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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.

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  • $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented Oct 18, 2011 at 23:31
  • $\begingroup$ Er, it's not on the arXiv, but it is on JStor. $\endgroup$
    – Ryan Budney
    Commented Oct 18, 2011 at 23:33
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    $\begingroup$ 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
    Commented Oct 19, 2011 at 0:09

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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.

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

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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...

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

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