Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

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.

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.

Is there an easy argument showing that this class is zero? Are there extra conditions required?

Edit: 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.

share|improve this question
    
How about Appendix A of math.princeton.edu/~rahulp/mpt.pdf ? –  Timo Schürg Dec 20 '11 at 8:40
    
Unfortunately (unless I am mis-reading something), this doesn't address the question that I have. The appendix in question talks about how to produce a reduced obstruction theory, but not why the original obstruction theory yields a trivial fundamental class. –  Simon Rose Dec 20 '11 at 17:58
    
An example: M a smooth scheme, E a vector bundle, s a section. Then the zero-set Z(s) carries a perfect obstruction theory, and a virtual class of dimension n-e, where n=dim(M), e=rk(E). Now add a trivial bundle to E. Then s+1 is a section of E+O, and Z(s+1)=Z(s). But the virtual dimension has dropped by 1, it is n-e-1. If you remove the factor, you get a virtual class of 1 dim larger. In the general case, it is not automatic that after removing the trivial factor, you still have an obstruction theory. Surjectivity on h^-1 can fail. If that holds, you'll increase the vir-dim by 1. Sound ok? –  Timo Schürg Dec 20 '11 at 20:33
    
I'm not sure I follow what you're saying. If you add a trivial bundle and look at the section $s + 1$ (which I interpret to mean $s$ on the $E$ part, and 1 on the trivial part, then $Z(s + 1) = \emptyset$, which was my intuitive description for the smooth case. Anyhow, I'm not concerned with whether or not the result when you remove the trivial factor is itself a perfect obstruction theory: in the linked paper it is made clear that that is not always the case. I just want my virtual fundamental class to be zero! –  Simon Rose Dec 20 '11 at 21:08
    
Sorry, my bad. The class is zero in my example, and should be zero in general too. Just that you can reduce the class to something non-zero doesn't imply it wasn't zero before. –  Timo Schürg Dec 20 '11 at 22:15

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.