Suppose that $G$ is a finite group, acting via homeomorphisms on $B^n$, the closed $n$-dimensional ball. Does $G$ have a fixed point?

A fixed point for $G$ is a point $p \in B^n$ where for all $g \in G$ we have $g\cdot p = p$. Notice that the answer is "yes" if $G$ is cyclic, by the Brouwer fixed point theorem. Notice that the answer is "not necessarily" if $G$ is infinite. If it helps, in my application I have that the action is piecewise linear.

First I thought this was obvious, then I googled around, then I read about Smith theory, and now I'm posting here.

  • $\begingroup$ Kwasik and Schultz have a paper on the arXiv where they give the example of $A_5$ acting on the Poincare Dodecahedral Space, and the action has precisely one fixed point. So the action on the punctured sphere (a homology ball) has a single fixed point. So maybe the answer to your question is yes, and perhaps even there's some homotopy spheres that have finite groups acting with a single fixed point. $\endgroup$ Oct 18, 2011 at 23:31
  • $\begingroup$ Er, it's not on the arXiv, but it is on JStor. $\endgroup$ Oct 18, 2011 at 23:33
  • 5
    $\begingroup$ If the answer were yes, then the Nielsen realization problem would have been much easier I think (given Thurston's action of Mod(g) on the compactification of Teichmuller space). en.wikipedia.org/wiki/Nielsen_realization_problem $\endgroup$
    – Ian Agol
    Oct 19, 2011 at 0:09

3 Answers 3


The answer is no.

A fixed point free action of the finite group $A_5$ on a $n$-cell was constructed by Floyd and Richardson in their paper An action of a finite group on an n-cell without stationary points, Bull. Amer. Math. Soc. Volume 65, Number 2 (1959), 73-76.

For some non-existence results, you can see the paper by Parris Finite groups without fixed-point-free actions on a disk, Michigan Math. J. Volume 20, Issue 4 (1974), 349-351.

  • 1
    $\begingroup$ You mean the answer is "no" (i.e. fixed point does not always exist). $\endgroup$
    – GH from MO
    Oct 18, 2011 at 23:32
  • $\begingroup$ Very nice. Thank you for the reference! $\endgroup$
    – Sam Nead
    Oct 19, 2011 at 16:04

Bob Oliver classified the finite groups that act without a global fixed point on some sufficiently high-dimensional disk. The conditions are somewhat complicated to state. But for finite abelian groups the conclusion is that such a group acts without fixed points on some disk if and only if it has three or more non-cyclic Sylow subgroups. Here's a link to the original announcement of his result.


The answer is "yes" (it has a fixed point) if the action is affine, i.e. if it satisfies for all $g \in G, x,y \in B^n$ and all $0 \leq t \leq 1$: $$g(tx+(1-t)y)=tgx+(1-t)gy$$.

In that case one can construct a fixed point by taking an $x \in B^n$ and averaging over its $G$-orbit: $$p:=\frac{1}{|G|}\Sigma_{g \in G}\ gx$$ By convexity of $B^n$ the point $p$ is again in $B^n$ and by the affineness of the action $p$ is indeed afixed point. Linear actions are of course affine, now with your piecewise linear action you have to see whether you can find an orbit which falls into a linear piece, for example.

The groups which allow the above kind of argument are called "amenable groups", as I just learned on monday...

  • 17
    $\begingroup$ Alternatively, one can take the center of the ball. $\endgroup$
    – user5810
    Oct 18, 2011 at 23:48
  • 1
    $\begingroup$ Oh my god, quite right :-) ! $\endgroup$ Oct 19, 2011 at 0:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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