11
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ @potentiallydense I may be the potentially dense one. :) Thanks for pointing that out. $\endgroup$ – Alex Youcis Jul 25 '15 at 14:44
  • 1
    $\begingroup$ I strongly suspect that the main question is false, but I'll need to think more about it. There is a typo in 2 by the way. 2 and 3 are open as far as I know. Although there exists smooth projective varieties with non residually finite fundamental groups (Toledo,...). $\endgroup$ – Donu Arapura Jul 25 '15 at 15:10
  • 2
    $\begingroup$ Question 2 is reminiscent of the Malcev-Grothendieck theorem : if the étale fundamental group of a smooth complex projective variety is trivial, then there is no non-trivial bundle with a flat connection. See the introduction of arxiv.org/abs/1112.4603 . $\endgroup$ – Niels Jul 25 '15 at 20:24
8
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It's kind of a shame that the name "good group" seems to have stuck. Given the large community of mathematicians here, I wonder if can vote to change it to something more else. How about "Serre group"? $\endgroup$ – Donu Arapura Jul 26 '15 at 12:20
  • $\begingroup$ I agree. I think I've seen "good group in the sense of Serre" somewhere before. $\endgroup$ – Piotr Achinger Jul 26 '15 at 12:24
  • $\begingroup$ I should've just looked in your thesis. :) $\endgroup$ – Alex Youcis Jul 26 '15 at 14:53
11
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think you need to use the orbifold fundamental group of the moduli space (fundamental group of the stack), correct? $\endgroup$ – Piotr Achinger Jul 26 '15 at 9:31
  • 1
    $\begingroup$ The point of going to level $\ge 3$ is that $\Gamma(n)$ acts without fixed points, so you can use the usual $\pi_1$. $\endgroup$ – Donu Arapura Jul 26 '15 at 11:37
  • $\begingroup$ Thanks for the nice explicit example Donu! By the way, you mentioned above that you thought that 2,3 were open. Do you still believe that? I haven't been able to find any explicit answers online, but that doesn't mean much. There is the book 'Fundamental Groups of Compact Kahler Manifolds', but I don't feel like reading hundreds of pages, and can't find a super relevant proposition. Thanks again! $\endgroup$ – Alex Youcis Jul 26 '15 at 14:54
  • 1
    $\begingroup$ Yes, I believe 2 and 3 are still open. Serre, in one of his books, asked explicitly whether the Higman group occurs as the fundamental group of a smooth projective variety. If so, it would provide a counterexample to 2. I'm not aware of any progress on this. $\endgroup$ – Donu Arapura Jul 26 '15 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.