TITLE: Does any surface of constant curvature admit a cocompact group action?
QUESTION [7 upvotes]: 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

REPLY [14 votes]: 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.