User 喻yuwei - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:40:59Z http://mathoverflow.net/feeds/user/19664 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110553/integral-hodge-classes-of-the-calabi-yau-3-fold integral hodge classes of the Calabi-Yau 3-fold 喻yuwei 2012-10-24T16:00:12Z 2012-10-24T17:21:11Z <p>I have been read many papers,But I don"t know a integral hodge class of the calabi-Yau 3-fold is algebraic or non-algebraic?Hope give some help and nice reference. Calabi-Yau 3-fold is a Kahler 3-fold with trival canonical bundle.Is it a open question？</p> http://mathoverflow.net/questions/108023/how-do-hodge-classes-for-calabi-yau-4-folds-compare-with-the-classes-for-tori How do Hodge classes for Calabi-Yau 4-folds compare with the classes for tori? 喻yuwei 2012-09-25T07:12:00Z 2012-09-25T09:01:40Z <p>Let $X$ be a Calabi-Yau 4-fold, i.e., a connected 4-dimensional compact Kahler manifold with $K_{X} \cong \mathscr{O}_{X}$ and $h^{i} (X,O_{X} )= 0$ for $0 \lt i \lt 4$. Given a general 4-dimensional Weil torus $T$, one has a Hodge class contained in $H^{2,2}( X )$. I would like to ask whether the following equality holds:</p> <p>$$Hg^{2} ( X)_{prim}= Hg^{2}\left ( T \right )_{prim}$$</p> <p>where $Hg^{2}\left ( X \right )_{prim}$ is a primitive Hodge class, and $Hg^2 ( \star )$ is defined as $H^4 ( \star ,\mathbb{Q} )\cap H^{2,2} ( \star )$ for $\star = X,T$. If the equality is faulty, I would like to know why. If Hodge class is not unique, how do we get the classification of Hodge classes？</p> http://mathoverflow.net/questions/106027/estimate-over-simply-connected-riemannian-manifold-with-non-positive-sectional-c estimate over simply-connected Riemannian manifold with non-positive sectional curvature 喻yuwei 2012-08-31T12:02:48Z 2012-09-11T19:19:44Z <p>Let<code>$M$</code> be a Complete simply-connected n-dimensional Riemannian manifold with nonpositive curvature,<code>$\Omega$</code>is a open subset of <code>$M$</code>,If <code>$n\geq 4$</code>Is anyone give a estimate of <code>$\frac{Vol_{n}\left ( \Omega \right )}{Vol_{n-1}\left ( \Omega \right )^{\frac{n}{n-1}}}$</code>?,If <code>$\Omega$</code>replace by <code>$B^{n}\left ( 1 \right )$</code>,it is a unit sphere,,estimate?</p> http://mathoverflow.net/questions/106086/isoperimetric-inequality-from-cartan-hadamard-manifold-extend-to-alexandrov-space Isoperimetric inequality from Cartan-Hadamard manifold extend to Alexandrov space 喻yuwei 2012-09-01T02:25:21Z 2012-09-01T13:35:43Z <p>We known that Croke proved a isoperimetric inequality for the four dimensional Cartan-Hadamard manifold,I want to ask extend to Alexandrov space that whether have same isoperimetric inequality ?Whether may be added to some condition,Is <code>$CD\left ( n,k \right )$</code>(curvature dimension),Is <code>$MCP\left ( n,k \right )$</code>(measure contraction property) and create to New condition?Of course,Maybe inequality is same.</p> http://mathoverflow.net/questions/106091/line-bundle-over-compact-simply-connected-topologically-regular-alexandrov-space Line bundle over compact simply-connected topologically regular Alexandrov space 喻yuwei 2012-09-01T04:01:34Z 2012-09-01T13:34:10Z <p>Let <code>$L$</code>be a line bundle over compact simply-connected topologically regular Alexandrov space<code>$X$</code>of demension 4 generated by fixed point set of smooth <code>$S^{1}-action$</code>,<code>$S^{1}-action$</code>is a section of <code>$L$</code>,Garcia proved that <code>$X$</code>with effective isometric action of <code>$S^{1}$</code>is Locally smmoth,For locally smooth topologically regular Alexandrov space<code>$X$</code>, I want to ask the following that whether isomorphism?<code>$H^{0}(X,L)\simeq H^{1}(X,X^{S^{1}})$</code> where <code>$X^{S^{1}}$</code>is smooth <code>$S^{1}-action$</code>.Do you have good idea?</p> http://mathoverflow.net/questions/106095/implication-question-about-mean-curvature-flow "implication" question about mean curvature flow 喻yuwei 2012-09-01T05:20:31Z 2012-09-01T13:28:39Z <p>About paper's problems: Let $H$ be a mean curvature vector field. We have the differential inequality</p> <p><code>$\frac{\partial }{\partial t} \left | H \right |^{2} \leq \Delta \left | H \right |^{2} + \beta \left | H \right |^{2}$</code></p> <p>where $\beta$ is a positive conatant. Putting $f=\left | H \right |^{2}$ and $B(R^{'}) = B_{g_{0}}(x,R^{'})$, and $q\geq 2$, then the above differential inequality apparently <strong><em>implies</em></strong> that</p> <p><code>$\frac{1}{q} \frac{\partial}{\partial t} \int_{B(R^{'})} f^{q} \eta^{2} d\mu_{t} \leq \int_{B(R^{'})} (\eta ^{2} f^{q-1} \Delta f d\mu _{t} + \beta f^{q} \eta ^{2}) d\mu_{t} + \int_{B(R^{'})} \frac{1}{q} f^{q} \eta^{2} \frac{\partial }{\partial t} d\mu _{t}$</code></p> <p>where<code>$\eta =0, 1$</code>, and <code>$d\mu _{t}$</code> satisfies <code>$\frac{\partial }{\partial t} d\mu_{t} = -\left | H \right |^{2} d\mu_{t}$</code>. My question is: "Why implies?，“How implies”，I don’t understand this implies？</p> http://mathoverflow.net/questions/95371/de-rham-vs-dolbeault-cohomology/95380#95380 Answer by 喻yuwei for de Rham vs Dolbeault Cohomology 喻yuwei 2012-04-27T17:09:41Z 2012-04-27T17:15:34Z <p>5.The cohomology of the projective space ，see akhil mathew's math bolg:<a href="http://amathew.wordpress.com/2010/11/22/the-cohomology-of-projective-space/" rel="nofollow">projective space</a></p> http://mathoverflow.net/questions/95371/de-rham-vs-dolbeault-cohomology/95375#95375 Answer by 喻yuwei for de Rham vs Dolbeault Cohomology 喻yuwei 2012-04-27T16:37:29Z 2012-04-27T16:37:29Z <p>1.De Rham cohomology is a cohomology of k-forms on complex manifold。 Dolbeault Cohomology is a cohomology of smooth sections of the vector bundle of complex differential forms of degree (p,q) on complex manifold。so Dolbeault Cohomology is subcohomology of the De Rham cohomology.2.when p+q=0 </p> http://mathoverflow.net/questions/29498/what-is-mordell-weil-lattice/95369#95369 Answer by 喻yuwei for What is Mordell-Weil lattice? 喻yuwei 2012-04-27T16:09:11Z 2012-04-27T16:15:46Z <p>[T Shioda ：Mordell-Weil lattice][1]</p> <p>[1]: <a href="http://www.rkmath.rikkyo.ac.jp/math/shioda/papers/mwl.pdf" rel="nofollow">http://www.rkmath.rikkyo.ac.jp/math/shioda/papers/mwl.pdf</a> more Basic，you also see the homepage of Chao Li about Elliptic Surfaces and Mordell-Weil Lattices in harvard university。</p> http://mathoverflow.net/questions/108023/how-do-hodge-classes-for-calabi-yau-4-folds-compare-with-the-classes-for-tori Comment by 喻yuwei 喻yuwei 2012-10-24T15:25:45Z 2012-10-24T15:25:45Z Is it a false question? http://mathoverflow.net/questions/110318/how-to-describe-the-splitting-of-alexandrov-space Comment by 喻yuwei 喻yuwei 2012-10-22T12:17:06Z 2012-10-22T12:17:06Z Please using Latex. http://mathoverflow.net/questions/109813/the-hodge-numbers-of-a-covering Comment by 喻yuwei 喻yuwei 2012-10-16T13:40:29Z 2012-10-16T13:40:29Z What is the Hodge rhomb? http://mathoverflow.net/questions/108688/how-to-type-this-strange-symbol-in-latex Comment by 喻yuwei 喻yuwei 2012-10-03T04:03:16Z 2012-10-03T04:03:16Z Should be close it. http://mathoverflow.net/questions/108152/topological-immersion Comment by 喻yuwei 喻yuwei 2012-09-26T12:38:28Z 2012-09-26T12:38:28Z It is not a real question.Who close it？ http://mathoverflow.net/questions/108023/how-do-hodge-classes-for-calabi-yau-4-folds-compare-with-the-classes-for-tori Comment by 喻yuwei 喻yuwei 2012-09-26T02:53:06Z 2012-09-26T02:53:06Z Explain ：the subspace$X^{'}$ of a Weil torus$X$ consists of the Hodge classes,and belongs to$H^{4}\left ( X,\mathbb{Q} \right )$ ,by Hodge decomposition，it contained in $H^{2,2}\left ( X \right )$ ，If it is not Hodge classes，maybe contained in$H^{1,3}\left ( X \right )$ ，$H^{3,1}\left ( X \right )$ ，$H^{0,4}\left ( X \right )$ or$H^{4,0}\left ( X \right )$ .Of course,the Hodge classes of$T$ whether exist in $X$ ,I hope have proof or counterexample .Prove its existence is a very diffficult. http://mathoverflow.net/questions/108049/what-is-projective-connection Comment by 喻yuwei 喻yuwei 2012-09-25T12:46:55Z 2012-09-25T12:46:55Z It is not real problem.Who close it http://mathoverflow.net/questions/108023/how-do-hodge-classes-for-calabi-yau-4-folds-compare-with-the-classes-for-tori Comment by 喻yuwei 喻yuwei 2012-09-25T10:18:40Z 2012-09-25T10:18:40Z Carnahan,thank you to fix English grammar fault，About the definition of Weil tori，see Voisin's papers and search in internet. http://mathoverflow.net/questions/107901/calabi-yau-and-ricci-flat-metrics Comment by 喻yuwei 喻yuwei 2012-09-24T16:09:48Z 2012-09-24T16:09:48Z Calabi-Yau theorem says that $Ric\left ( \omega \right )=\Omega$ ,and $X$ has Ricci-flat metric. http://mathoverflow.net/questions/107901/calabi-yau-and-ricci-flat-metrics Comment by 喻yuwei 喻yuwei 2012-09-24T02:20:24Z 2012-09-24T02:20:24Z Professor Yang，yuwei$\neq$ bruno，doesn't concern me http://mathoverflow.net/questions/107901/calabi-yau-and-ricci-flat-metrics Comment by 喻yuwei 喻yuwei 2012-09-23T18:14:24Z 2012-09-23T18:14:24Z Only show the canonical bundle $K_{X}$of $X$ is trivial. http://mathoverflow.net/questions/107901/calabi-yau-and-ricci-flat-metrics Comment by 喻yuwei 喻yuwei 2012-09-23T16:13:36Z 2012-09-23T16:13:36Z It should be $X$ be a compact n-dimensional kahler manifold, has a Kahler metric with vanishing Ricci curvature,vanishing first Chern number 。 http://mathoverflow.net/questions/106095/implication-question-about-mean-curvature-flow Comment by 喻yuwei 喻yuwei 2012-09-01T15:50:45Z 2012-09-01T15:50:45Z Professor Yang， Can you give some details？Thank you.