Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2015-05-03T17:06:03Zhttp://mathoverflow.net/feedshttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://mathoverflow.net/q/2045900Coefficients of the pull-backs of divisors by resolving morphism.JAPB1971http://mathoverflow.net/users/722842015-05-03T17:00:35Z2015-05-03T17:00:35Z
<p>Let $\varphi : X \dashrightarrow X$ be a rational map. By a theorem of Hironaka we can find a resolution of singularities $(\tilde{X}_\varphi,\pi)$ of $\varphi$, where $\tilde{X}_\varphi$ is a successive blow-up of $X$ and $\pi : \tilde{X}_\varphi \longrightarrow X$ is such that $\varphi \circ \pi : \tilde{X} \longrightarrow X$ extends to a morphism $\tilde{\varphi} : \tilde{X}_\varphi \longrightarrow X$. Suppose that the birrational map $\pi : X_r=\tilde{X}_\varphi \longrightarrow X=X_0$ is obtained as composition of monoidal transformations $\pi_i: X_i \longrightarrow X_{i-1} \hspace{2mm} (i=1 \dots r)$. If we denote by $F_i$ the exceptional divisor of $\pi_i : X_i \longrightarrow X_{i-1}$ and take $\pi_i'=\pi_{i+1} \circ \dots \circ \pi_r: \tilde{X} \longrightarrow X_i \hspace{2mm} (i=1 \dots r)$, the $Pic(\tilde{X}_\varphi)$ is the module
$$Pic(\tilde{X}_\varphi)=Pic(X) \oplus E_1 {\mathbb Z}\oplus \dots \oplus E_r {\mathbb Z},$$
where $E_i = (\pi'_i)^{\sharp} F_i$ be the proper transform of $F_i$ under $\pi'_i $. </p>
<p>For a divisor $D \in Pic(X)$ and
$\tilde{\varphi} D = D' + x_1 E_1 + \dots +x_r E_r , $
How to find the integers $x_i$'s?</p>
http://mathoverflow.net/q/2045882Why we study Geometric invariant theory?riu_sshttp://mathoverflow.net/users/722832015-05-03T15:57:27Z2015-05-03T15:57:27Z
<p>I am trying to learn <strong>Geometric invariant theory</strong> like it was introduced by Mumford. But I do not have a strong motivation and so I want to know the reason of studying Geometric invariant theory. I just know Geometric invariant theory plays an important role in the construction of moduli spaces.</p>
<p>What's a reason for studying Geometric invariant theory? And, what are related (big or small) problems with Geometric invariant theory in algebraic geometry?</p>
<p>Any answer will be very helpful for me. </p>
http://mathoverflow.net/q/204585-4Remainder is always a multiple of 9 [on hold]Emmanuel Garcíahttp://mathoverflow.net/users/642522015-05-03T13:49:12Z2015-05-03T13:58:00Z
<p>I am not a mathematician and certainly number theory is not my forte, but I have found the following pattern inspired on Kaprekar's routine (I am pretty sure this is already well-known although I have not seen anything similar on web, so I'd be really grateful if someone point out a reference for me). Can anyone explain why is that? Probably this is obvious. Thanks in advance.</p>
<p>Take any two-digit number, using at least two different digits. (Leading zeros are allowed.)</p>
<p>Arrange the digits in descending and then in ascending order to get a pair of two-digit numbers, adding leading zeros if necessary.</p>
<p>Subtract the smaller number from the bigger number.</p>
<p>Then, the remainder is always a multiple of 9.</p>
<p>Examples:</p>
<p>56: 65 - 56 = 09;
90: 90 - 09 = 81;
13: 31 - 13 = 18</p>
http://mathoverflow.net/q/2045810Is the following conjecture equivalent to the Second Hardy-Littlewood Conjecture? [on hold]user170039http://mathoverflow.net/users/574322015-05-03T13:18:43Z2015-05-03T13:38:49Z
<blockquote>
<p>Let $y$ be an arbitrary positive real number such that $y\ge 2$. Then if we can prove that, $$\lim_{x\to\infty}\dfrac{\pi(x)+\pi(y)}{\pi(x+y)}=1$$will it imply that for all sufficiently large $x$ and $y$ we will have $\pi(x)+\pi(y)\ge \pi(x+y)$ ?</p>
</blockquote>
<p>This conjecture was given to me by one of my friend who thought that this is not true but couldn't prove it. I also tried to disprove the problem but couldn't. How can one prove or disprove the claim?</p>
http://mathoverflow.net/q/2045802On closest unitary matrixOmid Hatamihttp://mathoverflow.net/users/187852015-05-03T12:54:01Z2015-05-03T12:54:01Z
<p>In this question $||A||_p$ is the normalized $p$-th Schatten norm which is defined to be $\left(\mathbb E_{i} \lambda_i^p\right)^{1/p}$, where $\lambda_i$ are singular values of matrix $A$.</p>
<p>Suppose that $A, B\in M_n(\mathbb C)$ are matrices with operator norm at most 1. Suppose that $||AB-I||_p < \varepsilon$. Can you prove that there is a unitary matrix $U$ such that $||A-U||_p < \varepsilon$?</p>
<p>I already know proof for this when $p=1$ or $2$ or $\infty$, but not for any other $p$.</p>
http://mathoverflow.net/q/2045770Uniqueness of a smooth function [on hold]Studenthttp://mathoverflow.net/users/722762015-05-03T11:51:16Z2015-05-03T15:13:04Z
<p>Let $x,y:[a,b]\to\mathbb{R},\ a<b, a,b\in\mathbb{R}$ be two smooth functions ($x,y\in C^{\infty}([a,b])$). How can I prove that there is a unique function $\theta:[a,b]\to\mathbb{R},\ \theta\in C^{\infty}([a,b])$ such that:</p>
<p>$$\begin{cases} x(t)\sin\theta(t)=y(t)\cos\theta(t),\ \forall\ t\in [a,b].\\ (x(t_0),y(t_0))=(x_0, y_0)\neq (0,0)\ \text{is given}\\ \theta(t_0)=\theta_0 \ \text{is given too such that}\ x(t_0)\sin\theta(t_0)=y(t_0)\cos\theta(t_0) . \end{cases}$$</p>
http://mathoverflow.net/q/204576-2Fermat's Last Theorem [on hold]mathxyzhttp://mathoverflow.net/users/722742015-05-03T11:19:50Z2015-05-03T11:37:40Z
<p>If there exist numbers $x,y,z\in{\mathbb N}$, such that $x^p+y^p= z^p $, $p$ is odd prime number. Prove or disprove that $x$ or $y$ or $z$ is prime.</p>
http://mathoverflow.net/q/2045750Coproduct Slice category [on hold]Silviohttp://mathoverflow.net/users/703772015-05-03T11:13:55Z2015-05-03T12:43:33Z
<p>In category theory, how can I prove that, if the category $C$ has co-product, also the slice category $C / I$ admit it?</p>
<p>Thanks a lot!</p>
http://mathoverflow.net/q/2045706Positive roots of a polynomialdimahttp://mathoverflow.net/users/238622015-05-03T08:23:09Z2015-05-03T09:00:45Z
<p>Let $a_i>0$, $i=1,\dots,n$, and put $\overline{a}:=\frac{1}{n}\sum_{i=1}^n a_i$. Assuming not all $a_i$'s are equal, take
$$
p(x):=\sum_{i=1}^n a_i (a_i-\overline{a})\prod_{k=1,\dots,n\;k\neq i} (x+a_k)^2.
$$
How to prove that $p(x)$ has exactly one positive root? (this is a conjecture, based on numerical experiments)</p>
http://mathoverflow.net/q/2045691Random graphs with boundary in a game (Tsuro)Stijnhttp://mathoverflow.net/users/494382015-05-03T08:00:31Z2015-05-03T12:39:13Z
<p>Suppose we have an $n \times n$ board and we have $n^2 - 1$ square tiles. These tiles consist of a 8 vertices, two on each edge, and every vertex is connected to precisely one other vertex. These tiles connect together to make paths.</p>
<p>At the start of the game there are 8 pieces belonging to 8 players which begin at random points on the edge of the board and players take turns placing tiles in any orientation they desire. The pieces then move along the newly created paths to the end and players must place tiles so that their piece moves. Pieces are removed if they collide with one another or move off the edge of the board. ( <a href="http://en.wikipedia.org/wiki/Tsuro" rel="nofollow">http://en.wikipedia.org/wiki/Tsuro</a> )</p>
<p>What is the probability that, with a randomly generated tileset, all pieces can survive 'til the end. I.e. that they're all clustered around the empty square? All players can see the entire tileset so they can play "perfectly". The original game is for a specific tile set and a $6 \times 6$ grid but I'm more interested in the general game.</p>
http://mathoverflow.net/q/2045673What kind of role has Functional Analysis played in Signal Processing? [on hold]Wfpiggiehttp://mathoverflow.net/users/26922015-05-03T07:03:05Z2015-05-03T11:01:40Z
<p>Does it serve mainly as a narration or is there any substantive consequence which might not be derived without tools of functional analysis? </p>
http://mathoverflow.net/q/204564-4Sequences and reflexivity [on hold]user2015http://mathoverflow.net/users/665582015-05-03T05:02:11Z2015-05-03T09:29:16Z
<p>Assume $X$ to be a real reflexive Banach space.
Why are sequential topological notions
topological notions ?
For ex :
sequentially closed set is closed,<br>
sequentially compact set is compact,
and so on. </p>
<p>Thanks.</p>
http://mathoverflow.net/q/2045394A question about simple closed curves in finite dimensional Euclidean spacesGarabed Gulbenkianhttp://mathoverflow.net/users/44232015-05-02T18:50:56Z2015-05-03T09:28:48Z
<p>Let n be a positive integer not less than 2. Does anyone know of a theorem stating that- for each n- there exists a simple closed curve c(n), which (1) is a subset of n-dimensional Euclidean space E(n) and (2) does not contain n+1 pairwise distinct points all belonging to the same (n-1)-dimensional Hyperplane of E(n)? For n=2 there is such a theorem. This theorem is almost "trivial". We have merely to picture the circumference of a circle. I am asking whether, and if so, to what extent this "trivial" theorem has been generalized. What is the situation when n is greater than 2?</p>
http://mathoverflow.net/q/2045360Redundancy of the Cantor Enumeration of the RationalsManfred Weishttp://mathoverflow.net/users/313102015-05-02T18:36:08Z2015-05-03T08:52:07Z
<p>What is the cardinality of the set of values corresponding to the first $n$ rationals generated in Cantor's enumeration scheme for proofing of their countability? </p>
<p>Edit:
following the suggestion of Todd, here I integrate the clarification from the comments here:
By "Cantor's enumeration scheme" I mean the one, that Cantor originally used (I am however not aware, which that was, so that would also be part of the question).<br>
In case of doubt, sort the fractions according to numerator+denominator or, in case of equal sum, accoding to numerator/denominator in ascending order in both cases.</p>
<p>As the rationals, whose value is encountered for the first time, are those for which numerator and denominator are relatively prime and because the probability that two natural numbers are relatively prime, is $\frac{6}{\pi^2}$, it would not be suprising if the number of different values encountered in the first $n$ fractions of the Cantor sequence would approach $n\frac{6}{\pi^2}$, so the interesting question is, whether the exact value of the factor for a given $n$ can be calculated or at least the asymptotic growth.</p>
http://mathoverflow.net/q/2045320Joint law of a standard Brownian motion and its local time at a nonzero levelAnandhttp://mathoverflow.net/users/368142015-05-02T17:25:10Z2015-05-03T16:37:42Z
<p>Let $B_t$ be the standard Brownian motion and $L_t^a$ be the local time at level $a$. It is known that the joint-density of $(L_t^0,B_t)$ is </p>
<p>$$
P\left(B_t\in d y, L_t^0\in d v\right) = \frac{|y|+v}{\sqrt{2\pi t^3}}\exp\left(-\frac{(|y|+v)^2}{2 t}\right) 1_{[0,\infty)}(v)1_{(-\infty,\infty)}(y) d y d v,
$$</p>
<p>see, e.g., p. 181 of <a href="http://rads.stackoverflow.com/amzn/click/0817633863" rel="nofollow">Chung & Williams</a>. </p>
<p>The problem is whether anyone knows the joint density of $(L_t^a,B_t)$ for $a\ne 0$?</p>
<p>Thank you very much for any hints.</p>
http://mathoverflow.net/q/2045221A table for irreducible integral representation of finite cyclic groupsuser2015http://mathoverflow.net/users/671402015-05-02T15:42:06Z2015-05-03T16:36:52Z
<p>Is there such a table where the irreducible integral representations of finite cyclic groups
are listed?</p>
<p>Edited:</p>
<p><em>Thanks for Todd Leason's comment.Acutally,i want to know all inequivalent indecomposable integral representations of cyclic groups. I think this is related to determining conjugacy classes of <strong>finite cyclic subgroups of $GL(n,Z)$</strong>.
I want to know such subgroups of $GL(n,Z)$ explicitely.Is there such a list?</em></p>
<p>For example,cyclic subgroups of $GL(3,Z)$ are determined explicitely in:</p>
<p><a href="http://projecteuclid.org/euclid.nmj/1118798212" rel="nofollow">ON THE FINITE SUBGROUPS OF GL (3, Z)</a></p>
http://mathoverflow.net/q/2045157Define "Mathematics Colloquium"?Luke Oedinghttp://mathoverflow.net/users/173862015-05-02T14:25:43Z2015-05-03T12:35:01Z
<p>I'm now a member of my department's colloquium committee. Our task is to make a great colloquium series. I thought that the first step would be to come up with an appropriate definition of "Mathematics Colloquium." Standard dictionaries are not too much help:</p>
<pre><code>Colloquium: n.
1. A usually academic meeting at which specialists deliver
addresses on a topic or related topics and then answer questions relating to them.
2. A conference at which scholars or other experts present papers on, analyze,
and discuss a specific topic.
</code></pre>
<p>So here's my question: What is the ideal definition of a mathematics colloquium? Of course I know a bunch of things that make for <strong>bad</strong> colloquium talks, so I'm interested in knowing the definition of a good or great mathematics colloquium. </p>
http://mathoverflow.net/q/20446431Do unit quaternions at vertices of a regular 4-simplex, one being 1, generate a free group?John Baezhttp://mathoverflow.net/users/28932015-05-01T17:06:45Z2015-05-03T14:56:44Z
<p>Choose unit quaternions $q_0, q_1, q_2, q_3, q_4$ that form the vertices of a regular 4-simplex in the quaternions. Assume $q_0 = 1$. Let the other four generate a group via quaternion multiplication. Is this a free group on 4 generators?</p>
<p>I heard from Adrian Ocneanu that the answer is <em>yes</em>, but I don't know a proof.</p>
<p>Here's why I care. As shown in this image by Greg Egan, you can inscribe a cube in a regular dodecahedron:</p>
<p><img src="http://i.stack.imgur.com/wqK0t.gif" alt="twin dodecahedra"></p>
<p>If you rotate the cube 90 degrees about an axis of 4-fold symmetry, the dodecahedron will be mapped to a <em>different</em> dodecahedron. Ocneanu calls this a <strong>twin</strong> of the original dodecahedron. For example, the red dodecahedron above has the blue one as a twin, and vice versa. Despite the name, a regular dodecahedron actually has 5 different twins, one for each cube that can be inscribed in it. </p>
<p>You can create a graph as follows. Start with a node for our original dodecahedron. Draw nodes for all the dodecahedra you can get from this one by repeatedly taking twins. Connect two nodes with an edge if and only if they are twins of each other.</p>
<p>Ocneanu claims the resulting graph is a tree! In other words, if you start at your original dodecahedron, and keep walking along edges of this graph by taking twins, youâ€™ll never get back to where you started except by undoing all your steps.</p>
<p>Ocneanu didn't tell me the proof, but he said the key to the proof was this: </p>
<p><strong>Claim</strong>: if you take unit quaternions at the vertices of a regular 4-simplex, one of them equal to 1, the remaining four are generators of a free group. </p>
<p>Indeed, Egan and I were able to use this claim to prove that the graph is a tree:</p>
<ul>
<li>John Baez, <a href="http://blogs.ams.org/visualinsight/2015/05/01/twin-dodecahedra/">Twin dodecahedra</a>, <em>Visual Insight</em>, 1 May 2015.</li>
</ul>
<p>So now I want to know why Ocneanu's claim is true --- and indeed, I want to know <em>that</em> it is true.</p>
<p>If it helps, you can assume the regular 4-simplex has these vertices:</p>
<p>$$ q_0 = 1 $$</p>
<p>$$q_1 = -\frac{1}{4} + \frac{\sqrt{5}}{4} i + \frac{\sqrt{5}}{4} j + \frac{\sqrt{5}}{4} k $$</p>
<p>$$ q_2 = -\frac{1}{4} + \frac{\sqrt{5}}{4} i -\frac{\sqrt{5}}{4} j -\frac{\sqrt{5}}{4} k $$</p>
<p>$$ q_3 = -\frac{1}{4} -\frac{\sqrt{5}}{4} i + \frac{\sqrt{5}}{4} j -\frac{\sqrt{5}}{4} k $$</p>
<p>$$ q_4 = -\frac{1}{4} -\frac{\sqrt{5}}{4} i -\frac{\sqrt{5}}{4} j +\frac{\sqrt{5}}{4} k $$</p>
http://mathoverflow.net/q/20446024How to prove this determinant is positive?user23765http://mathoverflow.net/users/712252015-05-01T16:45:03Z2015-05-03T14:05:16Z
<p>Given the matrices $ A_i=
\biggl(\begin{matrix}
0 & B_i \\
B_i^T & 0
\end{matrix} \biggr)
$, where $B_i$ are real matrices and $i=1,2,\ldots,N$, how to prove that $\det(I + e^{A_1}e^{A_2}\ldots e^{A_N}) \ge 0$ ? This seems to be true numerically. </p>
<p>Update: As was shown in below, the above inequality is related to another conjecture $\det(1+e^M)\ge 0$, given a $2n\times 2n$ real matrix $M$ that fulfills $\eta M \eta =-M^T$ and $\eta=diag(1_n, -1_n)$. The answers of Christian and Will, although inspiring, did not really disprove this conjecture as I understood. </p>
http://mathoverflow.net/q/2042701interchanging limits for doubly indexed random sequencesr_faszanatashttp://mathoverflow.net/users/569312015-04-29T13:40:01Z2015-05-03T09:57:06Z
<p>I've encountered the following problem which seems to be quite standard but for which I can't find any proper references (asking on mathematics SE didn't bring up any answers so I'm reposting my question here): </p>
<p>Suppose that for $n \in \mathbb{N}$ we have a sum of random variables $S(n)= \sum_{r=1}^{\infty} X_r(n)$, such that $X_r(n)$ and $X_s(n)$ are uncorrelated for $r \neq s$ and $S(n)$ is uniformly bounded in $L^2$. Denote by $S_k(n)$ the partial sums $\sum_{r=1}^k X_r(n)$. If we know that</p>
<ol>
<li>for each $k$,
$$S_k(n) \xrightarrow{d} N(\mu_k,\sigma^2_k), \qquad (n \to \infty),$$
where $\xrightarrow{d}$ is convergence is distribution and $N(\mu_k,\sigma^2_k)$ denotes a Gaussian with mean $\mu_k$ and variance $\sigma^2_k$</li>
</ol>
<p>and</p>
<ol start="2">
<li>the sequences $(\mu_k)$ and $(\sigma^2_k)$ both converge to some limits $\mu$ and $\sigma^2$ (so that in particular $N(\mu_k,\sigma^2_k) \xrightarrow{d} N(\mu,\sigma^2)$,</li>
</ol>
<p>what are sufficient conditions to conclude that $S(n) \xrightarrow{d} N(\mu,\sigma^2)$ as $n\to \infty$?
In other words, with some abuse of notation, i.e. writing $\lim$ for "limit in distribution", I'm looking for conditions which imply that
$$\lim_{n \to \infty} \lim_{k \to \infty} S_k(n) = \lim_{k \to \infty} \lim_{n \to \infty} S_k(n).$$
What about the more general situation where the partial sums are replaced by some doubly indexed random variables and the limiting Gaussians with some generic (continuous) random variable? </p>
<p>Thanks for your help!</p>
http://mathoverflow.net/q/2024662A Jordan Separation Theorem for Polyhedral SurfacesTranscendentalhttp://mathoverflow.net/users/506142015-04-09T17:53:34Z2015-05-03T07:03:59Z
<p>Let me begin by defining what a polyhedral surface is.</p>
<p>A <strong>path-connected</strong> subset $ P $ of $ \mathbb{R}^{3} $ is called a <em>polyhedral surface</em> iff it is the union of a <strong>finite</strong> collection $ \mathcal{C} $ of polygons (possibly non-convex) that satisfies the following three conditions:</p>
<ul>
<li>The intersection of any pair of distinct polygons in $ \mathcal{C} $ is exactly one of three things: (i) a common edge, (ii) a common vertex or (iii) $ \varnothing $. (This implies that the interiors of the polygons in $ \mathcal{C} $ are disjoint.)</li>
<li>Any edge of any polygon in $ \mathcal{C} $ is an edge of exactly one other polygon in $ \mathcal{C} $.</li>
<li>If an edge of a polygon in $ \mathcal{C} $ intersects an edge of another polygon in $ \mathcal{C} $ in a common vertex, then the two edges are also edges of a third polygon in $ \mathcal{C} $.</li>
</ul>
<p>I believe this to be a ubiquitously understood precise statement of what a polyhedral surface should be.</p>
<blockquote>
<p><strong>Question.</strong> Let $ P $ be a polyhedral surface. Is there an analogue of the Jordan Separation Theorem that states that $ P $ is the boundary of two path-connected open subsets of $ \mathbb{R}^{3} $, one of which is bounded and the other unbounded?</p>
</blockquote>
<p>The definition given above allows for a polyhedral surface that is homeomorphic to $ \mathbb{S}^{1} \times \mathbb{S}^{1} $, the boundary of a $ 2 $-torus, which is why I am imposing only path-connectedness and not simple-connectedness also.</p>
<p>I suspect that the way to proceed is to first prove that a polyhedral surface is indeed a topological surface (i.e., a topological $ 2 $-manifold) that is closed (i.e., compact and without boundary). Then one can apply the Classification Theorem for Surfaces to prove that it must be homeomorphic to either a $ 2 $-sphere or a finite connected sum of $ 2 $-tori. However, I feel that I may be missing something out.</p>
<p>Thank you very much for your help!</p>
http://mathoverflow.net/q/2018153A priori estimates for a nonlinear elliptic problem singular on the boundaryuserhttp://mathoverflow.net/users/700592015-04-02T16:55:57Z2015-05-03T14:08:20Z
<p>Let us consider the following elliptic problem
$$
\begin{cases}
-\Delta u = \frac{u^p}{|x|^2} \mbox{ in } \Omega \\
u >0 \mbox{ in } \Omega \\
u = 0 \mbox{ on } \partial \Omega.
\end{cases}
$$
with $N \geq 3$, $\Omega$ a bounded domain in $\mathbb{R}^N$ with smooth boundary,$1 < p < \frac{N+2}{N-2}$ and $0 \in \partial \Omega$.</p>
<p>We say that $u$ is an energy solution of the preceding problem if $u \in H^1_0(\Omega)$, $u >0$ such that
$$
\int_\Omega \nabla u \nabla \varphi \, dx = \int_\Omega \frac{u^p}{|x|^2} \varphi \, dx \quad \forall \varphi \in \mathcal{C}^\infty_0(\Omega).
$$</p>
<p>The authors proved the following Lemma.</p>
<p>If $u$ is an energy solution of the preceding problem then $u \in L^\infty(\Omega)$.</p>
<p>The proof, they said, is just a consequence of the a priori estimates of Gidas and Spruck, "A priori bounds for positive solutions of nonlinear elliptic equations", Comm. Partial Differential Equations 6 (1981), after suitable scaling.
Moreover they gave this hint. Suppose $x_0 \in \Omega$, $x\neq 0$. Let $r:= \frac{|x_0|}{2}$ and define the function
$$
v(y) := u(x_0 + ry) \quad \mbox{in } \frac{\Omega - x_0}{r}
$$
Then $v$ satisfies the equation
$$
- \Delta v = \frac{r^2}{|x_0 + ry|^2}v^p \quad \mbox{in } \frac{\Omega - x_0}{r}
$$
Now restrict the equation to $\frac{\Omega - x_0}{r} \cap B_1(0)$. In this region the weight $\frac{r^2}{|x_0 + ry|^2}$ is bounded and smooth. Now using the a priori estimates of Gidas and Spruck we deduce that $v(0) \leq C$ for some universal $C$.
Finally since $u(x_0) = v(0)$ we have the thesis, since $C$ is not depending on $x_0$.</p>
<p>Howewer I don't understand how the result of Gidas and Spruck could be useful in this setting. Indeed although in the region $\frac{\Omega - x_0}{r} \cap B_1(0)$ the weight is bounded and smooth, the boundary conditions of the problem for the function $v$ are "free" ( at least in some part of the boundary of that region). So, applying the Gidas and Spruck results we have that there exists a constant, but I don't get how can this constant be universal, since it seems, to me, depending on the boundary conditions and hence on the point $x_0$.</p>
http://mathoverflow.net/q/1990151Weak topology on a pre-Hilbert Spaceerzhttp://mathoverflow.net/users/531552015-03-04T06:45:24Z2015-05-03T12:07:50Z
<p>Since there was essentially no answers on my <a href="http://mathoverflow.net/questions/197887/recontruction-of-the-weak-topolgy-from-the-scalar-product-on-a-subset-of-a-hilbe?noredirect=1#comment491399_197887">previous</a> question, I will ask a partial case of it, which is very easy to state.</p>
<p>Let $\left(X,\left<\cdot,\cdot\right>\right)$ be an inner product (pre-Hilbert) space. Is it possible to describe the weak topology on $X$ explicetely in terms of $\left<\cdot,\cdot\right>$ as a function on $X\times X$? That means without mentioning such <strong>not-enough-constructive</strong> objects as "completion" and "the dual".</p>
<p>Thank you.</p>
http://mathoverflow.net/q/1921270What is the meaning of non-Hausdorff spaces in algebraic geometry [closed]truebaranhttp://mathoverflow.net/users/240782015-01-04T18:00:59Z2015-05-03T16:21:14Z
<p>At the beginning I should warn everybody reading this post: I don't know much about algebraic geometry so specialists in this subject may see my question as ignorant.</p>
<hr>
<p>As far I understood one on the main themes in algebraic geometry is to pursue as far as it is possible the duality between geometric objects and algebras: most basic result is the Hilbert Nullstellensatz but the theory goes much further-to the definition of general <em>schemes</em> due to Grothendieck. The notion of <em>space</em> has evolved through the history of mathematics but as far as some topological space was around, the minimal requirement (at least for me) was that the space should be Hausdorff. This is quite natural due to the following characterisation: each net has at most one limit. Moreover, when one is interested in compact or locally compact spaces, the assumpion of being Hausdorff automatically implies better behaviour (normality or complete regularity resp.).</p>
<p>Finally, there is the theory (which is close to my heart) of $C^*$-algebras: in this theory a fundamental result is the Gelfand-Najmark theorem which establishes the duality between compact Hausdorff spaces and commutative unital $C^*$-algebras. This is another <em>algebra-geometry</em> duality and allows one to think of the theory of general $C^*$-algebras as <em>noncommutative topology</em>: but there are plenty of situations when one has a "pathological" topological space (with some non Hausdorff topology) which is hard to deal with. Then one switches to the realm of algebras and tries to say something about this space using the associated algebra: in other words one doesn't stick to a geometric picture.</p>
<p>It seems that algebraic geometry goes the other way around and works very often with topological spaces which are non-Hausdorff. So my (rather vague) question is the following:</p>
<p><strong>Question.</strong> What is the <em>geometric</em> meaning and the intuition behind non-Hausdorff spaces in the realm of algebraic geometry? How to interpret such non Hausdorff topologies in this algebra-geometric context?</p>
<p>Let me give one example, which may clarify about what sort of things I'm asking: when one forms a quotient space one glues some points of the space to the another and in such a way one obtains a new set of points. In particular one can take some subset $A \subset X$ which is not closed and collapse it to the one point: then $X/A$ would be non Hausdorff and the special point in the quotient will be $\pi(a)$ where $a \in A$ is arbitrary and $\pi$ denotes the natural projection. My intuition behind this example is the following: the point $\pi(a) \in X/A$ was obtained from the richer set of data which was the set $A$ and the fact that $A$ was not closed. A more dramatic example would be $X/G$ where for each $g \in G$ its orbit is dense in $X$: then my intuition behind this example is the fact that the points in $X/G$ have some extra internal structure. So the operation of taking quotients very often gives a non-Hausdorff topology.</p>
http://mathoverflow.net/q/1860830Maximum of a mollified/convolution functionMolehttp://mathoverflow.net/users/612732014-11-03T10:24:37Z2015-05-03T13:08:20Z
<p>I have a function $f:{\mathbb R}\rightarrow {\mathbb R}_+$ which has a unique maximum at $x=0$. $f$ can be symmetric or asymmetric. I am interested on the mollified-f function</p>
<p>$$\tilde{f}(x)=\int_{-\epsilon}^{\epsilon}\varphi(y)f(x-y)dy,\,\,\, \epsilon>0$$</p>
<p>I wonder if there are general results on the <a href="http://en.wikipedia.org/wiki/Mollifier" rel="nofollow">Mollifier</a> $\varphi$, such that $\tilde{f}$ also has a unique maximum (not necessarily at $0$).</p>
http://mathoverflow.net/q/1229831Calculate channel capacity of general channel under constraintuser31757http://mathoverflow.net/users/317572013-02-26T14:35:35Z2015-05-03T07:07:50Z
<p>Hi!</p>
<p>Given a conditional distribution $P_{Y|X}$ I'd like to find the prior distribution $P_X$ that maximizes the mutual information $I(X;Y)$ with $P_Y(y)=\int P_{Y|X}(y|x)P_X(x)\text{dx}$ (this corresponds to finding the channel capacity $C(X;Y):=\max_{P_X}I(X;Y)$) subject to the constraint $E[-\log(X)]=a$.</p>
<p>In my particular case, $P_{Y|X}$ is a Bernoulli distribution with $X$ as its parameter and I'm looking for a distribution over the parameter.</p>
<p>My intuition would tell me this should be some Beta-distribution (something like $\text{Beta}(\frac{1}{a},1)$?!) in my particular case, but I don't know how to approach such a problem, much less in the general case.</p>
<p>Could anyone point me in the right direction?</p>
http://mathoverflow.net/q/102806-1Morphism in derived categoryMessihttp://mathoverflow.net/users/252572012-07-21T12:07:16Z2015-05-03T07:09:21Z
<p>We know that the morphisms between objects of derived category are roofs. But how to understand them,and how to compute them. For example, we consider the derived category $D(X)$ of a projective variety $X$, then $\Hom(O_X, E^.)=?$ for a complex $E^.$ and why $\Hom(A, B[1])=\Ext^1(A, B)$ for sheaves $A$ and $B$. Can we understand them only using the roofs.</p>
http://mathoverflow.net/q/892438Gluing Markov processesShawnDhttp://mathoverflow.net/users/172192012-02-23T00:39:23Z2015-05-03T11:07:50Z
<p>I am looking for a reference on the gluing together of strong Markov processes to get a new one.</p>
<p>Here is an example of what I have in mind. Let $B^1, B^2, \ldots $ be independent one-dimensional Brownian motions started at $0$ and $\tau_i$ be the first time $B^i$ hits $1$. Construct a new process $Y$ as follows. Let $Y(t)=B^1(t)$ for $t<\tau_1$, $Y(t)=1-B^2(t-\tau_1)$ for $\tau_1 \leq t<\tau_1+\tau_2$, $Y(t)=B^3(t-\tau_1-\tau_2)$ for $\tau_1+\tau_2\leq t <\tau_1+\tau_2+\tau_3$ and so on. I would like to conclude that the this construction gives a strong Markov process (in fact its a Brownian motion).</p>
<p>More generally, suppose I have a suitably nice topological space $E$ and open sets $U_1, \ldots, U_n$ such that $E=\cup U_i$. Suppose for each $\overline{U_i}$ I have a strong Markov process $X_i$ in $\overline{U_i}$ killed at $\overline{U_i}-U_i$ and that started at $x\in U_i \cap U_j$, $X_i$ and $X_j$ are the same up until the first time they leave $U_i \cap U_j$. Is it possible to "glue" these $X_i$'s together (analogous to the example above) to get a strong Markov process on all of $E$?</p>
<p>I think I can make sense of my example by looking at the infinitesimal generator for the process and observing that it is "nice" and therefore is the generator of a strong Markov process. Perhaps I can use the same sort of idea in general if the $X_i$ are "nice" (that is if the $X_i$ have "nice" generators operating on $C_0(U_i)$, I can use them to construct a nice generator on $C_0(E)$ and then use general theory to say there is a process with this generator), but I'm wondering if there is anything in the literature on this sort of construction.</p>
<p>EDIT: I slightly changed my example to try to emphasize that I don't want to a priori assume that I know anything about the process I get in the end.</p>
http://mathoverflow.net/q/7663110Vanishing of $\hat{A}$ genus and positive scalar curvatureshuhttp://mathoverflow.net/users/163262011-09-28T11:43:30Z2015-05-03T13:30:06Z
<p>Classicly, for a spin Riemannian manifold $M$, the $\hat{A}(M)$ genus will be $0$, if the scalar curvature is positive. </p>
<p>The proof is to use the Lichnerowicz formula. we have the index of the Dirac operator will be $0$, i.e.,
$$Ind(D_+)=0.$$
On the other hand, by the index theorem of Atiyah and Singer, we have
$$Ind(D_+)=\hat{A}(M).$$
So we get
$$\hat{A}(M)=0.$$</p>
<p>My question is "Can we have a different method to proof this result? without using the Lichnerowicz formula or without using the index theorem" Maybe a formal proof or explanation.</p>
http://mathoverflow.net/q/7602Does there exist a sequence of groups whose representation theory is described by plane partitions?Qiaochu Yuanhttp://mathoverflow.net/users/2902009-10-16T16:39:57Z2015-05-03T12:45:58Z
<p>More precisely, does there exist a sequence G<sub>1</sub> < G<sub>2</sub> < ... of finite groups such that the irreducible representations of G<sub>n</sub> are parameterized by the plane partitions of total size n?</p>