User zhang xiao - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:15:01Z http://mathoverflow.net/feeds/user/22727 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93522/two-approaches-to-compute-the-signature-of-a-kaehler-manifold Two approaches to compute the signature of a Kaehler manifold Zhang Xiao 2012-04-08T22:06:27Z 2013-01-08T18:33:51Z <p>Given a compact Kaehler manifold $M$ of complex dimension $2n$, there are essentially two ways to compute its signature $\sigma(M)$, i.e. the index of the intersection form on $H_{2n}(M,\mathbb{R})$:</p> <p>1.by Hodge index theorem $\sigma(M)=\sum_{p,q}(-1)^p h^{p,q}$, here $h^{p,q}$ stands for the Hodge numbers.</p> <p>2.by Hirzebruch signature theorem $\sigma(M)=L[M]$, here $L[M]$ stands for the $L$-genus, i.e. the characteristic number of the top $L$-class.This approach is more general since it works on any $4k$ dimensional real manifolds.</p> <p>My questions are</p> <p>1.Since these two approaches rest on different levels of cohomology theory, how are they interrelated? </p> <p>2.Of course, one possible way to answer Question 1 is to generalize both by the Hirzebruch-Riemann-Roch on Kaehler manifolds, a point already mentioned in Hirzebruch's <em>Neue topologische Methoden</em>. However, I am wondering if someone could relate these two approaches on a more fundamental level. </p> <p>To be precise,</p> <p>Is there a formula to express the Chern numbers/Pontryagin numbers out of the Hodge numbers on a compact Kaehler manifold $M$ of complex dimension $n$? Surely it is the case for $c_n[M]$ interpreted as the Euler characteristic number. </p> <p>Or, does anyone know such counterexamples that two Kaehler manifolds(notably, Kaehler surfaces, I guess) have the same Hodge numbers but different Chern numbers?</p> <p>Many thanks!</p> http://mathoverflow.net/questions/105166/how-markusyamabe-implies-jacobian How Markus–Yamabe implies Jacobian ? Zhang Xiao 2012-08-21T15:32:48Z 2012-08-21T15:58:27Z <p>To make myself precise, I would like to recall some backgrounds.</p> <p>(Markus-Yamabe, $\mathrm{MY}_n$) Given a $C^1$ map $f:\Bbb R^n \to \Bbb R^n$ with $f(0)=0$ and $Df$ everywhere Hurwitz stable (the real part of all eigenvalues of $Df$ are $&lt;0$), 0 is the global attractor of dynamical system $\dot{x}=f(x)$.</p> <p>$\mathrm{MY}_2$ is now a theorem (Fessler; Glutsyuk; Gutierrez), and a polynomial counterexample has been found for $\mathrm{MY}_3$ (Cima et al.)</p> <p>I would also like to distinguish 2 kinds of "real Jacobian conjecture":</p> <p>(real Jacobian, $\mathrm{RJ}_n$) Given a polynomial map $f:\Bbb R^n \to \Bbb R^n$, the Jacobian $J_f$ being some non-zero constant implies $f$ is a diffeomorphism. I believe a stronger statement can be made by taking $f$ to be $C^1$.</p> <p>(strong Jacobian, $\mathrm{SJ}_n$) Given a $C^1$ map $f:\Bbb R^n \to \Bbb R^n$, $J_f(x)>0$ implies $f$ is a diffeomorphism.</p> <p>Notice that Pinchuk has found a polynomial counterexample for $\mathrm{SJ}_2$.</p> <p>I learnt recently a very vague statement that the Markus-Yamabe conjecture implies the real Jacobian conjecture. I think these stuffs must be well-known among the experts, so could anyone </p> <p>(1)make this statement rigorous by indicating the relationship between $\mathrm{MY}_n$, $\mathrm{RJ}_n$ and $\mathrm{SJ}_n$?</p> <p>(2)or even better, show me how the argument goes,</p> <p>(3)or locate some reference?</p> <p>Thanks a lot!</p> http://mathoverflow.net/questions/103451/on-certain-decomposition-of-unitary-symmetric-matrices On certain decomposition of unitary symmetric matrices Zhang Xiao 2012-07-29T14:18:06Z 2012-07-30T19:29:38Z <p>This is by any means elementary, but since I have asked this question on Stark Exchange but received no satisfactory answers I decide to post it here.</p> <p>It is well known that a symmetric matrix over field $\Bbb F$ is congruent to a diagonal matrix, i.e., there exists some A s.t. $A^TUA=D$ with $U$ symmetric and $D$ diagonal. If $\Bbb F=\Bbb C$ then we can make $D=I$.</p> <p>Recently I learned that if $U$ is unitary that we can do one step further by requiring $A$ to be unitary too. A similar result holds for unitary skew matrices. But I fail to figure out a proof myself.</p> <p>Can anyone provide a proof of this or at least help me to locate some references? Many thanks!</p> http://mathoverflow.net/questions/97976/approachable-french-masters/99317#99317 Answer by Zhang Xiao for Approachable French Masters Zhang Xiao 2012-06-11T22:36:35Z 2012-06-11T22:36:35Z <p>For operator theory, Dixmier seems to be a good option. </p> http://mathoverflow.net/questions/95837/examples-of-theorems-with-proofs-that-have-dramatically-improved-over-time/96377#96377 Answer by Zhang Xiao for Examples of theorems with proofs that have dramatically improved over time Zhang Xiao 2012-05-08T21:31:25Z 2012-05-08T21:31:25Z <p>It occurs to me that Morse theory is a good example. At the time of Morse, algebraic topology (even the notion of CW complex or cell complex) is barely developed, which made his combinatorial arguments extremely difficult to read.</p> <p>Well, nowadays people can simply learn these topics by referring to the definite account of Milnor or Bott.</p> http://mathoverflow.net/questions/94149/polyas-conjecture-on-the-spectra-of-the-laplacians Pólya's conjecture on the spectra of the Laplacians Zhang Xiao 2012-04-15T20:46:18Z 2012-04-16T17:48:38Z <p>Recently I've learned something about the spectra of the Laplacians. Given a bounded domain $\Omega \subset \mathbb{R}^n$ with $\partial \Omega$ smooth, we can consider eigenfunctions of Dirichlet type, i.e. $u \in C^2(\Omega)\cap C(\partial \Omega)$ s.t. $-\triangle u=\lambda u$ and $u|_{\partial \Omega}=0$. By standard results in functional analysis, $-\triangle$ has a discrete spectrum $0&lt;\lambda_1 \leq \lambda_2 \leq \cdots$ with $\lambda_k \to +\infty$.</p> <p>A well-known asymptotic formula by Weyl says $\displaystyle \lambda_k \sim W_n (\frac{k}{V(\Omega)})^{2/n}$. We refer $W_n$ as the Weyl constant. And Pólya conjectured that $\displaystyle \lambda_k \geq W_n (\frac{k}{V(\Omega)})^{2/n}$ holds.</p> <p>As far as I know, the best known result is due to Li and Yau. They proved the conjecture in the sense of "average": $\displaystyle \sum_{j=1}^k \lambda_j \geq \frac{nW_n}{n+2}{k}^{(n+2)/n}{V(\Omega)}^{-n/2}$. </p> <p>I find their argument is elementary, only employing some standard Fourier tricks. And the big picture is quite clear if put into the quantum framework. My question is in some sense "soft", but it does make me feel absurd: what makes it so difficult to estimate the eigenvalues one by one while their average is so well understood? Does anyone work on this problem by carrying further Li &amp; Yau's analysis? I do know one instance: Kröger has transplanted their proof to Neumann settings, but how about the original Dirichlet problem? </p> http://mathoverflow.net/questions/93694/fundamental-theorem-of-algebra-theorems-of-brouwer-and-borsuk/93709#93709 Answer by Zhang Xiao for Fundamental Theorem of Algebra, Theorems of Brouwer and Borsuk Zhang Xiao 2012-04-10T21:12:24Z 2012-04-10T21:12:24Z <p>Since Ryan has discussed the relationship between Brouwer and FTOA, I would like to say something about Brouwer and Borsuk-Ulam（I prefer this name）. What I know is there are series of classical theorems on this direction, all closely related:</p> <p>Thm.1(Brouwer) Continuous map $f:B^{n+1} \to B^{n+1}$ has at least one fixed point.</p> <p>Thm.2(Hirsch) $\partial B^{n+1}$ is not a deformation retract of $B^{n+1}$.</p> <p>Thm.3 $\mathrm{id}:S^n \to S^n$ is not homotopic to a constant map. </p> <p>Thm.4 Continuous map $g:S^n \to S^n$ sending antipodal points to antipodal points is not homotopic to a constant map.</p> <p>Thm.5 There exists continuous map $h:S^n \to S^m$ sending antipodal points to antipodal points if and only if $n \leq m$.</p> <p>Thm.6(Borsuk-Ulam)For continuous map $k:S^n \to \mathbb{R}^n$，there exists $x \in S^n$ such that $k(x)=k(-x)$.</p> <p>Let me explain. Equivalence of 1 and 2 is quite standard, for example, see</p> <p>Milnor <em>Topology from the differentiable viewpoint</em></p> <p>By the way, it also contains a proof of FTOA, using the degree of maps.</p> <p>To see the relation of 2 and 3, simply glue the boundary to a point. </p> <p>4 is an obvious generalization of 3. You may prove it by studying $H_*(\mathbb{R}P^n)$. This study also leads to a proof of 5.</p> <p>Equivalence of 6 and (the "only if" part of) 5 is also not hard, you may take it as a exercise:)</p> <p>All in all, what I have shown is some knowledge of $H_*(\mathbb{R}P^n)$ is sufficient for both Brouwer and Borsuk-Ulam. Hope this helps.</p> http://mathoverflow.net/questions/93556/finding-topological-obstructions-for-a-complex-manifold-to-be-kaehler Finding topological obstructions for a complex manifold to be Kaehler Zhang Xiao 2012-04-09T09:24:26Z 2012-04-09T14:02:57Z <p>Well, it is of the "straightforward" questions one may ask. I propose it here to see if someone could tell me more on the recent status of this quite long-standing problem.</p> <p>To initiate, let me give a brief description of the "classical" invariants. Essentially they are provided by either the symplectic structure or the Hodge structure, most of which relating to vanishing theorems and integral theorems. As far as a compact manifold $M$ of complex dimension $n$ is concerned, we have the following:</p> <p>$b_2 \ne 0$ for the symplectic structure and many more;</p> <p>$b_{2k+1}$ are even and $b_{k-2} \leq b_k$ for $k \leq n$ for the Hodge structure, which date back to Lefschetz;</p> <p>Various integral results on Chern numbers by Hirzebruch-Riemann-Roch on Kaehler manifolds;</p> <p>and etc.</p> <p>To make the question precise, I would like to ask:</p> <p>Has there been any "higher" or essentially new invariants discovered so far? Particularly, one may observe that the above invariants are all torsion-free(as Yau did in his problem list), so the torsion invariants would be rather interesting, suppose they do exist. </p> <p>Any comments are welcomed and thanks a lot!</p> http://mathoverflow.net/questions/105220/wikipedia-story-about-bill-thurstons-death Comment by Zhang Xiao Zhang Xiao 2012-08-22T10:13:41Z 2012-08-22T10:13:41Z @Damian R&#246;ssler This is so sad. May he rest in peace. http://mathoverflow.net/questions/103451/on-certain-decomposition-of-unitary-symmetric-matrices/103488#103488 Comment by Zhang Xiao Zhang Xiao 2012-07-30T05:33:00Z 2012-07-30T05:33:00Z I like this down-to-ground proof, many thanks! http://mathoverflow.net/questions/103451/on-certain-decomposition-of-unitary-symmetric-matrices/103469#103469 Comment by Zhang Xiao Zhang Xiao 2012-07-30T05:32:27Z 2012-07-30T05:32:27Z Will and Paul: My guess is both Suvrit and Terry Loring interpret the question in the &quot;right&quot; way. And by &quot;right&quot; I mean the way I interpret it:) Anyway, thanks for the discussion. http://mathoverflow.net/questions/103451/on-certain-decomposition-of-unitary-symmetric-matrices/103473#103473 Comment by Zhang Xiao Zhang Xiao 2012-07-30T05:29:30Z 2012-07-30T05:29:30Z Thanks a lot Suvrit! I find your answer quite simple and illuminating. http://mathoverflow.net/questions/103451/on-certain-decomposition-of-unitary-symmetric-matrices Comment by Zhang Xiao Zhang Xiao 2012-07-29T15:53:44Z 2012-07-29T15:53:44Z Yuan: the formal statement is like this: for any unitary symmetry matrix (over $\Bbb C$), there exists some unitary matrix $A$ such that $A^TUA=I$. And for unitary skew matrix (again over $\Bbb C$), some unitary $A$ such that $A^TUA=S$, $S$ is the standard symplectic matrix. The crucial point is to make every matrix unitary. http://mathoverflow.net/questions/103451/on-certain-decomposition-of-unitary-symmetric-matrices Comment by Zhang Xiao Zhang Xiao 2012-07-29T14:41:24Z 2012-07-29T14:41:24Z It seems you do not read my statement carefully:) http://mathoverflow.net/questions/94149/polyas-conjecture-on-the-spectra-of-the-laplacians/94234#94234 Comment by Zhang Xiao Zhang Xiao 2012-04-17T17:15:51Z 2012-04-17T17:15:51Z I am reading the fascinating Sarnak now. Best regards Plm:) http://mathoverflow.net/questions/94149/polyas-conjecture-on-the-spectra-of-the-laplacians/94233#94233 Comment by Zhang Xiao Zhang Xiao 2012-04-17T17:08:30Z 2012-04-17T17:08:30Z Dear Denis: it seems to me that Polya tries to control the eigenvalues by some geometry properties of the domain while you seek for &quot;instinct&quot; restrictions. After looking into some references on the A.Horn's inequalities, I find this comment quite thought-provoking. Thanks. http://mathoverflow.net/questions/94149/polyas-conjecture-on-the-spectra-of-the-laplacians/94168#94168 Comment by Zhang Xiao Zhang Xiao 2012-04-16T02:49:26Z 2012-04-16T02:49:26Z To be frank I don't think the above notes make an attempt to attack the problem by detailed analysis on the &quot;quantum spectrum&quot;. But still many thanks! http://mathoverflow.net/questions/94149/polyas-conjecture-on-the-spectra-of-the-laplacians Comment by Zhang Xiao Zhang Xiao 2012-04-16T02:47:01Z 2012-04-16T02:47:01Z Some reference would be extremely appreciated, but thanks any way Mrc Plm! http://mathoverflow.net/questions/93556/finding-topological-obstructions-for-a-complex-manifold-to-be-kaehler/93569#93569 Comment by Zhang Xiao Zhang Xiao 2012-04-10T23:59:32Z 2012-04-10T23:59:32Z Thank you so much for your detailed explanation Misha! It seems I should take my time to expose to more rational homotopy theory:) http://mathoverflow.net/questions/93556/finding-topological-obstructions-for-a-complex-manifold-to-be-kaehler/93563#93563 Comment by Zhang Xiao Zhang Xiao 2012-04-10T23:57:53Z 2012-04-10T23:57:53Z I am not familiar with the kaehler cone but all these stuffs sound rather interesting. By the way, personally I regard $h^{1,1}=1$ as a rather strong restriction since they are &quot;immense&quot; only in the cardinality sense but only consist of a few of &quot;types&quot;:) http://mathoverflow.net/questions/93694/fundamental-theorem-of-algebra-theorems-of-brouwer-and-borsuk/93709#93709 Comment by Zhang Xiao Zhang Xiao 2012-04-10T23:44:39Z 2012-04-10T23:44:39Z @Stephan: You are correct. Although homology helps to make a unified picture, it contains strictly more information than Brouwer and Borsuk, hence of no use in setting the equivalence between two. http://mathoverflow.net/questions/93556/finding-topological-obstructions-for-a-complex-manifold-to-be-kaehler/93569#93569 Comment by Zhang Xiao Zhang Xiao 2012-04-10T01:23:54Z 2012-04-10T01:23:54Z Nice reference. My thanks to both Igor and Misha. Actually without knowing something has been done on the fundamental group, what I have in mind is the second homotopy group, which by Hurewicz a refinement of $H^2(M)$，the latter being much more important in the Kaehler case than $H^1(M)$.Has any work been done on this direction? http://mathoverflow.net/questions/93556/finding-topological-obstructions-for-a-complex-manifold-to-be-kaehler/93563#93563 Comment by Zhang Xiao Zhang Xiao 2012-04-10T01:17:19Z 2012-04-10T01:17:19Z Gunnar: I really appreciate your response. Since the criterion mentioned is rather differential-geometric than topological, I am wondering whether the seminegativeness of the sectional curvature of this &quot;moduli space&quot; has some implications on the topology of the original manifold:)