Can a finite group act freely (as homeomorphisms) on $\mathbb R^n$
I am asking if whether or not a finite group acts freely (as homeomorphisms) on $\mathbb R^n$.
To answer in the negative, it suffices to show: for any homeomorphism $f$ such that $f^d=\text{id}_{\mathbb R^n}$, then $f$ has a fixed point.
I am looking for a complete self-contained answer.
Solution 1:
Any continuous map, $f: \mathbb R^n \rightarrow \mathbb R^n$, such that $f^n = \text{ id}$ for any natural number $n$ must have a fixed point. The proof is not entirely trivial and there are two ways to do it: either using Smith theory or using algebraic topology (see Bredon, Geometry and Topology, for instance where a scheme for such a proof is laid out). In general, if a group acts freely and properly discontinuously on $\mathbb R^n$, it cannot have torsion. This is also the reason that classifying spaces of finite groups are infinite dimensional. For example, the classifying space of $Z/2$ is $R\mathbb P^\infty$.
Solution 2:
This is an elaboration of the answer by user44441. Note that none of these proofs is elementary, meaning both use material going beyond what is typically covered in a graduate topology class, although the proof using cohomological dimension is much simpler, you just need to know sheaf cohomology and some basic material of cohomology of groups.
- Via Smith Theory. Let $G$ be a finite nontrivial group of homeomorphisms of $R^n$. Then the action of $G$ extends continuously to its action on $S^n$, the 1-point compactification of $R^n$. Let $C< G$ be a nontrivial cyclic subgroup of $G$, it has order $k>1$. Let $p$ be a prime divisor of $k$. Then $C$ contains a subgroup $H$ of order $p$, isomorphic to ${\mathbb Z}/p$, cyclic group of order $p$. According to the Smith Theory, see Theorem 7.11 in
G.Bredon, "Introduction to Compact Transformation Groups", Elsevier, 1972.
the fixed point set $F$ of $H$ is a Chech cohomology sphere over $p$ of some dimension $r\in [-1, n]$. In other words, the Chech cohomology groups $H^i(F, {\mathbb Z}/p)$ are isomorphic to that of the sphere $S^r$, where $(-1)$-dimensional sphere is the empty set. In order to use this result, you do not need to know what Chech cohomology is (just for the record, it is the sheaf cohomology with respect to the constant sheaf $\underline{{\mathbb Z}/p}$ on $F$), as in our situation, if $H$ acts freely on $R^n$, the set $F$ is the singleton $\{\infty\}$ and its Chech cohomology is the same as any cohomology you are are likely familiar with (singular/cellular/simplicial/cubical). The cohomology of the singleton (with ${\mathbb Z}/p$ coefficients) is zero in all degrees except degree zero, and $H^0(F, {\mathbb Z}/p)\cong {\mathbb Z}/p$. But no sphere (empty or not) can have such cohomology groups: Again, you do not need to know what Chech cohomology is since for topological manifolds it is isomorphic to the cellular/singular/... cohomology. Hence, a free nontrivial finite group action on $R^n$ cannot exist.
- Via group cohomology. If $G\times R^n\to R^n$ is a free action of a finite group, the projection $R^n\to X=R^n/G$ is a covering map. Hence, the space $X$ is $K(G,1)$. Therefore, $H^*(X, {\mathcal F})\cong H^*(G, M)$ for any $ZG$-module $M$, where ${\mathcal F}$ is the sheaf on $X$ associated with the module $M$. (To construct this sheaf, take the locally trivial fiber bundle $E\to X$ associated with the $ZG$-module $M$ regarded as a set acted upon by $G$, and consider the sheaf of locally constant local sections of $E\to X$.)
Now the proof hinges upon two facts:
a. For every nontrivial finite group $G$ there is a $ZG$-module $M$ and an infinite sequence $k_i\to \infty$ such that $$ H^{k_i}(G, M)\ne 0. $$ You can find this in Ken Brown's book "Cohomology of Groups". In our setting, you only need to know this for finite cyclic groups $G$. In this situation, instead of reading Brown's book you can read his freely available lecture notes here, Exercise 1.5: You take $M=ZG$.
b. If $X$ is a, say, metrizable, topological space of covering dimension $\le n$, then for any sheaf ${\mathcal F}$ on $X$, the cohomology groups $H^k(X, {\mathcal F})=0$ for all $k>n$. Once you get through the definitions, this is actually a "soft" fact, since the covering dimension assumption gives you a cofinal sequence of coverings of $X$ whose nerves are of dimension $\le n$ and, hence, the cochain groups defined using these covering vanish (in degrees $>n$), which implies vanishing of cohomology groups. Assuming that you are familiar with simplicial cohomology, this proof is the same as of the fact that for an $n$-dimensional simplicial complex $X$, its simplicial cohomology vanish in degrees $>n$. Since $R^n$ has covering dimension $\le n$ (you only need to know that the covering dimension is $\le n$, which is proven by constructing an explicit cofinal sequence of covers ${\mathcal U}$), the space $X$ covered by $R^n$ also has dimension $\le n$, simply by projecting ${\mathcal U}$'s from $R^n$ to $X$.
Now, adding (a) and (b), you conclude that $H^k(R^n/G, {\mathcal F})=0$ for all $k>n$ and all sheaves ${\mathcal F}$ (from (b)) while $H^{k_i}(R^n/G, {\mathcal F})\ne 0$ for some sequence $k_i\to \infty$ (from (a)) and for some sheaf ${\mathcal F}$. This is a contradiction.
Solution 3:
If $G$ acts freely and continuously on $\mathbb{R}^n$ then the quotient map $\mathbb{R}^n \to \mathbb{R}^n/G=X$ is a covering map of degree $n=|G|$. Hence we have for the Euler characteristic (see https://en.wikipedia.org/wiki/Euler_characteristic#Covering_spaces), $$ \chi(X) \cdot n=\chi(\mathbb{R}^n)=1.$$ Since the Euler characterstic is an integer, this implies that $G$ is trivial.