Let $U$ be a smooth variety over $\mathbb{C}$. We know that there exists a smooth compactification $X$ such that $X-U$ is a normal crossings divisor $D$ and that the de Rham cohomology of $U$ can be computed using the complex of sheaves with logarithmic differentials on $X$:
$$
H_{dR}^\ast(U)=\mathbb{H}^\ast(X, \Omega^\bullet_X(\log D))
$$
In general, one needs hypercohomology here but I am wondering if, when $U$ is affine, one simply has
$$
H^\ast_{dR}(U)=H^\ast(\Gamma(\Omega^\bullet_X(\log D))
$$ I know that this is true if one uses instead the de Rham complex $\Omega^\bullet_U$ to compute cohomology and I know that $j_\ast \Omega_U^\bullet$ and $\Omega_X^\bullet(\log D)$ are quasi-isomorphic (here $j: U \hookrightarrow X$). Is this enough to conclude?
Best Answer
Unfortunately, no. Let $X$ be an elliptic curve, $D=p$ a point, $U=X-D$ then one sees from topology that $\dim H^1_{dR}(U)=2$. However, your proposed answer gives $$H^1(\Gamma(\mathcal{O}_X)\to \Gamma(\Omega_X^1(\log D)) = H^1(\mathbb{C} \stackrel{0}{\to} \mathbb{C})=\mathbb{C}$$ To understand what goes wrong, note from the spectral sequence (is this OK?) we get another term $H^1(\mathcal{O}_X)$ which needs to be there.
Here are a few additional comments, which might help. Given a bounded complex $\mathcal{F}^\bullet$, there is a spectral sequence $$E^{pq}= H^q(T,\mathcal{F}^p)\Rightarrow \mathbb{H}^{p+q}(T,\mathcal{F}^\bullet)$$