Is there some sort of classification of finite groups $G$ such that for at least one $n$ the group $G$ admit a free isometric action on the standard sphere $S^n $of curvature 1? Are there some simple criteria that permit to check (in some particular cases) if a given group has such an action (for at least on $n$) or not?
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$\begingroup$ Well, if the dimension of the sphere is even, then $\mathbb{Z}/2$ is the only nontrivial group acting freely. $\endgroup$– MauricioCommented May 9, 2012 at 14:02
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$\begingroup$ Mauricio, sure. I guess I formulated the question not quite precisely. $\endgroup$– aglearnerCommented May 9, 2012 at 14:11
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2$\begingroup$ @aglearner: The formulation was precise, I think. But the classification of such finite groups when n is even is quite simple. Anyhow you might be interested in these notes. $\endgroup$– MauricioCommented May 9, 2012 at 14:29
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$\begingroup$ Mauricio, thanks for the link, it is very interesting indeed. Yet, this note deals with a more general, topological question. It would be nice to have some stronger restrictions on groups G coming from linear algebra. I just saw that I forgot to say that I was interested in isometric actions on $S^n$, I corrected the question accordingly. $\endgroup$– aglearnerCommented May 9, 2012 at 14:46
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2$\begingroup$ This problem was solved by Joseph Wolf in his book "Spaces of Constant Curvature" (1967). (That is, if I understand the problem correctly -- so it's equivalent to the classification of complete manifolds of constant positive curvature.) I don't remember the answer -- its statement is quite involved $\endgroup$– macbethCommented May 9, 2012 at 15:56
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1 Answer
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After all these comments, a possible answer to your question goes as follows.
If $n$ is even, then the only group that can act freely and isometrically on $S^n$ is $\mathbb{Z}/2$. One way to see this is as in Max's answer above, i.e. by looking at the behavior of eingenvalues of orthogonal matrices. Here is another reason: an action of $G$ on $S^n$ means that there is a group homomorphism $G\rightarrow\mbox{Homeo}(S^n)$. But any homeomorphsim has degree $\pm 1$. So we get a homomorphism $G\rightarrow\mathbb{Z}/2$. But the action has no fixed points, then the degree of the image of every non trivial element in $G$ is $(-1)^{n+1}=-1$, therefore the map has a trivial kernel and therefore it is an isomorphism.
As macbeth already pointed out, when $n$ is odd, the problem is not so easy. In that case, we consider $S^n$ as the universal cover of a complete Riemannian manifold $M$ with constant sectional curvature. A simple argument using lifting properties of covering spaces shows that there is an isometry between $M$ and $S^n/G$ (whenever the action is free and properly discontinuous). So the problem of finding the subgrups with that particular action on the sphere is the same as the classification of complete Riemannian manifolds with constant sectional curvature (=1). That is done in Wolf's book Spaces of Constant Curvature.
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$\begingroup$ Mauricio, thank you for the answer. Unfortunately Wolf's book is not available online. Thanks to the link that you gave I found an online reference with an answer projecteuclid.org/DPubS/Repository/1.0/… I wonder if there is some user-friendly exposition of Wolf's classification. Or this is the "best" source to read. $\endgroup$ Commented May 9, 2012 at 23:07
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$\begingroup$ Thanks for that link. I honestly don't know a different source to read about that stuff, sorry. $\endgroup$– MauricioCommented May 9, 2012 at 23:52