Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2016-02-12T14:39:56Zhttp://mathoverflow.net/feedshttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://mathoverflow.net/q/2309690On Incidence structure of finite Projective planedasshttp://mathoverflow.net/users/619312016-02-12T14:31:28Z2016-02-12T14:31:28Z
<p>Consider a finite projective plane $\mathcal{P}$ over a finite field $F_q$, $q$ a prime power. Is it possible to define a map $f:\mathcal{P}\times \mathcal{P}\rightarrow \mathcal{P}$ such that</p>
<p>(i) $f$ is commutative,</p>
<p>(ii) $f(x,y)\neq x, y$ for all $x,y\in \mathcal{P}$,</p>
<p>(iii) $f(x,y),x$ and $y$ are not incident on the same line for any two distinct
points $x$ and $y$ in $\mathcal{P}$,</p>
<p>(iii) $f(x,y),f(y,z)$ and $f(z,x)$ are not incident on the same line for any three distinct points $x,y$ and $z$ in $\mathcal{P}$.</p>
http://mathoverflow.net/q/2309680Algebraic operations with memory hardness propertiesJeff Burdgeshttp://mathoverflow.net/users/141632016-02-12T14:19:42Z2016-02-12T14:19:42Z
<p>In cryptography, there are password hash functions like <code>scrypt</code> and <code>argon2</code> for which the fastest known algorithms employ large lookup tables, and more space efficient algorithms would require massive recalculation. </p>
<p>Are there any computable abelian groups whose operations appears to exhibit similar time-space trade offs? If not, are there any algebraic objects with such properties?</p>
<p>There are for example some formulations of elliptic curves in which fast point or scalar multiplication algorithms make use of several extra coordinates, but afaik never that many. I am unaware of any algebraic objects for which the algorithms benefit dramatically from thousands or millions of extra coordinates. There might however by constructions of abelian groups on Jacobian varieties or something. Or perhaps some non-abelian matrix groups with efficient sparse-ish representations that needed the full matrix during multiplication. </p>
<p>I'm asking because the current proposed post-quantum public-key systems require large key sizes and lack some properties familiar from elliptic curves, especially around blinding. A priori, one expects homomorphic properties to weaken any system's security, so it might take ages for post-quantum systems with desirable homomorphic properties.</p>
<p>A public-key system for which the key size is relatively small but the operations' need extreme space for performance could be viewed as post-quantum even if technically vulnerable to Shor's algorithm, such as by being based on an abelian hidden subgroup problem. In addition, if such a public-key system were based upon objects similar to elliptic curves, such as Jacobians, then cryptographers might discover optimizations and gain confidence in the system relatively quickly. </p>
http://mathoverflow.net/q/2309671Seeking more information regarding the "rigoidal category" of $\mathbb{N}$-graded setsgoblinhttp://mathoverflow.net/users/260802016-02-12T14:16:37Z2016-02-12T14:29:50Z
<p><strong>Definitions.</strong></p>
<ul>
<li><p>By an $\mathbb{N}$-<em>graded set</em>, I mean a set $X$ together with a function $|\Box|_X:\mathbb{N} \leftarrow X,$ called the <em>grading.</em> These will simply be called <em>graded sets</em> hereafter.</p></li>
<li><p>If $Y$ and $X$ are graded sets, then a function $f : Y \leftarrow X$ is said to be a <em>morphism of graded sets</em> iff it preserves the grading up to equality. Meaning that: $$\left(\mathop{\forall}_{x:X}\right)\;|f(x)|_Y = | x|_X$$</p>
<p>(There's a weaker version in which we only require $\leq$ in the above condition, but we won't be using that here.)</p></li>
<li><p>Write $\mathbf{Set}^\mathbb{N}$ for the category of graded sets (this is equivalent to the functor category $\mathbf{Set}^\mathbb{N}$).</p></li>
<li><p>If $Y$ and $X$ are graded sets, define $Y \oplus X$ as follows.</p>
<ul>
<li><p>Its underlying set is $Y \times X$.</p></li>
<li><p>The new grading is defined additively: $$\left(\mathop{\forall}_{y,x:X}\right)\; |(y,x)|_{Y \oplus X} = |y|_Y+|x|_X$$</p></li>
</ul></li>
<li><p>If $Y$ and $X$ are graded sets, define $Y \otimes X$ as follows.</p>
<p>$$Y \otimes X = \bigsqcup_{y:Y} X^{\oplus |y|}$$</p></li>
<li><p>Define a function $\mathbf{Set}^\mathbb{N} \leftarrow \mathbb{N}$ by assigning to each natural number $n$ a graded set $\underline{n}$ with precisely one element, whose grade equals $n.$</p></li>
</ul>
<p>We have: $$\left(\mathop{\forall}_{a,b:\mathbb{N}}\right)\; \underline{a+b} \cong \underline{a} \oplus \underline{b}, \qquad \left(\mathop{\forall}_{a,b:\mathbb{N}}\right)\;\underline{ab} \cong \underline{a} \otimes \underline{b}$$</p>
<p>I haven't checked the details, but this seems to make $\mathbf{Set}^\mathbb{N}$ into a kind of categorified rig; a "rigoidal category," if you will, with identity elements $\underline{0}$ and $\underline{1}$ respectively. My main interest in this structure is to give a description of operads; it seems to be the case that an <em>operad</em> can be defined as a monoid object in the monoidal category $(\mathbf{Set}^\mathbb{N}, \otimes, \underline{1}).$ Anyway, I'd like to get more information.</p>
<blockquote>
<p><strong>Questions.</strong></p>
<p><strong>Q0.</strong> Does this description of operads work? If so, can we describe symmetric operads in a similar way? What about cartesian operads? Where can I learn more?</p>
<p><strong>Q1.</strong> Does this "rigoidal category" misbehave in any unexpected ways? Failure of distributivity, etc? Further to that, supposing that we want $\mathbf{Set}^\mathbb{N}$ to form a "rigoidal category," what are the appropriate axioms of a "rigoidal category"? Is there somewhere I can learn more about such structures?</p>
</blockquote>
http://mathoverflow.net/q/2309631Is the solution of $-\Delta u + \lambda u =0$, $\partial_\nu u=\alpha$ in $L^\infty(\Omega)$?ACAhttp://mathoverflow.net/users/865882016-02-12T13:43:46Z2016-02-12T14:10:44Z
<p>Let $\lambda > 0$ be a constant and let $u$ be the weak solution on a bounded domain $\Omega$ of
$$-\Delta u + \lambda u = 0 \quad\text{in $\Omega$}$$
$$\partial_\nu u = \alpha \quad \text{on $\partial\Omega$}$$
where $\partial_\nu u$ is the normal derivative and $\alpha > 0$ is a constant. Actually $u$ satisfies
$$\int_\Omega \nabla u \nabla \varphi + \lambda u\varphi = \int_{\partial\Omega} \alpha \varphi$$
for all $\varphi \in H^1(\Omega)$.</p>
<p>In fact $u \in H^2(\Omega)$. </p>
<p>My question is, is $u \in L^\infty(\Omega)$, and what estimate is available on $\lVert u \rVert_{L^\infty(\Omega)}$? Can it bounded above by $\alpha$ and $\lambda$? I couldn't find anything after trying books by Grisvard, Wloka, Lions, ...</p>
http://mathoverflow.net/q/2309602Asymptotics for the number of abelian groups of order at most $x.$Igor Rivinhttp://mathoverflow.net/users/111422016-02-12T13:05:18Z2016-02-12T14:15:14Z
<p>The number of abelian groups of order $n$ (call it $a(n)$ is a studied subject (see <a href="http://oeis.org/A000688" rel="nofollow">http://oeis.org/A000688</a>), but I can't seem to find any asymptotic results. Obviously, there is no asymptotic for $a(n)$ itself (it is much too irregular), but there should an asymptotic for $\sum_{n\leq x} a(n),$ but I can't seem to find a reference.</p>
<p>Roberto answered the above, but another question is whether one has any distributional results (how high are the maxima, is there a limiting distribution, etc).</p>
http://mathoverflow.net/q/2309592Maximal TB number and slice genus relation of a knot in any 3-manifoldmrvscgnhttp://mathoverflow.net/users/865812016-02-12T12:50:13Z2016-02-12T12:50:13Z
<p>Lisca-Stipsicz say that if a knot $K\subset{S^{3}}$ satisfies $g_{s}(K)>{0}$ and $TB(K)=2g_{s}(K)-1$ then $S^{3}_r(K)$ carries positive, tight contact structures
for every $r\neq{TB(K)}$ where $S^{3}_r(K)$ is the new manifold by performing $r$-surgery to $K$ in $S^{3}$. Can we say this result is true for any 3-manifold not only for $S^{3}$?</p>
http://mathoverflow.net/q/230958-2Should I learn information technology? [on hold]H. Józsefhttp://mathoverflow.net/users/865832016-02-12T12:32:22Z2016-02-12T12:32:22Z
<p>I am 18 y.o., and I want to go to university. I will study
Software Engineering (Computer science), but I do not know yet what I want to be exactly.
I love math, phisycs, psychology, philosophy, literature, languages and I am interested in IT.
If I will study CS, in the future I will be able to go through another area (for example: math/phisycs/psychology/philosophy/languages or something else)? Of course, when I find my favourite profession, I will work hard and do anything to be the best! But now, I am just a confused 18 y. o. boy who does not know what to do in the future.
But I can feel, I want to be a scientist. </p>
<p>What do you think about it? Will I find my favourite profession in the future? </p>
<p>Sorry for the bad english! </p>
http://mathoverflow.net/q/2309560Cells in affine Weyl groupswkyhttp://mathoverflow.net/users/142262016-02-12T11:10:15Z2016-02-12T11:10:15Z
<p>This may sound like a very general question, which pretty much reflects my ignorance on the subject.</p>
<p>In the case of Weyl groups $W$, there is a notion of left/right/double cells, which is roughly some partitions on the elements of $W$. It turns out that these cells afford representations of $W$ (or $W \times W$ in the case of double cells).</p>
<p>The structure of these cells are well-known, for example in the case of $W = S_n$, the partition of cells is related to the Robinson-Schensted algorithm. </p>
<p>There are applications of cells in other aspects of mathematics. Due to my ignorance of the subject, I will simply refer what I mean by 'other aspects' to the notes <a href="https://www.google.com.hk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0ahUKEwi93rO7hfLKAhViOKYKHXMTBOsQFggeMAA&url=http%3A%2F%2Fwww.liegroups.org%2Fpapers%2Fsummer06%2Fcells.pdf&usg=AFQjCNGBhKai1Eb26tVvdg4vmxWIAgc2JA" rel="nofollow">here</a>. </p>
<p>On the other hand, Lusztig also defined 'cells' in affine Weyl groups in a series of papers under the title (Cells in affine Weyl groups), which I have literally zero knowledge about. Here is a couple of questions I am particularly interested in:</p>
<p>1) I read from Sommers-Gunnells (<a href="http://people.math.umass.edu/~esommers/dynkin_elements.pdf" rel="nofollow">here</a> p.7) that the double cells in affine Weyl group are parametrized by nilpotent orbits in the Langlands dual. This, to my knowledge, is quite different from the case of double cells in Weyl group, whose double cells are parametrized by <strong>special</strong> orbits in the Langlands dual (correct me if I am wrong). Can anyone give us some intuition on how the parameterization in the affine Weyl group case works?</p>
<p>2) There is a notion of canonical left cell in each double cell of affine Weyl group (<a href="http://www.sciencedirect.com/science/article/pii/000187088890031X" rel="nofollow">here</a>). Are there any results on how these canonical left cells are computed?</p>
http://mathoverflow.net/q/2309535Weak*-closure of finite rank operators on dual spaceHannes Thielhttp://mathoverflow.net/users/249162016-02-12T10:06:10Z2016-02-12T12:34:06Z
<p>Given a Banach space $X$, we consider the space $B(X^*)$ of bounded, linear operators on $X^*$ with the weak*-topology from its canonical predual $B(X^*)_*=X^*\hat{\otimes}X$. What is $\overline{F(X^*)}^{wk*}$, the weak*-closure of the finite rank operators on $X^*$? Since this is rather vague, here are some concrete questions:</p>
<blockquote>
<p>Q1: Do we always have $K(X^*)\subseteq\overline{F(X^*)}^{wk*}$, i.e., is every compact operator in the weak*-closure of finite-rank operators?</p>
<p>Q2: Is there a characterization when $B(X^*)=\overline{F(X^*)}^{wk*}$, i.e., when the finite-rank operators are weak*-dense?</p>
<p>Q3: Is there a characterization when $B(X^*)=\overline{K(X^*)}^{wk*}$, i.e., when the compact operators are weak*-dense?</p>
</blockquote>
<p>Considering $B(X)$ as a subalgebra of $B(X^*)$ in the usual way, we may ask related questions in connection with $F(X)$ and $K(X)$. The principle of local reflexivity implies $\overline{F(X)}^{wk*}=\overline{F(X^*)}^{wk*}$ in $B(X^*)$. However, it is not clear to me if we always have $\overline{K(X)}^{wk*}=\overline{K(X^*)}^{wk*}$ (I guess not). Therefore, we may also ask:</p>
<blockquote>
<p>Q4: Do we always have $K(X)\subseteq\overline{F(X^*)}^{wk*}$?</p>
</blockquote>
http://mathoverflow.net/q/2309511Continuous non-constant function with infinite intersections with horizontal line on a compact interval?Matija Sreckovichttp://mathoverflow.net/users/865732016-02-12T09:42:19Z2016-02-12T10:29:50Z
<p>The title might be misleading, but whether such a function exists is what boggles me about the following problem:</p>
<p>Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function such that for all $a<b $ satisfying $f(a)=f(b)$, there exists $c$ in $(a,b)$ such that $f(a)=f(c)=f(b)$.</p>
<p>Prove that $f$ is monotonous on $\mathbb{R}$.</p>
<p>What I'm intuiting about this problem is that for every such pair $a, b$ the function f is constant on $[a,b]$. Therefore, $f$ could have no local extremum. However, I'm not sure how to go about proving this. Is the set $X=\{x\in [a,b]| f(x)=f(a)\}$ be dense in $[a,b]$? If it is, how could I prove it? What also troubles me, though, is the existence of nowhere-monotone functions such as the Weierstrass function. Does the Weierstrass function satisfy the problem condition?</p>
<p>Furthermore, I'd like to be able to prove that an arbitrary horizontal line $g(x)=u, u \in \mathbb{R}$ either intersects f at a single point, or at a compact interval $[a_{1},b_{1}]$ $(a_{1}<b_{1})$. </p>
<p>I'm not sure if these two conditions are enough to prove that the function is monotonous. </p>
<p>What is the best approach towards proving this problem? Am I on the right track?</p>
http://mathoverflow.net/q/2309501Wave-like equation with 1st order time derivative and non-constant coefficientsNex_Friedrichhttp://mathoverflow.net/users/576992016-02-12T09:37:12Z2016-02-12T14:23:30Z
<p>We start with the following recurrence relation for complex coefficients $c_{n,m}$:
$$i\dot{c}_{n,m}(t) = \sqrt{(n+1)(n+2)(m-1)m}c_{n-2,m-2} + \sqrt{n(n-1)(m+1)(m+2)}c_{n+2,m-2}$$
where $\dot{c}_{n,m}$ denotes time derivative and $n,m = 0,1,2,...$.</p>
<p>In order to sole this equation for and find time evolution of $c_{n,m}$ (we assume that at $t=0$ we know values of $c_{n,m}$) defining generation function:
$$G(x,y,t) = \sum\limits_{n,m=0}^{\infty}x^{n}y^{m}\alpha_{n,m}c_{n,m}(t)$$
with some unknown coefficient $\alpha_{n,m}$ that can be adjusted manually. Recurrence relation can be rewritten in terms of differential equation for $G$:
$$i\partial_{t}G(x,y,t) = (y^2\partial_{x}^2 + x^2\partial_{y}^2)G(x,y,t)$$
if we set
$$\alpha_{n,m} = \alpha_{n+2,m-2}\sqrt{\frac{(n+1)(n+2)}{m(m-1)}}$$
We can assume that $G(x,y,0)$ is known and converge domain is $[0,1] \times [0,1]$ for any $t$.</p>
<p>Is there a way to find function $G(x,y,t)$ that is not represented by infinite series? This is a special kind of diffusion equation.</p>
http://mathoverflow.net/q/2309472Bott-Samelson theorem for simplicial setsghknbhxbfhdbyhttp://mathoverflow.net/users/865712016-02-12T08:36:40Z2016-02-12T10:33:57Z
<p>Let $X\in \mathrm{sSet}$ and $FX$ be the Milnor's construction (model for $\Omega\Sigma |X|$) - in each dimension $n$ this is the free group on $X_n$ with one relation $*=1$. I'm interested in $\mathbb Z FX$ - the level-wise group ring on this construction - which homotopy groups gives the integral homology of $FX$. From the classical Bott-Samelson theorem we know that (over the field) homology of $\Omega\Sigma |X|$ as Hopf algebra is isomorphic to the tensor algebra on reduced homology of $X$ (as far as I understand, restriction to field here is given only to ensure Hopf algebra structure on homology). I'm trying to trace this fact back to simplicial level. Note that $\mathbb Z FX$ is also a Hopf algebra (comultiplication is given by diagonal) and homotopy groups of the abelianization of Milnor's construction are exactly reduced homology of $X$. So my question here: is there a (canonical!) morphism of simplicial Hopf algebras $T(FX_{ab}) \to \mathbb Z FX$ (or in another direction) which is homotopy equivalence on the underlying simplicial sets?</p>
<p>Few observations here:</p>
<ol>
<li><p>By Magnus-Witt, associated graded $\bigoplus I^k/I^{k+1}$ of $\mathbb Z FX$ is isomorphic to the tensor algebra $T(FX_{ab})$ as algebras (here $I^k$ are powers of augmentation ideal). Also this augmentation ideal filtration give rise to the cobar spectral sequence $E^1_{p,q}=\pi_p(I^q/I^{q+1})\Rightarrow \pi_{p+q}\mathbb Z FX$. Curtis used the same spectral sequence for Kan's construction $GX$ to compute homology of loop spaces. So, if my question have positive answer, the corresponding spectral sequence should collapse on first page and all extensions should be trivial. Not sure, how to see this and how to extract homotopy equivalence from this spectral sequence. Also, Hopf algebra structure here is a mystery for me. Advantage of this approach is that we would have combinatorial control over elements like $g^{-1}$ in group ring and their image in tensor algebra, since $g^{-1}-1=-g-1 \ \mathrm{mod} \ I^2$.</p></li>
<li><p>From the short exact sequence of free abelian groups $I\to \mathbb Z FX\to \mathbb Z$ we can choose a splitting (even naturally with respect to $X$ since it serves as a basis) which will give the isomorphism of abelian groups $\mathbb Z FX\cong \mathbb Z\oplus I$. We can continue this process and get an isomorphism of abelian groups between the group ring and it's associated graded. Not sure if homotopy magic will help to turn this "map" to isomorphism of Hopf algebras.</p></li>
<li><p>Another point of view is an isomorphism of simplicial algebras $\mathbb Z JX\cong T(FX_{ab})$, here $JX$ denote free simplicial monoid on $X$ (aka James construction). Now, by theorem of Quillen, group completion $JX\to FX$ is a homotopy equivalence, so looks like we done. But in this approach the inverse of homotopy equivalence is untrackable, so one really can not understand images in tensor algebra of elements like $g^{-1}$ from group ring. </p></li>
</ol>
<p>So, all of these observations are different pieces of same puzzle which I can not put together. Will appreciate any help!</p>
http://mathoverflow.net/q/2309464Another interpretation of the $16$ dimensional Severi VairetySrinivasa Granujanhttp://mathoverflow.net/users/863972016-02-12T08:23:18Z2016-02-12T08:23:18Z
<p>I asked about an interpretation of this variety <a href="http://mathoverflow.net/questions/230876/looking-for-severi-varieties/230915#230915">here</a>. There is another one that could be easier. Let $K$ be an algebraically closed field of characteristic $0$. We denote the set of terns of $3\times 3$ matrices by
$$
V:=\mathcal{M}_{3}(K)\times \mathcal{M}_{3}(K)\times \mathcal{M}_{3}(K),
$$
and we define the cubic form
$$
F:V\rightarrow K, (A,B,C)\mapsto \det A+\det B+\det C-tr(ABC).
$$
We consider the algebraic group
$$
G:=\{g\in GL(V):F(g(v))=F(v)\text{ }\forall v\in V\},
$$
that acts on $V$.</p>
<p>Let $v\in V$ be a highest weight vector. We define
$$
X:=\mathbb{P}(G\cdot v)\subseteq\mathbb{P}(V)=\mathbb{P}^{26}.
$$</p>
<p>I would like to prove (I have not seen it explicitly in any text, so I could be wrong):</p>
<p>$i)$ The tern of matrices that we may choose as $v$. I don't know very much about representation theory, and I am not sure how to compute heighest weight vectors.</p>
<p>$ii)$ $X$ is a non-degenerate smooth algebraic variety of dimension $16$.</p>
<p>$iii)$ The secant variety of $X$ is
$$
SX=\{[v]\in\mathbb{P}^{26}:F(v)=0\}.
$$</p>
<p>If you know a book or article where this is proven I would also appreciate if you could tell me.</p>
http://mathoverflow.net/q/230943-3The line graph of a complete graph [on hold]Treesa Josehttp://mathoverflow.net/users/865682016-02-12T07:46:04Z2016-02-12T08:18:55Z
<p>Show that there exist a $\left\{P_{5},C_{4}\right\}$- decomposition of the graph $L(K_{9})$.</p>
http://mathoverflow.net/q/2309423Bounding and dominating numbers ${\frak b}, {\frak d}$ via ultrafiltersDominic van der Zypenhttp://mathoverflow.net/users/86282016-02-12T07:41:43Z2016-02-12T08:46:36Z
<p>Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$ and suppose that ${\cal U}$ is a free ultrafilter on $\omega$. We write $f \leq_{\cal U} g$ if $$\{n\in\omega: f(n) \leq g(n)\}\in{\cal U}.$$ </p>
<p>Similar to the <a href="https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum" rel="nofollow">usual bounding and dominating numbers</a>, we define $${\frak b}_{\cal U} = \min\{|B|: B\subseteq \omega^\omega \land \forall f\in\omega^\omega \exists g\in B(g\not \leq_{\cal U} f)\},$$
and
$${\frak d}_{\cal U} = \min\{|D|: D\subseteq \omega^\omega \land \forall f\in\omega^\omega \exists g\in D(f \leq_{\cal U} g)\}.$$</p>
<p>Do we have ${\frak b} = {\frak b}_{\cal U}$ and ${\frak d} = {\frak d}_{\cal U}$?</p>
http://mathoverflow.net/q/2309373Generic Smoothness Type of Results in Positive Characteristicuser80473http://mathoverflow.net/users/804732016-02-12T06:23:56Z2016-02-12T14:21:23Z
<p>Let $f:X\to Y$ be a surjective morphism between two projective varieties over a field of characteristic $p>0$. Also assume that $f_*\mathcal{O}_X=\mathcal{O}_Y$, and $X$ is smooth. </p>
<p>We know that the general fiber of $f$ is not smooth in general. But can we say that the general fiber is an integral scheme?</p>
<p>Maybe this is too much to ask for an arbitrary morphism $f$; are there any known classes of morphsims for which this property holds? What if $f$ is given by an Iitaka fibration of a nef line bundle, does it make any difference? </p>
<p>$\textbf{Note}:$ It is known that if the $\textit{generic}$ fiber of $f$ is $\textit{geometrically integral}$, then the general fiber is an integral scheme. So my question basically reduces to asking what are some known classes of morphisms with geometrically integral generic fibers? </p>
http://mathoverflow.net/q/2308860CLT for sums of an infinite sequence of rv with an asymptotic distributionArunhttp://mathoverflow.net/users/86182016-02-11T18:07:22Z2016-02-12T12:15:53Z
<p>Excuse me if the question is ill-posed. I'll do my best to explain the problem.I have a vector $(x^{(n)}_1, x^{(n)}_2, \ldots x^{(n)}_n),$ whose individual components can be shown to be asymptotically normally distributed as $n \to \infty,$ with some suitable scaling (here $\sqrt{n}.$) That is to say, $\sqrt{n}x_i \xrightarrow{\mathcal{D}} \mathcal{N}.$ I do not much about the joint distribution; I assume they are independent. Under the assumption that they are independent, is it possible to say something about the asymptotic distribution of $\sum_{i=1}^{n}f(x_i),$ for some function $f.$ Because here I only have access to the asymptotic mean and variance of $x_i,$ so a direct application of the CLT doesn't seem warranted. I know we can not say much about the distribution of sums of random variables that individually converge in distribution(unless we have Skorohod embedding), but here I have an infinite number of them. Any input is welcome. Thanks a lot.</p>
http://mathoverflow.net/q/2308681Frobenius series of Fuchsian PDEsMPTuitehttp://mathoverflow.net/users/46542016-02-11T15:33:44Z2016-02-12T12:25:08Z
<p>I'm interested in the analyticity of Frobenius-like series solutions to a PDE in $z=(z_1,\ldots ,z_N)\in\mathbb{C}^N$ with regular singular behavior at $z_\alpha=0$ for all $\alpha=1,\ldots, N$. </p>
<p>For $i=(i_1,\ldots,i_N)\in\mathbb{Z}_{+}$ define the order $|i|=\sum_{\alpha=1}^{N}i_\alpha$ differential operator $\mathcal{D}_{z}^{i}=\prod_{\alpha=1}^N z_\alpha^{i_\alpha}\frac{\partial^{i_\alpha}}{\partial z_{\alpha}^{i_{\alpha}}}$. Consider an order $n$ PDE of the form: </p>
<p>$(*)$ $\sum_{i}P^{i}(z)\mathcal{D}_{z}^{i}y=0$ </p>
<p>where $|i|\le n$ and where $P^{i}(z)$ is analytic in a polydisc $\Delta:\{|z_\alpha|<R_\alpha\}$. Furthermore, assume that $\frac{1}{P^{i}(z)}$ is also analytic in $\Delta$ for all $i$ of the form $(0,\ldots,n,\ldots,0)$ i.e. for each $\alpha=1,\ldots,N$ the multiplicative inverse of the coefficient of $z_\alpha^{n}\frac{\partial^{n}y}{\partial z_{\alpha}^{n}}$ in $(*)$ is also analytic in $\Delta$. </p>
<p>Consider a Frobenius series solution of $(*)$ of the form</p>
<p>$Y(z,\rho)=\sum_{\alpha=1}^N\sum_{k_\alpha\ge 0}y_k z^{\rho+k}$, $y_0=1$</p>
<p>for $z^k=\prod_{\alpha=1}^N z_\alpha^{k_\alpha}$ etc where $\rho\in\mathbb{C}^N$ satisfies a degree $n$ indicial polynomial equation. For a given root $\rho=r$ and assuming that $\rho+k$ is not an indicial root for all $k>0$, it is straightforward to show that a unique formal series solution $Y(z,r)$ exists.</p>
<p>My question is: is it true that $\sum_{\alpha=1}^N\sum_{k_\alpha\ge 0}y_k z^{k}=Y(z,r)z^{-r}$ is also analytic on $\Delta$?</p>
http://mathoverflow.net/q/2308659List of counting proofs instead of linear algebra method in combinatoricsdomotorphttp://mathoverflow.net/users/9552016-02-11T15:11:01Z2016-02-12T08:38:02Z
<p>I've just come across <a href="http://arxiv.org/abs/1007.1553" rel="nofollow">this</a> proof of the Graham-Pollak Theorem by Sundar Vishwanathan (thanks to Konrad Swanepoel's sporadic comments about it on this site), that must be called <em>beautiful</em> after its author. The paper mentions (after lemma 3) that certain other linear algebra proofs can also be replaced by similar pigeonhole principle applications. But I couldn't find any, so I thought that we should collect some similar <em>beautiful</em> proofs here. I don't mind if it also uses parity or other tricks, but linear algebra should be replaced! For example, can anyone prove the <a href="http://exwiki.org/mw/index.php?title=The_Oddtown_theorem" rel="nofollow">Oddtown theorem</a>?</p>
http://mathoverflow.net/q/2308444Can we always attain another prime via inserting digits between the digits of a fixed prime?jorohttp://mathoverflow.net/users/124812016-02-11T09:54:02Z2016-02-12T13:32:39Z
<p>The sequence <a href="https://oeis.org/A080437" rel="nofollow">OEIS A080437</a> is</p>
<blockquote>
<p>For n > 10, let m = n-th prime. If m is a k-digit prime then a(n) = smallest prime obtained by inserting digits between every pair of digits of m.</p>
</blockquote>
<p>I don't see why this sequence is defined for all $n$, i.e. why prime $a(n)$ must exist, that is why this process of digit-insertion always yields some prime.</p>
<p>For two digit primes, this appears an accident to me.</p>
<p>For sufficiently large primes, probabilistic arguments suggest
it is defined.</p>
<p><strong>Question:</strong> Is OEIS A080437 defined for all $n$?</p>
http://mathoverflow.net/q/230831-1A problem on finite Projective plane [on hold]dasshttp://mathoverflow.net/users/619312016-02-11T05:41:04Z2016-02-12T14:14:49Z
<p>Let Z=PG(q,2) be a finite Projective plane over a finite field Fq, q a prime power. Show that there exists a commutative binary operation * in Z such
that</p>
<p>(i) x*y is neither x nor y for any x and y, x not equal to y,</p>
<p>(ii) x,y and x*y are not incident on the same line for any two distinct
points x and y in the plane,</p>
<p>(iii) x<em>y, y</em>z and z*x are not incident on the same line for any three distinct</p>
<p>points x,y and z.</p>
<p>(Note: It is not necessary to define x*x!)</p>
http://mathoverflow.net/q/2308281Stochastic calculus in $L^1$CSAhttp://mathoverflow.net/users/368862016-02-11T04:41:27Z2016-02-12T14:03:11Z
<p>Does there exist a more general (than Malliavin or Itô) "Stochastic calculus" defined on $L^1$ space, or some Orlicz space between $L^2$ and $L^1$?</p>
<p>For examples: are there:</p>
<ul>
<li>Ito Isometry(-types) of results for L1 processes</li>
</ul>
http://mathoverflow.net/q/2303892Moduli space of log Calabi-Yau varieties exists?jolhttp://mathoverflow.net/users/215742016-02-06T16:14:01Z2016-02-12T12:50:41Z
<p>Let $\mathcal M^{(X,D)}$ be a moduli space of pair varieties $(X,D)$ which $K_X+D$ is trivial and $D$ is a divisor with conic singularities on Kaehler variety $X$. I am looking for a proof that such moduli space exists?</p>
<p>The log Weil-Petersson $\omega_{WP}^D$ correspounding to moduli space $\mathcal M^{(X,D)}$ is Kahler metric?(of course for Weil-Petersson metric it is known, but here we have instead Log Weil-Petersson metric) </p>
<p>A proof or freference for it is appreciated.</p>
<p>Here is some discussion of the moduli space of log Calabi-Yau varieties in case $dim X =2$ in Section 6 of the paper
<a href="http://arxiv.org/abs/1211.6367" rel="nofollow">http://arxiv.org/abs/1211.6367</a></p>
http://mathoverflow.net/q/23037314Unifying (& "justifying") the various definitions for differential operatorsSaal Hardalihttp://mathoverflow.net/users/228102016-02-06T13:10:43Z2016-02-12T10:49:26Z
<p>Reading about differential operators in different sources I've picked up several definitions which are not obviously equivalent (to me). Here they are:</p>
<blockquote>
<p><strong>Definition 1 ("naive")</strong>: Let $X$ be a (real) smooth manifold with structure sheaf
$\mathcal{O}_X$. A differential operator <strong>is an $\mathbb{R}$-linear
morphism of sheaves</strong> $D \in Hom_{\mathbb{R}}(\mathcal{O}_X ,\mathcal{O}_X)$ Which <strong>locally looks like a linear differential</strong>
operator. In other words every point $p \in X$ has a neighborhood $p
\in U$ for which $D |_U = \sum^{\infty}_{k =0} \sum_{|\alpha| \le k}f_{\alpha} \partial_{\alpha}$ for some collection $f_\alpha \in
\Gamma(U,\mathcal{O}_X)$.</p>
</blockquote>
<p>Letting $|\alpha| =r$ the notation above means: $\alpha = (\alpha_1,\dots,\alpha_{r})$ and $\partial_\alpha = \frac{\partial^{r}}{\partial x^{\alpha_1} \dots \partial x^{\alpha_r} }$.</p>
<blockquote>
<p><strong>Definition 2 (grothendiek):</strong> Let $X \to S$ be a locally ringed space over $S$ and let $\mathcal{I} = \ker \triangle^{\flat}$ where
$\triangle : X \to X \times_S X$ is the diagonal morphism. <strong>The sheaf
of $k$-jets of sections</strong> of $\mathcal{O}_X$ is then $\mathcal{J^k} :=
\mathcal{J^k}(\mathcal{O}_X) := \triangle^{*} ( \mathcal{O}_{X
\times_S X} / \mathcal{I}^{k+1})$. The sheaf of differential operators
of degree at most $k$ is then dual to the sheaf of $k$-jets
$\mathcal{D}^k := Hom_{\mathcal{O_X}}(\mathcal{J^k},\mathcal{O}_X)$.</p>
<p><strong>Definition 2.5 (possibly different definition for jets):</strong> For smooth manifolds one could do the same as definition (2) above only using the
familiar notion of a <strong>jet bundle</strong>. This is the bundle whose local
sections are discrete sections of $\bigcup_p J^r_p$ that locally agree
with a jet of some smooth function. </p>
<p><strong>Definition 3 (filtration):</strong> Let $(X, \mathcal{O}_X)$ be a locally ringed space whose structure sheaf is a sheaf of $k$-algebras. There's
a <strong>natural differential filtraion</strong> on
$Hom_k(\mathcal{O}_X,\mathcal{O}_X)$ which gives rise to the filtered
sub-algebra of differential operators. This is nicely explained in the
<a href="http://mathoverflow.net/questions/18203/is-there-a-categorical-description-of-grothendiecks-algebra-of-differential-o">following question</a></p>
<p><strong>Definition 4 (???):</strong> Let $(X,\mathcal{O}_X)$ be a smooth manifold. There is a topology on the sheaf of $\mathbb{R}$-vector
spaces $\mathcal{O}_X$ s.t. <strong>the sheaf of continuous
endomorphisms</strong> $ConEnd_{\mathbb{R}}(\mathcal{O}_X,\mathcal{O}_X)$ is
the sheaf of differential operators.</p>
</blockquote>
<p>Definition (4) is a theorem I've seen reffered to in several places yet it wasn't clear to me what's the correct topology there.</p>
<p><strong>Questions:</strong></p>
<ul>
<li><strong>Out of the definitions I've given what are the tautologies and what are the non-trivial equivalences/non-equivalences?</strong></li>
<li><strong>Does anything radically different happen when considering locally free sheaves?</strong> It seems to me that everything should stay the same as locally we still have just multi-valued sections of the structure sheaf. In particular, <strong>what's the correct definition and topology in the following sentence</strong>:</li>
</ul>
<blockquote>
<p><strong>Continuous $\mathbb{R}$-linear endomorphisms of a locally free sheaf $\mathcal{E}$ over a manifold $M$ are differential operators.</strong></p>
</blockquote>
http://mathoverflow.net/q/2301740If $N = qn^2$ is an odd perfect number, is it possible to have $q + 1 = \sigma(n)$?Kashitokiku Teshikiarihttp://mathoverflow.net/users/103652016-02-04T14:02:37Z2016-02-12T11:59:05Z
<p>The title says it all.</p>
<p><strong>Question</strong></p>
<blockquote>
<p>If $N = qn^2$ is an odd perfect number with Euler prime $q$, is it possible to have $q + 1 = \sigma(n)$?</p>
</blockquote>
<p><strong>Heuristic</strong></p>
<p>From the Descartes spoof, with quasi-Euler prime $q_1$:</p>
<blockquote>
<p>$$n_1 = 3003 < \sigma(n_1) = 5376 < q_1 = 22021$$</p>
</blockquote>
<p>So it appears that it might be possible to prove that $q + 1 \neq \sigma(n)$.</p>
<p><strong>Some Essential Estimates</strong></p>
<p><a href="http://www.worldscientific.com/doi/abs/10.1142/S1793042112500935" rel="nofollow">Acquaah and Konyagin</a> recently obtained the estimate $q < n\sqrt{3}$. We will use this estimate to obtain an upper bound for $\sigma(q)/n$.</p>
<p><a href="http://www.ams.org/journals/mcom/2012-81-279/S0025-5718-2012-02563-4/" rel="nofollow">Ochem and Rao</a> recently obtained the lower bound $N > {10}^{1500}$ for the magnitude of an odd perfect number. Using this bound, together with the inequality $n < q$, gives
$$I(q) < 1 + {10}^{-500}.$$</p>
<p><strong>Motivation</strong></p>
<p>We wish to prove the following proposition:</p>
<blockquote>
<p>If $N = {q^k}{n^2}$ is an odd perfect number with Euler prime $q$, then $3 \nmid N$ implies that $q < n$.</p>
</blockquote>
<p>If $q + 1 \neq \sigma(n)$, then it follows that
$$I(q) + I(n) \neq \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$
from which we obtain
$$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q},$$
since the reverse inequality
$$\frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < I(q) + I(n)$$
will violate the inequality $I(q) < \sqrt[3]{2} < I(n)$ (see this <a href="https://cs.uwaterloo.ca/journals/JIS/VOL15/Dris/dris8.html" rel="nofollow">paper</a>). (<strong>Edit February 8 2016</strong>: Assuming
$$\frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < I(q) + I(n)$$
is true, then
$$\left(q < n\right) \land \left(\sigma(n) < \sigma(q)\right)$$
is false. However, I am currently unable to rule out
$$\left(n < q\right) \land \left(\sigma(q) < \sigma(n)\right).$$
This particular case remains open.) </p>
<p>But the inequality
$$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$
implies that the biconditional
$$q < n \Longleftrightarrow \sigma(q) < \sigma(n)$$
holds.</p>
<p>This biconditional is then a key ingredient in the proof of the proposition mentioned earlier.</p>
<p>My method is able to rule out $\sigma(q) = q + 1 = \sigma(n)$ if $3 \nmid n$, since we obtain
$$2.799 \approx 1 + 2^{\frac{\log(6/5)}{\log(31/25)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(5))}{\log(I(5^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$
$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$
(where the smallest prime factor $u$ of $N$ satisfies $u \geq 5$), whence we arrive at a contradiction.</p>
<p><strong>Further Considerations</strong></p>
<p>If $\sigma(q) = q + 1 = \sigma(n)$ and $3 \mid n$, then the same method does not force a contradiction, because we then have
$$2.7199 \approx 1 + 2^{\frac{\log(4/3)}{\log(13/9)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(3))}{\log(I(3^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$
$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$
where $u$ is the smallest prime factor of $N$.</p>
<p><strong>Added February 7 2016</strong></p>
<p>If $q + 1 = \sigma(n)$, then $\sigma(n) \equiv 2 \pmod 4$, so that $n$ takes the form
$$n = {p^r}{m^2}$$
where $p$ is a prime with $p \equiv r \equiv 1 \pmod 4$ and $\gcd(p, m) = 1$.
If $3 \mid n$, then $p \neq 3$, so that $3 \mid m$.</p>
<p>So I have
$$\sigma(n) = \sigma(p^r)\sigma(m^2)$$
where $\sigma(p^r) \equiv r + 1 \equiv 2 \pmod 4$.</p>
<p>Since $\gcd(q, q+1) = 1$, then $\gcd(q, \sigma(n)) = 1$, so that
$$\gcd(q, \sigma(p^r)\sigma(m^2)) = 1.$$
Thus, $q \nmid \sigma(p^r)$ and $q \nmid \sigma(m^2)$.</p>
<p>However, note that $$q \mid \sigma(n^2) = \sigma(p^{2r})\sigma(m^4).$$</p>
<p>Alas here is where I get stuck.</p>
http://mathoverflow.net/q/2297260Do tori in a symplectic group always have invariant maximal isotropic subspaces?kneidellhttp://mathoverflow.net/users/144432016-01-31T00:48:08Z2016-02-12T08:45:03Z
<p>$\newcommand{\mbf}{\mathbf}$
Hi all,</p>
<p>I've been thinking about the following question for a while now, and got a little stuck trying to solve it. Hopefully, someone here might be able to help.</p>
<p>For starters, let $K$ be some field, and led $G$ be the group of $K$-points of the group $\mbf{G}=\mathrm{Sp}_{2n}$, for some $n\in\mathbb N$. That is to say- $G$ is the group of isometries of a $2n$-dimensional vector space $V$ over $K$, endowed with a non-degenerate alternating form $(\cdot,\cdot):V\times V\to K$. It is well-known that in this situation there exists a Darboux basis for $V$ over $K$, i.e. a basis $\lbrace e_1,\ldots,e_n,f_1,\ldots,f_n\rbrace$ where $(e_i,f_j)=\delta_{i,j}$ and $(e_i,e_j)=(f_i,f_j)=0$ for all $i,j=1,\ldots,n$. This allows us to identify $G$ with the matrix group
$$G=\lbrace\mbf{x}\in M_{2n}(K)\mid \mbf{x}^T\Omega\mbf{x}=\Omega\rbrace,$$
where $\Omega:=\left(\begin{matrix}\mbf{0}_n&\mbf{id}_n\\-\mbf{id}_n&\mbf{0}_n\end{matrix}\right)$.</p>
<p>A torus in $G$ is a connected abelian subgroup $T$ such that any $\mbf{y}\in T$ is diagonalizable in some field extension of $K$.</p>
<p>For example, if $K$ is algebraically closed, then any torus in $G$ is conjugate to the group $$\lbrace\left(\begin{smallmatrix}a_1\\&\ddots\\&&a_n\\&&&&a_1^{-1}\\&&&&&\ddots\\&&&&&&a_n^{-1}\end{smallmatrix}\right)\mid a_1,\ldots,a_n\in K^\times\rbrace.$$</p>
<p>My question is this-</p>
<p><strong>Is it true that for any torus $T\subseteq G$, there exists a maximal isotropic subspace $W\subseteq V$ such that $TW\subseteq W$.</strong> </p>
<p>Recall that a subspace $W\subseteq V$ is isotropic if it satisfies $(w_1,w_2)=0$ for all $w_1,w_2\in W$.</p>
<p>In the case where $K$ is algebraically closed, the answer to this question is obviously true. However, my main concern is with the case of non-a.c. fields, and in particular- finite and local fields.</p>
<p><em>What I've managed to show</em>- It is not very hard to show that the minimal polynomial $f(t)$ of any element of $G$ must satisfy the equality
$$\tilde{f}(t)=f(t),$$
where $\tilde{f}(t)$ is defined by "reversing the coefficients of $f$", i.e.
$$\tilde{f}(t):=\frac{t^{\deg(f)}}{f(0)}\cdot f(t^{-1}).$$
This implies that the minimal polynomial of any such elements can be written uniquely in the form \begin{equation}\tag{$*$}\prod_i f_i(t)^{r_i}\cdot\prod_j \left(g_j\cdot \tilde{g_j}(t)\right)^{q_i},\end{equation}
where the $f_i$ are irreducible and satisfy $\tilde{f_i}=f_i$ and he $g_j$'s are irreducible with $g_j\ne\tilde{g_j}$. </p>
<p>In the case where a torus $T\subseteq G$ contains an element $y$ whose minimal polynomial is of the form $g(t)\cdot\tilde{g}(t)$, where the polynomials $g$ and $\tilde{g}$ have no common irreducible factors, the existence of a $T$-invariant is true, and the subspace can be realized as $W=\ker\left(g(y)\right).$ </p>
<p>However, it is possible, to have tori which contain no such elements $y$ (for example, if $K=\mbf{F}_q$ is the field with $q$ elements and $G=\mathrm{Sp}_4(K)$, there is an emebedding of the group of elements of norm $1$ in the extension $\mbf{F}_{q^4}\mid \mbf{F}_{q^2}$ as a torus in $G$, and all elements of this group are either central, or have minimal polynomial whose irreducible factors satisfy $\tilde{f}=f$).</p>
<p>In any case, I would be very grateful if anyone could offer any insight for why this assertion is true or otherwise, point towards a counter example.</p>
<p>Thank you very much.</p>
http://mathoverflow.net/q/17576218Is formula valid for relating $\pi$ with ALL of its OEIS A002485(n)/A002486(n) convergents?Alexhttp://mathoverflow.net/users/558532014-07-10T13:10:23Z2016-02-12T14:10:12Z
<p>Could anyone try to prove that the below conjectured formula is valid for relating $\pi$ with ALL of its convergents - those, which are described in OEIS via A002485(n)/A002486(n)?</p>
<p><strong>$$(-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n))$$
$$=(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^{2(j+2)}(k+(i+k)x^2)\big)/(1+x^2)\; dx$$</strong></p>
<p>and in unformatted form:</p>
<p>(-1)^n*(Pi−A002485(n)/A002486(n))=(abs(i)*2^j)^(-1)<em>Int((x^l</em>(1-x)^(2*(j+2))*(k+(i+k)*x^2))/(1+x^2),x=0...1)</p>
<p>where integer $n = 0,1,2,3,...$ serves as the index for terms in OEIS A002485(n) and A002486(n), </p>
<p>and $\{i, j, k, l\}$ are some integers (to be found experimentally or otherwise), which are probably some functions of $n$.</p>
<p><strong>The "interesting" (I think) part of my generalization conjecture is that "i" is present in both denominator of the coefficient in front of the integral and in the body of the integral itself</strong></p>
<p>At this time it could be shown that the formula under question is applicable for some first few convergents (of the A002485(n)/A002486(n) type) </p>
<p>For example for $\frac{22}{7}$</p>
<p>$$\frac{22}{7} - \pi = \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,\mathrm{d}x$$</p>
<p>with $n=3, i=-1, j=0, k=1, l=4$ - with regards to my above suggested generalization. </p>
<p>In Maple notation</p>
<p>i:=-1; j:=0; k:=1; l:=4;Int(x^l*(1-x)^(2*j+2)*(k+(k+i)<em>x^2)/((1+x^2)</em>(abs(i)*2^j)),x= 0...1)</p>
<p>yields 22/7 - Pi</p>
<p>It also works for found by Lucas </p>
<p><a href="http://www.math.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf" rel="nofollow">http://www.math.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf</a></p>
<p>formula for $\frac{333}{106}$ </p>
<p>$$\pi - \frac{333}{106} = \frac{1}{530}\int_{0}^{1}\frac{x^5(1-x)^6(197+462x^2)}{1+x^2}\,\mathrm{d}x$$</p>
<p>with $n=4, i=265, j=1, k=197, l=5$ -with regards to my above suggested generalization. </p>
<p>In Maple notation
i:=265; j:=1; k:=197; l:=5;Int(x^l*(1-x)^(2*j+2)*(k+(k+i)<em>x^2)/((1+x^2)</em>(abs(i)*2^j)),x= 0...1)</p>
<p>yields Pi - 333/106</p>
<p>And it works for Lucas's formula for $\frac{355}{113}$</p>
<p>$$\frac{355}{113} - \pi = \frac{1}{3164}\int_{0}^{1}\frac{(x^8(1-x)^8(25+816x^2)}{(1+x^2)}$$</p>
<p>with $n=5, i=791, j=2, k=25, l=8$ -with regards to my above suggested generalization. </p>
<p>In Maple notation</p>
<p>i:=791; j:=2; k:=25; l:=8;Int(x^(2*j+2)<em>(1-x)^l</em>(k+(k+i)<em>x^2)/((1+x^2)</em>(abs(i)*2^j)),x= 0...1)</p>
<p>yields 355/113 - Pi</p>
<p>And it works as well for Lucas's formula for $\frac{103993}{33102}$</p>
<p>$$\pi - \frac{103993}{33102} = \frac{1}{755216}\int_{0}^{1}\frac{x^{14}(1-x)^{12}(124360+77159x^2)}{1+x^2}\,\mathrm{d}x$$</p>
<p>with $n=6, i= -47201, j=4, k=124360, l=14$ -with regards to my above suggested generalization. </p>
<p>In Maple notation</p>
<p>i:=-47201; j:=4; k:=124360; l:=14;Int(x^l*(1-x)^(2*j+2)*(k+(k+i)<em>x^2)/((1+x^2)</em>(abs(i)*2^j)),x= 0...1)</p>
<p>yields Pi - 103993/33102</p>
<p>And also it works Lucas's formula for $\frac{104348}{33215}$</p>
<p>$$\frac{104348}{33215} - \pi = \frac{1}{38544}\int_{0}^{1}\frac{x^{12}(1-x)^{12}(1349-1060x^2)}{1+x^2}\,\mathrm{d}x$$</p>
<p>with $n=7, i= -2409, j=4, k=1349, l=12$ - with regards to my above suggested generalization. </p>
<p>In Maple notation</p>
<p>i:=-2409; j:=4; k:=1349; l:=12;Int(x^l*(1-x)^(2*j+2)*(k+(k+i)<em>x^2)/((1+x^2)</em>(abs(i)*2^j)),x= 0...1)</p>
<p>yields 104348/33215 - Pi </p>
<p>And it works as well for $\frac{618669248999119}{196928538206400}$</p>
<p>which, by the way, is not part of A002485/A002486 OEIS sequences:</p>
<p>$$\frac{618669248999119}{196928538206400} - \pi = \frac{1}{755216}\int_{0}^{1}\frac{x^{14}(1-x)^{12}(77159+124360x^2)}{1+x^2}\,\mathrm{d}x$$</p>
<p>with $i= 47201, j=4, k=77159, l=14$ -with regards to my above suggested generalization. </p>
<p>In Maple notation</p>
<p>i:=47201; j:=4; k:=77159; l:=14;Int(x^l*(1-x)^(2*j+2)*(k+(k+i)<em>x^2)/((1+x^2)</em>(abs(i)*2^j)),x= 0...1)</p>
<p>yields
618669248999119/196928538206400 - Pi</p>
<p>This question relates to my answer given in
<a href="http://math.stackexchange.com/questions/1956/is-there-an-integral-that-proves-pi-333-106/127618#127618">http://math.stackexchange.com/questions/1956/is-there-an-integral-that-proves-pi-333-106/127618#127618</a> </p>
<p>Update:
Recently Thomas Baruchel (see his answer at <a href="http://math.stackexchange.com/questions/860499/seeking-proof-for-the-formula-relating-pi-with-its-convergents">http://math.stackexchange.com/questions/860499/seeking-proof-for-the-formula-relating-pi-with-its-convergents</a> ) has conducted extensive calculations and found that even the parametric formula (with four parameters) yields infinite number of solutions for each n.</p>
<p>Thomas shared with me his calculations results and supplied me with quite a few of valid combinations of i, j, k, l values - so now I have a lot of experimentally found five-tuples {n,i, j, k, l}, which satisfy above parameterization, where n varies in the range from 2 to 26.</p>
<p>Based on this data, of course, it would be nice to find how (if at all) i, j, k, l are inter-related between each other and with "n" - but such inter-relation (if exists) is not obvious and difficult to derive just by observation ... (though it is clearly seen that an absolute value of "i" is strongly increasing as "n" is growing from 2 to 26).</p>
<p>RHS could be reduced (after performing integration - please let me know if I made a mistake in doing this) to:</p>
<p>(abs(i)*2^j)^(-1)*Gamma(2*j+5)*((k+i)*Gamma(l+3)*HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2*j+l+8)+k*Gamma(l+1)*HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2*j+l+6))</p>
<p><strong>May be from discussed parametric identity one could derive irrationality measure for pi, if to assume that RHS in this identity holds true, when the rational fraction on the LHS is equal to 0?</strong></p>
<p><strong>Are there any {i,j,k,l}, which would satisfy such such condition?</strong></p>
<p>Pi = (abs(i)*2^j)^(-1)*Gamma(2*j+5)*((k+i)*Gamma(l+3)*HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2*j+l+8)+k*Gamma(l+1)*HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2*j+l+6))</p>
<p>Update #2:</p>
<p>Thanks to Jaume Oliver Lafont, at least one case, answering affirmatively to the last question, is identified: i=-1, j=-2, k=1, l=0</p>
<p>$$\pi = \int_{0}^{1}\frac{4}{1+x^2}\,\mathrm{d}x$$</p>
<p><strong>Should there be infinite number of such cases?</strong></p>
<p>P.S. Per discussion with Jaume Oliver Lafont, depending on the value of the polynomial x degree in the integral body's numerator (while denominator stays to be the same "1+x^2"), the result varies from "Pi" to "log(2)" and also to "+/- (Pi - p/q)" as well as to "+/-(log(2)-p/q)", so perhaps now one could produce two distinct families of parameterization: one for Pi and the differences between Pi and its convergents and another for log(2) and the differences between log(2) and its convergents.</p>
http://mathoverflow.net/q/1403376How to find or constrain "particularly good" (two-sided) spectral expanders?Robin Saundershttp://mathoverflow.net/users/43362013-08-24T23:57:30Z2016-02-12T09:15:56Z
<p>I'm new to graph theory, but a response to <a href="http://mathoverflow.net/questions/71451/cubic-graphs-which-are-difficult-to-navigate">a question</a> I asked a while ago introduced me to the concept of expander graphs.</p>
<p>A k-regular graph (henceforth "graph") on n nodes has eigenvalues k = λ<sub>1</sub> ≥ λ<sub>2</sub> ≥ ... λ<sub>n</sub> ≥ -k. According to the most widespread definition, the graph is considered to be a good expander if λ<sub>2</sub> is small. Random k-regular graphs (henceforth "random graphs") tend to be good expanders, and expanders have some of the important properties of random graphs, such as good connectivity.</p>
<p>However, there are many situations in which we would also like λ<sub>n</sub> to be small in size. For example, the expander mixing lemma states that the number of edges connecting two subsets of the nodes in an "expander" is roughly what you'd expect in a random graph, but here "expander" means that λ ≔ max(|λ<sub>2</sub>|,|λ<sub>n</sub>|) is small. Terry Tao, among others, <a href="http://terrytao.wordpress.com/2011/12/02/245b-notes-1-basic-theory-of-expander-graphs" rel="nofollow">has used</a> the term "two-sided expander" to differentiate these graphs from the usual "one-sided expanders" of which they are a subset.</p>
<hr>
<p>It is known that for any sequence of graphs with n increasing, lim inf λ ≥ 2√(k-1); and in fact λ converges almost-surely to this value for random graphs on n nodes.</p>
<p>My question concerns the "largest" (in terms of n) graph(s) for which λ ≤ x, given some x < 2√(k-1). How might we find these graphs, or at least constrain them (e.g. by finding a feasible range for n), possibly to a set which could be searched by computer? If the general problem is too difficult, what about the case k = 3, x = 2?</p>
<p><i>Edited the definitions in the first half and added some explanation, per comments on Igor Rivin's answer.</i></p>
http://mathoverflow.net/q/1377257Is the fundamental group of $II_{1}$ factors invariant under a relation?Sébastien Palcouxhttp://mathoverflow.net/users/345382013-07-25T12:23:25Z2016-02-12T12:59:18Z
<p>In order to define the equivalence relation, let's first recall the Tomita-Takesaki modular theory and conditional expectation for von Neumann algebras.</p>
<p>Let $H$ be a separable Hilbert space and $B(H)$ the algebra of bounded operators. </p>
<p><strong>Definition</strong>: A von Neumann algebra is a *-subalgebra $M \subset B(H)$ stable under bicommutant: $M^{*} = M$ and $M'' = M$. </p>
<p><strong>Modular theory</strong> : Let $M \subset B(H)$ be a von Neumann algebra. Let $\Omega \in H$ be a <em>cyclic</em> and <em>separating</em> vector (i.e., $M.\Omega$ and $M'.\Omega$ are dense in $H$). Let $S : H \to H$ be the closure of the anti-linear map $x\Omega \to x^{*}\Omega$. Then, $S$ admits a polar decomposition $S = J\Delta^{1/2}$, with $J$ anti-linear unitary and $\Delta$ positive. Then, $JMJ = M'$ and $\Delta^{it} M \Delta^{-it} = M$.<br>
Let $\sigma_{\Omega}^{t}(x) = \Delta^{it} x \Delta^{-it}$ the modular action of $\mathbb{R}$ on $M$. </p>
<p><strong>Conditional expectation</strong> (<a href="http://www.sciencedirect.com/science/article/pii/0022123672900043" rel="nofollow">Takesaki 1972</a>) : Let $N \subset M$ be an inclusion of von Neumann algebra, then there is a conditional expectation of $M$ onto $N$ with respect to $\Omega$ (cyclic and separating) if $N$ is invariant under the modular action (i.e., $\sigma_{\Omega}^{t}(N) = N)$.<br>
<strong>Notation</strong> : if $\exists \Omega$ verifying the previous conditions, we note $N \subset_{e} M$.</p>
<p><strong>Remark</strong> : The modular theory is trivial for $M = L(\Gamma) \subset B(H)$, with $\Gamma$ a discrete group and $H = l^{2}(\Gamma)$ (because $\Delta = I$). In particular, it's trivial for the abelian von Neumann algebras.<br>
As a consequence, in this case: $N \subset M$ $\Leftrightarrow$ $N \subset_{e} M$. </p>
<p><strong>Notation</strong> : Let $N$ and $M$ be two von Neumann algebras.<br>
If $\exists P \simeq N$ such that $ P \subset_{e} M$, we note $N \hookrightarrow_{e} M$. </p>
<blockquote>
<p><strong>Equivalence relation</strong> : $M \sim N$ if $N \hookrightarrow_{e} M \hookrightarrow_{e} N$.</p>
</blockquote>
<p><strong>Philosophy</strong> : $M \sim N$ could significate they are isomorphic as <em>noncommutative sets</em> (see <a href="http://mathoverflow.net/questions/137520/whats-a-noncommutative-set">here</a>).</p>
<p><strong>Examples</strong> : </p>
<ul>
<li>Among $l^{\infty}(\{1,2,...,n \})$, $l^{\infty}(\mathbb{N})$
and $L^{\infty}([0,1])$ none is equivalent to another. </li>
<li>$L^{\infty}([0,1])$, $L^{\infty}([0,1]\cup \{1,2,...,n \})$
and $L^{\infty}([0,1]\cup \mathbb{N})$ are pairwise equivalent,<br>
because $L^{\infty}([0,1]) \subset L^{\infty}([0,1] \cup \{2,3,...,n\}) \subset L^{\infty}([0,1] \cup \mathbb{N}_{\geq 2}) \hookrightarrow L^{\infty}(\mathbb{R})$<br>
and $L^{\infty}([0,1]) \simeq L^{\infty}(\mathbb{R})$</li>
<li>Obviously $L^{\infty}([0,1]) \not\sim B(H)$.</li>
<li>Let $R \subset B(H)$ be the hyperfinite $II_{1}$ factor, $R_{\infty} = R \otimes B(H)$ the hyperfinite $II_{\infty}$ factor. $ B(H) \hookrightarrow_{e} R_{\infty} \hookrightarrow_{e} B(H \otimes H)$ and $B(H) \simeq B(H \otimes H)$. So, $R \not\sim B(H) \sim R_{\infty}$.</li>
<li>Let $\Gamma$ be a non-amenable ICC discrete group. Then $L(\Gamma) \not\hookrightarrow_{e} B(H)$ and $L_{\infty}(\Gamma) = L(\Gamma) \otimes B(H) \not\hookrightarrow_{e} B(H \otimes H) $ so $L(\Gamma) \not\sim B(H) \not\sim L_{\infty}(\Gamma)$.</li>
<li>Let $\mathbb{F}_{2} = \langle a,b \vert \ \rangle $ and $\mathbb{F}_{\infty} = \langle a_{1},a_{2},... \vert \ \rangle $.<br>
Then $\mathbb{F}_{2} \hookrightarrow \mathbb{F}_{n} \hookrightarrow \mathbb{F}_{\infty} \hookrightarrow\mathbb{F}_{2} $ (the last injection is given by $a_{n} \to b^{-n}ab^{n}$).<br>
Consequence : $L(\mathbb{F}_{2}) \sim L(\mathbb{F}_{n}) \sim L(\mathbb{F}_{\infty}) $</li>
</ul>
<blockquote>
<p><strong>Fundamental group</strong> (see <a href="http://www.pnas.org/content/101/3/723.full" rel="nofollow">here</a>) : The fundamental group of a type $II_{1}$ factor is the set of numbers $t > 0$ for which its
<em>amplification</em> by $t$ is isomorphic to itself: $\mathcal{F}(M) = \{t>0 \ \vert \ M^{t}\simeq M \}$.</p>
</blockquote>
<p><strong>Examples</strong>:</p>
<ul>
<li>There is a semi-direct product $ \Gamma = \mathbb{Z}^{2} \rtimes SL(2,\mathbb{Z})$ such that $\mathcal{F}(L(\Gamma)) = \{1\}$</li>
<li>It's countable for $II_{1}$ factors with property (T).</li>
<li>$\mathcal{F}(R) = \mathcal{F}(L(\mathbb{F}_{\infty})) = \mathbb{R}_{+}^{*}$</li>
<li><strong>Open</strong> : $\mathcal{F}(L(\mathbb{F}_{2})) = \{1\}$ <strong>or</strong> $\mathbb{R}_{+}^{*}$, but we still do not know which it is.<br>
This is a reformulation of the free group factor isomorphism problem: $L(\mathbb{F}_{2}) \simeq L(\mathbb{F}_{\infty}) $ ?</li>
</ul>
<blockquote>
<p><strong>Question</strong>: Is the fundamental group $\mathcal{F}(M)$ of a $II_{1}$ factor $M$
invariant under $\sim$ ?</p>
</blockquote>
<p><strong>Remark</strong> : an affirmative answer would solve the free group factor isomorphism problem. </p>
<p>Because this problem is very difficult, if this question admits an affirmative answer, I do not expect that the proof will be given here without a colossal work, but I would be interested to know if (in your opinion) this way seems promising. If it admits a negative answer, then in addition to a possible counter-example, I would be interested to know if you see a manner to reformulate the question for becoming open. </p>
http://mathoverflow.net/q/13752018What's a noncommutative set?Sébastien Palcouxhttp://mathoverflow.net/users/345382013-07-23T17:19:12Z2016-02-12T13:06:24Z
<p>This issue is for logicians and operator algebraists (but also for anyone who is interested).</p>
<p>Let's start by short reminders on <a href="https://en.wikipedia.org/wiki/Von_Neumann_algebra">von Neumann algebra</a> (for more details, see <a href="http://www.math.berkeley.edu/~vfr/MATH20909/VonNeumann2009.pdf">[J]</a>, <a href="http://www.springer.com/mathematics/analysis/book/978-3-540-42248-8">[T]</a>, <a href="http://iml.univ-mrs.fr/~wasserm/OHS.ps">[W]</a>): </p>
<p>Let $H$ be a separable Hilbert space and $B(H)$ the algebra of bounded operators. </p>
<p><strong>Definition</strong>: A von Neumann algebra is a *-subalgebra $M \subset B(H)$ stable under bicommutant:<br>
$M^{*} = M$ and $M'' = M$. </p>
<p><strong>Theorem</strong>: The <a href="http://en.wikipedia.org/wiki/Abelian_von_Neumann_algebra">abelian von Neumann algebras</a> are exactly the algebras $L^{\infty}(X)$ with $(X, \mu)$ a standard measure space. They are isomorphic to one of the following: </p>
<ul>
<li>$l^{\infty}(\{1,2,...,n \})$, $n \geq 1$</li>
<li>$l^{\infty}(\mathbb{N})$ </li>
<li>$L^{\infty}([0,1])$</li>
<li>$L^{\infty}([0,1]\cup \{1,2,...,n \})$</li>
<li>$L^{\infty}([0,1]\cup \mathbb{N})$ </li>
</ul>
<p><strong>Noncommutative philosophy</strong>: There are various <em>schools</em> of noncommutative philosophy, here is the <em>school</em> close to operator algebras. This issue is not about philosophy, so I will explain it quickly (for more details see <em>for example</em> the introduction of this <a href="http://www.alainconnes.org/docs/book94bigpdf.pdf">book</a>). First an intuitive idea : in the same way as there are <em>classical physics</em> and <em>quantum physics</em>, there are <em>classical mathematics</em> and <em>quantum mathematics</em>. What does it mean in practice ? It means the following : in the <em>classical mathematics</em> there are many different structures, <em>for example</em>, the measurable, topological or Riemannian spaces. The point is to encode each structure by using the framework of commutative operator algebras. For the previous examples, it's the commutative von Neumann algebras, C$^{*}$-algebras and spectral triples. Now if we take these operator algebraic structures and if we remove the <em>commutativity</em>, we obtain what we call <em>noncommutative analogues</em> : <strong>noncommutative measurable, topological or Riemannian spaces</strong>.<br>
<em>This school</em> explores noncommutative analogues of more and more structured objects, it goes in one direction. My point is to question about the other direction (back to the <em>Source</em>) :<br>
What's the noncommutative analogue of a set (called a <strong>noncommutative set</strong>) ? </p>
<blockquote>
<p>What is a <em>noncommutative set</em>? </p>
</blockquote>
<p>The von Neumann algebras of the standard measure space $[0,1]$, $[0,1]\cup \{1,2,...,n \}$ and $[0,1]\cup \mathbb{N}$ are not isomorphic, but as sets, these spaces are isomorphic (i.e., same cardinal). </p>
<p>Is there a natural equivalence relation $\sim$ on the von Neumann algebras, forgetting the <em>measure space</em> but remembering the <em>set space</em>, on abelian von Neumann algebras? </p>
<p><strong>Remark</strong>: If $M \sim N$, we could say that they are isomorphic as <strong>noncommutative sets</strong>.<br>
The equivalence class could be called the <strong>quantum cardinal</strong> (a link with <a href="http://mathoverflow.net/questions/133050/the-cyclic-subfactors-theory-a-quantum-arithmetic">cyclic subfactor</a> theory?).</p>
<p>Are there <em>noncommutative analogues</em> of the ZFC axioms ?</p>
<p>What I'm looking for seems different of what is called <em>quantum set</em> in the literature...</p>