Consider the $n$-dimensional euclidean space $\mathbf{R}^n$. A self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ is said to be *of finite order* if $\phi^m = \mathrm{id}_{\mathbf{R}^n}$ for some positive integer $m$.

**Question**: Does every finite order self-homeomorphism $\phi:\mathbf{R}^n\to \mathbf{R}^n$ have a fixed point?

What I know about it:

If for every divisor $d | m$, the fixed-point set $\left(\mathbf{R}^n\right)^{\phi^d}$ of $\phi^d$ has its cohomology groups $H^*_c\left(\left(\mathbf{R}^n\right)^{\phi^d}, \mathbf{Z}\right)$ finitely generated over $\mathbf{Z}$, then the theorem of "Verdier, *Caractéristique d'Euler-Poincaré*, 1973" will be applicable. In particular, all the self-homeomorphisms of $\mathbf{R}^n$ of prime order has a fixed-point.

In that article Verdier derived a formula of the finite group representation on the alternating sum of the cohomology group with $\mathbf{Q}$-coefficients. This in particular implies a version of Lefschetz trace formula for finite-order self-homeomorphisms.

Unfortunately, I don't know if there can be some self-homeomorphism of non-prime order such that a certain power of it has its fixed-point set very complicated.

Haynes, Kwasik, Mast, Schultz, Periodic maps on R7 without fixed points; Math Proc Camb Phil Soc, 132, p. 131-136, 2002.by Elmonds. ams.org/mathscinet-getitem?mr=1866329 (continued) $\endgroup$ – YCor Aug 17 '17 at 22:024more comments