is there a detailed study of the opposite phenomenon, i.e., when the restriction maps are monomorphisms? For example, quasi-coherent sheaves which are locally torsion free have this property (over an integral scheme). </p>
<p>Has this property been studied in general: sheaves of $\mathcal O_X$-modules (not necessarily quasi-coherent) such that the restrictions are all monomorphisms? </p>
http://mathoverflow.net/q/2076111planes intersecting a convex polytopePawan Aurorahttp://mathoverflow.net/users/396632015-05-26T07:46:08Z2015-05-26T14:18:54Z
<p>We are given a $d$-dimensional convex polytope ${\cal P}$ in $N$-dimensional
space where $d<N-1$. Consider several planes $P_i$ corresponding to inequalities
$f_i(X)\ge 0$. We are given that each such plane intersects ${\cal P}$ such that there are points in ${\cal P}$ that violate the corresponding inequality and others that satisfy it. Let these intersections create $k$ regions $R_1,\ldots,R_k$ in polytope ${\cal P}$ such that $\cup_iR_i={\cal P}$ and each $R_i$ corresponds to the violated side of some inequality. Also, there must exist at least one point in each $R_i$ such that there is a unique $P_i$ whose inequality is violated by that point. We would like to get a non-trivial upper bound on the value of $k$. Any suggestions, ideas on how to arrive at such a bound
would be highly appreciated.</p>
http://mathoverflow.net/q/2075952When a smooth algebra is regular?user237522http://mathoverflow.net/users/722882015-05-26T01:43:02Z2015-05-26T11:57:47Z
<p>Let $A \subseteq B$ be noetherian integral domains, $A$ regular (=every localization at maximal ideal is a regular local ring) and $B$ is a smooth $A$-algebra. For the definition of a smooth algebra, please see the first page of</p>
<blockquote>
<p>Robert A Morris, Stuart Sui-Sheng Wang, <em>A Jacobian criterion for smoothness</em><br>
Journal of Algebra, Volume 69, Issue 2, April 1981, Pages 483–486<br>
<a href="http://dx.doi.org/10.1016/0021-8693(81)90217-9" rel="nofollow">doi:10.1016/0021-8693(81)90217-9</a></p>
</blockquote>
<p>which says that $B$ is a smooth $A$-algebra if the following two conditions are satisfied:</p>
<p>(1) For each $A$-algebra $C$, and each ideal $J$ in $C$ with $J^2=0$, the canonical homomorphism $Hom_{A-alg}(B,C) \to Hom_{A-alg}(B,C/J)$ is surjective.</p>
<p>(2) $B$ is finitely presented as an $A$-algebra.</p>
<p>My question: Is it true that $B$ must be regular too? If not, what additional conditions should we assume in order that $B$ will be regular?</p>
<p>I really apologize if this question is trivial; it's just that only recently I have started to study regular rings/smooth extensions.</p>
http://mathoverflow.net/q/20759013Origin of the term "Diophantine equation"Renéhttp://mathoverflow.net/users/179072015-05-25T23:04:37Z2015-05-26T10:45:17Z
<p>It seems that the term "Diophantine equation" has been around at least since the second half of the 19th century, since the historian Hermann Hankel writes (polemically) in the chapter on Diophantus in his <em>Zur Geschichte der Mathematik in Alterthum und Mittelalter</em> (p. 163):</p>
<blockquote>
At this point, there is a mistake to be corrected, a mistake that is reinforced by false nomenclature and therefore, or so I fear, impossible to weed out. In education, one designates linear equations of the form $ax+by=c$, that are to be solved in integers $x,y$, as <i>Diophantine</i>. Now, not only did Diophantus not know the solution method of these equations, that in the West was first obtained by his commentator Bachet, but the very problem would have been utterly alien to him, since he never fixes the condition that his solutions should be integral, but is completely satisfied with rational solutions.
</blockquote>
<p>(Italics are mine.) Now, in connection with this passage, I have the following questions: </p>
<ol>
<li>When did people start talking/writing about "Diophantine equations"? </li>
</ol>
<p>And also: </p>
<ol start="2">
<li>Were Diophantine equations originally considered as "polynomial equations to be solved in rational numbers", in accordance with Diophantus' own preference, or was the mistake that Hankel aims to correct made from the beginning?</li>
</ol>
http://mathoverflow.net/q/2075808Nontrivial finite group with trivial cohomology in prescribed degreeJens Reinholdhttp://mathoverflow.net/users/142332015-05-25T21:09:40Z2015-05-26T10:14:36Z
<p>For any non-trivial finite group $G$ there exists some $j > 0$ such that $H^{aj}(G) \neq 0$ for all $a = 1,2,3,\dots$, see e.g. this question: <a href="http://mathoverflow.net/questions/64688/non-vanishing-of-group-cohomology-in-sufficiently-high-degree">Non-vanishing of group cohomology in sufficiently high degree</a>.</p>
<p>Furthermore, it is not known whether there exists a positive $N$ such that $H^i(G) \neq 0$ for $0 < i \leq N$ implies $G = 1$, see this question <a href="http://mathoverflow.net/questions/52552/nontrivial-finite-group-with-trivial-group-homologies">Nontrivial finite group with trivial group homologies?</a>.</p>
<p>My question is: Given a positive integer $i$, does there always exist a non-trivial finite group $G$ with $H^i(G) = 0$? (All cohomology groups are meant to be with $\mathbb Z$ coefficients.)</p>
http://mathoverflow.net/q/2075702Maximizing entropy under constraintsBenoît Kloecknerhttp://mathoverflow.net/users/49612015-05-25T19:29:42Z2015-05-26T11:19:31Z
<p>This question is about an extension of the variational principle in thermodynamical formalism when one adds linear constraints to the measures.</p>
<p>Consider the one-sided shift $\sigma:\mathcal{A}^\mathbb{N}$ where $\mathcal{A}$ is a finite alphabet ($\{0,1\}$ say). Consider a test function $\varphi:\mathcal{A}^\mathbb{N}\to \mathbb{R}$, of Hölder regularity with respect to a usual metric, $d(x,y) = 2^{-\min\{i,x_i\neq y_i\}}$ say, and assume that $0$ is in the interior of the interval of values taken by $\int\varphi \,d\mu$ when $\mu$ runs over shift-invariant probability measures.</p>
<p>Consider the following question:</p>
<blockquote>
<p>How to find a shift invariant probability measure $\mu$ such that $\int\varphi \,d\mu=0$ and which maximizes the metric entropy $h(\sigma,\mu)$ under this constraint?</p>
</blockquote>
<p>Together with collaborators we developed tools about the thermodynamical formalism which allow us to answer such a question. What I would like to know is whether this should be considered a mere illustration of our methods or a true application.</p>
<blockquote>
<p><strong>My question:</strong> Is the answer to the above question known? If yes, could you give a reference? How difficult is it? If no, do you know a reference where this kind of question is explicitly asked?</p>
</blockquote>
<p>In fact, we can deal with multiple constraints of the same form, and optimize pressure $h(\sigma,\mu) + \int A \,d\mu$ of yet another function $A$, and we can deal with more general dynamical systems (we mostly need a spectral gap for the Ruelle operator).</p>
<p>Just to examplify the above, let me give the answer in two cases, using $\mathcal{A}=\{0,1\}$ in denoting by $w*$ the cylinder defined by a finite word $w$.</p>
<p><strong>Example 1:</strong> among shift-invariant measures $\mu$ such that $\mu(0*) =
.9$, the Bernoulli measure of parameter .9 (i.e. the law of the word
$\alpha_1\alpha_2\dots$ where the $\alpha_j$ are i.i.d. random variables taking the value $0$ with probability .9) maximizes entropy.</p>
<p><strong>Example 2:</strong> Among shift-invariant measures $\mu$ such that
$\mu(01*) = 2\mu(11*)$, the Markov measure associated
to the transition probabilities
\begin{align*}
\mathbb{P}(0\to 0) &= 1-a & \mathbb{P}(0\to1) &= a \\
\mathbb{P}(1\to0) &= \frac23 & \mathbb{P}(1\to1) &= \frac13
\end{align*}
where $a$ is the only real solution to
$$(1-a)^5=\frac{4}{27} a^2 \qquad (a\simeq 0.487803)$$
maximizes entropy.</p>
<p>The first example feels quite obvious (but I do not know if there is a really easy way to prove it) while the second surprised me quite a bot (hoping that I did not messed up the computations).</p>
http://mathoverflow.net/q/2075524Multi-podal pointsErel Segal-Halevihttp://mathoverflow.net/users/344612015-05-25T17:08:00Z2015-05-26T10:43:40Z
<p>Two points $x,y \in \mathbb{R}^n$ are called antipodal if $x = -y$.</p>
<p>Stated differently, $x,y$ are antipodal if:</p>
<ul>
<li>They have the same absolute value in each of their $n$ coordinates;</li>
<li>Each of their non-zero coordinates appears with a positive sign exactly once (either in $x$ or in $y$).</li>
</ul>
<p>The <a href="https://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem" rel="nofollow">Borsuk-Ulam theorem</a> is a well-known theorem about antipodal points.</p>
<hr>
<p>Let's call a set of points $\{x_i\}_{i=1}^k \subseteq \mathbb{R}^n$ <em>multipodal</em> if:</p>
<ul>
<li>They have the same absolute value in each of their $n$ coordinates;</li>
<li>Each of their non-zero coordinates appears with a positive sign exactly once.</li>
</ul>
<p>Every pair of antipodal points is a multipodal set of size 2. Here are some multipodal sets of size 3:</p>
<ul>
<li>$(+1, -2, -3); (-1, +2, -3); (-1, -2, +3)$</li>
<li>$(+1, -2, 0, -3, -2); (-1, +2, 0, -3, -2); (-1, -2, 0, +3, +2)$</li>
<li>$(+1,+2); (-1,-2); (-1,-2)$</li>
<li>$(-1,+2); (+1,-2); (-1,-2)$</li>
</ul>
<hr>
<p>Let's call a function $f$ <em>k-podal</em> if, for every multipodal set $X$ of size $k$:</p>
<p>$$\sum_{x\in X}f(x) = 0$$</p>
<p>In particular, a 2-podal function is just another name for an odd function: $f(-x)=-f(x)$.</p>
<p>I am looking for information about these "multipodal points" and "$k$-podal" functions. In particular:</p>
<ul>
<li>Is there another, standard term for them?</li>
<li>What do we know about the degree of "$k$-podal" functions, for $k\geq 3$? E.g, is there a theorem saying that every $k$-podal function from $S^n$ to $S^n$ has an odd degree? (this is known for $k=2$).</li>
<li>The following theorem is true for $k=2$ - it is the <a href="https://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theorem" rel="nofollow">Borsuk-Ulam theorem</a>:</li>
</ul>
<blockquote>
<p>Every continuous $k$-podal function $f: S^n \to \mathbb{R}^n$ has a zero.</p>
</blockquote>
<p>Is it correct for any $k\geq 3$?</p>
http://mathoverflow.net/q/20754013How many traces are there on Temperley-Lieb, Fuss-Catalan, Iwahori-Hecke, Birman-Wenzl-Murakami-Kauffman, ... algebras?მამუკა ჯიბლაძეhttp://mathoverflow.net/users/412912015-05-25T10:57:00Z2015-05-26T14:43:56Z
<p>There is a theorem (I believe by Ocneanu) that the Markov trace on the tower of <a href="http://en.wikipedia.org/wiki/Temperley%E2%80%93Lieb_algebra" rel="nofollow">Temperley-Lieb algebras</a> is (essentially) unique.</p>
<p>What about just traces on separate algebras? That is, take one of them, say $\mathbf{TL}_n$; what is the dimension of the space of linear functionals $\operatorname{tr}$ on it with the property $\operatorname{tr}(xy)=\operatorname{tr}(yx)$? (If one considers their versions with free parameters, i. e. as algebras over polynomials in parameters as free variables, then linearity is understood over these polynomials).</p>
<p>Equivalently, this is the question about the rank of the quotient of, say, $\mathbf{TL}_n$ by the subspace (resp. submodule over polynomials) spanned by all elements of the form $xy-yx$. If you want, about the rank of the 0th Hochschild homology (with itself as coefficients).</p>
<p>I could in fact restrict myself to the mother of them all - the group algebra of the braid group on $n$ strings. There, as Qiaochu Yuan says in a comment to the answer, there are as many independent traces as conjugacy classes in the group.</p>
<p>I've played with it a bit in lower dimensions, and got impression that the trace might be still essentially unique.</p>
<p>However as the answer below shows, this is certainly not so already for $\mathbf{TL}_4$ over $\mathbb C$.</p>
<p>Just for fun, here is the corresponding table of the (nontrivial cases of) "$xy=yx$" for $\mathbf{TL}_3$.</p>
<p><img src="http://i.stack.imgur.com/BxDBs.jpg" alt="enter image description here"></p>
http://mathoverflow.net/q/2075156Pseudodifferential operators on spaces with boundarystudenthttp://mathoverflow.net/users/740802015-05-25T01:31:59Z2015-05-26T10:07:15Z
<p>Consider the upper half space $\mathbb{R}^n_{+} = \{x = (x_1,..,x_n) \in \mathbb{R}^n : x_n \geq 0\}$. Consider the Laplacian on this space with either the Dirichlet boundary condition or the Neumann boundary condition. My question is: can the Laplacian $\Delta$ under such boundary conditions be treated as a pseudodifferential operator of order 2? I can understand that the usual trick of Fourier inversion does not work, but if we still forcibly write
$$-\Delta f(x) = \int p(x, \xi)\hat{f}(\xi)e^{ix.\xi}d\xi$$ and assume that $p(x, \xi)$ belongs to some symbol class $S^m_{\rho, \delta}$, what can go wrong?</p>
http://mathoverflow.net/q/2075112Can the epsilon induction condition be presented with successor case and limit case, similar to transfinite induction?Ioachim Drugushttp://mathoverflow.net/users/310352015-05-24T23:00:00Z2015-05-26T14:00:01Z
<p>The epsilon induction looks like this:
$\forall x \Big(\forall y (y \in x \rightarrow P(y)) \rightarrow P(x)\Big) \rightarrow \forall x P(x)$</p>
<p>Here, the quantifiers "run over" any sets and not only over ordinals, for which there are notions "successor" and "limit" used in transfinite induction (<a href="http://en.wikipedia.org/wiki/Transfinite_induction" rel="nofollow">http://en.wikipedia.org/wiki/Transfinite_induction</a>). </p>
<p>Thus, the question can be made more precise if notions similar to "successor" and "limit" for any sets are defined. There is some similarity between unary successor operation for ordinals and the binary operation over any sets called "adjunction" denoted as ";" (semicolon) and defined like this: $x;y = x \cup ${$y$}. But any non-empty set X is a successor of a set, since $X = (X \setminus ${x}) ; x, for any element x of $X$.</p>
<p>My interest in adjunction is due to a theory of inheritably finite sets based on adjunction operation with one induction principle by Laurence Kirby <a href="http://projecteuclid.org/euclid.ndjfl/1257862036" rel="nofollow">http://projecteuclid.org/euclid.ndjfl/1257862036</a></p>
<p>This theory has the negation of infinity axiom as an axiom as well as the axioms:</p>
<p>$ 0;x \ne 0, \ \ \ \ \ \ (1)$</p>
<p>$(x;y);y = x;y, \ \ \ \ (2)$</p>
<p>$(x;y);z = (x;z);y,\ \ \ (3)$</p>
<p>$(x;y);z = x;y \ \leftrightarrow \ x;z = x \vee z = y, \ (4)$</p>
<p>and the axiom scheme</p>
<p>$ P(0)\ \& \ \forall x y$ (P(x) & P(y) $\to P(x; y)) \to \forall x P(x). \ (5)$</p>
<p>Interestingly, in this theory, all regular set theoretic operations, including the unary union operation, are defined by induction. But probably, this principle cannot help defining this operation if the infinity axiom is postulated, and without the union operation a set theory sounds too poor.</p>
<p><strong>Addition 1</strong></p>
<p>To minimally modify the main text (since it was read by some participants of this forum), I just deleted its ending and labeled several axioms to make to them reference here, where I continue with comments clarifying the question. </p>
<p>The transfinite induction principle (<a href="http://en.wikipedia.org/wiki/Transfinite_induction" rel="nofollow">http://en.wikipedia.org/wiki/Transfinite_induction</a>) discusses about ordinals, and its condition is a conjunction of 3 formulas called "cases" - "zero case", "successor case", "limit case". These formulas discuss about three pairwise disjoint subclasses of the universe of discourse of set theory. By its form, the transfinite principle differs from the mathematical induction principle, which discusses about finite ordinals, without the limit case. One can say that Kirby extended the successor case so that it discusses about sets - hereditarily finite sets, and not only about finite ordinals.</p>
<p>Based on the above reasoning, the following hypothesis imposes as verisimilar: the condition of epsilon-induction which is about arbitrary sets (and not only about ordinals), can be also represented as a conjunction of 3 cases, discussing about 3 pairwise disjoint subclasses of the universe of discourse of set theory.</p>
<p>The induction principle is used in "proofs by induction", and such proofs employ different methods for each case. Therefore, if this hypothesis is a theorem in a set theory (in particular - an axiom), then this theorem will be a good contribution to proof theory for that set theory.</p>
<p>In search of a new form of induction principle, I proceed the manner described next. Suppose T is a theory with the axiom of infinity (and not its negation as in Kirby theory) and the axioms (1) - (4) of Kirby theory (but not also with the axiom scheme (5) of Kirby theory). For theory T, I present the induction principle with a condition which is the conjunction of three "cases" -- universal closures of these formulas:</p>
<p>Zero case: $P(0)$</p>
<p>Successor case: $\sigma(P)$ </p>
<p>Limit case: $\lambda(P)$,</p>
<p>Here, $\sigma(P)$ and $\lambda(P)$ are built of terms like "P(x)", where the expression "to be built of" has the same meaning as the expression "$ (P(x) \& P(y) \to P(x; y))$ is built of $P(x)$" in (5) -- I am not sure how they would say this in logic. </p>
<p>Now, I am looking for the formulas $\sigma(P)$ and $\lambda(P)$. "Unfortunately", "$ (P(x) \& P(y) \to P(x; y))$ cannot serve as $\sigma(P)$, because every non-empty set is described by this condition, so that this condition covers both the "successor case" and the "limit case". Thus, only a stronger condition than $ (P(x) \& P(y) \to P(x; y))$ can play the role of $\sigma(P)$, and then $\lambda(P)$ would be $\neg (P(0) \vee \lambda(P))$. But I did not find such a "successor case" stronger than Kerby's which would also make sense in set-theoretic conceptuality. </p>
<p>I am looking for a three case induction proceeding from Kirby theory only because this theory looks attractive to me due to its axioms which use an operation rather than the relation of membership (notice, that $x ; y = y \ \leftrightarrow \ x \ \epsilon \ y$, and thus the membership relation can be expressed in this theory). But my question is wider than the question which makes reference to Kirby theory. </p>
http://mathoverflow.net/q/2074593Earliest source for a Lie algebra constructionArnold Neumaierhttp://mathoverflow.net/users/569202015-05-24T11:26:59Z2015-05-26T14:01:00Z
<p>I am looking for the earliest reference to the fact that any associative algebra becomes a Lie algebra with bracket $AXB-BXA$, where $X$ is a fixed element of the algebra. This is observed in the following paper:</p>
<p>Yanovski, A. B. "Linear bundles of Lie algebras and their applications." Journal of Mathematical Physics 41.11 (2000): 7869-7882.</p>
<p>But surely this is a much older result, at least for the case of matrix algebras.</p>
http://mathoverflow.net/q/19595910An inequality for copulasClark Kimberlinghttp://mathoverflow.net/users/614262015-02-07T20:12:45Z2015-05-26T13:44:03Z
<p>Suppose that $f$ from $[0,\infty]$ onto $[0,1]$ is completely monotonic on $(0,\infty)$, and let $g$ be the inverse of $f$. For $(u,v)$ in $[0,1]^{2}$, define $C(u,v) = f(g(u)+g(v))$, and let $a = (u+v)/$2. Is $$C(a,a) - a^{2} \geq C(u,v) - uv$$ for all $(u,v)$ in $[0,1]^{2}$?</p>
<p>Some background: $C$ is an Archimedean copula, and if $f(x) = \mathrm{e}^{-x}$, then $C(u,v) = uv$. The complete monotonicity of $f$ is necessary and sufficient for $C$ to be extendible to a multivariate Archimedean copula, for the purpose of "joining" arbitrarily many distribution functions to form a joint distribution. Well-known (families of) copulas which apparently satisfy the inequality are the Clayton, Frank, and Gumbel families. A possibly new family (for $t>0$) results from $$f(x) = (\mathrm{e}^{(x+1)^{-t}}-1)/(e-1)$$.</p>
<p>"Completely monotonic" is not another term for "strictly monotonic"; $f$ is completely monotonic on (a,b) if $$(-1)^{k}f^{(k)}(x) \geq 0$$ for all x in (a,b). (Widder, <em>The Laplace Transform</em>, 1946, p. 145)</p>
<p>May 26 2015 - the question remains largely <strong>unanswered.</strong> The two "Answers" have proofs for simple examples, but the question is for a very large class of copulas (see, for example Roger Nelsen, <em>An Introduction of Copulas, 2nd ed.</em>, pp. 151-155). Can someone find a proof that uses the complete monotonicity of $f$ - or a counterexample?</p>
http://mathoverflow.net/q/1907912A surface on which all regular curves have nowhere vanishing curvatureAli Taghavihttp://mathoverflow.net/users/366882014-12-15T17:17:41Z2015-05-26T16:07:05Z
<p>Let $S$ be a surface in $\mathbb{R}^{3}$ such that every regular curve $\gamma\subset S$ has nowhere vanishing curvature, that is $\kappa(z)\neq 0$ for all $z\in \gamma$. Does this imply that $S$ is a part of a sphere?</p>
http://mathoverflow.net/q/1442467Seeking conceptual explanation of these nice bijections on roots of unityMichael Zievehttp://mathoverflow.net/users/304122013-10-08T02:56:24Z2015-05-26T16:08:52Z
<p>I proved the following facts by unenlightening calculations. Since the statements are quite clean, I think there should be a conceptual explanation for them, which my proof certainly is not.</p>
<p>Let $q$ be a prime power, and let $\mu_{q+1}$ be the set of $(q+1)$-th roots of unity in the finite field $\mathbf{F}_{q^2}$. If $b\in\mu_{q+1}$ and $c\in\mathbf{F}_{q^2}\setminus\mathbf{F}_q$ then
$$
x\mapsto \frac{cx-bc^q}{x-b}
$$
maps $\mu_{q+1}$ to $\mathbf{F}_q\cup\{\infty\}$. If $b\in\mu_{q+1}$ and $d\in\mathbf{F}_{q^2}\setminus\mu_{q+1}$ then
$$
x\mapsto \frac{x-bd^q}{dx-b}
$$
maps $\mu_{q+1}$ to itself. (It is also true that these are the only degree-one rational functions which map $\mu_{q+1}$ to either $\mathbf{F}_q\cup\{\infty\}$ or $\mu_{q+1}$, but I'm mainly interested in understanding the existence.)</p>
<p>I tagged this "group theory" because the first fact vaguely feels like a connection between orbits of a nonsplit torus and a split torus in $\textrm{PGL}_2(q)$. It's tempting to identify $\mathbf{F}_{q^2}$ with $\mathbf{F}_q\times\mathbf{F}_q$, and consider the resulting action of $\textrm{GL}_2(q)$ on $\mathbf{F}_{q^2}$, but I don't see how to go further in this way.</p>
<p>Any suggestions?</p>