This is something I've never understood.

Let me first recall the classical case, in which you start with a smooth algebraic variety $X$ over $\mathbb{C}$. One has the algebraic de Rham complex $\Omega^\bullet_{X/\mathbb{C}}$ and its analytification $\Omega^\bullet_{X^{an}}$ on $X^{an}$. Then:

Theorem (Grothendieck). The morphism of complexes $\Omega^\bullet_X \to \Omega^\bullet_{X^{an}}$ induces isomorphisms at the level of hypercohomology $$\mathbb{H}(X, \Omega^\bullet_{X/\mathbb{C}}) \to \mathbb{H}(X^{an}, \Omega^\bullet_{X^{an}}),$$ the first one being with respect to the Zariski topology while the second is taken in the usual cohomology of the complex analytic variety $X^{an}$.

Now assume you are given an integrable connection $$ \nabla: \mathcal{F} \to \mathcal{F} \otimes_{\mathcal{O}_X} \Omega^1_X $$ on some locally free $\mathcal{O}_X$-module. This defines a de Rham complex $DR(\mathcal{F}, \nabla)$ and one can also consider the analytic version $DR(\mathcal{F}^{an}, \nabla^{an})$. There is still a map $$ \mathbb{H}^\ast(X, DR(\mathcal{F}, \nabla)) \to \mathbb{H}^\ast(X^{an}, DR(\mathcal{F}^{an}, \nabla^{an}) $$ but now very easy examples show that this won't be an isomorphism in general. That's where the assumption of regular singularities comes in.

Fix a good compactification $j: X \hookrightarrow \bar{X}$ (so $\bar{X}$ is smooth and the complement is a normal crossings divisor $D$). The $(\mathcal{F}, \nabla)$ has regular singularities if it can be extended to some coherent $\mathcal{O}_{\bar{X}}$-module $\bar{\mathcal{F}}$ equipped with a logarithmic integrable connection $$ \overline{\nabla}: \bar{\mathcal{F}} \to \bar{\mathcal{F}} \otimes \Omega^1_X(\log D) $$ Now the advantage is that one has the GAGA theorem at disposal, so $$ \mathbb{H}^\ast(\bar{X}, DR(\bar{\mathcal{F}}, \bar{\nabla})) \simeq \mathbb{H}^\ast(\bar{X}^{an}, DR(\bar{\mathcal{F}}^{an}, \bar{\nabla}^{an})) $$ for any such extension. The problem is of course that the natural restriction map $$ \mathbb{H}^\ast(\bar{X}, DR(\bar{\mathcal{F}}, \bar{\nabla})) \to \mathbb{H}^\ast(X, DR(\mathcal{F}, \nabla)) $$ need not be an isomorphism. So the question is:

Question 1: Under what assumptions is this map an isomorphism? How do you prove it, and how do you put everything together to prove an analogue of Grothendieck's theorem?

Question 2: Does this story have a variant with compact support?


1 Answer 1


The morphism $\mathbb{H}^*(X,DR(\mathcal{F},\nabla))\rightarrow \mathbb{H}^*(X^\mathrm{an},DR(\mathcal{F}^\mathrm{an},\nabla^\mathrm{an}))$ is always an isomorphism when $\mathcal{F}$ is regular. This is Theoreme 6.2 of Deligne's book http://www.springer.com/mathematics/analysis/book/978-3-540-05190-9.

The reason (explained in Chapter 2) is that in the regular case, denoting by $j:X\rightarrow \bar{X}$ the inclusion, natural map $DR(\bar{\mathcal{F}},\bar{\nabla})\rightarrow \mathbb{R}j_*DR(\mathcal{F},\nabla)=j_*DR(\mathcal{F},\nabla)$ is a quasi-isomorphism of complexes, and so we get $\mathbb{H}^*(\bar{X},DR(\bar{\mathcal{F}},\bar{\nabla}))\cong \mathbb{H}^*(X,DR(\mathcal{F},\nabla))$ by taking hypercohomology.

You have a similar isomorphism $\mathbb{H}^*(\bar{X}^\mathrm{an},DR(\bar{\mathcal{F}}^\mathrm{an},\bar{\nabla}^\mathrm{an}))\cong \mathbb{H}^*(X^\mathrm{an},DR(\mathcal{F}^\mathrm{an},\nabla^\mathrm{an}))$ on the analytic side, and combining this with the GAGA isomorphism in the proper case gives you the version of Grothendieck's theorem you want.

I'm not sure how compact support fits into the picture, though.

  • $\begingroup$ Hi Chris. Thanks for your answer. I think it's wrong though. I don't think your map in line 5 is always a quasi-isomorphism. It depends on the extension and that was exactly my question: for which extensions is true and why? $\endgroup$
    – dmos
    Commented Jun 28, 2013 at 6:04

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