Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2014-09-02T12:05:47Zhttp://mathoverflow.net/feedshttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://mathoverflow.net/q/1799230Is Besove spaces $B^{s}_{p,q}$ invariant under Fourier transform?Inquisitivehttp://mathoverflow.net/users/330182014-09-02T11:50:08Z2014-09-02T11:56:01Z
<p>(This may be very easy question for MO; as I am just trying to understand Besov spaces)</p>
<p>Let $\phi \in C^{\infty}(\mathbb R^{n})$ with
$ \operatorname{supp} \phi \subset \{\xi \in \mathbb R^{n}: |\xi|\leq 2\} , \phi(\xi)=1$ if $|\xi|\leq 1.$
We put,
$$\phi_{j}(\xi)= \phi(2^{-j}\xi)- \phi(2^{-j+1}\xi), (\xi \in \mathbb R^{n}, j \in \mathbb N).$$
Then we have
$$\operatorname{supp} \phi_{j} \subset \{\xi\in \mathbb R^{n}: 2^{j-1}\leq |\xi| \leq 2^{j+1} \}, j\in \mathbb N $$
and, with $\phi_{0}=\phi,$
$$\sum_{k=0}^{\infty} \phi_{k}(\xi)=1, \text{if} \ \xi\in \mathbb R^{n}.$$</p>
<p>Perhaps there different ways to introduce <a href="http://en.wikipedia.org/wiki/Besov_space" rel="nofollow">Besov spaces</a>; we define in the following way.</p>
<p>Let $0<p\leq \infty, 0 <q \leq \infty$ and $s\in \mathbb R$ then
$$B^{s}_{p,q}(\mathbb R^{n})=\{f\in \mathcal{S'}(\mathbb R^{n}):\|f\|_{B^{s}_{p,q}}=\left(\sum_{k=0}^{\infty} 2^{ksq} \|(\phi_{k}\hat{f})^{\vee}\|_{L^{p}}^{q}\right)^{1/q}<\infty \}.$$</p>
<p>Examples. $B^{s}_{2,2}(\mathbb R^{n})= H^{s}(\mathbb R^{n})(=\text{Sobolev spaces}).$</p>
<p><strong>My naive questions are</strong>: </p>
<blockquote>
<p>(1) Is $B^{0}_{1,1}(\mathbb R^{n})$ can be embedded in $L^{1}(\mathbb R^{n})$ ? (or other $L^{p}$ for $1\leq p \leq \infty$)
(2) Is $B^{0}_{1,1}(\mathbb R^{n})$ is invariant under Fourier transform, that is, if $f\in B^{0}_{1,1}(\mathbb R^{n}),$ then $\hat{f} \in B^{0}_{1,1}(\mathbb R^{n})$ ? (3) What about $B^{s}_{p,q}(\mathbb R^{n})$ except for $p=q=2$ ? (4) What does this definition tells us intuitively ? (5) Is there some thing special about dyadic decompositions ?</p>
</blockquote>
<p>Thanks,</p>
http://mathoverflow.net/q/1799220integral curves and differential equations on arcsDima Sustretovhttp://mathoverflow.net/users/22342014-09-02T11:25:06Z2014-09-02T11:25:06Z
<p>I am trying to prove a statement that is obivious in analytic setting, but makes me feel at a loss in formal algebraic setting.</p>
<p>Let $M$ be a smooth curve over an algebraically closed field. Let $\mathcal D \subset TM^2$ be a distribution of one-dimensional subspaces on the tangent space of $M^2$. Suppose $X$ is a curve that passes through a point $P$ and let projection of $X$ on the first $M$ be \'etale. Choose a morphism $\iota: \mathrm{Spec}\ k[[x,y]] \to M^2$ such that the closed point of $\mathrm{Spec}\ k[[x]]$ is mapped to $P$. Let $\eta \in X(k[[x]])$ be defined (supposing $M$ affine for simplicity) as follows
$$
\eta^*: k[M \times M] \to k[[x]], \qquad g \mapsto \iota^*(g)(x,f)
$$
where $f$ is an element of the maximal ideal of $k[[x]]$ and the expression on the right is substitution of formal power series. (The choice of $\iota$ plays the role of the choice of local coordinates)</p>
<p>Suppose $X$ is an integral curve for the distribution, i.e. for every point $Q \in X$ $T_Q X = \mathcal{D}_Q$. How does one show then that $f$ satisfies the differential equation
$$
f'=h(x,f)
$$
where $h$ depends on the distribution.</p>
<p>In analytic setting $f$ would be a power series that define an actual section of the projection of a neighbourhood of $P$ in $X$ on $M$, and the differential equation would hold in some neigbourhood of $\pi_1(P)$. I am not sure how the derivative in formal power series is related to tangent spaces of points of $X$ close to $P$, although intuitively they should be related. </p>
http://mathoverflow.net/q/1799200Vector Quaternion multiplication [on hold]user3067395http://mathoverflow.net/users/577272014-09-02T10:17:34Z2014-09-02T10:17:34Z
<p>If I multiply two quaternions (representing rotations) Q1 * Q2, then the rotation of Q2 is performed on the local coordinate system of Q1, right? (And not at the world axis where x = (1, 0, 0), y = (0, 1, 0) and z = (0, 0, 1).)</p>
<p>If I multiply two vectors (also representing rotations) V1 * V2, then the rotation of V2 is performed on the global axes of the coordinate system.</p>
<p>What happens if I multiply a vector by a quaternion? Is the rotation performed on the global axes or on the local axes of the vector.
Also, how is the rotation performed if I multiply a Quaternion by a vector?</p>
http://mathoverflow.net/q/1799193(Smooth) Borel Conjecture for 4-dimensional torusLCC1http://mathoverflow.net/users/404842014-09-02T10:10:37Z2014-09-02T11:05:35Z
<p>Given an aspherical 4-dimensional closed manifold $M$ with fundamental group $\mathbb{Z}^4$, it is homotopy-equivalent to $T^4 = S^1 \times \ldots S^1$, the 4-dimensional torus. </p>
<p><strong>Question 1:</strong> Since I am no expert and could not dig out a reference I would be interested if it is open/known that under the circumstances above, $M$ must be homeomorphic (diffeomorphic) to $T^4$?</p>
<p><strong>Question 2:</strong> Can one say more if one knows that $M$ is smoothly covered by $\mathbb{R} \times T^3$?</p>
<p><em>Remark:</em> in all other dimensions it seems to be true (due to Wall et al.) that for dimensions $n \geq 5$ the manifold $M$ is finitely coverey by a manifold diffeomorphic to $T^n$ and for $n \leq 3$ it is even diffeomorphic to $T^3$.</p>
http://mathoverflow.net/q/1799182Characterization of externally definable setsPrimo Petrihttp://mathoverflow.net/users/574672014-09-02T09:51:23Z2014-09-02T09:51:23Z
<p>Let $\cal U$ be a saturated model of inaccessible cardinality $\kappa$. For arbitrary $\cal D\subseteq U$ denote by $\langle\cal U,D\rangle$ the expansion of $\cal U$ with a new predicate for $\cal D$. Write $e({\cal D}/A)$ for collection of subsets $\cal C\subseteq U$ such that $\langle{\cal U,C}\rangle\equiv_A\langle{\cal U, D}\rangle$.</p>
<blockquote>
<p><strong>Question:</strong> Is there some easy condition on $e(\cal D)$ that characterizes sets that are <em>externally definable</em>?</p>
</blockquote>
<p>Recall that $\cal D$ is <em>externally definable</em> if it has the form $\big\{a\ :\ \varphi(x;a)\in p\big\}$ for some global type $p\in S_x(\cal U)$ and some formula $\varphi(x;z)\in L$.</p>
<p>By the way of example, notice that we can easily characterize definablity. Namely, $\cal D$ is definable if and only if $e({\cal D}/A)$ is finite (or, for that matters, $=1$ or $<\kappa$) for some $A$.</p>
<p>Not relevant to the question (but motivates me): it can be shown that $\cal D$ is externally definable if for some $A$ the VC-dimension of $e({\cal D}/A)$ is finite. Clearly, the converse may fail unless we assume that $T$ is nip.</p>
http://mathoverflow.net/q/179914-4Roots of a quadratic formula [on hold]Asadhttp://mathoverflow.net/users/576912014-09-02T07:40:09Z2014-09-02T11:09:39Z
<p>I have a polynomial $ az^2 + bz +c = 0$, where z is a complex number. i.e. $ z = a +ib $ and a, b and c are the real numbers (or complex <em>e.g</em> $a = a+ i 0$) . I have manged to reach the $(z + b/a) = -b/2a + \sqrt{(b^2 - 4ac / 4a^2)}$. I need to know the conditions under which I can take the square root on numerator and denominator separately, which are complex in nature. </p>
http://mathoverflow.net/q/1799130A question for uniqueness of configuration theoremBobscotthttp://mathoverflow.net/users/531192014-09-02T07:16:58Z2014-09-02T11:40:29Z
<p>Recently I am reading a book of Katok and Hasselblatt.</p>
<p>I was confused by the proof for the following theorem:</p>
<p>If $f:R^n\times R^n\rightarrow\mathbb{R^n}$ is C^2, and for any $M>0$, there exists $\epsilon>0$ such that for $i=1,2$ we have $v_1\neq v_2\in R^n$ with the norm smaller that $M$ , $y_i: R\rightarrow R^n$. $y_i''=f(y_i,y_i'), y_i(0)=0, y_i'(0)=v_i$ for $t\in (0,\epsilon],$ then we have $y_1(t)\neq y_2(t)$.</p>
<p>They give a remark in their books:</p>
<p>This is not direct corollary of uniqueness of solution to an ordinary differential equation, since this proposition asserts uniquess in the configuration space. Notice that $\epsilon$ is independent of the initial conditions.</p>
<p>I do not understand their proof in the book (especially for the last line), I put it here.</p>
<p>Proof: Let $$g(t)=y_1''(t)-y_2''(t)$$. Since $f\in C^2$ there exist $k,l\in R$, independent of $v_1, v_2$, such that $$\|g(0)\|\leq k\|v_1-v_2\|$$
and
$$\|g'(0)\|\leq l\|v_1-v_2\|$$</p>
<p>Consequently there exist \epsilon>0, independent of $v_1$, and $v_2$, such that $\|g(t)\|\leq \frac{1}{\epsilon}\|v_1-v_2\|$, for $t\in (0,\epsilon)$.</p>
<p>we have $$\|\frac{d}{dt}(y_1(t)-y_2(t))\|>0$$.</p>
<p>Since $y_1(t)-y_2(0)=0$, we have $\|y_1(t)-y_2(t)\|>0$, for $t\in (0,\epsilon]$. (I do not know why?).</p>
<p>I know when the equation is about geodesic equation, this argument is obvious true, however, I did know a proof for the general case.</p>
http://mathoverflow.net/q/1799121Is every irreducible unitary class one representation induced?Corbennickhttp://mathoverflow.net/users/515082014-09-02T07:05:12Z2014-09-02T09:15:07Z
<p>Let $G$ be a connected semi simple Lie group with finite center.
Fix a maximal compact subgroup $K$.
An irreducible representation $(\pi,V)$ of $G$ is called a "class-one representation", if it contains non-zero $K$-fixed vectors. It is often treated as well-known, at least for splitrank one groups, that every irreducible unitary class-one representation is either trivial or infinitesimally equivalent to a representation induced from a minimal parabolic (and not only a sub representation of the latter as the sub representation theorem says). </p>
<p>Where can I find a proof of this "well-known"?</p>
http://mathoverflow.net/q/1799000A question on degree 4 binary formsStanley Yao Xiaohttp://mathoverflow.net/users/108982014-09-01T22:39:33Z2014-09-02T11:45:59Z
<p>Suppose that we have a binary form $f(x,y) \in \mathbb{Z}[x,y]$ of degree 4, and that we explicitly have</p>
<p>$$\displaystyle f(x,y) = a_0 x^4 + a_1 x^3 y + a_2 x^2 y^2 + a_3 xy^3 + y^4,$$
so that $(0,1)$ is a solution to the equation $f(x,y) = 1$ (most notably, we require that the equation $f(x,y) = 1$ have a solution in $\mathbb{Z}^2$).</p>
<p>Suppose we define the auxiliary form $g = u_0 x^4 + \cdots + u_4 y^4$ by $u_0 = a_0$,
$u_1 = a_1 + 4 a_0 p^{-k} (b-c)$, </p>
<p>$u_2 = a_2 + (b-c)p^{-k}(3a_1 + 6a_0 (b-c)p^{-k})$, $u_3 = a_3 + (b-c)p^{-k}(4a_0 (b-c)^2 p^{-2k} + 3a_1 (b-c)p^{-k} + 2a_2)$, and $u_4 = a_4 - a_3 \left(\frac{b-c}{p^k}\right) - a_2 \left(\frac{b-c}{p^k}\right)^2 - a_1 \left(\frac{b-c}{p^k}\right)^3 - a_0 \left(\frac{b-c}{p^k}\right)^4 $</p>
<p>where $p$ is a prime, $k \geq 1$ is an integer, and $b,c$ are such that each $u_i$ is an integer and $\gcd(u_0, \cdots, u_4) = 1$. Can one bound the number of choices of $b,c$ by an absolute constant for which $g(x,y) = 1$ also has a solution in $\mathbb{Z}^2$?</p>
<p>My motivation is the following situation. Stewart (my PHD advisor) conjectured in 1991 in "On the number of solutions of polynomial congruences and Thue equations", Journal of the American Mathematical Society, (4) Volume 4 (1991), 793-835 that there exists a constant $c_0$ such that for a given binary form $f(x,y) \in \mathbb{Z}[x,y]$ of degree $d \geq 3$, there exists a number $r(f) \in \mathbb{N}$ such that for all integers $h$ with $|h| \geq r(f)$, the number of <em>primitive</em> solutions to the equation $f(x,y) = h$ is bounded by $c_0$. A weaker form of the conjecture is to replace the number $c_0$ (which does not depend on the degree $d$) with a quantity $c(d)$ which depends on $d$. In the same paper, Stewart counted the solutions to $f(x,y) = h$ by considering a form $f(p^k x + by, y)$ with $p^k || h$ and where $b \pmod{p^k}$ is chosen so that the form $\tilde{f}(x,y) = f(p^k x + by, y)$ has content $p^k$. Then one obtains an equation of the form $f'(x,y) = h/p^k$, and continuing in this manner one eventually gets to a counting problem that one can solve. The above construction in my question 'reverses' this process, by first starting with a form $f(x,y)$ which manifestly has a solution to the equation $f(x,y) = 1$, then trying to lift the solution to one of the form $f'(x,y) = p^k$. My question is basically asking under what circumstances and two distinct forms (in my case, both degree 4) $f,g$ lift to the same form $f'$.</p>
http://mathoverflow.net/q/1798801Possible ways to create a graph representation from a distance matrix (through approximation)for3sthttp://mathoverflow.net/users/577062014-09-01T19:24:36Z2014-09-02T09:15:05Z
<p>Forgive me, Im not math professional, but a computer scientist at the beginning of my base research from my thesis, so bare with me if I miss something blatantly obvious.</p>
<p>I have a Euclidean distance matrix, which represent physical distances of nodes. </p>
<p>My goal is to create a geometrically correct undirected graph from this data. So basically I know the edges and their lengths and I want to know the locations of the nodes relative to each other. (So it can be drawn in a simple coordinate system). </p>
<p>Let's say we have the following distance matrix (did not check if this is valid)</p>
<pre><code> | A | B | C
A | 0 | 6 | 9
B | 6 | 0 | 3
C | 9 | 3 | 0
</code></pre>
<p>My naive approach would be to pick a random node and start at <code>(0,0)</code> in a coordinate system, lets say <code>A</code>. From <code>A</code> I can draw a circle with the radius 6 and know that <code>B</code> must be somewhere on this circle. The I would draw another circle with radius 9(<code>A</code>-><code>C</code>). I pick a random point on the <code>B</code>-circle and draw a circle with radius 3. Now this circle should have two intersection on the <code>C</code> circle. Now I should have 2 possible graphs that satisfy the distance matrix. But the problem is, I read in a paper about semidefinite graphs that this kind of problem is NP, so a realistic solution would be an approximation.</p>
<p>There is a phyton lib that can do exactly that (approximate a graphical represenation with a distance matrix): <a href="http://stackoverflow.com/questions/13513455/drawing-a-graph-or-a-network-from-a-distance-matrix">networkx</a></p>
<p>My question is:</p>
<p>Is this some kind of known problem? Are there known approximation algorithms for this or similiar problems? Is this realistically solveable in polynominal time or faster?</p>
<p><strong>Edit:</strong> thx to the comments, I know now this topic is refered to as <a href="http://en.wikipedia.org/wiki/Distance_geometry" rel="nofollow">distance geometry</a>, specificially Im interessted in Sensor Network Localization. And according to the wiki article this problem is called "Sensor Network Localization problem". I found a <a href="https://uwspace.uwaterloo.ca/bitstream/handle/10012/5093/Krislock_Nathan.pdf?sequence=1" rel="nofollow">PhD thesis about this here</a>.</p>
http://mathoverflow.net/q/1798450Repeatedly changing queue behaviorbryanjhttp://mathoverflow.net/users/312862014-09-01T12:23:29Z2014-09-02T11:09:41Z
<p>I'm not sure if this question is suited to MO. I will happily delete if not.</p>
<p><strong>Situation</strong></p>
<p>Consider a general queueing system $\mathscr{S}$, whose customer arrival times are independent, and whose service times are independent; both of these are allowed to have general distributions. No assumptions are made about whether not customers are served singly or in batches (or about the size of the batches in that case). The queue is also allowed, but not required, to have finite capacity.</p>
<p>The queue is explicitly assumed be a regenerative process.</p>
<p>Suppose that in certain situations as the queue evolves, the behavior of the service times or customer arrivals is allowed to be changed.<br>
For example, the modified queue might occasionally be allowed to turn away a customer (whereas the original system $\mathscr{S}$ would typically accept them). Or perhaps the modified queue in certain situations might be able to eliminate the remaining service time for a server which is currently busy, so as to immediately complete the service for a customer (whereas these customers in the original system $\mathscr{S}$ would still be left being served). </p>
<p><strong>EDIT: Example</strong></p>
<p>Let $\mathscr{U}$ be a finite capacity $M/G/K/K$ queue, and let $\mathscr{T}$ be a finite capacity $M/G'/K'/K'$ queue, which both <em>see the same identical customer demand realization</em> (you can imagine that customers are Poisson arrivals of pairs of customers, with one routed to each queue). The behavior of queue $\mathscr{T}$ is changed so that all customers are turned away whenever the queue $\mathscr{U}$ is at full capacity.</p>
<p><strong>Question</strong></p>
<p>If there are only finitely many epochs when these policy changes occur, then the regenerative nature of the queue guarantees there is no effect on the long run behavior of the queue.</p>
<p>But if on the other hand there is a policy (say for example driven either by the state of the queue itself, or maybe the state of some other process) which allows these "one off" behavior changes to become recurring, then there likely will be a change to the long run dynamics.</p>
<p>In particular, if the policy changes have the <em>short term</em> effect of decreasing service time for customers, or of reducing the number of customer arrivals, can one say anything about the effect on <em>long run</em> average queue length?</p>
<p>Deep thanks for any information.</p>
<p>(In my situation, the queue in question is the outstanding replenishment orders for a lost sales inventory system with constant lead time.)</p>
http://mathoverflow.net/q/1798105Are Anderson $T$-motives motives for the function field analogy?user40276http://mathoverflow.net/users/408832014-08-31T20:09:19Z2014-09-02T09:28:37Z
<p>this question is related to this one <a href="http://mathoverflow.net/questions/24282/geometry-for-andersons-motives/30315#30315">Geometry for Anderson's motives?</a>, though the previous one doesn't answer exactly my question. </p>
<p>Let $\mathbb{C}_{\infty}$ be the function field analog of $\mathbb{C}$ for $\mathbb{F}_q (\theta) = \mathbb{Q}_{\infty}$ and $A = \mathbb{C}_{\infty} [T, \tau]$ the Anderson ring ( i.e., $T$ is central and $\tau$ acts as the Frobenius endomorphism).</p>
<p>An Anderson $T$-motive is simply a left $A$-module $M$ which is free and finitely generated over $\mathbb{C}_{\infty} [\tau]$, and satisfies $$(T - \theta)^n M/\tau M = \{ 0 \}$$ for some $n > 0$. If, furthermore, $M$ is finitely generated over $\mathbb{C}_{\infty}[T]$ (or equivalently, free of finite rank), then it's called an abelian $T$-motive or $t$-motive.</p>
<p>First of all, why the word <strong>"motives"</strong> in "Anderson $T$-motives"?</p>
<p>Are $T$-motives analogous to motives of arithmetic schemes? What's its number field analogue?</p>
<p>Thanks in advance.</p>
http://mathoverflow.net/q/17979020Is fixed point property for posets preserved by products?M.Mirabihttp://mathoverflow.net/users/389662014-08-31T14:19:27Z2014-09-02T11:56:23Z
<p>Recall that a Partially Ordered Set (poset) $P$ has the fixed point property (FPP) if any order preserving function $f:P\longrightarrow P$ has a fixed point.</p>
<p><strong>Theorm :</strong> Suppose $P$ and $Q$ are posets with FPP and at least on of them is finite then $(P\times Q)$ has FPP.</p>
<p>Note : $(a,b)\le(c,d)$ if and only if $a\le c$ and $b\le d$.</p>
<p><strong>Question :</strong> Suppose $P$ and $Q$ are two infinite posets with FPP. Does $(P\times Q)$ have FPP ?</p>
http://mathoverflow.net/q/1796730How to find generators to Mordell weil groups of elliptic curves?MKJhttp://mathoverflow.net/users/576102014-08-29T15:53:30Z2014-09-02T10:12:26Z
<p>I am new to branch of elliptic curve and algebraic number theory .I want to find generators to Mordell Weil group of the Elliptic Curve $y^2=x^3-6321363052$ and class number of $\mathbb Q(\sqrt[3]{6321363052}) $. Some suggestions such as algorithm or softwares will be helpful.</p>
http://mathoverflow.net/q/1794599When does $Pr[vr_i=ur_i\mid \forall j < i: vr_j=ur_j] =O( 1/\sqrt n)$?Anushhttp://mathoverflow.net/users/400782014-08-26T18:05:58Z2014-09-02T08:36:30Z
<p>In <a href="http://mathoverflow.net/questions/176987/a-conjecture-about-the-entropy-of-matrix-vector-products/177415">A conjecture about the entropy of matrix vector products</a> I asked a conjecture relating to the entropy of a matrix-vector product. This conjecture is as yet unproven. domotorp then made another conjecture which may be simpler to solve and seems worth posing as a separate question. Here it is, lightly edited for content. The rotations of a vector are simply its cyclic permutations. That is the rotations of a vector $v = (v_1, v_2, v_3, \dots , v_n)$ are $(v_1, v_2, v_3, \dots , v_n)$, $(v_n, v_1, v_2, v_3, \dots , v_{n-1})$, $(v_{n-1}, v_n, v_1, v_2, v_3, \dots , v_{n-2})$ and so on. We assume the entries of the vectors we will define are chosen independently and uniformly from $\{0,1\}$</p>
<p>Suppose we want to compute the probability that for two different random vectors, denoted by $v$ and $u$, multiplying them with the rotations of a random vector $r$ we get the same values, i.e., if the first $k$ rotations of $r$ are denoted by $r_1,\ldots,r_k$, then what is the chance that for all $i$ we have $vr_i=ur_i$.
For any $i$, this is like a random walk, as $v$ and $u$ are also random, so $Pr[vr_i=ur_i]\approx 1/\sqrt n$.</p>
<p>Conjecture: this statement is true even in the following conditional form if $k$ is small enough: $$\forall i \leq k \; Pr[vr_i=ur_i\mid \forall j < i: vr_j=ur_j] =O( 1/\sqrt n).$$</p>
<p>How small is small enough? I am particularly interested in $k \approx n/\ln{n}$ for large $n$. Is this small enough?</p>
<hr>
<p><strong>Edit 1.</strong> Clarified the problem statement by adding the correct quantification and giving an example of rotations of a vector.</p>
http://mathoverflow.net/q/1789902Gaussian expectation of an exponentiated outer productMemminghttp://mathoverflow.net/users/149742014-08-20T23:36:58Z2014-09-02T05:10:26Z
<p>Given a normal random column vector $\mathbf{x} \sim N(\mu, \Sigma)$, I need the expectation,</p>
<p>$$ E\left[ \exp(\mathbf{xx}^\top)\right]$$</p>
<p>where $\exp(\cdot)$ is element-wise exponential function (not a matrix exponential).
<strong>Is there a closed form for this expression?</strong></p>
<p>I know that the inner product form has a closed form:</p>
<p>$$ E\left[ \exp(\mathbf{x}^\top A \mathbf{x})\right] = |I - 2A\Sigma|^{-\frac{1}{2}} \exp\left[ -\frac{1}{2} \mu^\top (I - (I - 2A\Sigma)^{-1})\Sigma^{-1}\mu \right]$$</p>
<p>for a real symmetric matrix $A$. Since each element in the resulting expectation is an exponentiated quadratic function, I feel like there should be a closed form solution, but my <em>Matrix-fu</em> is not strong enough.</p>
<p>(Context: this result is needed to derive a statistical estimator for a state-space model. Eventually, I need to numerically evaluate this expression.)</p>
<p><strong>EDIT</strong>:
Note that
$$ (\mathbf{xx^\top})_{ij} = \mathbf{x^\top}A\mathbf{x}$$
where $A = \frac{1}{2}(J_{(i,j)} + J_{(j,i)})$, and $J_{(i,j)}$ is a matrix with zeros except a 1 at $(i,j)$. So each entry is computable, but can it be simplified to allow matrix form evaluation?</p>
http://mathoverflow.net/q/1787115How to solve such an optimization problempenghttp://mathoverflow.net/users/572222014-08-17T07:22:46Z2014-09-02T05:57:45Z
<p>I encounter the following optimization problem, but I can't solve it.</p>
<p>Given $N$ variables satisfying $0 \leq x_1 \leq x_2 \leq x_3 \leq ... \leq x_N \leq 1$ and an integer $K$ no large than $N$, find the values of $\{x_i\}$ that maximize the following function.</p>
<p>$$\sum_{S \subset \{1,2,..., N\},\\ |S| = K} \prod_{i<j,\\ i,j \in S} (x_j - x_i)^2.$$</p>
<p>This problem is somehow related to Vandermonde matrix. Each additional term in the above target function is just the square of the determinate of Vandermonde matrix generated by the $K$ selected variables belonging set $S$. </p>
<p>\******************************************************************************************\</p>
<p>Many thanks for all who gave valuable comments and potential answers to this question. Based on all these responses, I'd like to summarize the current progress as follows.</p>
<p>The solution to this question may involve the following five steps.</p>
<p>Step 1. Prove that for general $N$ and $K$, the optimal values of all the $N$ $\{x_i\}$ can only take $K$ different numbers, i.e., they are divided into $K$ groups, and all the $\{x_i\}$ in the same group take the same value.</p>
<p>Status: Not proved</p>
<p>Step 2. Prove that the $K$ optimal values of $\{x_i\}$ are independent of the value of $N$.</p>
<p>Status: Can be proved if Step 1 is proved.</p>
<p>Step 3: Prove that the numbers $\{x_i\}$ in each group in Step 1 are almost the same, i.e., they differ by at most 1.</p>
<p>Status: Can be proved if Steps 1&2 are proved.</p>
<p>Step 4: Prove that the original question in the special case of $N = K$ has a unique solution.</p>
<p>Status: Can be proved.</p>
<p>Step 5: Find the closed-form expressions of these $K$ values.</p>
<p>Status: It has been known that these $K$ values are just the Fekete points. However, I still have not find the correct reference showing these closed-form expressions and the corresponding proof.</p>
<p>In summary, the remaining difficulties are Step 1 and Step 5. Step 1 requires more intelligent input, and Step 5 relies on finding the correct reference.</p>
<p>Thanks a lot for all your attention~! I will be greatly appreciated if someone can help me with Steps 1 and 5. </p>
http://mathoverflow.net/q/1787003Birkhoff Ergodic Theorem and Ergodic Decomposition Theorem for Continuous-Time Markov ProcessesJulian Newmanhttp://mathoverflow.net/users/155702014-08-16T22:58:49Z2014-09-02T06:46:53Z
<p>I have a couple of questions regarding ergodicity for Markov processes in continuous time. (In particular, the first question seems like it should be particularly basic, and yet I haven't managed to find a proof [or counterexample!].)</p>
<p>We have a family $(P_x^t)_{x \in \mathbb{R},t \geq 0}$ of Borel probability measures on $\mathbb{R}$ such that</p>
<ol>
<li>for all $A \in \mathcal{B}(\mathbb{R})$, the map $(x,t) \mapsto P_x^t(A)$ is Borel-measurable;</li>
<li><p>for all $x \in \mathbb{R}$, $P_x^0=\delta_x$;</p></li>
<li><p>for all $A \in \mathcal{B}(\mathbb{R})$ and $s,t \geq 0$, $P_x^{s+t}(A)=\int_\mathbb{R} P_y^t(A) \, P_x^s(dy)$.</p></li>
</ol>
<p>[We can refer to $(P_x^t)_{x \in \mathbb{R},t \geq 0}$ as a "measurable stochastic semigroup". In general, "stochastic semigroups" only need to be measurable in $x$ for each $t$.]</p>
<p>We will say that a probability measure $\rho$ on $\mathbb{R}$ is <em>stationary</em> if $\rho(A)=\int_\mathbb{R} P_x^t(A) \, \rho(dx)$ for all $A \in \mathcal{B}(\mathbb{R})$ and $t \geq 0$. We will say that a probability measure on $\mathbb{R}$ is <em>ergodic</em> if it is an extremal point of the convex set of stationary probability measures.</p>
<blockquote>
<p>Q1. Let $(\Omega,\mathcal{F},(\mathcal{F}_{t \geq 0}),\mathbb{P})$ be a filtered probability space, and let $(X_t)_{t \geq 0}$ be a progressively measurable real-valued homogeneous Markov process with transition probabilities given by $(P_x^t)_{x \in \mathbb{R},t \geq 0}$ -- that is to say, $P_{X_s(\cdot)}^t(A)$ is a conditional probability of $X_{s+t}^{-1}(A)$ with respect to $\mathcal{F}_s$ (for all $s,t,A$). Suppose also that $\rho:=X_{0\ast}\mathbb{P}$ is stationary. Fix any bounded measurable $f:\mathbb{R} \to \mathbb{R}$; is it the case that</p>
<p>$\hspace{5mm} \lim_{T \to \infty} \frac{1}{T} \int_0^T f(X_t(\omega)) \, dt$</p>
<p>exists for $\mathbb{P}$-almost all $\omega \in \Omega$?</p>
<p>(Please note that we do <em>not</em> assume any kind of continuity of $(X_t)$, but only that it is progressively measurable.)</p>
</blockquote>
<p>Now in terms of my motivation, what I am really after is an ergodic decomposition theorem for the setting that I'm currently working with; I think that a positive answer to Q1 will be enough for me to prove this. However, I would ideally like to know if ergodic decompositions exist more generally:</p>
<blockquote>
<p>Q2. Suppose $\rho$ is a stationary probability measure. Does there exist a probability measure $Q$ on the set $\mathcal{M}$ of probability measures on $\mathbb{R}$ (equipped with the usual $\sigma$-algebra, which is known to be standard) such that</p>
<ol>
<li><p>$Q$-almost every $\mu \in \mathcal{M}$ is ergodic;</p></li>
<li><p>for all $A \in \mathcal{B}(\mathbb{R})$, $\rho(A) = \int_\mathcal{M} \mu(A) \, Q(d\mu)$?</p></li>
</ol>
</blockquote>
<p>The following might be useful:</p>
<p><strong>Equivalent definitions of ergodicity</strong>: Given a stationary probability measure $\rho$, we will say that a set $A \in \mathcal{B}(\mathbb{R})$ is $\rho$-<em>almost stationary</em> if for all $t \geq 0$, $\rho(x \in A: P_x^t(A)=1)=\rho(A)$.</p>
<p>(1) In analogy to Proposition 7.2.4 of books.google.co.uk/books?isbn=0521515971 (p378) for deterministic systems, we have that a stationary probability measure $\rho$ is ergodic if and only if every $\rho$-almost stationary set has $\rho$-trivial measure: If $\rho(A) \in (0,1)$ and $A$ is $\rho$-almost stationary, then $\rho$ conditioned on $A$ and $\rho$ conditioned on $\mathbb{R} \setminus A$ are stationary probability measures which can be linearly combined in the obvious way to give $\rho$. In the other direction, it suffices to show that if every $\rho$-almost stationary set has trivial measure and $\tilde{\rho}$ is a stationary probability measure that is absolutely continuous with respect to $\rho$, then $\rho=\tilde{\rho}$. Take a density $g$ of $\tilde{\rho}$ with respect to $\rho$. For each $t$, define the probability measure $\rho_t$ on $\mathbb{R} \times \mathbb{R}$ by $\rho_t(A \times B) = \int_A P_x^t(B) \, \rho(dx)$. The stationarity of $\tilde{\rho}$ implies that</p>
<p>$\hspace{5mm} \int_{A \times (X \setminus A)} g(x_1) \, \rho_t(d(x_1,x_2)) \ = \ \int_{(X \setminus A) \times A} g(x_1) \, \rho_t(d(x_1,x_2))$</p>
<p>for any $A \in \mathcal{B}(\mathbb{R})$ and $t \geq 0$. Setting $A:=\{x \in X : g(x) \geq 1\}$, the above equation (combined with the stationarity of $\rho$) implies that $A$ is $\rho$-almost stationary, so $A$ has trivial measure. It follows that $\tilde{\rho}=\rho$.</p>
<p>(2) We will say that a set $A \in \mathcal{B}(\mathbb{R})$ is <em>invariant</em> if for all $t \geq 0$ and all $x \in A$, $P_x^t(A)=1$. Given a set $A$ that is $\rho$-almost stationary, there exists a set $A'$ that is invariant, with $\rho(A \triangle A')=0$. Namely, set</p>
<p>$\hspace{5mm} A' \ := \ \{ x \in X : \textrm{Leb}(t \geq 0 : P_x^t(A)<1) = 0 \}$</p>
<p>where $\textrm{Leb}$ denotes the Lebesgue measure. So a stationary probability measure $\rho$ is ergodic if and only if every invariant set has $\rho$-trivial measure.</p>
<p>(It is perhaps worth pointing out that (1) does not rely on the stochastic semigroup $(P_x^t)$ being a "measurable" stochastic semigroup, but the construction in (2) does rely on this.)</p>
http://mathoverflow.net/q/1693053Are there any new results on approximating Riemann $\Xi$ function by Polya-like Fourier transforms?mikehttp://mathoverflow.net/users/336722014-06-07T16:06:59Z2014-09-02T09:00:17Z
<p>I posted [this question][1] at math.stackexchange.com and was told that it is more appropriate to post this research related question here at mathoverflow.</p>
<p>So I re-post it below.</p>
<p>Riemann $\Xi(z)$ function is related to Riemann $\zeta(s)$ function via ($s=1/2+i z$):</p>
<p>$$\Xi(z)=\frac{1}{2}s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)$$</p>
<p>The functional equation for $\zeta(s)$ is equivalent to $\Xi(z)=\Xi(-z)$.</p>
<p>Riemann $\Xi(z)$ function can be expressed as a Fourier transformation:</p>
<p>$$\Xi(z)=2\int_0^{\infty}\Phi(u)\cos(z u){\rm d}u$$</p>
<p>where
$$\Phi(u)=\sum_{n=1}^{\infty}\left(4\pi^2n^4\exp(9u/2)-6\pi n^2\exp(5u/2)\right)\exp\left(-\pi n^2 \exp(2u)\right)=\Phi(-u)$$</p>
<p>(1) Polya approximated $\Phi(u)$ with $\Phi_{*}(u)$ and $\Phi_{**}(u)$:</p>
<p>$$\Phi_{*}(u)=8\pi^2\cosh(9u/2)\exp\left(-2\pi \cosh(2u)\right)$$</p>
<p>$$\Phi_{**}(u)=\left(8\pi^2\cosh(9u/2)-6\pi\cosh(5u/2)\right)\exp\left(-2\pi \cosh(2u)\right)$$</p>
<p>This is because he noticed that when $u\to\infty$, $\Phi(u)\to\Phi_{*}(u)$ and $\Phi(u)\to\Phi_{**}(u)$.</p>
<p>Polya proved that the resulting $\Xi_*(z)$ and $\Xi_{**}(z)$ have real zeros only.</p>
<p>(2) de Bruijn approximated $\Phi(u)$ with $\Phi_d(u)$:</p>
<p>$$\Phi_d(u)=2\cosh(5u/2)\left(2\pi^3-3\pi+4\pi^2\cosh(u)\right)\exp\left(-2\pi \cosh(2u)\right)$$</p>
<p>de Bruijn proved that the resulting $\Xi_d(z)$ has real zeros only.</p>
<p>(3) de Bruijn also approximated $\Phi(u)$ with $\Phi_\lambda(u)$:</p>
<p>$$\Phi_\lambda(u)=\exp(\lambda u^2)\Phi(u)$$</p>
<p>de Bruijn proved that when $\lambda\ge \frac{1}{8}$,the resulting $\Xi_\lambda(z)$ has real zeros only.</p>
<p>Newman showed that when $\lambda\lt 0$,$\Xi_\lambda(z)$ has non-real zeros as well.</p>
<p>(@SylvainJULIEN pointed out that) The so-called de Bruijn-Newman constant $\Lambda$ is defined in such a way that $4\lambda \ge \Lambda$ implies $\Xi_\lambda(z)$ has only real zeros. </p>
<p>(4) Hejhal approximated $\Phi(u)$ with $\Phi_N(u)$:</p>
<p>$$\Phi_N(u)=\sum_{n=1}^{\infty}\left(8\pi^2n^4\cosh(9u/2)-12\pi n^2\cosh(5u/2)\right)\exp\left(-2\pi n^2 \cosh(2u)\right)$$</p>
<p>Hejhal proved that almost all the zeros of the resulting $\Xi_N(z)$ are real.
However when $N\to\infty$ $\Phi_N(u) \not\to \Phi(u)$.</p>
<p>For more details please refer to two review papers by Dimitrov and Rusev [1] and Ki [2] and references therein.</p>
<p>[1]: Dimitrov and Rusev, The zeros of entire Fourier transforms, EAST JOURNAL ON APPROXIMATIONS Volume 17, Number 1 (2011), 1-108</p>
<p>[2]: Ki, The Zeros of Fourier Transforms </p>
<p>Question 1: Are there any new research results on approximating Riemann $\Xi(z)$ by Fourier transforms?</p>
<p>Question 2: Why approximating Riemann $\Xi(z)$ by Fourier transforms does not seem to be an active research field towards a possible proof of Riemann hypothesis$?</p>
<p>Best regards-
mike</p>
http://mathoverflow.net/q/1690135Strong Markov property for Poisson point processSinusxhttp://mathoverflow.net/users/410712014-06-04T13:31:56Z2014-09-02T06:02:37Z
<p>The question is thoroughly contained in the title. I just say that I would only like to find a reference for this question. I have searched in some books, to no avail.</p>
<p>Here is what I mean exactly. Let's say we have a process $N$ on $\mathbb{R} _+\times \mathbb{R}^d$, $\mathbb{R} _+$ is time, with the intensity measure being the Lebesgue measure on $\mathbb{R} _+\times \mathbb{R}^d$. Let $\mathscr{F} _t$ be the minimal $\sigma$ -algebra containing all random variable
$N(Q \times U)$, where $Q \in \mathscr{B}([0;T])$, $U \in \mathscr{B}(\mathbb{R}^d)$. Or we may take minimal complete right-continuous $\sigma$ - algebra with this property. Let $\tau$
be a stopping time with respect to $(\mathscr{F} _t)$. It seems reasonable to conjecture,
that the process $\bar N $ defined by </p>
<p>$$\bar N ([0;s] \times U) = N ([\tau;\tau + s] \times U), \ \ \ U \in \mathscr{B}(\mathbb{R}^d)$$</p>
<p>is a Poisson point process with the same intensity measure independent of $(\mathscr{F} _{\tau})$. I was not able however to find a reference to this statement.</p>
<p>In terms of random sets, it corresponds to the strong Markov property of the set
$[0;\tau] \times \mathbb{R}^d$, which is not compact. </p>
<p>In the book by Rozanov, <a href="http://link.springer.com/book/10.1007%2F978-1-4613-8190-7" rel="nofollow">link</a>, the strong Markov property is
considered for compact stopping sets.</p>
<p>I asked this question on math.stackexchange, <a href="http://math.stackexchange.com/questions/807773/strong-markov-property-for-poisson-point-process">link</a>.</p>
<p><strong>Update</strong></p>
<p>I would like to add two things. </p>
<p><strong>First</strong>, it looks like the strong Markov property for $\bar N$ follows from Theorem 20.9 <a href="http://www.cambridge.org/us/academic/subjects/statistics-probability/probability-theory-and-stochastic-processes/stochastic-processes-1" rel="nofollow">here</a>:</p>
<blockquote>
<p><strong>Theorem 20.9</strong> Suppose $(X_t , P^x)$ is a Markov process with respect to $\{\mathscr{F}_t \}$, that Assumption
20.1 holds, and that $T$ is finite stopping time. If Y is bounded and measurable with respect
to $\mathscr{F}_\infty$, then
$$
E^x[Y \circ \theta _T |\mathscr{F}_T ] = E^{X_T}Y, \ \ \ \ P^x-a.s.
$$</p>
</blockquote>
<p>Here $\{\mathscr{F}_t \}$ is the minimal right-continuous filtration, which contains all sets $N$ satisfying $P^x(N)=0$ for all $x$,
and such that
$(X_t)$ is adapted:
$$
\mathcal{F}_t^{00} = \sigma(X_s : s \leq t),
$$
$$ \mathcal{F}_t^{0} = \sigma \left(\mathcal{F}_t^{00} \cup \{ A \subset S : A \text{ is } \Bbb{P}^x\text{-null for all } x \in S\} \right) \quad \text{and} \quad \mathcal{F}_t = \mathcal{F}_{t+}^{0} = \bigcap_{\epsilon > 0} \mathcal{F}_{t+\epsilon}^{0},$$
$S$ is the state space and $P_t $ is associated with the process semigroup. Assumption 20.1:</p>
<blockquote>
<p><strong>Assumption 20.1</strong> Suppose $P_t f$ is continuous on $S$ whenever $f$ is bounded and continuous
on $S$.</p>
</blockquote>
<p>Assumption 20.1 is satisfied, if we consider a Poisson point process as a canonical process in the space $D_S [0;T_1]$, $T_1 >0$, $S =\Gamma $. Here $\Gamma$ is the space of all simple counting measures over $\mathbb{R}^d$, equipped with the vague topology, i.e. the smallest topology such that
for every $f \in C_K(\mathbb{R}^d)$ the mapping </p>
<p>$$
\Gamma \ni \gamma \to \int f d \gamma
$$
is continuous. With this topology, $Г$ is a Polish space, <a href="http://onlinelibrary.wiley.com/doi/10.1002/mana.200310392/abstract" rel="nofollow">link</a>. Then assumption 20.1 is equivalent to</p>
<p>\begin{equation}
E g(\gamma _n \cup N _t) \to E g(\gamma \cup N_t), \ \ \ \text{whenever} \ \gamma _n \to \gamma \ \ \ \ \ \ \ \ \ \ \ \ (1)
\end{equation}
where $g: \Gamma \to \mathbb{R}$ is a bounded and continuous function. Convergence
$\gamma _n \to \gamma$ in the vague topology implies $\gamma _n \cup N _t \to \gamma \cup N_t$ a.s. Therefore, (1) follows by the bounded convergence Theorem.</p>
<p>Also, one should prove that $(N_t)$ is a Markov process under $\{\mathscr{F}_t \}$ (the is an assumption of Theorem 20.1).</p>
<p><strong>Second</strong>, I have tried to prove the statement in the original question, using the idea suggested by Anthony Quas in his comment. To prove that $ N$ is a strong Markov, enough to show that</p>
<ol>
<li>for any $b>a>0$ and bounded open $U \subset \mathbb{R}^d$, $\bar N ((a;b),U)$ is an independent of $\mathscr{F} _\tau$ Poisson random variable with mean $(b-a)\lambda (U)$, $\lambda$ is the Lebesgue's measure on
$\mathbb{R}^d$, and</li>
<li>for any $b_k>a_k>0$, $k=1,...,m$, and any bounded open $U_k \subset \mathbb{R}^d$, such that $((a_i;b_i) \times U_i) \cap( (a_j;b_j) \times U_j) = \varnothing $, $i \ne j$,
random variables $\bar N ((a_k;b_k) \times U_k)$
are independent of each other.</li>
</ol>
<p>To do so, let $\tau _n$ be the sequence of stopping times taking only countably many values, $\tau _n \downarrow \tau$, $\tau _n - \tau \leq \frac{1}{2^n}$. Then $N$ satisfies strong Markov property for $\tau _n$, and the processes $\bar N _n$ defined by
$$
\bar N _n ([0;s] \times U) = N ([\tau_n;\tau_n + s] \times U),
$$
are Poisson point processes. To prove 1, note that $\bar N _n ((a;b) \times U) \to \bar N ((a;b) \times U) $ a.s. and all random variables $\bar N _n ((a;b) \times U)$ have the same distribution, therefore $\bar N ((a;b) \times U)$ is a Poisson random variable with mean
$(b-a)\lambda (U)$. Random variables $\bar N _n ((a;b) \times U)$
are independent of $\mathscr{F}_{\tau}$, hence $\bar N ((a;b) \times U)$ is independent of $\mathscr{F}_{\tau}$, too. Similarly, 2 follows.</p>
http://mathoverflow.net/q/1686436DG enhancements of $\ell$-adic derived categoriesReladenine Vakalwehttp://mathoverflow.net/users/239072014-05-30T17:48:07Z2014-09-02T06:54:40Z
<p>This question is similar in flavor to <a href="http://mathoverflow.net/questions/99689/existence-of-dg-realization-for-6-functors/100570#100570">Existence of dg realization for 6 functors</a></p>
<p>Let $X$ be a complex variety and $D(X)$ the bounded derived category of constructible sheaves (the Euclidean topology and field coefficients say) on $X$. Let $\mathcal{L}\in D(X)$. Then there exists a dg-algebra $\mathcal{E}$ whose cohomology is the shifted $Hom$-groups $Hom(L, L[i])$ in $D(X)$ and such that the derived category of (dg-modules of) $\mathcal{E}$ is equivalent to the triangulated subcategory of $D(X)$ generated by $\mathcal{L}$. (Strictly speaking I need some finiteness conditions but I am going to ignore those for now).</p>
<p><b>Question 1:</b> Does a similar result hold if $D(X)$ is replaced with the corresponding $\ell$-adic `derived category’?</p>
<p>If the answer to 1) is yes, then:</p>
<p><b>Question 2:</b> Assume that $X$ is defined over a finite field and $\mathcal{L}$ admits a mixed structure (a la Deligne). Then the shifted $Hom$-groups $Hom(\mathcal{L}, \mathcal{L}[i])$ in the ordinary (= non-mixed derived category) inherit mixed structures. Is it possible to lift the mixed structure to the dg-algebra $\mathcal{E}$ produced by an affirmative answer to 1?</p>
<p>In the circles I run in it is standard to assume that 1) is `morally' true but I have no real clue as to what a proof would look like. Not so sure about 2). They are both true for some particular $\mathcal{L}$ (such as if $\mathcal{L}$ is the direct sum of $IC$-complexes corresponding to a stratification by contractible strata). However I am looking for something more general. </p>
<p>A related question:</p>
<p><b>Question 3:</b> Does the category of <i>all</i> perverse sheaves on $X$ (say in the $\ell$-adic or complex algebraic setting) have enough projectives (or injectives)?</p>
<p>Question 2 has an analogue in the setting of mixed Hodge modules. I do not know the answer there either (except for some very special cases). My reason for asking it in the $\ell$-adic setting is my failure at answering it in the Hodge setting (which I am considerably more familiar with than the $\ell$-adic setup).</p>
<p><b>Added later:</b> The answer to Question 3 is no in general (look at $Ext$ with skyscrapers).</p>
<p><b>Added even later:</b> Let me try and explain what 2) is asking for via some "examples". Let $X$ be a complex variety, and $D(X)$ the usual derived category of constructible sheaves. Consider the constant sheaf $\mathcal{L}$ on $X$. Let $\mathcal{E}$ be the complex obtained by applying taking global sections of the Godement resolution of $\mathcal{L}$. Then $\mathcal{E}$ is a dg-algebra. Note:</p>
<p>a) The cohomology algebra of $\mathcal{E}$ is isomorphic to the $Ext$ algebra of $\mathcal{L}$ which is of course the same as the cohomology ring $H^*(X)$.</p>
<p>b) It is not quite obvious, but true (and well known) nonetheless, that the derived category of dg-modules of $\mathcal{E}$ is equivalent to the triangulated subcategory of $D(X)$ generated by $\mathcal{L}$. (To keep things simple I am omitting some finiteness adjectives here).</p>
<p>Now if we had been working with rational coefficients say, then $H^*(X)$ has a lot of additional structure such as a mixed Hodge structure. In particular $H^*(X)$ has an additional grading via weights (if you don't like Hodge theory, feel free to work with Gilet and Soule's motivic weight filtration in which case you can even work over the integers). What Question 2) is asking for is to lift this grading to the complex $\mathcal{E}$ so that the induced grading on $H^*(\mathcal{E})$ is exactly what you started with. Actually, the way it is stated Question 2) is asking to lift the full mixed structure, but for my purposes lifting the grading suffices.</p>
<p>With the Godement resolution it is not clear whether this can be done. However, say instead of the Godement resolution I had used the de Rham resolution (assume $X$ to be smooth), then both a) and b) are still true and now I can lift the weight grading.</p>
<p>My interest is in $\mathcal{L}$ more general than the constant sheaf: mainly some semisimple perverse sheaves. If one is working with the complex numbers, then basically using Godement resolutions (or any injective resolution) one can construct an $\mathcal{E}$ such that a) and b) hold (so this is my dg-algebra model of the category I am after). On the other hand mixed Hodge theory gives me a weight grading on the cohomology of $H^*(\mathcal{E}) = Ext^*(\mathcal{L}, \mathcal{L})$. And Question 2) asks to lift this to $\mathcal{E}$. </p>
<p>Now I simply don't know how to do this in the Hodge setting. But weights in $\ell$-adic cohomology (work over finite fields now) are in a sense much simpler (there's an actual Galois group giving the weights etc.) However, in the $\ell$-adic setup since the triangulated category that one is dealing with isn't quite a derived category, it is not clear if one can construct an $\mathcal{E}$ satisfying a) and b) in the first place. This is what Question 1) is asking to resolve, and then Question 2) is asking to lift weight gradings.</p>
<p>Let me point out that the same questions can be asked in the setting of motivic sheaves (say in the form given by Deglise-Cisinski), and if you believe in "Beilinson's world" of motivic sheaves (spruced up a bit to a dg-enhanced setting) then all of these questions must have positive answers. But the motivic theory as it stands has a deficiency (t-structure!) that makes it even harder to work with it directly for these things.</p>
<p>Finally, I think it's fair to ask why anyone would/should care about this? This is a long story (which I wont attempt to tell here). But it goes back to Deligne-Morgan-Sullivan's landmark result on the formality of the rational homotopy type of smooth projective varieties. These ideas transplanted into representation theory start having some highly non-trivial consequences (well, if one can prove formality). Let me point to:</p>
<p><a href="http://mathoverflow.net/questions/46521/formality-of-classifying-spaces?rq=1">Formality of classifying spaces</a></p>
<p><a href="http://arxiv.org/abs/1209.3760" rel="nofollow">http://arxiv.org/abs/1209.3760</a></p>
<p><a href="http://arxiv.org/abs/1404.6333" rel="nofollow">http://arxiv.org/abs/1404.6333</a></p>
<p>as starting points for why representation theorists might (or should?) care about these things.</p>
http://mathoverflow.net/q/1684700Natural integration constant for normal and discrete integration: is there a connection?Anixxhttp://mathoverflow.net/users/100592014-05-28T20:30:37Z2014-09-02T11:06:54Z
<p>It is often assumed that integration, unlike differentiation is defined only to an arbitrary constant. So the antiderivative function is often left undefined or postulated to be zero in zero.</p>
<p>But I tasked myself to derive an expression for a natural integration constant that will unambiguously define an antiderivative in the most natural way.</p>
<p>Let $f(x)$ be a function that is equal to its Newton's expansion.</p>
<p>The most obvious idea is to intepolate over consequtive derivatives with a Newton interpolation formula and if it converges to an analytic function, just take it at -1:</p>
<p>$$f^{(-1)}(0)=\sum_{m=0}^{\infty} \binom {-1}m \Delta^k g(0)$$</p>
<p>where $g(p)=f^{(p)}(0)$, or in full form,</p>
<p>$$f^{(-1)}(0)=\sum_{m=0}^{\infty} \binom {-1}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(0)$$</p>
<p>Unfortunately this method converges only for a very limited class of functions. One of the functions, for which it works is the exponent. Since all its derivatives in zero are equal to $1$, the Newton series converges to constant $1$ and as such, $f^{(-1)}(0)=1$ as well. Although this method does not converge for all power bases, it allows generalization to all exponents, and as such, to hyperbolic and trigonometric sine and cosine.</p>
<p>Lets designate the natural integration constant $$N_{h=0}[f] = f^{(-1)}(0)$$ now on.</p>
<hr>
<p>Now look at this expression for discrete integral (indefinite sum, antidifference), known as Faulhaber's formula:</p>
<p>$$\sum _x f(x)= \sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n(x) + C$$</p>
<p>The expression $\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n(x)$ gives the Ramanujan sum R[f] when taken at x=0. Maybe this can give us a clue for a natural integration constant for normal intergral? Lets write the expression in time scales and take a limit at $h\to0$:</p>
<p>$$R_h[f](x)=\lim_{h\to0} \sum_{n=1}^{\infty} \frac{h^nf^{(n-1)} (0)}{n!} B_n(x/h)=\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!}x^n$$</p>
<p>Which is just Taylor series for antidifference without the first term. It is always is equal to zero at $x=0$. Thus, the Ramanujan sum contributes nothing to the natural integration constant at $h=0$.</p>
<p>But if we look at the Faulhaber's formula (or any other formula for indefinite sum that gives Ramanujan's sum at zero), we will find that there is a tiny incompleteness in it: the sum starts from 1, thus missing the first term. So if we postulate the natural discrete integral to be the full summ starting from 0, we get, writing the zeroth term separately:</p>
<p>$$\Delta^{-1}f(x)=\sum_{n=0}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n(x)=f^{(-1)}(0)+\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n(x)$$</p>
<p>If $f^{(-1)}(x)$ is the natural integral, then we can express the natural integration constant for h=1 as follows (noting the rest of the terms is the Ramanujan's sum):</p>
<p>$$N_{h=1}[f]=N_{h=0}[f]+R[f]$$</p>
<p>or</p>
<p>$$N_{h=0}[f]=N_{h=1}[f]-R[f]$$</p>
<p>To find $N_{h=1}[f]$ we can employ the same technique as in the first part of this post, but interpolating over finite differences rather than derivatives:</p>
<p>$$f^{(-1)}(0)=\sum_{m=0}^{\infty} \binom {-1}m \sum_{k=0}^m\binom mk(-1)^{m-k}\Delta^k f(0)-R[f]$$</p>
<p>This should potentially work for a wider class of the functions.</p>
<p>The later expression can be simplified to</p>
<p>$$f^{(-1)}(0)=\sum_{m=-\infty}^{-1} f(m) -R[f]$$</p>
<hr>
<p>Thus my question is for a rigorous proof that the natural integration constants obtained via interpolation for normal and discrete integrals differ only by a value of the Ramanujan sum of the function.</p>
http://mathoverflow.net/q/1641070Collecting terms of a linear expression with nested sums and combinatorics in coefficientsGiovanni Ursinohttp://mathoverflow.net/users/499072014-04-23T10:19:59Z2014-09-02T09:59:36Z
<p>I need to collect the $\Pr(\cdot)$ terms of the following expression:</p>
<p>$\sum_{m=3}^{n}\frac{g_{m}\left( \cdot \right) }{\left( \sqrt{\theta \left(
1-\theta \right) }\right) ^{m}}\left[ \sum_{j=2}^{m-1}\sum_{i=0}^{j}%
\sum_{l=0}^{n-j-1}\left[ \frac{j!}{\left( j-i\right) !i!}\left( -\theta
\right) ^{j-i}\left( 1-\theta \right) ^{m-1-j+i}\Pr \left( \mathbf{0}%
_{j+1-i+l},\mathbf{1}_{i+n-j-1-l}\right) \right] \right] $</p>
<p>where $\Pr \left( \mathbf{0}%
_{j+1-i+l},\mathbf{1}_{i+n-j-1-l}\right)$ is the probability that $i+n-j-1-l$ out of $n$ Bernoulli variables take value $1$ while the remaining $j+1-i+l$ take value $0$ and $g_m(\cdot)$ is a polynomial which depends only on $m$ and other variables not relevant here (hence the $(\cdot)$ argument).</p>
<p>Each Bernoulli variable has probability equal to $\theta\in(0,1)$ of taking value 1 (identical marginals). However, variables are NOT independent and their joint distribution is only known to be symmetric (hence the notation above where it is only reported how many zeroes and ones are taken by the $n$ variables and not the exact sequence of all values). </p>
<p>For the question posed here the $\Pr(\cdot)$ terms are to be considered as $n+1$ parameters of an unknown distribution. I want to collect the $\Pr(\cdot)$ terms in an expression which, ideally, reads something like</p>
<p>$\Pr \left( \mathbf{0}_{1},\mathbf{1}_{n-1}\right) f_{1}\left( \theta
,n,g_{3}\left( \cdot \right) ,...,g_{n}\left( \cdot \right) \right) +\Pr
\left( \mathbf{0}_{2},\mathbf{1}_{n-2}\right) f_{2}\left( \theta
,n,g_{3}\left( \cdot \right) ,...,g_{n}\left( \cdot \right) \right) +...+\Pr
\left( \mathbf{0}_{n},\mathbf{1}_{0}\right) f_{n}\left( \theta ,n,g_{3}\left(
\cdot \right) ,...,g_{n}\left( \cdot \right) \right) $</p>
<p>where the $f_i(\cdot)$, $i=1,2,...,n$ are the coefficients of $\Pr(\mathbf{0}_i,\mathbf{1}_{n-i})$'s.</p>
<p>Hope the problem was stated sufficintly clearly. Thanks for any suggestion.</p>
http://mathoverflow.net/q/1625170properties of frequency-uniform decomposition operator $\square_{k}^{\sigma}$Inquisitivehttp://mathoverflow.net/users/330182014-04-05T11:39:50Z2014-09-02T09:50:04Z
<p>Let $\rho \in S(\mathbb R^{n})= \text{Schwartz space}, \ \rho:\mathbb R^{n}\to [0,1]$ be smooth radial function verifying $\rho(\xi)=1$ for $|\xi|_{\infty}\leq \frac{1}{2}$ and $\rho(\xi)=0$ for $|\xi|_{\infty}\geq 1$ [For $\xi=(\xi_{1},...,\xi_{n}), |\xi|_{\infty}:= \max_{i=1,2,...,n}|\xi_{i}|$ ]. Let $\rho_{k}$ be a translation of $\rho,$ $\rho_{k}(\xi):=\rho(\xi -k), k\in \mathbb Z^{n}.$
Let $Q_{k}$ be be the unit cube with center at $k,$ $\{Q_{k}\}_{k\in\mathbb Z^{n}}$ constitutes a decomposition of $\mathbb R^{n}.$ We notice, $\rho_{k}(\xi)=1$ in $Q_{k}$, and so $\sum_{k\in \mathbb Z^{n}}\rho_{k}(\xi)\geq 1$ for all $\xi \in \mathbb R^{n}.$ Denote,
$\sigma_{k}(\xi):= \rho_{k}(\xi)\left(\sum_{k\in\mathbb Z^{n}}\rho_{k}(\xi)\right)^{-1}, \ k\in \mathbb Z^{n}.$ Then we have, </p>
<p>(1) $|\sigma_{k}(\xi)|\geq c, \forall z \in Q_{k}$, </p>
<p>(2) $\text{supp} \ \sigma_{k} \subset \{\xi: |\xi-k|_{\infty}\leq 1 \},$ </p>
<p>(3) $\sum_{k\in \mathbb Z^{n}} \sigma_{k}(\xi)\equiv 1, \forall \xi \in \mathbb R^{n},$ </p>
<p>(4) $|D^{\alpha}\sigma_{k}(\xi)|\leq C_{|\alpha|}, \forall \xi \in \mathbb R^{n}, \alpha \in (\mathbb N \cup \{0\})^{n}.$</p>
<p>Hence, the set
$A_{n}:=\{ \{\sigma_{k}\}_{k\in \mathbb Z^{n}}:\{\sigma_{k}\}_{k\in \mathbb Z^{n}} \ \text{satisfies} (1) to (4) \}$ is non empty. </p>
<p>Let $\{\sigma_{k}\}\in A_{n}$ and define frequency- uniform decomposition operator
$$\square_{k}^{\sigma}:= \mathcal{F}^{-1}\sigma_{k}\mathcal{F}, k\in \mathbb Z^{n};$$
where $\mathcal{F}-$ denotes Fourier transform and $\mathcal{F}^{-1}-$ inverse Fourier transform.</p>
<blockquote>
<p><strong>My Questions</strong>: (I)Why this operator is known as frequency-uniform decomposition operator ? (II) Can we expect, $$\square_{k}^{\sigma}=\sum_{|\ell|_{\infty}\leq 1}\square_{k}^{\sigma}\square_{k+\ell}^{\phi};$$
for $\{\sigma_{k}\}_{k\in \mathbb Z^{n}}, \{\phi_{k}\}_{k\in \mathbb Z^{n}} \in A_{n}$; if yes, How ? </p>
</blockquote>
<p><strong>My Motivation</strong>:(Importance of frequency-uniform decomposition operator); It is well-known that $S(t)=e^{it\triangle}: L^{p}\to L^{p}$ if and only if $p=2.$ But the frequency-uniform decomposition has at least two advantages for the Shr\"odinger semi-group: (a) $\square_{k}e^{it\triangle}:L^{p'}\to L^{p}$ satisfies a uniform truncated decay, (b) $\square_{k}e^{it\triangle}$ is uniformly bounded on $L^{p}.$</p>
<p>Thanks,</p>
http://mathoverflow.net/q/1621720The Largest Root of Associated Laguerre PolynomialFederico Magallanezhttp://mathoverflow.net/users/113612014-04-02T08:07:25Z2014-09-02T11:59:36Z
<p>The <a href="http://mathworld.wolfram.com/LaguerrePolynomial.html" rel="nofollow">Laguerre polynomial</a> $L_n(x)$ is the solution to the <a href="http://mathworld.wolfram.com/LaguerreDifferentialEquation.html" rel="nofollow">Laguerre differential equation</a>
\begin{equation*}
x\,y'' + (1 - x)\,y' + n\,y = 0.
\end{equation*}</p>
<p>The <a href="http://mathworld.wolfram.com/AssociatedLaguerrePolynomial.html" rel="nofollow">associated Laguerre polynomial</a> $L_n^\alpha(x)$ is the solution to the more general Laguerre differential equation
\begin{equation*}
x\,y'' + (\alpha + 1 - x)\,y' + n\,y = 0.
\end{equation*}</p>
<p>One can easily see that $L_n(x) = L_n^0(x)$.</p>
<hr>
<p>My question is:</p>
<ol>
<li>Let $D$ be the derivative operator with respect to $x$. Some papers mention that iterating the operator $(I − \alpha D)$ for any $\alpha > 0$ for $x^n$ generates an associated Laguerre polynomial; that is,
\begin{equation*}
p_k(x) = (I − \alpha D)^k x^n,
\end{equation*}
But I'm not sure how the Laguerre differential equation is related to this differential operator notataion.</li>
<li>I'm wondering whether we can get the upper-bound of the largest root of $L_n^\alpha(x)$. Maybe it's hard to compute it. Is there any known result on this topic? </li>
</ol>
http://mathoverflow.net/q/1496749Coherence between different ranking methods of a graph's verticesFelix Goldberghttp://mathoverflow.net/users/220512013-11-22T21:51:49Z2014-09-02T09:34:22Z
<p>Given a (connected) graph $G$ it is natural to want to rank its vertices, with the more "central" vertices ranked higher.</p>
<p>Two natural ways of doing it are:</p>
<ol>
<li>By the degrees.</li>
<li>By the entries in a Perron eigenvector of the adjacency matrix.</li>
</ol>
<p>These two methods coincide for regular graphs and for so-called harmonic graphs (defined as graphs in which the degree vector is an eigenvector) - which is all a tad trivial. </p>
<p>What is more interesting is that for many random graphs I've checked the two orderings coincide as well and I am able to show that they coincide for threshold graphs.</p>
<blockquote>
<p>Have such graphs been studied?</p>
</blockquote>
<p>I did find in the mathematical sociology literature some work on the question when the most central vertex w.r.t both rankings is the same but nothing for the whole vector.</p>
<p>(If such graphs haven't been named yet, I propose to call them <strong>tranquil</strong>, since tranquility is a lesser form of harmony). </p>
<p>P.S.
Method 2 is essentially what Google does in its PageRank algorithm.</p>
http://mathoverflow.net/q/14604210Can the equation of motion with friction be written as Euler-Lagrange equation, and does it have a quantum version?semyon aleskerhttp://mathoverflow.net/users/161832013-10-27T13:17:53Z2014-09-02T08:35:30Z
<p>My (non-expert) impression is that many physically important equations of motion can be obtained as Euler-Lagrange equations. For example in quantum fields theories and in quantum mechanics quantum equations of motion are obtained from the classical ones only if a Lagrangian (or Hamiltonian) is known for the classical case. Is my understanding too oversimplified?</p>
<p><strong>Are there examples of physically important equations which are not Euler-Lagrange for any Lagrangian?</strong></p>
<p>More specifically, let us consider the classical motion of a particle in $\mathbb{R}^3$ with friction:
$$\overset{\cdot\cdot}{\vec x}=-\alpha \overset{\cdot}{\vec x},\, \, \alpha>0,$$
namely acceleration is proportional to velocity with negative coefficient. </p>
<p><strong>Is this equation Euler-Lagrange for an appropriate Lagrangian? Is there a quantum mechanical version of it?</strong></p>
<p><strong>Added later:</strong> As I mentioned in one of the comments below, I do not really know how to make formal what is "quantum mechanical version". As a first guess one could try to write a Schroedinger equation with general (time dependent?) Hamiltonian such that some version of the Ehrenfest theorem would be compatible with the classical equation of motion with friction.</p>
http://mathoverflow.net/q/1308272What structure has been found for functions with this relationship.Willhttp://mathoverflow.net/users/272152013-05-16T12:32:21Z2014-09-02T08:44:05Z
<p>Given $f$ and $g$</p>
<p>$\forall x y. f(x) = f(y) \Longrightarrow f(g(x)) = f(g(y))$</p>
<p>Or equivalently</p>
<p>$ker\ f \subseteq ker\ (f \circ g)$.</p>
<p>Note: if $f$ is injective then this holds for any $g$.</p>
<p>Explanation/motivation: I'm a CompSci who is trying to become more theoretical, and have been playing with recursive functions. I've discovered that a useful property about fixpoints is entailed by the above property, (namely that $\exists h. f(\mu g) = \mu h$,) and I'm wondering:</p>
<ol>
<li>What has been discovered around functions of this shape.</li>
<li>More generally, what area of maths should I be investigating to learn more.</li>
</ol>
http://mathoverflow.net/q/7690614A "dimension" for Tychonoff spacesAdrian Keethttp://mathoverflow.net/users/182052011-10-01T01:59:15Z2014-09-02T05:09:46Z
<p>It's well-known that any Tychonoff space $X$ can be embedded in $[0,1]^k$ for some cardinal $k$. It's natural to ask what the smallest such $k$ is (let's call it $k(X)$). However, this probably probably doesn't deserve to be called dimension, since it fails to satisfy some desirable properties. For instance, although $k(\text{point}) = 0$, we have $k(\text{2 points}) = 1$. This leads me to consider a local version:</p>
<p>If $X$ is Tychonoff and $x \in X$, let $D(X, x)$ be the smallest cardinal $k$ such that some neighbourhood of $x$ can be embedded in $[0,1]^k$, and let $D(X) = \sup_{x \in X} D(X, x)$. This satisfies some obvious properties:</p>
<ul>
<li>If $\{U_\alpha\}$ is an open cover of $X$, then $\dim(X) = \sup D(U_\alpha)$.</li>
<li>If $A$ is a subspace of $X$, then $D(A) \le D(X)$. Equality holds if, for instance, $A$ contains a neighbourhood of a point $x$ with $D(X,x) = D(X)$.</li>
<li>$D(X) = n$ if $X$ is a $n$-dimensional manifold.</li>
<li>$D(X \times Y) \le D(X) + D(Y)$.</li>
</ul>
<p>This last inequality may be strict; for instance, if $X$ is the Cantor set, then $D(X) = 1$ and $X \times X \cong X$.</p>
<p>If $\dim$ denotes the Lebesgue covering dimension, then for $X$ compact, we have $\dim(X) \le \dim([0,1]^{D(X)}) = D(X)$. I have no idea when equality holds (it would if $\dim(X) = n$ implied that $X$ could be <em>locally</em> embedded in $\mathbb{R}^n$, but I don't know if that's true).</p>
<p>Is there a name for this $D$, or has such an invariant been studied before? How is this related to other notions of dimension for a topological space? In particular, are there classes of nice spaces (for instance, compact metrizable) on which they agree?</p>
http://mathoverflow.net/q/7018923Proving that a function's image contains (1/n,...,1/n)Jennifer Gaohttp://mathoverflow.net/users/133632011-07-12T23:35:06Z2014-09-02T07:45:03Z
<p>This question is a follow-up to a previous question answered by Neil Strickland:</p>
<p><a href="http://mathoverflow.net/questions/67318/map-from-simplex-to-itself-that-preserves-sub-simplices">Map from simplex to itself that preserves sub-simplices</a></p>
<p>Let $B$ denote the closed unit ball in $\mathbb{R}^2$ and let $\Delta_{n-1}$ denote the $(n-1)$-simplex. I have a continuous function $f(x_1,\dots,x_n):B^n \rightarrow \Delta_{n-1}$ defined for all subsets $\lbrace x_1,\dots,x_n\rbrace \subset B$ of size $n$ that satisfy $x_i \neq x_j$ for all pairs $i,j$ (in other words, the function is only defined if all of the $n$ arguments are distinct). This function has the property that, if $\sigma$ denotes a permutation, then $f(\sigma(x_1,\dots,x_n)) = \sigma(f(x_1,\dots,x_n))$. In other words, permuting the arguments of the function merely permutes the output. My question is: are there non-trivial sufficient conditions on $f$ under which the point $(1/n , \dots, 1/n)$ lies in the image of this map? (or, even better, is this always the case?)</p>
<p>Here's one property of the map $f$ that I can add regarding the requirement that arguments be distinct: if $\lbrace \mathbf{x}_k \rbrace$ is a sequence of $n$-tuples (with distinct entries) in $B$ that converges to an $n$-tuple $\bar{\mathbf{x}}$ with (possibly) non-distinct entries, then the limit of $f(\mathbf{x}_k)$ exists if and only if, for each pair of entries $x_i^k$ and $x_j^k$ in the $n$-tuple, the unit direction vector from $x_i^k$ to $x_j^k$ (i.e. $\frac{x_i^k - x_j^k}{||x_i^k - x_j^k||}$) has a limit.</p>