# Which groups act freely on $S^n$?

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 $n > 1$?

 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.

• I have added some partial results with my answer below, as the question would be too long if I had placed them there. – Balarka Sen Dec 14 '15 at 20:27
• Not sure but isn't it more suitable to mathoverflow? Seems not like a standard textbook-exercise, but I may be wrong.. – Peter Franek Dec 14 '15 at 20:33
• @PeterFranek Could be. I'll let it stay here for a few days to see if it's hard. I'll cross-post to MO if I don't get any answers. – Balarka Sen Dec 14 '15 at 20:35
• @Peter: I think it's suitable on either site. The case for non-fixed $n$ ("what groups act on some sphere?") was famously resolved by a Wall in the 60s. Or at least, famous to the people it's famous to. – user98602 Dec 14 '15 at 20:36
• @BalarkaSen Finally you have posted a doubt... Although "out of topic comment" – Anubhav Mukherjee Dec 15 '15 at 11:28

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.

• Can you give a reference or proof of the fact that all abelian subgroups of $G$ are cyclic? +1. – Balarka Sen Dec 14 '15 at 21:06
• A reference is Theorem $2$ here. – Dietrich Burde Dec 14 '15 at 22:49
• The second paragraph is very close to being the classification of which groups act on (some) sphere: all it's missing is the condition that every subgroup of order $2p$ is cyclic. (That is, that the dihedral group $D_{2p}$ is not a subgroup.) – user98602 Dec 14 '15 at 23:25

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?