# Does a compact contractible metric space have a point that is fixed by all isometries?

Let $$(X,d)$$ be a compact and contractible metric space. Let $$\operatorname{Isom}(X)=\{\phi\colon X\to X\}$$ be its group of isometries.

Question: Is there a point $$x\in X$$ fixed by all $$\phi\in\operatorname{Isom}(X)$$?

I am happy to assume some additional niceness conditions for $$X$$, enough to ensure that $$X$$ satisfies some fixed-point theorem, guaranteeing that every continuous map $$\phi\colon X\to X$$ has a fixed point (e.g. triangulable, locally contractible; see A version of Brower's fixed point theorem for contractible sets? for details). The emphasize is therefore on whether all isometries have a common fixed point.

This post is a refinement/generalization of Symmetries of contractable subsets of $\Bbb R^n$.

• Duplicate of a question answered here 9 years ago. May 14 at 12:03

There are finite groups that act smoothly on a disk without a global fixed point. You can arrange the metric to be isometric, e.g. via the Mostow-Palais embedding theorem, which equivariantly and smoothly embeds the disk into a Euclidean space where the group acts orthogonally, and taking the metric induced by Euclidean one. The full isometry group is potentially bigger, so it cannot fix a point.

• Other way to arrange the metric so that the action is isometric is, if $G$ is the group and $d$ is the usual metric in the disk, define $d'(x,y)=\sum_{g\in G}d(gx,gy)$ May 13 at 18:21
• By “a disk” you mean a ball, right? What about the question of whether a metric space homeomorphic a ($2$-dimensional closed) disk has a point fixed by all isometries? May 13 at 22:22
• @Gro-Tsen Yes disk=ball. Any compact topological group action on a 2-disk is equivalent to a linear action (hence has a fixed point). Any smooth action of a compact Lie group on the 3-disk is smoothly equivalent to a linear action. Any smooth finite group action on a 4-disk has a fixed point. For $n>5$ there is a fixed point free smooth action of the alternating group $A_5$ on the n-disk. I think the case $n=5$ is open. References can be found e.g. in the introduction and appendix A of arxiv.org/pdf/1706.08135.pdf. May 13 at 22:32
• Of course you don't need Mostow-Palais to get an invariant metric here: just take the standard Euclidean metric and average it along the smooth finite group action to get an invariant Riemannian metric.
– YCor
May 14 at 7:51
• @YCor I only mentioned Mostow-Palais because the above also answers mathoverflow.net/questions/422384/…. May 14 at 10:45