most recent 30 from http://mathoverflow.net 2013-05-19T01:02:03Z http://mathoverflow.net/feeds/question/83906 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83906/trivial-obstructions-and-virtual-fundamental-classes Simon Rose 2011-12-19T22:25:38Z 2012-01-19T01:10:22Z

<p>Suppose $X$ is a DM stack, and let $E^\bullet$ be a perfect obstruction theory of $X$ such that the $E^{-1}$ term admits a trivial quotient/sub-bundle. Is it true that the virtual fundamental class $[X, E^\bullet]$ is zero?</p> <p>If $X$ is smooth, then this is true: In such a case, the virtual fundamental class is the top Chern class of the vector bundle $E^{-1}$, which is zero due to the trivial quotient/sub-bundle and the exact sequence $$0 \to \mathcal{O} \to E^{-1} \to coker \to 0$$ together with the multiplicative nature of $c_{top}$. Intuitively, you can use the trivial factor to "move" a section of this bundle away from the zero section.</p> <p>If $X$ is not smooth, then this argument doesn't work and we must use the intrinsic normal cone of Behrend and Fantechi to compute the virtual fundamental class. Instead of bundles, we obtain a cone $C(E^\bullet)$ contained in $(E^{-1})^\vee$, which we intersect with the zero section of $(E^{-1})^\vee$. In comparison with the smooth case, it seems like we should somehow be able to move the cone out of the zero section using the trivial portion of the bundle, but I don't see how to do this.</p> <p>Is there an easy argument showing that this class is zero? Are there extra conditions required?</p> <p><b>Edit</b>: The method which I have tried to no avail is to use Proposition 5.10 of Behrend-Fantechi. It states loosely that, given two obstruction theories $F$ and $F'$, and certain compatibility data between them, that $$ v^![X,F] = [X,F'] $$ However, I was not able to find a clear way to have this yield that my desired virtual class is zero.</p>

http://mathoverflow.net/questions/83906/trivial-obstructions-and-virtual-fundamental-classes/86057#86057 Simon Rose 2012-01-19T01:10:22Z 2012-01-19T01:10:22Z

<p>It turns out that this is true. In the paper "Localizing Virtual Cycles by Cosections" by Kiem and Li, they address the case where one has a surjection $Ob \to \mathcal{O}$. In the case of an injection $\mathcal{O} \to Ob$, one can produce via a diagram chase a corresponding surjection, which yields the claim.</p>