Which groups act freely on $S^n$?

Question

When $n$ is even, it is easy to classify groups which act freely on $S^n$ using degree theory: if $G$ acts on $S^n$, then associating to each element $g \in G$ the degree of the map obtained from multiplication by $g$, one gets a map $d : G \to \{\pm 1\}$. It is easy to verify this is a homomorphism. If $G$ acts freely, multiplication by any element $g$ is a fixed point free map, thus $d(g) = (-1)^{n+1} = -1$, making $d$ injective. The only nontrivial group which injects into $\Bbb Z/2$ is itself, so we're done.

However, a lot of groups act freely on $S^n$ for odd $n$. For example, $\Bbb Z/p$ acts on $S^3$ freely for all prime $p$ (so called lens space action). What do we know about such groups? Is it possible to classify them?

If $G$ is a finite group acting freely on $S^3$, then as free action of finite groups on Hausdorff spaces is properly discontinuous, quotient map $S^3 \to S^3/G$ is a covering projection. Thus $S^3/G$ is a closed 3-manifold with fundamental group $G$, hence $G$ is a closed 3-manifold group. On the other hand, if $G$ is a closed 3-manifold group, let $M$ be that manifold and $G$ must act on $\tilde{M}$ freely. But $\tilde{M}$ is a simply connected closed 3-manifold hence homeomorphic to $S^3$ by Poincare conjecture, so $G$ must act on $S^3$ freely.

Thus, finite groups acting freely on $S^3$ are precisely the finite closed 3-manifold groups. But what about infinite groups? Given an infinite group, how can we tell if it acts on $S^3$ or not? More generally, what about groups acting freely on $S^n$ for some fixed odd $n > 1$?

---

[edit] I am only interested in actions of discrete groups. Also, any sort of general remark (long enough to not fit as a comment) or partial answers (like the answers below) are welcome to me, you can post them as answers.

Answer 1

Here are some thoughts. If a finite group $G$ acts on $S^n$ freely, then it's automatically a covering space action as $S^n$ is Hausdorff. Thus, $\pi_1(S^n/G) \cong G$. Moreover, $\pi_i(S^n/G) \cong \pi_i(S^n) \cong 0$ for all $i < n$. $\pi_n(S^n/G) \cong \pi_n(S^n)$ which is infinite cyclic. Pick a map $S^n \to S^n/G$ representing the generator, and glue a $D^{n+1}$ along it to kill $\pi_n$. Kill $\pi_k$ for all $k > n$ similarly by gluing cells along maps representing generators of $\pi_k$ for each dimension $k$, to infinity if necessary.

One then obtains a $K(G, 1)$ from attaching just one $(n+1)$-cell to $S^n/G$. Thus, the group cohomology $H^{n+1}(G; \Bbb Z/p)$ is either $\Bbb Z/p$ or $0$ for any prime $p$ cellular cohomology.

As a corollary, for example, $\Bbb Z/2 \times \Bbb Z/2$ cannot act on any $S^n$ freely for any $n$, as by Kunneth formula

$$H^{n+1}(\Bbb Z/2 \times \Bbb Z/2; \Bbb Z/2) \cong \bigoplus_{i + j = n+1} H^i(\Bbb Z/2; \Bbb Z/2) \otimes H^j(\Bbb Z/2; \Bbb Z/2)$$

which is simply isomorphic to $(\Bbb Z/2)^{n+1}$, violating the previous obstruction. Similarly, $\Bbb Z/p \times \Bbb Z/p$ does not act on $S^n$ for any prime $p$. However, similar technique doesn't work for infinite groups. Also, can any obstruction be placed over higher group cohomologies?

Answer 2

You have already given answers yourself. I only have a two comments, with a recent reference, which you might have already seen.

If a finite group $G$ acts freely on a sphere then we know that all abelian subgroups of $G$ are cyclic, i.e., that $G$ has periodic cohomology, and that all elements of order $2$ are central. In particular, $G$ has at most one element of order $2$. For even-dimensional spheres there is only $C_2$, see also here.

For infinite groups, free actions of discrete groups have been studied a lot. A free action of a discrete group $G$ on an $n$-homotopy sphere $\Sigma(n)$ induces an action on $H^n(\Sigma(n),\mathbb{Z})$, i.e., an homomorphism $G\rightarrow Aut(H^n(\Sigma(n),\mathbb{Z}))$. For $G$ finite, and $n$ odd, this action is trivial. If the group $G$ is infinite there are more possibilities for the induced action of $G$, which makes it more difficult to characterise these induced actions. For a summary of some results and a certain classification see the recent preprint on Free and properly discontinuous actions of groups on homotopy $2n$-spheres.

Answer 3

This has almost but not quite been stated a few times, so to clear the air: the answer is known for finite groups, it is due to Madsen, Thomas, and Wall, and it says that a finite group $G$ acts freely on some sphere if and only if

1. all of the abelian subgroups of $G$ are cyclic; equivalently, the cohomology is periodic; equivalently, $\mathbb{Z}_p \times \mathbb{Z}_p$ does not occur as a subgroup for any prime $p$; and
2. every element of order $2$ is central.

The necessity of the first condition is due to Smith and the necessity of the second condition is due to Milnor. This is taken from the introduction to Alejandro Adem's Constructing and Deconstructing Group Actions.