Let $X$ be a connected qcqs scheme. We say that $X$ is a (étale) $K(\pi,1)$ if for every locally constant constructible abelian sheaf $\mathscr{F}$ on $X$ and every geometric point $\overline{x}$ the natural morphisms
$$\mathrm{H}^i(\pi_1(X,\overline{x}),\mathscr{F}_x) \to \mathrm{H}^i_\text{ét}(X,\mathscr{F})$$
are isomorphisms for all $i\geq 0$. (See section 4 on P. Achinger's Wild Ramification and $K(\pi,1)$ spaces for a quick review on this.)
There's a natural de Rham analog of this story. Let $X$ be a smooth algebraic variety over a characteristic zero field $k$. The category $\textsf{DE}(X/k)$ of vector bundles on $X$ with integrable connection is tannakian. In particular, a fiber functor $\omega$ induces an equivalence of categories $\omega:\textsf{DE}(X/k)\to \textsf{Rep}_\text{fd}(\Gamma)$ for some linear algebraic group $\Gamma$. (The index fd means finite-dimensional representations.) Then, we say that $X$ is a (de Rham) $K(\pi,1)$ if for every vector bundle with integrable connection $\mathscr{E}$ and every fiber functor $\omega$ we have an isomorphism
$$\mathrm{H}^i(\Gamma,\omega(\mathscr{E}))\simeq\mathrm{H}^i_\text{dR}(X,\mathscr{E}),$$
where the cohomology on the left is algebraic group cohomology (as in chapter 15 of Milne's book Algebraic Groups), for all $i\geq 0$.
I can prove that (as in the étale context) the isomorphism above always holds for $i=0$ but I know nothing more about this story. What are some classes of algebraic varieties that are de Rham $K(\pi,1)$'s? Is a variety an étale $K(\pi,1)$ if and only if it's a de Rham $K(\pi,1)$? (Or perhaps there's only one implication?)
In general, what's known about de Rham $K(\pi,1)$'s?
Best Answer
I had a derelict project (joint with Javier Fresan) trying to study this notion. We didn't prove much and we got stuck with the more interesting questions.
The isomorphism you wrote always holds for $i=1$. An easy way to see this is to compare both sides to extensions of $\mathcal{O}$ by $\mathcal{E}$.
If $X$ is proper (or if we restrict to regular connections), the comparison question becomes group-theoretic. Namely, we have a finitely presented group $\Gamma = \pi_1^{\rm top}(X)$, its profinite completion $\widehat{\Gamma} = \pi_1^{\rm et}(X)$, and its pro-algebraic completion $\Gamma_{\rm alg} = \pi_1^{\rm dR}(X)$. Cohomology comparison results (Deligne LNM 163 for de Rham, SGA4 for etale) give you isomorphisms $$ H^i_{\rm dR}(X, \mathcal{E}) \simeq H^i(X, \mathbf{V}) $$ and $$ H^i_{\rm et}(X, \mathcal{L}) \simeq H^i(X, \mathcal{L}) $$ where $\mathcal{E}$ is a regular connection corresponding to the local system $\mathbf{V}$, and where $\mathcal{L}$ is an etale local system of (pro)finite abelian groups (and also the corresponding Betti local system). The comparison maps from group cohomology to sheaf cohomology and the above maps, together with similar comparison maps on the level of group cohomology, give a $2\times 3$ commutative diagram that I don't remember how to draw on MO. You might be able to deduce something from this diagram. Recall that $\Gamma$ is "good" if its group cohomology (with coefficients in a finite module) agree with the group cohomology of $\widehat{\Gamma}$. By analogy, let us call $\Gamma$ "cute" if the same holds for every $\mathbf{C}$-representation, with $\widehat{\Gamma}$ replaced with $\Gamma_{\rm alg}$. With this you may be able to conclude something like: if $\Gamma$ is cute and good, then $X$ is an etale $K(\pi,1)$ if it is a de Rham $K(\pi, 1)$.
So, for example, I think abelian varieties and curves of genus $>0$ are $K(\pi,1)$ in all three senses.
If $X$ is not proper, and you allow irregular connections, then the de Rham fundamental group is nasty while the etale one is reasonable, and we expect no connection.
For $\mathbf{A}^n$, one can use the fibration trick (Deligne-Laumon) as in characteristic $p$, and I believe we can conclude that $\mathbf{A}^n$ is a de Rham $K(\pi, 1)$.
I have no idea what happens for arbitrary smooth affines. At least, the trick of reducing everything to $\mathbf{A}^n$ using finite etale maps completely fails.