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Suppose $\mathcal H$ is the upper half plane, and $\Gamma$ is an arithmetic subgroup of $\operatorname{PSL}_2(\mathbb Z)$, I want to ask can we interpret the fundamental group of $\mathcal H/\Gamma$ by some modification of $\Gamma$?

I know if we remove the elliptic points then the $\pi_1$ is exactly $\Gamma$. But I don't know anything (concrete) about $\pi_1(\mathcal H/\Gamma)$.

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  • $\begingroup$ There is a finite index, torsion-free subgroup $\Sigma$ of $\Gamma$ and $\pi_1(\mathcal{H}/\Sigma)$ is $\Sigma$. On the other hand, for $\Gamma=PSL_2(\mathbb{Z})$ the fundamental group of the quotient is trivial. The problem is about finite branched coverings. As you see, they might kill fundamental groups. $\endgroup$
    – Antonius
    Commented Nov 18 at 5:52
  • $\begingroup$ When you write $\pi_1$ you mean the fundamental group of the orbifold right? $\endgroup$
    – Asaf
    Commented Nov 18 at 12:07
  • $\begingroup$ Please use a high-level tag like "at.algebraic-topology". I added this tag now. Regarding high-level tags, see meta.mathoverflow.net/q/1075 $\endgroup$
    – GH from MO
    Commented Nov 18 at 14:00
  • $\begingroup$ @Asaf No, just the manifold. $\endgroup$
    – Richard
    Commented Nov 18 at 14:10
  • $\begingroup$ But when there are cone points, this quotient is not a manifold... $\endgroup$
    – Asaf
    Commented Nov 18 at 14:16

1 Answer 1

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  1. The statement

I know if we remove the elliptic points then the $\pi_1$ is exactly $\Gamma$.

is wrong. You can see this by considering $\Gamma=PSL(2,\mathbb Z)$.

What you probably meant to say is that if $\Gamma$ is torsion-free, then $\pi_1(H^2/\Gamma)\cong\Gamma$.

  1. It is a very general theorem due to Armstrong: Let $X$ be a path connected, simply connected, locally compact metrizable space and $\Gamma\times X\to X$ a proper discrete group action. Then $\pi_1(X/\Gamma)$ is isomorphic to the quotient of $\Gamma$ by its normal subgroup normally generated by elements which have fixed points in $X$. (In your case, these are elliptic elements of $\Gamma$.) From this, you get $\pi_1(H^2/\Gamma)$, where $\Gamma$ is any Fuchsian group.

Armstrong, M. A., The fundamental group of the orbit space of a discontinuous group, Proc. Camb. Philos. Soc. 64, 299-301 (1968). ZBL0159.53002.

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  • $\begingroup$ Sorry, what's your definition of $\pi_1(H^2/\Gamma)$? (What does superscript "2" mean) $\endgroup$
    – Richard
    Commented Nov 18 at 14:12
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    $\begingroup$ @Richard: $H^2$ (or $\mathbb H^2$) is the usual notation for the hyperbolic plane. $\endgroup$ Commented Nov 18 at 14:15

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