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 $H^0(X, \Omega^1_X /d\log \mathcal{O}_X^\ast)$?


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.

  • $\begingroup$ Thanks abx! So if $X$ is a curve what I wrote is correct, but in general you need to take closed forms. $\endgroup$
    – conn754
    Dec 24, 2013 at 7:36

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