Let $M$ be a Riemannian manifold such that its isometry group $G=\textrm{Iso}(M)$ is a Lie group, and let $\Gamma$ be a subgroup of $G$.

1) What does the phrase "$\Gamma$ is a cocompact group of isometries of $M$" mean? Does it mean that the quotient space $M/\Gamma$ is compact, or does it mean that the coset space $G/\Gamma$ is compact ?


2) Is there an example as above in which $M/\Gamma$ is compact and $G/\Gamma$ is not, or viceversa ?


3) Is there any difference between "$\Gamma$ is a (discrete) cocompact group of isometries of $M$" and "The (discrete) group $\Gamma$ acts on $M$ cocompactly by isometries" and "The action $\Gamma\times M \rightarrow M$ is cocompact, where $\Gamma$ is a (discrete) group of isometries" ?

  • $\begingroup$ G/Gamma is not necessarily a group. In any case there is no distinction if Gamma has compact stabilizers or something like that, right? $\endgroup$ – Qiaochu Yuan Feb 17 '11 at 15:09
  • $\begingroup$ Right: G/Gamma just homogeneus space. Edited accordingly. $\endgroup$ – Qfwfq Feb 17 '11 at 15:43
  • $\begingroup$ Ah, when I made the second statement I was assuming G acts transitively. $\endgroup$ – Qiaochu Yuan Feb 17 '11 at 16:30
  • $\begingroup$ "Quotient space is compact" is correct. Used in this way it is at least standard terminology in equivariant topology and is defined in this sense for example in nearly all books by, e.g., Lück and for example on p. 5 defined this way in Mislin-Valette's book "Proper group actions and the Baum-Connes conjecture". I'd say that all statements in 3) express the same thing. $\endgroup$ – Daniel Pape Feb 17 '11 at 17:08

1) from wikipedia "In mathematics, an action of a group G on a topological space X is cocompact if the quotient space X/G is a compact space."

2) Let $\bar{M}$ be a noncompact manifold, $\Gamma$ its fundamental group, and $M$ its universal cover. Given a riemannian metric on $\bar{M}$, we can lift the metric to $M$ and the group $\Gamma$ acts as a group of isometry on $M$. For a generic metric on $\bar{M}$, there is no other isometries on $\bar{M}$ than the one provided by $\Gamma$. In which case $G/\Gamma$ is trivial (hence compact) whereas $M/\Gamma$ is isometric to $\bar{M}$, hence noncompact. As an example, take an euclidean cylinder, give a good (asymetrical) kick in it, unroll it and you are done.

Commenting on Scatt Carnahan answer, the example of the hyperbolic upper half plane is misleading, because its isometry group is transitive. This is exceptional in the world of riemannian geometry. Just take some non compact quotient of the upper half plane, say $H/\Gamma(2)$, make a small bump on the quotient, lift the resulting metric, et voila you have killed most isometries on $H$ with respect to this new metric.


I believe the answer to question 1 is that "cocompact" refers to $M/\Gamma$. At least, this is what I see when $M$ is the complex upper half-plane.

For question 2, if $M$ is finite dimensional, the stabilizer of a point in $M$ is necessarily a compact group of automorphisms, since it is made up of orthogonal transformations. This means if $M/\Gamma$ is compact, then so is $G/\Gamma$. On the other hand, you can choose $M$ noncompact with trivial automorphism group as a case where $G/\Gamma$ is compact and $M/\Gamma$ is not. If you want an example where $G$ is nontrivial, take a hyperboloid with one sheet in $\mathbb{R}^3$, with $\Gamma$ a finite cyclic group of rotations.

For question 3, I think the three statements are equivalent. Again, my experience is restricted to the situation when $M$ is the complex upper half-plane.


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