# comparison theorem for connections with regular singularities

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?

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.