User botong wang - MathOverflowmost recent 30 from http://mathoverflow.net2013-06-19T19:45:07Zhttp://mathoverflow.net/feeds/user/4975http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/85908/stability-condition-for-vector-bundlesStability condition for vector bundlesBotong Wang2012-01-17T15:19:59Z2012-01-18T05:02:21Z
<p>Let $X$ be a smooth projective variety over $\mathbb{C}$, and fix $A$ an ample divisor as the polarization. We say a vector bundle $E$ to be (semi-)stable, if for any proper subsheaf $F$ of $E$, $\mu(F)<\mu(E)$ (resp. $\leq$). </p>
<p>I guess it is not sufficient to check just subbundles $F$ of $E$. But is there a counterexample? Or more precisely, is there an example of $(X, A, E, F)$ satisfying the following conditions?</p>
<p>(1)$X,A,E$ are as above, and $F$ is a proper subsheaf of $E$ breaking the stability condition, i.e., $\mu(F)\geq \mu(E)$.</p>
<p>(2)There exists no vector bundle which break the stability condition, i.e., $\mu(F')<\mu(E)$ for any subbundle $F'$ of $E$.</p>
http://mathoverflow.net/questions/19635/gauge-theory-construction-of-moduli-of-vector-bundlesGauge theory construction of moduli of vector bundlesBotong Wang2010-03-28T16:37:16Z2010-04-05T20:43:33Z
<p>Given a smooth projective complex variety $X$, instead of using Mumford's GIT to construct the moduli of rank $n$ topologically trivial vector bundles, we can also take the gauge theory approach. </p>
<p>To classify all topologically trivial vector bundles is same as to classify all possible complex structures on the topologically trivial vector bundle $X\times \mathbb{C}^n$ module the ambiguity from the choice of basis. And a complex structure is determined by $\bar\partial=\bar\partial_0+\eta$, where $\bar\partial_0$ corresponds to the holomorphic structure as a trivial vector bundle, and $\eta$ is an anti-holomorphic $End(\mathbb{C}^n)$ valued one form, such that $d\eta+\eta\wedge\eta=0$. Let $M$ denote the set of all such $\eta$, and $G$ denote the gauge group with fiber $GL(\mathbb{C}^n)$, then $G$ acts on $M$. Now, set-theoretically the orbits will correspond to the set of isomorphism classes of rank $n$ topologically trivial holomorphic vector bundles, but the quotient topology seems rather bad (non-Hausdorff).</p>
<p>My question is using this approach</p>
<p>(1)Whether stability condition is reflected in this construction?</p>
<p>(2)Can we obtain the same moduli space as using Mumford's GIT?</p>
<p>(3)I guess this is a natural approach. So is there any article or book about this?</p>
<p>The assumption of topologically trivial should not be necessary, just to simplify some notation. We only need to fix the underlying topological vector bundle.</p>
http://mathoverflow.net/questions/18271/what-out-of-print-books-would-you-like-to-see-re-printed/20310#20310Answer by Botong Wang for What out-of-print books would you like to see re-printed?Botong Wang2010-04-04T16:13:41Z2010-04-04T16:13:41Z<p>Kobayashi, "Differential geometry of complex vector bundles"</p>
http://mathoverflow.net/questions/9731/polynomial-representing-all-nonnegative-integers/19855#19855Answer by Botong Wang for Polynomial representing all nonnegative integersBotong Wang2010-03-30T17:41:18Z2010-03-30T17:41:18Z<p>How about an alternative question: does there exist a polynomial $f\in\mathbb{Q}[x,y]$ with integer values at lattice points, and of degree at least two on each variable, such that for any prime $p$, the map $f:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ composing with $\mathbb{Z}\to\mathbb{F}_p$ is surjective? Or more specifically, does a degree two such polynomial exist? The last part shouldn't be too hard, but I don't know how to solve it.</p>
http://mathoverflow.net/questions/19768/upper-half-plane-quotient-by-a-discrete-groupUpper half plane quotient by a discrete groupBotong Wang2010-03-29T21:05:37Z2010-03-30T11:57:43Z
<p>I was reading Mehta and Seshadri's paper "Moduli of vector bundles on curves with parabolic structures".</p>
<p>In the second paragraph, they wrote:</p>
<p>"Suppose that $H$ mod $\Gamma$ has finite measure ($H$ is the complex upper half plane, and $\Gamma$ is a discrete group). Let $X$ be the smooth projective curve containing $H$mod$\Gamma$ as an open subset and $S$ the finite subset of $X$ corresponding to parabolic and elliptic fixed points under $\Gamma$."</p>
<p>I am not sure about what parabolic and elliptic mean here. And why does such an $X$ exist? If I take a Riemann surface and remove several small balls from it, when is it biholomorphic to another Riemann surface removing several points?</p>
http://mathoverflow.net/questions/19505/riemannian-geometry-introductory-text/19639#19639Answer by Botong Wang for Riemannian Geometry Introductory TextBotong Wang2010-03-28T17:03:32Z2010-03-28T17:03:32Z<p>I like do Carmo's Riemannian geometry, which is more down-to-earth, and gives more intuition.</p>
<p>Jurgen Jost's Riemannian geometry and geometric analysis is also a good book, which covers many topics including Kahler metric. </p>
http://mathoverflow.net/questions/19635/gauge-theory-construction-of-moduli-of-vector-bundlesComment by Botong WangBotong Wang2010-04-02T20:52:32Z2010-04-02T20:52:32ZJoel, that's right. Thanks for pointing out.http://mathoverflow.net/questions/19635/gauge-theory-construction-of-moduli-of-vector-bundles/19637#19637Comment by Botong WangBotong Wang2010-04-02T20:51:23Z2010-04-02T20:51:23ZThanks, the book by Kobayashi is very helpful.http://mathoverflow.net/questions/19768/upper-half-plane-quotient-by-a-discrete-groupComment by Botong WangBotong Wang2010-03-31T21:28:33Z2010-03-31T21:28:33ZI am still not sure whether I understand the words elliptic and parabolic or not.
So an elliptic point means a point on $X$ with non-trivial stablizer, and a parabolic point means a point that is not in $X$, but comes from a natural completion of the quotient.
I am not familiar with the automorphisms of the complex unit disk. Is there a book that I can look at and get familiar with these basic concepts?http://mathoverflow.net/questions/19768/upper-half-plane-quotient-by-a-discrete-group/19799#19799Comment by Botong WangBotong Wang2010-03-31T21:19:11Z2010-03-31T21:19:11ZThanks for your answer too. I am not sure how is the Riemannian metric obtained? Do you just choose one which gives finite measure? Is there any relation with the conformal structure?http://mathoverflow.net/questions/19768/upper-half-plane-quotient-by-a-discrete-group/19771#19771Comment by Botong WangBotong Wang2010-03-31T21:17:19Z2010-03-31T21:17:19ZThanks for you answer, the third paragraph is really helpful. What will be the space of all structures with parabolic ends in $S^2$ minus four points?http://mathoverflow.net/questions/12892/why-cant-subvarieties-separate/12896#12896Comment by Botong WangBotong Wang2010-03-29T21:17:59Z2010-03-29T21:17:59ZIn Griffiths&Harris, page 21, it is proved that a variety is irreducible if and only if its smooth locus is connected.