Let $M$ be a complete Riemannian manifold. My question is, under what conditions does $M$ admit a discrete group of isometries $\Gamma$ which acts properly discontinuously and cocompactly on $M$, that is, $M / \Gamma$ is a compact Riemannian manifold? If needed, one can assume that $M$ has negative (and even constant negative) curvature. This is mainly a reference request.
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The question is phrased very generally, so I'm not sure if the following is what you're looking for. However, my answer to Does any surface of constant curvature admit a cocompact group action? might be useful to you, since you are interested in the case of constant negative curvature. The result can be summarised as follows:
Theorem A: If $M$ is a 2-dimensional, complete, non-compact hyperbolic manifold and $\pi_1M$ is finitely generated then $M$ admits a discrete, cocompact group action only if $M$ is simply connected.
Since a complete, simply connected, hyperbolic manifold is isometric to the hyperbolic plane, this necessary condition is of course also sufficient.
In higher dimensions, comparable results are much more difficult and delicate. For instance, using the tameness theorem of Agol and Calegari--Gabai, one can conclude the following:
Theorem B: If $M$ is a 3-dimensional, complete, non-compact hyperbolic manifold and $\pi_1M$ is finitely generated then $M$ admits a discrete, cocompact group action only if $M$ is simply connected or homotopy equivalent to a closed surface.
In this case, the necessary condition is no longer sufficient: for instance, the quotient of hyperbolic 3-space by a surface group preserving an isometrically embedded copy of the hyperbolic plane cannot admit a cocompact action.
Above dimension 3, very little is known. For instance, it is an open question whether or not a closed hyperbolic 4-manifold can have a normal surface subgroup. (Note that the above results follow from classification theorems for normal subgroups of hyperbolic manifold groups.) However, you might be interested in the following result, which can be deduced from deep work of Paulin and others:
Theorem C: Let $M$ be any non-compact manifold of negative sectional curvature. If $M$ is homotopy equivalent to a closed negatively curved manifold of dimension greater than 2 then $M$ does not admit a properly discontinuous, cocompact group action by isometries.
I'm happy to provide further details if these kinds of results are what you are looking for.
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$\begingroup$ I am really interested. Do you know of any more sufficient conditions? Your Theorem B and Theorem C are necessary conditions. $\endgroup$– SMSCommented Aug 20, 2022 at 18:43
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$\begingroup$ In particular, I was thinking of the following little problem: can one find a closed Riemannian surface $M$ and a complete non-compact Riemannian covering space $N$ such that $M = N /\Gamma$, and $N$ has prescribed number of ends? Intuitively this seems doable, but at the moment, I cannot come up with a construction. $\endgroup$– SMSCommented Aug 20, 2022 at 19:25
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$\begingroup$ @SMS: Regarding sufficient conditions, the first non-trivial (hyperbolic) case is in dimension 3, and by Theorem B we may assume that $\pi_1M$ is a surface group. The question is then whether or not $\pi_1M$ is a geometrically finite or geometrically infinite Kleinian group. I believe (perhaps someone else can confirm) that all geometrically infinite hyperbolic surface groups can be embedded as normal subgroups of lattices, whence the relevant group action will exist. I think nothing is known above dimension 3. $\endgroup$– HJRWCommented Aug 20, 2022 at 22:33
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$\begingroup$ @SMS: Regarding your second comment, the Svarc--Milnor lemma implies that $N$ is quasi-isometric to $\Gamma$ and so, by a well known result of Hopf, must have 0, 1, 2 or $\infty$ ends. (And each of these cases is fairly easy to realise.) $\endgroup$– HJRWCommented Aug 20, 2022 at 22:36