Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2014-11-26T04:39:24Zhttp://mathoverflow.net/feedshttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://mathoverflow.net/q/1880990Example of symplectic 4-manifolds with no Lefschetz fibration structure?Jie Minhttp://mathoverflow.net/users/222982014-11-26T02:26:22Z2014-11-26T04:24:22Z
<p>I just read about Donaldson's result on existence of Lefschetz pencil structure on symplectic manifolds (Donaldson 1999). However, one has to blow up the base locus to get a Lefschetz fibration structure.</p>
<p>So I wonder if every symplectic manifold (especially symplectic 4-manifold) admits a Lefschetz fibration instead of just a Lefschetz pencil? If not, are there any examples?</p>
http://mathoverflow.net/q/1880980Asymptotic analysis of a sum of complex summands using integralteaguthttp://mathoverflow.net/users/569132014-11-26T01:38:10Z2014-11-26T01:38:10Z
<p>I'm trying to find the exact asymptotics of a sum:</p>
<p>$$A = \sum^n_{i=0} \begin{pmatrix} 2n \\ i \end{pmatrix} x^{i} y^{2n-i} $$</p>
<p>as $n\rightarrow\infty$. Here $x,y$ are complex numbers, $|x|\leq1, |y|\leq1$. I also have $|x+y|<1$ and $|4xy|\leq1$. My strategy is to write the sum as an integral.
Rewrite this sum as:
$$ A = (x+y)^{2n} -
\sum^{n-1}_{i=0} \begin{pmatrix} 2n \\ i \end{pmatrix} x^{2n-i} y^i \quad\quad\quad\quad (1) $$</p>
<p>Note that this is a remainder for the $(n-1)$-order Taylor expansion so can use the integral formula for the remainder:</p>
<p>$$ A = \begin{pmatrix} 2n \\ n \end{pmatrix} y^n \int^1_0 (x+(1-t)y)^n nt^{n-1} dt $$</p>
<p>Changing variables and introducing $z=x/y$ get:</p>
<p>$$ A = \begin{pmatrix} 2n \\ n \end{pmatrix} (xy)^n \int^1_0 (1+(1-u^{\frac{1}{n}})z)^n du $$</p>
<p>In this integral the integrand converges to $u^{-z}$. Then for $Re(z)<1$ the integral converges and I get (using Stirling's formula) the following asymptotic expression:</p>
<p>$$ A \sim \frac{(4xy)^n}{\sqrt{n}}\frac{1}{1-z} $$</p>
<p>On the other hand, I can approximate the <strong>second</strong> term in $(1)$ as the Taylor remainder thus interchanging the roles of $x,y$ and then get for $Re(1/z)<1$ that </p>
<p>$$ A \sim (x+y)^{2n} - \frac{(4xy)^n}{\sqrt{n}}\frac{1}{1-1/z} $$</p>
<p>However, now I have something very strange: when both $Re(z)<1$ and $Re(1/z)<1$ are valid, both asymptotic expressions have to agree, which means that it must be $|x+y|^2 < |4xy|$. But it's clearly not true - take for example $x=-1/2$ and $y=1/100$ which satisfy the conditions on $Re(z)$ and $Re(1/z)$. Where have I made a mistake?</p>
http://mathoverflow.net/q/1880971"Automorphic simulteneous eigenvectors" under the set of weighted Laplacians of a Hilbert modular varietyHugo Chapdelainehttp://mathoverflow.net/users/117652014-11-26T00:35:13Z2014-11-26T03:33:36Z
<p>This question is closely related to the following MO question <a href="http://mathoverflow.net/questions/185126/characterizing-the-real-analytic-eisenstein-series">Characterizing the real analytic Eisenstein series</a></p>
<p>Let $\mathfrak{h}=\{z=x+iy\in\mathbf{C}\}$ be the Poincare upper half plane endowed with its Poincare metric. Let $w\in\mathbf{Z}$ be a weight and define the "weight $w$ Laplacian" on $\mathfrak{h}$ by
$$
\Delta[w]=-y^2(\partial_x^2+\partial_y^2)+i\cdot w\cdot y\cdot\partial_x.
$$
Here $i=\sqrt{-1}$. In general, $\Delta[w]$ will commute with the $|_{w}$ right action of $GL_2^+(\mathbf{R})$ on Maass forms as is explained on page 130 of Bump's book on automorphic forms and representations.</p>
<p>Let $g\in\mathbf{Z}_{\geq 1}$ be a fixed integer and consider the symmetric space $\mathfrak{h}^g$.
Choose (arbitrarily) a totally real field $K$ of degree $g$. Choose also (arbitrarily) a discrete subgroup $\Gamma\leq GL_2^+(K)$ which acts discontinuously on $\mathfrak{h}^g$ (by Moebius transformation through the distinct embeddings of $K$) and which has finite covolume. The specific choices of $K$ and $\Gamma$ <strong>are not important</strong> here.</p>
<p>Let $\underline{z}=(z_j)_{j=1}^g\in\mathfrak{h}^g$ where $z_j=x_j+iy_j$. We let
$D_j[w_j]$ be the weight $w_j$ Laplacian with respect to the variable $z_j$.
Let $\underline{w}=(w_1,w_2,\ldots,w_g)\in\mathbf{Z}^g$ be an integral weight vector. For
$\gamma=\left(\begin{matrix} a & b \\ c& d\end{matrix}\right)\in \Gamma$ and
$\underline{z}\in\mathfrak{h}^g$ we let
$$
j(\gamma,\underline{z})=(c^{(j)}z_j+d^{(j)})_{j=1}^g
$$
be the usual $1$-cocycle of $\Gamma$ taking values in the ring of analytic functions going from $\mathfrak{h}^g$ to $\mathfrak{h}^g$. We also let
$$
\omega_{\underline{w}}(j(\gamma,\underline{z})):=\prod_{j=1}^g |c^{(j)}z_j+d^{(j)}|^{2s-w_j}
$$
be a (convenient) automorphic factor of weight $\underline{w}$. Now consider a function
$$
(\star) \;\;\;\;\;\;\;\;\;\;\;\;\;\; F(Im(\underline{z}),s):=A(s)\prod_{j=1}^g y_j^{s-w_j/2}+B(s)\prod_{j=1}^g y_j^{1-s+w_j/2}
$$
where $A(s)$ and $B(s)$ are say (for the sake of being precise) holomorphic functions in $s$. Here $Im(\underline{z})=\underline{y}=(y_1,y_2,\ldots,y_g)$.</p>
<p>Then the function $F(Im(\underline{z}),s)$ satisfy the following two properties</p>
<p><strong>(1)</strong> For $1\leq j\leq g$, the linear differental operator
$$
D_j:=\Big(\Delta_j[w_j]-(s-w_j/2)(1-s+w_j/2)\Big)
$$
kill $F(Im(\underline{z}),s)$.</p>
<p><strong>(2)</strong> For all $\gamma\in \Gamma$, $F(Im(\gamma \underline{z}),s)=\omega_{\underline{w}}(\gamma,\underline{z})F(Im(\underline{z}),s)$.</p>
<p><strong>Question 1:</strong> Let $G(Im(\underline{z}),s)$ be a real analytic function in $\underline{y}$ and holomorphic function in $s$ which satisfy (1) and (2); Does it follow that $G(Im(\underline{z}),s)$ has the same form as the expression in $(\star)$?</p>
<p><strong>Question 2:</strong> Assuming that the answer to question 1 is positive, what is the natural context to phrase this question? It seems to me that similar questions could be phrased for other symmetric spaces acted on by arithmetic subgroups of finite covolume.</p>
<p><strong>Remark:</strong> </p>
<p>(a) We note that condition (1) is independent of the specific choices of $K$
and $\Gamma$. </p>
<p>(b) The answer to Question 1 is positive if $g=1$. In this case one has a homogenous linear ODE of order $2$ which can be easily solved. In fact, condition (1) alone is enough to guarantee that the solution is as in $(\star)$. </p>
<p>(c) The fact that $F(Im(\underline{z}),s)$ does not depend on the variables
$(x_1,x_2,\ldots,x_g)$ simplifies the way that $\Delta_j[w_j]$ acts on $F(Im(\underline{z}),s)$. But I thought it was more natural to keep $\Delta[w]$ as I originally defined it, i.e., as the weight $w$ Laplacian. </p>
<p>(d) If one only looks at condition (1), then $\bigcap_{j=1}^g\ker(\Delta_j[w_j])$ is infinite dimensional if $g\geq 2$. This was my motivation for adding the "artificial" condition (2) in the hope of obtaining uniqueness. But there are probably better ways of doing this.</p>
http://mathoverflow.net/q/1880930Absolutely continuous functionsJulio Valenciahttp://mathoverflow.net/users/622522014-11-25T23:33:50Z2014-11-26T02:36:11Z
<p>it is well known that if a function $f:[0,T]\to\mathbb{R}$ satisfies the inequality
$$\vert f(t)-f(s)\vert\leq \int_s^t{m(r) dr},$$
for $s<t$ and some $m\in L^1([0,T])$ then $f$ is absolutely continuous.
On the other hand, if the function satisfies the inequality
$$\vert f(t)-f(s)\vert\leq (g(s)+g(t))\vert t-s\vert,$$
for some $g\in L^1([0,T])$, then $f\in W^{1,1}([0,T])$, indeed both conclusions are the same. </p>
<p>My question is: if the inequality
$$\vert f(t)-f(s)\vert\leq (g(s)+g(t))\vert t-s\vert+\int_s^t{m(r) dr}$$
is valid, do we have the same conclusion? I think that it is true, but I don't know how to show it.</p>
http://mathoverflow.net/q/1880920abelian p- subgroups of E_6(q)daryahttp://mathoverflow.net/users/302522014-11-25T23:29:21Z2014-11-26T03:16:51Z
<p>Is there any result about maximal abelian p-subgroups of the exceptional group E_6(q), where q=p^a is prime power?</p>
http://mathoverflow.net/q/1880907Non-abelian freeness of SU_2Will Sawinhttp://mathoverflow.net/users/180602014-11-25T23:06:26Z2014-11-25T23:06:26Z
<p>The distribution of the trace of a random element of $SU_2$ is the Sato-Tate distribution.</p>
<p>The analogue of the Gaussian distribution in free probability theory is the Wigner semicircle distribution.</p>
<p>The Sato-Tate distribution and the Wigner semicircle distribution are the same.</p>
<p>Combinatorially, this comes from the fact that the moments of both those two distributions are the Catalan numbers.</p>
<p>We can categorify this as follows: If $A$ is a self-dual element of a monoidal abelian category over a field, then $Hom(1, A^{\otimes n})$ is a vector space. (By self-duality, this is also equal to $Hom(A^{\otimes a}, A^{\otimes b} )$ for any $a,b$ such that $a+b=n$.)</p>
<p>This categorifies the $n$th moment in the following way: If $G$ is a compact Lie group, and $A$ is a self-dual representation in the category of representations of $A$, then the dimension of $Hom(1, A^{\otimes n})$ is the $n$th moment of the trace of a random element of $G$ acting on $A$. So in the case $G=SU_2$, $A$ the standard representation, the dimension is the Catalan numbers.</p>
<p>On the other hand, if you take a free monoidal abelian category on one self-dual generator, the moments are the Catalan numbers. (One has to take dimensions, not just over the base field, but over the ring generated over the base field by the element of $Hom(1,1)$ defined by the canonical morphism $1 \to A \otimes A \to 1$.)</p>
<p>Since the category of representations of $SU_2$ is certainly a monoidal abelian category, we have a functor from the second category to the first that sends the free generator to the standard representation. So we have a morphism from the free moment vector spaces to the moment vector spaces of $SU_2$.</p>
<p>I believe I can show by an explicit combinatorial argument that this is an isomorphism, categorifying this identity between the two definitions of Catalan numbers.</p>
<p>Hence the free monoidal abelian category on one self-dual generator is equivalent to the category of finite-dimensional representations of $SU_2$.</p>
<blockquote>
<p>Is there some conceptual reason why this is so?</p>
</blockquote>
<p>Note that the limit of the category of representations of $USP_{2n}$ as $n$ goes to $\infty$ is the free symmetric monoidal abelian category on one self-dual generator.</p>
http://mathoverflow.net/q/1880880Dehn twist about an arbitrary curveasdfasdfgasdhttp://mathoverflow.net/users/314752014-11-25T22:50:02Z2014-11-25T22:50:02Z
<p>I need to know if there is an algortihm to write down a Dehn twist about an arbitrary curve on an orientable surface $S$ , as product of a set of generators of $MCG(S)$. </p>
<p>Since we have the conjugacy relation: $\phi^{-1}D_c\phi=D_{\phi(c)}$ ($D$ being the Dehn twist and equality understood up to isotopy), it will also help to know if for an arbitrary curve $C$ we can find a generator $a$(from any set of generators) and a diffeomorphism $\phi$ such that $\phi(a)=C$. </p>
http://mathoverflow.net/q/1880850Rank of negative definite even intersection formszygundhttp://mathoverflow.net/users/224312014-11-25T22:24:02Z2014-11-25T22:24:02Z
<p>I have a fairly basic question about the theory of symmetric bilinear integral forms. If an even unimodular form is negative definite, does its rank have to be a multiple of 8? </p>
<p>I read in Scorpan's book, "The Wild World of 4-Manifolds", that the analogous statement is true for <em>positive</em> definite even forms, a consequence of "van der Blij's lemma". Scorpan conspicuously omits negative definite forms from his analysis. Do there exist four-manifolds with negative definite even intersection forms whose rank is not divisible by 8?</p>
http://mathoverflow.net/q/188083-4finding points on elliptic curve over finite field [on hold]ugradmathhttp://mathoverflow.net/users/622502014-11-25T21:43:33Z2014-11-25T21:43:33Z
<p>Find the points on the elliptic curve y^2 = x^3 + 2x + 2 in F17 (field of prime 17).</p>
<p>Do I have to guess a first point and then use an algorithm to spit out all other points?</p>
http://mathoverflow.net/q/1880818A game of stonesLiviu Nicolaescuhttp://mathoverflow.net/users/203022014-11-25T21:23:38Z2014-11-25T23:10:37Z
<p>How I arrived at this question is a rather long story having to do with the honors calculus class I am teaching. At this point it's sheer curiosity on my part. Here is the game.</p>
<p>$\newcommand{\bZ}{\mathbb{Z}}$</p>
<p>We start with a finite collection of stones placed at random somewhere on the set of nodes $\newcommand{\eN}{\mathscr{N}}$ $$\eN=\{2,3,4,\dotsc\}. $$ We can view a distribution of stones as a function $s:\eN\to\bZ_{\geq 0} $ with finite support, $s(n)=$ the number of stones at $n$. Its <em>weight</em> is the nonnegative integer</p>
<p>$$|s|=\sum_{n\in\eN} s(n). $$</p>
<p>We say that a distribution $s$ is <em>overcrowded</em> if $s(n)> 1$ for some $n\in\eN$. A node $n$ is called <em>occupied</em> (with respect to $s$) if there is at least one stone at $n$, $s(n)>0$. </p>
<p>We are allowed the following moves: choose an occupied node $n$. Then you move one stone from location $n$ to location $n+1$ and add a new stone at location $n^2$. Note that such a move increases the weight by $1$.</p>
<p>Now comes the question.</p>
<blockquote>
<p>Is it true that for any initial distribution of stones $s:\eN\to\bZ_{\geq 0}$ and any positive integer $N$ there exists a finite sequence of allowable moves such that after these moves we obtain a new distribution of stones which (i) is <strong>not</strong> overcrowded, and (ii) no node $n<N$ is occupied. </p>
</blockquote>
<p>Empirical evidence leads me to believe that the answer to this question is positive. However, I have failed to find a conclusive argument.I'm hoping someone in the MO community will have more luck.</p>
http://mathoverflow.net/q/1880773Example of proof using the generic matrixAyman Moussahttp://mathoverflow.net/users/277672014-11-25T19:19:02Z2014-11-25T22:16:03Z
<p>There is a really nice proof of the Cayley-Hamilton Theorem using the generic matrix. I expose it briefly. </p>
<p>One defines the generic matrix $G:=(X_{ij})_{ij} \in\mathcal{M}_n(\mathbb{Z}[X_{ij}]_{ij})$. </p>
<p>The discriminant $\Delta_G$ of $\chi_G$ (characteristic polynomial of $G$) is an element of $\mathbb{Z}[X_{ij}]_{ij}$. </p>
<p>Now $\Delta_G = 0$ is impossible, since it would imply in particular (after specialization) that any element of $\mathcal{M}_n(\mathbb{C})$ has at least one double eigenvalue (which is clearly false).</p>
<p>Whence $\Delta_G \neq 0$. This means that $\chi_G$ has only simple roots (in some field of decomposition, say $\mathbb{K}$), which in turn means that $G$ is diagonalizable in $\mathcal{M}_n(\mathbb{K})$ and $\chi_G(G)$ follows easily. </p>
<p>We get then the Cayley-Hamilton for any matrix after specialization.</p>
<p>I am searching for other nice (and easy) use of this generic matrix. In fact any generic argument like this one, could be nice also.</p>
<p>I have to precise that I am <strong><em>not</em></strong> an algebraist, so I would really appreciate a <em>simple</em> example. </p>
<p>I find this proof a bit magic but at the same time very natural (meaning, after all, that $\chi_G(G)=0$ is just a general algebraic identity, just like $(a+b)^2 = a^2+ ab +ba + b^2$, and it is a way to compute it).</p>
<p>Thanks for your help !</p>
http://mathoverflow.net/q/18806111A funny factorization of the Jacobian coming from the lines on the Fermat cubicDavid Speyerhttp://mathoverflow.net/users/2972014-11-25T15:40:38Z2014-11-25T22:30:33Z
<p>Here is something which came up in my algebraic geometry class, and I'm wondering if it has a deeper explanation. Let $F(w,x,y,z) = w^3+x^3+y^3+z^3$ and let $X$ be the cubic surface in $\mathbb{P}^3$ where $F$ vanishes. As is well known, there are $27$ lines on $X$ (in characteristic $\neq 3$).</p>
<p>I had my students verify that the obvious equations on the Grassmannian $G(2,4)$ define the $27$ points corresponding to these lines as a reduced scheme. This is easiest to do in an affine cover. For example, consider the affine chart of lines of the form
$$\mathrm{RowSpan} \begin{pmatrix} 1 & 0 & p & q \\ 0 & 1 & r & s \end{pmatrix}.$$
This chart contains $18$ of the $27$ points in question. The obvious equations come from setting equal to $0$ the coefficients of $t^3$, $t^2 u$, $t u^2$ and $u^3$ in
$$F(t,u,pt+ru,qt+su).$$
I.e.
$$1 + r^3 + s^3 = 3 p r^2 + 3 q s^2= 3 p^2 r + 3 q^2 s= 1 + p^3 + q^3=0. \quad (\ast)$$</p>
<p>To verify that they define a radical ideal, one must check the Jacobian condition: I.e., that
$$\det \begin{pmatrix}
0 & 0 & 3 r^2 & 3 s^2\\
3 r^2 & 3 s^2 & 6pr & 6 qs \\
6pr & 6 qs & 3 p^2 & 3 q^2 \\
3 p^2 & 3 q^2 & 0 & 0 \\
\end{pmatrix} \neq 0$$
at each of the $18$ roots of $(\ast)$. (Actually, I only assigned them to do one root; then the large symmetry group of $X$ does the rest.)</p>
<p>So, here is the thing I can't explain. Out of curiosity, I computed the above determinant and factored it. It turns out that
$$\det \begin{pmatrix}
0 & 0 & 3 r^2 & 3 s^2\\
3 r^2 & 3 s^2 & 6pr & 6 qs \\
6pr & 6 qs & 3 p^2 & 3 q^2 \\
3 p^2 & 3 q^2 & 0 & 0 \\
\end{pmatrix} = - 81 \det\begin{pmatrix} p & q \\ r & s \end{pmatrix}^4 .$$</p>
<p>Is there any deep reason for this? Does that determinant even have any significance away from the $18$ points which describe lines on $X$?</p>
http://mathoverflow.net/q/1880590Powers of orthogonal matrices is closedSebastian Schlechthttp://mathoverflow.net/users/514782014-11-25T15:06:13Z2014-11-26T00:25:43Z
<p>This might be a basic question, nonetheless I cannot give a proof.</p>
<p>Given an orthogonal matrix $A$ with eigendecomposition $A = Q \Lambda Q^{-1}$ with only non-real eigenvalues. Given also a diagonal real matrix $\Phi$ with $\Phi_{ii} = \Phi_{jj}$ if $\Lambda_{ii} = \overline{\Lambda_{jj}}$. The following matrix power is defined as $[\Lambda^\Phi]_{ii} := \Lambda_{ii}^{\Phi_{ii}}$. </p>
<p>Why is $Q \Lambda^\Phi Q^{-1}$ orthogonal? (Unitarity is simple, but why is it real?)</p>
http://mathoverflow.net/q/1880490Distribution of the $\alpha$-parameter of a $2\times 2$ Haar-distributed, unitary matrixconfused guyhttp://mathoverflow.net/users/622372014-11-25T14:04:07Z2014-11-25T21:55:23Z
<p>It is well known that any $2\times 2$ unitary matrix $\mathbf{U}$ can be parametrized as
$$\mathbf{U}=\begin{pmatrix} 1 & 0 \\ 0 & \mathrm{e}^{\mathrm{j}\beta_1}\end{pmatrix} \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix}\begin{pmatrix} \mathrm{e}^{\mathrm{j}\beta_2}\ & 0 \\ 0 & \mathrm{e}^{\mathrm{j}\beta_3}\end{pmatrix}.$$</p>
<p>The question is, if $\mathbf{U}$ is Haar-distributed, what is the distribution of $\alpha$?</p>
http://mathoverflow.net/q/18804720Fermat's last theorem over larger fieldsPablohttp://mathoverflow.net/users/388892014-11-25T13:59:05Z2014-11-25T21:29:32Z
<p>Fermat's last theorem implies that the number of solutions of $x^5 + y^5 = 1$ over $\mathbb{Q}$ is finite. </p>
<p><strong><em>Is the number of solutions of $x^5 + y^5 = 1$ over $\mathbb{Q}^{\text{ab}}$ finite?</em></strong></p>
<p>Here $\mathbb{Q}^{\text{ab}}$ is the maximal abelian extension of $\mathbb{Q}$. By Kronecker-Weber, this is the field obtained from $\mathbb{Q}$ by adjoining all roots of unity.</p>
http://mathoverflow.net/q/1880420Continuity of a Functionaledithttp://mathoverflow.net/users/305102014-11-25T13:14:25Z2014-11-26T02:36:43Z
<p>A certain functional $T$ is defined as:
$$T(F)=\int_{(0,1)}F^{-1}(s)M(ds)$$
where $M$ is a probability measure.
The result that above functional is continuous at $F$, if the measure $M$ does not assign mass to the point where $F^{-1}$ is discontinuous is special case of Theorem 3.7. Robust Statistics by Huber and Ronchetti(2nd ed).</p>
<p>If $\frak{M}$ be a set of probability measures, is there a way to guarantee that
$$T_{sup}(F)=\sup_{\{M\in\frak{M}\}}\{\int_{(0,1)}F^{-1}(s)M(ds)\}$$ is continuous if none of $M$ assign mass to point where $F^{-1}$ is discontinuous? Is it false?</p>
<p>We can at least say that it is lower semi continuous. Can we say more?</p>
<p><strong>Details Added:</strong></p>
<p>Domain of $T$ is CDFs. $F^{-1}(x):=\inf\{y\in\mathbb{R}:F(y)\ge x\}$</p>
http://mathoverflow.net/q/1879840Eigenvalue of a linear map over finite fielduser3208http://mathoverflow.net/users/32082014-11-24T22:15:07Z2014-11-25T22:39:26Z
<p>Let $ F_q $ be a finite field with $ q $ elements.
Let $ g $ be a multiplicative generator of $ F_{q^2}^* $.
It implies that
$ <g^{q+1}> = F_q^* $.
Let $ l $ be a prime greater than $ q^2-1 $ and dividing
$ q^{2(q-1)}-1 $.</p>
<p>For any element $ b \in F_{q^2}^* $, there must exist
two unique integers $ 0\leq i(b) < q-1$ and $ 0\leq j(b) < q+1 $ such that
$ b = g^{ i(b) (q+1) + j(b)} $. Define the polynomial
$$ m(x) = \sum_{b \in g+F_q} q^{-2i(b)} x^{j(b)} $$</p>
<p>Identify $ F_l^{q+1} $ with $ R = F_l [x]/(x^{q+1}- q^{-2}) $.
Define two $ F_q $-linear maps on $ R $. The first one is $ M $, which is the
multiplication by $ m(x) $ in $ R $.
The second one is $ T $, which sends $ x^k $ to
$ q^{-2i(c)} x^{j(c)} $ where
$ c = (g^{kq+1}-g^k)/(g^{q}-g) $ for $ 0\leq k < q+1 $. </p>
<p>It can be proved that $ 1 $ is an eigenvalue of $ T M $.
We have verified that for many $ q $, the eigenvalue $ 1 $ has
multiplicity $ 1 $.
Is it true for all prime power $ q>3 $ ? Has the similar problem
been studied before?</p>
http://mathoverflow.net/q/1879616Exactness of an additive left Kan extensionSimone Virilihttp://mathoverflow.net/users/248912014-11-24T17:47:55Z2014-11-25T22:31:07Z
<p>Let $\phi:R\to S$ be a flat ring homomorphism and consider the induced adjoint pair
$$\phi_!:R-Mod\rightleftarrows S-Mod:\phi^*,$$
where $\phi_!=(S\otimes_R -)$. The right adjoint $\phi^*$ is easily described if we see $\phi$ as an additive functor between the one-object categories $R$ and $S$ and we view a left $S$-module $M$ as an additive functor $M:S\to Ab$. In this case, $\phi^*$ is just composition by $\phi$, $(M:S\to Ab)\mapsto (M\circ\phi:R\to Ab)$.</p>
<p>By definition of flat homomorphism, $\phi_!$ is an exact functor, while $\phi^*$ is exact for any ring homomorphism $\phi$.</p>
<p>Consider now the categories of finitely presented modules $fp(R)$ and $fp(S)$, and restrict the functor $\phi_!$ to an additive functor $F:fp(R)\to fp(S)$. Denote also by $fp(R)-Mod$ and $fp(S)-Mod$ the categories of additive functors $fp(R)\to Ab$ and $fp(S)\to Ab$, respectively. Of course, also here there is an induced exact functor
$$F^*:fp(S)-Mod\to fp(R)-Mod$$
and, using additive Kan extensions one may construct a left adjoint
$$F_!:fp(R)-Mod\to fp(S)-Mod.$$
Can we prove that the functor $F_!$ is exact?</p>
<p>Spelling this out, rember that, given a sequence of left $fp(R)$-modules
$$0\to M_1\to M_2\to M_3\to 0$$
this sequence is exact if and only if, for all $P\in fp(R)$, the sequence of abelian groups
$$0\to M_1(P)\to M_2(P)\to M_3(P)\to 0$$
is short exact. Thus one should prove (or disprove) that for any short exact sequence as above,
$$0\to F_!M_1(Q)\to F_!M_2(Q)\to F_!M_3(Q)\to 0$$
is short exact for all $Q\in fp(S)$.</p>
http://mathoverflow.net/q/1879385lower-bound for $Pr[X\geq EX]$LIUhttp://mathoverflow.net/users/622002014-11-24T12:44:44Z2014-11-26T01:07:37Z
<p>Given n random variables, $X_1, ..., X_n$, each takes value 0 or $a_i \in[0, 1]$. $X = \sum_{i=1}^n X_i$ and $EX \geq 1$ is the expected value of $X$. Can we get a lower-bound for $Pr[X \geq EX]$? It seems it can't be too small, may be constant or $\frac{1}{n}$.</p>
http://mathoverflow.net/q/1879281Density for Translated ProcessMatthias Ludewighttp://mathoverflow.net/users/167022014-11-24T09:57:07Z2014-11-26T02:37:20Z
<p>Let $M$ be a (compact) Riemannian manifold. Let $v$ be a smooth vector field on $M$ with flow $\Theta_t$. Let $L$ be an elliptic second order differential operator on $M$ that generates the Ito process $X_t$. </p>
<p>Define the process $Y_t := \Theta_t^* X_t$. I read that the generator of $Y_t$ is the time-dependent differential operator
$$L_s = \Theta_s^* L - v$$
I do not really know how to prove this though, and there is no explanation in the article I read.
Is the underlying probability measure of $Y_t$ absolutely continuous w.r.t. $X_t$? If yes, what is its density?</p>
<p>I know the Girsanov formula, which states e.g. that the process generated by $L-v$ has the density
$$\exp\left( \int_0^t v(B_s) \mathrm{d} B_s - \int_0^t |v(B_s)|^2 \mathrm{d} s\right)$$
However, in this situation, also the second-order part is altered and I don't know how to deal with this.</p>
http://mathoverflow.net/q/1878652Unitary factor in polar decompositionsM. Linhttp://mathoverflow.net/users/544582014-11-23T03:25:42Z2014-11-25T23:01:18Z
<p>Let $A, B$ be $n$-square (Hermitian) positive definite matrices. Let $AB=U|AB|$ be the polar decomposition of $AB$. So $U$ is unitary (called the unitary factor of $AB$). What is the optimal constant $c$ such that $\|I-U\|\le c$, where the norm is the usual spectral norm?</p>
<p>I want to have some understanding on the behaviour of the unitary factor for certain classes of matrices (e.g. matrices with real eigenvalues). Perhaps this is well known, any pointer to the existing papers is welcome. </p>
http://mathoverflow.net/q/1876672(Alternative) Presentation for the pure braid group of the spherePierrehttp://mathoverflow.net/users/370212014-11-20T21:31:35Z2014-11-25T22:42:30Z
<p>First I need some notation (it's all standard I think). For a manifold $M$, let $F_nM = F_{0,n}M$ be the space of $n$-tuples of distinct points on $M$ ; let $B_nM = B_{0,n}M = F_nM / \Sigma_n$. When $M= \mathbb{R}^2$ the fundamental group of $B_nM$ is the braid group $B_n$, and that of $F_nM$ is the pure braid group $P_n$.</p>
<p>Further, $F_{m,n} M$ is defined to be $F_n N$ where $N$ is $M$ with $m$ points removed. Likewise for $B_{m,n} M$.</p>
<p>Now this question is about $\pi_1 F_n S^2$. The fundamental group of $B_n S^2$(the "braid group of the sphere") is usually presented as a quotient of $B_n$, adding just one relation to Artin's usual ones ; so the fundamental group of $F_n S^2$ is a quotient of $P_n$.</p>
<p>However, a classical result asserts that there is a fibration</p>
<p>$$ F_{m+r, n-r}M \to F_{m,n}M \to F_{m,r}M$$</p>
<p>for all $m,n$ and $r\le n$. Taking $M=S^2$, $m=0$ and $r=1$ this becomes</p>
<p>$$ F_{n-1} \mathbb{R}^2 \to F_n S^2 \to S^2 $$</p>
<p>And the long exact sequence of homotopy groups gives in particular </p>
<p>$$ \mathbb{Z} \to P_{n-1} \to \pi_1 F_n S^2 \to 0$$</p>
<p>using that $\pi_1(S^2) = 0$ and $\pi_2(S^2) = \mathbb{Z}$.</p>
<p>This gives a presentation of $\pi_1 F_n S^2$ as a quotient of $P_{n-1}$ rather than $P_n$, adding just one relator (with a very simple proof indeed). The group $P_n$ can be generated by $n(n-1)/2$ generators and no fewer, and so $P_{n-1}$ can be generated by $(n-1)(n-2)/2$ generators, giving a much smaller set of generators for $\pi_1 F_n S^2$.</p>
<p>So my questions are: (EDITED)</p>
<blockquote>
<p>(1) Did I get something wrong in the above argument?</p>
<p>(2) Does someone know what the image of $\mathbb{Z}$ in $P_{n-1}$ is?</p>
<p>(3) Has this presentation been studied algebraically? Is it easier to work with the group $\pi_1 F_n S^2$ presented as a quotient of $P_{n-1}$ than with the presentation as a quotient of $P_n$ ? Any reference to a work in this direction?</p>
</blockquote>
<p>The answer by Ryan Budney below covers (1) and (2), I think. Any help with (3) appreciated.</p>
<p>Thanks !</p>
<p>Pierre</p>
http://mathoverflow.net/q/1870023Characterizing space that preserves positive-definiteness propertyuser2374357http://mathoverflow.net/users/617122014-11-13T04:13:42Z2014-11-26T00:25:41Z
<p>Given a symmetric positive-definite matrix $\Sigma$, consider the space $\mathcal{D}$ of diagonal matrices such that $\forall D\in\mathcal{D}$, the matrix $\Sigma-D\Sigma^{-1}D$ is positive definite. What is the connectedness of $\mathcal{D}$? Is it simply connected?</p>
http://mathoverflow.net/q/1853517Does Nelson try to prove PA inconsistent directly?Wojowuhttp://mathoverflow.net/users/301862014-10-25T10:09:28Z2014-11-26T00:17:17Z
<p>Edward Nelson is known for his serious attempts to show that Peano axioms, and sometimes even weaker theories, are inconsistent. I wasn't able to find Nelson's papers anywhere, so I wanted to ask a question about structure of his proofs:</p>
<blockquote>
<p>Did Nelson attempts try to, for some statement $\phi$, prove both $\phi$ and not $\neg\phi$?</p>
</blockquote>
<p>Some of you might ask a question "What other possibility could there be?", and here is one such possibility: PA might just prove the statement "PA is inconsistent". What is the difference? The difference is that supposed "proof" of contradiction might have nonstandard length. You can think of this in terms of different theory, namely $PA+\neg Con(PA)$. This theory shows that PA is inconsistent (because one of its axioms says so), and, as extension of PA, it can show itself inconsistent. However, the theory itself is still consistent.</p>
<p>What the reasoning above shows is that PA is not $\omega$-consistent. So, slightly restating the question,</p>
<blockquote>
<p>Did Nelson in his attempts really try to show PA inconsistent, or just $\omega$-inconsistent?</p>
</blockquote>
<p>Also, if anyone knows a source where I could find Nelson's papers, I'd be thankful; this is the reason I added "reference-request" tag.</p>
<p>Thanks for all feedback!</p>
http://mathoverflow.net/q/1684950Third order central moment of a positive linear combination of log-normal random variablesShravan Mohanhttp://mathoverflow.net/users/513912014-05-29T05:07:08Z2014-11-25T23:17:12Z
<p>What is the sign (+tive/-tive) of the third order central moment of a positive linear combination of log-normal random variables? </p>
<p>It seems to be a common notion that the skewness of random variables with longer tails to the right is positive. Is it correct? If so, how do you prove it? </p>
http://mathoverflow.net/q/1659201question about the optimal decomposition of supermartingaleCodeGolfhttp://mathoverflow.net/users/378502014-05-12T11:48:52Z2014-11-25T21:36:29Z
<p>Given a filtered probability space $(\Omega, \mathbb{F}, \{\mathcal{F}_t\}_{0\le t\le 1}, \mathbb{P})$, let $X$ be a cadlag martingale and $V$ be cadlag supermartingale. Suppose $V$ has the following decomposition: there exists a $X$-integrable predictable process $H$ s.t.</p>
<p>$$V_t=V_0+\int_0^tH_sdX_s-C_t,~ \forall t\in [0,1]$$</p>
<p>where $C$ is an adapted increasing process (not necessarily predictable!) with $C_0=0$. Now if we have another predictable process $H'$ s.t.</p>
<p>$$V_t\le V_{t-}+H'_t(X_t-X_{t-}), \forall t\in [0,1]$$</p>
<p>where $V_{t-}$ and $X_{t-}$ denote the left limit. Could we say that $H=H'$ a.s. or </p>
<p>$$\int_0^tH_sdX_s\le \int_0^tH'_sdX_s, \forall t\in [0,1]$$</p>
<p>Many thanks for the reply!</p>
http://mathoverflow.net/q/1633820Probability generating function zero implies random variable is infinitemathjungehttp://mathoverflow.net/users/496032014-04-14T22:00:59Z2014-11-25T21:17:17Z
<p>Let $V$ be a random variable supported on the nonnegative integers (including $\infty$) and $f(x) = \mathbf E x^V$ be the probability generating function. In our model $V$ is the number of visits to the root of a graph by an interacting particle system. We prove $V = \infty$ a.s. by showing that $f$ satisfies a recurrence relation</p>
<p>$$f(x) =\frac{x+2}{3}f\Bigl(\frac{x+1}{2}\Bigr)^2
+\frac{x+1}{3}f\Bigl(\frac{x}{2}\Bigr)\biggl(1
-f\Bigl(\frac{x+1}{2}\Bigr)\biggr)$$</p>
<p>which through analytic methods we prove can only be satisfied when $f \equiv 0$ on $[0,1)$. Though our technique works we are somewhat baffled and are hoping to, in our upcoming paper, give some context by providing examples of this type of argument occurring in probability literature.</p>
<p>So, the question is are there other examples of proving a r.v. is a.s. infinite by proving the generating function is identically zero?</p>
http://mathoverflow.net/q/1598970Homotopy with non piece-wise linear boundaryJuergenhttp://mathoverflow.net/users/480182014-03-09T23:52:21Z2014-11-26T01:17:18Z
<p>in the middle of a long proof I encounter the following problem.</p>
<p>Let $E$ be a closed and convex set in $\mathbb R^n$ such that for all $\vec x\in E$ it holds that $\sum_ix_i=1$. (We can understand $E$ as a set of probability functions.)</p>
<p>Let $x^\dagger$ be the probability function in $E$ which has maximum Shannon entropy:
i.e. $\{x^\dagger\}=\arg\sup_{\vec x\in E}-\sum_{i=1}^nx_i\cdot \log(x_i)$. This function is well-known to be unique. In the case I am interested in, I can assume that $x^\dagger_i>0$ for all $i$.</p>
<p>For all $k\in\mathbb N$ let $c_k(i)$ be an $n$-tuple of numbers such that for all $1\leq i\leq n$ it holds that $\lim_k c_k(i)=\log(x_i)$.</p>
<p>I need to show the following:
there exists a sequence $(q_k)_{k\in\mathbb N}$ with $q_k\in E$ such that</p>
<ol>
<li>$q_k \in \arg \sup_{\vec x\in E}-\sum_{i=1}^n x_i c_k(i)$ and</li>
<li>$\lim_k q_k = x^\dagger$.</li>
</ol>
<p>The main problem is that $\arg \sup_{\vec x\in E}-\sum_{i=1}^n x_i c_k(i)$ may contain more than one element.</p>
<p>So, if I replace $E$ by a closed, convex set with an open (i.e. non-empty) interior $U_\epsilon(E)\subset\mathbb R^n$ with a boundary which is nowhere piece-wise linear,
then $\arg \sup_{\vec x\in U_\epsilon(E)}-\sum_{i=1}^n x_i c_k(i)$
has a unique solution in $U_\epsilon(E)$. (This is a general fact about linear optimisation problems; or so I hope :))</p>
<p>These maxima will then obtain, in general, not for probability functions. But I think I can handle this.</p>
<p>What I need for my proof is the following: Given a fixed closed and convex set $E$ as above I need to construct sets $U_\epsilon(E)$
such that</p>
<ol>
<li>$U_\epsilon(E)$ is closed,</li>
<li>$U_\epsilon(E)$ varies continuously with $\epsilon>0$,</li>
<li>the interior of $U_\epsilon(E)$ is open for $\epsilon>0$,</li>
<li>$\{(1+\epsilon) x^\dagger\} = \arg \sup_{\vec x\in U_\epsilon(E)} -\sum_{i=1}^n ,x_i\log(x^\dagger_i)$,</li>
<li>the boundary of $U_\epsilon(E)$ is nowhere piece-wise linear and</li>
<li>$\lim_{\epsilon\rightarrow 0}U_\epsilon(E)=E$.</li>
</ol>
<p>The last limit is taken over all strictly positive $\epsilon$.</p>
<p>@4: It is well-known that $\{ x^\dagger\} = \arg \sup_{\vec x\in E} -\sum_{i=1}^n ,x_i\log(x^\dagger_i)$.</p>
<p>So, I can simply assume that there exists a homotopy which gives me what I need or do I have to/can I prove the existence of such a homotopy.</p>
<p>All help much appreciated.</p>
http://mathoverflow.net/q/6950943Small residue classes with small reciprocalTerry Taohttp://mathoverflow.net/users/7662011-07-05T01:47:19Z2014-11-26T02:12:01Z
<p>Let $p$ be a large prime. For any $m \in \{1,\ldots,p-1\}$, let $\overline{m} \in \{1,\ldots,p-1\}$ be the reciprocal in ${\bf Z}/p{\bf Z}$ (i.e. the unique element of $\{1,\ldots,p-1\}$ such that $m \overline{m} = 1 \hbox{ mod } p$).</p>
<p>I am interested in finding $m$ for which for which $m$ and $\overline{m}$ are both small compared with $p$, excluding the trivial case $m=1$ of course. For instance, using Weil's bound on Kloosterman sums and some Fourier analysis, it is not difficult to show that there exists nontrivial $m$ with $\max(m, \overline{m}) \ll p^{3/4}$. But this does not look sharp; probabilistic heuristics suggest that one should be able to get $\max(m, \overline{m})$ as small as $O(p^{1/2})$ or so (ignoring log factors), which would clearly be best possible. Is some improvement on the $O(p^{3/4})$ bound known? For my specific application I would like to reach $O(p^{2/3})$. (I would also be willing to do some averaging in $p$ if this improves the bounds. I'm actually more interested in asymptotics for the number of $m$ with $\max(m,\overline{m})$ bounded by a given threshold, but the existence problem already looks nontrivial.)</p>
<p>I tried playing around with Karatsuba's bounds for incomplete Kloosterman sums, but it was not clear to me how to use them to get both $m$ and $\overline{m}$ into intervals smaller than $p^{3/4}$.</p>
http://mathoverflow.net/q/372608Simplifying triangulations of 3-manifoldsMark Bellhttp://mathoverflow.net/users/31212010-08-31T12:56:24Z2014-11-26T01:07:04Z
<p>Throughout, by finite triangulation I mean a triangulation consisting of a finite number of triangles.</p>
<p>Suppose $T$ and $T'$ are finite triangulations of a 3-manifold $M$. We will say that $T'$ is simpler than $T$ iff $T'$ consists of the same number or fewer triangles than $T$ and that $T'$ is a simplest triangulation of $M$ iff $\forall$ triangulation $T$ of $M$, $T'$ is simpler than $T$.</p>
<p>Note: If a 3-manifold $M$ has a finite triangulation, then clearly it has a simplest triangulation.</p>
<p>By a theorem of Pachner (Theorem A.1.1. in 'The geometry of dynamical triangulations') any two triangulations of a manifold can be transformed from one to another by a finite number of stellar subdivisions. As we are only dealing with 3-manifolds, there are only 4 stellar subdivisions; known as the $1 \to 4$, $2 \to 3$, $3 \to 2$ and $4 \to 1$ moves as described in <a href="http://at.yorku.ca/t/a/i/c/45.pdf" rel="nofollow">http://at.yorku.ca/t/a/i/c/45.pdf</a> and hereafter called the Pachner moves. So clearly, there exists a finite sequence of Pachner moves from any finite triangulaiton $T$ of $M$ to $T'$, a simplest triangulation of $M$. </p>
<blockquote>
<p>If $T$ is a finite triangulation of $M$, does the greedy algorithm of just applying as many $4 \to 1$ and $3 \to 2$ Pachner moves to $T$ as possible always result in a simplest triangulation of $M$?</p>
</blockquote>
<p>Or alternatively,</p>
<blockquote>
<p>Is there a finite triangulation $T$ of a 3-manifold $M$ such that repeatedly applying only the $4 \to 1$ and $3 \to 2$ Pachner moves does not eventually result in a simplest triangulation of $M$?</p>
</blockquote>