Let $X$ be a quotient of the complex ball by an arithmetic group. How does the orbifold fundamental group of the complex points of $X$ compare to the étale fundamental group of $X$?
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4$\begingroup$ If the arithmetic group $\Gamma$ is torsion free, then the étale $\pi_1$ is the profinite completion $\hat\Gamma$. If you treat $X$ as a stack, then this would true in general. $\endgroup$– Donu ArapuraCommented May 3, 2021 at 16:46
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3$\begingroup$ Adding onto what @DonuArapura says, if you want the underlying normal projective variety and not the stack, you take the quotient $\Delta$ of $\Gamma$ by the subgroup generated by torsion associated with smooth components of the orbifold locus (it's standard that this is $\pi_1$ of the underlying space), then take the profinite completion of $\Delta$. (You need to be careful you aren't killing elements of finite order in the fundamental group of the underlying normal variety.) $\endgroup$– ToffeeCommented May 3, 2021 at 16:58
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2$\begingroup$ @Ramón For my first statement, I used the comparison between the usual topological fundamental group and etale group for a complex alg. variety, which is due to Grothendieck in SGA1. For the second, I'm not sure of a good reference. Perhaps you can look at papers of Noohi. $\endgroup$– Donu ArapuraCommented May 3, 2021 at 17:03
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2$\begingroup$ It's also maybe worth noting that the stack and underlying analytic space notions are very, very different. The profinite completion of a lattice in $\mathrm{PU}(n,1)$ is always very big and complicated, but there are examples of ball quotient orbifolds with underlying space $\mathbb{P}^2$, so the topological and algebraic fundamental groups of the underlying space are trivial. $\endgroup$– ToffeeCommented May 3, 2021 at 17:06
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1$\begingroup$ @Ramón It's the same as for the moduli space of elliptic curves as a quotient of the upper half plane by the modular group. The underlying analytic space is just $\mathbb{C}$, which is (topologically and algebraically) simply connected, but $\Gamma = \mathrm{PSL}_2(\mathbb{Z})$ is virtually a nonabelian free group, so it's profinite completion is huge. As a moduli space, it's the quotient of $\mathbb{P}^1$ minus $3$ points (elliptic curves with level $2$ structure), where there are no orbifold issues, by the action of the symmetric group $S_3 = \mathrm{PSL}_2(\mathbb{F}_2)$. $\endgroup$– ToffeeCommented May 3, 2021 at 19:52
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