Relationship between étale and topological $K(\pi,1)$s

Question

I was trying to find a proof, or a counterexample to the claim that if $X/\mathbb{C}$ is connected smooth projective, then $X$ is a $K(\pi^{\mathrm{\acute{e}t}},1)$ if and only if $X^\mathrm{an}$ is a $K(\pi,1)$.

For me, $X/\mathbb{C}$ should be a $K(\pi^{\mathrm{\acute{e}t}},1)$ if for all LCC $\mathcal{F}$ the natural map

$$H^i(\pi_1^{\mathrm{\acute{e}t}}(X,\overline{x}),\mathcal{F}_{\overline{x}})\to H^i(X_{\mathrm{\acute{e}t}},\mathcal{F})$$

is an isomorphism.

I believe this is equivalent to $X$ having vanishing higher étale homotopy groups. So, this question naturally led me to the following questions:

1. Does there exist $X/\mathbb{C}$ smooth projective and some $i>1$ such that $\pi_i^\mathrm{\acute{e}t}(X)=0$ but $\pi_i(X^\mathrm{an})\ne 0$?
2. Does there even exist $X/\mathbb{C}$ smooth projective with $\pi_1^{\mathrm{\acute{e}t}}(X)=0$ but $\pi_1(X^\mathrm{an})\ne 0$?
3. Does there even exist a connected compact Kähler manifold $X$ such that $\pi_1(X)\ne 0$ but $\widehat{\pi_1(X)}=0$ (profinite completion)?

Any help with these questions would be much appreciated!

Answer 1

Let $G$ be a group, $\iota: G\to \hat G$ its pro-finite completion. We call a $G$ a good group (cf. J.-P. Serre Cohomologie galoisienne, 2.6) if for every finite $\hat G$-module $M$, the maps $$ \iota^* : H^q(\hat G, M) \to H^q(G, M) $$ are isomorphisms for all $q\geq 0$.

Some examples of good groups: (1) finite groups, (2) finitely generated free groups and finitely generated free abelian groups, (3) iterated extensions of groups of type (1) and (2), (4) braid groups (a special case of (3)),

Arithmetic groups are not good in general, e.g., as Donu Arapura commented, ${\rm Sp}(2n,\mathbb{Z})$ is not a good group for $n > 1$. It is not known whether the mapping class groups $\Gamma_{g ,n}$ (the orbifold fundamental group of $\mathcal{M}_{g, n}$) are good groups (cf. Lochak–Schneps 2006, par. 3.4).

Now, say $X$ is a connected scheme of finite type over $\mathbb{C}$. Consider the following three conditions:

1. The scheme $X$ is a $K(\pi, 1)$ scheme.

2. The topological space $X^{\rm an}$ is a $K(\pi, 1)$ space.

3. The fundamental group $\pi_1(X^{\rm an})$ is a good group.

Then (1)+(2) implies (3) and (2)+(3) implies (1). To prove this, consider an lcc sheaf $\mathcal{F}$ on $X$ and look at the commutative squares $$ \begin{array}{c} H^q(\pi_1(X^{\rm an}, x), \mathcal{F}_x) & \to & H^q(X^{\rm an}, \mathcal{F}) \\ \uparrow & & \uparrow & \\ H^q(\pi_1(X, x), \mathcal{F}_x) & \to & H^q(X, \mathcal{F}). \end{array} $$

Note that we do not get that (1)+(3) implies (2), as neither (1) nor (3) gives us any information about the cohomology of $X^{\rm an}$ with non-torsion coefficients. Still, we see that (1)+(3) implies that $X^{\rm an}$ is a "$K(\pi, 1)$ for local systems of finite groups”.

The same should hold for fundamental groups of Deligne–Mumford stacks. In particular, the open question whether $\Gamma_{g ,n}$ is a good group would be equivalent to the question whether the stack $\mathcal{M}_{g, n}$ is a $K(\pi, 1)$ in the algebraic sense. Similarly (as in Donu Arapura's answer), as the orbifold fundamental group of $\mathcal{A}_g$ is ${\rm Sp}(2g ,\mathbb{Z})$, while the orbifold universal cover of $\mathcal{A}_g$ is the Siegel upper-half space (which is contractible), the stack $\mathcal{A}_g$ ($g > 1$) gives a probable example of a smooth Deligne–Mumford stack which is a $K(\pi, 1)$ in the analytic sense but not in the algebraic sense.

Answer 2

Let me expand my comment. I found the reference I had in mind: Ihara and Nakamura, Some examples for Anabelian geometry in high dimensions. The moduli space of $g$ dimensional principally polarized abelian varieties with level $n\ge 3$ structures is a $K(\pi, 1)$ because the universal cover, which is the Siegel upper half plane, is contractible. In particular, its fundamental group $\Gamma(n)\subset Sp_{2g}(\mathbb{Z})$ has finite cohomological dimension. However, when $g>1$, they show that the profinite completion $\widehat{\Gamma(n)}$ has infinite cohomological dimension. So it cannot be a $K(\pi^{et},1)$ in the sense you gave.

Added To avoid any confusion, $\Gamma(n)\subset Sp_{2g}(\mathbb{Z})$ is the congruence subgroup of full level $n$, i.e. the kernel of the map to $Sp_{2g}(\mathbb{Z}/n)$. Since $n\ge 3$, this is known to act without fixed points on the Siegel upper half plane, so there is no need to worry about orbifolds above.