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

Let $X$ be a smooth projective variety over a field $k$ of characteristic zero. Consider the hypercohomology in degree one of the logarithmic de Rham complex

$\mathbb{H}^1(X, \mathcal{O}_X^\ast \to \Omega^1_X \to \Omega^2_X \to \cdots)$

where the first arrow $d\log: \mathcal{O}_X^\ast \to \Omega^1_X$ sends $f$ to $\frac{f'}{f}$ (which is closed).

Question: Why is this isomorphic to $H^0(X, \Omega^1_X /d\log \mathcal{O}_X^\ast)$?

share|improve this question

1 Answer 1

I doubt that this is correct. What is true is that $\mathbb{H}^1$ is isomorphic to $H^0(X,\Omega ^1_{X,c}/d\log \mathcal{O}_X^*)$, where $\Omega ^1_{X,c}$ is the sheaf of closed 1-forms. I doubt that this is the same as what you write.

Here is why. For any complex of sheaves $K$ on $X$ there is a spectral sequence with $E^{pq}_2=H^p(X,\mathcal{H}^q(K))$ converging to the hypercohomology of $K$. This gives a low degree exact sequence $$0\rightarrow H^1(X,\mathcal{H}^0(K))\rightarrow \mathbb{H}^1(X,K)\rightarrow H^0(X,\mathcal{H}^1(K))\rightarrow H^2(X,\mathcal{H}^0(K))\ .$$ In your case $\mathcal{H}^0(K)$ is the constant sheaf $k^*$, hence it has trivial (Zariski) $H^i$ for $i\geq 1$; $\mathcal{H}^1(K)$ is $\Omega ^1_{X,c}/d\log \mathcal{O}_X^*$. This gives the above isomorphism.

share|improve this answer
    
Thanks abx! So if $X$ is a curve what I wrote is correct, but in general you need to take closed forms. –  conn754 Dec 24 '13 at 7:36

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.