19
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ see page 4 here for a cohomological criterion on G $\endgroup$ – Andres Mejia Mar 19 at 22:04
14
$\begingroup$

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?

| cite | improve this answer | |
$\endgroup$
13
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can you give a reference or proof of the fact that all abelian subgroups of $G$ are cyclic? +1. $\endgroup$ – Balarka Sen Dec 14 '15 at 21:06
  • 2
    $\begingroup$ A reference is Theorem $2$ here. $\endgroup$ – Dietrich Burde Dec 14 '15 at 22:49
  • $\begingroup$ 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.) $\endgroup$ – user98602 Dec 14 '15 at 23:25
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think Quaternions satisfies in these two criteria. So they act freely ..? $\endgroup$ – C.F.G Sep 5 at 12:29
  • 1
    $\begingroup$ @C.F.G Consider the $3$-manifold obtained from the cube $I^3$ by pasting opposing pairs of sides by a 90 degree twist; this in fact has fundamental group $Q_8$ (this is an exercise in Hatcher somewhere). So $Q_8$ is a finite $3$-manifold group, thus acts on $S^3$ freely. This manifold is a lot like the Poincare homology sphere, if you wish, except that's constructed with the dodecahedron instead of the cube. $\endgroup$ – Balarka Sen Sep 5 at 12:32
  • 1
    $\begingroup$ Alternatively one can construct it as follows; $V_4$ is a subgroup of $SO(3)$, and $Q_8$ is a central $\Bbb Z_2$-extension of $V_4$ so is a subgroup of $SU(2)$ (in fact it's $i, j, k$ acting on the unit quaternions). The action of $Q_8$ on $S^3 = SU(2)$ is by left multiplication. The manifold above is $SO(3)/V_4$, analogous to Poincare homology sphere, once again. $\endgroup$ – Balarka Sen Sep 5 at 12:37
  • $\begingroup$ @Qiaochuyuan: Hatcher (p.75) says that "$G$ contains at most one element of order 2". So together your second restriction it becomes: "$G$ contains at most one element of order 2 and if exist any it is unique and central." $\endgroup$ – C.F.G Oct 31 at 17:58
3
$\begingroup$

There are noncyclic finite groups that act freely as rotations of $\Bbb S^n$ for odd $n > 1$. These actions are classified quite explicitly in Joseph A. Wolf; Spaces of Constant Curvature 1984.

There is a theorem assert that

The group $\Bbb Z_p \times \Bbb Z_p$, where $p$ is prime, cannot act freely on any sphere.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I have given a proof of this theorem below in my answer; an alternative argument is that if $G$ is a finite group acting freely on $S^n$ then $G$ has periodic cohomology (although the converse is false, take $G = S_3$, see eg Dietrich Burde's answer). Good point that space forms of constant positive curvature are sphere quotients though, I will try to read the reference to see how a classification goes. One should note that there are wild topological actions of finite groups on spheres, eg, there is a free Z_2 action on S^7 not conjugate to the antipodal action, giving fake RP^7 quotients $\endgroup$ – Balarka Sen Apr 20 at 10:39
  • $\begingroup$ Sorry. I missed last paragraph of your answer. Anyway, the first part of your question is so interesting for me. (using degree theory). this is your argument or it is from a book? any reference? $\endgroup$ – C.F.G Apr 20 at 11:32
  • $\begingroup$ @BalarkaSen: $Z_2$ act on standard 7 sphere or exotic 7 sphere? $\endgroup$ – C.F.G Apr 20 at 11:34
  • $\begingroup$ That $\Bbb Z_2$ is the only group that can act on $S^n$ for even $n$ is standard. See for example Hatcher. The wild $\Bbb Z_2$-action on $S^7$ that I mentioned is constructed in Milnor-Hirsch, "Some curious involutions of spheres"; they do use Milnor's exotic $S^7$ to construct it, and the $\Bbb Z_2$-action is antipodal action on the $S^3$ fibers of the bundle $S^3 \to S^7 \to S^4$ in the construction of Milnor's 7-sphere $\endgroup$ – Balarka Sen Apr 20 at 12:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.