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Suppose $S$ is a non-compact and complete surface (2 dimensional smooth Riemannian manifold) of constant curvature. I am wondering if there exists a group $G$ which acts by isometries and properly discontinuously on $S$ such that $S/G$ becomes compact?! Are there maybe any reference where I can find results related to the above situation?

Best wishes

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    $\begingroup$ Such group does not exist quite often. For example, consider $\mathbb H^2$ - the hyperbolic plane. Suppose $\mathbb Z$ is acting on $\mathbb H^2$ without fixed points (i.e. it is parabolic, or hyperbolic). Then $S=\mathbb H^2/\mathbb Z$ does not admit an action that you would like to have. $\endgroup$
    – aglearner
    Dec 20, 2016 at 20:36
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    $\begingroup$ But that does not answer the question. For, on $\mathbb{H}^2$ there are indeed groups acting isometrically and properly discontinuously so that $\mathbb{H}^2/G$ is compact. Any compact, hyperbolic surface arises in this way. The question is asking, given $S$, does there exist a $G$ making $S/G$ compact. So exhibiting a $G$ such that $S/G$ is not compact does not answer the question. $\endgroup$
    – Paul Bryan
    Dec 20, 2016 at 20:53
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    $\begingroup$ Most open hyperbolic Riemann surfaces have only trivial automorphism. As isometries = conformal isomorphisms for such surfaces, the answer is no. $\endgroup$
    – Alexandre Eremenko
    Dec 20, 2016 at 21:10
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    $\begingroup$ This will happen iff $\pi_1(S)$, seen as a subgroup of the isometry group of $\tilde{S}\simeq \mathbb{H}^2$ (via the deck transformation action), has a discrete cocompact torsion free group which normalizes it. In this case, take $G$ to be the quotient group. Said differently, this is the case iff $\pi_1(S)$ is a normal subgroup in a surface subgroup of $\text{PSL}_2(\mathbb{R})$. $\endgroup$
    – Uri Bader
    Dec 20, 2016 at 21:10
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    $\begingroup$ constant curvature / isometries makes sense in a smooth Riemannian manifold, not in a smooth manifold. $\endgroup$
    – YCor
    Dec 20, 2016 at 22:41

1 Answer 1

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As Uri Bader says in the comments, covering-space theory implies that this happens if and only if $\pi_1S$ is a normal subgroup of $\pi_1\Sigma$, where $\Sigma$ is some compact surface.

The cases of positive and zero curvature are easy, so we may as well assume that $S$ and $\Sigma$ are of constant negative curvature. In this case, a theorem of Greenberg places strong restrictions on normal subgroups of surface groups.

Theorem (Greenberg): If $H$ is a finitely generated, normal subgroup of the fundamental group of a hyperbolic surface $\Sigma$, then $H$ is either trivial or of finite index

In particular, if $S$ is non-compact but not the hyperbolic plane, and $\pi_1S$ is finitely generated, then $S$ does not admit such a group action.

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  • $\begingroup$ Greenberg's theorem is apparently contained in [1960]. Or more directly Karass-Solitar [1973], in the form: "Fuchsian groups have the f.g.n. property". $\endgroup$
    – Francois Ziegler
    Dec 24, 2016 at 6:37
  • $\begingroup$ In fact, the proof of the theorem is very short: see YCor's comment on the answer to this MO question: mathoverflow.net/questions/193369/… . $\endgroup$
    – HJRW
    Dec 24, 2016 at 13:16
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    $\begingroup$ Very kapovich short! $\endgroup$
    – Francois Ziegler
    Dec 24, 2016 at 14:15

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