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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.

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  • $\begingroup$ How about Appendix A of math.princeton.edu/~rahulp/mpt.pdf ? $\endgroup$
    – Timo Schürg
    Dec 20, 2011 at 8:40
  • $\begingroup$ 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. $\endgroup$
    – Simon Rose
    Dec 20, 2011 at 17:58
  • $\begingroup$ 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? $\endgroup$
    – Timo Schürg
    Dec 20, 2011 at 20:33
  • $\begingroup$ 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! $\endgroup$
    – Simon Rose
    Dec 20, 2011 at 21:08
  • $\begingroup$ 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. $\endgroup$
    – Timo Schürg
    Dec 20, 2011 at 22:15

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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.

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