Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 13 / 5

We first prove that fixed-point-free sphere maps are homotopic to the antipodal map. If $f:S^d\to S^d$ has no fixed point, defineThe denominator could vanish only if $f(x)$ is a positive multiple of $x$. Both are unit vectors, so this would require $f(x)=x$ and $t=1/2$, excluded by the hypothesis. Thus $H$ is a continuous homotopy from $f$ to $x\mapsto -x$.

On $S^{2r}$, with $r\ge 1$, the antipodal map has degree $(-1{)}^{2r+1}=-1$, since it is the boundary map of the ambient linear transformation $-I$ on ${\mathbb{R}}^{2r+1}$. Homotopy invariance therefore gives degree minus one for every fixed-point-free action map. The degree of each homeomorphism is $\pm 1$, and functoriality makes degree a group homomorphismEvery nonidentity element in a free action has degree minus one, so this homomorphism has trivial kernel. HenceFor $S^0$ the same bound follows immediately from its two points. This proves that free actions on even-dimensional spheres have order at most two, without an Euler-characteristic argument requiring a cell structure on the quotient.

Identify $S^{2b-1}$ with the unit sphere in ${\mathbb{C}}^b$, and put $\zeta =e^{2\pi i/a}$. The generator of the cyclic group $\mathbb{Z}/a$ acts byIt preserves the sphere because $|\zeta |=1$. If its $k$th power fixes a point, then $({\zeta }^k-1)z_j=0$ for every $j$. Some coordinate is nonzero, so ${\zeta }^k=1$, meaning $k\equiv 0(\mathrm{mod}a)$. Thus the scalar root-of-unity action is free. The case $a=1$ is the trivial group and also satisfies freeness. The quotient is a standard lens space.

Suppose a nontrivial finite group acts freely on ${\mathbb{R}}^n$ by homeomorphisms. Choose a nonidentity element of finite order $m$, a prime $p$ dividing $m$, and the power of that element with exponent $m/p$. It has order $p$, so a subgroup $C_p$ also acts freely.

Let $M={\mathbb{R}}^n/C_p$. A finite free action is properly discontinuous: around any point choose mutually disjoint neighborhoods of the finitely many orbit points, then intersect their inverse translates to obtain an evenly covered neighborhood. Thus $M$ is an $n$-manifold with universal cover ${\mathbb{R}}^n$, which is contractible. The group-cohomology argument of Question 2 applies even if no cell structure has been chosen: singular simplices in $M$ lift to the cover, their lifts form free deck-group orbits, and the augmented singular chains of the contractible cover form an exact free resolution. Therefore

We calculate the right-hand side explicitly. Write $R=\mathbb{Z}[C_p]$, let $t$ generate $C_p$, and set $N=1+t+\cdots +t^{p-1}$. The periodic resolution of a finite cyclic group isIts exactness can be seen in coefficients. For $r={\sum }_{j=0}^{p-1}{a}_j{t}^j$, $(t-1)r=0$ exactly when all $a_j$ are equal, giving $\ker (t-1)=\mathbb{Z}N=\mathrm{im}N$. Also $Nr=\epsilon (r)N$, so $\ker N=\ker \epsilon$. If $\epsilon (r)=0$, thenHence $\ker N=\ker \epsilon =(t-1)R$, proving exactness everywhere.

Apply ${\mathrm{Hom}}_R(-,\mathbb{Z})$ with the trivial action. Multiplication by $t-1$ induces zero, and multiplication by $N$ induces multiplication by $p$. The cochain complex is thereforeIts positive even-degree cohomology iswhile its positive odd-degree cohomology is zero.

This contradicts the dimension of the quotient manifold. The noncompact form of Poincaré duality for noncompact manifolds, with the orientation local system when needed, isHere Borel-Moore homology is the homology of locally finite chains. Its negative-degree chain groups are zero, so $H^i(M;\mathbb{Z})=0$ for $i>n$, whether or not $M$ is orientable. Taking any $2j>n$ contradicts the displayed cyclic group calculation. ThusThis proves finite groups cannot act freely on Euclidean space for general homeomorphisms; it does not assume that the action consists of affine maps or isometries.