Recent Questions - MathOverflowmost recent 30 from mathoverflow.net2015-10-05T01:55:41Zhttp://mathoverflow.net/feedshttp://www.creativecommons.org/licenses/by-sa/3.0/rdfhttp://mathoverflow.net/q/2200040$d\bar d$-lemma on pair $(X,D)$Alonhttp://mathoverflow.net/users/563532015-10-05T00:56:16Z2015-10-05T00:56:16Z
<p>Let $X$ be a Kahler manifold with a simple normal crossing divisor $D$, i.e., pair $(X,D)$. Let $\omega$ and $\omega'$ be two Kahler forms in the same Kahler class then have we $d\bar d$-lemma on pair $(X,D)$?</p>
http://mathoverflow.net/q/2199991Did Erdős prove there are two primes $4a+1, 4b+3$ between between $n$ and $2n$?user236182http://mathoverflow.net/users/810522015-10-05T00:30:27Z2015-10-05T00:39:36Z
<p><a href="http://mathworld.wolfram.com/ChoquetTheory.html" rel="nofollow">http://mathworld.wolfram.com/ChoquetTheory.html</a></p>
<p>Is the claim in the link true? Here's the reference given there:</p>
<p><a href="https://www.renyi.hu/~p_erdos/1934-01.pdf" rel="nofollow">https://www.renyi.hu/~p_erdos/1934-01.pdf</a></p>
<blockquote>
<p>Erdős proved that there exist at least one prime $\equiv 1\pmod{4}$ and at least one prime $\equiv 3\pmod{4}$ between $n$ and $2n$ for all $n>6$.</p>
<p>References:
<em>Erdős, P. "A Theorem of Sylvester and Schur." J. London Math. Soc. 9, 282-288, 1934.</em></p>
</blockquote>
<p>Also: <a href="https://en.wikipedia.org/wiki/Talk:Bertrand%27s_postulate#Dubious_statement" rel="nofollow">https://en.wikipedia.org/wiki/Talk:Bertrand%27s_postulate#Dubious_statement</a></p>
<p>My question:</p>
<blockquote>
<p>Can you find the proof of this in the given pdf (maybe somewhere else)?</p>
</blockquote>
http://mathoverflow.net/q/2199980Showing a permutation module is reducibleHunterhttp://mathoverflow.net/users/810662015-10-05T00:00:40Z2015-10-05T00:00:40Z
<p>For a permutation module $V$, which is a permutation module if it has the basis $B = \{v_1,...,v_n\}$ such that the matrix of every $g \in G$ with respect to this basis is a permutation matrix. I need to show that a permutation module is always reducible.</p>
<p>I'm much more interested in how you approach solving this problem than what the answer is. </p>
<p>Is a permutation matrix a matrix which maps each basis vector to another basis vector, like a bijection? How would I better describe this property?</p>
<p>Thanks for any help you can give.</p>
http://mathoverflow.net/q/2199960A question on the integrability of eigenfunctions of the Laplacianuser38600http://mathoverflow.net/users/386002015-10-04T23:29:45Z2015-10-04T23:35:38Z
<p>Let $(M,g)$ be a closed Riemannian manifold. Let $\lambda$ and $u$ be (the $k$-th) eigenvalue and eigenfunction,
$$\Delta u=-\lambda u.$$</p>
<p>I was wondering under what condition (for example, spaces forms? Einstein manifolds? Kahler manifolds? with curvature constraints?) do we have
$$\int_M\frac{1}{u^2}dv<\infty,\quad\mathrm{or}\quad\int_M\frac{1}{u^{2k}}dv<\infty？$$
where $k$ is some positive real number.</p>
<p>I searched using google but didn't get any result. Thank you very much.</p>
http://mathoverflow.net/q/219995-4Problem on Shell Method [on hold]Henryhttp://mathoverflow.net/users/810652015-10-04T23:27:31Z2015-10-04T23:27:31Z
<p>I've been on this problem for awhile and have no idea how to figure it out. Some help would be greatly appreciated. </p>
<p>Thank you! <a href="http://i.stack.imgur.com/I4dP7.png" rel="nofollow">http://i.stack.imgur.com/I4dP7.png</a></p>
http://mathoverflow.net/q/2199941Veronese embeddings and locally free resolutionsKenny Wonghttp://mathoverflow.net/users/803372015-10-04T23:19:31Z2015-10-04T23:54:17Z
<p>Let $i : \mathbf P^1 \to \mathbf P^2$ be the second Veronese embedding. Clearly, $i_\star \mathcal O_{\mathbf P^1}$ has a locally free resolution of the form</p>
<p>$$
0 \to \mathcal O_{\mathbf P^2} (-2) \to \mathcal O_{\mathbf P^2} \to i_\star \mathcal O_{\mathbf P^1} \to 0
$$</p>
<p>More generally, for any integer $n$, we have $\mathcal O_{\mathbf P^1} (2n) \cong \mathcal O_{ \mathbf P^1} \otimes i^\star \mathcal O_{\mathbf P^2} (n)$, so $i_\star \mathcal O_{\mathbf P^1} (2n) $ has a resolution</p>
<p>$$
0 \to \mathcal O_{\mathbf P^2} (n-2) \to \mathcal O_{\mathbf P^2} (n) \to i_\star \mathcal O_{\mathbf P^1}(2n) \to 0
$$</p>
<p>My question is: how can I construct a locally free resolution for $i_\star O_{\mathbf P^1} (2n+1)$?</p>
<p>(I should add that it would make my life easier if the sheaves in this resolution are direct sums of invertible sheaves on $\mathbf P^2$.)</p>
http://mathoverflow.net/q/2199913Plurisubharmonic functions on Kähler manifolds, intuition?user81061http://mathoverflow.net/users/810612015-10-04T20:57:15Z2015-10-04T20:57:15Z
<p>As the question suggests, what is the intuition for working with plurisubharmonic functions on Kähler manifolds?</p>
http://mathoverflow.net/q/219988-4Lexicographic rank of an odd/even permutation [on hold]lassaendiehttp://mathoverflow.net/users/802052015-10-04T19:35:28Z2015-10-04T20:47:27Z
<p>Let $S_n$ be the set of all permutations of integers from $1$ to $n$. Let $P_1$ and $P_2$ be two partitions defined on $S_n$ as follows.</p>
<p>$P_1$ is the set of all those permutations which have even length cycles. $P_2$ is the set of all those permutations which have odd length cycles.
The sets $P_1$ and $P_2$ have a lexicographic ordering imposed on them.</p>
<p>Given a permutation $P$ </p>
<ol>
<li>Tell which partition it belongs to </li>
<li>What is the lexicographic rank($1$ based indexing) of this permutation in that partition. If this rank is $K$, you have to specify the number $K$ mod $M$ where $M = 1000000007 (1e9 + 7)$</li>
</ol>
<p>Here $N$ can be as large as $10^5$</p>
<p>My approach: Calculate the number of cycles in P using a simple DFS technique on my computer and decide which partition it belongs to. For calculating rank I first calculate its lexicographic rank in the set $S_n$. There is already an algorithm available for that. Let that rank be $K_1$($1$ based indexing) then the actual rank is coming out to be floor($\frac{K_1+1}{2}$). (I may also be wrong with this result.) I am having trouble applying modular arithmetic on this expression. Is there any other way to find the rank in context of $P_1$ or $P_2$ which does not involve taking floors ?</p>
<p>What if $P_1$ and $P_2$ are partitions containing odd and even permutations respectively. Will the answer change?</p>
<p>Note: By "cycle" I mean when the permutation is represented as a graph, the number of components or disjoint forests formed. </p>
http://mathoverflow.net/q/2199871Bertini's theorem in positive characteristic for smooth morphismsKalihttp://mathoverflow.net/users/543692015-10-04T19:31:18Z2015-10-04T19:44:29Z
<p>Let $f:X \to Y$ be a morphism of finite type of irreducible schemes over an algebraically closed field of characteristic $0$. Assume that $Y$ is non-singular. Let $x \in X$ be a closed point and $T_xf:T_xX \to T_{f(x)}Y$ the induced linear map. Then, Proposition III.$10.6$ of Hartshorne's "Algebraic geometry" implies that <em>for a general $x \in X$</em>, $\dim \mathrm{Im}(f) \le \mathrm{rk}(T_xf)$ (with notations same as in Hartshorne). </p>
<p>Does this result also hold if we drop the condition of zero characteristic and further assume that $f$ is a smooth at a general point $x \in X$? Is there any other condition on $f$, other than smoothness, under which this can hold true (in positive characteristic)? Any reference or hint on this question will be most welcome.</p>
<p>N.B. This is a continuation of a <a href="http://mathoverflow.net/questions/219937/bertini-type-theorem-in-positive-characteristic">question</a> I asked before.</p>
http://mathoverflow.net/q/2199842Finite orbits on an elliptic curve with two generic involutionsYankeeDoodleDandyhttp://mathoverflow.net/users/687222015-10-04T18:54:20Z2015-10-04T23:16:18Z
<p>Let $C$ be a (very) general genus 1 curve embedded in $\mathbb{CP}^1\times \mathbb{CP}^1$ as a (2,2)-divisor. </p>
<p>Each projection defines $C$ as a double cover of $\mathbb{CP}^1$ and induces an involution $\tau_i:C\to C$. Let $G\simeq \mathbb Z/2 * \mathbb Z/2$ be the subgroup of $Aut(C)$ generated by these. Note that $G$ is infinite.</p>
<p>Are there points on $C$ with finite orbit under $G$?</p>
http://mathoverflow.net/q/21996313Are there irreducible polynomials with all zeros on two concentric circles?Wolfganghttp://mathoverflow.net/users/297832015-10-04T12:51:18Z2015-10-05T00:39:59Z
<p>This is somewhat similar to <a href="http://mathoverflow.net/questions/219315/aligned-roots-of-irreducible-polynomials">this</a> recent question, but extending in a different direction.</p>
<p>Let $f(x)$ be an irreducible polynomial of degree $n$ with integer coefficients. Call such $f$ a <strong>bicycle polynomial</strong> if all its roots are located on two circles around $O$, i.e. all roots have one of two moduli. (Of course we'll exclude polynomials of cyclotomic type like $\Phi_n(mx)$ for $m\in\mathbb Z$, which have all their roots fit on <em>one</em> circle.)</p>
<p>For $n=2k$ or $n=3k$, examples of bicycle polynomials are $f(x)=g(x^k-1)$, where $g$ is irreducible of degree $2$ or $3$. Taking here a $g$ of degree $4$ with two pairs of complex roots yields still other bicycle polynomials for $n=4k$. </p>
<p>Now: except replacing the "$-1$" by any other nonzero integer, <em>those types of constructions already seem to be about all of it...</em> </p>
<blockquote>
<ul>
<li>Do bicycle polynomials of degree $n$ exist <strong>if $n$ has only prime factors $>3$?</strong></li>
</ul>
</blockquote>
<p>Assuming the existence of such a polynomial, it appears (?) to boil down to the existence of a degree $m$ polynomial ($m>3$ odd) with $m-1$ roots on one circle. This circle must presumably have an irrational radius because of what is known about <a href="http://mathoverflow.net/questions/114745/monic-polynomial-with-integer-coefficients-with-roots-on-unit-circle-not-roots/114748#114748">Salem polynomials</a>, but I'm stuck here. </p>
<p>Also related: <a href="http://mathoverflow.net/questions/21298/how-to-best-distribute-points-on-two-concentric-circles">How to best distribute points on two concentric circles?</a></p>
<p>Another question: </p>
<blockquote>
<ul>
<li>Does anything change if we allow complex integers as coefficients?</li>
</ul>
</blockquote>
http://mathoverflow.net/q/2199573Is a manifold generically real analytic (with generic real analytic metric)?Guesthttp://mathoverflow.net/users/810392015-10-04T10:23:33Z2015-10-04T20:20:28Z
<p>I have heard it said in some differential geometry talks that "the generic situation in such and such case is real analytic". My question is, is the generic smooth manifold also real analytic in some sense of genericity? I do not have a specific purpose in asking this question, just curious. Thanks!</p>
<p>Edit after Robert Bryant's comment: As an extension to my question, consider a Riemannian manifold with a smooth metric. Is this metric generically real analytic in a certain sense? That is, can we equip a smooth manifold with a real analytic metric generically (in a certain sense)?</p>
http://mathoverflow.net/q/2199455mod p cohomology ring of alternating groupshaohaizihttp://mathoverflow.net/users/658002015-10-04T06:13:12Z2015-10-04T18:08:51Z
<p>Let $A_n$ be the alternating group of $\{1,2,\cdots,n\}$. </p>
<p>(1). What is the cohomology ring
$$
H^*(A_4;\mathbb{Z}/3)
$$
and its Steenrod operation $P^i$'s?</p>
<p>(2). Are there general results about the cohomology ring
$$
H^*(A_{p+1};\mathbb{Z}/p)
$$
for general primes $p\geq 3$?</p>
<p>(3). What is the cohomology ring
$$
H^*(A_n;\mathbb{Z}_2)
$$
for $n\geq 4$ (we can impose conditions on $n$)?</p>
<p>Any references?</p>
http://mathoverflow.net/q/2199092A matching that covers vertices with maximum degreeMohemnisthttp://mathoverflow.net/users/810112015-10-03T14:42:25Z2015-10-04T23:28:14Z
<p>We have a graph G with maximum degree $\Delta$. The induced subgraph on vertices with degree equal to $\Delta$ is a bipartite graph (while the original graph is not).
Prove that G has a matching that covers all vertices with degree $\Delta$.</p>
<p>For example consider $K_{3,3}$ and add a vertex on one edge. So the graph has 7 vertices and 10 edges. The graph is not bipartite, but the induced subgraph on vertices with degree $\Delta(G) = 3$ is a bipartite subgraph. You can easily find a matching that covers all of the 6 vertices with degree 3.</p>
http://mathoverflow.net/q/2198641When does $R [x]/I $ has infinitely many idempotents?Es_Rohttp://mathoverflow.net/users/809842015-10-02T19:24:32Z2015-10-04T22:01:36Z
<blockquote>
<p>Let $R$ be a commutative ring with identity and $R[x] $ its polynomial ring. I am looking for a ring with finitely many idempotents and an unextended ideal $I$ in $R[x]$ such that $R[x]/I$ has infinitely many idempotents. </p>
</blockquote>
<p>Thank you for any help.</p>
http://mathoverflow.net/q/2198554Generalized density functions on the natural numbersJames Propphttp://mathoverflow.net/users/36212015-10-02T17:56:56Z2015-10-05T00:38:42Z
<p>If $a_1,a_2,\dots$ are IID random bits (correction as per Anthony Quas: these "bits" are $+1$ and $-1$ with equal probability), then with probability 1, the set of natural numbers $n$ such that $a_1+a_2+\dots+a_n \leq 0$ has lower density 0 and upper density 1, so it has no density in the ordinary sense. Still, I wonder if there is a principled way to generalize the manner in which we assign "densities" to subsets of the natural numbers in such a fashion that, with probability 1, the aforementioned set has generalized density 1/2 -- and, more generally, for every real $t$, the set of $n$ such that $(a_1+a_2+\dots+a_n)/\sqrt{n} \leq t$ has generalized density equal to the probability that the relevant Gaussian random variable has value less than $t$.</p>
http://mathoverflow.net/q/2195752Minimum of an apparently harmless function of two variablesPagliahttp://mathoverflow.net/users/411232015-09-29T14:18:33Z2015-10-04T20:26:38Z
<p>DISCLAIMER: I already posted this question on Mathematics a month ago, <a href="http://math.stackexchange.com/questions/1387569/minimum-of-an-apparently-harmless-function-of-two-variables">here</a>. However, since it has not been solved yet on that platform, I decided to ask it also here on mathoverflow. At a first glance, it looks like a straightforward calculus exercise, but it seems to hide some intrinsic difficulty... (at least, to me! :D)</p>
<hr>
<p>I would like to prove that the minimum of the function</p>
<p>$$
f(x,y):=\frac{(1-\cos(\pi x))(1-\cos (\pi y))\sqrt{x^2+y^2}}{x^2 y^2 \sqrt{(1-\cos(\pi x))(2+\cos(\pi y))+(2+\cos(\pi x))(1-\cos(\pi y))}}
$$</p>
<p>over the domain $[0,1]^2$ is $2\sqrt{2}$. Looking at the 2D plot of the function</p>
<p><a href="http://i.stack.imgur.com/QdfIB.png" rel="nofollow"><img src="http://i.stack.imgur.com/QdfIB.png" alt="Plot of f"></a></p>
<p>one immediately notices that the minimum is $f(1,1) = 2\sqrt{2}$. However, I can't figure out how to prove this in a rigorous way, even if the expression of $f$ seems to have a nice, "quasi-separable" structure...</p>
http://mathoverflow.net/q/2194894Loci in the moduli space of K3 surfaces associated to latticesYankeeDoodleDandyhttp://mathoverflow.net/users/687222015-09-28T18:33:49Z2015-10-04T22:25:47Z
<p>The moduli space of K3 surfaces forms a 20-dimensional family with countably many 19-dimensional components $M_d$ corresponding to the polarized K3s $(X,L)$ with $L^2=d$. The moduli space $M_d$ has a natural locus $M_d^W$ for each lattice $W$ with an embedding $W\subset H^{1,1}(X)\cap H^2(X,\mathbb Z)$. Is it known how these loci intersect? </p>
<p>For instance, is it true that $M_{d}^W\cap M_{e}\neq \emptyset$ for each $W,d,e$ provided $M_{d}^W\neq \emptyset$? </p>
http://mathoverflow.net/q/2187378Is every finitely generated flat modules projective over a commutative ring with a finite number of minimal primes?brunohhttp://mathoverflow.net/users/33332015-09-19T23:30:29Z2015-10-04T19:23:25Z
<p>Over a commutative ring $R$, a finite type locally free (weak sense) module for which the rank function is locally constant is projective.</p>
<p>If we notice that for each minimal prime $p$ of the ring, the rank function is constant on $V(p)$, the adherence of $p$ in the Zariski topology (because if $p\subset q$ the rank at $p$ is equals to the rank at $q$) on finite locally free (weak sense) modules, then if the ring has only a finite number of minimal primes, the rank function is always locally constant on finite flat modules. Am I right ?</p>
<p>I am asking this simple question because I read this great answer here
<a href="http://mathoverflow.net/a/33574/3333">http://mathoverflow.net/a/33574/3333</a>
where an important paper of Raynaud-Gruson is mentioned. It gives amongst a lot of generalizations the simple criteria that if $R$ has a finite number of associated primes then without any other hypothesis on $R$ every f.g. flat modules is actually projective. My reasoning above seems quite simple and gets a slightly more general result, but perhaps I am wrong ?</p>
<p>Edit: since nobody answered my question I dug around but I did not find any reference to this simple criteria. I only found the criteria about the finiteness of the number of associated primes. Is it equivalent ? Is there a counterexample with an infinite number of embedded primes but a finite number of minimal (isolated) primes ? </p>
<p>Edit 2: the reasoning is detailed here <a href="http://math.stackexchange.com/q/1450205/14860">http://math.stackexchange.com/q/1450205/14860</a></p>
http://mathoverflow.net/q/2186874Mixed norm estimate for the heat equationFan Zhenghttp://mathoverflow.net/users/371032015-09-19T11:40:47Z2015-10-04T19:40:26Z
<p>Consider the inhomogeneous linear heat equation</p>
<p>$$\partial_tu-\Delta u=F$$</p>
<p>on $\mathbb R^n\times [0,1]$ (say) with zero initial data. Assume $F$ is very nice (say Schwarz), so that we have a nice solution $u$. It is quite standard that</p>
<p>$$\|\nabla^2 u\|_{L_x^2L_t^2}\ll \|F\|_{L_x^2L_t^2}. $$</p>
<p>I'm now asking if there is a similar estimate</p>
<p>$$\|\nabla^2 u\|_{L_x^2L_t^p}\ll_p \|F\|_{L_x^2L_t^p} $$</p>
<p>for $1<p<\infty$.</p>
<p>Remark: It is false for $p=\infty$, because the $L^2\to L^2$ norm of second derivatives of the heat kernel $K_t$ (i.e. the $L^\infty$ norm of its spatial Fourier transform) blows up like $1/t$ as $t\to 0$, which is not integrable in $t$. Also by duality it is also false for $p=1$, so I'm interested in what happens in between.</p>
http://mathoverflow.net/q/2131220Ergodic automorphisms of a compact metric abelian group are BernoulliStéphane Laurenthttp://mathoverflow.net/users/213392015-08-05T20:00:47Z2015-10-04T23:10:36Z
<p>In the literature, such as <a href="http://link.springer.com/article/10.1007%2FBF02760804" rel="nofollow">in this article</a>, it is proved that every ergodic automorphism of a compact metric abelian group is Bernoulli. A rotation is not isomorphic to a Bernoulli shift because it has zero entropy. What does it mean ? Is it another notion of Bernoullicity ? (I do not have access to these papers).</p>
http://mathoverflow.net/q/2130762Generating function of alternating even terms in the Vandermonde Convolutionjimshttp://mathoverflow.net/users/767822015-08-05T07:00:07Z2015-10-05T00:10:36Z
<p>I have</p>
<p>\begin{equation}
G(x) = \sum_{i = 0}^{\infty} \sum_{r = 0}^{\infty} (-1)^i \binom{k}{2i} \binom{n-k}{r} x^{2i} x^{r} = \frac{1}{2} \left( (1 + x{\iota})^{k} + (1 - x \iota)^{k} \right)(1 + x)^{n-k}
\end{equation}</p>
<p>where $\iota$ is the imaginary unit, i.e., $\iota^2 = -1$. The $l$th term of the above is the alternating sum of even coefficients in the Vandermonde convolution, with $l \le k$. I am interested in obtaining a form of the $l$th term that involves a single binomial coefficient (as in the Vandermonde convolution) as opposed to the convolution of two binomial coefficients. Is there a way to do so? </p>
<p>Edit: Sorry, I needed to explain better. By the $l$th term I mean the coefficient of $x^l$. If we fix such an $l$, then we see that for $i = 0, 1, \ldots $, we have $r = l - 0, l - 2, \ldots $, and therefore fixing $r = l - 2i$, we obtain the following as the coefficient of $x^l$:</p>
<p>$\sum_{i = 0}^{\infty} (-1)^{i} \binom{k}{2i} \binom{n-k}{l - 2i}$</p>
<p>So, for instance, if $l = 3$, we get </p>
<p>$\binom{k}{0} \binom{n-k}{3 - 0} - \binom{k}{2} \binom{n-k}{3 - 2}$, where I have used the convention that $\binom{a}{b} = 0$ if $a < b$.</p>
<p>The Vandermonde convolution is:</p>
<p>$\sum_{i = 0}^{\infty} \binom{k}{i} \binom{n-k}{l - i} = \binom{n}{l}$. </p>
<p>I was hoping to get something like the RHS of the above, i.e., $\binom{n}{l}$, for the $l$th term of the main equation. Note that the inner sum in the main equation has the closed form $(1 + x)^{n-k}$, and the outer sum has the closed form $\frac{1}{2}\left((1 + x\iota)^k + (1 - x\iota)^k\right)$. Please bear in mind the difference between $i$ and $\iota$.</p>
http://mathoverflow.net/q/21017510Travelling Salesman Problem: Can the nearest neighbor algorithm be $n$ times longer than the optimal solution?Dominic van der Zypenhttp://mathoverflow.net/users/86282015-06-26T06:35:17Z2015-10-04T20:10:16Z
<p>This is inspired by a <a href="http://mathoverflow.net/questions/210049/length-of-nearest-neighbor-path-in-travel-salesman-problem">recent question</a>.</p>
<p>Given a positive integer $n\in\mathbb{N}$, is there a setting of finitely many points and a designated "starting point" $s$ in $\mathbb{R}^2$ such that the nearest-neighbor algorithm (described below) gives a tour that is $n$ times longer than the optimal solution starting at $s$?</p>
<blockquote>
<blockquote>
<p>Starting at $s$, pick the nearest neighbor not visited so far as the next node to visit.</p>
</blockquote>
</blockquote>
<p><strong>EDIT</strong>: If the answer is no, what is the maximum value that the ratio $r$ of "nearest neighbor trip" vs "best trip" can take?</p>
http://mathoverflow.net/q/20053545Continuous maps which send intervals of $\mathbb{R}$ to convex subsets of $\mathbb{R}^2$Abcdhttp://mathoverflow.net/users/694822015-03-20T14:16:55Z2015-10-04T22:10:31Z
<p>Let $f : \mathbb{R} \longrightarrow \mathbb{R}^2$ be a continuous map which sends any interval $I \subseteq \mathbb{R}$ to a convex subset $f(I)$ of $\mathbb{R}^2$. Is it true that there must be a line in $\mathbb{R}^2$ which contains the image $f(\mathbb{R})$ of $f$?</p>
<p>Yes, this question seems rather elementary, but I have already spent (or lost?) too much time on this devilish problem, and I have communicated this question to sufficiently many people to know that it is far from trivial...</p>
http://mathoverflow.net/q/1958923assumptions on local rademacher complexitiesuser3429697http://mathoverflow.net/users/666822015-02-06T22:40:33Z2015-10-04T21:10:18Z
<p>A lot of the work on Local Rademacher complexities of Koltchinskii, and Bartlett for fast rates of convergence is based on Bousquet's version of Talagrand's inequality [1] (Theorem 2.11). However the assumption used by Bousquet is that the space of functions used is countable. Nevertheless, all the results by Koltchinskii and Bartlett ignore this assumption and claim that it holds for arbitrary function spaces. </p>
<p>My question is does the general inequality follows from the countable inequality?
I have been trying to show this but it is not immediate without some continuity assumptions on the empirical process. </p>
<p>The proof of Bousquet also uses the fact that the set can be approximated with finite sets so I do believe it is necessary to have a countable space of functions. </p>
<p>[1] Concentration inequalities and Empirical Processes Theory Applied to the Analysis of Learning Algorithms. Olivier Bousquet, 2002.</p>
http://mathoverflow.net/q/1855851Hamiltonian potentials of holomorphic vector fields on modifications of Kahler manifoldsstudenthttp://mathoverflow.net/users/373702014-10-28T11:08:51Z2015-10-05T01:11:05Z
<p>let $(M,\omega)$ be a compact Kähler manifold. Let $\mathfrak{g}=H^{0}(M,T_{M})$ be the Lie algebra of holomorphic vector fields on $M$.We can decompose $\mathfrak{g}$ as
$$\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{a}$$
where $\mathfrak{h}$ is the space of holomorphic vector fields vanishing somewhere on $M$. Suppose $X\in \mathfrak{h}$ has a Hamiltonian potential $\varphi_{X}\in C^{\infty}(M,\mathbb{R})$
$$\overline{\partial}\varphi_{X}=-\frac{1}{2}i_{X}\omega$$
Now we blow up $M$ at some points $p_{1},\ldots,p_{n}$ and we denote with $\tilde{M}$ this blow up and with $\pi$ the canonical surjection on $M$. Suppose $X$ lifts to $\tilde{M}$ i.e. $X$ vanishes at $p_{1},\ldots,p_{n}$. Now we pick a Kähler metric $\omega_{\varepsilon}\in[\omega_{\varepsilon}]$</p>
<p>$$[\omega_{\varepsilon}]:=\pi^{*}[\omega]+\varepsilon\sum_{i=1}^{n}c_{1}(\mathcal{O}(-E_{i}))$$
with $E_{i}$ the <a href="http://en.wikipedia.org/wiki/Exceptional_divisor" rel="nofollow">exceptional divisors</a> at points. We DO NOT assume that $\omega_{\varepsilon}$ is invariant for the flow of the lift of $X$. </p>
<p>The question is the following: is there always a Hamiltonian potential $\tilde{\varphi}_{X}\in C^{\infty}(M,\mathbb{R})$ such that<br>
$$\overline{\partial}\tilde{\varphi}_{X}=-\frac{1}{2}i_{X}\omega_{\varepsilon}$$
or there is some obstruction?</p>
http://mathoverflow.net/q/14150116fixed point property for maps of compactsMishahttp://mathoverflow.net/users/216842013-09-07T13:04:08Z2015-10-04T18:08:55Z
<p>Definition. A topological space $X$ has the Fixed Point Property (FPP) if every continuous self-map $X\to X$ has a fixed point. </p>
<p>Question. If $X$ and $Y$ are homotopy-equivalent compact metrizable spaces and $X$ has the FPP, does it follow that $Y$ also has FPP? Another way to put it: Can one force a fixed point for a self-map of a compact by a "non-homological" argument?</p>
<p>I do not know an answer to this even for finite simplicial complexes, but my primary interest is in locally connected finite-dimensional compacts. </p>
http://mathoverflow.net/q/13583312What might extraterrestrial mathematics look like? [closed]K.B.http://mathoverflow.net/users/365612013-07-05T07:15:18Z2015-10-04T19:42:04Z
<p>In an extensive anthropological joint research project concerning the necessities in the development of life and civilisation my group is concerned with mathematics. This forum seems to be extremely well-suited to answer the following question:</p>
<blockquote>
<p>Given the case that a civilisation independent of mankind is or will be existing. How would their mathematics look like and <em>why</em>?</p>
</blockquote>
<p>Some teasers: Integers from simple counting to the Binomial Theorem, Geometry including Pythagoras' Theorem, Probability Theory, and Analysis seem to be dictated by practical requirements. Research on Prime Numbers may be a necessary by-product of calculating with integers. What about more advanced concepts like Grothendieck universes or Category Theory? Can even somewhat be said about branches that we don't yet have exploited ourselves?</p>
<p>(This question should be made wiki but I could not find the button. Seems that something has changed during the recent year.)</p>
http://mathoverflow.net/q/12679811What is the relationship between these two notions of "period"?Julian Rosenhttp://mathoverflow.net/users/52632013-04-07T18:44:40Z2015-10-04T23:57:54Z
<p>The motivation for this question is to understand a <a href="http://www.ihes.fr/~brown/MTZ.pdf" rel="nofollow">recent theorem of Francis Brown</a> which implies that all periods of mixed Tate motives over $\mathbb{Z}$ lie in $\mathcal{Z}[\frac{1}{2\pi i}]$, where $\mathcal{Z}$ is the $\mathbb{Q}$-span of the set of multiple zeta values (of positive integer arguments). My picture of mixed Tate motives is not very clear, and I would like to be able to relate their periods to something I understand better.</p>
<p>There is a survey article of Kontsevich and Zagier which defines a period as a complex number whose real and imaginary parts are given by convergent integrals of rational functions with rational coefficients, over domains in $\mathbb{R}^n$ cut out by finitely many polynomial inequalities with rational coefficients.</p>
<blockquote>
<p>What is the relationship between the set of periods of mixed Tate motives over $\mathbb{Z}$ and the set of periods in the sense of Kontsevich/Zagier? Does one of these sets contain the other?</p>
</blockquote>
<p>I would be interested to see examples of periods of one kind which are not periods of the other.</p>
http://mathoverflow.net/q/9373613What are the generalizations of the 27 lines on a cubic surface?David Feldmanhttp://mathoverflow.net/users/109092012-04-11T04:12:42Z2015-10-05T01:03:28Z
<p>The following doubtlessly naive heuristic suggests to me that there might be some generalizations. I don't know whether, at one extreme, the story is classical, or at the other extreme, the heuristic just fails.</p>
<p>Consider a generic hypersurface $S$ of degree $d$ in ${\Bbb P}^n$.
The intersection of $S$ with a generic plane $P$ should form a curve of degree $d$, hence a curve of genus $g=(d-1)(d-2)/2$.</p>
<p>The possible planes $P$ range over a Grassmannian ${\rm Gr}(n+1,3)$ of dimension $3(n-2)$.</p>
<p>One thus gets a morphism from ${\rm Gr}(n+1,3)$ to the moduli space of curves of genus $g$, which has dimension $3(d-1)(d-2)/2 - 3$ (unless $d=3$).</p>
<p>If the numbers work out right, one can try to make $n$ large enough, but not too large, so that one gets a 0-dimensional set of planes where the intersection gives rise to curves with some desired amount of degeneration. With enough degeneration perhaps, the original curves of genus $g$ will acquire components of smaller genus. So one might get interesting configurations of comparatively low genus curves (not necessarily all of the same genus).
For example, with $S$ of degree $4$ one might look for a configuration of genus 1 and 2 curves. (Personally, I don't know enough about compactifying moduli spaces even to guess the details at this point.)</p>
<p>In any case, with the 27-lines on the cubic surface, elliptic curves would seem to degenerate into a finite set of lines sharing common points, making this classic object an example of the heuristic above.</p>
<p>All that said, I'll make my question the broad one in the title. </p>