# Finite groups admitting free isometric actions on round spheres

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?

-
Well, if the dimension of the sphere is even, then $\mathbb{Z}/2$ is the only nontrivial group acting freely. –  Mauricio May 9 '12 at 14:02
Mauricio, sure. I guess I formulated the question not quite precisely. –  aglearner May 9 '12 at 14:11
@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. –  Mauricio May 9 '12 at 14:29
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. –  aglearner May 9 '12 at 14:46
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 –  macbeth May 9 '12 at 15:56

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.

-
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. –  aglearner May 9 '12 at 23:07
Thanks for that link. I honestly don't know a different source to read about that stuff, sorry. –  Mauricio May 9 '12 at 23:52

EDIT 2: My original and the following revised answer were nonsense; I'll just leave this edited, and very incomplete "answer" dealing with even $n$, just in case it helps somebody avoid to repeat my mistakes :(.

The isometry group of the standard sphere $S^n$ is the orthogonal group $G:=O_{n+1}(\mathbb{R})$. Each element of $G$ is either a rotation, a reflection or the product of a rotation and a reflection. Every reflection fixes a hyperplane, hence also a point on the sphere. Thus a group $H$ acting freely and isometrically cannot contain reflections.

Now if $n$ is even (and hence $n+1$ odd), then every rotation of $G$ fixes a point. (Its eigenvalues have absolute value 1, and it must have at least one real eigenvalue, hence 1 or -1. But the complex eigenvalues come in pairs $\lambda,\overline\lambda$. Since rotations have determinant 1, we conclude that the eigenvalue $-1$ must occur an even time. Hence there is an eigenvector with eigenvalue 1.)

This leaves products of a reflection and a rotation (such as the antipodal map $x\mapsto -x$). One can then show that at most one such non-trivial element can be contained in $H$ (as otherwise, we would get reflections or rotations inside $H$), and in fact, only the antipodal map can occur as non-trivial element of $H$.

Thus only the trivial group and the cyclic group of order 2 can act freely and isometrically on $S^n$ for even $n$.

On the other hand, for odd $n$, more possibilities arise, e.g. any cyclic group admits a free isometric action on $S^n$ for odd $n$.

-
Max, really sorry, but it is not true that only cyclic group can act freely on spheres... For example, $SU(2)=S^3$. $SU(2)$ has finite non-Abelian subgroups, they act on $S^3$ freely by multiplication on left –  aglearner May 9 '12 at 15:31
But is that an isometric action? –  Max Horn May 9 '12 at 15:35
Sure, it is an isometric action. –  aglearner May 9 '12 at 15:41
It is not true that if $n>1$ then rotations fix points. There are counterexamples in all odd dimensions. For instance, the element of $SO(2n)$ whose nonzero entries are $n$ $2\times 2$ blocks along the diagonal, each some nonzero rotation $\begin{pmatrix} \cos t & -\sin t \cr \sin t & \cos t\end{pmatrix}$. –  macbeth May 9 '12 at 15:45
$\mathbb{Z}$ does act freely on $S^{2k+1}$ in the same way. This doesn't contradict Wolf's theorem -- the point is that this action is not properly discontinuous, so the quotient by this action is not a manifold. –  macbeth May 9 '12 at 18:25