# Finite-order self-homeomorphisms of $\mathbf{R}^n$

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?

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.

• So 'automorphism' here doesn't mean anything about algebraic structure? That is, it's purely a topological question? – LSpice Aug 17 '17 at 21:34
• I mean topologically, yes. Maybe my vocabularies are too algebraic. – Wille Liou Aug 17 '17 at 21:35
• [Correcting here previous comment] Here's some info from the review of [HKMS] 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) – YCor Aug 17 '17 at 22:02
• Consider $f$ on $R^n$ with $f^m=id$. (1) "classical results of Smith": if $f$ is smooth, and if ($n\le 6$ or $m$ is prime-power) then answer is yes. (2) Kister, 1961-63 relying on Conner-Floyd 1959: construction of fixed-point-free smooth periodic maps on $R^8$. (3) [HKMS] for any $m$ non-prime-power and $n\ge 7$, existence of such smooth $f$. (4) Oliver 1979 fixed-point-free self-homeo on $R^n$ for $n\ge 6$ when $m$ is divisible by at least 3 primes. (5) (no ref): answer is yes for $n\le 4$ without smoothness. Open when $n=5$. – YCor Aug 17 '17 at 22:03
• In the polynomial case, the fixed point set of any power of $f$ is always homotopically equivalent to a finite CW-complex. So by Verdier's trace formula, $f$ always has a fixed point. – Wille Liou Aug 17 '17 at 22:11

Let T be a homeomorphism of period r on Euclidean space En. The classical result of P. A. Smith is that T has a fixed point if r is prime [Ann. of Math. (2) 35 (1934), 572–578] or a prime power [Amer. J. Math. 63 (1941), 1–8; MR0003199]. The author settles a question of long standing with the following theorem: If r is not a prime power, then there exists a triangulation μ of $E^{9r}$ and a homeomorphism T of period r on $E^{9r}$ without a fixed point, such that T is simplicial over μ. The construction is a modification of an example due to Conner and Floyd [Proc. Amer. Math. Soc. 10 (1959), 354–360; MR0105115].