Do finite groups acting on a ball have a fixed point? - MathOverflow

Do finite groups acting on a ball have a fixed point?

2011-10-18T22:46:04Z

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.

Answer by Francesco Polizzi for Do finite groups acting on a ball have a fixed point?

2011-10-18T23:27:39Z

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 on 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, see instead the paper by Parris Finite groups without fixed-point-free actions on a disk, Michigan Math. J. Volume 20, Issue 4 (1974), 349-351.

Answer by Allan Edmonds for Do finite groups acting on a ball have a fixed point?

2011-10-18T23:36:51Z

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.

Answer by Peter Arndt for Do finite groups acting on a ball have a fixed point?

2011-10-18T23:37:40Z

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...