MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am working on some basic of Gromov-Witten theory and stuck in understanding obstruction bundle. Recall that a perfect obstruction theory on a scheme or stack $M$ due to Behrend and Fantechi is a moprhism $\phi:\mathcal{E}\rightarrow \tau_{\ge -1}L_M$ in $D^{[-1,0]}(M)$ satisfying some conditions. Taking the first cohomology of the dual of the two-term complex $\mathcal{E}$, we get so-called obscturction sheaf $Ob=h^1(\mathcal{E}^{\vee})$.

Assume now that $M=\overline{M}_{g,n}(X)$ for some smooth variety $X$ with the usual perfect obstruction theory on it. My question is, why the obstruction sheaf forms a vector bundle on $M$ when the moduli space $M$ is non-singular?

I know that the modul space $M$ is intuitively obtained by cutting out the deformation space by $\dim Ob$ many equations, but I am not really convinced by this argument.

share|cite|improve this question
up vote 6 down vote accepted

The point is that if the moduli space is non-singular, then the tangent sheaf $h^0(\mathcal{E}^\vee)$ is locally free and so the map $E^{-1}\to E^0$ must be of constant rank. This implies that the cokernel is also locally free. The difference between the dimensions of fibers of the tangent sheaf and the obstruction sheaf is always constant --- it is the expected dimension of the moduli space. So if the dimension of the fibers of the tangent sheaf doesn't jump, then neither do the fibres of the obstruction sheaf.

share|cite|improve this answer
Thank you for the answer, Jim. That make perfect sense. – H Zheng Dec 28 '12 at 18:05

Your Answer


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.