User dung nguyen - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-24T14:46:45Zhttp://mathoverflow.net/feeds/user/4275http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/51290/calculating-normal-bundleCalculating normal bundleDung Nguyen2011-01-06T08:40:24Z2011-01-06T13:19:27Z
<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-morphismProving that a map is a morphismDung Nguyen2010-12-08T05:27:13Z2010-12-09T12:06:42Z
<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-morphismQuasi-finite + separated but not finite morphismDung Nguyen2010-09-24T04:47:50Z2010-09-24T22:32:05Z
<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#26238Answer by Dung Nguyen for Examples of the moduli space of X giving facts about a certain XDung Nguyen2010-05-28T07:31:36Z2010-05-28T07:31:36Z<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-2Cartier divisor on an open subscheme whose complement is of codim 2Dung Nguyen2010-03-28T14:42:31Z2010-03-28T15:12:13Z
<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-productsImplement intersection productsDung Nguyen2010-03-18T03:05:01Z2010-03-18T14:48:34Z
<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-moduleO_X module with support Z \subset X vs O_S module?Dung Nguyen2010-03-03T10:17:48Z2010-03-03T10:32:55Z
<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-curvesetale fundamental group and etale cohomology of curvesDung Nguyen2010-02-27T00:29:56Z2010-02-27T00:44:41Z
<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#43958Comment by Dung NguyenDung Nguyen2011-01-09T22:19:30Z2011-01-09T22:19:30ZThat's very funny!http://mathoverflow.net/questions/39822/quasi-finite-separated-but-not-finite-morphismComment by Dung NguyenDung Nguyen2010-10-09T02:44:02Z2010-10-09T02:44:02ZIt does not have to be a finite extension.http://mathoverflow.net/questions/39822/quasi-finite-separated-but-not-finite-morphism/39911#39911Comment by Dung NguyenDung Nguyen2010-10-09T02:41:45Z2010-10-09T02:41:45ZI should have added "surjective", but BCnrd's comment answered my question.