User dung nguyen - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:46:45Z http://mathoverflow.net/feeds/user/4275 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51290/calculating-normal-bundle Calculating normal bundle Dung Nguyen 2011-01-06T08:40:24Z 2011-01-06T13:19:27Z <p>I just realize that even though I know what normal bundes are, I dont know how to compute them. The main objective is to show that a ration curve C on a quintic threefold doesnt move. If C is a line, then its normal bundle in the ambient space is $\mathcal O^{\oplus 3}(1)$. If we know that C is rigid on the quintic threefold, then its normal bundle, being a subundle of $\mathcal O^{\oplus 3}(1)$, must be $\mathcal O^{\oplus 2}(-1)$. But how do we prove this? What about higher degree rational curves?</p> http://mathoverflow.net/questions/48607/proving-that-a-map-is-a-morphism Proving that a map is a morphism Dung Nguyen 2010-12-08T05:27:13Z 2010-12-09T12:06:42Z <p>Example : Consider the (open, not compactified) moduli space of stable maps $\mathcal M_g(1,d)$ of maps of smooth curves of genus $g$ to $\mathbb P^1$. To each map we associate its branch divisor, which is an element of $Sym^r(\mathbb P^1)$. Then we should have a morphism from $\mathcal M_g(1,d)$ to $Sym^r(\mathbb P^1)$. How do we prove this?</p> <p>In general, if we have a family of objects $A$ and for each $a \in A$ we can choose an element $b \in B$ that "depends continuously" on $a$. How do we prove that we have established a morphism $A \to B$? What is the general method to do this?</p> <p>Note about the example: Extending the morphism to the Kontsevich compactification of the moduli space was the main objective of a paper by Fantechi-Pandharipande. But I couldn't filter out the proof that the map is a morphism, or the proof wasn't there. </p> http://mathoverflow.net/questions/39822/quasi-finite-separated-but-not-finite-morphism Quasi-finite + separated but not finite morphism Dung Nguyen 2010-09-24T04:47:50Z 2010-09-24T22:32:05Z <p>What is an interesting example of that? Things like $Spec(K) \to Spec(L)$ do not count cause they are not interesting.</p> http://mathoverflow.net/questions/26112/examples-of-the-moduli-space-of-x-giving-facts-about-a-certain-x/26238#26238 Answer by Dung Nguyen for Examples of the moduli space of X giving facts about a certain X Dung Nguyen 2010-05-28T07:31:36Z 2010-05-28T07:31:36Z <p>Another example also in the flavor enumerative geometry: by considering the (Deligne- Mumford compactification of) moduli space of elliptic curves, we see that the point at infinity represents the genus one curve with one node, while each other point represents curve with fix j-invariant. The moduli space now look somewhat like $\mathbb{P}^1$ (except that it is not, for some stacky reasons, but for current purpose, we can pretend that it is ) and we can view the nodal curve as having $j = \infty$. Since any two points on $\mathbb{P}^1$ are equivalent as divisors , the nodal curve is "equivalent" to any other curve with fix j-invariant. ( except when j is 0,1,1728)</p> <p>This property translates well into the Kontsevich moduli space of stable maps of elliptic curves to say, $\mathbb{P}^2$. The locus of maps of nodal curves are equivalent to the locus of maps of any curves with fix j-invariant as divisors, hence they should have the same enumerative invariants, because after all, enumerative invariants are just intersection numbers on the moduli space of maps. Thus, the problem of counting elliptic curves with fix j-invariant is the same as counting nodal curves, and in $\mathbb{P}^2$, nodal curves are the same as rational curves ( as rational plane curves necessarily have nodes if the degree is more than 2 ). This argument was used by Pandharipande to compute the characteristic numbers of plane elliptic curves with fix j-invariant.</p> http://mathoverflow.net/questions/19622/cartier-divisor-on-an-open-subscheme-whose-complement-is-of-codim-2 Cartier divisor on an open subscheme whose complement is of codim 2 Dung Nguyen 2010-03-28T14:42:31Z 2010-03-28T15:12:13Z <p>1- If a Cartier divisor is defined away from a codim 2 closed subset, when can we say that this must be a restriction of a Cartier divisor on the whole scheme?</p> <p>2- If two Cartier divisor agree away from a codim 2 closed subset, when can we say that they must be the same?</p> http://mathoverflow.net/questions/18543/implement-intersection-products Implement intersection products Dung Nguyen 2010-03-18T03:05:01Z 2010-03-18T14:48:34Z <p>I am doing a counting problem, and it comes to compute intersection products ( Chow ring ) on some varieties. Is there any computer algebra that deals with this?</p> http://mathoverflow.net/questions/16963/o-x-module-with-support-z-subset-x-vs-o-s-module O_X module with support Z \subset X vs O_S module? Dung Nguyen 2010-03-03T10:17:48Z 2010-03-03T10:32:55Z <p>Given a $O_X$ module $\cal F$ whose support is a closed subscheme $Z \subset X$. Under what conditions can we say that $\cal F$ is an $O_S$ module ( how far off is $\cal F$ an $O_S$ module ? )</p> http://mathoverflow.net/questions/16566/etale-fundamental-group-and-etale-cohomology-of-curves etale fundamental group and etale cohomology of curves Dung Nguyen 2010-02-27T00:29:56Z 2010-02-27T00:44:41Z <p>Given a curve $C$. Is there any relation between the etale fundamental group $\pi_1(C)$ and the first etale cohomology of the constant sheaf , say $Z/nZ$, on $C$ ?</p> <p>For example, if $C$ is a complex curve, then the singular cohomology $H^1(C,Z)$ is the dual of the topological fundamental group divided by the commutators ( which is the same as Hom$(\pi_1(C),Z) )$.</p> <p>So it seems that there should be some relation between Hom$(\pi_1(C),Z/nZ)$ and $H^1(C,Z/nZ)$ in the etale case, but how?</p> http://mathoverflow.net/questions/43755/when-should-you-and-should-you-not-share-your-mathematical-ideas/43958#43958 Comment by Dung Nguyen Dung Nguyen 2011-01-09T22:19:30Z 2011-01-09T22:19:30Z That's very funny! http://mathoverflow.net/questions/39822/quasi-finite-separated-but-not-finite-morphism Comment by Dung Nguyen Dung Nguyen 2010-10-09T02:44:02Z 2010-10-09T02:44:02Z It does not have to be a finite extension. http://mathoverflow.net/questions/39822/quasi-finite-separated-but-not-finite-morphism/39911#39911 Comment by Dung Nguyen Dung Nguyen 2010-10-09T02:41:45Z 2010-10-09T02:41:45Z I should have added &quot;surjective&quot;, but BCnrd's comment answered my question.